Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
|---|---|---|
Inverse function of $f(x) = \frac{x+5}{x-2}$ Find the inverse of the function
$f(x) = \frac{x+5}{x-2}$
Here's what I have so far:
$y = \frac{x+5}{x-2}$
$x = \frac{y+5}{y-2}$
$(x)(y-2) = (y+5)$
but this seems to be a dead end.
How should I approach this?
Thank you!
| Almost there:
$$x = \frac{y+5}{y-2}$$
$$x(y-2) = y+5$$
$$xy - 2x = y+5$$
$$xy - y = 5 + 2x$$
$$y(x-1) = 5+2x$$
I'll let you finish.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1511766",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 1
} |
How to compute this gross limit.
How do I compute this limit?
$$
\lim_{n \to \infty}
\frac{\left(1 + \frac{1}{n} + \frac{1}{n^2}\right)^n -
\left(1 + \frac{1}{n} - \frac{1}{n^2}\right)^n
}{
2 \left(1 + \frac{1}{n} + \frac{1}{n^2}\right)^n -
\left(1 + \frac{1}{n} - \frac{1}{n^2 + 1}\right)^n -
\left(1 + \frac{1}{n} - \frac{1}{n^2 (n^2 +1)}\right)^n
}
$$
I think I got the correct limit by using fast converging limits to $e$.
In particular I used truncated Taylor series for the sqrt and 4th root.
Or squares and bisquares.
Example
$(1+1/2n)^{2n}$
Becomes
$(1 + 1/n + 1/4n^2)^n.$
In combination with l'hopital it gives me the answer.
But I guess that is not a very good (fast) method.
| Use the fact that $a^n-b^n = (a-b)(a^{n-1}+a^{n-2}b+\cdots+ab^{n-2}+b^{n-1})$. It is going to be quite lengthy so bear with me: The numerator is
$$
\left(1+\frac{1}{n}+\frac{1}{n^2}\right)^n-
\left(1+\frac{1}{n}-\frac{1}{n^2}\right)^n = \frac{2}{n^2}\sum_{m=0}^{n-1}
\left(1+\frac{1}{n}+\frac{1}{n^2}\right)^{n-1-m}
\left(1+\frac{1}{n}-\frac{1}{n^2}\right)^m
$$
Write the above as $2/n^2 f(n)$.
The denominator has two parts
$$
\left(1+\frac{1}{n}+\frac{1}{n^2}\right)^n-
\left(1+\frac{1}{n}-\frac{1}{n^2+1}\right)^n = \left(\frac{1}{n^2}+\frac{1}{n^2+1}\right)g(n)
$$
where
$$g(n) = \sum_{m=0}^{n-1}\left(1+\frac{1}{n}+\frac{1}{n^2}\right)^{n-1+m}\left(1+\frac{1}{n}-\frac{1}{n^2+1}\right)^m$$
and the second part
$$
\left(1+\frac{1}{n}+\frac{1}{n^2}\right)^n-
\left(1+\frac{1}{n}-\frac{1}{n^2(n^2+1)}\right)^n = \frac{1}{n^2}\left(1+\frac{1}{n^2+1}\right)k(n)
$$
where
$$k(n)=\sum_{m=0}^{n-1}\left(1+\frac{1}{n}+\frac{1}{n^2}\right)^{n-1+m}\left(1+\frac{1}{n}-\frac{1}{n^2(n^2+1)}\right)^m$$
so in the end you have
$$
\lim_{n\to\infty}\frac{\frac{2}{n^2}f(n)}{\frac{2n^2+1}{n^2(n^2+1)}g(n)+\frac{n^2+2}{n^2(n^2+1)}h(n)}=
\lim_{n\to\infty}\frac{2(n^2+1)f(n)}{(2n^2+1)g(n)+(n^2+2)h(n)}
$$
Now note that as $n\to \infty$ then $f(n)$ becomes approximately
$$f(n) = \sum_{m=0}^{n-1}
\left(1+\frac{1}{n}+\frac{1}{n^2}\right)^{n-1-m}
\left(1+\frac{1}{n}-\frac{1}{n^2}\right)^m\approx n\left(1+\frac{1}{n}\right)^n\approx ne$$
where $\approx$ means to order of $o(1)$. Actually $g(n)\approx ne\approx h(n)$ too.
So the limit reduces to
$$\to \lim_{n\to\infty}\frac{2(n^2+1)}{(2n^2+1)+(n^2+2)}=\frac{2}{3}$$
Hopefully I haven't made a mistake anywhere, but at least I think the method is clear.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1512063",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 6,
"answer_id": 1
} |
Greatest Common Divisor Proof: $\gcd(m^2-n^2, m^2+n^2) = 1$ Prove that if $\gcd(m,n) = 1$ and $m+n \equiv 1 (\text{mod} ~2)$ then $\gcd(m^2-n^2, m^2+n^2) = 1$
Workings:
Suppose that $\gcd(m,n) = 1$ and $m+n \equiv 1 (\text{mod} ~2)$
$\gcd(m^2-n^2, m^2+n^2)$
$= gcd((m-n)(m+n), (m-n)(m+n)+2n^2)$
Now I know that $m+n=1 (\text{mod} ~2)$ means that one of $m$ or $n$ is odd or even
But now I'm not too sure on what to do.
Any help will be appreciated.
| Euclid's algorithm gives:
\begin{align*}
\gcd(m^2 - n^2, m^2 + n^2)
&= \gcd(m^2 - n^2, (m^2 + n^2) - (m^2 - n^2)) \\
&= \gcd(m^2 - n^2, 2n^2)
\end{align*}
Since $m + n \equiv 1 \pmod{2}$, $m^2 - n^2 \equiv 1 \pmod{2}$, so we can forget about the $2$.
We get
\begin{align*}
&= \gcd(m^2 - n^2, n^2) \\
&= \gcd(m^2, n^2) \\
&= 1.
\end{align*}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1513549",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 0
} |
Find the limit as x approaches infinity $$\lim_{x\to \infty} {\cosh^{-1}(x^{3}) + \coth^{-1}(\sqrt{x^{2}+1}) - 3\sinh^{-1}(x)}$$
Honestly, I don't really know how to approach this. I know the logarithmic formulae for the inverse hyperbolic functions, but that gives me a very complicated expression. Is there an identity I can use or a rule? Any tips are welcome.
Thank you.
| We can directly use the expression of inverse hyperbolic functions namely
\begin{align}
\sinh^{-1}x &= \log(x + \sqrt{x^{2} + 1})\notag\\
\cosh^{-1}x &= \log(x + \sqrt{x^{2} - 1})\notag\\
\coth^{-1}x &= \frac{1}{2}\log\left(\frac{x + 1}{x - 1}\right)\notag
\end{align}
From the above we see that $\coth^{-1}x \to 0$ as $x \to \infty$. It follows by the same logic that $\coth^{-1}\sqrt{x^{2} + 1} \to 0$ so we need to consider the only first and last terms of the expression given in question. Clearly we have
\begin{align}
\cosh^{-1}x^{3} - 3\sinh^{-1}x &= \log(x^{3} + \sqrt{x^{6} - 1}) - 3\log(x + \sqrt{x^{2} + 1})\notag\\
&= \log\left(\frac{x^{3} + \sqrt{x^{6} - 1}}{(x + \sqrt{x^{2} + 1})^{3}}\right)\notag\\
&= \log\left(\frac{1 + \sqrt{1 - 1/x^{6}}}{(1 + \sqrt{1 + 1/x^{2}})^{3}}\right)\notag\\
&\to \log 1 = 0\notag
\end{align}
The desired limit is $0$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1513636",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Finding maximum $b$ in $x^5-20x^4+bx^3+cx^2+dx+e=0$ Let $b, c, d, e$ be real numbers such that the following equation
$$x^5-20x^4+bx^3+cx^2+dx+e=0$$
has real roots only. Find the largest possibe value of $b$.
What I have done is:
Let $x_1, x_2, x_3, x_4, x_5$ be the 5 real roots of the equation. Then we have
$$x_1+x_2+x_3+x_4+x_5=20$$ and
$$b=\sum_{0<i\le j \le 5} x_ix_j=\frac{1}{2}[(x_1+x_2+x_3+x_4+x_5)^2-(x_1^2+x^2_2+x_3^2+x_4^2+x_5^2)]$$
To find maximum $b$, we can find minimum $x_1^2+x^2_2+x_3^2+x_4^2+x_5^2$. Cauchy-Schwartz Inequality yields,
$$(x_1^2+x^2_2+x_3^2+x_4^2+x_5^2)(1+1+1+1+1)\ge(x_1+x_2+x_3+x_4+x_5)=20$$
Thus,
$$x_1^2+x^2_2+x_3^2+x_4^2+x_5^2\ge\frac{20}{5}=4$$
So,
$$b_{max}=\frac{1}{2}[20^2-4]=198$$
However, the answer was 160, yet I am pretty sure I am correct. Where did I go wrong?
Thanks in advance!
| Hint: Let $P(x) = x^5-20x^4+bx^3+cx^2+dx+e$. If $P$ has all real roots, then $P'''(x)$ must have two real roots..
$\implies 60x^2-480x+6b$ has real roots $\implies b \le 160$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1514076",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
Equation of locus Point P$(x, y)$ moves in such a way that its distance from the point $(3, 5)$ is proportional to its distance from the point $(-2, 4)$. Find the locus of P if the origin is a point on the locus.
Answer:
$$(x-3)^2 + (y-5)^2 = (x+2)^2 + (y-4)^2$$
or, $$10x+2y-14=0$$
or, $$5x+y-7=0$$
but answer given is $$7x^2+7y^2+128x-36y=0$$
| $$(x-3)^2 + (y-5)^2 = \lambda((x+2)^2 + (y-4)^2).$$
We express that the curve passes through the orgin:
$$(-3)^2 + (-5)^2 = \lambda((+2)^2 + (-4)^2),$$
hence
$$\lambda=\frac{17}{10}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1517502",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
Use integration by parts to find the integral $\int\frac{\sqrt {4x^2-9}}{x^2}dx$ $$\int\frac{\sqrt {4x^2-9}}{x^2}dx$$
I tried to solve this using integration by parts, but I come up with something that is much more difficult to solve. How can this be solved?
| $$\int \frac{\sqrt{4x^2-9}}{x^2}dx$$
$$=-\int \sqrt{4x^2-9} \,\ d(\frac{1}{x})$$
Using by parts, $$=-\left[\frac{\sqrt{4x^2-9}}{x}-\int \frac{4}{\sqrt{4x^2-9}}dx\right]$$
$$=-\left[\frac{\sqrt{4x^2-9}}{x}-2\int \frac{d(2x)}{\sqrt{(2x)^2-3^2}}\right]$$
$$=-\left[\frac{\sqrt{4x^2-9}}{x}-2 \ln \left|2x+\sqrt{4x^2-9}\right|\right]+c$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1518793",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 4,
"answer_id": 0
} |
Does the sequence $a_n = \frac{1 \cdot 3 \cdot 5 \cdots (2n-1)}{2 \cdot 4 \cdot 6 \cdots (2n)}$ converge?
Does the sequence $a_n = \frac{1 \cdot 3 \cdot 5 \cdots (2n-1)}{2 \cdot 4 \cdot 6 \cdots (2n)}$ converge?
This is a homework question for an analysis class.
Firstly, I know the sequence is comprised of odd numbers up to $2n-1$ over even numbers up to $2n$.
Secondly, I know the sequence decreases. I'm guessing it'll converge, but not specifically to $0$.
Where I am stuck is simplifying the series to something easier to work with. Any help would be greatly appreciated!
Edit: solved, thank you everyone. I ended up doing what was first suggested, not finding the actual limit but just proving the sequence converges. Thanks again!
| $$a^{ 2 }_{ n }={ \left( \frac { 1\cdot 3\cdot 5\cdot ...\cdot (2n-1) }{ 2\cdot 4\cdot 6\cdot ...\cdot (2n) } \right) }^{ 2 }=\frac { { 1 }^{ 2 }\cdot 3^{ 2 }\cdot 5^{ 2 }\cdot ...\cdot (2n-1)^{ 2 } }{ 2^{ 2 }\cdot 4^{ 2 }\cdot 6^{ 2 }\cdot ...\cdot (2n)^{ 2 } } =\\ =\frac { 1\cdot 3 }{ { 2 }^{ 2 } } \cdot \frac { 3\cdot 5 }{ 4^{ 2 } } \cdot \frac { \left( 2n-1 \right) \left( 2n+1 \right) }{ \left( 2n \right) ^{ 2 } } \cdot \frac { 1 }{ 2n+1 } <\frac { 1 }{ 2n+1 } \\ \\ $$
for every $n$ $ \epsilon$ $ \mathbb{N} $
$0<{ a }_{ n }<\frac { 1 }{ \sqrt { 2n+1 } } \quad $ so it converges and
$$ \lim _{ n\rightarrow \infty }{ { a }_{ n } } =0$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1519540",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
"answer_id": 2
} |
Express $\frac{1-x}{(x-1)^2+y^2}-\bigg(\frac{y}{(x-1)^2+y^2}\bigg)i$ in the form of $f(z)$ I need to express $f(z)$ from the form $\color{blue}{u+vi}$ to the form $\color{blue}z$ for example if: $g(z)=\frac{1}{x+yi}$ so $ =g(z)=\frac{1}{z}$
$$f(z)=\underbrace {\frac{1-x}{(x-1)^2+y^2}}_{=u(x,y)}+\underbrace {\bigg(-\frac{y}{(x-1)^2+y^2}\bigg)}_{=v(x,y)}i$$
My try:
$$f(z)=\frac{1-x-yi}{(x-1)^2+y^2}$$
$$=\frac{1-\overbrace{x-yi}^{=\bar z}}{\underbrace{x^2+y^2}_{=|z|^2}-2x+1^2}$$
$$-\frac{1-\bar z}{|z|^2-2x+1}$$
I'm stuck here
| Hint ( to prove the answer of @RonGordon):
$$
f(z)=\frac{1-x}{(x-1)^2+y^2}-\bigg(\frac{y}{(x-1)^2+y^2}\bigg)i=
$$
$$
=\frac{1}{(1-x)^2+y^2}\left(1-x-i y \right) = \frac{\bar u}{|u|^2}=\frac{1}{u}
$$
for $u=1-x +iy=1-(x-iy)=1-\bar z$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1522760",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
} |
Fermat's infinite descent for finding the squares that sum to a prime Fermat's theorem on sum of two squares states that an odd prime $p = x^2 + y^2 \iff p \equiv 1 \pmod 4$
Applying the descent procedure I can get to $a^2 + b^2 = pc$ where $c \in \mathbb{Z} \gt 1$
I want $c = 1$, so how do I proceed from here? How do I apply the procedure iteratively?
Example:
$$
p = 97
$$
$$97 \equiv 1 \pmod 4 \implies \left(\frac{-1}{97}\right) = 1 \implies x^2 \equiv -1 \pmod {97}$$ has a solution
$$x^2 + 1 \equiv 0 \pmod {97}$$
$$x^2 + 1 = 97m$$
We find an $x,m$ that solves the equation.
$$x = 75, m = 58$$
Now, we pick an $a,b$ such that $\frac{-m}{2} \leq a,b \leq \frac{m}{2}$
$$a \equiv x \pmod m = 17$$
$$b \equiv y \pmod m = 1$$
Observations:
*
*$ a^2 + b^2 \equiv x^2 + 1 \equiv 0 (\mod m)$
*$ (a^2 + b^2) = mc$
*$ (x^2 + 1) = mp$
Plugging in $a,b,m$ for 2, we get $c = 5$
By this identity, we know that
$(a^2 + b^2)(c^2 + d^2) = (ac + bd)^2 + (ad - bc)^2$
**$(a^2 + b^2)(x^2 + 1^2) = (ax + b)(a - bx) = m^2pc$
Dividing ** by $m^2$, $pc = (\frac{ax+b}{m})^2 + (\frac{a-bx}{m})^2$
Plugging in $a,b,m,p,c$ we get that $22^2 + (-1)^2 = 97*5$
So we have two squares that add up to 5 times our $p$. How do we turn the 5 into a 1? What is the next step in the descent?
| This is how we proceed.
We found out that $22^2+(−1)^2=97∗5$
Notice it's a sum of squares - so let's use the equation from before $x^2 + 1 = 97m$. In general, the $1$ might be something else. In this case, since $(-1)^2 = 1$, nothing really changes and we can use the exact same equation.
In the second iteration:
$m = 2$
$x = 22$
$y = 1$ (we can safely remove the negative)
Our $m$ is now $5$. So as before, we need to pick integers $a,b$ such that $a,b \in [-m/2,m/2] \implies a,b \in [-2.5, 2.5]$
$a \equiv x \equiv 2 \pmod 5$
$b \equiv y \equiv 1 \pmod 5$
Apply the Fibonacci identity from before:
** $(a^2 + b^2)(x^2 + y^2) = (ax+by)^2 + (ay-bx)^2 = (2*22+1*1)^2 + (2*1-1*22)^2 = (45)^2 + (-20)^2 = m^2pr = 25*97*c$
Again, solve for $c$. In this case $c = 1$.
Awesome! This is the last iteration.
Divide ** by $m^2$, $p =(\frac{45}{5})^2+(\frac{-20}{5})^2 = (9)^2 + (-4)^2$
So, $a=9, b=4$ solve $a^2 + b^2 = 97$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1525113",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 2
} |
If $x$ is a positive integer such that $x(x+1)(x+2)(x+3)+1=379^2$, find $x$
If $x$ is a positive integer such that $x(x+1)(x+2)(x+3)+1=379^2$, find $x$
This is a 1989 ARML problem. One, ugly way to solve this is:
Approximate this as $x^4=379^2$, so $x\approx \sqrt{379}\approx 19$ and guess and check around there to see that $18$ works.
What's a nicer way?
Hint
Difference of squares
| $$x(x+1)(x+2)(x+3) +1 =(x^2 +x)(x^2 +5x +6) +1 = x^4+5x^3 +6x^2 +x^3 +5x^2 +6x +1 =x^4 +6x^3 +11x^2 +6x +1 =x^2 (x^2 +6x +9) +2x(x+3)+1 =(x(x+3) +1)^2$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1528187",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 5,
"answer_id": 4
} |
What is the best way to solve an equation of the form $(f(x))^2-a(f(x))+b=x$? On a math contest I was told to solve the equation $$(x^2-3x+1)^2-3(x^2-3x+1)+1=x$$
For this particular problem I simplified by letting $$a\equiv x^2-3x+1$$
Then I continued with $$a^2-3a+1-x=0$$
$$a=\frac{3\pm\sqrt{9-4\left(1-x\right)}}{2}$$
$$3\pm\sqrt{5+4x}=2a=2x^2-6x+2$$
$$\pm\sqrt{5+4x}=2x^2-6x-1$$
I am not sure how to finish off this problem. Also, I have seen a bunch of problems like this in the past. What is the best way to approach a problem like this and also how could I finish solving this problem? Could someone also explain why f(f(x))=x has the same solutions as f(x)=x.
| Add $x^2-3x+1$ to both sides, giving
$$(x^2-3x+1)^2 - 2(x^2-3x+1) + 1 = x + x^2-3x+1 = x^2-2x+1,$$
or
$$((x^2-3x+1)-1)^2 = (x-1)^2.$$
Thus $(x^2-3x+1)-1 = \pm (x-1)$. The plus sign gives
$$x^2-4x+1=0\quad\Rightarrow\quad x = 2\pm\sqrt{3}.$$
The minus sign gives
$$x^2-2x-1=0\quad\Rightarrow\quad x = 1\pm\sqrt{2}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1528652",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 0
} |
Bound on difference between irrational square root and rational number I'm trying to show that if $\alpha$ is the square root of a non-perfect square $d$, then for any rational $p/q$ there's a positive real $x$ such that $|\alpha - p/q| > x/q^2$. I'm having trouble starting on the problem, could someone give a hint? Thanks
| Let $ d $ be a positive integer which isn't a square, and consider a rational $ \frac{p}{q} $ ($ p, q $ integers, $ q > 0 $).
Looking at $ \left| \sqrt{d} - \frac{p}{q} \right| $, it is $ \dfrac{1}{q} |q\sqrt{d} - p | $ $ = \dfrac{1}{q} \dfrac{|q^2 d - p^2|}{|q\sqrt{d}+p|} $. As $ \sqrt{d} $ is irrational, numerator $ |q^2 d - p^2 | $ is non-zero (and an integer), making $ | q^2 d - p^2 | \geq 1 $.
So $ \left| \sqrt{d} - \dfrac{p}{q} \right| \geq \dfrac{1}{q^2} \dfrac{1}{\left| \sqrt{d} + \frac{p}{q} \right|}$. Notice denominator $$\left| \sqrt{d} + \frac{p}{q} \right|= \left| 2\sqrt{d} -\sqrt{d} + \frac{p}{q} \right|\leq 2\sqrt{d} + \left| -\sqrt{d} + \frac{p}{q} \right|,$$ giving us
$$ \left| \sqrt{d} - \dfrac{p}{q} \right| \geq \dfrac{1}{q^2} \left(\dfrac{1}{2\sqrt{d} + \left| \sqrt{d} - \frac{p}{q} \right|} \right).$$
Writing $ t := \left| \sqrt{d} - \dfrac{p}{q} \right| $, this is $ t \geq \dfrac{1}{q^2 (2\sqrt{d} + t)} $. So $ q^2 t^2 + q^2 2\sqrt{d} t - 1 \geq 0 $.
Quadratic on LHS has roots $ \dfrac{-q \sqrt{d} \pm \sqrt{ q^2 d + 1 }}{q} $, so the inequality becomes $$ \left(t-\frac{-q\sqrt{d} - \sqrt{q^2 d+1}}{q} \right) \left( t - \frac{-q\sqrt{d}+\sqrt{q^2 d+1}}{q} \right) \geq 0.$$
The first factor is anyways $ > 0 $, giving $ t \geq \dfrac{-q\sqrt{d} + \sqrt{q^2 d + 1}}{q} $. Now RHS $$ \dfrac{-q\sqrt{d} + \sqrt{q^2 d + 1}}{q} = \dfrac{1}{q(q\sqrt{ d} + \sqrt{q^2 d + 1})}=\dfrac{1}{q^2\left(\sqrt{d} + \sqrt{d + \dfrac{1}{q^2}}\right)}\geq \dfrac{1}{q^2(\sqrt{d} + \sqrt{d+1})}= \dfrac{\sqrt{d+1}-\sqrt{d}}{q^2}$$
So finally, we have
$$ \left| \sqrt{d} - \dfrac{p}{q} \right| \geq \dfrac{\sqrt{d+1}-\sqrt{d}}{q^2} $$
for any integers $ p, q $ with $ q > 0 $.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1530454",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
When is EF longer than AC? (a generalization) ABC is an isosceles right triangle,
M is on AC,
and EMF is a straight line.
When is EF longer than AC?
]1
Note:
This is a generalization
of the following problem,
which has M in the
center of AC:
Prove that EF is longer than AC
In that case,
I showed that
EF is always longer than AC.
I can show the following:
If M is closer to C than A,
then EF is longer.
If M is closer to A than C,
then EF can be shorter,
but it will be longer when
EA is long enough.
If $|EB| \ge \sqrt{2}$,
then EF is always longer
no matter where M is.
If there are no answers in two days,
I will post mine.
| We may suppose that $$A(0,1),B(0,0),C(1,0),M(m,1-m),E(0,1-m-am),F\left(m+\frac{m-1}{a},0\right)$$
where $0\lt m\lt 1$ is the $x$ coordinate of $M$ and $a\lt -1$ is the slope of the line $EF$.
