Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
|---|---|---|
Best linear unbiased estimates Problem is : Model given $$Y_i=ib+ e \quad; \, i=1,2,3$$
Mean is $0$ and variance is $\sigma, 2\sigma, 3\sigma$. What will be best linear unbiased estimator for $b$?
Answer given in book is $(y_1+y_2+y_3)/6$? this is unbiased I know, but is it correct?
| (W)OLS estimator is the BLUE by the Gauss-Markov theorem, which is given by the normal equation (weighted by the inverse of the noise covariance matrix)
$\hat{b}=(X^T\Sigma^{-1}X)^{-1}X^T\Sigma^{-1}y$
Here we have
$X=
\left[ {\begin{array}{cc}
1 & 1 \\
1 & 2 \\
1 & 3 \\
\end{array} } \right]
$
,
$y=
\left[ {\begin{array}{c}
y_1 \\
y_2 \\
y_3 \\
\end{array} } \right]
$
and the noise covariance matrix
$\Sigma=
\left[ {\begin{array}{ccc}
\sigma & 0 & 0 \\
0 & 2\sigma & 0 \\
0 & 0 & 3\sigma \\
\end{array} } \right]$ (by homoscedasicity and independent noise assumption, all the off-diagonal entries are zeros)
s.t.,
$X^T\Sigma^{-1}X=
\left[ {\begin{array}{cc}
\frac{11}{6\sigma} & \frac{3}{\sigma} \\
\frac{3}{\sigma} & \frac{6}{\sigma} \\
\end{array} } \right]
$
$(X^T\Sigma^{-1}X)^{-1}=
\left[ {\begin{array}{cc}
3\sigma & -\frac{3\sigma}{2} \\
-\frac{3\sigma}{2} & \frac{11\sigma}{12} \\
\end{array} } \right]
$
so, we have
$(X^T\Sigma^{-1}X)^{-1}(X^T\Sigma^{-1}y)=
\left[ {\begin{array}{cc}
3\sigma & -\frac{3\sigma}{2} \\
-\frac{3\sigma}{2} & \frac{11\sigma}{2} \\
\end{array} } \right]
%
\left[ {\begin{array}{c}
\frac{y_1}{\sigma}+\frac{y_2}{2\sigma}+\frac{y_3}{3\sigma} \\
\frac{y_1}{\sigma}+\frac{y_2}{\sigma}+\frac{y_3}{\sigma} \\
\end{array} } \right]
= \left[ {\begin{array}{c}
\frac{1}{2}(3y_1 -y_3) \\
\frac{1}{12}(-7y_1+ 2y_2 + 5y_3) \\
\end{array} } \right]
$
s.t., $\hat{b_0}=\frac{1}{2}(3y_1 -y_3)$ (intercept) and
$\hat{b_1}=\frac{1}{12}(-7y_1+ 2y_2 + 5y_3)$.
The above derivation was for the WOLS coefficients in the general settings (with a bias term). If we drop the bias term in $X$, we shall have
$b = (X^T\Sigma^{-1}X)^{-1}(X^T\Sigma^{-1}y)=\frac{\sigma}{6}.\frac{y_1+y_2+y_3}{\sigma}=\frac{y_1+y_2+y_3}{6}$
Also, another way, if we want to fit WOLS without an intercept in 1D, we need to minimize the SSE (or MSE) $E = \sum\limits_i e_i^2 = \sum\limits_i\frac{(y_i-\hat{y_i})^2}{\sigma_i^2}=\sum\limits_i\frac{(y_i-ib)^2}{\sigma_i^2}$ w.r.t. $b$.
(here sum of errors $\sum\limits_i e_i = 0$ (with the assumption $\textbf{e} \sim N(0,\sigma^2)$)
Setting $\frac{\partial E}{\partial b} = 2\sum\limits_i\frac{(y_i-ib )(-i)}{\sigma_i^2}=0$,
we have $b=\frac{\sum\limits_{i=1}^{3}\frac{iy_i}{\sigma_i^2}}{\frac{\sum\limits_{i=1}^{3}i^2}{\sigma_i^2}}=\frac{\frac{y_1}{\sigma}+\frac{2y_2}{2\sigma}+\frac{3y_3}{3\sigma}}{\frac{1^2}{\sigma}+\frac{2^2}{2\sigma}+\frac{3^2}{3\sigma}}=\frac{y_1+y_2+y_3}{6}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4033950",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Proving that $a^{4}+b^{3}+c^{2}\ge a^{3}+b^{3}+c$ for positive real $a,b,c$, $abc\ge 1$ I have been able to simplify the problem. Since $a^{4}\ge a^{3}$ iff $a\ge 1$ (and analogously for $b,c$), $a^{4}+b^{3}+c^{2}\ge a^{3}+b^{3}+c$ holds iff $a+b+c\ge 3$. It is thus sufficient to prove $a+b+c\ge 3$.
I am not sure how to proceeds from there, however.
All insights are really appreciated.
| I think maybe the question should be $a^4+b^3+c^2\ge a^3+b^2+c$.
By AM-GM,
$a^4+a^4+a^4+1\ge 4a^3\Rightarrow a^4\ge \frac{4a^3-1}{3}$
Similarly, $b^3\ge \frac{3b^2-1}{2}$ , $c^2\ge 2c-1$
Hence, $a^4+b^3+c^2\ge \frac{1}{6}(8a^3+9b^2+12c-11)$
So we only need to prove that
$8a^3+9b^2+12c-11\ge 6a^3+6b^2+6c$ , which is equivalent to
$2a^3+3b^2+6c\ge 11$
By AM-GM, $2a^3+3b^2+6c=a^3+a^3+b^2+b^2+b^2+c+c+c+c+c+c\ge 11\sqrt[11]{a^6b^6c^6}\ge 11$
Hence, we are done.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4034683",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Find the expectation and variance of X A pdf is defined as
\begin{equation}
f(x)=
\begin{cases}
C(x+\frac{3}{2}),\quad0<x<2\\
0,\quad\quad\quad\quad\text{otherwise}
\end{cases}
\end{equation}
*
*Find the value of C.
*Find the expectation and variance of X.
*Find the expectation of random variable $Z=\frac{X}{2X+3}$
What I tried:
*
*Finding C is straightforward as we just need to make sure that $\int^\infty_{-\infty}f(x)dx=1$.
\begin{align*}
1&=\int^2_0C(x+\frac{3}{2})dx\\
&=\left[\frac{C}{2}x^2+\frac{3}{2}Cx\right]^2_0\\
&=\left[\frac{C}{2}(2)^2+\frac{3}{2}C(2)\right]-0\\
&=5C\\
&C=\frac{1}{5}
\end{align*}
*Using $\mathbb{E}[X]=\int^\infty_{-\infty}xf(x)dx$,
\begin{align*}
\mathbb{E}[X]&=\int^2_0x\left(\frac{1}{5}\left(x+\frac{3}{2}\right)\right)dx\\
&=\left[\frac{1}{15}x^3+\frac{3}{20}x^2\right]^2_0\\
&=\left[\frac{1}{15}(2)^3+\frac{3}{20}(2)^2\right]^2_0-0\\
&=\frac{17}{15}
\end{align*}
Variance is $$\text{Var}[x]=\mathbb{E}[X^2]-\mathbb{E}[X]^2=\mathbb{E}[X(X-1)]-\mathbb{E}[X]-\mathbb{E}[x]^2$$
$$\mathbb{E}[X(X-1)]=\int^2_0x(x-1)(\frac{x}{5}+\frac{3}{10})dx=\frac{7}{15}$$
Then, Var$[X]=\frac{7}{15}-\frac{17}{15}-(\frac{17}{15})^2=-\frac{439}{225}$
My concern is that the variance is negative, which should not be the case. Can I get some pointers on where I went wrong? Thank you.
*This one is simple. I just need to find $\int^2_0\frac{x}{2x+3}f(x)dx$
| Your $C=1/5$ and $E(x)=17/15$ are correct, then $$E(x^2)=\frac{1}{5}\int_{0}^{2} x^2(x+3/2)dx=\frac{8}{5}$$
$$Var=E(x^2)-(E(x))^2=\frac{8}{5}-(\frac{17}{15})^2=\frac{71}{225}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4040555",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Evaluate $\frac{4}{\sin^2 20^\circ} - \frac{4}{\sin^2 40^\circ} + 64\sin^2 20^\circ$ Evaluate the following expression:$$\frac{4}{\sin^2 20^\circ} - \frac{4}{\sin^2 40^\circ} + 64\sin^2 20^\circ$$
I tried combining the whole into a single fraction and using double-angle identity, product/sum to sum/product, but it didn't work.
| Formulae for $\sin4x,\,\sin5x$ provide another solution. Let $S:=\sin^220^\circ\ne\sin^260^\circ=\tfrac34$ so$$\begin{align}0&=\frac{\sin^2100^\circ-\sin^280^\circ}{S}\\&=(5-20S+16S^2)^2-16(1-S)(1-2S)^2\\&=256S^4-576S^3+432S^2-120S+9\\&=(4S-3)(64S^3-96S^2+36S-3)\\\implies64S^3-96S^2+36S-3&=0\\\implies\frac{4}{S}-\frac{1}{S(1-S)}+64S&=\frac{3-4S+64S^2-64S^3}{S(1-S)}\\&=32.\end{align}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4044022",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
} |
Determine the number of integer solutions of the equation $ x_{1}+x_{2}+x_{3}+x_{4}=20 $ Determine the number of integer solutions of the equation
$$
x_{1}+x_{2}+x_{3}+x_{4}=20
$$
under the restrictions:
$$
\begin{array}{l}
2 \leq x_{1} \leq 6 \\
0 \leq x_{2} \leq 5 \\
2 \leq x_{3} \leq 8 \\
2 \leq x_{4} \leq 6
\end{array}
$$
I did $x_1=y_1+2$ then $0\leq y_1 \leq4$
$x_2=y_2$ then $0\leq y_2 \leq5$
$x_3=y_3+2$ then $0\leq y_3 \leq6$
$x_4=y_4+2$ then $0\leq y_4 \leq4$
So $y_1+y_2+y_3+y_4 \leq 19$
So the number of solutions is
$\sum_{k=0}^{19}\left(\begin{array}{c}
k+3 \\
k
\end{array}\right)=\sum_{k=0}^{19}\left(\begin{array}{c}
k+3 \\
3
\end{array}\right)$
is right?
| The combined values of the maximum value for the $x$'s is 25. We can then take four part ordered partitions of $5$ (including zeros) and subtract each of those from the maximum.
The first set of partitions is $(5,0,0,0)$ with the $5$ being in four potential locations.
You cannot subtract $5$ from $x_1$ or $x_4$ so there are $2$ ways of doing this.
The second set of partitions is
$(4,1,0,0)$. There are $12$ ways of ordering these numbers.
The third set of partitions is $(3,2,0,0)$.There are $12$ ways of ordering these numbers.
The fourth set is $(3,1,1,0)$.There are $12$ ways of ordering these numbers.
The fifth set is $(2,2,1,0)$.There are $12$ ways of ordering these numbers.
Finally the sixth and last set is $(2,1,1,1)$. There are four locations for the $2$. Therefore there are $4$ ways of ordering these numbers.
$2+12+12+12+12+4=54$
There are $54$ solutions to the equation $x_1+x_2+x_3+x_4=20$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4047798",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Avoiding brute force: determining when a specific polynomial in $\mathbb{Q}[x]$ is an integer for any integer $x$ I have to prove that $\frac{1}{5}n^5+\frac{1}{3}n^3+\frac{7}{15}n$ is an integer for any $n$. I solved this by brute-force, exhausting all the possibilities methods. I was wondering if there was a way to solve this at-a-glance with some sort of theory? Because
Although my answer is correct, I was apparently supposed to use some theory to answer this question.
I started by putting everything on common denominator and factoring out $n$:
$$\frac{1}{5}n^5+\frac{1}{3}n^3+\frac{7}{15}n = \frac{n(3n^4+5n^2+7)}{15}$$
then I proceeded to plug in $\pm1 ,\pm2, \pm3 ,..., \pm7$ into $3n^4+5n^2+7 \pmod {3,5,\text{ or }15}$
$$3(\pm 1)^4 +5(\pm 1)^2 +7 \equiv 15 \equiv 0 \pmod{15}$$
$$3(\pm 2)^4 +5(\pm 2)^2 +7 \equiv 75 \equiv 5\cdot 15 \equiv 0 \pmod{15}$$
$$3(\pm 3)^4 +5(\pm 3)^2 +7 \equiv 295 \equiv 0 \pmod{5} \text{ while } n \equiv 0 \pmod 3$$
$$3(\pm 4)^4 +5(\pm 4)^2 +7 \equiv 855 \equiv 57\cdot 15\equiv 0 \pmod{15}$$
$$3(\pm 5)^4 +5(\pm 5)^2 +7 \equiv 2007 \equiv 0 \pmod{3}\text{ while } n \equiv 0 \pmod 5$$
$$3(\pm 6)^4 +5(\pm 6)^2 +7 \equiv 4075 \equiv 0 \pmod{5}\text{ while } n \equiv 0 \pmod 3$$
$$3(\pm 7)^4 +5(\pm 7)^2 +7 \equiv 7455 \equiv 497 \cdot 15\equiv 0 \pmod{15}$$
to conclude that if $n\equiv 0 \pmod {15}$ then $\frac{n}{15}$ is an integer from which $\frac{1}{5}n^5+\frac{1}{3}n^3+\frac{7}{15}n$ is an integer,
and if $n \not\equiv 0 \pmod {15}$ then either $3n^4+5n^2+7 \equiv 0 \pmod{15}$ or $n \equiv 0 \pmod 3$ and $ 3n^4+5n^2+7 \equiv 0 \pmod{5}$ or $n \equiv 0 \pmod 5$ and $ 3n^4+5n^2+7 \equiv 0 \pmod{3}$ from which the statement is clearly true.
Is there a less brute-force-ish way of concluding this? Is there some theory I should be using that would cause me to not be excessively lengthy in calculation if I were given different, larger numbers than $15$?
| To show that $15$ divides $n(3n^4+5n^2+7)$, show that $3$ and $5$ do.
$3$ does because either $3$ divides $n$ or $n^2\equiv1\pmod3$ by Fermat's little theorem,
in which case $3n^4+5n^2+7\equiv5n^2+7\equiv5+7=12\equiv0\pmod3.$
$5$ does because either $5$ divides $n$ or $n^4\equiv1\pmod5$ by Fermat's little theorem,
in which case $3n^4+5n^2+7\equiv3n^4+7\equiv3+7=10\equiv0\pmod5$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4049277",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 1
} |
Remainder When Divided By 70 : Remainder Theorem Problem If $x$ = $16^3$ + $17^3$ + $18^3$ + $19^3$ , then $x$ divided by $70$ leaves a remainder of?
I tried to solve this problem by using the remainder theorem which states that remainder of $$Rem[\frac{a+b+c+....}{x}] = Rem[\frac{a}{x}] + Rem[\frac{b}{x}] + Rem[\frac{c}{x}] + .... $$ where 'Rem' means 'Remainder Of'.
Using this same logic I can write $$Rem[\frac{16^3+17^3+18^3+19^3}{70}] = Rem[\frac{16^3}{70}] + Rem[\frac{17^3}{70}] + Rem[\frac{18^3}{70}] + Rem[\frac{19^3}{70}] = 18 +13+11+69 = 111 \Rightarrow Rem[\frac{111}{70}] = 41$$
So my answer comes up as 41. Can someone please explain to me what I might be doing wrong or what I have understood wrong?
| I made a mistake in my solution with the remainders. Below is the correct approach and answer :-
Using this same logic I can write $$Rem[\frac{16^3+17^3+18^3+19^3}{70}] = Rem[\frac{16^3}{70}] + Rem[\frac{17^3}{70}] + Rem[\frac{18^3}{70}] + Rem[\frac{19^3}{70}] = 36 +13+22+69 = 140 \Rightarrow Rem[\frac{140}{70}] = 0$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4049737",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 2
} |
Find the area of the space enclosed by points $(x, y)$ such that fullfill the following condition: $9x^3+9x^2y-45x^2=4y^3+4xy^2-20y^2$ Find the area of the space enclosed by points $(x, y)$ such that fullfill the following condition: $9x^3+9x^2y-45x^2=4y^3+4xy^2-20y^2$
My solution goes as follows:
Factorizing we have:
$(9x^2-4y^2)(x+y-5)=0$
$(3x+2y)(3x-2y)(x+y-5)=0$
Hence we get that the vertices of the triangle are points $(-10, 15), (0, 0)$ and $(2, 3)$ and from this we conclude that the area is $30$.
My question is how was I supposed to think of the particular factorization? Can you propose an intuitive method?
| Step by step, we have:
$$9(x^3+x^2y-5x^2)=4(y^3+xy^2-5y^2)$$
$$\Rightarrow 9x^2(x + y - 5) = 4y^2 (y + x - 5)$$
and moving everything to one side gives $(9x^2-4y^2)(x + y - 5) = 0$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4051731",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Question of rank of a matrix $4\times 6$ Let $a,b,c,d$ be complex numbers. Find the rank of the following matrix
\begin{align*}
\begin{pmatrix}
2& -3& 6& 0& -6& a\\
-1& 2& -4& 1& 8& b\\
1& 0& 0& 1& 6& c\\
1& -1& 2& 0& -1& d\\
\end{pmatrix}
\end{align*}
Idea: Proceed the elementary row operations and reduce to the raw Echelon form then the number of nonzero rows are the rank of the matrix.
Solution
\begin{align*}
&\begin{pmatrix}
2& -3& 6& 0& -6& a\\
-1& 2& -4& 1& 8& b\\
1& 0& 0& 1& 6& c\\
1& -1& 2& 0& -1& d\\
\end{pmatrix}\overset{r_1 \xrightarrow{} r_4} {}
\begin{pmatrix}
1& -1& 2& 0& -1& d\\
-1& 2& -4& 1& 8& b\\
1& 0& 0& 1& 6& c\\
2& -3& 6& 0& -6& a\\
\end{pmatrix} \underset{\overset{r_1(-2) + r_4 \xrightarrow{}r_4{}}{}}{\overset{r_1 + r_2 \xrightarrow{}r_2,\, (-1)r_1 + r_3 \xrightarrow{}r_3}{}}\\
&\begin{pmatrix}
1& -1& 2& 0& -1& d\\
0& 1& -2& 1& 7& b+d\\
0& 1& -2& 1& 7& c-d\\
0& -1& 2& 0& -4& a-2d\\
\end{pmatrix} \underset{\overset{r_2 + r_4 \xrightarrow{}r_4{}}{}}{\overset{r_2(-1) + r_3 \xrightarrow{}r_3}{}}\begin{pmatrix}
1& -1& 2& 0& -1& d\\
0& 1& -2& 1& 7& b+d\\
0& 0& 0& 0& 0& c-b-2d\\
0& 0& 0& 1& 3& a+b-d\\
\end{pmatrix}\overset{r_3 \xrightarrow{}\, r_4} {}\\
& \begin{pmatrix}
1& -1& 2& 0& -1& d\\
0& 1& -2& 1& 7& b+d\\
0& 0& 0& 1& 3& a+b-d\\
0& 0& 0& 0& 0& c-b-2d\\
\end{pmatrix}
\end{align*}
Answer: If $c-b-2d = 0$ the rank of the matrix is 3, If $c-b-2d \neq 0$ the rank of the matrix is 4
This is my answer, am I correct here? I feel it is not enough for an answer. Is there anything I have to do? Any advice would be appreciated.
| Guide:
Your idea is fine.
The question is what is the rank of the matrix, the answer should be a number, of which you have not provided.
Consider two cases:
*
*$c-b-2d=0$.
*$c-b-2d \ne 0$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4054464",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Does the square root function behave the same when calculating complex roots? This is incredibly basic, yet I cannot for the life of me confirm it. Let's say in the topic of complex numbers we are told to calculate the roots of:
$
z^4 = \sqrt 9
$
Are we to assume $ z^4 = 3 $ and calculate 4 roots or should we proceed with " $ z^8 = 9 $ ", and calculate 8 roots ? Since we are not operating in real numbers, the fact that a negative square root can exist in this instance confused me. Do we just take it as +3 since the square root function is specific to non-negative real numbers? I know the professor wanted us to calculate 4 roots, but I would like to make sure.
| $z^4=\sqrt9$ which implies: $z^4=3$ and $z^4=-3$
Solving: $z^4=3$
Let $u=z^2$, then $u^{2}=3 \rightarrow u=\pm \sqrt 3$
Now, $z^2=\sqrt3$ and $z^2=-\sqrt3$. Thus, there are four roots $$z=\pm\sqrt[4]{3}i \,\,\,\&\,\,\, z=\pm\sqrt[4]{3}$$
Solving $z^4=-3$
Let $u=z^2$, thus $u^{2}=-3 \rightarrow u=\sqrt3i \,\,\,\&\,\,\, u=-\sqrt3 i$
Hence, $z^2=\sqrt3i$ and $z^2=-\sqrt3 i$
Let $z=x+iy$, for $x,y \in \mathbb{R}$. Then $z^2=\left(x^2-y^2\right)+i(2xy)$.
Case 1: $z^2=\sqrt 3 i$
Now, the real part equals the real part, and the imaginary part equals the imaginary part. Hence, $$x^2-y^2=0\\ x=\pm y\\x=y$$ and $$2xy=\sqrt 3 \\2x^2=\sqrt 3 \\x=\pm\sqrt\frac{\sqrt 3}{2}\\ x=\pm\frac{\sqrt[4]{3}\sqrt{2}}{2}$$ and plugging back in these $x$-values into $2xy=\sqrt3$, we find that for $x=\frac{\sqrt[4]{3}\sqrt{2}}{2}$, $y=-\sqrt[4]{\frac{3}{4}}$ and for $x=-\frac{\sqrt[4]{3}\sqrt{2}}{2}$, $y=\sqrt[4]{\frac{3}{4}}$.
Hence writing them in terms of $z$, we get: $$z=-\frac{\sqrt[4]{3}\sqrt{2}}{2}+\sqrt[4]{\frac{3}{4}}i,\:z=\frac{\sqrt[4]{3}\sqrt{2}}{2}-\sqrt[4]{\frac{3}{4}}i$$
Case 2:
$z^2=-\sqrt 3 i$. Similarly, for $x,y \in \mathbb{R}$ $x^2-y^2=0\\ x=\pm y\\ x=-y$ and $2xy=-\sqrt 3\\-2x^2=-\sqrt 3\\ \therefore x= \pm \frac{\sqrt[4]{3}\sqrt{2}}{2}$ corresponding to $y=\pm \sqrt[4]{\frac{3}{4}}$. Hence the solution in terms of $z$ are: $$z=-\frac{\sqrt[4]{3}\sqrt{2}}{2}+\sqrt[4]{\frac{3}{4}}i,\:z=\frac{\sqrt[4]{3}\sqrt{2}}{2}-\sqrt[4]{\frac{3}{4}}i$$
Overall, there are $8$ roots to this equation: $$z=\pm\sqrt[4]{3}i \,\,\,\&\,\,\, z=\pm\sqrt[4]{3} \,\,\,\&\,\,\, z=-\frac{\sqrt[4]{3}\sqrt{2}}{2}+\sqrt[4]{\frac{3}{4}}i \,\,\,\&\,\,\,\:z=\frac{\sqrt[4]{3}\sqrt{2}}{2}-\sqrt[4]{\frac{3}{4}}i $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4054625",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
Solving a Polynomial Equation by Factoring. $n^3+12n^2+48n+64$
I know the sum of two cubes formula, $(a+b)(a^2-ab+b^2)$. I'm not sure how to apply it here? Any help would be appreciated.
| Differentiate $f(n) = n^3+12n^2+48n+64$ to get $f'(n) = 3n^2 + 24n + 48 = 3(n^2 + 8n + 16) = 3(n+4)^2$.
Hence $\int f'(n) \ dn = 3 \int (n+4)^2 \ dn = 3 \cdot \left(\frac{(n+4)^3}{3} + C \right) = (n+4)^3 + C' = f(n)$, by the fundamental theorem of calculus. Since $f(-4) = C' \Rightarrow 0 = C'$, $f(n) = (n+4)^3$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4060667",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 8,
"answer_id": 4
} |
Sequence queue - converges or diverges? I have the next sequence queue: $\sum_{n=1}^{\infty}{\frac{n^2-n-1}{n^4+n^2+1}}$. Does the queue converges or diverge?
My attempt:
I have tried to show that $\frac{1}{n^2}>|\frac{n^2-n-1}{n^4+n^2+1}|$ for any $n>0$, and then by the comparation test we get that since $\sum_{n=1}^{\infty}{\frac{1}{n^2}}$ converges, we have that $\sum_{n=1}^{\infty}{|\frac{n^2-n-1}{n^4+n^2+1}|}$ converges too, and by a theorem we have that since $\sum_{n=1}^{\infty}{|\frac{n^2-n-1}{n^4+n^2+1}|}$ $\implies$ $\sum_{n=1}^{\infty}{\frac{n^2-n-1}{n^4+n^2+1}}$ converges. Is that right?
| I continue from where you stopped,
Easy to show that :
$$\frac{n^2-n-1}{n^4+n^2+1} < \frac{n^2-1}{n^4+1} < \frac{n^2}{n^4} < \frac{1}{n^2} $$
When we know that $\sum_{n=1}^{\infty}{\frac{1}{n^2}}$ converges.
Therefore, by comparison test :
$$\underset{converges}{\underbrace{\frac{n^2-n-1}{n^4+n^2+1}}} < \underset{converges}{\underbrace{\frac{1}{n^2}}}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4064710",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
Integration using digamma functions I'm trying to prove that for $x,y,r>0$ the following identity (gotten via CAS) holds:
$$\int_0^x \frac{x^r}{x^r+y^r}\,dy=\frac{x}{2r}\left(\psi\left(\frac{1+r}{2r}\right)-\psi\left(\frac{1}{2r}\right)\right),$$
where $\psi(\cdot)=\frac{\Gamma'(\cdot)}{\Gamma(\cdot)}$ denotes the Digamma function. Does anyone has a clue, how integration can be done this way?
As an additional information following identity, holds also to be true:
$$\int_0^x \frac{y^r}{y^r+x^r}\,dy=\frac{x}{2r}\left(\psi\left(\frac{1}{2r}+1\right)-\psi\left(\frac{1}{2r}+\frac{1}{2}\right)\right),$$
Which was obtained using:
$$\psi(x)=H_{x-1} - \gamma,$$
where $\gamma$ denotes the Euler-Mascheroni constant and $H_x$ denotes the $x^{th}$ harmonic number.
| Just to add a bit to the evaluation provided by @giobrach and @Roman - another way to get Digamma function explicitly.
$$I=\int_0^x \frac{x^r}{x^r+y^r}\,dy\overset{z=\frac{y}{x}}=x\int_0^1 \frac{1}{1+z^r}\,dz=x\int_0^1 \frac{1-z^r}{(1+z^{r})(1-z^r)}\,dz=x\int_0^1 \frac{1-z^r}{1-z^{2r}}\,dz$$
Next we make regularisation:
$$I=lim_{\epsilon\to0} \,I_\epsilon $$ where
$$I_\epsilon=x\int_0^1 \frac{1-z^r}{(1-z^{2r})^{1-\epsilon}}\,dz\overset{t=z^{2r}}=\frac{x}{2r}\int_0^1 \frac{t^{(\frac{1}{2r}-1)}-t^{(\frac{1}{2r}-\frac{1}{2})}}{(1-t)^{1-\epsilon}}\,dt=\frac{x}{2r}\Bigr(B(\frac{1}{2r},\epsilon)-B(\frac{1}{2r}+\frac{1}{2},\epsilon)\Bigr)$$ $B(a,b)$ - Beta function: $B(a,b)=\int_0^1t^{a-1}(1-t)^{b-1}=\frac{\Gamma(a)\Gamma(b)}{\Gamma(a+b)}$
$$I_\epsilon=\frac{x}{2r}\biggl(\frac{\Gamma(\epsilon)\Gamma(\frac{1}{2r})}{\Gamma(\frac{1}{2r}+\epsilon)}-\frac{\Gamma(\epsilon)\Gamma(\frac{1}{2r}+\frac{1}{2})}{\Gamma(\frac{1}{2r}+\frac{1}{2}+\epsilon)}\biggr)$$
Taking into consideration that $\Gamma(\epsilon)=\frac{1}{\epsilon}+O(1)$ at $\epsilon\to0$ we get
$$I=lim_{\epsilon\to0} \,I_\epsilon = lim_{\epsilon\to0}\,\frac{x}{2r}\,\frac{1}{\epsilon}\biggl(\frac{\Gamma(\frac{1}{2r})}{\Gamma(\frac{1}{2r}+\epsilon)}-\frac{\Gamma(\frac{1}{2r}+\frac{1}{2})}{\Gamma(\frac{1}{2r}+\frac{1}{2}+\epsilon)}\biggr)$$ $$=lim_{\epsilon\to0}\,\frac{x}{2r}\,\frac{1}{\epsilon}\biggl(\frac{\Gamma(\frac{1}{2r})}{\Gamma(\frac{1}{2r})+\Gamma \,'(\frac{1}{2r})\epsilon + O(\epsilon^2)}-\frac{\Gamma(\frac{1}{2r}+\frac{1}{2})}{\Gamma(\frac{1}{2r}+\frac{1}{2})+\Gamma\,'(\frac{1}{2r}+\frac{1}{2})\epsilon)+O(\epsilon^2)}\biggr)$$
$$=lim_{\epsilon\to0}\,\frac{x}{2r}\,\frac{1}{\epsilon}\biggl(\frac{1}{1+\log'(\Gamma(\frac{1}{2r}))\epsilon + O(\epsilon^2)}-\frac{1}{1+\log'(\Gamma(\frac{1}{2r}+\frac{1}{2}))\epsilon+O(\epsilon^2)}\biggr)$$
$$=\frac{x}{2r}\,\biggl(\log'(\Gamma(\frac{1}{2r}+\frac{1}{2}))-\log'(\Gamma(\frac{1}{2r}))\biggr)$$
Finally
$$I=\frac{x}{2r}\,\biggl(\Psi(\frac{1}{2r}+\frac{1}{2})-\Psi(\frac{1}{2r})\biggr)$$
were $\Psi(t)=\frac{d}{dt}\log(\Gamma(t))$ - Digamma function
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4068558",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 2
} |
Harvard MIT math question Let P be a point selected uniformly at random in the cube $[0, 1]^3$. The plane parallel to x+y+z = 0
passing through P intersects the cube in a two-dimensional region R. Let t be the expected value of
the perimeter of R. If $t^2$ can be written as $\frac{a}{b}$, where a and b are relatively prime positive integers,compute 100a + b
Let me know if the solution below is right?
Let $x_1,y_1,z_1$ be any point P in the cube.
