Q
stringlengths 70
13.7k
| A
stringlengths 28
13.2k
| meta
dict |
|---|---|---|
Laurent series expansion with multiple taylor expansions Find the laurent series expansion for the following function: $$f(z)=\frac{e^z}{z^3-z^4}$$
What I've done is:
\begin{align}
f(z) & = \frac{e^z}{z^3-z^4} \\
& = \frac{e^z}{z^3(1-z)} \\
& = \frac{e^z}{z^3} .\frac{1}{1-z} \\
& = \frac{e^z}{z^3}\Big(1+z+z^2+\dots\Big) \\
& = e^z\Big(\frac{1}{z^3}+\frac{1}{z^2}+\frac{1}{z}+1+z+z^2+\dots\Big) \\
& = \Big(1+z+\frac{z^2}{2}+\frac{z^3}{6}+\frac{z^4}{24}+\frac{z^5}{60}+\dots \Big)\Big(\frac{1}{z^3}+\frac{1}{z^2}+\frac{1}{z}+1+z+z^2+\dots\Big) \\
\end{align}
Am I supposed to multiply this out? Surely there's a better way?
| Here's how I would do it. First of all, I would compute the Taylor expansion of $\dfrac{e^z}{1-z}$:\begin{align}\frac{e^z}{1-z}&=\left(1+z+\frac{z^2}{2!}+\frac{z^3}{3!}+\cdots\right)\left(1+z+z^2+z^3+\cdots\right)\\&=1+2z+\frac52z^2+\cdots+\left(\sum_{k=0}^n\frac1{k!}\right)z^n+\cdots\end{align}And then\begin{align}\frac{e^z}{z^3-z^4}&=\frac1{z^3}\cdot\frac{e^z}{1-z}\\&=\frac1{z^3}\left(1+2z+\frac52z^2+\cdots+\left(\sum_{k=0}^n\frac1{k!}\right)z^n+\cdots\right)\\&=\frac1{z^3}+\frac2{z^2}+\frac5{2z}+\cdots+\left(\sum_{k=0}^{n+3}\frac1{k!}\right)z^n+\cdots\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3362017",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Use coordinate method to solve a pretty hard geometry problem Here is a hard geometry problem for my homework.
Let $D$ be a point inside $\Delta ABC$ such that $\angle BAD=\angle BCD$ and $\angle BDC=90^\circ$. If $AB=5,BC=6$ and $M$ is the midpoint of $AC$, find the length of $DM$.
(Taken from HK IMO Prelim 2012)
The teacher wants us to use the coordinate geometry to solve this problem. It does not appear too hard, but there is a big problem.
I set the coordinates of $A,B,C,D,M$ as in the diagram. Then, we can get the following statement:
$$\begin{cases} b^2+c^2=36 \\ x^2+\left(b+y\right)^2=25 \\ \sin\angle BAD=\sin\angle BCD \end{cases}$$
We need to find the value of $\sin\angle BAD$ and $\sin\angle BCD$
$$\sin\angle BCD=\dfrac{BD}{BC}=\dfrac{b}{6}\\ \text{The area of }\Delta ABD=\dfrac{1}{2}\left(AB\right)\left(AD\right)\sin\angle BAD \\ \dfrac{bx}{2}=\dfrac{5\sqrt{x^2+y^2}\sin\angle BAD}{2} \\ \sin\angle BAD=\dfrac{bx}{5\sqrt{x^2+y^2}}\\ \dfrac{b}{6}=\dfrac{bx}{5\sqrt{x^2+y^2}} \\ \sqrt{x^2+y^2}=\dfrac{6}{5}x \\ x^2+y^2=\dfrac{36}{25}x^2 \\ y^2=\dfrac{11}{25}x^2 \\ y=\dfrac{\sqrt{11}}{5}x$$
Then, when I proceed further, the expression gets very complicated. I felt puzzled, so I stopped. I found a geometric solution, which gives the answer of $\dfrac{\sqrt{11}}{2}$.
I hope you guys can help me solve this question using coordinate method. Thank you!
| After your calculations we have the system:
$$\begin{cases} b^2+c^2=36, \\ x^2+\left(b+y\right)^2=25, \\ \sqrt{11}x = 5y, \end{cases}\tag{1}$$
where $x>0,\;y>0$.
Substituting $x=\dfrac{5}{\sqrt{11}}y\;$ into $2$nd equation of $(1)$, we get:
$$
\dfrac{25}{11}y^2+(b+y)^2=25;
$$
$$
\dfrac{36}{11}y^2+2by+(b^2-25)=0;\tag{2}
$$
and positive $y$-solution of quadratic equation $(2)$ is:
$$
y = -\dfrac{11}{36}b + \dfrac{5\sqrt{11}}{36} \sqrt{36-b^2};
$$
when apply $1$st equation of $(1)$, we have:
$$
y = \dfrac{-11b + 5\sqrt{11} c}{36};\tag{3}
$$
therefore
$$
x = \dfrac{-5\sqrt{11}b + 25c}{36}.\tag{3'}
$$
And
$$
DM = \dfrac{1}{2}\sqrt{(c-x)^2+y^2}=\\
\dfrac{\sqrt{(11c+5\sqrt{11}b)^2 + (-11b+5\sqrt{11}c)^2}}{2\cdot 36} = \\
\dfrac{\sqrt{(11^2+25\cdot 11)(b^2+c^2)}}{2\cdot 36} = \dfrac{\sqrt{36\cdot 11\cdot 36}}{2\cdot 36} = \dfrac{\sqrt{11}}{2}.
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3362125",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
} |
How many integral solutions does $2x + 3y + 5z = 900$ have when $ x, y, z \ge 0$? Solution: Let $2x + 3y = u.$ Then we must solve $\begin{align} u + 5z = 900 \tag 1 \\ 2x + 3y = u \tag 2 \end{align}$
For $(1),$ a particular solution is $(u_0, z_0) = (0, 180).$ Hence, all the integral solutions of $(1)$ are $\begin{cases} u = 5t \\ z = 180 - t \end{cases} (t \in \mathbb Z)$
Substituting $u = 5t$ into $(2)$ gives $2x + 3y = 5t$ whose particular solution is $(x_0, y_0) = (t, t).$ Hence all the integral solutions of $(2)$ are $\begin{cases} x = t - 3s \\ y = t + 2s \end{cases} (t \in \mathbb Z)$
Thus all the integral solutions of $2x + 3y + 5z = 900$ are given by $$\begin{cases} x = t - 3s \\ y = t + 2s \\ z = 180 - t \end{cases} (s,t \in \mathbb Z)$$
Now suppose $x, y, z \ge 0.$
Note, $180 - t \ge 0 \implies t \le 180$ and so $t + 2s \ge 0 \implies s \ge -90$ and $t - 3s \ge 0 \implies s \le 60.$ Thus we have $-90 \le s \le 60.$
Consider $0 \le s \le 60.$ Now $t \le 180, \ t \ge 3s \implies 3s \le t \le 180$. Thus in this range of $s$, there are $180 - 3s + 1 = 181 - 3s$ of $t$'s.
Consider $-90 \le s < 0.$ Now $t \le 180, \ t \ge -2s \implies -2s \le t \le 180$. Thus in this range of $s$, there are $180 + 2s + 1 = 181 + 2s$ of $t$'s.
Range $0 \le s \le 60$ has the following points:
$(0, 181 - 3(0)), \ (1, 181 - 3(1)), (2, 181 - 3(2), \ldots (60, 181 - 3(60))$ of which there are $61.$
The range $-90 \le s < 0$ must have $91$ points. In sum, we have $61 + 91 = 152$ points for $x, y, z \ge 0.$
My question:
According to the book the answer is $\displaystyle{\sum_{s = 0}^{60}(181 - 3s) + \sum_{s = -90}^{-1}(181 + 2s) = 13651.}$ I don't understand why they took the sum of all $t$'s in the range of $s$. That means some of my denotations and labels above must be incorrect. Where's the mistake? Thanks.
edit:
I think I see my mistake. The number $181 - 3s$ is the number of $t$'s, not necessarily the form of $t$. Given that, the number of ordered pairs (in the given range) must be $(181 - 3s)*61$ by the product rule.
| Since the generating function
$$ f(z)=\frac{1}{(1-z^2)(1-z^3)(1-z^5)} $$
has a triple pole at $z=1$ and simple poles at $9$ points of $S^1$, we may decompose it as
$$ f(z) = \frac{1}{30(1-z)^3}+\frac{7}{60(1-z)^2}+\sum_{k=1}^{8}\frac{R_k}{z-R_k} $$
and deduce that the coefficient of $z^n$ in $f(z)$, up to a bounded error term (bounded by $\sum_{k=1}^{8}|R_k|$) is given by
$$ [z^n]\left[\frac{1}{30(1-z)^3}+\frac{7}{60(1-z)^2}\right]=\frac{(n+1)(n+9)}{60} $$
(this follows from stars and bars). By evaluating the RHS at $n=900$ we get a bit more than $13650$, and the number of representations of $900$ as $2a+3b+5c$ is actually $13651$. In order to have a simultaneous control of the magnitude and the error term we may prove first that
$$ [x^n]\frac{1}{(1-x^2)(1-x^3)} = \frac{2n+5}{12}+3(-1)^n+\left\{\begin{array}{rcl}1&\text{if}&n\equiv 0\pmod{3}\\-1&\text{if}&n\equiv 1\pmod{3}\\0&\text{if}&n\equiv 2\pmod{3}\\\end{array}\right.$$
which equals $\left\lfloor\frac{n+4}{6}\right\rfloor$, increased by one if $n\equiv 0\pmod{6}$. It follows that the exact number of representations is given by
$$ 31+\sum_{k=0}^{180}\left\lfloor \frac{5k+4}{6}\right\rfloor = 13651. $$
The sum in the last line equals the number of lattice points in a trapezium. By Pick's theorem this is given by the area of the trapezoid and the number of lattice points on the line $y=\frac{5x+4}{6}$, for $x\in[0,180]$. This counting problem is analogous to the determination of
$$|[4,904]\cap6\mathbb{Z}|=180.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3362584",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 1
} |
How find the function $f(f(x)+xy)=f(x)+xf(y),\forall x,y\in R$ Find all function $f:R\to R$ and such
$$f(f(x)+xy)=f(x)+xf(y),\forall x,y\in R$$
Let $x=y=0$,then $$f(f(0))=f(0)$$
| Replace $x=0$ in the given identity, we get
$ f(f(0)) = f(0).$
If we let $f(0) = a$, then
$$f(a) = a\tag{1}.$$
Replace $y=0$ in the given identity, we get
$$f(f(x)) = f(x) + xf(0) = f(x) + ax.\tag{2}$$
Let $x=a$ in (2), we get
$$f(f(a)) = f(a) + a^2.\tag{3}$$
But from (2), $f(f(a)) = f(a) = a$. So (3) becomes
$$a = a + a^2,\text{or } f(0) = 0.$$
Therefore, (2) becomes
$$ f(f(x)) = f(x), \forall x\in \mathbb R.\tag{4}$$
Replace $x$ by $f(x)$ in the given identity, we have
$$f(f(f(x)) + f(x)y) = f(f(x)) + f(x)f(y).$$
Since $f(f(x)) = f(x)$, we have
$$f(f(x) (y+1)) = f(x)(1+f(y)), \forall x,y.\tag{5}$$
Let $y=-1$ in (5), we see that
$$ 0 = f(0) = f(x)(1+f(-1)), \forall x.$$
Since $f(x)\not\equiv 0$, $f(-1) = -1$. Now replace $x=-1, y=1$ in (5), we get
$$f(-2) = - 1 - f(1), \forall y.\tag{6}$$
We will now show that $f(1)\ne 0$. Indeed, assume that $f(1)=0$, then $f(-2)=-1$. Replace $x=-1, y=-2$ in (5), we get
$$f(f(-1)(1-2)) = f(-1)(1-2).$$
Since $f(-1)=-1$, the above becomes a contradiction $f(1) = 1$.
Thus we showed that $f(1) = b\ne 0$. Replace $1=1$ in the given identity, we have
$$f(b + y) = b + f(y).$$
This and (4) implies that
$$f(b+f(y)) = f(f(b+y)) = f(b+y) = b + f(y), \forall y.\tag{7}$$
Moreover, replace $y=b/x$ in the given identity we get
$$f(f(x) + b) = f(x) + xf(\frac{b}{x}),\forall x\ne 0.$$
The above and (7) gives
$$ b + f(x) = f(x) + xf(\frac{b}{x}), \forall x\ne 0,$$
or
$$ \frac{b}{x} = f(\frac{b}{x}), \forall x\ne 0.$$
Since $b\ne 0$, it follows that $f(x)=x$ for all $x\ne 0.$ Thus $f(x) = x$ for all $x\in \mathbb R$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3363483",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 5,
"answer_id": 0
} |
Prove that $\sqrt{x^{3} + y^{3}+1} + \sqrt{z^{3} + y^{3}+1} + \sqrt{x^{3} + z^{3}+1} \ge 2 + \sqrt{2(x^{3} + y^{3}+z^{3})+1}$ Given $x, y,z \ge 0$. Prove that
$$\sqrt{x^{3} + y^{3}+1} + \sqrt{z^{3} + y^{3}+1} + \sqrt{x^{3} + z^{3}+1} \ge 2 + \sqrt{2(x^{3} + y^{3}+z^{3})+1} $$
Attempt
Notice that $$(x^{3} + y^{3}+1) + (z^{3} + y^{3}+1) + (x^{3} + z^{3}+1) = 2 + 2(x^{3} + y^{3}+z^{3})+1 $$
So the sum of squares must be between two forms as below:
$$2 + \sqrt{2(x^{3} + y^{3}+z^{3})+1} \le \sqrt{x^{3} + y^{3}+1} + \sqrt{z^{3} + y^{3}+1} + \sqrt{x^{3} + z^{3}+1} \le 2 + 2(x^{3} + y^{3}+z^{3})+1$$
If I work backwards I will get, by letting the form in each root be $a,b,c$ respectively:
$$\sqrt{a} + \sqrt{b} + \sqrt{c} \ge 2 + \sqrt{a+b+c-2} $$
$$ a + b +c + 2(\sqrt{ab} + \sqrt{bc} + \sqrt{ca}) \ge 4 + (a+b+c-2) + 4\sqrt{a+b+c-2}$$
$$ (\sqrt{ab} + \sqrt{bc} + \sqrt{ca}) \ge 1 + 2\sqrt{a+b+c-2}$$
$$ ((\sqrt{ab} + \sqrt{bc} + \sqrt{ca}) -1))^{2} \ge 4(a+b+c-2)$$
$$ (ab + bc + ca + 2a^{2}\sqrt{bc} +2b^{2}\sqrt{ac} +2c^{2}\sqrt{ab}) - 2(\sqrt{ab} + \sqrt{bc} + \sqrt{ca}) + 1 \ge 4(a+b+c-2)$$
How to continue?
Apart from the attempt above, I have tried:
*
*AM-GM for the sum of the 3 squares
*AM-GM for the sum of the 3 squares, then using Holder's inequality from the GM
*AM-GM for each square
| It should be
$$ (ab + bc + ca + 2a\sqrt{bc} +2b\sqrt{ac} +2c\sqrt{ab}) - 2(\sqrt{ab} + \sqrt{bc} + \sqrt{ca}) + 1 \ge 4(a+b+c-2),$$
where $a\geq1$, $b\geq1$ and $c\geq1$, for which it's not so easy to find a proof.
By the way, the Karamata's inequality helps.
Indeed, let $f(x)=\sqrt{x+1},$ $x^3+y^3=c$, $x^3+z^3=b$ and $y^3+z^3=a$.
Thus, $f$ is a concave function and we need to prove that
$$\sum_{cyc}\sqrt{a+1}\geq2+\sqrt{a+b+c+1},$$ which is just Karamata because if $a\geq b\geq c$ so
$$(a+b+c,0,0)\succ(a,b,c)$$ and
$$f(a+b+c)+f(0)+f(0)\leq f(a)+f(a)+f(b)$$ and we are done!
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3363591",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
How to solve $x^{2}\equiv-7\:\left(128\right)$? How to solve $x^{2}\equiv-7\:\left(128\right)$ ?
I know that $x^{2}\equiv a\:\left(2^{l}\right)$ is solvable iff $a\equiv1\:\left(8\right)$
which is the case here, but how can we find the solution?
| The general method would be to first solve mod 2 and discover that $x \equiv 1 \pmod{2}$. Then "lift" to mod $2^2$, which means that you observe that if $x \equiv 1 \pmod{2}$, then $x\equiv 1$ or $3 \pmod{4}$. Both of these are solutions. So lift them again to mod $2^3$ and get that $x \equiv 1, 3, 5,$ or $7 \pmod{8}$. All of these are solutions. So lift again to mod $2^4$.
Now you get that $x \equiv 1,3, 5, 7, 9, 11, 13,$ or $15\pmod{16}$ but only 4 of these solve the congruence mod 16: 3, 5, 11, and 13.
lift these to mod 32: $3 \pmod{15}$ lifts to $3$ and $19 \pmod{32}$. 5 lifts to 5 and 21. 11 lifts to 11 and 27. 13 lifts to 13 and 29. Of these 8, only 5, 11, 21 and 27 solve the congruence mod 32.
Lift two more times to get all four solutions mod 128.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3364212",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 4
} |
Determinant of block matrix where all blocks are $n \times n$ I found the following formula on Wikipedia:
Suppose $A$, $B$, $C$, and $D$ are matrices of dimension $n \times n$, $n \times m$, $m \times n$, and $m \times m$, respectively.
$$\det \begin{pmatrix} A & B \\ C & D\end{pmatrix} = \det(D) \det \left(A - B D^{-1} C \right)$$
Hence, if $D$ is a zero matrix, the determinant should be zero. But when I calculate the determinant of the following matrix
$$\det \left[ \begin {array}{ccc|ccc} 0&0&0&-4&0&0\\ 0&-1&0
&0&-5&0\\ 0&0&-2&0&0&-6\\ \hline 7&0&0&0&0
&0\\ 0&8&0&0&0&0\\ 0&0&9&0&0&0
\end {array} \right] = 60480
$$
Is this formula above really only applicable, when $n\neq m$?
| Since all four blocks are square and the bottom ones do commute (the zero matrix commutes with everyone),
$$\det \left[ \begin {array}{ccc|ccc} 0&0&0&-4&0&0\\ 0&-1&0
&0&-5&0\\ 0&0&-2&0&0&-6\\ \hline 7&0&0&0&0
&0\\ 0&8&0&0&0&0\\ 0&0&9&0&0&0
\end {array} \right] = \det \left( \mathrm O_3 + \mbox{diag} (4 \cdot 7, 5 \cdot 8, 6 \cdot 9) \right) = \prod_{i=4}^9 i = \frac{9!}{3!} = \color{blue}{60480}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3365789",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
If $A \in \mathbb{R}^{n\times n}$ and $\textbf{x}\in \mathbb{R}^{n}$why $A = \frac{1}{2} A + \frac{1}{2} A^\top $ is not always true? If $A \in \mathbb{R}^{n\times n}$ and $\textbf{x}\in \mathbb{R}^{n}$,
It is possible to prove that
$$
\textbf{x}^\top A \textbf{x} = \textbf{x}^\top(\frac{1}{2} A + \frac{1}{2} A^\top)\textbf{x}
$$
That could let us think that
$$
A = \frac{1}{2} A + \frac{1}{2} A^\top ?
$$
However if we take
$$A = \begin{bmatrix}1&2\\ 3&4 \end{bmatrix}$$
and compute
$$B = \frac{1}{2} A + \frac{1}{2} A^\top$$
we get $$B = \begin{bmatrix}1&2.5\\ 2.5&4 \end{bmatrix}\ne A$$
Although
$$
\textbf{x}^\top A \textbf{x} = \textbf{x}^\top B \textbf{x}
$$
Here is my question:
How can we explain that the following is not always true?
$$
A = \frac{1}{2} A + \frac{1}{2} A^\top
$$
| Observe that
$A = \dfrac{1}{2}(A + A^T) \Longleftrightarrow A = A^T, \tag 1$
that is,
$A = \dfrac{1}{2}(A + A^T) \tag 2$
if and only if $A$ is a symmetric matrix. This is easily seen as follows: if
$A = A^T, \tag 3$
then
$A = \dfrac{1}{2}(2A) = \dfrac{1}{2}(A + A) = \dfrac{1}{2}(A + A^T); \tag 4$
going the other way, if
$A = \dfrac{1}{2}(A + A^T), \tag 5$
then
$A = \dfrac{1}{2}A + \dfrac{1}{2}A^T, \tag 6$
whence
$\dfrac{1}{2}A = A - \dfrac{1}{2}A = \dfrac{1}{2}A^T, \tag 7$
so
$A = A^T. \tag 8$
We thus see that (2) is false if $A$ is not symmetric.
Note we need not introduce $\mathbf x \in \Bbb R^n$ to establish this.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3368309",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 2
} |
Solve the equations and find $x,y,z$ Given:
$$x^3+y^3+z^3=x+y+z$$
And:
$$x^2+y^2+z^2=xyz$$
Find all real and positive solutions to these equations, if any.
So most probably, we'll factorise in this way:
$$x^3+y^3+z^3-3xyz=x+y+z-3xyz
\rightarrow (x+y+z)(x^2+y^2+z^2-xy-yz-zx)=(x+y+z)-3(x^2+y^2+z^2)$$
But I'm having trouble in proceeding further.
Kindly help.
| Denote by $\sigma_i$ the elementary symmetric functions and by $p_r$ the power sums of the nonnegative variables $x$, $y$, $z$. We are told that
$$p_3=\sigma_1,\qquad p_2=\sigma_3\ .\tag{1}$$
Of course $x=y=z=0$ is a solution of $(1)$. We therefore may assume $\sigma_1>0$ in the sequel. Using the inequalities $$\left({x+y+z\over3}\right)^2\leq{x^2+y^2+z^2\over3},\qquad \bigl(\sigma_3\bigr)^{1/3}\leq{\sigma_1\over3}$$
we find
$${1\over3}\sigma_1^2\leq p_2=\sigma_3\leq{1\over27}\sigma_1^3\ ,$$
so that $\sigma_1\geq9$. On the other hand
$${x^3+y^3+z^3\over3}\geq\left({x+y+z\over3}\right)^3$$
implies
$$\sigma_1=p_3\geq{1\over9}\sigma_1^3\ ,$$
hence $\sigma_1\leq3$. It follows that the given equations $(1)$ cannot be solved with nonnegative $x$, $y$, $z$ other than $(0,0,0)$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3368966",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
} |
Polynomial of 4th degree I would like to ask if someone could help me with the following equation.
\begin{equation}
x^4+ax^3+(a+b)x^2+2bx+b=0
\end{equation}
Could you first solve in general then $a=11$ and $b=28$.
I get it to this form but I stuck.
\begin{equation}
1+a\left(\frac{1}{x}+\frac{1}{x^2}\right)+b\left(\frac{1}{x^2}+\frac{1}{x^3}+\frac{1}{x^3}+\frac{1}{x^4}\right) = 0
\end{equation}
Thank you in advance.
| You are almost there !
$$1+a\left(\frac{1}{x}+\frac{1}{x^2}\right)+b\left(\frac{1}{x^2}+\frac{1}{x^3}+\frac{1}{x^3}+\frac{1}{x^4}\right) = 0$$
$$1+a\left(\frac{x+1}{x^2}\right)+b\left(\frac{x+1}{x^3}+\frac{x+1}{x^4}\right) = 0$$
$$1+a\left(\frac{x+1}{x^2}\right)+b\left(\frac{x+1}{x^2}\right)^2 = 0$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3369802",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
If $A+B+C=\pi$, prove that $\cos (A-B) \cos (B-C) \cos (C-A)\ge 8\cos A \cos B \cos C$ If $A+B+C=\pi$, prove that $\cos (A-B) \cos (B-C) \cos (C-A)\ge 8\cos A \cos B \cos C$
I know this is true for acute angle triangle.
I want to know whether it is true for every real $A,B,C$ such that $A+B+C=\pi.$
| Yes, this inequality is true for any reals $A,$ $B$ and $C$ such that $A+B+C=\pi.$
Indeed, let $\cos(A-B)=x$ and $\cos(A+B)=y.$
Thus, we need to prove that:
$$x(\cos(A+B-2C)+\cos(A-B))\geq8(\cos(A+B)+\cos(A-B))(-\cos(A+B))$$ or
$$x(\cos3(A+B)+x)+8y(x+y)\geq0$$ or
$$x(4y^3-3y+x)+8y(x+y)\geq0$$ or
$$x^2+(4y^3+5y)x+8y^2\geq0,$$ for which it's enough to prove that $f(x)\geq0,$ where
$$f(x)=x^2-(4y^3+5y)x+8y^2$$ and $\{x,y\}\subset[0,1].$
Now, if $$(4y^3+5y)^2-4\cdot8y^2\leq0$$ or
$$0\leq y\leq\frac{\sqrt{4\sqrt2-5}}{2},$$ so our inequality is proven.
But for $\frac{\sqrt{4\sqrt2-5}}{2}<y\leq1$ we see that $\frac{4y^3+5y}{2}>1,$ which says that $f$ decreases.
Id est, $$f(x)\geq f(1)=1-4y^3-5y+8y^2=(1-y)(2y-1)^2\geq0$$ and we are done!
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3371107",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Find all real matrices such that $X^{3}-4X^{2}+5X=\begin{pmatrix} 10 & 20 \\ 5 & 10 \end{pmatrix}$ The following question come from the 1998 Romanian Mathematical Competition:
Find all matrices in $M_2(\mathbb R)$ such that $$X^{3}-4X^{2}+5X=\begin{pmatrix} 10 & 20 \\ 5 & 10 \end{pmatrix}$$
Can you guy please help me? Thanks a lot!
| Let $X=\begin{pmatrix} a & b \cr c & d \end{pmatrix}$. Then the matrix equation immediately gives that $d=a$ and $b=4c$. Now it is easy to solve the last equations in $a$ and $c$. The only real solution is $a=2$ and $c=1$, and hence
$$
X=\begin{pmatrix} 2 & 4 \cr 1 & 2 \end{pmatrix}.
$$
For complex numbers there are several other solutions, e.g.,
$a=\frac{i + 6}{2},\; c= \frac{2- i }{4}$, or $a=-\frac{\sqrt{-5}}{2},\;c=-\frac{\sqrt{-5}}{4}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3372342",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 5,
"answer_id": 1
} |
Proof that $3^{10^n}\equiv 1\pmod{10^n},\, n\ge 2$ This should be rather straightforward, but the goal is to prove that
$$3^{10^n}\equiv 1\pmod{10^n},\, n\ge 2.$$
A possibility is to use
$$\begin{align*}3^{10^{n+1}}-1&=\left(3^{10^n}-1\right)\left(1+\sum_{k=1}^9 3^{10^n k}\right)\\&=\left(3^{10^n}-1\right)\left(3^{9\cdot 10^n}-1+3^{8\cdot 10^n}-1+\cdots +3^{10^n}-1+10\right),\end{align*}$$
but I don't see how this proves the equality in the question. I got only
$$3^{10^{n}}\equiv 1\pmod{3^{10^{n-1}}-1}.$$
Perhaps someone here can explain.
| The identity you used leads to a proof of the desired result by induction as follows. So suppose that $10^n$ divides $3^{10^n}-1$ for all $n=2,\ldots,N$. Applying the identity yields
$$
\frac{3^{10^{N+1}}-1}{3^{10^N}-1}=10 + \sum_{k=1}^9(3^{k\cdot 10^N}-1),
$$
and the right side is a multiple of $10$ since each term in the sum is a multiple of $3^{10^N}-1$.
