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Unsure if I have solved/proven this trigonometric inequality. This is my first post here. I apologize if it goes against any guidelines for posting. I study math as a hobby and am currently dealing with trigonometry on a high school level. I have so far learned the formulas for trigonometric addition and subtraction and double the angle, as well as what is in my language referred to as the ’trigonometric one’ - getting the radius of the unit circle by use of the pythagorean theorem. I have not yet gotten to deriving trigonometric functions. The following is a problem I could solve by plugging in a set of numbers, but in seeking a more elegant solution, perhaps, I found myself stuck and I don’t know what I am missing. I am appreciative of any help I get. The problem is as follows:
Show that if $A$ is an angle and $0^\circ<A<90^\circ$ then $\hspace{0.3cm}\left( 1+\dfrac {1}{\sin A}\right) \left( 1+\dfrac {1}{\cos A}\right)>5$
I began with the following assumption:
$$0^\circ<A<90^\circ\rightarrow0<{\sin A}<1\\0<{\cos A}<1\rightarrow\dfrac {1}{\sin A}\\\dfrac {1}{\cos A}>1$$
Given the above, it would follow that:
$$\begin{aligned}
\lim _{A\rightarrow 90^\circ}\dfrac {1}{\cos A}&=\infty \\
\lim _{A\rightarrow 0^\circ}\dfrac {1}{\sin A}&=\infty
\end{aligned}$$
This alone doesn’t seem like enough to show what is asked. I can show that at $A=45^\circ$ the product is still greater than 5, but I am not sure how any offset in degrees from there affects two trigonometric terms such that the product is still greater than 5. I also tried solving the inequality but ended up with fractioned terms I couldn’t add up or a cubic function if you will, that I couldn’t solve.
|
Use $AM-GM$ and $\sin^2 A + \cos^2 A = 1$.
.....
I start by simply doing.
$(1 + \frac {1}{\sin A})(1 + \frac 1{\cos A}) = $
$(1 + \frac 1{\sin A} + \frac 1{\cos A} + \frac 1{\sin A \cos A}$.
Now $0 < \sin A < 1$ so $ \frac 1{\sin A} > 1$ and $0 < \cos A < 1$ so $ \frac 1{\cos A} > 1$.
So $(1 + \frac 1{\sin A} + \frac 1{\cos A} + \frac 1{\sin A \cos A} > 3 + + \frac 1{\sin A \cos B}$.
Now common sense tells us $\sin A \cos A < 1*1$ so $\frac 1{\sin A \cos A} > 1$ and that gives us:
$(1 + \frac 1{\sin A} + \frac 1{\cos A} + \frac 1{\sin A \cos A}) > 4$ and that is not good enough.
So I need more. Now I know $\sin A$ and $\cos A$ aren't just any numbers less than one. I know $\sin^2 A + \cos^2 A = 1$ and I know that if gets close to $1$ or $0$ the other gets close to $0$ and $1$ and I know that $\sin 45 = \cos 45 = \frac{\sqrt {2}}{2}$ but other wise one is more and the other is less than $\frac{\sqrt{2}}{2}$.
This all triggers that I can probably use the AM-GM theorem to "boost" $\frac 1 {\sin A\cos b} > 1$ to $\frac 1{\sin A\cos B} > 2$. Maybe... I'll use AM-GM and see what happens.
So we need to use AM-GM to prove $\sin A \cos B < \frac 12$ and $\frac 1{\sin A \cos A}> 2$.
Now AM- GM states for $M,N > 0$ thatn $\frac {M+N}2 \ge \sqrt{MN}$ so $\sin A\cos A = \sqrt{\sin^2 A\cos^2 A} \le \frac {\sin^2 A + \cos^2 A}2 = \frac 12$.
ANd that's that.
===
Another way $\cos A = \sin (90 - A)$. Let $A = 45 + b$ then
$\sin A = \sin (45+b) = \sin 45 \cos b + \cos 45 \sin b=\frac {\sqrt 2}2(\cos b + \sin b)$ and $\cos A = \sin (45 - b) = \sin 45 \cos b - \cos 45 \sin b=\frac {\sqrt 2}2(\cos b - \sin b)$.
And $\sin A*\cos A = (\frac {\sqrt 2}2)^2(\cos b + \sin b)(\cos b - \sin b) = \frac 12(\cos^2 b - \sin^2 b)= \frac 12(1-2\sin^2 b) < \frac 12$
So $(1 +\frac 1{\sin A})(1 + \frac 1 {\cos A}) =$
$1 + \frac 1{\sin A} + \frac 1{\cos A} + \frac 1 {\sin A\cos B} > 1 + 1+1+2 =5$.
|
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"timestamp": "2023-03-29T00:00:00",
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|
Find the number of natural solutions of $5^x+7^x+11^x=6^x+8^x+9^x$
Find the number of natural solutions of $5^x+7^x+11^x=6^x+8^x+9^x$
It's easy to see that $x=0$ and $x=1$ are solutions but are these the only one? How do I demonstrate that?
I've tried to write them either:
$$5^x+7^x+11^x=2^x*3^x+2^{3x}+3^{2x}$$
or
$$5^x+7^x+11^x=(5+1)^x+(7+1)^x+(11-2)^x$$
and tried to think of some AM-GM mean inequality or to divide everything by $11^x$, but those don't seem like the way to go. Any hints?
|
Let $f(x)=x^k,$ where $k>1$ or $k<0$.
Thus, $f$ is a convex function and since $(11,7,5)\succ(9,8,6),$ by Karamata we obtain:
$$f(11)+f(7)+f(5)>f(9)+f(8)+f(6).$$
Also, for $0<k<1$ we see that $f$ is a concave function.
Thus, by Karamata again
$$f(11)+f(7)+f(5)<f(9)+f(8)+f(6).$$
Thus, it remains to check, what happens for $k\in\{0,1\}$.
|
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|
For $x\geq 0$, what is the smallest value of $\frac{4x^2+8x+13}{6(x+1)}$? I know that I have to use the AM-GM inequality.
I tried separating the fraction:
$$\frac{4x^2+2x+7}{6(x+1)} + \frac{6(x+1)}{6(x+1)}$$
However, it doesn't seem to make either side of the inequality into a number.
I would appreciate some help, thanks!
|
Solution
$$\frac{4x^2+8x+13}{6(x+1)}=\frac{4(x+1)^2+9}{6(x+1)}=\frac{2}{3}(x+1)+\frac{3}{2(x+1)}\geq 2$$
with the equality holding if and only if $\dfrac{2}{3}(x+1)=\dfrac{3}{2(x+1)}$,namely, $x=\dfrac{1}{2}.$
|
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|
How can these two equations be solved by elimination? In this question, the following two equations were solved using elimination. With a google crash course I sort of get how elimination works, but it seems like these are much too complex to add the left sides together in a way that cancels out x or y.
$$
\frac{3x(3x^2+9)}{2y}
-
\frac{(3x^2+9)}{8y^3}^3
-
y
=
6 \pmod {23}
$$
$$
\frac{(3x^2+9)^2}{(2y)^2}
-
2x
=
12 \pmod {23}
$$
Ultimately I'm trying to figure out any way (the easiest preferably) to reduce these such that x=18, y=10 or x=19, y = 3 in order to convert it to a function in a scripting language.
If elimination is the way to go, what were the steps involved in arriving at the following equation?
$$x^4-48 x^3-18 x^2+13968 x=-86481 \pmod {23}$$
|
If you want to implement the resultant yourself, you can use the Sylvester matrix (see wikipedia).
In M2:
R=ZZ/23[x,y]
f=3*x*(3*x^2+9)*(4*y^2)-(3*x^2+9)^3-(y+6)*8*y^3;g=(3*x^2+9)^2-(2*x+12)*(2*y)^2;
sylvesterMatrix(f,g,y)
yields
$${\tiny \begin{pmatrix}{-8}&
{-2}&
-10 x^{3}-7 x&
0&
-4 x^{6}+10 x^{4}+7 x^{2}+7&
0\\
0&
{-8}&
{-2}&
-10 x^{3}-7 x&
0&
-4 x^{6}+10 x^{4}+7 x^{2}+7\\
-8 x-2&
0&
9 x^{4}+8 x^{2}-11&
0&
0&
0\\
0&
-8 x-2&
0&
9 x^{4}+8 x^{2}-11&
0&
0\\
0&
0&
-8 x-2&
0&
9 x^{4}+8 x^{2}-11&
0\\
0&
0&
0&
-8 x-2&
0&
9 x^{4}+8 x^{2}-11\\
\end{pmatrix} }$$
determinant sylvesterMatrix(f,g,y) gives the same output as resultant(f,g,y).
As for factoring the output
$$-11 x^{16}-x^{15}-3 x^{13}-2 x^{12}-3 x^{11}+7 x^{10}-10 x^{9}+x^{8}+8 x^{7}-6 x^{4}+4 x^{3}+10 x^{2}+10 x+8$$
to $$(x+5) (x+4) ({x^{2}+3})^{6} (x^{2}-11 x-8) ({-11})$$
you probably don't need the whole functionality, just check on $x=-11,\ldots,11$ whether the value is divisible by $23$.
|
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|
Number of solutions of $\left\{x\right\}+\left\{\frac{1}{x}\right\}=1$ Find the number of solutions of $$\left\{x\right\}+\left\{\frac{1}{x}\right\}=1,$$ where $\left\{\cdot\right\}$ denotes Fractional part of real number $x$.
My try:
When $x \gt 1$ we get
$$\left\{x\right\}+\frac{1}{x}=1$$ $\implies$
$$\left\{x\right\}=1-\frac{1}{x}.$$
Letting $x=n+f$, where $n \in \mathbb{Z^+}$ and $ 0 \lt f \lt 1$, we get
$$f=1-\frac{1}{n+f}.$$
By Hint given by $J.G$, i am continuing the solution:
we have
$$f^2+(n-1)f+1-n=0$$ solving we get
$$f=\frac{-(n-1)+\sqrt{(n+3)(n-1)}}{2}$$ $\implies$
$$f=\frac{\left(\sqrt{n+3}-\sqrt{n-1}\right)\sqrt{n-1}}{2}$$
Now obviously $n \ne 1$ for if we get $f=0$
So $n=2,3,4,5...$ gives values of $f$ as
$\frac{\sqrt{5}-1}{2}$, $\sqrt{3}-1$, so on which gives infinite solutions.
|
Using continued fraction:
\begin{align}
f+\frac{1}{n+f} &= 1 \tag{$n>1$, $0<f<1$} \\[5pt]
n+f &= n+1-\frac{1}{n+f} \\[5pt]
x &= n+\frac{n+f-1}{n+f} \\[5pt]
&= n+\frac{1}{\dfrac{n+f}{n+f-1}} \\[5pt]
&= n+\frac{1}{1+\dfrac{1}{n+f-1}} \\[5pt]
x &= n+\frac{1}{1+\dfrac{1}{x-1}} \tag{$\star$} \\[5pt]
&= \left[ n;\overline{1,(n-1)} \right] \\[5pt]
\alpha &= \frac{n+1+\sqrt{(n-1)(n+3)}}{2} \tag{$\alpha=x$} \\[5pt]
\beta &= \frac{n+1-\sqrt{(n-1)(n+3)}}{2} \tag{$\beta=\frac{1}{x}$}
\end{align}
where $\alpha$, $\beta$ are the roots of $(\star)$.
Note that $\alpha \beta=1$, the symmetric roles for $\alpha$ and $\beta$.
|
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|
Plot $|\frac{z+i}{z-1}|<1$ Let $D=\{|\frac{z+i}{z-1}|<1\}$ plot $D$
$$\frac{z+i}{z-1}|<1\iff \frac{|z+i|}{|z-1|}<1\iff |z+i|<|z-1|\iff |x+(y+1)i|<|(x-1)+yi|\iff\\ \iff \sqrt{x^2+(y+1)^2}<\sqrt{(x-1)^2+y^2}\iff {x^2+(y+1)^2}<{(x-1)^2+y^2}\iff \\ \iff x^2+y^2+2y+1<x^2-2x+1+y^2\iff x<-y$$
Which is the are under the line $x=-y$
Is it correct?
|
$$\frac{|z+i|}{|z-1|}<1$$
$$|z-(-i)| < |z-1|$$
We are describing points that are closer to $(-i)$ than $1$.
The perpendicular bisector between the two points is $y=-x$ and the corresponding region is the area under $y=-x$.
|
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|
Is there a way to simplify $\prod_{i=1}^n\cos(a^i\theta)$, where $a<1$? I recently came upon the following expression in an attempt at getting a closed-form solution for a recursive relation:
$$\prod_{i=1}^n \cos(a^i\theta)$$
where $a<1$. Is there a way to make this product into a sum or otherwise make it simpler, or approximate it? In particular for the problem I was looking at, $a$ was $\frac{3}{4}$ and $\theta$ was $\frac{\pi}{4}$. Thanks!
|
from
$$
\cos z = {{e^{\,i\,z} + e^{\, - \,i\,z} } \over 2} = {1 \over 2}e^{\,i\,z} \left( {1 + e^{\, - \,i\,2z} } \right)
$$
we get
$$
\eqalign{
& \ln \cos z = \ln {1 \over 2} + i\,z + \ln \left( {1 + e^{\, - \,i\,2z} } \right) = \cr
& = \ln {1 \over 2} + i\,z + \ln \left( {1 + 1 - i2z - 2z^2 + O\left( {z^3 } \right)} \right) = \cr
& = \ln {1 \over 2} + i\,z + \ln \left( {2\left( {1 - iz - z^2 + O\left( {z^3 } \right)} \right)} \right) = \cr
& = \ln {1 \over 2} + i\,z - \ln 2 - iz - {{z^2 } \over 2} + O\left( {z^3 } \right) = \cr
& = - {{z^2 } \over 2} + O\left( {z^3 } \right) \cr}
$$
Therefore we can say that
$$
\eqalign{
& \ln P(x,a,n) = \ln \prod\limits_{k = 1}^n {\cos (a^{\,k} x)} = \sum\limits_{k = 1}^n {\ln \cos (a^{\,k} x)} = \cr
& = - {1 \over 2}\sum\limits_{k = 1}^n {\left( {x^{\,2} a^{\,2k} + O\left( {a^{\,3k} x^3 } \right)} \right)} = - {{x^{\,2} a^{\,2} } \over 2}{{1 - a^{2n} } \over {1 - a^{\,2} }} + O\left( {a^{\,3} x^3 } \right) \cr}
$$
|
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|
Prove there is no $x, y \in \mathbb Z^+ \text{ satisfying } \frac{x}{y} +\frac{y+1}{x}=4$ Prove that there is no $x, y \in \mathbb Z^+$ satisfying
$$\frac{x}{y} +\frac{y+1}{x}=4$$
I solved it as follows but I seek better or quicker way:
$\text{ Assume }x, y \in \mathbb Z^+\\ 1+\frac{y+1}{y}+\frac{x}{y} +\frac{y+1}{x}=1+\frac{y+1}{y}+4 \\
\Rightarrow \left(1+\frac{x}{y}\right)\left(1+\frac{y+1}{x}\right)=6+\frac{1}{y}\\
\Rightarrow (x+y)(x+y+1)=x(6y+1)\\
\Rightarrow x\mid (x+y) \;\text{ or }\; x\mid (x+y+1)\\
\Rightarrow x\mid y \;\text{ or }\; x\mid (y+1)\\
\text{Put}\; y=nx ,n \in \mathbb Z^+\;\Rightarrow\; \frac{x}{nx} +\frac{nx+1}{x}=4 \;\Rightarrow\; \frac{1}{n} +\frac{1}{x}=4-n \;\rightarrow\;(1)\\
\text{But}\; \frac{1}{n} +\frac{1}{x} \gt 0 \;\Rightarrow\; 4-n \gt 0 \;\Rightarrow\; n \lt 4\\
\text{Also}\; \frac{1}{n},\frac{1}{x}\le 1 \;\Rightarrow\; \frac{1}{n} +\frac{1}{x} \le 2 \;\Rightarrow\; 4-n \le 2 \;\Rightarrow\; n \ge 2\\
\;\Rightarrow\; n=2 \;\text{ or }\; 3, \;\text{substituting in eq. (1), we find no integral values for } x.\\ \text{The same for the other case.}\\
$
So is there any other better or intelligent way to get this result?
|
I include a proof for the original question. It is between the first two pictures.
There are infinitely many solutions with both $x,y \leq 0.$ This comes under the heading of Vieta Jumping. Given a solution to $x^2 - 4xy + y^2 + y = 0,$ we can create new solutions by alternately taking mappings on the hyperbola,
$$ (x,y) \mapsto (x, 4x-y - 1) \; \; \; , $$
$$ (x,y) \mapsto (4x-y,y) \; \; \; . $$
The solutions on the hyperbola branch that is (mostly) in the third quadrant begin
$$ (0,0); \; (0,-1); \; (-4,-1); \; (-4,-16); \; (-60,-16); \; (-60,-225); \; \ldots $$
There are infinitely many rational solutions with both $x,y > 0.$
$$ \left(\frac{1}{2},\frac{1}{2}\right); \; \; \left(\frac{3}{2},\frac{1}{2}\right); \; \; \left(\frac{3}{2},\frac{9}{2}\right); \; \; \left(\frac{33}{2},\frac{9}{2}\right); \; \; \left(\frac{33}{2},\frac{121}{2}\right); \; \; \left(\frac{451}{2},\frac{121}{2}\right); \; \; \left(\frac{451}{2},\frac{1681}{2}\right); \; \; \ldots $$
The $y$ values above are of the form $\frac{b_n^2}{2},$ where $b_{n+2} = 4 b_{n+1} - b_n \; , \;$ giving $1,3,11,41,153,571,...$
If there were any integer solutions in the first quadrant, these same mappings would take us to an integer solution with both $x,y$ fairly small obeying certain explicit inequalities.
The following diagram goes with the inequality part:
If there were an integer solution with $x,y > 0$ and $x+y \geq 10,$ we are going to show that one of the mappings given reduces $x+y$ by at least two; therefore a finite number of steps would take us to an integer solution with $x,y > 0$ and $x+y < 10.$ One can quickly inspect and find that there are no integer solutions with such small numbers. So that is it.
If $x+y \geq 10$ and both positive, one case, closer to the positive $x$ axis, has $x > y$ and $y > 2.$ From the quadratic formula and the inequalities we find $x = 2y + \sqrt{3y^2 - y}.$ Note that $3y^2 - y > 1.$ At this point, we have $x+y = 3y + \sqrt{3y^2 - y}.$ Now, after we apply the mapping $ \color{magenta}{ (x,y) \mapsto (4x-y,y) \; \; \; }, $ the replacement value for $x+y$ is $3y - \sqrt{3y^2 - y},$ so it has shrunk by at least $2.$
The second case is closer to the positive $y$ axis, $y > x$ and $x > 2.$ We find $y = \frac{4x-1 + \sqrt{12x^2 - 8x+1}}{2}.$ After applying the other mapping $ \color{magenta}{ (x,y) \mapsto (x,4x-1-y) \; \; \; }, $ The new $y$ value is $y = \frac{4x-1 - \sqrt{12x^2 - 8x+1}}{2}.$ Therefore the sum we keep calling $x+y$ has decreased by $\sqrt{12x^2 - 8x+1}$ which is larger than $2$ when $x > 2.$ Again, we have shrunken $x+y$ by at least $2.$
That's it. There cannot be any integer solutions with $x,y > 0$ because there are none with $x+y \leq 10.$
$$ \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc $$
I found out how to draw line segments in desmos. One can put in a short table of points and have it connect them in order. Here is the hyperbola in the first quadrant, showing the first four of the half integer points and how a stairway connects them, beginning an infinite process. The next point on this staircase, which goes up and to the right, would be $\left(\frac{33}{2},\frac{121}{2}\right);$
Alright, here is the hyperbola with the first few points in the third quadrant that have integer coordinates, joined up with green line segments, beginning with the origin and $(0,-1).$ The next point in this downward stairway would be $(-60,-16)$
|
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|
Show $\frac{1}{b+c+d} + \frac{1}{a+c+d} + \frac{1}{a+b+d} + \frac{1}{a+b+c} \ge \frac{16}{3(a+b+c+d)}$.
If $a,b,c,d > 0$ and distinct then show that
$$
\frac{1}{b+c+d} + \frac{1}{a+c+d} + \frac{1}{a+b+d} + \frac{1}{a+b+c}
\ge \frac{16}{3(a+b+c+d)}
$$
I tried using HM < AM inequality but am missing on $16$. Probably I am mistaken in solving.
|
I think the constant $16$ in the question should be $16/3$. Also, the AM-HM inequality implies
$$\frac{\frac{1}{b+c+d}+\frac{1}{c+d+a}+\frac{1}{d+a+b}+\frac{1}{a+b+c}}{4} \ge \frac{4}{3(a+b+c+d)}.$$
|
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|
Relationship between roots
Suppose we have the equation $x^2 + 2x + 1, x^2 + 3x + 1, \ldots, x^2 + (2 + k) x + 1, k \in \mathbb{N}$, consider all the roots less than $-1$. (see Appendix)
Is it possible to find the length of any one of the neighboring segments, knowing one of them? For example, if I know the length of a red segment, can I only use the knowledge of the equation and the length of the segment to find the blue segment?
|
Define
$$f_k(x):=x^2+(2+k)x+1$$
for $x<-1$.
When $f_k(x)=0$,
$$
x=\frac{-(2+k)\pm\sqrt{(2+k)^2-4}}{2}
$$
Since $k+2>2$, $$(2+k)^2-4>0$$ so we have two distinct roots. We can show that the case where $x<-1$ is when
$$x=-\frac{(2+k)+\sqrt{(2+k)^2-4}}{2}$$
Define $$r_k:=-\frac{(2+k)+\sqrt{(2+k)^2-4}}{2}$$
Define $$l_k:=r_{k}-r_{k+1}$$
as the lengths between the adjacent roots of $y=f_k$ and $y=f_{k+1}$.
We have
\begin{align}
l_k
&=\frac{(3+k)+\sqrt{(3+k)^2-4}}{2}-\frac{(2+k)+\sqrt{(2+k)^2-4}}{2}\\
&=\frac12\left(1+\sqrt{(3+k)^2-4}-\sqrt{(2+k)^2-4}\right)
\end{align}
which you can then use to calculate back your unknown $k$ from your known $l_k$ to then calculate $l_{k+1}$.
|
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|
Problems with Recurrence Relations as a form of Counting I have been having trouble trying to understand how to do the following problem
Solve by unfolding: $a_0=3$, and for $n\geq1$, $a_n=5a_{n-1}+3$. Hint: This will involve the geometric sum formula.
This is my work so far:
$$a_n=5a_{n-1}+3$$
$$a_n=5(5a_{n-2}+3)+3$$
$$a_n=(5(5(5a_{n-3}+3)+3)+3)$$
$$a_n=5^{n}*a_0+5^{n-1}*3+5^{n-2}*3+...+5*3+3$$
I am not sure if this is right, or how to really do this problem. Help, and hints would be much appreciated.
|
\begin{align}
a_n
&= 5a_{n-1}+3\\
&= 5(5a_{n-2}+3)+3\\
&= 5^2a_{n-2}+3(5^1+5^0)\\
&= 5^2(5a_{n-3}+3)+3(5^1+5^0)\\
&= 5^3a_{n-3}+3(5^2+5^1+5^0)\\
&= \dots\\
&= 5^na_0+3(5^{n-1}+\dots+5^0)\\
&= 5^n(3)+3\cdot\frac{5^n-1}{5-1}\\
&= 3\cdot5^n+\frac34(5^n-1)\\
&= \frac{15}4\cdot5^n-\frac34\\
&= \frac34(5^{n+1}-1)
\end{align}
|
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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|
Evaluating $\sum_{k=1}^{\infty} 2\ln{(2k)} - \ln{(2k-1)} - \ln{(2k+1)} $ I was trying to evaluate the following series, which I know converges:
$$\sum_{k=1}^{\infty} 2\ln{(2k)} - \ln{(2k-1)} - \ln{(2k+1)} \tag{1}\label{1} $$
In a telescoping fashion, I began writing out the terms in hopes to find a pattern:
$$= (2\ln{2} - \ln{1} - \ln{3}) + (2\ln{4} - \ln{3} - \ln{5}) + (2\ln{6} - \ln{5} - \ln{7}) + \ldots \tag{2}\label{2}$$
while nothing canceled out, I grouped terms together:
$$ = 2\ln{2} - 2\ln{3} + 2\ln{4} - 2\ln{5} + 2\ln{6} - 2\ln{7} + \ldots \tag{3}\label{3}$$
$$ = 2 \left[ \ln{2} - \ln{3} + \ln{4} - \ln{5} + \ln{6} - \ln{7} + \ldots \right] \tag{4}\label{4}$$
which left me with the following divergent series:
$$ = 2 \sum _{k=2} ^{\infty} (-1)^k \ln{k} \tag{5}\label{5}$$
Clearly, $\eqref{5}$ can't be equivalent to $\eqref{1}$.
I'm pretty new to calculus, and while I've covered telescoping series, it seems that this technique cannot be applied here. Though, I don't know why.
Where did I go wrong?
|
Here is the answer of the series. With expansion of $\ln$ we have
\begin{align}
\sum_{k=1}^{\infty} 2\ln(2k)-\ln(2k-1)-\ln(2k+1)
&= \sum_{k=1}^{\infty} -\ln\left(1-\dfrac{1}{2k}\right)-\ln\left(1-\dfrac{1}{2k}\right) \\
&= \sum_{k=1}^{\infty} \sum_{n\geqslant1} \dfrac{1}{n}\left(\dfrac{1}{2k}\right)^{2n} \\
&= \sum_{n\geqslant1} \dfrac{1}{n} \sum_{k=1}^{\infty} \left(\dfrac{1}{2k}\right)^{2n} \\
&= \sum_{n\geqslant1} \dfrac{\zeta(2n)}{n2^{2n}} \\
&= \color{blue}{\ln\dfrac{\pi}{2}}
\end{align}
the last step proved here.
|
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
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|
Using inverse Laplace transform to solve differential equation The differential equation is as follows-
$$\frac{d^2 x}{dt^2} + 5 \frac{dx}{dt} + 6x = e^t $$
I use laplace transform to make it to become - $$X(s) = \frac{1}{(s-1)(s+3)(s+2)}$$
where $X(s)$ is the Laplace transform of $X(t)$
So now I am trying to find $X(t)$ using inverse transform.
From partial fractions-
$X(s) = \frac{1}{(s-1)(s+3)(s+2)} = \frac{A}{s-1} + \frac{B}{s+3} + \frac{C}{s+2} $
Numerator - $ 1 = A(s+3)(s+2) + B(s-1)(s+2) + C(s-1)(s+3) $
I am stuck from here on how to carry on this partial fraction
Can I sub all s values to be 0 ?
