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Find equation of the circle whose diameter is the common chord of two other circles? Circle 1:
$$x^2 + y^2 +6x + 2y +6 = 0$$
Circle 2:
$$x^2 + y^2 + 8x + y + 10 = 0$$
My attampt:
From circle 1 and 2, I found
$$ y = 2x + 4 $$
which is the common chord.
Pluging that in equation 1 I got
$$5x^2 + 26x + 30 = 0$$
here I got stuck for too much complication if I use quadratic formula.Is there anything wrong?
| First, obtain the equations of the intersection points below for both $x$ and $y$,
$$5x^2 + 26x + 30= 0$$
$$5y^2 + 12y -8= 0$$
It may be more efficient not to solve for the intersection points explicitly. Rather, use the relationships from the above equations ,
$$x_1+x_2=-\frac{26}{5},\>\>\>x_1x_2=6$$
$$y_1+y_2=-\frac{12}{5},\>\>\>y_1y_2=-\frac 85 $$
Thus, the center of the circle is $\frac{x_1+x_2}{2}=-\frac{13}{5}, \frac{y_1+y_2}{2}=-\frac{6}{5}$ and its diameter squared is,
$$(x_1-x_2)^2 + (y_1-y_2)^2$$
$$ = (x_1+x_2)^2- 4x_1x_2 + (y_1+y_2)^2 - 4y_1y_2$$
$$= \left( \frac{26}{5} \right)^2 -4\cdot 6 + \left( \frac{12}{5}\right)^2 + 4\cdot \frac 85 = \frac{76}{5}$$
The equation of the circle is
$$\left( x+\frac{13}{5} \right)^2 + \left( y +\frac{6}{5}\right)^2 = \frac{19}{5}$$
| {
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"url": "https://math.stackexchange.com/questions/3363697",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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Find all natural numbers $n$ such that $n+1$ divides $3n+11$ Following the example of my teacher:
Find all natural numbers $n$ such that $n-2$ divides $n+5$.
$$n+5 = n-2+7$$
As $n-2|n-2$, $n-2$ will divide $n+5$ if and only if $n-2|7$. Yet, $7$ has the divisors $-7$, $-1$, $1$, $7$ hence the equations:
*
*$n-2=-7 \Leftrightarrow n = -5 $
*$n-2=-1 \Leftrightarrow n = 1 $
*$n-2=7 \Leftrightarrow n = 3 $
*$n-2=-1 \Leftrightarrow n = 9 $
So $S = \{ 1, 3, 9 \}$
I decompose $n+1$ the exact same way:
$$n+1 = 3n+11 - 2(n+5)$$
But I'd get stuck as $2(n+5)$ since only $-2$, $-1$, $1$ and $2$ are divisors, which don't satisify the equation as I'd hoped:
*
*$3n+11 = -2 \Leftrightarrow n =\frac{-13}{3}$
*$3n+11 = -1 \Leftrightarrow n = -4$
*$3n+11 = 1 \Leftrightarrow n =\frac{-10}{3}$
*$3n+11 = 1 \Leftrightarrow n = -3$
Any clues?
| $\!\! \bmod n\!+\!1\!:\,\ \color{#c00}{n\equiv -1}\,\Rightarrow\, 3\,\color{#c00}n+11\equiv 3(\color{#c00}{-1})+11\equiv 8\ $ by Congruence Sum & Product Rules $\ \ \ \ \ $
| {
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Prove inequalities $\frac 34 \le I(a,b) \le 1$ Given the expression,
$$ I(a,b) = \frac{a^2}{(1+a)(a+b)} + \frac{b^2}{(1+b)(a+b)} + \frac{1}{(1+a)(1+b)}$$
where $a\ge 0$ and $b\ge 0$, prove the following inequalities:
$$\frac 34 \le I(a,b) \le 1$$
I had trouble figuring it out. Tried some inequity techniques I am aware of and had no lack so far. I am not sure if it is wise to go brute force by examining the derivatives of $I(a,b)$, which I feel would be pretty messy due to its dual dimension.
Appreciate if anyone could offer a viable approach for the proof.
| I think, it means that $a+b>0$, otherwise your inequality is wrong for $a=b=0$.
Let $a+b=2x$.
Thus, by C-S and AM-GM we obtain:
$$I(a,b)=\frac{1}{a+b}\left(\frac{a^2}{1+a}+\frac{b^2}{1+b}\right)+\frac{1}{1+a+b+ab}\geq$$
$$\geq\frac{1}{2x}\cdot\frac{(a+b)^2}{1+a+1+b}+\frac{1}{1+2x+\left(\frac{a+b}{2}\right)^2}=$$
$$=\frac{x}{1+x}+\frac{1}{(x+1)^2}=\frac{3}{4}+\frac{1}{4}-\frac{1}{1+x}+\frac{1}{(1+x)^2}=$$
$$=\frac{3}{4}+\left(\frac{1}{2}-\frac{1}{1+x}\right)^2\geq\frac{3}{4}.$$
Also, $I(a,b)\leq1$ it's $$a^2(1+b)+b^2(1+a)+a+b\leq(1+a)(1+b)(a+b)$$ or $$ab\geq0$$ and we are done!
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3366373",
"timestamp": "2023-03-29T00:00:00",
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Parabolic mirror: why is $b^2-1/4 = \tan(2\tan^{-1}(2b)+{\pi}/{2})(b-0)$ for all $b$? Let us take the simple parabola $x^2$.
A ray of light will bounce on it making equal angles to both sides.
The derivative of $x^2$ is $2x$ and its normal is $-1/2x$.
Given light rays coming straight from above we get, given a variable b that represents the x value of the vertical ray we get that the reflected ray in general is:
$\left(b^2-y\right)=\tan\left(2\tan^{-1}\left(2b\right)+\frac{\pi}{2}\right)\left(b-x\right)$
Graph: https://www.desmos.com/calculator/tqf9a8fnsr
In fact we see that this always pass through a point (focal point) for all b by watching the graph while sliding b.
I want to prove all points pass there analitically.
I substitute in $b=1/2$ and $b=(\sqrt3)/2$ as examples and I get the intersection (0,1/4) (also clearly seen in the graph).
Now I substitute (x=0,y=1/4) back in $\left(b^2-y\right)=\tan\left(2\tan^{-1}\left(2b\right)+\frac{\pi}{2}\right)\left(b-x\right)$ and I get an equation that is true for all b as you can check by sliding the slider. $\left(b^2-\frac{1}{4}\right)-\tan\left(2\tan^{-1}\left(2b\right)+\frac{\pi}{2}\right)\left(b-0\right)$ This means that all reflected rays of light pass by it.
How can I prove this last statement analytically?
| Apply the following identities
$ tan(x+ \frac \pi 2)=-cot(x)$
$cot(2x) = \frac{ \cot^2(x)-1 }{ 2\cot(x) }$
$ \cot( \tan^{-1}(x) ) =\frac 1x $
Get $$ \tan\left(2\tan^{-1}\left(2b\right)+\frac{\pi}{2}\right)
\\ = -\cot( 2 \tan^{-1}(2b) )
\\ =\frac { 1-\cot^2( \tan^{-1}(2b) ) }{ 2\cot( \tan^{-1}(2b) }
\\ = \frac{ 1-\frac 1{4b^2} }{ \frac 1b }
\\ = b-\frac 1{4b}$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Using AM-GM inequality prove that $(1+\sqrt{2})(1+\sqrt{3})(1+\sqrt{5}) \lt 8\sqrt{30}$. It is trivial to prove that $(1+\sqrt{2})(1+\sqrt{3})(1+\sqrt{5}) \le
8\sqrt{30}$ using numeric methods. For example by multiplying
$(1+\sqrt{2}) \le 3 $
$(1+\sqrt{3}) \le 3 $
$(1+\sqrt{5}) \le 4 $
We get:
$(1+\sqrt{2})(1+\sqrt{3})(1+\sqrt{5}) < 36$
while $8\sqrt{30} \gt 40 $ because $\sqrt{30} \gt 5 $ . However for this
particular problem it is asked a solution using the AM-GM inequality and I am not able to find one. Can anyone help me?
| $$(1+\sqrt 2)(1+\sqrt 3)(1+\sqrt 5)=1+\sqrt 2 + \sqrt 3 + \sqrt 5 + \sqrt 6 + \sqrt {10} + \sqrt {15} + \sqrt {30}$$
and thus you can use the AM-GM inequality on this
Actually I think I messed up, the AM-GM inequality goes the wrong way and the exponents don't work out. Use the Root-Mean Square - Arithmetic Mean Inequality: https://artofproblemsolving.com/wiki/index.php/Root-Mean_Square-Arithmetic_Mean-Geometric_Mean-Harmonic_mean_Inequality
to get that
$$\frac{1+\sqrt 2 + \sqrt 3 + \sqrt 5 + \sqrt 6 + \sqrt {10} + \sqrt {15} + \sqrt {30}}{8}\leq \sqrt{\frac{72}{8}}=3\leq \sqrt{30}$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Infinite product $\Gamma(\tfrac14)=\mathrm{A}^3e^{-\mathrm{G}/\pi}2^{1/6}\sqrt{\pi}\prod_{k\ge1}\left(1-\frac1{2k}\right)^{(-1)^k k}$ I saw the following infinite product on Wikipedia:
$$\Gamma\left(\tfrac14\right)=\mathrm{A}^3e^{-\mathrm{G}/\pi}2^{1/6}\sqrt{\pi}\prod_{k\ge1}\left(1-\frac1{2k}\right)^{(-1)^k k}\tag{1}$$
where $\mathrm A$ is the Glaisher constant, and $\mathrm G$ is Catalan's constant. I am looking for a proof of this product.
I haven't gotten very far with this product, other than noting that if
$$\zeta_*(s)=\sum_{k\ge1}\frac1{(1-\frac1{2k})^{(-1)^k ks}}$$
then
$$\zeta_*'(0)=3\ln\mathrm A+\frac16\ln2+\frac12\ln\pi-\frac{\mathrm{G}}{\pi}-\ln\Gamma\left(\tfrac14\right),\tag2$$
of course, assuming that $(1)$ is true. Perhaps $(2)$ is easier to prove. Could I have some help? Thanks.
| A short proof.
$\displaystyle Q_0(x) :=\Gamma(x+1)=\lim_{n\to\infty}\frac{n^x}{\prod\limits_{k=1}^n\left(1+\frac{x}{k}\right)}~~ , ~~~~ Q_1(x) :=\lim_{n\to\infty}\frac{e^{xn}n^{-x^2/2}}{\prod\limits_{k=1}^n\left(1+\frac{x}{k}\right)^k}$
$\displaystyle \pi^{1/2} = Q_0\left(-\frac{1}{2}\right)~~ , ~~ A^3 = 2^{7/12}Q_1\left(-\frac{1}{2}\right)^2 ~~ , ~~ e^{G/\pi} = 2^{3/4}\left(\frac{ Q_0\left(-\frac{1}{2}\right) Q_1\left(-\frac{3}{4}\right) }{ Q_0\left(-\frac{3}{4}\right) Q_1\left(-\frac{1}{4}\right) }\right)^2$
It follows:
$\displaystyle \prod\limits_{k=1}^{2n}\left(1-\frac{1}{2k}\right)^{(-1)^k k} = \prod\limits_{k=1}^n \frac{\left(1-\frac{1}{4k}\right)^{2k}}{\left(1-\frac{1}{4k-2}\right)^{2k-1}} = \prod\limits_{k=1}^n \frac{\left(1-\frac{1}{2k}\right)^{2k-1} \left(1-\frac{1}{4k}\right)^{2k}}{ \left(1-\frac{3}{4k}\right)^{2k-1} } = $
$\displaystyle = \frac{ \prod\limits_{k=1}^n \left(1-\frac{3}{4k}\right) n^{-1/2} }{ n^{-3/4} \prod\limits_{k=1}^n \left(1-\frac{1}{2k}\right) } \left(\frac{ e^{-3n/4} n^{-9/32} \prod\limits_{k=1}^n \left(1-\frac{1}{2k}\right)^k \prod\limits_{k=1}^n \left(1-\frac{1}{4k}\right)^k }{\prod\limits_{k=1}^n \left(1-\frac{3}{4k}\right)^k e^{-n/2} n^{-1/8} e^{-n/4} n^{-1/32} }\right)^2$
$\displaystyle \to ~\frac{ Q_0\left(-\frac{1}{2}\right) }{ Q_0\left(-\frac{3}{4}\right) } \left(\frac{ Q_1\left(-\frac{3}{4}\right) }{ Q_1\left(-\frac{1}{2}\right) Q_1\left(-\frac{1}{4}\right) }\right)^2$
$\displaystyle = 2^{-1/6} \frac{ Q_0\left(-\frac{3}{4}\right) }{ Q_0\left(-\frac{1}{2}\right) } \cdot 2^{-7/12} Q_1\left(-\frac{1}{2}\right)^{-2} \cdot 2^{3/4} \left(\frac{ Q_0\left(-\frac{1}{2}\right) Q_1\left(-\frac{3}{4}\right) }{ Q_0\left(-\frac{3}{4}\right) Q_1\left(-\frac{1}{4}\right)}\right)^2$
$\displaystyle = \Gamma\left(\frac{1}{4}\right) 2^{-1/6} \pi^{-1/2} A^{-3} e^{G/\pi}$
| {
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A (possibly) difficult epsilon-delta proof So I'm struggling to find an epsilon-delta proof for why the function $f(h) = \dfrac{\sqrt[3]{1+h}-1}{h}$ approaches $\dfrac{1}{3}$ as $x \rightarrow 0$.
I'd like to know how one can show using an epsilon-delta proof or some other means that for $0 < \epsilon < \dfrac{1}{8}$ and for $0 < h < 4\epsilon$, we have that $1+(\dfrac{1}{3}-\epsilon)h < \sqrt[3]{1+h}$.
I can't seem to show that for $0 < h <4\epsilon$, $\dfrac{1}{3} - \dfrac{\sqrt[3]{1+h}-1}{h} < \epsilon$.
| $\lim_\limits {h\to 0} \frac{\sqrt [3]{1 + h} - 1}{h} = \frac{d}{dx} \sqrt [3] x$ evaluated at $1.$
First, let's write the $\epsilon - \delta$ definition of the limit
$\forall \epsilon > 0, \exists \delta > 0 : |h| < \delta \implies |\frac{(1 + h)^\frac 13 - 1}{h} - \frac 13| < \epsilon$
Use the generalized binomial theorem to expand out the radical.
$(1 + h)^{\frac 13} = 1 + \frac 13 h - \frac {1}{9} h^2 + \frac {5}{81} h^2 - \cdots$
The series converges when $|h| < 1$
We can use this to find establish an upper bound and a lower bound.
$ 1 + \frac 13 h - \frac 19 h^2 \le (1 + h)^{\frac 13} \le 1 + \frac 13 h$
and so
$- \frac 19 h\le \frac{(1 + h)^\frac 13 - 1}{h} - \frac 13 \le 0\\
|\frac{(1 + h)^\frac 13 - 1}{h} - \frac 13| < \frac {\delta}{9}$
let $\delta = \max(\frac {\epsilon}{9},1)$
| {
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Functional equation problem: $ f \left( y ^ 2 - f ( x ) \right) = y f ( x ) ^ 2 + f \left( x ^ 2 y + y \right) $ This functional equation problem is from the Latvian Baltic Way team selection competition 2019:
Find all functions $ f : \mathbb R \to \mathbb R $ such that for all real $ x $ and $ y $,
$$ f \left( y ^ 2 - f ( x ) \right) = y f ( x ) ^ 2 + f \left( x ^ 2 y + y \right) \text . \tag 1 \label {eqn1} $$
OK, so I think that the only answer is $ f ( x ) = 0 $.
I just want to see if my proof that it is the only solution is correct.
So we start off by plugging $ y = - y $. We get that
$$ f \left( y ^ 2 - f ( x ) \right) = - y f ( x ) ^ 2 + f \Big( - y \left( x ^ 2 + 1 \right) \Big) \text . $$
Then we add the two equations together getting that
$$ 2 f \left( y ^ 2 - f ( x ) \right) = f \Big( - y \left( x ^ 2 + 1 \right) \Big) + f \Big( y \left( x ^ 2 + 1 \right) \Big) \text . $$
From the above equations we get that
$$ \frac { f \Big( - y \left( x ^ 2 + 1 \right) \Big) + f \Big( y \left( x ^ 2 + 1 \right) \Big) } 2 = y f ( x ) ^ 2 + f \left( x ^ 2 y + y \right) \text. \tag 2 \label {eqn2} $$
Now, if we plug $ x = - x $ then we will get that the LHS is the same and that the RHS is $ y f ( - x ) ^ 2 + f \left( x ^ 2 y + y \right) $.
So we proceed by subtracting the two and getting that
$$ 0 = y f ( x ) ^ 2 - y f ( - x ) ^ 2 \text . $$
So, lets assume that $ y \ne 0 $ getting that
$$ 0 = \big( f ( x ) - f ( - x ) \big) \big( f ( x ) + f ( - x ) \big) \text . $$
Now we do a two case analysis, 1) the function is even and 2) the function is odd.
Lets start with the function being even then from \eqref{eqn2} we get that
$$ 0 = y f ( x ) ^ 2 \text , $$
which of course implies that the function is just $ 0 $.
OK, now the odd case. Since the function is odd, $ f ( 0 ) = 0 $. Then plugging $ x = 0 $ in \eqref{eqn1} we get that
$$ f \left( y ^ 2 \right) = f ( y ) \text , $$
which implies that $ f $ is also an even function. Since $ f $ is both even and odd, it can only be $ 0 $.
Since we got that $ f $ is zero in both cases, the only solution to the equation is $ f ( x ) = 0 $.
| Suppose we have $(x^2+1)^2+4f(x)\geq 0$ for some $x$. Then there exists some $y_0$, such that $$y_0^2-(x^2+1)y_0-f(x)=0.$$ We may also assume that $y_0\ne 0$, because the roots of the above quadratic can't both be $0$. Plugging $P(x, y_0)$ in the equation we get $$y_0f(x)^2=0,$$ and because $y_0\ne 0$, we must have $f(x)=0$.
This directly implies $f(x)\leq 0$ for all $x$. Suppose the function does not positive roots. This means that we must have $f(x)<\frac{-(x^2+1)^2}{4}$ for all positive $x$. If we plug $P(x, 0)$ in the equation, where $x$ is positive, we get $$f(0)=f(-f(x))<-\frac{(f(x)^2+1)^2}{4}<-\frac{\left(-\left(\frac{(x^2+1)^2}{4}\right)^2+1\right)^2}{4},$$ where the first inequality follows from $-f(x)>0$ and $f(x)<\frac{-(x^2+1)^2}{4}$, whereas the second one follows from the fact that squaring the inequality mentioned above implies $$f(x)^2>\left(\frac{-(x^2+1)^2}{4}\right)^2.$$
Seeing as the $RHS$ of the inequality is not bounded from below, we have reached a contradiction, therefore we must have a positive root $a$. Plugging $P\left(x, \frac{a}{x^2+1}\right)$ in the equation we get $$0\geq f\left(\left(\frac{a}{x^2+1}\right)^2-f(x)\right)=\frac{a}{x^2+1}f(x)^2\geq 0,$$ so $\boxed{f\equiv 0}$, which indeed is a solution.
| {
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What is the wrong step in the integral $ \int {\ln x \over x\sqrt{1-4\ln x -\ln^2 x}}dx $?
Evaluate the following integral:
$$
I= \int {\ln x \over x\sqrt{1-4\ln x -\ln^2 x}}dx
$$
I've started with a substitution: $t = \ln x$, then:
$$
dt = {dx \over x} \iff dx = xdt\\
I = \int {tdt\over \sqrt{1 - 4t - t^2}}
$$
Completing the square in the denominator I got:
$$
1-4t-t^2 = -(t^2 + 4t - 1 +5-5) = -(t+2)^2 + 5 = 5-(t+2)^2
$$
Then the integral becomes:
$$
\int \frac{tdt}{\sqrt{5-(t+2)^2}}
$$
Substitute $t+2 = s$, then $dt = ds$, and $t = s-2$:
$$
\int \frac{(s-2)ds}{\sqrt{5 - s^2}} = \int \frac{sds}{\sqrt{5 - s^2}} - \int\frac{2ds}{\sqrt{5 - s^2}} \tag1
$$
Then:
$$
I_1 = \int \frac{sds}{\sqrt{5 - s^2}}
$$
Substitute $p = s^2$, $dp = 2sds$, and $ds = {dp \over 2s}$:
$$
I_1 = \int
\frac{dp}{2\sqrt{5-p}} = {1\over 2}\arcsin{\sqrt{p}\over \sqrt5}+C = \\
{1\over 2}\arcsin{\ln x + 2\over \sqrt5}+C
$$
Going back to $(1)$:
$$
I_2 = \int\frac{2ds}{\sqrt{5 - s^2}} = 2\arcsin{s\over \sqrt5}+C= \\
2\arcsin{\ln x + 2\over \sqrt5}+C
$$
Which means:
$$
I = I_1 - I_2 = \boxed{{1\over 2}\arcsin{\ln x + 2\over \sqrt5} - 2\arcsin{\ln x + 2\over \sqrt5}+C}
$$
And that is not correct since the answer suggests:
$$
I = -\sqrt{1-4\ln x - \ln^2 x} - 2\arcsin{\ln x + 2\over \sqrt5}+C
$$
I've been trying to spot the error for a while without any success, where did it go wrong? Obviously my answer is wrong. By the way, I'm supposed to use substitution to solve the integral. Thank you in advance!
| Your computation of $I_1$ is wrong. $\int \frac 1 {\sqrt {5-p}} dp$ is $-2\sqrt {5-p}+C$
| {
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Solution to $x^2-y^{11}=23$ in positive integers? Are there positive integers that make the following equation true? $x^2-y^{11}=23$
I do not think it is possible because I have tried many numbers any none of them seem to work. I was able to solve it with an easier case: $x^2 - y^2 = 23$. We can factor to get: $(x+y)(x-y)=23$ Since 23 is prime, its prime factorization is $23 \cdot 1$. We can now solve for $x$ any by doing $(x+y) = 23$ and $(x-y)=1$. $x = 12$, $y = 11$. Although this works for this case, I don't know if it'll work for the original problem.
| Suppose a solution exists. First note that $x$ has to be even. Indeed, if it was odd, then $y^{11}=x^2-23$ would be even but not divisible by $4$ (look mod $4$), which is impossible.
It follows that $y$ has to be odd. But then, since $y^{11}$ is congruent to $y$ modulo $4$ for odd $y$, we find that $y\equiv 1\pmod 4$.
Now comes the trick. By adding $y^{11}+2025$ to both sides, we find
$$x^2+45^2=y^{11}+2^{11}=(y+2)(y^{10}-2y^9+2^2y^8\pm\dots+2^{10})=(y+2)A.$$
Observe that A is relatively prime to $y+2$ - we have $A=y^{10}-2y^9+2^2y^8\pm\dots+2^{10}\equiv 11(-2)^{10}\pmod{y+2}$, so since $y+2$ is odd, the only common factor could be $11$, which requires $y\equiv -2\pmod{11}$. But going back to the original equation, this would imply $-1$ is a square modulo $11$, which it certainly isn't.
To sum up, we have established that $y+2$ and $A$ are relatively prime numbers whose product is a sum of two squares. However, we can see that both of them are $3\pmod 4$, so both have prime factors which are $3\pmod 4$. One of those factors must not be equal to $3$, call it $p$. By considering the displayed equation modulo $p$, we find $x^2\equiv -45^2\pmod p$, and since $p\neq 3,5$, by taking multiplicative inverse of $45$ modulo $p$ we find $z^2\equiv -1\pmod p$ has an integer solution. However, it is well-known that this is impossible whenever $p\equiv 3\pmod 4$. This gives a contradiction, showing that the equation has no solution.
(Proof inspired by the proof of Theorem 2.1 here.)
| {
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Let $ a \in \mathbb{N}^\ast $, prove that $ \frac{1}{2a} - \frac{1}{2a^3} < \sqrt{a^2+1} - a < \frac{1}{2a} $ I have the following problem to solve. It's about convergent sequences.
Let $ a \in \mathbb{N}^\ast $, prove that:
$$ \frac{1}{2a} - \frac{1}{2a^3} < \sqrt{a^2+1} - a < \frac{1}{2a} $$
To solve it, I first solve:
$$ 0 < \sqrt{a^{2}+1} - a - \frac{1}{2a} + \frac{1}{2a^3} $$
To do so, I compute the result for $ a = 1 $ which is $ \sqrt{2} - \frac{11}{8} > 0$. Then I was thinking of studying the variation of the sequence with it's derivative but it makes weird equations:
$$ U_a = \sqrt{a^{2}+1} - a - \frac{1}{2a} + \frac{1}{2a^3} $$
$$ U_a' = \frac{a}{\sqrt{a^2 + 1}} - 1 + \frac{1}{2a^2} - \frac{3}{2a^4} $$
$$ U_a' = 0 \Leftrightarrow \frac{2a^5 \sqrt{a^2+1} + 2a^4}{7a^2+9} - 1 = 0 $$
I'm lost from here, someone can help me ?
If it helps, we are studying limited expansions.
Thank you.
| The Maclaurin series of $\sqrt{x^2+1}$:$$\sqrt{x^2+1}=1+\dfrac{1}{2}x^2-\dfrac{1}{8}x^4+\dfrac{1}{16}x^5-\cdots$$ Then, find $\dfrac{\sqrt{x^2+1}-1}{x}$,$$\dfrac{\sqrt{x^2+1}-1}{x}=\dfrac{1}{2}x-\dfrac{1}{8}x^3+\dfrac{1}{16}x^5-\cdots$$ Substitute $x=\dfrac{1}{a}$:$$\dfrac{\sqrt{x^2+1}-1}{x}=\dfrac{\sqrt{\frac{1}{a^2}+1}-1}{\frac{1}{a}}=\sqrt{a^2+1}-a=\dfrac{1}{2a}-\dfrac{1}{8a^3}+\dfrac{1}{16a^5}-\cdots$$
Then, obviously, $$\dfrac{1}{2a}-\dfrac{1}{2a^3}<\dfrac{1}{2a}-\dfrac{1}{8a^3}<\sqrt{a^2+1}-a<\dfrac{1}{2a}$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Find $\lim_{x \to 0} \frac{(\tan(\tan x) - \sin (\sin x))}{ \tan x - \sin x}$ Find $$\lim_{x\to 0} \dfrac{\tan(\tan x) - \sin (\sin x)}{ \tan x - \sin x}$$
$$= \lim_{x \to 0} \dfrac{\frac{\tan x \tan (\tan x)}{\tan x}- \frac{\sin x \sin (\sin x)}{\sin x}}{ \tan x - \sin x} = \lim_{x \to 0} \dfrac{\tan x - \sin x}{\tan x - \sin x} = 1$$
But the correct answer is $2$. Where am I wrong$?$
| This is a nice case for composition of Taylor series. Using
$$\tan(x)=x+\frac{x^3}{3}+\frac{2 x^5}{15}+O\left(x^7\right)$$
$$\sin(x)=x-\frac{x^3}{6}+\frac{x^5}{120}+O\left(x^7\right)$$
$$\tan(\tan(x))=x+\frac{2 x^3}{3}+\frac{3 x^5}{5}+O\left(x^7\right)$$
$$\sin(\sin(x))=x-\frac{x^3}{3}+\frac{x^5}{10}+O\left(x^7\right)$$ then
$$\frac{\tan(\tan( x)) - \sin (\sin( x)}{ \tan (x) - \sin (x)}=\frac {x^3+\frac{x^5}{2}+O\left(x^7\right) } {\frac{x^3}{2}+\frac{x^5}{8}+O\left(x^7\right) }=2+\frac{x^2}{2}+O\left(x^4\right)$$ which shows the limit and also how it is approached.
| {
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Show with the epsilon-delta definition that $\lim_{x \to 2} \frac{1}{x - 1} = 1$ I have an assignment about epsilon-delta proofs and I'm having trouble with this one. I have worked it through using some of the methods I've picked up for similar proofs but it's just something about this particular expression that doesn't sit right with me. Any feedback would be very helpful. This is how far I've come:
Let $\varepsilon > 0$. We want to find a $\delta$ so that $\left|\frac{1}{x - 1} - 1\right| < \varepsilon$ when $0 < |x - 2| < \delta$. We expand the expression:
\begin{align*}
\left|\frac{1}{x - 1} - 1\right| &< \varepsilon \\
\left|\frac{1}{x - 1} - \frac{x - 1}{x - 1}\right| &< \varepsilon \\
\left|\frac{2 - x}{x - 1}\right| &< \varepsilon \\
|{x - 1}| &< \frac{|x - 2|}{\varepsilon} \\
\end{align*}
We could let $\delta = \dfrac{|x - 2|}{\varepsilon}$ but $|x - 2|$ contains an unwanted variable. Since the limit is only relevant when $x$ is close to $a$ we'll restrict $x$ so that it's at most $1$ from $a$ or in other words, in our case, that $|x - 1| < 1$. This means $0 < x < 2$ and that $-2 < x - 2 < 0$. Looking at our previous inequality
\begin{align*}
|{x - 1}| &< \frac{|x - 2|}{\varepsilon}
\end{align*}
we see that the right-hand side is the smallest when $|x - 2|$ is the smallest which by the range above is when $x - 2 = -2$ and then we have that
\begin{align*}
|{x - 1}| &< \frac{|x - 2|}{\varepsilon} < \frac{2}{\varepsilon}
\end{align*}
We now have the two inequalities $|x - 1| < 1$ and $|x - 1| < \frac{2}{\varepsilon}$. Let $\delta = \textrm{min}(1, \frac{2}{\varepsilon})$ and by definition we have that for every $\varepsilon > 0$ there is a $\delta$ so that $|f(x) - A| < \varepsilon$ for every $x$ in the domain that satisfies $0 < |x - a| < \delta$. $\blacksquare$
| Assuming wlog $\frac32\le x\le \frac52$ we have
$$\left|\frac{2 - x}{x - 1}\right| < \varepsilon \iff \left|2 - x\right| < \varepsilon \left|x - 1\right|\le \frac 32 \varepsilon $$
then it suffices to take $\delta <\frac 32 \varepsilon $.
| {
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Use the relation $x^3=y$ to get $\alpha^3 + \beta^3+\gamma^3$ I have this equation
$$x^3-x^2-x+0.5=0$$
Now i have to use the relation $x^3=y$ to get $\alpha^3 + \beta^3+\gamma^3$ which will be the roots of the new equation obatained.
