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$\lim\limits_{x\to 0} \frac{\tan x - \sin x}{x^3}$? $$\lim_{x\to 0} \frac{\tan x - \sin x}{x^3}$$
Solution
\begin{align}\lim_{x\to 0} \frac{\tan x - \sin x}{x^3}&=\\&=\lim_{x\to 0} \frac{\tan x}{x^3} - \lim_{x\to 0} \frac{\sin x}{x^3}\\
&= \lim_{x\to 0}\frac{\tan x}{x}\lim_{x\to 0} \frac{1}{x^2} -\lim_{x\to 0} \frac{\sin x}{x}\lim_{x\to 0} \frac{1}{x^2}\\&= \lim_{x\to 0} \frac{1}{x^2} -\lim_{x\to 0} \frac{1}{x^2}\\
&= \lim_{x\to 0} \frac{1}{x^2} -\frac{1}{x^2}\\&=0 \end{align}
But the answer is $\dfrac{1}{2}$ by L'Hopital's Rule.
|
I don't know is there later mistakes or not, but I think there's a mistake at first equation. $ \lim\limits_{x \to 0}\big( f(x) - g(x)\big)$ is not always equal to $ \lim\limits_{x \to 0} f(x) - \lim\limits_{x \to 0} g(x)$.
|
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"language": "en",
"url": "https://math.stackexchange.com/questions/3008071",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 5,
"answer_id": 4
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|
Find the sum: $\sum_{n=2}^\infty \frac{1}{n^2-1}$
Evaluate : $$\sum_{n=2}^\infty \frac{1}{n^2-1}$$
I've tried to rewrite the questions as $$\sum _{n=2}^{\infty \:\:}\left(-\frac{1}{2\left(n+1\right)}+\frac{1}{2\left(n-1\right)}\right)$$ but still couldnt't get any answer as when I substitute numbers $(n)$ into the $Sn \left(-\frac{1}{6}+\frac{1}{2}-\frac{1}{8}+\frac{1}{4}-\frac{1}{10}\right)$ , $Sn$ just keeps going on and on. So how do I solve this problem and what does this series converge to?
|
As you correctly have: $$\frac{1}{n^2-1}=\frac{1}{(n-1)(n+1)}=\frac{1}{2(n-1)}-\frac{1}{2(n+1)}$$ Now, observe that $$\sum_{n=2}^{\infty}\frac{1}{n-1}=\sum_{n=1}^{\infty}\frac{1}{n} \qquad \text{and}\qquad\sum_{n=2}^{\infty}\frac{1}{n+1}=\sum_{n=3}^{\infty}\frac{1}{n}$$ Hence
$$\sum_{n=2}^{\infty}\frac{1}{n^2-1}=\frac12\sum_{n=2}^{\infty}\left(\frac{1}{n-1}-\frac{1}{n+1}\right)=\frac12\left(\sum_{n=1}^{\infty}\frac1n-\sum_{n=3}^{\infty}\frac1n\right)=\frac12\left(\frac11+\frac12\right)=\frac34$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3009920",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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|
roots of cubic equation complex
Given the cubic equation: $$x^3-2kx^2-4kx+k^2=0.$$
If one root of the equation is less than $1$, another root is in the interval $(1,4)$ and the third root is greater than $4$, then the value of $k$ lies in the interval $\left(a+\sqrt{b},b(a+\sqrt{6})\right)$ where $a,b\in \mathbb{N}$.
Find the value of $(a+b)^3+(ab+2)^2$.
My approach
Let the roots be $p,q,r$
So
$$p<1$$
$$1<q<4$$
$$r>4$$
$$p+q+r = 2k$$
$$pq+qr+rp = -4k$$
$$pqr = -k^2$$
|
Let $f(x)=x^3-2kx^2-4kx+k^2$. From the information given, $f$ has three distinct real roots. Also based on the leading term, we know $f \to \infty$ as $x \to \infty$ and $f \to -\infty$ as $x \to -\infty$.
For a root to be less than $1$, we should have $f(1) >0$. Likewise $f(4)<0$. This gives us
\begin{align*}
k^2-6k+1 & > 0\\
k^2-48k+64 & < 0
\end{align*}
The first inequality gives $k \in (-\infty, 3-2\sqrt{2}) \cup (3+2\sqrt{2}, \infty)$. The second inequality gives $k \in (24-16\sqrt{2}, \,\, 24+16\sqrt{2})$. The intersection of these intervals gives
$$k \in (3+2\sqrt{2}, \,\, 24+16\sqrt{2})=(3+\sqrt{8}, \,\, 8(3+\sqrt{8})).$$
Thus $a=3,b=8$.
Note: I think there is a possible typo in your question. It should be $k\in(a+\sqrt{b}, b(a+\color{red}{\sqrt{8}}))$
|
{
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|
How to prove that $(-18+\sqrt{325})^{\frac{1}{3}}+(-18-\sqrt{325})^{\frac{1}{3}} = 3$ How to prove that $\left(-18+\sqrt{325}\right)^{\frac{1}{3}}+\left(-18-\sqrt{325}\right)^{\frac{1}{3}} = 3$ in a direct way ?
I have found one indirect way to do so: Define $t=\left(-18+\sqrt{325}\right)^{\frac{1}{3}}+\left(-18-\sqrt{325}\right)^{\frac{1}{3}}$, and then observe $t$ is a root of $f(x) = x^3 + 3x - 36 = 0$. Also observe that $f(3) = 0$. Now $f(x)$ has only one real root, since $f'(x) = x^2+3$ has no real solution. So $t =3$, as desired.
But I couldn't find any direct/algebraic way of fiddling it for a while. I guess morever this is true: $\left(-\frac{t(t^2+3)}{2}+\sqrt{\left[\frac{t(t^2+3)}{2}\right]^2 + 1}\right)^{\frac{1}{3}}+\left(-\frac{t(t^2+3)}{2}-\sqrt{\left[\frac{t(t^2+3)}{2}\right]^2 + 1}\right)^{\frac{1}{3}} = t$, which can possibly be proved in the same indirect way but I don't know how to prove this by direct manipulation.
One way maybe is to write $z_+ = \left(-\frac{t(t^2+3)}{2}+\sqrt{\left[\frac{t(t^2+3)}{2}\right]^2 + 1}\right)^{\frac{1}{3}}$, $z_- = \left(-\frac{t(t^2+3)}{2}-\sqrt{\left[\frac{t(t^2+3)}{2}\right]^2 + 1}\right)^{\frac{1}{3}}$, and writing $t = z_+ + z_-$, observe $z_+z_- = -1$, so $(x-z_+)(x-z_-) = x^2 - tx -1$, and then with Fundamental Theorem of Symmetric Polynomials, maybe calculating $(z_+^3 - z_-^3)^2$ and $(z_+^3 + z_-^3)$ in terms of $t$ would help ?
|
It helps a lot that we know the answer in advance! Otherwise, some lucky guesswork (or knowledge of algebraic number theory) is required.
One can follow a quite similar line of reasoning to Rafael Bombelli (c.1526-1572). See Bombelli and the invention of complex numbers; or my answer to an earlier question here; or section 14.3, "Cubic Equations", of John Stillwell's book Mathematics and Its History (second edition 2002 - there is a third edition, which I don't have).
We know that if $a$ and $b$ are rational, and $(a + b\sqrt{13})^3 = c + d\sqrt{13}$, then $(a - b\sqrt{13})^3 = c - d\sqrt{13}$. So it is a natural idea - and a much less "radical" leap than Bombelli's (if you'll pardon the pun) - to look for rational $a, b$ such that $(a + b\sqrt{13})^3 = 18 + 5\sqrt{13}$. Then we will have $\sqrt[3]{18 + 5\sqrt{13}} + \sqrt[3]{18 - 5\sqrt{13}} = 2a$.
We know right away that $a = \frac{3}{2}$. (Honesty compels me to admit that I initially overlooked this obvious fact!) Equating coefficients of the irrational number $\sqrt{13}$ in the expanded version of either of the equations defining $a, b$, we have $a^3 + 39ab^2 = 18$, which simplifies to $\frac{9}{4} + 39b^2 = 12$. Taking $b > 0$, for definiteness, we have $b = \frac{1}{2}$. We now only
need check that $3a^2b + 13b^3 = 5$, i.e. $3a^2 + 13b^2 = 10$, to see that the
defining equations for $a, b$ are indeed satisfied.
|
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"url": "https://math.stackexchange.com/questions/3011230",
"timestamp": "2023-03-29T00:00:00",
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|
Limit of $\lim_{x \to 0} (\cot (2x)\cot (\frac{\pi }{2}-x))$ (No L'Hôpital)
$\lim_{x \to 0} (\cot (2x)\cot (\frac{\pi }{2}-x))$
I can't get to the end of this limit. Here is what I worked out:
\begin{align*}
& \lim_{x \to 0} \frac{\cos 2x}{\sin 2x}\cdot\frac{\cos(\frac{\pi }{2}-x )}{\sin(\frac{\pi }{2}-x )}
\lim_{x \to 0}\frac{\frac{\cos2x }{2x}}{\frac{\sin 2x}{2x}}\cdot \frac{\cos(\frac{\pi }{2}-x )}{\sin(\frac{\pi }{2}-x )}
= \lim_{x \to 0} \frac{\cos 2x}{2x} \cdot \frac{\cos(\frac{\pi }{2}-x )}{\sin(\frac{\pi }{2}-x )} \\
= & \lim_{x \to 0} \frac{{\cos^2 (x)}-{sin^2 (x)}}{2x}\cdot\frac{\cos(\frac{\pi }{2}-x )}{\sin(\frac{\pi }{2}-x )}
= \lim_{x \to 0} \left(\frac{\cos^2(x)}{2x}-\frac{sin^2 x}{2x}\right)\cdot \frac{\cos(\frac{\pi }{2}-x )}{\sin(\frac{\pi }{2}-x )} \\
= & \lim_{x \to 0} \left(\frac{1-\sin^2 x}{2x}-\frac{sin x}{2}\right)\cdot \frac{\cos(\frac{\pi }{2}-x )}{\sin(\frac{\pi }{2}-x )}
= \lim_{x \to 0} \left(\frac{1}{2x}-\frac{\sin^2 x}{2x}-\frac{sin x}{2}\right) \cdot \frac{\cos(\frac{\pi }{2}-x )}{\sin(\frac{\pi }{2}-x )} \\
= & \lim_{x \to 0} \left(\frac{1}{2x}-\frac{\sin x}{2}-\frac{sin x}{2}\right)\cdot \frac{\cos(\frac{\pi }{2}-x )}{\sin(\frac{\pi }{2}-x )}
= \lim_{x \to 0} \left(\frac{1}{2x}-2\frac{\sin x}{2}\right)\cdot \frac{\cos(\frac{\pi }{2}-x )}{\sin(\frac{\pi }{2}-x )} \\
= & \lim_{x \to 0} \left(\frac{1}{2x}-\sin x\right)\cdot \frac{\cos(\frac{\pi }{2}-x )}{\sin(\frac{\pi }{2}-x )}
\end{align*}
Here is where I can't seem to complete the limit, the 2x in the denominator is giving me a hard time and I don't know how to get rid of it. Any help would be appreciated. (In previous questions I got a really hard time because of my lack of context, I hope this one follows the rules of the site. I tried.)
|
HINT
We have that
$$\cot (2x)\cot\left(\frac{\pi }{2}-x\right)=\frac{\cos(2x)}{\sin(2x)}\frac{\sin x}{\cos x}$$
|
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"timestamp": "2023-03-29T00:00:00",
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|
Using a power series to show $\log(2) = \sum_{k = 1}^{\infty} \frac{1}{k \cdot 2^{k}}$ I want to show
$$\log(2) = \sum_{k = 1}^{\infty} \frac{1}{k \cdot 2^{k}}.$$
So, I start with the geometric series:
$$\frac{1}{1 - x} = \sum_{n = 0}^{\infty} x^{n} $$
Then take the integral to get
$$\log(1 - x) = -\sum_{n = 1}^{\infty} \frac{x^{n}}{n}.$$
Then plug in $x = -1$ to get
$$\log(2) = -\sum_{n = 1}^{\infty} \frac{(-1)^{n}}{n} = \sum_{n = 1}^{\infty} \frac{(-1)^{n + 1}}{n},$$
but this isn't equal to the original equality. Why not? How can I do the problem?
|
Since the series are convergent for $|x| < 1$ then the "trick" is to use $x=1/2$. This leads to
\begin{align}
\frac{1}{1 - x} &= \sum_{n = 0}^{\infty} x^{n} \\
\log(1 - x) &= -\sum_{n = 1}^{\infty} \frac{x^{n}}{n}
\end{align}
becoming
\begin{align}
2 &= \sum_{n=0}^{\infty} \frac{1}{2^{n}} \\
- \ln\left(1 - \frac{1}{2}\right) = \ln(2) &= \sum_{n=1}^{\infty} \frac{1}{2^{n} \, n}
\end{align}
|
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"timestamp": "2023-03-29T00:00:00",
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|
Proof using AM-GM inequality The questions has two parts:
Prove
(i) $ xy^{3} \leq \frac{1}{4}x^{4} + \frac{3}{4}y^{4} $
and
(ii) $ xy^{3} + x^{3}y \leq x^{4} + y^{4}$.
Now then, I went about putting both sides of $\sqrt{xy} \leq \frac{1}{2}(x+y)$
to the power of 4 and it left me with
$$-x^{3}y \leq \frac{1}{4}x^{4} + \frac{1}{4}y^{4} + xy^{3} + \frac{5}{2}x^{2}y^{2}. $$
Curiously squaring and multiplying $\sqrt{xy} \leq \frac{1}{2}(x+y)$ I've tried merging the results with my other inequality a few times to no avail - I just can't seem to get the signs right and nor can I seem to make the coefficients of the $x^{4}$ and $y^{4}$ different, as they are in (i). I feel like there's something I'm missing. Does anyone see a nice way about this problem?
|
(i) follows from Young's inequality for $p = 4$ and $q = \frac43$:
$$xy^3 \le |x||y|^3 \le \frac{|x|^p}{p} + \frac{\left(|y|^3\right)^q}q = \frac{|x|^4}4 + \frac{\left(|y|^3\right)^{4/3}}{4/3} = \frac{|x|^4+3|y|^4}4 = \frac{x^4+3y^4}4$$
(ii) follows from (i):
$$xy^3 + x^3y \le \frac{x^4+3y^4}4 + \frac{3x^4+y^4}4 = x^4+y^4$$
|
{
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"url": "https://math.stackexchange.com/questions/3020293",
"timestamp": "2023-03-29T00:00:00",
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|
If $\frac{x}{y^\frac{n-1}{n}}$ is constant, how do I prove $\frac{dx}{x}=\frac{n-1}{n}\frac{dy}{y}$? From:
$\frac{x}{y^\frac{n-1}{n}}=constant$
To:
$\frac{dx}{x}=\frac{n-1}{n}\frac{dy}{y}$
It's from a turbomachinery lecture. The text I'm studying says that binomial expansion is used, but I don't get it.
|
We want
$\dfrac{dy}{dx} = \dfrac{n}{n - 1} \dfrac{y}{x}; \tag 0$
from
$\dfrac{x}{y^{\frac{n - 1}{n}}} = \alpha = \text{constant}, \tag 1$
assuming of course $x$, $y$, and $n$ are all in range sufficient to legitimize our operations, we have:
$x = \alpha y^{\frac{n - 1}{n}}; \tag 2$
$x^n = \alpha^n y^{n - 1}; \tag 3$
$nx^{n - 1} = \alpha^n (n - 1)y^{n - 2} \dfrac{dy}{dx}; \tag 4$
$\alpha^n \dfrac{dy}{dx} = \dfrac{n}{n - 1} \dfrac{x^{n - 1}}{y^{n - 2}}; \tag 5$
from (3),
$\alpha^n = \dfrac{x^n}{y^{n - 1}}; \tag 6$
thus, dividing (5) by (6),
$\dfrac{dy}{dx} = \dfrac{n}{n - 1} \dfrac{x^{n - 1}}{y^{n - 2}} \dfrac{y^{n - 1}}{x^n} = \dfrac{n}{n - 1} \dfrac{y}{x}, \tag 7$
which we may write in differential form to obtain
$\dfrac{dx}{x} = \dfrac{n - 1}{n} \dfrac{dy}{y}; \tag 8$
as per request. $OE\Delta$
|
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"timestamp": "2023-03-29T00:00:00",
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|
Prove that $\int_0^{\frac{n\pi}{4}} \frac{1}{3\sin^4(x) + 3\cos^4(x) -1} dx = \frac{n\pi}{4} $
Prove that $$\int_0^\frac{n\pi}{4} \left(\frac{1}{3\sin^4(x) + 3\cos^4(x) -1}\right) dx = \frac{n \pi}{4} $$
is true for all integers $n$.