The plane that passes through the point parallel to x+y+z 0
is ($x+y+z=x_1+y_1+z_1 = t$)
This plane cuts the x axis at say (x,0,0) then x = t and the point is (t,0,0)
This plane cuts the y axis at say (0,y,0) then y = t and the point is (0,t,0)
This plane cuts the z axis at say (0,0,z) then z = t and the point is (0,0,t)
The region of intersection is an equilateral triangle whose perimeter is given to be $P = 3\sqrt{2t^2} = 3\sqrt{2}t$
Was it a fluke that i hit the correct answer or is the rationale correct?
$E(P) = 3\sqrt{2}\int_{0}^{1}\int_{0}^{1}\int_{0}^{\sqrt{3}-x_1-y_1} (x_1+y_1+z_1)dz_1dy_1dx_1 = \frac{11}{2\sqrt{2}}$. After extraction the answer is $12108$
| If you take a unit cube and try to cut it along a plane $x + y + z = r$, then you will notice that the two-dimensional region can be two different shapes.
More precisely, if $r \leq 1$ or $r \geq 2$ (which are symmetric via $x, y, z \mapsto 1 - x, 1 - y, 1 - z$), then the region $R$ is a triangle with side length $\sqrt 2 r$ or $\sqrt 2 (3 - r)$, respectively. Thus $R$ has perimeter $3\sqrt 2 r$ and area $\frac {\sqrt 3} 2 r^2$ when $r \leq 1$ (and symmetrically when $r \geq 2$).
If $1 < r < 2$, then the region $R$ will be a hexagon, whose angles are all $120^\circ$ and whose side lengths are $\sqrt 2(r - 1)$ and $\sqrt 2(2 - r)$, alternatively. In this case, $R$ has perimeter $3\sqrt 2$ and area $\frac{\sqrt 3}2 (-2r^2 + 6r - 3)$.
Since the point $P$ is chosen uniformly in the cube, the probability density of the plane being $x + y + z = r$ is proportional to the area of $R$. Therefore the expected value of the perimeter is calculated as
$$t = \left(2\int_0^1 3\sqrt 2 r \cdot \frac {\sqrt 3} 2 r^2 dr + \int_1^2 3\sqrt 2 \cdot \frac{\sqrt 3}2 (-2r^2 + 6r - 3)dr\right)/\left(2\int_0^1 \frac {\sqrt 3} 2 r^2 dr + \int_1^2 \frac{\sqrt 3}2 (-2r^2 + 6r - 3)dr\right) = \frac {11} 4 \sqrt 2.$$ Therefore $t^2 = \frac{121}{8}$ and the final answer is $12108$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4069687",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Quadratic inequality with Modulus of a Linear Equation : Solution Verification & Clarification Needed $x^2-|5x-3|-x<2$
For this problem I took two cases where in one case $5x-3 \geq 0$ and in second case $5x-3 \lt 0$.
$1st$ Case when $5x-3 \geq 0$ :-
For this case I got the solution set as $x \in [0.6,3+2\sqrt{2}]$.
$2nd$ Case when $5x-3 \lt 0$ :-
For this case I got the solution set as $x \in [-5,0.6]$.
Upon combining both the cases I am getting the solution set as $x \in [-5,3+2\sqrt{2}]$ but in the answer that has been provided for this problem is not including the points and thus the answer given is $x \in (-5,3+2\sqrt{2})$.
What am I doing wrong here that is making my endpoints of the solution inclusive whereas the the solution set provided has got endpoints exclusive. Please help me on this !
Thanks in advance !
| Let me forcus on when $5x-3<0$, we have $x < \frac35$ and
$$x^2-|5x-3|-x<2$$
becomes
$$x^2-(3-5x)-x<2$$
$$x^2+4x-5<0$$
$$(x+5)(x-1)<0$$
Hence we have $-5 < x < 1$ and $x < \frac35$, hence taking intersection, we have $-5<x < \frac35$.
To answer the question, why did we exclude $x=-5$, consider $(x+5)(x-1)<0$, if we substute $x=5$ inside, we get $0<0$ which is not true.
If $5x-3 \ge 0$, we have $x \ge \frac35$ and
$$x^2-|5x-3|-x < 2$$
becomes
$$x^2-5x+3-x < 2$$
$$x^2-6x+1 < 0$$
$$\frac{6-\sqrt{36-4}}{2} < x < \frac{6+\sqrt{36-4}}{2} $$
$$3-2\sqrt2< x < 3+2\sqrt2 $$
Taking intersection with $x \ge \frac35$, we have
$$\frac35 \le x < 3+2\sqrt2$$
Taking union, we have $-5 < x < 3+2\sqrt2$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4070725",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Determine whether the series $\sum_{n=1}^\infty \ln\left(\cos \frac{1}{n}\right)$ is convergent. Determine whether the series $\sum_{n=1}^\infty \ln\left(\cos \frac{1}{n}\right)$ is convergent.
Attempt:
Notice that the $p-$ series $\sum_{n=1}^\infty \frac{1}{n^2}$ with $p=2$ is convergent. Then, we have
\begin{align*}
\lim\limits_{n \to \infty} \frac{\ln\left(\cos \frac{1}{n} \right)}{\frac{1}{n^2}} &= -\frac{1}{2} \cdot \lim\limits_{n \to \infty} \frac{\sin \frac{1}{n}}{\frac{1}{n}} \cdot \frac{1}{\cos \frac{1}{n}} \\
&= -\frac{1}{2}.
\end{align*}
Hence, the limit is equal to $-\frac{1}{2} \ne 0$. Since the $p-$ series is convergent with $p=2$, by the limit comparison test, the series is convergent.
Am I true?
| For every $n\geqslant 1$, $\cos \dfrac{1}{n}$ gets closer and closer to $1$.
That is,
$$\cos \dfrac{1}{n}\approx 1-\dfrac{1}{2n^2}+\dfrac{1}{24n^4}-\dfrac{1}{720n^6}$$ and for large enough $n$, the larger terms go to zero, making the approximation more accurate; as $n$ approaches $\infty$, our polynomial, and the cosine, go to $1$. Thus in the sum
$$\sum_{n\geqslant 1}\ln\left(\cos\dfrac{1}{n}\right)$$ the terms
$$\ln\left(\cos\dfrac{1}{n}\right)\longrightarrow 0,$$ since $\ln 1 = 0$. The approximate value of $$\ln\cos \dfrac{1}{n}\approx -\dfrac{1}{2n^2} - \dfrac{1}{12n^4} - \dfrac{1}{45n^6} - \dfrac{17 }{2520n^8}$$ and the terms go to zero for sufficiently large $n$. Therefore your series converges and is (for fun) approximately
$$-\left[\frac{1}{2}\zeta(2)+\dfrac{1}{12}\zeta (4)+\dfrac{1}{45}\zeta(6)+\dfrac{17}{2520}\zeta(8)\right]$$ though this agrees only to two decimal places with the correct figure $-0.945369$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4071960",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
Finding maxima and minima of $f(x,y) = x^{4} + y^{4} - x^{3}$ Finding maxima and minima of $$f(x,y) = x^{4} + y^{4} - x^{3}$$
I tried solving this question but need help.
In this question, I calculated the partial derivatives of first and second order.
I found
$$f_{x} = 4x^{3} - 3x^{2}\\ f_{y}= 4y^{3} \\ f_{xx} = 12x^{2}-6x \\ f_{yy}= 0$$
On equating $$f_{x} = 0 \hspace{0.5cm} and\hspace{0.5cm} f_{y} = 0 $$ I get the critical points as (0,0) and (0.75,0) but I am not able to understand whether these points constitute a maxima or minima because $$f_{xx}f_{yy} - f_{xy}f_{yx}=0$$
Can somebody please guide?
| $$f(x,y)=x^4+y^4-x^3\\\qquad\quad ~~ ~=x^3(x-1)+y^4$$
Apply $x\to\ +\infty$ and $y=0$, you get $f(x,y)\to +\infty.$
So, global maxima doesn't exist.
Now, we will prove that global minimum exists and is equal to
$$\text{min}[x^4-x^3+y^2]=-\frac{27}{256}.$$
In this case , of course, $y=0$ must be. Then observe that,
$$\begin{align}x^4-x^3+\frac{27}{256}=\frac{1}{256}(256x^4-256x^3+27)=\frac{1}{256}(4 x - 3)^2 (16 x^2 + 8 x + 3)≥0. \end{align}$$
Because,
$\Delta_{\text{half}}=16-16\times 3=-32<0$.
This means, $16x^2+8x+3>0$ for all $x\in\mathbb R.$
Finally, we deduce that
$$\text{min}\left[x^4-x^3+\frac{27}{256}\right]=0$$
at $x=\dfrac 34.$
which follows
$$\text{min}\left[x^4-x^3+y^4\right]=-\frac{27}{256}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4074434",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 1
} |
Mathematical Induction: Can I assume that $P(k-1)$ is also true in the induction hypothesis? Suppose that $y \in \mathbb{R}$, $y \neq 0$ and $y + \frac{1}{y} \in \mathbb{Z}$. Prove using mathematical induction that $y^{n} + \frac{1}{y^{n}}$ is an integer for all $n \geq 1$.
Basis step: The statement is true for $n=1$ since: $y^{1} + \frac{1}{y^{1}} = y^{n} + \frac{1}{y^{n}} \in \mathbb{Z}$.
Induction hypothesis: Assume that $y^{k} + \frac{1}{y^{k}}$ and $y^{k-1} + \frac{1}{y^{k-1}}$ are integers for all $n \geq 1$.
Induction step: We show that $y^{k+1} + \frac{1}{y^{k+1}}$ is true.
$y^{k+1} + \frac{1}{y^{k+1}} = (y^{k} + \frac{1}{y^{k}})(y + \frac{1}{y}) - (\frac{y^{k}}{y} + \frac{y}{y^{k}})$
$y^{k+1} + \frac{1}{y^{k+1}} = (y^{k} + \frac{1}{y^{k}})(y + \frac{1}{y}) - (y^{k-1} + \frac{1}{y^{k-1}})$
Since it is given that $y + \frac{1}{y} \in \mathbb{Z}$ and by the induction hypothesis, $y^{k} + \frac{1}{y^{k}}$, $y^{k-1} + \frac{1}{y^{k-1}} \in \mathbb{Z}$:
$\Rightarrow (\mathbb{Z})(\mathbb{Z})-\mathbb{Z}=\mathbb{Z}$ (Closure property for integers)
Therefore, $y^{n} + \frac{1}{y^{n}}$ is an integer for all $n \geq 1$.
This is the only way I was able to do the proof. However, I want to know if I am allowed to assume that $P(k-1)$ is also true in the induction hypothesis since as far as I know we can only assume that $P(k)$ is true. Would also appreciate feedback and alternative solutions for my proof!
| Here's a way you can see it. Suppose $n$ is a counterexample, so that $y^n+\frac{1}{y^n} \notin \Bbb Z$ for some $n$. Then because every non-empty set of natural numbers has a least element, there must be some least $n$ that works as a counterexample. Therefore, for each $k \lt n, y^k+\frac{1}{y^k} \in \Bbb Z$. Moreover, $n \neq 1, 2$. From here, use your proof.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4076728",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 3
} |
Solving $|2x-5| + |7-2x| =2 $ $$|2x-5| + |7-2x| =2 $$
I approached this problem by making cases for the equation about the critical points i.e. $x=2.5$ and $x=3.5$ :-
*
*1st case, where $x\le2.5$ :-
$$5-2x+2x-7 = 2 \Rightarrow 2\neq 2$$
We can say that no solution exists in this range.
*2nd case, where $2.5<x\le3.5$ :-
$$2x-5+2x-7 = 2 \Rightarrow x=3.5$$
We can say that one solution exists in this range.
*3rd case, where $x>3.5$ :-
$$2x-5+7-2x = 2 \Rightarrow 2=2$$
We can say that infinite solutions exist in this case.
So, as per above calculations, what I can conclude that for all $x\geq2.5 $ solution exists.
But the solution provided in the book says that $|2x-5| + |7-2x| =2 $ will only be valid when $2.5 \leq x \leq 3.5$.
Can someone please help me out on how to decide the solutions and solution ranges for these type of questions?
Thanks in advance !
| You have the info about $7-2x$ completely backwards. I $x \le 3.5$ then $7-2x \ge 0$ and $|7-2x| = 7-2x$.
And if $x >3.5$ then $7-2x < 0$ and $|7-2x| = 2x - 7$.
So case 1: $x <2.5$ then $|2x - 5|+|7-2x| = (5-2x)+(7-2x) = 2$ and $12-4x = 2$ and $x =2.5$ but $x < 2.5$ so that isn't a solution.
Case 2: $2.5 \le x \le 3.5$ and $|2x-5|+|7-2x|=(2x-5)+(7-2x)= 2$ and $2=2$ has infinite solutions.
Case 3: $x > 3.5$ then $|2x-5|-|7-2x| = (5-2x) -(2x-7)= 12 - 4x = 2$ so $x=-2.5$ but $x > 3.5$ so that isn't a solution.
....
Note $|7-2x| = |2x-7|$ so the critical points don't have to mean $x < critical\implies (central stuff) < 0$ and $x > critical \implies (central stuff) > 0$.
It could be the exact opposite. One key thing to note $2x -7$ has a positive coefficient so it has $<> critical \implies <> 0$. But a negative coefficient would imply $<> critical \implies >< 0$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4076871",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 3
} |
How to prove the solution of an inequality is lower bounded? Let $x_{n,k}\in (0,1]$ be any solution of the following inequality
\begin{align}
n(1-x)^2+2kx^2(1-x)^2 -x^2-k(1-x)^4 \leq 0,
\end{align}
where $n, k$ are parameters such that $n\geq k\geq 1$.
I want to prove that there exists a strictly positive constant $c$ (i.e. $c>0$) such that
\begin{align}
x_{n,k} \geq c, \text{ for all } n\geq k \geq 1.
\end{align}
If you want to have a look at their graphs, they are available here thanks to Wolfram Alpha.
I tried a lot of estimations, but failed.
I guess that we should use something special here.
Could someone give me a hint? Thanks!
| Let $f(x) = n(1-x)^2+2kx^2(1-x)^2 -x^2-k(1-x)^4$.
We have, for all $x$ in $(0, 1]$,
\begin{align}
f(x) &\ge k(1-x)^2 + 0 - x^2 - k(1-x)^4 \\
&= k[(1-x)^2 - (1-x)^4] - x^2\\
&\ge (1-x)^2 - (1-x)^4 - x^2 \\
&\ge (1-x)^2 - (1-x)^3 - x^2 \\
&= x(x^2 - 3x + 1).
\end{align}
Thus, if $x\in (0, 1]$ satisfying $f(x) \le 0$, then
$x^2 - 3x + 1 \le 0$ which results in $\frac{3 - \sqrt{5}}{2} \le x \le 1$.
We are done.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4077879",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Representing all rational numbers between $\dfrac{1}{2}$ and $1$ How do I show that
$$\dfrac{2^{\left\lfloor\frac12 a_1\right\rfloor} + 2^{\left\lfloor\frac12 a_2\right\rfloor} + \ldots + 2^{\left\lfloor\frac12 a_n \right\rfloor}}{2^{\left\lceil\frac12 a_1\right\rceil} + 2^{\left\lceil\frac12 a_2\right\rceil} + \ldots + 2^{\left\lceil\frac12 a_n \right\rceil}}$$
represents all rationals between $\dfrac{1}{2}$ and $1$, where $a_1, a_2, \ldots, a_n \in \mathbb{N}$?
I have tried representing few rationals such as $\dfrac{7}{8}$ and $\dfrac{5}{8}$ and it seems like we can represent all of them. However I'm not completely sure why and don't have a rigorous proof.
For example
$$\dfrac{5}{8} = \dfrac{2^{\left\lfloor\frac12 3\right\rfloor} + 2^{\left\lfloor\frac12 4\right\rfloor} + 2^{\left\lfloor\frac12 5 \right\rfloor}}{2^{\left\lceil\frac12 3\right\rceil} + 2^{\left\lceil\frac12 4\right\rceil} + 2^{\left\lceil\frac12 5 \right\rceil}}$$
$$\dfrac{7}{8} = \dfrac{2^{\left\lfloor\frac12 1\right\rfloor} + 2^{\left\lfloor\frac12 2\right\rfloor} + 2^{\left\lfloor\frac12 4 \right\rfloor}}{2^{\left\lceil\frac12 1\right\rceil} + 2^{\left\lceil\frac12 2\right\rceil} + 2^{\left\lceil\frac12 4 \right\rceil}}$$
Any help would be highly appreciated. Thanks
| Some nice answers have been posted, but I'm still adding this answer since it gives a straightforward non-recursive way to get a representation.
That the expression is in $[{1 \over 2},1]$ is easy to see.
For the other way, consider ${x \over y} \in [{1 \over 2},1]$ and let $m=y-x$. Then $x-m \geq 0$.
Express $x-m=(a_k a_{k-1} \dots a_0)_2$ and $m=(b_l b_{l-1} \dots b_0)_2$ expressed in binary [$0 \notin \{a_k,b_l\}$].
Then, I claim that $\{2i : a_i = 1\} \cup \{2j+1 : b_j =1 \}$ gives our representation.
Indeed,
$$
{\sum_{a_i=1}2^{\lfloor{2i \over 2}\rfloor}+\sum_{b_j=1}2^{\lfloor{2j+1 \over 2}\rfloor} \over \sum_{a_i=1}2^{\lceil{2i \over 2}\rceil}+\sum_{b_j=1}2^{\lceil{2j+1 \over 2}\rceil}}
= {\sum_i a_i 2^i+\sum_j b_j 2^j \over \sum_i a_i 2^i+\sum_j b_j 2^{j+1}}
= {x-m+m \over x-m+2m} = {x \over y}
$$
The last line is due to the fact that in case of non-integers (i.e. ${2j+1 \over 2}$), the ceilings increase the powers of $2$ by $1$ and hence the number expressed is multiplied by $2$.
As an example, for $7 \over 10$, $x=7,y=10,m=3$.
Now, $x-m=4=(100)_2$ and $m=3=(11)_2$.
Using our algorithm, the representative numbers are $4,3,1$.
Indeed, ${2^{\lfloor{4 \over 2}\rfloor}+2^{\lfloor{3 \over 2}\rfloor}+2^{\lfloor{1 \over 2}\rfloor} \over 2^{\lceil{4 \over 2}\rceil}+2^{\lceil{3 \over 2}\rceil}+2^{\lceil{1 \over 2}\rceil}} = {2^2+2^1+2^0 \over 2^2+2^2+2^1} = {4+2+1 \over 4+4+2} = {7 \over 10}$.
I just noticed that @TheSimpliFire has done a similar thing in the comments, but I started writing it sometime ago, had some coffee, and then finished it. So I somehow missed his/her comment. Anyway, if you (@TheSimpliFire) decide to make it into an answer, please let me know and I'll remove mine.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4080410",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "21",
"answer_count": 4,
"answer_id": 1
} |
Trigonometric integral $\int^\pi_0\frac{d\theta}{1+a^2\sin^2\theta}$ The function
$f(x_1,x_2)=\frac{e^{-\sqrt{x^2_1 + x^2_2}}}{\sqrt{x^2_1+x^2_2}}$, $(x_1,x_2)\neq(0,0)$ is integrable in $\mathbb{R}^2\setminus\{(0,0\})$ as $f>0$ and
\begin{align}
\int_{\mathbb{R}^2}f(x_1,x_2)\,dx_1dx_2=2\pi\int^\infty_0e^{-r}\,dr
\end{align}
I am interested in its Fourier transform. Being radial, we have that
\begin{align}
\widehat{f}(t_1,t_2)&=\widehat{f}\big(0,\sqrt{t^2_1+t^2_2}\big)=\int_{(0,\infty)\times(-\pi,\pi)}e^{-r}e^{-2\pi i r\sqrt{t^2_1+t^2_2}\sin\theta}\,drd\theta\\
&=\int^\pi_{-\pi} \frac{d\theta}{1+ 2\pi i \sqrt{t^2_1+t^2_2}\sin\theta}=\int^\pi_{-\pi} \frac{1-2\pi i \sqrt{t^2_1+t^2_2}\sin\theta}{1+ 4\pi^2 (t^2_1+t^2_2)\sin^2\theta}\,d\theta\\
&=2\int^\pi_0\frac{d\theta}{1+ 4\pi^2 (t^2_1+t^2_2)\sin^2\theta}
\end{align}
Question:
At this point, the last integral escapes me. I would like to see if there is either a more direct way to estimate the Fourier transform of $f$, or whether the integral
\begin{align}
\int^\pi_0\frac{d\theta}{1+a^2\sin^2\theta}\tag{1}\label{one}
\end{align}
can be further evaluated.
Edit: As pointed out by @Koro, it turns out that \eqref{one} is rather trivial:
\begin{align}
\int^\pi_0\frac{d\theta}{1+a^2\sin^2\theta}&=2\int^{\pi/2}_0\frac{d\theta}{1+a^2-a^2\cos^2\theta}=2\int^{\pi/2}_0\frac{\sec^2\theta}{\sec^2\theta(1 +a^2)-a^2}\,d\theta\\
&=2\int^{\pi/2}_0\frac{\sec^2\theta}{1+(1+a^2)\tan^2\theta}\,d\theta
\stackrel{u=\tan\theta}{=}2\int^{\infty}_0\frac{du}{1+(1+a^2)u^2}\\
&=\frac{\pi}{\sqrt{1+a^2}}
\end{align}
Hence, the Fourier transform of $f$ is
$$\widehat{f}(t_1,t_2)=\frac{2\pi}{\sqrt{1+4\pi^2(t^2_1+t^2_2)}}$$
| \begin{aligned}\int_{0}^{\pi}{\frac{\mathrm{d}\theta}{1+a^{2}\sin^{2}{\theta}}}&=\frac{1}{2}\int_{0}^{\pi}{\frac{\mathrm{d}\theta}{1+a^{2}\sin^{2}{\theta}}}+\frac{1}{2}\int_{\pi}^{2\pi}{\frac{\mathrm{d}\theta}{1+a^{2}\sin^{2}{\left(2\pi -\theta\right)}}}\\ &=\frac{1}{2}\int_{0}^{2\pi}{\frac{\mathrm{d}\theta}{1+a^{2}\sin^{2}{\theta}}}\\ &=\frac{1}{2\,\mathrm{i}}\oint_{\left|z\right|=1}{\frac{\mathrm{d}z}{z\left(1-\frac{a^{2}}{4}\left(z-\frac{1}{z}\right)^{2}\right)}}\\ &=2\,\mathrm{i}\oint_{\left|z\right|=1}{\frac{z}{a^{2}z^{4}-\left(2a^{2}+4\right)z^{2}+a^{2}}\,\mathrm{d}z}\\ &=2\,\mathrm{i}\left(2\pi\,\mathrm{i}\,\mathrm{Res}\left(f,\sqrt{1+\frac{2}{a^{2}}-2\sqrt{\frac{1}{a^{2}}+\frac{1}{a^{4}}}}\right)+2\pi\,\mathrm{i}\,\mathrm{Res}\left(f,-\sqrt{1+\frac{2}{a^{2}}-2\sqrt{\frac{1}{a^{2}}+\frac{1}{a^{4}}}}\right)\right)\\ &=2\,\mathrm{i}\left(2\pi\,\mathrm{i}\times\left(-\frac{1}{8a^{2}\sqrt{\frac{1}{a^{2}}+\frac{1}{a^{4}}}}\right)+2\pi\,\mathrm{i}\times\left(-\frac{1}{8a^{2}\sqrt{\frac{1}{a^{2}}+\frac{1}{a^{4}}}}\right)\right)\\ \int_{0}^{\pi}{\frac{\mathrm{d}\theta}{1+a^{2}\sin^{2}{\theta}}}&=\frac{\pi}{\sqrt{1+a^{2}}}\end{aligned}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4082555",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Find $(a, b, c)$ that satisfies the system of equations $2a + 3b -4c = 7, a-b+2c=6$ $\begin{align}
2a + 3b -4c = 7, \\
a - b + 2c = 6
\end{align}$
I haven't seen a three-variable system of equations with only two equations before. I try to make the problem into a two-variable system.
Solving for $a$ in the first equation
$\begin{align}
a = \frac{-3b +4c+7}{2}
\end{align}$.
Solving for $a$ in the second equation:
$\begin{align}
a = b-2c+6
\end{align}$.
Then substitute the equations into $a$ for the original equations.
$\begin{align}
2(b-2c+6) + 3b -4c = 7 \\
2b-4c+12+3b-4c=7 \\
5b-8c = -5
\end{align}$
$\begin{align}
\frac{-3b +4c+7}{2} - b + 2c = 6 \\
-3b+4c+7-2b+4c=12 \\
-5b+8c=5
\end{align}$
This create a new system of equations:
$\begin{align}
5b-8c =-5 \\
-5b+8c=5
\end{align}$
This leads to the solution $0=0$. So I assume this means I can select any $b$ and $c$. This however isn't the case, I can't find $a$ that satisfies both equations with this premise.
The problem additionally asks to solve for $b$ in terms of $a$ and solve for $c$ in terms of $a$. I did so using elimination.
$\begin{align}
b = -4a +19 \\
c = \frac{-5a+25}{2}
\end{align}$
So we have both $b$ and $c$ depending on $a$. There is an infinite number of choices for $a$ so there are infinitely many solutions to this system.
Could I not find one or more ordered triples that satisfy both equations if I did not (following the instruction) create this final system of equations? Until now I thought the number of equations in the system was simply related to the number of variables.
| First, see my comment, following your query.
Since you are trying, not only to solve the equations, but also to express
$b$ and $c$ in terms of $a$, I would have approached it differently.
$$2a + 3b - 4c = 7. \tag1$$
$$a - b + 2c = 6. \tag2$$
Multiplying equation (2) above by $(3)$ gives
$$3a - 3b + 6c = 18.\tag3$$
Adding equations (3) and (1):
$$5a + 2c = 25 \implies c = \frac{25 - 5a}{2}. \tag4$$
Now, repeat the process to express $b$ in terms of $a$.
Multiplying equation (2) by $(2)$ gives
$$2a - 2b + 4c = 12. \tag5$$
Adding equations (5) and (1):
$$4a + b = 19 \implies b = 19 - 4a. \tag6$$
At this point, you are actually done. $(a)$ can be chosen to be any number.
Once $(a)$ is chosen, $(b)$ and $(c)$ are computed via equations (4) and (6) above.
Then, you have generated a solution $(a,b,c).$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4083679",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 2,
"answer_id": 0
} |
An upper bound for $\{r_n\}$ which satisfies $r_n = r_{n-1} + \frac{K}{r_{n-1}}$ I have a sequence $\{r_n\}$ of positive numbers defined recursively by
$$\tag{1} r_n = r_{n-1} + \frac{K}{r_{n-1}}.$$
for some positive $K$. The sequence is clearly strictly increasing and tends to $+\infty$ as $n\to \infty$. Also, by induction one can prove
$$\tag{2} r_n^2 \ge 2(n-1)K +r_1^2$$
for all $n$.
Question: Does there exists positive $K_2$ depending on $r_1, K$ such that
$$\tag{3} r_n^2 \le 2nK + K_2$$
for all large $n$?
Squaring (1) on both sides give
$$ r_n^2 = r_{n-1}^2 + 2K + \frac{K^2}{r_{n-1}^2},$$
If I try to prove (3) using induction, (2) gives
$$ \frac{K^2}{r_{n-1}^2} \le \frac{K^2}{ 2(n-2)K +r_1^2},$$
which is not very helpful, since $\sum \frac{1}{n-2}$ is not summable.
| First, reproving the lower bound: we have, for all $n\geq 1$,
$$
r_{n-1}^2 +K = r_n r_{n-1} \leq \frac{r_n^2+ r_{n-1}^2}{2} \tag{1}
$$
using the AM-GM inequality; from which
$$
r_n^2 \geq r_{n-1}^2 + 2K \tag{2}
$$
leading to $r_{n}^2 \geq 2K n + r_0^2$ for all $n\geq 0$.
Then, the upper bound: We know, using the lower bound, that there exists $N\geq 1$ such that, for all $n\geq N$,
$$
r_n^2 \geq 2Kn
$$
so we get now use this to get
$$
r_n = r_{n-1}+\frac{K}{r_{n-1}} \leq r_{n-1}+\frac{\sqrt{K}}{\sqrt{2(n-1)}} \tag{3}
$$
so that
$$
r_n - r_1 = \sum_{k=2}^n (r_k - r_{k-1}) \leq \sum_{k=2}^n \sqrt{\frac{K}{2(k-1)}}
= \sqrt{\frac{K}{2}}\sum_{k=1}^{n-1} \frac{1}{\sqrt{k}}
\leq \sqrt{\frac{K}{2}}\int_{0}^{n-1} \frac{dx}{\sqrt{x}} = \sqrt{2K(n-1)}
$$
and so $(r_n-r_1)^2 \leq 2Kn$. Now, since $r_n \to \infty$, for every $\varepsilon>0$ there exists some $n_\varepsilon\geq 1$ such that $(r_n-r_1)^2 \geq (1+\varepsilon/2)^{-1}r_n^2$ for all $n\geq n_\varepsilon$.
Conclusion. For all $\varepsilon>0$, there exists $n_\varepsilon$ such that, for $n\geq n_\varepsilon$,
$$
\boxed{ 2Kn + r_0^2 \leq r_n^2 \leq (2+\varepsilon)Kn + r_1^2 }
$$
(the lower bound actually holds for all $n\geq 0$).