Thus, since $10^N$ divides $3^{10^N}-1$ by the inductive hypothesis, we see that $10^{N+1}$ divides $3^{10^{N+1}}-1$ and this completes the inductive step. All that is left is to check the base case $n=2$, that $3^{100}-1$ is divisible by $100$. This is straightforward using standard modular arithmetic techniques, for instance by checking modulo $4$ and modulo $25$ separately.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3373359",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 0
} |
How can I find this limit applying squeeze theorem? How can I evaluate this limit applying squeeze theorem?
$$\lim_{x\to6}\frac{1-\sqrt{ 3-\sqrt{x-2}}}{x-6}$$
I found this limit with standart way:
$\lim_{x\to6}\frac{1-\sqrt{ 3-\sqrt{x-2}}}{x-6}=\lim_{x\to6}\frac{\left(1+\sqrt{ 3-\sqrt{x-2}} \right)\times \left( 1-\sqrt{ 3-\sqrt{x-2}}\right)}{(x-6)\times \left( 1+\sqrt{ 3-\sqrt{x-2}}\right)}=\frac 18$
I want to write this limit using squeeze theorem.
| First step let wlog $y=x-6\to 0^+$ thus
$$\lim_{x\to6^+}\frac{1-\sqrt{ 3-\sqrt{x-2}}}{x-6}=\lim_{y\to0^+}\frac{1-\sqrt{ 3-\sqrt{y+4}}}{y}$$
then by Bernoulli inequality
$$\sqrt{y+4} =2(1+y/4)^\frac12\le 2(1+y/8)=2+y/4$$
$$3-\sqrt{y+4}\ge 3-2-y/4=1-y/4$$
$$\sqrt{ 3-\sqrt{y+4}}\ge \sqrt{ 1-y/4}$$
$$1-\sqrt{ 3-\sqrt{y+4}}\le 1-\sqrt{ 1-y/4}$$
then
$$\frac{1-\sqrt{ 3-\sqrt{y+4}}}{y} \le \frac{1-\sqrt{ 1-y/4}}{y}$$
For the other inequality use that
$$\sqrt{y+4} =2(1+y/4)^\frac12\ge 2(1+y/8-y^2/128)=2+y/4-y^2/64$$
$$3-\sqrt{y+4}\le 3-2-y/4+y^2/64$$
$$\sqrt{ 3-\sqrt{y+4}}\le \sqrt{ 1-y/4+y^2/64}$$
$$1-\sqrt{ 3-\sqrt{y+4}}\ge 1-\sqrt{ 1-y/4+y^2/64}$$
Then finally
$$\frac{1-\sqrt{ 1-y/4+y^2/64}}{y} \le \frac{1-\sqrt{ 3-\sqrt{y+4}}}{y} \le \frac{1-\sqrt{ 1-y/4}}{y}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3373480",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
} |
Proof regarding the minimum of a function I want to show that the function
$$f(x)= \frac{2x}{1+(\frac{1}{k-x})^a} + \frac{k-2x}{1+2\cdot(\frac{1}{k-x})^a+ (\frac{1}{k-2x})^a}$$
with
$$ x \in [0,\frac{k}{2}], a \in [0,\infty], k \in [0,1] $$
has the minimum
$$min(f(x)) = \begin{cases}
\frac{k}{1+3\cdot(\frac{1}{k})^a},\, a\leq \frac{log(3)}{log(2)}
\\
\frac{k}{1+(\frac{1}{k/2})^a} ,\, else
\end{cases}$$
for all permited values of the parameters $a,k$. The minima can be obtained by setting either $x=0$ or $x=\frac{k}{2}$. I checked that this is in fact the minimum by simulating the function for all permitted parameter values.
My initial idea was to use the generalized version of the Binomial theorem
on the $1/(k-x)^a$ terms, but this only worked for $a\leq2$ because the elements of the series obtained from the theorem are not shrinking in absolute value.
Another idea was to show that every critical point of the function is never a global minimum, but I cannot find a solution for $f'(x)=0$.
Any ideas would be greatly appreciated. Thanks in advance.
| Update
We can also prove the case when $a > \frac{11}{4}$.
We will use Fact 1 and 2 whose proof is simple and thus omitted.
Fact 1: If $a > \frac{11}{4}$, we have
$$\frac{2x}{1+(\frac{1}{k-x})^a} \ge \frac{k}{1 + (\frac{2}{k})^a}$$
for $\frac{k}{4} \le x \le \frac{k}{2}$.
Fact 2: If $a > \frac{11}{4}$, we have
$$\frac{k^2}{k + 2(k-x)^{1-a} + (k-2x)^{1-a}} \ge \frac{k}{1+(\frac{2}{k})^a}$$
for $0\le x \le \frac{k}{4}$.
Now let us proceed.
If $0\le x\le \frac{k}{4}$, using the Cauchy-Bunyakovsky-Schwarz inequality and Fact 2, we have
$$f(x) \ge \frac{k^2}{k + 2(k-x)^{1-a} + (k-2x)^{1-a}} \ge \frac{k}{1+(\frac{2}{k})^a}.$$
If $\frac{k}{4} \le x \le \frac{k}{2}$, using Fact 1, we have
$$f(x) \ge \frac{2x}{1+(\frac{1}{k-x})^a} \ge \frac{k}{1 + (\frac{2}{k})^a}.$$
Previously written
Partial answer
I prove the case when $0 < a < 1$. (The case $a=1$ is easy.)
Let
\begin{align}
g(x) &= 2x \Big(1 + \Big(\frac{1}{k-x}\Big)^a\Big)
+ (k-2x)\Big(1 + 2\Big(\frac{1}{k-x}\Big)^a + \Big(\frac{1}{k-2x}\Big)^a\Big)\\
&= k + 2(k-x)^{1-a} + (k-2x)^{1-a}.
\end{align}
Using the Cauchy-Bunyakovsky-Schwarz inequality, we have
$$f(x) \ge \frac{(2x + k-2x)^2}{g(x)} = \frac{k^2}{g(x)}.$$
If $0 < a < 1$, then $g(x)$ is strictly decreasing for $x\in [0, k/2]$.
We have $f(x) \ge \frac{k^2}{g(0)} = \frac{k^2}{k+ 3k^{1-a}} = f(0)$.We are done.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3375808",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 1,
"answer_id": 0
} |
Find $\lim_{x \to 0} \frac{(\tan(\tan x) - \sin (\sin x))}{ \tan x - \sin x}$ Find $$\lim_{x\to 0} \dfrac{\tan(\tan x) - \sin (\sin x)}{ \tan x - \sin x}$$
$$= \lim_{x \to 0} \dfrac{\frac{\tan x \tan (\tan x)}{\tan x}- \frac{\sin x \sin (\sin x)}{\sin x}}{ \tan x - \sin x} = \lim_{x \to 0} \dfrac{\tan x - \sin x}{\tan x - \sin x} = 1$$
But the correct answer is $2$. Where am I wrong$?$
| Your way is wrong since
$${\tan(\tan x) - \sin (\sin x)}\neq { \tan x - \sin x}$$
indeed this step
$$\frac{\tan x \tan (\tan x)}{\tan x}=\tan x\cdot \frac{ \tan (\tan x)}{\tan x}\color{red}{=\tan x \cdot 1}=\tan x$$
and the similar for $\sin x$ are not allowed (in general we can't evaluate limit only for a part or factor of the whole expression; see also here).
We can avoid Tayor's expansion and use that (see here):
*
*$\lim_{x\to0}\frac{\tan x-x}{x^3}=\frac13$
*$\lim_{x\to0}\frac{\sin x-x}{x^3}=-\frac{1}6$
then
$$\dfrac{\tan(\tan x) - \sin (\sin x)}{ \tan x - \sin x}=\dfrac{\frac{\tan^3 x}{x^3}\frac{\tan(\tan x)-\tan x}{\tan^3x} - \frac{\sin^3 x}{x^3}\frac{\sin(\sin x)-\sin x}{\sin^3x}+\frac{\tan x-x}{x^3}-\frac{\sin x-x}{x^3}}{ \frac{\tan x-x}{x^3} - \frac{\sin x-x}{x^3}}\\\to\frac{1\cdot \frac13-1\cdot\left(-\frac16\right)+\frac13-\left(-\frac16\right)}{ \frac13-\left(-\frac16\right)}=2$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3376481",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
"answer_id": 2
} |
Variance of taking balls from a bag question A bag contains $2$ white balls and $2$ black balls.
Each instance a ball is taken from the bag, if it's white it's returned to the bag and if it's black it's replaced by a white ball. The game ends when no black balls remain in the bag. Let $Y$ be the number of instances of the game; calculate $\operatorname{var}(Y).$ Answer in the book: $14.$
I am not sure how to write the probability function of $Y;$ please help and thanks in advance!
| Let $Y$ be the number of black balls taken. The probability distribution table:
$$\begin{array}{c|l}
Y&P(Y)\\
\hline
2&\frac1{2^3}\\
3&\frac1{2^4}+\frac3{2^5}\\
4&\frac1{2^5}+\frac3{2^6}+\frac{3^2}{2^7}\\
5&\frac1{2^6}+\frac3{2^7}+\frac{3^2}{2^8}+\frac{3^3}{2^9}\\
\vdots
\end{array}$$
Note:
$$P(Y=n)=\sum_{i=1}^{n-1} \frac{3^{i-1}}{2^{n+i}}=\frac13\cdot \frac1{2^n}\sum_{i=1}^{n-1}\left(\frac32\right)^i=\frac{2\cdot 3^n-3\cdot 2^n}{3\cdot 4^n}.$$
Hence:
$$Var(Y)=\mathbb E(Y^2)-(\mathbb E(Y))^2=\\
\sum_{i=2}^\infty i^2\cdot \frac{2\cdot 3^i-3\cdot 2^i}{3\cdot 4^i}-\left(\sum_{i=2}^\infty i\cdot \frac{2\cdot 3^i-3\cdot 2^i}{3\cdot 4^i}\right)^2=\\
50-36=14.$$
WA answers: 1, 2, 3.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3376817",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
} |
$\lim_{x \rightarrow 4} \frac{\sqrt{x}-2}{x-4}$. Did I choose my $\delta$ Properly? Preliminaries:
We know that, the limit as $ x \rightarrow a $ of a function $ f(x) $ is $ L$ if for every $ \epsilon > 0 $ there exist a $ \delta > 0 $ such that $ 0 <|x-a|<\delta \implies |f(x) - L|<\epsilon$. Our function is $ f(x) = \frac{\sqrt{x} - 2}{x - 4}$.
Note that
$$
\lim_{x \rightarrow 4} \frac{\sqrt{x} - 2}{x - 4}
$$
cannot be solved as it is, for replacing $ x $ for $ 4 $ will render an undetermined answer. In order to fix this, we can rationalize the denominator by performing:
$$
\frac{\sqrt{x} - 2}{x - 4} \cdot \frac{\sqrt{x}+2}{\sqrt{x}+2} = \frac{(x-4)}{(x-4)(\sqrt{x}+2)} = \frac{1}{\sqrt{x}+2}
$$
Let the rationalized expression be $ g(x) $. as $ g(x) = f(x) $ we can compute the limit:
$$
\lim_{x \rightarrow 4}\frac{1}{\sqrt{x}+2} = \frac{1}{\sqrt{4}+2} = \frac{1}{2+2} = \frac{1}{4}
$$
The Process for Choosing the Delta:
We shall say, that the limit as $ x \rightarrow 4 $ of $ \frac{1}{\sqrt{x}+2}$ is $\frac{1}{4}$ if for every $ \epsilon > 0 $ there exist a $ \delta > 0$ such that $ 0 < | x - 4| < \delta \implies \left|\frac{1}{\sqrt{x}+2}-\frac{1}{4}\right|<\epsilon$
Thanks to the property of triangle inequality, we have that:
$$
\left|\frac{1}{\sqrt{x}+2}-\frac{1}{4}\right| \leq \left| \frac{1}{\sqrt{x}+2}\right| - \left|\frac{1}{4}\right|
$$
For instance, it is sufficient for
$$
\left| \frac{1}{\sqrt{x}+2}\right| - \frac{1}{4} < \epsilon
$$
which is equivalent to
$$
\left|\frac{1}{\sqrt{x}+2}\right| < \frac{4\epsilon+1}{4}
$$
Now, as per the definition of absolute value inequality, we have that:
$$
-\frac{4\epsilon + 1}{4} < \frac{1}{\sqrt{x}+2} < \frac{4\epsilon + 1}{4}
$$
By raising the expression to the power of minus one on each "side of the inequality" we can argue that,
$$
-\frac{4}{4\epsilon +1} < \frac{\sqrt{x}+2}{1} < \frac{4}{4\epsilon+1}
$$
which is equivalent to
$$
\left|\sqrt{x}+2\right| < \frac{4}{4\epsilon+1}
$$
by using the triangle inequality once again, we can argue that $\left|\sqrt{x}+2\right| \leq \left| \sqrt{x} \right| + |2|$
which means that it suffices for:
$$
\left|\sqrt{x}\right| + 2 < \frac{4}{4\epsilon + 1}
$$
$$
\left| \sqrt{x} \right| < \frac{4}{4\epsilon+1} - 2
$$
$$
\left|\sqrt{x}\right| < \frac{4-2(4\epsilon +1)}{4\epsilon+1}
$$
If we square both sides of the equation, we shall have
$$
\left|\sqrt{x}\right|^2 = \left|\sqrt{x}\right| \cdot \left|\sqrt{x}\right| = \left| x \right| < \left(\frac{4-2(4\epsilon +1)}{4\epsilon+1}\right)^2
$$
Now, using once again the triangle inequality, we have that
$$
\left|x-4\right| \leq \left|x\right| + \left|- 4\right|
$$
for instance, if
$$\left|x\right| + \left|- 4\right| < \left(\frac{4-2(4\epsilon +1)}{4\epsilon+1}\right)^2 + \left|-4\right| = \left(\frac{4-2(4\epsilon +1)}{4\epsilon+1}\right)^2 + 4$$
we will have
$$
\left| x-4 \right| < \left(\frac{4-2(4\epsilon +1)}{4\epsilon+1}\right)^2 + 4$$
For instance, if we choose
$$ \delta =\left(\frac{4-2(4\epsilon +1)}{4\epsilon+1}\right)^2 + 4$$
The Limit will exist.
question: Did I choose my Delta Correctly, and for instance is my proof correct?
|
Thanks to the property of triangle inequality, we have that:
$$
\left|\frac{1}{\sqrt{x}+2}-\frac{1}{4}\right| \leq \left| \frac{1}{\sqrt{x}+2}\right| \color{red}{\mathbf{-}} \left|\frac{1}{4}\right|
$$
This is not correct.
Instead, you could rewrite:
$$\left|\frac{1}{\sqrt{x}+2}-\frac{1}{4}\right|=\frac{1}{4}\left|\frac{2-\sqrt{x}}{2+\sqrt{x}}\right|=\frac{1}{4}\left|\frac{4-x}{\left(2+\sqrt{x}\right)^2}\right|=\frac{1}{4}\frac{\left|x-4\right|}{\left(2+\sqrt{x}\right)^2} \le \frac{\left|x-4\right|}{16}$$
And since $x \to 4$, you can make this arbitrarily small.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3385507",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
In how many ways can 10 blankets be given to 3 beggars such that each recieves at least one blanket? The question was to find the number of ways in which 10 identical blankets can be given to 3 beggars such that each receives at least 1 blanket. So I thought about trying the multinomial theorem...this is the first time I've tried it so I'm stuck at a point...
So $$x_1+x_2+x_3 = 10$$
Subject to the condition that :
$$1\leq x_1 \leq8$$
$$1\leq x_2 \leq8$$
$$1\leq x_3 \leq8$$
As each beggar can get at maximum 8 blankets and at minimum, 1.
So the number of ways must correspond to the coefficient of $x^{10}$ in:
$$(x^1+x^2+x^3+x^4+x^5+x^6+x^7+x^8)(x^1+x^2+x^3+x^4+x^5+x^6+x^7+x^8)(x^1+x^2+x^3+x^4+x^5+x^6+x^7+x^8)$$
= coeff of $x^{10}$ in $x^3(1+x^1+x^2+x^3+x^4+x^5+x^6+x^7)(1+x^1+x^2+x^3+x^4+x^5+x^6+x^7)(1+x^1+x^2+x^3+x^4+x^5+x^6+x^7)$
= coeff of $x^{10}$ in $x^3(1+x^1+x^2+x^3+x^4+x^5+x^6+x^7)^3$
= coeff of $x^{10}$ in $x^3(1-x^7)^3(1-x)^{-3}$
= coeff of $x^{10}$ in $x^3(1-x^{21}-3x^7(1-x^7))(1-x)^{-3}$
= coeff of $x^{10}$ in $(x^3-3x^{10})(1+\binom{3}{1}x + \binom{4}{2}x^2+...+ \binom{12}{10}x^{10})$
= $\binom{9}{7} - 3 = 33$
Is this right? From here I get the answer as $\binom{9}{7} - 3 = 33$ but the answer is stated as $36$. I don't understand where I'm making a mistake
| Your error is going from
$$(x^1+x^2+x^3+x^4+x^5+x^6+x^7+x^8) (x^1+x^2+x^3+x^4+x^5+x^6+x^7+x^8) \\(x^1+x^2+x^3+x^4+x^5+x^6+x^7+x^8)$$
to
$$x^3(x^1+x^2+x^3+x^4+x^5+x^6+x^7)(x^1+x^2+x^3+x^4+x^5+x^6+x^7) \\(x^1+x^2+x^3+x^4+x^5+x^6+x^7)$$
and you should have written an $x^0$ term as in
$$x^3(x^0+x^1+x^2+x^3+x^4+x^5+x^6+x^7)(x^0+x^1+x^2+x^3+x^4+x^5+x^6+x^7) \\(x^0+x^1+x^2+x^3+x^4+x^5+x^6+x^7)$$
getting you later to
coeff of $x^{10}$ in $x^3(1-x^8)^3(1-x)^{-3}$
and then giving you $36$ rather than $33$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3385830",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 0
} |
In multiplying 2 matrices, how do you know whether to operate on rows or columns? My understanding is that multiplying a matrix by a matrix on its left means operating on rows, and multiplying a matrix by a matrix on its right means operating on columns.
When there are 2 matrices next to each other to be multiplied, how can I know whether I'm supposed to operate on columns or on rows? These 2 operations appear to produce different results.
| I think this example will help, if we have any $3×3$ matrix $A$, and we multiply it by the permutation matrix $P_{2,3}$ from the right, here is what happens,
$$AP_{2,3}= \begin{pmatrix}
a & b & c \\
d & e & f \\
g & h & i
\end{pmatrix}
\begin{pmatrix}
1 & 0 & 0 \\
0 & 0 & 1 \\
0 & 1 & 0
\end{pmatrix}= \begin{pmatrix}
a & c & b \\
d & f & e \\
g & i & h
\end{pmatrix}$$
(The second and third columns are switched)
Now if we multiply it by the permutation matrix $P_{2,3}$ from the left, here is what happens,
$$P_{2,3}A= \begin{pmatrix}
1 & 0 & 0 \\
0 & 0 & 1 \\
0 & 1 & 0
\end{pmatrix} \begin{pmatrix}
a & b & c \\
d & e & f \\
g & h & i
\end{pmatrix} =
\begin{pmatrix}
a & b & c \\
g & h & i \\
d & e & f
\end{pmatrix}$$
(The second and third rows are switched)
So $P$ works on the columns of $A$ when multiplied by $A$ from the right, and $P$ works on the rows of $A$ when multiplied by $A$ from the left.
Note that $P_{2,3}$ is the matrix obtained from the identity by flipping the second and third row (or column).
Also, $$P_{2,3}P_{2,3}= \begin{pmatrix}
1 & 0 & 0 \\
0 & 0 & 1 \\
0 & 1 & 0
\end{pmatrix} \begin{pmatrix}
1 & 0 & 0 \\
0 & 0 & 1 \\
0 & 1 & 0
\end{pmatrix}= \begin{pmatrix}
1 & 0 & 0 \\
0 & 1 & 0 \\
0 & 0 & 1
\end{pmatrix}$$
You can say that the $P_{2,3}$ on the right switched the second and the third columns of the $P_{2,3}$ on the left, OR you can also say that the $P_{2,3}$ on the left switched the second and third rows of the $P_{2,3}$ on the right.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3388258",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Find all the values of the parameter 'a' for which the given inequality is satisfied for all real values of x. Find all the values of the parameter 'a' for which the inequality is satisfied for all real values of x.
$$a\cdot 9^x+4\cdot \left(a-1\right)\cdot 3^x+\left(a-1\right)>0$$
My attempt is as follows:-
$$a\cdot \left(9^x+4\cdot 3^x+1\right)-(4\cdot 3^x+1)>0$$
$$a\cdot \left(9^x+4\cdot 3^x+1\right)>4\cdot 3^x+1$$
$$a>\frac{4\cdot 3^x+1}{9^x+4\cdot 3^x+1}$$
Now if a is greater than the maximum value of $\frac{4\cdot 3^x+1}{9^x+4\cdot 3^x+1}$, then inequality will be true for all x.
So if we can find the range of $\frac{4\cdot 3^x+1}{9^x+4\cdot 3^x+1}$, then we can say a should be greater than the maximum value in the range.
Let's assume y=$\frac{4\cdot 3^x+1}{9^x+4\cdot 3^x+1}$ and substitute $3^x$ with $t$.
$$y=\frac{4t+1}{t^2+4t+1}$$
$$yt^2+4ty+y=4t+1$$
$$yt^2+4t(y-1)+y-1=0$$
We want to have real values of t satisfying the equation, so $D>=0$
$$16(y-1)^2-4*y*\left(y-1\right)>=0$$
$$4(y-1)(4y-4-y)>=0$$
$$4(y-1)(3y-4)>=0$$
$$y\in \left(-\infty,1 \right] \cup \left[\frac{4}{3},\infty\right)$$
So I am getting maximum value tending to $\infty$ for y=$\frac{4\cdot 3^x+1}{9^x+4\cdot 3^x+1}$
I am not able to understand where am I making mistake.
Official answer is $a\in \left[1,\infty\right) $
| Prove that $$\frac{4\cdot3^x+1}{9^x+4\cdot3^x+1}<1$$ for any real $x$.
Now, $$\lim_{x\rightarrow-\infty}\frac{4\cdot3^x+1}{9^x+4\cdot3^x+1}=1,$$ which gives $$\sup\frac{4\cdot3^x+1}{9^x+4\cdot3^x+1}=1$$ and the answer $a\geq1.$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3388670",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
Roots over $\mathbb{C}$ equation $x^{4} - 4x^{3} + 2x^{2} + 4x + 4=0 $. I need roots over $\mathbb{C}$ equation $$x^{4} - 4x^{3} + 2x^{2} + 4x + 4 = 0$$
From Fundamental theorem of algebra we have statement that the equation have 4 roots over complex.
But I prepare special reduction:
$$ \color{red}{ x^{4} - 4x^{3} + 2x^{2} + 4x + 4} = (x-1)^{4}-4(x-1)^{2} + 7 $$
for substitution $y = (x-1)^{2}$ we have:
$y^{2} - 4y + 7 = 0 $
$y_{0} = 2+i\sqrt{3}$
$y_1 = 2 - i\sqrt{3}$
and we have
$y_{0}^{1/2} + 1 = x_{0} $, $-y_{0}^{1/2} + 1 = x_{1}$,
$y_{1}^{1/2} + 1 = x_{2}$, $-y_{1}^{1/2} +1 = x_{3} $
I am not sure what is good results.
Please check my solution.
EDIT:
The LHS is not correct, I modify this equation. We should have $p(x) = x^{4} - 4x^{3} + 2x^{2} + 4x + 4 = (x-1)^{4}-4(x-1)^{2} + 7 $
EDIT2:
I need show that the $p(x)$ is reducible (or not) over $\mathbb{R}[x]$ for two polynomials of degrees 2.
But I am not sure how show that $\left(x-1-\sqrt{2-i\sqrt{3} }\right) \left(x-1+\sqrt{(2+i\sqrt3}\right)$ is (not) polynomial of degree 2.
| COMMENT.-These are the two real roots given by Wolfram. It is impossible that you can calculate them by simple means. The two non-real roots are equally complicated.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3390052",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 1
} |
Typical Olympiad Inequality? If $\sum_i^na_i=n$ with $a_i>0$, then $\sum_{i=1}^n\left(\frac{a_i^3+1}{a_i^2+1}\right)^4\geq n$
Let $\sum_i^na_i=n$, $a_i>0$. Then prove that $$ \sum_{i=1}^n\left(\frac{a_i^3+1}{a_i^2+1}\right)^4\geq n $$
I have tried AM-GM, Cauchy-Schwarz, Rearrangement etc. but nothing seems to work. The fourth power in the LHS really evades me, and I struggle to see what can be done.
My attempts didn’t lead me to any result ... Simply cauchy , where $a_i=x,$ $b_i=1$ to find an inequality involving $\sum x^2$ . I also tried finding an inequality involving $\sum x^3$ using $a_i=\frac{x^3}{2}$ and $b_i=x^{\frac{1}{2}}$
| You need to use another queue:
By Rearrangement, AM-GM and C-S we obtain: $$\sum_{i=1}^n\left(\frac{a_i^3+1}{a_i^2+1}\right)^4\geq \ \sum_{i=1}^n\left(\frac{a_i+1}{2}\right)^4\geq\sum_{i=1}^na_i^2=\frac{1}{n}\sum_{i=1}^n1^2\sum_{i=1}^na_i^2\geq\frac{1}{n}\left(\sum_{i=1}^na_i\right)^2=n. $$
I used the following.
$$\frac{x^3+1}{x^2+1}\geq\frac{x+1}{2}$$ it's
$$2(x^3+1)\geq(x^2+1)(x+1)$$ or
$$x^3+1\geq x^2+x$$ or
$$x^2\cdot x+1\cdot1\geq x^2\cdot1+x\cdot1,$$ which is true by Rearrangement because $(x^2,1)$ and $(x,1)$ have the same ordering.
Also, by AM-GM $$\left(\frac{x+1}{2}\right)^4\geq\left(\sqrt{x}\right)^4=x^2.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3390979",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 2,
"answer_id": 0
} |
Find all $x\in \mathbb R$ such that $\sqrt[3]{x+2}+\sqrt[3]{x+3}+\sqrt[3]{x+4}=0$ The question is as the title says:
Find all $x\in \mathbb R$ such that $\sqrt[3]{x+2}+\sqrt[3]{x+3}+\sqrt[3]{x+4}=0$.
I struggle to even start this question.
By inspection, I see that $x$ must be negative. Playing around yields $x=-3$ as a solution, though I do not know how to prove that there are no other solutions.
Upon differentiation, I obtain:
\begin{align}
\frac{d}{dx} (\sqrt[3]{x+2}+\sqrt[3]{x+3}+\sqrt[3]{x+4})&= \frac{1}{3}(\sqrt[3]{(x+2)^{-2}}+\sqrt[3]{(x+3)^{-2}}+\sqrt[3]{({x+4})^{-2}})\\
\end{align}
which is always positive for all real values of $x$, implying that the function defined as $f(x)=\sqrt[3]{x+2} + \sqrt[3]{x+3} + \sqrt[3]{x+4}$ is strictly increasing.
Is there a better way to solve this equation?
| let $t=x+3$ and work with the 'symmetric' form,
$$\sqrt[3]{t-1}+\sqrt[3]{t+1}=-\sqrt[3]{t}\tag{1}$$
Take the cubic power,
$$2t + 3(t^2-1)^{1/3}(\sqrt[3]{t-1}+\sqrt[3]{t+1})=-t$$
Simplify with (1),
$$(t^2-1)t=t^3$$
which leads to the only solution $t=0$, hence $x=-3$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3394661",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 4
} |
Pairs of perfect squares Two perfect squares are said to be friendly if one is obtained from the other by adding
the digit 1 on the left. For example, $1225 = 35 ^ 2$ and $225 = 15 ^ 2$ are friendly. Prove that there are
in finite pairs of friendly and odd perfect squares.
Maybe Any product of square-full numbers is square-full.
$4n(n+1)+1=4n^{2}+4n+1=(2n+1)^{2}$
I don't know if that would help: http://oeis.org/A060355
| Be $a, b$ belonging to integers, we want to show that for all $a, b$ belonging to integers, there is $ k $ belonging to the natural such that they are infinite.
$a ^ 2 - b ^ 2 = 10 ^ k$
$(a + b) (a-b) = 2^k. 5 ^ k $
$\begin{cases}
a + b = 2 ^ {k-1} \\
a - b = 2.5 ^ k
\end{cases}$
$2a = 2.5 ^ k + 2 ^ {k-1}$
$a = 5 ^ k + 2 ^ {k-1}$
$ 2b = 2 ^ {k-1} - 2.5 ^ k$
$b = 2 ^ {k-2} - 5 ^ k$
Taking $a = 2 ^ {k-2} + 5 ^ k$
$b = 5 ^ k - 2 ^ {k-2}$
Take: $a ^ 2 - b ^ 2 = 2 ^ {2k - 4} + 2.2 ^ {k-2} .5 ^ k + 5 ^ 2k - 5 ^ 2k + 2.2 ^ {k-2} .5 ^ k- 2 ^ {2k-4}$
$a ^ 2-b ^ 2 = 2 ^ k.5 ^ k$
$a ^ 2-b ^ 2 = 10 ^ k$
So we have k belonging to the natural such that $ k \geq 2$. We have infinite numbers
$a = 2 ^ {k-1} + 5 ^ k$
$b = 5 ^ k - 2 ^ {k-2}$ such that $a ^ 2$ and $b ^ 2$ are friendly and a, b are odd
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3397900",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Solve for equations involving floor function
Solve for real $x$: $$\frac{1}{\lfloor x \rfloor} + \frac{1}{\lfloor 2x \rfloor} = x - \lfloor x \rfloor + \frac{1}{3}$$
Hello! I hope everybody is doing well. I was not able to solve the above problem. And this problem becomes even more difficult with the fact that $x $ is not necessarily a positive real.
