Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
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Finding $x^3 + y^3$ when $y = \frac{3-\sqrt5}{3+\sqrt5} , x = \frac{3+\sqrt5}{3-\sqrt5} $ $y = \frac{3-\sqrt5}{3+\sqrt5} , x = \frac{3+\sqrt5}{3-\sqrt5} $
$x^3 + y^3 =?$
my answer =
$(3 + \sqrt5)^3 = 47 + 32\sqrt5$
$(3 - \sqrt5)^3 = 47 - 32\sqrt5$
$x^3 + y^3 = \frac{47 + 32\sqrt5}{47 - 32\sqrt5} + \frac{47 - 32\sqrt5}{47 + 32\sqrt5} = 2*7329/-2911$
why my answer is wrong? please help me
| Numerator:
$$(3-\sqrt5)^6+(3+\sqrt5)^6=(14-6\sqrt5)^3+(14+6\sqrt5)^3=8\,(7-3\sqrt5)^3+8(7+3\sqrt5)^3\\
=16\,(7^3+3\cdot7\cdot9\cdot5)=16\cdot7\cdot184.$$
Denominator:
$$(3-\sqrt5)^3(3+\sqrt5)^3=4^3.$$
Ratio:
$$322.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3350519",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 3
} |
find minimum of maximum of two functions
I need to find the angle $\theta$ so that: $$\max(\cos^2(\theta),1-\cos^2(45-\theta))$$ is minimized.
OK, so I wrote \begin{align*}f(\theta)&=\max(\cos^2(\theta),1-\cos^2(45-\theta))\\
&=\frac{\cos^2(\theta)+1-\cos^2(45-\theta)+|\cos^2(\theta)-1+\cos^2(45-\theta)|}{2}\end{align*}
Then I'm stuck.
| A little nudge in the right direction (hopefully!):
Now, using the required formula (which?), we can simplify: \begin{align} 1 - \cos^2(45^\circ - \theta) & = \sin^2(45^\circ - \theta) \\ & = \frac{1 - \sin 2\theta}{2}\end{align}
We are now required to find: $$\min(\max(\frac{1+\cos 2\theta}{2}, \frac{1-\sin 2\theta}{2}))$$
We can see that in the interval $[0, \pi]$, the solutions to $\cos 2\theta = - \sin 2\theta$ are: $\frac{3\pi}{8}, \frac{7\pi}{8}$. Also, the points $\frac{3\pi}{4}$ and $\pi$ may be of interest (why?)
Can you take it from here?
| {
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"timestamp": "2023-03-29T00:00:00",
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For positive real numbers $a,b,c$ prove that $ a^4 + b^4 + c^4 \ge abc(a+b+c)$ For positive reals $a,b,c$ prove that $$ a^4+b^4+c^4 \ge abc(a+b+c). $$
I tried to pls around trying to reorganize to get AM-GM but i couldn't
Thanks for the help in advance.
| \begin{eqnarray*}
&(x^2-y^2)^2+(y^2-z^2)^2+(z^2-x^2)^2 \\&+2(z^2-xy)^2 +2(x^2-yz)^2+2(y^2-zx)^2\geq 0
\end{eqnarray*}
now divide by $4$ and rearrange.
| {
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"timestamp": "2023-03-29T00:00:00",
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A hard inequality indian olympiad problem If $x,y,z$ are positive real numbers, prove that:
$\left(x+y+z\right)^2\left(yz+xz+xy\right)^2\le 3\left(y^2 + yz + z^2\right)\left(x^2 + xz + z^2\right)\left(x^2 + xy + y^2\right)$.
I have been stuck in it. It is an Indian Olympiad problem. Can you guys help me out, please?
| $\quad3\left(y^2 + yz + z^2\right)\left(x^2 + xz + z^2\right)\left(x^2 + xy + y^2\right)-\left(x+y+z\right)^2\left(yz+xz+xy\right)^2\\=\sum_{sym} \left(2x^4y^2z^0+0.5x^3y^3z^0+0.5x^4y^1z^1-2x^3y^2z^1-x^2y^2z^2\right)\\=2\sum_{sym} \left(x^4y^2z^0-x^3y^2z^1\right)+0.5\sum_{sym} \left(x^3y^3z^0-x^2y^2z^2\right)+0.5\sum_{sym} \left(x^4y^1z^1-x^2y^2z^2\right)$
By Muirhead's Inequality, $\left(4,2,0\right)\succ\left(3,2,1\right),\left(3,3,0\right)\succ\left(2,2,2\right),\left(4,1,1\right)\succ\left(2,2,2\right)$
$\because$ The expression $\ge 0$, equality holds when $x=y=z$
$\therefore 3\left(y^2 + yz + z^2\right)\left(x^2 + xz + z^2\right)\left(x^2 + xy + y^2\right)\ge\left(x+y+z\right)^2\left(yz+xz+xy\right)^2$
($\succ$ denotes majorization)
| {
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Prove that there are only 3 integer solutions to $2^a+3 ^b=5^c$ I have figured out that $a \pmod{2} = b\pmod{2} =c \pmod{2} $ by making a few modulo tables and using that both sides of the equation must be divisible by $5$, and I have found the three solutions, I'm just at a loss as to how to prove that those are the only three. I struggle specifically with how to move from something being divisible by $5$ to something being a power of $5$. I have also tried to use the binomial expansion by noting that $5=2+3$, but I quickly ran out of ideas what to do with it.
Any help would be appreciated!
| Let's try out some values of $a$. The tl, dr: we are forced $(a,b,c)=(1,1,1)$ when $a=1$ and to $(a,b,c)=(2,0,1)$ when $a=2$; then by using Pythagorean-triple-like solutions we fi n.v d these constraints eliminate all possibilities for larger $a$ except $(a,b,c)=(4,2,2)$.
$a=0$:
We can't have a power of $3$ and a power of $5$, both odd, differing by $1$. No solution.
$a=1$:
We have
$2+3^b=5^c$
For this to hold $\bmod 8$ we must have $b$ and $c$ odd. With that constraint, if $b\ge 2$ the above equation requires $b\equiv c\equiv 1\bmod 6$ to hold $\bmod 7$, whereas $c\equiv 5\bmod 6$ would be needed $\bmod 9$. These requirements are contradictory so $b\ge 2$ fails. With $b$ odd we accept $(a,b,c)=(1,1,1)$ as the only solution with $a=1$.
$a=2$:
This gives
$4+3^b=5^c$
For this to hold $\bmod 8$ we need $b$ even and $c$ odd. The latter causes failure $\bmod 3$ unless $b=0$, therefore this case admits only $(a,b,c)=(2,0,1)$.
$a\ge 3, \text{ odd}$
Now it gets meaty. When $a\ge 3$ the power of $2$ becomes fixed $\bmod 8$ and the only way for the equation to hold now is to render $b$ and $c$ both even.
We then have a primitive triple of the form
$2p^2+q^2=r^2, 2p^2=(r+q)(r-q)$
where $p=2^{(a-1)/2}q=3^{b/2}, r=5^{c/2}$. Then $r+q$ and $r-q$ must have a greatest common divisor of $2$, their product must be twice a square, and $q$ and $r$ must both be odd. We are forced to this solution
$p=2mn, q=|m^2-2n^2|, r=m^2+2n^2$
Since $p$ must be a power of $2$, the primitive solution must have $m=1$ or $n=1$; then we observe that $q$ (a power of $3$) must differ from $r$ (a power of $5$) by either $2$ or $4$. From the earlier cases this forces $r=5$, and then we must have $2^a+3^b=r^2=25$ therefore no solution with $a$ odd ($2^a$ being twice a square).
$a\ge 4, \text{ even}$
This goes similarly to the case just rendered above, except the power of $2$ is now a square instead of twice a square. We have the primitive Pythagorean triple
$p^2+q^2=r^2, p^2=(r+q)(r-q)$
with $p=2^{a/2}q=3^{b/2}, r=5^{c/2}$ and the solution
$p=2mn, q=|m^2-n^2|, r=m^2+n^2$
This time $q$ and $r$ must differ by $2$ forcing $q=3,r=5$, leading to $(p,q,r)=(4,3,5)$ and then $(a,b,c)=(4,2,2)$.
| {
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show this inequality $\sum_{cyc}\frac{a^3}{a^2+ab+b^2}\ge\sqrt{\sum a^3}$ let $a,b,c>0$.such $a+b+c=1$ show that
$$\sum_{cyc}\dfrac{a^3}{a^2+ab+b^2}\ge \sqrt{a^3+b^3+c^3}$$
I have show that not stronger inequality:
$$\sum\dfrac{a^3}{a^2+ab+b^2}=\sum_{cyc}\dfrac{a^4}{a^3+a^2b+ab^2}\ge \dfrac{(a^2+b^2+c^2)^2}{\sum (a^3+ab^2+a^2b)}=\dfrac{(a^2+b^2+c^2)^2}{(a+b+c)(a^2+b^2+c^2)}=\dfrac{a^2+b^2+c^2}{a+b+c}=a^2+b^2+c^2$$
But for $(1)$ I can't prove it
| Also, we can use SOS here.
Indeed, we need to prove that:
$$\sum_{cyc}\frac{a^3}{a^2+ab+b^2}-\frac{a^2+b^2+c^2}{a+b+c}\geq\sqrt{\frac{a^3+b^3+c^3}{a+b+c}}-\frac{a^2+b^2+c^2}{a+b+c}$$ or
$$\frac{(ab+ac+bc)\sum\limits_{cyc}(a^4b^2+a^4c^2-a^3b^2c-a^3c^2b)}{(a+b+c)\prod\limits_{cyc}(a^2+ab+b^2)}\geq$$
$$\geq\frac{(a^3+b^3+c^3)(a+b+c)-(a^2+b^2+c^2)^2}{(a+b+c)\left(\sqrt{(a^3+b^3+c^3)(a+b+c)}+a^2+b^2+c^2\right)}$$ or
$$\sum_{cyc}(a-b)^2\left(\frac{c^2(ab+ac+bc)}{(a^2+ac+c^2)(b^2+bc+c^2)}-\frac{ab}{\sqrt{(a^3+b^3+c^3)(a+b+c)}+a^2+b^2+c^2}\right)\geq0.$$
Now, by C-S $$\sqrt{(a^3+b^3+c^3)(a+b+c)}\geq a^2+b^2+c^2.$$
Thus, it's enough to prove that:
$$\sum_{cyc}(a-b)^2\left(\frac{c^2(ab+ac+bc)}{(a^2+ac+c^2)(b^2+bc+c^2)}-\frac{ab}{2(a^2+b^2+c^2)}\right)\geq0.$$
Now, let $a\geq b\geq c$.
Thus, $$\sum_{cyc}(a-b)^2\left(\frac{c^2(ab+ac+bc)}{(a^2+ac+c^2)(b^2+bc+c^2)}-\frac{ab}{2(a^2+b^2+c^2)}\right)\geq$$
$$\geq(a-b)^2\left(\frac{c^2(ab+ac+bc)}{(a^2+ac+c^2)(b^2+bc+c^2)}-\frac{ab}{2(a^2+b^2+c^2)}\right)+$$
$$+(a-c)^2\left(\frac{b^2(ab+ac+bc)}{(a^2+ab+b^2)(b^2+bc+c^2)}-\frac{ac}{2(a^2+b^2+c^2)}\right)\geq$$
$$\geq(a-b)^2\left(\frac{c^2(ab+ac+bc)}{(a^2+ac+c^2)(b^2+bc+c^2)}-\frac{ab}{2(a^2+b^2+c^2)}\right)+$$
$$+(a-b)^2\left(\frac{b^2(ab+ac+bc)}{(a^2+ab+b^2)(b^2+bc+c^2)}-\frac{ac}{2(a^2+b^2+c^2)}\right)\geq$$
$$\geq(a-b)^2\left(\frac{c^2(ab+ac+bc)}{(a^2+ac+c^2)(b^2+bc+c^2)}-\frac{ab+\frac{bc}{2}}{2(a^2+b^2+c^2)}\right)+$$
$$+(a-b)^2\left(\frac{b^2(ab+ac+bc)}{(a^2+ab+b^2)(b^2+bc+c^2)}-\frac{ac+\frac{bc}{2}}{2(a^2+b^2+c^2)}\right)=$$
$$=(a-b)^2(ab+ac+bc)\left(\tfrac{c^2}{(a^2+ac+c^2)(b^2+bc+c^2)}+\tfrac{b^2}{(a^2+ab+b^2)(b^2+bc+c^2)}-\tfrac{1}{2(a^2+b^2+c^2)}\right)\geq0.$$
and we are done!
| {
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"timestamp": "2023-03-29T00:00:00",
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Three real roots of $8x^3 – ax^2 + bx – 1 = 0$ in G.P. The equation $8x^3 – ax^2 + bx – 1 = 0$ has three real roots in G.P. If $λ_1 ≤ a ≤ λ_2$, then find ordered pair $(λ_1, λ_2)$.
My approach $f(x)=8x^3 – ax^2 + bx – 1 $
$f'(x)=24x^2 –2ax + b$
For real root
${(4a^2-96b)}>0$
Roots of $f'(x)$ are
$T=\frac{2a+\sqrt{(4a^2-96b)}}{48}$ &U= $\frac{2a-\sqrt{(4a^2-96b)}}{48}$
Now $f(T).f(U)<0$ then we have three real roots
If above condition is satisfied then
we need to frame another equation
Let the roots are $a',a'r,a'r^2$
$a'+a'r+a'r^2=\frac{a}{8}$
$a'^2(r+r^2+r^3)=\frac{b}{8}$
$a'^3r^3=\frac{1}{8}$
from here I am not able to proceed
| Suppose that the 3 roots are $k/r, k, kr$. Their product $k^3$ is equal to $1/8$. Therefore, $k = 1/2$.
Their sum is $a/8 = k(1/r+1+r)$. Hence $1/r+1+r = a/4$.
Finally, $k^2(1/r+1+r)= b/8$, which implies $(1/r+1+r)=b/2$.
If a solution exists, we must have $a =2b$ and $r^2 + (1-b/2) r+1=0$. If this last equation has real roots, its discriminant has to be positive
$$\Delta = (1-b/2)^2-4 \ge 0$$ which means
$$1-b/2 \in (-\infty , -2] \cup [2, \infty)$$ or
$$b \in (-\infty ,2] \cup [6, \infty)$$ with $a =2b$.
Conversely, if those conditions are met, the roots will form a geometric progression.
| {
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If $~\lim_{x\to\infty}\sqrt{64x^2 + ax + 7} - 8x + b = \frac32~ $, find $~ a+b=~?$ If $$~\lim_{x\to\infty}\sqrt{64x^2 + ax + 7} - 8x + b = \frac32~ ,$$ find $~ a+b=~?$
I get $a-16b = 24$,
But how to get $a+b$?
a,b is positive integer
| $$
\sqrt{64x^2 + ax + 7} =\sqrt{(8x)^2 + ax + 7} =\sqrt{(8x)^2 + 2\times(?)\times(8x)+ 7} \\
$$note that $$2\times(?)\times(8x)=ax \to ?=\frac{a}{16}$$so
$$\sqrt{(8x)^2 + 2(\frac{a}{16})(8x)+ 7}=\sqrt{(8x+\frac{a}{16})^2 -(\frac{a}{16})^2+ 7}\\x\to \infty \sqrt{(8x+\frac{a}{16})^2 -(\frac{a}{16})^2+ 7}\sim |8x+\frac{a}{16}| $$
so
$$lim_{x\to\infty}\sqrt{64x^2 + ax + 7} - 8x + b = \frac{3}{2}\\
lim_{x\to\infty}|8x+\frac{a}{16}| - 8x + b = \frac{3}{2}$$ now can you take over?
a,b are a positive integer so
$$\frac{a}{16}+b=\frac{3}{2}\\b=0\to \frac{a}{16}+0=\frac{3}{2} \to a=24 \to a+b=24\\b=1 \to \frac{a}{16}+1=\frac{3}{2} \to a=8 \to a+b=1+8=9\\b=2,3,... impossible $$
| {
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"timestamp": "2023-03-29T00:00:00",
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Can we represents $k$ in partition of consecutive positive integer? Let $a,b$ are non-negative integers.
Let define
$$S(a)=1+2+3+...+a=\sum_{i=0}^{a} i$$
and $$S(b)=1+2+3+...+b=\sum_{i=0}^{b} i$$
Question
Show that each $k\in \mathbb{N}$ and $k \ne 2^t$ $ \forall t\in \mathbb{N} $ can be represent as
$$ k = S(a)-S(b)$$ where $a \ne b+1$
My attempted
We can write $k$ as
$$ k = S(a)-S(b)$$
$$= n+(n+1)+...+(n+u)= \sum_{i=0}^{u}n+i$$
$$=\frac{(u+1)(2n+u)}{2}$$
Where $u\ge 1$
For $k>1$ and $k$ is odd number then $k$ can be represent as
$k= 2r+1 = r+(r+1)= S(r+1)-S(r-1)$
Now proof for, if $k= 2^t$ then $ k\ne S(a)-S(b)$
Proof
Let suppose $$k = n+(n+1)+...+(n+u)$$
$$=\frac{(u+1)(2n+u)}{2}= 2^t$$
So $$ (u+1)(2n+u)= 2^{t+1}$$
Case1
if $u$ is odd then $u+1=$ even and $2n+u =$ odd so even×odd $\ne 2^{t+1}$ because $ 2^{t+1}$ content only even multiples except $1$.
Case2
if $u$ is even then $u+1=$ odd and $2n+u =$ even so odd×even $\ne 2^{t+1}$ similarly as case 1
So both cases show complete proof for $k \ne 2^t$
Example
$6=1+2+3, 10=1+2+3+4,$
$ 12=3+4+5, 14=2+3+4+5, $
$18=5+6+7, 20=2+3+4+5+6$
$22=6+7+8, 24=7+8+9,$
$ 26=5+6+7+8,...$
You can check advanced similar problem here
| You have shown above that it is impossible for any number of the form $2^t$ to be expressed as $S(a)-S(b)$. You have also shown that it is possible for all odd numbers. All you have to do is to extend the same idea for even values.
Case 1: The number is of the form $2^{t-1}m$ where $m \geqslant 2^t+1$ is odd
$$\frac{2^t \cdot (2^t+1)}{2}=2^{t-1}(2^t+1)=1+2+\cdots+2^t$$
$$2^{t-1}({2^t+1}+2k)=(1+k)+(2+k)+\cdots+(2^t+k)=S(2^t+k)-S(k)$$
for all non-negative integers $k$. We set $k=m-2^t-1$
Case 2: The number is of the form $2^{t-1}m$ where $1<m \leqslant 2^t-1$ is odd
$$\frac{m(m+1)}{2}=1+2+\cdots+m$$
$$m(\frac{m+1}{2}+k)=(1+k)+(2+k)+\cdots+(m+k)=S(m+k)-S(k)$$
for all non-negative integers $k$. We set $k=2^t-1-m$
The reason why $m \neq 1$ is because we must have $a \neq b+1$ but at $m=1$, we have $a=k+1$ and $b=k$. Thus, using the two cases, any even number which is not a power of $2$ can be expressed as required.
| {
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Limit of $f(x, y)$ at $(0, 0)$ I need to show
$$\lim_{(x, y) \rightarrow (0,0)} \frac{xy(x^2-y^2)}{(x^2+y^2)^{3/2}} = 0
$$
Not really sure how to go about this without using the epsilon delta definition, which I would prefer not to. Any sort of help is appreciated.
Edit: I do have the inequality:
$$\frac{xy(x^2-y^2)}{(x^2+y^2)^{3/2}} \le \frac{xy(x^2+y^2)}{(x^2+y^2)^{3/2}} = \frac{xy}{(x^2+y^2)} $$
| Use polar coordinates: by letting $x=r\cos(\theta)$ and $y=r\sin(\theta)$, we have that
$$\frac{xy(x^2-y^2)}{(x^2+y^2)^{3/2}}= \frac{r^4\sin(2\theta)\cos(2\theta)}{2r^3}=\frac{r\sin(4\theta)}{4}.$$
Therefore
$$\left|\frac{xy(x^2-y^2)}{(x^2+y^2)^{3/2}}\right|\leq \frac{r}{4}$$
and the right-hand side goes to $0$ as $r\to 0$.
P.S. Your equality is too "weak" for the job: by taking $y=x$ the right-hand side is $\frac{1}{2}$.
| {
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a) Find a base for $W_1 \cap W_2$ and b) Find a base for $W_1 + W_2$ Let
$$W_1 = \left\{ \begin{pmatrix}
a & b & c \\
-b & a & b \\
c & b & a \\
\end{pmatrix} :\, a, b, c \in \Bbb R \right\}$$
$$W_2 = \left\{
\begin{pmatrix}
a & 0 & 0 \\
b & 0 & c \\
c & b & a \\
\end{pmatrix} :\, a, b, c \in \Bbb R \right\}$$
a) Find a basis for $W_1 \cap W_2$
b) Find a basis for $W_1 + W_2$
For a) I`m not sure if the set for $W_1 \cap W_2 = $ $ \{
\begin{pmatrix}
0 & 0 & 0 \\
0 & 0 & 0 \\
0 & 0 & 0 \\
\end{pmatrix}\}
$
so the base is only the zero matrix.
For b) I belive the set is: $W_1 + W_2 = $ $ \{
\begin{pmatrix}
2a & b & c \\
0 & a & b+c \\
2c & 2b & 2a \\
\end{pmatrix}
$:$ a, b, c \in R\}$
But don`t know how to give an
appropriate base for this set.
| For b), rewrite what you obtained as
$$a \begin{pmatrix}
2& 0& 0\\
0 & 1 & 0 \\
0 & 0& 2 \\
\end{pmatrix}+b \begin{pmatrix}
0 & 1 & 0\\
0 & 0 & 1 \\
0 & 2 & 0 \\
\end{pmatrix}+c \begin{pmatrix}
0 & 0 &1\\
0 & 0 & 1 \\
2 & 0 & 0 \\
\end{pmatrix}$$
Can you say whether the three above matrices are linearly independent?
| {
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How to solve $(x^2 - 11x + 29)^{(6x^2 + x - 2)}=1$? This question comes from PURE MATHEMATICS 1 for As and A levels. This question is part of exercise 1 and it has 5 Answers : $1/2, - 2/3, 4, 6 $ and $7$ . The first 2 values $1/2$ and $-
2/3 $ I am able to find but the rest I can't. I get this first 2 values by make 1 to $(x^2 - 11x + 29)^0$ and this solving by transposition as base of power are same so they get cancelled and make it $(6x^2 + x - 2)=0.$
| It's simple
This equation is true when either $x^2 -11x+29=1$ or $6x^2+x-2 =0$ or $x^2-11x+29=-1 \ \ and \ \ 6x^2+x-2= even$ because $1^k=1$ and $n^0=1$ and $(-1)^{even}=0$
Solve the equations
You have solved second equation so I'll leave that
For the first you will get $$x^2 -11x+28=0$$
$$(x-7)(x-4)=0$$
Solution is$ x=4,7 $as you desired
For third equation
$x^2-11x+30=0\ \ and \ \ 6x^2+x-2= even$
We get
$$x=5,6 \ \ and \ \ 6x^2+x-2=even$$
Plug in each value
For $x=5 6x^2+x-2=153$ which is odd
But for $x=6 6x^2+x-2= 220$ which is even
So x=6 is also a solution
This method is consistent with the As and A Lvl maths
| {
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"url": "https://math.stackexchange.com/questions/3368462",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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how to solve $\lim_{n\to\infty}{\left(\sum_{k=1}^{n}{\frac{1}{\sqrt{n^2+k}}}\right)^{n}}$? $\displaystyle\left(\sum_{k=1}^{n} \frac{1}{\sqrt{n^{2}+1}}\right)^{n}\ge\left(\sum_{k=1}^{n} \frac{1}{\sqrt{n^{2}+k}}\right)^{n}\ge\left(\sum_{k=1}^{n} \frac{1}{\sqrt{n^{2}+n}}\right)^{n}$
left=$\displaystyle\lim_{n\to\infty}{\left(\sum_{k=1}^{n}{\frac{1}{\sqrt{n^2+1}}}\right)^{n}}=e^{\displaystyle n \ln{\frac{n}{\sqrt{n^2+1}}} }=e^{0}=1$
right=$\displaystyle\lim_{n\to\infty}{\left(\sum_{k=1}^{n}{\frac{1}{\sqrt{n^2+n}}}\right)^{n}}=e^{\displaystyle n \ln{\frac{n}{\sqrt{n^2+n}}} }=e^{-\frac{1}{2}}$
left $\ne$ right ,what to do next?
$\displaystyle\lim_{n\to\infty}{\left(\sum_{k=1}^{n}{\frac{1}{\sqrt{n^2+k}}}\right)^{n}}=\lim _{n \rightarrow \infty} e^{\displaystyle n \ln \sum_{k=1}^{n} \frac{1}{\sqrt{n^{2}+k}}(1)}$
$(1)=\displaystyle \lim _{n \rightarrow \infty} n\left(\ln \frac{1}{n} \sum_{k=1}^{n} \frac{1}{\sqrt{1+k/n^{2}}}\right)$$=\lim _{n \rightarrow \infty} n\left(\ln \frac{1}{n} \sum_{k=1}^{n} \frac{1}{\sqrt{1+k/n \cdot 1/n}}\right)$$=\lim _{n \rightarrow \infty} n \ln \int_{0}^{1} \frac{1}{\sqrt{1+x/n}} d x$$=\lim _{n \rightarrow \infty} n \ln \int_{0}^{1} \frac{nd(x/n+1)}{\sqrt{1+x/n}}$$= \lim_{n\to\infty}{n\ln{n \cdot2 \left.\sqrt{1+\frac{x}{n} }\right|_{0}^{1}}}$$= \lim_{n\to\infty}{n\ln{n \cdot2 (\sqrt{1+\frac{1}{n}}-1)}}$$=\lim_{n\to\infty}{n\ln{n \cdot2 (\frac{1}{2n} -\frac{1}{2n^2} +o(\frac{1}{n^2}))}}$$=\lim_{n\to\infty}{n\ln{n \cdot2 (\frac{1}{2n} +\left(\frac{1}{2!}\cdot \frac{1}{2} \cdot \left(\frac{1}{2}-1\right) \right)\frac{1}{n^2} +o(\frac{1}{n^2}))}}=\lim_{n\to\infty}{n \ln{\left(1-\frac{1}{4n}\right)}}=-\frac{1}{4} $
so that $\displaystyle\lim_{n\to\infty}{\left(\sum_{k=1}^{n}{\frac{1}{\sqrt{n^2+k}}}\right)^{n}}=\lim _{n \rightarrow \infty} e^{(1)}=e^{-\frac{1}{4}}$
this solution is right.
| We can use the sandwich theorem twice for solving the limit
$$\text { Observe that }
\frac{n}{\sqrt{n^2 +n}}\le\alpha_n=\sum_{k=1}^{n}\frac{1}{\sqrt{n^2 +k}}\le\frac{n}{\sqrt{n^2+1}}, \text{ So, } \lim_{n\to\infty}\alpha_n=1.\space
\text { Proceeding as an usual case of } 1^{\infty}, $$
$$ \alpha=\lim_{n\to\infty}\left(\sum_{k=1}^{n}\frac{1}{\sqrt{n^2 +k}}\right) ^n=e^{\lim_{n\to\infty} n(\alpha_n -1)}=e^{\lim_{n\to\infty}\sum_{k=1}^{n}\left(\frac{n}{\sqrt{n^2+k}} -1\right)}=e^{\lim_{n\to\infty}\sum_{k=1}^{n}\left(\frac{-k}{(n+\sqrt{n^2+k})\sqrt{n^2+k}} \right)}$$
$$\text{ Now, use Sandwich theorem again and obtain that } \alpha=\frac{1}{\sqrt[4]{e}}$$
| {
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Factoring $3x^2+4x-4=0$ using the quadratic formula The correct answer is $(3x-2)(x+2)$ but I am getting $(x-\frac{2}{3})(x+2)$, why?
My calculations:
$x = \frac{-4\pm\sqrt{16-(-48)}}{6} = \frac{-4\pm8}{6}$, which gives the factors $(x-\frac{4}{6})(x+\frac{12}{6}) = (x-\frac{2}{3})(x+2)$.
| You are confusing factoring a quadratic with solving a quadratic.
The quadratic $ax^2 + bx + c=0$ will have the same solutions as $x^2 + \frac bax + ca = 0$ (in fact, that was the very first step in developing the quadratic formula) and will have the same as $dx^2 + \frac {db}ax + \frac {dc}a-0$ or any $akx^2 + bkx + ck =0$.
And you will get the same two solutions $r_1 = \frac {-b+\sqrt{b^2 -4ac}}{2a}$ and $r_2 = \frac {-b+\sqrt{b^2-4ac}}{2a}$ for all of those polynomials.
And $x^2 + \frac ba x + \frac ca$ will factor as $(x - r_1)(x - r_2)$. (If we expand $(x-r_1)(x-r_2)$ we get $x^2 - (r_1+r_2)x + r_1r_2$ and it's not hard to see that $-(r_1 + r_2)= \frac ba$ and $r_1r2 = \frac ca$.)
