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then maximum and minimum value of $x+y$ if $x,y\in R$ and $x^3+y^3=2\;,$ then maximum and minimum value of $x+y$
using $\displaystyle \frac{x^3+y^3}{2}\geq \left(\frac{x+y}{2}\right)^3$
So $(x+y)^3\leq 2^3$ so $x+y\leq 2$
could some help me to find minimum value, thanks
|
From
$$
2 = x^3 + y^3 = (x+y)(x^2-xy+y^2) = (x+y) \frac{(x-y)^2+x^2+y^2}2
$$
it follows that $x+y> 0$. On the other hand, for arbitrary $t > 0$
$$
x =-t \quad , \quad y = \sqrt[3]{2+t^3}
$$
satisfies $x^3+y^3 =2$ and
$$
x + y = \sqrt[3]{2+t^3} - t = \frac{2}{(\sqrt[3]{2+t^3})^2 + t \sqrt[3]{2+t^3} + t^2} \to 0
$$
for $t \to \infty$. Therefore the infimum is zero and a minimum
does not exist.
|
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|
Proving $\left(\frac{a}{a+b+c}\right)^2+\left(\frac{b}{b+c+d}\right)^2+\left(\frac{c}{c+d+a}\right)^2+\left(\frac{d}{d+a+b}\right)^2\ge\frac{4}{9}$ The inequality:
$$\left(\frac{a}{a+b+c}\right)^2+\left(\frac{b}{b+c+d}\right)^2+\left(\frac{c}{c+d+a}\right)^2+\left(\frac{d}{d+a+b}\right)^2\ge\frac{4}{9}$$
Conditions: $a,b,c,d \in \mathbb{R^+}$
I tried using the normal Cauchy-Scharwz, AM-RMS, and all such..
I think I can do it using some bash method like expanding the LHS and then maybe to go with homogeneity, normalization and then Muirhead, Jensen or something I dont know since I didn't go that way..
But can someone help me with a nice elegant solution. This is an olympiad question I was trying to solve, but couldn't manage an elegant solution.
|
By Holder inequality,we have
$$\left(\sum\dfrac{a^2}{(a+b+c)^2}\right)\left(\sum a(a+b+c)\right)^2\ge \left(\sum_{cyc} a^{\frac{4}{3}}\right)^3$$
therefore
$$\sum\dfrac{a^2}{(a+b+c)^2}\ge\dfrac{(a^{4/3}+b^{4/3}+c^{4/3}+d^{4/3})^3}{[(a+c)^2+(b+d)^2+(a+c)(b+d)]^2}$$
use Holder we have
$$a^{\frac{4}{3}}+c^{\frac{4}{3}}\ge 2\left(\dfrac{a+c}{2}\right)^{\frac{4}{3}}$$
$$b^{\frac{4}{3}}+d^{\frac{4}{3}}\ge 2\left(\dfrac{b+d}{2}\right)^{\frac{4}{3}}$$if setting
$$t^3=\dfrac{(a+c)}{b+d}$$,we have
$$\sum_{cyc}\dfrac{a^2}{(a+b+c)^2}\ge\dfrac{1}{2}\cdot\dfrac{(t^4+1)^3}{(t^6+t^3+1)^2}$$
it remain to show that
$$9(t^4+1)^3\ge 8(t^6+t^2+1)^2,t>0$$
or
$$9\left(t^2+\dfrac{1}{t^2}\right)^3\ge 8\left(t^3+\dfrac{1}{t^3}+1\right)^2$$
let $u=t+\dfrac{1}{t}\ge 2$ the above ineuqality becomes
$$9(u^2-2)^2\ge 8(u^3-3u+1)^2$$
or
$$(u-2)^2(u^4+4u^3+6u^2-8u-20)\ge 0$$
since
$$u^4+4u^4+6u^2-8u-20=u^4+4u^2(u-2)+4u(u-2)+10(u^2-2)\ge 0$$
|
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|
Solving Linear Equations with Modulo I'm having trouble approaching the following problem:
Solve the following system of equation -
$$3x+5y\equiv 14\mod 17$$
$$7x+3y\equiv 6\mod 17$$
Multiplying the top by $3$ and the bottom by $5$, I get $x = -3/5$ and $y = 51/15$, but I'm unsure how to apply (mod $17$) with $2$ variables
|
So what you do is elimination of variables, and then deal with it differently:
start by eliminating $y$: multiply the first equation by three, second equation by five, then you get:
$$
9x+15y \equiv 42 (\equiv 8) \mod 17 ; 35x+15y \equiv 30 (\equiv 13) \mod 17
$$
Now, subtract the first equation from the second, and you get:
$$
26x \equiv 5 \mod 17
$$
This has to be solved, but we can do this through the reverse-Euclid algorithm: we know that there exist integers $x,y$ such that $17x+26y=1$ by Bezout's theorem. So the question is what are these $x$ and $y$?
To do this, note that:
$$
26 = (17 \times 1)+9 ; 17 = (9 \times 1) + 8; 9 = (8 \times 1)+ 1
$$
So we cleverly reverse this procedure:
$$
1 = 9-(8 \times 1) = (26-17 \times 1) - (17-9 \times 1) = (26-17)- (17-(26-17)) = 26-17-17+26-17 = 2 \times 26-3 \times 17
$$
Therefore, $x=2$ satisfies the equation $26x \equiv 1 \mod 17$. Multiply by $5$ and we see that $x=2 \times 5=10$ satisfies $26x \equiv 5 \mod 17$.
Hence, $x=10$ is a solution. Substituting this in the second equation, we see that $70+3y \equiv 6 \mod 17$, which simplifies to $3y \equiv -64 \mod 17$, or $3y \equiv 4 \mod 17$. You can check again via reverse Euclid:
$$
17 = (3 \times 5)+2 ; 3 = (2 \times 1)+1
$$
Therefore
$$
1 = 3-(2 \times 1) = 3 - (17 - 3 \times 5) = 3 \times 6-17 \times 1
$$
So $y=6$ satisfies $3y \equiv 1 \mod 17$, so multiply by four to get $3y \equiv 4 \mod 17$ (for $y=24$, or $y=7$ when reducing mod $17$).
So the answer is $x=10,y=7 \mod 17$. Let's check this:
$$
3x+5y = 30+35 = 65 \equiv 14 \pmod{17} \\
7x+3y = 70+21 = 91 \equiv 6 \pmod{17}
$$
Hence the equation is solved.
|
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|
Solve the equation ${\sqrt{4x^2 + 5x+1}} - 2{\sqrt {x^2-x+1}} = 9x-3$ I tried factoring the expression inside the square root, but that does not seem to help. Squaring the equation makes it even more terrible.
Can anyone provide a hint about what should be done?
|
Notice that
$$\left({\sqrt{4x^2 + 5x+1}} - 2{\sqrt {x^2-x+1}}\right) \left({\sqrt{4x^2 + 5x+1}} + 2{\sqrt {x^2-x+1}}\right) = 9x-3.$$
Thus,
$$9x-3 = \left(9x-3\right)\left({\sqrt{4x^2 + 5x+1}} + 2{\sqrt {x^2-x+1}}\right) \implies $$
$$9x-3 = 0 \implies x=1/3$$ or
$${\sqrt{4x^2 + 5x+1}} + 2{\sqrt {x^2-x+1}} = 1.$$
But the minimal value of $x^2-x+1$ is $3/4$, which implies that
$$ 2{\sqrt {x^2-x+1}} \geq \sqrt{3} > 1.$$
Therefore $x =1/3$ is the only real solution.
|
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|
the value of $\frac{100^2}{100!}+\sum^{100}_{k=1}|(k^2-3k+1)S_{k}|$ is Let $S_{k},$ where $k=1,2,3,\cdots \cdots ,100$ denote the sum of the infinite geometric series whose first term is $\displaystyle \frac{k-1}{k!}$ and the common ratio is $\displaystyle \frac{1}{k},$ then the value of $\displaystyle \frac{100^2}{100!}+\sum^{100}_{k=1}|(k^2-3k+1)S_{k}|$ is
$\displaystyle S_{k} = \frac{a}{1-r} = \frac{(k-1)\cdot k}{k!\cdot (k-1)} = \frac{1}{(k-1)!}$
So $\displaystyle \sum^{100}_{k=1}\bigg|(k^2-3k+1)S_{k}\bigg| = \sum^{100}_{k=1}\bigg|\frac{(k-1)^2-k}{(k-1)!}\bigg| = \sum^{100}_{k=1}\bigg|\frac{(k-1)}{(k-2)!}-\frac{k}{(k-1)!}\bigg| $
wan,t be able to go further, could some help me with this, thanks
|
$$\frac{k-1}{(k-2)!}-\frac{k}{k!}=\frac{(k - 2) k^2}{k!}>0 \implies \bigg|\frac{(k-1)}{(k-2)!}-\frac{k}{(k-1)!}\bigg|=\frac{(k-1)}{(k-2)!}-\frac{k}{(k-1)!}$$
Then as @Archis said use telescoping sum.
|
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|
How can I prove this inequality in a triangle: $\frac{1}{(b+c-a)} +\frac{1}{(c+a-b)} +\frac{1}{(a+b-c)} \gt \frac{1}{a} +\frac{1}{b} +\frac{1}{c}$?
If $a,b,c$ are the sides of a triangle then show that-
$\frac{1}{(b+c-a)} +\frac{1}{(c+a-b)} +\frac{1}{(a+b-c)} \gt \frac{1}{a} +\frac{1}{b} +\frac{1}{c}$
I got this problem from an old book on algebra. I've been trying for a long time but can't reach the answer.
My try on the question-
We know that in a triangle, the sum of two sides is always greater than the third side
$b+c\gt a$ $\implies$ $b+c-a \gt 0$
$c+a\gt b$ $\implies$ $c+a-b\gt 0$
$a+b \gt c$ $\implies$ $a+b-c\gt 0$
Also,
$\frac {1}{(b+c)} \lt \frac{1}{a}$
$\frac{1}{(c+a)} \lt \frac{1}{b}$
$\frac{1}{(a+b)} \lt \frac{1}{c}$
Adding,
$\frac{1}{(b+c)} + \frac{1}{(c+a)} +\frac{1}{(a+b)} \lt \frac{1}{a} + \frac{1}{b} + \frac{1}{c}$
How do I proceed further? All genuine answers are welcome :)
|
Your inequality is wrong! Try $a=b=c$.
It should be
$$\frac{1}{b+c-a} +\frac{1}{c+a-b} +\frac{1}{a+b-c} \geq \frac{1}{a} +\frac{1}{b} +\frac{1}{c}$$
For the proof we can use the following way.
Let $a+b+c=3u$, $ab+ac+bc=3v^2$ and $abc=w^3$.
Since our inequality is fifth degree, we see that $f(w^3)\geq0,$ where $f$ is a linear function,
which says that it's enough to prove our inequality for an extreme value of $w^3$,
which happens in the following cases.
*
*$b=c=1$.
We obtain:
$$(a-1)^2\geq0;$$
*$c\rightarrow0^+$ and $b=1$.
We obtain:
$$(a+1)(a-1)^2\geq0;$$
*$a+b-c\rightarrow0^+$
In this case the inequality is obvious.
Done!
|
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|
How to derive this formula and provide a proof? I have a formula that "filters" out values of x where x mod 3 is 0:
$$f(x) := 3\cdot\left\lfloor \frac x2\right\rfloor\, +\,(x\mathrm{\,mod\,} 2)\,+\,1$$
e.g.
f(0) = 1
f(1) = 2
f(2) = 4
f(3) = 5
f(4) = 7
f(5) = 8
f(6) = 10
etc.
I've come up with this formula through intuition, but I'm trying to figure out how to create a proof and also how to derive the solution.
|
A recursive definition would be $f(0)=1$ and
$$
f(x+1)=\begin{cases}
f(x)+1 & \text{if $3\nmid f(x)+1$}\\[4px]
f(x)+2 & \text{if $3\mid f(x)+1$}
\end{cases}
$$
Let's examine your function $f$.
We should have
$$
0<f(x+1)-f(x)\le2
$$
and $f(x+1)-f(x)=2$ if and only if $f(x)+1$ is a multiple of $3$.
We have
$$
f(x+1)-f(x)=
3\left(\left\lfloor\frac{x+1}{2}\right\rfloor-
\left\lfloor\frac{x}{2}\right\rfloor
\right)+
((x+1)\bmod 2)-(x\bmod 2)
$$
If $x$ is even, then $\lfloor(x+1)/2\rfloor=\lfloor x/2\rfloor$ and $((x+1)\bmod 2)-(x\bmod 2)=1$, so $f(x+1)-f(x)=1$.
If $x$ is odd, say $x=2k+1$, we have
$$
3\left(\left\lfloor\frac{2y+2}{2}\right\rfloor-
\left\lfloor\frac{2y+1}{2}\right\rfloor
\right)=3(y+1)-3y=3
$$
and $((x+1)\bmod 2)-(x\bmod 2)=-1$, so $f(x+1)-f(x)=2$.
Now suppose that $f(x)+1$ is a multiple of $3$, so
$$
3\left\lfloor\frac{x}{2}\right\rfloor+
(x\bmod 2)+1+1=3k
$$
This means that $x\bmod2=1$, that is, $x$ is odd. In this case, $f(x+1)=f(x)+2$, as required.
|
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|
Find all reals $x$,$y$ satisfying the following equation:
Find all positive reals $x,y \in \mathbb{R}^+$ satisfying:
$$\frac{x^9}{y} + \frac{y^9}{x} = 10-\frac{8}{xy}$$
Since this involves higher exponents I am unable to tackle this problem. Please help me.
|
Just to give a non-AM-GM answer, let $u=xy$, which must be positive in order for $x^{10}+y^{10}=10xy-8$ to have a solution, and note that $x^{10}+y^{10}=(x^5-y^5)^2+2(xy)^5$. Thus
$$\begin{align}x^{10}+y^{10}-10xy+8
&=(x^5-y^5)^2+2(u^5-5u+4)\\
&=(x^5-y^5)^2+2(u-1)(u^4+u^3+u^2+u-4)\\
&=(x^5-y^5)^2+2(u-1)^2(u^3+2u^2+3u+4)\\
&\ge0
\end{align}$$
with equality if and only if $x^5=y^5$ and $xy=u=1$ (since $u\gt0$ implies $u^3+2u^2+3u+4\gt0$). From this we see that $x=y=\pm1$ are the only possibilities. In particular, $(x,y)=(1,1)$ is the only solution with $x,y\in\mathbb{R}^+$.
Remark: I showed the factorization of $u^5-5u+4$ in two steps for ease of checking.
|
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|
Computing $ \lim \limits_{x \rightarrow \infty} \left(1+\frac{1}{x} \right)^{\sqrt{ax^2 +bx+c}}$ For $a \geq 0 $, the following limit
$$ L = \lim_{x \rightarrow \infty} \left(1+\dfrac{1}{x} \right)^{\sqrt{ax^2 +bx+c}}$$
can be computed by applying L'Hopital rule as follows
$$L = \exp \left(\lim_{x \rightarrow \infty} \frac{\log \left(1+\dfrac{1}{x} \right)}{\frac{1}{\sqrt{ax^2 +bx+c}}}\right) = \exp \left(\lim_{x \rightarrow \infty} \dfrac{\frac{-1}{x(x+1)}}{\frac{-(2ax+b)}{2(ax^2 +bx+c)^{3/2}}}\right) =$$
$$ = \exp \left(\lim_{x \rightarrow \infty} \dfrac{2(ax^2 +bx+c)^{3/2}}{x(x+1)(2ax+b)}\right) = \exp \left(\lim_{x \rightarrow \infty} \dfrac{2\left(a+ \frac{b}{x}+\frac{c}{x^2}\right)^{3/2}}{1\cdot\left(1+\frac{1}{x}\right)\left(2a+\frac{b}{x}\right)}\right) = \exp(\sqrt{a}).$$
I am wondering if exists another method to compute this limit.
Thanks for any hint!
|
Hint
$$A=\left(1+\dfrac{1}{x} \right)^{\sqrt{ax^2 +bx+c}}\implies \log(A)={\sqrt{ax^2 +bx+c}}\log\left(1+\dfrac{1}{x} \right)$$ Now, use Taylor series $$\log\left(1+\dfrac{1}{x} \right)=\frac{1}{x}-\frac{1}{2 x^2}+O\left(\frac{1}{x^3}\right)$$ and you will arrive to the result (not only the limit but also how it is approached).
|
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|
Find the limit $\lim _{ x \to 0} \frac{\sqrt[3]{1+6x+3x^2+3x^3+3x^4}-\sqrt[4]{1+8x+4x^2+4x^3-2x^4}}{6x^2}$ I have been trying to find the limit,
$$\lim _{ x \to 0} \frac{\sqrt[3]{1+6x+3x^2+3x^3+3x^4}-\sqrt[4]{1+8x+4x^2+4x^3-2x^4}}{6x^2}$$
and sort of succeeded. But my $0$ answer doesn't converge with what Wolfram says which is $1/3$. Therefore, I would really appreciate it, if you could give me the right answer and drop a hint on the solution.
|
Another approach (apart from the one mentioned in my comments) is to use Binomial Theorem. Let the expression be denoted by $$\frac{u - v}{w}$$ and then multiply numerator and denominator by $$t = u^{11} + u^{10}v + \cdots + v^{11}$$ to get $$\frac{u^{12} - v^{12}}{wt}$$ and note that since $u, v$ both tend to $1$ as $x \to 0$ the variable $t \to 12$ and hence the desired limit is $$\frac{1}{72}\lim_{x\to 0}\frac{(1+6x+3x^2+3x^3+3x^4)^{4} - (1+8x+4x^2+4x^3-2x^4)^{3}}{x^{2}}$$ As can be seen the numerator can be written as $$(1 + 24x + 228x^{2} + o(x^{2})) - (1 + 24x + 204x^{2} + o(x^{2}))$$ or $$24x^{2} + o(x^{2})$$ and hence the answer is $24/72 = 1/3$.
On OP's request the solution by Taylor series is provided below.
We have the Taylor series for $(1 + x)^{n}$ for all real $n$ as $$(1 + x)^{n} = 1 + nx + \frac{n(n - 1)}{2!}x^{2} + o(x^{2})$$ as $x\to 0$. Hence $$(1+6x+3x^2+3x^3+3x^4)^{1/3} = 1 + 2x - 3x^{2} + o(x^{2})$$ and $$(1+8x+4x^2+4x^3-2x^4)^{1/4} = 1 + 2x - 5x^{2} + o(x^{2})$$ so the answer now comes easily as $1/3$.
|
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|
Evaluate $\lim_{x \to 1}\frac{1-x^2}{\sin\pi x}$ without L'Hopital I'm trying to evaluate the given limit without using L'hopitals rule. $$\lim_{x \to 1}\frac{1-x^2}{\sin(\pi x)}$$
Replacing $x$ by $1$ leads to $\frac{0}{0}$.
I have tried multiplying by $\frac{1+x^2}{1+x^2}$ and resoving the $\sin x$ factor by doing:
$$\frac{x}{\sin x} \cdot \frac{(\frac{1}{x} -x)}{\pi}$$
so $\frac{x}{\sin x} = 1$ but then the result still is $\frac{0}{\pi}$. All other options I see also lead to $\frac{0}{0}$ and I'm too sure how to proceed. Any hints are very welcome.
|
Write
$$\frac{1-x^2}{\sin\pi x}=\frac{1-x^2}{\sin(\pi-\pi x)}=\frac{1-x^2}{\sin\pi(1- x)}=\frac{(1-x)(1+x)}{2\sin\frac{\pi(1- x)}{2}\cos\frac{\pi(1- x)}{2}}=\frac12\times\frac{2}{\pi}\frac{\frac{\pi(1- x)}{2}}{\sin\frac{\pi(1- x)}{2}}\times\frac{1+x}{\cos\frac{\pi(1- x)}{2}}$$
so $t=1-x$ we have
$$\lim_{x\to1}\frac{1-x^2}{\sin\pi x}=\lim_{t\to0}\frac12\times\frac{2}{\pi}\frac{\frac{\pi t}{2}}{\sin\frac{\pi t}{2}}\times\frac{2-t}{\cos\frac{\pi t}{2}}=\frac12\times\frac{2}{\pi}\times\frac{2}{1}=\color{red}{\frac{2}{\pi}}$$
|
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|
If $2^{2017} + 2^{2014} + 2^n$ is a perfect square, find $n$. If $2^{2017} + 2^{2014} + 2^n$ is a perfect square, find $n$.
My first shot would be to assume the perfect square is $2^{2018}$, but how would I prove that? Even if it is, what is $n$? All help is appreciated.
|
$(2^x+2^y)^2=2^{2x}+2^{x+y+1}+2^{2y}$ so we can take $y=1007$ and $x=1009$ to conclude $2^{2018}+2^{2017}+2^{2014}$ is a square.
how to obtain all $n$:
$9(2^{2014})=2^{2017}+2^{2014}=k^2-2^n$.
Notice that we must have $k^2\equiv 1 \bmod 3$ implying $2^n\equiv 1\bmod 3$.
So $n=2m$.
From here we have $2^{2014}\times 9= (k+2^m)(k-2^m)=z(z+2^{m+1})$
there are now two cases, case one is when $z=9(2^j)$ and case two is when $z=2^j$
the first case is impossible because the number $9$ in binary is $1001$ and adding a power of two clearly wont make it a power of $2$. The other case is clearly only solved by taking $z=2^{1007}$ and taking $m=1009$ thus getting $z(z+2^{m+1})=2̣^{1007}(2^{1007}+2^{1010})=9(2^{2014})$.
So we need $n=2(1009)=2018$
|
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|
$a+b+c+ab+bc+ca+abc=1000$ find the minimum value of $a+b+c$ Given $a+b+c+ab+bc+ca+abc=1000$.
Find the minimum value of $a+b+c$.
Now we are considering $a$, $b$, $c$ to be integers and here in lies the pertinent problem as I could not get an answer in integers but in fractions.
|
Trial and error can be avoided. Note that minimizing $x+y+z$ where the positive real numbers $x,y,z$ have a specified product $1001$ is achieved by making $x,y,z$ as nearly equal as possible.
Here we minimize $(1+a)+(1+b)+(1+c)$ when we minimize $a+b+c$. Since $a,b,c$ are required to be (positive?) integers, the most equal we can make the factors is by assigning the primes $7,11,13$ to $(1+a),(1+b),(1+c)$ in some order. Although this assignment is not unique, the minimum sum $a+b+c = 6+10+12$ will be unique.
|
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Trigonometric Inequalities-Proof Verification. Prove the following inequalities:
If $x\in\Bbb R$, $x\ge0$ then,
(1) $-x\le \sin x\le x$. Moreover, does this inequality hold if $x<0$
(2) $1-\displaystyle\frac{x^2}{2}\le \cos x\le 1$
(3) $x-\displaystyle\frac{x^3}{6}\le \sin x\le x$
(4) $1-\displaystyle\frac{x^2}{2}\le \cos x\le 1-\displaystyle\frac{x^2}{2}+\displaystyle\frac{x^4}{24}$
(5) Also show that, $\left|\sin x\right|\le | x|$, $\forall x\in \Bbb R$
My Attempt:
(1)
Since, $\displaystyle sin^2t+cos^2t=1$, $ \forall \;\;t\in \displaystyle\Bbb R $
$\;\;\;\;\;\;\;\;\implies$ $\;-1\le \cos t\le1$ $ \forall \;\;t\in \displaystyle\Bbb R $
If $x\ge0$ then,
$$\int_{0}^x-dt\le\int_{0}^x \cos t\;dt\le\int_{0}^x dt\;\;\implies -x\le\sin x\le x$$
If $x<0$ we have, $x\le \sin x\le -x$.
(2)
On integrating the inequality in (1), we get
$$\int_{0}^x-t\;dt\le\int_{0}^x \sin t\;dt\le\int_{0}^x t\;dt\;\;\implies -\frac{x^2}{2} \le - \cos x +1 \le \frac{x^2}{2} \implies 1-\frac{x^2}{2} \le \cos x$$
Therefore we can conclude, $1-\displaystyle\frac{x^2}{2}\le \cos x\le 1$
On proceeding similarly, I have verified that (3) and (4) can also be obtained.
Is the above approach correct?
Furthermore I am curious to see what happens in (2), (3) and (4) for $x\le 0$.
In (2),
If $x \le 0$ $\implies $ $-x \ge 0$
Therefore, $1-\displaystyle\frac{(-x)^2}{2}\le \cos (-x)\le 1$ $\implies $ $1-\displaystyle\frac{x^2}{2}\le \cos x\le 1$
Thus, $$1-\displaystyle\frac{x^2}{2}\le \cos x\le 1, \;\; \forall x\in \Bbb R$$
In (3),
If $x \le 0$ $\implies $ $-x \ge 0$
Therefore, $(-x)-\displaystyle\frac{(-x)^3}{6}\le \sin(-x)\le (-x) \implies -x+\displaystyle\frac{x^3}{6}\le -\sin x\le -x \implies x \le \sin x\le x-\displaystyle\frac{x^3}{6}$
Thus, $$x-\displaystyle\frac{x^3}{6}\le \sin x\le x \;\;, \forall x\ge 0 $$
$$x \le \sin x\le x-\displaystyle\frac{x^3}{6}\;\;, \forall x\le 0$$
Similarly for part (4) , $1-\displaystyle\frac{x^2}{2}\le \cos x\le 1-\displaystyle\frac{x^2}{2}+\displaystyle\frac{x^4}{24}, \;\; \forall x\in \Bbb R$
How do I go about with proving (5)?