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4085112",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Advanced complex numbers/roots of unity Let $a, b, c, d$ be real numbers, none of which are equal to $-1$, and let $\omega$ be a complex number such that $\omega^3 = 1$ and $\omega \neq 1.$ Given that,
$$
\frac{1}{a + \omega} + \frac{1}{b + \omega} + \frac{1}{c + \omega} + \frac{1}{d + \omega} = \frac{2}{\omega},
$$
how can I deduce
$$
\frac{1}{a + 1} + \frac{1}{b + 1} + \frac{1}{c +1} + \frac{1}{d + 1}?
$$
I have tried clearing the denominators of the first equation, but that just results in a large mess. I don't know how to continue from there.
| A calculation by foot. We obtain from
\begin{align*}
\color{blue}{0}&\color{blue}{=\frac{1}{a+\omega}+\frac{1}{b+\omega}
+\frac{1}{c+\omega}+\frac{1}{d+\omega}-\frac{2}{\omega}}\tag{1}\\
\end{align*}
by multiplication with the common denominator $(a+\omega)(b+\omega)(c+\omega)(d+\omega)\omega$:
\begin{align*}
\color{blue}{0}&=(b+\omega)(c+\omega)(d+\omega)\omega\\
&\qquad+(a+\omega)(c+\omega)(d+\omega)\omega\\
&\qquad+(a+\omega)(b+\omega)(d+\omega)\omega\\
&\qquad+(a+\omega)(b+\omega)(c+\omega)\omega\\
&\qquad-2(a+\omega)(b+\omega)(c+\omega)(d+\omega)\\
&=bcd\omega+(bc+bd+cd)\omega^2+(b+c+d)\omega^3+\omega^4\\
&\qquad+acd\omega+(ac+ad+cd)\omega^2+(a+c+d)\omega^3+\omega^4\\
&\qquad+abd\omega+(ab+ad+bd)\omega^2+(a+b+d)\omega^3+\omega^4\\
&\qquad+abc\omega+(ab+ac+bc)\omega^2+(a+b+c)\omega^3+\omega^4\\
&\qquad-2abcd\\
&\qquad-2(abc+abd+acd+bcd)\omega\\
&\qquad-2(ab+ac+ad+bc+bd+cd)\omega^2\\
&\qquad-2(a+b+c+d)\omega^3\\
&\qquad-2\omega^4\\
&\,\,\color{blue}{=\left(2-(abc+abd+acd+bcd)\right)\omega-2abcd+a+b+c+d}\tag{2}
\end{align*}
In the last step (2) we observe that terms with $\omega^2$ cancel away and we also use the identities
\begin{align*}
\omega^3=1,\quad\omega^4=\omega
\end{align*}
We note from (1) and (2) we can write (1) as
\begin{align*}
\color{blue}{0=\frac{A\omega+B}{(a+\omega)(b+\omega)(c+\omega)(d+\omega)\omega}}\tag{3}
\end{align*}
and since $A,B\in\mathbb{R}$ and $\omega\in\mathbb{C}\setminus{\mathbb{R}}$ we conclude $A=B=0$, so that
\begin{align*}
\color{blue}{abc+abd+acd+bcd}&\color{blue}{=2}\tag{4}\\
\color{blue}{a+b+c+d}&\color{blue}{=2abcd}\\
\end{align*}
follows.
On the other hand we consider the expression
\begin{align*}
\color{blue}{\frac{1}{a+1}+\frac{1}{b+1}+\frac{1}{c+1}+\frac{1}{d+1}}&\color{blue}{=z}
\end{align*}
Multiplication of the LHS with the common denominator $(1+a)(1+b)(1+c)(1+c)$ gives
\begin{align*}
&(b+1)(c+1)(c+1)+(a+1)(c+1)(d+1)\\
&\qquad\quad+(a+1)(b+1)(d+1)+(a+1)(b+1)(c+1)\\
&\quad=1+(b+c+d)+(bc+bd+cd)+bcd\\
&\qquad\quad+1+(a+c+d)+(ac+ad+cd)+acd\\
&\qquad\quad+1+(a+b+d)+(ab+ad+bd)+abd\\
&\qquad\quad+1+(a+b+c)+(ab+ac+bc)+abc\\
&\quad=4+3(a+b+c+d)\\
&\qquad\quad+2(ab+ac+ad+bc+bd+cd)\\
&\qquad\quad+abc+abd+acd+bcd\\
&\,\,\color{blue}{=2\left(3+3abcd+ab+ac+ad+bc+bd+cd\right)}\tag{5}
\end{align*}
In the last line (5) we used the identities from (4).
Similarly, multiplication of the RHS with the common denominator gives
\begin{align*}
&z(a+1)(b+1)(c+1)(d+1)\\
&\qquad=z(1+(a+b+c+d)\\
&\qquad\quad+(ab+ac+ad+bc+bd+cd)\\
&\qquad\quad+(abc+abd+acd+bcd)+abcd)\\
&\,\,\color{blue}{\qquad=z(3+3abcd+ab+ac+ad+bc+bd+cd)}\tag{6}
\end{align*}
Again in the last line (6) we used the identities from (4) for simplification.
Comparing (5) and (6) we conclude
\begin{align*}
\color{blue}{\frac{1}{a+1}+\frac{1}{b+1}+\frac{1}{c+1}+\frac{1}{d+1}=2}
\end{align*}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4086078",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 3,
"answer_id": 0
} |
Solving $\tan^{-1}\left(\frac{2x+1}{x+1}\right)+\tan^{-1}\left(\frac{2x-1}{x-1}\right)=2\tan^{-1}\left(1+x\right)$
Find all $x$ that satisfy the equation $$\tan^{-1}\left(\frac{2x+1}{x+1}\right)+\tan^{-1}\left(\frac{2x-1}{x-1}\right)=2\tan^{-1}\left(1+x\right)$$
Attempt:
Taking tangent on both sides, and using the identities $\displaystyle \tan\left(\alpha+\beta\right)=\frac{\tan\alpha+\tan\beta}{1-\tan\alpha\tan\beta}$ and $\displaystyle \tan\left(2\alpha\right)=\frac{2\tan\alpha}{1-\tan^{2}\alpha}$,
We obtain $$\frac{\frac{\left(2x+1\right)}{x+1}+\frac{\left(2x-1\right)}{x-1}}{1-\frac{\left(4x^{2}-1\right)}{x^{2}-1}}=\frac{2\left(x+1\right)}{1-\left(x+1\right)^{2}}$$ and after removing the root $x=0$, we are left with a simple equation $(x^2-2)(2x+1)=0$. So, the possible roots are $x=0,\pm \sqrt{2}, -\dfrac{1}{2}$ , now my question is, how do I verify which roots satisfy the original equation, because one cannot see directly after plugging, whether it's true or not because it involves arctangents.
To avoid this situation, I re-wrote the original equation as $$\tan^{-1}\left(\frac{2x+1}{x+1}\right)-\tan^{-1}\left(1+x\right)=\tan^{-1}\left(1+x\right)-\tan^{-1}\left(\frac{2x-1}{x-1}\right)$$ and because while using the identity $\tan^{-1}\left(\alpha\right)-\tan^{-1}\left(\beta\right)=\tan^{-1}\left(\frac{\alpha-\beta}{1+\alpha\beta}\right)$, there are no restrictions on $\alpha,\beta$ I went ahead with this, and got the equation $$\frac{\frac{\left(2x+1\right)}{x+1}-\left(x+1\right)}{2\left(x+1\right)}=\frac{\left(x+1\right)-\frac{\left(2x-1\right)}{x-1}}{2x}$$ which simplified to $\left(x+1\right)^{2}\left(x-2\right)+x^{2}\left(x-1\right)=0$ and surprisingly from here, I am not even getting those roots!
I used this method to avoid the extraneous roots, what is going wrong here?
Edit: The immediately above issue has been resolved, however, the first issue remains unresolved.
Note that all calculations have to be performed by hand, since these types of questions are asked in exams in which the allotted time per question averages out to 2-3 minutes.
| This should be an exercise for high school students who just learned inverse trig function.
Denote $\alpha=\tan^{-1}((2x+1)/(x+1)), \beta=\tan^{-1}((2x-1)/(x-1))$ and $\gamma=\tan^{-1}(1+x)$. The original equation is
\begin{equation}
(1)\qquad \alpha+\beta=2\gamma.
\end{equation}
The equation you solved is equivalent to
\begin{equation}
(2)\qquad \tan(\alpha+\beta)=\tan(2\gamma).
\end{equation}
Equations $(1)$ and $(2)$ are clearly not equivalent. Actually, equation $(1)$ is equivalent to equation $(2)$ plus
\begin{equation}
(3)\qquad \alpha+\beta \textrm{ and } 2\gamma \textrm{ are in the same interval } ((k-\frac{1}{2})\pi, (k+\frac{1}{2})\pi), \textrm{ for some integer } k.
\end{equation}
It suffices to check whether your solutions of $(2)$ satisfy $(3)$ or not, which is just an easy exercise. For example, when $x=-\sqrt{2}$, then $\tan(\alpha)=1+\frac{x}{1+x}>1$, thus $\alpha=\tan^{-1}((2x+1)/(x+1))\in (\pi/4, \pi/2)$. Similarly, $\beta\in (\pi/4, \pi/2)$ and thus $\alpha+\beta\in (\pi/2,\pi)$. While $-1<1+x<0$, thus $\gamma=\tan^{-1}(1+x)\in (-\pi/4,0)$ and thus $2\gamma\in (-\pi/2,0)$. Thus $\alpha+\beta$ cannot be $2\gamma$. Instead, we should have $\alpha+\beta=2\gamma+\pi$ since $\tan(\alpha+\beta)=\tan(2\gamma)$ according your solution. The rest solutions can be checked easily. Among your solutions of $(2)$, only $x=-\sqrt{2}$ does not satisfy $(3)$ and hence does not satisfy $(1)$. All of the rest solutions satisfy (3) and thus (1), assuming your solutions of $(2)$ are correct.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4087642",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 2,
"answer_id": 0
} |
Calculating $\cos^{-1}{\frac{3}{\sqrt10}} + \cos^{-1}{\frac{2}{\sqrt5}}$ $$\cos^{-1}{\frac{3}{\sqrt{10}}} + \cos^{-1}{\frac{2}{\sqrt 5}}= ?$$
Let $\cos^{-1}{\frac{3}{\sqrt{10}}}=\alpha,
\cos^{-1}{\frac{2}{\sqrt 5}}=\beta$ then, $\cos\alpha=\frac{3}{\sqrt{10}}, \cos\beta=\frac{2}{\sqrt5}$
Therefore $$\cos\alpha=\frac{3\cdot2}{2\sqrt2\sqrt5}= \frac{3}{2\sqrt2}\cdot\cos\beta$$
This is all I did till now. Could you go further with this to answer?
| Like Proof for the formula of sum of arcsine functions $ \arcsin x + \arcsin y $,
using $\cos(A+B)$ and the definition of principal values
$$\cos^{-1}x+\cos^{-1}y=\begin{cases}\cos^{-1}(xy-\sqrt{(1-x^2)(1-y^2)}) &\mbox{if }\cos^{-1}x+\cos^{-1}y\le\pi \\
2\pi-\cos^{-1}(xy-\sqrt{(1-x^2)(1-y^2)}) & \mbox{otherwise } \end{cases}$$
Now $\cos^{-1}x+\cos^{-1}y\le\pi$ will happen
$\iff\cos^{-1}x\le\pi-\cos^{-1}y=\cos^{-1}(-y)$ using How do I prove that $\arccos(x) + \arccos(-x)=\pi$ when $x \in [-1,1]$?
$\iff\dfrac\pi2-\sin^{-1}x\le\dfrac\pi2-\sin^{-1}(-y)$
$\iff\sin^{-1}x\ge\sin^{-1}(-y)$
$\iff x\ge-y$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4087823",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 3
} |
$a_1Give that $a_1<a_2<a_3<a_4$ are positive integers such that $$\sum_{i=1}^4\frac{1}{a_i}=\frac{11}{6}$$, find the value of $a_4-a_2$.
My try:
Since $11$ is prime, atleast one of $a_1,a_2,a_3,a_4$ should be divisible by $6$.
Now we should express the fraction $\frac{11}{6}$ as $\frac{A+B+C+D}{E}$ such that $E$ is divisible by $A,B,C,D,E$.
I tried $$\frac{11}{6}=1+\frac{1}{2}+\frac{1}{3}$$ But fourth fraction is not possible.
Any hint?
| If $a_2>2$, then
$$ \sum_{k=1}^4\frac1{a_k}\le1+\frac13+\frac14+\frac15=\frac{107}{60}<\frac{11}{6}.$$
Hence we must have $a_2\le 2$ and therefore $a_1=1$, $a_2=2$.
Solving for $a_4$ in terms of $a_3$, we now find
$$ a_4=\frac1{\frac{11}6-1-\frac12-\frac1{a_3}}=\frac{3a_3}{a_3-3}$$
which forbids $a_3=3$, is no integer for $a_3=5$, and is $\le \frac{3a_3}3=a_3$ for $a_3\ge 6$. Remains only $a_3=4$ and $a_4=12$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4089436",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
$\angle DAB = 10^\circ$, $\angle CAD = 20^\circ$, $\angle CDB = 140^\circ$, $CD = BD$. Find $\angle ABD$ $\angle DAB = 10^\circ$
$\angle CAD = 20^\circ$
$\angle CDB = 140^\circ$
The triangle $\triangle BCD$ is isosceles. $CD = BD$
Find the angle $x$
The answer is $30º$.
Playing with GeoGebra, I found that $AC = AD$. But I don't know how to prove it.
| Construct an equilateral triangle $\triangle DCE$ inside of $\triangle ADC$ as follows:
We can see that $\angle ADB = 180^\circ - \angle DAB - \angle ABD = 170^\circ - x$ because $\angle DAB = 10^\circ$. Now, because $\angle CDB = 140^\circ$, we now know that $\angle ADC = 360^\circ - \angle ADB - \angle CDB = 50^\circ + x$. So $\angle EDA = \angle ADC - \angle EDC = x - 10^\circ$. (Note that because $\triangle DCE$ is equilateral, we must have $\angle EDC = \angle ECD = 60^\circ$ in particular.) Moreover, $\angle ACE = \angle ACD - \angle ECD = 180^\circ - \angle CAD - \angle ADC - \angle ECD = 50^\circ - x$.
Furthermore, $\angle EDB = \angle EDA + \angle ADB = 160^\circ$. As $|ED| = |DC| = |BD|$, $\triangle EDB$ is isosceles and therefore $\angle EBD = \angle EDB = 10^\circ$. But then $\angle DAB = 10^\circ = \angle DEB$. So $ABDE$ is a cyclic quadrilateral. From that we get that $\angle EAD = \angle EBD = 10^\circ$ and $\angle CAE = \angle CAD - \angle EAD = 10^\circ$ since $\angle CAD = 20^\circ$.
Note that $|AE| = |AE|$, $|ED| = |EC|$ and $\angle EAD = \angle EAC$ in $\triangle AED$ and $\triangle AEC$ respectively. To prove that $\triangle AEC \cong \triangle AED$, we therefore have to prove that it is impossible for one of the angles $\angle AED$ and $\angle AEC$ to be obtuse, and the other acute. This can be seen from the fact that $\angle AED + \angle AEC = 360^\circ - \angle CED = 300^\circ$: if one angle were indeed to be in the interval $(0^\circ, 90^\circ)$ and the other in the interval $(90^\circ, 180^\circ)$, then the sum would not be greater than $270^\circ$, resulting in a contradiction.
From $\triangle AEC \cong \triangle AED$ we at last see that $\angle ACE = \angle ADE$ and as such $50^\circ - x = x - 10^\circ$ or $x = 30^\circ$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4090521",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Find the number of solutions in nonnegative integers of the (strict) inequality $a+b+c+d+e<11$ when $a$ is at most $6$. Find the number of solutions in nonnegative integers of the (strict) inequality $a+b+c+d+e<11$.
We can see that $a+b+c+d+e \leq10$ and so there are $10+5 \choose 5$ solutions.
Now, what if $a$ is at most $6$? That is, $a\leq6$. This means $a\notin\{7,8,9,10\}$. I decided to use the subtraction principle. I let $a=7,8,9,10$ (four separate cases) and came to ${15 \choose 5}-{7 \choose 4}-{6 \choose 4}-{5 \choose 4}-{4 \choose 4}=3003-56=2947$ which is correct. Is there a better way to do this? Is there a quicker way to do this? The textbook solution is ${15 \choose 5}-{8 \choose 5}$ and I wonder how they came across that exact form. Any solutions or hints are greatly appreciated.
| The shortcut is that $a + b + c + d + e \leq 10$, where $a \geq 7$ bijects to $a + b + c + d + e \leq 3$, where $a,b,c,d,e$ are non-negative integers.
Therefore, the number of solutions to deduct is
$$\binom{7}{4} + \binom{6}{4} + \binom{5}{4} + \binom{4}{4} = \binom{8}{5}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4090946",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Understanding proof of $(m+n+2)^2 \neq 3(mn + 1)$ The problem is: Prove that there are no integers $m$ and $n$ such that:
$$(m+n+2)^2 = 3(mn + 1)$$
I found the explanation for the problem solution, but I still don't understand it.
Here's the solution I found:
If there are numbers $m$ and $n$ such that: $(m+n+2)^2 = 3(mn + 1)$, then $3
|(m+n+2)^2 \implies 9|(m+n+2)^2$ and $3 | (mn+1)$, that means that numbers $m$ and $n$ cannot be divisible by $3$ and they cannot have the same remainder after dividing by $3$. Let's rewrite $m$ and $n$ as $m = 3p + 1$ and $n = 3q + 2$, but then we have $m+n+2 = 3(p+q+1) + 2$ which is impossible since $3 | m+n+2$
The only thing I don't understand is why $3 | (mn+1)$?
Can someone explain this to me? Thanks.
| Alternative proof
$$(m+n+2)^2-3mn-3=0$$
$$\iff m^2 - m (n -4) + n^2 + 4 n + 1=0 $$
$$\begin{align}\Delta &=-3 n^2 - 24 n + 12\\
&=60 - 3 (n + 4)^2\end{align}$$
$$60 - 3 (n + 4)^2≥0, ~ 60 - 3 (n + 4)^2=T^2, T\in\mathbb Z$$
This means, such $n\in\mathbb Z$ doesn't exist.
Small supplement:
$$\begin{align}0&≤60 - 3 (n + 4)^2\\
&=60 - 3 (|n + 4|)^2\\
&=T^2,~T\in\mathbb Z \end{align}$$
All possible cases are as follows:
$$|n+4|:=1,2,3,4.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4091781",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Calculating $ \sum_{k=0}{n\choose 4k+1}$ Calculate $$\sum_{k=0}{n\choose 4k+1}$$
This should be an easy and short result but I'm messing up somewhere. What I've done so far is take $f(x)=(1+x)^n$ and with the binomial theorem expand $f(1), f(-1), f(i), f(-i)$ but it leads to taking cases for n's division rest to 4 (which is too long, $\displaystyle \sum_{k=0}{n\choose 4k}$ is much shorter to calculate). The sum should be, I think, $$\frac{1}{2}\left(2^{n-1}+2^{\frac{n}{2}}\sin\left(\frac{n\pi}{4}\right)\right)$$
| In case there is interest we can also do this one with residues, by way
of enrichment, and demonstrate some complex arithmetic. Start with
$$\sum_{k\ge 0} {n\choose 4k+1}
= \sum_{k\ge 0} {n\choose n-4k-1}
= [z^{n-1}] (1+z)^n \sum_{k\ge 0} z^{4k}
\\ = [z^{n-1}] (1+z)^n \frac{1}{1-z^4}
= \mathrm{Res}_{z=0} \frac{1}{z^n}
(1+z)^n \frac{1}{1-z^4}.$$
Now residues sum to zero and the residue at infinity is zero by
inspection. This leaves the residues at $\rho_m = \exp(m \pi i/2)$
where $0\le m\lt 4.$ We obtain for our sum (using the fact that the
poles are simple)
$$- \sum_\rho \mathrm{Res}_{z=\rho}
\frac{1}{z^n} (1+z)^n \frac{1}{1-z^4}
\\ = - \sum_\rho
\frac{1}{\rho^n} (1+\rho)^n
\lim_{z\rightarrow \rho}
\frac{z-\rho}{(1-z^4)-(1-\rho^4)}
\\ = - \sum_\rho
\frac{1}{\rho^n} (1+\rho)^n
\frac{1}{-4\rho^3}
= \frac{1}{4} \sum_\rho
\frac{1}{\rho^{n+3}} (1+\rho)^n
= \frac{1}{4} \sum_\rho
\frac{\rho}{\rho^{n}} (1+\rho)^n.$$
Expanding with $(1+\rho)/\rho = 1 + 1/\rho$, we get
$$\frac{1}{4} \times 1 \times 2^n
+ \frac{1}{4} \times i \times (1-i)^n
+ \frac{1}{4} \times -1 \times 0^n
+ \frac{1}{4} \times -i \times (1+i)^n
\\ = \frac{1}{4} 2^n +
\frac{1}{4} i \sqrt{2}^n
(\exp(-i\pi n/4)-\exp(i\pi n/4))
\\ = \frac{1}{4} 2^n
- \frac{1}{2} \sqrt{2}^n
\frac{\exp(-i\pi n/4)-\exp(i\pi n/4)}{2i}
\\ = \frac{1}{4} 2^n
- \frac{1}{2} \sqrt{2}^n \sin(-\pi n/4)
= \frac{1}{2} (2^{n-1} + 2^{n/2} \sin(\pi n/4)).$$
This is the claim.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4091959",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Given that $3\sin x + 2\sin y = 4$. Find the maximum value of $3\cos x + 2\cos y$.
Given that $3\sin x + 2\sin y = 4$. Find the maximum value of $3\cos x + 2\cos y$.
I wonder if calculus can be used to solve this problem? If not how to solve this? I am thinking about $(3\cos x + 2\cos y)^2=9\cos^2x+12\cos x\cos y + 4\cos^2y$ and work in this way but I don't quite see how.
| Hint:
$\begin{align}(3\cos x + 2\cos y)^2=&9\cos^2x+12\cos x\cos y + 4\cos^2y\\=&13+12\cos x\cos y +12\sin x\sin y -(9\sin^2x+4\sin^2y+12\sin x\sin y)\\=& 13+12\cos(x-y)-(3\sin x+2\sin y)^2\\=&-3+12\cos(x-y)\end{align}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4092309",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
Prove $f(x)$ is differentiable at a point.
Let $f(x)$ be a funciton defined over $(0,1)$ such that
$$f(x)=\begin{cases} \dfrac{1}{p^3},&x=\dfrac{q}{p},\text{where} ~p,q
\in \mathbb{N}, \text{and}~ {\rm GCD}(p,q)=1,\\ 0, &\text{otherwise}.
\end{cases}$$
Prove $f(x)$ is differentiable at $x=\dfrac{n\sqrt{2}}{m}$ for any
$m,n \in \mathbb{N}$.
Obviously, what we need to do is show that
$$\lim_{h \to 0}\dfrac{f\left(\dfrac{n\sqrt{2}}{m}+h\right)-f\left(\dfrac{n\sqrt{2}}{m}\right)}{h}$$
exists. Since $\dfrac{n\sqrt{2}}{m}$ is irrational, $f\left(\dfrac{n\sqrt{2}}{m}\right)=0$. Therefore
$$\lim_{h \to 0}\dfrac{f\left(\dfrac{n\sqrt{2}}{m}+h\right)-f\left(\dfrac{n\sqrt{2}}{m}\right)}{h}=\lim_{h \to 0}\dfrac{f\left(\dfrac{n\sqrt{2}}{m}+h\right)}{h}.$$
How to go on?
| The differential quotient
$$
\frac{f\left(x\right)-f\left(\frac{n\sqrt{2}}{m}\right)}{x - \frac{n\sqrt{2}}{m}}
$$
is zero for irrational $x$, therefore it remains to show that it tends to zero for rational numbers $x\to \frac{n\sqrt{2}}{m}$.
For $x=q/p$ is
$$
\left|\frac{f\left(x\right)-f\left(\frac{n\sqrt{2}}{m}\right)}{x - \frac{n\sqrt{2}}{m}} \right|
= \frac{1}{p^3\left| \frac qp - \frac{n\sqrt{2}}{m}\right|}
= \frac{m}{p^2|qm - pn \sqrt 2|} = \frac{m(qm+pn\sqrt 2)}{p^2|(qm)^2-2(pn)^2|} \, .
$$
Here we have expanded the fraction such that $|(qm)^2-2(pn)^2|$ in the denominator becomes a non-zero integer. It follows that
$$
\left|\frac{f\left(x\right)-f\left(\frac{n\sqrt{2}}{m}\right)}{x - \frac{n\sqrt{2}}{m}} \right|
\le \frac{m(qm+pn\sqrt 2)}{p^2} = \frac{m(xm+n\sqrt 2)}{p} \, .
$$
If $x_k = q_k/p_k$ converges to $\frac{n\sqrt{2}}{m}$ then necessarily $p_k \to \infty$, and therefore
$$
\frac{f\left(x_k\right)-f\left(\frac{n\sqrt{2}}{m}\right)}{x_k - \frac{n\sqrt{2}}{m}} \to 0 \, .
$$
This proves that $f'\left( \frac{n\sqrt{2}}{m}\right) = 0$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4092726",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 1,
"answer_id": 0
} |
If $a$,$b$ and $y$ are the roots of $3x^3+8x^2-1=0$ find $(b+1/y)(y+1/a)(a+1/b)$ If a b and y are the roots of $3x^3+8x^2-1=0$ find $(b+1/y)(y+1/a)(a+1/b)$
This is what I have done so far, but apparently it is incorrect. I want to know why.
$(b+1/y)(y+1/a)(a+1/b)$
$(by+1/y)(ay+1/a)(ab+1/b)$
$(aby^2+1/ay)(ab+1/b)$
which is equal to
$a^2b^2y^2 + 1/aby$
Using Vieta's formula I get:
$aby = -d/a$
$aby = 1/3$
Subbing in original:
$(1/9 + 1)/1/3$
which is...
10/27
The answer is supposedly 2/3, want to know what I did wrong.
| Write $(b+1/y)(y+1/a)(a+1/b)$ as $\frac{(by+1)(ya+1)(ab+1)}{yab}$ by finding a common denominator for each bracket. Expanding this gives:
$$\frac{(aby)^2+aby(a+b+y)+ab+by+ya+1}{aby}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4101608",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 1
} |
Find the volume of the region bounded by a sphere and a paraboloid using cylindrical and spherical coordinates. The sphere is $x^2 + y^2 + z^2 = 4$ and the paraboloid is $x^2 + y^2 = 3z$
I've already done for the cylindrical coordinates and got:
$$\int^{2\pi}_0\int_0^1\int_{\frac{r^2}{3}}^\sqrt{4-r^2}rdzdrd\theta = \frac{\pi(31-12\sqrt{3})}{6}$$
But I'm having troubles when building the integral for spherical coordinates. So far I've found the bounds of integration for $d\rho$ and $d\theta$.
$$\int^{2\pi}_0\int\int_0^2\rho ^2\sin\phi \,\,\,d\rho d\phi d\theta$$
What about the bounds for $d\phi$? Done a small search and had to solve $3\rho\cos\phi = \rho^2\sin^2\theta$, but this is the point where I'm stuck. Is this right? If yes, how to solve this?
| At the intersection of paraboloid and sphere,
$x^2 + y^2 + z^2 = 4, x^2+y^2 = 3z \implies z^2 + 3z - 4 = 0$.
i.e, $z = 1, x^2+y^2 = 3$.
So in cylindrical coordinates, $0 \leq r \leq \sqrt3$.
Now in spherical coordinates,
$x = \rho \cos\theta \sin\phi,y = \rho \sin\theta \sin\phi, z = \rho \cos\phi$
At the intersection of paraboloid and sphere,
$\rho = 2, z = \rho \cos\phi = 1, \implies \cos\phi = \frac{1}{2}$
So for $0 \leq \phi \leq \frac{\pi}{3}$, the region is bound by the sphere and for $\frac{\pi}{3} \leq \phi \leq \frac{\pi}{2}$, the region is bound by the paraboloid.
$3z = x^2 + y^2 \implies 3 \rho \cos\phi = \rho^2 \sin^2\phi$
$\rho = 3 \cos\phi \csc^2\phi$
So bounds for the second integral are,
$\frac{\pi}{3} \leq \phi \leq \frac{\pi}{2}, 0 \leq \rho \leq 3 \cos\phi \csc^2\phi $
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4104811",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Find the maximum of $\sum_{k=1}^n x_k^4$ Given $x_k\in\mathbb{R}$ satisfying
$$\sum_{k=1}^n x_k=0\quad\text{and}\quad\sum_{k=1}^n x_k^2=n^2-n$$
Find the maximum of
$$\sum_{k=1}^n x_k^4$$
I have notice that $\displaystyle\sum_{k=1}^n x_k=0$ implies that
$$x_m^2=\bigg(\sum_{\substack{k=1\\ k\not=m}}^n x_k\bigg)^2\leq (n-1)\sum_{\substack{k=1\\ k\not=m}}^n x_k^2=(n-1)(n^2-n-x_m^2)$$
using Chebyshev's inequality and $\displaystyle\sum_{k=1}^n x_k^2=n^2-n$. Therefore we obtain $nx_m^2\leq n(n-1)^2\Leftrightarrow x_m\leq n-1$. Hence we have
\begin{align*}\begin{split}&\sum_{k=1}^n (x_k-n+1)(x_k+1)^2\leq0\\ \Leftrightarrow \sum_{k=1}^n x_k^3 &\leq (n-3)\sum_{k=1}^n x_k^2+\sum_{k=1}^n (n-1)\quad\text{since}\;\sum_{k=1}^n x_k=0 \\ &= n(n-1)(n-2)=(n-1)^3-(n-1)\end{split}\end{align*}
since I notice that $(x_k+1)^2$ would result me with an optimal, but I have no idea why is it so. Now using a similar approach, we find that
\begin{align*}\begin{split}&\sum_{k=1}^n (x_k^2-(n-1)^2)(x_k+1)^2\leq0 \\ \Leftrightarrow \sum_{k=1}^n x_k^4 &\leq-2\sum_{k=1}^n x_k^3+(n^2-2n)\sum_{k=1}^n x_k^2+\sum_{k=1}^n (n^2-2n+1)\\ &=-2\sum_{k=1}^n x_k^3+(n^2-2n)(n^2-n)+n(n^2-2n+1)\end{split}\end{align*}
and I have no idea to how to proceed from here since I do not know the minimum of $\displaystyle\sum_{k=1}^n x_k^3$ but I would like to give a guess that the maximum of $\displaystyle\sum_{k=1}^n x_k^4$ is $(n-1)^4+(n-1)$.
| It looks smart at the first glance to use $(x_k^2-(n-1)^2)(x_k+1)^2\leq0$, but it would not get you to anywhere. In this case the equity holds at $x_1=1-n$, and $x_2=\cdots=x_n=1$, therefore we obtain $x_m\geq1-n$, hence we have
\begin{align*}\begin{split}&\qquad\qquad\sum_{k=1}^n (x_k+n-1)^2(x_k-1)^2\geq 0\\ \therefore \sum_{k=1}^n x_k^4 &\leq (4-2n)\sum_{k=1}^n x_k^3-(n^2-6n+6)\sum_{k=1}^n x_k^2-\sum_{k=1}^n (n^2-2n+1)\\ &\leq (4-2n)(-(n-1)^3+(n-1))-(n^2-6n+6)(n^2-n)-n(n^2-2n+1)\\ &= n^4-4n^3+6n^2-3n=(n-1)^4+(n-1)\end{split}\end{align*}
Not too sure if this is the solution since the inequality is flip yet I have no reason for that, maybe it is because of the $x_k^3$. Another thing to consider is that $x_m\geq1-n\Leftrightarrow x_m^3\geq(1-n)^3=-(n-1)^3$, one may also try using the fact that $x_m^3+(n-1)^3\geq0$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4108144",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 1,
"answer_id": 0
} |
Help with $ \lim \frac{n\pi}{4} - \left( \frac{n^2}{n^2+1^2} +\frac{n^2}{n^2+2^2} + \cdots + \frac{n^2}{n^2+n^2} \right) $ I have proved that $$\lim_{n\to \infty} \left( \frac{n}{n^2+1^2} +\frac{n}{n^2+2^2} + \cdots + \frac{n}{n^2+n^2} \right) = \frac{\pi}{4},$$
by using the Riemann's sum of $\arctan$ on $[0,1]$. Now I'm interested in computing the next limit
$$ L=\lim_{n\to \infty} \left[ \frac{n\pi}{4} - \left( \frac{n^2}{n^2+1^2} +\frac{n^2}{n^2+2^2} + \cdots + \frac{n^2}{n^2+n^2} \right) \right] $$
Note that
$$ L= \lim n \left[ \frac{\pi}{4} - \left( \frac{n}{n^2+1^2} +\frac{n}{n^2+2^2} + \cdots + \frac{n}{n^2+n^2} \right) \right] =[\infty \cdot 0].$$
I can't manage to calculate $L$. How can this indeterminate form be solved? Also, by using computer, I suspect that $L$ exists and is equal to $1/4$. Any idea or hint is welcome! (Actually, I dont need a complete solution, just a good start point). Have a nice day you all!
| By Taylor's formula
$$
\frac{1}{{1 + (x/n)^2 }} = \frac{1}{{1 + (k/n)^2 }} - \frac{1}{n}\frac{{2k/n}}{{(1 + (k/n)^2 )^2 }}(x - k) + \frac{{(3(\xi /n)^2 - 1)}}{{(1 + (\xi /n)^2 )^3 }}\frac{1}{{n^2 }}(x - k)^2 .
$$
for $0<x<n$, $0\leq k\leq n$ and some $0<\xi<n$ depending on $x$, $k$ and $n$. Since
$$
\left|\frac{{(3(\xi /n)^2 - 1)}}{{(1 + (\xi /n)^2 )^3}} \right|\leq 1
$$
for any $0<\xi<n$, we can write
$$
\frac{1}{{1 + (x/n)^2 }} = \frac{1}{{1 + (k/n)^2 }} - \frac{1}{n}\frac{{2k/n}}{{(1 + (k/n)^2 )^2 }}(x - k) + \mathcal{O}\left(\frac{1}{{n^2 }}\right)(x - k)^2
$$
uniformly for $0<x<n$ and $0\leq k\leq n$. Thus
$$
\int_{k - 1}^k {\frac{{dx}}{{1 + (x/n)^2 }}} = \frac{1}{{1 + (k/n)^2 }} + \frac{1}{n}\frac{{k/n}}{{(1 + (k/n)^2 )^2 }} + \mathcal{O}\!\left( {\frac{1}{{n^2 }}} \right),
$$
and so
$$
\frac{1}{n}\sum\limits_{k = 1}^n {\frac{1}{{1 + (k/n)^2 }}} = \frac{1}{n}\int_0^n {\frac{{dx}}{{1 + (x/n)^2 }}} - \frac{1}{{n^2 }}\sum\limits_{k = 1}^n {\frac{{k/n}}{{(1 + (k/n)^2 )^2 }}} + \mathcal{O}\!\left( {\frac{1}{{n^2 }}} \right),
$$
i.e.,
$$
\sum\limits_{k = 1}^n {\frac{{n^2 }}{{n^2 + k^2 }}} = \frac{\pi }{4}n - \frac{1}{n}\sum\limits_{k = 1}^n {\frac{{k/n}}{{(1 + (k/n)^2 )^2 }}} +\mathcal{O}\!\left( {\frac{1}{n}} \right).
$$
Finally,
$$
\frac{\pi }{4}n - \sum\limits_{k = 1}^n {\frac{{n^2 }}{{n^2 + k^2 }}} = \frac{1}{n}\sum\limits_{k = 1}^n {\frac{{k/n}}{{(1 + (k/n)^2 )^2 }}} + \mathcal{O}\!\left( {\frac{1}{n}} \right) \to \int_0^1 {\frac{x}{{(1 + x^2 )^2 }}dx} = \frac{1}{4}.
