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Prove that there are no such positive integers $a,b,c,d$ such that $a^2 + b^2 = 3(c^2 + d^2)$ Prove that there are no positive integers a, b, c, d such that $a^2 + b^2 = 3(c^2 + d^2)$. Hint: What can you say about divisibility of a and b by 3? Look at solution with smallest possible a.
|
Well, $A^2+B^2=1,0$ ( mod 4) so it couldn't be $A^2+B^2=4k+3$. But (4k+1)$(4k+3)=16k^2+16k+3=4(4k^2+4k)+3=4r+3$, while (4k+3)$(4k+3)=16k^2+24k+8+1=4(4k^2+6k+2)+1=4s+1$. So if a prime $p|A^2+B^2$ and p=4k+3, $p^2|A^2+B^2!$ But the exponent of number 3 is odd while it should be even. Hence there no positive integers A,B,C,D such that $A^2+B^2=3(C^2+D^2)$.
|
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|
How do I find the highest and lowest points made by the union of these two functions using Lagrange Multipliers?
Find the highest and lowest points made by the union of these two functions using Lagrange Multipliers.
$x^2+y^2+z^2 = 16$
$(x+1)^2+(y+1)^2+(z+1)^2 = 27$
I got the basics down, I used the first function as my constraint.
$2(x+1) = \lambda\times2x$
$2(y+1) = \lambda\times 2y$
$2(z+1) = \lambda\times2z$
$x^2+y^2+z^2 = 16$
and after solving I think it's all $x=y=z=\pm\dfrac{4}{\sqrt{3}}$
I just don't know how to interpret these results.
What does this mean? and how do I use this to answer
my problem?
Thank you for your time.
|
I think you are missing something. As @copper.hat noted you don't have nothing to optimize. I think that the two equations you uploaded are constrains, but the function you want to maximize/minimize isn't posted.
If that's not the case, add those two and you'll get that both minima and the maxima of the function is $27 + 16 = 43$.
If you want to find a solution for the system of equations, just subtract the first equation from the second and we have:
$$x + y + z = 4$$
$$x^2 + y^2 + z^2 = 16$$
Then there are infinite amount of solution. Set $x^3 + y^3 + z^3$ to any value you like and then using Newton's Identities you'll be able to obtain a cubic polynomial, where $x,y,z$ will be roots of that polynomial.
And from the thing you've posted I think you want to maximize/minimize the function
$$f(x,y,z) = (x+1)^2 + (y+1)^2 + (z+1)^2$$
under constrain:
$$g(x,y,z) = x^2 + y^2 + z^2 - 16 = 0$$
To make things even simplier expand the first function and you'll have:
$$f(x,y,z) = (x^2 + y^2 + z^2) + 2(x+y+z) + 3 = 2(x+y+z) + 19$$
Now apply Lagrange multipliers:
$$F(x,y,z,\lambda) = 2(x+y+z) + 19 - \lambda(x^2 + y^2 + z^2 - 16)$$
Now take partial derivatives and set them to 0.
$$F_x = 2 - 2x\lambda = 0$$
$$F_y = 2 - 2y\lambda = 0$$
$$F_z = 2 - 2z\lambda = 0$$
$$F_{\lambda} = x^2 + y^2 + z^2 - 16 = 0$$
Substituting $x = y = z \frac{1}{\lambda}$ from the first partial derivatives into the constarin we have:
$$\frac{3}{\lambda^2} = 16$$
$$\lambda = \pm \frac{\sqrt{3}}{4}$$
Substitute back and get that $ x = y = z = \pm \frac{4}{\sqrt{3}}$
Plug the values for $x,y,z$ into the initial function and you'll get that the maxima occurs at $f(\frac{\sqrt{3}}{4},\frac{\sqrt{3}}{4},\frac{\sqrt{3}}{4}) = 19 + 8\sqrt{3}$ and the minima at $f(\frac{- \sqrt{3}}{4}, - \frac{\sqrt{3}}{4}, - \frac{\sqrt{3}}{4}) = 19 - 8\sqrt{3}$
|
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|
Is there a base for each positive integer where this number can be represented a string of n (base - 1) digit? How are called positive integer numbers that have the following property of being represented as:
$$
N = \sum_{k=0}^n{(B-1)B^k} = (B - 1)\sum_{k=0}^n{B^k}
$$
with $N$ a positive integer number, $B$ the base in which $N$ is represented, $B \lt N$
For example, base 10:
$$
99_{10} = (10-1)10^1 + (10-1)10^0
$$
$$
999_{10} = (10-1)10^2 + (10-1)10^1 + (10-1)10^0
$$
$$
...
$$
base 2:
$$
3_{10} = 11_2 = (2 - 1) 2^1 + (2 - 1) 2^0
$$
$$
7_{10} = 111_2 = (2 - 1) 2^2 + (2 - 1) 2^1 + (2-1)2^0
$$
$$
15_{10} = 1111_2 = (2-1)2^3 + (2-1) 2^2 + (2-1)2^1 + (2-1)2^0
$$
$$
...
$$
base 3:
$$
4_{10} = 22_{3} = (3-1)3^1 + (3 - 1)3^0
$$
$$
26_{10} = 222_{3} = (3-1)3^2 + (3-1)3^1 + (3-1)3^0
$$
$$
..
$$
base 16:
$$
255_{10} = ff_{16} = (16-1)16^1 + (16-1)16^0
$$
$$
4095_{10} = fff_{16} = (16-1)16^2+(16-1)16^1 + (16-1)16^0
$$
$$
...
$$
In this other hand, does a base $B$ exist for any positive integer $N$ where it can be written
$$
N = \sum_0^n{(B - 1)B^n}
$$
With the additional condition that $B \lt N$
|
When it is $1$'s repeated, it is called a repunit. When it is $n$ repeated, it is called a repdigit.
We want $n=(b-1)\sum_{k=0}^{m-1}b^k=b^m-1 $
So whenever $n+1$ is an $m^{th}$ power, $n$ may be expressed as $m\ $ $b-1$'s
|
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|
Coordinate Geometry Oblique Coordinates Problem This is a elementary geometry problem which I have tried to solve using coordinate geometry but it is resulting in an impossible and impractical result. Maybe I have some misconceptions with oblique coordinates. So please help me with my misconceptions.
Suppose $\triangle ABC$ is a triangle in which $\angle B= 2 \angle C$. $D$ is a point on $BC$ such that $AD$ bisects $\angle BAC$ and $AB=CD$. Prove that $\angle BAC= 72^\circ$.
My solution
Let us construct the coordinate system in $\triangle BAC$ where the $X$-axis is along the side $BC$ and the $Y$ axis is along the side $AD$ with angle of origin=$\angle G$. Consider $D$ as the origin with coordinates $(0,0)$. Let the coordinates of $C$ be $(a,0)$ and that of $A$ be $(0,c)$. Let the coordinates of $B \equiv (b,0)$.
Then the equation of straight line $AC=\frac x a + \frac y c=1$ (by intercept form). That is, $cx + ay= ac$, i.e., $ cx+ay-ac=0$. Equation of straight line $\frac x{-b} + \frac y c =1$, i.e., $cx -by= -bc$, i.e., $cx-by+bc=0$.
As $AD$ bisects angle $\angle BAC$, the angle between $cx-by+bc= 0$ and $x=0$ = to the angle between $cx+ ay-ac=0$ and $x=0$. So their tangents are equal, i.e., $\frac{b \sin g}{c+b \cos g}= \frac{a \sin g}{c-a \sin g}$.
Cancelling $\sin g$ from both sides we have $- bc + ab \cos g =ac + ab \cos g $. Cancelling $ab \cos g$ from both sides we have $ac+ bc=0$ ie $a+b=0$
ie $a=-b$ This implies $DC= BD$. But by angle bisector theorem we have $\frac{AB}{AC}= \frac{BD}{DC}$ as $BD = DC$ we have $AB = AC$. Thus the $\triangle ABC$ is isosceles which is contradictory by the question.
How is this possible?
|
There are two difficulties here, as the post stands. One is that it appears as though there was an earlier version of the solution posted, which was then revised, leaving it in a form that is now internally inconsistent. Unless, for some reason, two coordinate axes are being used which are not perpendicular to one another (which would require significant modification to the trigonometric ratios shown), it cannot be the case that both point $ \ D \ $ is placed at the origin of the coordinate system and that point $ \ A \ $ lies on the $ y-$ axis. This is evidently the cause of the false conclusion that $ \ DC = BD \ . $ (There are isoceles triangles in this diagram: $ \ ABC \ $ is one of them, but not with the congruence of sides suggested.)
The other problem is with taking a purely analytical-geometric approach here. Since we wish to show that $ \angle BAC \ $ has a measure of 72º , we will eventually run up against the issue of whether we will recognize the exact value of a trigonometric ratio for that angle when we encounter it.
Application of trigonometry is a useful approach; I will lay out one particular route here.
I have marked the given information for the triangle in blue; we will derive the angles marked in red. The angle $ \ \angle BAC \ $ has measure $ \ 180º - 3 \theta \ , $ so we have $ \ m(\angle BAD) = m(\angle CAD) $ $ = 90º - \frac{3}{2} \theta \ . $ It follows by one or another line of argument that $ \ m(\angle ADC) = 90º + \frac{1}{2} \theta \ $ and $ \ m(\angle ADB) = 90º - \frac{1}{2} \theta \ . $
The specification that $ \ AB = CD \ $ (the lengths of which we shall call $ \ c \ $ ) is crucial to resolving this triangle. We may use the Law of Sines to write for triangle $ \ ABD \ , $
$$ \frac{\sin (90º - \frac{1}{2} \theta)}{c} \ = \ \frac{\sin 2 \theta}{d} \ , $$
and for triangle $ \ ACD \ , $
$$ \frac{\sin (90º - \frac{3}{2} \theta)}{c} \ = \ \frac{\sin \theta}{d} \ . $$
From these ratios and the use of the "co-relations", we obtain $ \ c \ \sin \theta \ = \ d \ \cos \frac{3}{2} \theta \ $ and $ \ c \ \sin 2 \theta \ = \ d \ \cos \frac{1}{2} \theta \ . $ Using the "double-angle formula" for sine and substituting the first equation into the second produces
$$ \ c \ \cdot 2 \ \sin \theta \ \cos \theta \ = \ d \ \cos \frac{1}{2} \theta \ \ \Rightarrow \ \ \ 2d \ \cos \frac{3}{2} \theta \ \cos \theta \ = \ d \ \cos \frac{1}{2} \theta \ , $$
from which we may cancel the factor $ \ d \ $ . The appropriate "product-to-sum formula" then leads us to
$$ 2 \ \left( \frac{1}{2} \ \left[ \ \cos (\frac{3}{2} \theta \ - \ \theta) \ + \ \cos (\frac{3}{2} \theta \ + \ \theta) \ \right] \ \right) \ = \ \cos \frac{1}{2} \theta $$
$$ \cos (\frac{1}{2} \theta ) \ + \ \cos (\frac{5}{2} \theta ) \ = \ \cos \frac{1}{2} \theta \ \ \Rightarrow \ \ \cos (\frac{5}{2} \theta ) \ = \ 0 \ \ . $$
We may restrict ourselves to the first quadrant for a solution, obtaining $ \ \frac{5}{2} \theta \ = \ 90º \ \Rightarrow \ \theta \ = \ 36º . $ At last, we conclude that
$$ \ m(\angle ACB) \ = \ 36º \ \ , \ \ m(\angle ABC) \ = \ 72º \ \ , \ \text{and} \ \ m(\angle BAC) \ = \ 72º \ . $$
$$ \ \ $$
Now for the more interesting aspect of this figure: with a little more work, we find that $ \ \ m(\angle BAD) \ = \ 36º \ \ \text{and} \ \ m(\angle ADB) \ = \ 72º \ . $ So, in fact, $ \ ACB \ $ and $ \ BAD \ $ are similar isosceles triangles. Therefore, we have a proportionality which tells us that
$$ \ \frac{c+x}{c} \ = \ \frac{c}{x} \ \ \Rightarrow \ \ x^2 \ + \ cx \ = \ c^2 \ , $$
which can be solved alternately to find $ \ c \ = \ x \ \left(\frac{1 + \sqrt{5}}{2} \right) \ = \ \phi \ $ or $ \ x \ = \ c \ \left(\frac{ \sqrt{5} - 1}{2} \right) \ = \ \frac{1}{\phi} \ . $ Sides $ \ BC \ ( \text{or} \ AC ) \ $ and $ \ AB \ $ and sides $ \ AB \ ( \text{or} \ AD ) \ $ and $ \ BD \ $ stand in proportions of the Golden Ratio!
Our triangle diagram is actually a portion of the so-called "Golden Pentagram", sacred to the Pythagoreans. (When the dominant culture of the region changed in subsequent centuries, the connotations associated with this figure followed the (somewhat simplistic) path, Greek $ \ \rightarrow \ $ pagan $ \ \rightarrow \ $ infidel $ \ \rightarrow \ $ evil.) This figure is rife with mathematical interest; see, for instance, the recent Posamentier and Lehmann, The Glorious Golden Ratio (2012), pp. 137-60.
|
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|
Putnam Series Question I'm studying for the Putnam Exam and am a bit confused about how to go about solving this problem.
Sum the series
$$
\sum_{m = 1}^{\infty} \sum_{n = 1}^{\infty} \frac{m^2n}{3^m(n3^m + m3^n)}.
$$
I've tried "splitting" the expression to see if a geometric sum pops up but that didn't get me anywhere. I've also tried examining the first few terms of the series for the first few values of $m$ to see if an inductive pattern emerged but no luck there either.
|
Let
$$
S = \sum_{m=1}^\infty \sum_{n=1}^\infty \frac{m^2n}{3^m(n3^m + m3^n)}= \sum_{m=1}^\infty \sum_{n=1}^\infty \frac{1}{\frac{3^m}{m}\left( \frac{3^m}{m} + \frac{3^n}{n} \right)}.
$$
Then we see by symmetry, that
$$
2S = \sum_{m=1}^\infty \sum_{n=1}^\infty \frac{1}{\left( \frac{3^m}{m} + \frac{3^n}{n} \right)}\left( \frac{1}{\frac{3^m}{m}} + \frac{1}{\frac{3^n}{n}}\right),
$$
or what is the same,
$$
2S = \sum_{m=1}^\infty\sum_{n=1}^\infty \frac{mn}{3^m 3^n} = \left( \sum_{m=1}^\infty \frac{m}{3^m} \right)^2.
$$
The problem reduces to calculating the last single sum.
To this end, recall that for $|x|< 1,$ geometric series yields
$$
\frac{1}{1-x} = 1 + x+ x^2 + \cdots,
$$
multiplying by $x,$ and differentiating (this is justified because the series on the right converges on compact subsets of $|x| < 1,$
$$
\frac{1}{(1-x)^2} = 1+ 2x + 3x^2 + \cdots,
$$
and multiplying by $x$ one more time,
$$
\frac{x}{(1-x)^2} = x + 2x^2 + 3x^3 + \cdots.
$$
Set $x = 1/3$ to evaluate the single sum on the right. We obtain (if I haven't messed up calculations)
$$
S = \frac{9}{32}.
$$
|
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|
How find this Fibonacci sequence sum $\sum_{k=0}^{\infty}\frac{1}{F_{2^k}}$ let sequence $\{F_{n}\}$ such $$F_{1}=1,F_{2}=1,F_{m+1}=F_{m}+F_{m-1},m\ge 2$$
Find this value
$$I=\sum_{k=0}^{\infty}\dfrac{1}{F_{2^k}}$$
My try: I know this
$$F_{n}=\dfrac{1}{\sqrt{5}}\left(\left(\dfrac{\sqrt{5}+1}{2}\right)^n-\left(\dfrac{\sqrt{5}-1}{2}\right)^n\right)$$
so
$$\dfrac{1}{F_{2^k}}=\dfrac{\sqrt{5}}{\left(\dfrac{\sqrt{5}+1}{2}\right)^{2^k}-\left(\dfrac{\sqrt{5}-1}{2}\right)^{2^k}}$$
so
we only find this sum
$$I=\sum_{k=0}^{\infty}\dfrac{\sqrt{5}}{\left(\dfrac{\sqrt{5}+1}{2}\right)^{2^k}-\left(\dfrac{\sqrt{5}-1}{2}\right)^{2^k}}$$
But I can't.Thank you
|
In fact, this is summable. This surprises me, because of results like those in this question at MO, which in short discusses how we don't know if the sum of the reciprocals of all the Fibonacci numbers is transcendental, and other related unknown results.
This question boils down to the following:
Claim: $\displaystyle \frac{1}{F_1} + \frac{1}{F_2} + \frac{1}{F_4} + \ldots + \frac{1}{F_{2^n}} = 3 - \frac{F_{2^n-1}}{F_{2^n}}$ for $n \geq 2$.
Proof: Induct, and use $F_{n}=\dfrac{1}{\sqrt{5}}\left(\left(\dfrac{\sqrt{5}+1}{2}\right)^n-\left(\dfrac{\sqrt{5}-1}{2}\right)^n\right)$ as you mentioned. $\diamondsuit$
Then take the limit as $n \to \infty$ to get that
$$\sum_{k=0}^{\infty}\dfrac{1}{F_{2^k}} = \frac{7-\sqrt 5}{2}.$$
|
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|
How prove this $\frac{a}{(bc-a^2)^2}+\frac{b}{(ca-b^2)^2}+\frac{c}{(ab-c^2)^2}=0$ let $a,b,c\in \mathbb{R}$, if such
$$\dfrac{1}{bc-a^2}+\dfrac{1}{ca-b^2}+\dfrac{1}{ab-c^2}=0$$
show that
$$\dfrac{a}{(bc-a^2)^2}+\dfrac{b}{(ca-b^2)^2}+\dfrac{c}{(ab-c^2)^2}=0$$
Does this problem has nice methods?
My idea:let
$$(ca-b^2)(ab-c^2)+(bc-a^2)(ab-c^2)+(bc-a^2)(ca-b^2)=0$$
then I can't. Thanks
|
Hint: if we have $$\dfrac{1}{bc-a^2}+\dfrac{1}{ca-b^2}+\dfrac{1}{ab-c^2}=0$$ Then let $$\dfrac{1}{bc-a^2} = x; \dfrac{1}{ca-b^2} =y; \dfrac{1}{ab-c^2} = z$$ So you have something more manageable, then note that if $x+y+z=0$, then $x = -(y+z)$.
Finally, note that $$\dfrac{a}{(bc-a^2)^2} = \dfrac{a}{bc-a^2}\cdot \dfrac{1}{bc-a^2} = \dfrac{a}{bc-a^2}\cdot x$$
Etc. From here, you should be able to make progress.
|
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|
Can $a^2+b+2$ and $b^2+4a$ both be perfect squares? Are there any positive integers $a$ and $b$ so that $a^2+b+2$ and $b^2+4a$ are both perfect squares?
|
Suppose that $a\le b$. Since $b^2$ is a perfect square, and $b^2+4a\lt (b+2)^2$, $b^2+4a$ cannot be a perfect square unless $b^2+4a=(b+1)^2$, which is impossible.
Now suppose that $a\gt b$. Then $a^2+b+2$ is too small to be a perfect square, since the smallest perfect square greater than $a^2$ is $a^2+2a+1$.
|
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|
Integration of trigonmetric function $(a\cos x + b\sin x + c)/(1 - d\cos x)^{2}$
Problem: Prove that the indefinite integral $$\int \frac{a\cos x + b\sin x + c}{(1 - d\cos x)^{2}}\,dx$$ is rational function of $\sin x, \cos x$ if and only if $ad + c = 0$.
My Try: Looking at the problem it is obvious that if $a = c = 0$ then the integral is a rational function of $\sin x, \cos x$. Another hope is that the integral if rational should be of the form $(A\cos x + B\sin x)/(1 - d\cos x)$ so that $$\left(\frac{A\cos x + B\sin x}{1 - d\cos x}\right)' = \frac{(B\cos x - A\sin x - Bd)}{(1 - d\cos x)^{2}}$$ Comparing we get $B = a, A = -b, c = -Bd$ so that $ad + c = 0$.
But the problem I see in this intuitive approach is to show that this is the only possible form of the integral if it is supposed to be a rational function of $\sin x, \cos x$. Any helpful ideas or a proper solution is welcome.
Source of the problem: A Course of Pure Mathematics, G. H. Hardy, 10th ed., Page 284, Problem 64.
|
Let
$$E=\frac{b}{d(d\cos x-1)}-\frac{2\left(cd+a\right)}{d^2-1} \cdot\frac{\tan \left(\frac x2\right)}{(d+1)\tan^2 \left(\frac x2\right)-d+1}
$$
$$F=\frac{ad+c}{(d^2-1)^{3/2}}\;\log \left|\frac{\sqrt{\frac{d+1}{d-1}}\tan\left(\frac x2\right)+1}{\sqrt{\frac{d+1}{d-1}}\tan\left(\frac x2\right)-1}\right|$$
Then, if $d^2>1$,
$$\int {{{a\,\cos x+b\,\sin x+c}\over{\left(1-d\,\cos x\right)^2}}
}{\;dx}=E+F$$
And if $d^2<1$, you have another form with arctan, with the same factor $ad+c$ (E is the same):
$$F=\frac{ad+c}{(1-d^2)^{3/2}}\;\arctan \left(\sqrt{\frac{d+1}{d-1}}\tan\left(\frac x2\right)\right)$$
Notice that $\tan \frac x2=\frac{\sin x}{1+\cos x}$.
Thus, the result is a rational function of sin, cos iff $ad+c=0$, but this comes from F.
By the way, if $d=0$, you get trivially a rational function of sin, cos whatever the values of a, b, c.
And if $d=1$, your whole primitive is
$$\frac a6 \cot^3 \left(\frac x2\right)\left(3\tan^2 \left(\frac x2\right)-1\right)-\frac c6 \cot^3 \left(\frac x2\right)\left(3\tan^2 \left(\frac x2\right)+1\right)+\frac{b}{\cos (x) - 1}$$
Thus it's again a rational function of sin, cos whatever the values of a, b, c.
The original statement should be "... iff $ad+c=0$ or $d=0$ or $d=1$".
Also, having written the primitive this way is not really a proof that it cannot be written with a rational function of sin, cos. I mean, it doesn't prove the log or arctan cannot be written otherwise. It looks obvious, but it's not proved as is.
|
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|
For how many integers $a$ is $\frac{2^{10} \cdot 3 ^8 \cdot 5^6}{a^4}$ an integer? In Mathleague $11316$ Target #$4$, the question is:
For how many integers $a$ is $$\frac{2^{10} \cdot 3 ^8 \cdot 5^6}{a^4}$$ an integer?
|
Hint: Factor $a$ into primes; it's evident that these primes must be $2$, $3$, and $5$ (why?). Then $a = 2^d \cdot 3^b \cdot 5^c$, so that
$$a^4 = 2^{4d} \cdot 3^{4b} \cdot 5^{4c}$$
What conditions on $d, b, c$ are necessary and sufficient to make this a divisor of $2^{10} \cdot 3^8 \cdot 5^6$?
|
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|
Can someone help me simplify this trig expression?
$$( \tan x+ \sec x )( \cot x-\cos x ) $$
I got stuck after a few steps of converting and adding and what not.
|
A simplification:
$$
\begin{align*}
\left( \tan x + \sec x \right) \left( \cot x - \cos x \right) &= \left(\tan x + \frac{1}{\cos x} \right)\left( \frac{1}{\tan x } - \cos x\right) \\
&=\frac{\tan x}{\tan x}-\cos x \tan x +\frac{1}{\cos x \tan x}-\frac{\cos x}{\cos x} \\
&=1-\cos x \tan x +\frac{1}{\cos x \tan x}-1 \\
&= \frac{1}{\cos x \tan x}-\cos x \tan x \\
&= \frac{\cos x}{\cos x \sin x}-\frac{\cos x \sin x}{\cos x} \\
&= \frac{1}{\sin x}-\sin x \\
&= \csc x - \sin x.
\end{align*}
$$
|
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|
Solution to a linear Ordinary Differential Equation (ODE) with variable coefficients. I need to solve a series of linear ODEs that are ``close'' to the Euler equation , in a certain sense. One example of my equations is given below:
\begin{equation}
8 \xi^3 \frac{d^3 r_\xi}{d \xi^3} + 36 \xi^2 \frac{d^2 r_\xi}{d \xi^2} + 2(12+\omega^2)\xi \frac{d r_\xi}{d \xi} + \omega^2(2 - \omega \sqrt{\xi}) r_\xi ==0
\end{equation}
As you can see, for small values of $\omega$ the equation is close to the Euler equation and thus is solved by applying an ansatz $r_\xi := \xi^r$. That solution reads:
\begin{equation}
r_\xi = C_\pm \xi^{-1/4} \xi^{\pm \imath \sqrt{\omega^2-1/4}} + C_3 \frac{1}{\xi}
\end{equation}
However, in general ,our equation is not an Euler equation so the ansatz does not work. Yet Mathematica gives the solution readily:
\begin{equation}
r_\xi = C_\pm \xi^{-1/4} \xi^{\pm\imath 1/2\sqrt{\omega^2-1/4}} F_{0,2}\left[{},{5/2\pm 1/2\sqrt{1-4\omega^2},1\pm\sqrt{1-4\omega^2}};\omega^3 \sqrt{\xi}\right] +
C_3 F_{0,2}\left[{},{-1/2+1/2\sqrt{1-4\omega^2},-1/2-1/2\sqrt{1-4\omega^2}};\omega^3 \sqrt{\xi}\right] \frac{1}{\omega^6 \xi}
\end{equation}
where $F_{0,2}$ is the hypergeometric function.