Then, we have
$$\begin{align}&|EF|\gt |AC|\\\\&\iff |EF|^2\gt |AC|^2\\\\&\iff \left(m+\frac{m-1}{a}\right)^2+(1-m-am)^2\gt 2\\\\&\iff m^2+\frac{2m(m-1)}{a}+\frac{(m-1)^2}{a^2}+1+m^2+a^2m^2-2m-2am+2am^2-2\gt 0\\\\&\iff (a+1)^2(a^2+1)m^2-2(a+1)(a^2+1)m+(a+1)(1-a)\gt 0\\\\&\iff (a+1)(a^2+1)m^2-2(a^2+1)m+1-a\lt 0\\\\&\iff \color{red}{\frac{-a^2-1+\sqrt{2a^2(a^2+1)}}{-(a+1)(a^2+1)}\lt m\lt 1}\tag1\end{align}$$
Here note that
$$0\lt f(a)=\frac{-a^2-1+\sqrt{2a^2(a^2+1)}}{-(a+1)(a^2+1)}\lt \color{blue}{\frac 12}$$
for $a\lt -1$ :
$$\begin{align}f(a)&=\frac{-a^2-1+\sqrt{2a^2(a^2+1)}}{-(a+1)(a^2+1)}\cdot\frac{-a^2-1-\sqrt{2a^2(a^2+1)}}{-a^2-1-\sqrt{2a^2(a^2+1)}}\\&=\frac{a-1}{-a^2-1-\sqrt{2a^2(a^2+1)}}\\&=\frac{1-\frac 1a}{-a-\frac 1a+\sqrt{2(a^2+1)}}\end{align}$$
is increasing for $a\lt -1$ with
$$\lim_{a\to -\infty}f(a)=0,\qquad\lim_{a\to -1^-}f(a)=\frac 12.$$
Added : Now I'm going to check if my answer agrees with what the OP wrote.
If M is closer to C than A, then EF is longer.
If $m\ge\frac 12$, then $(1)$ holds for any $(a,m)$.
If M is closer to A than C, then EF can be shorter, but it will be longer when EA is long enough.
If $m\lt\frac 12$, then EF can be shorter, but it will be longer when $a$ is small enough.
If $|EB|\ge \sqrt 2$, then EF is always longer no matter where M is.
If $1-m-am\ge \sqrt 2$, i.e. $m\ge\frac{1-\sqrt 2}{a+1}$, then $(1)$ holds for any $(a,m)$ because we have
$$\frac{-a^2-1+\sqrt{2a^2(a^2+1)}}{-(a+1)(a^2+1)}\lt \frac{1-\sqrt 2}{a+1}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1530698",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
An Algebraic Proof that $|y^3 - x^3| \ge |(y - x)|^3/4 $ I can prove this using calculus, but not by simple algebra: can anyone help ?
Calculus Proof:
Fix the separation of $x$ and $y$ so that $y = x + d$ with $d>0$ ($ \implies y > x \implies y^3 > x^3$) and now consider $ f(x) = y^3 - x^3 = (x+d)^3 - x^3$.
$ f(x) = (x+d)^3 - x^3$ = $3x^2d + 3d^2x +d^3$ which is a quadratic in $x$ with a minimum given by $f`(x) = 0 = 6xd + 3d^2$ giving $x = - d/2$ and therefore $y = d/2$. This is not surprising considering the geometry: it says that for a fixed separation of $x$ and $y$, $y^3 - x^3$ is minimised when $x$ and $y$ are symetrically placed around the inflexion point of the cubic.
So, $y^3 - x^3 = d^3/8 - (-d^3)/8 = d^3/4$ and since this is the minimum value it follows that for $x \in (-\infty, +\infty)$ and $y > x$ then $y^3 - x^3 \ge (y - x)^3/4 $, so that $|y^3 - x^3| \ge |(y - x)|^3/4 $ whether $y > x$ or $y < x$ (and clearly this is true for $y = x$.)
(The original interest in the inequality comes from this question: cauchy sequence on $\mathbb{R}$)
| Wlog. $y\ge x$. Then
$$\begin{align}y^3-x^3&\ge \frac{(y-x)^3}{4}&\iff\\
4(y^3-x^3)-(y-x)^3&\ge 0&\iff\\
3y^3+3y^2x-3x^3-3x^2y&\ge 0&\iff\\
3(y+x)^2(y-x)&\ge0\end{align}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1533407",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
$x^2+(a^2+b^2)x+ab=0$ where $a,b,x\in \mathbb Z$ Trivial solutions include $a=0, x=-b^2$ and $b=0,x=-a^2$. Then $a=b=1,x=-1$ or $a=b=-1, x=-1$. Is there any other? How could you prove there aren't? $(a^2+b^2)^2-4ab=(a^2+b^2)+2ab(ab-1)$ needs to be a square. I tried to look at it $\mod 4$ but it doesn't yield anything useful.
| Suppose $a,b\ne 0$ and $(a^2+b^2)^2-4ab = c^2$ for some integer $c\ge 0$. We can then write $c = a^2+b^2 - k$ for some $0<k\le a^2+b^2$. Then
$$ 4ab = (a^2+b^2) - c^2 = (a^2+b^2) - (a^2+b^2-k) = k(2a^2+2b^2-k) $$
Note that, for $0<k\le a^2+b^2$, the RHS is increasing in $k$. Hence, if $k\ge 2$, we have
$$ 4ab = k(2a^2+2b^2-k)\ge 4a^2+4b^2-4 \implies 2a^2+2b^2 - 4 + 2(a-b)^2\le 0. $$
This is false if $a^2+b^2> 2$, which occurs if at least one of $|a|$ or $|b|$ is greater than 1. In that case, $k = 1$, thus implying
$$ 4ab = 2a^2+2b^2 - 1\implies 2(a-b)^2 - 1 = 0 $$
which is also impossible. Hence, there are no nontrivial solutions, where nontrivial here means $|ab|>1$.
Edit: Starting from the equality $4ab = k(2a^2+2b^2-k)$, another way to limit the possibilities is to note the following: for $k\ge 1$, we have
$$ 4ab = k(2a^2+2b^2-k)\ge 2a^2+2b^2 - 1\implies 2(a-b)^2 - 1 \le 0.$$
Since $a$ and $b$ are integers, this can happen only if $a=b$, in which case we have
$$ 4a^2 = k(4a^2 - k)\implies k^2 - 4a^2k + 4a^2 = 0.$$
The discriminant of the above quadratic is
$$(4a^2)^2 - 4(4a^2) = 4a^2(4a^2-4)$$
which is positive if $a^2>1$, i.e. there are no solutions if $a^2>1$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1534809",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Compute the limit of the sequence of functions of $\lim_{n\to \infty} f_n(x) = \frac{x^2}{x^2+(1-nx)^2}$ Compute the limit of the sequence of functions of
$$
\lim_{n\to \infty} f_n(x) = \frac{x^2}{x^2+(1-nx)^2}.
$$
Attempt.
$$\lim_{n\to \infty} f_n(x) = \lim_{n\to \infty} \frac{x^2}{x^2+(1-nx)^2} = \lim_{n\to \infty} \frac{x^2}{x^2+n^2x^2-2nx + 1} $$
$$= \lim_{n\to \infty} \frac{x^2}{x^2+\frac{x^2}{n^2}-\frac{2x}{n} + \frac{1}{n}} = \lim_{n\to \infty} \frac{x^2}{x^2} = 1$$
| Well $$\lim \frac{x^2}{x^2+n^2x^2-2nx + 1} \ne \lim \frac{x^2}{x^2+\frac{x^2}{n^2}-\frac{2x}{n} + \frac{1}{n}} $$
Note that there is no need to do any algebraic manipulation here; you can just do it by inspection as $n \to \infty$ we immediately see that $$\lim_{n\to \infty} \frac{x^2}{x^2+(1-nx)^2}=0$$
as the denominator is tends to infinity the fraction must be zero.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1537927",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
Prove by induction $ \sin x + \sin 2x + ... + \sin nx = \frac {\sin (\frac {n + 1} {2} x)} {\sin \frac{x}{2}} \sin \frac{nx}{2} $
Prove by induction
$$ \sin x + \sin 2x + ... + \sin nx = \frac {\sin (\frac {n + 1} {2} x)} {\sin \frac{x}{2}} \sin \frac{nx}{2} $$
What I have for now:
$$ \frac {\sin (\frac {n + 1} {2} x)} {\sin \frac{x}{2}} \sin \frac{nx}{2} + \sin(n + 1)x = \frac {\sin (\frac {n + 2} {2} x)} {\sin \frac{x}{2}} \sin \frac{(n + 1)x}{2}$$
Letting $y = \frac {(n + 1)x} {2} $
$$\frac {\sin y} {\sin \frac{x}{2}} \sin (y - \frac{x}{2}) + \sin2y =
\frac {\sin (y + \frac {x} {2} )} {\sin \frac{x}{2}} \sin y $$
Then I used $\sin (\alpha + \beta)$ and $\sin (\alpha - \beta)$ formulas, bit they didn't help.
| You are almost there, but use $\sin 2y = 2\sin y\cos y$, considering your left hand side we get:
$$\frac {\sin y} {\sin \frac{x}{2}} \sin (y - \frac{x}{2}) + \sin2y =
\frac {\sin y} {\sin \frac{x}{2}} \sin (y - \frac{x}{2}) + 2\sin y\cos y = $$
$$\left(\frac {\sin (y-\frac{x}{2})} {\sin \frac{x}{2}} +2\cos y\right)\sin y = \left(\frac {\sin (y-\frac{x}{2}) +2\cos y\sin\frac{x}{2}} {\sin \frac{x}{2}}\right)\sin y =$$
Now use the subtraction of angles formula $\sin (\alpha-\beta)$:
$$=\left(\frac {\sin y\cos \frac{x}{2} - \cos y \sin \frac{x}{2} +2\cos y\sin\frac{x}{2}} {\sin \frac{x}{2}}\right)\sin y = \left(\frac {\sin y\cos \frac{x}{2} +\cos y\sin\frac{x}{2}} {\sin \frac{x}{2}}\right)\sin y =$$
And lastly use the addition of angles formula $\sin(\alpha+\beta)$:
$$=\left(\frac {\sin (y+ \frac{x}{2}) } {\sin \frac{x}{2}}\right)\sin y $$
Which is what you had on your right hand side, and thus the induction is done.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1538163",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 2,
"answer_id": 0
} |
How to prove that $\frac{1+4n^2}{2+2n^2}$ is a cauchy sequence? I know that the following sequence is a
$$\frac{1 + 4n^2}{2+2n^2}$$
How can I show that it is a cauchy sequence - well it has to cause its convergent but I want to understand it with the cauchy definition.
I know that i have to get the $|a_m - a_n | < \varepsilon$ meaning that the distance between two elements has to be smaller than epsilon. But how big is epsilon? I only know that there is an element called $N_0$ when this will become true.
This is where my problem starts - what will be the next step?
So I have $ \frac{1+4n^2+1}{2+2n^2+1} - \frac{1+4n^2}{2+2n^2} < \varepsilon $ but I don't know how much $\varepsilon$ is nor do I know where $N_0$ is.
| Let $a_n$ the $n$-th term of the sequence, since $$a_n=\frac{1+4n^2}{2+2n^2}=\frac{(4+4n^2)-3}{2+2n^2}=2-\frac{3}{2}\frac{1}{n^2+1}$$
it follows
$$|a_n-a_m|=\frac{3}{2}\left|\frac{1}{n^2+1}-\frac{1}{m^2+1}\right|\le\frac{3}{2}\left(\left|\frac{1}{n^2+1}\right|+\left|\frac{1}{m^2+1}\right|\right)\tag{1}$$
Given $\varepsilon>0$ by choosing $N_0$ such that $\frac{1}{1+N_0^2}<\varepsilon/3$ we get
$$n,m >N_0\qquad\implies\qquad |a_n-a_m|\le 3\max\left(\left|\frac{1}{n^2+1}\right|,\left|\frac{1}{m^2+1}\right|\right)<3\frac{1}{1+N_0^2}<\varepsilon$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1539786",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
} |
$\int\frac{(2x+3)dx}{(x^2+2x+3)\sqrt{x^2+2x+4}}$ $\int\frac{(2x+3)dx}{(x^2+2x+3)\sqrt{x^2+2x+4}}$
$I=\int\frac{(2x+3)dx}{(x^2+2x+3)\sqrt{x^2+2x+4}}=\int\frac{(2x+2)dx}{(x^2+2x+3)\sqrt{x^2+2x+4}}+\int\frac{1 dx}{(x^2+2x+3)\sqrt{x^2+2x+4}}$
Let first integral be called $I_1$ and the second integral be called $I_2$.
$I_1=\int\frac{dt}{t\sqrt{t+1}}$,where $x^2+2x+3=t$
$I_1=\int\frac{2dp}{p^2-1}$,where $p=\sqrt{t+1}$
$I_1=\log\frac{p-1}{p+1}=\log\frac{\sqrt{t+1}-1}{\sqrt{t+1}+1}=\log\frac{\sqrt{x^2+2x+4}-1}{\sqrt{x^2+2x+4}+1}+c$
But i could not find $I_2$,please help me.Thanks.
| Hint: the following substitutions reduce the original integral into a pair of very familiar ones.
$$\begin{align}
I
&=\int\frac{2x+3}{\left(x^2+2x+3\right)\sqrt{x^2+2x+4}}\,\mathrm{d}x\\
&=\int\frac{2y+1}{\left(y^2+2\right)\sqrt{y^2+3}}\,\mathrm{d}y;~~~\small{\left[x+1=y\right]}\\
&=\int\frac{2}{w^2-1}\,\mathrm{d}w;~~~\small{\left[\sqrt{y^2+3}=w\right]}\\
&~~~~~+\int\frac{\mathrm{d}t}{2+t^2};~~~\small{\left[\frac{y}{\sqrt{y^2+3}}=t\right]}.\\
\end{align}$$
I presume you can take it from there.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1544196",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Evaluating $\lim_{x\to 0}\frac{\sin(x)\arcsin(x)-x^2}{x^6}$ Step by Step Using L' Hopital Rule The limit to be found is
$$ \lim_{x\to 0}\frac{\sin(x)\arcsin(x)-x^2}{x^6}$$
I've tried l'hopital rule but it gets really messy. I've also tried splitting it into 2 limits but that doesn't work. I can't think of any meaningful substitution either.
PS: I know the answer $1/18$ but I'm interested in the method. Thank you.
| Some basic limits can be obtained by repeated use of
l'Hospital's Rule easily,
\begin{eqnarray*}
\lim_{x\rightarrow 0}\frac{\sin x-x}{x^{3}} &=&-\frac{1}{6},\ \ \ \ \ \ \ \
and\ \ \ \ \ \ \ \ \ \lim_{x\rightarrow 0}\frac{\arcsin x-x}{x^{3}}=\frac{1}{%
6} \\
\lim_{x\rightarrow 0}\frac{\sin x-x+\frac{1}{6}x^{3}}{x^{5}} &=&\frac{1}{120}%
,\ \ \ \ \ \ \ \ and\ \ \ \ \ \ \ \ \ \lim_{x\rightarrow 0}\frac{\arcsin x-x-%
\frac{1}{6}x^{3}}{x^{5}}=\frac{3}{40}
\end{eqnarray*}
Now, re-write the original expression as follows:
\begin{eqnarray*}
\frac{\sin x\arcsin x-x^{2}}{x^{6}} &=&\frac{(\left[ \sin x-x\right]
+x)([\arcsin x-x]+x)-x^{2}}{x^{6}} \\
&=&\frac{\left[ \sin x-x\right] [\arcsin x-x]+x\left[ \sin x-x\right]
+x[\arcsin x-x]+x^{2}-x^{2}}{x^{6}} \\
&=&\frac{\left[ \sin x-x\right] [\arcsin x-x]}{x^{6}}+\frac{x\left[ \sin x-x+%
\frac{1}{6}x^{3}\right] +x[\arcsin x-x-\frac{1}{6}x^{3}]}{x^{6}} \\
&=&\frac{\left[ \sin x-x\right] }{x^{3}}\frac{[\arcsin x-x]}{x^{3}}+\frac{%
\left[ \sin x-x+\frac{1}{6}x^{3}\right] }{x^{5}}+\frac{[\arcsin x-x-\frac{1}{%
6}x^{3}]}{x^{5}}
\end{eqnarray*}
It follows that
\begin{eqnarray*}
\lim_{x\rightarrow 0}\frac{\sin x\arcsin x-x^{2}}{x^{6}} &=&\lim_{x%
\rightarrow 0}\frac{\left[ \sin x-x\right] }{x^{3}}\lim_{x\rightarrow 0}%
\frac{[\arcsin x-x]}{x^{3}}+\lim_{x\rightarrow 0}\frac{\left[ \sin x-x+\frac{%
1}{6}x^{3}\right] }{x^{5}}+\lim_{x\rightarrow 0}\frac{[\arcsin x-x-\frac{1}{6%
}x^{3}]}{x^{5}} \\
&=&\left( -\frac{1}{6}\right) \left( \frac{1}{6}\right) +\left( \frac{1}{120}%
\right) +\left( \frac{3}{40}\right) \\
&=&\frac{1}{18}.
\end{eqnarray*}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1549725",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 3
} |
How does one get from $p+3=3k+2$ then $2^{p+3} \equiv 4 \pmod 7$, to $5 \cdot 2^{p+3} -31 \equiv 3 \pmod7$ How does one get from $p+3=3k+2$ then $2^{p+3}\equiv4 \pmod 7$, to $5 \cdot 2^{p+3} -31 \equiv 3 \pmod7$.
I am just starting with modular arithmetic so any help would be greatly appreaciated.
Credit to user236182 for original answer, just wanted to know how he got to it
| $2^3=8$, so the remainder when you divide $2^3$ by 7 is 1. That means $2^3=1\pmod7$.
$2^{3k}=(2^3)^k$ because of power laws. This equals $(7+1)^k$. Expand this using the Binomial Theorem, and all the terms are multiples of 7 except for the final term which is 1. So $2^{3k}=1\pmod7$.
Lastly, $2^{3k+2}=2^{3k}2^2=4.2^{3k}$. But $2^{3k}=7m+1$ for some number $m$, so $4.2^{3k}=4(7m+1)=7(4m)+4=4\pmod7$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1550027",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Locus of the point of intersection of the pair of perpendicular tangents to the circles $x^2+y^2=1$ and $x^2+y^2=7$ Locus of the point of intersection of the pair of perpendicular tangents to the circles $x^2+y^2=1$ and $x^2+y^2=7$ is the director circle of the circle with radius
$(A)\sqrt2\hspace{1cm}(B)2\hspace{1cm}(C)2\sqrt2\hspace{1cm}(D)4$
Since $x^2+y^2=1$ and $x^2+y^2=7$ are the concentric circles with center $O$(say).Let $P$ be any point on the required locus.Let $PT_1$ be the tangent from $P$ to the circle $x^2+y^2=7$ and let $PT_2$ be the tangent from $P$ to the circle $x^2+y^2=1$.Since $PT_1$ and $PT_2$ are perpendicular to each other.
But i dont know how to find the required locus.Please help me.Thanks.
| You know it's a circle with center $O$, so you just need a point to find the radius.
You can take a tangent the the first circle: $y=1$ and a perpendicular tangent to the other circle: $x=\sqrt{7}$ , the intersection $(1,\sqrt{7})$ is on the circle: the radius is
$$\sqrt{1^2+\sqrt{7}^2}=\sqrt{8}=2\sqrt{2}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1551153",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
If $\tan A + \tan B + \tan C = 3 \sqrt{3}$, then $\triangle ABC$ is equilateral.
Given $\triangle ABC$ such that
$$\tan A + \tan B + \tan C = 3 \sqrt{3}$$
prove that $\triangle ABC$ is equilateral.
The full process is needed.
| First we prove $\tan A + \tan B + \tan C = \tan A \tan B \tan C$.
We know that
$A + B + C = 180$
from the angle addition formula, so
$\tan((A + B) + C) = \frac{\tan(A + B) + \tan C}{1−\tan(A + B) \tan C}$
$= \frac{\tan A + \tan B + \tan C - \tan A \tan B \tan C}{1 - \tan A \tan B - \tan A \tan C - \tan B \tan C}$
We know that tan $180 = 0$, and since $A+B+C = 0$, we have
$\tan A + \tan B + \tan C - \tan A \tan B \tan C = 0$
$\tan A + \tan B + \tan C = \tan A \tan B \tan C$
Now we apply the AM-GM inequality, $\frac{a+b+c}3 \ge abc^{\frac13}$. ($x^{\frac13}$ is the cube root)
$\frac{\tan A + \tan B + \tan C}3 \ge (\tan A \tan B \tan C)^{\frac13}$
$\frac{\tan A \tan B \tan C}3 \ge (\tan A \tan B \tan C)^{\frac13}$
We substitute $x$ for $\tan A \tan B \tan C$.
$\frac x3 \ge x^{\frac13}$
$x \ge 3 x^{\frac13}$
$x^3 \ge 27 x$
$x^2 \ge 27$
$x \ge 3 \sqrt 3 $
$\tan A \tan B \tan C >= 3 \sqrt(3)$
$\tan A + \tan B + \tan C >= 3 \sqrt(3)$
The AM-GM inequality only becomes an equality when $x_1 = x_2 = x_3$, as can be seen from this page on wikipedia.
Thus, we know that $\tan A + \tan B + \tan C = 3 \sqrt 3$ only when $\tan A = \tan B = \tan C$. This will only be true when all angle are the same, so for $\tan A + \tan B + \tan C = 3 \sqrt 3$ to be true, the triangle has to be equilateral.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1551375",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Find the eigenvalues and associated eigenvectors for this matrix $$A = \begin{pmatrix} 1 & -1 & 4 \\ 3 & 2 & -1 \\ 2 & 1 & -1\end{pmatrix}$$
I solved for the determinant of matrix $A$ and got the following:
$\det\begin{bmatrix}
1-\lambda & -1 & 4\\
3 & 2-\lambda & -1 \\
2 & 1 & -1-\lambda
\end{bmatrix} = -\lambda^{3}+2\lambda^{2}+5\lambda-6$
How do I find the eigenvalues from this?
Once I have the eigenvalues, do I get the eigenvectors from the bases?
| To find eigenvalues, note that you want to find the solutions to the characteristic equation given by
$\det (A - \lambda I) = 0$
As you noted, we have the equation
$-\lambda^3 + 2\lambda^2 + 5 \lambda - 6 = 0$
Note that this is just
$-(\lambda - 3)(\lambda + 2)(\lambda -1) = 0$
so your eigenvalues are $\lambda_1 = 3, \lambda_2 = -2,$ and $\lambda_3 = 1$.
Now, to get the eigenvectors, you want to look for $v$ such that $Av = \lambda v$ or $(A- \lambda I) v = 0$.
For $\lambda_1$,
$A- \lambda I =
\begin{pmatrix}
-2 & -1 & 4 \\
3 & -1 & -1 \\
2 & 1 & -4
\end{pmatrix}$
so row reducing gives
$\begin{pmatrix}
1 & 0 & -1 \\
0 & 1 & -2 \\
0 & 0 & 0
\end{pmatrix}$
In other words, for your eigenvector $v_1 = (a_1,a_2,a_3)$,
\begin{align*}
a_1 - a_3 &= 0 \\
a_2 - 2a_3 &= 0
\end{align*}
Let $a_3 = t$, then $a_1 = t$ and $a_2 = 2t$ so a basis for the eigenspace corresponding to $\lambda_1$ is given by $\{\begin{pmatrix}
1 \\
2 \\
1 \\
\end{pmatrix}
\}$.
