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Nesbitt's Inequality $\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}\geq\frac{3}{2}$ I'm reading a book which focus in inequality.
I'm stuck in this question. Let $a,b,c$ be positive real numbers. (Nesbitt's inequality) Prove the inequality $$\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}\geq\frac{3}{2}$$
So the first step of solution given is $\frac{a+b}{b+c}+\frac{b+c}{a+b}+\frac{a+c}{c+b}+\frac{c+b}{a+c}+\frac{b+a}{a+c}+\frac{a+c}{b+a}\geq2+2+2=6$
I don't know how to proceed from the question to the first step of solution. Can anyone explain?
|
We have three expressions:
$$S=\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}$$
$$M=\frac{b}{b+c}+\frac{c}{c+a}+\frac{a}{a+b}$$
$$N=\frac{c}{b+c}+\frac{a}{c+a}+\frac{b}{a+b}$$
Obviously, we have $M+N=3$ According to AM-GM,we have:
$$M+S=\frac{a+b}{b+c}+\frac{b+c}{c+a}+\frac{c+a}{a+b}\geqslant 3$$
$$N+S=\frac{a+c}{b+c}+\frac{a+b}{c+a}+\frac{b+c}{a+b}\geqslant 3$$
Thus, $M+N+2S\geqslant 6$ so $2S\geqslant 3$ so $S\geqslant \dfrac{3}{2}$ q.e.d
|
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"language": "en",
"url": "https://math.stackexchange.com/questions/1589751",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 2
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|
Better proof for $\frac{1+\cos x + \sin x}{1 - \cos x + \sin x} \equiv \frac{1+\cos x}{\sin x}$ It's required to prove that
$$\frac{1+\cos x + \sin x}{1 - \cos x + \sin x} \equiv \frac{1+\cos x}{\sin x}$$
I managed to go about out it two ways:
*
*Show it is equivalent to $\mathsf{true}$:
$$\frac{1+\cos x + \sin x}{1 - \cos x + \sin x} \equiv \frac{1+\cos x}{\sin x}$$
$$\Longleftrightarrow\sin x(1+\cos x+\sin x)\equiv(1+\cos x)(1-\cos x+\sin x)$$
$$\Longleftrightarrow\sin x+\cos x\sin x+\sin^2 x\equiv1-\cos x+\sin x+\cos x-\cos^2 x+\sin x \cos x$$
$$\Longleftrightarrow\sin^2 x\equiv1-\cos^2 x$$
$$\Longleftrightarrow\cos^2 x +\sin^2 x\equiv1$$
$$\Longleftrightarrow \mathsf{true}$$
*Multiplying through by the "conjugate" of the denominator:
$${\rm\small LHS}\equiv\frac{1+\cos x + \sin x}{1 - \cos x + \sin x} $$
$$\equiv\frac{1+\cos x + \sin x}{1 - (\cos x - \sin x)} ~~\cdot ~~\frac{1+(\cos x - \sin x)}{1 +(\cos x - \sin x)}$$
$$\equiv\frac{(1+\cos x + \sin x)(1+\cos x - \sin x)}{1 - (\cos x - \sin x)^2}$$
$$\equiv\frac{1+\cos x - \sin x+\cos x + \cos^2 x - \sin x \cos x+\sin x + \sin x \cos x - \sin^2 x}{1 - \cos^2 x - \sin^2 x + 2\sin x \cos x}$$
$$\equiv\frac{1+ 2\cos x + \cos^2 x- \sin^2 x}{2\sin x \cos x}$$
$$\equiv\frac{1+ 2\cos x + \cos^2 x- 1 + \cos^2 x}{2\sin x \cos x}$$
$$\equiv\frac{2\cos x (1+\cos x)}{2\cos x(\sin x)}$$
$$\equiv\frac{1+\cos x}{\sin x}$$
$$~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~\equiv {\rm\small RHS}~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~\square$$
Both methods of proof feel either inelegant or unnecessarily complicated. Is there a simpler more intuitive way to go about this? Thanks.
|
alternatively, using abbreviations $c=\cos x$ and $s=\sin x$ we have
$$
s(1+c+s)=s(1+c) + s^2 = s(1+c) + 1-c^2=s(1+c)+(1-c)(1+c)=(1+s-c)(1+c)
$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1590059",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
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|
Prove that $xy+yz+zx+\frac{x^2y^2}{z^2}+\frac{y^2z^2}{x^2}+\frac{z^2x^2}{y^2}\ge 2x^2+2y^2+2z^2$ for every $x,y,z$ strictly positive I checked the inequality for $y=z$. In this particular case, after some simplifications, the inequality becomes:
$$
2x^3+y^3\ge 3x^2y,
$$
which is true, according to the arithmetic-geometric mean inequality applied to the numbers $x^3,\ x^3$ and $y^3$. I have no idea, at least for now, on how to proceed in the general case. Please give me a hint.
|
simplifying and factorizing the term $$xy+yz+zx+\frac{x^2y^2}{z^2}+\frac{y^2z^2}{x^2}+\frac{z^2x^2}{y^2}-2(x^2+y^2+z^2)$$ we get
$${\frac { \left( xy+zx+yz \right) \left( {y}^{3}{x}^{3}-{y}^{2}z{x}^{3
}-y{z}^{2}{x}^{3}+{z}^{3}{x}^{3}-{y}^{3}z{x}^{2}+3\,{z}^{2}{x}^{2}{y}^
{2}-y{z}^{3}{x}^{2}-{y}^{3}{z}^{2}x-{z}^{3}{y}^{2}x+{y}^{3}{z}^{3}
\right) }{{z}^{2}{x}^{2}{y}^{2}}}$$
it must be
$${y}^{3}{x}^{3}-{y}^{2}z{x}^{3}-y{z}^{2}{x}^{3}+{z}^{3}{x}^{3}-{y}^{3}z
{x}^{2}+3\,{z}^{2}{x}^{2}{y}^{2}-y{z}^{3}{x}^{2}-{y}^{3}{z}^{2}x-{z}^{
3}{y}^{2}x+{y}^{3}{z}^{3}\geq 0$$
assume that $x=\min(x,y,z)$ and set $y=x+u,z=x+y+z$ after simplification we obtain
$$\left( {u}^{2}+uv+{v}^{2} \right) {x}^{4}+ \left( 4\,{u}^{3}+6\,{u}^{
2}v+2\,u{v}^{2} \right) {x}^{3}+ \left( 6\,{u}^{4}+12\,{u}^{3}v+6\,{u}
^{2}{v}^{2} \right) {x}^{2}+ \left( 4\,{u}^{5}+10\,{u}^{4}v+8\,{u}^{3}
{v}^{2}+2\,{u}^{2}{v}^{3} \right) x+{u}^{6}+3\,{u}^{5}v+3\,{u}^{4}{v}^
{2}+{u}^{3}{v}^{3}
\geq 0$$ since $u,v\geq 0$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1590461",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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|
Solving $\lim \limits _{x \to \infty} (\sqrt[n]{(x+a_1) (x+a_2) \dots (x+a_n)}-x)$ $$\lim \limits _{x \to \infty}\bigg(\sqrt[n]{(x+a_1) (x+a_2) \dots (x+a_n)}-x\bigg)$$
We can see the limit is of type $\infty-\infty$. I don't see anything I could do here. I can only see the geometric mean which is the $n$-th root term. Can I do anything with it? Any tips on solving this?
|
It's more complicated to write in LaTeX, than to solve. Remember that
$$A-B = \frac {A^n - B^n} {A^{n-1} + A^{n-2} B + \dots + A B^{n-2} + B^{n-1}} .$$
Choosing $A = \sqrt[n]{(x+a_1) (x+a_2) \dots (x+a_n)}$ and $B = \sqrt[n] {x^n}$, note that the largest power of $x$ in $A^n - B^n$ is $x^{n-1}$, and it is multiplied by the coefficient $a_1 + \dots + a_n$.
Similarly, looking for the dominant part of $x$ in the denominator, write
$$A^{n-k} \cdot B^{k-1} = [ (x+a_1) (x+a_2) \dots (x+a_n) ] ^{\frac {n-k} n} \cdot (x^n) ^{\frac {k-1} n} = \\ (x^n) ^{\frac {n-k} n} \left[ \left( 1 + \frac {a_1} x \right) \dots \left( 1 + \frac {a_n} x \right)\right] ^{\frac {n-k} n} \cdot (x^n) ^{\frac {k-1} n} = x^{n-1} \left[ \left( 1 + \frac {a_1} x \right) \dots \left( 1 + \frac {a_n} x \right)\right] ^{\frac {n-k} n} .$$
Note that the part between square brackets tends to $1$ when $x \to \infty$ (because $n$, the number of factors, stays fixed and each factor tends to $0$), so $\dfrac {A^{n-k} \cdot B^{k-1}} {x^{n-1}} \to 1$ (i.e $A^{n-k} \cdot B^{k-1}$ behaves asymptotically like $x^{n-1}$ when $x \to \infty$). There are $n$ such terms in the denominator, so $\dfrac {n x^{n-1}} {\text{denominator}} \to 1$.
Putting everything together you get that
$$\sqrt[n]{(x+a_1) (x+a_2) \dots (x+a_n)} - \sqrt[n] {x^n} = A-B = \\ \frac {(a_1 + \dots a_n) x^{n-1} + \text{smaller powers of} \ x} {n x^{n-1}} \frac {n x^{n-1}} {\text{denominator}} \to \frac {a_1 + \dots + a_n} n .$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1591903",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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|
How to find critical numbers in awkward function I have problem to find critical numbers in this awkward function with Euler's constants.
$$f(x, y) = e^{2 x+3 y} \left(8 x^2-6 x y+3 y^2\right)$$
Task: Find critical numbers and determine if it's maximum or minimum
|
We find $f_x$ and $f_y$ set them equal to zero and find the critical points.
$$f_x = 2 e^{2 x+3 y} \left(8 x^2-6 x y+3 y^2\right)+e^{2 x+3 y} (16 x-6 y)=0$$
$$f_y = 3 e^{2 x+3 y} \left(8 x^2-6 x y+3 y^2\right)+e^{2 x+3 y} (6 y-6 x)=0$$
Clearly the exponential terms do not give zeros (divide those out), so we are left with solving:
$$2 \left(8 x^2-6 x y+3 y^2\right)+ (16 x-6 y)=0$$
$$3 \left(8 x^2-6 x y+3 y^2\right)+ (6 y-6 x)=0$$
Solving these leads to two critical points:
$$(x, y) = (0, 0), \left(-\dfrac{1}{4}, -\dfrac{1}{2}\right)$$
You can now try to classify them.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1592149",
"timestamp": "2023-03-29T00:00:00",
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|
How to solve the integral $\int\frac{x-1}{\sqrt{ x^2-2x}}dx $ How to calculate $$\int\frac{x-1}{\sqrt{ x^2-2x}}dx $$
I have no idea how to calculate it. Please help.
|
Substitute $u=x^2-2x,du=2(x-1)dx$. Then:
\begin{align*}
\int \frac{1}{2\sqrt{u}}du &=\frac{1}{2}\int u^{-0.5}du
\end{align*}
User the power rule:
$$\int x^adx=\frac{x^{a+1}}{a+1},\:\quad \:a\ne -1$$
$$\Rightarrow \frac{1}{2}\int u^{-0.5}du =\frac{1}{2}\frac{u^{-0.5+1}}{-0.5+1}$$
Back substitute and simplify:
$$\Rightarrow \frac{1}{2}\int u^{-0.5}du =\frac{1}{2}\frac{u^{-0.5+1}}{-0.5+1}=\frac{1}{2}\frac{\left(x^2-2x\right)^{-0.5+1}}{-0.5+1}=\frac{1}{2}\cdot 2\cdot\sqrt{x^2-2x}=\sqrt{x^2-2x},$$
and don't forget to add a constant $c\in\mathbb{R}$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1595026",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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|
If the inequality $\log_a(x^2-x-2)>\log_a(-x^2+2x+3)$ is known to be satisfied for $x=\frac{9}{4}$ in the interval $(x_1,x_2)$ If the inequality $\log_a(x^2-x-2)>\log_a(-x^2+2x+3)$ is known to be satisfied for $x=\frac{9}{4}$ in the interval $(x_1,x_2)$,then find the product $x_1x_2$.
Here $a$ is not specified .I know that $\log$ is an increasing function,so we can cancel it out from both the sides,
$\log_a(x^2-x-2)>\log_a(-x^2+2x+3)$
$(x^2-x-2)>(-x^2+2x+3)$
$2x^2-3x-5>0$
$(x+1)(2x-5)>0$
So $x\in(-\infty,-1)\cup(\frac{5}{2},\infty)$ but $\frac{9}{4}=2.25$ is not in this interval.
What wrong have i done in this?What is the correct way to solve this and what is the correct answer?
|
You have done interval for $x<0$ plug in $-\infty$ you will see why. So the correct interval is $(2,5/2)$ so the product is $5$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1595990",
"timestamp": "2023-03-29T00:00:00",
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|
Show $\frac{(p^d - 1)(p^{d-1} - 1)}{(p-1)(p^2 - 1)} \equiv 1 \pmod{p}$. Let $p$ be prime and $d \ge 2$. I want to show that
$$
\frac{(p^d - 1)(p^{d-1} - 1)}{(p-1)(p^2 - 1)} \equiv 1 \pmod{p}.
$$
I have a proof, but I think it is complicated, and the statement appears in a book as if it is very easy to see. So is there any easy argument to see it?
My proof uses
$$
\frac{p^n - 1}{p-1} = 1 + p + \ldots + p^{n-1}.
$$
So
$$
\frac{(p^d - 1)(p^{d-1} - 1)}{(p-1)(p^2 - 1)} = \frac{(1 + p + \ldots + p^{d-1})(1 + p + \ldots + p^{d-2})}{p+1}
$$
If $d - 2$ is odd, set $u := (1 + p + \ldots + p^{d-1})$ and then proceed
\begin{align*}
& \frac{u(1 + p + \ldots + p^{d-2})}{p+1} \\
& = \frac{u(1+p) + u(p^2 + \ldots + p^{d-4})}{p+1} \\
& = u + p^2\cdot \frac{u(1+p+\ldots + p^{d-4})}{p+1} \\
& = u + p^2\cdot \left( \frac{u(1+p) + u(p^2 + \ldots p^{d-4})}{p+1} \right) \\
& = u + p^2\cdot \left( u + p^2\cdot \frac{u(1+p+\ldots p^{d-6})}{p+1} \right) \\
& \quad \qquad\qquad \vdots \\
& = u + p^2\cdot \left( u + p^2 \left( u + p^2\left( u + \ldots + p^2 \frac{u(1+p)}{1+p} \right) \right) \right) \\
& = u + p^2\cdot ( u + p^2 \cdot ( u + p^2 ( u + \ldots + p^2 u ))) \\
& \equiv u \pmod{p} \\
& \equiv 1 \pmod{p}
\end{align*}
and similar if $d-1$ is odd then successively multiply $(1+p+\ldots + p^{d-1})$ out with $v := (1+p+\ldots p^{d-2})$. But as said, this seems to complicated for me, so is there another easy way to see this?
|
You need to prove that $p\mid\frac{(p^{d-1}-1)(p^d-1)}{(p-1)(p^2-1)}-1$
It is not difficult to see that $p\mid (p^{d-1}-1)(p^d-1)-(p-1)(p^2-1)$. Furthermore, we have that $(p-1)(p^2-1)\mid (p^{d-1}-1)(p^d-1)$ because either $d$ or $d-1$ is even and so either $p^{d}-1$ or $p^{d-1}-1$ is divisible by $p^2-1$ and the other factor is then divisible by $p-1$.
So to sum up, we have proven that:
$$
p\mid (p^{d-1}-1)(p^d-1)-(p-1)(p^2-1)
$$
and
$$
(p-1)(p^2-1)\mid (p^{d-1}-1)(p^d-1)-(p-1)(p^2-1)
$$
Now, because $\gcd(p,(p-1)(p^2-1))=1$, this implies:
$$
p(p-1)(p^2-1)\mid (p^{d-1}-1)(p^d-1)-(p-1)(p^2-1)\implies\\
p\mid \frac{(p^{d-1}-1)(p^d-1)}{(p-1)(p^2-1)}-1
$$
|
{
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"timestamp": "2023-03-29T00:00:00",
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|
Find all real numbers $x,y > 1$ such that $\frac{x^2}{y-1}+\frac{y^2}{x-1} = 8$
Find all real numbers $x,y > 1$ such that $$\frac{x^2}{y-1}+\frac{y^2}{x-1} = 8$$
Attempt
We can rewrite this as $x^2(x-1)+y^2(y-1) = 8(x-1)(y-1)$. Then I get a multivariate cubic, which I find hard to find all solutions to.
|
user236182 was on the right track by using Cauchy-Schwarz. We have from Cauchy-Schwarz that $$((y-1)+(x-1)) \left (\dfrac{x^2}{y-1}+\dfrac{y^2}{x-1} \right) \geq \left (\sqrt{y-1}\sqrt{\dfrac{x^2}{y-1}}+\sqrt{x-1}\sqrt{\dfrac{y^2}{x-1}} \right)^2$$ and letting $a = x+y$ we get $8 \geq \dfrac{a^2}{a-2} \implies a < 2$ or $a = 4$. But $a \not < 2$ and thus $a = 4 = x+y$. Plugging this into $x^2(x-1)+y^2(y-1) = 8(x-1)(y-1)$ for $y$ gives a quadratic in $x$ giving the solution $x =2 = y$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1596327",
"timestamp": "2023-03-29T00:00:00",
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|
How to evaluate $\lim\limits_{x \to +\infty} \frac{(x+2)!+4^x}{((2x+1)^2+\ln x)x!}$? I have a problem with this limit, I don't know what method to use. I have no idea how to compute it.
Can you explain the method and the steps used?
$$\lim\limits_{x \to +\infty} \frac{(x+2)!+4^x}{((2x+1)^2+\ln x)x!}$$
|
$$\lim_{ x\to +\infty } \frac { (x+2)!+4^{ x } }{ ((2x+1)^{ 2 }+\ln { x } )x! } =\lim_{ x\to +\infty } \frac { x!\left( x+1 \right) (x+2)+4^{ x } }{ ((2x+1)^{ 2 }+\ln { x } )x! } =\lim_{ x\to +\infty } \frac { x!\left( \left( x+1 \right) (x+2)+\frac { { 4 }^{ x } }{ x! } \right) }{ ((2x+1)^{ 2 }+\ln { x } )x! } =\\ =\lim_{ x\rightarrow \infty }{ \frac { { x }^{ 2 }+3x+2+\frac { { 4 }^{ x } }{ x! } }{ 4{ x }^{ 2 }+4x+1+\ln { x } } = } \lim_{ x\rightarrow \infty }{ \frac { { x }^{ 2 }\left( 1+\frac { 3 }{ x } +\frac { 2 }{ { x }^{ 2 } } +\frac { { 4 }^{ x } }{ { x }^{ 2 }x! } \right) }{ { x }^{ 2 }\left( 4+\frac { 4 }{ x } +\frac { 1 }{ { x }^{ 2 } } +\frac { \ln x }{ x } \right) } = } \frac { 1 }{ 4 } $$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1597093",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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|
Finding the limit of the sequence $T_1=0,T_2=1, T_n=\frac{T_{n-1}+T_{n-2}}{2}$ Given that $T_1=0$, $T_2=1$ and $T_n=\frac{T_{n-1}+T_{n-2}}{2}$, show that the sequence converges to $\frac{2}{3}$.
|
Start from the matrix representation of the problem
$$\begin{bmatrix}
T_{n} & T_{n-1}
\end{bmatrix}=\begin{bmatrix}
T_{n-1} & T_{n-2}
\end{bmatrix}
\begin{bmatrix}
\frac{1}{2}& 1\\
\frac{1}{2}& 0
\end{bmatrix}$$
$$\begin{bmatrix}
T_{2} & T_{1}
\end{bmatrix}=\begin{bmatrix}
1 & 0
\end{bmatrix}$$
Chaining it all you have
$$\begin{bmatrix}
T_{n} & T_{n-1}
\end{bmatrix}=\begin{bmatrix}
T_{2} & T_{1}
\end{bmatrix}
\begin{bmatrix}
\frac{1}{2}& 1\\
\frac{1}{2}& 0
\end{bmatrix}^{n-2}$$
So it all comes down to finding
$$\lim\limits_{n \to \infty}
\begin{bmatrix}
\frac{1}{2}& 1\\
\frac{1}{2}& 0
\end{bmatrix}^{n}$$
Now this part is really classical and you just find eigenvalues of the matrix $\lambda_{1}=-\frac{1}{2}$,$\lambda_{2}=1$ and diagonalize
$$
\begin{bmatrix}
\frac{1}{2}& 1\\
\frac{1}{2}& 0
\end{bmatrix}=\begin{bmatrix}
-1 & 2\\
1 & 1
\end{bmatrix}
\begin{bmatrix}
-\frac{1}{2}& 0\\
0 & 1
\end{bmatrix}
\begin{bmatrix}
-\frac{1}{3} & \frac{2}{3}\\
\frac{1}{3} & \frac{1}{3}
\end{bmatrix}$$
to have
$$\lim\limits_{n \to \infty}
\begin{bmatrix}
-1 & 2\\
1 & 1
\end{bmatrix}
\begin{bmatrix}
-\frac{1}{2}& 0\\
0 & 1
\end{bmatrix}^{n}
\begin{bmatrix}
-\frac{1}{3} & \frac{2}{3}\\
\frac{1}{3} & \frac{1}{3}
\end{bmatrix}=\lim\limits_{n \to \infty}
\begin{bmatrix}
-1 & 2\\
1 & 1
\end{bmatrix}
\begin{bmatrix}
(-\frac{1}{2})^{n}& 0\\
0 & 1
\end{bmatrix}
\begin{bmatrix}
-\frac{1}{3} & \frac{2}{3}\\
\frac{1}{3} & \frac{1}{3}
\end{bmatrix}=$$
$$\begin{bmatrix}
-1 & 2\\
1 & 1
\end{bmatrix}
\begin{bmatrix}
0 & 0\\
0 & 1
\end{bmatrix}
\begin{bmatrix}
-\frac{1}{3} & \frac{2}{3}\\
\frac{1}{3} & \frac{1}{3}
\end{bmatrix}=\frac{1}{3}\begin{bmatrix}
2 & 2\\
1 & 1
\end{bmatrix}$$
This is giving
$$\lim\limits_{n \to \infty}\begin{bmatrix}
1 & 0
\end{bmatrix}
\begin{bmatrix}
\frac{1}{2}& 1\\
\frac{1}{2}& 0
\end{bmatrix}^{n-2}=\begin{bmatrix}
1 & 0
\end{bmatrix}\frac{1}{3}\begin{bmatrix}
2 & 2\\
1 & 1
\end{bmatrix}=\begin{bmatrix}
\frac{2}{3} & \frac{2}{3}
\end{bmatrix}$$
which means that the series converges to $\frac{2}{3}$
|
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"timestamp": "2023-03-29T00:00:00",
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|
Decomposing $x^4-5x^2+6$ over some fields My book asks me to decompose
$$x^4-5x^2+6$$
over:
$K = \mathbb{Q},\\ K = \mathbb{Q[\sqrt{2}]},\\ K = \mathbb{R}$
For $K = \mathbb{Q}$, I substituted $x² = a$ to get:
$$a²-5a+6 = (a-3)(a-2)$$
So Getting back to $a = x²$ we get:
$$x^4-5x^2+6 = (x²-3)(x²-2)$$
Also, for $\mathbb{R}$ we can simply factor $x²-3 = (x-\sqrt{3})(x+\sqrt{3})$ and $(x²-2) = (x-\sqrt{2})(x+\sqrt{2})$ so we have:
$$x^4-5x^2+6 = (x-\sqrt{3})(x+\sqrt{3})(x-\sqrt{2})(x+\sqrt{2})$$
but what about
$$K = \mathbb{Q[\sqrt{2}]}\ ?$$
|
Your work is correct; for $\mathbb{Q}[\sqrt{2}]$, you surely have
$$
x^4-5x^2+6=(x-\sqrt{2})(x+\sqrt{2})(x^2-3)
$$
Is $x^2-3$ reducible over $\mathbb{Q}[\sqrt{2}]$? This would mean…
|
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|
Denoting the sum of $n$ odd numbers $1+3+5+\cdots+(2n-1) = 1+3+5+\cdots+(2n+1)$? Let's say we are denoting the sum of $n$ odd numbers.