For example
$1 = A(0+3)(0+2)$
$1= B(0-1)(0+2) $
$1 = C (0-1)(0+3) $
|
$$X(s) = \frac{1}{(s-1)(s+3)(s+2)}$$
Substitute $k=s+2$
$$F(k) = \frac{1}{(k-3)(k+1)k}$$
$$F(k) = \frac{1}{(k-3)}\left(\frac 1 {(k+1)k} \right)$$
$$F(k) = \frac{1}{(k-3)}\left(\frac 1 k-\frac 1 {k+1} \right)$$
$$F(k) = \frac{1}{(k-3)}\frac 1 k-\frac{1}{(k-3)}\frac 1 {(k+1)} $$
$$F(k) = \frac 13 \left (\frac{1}{(k-3)}-\frac 1 k\right )-\frac 14\left(\frac{1}{(k-3)}-\frac 1 {(k+1)} \right )$$
$$F(k) = \frac 1 {12} \frac{1}{(k-3)}-\frac 1{3k}+\frac 14\frac 1 {(k+1)} $$
Substitute back $k=s+2$
$$X(s)=-\frac 1{3(s+2)}+\frac 1{4(s+3)}+\frac 1 {12(s-1)}$$
$$x(t)=-\frac 1{3}e^{-2t}+\frac 1{4}e^{-3t}+\frac 1 {12}e^{t}$$
It's a bit longer but i find this method easier
|
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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|
Evaluating $\sum (-1)^{n+1} (n+1 + \frac{1}{n+1})/n!$ Let $a_{n}=n+\dfrac{1}{n}$ for $n \in \mathbb{N}$. Find the sum of series $$\sum_{n=1}^{\infty}(-1)^{n+1}\dfrac{a_{n+1}}{n!}.$$
This becomes:
$$\begin{align}
\sum_{n=1}^{\infty}(-1)^{n+1}\Big[\dfrac{(n+1)+\dfrac{1}{n+1}}{n!}\Big] &\implies
\sum_{n=1}^{\infty}(-1)^{n+1}\Big[\dfrac{(n+1)}{n!}+\dfrac{1}{(1+n)!}\Big] \\
&\implies \sum_{n=1}^{\infty}(-1)^{n+1}\dfrac{(n+1)}{n!}+\sum_{n=1}^{\infty}(-1)^{n+1}\dfrac{1}{(1+n)!}\\
& \implies \sum_{n=1}^{\infty}(-1)^{n+1}\dfrac{(n+1)}{n!}+e^{-1}-1
\end{align}$$
I don't know how to simplify further.
|
$\begin{array}\\
\sum_{n=1}^{\infty} (-1)^{n+1} (n+1 + \frac{1}{n+1})/n!
&=\sum_{n=2}^{\infty} (-1)^{n} \dfrac{n + \frac{1}{n}}{(n-1)!}\\
&=\sum_{n=2}^{\infty} (-1)^{n} \dfrac{n^2 + 1}{n!}\\
&=\sum_{n=2}^{\infty} (-1)^{n} \dfrac{n^2}{n!}+\sum_{n=2}^{\infty} (-1)^{n} \dfrac{1}{n!}\\
&=\sum_{n=2}^{\infty} (-1)^{n} \dfrac{n(n-1)+n}{n!}+e^{-1}\\
&=\sum_{n=2}^{\infty} (-1)^{n} \dfrac{n(n-1)}{n!}+\sum_{n=2}^{\infty} (-1)^{n} \dfrac{n}{n!}+e^{-1}\\
&=\sum_{n=2}^{\infty} (-1)^{n} \dfrac{1}{(n-2)!}+\sum_{n=2}^{\infty} (-1)^{n} \dfrac{1}{(n-1)!}+e^{-1}\\
&=\sum_{n=0}^{\infty} (-1)^{n+2} \dfrac{1}{n!}+\sum_{n=1}^{\infty} (-1)^{n+1} \dfrac{1}{n!}+e^{-1}\\
&=1+\sum_{n=1}^{\infty} ((-1)^{n+2}+(-1)^{n+1}) \dfrac{1}{n!}+e^{-1}\\
&=1+e^{-1}\\
\end{array}
$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2861212",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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|
Evaluating $\frac{1^2}{2^1}+\frac{3^2}{2^2}+\frac{5^2}{2^3}+\cdots$ using sigma notation This question can be solved by method of difference
but I want to solve solve it using sigma notation:
$$\frac{1^2}{2^1}+\frac{3^2}{2^2}+\frac{5^2}{2^3}+\cdots+\frac{(2r +1)^2}{2^r}+\cdots$$
I used the geometric progression summation for the $(1/2)^r$ part, and opened $(2r+1)^2$ to $(4r^2 + 1 + 4r)$. If I now express this in sigma notation, it becomes
$$\frac{4n(n+1)(2n+1)}{6} + n + \frac{4(n\cdot n+1)}{2}$$ but I getting problem while putting upper limit infinity.
Where did I go wrong? Please explain.
Answer = $17$
|
The general term is $\frac{(2k-1)^2}{2^k}$. Using that, we get
$$
\begin{align}
\sum_{k=1}^\infty\frac{(2k-1)^2}{2^k}
&=\sum_{k=1}^\infty\frac{4k(k-1)+1}{2^k}\tag1\\
&=\sum_{k=1}^\infty8\binom{k}{k-2}\left(\frac12\right)^k+\sum_{k=1}^\infty\binom{k-1}{k-1}\left(\frac12\right)^k\tag2\\
&=\sum_{k=1}^\infty8\binom{-3}{k-2}\left(-\frac12\right)^k-\sum_{k=1}^\infty\binom{-1}{k-1}\left(-\frac12\right)^k\tag3\\
&=2\left(1-\frac12\right)^{-3}+\frac12\left(1-\frac12\right)^{-1}\tag4\\[12pt]
&=17\tag5
\end{align}
$$
Explanation:
$(1)$: $(2k-1)^2=4k(k-1)+1$
$(2)$: $4k(k-1)=8\binom{k}{2}=8\binom{k}{k-2}$ for $k\ge2$ and $\binom{k-1}{k-1}=1$ for $k\ge1$
$(3)$: convert to negative binomial coefficients
$(4)$: Generalized Binomial Theorem
$(5)$: evaluate
|
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"timestamp": "2023-03-29T00:00:00",
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|
Trigonometry : Find the value of $\csc^2 \pi/7 + \csc^2 2\pi/7 + \csc^2 3\pi/7$ Find the value of $\csc^2 \pi/7 + \csc^2 2\pi/7 + \csc^2 3\pi/7$
My try : Converted it into Sin and then tried to apply series formula but failed
|
I provide one more solution. Judging by the question you asked, I'm assuming your manipulation of $\tan$ and $\cot$ need to reinforce.
First, some preparation.
We all know that
$$
\tan (x \pm y) = \frac {\tan (x) \pm \tan (y)} {1 \mp \tan (x)\tan (y)}.
$$
Then
$$
\cot (x - y) = \frac {\cot (x) \cot (y) + 1} { \cot (y) - \cot (x)},
$$
or
$$
\cot (x) \cot (y) = \cot(x-y) (\cot(y) -\cot(x)) - 1
$$
Especially,
$$
\cot (2x) = \frac {\cot^2 (x) -1} {2 \cot (x)},
$$
which yields
$$
\cot^2(x) - 1 = 2\cot(x)\cot(2x).
$$
Also,
$$
\csc^2(x) = \frac {\sin^2(x) + \cos^2(x)} {\sin ^2(x)} = 1 + \cot^2(x).
$$
Now let's start. Let $x = \pi /7$.
$$
P := \csc^2(x) + \csc^2(2x) + \csc^2(4x) = 3 + \cot^2(x) + \cot^2(2x) + \cot^2(3x).
$$
By the formula above,
$$
P = 6 + 2(\cot(x) \cot(2x) + \cot(2x) \cot (4x) + \cot (4x) \cot (8x)) =: 6 +2Q.
$$
Now change the angle: since $\cot(\pi \pm y) = \mp\cot (y)$, we have
$$
Q = \cot(x)\cot(2x) - \cot(2x) \cot (3x) - \cot(3x) \cot (x).
$$
Now,
\begin{align*}
Q &= 1+ \cot(x)(\cot(x) - \cot(2x)) - \cot(x) (\cot(2x) -\cot(3x)) - \cot(2x)(\cot(x) - \cot (3x))\\
&= 1+ \cot(x) (\cot(x) -3\cot(2x) + \cot(3x)) + \cot(2x) \cot(3x)\\
&= \cot(x) (\cot(x) -3\cot(2x) + \cot(3x)) + (\cot(2x)-\cot(3x))\cot(x) \\
&= \cot(x)(\cot(x) -2\cot(2x))\\
&= \cot^2(x) - 2\cot(2x) \cot(x)\\
&=1.
\end{align*}
Therefore $P = 6+2Q=8$.
|
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"timestamp": "2023-03-29T00:00:00",
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|
Arithmetic series problem Given $\left\{a_n\right\}$ arithmetic progression, $a_1=2$, $a_{n+1}=a_n+2n$ $\left(n\:\ge \:1\right)$. $a_{50}=?$
What i did:
$$a_n+d=a_n+2n$$
$$d=2n$$
$$a_{50}=2+d\left(n-1\right)$$
$$a_{50}=2+2\left(n^2-n\right)$$
$$a_{50}=2+2\cdot 2450$$
$$a_{50}=4902$$
But this is wrong. Answers:
$$A=2452,\:B=2450,\:C=2552,\:D=2500$$
|
\begin{align}a_1&=2\\a_{n+1}&=a_n+2n\\&=a_{n-1}+2[n+(n-1)]\\&=a_{n-2}+2[n+(n-1)+(n-2)]\\
&= \ldots\\
&=a_1+2[n+(n-1)+\ldots +1]\end{align}
Hence \begin{align}a_{50}&=a_1+2(49+\ldots + 1)\\&=2+2\cdot \frac{49(50)}{2}\\&=2+49(50)\\&=2+(50-1)(50)\\&=2+2500-50\\&=2452 \end{align}
Remark:
This is not an AP, if it is an AP, the difference between consecutive terms is a constant.
|
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|
Evaluation of $\lim_{x \to 0} \frac{\sin(x+a) -\sin(a)}{\sin2x}$. I am having trouble proving out the below:
$$\lim_{x \to 0} \frac{\sin(x+a) - \sin(a)}{\sin2x}$$
I have got as far as below using $\sin(a) + \sin(b)$ in numerator and $\sin(2a)$ in the denominator but am not sure how to expand further, or if these are the right rules to apply.
$$\lim_{x \to 0}\frac{\sin(x)\cos(a) + \sin(a)[\cos(x) - 1]}{2\sin(x)\cos(x)}$$
I need to simplify it down to apply $\displaystyle\lim_{x \to 0}\frac{\sin(x)}{x}$.
|
Use the sum to product formula
$$\sin(x)-\sin(y) = 2\cdot \cos\left(\frac{x+y}2\right)\sin\left(\frac{x-y}2\right)$$
Plugging in $x=x+a$ and $y=a$ leads to
$$\begin{align}
\sin(x+a)-\sin(a)~&=~2\cdot \cos\left(\frac{(x+a)+a}2\right)\sin\left(\frac{(x+a)-a}2\right)\\
&=~2\cdot\cos\left(\frac{x}2+a\right)\sin\left(\frac{x}2\right)
\end{align}$$
Subsitute this into the limit yields to
$$\begin{align}
\lim_{x\to 0}\frac{2\cdot\cos\left(\frac{x}2+a\right)\sin\left(\frac{x}2\right)}{2\cdot\sin(x)\cos(x)}~&=~\lim_{x\to 0}\frac{\cos\left(\frac{x}2+a\right)\sin\left(\frac{x}2\right)}{\sin(x)\cos(x)}\\
&=~\lim_{x\to 0}\frac{\cos\left(\frac{x}2+a\right)}{\cos(x)}\cdot\lim_{x\to 0}\frac{\sin\left(\frac{x}2\right)}{x}\\
&=~\cos(a)~\cdot\lim_{h\to 0}\frac{\sin\left(h\right)}{2h}\\
&=~\cos(a)~\cdot\frac12\lim_{h\to 0}\frac{\sin\left(h\right)}{h}\\
&=~\frac{\cos(a)}{2}
\end{align}$$
Where in the end $\frac{x}2$ was substituted by $h$.
|
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|
Showing that $\left|\frac{1}{z^4+1}\right|\leq\frac{1}{1-r^4}$
I am trying to show that if $|z|=r<1$, then
$$\left|\frac{1}{z^4+1}\right|\leq\frac{1}{1-r^4} \ \ \ \ \ \ \ \ \ \ \ \ \ \ (1)$$
I have shown the inequality $$\left|\frac{1}{z^3+1}\right|\leq\frac{1}{1-r^3} \ \ \ \ \ \ \ \ \ \ \ \ \ \ (2)$$ holds under the same conditions, but I am having showing the result of $(1)$.
I considered \begin{align}
|z^4+1|&\geq\left||z^4|-|-1|\right| \\
&=\left||z|^4-1\right| \\
&=\left|r^4-1\right|
\end{align}
Now, $r^4-1$ is positive $\forall r<1$\ $\{0\}$ $(r\neq -1)$. Ideally, like in $(2)$, if $r^4-1$ was strictly negative then the result would immediately follow.
A hint would be very helpful.
|
Recall that
$$\left|\frac{1}{z^4+1}\right|=\frac1{\left|z^4+1\right|}$$
and since
$$\left|z^4+1\right|\ge \left|r^4-1\right|=1-r^4\ge 0$$
the result follows.
|
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"timestamp": "2023-03-29T00:00:00",
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|
Polynomial division : finding the unknown parameters in polynomial via divisibility (Division statement) Here is a question that involves unknowns to be found out in a polynomial and in the divisor:
Find a relation between the constants $m$, $p$ and $q$, such that:
$x^4 + px^2 +q\space$ is divisible by $x^2+mx+1$.
This one I really don't know how approach, except maybe factoring the quadratic generally using the quadratic formula.
{Note that this problem is a pre-calculus problem, so one cannot make use of any Calculus}
|
If the quartic polynomial $x^4 + px^2 + q$ is divisible by a quadratic of the form $x^2 + mx + 1$, so that
$x^4 + px^2 + q = f(x)(x^2 + mx + 1) \tag 1$
for some polynomial $f(x)$, severe restrictions are placed upon $x^4 + px^2 + q$, $x^2 + mx + 1$, and $f(x)$, as is seen below:
First of all, the relation (1) implies that $f(x)$ must itself be a monic quadratic; clearly we have
$\deg f(x) = 2, \tag 2$
and furthermore, writing
$f(x) = lx^2 + ax + b, \tag 3$
we see that the leading coefficient $l$ of $f(x)$, must obey
$l = l(1) = 1, \tag 4$
by comparing coeffiecients $x^4$; therefore we may write
$f(x) = x^2 + ax + b, \tag 5$
and (1) becomes
$ x^4 + px^2 + q = (x^2 + mx + 1)(x^2 + ax + b)$
$= x^4 + ax^3 + bx^2 + mx^3 + am x^2 + bmx + x^2 + ax + b$
$= x^4 + (a + m)x^3 + (b + am + 1)x^2 + (bm + a)x + b; \tag 6$
so, comparing coeffiecients of powers of $x$ on either side we deduce that
$b = q, \tag 7$
$bm + a = 0, \tag 8$
$b + am + 1 = p, \tag 9$
$a + m = 0; \tag{10}$
from (10),
$a = -m; \tag{11}$
using this together with (7) in (9):
$q - m^2 + 1 = p, \tag{12}$
or
$m^2 = q - p + 1; \tag{13}$
from (8) and (11),
$m(b - 1) = mb - m = mb + a = 0; \tag{14}$
so, via (7),
$m(q - 1) = 0; \tag{15}$
we now branch according to the value of $q$; if
$q \ne 1, \tag{16}$
then
$m = 0, \tag{17}$
so by (11),
$a = 0; \tag{18}$
and from (13),
$p = q + 1, \tag{19}$
thus again invoking (7) we find
$x^4 + px^2 + q = x^4 + (q + 1)x^2 + q = (x^2 + 1)(x^2 + q); \tag{15}$
on the other hand, if
$q = 1, \tag{16}$
we have by (12) that
$p = 2 - m^2, \tag{17}$
so (9) and (11) lead us to
$b = p - am - 1 = 2 - m^2 + m^2 - 1 = 1, \tag{18}$
and finally,
$x^4 + px^2 + q = x^4 + (2 - m^2)x^2 + 1 = (x^2 + mx + 1)(x^2 -mx + 1). \tag{19}$
(15) and (19) give the only possible forms of $x^4 + px^2 + q$ and its factorization provided that $x^2 + mx + 1 \mid x^4 + px^2 + q$.
|
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"timestamp": "2023-03-29T00:00:00",
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|
[Integral][Please identify problem] $\displaystyle\int \cfrac{1}{1+x^4}\>\mathrm{d} x$ Here is my attempt. The result is not right. Please help identify the issue(s).
$\displaystyle f(x)=\int\cfrac{1}{x^4+1}\>\mathrm{d}x$, let $x=\tan t$, we have $ \mathrm{d}x = \sec^2 t\>\mathrm{d}t,\> t=\tan^{-1} x\in\left(-\cfrac{\pi}{2},\cfrac{\pi}{2}\right)$
\begin{align}
\displaystyle f(\tan t)&= \int\cfrac{\sec^2 t\> \mathrm{d}t}{1+\tan^4 t}=\int\cfrac{\cos^2 t\> \mathrm{d}t}{\cos^4 t+\sin^4 t}=\int\cfrac{\cfrac{1+\cos 2t}{2}\> \mathrm{d}t}{(\cos^2 t+\sin^2 t)^2-2\sin^2 t\cos^2 t} \notag\\
&=\int\cfrac{1+\cos 2t}{2-\sin^2 2t} \>\mathrm{d}t
=\int\cfrac{\mathrm{d}t}{2-\sin^2 2t} + \cfrac 12\int\cfrac{\mathrm{d}\sin 2t}{2-\sin^2 2t} \notag\\
&=\int\cfrac{\sec^2 2t \>\mathrm{d}t}{2\sec^2 2t-\tan^2 2t} + \cfrac {\sqrt{2}}8\int\cfrac{1}{\sqrt{2}-\sin 2t} + \cfrac{1}{\sqrt{2}+\sin 2t}\>\mathrm{d}\sin 2t \notag\\
&=\cfrac 12\int\cfrac{\mathrm{d}\tan 2t}{2+\tan^2 2t} +\cfrac{\sqrt{2}}{8}\ln \cfrac{\sqrt{2}+\sin 2t}{\sqrt{2}-\sin 2t} \notag\\
&=\cfrac {\sqrt{2}}4 \tan^{-1} \cfrac{\tan 2t}{\sqrt{2}} +\cfrac{\sqrt{2}}{8}\ln \cfrac{\sqrt{2}+\sin 2t}{\sqrt{2}-\sin 2t} \notag
\end{align}
As $\tan 2t=\cfrac{2\tan t}{1-\tan^2 t}=\cfrac{2x}{1-x^2}, \cfrac{\sqrt{2}+\sin 2t}{\sqrt{2}-\sin 2t}=\cfrac{\sqrt{2}\sec^2 t+\tan t}{\sqrt{2}\sec^2 t-\tan t}=\cfrac{\sqrt{2}(x^2+1)+x}{\sqrt{2}(x^2+1)-x},$
$f(x)=\cfrac {\sqrt{2}}4 \tan^{-1} \cfrac{\sqrt{2}x}{1+x^2} +\cfrac{\sqrt{2}}{8}\ln \cfrac{\sqrt{2}(x^2+1)+x}{\sqrt{2}(x^2+1)-x}+c$
If the above holds, $\displaystyle \int_0^{\infty} \cfrac{\mathrm{d} x}{1+x^4}$ would be $0$, which is impossible(Should be $\cfrac {\sqrt{2}\pi}{4}$).
|
Comparing to the method I used in the following, maybe the issue occurs when computing
$$
\int \frac {\mathrm dt} {2-\sin^2(2t)}.
$$
Then from now on $t$ cannot take the value $\pm \pi /4$ if we want to devide the numerator and the denominator by $\cos^2(2t)$. Now to compute the improper integral, we should take the limit $x \to 1^-$ and $x\to 1^+$ separately, since the result is discontinuous at $1$. Fundamental Theorem of Calculus may deduce the wrong result if we apply it to a discontinued antiderivative. So if we use the OP as the antiderivative, we should compute
$$
f(+\infty) - f(1^+) + f(1^-) - f(0),
$$
which would give the right result $\sqrt 2 \pi/4$.
Conclusion: the computation in the OP is right, but when apply it to compute definite integral, we should split the interval at the point $1$.
Appendix
I'm here to give another approach. We would introduce a conjugate pair. Assume $x \neq 0$.
\begin{align*}
\int \frac {\mathrm d x} {1+x^4}
&= \frac 12 \int \frac {1-x^2}{1+x^4} \mathrm dx + \int \frac {1+x^2}{1+x^4}\mathrm dx \\
&= \frac 12 \int \frac {x^{-2} - 1}{x^2+x^{-2 }} \mathrm dx + \frac 12 \int \frac {x^{-2} + 1}{x^2+x^{-2 }} \mathrm dx\\
&= -\frac 12 \int \frac {\mathrm d (x + x^{-1})} {(x+x^{-1} )^2 -2} \mathrm dx+\frac 12 \int \frac {\mathrm d(x-x^{-1})} {(x - x^{-1})^2 +2}\\
&= -\frac {\sqrt 2}8 \int \left( \frac 1 {x +x^{-1}-\sqrt 2} - \frac 1{x+x^{-1}+ \sqrt 2}\right)\mathrm d(x+x^{-1}) \\
&\phantom{==}+\frac {\sqrt2}4 \int \frac {\mathrm d(x - x^{-1})/\sqrt 2} {((x-x^{-1})/\sqrt 2)^2 +1} \\
&= \frac {\sqrt 2}8 \log \left( \frac {x + x^{-1}+\sqrt 2} {x +x^{-1}-\sqrt 2}\right) + \frac {\sqrt 2}4 \mathrm {arctan}\left( \frac {x-x^{-1}} {\sqrt 2}\right) + C \\
&= \frac {\sqrt 2}8 \log \left(\frac {x^2 +\sqrt 2 x + 1} {x^2 - \sqrt2 x +1}\right) +\frac {\sqrt 2}4 \mathrm{arctan} \left(\frac {x^2 -1}{\sqrt 2 x}\right) + C.
\end{align*}
If we use this as the result, then
$$
f(+\infty) - f(0) = \frac {\sqrt 2} 4 \left( \frac \pi 2 + \frac \pi 2\right) = \frac {\sqrt 2}4 \pi.
$$
Also note that when $x \neq 0$,
$$
\arctan(x) + \mathrm{arccot} (x) = \mathrm {sgn} (x)\frac \pi 2 \implies \arctan (x) = \mathrm {sgn} (x)\frac \pi 2 + \arctan \left(-\frac 1x\right),
$$
so the OP is correct.
|
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|
Finding $\lim_{x\to 0} \frac{1 + 1/x}{1 + 1/x^2}$ using definition of a limit I need to either compute $$\lim_{x\to 0} \dfrac{1 + 1/x}{1 + 1/x^2}$$ using definition of a limit or prove it doesn't exist.
My attempt:
Given $\epsilon > 0$, we wish to find $N$ such that $\forall x \geq N$,
$$\left|\dfrac{1 + 1/x}{1 + 1/x^2} - 0\right| < \epsilon.$$
We have
$$\left|\dfrac{1 + 1/x}{1 + 1/x^2} - 0\right| = \left|\dfrac{x^2 + x}{x^2 + 1}\right| \leq \dfrac{x^2 + x}{x} = x + 1 < \epsilon. $$
Thus, we can choose $N \geq \epsilon - 1$ and then whenever $x > N$, we have $\left|f(x)\right| < \epsilon \implies \lim_{x\to 0 } f(x) = 0$
|
Let $f:\mathbb{R}{\setminus}\{0\}\to \mathbb{R}$ be defined by
$$f(x)=\frac{1 + 1/x}{1 + 1/x^2}$$
Claim:$\;\lim_{x\to 0}f(x) = 0$.
Let $\epsilon > 0$, and let $\delta=\min\bigl(1,{\large{\frac{\epsilon}{2}}}\bigr)$.
Suppose $|x| < \delta$, $x\ne 0$. We want to show that $|f(x)| < \epsilon$.
Since $|x| < 1$, it follows that $|x+1| < 2$, hence
\begin{align*}
|f(x)|
&=\left|\frac{1 + 1/x}{1 + 1/x^2}\right|\\[4pt]
&=\left|\frac{x^2+x}{x^2 + 1}\right|\\[4pt]
&=\frac{\left|x^2+x\right|}{\left|x^2+1\right|}\\[4pt]
& < \left|x^2+x\right|\\[4pt]
&=\left|x(x+1)\right|\\[4pt]
&=|x||x+1|\\[4pt]
&< 2\left|x\right|\\[4pt]
& < 2\delta\\[4pt]
&< \epsilon\\[4pt]
\end{align*}
as was to be shown.
|
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|
if $a^2+b^2+c^2+2\rho(ab+bc+ca)\ge0$ then $\rho\ge-1/2$ This may be trivial but I am not able to prove that if $a^2+b^2+c^2+2\rho(ab+bc+ca)\ge0$ for $a,b,c\in\mathbb{R}$ then $\rho\ge-1/2$. Can anybody help me please?
Thanks!
|
Correct me if wrong:
Consider $(a,b,c)$ , then
1) $a^2+b^2+c^2=$
$ (a,b,c)\cdot(a,b,c)=||(a,b,c)||^2$.
2) $ab+bc +ca =$
$(a,b,c) \cdot (b,c,a) =$
$||(a,b,c)|| ||(a,b,c)|| \cos \phi =$
$||(a,b,c)||^2 \cos \phi$.
Combining:
$a^2+b^2+c^2 +2\rho (ab+bc+cd) \ge$
$0$, or
$||(a,b,c)||^2 + 2\rho ||(a,b,c)||^2 \cos \phi \ge 0.$
Assuming $(a,b,c)\not = (0,0,0)$:
$1+ 2\rho \cos \phi \ge 0$;
$\rho \cos \phi \ge -1/2$.
1) $0 \le \cos \phi \le 1$, then
$\rho \ge -1/2$.
2) $-1/2 \le \cos \phi \le 0$, then
$\rho \le 1$.
Note : $\cos \phi \ge -1/2$, since
$(a+b+c)^2 - (a^2+b^2+c^2) = $
$2(ab +ac+bc)$.
We have:
$-||(a,b,c)||^2 \le $
$(a+b+c)^2 -||(a,b,c)||^2 =$
$ 2||(a,b,c)||^2\cos \phi.$
Hence: $-1 \le 2\cos \phi $.
|
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|
You flip a coin $10$ times. How many ways can you get at least $7$ heads?
You flip a coin $10$ times. How many ways can you get at least $7$ heads?
My answer.
$$\binom{10}{10}+ \binom{10}9\cdot\binom{10}1 + \binom{10}8\cdot\binom{10}2+\binom{10}7\cdot\binom{10}3$$
You have $10$ Heads and $0$ tails $+$ $9$ Heads $\cdot$ $1$ Tail $+$ $8$ Heads $\cdot$ $2$ tails $+$ $7$ Heads $\cdot$ $3$ tails.