Have I done something wrong?
|
Everything looks fine. For (5) use (1) and the fact that $|a| \le |b|$ equivalent $-|b| \le a \le |b|$.
For $x \ge 0$ then, using (1) $|sin(x)|\le x$, therefore $|sin(x)|\le |x|$.
For $x \lt 0$ then, using (1) $x \le sin(-x) \le -x$ equivalent $x \le -sin(x) \le -x$ equivalent $x \le sin(x) \le -x$ equivalent $-|x| \le sin(x) \le |x|$
|
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|
Proof of convergence $\frac{n^2+3n+2}{6n^3+5}$
I need to prove that $a_n = \frac{n^2+3n+2}{6n^3+5}$ converges as $n\to\infty$. I think it will converge to $0$, so I want to prove it.
Let $\epsilon >0$. Then
side-work
$$\left|\frac{n^2+3n+2}{6n^3+5}\right|< \left| \frac{n^2+3n+2n}{6n^3+5}\right| < \left| \frac{n^2+5n}{6n^3}\right|$$ where I used $n \geq 1$ and then I continued $$\left| \frac{n^2+5n}{6n^3}\right| = \left|\frac{n+5}{6n^2}\right|<\left|\frac{6n}{6n^2}\right|$$ using again $n \geq 1$ and finally $$\left|\frac{6n}{6n^2}\right| = \left|\frac{1}{n}\right|=\frac{1}{n} < \epsilon$$
..Then take $n_0 = \max\left\{1, \frac{1}{\epsilon}\right\}$ and $\forall n > n_0$ we have $\left|\frac{n^2+3n+2}{6n^3+5}-0\right|<\epsilon$
Is this proof correct? Because my books give a different one
|
Unless you have some particular reason for an epsilon-delta proof:
$$\frac{n^2+3n+2}{6n^3+5} = \frac{n^2}{n^3} \cdot \frac{1 + 3/n + 2/n^2}{6+5/n^3} \;\;\to\;\; 0 \cdot \frac{1}{6} \;=\; 0$$
|
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|
if : $abc=8 $ then : $(a+1)(b+1)(c+1)≥27$ if : $$abc=8 :a,b,c\in \mathbb{R}_{> 0}$$
then :
$$(a+1)(b+1)(c+1)\ge 27.$$
My try :
$$(a+1)(b+1)(c+1)=1+(a+b+c)+(ab+ac+bc)+abc$$
$$(a+1)(b+1)(c+1)=1+(a+b+c)+(ab+ac+bc)+8, $$
then?
|
We have $$(a+1)(b+1)(c+1)=9+a+b+c+ab+bc+ca,$$
since $$abc=8.$$ We get $$\frac{a+b+c}{3}\geq \sqrt[3]{abc}=2$$ and $$\frac{ab+bc+ca}{3}\geq \sqrt[3]{(abc)^2}=4$$ Multplying by $3$ and adding up, we get the statement above.
|
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|
Quadratic equation with parameters If $a$, $b$, $c$ are real numbers such that $2a + 3b + 6c = 0$, prove that $ax^2+bx +c=0$ has a solution in the interval $[0, 1]$.
This should use high school maths or a little bit more then that.
|
\begin{align*}
0 &= 2a+3b+6c \\
f(x) &= ax^2+bx+c \\
f(0) &= c \\
f\left( \frac{1}{2} \right) &= \frac{a}{4}+\frac{b}{2}+c \\
&= -\frac{a}{12} \\
f(1) &= a+b+c \\
&= \frac{a}{3}-c
\end{align*}
Case I: $ac \ge 0$
$$f(0) f\left( \frac{1}{2} \right) = -\frac{ac}{12} \le 0$$
*
*$\exists x\in \left[ 0, \dfrac{1}{2} \right]$ such that $f(x)=0$.
Case II: $ac \le 0$
$$f\left( \frac{1}{2} \right) f(1)= -\frac{a^2}{36}+\frac{ac}{12} \le 0$$
*
*$\exists x\in \left[ \dfrac{1}{2}, 1 \right]$ such that $f(x)=0$.
Combining, $\exists x \in [0,1]$ such that $f(x)=0$
N.B.
When $0 <3ac <a^2$, there're two such roots, namely $\alpha \in \left( 0, \frac{1}{2} \right)$ and $\beta \in \left( \frac{1}{2}, 1 \right)$.
Updates
The value of $x=\dfrac{1}{2}$ can be inspired by plotting a family of curves by varying either $b$ or $c$. You can see a fixed point at $x=\dfrac{1}{2}$ by varying $c$ below.
Alternatively, $(x,y)=\left(\dfrac{1}{2},-\dfrac{a}{12} \right)$ is a solution of
$$ax^2-\left( \frac{2a}{3} + 2c \right)x+c-y=
\frac{\partial}{\partial c}
\left[
ax^2-\left( \frac{2a}{3} + 2c \right)x+c-y
\right]=0$$
|
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|
Converting a power series to a function
$$\sum (n+1)^2x^n = {1+x\over (1-x)^3}$$
I tried to find a similar power series and start from there. I see $ x^n $ and it reminds me of geometric series, but $(n+1)^2$ confuses me. How do I prove this?
|
Just another way to do it.
$$A=\sum_{n=0}^\infty (n+1)^2x^n =\sum_{n=0}^\infty (n^2+2n+1)x^n $$ Now, write $$n^2+2n+1=(n(n-1)+n)+2n+1=n(n-1)+3n+1$$ $$A=\sum_{n=0}^\infty n(n-1)x^n+3\sum_{n=0}^\infty nx^n+\sum_{n=0}^\infty x^n$$ $$A=x^2\sum_{n=0}^\infty n(n-1)x^{n-2}+3x\sum_{n=0}^\infty nx^{n-1}+\sum_{n=0}^\infty x^n$$ $$A=x^2\left(\sum_{n=0}^\infty x^n \right)''+3x\left(\sum_{n=0}^\infty x^n \right)'+\left(\sum_{n=0}^\infty x^n \right)$$ $$\left(\sum_{n=0}^\infty x^n \right)=\frac{1}{1-x}\implies\left(\sum_{n=0}^\infty x^n \right)'=\frac{1}{(1-x)^2}\implies\left(\sum_{n=0}^\infty x^n \right)''=\frac{2}{(1-x)^3}$$ $$A=\frac{2x^2}{(1-x)^3}+\frac{3x}{(1-x)^2}+\frac{1}{1-x}=\frac{1+x}{(1-x)^3}$$
Using the same steps, you could make the problem more general and show that $$B=\sum_{n=0}^\infty (an^2+bn+c)x^n =\frac{(a-b+c)x^2 + (a+b-2 c)x+c }{(1-x)^3}$$
|
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|
Is it possible to evaluate $\lim_{x\rightarrow 0}(-1+\cos x)^{\tan x}$? Let's say we have the following limit:
$$\lim_{x\rightarrow 0}(-1+\cos x)^{\tan x}$$
Would the following solution be correct?
The solution is incorrect, please see the correction of @YvesDaoust
\begin{align}
\lim_{x \rightarrow 0}(-1+\cos x)^{\tan x}
&= \lim_{x \rightarrow 0}\left((1-\cos x)^{\tan x}\cdot (-1)^{\tan x}\right) \\
&= \lim_{x \rightarrow 0}\left(1-\left(1-2\sin^2\left(\frac{x}{2}\right)\right)^{\tan x}\right) \cdot \lim_{x \rightarrow 0}(-1)^{\tan x} \\
&= \lim_{x \rightarrow 0}\left(2\sin^2 \left(\frac{x}{2}\right)\right)^{\tan x} \cdot 1\\
&= \lim_{x \rightarrow 0}2^{\tan x} \cdot \lim_{x \rightarrow 0}\sin \left(\frac{x}{2}\right)^{2\tan x} \\
&= 1 \cdot \lim_{x \rightarrow 0}\sin \left(\frac{x}{2}\right)^{2\tan x} \\
&= \lim_{x \rightarrow 0}\sin \left(\frac{x}{2}\right)^{2\frac{\sin x}{\cos x}} \\
&= \lim_{x \rightarrow 0}\sin \left(\frac{x}{2}\right)^{\frac{4\sin (\frac{x}{2})\cdot \cos (\frac{x}{2})}{\cos x}} \\
&= \lim_{x \rightarrow 0}\left(\sin \left(\frac{x}{2}\right)^{\sin \left(\frac{x}{2}\right)}\right)^{\frac{4\cos\frac{x}{2}}{\cos x}} \\
&= \left(\lim_{x \rightarrow 0}\sin \left(\frac{x}{2}\right)^{\sin \left(\frac{x}{2}\right)}\right)^{\lim_{x \rightarrow 0}\frac{4\cos\frac{x}{2}}{\cos x}} \\
&= \left(\lim_{u \rightarrow 0}u^u\right)^{4} \\
&= 1^4 \\
&= 1 \\
\end{align}
The result seems to be correct, but the way leading to it seems to be quite lengthy. Am I doing something redundant?
|
Quite long but it's correct to me.
To try with something else, you could:
1. Use Taylor Series
$$\tan(x)\approx x$$
$$\cos(x) \approx 1 - \frac{x^2}{2}$$
2. Use a different approach
For example the exponential representation
$$(\cos(x)-1)^{\tan(x)}
= \text{exp}\Big( \tan(x)\log\left(-2\sin^2\frac{x}{2}\right)\Big)$$
|
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|
Contour integration with semicircular arcs. I am trying to compute
$$\int_0^\infty \frac{\ln^2 x}{1+x^2} dx $$
using a complex contour integration of $f(z) = \ln^2 z / (1+z^2)$ around a closed contour with the segments:
$$C_1 : [-R,-\rho]\\C_2: \{z: |z| = \rho, 0 \leq \arg z \leq \pi\}\\ C_3: [\rho , R] \\ C_4: \{z: |z| = R, 0 \leq \arg z \leq \pi\}.$$
This contour encloses one singular point $z = i$ where I found $Res(f,i) = i\pi^2/8.$
My difficulty is in handling the $\int_{C_1}f(z) dz$ integral on the real interval $[-R,-\rho]$ which does not vanish as $R \to \infty$ and $\rho \to 0$.
Thank you.
|
By the residue theorem
$$\oint_{C_1 \cup C_2 \cup C_3 \cup C_4}f(z) \, dz = 2\pi i Res(f,i) = 2\pi i (i \pi^2/8) = -\frac{\pi^3}{4}.$$
It is straightforward to show that the integrals over $C_2$ and $C_4$ vanish in the limit as $R \to \infty$ and $\rho \to 0$.
Hence,
$$\tag{1} -\frac{\pi^3}{4} = \int_0^\infty \frac{\ln^2 x}{1 + x^2} \, dx + \lim_{R \to \infty, \rho \to 0} \int_{C_1} \frac{\ln^2 z}{1 + z^2} \, dz.$$
Now we can focus on the second integral on the RHS -- the source of your difficulty.
Note that
$$\begin{align}\lim_{R \to \infty, \rho \to 0} \int_{C_1} \frac{\ln^2 z}{1 + z^2} \, dz &= \int_{-\infty}^0 \frac{ \ln^2 x}{1 + x^2} \, dx \\ &= \int_{0}^\infty \frac{ \ln^2 (-x)}{1 + x^2} \, dx \\ &= \int_{0}^\infty \frac{ (\ln x + i \pi)^2}{1 + x^2} \, dx \\ &= \int_{0}^\infty \frac{ \ln^2 x }{1 + x^2} \, dx + 2\pi i\int_{0}^\infty \frac{ \ln x }{1 + x^2} \, dx - \pi^2\int_{0}^\infty \frac{ 1 }{1 + x^2} \, dx \\ &= \int_{0}^\infty \frac{ \ln^2 x }{1 + x^2} \, dx + 0 - \pi^2 \frac{\pi}{2} \\ &= \int_{0}^\infty \frac{ \ln^2 x }{1 + x^2} \, dx - \frac{\pi^3}{2}\end{align}.$$
The second of the three integrals on the RHS can be shown to be $0$ by showing that integrals over $[0,1]$ and $[1,\infty)$ must cancel (using a change of variables $u = 1/x)$.
Substituting into (1) we obtain
$$-\frac{\pi^3}{4} = 2\int_0^\infty \frac{\ln^2 x}{1 + x^2} \, dx - \frac{\pi^3}{2}.$$
Thus,
$$\int_0^\infty \frac{\ln^2 x}{1 + x^2} \, dx = \frac{\pi^3}{8}.$$
|
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|
System of two cubic equations: $x+y^2 = y^3$, $y+x^2=x^3$ I got stuck on this system of equations. Could you help and tell me how should I approach this problem?
\begin{align*}
x+y^2 &= y^3\\
y+x^2 &= x^3
\end{align*}
These are the solutions:
\begin{align*}
(0, 0); \\((1+\sqrt{5})/2, (1+\sqrt{5})/2); \\((1-\sqrt{5})/2, (1-\sqrt{5})/2);
\end{align*}
|
The solution set of the second equation is the cubic parabola
$$\gamma:\quad y=x^3-x^2$$
which intersects the symmetry line $y=x$ in the three points
$$\left({1-\sqrt{5}\over2},{1-\sqrt{5}\over2}\right),\quad(0,0),\quad\left({1+\sqrt{5}\over2},{1+\sqrt{5}\over2}\right)\ .\tag{1}$$
Otherwise $\gamma$ lies in the interior of the four shaded regions of the following figure (note that for $x>0$ one has $x^3-x^2+x=x(x^2-x+1)>x(x-1)^2>0$, hence $x^3-x^2>-x$):
The solution set of the first equation is the curve $\hat\gamma$ obtained by reflecting $\gamma$ in the symmetry line $y=x$. It contains the three points $(1)$ and is otherwise a subset of the reflected shaded regions. Since the latter are disjoint from the original regions it follows that the intersection $\gamma\cap\hat\gamma$ is given by $(1)$.
|
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|
Flipping a coin - expected value We have seven coins, two of them have tails on both sides. We choose (randomly) a coin and we flip a coin untill two tails appear (not necessarly in row). Calculate expected number of throws.
My attempt:
$\mathbf{P}\left( X=0\right) =\mathbf{P}\left( X=1\right) =0 \\
\mathbf{P}\left( X=2\right) =\frac{5}{7}\cdot\frac{1}{2^2} +\frac{2}{7} \\
\mathbf{P}\left( X=3\right) =\frac{5}{7}\cdot\frac{1}{2^3}\cdot {3\choose 2} \\
\mathbf{P}\left( X=n\right) =\frac{5}{7}\cdot\frac{1}{2^n}\cdot {n\choose 2}$
So the expected value is given by this formula:
$\mathbf{E}X=\lim_{n\to\infty}2\cdot\left(\frac{5}{7}\cdot\frac{1}{2^2}+\frac{2}{7}\right)+\sum_{k=3}^n \frac{5}{7}\cdot\frac{1}{2^k}\cdot{k \choose 2}\cdot k=\frac{13}{14}+\frac{5}{14}\sum_{k=3}^\infty\frac{k^2\cdot(k-1)}{2^k}$
Wolfram shows that this sum is equal about $4$, which seems to be fine, but I highly doubt that professor would give us such a sum on an exam.
Is there any other way to solve this?
|
You have $\frac 27$ chance of being guaranteed two tails in two throws. You have $\frac 57$ chance of expecting to take four throws, so the expected number of throws is $2 \cdot \frac 27+ 4\cdot \frac 57=\frac {24}7$ It should definitely be under $4$ as it would be four without the two two-tailed coins. Those will lower the expectation.
|
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|
Prove that; $8\cos^3 (\frac {\pi}{9})- 6\cos(\frac {\pi}{9})=1$ Prove that; $8\cos^3 (\frac {\pi}{9})- 6\cos(\frac {\pi}{9})=1$
My Attempt,
$$L.H.S=8\cos^3(\frac {\pi}{9}) - 6\cos(\frac {\pi}{9})$$
$$=2\cos(\frac {\pi}{9}) [4\cos^2(\frac {\pi}{9}) - 3]$$
$$=2\cos(\frac {\pi}{9}) [2+2\cos(\frac {2\pi}{9}) - 3]$$
$$=2\cos(\frac {\pi}{9}) [2\cos(\frac {2\pi}{9})-1]$$.
What should I do further?
|
$2(4 \cos^3 \frac π9 - 3 \cos \frac π9)$
As $\cos 3\theta = 4 \cos^3 \theta - 3\cos \theta$
= $2(\cos (3 × \frac π9))$
=$2 × \cos(\frac π3)$
=$2 × \frac 12 = 1$
|
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|
Find the integral of $\int \frac{x^4}{x-1}$ I was not happy with the responses so I went ahead and solved the integral to get feedback on the answer:
Dividing as the book shows us:
we get $x^4$ divided by $x-1$ which gives us the quotient plus the remainder over the divisor which is: $x^3 + \frac{x^3}{x-1}$
$\int x^3+\frac{x^3}{x-1} \to \frac{1}{4}x+x^3ln(\vert x-1 \vert)$
Instead of using shortcuts I want feedback on how it should be done with the process in the book.
|
$$\int\frac{x^4-1+1}{x-1}dx=\int\frac{x^4-1}{x-1}dx+\int\frac{1}{x-1}dx$$
note
$$x^4-1=(x-1)(x^3+x^2+x+1)$$
|
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$f(x)=3+2x+\alpha^2x^3+\beta x^4\in \mathbb{Z}_5[x]$ Find $\alpha,\beta,\lambda^{625}$ I have the following question :
$K=\frac{\mathbb{Z}_5[x]}{(f)}$ is a field with 125 elements.
$\lambda=x+(f)\in K$
$f(x)=3+2x+\alpha^2x^3+\beta x^4\in \mathbb{Z}_5[x]$ Find $\alpha,\beta,$ and $\lambda^{625}$ express $\lambda^{625}\in K$ in the following form $a_0+a_1 \lambda+a_2\lambda ^2$ while $a_0,a_1,a_2\in \mathbb{Z}_5$
I did manage to find $\beta$ which is $0$ since there's a theorem that say $q^n=125$ (125-number of elements) so the max power should be $3$, Yet I don't know how to find $\alpha$ and $\lambda^{625}$
Any help will be appreciated.
|
We have $\frac{\mathbb Z_5[x]}{(f)}\cong \mathbb Z_5(a)$ then number of the element in $\mathbb Z_5(a)=125=5^3$ then $[\mathbb Z_5(a):\mathbb Z_5]=3$ then $f(x)$ is a polynomial of degree $3$ and donot have roots in $\mathbb Z_5$, then $\beta=0$.
we have $\alpha\in \mathbb Z_5 $,then $\alpha^2=0,1,4 $
if $\alpha^2=0$ then $f(x)=3+2x$,then $deg(f)<3$ contradiction
if $\alpha^2=1$ then $f(x)=3+2x+x^3 $, then $2$ root of $f(x)$ contradiction
if $\alpha^2=4$ then $f(x)=3+2x+4x^3$ and donnot have a root in $\mathbb Z_5$
Let $a$ root of $f(x)$ in $ \mathbb Z_5(a)$ then $\lambda=x+(f)$ root in $\mathbb Z_5[x]/(f)$, for calculate $\lambda^{625}$ we must to calculate $a^{625}$
we have $4a^3+2a+3=0\Rightarrow a^3=2a+3\Rightarrow a^4=2a^2+3a \Rightarrow a^5=3a^2+4a+1\Rightarrow a^6=4a^2+2a+4 $, so $a^{10}=3a+3\Rightarrow a^{20}=4a^2+3a+4\Rightarrow a^{25}=2a^2+4\Rightarrow a^{50}=4a^2+2a+1 $, so $ a^{50}=a^6+2 \Rightarrow a^{100}=(a^6+2)^2=a^{12}+4a^6+4=4a^2+4a+4\Rightarrow a^{100}=a^6+2a $,so $a^{200}=(a^6+a)^2=4a+2 \Rightarrow a^{400}=a^2+a+4\Rightarrow a^{600}=a^2+a$.
Therefor $a^{625}=a^{600}a^{25}=3a^2+4a+1=a^5$
|
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What is the maximal value of the expression $\overline{abc}-(a^3+b^3+c^3)?$ Let $\overline{abc}$ be $3$-digits number. What is the maximal value of the expression $\overline{abc}-(a^3+b^3+c^3)?$
Computer gives the answer $396$ for the number $620$ but I have no idea how to do it by hand.
|
You can write the whole thing as $100a + 10b + c - a^3 - b^3 - c^3$.
Rearranging the terms gives $(100a - a^3) + (10b - b^3) + (c - c^3)$, which you can solve one summand at a time:
You maximize $100a - a^3$ for $a = 6$.
You maximize $10b - b^3$ for $b = 2$.
You maximize $c - c^3$ for $c = 0$ or $1$.
|
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Proving trigonometrical identity from given identity.
Given $$\dfrac{\sin^4 A}{a}+\dfrac{\cos^4 A}{b}=\dfrac{1}{a+b}$$ prove that:$$\dfrac{\sin^8 A}{a^3}+\dfrac{\cos^8 A}{b^3}=\dfrac{1}{(a+b)^3}$$
Using given, I proved: $b\sin^2 A=a\cos^2 A$ Help me proceeding after this.
|
You are almost finished. From where you left off, let us say $$b \sin^2 A =a \cos^2 A \iff \frac{\sin^2 A}{a}=\frac{\cos^2 A}{b}=c$$
For some $c$. Thus $$\sin^2 A+\cos^2 A=1 \iff ac+bc=1 \iff c=\frac{1}{a+b} \tag{1}$$
Now, note $$\dfrac{\sin^8 A}{a^3}+\dfrac{\cos^8 A}{b^3}=\frac{a^4c^4}{a^3}+\frac{b^4c^4}{b^3}=ac^4+bc^4=c^4(a+b)=\frac{1}{(a+b)^3}$$
From $(1)$. We are done.
|
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|
An expression for $f(2x)$ as a rational function of $f(x)$ A real-valued function $f$ defined on the $\mathbb{R}\setminus\{1\}$ by
\begin{equation*}
f(x) = \frac{3x + 1}{x - 1} = 3 + \frac{4}{x - 1}
\end{equation*}
is invertible, and
\begin{equation*}
f^{-1}(x) = \frac{x + 1}{x - 3} = 1 + \frac{4}{x - 3} .
\end{equation*}
Evaluate constants $A$, $B$, and $C$ such that
\begin{equation*}
f(2x) = \frac{Af(x) + B}{f(x) + C}
\end{equation*}
for every real number $x$ distinct from $1/2$ and 1.
This is a problem from a past high school competition given in a county - Monroe County - in New York. (I will post a solution to the problem.) Does every rational function that is equal to the quotient of two linear functions have such a property? Can $f(3x)$ be expressed analogously? How about $f(kx)$ for any positive integer $k$? Is there any advantage to such an expression over substituting "$2x$" in for "$x$"?
|
Rational functions of the form $x\mapsto \frac{ax+b}{cx+d}$ (with $ad\ne bc$) form a group under composition (Möbius transformations). You are given such an $f$ (and its inverse $f^{-1}$) and want some $g$ such that $g\circ f=f\circ 2$ (where the duplication function $2(x)=\frac{2x+0}{0x+1}$ is a Möbius transformation as well). We apply $\circ f^{-1}$ from the right to find $$ g=f\circ 2\circ f^{-1}.$$
In this particular case,
$$g(x)=f\left(2\cdot \frac{x+1}{x-3}\right) =\frac{3\cdot2\cdot \frac{x+1}{x-3}+1 }{2\cdot \frac{x+1}{x-3}-1}=\ldots =\frac{7x+3}{x+5}$$
As long as we stay within Möbius transformations , there is nothing special about the $2$ in this problem. It would not be more complicated to ask how to express $f(\frac{x+42}{2-x})$ in terms of $f(x)$.
|
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How to solve this problem using Jensen's inequality? This is the task to find the minimum value of $x^5 + y^5 + z^5 - 5xyz$
where $x,y$ and $z$ are any positive numbers.