Here is what I did:
Let $x=p+r, 0 \leq r <1$ Now we divide into cases whether $r $ is less than $0.5 $ or not and I reach here(1st part): $0 \frac{3}{2p} - \frac13 < \frac12$. Now I could have done that $\frac95<2\le [x]\le4<\frac92 $ but since $x$ is not necessarily positive, I guess this is not always true.
Any help would be appreciated. Thanks.
| I would approach the problem in this way.
Let's put
$$
{1 \over {\left\lfloor x \right\rfloor }} + {1 \over {\left\lfloor {2x} \right\rfloor }} = x - \left\lfloor x \right\rfloor + {1 \over 3} = \left\{ x \right\} + {1 \over 3} = t
$$
Since the fractional part of $x$ has values in the range $[0, \, 1)$ we have the the RHS is
$$
0 \le \left\{ x \right\} = t - {1 \over 3} < 1\quad \Rightarrow \quad {1 \over 3} \le t < {4 \over 3}
$$
At the same time, we know that
$$
x - 1 < \left\lfloor x \right\rfloor \le x
$$
thus for the LHS we get this set of inequalities
$$
\left\{ \matrix{
{1 \over 3} \le {1 \over {\left\lfloor x \right\rfloor }} + {1 \over {\left\lfloor {2x} \right\rfloor }} = t < {4 \over 3} \hfill \cr
{1 \over x} + {1 \over {2x}} \le {1 \over {\left\lfloor x \right\rfloor }} + {1 \over {\left\lfloor {2x} \right\rfloor }} < {1 \over {x - 1}} + {1 \over {2x - 1}} \hfill \cr} \right.
$$
Translating this into inequalities for $x$, we shall have that
$$
\left\{ \matrix{
{1 \over 3} < {1 \over {x - 1}} + {1 \over {2x - 1}} \hfill \cr
{1 \over x} + {1 \over {2x}} < {4 \over 3} \hfill \cr} \right.
$$
Solving the quadrics give
$$
\left\{ \matrix{
x \in \left( {1/2,\;3 - \sqrt {22} /2} \right) \cup \left( {1,\;3 + \sqrt {22} /2} \right) \hfill \cr
x \in \left( { - \infty ,\;0} \right) \cup \left( {9/8,\;\infty } \right) \hfill \cr} \right.\quad \Rightarrow \quad x \in \left( {9/8,\;3 + \sqrt {22} /2} \right) \approx \left( {1.1,\;5.34} \right)
$$
So we have just to find the possible solutions for
$$
\left\{ x \right\} = {1 \over {\left\lfloor x \right\rfloor }} + {1 \over {\left\lfloor {2x} \right\rfloor }} - {1 \over 3}
\quad \left| {\;x \in \left[ {1,\;3/2} \right)\,\; \cup \;\left[ {3/2,\;2} \right)\; \cup \; \cdots \,\; \cup \;\left[ {5,\;11/2} \right)} \right.
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3398962",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Show that $\sum_{k=0}^{n+1}\left(\binom{n}{k}-\binom{n}{k-1}\right)^2 = \frac{2}{n+1}\binom{2n}{n}$
Show that $$\displaystyle\sum_{k=0}^{n+1}\left(\dbinom{n}{k}-\dbinom{n}{k-1}\right)^2 = \dfrac{2}{n+1}\dbinom{2n}{n}$$ where $n \in \mathbb{N}$
Consider $\dbinom{n}{r}=0 $ for $r<0 $ and $r>n$. Using $\displaystyle\sum_{k=0}^{n}\dbinom{n}{k}^2 = \dbinom{2n}{n}$, we get $\displaystyle\sum_{k=0}^{n+1}\left(\dbinom{n}{k}-\dbinom{n}{k-1}\right)^2 = \displaystyle\sum_{k=0}^{n+1} \dbinom{n}{k}^2 + \displaystyle\sum_{k=0}^{n+1} \dbinom{n}{k}^2 - 2\displaystyle\sum_{k=0}^{n+1}\dbinom{n}{k}\dbinom{n}{k-1} \\ \hspace{47mm} = 2\dbinom{2n}{n}-2\displaystyle\sum_{k=1}^{n}\dbinom{n}{k}\dbinom{n}{k-1} \\ \hspace{47mm} = 2\dbinom{2n}{n}-2\displaystyle\sum_{k=1}^{n}\dfrac{1}{k(n-k+1)}\dbinom{n}{k-1}^2 \\ \hspace{47mm} = 2\dbinom{2n}{n}-\dfrac{2}{n+1}\displaystyle\sum_{k=1}^{n}\left(\dfrac{1}{k}+\dfrac{1}{n-k+1} \right)\dbinom{n}{k-1}^2 $
Now I am stuck at this point, how do I further progress in this proof? Is there a better way to solve this problem ?
| The given sum is equal to the constant term of $f(z)f(1/z)$, where $$f(z)=\sum_{k=0}^{n+1}\left[\binom{n}{k}-\binom{n}{k-1}\right]z^k=(1-z)(1+z)^n;$$ in other words, it is the coefficient of $z^{n+1}$ in $$z^{n+1}f(z)f(1/z)=-(1-z)^2(1+z)^{2n},$$ equal to $-\binom{2n}{n+1}+2\binom{2n}{n}-\binom{2n}{n-1}$ which reduces to the expected result.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3400677",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 1
} |
Find the distance travelled by an ant in the coordinate plane
An ant is standing on the origin. It begins by walking 1 unit in the positive $x$-direction and then turns $45$ degrees counterclockwise and walks $\dfrac{1}{2}$ units in that direction. The ant then turns another $45$ degrees and walks $\dfrac{1}{3}$ units in that direction. The ant keeps doing this endlessly. How far is the ant's final position from the initial position?
I think I should let the origin be $(0,0)$ and calculate the coordinates of the points that ant reaches each time it turns. I believe the $x$-coordinate at the $n$th step is $\sum_{k=1}^n\dfrac{cos(\dfrac{\pi}{4}(k-1))}{k}$ and the $y$-coordinate at the $n$th step is $\sum_{k=1}^n\dfrac{sin(\dfrac{\pi}{4}(k-1))}{k}$.
| The problem is amenable to some simple complex number analysis (you have to work in radian measure, note that $45^{\circ} = \frac{\pi}{4}$).
Each step can be characterised as a vector in the complex plane. The first step is $\displaystyle 1 = 1e^{i0}$, the second is $\displaystyle \frac 12e^{i\frac{\pi}{4}}$, and the $n$th step is $\displaystyle \frac 1n e^{i\frac{(n-1)\pi}{4}}$. The final position is the infinite sum of all of these complex numbers.
Let $\displaystyle z = e^{i\frac{\pi}{4}}$.
Then define a series sum $\displaystyle S(z) = 1 + \frac 12 z + \frac 13 z^2 + \frac 14 z^3 + ...$
Note that $zS(z) = z + \frac 12 z^2 + \frac 13 z^3 + \frac 14 z^4 + ...$
And now observe: $(zS(z))' = 1 + z + z^2 + z^3 + z^4 + ... = \frac 1{1-z}$, where convergence is assured for complex $|z| < 1$.
By integrating and rearranging, we find, $\displaystyle S(z) = - \frac 1z \ln(1-z)$. (The constant of integration is easily shown to be zero).
Now find $|\displaystyle S(e^{i\frac{\pi}{4}})|$, which is the distance of the final position of the ant from the origin.
$|\displaystyle S(e^{i\frac{\pi}{4}})| = |-e^{-\frac{i\pi}{4}}\ln(1-e^{i\frac{\pi}{4}})|$
Since $\displaystyle |z_1z_2| = |z_1||z_2|$, the above is equal to $\displaystyle |-e^{-\frac{i\pi}{4}}||\ln(1-e^{i\frac{\pi}{4}})| = |\ln(1-e^{i\frac{\pi}{4}})|$
(since $\displaystyle |-e^{-\frac{i\pi}{4}}| = 1$)
To find $\displaystyle |\ln(1-e^{i\frac{\pi}{4}})|$, we first express $\displaystyle 1-e^{i\frac{\pi}{4}}$ in the form $re^{i\theta}$.
$\displaystyle 1-e^{i\frac{\pi}{4}} = 1-\cos \frac{\pi}{4} - i\sin\frac{\pi}{4}$
Now $\displaystyle r = \sqrt{(1-\cos \frac{\pi}{4})^2 + (\sin\frac{\pi}{4})^2} = \sqrt{2 - 2\cos \frac{\pi}{4}} = \sqrt{2-\sqrt 2}$
and $\displaystyle \theta = \arctan{\frac{\sin\frac{\pi}{4}}{1-\cos \frac{\pi}{4}}} = \arctan (\sqrt 2 + 1)$
and $\displaystyle \ln(re^{i\theta}) = \ln r + i\theta$, giving:
$\displaystyle |\ln(1-e^{i\frac{\pi}{4}})| = |\frac 12 \ln{(2-\sqrt 2)} + i\arctan{(\sqrt 2 + 1)}| \\= \sqrt{[\arctan(\sqrt 2 + 1)]^2 + \frac 14[\ln(2-\sqrt 2)]^2} \\= 1.20806...$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3401038",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Prove the given inequality if $a+b+c=1$ Let $a,b,c$ be positive real numbers such that $a+b+c=1$, then prove that
$\frac{a}{a^2 +b^3 +c^3}+\frac{b}{a^3 +b^2 +c^3}+\frac{c}{a^3 +b^3 +c^2} \leq \frac{1}{5abc}$
Please provide some hint to proceed. I have used Arithmetic Mean-Geometric Mean inequality to proceed in such questions, but I am not getting how to use it if it is to be used. Kindly provide some directon.
Usually the questions I have dealt so far were like if $a+b+c=1$, find maximum value of $a^3b^4c^2$ which I find using weighted A.M-G.M, but here I am not able to gather the thoughts to proceed
| Also, by AM-GM
$$a^3+b^3+c^3+a^2b+a^2c\geq a^3+4\sqrt[4]{b^3\cdot c^3\cdot a^2b\cdot a^2c}=a^3+4abc.$$
Thus, it's enough to prove that:
$$\sum_{cyc}\frac{1}{a^2+4bc}\leq\frac{a+b+c}{5abc}$$ or
$$\sum_{cyc}\left(\frac{1}{a^2+4bc}-\frac{1}{4bc}\right)\leq\frac{a+b+c}{5abc}-\frac{a+b+c}{4abc}$$ or
$$\sum_{cyc}\frac{a^2}{bc(a^2+4bc)}\geq\frac{a+b+c}{5abc}$$ or
$$\sum_{cyc}\frac{a^3}{a^2+4bc}\geq\frac{a+b+c}{5},$$ which is true by Holder:
$$\sum_{cyc}\frac{a^3}{a^2+4bc}\geq\frac{(a+b+c)^3}{3\sum\limits_{cyc}(a^2+4ab)}\geq\frac{(a+b+c)^3}{3\sum\limits_{cyc}\frac {5}{3}(a^2+2ab)}=\frac{a+b+c}{5}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3401646",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Fourier series of a function in interval $[-\pi/2,\pi/2]$ I need to show that $$\frac{\cos x}{3}+\frac{\cos 3 x}{1 \cdot 3 \cdot 5}-\frac{\cos 5 x}{3 \cdot 5 \cdot 7}+\frac{\cos 7 x}{5 \cdot 7 \cdot 9}-\cdots=\frac{\pi}{8} \cos ^{2} x$$ on the interval $[-\pi/2,\pi/2]$. I have tried to find the Fourier coefficient $a_n$ with the formula $$a_n=\frac 2{\pi/2}\int_0^{\pi/2} \frac{\pi}{8}\cos^2(x)\cdot \cos \frac{n\pi x}{\pi/2}dx$$, but always end up getting something useless. How do I do it?
| The way to think about this problem is to notice that LHS has period $2\pi$ and satisfies $f(x+\pi)=-f(x)$, while RHS has period $\pi$ so it satisfies $g(x+\pi)=g(x)$. This means that you need to extend $\cos^2(x)$ to the full interval $[-\pi,\pi]$ in such a way that it satisfies the same property as LHS, so you consider:
$h(x)=\cos^2(x)=\frac{1+\cos 2x}{2}, -\frac{\pi}{2} \le x \le \frac{\pi}{2}$,
$h(x)=-\cos^2(x)=-\frac{1+\cos 2x}{2}, -\pi \le x \le-\frac{\pi}{2}, \frac{\pi}{2} \le x \le \pi$
(Since $h$ is obviously continuos being $0$ where the expressions join, you can use $\le$ in both)
Now $h$ is even hence it has a Fourier cosine series and it is obvious that the even cosine part (including the free term) is zero (by periodicity the plus and minus parts cancel out), while the odd part has coefficients - for $k \ge 1$ corresponding to $\cos {(2k-1)x}$:
$\frac{1}{\pi}(\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}\cos^2(x)\cos((2k-1)x)dx-\int_{-\pi}^{-\frac{\pi}{2}}\cos^2(x)\cos((2k-1)x)dx-\int_{\frac{\pi}{2}}^{\pi}\cos^2(x)\cos((2k-1)x)dx)=$
$\frac{4}{\pi}\int_{0}^{\frac{\pi}{2}}\cos^2(x)\cos((2k-1)x)dx=\frac{1}{\pi}((-1)^{k+1}\frac{2}{2k-1}+(-1)^{k}\frac{1}{2k-3}+(-1)^{k}\frac{1}{2k+1})$
and then for $k=1$ one gets $\frac{8}{3\pi}\cos x$, while for $k \ge 2$ we get $(-1)^k\frac{8}{(2k-3)(2k-1)(2k+1)\pi}\cos{(2k+1)x}$ which is the expected value
Putting all together we get that:
$\frac{\cos x}{3}+\frac{\cos 3 x}{1 \cdot 3 \cdot 5}-\frac{\cos 5 x}{3 \cdot 5 \cdot 7}+\frac{\cos 7 x}{5 \cdot 7 \cdot 9}-\cdots=\frac{\pi}{8} \cos ^{2} x, -\frac{\pi}{2} \le x \le \frac{\pi}{2}$
$\frac{\cos x}{3}+\frac{\cos 3 x}{1 \cdot 3 \cdot 5}-\frac{\cos 5 x}{3 \cdot 5 \cdot 7}+\frac{\cos 7 x}{5 \cdot 7 \cdot 9}-\cdots=-\frac{\pi}{8} \cos ^{2} x, -\pi \le x \le -\frac{\pi}{2}, \frac{\pi}{2} \le x \le \pi$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3402797",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Find real parameter $a$, such that the solution of the linear system lies in the second quadrant
For which real parameter $a$ lies the solution of the system of equations
$$\begin{aligned} \frac{x}{a+1} + \frac{y}{a-1} &= \frac{1}{a-1}\\ \frac{x}{a+1} - \frac{y}{a-1} &= \frac{1}{a+1} \end{aligned}$$
in the second quadrant?
I do not how to start to solve this system of equations. Any help?
| Adding and subtracting the equations we obtain
$$\frac{2x}{a+1} = \frac1{a-1}+\frac1{a+1}\implies x=\frac12 \frac{a+1}{a-1}+\frac12=\frac a{a-1}$$
$$\frac{2y}{a-1} = \frac1{a-1}-\frac1{a+1}\implies y = \frac12-\frac12 \frac{a-1}{a+1}=\frac 1{a+1}$$
then we need $y>0$ and $x<0$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3403343",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
Fixed Point Theorems and Contraction Mappings I am trying to solve this following exercise.
Let $F(x)$ be a continuously differentiable function defined on the interval $[a,b]$ such that $F(a) < 0$, $F(b) > 0$, and
\begin{align*}
0 < K_1 \leq F'(x) \leq K_2 \; \; \; (a \leq x \leq b).
\end{align*}
Use Theorem 1 (given below) to find the unique root of the equation $F(x) = 0$.
Hint: Introduce the auxiliary function $f(x) = x - \lambda F(x)$, and choose $\lambda$ such that the theorem works for the equivalent equation $f(x) = x$
(Theorem 1: Every contraction mapping $A$ defined on a complete metric space $R$ has a unique fixed point.)
Here is what I have so far:
Define the auxiliary function $f(x) = x - \lambda F(x)$. We first must show that $f$ is a contraction mapping, meaning that $|f(x) - f(y)| \leq K |x - y|$ where $x, y \in [a,b]$.
Thus, let $x, y \in [a,b]$. We have:
\begin{align*}
|f(x) - f(y)| & = |(x - \lambda F(x)) - (y - \lambda F(y))| & & \text{definition of $f(x)$} \\
& = |(x - y) - \lambda (F(x) - F(y))| & & \text{rearrange}
\end{align*}
Since $F(x)$ is continuous on $[a,b]$ and differentiable on $(a,b)$, we invoke the mean value theorem. Thus, $\exists c \in (x,y)$ such that $F'(c) = \frac{F(x) - F(y)}{x-y}$. (This requires us to impose the restriction that $x \neq y$, if $x = y$, then $f(x) = f(y)$, and $|f(x) - f(y)| = 0$, and the result trivially holds for any choice of $K$.) This implies that $F' (c) (x - y) = F(x) - F(y)$. Using this, we get:
\begin{align*}
& = |(x - y) - \lambda F'(c) (x-y)| \\
& = |(x-y)(1 - \lambda F'(c))| & & \text{take out factor of $(x-y)$} \\
& = |x-y||1 - \lambda F'(c)| & & \text{properties of abs. value.}
\end{align*}
Using our assumption, we have:
$K_1 \leq F'(c) \leq K_2$. For $\lambda \geq 0$, we have
\begin{align*}
\lambda K_1 \leq \lambda F'(c) \leq \lambda K_2 & \iff - \lambda K_1 \geq - \lambda F'(c) \geq - \lambda K_2 \\
& \iff - \lambda K_2 \leq - \lambda F'(c) \leq - \lambda K_1 \\
& \iff 1 - \lambda K_2 \leq 1 - \lambda F'(c) \leq 1 - \lambda K_1
\end{align*}
Now, set $\lambda - \frac{2}{K_2}$. Then, we have:
\begin{align*}
1 - \lambda K_2 = 1 - \frac{2}{K_2} K_2 = 1 - 2 = -1 \\
\end{align*}
Since $K_1 \leq K_2$, $\frac{1}{K_1} \leq \frac{1}{K_2}$, so $\frac{2}{K_2} \geq \frac{2}{K_1}$ and, hence, $- \frac{2}{K_2} \leq \frac{2}{K_1}$. Thus:
\begin{align*}
1 - \lambda K_1 = 1 - \frac{2}{K_2} K_1 \leq 1 - \frac{2}{K_1} K_1 = 1 - 2 = -1.
\end{align*}
Therefore, given this choice of $\lambda$, we have
\begin{align*}
-1 \leq 1 - \lambda F'(c) \leq 1.
\end{align*}
Thus,
\begin{align*}
|1 - \lambda F'(c)| \leq 1.
\end{align*}
Putting all of this together, we have:
\begin{align*}
|f(x) - f(y)| = |x-y||1 - \lambda F'(c)| \leq |x-y| \cdot 1 = |x-y|.
\end{align*}
Therefore, $f$ is a contraction mapping, meaning that, by Theorem 1, it has a unique fixed point. This implies that $\exists z \in [a,b]$ such that $f(z) = z$. By the definition of $f$, this implies that
\begin{align*}
z - \lambda F(z) = z,
\end{align*}
and then that
\begin{align*}
z - \frac{2}{K_2} F(z) = z.
\end{align*}
Subtracting $z$ from both sides gives
\begin{align*}
- \frac{2}{K_2} F(z) = 0.
\end{align*}
Finally, multiplying both sides by $- \frac{K_2}{2}$ gives:
\begin{align*}
F(z) = 0.
\end{align*}
Thus, there exists a unique root, $z$, to the equation $F(z) = 0$.
Thanks.
| Pick $\lambda$ so that $0<\lambda F'<1$. E.g. take $\lambda=\frac1{2K_2}$. Then $0<\frac{K_1}{2K_2}\le\frac {F'}{2K_2}\le\frac{K_2}{2K_2}=\frac12<1$. Using that $f'=1-\lambda F'$ we obtain $0<1-\frac12=\frac12\le f'=1-\frac {F'}{2K_2}\le1-\frac{K_1}{2K_2}<1$. If $\mu=1-\frac{K_1}{2K_2}$ then $0<\frac12\le f'\le\mu<1$ (so now you could invoke the mean value theorem for $f'$).
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3405134",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
$10000=abc$ with $a,b,c$ natural numbers don't have digit $0$. Smallest possible value of $a+b+c$ is? Let $10000=abc$ with $a,b,c$ natural numbers don't have digit $0$. If $a,b,c$ does not have to be all distinct, then the smallest possible value of $a+b+c$ is?
Attempt:
First write as prime factors: $10000 = 2^{4} 5^{4}$. The possible triples are:
$$ 2, 2^{3}, 5^{4} $$
$$ 2^{2}, 2^{2}, 5^{4} $$
$$ 2^{3}, 2, 5^{4} $$
$$ 2^{4}, 2^{4}, 5^{4} $$
$$ 5, 5^{3}, 2^{4}$$
$$ 5^{2}, 5^{2}, 2^{4}$$
$$ 5^{3}, 5, 2^{4}$$
The smallest sum is $5^{2} + 5^{2} + 2^{4}$. Are there better approaches?
| Clearly one of $a,b,c$ is $2^4$
We need to minimize $b+c$ with $bc=5^4$
Now $(b+c)^2-4bc=?\ge0$
So $(b+c)^2\ge4bc=?$ the equality occurs if $b=c$
Fortunately, that condition keeps $b,c$ natural
Else we need to resort to check for possible combinations
$$5^k,5^{4-k},0\le k\le4-k$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3405223",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 2
} |
Sum of floor function of general term While doing some solution to get a general term for some series I encountered at the end with this.
$S(n,i) = \left(\dfrac{n^2+n-2i}{n} - \lfloor\dfrac{n^2+n-2i}{2n}\rfloor \right) \left(\lfloor\dfrac{n^2+n-2i}{2n}\rfloor +1 \right)$
, where $i> \dfrac{n}{2}$. How can I get the general term from this summation by geting rid of the floor function.
| Assuming $n$ is even and $i<\frac{3}{2}n$,
$$\lfloor\frac{n^2+n-2i}{2n}\rfloor = \lfloor\frac{n^2-(n+2i)}{2n}\rfloor = \frac{n}{2}-\lfloor\frac{n+2i}{2n}\rfloor = \frac{n}{2}-1$$
So
$$S(n,i) = (\frac{n^2+n-2i}{n}-\frac{n}{2}+1)*(\frac{n}{2}) = (\frac{n}{2}-2\frac{i}{n}+2)*(\frac{n}{2})$$
So
$$S(n,i) = (\frac{n}{2})^2+n-i$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3405387",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
How to find x from the equation $\sqrt{x + \sqrt{x}} - \sqrt{x - \sqrt{x}} = m \sqrt{\frac{x}{x+\sqrt{x}}}$ For m are real number,
Find x from the equation
$$\sqrt{x + \sqrt{x}} - \sqrt{x - \sqrt{x}} = m \sqrt{\frac{x}{x+\sqrt{x}}}$$
I tried to multiply $\sqrt{x + \sqrt{x}}$ to the both sides and I get
$$x + \sqrt{x} - \sqrt{x^{2} - x} = m \sqrt{x}$$
What should I do now to get x? Can anyone show me a hint please?
| Squaring both sides we get $$x+\sqrt{x}+x-\sqrt{x}-2\sqrt{x^2-x}=\frac{m^2x}{x+\sqrt{x}}$$
Simplifying we get
$$2x-\frac{m^2x}{x+\sqrt{x}}=2\sqrt{x^2-x}$$
simplifying we get
$$m^4x^2=-4x(x^2+x+2x\sqrt{x})$$ diviniding by $$x\neq 0$$ $$4x+4+m^4=-8\sqrt{x}$$ squaring again
$$16x^2-32x+8xm^4+8m^4+m^8+16=0$$
Can you solve this equation?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3405511",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 1
} |
How to prove $\lim_{n \to \infty} \left(\frac{3n-2}{3n+1}\right)^{2n} = \frac{1}{e^2}$?
It's known that $\lim_{n \to \infty} \left(1 + \frac{x}{n} \right)^n = e^x$.
Using the above statement, prove $\lim_{n \to \infty} \left(\frac{3n-2}{3n+1}\right)^{2n} = \frac{1}{e^2}$.
My attempt
Obviously, we want to reach a statement such as $$\lim_{n \to \infty} \left(1 + \frac{-2}{n}\right)^n \quad \text{ or } \quad \lim_{n \to \infty} \left(1 + \frac{-1}{n}\right)^n \cdot \lim_{n \to \infty} \left(1 + \frac{-1}{n}\right)^n$$
in order to be able to apply the above condition. However, I was unable to achieve this. The furthest I've got was the following:
\begin{align}
\left(\frac{n-2}{3n+1}\right)^{2n} &= \left( \frac{9n^2 - 12n + 4}{9n^2 + 6n + 1} \right)^n\\
&= \left(1 + \frac{-18n+3}{9n^2+6n+1}\right)^n\\
f(n)&= \left(1 + \frac{-2 + \frac{3}{n}}{n+\frac{2}{3} + \frac{1}{9n}} \right)^n
\end{align}
It seems quite obvious that $\lim_{n \to \infty} \left(f(n)\right) = \left(1 + \frac{-2}{n + \frac{2}{3}}\right)^n$, however, this is not exactly equal to the statement given above. Are you able to ignore the constant and apply the condition regardness? If so, why? How would you go about solving this problem?
| $$L=\lim_{n \rightarrow} \left(\frac{3n-2}{3n+1} \right)^{2n}= \lim_{n \rightarrow \infty} \frac{\left([1-2/(3n)]^{3n/2}\right)^{4/3}}{\left([1+1/(3n)]^{3n}\right)^{2/3}}=\frac{ e^{-4/3}}{e^{2/3}}=e^{-2}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3405914",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 4
} |
Proving a much stronger version of AM-GM for three variables Here, @MichaelRozenberg stated the following inequality without proof:
Theorem. For all non-negative $a,b,c\in\mathbb R$,
$$a^6+b^6+c^6-3a^2b^2c^2\geq16(a-b)^2(a-c)^2(b-c)^2.$$
In my answer below I give a complete brute-force proof. However, more elegant proofs are welcome.
| I meant the following reasoning.
Let $a=\min\{a,b,c\},$ $b=a+u,$ $c=a+v$ and $u^2+v^2=2kuv.$
Thus, by AM-GM $k\geq1$ and
$$\sum_{cyc}(a^6-a^2b^2c^2)=\frac{1}{2}(a^2+b^2+c^2)\sum_{cyc}(a-b)^2(a+b)^2\geq$$
$$\geq\frac{1}{2}(u^2+v^2)(u^4+v^4+(u^2-v^2)^2)=(u^2+v^2)(u^4-u^2v^2+v^4).$$
Id est, it's enough to prove that:
$$(u^2+v^2)(u^4-u^2v^2+v^4)\geq16(u-v)^2u^2v^2$$ or
$$2k(4k^2-3)\geq16(2k-2)$$ or $$4k^3-19k+16\geq0,$$ which is true by AM-GM:
$$4k^3-19k+16\geq3\sqrt[3]{4k^3\cdot8^2}-19k=\left(12\sqrt[3]{4}-19\right)k>0.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3407096",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 0
} |
Complex Numbers Identity Let $a,b$ be non-zero complex numbers with:
$$a2^{|a|}+b2^{|b|} = (a+b)2^{|a+b|}$$
Show that $a^6 = b^6$.
I gave it a go, but to no avail.
| First note that $a+b=0$ always gives a solution for which $a^6=b^6$.
If $a$ and $b$ are real, we can assume without loss of generality that $a=1$. We get $2+b2^{|b|}=(1+b)2^{1+|b|}$, so $2^{1-|b|}+b=2(1+b)$, so $2^{1-|b|}-b=2$. By piecewise analysis, we find $b=0$ or $b=-1$. But $a$ and $b$ are required to be non-zero, so we have $a+b=0$.