My question is the following. How do I derive this solution? Is it not strange that the solution is expressed through hypergeometric functions which, as we know, are solutions to a second order ODE whereas our ODE is of the third order.
|
Let us use a ``method of variation of constants''. Since we know that for small values of $\xi$ the solution behaves as: $C_3/\xi$ why do we not make the constant $C_3$ a variable. Thus we conjecture:
\begin{equation}
r_\xi := \frac{f_\xi}{\xi}
\end{equation}
Substituting the above into the original ODE yields:
\begin{equation}
\frac{-1}{\sqrt{\xi}} \omega^3 f_\xi + 2 \omega^2 \frac{d f_\xi}{d \xi} + 12 \xi \frac{d^2 f_\xi}{d \xi^2} + 8 \xi^2 \frac{d^3 f_\xi}{d \xi^3} == 0
\end{equation}
Substituting for $u = \sqrt{\xi}$ yields an even simpler equation:
\begin{equation}
-\frac{1}{u} \omega^3 f_u + \frac{1}{u} \omega^2 \frac{d f_u}{du} + u \frac{d^3 f_u}{d u^3} == 0
\end{equation}
This equation is readily solved by series expansion methods $f_u := \sum\limits_{n=0}^\infty a_n u^{n+r}$. The details of this procedure are described in the literature so I will not reiterate them here. It is only worthwhile noting that the resulting iteration equation for the coefficients $a_n$ can indeed be solved since it is a recurrence based on two consecutive values only. In other words the values of $a_{n+1}$ will depend on $a_n$ only rather than on some previous values $a_{n-1},a_{n-2}$, etc. Plugging that final equation into Mathematica we get:
\begin{equation}
f_u = C_\pm u^{\frac{1}{2}(3\pm \sqrt{1-4 \omega^2})}F_{0,2}\left[{},1\pm\sqrt{1-4 \omega^2},\frac{5}{2} \pm \frac{1}{2}\sqrt{1-4 \omega^2}; \omega^3 u\right] + C_3 F_{0,2}\left[{},-\frac{1}{2}-\frac{1}{2}\sqrt{1-4 \omega^2}, -\frac{1}{2}+\frac{1}{2}\sqrt{1-4 \omega^2}; \omega^3 u\right]
\end{equation}
The above solution matches the answer given in the original question.
|
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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|
calculation of $\int\frac{1}{\sin^3 x+\cos^3 x}dx$ and $\int\frac{1}{\sin^5 x+\cos^5x}dx$
Solve the following indefinite integrals:
$$
\begin{align}
&(1)\;\;\int\frac{1}{\sin^3 x+\cos^3 x}dx\\
&(2)\;\;\int\frac{1}{\sin^5 x+\cos^5 x}dx
\end{align}
$$
My Attempt for $(1)$:
$$
\begin{align}
I &= \int\frac{1}{\sin^3 x+\cos ^3 x}\;dx\\
&= \int\frac{1}{\left(\sin x+\cos x\right)\left(\sin^2 x+\cos ^2 x-\sin x \cos x\right)}\;dx\\
&= \int\frac{1}{\left(\sin x+\cos x\right)\left(1-\sin x\cos x\right)}\;dx\\
&= \frac{1}{3}\int \left(\frac{2}{\left(\sin x+\cos x\right)}+\frac{\left(\sin x+\cos x \right)}{\left(1-\sin x\cos x\right)}\right)\;dx\\
&= \frac{2}{3}\int\frac{1}{\sin x+\cos x}\;dx + \frac{1}{3}\int\frac{\sin x+\cos x}{1-\sin x\cos x}\;dx
\end{align}
$$
Using the identities
$$
\sin x = \frac{2\tan \frac{x}{2}}{1+\tan ^2 \frac{x}{2}},\;\cos x = \frac{1-\tan ^2 \frac{x}{2}}{1+\tan^2 \frac{x}{2}}
$$
we can transform the integral to
$$I = \frac{1}{3}\int\frac{\left(\tan \frac{x}{2}\right)^{'}}{1-\tan^2 \frac{x}{2}+2\tan \frac{x}{2}}\;dx+\frac{2}{3}\int\frac{\left(\sin x- \cos x\right)^{'}}{1+(\sin x-\cos x)^2}\;dx
$$
The integral is easy to calculate from here.
My Attempt for $(2)$:
$$
\begin{align}
J &= \int\frac{1}{\sin^5 x+\cos ^5 x}\;dx\\
&= \int\frac{1}{\left(\sin x+\cos x\right)\left(\sin^4 x -\sin^3 x\cos x+\sin^2 x\cos^2 x-\sin x\cos^3 x+\cos^4 x\right)}\;dx\\
&= \int\frac{1}{(\sin x+\cos x)(1-2\sin^2 x\cos^2 x-\sin x\cos x+\sin^2 x\cos^2 x)}\;dx\\
&= \int\frac{1}{\left(\sin x+\cos x\right)\left(1-\sin x\cos x-\left(\sin x\cos x\right)^2\right)}\;dx
\end{align}
$$
How can I solve $(2)$ from this point?
|
For the second integral, decompose the integrand as follows
\begin{align}
&\frac52\int\frac{1}{\sin^5 x+\cos^5 x}dx\\
=&\int\frac{2}{\sin x+\cos x}-\frac{\sin x + \cos x}{2\sin x\cos x-\sec\frac{\pi}5}
-\frac{\sin x + \cos x}{2\sin x\cos x -\sec\frac{3\pi}5}\ dx\\
=&\sqrt2\tanh^{-1}\frac {\sin x- \cos x}{\sqrt2}
-\frac{1}{\sqrt{1-\sec\frac{3\pi}5}}\tanh^{-1}\frac{\sin x- \cos x}{\sqrt{1-\sec\frac{3\pi}5}}\\
&\hspace{4.5cm} +\frac{1}{\sqrt{\sec\frac{\pi}5-1}}\tan^{-1}\frac{\sin x- \cos x}{\sqrt{\sec\frac{\pi}5-1}}
\end{align}
Specifically
\begin{align}
&\int_0^{\pi/2}
\frac{1}{\sin^5 x+\cos^5 x}dx\\
=&\ \frac45\bigg(\sqrt2\coth^{-1}\sqrt2
-\frac{\coth^{-1}\sqrt{\sqrt5+2}}{\sqrt{\sqrt5+2}} +\frac{\cot^{-1}\sqrt{\sqrt5-2}}{\sqrt{\sqrt5-2}}\bigg)
\end{align}
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/595038",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "12",
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|
Finding $\int_{0}^{2\pi} \frac {(1+2\cos\theta)^n\cos(n\theta)}{3+2\cos\theta} \operatorname{d}\theta$ If $n$ is a positive integer find
$$
\int_{0}^{2\pi} \frac {(1+2\cos\theta)^n\cos(n\theta)}{3+2\cos\theta} \operatorname{d}\theta
$$
I know that I have to use contour integral with a circle of radius 1 centered at the origin, but I am having trouble converting the integral into the form $\int_{|z|}$
$$\int_{|z| = 1} \frac{(1+z+1/z)^n\cos(n\theta)}{3+(z+1/z)} \frac{1}{iz} \operatorname{d}z$$
I cant seem to find a way to expand $\cos(n\theta)$ into a function of $z$.
From the above equation, I can get that the poles of is at $z = -1.5 \pm \frac{\sqrt{5}}{2}$ and only the residual of $z = -1.5 + \frac{\sqrt{5}}{2}$ should be included.
|
Note that
$$\cos{n \theta} = \frac12 \left ( z^n + z^{-n}\right )$$
so upon your substitution, you get
$$-\frac{i}{2} \oint_{|z|=1} \frac{dz}{z} \frac{(1+z+z^{-1})^n (z^n+z^{-n})}{3+z+z^{-1}}$$
This can be rearranged to get
$$-\frac{i}{2} \oint_{|z|=1} \frac{dz}{z^{2 n}} \frac{(1+z+z^2)^n (1+z^{2 n})}{1+3 z+z^2} $$
There are poles where $z^2+3 z+1=0$, or
$$z_{\pm} = \frac{-3\pm \sqrt{5}}{2}$$
i.e., $z_- = -\phi^2$, $z_+=-1/\phi^2$, where $\phi=(\sqrt{5}+1)/2$. Thus only $z_+$ is within the unit circle and contributes to the integral.
For the pole at $z=0$, it may be easiest to find the coefficient of $z^{2 n-1}$ in the Maclurin expansion of
$$\frac{(1+z+z^2)^n (1+z^{2 n})}{1+3 z+z^2} $$
Not an easy task. Good luck.
|
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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|
Prove that $\sin^2(A) - \sin^2(B) = \sin(A + B)\sin(A -B)$ This is my attempt:
$$
\begin{align}
& \phantom{={}}\sin^2(A) - \sin^2(B) = \sin(A + B)\sin(A -B) \\[8pt]
& = (\sin(A)\cos(B)+\cos(A)\sin(B))(\sin(A)\cos(B) - \cos(A)\sin(B)) \\[8pt]
& = (\sin(A)+\sin(B))(\cos(B)+\cos(A))(\sin(A)-\sin(B))(\cos(B)-\cos(A)) \\[8pt]
& = (\sin(A)+\sin(B))^2(\cos(B)-\cos(A))^2
\end{align}
$$
But now I can't get rid of the cosines. How can I get rid of them?
|
$x^2-y^2=(x+y)(x-y)$, we have
$$\sin^2A-\sin^2B\\
=(\sin A+\sin B)(\sin A-\sin B)\\
=2\sin\frac{A+B}2\cos\frac{A-B}2\cdot2\cos\frac{A+B}2\sin\frac{A-B}2\\
=2\sin\frac{A+B}2\cos\frac{A+B}2\cdot2\sin\frac{A-B}2\cos\frac{A-B}2\\
=\sin(A+B)\sin(A-B)$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/600681",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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|
Help with hard complex numbers We had the topic of complex numbers for my senior math team meet this week, and I wasn't able to solve two of the problems.
1.) $z=i^{\displaystyle \left(i^{\displaystyle \left(i^{(2)}\right)}\right)}$ and $a$ is the real part of $z$, find the lowest positive value of $\ln(a)$
[ I know it comes to $i-i$ but I don't know why that is e^(pi/2)]
2.) $$\left[\cos \left(\frac{2\pi}{7}\right) + \cos \left(\frac{4\pi}{7}\right) + \cos \left(\frac{8\pi}{7}\right)\right]^2$$
[I think I can use de moivre's forumla, but I dont know how here]
It's non calculator and the answers are $\frac{\pi}{2}$ and \frac{1}{4}$ respectively. I just want to know how to solve them, thanks.
|
Okay, the second one took me some time but I solved it, sorry if I do not know how to write math in this program.
I will use 2 pieces of data, the first:
$(e^x)^i = cos(x)+i*sin(x)$ [Euler's identity]
The second [I don know the name, but is easy to prove]:
if you add the n nth roots of 1 you get 0.
$1+(-1) = 0$
$1+e^{\frac{2\pi i}{3}}+e^{\frac{-2\pi i}{3}}=0$
$1+i-1-i=0$
etc...
Now: $0=e^{\pi i \frac{0}{7}}+e^{\pi i \frac{2}{7}}+e^{\pi i \frac{4}{7}}+e^{\pi i \frac{6}{7}}+e^{\pi i \frac{8}{7}}+e^{\pi i \frac{10}{7}}+e^{\pi i \frac{12}{7}}$
(The 7 roots of 1 added up equal 0)
$S=\cos(\frac{2\pi}{7})+\cos(\frac{4\pi}{7})+\cos(\frac{6\pi}{7})=\cos(\frac{8\pi}{7})+\cos(\frac{10\pi}{7})+\cos(\frac{12\pi}{7})$
$S=\frac{1}{2}\big[\cos(\frac{2\pi}{7})+\cos(\frac{4\pi}{7})+\cos(\frac{6\pi}{7})=\cos(\frac{8\pi}{7})+\cos(\frac{10\pi}{7})+\cos(\frac{12\pi}{7})\big]=$
$\frac{1}{2}\big[\cos(\frac{2\pi}{7})+\cos(\frac{4\pi}{7})+\cos(\frac{6\pi}{7})=\cos(\frac{8\pi}{7})+\cos(\frac{10\pi}{7})+\cos(\frac{12\pi}{7})+\cos(0)-cos(0)\big]$
$=\frac{1}{2} Re(e^{\pi i \frac{0}{7}}+e^{\pi i \frac{2}{7}}+e^{\pi i \frac{4}{7}}+e^{\pi i \frac{6}{7}}+e^{\pi i \frac{8}{7}}+e^{\pi i \frac{10}{7}}+e^{\pi i \frac{12}{7}}+e^{\pi i \frac{0}{7}}-e^{\pi i \frac{0}{7}})$=
$Re(0-e^{\frac{0\pi}{7}})$
[sum of roots equals 0]
$S = \frac{Re(-e^0)}{2} = \frac{-1}{2}$
$S^2 = \frac{1}{4}$
So your answer is $\frac{1}{4}$, the problem is hard but solvable
|
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"timestamp": "2023-03-29T00:00:00",
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|
Which is bigger, $1+3\sqrt{2}$ or $3\sqrt{3}$? Which is bigger, $a = 1+3\sqrt2$ or $b = 3\sqrt3$?
To find out result, I am doing: $(3\sqrt3)^2-(1+3\sqrt2)^2=8-2\sqrt{18}$.
But how can I find if the value of $8-2\sqrt{18}$ is positive or negative?
Thank you.
|
Hint:
$$\color{blue}8-\color{red}{2\sqrt{18}}=\color{blue}{\sqrt{64}}-\color{red}{\sqrt{4}\sqrt{18}}=\color{blue}{\sqrt{64}}-\color{red}{\sqrt{72}}$$
|
{
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"url": "https://math.stackexchange.com/questions/607061",
"timestamp": "2023-03-29T00:00:00",
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|
Comparison of $S_{n}$ and $T_{n}$, where $S_{n} = \sum_{k=1}^{n}\frac{n}{n^2+nk+k^2}$ and $T_{n} = \sum_{k=1}^{n-1}\frac{n}{n^2+kn+k^2}$ Let $\displaystyle S_{n} = \sum_{k=1}^{n}\frac{n}{n^2+nk+k^2}$ and $\displaystyle T_{n} = \sum_{k=1}^{n-1}\frac{n}{n^2+kn+k^2}$ for $n=1,2,3,\dots$
Then which of the following options are Right.
Options:: $\displaystyle (a)\; S_{n}<\frac{\pi}{3\sqrt{3}}\;\;\;\;\;\;\;\;(b)\;S_{n}>\frac{\pi}{3\sqrt{3}}\;\;\;\;\;\;\;\;(c)\; T_{n}<\frac{\pi}{3\sqrt{3}}\;\;\;\;\;\;\;\;(d)\; T_{n}<\frac{\pi}{3\sqrt{3}}\;\;\;\;\;\;\;\;$
$\bf{My\; Try}::$ Given $\displaystyle S_{n} = \sum_{k=1}^{n}\frac{n}{n^2+kn+k^2}<\lim_{n\rightarrow \infty}\sum_{k=1}^{n}\frac{n}{n^2+kn+k^2} = \int_{0}^{1}\frac{1}{1+x+x^2}dx$
(By converting above infinite sum into Integral).
Now Let $\displaystyle I = \int_{0}^{1}\frac{1}{1+x+x^2}dx = \int_{0}^{1}\frac{1}{\left(x+\frac{1}{2}\right)^2+\left(\frac{\sqrt{3}}{2}\right)^2}dx = \frac{2}{\sqrt{3}}\cdot \frac{\pi}{6} = \frac{\pi}{3\sqrt{3}}$
So $\displaystyle S_{n} = \sum_{k=1}^{n}\frac{n}{n^2+kn+k^2}<\frac{\pi}{3\sqrt{3}}$
Now How can I solve for $\displaystyle T_{n} =\sum_{k=1}^{n-1}\frac{n}{n^2+kn+k^2}$
Help Required
Thanks
|
\begin{align*}
T_n&=\sum_{k=0}^{n-1}\frac{n}{n^2+kn+k^2}-\frac{1}{n}\\
&<\lim_{n\rightarrow \infty}\left(\sum_{k=0}^{n-1}\frac{n}{n^2+kn+k^2} -\frac{1}{n}\right)= \int_{0}^{1}\frac{1}{1+x+x^2}dx+0=\frac{\pi}{3\sqrt{3}}
\end{align*}
OR
\begin{align*}
T_n&=\sum_{k=1}^{n}\frac{n}{n^2+kn+k^2}-\frac{1}{3n}\\
&<\lim_{n\rightarrow \infty}\left(\sum_{k=1}^{n}\frac{n}{n^2+kn+k^2} -\frac{1}{3n}\right)= \int_{0}^{1}\frac{1}{1+x+x^2}dx+0=\frac{\pi}{3\sqrt{3}}
\end{align*}
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/607659",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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|
Limit of $\sum_{i=1}^n \left(\frac{{n \choose i}}{2^{in}}\sum_{j=0}^i {i \choose j}^{n+1}\right)$ I'm trying to calculate the limit for the sum of binomial coefficients:
$$S_{n}=\sum_{i=1}^n \left(\frac{{n \choose i}}{2^{in}}\sum_{j=0}^i {i \choose j}^{n+1} \right).$$
|
Simply multiplying the largest term by the number of terms in the sum yields
$$
\begin{align}
\frac{\binom{n}{i}}{2^{in}}\sum_{j=0}^i\binom{i}{j}^{n+1}
&\le(i+1)\binom{i}{i/2}\binom{n}{i}\left(\binom{i}{i/2}2^{-i}\right)^n\tag{1}
\end{align}
$$
using $\binom{i}{i/2}=\binom{i}{(i\pm1)/2}=\frac12\binom{i+1}{(i+1)/2}$ for odd $i$.
Since $\sum\limits_{j=0}^i\binom{i}{j}=2^i$, we know that
$$
\binom{i}{i/2}\le2^i\tag{2}
$$
Furthermore, $\binom{i}{i/2}2^{-i}$ is non-increasing:
$$
\begin{align}
\binom{2k-1}{k-1}2^{-2k+1}
&=\binom{2k}{k}2^{-2k}\\[6pt]
&=\frac{2k+2}{2k+1}\binom{2k+1}{k+1}2^{-2k-1}\\[6pt]
&\gt\binom{2k+1}{k+1}2^{-2k-1}\tag{3}
\end{align}
$$
Note that
$$
\sum\limits_{i=0}^n(i+1)\ 2^i\,\binom{n}{i}=\frac{2n+3}{3}3^n\tag{4}
$$
Combining $(1)$, $(2)$, $(3)$, and $(4)$ yields
$$
\begin{align}
\sum_{i=k}^n\frac{\binom{n}{i}}{2^{in}}\sum_{j=0}^i\binom{i}{j}^{n+1}
&\le\sum_{i=k}^n(i+1)\binom{i}{i/2}\binom{n}{i}\left(\binom{i}{i/2}2^{-i}\right)^n\\
&\le\sum_{i=k}^n(i+1)\ \quad2^i\quad\binom{n}{i}\left(\binom{i}{i/2}2^{-i}\right)^n\\
&\le\frac{2n+3}{3}3^n\left(\binom{k}{k/2}2^{-k}\right)^n\tag{5}
\end{align}
$$
Applying $(5)$ with $k=23$ gives
$$
\begin{align}
\sum_{i=23}^n\frac{\binom{n}{i}}{2^{in}}\sum_{j=0}^i\binom{i}{j}^{n+1}
&\le\frac{2n+3}{3}0.48354^n\\
&=o(2^{-n})\tag{6}
\end{align}
$$
For $i=1$,
$$
\frac{\binom{n}{i}}{2^{in}}\sum\limits_{j=0}^i\binom{i}{j}^{n+1}=\frac{2n}{2^n}\tag{7}
$$
For $i=2$,
$$
\frac{\binom{n}{i}}{2^{in}}\sum\limits_{j=0}^i\binom{i}{j}^{n+1}=\frac{n(n-1)/2}{4^n}\left(2+2^{n+1}\right)\sim\frac{n^2-n}{2^n}\tag{8}
$$
For $i\ge3$,
$$
\begin{align}
\frac{\binom{n}{i}}{2^{in}}\sum\limits_{j=0}^i\binom{i}{j}^{n+1}
&\le(i+1)\binom{i}{i/2}\binom{n}{i}\left(\binom{i}{i/2}2^{-i}\right)^n\\
&=O\left(n^4\left(\frac38\right)^n\right)\\[6pt]
&=o(2^{-n})\tag{9}
\end{align}
$$
because $\binom{n}{i}\le\frac{n^i}{i!}$ and for $i\ge5$, $\binom{i}{i/2}2^{-i}\lt\frac38$.
Thus, $(6)$, $(7)$, $(8)$, and $(9)$ show that
$$
S_n\sim\frac{n^2+n}{2^n}\tag{10}
$$
|
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|
Prove that, for $p > 3$, $(\frac 3p) = 1$ when $p \equiv 1,11 \pmod{12}$ and $(\frac 3p) = -1$ when $p \equiv 5,7 \pmod{12}$ $(\frac 3p)$ is the Legendre symbol here, not a fraction, if that wasn't clear.
This is what I have so far:
We know by Gauss's Lemma that $(\frac qp) = (-1)^v$ where
$$v = \#\left\{1 \le a \le \frac{p-1}{2} \mid aq \equiv k_a \pmod{12}),\space -p/2 < k_a < 0\right\}$$
And we also know that multiples of $(\dfrac{p-1}{2}) \times 3$ must be between $0$ and $p/2$, between $p/2$ and $p$, or between $p$ and $3p/2$.
Representing $p$ as $12k + r$ seems like the next step, but I'm not sure where to go from there.
|
By Quadratic Reciprocity $$\left(\frac{3}{p}\right)=\left(\frac{p}{3}\right)(-1)^{(p-1)(3-1)/4}=\left(\frac{p}{3}\right)(-1)^{(p-1)/2}=\left(\frac{p}{3}\right)\left(\frac{-1}{p}\right).$$ We will use the fact that $-1$ is a quadratic residue modulo $p$ if $p\equiv 1\pmod{4}$ and is not if $p\equiv -1\pmod{4}$. Since $3\nmid p$, $p\equiv 1,2\pmod{3}$. Thus, $\left(\frac{p}{3}\right)=\left(\frac{1}{3}\right)=1$ or $\left(\frac{p}{3}\right)=\left(\frac{2}{3}\right)=-1$. Now, for $\left(\frac{3}{p}\right)=1$ either $\left(\frac{p}{3}\right)=1$ and $\left(\frac{-1}{p}\right)=1$, or they are both equal to $-1$. In the former case, $p\equiv 1\pmod{4}$ and $p\equiv 1\pmod{3}$, so by CRT, $p\equiv 1\pmod{12}$. In the latter case $p\equiv 3\pmod{4}$ and $p\equiv 2\pmod{3}$, so again by CRT $p\equiv 11\equiv -1\pmod{12}$. The result follows from here.
|
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"timestamp": "2023-03-29T00:00:00",
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|
Divisibility by large powers Show that $2^{111} + 1$ divides $2^{555} + 1$ but does not divide $2^{444} + 1$.
My try:
$(2^{111}+1)^5 = 2^{555}+5*2^{444}+10*2^{333}+10*2^{222}+5*2^{111}+1$
$2^{555} + 1$ = $(2^{111}+1)^5$ - $(5*2^{444}+10*2^{333}+10*2^{222}+5*2^{111})$
= $(2^{111}+1)^5$ - ($5*2^{333}(2^{111}+1)+5*2^{222}(2^{111}+1)+5*2^{111}(2^{111}+1))$
Each term is divisible by $(2^{111} + 1)$ and hence divisible by $2^{111} + 1$.
Such a nice distribution is not possible for $2^{444} + 1$
I could answer the first part and the second part of indivisibility is not as clean as I would like. Could someone let me know how I could prove the second part.
|
$a^{odd}+b^{odd}$ is always divisible through $a+b$, and $a^{even}-b^{even}$ is always divisible through $a-b$.
$$a^{2n+1}+b^{2n+1}=(a+b)\cdot\sum_{k=0}^{2n}(-1)^k\cdot a^k\cdot b^{2n-k}$$
E.g., for $n=1$ we have the well-known formula $a^3+b^3=(a+b)(a^2-ab+b^2)$. Likewise, for $n=2$ we have $a^5+b^5=(a+b)(a^4-a^3b+a^2b^2-ab^3+b^4)$. In your case, $a=2^{111}$ and $b=1$
|
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"timestamp": "2023-03-29T00:00:00",
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|
How to prove that $(1+\cos a)/(\sin a)=(\sin a)/(1-\cos a)$? How can I prove this relation $(1+\cos a)/(\sin a)=(\sin a)/(1-\cos a)$ ?
I tried to start from relation $\cos^2a+\sin^2a=1$ but relation went crazy with lot of $\cos$ and $\sin$ and $\sin^2$.
|
$$\begin{align}\dfrac{1+\cos a}{\sin a} & = \dfrac{(1+\cos a)\cdot(1-\cos a)}{\sin a(1-\cos a)}\\ \, \\
&=\dfrac{1-\cos^2a}{\sin a(1-\cos a)}\quad\text{since: $(a+b)\cdot(a-b)=a^2-b^2$}\\\,\\
&=\dfrac{\cos^2a+\sin^2a-\cos^2a}{\sin a(1-\cos a)}\quad\text{since: $\cos^2a+\sin^2a=1$}\\\,\\
&=\dfrac{\sin^2a}{\sin a(1-\cos a)}\\\,\\&=\dfrac{\sin a\cdot\sin a}{\sin a(1-\cos a)}\\\,\\
&=\boxed{\dfrac{\sin a}{1-\cos a}}
\end{align}$$
|
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|
Probability that $5$ divides $a^2+b^2$ is $\frac9{25}$
Two positive integers $a$ and $b$ are randomly selected with replacement, then prove that the probability of $(a^2 +b^2)/5$ being positive integer is $9/25$.
I found out the pattern of the last digits of $a^2$ , $b^2$ but then the sample space isn't finite so I dropped idea of thinking of finding the a's and b's such that their square's last digit is either 0 or 5. Please help.
|
The idea of this question is that $a$ has $20\%$ chances of being each of $0$ or $1$ or $2$ or $3$ or $4$ modulo $5$, hence $a^2$ has $20\%$ chances of being $0$, $40\%$ to be $1$ and $40\%$ to be $4$ modulo $5$. The same holds for $b^2$ hence, assuming that $a$ and $b$ are independent, $a^2+b^2$ is $0$ modulo $5$ when $(a^2,b^2)$ is $(0,0)$ or $(1,4)$ or $(4,1)$ modulo $5$, that is, with probability
$$20\%\times20\%+40\%\times40\%+40\%\times40\%=36\%.
$$
This can be made fully rigorous by assuming that $a$ and $b$ are uniform between $1$ and $n$, and letting $n$ go to infinity. Then, for every $n$ multiple of $5$, the formal proof above becomes rigorous, hence the probability is exactly $36\%$, which implies that the limit when $n\to\infty$ ($n$ multiple of $5$ or not) indeed exists and is $36\%$.