Then in symbols $1+3+5+\cdots+(2n-1)$.
If we substitute $(k+1)$ for $n$. $2n-1=2k+1$
So $1+3+5+...+2k+1$
Then can we use $1+3+5+\cdots+(2n+1)$ instead of $1+3+5+\cdots+(2n-1)$?
Logically, I think they are the same, but when I think of the number of terms, $1+3+5+\cdots+(2n+1)$ has one more term than $1+3+5+\cdots+(2n-1)$.
I also think $1+3+5+\cdots+(2n+1)$ is equal to $1+3+5+\cdots+(2n-1)+(2n+1)$, so it would be contradiction. But I don't know how to explain $1+3+5+\cdots+(2n-1)=1+3+5+\cdots+2k+1$ is a contradiction.
|
Suppose your inductive hypothesis is $$1+3+5+...+(2n-1) = n^2.$$
You initially show the inductive hypothesis is true when $n=1$, e.g. by saying $1=1^2$ is indeed correct
Then you assume the inductive hypothesis is true for $n=k$ so $1+3+5+...+(2k-1) = k^2.$ You add $(2k+1)$ to both sides so the left hand side becomes $1+3+5+...+(2k+1)$ and the right hand side becomes $k^2 +2k +1 = (k+1)^2$. Since these are equal, this shows the inductive hypothesis would then be true for $n=k+1$.
So using the axiom of induction, this proves the inductive hypothesis.
There is no need for contradiction.
|
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|
Proving $\pi=2\sum_{n=0}^{\infty} \arctan \frac{1}{F_{2n+1}}$ How to prove that $$\pi=2\sum_{n=0}^{\infty} \arctan \frac{1}{F_{2n+1}}$$
Where $F_{n}$ is the Fibonacci Number.
|
The goal is to write $\arctan\left(\dfrac1{F_{2n+1}}\right)$ as $\arctan(a_{n+1}) - \arctan(a_{n})$. This means we need
$$\dfrac{a_{n+1}-a_n}{1+a_na_{n+1}} = \dfrac1{F_{2n+1}}$$
Recall that from Cassini/Catalan identity we have
$$F_{2n+1}^2 = 1+F_{2n+2}F_{2n}$$
Hence, let $a_n = F_{2n}$. We then have
$$\dfrac{a_{n+1}-a_n}{1+a_na_{n+1}} = \dfrac{F_{2n+2}-F_{2n}}{1+F_{2n+2}F_{2n}} = \dfrac{F_{2n+1}}{F_{2n+1}^2} = \dfrac1{F_{2n+1}}$$
Hence, we have
$$\arctan\left(\dfrac1{F_{2n+1}}\right) = \arctan(F_{2n+2})-\arctan(F_{2n})$$
I trust you can finish off from here.
Hence, we have $$\sum_{n=0}^m \arctan\left(\dfrac1{F_{2n+1}}\right) = \arctan(F_{2m+2}) \implies \sum_{n=0}^{\infty} \arctan\left(\dfrac1{F_{2n+1}}\right) = \lim_{m \to \infty}\arctan(F_{2m+2}) = \dfrac{\pi}2$$
|
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|
Prove that if $a$ and $b$ are nonnegative real numbers, then $(a^7+b^7)(a^2+b^2) \ge (a^5+b^5)(a^4+b^4)$ Prove that if $a$ and $b$ are nonnegative real numbers, then $(a^7+b^7)(a^2+b^2) \ge (a^5+b^5)(a^4+b^4)$
My try
My book gives as a hint to move everything to the left hand side of the inequality and then factor and see what I get in the long factorization process and to lookout for squares.
So that's what I have tried:
\begin{array}
((a^7+b^7)(a^2+b^2) &\ge (a^5+b^5)(a^4+b^4) \\\\
(a^7+b^7)(a^2+b^2)-(a^5+b^5)(a^4+b^4) &\ge 0 \\\\
(a+b)(a^6-a^5b+a^4b^2-a^3b^3+a^2b^4-ab^5+b^6)(a^2+b^2)-(a+b)(a^4-a^3b+a^2b^2-ab^3+b^4)(a^4+b^4) &\ge 0 \\\\
(a+b)\left[(a^6-a^5b+a^4b^2-a^3b^3+a^2b^4-ab^5+b^6)((a+b)^2-2ab)-(a^4+b^4)(a^4-a^3b+a^2b^2-ab^3+b^4)\right] &\ge 0
\end{array}
Now it's not clear what I have to do next.I am stuck.
Note: My book doesn't teach any advanced technique for solving inequality as AM-GM ,Cauchy inequality etc..
|
Multiplying out, this is equivalent to $a^7b^2 + a^2b^7 \geq a^5b^4 + a^4b^5$, or $a^5 + b^5 \geq a^3 b^2 + a^2b^3$ (the case $ab=0$ is easy).
We can prove this using weighted AM-GM:
$$\frac{3}{5}a^5 + \frac{2}{5}b^5 \geq a^3 b^2$$
$$\frac{2}{5}a^5 + \frac{3}{5}b^5 \geq a^2 b^3$$
(Sorry, I didn't notice that you specified no AM-GM, but I'll leave this up in case others are interested.)
|
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|
Prove that $a^2+b^2+c^2+3\sqrt[3]{a^2b^2c^2} \geq 2(ab+bc+ca).$
Let $a,b,c$ be three nonnegative real numbers. Prove that $$a^2+b^2+c^2+3\sqrt[3]{a^2b^2c^2} \geq 2(ab+bc+ca).$$
It seems that the inequality $a^2+b^2+c^2 \geq ab+bc+ca$ will be of use here. If I use that then I will get $a^2+b^2+c^2+3\sqrt[3]{a^2b^2c^2} \geq ab+bc+ca+3\sqrt[3]{a^2b^2c^2}$. Then do I use the rearrangement inequality similarly on $3\sqrt[3]{a^2b^2c^2}$?
|
Let $a+b+c=3u$, $ab+ac+bc=3v^2$ and $abc=w^3$.
Hence, our inequality it's $f(v^2)\geq0$, where $f(v^2)=3u^2-4v^2+w^2$.
Thus, $f$ is a linear function, which says that $f$ get's a minimal value for an extremal value of $v^2$,
which happens for equality case of two variables.
Let $b=a=x^3$ and $c=1$.
Hence, we need to prove that $x^6+2+3x^2\geq2(2x^3+1)$,
which is $(x-1)^2(x^2+2x+3)x^2\geq0$. Done!
|
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|
prove that $(x-y)^3+(y-z)^3+(z-x)^3 > 0.$
If $x<y<z$ are real numbers, prove that $$(x-y)^3+(y-z)^3+(z-x)^3 > 0.$$
Attempt
We have that $(x-y)^3+(y-z)^3+(z-x)^3=-3 x^2 y+3 x^2 z+3 x y^2-3 x z^2-3 y^2 z+3 y z^2$. Grouping yields $3(x^2z+xy^2+yz^2)-3(x^2y+xz^2+y^2z)$. Then I was thinking of using rearrangement but then I would get $\geq$ instead of $>$.
|
Hint:
If $a+ b +c=0, a^3+b^3+c^3=3abc$
|
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|
Show $\lim_{a \rightarrow \infty} \int_0^1 {f(x)x\sin(ax^2)}=0$.
Suppose $f$ is integrable on $(0,1)$, then show $$\lim_{a \rightarrow \infty} \int_0^1 {f(x)x\sin(ax^2)}=0.$$
I tried to write $$(0,1) = \bigcup _{k=0}^{{a-1}} \left(\sqrt{\frac{k}{a}},\sqrt{\frac{k+1}{a}}\right),$$ but cannot make the integral converge to $0$.
|
Suggestion:
Split $[0, 1]$
into the intervals where
$ax^2 = 2\pi n$,
$ax^2 = \pi(2 n+1)$,
$ax^2 = \pi(2 n+2)$.
These are
$I_{2n}
=[\sqrt{\frac{2\pi n}{a}}, \sqrt{\frac{\pi(2 n+1)}{a}})
$
and
$I_{2n+1}
=[\sqrt{\frac{\pi(2 n+1)}{a}}, \sqrt{\frac{\pi(2 n+2)}{a}})
$.
Since
$\sin(ax^2) > 0$
in
$I_{2n}$
and
$\sin(ax^2) < 0$
in
$I_{2n+1}$,
show that
the integral over
$I_{2n} \cup I_{2n+1}$
goes to zero.
|
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|
Solve equation on mathematical physics Show that
$$\gamma_+ - \gamma_-=\frac{2\beta_0\beta}{\sqrt{(1-\beta_0^2)(1-\beta^2)}}$$
where
$$\gamma_+=(1-\beta^2_+)^{-\frac{1}{2}} \ \mbox{and} \ \beta_+=\frac{\beta_0+\beta}{1+\beta_0\beta}$$
$$\gamma_-=(1-\beta^2_-)^{-\frac{1}{2}} \ \mbox{and} \ \beta_-=\frac{\beta_0-\beta}{1-\beta_0\beta}$$
|
I was moved $\gamma_-$ to other side equation:
$$\gamma_+=\frac{2\beta_0\beta}{\sqrt{(1-\beta_0^2)(1-\beta^2)}}+\gamma_-$$
So
$$\gamma_+=\frac{2\beta_0\beta}{\sqrt{(1-\beta_0^2)(1-\beta^2)}}+\frac{1}{\sqrt{1-\beta^2_-}}$$
Therefore
$$\gamma_-=\frac{2\beta_0\beta\sqrt{1-\beta^2}+\sqrt{(1-\beta^2_0)(1-\beta^2)}}{\sqrt{(1-\beta_0^2)(1-\beta^2)}\sqrt{1-\beta^2_-}}$$
So we come to the conclusion that
$$\gamma_-^2=\left(\frac{2\beta_0\beta\sqrt{1-\beta^2}+\sqrt{(1-\beta^2_0)(1-\beta^2)}}{\sqrt{(1-\beta_0^2)(1-\beta^2)}\sqrt{1-\beta^2_-}}\right)^2$$
$$\gamma_-^2=\frac{\left(2\beta_0\beta\sqrt{1-\beta^2}+\sqrt{(1-\beta^2_0)(1-\beta^2)}\right)^2}{\left(\sqrt{(1-\beta_0^2)(1-\beta^2)}\sqrt{1-\beta^2_-}\right)^2}$$
Well I think?
|
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|
Value of $X^4 + 9x^3+35X^2-X+4$ for $X=-5+2\sqrt{-4}$
Find the value of $X^4 + 9x^3+35X^2-X+4$ for $X=-5+2\sqrt{-4}$
Now the trivial method is to put $X=5+2\sqrt{-4}$ in the polynomial and calculate but this is for $2$ marks only and that takes a hell lot of time for $2$! So I was thinking may be there is some trick or other technique to get the result quicker . Can anybody help please $?$
Thank you .
|
It's not the that much work.
1) Just calculate $x, x^2, x^4 = (x^2)^2, x^3 = x^2 * x$ to the side first. and
2) consider every number as having two "parts"; a "normal" number part and a "square root of negative one part"
$x = -5 + \sqrt{-4} = -5 + 2\sqrt{-1}$
$x^2 = 25 + 4(-1) + 20\sqrt{-1} = 21 + 20\sqrt{-1}$
$x^4 = 21^2 + 20^2(-1) + 2*20*21\sqrt{-1} = 441 - 400 + 840\sqrt{-1}= 41 + 840\sqrt{-1}$
$x^3 = (21 + 20\sqrt{-1})(-5 + 2\sqrt{-1}) = -110 + 40(-1) + (-100 + 42)\sqrt{-1} = -150 - 58\sqrt{-1}$
So $X^4 + 9x^3+35X^2-X+4 = (41 + 9(-150) + 35(21) -(-5) + 4) + \sqrt{-1}(840 + 9(-58) + 35(2) - 2) = -565 + 386\sqrt{-1}$
This is a hint to complex numbers. We call $\sqrt{-1} = i$ and all numbers are of the form $(a, b*i)$. $(a,bi) + (c,di) = (a+c, (b+d)i)$ and $(a,bi)(c,di) = (ac - bd, (ad + bc)i)$.
|
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|
Prove that $\sqrt{(a+b+1)(c+2)}+\sqrt{(b+c+1)(a+2)}+\sqrt{(c+a+1)(b+2)}\ge{9}$. Let $a,b,c$ be non-negative real numbers such that $\sqrt{a}+\sqrt{b}+\sqrt{c}=3$. Prove that
$\sqrt{(a+b+1)(c+2)}+\sqrt{(b+c+1)(a+2)}+\sqrt{(c+a+1)(b+2)}\ge{9}$.
Here equality holds when $a=b=c=1$ so by using AM-GM on $c+1+1$ we get
$\sqrt{(a+b+1)(c+2)}+\sqrt{(b+c+1)(a+2)}+\sqrt{(c+a+1)(b+2)}\ge{9(abc)^{1/6}}$ but this is not right choice because by am-gm on given condition we have $(abc)^{1/6}\le{1}$. I tried Cauchy-Schwartz, too.
|
Cauchy Schwarz actually works:
$$
\sum_{cyc}\sqrt{(a+b+1)(1+1+c)}≥\sum_{cyc}\sqrt{\left(\sqrt a+\sqrt b+\sqrt c\right)^2}=3(\sqrt a+\sqrt b+\sqrt c)=9
$$
|
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|
Find all solutions to the diophantine equation $(x+2)(y+2)(z+2)=(x+y+z+2)^2$ Solve in postive integer the equation
$$(x+2)(y+2)(z+2)=(x+y+z+2)^2$$
It is rather easy to find several parametric solutions, (such $(a,b,c)=(2,1,1),(2,2,2)$).but it seems harder to find a complete enumeration of all the solutions.
and I have proved it for these some solutions :
since we have $$x^2+y^2+z^2=xyz+4$$
because we have
$$\left(a+\dfrac{1}{a}\right)^2+\left(b+\dfrac{1}{b}\right)^2+\left(c+\dfrac{1}{c}\right)^2=\left(a+\dfrac{1}{a}\right)\left(b+\dfrac{1}{b}\right)\left(c+\dfrac{1}{c}\right)+4,abc=1$$
so Let
$$x=a+\dfrac{1}{a},y=b+\dfrac{1}{b},z=c+\dfrac{1}{c},abc=1,a,b,c>0$$
It seems that there are no nontrivial solutions , but I cannot prove this.
|
For the equation.
$$(x+2)(y+2)(z+2)=(x+y+z+2)^2$$
If we use the solutions of Pell equations.
$$p^2-(z^2-4)s^2=1$$
Using my replacement.
$$a=p^2-2(z+2)ps+(z^2-4)s^2$$
$$b=p^2-2zps+(z^2-4)s^2$$
Decisions will be.
$$x=-2a^2+2(z+2)ab-z(z+2)b^2$$
$$y=-za^2+2(z+2)ab-2(z+2)b^2$$
|
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|
Value of $a^2 + b^2$ given $a^3 - 3ab^2 = 44$ and $b^3 - 3a^2 b = 8$ If $a$ and $b$ are real numbers such that $a^3-3ab^2=44$ and $b^3-3a^2b=8$ what is value of $a^2+b^2$? I have tried by adding and subtracting these equations, but can't find anything.
|
Notice $$(a+ib)^3 = (a^3-3ab^2) + (3a^2b-b^3)i = 44-8i$$
Multiply both sides by their complex conjugates, one find
$$(a^2+b^2)^3 = 44^2 + 8^2 = 2000\quad\implies\quad
a^2 + b^2 = \sqrt[3]{2000} = 10\sqrt[3]{2}$$
Update
If one really want to hide the use of complex numbers, one can
take the squares of both conditions, sum them and then employ the familiar $1,3,3,1$ pattern to factorize the result.
$$\begin{array}{rllll}
(a^3 - 3ab^2)^2 &= a^6 &- 6a^4b^2 &+ 9a^2b^4 & &= 44^2\\
(b^3 - 3a^2b)^2 &= &+ 9a^4b^2 &- 6a^2b^4 &+ b^6 &= 8^2\\
\end{array}\\
\Downarrow \rlap{\color{gray}{\leftarrow\text{ sum the 2 equations }}}\\
a^6 + 3a^4b^2 + 3a^2b^4 + b^6 = 44^2+8^2 = 2000\\
\Downarrow \rlap{\color{gray}{\leftarrow\text{ use the 1, 3, 3, 1 pattern on LHS }}}\\
(a^2 + b^2)^3 = 2000\\
\Downarrow\\
(a^2 + b^2 ) = \sqrt[3]{2000} = 10\sqrt[3]{2}
$$
|
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|
Show that the angles satisfy $x+y=z$ How can I show that $x+y=z$ in the figure without using trigonometry? I have tried to solve it with analytic geometry, but it doesn't work out for me.
|
Imagine that all the the squares are 1 by 1 and so the rectangle has base 3 and a height of $1$.
There are three right-angled triangles in the diagram. The one with angle $x$ has base 3, and height 1. The triangle with angle $y$ has base 2 and height 1. The triangle with angle $z$ has base 1 and height 1.
Using the standard trig' ratio $\tan \theta = \frac{\mathrm{opp}}{\mathrm{adj}}$, we get $\tan x = \frac{1}{3}$, $\tan y = \frac{1}{2}$ and $\tan z= \frac{1}{1}=1$.
There is a well-know formula for angle addition:
$$\tan(\alpha+\beta) = \frac{\tan \alpha + \tan \beta}{1-\tan \alpha \tan \beta}$$
Applying this formula to the case of $\alpha =x$ and $\beta = y$ gives:
$$\tan(x+y) = \frac{\tan x + \tan y}{1-\tan x \tan y}=\frac{\frac{1}{3}+\frac{1}{2}}{1-\frac{1}{3}\cdot\frac{1}{2}}=1$$
It follows that $\tan(x+y)=\tan z$. Since $0^{\circ} < x<y<z < 90^{\circ}$ it follows that
$$\tan(x+y) = \tan z \iff x+y = z$$
|
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|
Prove the identity $\binom{2n+1}{0} + \binom{2n+1}{1} + \cdots + \binom{2n+1}{n} = 4^n$ I've worked out a proof, but I was wondering about alternate, possibly more elegant ways to prove the statement. This is my (hopefully correct) proof:
Starting from the identity $2^m = \sum_{k=0}^m \binom{m}{k}$ (easily derived from the binomial theorem), with $m = 2n$:
$2^{2n} = 4^n = \binom{2n}{0} + \binom{2n}{1} + \cdots + \binom{2n}{2n-1} + \binom{2n}{2n}$
Applying the property $\binom{m}{k} = \binom{m}{m-k}$ to the second half of the list of summands in RHS above:
$4^n = \binom{2n}{0} + \binom{2n}{1} + \cdots + \binom{2n}{n-1} + \binom{2n}{n} + \underbrace{\binom{2n}{n-1} +\cdots \binom{2n}{1} + \binom{2n}{0}}_{\binom{m}{k} = \binom{m}{m-k} \text{ has been applied}}$
Rearranging the above sum by alternately taking terms from the front and end of the summand list in RHS above (and introducing the term $\binom{2n}{-1} = 0$ at the beginning just to make explicit the pattern being developed):
$4^n = (\binom{2n}{-1} + \binom{2n}{0}) + (\binom{2n}{0} + \binom{2n}{1}) + \cdots + (\binom{2n}{n-1} + \binom{2n}{n})$
Finally, using the property $\binom{m}{k} + \binom{m}{k-1} = \binom{m+1}{k}$ on the paired summands, we get the desired result:
$4^n = \binom{2n+1}{0} + \binom{2n+1}{1} + \cdots + \binom{2n+1}{n}$
|
Why not just
$$ \begin{align} 2^{2n+1} &=\binom{2n+1}{0}+\cdots+\binom{2n+1}{n}+\binom{2n+1}{n+1}+\cdots+\binom{2n+1}{2n+1} \\
&=\binom{2n+1}{0}+\cdots+\binom{2n+1}{n}+\binom{2n+1}{n}+\cdots+\binom{2n+1}{0} \\
&=2\left[\binom{2n+1}{0}+\cdots+\binom{2n+1}{n}\right] \end{align} $$
Then divide each extremity by 2.
|
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|
An inequality on three constrained positive numbers Assume $a,b,c$ are all positive numbers, and $2a^3b+2b^3c+2c^3a=a^2b^2+b^2c^2+c^2a^2$.
Prove that:
$$2ab(a-b)^2+2bc(b-c)^2+2ca(c-a)^2\ge(ab+bc+ca)^2$$
|
Let $a+b+c=3u$, $ab+ac+bc=3v^2$ and $abc=w^3$.
Hence, the condition gives $\sum\limits_{cyc}(a^3b+a^3c-a^2b^2)=\sum\limits_{cyc}(a^3c-a^3b)$
or $9u^2v^2-9v^4+uw^3=u\prod\limits_{cyc}(a-b)$
or $(9u^2v^2-9v^4+uw^3)^2=27u^2(3u^2v^4-4v^6-4u^3w^3+6uv^2w^3-w^6)$
or $28u^2w^6+18u(6u^4-8u^2v^2-v^4)w^3-54u^2v^6+81v^8=0$, which gives
$3(6u^4-8u^2v^2-v^4)^2-28(-2u^2v^6+3v^8)\geq0$
or $108u^8-288u^6v^2+156u^4v^4+104u^2v^6-81v^8\geq0$, which gives
$u^2\geq1.49098...v^2$.
By the way, we get $28u^2w^6+18u(6u^4-8u^2v^2-v^4)w^3=9v^4(6u^2v^2-9v^4)$.
In another hand, we need to prove that $2\sum\limits_{cyc}(a^3b+a^3c-2a^2b^2)\geq9v^4$ or
$6u^2v^2-9v^4+2uw^3\geq0$ or $9v^4(6u^2v^2-9v^4)+18uv^4w^3\geq0$
or $28u^2w^6+18u(6u^4-8u^2v^2-v^4)w^3+18uv^4w^3\geq0$
or $27u^3-36uv^2+7w^3\geq0$.
By Schur $w^3\geq4uv^2-3u^3$.
Hence, it remains to prove that $u^2\geq\frac{4}{3}v^2$, which is obvious.
|
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|
Coupon collector's problem using inclusion-exclusion Coupon collector's problem asks:
Given n coupons, how many coupons do you expect you need to draw with replacement before having drawn each coupon at least once?