So, for $\lambda_1 =3$, we get a corresponding eigenvector $(1,2,1)^T$. Try finding the eigenvectors for $\lambda_2$ and $\lambda_3$!
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1552375",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Express $\frac{\sin 7\theta}{\sin \theta}$ in powers of $\sin \theta$ only By using DeMoivre's theorm express
$$\frac{\sin 7\theta}{\sin \theta}$$
in the powers of Sine only
answer given in the book is
$$7-56\sin ^2\theta+112\sin ^4 \theta-64\sin^6 \theta$$
can any one help to solve the question
| $(\cos x+i\sin x)^7=\cos 7x+i\sin7x$.
We have
$$(r+t)^7=r^7+7r^6t+21r^5t^2+35r^4t^3+35r^3t^4+21r^2t^5+7rt^6+t^7$$
hence, making $\sin x=t$ and $\cos x=r$ and having in account the powers of $i$, we get
$$\sin 7x=7r^6t-35r^4t^3+21r^2t^5-t^7$$ so, because of $\cos^2x=1-\sin^2x$,
$$7t-21t^3+21t^5-7t^7-35t^3+70t^5-35t^7+21t^5-21t^7-t^7=-64t^7+112t^5-56t^3+7t$$
Now dividing by $t$, we finish.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1553047",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 2
} |
How to simplify an expression? I have tried to simplify this expression for quite a long time now but I can't find how to do it.
$$\left(\frac{1}{2+4m}-\frac{1-m}{8m^3+1}:\frac{1-2m}{2m^2-2m+1}\right)\cdot\frac{4m+2}{2m-1}-\frac{1}{1-4m+4m^2}$$
Can someone help me with it?
| $\require{cancel}$
Starting with the original expression:
$$\left(\frac{1}{2+4m}-\frac{1-m}{8m^3+1}:\frac{1-2m}{2m^2-2m+1}\right)\cdot\frac{4m+2}{2m-1}-\frac{1}{1-4m+4m^2}$$
Factoring where possible and changing the order of some terms:
$$\left(\frac{1}{2(2m+1)}+\frac{m-1}{(2m+1)(4m^2-2m+1)}:\frac{-(2m-1)}{2m^2-2m+1}\right)\cdot\frac{2(2m+1)}{2m-1}-\frac{1}{(2m-1)^2}$$
Converting division into multiplication by the reciprocal; distributing the outside fraction:
$$\left(\frac{1}{2(2m+1)}\frac{2(2m+1)}{2m-1}+\frac{m-1}{(2m+1)(4m^2-2m+1)}\frac{2m^2-2m+1}{-(2m-1)}\frac{2(2m+1)}{2m-1}\right)-\frac{1}{(2m-1)^2}$$
Some cancellations:
$$\hspace{-0.8mm}\left(\frac{1}{\cancel{2(2m+1)}}\frac{\cancel{2(2m+1)}}{2m-1}+\frac{m-1}{\cancel{(2m+1)}(4m^2-2m+1)}\frac{2m^2-2m+1}{-(2m-1)}\frac{2\cancel{(2m+1)}}{2m-1}\right)-\frac{1}{(2m-1)^2}$$
$$\implies\frac{1}{2m-1}-\frac{2(m-1)}{4m^2-2m+1}\frac{2m^2-2m+1}{(2m-1)^2}-\frac{1}{(2m-1)^2}$$
Finding common denominators:
$$\frac{8m^3-8m^2+4m-1}{(4m^2-2m+1)(2m-1)^2}-\frac{4m^3-8m^2+6m-2}{(4m^2-2m+1)(2m-1)^2}-\frac{4m^2-2m+1}{(4m^2-2m+1)(2m-1)^2}$$
Adding/subtracting:
$$\frac{4m^3-4m^2}{(4m^2-2m+1)(2m-1)^2}$$
Factoring one last time:
$$\frac{4m^2(m-1)}{(4m^2-2m+1)(2m-1)^2}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1554806",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Zero divided by zero must be equal to zero What is wrong with the following argument (if you don't involve ring theory)?
Proposition 1: $\frac{0}{0} = 0$
Proof: Suppose that $\frac{0}{0}$ is not equal to $0$
$\frac{0}{0}$ is not equal to $0 \Rightarrow \frac{0}{0} = x$ , some $x$ not equal to $0$ $\Rightarrow$ $2(\frac{0}{0}) = 2x$ $\Rightarrow$ $\frac{2\cdot 0}{0} = 2x$ $\Rightarrow$ $\frac{0}{0} = 2x$ $\Rightarrow$ $x = 2x$ $\Rightarrow$ $ x = 0$ $\Rightarrow$[because $x$ is not equal to $0$]$\Rightarrow$ contradiction
Therefore, it is not the case that $\frac{0}{0}$ is not equal to $0$
Therefore, $\frac{0}{0} = 0$.
Q.E.D.
Update (2015-12-01) after your answers:
Proposition 2: $\frac{0}{0}$ is not a real number
Proof [Update (2015-12-07): Part 1 of this argument is not valid, as pointed out in the comments below]:
Suppose that $\frac{0}{0}= x$, where $x$ is a real number.
Then, either $x = 0$ or $x$ is not equal to $0$.
1) Suppose $x = 0$, that is $\frac{0}{0} = 0$
Then, $1 = 0 + 1 = \frac{0}{0} + \frac{1}{1} = \frac{0 \cdot 1}{0 \cdot 1} + \frac{1 \cdot 0}{1 \cdot 0} = \frac{0 \cdot 1 + 1 \cdot 0}{0 \cdot 1} = \frac{0 + 0}{0} = \frac{0}{0} = 0 $
Contradiction
Therefore, it is not the case that $x = 0$.
2) Suppose that $x$ is not equal to $0$.
$x = \frac{0}{0} \Rightarrow 2x = 2 \cdot \frac{0}{0} = \frac{2 \cdot 0}{0} = \frac{0}{0} = x \Rightarrow x = 0 \Rightarrow$ contradiction
Therefore, it is not the case that $x$ is a real number that is not equal to $0$.
Therefore, $\frac{0}{0}$ is not a real number.
Q.E.D.
Update (2015-12-02)
If you accept the (almost) usual definition, that for all real numbers $a$, $b$ and $c$, we have $\frac{a}{b}=c$ iff $ a=cb $, then I think the following should be enough to exclude $\frac{0}{0}$ from the real numbers.
Proposition 3: $\frac{0}{0}$ is not a real number
Proof: Suppose that $\frac{0}{0} = x$, where $x$ is a real number.
$\frac{0}{0}=x \Leftrightarrow x \cdot 0 = 0 = (x + 1) \cdot 0 \Leftrightarrow \frac{0}{0}=x+1$
$ \therefore x = x + 1 \Leftrightarrow 0 = 1 \Leftrightarrow \bot$
Q.E.D.
Update (2015-12-07):
How about the following improvement of Proposition 1 (it should be combined with a new definition of division and fraction, accounting for the $\frac{0}{0}$-case)?
Proposition 4: Suppose $\frac{0}{0}$ is defined, so that $\frac{0}{0} \in \mathbb{R}$, and that the rule $a \cdot \frac{b}{c} = \frac{a \cdot b}{c}$ holds for all real numbers $a$, $b$ and $c$.
Then, $\frac{0}{0} = 0$
Proof: Suppose that $\frac{0}{0}=x$, where $x \ne 0$.
$x = \frac{0}{0} \Rightarrow 2x = 2 \cdot \frac{0}{0} = \frac{2 \cdot 0}{0} = \frac{0}{0} = x \Rightarrow x = 0 \Rightarrow \bot$
$\therefore \frac{0}{0}=0$
Q.E.D.
Suggested definition of division of real numbers:
If $b \ne 0$, then
$\frac{a}{b}=c$ iff $a=bc$
If $a=0$ and $b=0$, then
$\frac{a}{b}=0$
If $a \ne 0$ and $b=0$, then $\frac{a}{b}$ is undefined.
A somewhat more minimalistic version:
Proposition 5. If $\frac{0}{0}$ is defined, so that $\frac{0}{0} \in \mathbb{R}$, then $\frac{0}{0}=0$.
Proof: Suppose $\frac{0}{0} \in \mathbb{R}$ and that $\frac{0}{0}=a \ne 0$.
$a = \frac{0}{0} = \frac{2 \cdot 0}{0} = 2a \Rightarrow a = 0 \Rightarrow \bot$
$\therefore \frac{0}{0}=0$
Q.E.D.
| The first fault is in the very statement. You assume that there is a $0/0$ in the first place, you have to have that before you even assume that it's not equal to zero.
Then even if there were a $0/0$ you're assuming that there is a $2{0\over0}$ and then that it is equal to ${2\cdot 0\over0}={0\over0}$.
Finally you assume that if $2x=x$ you must have that $x=0$, even if you've got this far there's no reason to think that $x$ is a number and must obey the rules for numbers.
Your updated proposition is true, but your proof has some issues.
In the first case you use the identity ${a\over b}+{c\over d} = {ad+bc\over bd}$ which is only shown to be true if $bd\ne0$.
In the second case you use the identity $a{b\over c} = {ab\over c}$ which is only shown to be true if $c\ne 0$.
That is you use twice identities where the prerequisite is not fulfilled.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1554929",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "52",
"answer_count": 16,
"answer_id": 3
} |
Evaluate $\int e^{2\theta} \sin (3\theta)\ d\theta$
Evaluate $$\int e^{2\theta} \sin (3\theta)\ d\theta .$$
I am little stuck as to what I can do after this point. Please tell me if my method overall is flawed:
| You have some mistake:
$$
\int e^{2\theta}\sin(3\theta) d\theta=\frac{1}{2}e^{2\theta}\sin(3\theta)-\frac{3}{2}\int e^{2\theta}\cos(3\theta) d\theta=
$$
$$
=\frac{1}{2}e^{2\theta}\sin(3\theta)-\frac{3}{4} e^{2\theta}\cos(3\theta) -\frac{9}{4}\int e^{2\theta}\sin(3\theta) d\theta
$$
so:
$$
\left(1+\frac{9}{4} \right)\int e^{2\theta}\sin(3\theta) d\theta=\frac{1}{2}e^{2\theta}\sin(3\theta)-\frac{3}{4} e^{2\theta}\cos(3\theta)
$$
and
$$
\int e^{2\theta}\sin(3\theta) d\theta=\frac{e^{2\theta}}{13}\left(2\sin(3\theta)-3\cos(3\theta)\right)
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1556613",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 5,
"answer_id": 2
} |
is $\frac{a-b}{a+b} - \frac{c-d}{c+d}$ equal to $ \frac{a-c}{a+c}$? I am using values out of a cross-correlation analysis and my intuition tells me that the equation in the title is true, however I would like to prove it.
$$ \frac{a-b}{a+b} - \frac{c-d}{c+d} = \frac{a-c}{a+c}$$
Unfortunately my math skills are horrible and I could not find this equation on-line (which now makes me think that my intuition is faulty).
Could you at least point out what mental strategy you would follow to try to solve it?
Thanks for any help
PS= any suggestion to improve my way of asking questions is also welcome.
all the best
| $$\frac{a-b}{a+b}-\frac{c-d}{c+d}=$$
$$\frac{(a-b)(c+d)}{(a+b)(c+d)}-\frac{(c-d)(a+b)}{(c+d)(a+b)}=$$
$$\frac{(a-b)(c+d)-(c-d)(a+b)}{(a+b)(c+d)}=$$
$$\frac{2(ad-bc)}{(a+b)(c+d)}$$
If you would like to look if your identity is correct, set your variables:
Choose $a=2,b=3,c=4,d=5$:
$$ \frac{2-3}{2+3} - \frac{4-5}{4+5} = \frac{2-4}{2+4}$$
$$ \frac{-1}{5} - \frac{-1}{9} = \frac{-2}{6}$$
$$ -\frac{1}{5}+\frac{1}{9} = -\frac{1}{3}$$
$$ -\frac{4}{45} = -\frac{1}{3}$$
So you know that:
$$ -\frac{4}{45} \ne -\frac{1}{3}$$
So your identity is not correct
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1556923",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Convergence rate of the sequence $a_{n+1} = a_n-a_n^2, a_0=1/2$. The sequence converges to zero at a rate that seems to be slightly faster than $1/n$.
What are the best known results on the convergence rate of this sequence?
| The sequence is a monotonic decreasing one. Also, it is easy to see that it is always positive. Hence, $\{a_n\}$ converges.
Let $N$ be sufficiently large integer so that
$$
a_N - a_{N + 1} < \varepsilon_N,
$$
where $\varepsilon_n \to 0$. But we have
$$
a_N - a_{N + 1} = a_N (1 - 1 + a_N) = a_N^2 < \varepsilon_N.
$$
Hence, $a_N \to 0$ by squeeze theorem.
Now let $r_n = 2^{2^n}$, $b_n = b_{n - 1} (r_{n - 1} - b_{n - 1})$ and $b_0 = 1$. Then,
$$
a_n = \frac {b_n} {r_n}.
$$
Since
$$
\left(\frac {r_{k - 1}} {2} - b_{k - 1}\right)^2 = \frac {r_k} {4} - b_k
$$
for every nonnegative integer $k$, we deduce that
$$
\sum_{k = 1}^{n} \left(\frac {r_{k - 1}} {2} - b_{k - 1}\right)^2 = \sum_{k = 1}^{n} \frac {r_k} {4} - \sum_{k = 1}^{n} b_k,
$$
and, by Cauchy-Schwartz inequality, we have
$$
\sum_{k = 1}^{n} \frac {r_k} {4} - \sum_{k = 1}^{n} b_k > \frac {1} {n} \left (\sum_{k = 1}^{n} \frac {r_{k - 1}} {2} - \sum_{k = 1}^{n} b_{k - 1}\right)^2. \qquad \qquad (1)
$$
Denote
$$
R_n = r_1 + r_2 + \cdots + r_{n - 1}, \qquad B_n = b_1 + b_2 + \cdots + b_{n - 1}, \nonumber \\ \phi (n) = \sqrt {n \left (n + r_n - R_n + \frac {3} {2} - 4 b_n\right) - R_n}.
$$
After simplification, $(1)$ becomes
$$
\frac {2 R_n - 2n - 3} {4} - \frac {\phi (n)} {2} < B_n < \frac {2 R_n - 2n - 3} {4} + \frac {\phi (n)} {2}.
$$
Taking into account that $R_n = o (r_n)$ and $b_n = B_{n + 1} - B_n$, we have
$$
\frac {r_n} {n} < b_n < \frac {r_n} {n} (1 + o (1)).
$$
Hence, $a_n = \frac {1} {n} + o \left(\frac {1} {n}\right)$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1558592",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 1
} |
Define $a$ and $b$ rational numbers so they satisfy equation Define $a, b \in \mathbb{Q}$ so that
$$\frac{a}{\sqrt{7 + 4\sqrt{3}}} + \frac{b}{\sqrt{7 - 4\sqrt{3}}} = \sqrt{4 + 2\sqrt{3}}$$
Using $\sqrt{a \pm \sqrt{b}} = \sqrt{\frac{a + \sqrt{a^2 - b}}{2}} \pm \sqrt{\frac{a - \sqrt{a^2 - b}}{2}}$ I got $\frac{a}{2 + \sqrt{3}} + \frac{b}{2 - \sqrt{3}} = \sqrt{3}+1$ which results in $2(a+b)+\sqrt{3}(b-a) = \sqrt{3}+1$, I'm not sure how to proceed from that.
| Hint:
$$\frac{1}{\sqrt{7+4\sqrt{3}}}=2-\sqrt{3}$$
$$\frac{1}{\sqrt{7-4\sqrt{3}}}=2+\sqrt{3}$$
$$\sqrt{4+2\sqrt{3}}=1+\sqrt{3}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1558762",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
What are $\int\sqrt{a^2-x^2}\,\textrm{d}x, \int\sqrt{x^2+a^2}\,\textrm{d}x,\int\sqrt{x^2-a^2}\,\textrm{d}x$? Can someone confirm these equations below? I got it from my college textbook, unfortunately there are no proofs and more importantly I cannot seem to find any other sources that say have these equations.
$\displaystyle\int\sqrt{a^2-x^2}\,\textrm{dx}=\frac{x\sqrt{a^2-x^2}}{2}+\frac{a^2}{2}\sin^{-1}\left(\frac{x}{a}\right) + C$
$\displaystyle\int\sqrt{x^2+a^2}\,\text{dx}=\frac{x\sqrt{x^2+a^2}}{2}+\frac{a^2}{2}\ln\left(x+\sqrt{x^2+a^2}\right)+C$
$\displaystyle\int\sqrt{x^2-a^2}\,\textrm{dx}=\frac{x\sqrt{x^2-a^2}}{2}-\frac{a^2}{2}\ln\left(x+\sqrt{x^2-a^2}\right)+C$
| It's OK to me.
If you just want to confirm it, you may just differentiate each function on the right side to see if you get the integrand on the left hand side.
One may obtain these results by performing respectively the change of variable
$x:=a\sin t$, $x:=a\sinh t$ and $x:=a\cosh t$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1560912",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Values of quadratic form on unit circle We have the quadratic form $q(\begin{pmatrix}x\\y\end{pmatrix})=11x^2-16xy-y^2$.
Which values does $q$ take on the unit circle $x^2+y^2=1$?
I know that $q(x,y)$ is given by $q(x,y)=(x,y)\begin{pmatrix}11&-8\\-8&-1\end{pmatrix}\begin{pmatrix}x\\y\end{pmatrix}$ and that $\begin{pmatrix}11&-8\\-8&-1\end{pmatrix}=
\begin{pmatrix}-2&-1\\1&-2\end{pmatrix}
\begin{pmatrix}15&0\\0&-5\end{pmatrix}
\begin{pmatrix}-2&-1\\1&-2\end{pmatrix}^T$. But I don't know how I could use this further.
| From Joseph Curwen's hint:
$$\begin{align}q &= 11\cos^2\theta-16\cos \theta\sin \theta-\sin ^2\theta\\
&=12\cos^2\theta-16\cos \theta\sin \theta-1\end{align}$$
Now $$\sin 2\theta = 2\cos\theta\sin\theta\\\cos 2\theta = 2\cos^2 \theta - 1$$
So,
$$\begin{align}q &= 6\cos 2\theta - 8\sin 2\theta + 5\\&=6\cos t - 8\sin t + 5\end{align}$$
Where $t = 2\theta$ (since $\theta$ was arbitrary, there is no reason to stick with it). Now set $q' = 0$ and find the maximum and minimum values of $q$. Since $q$ is continuous, it will take on every value between them.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1560990",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Finding the nth derivative of $y=\frac4{(6x+8)^3}$ Find the $n$-th derivative of $y=\frac{4}{(6x+8)^3}$
\begin{align}
y' ={} & 4(-3)(6) \frac{1}{(6x+8)^4}
\\
y''={} & 4(-3)(-4)(6)^2 \frac{1}{(6x+8)^5}
\\
y'''={} & 4(-3)(-4)(-5)(6)^3\frac{1}{(6x+8)^6}
\end{align}
I recognise the pattern but can't interpret that into a formula.
| $y=\frac{4}{(6x+8)^3} y'$ so $\frac{\rm d}{\rm dx}(\ln y)=\frac{(6x+8)^3}{4}$.
Now you can find $\ln y$ and hence $y$, and then take the derivatives.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1562282",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 3
} |
Unitary transformation I have a matrix in following form
$$A=\begin{bmatrix}
1&0&0&0&0&0&0&0\\
0&-1&0&0&0&0&0&0\\
0&0&1&0&0&0&0&0\\
0&0&0&-1&0&0&0&0\\
0&0&0&0&1&0&0&0\\
0&0&0&0&0&-1&0&0\\
0&0&0&0&0&0&1&0\\
0&0&0&0&0&0&0&-1\\
\end{bmatrix}$$
I need some help to reduce the matrix in following form using unitary transformation
$$C=\begin{bmatrix}
1&0&0&0&0&0&0&0\\
0&1&0&0&0&0&0&0\\
0&0&1&0&0&0&0&0\\
0&0&0&1&0&0&0&0\\
0&0&0&0&-1&0&0&0\\
0&0&0&0&0&-1&0&0\\
0&0&0&0&0&0&-1&0\\
0&0&0&0&0&0&0&-1\\
\end{bmatrix}$$
I have used the unitary matrix transformation$UAU^*$ method but the result always come in following form
$$UAU^*=\begin{bmatrix}
1&0&0&0&0&0&0&0\\
0&1&0&0&0&0&0&0\\
0&0&-1&0&0&0&0&0\\
0&0&0&-1&0&0&0&0\\
0&0&0&0&1&0&0&0\\
0&0&0&0&0&1&0&0\\
0&0&0&0&0&0&-1&0\\
0&0&0&0&0&0&0&-1\\
\end{bmatrix}$$
| Denote by $e_i$ the standard basis for $\mathbb{C}^8$. You are given a matrix $A$ such that
$$ Ae_i = \begin{cases} e_i & i \text{ is odd}, \\
-e_i & i \text{ is even}. \end{cases} $$
and you look for a unitary matrix $U \in M_{8}(\mathbb{C})$ such that
$$ UAU^{*}(e_i) = \begin{cases} e_i & 1 \leq i \leq 4, \\
-e_i & 5 \leq i \leq 8. \end{cases} $$
We can achieve this by letting $U^{*}$ (and hence also $U$) be a permutation matrix that rearranges the $e_i$'s. If we define $U^{*}$ by
$$ U^{*}e_1 = e_1, U^{*}e_2 = e_3, U^{*}e_3 = e_5, U^{*}e_4 = e_7, U^{*}e_5 = e_2, U^{*}e_6 = e_4, U^{*}e_7 = e_6, U^{*}e_8 = e_8 $$
then you can check that $U^{*}$ is unitary and $U = U^{-1}$ acts on the standard basis vectors by
$$ Ue_1 = e_1, Ue_2 = e_5, Ue_3 = e_2, Ue_4 = e_6, Ue_5 = e_3, Ue_6 = e_7, Ue_7 = e_4, Ue_8 = e_8. $$
Calculating explicitly, we have
$$ UAU^{*}(e_i) = \begin{cases}
UAU^{*}(e_1) = UAe_1 = Ue_1 = e_1 & i = 1, \\
UAU^{*}(e_2) = UAe_3 = Ue_3 = e_2 & i = 2, \\
UAU^{*}(e_3) = UAe_5 = Ue_5 = e_3 & i = 3, \\
\ldots \\
UAU^{*}(e_8) = UAe_8 = U(-e_8) = -e_8 & i = 8.