The answer is $176$ though.
|
Since we need at least $7$ heads from $10$ trails
First we get $7$ heads and $3$ tails in $\dbinom{10}{7}$
Second we get $8$ heads and $2$ tails in $\dbinom{10}{8}$
Third we get $9$ heads and $1$ tail in $\dbinom{10}{9}$
Fourth we $10$ heads and $0$ tails in $\dbinom{10}{10}$
Now total number of permutations $=\dbinom{10}{7}+\dbinom{10}{8}+\dbinom{10}{9}+\dbinom{10}{10}=176$
|
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|
Find the value $\int_{0}^{1}f(x) \,\mathrm dx$ without words
Let $f(x)=ax^3+bx^2+cx+d$ such that: $$f(0)=1,\quad f(0.5)=5, \quad f(1)=15$$
Find: $$\int_{0}^{1}f(x) \,\mathrm dx$$
It is said that it can be solved without words.
|
$f(0) = 1\implies d = 1, f(0.5) = 5 \implies \dfrac{a}{8} +\dfrac{b}{4} + \dfrac{c}{2} = 4, f(1) = 15 \implies 2a+2b+2c = 28\implies \displaystyle \int_{0}^{1} f(x)dx = \dfrac{a}{4}+\dfrac{b}{3}+\dfrac{c}{2}+d= \dfrac{3a+4b+6c}{12}+1=\dfrac{32+28}{12}+1 = 6$
|
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|
How to proof the reason? I have this statement:
If $\frac{a}{b} = \frac{c}{d},$ prove that
$\frac{a+b}{a-b}=\frac{c+d}{c-d}$
I tried to add 1, multiply 1 and nothing.
My development was:
$\frac{a}{b} - \frac{b}{b} = \frac{c}{d} - \frac{d}{d}$
$\frac{a-b}{b} = \frac{c-d}{d}$
$\frac{b}{a-b} = \frac{d}{c-d}$ (I raised to $^{-1}$)
So far I have arrived, without much success.
How can I prove it? Thanks in advance.
|
$$\frac{a}{b} = \frac{c}{d}$$
write $$\frac{a}{b}+1 = \frac{c}{d}+1$$
$$\frac{a+b}{b} = \frac{c+d}{d}$$
also if $$\frac{a}{b} = \frac{c}{d}$$
then $$\frac{b}{a} = \frac{d}{c}$$
write $$1-\frac{b}{a} =1- \frac{d}{c}$$
$$\frac{a-b}{a} = \frac{c-d}{c}$$
we first obtained that $$\frac{a+b}{b} = \frac{c+d}{d}$$
$$\frac{(a+b)a}{(a-b)b} = \frac{(c+d)c}{(c-d)d}$$
also $$\frac{a}{b} = \frac{c}{d}$$
thus
$$\frac{a+b}{a-b}=\frac{c+d}{c-d}$$
|
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|
Natural parameterization of the following curve The given curve is: $ 27 x^{2} = 4 y^{3} $.
I have to find the natural parameterization.
First, I parameterized the curve:
$$ 27 x^2 = 4 y^3 $$
$$ x = \pm \sqrt{\frac{4}{27} y^3} $$
What should I do when I have two possibilities?
I took one possibility, for example: $$ x = \sqrt{\frac{4}{27} y^3} $$
Parameterization: $$ \alpha (t) = (\sqrt{\frac{4}{27} t^3},t) $$
The other steps:
$$ \alpha ' (t) = (\frac{\sqrt{t}}{\sqrt{3}},1) $$
$$ || \alpha ' (t) || = \frac{1}{\sqrt{3}} \sqrt{t+3} $$
$$s(t) = \frac{1}{\sqrt{3}} \int_{0}^{t} \sqrt{u+3} du = \frac{1}{\sqrt{3}}(\frac{2}{3}(t+3)^{\frac{3}{2}}- 2\sqrt{3})$$
Did I go wrong at some point? It became too complicated, so I need help. Thanks!
|
If we write $x=2t^3$ then we have $$\dfrac{y^3}{27}=\dfrac{x^2}{4}=t^6\implies y=3t^2.$$
Thus we get $$\alpha(t)=(2t^3,3t^2),$$ which is a nice parametrization to get the length of the curve.
|
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|
Is $\dfrac{a^2+b^2}{6}\leq \dfrac{a^2+b^2}{3}+\dfrac{ab}{3}\leq \dfrac{a^2+b^2}{2}$ for any $a,b\in\mathbb R$? Is $\dfrac{a^2+b^2}{6}\leq \dfrac{a^2+b^2}{3}+\dfrac{ab}{3}\leq \dfrac{a^2+b^2}{2}$ for any $a,b\in\mathbb R$ ?
For $a$ and $b$ are both positive or both negative,I proved this.
But I am not able to prove for all $a,b\in\mathbb R$ ?
Please help me !
|
Suppose $a \ge 0$
and $b < 0$.
Writing $-b$ for $b$,
this becomes
$\dfrac{a^2+b^2}{6}\leq \dfrac{a^2+b^2}{3}-\dfrac{ab}{3}\leq \dfrac{a^2+b^2}{2}
$
with $a \ge 0, b > 0$.
The right-hand inequality
is obviously true.
The left-hand one is
$ab/3 \le (a^2+b^2)/6$
or
$2ab \le a^2+b^2$
or
$0 \le (a-b)^2$
which is true.
|
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|
Prove that $\frac{1+2\sin{\theta}\cos{\theta}}{\cos^2{\theta}-\sin^2{\theta}}=\frac{1+\tan{\theta}}{1-\tan{\theta}}$ Prove that $$\frac{1+2\sin{\theta}\cos{\theta}}{\cos^2{\theta}-\sin^2{\theta}}=\frac{1+\tan{\theta}}{1-\tan{\theta}}$$
Here's my attempt
$$\require{cancel}\text{Left - Right} = \frac{(1+2\sin{\theta}\cos{\theta})(1-\tan{\theta})-(1+\tan{\theta})(\cos^2{\theta}-\sin^2{\theta})}{(\cos^2{\theta}-\sin^2{\theta})(1-\tan{\theta})}=
\frac{\cancel1\cancel{-\tan{\theta}}+2\sin^2{\theta}\cancel{+2\sin{\theta}\cos{\theta}}\cancel{-\cos^2{\theta}+\sin^2{\theta}}\cancel{-\sin{\theta}\cos{\theta}}\cancel{+\tan{\theta}}\cancel{-\sin{\theta}\cos{\theta}}}{(\cos^2{\theta}-\sin^2{\theta})(1-\tan{\theta})}
=\frac{2\sin^2{\theta}}{(\cos^2{\theta}-\sin^2{\theta})(1-\tan{\theta})}$$
It should be zero, but it isn't? Here's the proof:
$$\text{Left}=\frac{\sin^2{\theta}+\cos^2{\theta}+2\sin{\theta}\cos{\theta}}{\cos^2{\theta}-\sin^2{\theta}}=\frac{(\sin{\theta}+\cos{\theta})^2}{(\cos{\theta}+\sin{\theta})(\cos{\theta}-\sin{\theta})}=\frac{\sin{\theta}+\cos{\theta}}{\cos{\theta}-\sin{\theta}}
\\\text{Right}=\frac{1+\frac{\sin{\theta}}{\cos{\theta}}}{1-\frac{\sin{\theta}}{\cos{\theta}}}=\frac{\cos{\theta}+\sin{\theta}}{\cos{\theta}-\sin{\theta}}
\\{\therefore}\text{Left=Right}$$
|
Note that
$$
1+2\sin\theta\cos\theta=\cos^2\theta+\sin^2\theta+2\sin\theta\cos\theta=
(\cos\theta+\sin\theta)^2
$$
Therefore
\begin{align}
\frac{1+2\sin\theta\cos\theta}{\cos^2\theta-\sin^2\theta}
&=\frac{(\cos\theta+\sin\theta)^2}{(\cos\theta+\sin\theta)(\cos\theta-\sin\theta)}
\\[6px]
&=\frac{\cos\theta+\sin\theta}{\cos\theta-\sin\theta}
\\[6px]
&=\frac{\cos\theta(1+\tan\theta)}{\cos\theta(1-\tan\theta)}
\\[6px]
&=\frac{1+\tan\theta}{1-\tan\theta}
\end{align}
|
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|
Proving $\frac{\sin 30^\circ}{\sin 50^\circ}= \frac{\sin 40^\circ}{\sin 80^\circ}$
Prove that
$$\frac{\sin 30^\circ}{\sin 50^\circ}= \frac{\sin 40^\circ}{\sin 80^\circ}$$
I tried making $\sin 80^\circ=\sin(50^\circ+30^\circ)$, but it didn't go well. I also tried using, maybe, trig periodicities, but I still can't get it.
|
$\frac{\sin 30^\circ}{\sin 50^\circ}= \frac{\sin 40^\circ}{\sin 80^\circ}$
$\frac{\frac{1}{2}}{\sin 50^\circ} = \frac{\sin 40^\circ}{\sin (2(\cdot40^\circ))}$
$\frac{\frac{1}{2}}{\sin 50^\circ} = \frac{\sin 40^\circ}{2 \cdot \sin 40^\circ \cdot \cos 40^\circ }$
$\frac{\frac{1}{2}}{\sin 50^\circ} = \frac{1}{2\cdot \cos 40^\circ }$
$\cos 40^\circ = \sin 50^\circ $
and that's true because $\sin x = \cos(\frac{\pi}{2}-x)$.
|
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|
Evaluate the sum $\sum\limits_{k=1}^n {\binom{n}{k} f(k)}$.
Evaluate the following sum $:$
$$\sum\limits_{k=1}^n \binom{n}{k} f(k),$$ where $f(k)=0$ if $k$ is even and $f(k)=(-1)^{\frac {k-1} {2}}$ if $k$ is odd.
How do I proceed? Please help me in this regard.
Thank you very much.
|
As already shown, complex numbers will come very handy. Here's one trick:
$$(1+i)^{n} = \binom{n}{0} + i\binom{n}{1} -\binom{n}{2} - i \binom{n}{3}... + i^{n} \binom{n}{n}$$
$$(1-i)^{n} = \binom{n}{0} - i\binom{n}{1} -\binom{n}{2} + i \binom{n}{3}... + (-i)^{n} \binom{n}{n}$$
It is easy to check now, (by multiplying both series by $i$ and subtracting the second from the first):
$$\frac{1}{2}i\left ((1-i)^{n} - (1+i)^{n} \right ) = \binom{n}{1} - \binom{n}{3} +\binom{n}{5} - \binom{n}{7}... + ((-1)^n(i)^{n+1}-i^{n+1}) \binom{n}{n}$$
In terms of exponentials:
$$\frac{1}{2}i\left ((1-i)^{n} - (1+i)^{n} \right )=\frac{1}{2}i2^{\frac{n}{2}}(e^{-\frac{i n \pi }{4}}-e^{\frac{i n \pi }{4}})= 2^{\frac{n}{2}}\frac{e^{\frac{i n \pi }{4}}-e^{-\frac{i n \pi }{4}}}{2i}=2^{\frac{n}{2}}sin\left ( \frac{n\pi }{4} \right )$$
As given in the other answer.
|
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|
Maximum value of $n$ for which $8n^3+16n^2+72n+64$ is a perfect cube
What is the maximum value of $n$ for which
$$8n^3+16n^2+72n+64$$
is a perfect cube?
I know how this could be done for a quadratic but how to extend it for a cubic.
|
I shall find all integers (not necessarily positive) $n$ such that $$8n^3+16n^2+72n+64=2^3\,\left(n^3+2n^2+9n+8\right)$$
is the cube of an integer. Clearly, $p(n):=n^3+2n^2+9n+8$ must be a perfect cube. Note that
$$p(n)-(n-1)^3=5n^2+6n+9=\frac{(5n+3)^2+36}{5}>0$$
and
$$(n+3)^3-p(n)=7n^2+18n+19=\frac{(7n+9)^2+52}{7}>0\,.$$
Thus,
$$(n-1)^3<p(n)<(n+3)^3\,.$$
Therefore, $p(n)=n^3$, $p(n)=(n+1)^3$, or $p(n)=(n+2)^3$. In the first case, $p(n)=n^3$ iff $2n^2+9n+8=0$, or equivalently, $n=\dfrac{-9\pm\sqrt{17}}{4}$, none of whose values is an integer. In the second case, $p(n)=(n+1)^3$ iff $n^2-6n-7=0$, or equivalently, $n=-1$ or $n=7$. In the final case, $p(n)=(n+2)^3$ iff $4n^2+3n=0$, which is the same as saying that $n=0$ or $n=-\dfrac{3}{4}$. Consequently, all integer solutions are $n=-1$, $n=0$, and $n=7$.
|
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|
Let $a$, $b$ and $c$ be integers. Prove that if $a^2 \mid b$ and $b^3 \mid c$, then $a^4b^5 \mid c^3$.
Let $a$, $b$ and $c$ be integers. Prove that if $a^2 \mid b$ and $b^3 \mid c$, then $a^4b^5 \mid c^3$.
My (attempted) proof:
Suppose that $a^2 \mid b$ and $b^3 \mid c$, where $a$, $b$, and $c$ are integers.
Therefore, we have that $a^2 k_1 = b$ and $b^3 k_2 = c$ for some $k_1, k_2 \in \mathbb{Z}$.
I'm unsure of how to proceed from here. If I cube $a^2 k_1 = b$ to get $a^6 k_1 = b^3$, and then substitute this into $b^3 \mid c$, then I get $a^6k_1k_2 = c$, which isn't useful. The same happens if I do the other substitution.
I would greatly appreciate it if people could please take the time to clarify how I should be proceeding.
|
I keep your notations and write $a^2 k_1 = b$ and $b^3 k_2 = c$. Then we have
$$c^3=b^9k_2^3=b^2b^7k_2^3=(a^4k_1^2)b^7k_2^3$$
by squaring the first equation and replacing $b^2$. This can be written as
$$c^3=(a^4b^5)k_3$$
where $k_3=b^2k_1^2k_2^3$ is an integer. Hence, $a^4b^5 \mid c^3$.
|
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|
Prove that $1 + \frac{2}{3!} + \frac{3}{5!} + \frac{4}{7!} + \dotsb = \frac{e}{2}$
Prove that $1 + \frac{2}{3!} + \frac{3}{5!} + \frac{4}{7!} + \dotsb = \frac{e}{2}$.
This is problem 4 from page 303 of S.L.Loney's 'Plane Trigonometry'.
It seems fairly obvious that the series expansion $e^x$ will be used. However, I am unsure where to start. Should I consider other series?
|
Another approach is
$$\sinh x=x+\dfrac{x^3}{3!}+\dfrac{x^5}{5!}+\dfrac{x^7}{7!}+\cdots$$
then
$$(x\sinh x)'\Big|_{x=1}=2\left(1+\dfrac{2}{3!}+\dfrac{3}{5!}+\dfrac{4}{7!}+\cdots\right)$$
which gives the result.
|
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|
find angle between two lines between $y-\sqrt{3}x-5=0$ and $\sqrt{3}y-x+6$ how to find to angle between these two lines $y-\sqrt{3}x-5=0$ and $\sqrt{3}y-x+6=0$
i tried so far like this
$y-\sqrt{3}x-5=0$
$y=\sqrt{3}x+5$
in the form of $y=mx+b$
got the value for $m_1=\sqrt{3}$
and for
$\sqrt {3}y-x+6=0$
$y=\dfrac {x-6} {\sqrt {3}}$
$y=\dfrac {1} {\sqrt {3}}x-6$
$m_{2}=\dfrac {1} {\sqrt {3}}$
the formula to find angle between two lines
$\tan \theta =\dfrac {m_{2-}m_{1}} {1+m_{1}m_{2}}$
$\frac{\frac{1}{\sqrt{3}}-\sqrt{3}}{1+\sqrt{3}.\frac{1}{\sqrt{3}}}$
$\frac { - \frac { 2 } { \sqrt { 3 } } } { 1 + 1 }$
$\frac { - \frac { 2 } { \sqrt { 3 } } } { 2 }$
$- \frac { 2 } { 2 \sqrt { 3 } }$
$- \frac { 1 } { \sqrt { 3 } }$
is this right till now and how to find angle after wards . should i use tan inverse of $- \frac { 1 } { \sqrt { 3 } }$ or my algebra calculation is wrong . help me thank you
|
Your calculation is very good so far.
You have found the tangent of the angle between two line to be $$ \tan \theta = \frac {-1}{\sqrt 3}$$
Thus $$ \theta = \tan ^{-1} \frac {-1}{\sqrt 3}=-\pi /6$$
You want a positive value for your angle so you add $\pi$ to get $$ \pi -\pi/6 = 5\pi/6$$ which is $150$ degrees.
|
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"url": "https://math.stackexchange.com/questions/2896035",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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|
How to calculate $\int\frac{x}{x^2-x+1}\, dx$? $$\int \frac{x}{x^2-x+1}\, dx = \int \frac{x}{(x-\frac 1 2)^{2} + \frac 3 4}\, dx = \int \frac{x}{(x-\frac 1 2)^2 + (\frac {\sqrt{3}} {2})^2}$$
Substitute $u= \frac{2x-1}{\sqrt{3}}, du=\frac{2}{\sqrt{3}}dx$:
$$\frac {\sqrt{3}} 2 \int \frac{\frac{\sqrt{3}} {2}u + \frac 1 2}{(\frac{\sqrt{3}}{2}u)^2+(\frac {\sqrt{3}}{2})^2} = \int \frac{u}{u^2+1}du + \frac{1}{\sqrt{3}}\int\frac 1 {u^2+1}du.$$
This gives $$\frac 1 2\log({u^2+1})+\frac{1}{\sqrt{3}}\arctan{u}.$$
Substituting back in x yields $$\frac 1 2\log(\frac 4 3x^2-\frac 4 3 x+ \frac 4 3)+\frac 1 {\sqrt{3}}\arctan(\frac{2x-1}{\sqrt{3}})$$
However, according to Wolfram Alpha, the integral should evaluate to $$\frac 1 2 \log(x^2-x+1)+\frac{1}{\sqrt{3}}\arctan(\frac{2x-1}{\sqrt{3}})$$After working some time on the integral, I know how to reach this solution, but I don't understand why my first attempt didn't arrive at the correct answer. Do you see where I went wrong?
Thank you for any help!
|
$$\frac 1 2\log(\frac 4 3x^2-\frac 4 3 x+ \frac 4 3)+C$$
Can be written as
$$\frac 1 2 \log (\frac 4 3) +\frac 1 2\log( x^2-x+ 1)+C $$
Where first term can be added to the constant
$$\frac 1 2 \log (\frac 4 3) +C $$ write it as new constant $C_1$
|
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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|
Form a quadratic equation with the following details. If $\alpha, \beta$ are the roots of the equation
$
x^2 - px + q = 0
$
and $\alpha_1, \beta_1$ are the roots of the equation $x^2 - qx + p = 0$,
Form the quadratic equation whose roots are
$$
\frac{1}{\alpha_1 \beta} + \frac{1}{\alpha \beta_1} and \frac{1}{\alpha_1 \alpha} + \frac{1}{\beta_1 \beta}
$$
|
By Vieta's formulas:
$$\alpha+\beta=p; \alpha\beta=q;\\
\alpha_1+\beta_1=q; \alpha_1\beta_1=p.$$
The equation to be found: $X^2+BX+C=0$, whose roots must be:
$$\frac{1}{\alpha_1 \beta} + \frac{1}{\alpha \beta_1} \ \ \text{and} \ \ \ \frac{1}{\alpha_1 \alpha} + \frac{1}{\beta_1 \beta}$$
The sum:
$$-B=\frac{1}{\alpha_1 \beta} + \frac{1}{\alpha \beta_1} + \frac{1}{\alpha_1 \alpha} + \frac{1}{\beta_1 \beta}=\frac{1}{\alpha_1}\cdot \frac{\alpha+\beta}{\alpha\beta}+\frac{1}{\beta_1}\cdot \frac{\alpha+\beta}{\alpha\beta}=\frac{\alpha+\beta}{\alpha\beta}\cdot \frac{\alpha_1+\beta_1}{\alpha_1\beta_1}=\cdots=1.$$
The product:
$$C=\left(\frac{1}{\alpha_1 \beta} + \frac{1}{\alpha \beta_1}\right)\left(\frac{1}{\alpha_1 \alpha} + \frac{1}{\beta_1 \beta}\right)=\frac{1}{\alpha_1^2\alpha\beta}+\frac{1}{\beta_1^2\alpha\beta}+\frac{1}{\alpha^2\alpha_1\beta_1}+\frac{1}{\beta^2\alpha_1\beta_1}=\\
\frac{1}{\alpha\beta}\cdot \frac{(\alpha_1+\beta_1)^2-2\alpha_1\beta_1}{\alpha_1^2\beta_1^2}+\frac{1}{\alpha_1\beta_1}\cdot \frac{(\alpha+\beta)^2-2\alpha\beta}{\alpha^2\beta^2}=\cdots=\\
\frac{q^3-2pq+p^3-2pq}{p^2q^2}.$$
Note: The triple dots to be filled by you.
|
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|
Multiplicative inverse of cosets I wish to find the multiplicative inverse of the coset $(1 + x) + (x^2 + x + 1)Q[x]$ in the field $Q[x]/(x^2 + x + 1)Q[x]$ but am not sure how. Do I start with Euclidean Algorithm?
|
Note that every coset of $(x^2+x+1) Q[x]$ is of the form $(ax + b) + (x^2+x+1)Q[x]$ by the division algorithm.
The product of two cosets $p + (x^2+x+1)Q[x]$ and $q + (x^2+x+1)Q[x]$ ($p,q \in Q[x]$) is just $pq + (x^2+x+1)Q[x]$.
Therefore, your task is to find $a$ and $b$ such that $(ax + b) \times (x+1)$ leaves a remainder of $1$ when divided by $(x^2+x+1)Q[x]$, so that the product of those two cosets would be the identity of the field $Q[x]/(x^2+x+1)Q[x]$.
I will actually hide the solution below : you may see if you are stuck.
Now, $(ax+b)(x+1) = ax^2 + (a+b)x+ b = (ax^2+ax+a) + (bx + (b-a))$. By uniqueness of the remainder theorem, the remainder is $bx + (b-a)$ which we want equal to $1$. Consequently, we must have $b = 0$ and $b-a = 1$ so $a = -1$. Therefore , the multiplicative inverse is just the element $(-x) + (x^2+x+1)Q[x]$.
|
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|
Why is $y^2 = 1+x^4$ an elliptic curve? I saw in a document that $y^2 = 1+x^4$ is (the affine equation of) an elliptic curve. Why is it the case? Typically, SAGE tells me it is isomorphic to $y^2 = x^3 - 4x$, which is an elliptic curve with Weierstrass equation, but I don't know how to prove this.
Thank you!
|
A very concrete answer to your question can be found in Exercise 1.15 on page 31 of Silverman and Tate's Rational Points on Elliptic Curves (2nd Edition).
You ask in a comment to your question:
Do all equations of the form $y^2 = \text{quartic}$ give elliptic curves?
The answer is clearly no, since for example the curve $y^2 = x^4$ has a singularity at the origin. Nevertheless, if you add the assumption that the quartic has no repeated roots then the answer is yes, as has already been mentioned in another answer.
In particular, the content of the exercise mentioned above says that if $g(t) \in \mathbb{C}[t]$ is a quartic polynomial, if $\alpha \in \mathbb{C}$ is a root of $g(t)$, and if $\beta \neq 0$ is any number, then the equations
\begin{align*}
x = \frac{\beta}{t - \alpha} \quad \text{and} \quad y = x^2 u = \frac{\beta^2 u}{(t - \alpha)^2}
\end{align*}
give a birational transformation $\phi: \mathcal{Q} \dashrightarrow \mathcal{E}$ between the curve $\mathcal{Q}: u^2 = g(t)$ and the curve $\mathcal{E}: y^2 = f(x)$, where
\begin{align*}
\phi: \mathcal{Q} &\dashrightarrow \mathcal{E}\\
(t, u) &\mapsto (x, y) = \left(\frac{\beta}{t - \alpha}, \frac{\beta^2 u}{(t - \alpha)^2} \right)
\end{align*}
and $$f(x) = g'(\alpha) \beta x^3 + \dfrac{g''(\alpha)}{2!} \beta^2 x^2 + \dfrac{g'''(\alpha)}{3!} \beta^3 x + \dfrac{g^{''''}(\alpha)}{4!} \beta^4$$
is cubic. Moreover, the exercise asks to show that if all the complex roots of $g(t)$ are different, then also the roots of $f(x)$ are distinct, and hence that $\mathcal{Q}: u^2 = g(t)$ is an elliptic curve.
An Example
For instance we can apply this to the curve $u^2 = 1 - t^4$ (the roots of $1 - t^4$ are slightly easier to work with than the roots of $1 + t^4$). In this case $g(t)= 1 - t^4$ has as roots the fourth roots of unity $\pm 1, \pm i$. If we choose $\alpha = 1$ and $\beta = -\dfrac{1}{4}$, then the transformation
\begin{align*}
x = -\frac{1}{4} \frac{1}{t - 1} \quad \text{and} \quad y = \frac{1}{16} \frac{u}{(t - 1)^2}
\end{align*}
gives a birational transformation with the curve
$$
y^2 = f(x) = x^3 - \frac{3}{8}x^2 + \frac{1}{16} x - \frac{1}{256},
$$
which is already in Weierstrass form. Moreover, if you want you can depress the cubic by making the change $x \mapsto X - \dfrac{1}{3}\left( -\dfrac{3}{8} \right) = X + \dfrac{1}{8}$ and $y \mapsto Y$, which gives you the equation
$$
Y^2 = X^3 + \frac{1}{64} X.
$$
|
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|
Prove that if $a+b+c+d=4$, then $(a^2+3)(b^2+3)(c^2+3)(d^2+3)\geq256$
Given $a,b,c,d$ such that $a + b + c + d = 4$ show that $$(a^2 + 3)(b^2 + 3)(c^2 + 3)(d^2 + 3) \geq 256$$
What I have tried so far is using CBS:
$(a^2 + 3)(b^2 + 3) \geq (a\sqrt{3} + b\sqrt{3})^2 = 3(a + b)^2$
$(c^2 + 3)(d^2 + 3) \geq 3(c + d)^2$
$(a^2 + b^2)(c^2 + d^2) \geq (ac + bd)^2$
Then, we have:
$(a^2 + 3)(b^2 + 3)(c^2 + 3)(d^2 + 3) \geq 9(a + b)^2(c + d)^2$.
Thus, we have to prove that $9(a + b)^2(c + d)^2 \geq 256$.
Then, I used the following substitution:
$c + d = t$ and $a + b = 4 - t$.
We assume wlog that $a \leq b \leq c \leq d$.
Then, $4 = a + b + c + d \leq 2(c + d) = 2t$. Thus, $t \geq 2$.
Then, what we have to prove is:
$9t^2(4 - t)^2 \geq 256$.
We can rewrite this as:
$(3t(4 - t) - 16)(3t(4 - t) + 16) \geq 0$, or
$(3t^2 + 2t + 16)(3t^2 - 12t - 16) \geq 0$,
at which point I got stuck.
|
Minimize $f(a,b,c,d)=(a^2 + 3)(b^2 + 3)(c^2 + 3)(d^2 + 3)$ subject to $a+b+c+d=4$.
The Lagrange function: $L(a,b,c,d,t)=f(a,b,c,d)+t(4-a-b-c-d)$.