I know that i can use many other ways but how to solve by this method
I end up with this
$$y-y^{\frac {2}{3}}-y^{\frac {1}{3}}+0.5=0$$
I can't factor out $y^{\frac{1}{3}}$
Any help is appriciated
| We have $$2x^3+1=2x^2+2x$$
Cubing both sides $$8(x^3)^3+1^3+3(2x^3)(2x^3+1)=(2x^2+2x)^3=8(x^3)^2+8(x^3)+3(2x^2)(2x)(2x^2+2x)$$
Now as $x^3=y, 2x^2+2x=2x^3+1=2y+1$
$$8y^3+1+6y(2y+1)=8y^2+8y+12y(2y+1)$$
$$\implies8y^3+y^2(12-8-24)+(\cdots)y+(\cdots)=0$$ whose roots are
$$\alpha^3,\beta^3,\gamma^3$$
Using Vieta's formula $$\alpha^3+\beta^3+\gamma^3=-\dfrac{12-8-24}8$$
| {
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"timestamp": "2023-03-29T00:00:00",
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How to use radical rules for this?
$$\sqrt{3+\sqrt{2\sqrt{7}+1}}-\sqrt{3-\sqrt{2\sqrt{7}+1}}=?$$
Which one of the following is a true answer and why?
*
*$\sqrt{7}-7$
*$1-\sqrt{7}$
*$\sqrt{7}-1$
*$\sqrt{7}$
*$\sqrt{7}+7$
| Square it and take root:
\begin{align}
& \sqrt{3+\sqrt{2\sqrt{7}+1}}-\sqrt{3-\sqrt{2\sqrt{7}+1}} \\
= & \sqrt{\Big[\sqrt{3+\sqrt{2\sqrt{7}+1}}-\sqrt{3-\sqrt{2\sqrt{7}+1}}\Big]^2} \\
= & \sqrt{\big(3+\sqrt{2\sqrt{7}+1}\big) + \big(3-\sqrt{2\sqrt{7}+1}\big) - 2\sqrt{(3+\sqrt{2\sqrt{7}+1})(3-\sqrt{2\sqrt{7}+1})}} \\
= & \sqrt{6 - 2\sqrt{9-(2\sqrt{7}+1)}} \\
= & \sqrt{6 - 2\sqrt{8-2\sqrt 7}} \\
= & \sqrt{6 - 2(\sqrt 7 -1)} \\
= & \sqrt{8 - 2\sqrt 7} \\
= & \sqrt 7 -1
\end{align}
where I have used, twice, the fact that $\sqrt{8-2\sqrt 7} = \sqrt 7 -1$. This can be verified by, again, squaring then taking square root:
$$\sqrt 7 -1 = \sqrt{\big(\sqrt 7 -1\big)^2} = \sqrt{7 + 1 - 2\sqrt 7} = \sqrt{8 - 2\sqrt 7}$$
| {
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Tricky epsilon-delta proof $$\lim_{x\to -1}\frac{x-1}{x^2-x+1}=-\frac{2}{3}$$
What I've got so far is that:
$$\forall\epsilon>0,\exists\delta>0\text{ s.t. }0<|x-(-1)|< \delta\implies\left|\frac{x-1}{x^2-x+1}-(-\frac{2}{3})\right|<\epsilon\\
\forall\epsilon>0,\exists\delta>0\text{ s.t. }0<| x+1|< \delta\implies\left|\frac{(x+1)(2x-1)}{3x^2-3x+3}\right|<\epsilon$$
How do I go about finding a value for delta from here? Thanks.
| Note that you have $0<|x+1|<\delta$, and you have $(x+1)$ in your numerator in the second line. So we make this more explicit:
$$
\left|\frac{(x+1)(2x-1)}{3x^2-3x+3}\right|=|x+1|\cdot \left|\frac{2x-1}{3x^2-3x+3}\right|<\epsilon
$$
This is what we want to hold. Now, in order to get a handle on this, we have
$$
|x+1|\cdot \left|\frac{2x-1}{3x^2-3x+3}\right|<\delta\cdot \left|\frac{2x-1}{3x^2-3x+3}\right|
$$
and $\delta$ we can control. So we want to pick a $\delta$ small enough that the right-hand side here is less than $\epsilon$, since that will automatically make the left-hand side smaller than $\epsilon$. Thus we have to see how large that second factor on the right-hand side could possibly become. I claim that it's always smaller than $1$. Which in turn gives us:
$$
\delta\cdot \left|\frac{2x-1}{3x^2-3x+3}\right|<\delta
$$
So as long as we pick $\delta= \epsilon$, putting together all these inequalities gives us $\left|\frac{(x+1)(2x-1)}{3x^2-3x+3}\right|<\epsilon$, which is what we want, and we have proven our limit.
Proof of claim: We can either do calculus to find max and min, or we do some algebra and split into cases. I'll go with the algebra option. Note that the numerator $3x^2-3x+3$ is always positive, so we may remove the absolute value signs from it. This gives us
$$
\left|\frac{2x-1}{3x^2-3x+3}\right| = \frac{|2x-1|}{3x^2-3x+3}<1\\
|2x-1|<3x^2-3x+3
$$
For $x\leq \frac12$, this turns into $1-2x<3x^2-3x+3$, which is easily verified by the quadratic formula. For $x\geq\frac12$, it turns into $2x-1<3x^2-3x+3$, which is also easily verified with the quadratic formula. So we get that $\left|\frac{2x-1}{3x^2-3x+3}\right|<1$, and we are done.
| {
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Prove $\tan(\frac{x}{2}) = \frac{\sin x}{1 + \cos x} $ using the quadratic formula I am trying to prove the fact that $\tan \frac{x}{2} = \frac{1-\cos x}{\sin x}$ or alternatively $\tan \frac{x}{2} = \frac{1- \cos x}{\sin x}$. (I understand that it can be proved using the half-angle identities of $\sin$ and $\cos$ but I want to understand how to get to the solution from this specific method of derivation.)
\begin{align*}
\tan(2x) &= \frac{2\tan(x)}{1-\tan(x)^2} \\
\tan(x) &= \frac{2\tan(\frac{x}{2})}{1-\tan(\frac{x}{2})^2} \\
\end{align*}
I now let $A=\tan x$ and $B=\tan \frac{x}{2}$
\begin{align*}
A\cdot(1-B^2) &= 2B\\
AB^2+2B-A &= 0 \\
\end{align*}
Now I solve for B using the quadratic formula.
\begin{align*}
B &= \frac{-2\pm \sqrt{4+4A^2}}{2A} \\
B &= \frac{-1\pm \sqrt{1+A^2}}{A} \\
\tan(\frac{x}{2}) &= \frac{-1\pm \sqrt{1+\tan(x)^2}}{\tan(x)}\\
\tan(\frac{x}{2}) &= \frac{-1\pm \sqrt{(\sec x)^2}}{\tan(x)}\\
\tan(\frac{x}{2}) &= \frac{-1\pm |\sec x|}{\tan(x)}
\end{align*}
I am confused as to how to continue at this point (firstly, not sure how to deal with the absolute value, and secondly not sure how to deal with the plus-minus).
Any help is greatly appreciated, as I feel like I do not fully understand how to manipulate absolute values and the meaning of the plus-minus.
| Hint. Note that $\pm|x|=\pm x,$ without loss of any generality. Then split into two cases.
| {
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How can I find general formula of this sequence? Let be the sequence $(a_n)$ so that $a_1 = 1$, $a_{n+1}=\dfrac{20}{3 a_n+4}$, $\forall n \geqslant 1.$ I can prove that the sequence is an increasing sequence and convert to 2. But, I can find the general formula of this sequence. How can I find it?
| $$A_{n+1}=\frac{20}{3A_n+4} \implies A_{n+1}~(3A_n +4)= 20.~~~(1)$$
Let $$(3A_n+4)=\frac{B_n}{B_{n-1}} \implies A_n=\frac{1}{3} \left(\frac{B_n}{B_{n-1}}-4\right)~~~(2)$$
Then (1) becomes $$B_{n+1}-60 B_{n-1}-4 B_n =0~~~(3)$$
Now let us put $B_n=t^n$ in (3) to get
$$t^2-4t-60=0 \implies t=10,-6.$$
So the solution of (3) is $$B_n=P (10)^n +Q(-6)^n ~~~(4)$$
Inserting (4) in (2) we get $$A_n=\frac{1}{3} \left(\frac{6 R (10)^{n-1}-10 (-6)^{n-1}}{R (10)^{n-1}+ (-6)^{n-1}}\right), R=\frac{P}{Q}.$$
Using $A_1=1$ gives $$R=\frac{13}{3}.$$
Finally, we get $$A_n= \left( \frac{26~ (5)^{n-1}-10 ~ (-3)^{n-1}}{13~ (5)^{n-1} +3 ~(-3)^{n-1}} \right).$$
| {
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If volume of a sphere increases by 72.8%, what is change of its surface area? If the volume of a sphere is increased by 72.8% what would be the change in surface area?
I'm trying to solve the problem using application of derivatives. I noticed that on differentiating the formula for volume of a sphere we directly end up with that of the surface area. How to approach this link?
| Set up a formula that sets up Volume in terms of Area.
As Volume is determined by $V= \frac 43\pi r^3$ and Surface Area of a Sphere is $SA= 4 \pi r^2$ then $SA = 4\pi r^2 = 4\pi{\sqrt[3]{\frac {3V}{4\pi}}}^2$
So if volume is increased by $1.728$ then surface area is increased from $4\pi{\sqrt[3]{\frac {3r}{4\pi}}}^2$ to $4\pi{\sqrt[3]{\frac {3*1.728r}{4\pi}}}^2$
And the proportional increase is $\frac{4\pi{\sqrt[3]{\frac {3*1.728r}{4\pi}}}^2}{4\pi{\sqrt[3]{\frac {3*1.728r}{4\pi}}}^2}=1.782^{\frac 23}$
And percentage increase is $100(1.782^{\frac 23}-1)$
The real question, I suppose, is how to get the formulas for volumes and surface areas in the first place.
If we look at cross section circles of a sphere at points $x: -r\le x \le 4$ along the diameter of the sphere, these cross section circles have radii of $R = \sqrt{r^2 - x^2}$.
An area of one of these circles is $\pi R^2=\pi(r^2 - x^2)$ and the circumference of a circle is $2\pi R= 2\pi\sqrt{r^2 - x^2}$
So the volume of a sphere is $V= \int_{-r}^r \pi(r^2 - x^2)dx=\frac 43\pi r^3$ and the surface area of a sphere is $\int_{-r}^r2\pi\sqrt{r^2 - x^2}dx= 4\pi r^2$
.....
Well we are at it the area of circle is determine by $A=\int_{-r}^4 2\sqrt{r^2 - x^2}dx = \pi r^2$
| {
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Prove that the inequality $\frac{n^3}{3} < 3n-3$ applies to $n = 2$ and for all other $n \in \mathbb{N}$ the inequality does not hold I have been given the following task:
Prove that the following inequality applies to $n = 2$ and for all other $n \in \mathbb{N}$ the inequality does not hold:
$$\frac{n^3}{3} < 3n-3$$
My idea was to prove the statement by using induction.
For $n=1$ it follows:
$\frac{1^3}{3} = \frac{1}{3} \nless 0 = 3 * 1-3$
For $n=2$ it follows:
$\frac{2^3}{3} = \frac{8}{3} < 3 = 3 * 2-3$
For $n=3$ it follows:
$\frac{3^3}{3} = \frac{27}{3} = 9 \nless 6 = 3 * 3-3$
I now assume that the inequality does not hold for $n \in \mathbb{N} \backslash \{2\}.$
n $\rightarrow$ n+1
$\frac{(n+1)^3}{3} = \frac{n^3+3n^2+3n+1}{3} = \frac{n^3}{3}+ \frac{3n^2+3n+1}{3} < 3n-3 + \frac{3n^2+3n+1}{3}=\frac{3(3n-3)+3n^2+3n+1}{3}=\frac{3n^2+12n-8}{3}$
But how do I continue from this step or is this even the wrong approach?
| Show that $n^3/3 \ge 3n-3$ for $n\ge 3$, or equivalently
$n(n^2-9)\ge -9$, or
$n(n+3)(n-3) \ge -9$.
For $n \ge 3:$
$n(n+3)(n-3) \ge 0$ and we are done.
| {
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Find all $x\in \mathbb R$ such that $\sqrt[3]{x+2}+\sqrt[3]{x+3}+\sqrt[3]{x+4}=0$ The question is as the title says:
Find all $x\in \mathbb R$ such that $\sqrt[3]{x+2}+\sqrt[3]{x+3}+\sqrt[3]{x+4}=0$.
I struggle to even start this question.
By inspection, I see that $x$ must be negative. Playing around yields $x=-3$ as a solution, though I do not know how to prove that there are no other solutions.
Upon differentiation, I obtain:
\begin{align}
\frac{d}{dx} (\sqrt[3]{x+2}+\sqrt[3]{x+3}+\sqrt[3]{x+4})&= \frac{1}{3}(\sqrt[3]{(x+2)^{-2}}+\sqrt[3]{(x+3)^{-2}}+\sqrt[3]{({x+4})^{-2}})\\
\end{align}
which is always positive for all real values of $x$, implying that the function defined as $f(x)=\sqrt[3]{x+2} + \sqrt[3]{x+3} + \sqrt[3]{x+4}$ is strictly increasing.
Is there a better way to solve this equation?
| Let $\sqrt[3]{x+2}=a$ etc.
So, $a+b+c=0$
and $a^3+c^3=2b^3=2(-a-c)^3=-2(a^3+c^3+3a^2c+3ac^2)$
$$\iff0=a^3+ c^3+2a^2c+2ac^2=(a+c)(a^2-ac+c^2)+2ac(c+a)=(c+a)(a^2+ca+c^2)$$
But $a^2+ca+c^2=0$
$\implies$ either $a=c=0$
or $\dfrac ac=$ imaginary as $\left(\dfrac ac\right)^2+\dfrac ac+1=0$
$\implies a+c=0\implies a^3+c^3=0$
| {
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No solutions to $x^2+y^2+z^2 = 7t^2$ A problem in my book asks me to show that there are no solutions to
$$x^2+y^2+z^2 = 7t^2$$
in the integers apart from $(x,y,z,t)=(0,0,0,0)$.
The solution states that reducing modulo $4$ we see that $x,y,z,t$ must be even
and dividing through we get a smaller solution.
I don't understand how we can conclude that everything is even. If $(x,y,z,t)$ are all $1$ mod $4$ it still seems to hold.
| Hmm.... Well, if $x,y,z$ are even then $t$ is, of course and if, $x,y,z$ are all odd then $t$ is. and then $\mod 4$ we get $x^2 + y^2 + t^2 \equiv 3\mod 4$ and $7t^2 \equiv 3\pmod 4$ so that's not a problem. (Yet.)
Indeed It will always be the case that $(2m+1)^2 + (2n+1)^2 + (2k+1)^2 \equiv 3 \equiv 7(2j+1)^2$.
But we can rule out $1$ or $2$ odds.
If one of $x,y,z$ are odd and $t$ are odd but the rest are even we'd have $x^2 + y^2 + z^2\equiv 1 \pmod 4\not\equiv -1\equiv 7t^2$.
If two of $x,y,z$ are odd and $t$ is even we would have $x^2 + x^2 +z^2 \equiv 2\pmod 4$ and $7t^2 \equiv 0 \pmod 4$.
So $\mod 4$ yields either $x,y,z,t$ are all odd or all even.
$\mod 8$ we have $(8m\pm 1,3)^2 \equiv 1\pmod 8$ and so all odd would result in $x^2 + y^2 + z^2 \equiv 3 \not \equiv 7\equiv t^2$.
So that's it. If $x^2 + y^2 + z^2 = 7t^2$ then all $x,y,z,t$ are even but wolog we can assume $x,y,z,t$ have no factor $d$ in common be because $x^2 + y^2 + z^2 = 7t^2 \iff (\frac xd)^2 + (\frac yd)^2 + (\frac zd)^2=7(\frac td)^2$.
| {
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Series using partial sums... Determine if the following series converge by studying the partial sums. If it converges, compute its value.
$\sum_{n=1}^{\infty} \frac{2n+3}{(n+1)(n+2)}$.
We use partial fractions so
\begin{equation*}
\begin{split}
\frac{2n+3}{(n+1)(n+2)} &= \frac{A}{n+1}+\frac{B}{n+2} \\
\Longleftrightarrow2n+3 &= A(n+2)+B(n+1) \\
&= An+2A+Bn+B \\
&= (A+B)n+2A+B.
\end{split}
\end{equation*}
Equating terms tells us that $A+B = 2$ and $2A+B = 3$. We can find $A,B$ and $C$ by solving
\begin{equation*}
\begin{bmatrix}
1 & 1 \\
2 & 1
\end{bmatrix}\begin{bmatrix}
A \\
B
\end{bmatrix} = \begin{bmatrix}
2 \\
3
\end{bmatrix}.
\end{equation*}
Doing so yields $A = B = 1$. This means the partial sum is
\begin{equation*}
S_k = \sum_{n=1}^{k} \frac{2n+3}{(n+1)(n+2)} = \sum_{n=1}^{k} \left(\frac{1}{n+1}+\frac{1}{n+2}\right) = \left(\frac{1}{2}+\frac{1}{3}\right)+\left(\frac{1}{3}+\frac{1}{4}\right)+\ldots+\left(\frac{1}{k+1}+\frac{1}{k+2}\right).
\end{equation*}
Is this correct?
| Your idea of looking at the partial sum is good. So, as you wrote,
$$
S_k = \sum_{n=1}^{k} \frac{2n+3}{(n+1)(n+2)} = \sum_{n=1}^{k} \frac{1}{n+1}+\sum_{n=1}^{k} \frac{1}{n+2} $$ Now
$$\sum_{n=1}^{k} \frac{1}{n+a}=H_{a+k}-H_a$$ where appear harmonic numbers. So
$$S_k=H_{k+1}+H_{k+2}-\frac{5}{2}$$ Now, using the asymptotics
$$H_p=\gamma +\log \left({p}\right)+\frac{1}{2
p}+O\left(\frac{1}{p^2}\right)$$ apply it twice and continue with Taylor series to get
$$S_k=2 \log ({k})+2 \gamma
-\frac{5}{2}+\frac{4}{k}+O\left(\frac{1}{k^2}\right)$$ For example $S_{10}=\frac{12554}{3465}\approx 3.62309$ while the truncated expression would give $\approx 3.65960$.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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If $x^5-bx^3+cx^2+dx-e$ can be expressed as the product of a perfect square and a perfect cube then prove following If $x^5-bx^3+cx^2+dx-e$ can be expressed as the product of a perfect square and a perfect cube then prove that
$$\frac{12b}{5}=\frac{9d}{b}=\frac{5e}{c}=\frac{d^2}{c^2}$$
My attempt is as follows:
$$E=x^5-bx^3+cx^2+dx-e$$
$$E=(x-\alpha)^2(x-\beta)^3$$
$$E=x^5-\left(2\alpha+3\beta\right)x^4+\left(3{\alpha}^2+6\alpha\beta+{\beta}^2\right)x^3-\left({\alpha}^3+6{\alpha}^2\beta+3\alpha{\beta}^2\right)x^2+\left(2{\alpha}^3\beta+3{\alpha}^2{\beta^2}\right)x-{\alpha}^2{\beta}^3$$
$$x^5-bx^3+cx^2+dx-e=x^5-\left(2\alpha+3\beta\right)x^4+\left(3{\alpha}^2+6\alpha\beta+{\beta}^2\right)x^3-\left({\alpha}^3+6{\alpha}^2\beta+3\alpha{\beta}^2\right)x^2+\left(2{\alpha}^3\beta+3{\alpha}^2{\beta^2}\right)x-{\alpha}^2{\beta}^3$$
So we obtain five equations
$$\left(2\alpha+3\beta\right)=0$$
\begin{equation}
\frac{\alpha}{\beta}=\frac{-3}{2}\tag{1}
\end{equation}
$$3{\alpha}^2+6\alpha\beta+{\beta}^2=-b$$
$$\text {Dividing by ${\beta}^2$}$$
$$3\frac{\alpha^2}{\beta^2}+6\frac{\alpha}{\beta}+1=\frac{-b}{{\beta}^2}$$
\begin{equation}
\frac{b}{{\beta}^2}=\frac{5}{4}\tag{2}
\end{equation}
$${\alpha}^3+6{\alpha}^2\beta+3\alpha{\beta}^2=-c$$
$$\text {Dividing by ${\beta}^3$}$$
$$\frac{\alpha^3}{\beta^3}+6\frac{\alpha^2}{\beta^2}+3\frac{\alpha}{\beta}=\frac{-c}{{\beta}^3}$$
\begin{equation}
\frac{c}{{\beta}^3}=\frac{-45}{8}\tag{3}
\end{equation}
$$2{\alpha}^3\beta+3{\alpha}^2{\beta}^2=d$$
$$\text {Dividing by ${\beta}^4$}$$
$$2\frac{\alpha^3}{\beta^3}+3\frac{\alpha^2}{\beta^2}=\frac{d}{{\beta}^4}$$
\begin{equation}
\frac{d}{{\beta}^4}=0\tag{4}
\end{equation}
So $d$ is coming as $0$ and this is where I am stuck because for $$\frac{12b}{5}=\frac{9d}{b}=\frac{5e}{c}=\frac{d^2}{c^2}$$ $d$ should be non-zero because $\dfrac{12b}{5}=3\beta^2$ is non zero.
Please help me.
| Since $$\alpha=-\frac{3}{2}\beta,$$ we obtain:
$$E=\left(x+\frac{3}{2}\beta\right)^2(x-\beta)^3=x^5-\frac{15}{4}\beta^2x^3+\frac{5}{4}\beta^3x^2+\frac{15}{4}\beta^4x-\frac{9}{4}\beta^5,$$ which gives
$$b=\frac{15}{4}\beta^2,$$
$$c=\frac{5}{4}\beta^3,$$ $$d=\frac{15}{4}\beta^4$$ and $$e=\frac{9}{4}\beta^5.$$
Can you end it now?
For example, we need to prove that:
$$12bc^2=5d^2$$ or
$$12\cdot\frac{15}{4}\beta^2\cdot\left(\frac{5}{4}\beta^3\right)^2=5\cdot\left(\frac{15}{4}\beta^4\right)^2.$$
| {
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Formal proof function is continuous I'm a bit stuck in this formal proof that the functions is continuos in all of it's domain.
$f(x)= 3(x^2+1)^3$
$ \epsilon >0 , \delta > 0 $
$|x-c|<\delta \to |3(x^2+1)^3-3(c^2+1)^3| \to 3|x^6-c^6+3x^4+3x^2-3c^4-3c^2| $
that's where I get to by myself. Could somebody help me out :) . Thanks
| As I said in the comments: $a^3 - b^3 = (a - b)(a^2 + ab + b^2)$ so we can write this as
\begin{align}
&3|(x^2 + 1) - (y^2 + 1)| \cdot |(x^2 + 1)^2 + (x^2 + 1)(y^2 + 1) + (y^2 + 1)^2| \\
={} & 3|x - y| \cdot |x + y| \cdot |(x^2 + 1)^2 + (x^2 + 1)(y^2 + 1) + (y^2 + 1)^2|.
\end{align}
Then you just need a crude bound on $3|x + y| \cdot |(x^2 + 1)^2 + (x^2 + 1)(y^2 + 1) + (y^2 + 1)^2|$.
| {
"language": "en",
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} |
Finding more than one basis for a column space Let's say the matrix A is defined as the following:
$$A=\begin{pmatrix}6 & 3 & -1 & 0 \\ 1 & 1 & 0 & 4 \\ -2 & 5 & 0 & 2\end{pmatrix}$$
Whose RRE form is:
$$A_{RRE}=\begin{pmatrix}1 & 0 & 0 & * \\ 0 & 1 & 0 & * \\ 0 & 0 & 1 & *\end{pmatrix}$$
Then I can write a basis for the column space of matrix A using the three linearly independent vectors as such:
$$B_1=\left\{\begin{pmatrix}6\\ 1 \\ -2\end{pmatrix},\begin{pmatrix}3\\ 1 \\ 5\end{pmatrix},\begin{pmatrix}-1\\ 0 \\ 0\end{pmatrix}\right\}$$
However, what if I want to find a different basis for colA besides the one above, such that no vector in the basis is a scalar multiple of any of the vectors in B1? Is this method valid?
1. Re-arrange columns of matrix, which won't change the column span:
$$A'=\begin{pmatrix}0 & 6 & 3 & -1 \\ 4 & 1 & 1 & 0 \\ 2 & -2 & 5 & 0\end{pmatrix}$$
2. Put into RRE form:
$$A'_{RRE}=\begin{pmatrix}1 & 0 & 0 & * \\ 0 & 1 & 0 & * \\ 0 & 0 & 1 & *\end{pmatrix}$$
3. Preliminary basis:
$$B'_2=\left\{\begin{pmatrix}0\\ 4 \\ 2\end{pmatrix},\begin{pmatrix}6\\ 1 \\ -2\end{pmatrix},\begin{pmatrix}3\\ 1 \\ 5\end{pmatrix}\right\}$$
4. Take linear combinations of above basis:
$$B_2=\left\{\begin{pmatrix}0\\ 4 \\ 2\end{pmatrix},\begin{pmatrix}6\\ 5 \\ 0\end{pmatrix},\begin{pmatrix}9\\ 2 \\ 3\end{pmatrix}\right\}$$
I know a better method would be to just take linear combinations of the existing basis B1, but I want to know if this method works and if not, why.
| Yes it works fine and all the basis you have found are correct as any triple of linearly independent vectors.
A simpler choice would be: $v_1=(1,1,0)$, $v_2=(0,1,0)$, $v_3=(0,0,1)$.
| {
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For which values of $k$ does the equation $2\cos^{2}\theta +k\sin \theta + k = 2$ have real solutions? So I take A level maths and this question was in our textbook. We solved an inequality for when the discriminant is less than zero and this gave us the same answer that is in the textbook. The problem is that the solutions also have to give $\sin \theta$ as between one and minus one, and we didn't know how to solve that inequality.
Any help would be appreciated, thanks!
| Rewriting $2\cos^2\theta$ as $2-2\sin^2\theta$, we can substitute $t = \sin\theta$ and our equation becomes
$$
2t^2-kt-k = 0
$$
with solutions
$$
t = \frac{k\pm\sqrt{k^2+8k}}{4}
$$
Having a non-negative discriminant $k^2+8k$ forces either $k \leq -8$ or $k \geq 0$. But in addition, we require $-1 \leq t \leq 1$ for at least one of the solutions, so that $\theta$ can be real. This is equivalent to
$$
-4 \leq k\pm\sqrt{k^2+8k} \leq 4
$$
for at least one of the branches of the square root.