\end{cases} $$
as required and
$$ U = \begin{pmatrix} 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\
0 & 0 & 1 & 0 & 0 & 0 & 0 & 0 \\
0 & 0 & 0 & 0 & 1 & 0 & 0 & 0 \\
0 & 0 & 0 & 0 & 0 & 0 & 1 & 0 \\
0 & 1 & 0 & 0 & 0 & 0 & 0 & 0 \\
0 & 0 & 0 & 1 & 0 & 0 & 0 & 0 \\
0 & 0 & 0 & 0 & 0 & 1 & 0 & 0 \\
0 & 0 & 0 & 0 & 0 & 0 & 0 & 1
\end{pmatrix}. $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1565652",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Convergence of $\frac{1}{5} + \frac{1}{9} + \frac{1}{13} + ... = \sum_{i=1}^{\infty} \frac{1}{1+4i}.$ I've been working on an approximation for a problem in my numerical methods course, and I seemed to have run into the series
$$\frac{1}{5} + \frac{1}{9} + \frac{1}{13} + ... = \sum_{i=1}^{\infty} \frac{1}{1+4i}.$$
I'm trying to figure out the right test for this example. I haven't touched Calc II in a while, and I'm really not interested in getting a weird explanation from Wolfram. I was thinking that I might want to do some kind of ratio test between the $n$th and $n+1$th term.
| Notice that:
$$(\frac{1}{5}+\frac{1}{5}+\frac{1}{5}+\frac{1}{5})+(\frac{1}{9}+\frac{1}{9}+\frac{1}{9}+\frac{1}{9})+...\ge(\frac{1}{5}+\frac{1}{6}+\frac{1}{7}+\frac{1}{8})+(\frac{1}{9}+\frac{1}{10}+\frac{1}{11}+\frac{1}{12})+..$$
So $4\sum a_n$ diverges, hence also $\sum a_n$ diverges.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1566947",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 1
} |
How is this summation simplified? I'm taking EdX's probability class and one question asked to find the expectation of a uniform random variable (problem described here: https://youtu.be/vB6EKsX12hc).
There's one part of the calculation that I don't understand:
how does the numerator in the summation get reduced to a nice and simple k? Where did b+1go?
| $\sum_{i=1}^a 1 = a$. Sot there's that.
So we just need to show $\sum_{k=a + 1}^b \frac {b-k +1}{b-a - 1} = \frac{1}{b -a - 1}\sum_{k=1}^{b-a} k$.
Which is $\sum_{k=a + 1}^b \frac {b-k +1}{b-a - 1} =\frac{1}{b -a - 1}\sum_{k=a + 1}^{b} (b - k +1) = \frac{1}{b -a - 1}\sum_{k=1}^{b-a} (b - a - k +1) =\frac{1}{b -a - 1}\sum_{k=1}^{b-a} (b - a - k +1)$
Now for every $1 \le k \le b-a$ there is a unique $i; k = b -a + 1 - i; 1 \le i \le b - a$.
So $\sum_{k=1}^{b-a} (b - a - k +1)= \sum_{i=1}^{b-a} i = \sum_{k=1}^{b-a} k$ so $\sum_{k=a + 1}^b \frac {b-k +1}{b-a - 1} =\frac{1}{b -a - 1}\sum_{k=a + 1}^{b} (b - k +1) = \frac{1}{b -a - 1}\sum_{k=1}^{b-a} (b - a - k +1) =\frac{1}{b -a - 1}\sum_{k=1}^{b-a} k $
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1567032",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Polynomial $P(x)$ contains only terms of odd degree.When $P(x)$ is divided by $(x-3)$ ,the remainder is $6$. Polynomial $P(x)$ contains only terms of odd degree.When $P(x)$ is divided by $(x-3)$ ,the remainder is $6$.If $P(x)$ is divided by $(x^2-9)$ then remainder is $g(x).$Find the value of $g(2).$
As $P(x)$ is a polynomial containing only terms of odd degree.Therefore it should pass through origin.
$P(0)=0$
When $P(x)$ is divided by $(x-3)$ ,the remainder is $6$.
$P(x)=Q_1(x)(x-3)+6....................(1)$
If $P(x)$ is divided by $(x^2-9)$ then remainder is $g(x)$.
$P(x)=Q_2(x)(x^2-9)+ax+b............(2)$,where $ax+b=g(x)$
From $(1),P(3)=6$
From $(2),P(3)=6=3a+b$
But we need to find $g(2)=2a+b$
I dont know how to solve further.Please help me.Thanks.
| Keep in mind, $P(x)$ is an odd polynomial, so $P(-3)=-6=-3a+b$. So we adding that to $6=3a+b$, we get $0=2b$, so $b=0$. That leaves $a=2$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1567477",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
How to solve the equation of $\sqrt{x}+\sqrt{y}=\sqrt{2205}$ in integers? How to solve the equation of $\sqrt{x}+\sqrt{y}=\sqrt{2205}$ in integers?
How in general to solve the similar equations?
| Simple factorization yields $\sqrt{2205} = 21\sqrt{5} $
$$\sqrt{x}=\sqrt{2205}-\sqrt{y}$$
$$x=2205+y-2.21\sqrt{5}\sqrt{y}$$
$$42\sqrt{5y}=2205+y-x$$
$$y\text{ is in the form of } 5k^2$$
$$y=k\sqrt{5}\Rightarrow\sqrt{x}=l{5}$$
Then we have,
$\sqrt{x}=0,\sqrt{y}=21\sqrt{5}$
$\sqrt{x}=\sqrt{5},\sqrt{y}=20\sqrt{5}$ and so on...
For these type of problems, use the fact that $a+\sqrt{b}=c+\sqrt{d}$ where b,d are not perfect squares $\Rightarrow a=c \text{ and } b=d$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1575392",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Divisibility of $2^n-n^2$ by 7 How many positive integers $n<10^4$ are there such that $2^n - n^2$ is divisible by 7?
| Just write out periods:
$2^n:2,4,1,2,4,1,2,4,1,2,4,1,2,4,1,2,4,1,2,4,1,|2...$ with period $3$
$n^2:1,4,2,2,4,1,0,1,4,2,2,4,1,0,1,4,2,2,4,1,0,|1...$ with period $7$
So in every $21$ numbers there are $6$ numbers.
Now just calculate $10^4\over 21$$=476$ and since $10000$ itself is not included there are $3$ remainder left so totally $6\cdot476+1=2857$ numbers.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1576496",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Calculating $\lim_{n \to \infty}\frac{n^{3}}{(3+\frac{1}{n})^{n}}$ I need help to calculate this limit:
$$\lim_{n \to \infty}\frac{n^{3}}{(3+\frac{1}{n})^{n}}$$
| And this is why $\lim_{n\to \infty}\frac{n^3}{3^n} = 0$.
${3^n}=(1+2)^n=1+n2+\frac{n(n-1)}{2}2^2+\frac{n(n-1)(n-2)}{3!}2^3+\frac{n(n-1)(n-2)(n-3)}{4!}2^4+...+2^n>\frac{n(n-1)(n-2)(n-3)}{4!}$
And so,
$0 \le \lim_{n\to \infty}\frac{n^3}{3^n} \le \lim_{n\to \infty}\frac{n^3}{\frac{n(n-1)(n-2)(n-3)}{4!}}= 0$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1580904",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
} |
distinct roots of the equation $A\sin^3 x+B\cos^3 x+C =0$
The number of distinct roots of the equation $A\sin^3 x+B\cos^3 x+C =0$
no two of which differ by $2\pi$ is, Where $A,B,C\in \mathbb{R}$
$\bf{(a)}\;\;\;\;\;\; 3\;\;\;\;\;\; (b)\;\;\;\;\;\; 4\;\;\;\;\;\; (c)\;\;\;\;\;\; 6\;\;\;\;\;\; (d)\;\;\;\;\;\; infinite$
$\bf{My\; Try::}$ Using $$\sin 3x =3\sin x -4\sin^3 x\Rightarrow 4\sin^3 x=3\sin x-\sin 3x$$ and
$$\cos 3x = 4\cos^3 x-3\cos x\Rightarrow 4\cos^3 x=\cos 3x+3\cos x$$
So we get $$A(4\sin^3 x) +B(4\cos^3 x)+4C=0$$
So $$A(3\sin x-\sin 3x)+B(\cos 3x+3\cos x)+4C=0$$
I did not understand How can I solve that question, Help me
Thanks
| See after simplifying use double angle formula so $(2\sin x+\sin x-\sin 3x)=2\sin x-2\cos 2x \sin x=2 \sin x(1-\cos 2x)$ same goes for other so it becomes $2\cos x(1+\cos 2x)$ thus simplifying further you get $a \sin x(1-\cos 2x)+b\cos x(1+\cos 2x)+2c=0$ now you just need values of $cos,sin$ where $c$ has a considerable limit according to $\sin,\cos$ . If $a$ and $b$ are big too then its simply infinite values.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1580975",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
Find product of solutions of $x^6=-64$
If the six solutions of $x^6=-64$ are written in the form $a+bi$,where
$a$ and $b$ are real, then find the product of those solutions with $a>0$.
The answer in my book is given as $4$ but I don't see why.
Can you give me some hint on that?
| $$x^6=-64\Longleftrightarrow$$
$$x^6=64e^{\pi i}\Longleftrightarrow$$
$$x=\left(64e^{(2\pi k+\pi)i}\right)^{\frac{1}{6}}\Longleftrightarrow$$
$$x=2e^{\frac{1}{6}(2\pi k+\pi)i}\Longleftrightarrow$$
$$x=\begin{cases}2e^{\frac{1}{6}(2\pi\cdot0+\pi)i}\\
2e^{\frac{1}{6}(2\pi\cdot1+\pi)i}\\
2e^{\frac{1}{6}(2\pi\cdot2+\pi)i}\\
2e^{\frac{1}{6}(2\pi\cdot3+\pi)i}\\
2e^{\frac{1}{6}(2\pi\cdot4+\pi)i}\\
2e^{\frac{1}{6}(2\pi\cdot5+\pi)i}\end{cases}\Longleftrightarrow$$
$$x=\begin{cases}2e^{\frac{\pi i}{6}}\\
2i\\
2e^{\frac{5\pi i}{6}}\\
2e^{-\frac{5\pi i}{6}}\\
-2i\\
2e^{-\frac{\pi i}{6}}\end{cases}\Longleftrightarrow$$
$$x=\begin{cases}2e^{\pm\frac{\pi i}{6}}\\
\pm 2i\\
2e^{\pm\frac{5\pi i}{6}}
\end{cases}\Longleftrightarrow$$
$$x=\begin{cases}\sqrt{3}\pm1i\\
\pm 2i\\
-\sqrt{3}\pm1i
\end{cases}\Longleftrightarrow$$
$$x=\begin{cases}\sqrt{3}\pm i\\
\pm 2i\\
-\sqrt{3}\pm i
\end{cases}$$
With $k\in\mathbb{Z}$ and $k:0-5$
So only the first one is the one you need, notice that for $a,b\in\mathbb{R}$
$(a+bi)(a+bi)^*=(a+bi)(a-bi)=a^2+b^2$:
$$\left(\sqrt{3}+i\right)\left(\sqrt{3}-i\right)=\left(\sqrt{3}\right)^2+1^2=3+1=4$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1583626",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
Formula for $\sum \limits_{n=0}^{\infty} \frac{1}{(n+a)!}$ Is there a closed form for the infinite sum $$\sum \limits_{n=0}^{\infty} \frac{1}{(n+a)!} \mathrm{?}$$ where a is an integer greater than or equal to $0$.
When $a=0$, the sum is just the series expansion for $e$, $\sum \limits_{n=0}^{\infty} \frac{1}{n!}=e$
When $a=1$, the sum can be rewritten as $\sum \limits_{n=0}^{\infty} \frac{1}{(n+1)!}= \sum \limits_{n=1}^{\infty} \frac{1}{n!} =\frac{1}{1!}+\frac{1}{2!}+\frac{1}{3!}+
\dots=e-1$.
When $a=2$, the sum can be rewritten as $\sum \limits_{n=0}^{\infty} \frac{1}{(n+2)!}= \sum \limits_{n=2}^{\infty} \frac{1}{n!} =\frac{1}{2!}+\frac{1}{3!}+\frac{1}{4!}+\dots=e-2$.
When $a=3$, the sum can be rewritten as $\sum \limits_{n=0}^{\infty} \frac{1}{(n+3)!}= \sum \limits_{n=3}^{\infty} \frac{1}{n!} =\frac{1}{3!}+\frac{1}{4!}+\dots=e-\frac{5}{2}$.
$\vdots$
Thus, we can generalize that $$\sum \limits_{n=0}^{\infty} \frac{1}{(n+a)!}= \sum \limits_{n=a}^{\infty} \frac{1}{n!}.$$
Can this infinite sum be written in a closed form in terms of $a$? Following the pattern the general form would have some value, dependent on $a$, subtracted from $e$.
| Not clearly more convenient, but an alternative form can be built by first establishing that:
$$\int_0^1\frac{x^n}{n!}e^{-x}\,dx=1-\frac{1}{e}\sum_{k=0}^n\frac{1}{k!}$$
[This is best done by induction - if we call the integral on the left $I_n$, integrating by parts quickly gives us $I_n=I_{n-1}-\frac{1}{e\cdot n!}$.]
Noting that $\sum_{k=0}^{\infty}\frac{1}{k!}=e$, this can be recast as:
$$\int_0^1\frac{x^n}{n!}e^{-x}\,dx=\frac{1}{e}\sum_{k=n+1}^{\infty}\frac{1}{k!}$$
and shifting around the $n!$ and $e$ terms gives that
$$\sum_{n=a}^{\infty}\frac{1}{n!}=\int_0^1\frac{x^{a-1}}{(a-1)!}e^{1-x}\,dx$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1585822",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Solve the equation: $\cos^2(x)+\cos^2(2x)+\cos^2(3x)=1$ How to solve the equation
$$\cos^2(x)+\cos^2(2x)+\cos^2(3x)=1$$
Can anyone give me some hints?
| $\cos^2(x)+[2 \cos^2x-1]^2 +[4 \cos^3 x-3 \cos x]^2 = 1$, Simplify the equation to get: $ \cos^2 x [8 \cos^4 x - 10 \cos^2 x + 3] = 0$. Hope this will help.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1588032",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 5,
"answer_id": 4
} |
Let $k = 2008^2 + 2^{2008}$. What is the last digit of $k^2 + 2^k$? Let $k = 2008^2 + 2^{2008}$. What is the last digit of $k^2 + 2^k.$
I thought of this
$$2008^2+2^{2008}\pmod{10} ≡ {-2}^2+{2^4}^{502}\pmod{10} ≡ 4+{-4}^{502}\pmod{10} ≡ 4+6^{251} \pmod{10}$$
but I still cannot prove it. Maybe there is a clever solution but so far I have been unable to spot it. Can anyone help me?
| $k$ is a huge even number, hence $k^2+2^k$ is for sure an even number. To compute the last digit of $k^2+2^k$, we just need to understand what $k^2+2^k\pmod{5}$ is. We have $k\equiv 0\pmod{4}$ and
$$ k \equiv 3^2+4^{1004} \equiv -1+1\equiv 0\pmod{5}, $$
hence $k^2+2^k\equiv 0+1\equiv 1\pmod{5}$ by Fermat's little theorem, so the last digit of $k^2+2^k$ is $\color{red}{6}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1589634",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 1
} |
How to Simplify $-\frac14\sin\frac x4 + \frac{\sqrt{3}}4\cos\frac x4 = 0$? A problem requires finding the value of x, such that
$$-\frac14\sin\frac x4 + \frac{\sqrt{3}}4\cos\frac x4 = 0$$
The site reduces the problem to $\tan(\frac x4) = \sqrt{3}$.
How one should do this?
| $$-\frac{1}{4}\sin\left(\frac{x}{4}\right)+\frac{\sqrt{3}}{4}\cos\left(\frac{x}{4}\right)=0\Longleftrightarrow$$
$$\frac{1}{4}\left(\sqrt{3}\cos\left(\frac{x}{4}\right)-\sin\left(\frac{x}{4}\right)\right)=0\Longleftrightarrow$$
$$\sqrt{3}\cos\left(\frac{x}{4}\right)-\sin\left(\frac{x}{4}\right)=0\Longleftrightarrow$$
$$\sqrt{3}\cos\left(\frac{x}{4}\right)=\sin\left(\frac{x}{4}\right)\Longleftrightarrow$$
$$\sqrt{3}\cot\left(\frac{x}{4}\right)=1\Longleftrightarrow$$
$$\cot\left(\frac{x}{4}\right)=\frac{1}{\sqrt{3}}\Longleftrightarrow$$
$$\frac{x}{4}=\pi n+\text{arccot}\left(\frac{1}{\sqrt{3}}\right)\Longleftrightarrow$$
$$x=4\pi n+4\text{arccot}\left(\frac{1}{\sqrt{3}}\right)\Longleftrightarrow$$
$$x=4\pi n+\frac{4\pi}{3}\space\space\space\space\space\space\space\space\text{with}\space\space n\in\mathbb{Z}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1589875",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 6,
"answer_id": 5
} |
Solve the equation $27 \sin(x) \cdot \cos^2(x) \cdot \tan^3(x) \cdot \cot^4(x) \cdot \sec^5(x) \cdot \csc^6(x) = 256$.
Solve the equation $27 \sin(x) \cdot \cos^2(x) \cdot \tan^3(x) \cdot \cot^4(x) \cdot \sec^5(x) \cdot \csc^6(x) = 256$.
I was hoping some things would cancel out when I expanded this but nothing. I think using inequalities will help.
| since
\begin{align*}27\sin{x}\cos^2{x}\tan^3{x}\cot^4{x}\sec^5{x}\csc^6{x}&=27\sin{x}\cos^2{x}\cdot\dfrac{\sin^3{x}}{\cos^3{x}}\cdot\dfrac{\cos^4{x}}{\sin^4{x}}\cdot\dfrac{1}{\cos^5{x}}\cdot\dfrac{1}{\sin^6{x}}\\
&=\dfrac{27}{\cos^2{x}\sin^6{x}}
\end{align*}
so
$$\sin^6{x}\cos^2{x}=\dfrac{27}{256}$$
but on the other hand , we have
$$t(1-t)^3=\dfrac{27}{256}$$
since
$$ 256t(1-t)^3-27=-(4t-1)^2(16t^2-40t+27)$$
because $$16t^2-40t+27=(4t-5)^2+2>0$$
$$t=\dfrac{1}{4}$$
then
$$\cos{x}=\dfrac{1}{2}\rm{or}-\dfrac{1}{2}$$
where $t=\sin^2{x}\ge 0$
then you can do it
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1591141",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 5,
"answer_id": 4
} |
$\frac{1}{\sin 8^\circ}+\frac{1}{\sin 16^\circ}+....+\frac{1}{\sin 4096^\circ}+\frac{1}{\sin 8192^\circ}=\frac{1}{\sin \alpha}$,find $\alpha$ Let $\frac{1}{\sin 8^\circ}+\frac{1}{\sin 16^\circ}+\frac{1}{\sin 32^\circ}+....+\frac{1}{\sin 4096^\circ}+\frac{1}{\sin 8192^\circ}=\frac{1}{\sin \alpha}$ where $\alpha\in(0,90^\circ)$,then find $\alpha$(in degree.)
$\frac{1}{\sin 8^\circ}+\frac{1}{\sin 16^\circ}+\frac{1}{\sin 32^\circ}+....+\frac{1}{\sin 4096^\circ}+\frac{1}{\sin 8192^\circ}=\frac{1}{\sin \alpha}$
$\frac{2\cos8^\circ}{\sin 16^\circ}+\frac{2\cos16^\circ}{\sin 32^\circ}+\frac{2\cos32^\circ}{\sin 64^\circ}+....+\frac{2\cos4096^\circ}{\sin 8192^\circ}+\frac{1}{\sin 8192^\circ}=\frac{1}{\sin \alpha}$
$\frac{2^2\cos8^\circ\cos16^\circ}{\sin 32^\circ}+\frac{2^2\cos16^\circ\cos32^\circ}{\sin 64^\circ}+\frac{2^2\cos32^\circ\cos64^\circ}{\sin 128^\circ}+....+\frac{2\cos4096^\circ}{\sin 8192^\circ}+\frac{1}{\sin 8192^\circ}=\frac{1}{\sin \alpha}$
In this way this series is getting complicated at each stage,is there any way to simplify it?Please help me.Thanks.
| This is just the same idea as in lab bhattacharjee's answer, but using the identity from the Weierstrass Substitution
$$
\tan(x/2)=\frac{\sin(x)}{1+\cos(x)}
$$
we get
$$
\begin{align}
\frac1{\tan(x/2)}-\frac1{\tan(x)}
&=\frac{1+\cos(x)}{\sin(x)}-\frac{\cos(x)}{\sin(x)}\\
&=\frac1{\sin(x)}
\end{align}
$$
The rest is the same telescoping series
$$
\begin{align}
\sum_{k=0}^n\frac1{\sin\left(2^kx\right)}
&=\sum_{k=0}^n\left[\frac1{\tan\left(2^{k-1}x\right)}-\frac1{\tan\left(2^kx\right)}\right]\\
&=\frac1{\tan(x/2)}-\frac1{\tan\left(2^nx\right)}
\end{align}
$$
The question has $x=8^\circ$ and $n=10$, so we get
$$
\begin{align}
\sum_{k=0}^{10}\frac1{\sin\left(2^k8^\circ\right)}
&=\frac1{\tan(4^\circ)}-\frac1{\tan(8192^\circ)}\\
&=\frac1{\tan(4^\circ)}+\frac1{\tan(88^\circ)}\\
&=\frac1{\tan(4^\circ)}+\tan(2^\circ)\\
&=\frac{\cos(4^\circ)}{\sin(4^\circ)}+\frac{\sin(4^\circ)}{1+\cos(4^\circ)}\\
&=\frac{\cos(4^\circ)}{\sin(4^\circ)}+\frac{1-\cos(4^\circ)}{\sin(4^\circ)}\\
&=\frac1{\sin(4^\circ)}
\end{align}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1591220",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 2,
"answer_id": 0
} |
How to compute $\lim\limits_{x \to 0_+} \left(\frac{x^2+1}{x+1}\right)^{\frac{1}{\sin^2 x}}$? I have a problem with this limit, I dont know what method to use. After several attempts the eventually find was obtained ($\infty$). But in reality the result sought is $0$.
Can you explain the method and the steps used? Thanks.
$$\lim\limits_{x \to 0_+} \left(\frac{x^2+1}{x+1}\right)^{\frac{1}{\sin^2 x}}$$
| Note that $\frac{x^2+1}{x+1}=\frac{x+1 -(x-x^2)}{x+1}$. If $x$ is small positive, then $x-x^2\gt x/2$ and $x+1\lt 2$. It follows that if $x$ is small positive, then
$$0\lt \frac{x^2+1}{x+1}\lt 1-\frac{x}{4}.$$
Note also that if $x$ is small positive, then $\frac{1}{\sin^2 x}\gt \frac{1}{x^2}$. Thus if $x$ is small positive, then
$$0\lt \left(\frac{x^2+1}{x+1}\right)^{1/\sin^2 x}\lt \left(\left(1-\frac{x}{4}\right)^{1/x}\right)^{1/x}\tag{1}.$$
As $x\to0^+$, $(1-x/4)^{1/x}\to e^{-1/4}$, and therefore the right-hand side of (1) approaches $0$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1592141",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 6,
"answer_id": 2
} |
Prove $(1+\sec x)(1+\csc x) > 5$ if $x \in {\left]0,\frac{\pi}{2}\right[}$
Prove $(1+\sec x)(1+\csc x) > 5$ if $x \in {\left]0,\frac{\pi}{2}\right[}$.