The same reasoning shows that $a^2+b^2$ is $1$ or $2$ or $3$ or $4$ modulo $5$, each with probability $16\%$, in other words, the odds for $0:1:2:3:4$ are $9:4:4:4:4$.
|
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|
Why is $\left(\sum \limits_{n=1}^{\infty}\frac1{n^3}\right)^3 \gt \sum \limits_{n=1}^{\infty}\frac1{n^4} $?
$$\displaystyle \left(\sum \limits_{n=1}^{\infty}\frac1{n^3}\right)^3 \gt \sum_{n=1}^{\infty}\frac1{n^4} $$
Why is this true ?
The LHS can be written either as it is or like this: $\displaystyle \sum \limits_{n=1}^{\infty}\frac1{n^9}$ which is clearly smaller.
|
In general a power of a sum is different from the sum of powers, $(a+b)^3 = a^3 + b^3 + 3ab(a+b)$ coincides with $a^3 + b^3$ only if either (at least) one of $a,b$ is zero or $a+b = 0$. For sums with more terms or series, conditions for equality aren't so straightforward. Here we have
$$1 < \sum_{n=1}^\infty \frac{1}{n^4} < \sum_{n=1}^\infty \frac{1}{n^3} < \left(\sum_{n=1}^\infty \frac{1}{n^3}\right)^3,$$
the first inequality because the first term in the series is $1$ and all further terms positive, the second because for $n > 1$ we have $\dfrac{1}{n^4} < \dfrac{1}{n^3}$, and the last because $x < x^3$ for all $x > 1$.
|
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|
How prove this equation $A^2+B^2=C^2+D^2$ define:
plane $W:Ax+By+Cz+D=0$ and the
hyperboloid of one sheet $U:x^2+y^2-z^2=1$
if $$W\bigcap U=l_{1},W\bigcap U=l_{2}$$
where $l_{1},l_{2}$ be two straight lines
show that :$$A^2+B^2=C^2+D^2$$
My try: since
$$\begin{cases}
Ax+By+Cz+D=0\\
x^2+y^2-z^2=1
\end{cases}$$
then
$$(Ax+By+D)^2=C^2(x^2+y^2-1)$$
then
$$(A^2-C^2)x^2+(B^2-C^2)y^2+2ABxy+2BDy+2ADx+D^2+C^2=0$$
Follow is user44197 idea
$$(A^2-C^2)x^2+(2ABy+2AD)x+(B^2-C^2)y^2+D^2+C^2=0$$
$$\Delta (y)=(2ABy+2AD)^2-4(A^2-C^2)[(B^2-C^2)y^2+D^2+C^2]$$
$$\Delta=4A^2B^2y^2+8A^2BDy+4A^2D^2-4(A^2B^2-A^2C^2-B^2C^2+C^4)y^2-4(A^2D^2+A^2C^2-C^2D^2-C^4)$$
so
$$\Delta(y)=4[C^2(A^2+B^2-C^2)y^2+2A^2BDy+C^2(C^2+D^2-A^2)]$$
then I can't.Thank you very much!
|
Here's an answer with a bit of a geometric flavor.
Clearly, the hyperboloid contains no lines that are parallel to any of the coordinate axes; it also contains no points $(x,y,z)$ such that $x^2 + y^2 < 1$. Consequently, an embedded line $\ell$ must pass through a point $P$ on the $xy$-plane, which must in fact lie on the unit circle; we can write $P = (\cos\theta, \sin\theta, 0)$ for some $\theta$. Moreover, the projection of $\ell$ into the $xy$-plane must be tangent to the unit circle, so its direction vector, $v$, has the form $(\sin\theta,-\cos\theta,c)$ for some $c$. That is, the line has this parameterization:
$$\ell : (\cos\theta, \sin\theta, 0 ) + t (\sin\theta,-\cos\theta, c)$$
and we have
$$0 = x^2 + y^2 - z^2 - 1 = (\cos\theta+t\sin\theta)^2+(\sin\theta-t\cos\theta)^2+(ct)^2-1 =t^2(c^2-1)$$
which must hold for all $t$, so that $c=\pm 1$.
Since $P$ is on the plane:
$$A \cos\theta + B \sin\theta + D = 0 \qquad \implies \qquad D = -\left(A \cos\theta+ B \sin\theta\right)$$
Since $v \perp (A,B,C)$:
$$v\cdot(A,B,C) = A\sin\theta - B\cos\theta \pm C = 0 \qquad \implies \qquad C = \mp \left( A \sin\theta - B \cos\theta \right)$$
Therefore,
$$C^2 + D^2 = A^2 + B^2 \qquad (\star)$$
Note: We have shown that $(\star)$ holds whenever the plane meets the hyperboloid in even a single line. This is consistent with the fact that $(\star)$ removes only a single degree of freedom in the plane equation.
|
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|
show that $\int_0^1 (\sqrt[3]{1-x^7}-\sqrt[7]{1-x^3}) \, dx=0$ show that
$$I=\int_0^1 (\sqrt[3]{1-x^7}-\sqrt[7]{1-x^3}) \, dx=0$$
I find this is Nice equalition!
My try: let
$$\sqrt[3]{1-x^7}=t\Longrightarrow x=\sqrt[7]{1-t^3}$$
so
$$dx=-\dfrac{3}{7}t^2(1-t^3)^{-\dfrac{6}{7}} \, dt$$
so
$$I=\frac{3}{7}\int_0^1 \frac{t^3}{\sqrt[7]{(1-t^3)^6}} \, dt-\int_0^1 \sqrt[7]{1-x^3} \, dx$$
By parts,we have
$$\int_0^1 \sqrt[7]{1-x^3} dx=\dfrac{3}{7}\int_0^1 \frac{x^3}{\sqrt[7]{(1-x^3)^6}} \, dt$$
so
$$I=0$$
this problem maybe have more other nice methods!Thank you
|
The antiderivative of $(1-x^a)^{1/b}$, at least for integer values of $a$ and $b$, is given by $x{\rm Hypergeometric2F1}[1/a, -(1/b), 1 + 1/a, x^a]$.
The integral of $(1-x^a)^{1/b}$ between 0 and 1 is then given by
$(\Gamma[1 + 1/a] \Gamma[1 + 1/b]) / \Gamma[1 + 1/a + 1/b]$
Then, the two integrals are equal and your identity is proved.
|
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"timestamp": "2023-03-29T00:00:00",
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|
question on integrals Let $\displaystyle A=\int_0^1 \frac{dx}{1+x^8}$. Then which of the following are true:
1) $A\lt 1$,
2) $A\gt 1$,
3) $A\lt \frac{\pi}{4}$,
4) $A\gt\frac{\pi}{4}$.
|
If $0\lt x\lt 1$, we have
$$\frac{1}{1+x^2}\lt \frac{1}{1+x^8}\lt \frac{1}{1+0^2}.$$
Hence, we have
$$\frac{\pi}{4}=\int_0^1\frac{1}{1+x^2}dx\lt\int_0^1\frac{1}{1+x^8}dx\lt\int_0^1\frac{1}{1+0^2}dx=1.$$
That leads that the answer $4)$ and $1)$.
|
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|
If $n\in\mathbb N$ and $4^n+2^n+1$ is prime, prove that there exists an $m\in\mathbb N\cup\{0\}$ such that $n=3^m$. If $n\in\mathbb N$ and $4^n+2^n+1$ is prime, prove that there exists an $m\in\mathbb N\cup\{0\}$ such that $n=3^m$.
I.e. if $4^n+2^n+1$ is prime, prove that $n=3^m$, where $m\in\mathbb N\cup\{0\}$.
I don't know how I could solve this, I don't have any ideas. I know that $4^n+2^n+1$ is obviously odd. Assume $m=0$, i.e. $n=1$. Then $3$ is prime. So that doesn't disprove what we're trying to show.
So assume $m>0$. Then $4^n+2^n+1>3$, so $4^n+2^n+1$ is either $3k+1$ or $3k+2$, where $k\in\mathbb N$. So either $4^n+2^n$ or $4^n+2^n-1$ is divisible by 3. I think it's not something that could help out at solving this. Just a little observation. I don't have any ideas.
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Hint. For prime $p\not=3$ we have
$$(x^2+x+1)\mid (x^{2p}+x^p+1)$$
Proof.
\begin{align}x^{2p}+x^p+1&=\frac{x^{3p}-1}{x^p-1}\\&=\frac{(x^3-1)\left(\sum\limits_{k=0}^{p-1}x^{3k}\right)}{x^p-1}\\&=\frac{(x^2+x+1)\left(\sum\limits_{k=0}^{p-1}x^{3k}\right)}{(x^p-1)/(x-1)}\end{align} Now it suffices to prove \begin{align}\frac{x^p-1}{x-1}\big|\sum_{k=0}^{p-1}x^{3k}\end{align} Since $x^p\equiv 1\pmod{x^p-1}$, we see $x^{np+j}\equiv x^j\pmod{\frac{x^p-1}{x-1}}$. Now note that since $3\not\mid p$, for $k=0,1,\cdots,p-1$ the sequence $3k$ form a complete residue system of $p$. So \begin{align}\sum_{k=0}^{p-1}x^{3k}&\equiv\sum_{j=0}^{p-1}x^j\pmod{\frac{x^p-1}{x-1}}\\ & \equiv\frac{x^p-1}{x-1}\equiv 0\pmod{\frac{x^p-1}{x-1}}\end{align}
|
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|
Solving $\frac{\partial^2 u}{\partial x^2}+\frac{\partial^2 u}{\partial y^2}=c$ Question:
let $a,b,c$ be positive constants. Find $u=u(x,y)$ if is satisfies the partial differential equation
$$\dfrac{\partial^2 u}{\partial x^2}+\dfrac{\partial^2 u}{\partial y^2}=c$$
and the boundary condition
$$u=0,\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}=1.$$
my try: I know this is screened Poisson equation
$$\dfrac{\partial^2 u}{\partial x^2}+\dfrac{\partial^2 u}{\partial y^2}=c$$
I only find this poisson equation one of the solution
$$u(x,y)=-\dfrac{c}{2}\cdot\dfrac{a^2b^2}{a^2+b^2}\left(1-\dfrac{x^2}{a^2}-\dfrac{y^2}{b}\right)$$
because
$$\dfrac{\partial u}{\partial x}=-\dfrac{c}{2}\cdot\dfrac{a^2b^2}{a^2+b^2}\left(-\dfrac{2x}{a^2}\right)$$
$$\dfrac{\partial^2 u}{\partial x^2}=c\cdot\dfrac{b^2}{a^2+b^2}$$
and
$$\dfrac{\partial^2 u}{\partial y^2}=c\cdot\dfrac{a^2}{a^2+b^2}$$
so
$$\dfrac{\partial^2 u}{\partial x^2}+\dfrac{\partial^2 u}{\partial y^2}=c\cdot\left(\dfrac{a^2}{a^2+b^2}+\dfrac{b^2}{a^2+b^2}\right)=c$$
and when
$$\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}=1\Longrightarrow u=0$$
and this solution is uniqueness? and How prove it? Thank you
Thank you very much!
|
Hint. Try
$$
u(x,y)=\frac{x^2}{a^2}+\frac{y^2}{b^2}-1.
$$
Find a solution, and then show uniqueness.
|
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|
How find this $\left(\frac{1}{x^2+a^2}\right)^{(n)}$ Prove that
$$\left(\dfrac{1}{x^2+a^2}\right)^{(n)}=(-1)^{(n)}n!\dfrac{\sin{[(n+1)\cdot \mathrm{arccot}{(x/a)}]}}{a(x^2+a^2)^{(n+1)/2}}$$
my try:
since
$$\dfrac{1}{x^2+a^2}=\dfrac{1}{2ai}\left(\dfrac{1}{x-ai}-\dfrac{1}{x+ai}\right),i=\sqrt{-1}$$
so
$$\left(\dfrac{1}{x^2+a^2}\right)^{(n)}=\dfrac{(-1)^nn!}{2ai}\left(\dfrac{1}{(x-ai)^{n+1}}-\dfrac{1}{(x+ai)^{n+1}}\right)$$
so
let$$x=a\cot{\theta},0<\theta<\pi,$$
then
$$x\pm ai=a(\cos{\theta}\pm i\sin{\theta})/\sin{\theta}$$
so
$$\dfrac{1}{(x\pm ai)^{n+1}}=\dfrac{\sin^{n+1}{\theta}}{a^{n+1}}[\cos{(n+1)\theta}\mp i\sin{(n+1)\theta}]$$
so$$\left(\dfrac{1}{x^2+a^2}\right)^{(n)}=(-1)^{(n)}n!\dfrac{\sin{[(n+1)\cdot \mathrm{arccot}{(x/a)}]}}{a(x^2+a^2)^{(n+1)/2}}$$
Question:
Have other methods?
Because this is important reslut,so I think this have other methods? Thank you
|
First Method
Let $$\mathcal{L}\{f(t)\}=\int_0^\infty f(t)\operatorname{e}^{-xt}\operatorname{d}t=F(x)$$ be the Laplace transform of $f(t)$.
For $f(t)=\sin(at)u(t)$ then we have $$F(x)=\int_0^\infty \sin(at)\operatorname{e}^{-xt}\operatorname{d}t=\frac{a}{x^2+a^2}.$$
Recalling that $\mathcal{L}\{t^n f(t)\}=(-1)^n F^{(n)}(x)$ we have
$$\begin{align}
\frac{(-1)^nF^{(n)}(x)}{a}=\left(\frac{1}{x^2+a^2}\right)^{(n)}&=\frac{(-1)^n\mathcal{L}\{t^n f(t)\}}{a}=\frac{(-1)^n}{a}\int_0^\infty t^n \sin(at)\operatorname{e}^{-xt}\operatorname{d}t\\
&=\frac{(-1)^n}{2ia}\left[\int_0^\infty t^n \operatorname{e}^{-(x-ia)t}\operatorname{d}t-\int_0^\infty t^n \operatorname{e}^{-(x+ia)t}\operatorname{d}t\right]\\
&=\frac{(-1)^n}{2ia}\Gamma(n+1)\left[\frac{1}{(x-ia)^{n+1}}-\frac{1}{(x+ia)^{n+1}}\right]
\end{align}
$$
using the identity $\sin(at)=\frac{\operatorname{e}^{iat}-\operatorname{e}^{-iat}}{2i}$ and the Gamma Function.
For $x=a\cot{\theta},\,0<\theta<\pi,$ we have
$$
x\pm ia=\frac{a}{\sin\theta}\operatorname{e}^{\pm i\theta}={(x^2+a^2)^{1/2}}\operatorname{e}^{\pm i\theta}
$$
and finally
$$
\left(\frac{1}{x^2+a^2}\right)^{(n)}=\frac{(-1)^n}{2ia}\Gamma(n+1)\frac{\left[\operatorname{e}^{+i(n+1)\theta}-\operatorname{e}^{-i(n+1)\theta}\right]}{(x^2+a^2)^{\frac{n+1}{2}}}=(-1)^n n!\frac{\sin\left((n+1)\cot^{-1}\left(\frac{x}{a}\right)\right)}{a(x^2+a^2)^{\frac{n+1}{2}}}.
$$
Second Method
Observing that $$\frac{1}{x^2+a^2}=\frac{1}{x+ia}\cdot\frac{1}{x-ia}$$ and using the general Leibniz rule $$
(f \cdot g)^{(n)}=\sum_{k=0}^n {n \choose k} f^{(k)} g^{(n-k)} $$ with $f(x)=\frac{1}{x+ia}$ and $g(x)=\frac{1}{x-ia}$ we have $$\frac{\operatorname{d}^n}{\operatorname{d}x^n}(x\pm ia)^{-1}=(-1)^n n!(x\pm ia)^{-(n+1)}$$
and then
$$
\begin{align}
\left(\frac{1}{x^2+a^2}\right)^{(n)}&=\sum_{k=0}^n {n \choose k} (-1)^k k!(x+ ia)^{-(k+1)}(-1)^{n-k} (n-k)!(x- ia)^{-(n-k+1)}\\
&=(-1)^n n!\sum_{k=0}^n (x+ ia)^{-(k+1)}(x- ia)^{-(n-k+1)}.
\end{align}
$$
For $x=a\cot{\theta},\,0<\theta<\pi,$ we have
$$
x\pm ia=\frac{a}{\sin\theta}\operatorname{e}^{\pm i\theta}={(x^2+a^2)^{1/2}}\operatorname{e}^{\pm i\theta}
$$
and then
$$
\begin{align}
\left(\frac{1}{x^2+a^2}\right)^{(n)}
&=(-1)^n n!\left(\frac{\sin\theta}{a}\right)^{n+2}\sum_{k=0}^n \operatorname{e}^{+i(n-2k)\theta}\\
&= (-1)^n n!\left(\frac{\sin\theta}{a}\right)^{n+2}\operatorname{e}^{+i n\theta}\frac{1-\operatorname{e}^{-2i\theta(n+1)}}{1-\operatorname{e}^{-2i\theta}}
\end{align}
$$
using the geometric sum $ \sum_{k=0}^{n} z^k = \frac{1-z^{n+1}}{1-z} $ with $z=\operatorname{e}^{-2i\theta}$.
Using the Euler's identity $\operatorname{e}^{+i\varphi}-\operatorname{e}^{-i\varphi}=2i\sin\varphi$ and multiplying and dividing by $a\operatorname{e}^{i\theta}$ we obtain
$$
\begin{align}
\left(\frac{1}{x^2+a^2}\right)^{(n)}
&= (-1)^n n!\left(\frac{\sin\theta}{a}\right)^{n+2}\frac{a\operatorname{e}^{i\theta}}{a\operatorname{e}^{i\theta}}\frac{\operatorname{e}^{i\theta(n+1)}-\operatorname{e}^{-i\theta(n+1)}}{\operatorname{e}^{i\theta}-\operatorname{e}^{-i\theta}}\\
&=(-1)^n n!\left(\frac{\sin\theta}{a}\right)^{n+2}\frac{a}{\sin\theta}\frac{1}{a}\sin((n+1)\theta)\\
&=(-1)^n n!\frac{\sin\left((n+1)\cot^{-1}\left(\frac{x}{a}\right)\right)}{a(x^2+a^2)^{\frac{n+1}{2}}}.
\end{align}
$$
Third Method
Let be $$f(x)=\frac{1}{x^2+a^2}=\frac{1}{a^2}\frac{1}{1+t^2}=\frac{1}{a^2}\psi(t).$$ Observe that $$\frac{\operatorname{d}^n f(x)}{\operatorname{d}x^n}=\frac{1}{a^{n+2}}\frac{\operatorname{d}^{n+1} \psi(t)}{\operatorname{d}t^{n+1}}$$ where $\psi(t)=\arctan(t)$.
Putting $\sin\theta=\frac{1}{\sqrt{1+t^2}}$ the $n$th-derivative of $\psi(t)$ is
$$
\psi^{(n)}(t)=(-1)^{n-1}(n-1)!\sin^n\theta\sin(n\theta)\tag 1
$$
The existence of the derivatives follows from the analyticity of $\arctan t$ on the real line. The proof of formula (1) is by mathematical induction. Clearly, the (1) is true
for $n = 1$. Suppose the (1)is true for $n = k$; that is, suppose
$$
\psi^{(k)}(t)=(-1)^{k-1}(k-1)!\sin^k\theta\sin(k\theta)\tag 2
$$
We will show that the (1) is true for $n = k + 1$ whenever it is true for $n = k$.
Differentiating both sides of (2) with respect to $t$ and noting that $\frac{\operatorname{d} \theta}{\operatorname{d}t}=-\sin^2\theta$ gives
$$\frac{\operatorname{d}\psi^{(k)}(t)}{\operatorname{d}t}=(-1)^{k}k!\sin^{k+1}\theta[\cos\theta\sin(k\theta)+\cos(k\theta)\sin\theta]=(-1)^{k}k!\sin^{k+1}\theta\sin((k+1)\theta)$$
that is $$\psi^{(k+1)}(t)= (-1)^{k}k!\sin^{k+1}\theta\sin((k+1)\theta)$$ so the (1) is true for any $n\ge 1$.
Thus we have
$$
\left(\frac{1}{x^2+a^2}\right)^{(n)}=\frac{1}{a^{n+2}}\psi^{(n+1)}(t)=\frac{1}{a^{n+2}}(-1)^{n}n!\sin^{n+1}\theta\sin((n+1)\theta)
$$
and observing that $\frac{\sin\theta}{a}=\frac{1}{a\sqrt{1+t^2}}=\frac{1}{(x^2+a^2)^{1/2}}$ finally we obtain
$$
\left(\frac{1}{x^2+a^2}\right)^{(n)}=(-1)^n n!\frac{\sin\left((n+1)\cot^{-1}\left(\frac{x}{a}\right)\right)}{a(x^2+a^2)^{\frac{n+1}{2}}}.
$$
|
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|
Sum of an infinite series - obscure sum I am trying to show the following:
I can get to the following result:
I would appreciate any help with solving this problem.
Thanks in advance.
|
First thing to do is replace for $$x=\frac{s}{2}$$
$$\implies \frac{-q^2}{8\pi e} \left( \frac{1}{2\frac{s}{2}} + \sum_{n=1}^{\infty} \left(\frac{ns}{(ns)^2-(\frac{s}{2})^2} - \frac{1}{ns} \right) \right)$$
$$\frac{-q^2}{8\pi e} \left( \frac{1}{s} + \sum_{n=1}^{\infty} \left( \left(\frac{s}{s^2}\right)\frac{n}{(n)^2-(\frac{1}{2})^2} - \frac{1}{ns} \right) \right)$$
$$\frac{-q^2}{8\pi es}\left( 1 + \sum_{n=1}^{\infty} \left( \frac{n}{(n)^2-(\frac{1}{2})^2} - \frac{1}{n} \right) \right)$$
$$\frac{-q^2}{8\pi es}\left( 1 + \sum_{n=1}^{\infty} \frac{1}{n(2n-1)(2n+1)} \right)$$
The sum can be rewritten as:
$$\sum_{n=1}^{\infty} \frac{1}{n(2n-1)(2n+1)}= 2 \sum_{n=1}^{\infty} \frac{1}{(2n)(2n-1)(2n+1)}$$
Method 1:
If we let,
$$f(x)=\sum_{n=1}^{\infty} \frac{x^{2n+1}}{(2n)(2n-1)(2n+1)}$$
$$\implies f^{(3)}(x)=\sum_{n=1}^{\infty}x^{2n-2}=\sum_{n=0}^{\infty}x^{2n}=\frac{1}{1-x^2}$$
Now you must find $f(1)$
Since $$f(0)=f'(0)=f''(0)=0$$
$$\implies \int_0^{1} \int_0^{x} \int_0^{y}f'''(z)dzdydx=f(1)$$
$$\therefore 2f(1)=\int_0^1 (1-z)^2 f'''(z)dz=\int_0^1 (1-z)^2\frac{1}{1-z^2}dz=\int_0^1 \frac{1-z}{1+z}dz=2\ln(2)-1$$
So, $f(1)=\ln(2)-\frac{1}{2}$
$$\sum_{n=1}^{\infty} \frac{1}{n(2n-1)(2n+1)}= 2 \sum_{n=1}^{\infty} \frac{1}{(2n)(2n-1)(2n+1)}=2\ln(2)-1$$
So, the final sum is
$$E_i(x=s/2)=\frac{-q^2}{8\pi es} (1 + 2 \ln(2)-1)=\frac{-q^2}{8\pi es} \ln(4)$$
Method 2
Use partial fractions:
$$\frac{1}{(2n)(2n+1)(2n-1)}=\frac{A}{2n}+\frac{B}{2n+1}+\frac{C}{2n-1}$$
Solve for $A,B,C$
Proof: On how to prove $\int_0^x \int_0^y \int_0^z f(t) \ dt \ dz \ dy=\frac{1}{2} \int_0^x (x-t)^2 f(t) \ dt$
First imagine the triple integral as the integral of a region A,
$$\iiint_A f(t) dV$$
where $A=\{(t,z,y) \ | \ 0<t<z, \ 0<z<y, \ 0<y<x\}$
The objective is to change the order of integration to $dt$ being last because this avoids the complexity of having to integrate $f(t)$. So, now take $B$ as the projection on the $t\text{-} y$ plane then
$$B=\{(t,y) \ | \ 0<t<x, \ t<y<x \}$$
So, what is the new region $A$ with an integration order of $dz \ dy \ dt$?
Once you find the new region $A$ (which is best found by drawing it); it is simply:
$$A=\{(t,z,y)\ | \ t\leq z \leq y, \ t\leq y \leq x , \ 0\leq t \leq x \} $$
In this case, you can skip the drawing because the inequalities of region $A$ are simple and its just a matter of changing the function according to the order of integration. More specifically, $z$ is the first order, so its limits, from the first region of $A$, are $t<z$ and $z<y$ which is $t<z<y$. The other limits come from the projection, $B$
$$\therefore \int_0^x \int_0^y \int_0^z f(t) \ dt \ dz \ dy = \int_0^x \int_t^x \int_t^y f(t) \ dz \ dy \ dt=\int_0^x \left[\int_t^x (y-t)f(t) \ dy \right] \ dt$$
$$=\int_0^x \left(\frac{1}{2}x^2-tx-\frac{1}{2}t^2+t^2 \right) f(t) \ dt= \frac{1}{2} \int_0^x \left( x^2-2tx+t^2 \right) f(t) \ dt$$
$$=\frac{1}{2}\int_0^x \left(x-t\right)^2 f(t) \ dt$$
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|
Integration step need some explanation (Calculus) $\displaystyle{\frac15\int_0^1\left[v(1+v^2)^5-v^{11}\right]dv=\frac 15\left[\frac 12\cdot\frac 16(1+v^2)^6-\frac{1}{12}v^{12}\right]_0^1}$
Can someone explain how to get the RHS from LHS. It's the first term after the integral sign that I am confused about.
Thanks.