Here I've looked at the case where $n=1$ as $t=\tan(x)$ is injective in that domain. Then I could perhaps try to use some argument about periodicity and symmetry to extend it to all the integers.
$$
= \int_0^\frac{\pi}{4} \frac{\sec^4(x)}{3\tan^4(x)+3-\sec^4(x)} \text{d}x
= \frac12 \int_0^1 \frac{t^2+1}{t^4-t^2+1} \text{d}t
$$
Which looks a little tough but reminiscent of the well-known integral:
$$ \int_0^\infty \frac{1}{x^4 + 2x^2\cos(2a) + 1} \text{d}x = \frac12\int_0^\infty \frac{x^2+1}{x^4 + 2x^2\cos(2a) + 1} \text{d}x = \frac{\pi}{4\cos(a)} $$
Where we choose $a=\frac\pi3$. But I can't seem to adjust the bounds to get an answer. How would you evaluate this integral?
|
$$3(\sin^4x+\cos^4x)-1=3((\sin^2x+\cos^2x)^2-2\sin^2x\cos^2x)-1=2-\frac32\sin^22x=\frac{5+3\cos4x}4$$
and your integral is also
$$\int_0^{n\pi}\frac{dx}{5+3\cos x}=n\int_0^{\pi}\frac{dx}{5+3\cos x}$$ as the cosine is an even function.
Using the Weierstrass substitution, the last integral is shown to be $\dfrac\pi4$.
$$\int_0^\pi\frac{dx}{5+3\cos4x}=\int_0^\infty\frac{2\,dt}{(1+t^2)\left(5+3\dfrac{1-t^2}{1+t^2}\right)}=\int_0^\infty\frac{dt}{4+t^2}.$$
|
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|
$a(x-a)^2+b(x-b)^2=0$ has one solution; $a, b$ are not $0$. Prove $|a|=|b|$ $a(x-a)^2+b(x-b)^2=0$ has one solution; $a, b$ are not $0$. Prove $|a|=|b|$
simplifying the equation I got:
$$(a+b)x^2-2(a^2+b^2)x+(a^3+b^3)=0$$
solving for the Discriminant $D=0$, I got:
$$a(a^2b-2ab^2+b^3)=0$$
and since $a$ cannot equal $0$, I got:
$$a=b \implies |a|=|b|$$
In the same way I also got: $$b=a$$
But I don't need the absolute values of $a, b$ because I got $a=b$ (which means $|a|=|b|$ is not needed) and checking in the equation $a$ cannot equal $-b$ because then the equation will have infinitely many solutions instead of just one as specified in the task.
Now, because I need to prove that $|a|=|b|$ and not that $a=b$, I think that I am missing something. So, am I missing something or is my solution correct?
|
Oooh boy! Incorporating hamam_Abdallah's answer into your efforts:
Note: you want $(a+b)x^2-2(a^2+b^2)x+(a^3+b^3)=0$ to have one solution.
So you naturally did the quadratic equation to get the solutions are
$x= \frac {2(a^2 +b^2) \pm \sqrt{D}}{2(a+b)}$ and for that to have one unique solution you need $D= 0$.
That's fine and good up to the point that you assume $2(a+b) \ne 0$.
You forgot to take into account the possibility that $2(a+b) = 0$ or $a = -b$.
If $a = - b$ you get:
$(a+b)x^2-2(a^2+b^2)x+(a^3+b^3)=0\implies -2(a^2+b^2)x+(a^3+b^3)=0$ which is a linear equation and has exactly one solution (assuming $-2(a^2 +b^2) \ne 0$ which it doesn't if $a=-b \ne 0$).
So for $(a+b)x^2-2(a^2+b^2)x+(a^3+b^3)=0$ to have one solution EITHER:
1) $(a+b) = 0$ and the equation is a linear equation.
OR
2) $(a+b) \ne 0$ and $D = (-2(a^2 + b^2))^2 - 4(a+b)(a^3 + b^3) = 0$ and the equation is a quadratic with a double root.
If 1) then we get $a = -b$.
If 2) then we get $a = b$.
.....
(I didn't actually follow your calculations but mine got the same result:
$4(a^2 + b^2)^2 - 4(a+b)(a^3 + b^3) = 0$
$a^4 + 2a^2b^2 + b^4 = a^4 + ab^3 + a^3b + b^3$
$2a^2b^2 = ab^3 + a^3b$
$2ab = b^2 + a^2$
$(b -a)^2 = 0$
$b =a$.
)
===
FWIW.
If 1) $a = -b$ then solution is $x = -\frac {a^3 + b^3}{-2(a^2 + b^2)}=\frac {a^3 - a^3}{-2(a^2 + a^2)} = 0$.
And if 2)$a = b$ then the solution is $x = \frac {2(a^2 + b^2)\pm \sqrt{D}}{2(a+b)} = \frac {2a^2}{2a} = a$.
|
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"timestamp": "2023-03-29T00:00:00",
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|
Easier way to find eigenvalues of Matrices? I am trying to find eigenvalues for this matrix,
A =
$\begin{bmatrix}
3 & 2 & -3 \\
-3 & -4 & 9 \\
-1 & -2 & 5 \\
\end{bmatrix}$
I find the characteristic equation here:
$(\lambda I - A)
=
\begin{bmatrix}
\lambda - 3 & -2 & 3 \\
3 & \lambda + 4 & -9 \\
1 & 2 & \lambda - 5 \\
\end{bmatrix}$
The difficult part I am having is finding the determinant of the characteristic equation, mainly that it becomes insanely difficult for me to keep track of the factoring to get the eigenvalues...here I use rule of Sarrus to try to calculate the eigenvalues.
$\begin{vmatrix}
\lambda I - A \\
\end{vmatrix}
=
\begin{bmatrix}
\lambda - 3 & -2 & 3 & \lambda -3 & -2 \\
3 & \lambda + 4 & -9 & 3 & \lambda + 4 \\
1 & 2 & \lambda - 5 & 1 & 2 \\
\end{bmatrix}$
= $(1)(\lambda + 4)(3)$
$+ (2)(-9)(\lambda - 3)$
$+ (\lambda - 5)(3)(-2)$
$- (\lambda - 3)(\lambda + 4)(\lambda - 5)$
$- (-2)(-9)(1)$
$- (3)(3)(2)$
= $(3)(\lambda + 4) + (-18)(\lambda - 3) + (-6)(\lambda -5) - (\lambda - 3)(\lambda + 4)(\lambda - 5) - 18 - 18$
= $(3)(\lambda + 4) + (-18)(\lambda - 3) + (-6)(\lambda -5) - (\lambda - 3)(\lambda + 4)(\lambda - 5) - 36$
= $((\lambda + 4)(3) -(\lambda -3)(\lambda -5)) + (-18)(\lambda -3) + (-6)(\lambda - 5) - 36$
= $(\lambda - 3)((\lambda + 4)(3) -(\lambda - 5)) -18 + (-6)(\lambda - 5) - 36$
= $(\lambda - 5)(\lambda -3)((\lambda + 4)(3) - 1)) - 18 + (-6)(1) - 36$
= $(\lambda - 5)(\lambda -3)((\lambda + 4)(3) - 1)) - 60$
= $(\lambda - 5)(\lambda - 3)(3\lambda + 12 -1 - 60)$
= $(\lambda - 5)(\lambda - 3)(3\lambda - 49)$
I end up with 5, 3, and 16.3 as the eigenvalues (16.3 seems off). Obviously that was a ridiculous amount of simplification I had to do just to get eigenvalues and when I feel like I messed up (like here), it is pretty impossible to check my work in an effective manner. The amount of time it takes for me to calculate eigenvalues is unacceptable for my upcoming final exam. Do you guys have any tips or tricks that makes this process easier?
|
Here is a simple computation:
\begin{align}
\det(\lambda I - A)& = \begin{vmatrix}
\lambda - 3 & -2 & 3 \\
3 & \lambda + 4 & -9 \\
1 & 2 & \lambda - 5 \\
\end{vmatrix}=\begin{vmatrix}
\lambda - 2 & 0 & \lambda - 2 \\
3 & \lambda + 4 & -9 \\
1 & 2 & \lambda - 5 \\
\end{vmatrix}\\& =(\lambda - 2 )\begin{vmatrix}
1 & 0 & 1 \\
3 & \lambda + 4 & -9 \\
1 & 2 & \lambda - 5 \\
\end{vmatrix}=(\lambda - 2 )\biggl(\begin{vmatrix}
\lambda + 4 & -9 \\
2 & \lambda - 5 \\
\end{vmatrix}+\begin{vmatrix}
3 & \lambda + 4 \\
1 & 2 \\
\end{vmatrix}\biggl)\\
&=(\lambda - 2 )\Bigl(\bigl(
(\lambda + 4)(\lambda - 5)+18\bigr)
+ (6 -\lambda - 4) \Bigl)=(\lambda - 2 )(\lambda^2-2\lambda)\\
&=\lambda(\lambda - 2)^2.
\end{align}
|
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|
How to evaluate $\int\frac{1}{3+4x+4x^2}$?
How to evaluate
$$\int\frac{1}{3+4x+4x^2}\quad ?
$$
This is what I've done so far:
$$
\frac{1}{4} \int\frac{1}{x^2+x+\frac{3}{4}}
=\frac{1}{4} \int\frac{1}{(x+\frac{1}{2})^2 + \frac{1}{2}}
$$
$$
y = \frac{1}{a}\arctan\frac{u}{a},\quad \frac{dy}{du} = \frac{1}{a^2 + u^2}
$$
$$
a = \sqrt{\frac{1}{2}},\quad
u = (x+\frac{1}{2}),\quad \frac{du}{dx} = 1
$$
so
$$ \frac{dy}{dx} = \frac{dy}{du} \cdot 1
$$
I don't know how to proceed.
Could I also have some help with
$$
\int\frac{1}{\sqrt{-4x^2-4x+3}}\quad ?
$$
|
HINT:Complete the squares and use the results
$$\int\frac{1}{x^2 + a^2}dx= \frac{1}{a}\arctan \frac{x}{a}+C $$
and
$$\int\frac{1}{\sqrt{a^2-x^2}}dx=\arcsin \frac{x}{a}+C$$.
|
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|
Find $A$ given $A^2$ and $B$ and $(A^2)^{-1} = A^{-1}B$ This excersice took place in class I had today. The exercise was the following:
Let the regular matrix $A$,$B$:
$$A^2 = \left[ \begin{matrix} 2 &-2 & 2\\ -2 & 2 & 2 \\ 4&4&-4 \end{matrix} \right] \hspace{2cm} B = \left[ \begin{matrix} 0 &-1 & 2\\ 0 & 2 & 0 \\ 1&3&-1 \end{matrix} \right]$$
Knowing that $(A^2)^{-1} = A^{-1}B$, find A.
My Attempt
During the exam I tried the following:
$(A^2)^{-1} = A^{-1}B \Leftrightarrow A^2(A^2)^{-1} = A^2A^{-1}B \Leftrightarrow I_n = AB \Leftrightarrow A = B^{-1}$
And procedeed to find the inverse of B, which is:
$A = B^{-1} = \left[ \begin{matrix} \frac{1}{2} &\frac{-5}{4} & 1\\ 0 & \frac{1}{2} & 0 \\ \frac{1}{2}& \frac{1}{4}&0 \end{matrix} \right]$
But, as I tend to make silly mistakes, I figured out that it would be a better way to do find $A$ just by using matrix products. We had $I_n = AB \Leftrightarrow A = A^2B$. Using this method I got that
$A = A^2B = \left[ \begin{matrix} 2 & 0 & 2\\ 2 & 12 & -6 \\ -4& -8&12 \end{matrix} \right]$
But as you can see the $2$ matrix are different. Where is the mistake in my logic?
Another thing I noticed is that $I_n = AB \Leftrightarrow A*I_n*B = A*(AB)*B \Leftrightarrow I_n = AB = A^2B^2$, but $A^2B^2$ isn't the identity matrix, so maybe the exercise is wrong, but I rather not rush into that thinking.
|
Long story short, matrix multiplication is not commutative.
Te equation $A=A^2B$ is not correct. From the equation $$(A^2)^{-1} = A^{-1}B$$
you have to multiply it by $A$ from the left to get
$$A(A^2)^{-1} = B$$
and the multiply it by $A^2$ from the right and you get
$$A=B\cdot A^2$$
which is not the same as $A=A^2B$.
However, even using $A=BA^2$ results in two different calculations of what $A$ is, and this seems to me like a mistake in the exercise. Indeed, from $(A^2)^{-1} = A^{-1}B$, it should follow that $I=A^2A^{-1}B=AB$, and from that, it should follow that $A=B^{-1}$. However, it is not true that $(B^{-1})^2=A^2$ which means that $A=B^{-1}$ cannot hold, which is a contradiction.
|
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|
How to solve a recurrence relation with generating functions? I don't really understand how to solve (with generating functions) for the recurrence relation of $$a_n = a_{n-1}+2(n-1)$$ with initial conditions of $a_1 = 2$ when $n \geq 2$
This is what I was thinking,
\begin{align*}
g(x) &= a_0x^0+a_1x^1+a_2x^2+...+a_rx^r+...\\
g(x) &= a_0 + a_1x^1 + \sum^{\infty}_{n=2}a_nx^n\\
g(x) &= 1 + 2x + \sum^{\infty}_{n=2}{(a_{n-1}+2(n-1))x^n}\\
g(x) &= 1 + 2x + \sum^{\infty}_{n=2}a_{n-1}x^n + \sum^{\infty}_{n=2}(2n-2)x^n\\
g(x) &= 1 + 2x + x\sum^{\infty}_{n=1}a_{n-1}x^{n-1} + \sum^{\infty}_{n=2}2n x^n - \sum^{\infty}_{n=2}2x^n\\
g(x) &= 1 + 2x + x\sum^{\infty}_{m=1}a_{m}x^{m} + \sum^{\infty}_{n=2}2 \binom{n}{1}x^n - \sum^{\infty}_{n=2}2x^n\\
g(x) &= 1 + 2x + x(g(x) - a_0) + \sum^{\infty}_{n=2}2 \binom{n}{1}x^n - \sum^{\infty}_{n=2}2x^n\\
\end{align*}
|
You certainly do not need generating functions for solving $a_n-a_{n-1}=2(n-1)$: it is enough to sum both sides on $n=1,2,\ldots,N$ to get that $a_n$ depends on $\sum_{k=1}^{n}2(k-1)$. The only issue in your case is that the given values of $a_0$ and $a_1$ do not agree with the given recurrence relation. So, let us assume $a_0=\color{red}{2}$ and $a_{n}=a_{n-1}+2(n-1)$ for any $n\geq 1$. By denoting as $f(x)$ the following generating function
$$ f(x)=\sum_{n\geq 0} a_n x^n = 2+\sum_{n\geq 1}a_n x^n$$
we have
$$ x\cdot f(x) = \sum_{n\geq 0} a_n x^{n+1} = \sum_{n\geq 1} a_{n-1} x^n $$
hence by the recurrence relation
$$ (1-x) f(x) = 2+2\sum_{n\geq 1}(n-1)x^n = 2+\frac{2x^2}{(1-x)^2}$$
where the last identity follows from stars and bars in the form
$$ \frac{1}{(1-x)^{m+1}}=\sum_{n\geq 0}\binom{n+m}{m}x^n.$$
Since we have
$$ f(x) = \frac{2}{1-x}+\frac{2x^2}{(1-x)^3} $$
stars and bars also gives
$$ a_n = [x^n]f(x) = 2-n+n^2. $$
|
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|
Why do we automatically assume that when we divide a polynomial by a second degree polynomial the remainder is linear? Given question:
If a polynomial leaves a remainder of $5$ when divided by $x − 3$ and a remainder of $−7$ when divided by $x + 1$,
what is the remainder when the polynomial is divided by $x^2 − 2x − 3$?
Solution:
We observe that when we divide by a second degree polynomial the remainder will generally be linear. Thus
the division statement becomes
$p(x) = (x^2 − 2x − 3)q(x) + ax + b $
Can someone please explain at a PRE-CALCULUS level? Thanks
|
While José Carlos Santos's answer is correct, I think it might be useful to talk about why $r(x)$ is defined to have degree less than $p_2(x)$. (J.G. explains this, but at a higher level than I suspect someone wanting "pre-calculus" would follow.)