But that means $ax^2+bx + c = a(x^2 + \frac bax +\frac ca) = a(x-r_1)(x-r_2)$ will factor as $a(x-r_1)(x-r_2)$. In general $(x-r_1)(x-r_2)$ will expand to a polynomial with leading coefficient $1$. And $k(x-r_1)(x-r_2)$ will expand to a polynomial with leading coefficient $k$.
And we want leading coefficient $a=3$. So we need $3x^2 + 4x-4 = 3(x-r_1)(x-r_2)$
Now you figured out $r_1$ and $r_2$ perfectly correctly.
So we have $3x^2 +4x-4 = 3(x-\frac 23)(x+2) = (3x -2)(x+2)$.
....
you know... maybe this will be clearer if we complete the square:
$3x^2 + 4x - 4 =$
$3(x^2 + \frac 43x +\frac {-4}3)=$
$3([x^2 + 2\frac {4}{2*3}x + \frac {4^2}{4*3^2}] - \frac {4^2}{4*3^2}+\frac {-4}3) =$
$3([x+\frac 4{2*3}]^2 - \frac {4^2 -4*(-4)*3}{4*3^2})=$
$3((x+\frac 4{2*3} + \frac {\sqrt{4^2 - 4*(-4)*3}}{2*3})(x+\frac 4{2*3} - \frac {\sqrt{4^2 - 4*(-4)*3}}{2*3}) =$
$3(x- \frac {-4-\sqrt{4^2 -4*(-4)*3}}{2*3})(x-\frac{-4+\sqrt{4^2 -4*(-4)*3}}{2*3})=$.
$3(x-\frac 23)(x+2)= (3x-2)(x+2)$.
Notice how after the first line and until the very last line, the $3$ just hung around for the ride doing absolutely nothing? That's ... kind of the point. The leading coefficient $a$ doesn't really do much once you set up to solve (after all, if you set to $0$, which we didn't do this time, we can just divide it out) but it sticks around if we never actually set anything to zero (which in this case we didn't).
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Geometry Ratio in right triangle
In ABC, $\angle ACB=90$. Point O is chosen inside ABC such that [AOB]=[BOC]=[COA]. If OA=23 and OB=11, find $OC^2$.
Using the areas, we can find OD:OE:OF, where D, E, F are the feet of the altitudes from O to BC, AC, and AB, respectively. I have no idea how to proceed
| With $D$ and $E$ as stated, note that $OD=\frac{AC}3$, $OE=\frac{BC}3$, then $AE=\frac{2AC}3$, $BD=\frac{2BC}3$.
Applying Pythagorean theorem to triangles AEO and ODB gives $4AC^2+BC^2=9\times 23^2$ and $AC^2+4BC^2=9\times 11^2$
Adding these gives $5(AC^2+BC^2)=9\times 650$
Replacing $AC^2+BC^2$ with $AB^2$ and rearranging gives $\frac{AB^2}9=130$
Also note that $OC=\frac{AB}3$, therefore:
$$OC^2=\frac{AB^2}9=130$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Nice olympiad inequality :$\frac{xy^2}{4y^3+3}+\frac{yz^2}{4z^3+3}+\frac{zx^2}{4x^3+3}\leq \frac{3}{7}$ I have this to solve :
Let $x,y,z>0$ such that $x+y+z=3$ then we have :
$$\frac{xy^2}{4y^3+3}+\frac{yz^2}{4z^3+3}+\frac{zx^2}{4x^3+3}\leq \frac{3}{7}$$
I try to use Jensen's inequality but the function $f(x)=\frac{x^2}{4x^3+3}$ is neither concave or convex on the interval $[0,3]$
I can't use Karamata's inequality too .
Maybe brute force is the only way to solve it .
I try also to use the derivative but it becomes a little bit difficult .
In fact my idea was to use rearrangment inequality we have :
$$\frac{xy^2}{4y^3+3}+\frac{yz^2}{4z^3+3}+\frac{zx^2}{4x^3+3}\leq \frac{x^3}{4x^3+3}+\frac{y^3}{4y^3+3}+\frac{z^3}{4z^3+3}$$
An use the inequality of Jensen's on $[0.8,1.2]$ with $f(x)=\frac{x^3}{4x^3+3}$
So it's a partial answer .
My question is how to complete my answer or can you provide an other answer ?
Thanks a lot for sharing your knowledge and your time .
| Note that (tangent line trick)
$$\frac{5+2y}{49} - \frac{y^2}{4y^3+3} = \frac{(8y^2+36y+15)(y-1)^2}{49(4y^3+3)}.$$
Thus, we have
$$\frac{y^2}{4y^3+3} \le \frac{5+2y}{49}, \quad \forall y \ge 0.$$
Thus, we have
$$\sum_{\mathrm{cyc}} \frac{xy^2}{4y^3+3} \le \sum_{\mathrm{cyc}} \frac{x(5+2y)}{49}
= \frac{5(x+y+z) + 2(xy+yz+zx)}{49} \le \frac{3}{7}$$
where we have used the fact that $xy+yz+zx \le \frac{(x+y+z)^2}{3}$. We are done.
| {
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"timestamp": "2023-03-29T00:00:00",
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How to solve the following ordinary differential equation? Consider the ODE $$y' + x = \sqrt{x^2+y}.$$
I have no idea how to solve this equation. I have tried several ways but neither of them worked. Can anybody help me? Many thanks.
I have tried substituting $u=\sqrt{x^2+y}$, as well as squaring both sides of the ODE.
| This is a heuristic derivation of the complete solution showing a possible motivation chain. Notice that Jack D'Aurizio has given a special solution before.
How can we simplify the ODE?
$$y' + x = \sqrt{x^2+y}\tag{1}$$
The first thing to do would be to get rid the square root.
Hence the set $v^2 = x^2+y$ so that the r.h.s becomes just $v$.
But this means that $y=v^2-x^2$ and $y'(x) = 2 v v'(x) - 2 x$ so that the the ODE becomes
$$2 v v' - x = v \tag{2}$$
Here we would like to get rid of $x$, hopefully by cancelling a factor $x$. To this end we try the natural substitution $v\to u x$ which gives
$$ 2 u x (u x)' -x = u x \implies 2 u (u x)' -1 = u \implies 2 u^2 +2 x u u' = u+1 \\\implies x u' = -u +\frac{1}{2}+\frac{1}{2u} $$
Now we can separate variables:
$$ x \frac{du}{dx} = -u +\frac{1}{2}+\frac{1}{2u} \implies \frac{du}{-u +\frac{1}{2}+\frac{1}{2u}} =\frac{dx}{x} $$
Integrating is elementary on both sides. Hence we find the following relation betwee $x$ and $u$
$$2 \left(-\frac{1}{3} \log (1-u)-\frac{1}{6} \log (2 u+1)\right)=\log (x) + const\tag{3}$$
Multiplying by $-3$ and simplifying the $\log$ gives the cubic equation for $u$
$$(1-u)^2(1+2u) = \frac{c^3}{x^3} \tag{4}$$
Here $c$ is a constant of integration.
Now $(4)$ can be solved by standard methods giving three solution for $u(x)$. We skip the explicit expression here for the time being as Jack has already given a special solution.
Finally, we have found that the ODE $(1)$ has three solutions given by
$$y(x) = x^2(u(x)^2-1)$$
explicitly
$$y(x,c) = \left\{x^2 \left(\frac{1}{4} \left(-\frac{g}{c x}-\frac{c x}{g}+1\right)^2-1\right),\\x^2 \left(\frac{1}{4} \left(\frac{g}{c x}+\frac{c x}{g}+1\right)^2-1\right),\\x^2 \left(-1+\frac{1}{16} \left(\frac{2 (-1)^{2/3} g}{c x}+\frac{c \left(x-i \sqrt{3} x\right)}{g}+2\right)^2\right)\right\}
\tag{5a}$$
here
$$g = \sqrt[3]{c^3 x^3+2 \sqrt{1-c^3 x^3}-2}\tag{5b}$$
Here is a plot of the three solutions with three different parameters $c$
Notice the interesting cusp structure from which two of the three solutions emerge. The position of the cusp is given by $(x,y) = (\frac{1}{c} ,-\frac{1}{c^2})$
Discussion
*
*CAS
Mathematica solves the ODE with the command DSolve immediately, but it returns 6 solutions instead of 3. Three of these do not solve the ODE after explicit insertion. This confused my at first so I did not submit that solution. The reason is that Mathematica takes a an additional negative value of the square root into account, i.e. it simultaneously solves the ODE $y'+x = - \sqrt{x^2+y}$. The latter is obtained from $(1)$ by the substitution $x\to -x$.
*Modifications
We could simplify the ODE $(1)$ by subsequently modifying terms, as for instance
$$y'(x)=\sqrt{y(x)}\tag{m1}$$
$$y'(x)=\sqrt{y(x)+x}\tag{m2}$$
$$y'(x)=\sqrt{y(x)+x^2}\tag{m3}$$
$$y'(x)+x=\sqrt{y(x)}\tag{m4}$$
$$y'(x)+x=\sqrt{y(x)+x}\tag{m5}$$
All these ODEs can be transformed to a separable ODE which even can be integrated but the resulting implicit equations cannot be solved explicitly.
For example, the integral of $(m4)$ is
$$\frac{1}{2} \log \left(\frac{2 y(x)}{x^2}+\sqrt{\frac{y(x)}{x^2}}+1\right)-\frac{1}{\sqrt{7}}\text{arctan}\left(\frac{1+4 \sqrt{\frac{y(x)}{x^2}}}{\sqrt{7}}\right)=\log \left(\frac{c}{x}\right)\tag{i4}$$
| {
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How can I prove that $4^{n} + 5$ is divisible by $3$. I have trying to prove that $4^{n} + 5$.
I've already proved the base case, so I'm working on the inductive step.
I've done the following:
$4^{n} + 5$
$4^{n+1} + 5$
$4*4^{n} + 5$
But I am unsure where to go from here to prove that it is divisible by 3 since I am unsure how to get a $3$ or multiple of $3$ from this.
| You can prove that using induction
We suppose that $4^n + 5 \equiv 3$ is true
Induction base: $n = 1 \Longrightarrow 4^1 + 5 = 9 \equiv 3$ - true.
Induction transition: let $k = n+1 \Longrightarrow 4^k + 5 = 4^{n+1}+5 = 4 \cdot 4^n + 5 = 3\cdot4^n + 4^n + 5$
$3 \cdot 4^n \equiv 3, \ \ 4^n + 5 \equiv 3$
Hence $4^{n+1} + 5\equiv3 \Longrightarrow \ 4^n + 5 \equiv 3$
| {
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"timestamp": "2023-03-29T00:00:00",
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What is the principal root of $\sqrt{-i}$? Where is the mistake in this solution?
$$\sqrt{-i}=\sqrt {\cos \frac{3 \pi}2+i\sin \frac{3 \pi}2}=\cos \frac{3 \pi}4+i\sin \frac{3 \pi}4=-\frac{\sqrt 2}2+i \frac{\sqrt 2}2$$
WA gives me different result:
$$\frac{\sqrt 2}2-i \frac{\sqrt 2}2$$
Why the principal root must be equal to $\frac{\sqrt 2}2-i \frac{\sqrt 2}2$?
I used Moivre's formula. But I dont know. How can I must choose value of $k$ in Moivre's formula? I chose $k=0$.
$$z=r\left(\cos x+i\sin x\right)$$
$$z^{\frac 1n}=r^\frac1n \left( \cos \frac{x+2\pi k}{n} + i\sin \frac{x+2\pi k}{n}\right)$$
| Solve the quadratic equation
$$x^2=-i=e^{-i\pi/2}\implies x=e^{-i\pi/4+ik\pi},\; k=0,1$$
| {
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Prove the following triangle related inequality
Prove that $\dfrac32\leq \dfrac{a}{b+c}+\dfrac{b}{a+c}+\dfrac{c}{a+b} < 2.$ where $a$ ,$b$ ,$c$ are sides of Triangle
I tried using sine rule which makes up the the expression $\sum \dfrac{\sin{A}}{\sin{B}+\sin{C}}$. Now i dont how to find the range of $\sum \dfrac{\sin{A}}{\sin{B}+\sin{C}}$. Any ideas(hint) or solution would be really appreciated.
| Making $b = x a$ and $c=y a$ the conditions read
$$
\frac 32\le\frac{1}{x+y}+\frac{x}{1+y}+\frac{y}{1+x}\le 2 \ \ \text{s. t.}\ \ x + y\ge 1
$$
so making the lagrangian
$$
L(x,y,\lambda) = \frac{1}{x+y}+\frac{x}{1+y}+\frac{y}{1+x}+\lambda(x+y-1)
$$
We have two stationary points. One at $x^*=1, y^* = 1$ corresponding to a minimum $(\frac 32)$ and $x^* = \frac 12, y^* = \frac 12$ corresponding to a maximum $\frac 53$ so we have
$$
\frac 32\le \frac 32\le \frac 53\le 2
$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Calculate the value of $\frac{1}{2}+\frac{1}{4}+\frac{2}{8}+\frac{3}{16}+....+\frac{F_n}{2^n}+....$ The Fibonacci sequence starts with 1, 1, 2, 3, 5, 8, 13, ... .(Start from the 3rd term,
each term is the sum of the two previous terms). Let $F_n$ be the $n$th term of this sequence. $S$ is defined as $S=\frac{1}{2}+\frac{1}{4}+\frac{2}{8}+\frac{3}{16}+....+\frac{F_n}{2^n}+....$
Calculate the value of $S$
I have no idea how to solve this, hints aswell as solutions would be appreciated
Taken from the 2013 AITMO
| Hint: $$F_n{={\frac {\Phi ^{n}-\Psi ^{n}}{\sqrt {5}}}={\frac {1}{\sqrt {5}}}\left(\left({\frac {1+{\sqrt {5}}}{2}}\right)^{n}-\left({\frac {1-{\sqrt {5}}}{2}}\right)^{n}\right)}$$
$$S = \sum_{n=1}^{\infty}\dfrac{F_n}{2^n}=\dfrac{1}{\sqrt{5}}\sum_{n=1}^{\infty}\left[\left(\dfrac{\Phi}{2}\right)^n- \left(\dfrac{\Psi}{2}\right)^n\right]$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3379066",
"timestamp": "2023-03-29T00:00:00",
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Ramanujan's Nested Radical By noting Ramanujan's Nested Radical, we have
$3 = \sqrt{1+2\sqrt{1+3\sqrt{1+4\sqrt{1+\cdots}}}}$
On the other hand, we can manipulate the number $4$ by applying the similar principle. Here we have
$\begin{aligned} 4 & = \sqrt{16} \\ & = \sqrt{1+15} \\ & = \sqrt{1+2 \cdot \dfrac{15}{2}} \\ & = \sqrt{1+2\sqrt{\dfrac{225}{4}}} \\ & = \sqrt{1+2\sqrt{1+\dfrac{221}{4}}} \\ & = \sqrt{1+2\sqrt{1+3 \cdot \dfrac{221}{12}}} \\ & = \sqrt{1+2\sqrt{1+3\sqrt{\dfrac{44841}{144}}}} \\ & \vdots \\ & = \sqrt{1+2\sqrt{1+3\sqrt{1+4\sqrt{1+\cdots}}}} \end{aligned}$
How can it be? Something contradicts?
| The generating function is
$$
f(x) = \sqrt{1+(x+1)f(x+1)}\Rightarrow f^2(x) = 1+(x+1)f(x+1)
$$
so making $f(x) = a x + b$ we get $a = 1, b= 2$ or
$$
f(x) = x + 2
$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Substitution integral with $t=x^3$
$$
\int \frac{3}{t(3+\sqrt[3]{t})} \mathrm{d} t . \text { Substitution } t=x^{3}
$$
$$
\int \frac{3}{x^3(3+\sqrt[3]{x^3})}3x^2
$$
$$
\int \frac{9}{x(3+{x})}
$$
$$
9\int \frac{1}{3x+{x^2}}
$$
Now i'm stuck
my calculation so far
| To finish your calculation:
$$\int\frac{9}{x(3+x)}$$
Take partial fractions
$$=3\int \frac{1}{x+3}dx+2\int\frac{1}{x}dx$$
Let $u=x+3,du=dx$ have
$$=3\int\frac{1}{u}du+2\int\frac{1}{x}dx$$
Then exists $C\in\mathbb{R},s.t.$
$$=3\ln(u)+2\ln(x)+C$$
Substitute $u$ back, we have
$$=3\ln(x+3)+2\ln(x)+C$$
| {
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If $\alpha$, $\beta$, $\gamma$ and $\delta$ be the roots of the polynomial $x^4+px^3+qx^2+rx+s$, prove the following.
If $\alpha$, $\beta$, $\gamma$ and $\delta$ be the roots of the polynomial $x^4+px^3+qx^2+rx+s$, show that $$(1+\alpha^2)(1+\beta^2)(1+\gamma^2)(1+\delta^2)=(1-q+s)^2+(p-r)^2.$$
I have tried putting
\begin{align}P(1)&=(\alpha-1)(\beta-1)(\gamma-1)(\delta-1)=1+p+q+r+s\\P(-1)&=(\alpha+1)(\beta+1)(\gamma+1)(\delta+1)=1-p+q-r+s.\end{align}
Then \begin{align}P(1)P(-1)&=(\alpha^2-1)(\beta^2-1)(\gamma^2-1)(\delta^2-1)\\&=(1+p+q+r+s)(1-p+q-r+s)=(1+q+s)^2-(p+r)^2\end{align}
Somehow it does not match the statement given.
| Your approach ends up proving $(α^2 - 1)(β^2 - 1)(γ^2 - 1)(δ^2 - 1) = (1 - q + s)^2 - (p - r)^2$. This means we will have to do something with that minus sign and to do that we can use the identity $(x + ιy)(x - ιy) = x^2 + y^2$. Putting $P(ι)$, and $P(-ι)$ you can easily solve the problem with the same approach.
| {
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Find a such that $ax^{17}+bx^{16}+1$ is divisible by $x^2-x-1$.
Find $a$ such that $ax^{17}+bx^{16}+1$ is divisible by $x^2-x-1$.
I tried taking the roots of the polynomial which are $\frac{1±\sqrt{5}}{2}$
And I got the equation $a(\frac{1±\sqrt{5}}{2})^{17}+b(\frac{1±\sqrt{5}}{2})^{16}+1=0$
Now I don't know what to do next.
Any help would be appreciated.
| Hint. If $x=\frac{1±\sqrt{5}}{2}$ then $x^2=x+1$ and
\begin{align}
ax^{17}+bx^{16}+1 &=(ax+b)(x+1)^8+1 \\
&=(ax+b)(x^2+2x+1)^4+1\\
&=(ax+b)(3x+2)^4+1\\
&=(ax+b)(21x+13)^2+1.
\end{align}
Can you take it from here? Note that 3,2, 21,13 are all Fibonacci numbers.
| {
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"timestamp": "2023-03-29T00:00:00",
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Can we simply further the expression? We have the set $\left \{z\in \mathbb{C}\mid z\neq 1, \ \left |\frac{z}{z-1}\right |\leq 1\right \}$
Let $z=x+yi$.
We have that \begin{align*}\frac{z}{z-1}&=\frac{x+yi}{(x-1)+yi}=\frac{\left (x+yi\right )\left ((x-1)-yi\right )}{\left ((x-1)+yi\right )\left ((x-1)-yi\right )} \\ & =\frac{x(x-1)+y^2-yi}{(x-1)^2+y^2} =\frac{x(x-1)+y^2}{(x-1)^2+y^2}+\frac{-y}{(x-1)^2+y^2}i\end{align*}
Then we get \begin{align*}\left |\frac{z}{z-1}\right |&=\sqrt{\left (\frac{x(x-1)+y^2}{(x-1)^2+y^2}\right )^2+\left (\frac{-y}{(x-1)^2+y^2}\right )^2} \\ & =\sqrt{\frac{x^2(x-1)^2+2x(x-1)y^2+y^4}{[(x-1)^2+y^2]^2}+\frac{y^2}{[(x-1)^2+y^2]^2}} \\ & = \sqrt{\frac{x^2(x-1)^2+2x(x-1)y^2+y^4+y^2}{[(x-1)^2+y^2]^2}} \end{align*}
can we simplify that further?
Is not, we have to solve the inequality from that point. So we have from $\left |\frac{z}{z-1}\right |\leq 1$ the following:
\begin{align*}&\sqrt{\frac{x^2(x-1)^2+2x(x-1)y^2+y^4+y^2}{[(x-1)^2+y^2]^2}}\leq 1 \\ & \Rightarrow 0\leq \frac{x^2(x-1)^2+2x(x-1)y^2+y^4+y^2}{[(x-1)^2+y^2]^2}\leq 1 \\ & \Rightarrow 0\leq x^2(x-1)^2+2x(x-1)y^2+y^4+y^2\leq [(x-1)^2+y^2]^2 \end{align*}
From the second inequality we have \begin{align*}&x^2(x-1)^2+2x(x-1)y^2+y^4+y^2\leq (x-1)^4+2(x-1)^2y^2+y^4 \\ & \Rightarrow x^2(x-1)^2+2x(x-1)y^2+y^2\leq (x-1)^4+2(x-1)^2y^2\end{align*} How do we continue from here? I got stuck right now.
| It's much easier to rearrange to $|z|\le|z-1|$. Geometrically, the distance of $z$ from $0$ is less than or equal to the distance of $z$ from $1$ in the complex plane, so the complex numbers $z$ satisfying this are those with $\Re(z)\le\frac{1}{2}$, or $x\leq\frac{1}{2}$.
Alternatively, you can write $z=x+iy$ in $|z|\le|z-1|$ and squaring both sides gives $x^2+y^2\le (x-1)^2+y^2$, which simplifies to $0\le 1-2x$, which has the same solution.
| {
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How to calculate $lcm(2,3,4,5,6,7,8,9,10)$ solely by mental arithmetic? Denote $lcm(a,b)$ the lowest common multiple of $a$ and $b.$
Question: How to calculate
$$lcm(2,3,4,5,6,7,8,9,10)$$
solely by mental arithmetic?
I have a way to calculate, but it relies on writing on paper.
We can calculate the lcm using recursively, that is,
$$l = lcm( lcm(2,3),4,5,6,7,8,9,10 ) = lcm(lcm(6,4),5,6,7,8,9,10) = lcm(lcm(12,5),6,7,8,9,10)...$$
I know that my notations above are not correct but hopefully the idea can get through.
| Since we have a list of consecutive integers starting from $2$, just pick the largest powers of a prime:
$$\text{lcm}(2,3,4,5,6,7,8,9,10)=\text{lcm}(5,7,8,9)=5\cdot 7\cdot 8 \cdot 9=63\cdot 40=2520.$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Which is bigger 5^(√3) and 4^(√5)? How do you know which one is bigger between $5^\sqrt{3}$ and $4^\sqrt{5}$?
For my method I used in $2^\sqrt{3}$ and $3^\sqrt{2}$ I put both numbers in the function $f(x)=x^\sqrt{3}$ so $f(2^\sqrt{3})$ become $2^3$ which is equal to 8 and $f(3^\sqrt{2})$ become $3^\sqrt{6}$ and $3^\sqrt{6}>3^\sqrt{2}$ which is equal to 9 and 9>8 so $3^\sqrt{2}>2^\sqrt{3}$
But for this problem I don't know what I should multiply or is there any method besides that? Please kindly help. Thank you
| We make a power of $\sqrt{3}$ both expression:
$\left(5^\sqrt{3}\right)^\sqrt{3}=5^3=125$
$\left(4^\sqrt{5}\right)^\sqrt{3}=4^\sqrt{15}>4^{3.5}=4^3\times 2=128>125$
Therefore, $4^\sqrt{5}>5^\sqrt{3}$
| {
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Primitive instances where $c^3|(a^3+b^3)$ with $a,b,c\in\mathbb{N}$ It is known that by Fermat's Last Theorem there are no solutions to $a^3+b^3=c^3$ for $a,b,c\in\mathbb{N}$. I wondered about how multiplying the $c^3$ by a constant would change this fact.
Accordingly, I have been looking into instances where $c^3|(a^3+b^3)$ for $a,b\in\mathbb{N}$. In other words, solutions to the Diophantine Equation:
$a^3+b^3=dc^3$ where $a,b,$ and $c$ are pairwise co-prime and $a,b,c>0$
Obviously there are some trivial solutions. If $c=1$, for example, $a$ and $b$ can be any integers, and $d$ can be chosen to simply be $a^3+b^3$.
By requiring that $a,b,$ and $c$ are pairwise co-prime and that $c\not=1$, we eliminate the trivial solutions, and what remains is of much more interest.
For $a,b,c,d\le20$, there are 5 solutions:
$4^3+5^3=7*3^3$
$2^3+7^3=13*3^3$
$1^3+8^3=19*3^3$
$3^3+5^3=19*2^3$
$1^3+19^3=20*7^3$
Under $100$ there are $16$ solutions, as found by Mathematica.
My question about this equation: Has it been studied previously? Are there infinitely many primitive solutions (which it seems like there are)? If so, can they be parametrized?
| To show that there are infinitely many non-trivial solutions, it suffices to consider solutions of the following form, for $n=0,1,2\dots$
$\qquad a=3$
$\qquad b=24n+5$
$\qquad c=2$
$\qquad d=(3n+1)[9-3(24n+5)+(24n+5)^2]=1728n^3+1080n^2+225n+19$
Note that:
$$a^3+b^3 = (a+b)(a^2-ab+b^2)=(3+24n+5)[9-3(24n+5)+(24n+5)^2]$$
so that:
$$a^3+b^3=(3n+1)[9-3(24n+5)+(24n+5)^2]2^3=dc^3$$
and also:
$\gcd(a,b)=1$ since $24n+5\equiv2\pmod3$
$\gcd(a,c)=1$ and $\gcd(b,c)=1$ since $a,b$ odd.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Equilateral pentagon with increasing/equal angles
Suppose all sides of a convex pentagon ABCDE have the same size, and $\angle A \ge \angle B \ge \angle C \ge \angle D \ge \angle E $. Prove that this pentagon is a regular pentagon.
I know this should be proved using contradiction, but I'm not sure how to reach that point. Can anyone guide me to the right solution?
| WOLOG, we only need to consider the case where the side of the equilateral pentagon is $1$ and the vertices $A,B,C,D,E$ are ordered counterclockiwisely
on circumference of the pentagon.
Let $\alpha,\beta,\gamma,\delta,\epsilon$ be the external angles at $A, B, C, D, E$. We have
$$\angle A \ge \angle B \ge \angle C \ge \angle D \ge \angle E
\quad\implies\quad \alpha \le \beta \le \gamma \le \delta \le \epsilon
$$
Since $2\pi = \alpha + \beta + \gamma + \delta + \epsilon \le 5 \epsilon$,
we have $\epsilon \ge \frac{2\pi}{5}$. This leads to
$$\alpha + \beta + \gamma + \delta \le \frac{8\pi}{5}
\implies \alpha + \beta \le \frac{4\pi}{5}
\implies \alpha \le \frac{2\pi}{5}
$$
Choose a coordinate system so that $E = (-\frac12,0), A = (\frac12,0)$.
In this coordinate system, it is not hard to see the $x$-coordinate of $C$ is
$$\frac12 + \cos\alpha + \cos(\alpha+\beta)
\ge \frac12 + \cos\frac{2\pi}{5} + \cos\frac{4\pi}{5} = 0\tag{*1}$$
THis means $C$ is lying in the left-half plane and hence $|AC| \le |EC|$.
However
$$\begin{align} &|AC|^2 = |AB|^2 + |BC|^2 + 2|AB||BC|\cos\beta = 2(1+\cos\beta)\\
& |CE|^2 = |CD|^2 + |DE|^2 + 2|CD||DE|\cos\delta = 2(1+\cos\delta)
\end{align}
$$
Together with $\beta \le \delta$, we obtain $|AC| \ge |EC|$.
Combine with above, we get $|AC| = |EC|$. This forces $C$ to lies on the $y$-axis.