Case $1$ ($k \leq -8$). It is clear that $k - \sqrt{k^2+8k} < -4$, so we focus on $k + \sqrt{k^2+8k}$. But then
\begin{align}
k+\sqrt{k^2+8k} & < k+\sqrt{k^2+8k+16} \\
& = k+(-k-4) \\
& = -4
\end{align}
so Case $1$ yields no real solutions.
Case $2$ ($k \geq 0$). We see that $k - \sqrt{k^2+8k} \leq 0$. When is it at least $-4$?
\begin{align}
k - \sqrt{k^2+8k} & > k - \sqrt{k^2+8k+16} \\
& = k - (k+4) \\
& = -4
\end{align}
Hence there exists (at least) one real solution for all $k \geq 0$. The other solution is always non-negative but exceeds $1$ when
$$
k + \sqrt{k^2+8k} > 4 \\
4-k < \sqrt{k^2+8k} \\
k^2-8k+16 < k^2+8k \\
16k > 16 \\
k > 1
$$
Note: For $k > 4$, we can no longer go from the second line to the third line, but then $k + \sqrt{k^2+8k} > k > 4$ anyway.
| {
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"source": "stackexchange",
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} |
$x_1$ and $x_2$ are the solutiton of $\frac{2\cdot \sin(x) \cdot \cos(2x)}{\cos (x) \cdot \sin (2x)} - 5 \tan(x) + 5 = 0$ $x_1$ and $x_2$ are the solutiton of $\frac{2\cdot \sin(x) \cdot \cos(2x)}{\cos (x) \cdot \sin (2x)} - 5 \tan(x) + 5 = 0$
then, $\tan(x_1 + x_2) = ....$
i can do it by doing it
$\dfrac{2\cdot \sin(x) \cdot (\cos^2x - \sin^2x)}{\cos (x) \cdot 2 \sin x \cos x} - 5 \tan(x) + 5 = 0$
leads to $(\sin x - \cos x)(\sin x + 6 \cos x) = 0$
but it's complicated, do you know the less complicated way to solve it?
| Hint
Let $t=\tan(x)$ to make
$$2t \frac{1-t^2}{2t}-5t+5=0\implies t^2+5t-6=0$$
| {
"language": "en",
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"source": "stackexchange",
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Prove $\sin^3A-\cos^3A=\left(\sin^2A-\cos^2A\right)(1-2\sin^2A\cos^2A)$ Prove $\sin^3A-\cos^3A=\left(\sin^2A-\cos^2A\right)(1-2\sin^2A\cos^2A)$
My attempt is as follows:
Taking LHS:
$$\left(\sin A-\cos A\right)(1+\sin A\cos A)$$
$$\left(\sin^2A-\cos^2A\right)\frac{\left(1+\sin A\cos A\right)}{\left(\sin A+\cos A\right)}$$
$$\left(\sin^2A-\cos^2A\right)\frac{(\left(\sin A+\cos A\right)^2-\sin A\cos A)}{\sin A+\cos A}$$
$$\left(\sin^2A-\cos^2A\right)\left(\sin A+\cos A-\frac{\sin A\cos A}{\sin A+\cos A}\right)$$
I was not getting any breakthroughs from here.
So I tried RHS:
$$(\sin A-\cos A)(\sin A+\cos A)(1-2\sin^2A\cos^2A)$$
$$(\sin A-\cos A)(\sin A+\cos A)((\sin^2A+\cos^2A)^2-2\sin^2A\cos^2A)$$
$$(\sin A-\cos A)(\sin A+\cos A)(\sin^4A+\cos^4A)$$
Even from here I was not getting breakthroughs, what am i missing?
Please help me.
| $(\sin^2A-\cos^2A)(1-\sin^2A\cos^2A)$
$=(\sin A-\cos A)(\sin A+\cos A)(\sin^4A+\cos^4A).$
But I think the problem is wrong, just try with $A=\frac{\pi}{5}$.
| {
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"timestamp": "2023-03-29T00:00:00",
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Fixed Point Theorems and Contraction Mappings I am trying to solve this following exercise.
Let $F(x)$ be a continuously differentiable function defined on the interval $[a,b]$ such that $F(a) < 0$, $F(b) > 0$, and
\begin{align*}
0 < K_1 \leq F'(x) \leq K_2 \; \; \; (a \leq x \leq b).
\end{align*}
Use Theorem 1 (given below) to find the unique root of the equation $F(x) = 0$.
Hint: Introduce the auxiliary function $f(x) = x - \lambda F(x)$, and choose $\lambda$ such that the theorem works for the equivalent equation $f(x) = x$
(Theorem 1: Every contraction mapping $A$ defined on a complete metric space $R$ has a unique fixed point.)
Here is what I have so far:
Define the auxiliary function $f(x) = x - \lambda F(x)$. We first must show that $f$ is a contraction mapping, meaning that $|f(x) - f(y)| \leq K |x - y|$ where $x, y \in [a,b]$.
Thus, let $x, y \in [a,b]$. We have:
\begin{align*}
|f(x) - f(y)| & = |(x - \lambda F(x)) - (y - \lambda F(y))| & & \text{definition of $f(x)$} \\
& = |(x - y) - \lambda (F(x) - F(y))| & & \text{rearrange}
\end{align*}
Since $F(x)$ is continuous on $[a,b]$ and differentiable on $(a,b)$, we invoke the mean value theorem. Thus, $\exists c \in (x,y)$ such that $F'(c) = \frac{F(x) - F(y)}{x-y}$. (This requires us to impose the restriction that $x \neq y$, if $x = y$, then $f(x) = f(y)$, and $|f(x) - f(y)| = 0$, and the result trivially holds for any choice of $K$.) This implies that $F' (c) (x - y) = F(x) - F(y)$. Using this, we get:
\begin{align*}
& = |(x - y) - \lambda F'(c) (x-y)| \\
& = |(x-y)(1 - \lambda F'(c))| & & \text{take out factor of $(x-y)$} \\
& = |x-y||1 - \lambda F'(c)| & & \text{properties of abs. value.}
\end{align*}
Using our assumption, we have:
$K_1 \leq F'(c) \leq K_2$. For $\lambda \geq 0$, we have
\begin{align*}
\lambda K_1 \leq \lambda F'(c) \leq \lambda K_2 & \iff - \lambda K_1 \geq - \lambda F'(c) \geq - \lambda K_2 \\
& \iff - \lambda K_2 \leq - \lambda F'(c) \leq - \lambda K_1 \\
& \iff 1 - \lambda K_2 \leq 1 - \lambda F'(c) \leq 1 - \lambda K_1
\end{align*}
Now, set $\lambda - \frac{2}{K_2}$. Then, we have:
\begin{align*}
1 - \lambda K_2 = 1 - \frac{2}{K_2} K_2 = 1 - 2 = -1 \\
\end{align*}
Since $K_1 \leq K_2$, $\frac{1}{K_1} \leq \frac{1}{K_2}$, so $\frac{2}{K_2} \geq \frac{2}{K_1}$ and, hence, $- \frac{2}{K_2} \leq \frac{2}{K_1}$. Thus:
\begin{align*}
1 - \lambda K_1 = 1 - \frac{2}{K_2} K_1 \leq 1 - \frac{2}{K_1} K_1 = 1 - 2 = -1.
\end{align*}
Therefore, given this choice of $\lambda$, we have
\begin{align*}
-1 \leq 1 - \lambda F'(c) \leq 1.
\end{align*}
Thus,
\begin{align*}
|1 - \lambda F'(c)| \leq 1.
\end{align*}
Putting all of this together, we have:
\begin{align*}
|f(x) - f(y)| = |x-y||1 - \lambda F'(c)| \leq |x-y| \cdot 1 = |x-y|.
\end{align*}
Therefore, $f$ is a contraction mapping, meaning that, by Theorem 1, it has a unique fixed point. This implies that $\exists z \in [a,b]$ such that $f(z) = z$. By the definition of $f$, this implies that
\begin{align*}
z - \lambda F(z) = z,
\end{align*}
and then that
\begin{align*}
z - \frac{2}{K_2} F(z) = z.
\end{align*}
Subtracting $z$ from both sides gives
\begin{align*}
- \frac{2}{K_2} F(z) = 0.
\end{align*}
Finally, multiplying both sides by $- \frac{K_2}{2}$ gives:
\begin{align*}
F(z) = 0.
\end{align*}
Thus, there exists a unique root, $z$, to the equation $F(z) = 0$.
Thanks.
| You have
$$0 < \frac{K_1}{K_1+K_2} \leq \frac{F'(x)}{K_1+K_2} \leq \frac{K_2}{K_1+K_2} < 1$$
It follows
$$0 < 1- \frac{K_2}{K_1+K_2} \leq 1- \frac{1}{K_1+K_2}F'(x) \leq 1 - \frac{K_1}{K_1+K_2} < 1$$
So, $\boxed{\lambda = \frac{1}{K_1+K_2}}$ is a good choice.
| {
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"source": "stackexchange",
"question_score": "2",
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$10000=abc$ with $a,b,c$ natural numbers don't have digit $0$. Smallest possible value of $a+b+c$ is? Let $10000=abc$ with $a,b,c$ natural numbers don't have digit $0$. If $a,b,c$ does not have to be all distinct, then the smallest possible value of $a+b+c$ is?
Attempt:
First write as prime factors: $10000 = 2^{4} 5^{4}$. The possible triples are:
$$ 2, 2^{3}, 5^{4} $$
$$ 2^{2}, 2^{2}, 5^{4} $$
$$ 2^{3}, 2, 5^{4} $$
$$ 2^{4}, 2^{4}, 5^{4} $$
$$ 5, 5^{3}, 2^{4}$$
$$ 5^{2}, 5^{2}, 2^{4}$$
$$ 5^{3}, 5, 2^{4}$$
The smallest sum is $5^{2} + 5^{2} + 2^{4}$. Are there better approaches?
| Recognize that the numbers cannot both contain a multiple of $5$ and a multiple of $2$; otherwise they would include a zero. Since $10000=2^4\cdot 5^4$, we know that each of $a,b,c$ is of the form $2^r5^s,$ where $0\leq r,s\leq 4$. Using the above fact we only have three possibilities to consider, though two of them can be ruled out by intuition. $1$) $5^4$ is a term. Then the minimum sum must be greater than $625$. $2$) $5^3$ is a term. Then the minimum sum must be greater than $125$. $3$) $5^2$ is a term. Then another term must be $5^2$ and the remaining term is $2^4$. So the minimum sum is $5^2+5^2+2^4=66.$
| {
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How to prove $\lim_{n \to \infty} \left(\frac{3n-2}{3n+1}\right)^{2n} = \frac{1}{e^2}$?
It's known that $\lim_{n \to \infty} \left(1 + \frac{x}{n} \right)^n = e^x$.
Using the above statement, prove $\lim_{n \to \infty} \left(\frac{3n-2}{3n+1}\right)^{2n} = \frac{1}{e^2}$.
My attempt
Obviously, we want to reach a statement such as $$\lim_{n \to \infty} \left(1 + \frac{-2}{n}\right)^n \quad \text{ or } \quad \lim_{n \to \infty} \left(1 + \frac{-1}{n}\right)^n \cdot \lim_{n \to \infty} \left(1 + \frac{-1}{n}\right)^n$$
in order to be able to apply the above condition. However, I was unable to achieve this. The furthest I've got was the following:
\begin{align}
\left(\frac{n-2}{3n+1}\right)^{2n} &= \left( \frac{9n^2 - 12n + 4}{9n^2 + 6n + 1} \right)^n\\
&= \left(1 + \frac{-18n+3}{9n^2+6n+1}\right)^n\\
f(n)&= \left(1 + \frac{-2 + \frac{3}{n}}{n+\frac{2}{3} + \frac{1}{9n}} \right)^n
\end{align}
It seems quite obvious that $\lim_{n \to \infty} \left(f(n)\right) = \left(1 + \frac{-2}{n + \frac{2}{3}}\right)^n$, however, this is not exactly equal to the statement given above. Are you able to ignore the constant and apply the condition regardness? If so, why? How would you go about solving this problem?
| Let $m:=3n+1$. We have
$$\left(\frac{3n-2}{3n+1}\right)^{2n}=\left(1-\frac3m\right)^{2(m-1)/3}=\left(\left(1-\frac3m\right)^m\right)^{2/3}\left(1-\frac3m\right)^{-2/3}.$$
Hence the limit is $e^{-3\cdot2/3}\cdot1$.
| {
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Can someone explain this strange relationship between arithmetic, quadratic, and cubic means? Let $a,b,c$ be positive real numbers. Then, there exist unique positive real $x,y$ such that $\frac{a+b+c}{3}$=$\frac{x+y}{2}$, and $\sqrt{\frac{a^2+b^2+c^2}{3}}$=$\sqrt{\frac{x^2+y^2}{2}}$.
Strangely, if $c=(a+b)/2$, then $\sqrt[3]{\frac{a^3+b^3+c^3}{3}}$= $\sqrt[3]{\frac{x^3+y^3}{2}}$. Wolfram Alpha confirms this last claim, but it feels like there should be some proof of it besides ugly expansion.
Question: Is there an approach to prove this last statement which provides intuition as to why it is true?
| A slightly different approach: the system is equivalent to $$\left\{\begin{array}{c}x+y=\frac 2 3(a+b+c):=A\\
x^2+y^2=\frac 23(a^2+b^2+c^2):=B\end{array}\right.$$ By elimination, one clearly has $$xy=\frac 12(A^2-B):=C$$ which together with the first equation can be used to solve for $(x,y)$. Note that in general there are two solutions. To address your second problem, namely to obtain the following relation: $$x^3+y^3=\frac 23(a^3+b^3+c^3):=g(a,b,c),~{\rm if~}c=(a+b)/2,$$ consider $$x^3+y^3=(x+y)^3-3xy(x+y)=A^3-3CA:=f(a,b,c).$$ In order that $f(a,b,c)=g(a,b,c)$, consider the factorization $$f(a,b,c)-g(a,b,c)=\frac 2{27}(a+b-2c)(b+c-2a)(c+a-2b),$$ which equals zero if either of the following three conditions holds: $$c=(a+b)/2,a=(b+c)/2,~{\rm or~}b=(c+a)/2.$$
| {
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In the range $0\leq x \lt 2\pi$ the equation has how many solutions $\sin^8 {x}+\cos^6 {x}=1$ In the range $0\leq x \lt 2\pi$ the equation has how many solutions
$$\sin^8 {x}+\cos^6 {x}=1$$
What i did
$\cos^6 {x}=1-\sin^8 {x}=(1-\sin^4 {x})(1+\sin^4 {x})=(1-\sin^2 {x})(1+\sin^2 {x})(1+\sin^4 {x})$
$\cos^4 {x}=(1+\sin^2 {x})(1+\sin^4 {x}) , \cos^2{x}=0$
$(1-\sin^2{x})^2=(1+\sin^2 {x})(1+\sin^4 {x})$
$-3\sin^2{x}=\sin^6{x}$
Which is not possilbe
Is there a trick or something to solve this equation or to know how many solutions are there ?
| Let $\cos x=u\implies \sin x=\sqrt{1-u^2}$. Then $$\sin^8x+\cos^6x=(1-u^2)^4+u^6=(1-4u^2+6u^4-4u^6+u^8)+u^6=1$$ gives $$u^8-3u^6+6u^4-4u^2=0\implies u^2(u^2-1)(u^4-2u^2+4)=0$$ so $u=\cos x=0,\pm1$ as $\Delta_{v^2-2v+4}<0$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3410740",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 3
} |
Calculate $\lim_{x\to\infty}\frac{x^2}{x+1}-\sqrt{x^2+1}$ I am stuck on a limit of the indeterminate form $\infty-\infty$. I have tried many approaches, such as multiplying with conjugates etc. and I am unable to find a solution. I suspect that there is an elementary trick that I am plainly missing right here. Can anybody give me a hint or solution as to solve
$$\lim_{x\to\infty}\frac{x^2}{x+1}-\sqrt{x^2+1}$$
| If
$f(x)
=\dfrac{x^2}{x+1}-\sqrt{x^2+1}
$
then
$\begin{array}\\
f(x)
&=\dfrac{x^2}{x+1}-\sqrt{x^2+1}\\
&=\dfrac{x^2+x-x}{x+1}-\sqrt{x^2+1}\\
&=x-\dfrac{x}{x+1}-\sqrt{x^2+1}\\
&=x-\dfrac{x+1-1}{x+1}-\sqrt{x^2+1}\\
&=x-1+\dfrac{1}{x+1}-\sqrt{x^2+1}\\
\text{so}\\
f(x)
&\lt x-1+\dfrac{1}{x+1}-\sqrt{x^2}\\
&= -1+\dfrac{1}{x+1}\\
\text{and}\\
f(x)
&=-1+\dfrac{1}{x+1}+x-\sqrt{x^2+1}\\
&=-1+\dfrac{1}{x+1}+(x-\sqrt{x^2+1})\dfrac{x+\sqrt{x^2+1}}{x+\sqrt{x^2+1}}\\
&=-1+\dfrac{1}{x+1}-\dfrac{1}{x+\sqrt{x^2+1}}\\
&>-1+\dfrac{1}{x+1}-\dfrac{1}{2x}\\
&=-1+\dfrac{2x-(x+1)}{2x(x+1)}\\
&=-1+\dfrac{x-1}{2x(x+1)}\\
&=-1+\dfrac{x+1-2}{2x(x+1)}\\
&=-1+\dfrac{1}{2x}-\dfrac{1}{x(x+1)}\\
\end{array}
$
so
$f(x) \to -1$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3411076",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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If $x$ and $y$ are real number such that $x^2+2xy-y^2=6$, then find the minimum value of $(x^2+y^2)^2$ If $x$ and $y$ are real number such that $x^2+2xy-y^2=6$, then find the maximum value of $(x^2+y^2)^2$
My attempt is as follows:
$$(x-y)^2\ge 0$$
$$x^2+y^2\ge 2xy$$
$$2(x^2+y^2)\ge x^2+y^2+2xy$$
\begin{equation}
2(x^2+y^2)\ge (x+y)^2\tag{1}
\end{equation}
Solving the given equation:
$$x^2+2xy-y^2=6$$
$$(x+y)^2=2y^2+6$$
So putting the value of $(x+y)^2$ in equation $1$
$$2(x^2+y^2)\ge 2y^2+6$$
$$x^2+y^2\ge y^2+3$$
\begin{equation}
x^2\ge 3\tag{2}
\end{equation}
So $x\in \left(-\infty,-\sqrt{3}\right) \cup \left(\sqrt{3},\infty\right)$
But how to proceed from here?
| Perform a parametrization of the form $$x = r \cos \theta, \quad y = r \sin \theta.$$ Then we seek to minimize $(x^2 + y^2)^2 = r^4$ subject to the constraint $$\begin{align*}
6 & = r^2 \cos^2 \theta + 2 r^2 \cos \theta \sin \theta - r^2 \sin^2 \theta \\
&= r^2 \left( \cos^2 \theta - \sin^2 \theta + 2\cos \theta \sin \theta\right) \\
&= r^2 \left( \cos 2\theta + \sin 2\theta \right).
\end{align*}$$ Therefore, $$r^2 = \frac{6}{\cos 2\theta + \sin 2\theta},$$ and on $\theta \in [0, 2\pi)$, the LHS is minimized when the denominator is maximized. Rewriting this using an additional trigonometric identity, $$\cos 2\theta + \sin 2\theta = \sqrt{2} \left(\cos 2\theta \sin \frac{\pi}{4} + \sin 2\theta \cos \frac{\pi}{4}\right) = \sqrt{2} \sin \left( 2\theta + \frac{\pi}{4}\right).$$ So the maximum is attained whenever $2\theta + \frac{\pi}{4} = 2\pi k + \frac{\pi}{2}$ for some integer $k$; namely $$\theta \in \left\{\frac{\pi}{8}, \frac{9\pi}{8}\right\},$$ and the maximum value is $\sqrt{2}$; thus the minimum value is $$r^4 = \left( \frac{6}{\sqrt{2}} \right)^2 = 18.$$ It is a straightforward exercise to compute the set of $(x,y)$ values for which this minimum is attained.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Show $(a^2 + b^2 + c^2 + d^2)^3 \ge 3 (a^3 + b^3 + c^3 + d^3)^2$ Let $a,b,c,d$ be real numbers such that $a + b + c +d = 0$.
Show:
$(a^2 + b^2 + c^2 + d^2)^3 \ge 3 (a^3 + b^3 + c^3 + d^3)^2.$
I've applied AM-GM to the left hand side to find $ (a^2 + b^2 + c^2 + d^2) \ge 4 \cdot\sqrt[4]{a^2b^2c^2d^2}, $ so $ (a^2 + b^2 + c^2 + d^2)^3 \ge 64 \cdot (abcd)^{3/2}.$
And I've been trying to get the right hand side bounded by a lower value but haven't a way forward.
| We need to prove that $$(a^2+b^2+c^2+(a+b+c)^2)^3\geq3(a^3+b^3+c^3-(a+b+c)^3)^2$$ or
$$\left(\sum_{cyc}(a+b)^2\right)^3\geq27(a+b)^2(a+c)^2(b+c)^2,$$ which is just AM-GM:
$$\frac{\sum\limits_{cyc}(a+b)^2}{3}\geq\sqrt[3]{\prod_{cyc}(a+b)^2}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3412751",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
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} |
Trouble Calculating for $30 \cos(30x)+14=−16$ I am unable to see how,
$$30\cos(30x)+14=−16$$
is equal to
$$\frac{\pi}{30} + n \frac{\pi}{15}$$
I solved up to this
$$\cos(30x) = -1$$
$$\pi= 30x$$
$$\frac{\pi}{30} = x$$
But I am unsure where the $\frac{\pi}{15}$ came from. Can someone help me understand this?
| The thing is $\cos (x + 2n\pi ) = \cos x$ so if $\cos 30x = -1$ then $\cos 30(x+ \frac {2n\pi}{30})$ will also be equal to $-1$.
So when you got $\cos (30x ) = -1$ then
$30x = \pi$ is ONE of the possible values for $30x$ (because $\cos \pi = -1$).
$30x = 3\pi$ is another one (because $\cos 3\pi = -1)$ and $30x = -pi$ and $30x=5\pi$ and $30x = 2147\pi$ are some more (because $\cos -\pi = \cos 5\pi = \cos 2147 \pi = -1$).
Indeed for any integer, $n$, you have an infinite number of possibilities of $30x = (2n+1)\pi$ becase $\cos (2n+1)\pi = -1$.
So $x =\frac {\pi}{30}$ is one answer and $\frac {\pi}{10}$ or $-\frac {\pi}{30}$ and $\frac {\pi}{6}$ and $\frac {2147\pi}{30}$ are all also solutions because for all thus values of $x$ you'd have $30x = \{\pi,3\pi, -pi,5\pi, 2147\pi\}$ and $\cos 30x = -1$ for all of those.
In fact, there are an infinite number of solutions. For any $\frac{\pi}{30} + \frac n{15}\pi$ you will have $\cos 30x = \cos (30(\frac {\pi}{30} + \frac n{15}\pi))=\cos (\pi + 2n \pi) = \cos \pi = -1$.
So those are an infinite number of solutions where $\cos 30x = -1$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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} |
Eccentricity of conic $4x^2+4xy+4y^2+x-5=0$ is
Finding Eccentricity of conic $4x^2+4xy+4y^2+x-5=0$ is
what I tried:
let $S = 4x^2+4xy+4y^2+x-5$
$\dfrac{dS}{dx}=8x+4y+1$ and $\dfrac{dS}{dy}=4x+8y$
for center $\dfrac{dS}{dx}=0$ and $\dfrac{dS}{dy}=0$
getting center as $ x=-\dfrac{1}{6}$ and $y=\dfrac{1}{12}$
How do I solve it? Help me please
| The eccentricity only depends on the ratio of the axis so that only the quadratic terms matter.
By symmetry, it is obvious that one of the axis is parallel to $x=y$ and we apply the orthogonal transformation
$$x=u-v,y=u+v$$
which yields
$$3u^2+v^2.$$
Hence
$$e=\sqrt{1-\dfrac13}.$$
More generally, when the quadratic terms are $ax^2+2bxy+cy^2$, we compute the ratio of the Eigenvalues of
$$\begin{pmatrix}a&b\\b&c\end{pmatrix}$$ which are the roots of
$$x^2-(a+c)x+ac-b^2,$$
and that ratio is
$$\dfrac{a+c-\sqrt{(a-c)^2+4b^2}}{a+c+\sqrt{(a-c)^2+4b^2}}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3416401",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Ideal factorization in cubic extension Let $a:=\sqrt[3]3$
Factoring the ideal $(5)$ in $\Bbb Z[a]$, I used reduction of $X^3-3$ mod $5$ to find gives $(5)=(5,a-2)(5,a^2+2a+4)$
Checking my result under sage math gives $(5)=(a-2)(a^2+2a+4)$
I don't know why this difference?
We have $(5)=(5,a-2)(5,a^2+2a+4)=(5\Bbb Z[a]+(a-2)\Bbb Z[a])(5\Bbb Z[a]+(a^2+2a+4)\Bbb Z[a])$
$=25\Bbb Z[a]+5\Bbb Z[a](a-2+a^2+2a+4)+(a-2)(a^2+2a+4)$
Since $25\Bbb Z[a]=(25)\subset (5)$ and $5\Bbb Z[a](a-2+a^2+2a+4)\subset (5)$ we have
$(5)=(a-2)(a^2+2a+4)$
If there is no mistake, it means that we can just remove p all the time if we have a prime factorization: $(p)=(p,m_i)^{e_i}$ where the $m_i$ are the irreducible polynomials of factorization of the minimal polynomial of $a$ modulo $p$. Does this make sense?
Thank you for your help
| In a commutative ring, the product of two principal ideals $(a_1)(a_2)$ is equal to the ideal $(a_1a_2)$. So in this case, this indeed gives us $(a - 2)(a^2 + 2a + 4) = (a^3 - 8) = (-5) = (5)$.
However, we cannot remove the $p$ from your expression in general. For example, if we are working in $\mathbb{Z}[\sqrt[3]{5}]$ (let $b = \sqrt[3]{5}$), let's try to factorize $(3)$. By the same method you used, we get $(3) = (3, b + 1)(3, b^2 - b + 1)$. In this case, though, we have $(b + 1)(b^2 - b + 1) = (b^3 + 1) = (6) \neq (3)$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3418598",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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If $x=9$, then can we write $\sqrt{x}=\pm3$ or only $\sqrt{x}=3$ If $x=9$, then can we write $\sqrt{x}=\pm3$ or only $\sqrt{x}=3$.
I am confused as square root always gives positive number.
But the irony is that if we have $a^2=9$, then we write $a=\pm3 \text { where $a=\sqrt{a^2}$ }$
| Note that $x^2=9$ has two solutions namely $x=\sqrt 9=3$ and $x=-\sqrt 9 = -3$
When we write $x=\pm 3$ we mean that both $3$ and $-3$ are solutions and it does not mean that they are equal.
Thus we have $x=\pm \sqrt 9$ which means there are two solutions and they are opposite to each other.
In general for a real number $x$ we have $$\sqrt {x^2}= |x|$$ which results in $$-\sqrt {x^2}= -|x|$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3420992",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 0
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interpretation on $1^2+2^2+\ldots+n^2=\binom{n+1}{2}+2\binom{n+1}{3}$ and $1^3+2^3+\ldots+n^3=\binom{n+1}{2}+6\binom{n+1}{3}+6\binom{n+1}{4}$ How to interpret the results
$$
1^2+2^2+\ldots+n^2=\binom{n+1}{2}+2\binom{n+1}{3}
\\
1^3+2^3+\ldots+n^3=\binom{n+1}{2}+6\binom{n+1}{3}+6\binom{n+1}{4}
$$
I want to find a clear argument (combinatorial example,etc.) to prove this, other than induction or merely use the formula.
| The sum of cubes formula can also be derived by slightly generalizing this answer by Mike Earnest for the sum of squares case.
Consider counting ordered quadruplets of integers $(w,x,y,z)$ s.t.
*
*$0 \le w, x, y < z$
*$1 \le z \le n$
When $z=k$ the number of ways to choose $(w,x,y)\in \{0, 1, \dots, k-1\}^3$ is $k^3$, so the total is $\sum_{k=1}^n k^3$.