The well-known solution is $E(T)=n \cdot H_n$, where T is the time to collect all n coupons(proof).
I am trying to approach another way, by calculating possible arrangements of coupons using inclusion-exclusion(Stirling's numbers of the second kind) and that one coupon should only be collected at last and other coupons should be collected at least once:
$$P(T=k)=\frac{n!\cdot{k-1\brace n-1}}{n^k}\\
=\frac{\sum\limits_{i=1}^{n-1}(-1)^{n-i-1}\cdot{n-1\choose i}\cdot i^{k-1}}{n^{k-1}}\\
E(T)=\sum\limits_{k=n}^{\infty}k\cdot P(T=k)\\
=\sum\limits_{k=n}^{\infty}k\cdot\frac{\sum\limits_{i=1}^{n-1}(-1)^{n-i-1}\cdot{n-1\choose i}\cdot i^{k-1}}{n^{k-1}}\\
=\sum\limits_{i=1}^{n-1}(-1)^{n-i-1}\cdot{n-1\choose i}\cdot\sum\limits_{k=n}^{\infty}k\cdot (\frac i n)^{k-1}\\
=\sum\limits_{i=1}^{n-1}(-1)^{n-i-1}\cdot{n-1\choose i}\cdot(\frac i n)^{n-1}\cdot(\frac 1 {1-\frac i n})\cdot(n-1+\frac 1 {1-\frac i n})$$
Calculation of first 170 terms yields same results.
Are two formulas same?
|
By way of enrichment here is a proof using Stirling numbers of the
second kind which encapsulates inclusion-exclusion in the generating
function of these numbers.
First let us verify that we indeed have a probability distribution
here. We have for the number $T$ of coupons being $m$ draws that
$$P[T=m] = \frac{1}{n^m} \times
n\times {m-1\brace n-1} \times (n-1)!.$$
Recall the OGF of the Stirling numbers of the second kind which says
that
$${n\brace k} = [z^n] \prod_{q=1}^k \frac{z}{1-qz}.$$
This gives for the sum of the probabilities
$$\sum_{m\ge 1} P[T=m]
= (n-1)! \sum_{m\ge 1} \frac{1}{n^{m-1}} {m-1\brace n-1}
\\ = (n-1)! \sum_{m\ge 1} \frac{1}{n^{m-1}}
[z^{m-1}] \prod_{q=1}^{n-1} \frac{z}{1-qz}
\\ = (n-1)! \prod_{q=1}^{n-1} \frac{1/n}{1-q/n}
= (n-1)! \prod_{q=1}^{n-1} \frac{1}{n-q} = 1.$$
This confirms it being a probability distribution.
We then get for the expectation that
$$\sum_{m\ge 1} m\times P[T=m]
= (n-1)! \sum_{m\ge 1} \frac{m}{n^{m-1}} {m-1\brace n-1}
\\ = (n-1)! \sum_{m\ge 1} \frac{m}{n^{m-1}}
[z^{m-1}] \prod_{q=1}^{n-1} \frac{z}{1-qz}
\\ = 1 + (n-1)! \sum_{m\ge 1} \frac{m-1}{n^{m-1}}
[z^{m-1}] \prod_{q=1}^{n-1} \frac{z}{1-qz}
\\ = 1 + (n-1)! \sum_{m\ge 2} \frac{m-1}{n^{m-1}}
[z^{m-1}] \prod_{q=1}^{n-1} \frac{z}{1-qz}
\\ = 1 + \frac{1}{n} (n-1)! \sum_{m\ge 2} \frac{1}{n^{m-2}}
[z^{m-2}] \left(\prod_{q=1}^{n-1}
\frac{z}{1-qz}\right)'
\\ = 1 + \frac{1}{n} (n-1)!
\left.\left(\prod_{q=1}^{n-1}
\frac{z}{1-qz}\right)'\right|_{z=1/n}
\\ = 1 + \frac{1}{n} (n-1)!
\left. \left(\prod_{q=1}^{n-1}
\frac{z}{1-qz}
\sum_{p=1}^{n-1} \frac{1-pz}{z} \frac{1}{(1-pz)^2}
\right)\right|_{z=1/n}
\\ = 1 + \frac{1}{n} (n-1)!
\prod_{q=1}^{n-1} \frac{1/n}{1-q/n}
\left. \sum_{p=1}^{n-1} \frac{1}{z} \frac{1}{1-pz}
\right|_{z=1/n}
\\ = 1 + \frac{1}{n} (n-1)!
\prod_{q=1}^{n-1} \frac{1}{n-q}
\sum_{p=1}^{n-1} \frac{n}{1-p/n}
\\ = 1 + \frac{1}{n}
\sum_{p=1}^{n-1} \frac{n^2}{n-p}
= 1 + n H_{n-1} = n \times H_n.$$
What we have here are in fact two annihilated coefficient extractors (ACE) more of which may be found at this MSE link. Admittedly the EGF better represents inclusion-exclusion than the OGF and could indeed be used here where the initial coefficient extractor would then transform it into the OGF.
|
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|
Prove that $\frac{1}{a+ab}+\frac{1}{b+bc}+\frac{1}{c+ca} \geq \frac{3}{2}.$
Let $a,b,$ and $c$ be positive real numbers such that $abc = 1$. Prove that $$\dfrac{1}{a+ab}+\dfrac{1}{b+bc}+\dfrac{1}{c+ca} \geq \dfrac{3}{2}.$$
I thought about substituting in $abc = 1$ to get $$\dfrac{1}{a+\dfrac{1}{c}}+\dfrac{1}{b+\dfrac{1}{a}}+\dfrac{1}{c+\dfrac{1}{b}} = \dfrac{c}{ac+1}+\dfrac{a}{ab+1}+\dfrac{b}{bc+1}.$$ Then I am not sure what inequality to apply.
|
Let $a=\frac{x}{y}$, $b=\frac{y}{z}$, $c=\frac{z}{x}$.
$$\sum_{\text{cyc}}\frac{1}{\frac{x}{y}+\frac{x}{y}\frac{y}{z}}=\sum_{\text{cyc}}\frac{yz}{xz+xy}$$
Let $yz=c$, $xz=d$, $xy=e$. Then your inequality follows from Nesbitt's inequality.
|
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|
Prove that $ax+by+cz+2\sqrt{(xy+yz+xz)(ab+bc+ca)}\le{a+b+c}$ Let $a,b,c,x,y,z$ be positive real numbers such that $x+y+z=1$. Prove that
$$ax+by+cz+2\sqrt{(xy+yz+xz)(ab+bc+ca)}\le{a+b+c}$$.
my try:
$2\sqrt{(xy+yz+xz)(ab+bc+ca)}\le{\frac{2(a+b+c)}{3}}$
But this is not the right choice because
$ax+by+cz\le{\frac{a+b+c}{3}}$ is not always true.
|
Let $a+b+c=k$.
Hence, $(a-kx)^2+(b-ky)^2+(c-kz)^2\geq0$ gives
$(a-kx)^2+(b-ky)^2+(c-kz)^2\geq(a-kx+b-ky+c-kz)^2$ or
$(xy+xz+yz)k^2-(ay+bx+az+cx+bz+cy)k+ab+ac+bc\geq0$, which gives
$4(ab+ac+bc)(xy+xz+yz)\leq(ay+bx+az+cx+bz+cy)^2$ or
$ax+by+cz+2\sqrt{(ab+ac+bc)(xy+xz+yz)}\leq a+b+c$. Done!
|
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|
Factore $(x+1)(x+2)(x+3)(x+4)-35$ I need to factor
$$(x+1)(x+2)(x+3)(x+4)-35$$
I know that the answer will be
$$(x^2+5x+11)(x^2+5x-1)$$
I go out only
$$(x^2+5x)(x^2+5x+10)-11$$
Help me.
|
Put $t=x^2+5x$
Then the given expression will be $t(t+10)-11$
$\Rightarrow t^2+10t-11=t^2-t+11t-11=(t+1)(t-11)$
Now put the value of $t$
|
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|
A line is drawn through the point $A(1,2)$ to cut the line $2y=3x-5$ in $P$ and the line $x+y=12$ in $Q$. If $AQ=2AP$, find $P$ and $Q$. A line is drawn through the point $A(1,2)$ to cut the line $2y=3x-5$ in $P$ and the line $x+y=12$ in $Q$. If $AQ=2AP$, find the coordinates of $P$ and $Q$.
I found the lengths of the lines $AQ$ and $AP$ in terms of $x$ and $y$ and used $AQ=2AP$ to get this equation $11x^2-40x-16=0$
Working: $AP=\sqrt{(x-1)^2+(\frac{3}{2}x-\frac{9}{2})^2}=\sqrt{\frac{1}{4}(13x^2-62x+85)}$
$AQ=\sqrt{(x-1)^2+(-x+10)^2}=\sqrt{2x^2-22x+101}$
$AQ=2AP\Rightarrow \sqrt{2x^2-22x+101}=2\sqrt{\frac{1}{4}(13x^2-62x+85)}$
$\Rightarrow 2x^2-22x+101=13x^2-62x+85\Rightarrow 11x^2-40x-16=0$
Using the quadratic formula I found $x=4,x=-\frac{4}{11}$
I then substituted these values into the two equations in the question to get the coordinates $(4,\frac{7}{2}),(-\frac{4}{11},-\frac{15}{11})$ or $(4,8),(-\frac{4}{11},\frac{136}{11})$
But the answers in the book are $(4,\frac{7}{2}),(7,5)$ or $(\frac{2}{5},-\frac{19}{10}),(\frac{11}{5},\frac{49}{5})$
I don't know where I went wrong in my method.
Edit: Found some errors in my method and calculation.
Using $x=4$ in $2y=3x-5$ gives $P(4,\frac{7}{2})$
Line with points $AP$ has equation $y-2=\frac{\frac{7}{2}-2}{3}(x-1)\Rightarrow y=\frac{1}{2}x+\frac{3}{2}$
$Q$ lies on the same line. Therefore, the intersection between $y=\frac{1}{2}x+\frac{3}{2}$ and $y+x=12$ is the point $Q$. Working: $\frac{1}{2}x+\frac{3}{2}=-x+12\Rightarrow x=7$ Therefore, $y=-7+12=5$. Point Q has coordinates $(7,5)$.
But $x=-\frac{4}{11}$ is not the other $x$ coordinate of $P$ according to the answers in the book. I have no idea where I went wrong at this point.
|
Your initial equation setup is incorrect. Note that $P$ and $Q$ do not have the same $x$-coordinate! Because of this faulty assumption, anything you do after that will not lead to the right answer.
A good way to fix this is to give the coordinate different variable names. Let $P\left(a, \frac{3a-5}{2}\right)$ and $Q\left(b, 12-b \right)$ we have the distance formula
$$ \sqrt{(b-1)^2 + (10-b)^2} = 2\sqrt{(a-1)^2+\left(\frac{3a-5}{2}-2\right)^2} \tag{1} $$
Now use the fact that $A,\ P,\ Q$ are collinear to obtain a second equation
$$ \frac{10-b}{b-1} = \frac{\frac{3a-5}{2}-2}{a-1} \tag{2} $$
where the LHS is equal to the slope of $AQ$ and the RHS the slope of $AP$
Solve these two equations to obtain the $x$-coordinates of $P$ and $Q$
|
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|
How to show this fraction is equal to 1/2? I have the fraction:
$$\frac{\left(2 \left(\frac {a}{\sqrt{2}}\right) + a \right) a} {2(1 + \sqrt{2})a^2}$$
Using Mathematica, I've found that this simplifies to $\frac{1}{2}$, but how did it achieve the result? How can I simplify that fraction to $\frac12$?
|
Assume $a\neq 0$, we have \begin{align*}
\frac{\left(2\left(\frac {a}{\sqrt{2}}\right)+a\right)a} {2(1+\sqrt{2})a^2}&=\frac{\left(\frac{2a}{\sqrt 2}+a\right)}{2(1+\sqrt 2)a}\tag 1\\
&=\frac{\left(\frac{2}{\sqrt2}+1\right)a}{2(1+\sqrt 2)a}\tag2\\
&=\frac{\left(\frac{2}{\sqrt2}+1\right)}{(2+2\sqrt 2)}\\
&=\frac{\sqrt 2}{\sqrt 2}\cdot \frac{\left(\frac{2}{\sqrt2}+1\right)}{(2+2\sqrt 2)}\\
&=\frac{2+\sqrt 2}{2\sqrt 2+2\cdot 2}\\
&=\frac{2+\sqrt 2}{2\sqrt 2+4}\\
&=\frac{(2+\sqrt 2)}{2(2+\sqrt 2)}\\
&=\frac{1}{2}
\end{align*}
where in $(1)$ I cancelled the first $a$ and in $(2)$ I factored out the second $a$ to cancel it also.
|
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|
Probability of getting the same number three times. If I have a set of numbers $\{1 \ldots n\}$ where $n \ge 1$ and I pick $3$ numbers from the set independently and uniformly. Whats the probability I'll get all $2$'s, the probability I get all the same numbers and the probability that my first two numbers are the same.
I've I compared this to rolling a fair die three times. Wouldn't probability of getting three $2$'s and getting all the same numbers be $\left(\dfrac 16 \right)^3$ since its uniform. And the probability of getting two of the same numbers be $\left(\dfrac 16 \right)^2 \left(1-\dfrac{1}{6} \right)$??
Any help would be appreciated.
|
*
*You're right, the probability of all 2s is
$$P(222) = P(2)P(2)P(2) = \frac{1}{6}\cdot\frac{1}{6}\cdot\frac{1}{6} = \left(\frac{1}{6}\right)^3= \frac{1}{216},$$
where use the product by independence.
*When calculating probability that they are all the same is, you have to account for all the choices you have for a match, which is $\binom{6}{1}$. So,
$$P(\text{All match}) = \binom{6}{1}\left(\frac{1}{6}\right)^3 =\frac{1}{36}.$$
*For the first two to match you have again $\binom{6}{1} = 6$ choices for this pair. After, you will only have $\binom{5}{1} = 5$ choices for the remaining single. Lastly, you need to account for the order of the first two since for example $221$ is the same as if I switched the first two $221$, which is $\binom{3}{2,1} = 3$. So the probability of interest is
$$\binom{3}{2,1}\binom{6}{1}\left(\frac{1}{6}\right)^2\binom{5}{1}\left(\frac{1}{6}\right) = 90\left(\frac{1}{6}\right)^3 = \frac{5}{12}.$$
|
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|
How to find the complex roots of $y^3-\frac{1}{3}y+\frac{25}{27}$ I've been trying to solve this for hours and all found was the real solution by Cardano"s formula.
I vaguely remember that if $\alpha$ is a root of a complex number, the other roots are $\omega \alpha$ and $\omega ^2 \alpha$ where $\omega = - \frac{1}{2}+i \frac{\sqrt{3}}{2}$, $\omega ^2 = - \frac{1}{2}-i \frac{\sqrt{3}}{2}$.
The problem is, I don't know how to apply this to find my roots(if applicable at all).
Straightly put, how can I find other roots of a cubic given one root?(This time it happens to be around -0.75, for the real root). Given a single root, what is the relationship between others? Any formula to compute the remaining solutions at all?
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By $a^{3}+b^{3}+c^{3}-3abc=(a+b+c)(a+\omega b+\omega^{2} c)(a+\omega^{2} b+\omega c)$;
and replacing $a, b, c$ by $x, -u, -v$ respectively,
$x^{3}-3uvx-(u^{3}+v^{3})=0 \implies x_{k}=u\, \omega^{k}+ v\, \omega^{2k}$ for $k=0,1,2$.
$u, v$ are known as resolvents.
Let your reduced cubic be $y^{3}-3py-2q=0$.
(Will substitute $p=\frac{1}{9}, q=-\frac{25}{54}$ later.)
Now $uv=p$ and $2q=u^{3}+v^{3}$.
$\therefore 2q=u^{3}+\left( \frac{p}{u} \right)^{3}$
$\implies u^{6}-2qu^{3}-p^{3}=0$
$\implies u^{3}=q\pm \sqrt{q^{2}-p^{3}}$
$\implies u=\sqrt[3]{q\pm \sqrt{q^{2}-p^{3}}}$
$\displaystyle{
\therefore
v=\frac{p}{\sqrt[3]{q\pm \sqrt{q^{2}-p^{3}}}}
=\sqrt[3]{q\mp \sqrt{q^{2}-p^{3}}}
}$
$\therefore y=\sqrt[3]{q\pm \sqrt{q^{2}-p^{3}}}+\sqrt[3]{q\mp \sqrt{q^{2}-p^{3}}}$.
Note that $u, v$ are conjugates and symmetrical in roles.
By keeping the upper case,
$y=\sqrt[3]{q+\sqrt{q^{2}-p^{3}}}+\sqrt[3]{q-\sqrt{q^{2}-p^{3}}}$ is one of the root.
The other two roots are $\omega u+\omega^{2} v$ and $\omega^{2} u+\omega v$.
In this case,
$u+v=-\frac{1}{3} \left( \sqrt[3]{\frac{25+3\sqrt{69}}{2}}+\sqrt[3]{\frac{25-3\sqrt{69}}{2}} \right) $.
Careful manipulation in $\omega$ enables us to find the rest (complex roots).
P.S.: In case of the discriminant $\Delta = q^{2}-p^{3} <0$, there are 3 distinct real roots involving combination of $\cos(\frac{1}{3} \cos^{-1} (.))$ and not discussed here.
|
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|
an upper bound for number of primes in the interval $[n^2+n,n^2+2n]$ What is an upper bound for the number of primes in an interval of $n$ consecutive
numbers?
What is an upper bound for the number of primes in the interval $[n^2+n,n^2+2n]$?
|
According to the Prime-counting function
$$\text{number of primes below natural }n = \pi(n) \approx \frac{n}{\ln n}$$
so you just want an approximation of $\pi(n^2+2n) - \pi(n^2+n)$:
$$\begin{align}
\pi(n^2+2n) - \pi(n^2+n) &\approx \frac{n^2+2n}{\ln (n^2+2n)} - \frac{n^2+n}{\ln (n^2+n)}\\
&\approx \lim_{n \to \infty} \left( \frac{n^2+2n}{\ln (n^2+2n)} - \frac{n^2+n}{\ln (n^2+n)} \right)\\
&= \frac{n \left(-2 \ln \left(\frac{1}{n}\right)-1\right)}{4 \ln ^2\left(\frac{1}{n}\right)} - \frac{3 \left(\ln \left(\frac{1}{n}\right)+1\right)}{8 \ln ^3\left(\frac{1}{n}\right)}+O\left(\frac{1}{n}\right)\\
& \approx \frac{n \left(2 \ln n-1\right)}{4 \ln ^2n} - \frac{3 \left(-\ln n+1\right)}{-8 \ln ^3n}\\
& = \frac{n \left(2 \ln n-1\right)}{4 \ln ^2n} - \frac{3 \left(\ln n-1\right)}{8 \ln ^3n}
\end{align}$$
For example, let $n = 1000$, now you want to calculate the number of primes $\in [1000\cdot 1001, 1000\cdot 1002] \equiv [1001000, 1002000]$
and that's approximately
$$\frac{1000 \left(2 \ln 1000-1\right)}{4 \ln ^21000} - \frac{3 \left(\ln 1000-1\right)}{8 \ln ^31000} = 67.1365$$
while the real number is
$$78649 - 78572 = 77$$
so that's an error of about $13$% with $n = 1000$. Note that you'll be able to obtain better approximations either developing the series at $\infty$ to greater orders of magnitude, either taking $n$ very large.
|
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|
To evaluate the limits $\lim\limits_{n \to \infty} \{\frac{1}{1+n^3}+\frac{4}{8+n^3}+\ldots +\frac{n^2}{n^3+n^3}\}$ To me it seems like that we need to manipulate the given sum into the Riemann sum of some function. First writing in the standard summation form;
$$\{\frac{1}{1+n^3}+\frac{4}{8+n^3}+\ldots +\frac{n^2}{n^3+n^3}\}=\sum\limits_{k=0}^n \frac{k^2}{k^3+n^3}$$
I have tried factoring out $\frac{1}{n}$ from the above sum. But I cant seem to get into the right form. Any help!
|
Note that
$$\frac{k^2}{k^3+n^3}=\dfrac1n\cdot\dfrac{\left( \frac{k}{n} \right)^2}{\left( \frac{k}{n} \right)^3+1}$$
so you can rewrite the summation to be
$$\begin{align}\lim_{n\to \infty} \Big(\frac{1}{1+n^3}+\frac{4}{8+n^3}+\ldots +\frac{n^2}{n^3+n^3}\Big) &= \lim_{n\to \infty} \sum\limits_{k=0}^n \frac{k^2}{k^3+n^3}\\
&= \lim_{n\to \infty} \sum\limits_{k=0}^n \dfrac1n\cdot\dfrac{\left( \frac{k}{n} \right)^2}{\left( \frac{k}{n} \right)^3+1}\\
&= \lim_{h\to 0} \;h\sum_{k=1}^n \frac{{(kh)}^2}{1 + {(kh)}^3}\\
&= \int_0^1 \frac{x^2}{1+x^3} \mathrm{d}x\;\\
&= \frac{1}{3} \ln(1+x^3) \Big|_0^1\\
&= \frac{1}{3} \ln 2
\end{align}$$
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|
If $\omega$ is an imaginary fifth root of unity, then $\log_2 \begin{vmatrix} 1+\omega +\omega^2+\omega^3 -\frac{1}{\omega} \\ \end{vmatrix}$ =
If $\omega$ is an imaginary fifth root of unity, then $$\log_2 \begin{vmatrix}
1+\omega +\omega^2+\omega^3 -\frac{1}{\omega} \\
\end{vmatrix} =$$
My approach :
$$\omega^5 = 1 \\ \implies 1+\omega +\omega^2 +\omega^3 + \omega^4 =0$$
Therefore, \begin{align}\log_2 |1+\omega +\omega^2+ \omega^3 -\frac{1}{\omega}| &=\log_2 |1+\omega +\omega^2+ \omega^3 -\omega^4|\\& =\log_2|-2\omega^4|\\ &=\log_2 2 +\log_2 \omega^4 \end{align}
Now how to solve further; please suggest.
|
Notice:
$$\omega^5=1\Longleftrightarrow$$
$$\omega^5=e^{0i}\Longleftrightarrow$$
$$\omega=\left(e^{2\pi ki}\right)^{\frac{1}{5}}\Longleftrightarrow$$
$$\omega=e^{\frac{2\pi ki}{5}}$$
With $k\in\mathbb{Z}$ and $k:0-4$
So, the solutions are:
$$\omega_0=e^{\frac{2\pi\cdot0i}{5}}=e^{\frac{0}{5}}=e^0=1$$
$$\omega_1=e^{\frac{2\pi\cdot1i}{5}}=e^{\frac{2\pi i}{5}}=e^{\frac{2\pi i}{5}}$$
$$\omega_2=e^{\frac{2\pi\cdot2i}{5}}=e^{\frac{4\pi i}{5}}=e^{\frac{4\pi i}{5}}$$
$$\omega_3=e^{\frac{2\pi\cdot3i}{5}}=e^{\frac{6\pi i}{5}}=e^{-\frac{4\pi i}{5}}$$
$$\omega_4=e^{\frac{2\pi\cdot4i}{5}}=e^{\frac{8\pi i}{5}}=e^{-\frac{2\pi i}{5}}$$
So:
$$\omega=
\begin{cases}
1\\
e^{\pm\frac{2\pi i}{5}}\\
e^{\pm\frac{4\pi i}{5}}
\end{cases}$$
*
*When $\omega=1$:
$$\log_2\left|1+1+1^2+1^3-\frac{1}{1}\right|=\log_2\left|1+1+1+1-1\right|=\log_2\left|3\right|=\log_2(3)$$
*When $\omega=e^{\pm\frac{2\pi i}{5}}$:
$$\log_2\left|1+e^{\pm\frac{2\pi i}{5}}+\left(e^{\pm\frac{2\pi i}{5}}\right)^2+\left(e^{\pm\frac{2\pi i}{5}}\right)^3-\frac{1}{e^{\pm\frac{2\pi i}{5}}}\right|=1$$
*When $\omega=e^{\pm\frac{4\pi i}{5}}$:
$$\log_2\left|1+e^{\pm\frac{4\pi i}{5}}+\left(e^{\pm\frac{4\pi i}{5}}\right)^2+\left(e^{\pm\frac{4\pi i}{5}}\right)^3-\frac{1}{e^{\pm\frac{4\pi i}{5}}}\right|=1$$
|
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|
Solving $\sum_{i=0} ^{\log n} i2^i$
Solve (or simplify): $$\sum_{i=0} ^{\log n} i2^i$$ (without integrals)
Trying to change the parameter: $j=i2^i$, so since $ 0\le i \le \log n$, then the maximum value for $j$ is when $j=n\log n$. So we get $ \displaystyle \sum_{i=0} ^{\log n} i2^i = \sum_{j=0} ^{n\log n} j$
Now I can use the formula to get $ \displaystyle \sum_{j=0} ^{nlog n} j = \Theta ((n \log n)^2)$ but it doesn't seem right when I compare it with a series calculator.