Notice if either $\alpha < \frac{2\pi}{5}$ or $\alpha + \beta < \frac{4\pi}{5}$, the inequality in $(*1)$ becomes strict. For $C$ to lies on the $y$-axis, we need
$\alpha = \frac{2\pi}{5}$. This leads to
$$2\pi = 5\alpha \le \alpha + \beta + \gamma + \delta + \epsilon = 2\pi$$
Since $\alpha \le \beta \le \gamma \le \delta \le \epsilon$, this forces
$\alpha = \beta = \gamma = \delta = \epsilon$ and
$ABCDE$ is a regular pentagon.
| {
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"timestamp": "2023-03-29T00:00:00",
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help solving for delta in epsilon delta proof $$\lim_{x\to 9} \frac{1}{\sqrt{x}} = \frac{1}{3} $$
so the two statements are
$$0<|x-9|<\delta$$
$$\left|\frac{1}{\sqrt{x}}-\frac{1}{3}\right|<\epsilon$$
I've tried to multiply by the conjegate to get
$$\frac{\frac{1}{x}-\frac{1}{9}}{\sqrt{\frac{1}{x}}+\frac{1}{3}}<\epsilon$$
I'm unsure where to go from here to get that x-9 factor i need
| We have that
$$\left|\frac{1}{\sqrt{x}}-\frac{1}{3}\right|=\left|\frac{\sqrt{x}-3}{3\sqrt{x}}\right|=\left|\frac{(\sqrt{x}-3)(\sqrt{x}+3)}{3\sqrt{x}(\sqrt{x}+3)}\right|=\left|\frac{x-9}{3\sqrt{x}(\sqrt{x}+3)}\right|$$
then assuming wlog $|x-9|<1 \implies 3\sqrt{x}(\sqrt{x}+3)\ge 3\sqrt{8}(\sqrt{8}+3)$ therefore
$$\left|\frac{1}{\sqrt{x}}-\frac{1}{3}\right|=\left|\frac{x-9}{3\sqrt{x}(\sqrt{x}+3)}\right|\le\frac{|x-9|}{ 3\sqrt{8}(\sqrt{8}+3)}$$
and it suffices to assume
$$\delta<3\sqrt{8}(\sqrt{8}+3)\epsilon $$
to have
$$\left|\frac{1}{\sqrt{x}}-\frac{1}{3}\right|<\epsilon$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Prove that $\lim_{x\to0, y\to0}\frac{x^3+y^3}{x^2+y^2}=0$ Prove that $\lim_{x\to0, y\to0}\frac{x^3+y^3}{x^2+y^2}=0$
My Try
| HINT
Notice that
\begin{align*}
|x|^{3} = |x|\times x^{2} \leq |x|(x^{2} + y^{2}) \Rightarrow \left|\frac{x^{3}}{x^{2} + y^{2}}\right| \leq |x|
\end{align*}
Based on the same procedure, you can prove that
\begin{align*}
|y|^{3} = |y|\times y^{2} \leq |y|(x^{2} + y^{2}) \Rightarrow \left|\frac{y^{3}}{x^{2} + y^{2}}\right| \leq |y|
\end{align*}
Then apply the squeeze theorem.
Can you take from here?
| {
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formula for $\left(1\cdot2\cdot...\cdot k\right)+...+\left(n\left(n+1\right)...\left(n+k- 1\right)\right)$ proof if $k$ is a constant then for every $n$ natural numbers we have:
$$\left(1\cdot2\cdot...\cdot
k\right)+\left(2\cdot3\cdot...\cdot\left(k+1\right)\right)+...+\left(n\left(n+1\right)...\left(n+k-
1\right)\right)=\frac{n\left(n+1\right)\left(n+2\right)...\left(n+k\right)}{k+1}.$$
clearly $$\left(1\cdot2\cdot...\cdot
k\right)+\left(2\cdot3\cdot...\cdot\left(k+1\right)\right)+...+\left(n\left(n+1\right)...\left(n+k-
1\right)\right)=\sum_{m=1}^{n}\left(\prod_{j=0}^{k-1}\left(m+j\right)\right)$$
$$=\sum_{m=1}^{n}\left(\Gamma\left(m+j+1\right)\right)$$where$\left(0\le j\le k-1\right)$ and
$\Gamma(x)$ is Gamma function,
or equivalently $$=\sum_{m=1}^{n}\int_{0}^{∞}x^{\left(m+j\right)}e^{-x}dx$$
but I could not find the given formula.
| We can prove this by induction. The base case here is when $n=1.$ Clearly,
$n\cdot(n+1)\cdot(n+2)\cdot\dots\cdot(n+k-1)=1\cdot(2)\cdot(3)\cdot\dots\cdot(k)$
and
$\dfrac{(n)\cdot(n+1)\cdot(n+2)\cdot\dots\cdot(n+k)}{1+k}=(1)\cdot(2)\cdot(3)\cdot\dots\cdot(k)=n\cdot(n+1)\cdot(n+2)\cdot\dots\cdot(n+k-1)$
so the base case holds. Assume the statement is true for some $y\in\mathbb{N}.$ Then
$$
\begin{align}(1)\cdot(2)\cdot\dots\cdot(k)+(2)\cdot(3)\cdot\dots\cdot(k+1)+\dots+(y+1)\cdot(y+2)\cdot\dots\cdot(y+k)=[(1)\cdot(2)\cdot\dots\cdot(k)+(2)\cdot(3)\cdot\dots\cdot(k+1)\\+(y)\cdot(y+1)\cdot\dots\cdot(y+k-1)]+(y+1)\cdot(y+2)\cdot\dots\cdot(y+k)\\=\dfrac{(y)\cdot(y+1)\cdot(y+2)\cdot\dots\cdot(y+k)}{1+k}+(y+1)\cdot(y+2)\cdot\dots\cdot(y+k)=\dfrac{(y+1)\cdot(y+2)\dots(y+k+1)}{k+1},\end{align}
$$
so the equation holds by induction.
| {
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Find $\lim_{x \to \ 0}{\frac{1-\sqrt{\cos x}}{1-\cos\sqrt{x}}}$ without using the l'Hospital rule The task is to evaluate
$$\lim_{x \to \ 0}{\frac{1-\sqrt{\cos x}}{1-\cos \sqrt{x}}}.$$
I tried to use some trigonometric identities such as
$$\lim_{x \to \ 0}{\frac{1-\sqrt{\cos\left(x\right)}}{1-\cos\left(\sqrt{x}\right)}}= \lim_{x \to \ 0} \frac{1-
\sqrt{\cos\left(x\right)}}{1-
\cos\left(\sqrt{x}\right)}\cdot\frac{1+\sqrt{\cos\left(x\right)}}{1+\sqrt{\cos\left(x\right)}}$$$$= \lim_{x \to \ 0} \frac{1-
\cos\left(x\right)}{2\sin^{2}\left(\frac{\sqrt{x}}{2}\right)}\cdot\frac{1}{1+\sqrt{\cos\left(x\right)}}$$$$=
\lim_{x \to \ 0} \left(\frac{\sin\left(\frac{\sqrt{x^{2}}}{2}\right)}{\sin\left(\frac{\sqrt{x}}{2}\right)}
\right)^{2}\cdot\frac{1}{1+\sqrt{\cos\left(x\right)}}$$$$=\frac{1}{2}\lim_{x \to \
0}\left(\frac{\sin\left(\frac{\sqrt{x^{2}}}{2}\right)}{\sin\left(\frac{\sqrt{x}}{2}\right)}\right)^{2}$$
and this is where I have a problem.
| We have that
$$\frac{1-\sqrt{\cos\left(x\right)}}{1-\cos\left(\sqrt{x}\right)}=
\frac{1-\sqrt{\cos\left(x\right)}}{x}\frac{1+\sqrt{\cos\left(x\right)}}{1+\sqrt{\cos\left(x\right)}}
\frac{x}{1-\cos\left(\sqrt{x}\right)}=$$
$$=x\frac{1-\cos\left(x\right)}{x^2}\frac{1}{1+\sqrt{\cos\left(x\right)}}
\frac{x}{1-\cos\left(\sqrt{x}\right)}\to 0$$
By your way from here
$$\frac{1-
\cos\left(x\right)}{2\sin^{2}\left(\frac{\sqrt{x}}{2}\right)}=2x\cdot\frac{1-
\cos\left(x\right)}{x^2}\frac{\left(\frac{\sqrt{x}}{2}\right)^2}{\sin^{2}\left(\frac{\sqrt{x}}{2}\right)}\to 0$$
| {
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If $f(x) = (x-a)^3(x-b)^3$ then what is the nature of the roots of $f^{\prime\prime}(x) = f^\prime (x)$? If we try solving it by finding $f''(x)$ then it is very long and difficult to do, so my teacher suggested a way of doing it, he said find nature of all the roots of $f(x) =f'(x)$, and on finding nature of the roots we got them to be real(but not all distinct) and then he said as all the roots of $f(x) = f'(x)$ are real so all the roots of $f'(x)= f''(x)$ are real and distinct. I did not understand how to prove that if all the roots of $f(x) = f'(x)$ are real so all the roots of $f'(x)= f''(x)$ are real and distinct.Can anyone please help me to prove this?
Is this statement (all roots of $f(x) = f'(x)$ are real so all roots of $f'(x)= f''(x)$ are real and distinct) true only for this question or is it true in general for all function $f(x)$ whose all roots of $f(x) = 0$ are real?
If instead of $f(x) = (x-a)^3(x-b)^3$ we had $f(x) = (x-a)^4(x-b)^4$
then would we say that as all roots of $f(x) = f'(x)$ are real so all the roots of $f'(x)= f''(x)$ are real and distinct or rather we would say that as all roots of $f(x) = f'(x)$ are real so all roots of $f'(x)= f''(x)$ are real.
If anyone has any other way of solving this question
$f(x) = (x-a)^3(x-b)^3$ then what is the nature of the roots of $f''(x) = f'(x)$ please share it.
| Recall the Leibnitz differentiation formula
$$(fg)^{(n)}=\sum_{k=0}^n f^{(k)} g^{(n-k)}$$
Then
$$\left( (x-a)^3(x-b)^3 \right)'=3(x-a)^2(x-b)^3+3(x-a)^3(x-b)^2=3(x-a)^2(x-b)^3[x-a+x-b]$$
$$\left( (x-a)^3(x-b)^3 \right)''=6(x-a)(x-b)^3+18(x-a)^2(x-b)^2+6(x-a)^3(x-b)=6(x-a)(x-b)[(x-a)^2+3(x-a)(x-b)+(x-b)^2]
$$
Equating them, you get $x=a, x=b$ as solutions together with the roots of
$$(x-a)(x-b)[x-a+x-b]=2[(x-a)^2+3(x-a)(x-b)+(x-b)^2] $$
This is a cubic equation, for which the nature of the roots can be easily studied.
Note that if you have instead $(x-a)^n (x-b)^m$ you would still end up with acubic, after canceling $(x-a)^{n-2} (x-b)^{m-2}$. If I didn't make any misatke, your cubic would be
$$(x-a)(x-b)[m(x-a)+n(x-b)]=2[m(m-1)(x-a)^2+2mn(x-a)(x-b)+n(n-1)(x-b)^2] $$
| {
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Show that $\frac{3\> + \>\cos x}{\sin x}$ cannot have any value between $-2\sqrt2$ and $2\sqrt2$ Show that
$$\dfrac{3+\cos x}{\sin x}\quad \forall \quad x\in R $$
cannot have any value between $-2\sqrt{2}$ and $2\sqrt{2}$.
My attempt is as follows:
There can be four cases, either $x$ lies in the first quadrant, second, third or fourth:-
First quadrant: $\cos x$ will decrease sharply and sinx will increase sharply, so $y_{min}=3$ at $x=\dfrac{\pi}{2}$.
$y_{max}$ would tend to $\infty$ near to $x=0$
Second quadrant: $\cos x$ will increase in magnitude and sinx will decrease sharply, so $y_{min}=3$ at $x=\dfrac{\pi}{2}$.
$y_{max}$ would tend to $\infty$ near to $x=\pi$
Third quadrant: $\cos x$ will decrease in magnitude and sinx will increase in magnitude but negative, so $y_{min}$ would tend to $-\infty$ near to $x=\pi$
$y_{max}$ would be $-3$ at $x=\dfrac{3\pi}{2}$
Fourth quadrant: $\cos x$ will increase sharply and sinx will decrease in magnitude, so $y_{min}$ would tend to $-\infty$ near to $x=2\pi$
$y_{max}$ would be $-3$ at $x=\dfrac{3\pi}{2}$
So in this way I have proved that $\dfrac{3+\cos x}{\sin x}$ cannot lie between $-2\sqrt{2}$ and $2\sqrt{2}$, but is their any smart solution so that we can calculate quickly.
| Let the value of this expression at $x=p$ be $q$. Then we have
$${3+\cos p\over\sin p}= q$$
$$\implies\cos p-q\sin p = -3$$
Dividing both LHS and RHS by $\sqrt{q^2+1}$ we have
$$\cos p\cdot {1\over\sqrt{q^2+1}} - \sin p\cdot{q\over\sqrt{q^2+1}} = {-3\over\sqrt{q^2+1}}$$
Let $1\over\sqrt{q^2+1}$ be $\cos r$. So, we have $$\cos p\cos r - \sin p\sin r = {-3\over\sqrt{q^2+1}}$$
Or $$\cos{(p+r)} = \frac{-3}{\sqrt{q^2+1}}$$
For this to be a valid expression $\sqrt{q^2+1}$ must be greater than $3$ since $\cos x \in [-1,1]$. So, we have
$$q^2+1 \geq 9$$
$$\implies |q| \geq \sqrt 8$$
$$\implies q \in \left(-\infty,-2\sqrt 2\right]\cup\left[2\sqrt 2, \infty\right)$$
$$\implies\boxed{ {3+\cos x\over\sin x}\notin\left(-2\sqrt 2, 2\sqrt 2\right)}$$
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"language": "en",
"url": "https://math.stackexchange.com/questions/3411788",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 4,
"answer_id": 1
} |
Coincidence in Diophantine solution parametrization for Pythagorean triples etc Consider a triple of nonnegative integers $(a, b, c)$ such that $c^2 = a^2 + b^2$. This can be viewed as integer triangles with sides $(a, b, c)$ such that $c$ is the side opposite a $90°$ angle. Such triples are well-known as Pythagorean triples, and it is well-known (called Euclid's formula on Wikipedia) that all such primitive (i.e. $\gcd(a, b, c) = 1$) triples can be parametrized as:
\begin{align}
a &= m^2 - n^2 \cr
b &= 2mn \cr
c &= m^2 + n^2
\end{align}
I've always found it mildly amusing (and occasionally confusing) that we started trying to find a parametrization for triples where $c^2$ was a sum of two squares, and obtained a parametrization where $c$ itself is a sum of two squares, i.e. has the same form.
Today I encountered the problem of nonnegative triples $(a, b, c)$ such that $c^2 = a^2 + b^2 + ab$. This can be viewed as integer triangles with sides $(a, b, c)$ such that $c$ is the side opposite a $120°$ angle. Such triples are called 1-Pythagorean triples on OEIS, Eisenstein triples in this paper, and “Trythagorean” triples in this blog post. Whatever the name, it turns out that all such primitive triples can be parametrized (see this very nice page) as:
\begin{align}
a &= n^2 - m^2 \cr
b &= m^2 + 2mn \cr
c &= m^2 + mn + n^2
\end{align}
where $m < n$ such that $\gcd(m,n)=1$ and $m≢n \pmod 3$.
This is spooky: we looked for triples such that $c^2$ was of the form $a^2 + ab + b^2$, and it turns out that $c$ itself is of a similar form, $c = m^2 + mn + n^2$.
Question: Is this just a coincidence? If not, what's going on? What's the most general kind of problem for which this (whatever “this” is) is true?
There's a general method for homogeneous Diophantine equations of degree two, but I haven't yet tried other equations. Also, even when sometimes the form seems different, it is not really, for example, the same page parametrizes solutions to $c^2 = a^2 + b^2 - ab$ (corresponding to $60°$ angles) as $c = m^2 + n^2 + mn$ which would seem to be a counterexample, but replacing either $m$ with $-m$ or $n$ with $-n$ gives $m^2 + n^2 - mn$ so I'm not sure.
| The example I like to show is solving
$$ 2(x^2 + y^2 + z^2) - 113(yz + zx + xy)=0, $$ four "recipes," all made up of binary quadratic forms
$$
\left(
\begin{array}{r}
x \\
y \\
z
\end{array}
\right) =
\left(
\begin{array}{r}
37 u^2 + 51 uv + 8 v^2 \\
8 u^2 -35 uv -6 v^2 \\
-6 u^2 + 23 uv + 37 v^2
\end{array}
\right)
$$
$$
\left(
\begin{array}{r}
x \\
y \\
z
\end{array}
\right) =
\left(
\begin{array}{r}
32 u^2 + 61 uv + 18 v^2 \\
18 u^2 -25 uv -11 v^2 \\
-11 u^2 + 3 uv + 32 v^2
\end{array}
\right)
$$
$$
\left(
\begin{array}{r}
x \\
y \\
z
\end{array}
\right) =
\left(
\begin{array}{r}
38 u^2 + 45 uv + 4 v^2 \\
4 u^2 -37 uv -3 v^2 \\
-3 u^2 + 31 uv + 38 v^2
\end{array}
\right)
$$
$$
\left(
\begin{array}{r}
x \\
y \\
z
\end{array}
\right) =
\left(
\begin{array}{r}
29 u^2 + 63 uv + 22 v^2 \\
22 u^2 -19 uv -12 v^2 \\
-12 u^2 -5 uv + 29 v^2
\end{array}
\right)
$$
For all four recipes,
$$ x^2 + y^2 + z^2 = 1469 \left( u^2 + uv + v^2 \right)^2 $$
giving effective bounds on $u,v$ if given upper bound on $x^2 + y^2 + z^2$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3420653",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Solve : $\sin\left(\frac{\sqrt{x}}{2}\right)+\cos\left(\frac{\sqrt{x}}{2}\right)=\sqrt{2}\sin\sqrt{x}$ Solve : $\sin\left(\dfrac{\sqrt{x}}{2}\right)+\cos\left(\dfrac{\sqrt{x}}{2}\right)=\sqrt{2}\sin\sqrt{x}$
$$\dfrac{1}{\sqrt{2}}\cdot\sin\left(\dfrac{\sqrt{x}}{2}\right)+\dfrac{1}{\sqrt{2}}\cos\left(\dfrac{\sqrt{x}}{2}\right)=\sin\sqrt{x}$$
$$\sin\left(\dfrac{\sqrt{x}}{2}+\dfrac{\pi}{4}\right)=\sin\sqrt{x}$$
$$\sin\left(\dfrac{\sqrt{x}}{2}+\dfrac{\pi}{4}\right)-\sin\sqrt{x}=0$$
$$2\sin\left(\dfrac{\dfrac{\sqrt{x}}{2}-\dfrac{\pi}{4}}{2}\right)\cos\left(\dfrac{\dfrac{3\sqrt{x}}{2}+\dfrac{\pi}{4}}{2}\right)=0$$
$$\left(\dfrac{\sqrt{x}}{2}-\dfrac{\pi}{4}\right)=2n\pi \text { or }
\left(\dfrac{3\sqrt{x}}{2}+\dfrac{\pi}{4}\right)=(2n+1)\pi$$
$$\sqrt{x}=4n\pi+\dfrac{\pi}{2} \text { or } \sqrt{x}=\dfrac{4n\pi}{3}+\dfrac{\pi}{2}$$
$$x=\left(4n\pi+\dfrac{\pi}{2}\right)^2 \text { or } x=\left(\dfrac{4n\pi}{3}+\dfrac{\pi}{2}\right)^2 \text { where n $\in$ I}$$
But actual answer is $$x=\left(4n\pi+\dfrac{\pi}{2}\right)^2 \text { or } x=\left(\dfrac{4m\pi}{3}+\dfrac{\pi}{2}\right)^2 \text { where n,m $\in$ W}$$
I tried to find the mistake but didn't get any breakthroughs.
| First we square both sides to get
$$
\begin{align*}
\left(\sin\left(\frac{\sqrt{x}}{2}\right)+\cos\left(\frac{\sqrt{x}}{2}\right)\right)^2&=\left(\sqrt{2}\sin\left(\sqrt{x}\right)\right)^2,\\
\sin^2\left(\frac{\sqrt{x}}{2}\right)+\cos^2\left(\frac{\sqrt{x}}{2}\right)+2\sin\left(\frac{\sqrt{x}}{2}\right)\cos\left(\frac{\sqrt{x}}{2}\right)&=2\sin^2\left(\sqrt{x}\right),\\
1+2\sin\left(\frac{\sqrt{x}}{2}\right)\cos\left(\frac{\sqrt{x}}{2}\right)=2\sin^2\left(\sqrt{x}\right),\\
-2\sin^2\left(\sqrt{x}\right)+\sin\left(\sqrt{x}\right)+1=0.
\end{align*}$$
Let $y:=\sin(\sqrt{x})$ meaning we have the polynomial $-2y^2+y+1=0$. Factoring gives us $(2y+1)(y-1)=0$ meaning that $\sin(\sqrt{x})=1$ and $\sin(\sqrt{x})=-1/2$.
If $\sin(\sqrt{x})=1$ then $\sqrt{x}=\pi/2 + 2\pi n$ where $n \in \mathbb{N}$ which means that $x=\pi^2(4n+1)^2/4$.
If $\sin(\sqrt{x})=-1/2$ then the principle angle (i.e. angle in quadrant one), $\alpha$, is $\alpha=\pi/6+2\pi n$. Since $\sin$ is negative in quadrants three and four, $\sqrt{x}=7\pi/6 +2\pi n$ and $\sqrt{x}=11\pi/6 + 2\pi n$. This means that the solutions are $x=\pi^2(12n+7)^2/36$ and $\pi^2(12n+11)^2/36$.
Therefore, the solutions are
$$
\begin{align}
x&=\frac{\pi^2(4n+1)^2}{4},\\
x&=\frac{\pi^2\left(12n+7\right)^2}{36},\\
x&=\frac{\pi^2\left(12n+11\right)^2}{36}.
\end{align}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3420908",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
A general formula to generate functions of power series I have been playing with Maclaurin series lately, I have been able to come across this:
$\dfrac{1}{1+x}=1-x+x^2-x^3+x^4-x^5...$
$\dfrac{1}{(1+x)^2}=1-2x+3x^2-4x^3+5x^4-6x^5+7x^6...$
I found out by accident that:
$\dfrac{1-x}{(1+x)^3}=1-2^2x+3^2x^2-4^2x^3+5^2x^4+6^2x^5...$
I found in an old paper of Euler that this can continue on with these functions:
$\dfrac{1-4x+x^2}{(1+x)^4}=1-2^3x+3^3x^2-4^3x^3+5^3x^4+6^3x^5...$
$\dfrac{1-11x+11x^2-x^3}{(1+x)^5}=1-2^4x+3^4x^4-4^4x^3+5^4x^4+6^4x^5...$
$\dfrac{1-26x+66x^2-26x^3+x^4}{(1+x)^5}=1-2^5x+3^5x^4-4^5x^3+5^5x^4+6^5x^5...$
$\dfrac{1-57x+320x^2-302x^3+57x^4-x^5}{(1+x)^5}=1-2^6x+3^6x^4-4^6x^3+5^6x^4+6^6x^5...$
and so on...
Is there a general formula to generate the functions on the right hand side? How did Euler calculate these series? I have to say I deeply respect him since only he and Ramanujan know how to play with series.
| We are looking for polynomials $p_k(x)$ with
\begin{align*}
\frac{p_k(x)}{(1+x)^{k+1}}=\sum_{j=0}^\infty (-1)^j(j+1)^kx^j\qquad\qquad k\geq 0
\end{align*}
We can find $p_k(x)$ as follows:
\begin{align*}
\color{blue}{p_k(x)}&=\left(\sum_{j=0}^\infty(-1)^j(j+1)^kx^j\right)(1+x)^{k+1}\\
&=\left(\sum_{j=0}^\infty(-1)^j(j+1)^kx^j\right)\left(\sum_{l=0}^{k+1}\binom{k+1}{l}x^l\right)\\
&=\sum_{n=0}^\infty\left(\sum_{{j+l=n}\atop{j,l\geq 0}}\binom{k+1}{l}(-1)^j(j+1)^k\right)x^n\\
&\,\,\color{blue}{=\sum_{n=0}^\infty\left(\sum_{l=0}^{\min\{n,k+1\}}\binom{k+1}{l}(-1)^{n-l}(n-l+1)^k\right)x^n}\tag{1}
\end{align*}
OP's result indicates that for $k\geq 1$ we expect $p_k(x)$ to be a polynomial of degree less than or equal to $k-1$. We therefore want to show that for $k\geq 1$
\begin{align*}
p_k(x)=\sum_{n=0}^{\color{blue}{k-1}}\left(\sum_{l=0}^{\color{blue}{n}}\binom{k+1}{l}(-1)^{n-l}(n-l+1)^k\right)x^n\tag{2}
\end{align*}
In order to show (2) it is convenient to use the coefficient of operator $[x^n]$ to denote the coefficient of $x^n$. This way we can write for instance
\begin{align*}
n^k=k![x^k]e^{nx}\tag{3}
\end{align*}
We obtain for $n\geq k+1$
\begin{align*}
\color{blue}{\sum_{l\geq 0}}&\color{blue}{\binom{k+1}{l}(-1)^{n-l}(n-l+1)^k}\\
&=\sum_{l\geq 0}\binom{k+1}{l}(-1)^{n-l}k![x^k]e^{x(n-l+1)}\tag{4}\\
&=(-1)^nk![x^k]e^{x(n+1)}\sum_{l\geq 0}\binom{k+1}{l}(-1)^le^{-lx}\tag{5}\\
&=(-1)^nk![x^k]e^{x(n+1)}\left(1-e^{-x}\right)^{k+1}\tag{6}\\
&=(-1)^nk![x^k]e^{x(n-k)}\left(e^x-1\right)^{k+1}\tag{7}\\
&\,\,\color{blue}{=0}\tag{8}
\end{align*}
and the claim (2) follows for $n\geq k+1$. Similarly the claim can be shown for $n=k$.
Comment:
*
*In (4) we apply the coefficient of operator according to (3).
*In (5) we do some rearrangements.
*In (6) we apply the binomial theorem.
*In (7) we factor out $e^{-x(k+1)}$.
*In (8) we note that $(e^x-1)^{k+1}=x^{k+1}+\cdots$ consists of terms with powers of $x$ greater than $k$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3422095",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
how to factorize $a^3 - ab^3 + a^2 + b^2 + 1$ I've got the following question in a book about maths olympiad
factorize $a^3b - ab^3 + a^2 + b^2 + 1$
and the answer to this question is
$a^3b - ab^3 + a^2 + b^2 + 1$$= a^3b - a^2b^2 + ab + a^2b^2 - ab^3 + b^2 - ab + + a^2+ 1$ $= ab(a^2 - ab +1) + b^2 (a^2 - ab + 1) + (a^2 - ab + 1)$ $= (a^2 - ab + 1)(ab+b^2 + 1)$
I understand that $a^2b^2 + ab$ is added to the polynomials for further factorization. However, it seems like a magic to me. I didn't know how I can think of adding $a^2b^2 + ab$. Adding two terms is really hard to think of to me. If I am given this question in the future, I am quite sure that I still can't finish it.
My question is: How can I know what terms to add to the polynomails so that I can group things up and do factorization? Also, where can I find more factorization like this for me to practise?
| Another way to guesstimate is to try $b=1$ and $a=1$:
$$a^3b - ab^3 + a^2 + b^2 + 1=a^3+a^2-a+2=(a^2-a+1)(a+2)=(a^2-a+\color{red}1)(a+1+\color{red}1)\\
a^3b - ab^3 + a^2 + b^2 + 1=-b^3+b^2+b+2=(2-b)(b^2+b+1)=(1-b+\color{red}1)(b+b^2+\color{red}1)$$
Hence:
$$a^3b - ab^3 + a^2 + b^2 + 1=(a^2-ab+1)(ab+b^2+1)$$
Note: $\color{red}1$s must be fixed. One can guess the term $ab$ from the signs.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3422724",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
"answer_id": 3
} |
Determinant of a nondegenerate quadratic form Given a quadratic form $Q$ over a space $V$.
To put it simply, we say $Q$ is nondegenerate if for every $x\in V$ there exists at least one $y\in V$ such that $x.y\neq 0$ (where '$.$' is a generalization of the inner product, and $x.y=\frac{1}{2}(Q(x+y)-Q(x)-Q(y)$).
Equivalently, $Q$ is nondegenerate if $\det(A)\neq 0$, ($A$ is the matrix representation of $Q$).
$\det(A)\neq 0$ means the kernel of $Q$ is ${0}$, hence $Q(x)=0$ iff $x=0$. But it contradicts the fact that we have isotropic elements ($0\neq x\in V$ such that $Q(x)=0$) in nondegenerate spaces.
what am I missing?
Please try putting it as simple as you can.
| Let's put the focus on the matrix $A$. We have
$$Q(x) = x^T A x.$$
Suppose $Q(x) = 0$. This means $x^T (A x) = 0$, or equivalently, $x$ is orthogonal (under the standard inner product) to $Ax$. Note that this doesn't require that $Ax = 0$ - it only requires that $Ax$ be a vector which lives on the perpendicular subspace to $x$.
Example:
Let $$A = \begin{pmatrix}0 & 1 \\ 1 & 2\end{pmatrix}.$$
Then $$Q\begin{pmatrix}s \\ t \end{pmatrix} = \begin{pmatrix} s & t \end{pmatrix}\begin{pmatrix}0 & 1 \\ 1 & 2\end{pmatrix} \begin{pmatrix}s \\ t \end{pmatrix} = \begin{pmatrix} s & t \end{pmatrix} \begin{pmatrix} t \\ s + 2t\end{pmatrix} = 2(st+t^2).$$
Observe that $$Q\begin{pmatrix}1 \\ 0 \end{pmatrix} = 0,$$
however,
$$A \begin{pmatrix}1 \\ 0 \end{pmatrix} = \begin{pmatrix}0 \\ 1 \end{pmatrix} \neq 0.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3423084",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
$| \dfrac{-z+\sqrt{z^2-4}}{2}|\le 1$ Suppose $z=u+iv$, with $v>0$.
$$ m(z)=\dfrac{-z+\sqrt{z^2-4}}{2}$$ satisfies the equation $$m(z)+1/m(z)+z=0 $$ from which we found that $$|m(z)|=|m(z)+z|^{-1} $$
Take the branch of the square root so that the imaginary part of $m(z)$ is postive.