I have done the following:
Let $t=\tan \frac{x}{2}\in ]0,1[$, then using the $t$-formulae the LHS becomes
$$(1+\sec x)(1+\csc x) = \frac{2t+1+t^2}{t(1-t^2)}$$
I could simply it further but then I keep lowering the bound, now I continue as follows:
$$\frac{2t+1+t^2}{t(1-t^2)} = \frac{2}{1-t^2} + \frac{1}{t(1-t^2)}+\frac{t^2-1}{t(1-t^2)}+\frac{1}{t(1-t^2)}$$
Now since $1-t^2$ and $t \in ]0,1[$ then $\frac{1}{t(1-t^2)}> 1$ and so is $\frac{1}{1-t^2}$ then:
$$> 2+1+1-\frac{1}{t}>4$$
Close, but no cigar. Where could I improve the bounding?
| Let $$y=\left(1+\dfrac{1+t^2}{1-t^2}\right)\left(1+\dfrac{1+t^2}{2t}\right)=\dfrac{2(1+t)^2}{2t(1-t^2)}=\dfrac{1+t}{t(1-t)}$$ as $t+1\ne0$
$$\implies yt^2-t(y-1)+1=0$$
As $t$ is real, the discriminant $$(y-1)^2-4y\ge0$$
As the roots of $y^2-6y+1=0$ are $y=\dfrac{6\pm\sqrt{32}}2=3\pm2\sqrt2$
$y^2-6y+1\ge0\implies$ either $y\ge3+2\sqrt2$ or $y\le3-2\sqrt2$
As $0< x<\dfrac\pi2, \csc x,\sec x\ge1\implies y>3\implies y\not\le3-2\sqrt2$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1592203",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Find $\lim_{n\to \infty}r_n$ and $\lim_{n\to \infty}s_n$
A sequence of polynomials is given by $$p_n(x) =a_{n+2}x^2 +a_{n+1}x-a_n$$ for
$n\geq 0$, where $a_0=a_1=1$ and for $n\geq 0$ $$a_{n+2}=a_{n+1} +a_n$$ Denote by $r_n$ and $s_n$ the roots of $p_n(x)=0$, with $r_n\leq s_n$. Find $\lim_{n\to \infty}r_n$ and $\lim_{n\to \infty}s_n$.
This is a Fibonacci sequence because for $n\geq 0$,$a_{n+2}=a_{n+1} +a_n$ and $a_0=a_1=1$ is a definition of Fibonacci sequence.
It is also clear that $x=-1$ is a root of the equation $$p_n(x) =a_{n+2}x^2 +a_{n+1}x-a_n$$
So it is clear that $\lim_{n\to \infty}r_n=-1$. But I cannot find $\lim_{n\to \infty}s_n.$
What should I do? I am stuck here.
| We can use sum of roots and product of roots.
For a quadratic equation $ax^2+bx+c=0$ with roots $s, t$, we have $s+t=-\frac{b}{a}$ and $st=\frac{c}{a}$.
Since we already have $1$ root, we can substitute it into the product of roots equation: $s_nr_n=\frac{-a_n}{a_{n+2}}$
Hence $s_n=\frac{a_n}{a_{n+2}}$.
Since Fibonacci numbers increase exponentially, i.e.
$$\lim_{n\rightarrow\infty}\frac{a_{n+1}}{a_n}=\frac{1+\sqrt5}{2}$$
we have:
$$\lim_{n\rightarrow\infty}s_n=\lim_{n\rightarrow\infty}\frac{a_{n}}{a_{n+1}}\times\frac{a_{n+1}}{a_{n+2}}=\left(\frac{1+\sqrt5}{2}\right)^{-2}=\left(\frac{3+\sqrt5}{2}\right)^{-1}=\frac{2}{3+\sqrt5}\times\frac{3-\sqrt{5}}{3-\sqrt{5}}=\frac{2(3-\sqrt{5})}{4}=\frac{3-\sqrt{5}}{2}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1593711",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Find $\lfloor z \rfloor $ given that $z=(\{\sqrt{3}\}^2-2\{\sqrt{2}\}^2)/(\{\sqrt{3}\}-2\{\sqrt{2}\})$ Let $\lfloor x \rfloor$ denote the greatest integer function, and $\{x\}=x-[x]$ the fractional part of $x$. If $$z=\cfrac{\{\sqrt{3}\}^2-2\{\sqrt{2}\}^2}{\{\sqrt{3}\}-2\{\sqrt{2}\}}$$ find $\lfloor z \rfloor $
My attempt
$$z=\cfrac{\{\sqrt{3}\}^2-2\{\sqrt{2}\}^2}{\{\sqrt{3}\}-2\{\sqrt{2}\}}=\cfrac{(\sqrt{3}-1)^2-2(\sqrt{2}-1)^2}{\sqrt{3}-1-2\cdot (\sqrt{2}-1)}=\cfrac{3+1-2(3 -2\sqrt{2})}{\sqrt{3}-1-2\cdot \sqrt{2}+2}$$
After rationalizing I have $$z=\cfrac{4\sqrt{2}+4\sqrt{6}-2-2\sqrt{3}}{-2} $$ (I hope I haven't made any careless mistake now )
Now I am stuck as I don't know how I should take $\lfloor z \rfloor$ as I don't have that if $x=\cfrac{a}{b}$ then $\lfloor x \rfloor =\cfrac{\lfloor a \rfloor}{\lfloor b \rfloor}$ .
I also think I haven't noticed a more straightforward way to do the problem ...
| We have
\begin{align*}
z = \cfrac{(\sqrt{3}-1)^2-2(\sqrt{2}-1)^2}{\sqrt{3}-1-2\cdot (\sqrt{2}-1)} &= \frac{3+1-2\sqrt{3}-2(2+1-2\sqrt2)}{\sqrt{3}-2\sqrt2+1}\\
&= \frac{-2\sqrt3 + 4\sqrt2-2}{\sqrt{3}-2\sqrt2+1}\\
&= -2\frac{\sqrt{3}-2\sqrt2+1}{\sqrt{3}-2\sqrt2+1}\\
&= -2.
\end{align*}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1593951",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Trigonometry with Quadratic Equations If $\tan A$ and $\tan B$ are the roots of $x^2+px+q=0$, then prove that
$$\sin^2(A+B)+p \sin(A+B) \cos(A+B) + q \cos^2(A+B) = q$$
I tried the question but with $q$ other terms came associated.
| Hint : Let $E=\sin^2(A+B)+p \sin(A+B) \cos(A+B) + q \cos^2(A+B) $
Divide $E$ by $\cos^2(A+B)$.
Then you will get $$E\sec^2(A+B)= \tan^2(A+B)+p \tan(A+B) + q $$
Since $$\tan(A+B)=\frac{−p}{1−q} \Rightarrow \sec^2(A+B)=1+\frac{p^2}{(1−q)^2}$$, provided $q≠1$.
$$E\sec^2(A+B)= \frac{p^2}{(1−q)^2}-\frac{p^2}{1−q} + q $$
$$E\cdot \left(1+\frac{p^2}{(1−q)^2}\right) = \frac{p^2}{(1−q)^2}-\frac{p^2}{1−q} + q $$
Provided $q≠1$.$$\Rightarrow E((1-q)^2+p^2)=p^2-p^2(1-q)+q(1-q)^2$$
$$\Rightarrow E((1-q)^2+p^2)=p^2q+q(1-q)^2$$
$$\Rightarrow E((1-q)^2+p^2)=q(p^2+(1-q)^2)$$
$$\Rightarrow E=q$$ .
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1594061",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 0
} |
Simultaneous Equations (Stuck on the algebra)
Question: Solve the following simultaneous equations for real values of x and y
$$
\left\{
\begin{array}{l}
9^{2x+y} - 9^x \times 3^y = 6 \\
\log_{x+1}(y+3) + \log_{x+1}(y+x+4) = 3
\end{array}
\right.
$$
What I have attempted; for the first equation
$$
9^{2x+y} - 9^x \times 3^y = 6 \\
9^{2x+y} - 3^{2x} \times 3^y = 6 \\
9^{2x+y} - 3^{2x+y} - 6 = 0
$$
Let $z = 3^{2x+y}$. Then
$$
z^2 - z - 6 = 0
\iff (z-3)(z+2) = 0
\iff z = 3, z \ne -2
$$
where
$$
3^{2x+y} = 3 \iff y = 1-2x.
$$
Now using the second equation I get
$$
\log_{x+1}(y+3) + \log_{x+1}(y+x+4) = 3
$$
Substituting $y = 1-2x$
$$
\log_{x+1}(4-2x) + \log_{x+1}(5-x) = 3.
$$
Now this is the part I am stuck on , how do I solve for $x$ algebraically?
| we have $$(3^{2x+y})^2-3^{2x+y}=6$$ and $$log_{x+1}(y+3)(y+x+4)=3$$
the last equation can be written as $$(y+3)(y+x+4)=(x+1)^3$$
from the quadratic equation we get $$2x+y=1$$ can you proceed?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1595562",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 2
} |
The value of the polynomial at given point. Given that:
$f(x)=x^{10}+2x^9-2x^8-2x^7+x^6+3x^2+6x+1$.
Find the value of $f(x)$ at $x=\sqrt{2}-1$
Answer is an integer. I tried factorization but couldn't proceed towards anything promising.
| Just compute the value for each power $$x=\sqrt{2}-1$$ $$x^2=x \times x=3-2 \sqrt{2}$$ $$x^3=x\times x^2=5 \sqrt{2}-7$$ $$x^6=x^3 \times x^3=99-70 \sqrt{2}$$ $$x^7=x\times x^6=169 \sqrt{2}-239$$ $$x^9=x^7\times x^2=985 \sqrt{2}-1393$$ $$x^{10}=x^7\times x^3=3363-2378 \sqrt{2}$$ With this kind of problem, just work in a systematic manner.
Happy New Year !!
Edit
Parth Kohli's answer provides another interesting way to compute the different powers of $x$ (his/her solution is much more elegant and faster than my brute force based method).
Let us consider that the value for which we need to compute the polynomial is one of the roots of the quadratic $x^2+2x-1=0$. So $$x^2=-2x+1$$ $$x^3=-2x^2+x=-2(-2x+1)+x=5x-2$$ $$x^4=5x^2-2x=5(-2x+1)-2x=-12x+5$$ $$x^5=-12x^2+5x=-12(-2x+1)+5x=29x-12$$ $$x^6=29x^2-12x=29(-2x+1)-12x=-70x+29$$ $$x^7=-70x^2+29x=-70(-2x+1)+29x=169x-70$$ $$x^8=169x^2-70x=169(-2x+1)-70x=-408x+169$$ $$x^9=-408x^2+169x=-408(-2x+1)+169x=985x-408$$ $$x^{10}=985x^2-408x=985(-2x+1)-408x=-2378x+985$$ Please notice the nice pattern in the numbers. Compute now the expression and get the final result (and notice that you will not use at all the fact that $x=\sqrt 2-1$; this is explained in Parth Kohli's answer by the constant remainder).
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1596166",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
"answer_id": 0
} |
Probability of choosing positive 2-digit integer with 4 in either place GRE study exam guide has following
If an integer is randomly selected from all positive 2-digit integers,
what is the probability that the integer chosen has at least one 4 in
the tens place or the units place?
I understand probability of being in 10s place is $1/9$ and the probability of being in the units place is $1/10$
When I add $1/9$ + $1/10$ the answer is $19/90$. However, answer says $1/5$.
Please explain
| A = $4$ in a ten place,
B = $4$ in a unit place,
C = at least one $4$ in a ten or unit place.
$P(A) = \frac{1}{9}$, $P(B) = \frac{1}{10}$, $P(C) = \frac{1}{9} + (1-\frac{1}{9})*\frac{1}{10} = \frac{10 + 8}{90} = \frac{1}{5}$.
Here you go:
You have $10$ cases for event $A$: $40, 41, 42, 43,44,45,46,47,48,49$, and $9$ cases for event $B$: $14, 24, 34, 44, 45, 46, 47, 48, 49$. So when you get $A \cup B$ you have not $19$ cases, but 18. There are $90$ two-digit numbers, so $\frac{18}{90} = \frac{1}{5}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1596246",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 0
} |
Does $\sum_{n=1}^\infty \frac{2\cdot 4\cdot 6\cdot ...\cdot (2n)}{n^n}$ converge? Does $\sum_{n=1}^\infty \frac{2\cdot 4\cdot 6\cdot ...\cdot (2n)}{n^n}$ converges?
Well, I'm trying the approach using Cauchy's test for convergence. If $a_n >= 0$ and $\lim \sqrt[n]a_n = L $ exists. then, I need to check whether it is bigger or lesser than $1$. (if equal then we have a problem)
So.. $\sum_{n=1}^\infty \frac{2\cdot 4\cdot 6\cdot ...\cdot (2n)}{n^n}$ becomes $\sum_{n=1}^\infty \frac{\sqrt[n]{2\cdot 4\cdot 6\cdot ...\cdot (2n)}}{n}$ becomes $\frac{({2\cdot 4\cdot 6\cdot ...\cdot (2n)})^{1/n}}{n}$ becomes computing the limit of $1/n$ which is zero. Which means it converges?
| Note that $\frac{2*4*6*...*2n}{n^n}=\frac{n!2^n}{n^n}$.
By the ratio test, $\dfrac{\left(\frac{2}{n+1}\right)^{n+1}(n+1)!}{\left(\frac{2}{n}\right)^{n}(n)!}=\dfrac{2n^n}{(n+1)^n}=2\left(\dfrac{n}{n+1}\right)^n=2\left(1-\frac{1}{n+1}\right)^n\to\frac{2}{e}<1$.
Thus the serie is convergent.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1597356",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 1
} |
solving $\int \cos^2(2x)\sin(2x)\,dx$
$$\int \cos^2(2x)\sin(2x)\,dx$$
I have tried $\int \cos^2(2x)\sin(2x)\,dx=\int (1-\sin^2(2x))\sin(2x)\,dx$
And $\int \cos^2(2x)\sin(2x)\,dx=\int (\cos^2(2x))2\cos x\sin x\, dx$
But no substitution seems to work.
| $$\int \cos ^{ 2 } (2x)sin(2x)dx=-\frac { 1 }{ 2 } \int { \cos ^{ 2 }{ \left( 2x \right) d\left( \cos { \left( 2x \right) } \right) } } =-\frac { \cos ^{ 3 }{ \left( 2x \right) } }{ 6 } +C$$
substitution here is $u=\cos { \left( 2x \right) } $
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1598800",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
Help finding the determinant of a 4x4 matrix? Sorry for the lack of notation but the work should be easy to follow if you know what you are doing. Okay my problem is that the book says it can be done by expanding across any column or row but the only way to get what the book does in their practice example is to choose the row that they chose. This bothers me. As I should be able to do it as I see fit. I will post my work and someone point out the problem in my work. The matrix is as follows:
$$A = \left(
\begin{matrix}
5&-7&2&2\\
0&3&0&-4\\
-5&-8&0&3\\
0&5&0&-6\\
\end{matrix} \right)
$$
I decided to expand across row one and cross out columns as I found the minors. For the first minor obtaining:
$$
\begin{pmatrix}
3 & 0 & -4 \\
-8 & 0 & 3 \\
5 & 0 & -6 \\
\end{pmatrix}
$$
M1 being row one column one we attain $-1^2 = 1$. This is to be multiplied by the determinate of the minor. Now finding the determinant I did:
3 times $$
\begin{pmatrix}
0 & 3 \\
0 & -6 \\
\end{pmatrix}
$$
giving $3(0-0)= 0$
then:
0 times $$
\begin{pmatrix}
-8 & 3\\
5 & -6\\
\end{pmatrix}
$$
giving 0(48-15)=0
Then:
4 times
$$
\begin{pmatrix}
-8 & 0 \\
5 & 0 \\
\end{pmatrix}
$$
giving $4(0-0)=0$
adding the determinants we get $0+0+0=0$
So det M1 $= 0(1) = 0$
M2--> M(1,2)---> $-1^1+2= -1^3 = -1$
$$
\begin{pmatrix}
0 & 0 & -4 \\
-5 & 0 & 3 \\
0 & 0 & -6 \\
\end{pmatrix}
$$
o*
$$
\begin{pmatrix}
0 & 3 \\
0 & -6 \\
\end{pmatrix}
$$
giving $0(0-0)=0$
obviously the next matrix will look the same as the top term in column two is a zero so the determinant for that will be $0$. Now finally
4 times
$$
\begin{pmatrix}
-8 & 0 \\
5 & 0 \\
\end{pmatrix}
$$
giving 4(0-0)= 0
So the Determinant of Minor 2 is (0+0+0)(-1)= 0 Now on to Minor number 3
M3 --> $-1^4 = 1$
$$
\begin{pmatrix}
0 & 3 & -4 \\
-5 & -8 & 3 \\
0 & 5 & -6 \\
\end{pmatrix}
$$
for the determinant:
0 times
$$
\begin{pmatrix}
-8 & 3 \\
5 & -6 \\
\end{pmatrix}
$$
which gives $0(48-15)=0$
-3 times
$$
\begin{pmatrix}
-5& 3 \\
0 & -6 \\
\end{pmatrix}
$$
which gives $-3(30-0)= -90$
it is redundant to go on from here because after the final computation for this minor I get -100 and as a result get det M3 = -190 and get determinant of zeros for the following determinant of M4.
which gives: $0(5)+ 0(-7) + (-90)(2) + (0)(2)$ giving
Det Ax $= -380.$ The book says its $20$ and when I did it in a calculator it got 20 but the problem is that both the book and calculator expand across the row with the most zeros but theoretically speaking NO MATTER WHICH row or column you choose to expand across you should get the same answer. So what is it? Is my computation wrong or is my assumption that you can expand across any row or column wrong? Isn't it only important if the determinant doesn't equal zero? or does the exact value matter in more advanced cases?
| For M3 it is $-90+100$ rather than $-90-100$ which gives $10 \times 2=20$ as the final answer
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1599082",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
"answer_count": 7,
"answer_id": 2
} |
Integral $\int\left(\frac{1}{x^4+x^2+1}\right)dx$ Someone can halp me to solve this integral:
$$\int\left(\frac{1}{x^4+x^2+1}\right)$$
solution$$\frac{1}{4}\ln\left(\frac{x^2+x+1}{x^2-x+1}\right)+\frac{1}{2\sqrt3}\arctan\frac {x^2-1}{x\sqrt3}$$
I don't manage using partial fraction because $${x^4+x^2+1}$$ has $\Delta\lt 0$ and substituing $t=x^2,$ $2x$ appears.That's a problem.
| HINT: $$x^4+x^2+1=(x^2+1)^2-x^2=(x^2-x+1)(x^2+x+1)$$ then $$\dfrac
{1}{x^4+x^2+1}=\dfrac{1}{(x^2-x+1)(x^2+x+1)}=-\dfrac12\left(\dfrac{x-1}{x^2-x+1}\right)+\dfrac12\left(\dfrac{x+1}{x^2-x+1}\right).$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1599282",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 0
} |
The greatest integer less than or equal to the number $R=(8+3\sqrt{7})^{20}$
Given $$R=(8+3\sqrt{7})^{20}, $$ if $\lfloor R \rfloor$ is Greatest integer less than or equal to $R$, then which of the following option(s) is/are true?
*
*$\lfloor R \rfloor$ is an even number
*$\lfloor R \rfloor$ is an odd number
*$R-\lfloor R \rfloor=1-\frac{1}{R}$
*$R(R-\lfloor R \rfloor-1)=-1$
My try: I wrote $R$ as $$R=8^{20}\left(1+\sqrt{\frac{63}{64}}\right)^{20} \approx8^{20}\left(1+\sqrt{0.98}\right)^{20} \approx8^{20}\left(1.989\right)^{20} .$$
Now, $8^{20}\left(1.989\right)^{20}$ is slightly less than $8^{20} \times 2^{20}=2^{80}$,
$$\lfloor 2^{80}\rfloor=2^{80}$$
hence
$$\lfloor R \rfloor=2^{80}-1,$$
so option $2$ is correct.
How does one figure out whether options $3$ and $4$ are correct or wrong?
| Hints For (1), (2): Using the binomial expansion twice gives that \begin{align}A := R + (8 - 3 \sqrt{7})^{20} &= (8 + 3 \sqrt{7})^{20} + (8 - 3 \sqrt{7})^{20} \\ &= \sum_{k = 0}^{20} {20 \choose k} 8^{20 - k} (3 \sqrt{7})^k + \sum_{k = 0}^{20} {20 \choose k} 8^{20 - k} (-3 \sqrt{7})^k \\ &= \sum_{k = 0}^{20} {20 \choose k} 8^{20 - k} (1 + (-1)^k) (3 \sqrt{7})^k .\end{align}
The appearance of the factor $1 + (-1)^k$ means that summands with odd $k$ are zero, so only the even terms contribute, and we can rewrite the sum as
$$A = \sum_{j = 0}^{10} {20 \choose 2j} 8^{20 - 2j} (3 \sqrt{7})^{2j} = \sum_{j = 0}^{10} {20 \choose 2j} 64^{10 - j} 63^j .$$
In particular, $A$ is an integer. On the other hand, since $49 < 63 < 64$, we have $7 < 3 \sqrt{7} < 8$ and hence $0 < 8 - 3 \sqrt{7} < 1$.
For (3): Note that $(8 + 3 \sqrt{7})(8 - 3 \sqrt{7}) = 64 - 63 = 1.$
Additional hints For (1)-(2): So, the second summand of $A$ satisfies $0 < (8 - 3 \sqrt{7})^{20} < 1$. (In fact, it is very close to zero.) So, $A - 1 < R < A$, and in particular, $\lfloor R \rfloor = A - 1$. Since we can determine the parity of $A$ from the last summation expression, we can also determine that of $\lfloor R \rfloor$. For (3): So $$(8 + 3 \sqrt{7})^{20} (8 - 3 \sqrt{7})^{20} = 1 ,$$ hence $$(8 - 3 \sqrt{7})^{20} = \frac{1}{R} .$$
There appears to be a typo in the equation in (4).
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1600472",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 3,
"answer_id": 1
} |
calculating a limit of sequence with tan Calculating a limit
$$\lim_{n\longrightarrow +\infty} \dfrac{\tan(\frac{\pi n}{2n+1})}{\sqrt[3]{n^3+2n-1}}$$
thanks.
| Another way is based on the asymptotics of the numerator and denominator $$\tan \left(\frac{\pi n}{2 n+1}\right)=\frac{4 n}{\pi }+\frac{2}{\pi }-\frac{\pi }{12 n}+\frac{\pi }{24
n^2}+O\left(\left(\frac{1}{n}\right)^3\right)$$ $$\sqrt[3]{n^3+2 n-1}=n+\frac{2}{3 n}-\frac{1}{3 n^2}+O\left(\left(\frac{1}{n}\right)^3\right)$$ Making the long division,then $$\dfrac{\tan(\frac{\pi n}{2n+1})}{\sqrt[3]{n^3+2n-1}}=\frac{4}{\pi }+\frac{2}{\pi n}+O\left(\frac{1}{n^2}\right)$$ which shows the limit and how it is approached.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1601416",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 4
} |
Let $x,y \ge 0 $.Find the maximum value of $\cfrac{(3x+4y)^2}{x^2+y^2}$ Let $x,y \ge 0 $.Find the maximum value of $\cfrac{(3x+4y)^2}{x^2+y^2}$
This exercise is from the chapter of my book regarding the Cauchy-Schwarz inequality but I don't know if I've applied it correctly .
The problem asks for the maximum value of the expression so it's clear that it must be on the lesser side of the Cauchy Inequality ,i.e.
\begin{array}
RR&\ge \cfrac{(3x+4y)^2}{x^2+y^2} \\
R(x^2+y^2) &\ge (3x+4y)^2 \\
\left(\left(\sqrt{R}\right)^2 \right) (x^2+y^2) &\ge (3x+4y)^2 \\
\end{array}
Now this would be true iff
$$ \left(\left(\sqrt{R}\right)^2 +(4)^2\right) (x^2+y^2) \ge (3x+4y)^2 $$ where $R=9$ ,which would imply that this is the maximum.