FOC:
$$\begin{cases}L_a=2a(b^2 + 3)(c^2 + 3)(d^2 + 3)-t=0\\
L_b=2b(a^2 + 3)(c^2 + 3)(d^2 + 3)-t=0\\
L_c=2c(a^2 + 3)(b^2 + 3)(d^2 + 3)-t=0\\
L_d=2d(a^2 + 3)(b^2 + 3)(c^2 + 3)-t=0\\
L_t=4-a-b-c-d=0\\
\end{cases}$$
Consider the difference:
$$L_a-L_b=(c^2+3)(d^2+3)(2ab^2+6a-2ba^2-6b)=0 \iff \\
(2ab-6)(b-a)=0 \Rightarrow 1) \ a=b; \ \ 2) \ ab=3.$$
Similarly, other differences are found:
$$1)\ a=c; \ \ 2) \ ac=3\\
1)\ a=d; \ \ 2) \ ad=3\\
1)\ b=c; \ \ 2) \ bc=3\\
1)\ b=d; \ \ 2) \ bd=3\\
1)\ c=d; \ \ 2) \ cd=3\\$$
Cases:
$$\begin{align} \ &1) \ a=b=c\ne d \Rightarrow ad=bd=cd=3 \Rightarrow 4-\frac 3d-\frac 3d-\frac 3d-d=0 \Rightarrow \emptyset;\\
&2) \ a=b\ne c=d \Rightarrow ac=ad=bc=bd=3 \Rightarrow 4-\frac 3d-\frac3d-d-d=0 \Rightarrow \emptyset \\
&3) \ a=b\ne c\ne d\ne a \Rightarrow ac=bc=cd=ad=bd=3 \Rightarrow 4-\frac 3d-\frac3d-\frac3d-d=0 \Rightarrow \emptyset; \\
&4) \ a=b=c=d \Rightarrow a+b+c+d=4 \Rightarrow a=b=c=d=1.\end{align}$$
The only solution is: $(a,b,c,d)=(1,1,1,1)$.
Now we will check bordered Hessian (let $g(x)=a+b+c+d$):
$$\bar{H}=\begin{vmatrix}
0&g_a&g_b&g_c&g_d\\
g_a&L_{aa}&L_{ab}&L_{ac}&L_{ad}\\
g_b&L_{ba}&L_{bb}&L_{bc}&L_{bd}\\
g_c&L_{ca}&L_{cb}&L_{cc}&L_{cd}\\
g_d&L_{da}&L_{db}&L_{dc}&L_{dd}\\
\end{vmatrix}=\begin{vmatrix}
0&1&1&1&1\\
1&128&64&64&64\\
1&64&128&64&64\\
1&64&64&128&64\\
1&64&64&64&128\\
\end{vmatrix} \Rightarrow \\
\bar{H}_1=-1<0; \ \bar{H}_2=-128<0; \ \bar{H}_3=-12288<0; \ \bar{H}_4=-1048576<0,$$
which implies $f(1,1,1,1)=256$ is minimum.
|
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|
zero Jacobian matrix determinant and local inverse Consider the mapping $f: \mathbb R^2 \backslash \{(0,0)\} \to \mathbb R^2$ given by
$$\begin{aligned}
f(x,y) = \begin{pmatrix}
(x^2-y^2)/(x^2+y^2) \\
xy/(x^2+y^2)
\end{pmatrix}
\end{aligned}$$
Does $f$ have a local inverse at every point of $\mathbb R^2$?
Update: The interesting fact with this function is that we can not use the inverse function theorem to decide whether it is locally injective, since for all points $(x, y)$ we have $Jf (x, y) = 0$. Indeed,
\begin{align}
\det Jf(x,y) = &
\det \begin{pmatrix}
\frac{\partial}{\partial x} \frac{ (x^2-y^2)}{(x^2+y^2)}
&
\frac{\partial}{\partial y} \frac{ (x^2-y^2)}{(x^2+y^2)}
\\
\frac{\partial}{\partial x} \frac{ xy}{(x^2+y^2)}
&
\frac{\partial}{\partial y} \frac{xy}{(x^2+y^2)}
\end{pmatrix}
\\
=&
\det \begin{pmatrix}
\frac{ 2x(x^2+y^2)-(x^2-y^2)2x}{(x^2+y^2)^2}
&
\frac{-2y(x^2+y^2)-(x^2-y^2)2y}{(x^2+y^2)^2}
\\
\frac{y(x^2+y^2)-xy(2x) }{(x^2+y^2)^2}
&
\frac{ x(x^2+y^2)-xy(2y) }{(x^2+y^2)^2}
\end{pmatrix}
\\
=&
\det \begin{pmatrix}
\frac{ 4xy^2}{(x^2+y^2)^2}
&
\frac{-4x^2y}{(x^2+y^2)^2}
\\
\frac{-x^2y+y^3 }{(x^2+y^2)^2}
&
\frac{ x^3-xy^2 }{(x^2+y^2)^2}
\end{pmatrix}
\\
=&
\frac{ 4xy}{(x^2+y^2)^2}
\det \begin{pmatrix}
y
&
-x
\\
\frac{-x^2y+y^3 }{(x^2+y^2)^2}
&
\frac{ x^3-xy^2 }{(x^2+y^2)^2}
\end{pmatrix}
\\
=&
\frac{ 4xy}{(x^2+y^2)^2} \cdot \frac{ 1}{(x^2+y^2)^2}
\det \begin{pmatrix}
y
&
-x
\\
{-x^2y+y^3 }
&
{ x^3-xy^2 }
\end{pmatrix}
\\
=&
\frac{ 4xy}{(x^2+y^2)^2} \cdot \frac{ 1}{(x^2+y^2)^2}
[ y({ x^3-xy^2 }) +x({-x^2y+y^3 })]
\\
=&
\frac{ 4xy}{(x^2+y^2)^2} \cdot \frac{ 1}{(x^2+y^2)^2}
[ x^3y-xy^3-x^3y+xy^3]
\\
=&
\frac{ 4xy}{(x^2+y^2)^2} \cdot \frac{ 1}{(x^2+y^2)^2}
\cdot 0
\\
=&0
\end{align}
|
This map is not locally injective for all points in your domain. For $x = y=t$ we have
\begin{align}
f(t,t)=&
\begin{pmatrix}
0/(t^2+t^2) \\
t^2/(t^2+t^2)
\end{pmatrix}
=
\begin{pmatrix}
0 \\
\frac{1}{2}
\end{pmatrix}
\end{align}
Therefore, this function is not locally invertible in any neighborhood of points $ x = y $.
|
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|
Find angle UFO in the picture attached I sent this problem to Presh Talwalkar who suggested me to send it to this site.
I tried many things but was not able to find the correct solution.
*
*I made various segments trying to get an equilateral triangle similar to the Russian triangle problem, but no success.
*I also tried to flip the triangle UFO over side NO but again no success.
*I tried to find like triangles, but not enough.
Could you please give me a hint?
Thanks,
R. de Souza
|
Without loss of generality, let $OE=1$. By the Law of Sines on the triangle $ONE$, $ON=\dfrac{\sin(40^\circ)}{\sin(60^\circ)}$. Thus, using the Law of Sines on the triangle $ONF$, we get
$$NF=ON\,\left(\frac{\sin(20^\circ)}{\sin(100^\circ)}\right)=ON\,\left(\frac{\sin(20^\circ)}{\sin(80^\circ)}\right)=\frac{\sin(20^\circ)\,\sin(40^\circ)}{\sin(60^\circ)\,\sin(80^\circ)}\,.$$
Furthermore, the Law of Sines on the triangle $OUE$ gives
$$OU=\frac{\sin(35^\circ)}{\sin(65^\circ)}\,.$$
We also have $NE=\dfrac{\sin(80^\circ)}{\sin(60^\circ)}$ (applying the Law of Sines on the triangle $ONE$), which gives
$$NU=NE\,\left(\frac{\sin(5^\circ)}{\sin(115^\circ)}\right)=NE\,\left(\frac{\sin(5^\circ)}{\sin(65^\circ)}\right)=\frac{\sin(80^\circ)\,\sin(5^\circ)}{\sin(60^\circ)\,\sin(65^\circ)}\,,$$
using the Law of Sines on the triangle $UNE$.
Thus,
$$\frac{NU}{NF}=\frac{\sin^2(80^\circ)\,\sin(5^\circ)}{\sin(20^\circ)\,\sin(40^\circ)\,\sin(65^\circ)}\,.\tag{*}$$
Note that
$$\sin(65^\circ)\,\sin(25^\circ)=\frac{1}{2}\,\big(\cos(40^\circ)-\cos(90^\circ)\big)=\frac{\cos(40^\circ)}{2}\,,$$
where we use the identity $\sin(x)\,\sin(y)=\dfrac{1}{2}\,\big(\cos(x-y)-\cos(x+y)\big)$.
Thus, (*) becomes
$$\frac{NU}{NF}=\frac{2\,\sin^2(80^\circ)\,\sin(5^\circ)\,\sin(25^\circ)}{\sin(20^\circ)\,\sin(40^\circ)\,\cos(40^\circ)}\,.$$
From the identity $\sin(2x)=2\,\sin(x)\,\cos(x)$, we get
$$\frac{NU}{NF}=\frac{4\,\sin(80^\circ)\,\sin(5^\circ)\,\sin(25^\circ)}{\sin(20^\circ)}=\frac{4\,\cos(10^\circ)\,\sin(5^\circ)\,\sin(25^\circ)}{\sin(20^\circ)}\,.$$
That is,
$$\frac{NU}{NF}=\frac{4\,\cos(10^\circ)\,\cos(5^\circ)\,\sin(5^\circ)\,\sin(25^\circ)}{\sin(20^\circ)\,\cos(5^\circ)}\,.$$
As $\sin(4x)=2\,\sin(2x)\,\cos(2x)=4\,\sin(x)\,\cos(x)\,\cos(2x)$, we get
$$\frac{NU}{NF}=\frac{\sin(25^\circ)}{\cos(5^\circ)}=\frac{\sin(25^\circ)}{\sin(95^\circ)}\,.$$
Ergo, if $\theta:=\angle UFN$, then we have from the Law of Sines on the triangle $UNF$ that
$$\frac{\sin(\theta)}{\sin(120^\circ-\theta)}=\frac{NU}{NF}=\frac{\sin(25^\circ)}{\sin(120^\circ-25^\circ)}\,.$$
It follows immediately from the identity $\sin(x)\,\sin(y)=\dfrac{1}{2}\,\big(\cos(x-y)-\cos(x+y)\big)$ that
$$\cos(120^\circ+25^\circ-\theta)=\cos(120^\circ-25^\circ+\theta)\,.$$
That is,
$$25^\circ-\theta=n\cdot 180^\circ$$ for some integer $n$. As $0^\circ<\theta<100^\circ$ (because $\angle OFN=100^\circ$), we have $n=0$, whence $\theta=25^\circ$. That is,
$$\angle UFO=180^\circ-25^\circ-80^\circ=75^\circ\,.$$
|
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|
For $f(x,y) = \frac{xy-1}{x^2 y^2-1}$, what is the limit as $(x,y)$ goes to $(1,1)$?
For $f(x,y) = \frac{xy-1}{x^2 y^2-1}$, what is the limit as $(x,y)$ goes to $(1,1)$?
Since the denominator can be factored into $(xy-1)(xy+1)$ and then the $(xy-1)$'s in both the numerator and denominator can be cancelled, we are left with $f(x,y) = \frac{1}{xy+1}$ and substituting $x=1$ and $y=1$, the limit becomes $\frac{1}{2}$. Is this solution right? We had this question on our test todayin Multivariable Caclulus, and some people put $1/2$ while others put $DNE$. Who is right?
|
Consider $X=x-1$ and $Y=y-1$, then we have
$$
\begin{array}{c}
\lim_{(x,y)\to (1,1)}\frac{xy-1}{x^2 y^2-1}=
\lim_{(X,Y)\to (0,0)}\frac{XY-Y-X}{ \left( XY-Y-X+2 \right) \left( XY-Y-X \right)}
=\lim_{(X,Y)\to (0,0)}\frac{1}{ \left( XY-Y-X+2 \right)}=\frac{1}{2}
\end{array}
$$
|
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|
How to find integers $p$ and $q$ such that $(p\sqrt{2}+q)^2=34-24\sqrt{2}$ Find integers $p$ and $q$ such that $(p\sqrt{2}+q)^2=34-24\sqrt{2}$.
I approached this question first by expanding the the left-hand side to get:
$$2p^2 +2\sqrt{2}pq+q^2 = 34-24\sqrt{2}$$
The problem becomes intractable for me at this point
|
We want to find integers $p$ and $q$ such that
$$(p\sqrt{2} + q)^2 = 34 - 24\sqrt{2}$$
Expanding the expression on the left-hand side yields
$$2p^2 + 2pq\sqrt{2} + q^2 = 34 - 24\sqrt{2}$$
Matching rational and irrational parts yields
\begin{align*}
2p^2 + q^2 & = 34 \tag{1}\\
2pq\sqrt{2} & = -24\sqrt{2} \tag{2}
\end{align*}
Solving equation 2 for $q$ yields
$$q = -\frac{12}{p} \tag{3}$$
Substituting $-12/p$ for $q$ in equation 1 yields
\begin{align*}
2p^2 + \left(-\frac{12}{p}\right)^2 & = 34\\
2p^2 + \frac{144}{p^2} & = 34\\
2p^4 + 144 & = 34p^2\\
2p^4 - 34p^2 + 144 & = 0\\
p^4 - 17p^2 + 72 & = 0\\
(p^2 - 9)(p^2 - 8) & = 0\\
(p + 3)(p - 3)(p + 2\sqrt{2})(p - 2\sqrt{2}) & = 0
\end{align*}
which yields the solutions $p = -3, 3, -2\sqrt{2}, 2\sqrt{2}$. The corresponding values for $q$ can be obtained by substituting these values for $p$ into equation 3. Check that the results meet the stated requirements.
|
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}
|
How do I simplify $\frac{\log_7 32}{\log_7 8\cdot\sqrt2}$? So far I have got $\log_7 2^5 - \log_7 2^3 + \log_7 2^{1/2}$ and am unable to proceed. Am I, on the right track so far? and how do I proceed? thank you!
|
You can use the formula
$$\frac{\log_{a}b}{\log_{a}c}=\log_{c}b$$
$$\frac{\log_{7}32}{\log_{7}8}\cdot \sqrt{2}=\left(\log_{8}32\right)\cdot \sqrt{2}=\frac{5\sqrt{2}}{3}$$
Or If you are looking for this
$$\frac{\log_{7}32}{\log_{7}8\sqrt{2}} =\left(\log_{8\sqrt{2}}32\right)=\frac{5}{\frac{7}{2}}=\frac{10}{7}$$
|
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|
Show that the series converges and find its sum Show that
$$ \sum_{n=1}^\infty \left( \frac{1}{n(n+1)} \right) = \frac{1}{2}+\frac{1}{6}+\frac{1}{12}+ \;... $$
converges and find its sum.
My solution so far:
I am thinking about finding the partial sum first and show that the series converges since its finite partial sum converges.
Now
$$ S_N=\sum_{n=1}^N \left( \frac{1}{n(n+1)} \right)=\sum_{n=1}^N \left( \frac{1}{n}-\frac{1}{n+1} \right)=\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\;...= \left( \frac{1}{1}-\frac{1}{2} \right)+\left( \frac{1}{2}-\frac{1}{3} \right) + \;...+ \left( \frac{1}{N}-\frac{1}{N+1} \right)$$
but I don't know how to go on with this. Now $ \lim_{N \to\infty} \left( \frac{1}{N}-\frac{1}{N+1} \right)=0$ but the right answer should be $1$.
|
You can use the Cauchy condensation test to show convergence:
$$\sum_{n=1}^\infty \frac{2^n}{2^n(2^n+1)} = \sum_{n=1}^\infty \frac{1}{2^n+1} \le \sum_{n=1}^\infty \frac{1}{2^n} = 1$$
because the latter is a geometric series.
Hence $\displaystyle\sum_{n=1}^\infty \frac1{n(n+1)}$ also converges.
|
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|
Find the length of the curve y = $\sqrt{-x(x+1)} - \arctan \sqrt{\frac{-x}{x+1}}$ I used the formula for this example:
$\displaystyle L = \int_{a}^{b} \sqrt{1 + \left(\frac{dy}{dx}\right)^2} dx$
And I start by computing the derivative:
\begin{align*}
y' &= \left(\sqrt{-x^2-x}\right)' - \left(\arctan \sqrt{\frac{-x}{x+1}}\right)' \\
&= \frac{1}{2\sqrt{-x^2-x}} \cdot (-2x-1)-\frac{1}{\left(\sqrt{\frac{-x}{x+1}}\right)^2+1} \cdot \frac {1}{2\sqrt{\frac{-x}{x+1}}} \cdot \frac{-1}{(x+1)^2} \\
&= \frac{-2x-1}{2\sqrt{-x^2-x}} - \frac{1}{{\frac{-x}{x+1}}+1} \cdot \frac {1}{2\sqrt{\frac{-x}{x+1}}} \cdot \frac{-1}{(x+1)^2} \\
&= \frac{-2x-1}{2\sqrt{-x^2-x}}+ \frac{1}{2(x+1)^2 \sqrt{\frac{-x}{x+1}}\left(1-\frac{x}{x+1}\right)}
\end{align*}
but I stopped at this moment because it does not seem to me that this derivation was so complicated, how to do it then? Maybe I made a mistake somewhere?
|
I think that you had problems with the simplifications of derivatives. Let $y=A-B$
$$A=\sqrt{-x (x+1)}\implies A'=\frac{-2 x-1}{2 \sqrt{-x (x+1)}}$$
$$B=\tan ^{-1}\left(\sqrt{-\frac{x}{x+1}}\right)\implies B'=\frac{\frac{x}{(x+1)^2}-\frac{1}{x+1}}{2 \sqrt{-\frac{x}{x+1}}
\left(1-\frac{x}{x+1}\right)}=\frac{\sqrt{\frac{1}{x+1}-1}}{2 x}$$
$$y'=A'-B'=-\frac{x}{\sqrt{-x (x+1)}}\implies 1+(y')^2=\frac 1 {1+x}$$
I am sure that you can take it from here.
|
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|
Evaluate $ \lim_{x \to 0} \left( {\frac{1}{x^2}} - {\frac{1} {\sin^2 x} }\right) $ $$\lim_{x\to0}\left({\frac{1}{x^2}}-{\frac{1}{\sin^2x}}\right)$$
Using the L'Hospital Rule I obtained the value $-1/4$, but the answer is given to be $-1/3$. I can't find the mistake. Here's what I did; please point out the mistake.
\begin{align}
\lim_{x\to0}\left({\frac{1}{x^2}}-{\frac{1}{\sin^2x}}\right)&=\lim_{x\to0}\frac{(\sin x+x)(\sin x-x)}{(x\sin x)(x\sin x)} \\[1ex]
&=\lim_{x\to0}\left(\frac{\sin x+x}{x\sin x}\right)\lim_{x\to0}\left(\frac{\sin x-x}{x\sin x}\right) \\[1ex]
&=\lim_{x\to0}\left(\frac{\cos x+1}{\sin x+x\cos x}\right)\lim_{x\to0}\left(\frac{\cos x-1}{\sin x+x\cos x}\right) \\[1ex]
&=\lim_{x\to0}\:(\cos x+1)\,\lim_{x\to0}\left(\frac{\cos x-1}{(\sin x+x\cos x)^2}\right) \\[1ex]
&=\lim_{x\to0}\frac{-\sin x}{(\sin x+x\cos x)(2\cos x-x\sin x)} \\[1ex]
&=-\lim_{x\to0}\left[\frac{1}{1+\cos x\left(\frac{x}{\sin x}\right)}\right]\left(\frac{1}{2\cos x-x\sin x}\right) \\[1ex]
&=-\frac{1}{2}\left[\lim_{x\to0}\,\frac{1}{1+\cos x}\right] \\[1ex]
&=-\frac{1}{4}
\end{align}
|
As an alternative, following the idea by Count Iblis, we have that by Taylor expansion
$$\sin x = x-\frac16 x^3+o(x^3) \implies \frac1{\sin x}=\frac 1x\left(1-\frac16x^2+o(x^2)\right)^{-1}=\frac1x+\frac16x+o(x)$$
therefore
$$\left( {\frac{1}{x^2}} - {\frac{1} {\sin^2 x} }\right)
=\left( {\frac{1}{x}} + {\frac{1} {\sin x} }\right) \left( {\frac{1}{x}} - {\frac{1} {\sin x} }\right)=$$
$$=\left(\frac2x+\frac16x+o(x)\right) \left( -\frac16x+o(x)\right)
=-\frac13+o(1) \to -\frac13$$
|
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|
How to prove this inequality using AM-GM? Suppose $a,b,c$ are positive real numbers. Then prove that $$\Big(\frac{a+b}{2}\Big)\Big(\frac{b+c}{2}\Big)\Big(\frac{c+a}{2}\Big)\ge\Big(\frac{a+b+c}{3}\Big)\Big(abc\Big)^\frac{2}{3}\tag{*}$$
My approach: From AM-GM $$\Big(\frac{a+b}{2}\Big)\ge\Big(ab\Big)^\frac{1}{2}\tag{1}$$
Similarly, $\Big(\frac{b+c}{2}\Big)\ge\Big(bc\Big)^\frac{1}{2}\tag{2}$ and $\Big(\frac{c+a}{2}\Big)\ge\Big(ca\Big)^\frac{1}{2}\tag{3}$
Multiplying $(1), (2)$ and $(3)$, we get $$\Big(\frac{a+b}{2}\Big)\Big(\frac{b+c}{2}\Big)\Big(\frac{c+a}{2}\Big)\ge\Big(abc\Big)\tag{4}$$
Which can be written as $$\Big(\frac{a+b}{2}\Big)\Big(\frac{b+c}{2}\Big)\Big(\frac{c+a}{2}\Big)\ge\Big(abc\Big)^\frac{2}{3}\Big(abc\Big)^\frac{1}{3}\tag{4*}$$
Now, again from AM-GM $$\Big(\frac{a+b+c}{3}\Big)\ge\Big(abc\Big)^\frac{1}{3}\tag{5}$$
But now I am stuck. What should I do to get to $(*)$ ?
|
The inequality
$$ \sum_{sym} a^2 b^1 c^0 \geq \sum_{sym} a^1 b^1 c^1 $$
holds by bunching / Muirhead's inequality and it is equivalent to
$$ \frac{a+b}{2}\cdot\frac{a+c}{2}\cdot\frac{b+c}{2}\geq \frac{a+b+c}{3}\cdot\frac{ab+ac+bc}{3}.$$
The given inequality is weaker than the latter, since $\frac{ab+ac+bc}{3}\geq\left(abc\right)^{2/3}$ holds by AM-GM.
As pointed out in the comments, the given inequality is equivalent to
$$ R^3 r \geq \tfrac{8}{27}\Delta^2 $$
for triangles.
|
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|
Proving $\int_0^1\frac{(1-t)^n}{(1+t)^{n+1}}dt = \frac{1}{2n} + O\left( \frac{1}{n^2}\right)$
The integral $\int_0^1\frac{(1-t)^n}{(1+t)^{n+1}}dt$ pops up when estimating the remainder of the series $\sum_{k} \frac{(-1)^k}k$.
Indeed, by Taylor's theorem with integral remainder, $$\log(1+x) = \sum_{k=1}^n (-1)^{k+1}\frac{x^k}{k} + \int_0^x \frac{(x-t)^n}{n!}\frac{(-1)^nn!}{(1+t)^{n+1}}dt$$
For $x=1$, this yields $$\sum_{k=n+1}^\infty \frac{(-1)^{k+1}}k =(-1)^n\int_0^1\frac{(1-t)^n}{(1+t)^{n+1}}dt$$
Mathematica says that $\displaystyle \int_0^1\frac{(1-t)^n}{(1+t)^{n+1}}dt=\frac{1}{2n} + O\left( \frac{1}{n^2}\right)$. I'm wondering how this can be proved.
Since $t\mapsto \frac{(1-t)^n}{(1+t)^{n+1}}$ converges pointwise to $\mathbb 1_{\{0\}}(t)$, the integral can be estimated by splitting it as $\int_0^{\epsilon_n} + \int_{\epsilon_n}^1$, for some $\epsilon_n$. However choosing the right $\epsilon_n$ seems difficult. I don't have any other approach in mind.
|
Apply the substitution $u = \frac{1-t}{1+t}$ to notice that
$$ \int_{0}^{1} \frac{(1-t)^n}{(1+t)^{n+1}} \, dt
= \int_{0}^{1} \frac{u^n}{1+u} \, du. $$
Now integration by parts proves the desired estimates:
\begin{align*}
\int_{0}^{1} \frac{u^n}{1+u} \, du
&= \left[ \frac{u^{n+1}}{n+1} \cdot \frac{1}{1+u} \right]_{0}^{1}
+ \int_{0}^{1} \frac{u^{n+1}}{n+1} \cdot \frac{1}{(1+u)^2} \, du \\
&= \frac{1}{2(n+1)} + \mathcal{O}\left( \int_{0}^{1} \frac{u^{n+1}}{n+1} \, du \right) \\
&= \frac{1}{2n} + \mathcal{O}\left( \frac{1}{n^2} \right).
\end{align*}
Addendum. Generalizing this observation leads to the following convergent series representation:
\begin{align*}
\int_{0}^{1} \frac{u^n}{1+u} \, du
&= \frac{1}{2} \int_{0}^{1} \frac{u^n}{1 - \frac{1-u}{2}} \, du \\
&= \sum_{k=0}^{\infty} \frac{1}{2^{k+1}} \int_{0}^{1} u^n(1-u)^k \, du \\
&= \sum_{k=0}^{\infty} \frac{1}{2^{k+1}} \cdot \frac{k!}{(n+1)\cdots(n+k+1)}.
\end{align*}
|
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|
Solving an equation with Lambert's W function? Or by any other means? I am trying to solve the following equation for x in terms of $y$ and $c$ (with $x,y \in [0,1]$)
\begin{equation}
\log\left(\frac{x}{1-x-y}\right) + \frac{x}{1-x-y} + \frac{y}{1-x-y} = c
\end{equation}
I can solve this easier equation
\begin{equation}
\log\left(\frac{x}{1-x-y}\right) + \frac{x}{1-x-y} = c
\end{equation}
Let
\begin{equation}
z = \frac{x}{1-x-y}
\end{equation}
Then I can solve for $x$ using Lambert's W function
\begin{align}
\log(z) + z &= c \notag \\
z &= \exp(c)\exp(-z) \notag \\
z \exp(z) &= \exp(c) \notag \\
z &= W(\exp(c)) \notag \\
x &= \frac{(1-y)W(\exp(c))}{1+W(\exp(c))} \notag
\end{align}
Can anyone help me solve the harder equation? Is Lambert's W function helpful here?