From now on, assume $a+b\neq0$. By rotation, we may assume without loss of generality that $a+b$ is positive and real. Then $a2^{|a|}+b2^{|b|}=(a+b)2^{|a+b|}$ is also positive and real. Since $a$ and $b$ are not both real, neither is real. It follows that $2^{|a|}=2^{|b|}$, so $|a|=|b|$.
Write $a=re^{i\theta}$ and $b=re^{-i\theta}$. We get $a+b=2r\cos(\theta)>0$. So the equality becomes $$2r2^r\cos(\theta)=2r\cos(\theta)2^{2r\cos(\theta)},$$ so $r=2r\cos(\theta)$. This gives $\theta=\frac{\pi}3$, so $a^6=b^6$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3410350",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
If $x$ and $y$ are real number such that $x^2+2xy-y^2=6$, then find the minimum value of $(x^2+y^2)^2$ If $x$ and $y$ are real number such that $x^2+2xy-y^2=6$, then find the maximum value of $(x^2+y^2)^2$
My attempt is as follows:
$$(x-y)^2\ge 0$$
$$x^2+y^2\ge 2xy$$
$$2(x^2+y^2)\ge x^2+y^2+2xy$$
\begin{equation}
2(x^2+y^2)\ge (x+y)^2\tag{1}
\end{equation}
Solving the given equation:
$$x^2+2xy-y^2=6$$
$$(x+y)^2=2y^2+6$$
So putting the value of $(x+y)^2$ in equation $1$
$$2(x^2+y^2)\ge 2y^2+6$$
$$x^2+y^2\ge y^2+3$$
\begin{equation}
x^2\ge 3\tag{2}
\end{equation}
So $x\in \left(-\infty,-\sqrt{3}\right) \cup \left(\sqrt{3},\infty\right)$
But how to proceed from here?
|
Minimize $z=(x^2+y^2)^2$ s.t. $x^2+2xy-y^2=6$.
The contour curve is a circle with the center at the origin: $x^2+y^2=\sqrt z$.
To minimize $z$ implies to find the smallest radius of the circle. The smallest circle must touch the hyperbola. The slope of tangent to the hyperbola at $(x_0,y_0)$ is:
$$2x+2y+2xy'-2yy'=0 \Rightarrow y'=\frac{x_0+y_0}{y_0-x_0}$$
The normal line to the hyperbola at $(x_0,y_0)$ must pass through the origin:
$$y=\frac{x_0-y_0}{x_0+y_0}x$$
The intersection point of the hyperbola and the normal line is $(x_0,y_0)$:
$$\begin{cases}x_0^2+2x_0y_0-y_0^2=6 \\ y_0=\frac{x_0-y_0}{x_0+y_0}\cdot x_0\end{cases} \Rightarrow \begin{cases}x_0^2+2x_0y_0-y_0^2=6 \\ x_0^2-2x_0y_0-y_0^2=0\end{cases} \stackrel{+}\Rightarrow x_0=\sqrt{3+y_0^2} \Rightarrow \\
3+y_0^2+2\sqrt{3+y_0^2}\cdot y_0-y_0^2=6 \Rightarrow y_0^4+3y_0^2-\frac94=0 \Rightarrow \\
y_0^2=\frac{-3+\sqrt{18}}{2},x_0^2=\frac{3+\sqrt{18}}{2} \Rightarrow \\
z(x_0,y_0)=(x_0^2+y_0^2)^2=18 \ \text{(min)}$$
Desmos graph for reference.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3411531",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 4
} |
Express the polynomial in the form p(x) = (x+1) Q(x) +R where (x+1) is the divisor, Q(x) is the quotient and R is the remainder Express the polynomial in the form p(x) = (x+1) Q(x) +R where (x+1) is the divisor, Q(x) is the quotient and R is the remainder,
Hey I would just like to know how to solve this as the question had me confused I don’t know if I am suppose to do long division on write it out like in the question, thanks
| Any textbook would tell you to do polynomial division. Personally, I think polynomial division is too opaque of a technique to use before you know how to solve a question like this without it. So here is an example using what basically amounts to long division, but much less mysterious.
Take $p(x) = x^3+2x^2-3x+5$. We want to write this of the form $(x+1)Q(x) + R$ for some polynomial $Q$, and some polynomial $R$ with smaller degree than $x+1$ (which is to say, $R$ is a constant). We do this by building $Q$ term by term, reducing the degree of the remainder until we have our constant $R$.
The fact that the highest degree term of $p$ is $x^3$ means that the highest degree term of $Q$ must be $x^2$. So we add $(x+1)x^2$, and simultaneously subtract $x^3+x^2$ from the remainder. This doesn't change the value, but it lets us start on building our $Q$:
$$
p(x) = (x+1)x^2 - (x^3+x^2) + x^3+2x^2-3x+5\\
= (x+1)x^2 + x^2-3x+5
$$
We have now successfully gone from a third degree remainder (all of $p$) to a second degree remainder. We are one step closer to a constant remainder. Next step: The highest degree term of the remainder is $x^2$, so the next term to add to $Q$ must be $x$:
$$
p(x) = (x+1)x^2 - (x^2 + x) + (x+1)x + x^2-3x+5\\
= (x+1)(x^2+x) -4x+5
$$
And finally, in the last step, we see that the leading term of the current remainder is $-4x$, so the final term to add to $Q$ must be $-4$:
$$
p(x) = (x+1)(x^2+x) + (x+1)(-4) - (-4x-4) -4x+5\\
= (x+1)(x^2+x-4) + 9
$$
And there we have it: $Q(x) = x^2+x-4$ and $R = 9$. If you do the long division, you might recognize how each step in the long division corresponds to something I did here, and vice versa, but long division is something of a mystery if one doesn't know that this is the process that lies behind it.
Also, I have been extra verbose here. If I were to give this problem on a test, I wouldn't necessarily expect more from an answer than, say,
$$
p(x) = x^3+2x^2-3x+5\\
= (x+1)x^2 + x^2-3x+5\\
= (x+1)(x^2+x) -4x+5\\
= (x+1)(x^2+x-4) + 9
$$
All-in-all more transparent and easier to follow than polynomial long division.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3413435",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
$\sin 2x = 1/2, 2x = 150, 390$. But why not 30? $0<x<360$
So $\sin(2x) = \sin (30 + 360)$ or $\sin(2x) = \sin (180-30)$.
It derives from
\begin{align*}
\sin x - \cos x & =\frac{1}{\sqrt2}\\
\sin^2 x + \cos^2 x -2\sin x \cos x & = \frac{1}{2}\\
2\sin x \cos x & = \frac{1}{2}
\end{align*}
But why not $\sin (2x) = \sin (30)$ ? So $x = 15$.
The answer only $x = 75$ and $195$.
| We have that
$$\sin 2x = \frac12=\sin 30° \iff 2x=30°+k360° \quad \lor \quad2x=180°-30°+k360°$$
therefore all the solutions are in the form
*
*$x=15°+k180°=15°,15°\pm 180°,15°\pm360°,\ldots$
*$x=75°+k180°=75°,75°\pm 180°,75°\pm360°,\ldots$
and for the range $0°\le x <360°$ the solutions are $15°$, $75°$, $195°$, $255°$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3417822",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Can you simplify an Euler formula expression when you are only interested in the real part? I was trying to find a way to simplify $cos(x) + cos(x+a)$ using Euler's formula:
f(x) = $ e^{ix} + e^{i(x+a)} $
= $cos(x) + cos(x+a) +i(sin(x) +sin(x+a))$
= $ 2 cos(x+ \frac{a}{2} )cos(\frac{a}{2}) + i(2sin(x+\frac{a}{2})cos(\frac{a}{2})) $ {sum to product identities)
=$ 2cos(\frac{a}{2})(cos(x+ \frac{a}{2} )+isin(x+\frac{a}{2}))$
=$2cos(\frac{a}{2})e^{i(x+\frac{a}{2})}$ {Important step 1}
=$2cos(\frac{a}{2})e^{ix}e^{i\frac{a}{2}}$ {Important step 2}
At Important step 1 Re(f(x)) is $2cos(\frac{a}{2})cos{(x+\frac{a}{2})}$ which is a valid conclusion however at Important Step 2 Re(f(x)) is $2cos(\frac{a}{2})cos(x)cos({\frac{a}{2}})$ which is false (not equal to $cos(x) + cos(x+a)$). Why is it false when $e^{i(x+\frac{a}{2})}$ = $e^{ix}e^{i\frac{a}{2}}$ ?
| For the step 2 : $f(x) = 2cos(\frac{a}{2})e^{ix}e^{i\frac{a}{2}}$ is correct. However, the real part of this is not $2cos(\frac{a}{2})cos(x)cos({\frac{a}{2}})$ (did you forget the $i^2$ term?).
Let's compute the expression $z := e^{ix}e^{i\frac{a}{2}}$,
we have
$$\begin{eqnarray}z & = & \Big(cos(x) + i\ sin(x)\Big) \cdot \Big(cos(\frac{a}{2}) + i\ sin(\frac{a}{2})\Big) \\
\phantom{z} & = & \big(cos(x) cos(\frac{a}{2}) - sin(x) sin(\frac{a}{2}) \Big) + i \Big(
cos(x) sin(\frac{a}{2}) + sin(x) cos(\frac{a}{2}) \Big)
\end{eqnarray}
$$
Hence, $\operatorname{Re}(f(x)) = 2cos(\frac{a}{2})\operatorname{Re}(z) = 2cos(\frac{a}{2}) \Big(cos(x) cos(\frac{a}{2}) - sin(x) sin(\frac{a}{2})\Big)$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3420740",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Is this formula correct, series and Bernoulli numbers According to this page:
https://proofwiki.org/wiki/Definition:Bernoulli_Numbers
The series is :
$\dfrac{x}{e^x-1}=1-\dfrac{x}{2}+\dfrac{B_1x^2}{2!}+\dfrac{B_2x^4}{4!}+\dfrac{B_3x^6}{6!}...$
where $B_1=-\dfrac{1}{2}, B_2=\dfrac{1}{6}, B_3=0, B_4=-\dfrac{1}{30}$
How can I expand obtain the Bernoulli coefficient as indicated in the website?
Using the series above, I have:
$\dfrac{x}{e^x-1}=1-\dfrac{x}{2}+\dfrac{(-\frac{1}{2})x^2}{2!}+\dfrac{\frac{1}{6}x^4}{4!}+\dfrac{(0)x^6}{6!}...$
$=\dfrac{x}{e^x-1}=1-\dfrac{x}{2}+\dfrac{x^2}{4}+\dfrac{x^4}{144}+0...$
This looks wrong since the series expansion according to Wolfram is:
$\dfrac{x}{e^x-1}=1-\dfrac{x}{2}+\dfrac{x^2}{12}-\dfrac{x^4}{720}...$
But then the series of Wolfram doesn't show the Bernoulli number.
I am confused, can you help me here?
| Note, the generating function for the Bernoulli numbers is defined as
\begin{align*}
\frac{x}{e^x-1}=\sum_{n=0}^\infty\frac{B_nx^n}{n!}
\end{align*}
with $\frac{B_\color{blue}{n}}{n!}$ the coefficient of $x^\color{blue}{n}$.
Since we have $B_0=1, B_1=-\frac{1}{2}, B_2=\frac{1}{6}, B_3=0, B_4=-\frac{1}{30},\ldots$ we obtain
\begin{align*}
\color{blue}{\frac{x}{e^x-1}}&=\sum_{n=0}^\infty\frac{B_nx^n}{n!}\\
&=B_0+\frac{B_1x}{1!}+\frac{B_2x^2}{2!}+\frac{B_3x^3}{3!}+\frac{B_4x^4}{4!}+\cdots\\
&=1+\frac{\left(-\frac{1}{2}\right)x}{1}+\frac{\left(\frac{1}{6}\right)x^2}{2}+\frac{0x^3}{6}+\frac{\left(-\frac{1}{30}\right)x^4}{24}+\cdots\\
&\,\,\color{blue}{=1-\frac{x}{2}+\frac{x^2}{12}-\frac{x^4}{720}+\cdots}
\end{align*}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3420868",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Finding the complete Taylor expansion of $\frac{1}{1+z^2}$ around $z=0$. For an exercise, I need to find the complete Taylor expansion for $(1+z^2)^{-1}$ around $z=0$. I have tried decomposing first $(1+z^2)^{-1}$ into partial fractions. Since $1+z^2=0$ gives $z=\pm i$, the partial fractions are:
$$\frac{1}{1+z^2} = \frac{1}{(z+i)(z-i)} = \frac{i}{2(z+i)} - \frac{i}{2(z-i)} = \frac{i}{2} \Big{(}\frac{1}{z+i} - \frac{1}{z-i}\Big{)}$$
Therefore, my idea was to find the much easier Taylor expansion of both fractions, and add them together. I have worked out each Taylor expansion, since the $n$-th derivative of each of the fractions can be easily found:
$$\frac{1}{z+i}=\sum_{n=0}^\infty -i^{n+1}z^n, \hspace{25px} \frac{1}{z-i}=\sum_{n=0}^\infty \frac{-z^n}{i^{n+1}}$$
Typing each sum in WolframAlpha returns the original fraction, so I think they are okay. Now, I replace both sums into the partial fractions:
$$\frac{1}{1+z^2} = \frac{i}{2} \Big{(}\sum_{n=0}^\infty -i^{n+1}z^n - \sum_{n=0}^\infty \frac{-z^n}{i^{n+1}}\Big{)}$$
But I always get a sum that does not return the correct Taylor expansion for $1/(1+z^2)$. Where am I getting this wrong?
| Since $\frac{1}{1+z^2} = \frac{1}{1-(-z^2)}$, the Taylor expansion around $z = 0$ (i.e., the Maclaurin series) would be the sum of an infinite Geometric series, i.e.,
$$\frac{1}{1+z^2} = \sum_{n=0}^{\infty}\left(-z^2\right)^n \tag{1}\label{eq1A}$$
with this being convergent for $|z^2| \lt 1$.
Regarding your work, since $\frac{1}{z+i} = \frac{i}{iz - 1} = \frac{-i}{1 - iz} = -i\left(\frac{1}{1-(iz)}\right)$, you get
$$\frac{1}{z+i} = -i\sum_{n=0}^{\infty}(iz)^n = \sum_{n=0}^{\infty}-i^{n+1}z^n
\tag{2}\label{eq2A}$$
This matches what you got. Next, $\frac{1}{z-i} = \frac{i}{iz + 1} = \frac{i}{1 - (-iz)} = i\left(\frac{1}{1-(-iz)}\right)$, you get
$$\frac{1}{z-i} = i\sum_{n=0}^{\infty}(-iz)^n = \sum_{n=0}^{\infty}(-1)^n i^{n+1}z^n \tag{3}\label{eq3A}$$
With your terms, note
$$\begin{equation}\begin{aligned}
\frac{-1}{i^{n+1}} & = \frac{-i^{n+1}}{i^{2n+2}} \\
& = \frac{-i^{n+1}}{(i^{2})^{n+1}} \\
& = \frac{-i^{n+1}}{(-1)^{n+1}} \\
& = \frac{-(-1)^{n+1}i^{n+1}}{\left((-1)^{n+1}\right)^2} \\
& = -(-1)(-1)^{n}i^{n+1} \\
& = (-1)^{n}i^{n+1}
\end{aligned}\end{equation}\tag{4}\label{eq4A}$$
As you can see, it matches what I got, so there's nothing wrong there. However, I believe my version is easier to deal with. With the difference of \eqref{eq2A} and \eqref{eq3A}, note the even terms in \eqref{eq3A} have $i^{n+1}z^n$, so this doubles the term in \eqref{eq2A}, while the odd terms are the same and, thus, cancel out. In summary, you then get
$$\begin{equation}\begin{aligned}
\frac{1}{1+z^2} & = \frac{i}{2}\left(\sum_{n=0}^{\infty}-i^{n+1}z^n - \sum_{n=0}^{\infty}(-1)^n i^{n+1}z^n\right) \\
& = \frac{i}{2}\left(\sum_{n=0}^{\infty}(-1 - (-1)^n)i^{n+1}z^n\right) \\
& = \frac{i^2}{2}\left(\sum_{n=0}^{\infty}(-1 - (-1)^n)i^{n}z^n\right) \\
& = \frac{-1}{2}\left(\sum_{n=0}^{\infty}(-2)i^{2n}z^{2n}\right) \\
& = \sum_{n=0}^{\infty}i^{2n}z^{2n} \\
& = \sum_{n=0}^{\infty}(i^2z^2)^n \\
& = \sum_{n=0}^{\infty}(-z^2)^n
\end{aligned}\end{equation}\tag{5}\label{eq5A}$$
As you can see, this matches \eqref{eq1A}. As such, what you did was correct & I don't know why you think you weren't getting the correct result. If it was just due to the expression being different, as I show it simplifies to the same thing.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3424898",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Compute Taylor Series $\ln(x^2-x+1)$ Please share if you have a faster approach, but I’ll share what I have so far. I think it reduces to some combinatorial problem, I’d like to see a combinatorial argument if possible.
From computing for small $n$ in wolfram alpha, I think the answer is
\begin{align*}
\ln(x^2-x+1)&=\sum_{n=1}^\infty (-1)^n \frac{a_n}{n} x^n \\
&= -\frac{1}{1}x^1+\frac{1}{2}x^2-\frac{2}{3}x^3+\frac{1}{4}x^4-\frac{1}{5}x^5+\dots
\end{align*}
where $a_n=2$ if $n=3,9,27,81,\dots$ and $a_n=1$ otherwise.
We know
$$
\ln(1+x)=x-\frac{x^2}{2}+\frac{x^3}{3}-\frac{x^4}{4}+\dots
$$
So
$$
\ln(1+(x^2-x))=\frac{x}{1}(x-1)-\frac{x^2}{2}(x
-1)^2+\frac{x^3}{3}(x-1)^3-\dots$$
Now we compute the coefficient of $x^n$.
\begin{align*}
x^8 \to -\frac{1}{4}\binom{4}{4}+\frac{1}{5}\binom{5}{3}-\frac{1}{6}\binom{6}{2}+\frac{1}{7}\binom{7}{1}-\frac{1}{8}\binom{8}{8}=\frac{1}{8}\cdot8! \\
x^9 \to \frac{1}{5}\binom{5}{4}-\frac{1}{6}+\frac{1}{7}\binom{7}{2}-\frac{1}{8}\binom{8}{1}+\frac{1}{9}\binom{9}{0}=-\frac{2}{9}\cdot 9! \\
x^{10} \to \frac{1}{5}\binom{5}{5}-\frac{1}{6}\binom{6}{4}+\frac{1}{7}\binom{7}{3}-\frac{1}{8}\binom{8}{2}+\frac{1}{9}\binom{9}{1}-\frac{1}{10}\binom{10}{0}=\frac{1}{10} \cdot 10!
\end{align*}
In other words, I want this identity:
$$
\sum_{r+k=n} (-1)^{r+1} \frac{1}{r} \binom{r}{k} = (-1)^n \frac{a_n}{n} n!
$$
with $a_n$ defined as above. Any help to prove the identity would be appreciated. Any other faster approaches would also be appreciated. Thanks!
Also, this LaTeX was a pain to type. Is there a way to reduce it? (Like \frac to \f, \binom to \b)
| Tips:
$$\ln(1-x+x^2)=\ln(1+x^3)-\ln(1+x)$$
Answer:
$$\ln(1+x)=x-\dfrac{x^2}{2}+\dfrac{x^3}{3}-\dfrac{x^4}{4}+\cdots \\ \ln(1+x^3)=x^3-\dfrac{x^6}{2}+\dfrac{x^9}{3}-\dfrac{x^{12}}{4}+\cdots \\ \ln(1-x+x^2)=\ln(1+x^3)-\ln(1+x)=\sum_{k=1}^\infty a_kx^k \\ a_k=\begin{cases} \dfrac{2(-1)^{k+1}}{k} & \text{when }3|k \\ \dfrac{(-1)^k}{k} & \text{otherwise}\end{cases}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3425722",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Checking a newtonian binomial expansion I am trying to check that the following expansion of $\sqrt{1-x}$ using Newton's Binomial Theorem "appears correct by squaring both sides"
$$(1-x)^{1/2}=1-\frac{x}{2}-\frac{x^2}{8}-\frac{x^3}{16}-\frac{5x^4}{128}+...$$
then to do the check, I got the following:
$$(\sqrt{1-x})^2=(1-\frac{x}{2}-\frac{x^2}{8}-\frac{x^3}{16}-\frac{5x^4}{128}+...)^2$$
$$\Rightarrow 1=1+\frac{25x^8+80x^7+224x^6+896x^5-16384x}{16384}$$
| Your last line is wrong. Expanding the first few terms on the right-hand side:
$$\left(1-\frac{x}{2}-\frac{x^2}{8}-\frac{x^3}{16}-\ldots\right)^2=1-2\cdot\frac{x}{2}+\frac{x^2}{4}-2\cdot\frac{x^2}{8}+\frac{x^4}{64}+2\cdot\frac{x^3}{16}-2\cdot\frac{x^3}{16}+\ldots$$
Everything cancels apart from $1-2\cdot\frac{x}{2}=1-x$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3427819",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Can you go from the integral of x squared to the summation of x squared? To be specific, I was wondering if you could go from the indefinite integral of $x^2$, mainly
$\int$ $x^2$$dx = \frac{x^3}{3}$, to the summation of $x^2$, $\sum_{i=0}^n i^2 = \frac{(n^2+n)(2n+1)}{6}$?
| At first glance, $\frac{x^3}{3}$ and $\frac{x(x+1)(2x+1)}{6}$ seem like very different functions. Finding the sum of $n^2$ can be represented as a piecewise function: $f(x) = 0$ when $-1 ≤ x < 0$, $f(x) = 1$ when $0 ≤ x < 1$, $f(x) = 4$ when $1 ≤ x < 2$, $f(x) = 9$ when $2 ≤ x < 3$ and so on.
However, if we integrate a function that passes through the midpoints of each endpoint, such as $(-0.5, 0)$, $(0.5, 1)$, $(1.5, 4)$, the extreme values at either side will cancel out and produce a more accurate value. This is similar to using a midpoint rule to approximate an integral, only here we are using an integral to approximate a summation.
Therefore we have $f(x) = (x+0.5)^2$ and the area under it is:
$$\int (x+0.5)^2 \ \mathrm{d} x = \frac{(x+0.5)^3}{3} = \frac{1}{3}x^3 + \frac{1}{2}x^2+\frac{1}{4}x + \frac{1}{24}$$
which only differs by a linear term to:
$$\frac{x(x+1)(2x+1)}{6} = \frac{1}{3}x^3+\frac{1}{2}x^2+\frac{1}{6}x$$
The error comes from the fact that around $x = 0.5, 1.5, 2.5 \cdots$, $f(x)$ is not quite an odd function. Because the derivative of $f(x)$ is increasing, the right-hand side has slightly more area above the piecewise function compared to the left-hand side.
Now your task is to try to correct for the error.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3429184",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
} |
HCF $(x,y) = 16$ and LCM $(x,y) = 48000$. Then the possible number of pairs $(x,y)$ Let $S$ be the set of all ordered pairs $(x,y)$ of positive integers, with HCF $(x,y) = 16$ and LCM $(x,y) = 48000$. The
number of elements in $S$ is
My Attempt : $48000= 2^7. 3 . 5^3$ As the L.c.m contains $2^7$ as a factor and G.c.d contains $2^4$ as a factor , we can assure that one element contains $2^7$ and the other one contains $2^4$. And $3$ and $5^3$ will be divided between two numbers in a manner so that the L.C.M and G.C.D remain as given.
So the possible pairs are $(2^7 .3 . 5^3 , 2^4)$ , $(2^7 .3 , 5^3.2^4)$ , $(2^7 , 3.5^3.2^4)$ , $(2^7.5^3 , 3.2^4)$.
So I think the number of elements in the set $S$ is $4$. Have I gone wrong anywhere? Can anyone please help me ?
| If $$\gcd(x,y)=16=2^4$$ and $$\operatorname{lcm}(x,y)=48000=2^7\cdot3\cdot5^3$$ then $$xy=2^{11}\cdot3\cdot5^3$$
So we can let
$$x = 2^\alpha\cdot3^\beta\cdot5^\gamma \qquad \text{and} \qquad
y = 2^{11-\alpha}\cdot3^{1-\beta}\cdot5^{3-\gamma}$$
We need
$$\begin{align}
\min(\alpha, 11-\alpha) &= 4 \\
\max(\alpha, 11-\alpha) &= 7 \\
\min(\beta, 1-\beta) &= 0 \\
\max(\beta, 1-\beta) &= 1 \\
\min(\gamma, 3-\gamma) &= 0 \\
\max(\gamma, 3-\gamma) &= 3
\end{align}$$
So $$\begin{align}
\alpha &\in \{4,7\} \\
\beta &\in \{0,1\} \\
\gamma &\in \{0,3\}
\end{align}$$
So there are $2\cdot2\cdot2 = 8$ possible $(x,y)$ pairs.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3436574",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
} |
Help with algebra, rearrange equations I have:
$$
r=\frac{1-x^2-y^2}{(1-x)^2+y^2} \tag 1
$$
And want to write $(1)$ as:
$$
\Big (x-\frac{r}{1+r}\Big )^2+y^2=\Big (\frac{1}{1+r}\Big )^2 \tag 2
$$
First method, starting from $(1)$:
$$
r=\frac{1-x^2-y^2}{x^2-2x+1+y^2}\iff
$$
$$
r(x^2-2x+1+y^2)=1-x^2-y^2 \iff
$$
$$
rx^2-r2x+r+ry^2=1-x^2-y^2 \iff
$$
$$
rx^2+x^2-r2x+r+ry^2+y^2=1 \iff
$$
$$
x^2(r+1)-r2x+r+y^2(r+1)=1 \iff
$$
$$
x^2-x\frac{2r}{r+1}+\frac{r}{r+1}+y^2=\frac{1}{r+1} \tag 3
$$
Completing the square of $x^2-x\frac{2r}{r+1}$:
$$
x^2-x\frac{2r}{r+1}+\Big (\frac{r}{1+r} \Big )^2-\Big (\frac{r}{1+r} \Big )^2=\Big (x-\frac{r}{1+r}\Big )^2-\Big (\frac{r}{1+r} \Big )^2
$$
Inserting in $(3)$ gives:
$$
\Big (x-\frac{r}{1+r}\Big )^2-\Big (\frac{r}{1+r} \Big )^2+\frac{r}{r+1}+y^2=\frac{1}{r+1} \tag 4
$$
I'm stuck here, what is next?
Second method, starting from $(2)$:
$$
\Big (x-\frac{r}{1+r}\Big )^2+y^2=\Big (\frac{1}{1+r}\Big )^2 \iff
$$
$$
\Big (\frac{x(r+1)-r}{1+r}\Big )^2+y^2=\Big (\frac{1}{1+r}\Big )^2 \iff
$$
$$
(x(r+1)-r)^2+y^2(1+r)^2=1 \tag 5
$$
Expand $(x(r+1)-r)^2$:
$$
(x(r+1)-r)^2=x^2(1+r)^2-x2r(1+r)+r^2
$$
Inserting in $(5)$ gives:
$$
x^2(1+r)^2-x2r(1+r)+r^2+y^2(1+r)^2=1 \iff
$$
$$
x^2(1+2r+r^2)-2rx-2xr^2+r^2+y^2(1+2r+r^2)=1\iff
$$
$$
x^2+2rx^2+r^2x^2-2rx-2xr^2+r^2+y^2+2ry^2+r^2y^2=1\iff
$$
Collect $r^2$ and $r$:
$$
r^2(x^2-2x+1+y^2)+r(2x^2-2x+2y^2) = 1-x^2-y^2 \iff
$$
$$
r^2((x-1)^2+y^2)+r(2x(x-1)+2y^2) = 1-x^2-y^2 \tag 6
$$
I'm stuck here, I don't know how to go from $(6)$ to $(1)$.
| Use the componendo and dividendo rules, i.e.
$\frac ab = \frac cd \implies \frac{b-a}{b+a} = \frac{d-c}{d+c}$
$$r=\frac{1-x^2-y^2}{(1-x)^2+y^2}\implies
\frac{1-r}{1+r} =-x+\frac{y^2}{x-1}$$
Rearrange
$$y^2 = (x-1)\left(x+\frac{1-r}{1+r}\right)=
\Big (x-\frac{r}{1+r}\Big )^2-\Big (\frac{1}{1+r}\Big )^2
$$
where $u^2-v^2=(u+v)(u-v)$ is used in the last step.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3436993",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 3
} |
Prove that $4\tan^{-1}\left(\frac{1}{5}\right) - \tan^{-1}\left(\frac{1}{239}\right)= \frac{\pi}{4}$
Prove that $4\tan^{-1} \left(\dfrac{1}{5}\right) - \tan^{-1}\left(\dfrac{1}{239}\right)=\dfrac{\pi}{4}.$
I was wondering if there was a shorter solution than the method below?