However I am not really sure if I am correct.
| You are right till
$$R (x^2+y^2) \ge (3x+4y)^2$$
Now comparing with the CS inequality
$$(3^2+4^2)(x^2+y^2) \ge (3x+4y)^2$$
$\implies R = 3^2+4^2 = 25$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1601894",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
How to factor polynomials? I am wondering if there is a methodical, algorithmic, brain-dead way to factor polynomials. For example:
$x^6 - 14x^{5} + 73x^{4} - 188x^{3} + 256x^{2} - 176x^{1} + 48$
can be written as
$(x-1)^2 (x-2)^3 (x-6)$
But how do you actually get there?
| If you will accept a "brain-dead" method, why not just appeal to computer algebra, such as Mathematica?
Factor[x^6 - 14 x^5 + 73 x^4 - 188 x^3 + 256 x^2 - 176 x^1 + 48]
$(x-6) (x-2)^3 (x-1)^2$
or
Factor[-20284203008 + 12410729472 x + 2524112000 x^2 -
1341562640 x^3 - 334176180 x^4 + 12601652 x^5 + 12198347 x^6 +
1767405 x^7 + 127690 x^8 + 5150 x^9 + 111 x^10 + x^11]
$(x-4) (x-2)^2 (x+8)^3 (x+19)^5$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1602768",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
Limit $(n - a_n)$ of sequence $a_{n+1} = \sqrt{n^2 - a_n}$
Consider the sequence $\{a_n\}_{n=1}^{\infty}$ defined recursively by
$$a_{n+1} = \sqrt{n^2 - a_n}$$ with $a_1 = 1$. Compute $$\lim_{n\to\infty} (n-a_n)$$
I am having trouble with this. I am not even sure how to show the limit exists. I think if we know the limit exists, it is just algebra, but I'm not sure.
| First, we'll show that $0 \le n-a_n \le 2$ for all $n$. Then, we will show that $n-a_n \to \dfrac{3}{2}$ as $n \to \infty$.
Since, $a_1 = 1$, we have $1-a_1 = 0$, so $0 \le 1-a_1 \le 2$, as desired.
Now, suppose $0 \le n-a_n \le 2$ for some positive integer $n$. Then, we have:
$$\sqrt{n^2-n} \le \sqrt{n^2-a_n} \le \sqrt{n^2-n+2}$$
$$\sqrt{(n-\tfrac{1}{2})^2-\tfrac{1}{4}} \le a_{n+1} \le \sqrt{(n-\tfrac{1}{2})^2+\tfrac{7}{4}}$$
$$(n+1) - \sqrt{(n-\tfrac{1}{2})^2+\tfrac{7}{4}} \le (n+1)-a_{n+1} \le (n+1) - \sqrt{(n-\tfrac{1}{2})^2-\tfrac{1}{4}}$$
$$\tfrac{3}{2} + (n-\tfrac{1}{2}) - \sqrt{(n-\tfrac{1}{2})^2+\tfrac{7}{4}} \le (n+1)-a_{n+1} \le \tfrac{3}{2} + (n-\tfrac{1}{2}) - \sqrt{(n-\tfrac{1}{2})^2-\tfrac{1}{4}}$$
$$\dfrac{3}{2} - \dfrac{\tfrac{7}{4}}{(n-\tfrac{1}{2}) + \sqrt{(n-\tfrac{1}{2})^2+\tfrac{7}{4}}} \le (n+1)-a_{n+1} \le \dfrac{3}{2} + \dfrac{\tfrac{1}{4}}{(n-\tfrac{1}{2}) + \sqrt{(n-\tfrac{1}{2})^2-\tfrac{1}{4}}}.$$
For $n \ge 1$, we have $(n-\tfrac{1}{2}) + \sqrt{(n-\tfrac{1}{2})^2+\tfrac{7}{4}} \ge \tfrac{1}{2}+\sqrt{(\tfrac{1}{2})^2+\tfrac{7}{4}} = \tfrac{1}{2}+\sqrt{2} \ge \tfrac{7}{6}$,
as well as $(n-\tfrac{1}{2}) + \sqrt{(n-\tfrac{1}{2})^2-\tfrac{1}{4}} \ge \tfrac{1}{2}+\sqrt{(\tfrac{1}{2})^2-\tfrac{1}{4}} = \tfrac{1}{2}$.
Thus, the last equation implies $0 \le (n+1)-a_{n+1} \le 2$.
So by induction, $0 \le n-a_n \le 2$ for all positive integers $n$.
Then by repeating the above algebra, we have that $$\dfrac{3}{2} - \dfrac{\tfrac{7}{4}}{(n-\tfrac{1}{2}) + \sqrt{(n-\tfrac{1}{2})^2+\tfrac{7}{4}}} \le (n+1)-a_{n+1} \le \dfrac{3}{2} + \dfrac{\tfrac{1}{4}}{(n-\tfrac{1}{2}) + \sqrt{(n-\tfrac{1}{2})^2-\tfrac{1}{4}}}$$ holds for all positive integers $n$.
By using the squeeze theorem, we get $\displaystyle\lim_{n \to \infty}[(n+1)-a_{n+1}] = \dfrac{3}{2}$, and thus, $\displaystyle\lim_{n \to \infty}(n-a_n) = \dfrac{3}{2}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1602850",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 0
} |
Evaluating - $\int_{|z| =R,R > 0} z^2 \big( \sin\big(\frac{1}{z}\big) + \sin\big(\frac{2}{z}\big) + ...+ \sin\big(\frac{n}{z}\big) \big) dz $ Let:
$$ I_n =\int_{|z| =R , \text{ }R > 0} z^2 \left( \sin\left(\frac{1}{z}\right) + \sin\left(\frac{2}{z}\right) + ...+ \sin\left(\frac{n}{z}\right) \right)\, dz $$
Knowing that $n$ is a natural non-zero number, prove that $e^{12I_n}$ is an integer. This was a problem on local contest today and I have no idea how to solve this.
Can anyone help?
| \begin{eqnarray}
I_n&=&\sum_{k=1}^n\int_{|z|=R}z^2\sin\left(\frac{k}{z}\right)\,dz=\sum_{k=1}^n\sum_{m=0}^\infty\int_{|z|=R}z^2\frac{(-1)^m}{(2m+1)!}\frac{k^{2m+1}}{z^{2m+1}}\,dz\\
&=&\sum_{k=1}^n\sum_{m=0}^\infty\int_{|z|=R}\frac{(-1)^m}{(2m+1)!}\frac{k^{2m+1}}{z^{2m-1}}\,dz=\sum_{k=1}^n\int_{|z|=R}\frac{-k^3}{3!}\frac{1}{z}\,dz=-\frac13i\pi\sum_{k=1}^nk^3\\
&=&-\frac13i\pi\left(\sum_{k=1}^nk\right)^2=-\frac13i\pi\left[\frac{n(n+1)}{2}\right]^2=-i\frac{n^2(n+1)^2}{12}
\end{eqnarray}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1603404",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Show that $\ln\left(1+\frac1n\right)=\frac1n-\frac{\theta_n}{2n^2}$ I've been asked to show that for any natural number $n$, there exists $\theta_n\in (0,1)$ such that
$$\ln\left(1+\frac1n\right)=\frac1n-\frac{\theta_n}{2n^2}$$
I've tried to use integral identities in order to convert the logarithm into something easier to handle but it didn't work, so I would appreciate some hints. Thanks in advance.
| $$\ln\left(1+\frac1n\right) = \frac1n + \frac{\theta_n}{2n^2}$$
Write $f(x) = \ln(1+x)$. Then by definition of the derivative $$f(x) = f(0) + f'(0)x + \frac{f''(0)}{2}x^2 + o(x^2).$$ Where $o(x^2)/x^2 \to 0$ as $x \to 0$.
$f(0)= \ln(1+0) = 0$.
$f'(x) = \frac{1}{1+x}$ so $f'(0) = 1/(1+0) = 1$.
$f''(x) = \frac{-1}{(1+x)^2}$ and $f''(0) = -1$.
Thus $$f(1/n) = \frac{1}{n} - \frac{1}{2n^2} + o\left( \frac{1}{n^2} \right).$$
$o(1/n^2)$ represents a function for which $n^2o(1/n^2) \to 0$ as $n \to \infty$. Hence $o(1/n^2) \to 0$ as $n \to \infty$, and $$f(1/n) = \frac{1}{n} - \frac{1 - 2n^2o\left( \frac{1}{n^2} \right)}{2n^2}.$$
Now we must employ the taylor theorem to estimate the size of $o(1/n^2)$, henceforth write $E(x)=o(x^2)$.
If we constrain $f(x)$ to the interval $[0,1/n]$, $E(x)$ can be determined via bounds on $f'''(x)$. Note that $f'''(x) = \frac{2}{(1+x)^3}$. This function is decreasing and is bounded below by $0$. Thus $2/(1+1/n)^3 \le f'''(x) \le 2$ on $[0,1/n]$.
By the Remainder theorems for Taylor's formula (arising from the Mean Value Theorem), we have
$$0 < E(1/n) \le 2 \frac{1}{n^3 3!}.$$
Thus $$0< 2n^2 o\left(\frac{1}{n^2}\right) = 2n^2 E(1/n) \le \frac{4n^2}{n^3 3!} = \frac{2}{3n} < 1.$$
This yields $0 < \theta_n = 1-2n^2 E(1/n) < 1$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1603479",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 0
} |
Some equations from Russian maths book. Could you please help me with solving these equations. I would like to solve them in the most sneaky way. All of the exercises in this book can be solved in some clever way which I can't often find.
$$
\frac{(x-1)(x-2)(x-3)(x-4)}{(x+1)(x+2)(x+3)(x+4)} = 1
$$
$$
\frac{6}{(x+1)(x+2)} + \frac{8}{(x-1)(x+4)} = 1
$$
$$
\sqrt[7]{ (ax-b)^{3}} - \sqrt[7]{ (b-ax)^{3} } = \frac{65}{8}; a \neq 0
$$
| The first equation implies that the product of the distances from $x$ to $-1, -2, -3, -4$ is the same as the product of the distances from $x$ to $1, 2, 3, 4$. (This condition is equivalent to the quotient appearing in the equation being $\pm 1$.) If $x > 0$, the latter distances are smaller than the corresponding former ones, so the latter product is smaller. If $x < 0$, then the former is. So the only possibility is $x = 0$.
In the second equation, make the substitution $u = x^2 + 3x$.
For the third one, you were almost there with what you wrote in the comments. But remember that $\sqrt[7]{-A} = - \sqrt[7]{A}$. You get $(ax - b)^{3/7} = 65/16$. Raise both sides of the equation to the power of $7/3$.
Edit The third equation has a typo in it. It was supposed to read
$$\sqrt[7]{ (ax-b)^{3}} - \sqrt[7]{ (b-ax)^{-3} } = \frac{65}{8}.$$
In this case, Write $u = (ax-b)^{3/7}$. Then the equation becomes $u + 1/u = 65/8$. Since after clearing denominators this becomes a quadratic equation, there are at most two possibilities for $u$. Since $u = 8$ and $u = 1/8$ work, these must be the ones. We get $ax - b \in \{128, 1/128\}$, after which it's easy to solve for $x$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1604713",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 5,
"answer_id": 0
} |
Evaluate Integral Using Change of Variables Let $A = \{(x, y)\in \mathbb{R}^2 | x^2-xy+2y^2<1 \}$; define $f: \mathbb{R}^2 \to \mathbb{R}$ by $f(x, y) = xy$.
Express the integral
$$\int_A f$$
as an integral over the unit ball $B = \{(x, y)\in \mathbb{R}^2 | x^2+y^2<1 \}$.
Someone gave a solution here but it seems to be wrong.
My attempt:
Observe that $x^2-xy+2y^2=(x-\frac{1}{2}y)^2+(\frac{\sqrt{7}}{2}y)^2$.
Let $B^\ast = \{(r, \theta) | 0< r < 1, 0 < \theta < 2\pi\}$.
Let $A^\ast = \{(x, y)\in \mathbb{R}^2 | x^2-xy+2y^2<1$, and $x<0$ if $y=0\}$.
Then the function
$$g(r, \theta) = (r\cos\theta+\frac{1}{\sqrt{7}}r\sin\theta, \frac{2}{\sqrt{7}}r\sin\theta)$$
carries $B^\ast$ in a one-to-one fasion onto $A-\{(0, 0)\}$.
Now we obtain that
\begin{align*}
\text{det} Dg & =
\begin{vmatrix}\cos\theta+\frac{1}{\sqrt{7}}\sin\theta & -r\sin\theta+\frac{1}{\sqrt{7}}r\cos\theta \\ \frac{2}{\sqrt{7}}\sin\theta & \frac{2}{\sqrt{7}}r\cos\theta \\ \end{vmatrix} \\
& = \frac{2}{\sqrt{7}}r \\
& > 0
\end{align*}
if $(r, \theta)\in B^\ast$.
Since the non-negative $x$-axis has measure zero, using the formula of change of variables, we have that
\begin{align*}
\int_A f
& = \int_{A^\ast} f \\
& = \int_{B^\ast} f \circ g \cdot |\text{det} Dg| \\
& = \int_0^{2\pi} \int_0^1 \Big((r\cos\theta+\frac{1}{\sqrt{7}}r\sin\theta)\frac{2}{\sqrt{7}}r\sin\theta\frac{2}{\sqrt{7}}r \Big) drd\theta \\
& = \frac{2}{7}\int_0^{2\pi} \int_0^1 (2\cos\theta\sin\theta+\frac{2}{\sqrt{7}}\sin^2 \theta)r^3 drd\theta \\
& = \frac{2}{7}\int_0^{2\pi} \int_0^1 \Big( \sin2\theta+\frac{1}{\sqrt{7}}(1-\cos2\theta) \Big)r^3 drd\theta \\
& = \frac{2}{7}\int_0^{2\pi} \int_0^1 \Big( \sin2\theta-\frac{1}{\sqrt{7}}\cos2\theta+\frac{1}{\sqrt{7}} \Big)r^3 drd\theta \\
& = \frac{2}{7}\int_0^{2\pi} \Big[ \frac{1}{4}\Big( \sin2\theta-\frac{1}{\sqrt{7}}\cos2\theta+\frac{1}{\sqrt{7}} \Big)r^4 \Big]_{r=0}^{r=1} d\theta \\
& = \frac{2}{7}\int_0^{2\pi} \frac{1}{4}\Big( \sin2\theta-\frac{1}{\sqrt{7}}\cos2\theta+\frac{1}{\sqrt{7}} \Big) d\theta \\
& = \frac{1}{14} \Big[ -\frac{1}{2}\cos2\theta-\frac{1}{2\sqrt{7}}\sin2\theta+\frac{1}{\sqrt{7}}\theta \Big]_{\theta=0}^{\theta=2\pi} \\
& = \frac{\pi}{7\sqrt{7}}
\end{align*}
Am I making any mistake? Any help or suggestion is appreciated.
| Looks reasonable to me. As a check on your computations, here’s a solution using an approach similar to Ron Gordon’s from the original question.
Examine the matrix $Q$ associated with the quadratic form $x^2-xy+2y^2$: $$
Q = \pmatrix{1&-\frac12\\-\frac12&2}.
$$ $\det Q>0$ and $\operatorname{tr}Q>0$, so the eigenvalues are positive, which means that we have an ellipse centered on the origin and rotated through some angle $\phi$. Define the pullback $\beta:(\rho,\theta)\mapsto(x,y)$ by $$\begin{align}
\beta^*x &= \rho\,(a\cos\theta\cos\phi-b\sin\theta\sin\phi) \\
\beta^*y &= \rho\,(a\cos\theta\sin\phi+b\sin\theta\cos\phi),
\end{align}$$ where $a$ and $b$ are the half-axis lengths of the ellipse. (This is the polar equation of the ellipse multiplied by $\rho$, which can easily be derived by scaling and rotating the unit circle.) For this pullback, we also have $$\det{\partial(\beta^*x,\beta^*y)\over\partial(\rho,\theta)}=ab\rho.$$ So, if $C$ is the unit disk, $$\begin{align}
\int_Af\,dA &= \int_C\beta^*(f\,dA) \\
&=ab\int_0^{2\pi}\int_0^1\rho^3(a\cos\theta\cos\phi-b\sin\theta\sin\phi)(a\cos\theta\sin\phi+b\sin\theta\cos\phi)\,d\rho\,d\theta \\
&=\frac{ab}4\int_0^{2\pi}(a\cos\theta\cos\phi-b\sin\theta\sin\phi)(a\cos\theta\sin\phi+b\sin\theta\cos\phi)\,d\theta \\
&=\frac{ab}4\int_0^{2\pi}a^2\cos^2\theta\cos\phi\sin\phi - b^2\sin^2\theta\cos\phi\sin\phi + \cdots\,d\theta \\
&=\frac{\pi ab}4(a^2-b^2)\cos\phi\sin\phi.\tag{*}
\end{align}$$ Note that we can ignore terms that involve odd powers of $\cos\theta$ and $\sin\theta$ since we’re integrating from $0$ to $2\pi$.
To determine $a$, $b$ and $\phi$, go back to the matrix $Q$. If its eigenvalues are $\lambda_1 < \lambda_2$, then $a^2 = 1/\lambda_1$ and $b^2 = 1/\lambda_2$. It’s a bit more difficult to find an exact value for $\phi$ in general, but we don’t really need to since we can get its sine and cosine directly from the normalized eigenvectors (which is why I didn’t combine $\cos\phi\sin\phi$ into $\frac12\sin{2\phi}$). In this case, we have $$
a^2 = {2\over3-\sqrt2}, b^2 = {2\over3+\sqrt2} \\
\cos\phi = {\sqrt{2+\sqrt2}\over2}, \sin\phi = {\sqrt{2-\sqrt2}\over2}.
$$ Plugging these values into (*) and simplifying gives ${\pi\over7\sqrt7}$, which agrees with the value computed in the question above.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1604795",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
How to solve $\lim _{x\to \sqrt{2}}\left(\frac{e^{x^2}+e^2(1-x^2)}{[\ln(x^2-3\sqrt{2}x+5)]^2}\right)$? I have a problem with this limit, i have no idea how to compute it. Can you explain the method and the steps used(without Hopital if is possible)? Thanks
$$\lim_{x\to \sqrt{2}}\left(\frac{e^{x^2}+e^2(1-x^2)}{[\ln(x^2-3\sqrt{2}x+5)]^2}\right)$$
| Notice, one can easily apply L'Hospital's rule for $\frac 00$ form as follows $$\lim_{x\to \sqrt 2}\left(\frac{e^{x^2}+e^2(1-x^2)}{[\ln(x^2-3\sqrt 2 x+5)]^2}\right)$$
$$\lim_{x\to \sqrt 2}\left(\frac{2xe^{x^2}-2xe^2}{\frac{2\ln(x^2-3\sqrt 2 x+5)}{x^2-3\sqrt 2 x+5}\cdot (2x-3\sqrt 2)}\right)$$
$$\lim_{x\to \sqrt 2}\left(\frac{x(e^{x^2}-e^2)(x^2-3\sqrt 2 x+5)}{(2x-3\sqrt 2)\ln(x^2-3\sqrt 2 x+5)}\right)$$
$$\lim_{x\to \sqrt 2}\left(\frac{(e^{x^2}-e^2)(x^3-3\sqrt 2 x^2+5x)}{(2x-3\sqrt 2)\ln(x^2-3\sqrt 2 x+5)}\right)$$
applying L'Hospital's rule for $\frac 00$ form,
$$=\lim_{x\to \sqrt 2}\left(\frac{(e^{x^2}-e^2)(3x^2-6\sqrt 2 x+5)+(x^3-3\sqrt 2x^2+5x)(2xe^{x^2})}{\frac{(2x-3\sqrt 2)}{x^2-3\sqrt 2 x+5}\cdot(2x-3\sqrt 2)+\ln(x^2-3\sqrt 2 x+5)(2)}\right)$$
$$=\left(\frac{0+(\sqrt 2)(2\sqrt 2e^2)}{\frac{(-\sqrt 2)}{1}\cdot(-\sqrt 2)+0}\right)=\color{red}{2e^2}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1605491",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
Find $\lfloor 1000S \rfloor$ for $S =\sum_{n=1}^{\infty} \frac{1}{2^{n^2}} = \frac{1}{2^1}+\frac{1}{2^4}+\frac{1}{2^9}+\cdots.$
Let $$S = \displaystyle \sum_{n=1}^{\infty} \dfrac{1}{2^{n^2}} = \dfrac{1}{2^1}+\dfrac{1}{2^4}+\dfrac{1}{2^9}+\cdots.$$
Find $\lfloor 1000S \rfloor$
My attempt was to recognize that consecutive perfect squares increase in the form $3,5,7,9,\ldots$ and be able to form a geometric series this way. I couldn't figure out how to make that work, though.
| The sum of the first three terms is
$$1000\left[\frac{1}{2^{1}}+\frac{1}{2^{4}}+\frac{1}{2^{9}}\right] \simeq 564.453$$
The sum of the rest of the terms is bounded by (using a geometrical series)
$$1000\left[\frac{1}{2^{16}}+\frac{1}{2^{25}}+\frac{1}{2^{36}}+\ldots\right] < 1000\left(\frac{1}{2^{16}} + \frac{1}{2^{16+9}} + \frac{1}{2^{16+2\cdot 9}} + \ldots\right) \\= \frac{1000}{2^{16}}\frac{1}{1-\frac{1}{2^9}} < \frac{1000}{2^{15}} \simeq 0.031$$
From which we see that
$$564.453 < 1000S < 564.484 \implies \lfloor 1000S\rfloor = 564$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1606189",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
} |
Laurent expansion of $\frac{1}{z^2}$ I need to find a Laurent expansion of $\frac{1}{z^2}$ with centre in $z_0 = 1$ and $P(1, 2014, 2015)$.
If it was $\frac{1}{z}$, I'd rewrite the fraction like this:
$$
\frac{1}{(z-1) +1 }
$$
But with $z^2$ I'm not sure how to proceed...
My attempt:
According to the given recommendation, I've tried to do the expansion of $\frac{1}{z}$:
$$
\frac{1}{z} = \frac{1}{1+(z-1)}, a_{0} = 1, q = -(z-1) = 1-z
$$
$$
\sum^{\infty}_{n=0}(1-z)^{n} = 1 + (1-z) + (1-z)^2 + ...
$$
And after differentiation:
$$
0 + 1 + 2(1-z) + 3(1-z)^2 + ... = \sum^{\infty}_{n=0}n(1-z)^{n-1}
$$
| Another way is:
$\dfrac {1}{z^2}=\dfrac {\left(1-\dfrac{1}{z}\right)^2}{(z-1)^2}=
\dfrac {(1-z)^2+2(1-z)^3+3(1-z)^4+4(1-z)^5+ \ldots}{(1-z)^2}=1+2(1-z)+3(1-z)^2+4(1-z)^3+ \ldots$
using the product of convergent power series.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1609849",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
The sequence $\phi_n= 1+\frac{1}{2^4}+\frac{1}{3^4}+\ldots+\frac{1}{n^4}$ is bounded above . A sequence $(\phi_n)$ is defined as follows : $$\phi_n= 1+\frac{1}{2^4}+\frac{1}{3^4}+\ldots+\frac{1}{n^4}$$
Show that the sequence is convergent.
Because this sequence is monotonic, proving it is bounded above will be sufficient to prove that it is convergent.