Thanks!
|
\begin{align}
\ln\left(\frac{x}{1-x-y}\right)
+\frac{x}{1-x-y} + \frac{y}{1-x-y}
&= c
\tag{1}\label{1}
\end{align}
\begin{align}
\ln\left(\frac{x}{1-x-y}\right)
&= c+\frac{1-x-y-1}{1-x-y}
,\\
\ln\left(\frac{1-y}{1-x-y}-1\right)
&= c+1-\frac{1}{1-x-y}
,\\
\ln\left(\frac{1-y}{1-x-y}-1\right)
&= c-\left(\frac{1}{1-x-y}-1\right)
,\\
\ln\left(\frac{1-y}{1-x-y}-1\right)
&= c-\frac{1}{1-y}\left(\frac{1-y}{1-x-y}-(1-y)\right)
,\\
\ln\left(\frac{1-y}{1-x-y}-1\right)
&= c-\frac{1}{1-y}\left(\frac{1-y}{1-x-y}-(1-y)\right)
,\\
\ln\left(\frac{1-y}{1-x-y}-1\right)
&= c-\frac{y}{1-y}-\frac{1}{1-y}\left(\frac{1-y}{1-x-y}-1\right)
.
\end{align}
Let
\begin{align}
\frac{1-y}{1-x-y}-1&=z
,\\
c-\frac{y}{1-y}&=u
,\\
-\frac{1}{1-y}&=v
\end{align}
and we have an equation
\begin{align}
\ln z&=u+vz
,
\end{align}
which has a standard solution for $z$ in terms Lambert W function
\begin{align}
z&=-\frac{\operatorname{W}(-v\exp(u))}{v}
,\\
x&=(1-y)\left(1-\frac1{1+z}\right)
.
\end{align}
|
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"timestamp": "2023-03-29T00:00:00",
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|
Find this integral $I=\int\frac{x}{(1-3x)^{3/2}(x+1)^{3/2}}dx$ Find the integral
$$I=\int\dfrac{x}{(1-3x)^{3/2}(x+1)^{3/2}}dx$$
My try: $$(1-3x)(x+1)=-3x^2-2x+1=-3(x+1/3)^2+\dfrac{4}{3}$$
Thus
$$I=\int\dfrac{x}{(\frac{4}{3}-3(x+\frac{1}{3})^2)^{3/2}}dx$$
|
Since
$$\dfrac x{(1-3x)(1+x)} = \dfrac14\cdot\dfrac1{1-3x}-\dfrac14\cdot\dfrac1{1+x},$$
then
$$I(x)=\int\dfrac{x\,\mathrm dx}{(1-3x)^{^{^3/_2}}(1+x)^{^{^3/_2}}}
=I_1(x)+I_2(x),\tag1$$
where
$$I_1(x)=\dfrac14\int\dfrac{\mathrm dx}{(1-3x)^{^{^3/_2}}\sqrt{1+x}},\quad
I_2(x) = -\dfrac14\int\dfrac{\mathrm dx}{(1+x)^{^{^3/_2}}\sqrt{1-3x}}.$$
1.
Substitution
$$t=\dfrac1{\sqrt{1-3x}},\quad \mathrm dt=\dfrac32\dfrac{\mathrm dx}{(1-3x)^{^{^3/_2}}},\quad x=\dfrac{t^2-1}{3t^2},$$
allows to get
$$I_1=\dfrac{\sqrt3}6\int\dfrac{t\,\mathrm dt}{\sqrt{4t^2-1}}
=\dfrac{\sqrt3}{24} \sqrt{4t^2-1}+\mathrm{const}
=\dfrac{\sqrt3}{24} \sqrt{\dfrac4{1-3x}-1}+\mathrm{const},$$
$$I_1(x) = \dfrac18\sqrt{\dfrac{1+x}{1-3x}}+\mathrm{const}.\tag2$$
2.
Similarly, substituion
$$t=\dfrac1{\sqrt{1+x}},\quad \mathrm dt=-\dfrac12\dfrac{\mathrm dx}{(1+x)^{^{^3/_2}}},\quad x=\dfrac{1-t^2}{t^2},$$
allows to get
$$I_2 = \dfrac12\int\dfrac{t\,\mathrm dt}{\sqrt{4t^2-3}}
= \dfrac18\sqrt{4t^2-3}+\mathrm{const}
= \dfrac18\sqrt{\dfrac4{1+x}-3}+\mathrm{const},$$
$$I_2(x) = \dfrac18\sqrt{\dfrac{1-3x}{1+x}} + \mathrm{const}.\tag3$$
3.
From $(1)-(3)$ should
$$\color{brown}{\mathbf{I(x) = \dfrac18\sqrt{\dfrac{1+x}{1-3x}} + \dfrac18\sqrt{\dfrac{1-3x}{1+x}} + \mathrm{const}}}.$$
|
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|
Is the sequence $a_{n} = 1 + \frac14 + \frac{2^2}{4^2} + \cdots +\frac{n^2}{4^n}$ Cauchy? I think that it is Cauchy (but I am not sure of this) and this is my proof:
$$|a_{m} - a_{n}| = \left|\frac{n+1}{4^{n+1}} + \frac{n+2}{4^{n+2}} + ..... + \frac{m^2}{m}\right| =\sum_{k=n+1}^{m} \frac{k^2}{4^k}$$
And then knowing that $4^n \geq n^2$ for all $n \geq 4$. (which I will prove by induction)
Then $a_{m} - a_{n} =\sum_{k=n+1}^{m} \frac{k^2}{4^k} \le \sum_{k=n+1}^{m} \frac{k^2}{k^2} \text{(for $n \geq 4$)} = m-n < m$ and we want $a_{m} - a_{n} < \epsilon$, then we want $m<\epsilon$, but $m<\epsilon$ implies $n<\epsilon$ because $m\geq n$. So can I choose $N = \lfloor\epsilon\rfloor +1 $?
Is this a correct choice that shows that my sequence is Cauchy?
I hope that my question is following the instructions of your esteemed, educational and useful site
|
Prove by induction that $k^2 \le 2^k$ for $k \ge 4$ so
$$|a_{m} - a_{n}| \le \sum_{k=n+1}^m \frac{k^2}{4^k} \le \sum_{k=n+1}^\infty \frac1{2^k} = \frac1{2^{n}}\xrightarrow{m,n\to \infty} 0$$
Hence $(a_n)_n$ is Cauchy.
|
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|
Prove that $\sqrt[4]{xyzw} \leq \frac{x+y+z+w}{4}$ for any $x, y, z, w \geq 0$ Prove that
$\sqrt[4]{xyzw} \leq \frac{x+y+z+w}{4}$
For any $x, y, z, \geq 0$
And prove the AM-GM inequality for three numbers, $\sqrt[3]{xyz} \leq \frac{x+y+z}{3}$ where $x, y, z \geq 0$, by using the previous proof with $w= (xyz)^{1/3}$
|
You are on the good track, indeed let consider by $AM-GM$
$$u=\frac{x+y}{2}\ge \sqrt{xy} \quad v=\frac{z+w}{2}\ge \sqrt{zw}$$
then
$$\frac{u+v}{2}= \frac{x+y+z+w}{4}\ge \sqrt{uv}\ge \sqrt{\sqrt{xy}\sqrt{zw}}=\sqrt[4]{xyzw}$$
|
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|
Solution of a quadratic equation to satisfy a constraint I have the roots of a quadratic equation as
$$x = \frac{1 \pm \sqrt{1-4\theta^2}}{2\theta}$$ I know that $|\theta| < 1/2$. Among these two roots, I want to find the one which has a value $|x| < 1$.
My attempt is:
It can't be $\frac{1 + \sqrt{1-4\theta^2}}{2\theta}$ as numerator exceeds one and denominator $2|\theta| < 1$, hence the modulus of the root is greater than one.
If $x = \frac{1 - \sqrt{1-4\theta^2}}{2\theta}$:
\begin{equation}
0 < 1 - 4\theta^2 < 1 \implies 0 < \sqrt{1 - 4\theta^2} < 1.
\end{equation}
\begin{equation}
\left[(1 - \sqrt{1 - 4\theta^2}) - 2|\theta|\right]^2 > 0
\end{equation}
\begin{equation}
1 - \sqrt{1-4\theta^2} - 2|\theta|\sqrt{1-4\theta^2} > 0
\end{equation}
\begin{equation}
\frac{1 - \sqrt{1-4\theta^2}}{2|\theta|} > \sqrt{1-4\theta^2}
\end{equation}
Using the fact that $0 < \sqrt{1 - 4\theta^2} < 1$, left hand side of the inequality has to be less than one.
Is there a more elegant way of finding this?
|
You can use Vieta's formula. We can easly see that $x$ is a solution to $$\theta x^2-x+\theta=0$$ Since you figer out $|x_1|>1$ you can use now:$$x_1\cdot x_2 =1\implies |x_2| = {1\over |x_1|} <1$$
|
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|
For $a\geq2$, $b\geq2$ and $c\geq2$, prove that $\left(a^3+b\right)\left(b^3+c\right)\left(c^3+a\right)\geq125 abc$
For $a\geq2$, $b\geq2$ and $c\geq2$, prove that
$$(a^3+b)(b^3+c)(c^3+a)\geq 125 abc.$$
My try:
First I wrote the inequality as
$$\left(a^2+\frac{b}{a}\right) \left(b^2+\frac{c}{b}\right) \left(c^2+\frac{a}{c}\right) \geq 125. $$
Then I noted that
$$a^2+\frac{b}{a}\geq a^2+\frac{2}{a}, \\ b^2+\frac{c}{b}\geq b^2+\frac{2}{b}, \\ c^2+\frac{a}{c}\geq c^2+\frac{2}{c}. $$
But I don't know how this can help me.
|
$$a^3+b = {a^3\over 4}+{a^3\over 4}+{a^3\over 4}+{a^3\over 4}+b\geq 5\sqrt[5]{a^{12}b\over 2^8}$$
$$b^3+c = {b^3\over 4}+{b^3\over 4}+{b^3\over 4}+{b^3\over 4}+c\geq 5\sqrt[5]{b^{12}c\over 2^8}$$
$$c^3+a = {c^3\over 4}+{c^3\over 4}+{c^3\over 4}+{c^3\over 4}+a\geq 5\sqrt[5]{c^{12}a\over 2^8}$$
So $$\left(a^3+b\right)\left(b^3+c\right)\left(c^3+a\right) \geq 125abc\sqrt[5]{a^8b^8c^8\over 2^{24}}\geq 125abc$$
|
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|
Asymptotics of $\sum _{n \leq x}\sigma_{-2}(n)$ It is known that the sum of the squared inverses of the divisors satisfies $$\sum _{n \leq x} \sigma_{-2}(n) = \zeta(3)x + \mathcal O(1).$$ On the other hand, an alternate calculation gives me another answer:
$$\begin{align*}
\sum _{n \leq x} \sigma_{-2}(n) &= \sum _{a \leq x} \sum _{b \leq \frac xa} \frac 1{b^2}\\
& = \sum _{a \leq x} \left( -\frac ax + \zeta (2) + \mathcal O(\frac {a^2} {x^2}) \right) \\
&= -\frac 1x \left(\frac 12 x^2+ \mathcal O (x)\right) + \zeta(2)x + \frac 1 {x^2}\mathcal O (\sum _{a \leq x }a^2)\\
&= -\frac 12 x + \mathcal O(1) +\zeta(2) x + \frac 13 x + \mathcal O(1).
\end{align*}$$
And this is a contradiction. I cannot spot the error in my calculation. Where did it go wrong?
Proof of known asymptotic
$$\begin{align*}
\sum _{n \leq x} \sigma_{-2}(n) &= \sum _{a \leq x} \frac 1{a^2} \sum _{b \leq \frac xa} 1\\
& = \sum _{a \leq x} \frac 1 {a^2} \left(\frac xa + \mathcal O(1)\right) \\
&= x \sum_{a \leq x }\frac 1 {a^3} + \mathcal O(\sum _{a \leq x} \frac 1 {a^2}) \\&= \zeta(3)x + \mathcal O(1).
\end{align*}$$
|
You are tacitly assuming that a constant is $1$, while that is not the case.
$$ \sum_{b\leq N}\frac{1}{b^2} = \zeta(2)-\frac{1}{N}+\frac{1}{2N^2}-\frac{1}{6N^3}+\frac{1}{30N^5}+\ldots $$
hence by replacing $N$ with $\frac{x}{a}$ we get
$$ \sum_{b\leq\frac{x}{a}}\frac{1}{b^2} = \zeta(2)-\frac{a}{x}+\frac{a^2}{2x^2}-\frac{a^3}{6x^3}+\frac{a^5}{30x^5}+\ldots $$
and summing both sides over $a\leq x$ we have
$$ \sum_{a\leq x}\sum_{b\leq\frac{x}{a}}\frac{1}{b^2} = x\left(\zeta(2)-\frac{1}{2}+\frac{1}{6}-\frac{1}{24}+\frac{1}{180}+\ldots\right)+\mathcal{O}(1) $$
where by the Euler-Maclaurin summation formula the involved constant is
exactly $\zeta(3)$ and not $\zeta(2)-\frac{1}{6}$.
|
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|
Find $a$ so that a line is tangent, secant or external from a sphere I am given the following problem:
Given the line $$r \{ R = (1,0,a) + \lambda [a \quad a \quad 0]$$ and the sphere $$S \{ 8x^2 + 8y^2 +8z^2 - 16x +24y -8z + 19 = 0$$ find, relating to values of $a$, when the line is external, tangent and secant to the sphere.
I completed the square on the sphere's equation for some clarity
$$(x-1)^2 + \left( y + \frac{3}{2} \right)^2 + \left( z - \frac{1}{2} \right)^2 = \frac{9}{8}$$
which gave me a center $C = \left( 1 , - \frac{3}{2}, \frac{1}{2} \right)$ and a radius $r = \frac{3}{2 \sqrt{2}}$.
To evaluate when a line is external or not to a sphere one must relate the distance of the center of the sphere to the line and check if it is greater (or not) than the radius. And my question is: how can I do that given the fact that the line has two variables?
Is the problem not well-made?
|
The distance squared,
writing $c$ for $\lambda$, from
$R
= (1,0,a) + c [a \quad a \quad 0]
= (1+ca,ca,a)
$
to
$C = \left( 1 , - \frac{3}{2}, \frac{1}{2} \right)
$
is
$\begin{array}\\
d^2
&=(1+ca-1)^2+(ca+\frac32)^2+(a-\frac12)^2\\
&=c^2a^2+c^2a^2+3ca+\frac94+a^2-a+\frac14\\
&=2c^2a^2+3ca+\frac52+a^2-a\\
\end{array}
$
For fixed $a$,
since
$\frac{\partial d^2}{\partial c}
=4ca^2+3a
$
and
$\frac{\partial^2 d^2}{\partial c^2}
=4a^2
> 0
$,
the minimum,
unless $a = 0$,
is at
$4ca^2+3a = 0$
or
$c = -\frac{3}{4a}$
when
$\begin{array}\\
d^2
&=2c^2a^2+3ca+\frac52+a^2-a \\
&=2\frac{9}{16a^2}a^2-3\frac{3}{4a}a+\frac52+a^2-a\\
&=\frac98-\frac{9}{4}+\frac52+a^2-a\\
&=\frac98+\frac14+a^2-a\\
&=\frac98+(a-\frac12)^2\\
\end{array}
$
From this,
$d^2 \ge \frac98$
and,
since
$r^2 = \frac98$,
it seems that
the line is never
inside the sphere
for any $a$
and is tangent to the sphere
when $a = \frac12$.
When $a=0$,
the line is just the point
$(1, 0, 0)$
with distance
$d^2 = \frac52$.
|
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|
Any idea how to find $\lim_{x\to 0} \frac{\sqrt{1-\cos(x^2)}}{1-\cos(x)}$? $$\lim_{x\to 0} \frac{\sqrt{1-\cos(x^2)}}{1-\cos(x)}$$
I am trying to solve this limit for 2 days, but still cant find the solution which is $\sqrt{2}$ (that's what is written in the solution sheet)
I tried multiplying with the conjugate, tried with some identities but nothing much because of that $x^2$ in the $\cos$.
Then i tried L'Hopital because it is $\frac{0}{0}$ and still that $\cos$ in the square root is doing problems. I tried on symbolab calculator but it can't solve it.
So can someone help me how to solve this?
Thank you.
|
If you want to go beyond the limit it self, use Taylor series and binomial expansion
$$\cos(x^2)=1-\frac{x^4}{2}+\frac{x^8}{24}+O\left(x^{12}\right)$$
$$1-\cos(x^2)=\frac{x^4}{2}-\frac{x^8}{24}+O\left(x^{12}\right)$$
$$\sqrt{1-\cos \left(x^2\right)}=\frac{x^2}{\sqrt{2}}-\frac{x^6}{24 \sqrt{2}}+O\left(x^{10}\right)$$
$$1-\cos(x)=\frac{x^2}{2}-\frac{x^4}{24}+O\left(x^6\right)$$
$$\frac{\sqrt{1-\cos(x^2)}}{1-\cos(x)}=\frac{\frac{x^2}{\sqrt{2}}-\frac{x^6}{24 \sqrt{2}}+O\left(x^{10}\right) }{ \frac{x^2}{2}-\frac{x^4}{24}+O\left(x^6\right)}$$ Now, long division to get
$$\frac{\sqrt{1-\cos(x^2)}}{1-\cos(x)}=\sqrt{2}+\frac{x^2}{6 \sqrt{2}}+O\left(x^4\right)$$ which shows the limit and how it is approached.
For the fun, use your pocket calculator for $x=\frac \pi 6$. The exact value would be
$$2 \sqrt{2} \left(2+\sqrt{3}\right) \sin \left(\frac{\pi ^2}{72}\right)\approx 1.44244$$ while the above truncated series would give
$$\frac{432+\pi ^2}{216 \sqrt{2}}\approx 1.44652$$
|
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|
Solve this inequality with nested radicals (possibly by induction) I tried to solve this problem by induction but didn't succeed. Given the series
$$ a_n = \sqrt{1+\sqrt{2+\sqrt{3+\sqrt{... + \sqrt{n}}}}}$$
Prove that $a_n < 2 (\forall n \in \mathbb{N^*}) $
Now I thought that maybe I could find a reccurence formula. I haven't found one. Another way I tought of was squaring both sides and substracting the number before the radical n times but that made it more complicated. Can someone lend me a hand on this?
|
Fix $x \ge 0$, and define $(a_n)$ by
$$
a_n
=
\sqrt{x^2 +
\sqrt{x^4 +
\sqrt{x^8 +
\sqrt{\cdots +
\sqrt{x^{2^n}}
}
}
}
}
$$
Then for all $n > 1$, we have the relation
$$a_n^2=x^2+xa_{n-1}$$
Now let $x = \sqrt{5}-1$.
Claim:$\;a_n < 2$, for all $n$.
Proceed by induction on $n$.
For $n=1$, we have $a_1=\sqrt{x^2}=x = \sqrt{5}-1< 2$.
Suppose $a_n < 2$, for some positive integer $n$.
Then we get
$$a_{n+1}^2 = x^2+xa_n < x^2+2x = (\sqrt{5}-1)^2+2(\sqrt{5}-1)= 4$$
so $a_{n+1} < 2$, which completes the induction.
Next, compare $x^{2^n}$ and $n$ . . .
Claim:$\;x^{2^n} > n$, for all $n$.
Proceed by induction on $n$.
For $n=1$, we have $x^{2^1} = x^2 = (\sqrt{5}-1)^2 > 1$.
For $n=2$, we have $x^{2^2} = x^4 = (\sqrt{5}-1)^4 > 2$.
Suppose $x^{2^n} > n$, for some positive integer $n \ge 2$.
Then we get
$$x^{2^{n+1}}=\left(x^{2^n}\right)^2 > n^2 \ge 2n > n+1$$
so $x^{2^{n+1}} > n+1$, which completes the induction.
Hence, for all $n$, we have
$$
\sqrt{1 +
\sqrt{2 +
\sqrt{3 +
\sqrt{\cdots +
\sqrt{n}
}
}
}
} <
\sqrt{x^2 +
\sqrt{x^4 +
\sqrt{x^8 +
\sqrt{\cdots +
\sqrt{x^{2^n}}
}
}
}
} < 2
$$
|
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|
Application of Euler's totient function to find last digits Q:what are the last five digits of the number $2018^{2017^{.^{.^{.^{2^{1}}}}}}$.
My Approach:I know how to find the last two digits of $N=2018^{2017^{k}} $ by$N=2018^{2017^{k}\pmod{\phi(25)}}\pmod {25}\equiv 2018^{2017^{2016^b\pmod{8}}\pmod{20}}\pmod {25}\equiv 2018^{2017^0\pmod{20}}\pmod {25}\equiv 2018^1\pmod {25}\equiv18\pmod{25}$Again,$N=18+25n\equiv0\pmod4$ Suppose n is an arbitrary integer.Solving it to find n.$18+25n\equiv0\pmod4\Rightarrow n=2\pmod4 $ Therefore, we have:$N=25(2)+18\pmod{100}\equiv68\pmod{100}$ By chinese remainder theorem.
but now I couldn't think of how to go with it by $\pmod{100000}$.Any hints or solution will be appreciated.Thanks in advance.
|
Let $\psi(0)=1$ and $\psi (n)=n^{\psi (n-1)} $.
We have to find $\psi (2018)\pmod {10^5} $.
Clearly $\psi (2015)\equiv 0\pmod {5^2} $ and $\psi (2015)\equiv 1\pmod 4$, from which $\psi (2015)\equiv 25\pmod {4\cdot 5^2} $, by chinese remainder theorem.
Consequently, $\psi (2016)\equiv 2016^{25}\equiv 16^{25}\equiv 2^{100}\equiv 1\pmod {5^3}$.
On the other hand, $\psi (2016)\equiv 0\pmod 4$, from which $\psi (2016)\equiv 376\pmod {4\cdot 5^3} $, by chinese remainder theorem.
Consequently, $\psi (2017)\equiv 2017^{376}\equiv 142^{376}\equiv 406\pmod {5^4} $.
On the other hand $\psi (2017)\equiv 1\pmod 4$, from which $\psi(2017)\equiv 2281\pmod {4\cdot 5^4} $, by chinese remainder theorem.
Consequently, $\psi (2018)\equiv 2018^{2281}\equiv 11768\pmod {5^5} $.
On the other hand, $\psi (2018)\equiv 0\pmod {2^5} $, from which $\psi (2018)\equiv 36768\pmod {10^5} $, by chinese remainder theorem.
|
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|
$ \lim_{n \to \infty} \frac{(-1)^n(3-n)}{(3n-5)}$ and ... To find limits:
$(a) \lim_{n \to \infty} \frac{(-1)^n(3-n)}{(3n-5)}$
$(b) \lim_{n \to \infty} \frac{n^3}{n!}$
For the first one the sequence is oscillating so it does not converge.
For the 2nd one I used ratio test: Let $a_n = \frac{n^3}{n!} $, then $\frac{a_{n+1}}{a_n} = \frac{n! \times (n+1)^3}{n^3 \times (n+1)!} = (1+ \frac1n)^3 \times \frac{1}{1+n}$. Thus,
$|\frac{a_{n+1}}{a_n}| < 1$, by ratio test limit is $0$.
Is the solutions correct?
|
Another way for the second is
$$\dfrac{n^3}{n!}<\dfrac{n}{n}\dfrac{n}{n-1}\dfrac{n}{n-2}\dfrac{1}{n-3}<1\dfrac{n}{\frac{n}{2}}\dfrac{n}{\frac{n}{2}}\dfrac{1}{\frac{n}{2}}=\dfrac{8}{n}<\varepsilon$$
for $n>6$.
|
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|
Point of negative inflection
Find the range of values of a for which the function $f(x) = ax^3/3 +(a+2)x^2+(a-1)x + 2$ possesses a negative point of inflection.
My attempt
I differentiated it twice and equated it less than $0$ to get $x < \frac{-(a+2)}{a}$. This isn't giving me any range of $a$. Any hint?
t
|
$f(x) =\frac{a}{3}x^3 +(a+2)x^2+(a-1) x+2$
I think you are supposed to find the intervals for which the function is concave-down? Note that the first derivative of the given function will allow you to find local mins and maxes:
$\frac{df(x)}{dx} = ax^2 + 2(a + 2)x + (a-1) = 0$
$x = \frac{-(a+2) +- \sqrt{4a + 5}}{a}$
Check the concavity (sign of any point in $d^2 f/dx^2$) in the regions
$(-inf.,\frac{-(a+2) - \sqrt{4a + 5}}{a})$,
$(\frac{-(a+2) - \sqrt{4a + 5}}{a},\frac{-(a+2) + \sqrt{4a + 5}}{a})$,
$(\frac{-(a+2) + \sqrt{4a + 5}}{a}, inf)$
Second derivative is (as you found)
$\frac{df^2(x)}{dx^2} = 2ax + 2(a + 2)$
Check the sign of that function when evaluated at any convenient point in those intervals. You want - .
|
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|
Finding $ \int^{1}_{-1}\frac{x^3}{\sqrt{1-x^2}}\ln\bigg(\frac{1+x}{1-x}\bigg)dx$
Finding $\displaystyle \int^{1}_{-1}\frac{x^3}{\sqrt{1-x^2}}\ln\bigg(\frac{1+x}{1-x}\bigg)dx$
Try: Let $$\displaystyle I = \int^{1}_{-1}\frac{x^3}{\sqrt{1-x^2}}\ln\bigg(\frac{1+x}{1-x}\bigg)dx$$ (Function is even )
$$I = 2\int^{1}_{0}\bigg[\ln(1+x)-\ln(1-x)\bigg]\frac{x^3}{\sqrt{1-x^2}}dx$$
Using By parts
$$I = 2\bigg[\ln\bigg(\frac{1+x}{1-x}\bigg)\bigg(\sqrt{1-x^2}-\sqrt[3]{1-x^2}\bigg)\bigg]\bigg|^{1}_{0}+4\int^{1}_{0}\frac{1}{1-x^2}\cdot \bigg(\sqrt{1-x^2}-\sqrt[3]{1-x^2}\bigg)dx$$
Could some help me to solve second Integration, Thanks
|
Since $\int \frac{x^3}{\sqrt{1-x^2}} \, dx = -\frac{1}{3}(2+x^2)\sqrt{1-x^2}$, integration by parts yields
\begin{align*}
\int_{-1}^{1} \frac{x^3}{\sqrt{1-x^2}} \log\bigg(\frac{1+x}{1-x}\bigg) \, dx
&= \int_{-1}^{1} \frac{2(2+x^2)}{3\sqrt{1-x^2}} \, dx.