Below is my attempt using what I would call the standard approach to these kinds of problems.
The expression on the left hand side is equivalent to $$\tan^{-1}\left[\tan \left(4\tan^{-1}\left(\dfrac{1}{5}\right)\right)-\tan^{-1}\left(\dfrac{1}{239}\right)\right]\\
=\tan^{-1}\left(\dfrac{\tan(4\tan^{-1}(\frac{1}{5}))-\frac{1}{239}}{1+\frac{1}{239}\tan(4\tan^{-1}(\frac{1}{5}))}\right)\tag{1}.$$
We have that $$\tan\left(4\tan^{-1}\left(\frac{1}{5}\right)\right)=\dfrac{2\tan(2\tan^{-1}(\frac{1}{5}))}{1-\tan^2(2\tan^{-1}(\frac{1}{5})}\tag{2}$$
and that
$$\tan\left(2\tan^{-1}\left(\frac{1}{5}\right)\right)=\dfrac{2\cdot \frac{1}{5}}{1-(\frac{1}{5})^2}=\dfrac{5}{12}\tag{3}.$$
Plugging in the result of $(3)$ into $(2)$ gives
$$\tan\left(4\tan^{-1}\left(\frac{1}{5}\right)\right) = \dfrac{2\cdot \frac{5}{12}}{1-(\frac{5}{12})^2}=\dfrac{120}{119}\tag{4}.$$
Pluggin in the result of $(4)$ into $(1)$ gives that the original expression is equivalent to
$$\tan^{-1}\left(\dfrac{\frac{120}{119}-\frac{1}{239}}{1+\frac{1}{239}\cdot\frac{120}{119}}\right)=\tan^{-1}\left(\dfrac{\frac{119\cdot 239 + 239-119}{239\cdot 119}}{\frac{119\cdot 239+120}{119\cdot 239}}\right)=\tan^{-1}(1)=\dfrac\pi4,$$
as desired.
| Shortest proof:
$$(5+i)^4(239-i)=114244+114244i.$$
Taking the arguments,
$$4\arctan \frac15-\arctan\frac1{239}=\frac\pi4.$$
Note that the computation avoids the fractions and immediately generalizes to other Machin-like formulas (https://en.wikipedia.org/wiki/Machin-like_formula#More_terms).
To perform the computation by hand, consider
$$(5+i)^2=24+10i\propto12+5i,$$
$$(12+5i)^2=119+120i,$$
$$(119+120 i)(239-i)=(119\cdot239+120)+(120\cdot239-119)i\propto 1+i.$$
(After simplification by $119\cdot239$, we have $120=239-119$.)
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3438621",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "18",
"answer_count": 4,
"answer_id": 2
} |
prove that $5^{2n+1} - 3^{2n+1} - 2^{2n+1}$ is divisible by 30 for all integers n ≥ 0. Prove that $5^{2n+1} - 3^{2n+1} - 2^{2n+1}$ is divisible by 30 for all integers n ≥ 0.
I have tried induction as follows.
Step 1:
Try n = 0, we get: $5 - 3 - 2 = 0$, which is divisible by 30.
Try n = 1, we get: $5^{3} - 3^{3} - 2^{3} = 90$, which is also divisible by 30.
Step 2:
Assume it is true for n = k.
So we are assuming the following equality is true: $5^{2k+1} - 3^{2k+1} - 2^{2k+1} = 30M$, for some integer M.
Step 3:
Now we look at the next case: n = k + 1.
$5^{2(k+1)+1} - 3^{2(k+1)+1} - 2^{2(k+1)+1}$
= $5^{2k+3} - 3^{2k+3} - 2^{2k+3}$
= $25\times5^{2k+1} - 9\times3^{2k+1} - 4\times2^{2k+1}$
= $21\times5^{2k+1} + 4\times5^{2k+1} - 5\times3^{2k+1} - 4\times3^{2k+1} - 4\times2^{2k+1}$
= $21\times5^{2k+1} - 5\times3^{2k+1} + 4\times[5^{2k+1} - 3^{2k+1} - 2^{2k+1}]$
= $21\times5^{2k+1} - 5\times3^{2k+1} + 4\times30M$. (Assumed in step 2)
The last term is divisible by 30. But I cannot get a factor 30 out of the first two terms. I can show divisibility by 15 as follows:
= $7\times3\times5\times5^{2k} - 5\times3\times3^{2k} + 4\times30M$
= $7\times15\times5^{2k} - 15\times3^{2k} + 4\times15\times2M$
But how do I show divisibility by 30?
| You have shown divisibility by 15.
To show divisibility by 30,
just note that the expression is even
(odd-odd+even).
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3441310",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 6,
"answer_id": 0
} |
Limit of $\frac{e^{-xy}}{1+x^2+y^2}$ in $x,y ≥ 0$ I am trying to figure out if $f(x,y)=\frac{e^{-xy}}{1+x^2+y^2}$ goes to zero, in the area x,y ≥ 0, when $x^2+y^2$ goes to infinity.
So I rewrote the function as $\frac{1}{e^{xy}}\frac{1}{1+x^2+y^2}$.
The function $\frac{1}{1+x^2+y^2} $ definitley goes to zero. But what about
$\frac{1}{e^{xy}}$? Since $x,y ≥ 0$ it looks like to me that this function should also go to zero.
So if $x$ or $y$ is equal to zero then we have $\frac{1}{e^{0}}=1$
So in this case I would think you could argue that $f$ goes to $1\cdot0=0$ when $x^2+y^2$ goes to infinity.
When $x,y > 0$ I tried changing to polar coordinates then we have
$$0 <\frac{1}{e^{xy}}=\frac{1}{e^{r^2cos(t)sin(t)}}=\frac{1}{e^{r^2\frac{1}{2}sin(2t)}}$$
Since $- \frac{1}{2} ≤ \frac{1}{2}sin(2t) ≤ \frac{1}{2}$
I tried to use this to (maybe) see that $\frac{1}{e^{r^2\frac{1}{2}sin(2t)}}$ is a function that goes to zero. But with $\frac{1}{e^{r^2\frac{1}{2}sin(2t)}} ≤ \frac{1}{e^{-\frac{r^2}{2}}}$ we cant come to any conclusion since $\frac{1}{e^{-\frac{r^2}{2}}}$ goes to infinity when $r$ goes to infinity.
Could I prove that the limit is zero in some other way? If not, how is it possible that $\frac{1}{e^{xy}}$ $x,y > 0$ does not approach zero when $x^2+y^2$ goes to infinity?
| If $x=0,$ then $\frac{1}{1+y^2} \rightarrow 0$ as $y\rightarrow \infty$. If $x$ is finite, then both $e^{-xy}$ and $\frac{1}{1+x^2+y^2}$ go to zero, as $y\rightarrow \infty$, so their product goes to zero. If both $x$ and $y$ approach $\infty$, first let $y$ go to $\infty$ then take the limit as $x \rightarrow \infty$.
In all cases, $f(x,y) \rightarrow 0$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3443860",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Show that a function is negative over its domain I would like to demonstrate that the following function is negative
\begin{equation}
f(x)=-\frac{t}{4\sqrt{x}^3}\bigg[1-\bigg(1+\sqrt{x}\bigg)^{\frac{1}{t-1}}\bigg]+\bigg(\frac{1}{1-t}\bigg)\bigg(\frac{t}{2\sqrt{x}}\bigg)^2\bigg(1+\sqrt{x}\bigg)^{\frac{2-t}{t-1}}
\end{equation}
for $x>0$ and $1>t>0$.
I have graphed this in Desmos and confirmed that the function is negative given my restrictions on $x$ and $t$.
Any help to solve this analytically would be most appreciated!
Many thanks.
| Here is an attempt to answer my question:
The function $f(x)$ has the same sign as the function $g(x)$, where
\begin{equation*}
g(x)=\bigg(1+\sqrt{x}\bigg)^{\frac{1}{t-1}}\bigg[1+\frac{t}{1-t}\frac{\sqrt{x}}{(1+\sqrt{x})}\bigg]-1,
\end{equation*}
and so the problem is equivalent to showing that $g(x)<0$.
Note that $\lim_{x\rightarrow0}g(x)=0$.
Moreover, $g(x)$ is strictly decreasing.
The first-order condition is
\begin{multline*}
g'(x)=\frac{1}{t-1}\bigg(1+\sqrt{x}\bigg)^{\frac{1}{t-1}-1}\bigg(\frac{1}{2\sqrt{x}}\bigg)\\
+\frac{t}{1-t} \Bigg[\frac{\frac{1}{2\sqrt{x}}(1+\sqrt{x})-\sqrt{x}\big(\frac{1}{2\sqrt{x}}\big)}{(1+\sqrt{x})^2}\bigg(1+\sqrt{x}\bigg)^{\frac{1}{t-1}}\\
+\frac{1}{t-1}\bigg(1+\sqrt{x}\bigg)^{\frac{1}{t-1}-1}\bigg(\frac{1}{2\sqrt{x}}\bigg)\frac{\sqrt{x}}{(1+\sqrt{x})}\Bigg],
\end{multline*}
\begin{multline*}
=\frac{1}{2\sqrt{x}(t-1)}\bigg(1+\sqrt{x}\bigg)^{\frac{1}{t-1}-1}\\
+\frac{t}{1-t}\Bigg[\frac{1}{2\sqrt{x}(1+\sqrt{x})^2}\bigg(1+\sqrt{x}\bigg)^{\frac{1}{t-1}}\\
+\frac{\sqrt{x}}{2\sqrt{x}(t-1)(1+\sqrt{x})}\bigg(1+\sqrt{x}\bigg)^{\frac{1}{t-1}-1}\Bigg].
\end{multline*}
Simplifying further yields
\begin{equation*}
g'(x)=-\frac{1}{2\sqrt{x}(1-t)}\bigg(1+\sqrt{x}\bigg)^{\frac{1}{t-1}-1}+\bigg[\frac{t}{2\sqrt{x}(1-t)}-\frac{t\sqrt{x}}{2\sqrt{x}(1-t)^2}\bigg]\bigg(1+\sqrt{x}\bigg)^{\frac{1}{t-1}-2}.
\end{equation*}
Then, using
\begin{equation*}
\frac{1}{2\sqrt{x}(1-t)}=\frac{(1-t)(1+\sqrt{x})}{2\sqrt{x}(1-t)^2(1+\sqrt{x})} \ \text{ and } \ \frac{t}{2\sqrt{x}(1-t)}=\frac{t(1-t)}{2\sqrt{x}(1-t)^2},
\end{equation*}
implies that
\begin{align*}
g'(x)&=-\Bigg[\frac{(1-t)(1+\sqrt{x})}{2\sqrt{x}(1-t)^2(1+\sqrt{x})}\bigg(1+\sqrt{x}\bigg)^{\frac{1}{t-1}-1} -\bigg[\frac{t(1-t)}{2\sqrt{x}(1-t)^2}-\frac{t\sqrt{x}}{2\sqrt{x}(1-t)^2}\bigg]\bigg(1+\sqrt{x}\bigg)^{\frac{1}{t-1}-2}\Bigg],\\
&=-\bigg(1+\sqrt{x}\bigg)^{\frac{1}{t-1}-2}\bigg[\frac{1-t+\sqrt{x}-t\sqrt{x}-t+t^2+t\sqrt{x}}{2\sqrt{x}(1-t)^2}\bigg],\\
&=-\bigg(1+\sqrt{x}\bigg)^{\frac{1}{t-1}-2}\bigg[\frac{1-2t+\sqrt{x}+t^2}{2\sqrt{x}(1-t)^2}\bigg],\\
&=-\frac{(1+\sqrt{x})^{\frac{1}{t-1}}}{(1+\sqrt{x})^2}\bigg[\frac{(t-1)^2+\sqrt{x}}{2\sqrt{x}(1-t)^2}\bigg]<0.
\end{align*}
Therefore, $g(x)$ is a strictly decreasing function starting from a limit value of $0$,
which implies the original function is negative for $x>0$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3444158",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Riemann sum of $\int_1^2 {1\over x^2} dx$. I've spent quite a time solving the following problem:
Evaluate using Riemann's sum:
$$
I = \int_1^2{1\over x^2} dx
$$
I was first trying the following approach, which didn't work since the summation seems undoable to me:
$$
\Delta x = {1\over n}\\
I = \lim_{n\to\infty}\sum_{k=1}^nf\left(1+{k\over n}\right)\Delta x \\
= \lim_{n\to\infty}\sum_{k=1}^n{n^2\over (k+n)^2} {1\over n} \\
= \lim_{n\to\infty}\sum_{k=1}^n{n\over (k+n)^2}
$$
Wolfram evaluates this sum in terms of digamma function which is too advanced.
Several hours has passed before I decided to reconsider the point to choose in each partition. Let:
$$
\Delta x = {1\over n}\\
x_k = 1 + {k\over n}\\
\begin{align}
I &= \lim_{n\to\infty}\sum_{k=1}^nf\left(\sqrt{x_k x_{k-1}}\right)\Delta x \\
&= \lim_{n\to\infty}\sum_{k=1}^n{1 \over x_k x_{k-1}}\Delta x \\
&= \lim_{n\to\infty}\sum_{k=1}^n{1 \over \left(1+{k\over n}\right)\left(1+{k-1\over n}\right)}\Delta x \\
&= \lim_{n\to\infty}\sum_{k=1}^n{n^2 \over (n+k)(n+k-1)}{1\over n}\\
&=\lim_{n\to\infty}\sum_{k=1}^n{n \over (n+k)(n+k-1)} \\
&=\lim_{n\to\infty}\sum_{k=1}^n\left({n \over (n+k-1)} - {n \over (n+k)}\right)\\
&= {n\over n} - {n\over 2n}\\
&= \boxed{{1\over 2}}
\end{align}
$$
This sum telescopes nicely. Now I'm wondering whether the first approach is even doable. I've met some other questions but the first one lists a hint I don't really understand and the second one is closed as a duplicate.
What would be the way to finish the initial approach? In the first approach, the problem is actually reduced to finding the limit which I couldn't handle. Also is there some intuition in choosing the "right" points in the partitions?
| There's no easy closed form for
$$\sum_{k = 1}^{n} \frac{1}{(n+k)^2}\,,$$
but since we're interested in a limit we can achieve our goal by approximating the terms of the sum in such a way that the approximation has an easy closed form. A very good approximation is obtained by
\begin{align}
\sum_{k = 1}^{n} \frac{1}{(n+k)^2 - \frac{1}{4}} &= \sum_{k = 1}^{n} \frac{1}{\bigl(n+k - \frac{1}{2}\bigr)\bigl(n + k + \frac{1}{2}\bigr)} \\
&= \sum_{k = 1}^{n} \biggl(\frac{1}{n+k - \frac{1}{2}} - \frac{1}{n + k + \frac{1}{2}}\biggr) \\
&= \frac{1}{n + \frac{1}{2}} - \frac{1}{2n + \frac{1}{2}}
\end{align}
from which
$$\lim_{n \to \infty} \sum_{k = 1}^{n} \frac{n}{(n+k)^2 - \frac{1}{4}} = \frac{1}{2}$$
is easily read off.
It remains to verify that the error introduced by approximating the terms doesn't influence the result. One can argue that this is also a Riemann sum for the integral (choose the points $\xi_k = \frac{1}{n} \sqrt{(n+k)^2 - \frac{1}{4}}$ to evaluate the function at), but a direct estimate is more transparent:
$$0 < \frac{1}{(n+k)^2 - \frac{1}{4}} - \frac{1}{(n+k)^2} = \frac{1}{(n+k)^2\bigl(4(n+k)^2-1\bigr)} < \frac{1}{4n^4}\,,$$
so the total difference is
$$0 < n\sum_{k = 1}^n \biggl(\frac{1}{(n+k)^2 - \frac{1}{4}} - \frac{1}{(n+k)^2}\biggr) < n\cdot n\cdot \frac{1}{4n^4} = \frac{1}{4n^2}$$
and
$$\lim_{n \to \infty} \sum_{k = 1}^{n} \frac{n}{(n+k)^2} = \frac{1}{2}$$
is proved.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3445371",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
} |
Line $11x+3y-48 = 0$ tangent a graph $f(x) = \frac{4x + 3}{3x - 6}$ at $(a,b)$ Line $11x+3y-48 = 0$ tangent a graph $f(x) = \frac{4x + 3}{3x - 6}$ at (a,b) when $a<b$
$a-b = ...$
Find gradient of the line, df(x) / dx
$\frac{(4x+3)3 - (3x-6)(4)} {(3x-6)^2}$
Which the same as $ -11/3$
$(12x+9 - 12x + 24 ) 3= -11(3x-6)^2$
If i solve x, literally i solve $a $ right?
Is there less complicated way to solve it?
| The equation $$-\frac{11}{3}x+16=\frac{4x+3}{3x-6}$$ must have only on e solution.
Compute the discriminant! It is $x=3$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3447965",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Is there a better way to solve this equation? I came across this equation:
$x + \dfrac{3x}{\sqrt{x^2 - 9}} = \dfrac{35}{4}$
Wolfram Alpha found 2 roots: $x=5$ and $x=\dfrac{15}{4}$, which "coincidentally" add up to $\dfrac{35}{4}$. So I'm thinking there should be a better way to solve it than the naïve way of bringing the fractions together and then squaring. Is there any?
| Like Solve the equation $x^2+\frac{9x^2}{(x+3)^2}=27$
Clearly $x>0$
Let $\dfrac{3x}{\sqrt{x^2-9}}=y$ so that $x+y=\dfrac{35}4$
$$\implies9x^2=x^2y^2-9y^2\iff\dfrac19=\dfrac1{x^2}+\dfrac1{y^2}=\dfrac{(x+y)^2-2xy}{(xy)^2}$$
Use $x+y=\dfrac{35}4$ and $xy>0$ for find $xy=\dfrac{75}4$
So, $x,y$ are the roots of $$t^2-\dfrac{35}4t+\dfrac{75}4=0$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3449031",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 3
} |
Solve complex equation $\left(\frac{8}{z^3}\right) - i = 0$ from $$(a^3 + b^3) = (a+b)(a^2-ab+b^2)$$
I have $${(2/z)}^3 + i^3 =0$$
I have $$\left(\frac{2}{z} + i\right)\left(\left(\frac{2}{z}\right)^2-(2/z)(i)-1)\right) = 0$$
i.e.$ \left(\frac{2}{z}\right)+i = 0 $ or $ \left(\left(\frac{2}{z}\right)^2-(\frac{2}{z})(i)-1\right) = 0$
......
but I'm not sure that is the correct answer;
help me please.
Thank you.
| $\dfrac8{z^3}=i\implies z^3=\dfrac8i=-8i$.
Since $2^3=8$ and $i^3=-i$, it is apparent that $z=2i, 2i\omega$, or $2i\omega^2$,
where $\omega$ is a primitive cube root of $1$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3450012",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Find $\underset{x\rightarrow 0}\lim{\frac{1-\cos{x}\sqrt{\cos{2x}}}{x\sin{x}}}$
Find
$$\underset{x\rightarrow 0}\lim{\frac{1-\cos{x}\sqrt{\cos{2x}}}{x\sin{x}}}$$
My work.$$\underset{x\rightarrow 0}\lim\frac{1}{x\sin{x}}=\frac{\underset{x\rightarrow0}\lim{\;\frac{\sin{x}}{x}}}{\underset{x\rightarrow 0}\lim{\;x\sin{x}}}=\underset{x\rightarrow 0}\lim\frac{1}{x^2}$$
$$\underset{x\rightarrow 0}\lim{\frac{\cos{x}}{x\sin{x}}}=\underset{x\rightarrow 0}\lim{\frac{\sin{2x}}{2x\sin^2{x}}}=\underset{x\rightarrow 0}\lim{\frac{\sin{2x}}{2x}}\cdot\underset{x\rightarrow 0}\lim{\frac{1}{\sin^2{x}}}=\frac{1}{x^2}$$
$$\underset{x\rightarrow 0}\lim{\sqrt{\cos{2x}}}=\underset{x\rightarrow 0}\lim{\sqrt{1-2\sin^2{x}}}=\underset{x\rightarrow 0}\lim{\sqrt{1-2x^2}}$$
L' Hopital's rule:
$\underset{x\rightarrow 0}\lim{\frac{1-\cos{x}\sqrt{\cos{2x}}}{x\sin{x}}}=\underset{x\rightarrow 0}\lim{\frac{1-\sqrt{1-2x^2}}
{x^2}}=\underset{x\rightarrow 0}\lim{\frac{-4x}{x^3\sqrt{1-2x^2}}}$
What should I do next?
| By standard limits we have
$$\frac{1-\cos{x}\sqrt{\cos{2x}}}{x\sin{x}}=\frac{1-\cos x +\cos x-\cos{x}\sqrt{\cos{2x}}}{x\sin{x}}=$$
$$=\frac{1-\cos x }{x\sin{x}}+\cos x\frac{1-\sqrt{\cos{2x}}}{x\sin{x}}=$$
$$=\frac{1-\cos x }{x^2}\frac{x }{\sin{x}}+\cos x\frac{1-\sqrt{\cos{2x}}}{x\sin{x}}\frac{1+\sqrt{\cos{2x}}}{1+\sqrt{\cos{2x}}} \to \frac 32$$
indeed
$$\frac{1-\cos x }{x^2}\frac{x }{\sin{x}} \to \frac12$$
and
$$\cos x\frac{1-\sqrt{\cos{2x}}}{x\sin{x}}\frac{1+\sqrt{\cos{2x}}}{1+\sqrt{\cos{2x}}}=\frac{\cos x}{1+\sqrt{\cos{2x}}}\frac{1-\cos{2x}}{x\sin{x}}=$$
$$=\frac{\cos x}{1+\sqrt{\cos{2x}}}\frac{1-\cos{2x}}{4x^2}\frac{4x}{\sin x}\to \frac12\cdot \frac12\cdot 4=1$$
Refer also to the related
*
*How to prove that $\lim\limits_{x\to0}\frac{\sin x}x=1$?
*Limits of trig functions
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3450548",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 3
} |
$ 27 + ( 2 + \frac{a^{2}}{bc} ) (2 + \frac{b^{2}}{ac}) (2 + \frac{c^{2}}{ab}) \ge 6 (a+b+c)(\frac{1}{a} + \frac{1}{b} + \frac{1}{c}) $ Let $a,b,c$ be positive numbers. Prove that
$$ 27 + ( 2 + \frac{a^{2}}{bc} ) (2 + \frac{b^{2}}{ac}) (2 + \frac{c^{2}}{ab}) \ge 6 (a+b+c)(\frac{1}{a} + \frac{1}{b} + \frac{1}{c}) $$
Attempt:
The inequality is equivalent to
$$ 36 + 4 \left( \frac{ab}{c^{2}} + \frac{ac}{b^{2}} + \frac{bc}{a^{2}} \right) + 2 \left( \frac{c^{2}}{ab} + \frac{b^{2}}{ac} + \frac{a^{2}}{bc} \right) $$
$$ \ge 6 \left( 3 + \frac{a}{b} + \frac{b}{c} + \frac{c}{a} + \frac{b}{a} + \frac{c}{b} + \frac{a}{c} \right) $$
I have tried using $HM-GM$ but still does not get the result. Rewriting the above
$$ \underbrace{1+1+...+1}_{18} + \underbrace{ \frac{ab}{c^{2}} + \frac{ac}{b^{2}} + \frac{bc}{a^{2}} + ... + \frac{ab}{c^{2}} + \frac{ac}{b^{2}} + \frac{bc}{a^{2}} }_{4} + \left( \frac{c^{2}}{ab} + \frac{b^{2}}{ac} + \frac{a^{2}}{bc} + \frac{c^{2}}{ab} + \frac{b^{2}}{ac} + \frac{a^{2}}{bc} \right) $$
$$ \ge \underbrace{\left( \frac{a}{b} + \frac{b}{c} + \frac{c}{a} + \frac{b}{a} + \frac{c}{b} + \frac{a}{c} \right) + ... + \left( \frac{a}{b} + \frac{b}{c} + \frac{c}{a} + \frac{b}{a} + \frac{c}{b} + \frac{a}{c} \right) }_{6} $$
The left hand side and right hand side each has $36$ terms.
| A full expanding gives:
$$\sum_{cyc}(a^3b^3+2a^4bc-3a^3b^2c-3a^3c^2b+3a^2b^2c^2)\geq0$$ or
$$\sum_{cyc}(a^3b^3-a^3b^2c-a^3c^2b+a^2b^2c^2)+abc\sum_{cyc}(a^3-a^2b-a^2c+abc)\geq0,$$
which true by Schur twice.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3452662",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Prove $\frac{r_1}{bc}+\frac{r_2}{ac}+\frac{r_3}{ab}=\frac1r-\frac1{2R}$ Prove $\dfrac{r_1}{bc}+\dfrac{r_2}{ac}+\dfrac{r_3}{ab}=\dfrac{1}{r}-\dfrac{1}{2R}$ where $a , b, c$ are sides of a triangle; $r $is the inradius; $R$ is the circumradius; the $r_i$ are the exradii; $s$ is semiperimeter; and $\triangle$ is area
Solving L.H.S
$$\dfrac{1}{abc}\left(ar_1+br_2+cr_3\right)$$
$$\dfrac{\triangle}{abc}\left(\dfrac{a}{s-a}+\dfrac{b}{s-b}+\dfrac{c}{s-c}\right)$$
$$\dfrac{\triangle}{abc}\left(\dfrac{a-s+s}{s-a}+\dfrac{b-s+s}{s-b}+\dfrac{c-s+s}{s-c}\right)$$
$$\dfrac{\triangle}{abc}\left(-3+s\left(\dfrac{1}{s-a}+\dfrac{1}{s-b}+\dfrac{1}{s-c}\right)\right)$$
$$\dfrac{\triangle}{abc}\left(-3+s\left(\dfrac{2s-a-b}{(s-a)(s-b)}+\dfrac{1}{s-c}\right)\right)$$
$$\dfrac{\triangle}{abc}\left(-3+s\left(\dfrac{c}{(s-a)(s-b)}+\dfrac{1}{s-c}\right)\right)$$
$$\dfrac{\triangle}{abc}\left(-3+s\left(\dfrac{c(s-c)+(s-a)(s-b)}{(s-a)(s-b)(s-c)}\right)\right)$$
$$\dfrac{\triangle}{abc}\left(-3+s\left(\dfrac{cs-c^2+s^2-bs-as+ab}{(s-a)(s-b)(s-c)}\right)\right)$$
$$\dfrac{\triangle}{abc}\left(-3+s\left(\dfrac{(s-c)(s+c)-s(a+b-c)+ab}{(s-a)(s-b)(s-c)}\right)\right)$$
$$\dfrac{\triangle}{abc}\left(-3+s\left(\dfrac{(s-c)(s+c)-2s(s-c)+ab}{(s-a)(s-b)(s-c)}\right)\right)$$
$$\dfrac{\triangle}{abc}\left(-3+s\left(\dfrac{(s-c)(c-s)+ab}{(s-a)(s-b)(s-c)}\right)\right)$$
$$\dfrac{\triangle}{abc}\left(-3+s\left(\dfrac{2cs-s^2-c^2+ab}{(s-a)(s-b)(s-c)}\right)\right)$$
$$\dfrac{\triangle}{abc}\left(-3+s\left(\dfrac{c(a+b+c)-s^2-c^2+ab}{(s-a)(s-b)(s-c)}\right)\right)$$
$$\dfrac{\triangle}{abc}\left(-3+s\left(\dfrac{ab+bc+ca-s^2}{(s-a)(s-b)(s-c)}\right)\right)$$
Now how to proceed from here, I just want the way to continue from here. Please don't share other ways, I particularly want to know the way to continue from the last step as in multiple questions I get to such points and don't find the way to proceed from there.
| As $r_1=s\tan\dfrac A2$
$T(1,a)=\dfrac{r_1}{bc}=\dfrac{2Rs\sin A\tan\dfrac A2}{abc}=\dfrac{2Rs(1-\cos A)}{abc}=\dfrac{2R\triangle(1-\cos A)}{r\cdot4R\triangle}=\dfrac{1-\cos A}{2r}$
using $\triangle=rs=\dfrac{abc}{4R}$
$T(1,a)+T(2,b)+T(3,c) =\dfrac{3-(\cos A+\cos B+\cos C)}r$
Now use Prove trigonometry identity for $\cos A+\cos B+\cos C$
and from here, $r=4R\sin\dfrac A2\sin\dfrac B2\sin\dfrac C2\implies\cos A+\cos B+\cos C=1+\dfrac r{4R}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3453807",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
$k \in \mathbb Z^{+}$. Prove that for all $n ≥ 2k^2$, $n! ≥ k^{n}$
Proposition
$k \in \mathbb Z^{+}$. Then for all $n ≥ 2k^2$, $n! ≥ k^{n}$
My attempt:
Lemma 1. $k \in \mathbb Z^{+}$. For all $n \in \mathbb N$, $(k^2 + n)! ≥ k^{2n}$
Proof: By induction.