($m(z) $ here is the Stieljes Transformation of the Semicircle Law)
How do we show $$|m(z)|\le 1 $$ or equivalently $$|m(z)+z|\ge 1 $$
As pointed out by Conrad in the comment,$$|m(z)(z+m(z))|=1 $$If we can show $m(z)$ is the smaller of the two in norm, then we are done.
Noting that $$v>0,\qquad m(z)=\dfrac{-z+\sqrt{z^2-4}}{2},\qquad m(z)+z= \dfrac{z+\sqrt{z^2-4}}{2}$$
we have $$Im[m(z)+z] >Im[m(z)] $$ but how about the real part?
as shown in the first few lines of the accepted solution, by the constrains we have on the imaginary part of $z$ and $m(z)$. we have that the real and imaginary part of $$z$$ and $$\sqrt{z^2-4}$$ are of the of the sign. With this we can conclude that $$|m(z)+z|\ge |m(z)| $$
| Let $z=a+ib,\sqrt{z^2-r^2}=c+id$ where $a,b,c,d$ are real and $r>0$
$$r^2=z^2-(z^2-r^2)=(a+ib)^2-(c+id)^2=a^2+d^2-b^2-c^2+2i(ab-cd)$$
Equating the imaginary parts, $ab-cd=0\implies\dfrac ac=\dfrac db=k$(say)
$\implies a=ck,d=bk$
Equating the real parts, $$r^2=a^2+d^2-b^2-c^2=(b^2+c^2)(k^2-1)$$
$\implies k^2-1=\dfrac{r^2}{b^2+c^2}>0$
$\implies$ either $k>1$ or $k<-1$
$$|-z+\sqrt{z^2-r^2}|^2=|c+id-(a+ib)|^2=(c-a)^2+(d-b)^2=(b^2+c^2)(1-k)^2$$
$$=\dfrac{r^2(1-k)^2}{(k^2-1)}=\dfrac{r^2(k-1)}{k+1}$$ which will be $\le r^2$
$\iff\dfrac{k-1}{k+1}\le1\iff0\ge\dfrac{k-1}{k+1}-1=-\dfrac2{k+1}$
$\iff 0\le\dfrac2{k+1}\iff k+1>0\iff k>-1$
So, the proposition will be nullified if $k<-1$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3427116",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Compute $\int_0^{\pi/2} x^2\left(\sum_{n=1}^\infty (-1)^{n-1} \cos^n(x)\cos(nx)\right)dx$ How to prove
$$I=\int_0^{\pi/2} x^2\left(\sum_{n=1}^\infty (-1)^{n-1} \cos^n(x)\cos(nx)\right)dx=\frac16\left(\frac{\pi^3}{12}-\pi\operatorname{Li}_2\left(\frac13\right)\right)$$
This problem is proposed by Cornel which can be found here where he suggested that the problem can be solved with and without harmonic series.
Here is my approach but I got stuck at the blue integral:
Using the common identity
$$ \sum_{n=1}^{\infty}p^n \cos(nx)=\frac{p(\cos(x)-p)}{1-2p\cos(x)+p^2}, \ |p|<1$$
Set $p=-\cos(x)$ we get
$$ \sum_{n=1}^{\infty}(-1)^n \cos^n(x) \cos(nx)=-\frac{2\cos^2(x)}{1+3\cos^2(x)}=-\frac23+\frac23\frac1{1+3\cos^2(x)}$$
Multiply both sides by $-x^2$ then integrate from $x=0$ to $\pi/2$ we get
$$\int_0^{\pi/2} x^2\left(\sum_{n=1}^\infty (-1)^{n-1} \cos^n(x)\cos(nx)\right)dx=\frac23\int_0^{\pi/2} x^2dx-\frac23\color{blue}{\int_0^{\pi/2}\frac{x^2}{1+3\cos^2(x)}dx}\\=\frac{\pi^3}{36}-\frac23\left(\color{blue}{\frac{\pi^3}{48}+\frac{\pi}{4}\operatorname{Li}_2\left(\frac13\right)}\right)=\frac{\pi^3}{72}-\frac{\pi}{6}\operatorname{Li}_2\left(\frac13\right)$$
I have two Questions:
1) Can we evaluate $I$ in a different way?
2) How to finish the blue integral?
My try to the blue integral is using integration by parts
$$\int\frac{dx}{1+3\cos^2(x)}=\frac12\tan^{-1}\left(\frac{\tan(x)}{2}\right)=-\frac12\tan^{-1}\left(2\cot(x)\right)$$
which gives us
$$\int_0^{\pi/2}\frac{x^2}{1+3\cos^2(x)}dx=\frac{\pi^3}{16}-\int_0^{\pi/2}x\tan^{-1}\left(\frac{\tan(x)}{2}\right)dx$$
Or
$$\int_0^{\pi/2}\frac{x^2}{1+3\cos^2(x)}dx=\int_0^{\pi/2}x\tan^{-1}\left(2\cot(x)\right)dx$$
I also tried the trick $x\to \pi/2-x$ but got complicated
Proof of the identity:
\begin{align}
\sum_{n=0}^\infty p^ne^{inx}&=\sum_{n=0}^\infty\left(p e^{ix}\right)^n=\frac{1}{1-pe^{ix}},\quad |p|<1\\&=\frac{1}{1-p\cos(x)-ip\sin(x)}=\frac{1-p\cos(x)+ip\sin(x)}{1-2p\cos(x)+p^2}\\
&=\frac{1-p\cos(x)}{1-2p\cos(x)+p^2}+i\frac{p\sin(x)}{1-2p\cos(x)+p^2}
\end{align}
By comparing the real and imaginary parts, we get
$$\sum_{n=\color{blue}{0}}^\infty p^n \cos(nx)=\frac{1-p\cos(x)}{1-2p\cos(x)+p^2}\Longrightarrow \sum_{n=\color{blue}{1}}^\infty p^{n-1} \cos(nx)=\frac{\cos(x)-p}{1-2p\cos(x)+p^2}$$
and
$$\sum_{n=\color{red}{0}}^\infty p^n \sin(nx)=\frac{p\sin(x)}{1-2p\cos(x)+p^2}\Longrightarrow \sum_{n=\color{red}{1}}^\infty p^n \sin(nx)=\frac{p\sin(x)}{1-2p\cos(x)+p^2}$$
| We can use the following fourier series:$$\frac{1}{a+b\cos t}=\frac{1}{\sqrt{a^2-b^2}}+\frac{2}{\sqrt{a^2-b^2}}\sum_{n=1}^{\infty}\left(\frac{\sqrt{a^2-b^2}-a}{b}\right)^n\cos{(nt)},\ a>b$$
Plugging $a=5, b=3$ and $t=2x$ we get:
$$\frac{1}{1+3\cos^2 x}=\frac{2}{5+3\cos(2x)}=\frac{1}{2}+\sum_{n=1}^\infty (-1)^n\left(\frac{1}{3}\right)^n\cos(2nx)$$
$$\Rightarrow \int_0^\frac{\pi}{2}\frac{x^2}{1+3\cos^2 x}dx=\frac12\int_0^\frac{\pi}{2} x^2dx+\sum_{n=1}^\infty(-1)^n \left(\frac13\right)^n\int_0^\frac{\pi}{2}x^2 \cos(2nx)dx$$
$$=\frac{\pi^3}{48}+\frac{\pi}4\sum_{n=1}^\infty \left(\frac13\right)^n\frac{1}{n^2}=\frac{\pi^3}{48}+\frac{\pi}{4}\operatorname{Li}_2\left(\frac13\right)$$
Using the series obtain above, we can also conclude that:
$$\sum_{n=1}^{\infty}(-1)^n \cos^n(x) \cos(nx)=-\frac13+\frac23\sum_{n=1}^\infty \left(-\frac{1}{3}\right)^n\cos(2nx)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3428840",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 2,
"answer_id": 0
} |
Calculating the integral $\iint_D\frac{x}{\sqrt{x^2+y}}\,dx\,dy$ I wish to calculate the following integral:
$$\iint_D\frac{x}{\sqrt{x^2+y}} \, dx \, dy$$
where $D$ is the area where $y>0,0<x<1,y<x^2$
I was asked to calculate twice, first integrating by $x$ and then by $y$, and then again the by opposite order.
Here is what I did:
*
*If $u=x^2$:
$$\iint_D\frac{x}{\sqrt{x^2+y}} \, dx \, dy = \int_0^1 \int_y^1\frac{1}{2\sqrt{u+y}} \, du \, dy=\cdots=\bigg[\frac{2}{3} \big((1+y)^{\frac32}-\frac{(2y)^{\frac32}}{2}\big) \bigg]_0^1$$
2.
$$\iint_D\frac{x}{\sqrt{x^2+y}} \, dy \, dx = \int_0^1 \int_0^{x^2}\frac{x}{\sqrt{x^2+y}} \, dy \, dx=\cdots=\bigg[\frac{2}{3}(\sqrt2-1) x^3\bigg]_0^1$$
When I calculated both integrals I got two different results, I either made a mistake defining the integration intervals, or I did a calculation error, but checking myself I failed to find it.
| *
*gives you $$\frac{2}{3} \left(2^{3/2}- \frac{2^{3/2}}{2}\right) - \frac{2}{3}$$
It looks different, but can be simplified to $$\frac{2}{3}\left(\frac{2.2^{3/2} - 2^{3/2}}{2}\right) - \frac{2}{3} = \frac{2}{3} \frac{2^{3/2}}{2} - \frac{2}{3} = \frac{2}{3} \sqrt 2 - \frac{2}{3}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3431177",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Evaluate $\int_{0}^{\infty} \frac{x^4}{(1+2x^2)^4} dx$ Evaluate :$$ \int_{0}^{\infty} \frac{x^4}{(1+2x^2)^4} dx$$
using residue theorem. The integral is even and it can be written as :
$$ \frac{1}{2} \int_{-\infty}^{\infty} \frac{x^4}{(1+2x^2)^4} dx$$
The only pole with a positive imaginary part is $ \frac{i}{\sqrt2} $ and its degree is $4$. How can I evaluate this without differentiating 3 times when calculating the residue?
I also tried calculating the infinite residue but that didn't help much.
| You can instead find the laurent series, and pick up the coefficient of the $z^{-1}$ term.
i.e. $(1+2z^2)^{-4} = (1+2(z-\frac{i}{\sqrt2})^2-2(\frac{-1}{2})+4\frac{iz}{\sqrt2})^{-4}= (1+2(z-\frac{i}{\sqrt2})^2+1+2\sqrt2iz)^{-4} = (2\sqrt{2}i(z-\frac{i}{\sqrt2})+2\sqrt2i\frac{i}{\sqrt2}+2+2(z-\frac{i}{\sqrt2})^2)^{-4}=(2\sqrt2i(z-\frac{i}{\sqrt2})+2(z-\frac{i}{\sqrt2})^2)^{-4}$
Put $y=z-\frac{i}{\sqrt2}$, so you need to find the laurent series of $(2\sqrt{2}iy+2y^2)^{-4}=(2\sqrt2iy)^{-4}(1+\frac{-i}{\sqrt2}y)^{-4}$.
Writing $\frac{z^4}{(1+2z^2)^{4}}=\frac{(y+\frac{i}{\sqrt2})^4}{(2\sqrt2iy)^{4}(1+\frac{-i}{\sqrt2}y)^{4}}=\frac{(y+\frac{i}{\sqrt2})^4}{64y^4(1+\frac{-i}{\sqrt2}y)^{4}}$
Therefore you are only interested in the $y^{-1}$ coefficient which corresponds to $y^3$ coefficient of $\frac{(y+\frac{i}{\sqrt2})^4}{(1+\frac{-i}{\sqrt2}y)^{4}}$ scaled by $\frac{1}{64}$.
Looking at $(y+\frac{i}{\sqrt2})^4(1+\frac{-i}{\sqrt2}y)^{-4}=(1/4 - i \sqrt2 y - 3 y^2 + 2 i \sqrt2 y^3 + y^4)(1+2\sqrt2iy-5y^2-5\sqrt2iy^3+O(y^4))=....-\frac{i}{2\sqrt2}y^3+....$
Scaling that by $\frac{1}{64}$, we find the residue to be $\frac{-i}{128\sqrt2}$
| {
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Find $4x \equiv 7 \pmod{15}$ and $3x \equiv 5 \pmod{16}$ (different exercises) This is how I solved each:
$$4x \equiv 7\pmod {15} \\
4x - 7 \equiv 0\pmod{15} \\
4x-7 = 15k \Leftrightarrow 4x-15k= 7 \\
$$
$$15 = 4*3+3 \\
4=3*1+1\\
3=3*1+0$$
$$1 = 4-3*1 \\
3 = 15-4*3 \\
1 = 4 - (15-4*3)*1 \\
1 = 4-15+4*3 \\
1 = 4*4-15*1 \\
$$
$$7 = (7*4)*4-(7*1)*15$$
$7*4*4 = 112$ which is congruent with 7 in mod 15. Yet my book says the answer is 13. What went wrong?
The second one:
$$3x - 5 \equiv 0 \pmod{16} \\
3x-5=16k\\
3x-16k=5 \\$$
$$16=3*5+1\\
3=1*3+0$$
$$1 = 16-3*5 \\
5=16*5-3*5*5$$
Yet $3*5*5-5$ is not congruent with zero in mod 16. What went wrong?
| To correctly pattern-match $\,\color{#c00}x\,$ in the above $\,\rm\color{#0a0}{scaled}\,$ Bezout equations do as follows.
$\qquad\ \begin{align}
\ -7\cdot \color{#90f}{15}&\ +\ 4\cdot 7\cdot 4\,\ =\ \color{#0a0} 7\ \ [=\ \color{#0a0}{7\ {\rm times}} {\rm \ Bezout\ of\ } \gcd(15,4)=1\,]\\[.2em]
\Rightarrow\ \bmod \color{#90f}{15}&\!:\,\ \ \ \ 4\cdot\color{#c00}{ 7\cdot 4} \ \equiv\ \ 7\\
\rm so &\,\ \ \ \ \ \ \ 4\ \cdot\ \color{#c00} x\ \ \, \equiv\, \ \ 7\ \ \ {\rm for}\ \ \ \color{#c00}{x\equiv 7\cdot 4}\equiv 13 \\[.6em]
\ 5\cdot \color{#90f}{16}&\, + 3\cdot 5(-5)\, =\, \color{#0a0}5\ \ [=\ \color{#0a0}{5\ {\rm times}} {\rm \ Bezout\ of\ } \gcd(16,3)=1\,]\\[.2em]
\Rightarrow\ \bmod \color{#90f}{16}&\!:\,\ \ 3\cdot\color{#c00}{ 5(-5)}\, \equiv\ 5\\
\rm so &\,\ \ \ \ \ 3\ \cdot\ \color{#c00} x\ \ \ \,\equiv\ \ \ \ 5\ \ \ {\rm for}\ \ \ \color{#c00}{x\equiv 5(-5)}\equiv 7
\end{align}$
Simpler $\bmod 15\!:\,\ 4x\equiv7\iff x \equiv \dfrac{7}{4}\equiv \dfrac{-8}4\equiv -2\equiv 13$
Similarly $\bmod 16\!:\,\ 3x\equiv 5\iff x\equiv \dfrac{5}3\equiv\dfrac{21}3\equiv 7\ $ where we subtracted or added the modulus to the numerator to make the division exact, a special case of Inverse Reciprocity.
Beware $\ $ Modular fraction arithmetic is well-defined only for fractions with denominator coprime to the modulus. See here for further discussion.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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How to prove $\sum\limits_{k=0}^{N} \frac{(-1)^k {N \choose k}}{(k+1)^2} = \frac{1}{N+1} \sum\limits_{n=1}^{N+1} \frac{1}{n}$ In the process of proving a more complicated relation, I've come across the following equality that I'm having trouble proving:
$$
\sum\limits_{k=0}^{N} \frac{(-1)^k {N \choose k}}{(k+1)^2} = \frac{1}{N+1} \sum\limits_{n=1}^{N+1} \frac{1}{n}
$$
I was already able to prove the following similar equality:
$$
\sum\limits_{k=0}^N \frac{(-1)^k {N \choose k}}{k+1} = \frac{1}{N + 1}
$$
but I'm unsure how to proceed with the first one. I assume it has something to do with the fact that every term in the left hand side of the first equality is $\frac{1}{k+1}$ times a term in the left hand side of the second equality. Any help would be greatly appreciated.
| Introduce the function
$$f(z) = (-1)^n \times n! \times \frac{1}{(1+z)^2}
\prod_{q=0}^n \frac{1}{z-q}.$$
Then we have for $0\le k\le n$
$$\mathrm{Res}_{z=k} f(z)
= (-1)^n \times n! \times \frac{1}{(1+k)^2}
\prod_{q=0}^{k-1} \frac{1}{k-q}
\prod_{q=k+1}^{n} \frac{1}{k-q}
\\ = (-1)^n \times n! \times \frac{1}{(1+k)^2}
\frac{1}{k!} (-1)^{n-k} \frac{1}{(n-k)!}
= {n\choose k} (-1)^k \frac{1}{(1+k)^2}.$$
Now as residues sum to zero we get
$$\sum_{k=0}^n {n\choose k} (-1)^k \frac{1}{(1+k)^2}
+ \mathrm{Res}_{z=-1} f(z) + \mathrm{Res}_{z=\infty} f(z) = 0.$$
The residue at infinity is zero by inspection. We also have
$$\mathrm{Res}_{z=-1} f(z) =
(-1)^n \times n! \times
\left.\left(\prod_{q=0}^n \frac{1}{z-q}\right)'\right|_{z=-1}
\\ = (-1)^n \times n! \times
\left.\prod_{q=0}^n \frac{1}{z-q}
\sum_{q=0}^n \frac{1}{q-z} \right|_{z=-1}
\\ = (-1)^n \times n! \times
\prod_{q=0}^n \frac{1}{-1-q}
\sum_{q=0}^n \frac{1}{q+1}
= (-1)^n \times n! \times
\frac{(-1)^{n+1}}{(n+1)!}
\sum_{q=1}^{n+1} \frac{1}{q}.$$
This simplifies to
$$- \frac{1}{n+1} \sum_{q=1}^{n+1} \frac{1}{q}$$
so that
$$\bbox[5px,border:2px solid #00A000]{
\sum_{k=0}^n {n\choose k} (-1)^k \frac{1}{(1+k)^2} =
\frac{1}{n+1} \sum_{q=1}^{n+1} \frac{1}{q}
= \frac{1}{n+1} H_{n+1}.}$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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How to compute $\lim\limits_{x \to a} \frac{\sqrt{2 a^{3} x-x^{4}}-a \sqrt[3]{a^{2} x}}{a-\sqrt[4]{a x^{3}}}$?
How to compute $\lim\limits_{x \to a} \frac{\sqrt{2 a^{3} x-x^{4}}-a \sqrt[3]{a^{2} x}}{a-\sqrt[4]{a x^{3}}}$?
My process:
\begin{align}
\lim\limits_{x \to a} \frac{\sqrt{2 a^{3} x-x^{4}}-a \sqrt[3]{a^{2} x}}{a-\sqrt[4]{a x^{3}}}=\frac{\sqrt{2 a^{3} a-a^{4}}-a \sqrt[3]{a^{2} a}}{a-\sqrt[4]{a a^{3}}}=\frac{\sqrt{a^4}-a\cdot \sqrt[3]{a^3}}{a-\sqrt[4]{a^4}}=\frac{a^2-a\cdot a}{a-a}=\frac{a^2-a^2}{a-a}
\end{align}
which would be undefined, but Wolfram Alpha calculated this:
$\lim_{x\to a} \, \frac{\sqrt{2 a^3 x-x^4}-a \sqrt[3]{a^2 x}}{a-\sqrt[4]{a x^3}}=
\begin{array}{cc}
\{ &
\begin{array}{cc}
\frac{16 a}{9} & a\geq 0 \\
0 & (\text{otherwise}) \\
\end{array}
\\
\end{array}$
What have I done wrong? Is there a way to calculate the limit with L'Hospitals Rule?
| Similar to @user's answer, starting with $x=a y$, we have
$$A=\frac{\sqrt{2 a^{3} x-x^{4}}-a \sqrt[3]{a^{2} x}}{a-\sqrt[4]{a x^{3}}}=a\frac{ \sqrt[3]{y}-\sqrt{y \left(2-y^3\right)}}{y^{3/4}-1}$$ Now, let $y=1+t$ to get
$$A=a\frac{\sqrt[3]{1+t}-\sqrt{1-2 t-6 t^2-4 t^3-t^4}}{(1+t)^{3/4}-1}$$ Now, using Taylor series around $t=0$ or binomial expansion
$$A=a \left(\frac{16}{9}+\frac{128 }{27}t+O\left(t^2\right) \right)$$ which shows the limit and also how it is approached.
| {
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How big is the smallest triangle inscribed in a square A square is divided in 7 areas as show on the figure.
The dots show the corners of the square and the middle points on the edges.
How large a fraction is area $D$ and how do I work it out?
I have tried using trigonometry to calculate the area A. If we say each side of the square is $1$, and look at the triangle $ABC$ using Pythagoras, it's hypotenuse must be $ \sqrt {1 \cdot 1 + 0.5 \cdot 0.5} = 1.118$.
Then using the sine relation we see that the $ \hat A$ is $\sin A = 1/1.118 = 63.43°$. It then follows the other angles must be $90°$ and $71.57°$. If I use Heron's formula I can calculate the area of $A = \sqrt {p(p−a)(p−b)(p−c)} = \frac 1 {12}$.
I know $C$ is $ \frac1 {16}$ just by looking at the figure.
The area of $ABC$ is $ \frac 1 4 $, so $B$ must be $\frac 1 4 - C - A = \frac 5 {48}$.
Now the area of $BD$ must be $ \frac 1 8$. It therefore follows that $D = \frac 1 8 - \frac 5 {48} = \frac 1 {48}$.
The trigonometry part just seems too elaborate, and I was wondering if there is a much more simple solution I am missing?
| Coordinate geometry can also work out area $D$. If the bottom left corner is at the origin, and the square has a side length of $1$, the equations of the three lines are:
$$y = 1-x \tag{1}$$
$$y = \frac{1}{2} + \frac{1}{2}x \tag{2}$$
$$x = \frac{1}{2} \tag{3}$$
Solving $(1) = (2)$ gives the point $\left(\frac{1}{3}, \frac{2}{3} \right)$, solving $(2) = (3)$ gives $\left(\frac{1}{2}, \frac{3}{4} \right)$, and solving $(3) = (1)$ gives $\left(\frac{1}{2}, \frac{1}{2} \right)$.
Then the area of the triangle is $\frac{1}{2} \times \text{base} \times \text{height}$:
$$\frac{1}{2} \times \left(\frac{3}{4} - \frac{1}{2} \right) \times \left(\frac{1}{2} - \frac{1}{3} \right) $$
$$=\frac{1}{48}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3435395",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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condition for pair of straight line equation While determining the condition for the pair of straight line equation $$ax^2+2hxy+by^2+2gx+2fy+c=0$$ i.e $\quad$$ax^2+2(hy+g)x+(by^2+2fy+c)=0$
$$x=\frac{-2(hy+g)}{2a}\pm\frac{\sqrt{(hy+g)^2-a(by^2+2fy+c)}}{a}
$$$$x=\frac{-2(hy+g)}{2a}\pm\frac{\sqrt{(h^2-ab)y^2+2(hg-af)y+(g^2-ac)}}{a}$$
The terms inside square root need to be a perfect square. I understand this. What I do not understand is when the inside square root terms ,quadratic in y is taken to be zero. Because of which its determinant $4(hg-af)^2-4(h^2-ab)(g^2-ac)=0$ becomes the condition for the pair of straight line equation.
I am stuck here. Can someone please help. Thanks.
| If the condition
$$
a x^2 + 2 h x y + b y^2 + 2 g x + 2 f y + c = 0
$$
represents the product of two lines then the set of solutions for this conditions should be one solution point, void or infinite solution points associated to the cases in which we have the two lines intersection, two lines parallel and two lines coincident. With this idea we follow obtaining
$$
x = \frac{gh-af\pm2\sqrt{(g+hy)^2-a(b y^2+2fy+c)}}{2a}
$$
If the intersection point is unique then the condition is
$$
(g + h y)^2 - a (c + 2 f y + b y^2) = 0
$$
which gives
$$
y = \frac{gh-af\pm\sqrt{a^2 f^2 + a b g^2 - 2 a f g h + a c h^2-a^2 b c}}{ab-h^2}
$$
but again if the intersection point is unique we should have
$$
a^2 f^2 + a b g^2 - 2 a f g h + a c h^2-a^2 b c = 0
$$
or
$$
c = \frac{a f^2+b g^2-2 f g h}{a b-h^2}
$$
The condition for distinct parallel lines follow as
$$
a b -h^2 = 0
$$
Another approach:
Assuming $a \ne 0$ and dividing $ax^2+2hxy+by^2+2gx+2fy+c=0$ by $a$ we have
$$
x^2+b' y^2 + c' + 2 f' y + 2 g'x +2 h' xy = (x+c_1 y + c_2)(x+d_1 y + d_2)
$$
after equating coefficients and solving for $c_1,c_2,d_1,d_2$ we have the conditions
$$
\cases{h'^2-b' > 0\\
c' = -\frac{f'^2-b' g'^2-2f' g' h'}{h'^2-b'}}
$$
such that the two lines equivalence is feasible.
or equivalently
$$
\cases{
h^2-a b > 0\\
c = \frac{a f^2+b g^2-2f g h}{a b -h^2}
}
$$
| {
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"source": "stackexchange",
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Find x and y such that $xx...x6yy...4=k^2$. Find x and y such that $N=xx...x6yy...4=k^2$, where number of x and y is n.
Solution:
Supose n=1 and N=x6y4 and we must have:
$(x6y)\times 10 +4=k^2$
⇒ $(x6y)\times 10=(k-2)(k+2)$
That is the last digit of (k+2) or (k-2) must be zero, let $k+2 ≡0 \ mod 10$ ⇒ $k ≡8 \mod 10$
Let $k= a8$ , we can see that if $a=6$ then:
$68^2=4624$ ⇒ $x=4, y=2$ and $66...8^2=44...622...4$
Also if a=9 then:
$98^2=9609$ ⇒ $x=9, y=0$ and $99...8^2=99...600...4$
But this seems to be a sort of lucky solution. Can someone gives a more reasonable algorithm?
| This is an interesting problem but so far I only have a partial answer.
$k$ must be of the form $10l+2$ or $10l+8$. Let's look at the second of these possibilities.
$$(10l+8)^2=\frac {10^n-1}{9}10^{n+2}x+\frac {10^n-1}{9}10y +6\times 10^{n+1}+4$$
$$100l^2+160l+64=\left(\frac {10^n-1}{9}\right)\left (10^{n+2}x+10y+540\right )+64 $$
$$l(10l+16)=\left(\frac {10^n-1}{9}\right)\left (10^{n+1}x+y+54\right )$$
Now consider solutions where $l=\frac {10^n-1}{9}t$ for some integer $t$. (However there could be other solutions).
We now have $t(10l+16)=10^{n+1}x+y+54$ and so $$(10^n-1)10t^2+144t=10^{n+1}9x+9y+486$$
$$10^{n+1}(t^2-9x)y=10t^{2}-144t+9y+486.$$
Now $x$ and $y$ are digits and for reasonably large $n$ we must have $t^2=9x$. The possibilities are
$x=1,t=3$. Then $y$ is negative.
$x=4,t=6$. Then $y=2.$
$x=t=9$. Then $y=0.$
We thus have the two solutions that you know.
| {
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Factorial Digits Sum
A natural number is a factorion if it is the sum of the factorials of
each of its decimal digits. For example, $145$ is a factorion because
$145 = 1! + 4! + 5!$. Find every $3$-digit number which is a factorion.
My solution:
We can obviously only have factorials $1$–$6$.
*
*If we have a $6$ factorial, our number would need to start with $6$ to be the biggest possible, but that is still too small. Doesn't work.
*If we have a $5$ factorial, since that's $120$, we need our number to start with $1$. $5!+1!=121$. We need a $1\_\_$. If after the $1$ we have a $5$, we can do the casework.
*
*$151$ does not work.
*Neither does $152$.
*Neither does $153$.
*Neither does $154$.
*But we did the computations for $151$, $152$, $153$, and $154$ to find this, and we notice that $154$'s answer is $145$. So, $145$ works, and we quickly see there are no others, so only $145$ works.