Maybe did I get the upper bound wrong?
PS: Here: $\log n = \log _2 n$
|
Note that
\begin{align}
\sum_{i=1}^{n}i \cdot 2^{i} &= 2+(4+4)+(8+8+8)+\ldots+(\underbrace{ 2^{n}+2^{n}+\ldots+2^{n}}_{n\text{ times}})\\
&=(2+4+8+\ldots+2^{n-1}+2^{n})+(4+8+\ldots+2^{n-1}+2^{n})+(8+\ldots\\
&\quad+2^{n-1}+2^{n})+\ldots+(2^{n-2}+2^{n-1}+2^{n})+(2^{n-1}+2^{n})+2^{n}\\
&=2(1+2+4+\ldots+2^{n-2}+2^{n-1})+4(1+2+4+\ldots+2^{n-3}+2^{n-2})+8(1+\ldots\\
& \quad+2^{n-3})+\ldots+2^{n-2}(1+2+4)+2^{n-1}(1+2)+2^{n}\\
&=2(2^{n}-1)+4(2^{n-1}-1)+8(2^{n-2}-1)+\ldots+2^{n-2}(2^{3}-1)+2^{n-1}(2^{2}-1)+2^{n}\\
&= (n-1)2^{n+1}+2^{n}-2-4-\ldots-2^{n-1}=(n-1)2^{n+1}+2^{n}-2(2^{n-1}-1)\\
&= (n-1)2^{n+1}+2\\
&= 2(2^n(n-1) + 1)
\end{align}
thus, assuming that $\log n = \log_2 n$, you get
$$\sum_{i=0} ^{\log n} i2^i = 2(1+2^{\log n}(\log n-1)) = 2(1+n(\log n-1))$$
|
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|
Find the minimum value of $\sin^{2} \theta +\cos^{2} \theta+\sec^{2} \theta+\csc^{2} \theta+\tan^{2} \theta+\cot^{2} \theta$
Find the minimum value of
$\sin^{2} \theta +\cos^{2} \theta+\sec^{2} \theta+\csc^{2} \theta+\tan^{2} \theta+\cot^{2} \theta$
$a.)\ 1 \ \ \ \ \ \ \ \ \ \ \ \ b.)\ 3 \\
c.)\ 5 \ \ \ \ \ \ \ \ \ \ \ \ d.)\ 7 $
$\sin^{2} \theta +\cos^{2} \theta+\sec^{2} \theta+\csc^{2} \theta+\tan^{2} \theta+\cot^{2} \theta \\
=\sin^{2} \theta +\dfrac{1}{\sin^{2} \theta }+\cos^{2} \theta+\dfrac{1}{\cos^{2} \theta }+\tan^{2} \theta+\dfrac{1}{\tan^{2} \theta } \\
\color{blue}{\text{By using the AM-GM inequlity}} \\
\color{blue}{x+\dfrac{1}{x} \geq 2} \\
=2+2+2=6 $
Which is not in options.
But I am not sure if I can use that $ AM-GM$ inequality in this case.
I look for a short and simple way .
I have studied maths upnto $12$th grade .
|
Hint: $$\sin^{2} \theta +\cos^{2} \theta+\sec^{2} \theta+\csc^{2} \theta+\tan^{2} \theta+\cot^{2} \theta$$$$= \sec^2 \theta + \csc^2 \theta + \sec^2 \theta + \csc^2 \theta -1 = 2sec^2 \theta + 2\csc^2 \theta -1$$
|
{
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"url": "https://math.stackexchange.com/questions/1620239",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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|
Rolling a dice $5$ times
Rolling a dice $5$ times find the probability that we will get the continuum:
*
*A. $12345$
*B. $1234$
*C. $434$
My attempt:
*
*A: for each throwing there are $5$ possibilities so
$P(1)=P(2)=P(3)=P(4)=P(5)=\frac 1 6\Longrightarrow$ we are looking for $P(1)\cdot P(2)\cdot P(3)\cdot P(4)\cdot P(5)=(\frac{1}{6})^5=1/7776$
*B: $P(1)=P(2)=P(3)=P(4)=\frac 1 6\Longrightarrow$ we are looking for $P(1)\cdot P(2)\cdot P(3)\cdot P(4)=(\frac{1}{6})^4=1/1296$ now we will multiply the result by two because the series could be here $\underline{1}\underline{2}\underline{3}\underline{4}\underline{}$ or here $\underline{}\underline{1}\underline{2}\underline{3}\underline{4}$ so the final result will be $2/1296$
*C:$P(4)=P(3)=P(4)=\frac{1}{6}$ so $P(4)\cdot P(3)\cdot P(4)=(\frac{1}{6})^3=1/216$ now we will multiply the result by three because the series could be here $\underline{4}\underline{3}\underline{4}\underline{}\underline{}$ or here
$\underline{}\underline{4}\underline{3}\underline{4}\underline{}$ or here $\underline{}\underline{}\underline{4}\underline{3}\underline{4}$
This is correct?
|
Assuming it's a fair die and it has actually 6 faces then
*
*$P(1,2,3,4,5) = \frac{1}{6} \cdot \frac{1}{6} \cdot \frac{1}{6} \cdot \frac{1}{6} \cdot \frac{1}{6} = \left( \frac{1}{6} \right) ^ 5 = \frac{1}{6^5}$
*$P(1,2,3,4) = \frac{1}{6} \cdot \frac{1}{6} \cdot \frac{1}{6} \cdot \frac{1}{6} \cdot 2 = 2 \cdot \frac{1}{6^4}$
*$P(1,2,3,4) = \frac{1}{6} \cdot \frac{1}{6} \cdot \frac{1}{6} \cdot 3 - 1 (\text{the event } 43434)= 3 \cdot \frac{1}{6^3} - 1$
|
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|
Limit problem $\ln(x)$ and $1^\infty$ Can anyone help me with this limit problem without L'Hopital rule and Taylor series?
$$\lim_{x\rightarrow\ 1}\left(\frac{2^x+2}{3^x+1}\right)^{1/\ln(x)}$$
|
$$\frac{2^x+2}{3^x+1}=\left(1+(\frac{2^x+2}{3^x+1}-1)\right)=1+\frac{2^x-3^3+1}{3^x+1}$$ Puting $\frac{2^x-3^3+1}{3^x+1}=h$ one gets $$\left(\frac{2^x+2}{3^x+1}\right)^{\frac{1}{\ln x}}=(1+h)^{\frac{1}{\ln x}}=(1+h)^{{\frac{h}{h\ln x}}}$$
$$\lim_{x\rightarrow\ 1}\left(\frac{2^x+2}{3^x+1}\right)^{1/\ln(x)}=\lim_{h\rightarrow\ 0}\left((1+h)^{\frac 1h}\right)^{{\frac{h}{\ln x}}}$$
Now $$\frac{h}{\ln x}=\frac{2^x-3^x+1}{(3^x+1)\ln x}=\frac{1}{3^x+1}\cdot\frac{2^x-3^x+3-2}{\ln x}$$ it follows
$$\left(\frac{1}{3^x+1}\right)
\left(\frac{2(2^{x-1}-1)-3(3^{x-1}-1)}{\ln x}\right)$$ On the other hand, it is known that $$\lim_{x\rightarrow 1}\frac{a^{x-1}-1}{\ln x}=\ln a$$ Hence
$$\lim_{x\rightarrow 1}\left(\frac{1}{3^x+1}\right)
\left(\frac{2(2^{x-1}-1)-3(3^{x-1}-1)}{\ln x}\right)=\frac 14(2\ln 2-3\ln 3)$$
Hence we have as limit $$e^{-\frac 14 (\ln 27-\ln 4)}=e^{-\ln \sqrt[4]{\ln 27-\ln 4}}$$ i.e. our limit is equal to $$\color{red}{\sqrt[4]{\frac{4}{27}}}$$
|
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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|
Integration by parts - hint I'm stuck on a passage on my textbook:
$$ \int \frac{1}{(1+t^2)^3} dt = \frac{t}{4(t^2+1)^2}+\frac{3}{4} \int \frac{1}{(t^2+1)^2} dt$$
I know that it should be easy but I just can't figure out what is the product of functions considered in this integration by parts..
can you help me?
thanks a lot
|
Notice, in general, use integration by parts as follows $$\int\frac{1}{(1+x^2)^{n-1}}\ dx=\int \underbrace{\frac{1}{(1+x^2)^{n-1}}}_{I}\cdot \underbrace{1}_{II}\ dx$$
$$=\frac{1}{(1+x^2)^{n-1}}\int 1\ dx-\int\left(\frac{d}{dx}\frac{1}{(1+x^2)^{n-1}}\cdot \int 1\ dx\right)\ dx$$
$$=\frac{x}{(1+x^2)^{n-1}}-\int\left(-\frac{2(n-1)x^2}{(1+x^2)^{n}}\right)\ dx$$
$$=\frac{x}{(1+x^2)^{n-1}}+2(n-1)\int\left(\frac{1+x^2-1}{(1+x^2)^{n}}\right)\ dx$$
$$=\frac{x}{(1+x^2)^{n-1}}+2(n-1)\int \frac{1}{(1+x^2)^{n-1}}\ dx-2(n-1)\int \frac{1}{(1+x^2)^{n}}\ dx$$
$$\int \frac{1}{(1+x^2)^{n}}\ dx=\frac{x}{2(n-1)(1+x^2)^{n-1}}+\frac{2n-3}{2(n-1)}\int\frac{1}{(1+x^2)^{n-1}}\ dx$$
Above reduction formula which can be easily applied to find
$\int\frac{1}{(1+x^2)^{3}}\ dx$ as follows
$$\int\frac{1}{(1+x^2)^{3}}\ dx=\frac{x}{4(1+x^2)^{2}}+\frac 34\int \frac{1}{(1+x^2)^{2}}\ dx$$
$$=\frac{x}{4(1+x^2)^{2}}+\frac 34\left(\frac{x}{2(1+x^2)}+\frac 12\int \frac{1}{1+x^2}\ dx\right)$$
$$=\frac{x}{4(1+x^2)^{2}}+\frac{3x}{8(1+x^2)}+\frac{3}{8}\tan^{-1}(x)+C$$
|
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|
Proving Trig Identities (Complex Numbers)
Question: Prove that if $z = \cos (\theta) + i\sin(\theta)$, then
$$ z^n + {1\over z^n} = 2\cos(n\theta) $$
Hence prove that $\cos^6(\theta)$ $=$ $\frac 1{32}$$(\cos(6\theta)$ + $6\cos(4\theta)$ + $15\cos(2\theta)$ + $10$)
I learnt to prove the first part in another post linked here.
The second part is where I am confused because there is a 'hence'
so I thought of taking 2 approaches:
either $$ z^6 + \frac 1{z^6} $$
or $$ z^6 $$ and equating real parts.
I will start with my first approach
$$ z^6 + \frac 1{z^6} $$
$$ (\cos(x)+i\sin(x))^6 + \frac 1{(\cos(x)+i\sin(x))^6} $$
$$ (\cos(x)+i\sin(x))^6 + {(\cos(x)+i\sin(x))^{-6}} $$
$$ \cos(6x) + i\sin(6x) + \cos(-6x)+ i \sin(-6x) $$
$$ 2\cos(6x) $$
Which is no where near what I am suppose to prove..
So with my second approach (expanding and equation real parts)
$$ z^6 $$
$$ (\cos(x) + i \sin(x))^6 $$
Using pascals
$$ \cos^6(x) + i*6\cos^5(x)\sin(x) + i^2*15\cos^4(x)\sin^2(x) + i^3*20\cos^3(x)\sin(x) + i^4*15\cos^2(x)\sin^4(x) + i^5*6\cos(x)\sin^5(x) + i^6 * \sin^6(x) $$
Simplifying
$$ \cos^6(x) - 15\cos^4(x)\sin^2(x)+ 15\cos^2(x)\sin^4(x) - \sin^6(x) +i(6\cos^5(x)\sin(x)-20\cos^3(x)\sin(x) + 6cos(x)\sin^5(x)$$
Now considering only real
$$ \cos^6(x) - 15\cos^4(x)\sin^2(x)+ 15\cos^2(x)\sin^4(x) - \sin^6(x) $$
At this point I'm confused , am I on the right approach?
|
I don't think that the hint is really that useful. It would be better to do
$$
\cos\theta=\frac{e^{i\theta}+e^{-i\theta}}{2}
$$
that gives you
\begin{align}
64\cos^6\theta
&=(e^{i\theta}+e^{-i\theta})^6 \\[3px]
&=e^{6i\theta}+6e^{4i\theta}+15e^{2i\theta}
+20+15e^{-2i\theta}+6e^{-4i\theta}+e^{-6i\theta} \\[3px]
&=(e^{6i\theta}+e^{-6i\theta})+
6(e^{4i\theta}+e^{-4i\theta})+
15(e^{2i\theta}+e^{-2i\theta})+
20 \\[3px]
&=2\cos6\theta+12\cos4\theta+30\cos2\theta+20
\end{align}
This is no more than De Moivre’s formula in disguise, but much easier to manage.
|
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|
Finding $\lim_{(x,y)\to (0,0)} \frac{3x^2\sin^2y}{2x^4+2\sin y^4}$ I had following limit of two variables as a problem on my calculus test. How does one show whether the limit below exists or does not exist? I think it does not exist but I was not able to show that rigorously. There was a hint reminding that $\lim_{t\to 0}\sin t / t=1$.
$$\lim_{(x,y)\to (0,0)} \frac{3x^2\sin^2y}{2x^4+2\sin y^4}$$
|
$$\lim_{(x,y)\to(0,0)}\frac{3}{2}(\frac{x^2sin^2(y)}{x^4+siny^4})$$
If Dividing by $y^4$
$$\lim_{(x,y)\to(0,0)}\frac{3}{2}(\frac{\frac{x^2sin(y)sin(y)}{y^2yy}}{\frac{x^4}{y^4}+\frac {sin(y^4)}{y^4}})$$
Know:$\lim_{t\to0}\frac{sint}{t}=1$
$(y^4=t)$,$(y=t)$
$$\lim_{(x,y)\to(0,0)}\frac{3}{2}(\frac{\frac{x^2}{y^2}}{\frac{x^4}{y^4}+1})$$
$$\lim_{(x,y)\to(0,0)}\frac{3}{2}(\frac{x^2y^2}{x^4+y^4})$$
Way $y=x$
$$\lim_{(x,y)\to (0,0)}\frac{3x^4}{4x^4}=\frac{3}{4}$$
Way $Y=0$ limit hint to 0
Not exist
|
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|
A particle moves along the x-axis find t when acceleration of the particle equals 0 A particle moves along the x-axis, its position at time t is given by
$x(t)= \frac{3t}{6+8t^2}$, $t≥0$,
where t is measured in seconds and x is in meters.
Find time at which acceleration equals 0.
I got the answer 0.5 and 0 but apparently i got the answer wrong. I know I need to use derivative twice. Can someone please help me
Calculation
$velocity = \frac {-3(4t^2-3)}{2(4t^2+3)^2}$
$Acceleration = 36x(16x^4+8x^2-3)$
answer 0.5 and 0 was found through wolfram
|
The acceleration is expressed by the second derivative.
We have $$x(t)=\frac{3t}{6+8t^2}$$ $$x'(t)=\frac{3(6+8t^2)-3t\cdot 16t}{(6+8t^2)^2}=\frac{18+24t^2-48t^2}{(6+8t^2)^2}=\frac{18-24t^2}{(6+8t^2)^2}$$ $$ x''(t)=\frac{-48t(6+8t^2)^2-2(18-24t^2)(6+8t^2)(16t)}{(6+8t^2)^4}=\frac{(6+8t^2) \left [-48t(6+8t^2)-2(18-24t^2)(16t)\right ]}{(6+8t^2)^4}=\frac{ 16\left [-3t(6+8t^2)-2t(18-24t^2)\right ]}{(2(3+4t^2))^3}=\frac{ 16\left [-6t(3+4t^2)-6t(6-8t^2)\right ]}{8(3+4t^2)^3}=\frac{-12t\left [3+4t^2+6-8t^2\right ] }{(3+4t^2)^3}=\frac{-12t\left [9-4t^2\right ] }{(3+4t^2)^3}$$
Therefore, the acceleration of the particle equals $0$ at time $t=0$ seconds and $t=\frac{3}{2}$ seconds.
|
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|
Solving modulo equations with one variable Given the following equation:
$$10 = 4^x \pmod {18}$$
How can one know what are the correct values for $x$ ?
|
First, $x>0$, because $x=0$ is not a solution, and $4$ has no inverse mod $18$.
Also, $2$ is a primitive root mod $9$, because by Euler's Theorem (since $\phi(9)=6$ and $\gcd(2,9)=1$) we get $2^6\equiv 1\pmod{9}$, also $2^{6/2}\equiv 8\not\equiv 1\pmod{9}$ and $2^{6/3}\equiv 4\not\equiv 1\pmod{9}$.
$$4^x\equiv 10\pmod{18}\stackrel{:2}\iff 2^{2x-1}\equiv 5\equiv 2^5\pmod{9}$$
$$\iff 2x-1\equiv 5\pmod{6}\iff 2x\equiv 6\pmod{6}$$
$$\stackrel{:2}\iff x\equiv 3\equiv 0\pmod{3}\iff x=3k$$
for some $k\in\mathbb Z^+$. This is your solution.
|
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|
Solution to the equation $\sqrt{x^2 - 2x + 1} - 5 = 0$ I had this equation on my exam :
$$\sqrt{x^2-2x+1} - 5 = 0$$
My friends said the the solution could be :
$$|x-1| = -5$$
So the solution is nothing!
But I say the solution is:
$$x^2-2x+1 = 25 $$
so $$x = 6\ |\ x = -4$$
My Question here is which solution is right, and why.
Thanks in advance.
|
Your solution is definitely correct:
\begin{align}
&\sqrt{x^2 - 2x +1} - 5 = 0 \\
\Rightarrow \; &x^2 -2x+1 = 25 \\
\Rightarrow \; &(x-6)(x+4) = 0 \\
\Rightarrow \; &x = 6 \text{ or } x=-4
\end{align}
Your friend's approach is also correct, but he (or she) has a typo. After correcting this, you get:
\begin{align}
&\sqrt{x^2 - 2x +1} - 5 = 0 \\
\Rightarrow \; &|x-1| = 5 \; \text{ (not }-5)\\
\Rightarrow \; &x = 6 \text{ or } x=-4
\end{align}
|
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"timestamp": "2023-03-29T00:00:00",
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|
Series expansion of $\frac{1}{\sqrt{x^3-1}}$ near $x \to 1^{+}$ How can I arrive at a series expansion for $$\frac{1}{\sqrt{x^3-1}}$$ at $x \to 1^{+}$? Experimentation with WolframAlpha shows that all expansions of things like $$\frac{1}{\sqrt{x^y - 1}}$$ have $$\frac{1}{\sqrt{y}\sqrt{x-1}}$$ as the first term, which I don’t know how to obtain.
|
Set $x:=1+\epsilon$, with $\epsilon \to 0^+.$ Then, by the binomial theorem,
$$
x^3=(1+\epsilon)^3=1+3\epsilon+3\epsilon^2+\epsilon^3
$$ giving
$$
\sqrt{x^3-1}=\sqrt{3\epsilon+3\epsilon^2+\epsilon^3}=\sqrt{3}\:\sqrt{\epsilon}\:\sqrt{1+\epsilon+O(\epsilon^2)} \tag1
$$ Observe that, by the Taylor expansion, as $\epsilon \to 0^+$,
$$
\sqrt{1+\epsilon+O(\epsilon^2)}=1+O(\epsilon). \tag2
$$ From $(1)$ and $(2)$, one gets
$$
\frac1{\sqrt{x^3-1}}=\frac1{\sqrt{3}\:\sqrt{\epsilon}}\frac1{\sqrt{1+\epsilon+O(\epsilon^2)}}=\frac1{\sqrt{3}\:\sqrt{\epsilon}}\frac1{\left(1+O(\epsilon)\right)}=\frac1{\sqrt{3}\:\sqrt{\epsilon}}\left(1+O(\epsilon) \right)
$$ or, using $\epsilon=x-1$,
$$
\frac1{\sqrt{x^3-1}}=\frac1{\sqrt{3}\:\sqrt{x-1}}+O(\sqrt{x-1}).
$$
Similarly, one obtains, for $y>0$, as $x \to 1^+$,
$$
\frac1{\sqrt{x^y-1}}=\frac1{\sqrt{y}\:\sqrt{x-1}}+O(\sqrt{x-1}).
$$
|
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|
Prove that $a(x+y+z) = x(a+b+c)$ If $(a^2+b^2 +c^2)(x^2+y^2 +z^2) = (ax+by+cz)^2$
Then prove that $a(x+y+z) = x(a+b+c)$
I did expansion on both sides and got:
$a^2y^2+a^2z^2+b^2x^2+b^2z^2+c^2x^2+c^2y^2=2(abxy+bcyz+cazx) $
but can't see any way to prove $a(x+y+z) = x(a+b+c)$. How should I proceed?
|
By C-S inequality, $(ax+by+cz)^2\le (a^2+b^2+c^2)(x^2+y^2+z^2)$ with equality iff $(x,y,z)=\lambda(a,b,c)$ for some $\lambda$ or $(a,b,c)=(0,0,0)$. But, if $(a,b,c)=(0,0,0)$, the problem is trivially true. If it not the case, then $x=\lambda a$, $y=\lambda b$ and $z=\lambda c$.