Base case: $n = 0$.
$(k^2 + 0)! ≥ 1 = k^{2\cdot0} = k^{2n}$
Induction step:
Suppose
$$(k^2 + n)! ≥ k^{2n}$$
Then
$$\begin{align}(k^2 + n + 1)! = (k^2 + n)! \cdot (k^2 + n + 1) & ≥ k^{2n} \cdot (k^2 + n + 1) \\
& ≥ k^{2n} \cdot k \\
& = k^{2n+1}
\end{align}$$
Back to the proposition:
By induction.
Base case: $n = 2k^{2}$.
Applying lemma $1$, we conclude that $(2k^{2})! ≥ k^{2k^{2}}$
Induction step:
Suppose
$$n! ≥ k^{n}$$
Then
$\begin{align}(n+1)! = n! \cdot (n+1) & ≥ k^{n} \cdot (n+1)\\
& ≥ k^{n} \cdot k \space \space \space (\text{because } n > k)\\
& = k^{n+1}\\
\end{align}$
$\Box$
Is it correct?
| You have made a mistake at the end of the proof of the lemma
\begin{eqnarray*}
(k^2 + n + 1)! = (k^2 + n)! \cdot (k^2 + n + 1) & ≥ k^{2n} \cdot (k^2 + n + 1) \\
& ≥ k^{2n} \cdot \color{red}{k^2} \\
& = k^{2n+\color{red}{2}}.
\end{eqnarray*}
& in the base case of the proposition, you could be clearer by saying ... Using $n=k^2$ in the lemma gives $(2k^{2})! \geq k^{2k^{2}} $.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3454454",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Limit of power series, 2 variables In an assignment, I've run into the following problem.
$$\lim_{x\to\infty} \sum ^\infty_{n=3} \frac{n\cdot x^{n-2}\cdot(-1)^n\cdot(n-1)}{(2n)!}$$
I really hope someone can steer me the right way, thanks!
| We have that
$$\cos x=\sum ^\infty_{n=0} \frac{(-1)^n\cdot x^{2n}}{(2n)!}\implies \cos \sqrt x=\sum ^\infty_{n=0} \frac{(-1)^n\cdot x^{n}}{(2n)!}$$
$$-\frac{\sin \sqrt x}{2\sqrt x}=\sum ^\infty_{n=1} \frac{(-1)^n\cdot n x^{n-1}}{(2n)!}$$
$$\frac{\sin \sqrt x-\sqrt x \cos \sqrt x}{4\sqrt {x^3}}=\sum ^\infty_{n=2} \frac{(-1)^n\cdot n (n-1)x^{n-2}}{(2n)!}$$
therefore
$$\sum ^\infty_{n=3} \frac{(-1)^n\cdot n (n-1)x^{n-2}}{(2n)!}=\frac{\sin \sqrt x-\sqrt x \cos \sqrt x}{4\sqrt {x^3}}-\frac1{12}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3455139",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
What are the last two digits of 1^5 + 2^5 + 3^5 + ... +99^5? What are the last two digits of 1^5 + 2^5 + 3^5 + ... +99^5?
My work:
1^5 ends with 1.
2^5 ends with 2.
3^5 ends with 3.
And so on.
Do I simply add the ending digits to get my answer?
| Write the sum as
$$S=(1^5+99^5)+(2^5+98^5)+(3^5+97^5)+.....+ (49^5+51^5)+50^n$$ Then
$100=1+99,=2+98= 3+97,...$ is the common factor of all the baraketed terms. Because $(a^5+b^5)=(a+b)M(a,b)$. Next, $50^5$ is also divisible by 100. So the last two digits in $S$ are $00$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3455707",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 1
} |
Find derivative $\frac{dy}{dx}$, given $y(x)=\sin^{-1}\left(\frac{5\sin x+4\cos x}{\sqrt{41}}\right)$
Find $\dfrac{dy}{dx}$ if $y=\sin^{-1}\bigg[\dfrac{5\sin x+4\cos x}{\sqrt{41}}\bigg]$
My Attempt
Put $\cos\theta=5/\sqrt{41}\implies\sin\theta=4/\sqrt{41}$
$$
y=\sin^{-1}\big[\sin(x+\theta)\big]\implies\sin y=\sin(x+\theta)\\
y=n\pi+(-1)^n(x+\theta)\\
\boxed{\frac{dy}{dx}=(-1)^n}
$$
But my reference gives the solution $y'=1$, am I missing something here ?
| Rewriting,
$$\sin y=\dfrac{5\sin x+4\cos x}{\sqrt{41}}=\sin(x+\theta)\tag 1$$
with $\theta=\tan^{-1}\frac4{5}$ and recognize $y\in[-\frac\pi2,\frac\pi2]$, hence $\cos y \ge 0$,
$$\cos y= \text{sgn}[\cos(x+\theta) ]\cos(x+\theta) $$
Take the derivative of (1),
$$\> y' = \dfrac{\cos(x+\theta)}{\cos y}
= \dfrac{\cos(x+\theta)}{\text{sgn}[\cos(x+\theta) ]\cos(x+\theta)} \\
=\text{sgn}[\cos(x+\theta) ]=
\begin{cases}
1, & x+\theta -2k\pi \in \left(-\frac{\pi}2,\frac{\pi}2\right]\\
-1, & x+\theta -2k\pi\in \left(\frac{\pi}2,\frac{3\pi}2\right]
\end{cases}
$$
which, as seen in the graph below, alternates at either -1 or 1 depending on the value of $x$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3456149",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
} |
How to solve $2x^2-2\lfloor x \rfloor-1=0$ How do I solve $2x^2-2\lfloor x \rfloor-1=0$?
I have tried setting $x=\lfloor x \rfloor + \{x\}$, where $\{x\}$ is the fractional part of $x$. Then, I tried $$2x^2-2\lfloor x \rfloor-1=0$$ $$2(\lfloor x \rfloor + \{x\})^2-2\lfloor x \rfloor-1=0$$ $$2\lfloor x \rfloor^2 + 4\lfloor x \rfloor\{x\}+2\{x\}^2-2\lfloor x \rfloor-1=0$$ but now I am stuck. How should I proceed?
| $$2x^2-1=2[x]~~~~(1).$$ $[x]$ is GIF/ Floor function
We have $2[x]+1=2x^2 \ge 0 \implies [x] \ge 0 ~~~(2)$
Let $x=n+q$, $n$-is integer $n\ge 0$ and $0 \le q <1$, then we get
$$2n^2+2q^2+4nq-1=2n~~~~(3)$$
$$\implies 2n^2-2n-1=-2q^2-4nq \le 0 \implies 2n^2-2n-1 \le 0 \implies \frac{1-\sqrt{3}}{2} \le n \le \frac{1+\sqrt{3}}{2}.$$
$$\implies n=0,1$$
Now put $n=0$ in (3) to get $q=\frac{\sqrt{6}-2}{2} >0, \implies x =1+q =\sqrt{\frac{3}{2}}.$
Next put $n=0$ in (3) to get $2q^2-1=0 \implies q=\frac{1}{\sqrt{2}}>0 \implies x=0+q=\frac{1}{\sqrt{2}}.$
So only two solutions $x=\sqrt{\frac{3}{2}}, \frac{1}{\sqrt{2}}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3459421",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
sum of series $\log2-\sum^{100}_{n=1}\frac{1}{2^nn}$ The difference of
$$\log2-\sum^{100}_{n=1}\frac{1}{2^nn}$$
$1). $less than Zero
$2). $Greater than $1$
$3). $less than $\frac{1}{2^{100}101}$
$4). $greater than $\frac{1}{2^{100}101}$
Solution I tried: we know that $$\log2 = 1-\displaystyle \frac{1}{2}+\frac{1}{3}-\frac{1}{4}+....$$ hence the given series become $$1-\displaystyle \frac{1}{2}+\frac{1}{3}-\frac{1}{4}+.... \;\;\;- \left ( \frac{1}{2}+\frac{1}{8}+\frac{1}{24}+.. \right)$$
I have no idea how to proceed further; please provide a hint.
Thank you
| $$\log2-\sum^{100}_{n=1}\frac{1}{2^nn} = \sum^{\infty}_{n=101}\frac{1}{2^nn}$$
You can put bounds on this:
$$\frac{1}{2^{101} 101} \lt \sum^{\infty}_{n=101}\frac{1}{2^nn} \lt \sum^{\infty}_{n=101}\frac{1}{2^n 101}= \frac{1}{2^{100} 101}$$
so between about $3.9 \times 10^{-33}$ and about $7.9 \times 10^{-33}$; in fact about $7.7 \times 10^{-33}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3459869",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
For $f(x)=x^2+a\cdot x + b$ prove $b\leq -\frac{1}{4}$ Equation $f(f(x))=0$, for $f(x)=x^2+a\cdot x + b$, have four real solutions and sum of two of them is $-1$. Prove that $b\leq-\frac{1}{4}$.
My progress so far:
Let $x_{1}$ and $x_{2}$ be real solutions for $f(x)=0$. We know that discriminant of this equation is positive, so from $a^2-4b\geq 0$, we have $b<\frac{a^2}{4}$.
Now let $y_{1}$ and $y_{2}$ be real solutions for $f(x)=x_{1}$ and $y_{3}$ and $y_{4}$ be real solutions for $f(x)=x_{2}$. From Vieta's formulas we know $y_{1}+y_{2}=-a$ and $y_{3}+y_{4}=-a$. From here I don't know how to continue, had many ideas but none of them worked.
Can anyone help?
| We can bound the sum of roots and improve the bound to $b\le\frac{-2\phi^{2}-2\phi-1}{4\phi\left(\phi+2\right)}=−0.405$, without needing Vieta's formula.
By the quadratic formula, the roots of $f\circ f(x)=f^{2}+af+b$ are $\underbrace{f(x)}_{x^{2}+ax+b}=\frac{-a\pm\sqrt{a^2-4b}}{2}$. Then, by applying the formula again, the roots of $f\circ f(x)$ must be
$$x=\frac{-a\pm\sqrt{a^{2}-2a-4b\pm2\sqrt{a^{2}-4b}}}{2}\tag{*}$$
As $f\circ f$ is a composition of two quadratics, it - and by extension, its roots - are symmetric across the vertical line through its minimum, $x=\frac{-a}{2}$. So, $(*)$ can be considered as specifying the roots' $4$ distances from a centre, $\frac{-a}{2}$. It is given that these four roots exist (possibly up to multiplicity) and they are indexed by the $\pm$ signs in $(*)$. I will attempt to justify their order. As it is given that $a^2-4b\ge0$, we have $\sqrt{a^2-4b}\ge0$, so
$$\begin{align}\color{red}{-}\sqrt{a^{2}-4b}&\le\color{red}{+}\sqrt{a^{2}-4b}
\\
\frac{\sqrt{a^{2}-4b-2a\color{red}{-}2\sqrt{a^{2}-4b}}}2&\le \frac{\sqrt{a^{2}-4b-2a\color{red}{+}2\sqrt{a^{2}-4b}}}2\tag{**}\end{align}$$
The last LHS radical represents the distance from $\frac{-a}2$ of the two roots with an inner $\pm$ sign of $-$, so they must be closer to the centre as this is the smaller of the two. Conversely, the last RHS represents an inner $\pm$ sign of $+$, and these roots must be further away.
The argument of the outermost radical in $(**)$ is also nonnegative, so the order of the roots must be:
$$\frac{-a-\sqrt{+}}{2}\le \frac{-a-\sqrt{-}}{2}\le \frac{-a+\sqrt{-}}{2}\le \frac{-a+\sqrt{+}}{2}$$
Therefore, the sum of the roots of $f\circ f$ must be at least the sum of the two most negative roots and at most the sum of the two most positive roots. And it is given that the sum of two roots, $-1$ is within these bounds. The problem is therefore to show that the region of the $(a,b)$ plane that satisfies
$$\left(\frac{-a-\sqrt{+}}{2}+ \frac{-a-\sqrt{-}}{2}\right)\le -1 \le \left(\frac{-a+\sqrt{-}}{2}+ \frac{-a+\sqrt{+}}{2}\right)$$
is a subset of the region $b\leq \frac{-1}{4}$. The curves that bound the regions are shown below and are those where the inequality is an equality and those where the radicals arguments' are zero: $b=\frac{a^{2}}{4},\frac{a^{2}-2a-2\pm2\sqrt{2a+1}}{4}$.
For the lower bound, we start with $
\frac{-a-\sqrt{a^{2}-4b-2a-2\sqrt{a^{2}-4b}}}{2}+\frac{-a-\sqrt{a^{2}-4b-2a+2\sqrt{a^{2}-4b}}}{2}\le-1
$. With the substitution $c=a^2-4b$ and evaluating the case when both sides are equal, we have $2a-2+\sqrt{c-2a+2\sqrt{c}}+\sqrt{c-2a-2\sqrt{c}}$, which solves for $b=\frac{-2a^{2}+2a-1}{4a\left(a-2\right)}$. For the upper bound, the curve's boundary line can be treated similarly and solves for the same function, $b=\frac{-2a^{2}+2a-1}{4a\left(a-2\right)}$. When I tried solving this, it was algebraically inefficient and required the aid of a CAS. However, judging by the neat solution, it could probably be streamlined. From here, it is the trivial matter of showing that the permissible region is that below the minimum of $\frac{-2a^{2}+2a-1}{4a\left(a-2\right)}$ and $\frac{a^{2}-2a-2\pm2\sqrt{2a+1}}{4}$, where they are each defined, and that these curves have a maximum at $(-\phi,\frac{-2\phi^{2}-2\phi-1}{4\phi\left(\phi+2\right)})=(-1.62,−0.405)$ (red point on figure), which is indeed less than $-0.25$ (red line on figure).
What is even more interesting is that we can use this same method of bounding the plane to show that if the sum of two roots was a constant, $c$, then $b\le -\frac{\left(c+1\right)}{2}a$, of which we were given the degenerate case $c=-1$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3461586",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 2,
"answer_id": 1
} |
Volume of ellipsoid in Cartesian co ordinates ( w/o changing to spherical or cylindrical systems) I tried triple integrating over the $$\int_{-a}^{a} \int_{-b\sqrt{1-\frac{x^2}{a^2}}}^{b\sqrt{1-\frac{x^2}{a^2}}} \int_{-c\sqrt{1-\frac{x^2}{a^2}-\frac{y^2}{b^2}}}^{c\sqrt{1-\frac{x^2}{a^2}-\frac{y^2}{b^2}}} {d}z {d}y {d}x $$ but getting $4/3 π a^2 b c$. Checking the possibility of deriving it with Cartesian coordinates and not converting to spherical or cylindrical systems. The derivation is as below
$$ z = \pm c\sqrt{1-\frac{x^2}{a^2}-\frac{y^2}{b^2}} ; y = \pm b\sqrt{1-\frac{x^2}{a^2}} \{z=0\}; x = \pm \,\, a \{z=0,y=0\}$$
for simplicity let $z = \pm c_1 \, and \, y = \pm b_1$
$$ V_E = \int_{-a}^{a}\int_{-b_1}^{b_1} z\Big|_{-c_1}^{c_1}\,{d}y{d}x$$
evaluating for z
$$ V_E = \int_{-a}^{a}\int_{-b_1}^{b_1} c\sqrt{1-\frac{x^2}{a^2}-\frac{y^2}{b^2}}\,-\,\left[ -c\sqrt{1-\frac{x^2}{a^2}-\frac{y^2}{b^2}}\right] {d}y{d}x$$
$$ V_E = \int_{-a}^{a}\int_{-b_1}^{b_1} 2c\sqrt{1-\frac{x^2}{a^2}-\frac{y^2}{b^2}}\,{d}y{d}x$$
which is
$$V_E = \frac{2c}{ab} \int_{-a}^{a}\int_{-b_1}^{b_1} \sqrt{{a^2}{b^2}-{x^2}{b^2}-{a^2}{y^2}}\,{d}y{d}x$$
now we know that
$$\boxed {\int \sqrt{{\alpha^2}-{\chi^2}}\,{dx} = \frac{x}{2}\sqrt{{\alpha^2}-{\chi^2}} +\frac{\alpha^2}{2}sin^{-1}\left(\frac{\chi}{\alpha}\right) + C\,\,}$$
putting $\alpha = \sqrt{{a^2}{b^2}-{x^2}{b^2}}\, and \, \chi = \sqrt{{a^2}{y^2}}$ in above, we get
$$V_E = \frac{2c}{ab} \int_{-a}^{a}\, \frac{ay}{2}\sqrt{(ab)^2-(bx)^2-(ay)^2}+\frac{(ab)^2-(bx)^2}{2}sin^{-1}\left(\frac{ay}{\sqrt{(ab)^2-(bx)^2}}\right)\Big|_{-b_1}^{b_1}\,$$
Substituting the limits;
$$V_E = \frac{2c}{ab} \int_{-a}^{a}\,\left[\frac{a}{2}\left(b\sqrt{1-\frac{x^2}{a^2}}\right)\right]\sqrt{(ab)^2-(bx)^2-a^2\left(b^2\left(1-\frac{x^2}{a^2}\right)\right)} + \frac{(ab)^2-(bx)^2}{2}\,sin^{-1} \left(\frac{a}{\sqrt{(ab)^2-(bx)^2}}b\sqrt{1-\frac{x^2}{a^2}}\right) - \,\left[\frac{a}{2}\left(-b\sqrt{1-\frac{x^2}{a^2}}\right)\right]\sqrt{(ab)^2-(bx)^2-a^2\left(b^2\left(1-\frac{x^2}{a^2}\right)\right)} - \frac{(ab)^2-(bx)^2}{2}\,sin^{-1} \left(\frac{a}{\sqrt{(ab)^2-(bx)^2}}(-b)\sqrt{1-\frac{x^2}{a^2}}\right) dx$$
Simplifying
$$V_E = \frac{2c}{ab} \int_{-a}^{a}\,\frac{b}{2}\,\sqrt{a^2-x^2}\,(0) + \frac{b^2(a^2-x^2)}{2}\,sin^{-1}(1)\,-\,\,\frac{-b}{2}\,\sqrt{a^2-x^2}\,(0) - \frac{b^2(a^2-x^2)}{2}\,sin^{-1}(-1) dx$$
we know that $sin^{-1}(1) = \frac{\pi}{2}; sin^{-1}(-1) = \frac{-\pi}{2} $
$$V_E = \frac{2c}{ab} \int_{-a}^{a}\,0 + \frac{b^2(a^2-x^2)}{2}\,\frac{\pi}{2}-\,0 - \frac{b^2(a^2-x^2)}{2}\,\left(\frac{-\pi}{2}\right)dx$$
taking the constants out;
$$V_E = \frac{2\pi b^2c}{2ab} \int_{-a}^{a}(a^2-x^2)\,dx$$
Solving for x
$$V_E = \frac{\pi bc}{a}\left( a^2x - \frac{x^3}{3} \Big|_{-a}^{a} \right)$$
$$V_E = \frac{\pi bc}{a}\left(\left[ a^3 - \frac{a^3}{3}\right]-\left[- a^3 - \frac{-a^3}{3}\right]\right)$$
hence
$$V_E = \frac{4\pi a^2bc}{3}$$
Not sure what went wrong
| Let $V$ denote the volume. You can simplify things from the start by taking advantage of symmetry:
$$\begin{align*}
V&=\int_{-a}^a\int_{-b\sqrt{1-\frac{x^2}{a^2}}}^{b\sqrt{1-\frac{x^2}{a^2}}}\int_{-c\sqrt{1-\frac{x^2}{a^2}-\frac{y^2}{b^2}}}^{c\sqrt{1-\frac{x^2}{a^2}-\frac{y^2}{b^2}}}\,\mathrm dz\,\mathrm dy\,\mathrm dx\\[1ex]
&=8\int_0^a\int_0^{b\sqrt{1-\frac{x^2}{a^2}}}\int_0^{c\sqrt{1-\frac{x^2}{a^2}-\frac{y^2}{b^2}}}\mathrm dz\,\mathrm dy\,\mathrm dx
\end{align*}$$
The integral with respect to $z$ is trivial:
$$V=8c\int_0^a\int_0^{b\sqrt{1-\frac{x^2}{a^2}}}\sqrt{1-\frac{x^2}{a^2}-\frac{y^2}{b^2}}\,\mathrm dy\,\mathrm dx$$
Substitute $y=b\sqrt{1-\frac{x^2}{a^2}}\sin\theta$, treating $x$ as a constant, so that $\mathrm dy=b\sqrt{1-\frac{x^2}{a^2}}\cos\theta\,\mathrm d\theta$. Under this substitution, we have
$$y=0\implies\sin\theta=0\implies\theta=0$$
$$y=\sqrt{1-\frac{x^2}{a^2}}\implies\sin\theta=1\implies\theta=\frac\pi2$$
$$\begin{align*}
V&=8c\int_0^a\int_0^{\frac\pi2}\sqrt{\left(1-\frac{x^2}{a^2}\right)-\frac1{b^2}\left(b\sqrt{1-\frac{x^2}{a^2}}\sin\theta\right)^2}\,b\sqrt{1-\frac{x^2}{a^2}}\cos\theta\,\mathrm d\theta\,\mathrm dx\\[1ex]
&=8bc\underbrace{\left(\int_0^a1-\frac{x^2}{a^2}\,\mathrm dx\right)}_{\frac{2a}3}\underbrace{\left(\int_0^{\frac\pi2}\cos^2\theta\,\mathrm d\theta\right)}_{\frac\pi2}\\[1ex]
&=\frac{4\pi abc}3
\end{align*}$$
Not sure where your mistake occurs, but you're obviously missing a factor of $a$ in the leading fractions of the integral leading up to your result. I suspect it's because of the way you were rewriting the square root terms, which I'd suggest not doing if you don't have to.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3464055",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Difference of two sums of three squares I have proved that every integer is the difference of two sums of three squares, i.e.,
$n = (a^2 + b^2 + c^2) - (d^2 + e^2 + f^2)$
Is this result publishable?
| Any integer not congruent to $2$ modulo $4$ can be written as the difference of two integer squares. Hence the sum of two differences of two integer squares can always be made to represent any number. For example, let $n \ge 1$ be the number, and let $a \ge 1$ and $b \ge 1$ be any two numbers such that $a,b \not\equiv 2\!\pmod{4}$ and $a+b=n$. Then $a=a_1^2-a_2^2$ and $b=b_1^2-b_2^2$, and
$$n = a + b = (a_1^2-a_2^2)+(b_1^2-b_2^2) = (a_1^2+b_1^2)-(a_2^2+b_2^2).$$
This can be extended to any number of squares by choosing $n=a+b+c$ or $n=a+b+c+d$, and so on; your example is $n=a+b+c$.
So the answer to your question is basically "no"… although it might be a worthwhile paper if you showed the generalization(s), and found some interesting applications of the results.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3464975",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Find the infinite sum of the series $\sum_{n=1}^\infty \frac{n^2}{(n^2 +1)^2}$ WolframAlpha outputs the answer as
$$-\frac14 \pi\left(-\coth(\pi) + \pi \operatorname{csch}^2(\pi)\right)$$
But I have no idea how to get there. Tried partial fractions (by splitting into imaginary components), tried comparing with the Basel problem (turns out there's little similarities), nothing worked.
| First, we note that $$\frac{n^2}{(1+n^2)^2}=\color{blue}{\frac1{1+n^2}}-\color{orange}
{\frac{1}{(1+n^2)^2}}.$$
So $$\sum_{n=1}^\infty \frac{n^2}{(n^2 +1)^2}=\color{blue}{\sum_{n=1}^\infty \frac1{1+n^2}}-\color{orange}{\sum_{n=1}^\infty
\frac{1}{(1+n^2)^2}}.$$
Using the identities $$\color{blue}{\sum_{n=1}^\infty \frac1{1+n^2}}=\frac12\pi\coth(\pi)-\frac12$$
from Find the infinite sum of the series $\sum_{n=1}^\infty \frac{1}{n^2 +1}$ and
$$\color{orange}{\sum_{n=1}^\infty
\frac{1}{(1+n^2)^2}}=-\frac{1}{2}+\frac{1}{4} \pi \coth (\pi )+\frac{1}{4} \pi ^2 \text{csch}^2(\pi )$$ from How to calculate $\sum_{k=1}^{\infty}\frac{1}{(a^2+k^2)^2}$ after calculating $\sum_{k=1}^{\infty}\frac{1}{a^2+k^2}$ using Parseval identity?, we reach the final result $$\bbox[5px,border:2px solid #C0A000]{\sum_{n=1}^\infty \frac{n^2}{(n^2 +1)^2}=\frac{1}{4} \pi \coth (\pi )-\frac{1}{4} \pi ^2 \text{csch}^2(\pi )}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3467567",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Find the minimal polynomial for $\cos(\frac{2\pi}{5})$ and $\sin(\frac{2\pi}{5})$ Let $\omega$ be the primitive 5th root of $1$, then $\cos(\frac{2\pi}{5}) = \frac{w+w^{-1}}{2}$ and $\sin(\frac{2\pi}{5}) = \frac{w-w^{-1}}{2i}$. How to find the minimal polynomial of $\frac{w+w^{-1}}{2}$ then? (without using the Chebyshev polynomials)
Thanks.
| Below I'll show how to find
(1) the minimal polynomial for $\omega+\omega^{-1}$,
(2) the minimal polynomial for $\cos\left(\dfrac{2\pi}{5}\right)$, and
(3) the minimal polynomial for $\sin\left(\dfrac{2\pi}{5}\right)$.
(1) First note that $\omega$ is a root of $x^5-1$, but not of $x-1$. Hence $\omega$ is a root of $\dfrac{x^5-1}{x-1}=x^4+x^3+x^2+x+1$. So we have that
$$\omega^4+\omega^3+\omega^2+\omega+1=0.$$
Since $\omega\ne0$, we can divide through by $\omega^2$ to obtain
$$\omega^2+\omega+1+\omega^{-1}+\omega^{-2}=0.$$
Which we can rewrite as
$$\left(\omega^2+2+\omega^{-2}\right)+\left(\omega+\omega^{-1}\right)+1-2=0.$$
So that
$$\left(\omega+\omega^{-1}\right)^2+\left(\omega+\omega^{-1}\right)-1=0.$$
Hence $\omega+\omega^{-1}$ is a root of $x^2+x-1$. Since the roots of this polynomial are not rational, this polynomial is irreducible. Hence $x^2+x-1$ is the minimal polynomial of $\omega+\omega^{-1}$.
(2) $\cos\left(\dfrac{2\pi}{5}\right)=\dfrac{\omega+\omega^{-1}}{2}$, and $4\left(\dfrac{\omega+\omega^{-1}}{2}\right)^2+2\left(\dfrac{\omega+\omega^{-1}}{2}\right)-1=0$.
So the minimal polynomial for $\cos\left(\dfrac{2\pi}{5}\right)$ is $4x^2+2x-1$.
(3) The roots of $4x^2+2x-1$ are $\dfrac{-1\pm\sqrt{5}}{4}$, and $\cos\left(\dfrac{2\pi}{5}\right)>0$, so $\cos\left(\dfrac{2\pi}{5}\right)=\dfrac{\sqrt{5}-1}{4}$.
It follows that $\sin\left(\dfrac{2\pi}{5}\right)=\sqrt{\dfrac{\sqrt{5}+5}{8}}$.
For simplicity, let's let $\alpha=\sin\left(\dfrac{2\pi}{5}\right)=\sqrt{\dfrac{\sqrt{5}+5}{8}}$. So we have that
$$\alpha^2=\dfrac{\sqrt{5}+5}{8}.$$
$$8\alpha^2-5=\sqrt{5}.$$
$$64\alpha^4-80\alpha^2+25=5.$$
It follows that $\alpha$ is a root $16x^4-20x^2+5$, which is irreducible because of Eisenstein's criterion. So the minimal polynomial of $\sin\left(\dfrac{2\pi}{5}\right)$ is $16x^4-20x^2+5$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3471903",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 0
} |
Show that $2(\sin y + 1)(\cos y + 1) = (\sin y + \cos y + 1)^2$ The question states:
Show that: $$2(\sin y + 1)(\cos y + 1) = (\sin y + \cos y + 1)^2$$
This is what I have done
$2(\sin y + 1)(\cos y + 1) = 2(\sin y + \cos y + 1)^2$
L. H. S. = R. H. S.
From L. H. S.