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4111276",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 2,
"answer_id": 0
} |
Given $\sin{x} = \frac{1 + \sqrt{5}}{4}$, find $x$ If I have been given that
$$\sin{x} = \frac{1 + \sqrt{5}}{4}$$ is there any way using which I can find out what the value of $x$ is? Sure I can remember the values, but I was curious whether if I knew the values of $\sin{\frac{\pi}{2}}, \sin{\frac{\pi}{4}}, \sin{\frac{\pi}{3}}$ I could somehow figure out what $x$ would be? Is there any intuition behind it?
| Let $\cos y = \frac{\sqrt5+1}4$. Then
$$\cos 2y =2\cos^2 y-1= 2\left(\frac{\sqrt5+1}4\right)^2-1=\frac{\sqrt5-1}4$$
$$\cos 4y =2\cos^2 2y-1= -\frac{\sqrt5+1}4= -\cos y= \cos(\pi\pm y)$$
The $+$ sign leads to $y=\frac\pi3$ while the $-$ sign leads to $y=\frac\pi5$. But then the value above for $\cos 2y$ is positive so we must have $y<\frac\pi4$. Therefore, $y=\frac\pi5$ is selected and $x=\frac\pi2-\frac\pi5=\frac{3\pi}{10}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4111996",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 1
} |
Image under a Mobius Transformation I am studying the Mobius transformation
$$f(z) = \frac{z(1+i)}{z - i}$$
I need to find where the transformation maps the real axis, the imaginary axis and the unit circle. I know that Mobius transformations map lines and circles to lines or circles.
So I let $z = x + iy$, so we have $$f(x,y) = \frac{(x+iy)(1 + i)}{x + iy - i}$$ Now the real line is given by $y = 0$, and the imaginary axis by $x = 0$ and the unit circle by $x^2 + y^2 = 1$. From here, I substitute each of these values:
$$f(x, 0) = \frac{(x)(1+i)}{x - i}$$
Now I have seen that we only need three points to uniquely determine a line/circle so I tested several values to see if I could see any pattern
$$f(-1,0) = 1$$
$$f(0,0) = 0$$
$$f(1,0) = i$$
$$f(2,0) = \frac{2 + 6i}{5}$$
$$f(3,0) = \frac{3 + 6i}{5}$$
I graphed these values using Desmos, and this does not appear to be a line or a circle.
I did the same for $$f(0,y) = \frac{(iy)(1 + i)}{iy - i}$$ to get
$$f(0,-1) = \frac{1+i}{2}$$
$$f(0,0) = 0$$
$$f(0,1) = \infty$$
$$f(0,2) = 2+2i$$
$$f(0,3) = \frac{3 + 3i}{2}$$
Again, I do not see how this is a line or a circle.
Finally, I am not exactly sure what to do with the unit circle. Although I could plug in some values for the that lie on the unit circle, it does not seem that my approach is working.
I appreciate any help!
| Start with
$$
\frac{z}{z-i} = \frac{x+iy}{x+(y-1)i} = \frac{(x+iy)(x-(y-1)i)}{x^2+(y-1)^2} = \\
= \frac{x^2 + y(y-1) + (xy - x(y-1))i}{x^2+(y-1)^2} =
\frac{x^2 + y(y-1)}{x^2+(y-1)^2} + i\frac{x}{x^2+(y-1)^2}
$$
So, for $z = x + i 0$, $x\in\mathbb{R}$:
$$
f(z) = (1+i)\left(
\frac{x^2}{x^2+1} + i\frac{x}{x^2+1}
\right)
$$
Since $\frac{x^2}{x^2+1} = \frac{x^2}{x^2+1} - \frac{1}{2} + \frac{1}{2} = \frac{2x^2-x^2-1}{2x^2+2} +
\frac{1}{2} = \frac{x^2 -1}{2x^2+2} + \frac{1}{2} = \frac{1}{2}\frac{x^2-1}{x^2+1}+\frac{1}{2}$,
$$
f(z) = (1+i)\cdot\frac{1}{2}\cdot\left(\frac{x^2-1}{x^2+1} + i \frac{2x}{x^2+1} +
1\right)
$$
As $\left(\frac{x^2-1}{x^2+1}\right)^2 + \left(\frac{2x}{x^2+1}\right)^2 =
\frac{x^4 - 2x^2 + 1 + 4x^2}{x^4 + 2x^2 + 1} = 1$, one may guess that
$\left(\frac{x^2-1}{x^2+1}, \frac{2x}{x^2+1}\right), x\in\mathbb{R}$ curve is unit circle centered at zero, so $\frac{x^2}{x^2+1} + i\frac{x}{x^2+1}$ gives circle with radius $\frac{1}{2}$ centered at $(1, 0)$, and multiplying it with $1+i$ simply scales it and rotates, since $1+i = \sqrt{2}e^{i\pi/4}$.
Check it in Desmos. So, image of real line is cirlce with radius $\frac{\sqrt{2}}{2}$ and center in $(0.5, 0.5)$.
Similarly, for imaginary axis with $z = 0 + iy$, $y\in\mathbb{R}$:
$$
f(z) = (1+i)\left(\frac{y(y-1)}{(y-1)^2}\right) = (1+i)\frac{y}{y-1}
$$
So image of imaginary axis is line through $0$ and $(1, 1)$ points on a plane.
Finally, case of unit cirlce.
$$
\frac{z}{z-i} =
\frac{x^2 + y(y-1)}{x^2+(y-1)^2} + i\frac{x}{x^2+(y-1)^2} =
\frac{x^2 + y^2 - y}{x^2 + y^2 - 2y + 1} + i \frac{x}{x^2+y^2 -2y+1}
$$
For $z = e^{it}$, $t\in [0; 2\pi]$, $x^2 + y^2 = 1$, so
$$\frac{z}{z-i} = \frac{1-y}{2 - 2y} + i\frac{x}{2-2y} = \frac{1}{2} + i\cdot\frac{1}{2}\cdot\frac{\cos t}{1 - \sin{t}}$$
$\frac{\cos t}{1 - \sin{t}}$ is clearly unbounded, so it is equation of line $
\frac{1}{2} + i y$, $y\in\mathbb{R}$. Multiplying by $1+i$, you get
$\frac{1}{2} - y + i\left(\frac{1}{2} +y\right)$ - it's also a straight line, check it with Desmos.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4112715",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Solve the following differential equation by undetermined coefficients $(1-x^2)y''-2xy'+6y=0$ I'm trying to obtain a recursion formula relating $a_{n+2}$ to $a_n$ using the method of undetermined coefficients. Then determining $a_n$ explicitly for each $n$ and finding the sum of the series.
$$(1-x^2)y''-2xy'+6y=0$$
Taking the first and second derivative of the power series gives me:
$$(1)\space y = \sum_{n=0}^{\infty}a_nx^n $$
$$(2)y'\space = \sum_{n=1}^{\infty} na_nx^{n-1}$$
$$(3) \space y'' = \sum_{n=2}^{\infty}n(n-1)a_nx^{n-2}$$
Then plugging this into the differential equation:
$$ (a) \space \sum_{n=2}^{\infty}n(n-1)a_nx^{n-2}+\sum_{n=2}^{\infty}n(n-1)a_nx^{n}-2\sum_{n=1}^{\infty} na_nx^{n} + 6 \sum_{n=0}^{\infty}a_nx^n$$
Then simplifying the exponents:
$$(b) \space \sum_{n=0}^{\infty}(n+2)(n+1)a_{n+2}x^{n} + \sum_{n=0}^{\infty}n(n-1)a_{n}x^{n} - 2\sum_{n=0}^{\infty} na_nx^{n} + 6 \sum_{n=0}^{\infty}a_nx^n $$
Then simplifying with $x^n$:
$$(c) \space \left([(n+2)(n+1)a_{n+2}] + [n(n-1-2)+6]a_n\right)x^n$$
Finally, equating coefficients of $x^n$ to obtain the relation:
$$(d) \space [(n+2)(n+1)a_{n+2}] + [n(n-1-2)+6]a_n=0$$
Or what amounts to:
$$(4) \space a_{n+2} = \frac{n(n-1-2)a_n}{(n+2)(n+1)}$$
$$(5) \space a_n = \frac{(n+2)(n+1)}{n(n-1-2)}$$
I would really appreciate a check on my calculations and whether I missed something out or went wrong along the way. Additionally, how do I proceed from here?
| This is a special case of Legendre's differential equation at $\ell=2$.
The series expansion should be $$\sum_{n=2}^{\infty}n(n-1)a_nx^{n-2}\color{blue}-\sum_{n=2}^{\infty}n(n-1)a_nx^{n}-2\sum_{n=1}^{\infty} na_nx^{n} + 6 \sum_{n=0}^{\infty}a_nx^n=0$$ from which $$(n+2)(n+1)a_{n+2}-(n+3)(n-2)a_n=0.$$ Note that $a_2=-3a_0$ and $12a_4=0$ so $a_{2N}=0$ for all $N>1$. Iterating gives \begin{align}a_{n+2}&=\frac{(n+3)(n-2)}{(n+2)(n+1)}a_n=\frac{(n+3)(n-2)}{(n+2)(n+1)}\frac{(n+1)(n-4)}{n(n-1)}a_{n-2}=\cdots\\&=(n+3)(n+1)\frac{(n-1)!(n-2k)(n-2k-2)}{(n-2k+1)!}\frac{(n-2k)!}{(n+2)!}a_{n-2k}\\&=\frac{n+3}{(n+2)n}\frac{(n-2k)(n-2k-2)}{n-2k+1}a_{n-2k}.\end{align} When $n=2N+1$ choosing $k=N$ gives $$a_{2N+3}=-\frac{N+2}{(2N+3)(2N+1)}a_1$$ so the solution is $$y(x)=a_0\left(1-3x^2\right)+a_1x\left(1-\frac23x^2-\frac15x^4-\frac4{35}x^6-\cdots\right).$$ Indeed, taking $a_0=1/2$ and $a_1=0$ yields the Legendre polynomial of order two.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4114821",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Calculating $\lim_{(x,y)\rightarrow (\infty,\infty)}(x^2+y^2)e^{-(x+y)}$ Please help me calculate the:
$$\lim_{(x,y)\rightarrow (\infty,\infty)}(x^2+y^2)e^{-(x+y)}$$
My attempt is: First method: for $t>0$ we have $e^t\geq \frac{t^3}{3}$ now i have:
$$(x^2+y^2)e^{-(x+y)}\leq\frac{3(x^2+y^2)}{(x+y)}\rightarrow 0, (x,y)\rightarrow(\infty,\infty)$$
but i dint know how to continue
second method:the polar coord
$$\lim_{(x,y)\rightarrow (\infty,\infty)}(x^2+y^2)e^{-(x+y)}=\lim_{r\rightarrow \infty}(r^2)e^{-r(cos\theta+sin\theta)}=\lim_{r\rightarrow \infty}\frac{r^2}{e^{r(cos\theta+sin\theta)}}=\frac{1}{cos\theta+\sin\theta}\lim_{r\rightarrow \infty}\frac{2r}{e^r}=\frac{2}{cos\theta+\sin\theta}\lim_{r\rightarrow \infty}\frac{1}{e^r}=\frac{2}{cos\theta+\sin\theta}\cdot0=0$$ i use the L'Hopital rule,but i dint know its correct
| Using $ e^t > \frac{t^4}{4!}$, we have for $x,y >0$
$$e^{x+y} > \frac{(x+y)^4}{4!} = \frac{(x^2+y^2+2xy)^2}{4!} > \frac{(x^2+y^2)^2}{4!} $$
$$(x^2+y^2)e^{-(x+y)} < 4! \frac{x^2+y^2}{(x^2+y^2 )^2}=\frac{4!}{(x^2+y^2)} \to 0$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4117289",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
} |
Maximize $x^2y$ where $x^2+y^2=100$. (Looking for other approaches)
For positive numbers $x,y$ we have $x^2+y^2=100$. For which ratio of $x$ to $y$, the value of $x^2y$ will be maximum?
$1)\ 2\qquad\qquad2)\ \sqrt3\qquad\qquad3)\ \frac32\qquad\qquad4)\ \sqrt2$
It is a problem from a timed exam so I'm looking for alternative approaches to solve this problem quickly. Here is my approach:
We have $x^2=100-y^2$, so $x^2y=y(100-y^2)$. By differentiating with respect to $y$ and putting it equal to zero, we have:
$$(100-y^2)-2y^2=0\qquad\qquad y^2=\frac{100}3$$
Hence $x^2=\frac{200}3$ and $\frac{x^2}{y^2}=2$ therefore $\frac{x}{y}=\sqrt2$.
Is there another quick approach to solve it? (like AM-GM inequality or others)
| One approach could be to parameterize the curve $x^2+y^2=100$. Define the functions $x$ and $y$ by $x(t)=10\cos(t)$ and $y(t)=10\sin(t)$ on the interval $(0,\pi/2)$ (this constraint ensures that $x$ and $y$ are bijective and strictly positive, since $\cos(t)>0$ and $\sin(t)>0$ for all $t\in(0,\pi/2)$). The quantity $x^2y$ is then simply $\cos^2(t)\sin(t)$, and we can find its maximum by finding the maximum of $\cos^2(t)\sin(t)$ on $(0,\pi/2)$.
Differentiating and equating the resulting expression to $0$ gives
$$\frac{d}{dt}\left[\cos^2(t)\sin(t)\right]=-2\cos(t)\sin^2(t)+\cos^3(t)=0$$
$$\iff \cos^3(t)=2\cos(t)\sin^2(t)$$
$$\iff \frac{1}{2}=\tan^2(t)$$
$$\iff t=\tan^{-1}\left(\frac{1}{\sqrt{2}}\right)\text{, since tan(t) is bijective on }(0,\pi/2)$$
Once you've proven that the critical point $t=\tan^{-1}\left(\frac{1}{\sqrt{2}}\right)$ does indeed correspond to a maximum for $\cos^2(t)\sin(t)$, it follows that the ratio $\frac{x}{y}$ that maximizes $x^2 y$ is
\begin{align*}
\frac{\cos\left(\tan^{-1}\frac{1}{\sqrt{2}}\right)}{\sin\left(\tan^{-1}\frac{1}{\sqrt{2}}\right)} &= \frac{1}{\tan\left(\tan^{-1}\frac{1}{\sqrt{2}}\right)}\\
&= \frac{1}{1/\sqrt{2}}\\
&= \sqrt{2}
\end{align*}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4119234",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 0
} |
$\int_0^1\frac{\ln x\ln^2(1-x^2)}{\sqrt{1-x^2}}dx=\frac{\pi}{2}\zeta(3)-2\pi\ln^32$ I'm looking for proof of the following identity
$$\int_0^1\frac{\ln x\ln^2(1-x^2)}{\sqrt{1-x^2}}dx=\frac{\pi}{2}\zeta(3)-2\pi\ln^32$$
I have worked on this problem for quite some time, however since I'm not much comfortable with beta functions and stuff, I was unable to prove the required. I'm looking for an elementary approach, however, any detailed method (including beta function) is most welcomed. Thanks.
| $$I=\int_0^1\frac{\ln x\ln^2(1-x^2)}{\sqrt{1-x^2}}dx$$ $$=\frac{1}{4}\int_0^1\frac{\ln t\ln^2(1-t)}{\sqrt t\sqrt{1-t}}dt=\frac{1}{4}\frac{d^2}{db^2}\frac{d}{da}\int_0^1t^{a-1/2}(1-t)^{b-1/2}dt|_{a=b=0}$$
$$=\frac{1}{4}\frac{d^2}{db^2}\frac{d}{da}B(a+1/2;b+1/2)|_{a=b=0}$$
$$\frac{1}{4}B(a+1/2;b+1/2)=\frac{1}{4}\frac{\Gamma(a+1/2)\Gamma(b+1/2)}{\Gamma(a+b+1)}=\frac{\pi}{4}\frac{\Gamma(1+2a)}{\Gamma(1+a)}\frac{\Gamma(1+2b)}{\Gamma(1+b)}\frac{4^{-a}4^{-b}}{\Gamma(1+a+b)}$$
$$\Gamma(a)\Gamma\bigl(a+1/2\bigr)=2\sqrt{\pi}\,4^{-a}\,\Gamma(2a);\,\,a\Gamma(a)=\Gamma(1+a)=1-\gamma a+\Gamma''(1)\frac{a^2}{2}+...$$
$$\Psi(t)=\frac{\Gamma'(t)}{\Gamma(t)};\,\,\Psi(1)=-\gamma;\,\,\Gamma''(1)=\gamma^2+\frac{\pi^2}{6};\,\,\Psi'(1)=\zeta(2);\,\,\Psi''(1)=-2\zeta(3)$$
$$I=\frac{\pi}{4}\frac{d^2}{db^2}\frac{d}{da}\biggl(\frac{(1-a\gamma+...)(1-a\ln4+...)}{\Gamma(1+b)+\Gamma'(1+b)a+...}\frac{\Gamma(1+2b)}{\Gamma(1+b)}4^{-b}\biggr)|_{a=b=0}$$
$$=-\frac{\pi}{4}\frac{d^2}{db^2}\biggl(\bigl(1-b\ln4+\frac{b^2}{2}\ln^24-..\bigr)\frac{1-2\gamma b+2\Gamma''(1)b^2+..}{(1-\gamma b+\Gamma''(1)\frac{b^2}{2}+..)^2}\bigl(\gamma+\ln4+\Psi(1+b)\bigr)\biggr)|_{b=0}$$
$$=-\frac{\pi}{4}\frac{d^2}{db^2}\biggl(\bigl(1-b\ln4+\frac{b^2}{2}\ln^24+..\bigr)\bigl(1-2\gamma b+2\Gamma''(1)b^2+..\bigr)\bigl(1+2\gamma b-\Gamma''(1)b^2+3\gamma^2b^2+..\bigr)\bigl(\gamma+\ln4-\gamma+b\zeta(2)-b^2\zeta(3)\bigr)+...\Bigr)|_{b=0}$$
$$=-\frac{\pi}{2}\Bigl(\frac{\pi^2}{6}\ln4-\ln4\zeta(2)-\zeta(3)+\frac{1}{2}\ln^34\Bigr)$$
$$I=\frac{\pi}{2}\zeta(3)-2\pi\ln^32$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4120781",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 2,
"answer_id": 1
} |
Angle between tangents to the curve $x^2+3y^2=9$
Tangents drawn from the point $(\alpha,\alpha^2)$ to the curve $x^2+3y^2=9$ include an acute angle between them, then find $\alpha$.
My attempt is by using the equation for pair of tangents from an external point to the curve. If the curve is $S=0$ then the equation for pair of tangents is $SS_1=T^2$, where $S_1=0$ is obtained after putting the external point in $S=0$. And $T=0$ is obtained after changing $x^2$ to $xx_1$ and $y^2$ to $yy_1$ in $S=0$, where $(x_1,y_1)$ is the external point from which the tangents are being drawn.
So,
$$SS_1=T^2$$
$$\implies (x^2+3y^2-9)(\alpha^2+3\alpha^4-9)=(x\alpha+3y\alpha^2-9)^2$$
Now, if $\theta$ is the angle between the tangents then $\tan\theta=|\frac{2\sqrt{h^2-ab}}{a+b}|$, where, $2h=$ coefficient of $xy=6\alpha^3, a=$ coefficient of $x^2=3\alpha^4-9$ and $b=$ coefficient of $y^2=3\alpha^2-27$
Therefore, $\tan\theta=|\frac{2\sqrt{9\alpha^6-9(\alpha^4-3)(\alpha^2-9)}}{3\alpha^4+3\alpha^2-36}|=|\frac{2\sqrt{\alpha^6-(\alpha^6-9\alpha^4-3\alpha^2+27)}}{\alpha^4+\alpha^2-12}|=|\frac{2\sqrt{9\alpha^4+3\alpha^2-27}}{\alpha^4+\alpha^2-12}|$
Not able to proceed next.
| Hint: $\alpha^4+\alpha^2-12=(\alpha^2-3)(\alpha^2+4)$ making the angle $\pi/2$ for $\alpha=\pm\sqrt{3}.$
Now check for say $\alpha\approx 2.2$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4127113",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Double Integral of $\sqrt{x^2+y^2}$ I want to calcule the following integral,
$$\int_0^{\frac{1}{\sqrt{2}}}\int_0^{\frac{1}{\sqrt{2}}}\sqrt{x^2+y^2}dxdy$$
what I have is ...
Let $x=y\tan\theta$ where $\theta\in(-\pi/2,\pi/2)$, then, $dx=y\sec^2\theta d\theta$, so,
$$\int_0^{\frac{1}{\sqrt{2}}}\int_0^{\frac{1}{\sqrt{2}}}\sqrt{x^2+y^2}dxdy=\int_0^{\frac{1}{\sqrt{2}}}\int_0^{\arctan(\frac{y}{\sqrt{2}})}y^2\sec^3\theta \,d\theta\,dy$$
From here,
$$\int_0^{\arctan(\frac{y}{\sqrt{2}})}y^2\sec^3\theta \,d\theta=\frac{\sqrt{2y^2+1}}{4}+\frac{1}{2}y^2\ln\left(\frac{|\sqrt{2y^2+1}+1|}{\sqrt{2}|y|}\right)$$
If I continue with the main problem, $$\int_0^{\frac{1}{\sqrt{2}}}\int_0^{\arctan(\frac{y}{\sqrt{2}})}y^2\sec^3\theta \,d\theta\,dy=\int_0^{\frac{1}{\sqrt{2}}}\left[\frac{\sqrt{2y^2+1}}{4}+\frac{1}{2}y^2\ln\left(\frac{|\sqrt{2y^2+1}+1|}{\sqrt{2}|y|}\right)\right]dy$$
$$=\int_0^{\frac{1}{\sqrt{2}}}\frac{\sqrt{2y^2+1}}{4}dy+\frac{1}{2}\int_0^{\frac{1}{\sqrt{2}}}y^2\ln\left(\frac{|\sqrt{2y^2+1}+1|}{\sqrt{2}|y|}\right)dy$$
$$=\frac{\sqrt{2}+\ln(1+\sqrt{2})}{8\sqrt{2}}+\frac{1}{2}\int_0^{\frac{1}{\sqrt{2}}}y^2\ln\left(\frac{|\sqrt{2y^2+1}+1|}{\sqrt{2}|y|}\right)dy$$
Here I´m stuck, I don´t know how to keep going with the second integral.
| That integral is equal to$$2\int_0^{1/\sqrt2}\int_0^x\sqrt{x^2+y^2}\,\mathrm dy\,\mathrm dx.$$In polar coordinates, that is equal to$$2\int_0^{\pi/4}\int_0^{1/(\sqrt2\cos(\theta))}\rho^2\,\mathrm d\rho\,\mathrm d\theta=\frac{\sqrt{2}+\log \left(1+\sqrt{2}\right)}{6\sqrt{2}}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4128417",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Compute $\lim_{n\to\infty}\sqrt[3]{8n^3+4n^2+n+11}-{\sqrt{4n^2+n+9}}$ Compute $\lim_{n\to\infty}\sqrt[3]{8n^3+4n^2+n+11}-\sqrt{4n^2+n+9}$.
My first thoughts are the that I can remove a factor of $2n$ from each root, and obtain
$$ \lim_{n\to\infty}2n\sqrt[3]{1+\frac{1}{2n}+\frac{1}{8n^2}+\frac{11}{8n^3}}-2n\sqrt{1+\frac{1}{2n}+\frac{9}{4n^2}} $$
But factoring out $2n$ won't help evaluate the limit because of the indeterminate form... So then I thought of using L'Hop, but thinking about the chain rule there makes me think that its more effort than it's worth and there's a missing something I can't get to. Any suggestions?
| $$
\begin{align*}
&\lim_{n\rightarrow \infty} \left( \sqrt[3]{8n^3+4n^2+n+11}-\sqrt{4n^2+n+9} \right)
\\
&=\lim_{n\rightarrow \infty} 2n\left( \sqrt[3]{1+\frac{1}{2n}+\frac{1}{8n^2}+\frac{11}{8n^3}}-\sqrt{1+\frac{1}{4n}+\frac{9}{4n^2}} \right)
\\
&=\lim_{n\rightarrow \infty} 2n\left( \sqrt[3]{1+\frac{1}{2n}+\frac{1}{8n^2}+\frac{11}{8n^3}}-1 \right) +\lim_{n\rightarrow \infty} 2n\left( 1-\sqrt{1+\frac{1}{4n}+\frac{9}{4n^2}} \right)
\\
&=\lim_{n\rightarrow \infty} 2n\cdot \frac{1}{3}\left( \frac{1}{2n}+\frac{1}{8n^2}+\frac{11}{8n^3} \right) -\lim_{n\rightarrow \infty} 2n\cdot \frac{1}{2}\left( \frac{1}{4n}+\frac{9}{4n^2} \right)
\\
&=\frac{2}{3}\lim_{n\rightarrow \infty} \left( \frac{1}{2}+\frac{1}{8n}+\frac{11}{8n^2} \right) -\lim_{n\rightarrow \infty} \left( \frac{1}{4}+\frac{9}{4n} \right)
\\
&=\frac{1}{3}-\frac{1}{4}
\\
&=\frac{1}{12}
\end{align*}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4128775",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 4
} |
For $(1 + x + x^2)^6$, find the term which has $x^6$ in it.
For $(1 + x + x^2)^6$, find the term which has $x^6$ in it.
I tried to use Newton's binomial formula as:
$$
(1 + x + x^2)^6 = \sum_{k = 0}^{6}\left( \binom{6}{k}(1 + x)^{n-k} x^{2k}\right)
$$
and that's all I can think of, other then just to compute it.
| Apply the trinomial theorem: $$(1+x+x^2)^6=\sum_{a+b+c=6,\:0\le a,b,c\le 6} \binom{6}{a,b,c}(1)^a(x)^b(x^2)^c$$
To get the $x^6$ term, the possibilities are $(a,b,c)=(3,0,3),(2,2,2),(1,4,1),(0,6,0)$. Hence the $x^6$ term is $$\binom{6}{3,0,3}x^6+\binom{6}{2,2,2}x^6+\binom{6}{1,4,1}x^6+\binom{6}{0,6,0}x^6=141x^6$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4137533",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 9,
"answer_id": 0
} |
Summation over $a+b+c=5$ Let $a,b,c$ be positive integers. Compute
$$\sum_{a+b+c=5} (6-a)(6-b)(6-c).$$
The first thing I notice is symmetry, so that I can permute $3!=6$ ways, but i'm not really sure how that works with the condition $a+b+c=10.$ The other method is to fix $a$, but that is reall time-consuming and unfeasible if say $a+b+c=20.$ Is there a clever method to evaluate this sum?