So how to show that this sequence is bounded above?
| Since $t \mapsto \dfrac{1}{t^4}$ is decreasing, we have
$$\frac{1}{(k+1)^4}=\int_{k}^{k+1}\frac{dt}{(k+1)^4}\leq \int_{k}^{k+1}\frac{dt}{t^4},\,\,\, k\geq1,$$ which gives$$\phi_n= 1+\frac{1}{2^4}+\frac{1}{3^4}+\ldots+\frac{1}{n^4}=1+\sum_{k=1}^{n-1}\frac{1}{(k+1)^4}\leq 1+\int_{1}^{n}\frac{dt}{t^4}=\frac{4}{3}-\frac{1}{3n^3}<\frac{4}{3}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1612011",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 1
} |
Solving this limit $\lim_{x\to 3}\frac{x^2-9-3+\sqrt{x+6}}{x^2-9}$. The question is $\lim_\limits{x\to 3}\frac{x^2-9-3+\sqrt{x+6}}{x^2-9}$.
I hope you guys understand why I have written the numerator like that. So my progress is nothing but $1+\frac{\sqrt{x+6}-3}{x^2-9}$.
Now how do I rationalize the numerator?
It is giving the $\frac{0}{0}$ form after plugging in $3$.
| The solution goes as follows:
$$\lim_\limits{x\to 3}\frac{x^2-9-3+\sqrt{x+6}}{x^2-9}$$
$$=\lim_\limits{x\to 3}\left(1+\frac{\sqrt{x+6}-3}{x^2-9}\right)$$
$$=\lim_\limits{x\to 3}\left[1+\frac{(\sqrt{x+6}-3)(\sqrt{x+6}+3)}{(x^2-9)(\sqrt{x+6}+3)}\right]$$
$$=\lim_\limits{x\to 3}\left[1+\frac{x+6-9}{(x^2-9)(\sqrt{x+6}+3)}\right]$$
$$=\lim_\limits{x\to 3}\left[1+\frac{x-3}{(x-3)(x+3)(\sqrt{x+6}+3)}\right]$$
$$=\lim_\limits{x\to 3}\left[1+\frac{1}{(x+3)(\sqrt{x+6}+3)}\right]$$
$$=1+\frac{1}{(3+3)(\sqrt{3+6}+3)}$$
$$=\frac{37}{36}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1615527",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 1
} |
Let $f:[1,\infty)\to R$ be a monotonic and differentiable function and $f(1)=1.$If $N$ is the number of solutions of $f(f(x))=\frac{1}{x^2-2x+2}$. Let $f:[1,\infty)\to R$ be a monotonic and differentiable function and $f(1)=1.$If $N$ is the number of solutions of $f(f(x))=\frac{1}{x^2-2x+2}$.Find $N.$
$\frac{1}{x^2-2x+2}=\frac{1}{(x-1)^2+1}$.Its graph look like the graph of $\frac{1}{x^2+1}$ but peak at $(1,1)$ buti do not know how does the graph of $f(f(x))$ behave and hence i am not able to find the number of points of intersection of $f(f(x))$ and $\frac{1}{x^2-2x+2}$.
| Let $g(x)=f(f(x))$, then $g'(x)=f'(x)f'(f(x))\geq0$ since $f(x)$ is monotonic function.
Note that $x=1$ is the solution of $g(x)=f(f(x))=\frac{1}{x^2-2x+2}=\frac{1}{(x-1)^2+1}$
Now, since $\frac{1}{(x-1)^2+1}$ is strictly decresing function on interval $[1,\infty)$, and $g(x)$ is nondecreasing on that interval, the solution $x=1$ is unique. Hence, $N=1$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1617707",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Is there a way to prove vector triple product from quaternion multiplication? For pure imaginary quaternions $u, v, w$, is there a way to prove the vector triple product $u\times(v\times w) = v(u\cdot w) - w(u\cdot v)$ from the relation:
$$uv = -u\cdot v + u\times v \text{ for $u, v \in \mathbb{R}^3$ }$$
I tried many times but failed.
| The vector triple product identity follows from quaternion associativity.
A convenient tool for expressing quaternion statements in terms of dot and cross products is to write the quaternion as a real number and a 3-vector, as $a = (\alpha, \mathbf{a})$. Then quaternion multiplication of quaternion $b = (\beta, \mathbf{b})$ on the left by $a$ is given by
$$
a b =
\begin{bmatrix}
\alpha & -\mathbf{a}^T \\
\mathbf{a} & \alpha I + \mathbf{a} \times
\end{bmatrix}
\begin{pmatrix}
\beta \\
\mathbf{b}
\end{pmatrix}
=
\begin{pmatrix}
\alpha\beta - \mathbf{a} \cdot \mathbf{b} \\
\beta\mathbf{a} + \alpha \mathbf{b} + \mathbf{a} \times \mathbf{b}
\end{pmatrix} .
$$
This is much simplified for purely non-real quaternions
$a = (0,\mathbf{a})$, $b = (0,\mathbf{b})$ and $c = (0,\mathbf{c})$
$$
a b =
\begin{bmatrix}
0 & -\mathbf{a}^T \\
\mathbf{a} & \mathbf{a} \times
\end{bmatrix}
\begin{pmatrix}
0 \\
\mathbf{b}
\end{pmatrix}
=
\begin{pmatrix}
-\mathbf{a} \cdot \mathbf{b} \\
\mathbf{a} \times \mathbf{b}
\end{pmatrix} .
$$
And further,
$$
a b c =
\begin{pmatrix}
-(\mathbf{a} \times \mathbf{b}) \cdot \mathbf{c} \\
-(\mathbf{a} \cdot \mathbf{b}) \mathbf{c} + (\mathbf{a} \times \mathbf{b}) \times \mathbf{c}
\end{pmatrix} .
$$
To extract the identity from quaternion expressions, for three cyclic permutations of three generic quaternions $a$, $b$ and $c$, write equations expressing associativity, to form a system of equations:
$$
\left\{
\begin{array}{rcl}
( a b ) c - a ( b c ) &=& 0 \\
( b c ) d - b ( c d ) &=& 0 \\
( c a ) b - c ( a b ) &=& 0 .
\end{array}
\right.
$$
In case the three quaternions are purely non-real, write $a = (0,\mathbf{a})$, $b = (0,\mathbf{b})$ and $c = (0,\mathbf{c})$,
for which the above system implies
$$
\left\{
\begin{array}{rcl}
-( \mathbf{a} \cdot \mathbf{b} ) \mathbf{c}
+\mathbf{a} ( \mathbf{b} \cdot \mathbf{c} )
+ ( \mathbf{a} \times \mathbf{b} ) \times \mathbf{c}
- \mathbf{a} \times ( \mathbf{b} \times \mathbf{c} ) &=& 0 \\
-( \mathbf{b} \cdot \mathbf{c} ) \mathbf{a}
+\mathbf{b} ( \mathbf{c} \cdot \mathbf{a} )
+ ( \mathbf{b} \times \mathbf{c} ) \times \mathbf{a}
- \mathbf{b} \times ( \mathbf{c} \times \mathbf{a} ) &=& 0 \\
-( \mathbf{c} \cdot \mathbf{a} ) \mathbf{b}
+\mathbf{c} ( \mathbf{a} \cdot \mathbf{b} )
+ ( \mathbf{c} \times \mathbf{a} ) \times \mathbf{b}
- \mathbf{c} \times ( \mathbf{a} \times \mathbf{b} ) &=& 0 .
\end{array}
\right.
$$
The cross product terms can be compared more easily if the
anticommutativity of the cross product is applied:
$$
\left\{
\begin{array}{rcl}
-( \mathbf{a} \cdot \mathbf{b} ) \mathbf{c}
+\mathbf{a} ( \mathbf{b} \cdot \mathbf{c} )
- \mathbf{c} \times ( \mathbf{a} \times \mathbf{b} )
- \mathbf{a} \times ( \mathbf{b} \times \mathbf{c} ) &=& 0 \\
-( \mathbf{b} \cdot \mathbf{c} ) \mathbf{a}
+\mathbf{b} ( \mathbf{c} \cdot \mathbf{a} )
- \mathbf{a} \times ( \mathbf{b} \times \mathbf{c} )
- \mathbf{b} \times ( \mathbf{c} \times \mathbf{a} ) &=& 0 \\
-( \mathbf{c} \cdot \mathbf{a} ) \mathbf{b}
+\mathbf{c} ( \mathbf{a} \cdot \mathbf{b} )
- \mathbf{b} \times ( \mathbf{c} \times \mathbf{a} )
- \mathbf{c} \times ( \mathbf{a} \times \mathbf{b} ) &=& 0 .
\end{array}
\right.
$$
The identity results from adding the first two equations of the system, and subtracting the third.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1618508",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 1,
"answer_id": 0
} |
Complex number - wrong result at the end I need to solve this:
$$ \frac{i^4+3}{i-1}$$
On my book the result should be: $-2-2i$ but I get: $-1-2i$ and I do not understand where the error is.
My steps:
$$ \frac{i^4+3}{i-1} = \frac{i^4+3}{i-1} \cdot \frac{-1-i}{-1-i}$$
$$ \frac{(i^4+3)(-1-i) + (-1-i)(i-1)}{(i-1)(-1-i)}$$
$$i^4 = (i^2)^2 = 1$$
$$ \frac{(1+3)(-1-i) + (-1-i)(i-1)}{(i-1)(-1-i)}$$
$$ \frac{4(-1-i) + (-1-i)(i-1)}{(i-1)(-1-i)}$$
$$ \frac{4(-1-i) + (-i+1+1+i)}{(-i+1+1+i)}$$
$$ \frac{-4-4i + 2}{2}$$
$$ \frac{-2-4i}{2} = -1-2i$$
Where is the error?
| By the way, since $i^4=1$ the fraction simplifies as
$$
\frac{i^4+3}{i-1}=\frac{4}{i-1}=\frac{4(i+1)}{-2}=-2(i+1).
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1619360",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 1
} |
Prove that $\sqrt{6}-\sqrt{2}$ $> 1$. I'm trying to prove that $\sqrt{6}-\sqrt{2}$ $> 1$. I need to admit that I'm completely new to proof writing and I have completely no experience in answering that kind of questions. However, I came up with this:
Since $\sqrt{4}=2$ and $\sqrt{9}=3$ we have a following inequality: $2 < \sqrt{6} < 3$.
On the other hand we have $\sqrt{1}=1$ and $\sqrt{4}=2$ so we have $1 < \sqrt{2} < 2$.
Let $n = \sqrt{6}-\sqrt{2}$,
Therefore $ 1<n<3$, which implies that $n>1$.
Is this a correct proof?
Thank you.
| $$\sqrt{6}-\sqrt{2} = \frac{6-2}{\sqrt{6}+\sqrt{2}}= \frac{2}{\frac{\sqrt{6}+\sqrt{2}}{2}}> \frac{2}{\sqrt{\frac{6+2}{2}}}=1$$
$\bf{Added:}$
$$\sqrt{6} - \sqrt{2} > \sqrt{6.25} - \sqrt{2.25} = 2.5 - 1.5 = 1$$
since the function $x\mapsto \sqrt{x}$ is concave.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1620218",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 5,
"answer_id": 1
} |
Prove that $\sqrt{(1+x)}<1+\frac{1}{2}x$ $\forall x>0$ I was just practicing some real analysis for an upcoming quiz and I just wanted some feedback on if my proof was written sufficiently. The question's above, my answer's below. Thanks!
Pf: Let $x>0,$ $ x \in (0,x)$, and let $f(x) = \sqrt{(1+x)}-(1+\frac{1}{2}x)$. Observe that $f(0) = \sqrt{1+0} - (1 + \frac{1}{2}0) = 1 - 1 = 0.$ Also observe that $f'(x) = \frac{1}{2\sqrt{1+x}} - \frac{1}{2}.$ We know that $\forall x>0$, $\sqrt{1+x}>1$ implies that $\frac{1}{2\sqrt{1+x}} - \frac{1}{2} < 0.$ Thus $f(x)$ is decreasing by the Decreasing Criterion and $f(x) < 0 \Rightarrow$ $\sqrt{(1+x)}-(1+\frac{1}{2}x) < 0 \Rightarrow$ $\sqrt{(1+x)}<1+\frac{1}{2}x$ $\forall x>0.$
| Let $x>0$, then $\frac{1}{4}x^2>0$
By adding $1+x$ to both sides, we obtain:
$$1+x+\frac{1}{4}x^2>1+x$$It follows that:$$(1+\dfrac{1}{2}x)^2>1+x$$
Since both sides are positive for all $x>0$, then $$\sqrt{(1+x)}<1+\dfrac{1}{2}x$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1620274",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
On $4n+1 = x^2, 5n+1 = y^2$ and the Fibonacci numbers While answering this question, I decided to look at the particular case,
$$4n+1 = x^2\\5n+1 = y^2$$
to be solved simultaneously. The solutions for $n,x,y$ are A157459, A007805, and A049629, respectively,
$$\begin{aligned}
n_k &= 72, 23256, 7488432, 2411251920,\dots\\[2.5mm]
x_k &= \frac{F(6k+3)}{2}\\ &= 17, 305, 5473, 98209, 1762289, 31622993,\dots\\[2.5mm]
y_k &= \frac{-F(6k+1)+F(6k+5)}{4} = \frac{F(6k+2)+F(6k+4)}{4}\\ &= 19, 341, 6119, 109801, 1970299, 35355581,\dots
\end{aligned}$$
starting with $k=1$ and $F(k)$ are the Fibonacci numbers.
Trying to find an expression for the $n_k$ in terms of the Fibonacci numbers, the only remaining possibilities are $F(6k)$ and $F(6k+6)$. After some fiddling, I found,
$$n_k = \frac{-18+\phi^{12k+6}+\phi^{-(12k+6)}}{80} = \frac{F(6k)\,F(6k+6)}{16}\tag1$$
and where $\phi = \frac{1+\sqrt{5}}{2}$ is the golden ratio.
Q: How do we prove that the two expressions for $(1)$ are indeed equivalent?
| The Fibonacci numbers are given in terms of $\phi$ by
$$
F(n) = \frac{\phi^n - (-\phi)^{-n}}{\sqrt{5}}.
$$
Also note that for even $n$, $(-\phi)^{-n} = \phi^{-n}$. This implies that
\begin{align*}
F(6k) F(6k+6) &= \frac{1}{5} (\phi^{6k} - \phi^{-6k})(\phi^{6k+6} - \phi^{-6k - 6}) \\
&= \frac{1}{5} \left[ \phi^{12k + 6} + \phi^{-(12k + 6)} - \phi^6 - \phi^{-6} \right] \\
&= \frac{1}{5} \left[ \phi^{12k + 6} + \phi^{-(12k + 6)} - 18 \right],
\end{align*}
since $\phi^6 + \phi^{-6} = 18$. The given equality then follows.
EDIT: Just to prove that $\phi^6 + \phi^{-6} = 18$: note that
$$
(\phi^6 + \phi^{-6})(\phi^6 - \phi^{-6}) = \phi^{12} - \phi^{-12},
$$
from which it follows that
$$
\phi^6 + \phi^{-6} = \frac{F(12)}{F(6)} = \frac{144}{6} = 18.
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1621423",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
How can I integrate $ \int \frac{x^2}{1 + x^3} $?
$$\int \frac{x^2}{1 + x^3}$$
I found this problem in one of my past papers and didn't understand how the solution to this is $ \frac{1}{3} \ln (1+x^3) $. I would really appreciate it if someone guided me through this. I'm sure this is the answer of the integration because I checked it on WolframAlpha and in the mark scheme, but did not understand how to arrive at the answer. Thanks in advance!
| The most efficient solution has already been covered by Yagna, however it can never hurt to have another method. The method I will take here is without question an extremely overly complicated method for this question, but is good practice for integrals of rational functions as a general method.
Here you have a rational function of the form:
\begin{equation}
f(x) = \frac{P_n(x)}{Q_m(x)}
\end{equation}
Were $P_n(x)$ and $Q_m(x)$ are real valued polynomials of orders $n$ and $m$ respectively. The first step is to factor $Q_m(x)$. By the fundamental theorem of algebra we can factor $Q_m(x)$ is a product of constant, linear, and quadratic terms. For your question:
\begin{equation}
f(x) = \frac{x^2}{x^3 + 1} \rightarrow P_n(x) = P_2(x) = x^2,\quad Q_m(x) = Q_3(x) = x^3 + 1
\end{equation}
Here $Q_3(x)$ is factored quite easily:
\begin{equation}
Q_m(x) = Q_3(x) = x^3 + 1 = \left(x + 1\right)\left(x^2 - x + 1\right)
\end{equation}
Thus,
\begin{equation}
f(x) = \frac{x^2}{\left(x + 1\right)\left(x^2 - x + 1\right)}
\end{equation}
We now apply a Partial Fraction Decomposition to yield:
\begin{equation}
f(x) = \frac{x^2}{\left(x + 1\right)\left(x^2 - x + 1\right)} = \frac{1}{3}\cdot \frac{1}{x + 1} - \frac{1}{3}\cdot \frac{2x - 1}{x^2 - x + 1}
\end{equation}
And hence,
\begin{align}
\int \frac{x^2}{x^3 + 1}\:dx &= \frac{1}{3}\int \frac{1}{x + 1}\:dx + \frac{1}{3}\int \frac{2x - 1}{x^2 - x + 1}\:dx \\
&= \frac{1}{3}\ln\left|x + 1\right| + \frac{1}{3} \cdot \ln\left|x^2 - x + 1 \right| + C
\end{align}
Where $C$ is the constant of integration.
As before, this is overly complicated for this exact situation, but it is a process that is useful in tackling these types of integrals.
Edit I had an incorrect partial fraction decomposition which was picked by a commenter. Thank you to that person.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1623243",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
What is the range of $y$ if $x+y+z=4$ and $xy+yz+xz=5$ for $x, y, z \in\mathbb{R}_+$
What is the range of $y$ if $x+y+z=4$ and $xy+yz+xz=5$ for $x, y, z \in\mathbb{R}_+$
How to explain the following method?
Let $x=z$ then:
$$2x+y=4\quad;\quad 2xy+x^{2}=5$$
$$\implies \left( 4-y\right) y+\left( \dfrac {4-y} {2}\right) ^{2}=5 $$
Simplify: $(3y-2)(y-2)=0$. Then I get $\dfrac {2} {3}\leq y\leq 2$.
But why?
Thanks and sorry for my poor English
| $\bf{Another \; Way::}$
Given $x+y+z = 4\Rightarrow x+z=4-y$ and $xy+yz+zx = 5$
Using $$(x+y+z)^2 = x^2+y^2+z^2+2(xy+yz+zx)\Rightarrow x^2+y^2+z^2 = 6$$
So we get $x^2+z^2 = 6-y^2$
Now Using $\bf{Weighted\; A.M\geq G.M\;,}$ We get
$$\frac{x^2+z^2}{2}\geq \left(\frac{x+z}{2}\right)^2$$
So $$\frac{6-y^2}{2}\geq \left(\frac{4-y}{2}\right)^2\Rightarrow (4-y)^2\leq 12-2y^2$$
So $$16+y^2-8y\leq 12-2y^2\Rightarrow 3y^2-8y+4\leq 0$$
So $$3y^2-6y-2y+4\leq 0\Rightarrow \frac{2}{3}\leq y\leq 2$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1623373",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
How to solve this limit involving sine and log? I've tried L'Hopital's Rule but the differentiated numerator involves cos(1/x) which does not exist when x approaches 0.
$$ \lim_{x\to 0^+} \frac{x^2sin\frac{1}{x}}{\ln(1+2x)}$$
| We know
$$\lim_{x\rightarrow 0} \ x\sin\left(\frac{1}{x}\right) = 0$$
$$\ln(1+2x) = 2x - \dfrac{4x^2}{2} + \dfrac{8x^3}{3} ... $$
So the limit reduces to :
$$\lim_{x\rightarrow 0} \ \frac{x\sin\left(\frac{1}{x}\right)x}{2x - \dfrac{4x^2}{2} + \dfrac{8x^3}{3} ..}$$
$$ = \lim_{x\rightarrow 0} \ \frac{x\sin\left(\frac{1}{x}\right)}{2 - \dfrac{4x}{2} + \dfrac{8x^2}{3} ..} = 0$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1626179",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 3
} |
Evaluating the following limit: $\lim _{x\to \frac{\pi }{4}}\left(\tan\left(2x\right)\tan\left(\frac{\pi }{4}-x\right)\right)$ I don't find the right identities for this
$$\lim _{x\to \frac{\pi }{4}}\left(\tan\left(2x\right)\tan\left(\frac{\pi }{4}-x\right)\right)$$
Someone can help me ?
Thanks.
| Since
\begin{align}
\tan(2x)&=\frac{2\tan x}{1-\tan^2 x}&&&\text{and}&&\tan\left(\frac{\pi}{4}-x\right)&=\frac{\tan\frac{\pi}{4}-\tan x}{1+\tan\frac{\pi}{4}\tan x}=\frac{1-\tan x}{1+\tan x}
\end{align}
we have
\begin{align}
\lim_{x\to\frac{\pi}{4}}\left(\tan\left(2x\right)\tan\left(\frac{\pi }{4}-x\right)\right)&=\lim_{x\to\frac{\pi}{4}}\frac{(2\tan x)(1-\tan x)}{(1-\tan x)(1+\tan x)^2}\\
&=\lim_{x\to\frac{\pi}{4}}\frac{2\tan x}{(1+\tan x)^2}\\
&=\frac{2}{(1+1)^2}\\
&=\boxed{\color{blue}{\frac{1}{2}}}
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1626834",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 6,
"answer_id": 0
} |
Given $0This seems true, but I have so far failed to prove it. Any clever estimates I could use?
| Let $x=\sin^2 \alpha$ and $z=\cos^2\beta$, where $0<\alpha,\beta<\frac{\pi}{2}$. Then the condition $0\le z\le1-x$ is equivalent to $\alpha\le\beta$ and the inequality becomes
$$ \sqrt{\sin^2\alpha+\cos^2\beta}\le(1-\sin\alpha)\frac{\cos\beta}{\cos\alpha}+\sin\alpha $$
which is equivalent to
$$ \cos^2\beta\le \frac{(1-\sin\alpha)^2}{\cos^2\alpha}\cos^2\beta+2\frac{(1-\sin\alpha)\sin\alpha}{\cos\alpha}\cos\beta$$
or
$$ \frac{\cos^2\alpha-(1-\sin\alpha)^2}{\cos^2\alpha}\cos\beta\le2\frac{(1-\sin\alpha)\sin\alpha}{\cos\alpha} \tag{1}$$
Noting that
$$ \cos^2\alpha-(1-\sin\alpha)^2=2\sin\alpha-2\sin^2\alpha=2\sin\alpha(1-\sin\alpha)>0 $$
Thus (1) becomes
$$ \frac{2\sin\alpha(1-\sin\alpha)}{\cos^2\alpha}\cos\beta\le2\frac{(1-\sin\alpha)\sin\alpha}{\cos\alpha} $$
which is equivalent to
$$ \cos\beta\le\cos\alpha. $$
This is true because $\alpha\le\beta$. Thus the inequality is true. Done
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1627841",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
$X^4 - 4Y^4 = -Z^2$ has no solutions in non zero integers I am trying to prove that $X^4 - 4Y^4 = -Z^2$ has no solutions in non zero integers. I know there are similar questions on MS, but that minus signs before the $Z$ gives me a hard time.
For the moment, I converted the equation to the equivalent $X^4 + Z^2 = 4Y^2$, and I supposed that $(x,y,z)$ is a primitive solution for this equation. Thus we have $(x^2)^2 +z^2 = (2y^2)^2$.