$2(\sin y +1)(\cos y + 1) = 2(\sin y\cos y + \sin y + \cos y + 1)$
$= 2(\sin y\cos y + \sin y + \cos y + \sin^2 y + \cos^2 y) (\sin^2 y + \cos^2 y = 1)$
$= 2(\sin^2 y + \sin y\cos y + \sin y + \cos^2 y + \cos y)$
I got stuck here. I do not know what to do from here.
I have tried and tried several days even contacted friends but all to no avail.
| $$\sin y=s,\cos y=c$$
$$(c+s+1)^2=c^2+s^2+1+2c+2s+2cs=2\underbrace{(1+c)}+2s\underbrace{(1+c)}=?$$
Another way
As $s^2+c^2=1$
$$(s+(c+1))^2=s^2+(c+1)^2+2s(c+1)=1-c^2+(c+1)^2+2s(c+1)=(c+1)[1-c+c+1+2s]=?$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3472158",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 0
} |
How many positive integers have less than $90000$ have the sum of their digits equal to $17$? How many positive integers have less than $90000$ have the sum of their digits equal to $17$?
I tried to write the number as $ABCDE$ and use some math with that (so we need $A + B + C + D + E = 17$), and I tried to use stars and bars, but I got no progress.
Can someone please help me?
| You need to find the number of solutions of $$A+B+C+D+E=17$$
where $0\leq A\leq 8$ and all other variables lie between $0$ and $9$. Check that it is coefficient of $x^{17}$ in the following expression:
$$(1+x+x^2+x^3+x^4+x^5+x^6+x^7+x^8)$$
$$\times(1+x+x^2+x^3+x^4+x^5+x^6+x^7+x^8+x^9)^4$$
Now, apply GP formula and simplify, i.e.
$$Expression = \frac{1-x^9}{1-x}\times\Big(\frac{1-x^{10}}{1-x}\Big)^4$$
$$ = \frac{(1-x^9)(1-x^{10})^4}{(1-x)^5}$$
Now, apply binomial expansion in $(1-x^{10})^4$ and ignore terms which don't contribute to term of $x^{17}$, we get,
$$\frac{(1-x^9)(1-4x^{10})}{(1-x)^5}$$
Now, you have terms with powers of $x$ as $0,9\&10$ in numerator, find the coefficients of $x^{17}, x^8\& x^7$ in the power series expansion of denominator using formula and you are done.
Hope it helps:)
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3474620",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
$\int \frac{\tan x}{\cos^2x}dx$ $\int \frac{\tan x}{\cos^2x}dx=\int\frac{\frac{\sin x}{\cos x}}{\cos^2 x}dx=\int \frac{\sin x}{\cos^3x}dx$
$\cos x=t, -\sin xdx=dt \Rightarrow \sin xdx=-dt$
$\int \frac{-dt}{t^3}=-\int t^{-3}dt=\frac{1}{2}t^{-2}dt=\frac{1}{2\cos^2x}+C $
$\int \frac{\tan x}{\cos^2x}dx\\\tan x=t,\frac{dx}{\cos^2 x}=dt\\\int t dt=\frac{1}{2}t^2+C=\frac{1}{2}\tan^2x+C$
which solution is correct?
| Both are correct, recall $\tan^{2}(x)+1=\sec^{2}(x)$, so that your two answers differ by a constant.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3477880",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Finding $\int \frac{\cos(2x) dx}{\cos^4x+\sin^4x}$ $\int \frac{\cos(2x) dx}{\cos^4x+\sin^4x}$
I'm looking more into simplifying this than the solution itself (which I know involves using t=tan(x/2)).
I did:
$$\int \frac{\cos(2x) dx}{\cos^4x+\sin^4x} = \int\frac{\cos(2x)dx}{(\cos^2x+\sin^2x)^2-2\sin^2x\cos^2x} = \int\frac{\cos(2x)dx}{1-2\sin^2(2x)}$$
I tried applying the t=tan(y/2) (where y = 2x) but it became a mess so I figure this can be simplified further... help?
| You may just apply $t=\tan x$, $dx = \frac{dt}{1+t^2}$ and the resulting expression is quite manageable,,
$$I=\int \frac{\cos(2x) dx}{\cos^4x+\sin^4x} = \int\frac{\cos(2x)dx}{1-\frac12\sin^2(2x)}
= \int \frac {\frac{1-t^2}{1+t^2}\cdot\frac{dt}{1+t^2}}{1-\frac12\left(\frac{2t}{1+t^2}\right)^2}=\int \frac{1-t^2}{1+t^4}dt$$
Next, carry out the integral as follows,
$$I=-\int \frac{1 - \frac1{t^2}}{t^2+\frac1{t^2}}dt
=-\int \frac{d(t + \frac1{t})}{(t+\frac1{t})^2-2}
=\frac1 {\sqrt2} \coth^{-1} \left(\frac{t^2 + 1}{\sqrt2\>t}\right)+C $$
where $(\coth^{-1}t)' = \frac1{1-t^2}$ is used in the last step.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3480289",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 0
} |
Solving a first order ordinary differential equation that is not linear and not separable Problem:
Solve the following differential equations.
$$ ( 3x - y + 1 ) dx - ( 3x - y) dy = 0 $$
Answer:
I am going to use the substitution $z = 3x - y + 1$.
\begin{align*}
\frac{dz}{dx} &=3 - \frac{dy}{dx} \\
dz &=3 \, dx - dy \\
z + (z-1)( 3 - \frac{dz}{dx} ) &= 0 \\
z + 3(z-1) - (z-1)\frac{dz}{dx} &= 0 \\
4z - 1 - (z-1)\frac{dz}{dx} &= 0 \\
4z - 1 &= (z-1)\frac{dz}{dx} \\
\frac{dz}{dx} &= \frac{4z-3}{z-1} \\
dx &= \left( \frac{z-1}{4z-3} \right) \, dz
\end{align*}
Using long division, I find that:
\begin{align*}
\frac{z-1}{4z-3} &= \frac{1}{4} - \frac{ \frac{1}{4} } {4z-3} \\
\int \frac{z-1}{4z-3} \, dz &= \frac{z}{4} - \frac{1}{16} \ln{|4z-3|} + C_1 \\
x &= \frac{z}{4} - \frac{1}{16} \ln{|4z-3|} + C_1 \\
16x &= 4z - \ln{|4z-3|} + C_2 \text{ with } C_2 = 16C_1 \\
16x &= 12x - 4y + 4 - \ln{|12x - 4y + 4 - 3|} + C_2 \\
4x &= -4y - \ln{|12x - 4y + 1|} + C \text{ with } \,\, C = C_2 + 4
\end{align*}
Now, I am going to check the answer by differentiating it.
\begin{align*}
4 &= -4 \frac{dy}{dx}- \frac{12 - 4\frac{dy}{dx}}{12x - 4y + 1} \\
4( 12x - 4y + 1) &= -4( 12x - 4y + 1)\frac{dy}{dx} - 12 + 4 \frac{dy}{dx} \\
48x - 16y + 4 &= ( -48x + 16y - 4)\frac{dy}{dx} - 12 + 4 \frac{dy}{dx} \\
48x - 16y + 16 &= ( -48x + 16y - 4)\frac{dy}{dx} + 4 \frac{dy}{dx} \\
48x - 16y + 16 &= ( -48x + 16y )\frac{dy}{dx} \\
(3x - y + 1 ) \, dx &= ( -3x + y) \, dy \\
(3x - y + 1 ) \, dx - ( -3x + y) \, dy &= 0 \\
\end{align*}
Hence the answers do not check but I am close. It looks like a sign error to me but I cannot seem to find it.
| Making $u = 3x-y$ we have $\frac{du}{dx}=3-\frac{dy}{dx}$ and also
$$
u+1-u\frac{dy}{dx} = u+1-u\left(3-\frac{du}{dx}\right)=0
$$
and this DE now is separable
$$
\frac{u du}{2u-1} = dx
$$
or
$$
\frac 14\left(\ln(2u-1)+2u-1\right)=x + C_0
$$
or
$$
(2u-1)e^{2u-1}=C_1 e^{4x}
$$
and now using the Lambert function
$$
2u-1=W\left(C_1e^{4x}\right)
$$
and finally
$$
y = \frac 12\left(6x-1-W\left(C_1e^{4x}\right)\right)
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3484152",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Hard inequality for positive numbers The problem is to prove that for $a,b,c>0$ we have
$$\frac{a^2}{b^2}+\frac{b^2}{c^2}+\frac{c^2}{a^2}+\frac{9abc}{4(a^3+b^3+c^3)}\geq \frac{15}{4}.$$
I have tried to use Bergstrom/Engel inequality to write, for example, $\frac{a^2}{b^2}+\frac{b^2}{c^2}+\frac{c^2}{a^2}\geq \frac{(a+b+c)^2}{a^2+b^2+c^2}$, and then to use Muirhead's inequalities to prove the remaining inequality - but unsuccessfully, so far...
| The cyclically symmetric inequality is equivalent to:
$$\color{red}{\left(\frac{a^2}{b^2}+\frac{b^2}{a^2}-2 \right)} + \color{blue}{\left(\frac{b^2}{c^2}+\frac{c^2}{a^2}-\frac{b^2}{a^2}-1 \right)}+ \color{green}{\left(\frac{9abc}{4(a^3+b^3+c^3)}-\frac34\right)} \geqslant 0$$
$$\iff \color{red}{\frac{(a^2-b^2)^2}{a^2b^2}} + \color{blue}{\frac{(a^2-c^2)(b^2-c^2)}{a^2c^2}}+\color{green}{\frac{9abc-3(a^3+b^3+c^3)}{4(a^3+b^3+c^3)}}\geqslant 0$$
As $a^3+b^3+c^3-3abc=(a+b+c)((a-b)^2+(a-c)(b-c))$, we have above $\iff$
$$(a-b)^2\left(\color{red}{\frac{(a+b)^2}{a^2b^2}}-\color{green}{\frac{3(a+b+c)}{4(a^3+b^3+c^3)}} \right)+(a-c)(b-c)\left(\color{blue}{\frac{(a+c)(b+c)}{a^2c^2}} - \color{green}{\frac{3(a+b+c)}{4(a^3+b^3+c^3)}}\right)\geqslant 0$$
Now due to symmetry, we may assume $c=\min (a, b, c)$, hence it remains to show that under this condition, both
$$\color{red}{\frac{(a+b)^2}{a^2b^2}}-\color{green}{\frac{3(a+b+c)}{4(a^3+b^3+c^3)}} \geqslant 0, \qquad \color{blue}{\frac{(a+c)(b+c)}{a^2c^2}} - \color{green}{\frac{3(a+b+c)}{4(a^3+b^3+c^3)}} \geqslant 0$$
However $(a^3+b^3)(a+b)\geqslant (a^2+b^2)^2\geqslant 4a^2b^2 \implies$
$$\color{red}{\frac{(a+b)^2}{a^2b^2}}\geqslant 4\frac{a+b}{a^3+b^3}> 4\frac{a+b}{a^3+b^3+c^3}\geqslant \frac83\frac{a+b+c}{a^3+b^3+c^3}> \color{green}{\frac{3(a+b+c)}{4(a^3+b^3+c^3)}}$$
And $3(a^3+b^3+c^3)\geqslant (a+b+c)(a^2+b^2+c^2) \implies$
$$4(a^3+b^3+c^3)(a+c)(b+c)\geqslant \frac43(a+b+c)(a+c)(b+c)(a^2+b^2+c^2) \geqslant \frac43(a+b+c)(2c)(2c)(a^2)\geqslant 3(a+b+c)a^2c^2$$
$$\implies \color{blue}{\frac{(a+c)(b+c)}{a^2c^2}}\geqslant \color{green}{\frac{3(a+b+c)}{4(a^3+b^3+c^3)}}$$
Hence the inequality holds true, with equality when $a=b=c$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3485380",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "13",
"answer_count": 2,
"answer_id": 1
} |
Factoring $x^4-2x^3+2x^2+x+4$
I need to show that the polynomial is not irreducible and I am trying to factor the polynomial
$$x^4-2x^3+2x^2+x+4$$
I checked from a calculator that it has a factor but how do I get it by myself?
I tried grouping but It didnt work I got
$x^2(x^2-2x+2)+x+4$ And I dont know how should I proceed. My guts tell me that it should be of the form:
$(x^2-ax+k)(x^2+bx+l)$, should I just try to figure out the constants by trying out?
| Actually, if you just try to solve this like a general quartic equation you get a break: let $x=t+1/2$. Then
$$\begin{align}x^4-2x^4+2x^2+x+4&=\left(t+\frac12\right)^4-2\left(t+\frac12\right)^3+2\left(t+\frac12\right)^2+\left(t+\frac12\right)+4\\
&=t^4+\frac12t^2+2t+\frac{77}{16}\\
&=\left(t^2+at+b\right)\left(t^2-at+c\right)\\
&=t^4+\left(b-a^2+c\right)t^2-a(b-c)t+bc\end{align}$$
So we have
$$\begin{align}b+c&=\frac12+a^2\\
b-c&=-\frac2a\end{align}$$
With solutions
$$\begin{align}2b&=a^2+\frac12-\frac2a\\
2c&=a^2+\frac12+\frac2a\end{align}$$
Then
$$4bc=a^4+a^2+\frac14-\frac4{a^2}=\frac{77}4$$
Which simplifies to
$$a^6+a^4-19a^2-4=0$$
And here is our break: the resolvent cubic has a rational root: $a^2=4$. If we pick $a=2$, then $b=7/4$, $c=11/4$ and
$$\begin{align}x^4-2x^4+2x^2+x+4&=\left(t^2+2t+\frac74\right)\left(t^2-2t+\frac{11}4\right)\\
&=\left(\left(x-\frac12\right)^2+2\left(x-\frac12\right)+\frac74\right)\left(\left(x-\frac12\right)^2-2\left(x-\frac12\right)+\frac{11}4\right)\\
&=\left(x^2+x+1\right)\left(x^2-3x+4\right)\end{align}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3487085",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 4,
"answer_id": 2
} |
Choosing group of size 3 in a roll of 5 unique dice Suppose you have $5$ fair dice (with sides $1-6$) colored red, green, yellow, orange and blue.
In how many ways can you roll them such that the resulting set of numbers rolled is of size $3$? For instance, the case where we roll: $1,2,3,1,2$ is accepted but $1,2,3,4,1$ is not.
My attempt:
Choose the set of size $3$ to work with: ${6 \choose 3}$
Choose $3$ dice to insert those three unique numbers into: $5 \choose 3$
Assign the numbers in the set to colors (we have to use all three numbers): $3!$
In the two remaining dice we have three options: $3^2$
overall: ${6 \choose 3}\cdot{5 \choose 3}\cdot3!\cdot3^2$.
Is this correct?
| You can either have three of one number and two singletons or two of two numbers and one singleton. For the first, you have $6$ ways to choose the number there will be three of, $5 \choose 3$ ways to choose the dice with that number, $5$ ways to choose number on the first of the other dice and $4$ ways to choose number on the last, for $$6\cdot {5 \choose 3}\cdot 5 \cdot 4=1200$$
For two pairs plus one, you have $5 \choose 2$ ways to pick the first pair and $6$ ways to pick its number, $3 \choose 2$ ways to pick the second pair and $5$ ways to pick its number, then $4$ ways to pick the number for the last, but we have double counted because we can swap the two pairs.
$$\frac 12{5 \choose 2}6{3 \choose 2}5\cdot 4=1800$$
For a total of $3000$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3487500",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Finding the volume of the tetrahedron with vertices $(0,0,0)$, $(2,0,0)$, $(0,2,0)$, $(0,0,2)$. I get $8$; answer is $4/3$. The following problem is from the 7th edition of the book "Calculus and Analytic Geometry Part II". It can be found in section 13.7. It is
problem number 5.
Find the volume of the tetrahedron whose vertices are the given points:
$$ ( 0, 0, 0 ), ( 2, 0, 0 ), ( 0, 2, 0 ), ( 0, 0, 2 ) $$
Answer:
In this case, the tetrahedron is a parallelepiped object. If the bounds of such an object is given by the vectors $A$, $B$ and $C$ then
the area of the object is $A \cdot (B \times C)$. Let $V$ be the volume we are trying to find.
\begin{align*}
x^2 &= 6 - y^2 - z^2 \\[4pt]
A &= ( 2, 0, 0) - (0,0,0) = ( 2, 0, 0) \\
B &= ( 0, 2, 0) - (0,0,0) = ( 0, 2, 0) \\
C &= ( 0, 0, 2) - (0,0,0) = ( 0, 0, 2) \\[4pt]
V &= \begin{vmatrix}
a_1 & a_2 & a_3 \\
b_1 & b_2 & b_3 \\
c_1 & c_2 & c_3 \\
\end{vmatrix} =
\begin{vmatrix}
2 & 0 &0 \\
0 & 2 & 0 \\
0 & 0 & 2\\
\end{vmatrix} \\
&= 2 \begin{vmatrix}
2 & 0 \\
0 & 2\\
\end{vmatrix} = 2(4 - 0) \\
&= 8
\end{align*}
However, the book gets $\frac{4}{3}$.
| A tetrahedron is never a parallepiped. It is a pyramid with a triangular base.
All six faces of a parallelopiped are parallelograms but a tetrahedron has only four faces and all are triangles.
In short, what you are measuring is very unlike what you were asked to measure.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3488590",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 5,
"answer_id": 0
} |
Find $p$ and $q$ such that $x^2+px+q
Find $p$ and $q$ such that $$x^2+px+q<x$$ iff $$x \in (1,5)$$
I tried the following:
$$x^2+px+q = (x+\frac{p}{2})^2+q-\frac{p^2}{4}$$
where the global minimum is $$q-\frac{p^2}{4}$$ if $$x = \frac{-p}{2}$$
However this doesn't seem to help with the problem at hand...
| So $1$ and $5$ are the roots of $x^2+px + q = x$:
$$\begin{align*}
(x-1)(x-5) &= 0\\
x^2 - 6x + 5 &= 0\\
x^2 - 5x + 5 &= x\\
p &= -5\\
q&= 5
\end{align*}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3489588",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 0
} |
recursive succession $a_{n+1}=\frac{a_n+2}{3a_n+2}, a_0>0$ I'm given this recursive succession:
$a_{n+1}=\frac{a_n+2}{3a_n+2}, a_0>0$.
This is what I've done:
$L=\frac{L+2}{3L+2} \rightarrow L_1=\frac{2}{3}$ and $L_2=-1$
if $a_0 >0 $ then $a_n>0 \forall n \in N \rightarrow
$ the succession is positive $\forall n \in N $and $L_2=-1$ is impossible.
if $a_n >0 $ then $a_n+2<3a_n+2 \rightarrow a_{n+1}=\frac{a_n+2}{3a_n+2} <1$ and then all the succession is beween $0$ and $1 $ excluded.
the succession jumps back and forth the value $\frac{2}{3}$ because:
if $a_n<\frac{2}{3}$ then $a_{n+1}>\frac{2}{3}$
if $a_n>\frac{2}{3}$ then $a_{n+1}<\frac{2}{3}$
Once I arrived at this point I don't know how to conclude that the limit of the succesion is if $\frac{2}{3}$
Can someone help me to understand how to procede?
| Use Banach fixed-point theorem, for $a_{n+1}=f(a_n)$ where $f(x)=\frac{x+2}{3x+2}$ and $f'(x)= \frac{-4}{(3 x + 2)^2}$. In fact, for positive $x$
$$1>\frac{x+2}{3x+2} > \frac{1}{3}$$
and, starting with $n\geq 1$, $a_n\in\left(\frac{1}{3},1\right)$. As a result, using MVT
$$|f(x)-f(y)|=|f'(\varepsilon)|\cdot |x-y|=\frac{4}{(3\varepsilon+2)^2}\cdot|x-y|<
\frac{4}{9}\cdot|x-y|$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3491959",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
} |
minimum value of $\sum (x+1/x)^{10}$ expression
If $x,y,z>0$ . Then prove that $\displaystyle \bigg(x+\frac{1}{x}\bigg)^{10}+\bigg(y+\frac{1}{y}\bigg)^{10}+\bigg(z+\frac{1}{z}\bigg)^{10}\geq \frac{10^{10}}{3^{9}}.$
What i try
Let $\displaystyle f(x)=\bigg(x+\frac{1}{x}\bigg)^{10}.$ Then $\displaystyle f'(x)=10\bigg(x+\frac{1}{x}\bigg)^{9}\bigg(1-\frac{1}{x^2}\bigg)>0$ fir all $x$
So $$\frac{f(x)+f(y)+f(z)}{3}\geq f\bigg(\frac{x+y+z}{3}\bigg)$$
How do i find minimum of $f((x+y+z)/3)$ help me please
| If it means that $x+y+z=1$ so by Jensen for $f(x)=x^{10}$ and by C-S we obtain:
$$\sum_{cyc}\left(x+\frac{1}{x}\right)^{10}\geq\frac{1}{3^9}\left(\sum_{cyc}\left(x+\frac{1}{x}\right)\right)^{10}=\frac{1}{3^9}\left(1+(x+y+z)\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)\right)^{10}\geq$$
$$\geq\frac{1}{3^9}\left(1+9\right)^{10}=\frac{10^{10}}{3^9}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3494037",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Is $x^2 \geq \alpha(\alpha-1)$?
If $\alpha$ is a nonnegative real and $x$ is a real satisfying $(x+1)^2\geq \alpha(\alpha+1),$ is $x^2 \geq \alpha(\alpha-1)$?
The answer is yes. Consider two cases: $1) \, x < -1$ and $2)\, x\geq -1.$
In case $1,$ taking the square root of both sides of the inequality gives $-(x+1) \geq \sqrt{\alpha(\alpha+1)}\Rightarrow x \leq -\sqrt{\alpha(\alpha+1)}-1.$ Hence $x^2\geq \alpha(\alpha+1)+2\sqrt{\alpha(\alpha+1)}+1.$ Since $2\sqrt{\alpha(\alpha+1)}+1\geq 2\alpha+1 > -2\alpha, x^2>\alpha(\alpha+1)-2\alpha=\alpha(\alpha-1).$
Now in case $2,$ if $\alpha = 0, x^2 \geq 0,$ so we are done. Suppose $\alpha > 0$. Taking the square root of both sides gives $x+1 \geq \sqrt{\alpha(\alpha+1)}\Rightarrow x\geq \sqrt{\alpha(\alpha+1)}-1\Rightarrow x^2\geq \alpha(\alpha+1)-2\sqrt{\alpha(\alpha+1)}+1.$ $2\alpha-2\sqrt{\alpha(\alpha+1)}+1 =2\alpha\left(1-\sqrt{1+\frac{1}{\alpha}}\right)+1 =\dfrac{\sqrt{1+\frac{1}{\alpha}}-1}{1+\sqrt{1+\frac{1}{\alpha}}}>0.$ Hence $-2\sqrt{\alpha(\alpha+1)}+1>-2\alpha\Rightarrow x^2 > \alpha(\alpha+1)-2\alpha = \alpha(\alpha-1)$. I think this argument works, but I was wondering if there was a faster method?
| Here's another method.
Suppose $\alpha(\alpha+1) = (x+1)^2$ and $x\geq 0$. Then since $\left(x+\frac{1}{2}\right)\left(x+\frac{3}{2}\right)=(x+1)^2-\frac{1}{4},$ we have that $\alpha > x+\frac{1}{2}.$ Hence $\alpha(\alpha-1)=\alpha(\alpha+1)-2\alpha <(x+1)^2-2x-1=x^2.$
If $\alpha(\alpha+1) < (x+1)^2$ and $x\geq 0,$ then take $\beta > \alpha$ with $\beta(\beta + 1) = (x+1)^2.$ Since $\alpha - 1<\beta - 1$ and $\alpha \geq 0,$ $\alpha(\alpha-1) \leq \alpha(\beta-1) <\beta(\beta -1) x^2$.
When $x<0, -|x| -1 < x+1 < |x| + 1,$ so $(x+1)^2 < (|x|+1)^2,$ and $x^2 = |x|^2,$ so the result follows follows from the result for $\alpha, |x|$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3494612",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
Find the integral by first principle In my math course homework, I encountered this problem:
Find $$\frac{d}{dx}\int^{x^2}_0\frac{dt}{1+e^{t^2}}$$ by first principle
$$\begin{split}\frac{d}{dx}\int^{x^2}_0\frac{dt}{1+e^{t^2}}&=\lim_{h\rightarrow0}\frac{\int^{(x+h)^2}_0\frac{dt}{1+e^{t^2}}-\int^{x^2}_0\frac{dt}{1+e^{t^2}}}{h}\\&=\lim_{h\rightarrow0}\frac{\int^{(x+h)^2}_{x^2}\frac{dt}{1+e^{t^2}}}{h}\end{split}$$
Then I don't know how to do. Please help me!
| Compute $$\lim_{h\rightarrow0}\frac{\int^{(x+h)^2}_{x^2}\frac{dt}{1+e^{t^2}}}{h}
$$
without using the fundamental theorem of calculus or the mean value theorem.
If $x^2 < t < (x+h)^2$, then
$$
x^4 < t^2 < (x+h)^4
\\
1 + \exp(x^4) < 1+\exp(t^2) < 1+\exp((x+h)^4)
\\
\frac{1}{1 + \exp(x^4)} > \frac{1}{1+\exp(t^2)} > \frac{1}{1+\exp((x+h)^4)}
\\
\frac{(x+h)^2-x^2}{1 + \exp(x^4)} > \int_{x^2}^{(x+h)^2}\frac{dt}{1+\exp(t^2)} > \frac{(x+h)^2-x^2}{1+\exp((x+h)^4)}
\\
\frac{2hx+h^2}{h(1 + \exp(x^4))} > \frac{1}{h}\int_{x^2}^{(x+h)^2}\frac{dt}{1+\exp(t^2)} > \frac{2hx+h^2}{h(1+\exp((x+h)^4))}
\\
\frac{2x+h}{1 + \exp(x^4)} > \frac{1}{h}\int_{x^2}^{(x+h)^2}\frac{dt}{1+\exp(t^2)} > \frac{2x+h}{1+\exp((x+h)^4)}
$$
But
$$
\lim_{h \to 0} \frac{2x+h}{1 + \exp(x^4)} = \frac{2x}{1 + \exp(x^4)}\quad\text{and}
\quad
\lim_{h \to 0}\frac{2x+h}{1+\exp((x+h)^4)} = \frac{2x}{1 + \exp(x^4)}
$$
So our answer is squeezed between these:
$$
\lim_{h \to 0}\frac{1}{h}\int_{x^2}^{(x+h)^2}\frac{dt}{1+\exp(t^2)}= \frac{2x}{1 + \exp(x^4)} .
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3496096",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 1
} |
Multinomial theorem member that does not contain an irrational number Hey I am supposed to solve the following problem:
Specify a development member that does not contain an irrational number:
$$\left (\sqrt{5} - \sqrt[3]{2} +2 \right )^{6}.$$
So I used multinomial theorem:
$$\sum_{i=0}^{6}\binom{6}{n_{1},n_{2},n_{3}}\left ( \sqrt{5} \right )^{n_{1}}\left ( - \sqrt[3]{2} \right )^{n_{2}}\left ( 2 \right )^{n_{3}}$$
and then I know that :$ 2k+3l +n_{3}=6$ with $n_1=2k$ and $n_2=3l$.
Is that a correct answer?
$$\binom{6}{2,3,1}+\binom{6}{2,0,4}+\binom{6}{4,0,2}+\binom{6}{6,0,0}+\binom{6}{0,3,3}+\binom{6}{0,0,6}$$
| No, your answer is not correct. Since $n_1\in\{0,2,4,6\}$ and $n_2\in\{0,3,6\}$ with $n_1+n_2\leq 6$, we find that the required number is
$$N:=\sum_{(n_1,n_2)\in S}\binom{6}{n_{1},n_{2},6-n_1-n_2}\left ( \sqrt{5} \right )^{n_{1}}\left ( - \sqrt[3]{2} \right )^{n_{2}}\left ( 2 \right )^{6-n_1-n_2}$$
where $S:=\{(0,0), (0,3), (0,6), (2,0), (2,3),(4,0), (6,0)\}$. What is $N$?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3496197",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Solving Cauchy Schwarz inequality Taken from Titu Andreescu and Bogdan Enescu's Mathematical Olympiad Treasures on page 9 Problem 1.19, to prove,
$\frac{a}{b+c} + \frac{b}{a+c} + \frac{c}{a+b} \geqslant \frac{3}{2}$.