How can we generalize this?
| Brainstorming:
As you pointed out. If one of the digits of $N$ is $7$ then $7! > 1000> N$ which is impossible.
If one of the digits is $6$ then $N > 6! = 720$ so one of the digits is at least $7$ which we just showed was impossible.
Also if all three digits, call them $a,b,c$, are all $< 5$ then $a!, b!,c! \le 4! =24$ and $N = a! + b!+c! \le 3*24 = 72 < 100$ so that's impossible. So there must be at least one digit that is equal to $5$.
So one of the digits is $5$. Then if the other two are $a,b$ then $a,b \le 5$ and $N = a! + b! + 5! \le 5! + 5! + 5! = 360$. So the first digit is $3$ or less.
So $N = a! + b! +5! \le 3! + 5! + 5! = 246$ so the first digit is $2$ or less.
If the first digit, let's assume the first digit is $a$ is equal $2$ then $200 < N = 2! + b! + 5! = b! + 122$ so $b! > 78$ so $b >4$ but $b \le 5$ so $b=5$ and $N = 255$ but $2! + 5! + 5! = 242$ so $255$ is not factorial. So the first digit is less than $2$.
But $N > 100$ so the first digit is $1$.
So $N = 1! + b! + 5! = 126 +b!$. Where $b = 0,1,2,3,4,5$ and I guess at this point we can do trial and error.
But we have $100 < N < 200$ and $N = 126+b!$ so $b! < 74$ for $b \le 4$. So $N=126 +b! \le 126 + 24 = 150$. And $1! + 5! + 0! = 122\ne 150$ so $150$ is not factorial so $N \le 149$. but as one of the digits is $5$ that must be the last digit and $N = 100 + 10b + 5$.
So $N = 105 + 10b = 1! + b! + 5! = 121 + b!$ so $10b = b! + 16$. And we know $N>121$ so $b \ge 2$ and $b \le 4$ and $b! + 16 \equiv 0 \pmod {10}$ and.... how much further can I go without saying, look, $b$ has got to be $4$; that's the only option?
Well a little further.... just to be masochistic. $10b = 20,30$ or $40$ so $b! = 4,14$ or $24$ and $b = 2,3,4$ so either $2! = 4$, $3!=14$ or $4! = 24$ and ....
That's as far as I can go.
$b =4$ and $N = 100 + 40 + 5 = 1! + 4! +5!$.
| {
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Solve the non-linear congruence $x^3+2x^2+5x+4\equiv 0\pmod {60}$. Solve the non-linear congruence $x^3+2x^2+5x+4\equiv 0\pmod {60}$.
For this question, I think I might need to use the Chinese remainder Theorem to simplify the problem to several smaller moduli (e.g. possibly $\pmod 4,\pmod 5,\pmod 3$ and simplify the $x^3+2x^2+5x$ terms).
| Modulo $4$ we have $x^3+2x^2+x=x(x+1)^2\equiv0$,
which means $x\equiv0$ or $x+1\equiv2$ or $4$.
Modulo $5$ we have $x^3+2x^2+5x+4=(x+1)(x^2+x+4)\equiv(x+1)(x-2)^2\equiv0$,
which means $x\equiv 4$ or $2$.
Modulo $3$ we have $x^3+2x^2+5x+4=(x+1)(x^2+x+4)\equiv(x+1)(x-1)^2\equiv0$,
which means $x\equiv 2$ or $1$.
Can you take it from here with the Chinese remainder theorem
to find $12$ solutions modulo $60=4\times5\times3$?
| {
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Find all positive integers $x, y, z$ so that $(x+2y)(y+2z)(z+2x)$ is equal to prime power Find all positive integers $x, y, z$ so that $(x+2y)(y+2z)(z+2x)$ is equal to prime power.
My try
Since $x, y, z$ are positive integers it follows that $x+2y, y+2z, z+2x \geq 3$. We have following system of equations:
$$\begin{align}
x+2y &= p^r \tag{1}\label{eq1} \\
y+2z &= p^s \tag{2}\label{eq2} \\
z+2x &= p^t \tag{3}\label{eq3}
\end{align}$$
Where $p$ is prime number and $r, s, t$ are positive integers.
Solving this system of equations for $x, y, z$ gives us $x=\frac{4p^t-2p^s+p^r}{9}$, $y=\frac{4p^r-2p^t+p^s}{9}$, $z=\frac{4p^s-2p^r+p^t}{9}$. Since $x, y, z$ are integers following equations must hold:
$$\begin{align}
4p^t-2p^s+p^r &\equiv 0 \pmod 9 \\
4p^r-2p^t+p^s &\equiv 0 \pmod 9 \\
4p^s-2p^r+p^t &\equiv 0 \pmod 9
\end{align}$$
I don't know how to continue from there. Can somebody help?
| Suppose we have a solution $(x, y, z)$. Note that if $d = \gcd(x, y, z) > 1$, then $d$ must also be a power of $p$, and we get a smaller solution $(x', y', z') = (x/d, y/d, z/d)$ with $\gcd(x', y', z') = 1$.
Thus without loss of generality we can assume our solution has $\gcd(x, y, z) = 1$. Now, if $p \neq 3$, then by the formulas $x = (4p^t - 2p^s + p^r)/9$, etc., we must have that $x, y, z$ are divisible by $p$, contradicting $\gcd(x, y, z) = 1$. It follows that there can only be solutions in the case $p = 3$.
Now suppose without loss of generality that $z$ is the largest of $x, y, z$. Again write
$$x + 2y = 3^r \qquad y + 2z = 3^s \qquad z + 2x = 3^t.$$
Since $y \leq z$, we have $3^s \leq 3z$, and similarly since $x \leq z$, we have $3^t \leq 3z$. But then $3^{s-1} \leq z < z + 2x = 3^t$, so $s \leq t$, and conversely, $2(3^{t-1}) \leq 2z < y + 2z = 3^s$, so $t \leq s$, and thus we must have $s = t$. This gives $y + 2z = z + 2x$, hence $z = 2x - y \leq 2x$, so $3^t \leq 4x$, and $x \geq 3^t/4$. It follows that $3^r = x + 2y > 3^t/4 > 3^{t-2}$, so $r \geq t-1$. In the other direction, since $z-x=x-y$ and $z \geq x$, we have $x \geq y$, hence $3^r = x + 2y \leq z + 2x = 3^t$, so $r \leq t$.
Thus our only solutions for $(r, s, t)$ are of the forms $(t, t, t)$ and $(t-1, t, t)$. In the case $(t, t, t)$, it immediately follows that $x = y = z$, hence $(x, y, z) = (1, 1, 1)$. Solving the second case for $x, y, z$ gives
$$(x, y, z) = (3^{t-3}(12-6+1), 3^{t-3}(4-6+3), 3^{t-3}(12-2+3)) = (3^{t-3} \cdot 7, 3^{t-3}, 3^{t-3} \cdot 13)$$
and since we assumed $\gcd(x, y, z) = 1$, our solution is $(7, 1, 13)$.
Now relaxing the assumption that $z$ was the largest, so we allow for cyclic shifts, the solutions with $\gcd(x, y, z) = 1$ are exactly $(1, 1, 1), (7, 1, 13), (1, 13, 7), (13, 7, 1)$. Relaxing the other assumption that $\gcd(x, y, z) = 1$, the general solutions are exactly all scalings of these primitive solutions by a power of $3$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3448343",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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What are the values of the parameters $a,b \in \mathbb{R}$ such that $\lim\limits_{x \to \infty}(\sqrt{x^2+x+1}+\sqrt{x^2+2x+2}-ax-b)=0.$ I have to find the values of $a$ and $b$ (with $a,b \in \mathbb{R}$) such that the following is true:
$$\lim\limits_{x \to \infty} (\sqrt{x^2+x+1} + \sqrt{x^2+2x+2}-ax-b)=0$$
This is what I did:
$$\lim\limits_{x \to \infty} (\sqrt{x^2+x+1} + \sqrt{x^2+2x+2}-ax-b)=0$$
$$\lim\limits_{x \to \infty} \bigg (x\sqrt{1+\dfrac{1}{x}+\dfrac{1}{x^2}} + x\sqrt{1+\dfrac{2}{x}+\dfrac{2}{x^2}}-ax-b \bigg )=0$$
$$\lim\limits_{x \to \infty} x\bigg (\sqrt{1+\dfrac{1}{x}+\dfrac{1}{x^2}} + \sqrt{1+\dfrac{2}{x}+\dfrac{2}{x^2}}-a-\dfrac{b}{x} \bigg )=0$$
So, we'd have something like:
$$\infty \cdot(2-a) = 0$$
$(*)$If $a \in (-\infty, 2)$ $\Rightarrow$ $(2-a) > 0$, which means:
$$\infty \cdot(2-a) = \infty$$
$(*)$If $a\in (2, +\infty)$ $\Rightarrow$ $(2-a) < 0$, which means:
$$\infty \cdot(2-a) = -\infty$$
So the only option left for the limit to have any chance of being true is:
$$a=2$$
which would result in the indeterminate form:
$$\infty \cdot (2-a) = \infty \cdot 0$$
And now that we have $a=2$, we need to find the value of $b$ for the limit to be true. Substituting $a$ with $2$ in the initial limit we get:
$$\lim\limits_{x \to \infty} (\sqrt{x^2+x+1} + \sqrt{x^2+2x+2}-2x-b)=0$$
Since $b$ is a constant we can pull it out of the limit and get:
$$b=\lim\limits_{x \to \infty} (\sqrt{x^2+x+1} + \sqrt{x^2+2x+2}-2x)$$
And this is where I got stuck. I tried a bunch of methods and tricks for finding this limit and I got nowhere. That leads me to think that either I made some mistake/mistakes along the way, or I simply don't know how to solve this limit. So, how can I find $b$?
| Hint: Intuitively, for large $x$, we have $\sqrt{x^2+x+1} = \sqrt{(x+\tfrac{1}{2})^2+\tfrac{3}{4}} \approx x+\tfrac{1}{2}$ and $\sqrt{x^2+2x+2} = \sqrt{(x+1)^2+1} \approx x+1$. But how can we make these observations rigorous?
Using the identity $\sqrt{a}-\sqrt{b} = \dfrac{a-b}{\sqrt{a}+\sqrt{b}}$, we get that$$\sqrt{x^2+x+1} - (x+\tfrac{1}{2}) = \dfrac{(x^2+x+1)-(x+\tfrac{1}{2})^2}{\sqrt{x^2+x+1} + (x+\tfrac{1}{2})} = \dfrac{\tfrac{3}{4}}{\sqrt{x^2+x+1} + (x+\tfrac{1}{2})}$$ and $$\sqrt{x^2+2x+2} - (x+1) = \dfrac{(x^2+2x+2)-(x+1)^2}{\sqrt{x^2+2x+2} + (x+1)} = \dfrac{1}{\sqrt{x^2+2x+2} + (x+1)}.$$ Both of these expressions tend to $0$ as $x \to \infty$. Do you see how to use these results to solve the problem?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3448476",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Proving HM-GM inequality using AM-GM I want to prove
$$\frac{n}{\frac{1}{a_1} + \cdots + \frac{1}{a_n}} ≤ \sqrt[n]{a_1\cdots a_n}$$
Using AM-GM inequality.
My attempt:
Consider arbitrary $x_1,\cdots,x_n$ We know that
$$\sqrt[n]{x_1\cdots x_n} ≤ \frac{x_1 + \cdots + x_n}{n}$$
Replacing all $x$ with $\frac{1}{x}$
$$\begin{align}\sqrt[n]{\frac{1}{x_1}\cdots \frac{1}{x_n}} ≤ \frac{\frac{1}{x_1} + \cdots + \frac{1}{x_n}}{n} & \implies \frac{1}{x_1}\cdots \frac{1}{x_n} ≤ \biggl(\frac{\frac{1}{x_1} + \cdots + \frac{1}{x_n}}{n}\biggr)^{n} \\
& \implies \frac{1}{\frac{1}{x_1}\cdots \frac{1}{x_n}} ≥ \frac{1}{\biggl(\frac{\frac{1}{x_1} + \cdots + \frac{1}{x_n}}{n}\biggr)^{n}} \\
& \implies x_1\cdots x_n ≥ \frac{1}{\biggl(\frac{\frac{1}{x_1} + \cdots + \frac{1}{x_n}}{n}\biggr)^{n}} \\
& \implies \sqrt[n]{x_1\cdots x_n} ≥ \frac{1}{\frac{\frac{1}{x_1} + \cdots + \frac{1}{x_n}}{n}} \\
& \implies \sqrt[n]{x_1\cdots x_n} ≥ \frac{n}{\frac{1}{x_1} + \cdots + \frac{1}{x_n}}
\end {align}$$
$\Box$
Is it correct? Are there other (better) alternatives?
| a)Yes, it is correct.
b) A better method would be to take reciprocals in the first step, as both sides are positive.
$$\sqrt[n]{\frac{1}{x_1}\cdots \frac{1}{x_n}} \leq \frac{\frac{1}{x_1} + \cdots + \frac{1}{x_n}}{n}$$
$$\iff \sqrt[n]{x_1x_2\cdots x_n}\geq\frac{n}{\frac{1}{x_1} + \cdots + \frac{1}{x_n}}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3448960",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Triangle sides $a,b,c$ are in arithmetic progression. Show $\sin^2(A/2)\csc2A$, $\sin^2(B/2)\csc2B$, $\sin^2(C/2)\csc2C$ are in harmonic progression
If sides $a,b,c$ of the triangle ABC are in arithmetic progression, prove that $\sin^2\frac{A}{2}\mathrm{cosec}(2A),\sin^2\frac{B}{2}\mathrm{cosec}(2B),\sin^2\frac{C}{2}\mathrm{cosec}(2C)$ are in harmonic progression.
My attempt is as follows:-
$$T_1=\dfrac{1-\cos A}{2\sin2A}$$
$$T_1=\dfrac{1}{4}\dfrac{\sec A-1}{\sin A}$$
$$T_1=\dfrac{2R}{4}\left(\dfrac{\dfrac{2bc}{b^2+c^2-a^2}-1}{a}\right)$$
$$T_1=\dfrac{R}{2}\left(\dfrac{\dfrac{2bc-b^2-c^2+a^2}{b^2+c^2-a^2}}{a}\right)$$
$$T_1=\dfrac{R}{2}\left(\dfrac{\dfrac{a^2-(b-c)^2}{b^2+c^2-a^2}}{a}\right)$$
$$T_1=\dfrac{R}{2}\left(\dfrac{(a+c-b)(a+b-c)}{(b^2+c^2-a^2)a}\right)$$
By symmetry, we can say $T_2=\dfrac{R}{2}\left(\dfrac{(b+c-a)(b+a-c)}{(a^2+c^2-b^2)b}\right)$
$T_3=\dfrac{R}{2}\left(\dfrac{(c+a-b)(c+b-a)}{(a^2+b^2-c^2)c}\right)$
For $T_1,T_2,T_3$ to be in HP, $\dfrac{1}{T_1},\dfrac{1}{T_2},\dfrac{1}{T_3}$ should be in A.P
$$\dfrac{1}{T_1}+\dfrac{1}{T_3}-\dfrac{2}{T_2}=\dfrac{2}{R}\left(\dfrac{(b^2+c^2-a^2)a}{(a+c-b)(a+b-c)}+\dfrac{(a^2+b^2-c^2)c}{(c+a-b)(c+b-a)}-\dfrac{2(a^2+c^2-b^2)b}{(b+c-a)(a+b-c)}\right)$$
$$\dfrac{1}{T_1}+\dfrac{1}{T_3}-\dfrac{2}{T_2}=\dfrac{2}{R}\left(\dfrac{a(b^2+c^2-a^2)(b+c-a)+c(a^2+b^2-c^2)(a+b-c)-2b(a^2+c^2-b^2)(c+a-b)}{(b+c-a)(a+b-c)(c+a-b)}\right)$$
$$\dfrac{1}{T_1}+\dfrac{1}{T_3}-\dfrac{2}{T_2}=\dfrac{2}{R}\left(\dfrac{a(b^3+b^2c-ab^2+bc^2+c^3-ac^2-a^2b-a^2c+a^3)+c(a^3+a^2b-a^2c+b^2a+b^3-b^2c-c^2a-c^2b+c^3)-2b(a^2c+a^3-a^2b+c^3+ac^2-bc^2-b^2c-ab^2+b^3)}{(b+c-a)(a+b-c)(c+a-b)}-\right)$$
$$\dfrac{1}{T_1}+\dfrac{1}{T_3}-\dfrac{2}{T_2}=\dfrac{2}{R}\left(\dfrac{a^4+c^4-2b^4+ab^3-a^3b+ac^3-a^3c+a^3c+b^3c-ac^3-bc^3+ab^2c+abc^2+a^2bc+ab^2c-2a^2bc-2abc^2\cdot\cdot}{(b+c-a)(a+b-c)(c+a-b)}\right)$$
It was getting very difficult to solve from here, is there any other method in which we can solve this question?
| The arithmetic progression of sides $b-a = c-b$ leads to
$$\sin B - \sin A = \sin C - \sin B\tag 1$$
With $A+B+C=\pi$,
$$\sin\frac{C}2\sin\frac{B-A}2=\sin\frac{A}2\sin\frac{C-B}2\implies
\frac{\sin\frac{B-A}2}{\sin\frac A2}
= \frac{\sin\frac{C-B}2}{\sin\frac C2}\tag 2$$
Next, let $I_X = {\sin^2\frac{X}{2}\csc2X}$ to examine
$$\frac1{I_{B}}-\frac1{I_A}=\frac{\sin2B}{\sin^2\frac{B}{2}}-\frac{\sin2A}{\sin^2\frac{A}{2}}
=4\cdot\frac{\cos B\cos\frac{B}{2}\sin\frac{A}{2}-\cos A\cos\frac A2\sin\frac{B}{2}}{\sin\frac{B}{2}\sin\frac{A}{2}}$$
Use the identity $\cos x = 1-2\sin^2\frac x2$ to simplify,
$$\frac14 (\frac1{I_{B}}-\frac1{I_A})= \frac{\sin\frac{A-B}{2}}{\sin\frac{B}{2}\sin\frac{A}{2}}
-(\sin B - \sin A)\tag 3$$
Likewise,
$$\frac14 (\frac1{I_{C}}-\frac1{I_B})= \frac{\sin\frac{B-C}{2}}{\sin\frac{C}{2}\sin\frac{B}{2}}
-(\sin C - \sin B)\tag 4$$
Apply the results (1)-(2) to (3)-(4) to obtain,
$$ \frac1{I_{B}}-\frac1{I_A}=\frac1{I_{C}}-\frac1{I_B}$$
Hence, $I_A $, $I_{B}$ and $I_{C}$ are in harmonic progression.
| {
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"url": "https://math.stackexchange.com/questions/3450752",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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How many numbers from $1$ to $1000$ can be written as the sum of $4$s and $5$s?
I like the numbers $4$ and $5$. I also like any number that can be added together using $4$s and $5$s. Eg,
$$9 = 4+5 \qquad 40 = 5 + 5 + 5 + 5 + 5 + 5 + 5 +5$$
How many number have this property from 1 to 1000?
Multiples of $4$s and $5$s are easy, but how do I calculate the number of numbers from different combinations of adding $4$ and $5$? (And which ones are different from multiples of $4$ and $5$?)
| Google frobenius coin problem.
But notice. If $K = 4m + 5n$ is such number then $K+4 = 4(m+1)+5n$ is such a number and all $K + 4a = 4(m+a) + 5n$ will be such numbers.
$12 = 3*4$ and $13=2*4 + 5$ and $14=4+2*5$ and $15=3*5$. So every $12 + 4a$ and $13+4b$ and $14+4b$ and $15+4c$ will be such numbers And that is every number greater than or equal to $12$.
The hard part is finding out that $11$ is the largest that can't be done.
ANd then counting that the ones that are less than $11$ than can be done are $4; 8=2*4; 5;9=5+4;10=2*5$ and all the others $1,2,3,6,7,11$ cant be done.
So that is $6$ that can't be done and all the rest that can. So $994$.
.....
Another way.... harder, but more intuitive for me....
If $N = 4k + r$ where $r=0,1,2,3$ is the remainder. We can do $N = 4k+r = 4k-4r + 5r = 4(k-r)+5r$ provided that $k \ge r$.
So if $k=0$ then we can't do this. If $k=1$ then if $r \le 1$, i.e. if $N=4*1+0 =4$ or $N = 4*1 + 1 = 0*1 + 5$, we can do this. but we can't do this if $k=1$ and $r=2,3$ i.e. if $N =4*1 +2=6$ and $N = 4*1 + 3 = 7$.
In $k=2$ and $r\le 2$ we can do this. $N=4*2+ 0 =8; N=4*2 + 1=4*1 + 5 = 9$; and $N=4*2 +2 = 4*0 + 2*5$. But if $r=3$ we can not; $N=4*2+3=4*1 +5*1+2 = 4*0 +5*2+1$ can not be done.
But if $k \ge 3$ and $r \le 3$ we can do it and that is the case for all $N \ge 12$.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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prove: $\nabla \cdot \frac{\vec{r}}{r^3} = 0$ prove:
$\nabla \cdot \frac{\vec{r}}{r^3} = 0$
a few useful identities for proof:
$\vec{r} \cdot \vec{r} = 1$
$\nabla r^n = n r^{n-2} \vec{r}$
$\nabla \cdot (\vec{v} + \vec{w}) = \nabla \cdot \vec{v} + \nabla \cdot \vec{w}$
$\nabla \cdot (A \vec{v}) = (\nabla A) \cdot \vec{v} + A (\nabla \cdot \vec{v})$
$r = (x^2 + y^2 + z^2)^{-\frac{1}{2}}$
$\vec{r} = x\hat{\text{i}} + y\hat{\text{j}} + z\hat{\text{k}}$
ok...so i attempt this proof as follows:
$\nabla \cdot \frac{\vec{r}}{r^3} = \nabla ( r^{-3}~ \vec{r})$
by product rule:
$\nabla \cdot \frac{\vec{r}}{r^3} = (\nabla r^{-3}) \cdot \vec{r} + r^{-3}(\nabla \cdot \vec{r})$
by one of the identities that kind of looks like a derivative rule:
$\nabla \cdot \frac{\vec{r}}{r^3} = -3r^{-5}\vec{r} \cdot \vec{r} + r^{-3}(\nabla \cdot \vec{r})$
then because $\vec{r} \cdot \vec{r} = 1$:
$\nabla \cdot \frac{\vec{r}}{r^3} = -3r^{-5} + r^{-3}(\nabla \cdot \vec{r})$
At this point i get stuck trying to get this expression to equal zero...
| Solution : $$\nabla \cdot\frac{\vec r}{r^3}$$
$$=(\nabla \cdot \vec r)\frac{1}{r^3} + \vec r \cdot \left( \nabla \frac{1}{r^3} \right)$$
where, $\nabla \cdot(\phi \vec A)=(\nabla \cdot \vec A) \phi + \vec A \cdot (\nabla \phi)$
$$=\frac{3}{r^3}+\vec r \cdot [ (-3)r^{-3-2}\vec r ] $$
where , $ \nabla r^n=nr^{n-2} \vec r$
$$=\frac{3}{r^3}+\frac{-3}{r^3}=0$$
| {
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"url": "https://math.stackexchange.com/questions/3453680",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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$k=-\sqrt{3}(\sqrt{5+2\sqrt{6}}+\sqrt{5-2\sqrt{6}})$
Let $k$ be equal to:
$$k=-\sqrt{3} \left(\sqrt{5+2\sqrt{6}}+\sqrt{5-2\sqrt{6}} \right)$$
I am trying to simplify the expression and express $5+2\sqrt{6}$ and $5-2\sqrt{6}$ as squares. I don't think it's smart to multiply the brackets by $-\sqrt{3}$ at first. Is there any algorithm that I should know to express the expressions as squares?
| Squaring gives
\begin{eqnarray*}
k&=& -\sqrt{3} \left(\sqrt{5+ 2\sqrt{6}} + \sqrt{5- 2\sqrt{6}} \right) \\
k^2&=&3 \left(5+ 2\sqrt{6}+5- 2\sqrt{6} +2\sqrt{ (5+ 2\sqrt{6})( 5- 2\sqrt{6}) } \right) \\
k^2&=&3 \left(10 +2\sqrt{ 25- 4 \times{6} } \right) \\
k^2&=& 36 \\
\end{eqnarray*}
Now square root this and we recall we expect $k$ to be negative ... so $\color{red}{k=-6}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3456305",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Evaluate the following integral $\int \frac{2x^2+x\cos x+\sin^{-1}x}{1+\sqrt{1-x^2}}$ I change the integral $\int \frac{2x^2+x\cos x+\sin^{-1}x}{1+\sqrt{1-x^2}}$ into
$$\int \frac{2x^2}{1+\sqrt{1-x^2}} + \int \frac{x\cos x}{1+\sqrt{1-x^2}} + \int \frac{\sin^{-1}x}{1+\sqrt{1-x^2}}$$
and I figure out $\int \frac{2x^2}{1+\sqrt{1-x^2}} = -\sin^{-1}x + 2x-x\sqrt{1-x^2}$
, however I don't know how to calculate the remaining integrals...
I have tried trigonometric substitution but it seems not working.
Can anyone give me some tips? Thanks a lot!
| PARTIAL ANSWER
First solve
\begin{eqnarray}
\mathcal I_1 &=& \int \frac1{1+\sqrt{1-x^2}}dx=\\
&=&\int\frac{1-\sqrt{1-x^2}}{x^2} dx=\\
&=&-\frac1x-\int\frac{\sqrt{1-x^2}}{x^2}dx.
\end{eqnarray}
Using $x=\sin t$ we get
\begin{eqnarray}
\mathcal I_1 &=&-\frac1x - \int \cot^2 t dt=\\
&=&-\frac1x + \cot t+ t+C=\\
&=&-\frac1x +\frac{\sqrt{1-x^2}}{x}+\arcsin x + C.
\end{eqnarray}
Using this result we can deal with
\begin{eqnarray}
\mathcal I_2 &=& \int\frac{\arcsin x}{1+\sqrt{1-x^2}}dx=\\
&=& \int\arcsin x d\left(-\frac1x +\frac{\sqrt{1-x^2}}{x}+\arcsin x\right)=\\
&=&-\frac{\arcsin x}x +\frac{\arcsin x\sqrt{1-x^2}}{x}+\arcsin^2 x+\\
& &- \int\left(-\frac{1}{x\sqrt{1-x^2}}+\frac1x + \frac{\arcsin x}{\sqrt{1-x^2}}\right) dx=\\
&=&-\frac{\arcsin x}x +\frac{\arcsin x\sqrt{1-x^2}}{x}+\arcsin^2 x+\\
& &-\log |x|-\frac12 \arcsin^2 x+\int\frac1{x\sqrt{1-x^2}}dx.
\end{eqnarray}
The last term above can be dealt with again with a trigonometric substitution $x=\sin t$, yielding
\begin{eqnarray}
\mathcal I_3 &=& \int\frac1{x\sqrt{1-x^2}}dx = \\
&=& \int \csc t dt=\\
&=& -\log |\cot t + \csc t| + C=\\
&=& -\log\left|\frac{\sqrt{1-x^2}}{x}-\frac1x\right|+C=\\
&=&\log|x| - \log\left|\sqrt{1-x^2}-1\right|+C.
\end{eqnarray}
Thus the last term in you original sum is
\begin{eqnarray}
\mathcal I_2&=& -\frac{\arcsin x}x +\frac{\arcsin x\sqrt{1-x^2}}{x}+\\
& &+\frac12 \arcsin^2 x- \log\left|\sqrt{1-x^2}-1\right|+C.
\end{eqnarray}
| {
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"timestamp": "2023-03-29T00:00:00",
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Mean and variance of first order statistic from a uniform distribution - min(x) So here are the first two moments about the mean and the origin of the first order statistic of a uniform random variable in the interval $[0, 1]$. Maybe someone can derive the moment generating function and post it as a answer.