If the degree of $r(x)$ is the same size or greater than $p_2(x)$, then you can subtract off a multiple of $p_2(x)$ from $r(x)$ to get a new remainder of lower degree. I.e., we can continue the division process until $r(x)$ is of lower degree than $p_2(x)$. It is only then that we have to stop and call anything left over the remainder.
For example, dividing $3x^3 - 2x^2 + x - 5$ by $x^2 + 1$ follows these steps:
*
*Note that the ratio of the leading terms is $\frac {3x^3}{x^2} = 3x$
*Multiply $x^2 + 1$ by $3x$ and subtract the result ($3x^3 + 3x$) from $3x^3 - 2x^2 + x - 5$, leaving $-2x^2 - 2x - 5$. I.e.,$$3x^3 - 2x^2 + x - 5 = (3x)(x^2 + 1) + (-2x^2 - 2x - 5)$$
*Note that the ratio of the leading terms of the remainder and $x^2 + 1$ is $\frac {-2x^2}{x^2} = -2$.
*Multiply $x^2 + 1$ by $-2$ and subtract the result ($-2x^2 - 2$) from $-2x^2 - 2x - 5$, leaving $-2x - 3$. I.e., $$-2x^2 - 2x - 5 = (-2)(x^2 + 1) + (-2x-3)$$
*Note that the ratio of the leading terms is now $\frac {-2x}{x^2} = \frac {-2}x$, which is not a polynomial, so we cannot continue.
Combining the results from steps 2 and 4:
$$\begin{align}3x^3 - 2x^2 + x - 5 &= (3x)(x^2 + 1) + (-2)(x^2 + 1) + (-2x-3)\\&=(3x - 2)(x^2 + 1) + (-2x-3)\end{align}$$
If you stop at step 2, then $r(x)$ is indeed not linear. But at that stage, we could continue the division to get something smaller. It was only when the remainder was of lower degree that we had to stop.
|
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|
Evaluating $ \int \frac1{5S_{8} - 9S_{10} + 7S_{12} - 2S_{14}} {\rm d}x$, where $S_n = \sin^n(x) + \cos^n(x)$
Prove that
$$ \int \frac1{5S_{8} - 9S_{10} + 7S_{12} - 2S_{14}} {\rm d}x = 2x - \arctan \left( \frac{\tan2x}{2 + \tan^22x} \right) + C$$
where $S_n = \sin^n(x) + \cos^n(x)$.
Even differentiating the right doesn't end up with anything close to that monster integrand.
$$ \frac{{\rm d}}{{\rm d}x} \left( 2x - \arctan \left( \frac{\tan2x}{2 + \tan^22x} \right) + C \right)
\\ = 2 + \frac{(\tan^22x+1)(2\tan^22x-4)}{\tan^42x+5\tan^22x+4}
\\ = \frac{ 4 } {\sin^42x+ 5\sin^22x\cos^22x + 4\cos^42x}
\\ = \frac1{\sin^8x + \sin^2x\cos^6x +\cos^2x\sin^6x + \cos^8x} $$
|
Let $c=\cos2x$. Then applying the double angle identities,
$$\sin^2x=\frac{1-\cos2x}2,\quad\cos^2x=\frac{1+\cos2x}2,$$
to each occurrence of $\sin^nx$ and $\cos^nx$, we get
$$\begin{cases}
S_8=2^{-3}(1+6c^2+c^4)\\
S_{10}=2^{-4}(1+10c^2+5c^4)\\
S_{12}=2^{-5}(1+15c^2+15c^4+c^6)\\
S_{14}=2^{-6}(1+21c^2+35c^4+7c^6)
\end{cases}$$
Combining these expressions to match the denominator of the integrand, we find that it reduces to
$$5S_8-9S_{10}+7S_{12}-2S_{14}=2^{-2}(1+3c^2)$$
so that
$$\int\frac{\mathrm dx}{5S_8-9S_{10}+7S_{12}-2S_{14}}=\int\frac4{1+3\cos^22x}\,\mathrm dx=\int\frac8{5+3\cos4x}\,\mathrm dx$$
The antiderivative is $-\arctan(2\cot2x)+C$, which I don't think should be too hard to reconcile with the given solution.
|
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|
If $x_1,x_2,\ldots,x_n$ are the roots for $1+x+x^2+\ldots+x^n=0$, find the value of $\frac{1}{x_1-1}+\frac{1}{x_2-1}+\ldots+\frac{1}{x_n-1}$
Let $x_1,x_2,\ldots,x_n$ be the roots for $1+x+x^2+\ldots+x^n=0$. Find the value of
$$P(1)=\frac{1}{x_1-1}+\frac{1}{x_2-1}+\ldots+\frac{1}{x_n-1}$$
Source: IME entrance exam (Military Institute of Engineering, Brazil), date not provided (possibly from the 1950s)
My attempt:
Developing expression $P(1)$, replacing the 1 by $x$, follows
$$P(x)=\frac{(x_2-x)\cdots (x_n-x)+\ldots+(x_1-x)\cdots (x_{n-1}-x)}{(x_1-x)(x_2-x)\cdots (x_n-x)}$$
As $x_1,x_2,\ldots,x_n$ are the roots, it must be true that
$$Q(x)=(x-x_1)\cdots(x-x_n)=1+x+x^2+\ldots+x^n$$
and
$$Q(1)=(1-x_1)\cdots(1-x_n)=n+1$$
Therefore the denominator of $P(1)$ is
$$(-1)^{n} (n+1).$$ But I could not find a way to simplify the numerator.
Another fact that is probably useful is that
$$1+x^{n+1}=(1-x)(x^n+x^{n-1}+\ldots+x+1)$$
with roots that are 1 in addition of the given roots $x_1,x_2,\ldots,x_n$ for the original equation, that is
$$x_k=\text{cis}(\frac{2k\pi}{n+1}),\ \ k=1,\ldots,n.$$
This is as far as I could go...
Hints and full answers are welcomed.
|
We solve it using complex residues by way of enrichment. With the
function
$$f(z) = \frac{1}{z-1} \frac{(n+1)/z}{z^{n+1}-1}$$
we have for $\zeta_k = \exp(2\pi i k/(n+1))$ with $1\le k\le n$ that
$$\mathrm{Res}_{z=\zeta_k} f(z)
= \frac{1}{z-1}
\left.\frac{(n+1)/z}{(n+1)z^n}\right|_{z=\zeta_k}
= \frac{1}{\zeta_k-1}.$$
Residues sum to zero so we have
$$\sum_{k=1}^{n} \frac{1}{\zeta_k-1}
+ \mathrm{Res}_{z=0} f(z)
+ \mathrm{Res}_{z=1} f(z)
+ \mathrm{Res}_{z=\infty} f(z) = 0.$$
The residue at infinity is zero by inspection. The residue at $z=0$ is
$n+1$, also by inspection. We have for the residue at $z=1$
$$\mathrm{Res}_{z=1} f(z) =
\mathrm{Res}_{z=1} \frac{1}{(z-1)^2}
\frac{(n+1)/z}{1+z+\cdots +z^{n}}
\\ = \left.\left(
\frac{(n+1)/z}{1+z+\cdots +z^{n}}
\right)'\right|_{z=1}
\\ = \left.\left(
- \frac{(n+1)/z^2}{1+z+\cdots +z^{n}}
- \frac{(n+1)/z \times (1+2z+\cdots+nz^{n-1})}
{(1+z+\cdots +z^{n})^2}
\right)\right|_{z=1}
\\ = -1 - \frac{(n+1) \times \frac{1}{2} n (n+1)}{(n+1)^2}.$$
Now with
$$\sum_{k=1}^{n} \frac{1}{\zeta_k-1} =
- \mathrm{Res}_{z=0} f(z)
- \mathrm{Res}_{z=1} f(z)$$
we obtain
$$-(n+1) + 1 + \frac{1}{2} n.$$
and therefore our answer is
$$\bbox[5px,border:2px solid #00A000]{
\sum_{k=1}^{n} \frac{1}{\zeta_{k}-1}
= - \frac{n}{2}.}$$
|
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|
Find all Pythagorean triples $x^2+y^2=z^2$ where $x=21$ Consider the following theorem:
If $(x,y,z)$ are the lengths of a Primitive Pythagorean triangle, then $$x = r^2-s^2$$ $$y = 2rs$$ $$z = r^2+z^2$$ where $\gcd(r,s) = 1$ and $r,s$ are of opposite
parity.
According to the previous theorem,My try is the following:
since $x = r^2-s^2$, $x$ is difference of two squares implying that $x \equiv 0 \pmod 4$. But $x=21 \not \equiv 0 \pmod 4$. Hence, there are no triangles having such $x$.
Is that right?
Added:
My argument is false here. Please refer to the appropriate answer.
|
Since $A$ may be any odd number $\ge3$ and we can find one or more triples for any odd leg $A\ge 3$ using a function of $(m,A).$
We know that $A=m^2-n^2\implies n=\sqrt{m^2-A}.$
$$\text{We can let }n=\sqrt{m^2-A}\text{ where }\lceil\sqrt{A}\space\rceil\le m\le \frac{A+1}{2}$$
$\text{Note: }n\in \mathbb{R}\implies \sqrt{A}\lt m\land n<m\implies m\le\frac{a+1}{2}$
$$\text{For A=21}\quad m_{min}=\lceil\sqrt{21}\space\rceil=5\qquad \qquad m_{max}\frac{22}{2}=11$$
Testing for $5\le m\le 11$, we find $\mathbf{integers}$ for $(m,n)=(5,2)\text{ and }(11,10)$
$$f(5,2)=(21,20,29)\qquad f(11,10)=(21,220,221$$
But $21=3*7$
$$\text{For A=3}\quad m_{min}=\lceil\sqrt{3}\space\rceil=2\quad m_{max}=\frac{4}{2}=2\quad \quad 7f(2,1)=7(3,4,5)=21,28,35$$
$$\text{For A=7}\quad m_{min}=\lceil\sqrt{7}\space\rceil=3\quad m_{max}=\frac{8}{2}=4\quad \quad 3f(4,3)=3(7,24,25)=21,72,75$$
|
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|
Why is my proof for “if $0 \leqslant x \leqslant 2$, then $-x^3 + 4x + 1 > 0$” is false?
$$\text{if $0 \leqslant x \leqslant 2$, then $-x^3 + 4x + 1 > 0$}$$
$$x(4-x^2)>-1$$
$$x>\dfrac{1}{x^2-4}$$ if the last statement is smaller than $x$ then it is also smaller than $2$ because of first statement ($0 \leqslant x \leqslant 2$):
$$\dfrac{1}{x^2-4}<2$$
$$1<2x^2-8$$
$$4.5<x^2$$
$$-2.121<x<2.121$$
Which contradicts the first statement because $x$ cannot be greater than $2$. My proof is wrong but why? (I took that question from Math for Computer Science book p. 12)
|
You made two mistakes and recovered:
$$\dfrac{1}{x^2-4}<2 \not\Rightarrow 1<2(x^2-4) \ \ (\text{it must be} \ 1\color{red}>2(x^2-4), \text{because} \ x^2-4\le 0)\\
4.5<x^2 \not\Rightarrow -2.121<x<2.121 \ (\text{it must be} \ 4.5\color{red}>x^2 \Rightarrow -2.121<x<2.121)$$
Alternatively, you can prove it as follows:
$$-x^3 + 4x + 1 > 0 \iff x^3-4x-1<0 \iff x(x-2)(x+2)<1 \iff \\
x(x-2)(x+2)\le 0<1 \ \ \text{for} \ 0\le x\le 2 \ \text{(why?)}$$
|
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|
If $a_n > 0$ prove that $\sum_{n=1}^{\infty} \frac{a_n}{(a_1+1)(a_2+1)\cdots(a_n+1)}$ converges I have an interesting task: If $a_n > 0$, prove that $$\sum_{n=1}^{\infty} \frac{a_n}{(a_1+1)(a_2+1)\cdots(a_n+1)}$$ converges.
I thought that it will be simple because ratio test gives me:
$$\frac{u_{n+1}}{u_n}= \frac{a_{n+1}}{a_{n+1}+1}\cdot a_n^{-1} < 1 \cdot a_n^{-1} = \frac{1}{a_n}$$ and $a_n$ should be in $[0,1]$. But... In my opinion it can be over that... why need I assume that $ a_n \rightarrow g \in [0,1] $?
There is similar topic on this forum, but It was not solved there...
@edit
I saw that:
$$\sum_{n=1}^{N}\frac{a_n}{(1+a_1)(1+a_2)...(1+a_n)} = 1-\frac{1}{(1+a_1)(1+a_2)...(1+a_N)} < 1 $$
So if series of partial sum is bounded from up, the sum converges, that is right?
@edit2 but It is good? Look at that:
$$ \sum_{n=1}^{N}\frac{a_n+1-1}{(1+a_1)(1+a_2)...(1+a_n)} = \sum_{n=1}^{N}\frac{1}{(1+a_1)(1+a_2)...(1+a_{n-1})}-\frac{1}{(1+a_1)(1+a_2)...(1+a_n)} $$ why somebody changed first part into $1$?
@edit3 Ok, I think that I have understood, thanks for your time ;)
|
Hint:
$$
\eqalign{
& \sum\limits_{1\, \le \,n\,} {{{a_n } \over {\left( {a_1 + 1} \right)\left( {a_2 + 1} \right) \cdots \left( {a_n + 1} \right)}}} = \cr
& = \sum\limits_{1\, \le \,n\,} {{{\left( {a_n + 1} \right) - 1} \over {\left( {a_1 + 1} \right)\left( {a_2 + 1} \right) \cdots \left( {a_n + 1} \right)}}}
= \cdots \cr}
$$
(continuing)
$$
\eqalign{
& = {{a_1 } \over {\left( {a_1 + 1} \right)}} + {{a_2 } \over {\left( {a_1 + 1} \right)\left( {a_2 + 1} \right)}}
+ {{a_3 } \over {\left( {a_1 + 1} \right)\left( {a_2 + 1} \right)\left( {a_3 + 1} \right)}} + \cdots = \cr
& = {{a_1 } \over {\left( {a_1 + 1} \right)}} + {{a_2 + 1 - 1} \over {\left( {a_1 + 1} \right)\left( {a_2 + 1} \right)}}
+ {{a_3 + 1 - 1} \over {\left( {a_1 + 1} \right)\left( {a_2 + 1} \right)\left( {a_3 + 1} \right)}} + \cdots = \cr
& = {{a_1 } \over {\left( {a_1 + 1} \right)}} + {1 \over {\left( {a_1 + 1} \right)}} - {1 \over {\left( {a_1 + 1} \right)\left( {a_2 + 1} \right)}}
+ {1 \over {\left( {a_1 + 1} \right)\left( {a_2 + 1} \right)}} - {1 \over {\left( {a_1 + 1} \right)\left( {a_2 + 1} \right)\left( {a_3 + 1} \right)}} + \cdots = \cr
& = {{a_1 } \over {\left( {a_1 + 1} \right)}} + {1 \over {\left( {a_1 + 1} \right)}}
- {1 \over {\left( {a_1 + 1} \right)\left( {a_2 + 1} \right)\left( {a_3 + 1} \right)}} + \cdots = \cr
& = 1 - {1 \over {\left( {a_1 + 1} \right)\left( {a_2 + 1} \right)\left( {a_3 + 1} \right)}} + \cdots = \cdots \cr}
$$
|
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Proof verification that $\{x_n\} = 0,\underbrace{77\dots 7}_{\text{n times}}$ is a Cauchy sequence.
Given a sequence $\{x_n\}$:
$$
x_n = 0,\underbrace{77\dots 7}_{\text n\ times}
$$
Prove that $\{x_n\}$ is a Cauchy sequence.