I know that I may use this inequality:
$$(t_1\cdot t_2\cdot t_3\cdots t_n)^{\frac{1}{n}} \leq \frac{t_1+t_2+t_3+\cdots +t_n}{n}$$
|
By the inequality we have that $\frac{x^5 + y^5 + z^5 + 1 + 1}{5} \ge \sqrt[5]{x^5y^5z^5\cdot1 \cdot 1}$. Hence $x^5 + y^5 + z^5 + 2 \ge 5xyz$
$$x^5 + y^5 + z^5 - 5xyz = x^5 + y^5 + z^5 + 2 - 5xyz - 2 \ge 5xyz - 5xyz - 2 = -2$$
It's obtained for $x=y=z$
|
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|
Find $A^n$ for all $n \in \mathbb{N}$. I'm learning linear algebra and need help with the following problem:
Let $A = \begin{pmatrix}-2 & -3 & -3\\-1 & 0 & -1\\5 & 5 & 6\end{pmatrix} \in M_{3x3}(\mathbb{R}).$ Find $A^n$ for all $n \in \mathbb{N}$.
My first thought was to compute $A^2$, $A^3$, $A^4$ and see if a pattern emerge. I used Mathematica to compute the power of $A$ to save me some time. The computations gave
$A^2 = \begin{pmatrix}-8 & -9 & -9\\-3 & -2 & -3\\15 & 15 & 16\end{pmatrix}, \quad A^3 = \begin{pmatrix}-20 & -21 & -21\\-7 & -6 & -7\\35 & 35 & 36\end{pmatrix}, \quad A^4 = \begin{pmatrix}-44 & -45 & -45\\-15 & -14 & -15\\75 & 75 & 76\end{pmatrix}$.
The above results are clearly showing a pattern relating the first four powers of $A$, for instance $(a_{31}) = (a_{32}) = (a_{33}) + 1$, but I'm unable to write the general form of a matrix $A^{k}$.
$(Q1)$ How do I find/write the general form of a matrix $A^{k}$?
$(Q2)$ I only compute the first four powers of $A$. How do I know that this pattern won't fail for some integer $k$?
|
The method descirbed in this answer still works even when $A$ is not diagonalizable. The upside of this method is that you are not required to calculate eigenvectors or inverses of matrices.
The characteristic polynomial is $P_A(X)=(2-X)(X-1)^2$. The Cayley-Hamilton theorem says that $P_A(A)=(2Id-A)(A-Id)^2=0$. Consider the polynomial $X^n$. We can write $$X^n=Q(X)P_A(X)+r(X)$$ with $\deg(r(X))<3$. Hence we may write $r(X)=aX^2+bX+c$.
We can plug in the eigenvalue $2$ into the previous equation to get $2^n=r(2)=4a+2b+c$. (Here we used that $P_A(2)=0$).
Similarly $1=r(1)=a+b+c$. Last but not least we take the derivative w.r.t. $X$. This yields $$nX^{n-1}=Q'(X)P_A(X)+Q(X)P_A'(X)+2aX+bX.$$ Plug in $1$ to get $n=2a+b$. You can solve for $a,b$ and $c$. Then you know $r(X)$. Plug in $A$ in the original equation to get $$A^n=r(A)=aA^2+bA+cId.$$
|
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|
Simultaneous equation that turns into quadratic Solve for $x$ and $y$:
\begin{cases}
y &= 4x^2 - x - 6 \\
y &= 2 - x.
\end{cases}
I have tried rearranging to get $x + y = 2$ and then substituting $y$ into it but hit a dead end. I'm pretty sure it turns into a quadratic equation. Any help would be appreciated thanks.
|
You can substitute $y$ into the first equation to get:
$4x^2-x-6=2-x$
$4x^2-8=0$
Plus this into the quadratic formula where $a=4$, $b=0$, and $c=-8$
$x = \dfrac{-(0)\pm \sqrt{(0)^{2} - 4 (4)(-8)}}{2(4)}$
$x=\pm \sqrt {2}$
$y=2\pm \sqrt 2$
|
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|
All pair of $m,n$ satisfying $lcm(m,n)=600$
Find the number of pairs of positive integers $(m,n)$, with $m \le n$, such that
the ‘least common multiple’ (LCM) of $m$ and $n$ equals $600$.
My tries:
It's very clear that $n\le600$, always.
Case when $n=600=2^3\cdot 3\cdot 5^2$, and let $m=2^{k_1}\cdot 3^{k_2}\cdot 5^{k_3}$, all possible values of $k_1=3+1=4,\ k_2=1+1=2,\ k_3=2+1=3$. So number of $m$ which satisfy above will be $4\cdot 2 \cdot 3=24$
Help me analyzing when $n<600$.
|
As you have very nicely written, you have the factorisation
$$600=2^3\cdot 3\cdot 5^2.$$
Now recall that the lcm of two numbers, given their prime factorisation, is the product of their prime factors to the highest power. For example
$$lcm(2\cdot 3\cdot 5^7,2^2\cdot 5^6)=2^2\cdot3\cdot5^7$$
So you have to find all the possible numbers of the form $2^p3^q5^r$, $p\leq 3,q\leq 1,r\leq 2\ $ so that at least one of them contains $2^3$, $3^5$ and $5^2$ as factors.
|
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|
values of $a$ in inequality If $\sqrt{xy}+\sqrt[3]{xyz}<a(x+4y+4z)$ and $x,y,z>0$ then $a$ is
with the help of am gm inequality
$\displaystyle \sqrt{xy}\leq \left(\frac{x+y}{2}\right)$ and $\displaystyle \sqrt[3]{xyz}\leq \left(\frac{x+y+z}{3}\right)$
so $\displaystyle \sqrt{xy}+\sqrt[3]{xyz}\leq \frac{x+y}{2}+\frac{x+y+z}{3} = \frac{5x+5y+2z}{6}$
want be able to go further, could some help me
|
We'll prove that $$x+4y+4z\geq3\left(\sqrt{xy}+\sqrt[3]{xyz}\right).$$
Indeed, by AM-GM
$$4z+x+4y-3\sqrt{xy}=4z+2\cdot\frac{x+4y-3\sqrt{xy}}{2}\geq$$
$$\geq3\sqrt[3]{4z\left(\frac{x+4y-3\sqrt{xy}}{2}\right)^2}=3\sqrt[3]{z\left(x+4y-3\sqrt{xy}\right)^2}.$$
Thus, it remains to prove that $$\left(x+4y-3\sqrt{xy}\right)^2\geq xy,$$
which is $$(\sqrt{x}-2\sqrt{y})^2(x+4y-2\sqrt{xy})\geq0.$$
The equality occurs for $x=4y$ and $y=4z$.
Id est, $a>\frac{1}{3}$ are all values of $a$, for which the inequality
$$\sqrt{xy}+\sqrt[3]{xyz}<a(x+4y+4z)$$
is true for all positives $x$, $y$ and $z$.
|
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|
Power series expansion of a holomorphic function One must show that equality$$\sum_{n=0}^\infty n^2z^n = \frac{z(z+1)}{(1-z)^3} $$ holds for $z\in\mathbb C$ and $|z|<1$.
I tried to expand $f(z):=\frac{z(z+1)}{(1-z)^3}$ into power series where $ z_0=0$ is the center and coefficients given by $a_n=\frac{1}{2\pi i}\int_{|z|<1}\frac{f(z)}{(z-0)^{n+1}}dz$, because f is holomorphic on the open domain $D=${$z:|z|<1$}.
Is there a more easy way to show this equality?
|
We could also recall the binomial series expansion valid for $|z|<1$
\begin{align*}
\frac{1}{(1-z)^3}&=\sum_{n=0}^\infty\binom{-3}{n}(-z)^n=\sum_{n=0}^\infty\binom{n+2}{2}z^n\\
&=\frac{1}{2}\sum_{n=0}^\infty(n+2)(n+1)z^n \tag{1}
\end{align*}
Using the coefficient of operator $[z^n]$ to denote the coefficient of $z^n$ of a series, we obtain from (1)
\begin{align*}
[z^n]\frac{z+z^2}{(1-z)^3}&=\left([z^{n-1}]+[z^{n-2}]\right)\frac{1}{2}\sum_{k=0}^\infty(k+2)(k+1)z^k\\
&=\frac{1}{2}(n+1)n+\frac{1}{2}n(n-1)\\
&=n^2
\end{align*}
|
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|
Solve the inequality $\sin(x)\sin(3x) > \frac{1}{4}$
Find the range of possible values of $x$ which satisfy the inequation $$\sin(x)\sin(3x) > \frac{1}{4}$$
SOURCE : Inequalities (PDF)( Page Number 6; Question Number 306)
One simple observation is that both $x$ and $3x$ have to positive or negative simultaneously. I tried expanding $\sin(3x)$ by the regular indentity as :
$$\sin(x) \times \big(3\sin(x)-4\sin^3(x)\big) > \frac {1}{4}$$
$$\implies \sin^2(x)\times\big(3-4\sin^2(x)\big) >\frac{1}{4}$$
I do not find any way of proceeding. Wolfram Alpha gives 4 sets of answers. Do I have to observe this problem "case-by-case"? Can this question be solved without calculus ? Can anyone provide a hint to what should be done ?
Thanks in Advance ! :)
|
$$sin(x)sin(3x) = sin(2x-x) sin(2x+x) < \frac{1}{4}$$
$$ \text{or, } \frac{cos(x) - cos(2x)}{2} < \frac{1}{4}$$
$$ \text{or, } {cos(x) - cos(2x)} - \frac{1}{2} < 0$$
$$ \text{or, } {cos(x) - 2cos^2(x) + 1} - \frac{1}{2} < 0$$
Let $y = cos(x)$
$$ \text{or, } - 2y^2 + y + \frac{1}{2} < 0$$
Solving,
$$ \frac{1-\sqrt{5}}{4} < cos(x) < \frac{1+\sqrt{5}}{4}$$
Hence,
$$\frac{\pi}{5} < x < \frac{3 \pi}{5}$$ and other $2 \pi n$ difference of it.
|
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|
How can you prove $\frac{n(n+1)(2n+1)}{6}+(n+1)^2= \frac{(n+1)(n+2)(2n+3)}{6}$ without much effort?
I will keep it short and take only an extract (most important part) of
the old task.
$$\frac{n(n+1)(2n+1)}{6}+(n+1)^2= \frac{(n+1)(n+2)(2n+3)}{6}$$
What I have done is a lot work and time consuming, I have "simply" solved it. But I think with a lot less work, there would be an easier and faster way. It's just I cannot see it : /
If anyone wants see, here is my long solution which I'm not happy with:
$$\frac{n(n+1)(2n+1)+6(n+1)^2}{6}=\frac{(n^2+2n+n+2)(2n+3)}{6} \Leftrightarrow$$
$$\Leftrightarrow \frac{(2n^3+n^2+2n^2+n)+6n^2+12n+6}{6} = \frac{(n^2+3n+2)(2n+3)}{6} \Leftrightarrow$$
$$\Leftrightarrow \frac{2n^3+3n^2+n+6n^2+12n+6}{6}=\frac{2n^3+3n^2+6n^2+9n+4n+6}{6} \Leftrightarrow$$
$$\Leftrightarrow \frac{2n^3+9n^2+13n+6}{6}=\frac{2n^3+9n^2+13n+6}{6}$$
|
Note that
$$n(2n+1)+6(n+1)=2n^2+7n+6=(n+2)(2n+3)$$
Multiplying both sides by $\dfrac {n+1}6$
$$\frac {n(n+1)(2n+1)}6+(n+1)^2=\frac {(n+1)(n+2)(2n+3)}6$$
|
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|
To prove $\frac{1}{\sin 10^\circ}-\frac{\sqrt 3}{\cos 10^\circ}=4$ To prove:
$$\frac{1}{\sin 10^\circ}-\frac{\sqrt 3}{\cos 10^\circ}=4$$
I tried taking lcm but could not get to anything.
|
\begin{align*}
\frac {1}{\sin 10^\circ} - \frac {\sqrt 3}{\cos 10^\circ} &= \frac {\cos 10^\circ - \sqrt 3\sin 10^\circ}{\sin 10^\circ \cos 10^\circ} \\
&= 2\left(\frac {\frac 12\cos 10^\circ - \frac {\sqrt 3}{2}\sin 10^\circ}{\sin 10^\circ \cos 10^\circ}\right) \\
&= 2\left(\frac {\sin 30^\circ\cos 10^\circ - \cos 30^\circ\sin 10^\circ}{\sin 10^\circ \cos 10^\circ}\right) \\
(a) \ldots &= 2\left(\frac {\sin (30^\circ-10^\circ)}{\sin 10^\circ \cos 10^\circ}\right) \\
&= 2\left(\frac {\sin 20^\circ}{\sin 10^\circ \cos 10^\circ}\right) \\
(b) \ldots &= 2\left(\frac {2\sin 10^\circ \cos 10^\circ}{\sin 10^\circ \cos 10^\circ}\right) \\
&= 4
\end{align*}
$(a) \ldots \rightarrow \sin (X-Y)=\sin X \cos Y - \sin Y \cos X $
$(b) \ldots \rightarrow \sin (2X)=2\sin X \cos X $
|
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|
Sum of the squares of the sides of a triangle
I have solved this problem, but I had to use a calculator, how do I solve it without using a calculator(wont be allowed to use a calculator in the examination).
Here is my attempt:
|
$$a^2+b^2+c^2=4R^2(\sin^2\frac{\pi}{7}+\sin^2\frac{2\pi}{7}+\sin^2\frac{4\pi}{7})=$$
$$=2R^2\left(3-\cos\frac{2\pi}{7}-\cos\frac{4\pi}{7}-\cos\frac{8\pi}{7}\right)=2R^2\left(3+\frac{1}{2}\right)=7R^2$$
Because $$\cos\frac{2\pi}{7}+\cos\frac{4\pi}{7}+\cos\frac{8\pi}{7}=\frac{2\sin\frac{\pi}{7}\cos\frac{2\pi}{7}+2\sin\frac{\pi}{7}\cos\frac{4\pi}{7}+2\sin\frac{\pi}{7}\cos\frac{8\pi}{7}}{2\sin\frac{\pi}{7}}=$$
$$=\frac{\sin\frac{3\pi}{7}-\sin\frac{\pi}{7}+\sin\frac{5\pi}{7}-\sin\frac{3\pi}{7}+\sin\frac{9\pi}{7}-\sin\frac{7\pi}{7}}{2\sin\frac{\pi}{7}}=-\frac{1}{2}$$
|
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Is the product of two consecutive integers $1$ $\pmod n$? Is there a simple test for $n$ to determine if there exists an integer $x$ such that $x$($x+1$) $=$ $1$ $\pmod n$. For example, $n$ $=$ $3$ and $n$ $=$ $7$, there are no integers $x$ such that $x$($x+1$) $=$ $1$ $\pmod n$. But for $n$ $=$ $5$ and $n$ $=$ $11$, there are integers $x$ such that $x$($x+1$) $=$ $1$ $\pmod n$. There are namely,
$x = 2$,
$2*3$ $=$ $1$ $\pmod 5$
$x = 7$,
$7*8$ $=$ $1$ $\pmod {11}$
It is confusing to find out which integers $n$ have this property. Maybe there is a special (mathematical) property these numbers have. Help is appreciated. Thanks.
|
You are asking when the polynomial $x^2+x-1$ has a root mod $n$. Since $x(x+1)$ is always even, clearly $n$ must be odd for such an $x$ to exist. In that case $2$ is invertible mod $n$, and so by the quadratic formula $x^2+x-1$ has a root mod $n$ iff the discriminant $5$ has a square root mod $n$.
So you are asking for what odd integers $n$ is $5$ a square mod $n$. By the Chinese remainder theorem, $5$ is a square mod $n$ iff it is a square mod $p^k$ for each prime power $p^k$ appearing in the prime factorization of $n$. For $p\neq 2,5$, $n$ is a square mod $p^k$ iff $5$ is a square mod $p$ (for instance, by Hensel's lemma). By quadratic reciprocity, $5$ is a square mod $p$ for odd $p$ iff $p$ is a square mod $5$, i.e. iff $p$ is $0,1,$ or $4$ mod $5$. The case $p=2$ does not matter since $n$ must be odd, and $5$ is a square mod $5^k$ iff $k\leq 1$.
Putting it all together, we find that $x^2+x-1$ has a root mod $n$ iff every prime factor of $n$ is $0,1,$ or $4$ mod $5$ and $25$ does not divide $n$.
|
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|
Proving functions are linearly independent I'm currently going through Harvard's Abstract Algebra using Michael Artin's book, and have no real way of verifying my proofs, and was hoping to make sure that my proof was right.
The question reads:
Let $V$ be the vector space of functions on the interval $[0, 1]$. Prove that the functions $x^{3}$, $\sin(x)$, and $\cos(x)$ are linearly independent.
My proof goes as follows:
For these to be linearly dependent there must exist an $a_{i} \neq0$, where $ i = 1, 2, 3$ such that
$$a_{1}x^{3} + a_{2}\sin(x) + a_{3}\cos(x) = 0. $$
So, we'll do this in 3 cases:
Case 1: $x = 0$
In this case, $x^{3} = 0$, $\sin(x) = 0$ but $\cos(x) = 1$.
So, we have
$$0\times a_{1} + 0\times a_{2} + 1\times a_{3} = 0.$$
So, $a_{1}$ and $a_{2}$ could be anything but $a_{3}$ must be 0.
Case 2: $x \in (0,1)$
In this case, $x^{3} \neq 0$, $\sin(x) \neq 0$ and $\cos(x) \neq 0$.
So, for this to be true, $a_{1}$, $a_{2}$ and $a_{3}$ all must be $0$.
Case 3: $x = 1$
In this case, $x^{3} = 1$, $\sin(x) = .8...$ and $\cos(x) = .5...$.
So, we have
$$1\times a_{1} +.8\times a_{2} + .5\times a_{3} = 0.$$
So, $a_{3}$ could be any value, while $a_{1}$ and $a_{2}$ must be $0$.
So, if $a_{1} \neq 0$ then we have a problem in Case 3. If $a_{2} \neq 0$ we have a problem in Case 3. If $a_{3} \neq 0$ we have a problem in Case 1. So, we know that all of the $a$ values must be $0$ and we complete the proof.
|
Your case 2 and case 3 are wrong.
For case 2,
just because
$a_{1}x^{3} + a_{2}sin(x) + a_{3}cos(x) = 0$
does not imply that
the $a_i$ are all zero.
Actually,
for any $x \ne 0$,
and any $a_2, a_3$,
$a_1$ can always be chosen
so that
$a_{1}x^{3} + a_{2}sin(x) + a_{3}cos(x) = 0$
by setting
$a_{1}=-\dfrac{a_{2}sin(x) + a_{3}cos(x)}{x^{3}}$.
For your case 3,
$\sin(1) \ne 1$
and
$\cos(1) \ne 0$.
If you use
$\pi/2$,
then this works.
|
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solving a Diophantine system of two equations Find all triples $(a,b,c) \in \mathbb{N}$ satisfing the following equations:-
\begin{align}
a^2 + b^2 & = c^3 \\
(a + b)^2 & = c^4
\end{align}
Thank you for your help.
|
We have that
$$
c=\frac{(a+b)^2}{a^2+b^2}=\frac{a^2+2ab+b^2}{a^2+b^2}=1+\frac{2ab}{a^2+b^2},
$$
so $\dfrac{2ab}{a^2+b^2}$ must be an integer. But by Cauchy inequality $2ab \leq a^2+b^2$ and an integer can be only if $2ab=a^2+b^2$. It implies that that $a=b$ and $c=2$.
From the first equation we have $2 a^2=c^3=8 \implies a=2.$
Se we get the one solution $(2,2,2).$
If $0 \in \mathbb{N}$ then we have trivial solution $(0, 0,0).$ Also there are two more cases when $2ab \leq a^2+b^2$ is integer for $ a=0 \implies b=1$ and for $b=0 \implies a=1.$ So we get also two solutions $(0,1,1)$ and $(1,0,1).$
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Given $7$ marbles in a box, $3$ are white and numbered from $1$ to $3$ and other $4$ are grey and numbered from $4$ to $7$. Calculate Probabilities Question: You have $7$ marbles in a box, $3$ are white and are numbered from $1$ to $3$ and the other $4$ are grey and are numbered from $4$ to $7$. You extract them one by one till you empty the box. Calculate the following probabilities:
*
*Probability that the first $3$ marbles extracted are white
*Probability that the first three marbles extracted are white and are also in the sequence $[1, 2, 3]$
My Approach
For the first part, I simulated $3$ extractions and I said: first extraction, I have $3$ white marbles out of $7$ total, second extraction I have $2$ out of $6$ and third extraction $1$ out of $5$, so:
$P=\tfrac 3 7\cdot \tfrac 2 6\cdot\tfrac 1 5$
Another idea was using the binomial coefficient: total ways I can extract $3$ marbles out of $7$ is $7\choose 3$, while the way I can extract $3$ white from $3$ total white is $3\choose 3$, so
$$P=\frac{{3\choose 3}}{\binom 7 3}$$
The second part took me some troubles. I thought to think this probability as
$\begin{align}P(\text{First 3 are white & in sequence [1,2,3]}) & ={{P(\text{First White and Number 1})}\\\cdot{P(\text{Second White and Number 2})}\\\cdot{P(\text{Third White and Number 3})}}\\[1ex] & =\tfrac 3 7\cdot\tfrac 1 3\cdot\tfrac 2 6\cdot\tfrac 1 3\cdot\tfrac 1 5\cdot\tfrac 1 3\end{align}$.
Are this calculations correct?
|
From the first part, there are $3!$ sequences of having chosen all white balls, which are as follows: $$\begin{array}{ccc} 1 & 2 & 3\\ 1 & 3 & 2\\ 2 & 1 & 3\\ 2 & 3 & 1\\ 3 & 1 & 2\\ 3 & 2 & 1\end{array}$$
Out of this, only one sequence is in the order $1 2 3$, so, the probability is $$\frac{1}{6} \times P(\text{First Part}) = \frac{1}{6} \times \frac{6}{7\times 6 \times 5} = \frac{1}{210}$$ Hope it helps.
|
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How to deduce $\sin A+\sin B+\sin C=4\cos\frac A2\cos\frac B2\cos\frac C2$ from $A+B+C=\pi$?
If $A+B+C=\pi$,$$\sin A+\sin B+\sin C=4\cos\dfrac A2\cos\dfrac B2\cos\dfrac C2\tag1$$$$\sin A+\sin B-\sin C=4\sin\dfrac A2\sin\dfrac B2\cos \dfrac C2\tag2$$$$\cos A+\cos B+\cos C=4\sin\dfrac A2\sin\dfrac B2\sin\dfrac C2+1\tag3$$$$\cos A+\cos B-\cos C=4\cos\dfrac A2\cos\dfrac B2\sin\dfrac C2-1\tag4$$$$\tan A+\tan B+\tan C=\tan A\tan B\tan C\tag5$$$$\cot\dfrac A2+\cot\dfrac B2+\cot\dfrac C2=\cot\dfrac A2\cot\dfrac B2\cot\dfrac C2\tag6$$
Formulae $(1)$ through $(6)$ were given with the condition that $A+B+C=180^{\circ}$. I'm not sure how to arrive at them.
Question: How do you arrive at $(1)$ through $(6)$?
I need a place to start. I am well aware that$$\sin A+\sin B=2\sin\dfrac {A+B}2\cos\dfrac {A-B}2$$And$$\cos A+\cos B=2\cos\dfrac {A+B}{2}\cos\dfrac {A-B}2$$
However, I'm not sure how to get $\sin A\pm\sin B\pm\sin C$. I'm guessing it has something to do with the expansion of $\sin(A+B+C)$.
Note: In your answer, give a hint on where I can begin, then hide the rest of your answer.
|
hint for all of them: $\sin(C) = 2\sin\left(\frac{C}{2}\right)\cos\left(\frac{C}{2}\right)$, and also the little identities such as $\sin\left(\frac{A+B}{2}\right) = \cos\left(\frac{C}{2}\right)$. And the for $3)$, use $\cos(C) = 1- 2\sin^2\left(\frac{C}{2}\right)$.
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|
Functional equation $f\left(\frac{x+y}{2}\right)+f\left(\frac{2xy}{x+y}\right)=f(x)+f(y)$ implies $2f(\sqrt{xy})=f(x)+f(y)$ Prove that if the function $f$ is defined on the set of positive real numbers, its values are real, and $f$ satisfies the equation
$$f\left( \frac{x+y}{2}\right) + f\left(\frac{2xy}{x+y} \right) =f(x)+f(y)$$
for all positive $x,y$, then
$$2f(\sqrt{xy})=f(x)+f(y)$$
for every pair $x,y$ of positive numbers.