CDF:
$P\left(Y_{\left(1\right)}>y\right)=1-P\left(Y_{\left(1\right)}\le y\right)=P\left(Y_1>y\right)P\left(Y_2>y\right)\ldots=\left(1-y\right)^n$
$P\left(Y_{\left(1\right)}\le y\right)=1-\left(1-y\right)^n$
PDF:
$f\left(y\right)=n\left(1-y\right)^{n-1}
$
First moment about the origin:
$E\left(Y_{\left(1\right)}\right)=\int_{0}^{1}{ny\left(1-y\right)^{n-1}dy}=n\int_{0}^{1}{y\left(1-y\right)^{n-1}dy}=n\left[-y\frac{\left(1-y\right)^n}{n}+\frac{1}{n}\int{\left(1-y\right)^ndy}\right]_0^1=-\left[y\left(1-y\right)^n+\frac{\left(1-y\right)^{n+1}}{n+1}\right]_0^1=\frac{1}{n+1}$
Second moment about the origin:
$E\left({Y_{\left(1\right)}}^2\right)=\int_{0}^{1}{ny^2\left(1-y\right)^{n-1}dy}=n\int_{0}^{1}{y^2\left(1-y\right)^{n-1}dy}=n\left[-y^2\frac{\left(1-y\right)^n}{n}+\frac{2}{n}\int{y\left(1-y\right)^ndy}\right]_0^1=-\left[y^2\left(1-y\right)^n-2\int{y\left(1-y\right)^n}dy\right]_0^1=-\left[y^2\left(1-y\right)^n-2\int{y\left(1-y\right)^n}dy\right]_0^1=-\left[y^2\left(1-y\right)^n-2\left(-\frac{1}{n+1}\left(y\left(1-y\right)^{n+1}+\frac{\left(1-y\right)^{n+2}}{n+2}\right)\right)\right]_0^1=-\left[y^2\left(1-y\right)^n+\frac{2}{n+1}\left(y\left(1-y\right)^{n+1}+\frac{\left(1-y\right)^{n+2}}{n+2}\right)\right]_0^1=\frac{2}{(n+1)(n+2)}$
Second moment about the mean
$V\left(Y_{\left(1\right)}\right)=\frac{2}{\left(n+1\right)\left(n+2\right)}-\frac{1}{\left(n+1\right)^2}=\frac{2\left(n+1\right)-\left(n+2\right)}{\left(n+2\right)\left(n+1\right)^2}=\frac{n}{\left(n+2\right)\left(n+1\right)^2}$
EDIT: I have edited this post to make it more useful other people searching it in the future. The intention now is to illustrate the mean and variance of the first order statistic of a uniform random variable.
| The derivative with respect to $y$ of $1-(1-y)^n$ is $+n(1-y)^{n-1}$ not $-n(1-y)^{n-1}$, and densities should always be non-negative
Step by step:
*
*The derivative of $y^n$ is $+ny^{n-1}$
*The derivative of $(1-y)^n$ is $-n(1-y)^{n-1}$ by the chain rule
*The derivative of $-(1-y)^n$ is $+n(1-y)^{n-1}$
*The derivative of $1-(1-y)^n$ is $+n(1-y)^{n-1}$
| {
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"url": "https://math.stackexchange.com/questions/3460727",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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How to evaluate $\lim_{\left ( x,y \right )\rightarrow \left ( 0,0 \right )}\frac{x^{2}+y^{2}}{x+y}$ I am supposed to evaluate the following limit:
$$\lim_{\left ( x,y \right )\rightarrow \left ( 0,0 \right )}\frac{x^{2}+y^{2}}{x+y}$$
My solution:
$$\lim_{\left ( x,y \right )\rightarrow \left ( 0,0 \right )}\frac{x^{2}+y^{2}}{x+y}
=\lim_{\left ( x,y \right )\rightarrow \left ( 0,0 \right )}\frac{x^{2}}{x+y}
+\lim_{\left ( x,y \right )\rightarrow \left ( 0,0 \right )}\frac{y^{2}}{x+y}\\
=\lim_{\left ( x,y \right )\rightarrow \left ( 0,0 \right )}x\cdot\frac{x}{x+y}
+\lim_{\left ( x,y \right )\rightarrow \left ( 0,0 \right )}y\cdot\frac{y}{x+y}\\
=0+0=0$$
Can someone please tell me if it is correct?
| The answer is wrong, that limit doesn't exist.
$$\lim_{\left ( x,y \right )\rightarrow \left ( 0,0 \right )}\frac{x^{2}+y^{2}}{x+y}=\lim_{\left ( x,y \right )\rightarrow \left ( 0,0 \right )}\frac{x^{2}+y^{2}+2xy-2xy}{x+y}=\lim_{\left ( x,y \right )\rightarrow \left ( 0,0 \right )}(x+y)-2\lim_{\left ( x,y \right )\rightarrow \left ( 0,0 \right )}\frac{xy}{x+y}.$$
And
$$\lim_{\left ( x,y \right )\rightarrow \left ( 0,0 \right )}\frac{xy}{x+y}$$
dosen't exist as you can see from Does $\lim \frac{xy}{x+y}$ exist at (0,0)?.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3462835",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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How can I prove analytically that the golden ratio is less then $\frac{\pi^2}{6}$. Or stated in other terms, prove that
$$\frac{1+\sqrt{5}}{2} < \sum_{n=1}^{\infty}\frac{1}{n^2}$$
| $\sum_{n=1}^\infty\frac{(-1)^{n-1}}{n^2}=\frac12\sum_{n=1}^\infty \frac1{n^2}$ (see here). Now, the inequality is equivalent to $\frac{1+\sqrt5}4<\sum_{n=1}^\infty\frac{(-1)^{n-1}}{n^2}.$
\begin{equation}
\begin{split}
\sum_{n=1}^\infty\frac{(-1)^{n-1}}{n^2}&=\sum_{k=1}^\infty\left(\frac1{(2k-1)^2}-\frac1{(2k)^2}\right)\\
&>\left(\frac1{1^2}-\frac1{2^2}\right)+\left(\frac1{3^2}-\frac1{4^2}\right)+\left(\frac1{5^2}-\frac1{6^2}\right)\\
&>\frac{1+\sqrt5}4.\\
\end{split}
\end{equation}
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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} |
How to prove $\sqrt{(a-1)(b-1)}+\sqrt{(a-1)(c-1)}+\sqrt{(b-1)(c-1)}\geq a+b+c+\sqrt{ab}+\sqrt{ac}+\sqrt{bc}$? I am trying to prove this inequality:
$$\sqrt{(a-1)(b-1)}+\sqrt{(a-1)(c-1)}+\sqrt{(b-1)(c-1)}\geq a+b+c+\sqrt{ab}+\sqrt{ac}+\sqrt{bc}$$
if there is condition $a+b+c=1$. $a, b, c$ are positive rational numbers.
I have tried it in several ways, but I can't find a solution. (I think that it may be solved using AM-GM inequality.)
What have I done: (I have done much more, but it isn't useful in this example...)
$\sqrt{(a-1)(b-1)}+\sqrt{(a-1)(c-1)}+\sqrt{(b-1)(c-1)}=\sqrt{ab+c}+\sqrt{ac+b}+\sqrt{bc+a}$
I have also done some googling, but no success. I think that it may be solved using AM-GM inequality.
I also thought, that $\sqrt{a}+\sqrt{b}+\sqrt{c}\geq a+b+c$ if the condition is true. Then we write:
$$\sqrt{ab+c}+\sqrt{ac+b}+\sqrt{bc+a} \geq a+b+c+\sqrt{ab}+\sqrt{ac}+\sqrt{bc}$$
And somehow reduce to this:
$$\sqrt{a}+\sqrt{b}+\sqrt{c}\geq a+b+c$$
which is true.
Thank you for the answer and effort!
| We assume throughout that $a,b,c$ are positive. Using the condition $a+b+c = 1$, we may write $$\sqrt{(a-1)(b-1)} + \sqrt{(b-1)(c-1)} + \sqrt{(a-1)(c-1)} = \sqrt{(b+c)(a+c)} + \sqrt{(a+c)(a+b)} + \sqrt{(b+c)(a+b)}$$ Note that $$(b+c)(a+c) = c^2 + ab + c(a+b) \ge c^2 + ab + 2c\sqrt{ab} = (c+\sqrt{ab})^2$$ using the two-variable AM-GM inequality. Thus, $$\sqrt{(b+c)(a+c)} \ge c + \sqrt{ab}$$ and summing the analogous inequalities cyclically yields $$\sqrt{(b+c)(a+c)} + \sqrt{(a+c)(a+b)} + \sqrt{(b+c)(a+b)} \ge a+b+c+ \sqrt{ab}+\sqrt{bc}+\sqrt{ac}$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Maximum value of $\frac{x}{y+1}+\frac{y}{x+1}$ while $0\leq x,y \leq 1$ What is the maximum value of $\frac{x}{y+1}+\frac{y}{x+1}$ while $0\leq x,y \leq 1$?
Wolfram Alpha plots this expression on a 3d graph, but I want to solve it algebraicly, by modifying the expression
My Attempts
1) add and substract 2 at the equation and we get $\frac{x+y+1}{y+1}+\frac{x+y+1}{x+1}$ and the numerator is same
=>failed
2) use AM-GM or Cauchy-Schwarz inequality
=>also failed
| When $ 0\le x,y \le 1$, then
$$\frac{x}{y+1} \le \frac{x}{x+y}$$
$$\frac{y}{x+1} \le \frac{y}{x+y}$$
Adding them we have
$$\frac{x}{y+1}+\frac{y}{x+1} \le \frac{x+y}{x+y}=1.$$
Equality holds when $x=0$ and $y=1$ or $x=1$ and $y=0$ and the maximum of $1$ is attained.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3475979",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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If $I_m=\int \frac{x^m}{\sqrt{ax^2+c}}\, dx$, show that $amI_m=x^{m-1}\sqrt{ax^2+c}-(m-1)c I_{m-2}$. Problem: If $I_m=\int \dfrac{x^m}{\sqrt{ax^2+c}}\, dx$, show that $amI_m=x^{m-1}\sqrt{ax^2+c}-(m-1)c I_{m-2}$.
Observe
\begin{align*}
I_m= & \int \dfrac{x^m}{\sqrt{ax^2+c}}\, dx\\
= & \dfrac{1}{a}\int x^{m-1}d(\sqrt{ax^2+c}) \\
= & \dfrac{1}{a}\left[\dfrac{x^{m-1}}{\sqrt{ax^2+c}}-\int (m-1)x^{m-2} d( \sqrt{ax^2+c})\right] \\
=& \dfrac{1}{a}\left[x^{m-1}\sqrt{ax^2+c}-(m-1)I_{m-2}\right]
\end{align*}
This implies $aI_m=x^{m-1}\sqrt{ax^2+c}-(m-1)I_{m-2}$ which is different from required result. Which one is correct?
| Your third and fourth steps are wrong. The correct steps are
$$I_m = \frac{1}{a}\Bigg[x^{m-1}\sqrt{ax^2+c}-\int(m-1)x^{m-2}\sqrt{ax^2+c}\ dx\Bigg]$$
$$=\frac{1}{a}\Bigg[x^{m-1}\sqrt{ax^2+c}-(m-1)\int \frac{x^{m-2}(ax^2+c)}{\sqrt{ax^2+c}}dx\Bigg]$$
$$=\frac{1}{a}\Bigg[x^{m-1}\sqrt{ax^2+c}-a(m-1)I_m-c(m-1)I_{m-2}\Bigg]$$
and you get the desired result in next step.
Hope it helps:)
| {
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"timestamp": "2023-03-29T00:00:00",
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Proving by induction that strictly lower triangular matrix is nilpotent
Let matrix $A \in \mathbb{K}^{n \times n}$ have the property $a_{ij} = 0$ for $1 \leq i \leq j \leq n$. Show that $A^n = 0$.
Proof by induction:
Base Case:
for $n=2:
A=
\left[ {\begin{array}{cc}
0 & 0 \\
a_{21} & 0 \\
\end{array} } \right]$
So $A^2 = \left[ {\begin{array}{cc}
0 & 0 \\
a_{21} & 0 \\
\end{array} } \right] \cdot
\left[ {\begin{array}{cc}
0 & 0 \\
a_{21} & 0 \\
\end{array} } \right] =
\left[ {\begin{array}{cc}
0 & 0 \\
0 & 0 \\
\end{array} } \right]$
Inductive Hypothesis(IH):
Assume $A^n = 0$ holds true for some $n$.
Inductive Step:
$n \rightarrow n+1$, to show: $A^{n+1} = 0 $
$A^{n+1} = A^n \cdot A =^{IH} 0 \cdot A = 0$
It seems to be too simple. Is it correct to prove this by induction?
| If you try to deduce the $n=3$ case from $n=2$ that you describe, this happens:
$$
\begin{pmatrix}
0 & 0 & 0 \\
a_{21} & 0 & 0 \\
a_{31} & a_{32} & 0
\end{pmatrix}^{3} =
\begin{pmatrix}
0 & 0 & 0 \\
a_{21} & 0 & 0 \\
a_{31} & a_{32} & 0
\end{pmatrix}^{2}
\begin{pmatrix}
0 & 0 & 0 \\
a_{21} & 0 & 0 \\
a_{31} & a_{32} & 0
\end{pmatrix}
$$
You can't conclude that $A^2 = 0$ because this matrix is $3\times 3$ and not $2\times 2$ like your previous case.
| {
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"timestamp": "2023-03-29T00:00:00",
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} |
$ydx - (4x^2y + x)dy = 0$, where is my mistake? $ydx - (4x^2y + x)dy = 0$
I found two ways to solve this equation and they give two different answers, but I couldnt find where I did mistake.
Solution 1:
Write $x$ as a function of $y$
$y\frac{dx}{dy} - (4x^2y + x) = 0$
$x' - \frac{x}{y} - 4x^2 = 0$
It is a Bernoulli equation, so divide by $x^2$ and make substitution
$z' - \frac{z}{y} = 4$
Solve first $z' - \frac{z}{y} = 0$
We get $z = yC(y)$, $z' = C(y) + C'(y)y$
Plugging back to the equation:
$C(y) + C'(y)y - C(y) = 4$
$C(y) = 4ln(y) + C$
Finally $z = y(4ln(y)+C) = \frac{1}{x}$
Solution 2
$ydx - (4x^2y + x)dy = 0$
$ydx - xdy = 4x^2ydy$
Dividing by $-x^2$ both sides we get:
$d\frac{y}{x} + d2y^2 = 0$
Integrating we get:
$\frac{y}{x} + 2y^2 = C$
So I got two solutions:
$y(4ln(y)+C) = \frac{1}{x}$ and $\frac{y}{x} + 2y^2 = C$.
My question is where did I do a mistake?
| You made a sign mistake here since $z=\frac 1x \implies z'=-\frac {x'}{x^2}$
$$z' - \frac{z}{y} = 4$$
It should be
$$z' + \frac{z}{y} =- 4 \implies z' =- \frac{z}{y} \implies z=\frac {C(y)}{y}$$
$$\implies C(y)=-2y^2+K$$
For the first one. The Bernouilli equation gives the same answer:
$$yx'-4x^2y-x=0$$
$$x'-\frac x y=4{x^2}$$
$$\frac {x'}{x^2}-\frac 1 {xy}=4$$
$$y(-\frac {1}{x})'-\frac 1 {x}=4y$$
$$\left (-\frac {y}{x} \right )'=4y$$
Integration gives:
$$-\frac {y}{x}=2y^2+K$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Why should it be $\sqrt[3]{6+x}=x$? Find all the real solutions to:
$$x^3-\sqrt[3]{6+\sqrt[3]{x+6}}=6$$
Can you confirm the following solution? I don't understand line 3. Why should it be $\sqrt[3]{6+x}=x$?
Thank you.
$$
\begin{align}
x^3-\sqrt[3]{6+\sqrt[3]{x+6}} &= 6 \\
x^3 &= 6+ \sqrt[3]{6+\sqrt[3]{x+6}} \\
x &= \sqrt[3]{6+ \sqrt[3]{6+\sqrt[3]{x+6}}} \\
\sqrt[3]{6+x} &= x \\
x^3 &= 6+x \\
x^3-2x^2+2x^2-4x+3x-6 &= 0 \\
(x-2)(x^2+2x+3) &= 0 \\
x &= 2
\end{align}
$$
| Begin as in OP $\begin{align}
x^3-\sqrt[3]{6+\sqrt[3]{x+6}} &= 6 \tag{1}\\
x^3 &= 6+ \sqrt[3]{6+\sqrt[3]{x+6}} \tag{2}\\
x &= \sqrt[3]{6+ \sqrt[3]{6+\sqrt[3]{x+6}}} \tag{3}
\end{align}$
Setting $t=\sqrt[3]{6+x}$, the equation $(3)$ rewrites
$$t^3-6=\sqrt[3]{6+\sqrt[3]{6+t}},$$ equivalent to $(1).$ Hence $t=x.$
EDIT
The functions $f:x\to x$ and $g:x\to \sqrt[3]{6+ \sqrt[3]{6+\sqrt[3]{x+6}}}\;$ are
*
*continuous on $[0,\infty)$
*increasing ($f$ is increasing trivially, $g$ is composed from increasing functions)
*$f$ is linear, $g$ is concave
*$f(0)=0<g(0), f(21)=21>g(21)$
Hence there exists unique solution to $(3).$ If $t$ and $x$ both satisfy, then they are interchangeable.
| {
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "18",
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Can it be shown, $n^4+(n+d)^4+(n+2d)^4\ne z^4$? We know, $n^4+(n+d)^4= z^4$ has no solution in positive integers $n,d,z$.
Can it be shown, $n^4+(n+d)^4+(n+2d)^4= z^4$ has no solution in positive integers $n,d,z$?
I am check upto $1\le n, d, z\le 150$ without finding a counter example.
PARI/GP
for(n=1,150,for(d=1,150,for(p=1,150,if(sum(q=0,2,(n+q*d)^4)==p^4,print([n,d,p])))))
Generalization over problem
| First, assume there's at least one solution and then let $m = n + d$, with $m$ being the smallest positive integer which works. The equation then becomes
$$\begin{equation}\begin{aligned}
z^4 & = (m - d)^4 + m^4 + (m + d)^4 \\
& = m^4 - 4md^3 + 6m^2d^2 - 4m^3d + d^4 + m^4 + \\
& \; \; \; \; m^4 + 4md^3 + 6m^2d^2 + 4m^3d + d^4 \\
& = 3m^4 + 12m^2d^2 + 2d^4
\end{aligned}\end{equation}\tag{1}\label{eq1A}$$
Next, note $a^4 \equiv 1 \pmod 3$ for all integers $a$ which aren't a multiple of $3$. The right side of \eqref{eq1A} is congruent to $2d^4$ modulo $3$, i.e., $2$ if $d$ is not a multiple of $3$, which is not possible so $d$ must be a multiple of $3$.
Thus, the right side of \eqref{eq1A} has a factor $3$, so $z$ is also a multiple of $3$, which means the left side has at least $4$ factors of $3$. If $m$ is not a multiple of $3$, then the right side has only $1$ factor of $3$ (since $12m^2d^2 + 2d^4$ has at least $3$ factors of $3$), which is not possible. Thus, $m$ must also be a multiple of $3$. This means we can divide both sides of \eqref{eq1A} by $3^4 = 81$, which is equivalent to dividing $z$, $m$ and $d$ each by $3$ to, say, the $3$ integers $z_1 = \frac{z}{3}$, $m_1 = \frac{m}{3}$ and $d_1 = \frac{d}{3}$, which then results in
$$z_1^4 = (m_1 - d_1)^4 + m_1^4 + (m_1 + d_1)^4 \tag{2}\label{eq2A}$$
Note this is of the same form as \eqref{eq1A}, but with $m_1$ being a smaller positive integer than $m$ which works. However, initially, I set $m$ to be the smallest such integer. This is a contradiction, so there's no smallest positive value for $m$ and, thus, no solution at all.
| {
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"url": "https://math.stackexchange.com/questions/3481061",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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Find value of $a_1$ such that $a_{101}=5075$ Let $\{a_n\}$ be a sequence of real numbers where $$a_{n+1}=n^2-a_n,\, n=\{1,2,3,...\}$$
Find value of $a_1$ such that $a_{101}=5075$.
I have
$$a_2=1^2-a_1$$
$$a_3=2^2-a_2=2^2-1^2+a_1$$
$$a_4=3^2-2^2+1^2-a_1$$
$$a_5=4^2-3^2+2^2-1^2+a_1$$
$$\vdots$$
$$a_{101}=100^2-99^2+98^2-97^2+\ldots+2^2-1^2+a_1.$$
Therefore, $$a_{101}=\sum_{i=1}^{50}(2i)^2-\sum_{i=1}^{50}(2i-1)^2.$$
Thus, $$5075=\sum_{i=1}^{50}(4i^2-4i^2+4i-1)+a_1,$$
and, $$a_1=5075-4\sum_{i=1}^{50}(i)+\sum_{i=1}^{50}(1).$$
Hence, $$a_1=5075-4(\frac{50}{2})(51)+50=25,$$
and $a_1=25$.
Is it correct? Do you have another way? Please check my solution, thank you.
| In the most simple way:
$$A_{n+1}+A_n=n^2~~~(1)$$ is a non-homogeneous recurrence equation. Its homogeneous part
$$A_{n+1}+A_n=0 \implies A_{n+1}=-A_n \implies A_n= (-1)^n S~~~(2)$$ In the RHS of (1) being $n^2)$, we can take $A_n =P n^2+Q n+R$; inserting this in (1), we get
$2P=1,P+Q=0,P+Q+2R=0 \implies P=1/2, Q=1/2, R=0.$ Then the total finally solution of (1) is
$$A_n=\frac{n(n-1)}{2}+ (-1)^n S. $$ Given that $A_{101}=5075$, finally we get
$$A_n=\frac{n(n-1)}{2}+(-1)^{n+1}~ 25.$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Find the maximum of a sum Given $x + y + z = 0, x + 1 > 0, y + 1 > 0, z + 4 > 0$, find the maximum of $$Q=\frac{x}{x+1} + \frac{y}{y+1} + \frac{z}{z+4}$$
The answer is $\frac{1}{3}$
I've tried inverting each of the fractions then using AM-GM on them, but it doesn't give me the correct answer
| By C-S
$$\frac{x}{x+1}+\frac{y}{y+1}+\frac{z}{z+4}=3+\frac{x}{x+1}-1+\frac{y}{y+1}-1+\frac{z}{z+4}-1=$$
$$=3-\left(\frac{1}{x+1}+\frac{1}{y+1}+\frac{4}{z+4}\right)=$$
$$=3-\frac{1}{6}(x+1+y+1+z+4)\left(\frac{1}{x+1}+\frac{1}{y+1}+\frac{4}{z+4}\right)\leq$$
$$\leq3-\frac{1}{6}(1+1+2)^2=\frac{1}{3}.$$
The equality occurs for $$(x+1,y+1,z+4)||(1,1,2)$$ or for
$$(x,y,z)=\left(\frac{1}{2},\frac{1}{2},-1\right),$$ which says that we got a maximal value.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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How many ways are there to arrange students and teachers in two lines under the given conditions? There are $24$ children and $4$ teachers. There are $2$ pairs of twins within the 24 children. We want to arrange everyone in two lines. In each line, there will be $12$ children, their order of appearance (or just order) does not matter, but there will be $1$ pair of twins in each line, there will be $2$ teachers in each line, on the sides of the line, and their order DOES MATTER. How many possibilities are for arrangement are there?
What I did is pick $10$ out $20$ children, without/ rep & without/ order * pick $1$ out of $2$ couples of twins without/ rep & without/ order, pick $2$ out of $4$ teachers WITH order and without rep, and pick $2$ out of $2$ with order and without rep. This all leads to
$$\frac{20!}{10!10!} \cdot 2 \cdot 2 \cdot 3 \cdot 2 \cdot 2$$
The answer is $$\frac{20!}{10!10!} \cdot 2 \cdot 2 \cdot 3 \cdot 2$$
Can someone explain whats wrong with my solution? thanks
| Consider the line containing a particular teacher.
There are $\begin{pmatrix}20\\10\\\end{pmatrix}$ ways of choosing the children in this line and, for each of these choices, $\begin{pmatrix}2\\1\\\end{pmatrix}$ ways of choosing the pair of twins and $\begin{pmatrix}3\\1\\\end{pmatrix}$ ways of choosing the fellow teacher. The teachers can then be arranged in $2\times2$ ways giving the answer $$\begin{pmatrix}20\\10\\\end{pmatrix}\begin{pmatrix}2\\1\\\end{pmatrix} \begin{pmatrix}3\\1\\\end{pmatrix}\times2\times2=\frac{20!}{10!10!} \cdot 2 \cdot 2 \cdot 3 \cdot 2.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3485254",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Find min and max of $A = \sqrt{x-2} + 2\sqrt{x+1} + 2019 -x$
Find minimum and maximum of $$A = \sqrt{x-2} + 2\sqrt{x+1} + 2019 -x$$
How to solve this problem without using derivatives?
Michael Rozenberg's edit:
Because with a derivative it's not so easy:
$$A'(x)=\frac{1}{2\sqrt{x-2}}+\frac{1}{\sqrt{x+1}}-1=\frac{1}{2\sqrt{x-2}}-\frac{1}{2}+\frac{1}{\sqrt{x+1}}-\frac{1}{2}=$$
$$=\frac{1-\sqrt{x-2}}{2\sqrt{x-2}}+\frac{2-\sqrt{x+1}}{2\sqrt{x+2}}=$$
$$=(3-x)\left(\frac{1}{2(1+\sqrt{x-2})\sqrt{x-2}}+\frac{1}{2(2+\sqrt{x+1})\sqrt{x+2}}\right),$$
which gives $x_{max}=3.$
| By C-S and AM-GM we obtain:
$$\sqrt{x-2}+2\sqrt{x+1}-x+2019=\sqrt{x-2}+4\sqrt{\frac{x+1}{4}}-x+2019\leq$$
$$\leq\sqrt{(1+4)\left(x-2+4\cdot\frac{x+1}{4}\right)}-x+2019=\sqrt{5(2x-1)}-x+2019\leq$$
$$\leq\frac{5+2x-1}{2}-x+2019=2021.$$
The equality occurs for $x=3,$ which says that we got a maximal value.
The minimum does not exist. Try $x\rightarrow+\infty.$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3487710",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
let $f(x)=x+\frac{1}{x} \ \ x \geq 1$ and $g(x)=x^2+4x-6$ the find Min of $g(f(x))=?$
Let $f(x)=x+\frac{1}{x}$ for all $x \geq 1$ and $g(x)=x^2+4x-6$. Find minimum of $g\circ f$.
My try:
The domain of $g\circ f$ is $[1,+\infty)$ and we have
$$g(f(x))=\left(x+\frac{1}{x}\right)^2+4\left(x+\frac{1}{x}\right)-6.$$
At $x=-2$, the minimum of $g(f(x))=-10$. Is that right?
| 1)AM-GM:
$x>0$:
$x+1/x \ge 2 \sqrt{x(1/x)}=2$;
$f(1)=2=\min ${$f(x)| x \ge 1$}.
2) $g(x)=x^2+4x-6 =$
$(x+2)^2-10$ is increasing for $x \ge -2$.
3) $g(f(x))=(f(x)+2)^2-10 \ge $
$(2+2)^2-10=6$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3490211",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
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Po-Shen Loh's new way of solving quadratic equation Quadratic equation, $ax^2+bx+c=0$ and its solution is quadratic equation, $x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$
Now setting $a=1$ then we have $x^2+bx+c=0$
$$x=\frac{-b\pm \sqrt{b^2-4c}}{2}$$ rewrite as
$$x=-\frac{b}{2}\pm \sqrt{\left(\frac{b}{2}\right)^2-c}$$
In this new video Dr. Loh claims to discover a new way of solving the quadratic equation! How? It is the same as the above formula, by using the quadratic formula, the only thing I see different, is he rewrite it in the above form!
Can someone please explain to me how this is a new way?
| Suppose that $A$ is the arithmetic mean of the roots of the quadratic $P$, and $R$ is the geometric mean [where $P(x) = ax^2+bx+c = a(x-r_1)(x-r_2)$], then we have this equation:
$$\begin{align}
P(x) &= 0\\
ax^2+bx+c &= 0\\
x^2+\frac{b}{a}x+\frac{c}{a} &= 0\\
x^2+\beta x+\gamma &= 0\\
\end{align}$$
But we can also look at it from another point of view.
$$\begin{align}
a(x-r_1)(x-r_2) &= 0\\
(x-r_1)(x-r_2) &= 0\\
x^2 -(r_1+r_2)x+r_1r_2 &= 0\\
x^2 -2\bigg(\frac{r_1+r_2}{2}\bigg)x+\sqrt{r_1r_2}^2 &= 0\\
x^2 -2Ax+R^2 &= 0\\
\end{align}$$
which can be solved via completing the square.
$$\begin{align}
x^2-2Ax+R^2 &= 0\\
(x^2-2Ax+A^2)+(R^2-A^2) &= 0\\
(x-A)^2+(R^2-A^2) &= 0\\
(x-A)^2-(A+R)(A-R) &= 0\\
\dots\\
\end{align}$$
But isn't that just Dr. Loh's method?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3490752",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
"answer_count": 8,
"answer_id": 6
} |
Evaluate $\int \frac{\sqrt{\sin^4x+\cos^4x}}{\sin^3x\cos x}dx$ $$\int \dfrac{\sqrt{\sin^4x+\cos^4x}}{\sin^3x\cos x}$$
My multiple attempts are as follows:-
Attempt $1$:
$$\int \dfrac{\sqrt{1-2\sin^2x\cos^2x}\cdot\sin x}{\sin^4x\cos x}dx$$
$$\cos x=t$$
$$-\sin x=\dfrac{dt}{dx}$$
$$\int -\dfrac{\sqrt{1-2t^2(1-t^2)}}{t(1-t^2)^2}dt$$
$$\int -\dfrac{\sqrt{2t^4+1-2t^2}}{t(1+t^4-2t^2)}dt$$
$$\int -\dfrac{\sqrt{2t^2+\dfrac{1}{t^2}-2}}{1+t^4-2t^2}dt$$
$$\int -\dfrac{\sqrt{2t^2+\dfrac{1}{t^2}-2}}{\left(\dfrac{1}{t^2}+t^2-2\right)t^2}dt$$
$$\int -\dfrac{\sqrt{\dfrac{2}{t^2}+\dfrac{1}{t^6}-\dfrac{2}{t^4}}}{\left(\dfrac{1}{t^2}+t^2-2\right)}dt$$
Not finding the way to proceed from here.