OTOH all such quadruplets can be classified as follows:
*
*$w, x, y$ all distinct: For a particular ordering, say $w < x < y < z$, we have ${n+1 \choose 4}$ choices. There are $3!=6$ orderings among the $\{w,x,y\}$, so this case gives $6 \times {n+1 \choose 4}$.
*$w = x = y$: Then we're simply choosing two numbers, $w$ and $z$. So this case gives ${n+1 \choose 2}$.
*Exactly two of the three $\{w, x, y\}$ are equal: There are $3$ ways to choose the equal pair, then $2$ ways to decide if the remaining value is bigger or smaller. Once we decided that, say $x < w = y$, we are simply picking three numbers $x, w, z$. So this case gives $3 \times 2 \times {n+1 \choose 3}$.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Solve binomial coefficient equation My book asks me to solve this equation:
$$\begin{pmatrix} 6\\2 \end{pmatrix}+\begin{pmatrix} 6\\x \end{pmatrix}=\begin{pmatrix} 7\\x \end{pmatrix}$$
The solution is $x=3$ and the formula $$\begin{pmatrix} n-1\\k-1 \end{pmatrix}+\begin{pmatrix} n\\k \end{pmatrix}=\begin{pmatrix} n+1\\k \end{pmatrix}$$ is supposed to reason that solution. What I do not understand however is, if $n=6$ would $n-1$ not equal $5$?
| The formula should be
$$\binom{n}{k-1}+\binom{n}{k}=\binom{n+1}{k}.$$
It is the fundamental recurrence of the binomial coefficients.
Hence
$$\binom{6}{2}+\binom{6}{x}=\binom{7}{x}=\binom{6}{x-1}+\binom{6}{x}\implies \binom{6}{2}=
\binom{6}{x-1}.$$
and, by symmetry, it follows that we have TWO solutions: $x-1=2$ OR $x-1=6-2=4$.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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If $h(t)=-16t^2+80t$ then simplify $\frac{h(a)-h(1)}{a-1}$ I am to simplify $\frac{h(a)-h(1)}{a-1}$ given $h(t)=-16t^2+80t$. The solution provided is
$\frac{-64+80a-16a^2}{-1+a}$ = $-16a+64$
I cannot see how this was arrived at. Here's as far as I got:
$\frac{h(a)-h(1)}{a-1}$
$\frac{(-16a^2+80a)-(-16+80)}{a-1}$ # substitute in the function h(t)
$\frac{-16a^2+80a-64}{a-1}$ # simplify numerator
$\frac{16(-a^2+5a-4)}{1-a}$ # 16 is a common factor in the numerator, attempted to simplify
...
I was not able to factor $-a^2+5a-4$
How can I arrive at the provided solution? More granular baby steps appreciated.
| We have that
$$\frac{-16a^2+80a-64}{a-1}=\frac{-16(a^2-5a+4)}{a-1}$$
and
$$a^2-5a+4=(a-4)(a-1)$$
indeed by quadratic equation for $a^2-5a+4=0$
$$a_{1,2}=\frac{5\pm \sqrt{25-16}}{2}=4,1$$
that is
$$a^2-5a+4=(a-a_1)(a-a_2)=(a-4)(a-1)$$
therefore providing that $a\neq 1$
$$\frac{-16(a^2-5a+4)}{a-1}=\frac{-16(a-4)(a-1)}{a-1}=-16(a-4)=-16a+64$$
| {
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"timestamp": "2023-03-29T00:00:00",
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How to prove $n \ge 2 {\left ( \sum_{k=1}^n \frac{1}{k} \right)}^2$ for all $n \ge 2$? I'm trying to prove that $$n \ge 2 {\left ( \sum_{k=1}^n \frac{1}{k} \right)}^2$$ for all $n \ge 2$. I try induction on $n$ as follows:
My attempt:
The inequality holds for $n=2$. Let it holds for $n$. Our goal is to show that $$2 {\left ( \sum_{k=1}^n \frac{1}{k} \right)}^2+1 \ge 2 {\left ( \sum_{k=1}^{n+1} \frac{1}{k} \right)}^2$$
This is equivalent to $$1 \ge \frac{1}{n+1} \left ( \frac{1}{n+1}+ \sum_{k=1}^n \frac{1}{k} \right) =\frac{1}{(n+1)^2} + \frac{1}{n+1} \sum_{k=1}^n \frac{1}{k}$$
I'm unable to approximate the sum $\sum_{k=1}^n \frac{1}{k}$. Could you please shed me some light on the last step?
| We can use that bound for harmonic series
$$\sum_{k=1}^n\frac{1}{k} \leq \ln n + 1$$
therefore
$$\frac{1}{(n+1)^2} + \frac{1}{n+1} \sum_{k=1}^n \frac{1}{k}\le \frac{1}{(n+1)^2} + \frac{\ln n + 1}{n+1} \le 1$$
indeed
$$ \frac{1}{(n+1)^2} + \frac{\ln n + 1}{n+1} \le 1 \iff \frac{1}{n+1} + \ln n + 1 \le n+1 \iff \ln n \le n-\frac{1}{n+1}$$
which is true indeed
$$ \ln (1+(n-1)) \le n-1\le n-\frac{1}{n+1}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3433789",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Calculate the number of solutions Calculate the number of solutions presented by the equation.
$$\sqrt{1-x}+\sqrt{1-2x}+\sqrt{1-4x}=x^2+2$$
What I thought: The LHS is concave, the RHS is convex
| Hint : If $X= \sqrt{a}+ \sqrt{b}+\sqrt{c}$ ... keep squaring and rearranging ...
\begin{eqnarray*}
X &=& \sqrt{a}+ \sqrt{b}+\sqrt{c} \\
\frac{X^2-a-b-c}{2} &=& \sqrt{ab}+ \sqrt{bc}+\sqrt{ca} \\
\left(\frac{X^2-a-b-c}{2} \right)^2 &=& 2\sqrt{abc}(\sqrt{a}+ \sqrt{b}+\sqrt{c})=2\sqrt{abc}X. \\
\end{eqnarray*}
Now square one final time & what order is the polynomial ?
| {
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Help with algebra, rearrange equations I have:
$$
r=\frac{1-x^2-y^2}{(1-x)^2+y^2} \tag 1
$$
And want to write $(1)$ as:
$$
\Big (x-\frac{r}{1+r}\Big )^2+y^2=\Big (\frac{1}{1+r}\Big )^2 \tag 2
$$
First method, starting from $(1)$:
$$
r=\frac{1-x^2-y^2}{x^2-2x+1+y^2}\iff
$$
$$
r(x^2-2x+1+y^2)=1-x^2-y^2 \iff
$$
$$
rx^2-r2x+r+ry^2=1-x^2-y^2 \iff
$$
$$
rx^2+x^2-r2x+r+ry^2+y^2=1 \iff
$$
$$
x^2(r+1)-r2x+r+y^2(r+1)=1 \iff
$$
$$
x^2-x\frac{2r}{r+1}+\frac{r}{r+1}+y^2=\frac{1}{r+1} \tag 3
$$
Completing the square of $x^2-x\frac{2r}{r+1}$:
$$
x^2-x\frac{2r}{r+1}+\Big (\frac{r}{1+r} \Big )^2-\Big (\frac{r}{1+r} \Big )^2=\Big (x-\frac{r}{1+r}\Big )^2-\Big (\frac{r}{1+r} \Big )^2
$$
Inserting in $(3)$ gives:
$$
\Big (x-\frac{r}{1+r}\Big )^2-\Big (\frac{r}{1+r} \Big )^2+\frac{r}{r+1}+y^2=\frac{1}{r+1} \tag 4
$$
I'm stuck here, what is next?
Second method, starting from $(2)$:
$$
\Big (x-\frac{r}{1+r}\Big )^2+y^2=\Big (\frac{1}{1+r}\Big )^2 \iff
$$
$$
\Big (\frac{x(r+1)-r}{1+r}\Big )^2+y^2=\Big (\frac{1}{1+r}\Big )^2 \iff
$$
$$
(x(r+1)-r)^2+y^2(1+r)^2=1 \tag 5
$$
Expand $(x(r+1)-r)^2$:
$$
(x(r+1)-r)^2=x^2(1+r)^2-x2r(1+r)+r^2
$$
Inserting in $(5)$ gives:
$$
x^2(1+r)^2-x2r(1+r)+r^2+y^2(1+r)^2=1 \iff
$$
$$
x^2(1+2r+r^2)-2rx-2xr^2+r^2+y^2(1+2r+r^2)=1\iff
$$
$$
x^2+2rx^2+r^2x^2-2rx-2xr^2+r^2+y^2+2ry^2+r^2y^2=1\iff
$$
Collect $r^2$ and $r$:
$$
r^2(x^2-2x+1+y^2)+r(2x^2-2x+2y^2) = 1-x^2-y^2 \iff
$$
$$
r^2((x-1)^2+y^2)+r(2x(x-1)+2y^2) = 1-x^2-y^2 \tag 6
$$
I'm stuck here, I don't know how to go from $(6)$ to $(1)$.
| One:
$$r=\frac{1-x^2-y^2}{(1-x)^2+y^2} \Rightarrow y^2=\frac{1-x^2-(1-x)^2r}{1+r}$$
Two:
$$\Big (x-\frac{r}{1+r}\Big )^2+y^2=\Big (\frac{1}{1+r}\Big )^2 \Rightarrow \\
y^2=\left[\frac1{1+r}-x+\frac{r}{1+r}\right]\cdot \left[\frac1{1+r}+x-\frac{r}{1+r}\right]=\\
(1-x)\cdot\frac{(1+x)-(1-x)r}{1+r}=\frac{1-x^2-(1-x)^2r}{1+r}.$$
| {
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Prove that $4\tan^{-1}\left(\frac{1}{5}\right) - \tan^{-1}\left(\frac{1}{239}\right)= \frac{\pi}{4}$
Prove that $4\tan^{-1} \left(\dfrac{1}{5}\right) - \tan^{-1}\left(\dfrac{1}{239}\right)=\dfrac{\pi}{4}.$
I was wondering if there was a shorter solution than the method below?
Below is my attempt using what I would call the standard approach to these kinds of problems.
The expression on the left hand side is equivalent to $$\tan^{-1}\left[\tan \left(4\tan^{-1}\left(\dfrac{1}{5}\right)\right)-\tan^{-1}\left(\dfrac{1}{239}\right)\right]\\
=\tan^{-1}\left(\dfrac{\tan(4\tan^{-1}(\frac{1}{5}))-\frac{1}{239}}{1+\frac{1}{239}\tan(4\tan^{-1}(\frac{1}{5}))}\right)\tag{1}.$$
We have that $$\tan\left(4\tan^{-1}\left(\frac{1}{5}\right)\right)=\dfrac{2\tan(2\tan^{-1}(\frac{1}{5}))}{1-\tan^2(2\tan^{-1}(\frac{1}{5})}\tag{2}$$
and that
$$\tan\left(2\tan^{-1}\left(\frac{1}{5}\right)\right)=\dfrac{2\cdot \frac{1}{5}}{1-(\frac{1}{5})^2}=\dfrac{5}{12}\tag{3}.$$
Plugging in the result of $(3)$ into $(2)$ gives
$$\tan\left(4\tan^{-1}\left(\frac{1}{5}\right)\right) = \dfrac{2\cdot \frac{5}{12}}{1-(\frac{5}{12})^2}=\dfrac{120}{119}\tag{4}.$$
Pluggin in the result of $(4)$ into $(1)$ gives that the original expression is equivalent to
$$\tan^{-1}\left(\dfrac{\frac{120}{119}-\frac{1}{239}}{1+\frac{1}{239}\cdot\frac{120}{119}}\right)=\tan^{-1}\left(\dfrac{\frac{119\cdot 239 + 239-119}{239\cdot 119}}{\frac{119\cdot 239+120}{119\cdot 239}}\right)=\tan^{-1}(1)=\dfrac\pi4,$$
as desired.
| We can also use
$$\arctan(u) \pm \arctan(v) = \arctan\left(\frac{u \pm v}{1 \mp uv}\right)$$
to obtain in four steps
$$\frac{\frac15 - \frac1{239}}{1 + \frac1{5\cdot 239}}=\frac{239-5}{5\cdot 239+1}=\frac{234}{5\cdot 239+1}=\frac9{46} \to$$
$$\to \frac{\frac15 + \frac9{46}}{1 - \frac15\frac9{46}}=
\frac7{17} \\\to \frac{\frac15 + \frac7{17}}{1 - \frac15\frac7{17}}=
\frac2{3} \\\to \frac{\frac15 + \frac2{3}}{1 - \frac15\frac2{3}}=
1$$
| {
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"answer_count": 4,
"answer_id": 0
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Prove that $ \lim_{x \to 0} \frac{ \sqrt{x + \sqrt{ x + \sqrt{x}}} }{\sqrt[8]{x}}= 1$
Prove that $$ \lim_{x \to 0} \frac{ \sqrt{x + \sqrt{ x + \sqrt{x}}} }{\sqrt[8]{x}}= 1$$
Let $x=y^8$ with $t>0$. Then after we get
$$
\lim \frac{ \sqrt { t^8 + \sqrt{ t^8 + \sqrt{t^8}}}}{ t} = \lim \frac{ \sqrt{ t^8 + \sqrt{ t^8 + t^4}}}{ t}
$$
and can't go further than that
| We have that
$$ \frac{\sqrt{x + \sqrt{ x + \sqrt x}}}{\sqrt[8] x} = \frac{ \sqrt{x + \sqrt[4] x\sqrt{ \frac{x}{ \sqrt x}+1}}}{\sqrt[8] x} = \frac{\sqrt[8]x\sqrt{\frac{x}{\sqrt[4]x} + \sqrt{ \frac{x}{ \sqrt x}+1}}}{\sqrt[8] x} =$$
$$=\sqrt{\frac{x}{\sqrt[4]x} + \sqrt{ \frac{x}{ \sqrt x}+1}} \to \sqrt{0 + \sqrt{ 0+1}}=1$$
or in a similar way by your approach
$$\frac{ \sqrt{ t^8 + \sqrt{ t^8 + t^4}}}{ t}=\frac{ \sqrt{ t^8 + t^2\sqrt{ t^4 + 1}}}{ t}=\frac{ t\sqrt{ t^6 + \sqrt{ t^4 + 1}}}{ t}=\sqrt{ t^6 + \sqrt{ t^4 + 1}} \to 1$$
| {
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Is my approach accurate to find original position of boat? A boat goes upstream for $3$ hr $30$ min and then goes downstream for $2$ hr $30$ min. If the speed of the current and the speed of the boat
in still water are $\frac{10}{3}$ kmph and $\frac{15}{2}$ kmph respectively, how far
from its original position is the boat now?
Speed of boat in still water$=\frac{15}{2}$
Speed of stream $=\frac{10}{3}$
Upstream speed $=\frac{15}{2}-\frac{10}{3}$
Downstream speed $=\frac{15}{2}+\frac{10}{3}$
→ Downstream distance - Upstream distance = far from original.
$=\left(2+\frac{1}{2}\right)\left(\frac{15}{2}+\frac{10}{3}\right) - \left(3+\frac{1}{2}\right)\left(\frac{15}{2}-\frac{10}{3}\right)$
$=\left(\frac{5}{2}\right)\left(\frac{65}{6}\right) - \left(\frac{7}{2}\right)\left(\frac{25}{6}\right)$
$=\frac{325-175}{12}$
→ $12.5$ km downstream.
Is my approach accurate to find original position of boat?
| Your method looks great. As an alternative:
With reference to ground, the water travels downstream for $2.5+3.5=6$ hours.
With reference to water, the boat went upstream for $1$ hour. So the boat is
$$ \frac{10}{3} \times 6- \frac{15}{2}\times 1 = 12.5$$
$km$ downstream from the start.
| {
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Find the right positive definite block matrix Let $0<a<b$, and let $A, B \in \mathbb{R}^{n \times n}$ be two positive definite symmetric matrices (that do not commute).
My question is: are the matrices $M, P \in \mathbb{R}^{2n \times 2n}$ defined by
\begin{align*}
M = \begin{bmatrix}
bABA & aAB\\
aBA & bBAB
\end{bmatrix},\qquad P = \begin{bmatrix}
bA & aAB\\
aBA & bB
\end{bmatrix}
\end{align*}
positive definite or not? (In the sense $x^\intercal M x >0$ for any nonzero $x \in \mathbb{R}^{2n}$.)
Why this question: I am trying to find a positive definite symmetric matrix $Q \in \mathbb{R}^{2n \times 2n}$ such that the product
\begin{align*}
Q \begin{bmatrix}
0 & A^{-1}\\
B^{-1} & 0
\end{bmatrix}
\end{align*}
is a symmetric matrix. My candidates at the moment are M and P, which are such that
\begin{align*}
M \begin{bmatrix}
0 & A^{-1}\\
B^{-1} & 0
\end{bmatrix} = \begin{bmatrix}
aA & bAB\\
bBA & aB
\end{bmatrix}, \qquad P \begin{bmatrix}
0 & A^{-1}\\
B^{-1} & 0
\end{bmatrix} = \begin{bmatrix}
aA & bI_n\\
bI_n & aB
\end{bmatrix}
\end{align*}
but I don't know if they are positive definite.
What I tried: I tried to find an invertible matrix $K$ such that the product $K^\intercal P K$ (or $K^\intercal M K$ ) is easier to study, since $P$ is positive definite if and only if $K^\intercal P K$ is. For example, for
\begin{align*}
K = \begin{bmatrix}
A^{-1} & A^{-1}\\
B^{-1} & B^{-1}
\end{bmatrix}
\end{align*}
we have
\begin{align*}
K^\intercal P K = \begin{bmatrix}
bA^{-1} + 2a I_n + bB^{-1} & bA^{-1} - bB^{-1} \\
bA^{-1} - bB^{-1} & bA^{-1} - 2a I_n + bB^{-1}
\end{bmatrix}.
\end{align*}
But I don't know how to proceed for this matrix either.
| First of all, we must assume that $b>0$ for either matrix to be positive definite.
For $M$, we compute the Schur complement
$$
bABA - \frac{a^2}{b}(AB)(BAB)^{-1}(BA)=
bABA - \frac{a^2}{b}A.
$$
Since $bABA$ is positive definite, $M$ will be positive definite if and only if $bABA - \frac{a^2}{b}A$ is positive definite. With the Loewner order, we can write this condition as
$$
bABA - \frac{a^2}{b}A > 0 \iff
bABA > \frac{a^2}{b}A \iff \\
ABA > \frac{a^2}{b^2} A \iff
A^{-1/2}(ABA)A^{-1/2} > A^{-1/2}[\frac{a^2}{b^2} A]A^{-1/2} \iff\\
A^{1/2}BA^{1/2} > \frac{a^2}{b^2}I.
$$
That is, $M$ will be positive definite if and only if the eigenvalues of $A^{1/2}BA^{1/2}$ are greater than $a^2/b^2$. We see that the matrix $AB$ is similar since
$$
AB = A^{1/2}(A^{1/2}BA^{1/2})A^{-1/2},
$$
so $M$ will be positive definite if and only if the eigenvalues of $AB$ are greater than $a^2/b^2$.
For $P$, we compute the Schur complement to be
$$
bA - \frac{a^2}{b} ABA
$$
and a similar analysis can be applied. We find that $P$ will be positive definite if and only if either $a=0$ or the eigenvalues of $AB$ are less than $b^2/a^2$.
In summary, $M$ and $P$ will be positive definite if and only if $a=0$ or the eigenvalues of $AB$ lie inside the interval $(a^2/b^2,b^2/a^2)$.
A potentially helpful insight: we can reduce your original problem to a possibly simpler case by considering a congruent matrix. For instance:
$$
\pmatrix{A^{1/2}\\&B^{1/2}} \pmatrix{0&A^{-1}\\B^{-1}&0} \pmatrix{A^{1/2}\\&B^{1/2}} =
\pmatrix{0&A^{-1/2}B^{1/2}\\B^{-1/2}A^{1/2} & 0}\\
%
\pmatrix{A^{1/2}\\&A^{1/2}} \pmatrix{0&A^{-1}\\B^{-1}&0} \pmatrix{A^{1/2}\\&A^{1/2}} =
\pmatrix{0&I\\A^{1/2}B^{-1}A^{1/2} & 0}
$$
| {
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"source": "stackexchange",
"question_score": "2",
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Where am I going wrong in calculating the projection of a vector onto a subspace? I am currently working my way through Poole's Linear Algebra, 4th Edition, and I am hitting a bit of a wall in regards to a particular example in the chapter on least squares solutions. The line $y=a+bx$ that "best fits" the data points $(1,2)$, $(2,2)$, and $(3,4)$ can be related to the (inconsistent) system of linear equations $$a+b=2$$ $$a+2b=2$$ $$a+3b=4$$ with matrix representation $$A\mathbf{x}=\begin{bmatrix}1&1\\1&2\\1&3\\\end{bmatrix}\begin{bmatrix}a\\b\\\end{bmatrix}=\begin{bmatrix}2\\2\\4\\\end{bmatrix}=\mathbf{b}$$ Using the least squares theorem, Poole shows that the least squares solution of the system is $$\overline{\mathbf{x}}=\left(A^T A \right)^{-1} A^T \mathbf{b}=\left(\begin{bmatrix}3&6\\6&14\\\end{bmatrix}\right)^{-1}\begin{bmatrix}8\\18\\\end{bmatrix}=\begin{bmatrix}\frac{7}{3}&-1\\-1&\frac{1}{2}\\\end{bmatrix}\begin{bmatrix}8\\18\\\end{bmatrix}=\begin{bmatrix}
\frac{2}{3}\\1\\\end{bmatrix}$$ so that the desired line has the equation $y=a+bx=\frac{2}{3} +x$. The components of the vector $\overline{\mathbf{x}}$ can also be interpreted as the coefficients of the columns of $A$ in the linear combination of the columns of $A$ that produces the projection of $\mathbf{b}$ onto the column space of $A$ [which the Best Approximation Theorem identifies as the best approximation to $\mathbf{b}$ in the subspace $\mathrm{col}(A)$]. In other words, the projection of $\mathbf{b}$ onto $\mathrm{col}(A)$ can be found from the coefficients of $\overline{\mathbf{x}}$ by $$\mathrm{proj}_{\mathrm{col}(A)}(\mathbf{b})=\frac{2}{3}\begin{bmatrix}1\\1\\1\\\end{bmatrix}+1\begin{bmatrix}1\\2\\3\\\end{bmatrix}=\begin{bmatrix}\frac{5}{3}\\\frac{8}{3}\\\frac{11}{3}\\\end{bmatrix}$$ But when I try to calculate $\mathrm{proj}_{\mathrm{col}(A)}(\mathbf{b})$ directly [taking $\mathbf{a}_{1}$ and $\mathbf{a}_{2}$ to be the first and second columns of $A$, respectively], I get $$\mathrm{proj}_{\mathrm{col}(A)}(\mathbf{b})=\left(\frac{\mathbf{a}_{1}\cdot\mathbf{b}}{\mathbf{a}_{1}\cdot\mathbf{a}_{1}}\right)\mathbf{a}_{1}+\left(\frac{\mathbf{a}_{2}\cdot\mathbf{b}}{\mathbf{a}_{2}\cdot\mathbf{a}_{2}}\right)\mathbf{a}_{2}=\left(\frac{\begin{bmatrix}1\\1\\1\\\end{bmatrix}\cdot\begin{bmatrix}2\\2\\4\\\end{bmatrix}}{\begin{bmatrix}1\\1\\1\\\end{bmatrix}\cdot\begin{bmatrix}1\\1\\1\\\end{bmatrix}}\right)\begin{bmatrix}1\\1\\1\\\end{bmatrix}+\left(\frac{\begin{bmatrix}1\\2\\3\\\end{bmatrix}\cdot\begin{bmatrix}2\\2\\4\\\end{bmatrix}}{\begin{bmatrix}1\\2\\3\\\end{bmatrix}\cdot\begin{bmatrix}1\\2\\3\\\end{bmatrix}}\right)\begin{bmatrix}1\\2\\3\\\end{bmatrix}$$ $$=\frac{8}{3}\begin{bmatrix}1\\1\\1\\\end{bmatrix}+\frac{18}{14}\begin{bmatrix}1\\2\\3\\\end{bmatrix}=\begin{bmatrix}\frac{8}{3}\\\frac{8}{3}\\\frac{8}{3}\\\end{bmatrix}+\begin{bmatrix}\frac{9}{7}\\\frac{18}{7}\\\frac{27}{7}\\\end{bmatrix}=\begin{bmatrix}\frac{83}{21}\\\frac{110}{21}\\\frac{137}{21}\\\end{bmatrix}$$ I am quite confident that my calculation is incorrect, for a number of reasons. For example, when I take the component of $\mathbf{b}$ orthogonal to $\mathrm{col}(A)$ $$\mathrm{perp}_{\mathrm{col}(A)}(\mathbf{b})=\mathbf{b}-\mathrm{proj}_{\mathrm{col}(A)}(\mathbf{b})=\begin{bmatrix}2\\2\\4\\\end{bmatrix}-\begin{bmatrix}\frac{83}{21}\\\frac{110}{21}\\\frac{137}{21}\\\end{bmatrix}=\begin{bmatrix}-\frac{41}{21}\\-\frac{68}{21}\\-\frac{53}{21}\\\end{bmatrix}$$ I get a vector that is not perpendicular to either $\mathbf{a}_{1}$ or $\mathbf{a}_{2}$, indicating that this vector is not in the orthogonal complement of $\mathrm{col}(A)$. Can somebody help me identify where I'm going wrong in my attempt to calculate the projection of $\mathbf{b}$ onto $\mathrm{col}(A)$?
| The column space of $A$, namely $U$, is the span of the vectors $\mathbf{a_1}:=(1,1,1)$ and $\mathbf{a_2}:=(1,2,3)$ in $\Bbb R ^3$, and for $\mathbf{b}:=(2,2,4)$ you want to calculate the orthogonal projection of $\mathbf{b}$ in $U$; this is done by
$$
\operatorname{proj}_U \mathbf{b}=\langle \mathbf{b},\mathbf{e_1} \rangle \mathbf{e_1}+\langle \mathbf{b},\mathbf{e_2} \rangle \mathbf{e_2}\tag1
$$
where $\mathbf{e_1}$ and $\mathbf{e_2}$ is some orthonormal basis of $U$ and $\langle \mathbf{v},\mathbf{w} \rangle:=v_1w_1+v_2w_2+v_3 w_3$ is the Euclidean dot product in $\Bbb R ^3$, for $\mathbf{v}:=(v_1,v_2,v_3)$ and $\mathbf{w}:=(w_1,w_2,w_3)$ any vectors in $\Bbb R ^3$.
Then you only need to find an orthonormal basis of $U$; you can create one from $\mathbf{a_1}$ and $\mathbf{a_2}$ using the Gram-Schmidt procedure, that is
$$
\mathbf{e_1}:=\frac{\mathbf{a_1}}{\|\mathbf{a_1}\|}\quad \text{ and }\quad \mathbf{e_2}:=\frac{\mathbf{a_2}-\langle \mathbf{a_2},\mathbf{e_1} \rangle \mathbf{e_1}}{\|\mathbf{a_2}-\langle \mathbf{a_2},\mathbf{e_1} \rangle \mathbf{e_1}\|}\tag2
$$
where $\|{\cdot}\|$ is the Euclidean norm in $\Bbb R ^3$, defined by $\|\mathbf{v}\|:=\sqrt{\langle \mathbf{v},\mathbf{v} \rangle}=\sqrt{v_1^2+v_2^2+v_3^2}$.
Your mistake is that you assumed that
$$
\operatorname{proj}_U\mathbf{b}=\frac{\langle \mathbf{b},\mathbf{a_1} \rangle}{\|\mathbf{a_1}\|^2}\mathbf{a_1}+ \frac{\langle \mathbf{b},\mathbf{a_2} \rangle}{\|\mathbf{a_2}\|^2}\mathbf{a_2}\tag3
$$
however this is not true because $\mathbf{a_1}$ and $\mathbf{a_2}$ are not orthogonal.
| {
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Prove $4^n+5^n+6^n$ is divisible by 15 Prove by induction:
$4^n+5^n+6^n$ is divisible by 15 for positive odd integers
For $n=2k-1,n≥1$ (odd integer)
$4^{2k-1}+5^{2k-1}+6^{2k-1}=15N$
To prove $n=2k+1$, (consecutive odd integer)
$4^{2k+1}+5^{2k+1}+6^{2k+1}=(4)4^{2k}+(5)5^{2k}+(6)6^{2k}$,
How do I substitute the statement where $n=2k-1$ to the above, to factor out 15 in order to prove divisibility? Would it be easier to assume $n=k$ is odd and prove $n=k+2$ is divisible by 15?
| $4^{2k+1}+5^{2k+1}+6^{2k+1}=(4)4^{2k}+(5)5^{2k}+(6)6^{2k}$
How do I substitute the statement where n=2k−1 to the above
By factoring one more power out...