Then $x+y+z=\lambda(a+b+c)$. Multiplying by $a$ both sides and remembering $x=\lambda a$ yield us the proof.
|
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|
Number of integer solutions of $\frac{1}{x}+\frac{1}{y}=\frac{1}{2016}$ How can we find number of integer solutions of
$\frac{1}{x}+\frac{1}{y}=\frac{1}{2016}$
I want to ask what approach in general should be followed in such types of question?
|
$$\frac{1}{x}+\frac{1}{y}=\frac{1}{2016}\Longleftrightarrow$$
$$\frac{y}{xy}+\frac{x}{xy}=\frac{1}{2016}\Longleftrightarrow$$
$$\frac{x+y}{xy}=\frac{1}{2016}\Longleftrightarrow$$
$$xy=2016(x+y)\Longleftrightarrow$$
$$xy-2016x=2016y\Longleftrightarrow$$
$$x\left(y-2016\right)=2016y\Longleftrightarrow$$
$$x=\frac{2016y}{y-2016}$$
With $y\ne2016$ and $y\ne0$.
|
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|
How to simplify $\sqrt[3]{7+5\sqrt{2}}+\sqrt[3]{7-5\sqrt{2}}$ The answer is 2. But I want to learn how to simplify this expression without the use of calculator.
|
By direct calculation we have the identity
$a^3+b^3+c^3-3abc=(a+b+c)(a^2+b^2+c^2-ab-bc-ac).$
Then set $a=\sqrt[3]{7+5\sqrt{2}},b=\sqrt[3]{7-5\sqrt{2}}$ and $c=-2$. Note that $ab=-1$ and $a^3+b^3=14$, so $a^3+b^3+c^3=3abc$.
Since $(a-b)^2+(b-c)^2+(b-c)^2>0\Longleftrightarrow a^2+b^2+c^2-ab-bc-ac>0$ holds for any three distinct real numbers, we have that $a+b+c=0$, which means $$\sqrt[3]{7+5\sqrt{2}}+\sqrt[3]{7-5\sqrt{2}}=2.$$
|
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|
Trigonometric Equation Simplification $$3\sin x + 4\cos x = 2$$
To solve an equation like the one above, we were taught to use the double angle identity formula to get two equations in the form of $R\cos\alpha = y$ where $R$ is a coefficient and $\alpha$ is the second angle being added to $x$ when using the double angle identity.
Why can't we use the identity $\sin(x) = \cos(x-90)$ to get $3\cos(x-90) + 4\cos x = 2$? Is this equation difficult to simplify further?
Additionally, why was the relationship between $sinx$ and $cosx$ in the pythagorean theorem (modified for the unit circle) not put to use? I did the following:
$$\sin^2x = 1 - \cos^2x$$
$$\therefore \sin x = ±\sqrt{1-\cos^2x}$$
If:
$$\sin x = y$$
Then,
$$y = ±\sqrt{1-\cos^2x}$$
Meaning that,
$$\cos x = ±\sqrt{1-y^2}$$
Inputting this into the original equation,
$$3y + 4\sqrt{1-y^2} = 2$$
We see, $$3y-2=-4\sqrt{1-y^2}$$
So, $$(3y-2)^2=16-16y^2$$
Therefore, $$9y^2-12y+4=16-16y^2$$
Rearranging gives, $$25y^2-12y-12=0$$
And so the solutions are, $$y_0,y_1=\frac{12}{50}\pm\frac{1}{50}\sqrt{144+1200}$$
And simplifying yields, $$y_0,y_1=\frac{6\pm4\sqrt{21}}{25}$$
Checking these solutions will give us the unique solution: $$y_0=\frac{6-4\sqrt{21}}{25}$$
$$q.e.d.$$
Can the above method be generalised? Has it been generalised?
|
A different approach:
You can trade the trigonometric functions for a rational expression with
$$3\sin x+4\cos x=3\frac{2t}{1+t^2}+4\frac{1-t^2}{1+t^2}=2.$$
Thus, the quadratic equation
$$6t^2-6t-2=0,$$ which you can readily solve.
Then
$$\tan x=\frac{2t}{1-t^2}$$
and
$$x=\arctan\frac{-12\pm2\sqrt{21}}5+k\pi.$$
To determine $k$, you must ensure that the angle lies in the quadrant of $(1-t^2,2t)$.
To avoid the computation of $\dfrac{2t}{1-t^2}$, you can also use
$$\tan x=\tan2\frac x2=\frac{2\tan\frac x2}{1-\tan^2\frac x2}=\frac{2t}{1-t^2}$$ or
$$\tan\frac x2=t,$$
$$x=2\arctan t+2k\pi.$$
|
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|
Calculate $\sqrt{x^2+y^2+2x-4y+5} + \sqrt{x^2+y^2-6x+8y+25}$, if $3x+2y-1=0$ As the title says, given $x,y \in \mathbb{R}$ where $3x+2y-1=0$ and $x \in [-1, 3]$, calculate $A = \sqrt{x^2+y^2+2x-4y+5} + \sqrt{x^2+y^2-6x+8y+25}$.
I tried using the given condition to reduce the complexity of the roots, but couldn't get rid of them.
|
$-1\le x\le3\iff-2\le x-1\le2$
WLOG $x-1=2\cos u\iff y=-1-3\cos u$
$x^2+y^2-6x+8y+25=(x-3)^2+(y+4)^2=(2^2+3^2)(1-\cos u)^2$
$\implies\sqrt{x^2+y^2-6x+8y+25}=\sqrt{13}(1-\cos u)$ as $1-\cos u\ge0$
Can you take it from here?
|
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|
Probability of getting $5$ heads on $10$ (fair) coin flips? Even before attempting the problem, I immediately defaulted to an answer: $\frac{1}{2}$.
I thought that this was a possible answer since the probability of flipping a head on one flip is definitely $\frac{1}{2}$.
I then worked through the problem:
Let E be the event in the problem statement.
The total number of possible outcomes of $10$ coin flips: $N = 2^{10}$
The total number of ways in which E can result:for $n=10,r=5, nCr= 252$
$P(E) =\frac{n}{N} = .246$
How do I correct my intuition?
|
Let's imagine a case with $4$ rolls where we desire $2$ heads (MathJax diagrams only go so far...)
$$\newcommand{\mychoose}[2]{\bigl({{#1}\atop#2}\bigr)}$$
$$\begin{array}{ccccccccccc} & & & & & & & H & & & & & \\
& & & & & & \swarrow & & \searrow & \\
& & & & & H & & & & T & \\
& & & & \swarrow & \downarrow & & & & \downarrow & \searrow \\
& & & H & & T & & & & H & & T \\
& & \swarrow & \downarrow & & \downarrow & \searrow & & \swarrow & \downarrow & & \downarrow & \searrow \\
& H & & T & & H & \color{red}{T} & & H & \color{red}{T} & & \color{red}{H} & & T\end{array}$$
$$ $$
$$\begin{array}{ccccccccccc} & & & & & & & T & & & & & \\
& & & & & & \swarrow & & \searrow & \\
& & & & & H & & & & T & \\
& & & & \swarrow & \downarrow & & & & \downarrow & \searrow \\
& & & H & & T & & & & H & & T \\
& & \swarrow & \downarrow & & \downarrow & \searrow & & \swarrow & \downarrow & & \downarrow & \searrow \\
& H & & \color{red}{T} & & \color{red}{H} & T & & \color{red}{H} & T & & H & & T\end{array}$$
Thus, we have a $\frac{6}{16} = \frac{3}{8}$ chance of getting two heads from four rolls by analyzing our diagram; this can also be found by taking $\frac{\mychoose{2}{4}}{2^4} = \frac{6}{16} = \frac{3}{8}$. As an addendum, note that your intuition calculation gives the chance that you get either $2$ or $1$ head from $4$ rolls ( $= \frac{8}{16} = \frac{1}{2}$ which can be easily identified from the diagram). Now just apply the logic from this example to your $10$ roll case. Also, pushing MathJax diagrams this far is a pain. You have been warned.
|
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|
Evaluate this limit of inverse trigonometric and radical functions without l'Hospital How can I solve this using only 'simple' algebraic tricks and asymptotic equivalences? No l'Hospital.
$$\lim_{x \rightarrow0}
\frac
{\sqrt[3]{1+\arctan{3x}} - \sqrt[3]{1-\arcsin{3x}}}
{\sqrt{1-\arctan{2x}} - \sqrt{1+\arcsin{2x}}}
$$
Rationalizing the numerator and denominator gives
$$
\lim_{x \rightarrow0}
\frac
{A(\arctan{3x}+\arcsin{3x})}
{B(\arctan{2x} + \arcsin{2x})}
$$
where $\lim_{x \rightarrow 0} \frac{A}{B} = -\frac{2}{3} $
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We will use the following standard limits $$\lim_{x \to 0}\frac{\arctan x}{x} = 1 = \lim_{x \to 0}\frac{\arcsin x}{x},\,\lim_{x \to a}\frac{x^{n} - a^{n}}{x - a} = na^{n - 1}\tag{1}$$ We have
\begin{align}
L &= \lim_{x \to 0}\frac{\sqrt[3]{1 + \arctan 3x} - \sqrt[3]{1 - \arcsin 3x}}
{\sqrt{1 - \arctan 2x} - \sqrt{1 + \arcsin 2x}}\notag\\
&= \lim_{x \to 0}\dfrac{\dfrac{\sqrt[3]{1 + \arctan 3x}}{x} - \dfrac{\sqrt[3]{1 - \arcsin 3x}}{x}}
{\dfrac{\sqrt{1 - \arctan 2x}}{x} - \dfrac{\sqrt{1 + \arcsin 2x}}{x}}\notag\\
&= \lim_{x \to 0}\dfrac{\left(\dfrac{\sqrt[3]{1 + \arctan 3x} - 1}{x}\right) - \left(\dfrac{\sqrt[3]{1 - \arcsin 3x} - 1}{x}\right)}
{\left(\dfrac{\sqrt{1 - \arctan 2x} - 1}{x}\right) - \left(\dfrac{\sqrt{1 + \arcsin 2x} - 1}{x}\right)}\notag\\
&= \frac{A - B}{C - D}\notag
\end{align}
The limit of each of the 4 bracketed expressions can be evaluated easily. These limits I have denoted by $A, B, C, D$ respectively. I will illustrate the complete evaluation for the first expression. We have
\begin{align}
A &= \lim_{x \to 0}\frac{\sqrt[3]{1 + \arctan 3x} - 1}{x}\notag\\
&= \lim_{x \to 0}\frac{\sqrt[3]{1 + \arctan 3x} - 1}{\arctan 3x}\cdot\frac{\arctan 3x}{3x}\cdot 3\notag\\
&= 3 \lim_{x \to 0}\frac{\sqrt[3]{1 + \arctan 3x} - 1}{\arctan 3x}\notag\\
&= 3 \lim_{t \to 1}\frac{t^{1/3} - 1}{t - 1}\text{ (putting }t = 1 + \arctan 3x)\notag\\
&= 3\cdot\frac{1}{3}\notag\\
&= 1\notag
\end{align}
Similarly $B = -1, C = -1, D = 1$ and hence the desired limit $L = (A - B)/(C - D) = -1$.
Rationalization of expressions involving radicals makes sense mostly if we are dealing with square roots. In cases where we have cube roots (and higher roots) it is difficult to write/type the long expressions (obtained during rationalization process) and it is best to make use of the standard limit formula $\lim\limits_{x \to a}\dfrac{x^{n} - a^{n}}{x - a} = na^{n - 1}$.
|
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|
Calculate $\frac{1}{\sin(x)} +\frac{1}{\cos(x)}$ if $\sin(x)+\cos(x)=\frac{7}{5}$ If
\begin{equation}
\sin(x) + \cos(x) = \frac{7}{5},
\end{equation}
then what's the value of
\begin{equation}
\frac{1}{\sin(x)} + \frac{1}{\cos(x)}\text{?}
\end{equation}
Meaning the value of $\sin(x)$, $\cos(x)$ (the denominator) without using the identities of trigonometry.
The function $\sin x+\cos x$ could be transformed using some trigonometric identities to a single function. In fact, WolframAlpha says it is equal to $\sqrt2\sin\left(x+\frac\pi4\right)$ and there also are some posts on this site about this equality. So probably in this way we could calculate $x$ from the first equation - and once we know $\sin x$ and $\cos x$, we can calculate $\dfrac{1}{\sin x}+\dfrac{1}{\cos x}$. Is there a simpler solution (perhaps avoiding explicitly finding $x$)?
|
$$ x+y= p\tag1$$
Square, since $( x^2+y^2=1 )$
$$ 1+ 2 x\;y = p^2, \; x y= \dfrac{p^2-1}{2} \tag2$$
From (1) and (2)
$$ \dfrac{1}{x}+ \dfrac{1}{y} = \dfrac{x+y}{x y}= \dfrac{2p}{p^2-1} $$
$$ = \dfrac{35}{12},\;$$
if
$$\;p= \dfrac{7}{5} $$
|
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|
Minimizing distance between line and point So I am given the problem where a line of the form $ax+by+c=0$ and point $(x_0,y_0)$ are given and I have to find the minimum distance between these two. I was able to do it with the projection of a vector normal to the line and found the answer to be $$\frac{|ax_0+by_0+c|}{\sqrt{a^2+b^2}}$$ but I am supposed to do it using minimization techniques (setting partials equal to 0, solving for critical points and using the Hessian to see whether it is a min) but I have no idea how to even start with this method? Any help at all would be greatly appreciated especially if they are hints rather than the actual answer. Thanks!!
|
To find the minimum distance between a point $P(x_0,y_0)$ and a line $ax + by + c = 0$, we will use the distance formula $$f(x) = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$$
to create a function that represents the distance between $P(x_0,y_0)$ and the line $ax + by + c = 0$ as a function of $x$.
First, let's rewrite $ax + by + c = 0$ as $$y = \frac{-ax-c}{b} \tag{1}$$
So our distance function will calculate the distance between $(x_0,y_0)$ and $(x,y)$.
Let $d$ be the distance function between the point and the line:
$$d(x) = \sqrt{(x - x_0)^2 + (y - y_0)^2}\tag{2}$$
Substituting $(1)$ into $(2)$ we get $$d(x) = \sqrt{(x - x_0)^2 + \left(\frac{-ax-c}{b} - y_0\right)^2}\tag{3}$$
Now we need to go about calculating the absolute minimum of $(3)$ by calculating the derivative of $(3)$, equating it to zero, and then solving it for $x$ which will be the value at which $(3)$ has its minimum(shortest) value(distance).
To simplify the process somewhat, we will take the derivative of the square of the distance in order to get rid of the square root when we're calculating the derivative. Note that in doing so will not alter the solution.
So we let D be the function that represents the square of the distance, so that
$$D(x) = (x - x_0)^2 + \left(\frac{-ax-c}{b} - y_0\right)^2\tag{4}$$
So now we take the derivative of $(4)$,
$$D'(x) = \frac{2b^2x - 2b^2x_0 + 2a^2x+2ac + 2aby_0}{b^2}\tag{5}$$
Now we set $(5)$ equal to zero and solve for $x$, which is where the minimum occurs. We do this by setting the numerator of $(5)$ to zero
$$
2b^2x - 2b^2x_0 + 2a^2x+2ac + 2aby_0 = 0
$$
so,
$$
x = \frac{b^2x_0 - ac - aby_0}{a^2+b^2}\tag{6}
$$
Plugging $(6)$ into $(3)$ we get
$$
d(x) = \sqrt{\left(\frac{b^2x_0 - ac - aby_0}{a^2+b^2} - x_0\right)^2 + \left(\frac{-a\frac{b^2x_0 - ac - aby_0}{a^2+b^2}-c}{b} - y_0\right)^2}
$$
$$
= \sqrt{\left(\frac{-a^2x_0-aby_0-ac}{a^2+b^2}\right)^2 + \left(\frac{-ab^2x_0-y_0b^3-b^2c}{b(a^2+b^2)}\right)^2}
$$
$$
= \sqrt{\frac{(a^2)(ax_0 + by_0 + c)^2}{(a^2+b^2)^2} + \frac{(b^2)(ax_0 + by_0 + c)^2}{(a^2+b^2)^2}}
$$
$$
=\bbox[5px,border:2px solid red]
{
\frac{\lvert ax_0 + by_0 + c \rvert}{\sqrt{a^2 + b^2}}
}
$$
|
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|
Solving for an inverse diophantine equation $\frac{1}{x_1} +\frac{1}{x_2} + ... + \frac{1}{x_n} +\frac{1}{x_1 x_2 ... x_n} = 1$ How can I prove that the Diophantine equation $$\frac{1}{x_1} +\frac{1}{x_2} + ... + \frac{1}{x_n} +\frac{1}{x_1 x_2 ... x_n} = 1$$ has at most one solution? All $x_i$ and $n$ are natural numbers.
My attempt was:
For example consider equation for $n=3$:
$$\frac{1}{x_1} + \frac{1}{x_2} + \frac{1}{x_3} + \frac1{x_1x_2x_3} =1$$
then
$$x_1x_2 + x_1x_3 + x_2x_3 +1 = x_1x_2x_3 \tag{1} $$
and
\begin{cases}
x_2x_3+1\equiv 0\pmod {x_1} \\
x_1x_3+1\equiv 0\pmod {x_2} \\
x_1x_2+1\equiv 0\pmod {x_3} \\
\end{cases}
so
\begin{cases}
x_2x_3=k_1x_1-1 \\
x_1x_3=k_2x_2-1 \\
x_1x_2=k_3x_3-1 \\
\end{cases}
substitute in (1)
gives this
$$k_1x_1 + k_2x_2 + k_3x_3 = x_1x_2x_3 + (3-1) $$
general form will be
$$k_1x_1 + k_2x_2 + k_3x_3 + ... + k_nx_n = x_1x_2x_3...x_n + (n-1) $$
I know to solve this but $x_1x_2x_3...x_n$ term is the problem.
|
Assuming that you want the number of solutions (you have referred to as root so it is a bit confusing).
This is NOT true, for example if $n=5$, then there are $3$ solutions.
\begin{align*}
1 & = \frac{1}{2}+\frac{1}{3}+\frac{1}{7}+\frac{1}{43}+\frac{1}{1807}+\frac{1}{2 \cdot 3 \cdot 7 \cdot 43 \cdot 1807}\\
& = \frac{1}{2}+\frac{1}{3}+\frac{1}{7}+\frac{1}{47}+\frac{1}{395}+\frac{1}{2 \cdot 3 \cdot 7 \cdot 47 \cdot 395}\\
& = \frac{1}{2}+\frac{1}{3}+\frac{1}{11}+\frac{1}{23}+\frac{1}{31}+\frac{1}{2 \cdot 3 \cdot 11 \cdot 23 \cdot 31}\\
\end{align*}
|
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|
How can I generalize this? I'm not sure what to tag this question as or whether its a bit nonsensical, but I'm a bit curious. I asked a question on a pretty (turned out to be) easy question about the Collatz Sequence here:
Collatz $4n+1$ rule?
My question is observe the following sequences (next term is $4n+1$):
$$3\ 13\ 53, \cdot\cdot\cdot\\5\ 21\ 85, \cdot\cdot\cdot\\7\ 29\ 117, \cdot\cdot\cdot\\9\ 37\ 149, \cdot\cdot\cdot\\11\ 45\ 181, \cdot\cdot\cdot\\15\ 61\ 245, \cdot\cdot\cdot\\17\ 69\ 277, \cdot\cdot\cdot\\19\ 77\ 309, \cdot\cdot\cdot$$
Basically take some odd number and multiply it by $4$ and add $1$ and you get the next number in the series. I am trying to find all the starting numbers of all the sequences that don't collide with each other. The sequences above do not repeat each other (I think?) and you can see the starting numbers are (if you expand it)
$$3, 5, 7, 9, 11, 15, 17, 19, 23, 25, 27, 31, 33, 35, 39, 41, 43, 47, 49, 51, 55, 57, 59, ...$$
With the numbers $13, 37, 53$ colliding. How can I generalize which numbers will not collide or have been covered by others as these are the true "unknown" values of the odd numbers of the Collatz Sequence?
|
You could notice that the formula describing each of those sequences is of the form $\frac{1}{3}(-1+4^n+3\cdot 4^n\cdot(2j-1))$, where $2j-1$ is the initial odd value.
Then the numbers $2k-1$ which have already been covered in a sequence must be of the form $2k-1=\frac{1}{3}(-1+4^n+3\cdot 4^n\cdot(2j-1))$ for some $n,j\in\mathbb{N}$. Rearranging this yields: $$k=\frac{\frac{1}{3}(-1+4^n+3\cdot 4^n\cdot(2j-1))+1}{2}$$ By varying $j$ and $n$, we then have a formula for all $k$ so that $2k-1$ already appears in a sequence.
EDIT: As requested, to find the formula describing each of the sequences, we note that given an odd number $c=2j-1$, any of the sequences may be described by the recursive relation $a_0=c$; $a_n=4a_{n-1}+1$, for $n\geq 1$. Then, we form the generating function $f(x)=\sum\limits_{n=0}^{\infty}a_nx^n$. From this, we obtain: $$f(x)=\sum\limits_{n=0}^{\infty}a_nx^n=c+\sum\limits_{n=1}^{\infty}a_nx^n=c+\sum\limits_{n=1}^{\infty}(4a_{n-1}+1)x^n$$ Expanding the last term and pulling out an $x$ yields: $$f(x)=c+x(4\sum\limits_{n=1}^{\infty}a_{n-1}x^{n-1}+\sum\limits_{n=1}^{\infty}x^{n-1})$$ Reindexing yields: $$f(x)=c+x(4\sum\limits_{n=0}^{\infty}a_{n}x^{n}+\sum\limits_{n=0}^{\infty}x^{n})$$ Substituting $f(x)=\sum\limits_{n=0}^{\infty}a_nx^n$ and using $\sum\limits_{n=0}^{\infty}x^{n}=\frac{1}{1-x}$ gives us: $$f(x)=c+x(4\cdot f(x)+\frac{1}{1-x})$$ Rearranging and solving for $f(x)$: $$f(x)=\frac{c}{1-4x}+\frac{x}{(1-x)(1-4x)}=c\cdot\frac{1}{1-4x}-\frac{1}{3}\cdot\frac{1}{1-x}+\frac{1}{3}\cdot\frac{1}{1-4x}$$ Then using the power series expansion $\frac{1}{1-rx}=\sum\limits_{n=0}^{\infty}r^nx^n$, we finally obtain: $$f(x)=c\cdot\sum\limits_{n=0}^{\infty}4^nx^n-\frac{1}{3}\cdot\sum\limits_{n=0}^{\infty}x^n+\frac{1}{3}\cdot\sum\limits_{n=0}^{\infty}4^nx^n$$ Combining these: $$f(x)=\sum\limits_{n=0}^{\infty}\frac{1}{3}(3\cdot 4^n\cdot c-1+4^n)x^n$$ Then comparing coefficients on both sides gives us the desired result of: $$a_n=\frac{1}{3}(3\cdot 4^n\cdot c-1+4^n)$$
|
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|
Proving for all integer $n \ge 2$, $\sqrt n < \frac{1}{\sqrt 1} + \frac{1}{\sqrt 2}+\frac{1}{\sqrt 3}+\cdots+\frac{1}{\sqrt n}$ Prove the following statement by mathematical induction:
For all integer $n \ge 2$, $$\sqrt n < \frac{1}{\sqrt 1} + \frac{1}{\sqrt 2}+\frac{1}{\sqrt 3}+\cdots+\frac{1}{\sqrt n}$$
My attempt: Let the given statement be $p(n)$ .