\end{align*}
Now the substitution $x = \sin\theta$ quickly yields the answer $\frac{5}{3}\pi$.
|
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|
first order Ordinary differential equation help $T''+T'+\lambda T=0$ I stuck with these ODE question.
solve the ODE
$T$ is function of $t$
$T''+T'+\lambda T=0$
$T(0)=0,T'(0)=0,\lambda=n^2 $
when $n\in \Bbb N$
My sulution So far:
Characteristic polynomial
$r^2+r+\lambda=0$
sol:
$r_{1,2}= \frac{-1\pm \sqrt {1-4\lambda} }{2} $
therefor
$T(t)=c_1e^{t\frac{-1+ \sqrt {1-4\lambda} }{2}}+c_2e^{t\frac{-1-\sqrt {1-4\lambda}}{2}}$
$T(0)=0 \rightarrow T(0)=c_1 +c_2 =0 \rightarrow c_1=-c_2 $
$T'(0)=c_1(-\frac{1}{2}+\frac{\sqrt{1-4 \lambda }}{2})+c_2(-\frac{1}{2}-\frac{\sqrt{1-4 \lambda }}{2}) =-1$
remember
$c_1=-c_2 $
$c_1(-\frac{1}{2}+\frac{\sqrt{1-4 \lambda }}{2})-c_1(-\frac{1}{2}-\frac{\sqrt{1-4 \lambda }}{2}) =-1$
$c_1(\sqrt{1-4 \lambda })=-1$
therefor
$c_1=\frac {-1}{\sqrt {1-4 \lambda }}$
and
$c_2=\frac {1}{\sqrt {1-4 \lambda }}$
plug in $c_1,c_2$ at$T(t)$:
$T(t)=\frac {-1}{\sqrt {1-4 \lambda }}e^{t\frac{-1+ \sqrt {1-4\lambda} }{2}}+\frac {1}{\sqrt {1-4 \lambda }}e^{t\frac{-1-\sqrt {1-4\lambda}}{2}}$
plug in $\lambda=n^2$
and we got
$$T(t)=\frac {-1}{\sqrt {1-4 n^2 }}e^{t\frac{-1+ \sqrt {1-4n^2} }{2}}+\frac {1}{\sqrt {1-4 n^2 }}e^{t\frac{-1-\sqrt {1-4n^2}}{2}}$$
how do i get to the solution:
$e^{-\frac{t}{2}}[Asin(\frac{\sqrt {4n^2-1}}{2})+Bcos(\frac{\sqrt {4n^2-1}}{2})]$
from here
thank's
|
I would say, if
$T(t)=C_1e^{t\frac{-1+ \sqrt {1-4n^2} }{2}}+C_2e^{t\frac{-1-\sqrt {1-4n^2}}{2}}$ then is for $n=1,2,3\cdots$
$T(t)=e^{-t/2}\left(C_1e^{\frac{it\sqrt {4n^2-1
} }{2}}+C_2e^{\frac{-it\sqrt {4n^2-1}}{2}}\right), $ i is the imaginary unit
$=e^{-t/2}\left(C_1\cos \frac{t\sqrt {4n^2-1
} }{2}+iC_1\sin \frac{t\sqrt {4n^2-1
} }{2}+C_2\cos \frac{t\sqrt {4n^2-1
} }{2}-iC_2\sin \frac{t\sqrt {4n^2-1
} }{2}\right)=e^{-t/2}\left((C_1+C_2)\cos \frac{t\sqrt {4n^2-1
} }{2}+i(C_1-C2)\sin \frac{t\sqrt {4n^2-1
} }{2}\right)=e^{-t/2}\left(A\cos \frac{t\sqrt {4n^2-1
} }{2}+B\sin \frac{t\sqrt {4n^2-1
} }{2}\right)$
|
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|
A choice of Lyapunov function for this 2D system? I am thinking of a choice of a suitable Lyapunov function$V(x_{1},x_{2})$ which can make the system stable around the fixed point $x_{1}=1,x_{2}=1$
$\dot{x_{1}} = x_{1}x_{2} - x_{1}^2 $
$\dot{x_{2}} = x_{2} - x_{1}x_{2} + 2 -2x_{2}^2$
I thought of using $V(x_{1},x_{2}) = a(x_{1}-1)^2 + b(x_{2}-1)^2$, but then I am unable to show the $\frac{dV}{dt} <0$ but still we have $V(1,1) =0$, which I asked here - Determining $a$ and $b$ such that the expression is negative? and it seems I cannot take such a $V$? any help with the expression of $V(x_{1},x_{2})$?
|
You only need that $\dot V<0$ close to the sink $(1,1)$, so check the expression you originally tried, $V=a(x-1)^2+b(y-1)^2$ with
\begin{align}
\dot V&=2a(x−1)(xy−x^2)+2b(y−1)(y−xy+2−2y^2)\\
&=2a(x−1)x(y-x)+2b(y−1)(y(1-x)+2(1+y)(1-y)\\
&=-2ax(x-1)^2-2b(1+y)(y-1)^2~+~(2ax-2by)(x-1)(y-1)
\end{align}
Thus for $a=b=1$ you get for $(x,y)\approx(1,1)$ that $$\dot V\approx-2(x-1)^2-4(y-1)^2$$
which is negative. The remainder is of size $V^{3/2}$, thus smaller than the quadratic contributions close to $(1,1)$.
In some rough but exact estimate,
\begin{align}
\dot V&=-2(x-1)^2-4(y-1)^2~-~2(x-1)^3-2(y-1)^3+2(x-1)^2(y-1)-2(x-1)(y-1)^2\\
&\le-2V+8V^{3/2}
\end{align}
which is negative for $4V^{1/2}<1$ or $V<\frac1{16}$. Using better estimates that combine some of the third order terms one might get larger bounds,
\begin{align}
\dot V&=-2(x-1)^2-4(y-1)^2~-~2(x-1)^3-2(y-1)^3+2(x-1)^2(y-1)-2(x-1)(y-1)^2\\
&\le-2V+2\Bigl[(x-1)^2+(y-1)^2\Bigr]\bigl(|x-1|+|y-1|\bigr)\\
&\le-2V+2\sqrt2V^{3/2},
\end{align}
which means that the admissible region is $V<\frac12$.
|
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|
Finding the complex square roots of a complex number without a calculator
The complex number $z$ is given by $z = -1 + (4 \sqrt{3})i$
The question asks you to find the two complex roots of this number in the form $z = a + bi$ where $a$ and $b$ are real and exact without using a calculator.
So far I have attempted to use the pattern $z = (a+bi)^2$, and the subsequent expansion $z = a^2 + 2abi - b^2$. Equating $a^2 - b^2 = -1$, and $2abi = (4\sqrt{3})i$, but have not been able to find $a$ and $b$ through simultaneous equations.
How can I find $a$ and $b$ without a calculator?
|
Let$$
z^2=(x+yi)^2=−1+4\sqrt3i,
$$
i.e.$$
(x^2-y^2)+2xyi=−1+4\sqrt3i.
$$
Compare real parts and imaginary parts, $$
\begin{cases}
x^2 - y^2 = -1&\qquad\qquad(1)\\
2xy = 4\sqrt3&\qquad\qquad(2)
\end{cases}
$$
Now, consider the modulus: $|z|^2 =|z^2|$, then
$$x^2 + y^2 = \sqrt{\smash[b]{(-1)^2+(4\sqrt3)^2}} = 7\tag3$$
Solving $(1)$ and $(3)$, we get $x^2 = 3\Rightarrow x = \pm\sqrt3$ and $y^2 = 4\Rightarrow y = \pm2$.
From $(2)$, $x$ and $y$ are of same sign,
$$\begin{cases}
x = \sqrt3\\
y = 2
\end{cases}\text{ or }
\begin{cases}
x = -\sqrt3\\
y = -2
\end{cases}
$$
then$$z = \pm(\sqrt3 + 2i).$$
|
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Find $xyz$ if $x^2+2y^2+2z^2-2x-6y-10z+2xy+2yz+14=0$ The question is
Find the sum of all possible values of $xyz$ given that $x, y, z\in \Bbb Z$ satisfying
$$x^2+2y^2+2z^2-2x-6y-10z+2xy+2yz+14=0.$$
Some thought so far:
Obviously $x$ is even. I assume $x=2k$, but I don't know how to proceed. I tried to complete squares and it became more complicated:
$$(x+y-1)^2+(y+z+4)^2+(z-9)^2=82.$$
WolframAlpha gives $48$ set of solutions, so I think there are ways calculating sum of all possible values of $xyz$ without finding all solutions.
|
Completing the squares:
$$(x+y-1)^2+(y+z-2)^2+(z-3)^2=0 \Rightarrow \\
\begin{cases}x+y-1=0\\ y+z-2=0\\ z-3=0\end{cases} \Rightarrow \\
(x,y,z)=(2,-1,3) \Rightarrow xyz=-6.$$
|
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|
First year college question on divisibility of integers I'm having a hard time with a practice question.
Given $n$ is an integer, prove $2$ divides $(n^4 -3)$ iff $4$ divides $(n^2 +3)$.
So I know since it's an iff statement, I have to show the implication going both ways. Let's start with the left side first.
There exists an integer $r$ such that $2 r = n ^ 4 - 3 $.
Here, I'm thinking in my head how can I get the equation to look like the conclusion, that is $4a = (n^2 + 3)$ for an integer $a$. I see that we can play with the $3$ on both sides.
$$3 = n^4 - 2r$$
$$3 + n^2 = n^4 + n^2 - 2r$$
So here's where I'm stuck. How can I show I can factor out a $4$ out if this right side? Thanks.
|
Proof:
Since the statement is biconditional we must prove the following two statements:
$(a)$ If $2|(n^4 − 3)$ then $4|(n^2 + 3)$
$(b)$ If $4|(n^2 + 3)$ then $2|(n^4 − 3)$
We will begin with statement $(b) ($since it should be the easiest to prove$)$
We will prove this statement directly. Assume that $4|(n^2 + 3)$. This implies that there is an integer $x$ such that
$n^2 + 3 = 4x$. Rearranging we get $n^2 = 4x − 3$. Now if we evaluate $n^4 − 3$ we get
$$n^4=(n^2)^2-3$$$$=(4x-3)^2-3$$$$=(16x^2-24x+9)-3$$$$2(8x^2-12x+3)$$
Notice that $8x^2 − 12x + 3$ is an integer. This implies that $2|(n^4 − 3)$
To prove statement (a) we could approach this directly or by contrapositive. Both directions could prove informative
so both will be presented here.
Direct: Assume that $2|(n^4 − 3)$. This means that there is some integer $y$ such that $n^4 − 3 = 2y$. We wish to prove something about $n^2$ so we will have to (somehow) reduce the power on $n$. To do this, notice that $n^4 = 2y+3 = 2(y+1)+1$.
Since $y + 1$ is an integer, we see that $n^4$ is odd.
If $n^4$ is odd then $n^2$ is odd. We see that $n$ must be odd. Therefore there is an integer $k$ such that $n = 2k + 1$.
Now we can write $$n^2+3=(2k+1)^2+3$$$$=4k^2+4k+1+3$$$$=4(k^2+k+1)$$
Since $k^2+k+1$ is an integer, we see that $4|(n^2+3)$
|
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|
Why does Wolfram|Alpha make a mistake here? We want to evaluate $$\lim_{x \to -8}\frac{\sqrt{1-x}-3}{2+\sqrt[3]{x}}.$$The solving process can be written as follows:\begin{align*}\lim_{x \to -8}\frac{\sqrt{1-x}-3}{2+\sqrt[3]{x}}&=\lim_{x \to -8}\left[\frac{(\sqrt{1-x}-3)(\sqrt{1-x}+3)}{(2+\sqrt[3]{x})(4-2\sqrt[3]{x}+\sqrt[3]{x^2})}\cdot \frac{4-2\sqrt[3]{x}+\sqrt[3]{x^2}}{\sqrt{1-x}+3}\right]\\&=\lim_{x \to -8}\left[\frac{-(x+8)}{x+8}\cdot \frac{4-2\sqrt[3]{x}+\sqrt[3]{x^2}}{\sqrt{1-x}+3}\right]\\&=-\lim_{x \to -8} \frac{4-2\sqrt[3]{x}+\sqrt[3]{x^2}}{\sqrt{1-x}+3}\\&=-2.\end{align*}
But when I input this
lim\frac{\sqrt{1-x}-3}{2+\sqrt[3]{x}} as x to -8
into Wolfram|Alpha, it gives the limit $0$.
Why is Wolfram|Alpha making a mistake here?
|
WolframAlpha understands the expression $\sqrt[3]{x}$ for negative x in a different way than you expect.
Try this: lim\frac{\sqrt{1-x}-3}{2+surd(x,3)} as x to -8
|
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|
$\frac{a^{3}+1}{b+1}+\frac{b^{3}+1}{a+1}$ an integer $\Rightarrow \frac{a^{3}+1}{b+1}$ and $\frac{b^{3}+1}{a+1}$ are integers. I want to show that if the natural numbers $a,b \in \mathbb{N}$ are such that $\frac{a^3+1}{b+1}+\frac{b^3+1}{a+1} \in \mathbb{N}$, then, necessarily, $\frac{a^3+1}{b+1} \in \mathbb{N}$ and $\frac{b^3+1}{b+1} \in \mathbb{N}$.
I have thought the following.
We are given that $\frac{a^3+1}{b+1}+\frac{b^3+1}{a+1} \in \mathbb{N}$.
This means that $\frac{a^3+1}{b+1}+\frac{b^3+1}{a+1}=k, \text{ for some } k\in \mathbb{N}$.
We also have that $b+1 \mid a^3+1$ and $a+1 \mid b^3+1$, right?
So does it suffice to reject the cases that $a^3+1=k_1(b+1)$ and $b^3+1=k_2 (a+1)$ for negative $k_1, k_2$ ? If so, then we pick all the possible combinations and want to get a contradiction from the fact that $\frac{a^3+1}{b+1}+\frac{b^3+1}{a+1} \in \mathbb{N}$, or not?
Or do we show somehow else that $\frac{a^3+1}{b+1} \in \mathbb{N}$ and $\frac{b^3+1}{b+1} \in \mathbb{N}$ ?
|
Hint $\ r\! +\! s,\, rs \in \Bbb Z\,\Rightarrow\, r,s \in \Bbb Z\ $ by applying the Rational Root Test to $\,(x\!-\!r)(x\!-\!s)\in\Bbb Z[x]$
|
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|
Prove central binomial coefficient upper bound I am trying to prove that $\binom{2n}{n} < \frac{4^n}{\sqrt{2n}}$. I tried induction, but with no effect (all I can get to is $(2n+1)(2n+2) < 4\sqrt{n(n+1)}$ which is false)
|
$ n^n e^{-n}\sqrt{2\pi n} < n! < n^n e^{-n} \sqrt{2\pi n} (1+\frac{1}{8n}) $ this is Stirling approximation
So $\binom{2n}{n} \leq \frac{(2n)^{2n} e^{-2n} \sqrt{4\pi n} (1+\frac{1}{16n})}{(n^n e^{-n} \sqrt{2\pi n})^2} = \frac{4^n (n)^{2n} e^{-2n} \sqrt{4\pi n} (1+\frac{1}{16n})}{n^{2n} e^{-2n}* 2\pi n} = \frac{4^n (1+\frac{1}{16n})}{\sqrt{\pi n}} \leq \frac{4^n}{\sqrt{2n}}$.
|
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|
Proving that the sequence space is a metric space in $\mathbb{R}$ I need to show that $(X, d)$ is a metric space:
$X$ is the sequence space where its elements are sequences of real numbers, and $d : X \times X \rightarrow \mathbb{R}$ where d is defined as:
$$d(x,y) = \sum_{j=1}^\infty\frac{1}{2^j} \frac{|x_j - y_j|}{1 + |x_j - y_j|}$$
I think that I need to prove the right $\frac{|x_j - y_j|}{1 + |x_j - y_j|}$ is montone increasing but I have no clue how to start with proving the triangle inequality for this expression. Any help is appreciated.
|
Let $f(x) = \frac{x}{1 + x}$. We wish to show that, for $x, y \ge 0$, we have $f(x + y) \le f(x) + f(y)$ (i.e. the function is subadditive over $[0, \infty)$). I hope you can see how this would prove triangle inequality for the proposed metric.
Suppose $x, y \ge 0$. Consider
\begin{align*}
f(x + y) \le f(x) + f(y) &\iff \frac{x + y}{1 + x + y} \le \frac{x}{1 + x} + \frac{y}{1 + y} \\
&\iff 1 - \frac{1}{1 + x + y} \le 2 - \frac{1}{1 + x} - \frac{1}{1 + y} \\
&\iff \frac{1}{1 + x} + \frac{1}{1 + y} \le 1 + \frac{1}{1 + x + y} \\
&\iff (1 + x + y)(1 + y) + (1 + x + y)(1 + x) \le (2 + x + y)(1 + x)(1 + y) \\
&\iff 1 + x + y \le (1 + x)(1 + y) \\
&\iff xy \ge 0,
\end{align*}
which is true.
You can also use the representation $f(x) = 1 - \frac{1}{1 + x}$ to quickly conclude that $f$ increases over $[0, \infty)$.
|
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|
Maximum and minimum absolute value of a complex number
Let, $z \in \mathbb C$ and $|z|=2$. What's the maximum and minimum value of $$\left| z - \frac{1}{z} \right|? $$
I only have a vague idea to attack this problem.
Here's my thinking :
Let $z=a+bi$
Exploiting the fact that, $a^2+b^2=4$
We get $z-\dfrac{1}{z}=a-\dfrac{a}{4}+i\left(b+\dfrac{b}{4}\right)$
So
$$
\begin{split}
\left|z-\frac{1}{z}\right|
&=\sqrt{\left(a-\dfrac{a}{4}\right)^2+\left(b+\dfrac{b}{4}\right)^2}\\
&=\sqrt{4+\dfrac{1}{4}-\dfrac{a^2}{2}+\dfrac{b^2}{2}}\\
&=\sqrt{4+\dfrac{1}{4}+\dfrac{1}{2}(b^2-a^2)}
\end{split}
$$
The minimum value can be obtained if we can minimize $b^2-a^2$.
Setting $b=0$ gives
the minimum value $\sqrt{2+\dfrac{1}{4}}=\dfrac{3}{2}$
Now, comes the maximum value.
We can write $$\sqrt{4+\dfrac{1}{4}+\dfrac{1}{2}(b^2-a^2)}$$
$$=\sqrt{4+\dfrac{1}{4}+\dfrac{1}{2}(4-2a^2)}$$
$$=\sqrt{4+\dfrac{1}{4}+2-a^2}$$
$$=\sqrt{6+\dfrac{1}{4}-a^2}$$
Setting $a=0$ gives the maximum value $\sqrt{6+\dfrac{1}{4}}=\dfrac{5}{2}$.
I don't know if it's okay to set $b=0$ since $z$ would become a real number then.
|
You could try like this (with help of triangle inequality):
$$|z-{1\over z}|= |{z^2-1\over z}| = {|z^2-1|\over 2} \geq {|z^2|-1\over 2} ={3\over 2}$$
clearly this can be achieved at $z = 2$ and $$ {|z^2-1|\over 2} \leq {|z^2|+1\over 2} = {5\over 2}$$ which can be achieved at $z=2i$.
|
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|
How to find inverse of general curvilinear coordinates Lets say I have a curvilinear coordinate system
$A=A(x,y,z) = \frac{x^2+y^2+z^2}{2z} $, $B=B(x,y,z)= \frac{x^2+y^2+z^2}{2\sqrt{x^2+y^2}}$, $C=C(x,y,z)=\tan^{-1}(y/x)$
How do I find the inverse of those i.e
$x=x(A,B,C)$, $y=y(A,B,C)$, $z=z(A,B,C)$
I know that in cylindrical or spherical coordinates I could do it based on the geometry, but I dont have the geometry, I just have the equations for A, B, and C in terms of x, y, and z
|
I will assume that $x,y,z>0$. You will need to check if it make sense to have them negative.
The first equation we are going to use is $$\tan C=\frac{y}{x}$$
Since everything else we have squares, I will write this as $$y^2=x^2\tan^2C\tag{1}$$
The second equation can be written from the ratio $A/B$ or in fact we can use the square of that:
$$\frac{A^2}{B^2}=\frac{x^2+y^2}{z^2}$$
This yields $$z^2=\frac{B^2}{A^2}(x^2+y^2)\tag{2}$$
The third equation is $$\frac{1}{A^2}+\frac{1}{B^2}=\frac{4(x^2+y^2+z^2)}{(x^2+y^2+z^2)^2}=\frac{4}{x^2+y^2+z^2}$$
I will rewrite this as $$x^2+y^2+z^2=\frac{4}{\frac{1}{A^2}+\frac{1}{B^2}}\tag{3}$$
Now it should be trivial to find $x^2$, $y^2$, and $z^2$ from $(1)$, $(2)$, and $(3)$.
|
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How do I integrate $\int \frac{x^2}{(x^2+9)^2}\, dx$? How do I integrate $\displaystyle \int \frac{x^2}{(x^2+9)^2}\, dx$
?
I tried doing using algebra and solving the question a bit. But it didn't become something that looks solvable.
How should I do this?
|
I think the best thing to do in such a question, should be suitable substitution.
So let
$x=3\tan{\theta}$
$dx=3\sec^2(\theta) d\theta$
$\begin{align} \displaystyle \int \frac{x^2}{(x^2+9)^2}\, dx & = \int \frac{9 \tan^2(\theta) }{(9 \tan^2(\theta) + 9)^2} 3\sec^2(\theta)\, d\theta = \int \frac{27 \tan^2(\theta)\sec^2(\theta)}{81 \sec^4(\theta)} \, d\theta \\ & = \int \frac{1}{3} \frac{ \tan^2(\theta)}{ \sec^2(\theta)}\, d\theta = \frac{1}{3} \int \sin^2(\theta) \, d\theta \\ & = \frac{1}{3} \int \frac{1}{2} - \frac{\cos(2\theta)}{2} \, d\theta = \frac{1}{3} (\frac{1}{2}\theta - \frac{\sin(2\theta)}{4}) + C\\ & = \frac{1}{6} \theta - \frac{2\sin\theta\cos\theta}{12} + C = \frac{1}{6} \theta - \frac{1}{6} \sin{\theta}\cos{\theta} + C \end{align}$
And since $x=3\tan \theta$, we get that:
*
*$\displaystyle \theta = \arctan(\frac{x}{3})$
*$\displaystyle \cos \theta = \frac{3}{\sqrt{x^2+9}}$
*$\displaystyle \sin \theta = \frac{x}{\sqrt{x^2+9}}$
Therefore, turning $\theta$ back into $x$, we get that:
$\displaystyle \begin{align}\int \frac{x^2}{(x^2+9)^2} \, dx & = \frac{1}{6} \theta - \frac{1}{6} \sin{\theta}\cos{\theta} + C \\ & = \frac{1}{6} \arctan(\frac{x}{3}) - \frac{1}{6} \frac{x}{\sqrt{x^2+9}}\frac{3}{\sqrt{x^2+9}} + C \\ & = \frac{1}{6} \arctan(\frac{x}{3}) - \frac{1}{2} \left(\frac{x}{x^2+9}\right)+C\end{align}$
And that should be your final answer.
|
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|
Roots of polynomial equation $x^6+2x^5+4x^4+8x^3+16x^2+32x+64=0$
If $x_1,x_2,...,x_6$ be the roots of $x^6+2x^5+4x^4+8x^3+16x^2+32x+64=0$ then I have to show that $|x_j|=2\space\space\space\forall j\in\{1,2,3,4,5,6\}$
I get that the roots of the equation must be complex, of the form $a+ib$ where $|a+ib|=\sqrt{a^2+b^2}=2$ for all $|x_i|$. I don't get how to find the value of $a+ib$. How do I find the roots?
|
$$x^6+2x^5+4x^4+8x^3+16x^2+32x+64=0$$
Let $z=x/2$, so
$$z^6+z^5+z^4+z^3+z^2+z+1=0$$
from $z \neq 1$
Then $\frac{1-z^7}{1-z}=0 $
so $1-z^7=0$
so $z$ is the $n$ th root of $7$.
from $x=2z$ ,so $|x|=|2z|$ so $|x|=|2z|=2$
|
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|
Sum of $\frac{x^2}{2*1} - \frac{x^3}{3*2} + \frac{x^4}{4*3} - ...$ I have to find the sum of :
$$\frac{x^2}{2*1} - \frac{x^3}{3*2} + \frac{x^4}{4*3} - \frac{x^5}{5*4} +\cdots$$
So far I have :
$$\sum_{n=1}^{\infty} \frac{(-1)^{n-1} \, x^{n+1}}{(n+1)(n)}$$
which is very close to $\ln(1+x)$... but I just can't figure out what I have to do from there.
|
Note that $$\frac {\mathrm d^2}{\mathrm dx^2}\frac{(-1)^nx^n}{n(n-1)}=\frac {\mathrm d}{\mathrm dx}\frac{(-1)^nx^{n-1}}{n-1}=(-1)^nx^{n-2}=(-x)^{n-2}$$
so that we expect $f(x)_=\sum_{n=2}^\infty\frac{(-1)^nx^n}{n(n-1)}$ to be a function with $f''(x)=\sum_{n=0}^\infty (-x)^{n} =\frac1{1+x}$.
|
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|
Problems with an exact differential eqution Consider the following differential equation
$$
\left(\frac{1}{x}-\frac{y^2}{(x-y)^2}\right)dx=\left(\frac{1}{y}-\frac{x^2}{(x-y)^2}\right)dy
$$
I want to find its general solution. I get that this equation is exact but, I trying to solve it for this method and it seems not work. Can someone give me a hit?
|
We have an exact differential equation in the form $Mdx +Ndy = 0$, with
$$M \equiv \frac{1}{x} - \left(\frac{y}{x-y}\right)^2, N \equiv \left(\frac{x}{x-y}\right)^2 - \frac{1}{y}.$$
If we let the implicit solution be $F(x, y) = c$, since the equation is exact, it is known that $F_x = M, F_y = N.$
$$F_x = M \Rightarrow, F = \int \left ( \frac{1}{x} - \left(\frac{y}{x-y}\right)^2 \right )dx,$$
$$\Rightarrow F = \ln|x| + \frac{y^2}{x-y} + \phi(y),$$
where $\phi(y)$ is a function of y.
$F_y = N,$
$$\Rightarrow \frac{y(2x-y)}{(x-y)^2} +\phi'(y) = \frac{x^2}{(x-y)^2} - \frac{1}{y},$$
$$\Rightarrow \phi'(y) = \frac{x^2-2xy + y^2}{(x-y)^2} - \frac{1}{y},$$
$$\Rightarrow \phi'(y) = 1 - \frac{1}{y},$$
$$\phi(y) = y - \ln|y|+C.$$
Thus, the implicit solution is $$F(x, y) = \ln\left|\frac{x}{y}\right| + \frac{xy}{x-y} = c.$$
|
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|
Find polynomial $p(n)$ such that $\displaystyle \sum_{n=1}^{\infty} \big(\sqrt[4]{n^4+2n^2}-\sqrt[3]{p(n)}\big)$ converges
Find a polynomial $p(n)$ such that series:
$$\sum_{n=1}^{\infty} \big(\sqrt[4]{n^4+2n^2}-\sqrt[3]{p(n)}\big)$$
converges.
Attempt. If $p(n)=q^3(n)$ for some polynomial $q(n)$ then
$$\sqrt[4]{n^4+2n^2}-\sqrt[3]{p(n)}=
\sqrt[4]{n^4+2n^2}-q(n)=\frac{n^4+2n^2-q^4(n)}{(\sqrt[4]{n^4+2n^2}+q(n))\,(\sqrt{n^4+2n^2}+q^2(n))}$$
and I thought of getting $q(n)=n$, but in that case
$$\sqrt[4]{n^4+2n^2}-\sqrt[3]{p(n)}\sim \frac{1}{n}$$
and in that case we get divergence, by limit comparison test.