I would like to have a generalized method, please.
| $$
\begin{array}{l}
a + b + c = 5\quad \left| {\,1 \le a,b,c\left( { \le 3} \right)} \right.\quad \Rightarrow \\
\Rightarrow \quad \left( {6 - a} \right) + \left( {6 - b} \right) + \left( {6 - c} \right)
= x + y + z = 13\quad \left| {\,3 \le x,y,z, \le 5} \right.\quad \Rightarrow \\
\Rightarrow \quad \left( {x,y,z} \right) \in S = \left\{ {\left( {3,5,5} \right) \vee \left( {4,4,5} \right)
\vee permut.} \right\}\quad \Rightarrow \quad \\
\Rightarrow \quad \sum\limits_{\left\{ {\begin{array}{*{20}c}
{\,1 \le a,b,c\left( { \le 3} \right)} \\
{a + b + c = 5} \\
\end{array}} \right.} {\left( {6 - a} \right)\left( {6 - b} \right)\left( {6 - c} \right)} = \\
= \sum\limits_{\left( {x,y,z} \right) \in S} {xyz} = 3\left( {3 \cdot 5 \cdot 5} \right)
+ 3\left( {4 \cdot 4 \cdot 5} \right) = 225 + 240 = 465 \\
\end{array}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4138871",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 1
} |
Infinite fraction's derivative
$f(x)=x+\dfrac{1}{x+\dfrac{1}{x+\dfrac{1}{x+\ldots}}}$
$f'\left(\dfrac{3}{2}\right)=?$
I tried to make equation like $y^2=xy+1$ but I can't made it clearly.
| Using implicit differentiation,
$f^2-xf-1=0
$
gives
$0
=2ff'-f-xf'
=(2f-x)f'-f
$
so
$f'
=\dfrac{f}{2f-x}
$.
Since
$f(x)
=\dfrac{x+\sqrt{x^2+4}}{2}
$
(need the positive root),
$f(\frac32)
=\dfrac{\frac32+\sqrt{\frac94+4}}{2}
=\dfrac{\frac32+\sqrt{\frac{25}{4}}}{2}
=\dfrac{\frac32+\frac52}{2}
=2
$
so
$f'(\frac32)
=\dfrac{2}{4-\frac32}
=\frac45
$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4140266",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
What is the limit of the $a_{n+1}=\frac{1+a_n+{a_{n-1}}^3}{3}$? $a_{n+1}=\frac{1+a_n+a_{n-1}^3}{3}$ with $n \geq 2$ ( $a_1 =0$, $a_2 = 1/2$)
By mathematical induction, the $0 \leq a_n \leq 1$ and the $a_n$ is monotonically increasing. If the limit exists, then $1$ or $\frac{-1 + \sqrt{5}}{2}$ would be value.
What is the exact limit of this recurrence relation between two values?
| If the limit exists, then let $L = \lim\limits_{n \to \infty} a_n$. We must have $L = \lim\limits_{n \to \infty} a_n = \lim\limits_{n \to \infty} a_{n + 1} = \lim\limits_{n \to \infty} \frac{1 + a_n + a_{n - 1}^3}{3} = \frac{1 + (\lim\limits_{n \to \infty} a_n) + (\lim\limits_{n \to \infty} a_{n - 1})^3}{3} = \frac{1 + L + L^3}{3}$.
Then we must have $L^3 - 2L + 1 = 0$. Note that $L^3 - 2L + 1 = (L - 1)(L^2 + L - 1)$. Then the three possible values are $L = 1$ and $L = \frac{-1 \pm \sqrt{5}}{2}$.
Note that we have $a_3 = \frac{1 + 1/2}{3} = 1/2$. Then it's easy to show that for all $n$, we have $a_{n + 1} \geq a_n$ using strong induction, since the base cases of $n = 1, 2$ are satisfied and for larger $n$, we have $a_{n + 1} = \frac{1 + a_n + a_{n - 1}^3}{3} \geq \frac{1 + a_{n - 1} + a_{n - 2}^3}{3} = a_n$.
So the sequence is increasing. Now, we show that the sequence is bounded. In fact, we see that for all $n$, we have $a_n \leq \frac{-1 + \sqrt{5}}{2}$. Let $w = \frac{-1 + \sqrt{5}}{2}$. This can be shown directly for $n = 1, 2$. To show it for other values of $n$, we use strong induction and note that $a_{n + 1} = \frac{1 + a_n + a_{n - 1}^3}{3} = \frac{1 + w + w_{n - 1}^3}{3} = w$.
The sequence is bounded and increasing, so it must have a limit. Since the sequence is increasing and $a_2 > \frac{-1 - \sqrt{5}}{2}$, that can't be the limit. And since $w < 1$ and the sequence is bounded by $w$, $1$ cannot be the limit either. Thus, the only possibility for the limit is $w$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4143849",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
Finding the bounds to solve for the volume using cylindrical coordinates My goal is to set up the triple integral that will solve for the volume inside the sphere $x^2 + y^2 + z^2 = 2z$ and above the paraboloid $x^2 + y^2 = z$ using cylindrical coordinates.
Upon solving, I know that the intersection of the sphere and the paraboloid is r = 1 and the bounds for $\theta$ is $0 \leq \theta \leq 2\pi$.
However, I am uncertain for the bounds of the z variable. I solve for the value of z in converted equation of the sphere $r^2 + z^2 = 2z$. By quadratic formula, I got the value of
$$ z = \frac{2 \pm \sqrt{4 - 4r^2}}{2}$$
Should I only include $z = \frac{1}{2} (2 + \sqrt{4-4r^2})$ as one of the bounds in the z variable? Or there will be a split integral?
My yielding triple integral in this case if I only include $z = \frac{1}{2} (2 + \sqrt{4-4r^2})$ would be
$$ \int_0^{2\pi} \int_0^1 \int_{r^2}^{\frac{1}{2}(2+\sqrt{4-4r^2})} r dzdrd\theta$$
| Yes you are correct. Paraboloid $z = x^2 + y^2 = r^2$ and Sphere $r^2 + z^2 = 2z$
To find $z$ at intersection, plug in $r^2 = z$ from equation of paraboloid into the equation of sphere. We get $z^2 - z = 0 \implies z = 0, 1$. So at intersection above origin, $z = 1$. We also find that at intersection $r = 1$ which is the radius of the sphere too.
Now as you mentioned, from the equation of the sphere, $ r^2 + (z-1)^2 = 1 \implies z = 1 \pm \sqrt{1-r^2}$
As the sphere is centered at $(0, 0, 1)$, for any $0 \leq r \lt 1$, there are two $z$ values on sphere - one below $z = 1 \ , (z = 1 - \sqrt{1-r^2})$ and one above $z = 1 \ , (z = 1 + \sqrt{1-r^2})$.
The intersection of paraboloid and sphere is also at $z = 1$, $z$ is bound below by paraboloid and above by sphere for $0 \leq r \leq 1$. So the point that you should take on sphere as upper bound is $z = 1 + \sqrt{1-r^2}$. Hence it should be,
$ \displaystyle \int_0^{2\pi} \int_0^1 \int_{r^2}^{1 + \sqrt{1-r^2}} r \ dz \ dr \ d\theta$
Edit: Now knowing that intersection is at $r = 1, z = 1$, the combined volume is volume of paraboloid between $0 \leq z \leq 1$ and half of the unit sphere above $z = 1$. So if you are allowed to use basic formula for volume, you can use the fact that volume of half unit sphere is $\frac{2 \pi }{3}$. In that case, the volume can be obtained as,
$\displaystyle \int_0^{2\pi} \int_0^1 \int_{r^2}^1 r \ dz \ dr \ d\theta + \frac{2 \pi}{3}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4148408",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Solve the inequality $\frac 1{x+1} - \frac 1{x} \le \frac 1{x-1} - \frac 1{x-2}$ $$\frac 1{x+1} - \frac 1{x} \le \frac 1{x-1} - \frac 1{x-2}$$
The answer for this inequality is given as $x ∈ (- \infty , -1) \cup(0, 1) \cup (2, \infty)$ but when I solve it, I am only able to get $x ∈ ( - \infty, -1)$.
How should I solve it to get the complete answer?
| Go step by step:
$$\frac 1{x+1} - \frac 1{x} \le \frac 1{x-1} - \frac 1{x-2} \\
\frac{-1}{x(x+1)} \le \frac{-1}{(x-2)(x-1)} \\
\frac{1}{(x-2)(x-1)} - \frac{1}{x(x+1)} \le 0 \\
\frac{x(x+1) - (x-2)(x-1)}{(x-2)(x-1)x(x+1)} \le 0 \\
\frac{x^2+x - (x^2-3x+2)}{(x-2)(x-1)x(x+1)} \le 0 \\
\frac{4x-2}{(x-2)(x-1)x(x+1)} \le 0$$
Can you finish this now?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4149304",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
Prove that $\sum_{n=1}^{\infty}\frac{1}{n^{2}} = \frac{\pi^{2}}{6}$
The goal is to prove for equality: \begin{equation*}
\sum_{n = 1}^{\infty} \frac{1}{n^{2}}=\frac{\pi^{2}}{6} \end{equation*} To do this, follow the steps below:
*
*Study the poles of the function $\phi(z) = \frac{\pi\cos(\pi z)}{\sin(\pi z)}$.
*Find an expression for the integral $$\int_{C_{N}} \frac{\pi\cos(\pi z)}{z^{2}\sin(\pi z)}$$, where $C_{N}$ is a curve
that encloses the integer up until $\pm N$.
*Considering $C_{N}$ as the rectangle with vertices $\pm(N+1/2)\pm(N+1/2)i$, prove that
$$\lim_{N\to\infty}\int_{C_{N}}\frac{\phi(z)}{z^{2}}dz=0$$
*Conclude.
What i try is
The dist thing is to note that the set of poles on $C_{N}$ is:
\begin{equation*}
\mathcal{P}(N) = \{k : |k|\leq N\}
\end{equation*}
then I note that the residue on $k=0$ is:
\begin{equation*}
\begin{split}
\text{Res}\left(\phi,k=0\right)
&= \lim_{z\to 0}(z-0)\phi(z)\\
&= \lim_{z\to0}(z-0)\cdot\frac{\pi\cos(\pi z)}{\sin(\pi z)}\\
&= \lim_{z\to0}\cos(\pi z)\cdot\frac{\pi x}{\sin(\pi z)}=1\\
\end{split}
\end{equation*}
To calculate the integral
\begin{equation*}
\int_{C_{N}}\frac{\phi(z)}{z^{2}}dz = \int_{C_{N}}\frac{\pi\cos(\pi z)}{z^{2}\sin(\pi z)}dz
\end{equation*}
whe can see that the pole is of order, meanwhile the other poles are simple, then:
\begin{eqnarray*}
\frac{1}{2\pi i}\int_{C_{N}}\frac{\pi\cos(\pi z)}{z^{2}\sin(\pi z)}dz
& = & \sum_{k=-N}^{N}\text{Res}\left(\frac{\phi(z)}{z^{2}},k\right)\\
& = & \text{Res}\left(\frac{\phi(z)}{z^{2}},0\right)+\sum_{k=-N,k\neq0}^{N}\text{Res}\left(\frac{\phi(z)}{z^{2}},k\right)\\
& = & \lim_{z\to0}\frac{1}{2!}\frac{d^{2}}{dz^{2}}\left[(z-0)^{3}\frac{\pi\cos(\pi z)}{z^{2}\sin(\pi z)}\right]+\sum_{k=-N,k\neq0}^{N}\text{Res}\left(\frac{\phi(z)}{z^{2}},k\right)
\end{eqnarray*}
\begin{eqnarray*}
\text{Res}\left(\frac{\phi(z)}{z^{2}},k\right)
& = & \lim_{z\to k}(z-k)\frac{\pi\cos(\pi z)}{z^{2}\sin(\pi z)}\\
& = & \frac{1}{k^2}\lim_{z\to k}(z-k)\frac{\pi\cos(\pi z)}{\sin(\pi z)}\\
& = & \frac{1}{k^2}\lim_{z\to k}\frac{\pi\cos(\pi z) - (z-k)\pi^{2}\sin(\pi z)}{\pi\cos(\pi z)}\\
& = & \frac{1}{k^2}\lim_{z\to k} \left(\frac{\pi\cos(\pi z)}{\pi\cos(\pi z)}-\frac{(z-k)\pi^{2}\sin(\pi z)}{\pi\cos(\pi z)}\right) = \frac{1}{k^{2}}\\
\end{eqnarray*}
Entonces
\begin{equation*}
\frac{1}{2\pi i}\int_{C_{N}}\frac{\pi\cos(\pi z)}{z^{2}\sin(\pi z)}dz = \text{Res}\left(\frac{\phi(z)}{z},0\right) + \sum_{\begin{subarray}{c}k=-N\\k\neq0\end{subarray}}^{N}\frac{1}{k^{2}}
\end{equation*}
and
\begin{equation*}
\cot(z) = \frac{1}{z} - \frac{z}{3}-\frac{z^{3}}{45}+o(z^{5})
\end{equation*}
with this we calculate
$$\text{Res}\left(\frac{\phi(z)}{z^{2}},0\right)=\text{Res}\left(\frac{\pi\cot(z)}{z^{2}},0\right)$$
On the other side:
\begin{equation*}
\cot(\pi z) = \frac{\cos(\pi z)}{\sin(\pi z)} = i\frac{e^{\pi i z} + e^{-\pi i z}}{e^{\pi i z} - e^{-\pi i z}} = i\frac{e^{2\pi i z} + 1}{e^{2\pi i z} - 1}
\end{equation*}
as $|\cot(\pi z)|<2$, then
\begin{eqnarray*}
\left|\int_{C_{N}}\frac{\phi(z)}{z^{2}}dz\right|
& \leq & \int_{C_{N}}\left|\frac{\phi(z)}{z^{2}}\right|dz\\
& \leq & \int_{C_{N}}\left|\frac{2\pi}{z^{2}}\right|dz\\
& \leq & |2\pi|\int_{C_{N}}\frac{1}{\left|z^{2}\right|}dz\\
& \leq & 2\pi\max_{z\in C_{N}}\left(\frac{1}{z^{2}}\right)\cdot L(C_{N})\\
\end{eqnarray*}
Could someone help me with this part (3)?
| You had the right idea using the series expansion to calculate the residue at 0. You can do the following:
In our case we have that $\cot(\pi z) = \frac{1}{\pi z} - \frac{\pi z}{3} - \frac{(\pi z)^3}{45} + o(z^5)$. Then:
\begin{eqnarray*}
\text{Res}\left(\frac{\phi(z)}{z^{2}},0\right)
& = & \lim_{z\to0}\frac{1}{2!}\frac{d^{2}}{dz^{2}}\left[(z-0)^{3}\frac{\cot(\pi z)}{z^{2}}\right] \\
& = & \lim_{z\to0}\frac{1}{2!}\frac{d^{2}}{dz^{2}}\left[z\pi\cot(\pi z)\right] \\
& = & \lim_{z\to0}\frac{1}{2}\frac{d^{2}}{dz^{2}}\left[1 - \frac{(\pi z)^2}{3} - \frac{(\pi z)^3}{45} + o(z^5) \right] \\
& = & \lim_{z\to0}\frac{1}{2}\frac{d}{dz}\left[\frac{2 \pi^2 z}{3} - \frac{3 \pi z^2}{45} \right] \\
& = & \lim_{z\to0}\frac{1}{2}\left[\frac{ 2\pi}{3} - \frac{6 \pi z}{45}\right] \\
& = & \frac{- \pi }{3}
\end{eqnarray*}
Note that if you write out the expansion with higher orders, they'll go to zero as $z \to 0$ just like the term $\frac{(\pi z)^3}{45}$.
Now we can see that $C_{N}\subseteq B(0,kN)$ for some $k\in\mathbb{R}$, so we have that $L(C_{N})\leq 2\pi kN$, then we have the upper bound:
\begin{equation*}
\frac{1}{z^{2}} \leq \frac{2\pi k}{N^{2}}
\end{equation*}
Now we can say that:
\begin{equation*}
2\pi\max_{z\in C_{N}} \left(\frac{1}{z^{2}}\right) \cdot L(C_{N})\leq2\pi\max_{z\in C_{N}}\left(\frac{2\pi k}{N^{2}}\right)\cdot L(C_{N})\to0\text{ when }N\to\infty
\end{equation*}
Then it's easy to see the next limit:
\begin{equation*}
\left|\int_{C_{N}}\frac{\phi(z)}{z^{2}}dz\right|\to0
\end{equation*}
Finally, considering we have:
\begin{eqnarray*}
0=\lim_{N\to\infty}\int_{C_{N}}\frac{\phi(z)}{z^{2}}dz
& = & 2\pi i\lim_{N\to\infty} \left( -\frac{\pi^{3}}{3} + \sum_{\begin{subarray}{c}k=-N\\k\neq0\end{subarray}}^{N}\frac{\pi}{k^{2}} \right)\\
& = & 2\pi i\lim_{N\to\infty} \left( -\frac{\pi^{3}}{3} + \sum_{k=1}^{N}\frac{2\pi}{k^{2}} \right)\\
\end{eqnarray*}
We can conclude that
\begin{equation*}
\sum_{k=1}^{\infty}\frac{1}{k^{2}}=\frac{\pi^{2}}{6}
\end{equation*}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4150349",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 1,
"answer_id": 0
} |
Maximum value of $x+2 x y+4 x y z$ Given that $x,y,z \in R^+$ such that $x+y+z=1.5$. Without using calculus find the maximum value of $x+2 x y+4 x y z $
My try:
We can write $S=x+2 x y+4 x y z =x(1+2y+4yz)$
$\implies$
$$S=x((1+2y)(1+2z)-2z)$$
$\implies$
$$S=\frac{1}{2}(1+2 x-1)(1+2 y)(1+2 z)-2 x z$$
$\implies$
$$S=\frac{1}{2}(1+2 x)(1+2 y)(1+2 z)-\frac{1}{2}(1+2 y)(1+2z)-2xz$$
By $AM-G.M$, we have
$$(1+2 x)(1+2 y)(1+2 z) \leq\left(\frac{3+2(x+y+z)}{3}\right)^{3}$$
$\implies$
$$(1+2 x)(1+2 y)(1+2 z) \leq 8$$
Any hint from here?
| The idea is to reduce the number of variables.
EG We could use the substitution $ z = 3/2 - y - x $ to make this a 2 variable inequality.
How can we make this a 1 variable inequality?
Work through the following. If you're stuck, explain what you've tried.
*
*(In hindsight, because) The maximum occurs at $ (1, 0.5, 0)$, so we will allow for non-negative reals (or deal with finding the supremum instead, your choice).
*We use a substitution $ 2x = a, 2y = b, 2c = z$ to simplify the problem
$a+b+c = 3$ are non-negative reals, what is the maximum of $a+ab+abc$?
*Suppose that $ b + c = k$, how can we maximize $ b+ bc$? What is it's value in terms of $k$?
Since $ b +bc = b ( 1+c ) = b ( 1 + k - b) \leq (\frac{ 1 + k}{2} )^2 $, hence the maximum occurs when $ b = 1+c = \frac{k+1}{2}$.
The max value is $\frac{1}{4} (k+1)^2$.
*Hence, what is the maximum of $ a ( 1 + b+bc) $, subject to $ b+c = 3-a$?
We want to find the maximum of $ a ( 1 + \frac{1}{4} ( 4 - a ) ^2 ) $, subject to $ 0 \leq a \leq 3$.
Use your favorite approach to maximize the above cubic in a restricted range.
The goal is to find the turning points of the cubic. Once we have those, we can check against the end points to find the max/min.
I personally would use calculus here, but that's not absolutely necessary:
Consider $Y(a) = a^3 - 8a^2 +20a $, which is our expanded cubic after multiplying by 4.
We want to find a $D$ such that $Y - D = a^3 - 8a^2 + 20a - D$ can be written in the form $ (a - p)(a-q)^2 $, then $ a = q$ will be a turning point.
Expanding and equating terms, we have $ p+2q = 8, q(2p+q) = 20, pq^2 = D$.
Solving the first 2 equations and substituting it into the 3rd, we have 2 solutions: $ ( p,q, D ) = (4, 2, 16), ( 4/3, 10/3, 400/27)$.
So the local maximum is at $a = 2$, the local minimum is at $ a = 10/3$.
Finally, the maximum of $ Y(a)$ on $[0,3]$ is the max of $ Y(0), Y(2), Y(3)$, which is 16.
Hence, the maximum occurs at $ a = 2, b = 1, c = 0$.
Substituting this back, we get $ x + 2xy + 4xyz \leq 2$ with equality at $ (1, 0.5, 0)$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4151240",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 1,
"answer_id": 0
} |
What values of $n$ is $A=\frac{a_1}{a_2}+\frac{a_2}{a_3}+\cdots +\frac{a_{n-1}}{a_n}+\frac{a_n}{a_1}$ an integer?
Suppose $a_i\in \mathbb{N} $ such that $a_i\ne a_j$ and $\displaystyle\frac{a_i}{a_{i+1}}\ne \frac{a_j}{a_{j+1}}~$ for $i \ne j$ $($take $a_{n+1}=a_1)$ and $\displaystyle A=\frac{a_1}{a_2}+\frac{a_2}{a_3}+\cdots +\frac{a_{n-1}}{a_n}+\frac{a_n}{a_1}$.
Then for what values of $n\ge 2$ can $A$ be an integer?
I found that if $a_i=(n-1)^i$ then $A$ is a integer but I realised that this violates the condition $\displaystyle\frac{a_i}{a_{i+1}}\ne \frac{a_j}{a_{j+1}}~$. It was easy to prove that $A$ is not an integer for $n=2$, but I was stuck for the cases where for $n>2$. I then I plugged in some values for $a_i$ but for $n>2$ but I found no solution.
Intuitively I think there is no solution but I am unable to prove it. Any help would be appreciated.
Edit-1: As pointed out by @WillJagy it is possible for $n=3,4,5$. So can we actually prove it is true for $n>6$ or are there some $n>6$ for which it isn't true?
Edit-2: I was able to find a solution for $n=5$ from a solution for $n=4$.
$$~~~~~~~~~~\frac{6}{4} + \frac{4}{3} + \frac{3}{1} + \frac{1}{6} = 6$$
$$\implies \frac{2}{4} + \frac{4}{3} + \frac{3}{1} + \frac{1}{6} = 5$$
$$\implies \frac{6}{2} +\frac{2}{4} + \frac{4}{3} + \frac{3}{1} + \frac{1}{6} = 8$$
But I could not find a solution for $n=6$ from a solution for $n=5$ because when I rewrite $\frac{4}{3} =1+\frac{1}{3}$, the $1$ in the numerator of $\frac{1}{3}$ is used by another $a_i$. I don't know whether we can prove by induction. Any suggestions is welcome.
| $$ \frac{81}{2} + \frac{36}{81} + \frac{2}{36} = 41 $$
$$ \frac{98}{12} + \frac{63}{98} + \frac{12}{63} = 9 $$
$$ \frac{6}{4} + \frac{4}{3} + \frac{3}{1} + \frac{1}{6} = 6 $$
$$ \frac{6}{5} + \frac{5}{3} + \frac{3}{10} + \frac{10}{12} + \frac{12}{6} = 6 $$
$$ \frac{3}{12} + \frac{12}{9} + \frac{9}{10} + \frac{10}{8} + \frac{8}{5} + \frac{5}{3} = 7 $$
$$ \frac{2}{8} + \frac{8}{10} + \frac{10}{6} + \frac{6}{5} + \frac{5}{4} + \frac{4}{3} + \frac{3}{2}= 8 $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4157159",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 0
} |
Find the area of an isosceles trapezoid $ABCD$ $(AB\parallel CD)$ with height $m$ and perpendicular diagonals $(AC\perp BD)$ Find the area of an isosceles trapezoid $ABCD$ $(AB\parallel CD)$ with height $m$ and perpendicular diagonals $(AC\perp BD)$.
The area of $ABCD$ is given by $$S_{ABCD}=\dfrac{a+b}{2}.h=\dfrac{a+b}{2}.m$$
where $a$ and $b$ are the two bases $(a>b)$. So we have to find $a+b$ in terms of $m$ or maybe even $\dfrac{a+b}{2}$. Any help would be appreciated. Thank you in advance!
| The trapezoid is isoceles, hence $\,\overline{AC} = \overline{BD}\,$ in:
Given the above coordinate system, we have for the diagonals, with a bit of analytic geometry:
$$
\overline{AD} \; : \; \begin{cases} x = -a/2 + (b/2+a/2)t \\ y = mt \quad ; \quad 0 \le t \le 1 \end{cases}
\quad \Longrightarrow \quad x-(a/2+b/2)y/m=-a/2 \\
\overline{BC} \; : \; \begin{cases} x = +a/2 - (b/2+a/2)t \\ y = mt \quad ; \quad 0 \le t \le 1 \end{cases}
\quad \Longrightarrow \quad x+(a/2+b/2)y/m=+a/2
$$
The diagonals are perpendicular to each other if the
dot product
of the normals of the line segments $\overline{AD}$ and $\overline{BC}$ is zero:
$$
\left(1,-\frac{a/2+b/2}{m}\right)\cdot\left(1,+\frac{a/2+b/2}{m}\right) = 1-\frac{(a/2+b/2)^2}{m^2} = 0
\\ \Longrightarrow \quad m = \frac{a+b}{2}
$$
$$
\mbox{Area} = m^2 = \left(\frac{a+b}{2}\right)^2
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4158287",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Prove $7^{2n}-1$ is divisible by 48 Trying to prove $7^{2n}-1$ is divisible by 48.
I began by taking the base case $n=1$ so $7^{2}-1=48$ which is true.
Now from here taking the inductive step to get
$7^{2(n+1)}-1=48k$
$7^{2n+2}-1=48k$
$7^{2}*7^{2n}-1=48k$
Not quite sure where to go from here. Thanks.
| Let S = $1 + 49 + 49^2 + \cdots + 49^{n-1} $
\begin{align}
S &= 1 + 49 + 49^2 + \cdots + 49^{n-1} + 49^n - 49^n\\
S &= 1 + 49(1 + 49 + 49^2 + \cdots + 49^{n-1}) - 49^n \\
S &= 1 + 49S - 49^n \\
49^n - 1 &= 48S\\
48 &\mid 49^n - 1 \\
48 &\mid 7^{2n} - 1 \\
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4162577",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 1
} |
Rate of change of radius of spherical balloon
A spherical balloon is filled with gas at a rate of $4 \text{ cm}^3/\text{s}$. What rate is the radius $r$ changing with respect to the time when the vol $V=36π \text{ cm}^3$?
[ans:$\frac 19, \frac{\pi}{9} $or 9$\pi$]
I tried this one with implicit differentiation:
$\frac{dV}{dt} = \frac {4}{3} \pi \cdot \frac{d}{dt}[r^2]$
*
*$4 = \frac {4}{3} \pi \cdot 2r \frac{dr}{dt}$
Volume formula: $36 π = \frac43 π r^2$
*
*$r^2= 27$
*$r= \pm 3 \sqrt{3}$
Substituting r into the differentiation to find $\frac{dr}{dt}$:
$4 = \frac {4}{3} \pi \cdot 2r \frac{dr}{dt}$
$4 = \frac {4}{3} \pi \cdot 2(3 \sqrt{3}) \frac{dr}{dt}$
$\frac{dr}{dt} = \frac{1}{2\sqrt{3}\pi}$ , which seems wrong. Please can someone tell me where my knowledge gap is?
| $\frac{dV}{dt} = \frac {4}{3} \pi \cdot \frac{d}{dt}[r^3]$
*
*$4 = \frac {4}{3} \pi \cdot 3r^2 \frac{dr}{dt}$
Volume formula: 36 π = 4/3 π r^3
*
*$r^3= 27$
*$r= 3$
Substituting r into the implicit differentiation to find $\frac{dr}{dt}$:
$4 = \frac {4}{3} \pi \cdot 3r^2 \frac{dr}{dt}$
$4 = \frac {4}{3} \pi \cdot 2(9) \frac{dr}{dt}$
$\frac{dr}{dt} = \frac{1}{9\pi}$
Many thanks @alessandro!
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4163944",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Calculating $ \tan w$ for $ w = \dfrac{1} {i} \log\dfrac{1-iz} {1+iz}$ Let $ w = \dfrac{1} {i} \log\left(\dfrac{1-iz} {1+iz}\right)$ and calculate $\tan w$.
I begin by using the fact that
$$ \tan w = \frac{\sin w}{\cos w} = \frac1i\frac{e^{iw} - e^{-iw}}{e^{iw}+e^{-iw}}$$
I now replace $ w = \dfrac{1} {i} \log(\dfrac{1-iz} {1+iz})$ and get
$$\tan w = \frac{1} {i} \left(\frac{{e^{\log(\frac{1-iz} {1+iz})} - e^{-{\log(\frac{1-iz} {1+iz})}}}} {e^{\log(\frac{1-iz} {1+iz})} + e^{-\log(\frac{1-iz} {1+iz})}}\right). $$
It is at this point that i run into troubles. The next step should be to use the fact that
$$e^{-{\log(\frac{1-iz} {1+iz})}} = e^{{\log(\frac{1-iz} {1+iz})^{-1}}} = e^{{\log(\frac{1+iz} {1-iz})}}. $$
The thing that i have a hard time grasping is why
$$e^{-{\log(\frac{1-iz} {1+iz})}} = e^{{\log(\frac{1-iz} {1+iz})^{-1}}} $$ is true for complex numbers aswell, since i've always worked under the assumption that the normal logarithmical rules doesn't always apply when it comes to complex numbers. How come it applies here?
| Use
$$e^{\log (a)} = a,$$
and
$$e^{-\log(a)} = \frac{1}{a}.$$
Then:
$$\tan w = \frac{1} {i} \left(\frac{{e^{\log\left(\frac{1-iz} {1+iz}\right)} - e^{-{\log\left(\frac{1-iz} {1+iz}\right)}}}} {e^{\log\left(\frac{1-iz} {1+iz}\right)} + e^{-\log\left(\frac{1-iz} {1+iz}\right)}}\right) = \\
= \frac{1}{i} \frac{\frac{1-iz}{1+iz} - \frac{1+iz}{1-iz}}{\frac{1-iz}{1+iz} + \frac{1+iz}{1-iz}} = \\
= \frac{1}{i} \frac{\frac{(1-iz)^2 - (1+iz)^2}{(1+iz)(1-iz)}}{\frac{(1-iz)^2 + (1+iz)^2}{(1+iz)(1-iz)}} = \\
= \frac{1}{i} \frac{(1-iz)^2 - (1+iz)^2}{(1-iz)^2 + (1+iz)^2} = \\
= \frac{1}{i} \frac{1-z^2-2iz - 1 + z^2 - 2iz}{1-z^2-2iz + 1 -z^2 + 2iz} = \\
= -\frac{1}{i} \frac{4iz}{2(1-z^2)} = \\
= \frac{2z}{z^2-1}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4166241",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
If $b$ is a divisor of $a^2 - a + 1$, can $a$ be a divisor of $b^2 - b + 1$ I observed that for $a > 1$, if $b$ is a divisor of $a^2 - a + 1$ then $a$ is not a divisor of $b^2 - b + 1$. This implies that their divisors are mutually exclusive. Is this true in general?
$$
b \mid a^2 - a + 1 \implies a \nmid b^2 - b + 1
$$
I have verified this for $a \le 10^9$.
| We restrict our attention to positive integers.