1.\begin{align*} \frac{1}{\sqrt 1} + \frac{1}{\sqrt 2} & =\frac{\sqrt 2 +1}{\sqrt 2} \\
2 &< \sqrt 2 +1 \\
\sqrt 2 &< \frac{\sqrt 2 +1}{\sqrt 2}=\frac{1}{\sqrt 1} + \frac{1}{\sqrt 2} \end{align*}
Hence, $p(2)$ is true.
2.For an arbitrary integer $k \ge 2$, suppose $p(k)$ is true.
That is, $$\sqrt k < \frac{1}{\sqrt 1} + \frac{1}{\sqrt 2}+\frac{1}{\sqrt 3}+\cdots+\frac{1}{\sqrt k}$$
Then we must show that $p(k+1)$ is true.
We're going to show that $$\sqrt {k+1} < \frac{1}{\sqrt 1} + \frac{1}{\sqrt 2}+\frac{1}{\sqrt 3}+\cdots+\frac{1}{\sqrt k}+\frac{1}{\sqrt {k+1}}$$
I'm stuck on this step. I can't develop it further. How can I complete this proof?
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$$\frac{1}{\sqrt 1} + \frac{1}{\sqrt 2}+\frac{1}{\sqrt 3}+\cdots+\frac{1}{\sqrt k}+\frac{1}{\sqrt k+1} > \sqrt{k} + \frac{1}{\sqrt {k+1}} =
\frac{\sqrt{k(k+1)} + 1}{\sqrt{k+1}} > \frac{\sqrt{k^2} + 1}{\sqrt{k+1}} = \sqrt{k+1}$$
|
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|
If $x,y,z$ are positive real number number, Then minimum value of $\frac{x^4+y^4+z^2}{xyz}$
If $x,y,z$ are positive real number number, Then minimum value of $\displaystyle \frac{x^4+y^4+z^2}{xyz}$
$\bf{My\; Try::}$ Given $x,y,z>0.$ So Using $\bf{A.M\geq G.M\;,}$ We get
$$\displaystyle x^4+y^4\geq 2x^2y^2$$ and then Using $$2x^2y^2+z^2\geq 2\sqrt{2}|xyz|\geq 2\sqrt{2}xyz$$
So we get $$x^4+y^4+z^4\geq 2x^2y^2+z^2 \geq 2\sqrt{2}xyz$$
and equality hold when $x^4=y^4$ and $2x^2y^2=z^2$
So we get $x=y$ and $z=\sqrt{2}x^2$, Now Minimum occur at $\left(x,x,\sqrt{2}z^2\right)$
So at that point we get $\displaystyle \left[\frac{x^4+y^4+z^4}{xyz}\right]_{\bf{\min}} = \sqrt{2}+2x^2$
and Answer given is $2\sqrt{2}$, I did not understand how this can happen
olz explain me, Thanks
My Question is
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$$x^4+y^4+z^2=x^4+x^4+(\sqrt{2}x^2)^2=4x^4\\
xyz=xx\sqrt{2}x^2=\sqrt{2}x^4$$
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"timestamp": "2023-03-29T00:00:00",
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|
$\lim_{x\to 0} (2^{\tan x} - 2^{\sin x})/(x^2 \sin x)$ without l'Hopital's rule; how is my procedure wrong?
please explain why my procedure is wrong i am not able to find out??
I know the property limit of product is product of limits (provided limit exists and i think in this case limit exists for both the functions).
The actual answer for the given question is $\frac{1}{2}\log(2)$.
My course book has shown that don't use this step but has not given the reason.
AND Please TELL why i am WRONG
|
The basic limits needed are
$\frac{e^x-1}{x}
\to 1
$,
$\frac{\sin x}{x}
\to 1
$,
and
$\cos x
\to 1
$ as
$x \to 0$.
First,
as $x \to 0$,
$\begin{array}\\
\tan x -\sin x
&=\frac{\sin x}{\cos x}-\sin x\\
&=\sin x(\frac{1}{\cos x}-1)\\
&=\sin x(\frac{1-\cos x}{\cos x})\\
&=\sin x(\frac{2\sin^2(x/2)}{\cos x})\\
&\to x(\frac{2(x/2)^2}{1})\\
&=x^3/2\\
\end{array}
$
Then
$\begin{array}\\
\frac{2^{\tan x}-2^{\sin x}}{x^2\sin x}
&=2^{\sin x}\frac{2^{\tan x-\sin x}-1}{x^2\sin x}\\
&\to \frac{2^{\tan x-\sin x}-1}{x^2\sin x}
\qquad\text{since } 2^{\sin x} \to 1\\
&\to \frac{2^{x^3/2}-1}{x^3}\\
&= \frac{e^{\ln 2 x^3/2}-1}{x^3}\\
&\to \frac{\ln 2 x^3/2}{x^3}\\
&= \ln 2/2\\
\end{array}
$
|
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|
Probability of combinations so that only $2$ of the $3$ types remains. $15$ telephones have just been received at an authorized service center. $5$ of these telephones are cellular, $5$ are cordless, and the other $5$ are corded phones. Suppose these components are randomly allocated the number $1,2,3, \dots$.
What is probability that after servicing $10$ of these phones, phones of only $2$ of the $3$ types remain to be serviced?
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To start, lets fix a telephone type, solve the problem for that type and then multiply with $3$ the obtained probability to obtain the correct answer (due to symmetry). We need to service all $5$ phones of this type and at least $1$ phone of each of the other $2$ types (in order to exhaust exactly this type and not $2$ types). The probability for this is (hypergeometric) $$\dfrac{\dbinom{5}{5}\sum_{k=1}^{4}\dbinom{5}{k}\dbinom{5}{5-k}}{\dbinom{15}{10}}=\dfrac{\sum_{k=1}^{4}\dbinom{5}{k}\dbinom{5}{5-k}}{\dbinom{15}{10}}$$ Since there are $3$ types it remains to multiply this probability with $3$ to obtain the answer $$3\cdot\dfrac{\sum_{k=1}^{4}\dbinom{5}{k}\dbinom{5}{5-k}}{\dbinom{15}{10}}=3\cdot\frac{2\dbinom{5}{1}^2+2\dbinom{5}{2}^2}{\dbinom{15}{10}}=\frac{750}{3003}\approx 0,25$$
|
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|
Infinity Series to Approximate a fraction I have observed that the following series is a good approximation for
$\frac{1}{10}$.
$\frac{1}{8}- \frac{1}{16} + \frac{1}{32} + \frac{1}{64} - \frac{1}{128} - \frac{1}{256} + \frac{1}{512} + \frac{1}{1024} - \frac{1}{2048} -
\frac{1}{(2\cdot2048)} + \frac{1}{(4\cdot2048)} + \frac{1}{(8\cdot2048)} - \frac{1}{(16\cdot2048)} - \frac{1}{(32\cdot2048)} + ...$
I believe there is a pattern to the series. However, I cannot prove that pattern holds. Does it? Can it be proved? I am thinking some Taylor series might be the way to go but that is just a hunch.
Bob
|
$$\frac{1}{8}-\frac{1}{16}+\sum \limits_{n=2}^{\infty} \frac{(-1)^n}{2^{2n+1}}+\sum \limits_{n=2}^{\infty} \frac{(-1)^n}{2^{2n+2}}=\dots$$
|
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|
State the range of this How do I state the range of the following equation:
$$8x-4x^2$$ using $$c-\frac{b^2}{4a}$$
I'm again unsure of what $a$, $b$, and $c$ are.
|
Generally, $f(x)=ax^2+bx+c$ has its extreme value at $f'(x)=0$, i.e.
$$2ax+b=0\implies x=-\frac{b}{2a}.$$
Substituting $x=-\frac{b}{2a}$ into $f(x)$ gives:
$$f\left(-\frac{b}{2a}\right)=a\left(-\frac{b}{2a}\right)^2+b\left(-\frac{b}{2a}\right)+c=\frac{ab^2}{4a^2}-\frac{b^2}{2a}+\frac{2ac}{2a}$$
$$=\frac{ab^2}{4a^2}-\frac{2ab^2}{4a^2}+\frac{4a^2c}{4a^2}=\frac{4a^2c-ab^2}{4a^2}=c-\frac{b^2}{4a}.$$
So $f$ has its extremum at the point $\left(-\frac{b}{2a},c-\frac{b^2}{4a}\right).$
Given the extreme value, do you know how to continue in order to find the whole range of $f$? To begin with, do you know what $f$ looks like? What kind of function is $f$?
|
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|
Prove that $f(x)=8$ for all natural numbers $x\ge{8}$ A function $f$ is such that $$f(a+b)=f(ab)$$ for all natural numbers $a,b\ge{4}$ and $f(8)=8$. Prove that $f(x)=8$ for all natural numbers $x\ge{8}$
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$\>$$\>$$\>$$\>$$\>$SS_C4's answer has two problems; the untrue claim that $2y>y+6$, for $y\geq 6$, and not demonstrating that $f(x)=f(8)$, for $9\leq x<17$, but SS_C4's proof can be saved as follows. The basic ideas belong to SS_C4. The value of $f(8)$ is immaterial. Lower-case Latin letters, except $f$, denote positive integers.
(A) $f(9)=f(4+5)=f(20)=f(4+16)=f(64)=f((8)(8))=f(16)=f((4)(4))=f(8)$.
(B) $f(10) = f(5+5)=f(25)=f(5+20)=f(100)=f((10)(10))=f(20)=f(8)$, by (A).
(C) $f(11)=f(4+7)=f(28)=f(4+24)=f(96)=f((8)(12))=f(20)=f(8)$, by (A).
(D) $f(12)=f(5+7)=f(35)=f(5+30)=f(150)=f((10)(15))=f(25)=f(8)$, by (B).
(E) $f(13)=f(5+8)=f(40)=f((4)(10))=f(14)=f(6+8)= f(48)=f((4)(12))=$
$\>$$\>$$\>$$\>$$\>$$\>$$\>$$\>$$\>$$\>$$f(16)=f(8)$, by (A).
(F) $f(15)=f(4+11)=f(44)=f(4+40)=f(160)=f((8)(20))=f(28)=f(8)$, by (C).
$\>$$\>$$\>$$\>$$\>$$\>$$\>$$\>$$\>$$\>$$\>$$\>$$\>$$\>$$\>$$\>$$\>$$\>$$\>$$\>$$\>$$\>$$\>$$\>$$\>$$\>$$\>$$\>$$\>$$\>$$\>$$\>$$\>$$\>$$\>$$\>$$\>$$\>$$\>$$\>$$\>$$\>$$\>$$\>$$\>$$\>$$\>$$\>$$\>$$\>$$\>$$\>$$\>$$\>$$\>$$\>$$\>$$\>$$\>$$\>$$\>$$\>$$\>$$\>$$\>$$\>$$\>$$\>$$\>$$\>$$\>$$\>$$\>$$\>$$\>$$\>$$\>$$\>$$\>$$\>$$\>$$\>$$\>$$\>$$\>$$\>$$\>$$\>$$\>$$\>$$\>$$\>$$\>$$\>$$\>$$\>$$\>$$\>$$\>$$\>$$\>$$\>$$\>$$\>$$\>$$\>$$\>$$\>$$\>$$\>$$\>$$\>$$\>$$\>$$\>$$\>$$\>$$\>$$\>$$\>$$\>$$\>$$\>$$\>$$\>$$\>$$\>$$\>$$\>$$\>$$\>$$\>$$\>$$\>$$\>$$\>$$\>$$\>$$\>$$\>$$\>$$\>$$\>$$\>$$\>$$\>$$\>$$\>$$\>$$\>$$\>$$\>$$\>$$\>$$\>$$\>$$\>$$\>$$\>$$\>$$\>$$\>$$\>$$\>$$\>$$\>$$\>$$\>$$\>$$\>$$\>$$\>$$\>$$\>$$\>$$\>$$\>$$\>$$\>$$\>$$\>$$\>$$\>$$\>$$\>$$\>$Suppose $f(x) \neq f(8)$, for some $x\geq9$. Let $m=$Min$\{x|x\geq9$ and $f(x)\neq f(8)\}$.
$m \geq 17$, by (A) - (F).
$\>$$\>$$\>$$\>$$\>$$\>$Suppose $m$ is even. Then, $m=2y$. Thus, $y\geq 9, y-2\geq7, 2y-4\geq14$ and
$15\leq y+6<2y=m$. Therefore, by the definition of $m$,
$\>$$\>$$\>$$\>$$\>$$\>$$\>$$f(8)=f(y+6)=f(8+(y-2))=f(8(y-2))=f(4(2y-4))=f(2y)=f(m)$,
contradicting the definition of $m$. Thus, $m$ is odd. So, $m=2y+1$. Therefore,
$y\geq 8,y-2\geq6, 2y-4\geq12$ and $16\leq y+8<2y+1=m$. Hence, by the definition of $m$,
$\>$$\>$$\>$$\>$$\>$$\>$$f(8)=f(y+8)=f(10+(y-2))=f(10(y-2))=f(5(2y-4))=$
$\>$$\>$$\>$$\>$$\>$$\>$$f(5+(2y-4))=f(2y+1)=f(m),$
again contradicting the definition of $m$. Thus, $f(x)=f(8)$, for $x\geq9$.
|
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|
How can I solve $7^{77}\mod 221$ How is it possible to solve this without calculator etc.: $$7^{77} \mod 221$$
I started with:
\begin{align}
7&\equiv 7 \mod 221 \\
7^2 &\equiv 49 \mod 221 \\
7^4 &\equiv \ ? \mod 221
\end{align}
Sure i can calculate this by my own, but is there a trick to calclulate this with any tools?
|
Without the Chinese Remainder Theorem :
$$
\eqalign{
7^5 &\equiv 11 \pmod {221} \cr
7^{75} &\equiv 11^{15} \pmod {221} \cr
7^{77} &\equiv 7^2 \cdot 11^{15} \pmod {221} \cr
}
$$
Also
$$
\eqalign{
11^3 &\equiv 5 \pmod {221} \cr
11^{15} &\equiv 5^{5} \pmod {221} \cr
}
$$
To conclude
$$
7^{77} \equiv 7^2 \cdot 11^{15} \equiv 7^2 \cdot 5^5 \equiv 49 \cdot 31 \equiv 193
$$
|
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|
Solving recurrence relation? Consider the recurrence relation $a_1=8,a_n=6n^2+2n+a_{n−1}$. Let $a_{99}=K\times10^4$ .The value of $K$ is______ .
My attempt :
$a_n=6n^2+2n+a_{n−1}$
$=6n^2+2n+6(n−1)^2+2(n−1)+a_{n−2}$
$=6n^2+2n+6(n−1)^2+2(n−1)+6(n−2)^2+2(n−2)+......+a_1$
$=6n^2+2n+6(n−1)^2+2(n−1)+6(n−2)^2+2(n−2)+......+6.1^2+2.1$
$=6(n^2+(n−1)^2+...+2^2+1^2)+2(n+(n−1)+...+2+1)$
$=6×n(n+1)(2n+1)/6+2×n(n+1)/2$
$=n(n+1)(2n+1+1)$
$=2n^3+2n^2+2n^2+2n$
$=2n(n^2+n+n+1)$
$=2n(n^2+2n+1)$
$a_n=2n(n+1)^2$
for $n=99, a_{99}=2×99×(99+1)^2=198×10^4$
I'm looking for short trick or alternative way, can you explain please?
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$$a_n-a_{n-1}=6n^2+2n$$
Now sum both sides from $n=2$ to $x$. You end up with a telescoping sum on the left hand side which will drastically make things easier and help you find the closed form, assuming you know some summation formulas.
$$\sum_{n=2}^x (a_n-a_{n-1})= \sum_{n=2}^x (6n^2+2n)$$
$$a_x-a_1=\sum_{n=2}^x (6n^2+2n)$$
$$a_x=a_1+\sum_{n=2}^x (6n^2+2n)$$
$$a_x=a_1-8+\sum_{n=1}^x (6n^2+2n)$$
$$a_x=\sum_{n=1}^x (6n^2+2n)$$
Therefore $a_{99}$ would be:
$$a_{99}=\sum_{n=1}^{99} (6n^2+2n)$$
Where now you can apply summation formulas.
|
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|
Integral $\int \frac{\sqrt{x}}{(x+1)^2}dx$ First, i used substitution $x=t^2$ then $dx=2tdt$ so this integral becomes $I=\int \frac{2t^2}{(1+t^2)^2}dt$ then i used partial fraction decomposition the following way:
$$\frac{2t^2}{(1+t^2)^2}= \frac{At+B}{(1+t^2)} + \frac{Ct + D}{(1+t^2)^2} \Rightarrow 2t^2=(At+B)(1+t^2)+Ct+D=At+At^3+B+Bt^2 +Ct+D$$
for this I have that $A=0, B=2, C=0, D=-2$
so now I have
$I=\int \frac{2t^2}{(1+t^2)^2}dt= \int\frac{2}{1+t^2}dt - \int\frac{2}{(1+t^2)^2}dt$
Now,
$$ \int\frac{2}{1+t^2}dt = 2\arctan t$$
and
$$\int\frac{2}{(1+t^2)^2}dt$$
using partial integration we have:
$$u=\frac{1}{(1+t^2)^2} \Rightarrow du= \frac{-4t}{1+t^2}$$
and $$dt=dv \Rightarrow t=v$$
so now we have:
$$\int\frac{2}{(1+t^2)^2}dt =\frac{t}{(1+t^2)^2} + 4\int\frac{t^2}{1+t^2}dt = \frac{t}{(1+t^2)^2} + 4\int\frac{t^2 + 1 -1}{1+t^2}dt = \frac{t}{(1+t^2)^2} + 4t -4\arctan t$$
so, the final solution should be:
$$I=2\arctan t - \frac{t}{(1+t^2)^2} - 4t +4\arctan t$$
since the original variable was $x$ we have
$$I= 6\arctan \sqrt{x} - \frac{\sqrt{x}}{(1+x)^2} - 4\sqrt{x} $$
But, the problem is that the solution to this in my workbook is different, it says that solution to this integral is $$I=\arctan \sqrt{x} - \frac{\sqrt{x}}{x+1}$$
I checked my work and I couldn't find any mistakes, so i am wondering which solution is correct?
|
I'd try the following: substitute $\;t^2=x\implies 2t\,dt=dx\;$, and your integral becomes
$$I=\int\frac{2t^2dt}{(t^2+1)^2}$$
and already here integrate by parts:
$$\begin{cases}u=t&u'=1\\{}\\v'=\frac{2t}{(t^2+1)^2}&v=-\frac1{t^2+1}=\end{cases}\;\;\;\implies$$
$$I=-\frac t{t^2+1}+\int\frac1{1+t^2}dt=-\frac t{1+t^2}+\arctan t+C$$
and going back to the original variable
$$I=-\frac{\sqrt x}{x+1}+\arctan\sqrt x+C$$
so I think the book's right.
|
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|
Solving $3^x \cdot 4^{2x+1}=6^{x+2}$ Find the exact value of $x$ for the equation $(3^x)(4^{2x+1})=6^{x+2}$
Give your answer in the form $\frac{\ln a}{\ln b}$ where a and b are integers.
I have tried using a substitution method, i.e. putting $2^2$ to be $y$, but I have ended up with a complicated equation that hasn't put me any closer to the solution in correct form. Any hints or guidance are much appreciated.
Thanks
|
$$\color{violet}{(3^x)(4^{2x+1})=6^{x+2}}$$
$$\color{indigo}{3^x\cdot 2^{2(2x+1)}=2^{x+2}\cdot 3^{x+2}}$$
$$\color{blue}{2^{2(2x+1)}\cdot 3^x=2^{x+2}\cdot 3^{x+2}}$$
$$\color{green}{\frac{2^{2(2x+1)}}{2^{x+2}}=\frac{3^{x+2}}{3^x}}$$
$$\color{brown}{2^{2(2x+1)-(x+2)}=3^{(x+2)-x}}$$
$$\color{orange}{2^{3x}=3^2}$$
$$\color{red}{8^x=9}$$
Taking natural logarithm on both sides, we get
$$\ln 8^x= \ln 9$$
$$\color{blue}{x=\frac{\ln 9}{\ln 8}}$$
To be a bit more precise i.e. simplifying a bit more we can have,
$$\color{red}{x=\frac{2\ln 3}{3\ln 2}}$$
|
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|
How to prove the trigonometric identity $\frac{1-\sin\varphi}{1+\sin\varphi} = (\sec\varphi - \tan\varphi)^2$ $$\frac{1-\sin\varphi}{1+\sin\varphi}$$
I have no idea how to start this, please help. This problem is essentially supposed to help me solve the proof of
$$\frac{1-\sin\varphi}{1+\sin\varphi} = (\sec\varphi - \tan\varphi)^2$$
|
$$(\sec\varphi - \tan\varphi)^2 = \left(\frac{1}{\cos \varphi}-\frac{\sin \varphi}{\cos \varphi}\right)^2 = \frac{(1 - \sin \varphi)^2}{\cos^2 \varphi}$$
$$ = \frac{(1 - \sin \varphi)^2}{ 1 - \sin^2 \varphi} = \frac{(1 - \sin \varphi)^2}{(1 - \sin \varphi)(1+\sin \varphi)} = \frac{1-\sin \varphi}{1+\sin\varphi}$$
|
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|
What is $\,a\bmod 63\,$ if $\,a\,$ is both a square and a cube? Let there be integers $a,m,n$ such that $a=m^2=n^3$. Show that $a\equiv 0,1,28,36\pmod{63}$.
I have established that $a$ must be a sixth power of an integer:
$$a=\frac{a^3}{a^2} = \frac{m^6}{n^6} =\left (\frac{m}{n}\right )^6$$
$n^6\mid m^6$ implies $n\mid m$ so $\frac{m}{n}=:b\in\mathbb{Z}$.
My objective is to show that $b\equiv 0,1,28,36\pmod{63}$. Then for every $f\in\mathbb{Z}[X]$:
$$b\equiv r\pmod{n}\Longrightarrow f(b)\equiv f(r)\pmod{n} $$
(just pick $f(x)=x^6$). Trouble is, I have no idea how to show $b$'s congruences. How to proceed?
Alternatively, I could also state that $a = 63t+r, 0\leq r<63$, but I don't know how to tie in the fact that $a$ is a perfect cube/square at the same time. Advice?
|
You must show $x^6\equiv 0,1,28,36\pmod{63}$ For $ 0 \leq x \leq 62$. By Fermat's little Theorem we conclude $x^6 \equiv 1 \pmod{7}$ if $x\neq 7k$ and it's easy to check $x^6 \equiv 1 \pmod{9}$ if $x\neq 3k$.
Now because of theorem :
$a\equiv b \pmod {m_1}$ and $a\equiv b \pmod {m_2}$ then $a\equiv b \pmod {lcm (m_1,m_2)}$
It remains to show for $x=3 , 7$:
$3^6 \equiv 36\pmod{63}$ and $7^6 \equiv 28\pmod{63}$.