Thanks in advance.
|
$$\sqrt[4]{n^4+2n^2} = n\sqrt[4]{1+\frac{2}{n^2}}= n\left(1+\frac{1}{2n^2}-\frac{3+o(1)}{8n^4}\right) = n+\frac{1}{2n}-\frac{3+o(1)}{8n^3} $$
and $\left(n+\frac{1}{2n}\right)^3 = n^3+\frac{3n}{2}+o(1)$, hence by taking $\color{red}{p(n)=n^3+\frac{3n}{2}}$ we are fine, since
$$\sqrt[3]{n^3+\frac{3n}{2}}=n\sqrt[3]{1+\frac{3}{2n^2}}=n\left(1+\frac{1}{2n^2}-\frac{1+o(1)}{4n^4}\right) = n+\frac{1}{2n}-\frac{1+o(1)}{4n^3}$$
and $\sum_{n\geq 1}\frac{1}{n^3}=\zeta(3)$ is finite, close to $\frac{6}{5}$.
|
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|
Solving $2 \sin x \left( \sqrt{3} \cos x- \sin x \right)= \sqrt{2} - 1$ I wanted to know if this equation could be solved any further please.
$$2 \sin x \left( \sqrt{3} \cos x- \sin x \right)= \sqrt{2} - 1$$
I have gone this far:
$$4 \sin x \sin(60^\circ-x)= \sqrt{2}- 1$$
Thank you
|
Hint:
$$2\sqrt{3} \sin x \cos x- 2\sin^2x = \sqrt3\sin 2x+\cos2x-1=\sqrt{2} - 1.$$
Then
$$3\sin^22x=3(1-\cos^22x)=(\sqrt2-\cos 2x)^2.$$
This is a quadratic equation in $\cos 2x$.
|
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}
|
Prove or disprove each of the follow function has limits $x \to a$ by the definition $\lim_{(x, y) \to (0, 0)} \frac{xy^2}{x^2 + y^2}$ Prove or disprove each of the follow function has limits $x \to a$ by the definition
$\lim_{(x, y) \to (0, 0)} \frac{xy^2}{x^2 + y^2}$
Given $\epsilon > 0$. Choose $\delta = \epsilon$, then $||(x, y) - (0, 0)|| = \sqrt{x^2+y^2}$, and so $\|(x, y) - (0,0)\| < \delta$ implies that
$$\left| \frac{xy^2}{x^2-y^2} - 0\right| = \frac{|xy^2|}{x^2+y^2} \leq \frac{\sqrt{x^2+y^2} \cdot (x^2+y^2)}{x^2 + y^2} = \sqrt{x^2+y^2} < \delta = \epsilon$$
Therefore the limit does exist.
Is this right?
|
You can also use this.
$$\left|{xy^2\over x^2 + y^2 }\right| = |x| {y^2\over x^2 + y^2 } \le |x|. $$
|
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"language": "en",
"url": "https://math.stackexchange.com/questions/2960120",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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|
Determine whether $\sum_{k=1}^{\infty}\frac{k+2}{\sqrt{k^5+4}}$ converges or diverges Question
Determine if
$$\sum_{k=1}^{\infty} \frac{k+2}{\sqrt{k^5+4}}$$
converges or diverges.
I'm looking for a proof that this sum converges in a simpler way than I've shown.
My (ugly) work
I have the following chain of implications for $k\ge 1$:
*
*$k^3(3k^2 - 4k -4) \ge -16 \Longrightarrow$
*$3k^5 + 16 \ge 4k^4 + 4k^3 \Longrightarrow$
*$4k^5 + 16 \ge k^5 + 4k^4 + 4k^3 \Longrightarrow$
*$2(k^5+4)^{1/2} \ge k^{5/2} + 2k^{3/2} \Longrightarrow$
*$2k(k^5+4)^{1/2} \ge (k+2)k^{5/2} \Longrightarrow$
*$\frac{2k}{k^{5/2}} \ge \frac{k+2}{\sqrt{k^5+4}}\Longrightarrow$
*$2\sum_{k=1}^{\infty} k^{-3/2} \ge \sum_{k=1}^{\infty} \frac{k+2}{\sqrt{k^5+4}}$.
Since the sum on the left-hand side converges, the given sum also converges.
Gross!
|
$$\sum_{k=1}^{\infty} \frac{k+2}{\sqrt{k^5+4}} \le \sum_{k=1}^{\infty} \frac{k+2}{\sqrt{k^5}} \le 3+\sum_{k=2}^\infty\frac{2k}{k^{2.5}}=3+2\sum_{k=2}^\infty\frac1{k^{1.5}}$$
Hence it converges.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2960292",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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|
find the answer in terms of $a$ and $b$ only ($a, b$ are roots of $\ x^4 + x^3 - 1 = 0$ If $a$ and $b$ are the two solutions of $\ x^4 + x^3 - 1 = 0$ , what is the solution of $\ x^6 + x^4 + x^3 - x^2 - 1 = 0$ ?
Well I am not able to eliminate or convert $\ x^6$. Please help.
|
Let $f(x) = x^4+x^3-1$ and $g(x) = x^6+x^4 + x^3 - x^2 - 1$.
Let $a, b, c, d$ be the roots of $f(x)$. It is easy to see these four roots are
distinct and differ from zero. If we set $\lambda = a + b$ and $\mu = ab$, we will have $\mu \ne 0$.
Since $a \ne b$ are roots of $f(x)$, $b$ is a root of
$$\begin{align}A(x,a) \stackrel{def}{=}
\frac{f(x)-f(a)}{x-a} &= x^3 + (a+1) x^2 + (a^2+a) x + (a^3+a^2)\\
&= x^3 + \frac{x^2}{a^3} + \frac{x}{a^2} + \frac{1}{a}
\end{align}$$
This implies $\mu$ is a root of
$$
B(x,a) \stackrel{def}{=} a^3 A\left(\frac{x}{a},a\right)
= x^3 + \frac{x^2}{a^2} + x + a^2 = (x^2+1)\left(x+\frac{1}{a^2}\right) - a
$$
Exchange the role of $a,b$, we find $\mu$ is also a root of $B(x,b)$. As a result, $\mu$ is a root of
$$\left(x + \frac{1}{b^2}\right)B(x,a) - \left(x + \frac{1}{a^2}\right)B(x,b)
= -a\left(x + \frac{1}{b^2}\right)
+ b\left(x + \frac{1}{a^2}\right)\\
= \frac{b-a}{a^2b^2}\left[a^2b^2x + (a^2 + ab + b^2)\right]
$$
In terms of $\lambda$ and $\mu$, this leads to
$$\mu^3 + (\lambda^2 - \mu) = 0\quad\iff\quad \lambda^2 = \mu - \mu^3$$
Since $f(a) = f(b) = 0$, we also have
$$\frac{f(a)-f(b)}{a-b} =
\frac{a^4 - b^4 + a^3-b^3}{a-b} =
(a^2+b^2)(a+b) + (a^2+ab + b^2) = 0$$
In terms of $\lambda$ and $\mu$, this is equivalent to
$$(\lambda^2 - 2\mu)\lambda + \lambda^2 - \mu = 0
\quad\iff\quad (\mu + \mu^3)\lambda + \mu^3 = 0
$$
Since $\mu \ne 0$, this leads to
$$\mu^4 = (-\mu^2)^2 = (1+\mu^2)^2\lambda^2
= (1+\mu^2)^2(\mu - \mu^3)$$
Get rid of a non-zero $\mu$ from both sides, we get
$$g(\mu) = \mu^6 + \mu^4 + \mu^3 - \mu^2 - 1 = (\mu^2+1)^2(\mu^2 - 1) + \mu^3 = 0$$
This means $ab = \mu$ is a root of the polynomial $g(x)$.
Swapping the roles of $a,b, c, d$ in suitable order, we can deduce
$ac, ad, bc, bd, cd$ are the other roots of $g(x)$. In short, $g(x)$ has following decomposition:
$$g(x) = (x-ab)(x-ac)(x-ad)(x-bc)(x-bd)(x-cd)$$
|
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"answer_id": 0
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|
How much bigger is 3↑↑↑↑3 compared to 3↑↑↑3? 3↑↑↑3 is already mind-bogglingly large, but how much larger is 3↑↑↑↑3? Is it so large that it is simply around 3↑↑↑↑3 times larger than 3↑↑↑3? Or is there another way to express its magnitude in terms of 3↑↑↑3?
|
For any integers $B>A>1,$ to gauge "how much bigger is $B$ compared to $A,$" consider the following questions :
*
*How many $A$s do we need to add together to reach $B$?
Ans: The least $x$ such that $\underbrace{A+A+...+A}_{x\ A\text{s}}\ge B$, i.e., $A\times x \ge B: \quad x=\left\lceil{B\over A}\right\rceil.$
*How many $A$s do we need to multiply together to reach $B$?
Ans: The least $x$ such that $\underbrace{A\times A\times...A\times A}_{x\ A\text{s}}\ge B$, i.e., $A\uparrow x\ge B:\quad x=\left\lceil{\log B\over \log A}\right\rceil.$
*More generally, for any operation $@$ in the hyperoperation sequence $(+,\times,\uparrow,\uparrow^2,...)$, we can ask for the least $x$ such that $\underbrace{A@ A@...A@A}_{x\ A\text{s}}\ge B$, i.e., $A@'x\ge B,$ where $@'$ is the next hyperoperation after $@$.
Now in the case of $A=3\uparrow^3 3$ and $B=3\uparrow^4 3=3\uparrow^3 A,$ note the following:
$$\begin{align}B &=3\uparrow^4 3\\
&=3\uparrow^3 3\uparrow^3 3\\
&=3\uparrow^3 A\\
&=3\uparrow^2 3\uparrow^3 (A-1)\\
&=3\uparrow^1 3\uparrow^2 (3\uparrow^3 (A-1) - 1)\\
&=3\uparrow^1 3\uparrow^1 3\uparrow^2 (3\uparrow^3 (A-1) - 2)\\
&\\
A&=3\uparrow^3 3\\
&=3\uparrow^2 3\uparrow^3 2\\
&=3\uparrow^1 3\uparrow^2 (3\uparrow^3 2 - 1)\\
&=3\uparrow^1 3\uparrow^1 3\uparrow^2 (3\uparrow^3 2 - 2)\\
\end{align}$$
Then the answer to (1) is $x$ such that $A\times x=B$:
$$\begin{align}x={B\over A}={3^{3\uparrow^2 (3\uparrow^3 (A-1) - 1)} \over 3^{3\uparrow^2 (3\uparrow^3 2 - 1)}}=3^{3\uparrow^2 (3\uparrow^3 (A-1) - 1) - 3\uparrow^2 (3\uparrow^3 2 - 1)}.
\end{align}$$
Similarly, the answer to (2) is $x$ such that $A\uparrow x=B$:
$$\begin{align}x&= {\log B\over \log A}={3^{3\uparrow^2 (3\uparrow^3 (A-1) - 2)} \over 3^{3\uparrow^2 (3\uparrow^3 2 - 2)}}=3^{3\uparrow^2 (3\uparrow^3 (A-1) - 2) - 3\uparrow^2 (3\uparrow^3 2 - 2)}.
\end{align}$$
Furthermore, we can give a lower bound on the least $x$ such that $A\uparrow^3 x\ge B$, as we have by Saibian's inequality,
$$(3\uparrow^3 3)\uparrow^3 (3\uparrow^3 3\,-\,3)\ < \ (3)\uparrow^3 (3\uparrow^3 3\,-\,3\,+\,3)\ =\ 3\uparrow^4 3;$$ that is, $$A\uparrow^3 (A-3)=\underbrace{A\uparrow^2 A\uparrow^2 ...A\uparrow^2 A}_{(A-3)\ A\text{s}}< B.$$
Therefore, $A\uparrow^2 A\uparrow^2 ...A\uparrow^2 A\ge B$ requires more than $(A-3)$ $A$s on the left-hand side.
(Of course, this is also a lower bound on the least $x$ such that $\underbrace{A\uparrow A\uparrow ...A\uparrow A}_{x\ A\text{s}}=A\uparrow^2 x\ge B.$)
|
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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|
Compute : $1 + 1/3 - 1/2 + 1/5 + 1/7 - 1/4 + 1/9 + \cdots$
Compute : $1 + 1/3 - 1/2 + 1/5 + 1/7 - 1/4 + 1/9 + \cdots$
My try
It can be verified that $\lim_{k \to \infty} S_{3k} < + \infty$ and $\lim_{k \to \infty} S_{3k} = \lim_{k \to \infty} S_{3k+1} = \lim_{k \to \infty} S_{3k+2}$.
So letting $a_n := S_{3n}$, $a_{n+1} - a_n = \frac{1}{4n+1} + \frac{1}{4n+3} - \frac{1}{2n + 2} = \frac{8k+5}{(4k+1)(4k+3)(2k+2)}$.
Since $\lim (a_{n+1} - a_n) = \lim S_n - a_1$, suffice to compute $\lim_{n\to\infty}(a_{n+1} - a_n)$.
$$
\begin{aligned}
\lim_{n \to \infty} (\frac{1}{4n+1} + \frac{1}{4n+3} - \frac{1}{2n + 2}) &=\lim_{n \to \infty} (\frac{1}{4n+1}) + \lim_{n \to \infty} (\frac{1}{4n+3}) - \lim_{n \to \infty} (\frac{1}{2n+2}) \\
&= \frac{5}{6}
\end{aligned}
$$
And $a_1 = S_3 = 5/6$, thus $\lim S_n = 5/3$. Am I right?
|
The sum in the question can be re-written as
$$
\begin{align}
\sum_{k=1}^\infty\left(\frac1{4k-3}+\frac1{4k-1}-\frac1{2k}\right)
&=\lim_{n\to\infty}\sum_{k=1}^n\left(\frac1{4k-3}+\frac1{4k-1}-\frac1{2k}\right)\\
&=\lim_{n\to\infty}\left(\sum_{k=1}^{4n}\frac1k-\sum_{k=1}^{2n}\frac1{2k}-\sum_{k=1}^n\frac1{2k}\right)\\
&=\frac12\lim_{n\to\infty}\left(\sum_{k=1}^{4n}\frac1k-\sum_{k=1}^{2n}\frac1k\right)+\frac12\lim_{n\to\infty}\left(\sum_{k=1}^{4n}\frac1k-\sum_{k=1}^n\frac1k\right)\\
&=\frac12\lim_{n\to\infty}\sum_{k=2n+1}^{4n}\frac nk\frac1n+\frac12\lim_{n\to\infty}\sum_{k=n+1}^{4n}\frac nk\frac1n\\
&=\frac12\int_2^4\frac1x\,\mathrm{d}x+\frac12\int_1^4\frac1x\,\mathrm{d}x\\[6pt]
&=\frac12\log(2)+\frac12\log(4)\\[9pt]
&=\frac32\log(2)
\end{align}
$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2967089",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 1
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|
Finding dependencies such that $0 > \frac{2b^2r^2}{z}-\left(2r ^2-2br\sqrt{1-\frac{b^2}{z^2}}\right)z$ I'm trying to solve an inequality with 3 variables.
$$0 > \frac{2 b^2 r^2}{z} - \left(2 r ^2 - 2 b r \sqrt{1 - \frac{b^2}{z^2}}\right) z$$
Basically, I want to know under which dependencies the formula is less than zero.
I tried to transform it in many ways, but it seems I cannot get a nice result. Especially the root seems to make problems:
$$2r^2z - 2brz \sqrt{1 - \frac{b^2}{z^2}} > \frac{2b^2r^2}{z} \tag{1}$$
$$rz - bz \sqrt{1 - \frac{b^2}{z^2}} > \frac{b^2r}{z} \tag{2}$$
$$r - b \sqrt{1 - \frac{b^2}{z^2}} > \frac{b^2r}{z^2} \tag{3}$$
I know that all variables are > 0.
So: $$r > 0 \qquad b > 0 \qquad z > 0$$ I also know that $$b \leq z$$
Do you have a hint, what I can try?
Do you think it is possible to calculate a nice solution, in which relation the 3 variables have to be, such that the formula is negative?
Thank you very much.
|
The final solution comes out to be very nice, so I suspect this is a homework problem. As such, I will just give you an outline.
$$0 < \frac{2 b^2 r^2}{z} - \left(2 r ^2 - 2 b r \sqrt{1 - \frac{b^2}{z^2}}\right) z$$
Simplifying terms, dividing both sides by $z$, letting $\frac{b}{z}=\alpha$ and pulling the square root to one side,
$$b \sqrt{1 - \alpha^2} > r-r\alpha^2$$
Now just factor the square root from both sides and then square both sides. You should end up with
$$a^2+b^2>r^2$$
|
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"timestamp": "2023-03-29T00:00:00",
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|
Show that $\left(1+\frac{1}{1^3}\right)\left(1+\frac{1}{2^3}\right)\left(1+\frac{1}{3^3}\right)\cdots\left(1+\frac{1}{n^3}\right) < 3$ I have this problem which says that for any positive integer $n$, $n \neq 0$ the following inequality is true: $$\left(1+\frac{1}{1^3}\right)\left(1+\frac{1}{2^3}\right)\left(1+\frac{1}{3^3}\right)\cdots\left(1+\frac{1}{n^3}\right) < 3$$
This problem was given to me in a lecture about induction but any kind of solution would be nice.And also I'm in 10th grade :)
|
Less Than $\boldsymbol{3}$
The inequality
$$
1+\frac1{n^3}\lt\frac{1+\frac1{2(n-1)^2}}{1+\frac1{2n^2}}\tag1
$$
can be verified by cross-multiplying and then multiplying both sides by $2n^5(n-1)^2$; that is,
$$
2n^7-4n^6+3n^5\underbrace{-3n^3+3n^2-2n+1}_\text{$-(3n^2+2)(n-1)-1\lt0$ for $n\ge1$}\lt2n^7-4n^6+3n^5\tag2
$$
Therefore, employing a telescoping product,
$$
\begin{align}
\prod_{n=1}^\infty\left(1+\frac1{n^3}\right)
&\lt2\prod_{n=2}^\infty\frac{1+\frac1{2(n-1)^2}}{1+\frac1{2n^2}}\\
&=2\cdot\frac32\\[9pt]
&=3\tag3
\end{align}
$$
Actual Value
$$
\begin{align}
\lim_{n\to\infty}\prod_{k=1}^n\frac{k^3+1}{k^3}
&=\lim_{n\to\infty}\frac{\Gamma(n+2)\,\Gamma\!\left(n+\frac12+i\frac{\sqrt3}2\right)\Gamma\!\left(n+\frac12-i\frac{\sqrt3}2\right)}{\Gamma(2)\,\Gamma\!\left(\frac12+i\frac{\sqrt3}2\right)\Gamma\!\left(\frac12-i\frac{\sqrt3}2\right)\Gamma(n+1)^3}\tag4\\
&=\frac1{\Gamma\!\left(\frac12+i\frac{\sqrt3}2\right)\Gamma\!\left(\frac12-i\frac{\sqrt3}2\right)}\\
&\times\lim_{n\to\infty}\frac{\Gamma(n+2)\,\Gamma\!\left(n+\frac12+i\frac{\sqrt3}2\right)\Gamma\!\left(n+\frac12-i\frac{\sqrt3}2\right)}{\Gamma(n+1)^3}\tag5\\
&=\frac{\sin\left(\frac\pi2+i\frac{\pi\sqrt3}2\right)}{\pi}\times1\tag6\\[6pt]
&=\frac{\cosh\left(\frac{\pi\sqrt3}2\right)}{\pi}\tag7
\end{align}
$$
Explanation:
$(4)$: $\prod\limits_{k=1}^n(k+x)=\frac{\Gamma(n+1+x)}{\Gamma(1+x)}$ and $k^3+1=(k+1)\left(k-\frac12+i\frac{\sqrt3}2\right)\left(k-\frac12-i\frac{\sqrt3}2\right)$
$(5)$: pull out the constant factor using $\Gamma(2)=1$
$(6)$: apply Euler's Reflection Formula $\Gamma(x)\,\Gamma(1-x)=\frac\pi{\sin(\pi x)}$
$\phantom{(6)\text{:}}$ and Gautschi's Inequality, which implies $\lim\limits_{n\to\infty}\frac{\Gamma(n+x)}{\Gamma(n)\,n^x}=1$
$(7)$: $\cos(ix)=\cosh(x)$
|
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"timestamp": "2023-03-29T00:00:00",
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|
Proof for a formula of a special real determinant For some $x_1,\cdots,x_n$ in $\mathbb{R}$ we define:
$$A_n:=
\left(\begin{matrix}
1 &x_1 &x_1^2 &\cdots &x^{n-1}_1\\
1 &x_2 &x_2^2 &\cdots &x_2^{n-1}\\
\vdots &\vdots &\vdots & &\vdots\\
1 &x_n &x_n^2 &\cdots &x^{n-1}_n
\end{matrix}\right)
\in \mathbb{R}^{n\times n}
$$
I now need to show that:
$$\det(A_n)=\prod_{1 \le i < j \le n}(x_j-x_i)$$
I tried subtracting the $n$-th row from all others to get:
$$\det(A_n)=\det\left(\begin{matrix}
0 &x_1-x_n &x_1^2-x_n^2 &\cdots &x^{n-1}_1-x_n^{n-1}\\
0 &x_2-x_n &x_2^2-x_n^2 &\cdots &x_2^{n-1}-x_n^{n-1}\\
\vdots &\vdots &\vdots & &\vdots\\
0 &x_{n-1}-x_n &x_{n-1}^2-x_n^2 &\cdots &x_{n-1}^{n-1}-x_n^{n-1}\\
1 &x_n &x_n^2 &\cdots &x^{n-1}_n
\end{matrix}\right)$$
With the usage of Laplaces Expansion utilising the numerous $0$s I was able to write:
$$
\begin{align*}
\det(A_n)
&=(-1)^{n+1}\cdot1\cdot\det
\left(\begin{matrix}
x_1-x_n &x_1^2-x_n^2 &\cdots &x^{n-1}_1-x_n^{n-1}\\
x_2-x_n &x_2^2-x_n^2 &\cdots &x_2^{n-1}-x_n^{n-1}\\
\vdots &\vdots & &\vdots\\
x_{n-1}-x_n &x_{n-1}^2-x_n^2 &\cdots &x_{n-1}^{n-1}-x_n^{n-1}\\
\end{matrix}\right)\\
&=(-1)^{n+1}\det\left(x_i^j-x_n^j\right)_{\substack{1\le i \le n-1\\1\le j \le n-1}}
\end{align*}
$$
Because I'm allowed to use the fact that
$$x^n-y^n=(x-y)\sum_{k=0}^{n-1}x^ky^{n-1-k}$$
I then used the multilinearity of the determinant ($(x_i-x_n)$ is independant of $j$):
$$
\begin{align*}
\det(A_n)
&=(-1)^{n+1}\det\left((x_i-x_n)\sum_{k=0}^{j-1} x_i^k\,x_n^{j-1-k}\right)_{i,j}\\
&=(-1)^2(-1)^{n-1}\prod_{1\le i < n} (x_i-x_n) \det\left(\sum_{k=0}^{j-1} x_i^k\,x_n^{j-1-k}\right)_{i,j}\\
&=\prod_{1\le i < n} (x_n-x_i)\det\left(\sum_{k=0}^{j-1} x_i^k\,x_n^{j-1-k}\right)_{i,j}
\end{align*}
$$
This reminds me an awful lot of an induction; for that I'd need to show that:
$$\det(A_{n-1})=\det\left(\sum_{k=0}^{j-1} x_i^k\,x_n^{j-1-k}\right)_{i,j}$$
But unfortunately at this point I do not know how to proceed.
I probably got stuck, trying to solve it this way, with the consequence of overlooking an easier proof :P Does anyone see an/the solution and is willing to guide me in the right direction?
Thanks for checking in and reading all that above :)
~Cedric
|
Take the case $n = 4$ as an example. Let $(x_1,x_2,x_3,x_4) = (x,y,z,t)$, we have
$$\begin{align}
\left|\begin{matrix}
1 & x & x^2 & x^3 \\
1 & y & y^2 & y^3 \\
1 & z & z^2 & z^3 \\
1 & t & t^2 & t^3 \\
\end{matrix}\right|
&\stackrel{\color{blue}{[1]}}{=}
\left|\begin{matrix}
0 & x-t & x^2-t^2 & x^3-t^3 \\
0 & y-t & y^2-t^2 & y^3-t^3 \\
0 & z-t & z^2-t^2 & z^3-t^3 \\
1 & t & t^2 & t^3 \\
\end{matrix}\right|\\
&\stackrel{\color{blue}{[2]}}{=} (-)^{n-1}\left|\begin{matrix}
x-t & x^2-t^2 & x^3-t^3 \\
y-t & y^2-t^2 & y^3-t^3 \\
z-t & z^2-t^2 & z^3-t^3
\end{matrix}\right|\\
&\stackrel{\color{blue}{[3]}}{=} (t-x)(t-y)(t-z)\left|\begin{matrix}
1 & x+t & x^2 + tx + t^2\\
1 & y+t & y^2 + ty + t^2\\
1 & z+t & z^2 + tz + t^2
\end{matrix}\right|\\
&\stackrel{\color{blue}{[4]}}{=}
(t-x)(t-y)(t-z) \left|\begin{matrix}1 & x & x^2\\ 1 & y & y^2 \\ 1 & z & z^2\end{matrix}\right|
\end{align}$$
*
*$\color{blue}{[1]}$ - subtract $n^{th}$ row from $k^{th}$ row for $k = 1,\ldots,n-1$.
*$\color{blue}{[2]}$ - Laplace expand against column $1$.
*$\color{blue}{[3]}$ - extract common factors from each row.
*$\color{blue}{[4]}$ - substract $x_n$ times $(k-1)^{th}$ column form $k^{th}$ column for $k = 2, \ldots, n - 1$.
Can you see the pattern?
|
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|
How to prove Fibonacci recurrence holds mod p? Let $$J_n \equiv c^{-1} \Big( \Big( \dfrac{1+c}{2} \Big)^n - \Big( \dfrac{1-c}{2} \Big)^n \Big) \ \text{(mod p)},$$ with $c$ and $c^{-1}$ integers such that $c^2 \equiv 5 \ \text{(mod p)}$ and $cc^{-1} \equiv 1 \ \text{(mod p)}$. And $c$ is an odd integer.
It is easy to check that $J_1 \equiv J_2 \equiv 1 \ \text{(mod p)}$. However, I can't prove that $J_{n} \equiv J_{n-1} + J_{n-2} \ \text{(mod p)}$.
My attempt: $$J_{n-1} + J_{n-2} = c^{-1} \Big( \Big( \dfrac{1+c}{2} \Big)^{n-2} - \Big( \dfrac{1-c}{2} \Big)^{n-2}\Big) \Big( \dfrac{6+2c}{4} \Big) + k_3 p = c^{-1} \Big( \Big( \dfrac{1+c}{2} \Big)^n - \Big( \dfrac{1-c}{2} \Big)^n\Big) + c^{-1} \Big( \dfrac{1+c}{2} \Big)^{n-2} \dfrac{k_1p}{4} - c^{-1} \Big( \dfrac{1-c}{2} \Big)^{n-2} \Big(1+ \dfrac{k_2p}{4} \Big) + k_3 p$$ which after a long try still I can't reduce it to the desired result because of factor 4 in the denominator.
PS I always keep $k_ip$'s to avoid mistakes working in mod p.
Please help!
Edit. If $\dfrac{c^{-1}}{2^2} ((1+c)^{n-2} - (1-c)^{n-2})$ were an integer for any $n$ then we are done! But that also I couldn't prove.