Now $(2y^2)^2$ is even, it tells us that $x$ and $z$ must have the same parity. But we know by Diophante's equation that there exist $a$ and $b$ with $a > b$, $(a, b) = 1$. If $a \not \equiv b \pmod 2$, we have $x^2 = a^2 - b^2, z = 2ab$ and $2y^2 = a^b + b^2$. But this is impossible, because of $2y^2 = a^2 + b^2$, $a$ and $b$ should have the same parity, which is not our hypothesis.
If $a$ and $b$ have the same parity, they are both odd (otherwise they are not coprime) and we have $x^2 = \frac{a^2 - b^2}{2}$, $z = ab$ and $2y^2 = \frac{a^2 + b^2}{2}$. But once again, as $(a, b) = 1$ and are both odd, they have no factor "2" in common thus $2y^2 = \frac{a^2 + b^2}{2}$ is impossible.
Therefore the equation $X^4 - 4Y^4 = -Z^2$ has no solution in $\mathbb{N} \setminus \{0\}$.
What do you think of this argument? My first idea was to use the infinite descent, but in this case I did not manage to do it.
My second idea was to prove that it was equivalent to another equation such as $X^4 + Y^4 = Z^2$ which we know has no solutions such that $XYZ \neq 0$ but I did not have more success.
| Your argument is correct, and no quicker method is springing to mind to solve this.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1628270",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Find the last $2$ digits of $7^{7^{7^{10217}}}$
Find the last $2$ digits of $$\large7^{7^{7^{10217}}}$$
So far I have: $17\cdot 601=10217$ and $7=7 \pmod{10}$. Any help greatly appreciated.
| Note that $100=4\cdot 25$. We can find $7^{7^{7^{10217}}}$ mod $4$, $25$, then combine this to get the value mod $100$ using the Chinese Remainder Theorem.
$7^4\equiv \left(7^2\right)^2\equiv (-1)^2\equiv 1\pmod{25}$
Therefore $$7^{7^{7^{10217}}}\equiv 7^{7^{7^{10217}}\pmod 4}\equiv 7^{(-1)^{7^{10217}}\pmod 4}\pmod{25}$$
$$\equiv 7^{-1\pmod{4}}\equiv 7^{3}\equiv \left(7^2\right)\cdot 7\equiv (-1)\cdot 7\equiv 18\pmod{25}$$
Also
$$7^{7^{7^{10217}}}\equiv (-1)^{7^{7^{10217}}}\equiv -1\equiv 3\pmod{4}$$
Using the Chinese Remainder theorem, $7^{7^{7^{10217}}}\equiv 43\pmod{100}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1631133",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Show that $\int_{-\infty}^{\infty}\frac{dx}{\sqrt{x^4+1}}$ converges. Show that
$$
\int_{-\infty}^{\infty}\frac{dx}{\sqrt{x^4+1}}
$$
converges.
I recognized that that since the integrand is even then
$$
\int_{-\infty}^{\infty}\frac{dx}{\sqrt{x^4+1}} = 2\int_{0}^{\infty}\frac{dx}{\sqrt{x^4+1}}
$$
Then, I started with the following statement and manipulated it to match the integrand.
$$
\begin{align*}
x^4 +1 &\gt x^4 \\ \sqrt{x^4+1} &\gt x^2 \\ \frac{1}{\sqrt{x^4 +1}} &\lt\frac{1}{x^2}
\end{align*}
$$
It is a well-known result that
$$
\int_1^{\infty}\frac{dx}{x^2}
$$
converges. So would it be correct to say that, by comparison, that
$$
\int_{-\infty}^{\infty}\frac{dx}{\sqrt{x^4+1}}
$$
converges as well?
| In the regime $|x| \ge 1$, you can compare $(x^4+1)^{-1/2}$ against $x^{-2}$, but in the neighborhood of $0$, the latter is unbounded so you are best off noting that $$\int_{x=-\infty}^\infty (x^4+1)^{-1/2} \, dx = 2 \int_{x=0}^\infty (x^4+1)^{-1/2} \, dx < 2\left( \int_{x=0}^1 (x^4+1)^{-1/2} \, dx + \int_{x=1}^\infty x^{-2} \, dx \right).$$ The first term is obviously finite since $0 < (x^4 + 1)^{-1/2} \le 1$ on $[0,1]$; and the second is also finite as you stated.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1631309",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 6,
"answer_id": 0
} |
How could we define the factorial of a matrix? Suppose I have a square matrix $\mathsf{A}$ with $\det \mathsf{A}\neq 0$.
How could we define the following operation? $$\mathsf{A}!$$
Maybe we could make some simple example, admitted it makes any sense, with
$$\mathsf{A} =
\left(\begin{matrix}
1 & 3 \\
2 & 1
\end{matrix}
\right)
$$
| I use the well known (and simple) definitions
$$n!=\Gamma (n+1)$$
and
$$\Gamma (A+1)=\int_0^{\infty } \exp (-t) \exp (A \log (t)) \, dt$$
Now, if A is a (square) matrix, all we need is to define the exponential function for a matrix.
This can always be done (in principle) via the power series which only requires to calculate powers of the matrix and adding matrices.
We skip here a possible diagonalization procedure (which was shown by others before) and use the function MatrixExp[] of Mathematica. For the Matrix given in the OP
$$A=\left(
\begin{array}{cc}
1 & 3 \\
2 & 1 \\
\end{array}
\right);$$
we have
Ax = MatrixExp[A Log[t]]
which gives
$$\left(
\begin{array}{cc}
\frac{t^{1-\sqrt{6}}}{2}+\frac{t^{1+\sqrt{6}}}{2} & \frac{1}{2} \sqrt{\frac{3}{2}} t^{1+\sqrt{6}}-\frac{1}{2} \sqrt{\frac{3}{2}} t^{1-\sqrt{6}} \\
\frac{t^{1+\sqrt{6}}}{\sqrt{6}}-\frac{t^{1-\sqrt{6}}}{\sqrt{6}} & \frac{t^{1-\sqrt{6}}}{2}+\frac{t^{1+\sqrt{6}}}{2} \\
\end{array}
\right)$$
We observe that in some cases the exponent of t is less than -1 ($1-\sqrt{6}=-1.44949$). This leads to a divergent integral which will then be understood as the analytic continuation.
This is accomplished simply by replacing each $t^q$ by $\Gamma (q+1)$.
fA = Ax /. t^q_ -> Gamma[q + 1]
giving
$$\left(
\begin{array}{cc}
\frac{\Gamma \left(2-\sqrt{6}\right)}{2}+\frac{\Gamma \left(2+\sqrt{6}\right)}{2} & \frac{1}{2} \sqrt{\frac{3}{2}} \Gamma \left(2+\sqrt{6}\right)-\frac{1}{2} \sqrt{\frac{3}{2}} \Gamma \left(2-\sqrt{6}\right) \\
\frac{\Gamma \left(2+\sqrt{6}\right)}{\sqrt{6}}-\frac{\Gamma \left(2-\sqrt{6}\right)}{\sqrt{6}} & \frac{\Gamma \left(2-\sqrt{6}\right)}{2}+\frac{\Gamma \left(2+\sqrt{6}\right)}{2} \\
\end{array}
\right)$$
Numerically this is
$$\left(
\begin{array}{cc}
3.62744 & 8.84231 \\
5.89488 & 3.62744 \\
\end{array}
\right)$$
This is in agreement with the result of hans and travis.
Discussion
(1) Let me point out that the definitions presented here do not need any specific property of the matrix. For example, is does not matter whether the eigenvalues and eigenvectors are singular or not, the matrix might well be deficient.
(2) I have used Mathematica here just to facilitate things. In the end we all use some tools at a certain stage to come to terms. The main ideas are idependent.
(3) The procedure desribed here obviously generalizes to other more or less complicated analytic functions.
As a more exotic example let us take the harmonic number H(A) of a matrix A. This function can be defined using the integral representation (see e.g. Relation between binomial coefficients and harmonic numbers)
$$H(\text{A})\text{=}\int_0^1 \frac{1-(1-x)^A}{x} \, dx$$
This definition also needs only the exponential function of the matrix.
The result for our matrix A is (after some analytic continuation)
$$\left(
\begin{array}{cc}
\frac{1}{2} \left(H_{-\sqrt{6}}+H_{\sqrt{6}}+\frac{1}{1-\sqrt{6}}+\frac{1}{1+\sqrt{6}}\right) & \frac{1}{2} \sqrt{\frac{3}{2}} \left(-H_{-\sqrt{6}}+H_{\sqrt{6}}-\frac{1}{1-\sqrt{6}}+\frac{1}{1+\sqrt{6}}\right) \\
\frac{-H_{-\sqrt{6}}+H_{\sqrt{6}}-\frac{1}{1-\sqrt{6}}+\frac{1}{1+\sqrt{6}}}{\sqrt{6}} & \frac{1}{2} \left(H_{-\sqrt{6}}+H_{\sqrt{6}}+\frac{1}{1-\sqrt{6}}+\frac{1}{1+\sqrt{6}}\right) \\
\end{array}
\right)$$
Numerically,
$$\left(
\begin{array}{cc}
1.51079 & 0.542134 \\
0.361423 & 1.51079 \\
\end{array}
\right)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1634488",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "204",
"answer_count": 7,
"answer_id": 0
} |
If $ A=\frac{1}{2\sqrt{1}}+\frac{1}{3\sqrt{2}}+\frac{1}{4\sqrt{3}}+.........+\frac{1}{100\sqrt{99}}\;,$ Then $\lfloor A \rfloor =$
If $\displaystyle A=\frac{1}{2\sqrt{1}}+\frac{1}{3\sqrt{2}}+\frac{1}{4\sqrt{3}}+.........+\frac{1}{100\sqrt{99}}\;,$ Then $\lfloor A \rfloor =$
Where $\lfloor x \rfloor$ represent floor function of $x$.
$\bf{My\; Try:}$ For lower bound $$\sum^{99}_{k=1}\frac{1}{(k+1)\sqrt{k}}>\sum^{99}_{k=1}\frac{1}{(k+1)k}=\sum^{99}_{k=1}\left[\frac{1}{k}-\frac{1}{k+1}\right]=1-\frac{1}{99}$$
Now I didn't understand how can I solve it, any help?
Thanks
| Let
$$A = \sum_{k=1}^{99} \frac1{(k+1) \sqrt{k}} $$
You can show that the sum is greater than $1$. First we can lower bound the sum by
$$\sum_{k=1}^{99} \frac1{k (k+1)} = \frac{99}{100} $$
which, while less than one, may be corrected by its second term:
$$\frac{99}{100} + \frac13 \left (\frac1{\sqrt{2}} - \frac12 \right ) \gt 1$$
i.e., $A \gt 1$. Next, consider that
$$A = \sum_{k=1}^{99} \frac1{(k+1) \sqrt{k}} = \sum_{k=1}^{99} \left (\frac{\sqrt{k}}{k} - \frac{\sqrt{k+1}}{k+1} \right ) + \sum_{k=1}^{99} \frac{\sqrt{k+1}-\sqrt{k}}{k+1}$$
The first sum on the RHS is just $9/10$. The second sum, however, is
$$ \sum_{k=1}^{99} \frac1{(k+1) (\sqrt{k+1}+\sqrt{k})} \le \frac12 \sum_{k=1}^{99} \frac1{(k+1) \sqrt{k}} = \frac12 A$$
Thus $A \le \frac9{10} + \frac12 A$, or
$$A = \sum_{k=1}^{99} \frac1{(k+1) \sqrt{k}} \le \frac95$$
and is also greater than one. Thus, the floor of the sum is $1$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1634645",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 4,
"answer_id": 2
} |
Prove inequality $1 < \frac{1}{n} + \frac{1}{n+1} + \ldots + \frac{1}{3n-1} < 2$ Prove the inequality $1 < \frac{1}{n} + \frac{1}{n+1} + \ldots + \frac{1}{3n-1} < 2$ For all $n \in \mathbb{N}$
I've done the right hand side, but can't do the left side of the inequality. For the right:
$\frac{1}{n} + \ldots + \frac{1}{3n-1} < \underbrace{\frac{1}{n} + \ldots + \frac{1}{n}}_{2n} = 2$
Now I haven't been able to make progress with the left.
| Hint:
Show $\dfrac{1}{n} \lt \dfrac{1}{n+k}+\dfrac{1}{3n-1-k} \lt \dfrac{2}{n} $ for $0 \le k \lt n$
Then sum over $k$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1635728",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 1,
"answer_id": 0
} |
How do I solve the following equation: $z^4+z^3+z+1=0$ Is there an existing method to solve the following equation:
$z^4+z^3+z+1=0$?
| Hint:
$$z^4 + z^3 + z + 1 = z^4 + z + z^3 + 1 = z(z^3+1) + z^3 + 1 = (z+1)(z^3 + 1)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1637779",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 0
} |
Find all incongruent solutions to $21x \equiv 14 \pmod{91}$ Find all incongruent solutions to $21x \equiv 14 \pmod{91}$.
I am able to work out the solution using Euclidean algorithm techniques, but the signs on the expression do not match up with the initial expression when I check my work. So by the linear congruence theorem, my solution has to satisfy: $$21x - 91y = 14$$ but after going through the process with the $\gcd(21, 91)$ my expression ends up as $$91 - 21(4) = 7$$ which I multiply by $2$ to get: $$91(2) - 21(8) = 14$$
Which would mean my solution has to have a negative somewhere in it. I can "put" a negative on one of my values and the original expression would be satisfied but that is not what I obtained through the work I did. Is the confusion in signs occurring on purpose or am I treating something wrong?
| By definition, the congruence
$$21x \equiv 14 \pmod{91} \tag{1}$$
is equivalent to the equation
$$21x = 14 + 91t, t \in \mathbb{Z} \tag{2}$$
If we divide each term of equation 2 by $7$, we obtain the equivalent equation
$$3x = 2 + 13t, t \in \mathbb{Z}$$
which is equivalent to the congruence
$$3x \equiv 2 \pmod{13} \tag{3}$$
Hence,
$$21x \equiv 14 \pmod{91} \Longleftrightarrow 3x \equiv 2 \pmod{13}$$
Since $\gcd(3, 13) = 1$, the congruence $3x \equiv 2 \pmod{13}$ has a solution. We can find it by applying the extended Euclidean algorithm.
\begin{align*}
13 & = 4 \cdot 3 + 1\\
3 & = 3 \cdot 1
\end{align*}
Solving for $1$ in terms of $3$ and $13$ yields
$$1 = 13 - 4 \cdot 3$$
Thus,
$$1 \equiv -4 \cdot 3 \pmod{13} \implies -4 \equiv 3^{-1} \pmod{13}$$
Therefore, if we multiply both sides of congruence 3 by $-4$, we obtain
$$x \equiv -8 \pmod{13}$$
To find all the solutions of congruence 1, we must find all the solutions of the inequality
$$0 \leq -8 + 13t < 91$$
in the integers.
\begin{align*}
0 & \leq -8 + 13t < 91\\
8 & \leq 13t < 99\\
\end{align*}
Hence, $1 \leq t \leq 7$. Therefore, the solutions of the congruence $21x \equiv 14 \pmod{91}$ are
\begin{align*}
x & \equiv 5 \pmod{91}\\
& \equiv 18 \pmod{91}\\
& \equiv 31 \pmod{91}\\
& \equiv 44 \pmod{91}\\
& \equiv 57 \pmod{91}\\
& \equiv 70 \pmod{91}\\
& \equiv 83 \pmod{91}
\end{align*}
which you can check by direct computation.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1638266",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
} |
Reasoning Aptitude CSIR 15 A single-celled spherical organism contains $70$% water by volume. If it loses $10$% of its water content, how much would its surface area change by approximately?
*
*$3\text{%}$
*$5\text{%}$
*$6\text{%}$
*$7\text{%}$
| Let $r$ be the original radius the original volume of water is $$V_1=\frac{70}{100}\times \frac{4}{3}\pi r^3=\frac{14}{15}\pi r^3$$ & volume of organic material
$$V_2=\frac{30}{100}\times \frac{4}{3}\pi r^3=\frac{2}{5}\pi r^3$$
& surface area $S_0=4\pi r^2$
when it loses $10$% of water then remaining volume of water $$V_1'=\frac{90}{100}\times V_1= \frac{90}{100}\times\frac{70}{100} \times \frac{4}{3}\pi r^3=\frac{21}{25}\pi r^3$$
If $r'$ is the radius of remaining cell then new volume of cell $$V_1'+V_2=\frac{21}{25}\pi r^3+\frac{2}{5}\pi r^3$$
$$r'=\left(0.93 \right)^{1/3}r$$
hence the new surface area of cell $$S'=4\pi r'^2=4\pi \left(\left(0.93\right)^{1/3}r\right)^2=4\pi \left(0.93 \right)^{2/3}r^2$$
hence % change in surface area of the cell $$\frac{S'-S_0}{S_0}\times 100$$
$$=\frac{4\pi \left(0.93 \right)^{2/3}r^2-4\pi r^2}{4\pi r^2}\times 100=4.722877503\approx 5\ \text{%}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1640285",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Basic algebra problem: $ \frac{\frac{1}{x}+\frac{1}{y}}{\frac{1}{x^2}-\frac{1}{y^2}} $ Basic algebra problem I can't seem to figure out: $$ \frac{\frac{1}{x}+\frac{1}{y}}{\frac{1}{x^2}-\frac{1}{y^2}} $$ $x,y \in \mathbb{R}, x^2 \neq y^2, xy\neq0$.
Now I know the result is: $\frac{xy}{y-x}$, but I am not sure how to get it, I get into a mess like this: $=x+\frac{x^2}{y}-\frac{y^2}{x}-y=\frac{x(xy)+x^3-y^3-y(xy)}{xy}=?$ which doesn't seem to help me much. Halp please.
| $$a=\frac1x+\frac1y=\frac{x+y}{xy}$$
$$b=\frac1{x^2}-\frac1{y^2}=\frac{y^2-x^2}{(xy)^2}$$
$$\frac{a}b=\frac{(x+y)xy}{x^2-y^2}=\frac{(x+y)xy}{(y-x)(y+x)}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1642208",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 3
} |
$p = \sqrt{1+\sqrt{1+\sqrt{1 + \cdots}}}$; $\sum_{k=2}^{\infty}{\frac{\lfloor p^k \rceil}{2^k}} = ? $
Let $p = \sqrt{1+\sqrt{1+\sqrt{1 + \cdots}}}$
The sum $$\sum_{k=2}^{\infty}{\dfrac{\lfloor p^k \rceil}{2^k}}$$
Can be expressed as $\frac{a}{b}$. Where $\lfloor \cdot \rceil$ denotes the nearest integer function. Find $a+b$.
My Work
$p^2 = 1+p \implies p = \dfrac{1+\sqrt{5}}{2} = \phi$
And also $\phi^n = \dfrac{L_n + F_n\sqrt{5}}{2}$. Where $L_n$ is the $n-th$ Lucas-Number.
What to do with the next part?
Note: Problem Collected from Brilliant.org
| For $n \ge 2$, the numbers $\lfloor \phi^k \rceil$ are equal to the Lucas numbers, which satisfy the Fibonacci recurrence but in which $L_1=1$ and $L_2=3$. The Lucas numbers satisfy
$$L_k = \phi^k + (-1)^k \phi^{-k}$$
and thus have the generating function
$$L(x) = \frac{2-x}{1-x-x^2} $$
so the sum we want is
$$L \left (\frac12 \right ) - L_0 - \frac12 L_1 = \frac72$$
Note that $\phi^{-1}= \phi-1 = 0.618\ldots$ and exponentiation to the power of $k$ does not affect a rounding for $k \ge 2$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1644523",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 2,
"answer_id": 1
} |
proving an inequality related to $AM\ge GM$ $$a^2+ab+b^2\ge 3(a+b-1)$$
$a,b$ are real numbers
using $AM\ge GM$
I proved that
$$a^2+b^2+ab\ge 3ab$$
$$(a^2+b^2+ab)/3\ge 3ab$$
how do I prove that $$3ab\ge 3(a+b-1)$$
if I'm able to prove the above inequality then i'll be done
| Follow @BarryCipra's hint of letting $a=x+1$ and $b=y+1$:
$$(x+1)^2+(x+1)(y+1)+(y+1)^2 \geq 3((x+1)+(y+1)-1)$$
Expand the left side and simplify the right side:
$$x^2+2x+1+xy+x+y+1+y^2+2y+1 \geq 3(x+y+1)$$
Simplify the left side and distribute the $3$ on the right side:
$$x^2+xy+y^2+3x+3y+3 \geq 3x+3y+3$$
Subtract both sides by $3x+3y+3$:
$$x^2+xy+y^2 \geq 0$$
Now, to prove this. we need to make some changes. We know that $xy \geq \lvert xy \rvert$. We also know that $\lvert xy \rvert \geq -\lvert xy \rvert$ because $\lvert xy \rvert$ is non-negative and non-negative numbers are always greater than or equal to their negatives. Finally, we have $-\lvert xy \rvert \geq -2\lvert xy \rvert$ because a non-positive number added to itself is always less than or equal to the original (e.g. $-2+(-2) \leq -2$, $-1+(-1) \leq -1$, $0+0 \leq 0$). Therefore, by the Transitive Property of Equality, we have $xy \geq -2\lvert xy \rvert$ and by adding $x^2+y^2$ by both sides of the inequality, we have $x^2+xy+y^2 \geq x^2-2\lvert xy \rvert+y^2$. This means we only need to prove the latter is non-negative to prove the former is non-negative, so we've reduced the problem to:
$$x^2-2\lvert xy \rvert+y^2 \geq 0$$
Now, we know that $x^2=\lvert x \rvert^2$, $y^2=\lvert y\rvert^2$, and $\lvert xy \rvert=\lvert x \rvert \lvert y \rvert$, so we can substitute to get everything in terms of $\lvert x \rvert$ and $\lvert y \rvert$:
$$\lvert x \rvert^2-2\lvert x \rvert \lvert y \rvert+\lvert y \rvert^2 \geq 0$$
Clearly, this is the same as $(\lvert x \rvert - \lvert y \rvert)^2$:
$$(\lvert x \rvert - \lvert y \rvert)^2 \geq 0$$
This is obviously true because the square of a real number is always non-negative. Thus, by proving this is non-negative, we have proven that $x^2+xy+y^2$ is non-negative by our previous argument, which proves the whole inequality.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1644694",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Prove that one of the lines represented by $ax^2+2hxy+by^2=0$ will bisect the angle between the coordinate axes if $(a+b)^2=4h^2$. Prove that one of the lines represented by $ax^2+2hxy+by^2=0$ will bisect the angle between the coordinate axes if $(a+b)^2=4h^2$.
Solution
I calculated the two lines represented by $ax^2+2hxy+by^2=0$ as follows;
here.
$$ax^2+2hxy+by^2=0$$
Multiplying by $a$ on both sides and adding $h^2y^2$ to both sides :
$$ax+hy=\pm y\sqrt{h^2-ab}.$$
What should I do next?
| Let $y=mx$ be the required line. Then we can rewrite the given pair of lines as $$bm^2+2hm+a=0$$.
$$\implies m=\frac{-2h\pm\sqrt{4h^2-4ab}}{2b}$$
$$(a+b)^2=4h^2 \implies 4h^2-4ab=(a-b)^2$$
Substituting in the value of $m$, $$m=\frac{-\vert a+b\vert\pm\vert a-b\vert}{2b}$$
For all values of $a,b$, the above will always gives at least $1$ or $-1$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1647804",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 6,
"answer_id": 3
} |
Subsets and Splits
Fractions in Questions and Answers
The query retrieves a sample of questions and answers containing the LaTeX fraction symbol, which provides basic filtering of mathematical content but limited analytical insight.