Source: Miklos Schweitzer Memorial Competition 2001
I can see how the repeated application of the functional equation condition upon itself forms a bound, but how can I formally prove this?
|
Solution modified from here
By repeated application of the function property for some positive reals $a,b,c,d$ as you suggested:
$$\begin{align*}f(a)+f(b)+f(c)+f(d)&= f\left({a+b\over2}\right)+f\left({2ab\over a+b}\right)+f\left({c+d\over2}\right)+f\left({2cd\over c+d}\right) \\
&=f\left({{a+b\over2}+{c+d\over2}\over2}\right)+f\left({2{a+b\over2}\cdot{c+d\over2}\over{a+b\over2}+{c+d\over2}}\right)+f\left({{2ab\over a+b}+{2cd\over c+d}\over2}\right)+f\left({2{2ab\over a+b}\cdot{2cd\over c+d}\over{2ab\over a+b}+{2cd\over c+d}}\right) \\ &= f\left({a+b+c+d\over4}\right)+f\left({(a+b)(c+d)\over a+b+c+d}\right)+f\left({abc+abd+acd+bcd\over(a+b)(c+d)}\right)+f\left({4abcd\over abc+abd+acd+bcd}\right) \end{align*}$$Now if we 'swap' $b$ and $c$ and repeat a similar process, one finds $$f\left({(a+b)(c+d)\over a+b+c+d}\right)+f\left({abc+abd+acd+bcd\over(a+b)(c+d)}\right)=f\left({(a+c)(b+d)\over a+b+c+d}\right)+f\left({abc+abd+acd+bcd\over(a+c)(b+d)}\right)\tag{1}$$ Now substitute $a=c$, $b=\frac{a^2}{d}$ and $t=\frac{a}{b}+\frac{b}{a}$ so that $$\frac{(a+b)(c+d)}{ a+b+c+d}=\frac{abc+abd+acd+bcd}{(a+b)(c+d)}=a$$ $${(a+c)(b+d)\over a+b+c+d}=a\cdot{2t\over2+t}$$ and $${abc+abd+acd+bcd\over(a+c)(b+d)}=a\cdot{2+t\over2t}$$ If we substitute these results into $(1)$, we find $$2f(a)=f\left(a\cdot{2t\over2+t}\right)+f\left(a\cdot{2+t\over2t}\right)\tag{2}$$ It can be seen that $t\geq 2$ by AMGM and thus $\frac{2t}{2+t}\geq 1$. Hence for all pairs $x\geq y$ there exist suitable $a$ and $t$ so that $$a\cdot{2t\over2+t}=x,a\cdot{2+t\over2t}=y\text{ and } \sqrt{xy}=a$$ so that $(2)$, and thus the required property, holds.
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How to find parameterization if 2 points are given? It is given $a=\begin{pmatrix} y-z \\ 3x \\ 2x \end{pmatrix} $ and points $A=(3,1,-2)$ and $D=(4,2,1)$. Give parameterization $t\in \left[0,1\right]$
$x=?$ Solution: $t+3$
$y=?$ Solution: $t+1$
$z=?$ Solution: $3t-2$
|
If you're trying to find a parametrization between $A$ and $D$, we can create ${\bf v} = \vec{D} - \vec{A}= (4,2,1) - (3,1,-2) = (1,1,3)$. So, if we want to make a line in $3D$ passing through $A$ and $D$, we need the vector parallel to the line and an initial point. Let's take point $A$.
Then, $$\begin{align}\vec{r}(t) & = \vec{A} + t{\bf v} \\ &= (3,1,-2) + t(1,1,3) \\ &=(3 + t, 1 + t, -2+3t) \\&=(\underbrace{t+3}_{x(t)}, \underbrace{t+1}_{y(t)}, \underbrace{3t-2}_{z(t)}). \tag{Rearrange terms}\end{align}$$
Therefore, we get that $$\begin{align}a &= \pmatrix{y-z\\ 3x\\ 2x} \\&= \pmatrix{(t+1) - (3t-2)\\3(t+3)\\2(t+3)} \tag{Substitute in $x,y,z$}\\ &=\pmatrix{-2t+3 \\ 3t+9 \\ 2t+6} \tag{Combine like terms}\\ &= \pmatrix{-2\\3\\2}t + \pmatrix{3\\9\\6}\tag{Simplify}\end{align}$$
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Definite integration with variable substitution gives wrong result I have the following definite integral:
$$\int^{5}_{-5}{\sqrt{25-x^2} dx}$$
I do a variable substitution:
$$y = 25 - x^2$$
$$x = \sqrt{25 - y}$$
$$dy = -2x~dx$$
$$dx = \frac{dy}{-2x} = -\frac{1}{2\sqrt{25-y}}$$
I get the new integral:
$$-\frac{1}{2}\int^{0}_{0}{\sqrt{\frac{y}{25-y}} dy}$$
From Integral table I get that this integral (24) may be solved as:
$$-\frac{1}{2}(-\sqrt{y(25-y)}-25\tan^{-1}\frac{\sqrt{y(25-y)}}{y-25})\biggr\rvert^0_{0}$$
Going back to $x$ I get:
$$-\frac{1}{2}(-\sqrt{(25-x^2)x^2}-25\tan^{-1}\frac{\sqrt{(25-x^2)x^2}}{-x^2})\biggr\rvert^5_{-5}$$
Then:
$$\frac{1}{2}x\sqrt{25-x^2}+\frac{25}{2}\tan^{-1}\frac{-\sqrt{25-x^2}}{x}\biggr\rvert^5_{-5}$$
But I get $0$ as result, while the right answer is $\frac{25\pi}{2}$.
Where is the mistake?
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$\sqrt{25-x^2}$ is an even function.
So $$\boxed{\int_{-5}^5 \sqrt{25-x^2} dx=2\cdot \int_{0}^5 \sqrt{25-x^2} dx}$$
Now proceed the way you have, by substitution with $y$ and so on. You will get the desired result.
|
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Prove that $(n^3 - 1)n^3(n^3 + 1)$ is divisible by $504$ How to prove that $(n^3 - 1)n^3(n^3 + 1)$ is divisible by $504$? Factoring $504$ yields $2^3 \cdot 3^2 \cdot 7$. Divisibility by $8$ can be easily seen, because if $n$ is even then $8 | n^3$, else $(n^3 - 1)$ and $(n^3 + 1)$ are both even and one of these is divisible by $4$ so $8|(n^3 - 1)(n^3 + 1)$. I'm stuck at proving divisibility by $9$ and $7$
|
Hints
*
*For $7$, what are the possible values of $n^3 \pmod 7$? (It suffices to check $-3, \ldots 3$.)
*For $9$, what are the values of $n^3 - 1, n^3, n^3 + 1 \pmod 9$? For example, for $n = 4 \pmod 9$ we have $n^3 - 1 = 0 \pmod 9$, so the given product is also divisible by $9$.
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If $f_k(x)=\frac{1}{k}\left (\sin^kx +\cos^kx\right)$, then $f_4(x)-f_6(x)=\;?$ I arrived to this question while solving a question paper. The question is as follows:
If $f_k(x)=\frac{1}{k}\left(\sin^kx + \cos^kx\right)$, where $x$ belongs to $\mathbb{R}$ and $k>1$, then $f_4(x)-f_6(x)=?$
I started as
$$\begin{align}
f_4(x)-f_6(x)&=\frac{1}{4}(\sin^4x + \cos^4x) - \frac{1}{6}(\sin^6x + \cos^6x) \tag{1}\\[4pt]
&=\frac{3}{12}\sin^4x + \frac{3}{12}\cos^4x - \frac{2}{12}\sin^6x - \frac{2}{12}\cos^6x \tag{2}\\[4pt]
&=\frac{1}{12}\left(3\sin^4x + 3\cos^4x - 2\sin^6x - 2\cos^6x\right) \tag{3}\\[4pt]
&=\frac{1}{12}\left[\sin^4x\left(3-2\sin^2x\right) + \cos^4x\left(3-2\cos^2x\right)\right] \tag{4}\\[4pt]
&=\frac{1}{12}\left[\sin^4x\left(1-2\cos^2x\right) + \cos^4x\left(1-2\sin^2x\right)\right] \tag{5} \\[4pt]
&\qquad\quad \text{(substituting $\sin^2x=1-\cos^2x$ and $\cos^2x=1-\sin^2x$)} \\[4pt]
&=\frac{1}{12}\left(\sin^4x-2\cos^2x\sin^4x+\cos^4x-2\sin^2x\cos^4x\right) \tag{6} \\[4pt]
&=\frac{1}{12}\left[\sin^4x+\cos^4x-2\cos^2x\sin^2x\left(\sin^2x+\cos^2x\right)\right] \tag{7} \\[4pt]
&=\frac{1}{12}\left(\sin^4x+\cos^4x-2\cos^2x\sin^2x\right) \tag{8} \\[4pt]
&\qquad\quad\text{(because $\sin^2x+\cos^2x=1$)} \\[4pt]
&=\frac{1}{12}\left(\cos^2x-\sin^2x\right)^2 \tag{9} \\[4pt]
&=\frac{1}{12}\cos^2(2x) \tag{10}\\[4pt]
&\qquad\quad\text{(because $\cos^2x-\sin^2x=\cos2x$)}
\end{align}$$
Hence the answer should be ...
$$f_4(x)-f_6(x)=\frac{1}{12}\cos^2(2x)$$
... but the answer given was $\frac{1}{12}$.
I know this might be a very simple question but trying many a times also didn't gave me the right answer. Please tell me where I am doing wrong.
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HINT:
$$\sin^6x+\cos^6x=(\sin^2x+\cos^2x)^3-3\sin^2x\cos^2x(\sin^2x+\cos^2x)=1-3\sin^2x\cos^2x$$
$$\sin^4x+\cos^4x=(\sin^2x+\cos^2x)^2-2\sin^2x\cos^2x=1-2\sin^2x\cos^2x$$
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What will happen to the roots of $ax^2 + bx + c = 0$ if the $a \to 0$? Exercise:
What will happen to the roots of the quadratic equation
$$ax^2 + bx + c = 0$$
if the coefficient $a$ approaches zero while the coefficients $b$ and $c$ are constant, and $b \neq 0$?
Attempt:
$\lim\limits_{a \to 0}{(ax^2 + bx + c)} = bx + c = 0 \longrightarrow x = -\frac{c}{b}$
However, I don't think my solution is complete; shouldn't I end up with $2$ roots? (I've only found $1$.)
Request:
Is there indeed another root to find? If so, how to I find it?
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Quadratic equation $ax^x+bx+c=0$ has two roots, $x_1=\frac{-b+\sqrt{b^2-4ac}}{2a}$ and $x_2=\frac{-b-\sqrt{b^2-4ac}}{2a}$.
We can investigate their behavior when $a \to 0$ by calculating their limits. We assume $b>0$ (we can always mupltiply the equation by -1):
\begin{split}
\lim_{a\to 0}\frac{-b+\sqrt{b^2-4ac}}{2a}=&\lim_{a\to 0}\frac{-b+\sqrt{b^2-4ac}}{2a} \cdot\frac{\sqrt{b^2-4ac}+b}{\sqrt{b^2-4ac}+b} \\
= & \lim_{a\to 0}\frac{b^2-4ac-b^2}{2a(\sqrt{b^2-4ac}+b)} \\
= & \lim_{a \to 0}\frac{-4ac}{2a(\sqrt{b^2-4ac}+b)} \\
= & \lim_{a \to 0}\frac{-4c}{2(\sqrt{b^2-4ac}+b)} \\
= & -\frac{c}{b}, \\
\lim_{a\to 0}\frac{-b-\sqrt{b^2-4ac}}{2a}=&\lim_{a\to 0}\frac{-2b}{2a} \\
= &\lim_{a\to 0}\left( -\frac{b}{a}\right).
\end{split}
The last expression doesn't only depend on the sign of $b$ but also on the sign of $a$, i.e. the direction from which we're approaching zero, so the limit does not exist. The one-sided limits are equal to $\pm \infty$.
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Showing $3^n=\sum_{k=0}^n{(-1)^k\binom{n}{k}4^{n-k}}$. This is Exercise 1.9.3 of F. M. Goodman's "Algebra: Abstract and Concrete".
Show $$3^n=\sum_{k=0}^n{(-1)^k\binom{n}{k}4^{n-k}}.$$
My Attempt:
I tried induction on $n$ as follows.
Base: Try when $n=1$. Then $LHS=3^1=3$ and
$$\begin{align}
RHS&=\sum_{k=0}^1{(-1)^k\binom{1}{k}4^{1-k}} \\
&=(-1)^0\binom{1}{0}4^1+(-1)^1\binom{1}{1}4^0 \\
&=4-1 \\
&=3
\end{align}$$ so the result holds for $n=1$.
Assume the result for $n=r\in\mathbb{N}$. Then $$3^r=\sum_{k=0}^r{(-1)^k\binom{r}{k}4^{r-k}}.$$
Consider when $n=r+1$: I have $3^{r+1}=3\cdot 3^r=3\sum_{k=0}^r{(-1)^k\binom{r}{k}4^{r-k}}$ and have considered
$$\sum_{k=0}^{r+1}{(-1)^k\binom{r+1}{k}4^{r+1-k}}$$ by writing $\binom{r+1}{k}=\binom{r}{k}+\binom{r}{k-1}$ but to no avail.
|
$$\begin{align}
\sum_{k=0}^n(-1)^k\binom nk 4^{n-k}
&=4^n\sum_{k=0}^n \binom nk \left(-\frac14\right)^k\\
&=4^n\left(1-\frac 14\right)^n\\
&=4^n\left(\frac 34\right)^n\\
&=3^n\end{align}$$
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Finding $\lim_{n\to\infty}(\frac{1}{\sqrt{n^2+1}} + \frac{1}{\sqrt{n^2+2}} + ... + \frac{1}{\sqrt{n^2+n}})$ I'm trying to find $\lim_{n\to\infty}(\frac{1}{\sqrt{n^2+1}} + \frac{1}{\sqrt{n^2+2}} + ... + \frac{1}{\sqrt{n^2+n}})$.
*
*I tried to use the squeeze theorem, failed.
*I tried to use a sequence defined recursively: $a_{n+1} = {a_n} + \frac{1}{\sqrt{(n+1)^2 +n+1}}$. It is a monotone growing sequence, for every $n$, $a_n > 0$. I also defined $f(x) = \frac{1}{\sqrt{(x+1)^2 +x+1}}$. So $a_{n+1} = a_n + f(a_n)$. But I'm stuck.
How can I calculate it?
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If $k\in[1,n]$ then the difference between $\frac{1}{\sqrt{n^2+k}}$ and $\frac{1}{n}$ is rather small:
$$ 0\leq \frac{1}{n}-\frac{1}{\sqrt{n^2+k}} = \frac{k}{n\sqrt{n^2+k}(n+\sqrt{n^2+k})}\leq \frac{1}{2n^2} $$
hence $\sum_{k=1}^{n}\frac{1}{\sqrt{n^2+k}}$ tends to $1$ as $n\to +\infty$, since $\sum_{k=1}^{n}\frac{1}{n}=1$ and $0\leq\sum_{k=1}^{n}\frac{1}{2n^2}=\frac{1}{2n}$.
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Prove that the sum of pythagorean triples is always even Problem: Given $a^2 + b^2 = c^2$ show $a + b + c$ is always even
My Attempt, Case by case analysis:
Case 1: a is odd, b is odd. From the first equation,
$odd^2 + odd^2 = c^2$
$odd + odd = c^2 \implies c^2 = even$
Squaring a number does not change its congruence mod 2.
Therefore c is even
$ a + b + c = odd + odd + even = even$
Case 2: a is even, b is even. Similar to above
$even^2 + even^2 = c^2 \implies c$ is even
$a + b + c = even + even + even = even$
Case 3:
One of a and b is odd, the other is even
Without loss of generality, we label a as odd, and b as even
$odd^2 + even^2 = c^2 \implies odd + even = c^2 = odd$
Therefore c is odd
$a + b + c = odd + even + odd = even$
We have exhausted every possible case, and each shows $a + b + c$ is even. QED
Follow Up:
Is there a proof that doesn't rely on case by case analysis?
Can the above be written in a simpler way?
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Note that $x^2\equiv x\pmod 2$ and thus $a^2+b^2=c^2$ implies $$a+b+c\equiv a^2+b^2+c^2\equiv 2c^2\equiv 0\pmod 2$$
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Calculate :$\int_{0}^{\infty}{ }\frac{dx}{a^6+(x-\frac{1}{x})^6}$ Find the:
$$\int_{0}^{\infty}{ }\frac{dx}{a^6+(x-\frac{1}{x})^6}:a>0$$
My Try:
$$\frac{1}{a^6}\int_{0}^{\infty}{ }\frac{dx}{1+(\frac{x-\frac{1}{x}}{a})^6}$$
$$\int_0^\infty{dx\over1+u^6}={1\over2}\left({\pi\over2}+{\pi\over6}+0\right){\pi\over3}$$
Is it right?
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Let $$I = \int^{\infty}_{0}\frac{1}{a^6+\bigg(x-\frac{1}{x}\bigg)^6}dx$$
substitute $\displaystyle x= \frac{1}{t}$ Then $$I = \int^{\infty}_{0}\frac{1}{a^6+\bigg(t-\frac{1}{t}\bigg)^6}\cdot \frac{1}{t^2}dt = \int^{\infty}_{0}\frac{1}{a^6+\bigg(x-\frac{1}{x}\bigg)^6}\cdot \frac{1}{x^2}dx$$
So $$2I = \int^{\infty}_{0}\frac{1}{a^6+\bigg(x-\frac{1}{x}\bigg)^6}\cdot \bigg(1+\frac{1}{x^2}\bigg)dx$$
Substitute $\displaystyle \bigg(x-\frac{1}{x}\bigg)=u,$ then $\displaystyle \bigg(1+\frac{1}{x^2}\bigg)dx = du$
so $$2I = \int^{\infty}_{-\infty}\frac{1}{a^6+u^6}du = 2\int^{\infty}_{0}\frac{1}{a^6+u^6}du=\frac{2}{a^6}\int^{\infty}_{0}\frac{1}{1+\left(\frac{u}{a}\right)^6}du$$
put $\displaystyle \frac{u}{a} = v\;,$ then $\displaystyle I = \frac{1}{a^5}\int^{\infty}_{0}\frac{1}{1+v^6}dv = \frac{\pi}{6a^5}\cdot \frac{1}{\sin \frac{\pi}{6}} = \frac{2\pi}{6a^5} = \frac{\pi}{3a^5}$
Show that $\int_0^ \infty \frac{1}{1+x^n} dx= \frac{ \pi /n}{\sin(\pi /n)}$ , where $n$ is a positive integer.
|
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Most efficient way of calculating primitive cube roots of unity I understand the definition of a primitive cube root of unity in a finite field $\mathbb{F}_p$ to be all those numbers $x$ such that $x^3=1$ but $x\neq 1$ and $x^2 \neq 1$
When we have a small $p$, say $p=7$, we can compute these through 'brute force' - that is filling in the below table:
\begin{array}{|c|cccccc|}
\hline
x & x^1 & x^2 & x^3 & x^4 & x^5 & x^6\\
\hline
1 & \underline{\bf{1}} & 1 & 1 & 1 & 1 & 1\\
2 & 2 & 4 & \bf{\underline{1}} & 2 & 4 & 1\\
3 & 3 & 2 & 6 & 4 & 5 & \underline{\bf{1}}\\
4 & 4 & 2 & \underline{\bf{1}} & 4 & 2 & 1\\
5 & 5 & 4 & 6 & 2 & 3 & \underline{\bf{1}}\\
6 & 6 & \underline{\bf{1}} & 6 & 1 & 6 & 1\\
\hline
\end{array}
We then look along all the rows for any value of $x$ which has the first value of $1$ in the $x^3$ column; in this example we have the primitive cubed roots of unity as $2$ and $4$ (first value of $1$ in each row is bold and underlined in the table above)
However this becomes unfeasible when $p$ becomes very big.
Can someone point me towards an easy method for computing the primitive cube roots of unity which requires as little computation as possible (eventually I will be implementing this in Python using values of $p$ several hundred digits long)
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Perhaps I miss something here, but it all sums up to calculate the roots of $\;x^2+x+1=0\,\pmod p\;$ , and thus we have to know whether $\;-3\;$ is a quadratic residue, and then
$$\binom{-3}p=1\iff p=1\pmod3$$
Thus, for example in $\;\Bbb F_{17}\;$ there are no primitive roots cube roots of unity, but in $\;\Bbb F_{19}\;$ there are : $\;7\;,\;\;11\;$, obtained from the usual quadratic formula for $\;x^2+x+1=0\;$ :
$$\Delta=1-4=-3=16\pmod{19}\implies $$
$$x_{1,2}=\frac{-1\pm\sqrt{16}}2=\frac{-1\pm4}2=\begin{cases}-\frac52=-5\cdot10=-50=-12=7\pmod{19}\\{}\\\frac32=3\cdot10=30=11\pmod{19}\end{cases}$$
Observe that we also know, in general, that if $\;\omega\;$ is primitive cube root then also $\;\omega^2\;$ is, so if we know $\;7\pmod{19}\;$ is , then also $\;7^2=11\pmod{19}\;$ is
|
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|
$5x^2−10x+7$ in completed square form? I was studying quadratic equations and practicing to solve them using the technique of completing the squares.
My answer was as follows:
$$5x^2−10x=−7$$
$$5x^2−10x+25=18$$
$$x(x-5)^2−5(x-5)=18$$
$$\boldsymbol{(x-5)^2-18=0}$$
The answer in the book is:
$$\boldsymbol{5(x-1)^2+2=0}$$
I'm confused. I just started with these sums today. Could you please tell me where I've gone wrong?
|
Generally we look for a zero on the right side of the equation. This is standard form and allows everyone to work in the same way, consistently.
If we do not have an equation but just a quadratic polynomial, put it in normal descending order.
$5x^2−10x+7 = 0 \ \ $ is the standard equation form.
$P(x) = 5x^2−10x+7 \ \ $ is the standard polynomial form
Completing the square is the reverse of squaring a binomial. You should remember that $(x-h)^2 = x^2 - 2hx + h^2$ and so $a(x-h)^2 = ax^2 - 2ahx + ah^2$
We want to work backwards from the right side to the left side. We start by identifying $a$, which is the coefficient of $x^2$.
In your problem, $5x^2−10x+7 = 0 \ \ $, $a = 5$
In the multiplication we multiplied by $a$ so here, working backwards we divide or factor out the $a$. It is a matter of choice whether you divide the constant term or not -- it needs to be adjusted later anyway.
$5(x^2 - 2x) + 7 = 0 \ \ $ OR $5(x^2 -2x + 7/5) = 0$
Now we need to identify the $h$. In the multiplication above the coefficient of the $x$ term (after removing $a$) was $-2h$. So $h$ = coefficient of x divided by $-2$. In your example, $h = (-2/-2) = 1
Finally in the perfect square multiplication above, there is an $h^2$ term. We need to add this in.
But we need to balance the books. We can't just go sticking in a term out of nowhere. There are two general methods:
(a) If you have a quadratic equation $= 0 \ \ $ then you can add the same thing both sides, $h^2 = h^2$.
(b) In any form you can always add zero and the value will not change. The expression $h^2 - h^2$ is a fancy form of zero.
We found $h = 1$ so we add either $1^2 = 1^2$ or zero $= 1^2 - 1^2$. We add this inside the parenthesis because it will be part of the $(x - h)^2$ (If you add both sides be careful to have a factor of $a$ on the right to match the left.)
This adding $h^2$ is completing the square.
$5(x^2 - 2x + 1 - 1) + 7 = 0 \ \ $ OR $5(x^2 -2x + 1 - 1 + 7/5) = 0$
Now we have a perfect square. We know it factors as a square because we built it that way.
We separate out the remaining constant term and simplify.
$5(x^2 - 2x + 1) + 5( - 1) + 7 = 0 \ \ $ OR $5(x^2 -2x + 1) + 5( - 1 + 7/5) = 0$
$5(x- 1)^2 -5 + 7 = 0 \ \ $ OR $5(x-1)^2 + 5( - 5/5 + 7/5) = 0$
$5(x- 1)^2 + 2 = 0 \ \ $
There is the same answer as your text and arrived at logically :-)
Study and learn this system.
|
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|
$\int \sec^3x dx$ in disguise I found this integral,
$$\int \sqrt{x^2+1}dx$$
on a problem list and I think it is a sneaky way of hiding a $\int \sec^3xdx$ problem but I am not sure if what I did was correct though, because of what happens at the end.