Attempt $2$:
$$\int \dfrac{\sqrt{\tan^4x+1}\cos x}{\sin^3 x}dx$$
$$\int \dfrac{\sqrt{\tan^4x+1}}{\tan^3 x\cos^2x}dx$$
$$\tan x=t$$
$$\int \dfrac{\sqrt{t^4+1}}{t^3}dt$$
$$t=\sqrt{\tan\theta}$$
$$\dfrac{dt}{d\theta}=\dfrac{1}{2\sqrt{\tan\theta}}\sec^2\theta$$
$$\int \dfrac{\sec\theta\sec^2\theta}{2\tan^2\theta}d\theta$$
$$\int \dfrac{\sec\theta}{2\sin^2\theta}d\theta$$
$$\int \dfrac{\cos\theta}{2\sin^2\theta\cos^2\theta}d\theta$$
$$y=\sin\theta$$
$$\int \dfrac{dy}{2y^2(1-y^2)}$$
$$-\dfrac{1}{2}\cdot\int\dfrac{y^2-(y^2-1)}{y^2(y^2-1)}dy$$
$$-\dfrac{1}{2}\cdot\left(\dfrac{1}{2}\cdot\ln\left|\dfrac{y-1}{y+1}\right|+\dfrac{1}{y}\right)+C$$
$$-\dfrac{1}{4}\cdot\ln\left|\dfrac{\sin\theta-1}{\sin\theta+1}\right|-\dfrac{1}{2\sin\theta}+C$$
$$-\dfrac{1}{4}\cdot\ln\left|\dfrac{t^2-\sqrt{1+t^4}}{t^2+\sqrt{1+t^4}}\right|-\dfrac{\sqrt{1+t^4}}{2t^2}+C$$
$$-\dfrac{1}{4}\cdot\ln\left|\dfrac{\tan^2x-\sqrt{1+\tan^4x}}{\tan^2x+\sqrt{1+\tan^4x}}\right|-\dfrac{\sqrt{1+\tan^4x}}{2\tan^2x}+C$$
$$-\dfrac{1}{4}\cdot\ln\left|\dfrac{1-\sqrt{\cot^4x+1}}{1+\sqrt{\cot^4x+1}}\right|-\dfrac{\sqrt{1+\cot^4x}}{2}+C$$
But real answer is $-\dfrac{\mathrm{cosec}x}{2}-\dfrac{1}{4}\cdot\ln\left|\dfrac{\mathrm{cosec}x-1}{\mathrm{cosec}x+1}\right|+C$
What am I missing here?
| Claim. The "real answer" is incorrect.
Proof. Let $\csc(x)=\frac1{\sin(x)}$ and $\sec(x)=\frac1{\cos(x)}$. Then, for the given solution at points where $\frac{\csc(x)-1}{\csc(x)+1}>0$, $$\frac{\mathrm d}{\mathrm dx}\bbox[10px,#ffd]{\frac{1}{4} \big(-2\csc (x)-\log (\csc (x)-1)+\log (\csc (x)+1)\big)}=\frac{\csc(x)}{4} \left(2 \cot (x)+\underbrace{\frac{\cot(x)}{\csc (x)-1}-\frac{\cot
(x)}{\csc (x)+1}}_{\large =2\tan(x)}\right)=\frac{\csc(x)}2(\tan(x)+\cot(x))=\csc(x)\csc(2x)$$
which does not equal your original function. $\square$
Claim. Your solution is correct.
(Constructive) proof. After $$\int \frac{\sqrt{t^4+1}}{t^3}\,\mathrm dt$$ we can continue with the substitution $u=t^{-4}$ and $\,\mathrm du=-4t^{-5}\,\mathrm dt$ to obtain $$-\frac14\int \frac{\sqrt{u+1}}u\,\mathrm du=-\frac12\int\frac{u+1}{2u\sqrt{u+1}}\,\mathrm du\overset{s=\sqrt{u+1}}=-\frac12\int\frac{s^2}{s^2-1}\,\mathrm ds=-\frac s2+\frac14\ln\left|\frac{s+1}{s-1}\right|+C$$
Since $s=\sqrt{1+t^{-4}}=\sqrt{1+\cot^4(x)}$, we get exactly your expression. $\square$
Remark. The integral can also be written as $$\frac{1}{2} \operatorname{arcsinh}\left(\tan ^2(x)\right)-\frac{1}{2} \sqrt{\tan ^4(x)+1} \cot
^2(x)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3491384",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 0
} |
Why does this pattern occur: $123456789 \times 8 + 9 = 987654321$ I came across the following:
$\begin{align} 1 \times 8 + 1 &= 9 \\ 12 \times 8 + 2 & = 98 \\ 123 \times 8 + 3 & = 987 \\ 1234 \times 8 + 4 & = 9876 \\ 12345 \times 8 + 5 & = 98765 \\ 123456 \times 8 + 6 & = 987654 \\ 1234567 \times 8 + 7 & = 9876543 \\ 12345678 \times 8 + 8 & = 98765432 \\ 123456789 \times 8 + 9 & = 987654321. \\ \end{align}$
I'm looking for an explanation for this pattern. I suspect that there is some connection to the series $\frac{1}{(1 - x)^2} = 1 + 2x + 3x^2 + \cdots$.
This post asks the same question but has no answers posted.
| This is something I noticed but I'm still thinking about if it means anything:
$$\boxed{1\cdot8+1=9}\\\downarrow$$
$$10\cdot8+10=90$$
$$10\cdot8+18=98$$
$$(10+2)\cdot8+2=98$$
$$\boxed{12\cdot8+2=98}\\\downarrow$$
$$120\cdot8+20=980$$
$$120\cdot8+27=987$$
$$(120+3)\cdot8+3=987$$
$$\boxed{123\cdot8+3=987}\\\downarrow$$
$$1230\cdot8+30=9870$$
$$1230\cdot8+36=9876$$
$$(1230+4)\cdot8+4=9876$$
$$\boxed{1234\cdot8+4=9876}\\\downarrow\\\cdot\\\cdot\\\cdot$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3492701",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "15",
"answer_count": 5,
"answer_id": 3
} |
If $a_k:=\tan( \sqrt{2} + \frac{k\pi}{2011})$, then evaluate $ \frac{a_1+a_2+\cdots+a_{2011}}{a_1a_2~\cdots~a_{2011}}$
Set $$
a_k = \tan{\left( \sqrt{2} + \frac{ k \pi }{2011} \right)}\quad\text{for}~k=1,2,3,\ldots, 2011
$$
Find the value of
$$
\frac{a_1 + a_2 + \ldots + a_{2011}}{a_1 a_2 \ldots a_{2011}}
$$
Can anyone help me in this? It's going too long.
| Using Sum of tangent functions where arguments are in specific arithmetic series,
$$\tan(2m+1)x=\dfrac{\binom{2m+1}1 t-\binom{2m+1}3t^3+\cdots+(-1)^m\binom{2m+1}{2m+1}t^{2m+1}}{\binom{2m+1}0-\binom{2m+1}2t^2+\cdots++(-1)^m\binom{2m+1}{2m}t^{2m}}$$
So, if $\tan(2m+1)x=\tan y$
$(2m+1)x=k\pi+y$ where $k$ is any integer
$x=\dfrac{k\pi+y}{2m+1}; k=1,2,\cdots,2m+1$
So, the roots of $$\tan y=\dfrac{\binom{2m+1}1 t-\binom{2m+1}3t^3+\cdots+(-1)^m\binom{2m+1}{2m+1}t^{2m+1}}{\binom{2m+1}0-\binom{2m+1}2t^2+\cdots+(-1)^m\binom{2m+1}{2m}t^{2m}}$$
$$\iff(-1)^mt^{2m+1}-\tan y(-1)^m(2m+1)t^{2m}+\cdots+\tan y=0$$ are $\tan x;x=\dfrac{k\pi+y}{2m+1}, k=1,2,\cdots,2m+1$
Can apply Vieta's formula?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3492939",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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} |
Find $\arctan x_1 \cdot \arctan x_2$, where $x_1$ and $x_2$ are roots of $x^2 - 2\sqrt{2}x + 1 = 0$. I am told that $x_1$ and $x_2$ are the roots of the following equation:
$$ x^2 - 2\sqrt{2}x + 1 = 0 $$
And I have to find the following:
$$\hspace{6cm} \arctan x_1 + \arctan x_2 \hspace{5cm} (1)$$
$$\hspace{6.cm} \arctan x_1 \cdot \arctan x_2 \hspace{5.3cm} (2)$$
Now I know that, in general:
$$\arctan(a) + \arctan(b) = \arctan \bigg ( \dfrac{a + b}{1 - ab} \bigg )$$
So $(1)$ is not that difficult to find, it actually equals $\dfrac{\pi}{2}$.
The problem is that I don't know how to find $(2)$.
| The roots of the equation are $ x_1 = \sqrt{2} + 1$ and $ x_2 =\sqrt{2} - 1 $. So we can explicitly work out $ \arctan(x_1) \, \cdot \, \arctan(x_2) = \frac{\pi}{8} \, \cdot \, \frac{3\pi}{8} = \frac{3\pi^2}{64} $
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3494497",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 3
} |
$4$ by $4$ Determinant with variables I was given this $4$ by $4$ determinant
$$
\begin{vmatrix}
x & a & a & a \\
a & x & a & a \\
a & a & x & a \\
a & a & a & x
\end{vmatrix}
= 0 $$
Clearly one of the answers is $x=a,$ how do i find the other answer?
I've tried splitting it into $4$ "$3$ by $3$" determinants but that didnt work well...
| Subtracting the 2nd row from the first, the 3rd from the second, the 4th from the third, tou obtain the determinant
$$\begin{vmatrix}
x-a&a-x&0&0\\0&x-a&a-x&0\\0&0&x-a&a-x\\ a&a&a&x
\end{vmatrix}=(x-a)^3\begin{vmatrix}
1 & -1 & 0 & 0\\ 0 & 1 & -1 & 0\\ 0 & 0 & 1 & -1\\ a&a&a&x
\end{vmatrix}$$
Then, adding successively the 1st column to the first, the 2nd to the third, the 3rd to the fourth, we get an upper triangular determinant:
$$\begin{vmatrix}
1 & -1 & 0 & 0\\ 0 & 1 & -1 & 0\\ 0 & 0 & 1 & -1\\ a&a&a&x
\end{vmatrix}=\begin{vmatrix}
1 & 0 & 0 & 0\\ 0 & 1 & 0 & 0\\ 0 & 0 & 1 & 0\\ a&2a&3a&x+3a
\end{vmatrix}=x+3a.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3495175",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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How to factorize $X^5-1$ into irreductible factors? I have to factorize the polynomial $P(X)=X^5-1$ into irreducible factors in $\mathbb{C}$ and in $\mathbb{R}$, this factorisation happens with the $5$th roots of the unity.
In $\mathbb{C}[X]$ we have $P(X)=\prod_{k=0}^4 (X-e^\tfrac{2ki\pi}{5})$.
In $\mathbb{R}[X]$ the solution states that by gathering all complex conjugate roots we find that $P(X)=(X-1)(X^2-2\cos(\frac{2\pi}{5})+1)(X^2-2\cos(\frac{4\pi}{5})+1)$, but I can't figure out how. Another problem I ran into was trying to figure out where the $2\cos(\frac{2\pi}{5})$ and $2\cos(\frac{4\pi}{5})$ come from so I tried these two methods:
The sum of the roots of unity is zero so we have:
$1+e^\tfrac{2i\pi}{5}+e^\tfrac{4i\pi}{5}+e^\tfrac{6i\pi}{5}+e^\tfrac{8i\pi}{5}=0$
In the $5$th roots of unity circle P3 and P4 are images according to the x-axis the same goes to P2 and P5, therefore $e^\tfrac{6i\pi}{5}=e^\tfrac{-4i\pi}{5}$ and $e^\tfrac{8i\pi}{5}=e^\tfrac{-2i\pi}{5}$ afterwards by using Euler's formula we find $1+2\cos(\frac{2\pi}{5})+2\cos(\frac{4\pi}{5})=0$.
Another method is that $\cos(6\pi/5) = \cos(-6\pi/5) = \cos(-6\pi/5 + 2\pi) = \cos(4\pi/5)$ $\cos(8\pi/5) = \cos(-8\pi/5) = \cos(-8\pi/5 + 2\pi) = \cos(2\pi/5)$
therefore $1 + \cos(2\pi/5) + \cos(4\pi/5) + \cos(4\pi/5) + \cos(2\pi/5) = 0$ and we find $1+2\cos(\frac{2\pi}{5})+2\cos(\frac{4\pi}{5})=0$
I don't know if both of these methods are correct on their own and I don't know if they will help in the factorisation since I don't know how to go from there and find $P(X)=(X-1)(X^2-2\cos(\frac{2\pi}{5})+1)(X^2-2\cos(\frac{4\pi}{5})+1)$
| There are various ways to render separate values of $\cos(2\pi/5)$ and $\cos(4\pi/5)$ given the equation
$1+2\cos(2\pi/5)+2\cos(4\pi/5)=0$
I demonstrate one approach. Separate out the factor of $2$ from the last two terms and apply the trigonometric sum-product relations:
$0=1+2(\cos(2\pi/5)+\cos(4\pi/5))=1+4\cos(2\pi/5)\cos(6\pi/5)$
Then $\cos(6\pi/5)=\cos(4\pi/5)$ because the arguments sum to $2\pi$. So now we have a sum and a product for $\cos(2\pi/5)$ and $\cos(4\pi/5)$:
$(\cos(2\pi/5))+(\cos(4\pi/5))=(-1/2)$
$(\cos(2\pi/5))\cdot(\cos(4\pi/5))=(-1/4)$
By Vieta's formulas for sums and products of roots these two quantities must then he roots of a quadratic equation
$4x^2+2x-1=0$
which can be solved by the usual methods, and out pop the individual cosine values you need to complete the factorization.
Incidentally, an iterated version of this technique gives radical expressions for $\cos(2\pi/p)$ for any Fermat prime $p$. The existence of this splitting process is what guarantees the constructibility of regular Fermat-prime sided polygons (but you might need a planet bigger than Earth to distinguish a regular $65537$-gon from a circle with standard drafting equipment).
| {
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"url": "https://math.stackexchange.com/questions/3496594",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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How to derive this series
The function $\dfrac1{1-x}$, equal to $$1 + x + x^2 + x^3 + \cdots,$$ can also be developed according to the series $$1 + \frac{x}{1 + x} + \frac{1\cdot2\cdot x^2}{(1 + x)(1 + 2x)} + \frac{1\cdot2\cdot3\cdot x^3}{(1 + x)(1 + 2x)(1 + 3x)} + \cdots $$ when $x$ is positive and smaller than $1$.
I know the first series and it is easy to obtain it. But the second series is strange. It is not a power series, not a Taylor series. How does one obtain this series?
| Proof of the Formula
Below, we show inductively
$$
\frac1{1-x}=\sum_{k=0}^{n-1}\frac{k!\,x^k}{\prod_{j=1}^k(1+jx)}+\frac{n!\,x^n}{\prod_{j=1}^n(1+jx)}\frac{1+nx}{1-x}\tag1
$$
where the empty sum is $0$ and the empty product is $1$.
Gautschi's Inequality says that
$$
\begin{align}
\frac{n!\,x^n}{\prod_{j=1}^n(1+jx)}
&=\frac{\Gamma(n+1)\,\Gamma\!\left(1+\frac1x\right)}{\Gamma\!\left(n+1+\frac1x\right)}\\
&\sim\frac{\Gamma\!\left(1+\frac1x\right)}{(n+1)^{1/x}}\tag2
\end{align}
$$
Thus, for $0\lt x\lt1$, the series
$$
\sum_{k=0}^\infty\frac{k!\,x^k}{\prod_{j=1}^k(1+jx)}\tag3
$$
converges and the remainder term
$$
\frac{n!\,x^n}{\prod_{j=1}^n(1+jx)}\frac{1+nx}{1-x}\tag4
$$
vanishes as $n\to\infty$. Therefore, for $0\lt x\lt1$,
$$
\bbox[5px,border:2px solid #C0A000]{\sum_{k=0}^\infty\frac{k!\,x^k}{\prod_{j=1}^k(1+jx)}=\frac1{1-x}}\tag5
$$
Inductive Proof of $\bf{(1)}$
Trivially, we have that $(1)$ is true for $n=0$.
Suppose that we have $(1)$ for some $n$. Then
$$
\begin{align}
\frac1{1-x}
&=\sum_{k=0}^{n-1}\frac{k!\,x^k}{\prod_{j=1}^k(1+jx)}+\frac{n!\,x^n}{\prod_{j=0}^n(1+jx)}\frac{1+nx}{1-x}\\
&=\sum_{k=0}^{n-1}\frac{k!\,x^k}{\prod_{j=1}^k(1+jx)}+\frac{n!\,x^n}{\prod_{j=0}^n(1+jx)}\left(\color{#C00}{1}+\color{#090}{\frac{(n+1)x}{1-x}}\right)\\
&=\sum_{k=0}^{\color{#C00}{n}}\frac{k!\,x^k}{\prod_{j=1}^k(1+jx)}+\color{#090}{\frac{(n+1)!\,x^{n+1}}{\prod_{j=1}^{n+1}(1+jx)}\frac{1+(n+1)x}{1-x}}\tag6
\end{align}
$$
Thus, $(1)$ holds for $n+1$.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "19",
"answer_count": 4,
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Calculate the maximum value of $\lfloor x\lfloor x \rfloor \rfloor + \lfloor y\lfloor y \rfloor \rfloor$.
Given negatives $x$ and $y$ such that $\left(\dfrac{x}{2} - 1\right)^2 + \left(\dfrac{y}{2} - 1\right)^2 \le \dfrac{125}{2}$. Calculate the maximum value of $$\large \lfloor x \lfloor x \rfloor \rfloor + \lfloor y \lfloor y \rfloor \rfloor$$
We could solve for the maximum value of $x^2 + y^2$.
We have that $$\left(\frac{x}{2} - 1\right)^2 + \left(\frac{y}{2} - 1\right)^2 \le \frac{125}{2} \iff \frac{x^2 + y^2}{4} - (x + y) + 2 \le \frac{125}{2}$$
$$\iff x^2 + y^2 \le 2[121 - 2(x + y)]$$
Moreover, $$\frac{(x + y)^2}{8} - (x + y) - \frac{121}{2} \le 0 \implies x + y \in (4 - 10\sqrt5, 0)$$ since $x, y < 0$.
$$\implies x^2 + y^2 \le 2[121 - 2(4 - 10\sqrt5)] = 2(113 + 20\sqrt5)$$
But I'm uncertain about the case for $x \lfloor x \rfloor + y \lfloor y \rfloor$ or even $\lfloor x \lfloor x \rfloor \rfloor + \lfloor y \lfloor y \rfloor \rfloor$.
| Well, we have the following problem:
$$\left(\frac{x}{\Delta}+\alpha\right)^2+\left(\frac{\text{y}}{\Delta}+\alpha\right)^2\le\frac{\rho}{\Delta}\tag1$$
Where all the constants are real numbers.
Now, if we multiply the LHS and the RHS with $\Delta^2$ we get:
$$\Delta^2\cdot\left(\left(\frac{x}{\Delta}+\alpha\right)^2+\left(\frac{\text{y}}{\Delta}+\alpha\right)^2\right)\le\Delta^2\cdot\frac{\rho}{\Delta}\tag2$$
Which we can rewrite as:
$$\left(x+\Delta\alpha\right)^2+\left(\text{y}+\Delta\alpha\right)^2\le\Delta\rho\tag3$$
This represents a circle with centre coordinates $\left(-\alpha,-\alpha\right)$ and radius $\sqrt{\Delta\rho}$.
| {
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"timestamp": "2023-03-29T00:00:00",
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Without using characteristic equation, how can I show genenal formula of this sequence? I know that, solution of the sequence $x_n$ so that
$$x_{n+2}-2 x_{n+1}= 2 x_{n+1} - 4 x_{n}$$
is $x_n = (A+ Bn)2^n$. I want to show this without using characteristic equation. I tried
\begin{align*}
x_{n+2}-2 x_{n+1} &= 2 x_{n+1} - 4 x_{n} \\
&= 2(x_{n+1}-2x_{n}) \\
&= 2^2(x_{n}-2x_{n-1}) \\
&= \ldots \ldots \\
&= 2^n (x_2 - 2x_1)
\end{align*}
From here, I cannot get the solution. Without using characteristic equation, how can I show general formula of this sequence?
| Use induction.
We are given a sequence $x_1,x_2,\dots$ satisfying the recursion
$$
x_{n+2}-2 x_{n+1}= 2 x_{n+1} - 4 x_{n},
$$
and we want to show that for some $A,B$ we have $x_n = (A + Bn)2^n$ for all $n$.
Select $A,B$ such that $x_n = (A + Bn)2^n$ holds for $n = 1,2$ (or note that such an $A,B$ necessarily exist). The base case of our induction (that $x_n = (A + Bn)2^n$ holds for $n = 1,2$) is now trivial. For the inductive step: suppose that $x_n = (A + Bn)2^n$ holds for all $n \leq k$ (where $k \geq 2$). We find that
$$
\begin{align}
x_{k+1} &= 2x_{k} + (x_{k+1} - 2x_{k}) =
2x_{k} + (2x_{k} - 4x_{k-1})
\\ & =
2\cdot (A + Bk)2^k + (2[(A + Bk)2^k] - 4[(A + B(k-1))2^{k-1}])
\\ &=
(A + Bk)2^{k+1} + (2[(A + Bk)2^k] - 4[(A + B(k-1))2^{k-1}])
\\ & = (A + Bk) 2^{k+1} + B\cdot 2^{k+1}
\\&= (A + B(k+1))2^{k+1}.
\end{align}
$$
We have now completed the inductive proof and reached the desired conclusion.
Using what you tried: you showed that the $x_n$ satisfies the recursion
$$
x_{n+2} = 2x_{n+1} + b\cdot 2^n, \quad n \geq 0
$$
where $b = (x_2 - 2x_1)$. By solving this first order linear recurrence (method of your choice), we find that the general solution is
$$
x_n = 2^{n-2}(bn + C)
$$
where $C$ is an arbitrary constant. The conclusion follows.
Another approach: suppose that $x_1,x_2,\dots$ is a sequence satisfying
$$
x_{n+2}-2 x_{n+1}= 2 x_{n+1} - 4 x_{n}.
$$
Define the sequence $y_1,y_2,\dots$ by $y_n = 2^{-n} x_n$. It's easy to see that
$$
y_{n+2} - y_{n+1} = y_{n+1} - y_n.
$$
Solve this to find that $y_n$ has the form $y_n = A + Bn$. The conclusion follows.
| {
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Area of the curve $y=\sin\theta \cos^2\theta$; $x=\sin^2\theta \cos\theta$ bounded by tangent parallel to axes Area bounded by tangents of the curve given by $y=\sin\theta \cos^2\theta$; $x=\sin^2\theta \cos\theta$, which are parallel to co-ordinate axes(other than axes) is
(1)$\frac{4}{27}$
(2) $\frac{27}{4}$
(3)$\frac{16}{27}$
(4) $\frac{27}{16}$
My approach is as follow $\frac{x}{y}=tan\theta$, putting the value we get $(x^2+y^2)^{\frac{3}{2}}=xy$ after this step I am not able to approach further.
| No need to eliminate the parameter. Tangents to $(x(t),y(t))$ have slope $\frac{\mathrm dy}{\mathrm dx}$. By the chain rule,
$$\frac{\mathrm dy}{\mathrm dx}=\frac{\mathrm dy}{\mathrm d\theta}\frac{\mathrm d\theta}{\mathrm dx}=\frac{\frac{\mathrm dy}{\mathrm d\theta}}{\frac{\mathrm dx}{\mathrm d\theta}}=\frac{\cos^3\theta-2\sin^2\theta\cos\theta}{2\sin\theta\cos^2\theta-\sin^3\theta}=\frac{1-2\tan^2\theta}{2\tan\theta-\tan^3\theta}$$
The tangent line is parallel to the $x$-axis when the slope is $0$:
$$\frac{1-2\tan^2\theta}{2\tan\theta-\tan^3\theta}=0\implies1-2\tan^2\theta=0\implies\tan\theta=\pm\frac1{\sqrt2}$$
$$\implies\theta=\pm\tan^{-1}\left(\frac1{\sqrt2}\right)+n\pi$$
where $n$ is any integer.
The tangent is parallel to the $y$-axis when the slope approaches $\pm\infty$, or when $\frac{\mathrm dx}{\mathrm dy}=0$:
$$\frac{2\tan\theta-\tan^3\theta}{1-2\tan^2\theta}=0\implies\tan\theta(2-\tan^2\theta)=0\implies\tan\theta=0\text{ or }\tan\theta=\pm\sqrt2$$
$$\implies\theta=2n\pi\text{ or }\theta=\pm\tan^{-1}(\sqrt2)+n\pi$$
The curve has $4$ unique tangents that are either horizontal or vertical when $\theta=\pm\tan^{-1}\left(\frac1{\sqrt2}\right)$ and $\theta=\pm\tan^{-1}(\sqrt2)$, and they have equations $y=\pm\frac2{3\sqrt3}$ and $x=\pm\frac2{3\sqrt3}$. These tangents form the boundary of a square with side lengths $\frac4{3\sqrt3}$, i.e. a square with area $\left(\frac4{3\sqrt3}\right)^2=\color{red}{\frac{16}{27}}$.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Finding the first three terms of the Laurent series near $z=0$
Let $$f(z) = \frac{1-e^z}{\sin^2z}$$
Find the first three terms of the Laurent series near $z=0$ and then evaluate $$I =\oint_{|z| = 1}f(z)dz$$
If we write $\sin^2 z = \frac{1-\cos2z}{2}$ and then try to use Taylor expansions of $e^z$ and $\cos z$, we should then divide two power series and this is really difficult by hand(this question was in an exam). Also if we do this, what's the radius of convergence? Is there any easier method for finding the first three terms of the Laurent series?
| $$e^z=1+z+\frac{z^2}2+\frac{z^3}6+\ldots\,,\;\sin z=z-\frac{z^3}6+\frac{z^5}{120}-\ldots\implies$$
$$\implies\sin^2z=z^2-\frac13z^4+\frac{2z^4}{45}\ldots\implies$$
$$\frac1{\sin^2z}=\frac1{z^2\left(1-\frac13z^2+\ldots\right)}=\frac1{z^2}\left(1+\frac{z^2}3+\frac{z^4}9+\ldots\right)\implies$$
$$\frac{1-e^z}{\sin^2z}=\left(-z-\frac{z^2}2-\frac{z^3}6-\ldots\right)\frac1{z^2}\left(1+\frac{z^2}3+\frac{z^4}{49}\ldots\right)=$$
$$=\frac1{z^2}\left(-z-\frac{z^2}2-\frac{z^3}3+\ldots\right)=-\frac1z-\frac12-\frac z2+\ldots$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3504053",
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Number of solutions via generating function How to find generating function for number of solutions
$x_1+x_2+x_3=n$ in set of positive integers such that $x_1 \ge x_2 \ge x_3$ and $x_1<x_2+x_3$.
| Hint: First rewrite $x_2 = x_3 + d_{2,3}$, with $d_{2,3} \geq 0$ and $x_1 = x_2 + d_{1,2} = x_3 + d_{2,3} + d_{1,2}$ with $d_{1,2} \geq 0$. In terms of $d_{1,2} \geq 0$, $d_{2,3} \geq 0$, $x_3 \geq 1$, we have arranged for all three variables to be (so far) independent.
Now we need to implement $x_1 < x_2 + x_3$, or, what is the same thing, $x_3 + d_{2,3} + d_{1,2} < x_3 + d_{2,3} + x_3$, which is just $d_{1,2} < x_3$. Set $x_3 = d_{1,2} + e$ with $e \geq 1$. Now we have arranged for the problem to be in terms of the independent variables $d_{1,2} \geq 0$, $d_{2,3} \geq 0$, and $e \geq 1$.
Rolling back, we see that \begin{align*}
n &= x_1 + x_2 + x_3 \\
&= (x_3 + d_{2,3} + d_{1,2}) + (x_3 + d_{2,3}) + x_3 \\
&= 3 x_3 + 2 d_{2,3} + d_{1,2} \\
&= 3(d_{1,2} + e) + 2 d_{2,3} + d_{1,2} \\
&= 4d_{1,2} + 2d_{2,3} + 3e \text{.}
\end{align*}
So the given problem is equivalent to finding a partition of $n$ into zero or more $2$s, one or more $3$s, and zero or more $4$s.
| {
"language": "en",
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"source": "stackexchange",
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If $\frac{\sin(b+\theta)\sin a}{\sin b}=\frac{\sin(c+\theta)\sin b}{\sin c}=\frac{\sin(a+\theta)\sin c}{\sin a}$, for arbitrary $\theta$, then $a=b=c$
If $$\frac{\sin(b+\theta)\sin a}{\sin b}= \frac{\sin(c+\theta)\sin b}{\sin c}= \frac{\sin(a+\theta)\sin c}{\sin a}$$ for some arbitrary constant $\theta$, then prove $a=b=c$.