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Remember that integration is linear so:
$$\int x(1+x^2)^5-x^{11} \, \operatorname{d}\!x = \int x(1+x^2)^5 \, \operatorname{d}\!x - \int x^{11} \, \operatorname{d}\!x$$
Hopefully you already know that
$$\int x^{11} \, \operatorname{d}\!x = \frac{1}{12}x^{12}+C$$
You need to show that
$$\int x(1+x^2)^5 \, \operatorname{d}\!x = \frac{1}{12}(1+x^2)^6 + C$$
We can perform this integration by substitution. If $u=1+x^2$ then $\operatorname{d}\!u = 2x\,\operatorname{d}\!x$. Hence
$$\int x(1+x^2)^5 \, \operatorname{d}\!x = \int u^5 \cdot \tfrac{1}{2}\,\operatorname{d}\!u = \frac{1}{2}\int u^5 \, \operatorname{d}\!u = \frac{1}{2}\cdot\frac{1}{6}\cdot u^6 + C$$
Recalling that $u=1+x^2$, we may conclude that
$$\int x(1+x^2)^5 \, \operatorname{d}\!x = \frac{1}{2}\cdot \frac{1}{6} \cdot (1+x^2)^6 + C$$
|
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Why is it that $2^{10} + 2^{9} + 2^{8} + \cdots + 2^{3} + 2^2 + 2^1 = 2^{11} - 2$? Why is it that $$2^{10} + 2^{9} + 2^{8} + \cdots + 2^{3} + 2^2 + 2^1 = 2^{11} - 2?$$
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Try using the fact that $x^n + x^{n - 1} + \cdots + x + 1 = \frac{x^{n + 1} - 1}{x - 1}$.
|
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|
Find $\int_0^{\frac{\pi}{4}}e^{\sec^2 x}dx$ How can we find
$$\int_0^{\frac{\pi}{4}}e^{\sec^2 x}dx$$
I tried $t=\frac{\pi}{4}-x$ but this seems not work. Any hints?
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$\int_0^\frac{\pi}{4}e^{\sec^2 x}~dx$
$=\int_0^\frac{\pi}{4}\sum\limits_{n=0}^\infty\dfrac{\sec^{2n}x}{n!}~dx$
$=\int_0^\frac{\pi}{4}dx+\int_0^\frac{\pi}{4}\sum\limits_{n=1}^\infty\dfrac{\sec^{2n}x}{n!}~dx$
$=[x]_0^\frac{\pi}{4}+\int_0^\frac{\pi}{4}\sum\limits_{n=1}^\infty\dfrac{\sec^{2n-2}x}{n!}~d(\tan x)$
$=\dfrac{\pi}{4}+\int_0^\frac{\pi}{4}\sum\limits_{n=1}^\infty\dfrac{(1+\tan^2x)^{n-1}}{n!}~d(\tan x)$
$=\dfrac{\pi}{4}+\int_0^\frac{\pi}{4}\sum\limits_{n=1}^\infty\sum\limits_{k=1}^n\dfrac{C_{k-1}^{n-1}\tan^{2k-2}x}{n!}~d(\tan x)$
$=\dfrac{\pi}{4}+\left[\sum\limits_{n=1}^\infty\sum\limits_{k=1}^n\dfrac{(n-1)!\tan^{2k-1}x}{n!(n-k)!(k-1)!(2k-1)}\right]_0^\frac{\pi}{4}$
$=\dfrac{\pi}{4}+\sum\limits_{n=1}^\infty\sum\limits_{k=1}^n\dfrac{1}{n(n-k)!(k-1)!(2k-1)}$
$=\dfrac{\pi}{4}+\sum\limits_{k=1}^\infty\sum\limits_{n=k}^\infty\dfrac{1}{n(n-k)!(k-1)!(2k-1)}$
$=\dfrac{\pi}{4}+\sum\limits_{k=0}^\infty\sum\limits_{n=0}^\infty\dfrac{1}{(n+k+1)n!k!(2k+1)}$
$=\dfrac{\pi}{4}+\int_0^1\sum\limits_{k=0}^\infty\sum\limits_{n=0}^\infty\dfrac{t^{n+k}}{n!k!(2k+1)}~dt$
$=\dfrac{\pi}{4}+\int_0^1\sum\limits_{k=0}^\infty\dfrac{t^ke^t}{k!(2k+1)}~dt$
$=\dfrac{\pi}{4}+\left[\sum\limits_{k=0}^\infty\sum\limits_{n=0}^k\dfrac{(-1)^{k-n}t^ne^t}{n!(2k+1)}\right]_0^1$ (according to https://en.wikipedia.org/wiki/List_of_integrals_of_exponential_functions#Indefinite_integral)
$=\dfrac{\pi}{4}+\sum\limits_{k=0}^\infty\sum\limits_{n=0}^k\dfrac{(-1)^{k-n}e}{n!(2k+1)}-\sum\limits_{k=0}^\infty\dfrac{(-1)^k}{2k+1}$
$=\dfrac{\pi}{4}+\sum\limits_{n=0}^\infty\sum\limits_{k=n}^\infty\dfrac{(-1)^{k-n}e}{n!(2k+1)}-\dfrac{\pi}{4}$
$=\sum\limits_{n=0}^\infty\sum\limits_{k=0}^\infty\dfrac{(-1)^ke}{n!(2n+2k+1)}$
$=\dfrac{e}{2}\Phi_1\left(\dfrac{1}{2},1,\dfrac{3}{2};-1,1\right)$ (according to http://en.wikipedia.org/wiki/Humbert_series)
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|
Inequality $\cos x+\cos y+\cos z=0$ and $\cos{3x}+\cos{3y}+\cos{3z}=0$ prove that $\cos{2x}\cdot \cos{2y}\cdot \cos{2z}\le 0$ Let $x,y,z$ be real numbers such that $\cos x+\cos y+\cos z=0$ and $\cos{3x}+\cos{3y}+\cos{3z}=0$ prove that $\cos{2x}\cdot \cos{2y}\cdot \cos{2z}\le 0$.
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Setting $\cos x=a$ etc.
We have
$\displaystyle a+b+c=0\ \ \ \ (1)$
$\displaystyle\implies a^3+b^3+c^3=(a+b)^3-3ab(a+b)+c^3=(-c)^3+-3ab(-c)+c^3$
$\displaystyle\implies a^3+b^3+c^3=3abc\ \ \ \ (2)$
Again as $\displaystyle\cos3x=4\cos^3x-3\cos x\implies,$
$\sum\cos3x=0\implies 4(a^3+b^3+c^3)=3(a+b+c)=0\ \ \ \ (3)$
From $\displaystyle (2),(3) 3abc=0$
$\displaystyle\implies $ at least one of $a,b,c$ is zero
If $c=0,$ from $(1), a+b=0\iff b=-a$
$\displaystyle\implies\cos2x\cos2y\cos2z=\prod(2\cos^2x-1)=-(2\cos^2x-1)^2\le0$ as $\cos x=-\cos y\implies \cos^2x=\cos^2y$
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Let $n=x^2+y^2$; $n=2^{2k}m$ or $n=2^{2k+1}m$ with $m$ odd. Prove that $2^{k}$ divides both $x$ and $y$. Let $n=x^2+y^2$ where x,y are integers, be one of the forms $n=2^{2k}m$ respectively $n=2^{2k+1}m$ with m odd. Prove that $2^{k}$ divides both x and y.
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We prove something a little stronger. Let $2^a$ be the highest power of $2$ that divides both $x$ and $y$. We show that if $2^{2k}$ or $2^{2k+1}$ is the highest power of $2$ that divides $x^2+y^2$, then $a=k$.
Since $2^a$ is the highest power of $2$ that divides both $x$ and $y$, we have $x=2^a s$ and $y=2^at$, where $s$ and $t$ are not both even.
If $s$ and $t$ are of opposite parity, then $x^2+y^2=2^{2a}(s^2+t^2)$, and $s^2+t^2$ is odd. So $2^{2a}$ is the highest power of $2$ that divides $x^2+y^2$, and therefore $a=k$.
If $s$ and $t$ are both odd, then $s^2\equiv 1\pmod{4}$ and $t^2\equiv 1\pmod{4}$. It follows that $s^2+t^2\equiv 2\pmod{4}$. So the highest power of $2$ that divides $s^2+t^2$ is $2^1$. It follows that the highest power of $2$ that divides $x^2+y^2$ is $2^{2a+1}$. So again $a=k$.
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Prove that $ \lim_{n \rightarrow \infty } \frac{n+6}{n^2-6} = 0 $. My attempt:
We prove that $ \lim\limits_{n \rightarrow \infty } \dfrac{n+6}{n^2-6} =0$.
It is sufficient to show that for any $ \epsilon \in\textbf{R}^+ $, there exists an $ K \in \textbf{R}$ such that for any $ n > K $
$ \displaystyle \left| \frac{n+6}{n^2-6} - 0 \right| < \epsilon $.
Suppose $ n>4 $. Then $ n+6 < 7n $ and $ n^2 -6 > \frac{1}{2} n^2 $. So
$ \displaystyle \left| \frac{n+6}{n^2-6} - 0 \right| = \frac{n+6}{n^2-6 } < \frac{7n}{\frac{1}{2}n^2} = \frac{14}{n} $.
Consider $ K = \max\{4,\displaystyle \frac{14}{\epsilon}\} $ and suppose $ n> K $. Then $ n > \displaystyle \frac{14}{\epsilon} $. This implies that $ \epsilon > \displaystyle \frac{14}{n} $. Therefore
$ \displaystyle \left| \frac{n+6}{n^2-6} - 0 \right| = \frac{n+6}{n^2-6 } < \frac{7n}{\frac{1}{2}n^2} = \frac{14}{n} < \epsilon $.
Thus $ \lim\limits_{n \rightarrow \infty } \displaystyle \frac{n+6}{n^2-6}=0 $.
Is this proof correct? What are some other ways of proving this? Thanks!
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It is sufficient to use the fact that $\displaystyle \frac{n+6}{n^2-6}\sim_{\infty}\frac{n}{n^2}=\frac{1}{n}$ then you have the result.
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Solving $2\cos(x) = \sin(x)$ How would you solve equations of the form $ a \sin (x+b) = \sin (x)$?
Eg. $ 2 \cos(x) = \sin(x) $
I realy have no idea how I would solve this kind of equations.
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For your initial question, you can use the angle-sum rule for $\sin(\alpha + \beta):$ $$\sin(\alpha \pm \beta) = \sin \alpha \cos \beta \pm \cos \alpha \sin \beta$$
So $$a \sin (x+b) = \sin (x) \iff a (\sin x \cos b + \cos x \sin b) = \sin x$$
We can divide through by $\cos x$, provided $\cos x \neq 0$, noting that $a, \cos b, \sin b$ are all constants.
For your example:
Note that if $\cos x \neq 0$, then $2 \cos x = \sin x \iff 2 = \dfrac {\sin x}{\cos x} = \tan x \iff \tan^{-1}(2) = x$.
And as @Alex notes, $2\cos x = \sin x \implies \cos x \neq 0$.
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|
Prove that: $\cos(x) -1 < -\frac{x^2}{2} + \frac{x^4}{24}$
Prove that:
$$\cos(x) -1 < -\frac{x^2}{2} + \frac{x^4}{24}$$ for $x \ne 0$
I need to prove this using Cauchy's mean value theorem.
What I did:
$f(x) = \cos(x) -1$
$$g(x) = -\frac{x^2}{2} + \frac{x^4}{24}$$
If I can show that:
$\frac{f(x)}{g(x)}\lt 1$ that would prove the original inequality
so,
$$\frac{f(x)}{g(x)} = \frac{f(x) - f(0)}{g(x) - g(0)} = \frac{f'(c)}{g'(c)}$$ for $0\lt c \lt x$
$$\dfrac{f'(c)}{g'(c)} = \dfrac{-\sin(c)}{-c + \frac{c^3}{6}}$$
My problem is the following expression is not always less then 1! (for example $c=2.4$)
What am I doing wrong?
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I think it will be easier to show that: $$ \cos x - 1 + \frac{x^2}2 < \frac{x^4}{24} $$
Define $f(x) = \cos x - 1 + \frac{x^2}2$, $g(x) = \frac{x^4}{24}$.
Now using the Cauchy MVT: $$ \frac{f(x)-f(0)}{g(x)-g(0)} = \frac{6(c-\sin c)}{c^3} $$
Now just show that $ h(x) = \frac{6(x-\sin x)}{x^3} $ is bounded above globably by $1$, and you'll get what you need.
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What is the integrating factor of $2xy'+x^{2}e^{1-x^2}y=2$ I know how to start solving it, its dividing everything by $2x$, but I can't solve
$$\int \dfrac{x^2e^{1-x^2}}{2x}\,dx$$
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$$\int \frac{x^2 e^{1-x^2}}{2x}dx=\int \frac{xe^{1-x^2}}{2}dx=-\frac{1}{2}\int\frac{-2xe^{1-x^2}}{2}dx$$
Let $u=1-x^2$ then $du=-2xdx$
$$=-\frac{1}{4}\int-2xe^{1-x^2}dx=-\frac{1}{4}\int e^udu =-\frac{1}{4}e^u$$
But $u=1-x^2$
$$\int \frac{x^2 e^{1-x^2}}{2x}dx=-\frac{1}{4}e^{1-x^2}$$
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|
Solving a challenging differential equation How would one go about finding a closed form analytic solution to the following differential equation?
$$\frac{d^2y}{dx^2} +(x^4 +x^2+x+c)y(x) =0 $$
where $c\in\mathbb{R}$
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This is the way I would do it.
Let $p(x)=x^4+x^2+x+c$. Put $z=y'$. Then the equation is equivalent to the system
$$
\begin{bmatrix}y\\ z\end{bmatrix}'=\begin{bmatrix}0&1\\-p(x)&0\end{bmatrix}\,\begin{bmatrix}y\\ z\end{bmatrix}.
$$
Let us write $A(x)$ for the $2\times 2$ matrix above. For a given vector $\begin{bmatrix}c_1\\ c_2\end{bmatrix}$, a solution will be given by
$$
e^{B(x)}\,\begin{bmatrix}c_1\\ c_2\end{bmatrix},
$$
where $B'(x)=A(x)$. That is $B(x)=\begin{bmatrix}0&x\\-q(x)&0\end{bmatrix}$, with $q(x)=x^5/5+x^3/3+x^2/2+cx$. If I didn't make a mistake, the exponential is
$$
e^{B(x)}=\begin{bmatrix}\cos\sqrt{xq(x)}&\sqrt\frac{x}{q(x)}\,\sin\sqrt{xq(x)}\\
-\sqrt\frac{q(x)}{x}\,\sin\sqrt{xq(x)}&\cos\sqrt{xq(x)}\end{bmatrix}.
$$
As $y$ will be the first coordinate of $e^{B(x)}\begin{bmatrix}c_1\\ c_2\end{bmatrix}$, we get
$$
y(x)=c_1\,\cos\sqrt{\frac{x^6}5+\frac{x^4}3+\frac{x^3}2+cx^2}+c_2\,\frac{\sin\sqrt{\frac{x^6}5+\frac{x^4}3+\frac{x^3}2+cx^2}}{\sqrt{\frac{x^5}5+\frac{x^3}3+\frac{x^2}2+cx}}.
$$
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If $a$ and $b$ are cathets a right triangle whose hypotenuse is $1$ determine the highest value of $2a + b$ Can some one help me out on where to go?
If $a$ and $b$ are cathets a right triangle whose hypotenuse is $1$ determine the highest value of $2a + b$ ?
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Let the two sides be $\sin(x)$ and $\cos(x)$.
So we want to maximize $2 \cos(x) + \sin(x)$.
We can write (2,1) in polar form as
$$
2 = R\cos(y),~ 1 = R\sin(y)~~~~\Rightarrow R = \sqrt{5}$$
Hence
$$
2 \cos(x) + \sin(x) = R (\cos(x) \cos(y) + \sin(x) \sin(y)) =\sqrt{5} \cos(x-y)$$
We really do not need to find $x$ or $y$ since
$$
\text{Maximum value of } \sqrt{5} \cos(x-y) = \sqrt{5}$$
Hence answer is $\sqrt{5}$
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|
Positive recurrent, zero recurrent and transient states How do we prove that a state in a Markov chain process is positive recurrent, zero recurrent or transient? For example, if we have a transition matrix
$$P=\left(\begin{array}{ccc}
\frac{1}{3} & \frac{1}{3} & \ 0 & \frac{1}{3} \\
0 & \frac{1}{2} & \frac{1}{2} & 0\\
\frac{1}{2} & \ 0 & 0 & \frac{1}{2} \\ 0 & \frac{1}{3} & \frac{1}{3} & \frac{1}{3}
\end{array}\right)$$
how do we do it??
Thanks a lot!!
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The path $1\to4\to2\to3\to1$ enumerates the state space and has positive probability hence every state is positive recurrent.
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minimum value of $\cos(A-B)+\cos(B-C) +\cos(C-A)$ is $-3/2$ How to prove that the minimum value of $\cos(A-B)+\cos(B-C) +\cos(C-A)$ is $-3/2$
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As Berci suggested ,$x=A-B,y=B-C \to f=\cos {x} +\cos{y}+cos{(x+y)}$
$f=2\cos{\dfrac{x+y}{2}} \cos{\dfrac{x-y}{2}}+2\left(\cos{\dfrac{x+y}{2}}\right)^2-1 \ge 2\cos{\dfrac{x+y}{2}}+2\left(\cos{\dfrac{x+y}{2}}\right)^2-1 =2\left(\cos{\dfrac{x+y}{2}}+\dfrac{1}{2}\right)^2-\dfrac{3}{2} \ge -\dfrac{3}{2}$
first " $\ge$ " : $\cos{\dfrac{x+y}{2}}<0,x=y$
second " $\ge$ " : $\cos{\dfrac{x+y}{2}}=-\dfrac{1}{2}$
so the "=" will hold when $x=y= \dfrac{2\pi}{3}(or \dfrac{4\pi}{3})+2k\pi $
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|
Find all solutions of $z^5+a^5=0$ The task is as follows:
Find all solutions of $z^5+a^5=0$, where $a$ is a positive real number.
My initial attempt
(which leads nowhere)
My guess is that i'll have to find the 5 5th roots of $-z^5$:
$w_1 = |-z^5|^5(cos(\frac{\theta}{5})+isin(\frac{\theta}{5})) \\
w_1 = |z|(cos(\frac{\theta}{5})+isin(\frac{\theta}{5})$
Then, the other roots are:
$w_2 = |z|(cos(\frac{\theta+2\pi}{5})+isin(\frac{\theta+2\pi}{5}) \\
w_3 = |z|(cos(\frac{\theta+4\pi}{5})+isin(\frac{\theta+4\pi}{5}) \\
w_4 = |z|(cos(\frac{\theta+6\pi}{5})+isin(\frac{\theta+6\pi}{5}) \\
w_5 = |z|(cos(\frac{\theta+8\pi}{5})+isin(\frac{\theta+8\pi}{5})$
but i'm not sure what to do with this new information. Also, I guess $z^5$ have to be a negative real number, since added to $a^5$ its $0$. Again, not sure what to do with this either.
|
To make our life a bit easier, start by writing $z=r(\cos\theta+i\sin\theta)$ for some $r>0$ and some real $\theta,$ so that $$-a^5=z^5\\-a^5=r^5(\cos\theta+i\sin\theta)^5\\-a^5=r^5(\cos5\theta+i\sin5\theta)$$ Now, taking the modulus of both sides gives us $$a^5=r^5,$$ so we'll need $r=a,$ and so $z=a(\cos\theta+i\sin\theta).$ Moreover, since $$-a^5=r^5(\cos5\theta+i\sin5\theta)\\-a^5=a^5(\cos5\theta+i\sin5\theta),$$ then $$-1=\cos5\theta+i\sin5\theta,$$ meaning that we need $\theta$ such that $$-1=\cos5\theta\\0=\sin5\theta.$$ This is true if and only if $5\theta$ is an odd integer multiple of $\pi.$ Can you take it from there to find appropriate values of $\theta$ to give $5$ distinct solutions $z$?
|
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"timestamp": "2023-03-29T00:00:00",
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|
Diophatine equation $x^2+y^2+z^2=t^2$ Probably duplicate but I don't find: I'd like to solve the diophantine equation
$$x^2+y^2+z^2=a^2$$
which has solutions, by exemple $1^2+2^2+2^2=3^2$ or $2^2+3^2+6^2=7^2$. Every such solution gives a rational point on the unit sphere.
Is there a complete description of the solutions such as for pythagorician triplet?
|
We'll start with Pythagorean triples to see the pattern. The complete rational solution to $x_1^2+x_2^2 = y_1^2$ has the form,
$$((a^2-b^2)t)^2+(2abt)^2 = ((a^2+b^2)t)^2\tag{1}$$
where $t$ is a scaling factor.
Proof: For any solution where $x_1+y_1 \neq 0$ , one can always find rational {$a,b,t$} using the formulas $a,b,t = x_1+y_1,\; x_2,\; \frac{1}{2(x_1+y_1)}$.
Similarly, for $x_1^2+x_2^2+x_3^2 = y_1^2$, it is,
$$((a^2-b^2-c^2)t)^2+(2abt)^2+(2act)^2 = ((a^2+b^2+c^2)t)^2\tag{2}$$
Proof: One can always find rational {$a,b,c,t$} using $a,b,c,t = x_1+y_1,\; x_2,\; x_3,\; \frac{1}{2(x_1+y_1)}$.
and so on for $n$ squares. See also Sums of Three Squares for more.
|
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"url": "https://math.stackexchange.com/questions/646255",
"timestamp": "2023-03-29T00:00:00",
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|
Solve the system using Gaussian elimination with back-substitution or Gauss-Jordan elimination Here is the system:
$$
\left\{
\begin{aligned}
x_1-3x_3&=-2 \\
3x_1+x_2-2x_3&=5 \\
2x_1+2x_2+x_3&=4
\end{aligned}
\right.
$$
This is my very first problem actually using a matrix so here is my attempt
First I setup the augmented matrix:
$$ \left[
\begin{array}{ccc|c}
1&0&-3&-2\\
3&1&-2&5 \\
2&2&1&4
\end{array}
\right]$$
Then I did $(-3)R_1+R_2->R_2$ which produced:
$$ \left[
\begin{array}{ccc|c}
1&0&-3&-2\\
0&1&7&11 \\
2&2&1&4
\end{array}
\right]$$
And then I did $(-2)R_1+R_3->R_3$:
$$ \left[
\begin{array}{ccc|c}
1&0&-3&-2\\
0&1&7&11 \\
0&2&7&8
\end{array}
\right]$$
And then $(1/2)R_3->R_3$:
$$ \left[
\begin{array}{ccc|c}
1&0&-3&-2\\
0&1&7&11 \\
0&1&\frac{7}{2}&4
\end{array}
\right]$$
And then $(-1)R_2+R_3->R_3$
$$ \left[
\begin{array}{ccc|c}
1&0&-3&-2\\
0&1&7&11 \\
0&0&\frac{-7}{2}&7
\end{array}
\right]$$
And finally $\left(\frac{-1}{7}\right)R_3->R_3$
$$ \left[
\begin{array}{ccc|c}
1&0&-3&-2\\
0&1&7&11 \\
0&0&1&-1
\end{array}
\right]$$
But this is not correct according to the solution. Can someone tell me where I went wrong? Clearly I did something really wrong here.
|
From your step: And then $(-1)R_2+R_3->R_3$
If you look at where you have $−11+4$.
|
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|
Induction: show that $\sum\limits_{k=1}^n \frac{1}{\sqrt{k}} < 2 \sqrt{n}$ for all n $\in Z_+$ The question:
show by using induction that $\sum\limits_{k=1}^n \frac{1}{\sqrt{k}} < 2 \sqrt{n}$ for all n $\in Z_+$
My attempt at a solution:
The base case $n = 1$ is true.
First we use the inductive assumption that the statement holds for some $k = n$
This gives us:
$\frac{1}{\sqrt{1}} + ... + \frac{1}{\sqrt{n}} < 2 \sqrt{n}$
Then we have to prove that it holds true for $k = n+1$
$\frac{1}{\sqrt{1}} + ... + \frac{1}{\sqrt{n+1}} < 2 \sqrt{n+1}$
This means that:
$2 \sqrt{n+1} - 2 \sqrt{n} > \frac{1}{\sqrt{n+1}}$
We can start by factoring out $2$:
$2 (\sqrt{n+1} - \sqrt{n}) > \frac{1}{\sqrt{n+1}}$
We can then multiply with ${\sqrt{n+1}}$:
$2 ((n+1) - \sqrt{n}\sqrt{n+1}) > 1$
But I am stuck here... Please help me out with the last steps!
Thank you very much!
|
Note that, by induction hypothesis
$$
\sum_{k=1}^{n+1} \frac{1}{\sqrt{k}} < 2\sqrt{n} + \frac{1}{\sqrt{n+1}}
$$
The RHS is $<2\sqrt{n+1}$ iff
$$
\frac{1}{\sqrt{n+1}} < 2(\sqrt{n+1} - \sqrt{n})
$$
$$
\Leftrightarrow \frac{\sqrt{n+1} + \sqrt{n}}{\sqrt{n+1}} < 2(n+1-n) = 2
$$
$$
\Leftrightarrow 1 + \sqrt{\frac{n}{n+1}} < 2
$$
$$
\Leftrightarrow \sqrt{n} < \sqrt{n+1}
$$
which is true.
|
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"timestamp": "2023-03-29T00:00:00",
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|
Identifying a quotient group (NBHM-$2014$)
Let $\mathbb C^*$ denote the multiplicative group of non-zero complex numbers and let $P$ denote the subgroup of positive real numbers. Identify the quotient group.
My thought $$\frac{\mathbb C^*}{P}=\{P,-P,iP,-iP\}.$$ Is it right?
|
I am not sure if this is in any way different than B.S's way. i am just omitting maps and trying to see just as a quotient.