Recall the definition of a fundamental sequence:
$$
x_n\ \text{is fundamental} \ \iff \forall \epsilon>0 \exists N\in \Bbb N: \forall n, m >N\implies |x_n - x_m| < \epsilon
$$
Rewrite $x_n$:
$$
x_n = {7\over 10^1} + {7\over 10^2} + \cdots + {7\over 10^n} = \sum_{k=1}^n \frac{7}{10^k}
$$
By geometric series sum:
$$
x_n = \sum_{k=1}^n \frac{7}{10^k} = \frac{7}{9}\left(1 - {1\over 10^n}\right) \\
x_m = \sum_{k=1}^m \frac{7}{10^k} = \frac{7}{9}\left(1 - {1\over 10^m}\right) \\
$$
Suppose $m > n$:
$$
\begin{align}
|x_n - x_m| &= |x_m - x_n| = \\
&= \left|\frac{7}{9}\left(1 - {1\over 10^m}\right) - \frac{7}{9}\left(1 - {1\over 10^n}\right)\right| = \\
&= \left|\frac{7}{9}\left(1 - {1\over 10^m} - 1 + {1\over 10^n}\right)\right| =
\\
&= \left|\frac{7}{9}\left({1\over 10^n} - {1\over 10^m}\right)\right| \le \left|\frac{7}{9}{1\over 10^n}\right| \le \frac{7}{9\cdot 10^N} < \epsilon
\end{align}
$$
This shows we've found $N$ which depends on $\epsilon$ and satisfies the definition of a Cauchy sequence.
This is the first time I'm dealing with proving a sequence is fundamental, could someone please verify whether my proof is valid?
|
Since my proof in OP is correct I would like to add a generalization also. Consider the following $x_n$:
$$
x_n = a + aq + aq^2 + \cdots + aq^{n-1}
$$
Using geometric series sum for $|q| < 1$:
$$
x_n = \frac{a(1-q^n)}{1-q}
$$
Since $|q| < 1$ we may rewrite it as:
$$
q = \frac{1}{1+r},\ r \in \Bbb R_{>0}
$$
Then for $m > n$:
$$\begin{align}
|x_m - x_n| &= \left|\frac{a}{1-q} \left(q^n - q^m\right)\right|\\ &= \left|\frac{a}{1-q} \left(\frac{1}{(1+r)^n} - \frac{1}{(1+r)^m}\right)\right| \\&\le \left|\frac{a}{1-q} \left(\frac{1}{(1+r)^n}\right)\right| \\ &
\le \frac{a}{1-q} \left(\frac{1}{(1+r)^N}\right) < \epsilon\end{align}
$$
Which shows any sequence of such kind is Cauchy. Or with direct statement:
$$
\frac{1-q}{a} \left((1+r)^N\right) > {1\over \epsilon}\\
(1+r)^N > \frac{a}{(1-q)\epsilon} \\
\exists N >\log_{1+r}\frac{a}{(1-q)\epsilon}, \forall m>n>N \implies |x_m - x_n| < \epsilon
$$
|
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|
Find the value of $\cot^{-1}21+\cot^{-1}13+\cot^{-1}(-8)$
Find the value of $\cot^{-1}21+\cot^{-1}13+\cot^{-1}(-8)$
My Attempt
\begin{align}
\cot^{-1}21+\cot^{-1}13+\cot^{-1}(-8)&=\pi-\tan^{-1}\frac{1}{8}+\tan^{-1}\frac{1}{13}+\tan^{-1}\frac{1}{21}\\
&=\pi-\tan^{-1}\frac{1}{8}+\tan^{-1}\frac{1}{13}+\tan^{-1}\frac{1}{21}\\
&=\pi-\tan^{-1}\frac{1}{8}+\tan^{-1}\frac{4-3}{1+4.3}+\tan^{-1}\frac{5-4}{1+5.4}\\
&=\pi-\tan^{-1}\frac{1}{8}+\tan^{-1}5-\tan^{-1}3\\
&=\pi-\tan^{-1}\frac{1}{8}+\tan^{-1}\frac{1}{8}=\pi
\end{align}
My reference gives the solution $0$, so what's going wrong with my attempt ?
|
Hint.
$$
\tan(\mbox{arccot}(a)) = \frac 1a
$$
and
$$
\tan(a+b) = \frac{\tan(a)+\tan(b)}{1-\tan(a)\tan(b)}
$$
so
$$
\tan(\mbox{arccot}(a)+\mbox{arccot}(b)) = \frac{a+b}{a b -1}
$$
etc.
NOTE
You can try now
$$
\tan(\mbox{arccot}(a)+\mbox{arccot}(b)+\mbox{arccot}(c)) = d = \frac{a (b+c)+b c-1}{abc-(a+b+c)}
$$
and then
$$
\mbox{arccot}(a)+\mbox{arccot}(b)+\mbox{arccot}(c) = \mbox{arctan}(d)
$$
after substitution we have $d=0$ and $\arctan(0) = 0$
|
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|
Integer Solutions of the Equation $u^3 = r^2-s^2$ The question says the following:
Find all primitive Pythagorean Triangles $x^2+y^2 = z^2$ such that $x$ is a perfect cube.
The general solution for each variable are the following:
$$x=r^2-s^2$$
$$y=2rs$$
$$z=r^2+s^2$$
such that $\gcd(r,s) = 1$and $r+s \equiv 1 \pmod {2}$
In order to make $x$ a perfect cube, I shall have the equation $x=u^3=r^2-s^2$. However, I am stuck to find a general formula for such cubes.
I know that a subset of the solutions might be the difference between two consecutive squares. This difference is always an odd integer. I can collect some examples such $14^2-13^2 = 27$ but I cannot give a formula for such type either.
Any ideas?
|
$1^2=1^3$
$3^2=(1+2)^2=1^3+2^3$
$6^2=(1+2+3)^2=1^3+2^3+3^3$
...
The difference between two consecutive squares on the left will give you a cube:
$1^3=1^2-0^2$
$2^3=3^2-1^2$
$3^3=6^2-3^2$
...
Which means the solutions are pairs of this form: $(\frac{n(n-1)}{2}, \frac{n(n+1)}{2})$
$1^3=1^2-0^2$
$2^3=3^2-1^2$
$3^3=6^2-3^2$
...
|
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|
Can't solve a quartic equation I'm trying to solve an algebraic question.The question wants me to solve $n^4+2n^3+6n^2+12n+25=m^2$.The question also states that n is a positive integer and the answer for $n^4+2n^3+6n^2+12n+25$ is a square number. Here's how I tried to solve it:
$$n^4+2n^3+6n^2+12n+25=\\n^4+6n^2+2n^3+12n+25=\\
n^2(n^2+6)+2n(n^2+6)+5^2=\\
(n\sqrt {n^2+6})^2+2n(n^2+6)+5^2.\\$$
Because $a^2+2ab+b^2=(a+b)^2$,so
$\sqrt {(n\sqrt {n^2+6})^2}\cdot\sqrt5^2=n(n^2+6)$
Then:
$$n\sqrt {n^2+6}\cdot 5=n(n^2+6)\\
\sqrt {n^2+6}\cdot 5=n^2+6\\
25(n^2+6)=n^4+12n^2+36\\
n^4+12n^2+36=25n^2+150\\
n^4-13n^2-114=0\\
(n^2+6)(n^2-19)=0\\
n^2=19\\
n=\sqrt 19$$
But $n$ is a positive integer.
Can anyone help?
|
Above equation shown below:
$n^4+2n^3+6n^2+12n+25=m^2$
As pointed out by Will Jagy the only positive
integer solution to above equation is
$(n,m)=(8,75)$
Also $m$ is a multiple of five as mentioned by Rhyes hughes
|
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|
AMC 1834 - If $a, b $ and $c$ are nonzero numbers such that (a+b-c)/c=(a-b+c)/b=(-a+b+c)/a and x=((a+b)(b+c)(c+a))/abc and x<0, then x=? I am getting stuck on this question:
If a, b, and c, are nonzero numbers such that $\frac{a+b-c}{c}$=$\frac{a-b+c}{b}$=$\frac{-a+b+c}{a}$ and x=$\frac{(a+b)(b+c)(c+a)}{abc}$ and x<0, then x=?
What I have done so far: Consider, if $\frac{a}{b}$=$\frac{c}{d}$, and (b+d) is non-zero, then $\frac{a+c}{b+d}$=$\frac{a}{b}$=$\frac{c}{d}$. So with that fact, I did... $\frac{a+b-c}{c}$=$\frac{a-b+c}{b}$ $\implies$
$\frac{2a}{c+b}$ and $\frac{2a}{c+b}$=$\frac{-a+b+c}{a}$$\implies$ $\frac{(a+b+c)}{a+b+c}$ = 1. Then I took the original equations and did $\frac{a+b-c}{c}$=1, so $a+b=2c$ , and similarly, $(a+c)=2b$ and $b+c=2a$. So plugging that into x=$\frac{(a+b)(b+c)(c+a)}{abc}$ we have $x=\frac{(2a)(2b)(2c)}{abc}$, which is $x=\frac{8(abc)}{abc}=8$. This is wrong though, because it says in the problem x<0. So how do I do this?
|
You correctly derived:
$$\frac{2a}{c+b}=\frac{-a+b+c}{a}$$
Your mistake is here:
Then I took the original equations and did $\frac{a+b-c}{c}=1$
Instead, denote: $\frac{b+c}{a}=t$, then the above equation becomes:
$$\frac2{t}=-1+t \Rightarrow t^2-t-2=0 \Rightarrow t_1=-1 \ \ \text{and} \ \ t_2=2.$$
Also note that:
$$\frac{a+b-c}{c}=\frac{a-b+c}{b}=\frac{-a+b+c}{a}\iff \\
\frac{a+b}{c}-1=\frac{a+c}{b}-1=\frac{b+c}{a}-1 \iff \\
\frac{a+b}{c}=\frac{a+c}{b}=\frac{b+c}{a}=t.$$
Hence:
$$x=\frac{(a+b)(b+c)(c+a)}{abc}=t^3=(-1)^3=-1;\\
x=\frac{(a+b)(b+c)(c+a)}{abc}=t^3=2^3\not< 0.$$
|
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|
How would I go about solving for $x$ in $\frac{(x-a)\sqrt{x-a}+(x-b)\sqrt{x-b}}{\sqrt{x-a}+\sqrt{x-b}}=a-b$? The question
This is a homework question. Given the following, I am to solve for $x$ in terms of $a$ and $b$:
$$\frac{(x-a)\sqrt{x-a}+(x-b)\sqrt{x-b}}{\sqrt{x-a}+\sqrt{x-b}}=a-b;a>b.$$
My attempt
Although I see the pattern of multiple occurrences of $(x-a)$, $(x-b)$ I can't see any way to simplify the fraction further, so I go on to simplify the expression by multiplying by $\sqrt{x-a}+\sqrt{x-b}$:
$$\begin{align*}
(x-a)\sqrt{x-a}+(x-b)\sqrt{x-b}&=(a-b)(\sqrt{x-a}+\sqrt{x-b})\\
&=a\sqrt{x-a}+a\sqrt{x-b}-b\sqrt{x-a}-b\sqrt{x-b}
\end{align*}$$
Now I have the following:
$$(x-a)\sqrt{x-a}+(x-b)\sqrt{x-b}=a\sqrt{x-a}+a\sqrt{x-b}-b\sqrt{x-a}-b\sqrt{x-b}$$
Simplifying the RHS as I was out of ideas at that point:
$$x\sqrt{x-a}-a\sqrt{x-a}+x\sqrt{x-b}-b\sqrt{x-b}=a\sqrt{x-a}+a\sqrt{x-b}-b\sqrt{x-a}-b\sqrt{x-b}$$
I noticed that all one of the common factors $\sqrt{x-a},\sqrt{x-b}$ so I tried to isolate them and factor them out -- that is, all $\sqrt{x-b}$ terms on one side and $\sqrt{x-a}$ terms on the other.
$$\sqrt{x-b}(x-a)=\sqrt{x-a}(2a-b-x)$$
I tried to then square both sides, but that led to quite a mess.
$$(x-b)(x^2-2ax+a^2)=(x-a)(4a^2-4ab+2bx-4ax+b^2+x^2)$$
I'm afraid to even begin trying to simplifying this. I'm convinced I'm going about it in the wrong way.
The $a>b$ hint is interesting, but I have no clue what implication it may have here.
I think the $(x-a)\sqrt {x-a}$ patterns may mean something, perhaps I could do something with $a\sqrt a=\sqrt{a^3}$, but at this point it is probably a dead end.
I appreciate any help.
|
Use the formula: $a^3+b^3=(a+b)(a^2-ab+b^2)$.
We get:
$$\frac{(x-a)\sqrt{x-a}+(x-b)\sqrt{x-b}}{\sqrt{x-a}+\sqrt{x-b}}=\\
\frac{(\sqrt{x-a}+\sqrt{x-b})((x-a)-\sqrt{(x-a)(x-b)}+(x-b))}{\sqrt{x-a}+\sqrt{x-b}}=\\
2x-a-b-\sqrt{(x-a)(x-b)}=a-b \Rightarrow \\
(x-a)(x-b)=(2x-2a)^2 \Rightarrow \\
3x^2+(b-7a)x+4a^2-ab=0 \Rightarrow \\
x=\frac{(7a-b)\pm \sqrt{(b-7a)^2-12(4a^2-ab)}}{6}=\\
\frac{7a-b\pm (a-b)}{6}=\\
\frac{4a-b}{3}; a.$$
|
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|
Find the limit of $\lim_{n\to\infty}((n^3+n^2)^{1/3}-(n^3+1)^{1/3})$ without using the identity $a^3-b^3=(a-b)(a^2+ab+b^2)$ Find the limit of the sequence $$\lim_{n\to\infty}((n^3+n^2)^{1/3}-(n^3+1)^{1/3})$$
I showed that the limit is $1/3$, using the identity $$a^3-b^3=(a-b)(a^2+ab+b^2)$$
we get the sequence is equal to
$$\frac{n^3+n^2-n^3-1}{(n^3+n^2)^{2/3}+(n^6+n^5+n^3+n^2)^{1/3}+(n^3+1)^{2/3}}\longrightarrow\frac{1}{3}$$
but I couldn't manage to solve it without using the above identity, and I was wondering if it is possible.
|
My favorite way is to consider
$$
f(x)=\left(\frac{1}{x^3}+\frac{1}{x^2}\right)^{1/3}-\left(\frac{1}{x^3}+1\right)^{1/3}
=\frac{\sqrt[3]{1+x}-\sqrt[3]{1+x^3}}{x}
$$
so that your sequence is $f(1/n)$ and so you can compute
$$
\lim_{x\to0^+}f(x)=\lim_{x\to0}\frac{(1+x/3+o(x)-(1+x^3/3+o(x^3))}{x}=\frac{1}{3}
$$
Without Taylor expansion, the sought limit is the derivative at $0$ of $g(x)=\sqrt[3]{1+x}-\sqrt[3]{1+x^3}$, so
$$
g'(x)=\frac{1}{3\sqrt[3]{(1+x)^2}}-\frac{3x^2}{3\sqrt[3]{(1+x^3)^2}}
$$
and $g'(0)=1/3$.
|
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|
How to factor this quadratic expression? A bit confused on how to factor $2x^2 + 5x − 3 = 0$. Firstly, I multiplied $a \cdot c$, so $2(-3)=-6$, however couldn't find two numbers that will add up to $5$. Then, I thought of the factors could be $6(-1)=-6$, and $6+-1=5$. However, shouldn't the $a$ and $c$ values always be multiplied together, (the $(2)(-3)$, rather than $6(-1)$?
When following the factors $-1$ and $6$. I have
$(2x^2-1x)(6x-3)$
$x(2x-1)+3(2x-1)$
Is this correct, if not; what is the best way to solve a leading coefficient when factoring?
|
We have that $(2x+a)(x+b)=2x^2+5x-3$, where $x=-\frac a2$ and $x=-b$ are the solutions you require.
Then $$(2x^2+(a+2b)x+ab=2x^2+5x-3$$
leads to $$a+2b=5; ab=-3$$
We set $b=\frac12(5-a)\to\frac12a(5-a)=-3 \to a^2-5a-6=0$
$$\to a=-1, 6$$
$$\to b= 3, -\frac 12$$
So $$2x^2+5x-3=(2x-1)(x+3)=(2x+6)(x-\frac12)$$
|
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|
Evaluating $\int_0^1\arctan x\ln(1+x)\left(\frac2x-\frac3{1+x}\right)dx$
How can we find the value of $$\int_0^1\arctan x\ln(1+x)\left(\frac2x-\frac3{1+x}\right)dx$$ using elementary methods?
With some help of calculator I get the result: $\displaystyle{\frac3{128}\pi^3-\frac9{32}\pi\ln^22}$.
Thoughts of this integral
Since I have asked this question and Pisco gave a brilliant answer, I tried to convert $$I_1=\int_0^1\arctan x\ln(1+x)\frac{dx}x\text{ and }I_2=\int_0^1\arctan x\ln(1+x)\frac{dx}{1+x}$$ into the form of integral Pisco gave.