I’ve tried everything, from product to sum manipulation, setting them to some $k$, to brute force expansion. But nothing seems to work.
| First, let's combine the two equations into one and eliminate the $\theta$. Take the first equation, expand $\sin(b+\theta)$ and $\sin(c+\theta)$.
$$(\sin \theta \cot b + \cos \theta)\sin a = (\sin \theta \cot c + \cos \theta)\sin b$$
$$\sin \theta (\cot b \sin a - \cot c \sin b) = -\cos \theta (\sin a - \sin b)$$
Do the same with the second equation:
$$\sin \theta (\cot c \sin b - \cot a \sin c) = -\cos \theta (\sin b - \sin c)$$
If $\sin \theta \neq 0$ and $\cos \theta \neq 0$, flip the second equation, multiply them together and cancel the $-\sin \theta \cos \theta$:
$$(\cot b \sin a - \cot c \sin b)(\sin b - \sin c) = (\cot c \sin b - \cot a \sin c)(\sin a - \sin b)$$
If $\sin \theta = 0$, then the terms without the cotangents are both zero, and if $\cos \theta = 0$, then the terms with the cotangents are both zero. So this equation is true no matter what $\theta$ is.
Expand and cancel the two like terms. The result has symmetry when the variables are cycled.
$$\sum_{a,b,c} \cos a \sin c = \sum_{a,b,c} \cot a \sin b \sin c $$
Now use the fact that $a+b+c=\pi$ to simplify this. Replace the $\sin b$ with an expression in terms of $a$ and $c$.
$$\sum_{a,b,c} \cos c \sin b = \sum_{a,b,c} \frac{\cos a \sin b \sin c}{\sin a} $$
$$\sum_{a,b,c} \cos c (\sin a \cos c + \cos a \sin c) = \sum_{a,b,c} \frac{\cos a (\sin a \cos c + \cos a \sin c) \sin c}{\sin a} $$
$$\sum_{a,b,c} (\sin a \cos^2 c + \cos a \sin c \cos c) = \sum_{a,b,c} \frac{(\sin a \cos a \sin c \cos c + \cos^2 a \sin^2 c) }{\sin a} $$
$$\sum_{a,b,c} \sin a \cos^2 c = \sum_{a,b,c} \frac{\cos^2 a \sin^2 c}{\sin a} $$
$$\sum_{a,b,c} \sin a (1 - \sin^2 c) = \sum_{a,b,c} \frac{(1 - \sin^2 a) \sin^2 c}{\sin a} $$
$$\sum_{a,b,c} \frac{\sin^2 a}{\sin a} = \sum_{a,b,c} \frac{\sin^2 c}{\sin a} $$
Let's use the Rearrangement inequality. Let the $x_i$'s be $\sin^2 a$, etc., and let the $y_i$ be $-1/\sin a$, etc. If you arrange the $x_i$'s so they increase, the corresponding $y_i$'s will increase too (because the sines are all positive). So we have equality if and only if $\sin a = \sin b = \sin c$. Because $a$, $b$ and $c$ are nonnegative and sum to $\pi$, this forces $a=b=c$.
Edit: Here’s some more detail about the Rearrangement inequality. If we have equality, then there must be at least one matching element among either the $x_i$’s or the $y_i$’s. In either case, there two variables with the same sine value. Assume that these are $b$ and $c$. So replace $c$ with $b$ in the equation, and one of the terms on each side will cancel:
$$ \frac{\sin^2 a}{\sin a} + \frac{\sin^2 b}{\sin b}
= \frac{\sin^2 b}{\sin a} + \frac{\sin^2 a}{\sin b} $$
Now apply the Rearrangement inequality again for this case, and we have $\sin a = \sin b$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3506797",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Solving differential equation by separating variables Solve the following differential equation:
$$\frac yx \frac {dy}{dx}=\sqrt {1+x^2+y^2+x^2y^2}$$
I have tried like this:
$$\frac yx \frac{dy}{dx}=\sqrt {1+x^2+y^2(1+x^2)}\\
\implies \frac yx \frac {dy}{dx}=\sqrt {(1+x^2)(1+y^2)}\\
\implies \int \frac {ydy}{\sqrt {1+y^2}}=\int \frac {xdx}{\sqrt{1+x^2}}\\
\implies \log \sqrt{1+y^2}=\log \sqrt {1+x^2}+c_1\\
\implies \frac {1+y^2}{1+x^2}=c\\$$
But in my book the answer is:
$$(1+x^2)^{3/2}-3\sqrt {1+y^2}=c$$
I can't understand how they did it,please check this..
| Hint:
Rewrite as below and integrate.
$$\left(\sqrt{1+y^2}\right)^{-1/2}d(y^2)-\left(\sqrt{1+x^2}\right)^{1/2}d(x^2)=0.$$
Mind the asymmetry in the exponents.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3508437",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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solution of fuctional equation: $f(g(x))=x^2$. For any $x \in (1,+\infty)$, $f(g(x))=x^2$, $g(f(x))=x^4$.
How to find the $f(x)$， $g(x)$ statisfying above functional equation.
Suppose inverse function existed.
The problem can be simplified to $f^2(x)=f(x^4)$.
How to find $f(x)$?
| Solving
$$
f(x)^2 = f(x^4) .
\tag{1}$$
We do not require that $f$ be continuous!
Let $f$ be arbitrary $[2,2^4) \to (1,+\infty)$. Extend recursively:
For $x \in [2^4,2^{4^2})$, let $f(x) = f(\sqrt[4]{x}\;)^2$. This means that $f$ satisfies $(1)$ for $x \in [2,2^{4})$.
For $x \in [2^{4^2}, 2^{4^3})$, let $f(x) = f(\sqrt[4]{x}\;)^2$. This means that $f$ satisfies $(1)$ for $x \in [2^4,2^{4^2})$.
Continue in this way: for $x>2^4$ define $f(x) = f(\sqrt[4]{x}\;)^2$.
In the other direction: For $x \in [2^{4^{-1}},2)$, let
$f(x) = \sqrt{f(x^4)}$. This means that $f$ satisfies $(1)$ for $x \in [2^{4^{-1}},2)$.
For $x \in [2^{4^{-2}},2^{4^{-1}})$, let
$f(x) = \sqrt{f(x^4)}$. This means that $f$ satisfies $(1)$ for $x \in [2^{4^{-2}},2^{4^{-1}})$.
Continue in this way: For $1 < x < 2$ define $f(x) = \sqrt{f(x^4)}$.
| {
"language": "en",
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"question_score": "1",
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Find $a, b$ so polynomial is divisible
Find values $a,b\in\Bbb{R}$ so the polynomial $$P(x)=6x^4-7x^3+ax^2+3x+2$$
is divisible by the polynomial
$$Q(x)=x^2-x+b$$
So what I know and how I do these problems most of the time, is since:
$$P(x)=Q(x)D(x)$$
I would know that by plugging the roots of $Q(x)$ in $P(x)$ should give me enough equations for me to solve this.
So I tried finding the roots of $Q(x)$:
$$x^2-x+b=0\\x_{1/2}={1\pm\sqrt{1-4b}\over2}$$
Okay so this $b$ value is giving me a headache here. The only thing I gathered from this is that (probably) $b\le0$. I tried now plugging this in $P(x)$, as I know that
$P(x_1)=0$ and $P(x_2)=0$
The exponents on everything made this a real pain and I'm pretty certain that it shouldn't be done this way. I'm stuck.
| Let $D(x) = 6x^2+cx+d$ (to fix the coefficient of $x^4$). Then multiply out:
$$Q(x)D(x) = (x^2-x+b)(6x^2+cx+d) = 6x^4+(c-6)x^3+(6b+d-c)x^2+(bc-d)x+bd = 6x^4-7x^3+ax^2+3x+2$$
Equate coefficients:
$$c-6=-7 \Longrightarrow c=-1$$
$$6x^4-7x^3+(6b+d+1)x^2-(b+d)x+bd = 6x^4-7x^3+ax^2+3x+2$$
$$bd = 2 \Longrightarrow d=\dfrac{2}{b} \\ b+d=-3 \Longrightarrow b+\dfrac{2}{b}=-3 \Longrightarrow b^2+3b+2=0 \Longrightarrow b\in \{-1,-2\}$$
Case 1: $b=-1$
$$d=\dfrac{2}{b}=\dfrac{2}{-1}=-2 \\ 6b+d+1 = 6(-1)+(-2)+1 = -7 = a$$
Case 2: $b=-2$
$$d=\dfrac{2}{b} = -1 \\ 6b+d+1 = 6(-2)+(-1)+1=-12 = a$$
So, the options are $(a,b) \in \{(-7,-1),(-12,-2)\}$.
| {
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"timestamp": "2023-03-29T00:00:00",
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Find the volume cut off from the sphere $x^2+y^2+z^2=a^2$ by the cylinder $x^2+y^2=ax$
Find the volume cut off from the sphere $x^2+y^2+z^2=a^2$ by the cylinder $x^2+y^2=ax$
Attempt: The projection of the Cylinder ( denoted $D$) on the $xy$ plane is a circle which has the equation: $x^2+y^2=ax ~~~\equiv~~~(x-a/2)^2+y^2=(a/2)^2 ~~~\equiv~~~~r=a \cos \theta$ ( In Polar Coordinates)
The circle has centre $(a/2,0)$ and radius $a/2$. Hence, the $y$ axis is a tangent to it.
Hence, the required volume $V = \int \int \int dv $
$= \int \int_D \int_{-\sqrt {a^2-x^2-y^2}} ^{\sqrt {a^2-x^2-y^2}}dz~~dx~dy$
$=2\int \int_D \sqrt {a^2-x^2-y^2}~~ dx dy$
Switching to Polar Coordinates :
$V=2\int_{-\pi/2}^{\pi/2} \int_{0}^{a \cos \theta} \sqrt {a^2-r^2}~~ r~dr~ d\theta=\dfrac {2}{3}a^3 \pi$
Although, the limits of $\theta$ look conceptually fine to me, my textbook uses the limits $0$ to $\pi$ and gives the result $= \dfrac {2}{3}a^3 (\pi-\dfrac{4}{3})$.
Could someone please clarify why $-\pi/2$ to $\pi/2$ usage in my expression might be incorrect?
Thanks a lot for your help!
| $V=4\displaystyle\int_{0}^{\pi/2} \displaystyle\int_{0}^{a\cos\theta}r\sqrt{{a^2}-r^2}\ drd\theta$
$=\dfrac{4a^3}{3}\displaystyle\int_{0}^{\pi/2} (1-\sin^3\theta)d\theta$
$=\dfrac{4a^3}{3}\left(\dfrac{\pi}{2}-\dfrac{2}{3}\right)=\dfrac{2a^3}{3}\left({\pi}-\dfrac{4}{3}\right)$
note: the factor of $4$ accounts for symmentry along $z$ -axis(into & outside page) and $y$-axis.see fig. below. $\theta$ is Polar angle taken from horizontal x-axis
| {
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Finding the next $x \in \mathbb{Z}^+$ such that $x^2 \equiv 17 \pmod{2^{n+1}}$ Only parts 4. and 5. are relevant, but for completion's sake, I'll just include the whole problem.
Let $N$ be any positive and let $S(N)$ be the set of remainders when the square numbers $0,1,4,9,\dots$ are divided by $N$.
*
*Prove that all the elements of $S(12)$ are square numbers.
*Find an odd integer $N$ and $x \in S(N)$ such that $x$ is not a square number.
*For all positive integers $N$, prove that $S(N)$ has at least $\sqrt{N}$ elements.
*It is given that there are integers $x,\lambda,n$, where $n \geq 5$, such that $x^2 = 17 + 2^n\lambda$. Prove that $17 \in S(2^{n+1})$.
*For $n \geq 5$, prove that $S(2^n)$ has at least $1 + \sqrt{2^n}$ elements.
I don't think part 1. to 3. are related to 4. at all, but I may be missing something. Firstly, part 1. to 3. are easy. For 1., we just need to show that $0,1,4,9,16,25$ have square remainders, as $(n + 6)^2 \equiv n^2 \pmod{12}$. For 2., we have $N = 7$, then $2 \in N$ as $9 \equiv 2 \pmod{7}$. For 3., simply observe that $\{0^2,1^2,\dots,(\lceil\sqrt{N}\rceil - 1)^2\} \subseteq S(N)$.
My main issue is part 4., as part 5. should follow immediately by induction and by part 3., where all of those $\sqrt{N}$ elements are perfect squares but not $17$.
Attempt: The first thing to observe for part 4. is that if $2 \mid \lambda$, then we are done. Thus, we just need to prove for the case of $\lambda$ is odd, which implies that $x^2 \equiv 17 + 2^n \pmod{2^{n+1}}$. My approaches revolve around constructing a square number that is $17 \pmod{2^n}$ from $x^2$.
*
*If $k$ is even, then $kx^2 \equiv 17k \pmod{2^{n+1}}$. Furthermore, $x^4 \equiv 17^2 \pmod{2^{n+1}}$. I observe that $x^4 - 16x^2 \equiv 17 \pmod{2^{n+1}}$, but of course we don't know if $x^4 - 16x^2$ is a square number.
*I conjectured that if $x^2 \equiv 17 \pmod{2^n}$ then $x^{2m} \equiv 17 \pmod{2^{n+1}}$ for some $m$. Taking the "base case" as $49 = 17 + 2^5$ and iterating it, it appears to be false.
*I tried to consider $(x + k)^2$ for some $k$, such as $(x + 2^n)^2 = x^2 + 2^{n+1}x + 2^{2n} \equiv x^2 \pmod{2^{n+1}}$. If this is indeed the approach, then we will have the carefully pick which $k$ to use. The main problem with this apprach is that we don't know what $x \pmod{2^n}$ looks like.
Any help would be appreciated.
| If $\lambda$ is even, we're obviously done. If not, then notice that $x^2 \equiv 17 + 2^n \pmod{2^{n+1}}$, and so $\left( x - 2^{n-1} \right)^2 \equiv x^2 - 2^n + 2^{2n-2} \equiv \left( 17 + 2^n \right) - 2^n + 0 \equiv 17 \pmod{2^{n+1}}$. This is not too different conceptually from the standard proof that the multiplicative group of integers mod $p^2$ for $p$ prime has a generator.
EDIT: Oops, bad factors. Clearly $x$ is odd, so we may invert it $\pmod{2^{n+1}}$. Then expanding $\left( x - 2^{n-1} \mathbf{x^{-1}} \right)^2$ gives $x^2 - 2^n - \frac{2^{2n-2}}{x^2}$, but $x^2$ is nonzero and $2^{2n-2}$ is zero, so the last term is still $0$ and we're fine.
| {
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"timestamp": "2023-03-29T00:00:00",
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On the Diophantine equation $m^2 - p^k = 4z$, where $z \in \mathbb{N}$ and $p$ is a prime satisfying $p \equiv k \equiv 1 \pmod 4$ The title says it all:
Do there always exist $m, p, k \in \mathbb{N}$ such that $m^2 - p^k = 4z$, $z \in \mathbb{N}$ and $p$ is prime satisfying $p \equiv k \equiv 1 \pmod 4$?
MY ATTEMPT FOR $z=1$
When $z = 1$, I get
$$m^2 - p^k = 4$$
$$p^k = m^2 - 4 = (m+2)(m-2)$$
Hence, we have the simultaneous equations $p^{k-x} = m+2$ and $p^x = m-2$ where $k \geq 2x + 1$. Consequently, we have $p^{k-x} - p^x = 4$, which implies that
$$p^x (p^{k-2x} - 1) = 4$$
where $k-2x$ is odd. Since $(p-1) \mid (p^y - 1) \hspace{0.05in} \forall y \geq 1$, this last equation implies that $(p-1) \mid 4$. Likewise, the congruence $p \equiv 1 \pmod 4$ implies that $4 \mid (p-1)$. These two divisibility relations imply that $p-1=4$, or $p=5$. Hence,
$$5^x (5^{k-2x} - 1) = 4.$$
Since $5$ does not divide $4$, $x=0$ and thus $5^k - 1 = 4$, which means that $k=1$.
Therefore, $p=5$, $k=1$, and $x=0$. Consequently, $p^{k-x} = 5^{1-0} = 5 = m + 2$ and $p^x = 5^0 = 1 = m - 2$. Either way, $m=3$.
CONCLUSION
Thus, for the equation $m^2 - p^k = 4$, I get the solution $m=3$, $p=5$, and $k=1$.
QUESTION
How does one treat the case for general $z \in \mathbb{N} \setminus \{1\}$?
For one, I cannot seem to wrap my head around $m^2 - p^k = 8$.
I could, of course, rewrite it as
$$(m+3)(m-3) = m^2 - 9 = p^k - 1 = (p-1)\sigma(p^{k-1})$$
(where $\sigma$ is the classical sum-of-divisors function), but then I am no longer able to apply Euclid's Lemma.
UPDATE TO QUESTION (January 20, 2020)
The equation $m^2 - p^k = 4z$ (under the given constraints) does not have a solution when $z=4$.
Here is my additional question:
When is the equation $m^2 - p^k = 4z$, where $z \in \mathbb{N}$ and $p$ is a prime satisfying $p \equiv k \equiv 1 \pmod 4$, guaranteed to have a solution?
| This is a partial answer.
Your idea works for $z=2^{2n}$.
Claim : For $z=2^{2n}$, if $2^{n+2}+1$ is prime, then $(m,p,k)=(2^{n+1}+1,2^{n+2}+1,1)$. If $2^{n+2}+1$ is not prime, then there is no solution.
Proof :
For $z=2^{2n}$, we have
$$m^2 - p^k = 2^{2n+2}\iff p^k=(m-2^{n+1})(m+2^{n+1})$$
So, there is a non-negative integer $a$ such that
$$m+2^{n+1}=p^{k-a},m-2^{n+1}=p^a\implies 2^{n+2}=p^a(p^{k-2a}-1)$$
Since $\gcd(2,p)=1$, we get $a=0$ to have
$$ p^{k}-2^{n+2}=1$$
If $2^{n+2}+1$ is prime, then $k=1$.
If $2^{n+2}+1$ is not prime, then $k\ge 2$. By Catalan's conjecture (or Mihăilescu's theorem), there is no solution.
| {
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Solve inequality using AM-GM / Cauchy-Schwarz Given positive real numbers $a$, $b$, $c$ satisfying $a^2+b^2+c^2=3$. Prove that
$$\dfrac{1}{a^2+7}+\dfrac{1}{b^2+7}+\dfrac{1}{c^2+7} \leq \dfrac{1}{4} \left(\dfrac{1}{a+b}+\dfrac{1}{b+c}+\dfrac{1}{c+a}\right).$$
I have tried using $a+b\leq \sqrt{2\left(a^2+b^2\right)}=\sqrt{2\left(3-c^2\right)}$ but it goes to a wrong inequality. I know this problem can be solved by using AM-GM inequality (Cauchy). Please help me with this, thanks.
| Let $a+b+c=3u.$
Thus, $u\leq1$ and by C-S $$\sum_{cyc}\frac{1}{a^2+7}=\sum_{cyc}\frac{b^2+c^2+6}{(a^2+1+1+5)(1+b^2+c^2+5)}\leq$$
$$\leq\sum_{cyc}\frac{b^2+c^2+6}{(a+b+c+5)^2}=\frac{24}{(a+b+c+5)^2}=\frac{24}{(3u+5)^2}.$$
In another hand, by C-S again:
$$\sum_{cyc}\frac{1}{a+b}=\frac{1}{2(a+b+c)}\sum_{cyc}(a+b)\sum_{cyc}\frac{1}{a+b}\geq\frac{9}{2(a+b+c)}=\frac{3}{2u}.$$
Id est, it's enough to prove that
$$\frac{3}{8u}\geq\frac{24}{(3u+5)^2}$$ or
$$(1-u)(25-9u)\geq0,$$ which is obvious.
| {
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Cyclic Inequality $(x^8+1)(x^4+1)(y^8+1)(y^4+1)(z^8+1)(z^4+1)≥((x^2+1)(y^2+1)(z^2+1))^2$ I have been challenged to prove the following:
If $xy+yz+zx=3$ for positive $x,y,z$, then $(x^8+1)(x^4+1)(y^8+1)(y^4+1)(z^8+1)(z^4+1)≥((x^2+1)(y^2+1)(z^2+1))^2$.
Initial attempts at AM-GM are not very helpful for me, I don't think.
$x^8+1≥2x^4,$ and $x^4+1≥2x^2$, so $(x^8+1)(x^4+1)≥4x^2(2x^2-1)$. It is easy to show $(x^8+1)(x^4+1)≥4x^2(2x^2-1)≥(x^2+1)^2$ if $x≥\sqrt{\frac{3}{7}},$ but that's insufficient.
I doubt that this problem turns into a Vieta bash due to the nature of having $x^8$ on the LHS. Initial attempts at Newton's sums were giving what looked like annoying numbers and something more bashy than what I would guess is a more elegant solution.
| Note: This answer is very similar with Michael Rozenberg's, maybe a little more direct, but I think it's worth posting.
Using Holder's inequality:
$$8(x^8+1)=(1^4+1^4)(1^4+1^4)(1^4+1^4)[(x^2)^4+1] \geq (x^2+1)^4$$
and from Cauchy-Schwarz:
$$2(x^4+1) = (1^2+1^2)((x^2)^2+1) \geq (x^2+1)^2$$
Multiplying with similar inequalities, we get:
$$(x^8+1)(x^4+1)(y^8+1)(y^4+1)(z^8+1)(z^4+1) \geq \frac{1}{2^{12}} \left[(x^2+1)(y^2+1)(z^2+1)\right]^6$$
Therefore, we are left to prove:
$$\frac{1}{2^{12}} \left[(x^2+1)(y^2+1)(z^2+1)\right]^6\geq \left[(x^2+1)(y^2+1)(z^2+1)\right]^2$$
or equivalent:
$$(x^2+1)(y^2+1)(z^2+1) \geq 8$$
This follows from the Cauchy-Schwarz inequality again:
$$2(x^2+1)(y^2+1)(z^2+1) = (1^2+x^2+x^2+1^2)(1^2+y^2+z^2+y^2z^2)$$
$$\geq (1+xy+zx+yz)^2 = 16$$
and the inequality is proven. Equality occurs when $x = y= z = 1$.
| {
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Find volume of body given by $x^2+y^2+z^2 \leq a^2$ and $x^2+y^2 \geq a(a-2z)$, $a>0$. Is this answer right? What are mistakes in my solution? It appears to be an upside-down 'bowl', cut out of a sphere by a paraboloid. Moving to cylindrical coordinates,
\begin{align}
4 \int_0^{\pi/2}\ d\phi \int_{0}^{a}\ r \ dr \int_{(a^2-r^2)/2a}^{\sqrt{a^2-r^2}} \ dz
&=
4 \int_0^{\pi/2}\ d\phi \int_{0}^{a}\ \left( \frac{a^2-r^2}{2}-\frac{(a^2-r^2)^2}{2a\cdot\ 2a\cdot2} \right) r \ dr
\\ \ \\
&=
2 \int_0^{\pi/2}\ d\phi \int_{0}^{a}\ \left( \frac{a^2-r^2}{1}-\frac{(a^2-r^2)^2}{4a^2} \right) r \ dr
\\ \ \\
&=
\int_0^{\pi/2}\ d\phi \int_{0}^{a}\ \left( \frac{3a^2}{2}r-r^3-\frac{r^5}{2a^2} \right)\ dr
\\ \ \\
&=
\int_0^{\pi/2}\ \frac{3a^2}{2}\cdot \frac{a^2}{2}-\frac{a^4}{4}-\frac{1}{2a^2}\cdot\frac{a^6}{6} \ d\phi
\\ \ \\
&=
\frac{5\pi a^4}{24}
\end{align}
Was this correct? If not, where?
I'd also love to see alternative ways of doing this.
| *
*Your setup is the canonical one, with the exception of taking $0\leq \phi\leq \pi/2$ and then multiplying by $4$: you could have just integrated from $0$ to $2\pi$.
*You are assuming that $a>0$, which is probably the case but you don't say.
*From the first to the second line: instead of $\sqrt{a^2-r^2}-\frac{a^2-r^2}{2a}$ you wrote $\frac{a^2-r^2}2-\frac{(a^2-r^2)^2}{2a\cdot2a\cdot2}$. I can't imagine why. That's where things derail.
What you should have done:
\begin{align}
\int_0^{2\pi}\int_0^a \int_{\frac{a^2-r^2}{2a}}^{\sqrt{a^2-r^2}}\,1\,dz\,r\,dr\,d\theta
&=\int_0^{2\pi}\int_0^a\left({\sqrt{a^2-r^2}}-{\frac{a^2-r^2}{2a}}\right)\,r\,dr\,d\theta\\ \ \\
&={2\pi}\int_0^a\left({\sqrt{a^2-r^2}}-{\frac{a^2-r^2}{2a}}\right)\,r\,dr\\ \ \\
&={2\pi}\left.\vphantom{\int}\left(-\frac23(a^2-r^2)^{3/2}+\frac{(a^2-r^2)^2}{4a}\right)\right|_0^a\\\ \ \\
&=2\pi\,\left(\frac13(a^2)^{3/2}-\frac{(a^2)^2}{8a}\right)\\ \ \\
&=\frac{5\pi a^3}{12}
\end{align}
As a way to confirm the result, if you calculate the bottom half, the limits of the inner integral would have been
$$
\int_{\sqrt{a^2-r^2}}^{\frac{a^2-r^2}{2a}},
$$
and a very similar calculation as above yields $\frac{11\pi a^3}{12}$. Not surprisingly, if you add both volumes you get
$$
\frac{5\pi a^3}{12}+\frac{11\pi a^3}{12}=\frac{16\pi a^3}{12}=\frac{4\pi a^3}3,
$$
the volume of the whole interior of the sphere.
| {
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Trisected sides of a scalene triangle Scalene triangle $\bigtriangleup ABC$ has area 45. Points $P_1$ and $P_2$ are located on side $AB$ such that $AP_1 = P_1P_2 = BP_2$. Additionally, the points $Q_1$ and $Q_2$ are located on side $AC$ such that $AQ_1 = Q_1Q_2 = CQ_2$. The area of the intersection of triangles $BQ_1Q_2$ and $CP_1P_2$ can be expressed as a common fraction $\frac{m}{n}$, where $m$ and $n$ are relatively prime positive integers. What is $m+n$?
$\textbf{(A) } 15 \qquad \textbf{(B) } 47 \qquad \textbf{(C) } 79 \qquad \textbf{(D) } 95 \qquad \textbf{(E) } 257 $
So to solve this, I assumed that an equilateral triangle wouldn't change the answer (I'm very lazy), so I fakesolved with an equilateral triangle.
I scaled it down so that the area was $\sqrt{3}$.
Then, the coordinates of the kite in the middle (in counter-clockwise order) are:
$(1,\frac{\sqrt{3}}{5})$
$(\frac{4}{7},\frac{2\sqrt{3}}{7})$
$(1,\frac{\sqrt{3}}{2})$
$(\frac{10}{7},\frac{2\sqrt{3}}{7})$
The product of the two diagonals divided by 2 is: $\frac{3}{7} \cdot \frac{3\sqrt{3}}{10}=\frac{9\sqrt{3}}{70}$. Multiplying by $\frac{45}{\sqrt{3}}=15\sqrt{3}$ gives $\frac{9 \cdot 15 \cdot 3}{70}=\frac{405}{70}=\frac{81}{14} \implies 95$.
No answer, can you guys check this? Thanks.
|
Let [.] denote areas and I = [ABC] = 45. Observe that the intersection area [DEGF] is equal to
$$[DEGF] = [DBC] - [EBC] - [FBC] + [GBC]\tag 1$$
Evaluate [EBC], one of the four triangle areas on the RHS, as follows.
$$\frac{[EBC]}{[P_2BC]}=\frac{EC}{P_2C}
= \frac{[Q_1BC]}{[Q_1P_2BC]} = \frac{[Q_1BC]}{I - [AP_2Q_1]}
= \frac{\frac23I}{I - \frac13\cdot\frac23I}=\frac67 $$
Then,
$$[EBC] = \frac67[P_2BC] = \frac67\cdot \frac13I = \frac27I$$
Following the same procedure to obtain the areas of the other three triangles,
$$[FBC] = \frac27I, \>\>\>\>\> [DBC] = \frac12I, \>\>\>\>\> [FBC] = \frac15I$$
Substitute the four areas and $I=45$ into (1),
$$[DEGF] = \left(\frac12 - \frac27- \frac27 + \frac15\right)I=\frac9{70}\cdot 45 = \frac{81}{14}$$
| {
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