We see $x+iy\in \mathbb{C}$ as $(\sqrt{x^2+y^2})(\dfrac{x}{\sqrt{x^2+y^2}}+i\dfrac{y}{\sqrt{x^2+y^2}})$
Now, As $\sqrt{x^2+y^2}\in \mathbb{R}$ when we see $(\sqrt{x^2+y^2})(\dfrac{x}{\sqrt{x^2+y^2}}+i\dfrac{y}{\sqrt{x^2+y^2}})$ modulo real numbers we would get :
$$(\sqrt{x^2+y^2})(\dfrac{x}{\sqrt{x^2+y^2}}+i\dfrac{y}{\sqrt{x^2+y^2}})\equiv (\dfrac{x}{\sqrt{x^2+y^2}}+i\dfrac{y}{\sqrt{x^2+y^2}}) \text { mod }\mathbb{R}$$
And then we have $|(\dfrac{x}{\sqrt{x^2+y^2}}+i\dfrac{y}{\sqrt{x^2+y^2}})|=\sqrt{(\dfrac{x}{\sqrt{x^2+y^2}})^2+(\dfrac{y}{\sqrt{x^2+y^2}})^2}=1$
So,for any $z\in \mathbb{C}$ we have $z=|z|(z')$ for $|z|\in \mathbb{R}$ and $|z'|=1$.
Thus $\mathbb{C}^*/P=\{z\in \mathbb{C} : |z|=1\}$
|
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|
Solve $2000x^6+100x^5+10x^3+x-2=0$ One of the roots of the equation $2000x^6+100x^5+10x^3+x-2=0$ is of the form $\frac{m+\sqrt{n}}r$, where $m$ is a non-zero integer and $n$ and $r$ are relatively prime integers.Then the value of $m+n+r$ is?
Tried to use the fact that another root will be $\frac{m-\sqrt{n}}r$ as coefficients are rational but there are six roots and using sum and product formulas would allow many variables in the equations.
|
$$2000x^6+100x^5-200x^4+200x^4+10x^3-20x^2+20x^2+x-2$$
$$(2000x^6+200x^4+20x^2)+(100x^5+10x^3+x)-(200x^4+20x^2+2)$$
$$x^2(2000x^4+200x^2+20)+\frac{x}{20}(2000x^4+200x^3+20x)-\frac{1}{10}(2000x^4+200x^2+20)$$
$$=(x^2+\frac{x}{20}-\frac{1}{10})(2000x^4+200x^2+20)=0$$
|
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"url": "https://math.stackexchange.com/questions/651024",
"timestamp": "2023-03-29T00:00:00",
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|
Showing $\frac{x}{1+x}<\log(1+x)0$ using the mean value theorem I want to show that $$\frac{x}{1+x}<\log(1+x)<x$$ for all $x>0$ using the mean value theorem. I tried to prove the two inequalities separately.
$$\frac{x}{1+x}<\log(1+x) \Leftrightarrow \frac{x}{1+x} -\log(1+x) <0$$
Let $$f(x) = \frac{x}{1+x} -\log(1+x).$$ Since $$f(0)=0$$ and $$f'(x)= \frac{1}{(1+x)^2}-\frac{1}{1+x}<0$$ for all $x > 0$, $f(x)<0$ for all $x>0$. Is this correct so far?
I go on with the second part:
Let $f(x) = \log(x+1)$. Choose $a=0$ and $x>0$ so that there is, according to the mean value theorem, an $x_0$ between $a$ and $x$ with
$f'(x_0)=\frac{f(x)-f(a)}{x-a} \Leftrightarrow \frac{1}{x_0+1}=\frac{ \log(x+1)}{x}$.
Since $$x_0>0 \Rightarrow \frac{1}{x_0+1}<1.$$ $$\Rightarrow 1 > \frac{1}{x_0+1}= \frac{ \log(x+1)}{x} \Rightarrow x> \log(x+1)$$
|
As a consequence of MVT, there is a $\xi\in(0,x)$, such that
$$
\log(1+x)=\log(1+x)-\log 1=x\cdot \left(\log(1+x)\right)'_{x=\xi}=x\cdot\frac{1}{1+\xi}<x.
$$
Let $y=\frac{x}{1+x}$. Then there is $\xi\in\big(0,y\big)$, such that
\begin{align}
\log(1+x)&=-\log\left(\frac{1}{1+x}\right)=\log 1-\log\left(1-\frac{x}{1+x}\right) \\&=
\log 1-\log\left(1-y\right) =
y\left(\log(1-y)\right)'_{y=\xi}=y\cdot\frac{1}{1-\xi}>y=\frac{x}{x+1.}
\end{align}
|
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|
Maximum and minimum distance from the origin
Find the maximum and minimum distances from the origin to the curve $5x^3+6xy+5y^2-8=0$
My attempt:
We have to maximise and minimise the following function $x^2+y^2$ with the constraint that $5x^3+6xy+5y^2-8=0$.
Let
$$F(x,y)=x^2+y^2+\lambda(5x^2+6xy+5y^2-8)$$
$$\frac{\delta F(x,y)}{\delta x}=2x+\lambda(10x+6y)$$
and
$$\frac{\delta F(x,y)}{\delta y}=2y+\lambda(6x+10y)$$
Multiplying the 2 equations by y,x respectively and subtracting I get
$$\lambda(y^2-x^2)=0$$
Hence $$y=x$$
Substituting $x=y$ in $5x^3+6xy+5y^2-8=0$, I get the $x=\pm \frac{1}{\sqrt2}$ and $y=\pm \frac{1}{\sqrt2}$ . Now I am stuck. Both the points corresponds to only one distance. Did I do something wrong?
|
You have found the points that, under the contraint $5x^2+6xy+5y^2−8=0$, minimise $x^2+y^2$and therefore minimise the euclidean distance between the origin and the point $(x,y)$ that is
$$d_{euclidean}\left( (0,0), (x,y) \right) \doteq \sqrt{(x-0)^2+(y-0)^2} = \sqrt{x^2+y^2}$$
So your minimum distance is the distance of $\left(+\frac{1}{\sqrt{2}}, +\frac{1}{\sqrt{2}}\right)$ from the origin (the same with $\left(-\frac{1}{\sqrt{2}},-\frac{1}{\sqrt{2}}\right)$. You have found the minimum distance: $$\sqrt{x^2+y^2}=\sqrt{\left( -\frac{1}{\sqrt{2}} \right)^2+\left( -\frac{1}{\sqrt{2}} \right) ^2}=\sqrt{\frac{1}{2}+\frac{1}{2}}=\sqrt{1}= 1$$
As lsp said, from $\lambda (y^2-x^2)$ you find an other possibility: $y=-x$ from which you get the points $\left(-\sqrt{2}, +\sqrt{2}\right)$ and $\left(+\sqrt{2}, -\sqrt{2}\right)$. Those points are the furthest from the origin as you will see when computing the distance to the origin: $$\sqrt{x^2+y^2}=\sqrt{\sqrt{2}^2+\sqrt{2}^2}=\sqrt{2+2}=\sqrt{4}= 2$$
|
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|
Solve the difference quotient
So I tried a variety of answers and they all came up incorrect. I guess I don't understand difference quotient because I feel like it's pointless and a slow way to do the problem. That being said... can't get the answer
|
You are expected to write
$$\frac{f(x)-f(4)}{x-4}=\frac{\frac{x+6}{x+4}-\frac{10}{8}}{x-4},\tag{1}$$
and then simplify.
We first simplify the numerator. Bringing it to the common denominator $(x+4)(8)$, we get
$$ \frac{x+6}{x+4}-\frac{10}{8}=\frac{(x+6)(8)-(x+4)(10)}{(x+4)(8)}=\frac{-2x+8}{(x+4)(8)}.$$
This simplifies to $\frac{-x+4}{(x+4)(4)}$.
Dividing by $x-4$, we find that if $x\ne 4$, then
$$\frac{f(x)-f(4)}{x-4}=-\frac{1}{(x+4)(4)}.$$
Remarks: The algebra is somewhat simpler if first we rewrite $\frac{x+6}{x+4}$ as $1+\frac{2}{x+4}$.
From the answer you got, it looks as if you evaluated the difference quotient correctly at $x=4$. That is not what the question asked for: you did extras work!
|
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"timestamp": "2023-03-29T00:00:00",
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|
Prove trig identity: $\tan(x) + \cot(x) = \sec(x) \csc(x)$ wherever defined I appreciate the help.
My attempt:
$$
\begin{align}
\tan(x) + \cot(x) &= \frac{\sin(x)}{\cos(x)} + \frac{\cos(x)}{\sin(x)} \\
&= \frac{\sin^2(x)}{\cos(x) \sin(x)}+\frac{\cos^2(x)}{\cos(x) \sin(x)} \\
&= \frac{\sin^2(x)+\cos^2(x)}{\cos(x) \sin(x)}\\
&= \frac{1}{\cos(x) \sin(x)}\\
&= \frac{1}{\frac{1}{\sec(x)}\frac{1}{\csc(x)}}\\
&=\frac{1}{\frac{1}{\sec \csc}}\\
&=\frac{1}{1}\cdot \frac{\sec(x) \csc(x)}{1}\\
&= \sec(x) \csc(x)
\end{align}
$$
|
An alternative approach writes $t=\tan x/2$ so $$\tan x+\cot x=\frac{2t}{1-t^2}+\frac{1-t^2}{2t}=\frac{1+t^2}{1-t^2}\frac{1+t^2}{2t}=\sec x\csc x.$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/661439",
"timestamp": "2023-03-29T00:00:00",
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"question_score": "10",
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|
Computing the limit of $n \cdot \arccos \left( \left(\frac{n^2-1}{n^2+1}\right)^{\cos (1/n)} \right)$ I need to solve this earth's wonder:
$$\lim_{n \rightarrow \infty} \left[n \; \arccos
\left( \left(\frac{n^2-1}{n^2+1}\right)^{\cos \frac{1}{n}} \right)\right]$$
I have tried to write down it using $e^{v \ln u}$,and then used L'Hôpital's rule, but with no luck, i'm constantly getting indeterminate form like $\infty-\infty$ inside $\ln$ which makes another application of L'Hôpital impossible.
My professor told me (with great smile on its face) that if i use Taylor expansion, it will lead me into the abyss...
any hints about possible rewriting this limit and what i should use would be VERY helpful. Thanks in advance.
|
The function given here is really bit complex (at least in typing). We first need to check if we know any fundamental limit associated with $\arccos$ function. Clearly there isn't any, but we do have the standard limit $$\lim_{x \to 0}\frac{1 - \cos x}{x^{2}} = \lim_{x \to 0}\dfrac{2\sin^{2}(x/2)}{x^{2}} = \frac{1}{2}$$ Putting $\cos x = t$ we can easily see that $$\lim_{t \to 1^{-}}\dfrac{\arccos^{2}t}{1 - t} = 2$$ and further putting $1 - t = x$ we get $$\lim_{x \to 0^{+}}\dfrac{\arccos^{2}(1 - x)}{x} = 2$$ or $$\lim_{x \to 0^{+}}\frac{\arccos(1 - x)}{\sqrt{x}} = \sqrt{2}\tag{1}$$ Next we focus on the complicated function given in the question. If we put $x = 1/n$ then we can see that the function is given by $$f(x) = \frac{1}{x}\cdot\arccos\left(\left(\frac{1 - x^{2}}{1 + x^{2}}\right)^{\cos x}\right)$$ If we put $y = ((1 - x^{2})/(1 + x^{2}))^{\cos x}$ then $y \to 1^{-}$ and hence $z = 1 - y \to 0^{+}$. And we have
\begin{align}
L &= \lim_{x \to 0^{+}}f(x)\notag\\
&= \lim_{x \to 0^{+}}\frac{1}{x}\cdot\arccos(1 - z)\notag\\
&= \lim_{x \to 0^{+}}\frac{1}{x}\cdot\sqrt{z}\cdot\frac{\arccos(1 - z)}{\sqrt{z}}\notag\\
&= \sqrt{2}\lim_{x \to 0^{+}}\sqrt{\dfrac{z}{x^{2}}}\text{ (using (1))}\\
&= \sqrt{2}\lim_{x \to 0^{+}}\sqrt{\dfrac{1 - y}{x^{2}}}\tag{2}
\end{align}
Next we can see that
\begin{align}
1 - y &= 1 - \left(\frac{1 - x^{2}}{1 + x^{2}}\right)^{\cos x}\notag\\
&= 1 - \left(1 - \frac{2x^{2}}{1 + x^{2}}\right)^{\cos x}\notag\\
&\geq 1 - \left(1 - \frac{2x^{2}\cos x}{1 + x^{2}}\right)\notag\\
&= \frac{2x^{2}\cos x}{1 + x^{2}}\tag{3}
\end{align}
and
\begin{align}
1 - y &= 1 - \left(\frac{1 - x^{2}}{1 + x^{2}}\right)^{\cos x}\notag\\
&= 1 - \left(1 - \frac{2x^{2}}{1 + x^{2}}\right)^{\cos x}\notag\\
&\leq 1 - (1 - 2x^{2})^{\cos x}\notag\\
&\leq 1 - (1 - 2x^{2})\notag\\
&= 2x^{2}\tag{4}
\end{align}
It follows from $(3)$ and $(4)$ that $$\frac{2\cos x}{1 + x^{2}}\leq \frac{1 - y}{x^{2}} \leq 2$$ and taking limits as $x \to 0^{+}$ and using Squeeze theorem we get $$\lim_{x \to 0^{+}}\frac{1 - y}{x^{2}} = 2$$ Using this limit in equation $(2)$ we get the desired limit $L = \sqrt{2}\cdot\sqrt{2} = 2$.
The above solution avoids the use of Taylor's series and instead relies on fundamental limit theorems. Using Taylor's series would be a challenge if we proceed directly. Did does it so smartly in his answer and I am at loss of words to appreciate his technique. So whenever we are asked to use Taylor's we must make certain optimizations (regarding no of terms of the series to be used) and this requires bit of experience to get the right answer in the most efficient manner. On the other hand avoiding Taylor's series or LHR almost always requires various algebraical/trigonometric manipulations and some ingenious use of inequalities and the use of Squeeze theorem (like I have done above).
Update: The limit of $(1 - y)/x^{2}$ can also be evaluated by expressing $y$ as $y = \exp(\log y)$ and this appears to be simpler than the approach via Squeeze theorem presented above. We have
\begin{align}
A &= \lim_{x \to 0^{+}}\frac{1 - y}{x^{2}}\notag\\
&= \lim_{x \to 0^{+}}\frac{1 - \exp(\log y)}{\log y}\cdot\frac{\log y}{x^{2}}\notag\\
&= -\lim_{x \to 0^{+}}\frac{\cos x\log((1 - x^{2})/(1 + x^{2}))}{x^{2}}\notag\\
&= \lim_{x \to 0^{+}}\frac{\log((1 + x^{2})/(1 - x^{2}))}{x^{2}}\notag\\
&= \lim_{x \to 0^{+}}\frac{\log(1 + x^{2})}{x^{2}} - \frac{\log(1 - x^{2})}{x^{2}}\notag\\
&= 1 - (-1) = 2\notag
\end{align}
and thus $L = \sqrt{2} \sqrt{A} = 2$.
|
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|
How prove $3(a^4+b^4+c^4)+2(a+b+c)abc\ge 5(a^2b^2+b^2c^2+a^2c^2)$ let $a,b,c>0$ and such $abc=1$,show that
$$3(a^4+b^4+c^4)+2(a+b+c)\ge 5\left(\dfrac{1}{a^2}+\dfrac{1}{b^2}+\dfrac{1}{c^2}\right)$$
my idea: maybe can use AM-GM inequality,
$$3(a^4+b^4+c^4)a^2b^2c^2+2(a+b+c)(a^2b^2c^2)\ge 5(a^2b^2+b^2c^2+a^2c^2)$$
$$\Longleftrightarrow 3(a^4+b^4+c^4)+2(a+b+c)abc\ge 5(a^2b^2+b^2c^2+a^2c^2)$$
then I can't,Thank you very much
|
Consider Schur's inequality for $r = 2$:
$$a^4 + b^4 + c^4 + (a + b + c)abc \geqslant a^3b + a^3c + b^3a + b^3c + c^3a + c^3b$$
plus three AM-GMs:
$$
\begin{aligned}
a^3b + b^3a &\geqslant 2a^2b^2\\
b^3c + c^3b &\geqslant 2b^2c^2\\
a^3c + c^3a &\geqslant 2a^2c^2
\end{aligned}
$$
This gives you, after multiplying by $2$:
$$2(a^4 + b^4 + c^4) + 2(a + b + c)abc \geqslant 4(a^2b^2 + b^2c^2 + a^2c^2).$$
Three more AM-GMs:
$$
\begin{align}
\frac{a^4 + b^4}{2} &\geqslant a^2b^2,\\
\frac{b^4 + c^4}{2} &\geqslant b^2c^2,\\
\frac{a^4 + c^4}{2} &\geqslant a^2c^2.
\end{align}$$
Sum up 4 last inequalities and you're done.
Equality, of course, holds only if $a = b = c$, which can be seen from all those AM-GMs.
|
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|
Solve $t^4+4 t^3+6 t^2+4 t-32 t^{1/4}+1 = -16 $ I'm trying to solve the following equation: $$(t+1)^4 - 32 t^{\frac{1}{4}}=-16 $$ where t $\geq 0$,
which is equivalent to $$t^4+4 t^3+6 t^2+4 t-32 t^{\frac{1}{4}}+1 = -16 $$
Wolfram Alpha tells that the equation is equivalent to $$t+1=2t^{\frac{1}{4}} $$ if we assume that t>0. How to prove this? The hard part is how to factorise... Is there another solution? (BTW, without to obtaina 3-rd degree equation and to use Cardano)
|
$u^{12}+2u^9+3u^8+4u^6+4u^5+3u^4+8u^3+4u^2+2u+17=u^8(u^4+2u+3)+(x+1)^2(4x^4-4x^3+7x^2-2x+1)+2(x+1)+14$
$u^4+2u+3 \ge 2u^2+2u+2 \ge \dfrac{3}{2}$
$4x^4-4x^3+7x^2-2x+1=x^2(2x-1)^2+6(x-\dfrac{1}{6})^2+\dfrac{5}{6} > \dfrac{5}{6}$
$(x+1)^2(4x^4-4x^3+7x^2-2x+1)+2(x+1)+14>\dfrac{5}{6}(x+1)^2+2(x+1)+14=\dfrac{5}{6}(x+1+\dfrac{6}{5})^2+12\dfrac{4}{5}>12\dfrac{4}{5}$
$u^{12}+2u^9+3u^8+4u^6+4u^5+3u^4+8u^3+4u^2+2u+17>12.8$
|
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|
Prove that $\left(\frac12(x+y)\right)^2 \le \frac12(x^2 + y^2)$ Prove that $$\left(\frac12(x+y)\right)^2 \leq \frac12(x^2 + y^2)$$
I've gotten that
$$\left(\frac12(x+y)\right)^2 \ge 0 $$
but stumped on where to go from here...
|
let investigate $$\frac{(x+y)^2}{4}-\frac{x^2+y^2}{2}=\frac{x^2+y^2+2xy-2x^2-2y^2}{4}=\frac{-(x-y)^2}{4}$$ then it is obvious that we have $$\frac{(x+y)^2}{4}-\frac{x^2+y^2}{2}\leq 0.$$ So $$\frac{(x+y)^2}{4}\leq\frac{x^2+y^2}{2}.$$
|
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|
How to use the generating function $F(x) =x/(1-x-x^2).$ The generating function for the Fibonacci sequence is
$$F(x) =x/(1-x-x^2).$$
To work out the 20th value of the sequence I understand you somehow expand this and look at the coefficient of $x^{20}$. How exactly do you do this?
|
You can do the following:
*
*Solve the equation $1-x-x^2=0$. The roots are $\frac{-1-\sqrt{5}}{2}$ and $\frac{-1+\sqrt{5}}{2}$.
*Then you have $1-x-x^2=-(x-\frac{-1-\sqrt{5}}{2})(x-\frac{-1+\sqrt{5}}{2})=-(x+\frac{1+\sqrt{5}}{2})(x+\frac{1-\sqrt{5}}{2})$
*$\frac{-x}{(x+\frac{1+\sqrt{5}}{2})(x+\frac{1-\sqrt{5}}{2})}=\frac{A}{(x+\frac{1+\sqrt{5}}{2})}+\frac{B}{(x+\frac{1-\sqrt{5}}{2})}$
*Find A and B then you can expand both fractions in geometric series.
$\frac{-x}{(x+\frac{1+\sqrt{5}}{2})(x+\frac{1-\sqrt{5}}{2})}=\frac{Ax+Bx+A\frac{1-\sqrt{5}}{2}+B+\frac{1+\sqrt{5}}{2}}{(x+\frac{1+\sqrt{5}}{2})(x+\frac{1-\sqrt{5}}{2})}=\frac{(A+B)x+A\frac{1-\sqrt{5}}{2}+B\frac{1+\sqrt{5}}{2}}{(x+\frac{1+\sqrt{5}}{2})(x+\frac{1-\sqrt{5}}{2})}$
Then $(A+B)=-1$ and $A\frac{1-\sqrt{5}}{2}+B\frac{1+\sqrt{5}}{2}=0$
This s system gives $A$ and $B$.
*
*Add the two series and you are done.
|
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|
Show if $a^2+b^2 \le 2$ then $a+b \le 2$ If $a^2+b^2 \le 2$ then show that $a+b \le2$
I tried to transform the first inequality to $(a+b)^2\le 2+2ab$ then $\frac{a+b}{2} \le \sqrt{1+ab}$ and I thought about applying $AM-GM$ here but without result
|
Suffices to show if $a^2+b^2 = 2$ then $a+b \leq 2$. From the constraint consider
$$
f(a) = a + \sqrt{2-a^2}
$$
and we need to prove $f(a) \leq 2$ over $[0,\sqrt{2}]$.
$$
f'(a) = 1 - \frac{a}{\sqrt{2-a^2}} \Leftrightarrow a = 1
$$
which is a maximum by 1st derivative test.
Since $f(0), f(1), f(\sqrt{2})$ are $\sqrt{2}, 2, \sqrt{2}$ we have $f(a) \leq 2$ as desired.
|
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|
How to show that $ 0< a\leq\cos^2(\theta)\leq b<1$ in this problem? The inequality $2\cos^4(\theta/2)-2\cos^2(\theta/2)+1/4\leq 0$ means that $\cos^2(\theta/2)$ lies between the roots of $2x^2-2x+1/4$ i. e., we can conclude that
$$
\frac{2-\sqrt{2}}{4}\leq\cos^2(\theta/2)\leq\frac{2+\sqrt{2}}{4}.
$$
The question is: from this, can we conclude that
$$
0< a\leq\cos^2(\theta)\leq b<1
$$
for some numbers $a$ and $b$ or, at least $0< a\leq\cos^2(\theta)$?
|
You can use the "double angle formula"
$$
\cos(\theta) = 2\cos^2(\theta / 2) - 1
$$
so
$$
\cos^2(\theta) = (2\cos^2(\theta / 2) - 1)^2
$$
Then we write
\begin{align*}
\frac{2-\sqrt{2}}{4} &\leq \cos^2(\theta/2)\leq\frac{2+\sqrt{2}}{4} \\
\frac{2-\sqrt{2}}{2} &\leq 2\cos^2(\theta/2)\leq\frac{2+\sqrt{2}}{2} \\
\frac{-\sqrt{2}}{2} &\leq 2\cos^2(\theta/2) - 1\leq\frac{\sqrt{2}}{2} \\
\end{align*}
which implies
\begin{align*}
0 &\leq (2\cos^2(\theta/2) - 1) ^2 \leq \left( \frac{\sqrt{2}}{2} \right)^2 = \frac{1}{2} \\
0 &\leq \cos(\theta) ^2 \leq \frac{1}{2} \\
\end{align*}
You can't do better than $0$ for a lower bound, though, because it is possible that $\theta = \frac{\pi}{2}$. Then $\cos(\theta) = 0$, and $\theta$ satisfied the required inequality
$2\cos^4(\theta/2)-2\cos^2(\theta/2)+1/4\leq 0$.
|
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|
if $(a+\sqrt{a^2+1})$ and $(b+\sqrt{b^2+1})$ are converse then prove that a and b are opposites $(a+\sqrt{a^2+1})\,(b+\sqrt{b^2+1})=1$ is supposed to equal: $b = -a$ but how do i get that? I've been trying to solve for like 2 days now.
|
First approach. You can do it quite directly. This is not an elegant path, but it's easy to grasp and it is reliable. Regard your equality as an equation, where $b$ is fixed and $a$ is an unknown. How would you go about solving it?
It seems that the natural thing to do is to get $\sqrt{a^2+1}$ on one side of the equation, and then square everything to get rid of the radicals. Like this:
$$
\begin{align}
a + \sqrt{a^2+1} &= \frac{1}{\sqrt{b^2+1} + b} \\
\sqrt{a^2 + 1} &= \frac{1}{\sqrt{b^2+1} + b} - a \\
a^2 + 1 &= \left(\frac{1}{\sqrt{b^2+1} + b} - a\right)^2 = a^2 - \frac{2a}{\sqrt{b^2+1} + b} + \frac{1}{\left(\sqrt{b^2+1} + b\right)^2} \\
\frac{2a}{\sqrt{b^2+1}+b} &= \frac{1}{\left(\sqrt{b^2+1} + b\right)^2} - 1 \\
2a &= \frac{1}{\sqrt{b^2+1}+b} - \sqrt{b^2+1} - b = \frac{1 - (b^2 + 1) - 2b\sqrt{b^2+1} - b^2}{\sqrt{b^2+1} + b} =
-2b
\end{align}
$$
And therefore $a=-b$.
Second approach. Another way to go about the problem is to exercise your curiosity. You might try to look at the converse statement, just to see what comes out of it: is it true that whenever $a=-b$, the equality $(a + \sqrt{a^2+1})(b + \sqrt{b^2+1})=1$ holds? It this is so, then it would mean that $(a + \sqrt{a^2+1})(-a + \sqrt{a^2+1})=1$ is true for any $a$. Does it look like that's indeed the case? Yes it does, by the formula for the difference of two squares:
$$
(\sqrt{a^2+1}+a)(\sqrt{a^2+1}-a) = (\sqrt{a^2+1})^2 - a^2 = 1.
$$
So, this is always true. Can we use this equality somehow to prove the original statement? We can, but I won't do it, because it's already been done in other answers.
|
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|
Three Surface Integrals Could someone assist with the following three surface integrals?
Q1 The portion of the cone $z=\sqrt{x^2+y^2}$ that lies inside the cylinder $x^2+y^2 =2x$.
Q2 The portion of the paraboloid $z=1-x^2-y^2$ that lies above the $xy$-plane.