Integrating by parts to the second integral converts $I_2$ into $\int_0^1\frac{\ln^2(1+x)}{1+x^2}dx$.
But for $I_1$? Integrating by parts gives a dilog function and I tried substitution $x=\frac{1-t}{1+t}$ and got $$\frac{\ln\frac{2}{t+1} \arctan\frac{1-t}{1+t}}{1-t^2}$$ which is not what I want.
|
Here is an elementary approach, although it turned into a crossover with FDP's answer.
First note that from here we have:
$$\color{blue}{\int_0^1 \frac{\arctan x \ln(1+x)}{x}dx}=\frac{3}{2}\int_0^1 \frac{\arctan x\ln(1+x^2)}{x}dx$$
$$\overset{IBP}=\frac32 \underbrace{\ln x\arctan x\ln(1+x^2)\bigg|_0^1}_{=0}-\frac32 \left(\int_0^1 \frac{\ln x\ln(1+x^2)}{1+x^2}dx+2\int_0^1 \frac{x\arctan x\ln x}{1+x^2}dx\right) $$
Back to the original integral, we have:
$$I=\color{blue}{2\int_0^1 \frac{\arctan x \ln(1+x)}{x}dx}-\color{red}{3\int_0^1 \frac{\arctan x \ln(1+x)}{1+x}dx} $$
$$=\color{blue}{-3\left(\int_0^1 \frac{\ln x\ln(1+x^2)}{1+x^2}dx+2\int_0^1 \frac{x\arctan x\ln x}{1+x^2}dx\right)}-\color{red}{3\int_0^1\frac{\arctan x\ln(1+x)}{1+x}dx}$$
$$\Rightarrow I=-3(B+2A+J)\quad \quad (1)$$
Where I kept the notation like in FDP's answer. Namely:
$$\begin{align*}
\displaystyle A&=\int_0^1 \dfrac{x\arctan x\ln x}{1+x^2}dx\\
\displaystyle B&=\int_0^1 \dfrac{\ln x \ln(1+x^2)}{1+x^2}dx\\
\displaystyle J&=\int_0^1\dfrac{\arctan x\ln(1+x)}{1+x}dx
\end{align*}$$
Also another two identities follows from that post, see $(8)$ and $(9)$: $$J=\dfrac{5}{3}G\ln 2-\dfrac{\pi^3}{128}+\dfrac{3\pi\left(\ln 2\right)^2}{32}+B+\dfrac{2}{3}\left(\dfrac{G\ln 2}{2}-\dfrac{\pi^3}{64}\right)-\dfrac{2}{3}\cdot\frac{\pi^3}{32} $$
$$\Rightarrow \color{purple}{J=2G\ln 2 -\frac{5\pi^3}{128}+\frac{3\pi}{32}\ln^2 2 +B} \tag 2$$
$$\color{magenta}{A=\dfrac{1}{64}\pi^3-B-G\ln 2} \tag 3$$
Now plugging $(2)$ and $(3)$ in $(1)$ yields:
$$I=-3\left(B+2\left(\color{magenta}{\dfrac{1}{64}\pi^3-B-G\ln 2}\right)+ \color{purple}{2G\ln 2 -\frac{5\pi^3}{128}+\frac{3\pi}{32}\ln^2 2 +B}\right)$$
$$\Rightarrow I=-3\left(-\frac{\pi^3}{128}+\frac{3\pi}{32}\ln^2 2\right)=\boxed{\frac{3\pi^3}{128}-\frac{9\pi}{32}\ln^2 2}$$
Credits to FDP for his amazing answer there!
|
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Solving : $2n^{\frac{1}{2}}-1.5n^{\frac{1}{3}}+n^{\frac{1}{6}}=y$ for $n$ in terms of $y$ I have been trying to solve this type of equation for $n$ in terms of $y$: $an^{\frac{1}{2}}-bn^{\frac{1}{3}}+cn^{\frac{1}{6}}=y$ but I have yet to find a way that works. I have also tried several online equation solvers with no avail.
Here is what I have tried so far in solving the following equation for $n$ in terms of $y$:
$2n^{\frac{1}{2}}-1.5n^{\frac{1}{3}}+n^{\frac{1}{6}}=y$
I first tried to rewrite it as a quadratic but after pulling out $n^{\frac{1}{6}}$ like this:
$n^{\frac{1}{6}}(2n^{\frac{1}{3}}-1.5n^{\frac{1}{6}}+1)=y$
But then there is nothing further I can do to factor this.
I couldn't find any power that I could raise $n$ to multiply both sides either.
Any advice as how to proceed in solving this would be greatly appreciated.
Thank you.
|
Let $x=n^{1/6}$ (i.e. $n=x^6$):
$$2x^3-\frac 3 2x^2+x=y$$
Subtract $y$ and divide both sides by $2$:
$$x^3-\frac 3 4x^2+\frac 1 2x-\frac 1 2y=0$$
In order to reduce this to a depressed cubic, let $x=z+\frac 1 4$:
$$z^3+\frac{5z}{16}+\frac{3}{32}-\frac 1 2y=0$$
In order to turn this into a quadratic, let $z=w+\frac{5}{48w}$:
$$w^3+\frac{3}{32}-\frac 1 2y-\frac{125}{110592w^3}=0$$
Multiply by $w^3$:
$$w^6+\left(\frac{3}{32}-\frac 1 2y\right)w^3-\frac{125}{110592}=0$$
Time to apply the quadratic formula:
$$w^3=\frac{\frac 1 2 y-\frac{3}{32}\pm\sqrt{\left(\frac{3}{32}-\frac 1 2 y\right)^2-\frac{125}{27649}}}{\frac{125}{55296}}$$
Now, there are three complex solutions to this equation. To represent this, I will use $\omega$ to represent a cube root of unity. Thus, the following equation represents three solutions for $w$: One for $\omega=1$, one for $\omega=\frac{-1+\sqrt{-3}}{2}$, and one for $\omega=\frac{-1-\sqrt{-3}}{2}$:
$$w=\omega\sqrt[3]{\frac{\frac 1 2 y-\frac{3}{32}\pm\sqrt{\left(\frac{3}{32}-\frac 1 2 y\right)^2-\frac{125}{27649}}}{\frac{125}{55296}}}$$
From here, I leave the rest to you: Solve for $z$ using $z=w-\frac{5}{48w}$, solve for $x$ using $x=z+\frac 1 4$, and, finally, solve for $n$ using $n=x^6$. Good luck finishing the problem!
|
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Sum of Infinity of Trigo to Pi I am currently working on a proof with a good friend of mine that involves adding more and more triangles to the sides of a regular polygon but keeping the longest diagonal constant until eventually, it becomes a circle. And we ended up with this formula.
$4$-sided regular$ \to 8$-sided regular$\to 16$-sided regular$\to 32$-sided regular\to \ldots \to n$-sided regular
(When $n$ tends to infinity, the area will be equal to that of a circle with the longest diagonal as diameter)
Moreover, we have used our calculator to input the numbers and we get the value of $3.140\ldots$ which is very close to π But we can't be completely sure that the infinity sum really equals to π.
That is why we really need your knowledge of Maths to solve this.
Thanks in advance and Happy Holidays Everyone! :D
So the question is:
Is there a way to legitly prove that
$$\sum_{n=0}^\infty 2^n \left(2\cdot \sin\frac{90^\circ}{2^n}-\sin \frac{180^\circ}{2^n}\right)=\pi$$
(please kindly refer to Appendix 1)
P.S. I apologize for my poor penmanship...
P.P.S. We tried using Wolfram Alpha, it didn't show us the step-by-step calculations
|
\begin{align}
\sum_{n=0}^m 2^n \left(2 \sin \frac{90^\circ}{2^n} - \sin \frac{180^\circ}{2^n} \right) &=\sum_{n=0}^m 2^n \left(2 \sin \frac{\pi}{2^{n+1}} - \sin \frac{\pi}{2^n} \right)\\
&=\sum_{n=0}^m \left(2^{n+1} \sin \frac{\pi}{2^{n+1}} - 2^n\sin \frac{\pi}{2^n} \right)\\
&=\left(2^1 \cdot \sin \frac{\pi}{2} -2^0\cdot \sin \frac{\pi}{2^0}\right)\\
&+\left(2^2 \cdot \sin \frac{\pi}{2^2} -2^1\cdot \sin \frac{\pi}{2^1}\right)\\
&+\left(2^3 \cdot \sin \frac{\pi}{2^3} -2^2\cdot \sin \frac{\pi}{2^2}\right)\\
&\vdots\\
&+\left(2^{m+1} \sin \frac{\pi}{2^{m+1}} - 2^m\sin \frac{\pi}{2^m} \right)\\
&= 2^{m+1}\sin \frac{\pi}{2^{m+1}}-2^0 \sin \pi \\
&= 2^{m+1}\sin \frac{\pi}{2^{m+1}}
\end{align}
Hence
\begin{align}\sum_{n=0}^\infty 2^n \left(2 \sin \frac{90^\circ}{2^n} - \sin \frac{180^\circ}{2^n} \right)&= \lim_{m \to \infty}\sum_{n=0}^m 2^n \left(2 \sin \frac{90^\circ}{2^n} - \sin \frac{180^\circ}{2^n} \right) \\
&=\lim_{m \to \infty} 2^{m+1}\sin \frac{\pi}{2^{m+1}}\\
&=\lim_{m \to \infty} 2^{m+1}\cdot \frac{\pi}{2^{m+1}}\cdot \frac{\sin \frac{\pi}{2^{m+1}}}{\frac{\pi}{2^{m+1}}}\\
&=\lim_{m \to \infty} 2^{m+1}\cdot \frac{\pi}{2^{m+1}}\cdot \lim_{m \to \infty}\frac{\sin \frac{\pi}{2^{m+1}}}{\frac{\pi}{2^{m+1}}}\\
&= \pi \cdot 1\\
&= \pi\end{align}
|
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|
Proving that a triangle is isosceles. In a triangle $ABC$, let $D$ be a point on the segment $BC$ such that $AB + BD = AC + CD$. Suppose that the points $B,C$ and the centroids of triangles $ABD$ and $ACD$ lie on a circle. Prove that $AB = AC$.
I began this way: Denote the centroid of triangle $ABD$ by $G_1$ and the centroid of triangle $ACD$ by $G_2$. Then by using $Basic$ $Proportionality$ $Theorem$, I got that $BG_1G_2C$ is an isosceles trapezium. After that I am stuck. Please help.
|
Now, in the standard notation $$c+BD=b+a-BD,$$ which gives
$$BD=\frac{a+b-c}{2}$$ and $$CD=\frac{a+c-b}{2}.$$
Thus, $$CG_2=\frac{1}{3}\sqrt{2b^2+2\left(\frac{a+c-b}{2}\right)^2-AD^2}$$ and
$$BG_1=\frac{1}{3}\sqrt{2c^2+2\left(\frac{a+b-c}{2}\right)^2-AD^2},$$ which gives
$$2b^2+2\left(\frac{a+c-b}{2}\right)^2-AD^2=2c^2+2\left(\frac{a+b-c}{2}\right)^2-AD^2$$ or
$$(b-c)(b+c-a)=0$$ or $$b=c.$$
|
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|
Find all functions satisfying $f(x+1)=\frac{f(x)-5}{f(x)-3}$ Find all functions satisfying
$$f(x+1)=\frac{f(x)-5}{f(x)-3}$$
My try:
We have $$f(x+1)=1-\frac{2}{f(x)-3}$$
Letting $g(x) =f(x+1)-3$
We get $$g(x+1)=-2-\frac{2}{g(x)}$$
Any clue here?
|
Let $f(x)=g(x)+3$ ,
Then $g(x+1)+3=\dfrac{g(x)+3-5}{g(x)+3-3}$
$g(x+1)+3=\dfrac{g(x)-2}{g(x)}$
$g(x+1)+3=1-\dfrac{2}{g(x)}$
$g(x+1)=-2-\dfrac{2}{g(x)}$
Let $g(x)=\dfrac{h(x+1)}{h(x)}$ ,
Then $\dfrac{h(x+2)}{h(x+1)}=-2-\dfrac{2h(x)}{h(x+1)}$
$\dfrac{h(x+2)}{h(x+1)}=-\dfrac{2h(x+1)+2h(x)}{h(x+1)}$
$h(x+2)+2h(x+1)+2h(x)=0$
$h(x)=\theta_1(x)(-1+i)^x+\theta_2(x)(-1-i)^x$ , where $\theta_1(x)$ and $\theta_2(x)$ are arbitrary periodic functions with unit period
$h(x)=\theta_1(x)e^{x\ln(-1+i)}+\theta_2(x)e^{x\ln(-1-i)}$ , where $\theta_1(x)$ and $\theta_2(x)$ are arbitrary periodic functions with unit period
$h(x)=\theta_1(x)e^{\frac{x\ln2}{2}+\frac{3i\pi x}{4}}+\theta_2(x)e^{\frac{x\ln2}{2}-\frac{3i\pi x}{4}}$ , where $\theta_1(x)$ and $\theta_2(x)$ are arbitrary periodic functions with unit period
$h(x)=\Theta_1(x)2^\frac{x}{2}\sin\dfrac{3\pi x}{4}+\Theta_2(x)2^\frac{x}{2}\cos\dfrac{3\pi x}{4}$ , where $\Theta_1(x)$ and $\Theta_2(x)$ are arbitrary periodic functions with unit period
$\therefore f(x)=\dfrac{\Theta_1(x+1)2^\frac{x+1}{2}\sin\dfrac{3\pi(x+1)}{4}+\Theta_2(x+1)2^\frac{x+1}{2}\cos\dfrac{3\pi(x+1)}{4}}{\Theta_1(x)2^\frac{x}{2}\sin\dfrac{3\pi x}{4}+\Theta_2(x)2^\frac{x}{2}\cos\dfrac{3\pi x}{4}}+3$ , where $\Theta_1(x)$ and $\Theta_2(x)$ are arbitrary periodic functions with unit period
$f(x)=\dfrac{\sqrt2\sin\dfrac{3\pi(x+1)}{4}+\Theta(x)\sqrt2\cos\dfrac{3\pi(x+1)}{4}}{\sin\dfrac{3\pi x}{4}+\Theta(x)\cos\dfrac{3\pi x}{4}}+3$ , where $\Theta(x)$ is an arbitrary periodic function with unit period
|
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|
How do I find out that the following two matrices are similar? How do I find out that the following two matrices are similar?
$N =
\begin{pmatrix}
0 & 1 & 0 & 0 \\
0 & 0 & 0 & 0 \\
0 & 0 & 0 & 0 \\
0 & 0 & 0 & 0
\end{pmatrix}$
and $M=
\begin{pmatrix}
0 & 0 & 0 & 0 \\
0 & 0 & 0 & 0 \\
0 & 0 & 0 & 1 \\
0 & 0 & 0 & 0
\end{pmatrix}$
I initially tried to think of the left multiplication of a matrix $P$ as a row operation and tried
$P=
\begin{pmatrix}
0 & 0 & 1 & 0 \\
0 & 1 & 0 & 0 \\
1 & 0 & 0 & 0 \\
0 & 0 & 0 & 1
\end{pmatrix}$
such that $PN = \begin{pmatrix}
0 & 0 & 0 & 0 \\
0 & 0 & 0 & 0 \\
0 & 1 & 0 & 0 \\
0 & 0 & 0 & 0
\end{pmatrix}$ but then $PNP^{-1} \neq M$.
My linear algebra is a bit rusty. Is there a more elaborate way to do this?
|
Let $P_{ij}$ be a matrix that switches the $i$th and $j$th rows. Then $P^T_{ij}$ switches the $i$th and $j$th columns. Thus, $M=P^T_{24}P_{13}N$. If you take $P=P^T_{24}P_{13}$, then $PNP^{-1}$= $P^T_{24}P_{13}N (P^T_{24}P_{13})^{-1}$=$M(P^T_{24}P_{13})^{-1}$ = $MP_{13}^{-1}(P^T_{24})^{-1}$. Switching rows and columns is its own inverse, so this is $MP_{13}P^T_{24}$. If the left-action of a matrix is to switch rows, then the right-action of that matrix is to switch columns. So $MP_{13}P^T_{24}$ means switch the 1st and 3rd columns of $M$, then switch the 2nd and 4th rows. But that doesn't affect $M$. So $PNP^{-1}$=$M$.
|
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|
On calculating the limit of the infinite product $\prod_{k=3}^n (1-\tan^4\frac{\pi}{2^k})$
Let $S_n=\prod_{k=3}^n (1-\tan^4\frac{\pi}{2^k})$. What is the value of $\lim_{n \to \infty} S_n$ ?