For example :
$12^6 \equiv 9k \pmod{63}$ we have : $4^6 . 3^4 \equiv k \pmod{7} $ (Because of theorem says : $ak=bk \pmod m $ then $a=b \pmod {\frac{m}{\gcd(k,m)}}$ ) and because $ 4^6 \equiv 1 \pmod 7 , 81 \equiv 4 \pmod 7$ then $k=4$ and $12^6 \equiv 3^6 \equiv 36 \pmod {63}$.
|
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The indefinite integral $\int x^{13}\sqrt{x^7+1}dx$ I used integration by parts technique to solve this problem
so $u = \sqrt{x^7+1}$ ,$du = \frac{1}{2}\frac{7x^6}{\sqrt{x^7+1}}dx$
$v =\frac{x^{14}}{14}$ $dv=x^{13}dx$
then it becomes
$\frac{x^{14}}{14}\sqrt{x^7+1}-\int\frac{x^{14}}{28}\frac{7x^6}{\sqrt{x^7+1}}dx$
and this is where i got stuck at. I tried to substitute $u =x^7$
but then the integral become $\int \frac{u^2}{28\sqrt{u+1}}du$
the final answer that I found using wolffram calculator is $\frac{2}{105}(x^7+1)^{3/2}(3x^7-2)$
|
Hint: Try $u=x^7+1$. $\,\,\,\,\,$
|
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|
Antiderivative problem What is the antiderivative of $(2x+7)^{1/2}$?
My understanding is that it would be $\frac 23 \times(2x+7)^{3/2}$ but according to the source I am working from the answer is $\frac 13 \times (2x+7)^{3/2}$.
|
Notice:
$$\int x^n\space\text{d}x=\frac{x^{1+n}}{1+n}+\text{C}$$
$$\int\sqrt{2x+7}\space\text{d}x=$$
Substitute $u=2x+7$ and $\text{d}u=2\space\text{d}x$:
$$\frac{1}{2}\int\sqrt{u}\space\text{d}u=\frac{1}{2}\int u^{\frac{1}{2}}\space\text{d}u=\frac{1}{2}\cdot\frac{u^{1+\frac{1}{2}}}{1+\frac{1}{2}}+\text{C}=\frac{u^{\frac{3}{2}}}{3}+\text{C}=\frac{(2x+7)^{\frac{3}{2}}}{3}+\text{C}$$
|
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|
Maximum value of $\frac{3x^2+9x+17}{x^2+2x+9}$ If $x$ is real, the maximum value of $\frac{3x^2+9x+17}{x^2+2x+9}$ is?
Is it necessary that this function will attain maximum when the denominator will be minimum?
|
You need to solve:
$$\frac{\text{d}}{\text{d}x}\left[\frac{3x^2+9x+17}{x^2+2x+9}\right]=0\Longleftrightarrow$$
$$\frac{47+x(20-3x)}{(9+x(2+x))^2}=0\Longleftrightarrow$$
$$47+x(20-3x)=0\Longleftrightarrow$$
$$x(20-3x)=-47\Longleftrightarrow$$
$$-3x^2+20x=-47\Longleftrightarrow$$
$$3x^2-20x=47$$
If you look to it graphicly you see that the maximum is about $8.5$ so finding the solutions of the equation we get:
$$x=\frac{10+\sqrt{241}}{3}\approx8.5081$$
So:
$$\text{max}\left[\frac{3x^2+9x+17}{x^2+2x+9}\right]=\frac{35+\sqrt{241}}{16}\space\text{at}\space x=\frac{10+\sqrt{241}}{3}$$
|
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|
Understanding a solution to $m^3-n^2=2$ I am reading through a proof that the only integer solutions to $m^3-n^2=2$ are $(m,n)=(3,\pm 5)$.
Working in $\mathbf{Z}[\sqrt{-2}]:$ we have $m^3=n^2+2=(n+\sqrt{-2})(n-\sqrt{-2})$. Also: $$\text{gcd}\,\big(n+\sqrt{-2},n-\sqrt{-2}\big)\,\big|\,2\sqrt{-2}$$
That's all fine. The next step says that from unique factorisation, for some $x\in\mathbf{Z}[\sqrt{-2}]:$
$$n+\sqrt{-2}=x^3\quad\text{or}\quad 2x^3\quad\text{or}\quad \sqrt{-2}\,x^3$$
I'm a bit confused by this assertion, maybe I am missing something simple. I see that the $\gcd$ is $\pm1,\,\pm2,\,\pm\sqrt{-2}$ or $\pm2\sqrt{-2}$, but how does the claim about $x$ follow?
Thanks for any help.
|
First: $2$ can't divide $n+\sqrt{-2}$, and thus neither can $2\sqrt{-2}$. So $\gcd(n+\sqrt{-2},n-\sqrt{-2})\mid \sqrt{-2}$.
If two numbers are relatively prime and their product is a cube in a unique factorization domain, then both numbers are cubes times a unit. Since the only units here are also cubes, this means if $\gcd(n+\sqrt{-2},n-\sqrt{-2})=1$ then $n+\sqrt{-2}=x^3$ for some $x\in\mathbb Z[\sqrt{-2}]$.
If $a$ and $b$ have a GCD of $\sqrt{-2}$ and $ab=m^3$for $m\in\mathbb Z$, then $m^3$ must be even, so $m$ must be even, and $ab=m^3$ is divisible by $(\sqrt{-2})^6$. So one of $a$ or $b$ must be divisible by $4\sqrt{-2}$. That isn't possible for $a=n+\sqrt{-2}$ or $b=n-\sqrt{-2}$.
|
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|
Nested radical $\sqrt{x+\sqrt{x^2+\cdots\sqrt{x^n+\cdots}}}$ I am studying the $f(x) = \sqrt{x+\sqrt{x^2+\cdots\sqrt{x^n+\cdots}}}$ for $x \in (0,\infty)$ and I am trying to get closed form formula for this, or at least some useful series/expansion. Any ideas how to get there?
So far I've got only trivial values, which are
$$f(1)=\sqrt{1+\sqrt{1+\cdots\sqrt{1}}}=\frac{\sqrt{5}+1}{2}$$
$$f(4) = 3$$
Second one follows from
$$2^n+1 = \sqrt{4^n+(2^{n+1}+1)} = \sqrt{4^n+\sqrt{4^{n+1}+(2^{n+2}+1)}} = \sqrt{4^n+\sqrt{4^{n+1}+\cdots}}$$
I have managed to compute several derivatives in $x_0=1$ by using chain rule recursively on $f_n(x) = \sqrt{x^n + f_{n+1}(x)}$, namely:
\begin{align*}
f^{(1)}(1) &= \frac{\sqrt{5}+1}{5}\\
f^{(2)}(1) &= -\frac{2\sqrt{5}}{25}\\
f^{(3)}(1) &= \frac{6\sqrt{5}-150}{625}\\
f^{(4)}(1) &= \frac{1464\sqrt{5}+5376}{3125}\\
\end{align*}
These gave me Taylor expansion around $x_0=1$
\begin{align*}
T_4(x) &= \frac{\sqrt{5}+1}{2} + \frac{\sqrt{5}+1}{5} (x-1) - \frac{\sqrt{5}}{25} (x-1)^2 + \frac{6\sqrt{5}-150}{3750} (x-1)^3 \\
&\ \ \ \ \ + \frac{61\sqrt{5}+224}{3125} (x-1)^4
\end{align*}
However this approach seems to be useful only very closely to the $x=1$. I am looking for something more general in terms of any $x$, but with my limited arsenal I could not get much further than this. Any ideas?
This was inspiring but kind of stopped where I did
http://integralsandseries.prophpbb.com/topic168.html
Edit:
Thanks for the answers, i will need to go through them, looks like the main idea is to divide by $\sqrt{2x}$, so then I am getting
$$\frac{\sqrt{x+\sqrt{x^2+\cdots\sqrt{x^n+\cdots}}}}{\sqrt{2x}} = \sqrt{\frac{1}{2}+\sqrt{\frac{1}{4}+\sqrt{\frac{1}{16x}+\sqrt{\frac{1}{256x^4}+...}}}}$$
Then to make expansion from this. This is where I am not yet following how to get from this to final expansion.
|
You can obtain an expansion in negative powers of $x$, valid for large $x$, by trying to find $$\frac{f(x)}{\sqrt{2x}}$$. If you pull the $\frac{1}{\sqrt{2x}}$ into the square roots (note that it gets raised to progressively higher powers as it is pulled into more inner square roots) you get an expansion of the form $$f(x) = \sqrt{2x} + \frac{\sqrt{2}}{8} + \frac{\kappa}{\sqrt{x}} + \cdots$$
where calculating $\kappa$ is straightforward but a bit messy (the value I get is $-\frac{3\sqrt{2}}{32}$ but numerically it looks to be about $-\frac{1}{18}$). Even the first order expression (without the $\frac{\kappa}{\sqrt{x}}$ term) is already accurate to an error of smaller than $0.0052$ for all $x>4$.
|
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For what values of $k$ does the integral $\int_{0^+}^{\infty} \frac{dx}{x^k \sqrt{x+x^2}}$ converge? For what values of $k$ does the integral $$\int_{0^+}^{\infty} \frac{dx}{x^k \sqrt{x+x^2}}$$ converge?
I believe it does not converge for any $k$.
|
$\int_{0^+}^{\infty} \frac{dx}{x^k \sqrt{x+x^2}}=\int_{0^+}^{1} \frac{dx}{x^k \sqrt{x+x^2}}+\int_{1}^{2} \frac{dx}{x^k \sqrt{x+x^2}}+\int_{2}^{\infty} \frac{dx}{x^k \sqrt{x+x^2}}$
If $0<x<1$, $\sqrt{x+x^2}\ge \sqrt{x}$ hence $\int_{0^+}^{1} \frac{dx}{x^k \sqrt{x+x^2}}\le \int_{0^+}^{1} x^{-k-\frac{1}{2}} dx<\infty$ if $-k+\frac{1}{2}>0$, that is $k<\frac{1}{2}$.
If $x>2$, $\sqrt{x+x^2}\ge x$ hence $\int_{2}^{\infty} \frac{dx}{x^k \sqrt{x+x^2}}\le \int_{2}^{\infty} x^{-k-1} dx<\infty$ if $-k<0$, that is $k>0$.
So it converges for $0<k<\frac{1}{2}$.
If $k\ge \frac{1}{2}$,$\int_{0^+}^{1} \frac{dx}{x^k \sqrt{x+x^2}}=\infty$ while if $k\le 0$, $\int_{2}^{\infty} \frac{dx}{x^k \sqrt{x+x^2}}=\infty$.
|
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|
Finding Maximum Area of a Rectangle in an Ellipse
Question: A rectangle and an ellipse are both centred at $(0,0)$.
The vertices of the rectangle are concurrent with the ellipse as shown
Prove that the maximum possible area of the rectangle occurs when the x coordinate of
point $P$ is $x = \frac{a}{\sqrt{2}} $
What I have done
Let equation of ellipse be
$$ \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$$
Solving for y
$$ y = \sqrt{ b^2 - \frac{b^2x^2}{a^2}} $$
Let area of a rectangle be $4xy$
$$ A = 4xy $$
$$ A = 4x(\sqrt{ b^2 - \frac{b^2x^2}{a^2}}) $$
$$ A'(x) = 4(\sqrt{ b^2 - \frac{b^2x^2}{a^2}}) + 4x\left( (b^2 - \frac{b^2x^2}{a^2})^{\frac{-1}{2}} \times \frac{-2b^2x}{a^2} \right) $$
$$ A'(x) = 4\sqrt{ b^2 - \frac{b^2x^2}{a^2}} + \frac{-8x^2b^2}{\sqrt{ b^2 - \frac{b^2x^2}{a^2}}a^2} = 0 $$
$$ 4a^2\left(b^2 - \frac{b^2x^2}{a^2} \right) - 8x^2b^2 = 0 , \sqrt{ b^2 - \frac{b^2x^2}{a^2}a^2} \neq 0 $$
$$ 4a^2\left(b^2 - \frac{b^2x^2}{a^2} \right) - 8x^2b^2 = 0 $$
$$ 4a^2b^2 - 4b^2x^2 - 8x^2b^2 = 0 $$
$$ 4a^2b^2 - 12x^2b^2 = 0 $$
$$ 12x^2b^2 = 4a^2b^2 $$
$$ x^2 = \frac{a^2}{3} $$
$$ x = \frac{a}{\sqrt{3}} , x>0 $$
Where did I go wrong?
edit:The duplicate question is the same but both posts have different approaches on how to solve it so I don't think it should be marked as a duplicate..
|
Your mistake is here
$$
A'(x) = 4(\sqrt{ b^2 - \frac{b^2x^2}{a^2}}) +\color{red}{\frac12} \times4x\left( (b^2 - \frac{b^2x^2}{a^2})^{\frac{-1}{2}} \times \frac{-2b^2x}{a^2} \right).
$$
|
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|
$x^{2}+2x \cdot \cos b \cdot \cos c+\cos^2{b}+\cos^2{c}-1=0$ Solve this equation :
$$x^{2}+2x \cdot \cos b \cdot \cos c+\cos^2{b}+\cos^2{c}-1=0$$
Such that $a+b+c=\pi$
I don't have any idea. I can't try anything.
|
Not as elegant as Roman83's solution but here you go:
Solve the quadratic equation in $x$:
$$ x = \frac{-B\pm \sqrt{B^2 - 4AC}}{2A}$$
$A = 1, \ B = 2\cos b\cos c, \ C =\cos^2 b +\cos^2 c - 1 $
$$ x = \frac{-2\cos b\cos c\pm \sqrt{(2\cos b\cos c)^2 - 4(\cos^2 b +\cos^2 c - 1)}}{2}$$
$$ x = \frac{-2\cos b\cos c\pm \sqrt{4(\cos^2 b\cos^2 c - \cos^2 b -\cos^2 c + 1)}}{2}$$
Using $\cos^2\theta + \sin^2\theta = 1$:
$$ x = \frac{-2\cos b\cos c\pm 2\sqrt{(\cos^2 b -1)(\cos^2 c -1)}}{2}$$
$$ x = \frac{-2\cos b\cos c\pm 2\sqrt{\sin^2b\sin^2c}}{2}$$
$$ x = \frac{-2\cos b\cos c\pm 2\sin b\sin c}{2} = -\cos b\cos c\pm \sin b\sin c $$
$$ x = (-\cos b\cos c+ \sin b\sin c) \ \text{or} \ (-\cos b\cos c- \sin b\sin c) $$
$$ x = -\cos(b+c)= \cos a \ \text{or} -\cos(b-c)$$
|
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|
How to find the limit of $\lim_{x\to 0} \frac{1-\cos^n x}{x^2}$ How can I show that
$$
\lim_{x\to 0} \frac{1-\cos^n x}{x^2} = \frac{n}{2}
$$
without using Taylor series $\cos^n x = 1 - \frac{n}{2} x^2 + \cdots\,$?
|
I have seen several limit problems on MSE where people don't use the standard limit $$\lim_{x \to a}\frac{x^{n} - a^{n}}{x - a} = na^{n - 1}\tag{1}$$ whereas frequent use is made of other standard limits like $$\lim_{x \to 0}\frac{\sin x}{x} = \lim_{x \to 0}\frac{\log(1 + x)}{x} = \lim_{x \to 0}\frac{e^{x} - 1}{x} = 1\tag{2}$$ and this question is also an instance where the limit $(1)$ should be used.
We have
\begin{align}
L &= \lim_{x \to 0}\frac{1 - \cos^{n}x}{x^{2}}\notag\\
&= \lim_{x \to 0}\frac{1 - \cos^{n}x}{1 - \cos x}\cdot\frac{1 - \cos x}{x^{2}}\notag\\
&= \lim_{x \to 0}\frac{1 - \cos^{n}x}{1 - \cos x}\cdot\lim_{x \to 0}\frac{1 - \cos x}{x^{2}}\notag\\
&= \lim_{t \to 1}\frac{1 - t^{n}}{1 - t}\cdot\lim_{x \to 0}\frac{1 - \cos^{2}x}{x^{2}(1 + \cos x)}\text{ (putting }t = \cos x)\notag\\
&= \lim_{t \to 1}\frac{t^{n} - 1}{t - 1}\cdot\lim_{x \to 0}\frac{\sin^{2}x}{x^{2}}\cdot\frac{1}{1 + \cos x}\notag\\
&= n\cdot 1\cdot\frac{1}{1 + 1}\notag\\
&= \frac{n}{2}\notag
\end{align}
|
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Prove that largest root of $Q_k(x)$ is greater than that of $Q_j(x)$ for $k>j$. Consider the recurrence relation:
$P_0=1$
$P_1=x$
$P_n(x)=xP_{n-1}-P_{n-2}$;
1). What is closed form of $P_n$?
2). Let $k,j.\in\{1,2,...,n+1\}$ and $Q_k(x)=xP_{2n+1}-P_{k-1}.P_{2n+1-k}$. Then prove that largest root of $Q_k$ is greater than that of $Q_j$ for $k>j$. Another observation is taking $k,j$ to be even and $k>j$, smallest positive root of $Q_k$ is greater than that of $Q_j$. Without explicitly calculating the roots, is there any way to prove these observations.
|
The expression for $P_n$ in terms of $n$ is pretty ugly,
$$P_n=xP_{n-1}-P_{n-2}$$
Exact form for $P_n$ can be solved by using characteristic equation:
It comes from guessing $P_n=ar^n$ might be an answer for $P_n$, you can search "recurrence relation" online to see how it works.
$ar^n=x(ar^{n-1})-ar^{n-2}$
$r^2=xr-1$
Here $x$ can be treated as a constant since it wont affect our result,
$$r=\frac{x\pm\sqrt{x^2-4}}2$$
So that means the general form of $P_n$ is:
$$P_n=A(\frac{x+\sqrt{x^2-4}}2)^n + B(\frac{x-\sqrt{x^2-4}}2)^n$$
A and B are constant (with respect to $n$) that need to satisfy initial conditions.
$$P_0=A+B=1$$
And $$P_1= A\frac{x+\sqrt{x^2-4}}2 + B\frac{x-\sqrt{x^2-4}}2=x$$
So $$(A+B)\frac x2 +(A-B)\frac{\sqrt{x^2-4}}2=x$$
$$A-B=\frac{x}{ \sqrt{x^2-4}}$$
So $$A=\frac{\sqrt{x^2-4}+x}{2\sqrt{x^2-4}}$$
And
$$B=\frac{\sqrt{x^2-4}-x}{2\sqrt{x^2-4}}$$
Finally,
$$P_n=2^{-n}[\frac{x+\sqrt{x^2-4}}{2\sqrt{x^2-4}}(x+\sqrt{x^2-4})^n+\frac{\sqrt{x^2-4}-x}{2\sqrt{x^2-4}}(x-\sqrt{x^2-4})^n]$$
|
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|
Given $\tan(x) = 2\sqrt2 $ , $ x\in[ \pi , \frac{3\pi}{2}] $ , What is the exact value of $\sin(3x)$?
Given $\tan(x) = 2\sqrt2 $ , $ x\in[ \pi , \frac{3\pi}{2}] $
What is the exact value of $\sin(3x)$?
What I have done:
Given $\tan(x) = 2\sqrt2 $ , I drew a right angled triangle and found the hypotenuse to be $3$ so $\sin(x) = \frac{2\sqrt2}{3}$
Recalling that $$\sin(3x) = 3\sin(x) - 4\sin^3(x)$$
Subbing in $\sin(x) = \frac{2\sqrt2}{3}$ and $\sin^3(x) = \frac{16\sqrt2}{27}$
$$\sin(3x) = 3(\frac{2\sqrt2}{3}) - 4(\frac{16\sqrt2}{27})$$
$$ \sin(3x) = {2\sqrt2} - \frac{64\sqrt2}{27} $$
$$ \sin(3x) = \frac{-10\sqrt2}{27} $$
Is this correct? If so would there be any other way I could have gone about solving this?
|
$$\sin^2x=\frac1{1+\cot^2x}=\frac1{1+\frac1{8}}=\frac89\implies \sin x=-\frac{2\sqrt2}3\,,\,\,\text{since}\;\;x\in[\pi,\,3\pi/2]\implies$$
$$\sin3x=3\sin x-4\sin^3x=-2\sqrt2+\frac{64\sqrt2}{27}=\frac{10\sqrt2}{27}$$
since
$$\pi\le x\le\frac{3\pi}2\implies3\pi\le 3x\le4\pi+\frac\pi2$$
|
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|
Prove that $\frac{\sqrt{3}}{8}<\int_{\frac{\pi}{4}}^{\frac{\pi}{3}}\frac{\sin x}{x}\,dx<\frac{\sqrt{2}}{6}$
Prove that $$\frac{\sqrt{3}}{8}<\int_{\frac{\pi}{4}}^{\frac{\pi}{3}}\frac{\sin x}{x}\,dx<\frac{\sqrt{2}}{6}$$
My try: Using $$\displaystyle \sin x<x$$ and $$\frac{\sin x-0}{x-0}>\frac{1-0}{\frac{\pi}{2}-0}=\frac{2}{\pi}$$
So we get $$\frac{2}{\pi}<\frac{\sin x}{x}<1$$
So we get $$\frac{2}{\pi}\int_{\frac{\pi}{4}}^{\frac{\pi}{3}}1\,dx<\frac{\sin x}{x}<1\cdot \int_{\frac{\pi}{4}}^{\frac{\pi}{3}}\,$$
But this is not what I have to prove here.
|
This is not an answer but it is too long for a comment.
I cannot resist the pleasure of reporting again the magnificent approximation proposed by Mahabhaskariya of Bhaskara I, a seventh-century Indian mathematician.$$\sin(x) \simeq \frac{16 (\pi -x) x}{5 \pi ^2-4 (\pi -x) x}\qquad (0\leq x\leq\pi)$$ (see here).
So, as an approximation $$\int\frac{\sin(x)}x dx\approx-2 \left(\log \left(4 x^2-4 \pi x+5 \pi ^2\right)+\tan
^{-1}\left(\frac{1}{2}-\frac{x}{\pi }\right)\right)$$ which gives $$\int_{\frac{\pi}{4}}^{\frac{\pi}{3}}\frac{\sin x}{x}dx\approx 2 \log \left(\frac{153}{148}\right)+\tan ^{-1}\left(\frac{100}{621}\right)\approx 0.226111$$ while the exact solution is $$\int_{\frac{\pi}{4}}^{\frac{\pi}{3}}\frac{\sin x}{x}dx=\text{Si}\left(\frac{\pi }{3}\right)-\text{Si}\left(\frac{\pi }{4}\right)\approx 0.226483$$ your bounds being $\approx 0.216506$ and $\approx 0.235702$
|
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|
Limits and continuity: $\lim_{ x\rightarrow0} {\frac{ax+bx^3}{x^2}} = 2$. Find the values of $a$ and $b$ Please help
$$\lim_{ x\rightarrow0} {\frac{ax+bx^3}{x^2}} = 2$$
Find the values of $a$ and $b$.
|
\begin{align}
\lim_{ x\rightarrow0} \frac{ax+bx^3}{x^2}
&=\lim_{ x\rightarrow0} \frac{x^2(\frac{a}{x}+bx)}{x^2}\\[6px]
&=\lim_{ x\rightarrow0} \left(\frac{a}{x}+bx\right)\\[6px]
&=\begin{cases}
\pm\infty&\text{if $a\neq 0$ (*)}\\[6px]
0 &\text{if $a=0$}
\end{cases}
\end{align}
(*) Depending on the sign of $a$ and if it is $x\to 0^+$ or $x\to 0^-$.
Then, if $a\neq 0$, the limit doesn’t exist. Therefore $a=0$, but in this case, the limit is $0$. The limit is never $2$.
|
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|
Inequality involving four numbers Show that, if $a$, $b$, $c$ and $d$ are four positive numbers with sum $1$, then $$\frac 3 {1-a} + \frac 3 {1-b} + \frac 3 {1-c} + \frac 3 {1-d} \ge \frac 5 {1+a} + \frac 5 {1+b} + \frac 5 {1+c} + \frac 5 {1+d}.$$ I tried to subtract the fractions, but I didn't get to any result.