Q3 The portion of the paraboloid $2z = x^2+y^2$ that is inside the cylinder $x^2+y^2=8$.
Any assistance will be greatly appreciated.
|
You have $z=f(x,y)$ and the surface area is $$\int\int_A\sqrt{1+f_x^2+f_y^2}dxdy$$ where $A$ is the projection of the surface area on the $xy$-plane.
Use polar coordinates and so $dA=rdrd\theta$
*
*$r=0\to r=2\cos\theta,\theta=-\pi/2\to\pi/2$
*$r=0\to r=1,\theta=0\to2\pi$
*$r=0\to r=\sqrt8,\theta=0\to2\pi$
For the first one, $z=\sqrt{x^2+y^2}$ and so
$$\int\int_A\sqrt{1+\frac{x^2}{x^2+y^2}+\frac{y^2}{x^2+y^2}}dxdy
\\=\int_{-\pi/2}^{\pi/2}\int_0^{2\cos \theta}\sqrt{2}rdrd\theta
\\=\frac{\sqrt{2}}{2}\int_{-\pi/2}^{\pi/2}r^2|_0^{2\cos \theta}d\theta$$
|
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|
Prove for any integer $a$, one of the integers $a$, $a+2$, $a+4$ is divisible by $3$. Prove for any integer $a$, one of the integers $a$, $a+2$, $a+4$ is divisible by $3$.
I know I would use the division algorithm but I am really confused how to go about this. Step by step explanation please? thank you so much!
|
For every number $a$: $a \equiv {0, 1, 2} \pmod{3}$
In the first case: $a \equiv 0 \pmod{3}$
For the second one:
$a \equiv 1 \pmod{3}$ =>
$a+2 \equiv 3 \pmod{3} \equiv 0 \pmod{3}$
For the third one:
$a \equiv 2 \pmod{3}$ =>
$a+4 \equiv 6 \pmod{3} \equiv 0 \pmod{3}$
|
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|
Proves of identities in inverse trigonometry Can someone please help me prove the following results from inverse trigonometry?
$$\tan^{-1}x + \tan^{-1}y = \pi + \tan^{-1}\frac{x+y}{1-xy}( x>0, y>0, xy>1)$$
and
$$\tan^{-1}x + \tan^{-1}y = -\pi + \tan^{-1}\frac{x+y}{1-xy} ( x<0, y< 0, xy>1)$$
I know the prove for $\tan^{-1}x + \tan^{-1}y =\tan^{-1}\frac{x+y}{1-xy} ( xy<1)$ but cant prove the other two. Please do help.
Thanks in advance :)
|
Hint: Let $\tan^{-1} x = a$ and $\tan^{-1} y = b$. Then $x = \tan a$ and $y = \tan b$.
$$\frac{x+y}{1-xy} = \frac{\tan a + \tan b}{1- \tan a \tan b}$$
$$\frac{x+y}{1-xy} = \tan {(a+b)}$$
|
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|
How to find the determinant of this $3 \times 3$ Hankel matrix? Today, at my linear algebra exam, there was this question that I couldn't solve.
Prove that
$$\det \begin{bmatrix}
n^{2} & (n+1)^{2} &(n+2)^{2} \\
(n+1)^{2} &(n+2)^{2} & (n+3)^{2}\\
(n+2)^{2} & (n+3)^{2} & (n+4)^{2}
\end{bmatrix} = -8$$
Clearly, calculating the determinant, with the matrix as it is, wasn't the right way. The calculations went on and on. But I couldn't think of any other way to solve it.
Is there any way to simplify $A$, so as to calculate the determinant?
|
Recall that $a^2-b^2=(a+b)(a-b)$. Subtracting $\operatorname{Row}_1$ from $\operatorname{Row}_2$ and from $\operatorname{Row}_3$ gives
$$
\begin{bmatrix}
n^2 & (n+1)^2 & (n+2)^2 \\
2n+1 & 2n+3 & 2n+5 \\
4n+4 & 4n+8 & 4n+12
\end{bmatrix}
$$
Then subtracting $2\cdot\operatorname{Row}_2$ from $\operatorname{Row}_3$ gives
$$
\begin{bmatrix}
n^2 & (n+1)^2 & (n+2)^2 \\
2n+1 & 2n+3 & 2n+5 \\
2 & 2 & 2
\end{bmatrix}
$$
Now, subtracting $\operatorname{Col}_1$ from $\operatorname{Col}_2$ and $\operatorname{Col}_3$ gives
$$
\begin{bmatrix}
n^2 & 2n+1 & 4n+4 \\
2n+1 & 2 & 4 \\
2 & 0 & 0
\end{bmatrix}
$$
Finally, subtracting $2\cdot\operatorname{Col}_2$ from $\operatorname{Col}_3$ gives
$$
\begin{bmatrix}
n^2 & 2n+1 & 2 \\
2n+1 & 2 & 0 \\
2 & 0 & 0
\end{bmatrix}
$$
Expanding the determinant about $\operatorname{Row}_3$ gives
$$
\det A
=
2\cdot\det
\begin{bmatrix}
2n+1 & 2\\
2 & 0
\end{bmatrix}
=2\cdot(-4)=-8
$$
as advertised.
|
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|
How find this maximum $f(x)=\sqrt{\frac{1+x^2}{2}}+\sqrt{x}-x$ let $x\ge 0$,Find the follow maximum
$$f(x)=\sqrt{\dfrac{1+x^2}{2}}+\sqrt{x}-x$$
I find
$$f'(x)=\dfrac{1}{2}\left(\dfrac{\sqrt{2}x}{\sqrt{x^2+1}}+\dfrac{1}{\sqrt{x}}-2\right)$$
so
$$f'(x)=0\Longrightarrow x=1$$
so I think
$$f(x)_{max}=f(1)$$
But I this methods is ugly, and I think this have AM-GM inequality or Cauchy-Schwarz inequality to solve this problem
$$f(x)=\sqrt{\dfrac{1+x^2}{2}}+\sqrt{x}-x\le\sqrt{\dfrac{1+x^2}{2}}+\dfrac{1}{2}(1+x)-x=\sqrt{\dfrac{1+x^2}{2}}+\dfrac{1}{2}(1-x)$$
Then I can't,Thank you
|
Simply consider that $\sqrt{x}$ is a concave function, hence, for any $a,b\geq 0$, $\sqrt{a}+\sqrt{b}\leq \sqrt{2(a+b)}$. By taking $a=x$ and $b=\frac{1+x^2}{2}$ you get $f(x)\leq 1$ with equality only when $x=\frac{1+x^2}{2}$, i.e. $x=1$, as wanted.
|
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|
If $x+\frac1x=5$ find $x^5+\frac1{x^5}$. If $x>0$ and $\,x+\dfrac{1}{x}=5,\,$ find $\,x^5+\dfrac{1}{x^5}$.
Is there any other way find it?
$$
\left(x^2+\frac{1}{x^2}\right)\left(x^3+\frac{1}{x^3}\right)=23\cdot 110.
$$
Thanks
|
Hint: Expand (binomial formula) $\left(x+\frac{1}{x}\right)^3$ and then $\left(x+\frac{1}{x}\right)^5$. (No need for even powers.)
|
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|
Let $\theta=\frac{2\pi}{5}$. Show $2\cos(2\theta)+2\cos(\theta)+1=0$. I have been at this for a while. Any ideas?
|
I think this is a simple way. By using De Moivre’s formula, we obtain the fifth time angle formula:
\begin{equation}
\cos(5\theta)=16\cos^5\theta-20\cos^3\theta+5\cos\theta
\end{equation}
Set $\theta=2\pi/5$, we have:
\begin{equation}
\cos(2\pi)=16\cos^5\theta-20\cos^3\theta+5\cos\theta\\
16\cos^5\theta-20\cos^3\theta+5\cos\theta-1=0
\end{equation}
By using factorization, we have:
\begin{equation}
(\cos\theta-1)(4\cos^2\theta+2\cos\theta-1)^2=0
\end{equation}
Obviously, $\cos\theta-1\neq0$, then the equality have been prove.
|
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|
Show that the sequence $a_{n+1}=a_{n}+\dfrac{a^2_{n}}{n^2}$ is upper bounded Let $\{a_{n}\}$ be defined with $a_{1}\in(0,1)$, and
$$a_{n+1}=a_{n}+\dfrac{a^2_{n}}{n^2}$$
for all $n\gt 0$. Show that the sequence is upper bounded.
My idea: since
$$a_{n+1}=a_{n}\left(1+\dfrac{a_{n}}{n^2}\right)$$
then
$$\dfrac{1}{a_{n+1}}=\dfrac{1}{a_{n}}-\dfrac{1}{a_{n}+n^2}$$
then
$$\dfrac{1}{a_{n}}-\dfrac{1}{a_{n+1}}=\dfrac{1}{a_{n}+n^2}$$
so
$$\dfrac{1}{a_{1}}-\dfrac{1}{a_{n+1}}=\sum_{i=1}^{n}\dfrac{1}{a_{i}+i^2}$$
since
$$a_{n+1}>a_{n}\Longrightarrow \dfrac{1}{a_{i}+i^2}<\dfrac{1}{a_{1}+i^2}$$
so
$$\dfrac{1}{a_{n+1}}>\dfrac{1}{a_{1}}-\left(\dfrac{1}{1+a_{1}}+\dfrac{1}{2^2+a_{1}}+\cdots+\dfrac{1}{a_{1}+n^2}\right)$$
But the RHS might be $\lt0$ for a sufficiently large starting value; for instance, with $a_{1}=\dfrac{99}{100}$
then
$$\dfrac{1}{a_{1}}-\left(\dfrac{1}{1+a_{1}}+\dfrac{1}{2^2+a_{1}}+\cdots+\dfrac{1}{a_{1}+n^2}\right)<0,n\to\infty$$
see:
so this method won't let me bound the series and I don't know what else to do.
|
Here is a solution: Since $(a_{n})$ is monotone increasing, either it remains bounded or it diverges to $+\infty$. Assume that $a_{n} \nearrow +\infty$.
Observation 1. Referring to OP's identity
$$ \frac{1}{a_{n}} - \frac{1}{a_{n+1}} = \frac{1}{a_{n} + n^{2}}, $$
we have
$$ \frac{1}{a_{n}} - \frac{1}{a_{m+1}} = \sum_{k=n}^{m} \frac{1}{a_{k} + k^{2}}. $$
Taking $m \to \infty$, it follows that
\begin{align*}
\frac{1}{a_{n}}
= \sum_{k=n}^{\infty} \frac{1}{a_{k} + k^{2}}
\leq \sum_{k=n}^{\infty} \frac{1}{k^{2} - k}
= \frac{1}{n-1}.
\end{align*}
Thus we must have $n-1 \leq a_{n}$ for any $n \geq 2$.
Observation 2. Now we prove that $a_{n} \leq a_{1} n$ for any $n$. Indeed, this is trivial when $n = 1$. Also, if it holds for $n$ then
$$ a_{n+1} = a_{n} \left( 1 + \frac{a_{n}}{n^{2}} \right) \leq n a_{1} \left( 1 + \frac{1}{n} \right) = (n+1)a_{1}. $$
Therefore by induction, we have the desired estimate.
Conclusion. Combining two observation, it follows that
$$n - 1 \leq a_{n} \leq na_{1} \quad \Longrightarrow \quad 1 - \frac{1}{n} \leq a_{1}.$$
for $n \geq 2$. Taking $n\to\infty$, we have $1 \leq a_{1}$, a contradiction! Therefore $(a_{n})$ must remain bounded.
Addendum. Now let us investigate an asymptotic behavior of the limit $f(a_{1}) = \lim_{n\to\infty} a_{n}$. We have
\begin{align*}
\frac{1}{a_{1}} - \frac{1}{f(a_{1})}
&= \sum_{n=1}^{\infty} \frac{1}{n^{2} + a_{n}}
\geq \sum_{n=1}^{\infty} \frac{1}{n^{2} + n}
= 1,
\end{align*}
So we have a lower bound.
$$ f(a_{1}) \geq \frac{a_{1}}{1 - a_{1}}. $$
|
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|
Sum of four squares not a prime Let $ a, b, c, d $ be natural numbers such that $ ab=cd $. Prove that $ a^2+b^2+c^2+d^2 $ is not a prime.
I am clueless on this one. I tried contradiction, but didn't get anywhere.
Can you help?
Edit: I understand natural numbers to be strictly positive, excluding $0 $.
|
Let $N := a^2 + b^2 + c^2 + d^2$
Note that $bd(a^2 + c^2) = ac(b^2 + d^2)$, while $ac < a^2+c^2$, so that $$lcm(a^2 + c^2, b^2 + d^2) < (a^2 + c^2)(b^2 + d^2)$$ This tells us that $a^2 + c^2$ and $b^2 + d^2$ are not coprime, and so $$ 1 < gcd(a^2 + c^2, b^2 + d^2)< N $$
Finally this $gcd|N$, which completes the proof.
|
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|
Have I done something wrong in solving the following pair of equations? Question given:
Solve,$3x^2-5y^2-7=0\\3xy-4y^2-2=0$
What I have done so far:
$$
3xy-4y^2-2=0\\\frac{3xy}{y}-\frac{4y2}{y}-\frac{2}{y}=0\\3x-4y-\frac{2}{y}=0\\3x=4y+\frac{2}{y}\\x=\frac{(4y+\frac{2}{y})}{3}
$$By substituting $x=\frac{(4y+\frac{2}{y})}{3}$ in the first equation,$$3x^2-5y^2-7=0\\3(\frac{4y+\frac{2}{y}}{3})^2-5y^2-7=0\\3(\frac{16y^2+\frac{4}{y^2}+16}{9})-5y^2-7=0\\16y^2+\frac{4}{y^2}+16-15y^2-21=0\\y^2+\frac{4}{y^2}-5=0\\y^4-5y^2+4=0$$Let $a=y^2$,$$(y^2)^2-5(y^2)+4=0\\a^2-5a+4=0\\a=\frac{5\pm\sqrt{25-4.1.4}}{2}\\a=\frac{5\pm\sqrt{25-16}}{2}\\a=\frac{5\pm3}{2}\\y^2=\frac{5\pm3}{2}\\y=\pm\sqrt{\frac{5\pm3}{2}}$$
Have I done something wrong when solving these pair of equations ? If so, please correct me.
Best Regards !
|
What you did is perfectly correct. You only missed to continue the simplifcations since $a$ is either $4$ or $1$ and then $y$ is ...
Good job !
|
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|
Find the value of $a+b+c$ If $$3.\sqrt{5.\sqrt[3]{37}-16}=\sqrt[3]a-\sqrt[3]b-c$$
What is the value of $a+b+c$?
EDIT: I forgot to mention that $a$, $b$ and $c$ are positive integers.
I tried squaring and then cubing but it got very lengthy. Is there some elegant method to do it?
|
Let $\theta=\sqrt[3]{37}$. If we put $\alpha=\theta^2-2\theta-2 \approx 2.4$, then
$$
\begin{array}{lcl}
\alpha^2 &=& (\theta^2-2\theta-2)^2 \\
&=& \theta^4 - 4 \theta^3 + 8 \theta + 4 \\
&=& 37\theta -4\times 37+8\theta+4 \\
&=& 45\theta -144 \\
&=& 9(5\theta-16)
\end{array}
$$
It follows that $3\sqrt{5\theta-16}=\alpha$, so
$a=37^2,b=8\times 37,c=2$, and hence $a+b+c=1667$.
|
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|
Partial Fraction Decomposition of $\frac{x^2}{(x-1)(x+1)^2}$ I am suppose to find the Partial Fraction Decomposition of
$$\dfrac{x^2}{(x-1)(x+1)^2}$$
My work so far:
$$\dfrac{x^2}{(x-1)(x+1)^2} = \dfrac{A}{x-1} + \dfrac{B}{x+1} + \dfrac{C}{(x+1)^2}$$
Thus,
$$ x^2 = A(x+1)^3 + B(x-1)(x+1)^2 + C(x+1)(x-1)$$
With x = 0 we get,
$$ 0 = A - B - C $$
I am not sure what to do after this, as plugging in the remaining factors (x = -1, 1) Does not help me find the unknown coefficients.
Any help would be appreciated. Thanks
Edit. With $x = 1$,
$$1 = A + 0B + 0C$$
$$A = 1$$
With $x = -1$
$$1 = 0A + 0B + 0C$$
$$1 = 0$$
|
After your "thus" it must be
$$x^2=A(x+1)^2+B(x-1)(x+1)+C(x-1)$$
and not what you wrote there. Now substitute $\;x=1\;$ in both sides and get
$$1^2=A\cdot2^2\implies A=\frac14$$
and then substitute $\;x=-1\;$ and get ...etc.
|
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|
Difference of consecutive cubes never divisible by 5. This is homework from my number theory course.
Since $(x+1)^3-x^3=3x^2+3x+1$ and $x^3-(x+1)^3=-3x^2-3x-1$, to say that the difference of two cubes is divisible by 5 is the same as saying that $3x^2+3x+1\equiv 0\mod 5$ or $-3x^2-3x-1\equiv 0\mod 5$. Both of these statements imply that $x(x+1)\equiv 3\mod 5$. Thus I can finish this by showing that there are no such integers which satisfy $x(x+1)\equiv 3\mod 5$.
I want to say that it is sufficient to check by hand for the values 0,1,2,3, and 4 (for which it is not true), but other than following this by a messy induction I was wondering if there is an easier way to show that there are no integers such that $x(x+1)\equiv 3\mod 5$?
|
$$x^2+x\equiv 3\pmod 5\\
\implies x^2+x+4^{-1}\equiv 3+4^{-1}\pmod 5\\
\implies (x+2^{-1})^2=(x+3)^2\equiv 2\pmod 5$$
The quadratic residues modulo $5$ are $0,1,4$ and this list does not include $2$. QED
|
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|
Find limit of this decreasing sequence $$a_n=\left(1-\frac{1}{2^2}\right)\left(1-\frac{1}{3^2}\right)\left(1-\frac{1}{4^2}\right) \cdots
\left(1-\frac{1}{n^2}\right)
$$
I have proved that this sequence is decreasing. However I am trying to figure out how to find its limit.
|
Answer:
$$\left(\frac{3}{2}\frac{4}{3}\frac{5}{4}\cdots\frac{n}{n-1}\frac{n+1}{n}\right)\cdot\left(\frac{1}{2}\frac{2}{3}\frac{3}{4}\cdots\frac{n-2}{n-1}\frac{n-1}{n}\right)$$
After mass cancellations, pull the $$\frac{n+1}{2}\text{ and }\frac{1}{n}$$
$$\frac{n+1}{2}\cdot\frac{1}{n}$$
Limit of this function tending to infinity $= 1/2$.
|
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|
Asymptotic solution of recurrence equation I need help to find asymptotic solution of this recurrence equation
$T(n)=\sqrt{n}T(\sqrt{n})+cn$ where $c$ is constant.
|
Here are two thoughts, either one may help.
$$\begin{split}
T(n) &= n^{1/2} T\left(n^{1/2}\right) + cn \\
&= n^{1/2} \left[ n^{1/4} T\left(n^{1/4}\right)+cn^{1/2} \right] + cn\\
&= n^{1/2+1/4} T\left(n^{1/4}\right) + cn^{1/2+1/2} + cn\\
&= n^{1/2+1/4} \left[ n^{1/8} T\left(n^{1/8}\right)+cn^{1/4} \right] + 2cn\\
&= n^{1/2+1/4+1/8} T\left(n^{1/8}\right) + 3cn \\
&= \ldots \\
&= n^{1-1/2^k} T\left(n^{1/2^k}\right) + kcn \\
&= \ldots \text{ let } k = \log \log n
\text{ so } 2^k = \log n
\text{ and } n^{2^{-k}} = n^{1/\log n} = 2
\ldots\\
&= \frac{n}{2} T\left(2\right) + cn\log \log n \\
&= \Theta (n \log \log n)
\end{split}$$
A different approach is to divide by $n$:
$$
\frac{T(n)}{n} = \frac{T \left( \sqrt{n} \right)}{\sqrt{n}}+c
$$
and changing variables to $S(n) = T(n)/n$ yields $S(n) = S\left(n^{1/2}\right)+c$, which should be easier to solve that the one above directly.
|
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|
Integrate $\frac{x^2-1}{x^2+1}\frac{1}{\sqrt{1+x^4}}dx$ An antiderivative from Spivak
$$\int \frac{x^2-1}{x^2+1}\frac{1}{\sqrt{1+x^4}}dx$$
The idea I had was to write the first factor as $\left(1-\dfrac{2}{x^2+1}\right)$, but I don't see how that's helping!
|
$$
\begin{align}
\int\frac{x^2 - 1}{x^2 + 1}\frac{1}{\sqrt{x^4+1}}\,\mathrm{d}x &=-\int\frac{1-x^2}{1+x^2}\frac{\mathrm{d}x}{\sqrt{\dfrac{\left(1+x^2\right)^2 + \left(1-x^2\right)^2}{2}}}\\
&=-\frac{\sqrt{2}}{2}\int\frac{1-x^2}{1+x^2}\frac{1}{\sqrt{1 + \left(\dfrac{1-x^2}{1+x^2}\right)^2}}\frac{2}{1 + x^2}\,\mathrm{d}x
\end{align}
$$
Now, let's use a "reverse" Weierstrass substitution, setting up $x = \tan\left(\dfrac{\theta}{2}\right) $ and $\dfrac{2}{1+x^2}\,\mathrm{d}x = \mathrm{d}\theta $. Also, we have
$$\cos\theta = \dfrac{1-x^2}{1+x^2}\qquad\qquad\sin\theta=\frac{2x}{x^2+1} $$ Then,
$$\begin{aligned}-\frac{\sqrt{2}}{2}\int\frac{\cos\theta}{\sqrt{1 + \cos^2\theta}}\,\mathrm{d}\theta&=-\frac{\sqrt{2}}{2}\int\frac{\cos\theta}{\sqrt{2 - \sin^2\theta}}\,\mathrm{d}\theta\\
&=-\frac{1}{\sqrt{2}}\arcsin\left(\frac{\sin\theta}{\sqrt{2}}\right) + C\\
&=-\frac{1}{\sqrt{2}}\arcsin\left(\frac{x\sqrt{2}}{x^2 + 1}\right)+C \end{aligned}$$
|
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|
Find square roots upto infinte times Evaluate :
$\sqrt{1+ 2 \sqrt{1+3 \sqrt{1+\dots\infty}}}$
Is it possible to solve in the following way :
Let
$x=\sqrt{1+ 2 \sqrt{1+3 \sqrt{1+\dots\infty}}}$
$x^2= 1+ 2 \sqrt{1+3 \sqrt{1+\dots\infty}}$
next ??????
I don't know
|
Note that $(x+1)^2=1+x(x+2)$, now taking the principal square root of both sides and re-substituting the resulting expression for $x+1$ multiple times gives us that:
$$(x+1)=\sqrt{1+x(x+2)}$$
$$(x+1)=\sqrt{1+x\sqrt{1+(x+1)(x+3)})}$$
$$(x+1)=\sqrt{1+x\sqrt{1+(x+1)\sqrt{1+(x+2)(x+4)}}}$$
$$(x+1)=\sqrt{1+x\sqrt{1+(x+1)\sqrt{1+(x+2)\sqrt{1+(x+3)\sqrt{1+(x+4)....}}}}}$$
$$3=\sqrt{1+2\sqrt{1+3\sqrt{1+4\sqrt{1+5\sqrt{1+6\sqrt{1+7}}...}}}}$$
You can use the monotone convergence theorem to prove that this resulting expression converges.
|
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|
Fibonacci proof by Strong Induction Prove by strong induction that for a ∈ A we have
$F_a + 2F_{a+1} = F_{a+4} − F_{a+2}.$
$F_a$ is the $a$'th element in the Fibonacci sequence
|
Do you consider the sequence starting at 0 or 1? I will assume 1.
If that is the case, $F_{a+1} = F_a + F_{a-1}) $ for all integers where $a \geq 3$.
The original equation states $F_{a+1} = (F_a) + F_{a-1} $.
.
$F_{a+1} = (F_a) + F_{a-1} $
$-(F_a) = -F_{a+1}+ F_{a-1} $
$F_a = F_{a+1}- F_{a-1}$. This equation is important.
.
$F_{a+3} = F_{a+4} - F_{a+2}$
after subtracting and dividing by -1 we have
$F_{a+4} = F_{a+3} + F_{a+2}$. This equation is important too.
.
By shifting we have $F_{a+3} = F_{a+2} + F_{a+1}$ and $F_{a+2} = F_{a+1} + F_{a}$. These formulas will be used to "reduce the power," in a sense.