What I attempted:-
$\log S_n=\sum_{k=3}^n \log (1-\tan^4\frac{\pi}{2^k})$.
Since $\lim_{x \to 0} \frac{\tan x}{x}=1$, $\tan^4\frac{\pi}{2^k}\approx \left(\frac{\pi}{2^k}\right)^4$
Thus, $\log S_n=\sum_{k=3}^n \log (1-\frac{\pi^4}{2^{4k}})\approx \sum_{k=3}^n\left( -\frac{\pi^4}{2^{4k}}\right) $
Taking limit as $n \to \infty$, $\lim_{n \to \infty} \log S_n=\frac{-\pi^4}{3840}$.
Finally, $\lim_{n \to \infty}S_n=e^{\frac{-\pi^4}{3840}}\approx 1-\frac{\pi^4}{3840}$
Am I correct? Is there any other better method which yield the accurate limit? I was told that the exact limit should be one of the $4$ options:- $\frac{\pi^3}{4},\frac{\pi^3}{16}, \frac{\pi^3}{32},\frac{\pi^3}{256}$. I guess the third option is correct as it is giving almost the same value that I have obtained.
|
What you're describing is how to arrive at an approximation of the actual product.
But you can compute the exact value for the product using trigonometric identities.
$$\begin{split}
S_n&=\prod_{k=3}^n (1-\tan^4\frac{\pi}{2^k})\\
&=\prod_{k=3}^n (1-\tan^2\frac{\pi}{2^k})(1+\tan^2\frac{\pi}{2^k})\\
&=\prod_{k=3}^n (\frac {\cos^2(\frac{\pi}{2^k}) - \sin^2(\frac{\pi}{2^k})}{\cos^2 \frac{\pi}{2^k}})(\frac 1 {\cos^2\frac{\pi}{2^k}})\\
&=\prod_{k=3}^n \frac {\cos(\frac {\pi}{2^{k-1}})}{\cos^4 \frac{\pi}{2^k}}\\
&= \frac{\cos\frac \pi 4}{\cos^4 \frac \pi 8}\frac{\cos\frac \pi 8}{\cos^4 \frac \pi {16}}\frac{\cos\frac \pi {16}}{\cos^4 \frac \pi {32}}...\\
&=\cos\frac \pi 4 \prod_{k=3}^n \frac 1 {\cos^3 \frac \pi {2^k}}
\end{split}$$
Now because $$\frac 1 {\cos \theta}=\frac {2 \sin \theta}{\sin 2\theta}$$
you can verify that, for any $0<\theta<\pi$,
(Viète's formula)
$$
\prod_{k=1}^{+\infty} \frac 1 {\cos \frac \theta {2^k}}=\frac \theta {\sin \theta}
$$
Which yields, for $\theta=\frac \pi 4$,
$$
\prod_{k=3}^{+\infty} \frac 1 {\cos \frac \pi {2^k}}=\frac \pi {4\sin \frac \pi 4}
$$
So finally,
$$\lim_{n\rightarrow+\infty}S_n=\left(\cos{\frac \pi 4}\right)\frac {\pi^3} {4^3\sin^3 \frac \pi 4}$$
In other words,
$$\lim_{n\rightarrow+\infty}S_n=\frac {\pi^3}{32}$$
|
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|
Prove that $\frac{1}{2}(x+\frac{a}{x}) \ge \sqrt{a}$ Let x and a be real numbers > 0. Prove that $\frac{1}{2}(x+\frac{a}{x}) \ge \sqrt{a}$
My idea is that I'm going to use $a>b \iff a^2>b^2$ since we are only dealing with postive real numbers we won't run into problems with the root.
$\frac{1}{2}(x+\frac{a}{x}) \ge \sqrt{a} \iff (\frac{1}{2}(x+\frac{a}{x}))^2 = \frac{1}{4}(x+\frac{a}{x})(x+\frac{a}{x})= \frac{1}{4}(x^2+2a+\frac{a^2}{x^2}) \ge \sqrt{a}^2=a$
And from that we get:
$\frac{1}{4}x^2 +\frac{1}{2}a+\frac{a^2}{4x^2} \ge a$
I don't really know how to proceed from here. What about looking at different possibilities for the values of a and x?
Let x>a. Then:
$\frac{1}{4}x^2 +\frac{1}{2}a+\frac{a^2}{4x^2} \ge \frac{1}{4}a^2 +\frac{1}{2}a+\frac{a^2}{4x^2} \ge a \iff \frac{1}{4}a^2x^2 +\frac{1}{2}ax^2+\frac{a^2}{4} \ge ax^2$
And now I don't know how to prcoeed from here and doubt that my approach is correct. Can someone please help me out here? Tanks in advance.
|
Because by AM-GM $$x+\frac{a}{x}\geq2\sqrt{x\cdot\frac{a}{x}}=2\sqrt{a}.$$
|
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|
Show that $CD\perp AB $. Let $\triangle ABC $ with all angles $<90°$. Let $A_1$ the middle of $ [BC] $.
If $\angle BAA_1=30°$ and $D\in [AB] $ s.t. $CD=AB $ show that $CD\perp AB $.
My idea : I draw $A_1T\perp AB $, $T\in [AB]$.
I have to show that $A_1T $ is the middle line in $\triangle CBD $ $\Leftrightarrow$ $A_1T=\frac {CD}{2}=\frac {AB}{2} $.
But $A_1T=\frac{AA_1}{2} $. So I need to prove that $ AA_1=AB $. Now I am stuck.
|
Let $A = (1,0),$ $B=(0,0),$ and $C = \left(2 - \frac2{\sqrt3},\frac23\right).$
Since $2 - \frac2{\sqrt3} \approx 0.845299,$
angles $\angle CAB$ and $\angle CBA$ are acute,
and since the distance from $C$ to $\left(\frac12,0\right)$ is greater than
$\frac12,$ the angle $\angle ABC$ also is acute.
Then $A_1 = \left(1 - \frac1{\sqrt3},\frac13\right).$
Line $AA_1$ has the equation $y = \frac1{\sqrt3}(1 - x),$
which you can confirm by plugging in the coordinates of $A$ and $A_1,$
and the slope of $AA_1$ is
$-\frac1{\sqrt3} = \tan(-30^\circ),$
so $\angle BAA_1 = 30^\circ.$
The circle of radius $AB = 1$ around $C$ has equation
$$\left(x + \frac2{\sqrt3} - 2\right)^2 + \left(y - \frac23\right)^2 = 1.$$
Put $y = 0$ in this equation and we find that
$$\left(x + \frac2{\sqrt3} - 2\right)^2 + \frac49 = 1,$$
$$x + \frac2{\sqrt3} - 2 = \pm \sqrt{\frac59},$$
$$x = 2 - \frac2{\sqrt3} \pm \frac{\sqrt5}3.$$
We have $\frac{\sqrt5}3 \approx 0.745356,$
so one of the solutions for $x$ is
$x_1 = 2 - \frac2{\sqrt3} - \frac{\sqrt5}3 \approx 0.099943.$
Let $D = (x_1, 0).$ Then $D$ lies on the segment $AB$ and $CD = AB = 1,$
but $CD$ is not perpendicular to $AB.$
In fact, if $C$ is any point $(x,y)$ that satisfies $y = \frac1{\sqrt3}(2 - x)$
and $0 < x < 1,$ then all three angles of the triangle are acute and
$\angle BAA_1 = 30^\circ.$
But the only way we have $D$ on the segment $AB$ with $CD = AB$ and $CD \perp AB$
is if $C = \left(1 - \frac{\sqrt3}2,1\right).$
It seems some necessary condition is missing from your problem statement.
|
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|
Compute the given limit as $x$ approaches $\infty$
If $f(x) = 8x^3+3x$ then, $$\lim_{x \to \infty} \frac{f^{-1}(8x)-f^{-1}(x)}{x^{1/3}}$$ is?
My attempt:
It is clear that the function cannot be easily inverted. So, there must be something in the limit given itself that may simplify the problem.
Honestly, I have no clue what to do here.
There are a few things which I could see is that the function has only $1$ root(i.e $0$) and is bijective on $x \in \mathbb R$. But that gave no benefit except showing that the inverse of the function exists.
Any help would be appreciated.
|
Consider the equation
$$y=8x^3+3x$$ As you say, there is only one real root which is given by
$$x(y)=\frac{1}{2} \left(\frac{\sqrt[3]{\sqrt{2} \sqrt{2 y^2+1}+2
y}}{2^{2/3}}-\frac{1}{\sqrt[3]{2} \sqrt[3]{\sqrt{2} \sqrt{2 y^2+1}+2 y}}\right)$$
Now, tedious but doable,for each piece, use Taylor expansion for infinitely large values of $y$ to get
$$x(8y)=\sqrt[3]{y}-\frac{\sqrt[3]{\frac{1}{y}}}{8}+O\left(\frac{1}{y^{4/3}}\right) $$
$$x(y)=\frac{\sqrt[3]{y}}{2}-\frac{\sqrt[3]{\frac{1}{y}}}{4}+O\left(\frac{1}{y^{4/3}}\right)$$
$$x(8y)-x(y)=\frac{\sqrt[3]{y}}{2}+\frac{\sqrt[3]{\frac{1}{y}}}{8}+O\left(\frac{1}{y^{4/3}}\right)$$ will show the limit and also how it is approached.
Edit
May be tricky but without Taylor series for the root of the cubic equation. Let us consider that, for large values of $x$
$$8x^3+3x=8 x^3+3 x+O\left(\frac{1}{x}\right)$$ Now, series reversion gives
$$x(y)=\frac{\sqrt[3]{y}}{2}-\frac{\sqrt[3]{\frac{1}{y}}}{4}+O\left(\frac{1}{y}\right)$$ and continue for the same result.
|
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|
Trying to simplify $\frac{\sqrt{8}-\sqrt{16}}{4-\sqrt{2}} - 2^{1/2}$ into $\frac{-5\sqrt{2}-6}{7}$ I'm asked to simplify $\frac{\sqrt{8}-\sqrt{16}}{4-\sqrt{2}} - 2^{1/2}$ and am provided with the solution $\frac{-5\sqrt{2}-6}{7}$
I have tried several approaches and failed. Here's one path I took:
(Will try to simplify the left hand side fraction part first and then deal with the $-2^{1/2}$ later)
$\frac{\sqrt{8}-\sqrt{16}}{4-\sqrt{2}}$
The root of 16 is 4 and the root of 8 could be written as $2\sqrt{2}$ thus:
$\frac{2\sqrt{2}-4}{4-\sqrt{2}}$
Not really sure where to go from here so I tried multiplying out the radical in the denominator:
$\frac{2\sqrt{2}-4}{4-\sqrt{2}}$ = $\frac{2\sqrt{2}-4}{4-\sqrt{2}} * \frac{4+\sqrt{2}}{4+\sqrt{2}}$ = $\frac{(2\sqrt{2}-4)(4+\sqrt{2})}{16-2}$ =
(I become less certain in my working here)
$\frac{8\sqrt{2}*2(\sqrt{2}^2)-16-4\sqrt{2}}{14}$ = $\frac{8\sqrt{2}*4-16-4\sqrt{2}}{14}$ = $\frac{32\sqrt{2}-16-4\sqrt{2}}{14}$ = $\frac{28\sqrt{2}-16}{14}$
Then add back the $-2^{1/2}$ which can also be written as $\sqrt{2}$
This is as far as I can get. I don't know if $\frac{28\sqrt{2}-16}{14}-\sqrt{2}$ is still correct or close to the solution. How can I arrive at $\frac{-5\sqrt{2}-6}{7}$?
|
$$\begin{align}
\frac{\sqrt{8}-\sqrt{16}}{4-\sqrt{2}}-\sqrt{2}&=\frac{2\sqrt{2}-4}{4-\sqrt{2}}-\sqrt{2}\\
&=\frac{2\sqrt{2}-4}{4-\sqrt{2}}-\frac{4\sqrt{2}-2}{4-\sqrt{2}}\\
&=\frac{-2\sqrt{2}-2}{4-\sqrt{2}}\\
&=\frac{-2\sqrt{2}-2}{4-\sqrt{2}}~\cdot~\frac{4+\sqrt{2}}{4+\sqrt{2}}\\
&=\frac{-10\sqrt{2}-12}{14}\\
&=\frac{-5\sqrt{2}-6}{7}
\end{align}$$
|
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|
Polynomial Long Division Confusion (simplifying $\frac{x^{5}}{x^{2}+1}$) I need to simplify \begin{equation}
\frac{x^{5}}{x^{2}+1}
\end{equation}
by long division in order to solve an integral.
However, I keep getting an infinite series:
\begin{equation}
x^{3}+x+\frac{1}{x}-\frac{1}{x^{3}}+...
\end{equation}
|
When dividing one polynomial by another, you expect a polynomial quotient. For example, when you divide $x^3+x+1$ by $x+1$,
*
*First multiply $x+1$ by $x^2$ and subtract the result, $x^3+x^2$, from $x^3+x+1$, giving the remainder $-x^2+x+1$;
*Now, multiply $x+1$ by $-x$ and subtract the result, $-x^2-x$, from $-x^2+x+1$, giving the remainder $2x+1$;
*As a final step, multiply $x+1$ by $2$ and subtract the result, $2x+2$, from $2x+1$, giving you the remainder $-1$. Thus,$$x^3+x+1=(x+1)(x^2-x+2)-1$$
Note that you don't go beyond the last step, because the degree of the remainder, $2,(0)$ is smaller than the degree of the divisor, $x+1, (1)$. To go any further, you will have to multiply $x+1$ by terms containing negative powers of $x$, which would mean the quotient will no longer be a polynomial in $x$. Moreover, there is no guarantee that this procedure will terminate, as you have seen.
The essence of performing long-division to solve the integral $$\int\frac{x^5}{x^2+1}dx$$ is to express $x^5=(x^2+1)Q(x)+R(x)$, where $Q(x),R(x)$ are polynomials in $x$ with degree of $R(x)<2$, which is the degree of the divisor, $x^2+1$.
This gives $$\frac{x^5}{x^2+1}=Q(x)+\frac{R(x)}{x^2+1}$$where $Q(x)$ can be easily integrated because it is a polynomial, and $\displaystyle\frac{R(x)}{x^2+1}$ can be integrated using partial fractions or similar techniques.
As other answers have pointed out, $$\int\frac{x^5}{x^2+1}dx=\int x^3-x+\frac x{x^2+1}dx=\frac{x^4}4-\frac{x^2}2+\frac12\ln(x^2+1)+C$$
|
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|
Find all polynomials $p(x)$ such that $(p(x))^2 = 1 + x \cdot p(x + 1)$ for all $x\in \mathbb{R}$
Find all polynomials $p(x)$ such that $(p(x))^2 = 1 + x \cdot p(x + 1)$ for all $x\in \mathbb{R}$.