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Write $a=\frac12(1+c)$ and $b=\frac12(1-c)$. Then $a+b=1$ and
$$ab=\frac14(1-c^2)\equiv\frac14(1-5)\equiv-1\pmod p.$$
Therefore
$$a^2-a-1\equiv a^2-(a+b)a+ab\equiv0\pmod p$$
and
$$b^2-b-1\equiv b^2-(a+b)b+ab\equiv0\pmod p.$$
Then
$$c(J_n-J_{n-1}-J_{n-2})\equiv a^n-b^n- a^{n-1}+b^{n-1}-a^{n-2}+b^{n-2}
\equiv(a^2-a-1)a^{n-2}+(b^2-b-1)b^{n-2}\equiv0\pmod p.
$$
It follows that $J_n\equiv J_{n-1}+J_{n-2}$ (mod $p$).
|
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|
Gaussian curvature of a given metric The exercise is from do Carmo, Differential Forms and Application, p.97. Consider $\mathbb{R}^2$ with the following inner product: if $p=(x,y)\in\mathbb{R}^2$ and $u,v\in T_p\mathbb{R}^2$, then $$\langle u,v\rangle_p=\frac{u\cdot v}{(g(p))^2},$$where $u\cdot v$ is the canonical inner product of $\mathbb{R}^2$ and $g:\mathbb{R}^2\rightarrow\mathbb{R}$ is a differentiable positive function. Prove that the Gaussian Curvature of this metric is $$K=g(g_{xx}+g_{yy})-(g_x^2+g_y^2)$$ I need some help to solve it. I guess all the details are given in the exercise.
|
I'm not sure if there's a simpler way to do it, but here's a way that only involves the basic theory of surfaces.
Your first fundamental form is $E = G = \frac{1}{g^2}, F=0.$ Hence you can compute the Christoffel symbols by solving
\begin{equation}
\begin{pmatrix}
1/g^2 & 0\\
0 & 1/g^2
\end{pmatrix}
\begin{pmatrix}
\Gamma_{11}^1\\
\Gamma_{11}^2
\end{pmatrix}
= \begin{pmatrix}
\frac{1}{2}E_x\\
F_x-\frac{1}{2}E_y
\end{pmatrix}
= \begin{pmatrix}
-g_x/g^3\\
g_y/g^3
\end{pmatrix}
\quad
\Rightarrow
\quad
\begin{pmatrix}
\Gamma_{11}^1\\
\Gamma_{11}^2
\end{pmatrix}
= \begin{pmatrix}
-g_x/g\\
g_y/g
\end{pmatrix}
\end{equation}
\begin{equation}
\begin{pmatrix}
1/g^2 & 0\\
0 & 1/g^2
\end{pmatrix}
\begin{pmatrix}
\Gamma_{12}^1\\
\Gamma_{12}^2
\end{pmatrix}
= \begin{pmatrix}
\frac{1}{2}E_y\\
\frac{1}{2}G_x
\end{pmatrix}
= \begin{pmatrix}
-g_y/g^3\\
-g_x/g^3
\end{pmatrix}
\quad
\Rightarrow
\quad
\begin{pmatrix}
\Gamma_{12}^1\\
\Gamma_{12}^2
\end{pmatrix}
= \begin{pmatrix}
-g_y/g\\
-g_x/g
\end{pmatrix}
\end{equation}
\begin{equation}
\begin{pmatrix}
1/g^2 & 0\\
0 & 1/g^2
\end{pmatrix}
\begin{pmatrix}
\Gamma_{22}^1\\
\Gamma_{22}^2
\end{pmatrix}
= \begin{pmatrix}
F_y-\frac{1}{2}G_x\\
\frac{1}{2}G_y
\end{pmatrix}
= \begin{pmatrix}
g_x/g^3\\
-g_y/g^3
\end{pmatrix}
\quad
\Rightarrow
\quad
\begin{pmatrix}
\Gamma_{22}^1\\
\Gamma_{22}^2
\end{pmatrix}
= \begin{pmatrix}
g_x/g\\
-g_y/g
\end{pmatrix}
\end{equation}
Furthermore, $(\Gamma_{12}^2)_x = -g_{xx}/g+g_x^2/g^2,$ and $(\Gamma_{11}^2)_y = g_{yy}/g-g_y^2/g^2$, so that the Gauss formula (see do Carmo's Differential Geometry of Curves and Surfaces, chapter 4-3, eq. 5) gives
\begin{align}
K &= -\frac{1}{E}\big( (\Gamma_{12}^2)_x - (\Gamma_{11}^2)_y + \Gamma_{12}^1\Gamma_{11}^2-\Gamma_{11}^2\Gamma_{22}^2-\Gamma_{11}^1\Gamma_{12}^2\big)\\
&= -g^2 \big( -g_{xx}/g+g_x^2/g^2 - g_{yy}/g + g_y^2/g^2 +g_y^2/g^2 +g_x^2/g^2 -g_y^2/g^2 -g_x^2/g^2\big)\\
&= - \big( -g_{xx}g-g_{yy}g + g_x^2 + g_y^2 \big)\\
&= g(g_{xx}+g_{yy}) - (g_x^2+g_y^2) .
\end{align}
|
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|
Two different series for $\frac{1}{x + 2}$
Find the appropriate power series for $f(x) = \frac{1}{x + 2} $
First method : $\frac{1}{x + 2} = \frac{1}{2(1 - (-\frac{x}{2}))} = \sum_{n=0}^\infty \frac{(-1)^n}{2^{n+1}}x^n$ , $|x|\lt2$
Second method : $\frac{1}{x + 2} = \frac{1}{1 - (-x-1)} = \sum_{n=0}^\infty (-x-1)^n$ , $-2 \lt x \lt 0$
Is there any problem in my answers ? Why these power series have different radii of convergence ?
|
The first one is the series centered at $x=0$ while the second one is the series centered at $x=-1$ therefore the radius of convergence are different.
|
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|
Show that $\int_0^{\pi} \dfrac{\cos2\theta \ d \theta}{1-2 a \cos {\theta} +a^2}=\dfrac{a^2 \pi}{1-a^2}; \ \ (-1For $a=0$ the equality holds. For $a \ne 0$, by change of variable and letting $C_0$ be the unit circle about the origin, we have:
\begin{align*}
\int_{-\pi}^{\pi} \dfrac{\cos2\theta \ d \theta}{1-2 a \cos {\theta} +a^2} &= \int_{C_0} \dfrac{\frac12 (z^2+z^{-2}) dz/zi}{1- 2a \frac12 (z+z^{-1})+ a^2} \\
& = \int_{C_0} \dfrac{(1+z^4)dz}{2iz^2(-a)(z+a)(z+\frac1a)}\\
& = \dfrac{\pi (a^4+1)}{a^2(a^2-1)} \\
& \ne 2\dfrac{a^2 \pi}{1-a^2}
\end{align*}
Where did I make a mistake?
PS This post couldn't help to find my mistake.
|
From the second line, it should be
$$\int_{C_0} \dfrac{(1+z^4)dz}{2iz^2(z-a)(1-az)}
=\pi\left(\dfrac{ 1+a^4}{a^2(1-a^2)}-\frac{1+a^2}{a^2}\right)=\dfrac{2a^2 \pi}{1-a^2}$$
where we evaluated the residues at $a$ AND at $0$.
|
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|
Find the interpolating polynomial of degree $3$ that interpolates $f(x) = x^3$
Find the interpolating polynomial of degree $3$ that interpolates $f(x) = x^3$ at the nodes $x_0=0, x_1 = 1, x_2=2, x_3 = 3$.
Here are my workings below
The basic Lagrange polynomials are:
$$L_0(x) = \frac{(x-1)(x-2)(x-3)}{(0-1)(0-2)(0-3)}$$
$$L_1(x) = \frac{(x-0)(x-2)(x-3)}{(1-0)(1-2)(1-3)}$$
$$L_2(x) = \frac{(x-0)(x-1)(x-3)}{(2-0)(2-1)(2-3)}$$
$$L_3(x) = \frac{(x-0)(x-1)(x-2)}{(3-0)(3-1)(3-2)}$$
Then the interpolating polynomial is:
$$P(x) = L_0(x)+(1)^3L_1(x)+(2)^3L_2(x)+(3)^3L_3(x)$$
Am I allowed to find the interpolating polynomial using these basic lagrange polynomials? and is my $P(x)$ correct? I wasn't sure if the first term should be $L_0(x)$?
|
Your calculations are correct.
The final result is
$P_3(x) = 0.0 + 0.5 (x-3) (x-2) (x+0)-4 (x-3) (x-1) (x+0)+4.5 (x-2) (x-1) (x+0) = x^3$
The formula for the error bound is given by:
$$E_n(x) = {f^{(n+1)}(\xi(x)) \over (n+1)!} \times (x-x_0)(x-x_1)...(x-x_n)$$
So, we have
$$E_3(x) = {f^{(4)}(\xi(x)) \over 4!} \times (x-0)(x-1)(x-2)(x-3)$$
The fourth derivative of $f(x) = x^3$ is zero, so $E_3(x) = 0$.
The reason for this is if $f(x) = $ polynomial of degree $M$ where $M \le N$, then
$$f^{(n)}(x) = 0 \implies E_n(x) = 0 ~\forall~ x$$
Therefore $P_3(x)$ is an exact representation of $f(x) = x^3$.
|
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|
How can one calculate $342342^{1001}$ mod $5$? How can one calculate $342343^2$ mod $3$? I know that the answer is $1$.
And $342342^{1001}$ mod $5$.
I know that
$
3^0 \mod 5 = 1 \\
3^1 \mod 5 = 3 \\
3^2 \mod 5 = 4 \\
3^3 \mod 5 = 2 \\\\
3^4 \mod 5 = 1 \\
3^5 \mod 5 = 3 \\
3^6 \mod 5 = 4 \\
$
So 1001 = 250 + 250 + 250 + 250 + 1, which is why the answer is also 1?
|
First, Note that
$$342342 = 34234*10+2= 34234*2*5 +2$$
So you have
$$342342 \equiv 2 \pmod 5$$
Then, remember that
$$\forall a,b,c,n \in \mathbb N, a \equiv b \pmod n \implies a^c \equiv b^c \pmod n$$
Therefore,
$$342342^{1001} \equiv 2^{1001} \pmod 5$$
Finaly, note that $1001 = 1000+1 = 4*250 +1$ and try to conclude
|
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|
Elementary Level Algebra Question I'm trying to solve the following homework problem:
If $a\neq b$, $a^3-b^3 = 19x^3$ and $a-b=x$, which of the following conclusions is correct?
\begin{align}
\text{(1) }& a = 3x \\
\text{(2) }& a = 3x \text{ or } a = -2x \\
\text{(3) }& a = -3x \text{ or } a = 2x \\
\text{(4) }& a = 3x \text{ or } a = 2x
\end{align}
I am getting option (4): $a = 3x$ or $a=2x$ as the answer, which is incorrect. The correct answer is $a=3x$ or $a=-2x.$
My work:-
$$ (a-b)^3 + 3ab(a-b)=19x^3$$
$$x^3 + 3ab(x) = 19x^3$$
$$ab=6x^2$$
Hence by comparing we have,
$$a * b = 3x * 2x \text{ or } a * b = 2x * 3x$$
So a can be either $2x$ or $3x$.
Why is my answer wrong?
|
Use difference of cubes.
$$a^3-b^3 = (a-b)(a^2+ab+b^2)$$
$$\implies (a-b)(a^2+ab+b^2) = 19x^3$$
Set $\color{purple}{a-b = x}$.
$$\implies \color{purple}{x}(a^2+ab+b^2) = 19x^3 \implies a^2+ab+b^2 = 19x^2$$
Set $\color{blue}{b = a-x}$.
$$\implies a^2+a\color{blue}{(a-x)}+\color{blue}{(a-x)}^2 = 19x^2$$
Move $19x^2$ to the LHS, expand, and simplify.
$$\implies a^2+a^2-ax+a^2-2ax+x^2-19x^2 = 0$$
$$\implies 3a^2-3ax-18x^2 = 0$$
$$\implies a^2-ax-6x^2 = 0$$
Factor the trinomial.
$$\implies (a-3x)(a+2x) = 0$$
Set either factor equal to $0$.
$$a = 3x \text{ or } a = -2x$$
Edit: You’ve asked where your error is. Your comparison part wasn’t correct.
$$ab = 6x^2$$
You can’t just jump to conclusions on what $a$ and $b$ can be. Here, make the substitution $\color{blue}{b = a-x}$.
$$a\color{blue}{(a-x)} = 6x^2$$
$$a^2-ax = 6x^2 \implies a^2-ax-6x^2 = 0$$
This leads to the same answer.
|
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|
How to evaluate $\int_{0}^{2\pi}x^2\ln (1-\cos x)dx$? Wolfram Alpha shows that $$\int_{0}^{2\pi}x^2\ln (1-\cos x)dx = -\frac{8}{3} \pi (\pi^2 \ln(2) + 3 \zeta(3))$$
I tried to use the Fourier series
$$\ln (1-\cos x)=-\sum_{n=1}^{\infty} \frac{\cos^nx}{n}.$$
I am not sure how to continue from this point. I need some help.
|
Thanks mrtaurho, I am able to finish it.
$$\begin{align}
\int_{0}^{2\pi}\pi^2\ln(1-\cos x)dx
&=-\frac{8\pi^3}3\ln(2)-\sum_{n=1}^{\infty}\frac{16}n\int_0^{\pi}x^2\cos(2nx)~dx\\
&=-\frac{8\pi^3}3\ln(2)-\sum_{n=1}^{\infty}\frac{16}n\left[-x^2\frac{\sin(2nx)}{2n}+\frac{2x\cos(2nx)}{4n^2}-\frac{2\sin(2nx)}{8n^3}\right]_{0}^{\pi}\\
&=-\frac{8\pi^3}3\ln(2)-\sum_{n=1}^{\infty}\frac{16}{n}\frac{2\pi}{4n^2}\\
&= -\frac{8\pi^3}3\ln(2)-8\pi\sum_{n=1}^{\infty}\frac{1}{n^3}\\
&=-\frac{8}{3}\pi(\pi^2\ln2+3\zeta(3))
\end{align}$$
|
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|
How to show the existence of the limit $\lim_{n\to \infty}\frac{x_n}{n}$ if $x_n$ satisfy $x^{-n}=\sum_{k=1}^\infty (x+k)^{-n}$? Suppose $x_n$ is the only positive solution to the equation $x^{-n}=\sum\limits_{k=1}^\infty (x+k)^{-n}$,how to show the existence of the limit $\lim_{n\to \infty}\frac{x_n}{n}$?
It is easy to see that $\{x_n\}$ is increasing.In fact, the given euation equals
$$1=\sum_{k=1}^\infty(1+\frac{k}{x})^{-n} \tag{*}$$
If $x_n\ge x_{n+1}$,then notice that for any fixed$ k$,$(1+\frac{k}{x})^{-n}$ is increasing,thus we can get
$$\frac{1}{(1+\frac{k}{x_n})^n}\ge \frac{1}{(1+\frac{k}{x_{n+1}})^n}>\frac{1}{(1+\frac{k}{x_{n+1}})^{n+1}}$$
By summing up all k's from 1 to $\infty$,we can see
$$\sum_{k=1}^\infty\frac{1}{(1+\frac{k}{x_n})^n}>\sum_{k=1}^\infty\frac{1}{(1+\frac{k}{x_{n+1}})^{n+1}}$$
then from $(*)$ we see that the two series in the above equality are all equals to $1$,witch is a contradiction!
But it seems hard for us to show the existence of $\lim_{n\to \infty}\frac{x_n}{n}$.What I can see by the area's principle is
$$\Big|\sum_{k=1}^\infty\frac{1}{(1+\frac{k}{x_n})^n}-\int_1^\infty \frac{1}{(1+\frac{x}{x_n})}dx\Big|<\frac{1}{(1+\frac1{x_n})^n}$$
or
$$\Big|1-\frac{x_n}{n-1}(1+\frac{1}{x_n})^{1-n}\Big|<\frac{1}{(1+\frac1{x_n})^n}$$
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For any $n \ge 2$, consider the function $\displaystyle\;\Phi_n(x) = \sum_{k=1}^\infty \left(\frac{x}{x+k}\right)^n$.
It is easy to see $\Phi_n(x)$ is an increasing function over $(0,\infty]$.
For small $x$, it is bounded from above by $x^n \zeta(n)$ and hence decreases to $0$ as $x \to 0$.
For large $x$, we can approximate the sum by an integral and $\Phi_n(x)$ diverges like $\displaystyle\;\frac{x}{n-1}$ as $x \to \infty$. By definition, $x_n$ is the unique root for $\Phi_n(x_n) = 1$. Let $\displaystyle\;y_n = \frac{x_n}{n}$.
For any $\alpha > 0$, apply AM $\ge$ GM to $n$ copies of $1 + \frac{\alpha}{n}$ and one copy of $1$, we obtain
$$\left(1 + \frac{\alpha}{n}\right)^{n/n+1} > \frac1{n+1} \left[n\left(1 + \frac{\alpha}{n}\right) + 1 \right] = 1 + \frac{\alpha}{n+1}$$
The inequality is strict because the $n+1$ numbers are not identical. Taking reciprocal on both sides, we get
$$\left( \frac{n}{n + \alpha} \right)^n \ge \left(\frac{n+1}{n+1 + \alpha}\right)^{n+1}
$$
Replace $\alpha$ by $\displaystyle\;\frac{k}{y_n}$ for generic positive integer $k$, we obtain
$$\left( \frac{x_n}{x_n + k} \right)^n = \left( \frac{n y_n}{n y_n + k} \right)^n > \left(\frac{(n+1)y_n}{(n+1)y_n + k}\right)^{n+1}$$
Summing over $k$ and using definition of $x_n$, we find
$$\Phi_{n+1}(x_{n+1}) = 1 = \Phi_n(x_n) > \Phi_{n+1}((n+1)y_n)$$
Since $\Phi_{n+1}$ is increasing, we obtain $x_{n+1} > (n+1)y_n \iff y_{n+1} > y_n$.
This means $y_n$ is an increasing sequence.
We are going to show $y_n$ is bounded from above by $\frac32$
(see update below for a more elementary and better upper bound).
For simplicity, let us abberivate $x_n$ and $y_n$ as $x$ and $y$. By their definition, we have
$$\frac{2}{x^n} = \sum_{k=0}^\infty \frac{1}{(x+k)^n}$$
By Abel-Plana formula, we can transform the sum on RHS to integrals. The end result is
$$\begin{align}\frac{3}{2x^n} &= \int_0^\infty \frac{dk}{(x+k)^n} +
i \int_0^\infty \frac{(x+it)^{-n} - (x-it)^{-n}}{e^{2\pi t} - 1} dt\\
&=\frac{1}{(n-1)x^{n-1}}
+ \frac{1}{x^{n-1}}\int_0^\infty \frac{(1+is)^{-n} - (1-is)^{-n}}{e^{2\pi x s}-1} ds
\end{align}
$$
Multiply both sides by $nx^{n-1}$ and replace $s$ by $s/n$, we obtain
$$\begin{align}\frac{3}{2y} - \frac{n}{n-1} &=
i \int_0^\infty \frac{(1 + i\frac{s}{n})^{-n} - (1-i\frac{s}{n})^{-n}}{e^{2\pi ys} - 1} ds\\
&= 2\int_0^\infty \frac{\sin\left(n\tan^{-1}\left(\frac{s}{n}\right)\right)}{\left(1 + \frac{t^2}{n^2}\right)^{n/2}} \frac{ds}{e^{2\pi ys}-1}\tag{*1}
\end{align}
$$
For the integral on RHS, if we want its integrand to be negative, we need
$$n\tan^{-1}\left(\frac{s}{n}\right) > \pi
\implies \frac{s}{n} > \tan\left(\frac{\pi}{n}\right) \implies s > \pi$$
By the time $s$ reaches $\pi$, the factor $\frac{1}{e^{2\pi ys} - 1}$ already drops to very small. Numerically, we know $y_4 > 1$, so for $n \ge 4$ and $s \ge \pi$, we have
$$\frac{1}{e^{2\pi ys} - 1} \le \frac{1}{e^{2\pi^2} - 1} \approx 2.675 \times 10^{-9}$$
This implies the integral is positive. For $n \ge 4$, we can deduce
$$\frac{3}{2y} \ge \frac{n}{n-1} \implies y_n \le \frac32\left(1 - \frac1n\right) < \frac32$$
Since $y_n$ is increasing and bounded from above by $\frac32$, limit
$y_\infty \stackrel{def}{=} \lim_{n\to\infty} y_n$ exists and $\le \frac32$.
For fixed $y > 0$, with help of DCT, one can show the last integral of $(*1)$
converges.
This suggests $y_\infty$ is a root of following equation near $\frac32$
$$\frac{3}{2y} = 1 + 2\int_0^\infty \frac{\sin(s)}{e^{2\pi ys} - 1} ds$$
According to DLMF,
$$\int_0^\infty e^{-x} \frac{\sin(ax)}{\sinh x} dx = \frac{\pi}{2}\coth\left(\frac{\pi a}{2}\right) - \frac1a\quad\text{ for }\quad a \ne 0$$
We can transform our equation to
$$\frac{3}{2y} = 1 + 2\left[\frac{1}{4y}\coth\left(\frac{1}{2y}\right) - \frac12\right]
\iff \coth\left(\frac{1}{2y}\right) = 3$$
This leads to $\displaystyle\;y_\infty = \frac{1}{\log 2}$.
This is consistent with the finding of another answer (currently deleted):
If $L_\infty = \lim_{n\to\infty}\frac{n}{x_n}$ exists, then $L_\infty = \log 2$.
To summarize, the limit $\displaystyle\;\frac{x_n}{n}$ exists and should equal to $\displaystyle\;\frac{1}{\log 2}$.
Update
It turns out there is a more elementary proof that $y_n$ is bounded from above by the optimal bound $\displaystyle\;\frac{1}{\log 2}$.
Recall for any $\alpha > 0$. we have $1 + \alpha < e^\alpha$. Substitute
$\alpha$ by $\frac{k}{n}\log 2$ for $n \ge 2$ and $k \ge 1$, we get
$$\frac{n}{n + k\log 2} = \frac{1}{1 + \frac{k}{n}\log 2} > e^{-\frac{k}{n}\log 2} = 2^{-\frac{k}{n}}$$
This leads to
$$\Phi_n\left(\frac{n}{\log 2}\right)
= \sum_{k=1}^\infty \left(\frac{n}{n + \log 2 k}\right)^n
> \sum_{k=1}^\infty 2^{-k}
= 1 = \Phi_n(x_n)
$$
Since $\Phi_n(x)$ is increasing, this means
$\displaystyle\;\frac{n}{\log 2} > x_n$ and $y_n$ is bounded from above by $\displaystyle\;\frac{1}{\log 2}$.
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{
"language": "en",
"url": "https://math.stackexchange.com/questions/2982897",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 4,
"answer_id": 0
}
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For any positive integer $n$, prove the following inequality. Prove that for any positive integer n, the following inequality is true.
$$\left(1+ \frac{1}{n}\right)^n < \left(1+\frac{1}{n+1}\right)^{n+1}$$
Attempt
Not a good attempt but this is my thinking
It will not be a problem when checking whether this is true for smaller integers. Only for larger integers, it will be difficult to prove. So, I took the equation LHS-RHS < $0$. I took limit for n tending to infinity and then applied L'Hospital Rule( though the equation was difficult to handle) & therefore could not do anything further.
Edit: I, now, want to know about @Rebellos' last equation and his answer (though at first my intention was to know about the question). Please see the comments of @Rebellos ' answer.
Adding graph for @Rebellos 's last equation
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I know this question is already answered, but I want to give an answer that only uses the binomium of Newton. This is certainly not the easiest way to show the inequality, but it is an "artisanal" way.
By the binomium we get
$$
\begin{align*}
\left(1+ \frac{1}{n}\right)^n &= \sum_{k=0}^n \binom{n}{k} \frac{1}{n^k} \\
&= 1 + \frac{n}{1!}\frac{1}{n} + \frac{n(n-1)}{2!}\frac{1}{n^2}
+ \frac{n(n-1)(n-2)}{3!}\frac{1}{n^3}+ \ldots
+ \frac{n(n-1)(n-2)\cdots 1}{n!}\frac{1}{n^n} \\
&= 2 + \frac{1}{2!}\left(1-\frac{1}{n}\right)+ \frac{1}{3!}\left(1-\frac{1}{n}\right)\left(1-\frac{2}{n}\right)+ \cdots +
\frac{1}{n!}\left(1-\frac{1}{n}\right)\left(1-\frac{2}{n}\right)\cdots\left(1-\frac{n-1}{n}\right).
\end{align*}
$$
Similarly
$$
\begin{align*}
\left(1 + \frac{1}{n+1}\right)^{n+1}
&= 2 + \frac{1}{2!}\left(1-\frac{1}{n+1}\right)+ \frac{1}{3!}\left(1-\frac{1}{n+1}\right)\left(1-\frac{2}{n+1}\right)+ \cdots \\
& + \frac{1}{n!}\left(1-\frac{1}{n+1}\right)\left(1-\frac{2}{n+1}\right)\cdots\left(1-\frac{n-1}{n+1}\right) \\
&+ \frac{1}{(n+1)!}\left(1-\frac{1}{n+1}\right)\left(1-\frac{2}{n+1}\right)\cdots\left(1-\frac{n}{n+1}\right).
\end{align*}
$$
For every term in the expression for $\left(1+\tfrac{1}{n}\right)^n$ there is a similar term in the expression for $\left(1+\tfrac{1}{n+1}\right)^{n+1}$ that is equal of bigger. Furthermore, the last expression contains one positive term more. Hence the latter expression is bigger.
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{
"language": "en",
"url": "https://math.stackexchange.com/questions/2986097",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 2
}
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Algebraic manipulation with indices The question is:
For a>0 and $\sqrt{a}+\frac{1}{\sqrt{a}}=3$, find the value of $a\sqrt{a}+\frac{1}{a\sqrt{a}}$
So I first squared the given equation and got:
$$a+\frac{1}{a}+2=9$$
$$a+\frac{1}{a}=7$$
Then to get the form of $a\sqrt{a}+\frac{1}{a\sqrt{a}}$:
$$(a+\frac{1}{a})(\sqrt{a}+\frac{1}{\sqrt{a}})=a\sqrt{a}+\frac{1}{a\sqrt{a}}+\frac{\sqrt{a}}{a}+\frac{a}{\sqrt{a}}$$
And I just got stuck right here because I didn't really know what to do with the $\frac{\sqrt{a}}{a}+\frac{a}{\sqrt{a}}$. So I looked into the solutions and apparently it's
$$(a+\frac{1}{a})(\sqrt{a}+\frac{1}{\sqrt{a}})=a\sqrt{a}+\frac{1}{a\sqrt{a}}+\sqrt{a}+\frac{1}{\sqrt{a}}$$
I'm not sure how those two are equal...
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Let $x=\sqrt{a}$ then $x+\frac{1}{x}=3$ and
$$
\\a\sqrt{a}+\frac{1}{a\sqrt{a}}=x^3+\frac{1}{x^3}=(x+\frac{1}{x})^3-3x\cdot\frac{1}{x}(x+\frac{1}{x})=27-3\cdot3=18
$$
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{
"language": "en",
"url": "https://math.stackexchange.com/questions/2988410",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
}
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