$F_{a+4} - F_{a+2} = F_{a+2} + F_{a+1} + F_{a+2} - F_{a+2}$
$F_{a+4} - F_{a+2} = F_{a+2} + F_{a+1}$
By using the substitution $F_{a+2} = F_{a+1} + F_{a}$ we have $F_{a+4} - F_{a+2} = (F_{a} + F_{a+1}) + F_{a+1}$
Therefore $F_{a+4} - F_{a+2} = F_{a} + 2F_{a+1}$
|
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|
If $a,b,c>0$ and $a+b+c=1$ prove inequality: $\frac a{b^2 +c} + \frac b{c^2+a} + \frac c{a^2+b} \ge \frac 94$ If $a,b,c>0$ and $a+b+c=1$ prove inequality: $$\frac a{b^2 +c} + \frac b{c^2+a} + \frac c{a^2+b} \ge \frac 94$$
|
Answer:
Multiplying a in the numerator and the denominator, you get
$$LHS = \frac{\sum a^{2}}{\sum (ac+b^{2}a)}$$
Applying the Cauchy-Schwarz inequality in the below steps
\begin{align}
&\geqslant\frac{(\sum a)^{2}}{\sum ac + \sum b^{2}a}\\
&\geqslant \frac{1}{\sum ac + \frac{1}{3}\sum a \sum a^2} = \frac{1}{3\sum ac +\sum a (\sum a)^2}\\
&\geqslant \frac{3}{\sum ac + (\sum a)^{2}}
\end{align}
Applying the inequality $$\sum ac \leqslant \frac{1}{3} (\sum a)^{2}$$
$$LHS \geqslant \frac{9}{4}$$
Edit: @9rm, Apply the Cauchy-Schwarz inequlality to $$\sum b^{2}a \leqslant (\sum b^4)^\frac{1}{2}\cdot (\sum a^2)^\frac{1}{2}.$$ This will reduce to $$\leqslant (abc)^\frac{1}{2}\cdot \sum a\cdot \sqrt{\frac{1}{3}}({\sum a})^{2}.$$ This will further reduce to $$\sqrt{\frac{1}{3}}\cdot \sum a\cdot \sqrt{\frac{1}{3}}\cdot {\sum a}^{2} = \frac{1}{3}\cdot \sum a(\sum a)^{2}.$$
|
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|
Proof this curious trigonometric identity Proof that $$\cos^2{10^\circ} + \cos^2{50^\circ} - \sin{40^\circ}\sin{80^\circ} = \frac{3}{4}$$
I notice that $10^\circ + 80^\circ = 90^\circ$, and $50^\circ +40^\circ = 90^\circ$. I tried doing some manipulation but my efforts were futile. Any hints?
|
Convert into a polynomial of $\cos(40^\circ)$:
$$
\begin{align}
&\cos^2(10^\circ)+\cos^2(50^\circ)-\sin(40^\circ)\sin(80^\circ)\\
&=\sin^2(80^\circ)+\sin^2(40^\circ)-\sin(40^\circ)\sin(80^\circ)\tag{1}\\
&=4\sin^2(40^\circ)\cos^2(40^\circ)+\sin^2(40^\circ)-2\sin^2(40^\circ)\cos(40^\circ)\tag{2}\\
&=1-2\cos(40^\circ)+3\cos^2(40^\circ)+2\cos^3(40^\circ)-4\cos^4(40^\circ)\tag{3}
\end{align}
$$
Explanation:
$(1)$: $\cos(10^\circ)=\sin(80^\circ)$ and $\cos(50^\circ)=\sin(40^\circ)$
$(2)$: $\sin(80^\circ)=2\sin(40^\circ)\cos(40^\circ)$
$(3)$: $\sin^2(40^\circ)=1-\cos^2(40^\circ)$
Since $\cos(3x)=4\cos^3(x)-3\cos(x)$, we have
$$
4\cos^3(40^\circ)-3\cos(40^\circ)=\cos(120^\circ)=-\tfrac12\tag{4}
$$
Thus, we write the polynomial from $(3)$, with $x=\cos(40^\circ)$, as
$$
\underbrace{1-2x+3x^2+2x^3-4x^4}_{\text{from }(3)}=\tfrac34+\left(\tfrac12-x\right)\underbrace{\left(\tfrac12-3x+4x^3\right)}_{0\text{ from }(4)}\tag{5}
$$
Therefore, $(3)$, $(4)$ and $(5)$ say
$$
\cos^2(10^\circ)+\cos^2(50^\circ)-\sin(40^\circ)\sin(80^\circ)=\frac34\tag{6}
$$
|
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|
Product of reflections is a rotation, by elementary vector methods Let $\mathbf{u}$ and $\mathbf{v}$ be two 3D unit vectors. The transform that performs reflection in the plane normal to $\mathbf{u}$ is given by
$$
T_{\mathbf{u}}(\mathbf{x}) = \mathbf{x} - 2(\mathbf{x} \cdot \mathbf{u})\mathbf{u}
$$
and similarly, reflection in the plane normal to $\mathbf{v}$ is performed by
$$
T_{\mathbf{v}}(\mathbf{x}) = \mathbf{x} - 2(\mathbf{x} \cdot \mathbf{v})\mathbf{v}
$$
Let $\theta$ be the angle between $\mathbf{u}$ and $\mathbf{v}$, and let $\mathbf{n}$ be the unit vector in the direction of $\mathbf{u} \times \mathbf{v}$. So, then we know that $\cos\theta = \mathbf{u} \cdot \mathbf{v}$, and $\mathbf{u} \times \mathbf{v} = (\sin\theta)\mathbf{n}$.
The composition of these two reflections is a rotation around $\mathbf{n}$ by an angle of $2\theta$ (I believe), and that rotation is given by Rodrigues' formula:
$$
R(\mathbf{x}) = (\cos2\theta)\mathbf{x} +
(1 - \cos2\theta)(\mathbf{x} \cdot \mathbf{n})\mathbf{n} +
(\sin 2\theta)(\mathbf{x} \times \mathbf{n})
$$
It seems to me that we ought to be able to prove from first principles that
$$
T_{\mathbf{u}}\big( T_{\mathbf{v}}(\mathbf{x}) \big) = R(\mathbf{x})
$$
I've slogged through pages of vector algebra for a few hours, but to no avail. It's depressing -- I used to be good at this stuff, but apparently not any more. I'd like a proof that uses nothing but elementary vector arithmetic, and I'd like it to be coordinate-free, please.
Edit
As a couple of people have mentioned, it seems sensible to work in the $\mathbf{u}\text{-}\mathbf{v}\text{-}\mathbf{n}$ coordinate system. This doesn't violate my "coordinate free" requirement as long as we don't start writing out explicit coordinates for $\mathbf{u}$, $\mathbf{v}$ and $\mathbf{n}$. The vector $R(\mathbf{x}) - \mathbf{x}$ should be entirely in the
$\mathbf{u}\text{-}\mathbf{v}$ plane, so all of its $\mathbf{n}$ terms must vanish, and we should be left with an expression that involves only $\mathbf{u}$ and $\mathbf{v}$, which (I hope) will give us the link to the reflections. The algebraic grunt-work involved is what's giving me trouble.
|
Write $c := \cos\theta$, $s := \sin\theta$, $\mathbf{w} := \mathbf{u}\times\mathbf{v} = s\mathbf{n}$, and $T := T_\mathbf{u}\left(T_\mathbf{v}\right)$, so that we have ...
$$\begin{align}
T(\mathbf{x}) &=\mathbf{x}-2(\mathbf{x}\cdot\mathbf{u})\mathbf{u}-2(\mathbf{x}\cdot\mathbf{v})\mathbf{v} + 4c(\mathbf{x}\cdot\mathbf{v})\mathbf{u}\\
R(\mathbf{x}) &= (2c^2-1)\mathbf{x} + 2 s^2 (\mathbf{x}\cdot\mathbf{n}) \mathbf{n} + 2 s c (\mathbf{x}\times\mathbf{n}) \\
&= (2c^2-1)\mathbf{x} + 2 (\mathbf{x}\cdot\mathbf{w})\mathbf{w} + 2 c (\mathbf{x}\times\mathbf{w}) \\
\end{align}$$
Decomposing $\mathbf{x}$ as $p\mathbf{u} + q\mathbf{v} + r \mathbf{w}$, we can get fairly directly ...
$$\mathbf{x}\cdot\mathbf{u} = p + q c \qquad \mathbf{x}\cdot\mathbf{v}=pc+q \qquad \mathbf{x}\cdot\mathbf{w}=rs^2 \qquad (\star)$$
$$\mathbf{x}\times\mathbf{w} = \mathbf{x}\times \left(\mathbf{u}\times\mathbf{v}\right) = (\mathbf{x}\cdot\mathbf{v})\mathbf{u}-(\mathbf{x}\cdot\mathbf{u})\mathbf{v} \qquad (\star\star)$$
Then it's straightforward to show that the difference of the transformations vanishes:
$$\begin{align}
T(\mathbf{x}) - R(\mathbf{x}) &=\mathbf{x}-2(p+qc)\mathbf{u}-2(pc+q)\mathbf{v} + 4c(pc+q)\mathbf{u}\\
&-\left( (2c^2-1)\mathbf{x} + 2 r s^2 \mathbf{w} + 2 c \left( (pc+q)\mathbf{u} - (p+qc)\mathbf{v} \right) \right) \\[6pt]
&= (2-2c^2)\;\mathbf{x} + 2 \left(-p-qc+2pc^2+2qc-pc^2-qc\right)\;\mathbf{u} \\
&+ 2\left(-pc-q+pc+qc^2\right)\mathbf{v} - 2 r s^2 \mathbf{w} \\[6pt]
&= 2 s^2 \left( \mathbf{x} - p\mathbf{u} - q \mathbf{v} - r\mathbf{w} \right) \\[6pt]
&= 0
\end{align}$$
and we conclude that the transformations are equivalent. $\square$
Edit. Without jumping immediately to the decomposition of $\mathbf{x}$, we can use the expansion in $(\star\star)$ to write
$$\begin{align}
\frac{T(\mathbf{x})-R(\mathbf{x})}{2s^2} \;\;&=\;\; \mathbf{x}
\;-\; \left( \; \frac{\mathbf{x}.( \mathbf{u} - c \mathbf{v} )}{s^2}\;\mathbf{u}
\;+\; \frac{\mathbf{x}.( \mathbf{v} - c \mathbf{u} )}{s^2}\;\mathbf{v}
\;+\; \frac{\mathbf{x}.\mathbf{w}}{s^2}\;\mathbf{w} \;\right)
\end{align}$$
If you can "see" that the coefficients of $\mathbf{u}$, $\mathbf{v}$, $\mathbf{w}$ are the components of $\mathbf{x}$ ---which would be clear for orthogonal $\mathbf{u}$ and $\mathbf{v}$, for which $c=0$ and $s=1$--- then you're done. If not, note that you can arrive at this insight by solving the dot-product equations $(\star)$ for $p$, $q$, $r$.
|
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|
Integral $\int_0^\infty F(x)\,F\left(x\,\sqrt2\right)\frac{e^{-x^2}}{x^2} \, dx$ involving Dawson's function I need your help evaluating this integral:
$$I=\int_0^\infty F(x)\,F\left(x\,\sqrt2\right)\frac{e^{-x^2}}{x^2} \, dx,\tag1$$
where $F(x)$ represents Dawson's function/integral:
$$F(x)=e^{-x^2}\int_0^x e^{y^2} \, dy = \frac{\sqrt{\pi}}{2} e^{-x^{2}} \operatorname{erfi}(x).\tag2$$
Dawson's function can also be represented by the infinite integral $$F(x) = \frac{1}{2} \int_{0}^{\infty} e^{-t^{2}/4} \sin(xt) \, dt.$$
Since $F(x)$ behaves like $x$ near $x=0$ and like $\frac{1}{2x}$ for large values of $x$, we know that integral $(1)$ converges.
|
Notice that for $a>0$, we have $$F(ax) = e^{-a^{2}x^{2}}\int_{0}^{ax} e^{y^{2}} \mathrm dy = e^{-a^{2}x^{2}} \int_{0}^{a} u e^{x^{2}u^{2}} \, \mathrm du . \tag{1}$$
Then using $(1)$, we get
$$ \begin{align} I &= \int_{0}^{\infty} F(x) F(x \sqrt{2}) \, \frac{e^{-x^{2}}}{x^{2}} \, \mathrm dx \\&= \int_{0}^{\infty} \int_{0}^{\sqrt{2}} \int_{0}^{1} x e^{-x^{2}} e^{x^{2} y^{2}} x e^{-2x^{2}} e^{x^{2}z^{2}} \, \frac{e^{-x^{2}}}{x^{2}} \, \mathrm dy \, \mathrm dz \, \mathrm dx \\ &= \int_{0}^{\sqrt{2}} \int_{0}^{1} \int_{0}^{\infty} e^{-(4-y^{2}-z^{2})x^{2}} \, \mathrm dx \, \mathrm dy \, \mathrm dz \\ &= \frac{\sqrt{\pi}}{2} \int_{0}^{\sqrt{2}} \int_{0}^{1} \frac{1}{\sqrt{4-y^{2}-z^{2}}} \, \mathrm dy \, \mathrm dz \\&= \frac{\sqrt{\pi}}{2} \int_{0}^{\sqrt{2}} \int_{0}^{\arcsin ( \frac{1}{\sqrt{4-z^{2}}})} \, \mathrm d \theta \, \mathrm d z \tag{2} \\ &= \frac{\sqrt{\pi}}{2} \int_{0}^{\sqrt{2}} \arcsin \left( \frac{1}{\sqrt{4-z^{2}}} \right) \, \mathrm dz \\ &= \frac{\sqrt{\pi}}{2} \left( \frac{ \sqrt{2} \pi}{4} - \int_{0}^{\sqrt{2}} \frac{z^{2}}{\sqrt{3-z^{2}} (4-z^{2})} \, \mathrm dz \right) \tag{3} \\ &= \frac{\sqrt{\pi}}{2} \left( \frac{\sqrt{2} \pi}{4} - \int_{1 / \sqrt{2}}^{\infty} \frac{1}{\sqrt{3u^{2}-1} (4u^{2}-1)} \frac{\mathrm du}{u}\right) \tag{4} \\ &= \frac{\sqrt{\pi}}{2} \left( \frac{\sqrt{2} \pi}{4} - 3 \int_{1 /\sqrt{2}}^{\infty} \frac{1}{ (4w^{2}+1)(w^{2}+1)} \, \mathrm dw\right) \tag{5}\\ &=\frac{\sqrt{\pi}}{2} \left( \frac{\sqrt{2} \pi}{2} - 4 \int_{1/ \sqrt{2}}^{\infty} \frac{1}{4w^{2}+1} + \int_{1/ \sqrt{2}}^{\infty} \frac{1}{w^{2}+1} \, \mathrm dw\right) \\ &= \frac{\sqrt{\pi}}{2} \left[ \frac{\sqrt{2} \pi}{4} - \pi +2 \arctan \left( \sqrt{2} \right) +\frac{\pi}{2} - \arctan \left( \frac{1}{\sqrt{2}} \right) \right] \\ &= \frac{\sqrt{\pi}}{2} \left( \frac{\sqrt{2} \pi}{4} - \pi + 3 \arctan{\sqrt{2}} \right). \end{align}$$
$(2)$ Let $y=\sqrt{4-z^{2}}\sin \theta$.
$(3)$ Integrate by parts.
$(4)$ Let $z = \frac{1}{u}$.
$(5)$ Let $w^{2}=3u^2-1$.
EDIT:
Using the same approach, I get
$$ \int_{0}^{\infty} F(ax) F(bx) \, \frac{e^{-p^{2}x^{2}}}{x^{2}} \, \mathrm dx $$
$$ = \frac{\sqrt{\pi}}{2} \left[b \arcsin \left( \frac{a}{\sqrt{a^{2}+p^{2}}} \right) - \sqrt{a^{2}+b^{2}+p^{2}} \arctan \left(\frac{ab}{p \sqrt{a^{2}+b^{2}+p^{2}}} \right) + a \arctan \left( \frac{b}{p} \right)\right]$$
where $a, b,$ and $p$ are all positive parameters.
|
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|
How to show $\sum_{i=1}^{n} \binom{i}{2}=\binom{n+1}{3}$?
Show that $\,\displaystyle\sum_{i=1}^{n} \binom{i}{2}=\binom{n+1}{3}$.
I'm thinking right now (though not getting anywhere with it) that I want to expand out the summation portion to $i!/2!(i-2)!$ and simplify from there? Not sure if that will help, not to mention if I put $1$ in for $i$ I get $1/-2$ which I don't think is right.
Anyone care to shed some light on the subject?
Thanks.
|
$$\sum_{i=2}^n\binom{i}{2}=\binom{2}{2}+\binom{3}{2}+\binom{4}{2}+\binom{5}{2}+\cdots+\binom{n}{2}$$
Since $\binom22=\binom33$
$$\sum_{i=2}^n\binom{i}{2}=\binom{3}{3}+\binom{3}{2}+\binom{4}{2}+\binom{5}{2}+\cdots+\binom{n}{2}$$
Pascal's Triangle Identity states that
$$\binom{n-1}k+\binom{n-1}{k-1}=\binom{n}{k}$$
So,
$$\sum_{i=2}^n\binom{i}{2}=\color{red}{\binom{3}{3}+\binom{3}{2}}+\binom{4}{2}+\binom{5}{2}+\cdots+\binom{n}{2}$$
$$\sum_{i=2}^n\binom{i}{2}=\color{red}{\binom{4}{3}}+\binom{4}{2}+\binom{5}{2}+\cdots+\binom{n}{2}$$
$$\sum_{i=2}^n\binom{i}{2}=\color{green}{\binom{4}{3}+\binom{4}{2}}+\binom{5}{2}+\cdots+\binom{n}{2}$$
$$\sum_{i=2}^n\binom{i}{2}=\color{green}{\binom{5}{3}}+\binom{5}{2}+\cdots+\binom{n}{2}$$
So on and so forth, until we have
$$\sum_{i=2}^n\binom{i}{2}=\binom{n}{3}+\binom{n}{2}=\binom{n+1}{3}$$
|
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"timestamp": "2023-03-29T00:00:00",
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|
the probability that exactly 4 dice are sixes given that none are fives. My answer key tells me the answer is (10 Choose 4)(1/5)^4 (4/5)^6.
But isn't conditional probability of P(A|B) = P(A and B)/P(B)? Why isn't this answer divided by (5/6)^10? Any help with my intuition? Thanks!
Edit: Sorry, left out some information. It is rolled ten times.
|
The most straight forward way to interpret that none are fives is simply consider the possible outcomes reduced. That is, you know each roll was one of $\{1,2,3,4,6\}$. So knowing this, the probability that exactly four were $6$ is given as
$$ {10 \choose 4} \left( \frac{1}{5} \right) ^4 \left( \frac{4}{5} \right) ^6 $$
If you'd like to use conditional probability, you can write
$$ P(A|B) = \frac{P(AB)}{P(B)} $$
where $P(B) = \left( \frac{5}{6} \right)^{10}$ and $P(AB) = \left( \frac{5}{6} \right)^{10} {10 \choose 4} \left( \frac{1}{5} \right) ^4 \left( \frac{4}{5} \right) ^6 $ which will cancel the factor of $\left( \frac{5}{6} \right)^{10}$ out to give you the answer above :)
|
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|
Integral $\int_0^1 \left(\arctan x \right)^2\,dx$ Evaluate
$$\int_0^1 \left(\arctan x \right)^2\,dx$$
The answer should be
$${\pi^2\over16} + \frac{\pi\ln(2)}{4} -C$$
where $C$ is Catalan's constant.
How do I proceed?
I tried doing integration by parts twice and got stuck at $$\int_0^1{\frac{\log\left(\frac1{x}+x\right)}{1+x^2}}\,dx$$
|
To get rid of the inverse, we let $y=\arctan x$ and transform the integral into
$$
\begin{aligned}
I &=\int_{0}^{\frac{\pi}{4}} y^{2} \sec ^{2} y d y \\
&=\int_{0}^{\frac{\pi}{4}} y^{2} d(\tan y) \\
& \stackrel{IBP}{=} \left[y^{2} \tan y\right]_{0}^{\frac{\pi}{4}}-2 \int_{0}^{\frac{\pi}{4}} y \tan y d y \\
&=\frac{\pi^{2}}{16}+2 \int_{0}^{\frac{\pi}{4}} y d(\ln (\cos y)) \\
& \stackrel{IBP}{=}\frac{\pi^{2}}{16}+2[y \ln (\cos y)]_{0}^{\frac{\pi}{4}}-2 \int_{0}^{\frac{\pi}{4}} \ln (\cos y) d y \\
&=\frac{\pi^{2}}{16}+2\left[\frac{\pi}{4} \ln \left(\frac{1}{\sqrt{2}}\right)\right]-2\left(\frac{G}{2}-\frac{\pi}{4} \ln 2\right) \\
&=\frac{\pi^{2}}{16}+\frac{\pi}{4} \ln 2-G
\end{aligned}
$$
where the last integral comes from my post.
|
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|
Solve for $x$ in the given determinant. Solve for $x$.
$$
\begin{vmatrix}
x^2-a^2&x^2-b^2&x^2-c^2\\
(x-a)^3&(x-b)^3&(x-c)^3\\
(x+a)^3&(x+b)^3&(x+c)^3\\
\end{vmatrix}=0.
$$
I could factorise each term, but could not find three factors in a single row or column. Atmost, there were $2$. I cannot think of any manipulations of the rows or columns. Please help.
|
I used brute force to solve this.
When we calculate:
$$
\begin{vmatrix}
x^2-a^2&x^2-b^2&x^2-c^2\\
(x-a)^3&(x-b)^3&(x-c)^3\\
(x+a)^3&(x+b)^3&(x+c)^3\\
\end{vmatrix} = 0
$$
We get:
$$-8 a^3 b^2 x^3+8 a^3 c^2 x^3+8 a^2 b^3 x^3+24 a^2 b x^5-8 a^2 c^3 x^3-24 a^2 c x^5-24 a b^2 x^5+24 a c^2 x^5-8 b^3 c^2 x^3+8 b^2 c^3 x^3+24 b^2 c x^5-24 b c^2 x^5 = 0$$
We see that we can divide by $-8 x^3$ and eliminate a triple root of $x = 0$, which reduces to:
$$(a-b) (a-c) (b-c) \left(-3x^2 + a (b+c)+b c\right) = 0$$
Using the quadratic, we find the other two roots as:
$$x=\pm ~\dfrac{\sqrt{a b+a c+b c}}{\sqrt{3}}$$
|
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|
$x$-coordinates of points where tangent line is horizontal or vertical. Using implicit differentiation For the implicit equation $x^3 + y^3 - xy^2 = 8$
Determine the exact x-coordinates of all points where the tangent line is horizontal or vertical
I figured out $\dfrac{dy}{dx} =\dfrac{y^2 - 3x^2}{ 3y^2 - 2xy}$
I know for it to be horizontal $y^2 - 3x^2 =0$ (and the denominator does not $= 0$)
For it to be vertical $3y^2 - 2xy = 0$ (and the numerator does not $= 0$)
But I have no clue how to solve for that. Any advice?
|
As OP already found, the expression for the slope of a tangent line to a point on the curve in question is
$$ y \ ' \ = \ \frac{y^2 \ - \ 3x^2}{3y^2 \ - \ 2xy} \ = \ \frac{y^2 \ - \ 3x^2}{y \ (3y \ - \ 2x)} \ \ . $$
Something useful to check for immediately is whether this rational function can be "indeterminate", that is, whether both numerator and denominator can be zero at the same time for a point on the curve. There are two cases to consider, due to the two factors in the denominator:
I -- $ \ y^2 \ = \ 3x^2 \ \ , \ \ y \ = \ 0 \ \ \Rightarrow \ \ x \ = \ 0 \ $ ; however, the point $ \ ( 0 , 0 ) \ $ does not lie on this curve
II -- $ \ y^2 \ = \ 3x^2 \ \ , \ \ 3y \ = \ 2x \ \ \Rightarrow \ \ \left( \frac{2}{3}x \right)^2 \ = \ 3 x^2 \ \ \Rightarrow \ \ x \ = \ 0 \ \ , \ \ y \ = \ 0 \ $ ; again, this is not a solution to the curve equation.
So the issue of "slope indeterminancy" does not occur for this curve, as it does for one of its "cousins", the folium of Descartes (discussed, for instance, here).
We can then locate the points at which horizontal tangents occur by inserting the relation $ \ y^2 \ = \ 3x^2 \ \Rightarrow \ y \ = \ \pm \ 3^{1/2}x \ $ into the equation for the curve (as mentioned by Graham Kemp), thus:
$$ x^3 \ + \ y^3 \ - \ xy^2 \ = \ 8 \ \ \Rightarrow \ \ x^3 \ + \ \left( \pm \ 3^{1/2}x \right)^3 \ - \ x \ \left( \pm \ 3^{1/2}x \right)^2 \ = \ 8 $$
$$ \Rightarrow \ \ \left( 1 \ \pm \ 3^{3/2} \ - \ 3 \right) \ x^3 \ = \ 8 $$
$$ \Rightarrow \ \ x \ = \ \left[ \frac{8}{-2 \ \pm \ 3 \sqrt{3}} \right]^{1/3} = \ \frac{2}{( -2 \ \pm \ 3 \sqrt{3} \ )^{1/3}} \ \approx \ -1.036 \ , \ 1.358 $$
$$ y \ = \ \pm \ \sqrt{3} \ \cdot \ x \ \rightarrow \ 1.794 \ , \ 2.352 \ \ . $$
[Upon insertion into the original curve equation, we see that the negative values for $ \ y \ $ are not solutions.] Thus, the horizontal tangent lines are found at
$$ \left( \frac{2}{( -2 \ - \ 3 \sqrt{3} \ )^{1/3}} \ , \ \frac{2 \ \sqrt{3}}{( -2 \ - \ 3 \sqrt{3} \ )^{1/3}} \right) \ \ , \ \ \left( \frac{2}{( -2 \ + \ 3 \sqrt{3} \ )^{1/3}} \ , \ \frac{2 \ \sqrt{3}}{( -2 \ + \ 3 \sqrt{3} \ )^{1/3}} \right) \ \ . $$
Vertical tangents occur at the points for which one of the factors in the denominator of the expression for $ \ y \ ' \ $ equals zero:
$$ y \ = \ 0 \ \ \Rightarrow \ \ x^3 \ + \ 0^3 \ - \ x \cdot 0^2 \ = \ 8 \ \ \Rightarrow \ \ x \ = \ 2 \ \ ; $$
$$ y \ = \ \frac{2}{3}x \ \ \Rightarrow \ \ x^3 \ + \ \left( \frac{2}{3}x \right)^3 \ - \ x \cdot \left( \frac{2}{3}x \right)^2 \ = \ 8 $$
$$ \Rightarrow \ \ \left( 1 \ + \ \frac{8}{27} \ - \ \frac{4}{9} \right) \ x^3 \ = \ \left( \frac{27 \ + \ 8 \ - 12}{27} \right) x^3 \ = \ 8 $$
$$ \Rightarrow \ \ x \ = \ \left( \frac{ 8 \ \cdot \ 27}{23} \right)^{1/3} \ = \ \frac{6}{23^{1/3}} \ \approx \ 2.110 \ \ \Rightarrow \ \ y \ = \ \frac{2}{3} \cdot \ \frac{6}{23^{1/3}} \ = \ \frac{4}{23^{1/3}} \ \approx \ 1.407 \ \ . $$
So the points where vertical tangents are found are
$$ ( \ 2, 0 \ ) \ \ \text{and} \ \ \left( \frac{6}{23^{1/3}} , \frac{4}{23^{1/3}} \right) \ \ . $$
A graph of the curve is presented below.
horizontal tangents are marked in green, vertical tangents, in red
|
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|
evaluation of $\int \cos (2x)\cdot \ln \left(\frac{\cos x+\sin x}{\cos x-\sin x}\right)dx$
Compute the indefinite integral
$$
\int \cos (2x)\cdot \ln \left(\frac{\cos x+\sin x}{\cos x-\sin x}\right)\,dx
$$
My Attempt:
First, convert
$$
\frac{\cos x+\sin x}{\cos x-\sin x} = \frac{1+\tan x}{1-\tan x} = \tan \left(\frac{\pi}{4}+x\right)
$$
This changes the integral to
$$
\int \cos (2x)\cdot \ln \left(\tan \left(\frac{\pi}{4}+x\right)\right)\,dx
$$
Now let $t=\left(\frac{\pi}{4}+x\right)$ such that $dx = dt$. Then the integral with changed variables becomes
$$
\begin{align}
\int \cos \left(2t-\frac{\pi}{2}\right)\cdot \ln (\tan t)dt &= \int \sin (2t)\cdot \ln (\tan t)dt\\
&= -\ln(\tan t)\cdot \frac{\cos (2t)}{2}+\frac{1}{2}\int \frac{\sec^2(t)}{\tan t}\cdot \cos (2t)\\
&= -\ln(\tan t)\cdot \frac{\cos (2t)}{2}+\frac{1}{2}\int \cot (2t)dt\\
&= -\ln(\tan t)\cdot \frac{\cos (2t)}{2}+\frac{1}{2}\ln \left|\sin (2t)\right|
\end{align}
$$
where $t=\displaystyle \left(\frac{\pi}{4}+x\right)$.