This is a Vieta jumping solution, where we use the existing solution set to find another solution set, and hope to reach the base case or find a contradiction.
We call $(a, b)$ a valid set if $ a \geq b > 0$, $a \mid b^2 - b + 1, b \mid a^2 - a + 1$, and $ a > 1$.
There is a solution $ (1,1)$, which isn't a valid set because $ a \not > 1 $.
We will show that there are no valid sets.
Proof by contradiction. Suppose we have a valid set. Let $ ac = b^2 - b + 1 $.
We will show that $(b, c)$ is also a valid set by checking that it satisfies the conditions.
*
*If $b = 1$, then $ a \mid b^2 - b + 1 = 1$ gives us $ a =1$, so $ b > 1$.
*$ b^2 - b + 1 = ac \equiv 0 \pmod {c}$.
*Since $ac = b^2 - b + 1 > 0$, hence $ c > 0$.
*Since $b^2 > b^2 -b + 1 = ac $, and $ a \geq b$, hence $ b > c $. (This is stronger than the $ b \geq c$ that we need, and we will use this subsequently.)
*$ ac \equiv 1 \pmod{b}$, so $ a^2(c^2 - c + 1 ) \equiv 1 - a + a^2 \equiv 0 \pmod{b}$. Dividing out by $a^2 $ (Since $ \gcd(a, b) = 1$) gives us $ c^2 - c + 1 \equiv 0 \pmod{b}$,
We can repeat the above process to get strictly smaller valid sets (Since $ b > c$).
Let this chain be $ a_1 = a, a_2 = b, a_3 = \frac{a_2^2 - a_2 + 1 } { a_1} , \ldots, a_{i+1} = \frac{ a_i ^ 2 - a_i + 1 }{ a_{i-1}} $. We have:
*
*$ a_i > a_{i+1} > a_{i+2} $
*$a_{i} a_{i+2} = a_{i+1}^2 - a_{i+1} + 1 $
However, there is no strictly decreasing chain of positive integers. Hence, we have a contradiction.
It's not immediately clear to me how to deal with the negative integers. Note that several of the inequalities need not be true.
We have the 3 chains
*
*$1, 1, 1, \ldots$
*$ 1, -1, 1, -1, \ldots $
*$ - 1, -1, -1, \ldots $
I'm not sure if there are other solutions.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4166682",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 1,
"answer_id": 0
} |
How to solve $\iiint_T \frac{x}{\sqrt{x^2+y^2}}\,dx\,dy\,dz$? I need to solve the integral $$\iiint_T \frac{x}{\sqrt{x^2+y^2}}\,dx\,dy\,dz\,,$$ where $T$ is the domain blocked by the cylinder $x^2+y^2=2x$, the cone $z=\sqrt{x^2+y^2}+1$ and plane $z=0$.
The solution:
In order to understand exactly the relationship between $\theta $ and $ r $, we will look at the equation of the cylinder. The domain $T$ is blocked by a cylinder, so in $T$ there is $(x-1)^2 + y^2 \leq1 $. We will place the change of variables and we will get:
$$
\geq (r\cos\theta -1)^2+r^2\sin^2\theta =1+r^2-2r\cos\theta
$$
So $r\leq 2\cos\theta$.
I really struggle to understand the solution. Why, in order to find the constraint on $r$, did they have to perform inequality on the cylinder equation? Is it possible to explain the solution?
| In $T$, you have $0<z<\sqrt{x^2+y^2}+1$ and then
$$\iiint_T \frac{x}{\sqrt{x^2+y^2}}\,dx\,dy\,dz=\iint_D\int_0^{\sqrt{x^2+y^2}+1}\frac{x}{\sqrt{x^2+y^2}}\,dz\,dx\,dy=$$
$$\iint_D\frac{x(\sqrt{x^2+y^2}+1)}{\sqrt{x^2+y^2}}\,dx\,dy=\iint_D x+\frac{x}{\sqrt{x^2+y^2}}\,dx\,dy$$
where $D$ is the projection of $T$ over $XY$ plane, that is, $x^2+y^2-2x\leq0$ or $(x-1)^2+y^2\leq1$. Working on polars coordinates
$$\int_{-\pi/2}^{\pi/2}\int_0^{2\cos(\theta)}\left(r\cos(\theta)+\cos(\theta)\right)r\,dr\,d\theta=\int_{-\pi/2}^{\pi/2} \frac{8 \cos ^4(\theta)}{3}+2 \cos ^3(\theta)\,d\theta=\boxed{\frac{8}{3}+\pi}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4168973",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Proving $\int_a^{b}\frac{\ln (1+x)}{a^2b^2(1+x)^2+({a+b})^2}\ dx=\frac{\ln (a+b)-\ln (ab)}{ab(a+b)}\left(\frac{\pi}{2}-2\tan^{-1}\frac{a}{b}\right)$
How can I prove this?
$$\int_a^{b}\frac{\ln (1+x)}{a^2b^2(1+x)^2+({a+b})^2}\ dx=\frac{\ln (a+b)-\ln (ab)}{ab(a+b)}\left(\frac{\pi}{2}-2\tan^{-1}\frac{a}{b}\right)$$
Here is my attempt
$$ \frac{1}{a^2b^2}\int_a^{b}\frac{\ln (1+x)}{(1+x)^2+(\frac{a+b}{ab})^2}\ dx$$
Applying the substitution $1+x=t \left (\dfrac{a+b}{ab}\right )$
$$ \frac{\frac{a+b}{ab}}{a^2b^2(\frac{a+b}{ab})^2}\int_{\frac{ab(a+1)}{a+b}}^{\frac{ab(b+1)}{a+b}}\frac{\ln (t)+\ln \left (\frac{a+b}{ab}\right )}{t^2+1}\ dt\\\\$$
$$ \frac{1}{ab(a+b)}\int_{\frac{ab(a+1)}{a+b}}^{\frac{ab(b+1)}{a+b}}\frac{\ln (t)+\ln \left (\frac{a+b}{ab}\right )}{t^2+1}\ dt\\\\$$
Again applying substitution $t\ =\dfrac{1}{u}$
$$ \frac{-1}{ab(a+b)}\int_{\frac{a+b}{ab(a+1)}}^{\frac{a+b}{ab(b+1)}}\frac{-\ln (u)+\ln \left (\frac{a+b}{ab}\right )}{u^2+1}\ du\\\\$$
| By using integration by parts,
Differentiating ln(1+x) and integrating the remaining function gives an inverse tangent function integral in second part ,now set 1+x =z×(a+b)/ab this substitution reduces the second integral to integration of arctan(z)/z under respective limits and observe that this integral is solvable by feynman parameter or integral under differential sign.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4172313",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
How can you quickly use the graph of $3x-6y=4$ to produce the graph of $3x-6y=10$? To start with, I am asked to graph the lines $3x-2y=k$ for $k=1,2,3$ and $4$.
I find that the equations produce different parallel lines.
For each line where $k=2,3$ and $4$, the coordinates points are $0.5k$ below the coordinates points of the line $3x-2y=1$.
If we let $x=0$, we can find the $y$-intercept of the lines.
\begin{align*}
-2y&=k\\
y &=\frac{k}{-2}
\end{align*}
and if we increase $k$ by $1$ we decrease $y$ by $0.5$.
\begin{align*}
-2y&=k+1\\
y &=\frac{k}{-2}-\frac{1}{2}
\end{align*}
I'm not certain the above logic suffices in quickly using the graph of $3x -6y = 4$ to produce the graph of $3x-6y=10$.
\begin{align*}
-6y&=4+6\\
y&=-\frac{2}{3}-1 \\
y&=-\frac{5}{3}
\end{align*}
Can my answer be improved?
| Since the goal is to describe the relationship between the two lines quickly, I will give a one-sentence way to do this:
The difference between the graphs of $3x-2y=4$ and $3x-2y=10$ is that the graph of the latter equation is a dilation, centered at the origin, of the graph of the former by a factor of $10/4$.
That means that every point on the original graph "moves" to a point that is $2.5$ times farther away from the origin.
(The detailed explanation for why this is true is contained in Amaan M's answer.)
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4173812",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 1
} |
Why does the triangle inequality seem to be false when rewritten based on $|x| = \sqrt{x^{2}}$? The triangle inequality is $$|x + y| \leq |x| + |y|.$$
Also, we know that $|x| = \sqrt{x^{2}}$. Then,
\begin{align*}
\sqrt{x + y} &\leq \sqrt{x} + \sqrt{y} \\
x + y &\leq x + y + 2\sqrt{xy} \\
0 &\leq 2\sqrt{xy} \\\\
\sqrt{xy} &\geq 0
\end{align*}
Now, we can see that the inequality $|x + y| \le |x| + |y|$ holds for real $x$ and $y$, but $\sqrt{xy} \geq 0$ does not for $x < 0, y > 0$ or $x > 0, y < 0$.
What seems to be the problem? Is it the statement $|x| = \sqrt{x^{2}}$ or is it much more than that?
| The problem is not with writing $|x|=\sqrt{x^2}$: that statement is true as long as $x$ is a real number. But the correct translation of $|x+y|\le|x|+|y|$ is
$$
\sqrt{(x+y)^2} \le \sqrt{x^2} + \sqrt{y^2} \tag{*}\label{*} \, ,
$$
not what you wrote. To prove that $\eqref{*}$ is true, we can use the fact that if $a$ and $b$ are nonnegative, then $a\leq b \iff a^2\le b^2$. Hence,
\begin{align}
& \sqrt{(x+y)^2} \le \sqrt{x^2} + \sqrt{y^2} \\[4pt]
\iff & (x+y)^2 \le x^2+y^2+ 2\sqrt{x^2}\sqrt{y^2} \\[4pt]
\iff & x^2+y^2+2xy \le x^2+y^2 + 2|x||y| \\[4pt]
\iff & 2xy \le 2|x||y| \\[4pt]
\iff & xy \le |x||y| \\[4pt]
\iff & xy \le |xy| \, .
\end{align}
Since the final line is true for all $x,y\in\Bbb{R}$, the first line must also be true for all $x,y\in\Bbb{R}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4178104",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Is a symmetric matrix a subspace of nxn matrices? Let $M$ be a vector space for all 2x2 matrices.
Show that the set of all the symmetric matrices $M2$={$B ∈ M2 : B=B^t$} is a subspace of M.
My solution:
*
*The null matrix is symmetric
$$B=\begin{pmatrix}0&0\\0&0\end{pmatrix}$$
*Let $B$, $C$ $∈$ $M2$, where $B = B^t$, $C=C^t$. Then, $B+C=B^t + C^t$.
$$B = \begin{pmatrix}a&c\\c&b\end{pmatrix}$$
$$C = \begin{pmatrix}d&f\\f&e\end{pmatrix}$$
$$B+C = \begin{pmatrix}a+d&c+f\\c+f&b+e\end{pmatrix}$$
$$B^t = \begin{pmatrix}a&c\\c&b\end{pmatrix}$$
$$C^t = \begin{pmatrix}d&f\\f&e\end{pmatrix}$$
$$B^t+C^t = \begin{pmatrix}a+d&c+f\\c+f&b+e\end{pmatrix}$$
Therefore, $B + C ∈ M2$.
*Let $B ∈ M2$, $θ ∈ R$. Then, $θB = θB^t$
$$B = \begin{pmatrix}a&c\\c&b\end{pmatrix}$$ $$θB = \begin{pmatrix}θa&θc\\θc&θb\end{pmatrix}$$
$$B^t = \begin{pmatrix}a&c\\c&b\end{pmatrix}$$ $$θB^t = \begin{pmatrix}θa&θc\\θc&θb\end{pmatrix}$$
Therefore, $θB ∈ M2$.
So, $M2$ is, indeed, a subspace of $M$.
I am just learning linear algebra, so I apologize if I've made any mistakes.
Also, I just got into this community, so I am still learning the formatting, so I apologize for any mistakes as well.
Appreciate all the answers.
| Your answer is fine.
Or you could have noticed that there is a basis for your space:
$$m_1=\begin{pmatrix}1&0\\0&0\end{pmatrix}, m_2=\begin{pmatrix}0&1\\1&0\end{pmatrix},
m_3=\begin{pmatrix}0&0\\0&1\end{pmatrix}$$
Show a $2\times 2$ matrix is symmetric if and only if it can be written as $$am_1+bm_2+cm_3$$
for some real $a,b,c.$
But this answer requires knowledge of bases. Your answer is more direct.
But the general rule makes it unnecessary to prove many individual case:
If $V$ is a real vector space, and $v_1,v_2,\cdots,v_k\in V$ then:
$$W=\{v\in V\mid \exists a_1,\dots,a_k\in \mathbb R(v=a_1v_1+a_2v_2+\cdots+a_kv_k\}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4180505",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
} |
Solve the system of equations $x + y^2 + z^3 = 3; y + z^2 + x^3 = 3; z + x^2 + y^3 = 3$ (where x,y,z are positive real numbers) It seems that $x=y=z=1$ is the only solution, but I can't find how to prove it. I've tried defining $a=x-1; b=y-1; c=z-1$, but with no success. Any ideas?
Geometrically, each equation defines a 3D surface. All 3 surfaces are identical, except they are oriented along different axes. We're looking for their intersection.
| The system can be written as:
$$
\begin{align}
\begin{cases}
\;\; (x-1) &+\;\; (y-1)\,(y+1) &+\;\; (z-1)\,(z^2+z+1) &= 0
\\ \;\; (x-1)\,(x^2+x+1) &+\;\;(y-1) &+\;\; (z-1)\,(z+1) &= 0
\\ \;\; (x-1)\,(x+1) &+\;\;(y-1)\,(y^2+y+1) &+\;\; (z-1) &= 0
\end{cases}
\end{align}
$$
Regarding it as a linear homogeneous system in $x-1, y-1, z-1$, its determinant is:
$$
\Delta =
\begin{vmatrix}
1 &y+1 &z^2+z+1
\\ x^2+x+1 &1 &z+1
\\ x+1 &y^2+y+1 &1
\end{vmatrix}
$$
After routine calculations courtesy WA, and using that $x,y,z \gt 0\,$:
$$
\begin{align}
\Delta &= x^2 y^2 z^2 + x^2 y^2 z + x^2 y^2 + x^2 y z^2 + x^2 y z + x^2 z^2 + x^2 z + x y^2 z^2
\\ &\;\;\;\; + x y^2 z + x y^2 + x y z^2 + 2 x y z + x y + x z + y^2 z^2 + y z^2 + y z
\\ &\gt 0
\end{align}
$$
It follows that the only solution is $x-1=y-1=z-1=0\,$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4181543",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
} |
Solutions of $ 5^x+5^{x^2}=4^x+6^{x^2} \quad \left(x\in \mathbb{R}\right)$ For equation $$ 5^x+5^{x^2}=4^x+6^{x^2} \quad \left(x\in \mathbb{R}\right)$$
is there any nontrivial solution?
Easy to find $x=0,1$ are the trivial solutions, and also, easy to figure out that $ x>1 $ or $ x<0 $ would not satisfy the equation.
But I found it hard to deal with the $ \left(0,1\right) $ case, and I haven't got any idea yet.
Any Advice Would Be Greatly Appreciated.
| We prove there is no solution in $(0,1)$. The equation is equivalent to
$$\iff 5^x\left(1-\left( \frac{4}{5} \right)^x \right) = 5^{x^2}\left(\left( \frac{6}{5} \right)^{x^2} -1\right)$$
$$\iff 5^{x-x^2}\left(1-\left( \frac{4}{5} \right)^x \right) = \left( \frac{6}{5} \right)^{x^2} -1 \tag{1}$$
We observe that, for $x \in (0,1)$
*
*The LHS of $(1)$ is greater than $1-\left( \frac{4}{5} \right)^x$
*The RHS of $(1)$ is smaller than $\left( \frac{6}{5} \right)^{x} -1$
Then from $(1)$, we deduce that
$$(1) \implies 1-\left( \frac{4}{5} \right)^x < \left( \frac{6}{5} \right)^{x} -1 \implies \left( \frac{6}{5} \right)^{x}+\left( \frac{4}{5} \right)^x > 2 \tag{2} $$
(The equality doesn't occur for $x \in (0,1)$)
Study the function $f(t) = t^x$ for $t\in \Bbb R^+$. We have $f''(t) = x(x-1)t^{x-2}<0$ for $x\in (0,1)$. Applying the Jensen's inequality, we obtain that
$$\frac{1}{2} \left( f\left(\frac{4}{5} \right)+f\left(\frac{6}{5} \right) \right) \le f\left(\frac{1}{2} \left(\frac{4}{5} +\frac{6}{5} \right) \right) =1$$
or
$$\left( \frac{6}{5} \right)^{x}+\left( \frac{4}{5} \right)^x \le 2 \tag{3}$$
From $(2)$ and $(3)$, we deduce that there is no solution for $x \in (0,1)$.
Q.E.D
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4182737",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
How would one solve $z^6+z^4+z^3+z^2+1=0$? I recently came upon the following equation, which is supposed to be solvable using complex numbers:
$$z^6+z^4+z^3+z^2+1=0$$
The problem is that it is a degree $6$ polynomial with many other exponential terms, which I don't know how to deal with. I try substituting $w=z^2$ to try and make the equation simplier:$$w^3+w^2+wz+w+1=0$$but I can't make any progress with this either. Only thing that I see that could be used is that $w^3,w^2,w,1$ is a geometric series, not sure how that helps though.
| Note that neither $z = 1$ nor $z = -1$ are roots. Then our equation becomes equivalent to
$$0 = (z^2 - 1)(z^6 + z^4 + z^3 + z^2 + 1) \text{ and } z \neq \pm 1.$$
This equation simplifies to
$$z^8 + z^5 - z^3 - 1 = 0.$$
We can factor this by:
$$z^8 + z^5 - z^3 - 1 = z^5(z^3 + 1) - 1(z^3 + 1) = (z^5 - 1)(x^3 + 1) = 0.$$
From this, and the fact that $z \neq \pm 1$, you can get the result easily.
How did I get this answer? It was mostly inspired by jlammy's answer, but one of the things that jumped out at me was that $z^3$ was an odd term out of a geometric series $z^6 + z^4 + z^2 + 1$. This meant that multiplying by $z^2 - 1$ would achieve some kind of simplification. I thought it was worth a shot.
In general, if you have a polynomial, whose coefficients are $\pm 1$, and you suspect that some roots of unity are involved, it's probably not a bad idea to try multiplying by $z^2 - 1$ anyway. It should take any of the cylotomic factors like $z^4 + z^3 + z^2 + z + 1$ or $z^2 - z + 1$, and turn them into simpler binomial factors like $z^5 - 1$ or $z^3 + 1$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4184696",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Fraction/Binomial Coefficient Estimate Let $k \leq k+x <n.$ Prove
$$(1)\ \ \ \ \ \ \ \ \ \ \ \ \left( \frac{n-k-x}{n-x}\right)^x \leq \frac{{n-x \choose k}}{{n \choose k}} \leq \left(\frac{n-k}{n}\right)^x.$$
(This is part of formula 1.23 in Stasys Jukna's extremal combinatorics.)
What I have tried:
(A) I tried to apply Stirling's formula to the factorials. This version of Stirling states
$$\Gamma(n+1) = \sqrt{\tau}\frac{n^{n+1/2}}{e^n} e^{\varepsilon_n},$$
with some kind of error bounds on $\varepsilon_n,$ e.g. $0 < \varepsilon_n < \frac{1}{12n}$ if $n \geq 1.$ Applying Stirling's formula to the first inequality in (1), we get an equivalent inequality
$$\text{ERROR}\cdot \left(\frac{n-k}{n-k-x}\right)^{n-k+1/2} \leq \left(\frac{n}{n-x}\right)^{n+1/2},$$
where the ERROR term is from Stirling's approximation. I don't know what to do from here.
(B) I tried expanding and applying a logarithm. Using the power series for $\ln(1+x),$ I found that (1) is equivalent to
$$\frac{k}{n}x + \left(\frac{k}{n}\right)^2 x + ... \leq \left(\frac{1}{n} + \frac{1}{n-1} + ... + \frac{1}{n-k+1} \right)x + \left(\frac{1}{n^2} + ... + \frac{1}{(n-k+1)^2} \right)x^2 + ... \leq \frac{k}{n-x}x + \left(\frac{k}{n-x}\right)^2 x + ....$$
I don't know what to do from here either.
| It may be helpful to adjust the $\dfrac{{n-x \choose k}}{{n \choose k}}$ expression so each of the three items is the product of $x$ terms:
*
*$\dfrac{{n-x \choose k}}{{n \choose k}}= \frac{n-x}{n}\frac{n-x-1}{n-1}\frac{n-x-2}{n-2}\cdots \frac{n-x-k+1}{n-k+1} = \frac{n-k}{n}\frac{n-k-1}{n-1}\frac{n-k-2}{n-2}\cdots \frac{n-k-x+1}{n-x+1} = \dfrac{{n-k \choose x}}{{n \choose x}}$ with the second product having $x$ terms
*$\frac{n-k-x}{n-x} = 1- \frac{k}{n-x}\le \frac{n-k-i}{n-i} = 1- \frac{k}{n-i}\le \frac{n-k}{n} = 1- \frac{k}{n}$ when $0 \le i \le x \le n-k$
*so $\left(\frac{n-k-x}{n-x}\right)^x \le \frac{n-k}{n}\frac{n-k-1}{n-1}\frac{n-k-2}{n-2}\cdots \frac{n-k-x+1}{n-x+1} \le \left(\frac{n-k}{n}\right)^x$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4186913",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
How to evaluate $\int_0^{\infty}\frac{\arctan{x}}{x^{1/4}(1+x)}\mathrm{d}x$? I want to evaluate
$ \displaystyle \int_{0}^{\infty} \frac{\arctan{x}}{x^{1/4}(1+x) } \mathrm{d}x \tag*{} $
I know that it equals
$\displaystyle \frac{\pi^2}{2 \sqrt{2} } + \frac{\pi}{\sqrt{2}} \ln \left( \frac{2+\sqrt{2-\sqrt{2}} }{2-\sqrt{2-\sqrt{2}}}\right) \tag*{} $
But I can't seem to find a way to prove it.
After substituting $x=t^4$ , the integral is transformed to
$ \displaystyle I= 4\int_{0}^{\infty} \frac{t^2 \arctan(t^4) }{1+t^4} \mathrm{d}t \tag*{} $
After this I tried differentiating under the integral, but the resulting integrand is very complicated, and I couldn't make much progress futher on.
I'm not too familiar with contour integration, maybe that can help here?
| Let $I(a)=\int_{0}^{\infty} \frac{\arctan{(ax)}}{x^{1/4}(1+x) } dx $ and evaluate
\begin{align}
I’(a)=& \int_{0}^{\infty} \frac{x^{3/4}}{(1+x)(1+a^2 x^2) } dx \\
=& \>\frac1{1+a^2}\int_0^\infty \left(\frac{a^2 x^{3/4}}{1+a^2 x^2} + \frac{x^{-1/4}}{1+a^2 x^2} - \frac{x^{-1/4}}{1+x}\right)dx\\
=& \>\frac1{1+a^2}\left(\frac\pi2 \csc\frac\pi8\>a^{1/4}+ \frac\pi2 \sec\frac\pi8\>a^{-3/4} - \sqrt2\pi \right)dx\\
\end{align}
Then
\begin{align}
\int_{0}^{\infty} \frac{\arctan{x}}{x^{1/4}(1+x) } dx
=& \>I(1) =\int_0^1\overset{a=y^4}{I’(a)\>da}\\
= &\>2\pi\int_0^1 \frac{\csc\frac\pi8 \>y^4 +\sec\frac\pi8 }{1+y^8}\>dy- \frac{\pi^2}{2\sqrt2}\\
= &\>2\pi\cdot \left(\frac\pi{2\sqrt2}-\frac1{\sqrt2}\ln \tan\frac{3\pi}{16}\right)- \frac{\pi^2}{2\sqrt2}\\
= &\> \frac{\pi^2}{2\sqrt2} -\sqrt2\pi\ln \tan\frac{3\pi}{16} \
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4189454",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Integer solution of $abc=a+b+c+2$. Let $a,b,c$ be integers greater than $1$. I am trying to prove that $$abc\geq a+b+c+2$$
with equality if and only if $a=b=c=2$.
I can prove the inequality by using the fact that $ab\geq a+b$. Since $ab\geq 4$ and $c\geq 2$, it follows that $abc\geq 4c$ and $abc\geq 2ab$. Therefore $abc\geq ab+2c\geq ab+c+c\geq a+b+c+c\geq a+b+c+2$.
The main problem I face is to justify that $$abc=a+b+c+2\implies a=b=c=2.$$
My idea is to assume that $a>2$ and try to get a contradiction.
| $bc > 4 \implies abc > 4a = a+a+a+a > a+b+c+2$ since wlog $a \geq b \geq c \geq 2$.
$bc = 4 \implies b,c = 2 \implies a=2 \implies abc=a+b+c+2 \implies bc=4$ using the contrapositive of the above statement and all is proved here.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4189755",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 7,
"answer_id": 6
} |
Prove if number is rational or irrational I've been asked to prove if $\frac{\sqrt{3+\sqrt5}}{\sqrt{2} + \sqrt {10}}$ is a rational number. I've tried a proof as follows:
Suppose the number is rational, so it can be written as the quotient of 2 numbers $a$ and $b$
\begin{align*}
\frac{\sqrt{3+\sqrt{5}}}{\sqrt{2} + \sqrt{10}} = \frac{a}{b} \\
\frac{3+\sqrt{5}}{12 + 4\sqrt{5}} = \frac{a^2}{b^2} \\
\frac{3+\sqrt{5}}{4(3+\sqrt{5})} = \frac{a^2}{b^2} \\
\frac{1}{4} = \frac{a^2}{b^2}
\end{align*}
And because we get $\frac{1}{4}$ which is rational, we can conclude that the proof is right and there aren't any contradictions. Hence $\frac{\sqrt{3+\sqrt{5}}}{\sqrt{2} + \sqrt{5}}$ is a rational number.
I guess my proof lacks something and I don't feel it's complete yet. I would appreciate any recommendations on how to improve my answer. Thanks in advance.
| Consider
$$
x=\dfrac{\sqrt{3+\sqrt{5}}}{\sqrt{2}+\sqrt{10}}
$$
Then you know that $x>0$. Then
$$
x^2(\sqrt{2}+\sqrt{10})^2=3+\sqrt{5}
$$
that becomes
$$
x^2(2+10+2\sqrt{20})=3+\sqrt{5}
$$
and therefore, owing to $2\sqrt{20}=4\sqrt{5}$,
$$
4x^2=1
$$
so $x=1/2$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4190391",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
"answer_id": 1
} |
Find the values of $a$ for which the function $f(x)=\sin (2x)-8(a+1) \sin x+(4a^2+8a-14)x$ increases for all $x\in R$.
Find the values of $a$ for which the function $f(x)=\sin (2x)-8(a+1) \sin x+(4a^2+8a-14)x$ increases for all $x\in R$ and has no critical points.
For increasing function, $$f'(x)\gt0\\2\cos(2x) - 8(a+1)\cos x + (4a^2 + 8a -14)\gt0$$
Using $\cos(2x)=2\cos^2x-1$, $$4\cos^2x-8(a+1)\cos x+4a^2+8a-16\gt0\\\cos^2x-2(a+1)\cos x+a^2+2a-4\gt0\\(\cos x-(a+1))^2-5\gt0\\(\cos x-a-1+\sqrt5)(\cos x-a-1-\sqrt5)\gt0\\\cos x\lt a+1-\sqrt5 \text{ or }\cos x\gt a+1+\sqrt5$$
Now, I know that $\cos x$ lies between $-1$ and $1$. Not able to use that to solve the inequality.
Also, the same question has been discussed here, but the approach is different. Also, the final answer posted there is not correct.
| By your work we need $$\cos^2x-2(a+1)\cos x+a^2+2a-4>0.$$
We have three cases:
*
*$a+1>0$. In this case we need $$1^2-2(a+1)\cdot1+a^2+2a-4>0.$$
*$a+1<-1$. In this case we need $$(-1)^2-2(a+1)\cdot(-1)+a^2+2a-4>0.$$
*$$(a+1)^2-(a^2+2a-4)<0.$$
Can you end it now?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4190897",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Find $\sin^5 x + \cos^5 x$
Given $$\cos \left[\sqrt{\left(\sin x + \cos x\right)\left(1 - \sin x \cos x \right)}\right] \\{}= \sqrt{\cos \left(\sin x + \cos x \right) \cos \left(1 - \sin x \cos x\right)}.$$
Find $\sin^5 x + \cos^5 x.$
My try:
\begin{align*}
\sin^5x+\cos^5x
&=(\sin x+\cos x)(\sin^4x-\sin^3x\cos x+\sin^2x\cos^2x-\sin x \cos^3x+\sin^4x)\\
&=(\sin x+\cos x)[(\sin^2x+\cos^2x)^2-(\sin^2x+\cos^2x)\sin x \cos x-\sin^2x\cos^2x]\\
&=(\sin x+\cos x)[1-\sin x\cos x-\sin^2x\cos^2x]\\
&=\frac{1}{4}(\sin x+\cos x)(4-2\sin 2x- \sin^2 2x)\\
\end{align*} I don't know what next?
| This is my try.. I'm pretty sure that there are shorter methods to solve this problem. And it's not complete as I have ignored some tedious cases.
Let
$$A=\sin x + \cos x=\sqrt{2}\cos(\pi/4-x)\tag{1}$$
$$B=1-\sin x\cos x=1-\frac{\sin 2x}{2}=1-\frac{\cos[2(\pi/4-x)]}{2}=\frac{3}{2}-\cos^2(\pi/4-x) \tag{2}$$
Multiplying $(1)$ and $(2)$
$$AB=\sin^3{x} + \cos^3{x}=\frac{3\sqrt{2}}{2}\cos(\pi/4-x)-\sqrt{2}\cos^3(\pi/4-x)$$
Given,
$$\cos(\sqrt{AB})=\sqrt{\cos A\cos B}$$
$$\implies2\cos^2(\sqrt{AB})=\cos(A+B)+\cos(A-B)$$
$$2\cos^2(\sqrt{AB})-1=\cos(A+B)+\cos(A-B)-1$$
$$\implies \cos(2\sqrt{AB})=\cos(A+B)+\cos(A-B)-1$$
$$\cos(2\sqrt{AB})-\cos(A-B)=\cos(A+B)-\cos(0)$$
$$-2\sin\bigg(\frac{2\sqrt{AB}+A-B}{2}\bigg)\sin\bigg(\frac{2\sqrt{AB}-A+B}{2}\bigg)=-2\sin^2\bigg(\frac{A+B}{2}\bigg)$$
$$\implies\sin\bigg(\frac{2\sqrt{AB}+A-B}{2}\bigg)\sin\bigg(\frac{2\sqrt{AB}-A+B}{2}\bigg)=
\sin\bigg(\frac{A+B}{2}\bigg)\sin\bigg(\frac{A+B}{2}\bigg)\tag{$\star$}$$
$$\implies\sin\bigg(\frac{2\sqrt{AB}+A-B}{2}\bigg)=\sin\bigg(\frac{2\sqrt{AB}-A+B}{2}\bigg)$$
$$\implies\sin\bigg(\frac{2\sqrt{AB}+A-B}{2}\bigg)-\sin\bigg(\frac{2\sqrt{AB}-A+B}{2}\bigg)=0$$
$$\implies2\cos\big(2\sqrt{AB}\big)\sin\big(A-B\big)=0$$
Note that, for $(\star)$,the graph looks like this.I have only taken the simplest case.