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I’ll show that the function $$g(a,b,c,d)=\frac 3 {1-a} + \frac 3 {1-b} + \frac 3 {1-c} + \frac 3 {1-d} - \left(\frac 5 {1+a} + \frac 5 {1+b} + \frac 5 {1+c} + \frac 5 {1+d}\right)$$ is minimized over positive $a,b,c,d$ with sum $1$ when $a=b=c=d=\frac{1}{4}$. Its minimum is then $g(\frac14,\frac14,\frac14,\frac14)=0$, proving the inequality.
Consider the function $$f(x,y)=\frac 3 {1-x} + \frac 3 {1-y} - \left(\frac 5 {1+x} + \frac 5 {1+y} \right).$$
Its directional derivative in the direction $(1,-1)$ is $$3\left(\frac{1}{(1-x)^2}-\frac{1}{(1-y)^2}\right)+5\left(\frac{1}{(1+x)^2}-\frac{1}{(1+y)^2}\right),$$ which is negative when $x<y$ and positive when $x>y$. Consequently, if $a<b$ and $0<\epsilon<\frac{b-a}{2}$, $$f(a+\epsilon,b-\epsilon)< f(a,b),$$ and therefore $g$ can be made smaller by averaging (or making closer) any two of its parameters that are unequal (which leaves $a+b+c+d$ unchanged).
|
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|
What is the correct u-substitution for this integral? I'm wondering how I can solve the following indefinite integral:
$$\int\frac{x}{x^2-2x+2}dx$$
The $u$-substitution I did was $x^2-2x+2$, but I've gotten a little stuck. I've shown my steps below for clarification on the problem.
$$u=x^2-2x+2$$
$$du = (2x-2)dx$$
$$du = 2(x-1)dx$$
$$\frac{1}{2}du = (x-1)dx$$
I'm not sure how to continue because the numerator of the function is $x$, but I got $x-1$ after continuing with the $u$-substitution. Is my $u$-substitution correct, or are there additional steps I need to perform in order to reach the final answer?
|
$$
\int\frac{x}{x^2-2x+2}dx = \frac{1}{2}\int\frac{2x-2+2}{x^2-2x+2}dx\\
= \frac{1}{2}\int\frac{2x-2}{x^2-2x+2}dx + \int\frac{1}{(x-1)^2+1}dx\\
= \frac{1}{2} \log(x^2-2x+2) + \tan^{-1}(x-1) + c
$$
|
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|
Prove that for $n\ge 2$, $2{n \choose 2}+{n \choose 1} = n^2$ Problem : Prove that for $n\ge 2$, $2{n \choose 2}+{n \choose 1} = n^2$
My Approach : I would assume that we can prove by induction.
Base case $n=2$.
$$=2{2 \choose 2}+ {2 \choose 1}= (2\cdot 1)+2 =4$$
$$n^2 =2^2 = 4.$$
Assume for $n\ge 2$, $n\ge 2$, $2{n \choose 2}+{n \choose 1} = n^2$ for $n \le k$.
Let $n=k+1$
$$2{k+1 \choose 2}+{k+1 \choose 1} = (k+1)^2$$
And, that's as far as I got.
I get stuck with the ${k+1 \choose 2}$ part.
|
It's much simpler to give a direct proof:
$$
2 \binom{n}{2} + \binom{n}{1}
=
2 \cdot \frac{n(n-1)}{2} + n
=
n^{2} - n + n
=
n^{2}.
$$
|
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How can we find a condition on $a$ so that there exists a regular matrix $P$ s.t. $B=P^{-1}AP$ $$ A=\begin{pmatrix} 4 & 1 & 1 \\ -1 & 1 & 0 \\ -2 & -1 & 0 \\ \end{pmatrix}, \, \, B=\begin{pmatrix} 0 & 1 & 0 \\ -3 & 4 & 0 \\ a & 2 & 1 \\ \end{pmatrix} $$
How can we find a condition on $a$ so that there exists a regular matrix $P$ s.t. $B=P^{-1}AP$
Answer:
$$(1- \lambda)^2( \lambda -3)=0$$
Eigenvalues are
$ \lambda_1 =1$ and $ \lambda_2= 3$ .
Corresponding eigenvectors are
$$v_1=\begin{pmatrix} 0 \\ 1 \\ -1 \\ \end{pmatrix}, v_2=\begin{pmatrix} 0 \\ 0 \\ 0 \\ \end{pmatrix} , v_3=\begin{pmatrix} -2 \\ 1 \\ 1\\ \end{pmatrix} $$
$B=P^{-1}AP$ implies the similarity of $A$ and $B$.
Since $A$ and $B$ are similar, $B$ has the same eigenvalues as $A$
Then how can we find the condition on $a$?
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$(A-I)(A-3I)\not=0,(A-I)^2(A-3I)=0$ and $(B-I)(B-3I)=(a+2)\begin{pmatrix}0&0&0\\0&0&0\\-3&1&0\end{pmatrix},(B-I)^2(B-3I)=0$.
Thus $A,B$ are similar iff $a\not=-2$.
|
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|
Prove that the determinant is a multiple of $17$ without developing it Let, matrix is given as :
$$D=\begin{bmatrix}
1 & 1 & 9 \\
1 & 8 & 7 \\
1 & 5 & 3\end{bmatrix}$$
Prove that the determinant is a multiple of $17$ without developing it?
I saw a resolution by the Jacobi method , but could not apply the methodology in this example.
|
$$
|D|
=
\begin{vmatrix}
1 & 1 & 9 \\
1 & 8 & 7 \\
1 & 5 & 3\end{vmatrix}
=
\begin{vmatrix}
1 & 1 & 9 \\
0 & 7 & -2 \\
0 & 4 & -6\end{vmatrix}
=
\begin{vmatrix}
1 & 1 & 9 \\
0 & 7 & -2 \\
0 & 0 & -\dfrac{34}{7}\end{vmatrix}
=
1 \times 7 \times -\dfrac{34}{7}
=
-34
=-2 \times 17
$$
|
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|
divergence/convergence $\sum_{k=1}^{\infty} \frac{1}{5k+2}$
$$\sum_{k=1}^{\infty} \frac{1}{5k+2}$$
Can I say that $$\sum_{k=1}^{\infty} \frac{1}{5k+2}=\sum_{k=1}\frac{1}{5k}+\sum_{k=1}^{\infty} \frac{1}{2}=\frac{1}{5}\sum_{k=1}^{\infty}\frac{1}{k}+\frac{1}{2}$$
and we know that $\\sum_{k=1}^{\infty} \frac{1}{k}$ diverges so the expression diverges too?
Or should I use the integral test $$\int_{1}^{\infty} \frac{1}{5k+2}=\frac{ln|5k+2|}{5}$$ from 1 to $\infty$= $\infty$
|
I would say that
$$ \sum_{k=1}^\infty\frac{1}{10k}<\sum_{k=1}^\infty \frac{1}{5k+2}$$
which is true because
$$ \frac{1}{10k}<\frac{1}{5k+2}\:\:\forall k\in \mathbb{N}.$$
$$ \sum_{k=1}^\infty \frac{1}{10k}=\frac{1}{10}\sum_{k=1}^\infty \frac{1}{k}$$
which implies that it diverges. So, because
$$ \sum_{k=1}^\infty\frac{1}{5k+2}$$
is greater term by term than a series which diverges, it too diverges.
|
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|
Find the minimum length of a line segment with endpoints on the coordinate axes that passes through the point $(1, 1)$ A line passes through the point $P=(1,1)$ and through the $x$ and $y$-axes at points $A$ and $B$ respectively. Find the minimum length of the line segment $AB$.
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$\frac{y-1}{x-1}=m$ defines the line.
$A$ is the $x$ intercept which is $(1-\frac{1}{m}, 0)$ and $B$ is the $y$ intercept which is $(0,1-m)$.
By minimizing the distance between these two points we can calculate $m$ which will then give us the distance. To make the algebra simple we will actually minimize the distance squared, call it d.
Then $d=(\frac{1}{m}-1)^2+(1-m)^2$ and taking the derivative with respect to $m$ gives $\frac{dd}{dm}=-\frac{2}{m^2}(\frac{1}{m}-1)-2(1-m)=\frac{-2+2m}{m^3}-\frac{2m^3(1-m)}{m^3}=\frac{-2+2m-2m^3+2m^4}{m^3}$
Setting this equal to $0$ gives $-2+2m-2m^3+2m^4=0\implies -1+m-m^3+m^4\implies m = 1$ or $m=-1$.
If $m=1$ then both the $x$ and $y$ intercepts will be $0$ so $A=B$ and the distance is $0$.
If $m=-1$ then $A=(2,0)$ and $B=(0, 2)$ so the distance is $\sqrt{2^2+2^2}=\sqrt{8}=2\sqrt{2}$
Therefore, the minimum distance is $0$ when $A=B$. If we change the problem to say that $A$ and $B$ are distinct then the minimum distance is $2\sqrt{2}$.
|
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|
How to calculate this integral without any integration techniques?
Differentiate $f(x) = (5x+2)\ln(2x+1)$ with respect to $x$. Hence, find $\int \ln(2x+1)^3dx$.
Because of the word "Hence" I'm assuming that the question doesn't allow integration techniques such as integration by parts or substitution.
The first part is trivial. The derivative is $5\ln(2x+1) + \dfrac{2(5x+2)}{2x+1}$. Now my line of thought so far has been to somehow get this derivative to the desired result: $\ln(2x+1)^3$ and then use $f(x)$ to get the integral. However I don't see any straight way to do this.
So I decided to experiment a little. Firstly I differentiated $(3x+2)\ln(2x+1)$ and got $$3\ln(2x+1) + \dfrac{2(3x+2)}{2x+1}$$ Now at least I have the $\ln(2x+1)^3$ term but I have another complication: $\dfrac{2(3x+2)}{2x+1}$.
I decided to take a look at the answer: $3x\ln(2x+1) + \dfrac{3}{2}\ln(2x+1) - 3x + C$. I thought that if I differentiated this, I would have some idea of where I should go, and how to proceed. So that's what I did, and I got:
$$3\ln(2x+1) + \dfrac{2(3x)}{2x+1} + \dfrac{3}{2x+1} - 3$$ This is perfect because $\dfrac{2(3x)}{2x+1} + \dfrac{3}{2x+1} - 3$ cancels out perfectly. But I have no idea how to get here, starting from $f'(x)$. But I'd say that my first experiment was pretty close. Any help would be highly appreciated.
|
Alright so from my first experiment we have the derivatie $3\ln(2x+1) + \dfrac{2(3x+2)}{2x+1}$. To get this to just $3\ln(2x+1)$ we subtract $3$ and $\dfrac{1}{2x+1}$. This gives us the expression $$\bigg(3\ln(2x+1) + \dfrac{6x}{2x+1} + \dfrac{4}{2x+1} \bigg) - 3 - \dfrac{1}{2x+1}$$ Integrating gives us $(3x+2)\ln(2x+1) - 3x - \dfrac{1}{2}\ln(2x+1) + C$
|
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|
Evaluate $\iint_{D}\sqrt{1-\frac{x^2}{a^2}-\frac{y^2}{b^2}}\text{d}x\text d y$
Evaluate using polar coordinates:$$\iint_{D}\sqrt{1-\frac{x^2}{a^2}-\frac{y^2}{b^2}}\text{d}x \, \text d y\qquad D=\bigg\{\frac{x^2}{a^2}+\frac{y^2}{b^2}\leqslant 1\bigg\}$$
What I did:
I found that the Jacobian matrix equal to $a\cdot b\cdot r$ I am sure about that and then I tried this:
$$\int_0^{2\pi}\int_0^1 \sqrt{1-x^2/a^2-y^2/b^2} \cdot (abr) \, \text{d} r \, \text d\theta$$ which leads to wrong answer,
a small bird whisper in my ear that I should get this:
$$\int_0^{2\pi}\int_0^1 \left(\sqrt{1-r^2}abr\right) \, \text{d}r \, \text d\theta$$
I wanted to tell her that I don't understed how to come to this result but she fly away, can you help me with that?
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If the change is $x = a r\cos\theta$, $y = b r\sin\theta$:
$$\frac{x^2}{a^2} + \frac{y^2}{b^2} = r^2\cos^2\theta + r^2\sin^2\theta = r^2.$$
|
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|
Proving $(1)\;\;\left(\frac n3\right)^n
Proving $$(1)\;\;\;\;\left(\frac n3\right)^n<n!<\left(\frac n2\right)^n\forall\,\,n\ge 6,n\in\mathbb{N}$$
and $$(2)\;\; \sqrt{n^n}\le n!\,\,\forall n\in\mathbb{N}$$ without induction
$\bf{My\; Try::}$ For $\bf{R.H.S}$ Inequality.
Using $\bf{A.M\geq G.M\;,}$ We get $$\frac{2+3+4+5+......+n}{n-1}>(2\cdot 3\cdot 4\cdot 5\cdot.......n)^{n-1}$$
So we get $$\frac{(n-1)(n+2)}{2(n-1)}>(n!)^{n-1}\Rightarrow n!<\left(\frac{n+2}{2}\right)^{n-1}$$
Now How can I solve after that also How can I prove Second one
Thanks
|
For (2), note that if $n$ is even we have:
$$\begin{align*}\dfrac{n^n}{(n!)^2}&=\dfrac{n\cdot n\cdot n\cdot n\cdots n\cdot n}{n\cdot n\cdot (n-1)\cdot (n-1)\cdots2\cdot 2\cdot 1\cdot 1}\\&=\dfrac{n\cdot n}{n\cdot n}\cdot \dfrac{n\cdot n}{(n-1)(n-1)}\cdots \dfrac{n\cdot n}{(n/2+1)(n/2+1)}\cdot \dfrac{1}{(n/2)(n/2)}\cdots \dfrac{1}{1\cdot 1}\\
&=\prod_{k=1}^{n/2}\dfrac{n^2}{(n-k+1)^2}\cdot \prod_{k=1}^{n/2} \dfrac{1}{k^2}=\prod_{k=1}^{n/2}\left(\dfrac{n}{n-k+1}\right)^2\cdot \prod_{k=1}^{n/2} \dfrac{1}{k^2}\\
&=\prod_{k=1}^{n/2}\left(1+\dfrac{k-1}{n-k+1}\right)^2\cdot \prod_{k=1}^{n/2} \dfrac{1}{k^2}=\prod_{k=1}^{n/2} \left[ \dfrac{1}{k^2}\left(1+\dfrac{k-1}{n-k+1}\right)^2\right]\\
&=\prod_{k=2}^{n/2} \left[ \dfrac{1}{k^2}\left(1+\dfrac{k-1}{n-k+1}\right)^2\right] \hspace{2cm}\text{(for $k=1$, the factor is $1$)}\\
&=\prod_{k=2}^{n/2} \left[\dfrac{1}{k}\left(1+\dfrac{k-1}{n-k+1}\right)\right]^2
\end{align*}$$ In the case when $n$ is odd, the limits in the product change from $1$ to $\lfloor n/2\rfloor$ and an extra factor of $\dfrac{n}{\left(n-\lfloor n/2 \rfloor \right)^2}\leq \dfrac{n}{(n/2)^2}\leq \dfrac{4}{n}$ must be multiplied at the end.
In any case, since $\dfrac{1}{k}\leq \dfrac{1}{2}$ and $1+\dfrac{k-1}{n-k+1}\leq 2$ for each $k$, we have that each factor is less than or equal to $1$, and so $$\dfrac{n^n}{(n!)^2}\leq 1 \Rightarrow \sqrt{n^n}\leq n!$$
|
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|
If possible, solve $7x^2 - 4x + 1 \equiv 0 $ (mod 11) If possible, solve $7x^2 - 4x + 1 \equiv 0 $ (mod 11)
Not sure if I'm approaching this problem correctly, any help is appreciated.
So far I have:
$7x^2 - 4x + 1 \equiv 0 $ (mod 11)
$21x^2 - 12x + 3 \equiv 0 $ (mod 11)
$-x^2 - x + 3 \equiv 0 $ (mod 11)
$-x^2 - x \equiv -3 $ (mod 11)
$-1(x^2 + x) \equiv -3 $ (mod 11)
$(x^2 + x) \equiv 3 $ (mod 11)
-Not sure if I am heading the right way, once again any help is appreciated.
|
In $\mathbb F_{11}$, one has $$7x^2 - 4x + 1= 0\iff-4x^2-4x+1=0\iff x(x+1)=\frac {1}{2^2}=2^8=3$$
For all $x\in\mathbb F_{11}$ we have $x(x+1)\ne 3$ (successively one has $0,2,6,1,9,8,9,1,6,2,0$)
Hence there is not solution.
|
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|
Let $x,y,z>0$ and $x+y+z=1$, then find the least value of ${{x}\over {2-x}}+{{y}\over {2-y}}+{{z}\over {2-z}}$ Let $x,y,z>0$ and $x+y+z=1$, then find the least value of
$${{x}\over {2-x}}+{{y}\over {2-y}}+{{z}\over {2-z}}$$
I tried various ways of rearranging and using AM > GM inequality. But I couldn't get it. I am not good at inequalities. Please help me.
I wrote $x$ as $1-(y+z)$ and I took $x+y$ as $a$ and the others as $b$ and $c$. And I am trying.
|
Here we use the AM-HM (harmonic mean) inequality.
\begin{align*}
{{x}\over {2-x}}+{{y}\over {2-y}}+{{z}\over {2-z}}
&=-3+2\left(\frac{1}{2-x}+\frac{1}{2-y}+\frac{1}{2-z}\right)\\
&\geq
-3+2\left[\frac{3^2}{(2-x)+(2-y)+(2-z)}\right]\\
&=-3+\frac{18}{5}=\frac{3}{5}.
\end{align*}
Also, when $x=y=z=\frac{1}{3}$, the above equality holds. Therefore the least value is $\frac{3}{5}$.
|
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|
Finding the error in an evaluation of the limit $\lim_{x\to0} \frac{e^x-x-1}{x^2} $ \begin{align}
\lim_{x\to0} \frac{e^x-x-1}{x^2}
&= \lim_{x\to0} \frac{e^x-1}{x^2} - \lim_{x\to0} \frac{1}{x} \\
&= \lim_{x\to0} \frac{e^x-1}{x}\lim_{x\to0} \frac{1}{x} - \lim_{x\to0} \frac{1}{x} \\ &= \lim_{x\to0} \frac{1}{x} - \lim_{x\to0} \frac{1}{x} \\
&= 0
\end{align}
|
The first claimed equality,
$$\lim_{x\to0} \frac{e^x-x-1}{x^2} \color{red}{\stackrel{\textrm{(false)}}{=}} \lim_{x\to0} \frac{e^x-1}{x^2} - \lim_{x\to0} \frac{1}{x} ,$$ already doesn't hold: Neither of the limits on its r.h.s. exist.
To evaluate the limit:
Hint Apply l'Hopital's Rule (twice) or expand $e^x$ in a Maclaurin series.
|
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|
To prove $X^2$ and $Y^2$ are independent for a given joint pdf $f(x,y)=\dfrac{1}{4}(1+xy)$ To prove $X^2$ and $Y^2$ are independent for a given joint pdf $f(x,y)=\dfrac{1}{4}(1+xy)$
$f(x,y) =
\begin{cases}
\dfrac{1}{4}(1+xy), & |x|<1,|y|<1 \\
0, & \text{otherwise}
\end{cases}$
The given solution starts with trying to prove that
\begin{align}
$P(X^2\leq x \cap Y^2\leq y) & = P(|X|\leq \sqrt{x} \cap |Y|\leq \sqrt{y}) \\
& = \int\limits_{-\sqrt{x}}^{\sqrt{x}}\int\limits_{-\sqrt{y}}^{\sqrt{y}}f(u,v)\,du\,dv \\
& = \int\limits_{-\sqrt{x}}^{\sqrt{x}}\int\limits_{-\sqrt{y}}^{\sqrt{y}}\dfrac{1}{4}\,du\,dv
\end{align}
I did not understand how we got to $f(u,v)=\dfrac{1}{4}$ rather that $\dfrac{1}{4}(1+xy)$
I tried with $X^2=u$ and $Y^2=v$, does not work out.
Please help.
|
The answer is rather simple. Just verify by a simple calculation that
$\frac{1}{4} \int\limits_{-\sqrt{x}}^{-\sqrt{x}}\int\limits_{-\sqrt{y}}^{-\sqrt{y}} uv \, du dv = 0$
and you're done.
Edit: here all steps:
$\int\limits_{-\sqrt{x}}^{-\sqrt{x}}\int\limits_{-\sqrt{y}}^{-\sqrt{y}} f(u,v) \, du dv = \int\limits_{-\sqrt{x}}^{-\sqrt{x}}\int\limits_{-\sqrt{y}}^{-\sqrt{y}} \frac{1}{4} \, du dv + \frac{1}{4} \int\limits_{-\sqrt{x}}^{-\sqrt{x}}\int\limits_{-\sqrt{y}}^{-\sqrt{y}} uv \, du dv = \int\limits_{-\sqrt{x}}^{-\sqrt{x}}\int\limits_{-\sqrt{y}}^{-\sqrt{y}} \frac{1}{4} \, du dv + \int\limits_{-\sqrt{x}}^{-\sqrt{x}} \left[ \frac{1}{2} uv^2 \right] _{-\sqrt{y}}^{-\sqrt{y}} \, du = \int\limits_{-\sqrt{x}}^{-\sqrt{x}}\int\limits_{-\sqrt{y}}^{-\sqrt{y}} \frac{1}{4} \, du dv + \int\limits_{-\sqrt{x}}^{-\sqrt{x}} 0 \, du = \int\limits_{-\sqrt{x}}^{-\sqrt{x}}\int\limits_{-\sqrt{y}}^{-\sqrt{y}} \frac{1}{4} \, du dv$
|
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Some uncommon improper integral. Convergence. I bet the integral $\int_1^3\frac{dx}{\sqrt{\tan(x^3-7x^2+15x-9)}}$ converges, but have no idea about proof, except expansion of tan in a series near zeros of its argument (limits of the integration).
|
Factoring the polynomial in the function gives us
$$\sqrt{\tan(x^3-7x^2+15x-9)}=\sqrt{\tan((x-1)(x-3)^2)}$$
The tangent function can be bounded above as the following: $$\tan(y)=\frac{\sin(y)}{\cos(y)} < \frac{y}{\cos(y)} < 2y$$ when $y$ is sufficiently close to zero. Therefore $$\sqrt{\tan((x-1)(x-3)^2)} < \sqrt{2(x-1)(x-3)^2} < \sqrt{6(x-3)^2} < \sqrt{6}|x-3|$$ for $x$ sufficiently close to 3. Therefore $$\frac{1}{\sqrt{\tan((x-1)(x-3)^2)}} > \frac{1}{\sqrt{6}|x-3|}$$
for $x$ sufficiently close to 3. This can you use to prove that the integral does not converge.
|
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"question_score": "1",
"answer_count": 2,
"answer_id": 0
}
|
Subsets and Splits
Fractions in Questions and Answers
The query retrieves a sample of questions and answers containing the LaTeX fraction symbol, which provides basic filtering of mathematical content but limited analytical insight.