Is this solution correct? Is there another method for finding the solution?
|
Let
\begin{equation*}
I=\int \cos 2x\cdot\ln \left|\frac{\cos x+\sin x}{\cos x-\sin x}\right|\,dx.
\end{equation*}
Using the following identity
\begin{equation*}
\cos 2x=2\cos ^{2}x-1
\end{equation*}
and the substitution
\begin{equation*}
u=\cos x,
\end{equation*}
we get
\begin{equation*}
I=\int \frac{1-2u^{2}}{\sqrt{1-u^{2}}}\cdot\ln \left|\frac{u+\sqrt{1-u^{2}}}{u-\sqrt{1-u^{2}}}\right|\,du.
\end{equation*}
$I$ is integrable by parts, differentiating the factor $\ln \left|\frac{u+\sqrt{1-u^{2}}}{u-\sqrt{1-u^{2}}}\right|$ and integrating the factor $\frac{1-2u^{2}}{\sqrt{1-u^{2}}}$. After simplifying, we obtain
\begin{eqnarray*}
I &=&u\sqrt{1-u^{2}}\cdot\ln \left|\frac{u+\sqrt{1-u^{2}}}{u-\sqrt{1-u^{2}}}\right|+2\int
\frac{u}{2u^{2}-1}du \\[2ex]
&=&u\sqrt{1-u^{2}}\cdot\ln \left|\frac{u+\sqrt{1-u^{2}}}{u-\sqrt{1-u^{2}}}\right|+\frac{1}{2}
\ln \left| 2u^{2}-1\right| +C \\[2ex]
&=&\left( \cos x\cdot\sin x\right)\cdot \ln \left|\frac{\cos x+\sin x}{\cos x-\sin x}\right|+\frac{1
}{2}\ln \left| 2\cos ^{2}x-1\right| +C\\[2ex]
&=&\frac{\sin 2x
}{2} \ln \left|\frac{\cos x+\sin x}{\cos x-\sin x}\right|+\frac{\ln \left| \cos 2x\right|
}{2} +C.
\end{eqnarray*}
|
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|
Inequality $\sum_{cyc}\frac{(a+b)^{3}}{\sqrt[3]{2(a+b)(a^{2}+b^{2})}}\ge{12}$ when $ab+bc+ca=6$ Let $a$, $b$ and $c$ be non-negative reals such that $ab+bc+ca=6$.
Prove that:
$$\frac{(a+b)^{3}}{\sqrt[3]{2(a+b)(a^{2}+b^{2})}}+\frac{(b+c)^{3}}{\sqrt[3]{2(b+c)(b^{2}+c^{2})}}+\frac{(c+a)^{3}}{\sqrt[3]{2(c+a)(c^{2}+a^{2})}}\ge{12}$$
|
A simple and an elegant solution:
first,
$$\sqrt{\frac{a^2+b^2}{2}} \leq \frac{a+b}{2}$$ or $$a^2+b^2 \leq 2 \cdot \frac{(a+b)^2}{4}=\frac{(a+b)^2}{2}$$
So $$2(a+b)(a^2+b^2) \leq 2(a+b)\frac{(a+b)^2}{2}=(a+b)^3$$
now: $$\frac{1}{\sqrt[3]{2(a+b)(a^2+b^2)}} \geq \frac{1}{\sqrt[3]{(a+b)^3}}=\frac{1}{a+b}$$
Our inequality becomes:
$$\frac{(a+b)^3}{(a+b)}+\frac{(b+c)^3}{(b+c)}+\frac{(c+a)^3}{(c+a)} =(a+b)^2+(b+c)^2+(c+a)^2=$$
$$2(a^2+b^2+c^2)+2(ab+bc+ca)=2(a^2+b^2+c^2)+12 \geq 12.$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/719523",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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|
Limit question involving L'Hospitals rule I need help in solving the below question using L'Hospitals rule:
$$ \lim_{x \rightarrow 0} \left( \frac{1}{x^2}-\cot^2 x \right) $$
I'm getting infinity as the final answer. Thanks in advance.
|
$$
\lim_{x \rightarrow 0} \left( \frac{1}{x^2}-\cot^2 x \right) \\
= \lim_{x \rightarrow 0} \left( \frac{1}{x^2}-\frac{\cos^2 x}{\sin^2 x} \right) \\
= \lim_{x \rightarrow 0} \left( \frac{\sin^2 x}{x^2 \sin^2 x}-\frac{x^2 \cos^2 x}{x^2 \sin^2 x} \right) \\
= \lim_{x \rightarrow 0} \left( \frac{\sin^2 x - x^2 \cos x}{x^2 \sin^2 x} \right) \\
$$
Now you have a fraction where both numerator and denominator approach $0$ as $x \rightarrow 0$, so you can invoke L'Hospital's rule by taking the derivative of both top and bottom.
|
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|
Evaluate the integral. $\int sin^3 (5x) dx$ The bounds are from $(\pi/5)$ to $0$. I know we use a pythagorean identity for this. So $u=\sin^2 (x)$ and $du = \sin (2x) dx$. But I'm helping trouble solving this problem.
|
\begin{align*}
\int \sin^3(5x) \, dx &= \int \sin^2(5x) \cdot \sin(5x) \, dx \\
&= \int \left( 1-\cos^2(5x) \right)\sin(5x) \, dx \\
&= \int \left( \sin(5x)-\cos^2(5x)\sin(5x) \right) \, dx \\
&= -\frac{\cos(5x)}{5}-\int \cos^2(5x)\sin(5x) \, dx \\
\end{align*}
For the second integral, let $u=\cos(5x)$.
\begin{align*}
u &=\cos(5x) \\
\Rightarrow du &=-5\sin(5x) \, dx \\
\Rightarrow -\frac{1}{5} du &= \sin(5x) \, dx.
\end{align*}
Hence,
\begin{align*}
-\int \cos^2(5x)\sin(5x) \, dx &=\frac{1}{5}\int u^2 \, du \\
&=\frac{u^3}{15}+c \\
&=\frac{\cos^3(5x)}{15}+c.
\end{align*}
Putting it all together,
$$\int \sin^3(5x) \, dx =-\frac{\cos(5x)}{5}+\frac{\cos^3(5x)}{15}+c.$$
I leave you to evaluate your limits.
|
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|
Solving the functional equation $f\left(x^2+f(y)\right)=f(x)^2+y^4+2f(xy)$ Problem: find all functions $f:\mathbb{R}\to\mathbb{R}$ such that
$$f\left(x^2+f(y)\right)=f(x)^2+y^4+2f(xy),\ \ \ \forall x,y\in\mathbb{R}\text.$$
|
It can be shown that the only function $ f : \mathbb R \to \mathbb R $ satisfying
$$ f \left( x ^ 2 + f ( y ) \right) = f ( x ) ^ 2 + y ^ 4 + 2 f ( x y ) \tag 0 \label 0 $$
for all $ x , y \in \mathbb R $ is $ f ( x ) = x ^ 2 $. It's straightforward to check that this is indeed a solution. We prove that it's the only one.
General Properties of $ f $:
Let $ a = f ( 0 ) $ and $ b = f ( 1 ) $. By letting $ y = 0 $ and $ y = 1 $ in \eqref{0}, we respectively get the following:
$$ f \left( x ^ 2 + a \right) = f ( x ) ^ 2 + 2 a \text ; \tag 1 \label 1 $$
$$ f \left( x ^ 2 + b \right) = f ( x ) ^ 2 + 2 f ( x ) + 1 \text . \tag 2 \label 2 $$
Substituting $ - x $ for $ x $ in \eqref{1} and comparing with \eqref{1} itself, we have $ f ( - x ) ^ 2 = f ( x ) ^ 2 $. Then, substituting $ - x $ for $ x $ in \eqref{2} and comparing with \eqref{2} itself we get $ f ( - x ) = f ( x ) $; i.e. $ f $ is an even function. Letting $ x = 0 $ in \eqref{0} we have
$$ f \big( f ( y ) \big) = y ^ 4 + a ^ 2 + 2 a \text . \tag 3 \label 3 $$
Showing $ f ( 0 ) = 0 $ and $ f ( 1 ) = 1 $:
We can use \eqref{1} and \eqref{3} together to get
$$ f \left( x ^ 4 + 2 a x ^ 2 + a ^ 2 + a \right) = f \left( x ^ 2 + a \right) ^ 2 + 2 a \\
= f ( x ) ^ 4 + 4 a f ( x ) ^ 2 + 4 a ^ 2 + 2 a = f \Big( f \big( f ( x ) \big) \Big) + 4 a f ( x ) ^ 2 + 3 a ^ 2 \\
= f \left( x ^ 4 + a ^ 2 + 2 a \right) + 4 a f ( x ) ^ 2 + 3 a ^ 2 \text , $$
which by letting $ x = \sqrt { \frac 1 2 } $ yields $ 4 a f \left( \sqrt { \frac 1 2 } \right) ^ 2 + 3 a ^ 2 = 0 $, and that implies $ a \le 0 $. We can then use \eqref{1} and the fact that $ f $ is even to get
$$ f \left( \sqrt { - \frac a 2 } \right) ^ 2 + 2 a = f \left( \frac a 2 \right) = f \left( - \frac a 2 \right) = f \left( \sqrt { - \frac { 3 a } 2 } \right) ^ 2 + 2 a \text . $$
On the other hand, substituting $ f ( x ) $ for $ x $ in \eqref{2} we can use \eqref{3} and see that
$$ f \left( f ( x ) ^ 2 + b \right) = \Big( f \big( f ( x ) \big) + 1 \Big) ^ 2 = x ^ 8 + 2 ( a + 1 ) ^ 2 x ^ 4 + ( a + 1 ) ^ 4 \text . $$
As we know that $ f \left( \sqrt { - \frac a 2 } \right) ^ 2 = f \left( \sqrt { - \frac { 3 a } 2 } \right) ^ 2 $, we can conclude that
$$ \left( \sqrt { - \frac a 2 } \right) ^ 8 + 2 ( a + 1 ) ^ 2 \left( \sqrt { - \frac a 2 } \right) ^ 4 = \left( \sqrt { - \frac { 3 a } 2 } \right) ^ 8 + 2 ( a + 1 ) ^ 2 \left( \sqrt { - \frac { 3 a } 2 } \right) ^ 4 \text , $$
which can be simplified to $ a ^ 2 \left( 5 a ^ 2 + 4 ( a + 1 ) ^ 2 \right) = 0 $, and thus $ a = 0 $. Now, putting $ x = 1 $ in \eqref{1} we have $ b ^ 2 = b $. But if $ b = 0 $, putting $ x = 0 $ in \eqref{2} leads to a contradiction. Hence $ b $ can't be equal to $ 0 $, and we must have $ b = 1 $.
Proving $ f ( x ) = x ^ 2 $:
Using \eqref{0}, \eqref{1}, \eqref{3} and $ a = 0 $, we have
$$ f \left( x ^ 4 + y ^ 4 \right) = f \Big( x ^ 4 + f \big( f ( y ) \big) \Big) = f \left( x ^ 2 \right) ^ 2 + f ( y ) ^ 4 + 2 f \big( x ^ 2 f ( y ) \big) \\
= f ( x ) ^ 4 + f ( y ) ^ 4 + 2 f \big( x ^ 2 f ( y ) \big) \text , $$
which by symmetry yields
$$ f \big( x ^ 2 f ( y ) \big) = f \big( y ^ 2 f ( x ) \big) \text . \tag 4 \label 4 $$
Now, putting $ y = 1 $ in \eqref{4} and using \eqref{3}, $ a = 0 $ and $ b = 1 $ we get $ f \left( x ^ 2 \right) = f \big( f ( x ) \big) = x ^ 4 $, and thus $ f ( x ) = x ^ 2 $ for all $ x \ge 0 $. As $ f $ is even, we conclude that $ f ( x ) = x ^ 2 $ for all $ x \in \mathbb R $.
|
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"language": "en",
"url": "https://math.stackexchange.com/questions/722789",
"timestamp": "2023-03-29T00:00:00",
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|
standard Taylor series using substitution Find Taylor series using substitution about $0$ for $f(x)=\frac{125}{(5+4x)^3}$ by writing $\frac{125}{(5+4x)^3}=\frac{1}{(1+\frac{4}{5}x)^3}$?
Determine a range of validity for this series.
|
As Claude did,
$$\frac{1}{1+y}=1-y+y^2-y^3+y^4-y^5+...+(-1)^ny^n...$$ and take second derivative of this expension to get,
$$ \frac{2}{(1+y)^3}=(\frac{1}{1+y})^{''}=2-6y+12y^2-20y^3...n(n-1)(-1)^ny^{n-2}...$$
$$\frac{1}{(1+y)^3}=1-3y+6y^2-10y^3...{n\choose2}(-1)^{n+1}y^{n-2}... $$
Then replace $y$ by $4x\over5$ then you will get your series.
Since above expension convergent for $|y|<1$ your expension is convergent for $|{4x\over5}|<1$
|
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"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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|
Matrix Exponential using the Cayley-Hamilton theorem For the matrix $$P=\left( \begin{matrix} 0&1&0 \\ 0&0&1 \\ 0&1&0 \end{matrix} \right)$$ how do you find $e^{Pt}$ using the Cayley-Hamilton theorem?
I have found it by diagonalising $P$, but the question states to use the Cayley-Hamilton theorem.
|
If you calculate the characteristic polynomial $\chi_A(x)=x^3-x$, you know from Cayley-Hamilton theorem that $P^3-P=0$, i.e. $P^3=P$.
This implies that $P=P^3=P^5=\dots$ and $P^2=P^4=P^6=\dots$. Hence
$$
\begin{align}
e^{Pt}
&= I+Pt + \frac{P^2t^2}2 + \frac{P^3t^3}{3!} + \dots = \\
&= I + P\left(t+\frac{t^3}{3!}+\frac{t^5}{5!}+\dots\right) + P^2\left(\frac{t^2}2+\frac{t^4}{4!}+\frac{t^6}{6!}+\dots\right) = \\
&= I+ P \frac{e^t-e^{-t}}2 + P^2 \left(\frac{e^t+e^{-t}}2-1\right) = \\
&= I+ P\sinh t + P^2(\cosh t-1)
\end{align}
$$
Using
$P=\begin{pmatrix}0&1&0\\0&0&1\\0&1&0\end{pmatrix}$
and
$P^2=\begin{pmatrix}0&0&1\\0&1&0\\0&0&1\end{pmatrix}$
I get
$$e^{Pt}=
\begin{pmatrix}
1&\sinh t&\cosh t-1\\
0&\cosh t&\sinh t\\
0&\sinh t&\cosh t
\end{pmatrix}.$$
If I did not miss something, this seems to be the same result as WolframAlpha returns. (Up to some algebraic manipulation.)
|
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|
How to prove that the angle between two sides of that triangle is less than 60 degree? The product of two sides of triangle is equal to 8*(R*r) where R is circumradius of this triangle, and r is inradius of this triangle.
How to prove that the angle between two sides of that triangle is less than 60 degree?
|
Suppose $ab = 8Rr$. We have $abc = 4RA$ where $A$ is the area of the triangle, hence $2rc = A$. The area of the triangle is also given by $rs$ where $2s=a+b+c$. It follows that $2rc = rs$ thus $4c=a+b+c$ or $3c=a+b$. It follows that $$ \cos C = \frac{a^2+b^2-c^2}{2ab} = \frac{9a^2+9b^2-(a+b)^2}{18ab} = \frac{8a^2+8b^2-2ab}{18ab} \geq \frac{14}{18}, $$
(using the fact that $a^2+b^2 \geq 2ab$)
implying that $C \leq \arccos(\frac{14}{18}) = 38.9^\circ$.
|
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|
Volume within the sphere Find the volume of the solid that lies within the sphere ,
above the xy plane, and outside the cone
My problem is finding the integral function and the limits
|
First, here is a sketch showing that the cone cuts out washers at each value of $z$. The inner radius is given by the cone and the outer radius by the sphere. In cylindrical coordinates, the cone is given by $z = 7r$ and the sphere by $r^2 + z^2 = 1 \rightarrow r^2 = 1 - z^2$. The $z$ coordinate goes from $z = 0$ to whatever value of $z$ makes the radius of the cone equal to the radius of the sphere (at that $z$ value):
$$
\text{cone}: r = \frac{z}{7} \text{ plug into sphere equation}\\
\left(\frac{z}{7}\right)^2 + z^2 = 1 \rightarrow z^2 = \frac{1}{1 + \frac{1}{49}} = \frac{49}{50} \\
z_{top} = \frac{7}{\sqrt{50}}
$$
Now you just need to find the volume of each differential washer:
$$
dV = Adh = \pi(R^2 - r^2)dz = \pi\left((1 - z^2) - \left(\frac{z}{7}\right)^2\right)dz
$$
The outer radius, $R$, is from the sphere and the inner radius, $r$, is from the cone. So finally, you get the volume:
$$
V = \int dV = \pi\int\limits_0^{\frac{7}{\sqrt{50}}}\left(1 - \frac{50}{49}z^2\right)dz = \pi\left.\left(z - \frac{50}{147}z^3\right)\right|_0^{\frac{7}{\sqrt{50}}} \\
V = \pi \left(\frac{7}{\sqrt{50}} - \frac{50}{3\cdot49}\cdot\frac{7^3}{50\sqrt{50}}\right) \\
V = \pi\left(\frac{7}{\sqrt{50}} - \frac{7}{3\sqrt{50}}\right) = \frac{7\pi}{\sqrt{50}}\cdot \frac{2}{3} = \frac{14\pi}{3\sqrt{50}}
$$
edit
If you absolutely must, the "full" integral, in cylindrical coordinates, would be something like this:
$$
V = \int\limits_0^\frac{7}{\sqrt{50}}dz\int\limits_{\frac{z}{7}}^{\sqrt{1 - z^2}}dr\int\limits_0^{2\pi} rd\phi
$$
...but it's easier to visualize it in this case and just use high school geometry to create a single integral.
|
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|
How find this value $\frac{S_{\Delta ABC}}{\overrightarrow{OA}\cdot\overrightarrow{OB}}$
In the plane have $4$ point $O,A,B,C$ such
$$\overrightarrow{OA}+\overrightarrow{OB}+\overrightarrow{OC}=\overrightarrow{0},2|\overrightarrow{OA}|=2|\overrightarrow{OB}|+\overrightarrow{OC}$$,and let
$$\theta=<\overrightarrow{AB},\overrightarrow{OC}>,\tan{\theta}=\dfrac{3}{2}$$
Find the value $$\dfrac{S_{\Delta ABC}}{\overrightarrow{OA}\cdot\overrightarrow{OB}}$$
My try: since
$$\overrightarrow{OA}+\overrightarrow{OB}+\overrightarrow{OC}=\overrightarrow{0}$$
then $O$ is The center of gravity with $\Delta ABC$.
so
$$2|\overrightarrow{OA}|=2|\overrightarrow{OB}|+\overrightarrow{OC}\Longrightarrow \dfrac{4}{3}|AD|=\dfrac{4}{3}|BE|+\dfrac{2}{3}|CF|$$
$$\Longrightarrow |AD|=|BE|+2|CF|$$
then I can't.Thank you
|
The answer as given by
orion
is repeated here for convenience.
This answer is certainly incomplete, but let it be assumed that it is correct so far.
$$
c^2+(a-b)^2-2c(a-b)\cos\theta=a^2 \\
c^2+(a-b)^2+2c(a-b)\cos\theta=b^2 \\
(2c)^2=a^2+b^2-2ab\cos\alpha
$$
Quote: "but $a,b,c$ can be scaled without affecting the outcome".
Hence, without loss of generality we can put (WLOG equation):
$$
a^2+b^2+c^2 = 1
$$
Orion's equations can be simplified a great deal.
Subtracting the first two and using $\;\tan{\theta}=3/2$ :
$$
-4c(a-b)\cos\theta=a^2-b^2 \quad \Longrightarrow \quad -4c\cos\theta=a+b
\quad \Longrightarrow \quad a+b=-4c\frac{2}{\sqrt{13}}
$$
Adding the first two equations:
$$
2c^2+2(a-b)^2=a^2+b^2 \quad \Longrightarrow \quad c^2+(a^2+b^2+c^2)-4ab=0
\quad \Longrightarrow \quad c^2-4ab=-1
$$
And orion's last equation:
$$
(2c)^2+c^2=(a^2+b^2+c^2)-2ab\,\cos\alpha \quad \Longrightarrow \quad 5c^2+2\cos\alpha \,ab=1
$$
Hence we can solve $c^2$ and $ab$ from:
$$
\left[ \begin{array}{cc} 1 & -4 \\ 5 & 2\cos\alpha \end{array} \right]
\left[ \begin{array}{c} c^2 \\ ab \end{array} \right] =
\left[ \begin{array}{c} -1 \\ 1 \end{array} \right] \quad \Longrightarrow \quad
\left[ \begin{array}{c} c^2 \\ ab \end{array} \right] =
\frac{1}{10+\cos\alpha} \left[ \begin{array}{c} 2-\cos\alpha \\ 3 \end{array} \right]
$$
We also have:
$$
a+b=-4c\frac{2}{\sqrt{13}} \qquad \Longrightarrow \qquad (a+b)^2 = \frac{64}{13}\frac{2-\cos\alpha}{10+\cos\alpha}
$$
Hence our WLOG equation can be rewritten as:
$$
a^2+b^2+c^2 = (a+b)^2 - 2ab + c^2 = 1 \qquad \Longrightarrow \\
\frac{64}{13}(2-\cos\alpha)-2\cdot3+(2-\cos\alpha)=10+\cos\alpha \quad \Longrightarrow \quad \cos\alpha = -\frac{3}{5}
$$
Now $(3,4,5)$ is a famous rectangular triangle. So what we finally get is an incredibly simple answer:
$$ \frac{S}{\vec{OA}\cdot\vec{OB}}=\frac32\tan \alpha \quad \wedge \quad
\tan\alpha = -\frac{4}{3} \qquad \Longrightarrow \qquad
\frac{S}{\vec{OA}\cdot\vec{OB}}= -2
$$
Where the minus sign must be due to the fact that the angle AOB is obtuse.
Bonus. Calculating $\;a,b,c\;$ too.
$$
c = \pm \sqrt{\frac{2-\cos\alpha}{10+\cos\alpha}} = \pm \sqrt{\frac{13}{47}} \\
a+b = -\frac{8c}{\sqrt{13}} > 0 \quad \Longrightarrow \quad \frac{a+b}{2} = \frac{4}{\sqrt{47}} \qquad ; \qquad ab = \frac{3}{10+\cos\alpha}=\frac{15}{47} \\ \Longrightarrow \qquad
\{a,b\} = \frac{a+b}{2} \pm \sqrt{\left(\frac{a+b}{2}\right)^2-ab} \; = \; \frac{4 \pm 1}{\sqrt{47}}
$$
Without the WLOG restriction (or $\;a^2+b^2+c^2=47\,t^2$ ) with $t$ an arbitrary positive real constant:
$$
c = \sqrt{13}\,t \quad ; \quad a = 5\,t \quad ; \quad b = 3\,t
$$
Where it is noted that $\,a\,$ and $\,b\,$ can be swapped.
|
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|
Find$f$ s.t. $f(1)=2$, $f(2)= 4$, $f(3)= 6$ and $f(4)= \pi$. Find a function where $f(1)=2$, $f(2)= 4$, $f(3)= 6$ and $f(4)= \pi$.
I got $\dfrac16(x-3)(x-2)(x-1)\pi$ as a start to get rid of $\pi$.
|
Your way of dealing with $\pi$ is very smart, but apparently you only took it because $\pi$ is irrational, and you are expecting to get the "nicer" values of $2,4,6$ in different ways. Here's something you can do, use seperate terms that are zero at $1,2,\pi$ and $6$ at $3$. Try it for yourself. Add all the terms. You already have the requisite insight.
$$\frac{(x-3)(x-2)(x-1)\pi}{6}+\frac{(x-\pi)(x-1)(x-2)6}{2(3-\pi)}+\frac{(x-\pi)(x-1)(x-3)4}{(2-\pi)\times-1}\\+\frac{(x-2)(x-3)(x-\pi)2}{(-1)(-2)(1-\pi)}$$
Simplyfying,
$$\frac{(x-3)(x-2)(x-1)\pi}{6}+\frac{3(x-\pi)(x-1)(x-2)}{(3-\pi)}+\frac{4(x-\pi)(x-1)(x-3)}{(\pi-2)}+\\\frac{(x-2)(x-3)(x-\pi)2}{2(1-\pi)}$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/731601",
"timestamp": "2023-03-29T00:00:00",
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|
Subsets and Splits
Fractions in Questions and Answers
The query retrieves a sample of questions and answers containing the LaTeX fraction symbol, which provides basic filtering of mathematical content but limited analytical insight.