For the sake of simplicity, we will consider $A=B$ as of now and ignore the other cases..
We have,
$$\sqrt{2}\cos(\pi/4-x)=\frac{3}{2}-\cos^2(\pi/4-x)$$
Let $\cos(\pi/4-x)=k$
$$\implies k^2 +\sqrt{2}k-\frac{3}{2}=0$$
$$k=\frac{-3\sqrt{2}}{2},k=\frac{1}{\sqrt{2}}$$
$$\cos(\pi/4-x)=\cos(\pi/4)$$
$$\implies x=2n\pi +\frac{\pi}{2}\text{ or } x=2n\pi\text{ (where }n\in\mathbb{Z})$$
Hence $\sin^5x+\cos^5x=1$ provided that the condition given in the question is true
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4191572",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
Find all possible solutions of $\sqrt[3]{(x^2 + x - 1)} + \sqrt[3]{(10 - x - x^2)} = 3$ The Question: Find all possible solutions $x$ to the equation $\sqrt[3]{(x^2 + x - 1)} + \sqrt[3]{(10 - x - x^2)} = 3$.
My answer:
Let $x^2 + x - 1 = y$.
Then, $\sqrt[3]{y} + \sqrt[3]{9 - y} = 3.$
So, $\sqrt[3]{9 - y} = 3 - \sqrt[3]{y}$.
Cubing both sides, we get, $9 - y = 27 - 27\sqrt[3]{y} + 9(\sqrt[3]{y})^2 - y.$
Simplifying and dividing by $9$, we get $(\sqrt[3]{y})^2 - 3\sqrt[3]{y} + 2 = 0.$
Factoring, we get: $(\sqrt[3]{y} - 1)(\sqrt[3]{y} - 2) = 0.$
Therefore, we get $y = 1, \sqrt[3]{2}$.
Substituting this back into the equation $x^2 + x - 1 = y$, we get $\boxed{x = 1, -2,}$ and $\boxed{{\frac{-1 \pm \sqrt{37}}{2}}}$.
| Alternatively, let $x^2+x=t+\frac{11}2$ to get the symmetric equation in $t$
$$\sqrt[3]{\frac92-t} + \sqrt[3]{\frac92+t} = 3 \implies
\sqrt[3]{\left(\frac{9}2\right)^2-t^2}=2\implies t=\pm \frac72
$$
which leads to $x^2+x=9$ and $x^2+x=2$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4192015",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Patterns in indefinite integral (1) So far, I've found out this:
$\int \arcsin x \mathrm d x = x \arcsin x + \sqrt{1-x^2}+C$
$\int x\arcsin x \mathrm d x = \dfrac{x^2}{2}\arcsin x +\dfrac {x}{4}\sqrt{1-x^2} -\dfrac{1}{4}\arcsin x + C$
$\int x^2 \arcsin x \mathrm d x = \dfrac {x^3}{3}\arcsin x+\dfrac{x^2}{9}\sqrt{1-x^2}+\dfrac{2}{9}\sqrt{1-x^2} + C$
$\int x^3 \arcsin x \mathrm d x = \dfrac{x^4}{4}\arcsin x+\dfrac{x^3}{4\times 4}\sqrt{1-x^2}+\dfrac{3x}{4\times 4\times 2}\sqrt{1-x^2}-\dfrac{3\times 1}{4\times 4\times 2}\arcsin x + C$
$\int x^4 \arcsin x \mathrm d x = \dfrac {x^5}{5}\arcsin x +\dfrac {x^4}{5 \times 5}\sqrt{1-x^2}+\dfrac{4x^2}{5\times 5\times 3}\sqrt{1-x^2} + \dfrac{4\times2}{5\times 5\times 3}\sqrt{1-x^2}+C$
$\int x^5 \arcsin x \mathrm d x =
\dfrac {x^6}{6}\arcsin x
+\dfrac {x^5}{6\times6}\sqrt{1-x^2}
+\dfrac{5x^3}{6\times6\times4}\sqrt{1-x^2}
+\dfrac{5\times3x}{6\times6\times4\times2}\sqrt{1-x^2}
-\dfrac{5\times3\times1}{6\times6\times4\times2}\arcsin x + C$
There seems to be a pattern. It seems like for any positive whole number $n$,
$\int x^n \arcsin x \mathrm d x = \mathrm{(polynomial)}\arcsin x + \mathrm{(polynomial)}\sqrt{1-x^2} + C$
So, what is the pattern? What should the above polynomial be?
| Here is what Wolfram alpha thinks about the solution of
$$\int x^n \arcsin (x) \;\text d x$$
Now, the hypergeometric function is of course a polynomial. So, your guess for the pattern was quite right.
Now, Wolfram doesn't give you the steps. But, once you have the solution, it's very easy to prove that using induction. Note that in the induction step, you just need to do
$$\int x^{n+1} \arcsin (x) \;\text{d}x = \int x\cdot x^n \arcsin (x) \;\text
{d}x$$
and integrate by parts.
Scroll down the link I embedded to have the formula of differentiating the hypergeometric series. You will need it in the by parts method.
I hope, that helps.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4194159",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Finding $x_1+x_2$, where the $x_i$, and $y$ and $z$, are positive integers satisfying $x^2+y^2+z^2=2(xy+1)$ and $x+y+z=2022$
Let $x$, $y$, and $z$ be positive integers satisfying
$$(x)^2+(y)^2+(z)^2=2(xy+1) \quad\text{and}\quad x+y+z=2022$$
If $x_1$ and $x_2$ are 2 distinct solutions for $x$, what is the value of $x_1+x_2$?
Attempted to express $(x)^2+(y)^2+(z)^2=2(xy+1)$ as $(x-y)^2+(z)^2=2$, then I'm stuck.
Kindly advise! Thanks!
| So we have , $$(x-y)^2+z^2=2$$
Since $(x,y,z)\in\mathbb Z^+$ we can say $|x-y|=1$ and $z=1$ [ It is because only $1^2 ,1^2$ are the positive integers that sum upto $2$ ]
We have also given that , $$x+y+z=2022$$ $$x+y=2021$$
Now : $$x+y=2021$$ $$x-y=1$$
From these two linear equations in two variable you'll get , $$(x,y)\in\{1010,1011\}$$
Here you go you have two distinct values of $x$ .
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4194786",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Prove that the following improper integral $\int_2^\infty x^5 \ln(x^5-1)-5x^5\ln x$ diverges
Prove that the following improper integral $\int_2^\infty x^5 \ln(x^5-1)-5x^5\ln x$ diverges.
$Proof.$ by simplifying the integral we would get: $$\int _{2}^{\infty }\left( x^{5} \cdot \ln\left( x^{5} -1\right) -5x^{5}\ln( x)\right) dx=\lim _{M\rightarrow \infty }\int _{2}^{M}\left( x^{5}\left(\ln\left( x^{5} -1\right) -5\ln( x)\right)\right) dx\\
\\
=\lim _{M\rightarrow \infty }\int _{2}^{M}\left( x^{5}\left(\ln\left( x^{5} -1\right) -\ln\left( x^{5}\right)\right)\right) dx=\lim _{M\rightarrow \infty }\int _{2}^{M}\left( x^{5}\ln\left(\frac{x^{5} -1}{x^{5}}\right)\right) dx\\
\\
=\lim _{M\rightarrow \infty }\int _{2}^{M}\ln\left(\left(\frac{x^{5} -1}{x^{5}}\right)^{x^{5}}\right) dx=\lim _{M\rightarrow \infty }\int _{2}^{M}\ln\left(\left( 1-\frac{1}{x^{5}}\right)^{x^{5}}\right) dx\\
\\
\xrightarrow[x\rightarrow \infty ]{}\lim _{M\rightarrow \infty }\int _{1}^{M}\ln\left(\left(\frac{1}{e}\right)\right) dx=\lim _{M\rightarrow \infty }\int _{1}^{M} -1\\
\\
=\lim _{M\rightarrow \infty } -x\Bigl|_{2}^{M} =\lim _{M\rightarrow \infty } -M+2=-\infty $$
Now, I know that it easily diverges because the function has limit $L= \frac{1}{e}$ and it isn't zero. However, is what I have done still valid? or it is a critical mistake? Thanks!
| As you started writing it $$x^5 \log(x^5-1)-5x^5\log( x)=x^5 \log(x^5-1)-x^5\log( x^5)=x^5\log \left(1-\frac{1}{x^5}\right)$$ When $x$ is large
$$x^5\log \left(1-\frac{1}{x^5}\right)=-1-\frac{1}{2 x^5}+O\left(\frac{1}{x^{10}}\right)$$
$$\int x^5\log \left(1-\frac{1}{x^5}\right)\,dx=-x+\frac{1}{8 x^4}+O\left(\frac{1}{x^9}\right)$$
What does happen if you integrate for a large upper bound ?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4195888",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
modulo in a 24 h clock Suppose we have a clock with $24$ hours and it's $12$ o'clock. I need to calculate what time it will be after $100^{100}$ hours.
Obviously I just have to calculate $100^{100}=4 \pmod {24}$
So I started by $100=4 \pmod {24}$ but I cannot go anywhere from here.
I also tried typing it as $100=-20 \pmod {24}$ which didn't prove to be helpful either. Could someone help me with this one?
| Many ways to approach this sort of problem. Mine may be a bit long, but it works.
You want $4^{100} = 2^{200} \pmod{24}$.
First note that $2^7 = 128 \equiv 8 = 2^3 \pmod{24}$
This allows you to write $2^{7k} \equiv 2^{3k} \pmod{24}$.
Now $200 = 7(28)+4$ so you can write $2^{200} \equiv 2^{3(28) + 4} \\= 2^{7(12)+4} \equiv 2^{3(12)+4} \\= 2^{40} \\= 2^{7(5)+5} \equiv 2^{3(5)+5} \\= 2^{20} \\= 2^{7(2)+ 6}\equiv 2^{12} \\= 2^{7+5} \equiv 2^{3+5} \\= 2^8 \\= 2^{7+1} \equiv 2^{3+1} \\= 16 \pmod{24}$
So the answer is $16$ hours after your start point.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4201399",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Proving on the equation $\frac1{x+1}+\frac1{2(x+2)}+\frac1{3(x+3)}+\cdots+\frac1{n(x+n)}=1$ Consider the equation:
\begin{gather*}
\frac{1}{x+1} + \frac{1}{2(x+2)} + \frac{1}{3(x+3)} + \cdots + \frac{1} {n(x+n)} = 1.\tag1
\end{gather*}
Prove that:
*
*For every $n \geq 2$, equation $(1)$ always has a unique positive solution $x_n$.
*The sequence $(x_n)$ has a finite limit with $n \to \infty+$. Find that limit.
Here's all I did:
$\frac {1}{x+1} + \frac{1}{2(x+2)} + \frac{1}{3(x+3)} + .... + \frac {1} {n(x+n)} = 1.$
$\Leftrightarrow 2.3...n(x+1)(x+2)...(x+n) + 1.3...n(x+1)(x+3)...n + ... + 1.2...(n-1)(x+1)(x+2)...(x+n-1) = n! (x+1)(x+2)...(x+n) $
Consider the polynomial $P(x) = 2.3...n(x+1)(x+2)...(x+n) + 1.3...n(x+1)(x+3)...n + ... + 1.2...(n-1)(x+1)(x+2)...(x+n-1) - n! (x+1)(x+2)...(x+n) $
Consider the highest coefficient of the polynomial:$ -n! < 0 $
Consider the lowest coefficient of the polynomial:$ (2.3...n)^2+(1.3...n)^2+...(1.2...(n-1))^2 - n! = (1.3...n)^2+...(1.2...(n-1))^2 > 0$
Let $x_1,x_2,x_3....$ be the solutions of$ P(x) $
Assume that equation (1) has no positive solution.
According to Viette's theorem: $x_1 x_2 x_3....x_n = (-1)^n \frac{a_0}{a_n} $
$\frac {a_0}{a_n} < 0 \Rightarrow $if $n$ is an odd number , then $x_1 x_2 x_3....x_n < 0$ , but
$(-1)^n \frac{a_0}{a_n} >0 \Rightarrow $ (nonsense) . Same with $n $being an even number.
So the polynomial must have at least one integer root.
Suppose the polynomial has at least 2 positive roots $x$ and $y$ .
$\frac {1}{y+1} + \frac{1}{2(y+2)} + \frac{1}{3(y+3)} + .... + \frac {1} {n(y+n)} =\frac {1}{x+1} + \frac{1}{2(x+2)} + \frac{1}{3(x+3)} + .... + \frac {1} {n(x+n)} .$
$\Rightarrow (y-x)( \frac {1}{(x+1)(y+1)}+ \frac {1}{2(x+2)(y+2)} +...+\frac {1}{n(x+n)(y+n)} ) = 0 $
$\Rightarrow x=y \Rightarrow$ (no sense)
So for every $n \geq 2$ , equation $( 1 )$ always has a unique positive solution $x_n$
That's all I did , and I have no idea what to do next , I want to establish a relationship between $x_n$ but for $n \geq 3$ the polynomial's solution is a " bad " solution . " so it's very difficult to build. Hope to get help from everyone. Thanks very much .
| Let $$f_n(x) = \frac{1}{x+1} + \cdots + \frac{1}{n(x+n)}.$$
Observe that $f_n(x)$ is strictly decreasing in $x$ for $x\ge 0$. Further, for $n\ge 2$, $f_n(0) > 1$ and $\lim_{x\to\infty} f_n(x) = 0$. It follows that there is a unique positive solution $x_n$ to $f_n(x) = 1$.
Let $f$ be defined by
$$f(x) = \sum_{i=1}^\infty \frac{1}{i(x+i)}.$$
Note that $(f_n)$ uniformly converges to $f$ (for $x\ge 0$, $|(f-f_N)(x)| \le \sum_{i=N}^\infty 1/i^2$, which can be made arbitrarily small for sufficiently large $N$).
Let the $x_n$ converge to $l$. Then, uniform convergence implies that $1 = \lim_{n\to\infty} f_n(x_n) = f(l)$. So,
$$
\begin{align*}
1 &= \sum_{i=1}^\infty \frac{1}{i(l+i)} \\
l &= \sum_{i=1}^\infty \frac{1}{i} - \frac{1}{l+i}.
\end{align*}
$$
Equality is seen to hold for $l=1$ since then, the right hand side becomes a telescoping sum which is evaluated to be $1$, and the left is $1$.
Thus, $x_n \to 1$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4203911",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
} |
Find the point on the curve $x^{2} - yz=1$ that is closest to the origin The problem statement is as follows:
Find the point on the curve $x^{2} - yz=1$ that is closest to the origin.
To do this we minimize $D^{2} = x^{2} + y^{2} + z^{2}$ with the above constraint.
Attempted solution:
Isolate $z$ in the constraint:
$$
x^{2} - yz = 1 \implies z = \frac{x^{2}-1}{y}
$$
Substitute into distance formula:
$$
x^{2} + y^{2} + \left(\frac{x^{2}-1}{y}\right)^{2}
$$
Differentiate in terms of x and y and set to zero
$$
\frac{\partial D^{2}}{\partial x} = 2x+4x\left(\frac{x^{2}-1}{y^{2}}\right)=0,\frac{\partial D^{2}}{\partial y} = 2y-2\left(\frac{(x^{2}-1)^{2}}{y^{3}}\right)=0
$$
which has real solutions $x=0, y=+1$, giving the critical point as $x=0, y=1, z=-1$.
The correct answer is $x=\pm 1, y=0, z=0$.
What am I doing wrong?
| An alternative approach is using Lagrange multipliers. Construct the Lagrangian auxiliary function
$$\mathcal L=x^{2} + y^{2} + z^{2}+\lambda\left[1-\left(x^{2} - yz\right)\right]$$
and solve the first order conditions
$$\begin{aligned}
\frac{d\mathcal L}{d x}=2x-2x\lambda=0\\
\frac{d\mathcal L}{d y}=2y+z\lambda=0\\
\frac{d\mathcal L}{d z}=2z+y\lambda=0\\
\frac{d\mathcal L}{d \lambda}=1-x^2+yz=0\\
\end{aligned}$$
that, of course, result in $x=\pm 1$, $y=0$ and $z=0$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4204521",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
Given $a,b,c$ are sides of a triangle, Prove that :- $\frac{1}{3} \leq \frac{a^2+b^2+c^2}{(a+b+c)^2} < \frac{1}{2}$
Given $a,b,c$ are sides of a triangle, Prove that :-
$$\frac{1}{3} \leq \frac{a^2+b^2+c^2}{(a+b+c)^2} < \frac{1}{2}$$
What I Tried:- I was able to solve the left hand side inequality. From RMS-AM Inequality on $a,b,c$ :-
$$\sqrt{\frac{a^2+b^2+c^2}{3}} \geq \frac{a+b+c}{3}$$
$$\rightarrow \frac{a^2+b^2+c^2}{3} \geq \frac{(a+b+c)^2}{9}$$
$$\rightarrow \frac{1}{3} \leq \frac{a^2+b^2+c^2}{(a+b+c)^2}$$
I got no progress for the second part. I also have a clue, that $a,b,c$ are sides of a triangle, which I have not used yet. So maybe that should be used somehow, but I am not getting it.
Can anyone help me? Thank You.
| Using triangle inequality,
$a(b+c)+b(c+a)+c(a+b) \gt a^2 + b^2 + c^2$
$a^2+b^2+c^2+a(b+c)+b(c+a)+c(a+b) \gt 2 (a^2 + b^2 + c^2)$
$(a+b+c)^2 \gt 2(a^2 + b^2 + c^2)$
$\cfrac{a^2 + b^2 + c^2}{(a+b+c)^2} \lt \cfrac{1}{2}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4206765",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 1
} |
show that series $\sum_{n=1}^\infty \frac {(-1)^{n-1}}{n+x^2}$ is uniformly convergent for all real values of x I tried using weirestrass M test .
My method : let $$\ f(x)=\frac {1}{(n+x^2)}\ $$
$$f'(x)= \frac {-2x}{(n+x^2)^2}=0 $$
this gives $x=0 $
$$f''(x)= -2[\frac{(n+x^2)^2 -2(n+x^2)(2x.x)}{(n+x^2)^4}]\ = \frac {-2(n-3x^2)}{(n+x^2)^3}$$ $$f''(x)= \frac {-2}{n^2} \lt 0$$
Thus, f(x) has a maxima at x=0 $$f(x)_{max} = f(0)= \frac 1n$$
Let $M_{n}$(weirestress M sequence) be $$\frac {1}{n}$$ thus $$f(x) \lt \frac 1n $$
But $$\frac 1n$$ is a diverging sequence thus the series is non uniformly convergent which is contrary to correct answer(i.e series is uniformly convergent) .
tell my what i'm doing wrong . have i taken $M_{n}$ series wrong ?
| You can rewrite your sum as follows:
\begin{align*}
\sum_{n=1}^{\infty}\frac{1}{2n-1 + x^2} - \frac{1}{2n + x^2} & = \sum_{n=1}^{\infty} f(2n-1) - f(2n) \\
\end{align*}
where $$f(t) = \frac{1}{t+x^2}$$
The Mean Value Theorem says there is some $c \in (t-1,t)$ for which we have $$f(t-1) - f(t) = (\frac{d}{dt}f)(c) = - \frac{1}{(c+x^2)^2}$$
For $c \in (t-1,t)$ we have that
$$\bigg|- \frac{1}{(c+x^2)^2} \bigg| \leq \frac{1}{(t-1)^2} $$
Let $t = 2n$ and apply the test there.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4206870",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Determine $\sin(x + y)$ and $\cos(x - y)$, if you know $\cos(x) = \frac{5}{7}, \quad x\in(0,\pi/2), \quad \sin(y) = \frac{1}{5},\quad y\in(\pi/2,\pi)$ I need advice on this example.
Determine $\sin (x + y)$ and $\cos (x - y)$, if you know that
$\cos(x) = \frac{5}{7}, \quad x\in(0,\pi/2), \quad \sin(y) = \frac{1}{5}, \quad y\in(\pi/2,\pi).$
I used this formula to express $\sin (x)$ and $\cos (y)$
$$\sin^2(x)+\cos^2(x)=1$$
I recieved:
$$\sin(x)=\sqrt{24}/7\quad\text{and}\quad\cos(y)=\sqrt{24}/5.$$
However, if I now substitute into these formulas I get a different result than I have in the assignment
$$
\begin{align*}
\sin(x+y)&=\sin(x)\cdot \cos(y)+\cos(x)\cdot \sin(y)\\
\cos(x-y)&=\cos(x)\cdot \cos(y)+\sin(x)\cdot \sin(y)
\end{align*}
$$
The correct result should be
$$\cos(x-y) = -8\sqrt{6}/35\quad\text{and}\quad\sin(x+y) = -19/35.$$
| You have a sign error.
$y$ is in the second quadrant, hence $\cos(y) = -\frac{\sqrt{24}}{5}$.
\begin{align}
\sin(x+y) = -\frac{\sqrt{24}}{7} \cdot \frac{\sqrt{24}}{5} + \frac57 \cdot \frac15 = -\frac{19}{35}
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4211864",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Comparing $\sqrt{5} + \sqrt{6} + \sqrt{11}$ and $8$ without calculating the values
I want to compare $\sqrt{5} + \sqrt{6} + \sqrt{11}$ and $8$ without calculating the actual value of square roots.
I tried to apply square on both side but it still carries the root terms.
Any trick I was missing here?
| The following is a somewhat more laborious, but also more generic, way to solve the problem.
*
*The minimal (rational) polynomial of $\,x_0=\sqrt{5}+\sqrt{6}+\sqrt{11}\,$ is
$$
P(x)=x^8 - 88 x^6 + 1696 x^4 - 10560 x^2 + 14400
$$
which can be determined similar to Minimal Polynomial of $\sqrt{2}+\sqrt{3}+\sqrt{5}$.
*With the substitution $\,x \to x+4\,$:
$$P(x+4) = x^8 + 32 x^7 + 360 x^6 + 1472 x^5 - 1504 x^4 - 28160 x^3 - 70976 x^2 - 59904 x - 15296
$$
By Descartes' rule of signs $A(x)=P(x+4)$ has exactly one real positive root, since there is one change of signs and $A(0) \lt 0\,$. It follows that $P(x)$ has exactly one real root larger than $\,4\,$ and that root must be $\,x_0\,$ since $\,x_0=\sqrt{5}+\sqrt{6}+\sqrt{11} \gt \sqrt{4}+\sqrt{4} = 4\,$.
*With the substitution $\,x \to x+8\,$:
$$P(x+8) = x^8 + 64 x^7 + 1704 x^6 + 24448 x^5 + 203936 x^4 + 988160 x^3 + 2574016 x^2 + 2780160 x - 6080
$$
Again by Descartes' rule of signs $B(x)=P(x+8)$ has exactly one real positive root, since there is one change of signs and $B(0) \lt 0\,$. It follows that $P(x)$ has exactly one real root larger than $\,8\,$, and that root must be the unique root larger than $\,4\,$ as established at the previous step, so in the end $\,x_0 \gt 8\,$.
As a side note, the middle step and related calculations could have been avoided by noting that the roots of $\,P(x)\,$ are the rational conjugates $\,\pm\sqrt{5}\pm\sqrt{6}\pm\sqrt{11}\,$, so $\,x_0\,$ is the largest real root.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4212106",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 2
} |
solve the equation: $\cos (z) = 1 + i$ , $z \in ℂ$. I want to solve the equation: $\cos (z) = 1 + i$ , $z \in ℂ$.
I started by saying $\cos (z) = \frac {e^{iz} + e^{-iz}}{2}$.
I am therefore led to solve:
$$e^{iz} + e^{-iz} = 2 + 2i $$ this implies that $$e^{2iz}+1 =(2 + 2i) e^{iz}$$
By setting $x = e^{iz}$
Consider the equation:
$$x^2-(2 + 2i) x + 1 = 0$$
which is equivalent to solving the equation $$ x^2-2 \sqrt{2} e^{i \frac {\pi}{4}} x + 1 = 0$$
then $$ (x- \sqrt{2} e^{i \frac{\pi}{4} })^2 + 1 + 2i = 0$$
We set $$w = x- \sqrt {2} e^{i \frac{\pi}{4}}$$ so $$w^2 = -1-2i$$ But I can't seem to continue. An idea please.
| At this point
$x^2-(2 + 2i) x + 1 = 0.$
Applying the quadratic formula:
$x = 1+i \pm \sqrt {2i - 1}:$
$(1+i + \sqrt {2i - 1})(1+i - \sqrt {2i- 1}) = 2i - (2i-1) = 1$
This means we can let $e^{iz} = 1+i + \sqrt {2i - 1}$ and $e^{-iz} = 1+i - \sqrt {2i - 1}$
$iz = \ln (1+i + \sqrt {2i - 1})\\
z = -i\ln (1+i + \sqrt {2i - 1})$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4212296",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Evaluating integral $\int_0^{2\pi}\frac{1}{2-\sin(x)}\,dx$ I come across a problem, which is to evaluate: $$\int_0^{2\pi}\frac{1}{2-\sin(x)}\,dx$$
My attempt so far is:
$$\begin{align}\int_0^{2\pi}\frac{1}{2-\sin(x)}\,dx
&= \int_0^{2\pi}\frac{1}{2-\frac{2\tan(\frac{x}{2})}{1+\tan^2(\frac{x}{2})}}\,dx\\
&= \int_0^{2\pi}\frac{1+\tan^2(\frac{x}{2})}{2+2\tan^2(\frac{x}{2})-2\tan(\frac{x}{2})}\,dx\\
&= \int_0^{2\pi}\frac{\frac{1}{2} \sec^2(\frac{x}{2})}{1+\tan^2(\frac{x}{2})-\tan(\frac{x}{2})}\,dx\end{align}$$
I'm wondering whether I'm in the right direction and how can I proceed?
| Note that because of periodicity, $\displaystyle \int_0^{2\pi} = \int_{-\pi}^{+\pi}.$
If you let $u = \tan\tfrac x2$ so that $\sin x= \dfrac{2u}{1+u^2}$ and $du = \dfrac{2\,du}{1+u^2},$ then as $x$ goes from $-\pi$ to $+\pi,$ $u$ goes from $-\infty$ to $+\infty.$
\begin{align}
& \int_{-\pi}^{+\pi} \frac 1 {1+\tan^2(\frac{x}{2})-\tan(\frac{x}{2})} \big(\tfrac 1 2\sec^2\tfrac x 2 \,dx\big) \\[8pt]
= {} & \int_{-\infty}^{+\infty} \frac 1 {1+u^2 - u} \, du \\[12pt]
= {} & \int_{-\infty}^{+\infty} \frac 1 {(u-\frac 1 2 )^2 + \frac 3 4} \, du \quad \left( \begin{array}{c} \text{This is complet-} \\ \text{ing the square.} \end{array} \right)\\[10pt]
= {} & \frac 4 3 \int_{-\infty}^{+\infty} \frac 1 {\left( \frac{2u-1}{\sqrt3} \right)^2 + 1} \, du \quad \left( \begin{array}{c} \text{Here we multiplied the} \\ \text{top and bottom by }4/3. \end{array} \right) \\[10pt]
= {} & \frac 2 {\sqrt3}\int_{-\infty}^{+\infty} \frac 1 {\left( \frac{2u-1}{\sqrt3} \right)^2+1} \left( \tfrac 2 {\sqrt3} \, du \right) \\[10pt]
= {} & \frac 2 {\sqrt3} \int_{-\infty}^{+\infty} \frac 1 {w^2+1} \, dw = \frac{2\pi}{\sqrt3}.
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4213093",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
Irreducibilty of polynomial $x^9-6x^6+282x^3-8$ over $\mathbb {Q} $ While trying to deal with the final parts of this answer I found that one needs to establish $$a=\sqrt[3]{1+\sqrt{11}}\notin\mathbb{Q} (\sqrt{11},\sqrt[3]{10})=F$$ Since both $a, F$ are of degree $6$ over $\mathbb {Q} $ the tower theorem does not help much.
Then I reasoned via contradiction. If $a\in F$ then $b=\sqrt[3]{1-\sqrt{11}}=-\sqrt[3]{10}/a\in F$ and hence $c=a+b\in F$.
One can check that $$c^3=a^3+b^3+3abc=2-3\sqrt[3]{10}c$$ or $$(c^3-2)^3=-270c^3$$ or $c$ is a root of $$f(x) =x^9-6x^6+282x^3-8$$ I checked using pari/gp that the polynomial $f(x) $ is irreducible over $\mathbb {Q} $. Further it can be observed that $f(289)$ is prime (also checked via pari/gp) so that the polynomial $f(x) $ is irreducible by Murty's criterion.
This shows that $c$ is of degree $9$ over $ \mathbb {Q} $ and hence $c\notin F$.
It can be observed that $f(x) $ can not be handled by Eisenstein as $2$ is the only prime which divides all non-leading coefficients and $4\mid 8$. I also checked the reducibility mod $3$ and the polynomial is reducible mod 3. So even that approach does not work. I don't know if using reducibility modulo other primes would help.
Is there is any other simpler way to prove the irreducibilty of $f(x) $ using hand computation?
Update: We have $$f(2x)=8(64x^9-48x^6+282x^3-1)=8g(x)$$ and it is somewhat easier to apply Murty's criterion on $g(x) $ with $g(8)=8577496063$ being prime. But I still find this unsuitable for hand calculation.
| The polynomial $P(x)$ factors $\!\!\!\mod \!\!13$ as $(x^3+6) ( x^6+ x^3 + 3) $ (link).
Assume that $P(x)$ is reducible. We conclude that $P(x)$ is a product of two polynomials $P_1(x)$, $P_2(x)\in \mathbb{Z}[x]$. with $P_1(x)\equiv x^3 + 6 \!\!\!\mod \!\!13$, $P_2(x)\equiv x^6 + x^3 + 3 \!\!\!\mod\!\! 13$.