$4^{2k+1}+5^{2k+1}+6^{2k+1}=(4)4^{2k}+(5)5^{2k}+(6)6^{2k}=(16)4^{2k-1} + (25)5^{2k-1} + (36)5^{2k-1}$
So this is $[16(4^{2k-1} + 5^{2k-1}+6^{2k-1})] + 9*5^{2k-1} + 20*6^{2k-1}$.
And it's easy to finish:
$=[16*15N] + 3*15*5^{2k-2} + 4*15*2*6^{2k-2}$.
====
But if you know modulo arithmetic this is CUTE!
$4^{n} + 5^n + 6^n = (3+1)^n + (6-1)^n + 6^n \equiv 1^n+(-1)^n + 0^n \equiv 0 \pmod 3$ so $3|4^n + 5^n +6^n$.
And $4^n + 5^n + 6^n = (5-1)^n + 5^n + (5+1)^n\equiv (-1)^n + 0^n + 1^n \equiv 0 \pmod 5$ so $5|4^n + 5^n +6^n$.
So $15|4^n + 5^n +6^n$.
....
If you don't know modulo arithmetic the you can use binomial theorem.
$4^n + 5^n + 6^n =(5-1)^n + 5^n + (5+1)^n =$
$(5^n - n*5^{n-1}+ C_{n,2} 5^{n-2} -..... +n*5 - 1) + 5^n +(5^n - n*5^{n-1}+ C_{n,2} 5^{n-2} -..... -n*5 + 1)=$
$(5^n - n*5^{n-1}+ C_{n,2} 5^{n-2} -..... +n*5) + 5^n +(5^n - n*5^{n-1}+ C_{n,2} 5^{n-2} -..... -n*5)$
Which is divisible by $5$.
Do the same for $4^n + 5^n + 6^n = (3+1)^n + (6-1)^n + 6^n$ to show it is divisible by $3$.
| {
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Prove that $3^{2015}-2^{2015}>2016^2$ I have to prove:$$3^{2015}-2^{2015}>2016^2$$
by induction.
I wanted to use the standard factorizing formula:
$$a^{mn}−b^{mn}=(a^n−b^n)(a^{n(m−1)}+a^{n(m−2)}b^n+...+a^nb^{n(m−2)}+b^{n(m−1)}).$$
$2015=403\cdot 5$
What would be the next step?
| We will prove for a more general case , i.e :
$$3^x - 2^x > (x+1)^2 \quad\quad \text{ For x > 3 }$$
For base case , $3^3 - 2^3 = 19 > 16$
Now let be true for a value $n$ , such that $3^n - 2^n > (n+1)^2$
$$\begin{align}3^n - 2^n + 2n+3& > (n+1)^2 + 2n+3\\
3^n - 2^n + 2n+3& > (n+2)^2 \end{align}$$
Now it is left to show that $3^{n+1} - 2^{n+1} >3^n - 2^n + 2n+3 $.We show it by assuming it is true.
$$\begin{align}3^{n+1} - 2^{n+1} &>3^n - 2^n + 2n+3 \\
3^{n+1} - 3^n + 2^n - 2^{n+1} &> 2n+3 \\
2.3^n - 2^n &> 2n+3\end{align}$$
Which follows from the fact that $3^n - 2^n > (n+1)^2 > 2n+3$.
Hence our induction is complete.
| {
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Evaluate $\lim\limits_{n \to \infty}\sum\limits_{k=1}^{n}\frac{\sqrt[k]{k}}{\sqrt{n^2+n-nk}}$
$$\lim_{n \to
\infty}\sum_{k=1}^{n}\frac{\sqrt[k]{k}}{\sqrt{n^2+n-nk}}$$
How to consider it?
| Here's my solution. Please correct me if I'm wrong.
First, let's introduce and prove a result, which will be applied soon. That is
$$\lim_{n \to \infty}\sum_{k=1}^{n}\frac{1}{\sqrt{n^2-n+nk}}=2.$$
Consider using the squeeze theorem. Notice that
\begin{align*} \frac{1}{\sqrt{n+1-k}+\sqrt{n-k}}\geq\frac{1}{2\sqrt{n+1-k}}&\geq\frac{1}{\sqrt{n+1-k}+\sqrt{n+2-k}}, \end{align*}
which implies
\begin{align*} \sqrt{n+1-k}-\sqrt{n-k}\geq\frac{1}{2\sqrt{n+1-k}}&\geq\sqrt{n+2-k}-\sqrt{n+1-k}. \end{align*}
Therefore
$$2\geq\sum_{k=1}^{n}\frac{1}{\sqrt{n^2+n-nk}}\geq\frac{2\left(\sqrt{n+1}-1\right)}{\sqrt{n}}\to 2(n \to \infty)$$
As per the squeeze theorem, the conclusion is followed.
Now, let's manage to solve the present problem. Since $\sqrt[k]{k}$ is decreasing for $k\geq 2$, and converges to $1$ as $k \to \infty$. Thus
$$\forall \varepsilon>0,\exists N>0,\forall n>N,\sqrt[k]{k}\leq1+\varepsilon.$$
Hence
\begin{align*}
\sum_{k=1}^{n}\frac{1}{\sqrt{n^2-n+nk}}&\leq\sum_{k=1}^{n}\frac{\sqrt[k]{k}}{\sqrt{n^2-n+nk}}\\&=\sum_{k=1}^{N}\frac{\sqrt[k]{k}}{\sqrt{n^2-n+nk}}+\sum_{k=N+1}^{n}\frac{\sqrt[k]{k}}{\sqrt{n^2-n+nk}}\\
&\leq\sum_{k=1}^{N}\frac{\sqrt[k]{k}}{\sqrt{n^2-n+nk}}+(1+\varepsilon)\sum_{k=N+1}^{n}\frac{1}{\sqrt{n^2-n+nk}}\\
&\leq\sum_{k=1}^{N}\frac{\sqrt[k]{k}}{\sqrt{n^2-n+nk}}+(1+\varepsilon)\sum_{k=1}^{n}\frac{1}{\sqrt{n^2-n+nk}}.
\end{align*}
Take the limits as $n \to \infty$ of both sides. We obtain
$$2\leq \lim_{n \to \infty}\sum_{k=1}^{n}\frac{\sqrt[k]{k}}{\sqrt{n^2-n+nk}}\leq 2(1+\varepsilon).$$
Since $\varepsilon$ is arbitary, we can conclude
$$\lim_{n \to \infty}\sum_{k=1}^{n}\frac{\sqrt[k]{k}}{\sqrt{n^2-n+nk}}=2,$$
which is what we want.
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"answer_id": 0
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In $\triangle ABC$, if length of medians $BE$ and $CF$ are $12$ and $9$ respectively. Find $\triangle_{max}$ In $\triangle ABC$, if length of medians $BE$ and $CF$ are $12$ and $9$ respectively. Find $\triangle_{max}$
My attempt is as follows:-
$$\triangle=\dfrac{1}{2}bc\sin A$$
For having the maximum area,
$$A=90^{\circ}$$
$$\triangle_{max}=\dfrac{1}{2}bc$$
According to the above figure,
$$BE=12$$
$$AC=b$$
$$AE=EC=\dfrac{b}{2}$$
$$CF=9$$
$$BA=c$$
$$BF=FA=\dfrac{c}{2}$$
For $\triangle AEB$,
$$AB^2+AE^2=BE^2$$
$$c^2+\left(\dfrac{b}{2}\right)^2=12^2$$
$$4c^2+b^2=144\cdot4\tag{1}$$
For $\triangle FAC$
$$AC^2+AF^2=CF^2$$
$$b^2+\left(\dfrac{c}{2}\right)^2=9^2$$
$$4b^2+c^2=81\cdot4\tag{2}$$
Solving equations $(1)$ and $(2)$
$$16b^2-b^2=81\cdot4\cdot4-144\cdot4$$
$$15b^2=144(9-4)$$
$$b^2=48$$
$$b=4\sqrt{3}$$
Putting the value of $b$ in $(1)$
$$4\cdot48+c^2=324$$
$$c^2=132$$
$$c=2\sqrt{33}$$
Hence $$\triangle_{max}=\dfrac{1}{2}8\sqrt{99}$$
$$\triangle_{max}=12\sqrt{11}$$
But actual answer is $72$. But this method seems to be correct, why am I not getting the correct answer from this method? Please help me in this.
| It's true that a triangle with given sides $b$ and $c$ attains it's maximum area when $\angle A = 90^{\circ}$.
But it's not true that a triangle with given medians attains it's maximum area when one of it's angle is a right angle which you assume in your attempt.
Here's one useful lemma to tackle the problem :
Given a triangle $\triangle ABC$ with area $S$ and median $m_a,m_b,m_c$. The area of triangle $\triangle XYZ$ whose sides lengths are equal to $m_a,m_b,m_c$ is equal to $\frac{3}{4}S$.
You could find the proof here or in Problems in Plane and Solid Geometry by V. Prasolov at page 26 problem 1.36.
Using this lemma, we construct another triangle $\triangle XYZ$ whose sides length area equal to the median lengths of $\triangle ABC$. Now, we are already given that two sides length of $\triangle XYZ$ are $12$ and $9$. Since $[ABC] = \frac{4}{3}[XYZ]$, in order to maximize $[ABC]$, it suffices to maximize $[XYZ]$. But clearly, the maximum area of $[XYZ]$ is $\frac{1}{2} \times 12 \times 9 = 54$. Thus, the maximum area of $ABC$ is $\frac{4}{3} \times 54 = 72$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3448752",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Is there a better way to solve this equation? I came across this equation:
$x + \dfrac{3x}{\sqrt{x^2 - 9}} = \dfrac{35}{4}$
Wolfram Alpha found 2 roots: $x=5$ and $x=\dfrac{15}{4}$, which "coincidentally" add up to $\dfrac{35}{4}$. So I'm thinking there should be a better way to solve it than the naïve way of bringing the fractions together and then squaring. Is there any?
| We see that we need $x>3$ then let $x=\frac3{\cos y}$ with $y\in\left(0,\frac \pi 2\right)$
$$x + \dfrac{3x}{\sqrt{x^2 - 9}} = \dfrac{35}{4} \iff \frac1{\cos y}+\frac1{\sin y}=\dfrac{35}{12}$$
and by half tangent identities by $t=\tan \frac y2$ we obtain
$$\frac{1+t^2}{1-t^2}+\frac{1+t^2}{2t}=\dfrac{35}{12} \iff (3t-1)(2t-1)(t^2-7t-6)=0$$
and $t=\frac 13, \frac12$ lead to the answer indeed
*
*$\dfrac3{\cos(2\cdot \arctan\left(\frac13\right)}=5$
*$\dfrac3{\cos(2\cdot \arctan\left(\frac12\right)}=\frac{15}4$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3449031",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 1
} |
Find $\underset{x\rightarrow 0}\lim{\frac{1-\cos{x}\sqrt{\cos{2x}}}{x\sin{x}}}$
Find
$$\underset{x\rightarrow 0}\lim{\frac{1-\cos{x}\sqrt{\cos{2x}}}{x\sin{x}}}$$
My work.$$\underset{x\rightarrow 0}\lim\frac{1}{x\sin{x}}=\frac{\underset{x\rightarrow0}\lim{\;\frac{\sin{x}}{x}}}{\underset{x\rightarrow 0}\lim{\;x\sin{x}}}=\underset{x\rightarrow 0}\lim\frac{1}{x^2}$$
$$\underset{x\rightarrow 0}\lim{\frac{\cos{x}}{x\sin{x}}}=\underset{x\rightarrow 0}\lim{\frac{\sin{2x}}{2x\sin^2{x}}}=\underset{x\rightarrow 0}\lim{\frac{\sin{2x}}{2x}}\cdot\underset{x\rightarrow 0}\lim{\frac{1}{\sin^2{x}}}=\frac{1}{x^2}$$
$$\underset{x\rightarrow 0}\lim{\sqrt{\cos{2x}}}=\underset{x\rightarrow 0}\lim{\sqrt{1-2\sin^2{x}}}=\underset{x\rightarrow 0}\lim{\sqrt{1-2x^2}}$$
L' Hopital's rule:
$\underset{x\rightarrow 0}\lim{\frac{1-\cos{x}\sqrt{\cos{2x}}}{x\sin{x}}}=\underset{x\rightarrow 0}\lim{\frac{1-\sqrt{1-2x^2}}
{x^2}}=\underset{x\rightarrow 0}\lim{\frac{-4x}{x^3\sqrt{1-2x^2}}}$
What should I do next?
| Note that, as $x\to 0$,
$$\begin{align}\frac{1-\cos{x}\sqrt{\cos{2x}}\pm\sqrt{\cos{2x}}}{x\sin{x}}&=
\sqrt{\cos{2x}}\cdot \frac{1-\cos{x}}{x^2}\cdot\frac{x}{\sin(x)}\\&\quad\qquad+2\cdot\frac{\sqrt{1-2\sin^2(x)}-1}{-2\sin^2(x)}\cdot\frac{\sin(x)}{x}\\
&\to1\cdot \frac{1}{2}\cdot 1+2\cdot\frac{1}{2}\cdot 1=\frac{3}{2}\end{align}$$
where we used $\cos(2x)=1-2\sin^2(x)$ and the stardard limits:
$$\lim_{t\to 0}\frac{\sin(t)}{t}=1\quad
\lim_{t\to 0}\frac{1-\cos(t)}{t^2}=\frac{1}{2}\quad
\lim_{t\to 0}\frac{\sqrt{1+t}-1}{t}=\frac{1}{2}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3450548",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 1
} |
Calculation of the limit $\lim_{x\to 0} (\cos x)^{1/x^2}$ without De l'Hospital/Landau's symbols/asymptotic comparison I have calculate this limit
$$\lim_{x\to 0}\ (\cos x)^{1/x^2}$$
with these steps. I have considered that:
$$(\cos x)^{1/x^2}=(\cos x -1+1)^{1/x^2}=\left(\frac{1}{\frac{1}{\cos x -1}}+1\right)^{\frac{1}{x^2}}$$
I remember that $1/(\cos x -1)$ when $x\to 0$ the limit is $\infty$. Hence
$$\left(\frac{1}{\frac{1}{\cos x -1}}+1\right)^{\frac{1}{x^2}}=\left[\left(\frac{1}{\frac{1}{\cos x -1}}+1\right)^{\frac{1}{x^2}}\right]^{\frac{\frac{1}{\cos x -1}}{\frac{1}{\cos x -1}}}=\left[\left(\frac{1}{\frac{1}{\cos x -1}}+1\right)^{\frac{1}{\cos x-1}}\right]^{\frac{\frac{1}{x^2}}{\frac{1}{\cos x -1}}} \tag{1}$$
But if I take
$$p=\frac{1}{\cos x -1}\xrightarrow{x\to 0}p\to \infty$$
therefore I consider the
$$\lim_{p\to \infty}\left(1+\frac 1p\right)^p=e$$
Consequently for the $(1)$,
$$\left(\frac{1}{\frac{1}{\cos x -1}}+1\right)^{\frac{1}{\cos x-1}}\xrightarrow{p\to \infty} e$$
and the exponent
$$\lim_{x\to 0}\frac{\frac{1}{x^2}}{\frac{1}{-(-\cos x +1)}}=-\frac 12\tag{2}$$
At the end $\displaystyle \lim_{x\to 0}\ (\cos x)^{1/x^2}=e^{-\frac 12}$.
I have followed this strategy in my classroom with my students. Is there a shorter solution to the exercise than the one I have given?
| $$
\cos x = 1 - \frac{x^2} 2 + \frac{x^4}{24} - \frac{x^6}{720} + \cdots
$$
So for $x$ near $0$ we have
\begin{align}
(\cos x)^{1/x^2} & \ge \left( 1 - \frac{x^2} 2 \right)^{1/x^2} \\[10pt]
& = \left( 1 + \frac{-1/2}{u} \right)^u \\[8pt]
& \to e^{-1/2} \quad \text{as } u \to+\infty. \\[12pt]
(\cos x)^{1/x^2} & \le \left( 1 - \frac{x^2}{2+\varepsilon} \right)^{1/x^2} \text{ Why is this true? See below.} \\[8pt]
& = \left( 1 + \frac{-1/(2+\varepsilon)}{u} \right)^u \\[8pt]
& \to e^{-1/(2+\varepsilon)} \quad \text{as } u\to+\infty.
\end{align}
If the limit is $\ge e^{-1/2}$ and is $\le e^{-1/(2+\varepsilon)}$ for EVERY sufficiently small $\varepsilon>0,$ then the limit is $\le\lim_{\varepsilon\,\downarrow\,0} e^{-1/(2+\varepsilon)} = e^{-1/2}.$
$\text{“Why is this true? See below.''}$
$$
\cos x \le \underbrace{ 1 - \frac{x^2} 2 + \frac{x^4}{24} \le 1 - \frac{x^2}{2+\varepsilon} }
$$
The inequality over the $\underbrace{\text{underbrace}}$ holds whenever $x^2\le 12\varepsilon/(2+\varepsilon),$ and thus holds in the limit as $x\to0.$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3453794",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 2
} |
Problem with derivative of generating function and sequence of Catalan number
$Q^1$: Given that
\begin{align}
a_0 &= 0\\
a_{n+1} &= na_n+1\\
\end{align}
find the closed form$?$
My attempt:
\begin{align}
\sum_{n\geq0}^{\infty}a_{n+1}x^{n+1}&=\sum_{n\geq0}^{\infty}na_nx^{n+1}+\sum_{n\geq0}^{\infty}1.x^{n+1}\\
(G(z)-a_0)&=G'(z)x^2+\frac{x}{1-x} \qquad\text{Using }G'(z)x=\sum_{n\geq0}^{\infty}na_nx^n
\end{align}
But how to deal with $G'(z)?$
$Q^2$: Let $\{C_n\}$ be the sequence of Catalan
numbers, that is, the solution to the recurrence relation $C_n=\sum_{k=0}^{n-1}C_kC_{n-k-1}$ with $C_0=C_1=1$
$(a)$ Show that if $G(x)$ is the generating function for the sequence of Catalan numbers, then $xG(x)^2-G(x)+1=0$ concluding $G(x)=\frac{1-\sqrt{1-4x}}{2x}$.
$(b)$ Conclude that $G(x)=\sum_{n=0}^{\infty}\frac{1}{n+1}\binom{2n}{n}x^n$ so that $C_n=\frac{1}{n+1}\binom{2n}{n}$
My attempt:
I solve $(a)$ but stuck in $(b)$. The solution provided said that
Using the extended binomial theorem: $$(1-4x)^{-\frac{1}{2}}=\sum_{k=0}^{\infty}\binom{2k}{k}x^k$$
$\color{red}{\text{Note that}(1-4x)^{-\frac{1}{2}}\text{is the derivative of}G(x)=\frac{1-\sqrt{1-4x}}{2x},\text{thus }xG(x)\text{is the integral of}(1-4x)^{-\frac{1}{2}}}$ $$xG(x)=\int_0^x(1-4x)^{-\frac{1}{2}}dx=\cdots=\sum_{n=0}^{\infty}\frac{1}{n+1}\binom{2n}{n}x^n$$ Hence, $C_n=\frac{1}{n+1}\binom{2n}{n}$.
I didn't understood the red line. I know asking one question at a time but since these two question related with derivative and integration hence putted on a single thread.Thanks in advance and thanks for your time .
| $Q^1$: I've played around a bit with initial conditions and indexing and arrived at the following solution. This is just the OEIS sequence A000522 with a $0$ tacked on in the front.
We will use exponential generating functions. First let $b_n = a_{n+1}$. Then we have
$$b_{n} = nb_{n-1} + 1; \qquad b_0 = 1$$
Let $B(x) = \sum_{n\ge 0} b_n \frac{x^n}{n!}$, then
$$\sum_{n\ge 0} b_{n}\frac{x^n}{n!} = \sum_{n \ge 0 }nb_{n-1} \frac{x^n}{n!} + \sum_{n \ge 0} \frac{x^n}{n!}$$
which yields
\begin{align}
B(x) &= x\sum_{n \ge 0 }b_{n-1} \frac{x^{n-1}}{(n-1)!} + e^x = xB(x) + e^x
\end{align}
and so we obtain the g.f. $B(x) = \frac{e^x}{1-x}$. Using the series expansions for $e^x$ and $(1-x)^{-1}$ we have
\begin{align}
\frac{e^x}{1-x} &= \left(\sum_{n\ge 0} \frac{x^n}{n!}\right)\left(\sum_{n\ge 0} x^n\right)\\
&= \sum_{n\ge 0} \left(\sum_{j=0}^{n} \frac{1}{j!} \cdot 1\right) x^n\\
&= \sum_{n\ge 0} \left(\sum_{j=0}^{n} \frac{n!}{j!} \right) \frac{x^n}{n!}.
\end{align}
(The product of $2$ polynomials $\left(\sum_{n\ge 0} s_n x^n\right)\left(\sum_{n\ge 0} t_n x^n\right) = \sum_{n\ge 0} \left(\sum_{j=0}^{n} s_jt_{n-j}\right)x^n$)
So our closed formula is $b_n = \sum_{j=0}^n \frac{n!}{j!}$, and to go back to our original problem we have
$$a_n = \begin{cases}
0 &n=0\\
\sum_{j=0}^n \frac{n!}{j!} &n\ge 1
\end{cases}$$
Well, it depends on what you really mean by closed formula. If someone has a less hacky solution I'm all ears.
$Q^2$: For other answers to this, you may want to see this the stack exchange question (specifically rogerl's answer for a more algebraic solution).
I believe the solution (or you) may have a typo. Note that $(1-4x)^{-\frac{1}{2}}$ is the derivative of $xG(x) = \frac{1-(1-4x)^{\frac{1}{2}}}{2}$, or in other words by the Fundamental Theorem of Calculus:
$$xG(x) = \frac{1-(1-4x)^{\frac{1}{2}}}{2} = \int_0^x(1-4t)^{-\frac{1}{2}} dt$$
which then allows us to integrate $\sum_{k=0}^\infty \binom{2k}{k}x^k$ term by term to obtain $$xG(x) = \sum_{k=0}^\infty \binom{2k}{k}\frac{1}{k+1}x^{k+1}$$ and dividing both sides by $x$ gives us our desired solution.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3454808",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Find derivative $\frac{dy}{dx}$, given $y(x)=\sin^{-1}\left(\frac{5\sin x+4\cos x}{\sqrt{41}}\right)$
Find $\dfrac{dy}{dx}$ if $y=\sin^{-1}\bigg[\dfrac{5\sin x+4\cos x}{\sqrt{41}}\bigg]$
My Attempt
Put $\cos\theta=5/\sqrt{41}\implies\sin\theta=4/\sqrt{41}$
$$
y=\sin^{-1}\big[\sin(x+\theta)\big]\implies\sin y=\sin(x+\theta)\\
y=n\pi+(-1)^n(x+\theta)\\
\boxed{\frac{dy}{dx}=(-1)^n}
$$
But my reference gives the solution $y'=1$, am I missing something here ?
| As you correctly pointed out we have
$$y = \arcsin\left[\sin\left(x +\theta\right) \right],$$
where $\theta = \arcsin\frac4{\sqrt{41}}$.
Now observe that ($k\in \Bbb Z$)
$$
\arcsin\sin \alpha = \begin{cases}\alpha-2k\pi & \left(2k\pi-\frac{\pi}2\leq \alpha < 2k\pi+\frac{\pi}2\right)\\
- \alpha - (2k-1)\pi & \left(2k\pi+\frac{\pi}2\leq \alpha < 2k\pi+\frac{3\pi}2\right).\end{cases}
$$
Thus your function is the triangular wave (see Figure below)
$$
y =
\begin{cases}
x +\theta- 2k \pi & \left(2k\pi-\frac{\pi}2-\theta\leq x < 2k\pi+\frac{\pi}2-\theta\right)\\
-x-\theta-(2k-1)\pi-\theta & \left(2k\pi+\frac{\pi}2-\theta\leq x< 2k\pi+\frac{3\pi}2-\theta\right)
\end{cases}
$$
whose derivative is
$$
y =
\begin{cases}
1 & \left(2k\pi-\frac{\pi}2-\theta< x < 2k\pi+\frac{\pi}2-\theta\right)\\
-1 & \left(2k\pi+\frac{\pi}2-\theta< x<2 k\pi+\frac{3\pi}2-\theta\right)
\end{cases}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3456149",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
} |
Show that $2^{y-1}|(\alpha_{0}+\alpha_{1}+...+\alpha_{2^{y-1}-1})$
Problem show that
Let $\alpha_{0},\alpha_{1},\alpha_{2},...,\alpha_{2^{y-1}-1}$ are smallest non negative integers such that
$$2^y|(3^{(2^{y-1}+i)}-\alpha_{i})$$
For $0\le i \le 2^{y-1}-1$
Then $$2^{y-1}|(\alpha_{0}+\alpha_{1}+...+\alpha_{2^{y-1}-1})$$
Where $y$ is given positive integer.
Example
Let $y=3 \implies \alpha_{0}=1,\alpha_{1}=3,\alpha_{2}=1,\alpha_{3}=3$
and $2^{y-1}=4|(\alpha_{0}+\alpha_{1}+\alpha_{2}+\alpha_{3})=8$
| Let $m=2^{y-1}$.
Then $\alpha_i\equiv 3^{m+i}\pmod{2m}$.
Since $\varphi(2m)=m$, we have $3^m\equiv 1\pmod{2m}$.
Consequently:
\begin{align}
s
&=\sum_{i=0}^{m-1}\alpha_i\\
&\equiv\sum_{i=0}^{m-1}3^{m+i}\\
&\equiv\sum_{i=0}^{m-1}3^{i}
\pmod{2m}
\end{align}
Then
\begin{align}
3s
&\equiv\sum_{i=0}^{m-1}3^{i+1}\\
&\equiv s\pmod{2m}
\end{align}
hence $2s\equiv 0\pmod{2m}$ from which $s\equiv 0\pmod m$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3456259",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Determinant with ones on diagonal and increasing powers of a constant in rows and columns I've found a few determinants of this form and I would like to reach a simple expression for this determinant in terms of a and n.
$$
D_{a,n}=\begin{vmatrix}
1 & a & a^2 &\cdots & a^{n-1} & a^n\\
a & 1 & a &\cdots & a^{n-2} & a^{n-1}\\
a^2 & a & 1 &\cdots & a^{n-3} & a^{n-2}\\
\vdots & \vdots & \vdots & \ddots &\vdots & \vdots \\
a^{n-1} & a^{n-2} & a^{n-3} & \cdots & 1 & a \\
a^n & a^{n-1} & a^{n-2} & \cdots & a & 1
\end{vmatrix}_{(n+1) \times (n+1)}
$$
My idea was to subtract the first row multiplied by the leading coefficient (of the other rows) from all others. This will put $ (1-a^2)$ on the main diagonal if repeated n times, while also making $D_{a,n}$ the determinant of an upper triangular matrix.
I.e:
$$
D_{a,n}=\begin{vmatrix}
1 & a & a^2 &\cdots & a^{n-1} & a^n\\
0 & 1-a^2 & a(1-a^2) &\cdots & a^{n-2}(1-a^2) & a^{n-1}(1-a^2)\\
0 & a(1-a^2) & 1-a^4 &\cdots & a^{n-3}(1-a^4) & a^{n-2}(1-a^4)\\
\vdots & \vdots & \vdots & \ddots &\vdots & \vdots \\
0 & a^{n-2}(1-a^2) & a^{n-3}(1-a^2) & \cdots & 1-a^{2(n-1)} & a(1-a^{2(n-1)}) \\
0 & a^{n-1}(1-a^2) & a^{n-2}(1-a^2) & \cdots & a(1-a^{2(n-1)}) & 1-a^{2n}
\end{vmatrix}_{(n+1) \times (n+1)}
\\
$$
$$
D_{a,n}=\begin{vmatrix}
1 & a & a^2 &\cdots & a^{n-1} & a^n\\
0 & 1-a^2 & a(1-a^2) &\cdots & a^{n-2}(1-a^2) & a^{n-1}(1-a^2)\\
0 & 0 & 1-a^2 &\cdots & a^{n-3}(1-a^4) & a^{n-2}(1-a^4)\\
\vdots & \vdots & \vdots & \ddots &\vdots & \vdots \\
0 & 0 & a^{n-3}(1-a^2)^2 & \cdots & 1-a^{2(n-2)} & a(1-a^{2(n-2)}) \\
0 & 0 & a^{n-2}(1-a^2) & \cdots & a(1-a^{2(n-2)}) & 1-a^{2(n-1)}
\end{vmatrix}_{(n+1) \times (n+1)}
\\ \vdots
$$
$$
D_{a,n}=\begin{vmatrix}
1 & a & a^2 &\cdots & a^{n-1} & a^n\\
0 & 1-a^2 & a(1-a^2) &\cdots & a^{n-2}(1-a^2) & a^{n-1}(1-a^2)\\
0 & 0 & 1-a^2 &\cdots & a^{n-3}(1-a^4) & a^{n-2}(1-a^4)\\
\vdots & \vdots & \vdots & \ddots &\vdots & \vdots \\
0 & 0 & 0 & \cdots & 1-a^2 & something \\
0 & 0 & 0 & \cdots & 0 & 1-a^2
\end{vmatrix}_{(n+1) \times (n+1)}
$$
$$
\implies D_{a,n}= (1-a^2)^{n}
$$
However, I don't believe this is rigorous enough proof to be used on an exam.