It is easy to see why $LHS$ may yield
$\geqslant\frac{(a+b+c)^2}{2(ab+bc+ca)}$ from Cauchy.
Yet how do one bring in the '$3$' part from the $RHS$ knowing that $a^2+b^2+c^2 \geqslant ab+bc+ca $?
Any help would be appreciated.
| Because $$\frac{(a+b+c)^2}{2(ab+ac+bc)}\geq\frac{3}{2}$$ it's
$$(a+b+c)^2\geq3(ab+ac+bc)$$ or
$$a^2+b^2+c^2+2ab+2ac+2bc\geq3(ab+ac+bc)$$ or
$$a^2+b^2+c^2\geq ab+ac+bc.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3496637",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Estimation of sum of series with cosine Prove the following:
$$
\sum_{{\large j = 1} \atop {\large j \neq k}}^{n}
{1 \over \left\vert\cos\left(k\pi/ n\right) -
\cos\left(j\pi/n\right)\right\vert} \leq cn^{2}
\qquad\mbox{where}\quad 1 \leq k \leq n\quad\mbox{is fixed.}
$$
I was able to get the upper bound to be $cn^{3}$.
Any method can be used for getting the desired bound.
I also tried taking the upper bound in terms of an integral.
I will appreciate any suggestions.
| By sum-to-product formulae, the sum $S(n)$ equals
$$S(n)=\frac12\sum^n_{j=1}\left|\csc\frac{(j+k)\pi}{2n}\csc\frac{(j-k)\pi}{2n}\right|$$
By Cauchy-Schwarz inequality,
$$S(n)\le\frac12\sqrt{
\underbrace{\sum^n_{j=1}\csc^2\frac{(j+k)\pi}{2n}}_{S_1}
\cdot\underbrace{\sum^n_{j=1}\csc^2\frac{(j-k)\pi}{2n}}_{S_2}}$$
Regarding $S_1$,
$$
\begin{align}
S_1
&=\sum^n_{j=1}\csc^2\frac{(j+k)\pi}{2n} \\
&=\sum^{n+k}_{j=k+1}\csc^2\frac{\pi}{2}\frac jn \\
&=\sum^{n}_{j=k+1}\csc^2\frac{\pi}{2}\frac jn +\sum^{n+k}_{j=n+1}\csc^2\frac{\pi}{2}\frac jn\\
&=\sum^{n}_{j=k+1}\csc^2\frac{\pi}{2}\frac jn +\sum^{k}_{j=1}\csc^2\frac{\pi}{2}\left(1+\frac jn\right)\\
&=\sum^{n}_{j=k+1}\csc^2\frac{\pi}{2}\frac jn +\sum^{k}_{j=1}\sec^2\frac{\pi}{2}\frac jn\\
&\le \sum^{n}_{j=k+1}\csc^2\frac{\pi}{2}\frac jn+k \sec^2\frac{k\pi}{2n}\\
&\le \sum^{n}_{j=k+1}\frac{n^2}{j^2}+k \sec^2\frac{k\pi}{2n} \qquad\text{NB: }x\csc\frac{\pi x}{2}\le1 \text{ for }|x|\le 1\\
&\le\frac{\pi^2}{6}n^2+k \sec^2\frac{k\pi}{2n}
\end{align}
$$
Thus, $S_1=O(n^2)$.
Regarding $S_2$,
$$
\begin{align}
S_2
&=\sum^{k-1}_{j=1}\csc^2\frac{(k-j)\pi}{2n}+\sum^n_{j=k+1}\csc^2\frac{(j-k)\pi}{2n} \\
&=\sum^{k-1}_{j=1}\csc^2\frac{\pi j}{2n}+\sum^{n-k}_{j=1}\csc^2\frac{\pi j}{2n} \\
&\le\sum^{k-1}_{j=1}\frac{n^2}{j^2}+\sum^{n-k}_{j=1}\frac{n^2}{j^2} \\
&\le \frac{\pi^2}{3}n^2
\end{align}
$$
Hence, $S_2=O(n^2)$. Therefore,
$$S(n)=\sqrt{O(n^2)\cdot O(n^2)}=O(n^2)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3496886",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Prove a trigonometric inequality Let $a,b,c$ reals with sum 0. Prove that
$\left|\sin{a}+\sin{b}+\sin{c}\right| \leq \frac{3\sqrt{3}}{2}.$
My idea was to set $z_1=\cos{a}+i\sin{a}$ and also $z_2,z_3$. Then I tried replacing the sines in the inequality with these complex numbers. However, I didn’t get to anything interesting that could help me prove this.
Thanks in advance!
| By the triangle inequality, C-S and AM-GM we obtain:
$$|\sin{a}+\sin{b}+\sin{c}|=|\sin{a}+\sin{b}-\sin{a}\cos{b}-\cos{a}\sin{b}|=$$
$$=|(1-\cos{b})\sin{a}-\sin{b}\cos{a}+\sin{b}|\leq|(1-\cos{b})\sin{a}-\sin{b}\cos{a}|+|\sin{b}|\leq$$
$$\leq\sqrt{((1-\cos{b})^2+(-\sin b)^2)(\sin^2a+\cos^2a)}+|\sin{b}|=$$
$$=\sqrt{2-2\cos{b}}+|\sin{b}|=2|\sin\frac{b}{2}|+|\sin{b}|=2|\sin\frac{b}{2}|\left(1+|\cos\frac{b}{2}|\right)=$$
$$=2\sqrt{\left(1-|\cos\frac{b}{2}|\right)\left(1+|\cos\frac{b}{2}|\right)^3}=\frac{2}{\sqrt3}\sqrt{\left(3-3|\cos\frac{b}{2}|\right)\left(1+|\cos\frac{b}{2}|\right)^3}\leq$$
$$\leq\frac{2}{\sqrt3}\sqrt{\left(\frac{3-3|\cos\frac{b}{2}|+3\left(1+|\cos\frac{b}{2}|\right)}{4}\right)^4}=\frac{3\sqrt3}{2}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3502090",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 2,
"answer_id": 1
} |
Question on Integration solution I am looking through the solution of integrating: $$\int_{}^{} \frac{x^4+1}{x^3+1} dx$$
I understand it but for the following part:
My problem is limited to the first and penultimate column. I would very much appreciate it if someone could explain to me these, especially the penultimate one.
EDIT 0: Clarification
I have a problem with this line:
How do I get those thetas?
My main problem is with this part:
Why 1 = ..., -1 = ... and 0 = ...?
| Solution of integral will be by partial fraction method
$$I=\int (x+\frac{1-x}{x^3+1})dx$$
$$I = \frac{x^2}{2}+\int (\frac{1-x}{(x+1)(x^2-x+1)})dx$$
Where you can solve for
$$I_1=\int (\frac{1-x}{(x+1)(x^2-x+1)})dx$$
Using partial fraction like done here
https://www.mathsdiscussion.com/discussion-forum/topic/indefinite-integral/?part=1#postid-77
$$I=\frac{x^2}{2}-\frac{1}{3}ln(\frac{x^2-x+1}{(x+1)^2})+C$$
Problem what you say in your solution is when let
$$\frac{1-x}{(x+1)(x^2-x+1)}=\frac{a}{x+1}+\frac{bx+c}{x^2-x+1}$$
Multiplying both sides by $(x+1)(x^2-x+1)$ we ge
$$1-x=a(x^2-x+1)+(bx+c)(x+1)$$
Now comparing coefficients of $x^2$ , x and constant terms on both sides we get
$$ $$
Coefficient of$x^2$ on both sides 0=a+b
$$. $$
Coefficient of x on both sides -1=-a+b+c
$$. $$
Constant terms on both sides 1=a+c
$$ $$ This how we get those 3 equations
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3503714",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Calculating an integral using Leibniz parametric integration formula Let there be $a, b > 0$, find:
$\int_0^{\pi/2}$$\frac{dx}{(a\cos^2(x)+b\sin^2(x))^2}$
Hints given:
1. Remember: $\cos^2(x) + \sin^2(x) = 1$, use it to make the integral a sum of two parts.
2. Define $f(x,y)=\frac{-1}{(y\cos^2(x)+b\sin^2(x))^2}$, and calculate:
$F(y) = \int_0^{\pi/2}f'y(x,y)\,dx$, and use Leibniz rule for parametric integrals while $y=a$.
what I did so far(at the image. im stuck):
any help/hints/anything would be much appreciated. thanks in advance!
i
| $$f(a,b)=\int_0^{\pi/2}\frac{dx}{a\cos^2(x)+b\sin^2(x)}=\int_0^{\pi/2}\frac{sec^2(x)}{a+b \tan^2(x)}dx=\frac{1}{\sqrt{ab}}arctan\left(\sqrt{\frac{b}{a}}tan(x)\right)_0^{\pi/2}$$
$$\color{red}{f(a,b)=\int_0^{\pi/2}\frac{dx}{a\cos^2(x)+b\sin^2(x)}=\frac{\pi}{2\sqrt{ab}}}$$
$$\frac{\partial f(a,b)}{\partial a}+\frac{\partial f(a,b)}{\partial b}=-\int_0^{\pi/2}\frac{cos^2(x)}{(a\cos^2(x)+b\sin^2(x))^2}dx-\int_0^{\pi/2}\frac{sin^2(x)}{(a\cos^2(x)+b\sin^2(x))^2}dx$$
$$\frac{\partial f(a,b)}{\partial a}+\frac{\partial f(a,b)}{\partial b}=\frac{\pi}{2}\left(-\frac{1}{2\sqrt{a^3b}}-\frac{1}{2\sqrt{ab^3}}\right)$$
Hence:
$$\int_0^{\pi/2}\frac{dx}{(a\cos^2(x)+b\sin^2(x))^2}=\frac{\pi}{4\sqrt{ab}}\left(\frac{1}{a}+\frac{1}{b}\right)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3507989",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Given positives $a, b, c$, prove that $\frac{a}{(b + c)^2} + \frac{b}{(c + a)^2} + \frac{c}{(a + b)^2} \ge \frac{9}{4(a + b + c)}$.
Given positives $a, b, c$, prove that $$\large \frac{a}{(b + c)^2} + \frac{b}{(c + a)^2} + \frac{c}{(a + b)^2} \ge \frac{9}{4(a + b + c)}$$
Let $x = \dfrac{b + c}{2}, y = \dfrac{c + a}{2}, z = \dfrac{a + b}{2}$
It needs to be sufficient to prove that $$\sum_{cyc}\frac{\dfrac{a + b}{2} + \dfrac{b + c}{2} - \dfrac{c + a}{2}}{\left(2 \cdot \dfrac{c + a}{2}\right)^2} \ge \frac{9}{\displaystyle 4 \cdot \sum_{cyc}\dfrac{c + a}{2}} \implies \sum_{cyc}\frac{z + x - y}{y^2} \ge \frac{9}{y + z + x}$$
According to the Cauchy-Schwarz inequality, we have that
$$(y + z + x) \cdot \sum_{cyc}\frac{z + x - y}{y^2} \ge \left(\sum_{cyc}\sqrt{\frac{z + x - y}{y}}\right)^2$$
We need to prove that $$\sum_{cyc}\sqrt{\frac{z + x - y}{y}} \ge 3$$
but I don't know how to.
Thanks to Isaac YIU Math Studio, we additionally have that $$(y + z + x) \cdot \sum_{cyc}\frac{z + x - y}{y^2} = \sum_{cyc}(z + x - y) \cdot \sum_{cyc}\frac{z + x - y}{y^2} \ge \left(\sum_{cyc}\frac{z + x - y}{y}\right)^2$$
We now need to prove that $$\sum_{cyc}\frac{z + x - y}{y} \ge 3$$, which could be followed from Nesbitt's inequality.
I would be greatly appreciated if there are any other solutions than this one.
| Hint : Using homogeneity, WLOG we may set $a+b+c=3$, then note $f(x) = \dfrac{x}{(3-x)^2}$ is convex and use Jensen’s inequality.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3508789",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 6,
"answer_id": 0
} |
Find $\lim\limits_{n \to \infty} \left ( n - \sum\limits_{k = 1} ^ n e ^{\frac{k}{n^2}} \right)$. I have to find the limit:
$$\lim\limits_{n \to \infty} \bigg ( n - \displaystyle\sum_{k = 1} ^ n e ^{\frac{k}{n^2}} \bigg)$$
This is what I managed to do:
$$ e^{\frac{1}{n^2}} + e^{\frac{1}{n^2}} + ... + e^{\frac{1}{n^2}}
\le e^{\frac{1}{n^2}} + e^{\frac{2}{n^2}} + ...e^{\frac{n}{n^2}} \le
e^{\frac{n}{n^2}} + e^{\frac{n}{n^2}} + ... e^{\frac{n}{n^2}}$$
$$ n e^{\frac{1}{n^2}}
\le \displaystyle\sum_{k = 1} ^ n e ^{\frac{k}{n^2}} \le
ne^{\frac{1}{n}}$$
$$ -n e^{\frac{1}{n}}
\le - \displaystyle\sum_{k = 1} ^ n e ^{\frac{k}{n^2}} \le
- n e^{\frac{1}{n^2}}$$
$$ n - n e^{\frac{1}{n}}
\le n - \displaystyle\sum_{k = 1} ^ n e ^{\frac{k}{n^2}} \le
n - n e^{\frac{1}{n^2}}$$
Here I found that the limit of the left-hand side is equal to $-1$, while the limit of the right-hand side is $0$. So I got that:
$$-1 \le n - \displaystyle\sum_{k = 1} ^ n e ^{\frac{k}{n^2}} \le 0$$
And I cannot draw a conclusion about the exact limit. What should I do?
| As @Andrei already answered, you face a geometric sum, that is to say that
$$\sum_{k = 1} ^ n e ^{\frac{k}{n^2}}=\sum_{k = 1} ^ n \left(e ^{\frac{1}{n^2}}\right)^k=\frac{e^{\frac{1}{n^2}} \left(e^{\frac{1}{n}}-1\right)}{e^{\frac{1}{n^2}}-1}$$ Now, expanding the exponentials as Taylor series, multiplying for the numerator and then long division, you should get
$$\sum_{k = 1} ^ n e ^{\frac{k}{n^2}}=n+\frac{1}{2}+\frac{2}{3 n}+O\left(\frac{1}{n^2}\right)$$
$$ n - \displaystyle\sum_{k = 1} ^ n e ^{\frac{k}{n^2}}=-\frac{1}{2}-\frac{2}{3 n}+O\left(\frac{1}{n^2}\right) $$ which shows the limit and also how it is approached.
Edit
In comments, @marty cohen made a good point.
If we consider
$$n-\sum_{k = 1} ^ {n^a} e ^{\frac{k}{n^2}}=n-\frac{e^{\frac{1}{n^2}} \left(e^{n^{a-2}}-1\right)}{e^{\frac{1}{n^2}}-1}$$ it will not converge for any $a \neq 1$ since it will be
$(n-n^a+\cdots)$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3509406",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 3,
"answer_id": 2
} |
Find the limit of $a_n = n \sqrt{n}(\sqrt{n + 1} - a\sqrt{n} + \sqrt{n - 1})$ given that the sequence $(a_n)$ is convergent. I am given the sequence:
$$a_n = n\sqrt{n}(\sqrt{n + 1} - a\sqrt{n} + \sqrt{n - 1})$$
with $n \in \mathbb{N}^*$ and $a \in \mathbb{R}$. I have to find the limit of $(a_n)$ given that the sequence $(a_n)$ is convergent.
This is what I did:
We have:
$$a_n = n\sqrt{n}(\sqrt{n + 1} - a\sqrt{n} + \sqrt{n - 1})$$
$$a_n = n\sqrt{n^2+n}-an^2+n\sqrt{n^2-n}$$
$$a_n = n^2\sqrt{1+\dfrac{1}{n}} + n^2 \sqrt{1 - \dfrac{1}{n}} - an^2$$
$$a_n = n^2 \bigg ( \sqrt{1 + \dfrac{1}{n}} + \sqrt{1-\dfrac{1}{n}} - a \bigg )$$
The only way we could have $a_n$ convergent is if we would have the limit result in an indeterminate form. In this case, we need $\infty \cdot 0$. So we have:
$$\lim_{n \to \infty} \bigg ( \sqrt{1 + \dfrac{1}{n}} + \sqrt{1-\dfrac{1}{n}} - a \bigg ) = 0$$
And from that we can conclude that:
$$a = 2$$
So now that I found $a$, I must find the limit of the sequence. So this limit:
$$\lim_{n \to \infty} a_n = \lim_{n \to \infty} n\sqrt{n}(\sqrt{n + 1} - 2\sqrt{n} + \sqrt{n - 1})$$
I tried this:
$$\lim_{n \to \infty} a_n = \lim_{n \to \infty} n\sqrt{n^2+n}-2n^2+n\sqrt{n^2-n}$$
$$= \lim_{n \to \infty}( n\sqrt{n^2+n} - n^2) + \lim_{n \to \infty} (n\sqrt{n^2-n} -n^2)$$
And then I multiplied both of those limits with its respective conjugate, but after my calculations, it still results in an indeterminate form, only it's $\infty - \infty$ this time.
So, if my previous calculations aren't wrong, my question is how can I find this limit:
$$\lim_{n \to \infty} n\sqrt{n}(\sqrt{n + 1} - 2\sqrt{n} + \sqrt{n - 1})$$
| You could have done everything a bit faster. Starting from what you wrote
$$a_n = n^2 \bigg ( \sqrt{1 + \dfrac{1}{n}} + \sqrt{1-\dfrac{1}{n}} - a \bigg )$$ let $x=\frac 1 n$, use the binomial expansion or Taylor series, replace $x$ by $\frac 1n$ to get
$$a_n=(2-a) n^2-\frac{1}{4}-\frac{5}{64 n^2}+O\left(\frac{1}{n^4}\right)$$ which gives the results $a=2$, the limit and how the limit is approached.
For your curiosity, using @Peter Foreman's result for $a=2$
$$a_n =-\sum_{k=0}^\infty\frac{\binom{4k+1}{2k}}{4^{2k+1}(k+1)n^{2k}}=\left( \sqrt{2\sqrt{1-\frac{1}{n^2}}+2}-2\right) n^2$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3510658",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 3
} |
Find $a$, $b$ for $z = \frac{1}{(a + ib)^3}$ to be on negative real axis
For what values of $a$ and $b$ does
\begin{equation}
z = \frac{1}{(a + ib)^3}
\end{equation}
lie on the negative real axis. Hence, or otherwise, find an expression in terms of $a$ only for $\lvert z \rvert$ when $\operatorname{arg}(z) = - \pi$.
How would I approach this question?
| For $z$ to lie on the negative real axis, we have
$$z = \frac{1}{(a + ib)^3}=-r=re^{i(1+2n)\pi}$$
where $r>0$. Rearrange to get
$$a+bi= \frac1{\sqrt[3]r} e^{-i\frac{(1+2n)\pi}3}
=\frac1{\sqrt[3]r} \left( \cos\frac{1+2n}3\pi - i\sin\frac{1+2n}3\pi \right)
$$
with $n=0,1,2$. Then,
$$a= \frac1{\sqrt[3]r} \cos\frac{1+2n}3\pi,\>\>\>\>\>
b= -\frac1{\sqrt[3]r} \sin\frac{1+2n}3\pi$$
Evalute at $n=0,1,2$ to obtain the values for $(a,b)$
$$(a.b)=( -c,0),\>(c,\pm\sqrt3c)$$
where $c$ represents any positive real number.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3511850",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
What is the error calculating the sum of this series? I need to determine if $$\sum_{n=1}^\infty \frac{2^n-1}{4^n}$$ converges, in which case I must also find its sum, or diverges. This is my approach:
$i$) $$\frac{2^n-1}{4^n}=\frac{2^n}{4^n}-\frac{1}{4^n}=\frac{2^n}{2^{2n}}-\frac{1}{4^n}=\frac{1}{2^n}-\frac{1}{4^n}$$
Then $$\sum_{n=1}^\infty \frac{2^n-1}{4^n}=\sum_{n=1}^\infty \left(\frac{1}{2^n}-\frac{1}{4^n}\right)=\sum_{n=1}^\infty \frac{1}{2^n}-\sum_{n=1}^\infty \frac{1}{4^n}$$
$$=\sum_{n=1}^\infty \frac{1}{4} \left(\frac{1}{2}\right)^{n-1}-\sum_{n=1}^\infty \frac{1}{16}\left(\frac{1}{4}\right)^{n-1}$$
$ii$) We found our series can be written as the difference of two geometric series with $|r|<1$, and the solution is
$$=\sum_{n=1}^\infty \frac{1}{4} \left(\frac{1}{2}\right)^{n-1}-\sum_{n=1}^\infty \frac{1}{16}\left(\frac{1}{4}\right)^{n-1}=\frac{\frac{1}{4}}{1-\frac{1}{2}}-\frac{\frac{1}{16}}{1-\frac{1}{4}}=\frac{5}{12}$$
I have given this series to an online series calculator, and it claims the result is not $\frac{5}{12}$ but $\frac{2}{3}$, and yet I can not find the mistake in my procedure. Can anyone point it out to me?
EDIT: Please remember, I'm asking more than "what is the right way to solve this problem"; I'm asking precisely what was the mistake I made. Correlated, but not the same.
| My short solution:
$$\sum_{n=1}^\infty \frac{2^n-1}{4^n}=\sum_{n=1}^\infty \left[\left(\frac{1}{2}\right)^n-\frac{1}{4^n}\right]=1-\frac13=\frac 23$$
PS: I have applied a criteria of a series geometric when $|x|<1$. See the link http://mathworld.wolfram.com/GeometricSeries.html formula number (9).
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3512982",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
Finite Derivative of $\frac{x^3-1}{x+1}$ I have the following qeustion:
Evaluating the front, back and central derivative I just need to plug it into the corresponding:
$$D_{+}(x)=\frac{f(x+h)-f(x)}{h},D_{-}(x)=\frac{f(x)-f(x-h)}{h},D_{c}(x)=\frac{f(x+h)-f(x-h)}{2h}?$$
Without choosing some point or value for $h$?
For the least square: I first find data points $$(0,-1),(1,0),(2,2.33),(3,6.5),(4,12.6),(5,20.66),(6,30.71)$$
And find $ax^3+bx^2+cx+d$ using least square and then derive it using front/back/central derivative?
| You do the calculations indicated, as tedious as they might be!
Since $f(x)= \frac{x^3- 1}{x+ 1}$, $f(x+ h)= \frac{(x+ h)^3- 1}{x+ h+ 1}= \frac{x^3+ 3x^2h+ 3xh^2+ h^3- 1}{x+ h+ 1}$.
So $f(x+ h)- f(x)= \frac{(x+ h)^3- 1}{x+ h+ 1}- \frac{x^3- 1}{x+ 1}= \frac{x^3+ 3x^2h+ 3xh^2+ h^3- 1}{x+ h+ 1}- \frac{x^3- 1}{x+ 1}$.
To do that subtraction, we need to get the "common denominator", $(x+ h+ 1)(x+ 1)$. Multiply both numerator and denominator of the first fraction by $x+ 1$ and multiply both numerator and denominator of the second fraction by $x+ h+ 1$:
\begin{align}
&\frac{(x^3+ 3x^2h+ 3xh^2+ h^3- 1)(x+ 1)}{(x+ h+ 1)(x+ 1)}- \frac{(x^3- 1)(x+ h+ 1)}{(x+ h+ 1)(x+ 1)}
\\
&= \frac{x^4+ 3x^3h+ 3x^2h^2+ xh^3- x+ x^3+ 3x^2h+3xh^2+ h^3- 1 }{(x+ h+ 1)(x- 1)}- \frac{(x^4+hx^3+x^3- x- h- 1)}{(x+ h+ 1)(x+ 1)}
\\
&= \frac{2x^3h+ 3x^2h+h+ 3x^2h^2+ 3xh^2+ xh^3+ h^3}{(x+ h- 1)(x- 1)}
\end{align}
Notice that every term in the numerator has an "h" factor so dividing by h cancels:
$$\frac{f(x+h)- f(x)}{h}= \frac{2x^3+ 3x^2+1+ 3x^2h+ 3xh+ xh^2+ h^2}{(x+ h+ 1)(x- 1)}$$
And now we can take the limit as goes to $0$ by just setting $h$ equal to $0$: $$\frac{2x^3+ 3x^2+1}{(x+ 1)^2}=2x+\frac{-x^2-2x+1}{(x+ 1)^2}=2x-1+\frac2{(x+ 1)^2}.$$
Test this with the alternative form $f(x)=\frac{(x^3+1)-2}{x+1}=x^2-x+1-\frac2{x+1}$, so that $f'(x)=2x-1+\frac{2}{(x+1)^2}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3514500",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Find sides of triangle, and radius of inscribed circle
$AD = x, \angle A = 60^{\circ}$
I was asked to find $BC$ when $BD=4, CF=2$
I tried using law of cosines,
$(4+x)^2 + (2+x)^2 - 2(4+x)(2+x) \cos 60^{\circ} = BC^2$
I get $x^2+6x+12=BC^2$
But the answer is $BC= 6, x^2+6x-24=0$
Also i was asked to find $\frac{\triangle ADF}{AG.AE}$.
I found $\triangle ADF = \frac{\sqrt3}{4}x^2$. But how to find $AG.AE$? Some hints anyone?
| In the standard notation: $$BD=\frac{a+c-b}{2}$$ and $$CF=\frac{a+b-c}{2}.$$
Thus,
$$BC=BD+CF=6.$$
Also, $$\frac{S_{\Delta ADF}}{AG\cdot AE}=\frac{\frac{1}{2}AD\cdot AF\cdot\sin60^{\circ}}{AF^2}=\frac{\sqrt3}{4}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3515619",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Equations reducible to homogeneous form Solve the following differential equation:
$$(2x+y-3)dy=(x+2y-3)dx$$
I have tried this & I found:
$$(y-x)^3=c(y+x+2)$$
But in my book the answer is:
$$(y-x)^3=c(y+x-2)$$
Please tell which one is correct..
| $$ (x+2y-3)dx=(2x+y-3)dy$$
The two constants of $-3$ are keeping the equation from being homogeneous. We can eliminate the constants by an appropriately chosen substitution:
$$X=x+a,\quad Y=y+b$$and let's see if we can choose suitable values for $a$ and $b$.
$$ (X - a + 2Y - 2b - 3)dX = (2X-2a + Y - b - 3)dY $$
It's easy to find what the "suitable" values should be, since we just want them to cancel out the constants! So we solve the following system of equations:
$$\begin{align*}
3 &= -a - 2b\\
3 &= -2a - b
\end{align*}$$
I'll let you verify that the correct values are $a=-1$ and $b=-1$. This leaves us with
$$(X + 2Y)dX = (2X+Y)dY$$
Now let $Y=VX$, and hence $dY = VdX + XdV$.
$$\begin{align*}
X(1 + 2V)dX &= X(2+V)(VdX+XdV) \\
X(1-V^2)dX &= X^2(2+V)dV\\
\dfrac{1}{X}dX &= \dfrac{2+V}{1-V^2}dV\\
\log\left|X\right| + C_1 &= \dfrac{1}{2}(-3 \log\left|1-V\right|+\log\left|1+V\right|)\\
C_2X^2 &= \dfrac{1+V}{(1-V)^3}\\
C_2X^2(1-V)^3 &= 1+V\\
C_2X^2(1-\dfrac{Y}{X})^3 &= 1+\dfrac{Y}{X}\\
C_2\dfrac{(X-Y)^3}{X}&= 1+\dfrac{Y}{X}\\
C_2(X-Y)^3 &= X + Y\\
C_2(x-y)^3 &= x + y - 2
\end{align*}$$
So your book has the correct solution.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3515988",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Integrate $\int \frac{x}{x^4+4}dx$ I'm having trouble solving the integral
$\int \frac{x}{x^4+4}dx$
I already tried the following but, I'm getting stuck after step 3.
$\int \frac{x}{(x^2)^2+4}dx $
$u = x^2 , du = 2x dx, \frac{du}{2}=xdx$
$\frac{1}{2}\int \frac{1}{u^2+4}du$
| $$I=\int \frac{x}{x^4+4}dx= \frac{1}{2} \int \frac{dt}{4+t^2}=\frac{1}{2} \tan^{-1} t/2+C=\frac{1}{4} \tan^{-1} x^2/2+C$$ Here, we took $x^2=t \implies 2xdx=dt$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3516605",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Subsets and Splits