So what I did was use trig-substitution and let $u=\tan\theta$ and $du=\sec^2\theta d\theta$
$$\int \sqrt{x^2+1}dx=\int\sqrt{\tan^2\theta+1}\sec^2\theta d\theta=\int \sec^3\theta d\theta$$
$$= \int (\tan^2\theta+1)\sec\theta d\theta = \int \tan^2\theta \sec\theta +\int \sec\theta d\theta$$
I will label $I_s=\int \sec\theta d\theta = \ln|\sec\theta+\tan\theta|$ so I have
$$\int \sec^3xdx=\int\tan^2x\sec x dx+I_s$$
and use integration by parts to solve $\int tan^2\theta \sec\theta d\theta$ that remains. I let $u=\sec\theta$ so $du=\sec\theta\tan\theta d\theta$ and $dv=\tan^2\theta\, d\theta$ such that $v=\tan\theta -\theta$
$$\int\tan^2x\sec x dx =\sec\theta (\tan\theta -\theta) - \int (\tan\theta-\theta)\sec\theta\tan\theta d\theta$$
$$= \frac 12 \sec\theta (\tan\theta -\theta) + \frac 12\int \theta \sec\theta\tan\theta \, d\theta$$
Again I use integration by parts, this time on $\frac 12\int \theta \sec\theta\tan\theta \, d\theta$ and let $u=\theta$ so $du=d\theta$ and $dv=\sec\theta\tan\theta \,d\theta$ so $v=\sec\theta$
$$\frac 12\int \theta \sec\theta\tan\theta \, d\theta = \frac 12 \theta \sec\theta - \frac 12 I_s$$
$$= \frac 12(\theta\sec\theta - I_s)$$
and use this result to get
$$\int\tan^2x\sec x dx = \frac 12 \sec\theta (\tan\theta -\theta) + \frac 12(\theta\sec\theta - I_s)$$
$$= \frac 12 \sec\theta\tan\theta - \frac 12 I_s$$
Putting everything back together and including an arbitrary constant $C$ i get
$$\int \sec^3\theta\,d\theta = \int \tan^2\theta\sec\theta d\theta + I_s$$
$$= \frac 12 \sec\theta\tan\theta - \frac 12 I_s + I_s + C$$
$$=\frac 12\sec\theta\tan\theta +\frac 12 \ln|\sec\theta+\tan\theta|+C$$
and substituting back into $x$-variable terms using $\tan\theta = x$ I have $\sec\theta = \sqrt{x^2+1}$ so
$$\int \sqrt{x^2+1}dx =\frac 12x\sqrt{x^2+1} +\frac 12 \ln|\sqrt{x^2+1}+x|+C$$
The reason why I am not sure about this is because when substituting back I have $\sec\theta = \sqrt{x^2+1}$ which is the same as the integrand $\sqrt{x^2+1}$ that became $\sec^3\theta$ in the initial substitutions. What is going on here?
Thank you all for the precious help!
|
From your trig sub $x=\tan\theta$,
\begin{align}
\int \sqrt{x^2+1}dx&=\int \sec\theta(\sec^2\theta d\theta)
\end{align}
That is, $\sqrt{x^2+1}=\sec\theta$ and $dx=\sec^2\theta d\theta$. The integrand only changed to $sec^3\theta$ after both of these pieces are considered.
Reassuringly, $\sqrt{x^2+1}\ne\sec^3\theta$, because $dx\ne d\theta$.
|
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|
Show that if for prime $p$ there exists $m,n \in \mathbb{N}$ such that $p^2=2^n 3^m + 1$, then $p\leq 17$
Show that if for prime $p$ there exists $m,n \in \mathbb{N}$ such that $p^2=2^n 3^m + 1$, then $p\leq 17$.
I have no idea even how to begin approaching this problem.
|
Suppose $p$ is a prime with the property that the only prime factors of $p-1$ and $p+1$ are $2$ and $3$
We assume $p>17$
Then, one of the numbers $p-1$ and $p+1$ contain only the prime factor $2$, hence is a power of $2$.
Case $1$ : $p-1$ is a power of $2$. Then $p=2^n+1$ and $p+1=2^n+2=2\cdot (2^{n-1}+1)$.
Case $2$ : $p+1$ is a power of $2$. Then $p=2^n-1$ and $p-1=2^n-2=2\cdot (2^{n-1}-1)$
We have $n>4$ because of $p>17$
One of the numbers $2^{n-1}+1$ and $2^{n-1}-1$ has no prime factor other than $3$, and because of $n>4$ . we have $2^{n-1}-1\ge 15>9$.
So, one of the numbers $2^{n-1}+1$ and $2^{n-1}-1$ must be a non-trivial power of $3$. Hence, we get a solution of the equation $2^m-3^n=\pm1$.
Since it is well known that the only solution with $m,n>1$ is $m=3$ and $n=2$ (Catalan's conjecture, now proven) , we get a contradiction.
|
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|
Determine value of $m$ Consider $m = \tan x + \frac{\sin x + \cos x}{\sin x - \cos x}$ . Now determine $m$ so that expression has solution .
I used Wolfram to find range of $f(x) = \tan x + \frac{\sin x + \cos x}{\sin x - \cos x}$ but it was unable to find it!
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$m = \frac{\sin x}{\cos x} + \frac{\sin x + \cos x}{\sin x - \cos x}$
$m = \frac{\sin x(\sin x - \cos x) +\cos x(\sin x + \cos z)}{\cos x(\sin x - \cos x)}$
$m = \frac{\sin^{2}x + \cos^{2}x}{\sin x \cos x - \cos^{2}x}$
$(\sin x \cos x - \cos^{2}x)m - 1 = 0$
$(\tan x - 1)m - (1+\tan^{2}x) = 0$
$tan^{2}x - m\tan x +m+1 = 0$
This has real solutions when the discriminant is greater than or equal to $0$:
$m^{2} - 4(m+1) \geq 0 \Leftrightarrow (m+2)^{2} \geq 0$
So it has solutions $\forall m \in \mathbb{R}$
The range is $\tan x \in \mathbb{R}$
|
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|
Find the value of $\left[\frac{1}{\sqrt 2}+\frac{1}{ \sqrt 3}+......+\frac{1}{\sqrt {1000}}\right]$
Find the value of $$\left[\frac{1}{\sqrt 2}+\frac{1}{\sqrt 3}+.....+\frac{1}{\sqrt {1000}}\right]$$.Where [•] denote the greatest integer function.
I am very confused about this problem. I tried to find the upper and lower bound of the function. But I can't find any formulas to find the bounds. Somebody please help me.
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This is all about providing an accurate approximation for the involved generalized harmonic sum. I will use a technique (creative telescoping) clearly outlined in the first chapter of these course notes.
We may notice that
$$ \sqrt{n+\frac{1}{2}}-\sqrt{n-\frac{1}{2}} = \frac{1}{\sqrt{n+\frac{1}{2}}+\sqrt{n-\frac{1}{2}}} $$
is a telescopic term and it is, additionally, pretty close to $\frac{1}{2\sqrt{n}}$. In particular
$$ \frac{1}{\sqrt{n}}-2\left(\sqrt{n+\frac{1}{2}}-\sqrt{n-\frac{1}{2}}\right)= d_n\\=-\frac{1}{2\sqrt{n}\left(\sqrt{n+1/2}+\sqrt{n-1/2}\right)^2\left(\sqrt{n+1/2}+\sqrt{n}\right)\left(\sqrt{n-1/2}+\sqrt{n}\right)}$$
is a negative term that behaves like $-\frac{1}{32 n^{5/2}}$ for large $n$s. It follows that
$$ \sum_{k=2}^{1000}\frac{1}{\sqrt{k}} = 2\sum_{k=2}^{1000}\left(\sqrt{k+\frac{1}{2}}-\sqrt{k-\frac{1}{2}}\right)+\sum_{k=2}^{1000}d_k $$
has a distance from
$$ 2\sum_{k=2}^{1000}\left(\sqrt{k+\frac{1}{2}}-\sqrt{k-\frac{1}{2}}\right) = 2\left(\sqrt{1000+\frac{1}{2}}-\sqrt{2-\frac{1}{2}}\right) $$
that is less$^{(*)}$ than $\frac{8}{1000}$. By computing the last quantity it follows that the answer is $\color{red}{60}$.
$(*)$ Proof: we have
$$ |d_k|\leq \frac{1}{48 k^{3/2}}-\frac{1}{48(k+1)^{3/2}} $$
hence
$$ \sum_{k=2}^{1000}|d_k|\leq \sum_{k\geq 2}|d_k|\leq \frac{1}{48\cdot 2^{3/2}}<\frac{8}{1000}.$$
|
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Infinitely many $n$ such that $n, n+1, n+2$ are each the sum of two perfect squares. Prove that there exist infinitely many integers $n$ such that $n$, $n+1$, $n+2$ are all the sum of two perfect squares. Induction does not seem to be yielding any results.
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If we can find $y$ such that $y^2-1$ is a sum of two squares, say $y^2- 1 = a^2+b^2$, then take $n = y^2-1$ and we'll have $n = y^2-1 = a^2+b^2$, $n+1 = y^2-1 +1 = y^2 +0$, $n+2 = y^2 + 1$. So, we want to show the existence of infinitely many $y$ such that $y^2-1$ is a sum of two squares.
Look at $3^{2^k}$ for integers $k \geq 1$. For $k = 1$, $3^2 - 1= 8 = 2^2 + 2^2$.
Now, $3^{2^2} -1 = 3^4 -1 = (3^2-1)(3^2+1) = (2^2 +2^2)(3^2+1^2) $
Note the identity $(p^2+q^2)(c^2+d^2) = (pc+qd)^2 + (pd-qc)^2$. So we know that the product of two sums of two squares is again a sum of two squares. And hence, without explicitly calculating, we can say $3^{2^2}-1$ is a sum of two squares.
Now suppose for the sake of induction that $3^{2^k} - 1$ is a sum of two squares. Then $3^{2^{k+1}} -1 = (3^{2^k}-1)((3^{2^{k-1}})^2 +1^2)$. Thus from the induction hypothesis and our neat identity, $3^{2^{k+1}} -1$ is a sum of two squares.
So, for all integers $k \geq 1$, taking $y = 3^{2^k}$ and $n = y^2-1$gives us infinitely many integers satisfying the problem's requirement.
|
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|
Determine whether $\int^{\infty}_{2} \frac{1}{x\sqrt{x^2-4}} dx$ converges. Determine whether $$\int^{\infty}_{2} \frac{1}{x\sqrt{x^2-4}} dx$$ converges.
Doing some rough work, I realize that this function near $\infty$, behaves like $$\frac{1}{x\sqrt{x^2}} = \frac{1}{x^2}$$
I know this function converges, but I am having a hard time finding a $larger$ function that converges too.
I know that:
$$\frac{1}{x\sqrt{x^2-4}} \leq \frac{1}{\sqrt{x^2-4}}$$
but the bigger function diverges, so it doesn't work.
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Easier and more accurate to evaluate directly. I took a look at that integral and it reminded me of something to do with secant or was that inverse secant? Tried an integral table quick search but failed to find the one needed. Paper and pen a couple of minutes gives:
By trig substitution
Let $x/2 = \sec u;\ \ x = 2 \sec u$
$dx = 2 \sec u \tan u \ du; \ \ \sqrt{x^2 - 4} = \sqrt{4\sec^2 u -4} = 2 \tan u $
$$ \int\frac{dx}{x\sqrt{x^2 - 4}} = \int \frac{2 \sec u \tan u \ du}{ 2 \sec u \ 2 \tan u} = \frac{1}{2}\int du = u = \frac{1}{2}\sec^{-1}(\frac{x}{2}) $$
$$ \int_2^{\infty}\frac{dx}{x\sqrt{x^2 - 4}}=\lim_{b \rightarrow \infty} \frac{1}{2}\sec^{-1}(\frac{x}{2})]_{x=2}^b $$
$$ = \frac{1}{2}(\lim_{b \rightarrow \infty} \sec^{-1}(\frac{b}{2}) - \sec^{-1}(\frac{2}{2})) $$
$$ = \frac{1}{2} (\pi/2 - 0) = \pi/4$$
So yes it converges and we have the limit.
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Prove that $a^2+\dfrac{1}{a^a-a+1}\ge a+1$, for any number $a$
Prove that $a^2+\dfrac{1}{a^a-a+1}\ge a+1$, for any number $a$
I think I can solve the problem by putting all the variables on one side and then $0$ on the other side, then factoring the side with the variables to become a square. This proves the inequality because anything squares is positive. Can I solve this problem using algebra? Obviously you can solve it using induction.
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Let $f:[1,\infty) \to \mathbb{R}$ be defined by
$$f(a) = a^2 - a - 1 + \frac{1}{a^a - a + 1}$$
The goal is to show $f(a) \ge 0$ for all $a \in [1,\infty)$.
First suppose $a \ge 2$. Then
\begin{align*}
f(a) &= a^2 - a - 1 + \frac{1}{a^a - a + 1}\\[6pt]
&> a^2 - a - 1\\[6pt]
&\ge 2^2 - 2 - 1 = 1\\[6pt]
&> 0\\
\end{align*}
Next suppose $1 \le a < 2$. Then
\begin{align*}
f(a) &= a^2 - a - 1 + \frac{1}{a^a - a + 1}\\[6pt]
&\ge a^2 - a - 1 + \frac{1}{a^2 - a + 1}\\[6pt]
&= \left(a^2 - a + 1 + \frac{1}{a^2 - a + 1}\right) - 2\\[6pt]
&\ge 2 - 2\qquad\text{[by AM-GM]}\\[6pt]
&=0\\
\end{align*}
Therefore $f(a) \ge 0$ for all $a \in [1,\infty)$, as required.
|
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Proving: $\left(a+\frac1a\right)^2+\left(b+\frac1b\right)^2\ge\frac{25}2$
For $a>0,b>0$, $a+b=1$ prove:$$\bigg(a+\dfrac{1}{a}\bigg)^2+\bigg(b+\dfrac{1}{b}\bigg)^2\ge\dfrac{25}{2}$$
My try don't do much, tough
$a+b=1\implies\dfrac{1}{a}+\dfrac{1}{b}=\dfrac{1}{ab}$
Expanding: $\bigg(a+\dfrac{1}{a}\bigg)^2+\bigg(b+\dfrac{1}{b}\bigg)^2\\=a^2+b^2+\dfrac{1}{a^2}+\dfrac{1}{b^2}+4\\=(a+b)^2-2ab+\bigg(\dfrac{1}{a}+\dfrac{1}{b}\bigg)^2-\dfrac{2}{ab}+4\\=4-2ab-\dfrac{2}{ab}+1+\dfrac{1}{a^2b^2}\\=4-2\bigg(\dfrac{a^2b^2+1}{ab}\bigg)+\dfrac{a^2b^2+1}{a^2b^2}\\=4-\bigg(\dfrac{a^2b^2+1}{ab}\bigg)\bigg(2-\dfrac{1}{ab}\bigg)\\=4-\bigg(ab+\dfrac{1}{ab}\bigg)\bigg(2-\dfrac{1}{ab}\bigg)$
Seems to using Cauchy-Schwartz. Please help.
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Ok, so we have that $a+b=1$.
From AM-GM, this immediately tells us that $ab \leq 0.25$. Furthermore, since you said that $\frac{1}{a} + \frac{1}{b} = \frac{1}{ab}$, this shows that $\frac{1}{a} + \frac{1}{b} \geq 4$.
Now, we are given two numbers $a + \frac 1a$ and $b+ \frac 1b$. Their sum is greater than or equal to $5$, since $a+b=1$. Let their sum be $c$.
Then, by AM-GM, $(a+\frac{1}{a})(b+\frac 1b) \leq \frac{c^2}{4}$, whose minimum value is $6.25$, since $c \geq 5$.
Hence, it follows that $(a+\frac 1a)^2 + (b + \frac 1b)^2 = c^2 - 2(a+\frac{1}{a})(b+\frac 1b) \geq 25 - 12.5 \geq 12.5$, since $c \geq 5$.
This proof required nothing more than just ordinary AM-GM. Sometimes it's nice to think simple.
|
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Prove by induction that $n^4+2n^3+n^2$ is divisible by 4
I'm trying to prove by induction that $n^4+2n^3+n^2$ is divisible by 4.
I know that P(1) it's true. Then $ n=k, P(k):k^4+2k^3+k^2=4w$ it's true by the hypothesis of induction.
When I tried to prove $n=k+1, P(k+1):(k+1)^4+(k+1)^3+(k+1)^2 = 4t$,
$$k^4+4k^3+6k^2+4k+1+2k^3+6k^2+6k+2+k^2+2k+1 = 4t.$$
$$(k^4+2k^3+k^2)+4k^3+6k^2+4k+1+6k^2+6k+2+2k+1 = 4t.$$
Can I replace $(k^4+2k^3+k^2)$ of the previous expression by $4w$ and then continue the proof? Is this correct?
Thanks.
|
That method will work fine, but you can simplify it by noting that $ f(n) = n^2(n+1)^2$ therefore $\,n,$ or $n+1$ is even, so $4$ divides its square, so also $f(n)$ (provable directly or by induction if need be).
|
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Is this elementary proof of FLT correct? Consider the classic FLT for $n=3$, $x^3+y^3=z^3 $
Without loss of generality,we can rewrite as $a^3+(b-a)^3=c^3…Equation (1)$
We can also assume there is no common factor between $a,b-a$ and $c …Assumption (1)$
Since $m+n$ is a factor of $m^3+n^3$, $b$ divides $c^3$.
Let $p$ be a common prime factor of $b$ and $c … Assumption (2)$
$p$ does not divide $a$ as if it did, $p$ divides $b-a$ and hence $p$ divides $a$ violating $Assumption (1)$
Rewriting Equation(1) we get, $a^3+b^3-a^3-3b^2 a+3ba^2=c^3…Equation (2)$
rearranging we get, $b^3-c^3=3b^2 a-3ba^2 …Equation (3)$
Note that $p^3$ divides the LHS of $Equation (3)$
Two cases:
Case I: $p \ne $n
On the RHS of $Equation (3)$, $p^2$ divides the first term $3b^2 a$ but not $3ba^2$
Case II: $p = $n
Now $p^3$ divides $3b^2a$ but not $3ba^2$.
In both cases, LHS of $Equation (3) \ne RHS $
The General case of FLT for any odd prime n:
Without loss of generality,we can rewrite FLT as $a^n+(b-a)^n=c^n…Equation (4)$
Expanding, we get
$a^n+b^n-nC_1 b^{n-1} a+nC_2 b^{n-2} a^2-⋯-nC_{n-2} b^2 a^{n-2}+nC_{n-1} ba^{n-1}-a^n=c^n…Equation (5)$
rearranging we get,
$b^n-c^n=nC_1 b^{n-1} a- nC_2 b^{n-2} a^2+ ⋯+nC_{n-2} b^2 a^{n-2}-nC_{n-1} ba^{n-1}…Equation (6)$
Therefore $p^n$ divides the LHS of $Equation (6)$
Two cases.Case I: $p \ne n$
On RHS of Equation (6), $p^2$ divides all the terms except the last term $nC_{n-1} ba^{n-1}$
Case II: $p=n$
Now $p^3$ divides all terms of RHS except the last term $nC_{n-1} ba^{n-1}$
In both cases, LHS of $Equation (6) \ne RHS $
|
No. What makes you think $b$ and $c$ have to have a common prime factor?
|
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Does there exist $a \in \mathbb{Q}$ such that $a^2 - a + 1$ is a square? Does there exist $a \in \mathbb{Q}$ such that $a^2 - a + 1$ is a square?
Context for this question: For pedagogical purposes I was trying to create an example of a cubic polynomial with both its critical values and its zeros all at integers. After fumbling around with it unsuccessfully for a while, I realized that the problem of creating such an example was equivalent to finding a rational number $a$ ($a \ne 0,1$) with the property that $a^2 - a + 1$ is a perfect square. (Equivalently, find two integers $m,n$ such that $m^2 - mn + n^2$ is a perfect square.) I haven't been able to find an example and strongly suspect that it's not possible, but can't see a simple proof of why that should be so.
Any proof or counterexample to the conjecture that no such $a$ exists?
EDIT: I may have found the beginning of an argument, within a few minutes of posting the question. I'd appreciate feedback.
Suppose $b$ is a positive integer such that $a^2-a+1=b^2$. Then $a-1 = a^2-b^2$, so $a-1= (a-b)(a+b)$. Now if $a-b > 1$ then we have
$$a-1 = (a-b)(a+b) > a+b$$ whence $b < -1$, contradicting the assumption that $b$ is a positive integer.
So if such a $b$ exists, it must be that $a-b \le 1$. Then.... (?)
|
Here is an alternative solution that I came up with shortly after Ahmed S. Attaalla posted his answer.
If $a^2 - a + 1 = b^2$
for some $b \in \mathbb{Q}$ then $a^2 - a + (1-b^2) = 0$, whence by the quadratic formula we have
$$a = \frac{1 \pm \sqrt{1 - 4\left(1-b^2\right)}}{2} = \frac{1 \pm \sqrt{4b^2-3}}{2}$$
If this is to be rational, then we need $4b^2 - 3$ to be a rational square. In other words we need to find $c$ such that $4b^2 - 3 = c^2$, or equivalently such that $\left(2b\right)^2 - c^2 = 3$.
Now the problem has been reduced to: "Find two rational numbers (in this base $2b$ and $c$) whose squares differ by a given number (in this case $3$)." This is a classical problem whose solution is given in Diophantus's Arithmetica (Book II, Problem 10). The solution given there (generalized and in modern language and notation) runs more or less as follows:
Choose any nonzero rational parameter $k$, and set $2b = c + k$. Then the condition $\left(2b\right)^2 - c^2 = 3$ reduces to
$$(c+k)^2 - c^2 = 3$$
$$c^2 + 2ck + k^2 - c^2 = 3$$
$$c = \frac{3-k^2}{2k}$$
Now follow the breadcrumbs back to the original problem: We have
$$2b = \frac{3-k^2}{2k} + k = \frac{3+k^2}{2k}$$
$$b= \frac{3+k^2}{4k}$$
So
$$\begin{align*}a &= \frac{1 \pm \sqrt{4\left(\frac{3+k^2}{4k}\right)^2-3}}{2}\\
&= \frac{1 \pm \sqrt{4\left(\frac{9+6k^2+k^4}{16k^2}\right)-3}}{2}\\
&=\frac{1\pm\sqrt{\frac{9-6k^2+k^4}{4k^2}}}{2}\\
&=\frac{1 \pm \frac{3-k^2}{2k}} {2}\\
&=\frac{2k \pm (3 - k^2)}{4k}\\
\end{align*}$$
To summarize: for any rational $k\ne 0$, we have $a = \frac{2k \pm (3 - k^2)}{4k}$ are two rational numbers satisfying the conditions of the problem.
For example, with $k$ = 1, we get $a = \frac{2\pm(2)}{4}$, which produces the two trivial solutions $a=0,1$. (It's interesting to note that $k=3$ also produces the same trivial solution.)
For a nontrivial example, with $k=4$, we get $a = \frac{8\pm 13}{16}$. We can confirm that
$$\left(\frac{21}{16} \right)^2 - \left(\frac{21}{16} \right) + 1 =\frac{441-336+256}{256} = \frac{361}{256} = \left(\frac{19}{16}\right)^2 $$
and
$$\left(\frac{-5}{16} \right)^2 - \left(\frac{-5}{16} \right) + 1 =\frac{25+80+256}{256} = \frac{361}{256} = \left(\frac{19}{16}\right)^2 $$
|
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Infinite series of formula How do I compute this infinite series
$$ \sum_{k=0}^{\infty}{\frac{(7n+32)(3^n)}{n(n+2)(4^n)}}$$
I believe partial fraction decomposition is part of the solving method but am a little stuck because the $3^n$ and the $4^n$ make the decomposition method a bit strange:
${\frac{(7n+32)(3^n)}{n(n+2)(4^n)}} = {\frac{A}{n}} + {\frac{B}{n+2}} + {\frac{C}{4^n}}$
$(7n+32)(3^n) = A(n+2)(4^n) + B(n)(4^n) + C(n)(n+2)$
$7n(3^n) + 32(3^n) = An (4^n) + 2A(4^n) + B(n)(4^n) + C(n^2 + 2n)$
$7n(3^n) + 32(3^n) = (An +2A + Bn)(4^n) + C(n^2 + 2n)$
..Stuck on how to proceed.
|
I would do it as follows.
$$ \sum_{n=0}^{\infty}{\frac{(7n+32)(3^n)}{n(n+2)(4^n)}}=
7\sum_{n=0}^{\infty}{\frac{1}{n+2}{\left(\frac34\right)}^n} +
32\sum_{n=0}^{\infty}{\frac{1}{n(n+2)}{\left(\frac34\right)}^n}$$
Now, we have that
$$\sum_{n=0}^{\infty}{\frac{1}{n+2}{\left(\frac34\right)}^n}=
{\left(\frac34\right)}^{-2}\cdot\sum_{n=0}^{\infty}{\frac{1}{n+2}{\left(\frac34\right)}^{n+2}}\\
={\left(\frac34\right)}^{-2}\cdot\sum_{n=2}^{\infty}{\frac{1}{n}{\left(\frac34\right)}^{n}}$$
Now, for $|r|<1$, we have that
$${(1-r)}^{-1}=\sum_{n\geq0}r^n,$$
so that integrating both sides by $r$ yields
$$-\ln(1-r)=\sum_{n\geq0}\frac1{n+1}r^{n+1}=\sum_{n\geq1}\frac1nr^n.$$
It follows that
$$\sum_{n=2}^{\infty}{\frac{1}{n}{\left(\frac34\right)}^{n}}=-\ln\left(1-\frac34\right)-\frac34=2\ln(2)-\frac34.$$
Finally, we use a partial fraction decomposition on $\frac1{n(n+2)}=\frac12\frac1n-\frac12\frac1{n+2}$, and this splits the final series into two series that can be solved like the one we just did. Do you think you can finish it?
EDIT:
Well, I just noticed that in your opening question there is a division by $0$ in the $n=0$ term, but this shouldn't change your approach to solving the question.