I'm hoping someone could point me in the right direction.
(I've thought of using induction, but it doesn't feel like the simplest approach)
(Also, what would "something" be if done the way I started?)
Edit: I just found out this is the determinant of a Toeplitz matrix, which has already been answered before here.
| Denote by $M$ the matrix in question (note that there are $n+1$ entries on the first row of $M$; thus $M$ is $(n+1)\times(n+1)$, not $n\times n$). Then
$$
\pmatrix{1\\ -a&1\\ &\ddots&\ddots\\ &&-a&1}M=\pmatrix{1&\ast&\cdots&\ast\\ &1-a^2&\ddots&\vdots\\ &&\ddots&\ast\\ &&&1-a^2}.
$$
Therefore $D_{a,n}=\det M=(1-a^2)^n$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3456479",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Prove that $W(n,\sum_{k=1}^{n}k^{2})=2(n-1)$ Let $F_{a}(n)$ be the digit sum of $n$ in base $a$,
define $W(a,b)=F_{a}(a^{\lceil\frac{\log{b}}{\log{a}} \rceil}-b)$,
prove that $\displaystyle\ W(n,\sum_{k=1}^{n}k^{2})=2(n-1)$ if $n−1 \in 6\mathbb{N} \pm 1$.
| For positive integers $n,k$, let
$$S(n,k)=\sum_{i=1}^{n}i^k$$
and for positive integers $m,b$, with $b>1$, let $D(b,m)$ be the sum of the base-$b$ digits of $m$.
Let $k=2$.
Thus, suppose $a$ is a positive integer such that $a \mid S(a,2)$, and let $b=a+1$.
Identically, we have
$$
S(n,2)
=
\sum_{i=1}^n i^2
=
\frac{n(n+1)(2n+1)}{6}
$$
hence
\begin{align*}
&a \mid S(a,2)\\[4pt]
\implies\;&a{\;|}\left(
\frac{a(a+1)(2a+1)}{6}
\right)\\[4pt]
\implies\;&6 \mid \left((a+1)(2a+1)\right)\\[4pt]
\implies\;&6 \mid \left(b(2b-1)\right)\\[4pt]
\implies\;&6 \mid b\;\;\text{or}\;\;\Bigl(2 \mid b\;\;\text{and}\;\;3 \mid (2b-1)\Bigr)\\[4pt]
\end{align*}
If $6 \mid b$, then
\begin{align*}
S(b,2)&=\frac{b(b+1)(2b+1)}{6}\\[4pt]
&=\frac{b^3}{3}+\frac{b^2}{2}+\frac{b}{6}\\[4pt]
&=
\left({\small{\frac{b}{3}}}\right)\!{\cdot}\,b^2
+
\left({\small{\frac{b}{2}}}\right)\!{\cdot}\,b^1
+
\left({\small{\frac{b}{6}}}\right)\!{\cdot}\,b^0
\end{align*}
hence
$$
D(b,S(b,2))
=
\left({\small{\frac{b}{3}}}\right)
+
\left({\small{\frac{b}{2}}}\right)
+
\left({\small{\frac{b}{6}}}\right)
=
b
$$
If $2 \mid b$ and $3 \mid (2b-1)$, then $b\equiv 2 \pmod3$, so
\begin{align*}
S(b,2)&=\frac{b(b+1)(2b+1)}{6}\\[4pt]
&=\frac{b^3}{3}+\frac{b^2}{2}+\frac{b}{6}\\[4pt]
&=
\left({\small{\frac{b+1}{3}}}\right)\!{\cdot}\,b^2
+
\left({\small{\frac{b-2}{6}}}\right)\!{\cdot}\,b^1
+
\left({\small{\frac{b}{2}}}\right)\!{\cdot}\,b^0
\end{align*}
hence
$$
D(b,S(b,2))
=
\left({\small{\frac{b+1}{3}}}\right)
+
\left({\small{\frac{b-2}{6}}}\right)
+
\left({\small{\frac{b}{6}}}\right)
=
b.
$$
Thus, for all subcases, we have $D(b,S(b,2))=b$.
So we are considering the summation $\sum_{i=1}^a i^2$ in base $a$. A quick observation yields
$$
a^{2+1} = \sum_{i=1}^a a^2 \geq \sum_{i=1}^a i^2 \geq a^2.
$$
See $W(b,S(b,2))=D(b,b^3-S(b,2))$
$\implies W(b,S(b,2))=3a+1-D(b,S(b,2))=2a$
and also note $a\in \{6t\pm 1\}$ then $a|S(a,2)$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3456752",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
} |
Asymptotic evaluations of a limit: $\lim_{x\to 0}\frac{\ln(1+x^3\cos 3x)}{2x^3(1-\sqrt{1+\cos 3x)}}$ I have this simple limit
$$\lim_{x\to 0}\frac{\ln(1+x^3\cos 3x)}{2x^3(1-\sqrt{1+\cos 3x})}$$
I have solved this with these steps:
$$\frac{\ln(1+x^3\cos 3x)}{2x^3(1-\sqrt{1+\cos 3x})}=\frac{\ln(1+x^3\cos 3x)(1+\sqrt{1+\cos 3x})}{2x^3(1-\sqrt{1+\cos 3x})\cdot(1-\sqrt{1+\cos 3x})}$$
$$=-\frac 12 \frac{\ln(1+x^3\cos 3x)}{x^3\cos 3x}\cdot (1+\sqrt{1+\cos 3x})\stackrel{x\to 0}{=}-\frac{1+\sqrt 2}{2}$$
My actual problem is to use the asymptotic evaluations: if I use $$\ln(1+x^3\cos 3x)\sim (1+x^3\cos 3x)$$
and
$$\sqrt[n]{1+\psi(x)}-1\sim \frac{\psi(x)}{n}$$
where $\psi(x)=\cos 3x$, I obtain
$$\frac{\ln(1+x^3\cos 3x)}{2x^3(1-\sqrt{1+\cos 3x})}\asymp\frac{(1+x^3\cos 3x)}{2x^3\left(-\frac 12\cos 3x\right)}$$
but is $x\to 0$ this limit diverges. I do not often use asymptotic evaluations but, it may be my fatigue, but I do not find the error at the moment. Thank you for your help.
| We have
*
*$\ln(1+x^3\cos 3x) \sim x^3\cos 3x$
moreover we have $\psi(x) \to 1$ and binomial expansion doesn't apply.
Therefore we obtain
$$\frac{\ln(1+x^3\cos 3x)}{2x^3(1-\sqrt{1+\cos 3x)}}\sim \frac{\cos 3x}{2(1-\sqrt{1+\cos 3x)}} \to\frac1{2(1-\sqrt 2)}$$
Anyway I suggest to proceed with caution for the asymptotic evaluation of limits since it can easily leads to some mistake.
In general it is better use little-o or big-O notation to avoid mistakes.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3457599",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Solve intial value problem using power series $xy''+y'+2y=0$ with y(1) =2, y'(1) =4.
What I tried:
\begin{equation}
xy'' +y'+2y = 0
\end{equation}
Let $y=\sum_{k=0}^{\infty}c_kx^k$, $y^\prime=\sum_{k=1}^{\infty}kc_kx^{k-1}$, $y^{\prime\prime}=\sum_{k=2}^{\infty}k(k-1)c_kx^{k-2}$
Then
\begin{equation}
x\sum_{k=2}^{\infty}k(k-1)c_kx^{k-2}+\sum_{k=1}^{\infty}kc_{k+1}x^{k-1}+2\sum_{k=0}^{\infty}c_kx^k=0
\end{equation}
\begin{equation}
\sum_{k=2}^{\infty}k(k+1)c_kx^{k-1}+\sum_{k=1}^{\infty}kc_kx^{k-1}+\sum_{k=0}^{\infty}2c_kx^k=0
\end{equation}
\begin{equation}
\sum_{k=0}^{\infty}k(k+1)c_{k+1}x^{k}+\sum_{k=0}^{\infty}(k+1)c_{k+1}x^{k}+\sum_{k=0}^{\infty}2kc_kx^{k}=0
\end{equation}
\begin{equation}
\sum_{k=0}^{\infty}[k(k+1)c_{k+1}+(k+1)c_{k+1}+2kc_k]x^k=0
\end{equation}
Therefore
\begin{equation}
k(k+1)c_{k+1}+(k+1)c_{k+1}+2kc_k=0 \text{ for }k\ge0
\end{equation}
So
\begin{equation}
c_{k+1}=\frac{-2k}{(k+1)^2}c_k
\end{equation}
This is how much I have tried. The answer is given to be:
$y=2+4(x-1)-4(x-1)^2+\frac{4}{5}(x-1)^3-\frac{1}{3}(x-1)^4+\frac{2}{15}(x-1)^5+.... $
How do I obtain this form. Pls help.
Question link: http://imgur.com/a/gTdCYbk
| We propose to solve $$xy''+y'+2y=0 ~~~(1)$$ by taylor series about $x=1$ as
$$y(x)=\sum_{k=0}^{\infty}~ y^{(k)}(1) \frac{(x-1)^k}{k!}, y^{(k)}(1)~ \text{denotes
$k$th derivative of $y(x)$ at $x=1$}~~~~(2)$$
solution where we are
given that $y(1)=2, y'(1)=4$. We differentiate (1) $n$ times as
$$xy^{(n+2)}+n y^{(n+1)}+2y^{(n)}=0 \implies y^{(n+2)}(1)=-(n+1)y^{(n+1)}(1)-2y^{(n)}(1)=0$$
$$\implies y^{(n)}(1)=-(n-1)y^{(n-1)}(1)-2y^{(n-2)}(1)$$
$$\implies y^{(2)}(1)=-8, y^{(3)}(1)=8, y^{(4)}(1)=-8, y^{{5}}(1)=16,...$$
Inserting these values in (2), we get
$$y(x)=2+4(x-1)-8\frac{(x-1)^2}{2!}+8\frac{(x-1)^3}{3!}-8\frac{(x-1)^{4}}{4!}+16 \frac{(x-1)^5}{5!}+.....$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3457927",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
prove that the series convereges and find its value $\frac{5^{k+1}+(-3)^k}{7^{k+2}}$ $$\sum_{k=0}^{\infty}\frac{5^{k+1}+(-3)^{k}}{7^{k+2}}$$
My first inclination is to use the divergence test.
$$\lim_{k\to\infty} \frac{5^{k+1}+(-3)^{k}}{7^{k+2}}$$
$$ \frac{5^{\infty}+(-3)^{k}}{7^{\infty}} \to \frac{5^{\infty}+(-3)^{\infty}}{7^{\infty}}$$
$$\frac{5^{\infty}+(-3)^{k}}{7^{\infty}} \to \frac{\infty+\infty}{\infty}$$
Since the sequence $a_k$ does not converge the series diverges by the diveregence test
| hint
We have
$$|5^{k+1}+(-3)^k | \le 5^{k+1}+3.5^{k+1}\le 5^{k+2}$$
thus
$$|u_n| \le (\frac 57)^{k+2}$$
Its sum is
$$\frac{5}{49}\sum_{k=0}^\infty (\frac 57)^k+\frac{1}{49}\sum_{k=0}^\infty(\frac{-3}{7})^k=$$
$$\frac{5}{49}\frac{1}{1-\frac 57}+\frac{1}{49}\frac{1}{1+\frac{3}{7}}=\frac{13}{35}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3458663",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Find the limit of $\sqrt[n]{n^2 + n}$ To find the limit I got the $\sqrt[3n]{n^2+n}$
Particularly, $\sqrt[3n]{n^2+n} \ge 1 \rightarrow \sqrt[3n]{n^2+n} = 1 + d_n$ where $d_n\ge 0$.
According to the Bernoulli's rule
$\sqrt{n^2+n} = (1+d_n)^n \ge d_n\cdot n \rightarrow d_n \le \frac{\sqrt{n^2+n}}{n}$
The $\frac{\sqrt{n^2+n}}{n} \rightarrow 1$, so $\lim d_n=1 $
So, $\lim\sqrt[n]{n^2+n} = \lim (1+d_n)^3 = \lim(1+3d_n^2+3d_n+d_n^3) =8$
However, $\sqrt[n]{n^2+n}$ tends to $1$. Where is the problem of my solution ? Can you give me a hint of how can I solve it with Bernoulli's rule ?
| How to fix your approach
You define $d_n=\sqrt[3n]{n^2+n}-1$. This then implies that $\sqrt[\large\color{#C00}{3}]{n^2+n}=(1+d_n)^n\ge nd_n$ (the $3$ was left out). Thus, $d_n\le\frac{\sqrt[3]{n^2+n}}n=\sqrt[\large3]{\frac{n+1}{n^2}}\to0$.
A Different Bernoulli Approach
Bernoulli's Inequality says
$$
\begin{align}
n^2+n
&\le\left(1+\sqrt{n}\right)^4\tag{expand}\\
&\le\left(1+\frac1{\sqrt{n}}\right)^{4n}\tag{Bernoulli}
\end{align}
$$
Therefore,
$$
\sqrt[\large n]{n^2+n}\le\left(1+\frac1{\sqrt{n}}\right)^4
$$
which tends to $1$ as $n\to\infty$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3460379",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 7,
"answer_id": 6
} |
Evaluate the following integral: $\int_{0}^{\pi/4} \frac{\sin\left(x\right)+\cos\left(x\right)}{9+ 16\sin\left(2x\right)} \, dx$ This is a question for the 2006 MIT Integration Bee that went unanswered by the contestants. I am not sure how to solve it either. I was only able to use the double angle formula to simplify the integral: $\sin\left({2x}\right) = 2 \sin\left(x\right) \cos\left(x\right)$
The final answer given was: $\frac{1}{20} \ln \left({3}\right)$
$$\int_{0}^{\frac{\pi}{4}} \dfrac{\sin\left(x\right)+\cos\left(x\right)}{9+ 16\sin\left(2x\right)} \, dx = $$
$$\int_{0}^{\frac{\pi}{4}} \dfrac{\sin\left(x\right)+\cos\left(x\right)}{9+ 32 \sin\left(x\right) \cos\left(x\right)} \, dx $$
| Note that \begin{align}
9+ 16\sin\left(2x\right)&=25+32\sin(x)\cos(x)-16\\
&=25-16\sin^2x-16\cos^2x+32\cos x\sin x\\
&=5^2-4^2(\sin x - \cos x)^2\\
&=(5-4\cos x+4\sin x)(5+4\cos x-4\sin x)
\end{align}
hence
\begin{align}
\frac{\sin x+\cos x}{(5-4\cos x+4\sin x)(5+4\cos x-4\sin x)}&=-\frac{1}{40}\frac{-4\cos x-4\sin x}{(5+4\cos x-4\sin x)}\\
&+\frac{1}{40}\frac{4\cos x+4\sin x}{(5-4\cos x+4\sin x)}
\end{align}
... the rest is shall be manageable: for each fraction the numerator is the derivative of the denominator, hence you need to recall $\int du / u = \ln(u)+k$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3461475",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 1
} |
Solution to the following Non linear Ode We have the following ODE:
$x'(t) = \frac{3x(t)ln(x(t))}{t^2-3t}$
$x(t_0) = x_0 ∈ R$
My professor claims that the solution is, for $t_0 = 6$ and $x_0 = 3$, $x(t) = 9e^{\frac{− 6 ln 3}{t}}$. Its derivative is $9e^{\frac{− 6 ln 3}{t}}\frac{6 ln 3}{t^2}$. But if we plug our solution $x(t)$ into $x'(t) = \frac{3x(t)ln(x(t))}{t^2-3t}$, we get a different function. How is this possible?
| There is at least an error in the work for the Laplace transform of the integral in $t$. Since both of these integrals are fairly simple, let's just do them.
\begin{align*}
\int \frac{\mathrm{d}x}{3 x \ln x} &= \frac{1}{3} \int \frac{\mathrm{d}u}{u} & &\left[u = \ln x, \mathrm{d}u = \frac{\mathrm{d}x}{x} \right] \\
&= \frac{1}{3} \ln |u| + C_1 \\
&= \frac{1}{3} \ln |\ln x| + C_1
\end{align*}
and
\begin{align*}
\int \frac{\mathrm{d}t}{t^2 - 3t} &= \int \frac{1/3}{t-3} - \frac{1/3}{t} \,\mathrm{d}t \\
&= \frac{1}{3} \ln |t-3| - \frac{1}{3}\ln |t| + C_2 \\
&= \frac{1}{3} \ln \left| \frac{t-3}{t} \right| + C_2
\text{.}
\end{align*}
Equating these two results, combining constants of integration, and multiplying through by $3$, we arrive at
\begin{align*}
\ln |\ln x(t)| &= \ln \left| \frac{t-3}{t} \right| + C \\
&= \ln \left| \frac{t-3}{t} \right| + \ln \mathrm{e}^C \\
&= \ln \left( \left| \frac{t-3}{t} \right| \mathrm{e}^C \right) \text{.}
\end{align*}
The real valued logarithm is injective (also called one-to-one), so
$$ |\ln x(t)| = \left| \frac{t-3}{t} \right| \mathrm{e}^C \text{.} $$
As long as $t > 3$, the fraction on the right is positive, so we may remove the absolute value bars. Since we are working on the connected part of the domain containing the initial condition $(t_0,x_0) = (6,3)$, we must have $t > 3$.
$$ |\ln x(t)| = \frac{t-3}{t} \mathrm{e}^C \text{.} $$
This right-hand side is always positive, so, since $x(t)$ is continuous, the sign of $\ln x(t)$ is always positive or always negative. At the given initial condition, $\ln(x_0) = \ln 3 > 0$, so $\ln x(t) > 0$ for the particular solution we are seeking. Thus, we may remove the absolute value bars from the left-hand side.
$$ \ln x(t) = \frac{t-3}{t} \mathrm{e}^C \text{.} $$
Applying initial conditions immediately, we have
$$ \ln 3 = \frac{6-3}{6} \mathrm{e}^C $$
so
$$ 2 \ln 3 = \mathrm{e}^C $$
and we obtain
$$ x(t) = \exp \left( \frac{t-3}{t} 2 \ln 3 \right) \text{.} $$
Now we check. Using that $t > 3$ during simplification, we find
$$ x'(t) = \frac{27^{1 - 2/t} \ln 9}{t^2} $$
and
$$ \frac{3 x(t) \ln x(t)}{t^2 - 3t} = \frac{27^{1 - 2/t} \ln 9}{t^2} \text{.} $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3461976",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Riemann sums for $x^3$ on [a,b] So far, I have this for the n'th right Riemann sum: $\sum_{i=1}^{n} \left(a+\frac{(b-a) i}{n}\right)^3 \left(\frac{b-a}{n}\right)$
The first component is the height of the rectangle and the second is the width
Obviously I know this should eventually become $\frac{b^4}{4}-\frac{a^4}{4}$ since that's what the integration rules will tell us, but I'm not sure how the simplification can work.
What I'm doing right now is summations of $\sum_{i=1}^{n}1$, $i$, $i^2$ and $i^3$ and multiplying that by various $\frac{(b-a)i}{n}$ terms when taken to similar powers.
The strategy is that if I take the $\lim n \to \infty$ of this products, all those summation formulas would have $n$ taken to powers that would cancel when the $\frac{(b-a)i}{n}$ terms are taken to similar powers and I'd just be left with numbers.
What I have is $\lim_{n \to \infty} \left( \frac{b-a}{n}(a^3)(n)+ 3\left(\frac{b-a}{n}\right)^2\left(\frac{n(n+1)}{2}\right) + 3\left(\frac{b-a}{n}\right)^3\left(\frac{n(n+1)(2n+1)}{6}\right) + \left(\frac{b-a}{n}\right)^4 \left(\frac{n^2(n+1)^2}{4}\right) \right)$ and I'm hoping that this could simplify into the correct $\frac{b^4}{4}-\frac{a^4}{4}$
Still by using Riemann sums, is there a more elegant way of doing this? (Or am I making it too complicated)
| First, we divide the area from a to b into n-equal subintervals. So, $\Delta x_i = \frac{(b-a)}{n}$ . From now on, let k = (b-a). So, as we know, the Riemann sum will be found by the summation $\sum_{i=1}^n f(x_i) \Delta x_i$, we need to find $x_i$. See $$x_0 = a$$ $$x_1 = a + \frac{k}{n}$$ $$x_2 = a + \frac{2k}{n}$$. So, one sees that the pattern illuminates $$x_i = a + \frac{ik}{n}$$ and that $$f(x_i) = (x_i)^3 = a^3 + \frac{3a^2ik}{n}\ + \frac{3ai^2k^2}{n^2}\ + \frac{i^3k^3}{n^3}.$$ Going back to our summation, we have $$\sum_{i=1}^n \frac{a^3k}{n}\ + \frac{3a^2ik^2}{n^2}\ + \frac{3ai^2k^3}{n^3}\ + \frac{i^3k^4}{n^4}.$$ This is equivalent to $$a^3k + \frac{3a^2k^2 (n+1)}{2n}\ + \frac{ak^3 (n+1)(2n+1)}{2n^2}\ + \frac{k^4(n+1)^2}{4n^2}$$ Now taking the limit as n approaches infinity, we have $$a^3k + \frac{3a^2k^2}{2}\ + ak^3 + \frac{k^4}{4}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3463880",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Solve: $\int \frac{\cos x}{x} dx$ Please help me solve this question. I have tried using series expansion, but I am not getting an finite answer.
| There is no closed form of this integral using elementary functions,but we can show an infinite series using Taylor expansion as an answer. In case you want to know more about the integral, you can find it here
$\int \frac{\cos(x)}{x} dx= \int( \frac{1-\frac{x^2}{2!}+\frac{x^4}{4!}-\frac{x^6}{6!}+\cdots}{x})dx = \int (\frac{1}{x}-\frac{x}{2!}+\frac{x^3}{4!}-\frac{x^6}{6!}+\cdots) dx= \ln(x) - \frac{x^2}{2\cdot2!}+\frac{x^4}{4\cdot4!}-\frac{x^6}{6\cdot6!}+\cdots$
So,
$\int \frac{\cos(x)}{x} dx = \ln(x) - \frac{x^2}{2\cdot2!}+\frac{x^4}{4\cdot4!}-\frac{x^6}{6\cdot6!}+\cdots$
Or more generally, $\int \frac{\cos(x)}{x} dx = \ln(x) + \Sigma_{r=1}^{\infty} (-1)^{r}\frac{x^{2r}}{2r\cdot(2r)!}$
Hope this helps !
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3464200",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Limit $\lim\limits_{n\to\infty}\left(\frac{n}{n^2+1}+\frac{n}{n^2+2}+\frac{n}{n^2+3}+\cdots+\frac{n}{n^2+n}\right)$ $\lim\limits_{n\to\infty}\left(\dfrac{n}{n^2+1}+\dfrac{n}{n^2+2}+\dfrac{n}{n^2+3}\cdots\cdots+\dfrac{n}{n^2+n}\right)$
Can we write it as following
$E=\lim\limits_{n\to\infty}\left(\dfrac{n}{n^2+1}\right)+\lim\limits_{n\to\infty}\left(\dfrac{n}{n^2+2}\right)+\lim\limits_{n\to\infty}\left(\dfrac{n}{n^2+3}\right)\cdots\cdots+\lim\limits_{n\to\infty}\left(\dfrac{n}{n^2+n}\right)\tag{1}$
Let's see what happens:-
$$\lim\limits_{n\to\infty}\left(\dfrac{n}{n^2+1}\right)$$
$$\lim\limits_{n\to\infty}\dfrac{\dfrac{1}{n}}{1+\dfrac{1}{n^2}}=0$$
In the same way for further terms, we will get $0$
Let's also confirm for general term
$$\lim\limits_{n\to\infty}\left(\dfrac{n}{n^2+n}\right)$$
$$\lim\limits_{n\to\infty}\left(\dfrac{\dfrac{1}{n}}{1+\dfrac{1}{n}}\right)=0$$
So the whole expression $E$ will be zero
But actual answer is $1$
Let's see what happens if we evaluate the original expression $OE=\lim\limits_{n\to\infty}\left(\dfrac{n}{n^2+1}+\dfrac{n}{n^2+2}+\dfrac{n}{n^2+3}\cdots\cdots+\dfrac{n}{n^2+n}\right)$
$OE=\lim\limits_{n\to\infty}\left(\dfrac{\dfrac{1}{n}}{1+\dfrac{1}{n^2}}+\dfrac{\dfrac{1}{n}}{1+\dfrac{2}{n^2}}+\dfrac{\dfrac{1}{n}}{1+\dfrac{3}{n^2}}\cdots\cdots+\dfrac{\dfrac{1}{n}}{1+\dfrac{1}{n}}\right)$
Now we can easily see that each term inside the bracket is tending to $0$, so can we say that sum of all terms upto infinity as well tends to zero?
I think we cannot because the quantity is not exactly zero, it is tending to zero, so when we add the values tending to zero upto infinity, we may not get zero.
But I got the following counter thought:-
$\lim\limits_{x\to0}\dfrac{(1+x)^\frac{1}{3}-1}{x}$
As we know $(1+x)^n=1+nx+\dfrac{n(n-1)}{2}x^2+\dfrac{n(n-1)(n-2)}{6}x^3\cdots\cdots\infty$ where $|x|<1$
$\lim\limits_{x\to0}\dfrac{\left(1+\dfrac{1}{3}x-\dfrac{1}{3}\cdot\dfrac{2}{3}\cdot\dfrac{1}{2}x^2+\dfrac{1}{3}\cdot\dfrac{2}{3}\cdot\dfrac{5}{3}\cdot\dfrac{1}{6}x^3\cdots\cdots\right)-1}{x}$
$\lim\limits_{x\to0}\dfrac{1}{3}-\dfrac{1}{3}\cdot\dfrac{2}{3}\cdot\dfrac{1}{2}x+\dfrac{1}{3}\cdot\dfrac{2}{3}\cdot\dfrac{5}{3}\cdot\dfrac{1}{6}x^2\cdots\cdots$
Now here also all the terms except $\dfrac{1}{3}$ are tending to $0$. So here also we can say that the whole quantity may not turn out to be zero as we are adding all terms upto infinity.
But surprisingly $\dfrac{1}{3}$ is the correct answer.