I assume that $O(p(x)=n)$ (Where O(p) denotes the order of p)
let $p(x)=a_nx^n+a_{n-1}x^{n-1}+a_{n-2}x^{n-2} \dots \dots +a_{1}x^{1}+a_{n-1}$ also $a_n \neq 0$
Then we have $a_n^2x^{2n} +\dots +a_0^2=1+a_nx^{n+1}+ \dots a_0x$
since $a_n \neq 0$ we must have $2n=n+1$ and hence $n=1$ hence the polynomial must be linear and hence there doesn't exist any solution(I proved assuming $p(x)=ax+b$)
Is it correct?
|
Our OP SunShine has correctly deduced that $\deg p(x) = 1$, so that
$p(x) = ax + b; \tag 1$
we compute:
$p^2(x) = a^2x^2 + 2ab x + b^2; \tag 2$
$p(x + 1) = ax + a + b; \tag 3$
$xp(x + 1) = ax^2 + ax + bx; \tag 4$
$xp(x + 1) + 1 = ax^2 + ax + bx + 1; \tag 5$
$a^2 x^2 + 2ab x + b^2 = ax^2 + ax + bx + 1; \tag 6$
$(a^2 - a)x^2 + 2abx + b^2 = ax + bx + 1; \tag 7$
$a^2 = a, \; b^2 = 1, \; 2ab = a + b; \tag 8$
$a = 0, 1; \; b = \pm 1; \tag 9$
$b = 1 \Longrightarrow 2a = a + 1 \Longrightarrow a = 1 \in \{0, 1 \}; \tag{10}$
$b = -1 \Longrightarrow -2a = a - 1 \Longrightarrow a = \dfrac{1}{3} \notin \{0, 1 \}; \tag{11}$
thus the only solution is $a = b = 1$, so that
$p(x) = x + 1. \tag{12}$
CHECK:
$p^2(x) = (x + 1)^2 = x^2 + 2x + 1$
$= x(x + 2) + 1 = x((x + 1) + 1) + 1 = xp(x + 1) + 1; \tag{13}$
it thus appears that the only solution is
$p(x) = x + 1. \tag{14}$
|
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|
Minimum value of $\frac{4}{4-x^2}+\frac{9}{9-y^2}$ Given $x,y \in (-2,2)$ and $xy=-1$
Minimum value of $$f(x,y)=\frac{4}{4-x^2}+\frac{9}{9-y^2}$$
My try:
Converting the function into single variable we get:
$$g(x)=\frac{4}{4-x^2}+\frac{9x^2}{9x^2-1}$$
$$g(x)=\frac{4}{4-x^2}+1+\frac{1}{9x^2-1}$$
Using Differentiation we get:
$$g'(x)=\frac{8x}{(4-x^2)^2}-\frac{18x}{(9x^2-1)^2}$$
$$g'(x)=2x\left(\frac{4(9x^2-1)^2-9(4-x^2)^2}{(4-x^2)^2(9x^2-1)^2}\right)$$
$$g'(x)=70x\frac{9x^4-4}{(4-x^2)^2(9x^2-1)^2}=0$$
So the critical points are:
$x=0, x=\pm \sqrt{\frac{2}{3}}$
But $x \ne 0$ since $xy=-1$
$$g'(x)=70x \frac{(3x^2-2)(3x^2+2)}{(4-x^2)^2(9x^2-1)^2}$$
By using derivative test we get Minimum occurs when $x=\pm \sqrt{\frac{2}{3}}$
Hence $$x^2=\frac{2}{3}, y^2=\frac{3}{2}$$
Min value is $$\frac{4}{4-\frac{2}{3}}+\frac{9}{9-\frac{3}{2}}=\frac{12}{5}$$
Is there any other approach?
|
With the change of variable $t=x^2$, and differentiating the logarithm of the function, things should be a bit easier:
$$h(t)=\frac4{4-t}+\frac{9t}{9t-1},t\in\left(\frac14,4\right)$$
Then
$$h(t)=\frac{-9t^2+72t-4}{(4-t)(9t-1)}$$
$$\log h(t)=\log(-9t^2+72t-4)-\log(4-t)-\log(9t-1)$$
$$(\log h(t))'=\frac{-18t+72}{-9t^2+72t-4}+\frac1{4-t}-\frac9{9t-1}=\frac{315t^2-140}{\text{Irrelevant}}$$
This is $0$ for $t=\frac23$
|
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|
A verification request upon a combinatorial problem An urn contains $100$ balls numbered from $1$ to $100$. Four are removed at random without being replaced. Find the probability that the number on the last ball is smaller than the number on the first
ball.
MY ATTEMPT
If the first removed ball is $2$, then there is $1$ possibility for the last ball and $98\cdot 97$ for the other two. Analogously, if the first removed ball is $3$, then there are $2$ possibilities for the last ball and $98\cdot 97$ for the other two. According to this reasoning, there are
\begin{align*}
1\cdot 1\cdot 98\cdot 97 + 1\cdot 2\cdot 98\cdot 97 + \ldots + 1\cdot 99\cdot 98\cdot 97 = \left[\frac{(1+99)\cdot99}{2}\right]\cdot98\cdot 97
\end{align*}
possible results which satisfy the given restriction. Consequently, the sought probability is
\begin{align*}
\textbf{P}(E) = \frac{50\cdot 99\cdot 98\cdot 97}{100\cdot 99\cdot 98 \cdot 97} = \frac{1}{2}
\end{align*}
Could someone please tell me if I am right? Thanks in advance.
|
Here is a simple argument:
Either the number on the last ball is greater than the number on the last ball or less than the first ball's number. Hence the required probability is $\frac12$.
|
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|
If $x \geq y>x/2$ then is it true that $x \pmod y < x/2$? I'm not sure how to prove this statement, which I believe is true:
given $x,y \in Z$ such that $x \geq y > \frac {x}{2}$ then $x$ (mod $y$) < $\frac {x}{2}.$
Edit:
Would this be an acceptable sketch of the proof?
Suppose that $x \geq y \geq \frac{x}{2};$ consider the bounds where $y = x$ or $y = \frac{x}{2}$.
In the case in which $x=y,$ then $x$ (mod $y$) = $x$ (mod $x) = 0$.
On the other hand, if $y = \frac{x}{2},$ then $x$ (mod $y$) = $x$ (mod $\frac {x}{2}) = \frac {x}{2}$.
Therefore, $0 \leq x$ (mod $y$) < $\frac {x}{2}$.
|
This is true. Suppose that $0\leq \frac{x}{2} < y \leq x$. Long divide $x$ by $y$ by the Division Algorithm to write $x=yq+r$ where $0 \leq r <y$. Here, $r$ is your $x \bmod y$. We claim that $r < \frac{x}{2}$.
Suppose not, so that $r \geq \frac{x}{2}$. Then $x=yq+r \geq yq+\frac{x}{2}$. Subtracting $\frac{x}{2}$ from each side gives $\frac{x}{2} \geq yq$. But $y >\frac{x}{2}$ so this gives $\frac{x}{2} > \frac{x}{2}q$ so $q < 1$. Hence $q=0$ and $x=r < y \leq x$, a contradiction.
|
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|
How to find the range of $y=\frac{x^2+1}{x+1}$ without using derivative? The only thing I know with this equation is $y=\frac{x^2+1}{x+1}=x+1-\frac{2x}{x+1}$.
Maybe it can be solved by using inequality.
|
Out of the given answers, one simple way is to get the $x$ in terms of $y$ and solve it :)
$y=\frac{x^2+1}{x+1}$
$$\Rightarrow yx + y = x^2 +1 \Rightarrow -x^2 + yx +(y-1)=0 \Rightarrow x = \frac{-y \pm\sqrt{y^2 + 4(y-1)}}{-2} \Rightarrow \frac{-y \pm \sqrt{(y-(-2-\sqrt5))(y - (\sqrt5-2))}}{-2}$$
The Lowest Value so will be obtained will be when the value under square root = 0, then in that case
$$ = \frac{-\sqrt5+2}{-2}$$
Now plug that value in the expression for $x$ Gives the minimal value of $y$:
|
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|
Solving a Cauchy problem, differential equation I have the following Cauchy problem
\begin{cases} y'(x) + \frac{1}{x^2-1}y(x) = \sqrt{x+1} \\ y(0) = 0 \end{cases}
I proceed by finding $e^{A(x)} $ where $A(x)$ is the primitive of $a(x)= \frac{1}{x^2-1}$ :
$$\int A(x)dx=\int \frac{1}{x^2-1}dx= \frac{1}{2} \log\Big(\frac{|x-1|}{|x+1|}\Big)+c $$
then I obtain : $$e^{A(x)}=e^{\frac{1}{2} \log\Big(\frac{|x-1|}{|x+1|}\Big)}=\Big(\frac{|x-1|}{|x+1|}\Big)^{\frac{1}{2}}=\sqrt{\frac{|x-1|}{|x+1|} }$$
I have attempted to solve it in this way:
$$ \sqrt{\frac{|x-1|}{|x+1|} }\cdot y'(x) + \frac{1}{x^2-1}\cdot \sqrt{\frac{|x-1|}{|x+1|} }y = \sqrt{x+1}\cdot\sqrt{\frac{|x-1|}{|x+1|} }$$
$$\sqrt{\frac{|x-1|}{|x+1|} }*y(x) =\int \sqrt{\frac{x+1}{|x+1|}}\cdot\sqrt{|x-1|}dx$$
$$y(x) =\Big(\sqrt{\frac{|x-1|}{|x+1|} }\Big)^{-1}\cdot\int \sqrt{\frac{x+1}{|x+1|}}\cdot\sqrt{|x-1|}dx$$
Is it correct doing this? From here I am not sure how to proceed. Thanks in advance for any help.
|
Note that since you have on the right side $\sqrt{x+1}$, we may assume that $x\geq -1$. Moreover the coefficient $\frac{1}{x^2-1}$ implies that $x\not=\pm 1$. Since the initial point is given at $x=0$, the interval $I$ of existence of your solution is contained in $(-1,1)$. Hence you may decide the sign of the arguments of the absolute values and, according to your attempt (which is correct),
$$\sqrt{\frac{1-x}{1+x}}\cdot y(x) =\int \sqrt{1-x}\,dx.$$
Can you take it from here? ... and do not forget the constant of integration!
|
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|
Problem on characteristic polynomial of a matrix Without computing $\det(xI-A)$, how to find the characteristic polynomial of $A$ where $A$ is a $4 \times 4$ matrix given by:
$$A = \begin{bmatrix} 0 & 0 & 0 & -4 \\ 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 5 \\ 0 & 0 & 1 & 0 \end{bmatrix}$$
|
To compute the characteristic polynomial one generally use the Faddeev-LeVerrier algorithm
In your peculiar case however, your matrix has the form of a companion matrix:
$$
A=\begin{bmatrix}0&0&\dots &0&-c_{0}\\1&0&\dots &0&-c_{1}\\0&1&\dots &0&-c_{2}\\\vdots &\vdots &\ddots &\vdots &\vdots \\0&0&\dots &1&-c_{{n-1}}\end{bmatrix}
$$
and by identification the characteristic polynomial is
$$
P(\lambda)=\sum_{k=0}^m c_k \lambda^k=4-5\lambda^2+\lambda^4
$$
In the general case if you want to play the game of not using $\det(\lambda I -A)$ you can use the Jacobi formula which can be obtained by "unfoding" the recurrence relation of the Faddeev-LeVerrier algorithm:
For a $A$ a $n\times n$ matrix, you have
$$
c_{n-m}={\frac {(-1)^{m}}{m!}}{\begin{vmatrix}\text{tr} A&m-1&0&\cdots \\\ \text{tr} A^{2}&\text{tr} A&m-2&\cdots \\\ \vdots &\vdots &&&\vdots \\\ \text {tr} A^{m-1}&\operatorname {tr} A^{m-2}&\cdots &\cdots &1\\\ \text {tr} A^{m}&\operatorname {tr} A^{m-1}&\cdots &\cdots &\operatorname {tr} A\end{vmatrix}}
$$
However there are a lot of computations and it is easier to directly compute $\det(\lambda I -A)$.
Anyway, here is the details for your example:
$$
A=\left(
\begin{array}{cccc}
0 & 0 & 0 & -4 \\
1 & 0 & 0 & 0 \\
0 & 1 & 0 & 5 \\
0 & 0 & 1 & 0 \\
\end{array}
\right)
$$
*
*$m=0$ you have $c_4=1$ (the domimant coefficient of the polynomial is always $1$)
*$m=1$, $\text{tr}(A)=0$, hence $c_{4-1}=c_3=0$
*$m=2$, $\text{tr}(A^2)=10$
$$
c_{4-2}=\frac{(-1)^2}{2!}\left|
\begin{array}{cc}
0 & 2-1 \\ 10 & 0
\end{array}
\right|=\frac{1}{2}(-10)=-5
$$
*$m=3$, $\text{tr}(A^3)=0$
$$
c_{4-3}=\frac{(-1)^3}{3!}\left|
\begin{array}{ccc}
0 & 3-1 & 0 \\ 10 & 0 & 3-2 \\
0 & 10 & 0
\end{array}
\right|=\frac{-1}{6}(0)=0
$$
*$m=4$, $\text{tr}(A^4)=34$
$$
c_{4-4}=\frac{(-1)^4}{4!}\left|
\begin{array}{cccc}
0 & 4-1 & 0 & 0 \\
10 & 0 & 4-2 & 0 \\
0 & 10 & 0 & 4-3 \\
34 & 0 & 10 & 0 \\
\end{array}
\right|=\frac{1}{24}(96)=4
$$
Your characteristic polynomial is
$$
P(\lambda)=4-5\lambda^2+\lambda^4
$$
which is hopefully the same answer as the "companion matrix" identification trick.
|
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|
If $\cos^4 \alpha+4\sin^4 \beta-4\sqrt{2}\cos \alpha \sin \beta +2=0$, then find $\alpha$, $\beta$ in $(0,\frac\pi2)$
If $\cos^4 \alpha+4\sin^4 \beta-4\sqrt{2}\cos \alpha \sin \beta +2=0$,
where $\displaystyle \alpha, \beta \in \bigg(0,\frac{\pi}{2}\bigg)$. Then value of $\alpha,\beta$ are
Try: I am trying to convert it into sum of square of quantity
like $$(\cos^2 \alpha)^2+(2\sin^2 \beta)^2-2\cdot \cos^2 \alpha \cdot 2\sin^2 \beta-4\sqrt{2}\cos \alpha \sin \beta +2+4\cos^2 \alpha \cdot \sin^2 \beta$$
$$(\cos^2 \alpha--2\sin^2 \beta)^2-4\sqrt{2}\cos \alpha \sin \beta+4\cos^2 \alpha \cdot \sin^2 \beta$$
Now i did not know how to solve it, could some help me
|
Hint:
For real $\cos^2\alpha,\sin^2\beta$
$$\dfrac{\cos^4\alpha+4\sin^4\beta+1+1}4\ge\sqrt[4]{\cos^4\alpha\cdot4\sin^4\beta}$$
The equality will occur if $\cos^4\alpha=4\sin^4\beta=1$
|
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|
How to see $x^6-1=(x^2−1)(x^2+x+1)(x^2−x+1)$? I was reading an example where the purpose was to compute a certain Galois group. Along the way, the writer says : note $x^6-1=(x^2−1)(x^2+x+1)(x^2−x+1)$. But how do I note this? I understand you can factorize by $x^2-1$, since when I draw on the unit circle I see that $-1$ and $+1$ are roots. But for the rest?
Edit :
I see you can then factor $(x^2-1)(x^4+x^2+1)$ and than substitute $x^2=y$ and solve quadratic equation but can you actually see the solution visually?
|
Also, by using your idea we obtain:
$$x^6-1=(x^2-1)(x^4+x^2+1)=(x^2-1)(x^4+2x^2+1-x^2)=$$
$$=(x^2-1)((x^2+1)^2-x^2)=(x^2-1)(x^2-x+1)(x^2+x+1).$$
|
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|
Find all positive triples of positive integers a, b, c so that $\frac {a+1}{b}$ , $\frac {b+1}{c}$, $\frac {c+1}{a}$ are also integers. Find all positive triples of positive integers a, b, c so that $\frac {a+1}{b}$ , $\frac {b+1}{c}$, $\frac {c+1}{a}$ are also integers.
WLOG, let a$\leqq b\leqq c$,
|
Hint: given $a \le b$ and $\frac {a+1}b$ is an integer, you must have $b=a+1$ or $a=b=1$.
|
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|
Verifying $ \sum_{k=0}^{\infty}{3k \choose k}\frac{9k^2-3k-1}{(3k-1)(3k-2)}\left(\frac{2}{27}\right)^k=\frac{1}{4}$ $$
\sum_{k=0}^{\infty}{3k \choose k}\dfrac{9k^2-3k-1}{(3k-1)(3k-2)}\left(\dfrac{2}{27}\right)^k=\dfrac{1}{4}
$$
After some simplification, I got the following result:
$$
\sum_{k=0}^{\infty}\left\{{3k \choose k}+\dfrac{27}{2k}{3k-4 \choose 2k-3}\right\}\left(\dfrac{2}{27}\right)^k.
$$
Now, how I can proceed?
|
By Lagrange's inversion theorem (see Bring radical for the quintic analogue)
$$ \sum_{k\geq 0}\binom{3k}{k}\left(\frac{4}{27}\right)^k x^{2k} = \frac{\cos\left(\frac{1}{3}\arcsin x\right)}{\sqrt{1-x^2}} \tag{1}$$
and by partial fraction decomposition
$$ \frac{9k^2-3k-1}{(3k-1)(3k-2)} = 1+\frac{1}{3k-1}+\frac{1}{3k-2}\tag{2} $$
hence by evaluating $(1)$ at $x=\frac{1}{\sqrt{2}}$ we immediately get
$$ \sum_{k\geq 0}\binom{3k}{k}\left(\frac{2}{27}\right)^k = \frac{1+\sqrt{3}}{2}. \tag{3}$$
By reindexing the original series equals
$$-\frac{1}{2}+\sum_{k\geq 0}\binom{3k}{k}\frac{3(3k+2)(3k+1)}{(2k+2)(2k+1)}\left[1+\frac{1}{3k+1}+\frac{1}{3k+2}\right]\left(\frac{2}{27}\right)^k \tag{4}$$
hence it is enough to compute the integrals over $(0,1/\sqrt{2})$ of the RHS of $(1)$ and of the RHS of $(1)$ multiplied by $x$. They are $\frac{3}{2\sqrt{2}}(\sqrt{3}-1)$ and $\frac{3}{16}(5-2\sqrt{3})$ by the substitution $x\mapsto\sin\theta$. By re-combining these pieces the claim is proved.
|
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|
Creating an integral to represent the volume of the intersection of two balls in cartesian coordinates The question states:
Let $A$ be the intersection of the balls
$x^2+y^2+z^2\leq 9$ and $x^2+y^2+(z-8)^2\leq 49$
I am asked to just set up the iterated triple integral that represents the volume of $A$ in cartesian coordinates.