Assuming the summation is carried out for $n\geq1$, you should obtain $\frac{33}2$ in the end.
|
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|
Determine the eigenvector and eigenspace and the basis of the eigenspace The yellow marked area is correct, so don't check for accuracy :)
$A=\begin{pmatrix} 0 & -1 & 0\\ 4 & 4 & 0\\ 2 & 1 & 2
\end{pmatrix}$ is the matrix.
Characteristic polynomial is $-\lambda^{3}+6\lambda^{2}-12\lambda+8=0$
The (tripple) eigenvalue is $\lambda=2$.
Calculate the eigenvectors now:
$\begin{pmatrix}
-2 & -1 & 0\\ 4 & 2 & 0\\ 2 & 1 & 0 \end{pmatrix} \begin{pmatrix} x\\ y\\ z \end{pmatrix}= \begin{pmatrix} 0\\ 0\\ 0
\end{pmatrix}$
We get the equations:
$I: -2x-y=0 \Leftrightarrow y = -2x$
$II: 4x+2y=0$
$III: 2x+y=0 \Leftrightarrow 2x-2x=0 \Leftrightarrow 0=0$
We see that in every eequation $z$ is unknown, so we can choose an arbitrary $z$.
$x\begin{pmatrix}
1\\
-2\\
z
\end{pmatrix}$ and this is the eigenspace...?
And what is the basis of this eigenspace? Can I just set $x=1$ and some value for $z$? So this would be a correct basis of the eigenspace: $\begin{pmatrix}
1\\
-2\\
3
\end{pmatrix}$?
Now we need three linearly independent eigenvectors but I couldn't find them as I always got linearly dependent vectors...
I need a detailled, not too complicated answer that explains it well and I will give that answer a nice bounty (up to 200 rep) because I couldn't find another site explaining this correctly to me and I'm really in need of it.
|
The $x$ shouldn't be outside the vector. The solution to equations I,II, and III is
\begin{pmatrix}
x\\
-2x\\
z
\end{pmatrix}
where $x$ and $z$ are arbitrary. Every vector of this form is an eigenvector for $A$. You can write each such vector as a linear combination of two vectors $e_1$ and $e_2$ defined by
$$e_1:=
\begin{pmatrix}
1 \\ -2 \\ 0
\end{pmatrix}$$
and
$$e_2:=
\begin{pmatrix}
0 \\ 0 \\ 1
\end{pmatrix}.$$
More concretely, we have
$$
\begin{pmatrix}
x \\ -2x \\ z
\end{pmatrix}
=
\begin{pmatrix}
x \\ -2x \\ 0
\end{pmatrix}
+
\begin{pmatrix}
0 \\ 0 \\ z
\end{pmatrix}
=
x\begin{pmatrix}
1 \\ -2 \\ 0
\end{pmatrix}
+
z\begin{pmatrix}
0 \\ 0 \\ 1
\end{pmatrix}
=
xe_1 + ze_2$$
A basis for the eigenspace is the two vectors $e_1$ and $e_2$, since every vector in the eigenspace can be written uniquely as a linear combination of those two vectors.
There's no reason it should have 3 linearly independent eigenvectors.
|
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|
Is this matrix diagonalisable or not? We will only be looking at one specific eigenvalue, eigenspace (there are three eigenvalues in total, I know two of them are fine and last one seems not but as I'm not sure I need to ask you).
We have matrix $A=\begin{pmatrix}
3 & -1 & 0\\
2 & 0 & 0\\
-2 & 2 & -1
\end{pmatrix}$
The eigenvalues are $\lambda_{1}=1, \lambda_{2}=2, \lambda_{3}=-1$
We will only concentrate on $\lambda_{3}=-1$ because I have checked the others already and I'm very sure they are fine (they show that the matrix is diagonalisable).
We insert $\lambda_{3}=-1$ here $\begin{pmatrix}
3-\lambda & -1 & 0\\
2 & -\lambda & 0\\
-2 & 2 & -1-\lambda
\end{pmatrix} \begin{pmatrix}
x\\
y\\
z
\end{pmatrix}=\begin{pmatrix}
0\\
0\\
0
\end{pmatrix}$
Then we get $\begin{pmatrix}
4 & -1 & 0\\
2 & 1 & 0\\
-2 & 2 & 0
\end{pmatrix} \begin{pmatrix}
x\\
y\\
z
\end{pmatrix}=\begin{pmatrix}
0\\
0\\
0
\end{pmatrix}$
$I: 4x-y=0 \Leftrightarrow y=4x$
$II: 2x+y=0$
$III: -2x+2y=0$
We see that there is no value for $z$, so we can choose an arbitrary $z$. We get the eigenspace $E_{A}(-1)=\left\{ \left. \begin{pmatrix}
x\\
4x\\
z
\end{pmatrix} \right | x,z \in \mathbb{R} \right\}$
Two variables mean the basis is made up by $2$ vectors and this means the dimension of the eigenspace is $2$ and this means the matrix is not diagonalisable because $\lambda_{3}=-1$ is just a single eigenvalue?
Can you please tell me if this is correct?
|
From I and II, we have $y = 4x$ and $y = -2x$. The only way these can simultaneously hold is if $x = y = 0$. However, $z$ is indeed a free variable. We find that
$$
E_A(-1) = \{(0,0,z)^T : z \in \Bbb R\}
$$
|
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|
$p^{p+1}+(p+1)^p-1$ a perfect square For which primes $p$ is $$p^{p+1}+(p+1)^p-1$$ a perfect square?
Context:
Once again a modified problem, this time from $p^{p+1}+(p+1)^p$ a perfect square, for which the answer is no such $p$ exist. For the modified form above, however, $p=2,3$ work, the main trouble I've had being that one can't easily find prime divisors of the above number that aren't already divisors of $p$ or $p+1$.
Edit:
I forgot to say that I have already tested this statement up to $p=19$ or as far as the calculators on a computer will go without finding any other examples apart from $2$ and $3$
|
This is my experience :
$A=P^{p+1}+(p+1)^p -1$
$p^{p+1}=(p^2)^\frac{p+1}{2}=k_1 p^2$
$(p+1)^p=∑ C^p_r p^r +1= k_2 p^2 +1$
$⇒A=P^{p+1}+(p+1)^p -1=(k_1+k_2)p^2 $
Also:
$P^{p+1}=(p+1-1)^{p+1}=∑ C^{p+1}_r (p+1)^r +1=t_1 (p+1)^2+1$
$(p+1)^p = (p+1)^2 (p+1)^{p-2}= t_2(p+1)^2$
$⇒A=P^{p+1}+(p+1)^p -1=(t_1+t_2)(p+1)^2 $
$⇒A=P^{p+1}+(p+1)^p -1=m p^2(p+1)^2 $
Where m is a composite integer; for example:
$7^8+8^7-1=23 . 109 . 56^2$
It can be seen that for Fermat primes a power of $2$ is also among the factors of m:
$(2^{2^n}+1)^{2^{2^n}+2}+(2^{2^n}+2)^{2^{2^n}+1}-1= k . (2^{2^n}+1)^2. (2^{2^n}+2)^2.2^{2^n}$
For example:
$17^{18} +18^{17}-1= k .2^4 . 17^2 . 18^2 $
There may exist primes of form $(x^{2^n}-1)x^{2^n}+1$ which result in A such that a composite factor like $m$, composed of even powers of some prime factors of $x$, arises; for example:
$p=15 . 16 + 1= 241$; $ 241^{241}+242^{241}-1=k . 2^4 . 241^2 . 242^2$
Conclusion:
We may conclude that;
a : The solutions are only $p=2$ and $p=3$ because m has square free prime factors.
b: Conjecture: There are primes other than 2 and 3 so that $A=P^{p+1}+(p+1)^p -1$ is perfect square.
|
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|
Prove the inequality, power. $\{ x,y \in\Bbb R\ \}$ If $x+y = 2$ then prove the inequality:
$x^4 + y^4 \ge 2$
How I started
*
*$(x+y)^2 = 4$
*$x^2 + y^2 = 4 - 2xy$
*$(x^2+y^2)^2 - 2(xy)^2 \ge 2$
*$(4-2xy)^2 - 2(xy)^2 \ge 2$
*$16-16xy + 4(xy)^2 -2(xy)^2 - 2 \ge 0$
*$2(xy)^2 - 16xy + 14 \ge 0$
*for $t=xy$
*$2t^2 - 16t + 14 \ge 0$
It isn't always true, I think that I should have a assumption for $t$, but I don't know how should I do this.
|
Also we can use C-S twice:
$x^4+y^4=\frac{1}{2}(1+1)(x^4+y^4)\geq\frac{1}{2}(x^2+y^2)^2=\frac{1}{8}((1+1)(x^2+y^2))^2\geq\frac{1}{8}\left((x+y)^2\right)^2=2$
|
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|
Noob question about $\int \frac{1}{\sin(x)}dx$ I manually integrate $\int \frac{1}{\sin(x)}dx$ as $$\int \frac{\sin(x)}{\sin^{2}(x)}dx = -\int \frac{1}{\sin^{2}(x)}d\cos(x) = \int \frac{1}{\cos^{2}(x) - 1}d\cos(x).$$
After replacing $u = \cos(x)$,
$$\int \frac{1}{u^{2} - 1}du = \int \frac{1}{u^{2} - 1}du = \frac {1} {2} \int \left(\frac {1} {u - 1} - \frac {1} {u+1}\right) du = \frac {1} {2} \ln\left(\frac {u-1} {u+1}\right) + C.$$
Substitute back to obtain
$$\frac {1} {2} \ln\left(\frac {\cos(x)-1}{\cos(x)+1}\right) + C.$$
The problem is that this solution is incorrect (I guess) because for example http://www.integral-calculator.com/ gives another solution
$$\frac {1} {2} \ln\left(\frac {1 - \cos(x)}{1 + \cos(x)}\right) + C.$$
And all other online solvers gives equivalent solution to
$$\frac {1} {2} \ln\left(\frac {1 - \cos(x)}{1 + \cos(x)}\right) + C.$$
The question is there I made a mistake?
Update: some of you may say that in complex space my answer is right but not so fast:
Take wolfram solver: integrate 1/sinx
The we get: $-ln(cot(x) + csc(x)) + C$ It is easy to see that it is equvalent to $$\frac {1} {2} \ln\left(\frac {1 - \cos(x)}{1 + \cos(x)}\right) + C.$$
$-\ln(\cot(x) + \csc(x)) + C = -\ln(\frac {\cos(x)} {\sin(x)} + \frac {1} {\sin(x)})$
then
$-\ln(\frac {\cos(x)} {\sin(x)} + \frac {1} {\sin(x)}) = -\frac {1}{2} \ln(\frac {(1+\cos(x))^{2}} {\sin^{2}x}) = -\frac {1}{2} \ln(\frac{1+\cos(x)+\cos(x)+\cos^{2}(x)} {1-\cos^{2}x}) = -\frac {1}{2} \ln(\frac {(1+\cos(x))(\cos(x)+\cos^{2})(x)} {1-\cos^{2}x}) = -\frac {1}{2} \ln(\frac {1+\cos(x)} {1-\cos(x)}) = \frac {1}{2} \ln(\frac {1-\cos(x)} {1+\cos(x)})$
|
Where did I make a mistake?
First, you rather have
$$
\frac {1} {2} \int \left(\frac {1} {u - 1} - \frac {1} {u+1}\right) du = \frac {1} {2} \ln\left|\frac {u-1} {u+1}\right| + C
$$then observe that
$$
\left|\frac {\cos(x)-1}{\cos(x)+1}\right|=\frac {1-\cos(x)}{1+\cos(x)}
$$ since
$$
1 - \cos(x)\ge 0,\quad 1+ \cos(x)\ge0,
$$ giving that
$$
\ln\left|\frac {\cos(x)-1}{\cos(x)+1}\right|=\ln\left(\frac {1-\cos(x)}{1+\cos(x)}\right).
$$
|
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|
How to solve this limit? $ \lim_{x \to +\infty} \frac{(\int^x_0e^{x^2}dx)^2}{\int_0^xe^{2x^2}dx}$ There's a limit which confuses me:
$$ \lim_{x \to +\infty} \dfrac{(\int^x_0e^{x^2}dx)^2}{\int_0^xe^{2x^2}dx}$$
Is it possible to use L'Hôpital's rule here?
|
An asymptotic way:
Let $\displaystyle F:x\mapsto \int_0^x e^{t^2} dt$.
Note that $$\frac{(\int^x_0e^{t^2}dt)^2}{\int_0^xe^{2t^2}dt} = \sqrt 2\frac{(F(x))^2}{F(\sqrt 2 x)}$$
Moreover, $$\begin{align}
F(x)&=\int_0^1 e^{t^2} dt + \int_1^x \frac{1}{2t}\cdot 2te^{t^2}dt \\
&= \int_0^1 e^{t^2} dt + \frac{e^{x^2}}{2x}-\frac e2 +\frac 12 \int_1^x \frac{e^{t^2}}{t^2}dt
\end{align}$$
The same trick performed on $\displaystyle \int_1^x \frac{e^{t^2}}{t^2}dt$ yields
$$\begin{align}
F(x)&= \int_0^1 e^{t^2} dt + \frac{e^{x^2}}{2x}-\frac e2 + \frac{e^{x^2}}{4x^3} - \frac e4 + \frac 34 \int_1^x \frac{e^{t^2}}{t^4}dt
\end{align}$$
Note that for large enough $x$, $\displaystyle \int_1^x \frac{e^{t^2}}{t^4}dt\leq (x-1)\frac{e^{x^2}}{x^4}\leq \frac{e^{x^2}}{x^3}$.
Therefore, $$F(x)=\frac{e^{x^2}}{2x} + \frac{e^{x^2}}{4x^3} + O\left(\frac{e^{x^2}}{x^3}\right)$$
For the purpose of the question, it is enough to write $$F(x)=\frac{e^{x^2}}{2x} + o\left(\frac{e^{x^2}}{2x}\right)$$
which is the same as $\displaystyle F(x)\sim \frac{e^{x^2}}{2x}$. This implies $\displaystyle F(\sqrt 2 x)\sim \frac{e^{2x^2}}{2\sqrt 2x}$
Therefore, $$\frac{(\int^x_0e^{t^2}dt)^2}{\int_0^xe^{2t^2}dt} = \sqrt 2\frac{(F(x))^2}{F(\sqrt 2 x)}\sim \frac 1x\xrightarrow[x\to \infty]{}0$$
|
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|
Incorrectly solving the determinant of a matrix Compute $det(B^4)$, where $B =
\begin{bmatrix}
1 & 0 & 1 \\
1 & 1 & 2 \\
1 & 2 & 1
\end{bmatrix}
$
I created
$C=\begin{bmatrix}
1 & 2 \\
1 & 1 \\
\end{bmatrix}
$ and
$
D= \begin{bmatrix}
1 & 1 \\
1 & 2 \\
\end{bmatrix}
$
$det(B) = -det(C) -2det(D) = -(1-2) - 2(2-1) = -(-1)-2(1)=1-2=1, 1^4=1$
However, the correct answer is 16. I'm confused on where I made my wrong turn
|
We have
$$
\det(B) = 1 \times \det(E) + 1 \times \det(D) = (1 - 4) + (2 - 1) = -2
$$
where $ E $ is
$$
\begin{bmatrix}
1 & 2 \\ 2 & 1
\end{bmatrix}
$$
Then
$$
\det(B^4) = \det(B)^4 = (-2)^4 = 16
$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2184156",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
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|
Evaluating limits using taylor expansion $$\lim_{x\to 0^{+}} (\ln x)^3\left(\arctan\left(\ln\left(x+x^2\right)\right) + \frac{\pi}{2}\right) + (\ln x)^2$$
I have this limit in my sheet and the answer is $\frac 13$.
But I don't know how to approach any step using taylor expansion near zero (this is our lesson by the way). Please help me with the simplest way.
|
As $x \to 0^+$, one has $\ln x<0$, giving
$$
\arctan\left(\ln\left(x+x^2\right)\right)=-\frac \pi2-\arctan\left(\frac1{\ln\left(x+x^2\right)}\right),
$$ and using $\dfrac1{\ln\left(x+x^2\right)} \to 0$ one gets, by applying a standard taylor series expansion,
$$
\arctan\left(\frac1{\ln\left(x+x^2\right)}\right)=\frac1{\ln\left(x+x^2\right)}-\frac1{3\ln^3\left(x+x^2\right)}+O\left( \frac1{\ln^5\left(x+x^2\right)}\right)
$$ and
$$
\begin{align}
&(\ln x)^3\left(\arctan\left(\ln\left(x+x^2\right)\right) + \frac{\pi}{2}\right) + (\ln x)^2
\\\\&= (\ln x)^2-\frac{(\ln x)^3}{\ln\left(x+x^2\right)}+\frac{(\ln x)^3}{3\ln^3\left(x+x^2\right)}+O\left( \frac{(\ln x)^3}{\ln^5\left(x+x^2\right)}\right)
\\\\&= (\ln x)^2-\frac{(\ln x)^3}{\ln x+O(x)}+\frac{(\ln x)^3}{3\ln^3x+O(x)}+O\left( \frac{(\ln x)^3}{\ln^5x+O(x)}\right)
\\\\&= (\ln x)^2-(\ln x)^2+\frac13+O\left( \frac1{\ln^2x}\right)+O(x)
\\\\& \to \frac13
\end{align}
$$ as announced.
|
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"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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|
Menelaus theorem & collinear points From vertex C of the right triangle ABC height CK is dropped and in triangle ACK bisector CE is drawn. Line that passes through point B parallel to CE meets CK at point F. Prove that line EF divides segment AC in halves.
So far I have:
Construct point M on AC such that AM=MC. WE want to show that M, E, and F are collinear. So we want to prove using Menelaus' that
$\displaystyle\frac{AM}{CM} \cdot \displaystyle\frac{CF}{KF} \cdot \displaystyle\frac{KE}{AE} = 1$.
We know AM=MC so that cancels out to 1. From the angle bisector theorem in triangle ACK we know that $\displaystyle\frac{KE}{AE} = \displaystyle\frac{CK}{CA}$.
So from this we have
$\displaystyle\frac{CF}{KF} \cdot \displaystyle\frac{CK}{CA}=1$. Using the fact that $\triangle CKE$ is similar to $\triangle BFK$ we know $\displaystyle\frac{CK}{KF}=\displaystyle\frac{EK}{KB}$
Now we obtain that $\displaystyle\frac{CF}{CA} \cdot \displaystyle\frac{EK}{KB} = 1$
FROM HERE I AM STUCK.
|
You have to check that by choosing $M$ as the midpoint of $AC$ then
$$ \frac{AM}{CM}\cdot\frac{CF}{KF}\cdot\frac{KE}{AE} = 1 $$
holds, but $\frac{AM}{CM}=1$ and by the bisector theorem $\frac{KE}{AE}=\frac{CK}{CA}$, so it is enough to prove that
$$ CF\cdot CK = AC\cdot KF $$
or
$$ \frac{CA}{CK} = \frac{FC}{FK} = 1+\frac{CK}{FK}=1+\frac{EK}{KB}. $$
If $a,b$ are the lengths of the legs of $ABC$, we have $CK=\frac{ab}{\sqrt{a^2+b^2}},CA=b$, $AK=\frac{b^2}{\sqrt{a^2+b^2}}$, $BK=\frac{a^2}{\sqrt{a^2+b^2}}$ and
$$AE=\frac{b}{b+\frac{ab}{\sqrt{a^2+b^2}}}\cdot\frac{b^2}{\sqrt{a^2+b^2}}=\frac{b^2}{a+\sqrt{a^2+b^2}}$$
$$EK=\frac{\frac{ab}{\sqrt{a^2+b^2}}}{b+\frac{ab}{\sqrt{a^2+b^2}}}\cdot\frac{b^2}{\sqrt{a^2+b^2}}=\frac{ab^2}{a^2+b^2+a\sqrt{a^2+b^2}}$$
The problem boils down to tedious&straightforward algebra.
|
{
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"timestamp": "2023-03-29T00:00:00",
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|
Prove that$ k^3n-kn^3$ is divisible by $6 $ for all n∈N. Hello I have problem with solution of task.
Prove that $k^{3}n-kn^3$ is divisible by $6$ for all $n∈N$, $k∈N$ .
Help me, please.
I know, when $n^3-n$ is divisible by 6.
$n^3-n= (n-1)(n)(n+1)$ and is divisible.
I having similar idea $(kn)^3-(kn)= (kn-1)(kn)(kn+1)$
and my 2 idea is $k^3n−kn^3 =(k^3-k)(n-n^3)+(kn)^3+kn$
|
$k^{3}n-kn^3 = kn(k+n)(k-n)$
If this is divisible by both $2$ and $3$, then it is divisible by $6$.
Assuming $k$ is an integer, if either $k$ or $n$ is even, then the expression is even, and if they are both odd, then $k+n$ is even, so the expression is still even.
If either $k$ or $n$ are divisible by $3$, then the expression is divisible by $3$.
If $k$ and $n$ are either both $\equiv 1\pmod 3$ or both $\equiv 2\pmod 3$, then $k-n$ will be divisible by $3$.
If one of the two variables is $\equiv 1\pmod 3$ and the other is $\equiv 2\pmod 3$ then $k+n$ will be divisible by $3$.
So in all scenarios, the expression is divisible both by $2$ and $3$ and therefore divisible by $6$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2187852",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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|
Relations similar to $\sin(\pi/18) + \sin(5\pi/18) = \sin(7\pi/18)$ This I checked numerically (and can be proven analytically easily). I guess there are many similar relations between the numbers
$$\{ \sin\frac{m\pi}{q}, 1\leq m \leq q \} , $$
where $q$ is some integer.
Can anyone give some reference?
|
I have studied this problem and found two kinds of solutions.
[Solution 1]
$ \cos\left(\frac{\pi k}{3 n}\right)=\cos \left(\frac{\pi (n-k)}{3 n}\right)+\cos \left(\frac{\pi (k+n)}{3 n}\right) $
Example:
$ \cos\left(\frac{\pi }{15}\right)=\cos\left(\frac{4 \pi }{15}\right)+\cos \left(\frac{6 \pi }{15}\right) $
$ \cos \left(\frac{2 \pi }{15}\right)=\cos \left(\frac{3 \pi }{15}\right)+\cos \left(\frac{7 \pi }{15}\right) $
$ \cos \left(\frac{3 \pi }{15}\right)=\cos \left(\frac{5 \pi }{15}\right)+\cos \left(\frac{6 \pi }{15}\right) $
$ \cos \left(\frac{ \pi }{21}\right)=\cos \left(\frac{6 \pi }{21}\right)+\cos \left(\frac{8 \pi }{21}\right) $
$ \cos \left(\frac{2 \pi }{21}\right)=\cos \left(\frac{5 \pi }{21}\right)+\cos \left(\frac{9 \pi }{21}\right) $
$ \cos \left(\frac{3 \pi }{21}\right)=\cos \left(\frac{4 \pi }{21}\right)+\cos \left(\frac{10 \pi }{21}\right) $
$ \cos \left(\frac{ \pi }{27}\right)=\cos \left(\frac{8 \pi }{27}\right)+\cos \left(\frac{10 \pi }{27}\right) $
$ \cos \left(\frac{2 \pi }{27}\right)=\cos \left(\frac{7 \pi }{27}\right)+\cos \left(\frac{11 \pi }{27}\right) $
$ \cos \left(\frac{3 \pi }{27}\right)=\cos \left(\frac{6 \pi }{27}\right)+\cos \left(\frac{12 \pi }{27}\right) $
$ \cos \left(\frac{4 \pi }{27}\right)=\cos \left(\frac{5 \pi }{27}\right)+\cos \left(\frac{13 \pi }{27}\right) $
[Solution 2]
$ \cos \left(\frac{\pi (6 k-3)}{5 (6 k-3)}\right)=\cos \left(\frac{\pi (10 k-5)}{3 (10 k-5)}\right)+\cos \left(\frac{(2 \pi ) (6 k-3)}{5 (6 k-3)}\right) $
Example:
$ \cos \left(\frac{ \pi }{5}\right)=\cos \left(\frac{ \pi }{3}\right)+\cos \left(\frac{2 \pi }{5}\right) $
$ \cos \left(\frac{3 \pi }{15}\right)=\cos \left(\frac{ 5 \pi }{15}\right)+\cos \left(\frac{6 \pi }{15}\right) $
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2188715",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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|
Limit without l'Hopital or Taylor series: $\lim\limits_{x \to 0} \frac{x\cos x- \sin x}{x^3}$ find the limit without l'Hôpital and Taylor rule :
$$\lim\limits_{x \to 0} \frac{x\cos x- \sin x}{x^3}=?$$
My Try :
$$\lim\limits_{x \to 0} \frac{x\cos x- \sin x}{x^3}\\=\lim\limits_{x \to 0}\frac{x\cos x \sin x- \sin x\sin x}{x^3\sin x}=\\\lim\limits_{x \to 0}\frac{x\sin 2x- \sin^2 x}{2x^3\sin x}=$$?
what now ?
|
Write
$$\frac{x\cos x-\sin x}{x^3}=\frac{\cos x-1}{x^2}+\frac{x-\sin x}{x^3} $$
The first fraction goes to $-\frac{1}{2}$ (it follows from $\frac{\sin x}{x} \to 1$, no de l'Hopital or Taylor needed), while the second goes to $\frac{1}{6}$ (see Solving $\lim\limits_{x\to0} \frac{x - \sin(x)}{x^2}$ without L'Hospital's Rule. ). Overall
$$\lim_{x \to 0} \frac{x\cos x-\sin x}{x^3} = -\frac{1}{3}. $$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2189466",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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|
Does the sequence $x_{n+1}=x_n+\frac{1}{x_n}$ converge or diverge? Set $x_1=a$, where $a > 0$ and let $x_{n+1}= x_n + \frac{1}{x_n}$. Determine if the sequence $ \lbrace X_n \rbrace $ converges or diverges.