Now, $P(1) =269$, a prime number. We conclude $P_1(1)=\pm 1$, or $P_2(1)=\pm 1$. But $P_1(1)\equiv 7 \!\!\!\mod \!\!13$, while $P_2(1)\equiv 5 \!\!\!\mod \!\!13$, contradiction.
..........................................................
$\bf{Added:}$ About the factorization $\mod 13$. Our polynomial is $P(x) = Q(x^3)$, with $Q(x)= x^3 - 6 x^2 + 282 x - 8$. Now $Q(x)$ has root $7$ $\mod 13$, so we have
$Q(x) = (x-7)(x^2 + x + 3)$, and so
$P(x) = (x^3 - 7)(x^6+x^3 + 3)$. Now $x^3-7$ has no root $\mod 13$, because $7^{\frac{12}{3}}= 7^4 \not \equiv 1 \mod 13$. Also, $x^2 + x + 3 = x^2 + 2 \cdot 7 x + 7^2 + 3 - 7^2= (x+7)^2 - 7$, and $7$ is not a quadratic residue $\mod 13$, so $x^2 + x + 3$ is irreducible.
Based on Jyrki's idea: Note that since $13\equiv 1 \mod 3$, there exists $\omega \in \mathbb{F}_{13}$, such that $\omega^3 = 1$, and $\omega\ne 1$ ( a primitive root of $1). \mod 13$). Now, consider the action of the group of roots of $1$ on monic polynomials $f(x) \mapsto \frac{1}{\omega^{\deg f}} f(\omega x)$. The polynomials in $x^3$ are invariated by the action ( the fixed points are $x^m \cdot $ pol in $x^3$). Therefore, the a polynomial in $x^3$, its irreducible factors will be either polynomials in $x^3$, or group in orbits of size $3$ under this action. Now, if the polynomial $x^6 + x^3 + 3$ were reducible, it would equal $R(x) R(\omega x) R(\omega^2 x)$, where $R$ is a polynomial of degree $2$. But that would imply that the free term $3$ is a cube. However, we have $3^{\frac{12}{3}} = 3^4 =3 \mod 13$. Therefore, $x^6 + x^3 + 3$ is irreducible $\mod 13$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4214793",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 3,
"answer_id": 1
} |
Contradiction in arranging words in two different ways
In how many ways can the letters in "WONDERING" be arranged with exactly two consecutive vowels ?
The solution is $18 \times 7!$ by classical counting technique
(because firstly choose $2$ adjacent vowels out of $3$ vowels by $3$ ways.Lets say we chose $O,E$ After that , arrange them by $2!\times (8!/2!)$ ways because there are double N. Result = $3 \times 8!$ . However , when we arrange $2$ vowels ,there will be cases where the third vowel $I$ is adjacent to them , so we must subtract the cases where $3$ vowels are adjacent. The result is $3 \times 2 \times 2 \times (7!/2!)$ , so the main result is $3 \times 8! - 3 \times 2 \times 2 \times (7!/2!) =18 \times 7!$) .
However , when i wanted to solve it by using formula , i encounter with different result.
My formula is $E_m=S_m - (\frac{m+1}{1})S_{m+1} + (\frac{m+2}{1})S_{m+2} - ...$ where $m$ represent the number of elements in $S$ that satisfy exactly $m$ of the conditions. Here , $S_m$ means that the number of the arrangements of $m$ adjacent vowels. For example $S_2= 3\times 8!$
Then , $E_2=S_2 - (\frac{2+1}{1})S_{2+1} =C(3,2) \times 2! \times \frac{8!}{2!} - 3 \times 3! \times \frac{7!}{2!} =15 \times 7!$
What am i missing ? This formula is working when the letters are CORRESPONDENTS , in that case , we write $C(5,2)[12! / (2!)^3]-C(3,1)C(5,3)[11! / (2!)^2]+C(4,2)C(5,4)[10! / (2!)]-C(5,3)C(5,5)[9!]$ for the case where there are exactly 2 pairs of consecutive identicsl letters.
However , the formula does not work for WONDERING , WHY ??
| Your first calculation is correct.
Method 1: First, arrange the six consonants N, N, D, G, R, W. This can be done in
$$\binom{6}{2}4! = \frac{6!}{4!2!} \cdot 4! = \frac{6!}{2!}$$
ways since we must select two of the six positions for the Ns, then arrange the remaining four distinct letters in the remaining six positions. This creates seven spaces in which to place the vowels, five between successive consonants and two at the ends of the row.
$$\square C_1 \square C_2 \square C_3 \square C_4 \square C_5 \square C_6 \square$$
Choose one of those seven spaces in which to place a block of two vowels. To ensure that there are exactly two consecutive vowels, choose one of the six remaining spaces in which to place the remaining vowel. Arrange the three distinct vowels from left to right in the selected spaces in $3!$ ways. Hence, there are
$$\frac{6!}{2!} \cdot 7 \cdot 6 \cdot 3! = 6! \cdot 7 \cdot 6 \cdot 3 = 18 \cdot 7!$$
distinguishable arrangements of the letters of the word WONDERING in which exactly two of the three vowels are consecutive.
Method 2: We apply the Inclusion-Exclusion Principle.
A pair of consecutive vowels: Choose which two of the three vowels will form a block. We now have eight objects to arrange, the six consonants, the block of two vowels, and the remaining vowel. Choose two of those eight positions for the Ns, arrange the remaining six distinct objects in the remaining six positions, then arrange the two vowels within the block. This can be done in
$$\binom{3}{2}\binom{8}{2}6!2! = \frac{3!}{2!1!} \cdot \frac{8!}{2!6!} \cdot 6!2! = 3 \cdot 8!$$
ways, which agrees with your calculation.
However, we have counted each case in which there are two pairs of consecutive vowels twice, once for each pair of consecutive vowels. We only want to count such arrangements once, so we must subtract them from the total.
Two pairs of consecutive vowels: Since there are only three vowels, this can only occur if the three vowels are consecutive since the two pairs of consecutive vowels must overlap. Thus, we have seven objects to arrange, the six consonants and the block of three consecutive vowels. The objects can be arranged in $7!$ ways. The three vowels can be arranged within the block in $3!$ ways. Hence, there are
$$7!3!$$
arrangements with two pairs of consecutive vowels.
Thus, by the Inclusion-Exclusion Principle, the number of distinguishable arrangements of the letters of the word WONDERING with exactly two adjacent vowels is
$$3 \cdot 8! - 7!3! = 3 \cdot 8 \cdot 7! - 6 \cdot 7! = (24 - 6) \cdot 7! = 18 \cdot 7!$$
as we found above.
Notice that I subtracted the number of arrangements with two pairs of adjacent vowels from cases with one pair of adjacent vowels rather than subtracting arrangements with three consecutive vowels from arrangements with two consecutive vowels. This is where you made your error.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4215554",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
} |
How to prove that addition modulo 2 is associative I learned that |a⊕(a⊕b)⟩=|(a⊕a)⊕b⟩ as addition modulo 2 is associative. However, I am wondering why addition modulo 2 is associative?
| *
*Try them all; there are only eight cases
a) $(0\oplus0)\oplus0=0\oplus0=0;0\oplus(0\oplus0)=0\oplus0=0$
b) $(0\oplus0)\oplus1=0\oplus1=1;0\oplus(0\oplus1)=0\oplus1=1$
c) $(0\oplus1)\oplus0=1\oplus0=1;0\oplus(1\oplus0)=0\oplus1=1$
d) $(0\oplus1)\oplus1=1\oplus1=0;0\oplus(1\oplus1)=0\oplus0=1$
e) $(1\oplus0)\oplus0=1\oplus0=1;1\oplus(0\oplus0)=1\oplus0=1$
f) $(1\oplus0)\oplus1=1\oplus1=0;1\oplus(0\oplus1)=1\oplus1=0$
g) $(1\oplus1)\oplus0=0\oplus0=0;1\oplus(1\oplus0)=1\oplus1=0$
h) $(1\oplus1)\oplus1=0\oplus1=1;1\oplus(1\oplus1)=1\oplus0=1$
*Reason it out. But handwave a little bit.
Let $(a \oplus b) \oplus c \equiv k \pmod 2$. THen by definition $[(a+b)+c]-k$ is divisible by $2$ where $a,b,c,k$ are integers and $+,-$ are integer operations. As $+$ is associative over $\mathbb Z$ we have that $[(a+b) + c] -k = [a+(b+c)]-k$ and $[a+(b+c)] -k$ is divisible by $2$. So $a\oplus (b\oplus c)\equiv k \equiv a\oplus (b\oplus c)\pmod 2$.
In other words. Addition modulo $2$ "inherits" associativity from addition over the integers.
*Reason it out with technical overkill
Let $a\oplus b = k$. Then by definition that means $a+b - k$ is divisible by $2$ ($a,b,k$ are regular old integers and $+,-$ is addition and subtraction. $k$ is either $0$ or $1$).
Let $(a+b)-k = 2w$ where $w$ is some integer. Then $k = (a+b)-2w$
Let $(a\oplus b)+c = k\oplus c = m$ ($m$ is either $0$ or $1$). Then by definition $k+c -m$ is even. Replace $k$ with $a+b-2w$ and we get $[(a+b)-2w]+c-m = ((a+b)+c) -2w -m$ is even. Let $((a+b)+c)-2w - m = 2\omega$. And that means $m = (a+b+c)-2(w+\omega)$. (Note, integer addition is just addition and it is commutative and associative.)
Now $m$ is either $0$ or $1$. $2(w+\omega)$ is even. So if $a+b+c$ is odd, then $m =1$. If $a+b+c$ is even then $m = 0$.
Now let $b\oplus c = j$. $a\oplus(b\oplus c)=a\oplus j= h$. That means $(b+c)-j$ is an even number. Let $(b+c)-j=2v$ so $j=(b+c)-2v$. And it means $(a+j)-h=(a+(b+c)-2v)-h$ is even. Let $((a+(b+c)-2v)-h = 2\alpha$ and that means $h = (a+b+c)-2(v-\alpha)$.
Again $h$ is either equal to $0$ or $1$ and if $a+b+c$ is odd then $h=1 = m$. Or if $a+b+c$ is even then $h = 0 = m$.
Either way $ a\oplus(b\oplus c) =h = m = (a\oplus b)\oplus c$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4217501",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
$x_1$ , $x_2$ are roots of $x=5-x^2$. Find the equation with roots $\frac1{(x_1+1)^3}$ and $\frac1{(x_2+1)^3}$.
Suppose $x_1$ , $x_2$ are the roots of the equation $x=5-x^2$.Then
$\dfrac1{(x_1+1)^3}$ and $\dfrac1{(x_2+1)^3}$ are roots of which
equation?
$1)125x^2+16x=1$
$2)125x^2=16x+1$
$3)125x^2=12x+1$
$4)125x^2+12x=1$
I solved this problem with the following approache,
I've denoted the roots of the original equation by $\alpha$ and $\beta$ rather than $x_1$, $x_2$ ,
$S=\alpha+\beta=-1$ and $P=\alpha\beta=-5$. We find $S'$ and $P'$ of the new equation,
$$P'=\dfrac1{[(\alpha+1)(\beta+1)]^3}=\dfrac1{(S+P+1)^3}=-\frac1{125}$$
$$S'=\dfrac{(\beta+1)^3+(\alpha+1)^3}{[(\alpha+1)(\beta+1)]^3}=-\frac1{125}\times\left((\alpha^3+\beta^3)+3(\alpha^2+\beta^2)+3(\alpha+\beta)+2\right)$$
$$=-\dfrac{(S^3-3PS)+3(S^2-2P)+3S+2}{125}=-\dfrac{-16+33-3+2}{125}=-\frac{16}{125}$$
Hence the new equation is $125x^2+16x-1=0$. And the answer is first choice.
This was a problem from a timed exam. So can you solve it with other approaches (preferably quicker one) ?
| Using that :
*
*$x^2=-x+5$
*$x^3=x\cdot x^2=-x^2+5x=x-5+5x=6x-5$
Then:
$$
\require{cancel}
y = \frac{1}{(x+1)^3}=\frac{1}{x^3+3x^2+3x+1}=\frac{1}{(6x-5)+3(-x+5)+3x+1} = \frac{1}{6x+11}
\\ \iff\;\;\;\; 6x = \frac{1}{y}-11
$$
Substituting the latter in $36x^2 + 36x - 180 = 0$ and multiplying by $y^2$ gives:
$$
(1-11y)^2 + 6y(1-11y) - 180y^2 = 0 \;\;\;\;\iff\;\;\;\; 125 y^2 + 16 y - 1 = 0
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4218287",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 0
} |
Find all primes $r$ and $s$ such that $ r^7+s=s^3+r $
Find all primes $r$ and $s$ such that $\;r^7+s=s^3+r\;.$
I have tried to factorise the expression since we are dealing with primes so:
$$ s(s^2-1)= r(r^6-1)=r(r^2-1)(r^4+r^2+1).$$
Since $ s^3>r^7$, then $\;s^2-1>r^4-1$.
That is all what I have found, thank you in advance for your precious help!
| Trying to milk more out of the factorization
$$
s(s-1)(s+1)=s^3-s=r^7-r=r(r-1)(r+1)(r^2-r+1)(r^2+r+1).\qquad(*)
$$
The OP already observed that $s^3>r^7$. Therefore $s>r^2$. Let's remember that $s$ is a prime. Staring at $(*)$ for a few seconds tells us that the only possible prime factor of the RHS larger than $r^2$ is $r^2+r+1$.
Therefore we must have $s=r^2+r+1$.
Plugging this into $(*)$ gives us
$$
\begin{aligned}0&=r^7-r-(r^2+r+1)^3+(r^2+r+1)\\
&=r(r+1)(r-3)(r^2+1)(r^2+r+1).
\end{aligned}
$$
Obviously $r=3$ is the only relevant solution and, luckily, $s=r^2+r+1=13$ turns out to be a prime.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4222576",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 2,
"answer_id": 0
} |
Evaluate by polynomial expansion the following integral with an error less than 0.000001 My work till now:
$\int_0^{0.2} \frac{1}{1+x^5} dx=\int_0^{0.2}(1+x^5)^{-1} dx$
$(1+x)^n=1+nx+ \frac{n(n-1)x^2}{2}...$
$\int_0^{0.2} \bigl( 1+(-1)x^5+ \frac{(-1)(-2)x^{10}}{2}- \frac{1(-2)(-3)x^{15}}{3!}+...\bigr)dx$
Now, I could integrate the above series but how can I find the result within that error?
Any tip/help is appreciated. Thanks!
| I think that the approximation works well: since
$$1+x^{15}=(1+x^5)(1-x^5+x^{10})$$
it follows that for any $x\in [0,1]$
$$\frac{1}{1+x^5}=1-x^5+x^{10}-\frac{x^{15}}{1+x^5}.$$
Hence
$$\begin{align}
\left|\int_{0}^{0.2}\frac{1}{1+x^5}\,dx-\int_{0}^{0.2}(1-x^5+x^{10})\,dx\right|&\leq\int_{0}^{0.2} \frac{x^{15}}{1+x^5}\,dx\\
&\leq \int_{0}^{0.2} x^{15}\,dx=\frac{(0.2)^{16}}{16}<10^{-12}.
\end{align}$$
Note that
$$\int_{0}^{0.2}\frac{1}{1+x^5}\,dx
\approx 0.199989335194742$$
and
$$\int_{0}^{0.2}(1-x^5+x^{10})\,dx\approx 0.1999893351951515.$$
P.S. Actually for an error less that $0.000001=10^{-6}$ you may consider less terms:
$$\frac{1}{1+x^5}=1-x^5+\frac{x^{10}}{1+x^5}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4222914",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
Solving $\sqrt[3]{4 - x^{2}} + \sqrt{x^{2} - 3} = 1$ I’m having a hard time with this radical equation.
$$\sqrt[3]{4 - x^{2}} + \sqrt{x^{2} - 3} = 1$$
I subtracted square root of x^2-3 then I got ^3 square root of -x+4 + square root of x^2-3 = 1. Next I cube both sides which I got -x^2-3= (1- square root of x^2-3) ^3. Would o need to add that back to the other side and use difference of cubes ?
| HINT
We can let $u = x^{2} - 3$, where $x^{2} - 3 \geq 0$. Then we get that
\begin{align*}
\sqrt[3]{1 - u} + \sqrt{u} = 1 & \Longleftrightarrow \sqrt[3]{1-u} = 1 - \sqrt{u}\\\\
& \Longleftrightarrow 1 - u = 1 - 3\sqrt{u} + 3u - \sqrt{u^{3}}\\\\
& \Longleftrightarrow \sqrt{u^{3}} - 4u + 3\sqrt{u} =0\\\\
& \Longleftrightarrow (u - 4\sqrt{u} +3)\sqrt{u} = 0
\end{align*}
Can you take it from here?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4223355",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Evaluate a coordinate of the center of mass of the solid $A$ with constant density $1$ Find the $y_G$ coordinate of the center of mass $G$ of the solid $A=\{(x,y,z) \in \mathbb{R}^3 \ | \ 3x^2 \leq y^2+z^2 \leq 3-x^2, yz \leq 0, z \geq |y|\}$ with constant density $\rho(x,y,z)=1$.
Can someone help me with the integration intervals for $y_G$? I'm not sure they are correct.
My try is the following: since $z \geq |y|$, it is $z \geq 0$ and so $yz \leq 0 \iff y \leq 0$, from this it follows that $|y|=-y$; so the set $A$ is equivalent to
$$A=\{(x,y,z) \in \mathbb{R}^3 \ | \ 3x^2 \leq y^2+z^2 \leq 3-x^2, y \leq 0, z \geq -y\}$$
Using cylindrical coordinates $x=x$, $y=r \cos \theta$ and $z=r \sin \theta$ the set $A$ becomes
$$B=\{(x,r,\theta) \in \mathbb{R}^3 \ | \ 3x^2 \leq r^2 \leq 3-x^2, \cos \theta \leq 0, \sin \theta \geq -\cos \theta\}$$
It is $\cos \theta \leq 0 \iff \frac{\pi}{2}+2k\pi \leq \theta \leq \frac{3\pi}{2}+2k\pi$, intersected with $0 \leq \theta < 2\pi$ it gives $\frac{\pi}{2} \leq \theta \leq \frac{3\pi}{2}$.
It is
$$\sin \theta \geq - \cos \theta \iff \cos \left(\theta-\frac{\pi}{4}\right) \geq0 \iff \frac{\pi}{2}+2k\pi \leq \theta-\frac{\pi}{4} \leq \frac{3\pi}{2}+2k\pi$$
$$\iff \frac{3\pi}{4}+2k\pi \leq \theta \leq \frac{7\pi}{4}+2k\pi$$
Intersected with $0 \leq \theta < 2\pi$ it is $\frac{3\pi}{4} \leq \theta \leq \frac{7\pi}{4}$.
Intersecting the two conditions on $\theta$, it is $\frac{3\pi}{4} \leq \theta \leq \frac{3\pi}{2}$.
Finally, it is $3x^2 \leq r^2 \leq 3-x^2 \iff \sqrt{3}|x| \leq r \leq \sqrt{3-x^2}$; moreover, this implies that $\sqrt{3}|x| \leq \sqrt{3-x^2} \iff 3x^2 \leq 3-x^2 \iff -\sqrt{3}/2 \leq x \leq \sqrt{3}/2$ and so it is
$$B=\left\{(x,r,\theta) \in \mathbb{R}^3 \ | \ -\frac{\sqrt{3}}{2} \leq x \leq \frac{\sqrt{3}}{2}, \frac{3\pi}{4} \leq \theta \leq \frac{3\pi}{2}, \sqrt{3}|x| \leq r \leq \sqrt{3-x^2} \right\}$$
So it is
$$\int_A y dxdydz=\int_B r^2 \cos \theta dxdrd\theta = \int_{-\sqrt{3}/2}^{\sqrt{3}/2}dx\int_{3\pi/4}^{3\pi/2}d\theta \int_{\sqrt{3}|x|}^{\sqrt{3-x^2}}r^2 \cos \theta dx$$
Is this correct?
Edit: Found my mistake, I've solved incorrectly $\cos \left(\theta-\frac{\pi}{4}\right) \geq 0$; solving it correctly, it leads to $-\frac{\pi}{2} \leq \theta -\frac{\pi}{4} \leq \frac{\pi}{2}$ and so I agree with the answer of @Math Lover. However, my questions remain:
Questions:
i) I was thinking about this: since $z \geq 0$ it is $z=|z|$, hence $A$ seems even in $z$ too and so I would say that $z_G=0$: however I'm suspicious about this reasoning, and I think it doesn't work because for the symmetry it is required that $(x,y,-z) \in A$ and if I specify that $z \geq 0$ and so $z=|z|$, while it's true that $|-z|=|z|$ but from $-z \geq 0$ follows $z \leq 0$ and so the set isn't even in $z$ because the conditions $z \geq 0$ and $z \leq 0$ are different. Hence $(x,y,-z) \notin A$ if I write $A$ equivalently with $z \geq 0$ and $z=|z|$; is this the reason why it is wrong to say that $z_G=0$ with a symmetry argument?
ii) From the fact that $\sqrt{3}|x| \leq r \leq \sqrt{3-x^2}$ I have deduced that $\sqrt{3}|x| \leq \sqrt{3-x^2}$ and found a stricter bound on $x$ (because if I only considered the condition on $\sqrt{3-x^2}$ to be real it would've been $-\sqrt{3} \leq x \leq \sqrt{3}$); when I have these chained inequalities of the form "$E_1 \leq E_2 \leq E_3$", I must always consider the condition $E_1 \leq E_3$ too because, if I don't, there could be values of $E_3$ "that go below $E_1$" and so I'm integrating on a bigger set, overestimating the value of the integral?
Thank you.
| We are talking about solid bound outside cone $3x^2 = y^2 + z^2$ and inside sphere $x^2+y^2+z^2 = 3$ along with constraints, $y z \leq 0$ and $z \geq |y|$.
First of all there is symmetry about $x = 0$, so $\overline x = 0$.
We are using $y = r \cos\theta, z = r \sin\theta$ and as we have $y z \leq 0$ and $z \geq |y|$,
we get $\frac{\pi}{2} \leq \theta \leq \frac{3\pi}{4}$
As the mass density is $1$, the integral to find mass should be,
$\displaystyle m = \int_{-\sqrt3/2}^{\sqrt3/2} \int_{\pi/2}^{3\pi/4} \int_{\sqrt3|x|}^{\sqrt{3-x^2}} r \ dr \ d\theta \ dx$
Now $ \displaystyle \overline y = \frac{1}{m} \iiint y \ dV, \ \overline z = \frac{1}{m} \iiint z \ dV \ $ (as $dm = dV$)
Please note that due to symmetry about $x = 0$, we could have as well considered the solid only for $x \geq 0$. We would get same values of $\overline y$ and $\overline z$.
Last but not least, this can be much simpler in spherical coordinates,
$y = \rho \cos\theta \sin\phi, z = \rho \sin\theta \sin\phi, x = \rho \cos\phi$
$0 \leq \rho \leq \sqrt3, \frac{\pi}{3} \leq \phi \leq \frac{2 \pi}{3}, \frac{\pi}{2} \leq \theta \leq \frac{3 \pi}{4}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4224563",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Finding the sum of the series $1^2 + 2 × 2^2 + 3^2 + 2 × 4^2 + 5^2 + 2 × 6^2 + . . . + 2(n − 1)^2 + n^2 ,$ The following is a question which has been bugging me for quite a while,
Find the sum of the series
$$1^2 + 2 × 2^2 + 3^2 + 2 × 4^2 + 5^2 + 2 × 6^2 + . . . + 2(n − 1)^2 + n^2 ,$$
Where $n$ is odd
I started by denoting this entire series with $S$, from there it is apparent that $S$ infact consists of two lesser series I shall call;
*
*$S_A$ The sum of the squares of odd natural numbers up till $n^2$
*$S_B$ Twice the sum of the squares of even natural numbers up till $(n-1)^2$
I resolved this would be easier to tackle by noting that
$$S_A + \frac{1}{2}S_B = \sum_{r=1}^{n} r^2$$
Which is just the sum of the squares of the first $n$ natural numbers
$$\therefore S_A +\frac{1}{2}S_B=\frac{n}{6}(n+1)(2n+1)$$
Leaving only the value of $\frac{1}{2}S_B$ to be found, this is where I am currently facing difficulty as I am unsure on whether my working is correct;
For the sum of the squares of the first n even natural numbers;
$$2^2 + 4^2 .... (2n)^2=2^2\sum_{r=1}^{n} r^2$$
$$\implies \frac{2}{3}n(n+1)(2n+1) $$
Hence the sum of the first $n-1$ even natural numbers should be
$$ \frac{2}{3}n(n-1)(2n-1)$$
And
$$S= \frac{n}{6}(n+1)(2n+1) + \frac{2}{3}n(n-1)(2n-1)$$
$$\therefore S= \frac{1}{6}n(10n^2 -9n + 5n) $$
However the correct answer is
$$\frac{1}{2}n^2(n+1)$$
Where has my working gone wrong and how would I arrive at the correct answer?
| As implicitely pointed out in the given answer, your mistake is due to the fact that for the sum of the squares of even natural numbers up to $n-1$ we need to consider the following:
$$2^2 + 4^2 .... (n-1)^2=2^2\left(1+2^2+\dots+\left(\frac{n-1}2\right)^2\right) $$
and not
$$2^2 + 4^2 .... (2n)^2=2^2\sum_{r=1}^{n} r^2$$
which is the sum of the squares of even natural numbers up to $2n$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4225909",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
Proving $\sum_k\binom{n}{2k}\frac{1}{2k+1}=\frac{2^n}{n+1}$ using the extraction/absorbtion identity. I want to use the extraction/absorbtion identity to prove the equality
$$
\sum_k\binom{n}{2k}\frac{1}{2k+1}=\frac{2^n}{n+1}
$$
Followed from this identity, we could directly obtain
$$
\binom{n+1}{2k+1}=\frac{n+1}{2k+1}\binom{n}{2k}
$$
Then we have
$$
\sum_k\binom{n}{2k}\frac{1}{2k+1}=\frac{1}{n+1}\sum_k\binom{n+1}{2k+1}
$$
Then how can I prove that $\sum_k\binom{n+1}{2k+1}=2^n$? Thanks!
| Since $\binom{n}{k} = \binom{n-1}{k-1} + \binom{n-1}{k}$, we have
$$\binom{n+1}{1} + \binom{n+1}{3} + \binom{n+1}{5} + \dots = \left[\binom{n}{0} + \binom{n}{1}\right] + \left[\binom{n}{2} + \binom{n}{3}\right] + \cdots$$
which is $2^n$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4227114",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 1,
"answer_id": 0
} |
Deducing an inequality from a Taylor series expansion Consider the inequality
$$
1-\frac{x}{2}-\frac{x^2}{2} \le \sqrt{1-x} < 1-\frac{x}{2}
$$
for $0 < x < 1$. The upper bound can be read off the Taylor expansion for $\sqrt{1-x}$ around $0$,
$$
\sqrt{1-x} = 1 - \frac{x}{2} - \frac{x^2}{8} - \frac{x^3}{16} - \dots
$$
by noting that all the non linear terms are negative. Can the left side inequality be read-off the expansion by a similar reasoning? Please do not try to prove the left side inequality by other means (such as minimizing $\sqrt{1-x} - 1 + \frac{x}{2} + \frac{x^2}{2}$ using derivatives).
| We have a function
$$f(x) =\sqrt{1-x} = 1 - \frac{x}{2} - \frac{x^2}{8} - \frac{x^3}{16} - \frac{5 x^4}{128 }- \frac{7 x^5}{256}- \cdots = \\=1 - a_1 x - a_2 x^2 -a_3 x^3 \cdots $$
The series converges for every $x\in [0,1]$, so we have
$$a_1 + a_2 + \cdots = 1$$
Therefore we can write
$$1 - \frac{x}{2} - \frac{x^2}{2} = 1 - a_1 x - (a_2 + a_3 + \cdots ) x^2< \\ <1 - a_1 x - a_2 x^2 - a_3 x^3 -\cdots = f(x)$$ for $x\in (0,1)$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4228826",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 2,
"answer_id": 0
} |
Prove the inequality $\sum_{cyc}\frac{a^2}{\sqrt{(b+c)(b^3+c^3)}}\geq\frac 32$ where $a,b,c$ are positive reals.
Let $a,b,c$ be positive real numbers, prove that $$\sum_{cyc}\frac{a^2}{\sqrt{(b+c)(b^3+c^3)}}\geq\frac 32.$$
The problem is from an inequality handout. Here is my attempt to solve the problem:
I first rewrote the inequality as $$\sum_{cyc}\frac{a^2}{(b+c)\sqrt{b^2-bc+c^2}}\geq\frac 32.$$
Here I noticed that the inequality has some similarities with Nesbitt's inequality, because that says $$\sum_{cyc}\frac{a}{b+c}\geq\frac 32.$$
But I don't know how to use this to reach the given inequality.
I also used Cauchy-Schwarz and got $$\sum_{cyc}\frac{a^2}{(b+c)\sqrt{b^2-bc+c^2}}\geq\frac{(a+b+c)^2}{\sum_{cyc}(b+c)\sqrt{b^2-bc+c^2}}.$$
But this also makes the problem too complicated that involves too many square roots. I think the square root on the denominator should be removed first to prove the inequality. But I am unable to do that.
So, I need a solution to the problem. Also it would be helpful for me if the answerer provides some motivation for the solution.
| By AM-GM, C-S, Schur and Muirhead we obtain:
$$\sum_{cyc}\frac{a^2}{\sqrt{(b+c)(b^3+c^3)}}=\sum_{cyc}\frac{4a^2}{2\sqrt{(b+c)^2(4b^2-4bc+4c^2)}}\geq\sum_{cyc}\frac{4a^2}{(b+c)^2+4b^2-4bc+4c^2}=$$
$$=\sum_{cyc}\frac{4a^2}{5b^2-2bc+5c^2}=\sum_{cyc}\frac{4a^4}{5a^2b^2-2a^2bc+5a^2c^2}\geq\frac{4(a^2+b^2+c^2)^2}{\sum\limits_{cyc}(10a^2b^2-2a^2bc)}=$$
$$=\frac{\sum\limits_{cyc}(4a^4+8a^2b^2)}{\sum\limits_{cyc}(10a^2b^2-2a^2bc)}\geq\frac{\sum\limits_{cyc}(a^4+3a^3b+3a^3c+8a^2b^2-3a^2bc)}{\sum\limits_{cyc}(10a^2b^2-2a^2bc)}\geq\frac{3}{2}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4230355",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 2,
"answer_id": 1
} |
Subsets and Splits
Fractions in Questions and Answers
The query retrieves a sample of questions and answers containing the LaTeX fraction symbol, which provides basic filtering of mathematical content but limited analytical insight.