I am feeling very confused in these two things. Please help me.
| $\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
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$\ds{\bbox[5px,#ffd]{}}$
\begin{align}
&\bbox[5px,#ffd]{\sum_{k = 1}^{n}{n \over n^{2} + k}} =
n\sum_{k = 0}^{n - 1}{1 \over k + 1 + n^{2}}
\\[5mm] = &\
n\sum_{k = 0}^{\infty}\pars{{1 \over k + 1 + n^{2}} -
{1 \over k + n + 1 + n^{2}}} = n\pars{H_{n^{2} + n} - H_{n^{2}}}
\\[5mm] \stackrel{\mrm{as}\ n\ \to\ \infty}{\sim}\,\,\,&
n\braces{\bracks{\ln\pars{n^{2} + n} + \gamma +
{1 \over 2n^{2} + 2n}} -
\bracks{\ln\pars{n^{2}} + \gamma + {1 \over 2n^{2}}}}
\\[5mm] \stackrel{\mrm{as}\ n\ \to\ \infty}{\sim}\,\,\, &\
n\ln\pars{1 + {1 \over n}}
\stackrel{\mrm{as}\ n\ \to\ \infty}{\to}\,\,\,
\bbx{1} \\ &
\end{align}
$\ds{H_{z}}$ is a Harmonic Number and $\ds{\gamma}$ is the Euler-Mascheroni Constant.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3467697",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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For positive $a$, $b$, $c$, show $\frac{a}{a+b}+\frac{b}{b+c}+\frac{c}{c+a} \ge \frac{a+b+c}{a+b+c+\sqrt[3]{abc}}$
Prove that for every three positive numbers $a, b, c$:
$$\frac{a}{a+b}+\frac{b}{b+c}+\frac{c}{c+a} \ge \frac{a+b+c}{a+b+c+\sqrt[3]{abc}}$$
I tried using $\sum_{\mathrm{cyc}}$ but I haven't got far. I also tried:
$$\sqrt[3]{abc}=\frac{1}{3}\sqrt[3]{3a \cdot 3b \cdot 3c}$$
and according to the HM-GM inequality we can replace the root by:
$$\frac13 \frac{3}{\frac1{3a} +\frac1{3b}+\frac1{3c}}=\frac{1}{\frac13\left(\frac1a+\frac1b+\frac1c\right)}=\frac{3}{ \frac1a+\frac1b+\frac1c}$$
And that's it. Can you give me a hint or a solution for the question?
| I'll prove a stronger inequality:
Let $a$, $b$ and $c$ be positive numbers. Prove that:
$$\frac{a}{a+b}+\frac{b}{b+c}+\frac{c}{c+a}\geq\frac{a+b+c}{a+b+c-\sqrt[3]{abc}}.$$
Indeed, by C-S
$$\sum_{cyc}\frac{a}{a+b}=\sum_{cyc}\frac{a^2}{a^2+ab}\geq\frac{(a+b+c)^2}{\sum\limits_{cyc}(a^2+ab)}.$$
Thus, it's enough to prove that
$$\frac{(a+b+c)^2}{\sum\limits_{cyc}(a^2+ab)}\geq\frac{a+b+c}{a+b+c-\sqrt[3]{abc}}$$ or
$$\frac{a+b+c}{(a+b+c)^2-ab-ac-bc}\geq\frac{1}{a+b+c-\sqrt[3]{abc}}$$ or
$$ab+ac+bc\geq(a+b+c)\sqrt[3]{abc},$$ which says that in the last case our inequality is proven.
Now, let $$ab+ac+bc\leq(a+b+c)\sqrt[3]{abc}.$$
Thus, by C-S again we obtain:
$$\sum_{cyc}\frac{a}{a+b}=\sum_{cyc}\frac{a^2c^2}{a^2c^2+c^2ab}\geq\frac{(ab+ac+bc)^2}{\sum\limits_{cyc}(a^2b^2+a^2bc)}$$ and it remains to prove that
$$\frac{(ab+ac+bc)^2}{\sum\limits_{cyc}(a^2b^2+a^2bc)}\geq\frac{a+b+c}{a+b+c-\sqrt[3]{abc}}$$ or
$$\frac{(ab+ac+bc)^2}{(ab+ac+bc)^2-(a+b+c)abc}\geq\frac{a+b+c}{a+b+c-\sqrt[3]{abc}},$$ which is
$$ab+ac+bc\leq(a+b+c)\sqrt[3]{abc}$$ exactly.
Done!
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3472385",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
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Prove that the sum of the squares of three consecutive odd numbers plus one is divisible by 12 but not by 24. I tried to find a similar problem to help me with this question, but I couldn’t find anything. The only relevant thing I may know is that this may be a quadratic word problem; any ideas how to solve this?
| Let the numbers be $x-2$ , $x$ and $x+2$. Now ,
$$\begin{align}S &= (x-2)^2 + x^2 + (x+2)^2+1\\ &= x^2 +4 -4x+x^2+x^2+4+4x +1\\ & = 3x^2 + 9 \end{align}$$
Since $x$ is odd , we have $x=2k+1$
$$S = 3(2k+1)^2 + 0 \implies \color{#d05}{S = 12k^2 + 12k+12}$$
Clearly $S $ is divisible by $12$ .
Since $(k^2+k)$ is always divisible by $2$ , we take $(k^2+k) = 2\lambda$.So ,
$$S = 12(2\lambda) + 12 \implies \color{#2bf}{S = 24\lambda + 12}$$
So $S $ always leaves $12$ as the remainder when divided by $24.$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3472685",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Solution for $I = \int_1^2 \int_0^\sqrt{1-(1-x)^2} x/(x^2+y^2) \ \mathrm dy\ \mathrm dx$ The given integration is:
$$I = \int_1^2 \int_0^\sqrt{1-(1-x)^2} \dfrac{x}{x^2+y^2} \ \mathrm dy\ \mathrm dx$$
After substituting to polar coordinate, I get:
$$I = \int_0^{\pi/2} \int_0^{2\cos\theta} \cos\theta\ \mathrm dr\ \mathrm d\theta$$
And finally, I got $I = \pi/2$.
But the answer is 1/2.
Where I wrong is substitution to the second integration, which is surely $\pi/2$.
But I can't find the details.
Please let me find where I missed.
**Added : the region is upper right quarter of the circle of radius 1 centered at (1,0).
| This is an attempt, I can't really find a simple way with polar coordinates.
Assuming your domain is the circle of radius $1$ with center at $(1,0)$ your integral can be written as
$$
\int_{-1}^1 \int_{1 - \sqrt{1-y^2}}^{1 + \sqrt{1-y^2}} \frac{x}{x^2+y^2}dxdy =
2 \int_{0}^1 \int_{1 - \sqrt{1-y^2}}^{1 + \sqrt{1-y^2}} \frac{x}{x^2+y^2}dxdy = \\
\int_{0}^1 \left( \int_{1 - \sqrt{1-y^2}}^{1 + \sqrt{1-y^2}} \frac{1}{x^2+y^2}dx^2 \right) dy =
\int_{0}^1 \log \left( \frac{1 + \sqrt{1 - y^2}}{1 - \sqrt{1 - y^2}}\right) dy
$$
Check the computations... note that
$$
\log \left( \frac{1 + \sqrt{1 - y^2}}{1 - \sqrt{1 - y^2}} \right) = 2 \tanh^{-1} \left(\sqrt{1 - y^2} \right)
$$
So we have the integral
$$
2 \int_0^1 \tanh^{-1} \left(\sqrt{1 - y^2} \right)dy = \left. 2y \tanh^{-1} \left(\sqrt{1 - y^2} \right) \right|_{0}^1 - 2 \int_{0}^1 y d \left(\tanh^{-1} \left(\sqrt{1 - y^2} \right)\right)
$$
The boundary term goes to 0, so we are left with
$$
2 \int_0^1 \tanh^{-1} \left(\sqrt{1 - y^2} \right)dy = - 2 \int_{0}^1 y d \left(\tanh^{-1} \left(\sqrt{1 - y^2} \right)\right)
$$
The derivative of the inverse tangent should be
$$
\frac{d}{dy} \left(\tanh^{-1} \left(\sqrt{1 - y^2} \right)\right) = -\frac{1}{y^2} \frac{y}{\sqrt{1 - y^2}}
$$
Substituing this in my last integral gives you the arcsin function evaluated between 0 and 1, at 0 you get 0 while at 1 you'll get your result.
Update
If you want to use the polar coordinates, given your domain define
$$
\left\{
\begin{array}{l}
x - 1 = r \cos \theta \\
y = r \sin \theta \\
\end{array}
\right.
$$
The differential won't change and your integral can be written as
$$
\int_{0}^1 \int_{0}^{2\pi} \frac{1 + r \cos \theta}{r^2 +2r \cos \theta + 1} dr d\theta
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3476045",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 3
} |
$ABC$ rectangle in $A$ or $C$ iff $\frac{\sin(\alpha)+\sin(\gamma)}{\sin(\beta)}=\cot\left(\frac{\beta}{2}\right)$ The triangle $ABC$ is rectangle in $A$ or $C$ if and only if $\frac{\sin(\alpha)+\sin(\gamma)}{\sin(\beta)}=\cot\left(\frac{\beta}{2}\right)$, where $\alpha$ is the angle in $A$, $\beta$ is the angle in $B$ and $\gamma$ is the angle in $C$.
I have already managed to prove the ‘only if’ direction by putting $\alpha=\frac{\pi}{2}$ respectively $\gamma=\frac{\pi}{2}$ and using trigonometric formulas. However, I’m stuck for the ‘if’ direction. How could I proceed ?
Thanks for your help !
| $$\cot\frac{\beta}{2}=\frac{\cos\frac{\beta}{2}}{\sin\frac{\beta}{2}}=\frac{2\cos^2\frac{\beta}{2}}{\sin\beta}$$
so if you assume that equality, you get
$$\sin\alpha+\sin\gamma=2\cos^2\frac{\beta}{2}\Leftrightarrow \\
2\sin\frac{\alpha+\gamma}{2}\cos\frac{\alpha-\gamma}{2}=2\cos^2\frac{\beta}{2}\Leftrightarrow \\
2\cos\frac{\beta}{2}\cos\frac{\alpha-\gamma}{2}=2\cos^2\frac{\beta}{2} \\
\cos\frac{\alpha-\gamma}{2}=\cos\frac{\beta}{2} $$
since $\cos\frac{\beta}{2}>0$, and then note the equality of the cosines implies
$$\frac{\alpha-\gamma}{2}=\pm\frac{\beta}{2} \Leftrightarrow \\
\alpha-\gamma=\pm\beta $$
hence $\alpha$ or $\gamma$ is a right angle.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3476298",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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} |
$\begin{array}{|l}x^2-y^2=7 \\ x^2+xy+y^2=37\end{array}$ Any suggestions how to solve: $$\begin{array}{|l}x^2-y^2=7 \\ x^2+xy+y^2=37\end{array}$$
I can simplify the system and get a homogeneous polynomial of degree $2$, but I think it must have an easier way.
| Add the two to get:
$$y=\frac{44-2x^2}{x}$$
Substitute to the first:
$$x^2-\frac{(44-2x^2)^2}{x^2}=7 \Rightarrow \\
3x^4-169x^2+44^2=0 \Rightarrow \\
x^2=\frac{169\pm \sqrt{169^2-12\cdot 44^2}}{6}=\frac{169\pm 73}{6}=16;\frac{121}{3} \Rightarrow \\
x_{1,2}=\pm 4,x_{3,4}=\pm \frac{11}{\sqrt{3}}\\
y_{1,2}=\pm 3,y_{3,4}=\mp \frac{10}{\sqrt{3}}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3476644",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 6,
"answer_id": 5
} |
Show that $\lim\limits_{x\to 0} \frac{\sin x\sin^{-1}x-x^2}{x^6}=\frac1{18}$ Question: Show that $\lim\limits_{x\to 0} \dfrac{\sin x\sin^{-1}x-x^2}{x^6}=\dfrac{1}{18}$
My effort: $\lim\limits_{x\to 0} \dfrac{\sin x\sin^{-1}x-x^2}{x^6}=\lim\limits_{x\to 0} \dfrac{\dfrac{\sin x}{x} x\sin^{-1}x-x^2}{x^6}=\lim\limits_{x\to 0} \dfrac{\sin^{-1}x-x}{x^5}=\lim\limits_{x\to 0} \dfrac{\frac{1}{\sqrt{1-x^2}}-1}{5x^4}$ .
Is my approach correct?
| When $x$ is small, $\sin x \approx x-x^3/6+x^5/120$ and $\sin^{-1} x\approx x+x^3/6+3x^5/40$
Then $$\lim_{x \rightarrow 0} \frac{(x-x^3/6+x^5/120)(x+x^3/6+3x^5/40)-x^2}{x^{6}}=
\lim_{x\rightarrow 0}\frac{x^6/18+(.)x^8}{x^6}=\frac{1}{18}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3476795",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
} |
Formula for the inverse of a block-matrix Let $A, B, C, D ∈ R^{n×n}$. Show that if $A, B, C − D(B^{−1})A$, and $D − C(A^{−1}B$ are nonsingular then
$$\begin{bmatrix}
\mathbf{A} & \mathbf{B} \\
\mathbf{C} & \mathbf{D}
\end{bmatrix}^{-1} = \begin{bmatrix}
\mathbf{A}^{-1} + \mathbf{A}^{-1}\mathbf{B}\left(\mathbf{D} - \mathbf{CA}^{-1}\mathbf{B}\right)^{-1}\mathbf{CA}^{-1} &
-\mathbf{A}^{-1}\mathbf{B}\left(\mathbf{D} - \mathbf{CA}^{-1}\mathbf{B}\right)^{-1} \\
-\left(\mathbf{D}-\mathbf{CA}^{-1}\mathbf{B}\right)^{-1}\mathbf{CA}^{-1} &
\left(\mathbf{D} - \mathbf{CA}^{-1}\mathbf{B}\right)^{-1}
\end{bmatrix}$$
| Let $\begin{bmatrix}\mathbf{X}_1 & \mathbf{X}_2 \\\mathbf{X}_3 & \mathbf{X}_4 \end{bmatrix} \in \mathbb{R}^{2n\times2n}$ be a matrix such that,
$$\begin{bmatrix}\mathbf{A} & \mathbf{B} \\\mathbf{C} & \mathbf{D} \end{bmatrix} \begin{bmatrix}\mathbf{X}_1 & \mathbf{X}_2 \\\mathbf{X}_3 & \mathbf{X}_4 \end{bmatrix} = \mathbf{I}_{2n\times2n}.$$
Then, we have four matrix equations in four matrix unknowns,
\begin{align}
\mathbf{A}\mathbf{X}_1 + \mathbf{B}\mathbf{X}_3 &= \mathbf{I}_{n\times n}, \tag{1}\\
\mathbf{A}\mathbf{X}_2 + \mathbf{B}\mathbf{X}_4 &= \mathbf{0}_{n\times n}, \tag{2}\\
\mathbf{C}\mathbf{X}_1 + \mathbf{D}\mathbf{X}_3 &= \mathbf{0}_{n\times n},\tag{3}\\
\mathbf{C}\mathbf{X}_2 + \mathbf{D}\mathbf{X}_4 &= \mathbf{I}_{n\times n} \tag{4}.
\end{align}
Solve for $\mathbf{X}_1, \mathbf{X}_2, \mathbf{X}_3$ and $\mathbf{X}_4$, and use the Woodbury identity,
$$\left(\mathbf{A} + \mathbf{U}\mathbf{B}\mathbf{V} \right)^{-1} = \mathbf{A}^{-1} - \mathbf{A}^{-1}\mathbf{U}\left(\mathbf{B}^{-1}+\mathbf{V}\mathbf{A}^{-1}\mathbf{U}\right)^{-1}\mathbf{V}\mathbf{A}^{-1}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3478260",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Doubts in solving the integral $\int \frac{1}{\sqrt{1+\sin2x}}dx$ I tried to solve the integral $\int \frac{1}{\sqrt{1+\sin2x}}dx$ by using $1+\sin2x=(\sin x+\cos x)^2$ but got stuck. So I referred the solution in my book which is given below:
$$
I=\int \frac{1}{\sqrt{1+\sin2x}}dx
$$
$$
= \int \frac{1}{\sqrt{1-\cos(\frac \pi 2+2x)}}dx\tag1\\
= \int \frac{1}{\sqrt{2\sin^2(\frac \pi 4+x)}}dx\\
=\frac 1 {\sqrt2} \int \csc\left(\frac \pi 4+x\right)dx\\
= \color{red}{\frac 1 {\sqrt2} \log\left|\tan\left(\frac \pi 8+\frac x 2\right)\right|+C}(*)
$$
I wondered what if I replaced $\sin 2x$ by $\cos\left(\frac \pi 2 -2x\right)$ instead of $-\cos\left(\frac \pi 2 +2x\right)$ in step $(1)$. So I proceeded as follows:
$$
I=\int \frac{1}{\sqrt{1+\sin2x}}dx
$$
$$
= \int \frac{1}{\sqrt{1+\cos(\frac \pi 2-2x)}}dx\\
= \int \frac{1}{\sqrt{2\cos^2(\frac \pi 4-x)}}dx\\
=\frac 1 {\sqrt2} \int \sec\left(\frac \pi 4-x\right)dx\\
= \color{red}{\frac 1 {\sqrt2} \log\left|\tan\left(\frac {3\pi} 8-\frac x 2\right)\right|+C}(**)
$$
But I got a different result. Could you please explain the reason for this anomaly? Is it wrong to do a different replacement in step $(1)$? I think it shouldn't make any difference.
Further, could you please explain how to think we must be doing the replacement instead of using $1+\sin2x=(\sin x+\cos x)^2$ to solve this integral? I got this idea only after looking the solution.
*Using $\int \csc x dx=\log\left|\tan\left(x/2\right)\right|+C$
**Using $\int \sec x dx=\log\left|\tan\left(\frac \pi 4 +\frac x 2\right)\right|+C$
| $\sqrt{1+sin2x}=\sqrt{(sinx+cosx)^{2}}=\sqrt{2}\left| sin(x+\frac{\pi}{4}) \right|\\
z=x+\frac{\pi}{4}\Rightarrow dz=dx\\
\displaystyle\int\frac{dx}{\sqrt{1+sin2x}}=\frac{1}{\sqrt{2}}\int\frac{dz}{\left| sinz \right|}\\
1)sinz\gt 0\\
\displaystyle\int\frac{dz}{\left| sinz \right|}=\int\frac{sinzdz}{sin^{2}z}=\int\frac{sinzdz}{1-cos^{2}z}\\
(y=cosz)\to \displaystyle \int\frac{dz}{\left| sinz \right|}=-\int\frac{dy}{1-y^{2}}=-\frac{1}{2}\int(\frac{1}{1+y}+\frac{1}{1-y})dy=-ln\sqrt{1-y^{2}}+c\\
\displaystyle\int\frac{dx}{\sqrt{1+sin2x}}=-\frac{1}{\sqrt{2}}ln\left| sin(x+\frac{\pi}{4}) \right|+c\\
2)sinz\lt 0\\
\displaystyle\int\frac{dz}{\left| sinz \right|}=-\int\frac{dz}{sinz}=\frac{1}{\sqrt{2}}ln\left| sin(x+\frac{\pi}{4}) \right|+c$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3480588",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Getting co-ordinates from equation of a circle I am given the question:
How many points with integer coordinates lie on the circumference of circle $x^2 + y^2 = 5^2$
I know the general equation of a circle so the radius is 5. How do I go about it now ?
| We want to find the integer solutions of the equation $x^2+y^2=5^2$ .
Observe that we must have $-5\le x,y\le 5$ . WLOG assume $x\lt y.$
We get : $x^2+x^2\lt x^2+y^2\implies 2x^2 \lt 25 \implies x\le3$
Setting $x=0$ , we get $y = \pm5$.
Setting $x=\pm1$ , we get $y = \pm \sqrt{24}$ , which is not a solution.
Setting $x=\pm2$ , we get $y = \pm \sqrt{21}$ , which is not a solution .
Setting $x=\pm3$ , we get $y = \pm4$ .
Now similarly take the case of $y\lt x$ and find all coordinates.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3484265",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 0
} |
For $a$, $b$, $c$ the sides of a triangle, show $ 7(a+b+c)^3-9(a+b+c)\left(a^2+b^2+c^2\right)-108abc\ge0$
If $a$, $b$, and $c$ are the three sidelengths of an arbitrary triangle, prove that the following inequality is true, with equality for equilateral triangles.
$$ 7\left(a+b+c\right)^3-9\left(a+b+c\right)\left(a^2+b^2+c^2\right)-108abc\ge0 \tag{1}$$
In expanded form:
$$ 6\left(a^2b+a^2c+b^2a+b^2c+c^2a+c^2b \right)-\left(a^3+b^3+c^3\right)-33abc\ge0 \tag{2}$$
This a part of an ongoing research in triangle geometry and related to solving a cubic equation.
| By your work $$7(a+b+c)^2-9(a+b+c)(a^2+b^2+c^2)-108abc=$$
$$=\sum_{cyc}(12a^2b+12a^2c-2a^3-22abc)=\sum_{cyc}(a^2b+a^2c-2a^3+11(a^2b+a^2c-2abc))=$$
$$=\sum_{cyc}a^2(c-a-(a-b))+11(b^2c+a^2c-2abc))=$$
$$=\sum_{cyc}((a-b)(b^2-a^2)+11c(a-b)^2))=$$
$$=\sum_{cyc}(a-b)^2(11c-a-b)\geq\sum_{cyc}(a-b)^2(3c-a-b).$$
Now, let $a\geq b\geq c$.
Thus, since $3a-b-c>0$, $3b-a-c\geq b+c-a>0$ and $a-c\geq a-b$, we obtain:
$$\sum_{cyc}(a-b)^2(3c-a-b)\geq(a-b)^2(3c-a-b)+(a-c)^2(3b-a-c)\geq$$
$$\geq(a-b)^2(3c-a-b)+(a-b)^2(3b-a-c)=2(a-b)^2(b+c-a)\geq0.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3484343",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 3,
"answer_id": 2
} |
$\frac5x-\frac3y=-4$ and $\frac3{x^2}+\frac6{y^2}=\frac{11}3$
Solve the system:
$$\begin{array}{|l} \dfrac{5}{x}-\dfrac{3}{y}=-4 \\ \dfrac{3}{x^2}+\dfrac{6}{y^2}=\dfrac{11}{3} \end{array}$$
First, we have $x,y \ne 0$. If we multiply the first equation by $xy$ and the second by $3x^2y^2$, we get $5y-3x=-4xy$ and $9y^2+18x^2=11x^2y^2$. It doesn't seem I'm on the right track. What else can I try?
| $$\dfrac{5}{x}-\dfrac{3}{y}=-4 \tag 1$$
$$\dfrac{3}{x^2}+\dfrac{6}{y^2}=\dfrac{11}{3}\tag 2$$
Substitute $\frac1y = \frac13(\frac5x+4)$ from (1) into (2) to obtain,
$$\frac{59}{x^2} +\frac{80}x+21=0\implies\left(\frac1x+1\right)\left(\frac{59}{x}+21\right)=0$$
Solve to obtain the solutions $(-1,-3)$ and $(-\frac{59}{21}, \frac{177}{131})$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3484812",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 1
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Evaluate $\prod_{n=1}^{\infty}\left(\frac{\Gamma\left(2^n+\frac{1}{2}\right)}{a^n\Gamma(2^n)}\right)^{2^{-n}}$ An interesting infinite product with the closed form,
How can we show show that the formula is correct
$$\prod_{n=1}^{\infty}\left(\frac{\Gamma\left(2^n+\frac{1}{2}\right)}{a^n\Gamma(2^n)}\right)^{2^{-n}}=\frac{8}{a^2}\cdot \frac{\sqrt{\pi}}{e^2}$$
Let
$$\prod_{n=1}^{\infty}\left(\frac{\Gamma\left(2^n+\frac{1}{2}\right)}{a^n\Gamma(2^n)}\right)^{2^{-n}}=X$$
take the log
$$\sum_{n=1}^{\infty}\frac{1}{2^n}\log\frac{\Gamma\left(2^n+\frac{1}{2}\right)}{a^n\Gamma(2^n)}=\log X$$
$$\sum_{n=1}^{\infty}\frac{1}{2^n}\log\Gamma\left(2^n+\frac{1}{2}\right)-\sum_{n=1}^{\infty}\frac{1}{2^n}\log\Gamma\left(2^n\right)-\sum_{n=1}^{\infty}\frac{n}{2^n}\log a=\log X$$
$$\sum_{n=1}^{\infty}\frac{1}{2^n}\log\Gamma\left(2^n+\frac{1}{2}\right)-\sum_{n=1}^{\infty}\frac{1}{2^n}\log\Gamma\left(2^n\right)-\log a^2=\log X$$
$$\sum_{n=1}^{\infty}\frac{1}{2^n}\log\Gamma\left(2^n+\frac{1}{2}\right)-\sum_{n=1}^{\infty}\frac{1}{2^n}\log\Gamma\left(2^n\right)=\log(a^2X)$$
$$\sum_{n=1}^{\infty}\frac{1}{2^n}\log\Gamma\left(2^n+\frac{1}{2}\right)-\sum_{n=1}^{\infty}\frac{\log(2^n-1)!}{2^n}=\log(a^2X)$$
...
| As I said in the comments, evaluate the finite product and then take the limit as the number of terms, $N \to \infty$.
First let's state up front that
$$\frac{\Gamma{\left ( n+\frac12 \right )}}{\Gamma{(n)}} = \sqrt{\pi} n \frac{(2 n)!}{2^{2 n} (n!)^2} $$
so that we can write the product out to $N$ terms as follows:
$$\left (\sqrt{\pi}
\left ( \frac{2}{a^2} \right)^1 \frac{\left (2^{2} \right )!}{2^{2^{2}} \left ( 2^1 \right )!^2} \right )^{1/2^1} \left (\sqrt{\pi}
\left ( \frac{2}{a^2} \right)^2 \frac{\left (2^{3} \right )!}{2^{2^{3}} \left ( 2^2 \right )!^2} \right )^{1/2^2} \cdots \left (\sqrt{\pi}
\left ( \frac{2}{a^2} \right)^N \frac{\left (2^{N+1} \right )!}{2^{2^{N+1}} \left ( 2^N \right )!^2} \right )^{1/2^N}$$
which can be simplified to a single term as follows. First, the power of the $\sqrt{\pi}$ term is a geometric series:
$$\frac12 + \frac1{2^2} + \cdots +\frac1{2^N} = 1-\frac1{2^N}$$
Then the power of the $2/a^2$ term is another well-known series:
$$\frac12 + \frac{2}{2^2} + \cdots +\frac{N}{2^N} = 2-\frac{N+2}{2^N}$$
Finally, we get cancellations of the $\left ( 2^{n+1} \right )!$ terms in the denominator, leaving one of the corresponding $\left ( 2^{n} \right )!$ in the denominator. The product is then, exactly,
$$\left ( \sqrt{\pi} \right )^{1-2^{-N}} \left (\frac{2}{a^2} \right )^{2-(N+2) 2^{-N}} 2^{-(2 N+1)} \left ( \left ( 2^{N+1} \right )! \right )^{2^{-N}}$$
Now we consider the limit as $N \to \infty$. Here we use Stirling and find that
$$\begin{align} \left ( \left ( 2^{N+1} \right )! \right )^{2^{-N}} &\approx (2 \pi)^{2^{-(N+1)}} 2^{2 N+2} 2^{(N+1) 2^{-(N+1)}} e^{-2} \end{align} $$
so that, when plugged into the exact expression, and the limit as $N \to \infty$ is taken, produces as the product $8 \sqrt{\pi}/e^2 1/a^2$, as asserted.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3485371",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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} |
Finding a formula for a triangle similar to Pascal's. I need to find a formula expressed in binomial coefficients of the following triangle:
$$D_n^k=\begin{cases}
n \quad\quad\quad\quad\quad\text{ if } n=k \text{ or } k=0 \\
D_{n-1}^k+D_{n-1}^{k-1} \text{ otherwise}
\end{cases}$$
This triangle is the same as the Pascal's triangle, except for the base value ($n$ instead of $1$).
| As darij grinberg has pointed out, the formula is $$D_n^k=\binom{n+2}{k+1}-2\binom{n}{k}$$ Here is a simple proof that it fits the definition:
$$D_n^n=\binom{n+2}{n+1}-2\binom{n}{n}=n+2-2=n$$
$$D_n^0=\binom{n+2}{1}-2\binom{n}{0}=n+2-2=n$$
$$D_n^k=\binom{n+2}{k+1}-2\binom{n}{k}\stackrel{(*)}{=}\binom{n+1}{k+1}+\binom{n+1}{k}-2(\binom{n-1}{k}+\binom{n-1}{k-1})\\=\binom{n+1}{k+1}-2\binom{n-1}{k}+\binom{n+1}{k}-2\binom{n-1}{k-1}=D_{n-1}^k+D_{n-1}^{k-1}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3486179",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Alterative Sum of Squared Error formula proof The well-known formula of calculating Sum of Squared Error for a cluster is this:
SSE formula
where "c" is the mean and "x" is the value of an observation.
But this formula also brings the same result:
Alternative SSE formula
where "m" is the number of the observations and "y" takes in every iteration, values of the observations.