I think this sequence diverges since Let $x_1= a > 0$ and $x_{n+1}= x_n + (1/x_n)$ be given. Let $x_1=2$ since $a >0$. Then...
$x_2 = 2 + \frac{1}{2} = 2.5$
$x_3 = 2.5 + \frac{1}{2.5} = 2.9$
$x_4 = 2.9 + \frac{1}{2.9} = 3.2448$
$x_5 = 3.2448 + \frac{1}{3.2448} = 3.5530$
$x_6 = 3.5530 + \frac{1}{3.5530} = 3.8344$
$x_7 = 3.8344 + \frac{1}{3.8344} = 4.0952$
$x_8 = 4.0952 + \frac{1}{4.0952} = 4.3394$
$x_9 = 4.3394 + \frac{1}{4.3394} = 4.5698$
$x_{10} = 4.5698 + \frac{1}{4.5698} = 4.7887$
but others have said it converges so I'm confused on whether it converges or diverges? Can someone please explain.
|
We can see the series $x_{n+1}$ is monotone increasing as $n\rightarrow \infty$ using the Monotone Convergence Theorem:
Monotone Increasing $\Rightarrow x_{n+1}>x_{n}, \forall n$
$\Rightarrow x_n+\frac{1}{x_n}>x_n$
$\Rightarrow \frac{1}{x_n} > 0$
Which is true since $x_1=a$ and $a>0$.
So now we know we have an increasing sequence. Lets assume some limit $x$ exists, which implies $x_{n+1}=x_n$, so we solve to 'find' our limit:
$x=x+\frac{1}{x}\Rightarrow x^2=x^2+1$, leading to a contradiction.
Therefore the limit does not exist and the sequence diverges.
|
{
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"timestamp": "2023-03-29T00:00:00",
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|
Expectation of a random variable Suppose $Y$ is normal random variable.
Show that for any integer $n \ge 1$,
$$
E[(Y- \mu)^{2n}] = \sigma^{2n} \frac{2n!}{2^{n}n!}.
$$
I've mangaged to show that that $f(\mu+y)$ = $f(\mu-y)$ and that therefore $(Y-\mu)$ and $(\mu - Y)$ have the same distribution. I think I have to somehow use this information to help compute the expectation, but I still do not know how to deal with the integral.
Can someone offer some help?
|
Assume $Y\sim N(\mu,\sigma)$ is a random variable with a normal distribution. Then for any $n\geq 1$,
$$
E[(Y-\mu)^{2n}] = E\left[\left(\sigma\cdot \dfrac{Y-\mu}{\sigma}\right)^{2n}\right] =\sigma^{2n}E\left[\left( \dfrac{Y-\mu}{\sigma}\right)^{2n}\right].
$$
Let $X=\dfrac{Y-\mu}{\sigma}$. Then $X\sim N(0,1)$ has the standard normal distribution. So we need to calculate
$$
E[X^{2n}]=E\left[\left( \dfrac{Y-\mu}{\sigma}\right)^{2n}\right].
$$
Since
$$
\begin{aligned}
\left( \int_{-\infty}^{\infty} \dfrac{1}{\sqrt{2\pi}} e^{-cx^2} dx\right)^2 &=
\int_{-\infty}^{\infty} \int_{-\infty}^{\infty} \dfrac{1}{2\pi} e^{-c(x^2+y^2)}dx\, dy \\
&= \int_{0}^{2\pi} \int_{0}^{\infty}\dfrac{1}{2\pi} e^{-c r^2}r\, dr\, d\theta \\
&= \frac{1}{2} \int_{0}^{2\pi} \int_{0}^{\infty}\dfrac{1}{2\pi} e^{-c u}du \, d\theta \hspace{4mm}\mbox{ where }u=r^2\\
&= \dfrac{1}{2}\dfrac{2\pi}{2\pi}\left( -\frac{1}{c}\right) e^{-cu}\Big|_{u=0}^{\infty} \\
&= \left( -\dfrac{1}{2c} \right) \left( 0-1\right) =\dfrac{1}{2c},
\end{aligned}
$$
we conclude
\begin{equation}
\int_{-\infty}^{\infty} \dfrac{1}{\sqrt{2\pi}} e^{-cx^2} dx = \sqrt{\dfrac{1}{2c}}. \hspace{16mm} (\dagger)
\end{equation}
This is called the $\textbf{Gaussian integral}$, whose technique is often used in mathematical physics and complex analysis.
Now, take the derivative of both sides of $(\dagger)$ with respect to $c$ $n$-times to obtain:
$$
\int_{-\infty}^{\infty} \dfrac{1}{\sqrt{2\pi}}\left( -x^2 \right)^n e^{-cx^2}dx
= \dfrac{1}{\sqrt{2}}\left(-\dfrac{1}{2} \right) \left(-\dfrac{3}{2} \right) \cdots \left(-\left( \dfrac{2n-1}{2}\right) \right) c^{-\frac{2n+1}{2}}.
$$
Now let $c=\frac{1}{2}$. Then
$$
\begin{aligned}
(-1)^n E[X^{2n}] &= 2^{\frac{2n+1}{2}-\frac{1}{2}}(-1)^n \dfrac{1}{2} \dfrac{3}{2} \cdots \dfrac{2n-1}{2} \\
&= 2^n (-1)^n \dfrac{1\cdot 3 \cdot 5 \cdots (2n-1)}{2^n} \\
&= (-1)^n 1\cdot 3 \cdot 5 \cdots (2n-1) \\
&= (-1)^n \dfrac{1\cdot 2\cdot 3 \cdot 4\cdots 2n}{2\cdot 4\cdot 6 \cdot 8\cdots 2n} \\
&= (-1)^n \dfrac{(2n)!}{2^n(1\cdot 2 \cdot 3\cdot 4\cdots n)} \\
&= (-1)^n \dfrac{(2n)!}{2^n \cdot n!}.
\end{aligned}
$$
Since
$$
E[X^{2n}] = \dfrac{(2n)!}{2^n \cdot n!},
$$
we conclude that
$$
E[(Y-\mu)^{2n}] = \sigma^{2n} \dfrac{(2n)!}{2^n \cdot n!}.
$$
|
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"timestamp": "2023-03-29T00:00:00",
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|
Proving $x^{-1/2}\leq x^{n}y\leq x^{1/2}$ Let $x$ and $y$ be positive real numbers. Prove that there exists an integer $n$ such that $x^{-1/2}\leq x^{n}y\leq x^{1/2}$.
I have no idea how to do this.
|
For $x,y>0$ the required inequality is equivalent to write
$$1\le x^{n+\frac{1}{2}}y\le x.$$
The right hand side:
Let $f(x,y)=x-x^{n+\frac{1}{2}}y$ we want to show that $f(x,y)\ge 0$. So that
\begin{align}
x-x^{n+\frac{1}{2}}y = x \left(1-x^{n-\frac{1}{2}}y\right) \ge 0 \qquad(?)
\end{align}
But since $x>0$ so we need $1-x^{n-\frac{1}{2}}y \ge 0$ which means
\begin{align}
x^{n-\frac{1}{2}}y \le 1
\end{align}
Taking $\ln(\cdot)$ for both sides (an increasing function), we get
\begin{align}
\ln x^{n-\frac{1}{2}}+ \ln y \le 0
\end{align}
i.e.,
\begin{align}
\left\{ \begin{array}{l}
n\le \frac{1}{2}- \frac{\ln y}{\ln x}, \qquad x>1,y>0 \\
\\
n\ge\frac{1}{2}+ \frac{\ln y}{\ln x}, \qquad 1>x>0,y>0 \\
\end{array} \right.
\qquad(*)
\end{align}
So that the value of such $n$ must chosen taking into account the above condition. The left hand side goes similarly so we can obtain
\begin{align}
\left\{ \begin{array}{l}
n\ge -\frac{1}{2}- \frac{\ln y}{\ln x}, \qquad x>1,y>0 \\
\\
n\le-\frac{1}{2}+ \frac{\ln y}{\ln x}, \qquad 1>x>0,y>0 \\
\end{array} \right.
\qquad(**)
\end{align}
Combining the inequalities (*) and (**) we can say that such integer $n$ satisfies
\begin{align}
\left\{ \begin{array}{l}
\left| {n + \frac{{\ln y}}{{\ln x}}} \right| \le \frac{1}{2},\qquad x>1,y>0 \\
\\
\left| {n- \frac{{\ln y}}{{\ln x}}} \right| \le \frac{1}{2}, \qquad 1>x>0,y>0
\end{array} \right.
\end{align}
|
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"timestamp": "2023-03-29T00:00:00",
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|
Surds question grade 9 I am a student and I need help answering this question.
Simplify:
$\frac{6}{\sqrt{28}}$-$\frac{9}{\sqrt{63}}$
What I did:
$\frac{6}{\sqrt{28}}$- $\frac{9}{\sqrt{63}}$
=$\frac{6}{\sqrt{7×4}}$ - $\frac{9}{\sqrt{7×4}}$
= $\frac{6}{2\sqrt{7}}$ - $\frac{9}{3\sqrt{7}} $
$\frac{6}{2\sqrt{7}}$ × $\frac{2\sqrt{7}}{2\sqrt{7}}$ = $\frac{12\sqrt{7}}{14}$
$\frac{9}{3\sqrt{7}}$ × $\frac{3\sqrt{7}}{3\sqrt{7}}$ = $\frac{27\sqrt{7}}{63}$
($\frac{12\sqrt{7}}{14}$)/2 = $\frac{6\sqrt{7}}{7}$
($\frac{27\sqrt{7}}{63}$)
/2 = $\frac{3\sqrt{7}}{7}$
$\frac{6\sqrt{7}}{7}$ - $\frac{3\sqrt{7}}{7}$
= $\frac{3\sqrt{7}}{7}$
Thankyou and help is appreciated.
|
Simply note that
$$\frac{6}{\sqrt{28}}-\frac{9}{\sqrt{63}}=\frac{6}{2\sqrt7}-\frac{9}{3\sqrt7}=\frac{3}{\sqrt7}-\frac{3}{\sqrt7}=0$$
|
{
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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|
proof by induction that $3^{2n-1} + 2^{2n-1}$ is divisible by $5$ I am supposed to proof by induction that $3^{2n-1} + 2^{2n-1}$ is divisible by $5$ for $n$ being a natural number (integer, $n > 0$).
What I have so far:
Basis: $n = 1$
\begin{align}
3^{2 \cdot 1-1} + 2^{2 \cdot 1-1} & = 3^1 + 2^1\\
& = 5
\end{align}
Assumption: $3^{2n-1} + 2^{2n-1}$ is divisible by $5$ for $n = k \in \mathbb{N}$.
$5 \mid (3^{2n-1} + 2^{2n-1}) \implies 3^{2n-1} + 2^{2n-1} = 5m$, $m \in \mathbb{Z}$
Proof: Let $n = k + 1$
\begin{align}
3^{2 \cdot (k+1)-1} + 2^{2 \cdot (k+1)-1} & = 3^{2k+2-1} + 2^{2k+2-1}\\
& = 3^{2k+1} + 2^{2k+1}\\
& = 3^{2k} \cdot 3^1 + 2^{2k} \cdot 2^1\\
& = 3^{2k} \cdot 3 + 2^{2k} \cdot 2
\end{align}
And here I got stuck. I don't know how to get from the last line to the Assumption. Either I am overlooking a remodeling rule or I have used a wrong approach.
Anyway, I am stuck and would be thankful for any help.
|
The sequence $a_n = 3^{2n-1}+2^{2n-1}$ fulfills the recurrence relation
$$ a_{n+2} = 13 a_{n+1} - 36 a_n $$
and since both $a_1$ and $a_2$ are $\equiv 0\pmod{5}$, by the previous recurrence relation every term of the sequence is $\equiv 0\pmod{5}$. Anyway, it is probably easier to notice that if $d$ is a positive odd integer, $(x+y)$ is a divisor of $x^d+y^d$, hence $5=(2+3)$ is for sure a divisor of $2^{2n-1}+3^{2n-1}$.
|
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "4",
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|
Compute : $\int\frac{x+2}{\sqrt{x^2+5x}+6}~dx$ Question: Compute this integral
$$\int\frac{x+2}{\sqrt{x^2+5x}+6}~dx$$
My Approach:
$$\int\frac{x+2}{\sqrt{x^2+5x}+6}~dx$$
$$=\int\frac{x+2}{\sqrt{x^2+5x}+6}\times \frac{{\sqrt{x^2+5x}-6}}{{\sqrt{x^2+5x}-6}}~dx$$
$$\int\frac{(x+2)(\sqrt{x^2+5x})}{x^2+5x-36}~dx~~- \underbrace {~\int\frac{(6x+12)}{x^2+5x-36}~dx~}_{\text{This one I know how to deal with} }$$
$$\text{Now:} ~\int\frac{(x+2)(\sqrt{x^2+5x})}{x^2+5x-36}~dx$$
$$=\frac{1}{2}\int\frac{(2x+5-1)(\sqrt{x^2+5x})}{x^2+5x-36}~dx$$
$$=\frac{1}{2}\int\frac{(2x+5)(\sqrt{x^2+5x})}{x^2+5x-36}~dx~~- \frac{1}{2}\int\frac{(\sqrt{x^2+5x})}{x^2+5x-36}~dx$$
$$\Big( \text{Let} ~ x^2+5x=t \implies (2x+5)~dx = dt \Big)$$
$$ \underbrace{\frac{1}{2}\int \frac{\sqrt{t}}{t-36}~dt}_{\text{I can deal with this}} ~~- \frac{1}{2}\int \frac{\sqrt{x^2+5x}}{x^2+5x-36}~dx$$
Now I'm stuck. I am unable to calculate: $$ \int \frac{\sqrt{x^2+5x}}{x^2+5x-36}~dx$$
P.S.: I am high school student so please try to use elementary integrals only; i.e. integration by parts and substitution. I don't know how to use complex numbers in integration, multiple integrals, error function, etc. (I don't know if it can be used here or not, just clarifying.)
As answered by @Kanwaljit Singh: Finally I have to compute:
$$\int \frac{1}{\sqrt{x^2+5x}-6}$$
But if I was able to compute it, I would have done it in the very first step, id est ;
$$\int \frac{x+2}{\sqrt{x^2+5x}+6}~dx =
\frac{1}{2}\int \frac{2x+5-1}{\sqrt{x^2+5x}+6}~dx
\\ \frac{1}{2}\int \frac{2x+5}{\sqrt{x^2+5x}+6}~dx ~- \frac{1}{2}\int \frac{1}{\sqrt{x^2+5x}+6}~dx
\\ \Big( \text{Let} ~ x^2+5x=t \implies (2x+5)~dx = dt \Big)
\\ \underbrace{\frac{1}{2}\int \frac{1}{t+6}~dt}_{\text{Doable}} ~-~\frac{1}{2}\int \frac{1}{\sqrt{x^2+5x}+6}~dx
\\ \int \frac{1}{\sqrt{x^2+5x}+6}~dx $$
Reached to a similar step by a short path.
But how do I compute this one?
A screenshot of this question:
|
Introduce the Euler substitution:
Let $u=\dfrac{\sqrt{x^2+5x}}{x}$ ,
Then $x=\dfrac{5}{u^2-1}$
$dx=-\dfrac{10u}{(u^2-1)^2}~du$
$\therefore\int\dfrac{x+2}{\sqrt{x^2+5x}+6}~dx$
$=-\int\dfrac{\dfrac{5}{u^2-1}+2}{\dfrac{5u}{u^2-1}+6}\dfrac{10u}{(u^2-1)^2}~du$
$=-\int\dfrac{(2u^2+3)10u}{(6u^2+5u-6)(u^2-1)^2}~du$
$=-\int\dfrac{(2u^2+3)10u}{(3u-2)(2u+3)(u+1)^2(u-1)^2}~du$
$=\int\left(-\dfrac{5}{2(u+1)^2}-\dfrac{5}{2(u-1)^2}+\dfrac{11}{2(u+1)}+\dfrac{13}{2(u-1)}-\dfrac{144}{13(2u+3)}-\dfrac{252}{13(3u-2)}\right)~du$ (according to http://www.wolframalpha.com/input/?i=-((2u%5E2%2B3)10u)%2F((3u-2)(2u%2B3)(u%2B1)%5E2(u-1)%5E2))
$=\dfrac{5}{2(u+1)}+\dfrac{5}{2(u-1)}+\dfrac{11\ln(u+1)}{2}+\dfrac{13\ln(u-1)}{2}-\dfrac{72\ln(2u+3)}{13}-\dfrac{84\ln(3u-2)}{13}+C$
$=\dfrac{5u}{u^2-1}+\dfrac{11\ln(u+1)}{2}+\dfrac{13\ln(u-1)}{2}-\dfrac{72\ln(2u+3)}{13}-\dfrac{84\ln(3u-2)}{13}+C$
$=\sqrt{x^2+5x}+\dfrac{11}{2}\ln\dfrac{x+\sqrt{x^2+5x}}{x}+\dfrac{13}{2}\ln\dfrac{\sqrt{x^2+5x}-x}{x}-\dfrac{72}{13}\ln\dfrac{3x+2\sqrt{x^2+5x}}{x}-\dfrac{84}{13}\ln\dfrac{3\sqrt{x^2+5x}-2x}{x}+C$
|
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|
What is the dimension of the following function space? Construct a function space out of the following:
$v_1 = x^4+x^3+x^2+x+1$
$v_2 = x^4+x^2+1$
$v_3 = x^3+x$
$v_4 = x+1$
Where the space S is defined as linear product of the $v$'s:
$S=lin(v_1,v_2,v_3,v_4)$
What is the dimension of S? Since $v_1$ is the linear product of $v_2$ and $v_3$ and the rest of the "vectors" appear to not be linearly dependent, is the correct answer 3? (Almost seems too easy to be the correct answer).
|
Pass from polynomials to coordinate vectors:
$$v_1\to(1,1,1,1,0)\;,\;\;v_2\to(1,0,1,0,1)\;,\;\;v_3\to(0,1,0,1,0)\;,\;\;v_4\to(0,0,0,1,1)$$
Now form the corresponding matrix with the above as rows, and reduce the matrix (Gauss elementary operations):
$$\begin{pmatrix}
1&1&1&1&0\\
1&0&1&0&1\\
0&1&0&1&0\\
0&0&0&1&1\end{pmatrix}\stackrel{R_2-R_1}\longrightarrow\begin{pmatrix}
1&1&1&1&0\\
0&\!-1&0&\!-1&1\\
0&1&0&1&0\\
0&0&0&1&1\end{pmatrix}\stackrel{R_3-R_2}\longrightarrow\begin{pmatrix}
1&1&1&1&0\\
0&\!-1&0&\!-1&1\\
0&0&0&0&1\\
0&0&0&1&1\end{pmatrix}\stackrel{R_3\leftrightarrow R_4}\longrightarrow$$$${}$$
$$\longrightarrow\begin{pmatrix}
1&1&1&1&0\\
0&\!-1&0&\!-1&1\\
0&0&0&1&1\\
0&0&0&0&1\end{pmatrix}$$$${}$$
We see we get a matrix of rank $\;4\;$ and then the four vectors above, and thus the four original ones (the polynomials) as well, are linearly independent, and this means
$$\dim\text{ Span}\,\{v_1,v_2,v_3,v_4\}=4$$
Added after correction by the OP:
$$\begin{pmatrix}
1&1&1&1&1\\
1&0&1&0&1\\
0&1&0&1&0\\
0&0&0&1&1\end{pmatrix}\stackrel{R_2-R_1}\longrightarrow\begin{pmatrix}
1&1&1&1&0\\
0&\!-1&0&\!-1&0\\
0&1&0&1&0\\
0&0&0&1&1\end{pmatrix}\stackrel{R_3-R_2}\longrightarrow\begin{pmatrix}
1&1&1&1&0\\
0&\!-1&0&\!-1&0\\
0&0&0&0&0\\
0&0&0&1&1\end{pmatrix}\stackrel{R_3\leftrightarrow R_4}\longrightarrow$$$${}$$
and thus we get at once the dimension is three...
|
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|
Find the value of $\tan^6(\frac{\pi}{7}) + \tan^6(\frac{2\pi}{7}) + \tan^6(\frac{3\pi}{7})$ I wish to find the value of $\tan^6(\frac{\pi}{7}) + \tan^6(\frac{2\pi}{7}) + \tan^6(\frac{3\pi}{7})$ as part of a larger problem. I can see that the solution will involve De Moivre's theorem somehow, but I cannot see how to apply it. I have looked at solutions of $z^7 - 1 = 0$ but to no avail. Can anyone suggest a method for solving this problem?
|
Misha beat me on time, so I will give a slightly different point of view. It is well-known that $\cos\frac{2\pi}{7}$, $\cos\frac{4\pi}{7}$ and $\cos\frac{6\pi}{7}$ are algebraic conjugates, roots of the Chebyshev polynomial
$$ p(x)=8x^3+4x^2-4x-1.$$
On the other hand
$$ \tan^6\frac{\pi}{7}=\left(\frac{1}{\cos^2\frac{\pi}{7}}-1\right)^3 = \left(\frac{1-\cos\frac{2\pi}{7}}{1+\cos\frac{2\pi}{7}}\right)^2$$
hence $\tan^2\frac{\pi}{7}$, $\tan^2\frac{2\pi}{7}$, $\tan^2\frac{3\pi}{7}$ are algebraic conjugates as well, roots of the polynomial
$$ q(x)=x^3-21x^2+35x-7 $$
with companion matrix
$$ M = \begin{pmatrix} 0 & 0 & 7 \\
1 & 0 & -35 \\
0 & 1 & 21 \end{pmatrix}$$
whose third power is
$$ M^3 = \left(
\begin{array}{ccc}
7 & 147 & 2842 \\
-35 & -728 & -14063 \\
21 & 406 & 7798 \\
\end{array}
\right)$$
The wanted sum is just $\text{Tr}(M^3)$, hence it equals $7-728+7798=\color{red}{7077}$.
|
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|
Determine the acceleration given the position function of a particle
The position function of a particle in a test laboratory is $s(t) = \frac{10t}
{t^2+3}$.
Determine the acceleration of the particle of particle after 4 seconds.
The speed $v(t)$ of the particle is the derivative of the position $\frac{ds(t)}{dt}$.
I got $v(t) = \frac{10(-t^2+3)}{(t^2+3)^2}$
For the acceleration $a(t)$ I tried taking the derivative of $v(t)$ but got the wrong answer at $a(4)$.
$$a(t)=\frac{-2t\cdot (t^3+3)^2-10(-t^2+3)\cdot 2(t^2+3)\cdot(2t)}{(t^2+3)^4}$$
|
$$s(t) = \frac{10t}{t^2+3}$$
$$\implies v(t) = \frac{ds}{dt} = \frac{(t^2+3)\cdot 10 - 10t\cdot 2t}{(t^2+3)^2} $$
$$v(t) = \frac{10t^2+30-20t^2}{(t^2+3)^2} = \frac{30-10t^2}{(t^2+3)^2}$$
$$\implies a(t) = \frac{dv}{dt} = \frac{(t^2+3)^2\cdot -20t -(30-10t^2)\cdot 2(t^2+3)\cdot 2t}{(t^2+3)^4}$$
$$a(t) = \frac{-20t(t^2+3)+4t(10t^2-30)}{(t^2+3)^3}$$
$$\implies a(4) = \frac{-80\cdot 19 +16\cdot 130}{19^3} = \frac{560}{6859}$$
|
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|
If $\dfrac {1}{a+b} +\dfrac {1}{b+c}=\dfrac {3}{a+b+c}$, prove that $\angle B=60^\circ $ If $\dfrac {1}{a+b}+\dfrac {1}{b+c}=\dfrac {3}{a+b+c}$, prove that $\angle B=60^\circ$.
My Attempt
$$\dfrac {1}{a+b}+\dfrac {1}{b+c}=\dfrac {3}{a+b+c}$$
$$\dfrac {a+2b+c}{(a+b)(b+c)}=\dfrac {3}{a+b+c}$$
$$a^2-ac-b^2+c^2=0$$.