For example, if we have {3, 7, 8} , our mean "c" = 6 and:
Using the usual formula: (6-3)² + (6-7)² + (6-8)² = 14
Using the alternative formula: [ 1∕(2*3) ] × [ (3-3)² + (3-7)² + (3-8)² + (7-3)² + (7-7)² + (7-8)² + (8-3)² + (8-7)² + (8-8)²] = 14
Starting from the first formula, I 'm trying to prove the alternative, but I 'm lost. Can someone help me with the proof?
| Working with your numerical example:
$$14=SSE=(6-3)² + (6-7)² + (6-8)² = \\
\left(\frac{3+7+8}{3}-3\right)^2+\left(\frac{3+7+8}{3}-7\right)^2+\left(\frac{3+7+8}{3}-8\right)^2=\\
\frac{(7-3+8-3)^2}{3^2}+\frac{(3-7+8-7)^2}{3^2}+\frac{(3-8+7-8)^2}{3^2}=\\
\frac{2[(7-3)^2+(8-3)^2+(7-8)^2]+2[(7-3)(8-3)+(3-7)(8-7)+(3-8)(7-8)]}{3^2}=\\
\frac{3[(7-3)^2+(8-3)^2+(7-8)^2]}{3^2}-\frac{[(7-3)+(3-8)+(8-7)]^2}{3^2}=\\
\frac{2[(7-3)^2+(8-3)^2+(7-8)^2]}{2\cdot 3}-0=\\
\frac{\sum_{i}\sum_{j}dist(x_i,y_j)^2}{2\cdot3}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3488587",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Finding the remainder when $5^{55}+3^{55}$ is divided by $16$
Find the remainder when $5^{55}+3^{55}$ is divided by $16$.
What I try
$a^{n}+b^{n}$ is divided by $a+b$ when $n\in $ set of odd natural number.
So $5^{55}+3^{55}$ is divided by $5+3=8$
But did not know how to solve original problem
Help me please
| We can simplify the expression as follows: $5^{55} + 3^{55} \equiv (5^4)^{13}\cdot 5^3 + (3^4)^{13}\cdot 3^3 \pmod {16}\\
\equiv (9^2)^{13}\cdot 5^2\cdot 5 + 1^{13}\cdot 11\pmod {16}\\
\equiv 1^{13}\cdot9\cdot 5 + 11\pmod {16}\\
\equiv 8\pmod {16}$.
So the remainder is $8$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3490156",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 8,
"answer_id": 7
} |
recursive succession $a_{n+1}=\frac{a_n+2}{3a_n+2}, a_0>0$ I'm given this recursive succession:
$a_{n+1}=\frac{a_n+2}{3a_n+2}, a_0>0$.
This is what I've done:
$L=\frac{L+2}{3L+2} \rightarrow L_1=\frac{2}{3}$ and $L_2=-1$
if $a_0 >0 $ then $a_n>0 \forall n \in N \rightarrow
$ the succession is positive $\forall n \in N $and $L_2=-1$ is impossible.
if $a_n >0 $ then $a_n+2<3a_n+2 \rightarrow a_{n+1}=\frac{a_n+2}{3a_n+2} <1$ and then all the succession is beween $0$ and $1 $ excluded.
the succession jumps back and forth the value $\frac{2}{3}$ because:
if $a_n<\frac{2}{3}$ then $a_{n+1}>\frac{2}{3}$
if $a_n>\frac{2}{3}$ then $a_{n+1}<\frac{2}{3}$
Once I arrived at this point I don't know how to conclude that the limit of the succesion is if $\frac{2}{3}$
Can someone help me to understand how to procede?
| Note that if $a_0>0$ then $a_n>0$ for all $n\geq 1$ and
$$|a_{n+1}-2/3|=\frac{|a_n-2/3|}{|3a_n+2|}\leq \frac{|a_n-2/3|}{2}.$$
Hence,
$$|a_{n+1}-2/3|\leq \frac{|a_n-2/3|}{2}\leq
\frac{\frac{1}{2}|a_{n-1}-2/3|}{2}=\frac{|a_{n-1}-2/3|}{2^2}\leq \dots\leq \frac{|a_0-2/3|}{2^{n+1}}.$$
Therefore, we have that $\frac{|a_0-2/3|}{2^{n+1}}\to 0$ as $n\to \infty$, and we may conclude that $a_{n+1}\to 2/3$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3491959",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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If n approaches infinity, will the descending numbers in n number system divided by the ascending numbers of n number system be closer to an integer? We know that $$\frac{987654321}{123456789}$$ is very close to $8$. If $n$ is the total amount of numbers in the number system ($10$ being the total amount of numbers in the decimal number system), will the descending numbers in $n$ number system divided by the ascending numbers of $n$ number system be even closer to an integer (e.g. $$\frac{fedcba987654321}{123456789abcdef}$$ in the hexadecimal number system)?
| We have that in a base $b$ number system, the numerator is
$$1\cdot b^0 +2\cdot b^1+3\cdot b^2+\cdots +(b-1)b^{b-2}$$
while the denominator is
$$(b-1)\cdot b^0 +(b-2)\cdot b^1+(b-3)\cdot b^2+\cdots +b^{b-2}$$
Thus, you are asking if
$$\frac{\sum_{i=1}^{b-1}i b^{i-1}}{\sum_{i=1}^{b-1}(b-i) b^{i-1}}$$
gets closer to an integer as $b$ goes to infinity? It does, and in fact converges to $b-2$. First, break apart the numerator into two parts:
$$\sum_{i=1}^{b-1}i b^{i-1}=(b-1)b^{b-2}+(b-2)b^{b-3}+\sum_{i=1}^{b-3}i b^{i-1}$$
Now, this sum has a closed form:
$$\sum_{i=1}^{b-3}i b^{i-1}=\frac{((b-4) b+2) b^b+b^3}{(b-1)^2 b^3}=\frac{-4 b^{b+1}+b^{b+2}+2 b^b+b^3}{b^5-2 b^4+b^3}=O(b^{b-3})$$
This notation simply means that as $b$ gets large, the expression behaves like $b^{b-3}$ since the largest power on the top is $b^{b+2}$ while the largest power on the bottom is $b^5$. Thus, there exists a constant $C\in\mathbb{R^{+}}$ such that
$$\left|\sum_{i=1}^{b-3}i b^{i-1}\right|\leq C b^{b-3}$$
Now, we can bound the denominator by
$$\sum_{i=1}^{b-1}(b-i) b^{i-1}\geq b^{b-2}$$
Thus
$$\lim_{b\to\infty} \left|\frac{\sum_{i=1}^{b-3}i b^{i-1}}{\sum_{i=1}^{b-1}(b-i) b^{i-1}}\right|\leq \lim_{b\to\infty} \left|\frac{Cb^{b-3}}{b^{b-2}}\right|=\lim_{b\to\infty} \frac{C}{b}=0$$
Thus, we only have to care about
$$(b-1)b^{b-2}+(b-2)b^{b-3}$$
in the numerator. In fact, we can bound the denominator by $b^{b-2}+2b^{b-3}$. If we can show
$$\lim_{b\to\infty} \left|\frac{(b-1)b^{b-2}+(b-2)b^{b-3}}{\sum_{i=1}^{b-1}(b-i) b^{i-1}}-b+2\right|\leq\lim_{b\to\infty}\left|\frac{(b-1)b^{b-2}+(b-2)b^{b-3}}{b^{b-2}+2b^{b-3}}-b+2\right|=0$$
we are done. However, this simplifies incredibly well to
$$\lim_{b\to\infty}\left|\frac{(b-1)b^{b-2}+(b-2)b^{b-3}}{b^{b-2}+2b^{b-3}}-b+2\right|=\lim_{b\to\infty}\left|\frac{2}{2+b}\right|=0$$
and the conjecture is proved.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Can someone please explain why the roots are real as well as imaginary? I have this polynomial $x^3+x^2(3-4n)+x(4n^2-6n)-2n^2+6n-4=0$
If I put $n=3$, I get it as $x^3-9x^2+18x-4$
On computing the roots of $x^3+x^2(3-4n)+x(4n^2-6n)-2n^2+6n-4=0$ for any $n$ using Wolfram Alpha I am getting imaginary roots, here.
But if I put $n=3$ I get roots as real, see here.
Can someone please explain why the roots are real as well as imaginary?
| Let $x_1$, $x_2$ and $x_3$ be roots of our polynomial.
Easy to see that if all roots are reals so $$(x_1-x_2)^2(x_1-x_3)^2(x_2-x_3)^2\geq0,$$ while if there are two complex roots, so
$$(x_1-x_2)^2(x_1-x_3)^2(x_2-x_3)^2<0.$$
Now, let $x_1+x_2+x_3=3u,$ $x_1x_2+x_1x_3+x_2x_3=3v^2,$ where $v^2$ can be negative and $x_1x_2x_3=w^3.$
Thus, $$u=\frac{4n-3}{3},$$
$$v^2=\frac{4n^2-6n}{3},$$ $$w^3=2n^2-6n+4$$ and
$$(x_1-x_2)^2(x_1-x_3)^2(x_2-x_3)^2=27(3u^2v^4-4v^6-4u^3w^3+6uv^2w^3-w^6)=$$
$$=4n(16n^4-75n^3+140n^2-126n+54).$$
Now, we can understand for which value of $n$ we'll get all real roots.
I got that for any $n\geq0$ our equation has three real roots and for any $n<0$ our equation has an unique real root.
It follows from:
$$16n^4-75n^3+140n^2-126n+54=$$
$$=\left(4n^2-\frac{75}{8}n+6\right)^2+\frac{1}{64}(263n^2-864n+1152)>0.$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Find a function $f$ such that $\gcd(f(x)-f(y),x-y)\mid 2$ For all integers $x,y$. Find a function $f$ such that $\gcd(f(x)-f(y),x-y)\mid 2$ For all integers $x,y$.
This was a question proposed to me by my teacher to attempt,
I tried to find a function such that $f(x)-f(y)=x-y+1 $ but there are no solutions for that because$f(y)-f(z)=y-z+1$ therefore $f(x)-f(z)=x-z+2$ which is a contradiction
Excuse any bad english please, it is not my first language
| I claim that no such function exists.
Suppose that $f$ is such a function, and consider the values
$$
\begin{align*}
a & = f(4) - f(1) & b & = f(7) - f(4) & c & = f(10) - f(7).
\end{align*}
$$
If $3 \mid a$ then we would have that $\gcd(f(4) - f(1), 4 - 1) \geq 3$, which would be a contradiction. Thus $3 \nmid a$. Similarly, $3 \nmid b$ and $3 \nmid c$.
So $a$, $b$, and $c$ all leave a remainder of $1$ or $2$ when divided by $3$. Suppose that two consecutive values, say $a$ and $b$ have different remainders when divided by $3$. Then we have that $3 \mid a + b$, and so $\gcd(f(7) - f(1), 7 - 1) = \gcd(a + b, 6) \geq 3$. Thus we must have that $3 \nmid a + b$, and so $a$ and $b$ have the same remainder when divided by $3$. Similarly, $b$ and $c$ also have the same remainder when divided by $3$.
But then we have that $a$, $b$, and $c$ all have the same remainder when divided by $3$, and so $3 \mid a + b + c$. We thus have that $\gcd(f(10) - f(1), 10 - 1) = \gcd(a + b + c, 9) \geq 3$, a contradiction.
edit: We can use a similar idea to show that for any fixed positive integer $n$, there are no functions $f : \mathbb{Z} \to \mathbb{Z}$ such that
$$
\gcd(f(x) - f(y), x - y) < n
$$
for all integers $x$ and $y$.
Consider the values $f(0), f(n), f(2n), \dots, f(n^2)$. There are $n + 1$ of them, and so by the pigeonhole principle, two of them have the same remainder when divided by $n$. Let these be $f(kn)$ and $f(mn)$ so that $n \mid f(kn) - f(mn)$. We then have that
$$
\gcd(f(kn) - f(mn), kn - mn) \geq n,
$$
a contradiction.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Proving recursive function outputs $2^x \cdot {x \choose y}$ Consider following function
$f: \mathbb{N}\times\mathbb{N} \rightarrow \mathbb{N}$:
\begin{align*}
f(x ,y) =
\begin{cases}
0 & \text{if } x < y\\
2^x & \text{if } y = 0\\
2 \cdot (f(x - 1, y - 1) + f(x - 1, y)) & \text{else.}\\
\end{cases}
\end{align*}
I want to prove that $f(x, y) = 2^x \cdot {x \choose y}$, $x, y \in \mathbb{N}$, $x \geq y$.
I do see that $f(x - 1, y - 1)$ resembles ${x - 1 \choose y - 1}$ and $f(x - 1, y)$ resembles ${x - 1 \choose y}$, $\left({x \choose y} = {x - 1 \choose y - 1} + {x - 1 \choose y}\right)$ but I don't know where to go from this.
| Going directly, I would try induction on $x$.
If $x=0$, $$f(0,y) = \begin{cases} 1 & y = 0 \\ 0 & \text{otherwise}\end{cases}$$
as required.
Now fix $x$, and let $F(y) = f(x,y)$. You have $F(0) = 2^{x}$, and (by assumption), $$F(y) = 2\left(2^{x-1}\binom{x-1}{y-1} + 2^{x-1}\binom{x-1}{y}\right) = 2^{x}\binom{x}{y}$$
which works up until $F(x) = 2^x$, and then the rest are $0$.
| {
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"timestamp": "2023-03-29T00:00:00",
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Determine all value of $p,q\in\mathbb{N}$ such that : $2^{5}5^{3}=(p+1)(2q+p)$ Problem :
Determine all value of $p,q\in\mathbb{N}$ such that :
$$2^{4}5^{3}=(p+1)(2q+p)$$
My try :
$$2q+p-p-1=2q$$
So $2p+q$ odd or $p+1$ odd
I'm going to try all divisible :
$$1,4,5,8,25,125,10,50,250,20,200,500,40,200,1000,16,80,400,2000$$
So : we see that :
$p=0,q=1000$ is a solution
$p+1=16$ and $2q+p=125$ we find $(p,q)=(15,55)$
Also : $250=p+1$ and $2q+p=8$
Is my solution correct?
| Observe that $p+1 \mid 2^{4}5^{3}$, so we can write $p+1 = 2^{a}5^{b}$, where $a$ and $b$ are integers such that $0\le a\le 4$, $0\le b\le 3$.
Now, plug this into the original equation, we then have $2q+p = 2^{4-a}5^{3-b} \Rightarrow 2q = 2^{4-a}5^{3-b} - p \Rightarrow 2q = 2^{4-a}5^{3-b} - 2^a5^b +1$. Now observe that if neither $4-a$ or $a$ were zero, then $2q$ will become an odd number, a contradiction. Thus we must have either one of $4-a$ or $a$ that is equal to zero. Checking the two cases, we have the solutions $(p,q) = (15,55), (4,198), (24,28)$.
| {
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Find $\lim\limits_{n \to \infty} \left ( n - \sum\limits_{k = 1} ^ n e ^{\frac{k}{n^2}} \right)$. I have to find the limit:
$$\lim\limits_{n \to \infty} \bigg ( n - \displaystyle\sum_{k = 1} ^ n e ^{\frac{k}{n^2}} \bigg)$$
This is what I managed to do:
$$ e^{\frac{1}{n^2}} + e^{\frac{1}{n^2}} + ... + e^{\frac{1}{n^2}}
\le e^{\frac{1}{n^2}} + e^{\frac{2}{n^2}} + ...e^{\frac{n}{n^2}} \le
e^{\frac{n}{n^2}} + e^{\frac{n}{n^2}} + ... e^{\frac{n}{n^2}}$$
$$ n e^{\frac{1}{n^2}}
\le \displaystyle\sum_{k = 1} ^ n e ^{\frac{k}{n^2}} \le
ne^{\frac{1}{n}}$$
$$ -n e^{\frac{1}{n}}
\le - \displaystyle\sum_{k = 1} ^ n e ^{\frac{k}{n^2}} \le
- n e^{\frac{1}{n^2}}$$
$$ n - n e^{\frac{1}{n}}
\le n - \displaystyle\sum_{k = 1} ^ n e ^{\frac{k}{n^2}} \le
n - n e^{\frac{1}{n^2}}$$
Here I found that the limit of the left-hand side is equal to $-1$, while the limit of the right-hand side is $0$. So I got that:
$$-1 \le n - \displaystyle\sum_{k = 1} ^ n e ^{\frac{k}{n^2}} \le 0$$
And I cannot draw a conclusion about the exact limit. What should I do?
| Hint: $$a+a^2+...+a^n=\frac{a(a^n-1)}{a-1}$$
| {
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Find the limit of $a_n = n \sqrt{n}(\sqrt{n + 1} - a\sqrt{n} + \sqrt{n - 1})$ given that the sequence $(a_n)$ is convergent. I am given the sequence:
$$a_n = n\sqrt{n}(\sqrt{n + 1} - a\sqrt{n} + \sqrt{n - 1})$$
with $n \in \mathbb{N}^*$ and $a \in \mathbb{R}$. I have to find the limit of $(a_n)$ given that the sequence $(a_n)$ is convergent.
This is what I did:
We have:
$$a_n = n\sqrt{n}(\sqrt{n + 1} - a\sqrt{n} + \sqrt{n - 1})$$
$$a_n = n\sqrt{n^2+n}-an^2+n\sqrt{n^2-n}$$
$$a_n = n^2\sqrt{1+\dfrac{1}{n}} + n^2 \sqrt{1 - \dfrac{1}{n}} - an^2$$
$$a_n = n^2 \bigg ( \sqrt{1 + \dfrac{1}{n}} + \sqrt{1-\dfrac{1}{n}} - a \bigg )$$
The only way we could have $a_n$ convergent is if we would have the limit result in an indeterminate form. In this case, we need $\infty \cdot 0$. So we have:
$$\lim_{n \to \infty} \bigg ( \sqrt{1 + \dfrac{1}{n}} + \sqrt{1-\dfrac{1}{n}} - a \bigg ) = 0$$
And from that we can conclude that:
$$a = 2$$
So now that I found $a$, I must find the limit of the sequence. So this limit:
$$\lim_{n \to \infty} a_n = \lim_{n \to \infty} n\sqrt{n}(\sqrt{n + 1} - 2\sqrt{n} + \sqrt{n - 1})$$
I tried this:
$$\lim_{n \to \infty} a_n = \lim_{n \to \infty} n\sqrt{n^2+n}-2n^2+n\sqrt{n^2-n}$$
$$= \lim_{n \to \infty}( n\sqrt{n^2+n} - n^2) + \lim_{n \to \infty} (n\sqrt{n^2-n} -n^2)$$
And then I multiplied both of those limits with its respective conjugate, but after my calculations, it still results in an indeterminate form, only it's $\infty - \infty$ this time.
So, if my previous calculations aren't wrong, my question is how can I find this limit:
$$\lim_{n \to \infty} n\sqrt{n}(\sqrt{n + 1} - 2\sqrt{n} + \sqrt{n - 1})$$
| Hint
Rationalizing you get
$$\sqrt{n + 1} - 2\sqrt{n} + \sqrt{n - 1}= (\sqrt{n + 1} - \sqrt{n})-(\sqrt{n} - \sqrt{n - 1})\\=\frac{1}{\sqrt{n+1}+\sqrt{n}}- \frac{1}{\sqrt{n}+\sqrt{n-1}}\\
=\frac{\sqrt{n-1}-\sqrt{n+1}}{(\sqrt{n+1}+\sqrt{n})(\sqrt{n}+\sqrt{n-1})}$$
Repeat:
$$\sqrt{n-1}-\sqrt{n+1}=\frac{-2}{\sqrt{n-1}+\sqrt{n+1}}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3510658",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 1
} |
Determine the conditions for $n$ and $\theta\neq0+2k\pi$ such that $(\frac{1+\cos\theta+i\sin\theta}{1-\cos\theta+i\sin\theta})^n \in\mathbb{R}$ Let $z=1+\cos\theta+i\sin\theta=|z|(\cos\alpha+i\sin\alpha)$ and $z'=1-\cos\theta+i\sin\theta=|z'|(\cos\alpha'+i\sin\alpha')$.
My line of reasoning is to convert the numerator and denominator to polar form, then knowing that in order for $(\frac{z}{z'})^n$ to be real, we must have $\arg(\frac{z}{z'})^n \equiv 0 \pmod{\pi}$ or more succinctly $$n(\arg(z)-\arg(z')) \equiv 0 \pmod{\pi}$$.
Determining $\arg(z) = \alpha$ and $\arg(z') = \alpha'$: $$\tan{\alpha}=\frac{\sin{\theta}}{1+\cos{\theta}}=\frac{2\sin{\frac{\theta}{2}\cos{\frac{\theta}{2}}}}{2\cos^2{\frac{\theta}{2}}}=\frac{\sin{\frac{\theta}{2}}}{\cos{\frac{\theta}{2}}}=\tan{\frac{\theta}{2}}$$ thus $\alpha = \frac{\theta}{2}$. $$\tan{\alpha'}=\frac{\sin{\theta}}{1-\cos{\theta}}=\frac{2\sin{\frac{\theta}{2}\cos{\frac{\theta}{2}}}}{2\sin^2{\frac{\theta}{2}}}=\frac{\cos{\frac{\theta}{2}}}{\sin{\frac{\theta}{2}}}=\frac{\sin{\alpha'}}{\cos{\alpha'}}$$ thus $$\sin{\alpha'}\sin{\frac{\theta}{2}}=\cos{\alpha'}\cos{\frac{\theta}{2}} \Leftrightarrow \frac{1}{2}(\cos(\alpha'-\frac{\theta}{2})-\cos(\alpha'+\frac{\theta}{2}))=\frac{1}{2}(\cos(\alpha'-\frac{\theta}{2})+\cos(\alpha'+\frac{\theta}{2}))$$ therefore $$-\cos(\alpha'+\frac{\theta}{2})=\cos(\alpha'+\frac{\theta}{2})$$. Knowing that $$\cos(\pi+\alpha'+\frac{\theta}{2})=\cos(\alpha'+\frac{\theta}{2})$$ is impossible, we take $$\cos(\pi-\alpha'-\frac{\theta}{2})=\cos(\alpha'+\frac{\theta}{2})$$ hence $$\pi-\alpha'-\frac{\theta}{2} = \alpha'+\frac{\theta}{2} \Leftrightarrow \alpha' = \frac{\pi-\theta}{2}$$.
So now we have $$\frac{1+\cos\theta+i\sin\theta}{1-\cos\theta+i\sin\theta} = \frac{|z|(\cos{\frac{\theta}{2}}+i\sin{\frac{\theta}{2}})}{|z'|(\cos{\frac{\pi-\theta}{2}}+i\sin{\frac{\pi-\theta}{2}})}$$
Here's where I have some difficulties. Supposing
$$\frac{\theta}{2}-\frac{\pi-\theta}{2} \equiv 0 \pmod{\pi}$$ therefore $$\frac{2\theta - \pi}{2} \equiv 0 \pmod{\pi} \Leftrightarrow 2\theta - \pi \equiv 0 \pmod{2\pi}$$ thus $$\theta \equiv \frac{\pi}{2} \pmod{\pi}$$.
However, for all $n$ and using the above condition, $$(\frac{1+\cos\theta+i\sin\theta}{1-\cos\theta+i\sin\theta})^n = 1$$
It can't be for all $n$, there's something here I'm missing.
| Use the exponential form for complex numbers, it will be much shorter:
$$\frac{1+\cos\theta+i\sin\theta}{1-\cos\theta+i\sin\theta}=\frac{1+\mathrm e^{i\theta}}{1-\mathrm e^{-i\theta}}=\frac{\mathrm e^{\tfrac{i\theta}2}}{\mathrm e^{-\tfrac{i\theta}2}}\,\frac{\mathrm e^{\tfrac{i\theta}2}+\mathrm e^{-\tfrac{i\theta}2}}{\mathrm e^{\tfrac{i\theta}2}-\mathrm e^{-\tfrac{i\theta}2}}=\mathrm e^{i\theta}\,\frac{\cos\cfrac\theta2}{i\sin\cfrac\theta 2},$$
which shows that
$\arg\Bigl(\dfrac{1+\cos\theta+i\sin\theta}{1-\cos\theta+i\sin\theta}\Bigr)=\theta-\frac\pi 2$.
The $n$-th power of this number will be a real number if and only if
$$n\Bigl(\theta-\frac\pi 2\Bigr)\equiv 0\pmod \pi\iff \theta-\frac\pi 2 \equiv 0\pmod {\frac\pi n}\iff \theta\equiv \frac\pi 2 \pmod{\frac\pi n}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3515051",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Bernoulli's equation(differential equations) Solve the following equation:
$$\sec^2y\frac{dy}{dx}+x \tan y =x^3$$
I have tried like this:
Let, tan y =z
$$\implies \sec^2 y\frac{dy}{dx}=\frac{dz}{dx}\\$$
$$\therefore \frac{dz}{dx}+zx=x^3\\$$
$$I.F.=e^{\int{xdx}}=e^{\frac{x^2}{2}}\\$$
$$\therefore ze^{\frac{x^2}{2}} =\int{x^3 e^{\frac{x^2}{2}}dx}\\$$
$$=2(\frac{x^2}{2} e^{\frac{x^2}{2}}-e^{\frac{x^2}{2}})+c\\$$
$$\therefore \tan y e^{\frac{x^2}{2}}=e^{\frac{x^2}{2}}(x^2-2)+c\\$$
$$\implies \tan y=x^2-2+ce^{\frac{-x^2}{2}}$$
But in my book the answer is:
$$\tan y=x^3-3x^2+6x-6+ce^{-x}$$
I can't understand where is my mistake..please check this..
| $$(\tan y )'+x \tan y =x^3$$
Substitute $z=\tan y$
$$z'+xz=x^3$$
$$(ze^{x^2/2})'=x^3e^{x^2/2}$$
$$ze^{x^2/2}=\int x^3e^{x^2/2} dx$$
Substitute $u=\dfrac {x^2}2$
$$ze^{x^2/2}=2\int ue^{u} du$$
$$ze^{x^2/2}=2( ue^{u}.-e^u)+C $$
$$z=2(\frac {x^2}2-1)+Ce^{-x^2/2}$$
Finally:
$$ \boxed {\tan y= {x^2}-2+Ce^{-x^2/2}}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3515424",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
find the limit for this $(a_{n})_{n \in \mathbb{N}}$, and $a_{0}>0 \\$ and
$a_{n+1}=a_{n}+\frac{1}{a_{n}^2+a_{n}+1}$
Find the limit of: $\lim_{n \to \infty} \frac{a_{n}^3}{n}$
please help me! :D
Can i solve this whith Cesaro-Stolz theorem or what else?
| $$a_{n+1}^3-a_n^3 = (a_{n+1}-a_n)(a_{n+1}^2+a_{n+1} a_{n}+a_n^2) = \frac{a_{n+1}^2+a_{n+1}a_n+a_{n}^2}{a_n^2+a_n+1}$$
and by letting $\delta_n=a_{n+1}-a_n$
$$\begin{eqnarray*} a_{n+1}^2+a_{n+1}a_n+a_n^2 &=& (a_n+\delta_n)^2+(a_n+\delta_n)a_n+a_n^2\\&=&3a_n^2+3\delta_n a_n+\delta_n^2\end{eqnarray*} $$
such that
$$ a_{n+1}^3-a_n^3 = \frac{3a_n^2+\frac{3a_n}{a_n^2+a_n+1}+\frac{1}{(a_n^2+a_n+1)^2}}{a_n^2+a_n+1}.$$
If you manage to prove that $a_n\to +\infty$ the previous line gives $a_{n+1}^3-a_n^3\to 3$, then Stolz-Cesàro proves $a_n\sim\sqrt[3]{3n}$.
Since $x^2+x+1$ has a negative discriminant we have that $a_n^2+a_n+1$ is strictly positive, so the sequence $\{a_n\}_{n\geq 0}$ is increasing and the limit $\lim_{n\to +\infty} a_n$ is either a positive constant $C$ or $+\infty$. On the other hand there is no positive constant $C$ such that
$$ C = C + \frac{1}{C^2+C+1}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3515974",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Can the quadratic formula be explained intuitively? Most people know what the quadratic formula is so I won’t post it here (Besides, I don’t know how to properly post formulas in general).
I was wondering if there is an intuitive explanation as to why the quadratic formula is structured the way it is.
| The development of the quadratic formula is based on solving the quadratic equation in the form $$ax^2+bx+c=0.$$
By completing the square, we have
\begin{align}
ax^2+bx+c&=0\\
ax^2+bx&=-c\\
x^2+\frac{b}{a}x&=-\frac{c}{a}\\
x^2+\frac{b}{a}x\color{blue}{+\left(\frac{b}{2a}\right)^2}&=\color{blue}{\left(\frac{b}{2a}\right)^2}-\frac{c}{a}\\
\left(x+\frac{b}{2a}\right)^2&=\frac{b^2-4ac}{4a^2}\\
x+\frac{b}{2a}&=\pm\sqrt{\frac{b^2-4ac}{4a^2}}\\
x&=-\frac{b}{2a}\pm\frac{\sqrt{b^2-4ac}}{2a}\\
x&=\frac{-b\pm\sqrt{b^2-4ac}}{2a}
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3516253",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 1
} |
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