How to prove further?
|
Here's a way to prove it without using trigonometry. For sake of discussion, in the pictured triangle assume $∠B$ is lower left corner and side $a$ is the bottom (side $c$ is left and $b$ is right). Now look at the three cases where $∠C$ is $<90°$, $=90°$, and $>90°$.
The easy case is $∠C =90°$. In this case $c=2a$ and $b=\sqrt{3}a$. Plug those values into $a^2−ac−b^2+c^2=0$ to get $a^2 - 2a^2 - 3a^2 +4a^2 = 0$ which confirms the equality.
Take the case $∠C <90°$. In this case let $h$ be the height of the triangle from $a$ up to $∠A$. Let $d$ be the length of the line segment from the lower left corner to where the height intercepts $a$. And let $e=a-d$.
Calculate the area of the triangle using Heron's formula $A=\sqrt{s(s-a)(s-b)(s-c)}$ where $S=(a+b+c)/2$. Since $A=ah/2$ calculate $$h=2A/a=\frac{\sqrt{2(c^2a^2+c^2b^2+b^2a^2)+a^4+b^4+c^4}}{2a}$$ Knowing $h$ and $c$ allows you to calculate $d$ by Pythagorean. Then calculate $e=a-d$. Knowing $e$ and $h$ allows you to calculate $b=\sqrt{a^2-ac+c^2}$.
Plugging that into your equation gives $a-ac-(a^2-ac+c^2)+c^2=0$, again confirming the equality.
I'll leave the third case as an exercise.
|
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|
Using 3 equations, find $2a-b+c$. From the problem, I extract the following 3 equations.
*
*$a+b+c=10$
*$ab+bc+ca=31$
*$abc=30$
The question is to find $(2a-b+c)$.
Using the equation $(a+b+c)^2 = a^2 + b^2 + c^2 +2 (ab+bc+ca)$, I found $a^2+b^2+c^2=38$.
Using the equation $a^3+b^3+c^3 - 3abc = (a+b+c)(a^2 + b^2+c^2-ab-bc-ca)$, I found $a^3+b^3+c^3=160$.
However, I was stuck here. Please let me know some directions. Thank you.
|
the first equation gives:
$$a+b=10-c$$
the second one:
$$c(a+b)=31-ab$$ with the first
$$c(10-c)=31-ab$$ and with $$c=\frac{30}{ab}$$ we obtain
$$c(10-c)=31-\frac{30}{c}$$
can you proceed?
|
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|
Find det(A) for matrix, find C and adj(A) I am asked to find the determinant of matrix A by using cofactors method. I understand how to do that portion, but then it asks to also find matrix C, and adj(A). How can I go about that?
for det(A) I got -20.
$ A = \begin{bmatrix}
1 & 3 & -3 \\
-3 & -3 & 2 \\
-4 & 4 & -6 \\
\end{bmatrix}$
|
$A = \begin{bmatrix}
1 & 3 & -3 \\
-3 & -3 & 2 \\
-4 & 4 & -6 \\
\end{bmatrix}$
$\det A = 4$.
We know $A^{-1}=\frac{1}{\det A} adj (A)$, where $adj$ is adjoint of matrix $A$.
$adj(A)= \left( \begin{array}{ccc} -26 & -30 & -3 \\ 10 & -6 & 7 \\ -24 & -16 & 6 \end{array} \right)$.
$\therefore A^{-1}= \frac{1}{4}\left( \begin{array}{ccc} -26 & -30 & -3 \\ 10 & -6 & 7 \\ -24 & -16 & 6 \end{array} \right)$
$A^{-1}= \left( \begin{array}{ccc} -26/4 & -30/4 & -3/4 \\ 10/4 & -6/4 & 7/4 \\ -24/4 & -16/4 & 6/4 \end{array} \right)$
|
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|
How to prove that $\frac{1}{n+1}+\frac{1}{n+2}+...+\frac{1}{2n} \le \frac{n}{n+1}$? I have a series $$a_n = \frac{1}{n+1}+\frac{1}{n+2}+...+\frac{1}{2n} \quad n \ge 1$$
For example, $a_3 = \frac{1}{4}+\frac{1}{5}+\frac{1}{6}$.
I need to prove that for $n \ge 1$:
$$a_n = \frac{1}{n+1}+\frac{1}{n+2}+...+\frac{1}{2n} \le \frac{n}{n+1}$$
I guess one could say that:
$$
\sum_{i=1}^n\frac{1}{n+i} \le \sum_{i=1}^n\frac{n}{n+1}
$$
However, I'm not sure this is rigorous enough (for example, in $\sum_{i=1}^n\frac{1}{n+i}$ how do we really know that the index goes from $1$ to $n$) and I think this needs to be proven via induction.
So the base case is:
$$a_1 = \frac{1}{2} \le \frac{1}{2} = \frac{n}{n+1}$$
The step: suppose $a_n \le \frac{n}{n+1}$ then let's prove that $$a_{n+1} \le \frac{n+1}{n+2}$$
The above can be developed as:
$$
\frac{1}{n+1}+\frac{1}{n+3}+...+\frac{1}{2(n+1)} \le \frac{n}{n+1}+\frac{1}{n+2}
$$
This is where I get stuck. If I could somehow prove that the number of terms to the left $\le$ the terms to the left I would be golden. Or maybe there's another way.
|
HINT:
$$\dfrac1{n+1}>\dfrac1{n+r}$$ for $2\le r\le n$
|
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|
Roundest ellipse with specified tangents Suppose you are given two points $\mathbf p_1$ and $\mathbf p_2$ in the plane with associated vectors $\mathbf v_1$ and $\mathbf v_2$. You want to find an ellipse passing through $\mathbf p_1$ tangent to $\mathbf v_1$ and through $\mathbf p_2$ tangent to $\mathbf v_2$. There are infinitely many solutions,
but a natural choice is to take the one that is the closest to a circle, i.e. the one with the minimum eccentricity. Is there an elegant way to compute such an ellipse given $(\mathbf p_1, \mathbf v_1)$ and $(\mathbf p_2, \mathbf v_2)$?
|
At the moment I can offer an analytical solution, which is quite cumbersome but manageable. I hope someone will be able to give a simpler solution.
To reduce the number of parameters, we can set up a coordinate system such that $\mathbf{p}_1=(\alpha,0)$, $\mathbf{p}_2=(0,\beta)$, $\mathbf{v}_1=(0,1)$, and define then $m=v_{2y}/v_{2x}$.
Let's start with the generic equation for a conic section, where we can set to $1$ the coefficient of $x^2$ because we are dealing with an ellipse:
$$
x^2+By^2+Cxy+Dx+Ey+F=0.
$$
We have four conditions: the conic passes through $\mathbf{p}_1$, $\mathbf{p}_2$ and is there tangent to $\mathbf{v}_1$, $\mathbf{v}_2$. These conditions translate into four equations:
$$
\begin{align}
\cases{
\alpha^2+\alpha D+F=0\\
\beta^2B+\beta E+F=0\\
2m\beta B+\beta C+D+mE=0\\
\alpha C+E=0
}
\end{align}
$$
From there, one can find expressions for $B$, $C$, $D$, $E$ as a function of $F$. Plugging these into the formulas for the semiaxes $a$ and $b$ we can then compute the eccentricity $\epsilon=\sqrt{1-b^2/a^2}$. The smaller the eccentricity, the rounder the ellipse, so we must find the minimum of $\epsilon$ as a function of $F$.
I computed $d\epsilon/dF$ with Mathematica and found that it vanishes for
$$
F=-\frac{\alpha ^2 \beta \left(\alpha ^2 \beta +\beta ^3-2 \alpha ^2 \beta m^2+2 \alpha ^3
m\right)}{\beta ^2 \left(\alpha ^2+\beta ^2\right)+2 m^2 \left(\alpha ^4+2 \alpha ^2 \beta
^2\right)+2 \alpha \beta m \left(\alpha ^2+2 \beta ^2\right)}.
$$
Once $F$ is known, one can compute the other coefficients and find the equation of the "roundest" ellipse:
$$
\begin{align}
\cases{
B=\frac{\displaystyle\alpha ^2 \left(\beta ^2+\alpha ^2 \left(2 m^2+1\right)+2 \alpha \beta m\right)}{\displaystyle\beta ^2\left(\alpha ^2+\beta ^2\right)+2 m^2 \left(\alpha ^4+2 \alpha ^2 \beta ^2\right)+2 \alpha
\beta m \left(\alpha ^2+2 \beta ^2\right)}\\
\\
C=\frac{\displaystyle2 \alpha ^2 m \left(-\alpha ^2+\beta ^2+2 \alpha \beta m\right)}{\displaystyle\beta ^2 \left(\alpha
^2+\beta ^2\right)+2 m^2 \left(\alpha ^4+2 \alpha ^2 \beta ^2\right)+2 \alpha \beta m
\left(\alpha ^2+2 \beta ^2\right)}\\
\\
D=-\frac{\displaystyle2 \alpha ^2 m \left(2 \beta ^3+m \left(\alpha ^3+3 \alpha \beta ^2\right)\right)}{\displaystyle\beta ^2
\left(\alpha ^2+\beta ^2\right)+2 m^2 \left(\alpha ^4+2 \alpha ^2 \beta ^2\right)+2 \alpha
\beta m \left(\alpha ^2+2 \beta ^2\right)}\\
\\
E=-\frac{\displaystyle2 \alpha ^3 m \left(-\alpha ^2+\beta ^2+2 \alpha \beta m\right)}{\displaystyle\beta ^2 \left(\alpha
^2+\beta ^2\right)+2 m^2 \left(\alpha ^4+2 \alpha ^2 \beta ^2\right)+2 \alpha \beta m
\left(\alpha ^2+2 \beta ^2\right)}
}
\end{align}
$$
A check with GeoGebra confirms that this is indeed the value of $F$ giving a minimum eccentricity.
EDIT.
Following the suggestion by Rahul, we can get a much simpler result. Choose a coordinate system such that
$\mathbf{p}_1=(\alpha,0)$, $\mathbf{p}_2=(-\alpha,0)$,
and define $m=v_{1y}/v_{1x}$, $n=v_{2y}/v_{2x}$.
With those choices, the equation of the ellipse can be written as:
$$
x^2+By^2-{m+n\over mn}xy+\alpha{m-n\over mn}y-\alpha^2=0.
$$
The eccentricity $\epsilon$ is then a function of $B$, and
$d\epsilon/dB=0$ for
$$
B=1+\frac{(m+n)^2}{2 m^2 n^2}.
$$
In summary, the equation of the roundest ellipse turns out to be:
$$
x^2+\left(1+\frac{(m+n)^2}{2 m^2 n^2}\right)y^2-{m+n\over mn}xy+\alpha{m-n\over mn}y-\alpha^2=0.
$$
Added by Rahul:
Things become even simpler if we use the slope of the normals, $\mu=-1/m$ and $\nu=-1/n$:
$$
x^2+\left(1+\tfrac12(\mu+\nu)^2\right)y^2+(\mu+\nu)xy+\alpha(\mu-\nu)y-\alpha^2=0.
$$
|
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|
Let $n$ be an even perfect number greater than $ 6$. Prove that $n \equiv 4 \pmod 6$. Things I've gotten so far:
Proven that $2^p-1\equiv \pmod 3$ (for some odd prime)
$n$ is in the form $n =( 2^p-1)(2^{p - 1})$ where $(2^p - 1)$ is prime.$ n > 6$, so $p > 2$.
So I have the congruence $2^p-1 \equiv 1 \pmod 3$,
and I think I have to prove that $2^p -1 \equiv 4 \pmod 6$ or $-2 \pmod 6$,
but I'm not sure how to accomplish that step.
|
Observe that $$x^{ab}-1=(x^a-1)(x^{a(b-1)}+x^{a(b-2)}+\cdots+x^a+1).$$ Hence a number of this form could be prime if and only if $x=2$ and $a=1$ and $b$ is a prime. Here it has shown that an even perfect number is of the form $2^{p-1}(2^p-1),$ where $2^p-1$ is a prime. Hence we only need to consider about prime powers!!
Consider $p=2$ separately (which gives you the prefect number $6$) and note that all the other primes are odd. Let $p=2m+1.$ Binomial theorem gives us,
$$2^{2m+1}=2(3+1)^m=2(3^m+\cdots+3m+1)=6.3^{m-1}+\cdots+6m+2.$$ Now $n=2^{2m}(2^{2m+1}-1)=2^{2m}(6.3^{m-1}+\cdots+6m+1).$
Hence $n \equiv 2^{2m} \pmod 6$ and you can easily prove that $2^{2m} \equiv 4 \pmod 6$ by mathematical induction.
|
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|
Find $\displaystyle\lim_{x\to\infty}x - \sqrt{x+1}\sqrt{x+2}$ using squeeze theorem
Find $$\lim_{x\to\infty}x - \sqrt{x+1}\sqrt{x+2}$$ using squeeze theorem
Tried using binomial expansion, but have no idea on how to continue.
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Your idea of a use of the binomial theorem was good, writing$$A=x - \sqrt{x+1}\times\sqrt{x+2}=x-\sqrt x \times\sqrt{1+\frac{1}{x}}\times\sqrt x \times\sqrt{1+\frac{2}{x}}$$ that is to say $$A=x\left(1-\sqrt{1+\frac{1}{x}}\times \sqrt{1+\frac{2}{x}}\right)$$ Now consider, using the binomial theorem with $t=\frac 1x$$$\sqrt{1+t}=1+\frac{t}{2}-\frac{t^2}{8}+\cdots$$ $$\sqrt{1+2t}=1+t-\frac{t^2}{2}+\cdots$$ $$\sqrt{1+t}\times\sqrt{1+2t}=1+\frac{3 t}{2}-\frac{t^2}{8}+\cdots$$ Replace $t$ by $\frac 1x$ to get $$A=x\left(1-1-\frac{3 }{2x}+\frac{1}{8x^2}+\cdots\right)=-\frac{3 }{2}+\frac{1}{8x}+\cdots$$
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"timestamp": "2023-03-29T00:00:00",
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Identify $\lim\limits_{x \to +\infty } x^2 \left(\sqrt{x^4+x+1}-\sqrt{x^4+x+5}\right)$
Identify $$\lim\limits_{x \to +\infty } x^2 \left(\sqrt{x^4+x+1}-\sqrt{x^4+x+5}\right)$$
My Try :
$$\sqrt{x^4+x+1}=\sqrt{x^4(1+\frac{1}{x^3}+\frac{1}{x^4})}=x^2\sqrt{(1+\frac{1}{x^3}+\frac{1}{x^4})}$$
Now : $$\frac{1}{x^3}+\frac{1}{x^4}=z$$
$$(1+z)^{\frac{1}{n}}= 1 + \frac1n x + \frac{1 - n}{2n^2}x^2 + \frac{2n^2 - 3n + 1}{6n^3}x^3 + O(x^4)$$
$$(1+(\frac{1}{x^3}+\frac{1}{x^4}))^{\frac{1}{2}}= 1 + \frac12 z - \frac{1}{8}z^2 + + O(z^3)$$
$$(1+(\frac{1}{x^3}+\frac{1}{x^4}))^{\frac{1}{2}}= 1 + \frac12 (\frac{1}{x^3}+\frac{1}{x^4}) - \frac{1}{8}(\frac{1}{x^3}+\frac{1}{x^4})^2 + O(z^3)$$
$$(1+(\frac{1}{x^3}+\frac{1}{x^4}))^{\frac{1}{2}}= 1 + \frac{1}{2x^3}+\frac{1}{2x^4} - \frac{x^2+2x+1}{8x^8} + O(z^3)$$
And :
$$\sqrt{x^4+x+1}=\sqrt{x^4(1+\frac{1}{x^3}+\frac{5}{x^4})}=x^2\sqrt{(1+\frac{1}{x^3}+\frac{5}{x^4})}$$
So :
$$(1+(\frac{1}{x^3}+\frac{5}{x^4}))^{\frac{1}{2}}= 1 + \frac{1}{2x^3}+\frac{5}{2x^4} - \frac{x^2+10x+25}{8x^8}+ O(z^3)$$
$$\lim\limits_{x \to +\infty }=x^4( 1 + \frac{1}{2x^3}+\frac{1}{2x^4} - \frac{x^2+2x+1}{8x^8} + O(z^3)-( 1 + \frac{1}{2x^3}+\frac{5}{2x^4} - \frac{x^2+10x+25}{8x^8}+ O(z^3)))$$
$$\lim\limits_{x \to +\infty }=x^4(\frac{1}{2x^4}-\frac{5}{2x^4})=x^4(\frac{-4}{2x^4})=-2$$
is it right ?
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Write
\begin{align}
\sqrt{x^4+x+1}-\sqrt{x^4+x+5}
=\frac{(\sqrt{x^4+x+1}-\sqrt{x^4+x+5})(\sqrt{x^4+x+1}+\sqrt{x^4+x+5})}{\sqrt{x^4+x+1}+\sqrt{x^4+x+5}}
\end{align}
and use
\begin{align}
\frac{1}{2\sqrt{x^4+x+5}}\leq\frac{1}{\sqrt{x^4+x+1}+\sqrt{x^4+x+5}}\leq \frac{1}{2\sqrt{x^4+x+1}}
\end{align}
and
\begin{align}
\sqrt{x^4+x+1}&=\sqrt{x^4(1+x^{-3}+x^{-4})}=x^2\sqrt{1+x^{-3}+x^{-4}}\\
\sqrt{x^4+x+5}&=\sqrt{x^4(1+x^{-3}+5x^{-4})}=x^2\sqrt{1+x^{-3}+5x^{-4}}
\end{align}
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"url": "https://math.stackexchange.com/questions/2211614",
"timestamp": "2023-03-29T00:00:00",
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Solve $\frac{2}{\sin x \cos x}=1+3\tan x$ Solve this trigonometric equation given that $0\leq x\leq180$
$\frac{2}{\sin x \cos x}=1+3\tan x$
My attempt,
I've tried by changing to $\frac{4}{\sin 2x}=1+3\tan x$, but it gets complicated and I'm stuck. Hope someone can help me out.
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Let $\tan\frac{x}{2}=t$.
Hence, we need to solve
$$\frac{2}{\frac{2t}{1-t^2}\cdot\frac{1-t^2}{1+t^2}}=1+\frac{6t}{1-t^2}$$ or
$$t^4+t^3-4t^2-t+1=0$$ or
$$t^4-t^3-t^2+2t^3-2t^2-2t-t^2-t+1=0$$ or
$$(t^2-t-1)(t^2+2t-1)=0,$$
which gives
*
*$\tan\frac{x}{2}=\sqrt2-1$ or $\frac{x}{2}=22.5^{\circ}+180^{\circ}k$, where $k=0$ or $x=45^{\circ}$;
*$\tan\frac{x}{2}=-\sqrt2-1$, which is impossible;
*$\tan\frac{x}{2}=\frac{1+\sqrt5}{2}$ or $x=2\arctan\frac{1+\sqrt5}{2}$;
*$\tan\frac{x}{2}=\frac{1-\sqrt5}{2}$, which is impossible.
Id est, the answer is $\left\{\frac{\pi}{4}, 2\arctan\frac{1+\sqrt5}{2}\right\}$.
Done!
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"timestamp": "2023-03-29T00:00:00",
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Find the minimum value of $f$ on $\{(a,b,c)\in \mathbb{R^3}\mid a+b+c=1\}$ where $f(a,b,c)=\int_{0}^1 (a+bx+cx^2)^2dx$. Find the minimum value of $f$ on $\{(a,b,c)\in \mathbb{R^3}\mid a+b+c=1\}$ where $f(a,b,c)=\int_0^1 (a+bx+cx^2)^2 \, dx$.
What I did is
$$
\int_0^1 (a+bx+cx^2)^2 \, dx=a^2+a \left( b+\frac{2c}{3} \right) + \frac{b^2}{3} + \frac{bc}{2}+\frac{c^2}{5}.
$$
Then we consider the function $F(a,b,c)=a^2+a(b+\frac{2c}{3})+\frac{b^2}{3} + \frac{bc}{2} + \frac{c^2}{5}-\lambda(a+b+c-1)$. Thus, we can consider $\nabla F=0$ with $a+b+c=1$ and try to find $a,b$ and $c$.
However, since the function and the constraint seem to be related, I am curious whether there is some other way to see the minimum without computing Lagrange multiplier. Thank you.
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Just substitute $a=1-b-c$ and minimize the function of two variables
$$g(b,c)=f(1-b-c,b,c)=\int_0^1(1-b-c+bx+cx^2)^2 dx.$$
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"url": "https://math.stackexchange.com/questions/2213882",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Finding solution of trigonometric equation How to find the solutions of this trigonometric equation $$\sum_{m=1}^6 cosec(\theta + \frac{(m-1)\pi}{4}) cosec(\theta + \frac{m\pi}{4}) =4\sqrt{2}$$ if $0<\theta<\pi/2$.
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HINT:
$\csc\left(\theta+\dfrac{(m-1)\pi}4\right)\cdot\csc\left(\theta+\dfrac{m\pi}4\right)=\dfrac1{\sin\left(\theta+\dfrac{(m-1)\pi}4\right)\cdot\sin\left(\theta+\dfrac{m\pi}4\right)}$
$\dfrac{\sin(A-B)}{\sin A\sin B}=\dfrac{\sin A\cos B-\cos A\sin B}{\sin A\sin B}=\cot B-\cot A$
Here $A=\theta+\dfrac{m\pi}4$ and $B=\theta+\dfrac{(m-1)\pi}4, A-B=\dfrac\pi4$
$\implies\dfrac1{\sin\left(\theta+\dfrac{(m-1)\pi}4\right)\cdot\sin\left(\theta+\dfrac{m\pi}4\right)}$
$=\dfrac1{\sin\dfrac\pi4}\cdot\dfrac{\sin\left\{\theta+\dfrac{m\pi}4-\left(\theta+\dfrac{(m-1)\pi}4\right)\right\}}{\sin\left(\theta+\dfrac{(m-1)\pi}4\right)\cdot\sin\left(\theta+\dfrac{m\pi}4\right)}$
$=\dfrac{\cot\left(\theta+\dfrac{(m-1)\pi}4\right)-\cot\left(\theta+\dfrac{m\pi}4\right)}{\sin\dfrac\pi4}$
Set $m=1,2,3,4,5,6$ and add and finally simplify.
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"timestamp": "2023-03-29T00:00:00",
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If $a+b+c=ab+ac+bc$ then $\sum\limits_{cyc}\frac{a}{a^2+2b}\leq1$.
Let $a$, $b$ and $c$ be positive numbers such that $a+b+c=ab+ac+bc$. Prove that:
$$\frac{a}{a^2+2b}+\frac{b}{b^2+2c}+\frac{c}{c^2+2a}\leq1$$
We can try the following.
Replace $a\rightarrow\frac{1}{a}$, $b\rightarrow\frac{1}{b}$ and $c\rightarrow\frac{1}{c}$.
Hence, the condition it's the same and we need to prove that $\sum\limits_{cyc}\frac{ab}{2a^2+b}\leq1$.
But by C-S
$$\sum\limits_{cyc}\frac{ab}{2a^2+b}=a+b+c+\sum\limits_{cyc}\left(\frac{ab}{2a^2+b}-a\right)=$$
$$=a+b+c-2\sum\limits_{cyc}\frac{a^3}{2a^2+b}=a+b+c-2\sum\limits_{cyc}\frac{a^4}{2a^3+ab}\leq$$
$$\leq a+b+c-\frac{2(a^2+b^2+c^2)^2}{2(a^3+b^3+c^3)+ab+ac+bc}$$
and it remains to prove that
$$a+b+c-\frac{2(a^2+b^2+c^2)^2}{2(a^3+b^3+c^3)+ab+ac+bc}\leq1,$$
which is wrong for $b=c$.
Thank you!
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The BW helps.
Let $a=\min\{a,b,c\}$, $b=a+u$ and $c=a+v$.
Hence, we need to prove that
$$\sum_{cyc}\frac{a}{a^2+\frac{2b(ab+ac+bc)}{a+b+c}}\leq\frac{a+b+c}{ab+ac+bc}$$ or
$$\sum_{cyc}\frac{a}{a^2(ab+ac+bc)+2b(a+b+c)}\leq\frac{1}{ab+ac+bc}$$ or
$$63(u^2-uv+v^2)a^7+(149u^3-11u^2v+105uv^2+149v^3)a^6+$$
$$+(145u^4-59u^3v-30u^2v^2+373uv^3+145v^4)a^5+$$
$$+(67u^5+33u^4v-164u^3v^2+268u^2v^3+357uv^4+67v^5)a^4+$$
$$+(12u^6+42u^5v-88u^4v^2+7u^3v^3+280u^2v^4+154uv^5+12v^6)a^3+$$
$$+2(5u^5-3u^4v-19u^3v^2+33u^2v^3+57uv^4+13v^5)uva^2+$$
$$+(2u^4-5u^3v-2u^2v^2+27uv^3+18v^4)u^2v^2a+4u^3v^6\geq0,$$
which is obvious.
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"timestamp": "2023-03-29T00:00:00",
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Fractions in Questions and Answers
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