What I am able to determine so far is that for the equation
$x^2+y^2+z^2\leq 9 $:
$-\sqrt{9-x^2-y^2}\leq z \leq \sqrt{9-x^2-y^2}$
$-\sqrt{9-x^2}\leq y \leq \sqrt{9-x^2}$
$-3\leq x \leq 3$
If I were to set up this integral it would be:
$\int_{-3}^{3}\int_{-\sqrt{9-x^2}}^{\sqrt{9-x^2}} \int_{-\sqrt{9-x^2-y^2}}^{\sqrt{9-x^2-y^2}} dzdydx$
But I don't know how I'm supposed to set up the intersection of the two balls?
I was thinking of setting the two equations equal to each other so that
$x^2+y^2\leq 9-z^2$ and
$x^2+y^2\leq 49-(z-8)^2$
so then I have $9-z^2=49-(z-8)^2$
Solving for $z$ I get $z=3/2$ but I don't know what to do with this information.
|
\begin{align}
V &= \int_{8-7}^{\frac32}\left(\int\int_{x^2+y^2 \le 49-(z-8)^2}\mathrm dx\mathrm dy\right)\mathrm dz + \int_{\frac32}^{3}\left(\int\int_{x^2+y^2 \le 9-z^2}\mathrm dx\mathrm dy\right)\mathrm dz\\
&=\int_{1}^{\frac32}\int_{-\sqrt{49-(z-8)^2}}^{\sqrt{49-(z-8)^2}}\int_{-\sqrt{49-(z-8)^2-y^2}}^{\sqrt{49-(z-8)^2-y^2}}\mathrm dx\mathrm dy\mathrm dz + \int_{\frac32}^{3}\int_{-\sqrt{9-z^2}}^{\sqrt{9-z^2}}\int_{\sqrt{9-z^2-y^2}}^{\sqrt{9-z^2-y^2}}\mathrm dx\mathrm dy\mathrm dz.
\end{align}
The idea is similar to what you mentioned when you wanted to compute each ball separately. The first term is the part of the intersection that is clipped by the second ball and the second term is the part clipped by the first ball.
|
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|
What's the answer to $\int \frac{\cos^2x \sin x}{\sin x - \cos x} dx$? I tried solving the integral $$\int \frac{\cos^2x \sin x}{\sin x - \cos x}\, dx$$ the following ways:
*
*Expressing each function in the form of $\tan \left(\frac{x}{2}\right)$, $\cos \left(\frac{x}{2}\right)\,$ and $\,\sin \left(\frac{x}{2}\right)\,$ independently, but that didn't go well for me.
*Multiplying and dividing by $\cos^2x$ or $\sin^2x$.
*Expressing $\cos^2x$ as $1-\sin^2x$ and splitting the integral, and I was stuck with $\int \left(\frac{\sin^3x}{\sin x - \cos x}\right)\, dx$ which I rewrote as $\int \frac{\csc^2x}{\csc^4x (1-\cot x) } dx,\,$ and tried a whole range of substitutions only to fail.
*I tried to substitute $\frac{1}{ \sin x - \cos x}$, $\frac{\sin x}{ \sin x - \cos x}$, $\frac{\cos x \sin x}{ \sin x - \cos x}$ and $\frac{\cos^2x \sin x}{\sin x - \cos x},$ independently, none of which seemed to work out.
*I expressed the denominator as $\sin\left(\frac{\pi}{4}-x\right)$ and tried multiplying and dividing by $\sin\left(\frac{\pi}{4}+x\right)$, and carried out some substitutions. Then, I repeated the same with $\cos\left(\frac{\pi}{4}+x\right)$. Neither of them worked.
|
Let $\displaystyle I =\frac{1}{2}\int\frac{2\cos^2 x\cdot \sin x}{\sin x-\cos x}dx=\frac{1}{2}\int\frac{(1+\cos 2x)\cdot \sin x}{\sin x-\cos x}dx$
So $\displaystyle I =\frac{1}{4}\int \frac{2\sin x}{\sin x-\cos x}dx+\frac{1}{2}\int\frac{\cos 2x\cdot \sin x}{\sin x-\cos x}dx$
Now writting
$2\sin x=(\sin x+\cos x)+(\sin x-\cos x)$
and $\cos (2x)=\cos^2 x-\sin^2 x.$
|
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|
Write a Limit to calculate $f'(0)$ Let $f(x) = \frac {2}{1+x^2} $
I need to write a limit to calculate $f'(0)$.
I think I have the basic understanding. Any help would be greatly appreciated.
d=delta and so far what I have is
$f'(x)$= lim (f(x+dx)-f(x))/dx
(dx)->0
((2/1+(x+dx)^2)-(2/1+x^2))/dx
((2/1+x^2+2xdx+dx^2)-(2/1+x^2))/dx
((2(1+x^2)-2(1+x^2+2xdx+dx^2))/(1+x^2+2xdx+dx^2)(1+x^2))/dx
((-4xdx-2dx^2)/(1+x^2+2xdx+dx^2)(1+x^2))/dx
(-4xdx-2dx^2)/(1+x^2+2xdx+dx^2)(1+x^2)(dx)
(-2dx(2xdx+dx)/(1+x^2+2xdx+x^2)(1+x^2)(dx)
(-2(2xdx+dx)/(1+x^2+2xdx+dx^2)(1+x^2)
that's as far as I have gotten. Any input would be great.
|
$f'(0) = \displaystyle \lim_{h \to 0} \dfrac{f(h) - f(0)}{h} = \lim_{h \to 0} \left (\dfrac{1}{h} \right ) \left ( \dfrac{2}{1 + h^2} - 2 \right )$
$= \lim_{h \to 0} \left (\dfrac{1}{h} \right ) \left ( -\dfrac{2h^2}{1 + h^2} \right ) = \lim_{h \to 0} -\dfrac{2h}{1 + h^2} = 0. \tag 1$
|
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|
Fourier transform of the Klein group Obviously we have the elements of $G = \lbrace e, a, b, c\rbrace\;$ where $\;c=ab$
Define:
$$\chi : G \rightarrow \mathbb{Z}$$
How would I construct the transform for each of the elements of $G$?
What I mean is, for example take another group $G$ as $G = \mathbb{Z_5} = \lbrace0,1,2,3,4\rbrace$
As defined above, let $\chi : G \rightarrow \mathbb{Z} $
Define:
$$\chi(0)=1$$
$$\chi(1)=a$$
$$\chi(2)=a^2$$
$$\chi(3)=a^3$$
$$\chi(4)=a^4$$
Let:
\begin{align*}
\chi_0(x) &= 1 \\[1ex]
\chi_1(x) &=
\begin{cases}
1, & \text{if } x = 0, \\
a, & \text{if } x = 1, \\
a^2, & \text{if } x = 2, \\
a^3, & \text{if } x = 3, \\
a^4, & \text{if } x = 4,
\end{cases}
&
\chi_3(x) &=
\begin{cases}
1, & \text{if } x = 0, \\
a^3, & \text{if } x = 1, \\
a, & \text{if } x = 2, \\
a^4, & \text{if } x = 3, \\
a^2, & \text{if } x = 4,
\end{cases}
\\[1ex]
\chi_2(x) &=
\begin{cases}
1, & \text{if } x = 0, \\
a^2, & \text{if } x = 1, \\
a^4, & \text{if } x = 2, \\
a, & \text{if } x = 3, \\
a^3, & \text{if } x = 4,
\end{cases}
&
\chi_4(x) &=
\begin{cases}
1, & \text{if } x = 0, \\
a^4, & \text{if } x = 1, \\
a^3, & \text{if } x = 2, \\
a^2, & \text{if } x = 3, \\
a, & \text{if } x = 4.
\end{cases}
\end{align*}
Also define:
$\hat f(\chi) = \sum {f}(a)\bar{\chi}(a)$ - The Fourier transform
$f(a)=\frac{1}{\rule{0pt}{0.65em} |G|}\sum \hat f(\chi)\chi_i(a)$ - The inverse Fourier transform
Then we can compute the transforms for each element of $G$.
Now my question is how would I do the same with the Klein group? I mean, can I proceed by doing this:
$$\chi(e)=1$$
$$\chi(a)=\alpha$$
$$\chi(b)=\alpha^2$$
$$\chi(c)=\alpha^3$$
Then proceed like the group $\mathbb{Z}_5$?
EDIT: I just checked and this is not possible as the Cayley table doesn't not match
EDIT: I have solved this problem now.
|
Since $K_4$ is abelian of order $4$, there are exactly four irreducible representations, all of dimension one. Since every element has order two, all values of the characters are $\chi(g)= \pm 1$. This implies the character table will take the form:
\begin{align*}
\begin{array}{c | c c c c }
& e & a & b & c\\
\hline
\chi_1 & 1 & 1 & 1 & 1\\
\chi_2 & 1 & \pm 1 & \pm 1 & \pm 1\\
\chi_3 & 1 & \pm 1 & \pm 1 & \pm 1\\
\chi_4 & 1 & \pm 1 & \pm 1 & \pm 1\\
\end{array}
\end{align*}
The only way to make all the rows of this table orthogonal, is by taking:
\begin{align*}
\begin{array}{c | c c c c }
& e & a & b & c\\
\hline
\chi_1 & 1 & 1 & 1 & 1\\
\chi_2 & 1 & 1 & -1 & -1\\
\chi_3 & 1 & -1 & 1 & -1\\
\chi_4 & 1 & -1 & -1 & 1\\
\end{array}
\end{align*}
Now we are in a position to construct the Fourier transform of the Klein group. Using the character table above and definition of Fourier transform, we have:
$$\hat{f}(\chi_1) = f(e)+f(a)+f(b)+f(c)$$
$$\hat{f}(\chi_2) = f(e)+f(a)-f(b)-f(c)$$
$$\hat{f}(\chi_3) = f(e)-f(a)+f(b)-f(c)$$
$$\hat{f}(\chi_4) = f(e)-f(a)-f(b)+f(c)$$
Computing the inverse Fourier transform using definition of inverse Fourier transform, we have:
$$f(e) = \frac{1}{4}[f(e)+f(e)+f(e)+f(e)] = f(e)$$
$$f(a) = \frac{1}{4}[f(a)+f(a)+f(a)+f(a)] = f(a)$$
$$f(b) = \frac{1}{4}[f(b)+f(b)+f(b)+f(b)] = f(b)$$
$$f(c) = \frac{1}{4}[f(c)+f(c)+f(c)+f(c)] = f(c)$$
as expected.
|
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|
Finding real $a$, $b$, $c$ such that $x^5 - 2x^4 + ax^3 + bx^2 - 2x + c$ has $1+i$ as a zero, and one negative integer as a zero with multiplicity $2$
Find $a, b, c \in \mathbb{R}$, if one zero of $p$ is $1+i$ and $p$ has one negative integer zero with multiplicity $2$, where:
$$p(x) = x^5 - 2x^4 + ax^3 + bx^2 - 2x + c$$
I know that one zero of $p$ is $1-i$, so I can reduce it one degree down after dividing with $x-2$, but after that I'm stuck.
I tried using Vieta's formulas, but with no result.
|
If $ 1 + i $ is the root of the polynomial
$$
p(x) = x^5 - 2x^4 + ax^3 + bx^2 - 2x + c
$$
and $a,b,c\in\mathbb{R}$ then $1-i$ is root of $p(x)$. Soon $q(x)=(x-1+i)\cdot(x-1-i)=(x^2-2x+2)$ divide $p(x)$ and
\begin{align}
x^5 - 2x^4 + ax^3 + bx^2 - 2x + c
=&
(x^2-2x+2)\big( x^3 -(2r+s)x^2+(r^2+2rs)x-r^2s\big)
\\
=&x^5
\\
&\hspace{0.5cm} -2x^4-(2r+s)x^4
\\
&\hspace{1.0cm} +2x^3+2(2r+s)x^3+(r^2+2rs)x^3
\\
&\hspace{1.5cm} -2(2r+s)x^2-2(r^2+2rs)x^2-r^2sx^2
\\
&\hspace{2.0cm} +2(r^2+2rs)x+2r^2sx
\\
&\hspace{2.5cm} +2(r^2+2rs)x+2r^2sx
\\
&\hspace{3.0cm} -2r^2s
\end{align}
Note that
\begin{align}
-2-(2r+s)=&-2\quad \implies \quad 2r+s=0\\
+2+2(2r+s)+(r^2+2rs)=&a\\
-2(2r+s)-2(r^2+2rs)-r^2s=&b\\
+2(r^2+2rs)+2r^2s=&-2 \quad \implies \quad 2r^2+4rs+2r^2s=-2\\
-2r^2s=&c\\
\end{align}
|
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|
Finding out the value of $\angle DQC$ in a trapezium $ABQD$ where $\angle DCB$ = 30$^\circ$ In this below diagram, $\angle ABC=60^\circ, \angle DCB=30^\circ $, $AD$ is parallel to $BC$ and $AP$ is perpendicular to $BC$. Both the area and perimeter of $ABCD$ and $APQD$ are equal . What is the value of /angle $DQC$?
At first, I started with the condition of area of the both the trapezium (mentioned in the question) being equal. As $AD$ is parallel to $BC$, So we can show that,
$\frac{1}{2}(AD+BC)$ ×$AP$ = $\frac{1}{2}(AD+PQ)$ ×$AP$
$AD$+$BC$ = $AD$+$PQ$
$BC$ = $PQ$ .................. (1)
And now, we can show their perimeter are equal with the below equation:
$AB+BC+CD+AD$ = $AP+PQ+DQ+AD$
$AB+BC+CD$ = $AP+PQ+DQ$
$AB+PQ+CD$ = $AP+PQ+DQ$............(from 1)
$CD+AB$ = $AP+DQ$.............(2)
I drew a vertical line $DE$ which is parallel to $AP$. $DE$ is equal to $AP$ as $AD$ and $BC$ are parallel to each other. Here /angle $APB$ and /angle $DEQ$ are respectively right-angle. So, we get two right-angled triangle $APB$ and $DEQ$.
By trigonometry and from the traingle $APB$, we can write that:-
$\frac{AB}{AP}$ = $\mathrm{cosec} 60^\circ$
$AB$ = $\frac{2AP}{\sqrt 3}$..........(3)
Similarly, from $DEC$, we can write that:-
$\frac{DC}{DE}$ = $\mathrm{cosec} 30^\circ$
$DC$ = 2$DE$
$CD$ = 2$DE$..........(4)
Let us denote the $\angle DQE = $$\theta$
So, now from the equation (2), we can get:-
$2AP$ + $\frac{2AP}{\sqrt 3}$ = $AP+ DQ$.........(from 3 and 4)
$\frac{2\sqrt 3AP + 2AP - \sqrt 3AP}{\sqrt 3} $ = $DQ$
$AP × \frac{2+ \sqrt 3}{\sqrt 3} $ = $DQ$
$\frac{2+ \sqrt 3}{\sqrt 3} $ = $\frac{DQ}{AP} $
$\frac{DQ}{DE} $ = $\frac{2+ \sqrt 3}{\sqrt 3} $......($AP = DE$ according to the diagram)
$\frac{DE}{DQ} $ = $\frac{\sqrt 3}{2 +\sqrt 3} $
$\sin \theta$ = $\frac{\sqrt 3}{2} + 1 $.................(5)
We know that if function of x is described as f(x) = $\sin^\text{-1} $x, than function of x will be real and valid if and only if its domain is [-1,1]. But there is a little bit problem in my calculation.
(5) is invalid because the real value of $\sin \theta$ is above 1 which is impossible in this case. For this reason, the value of $\theta$ is unreal. So, there is an extensive mistake either in my calculation or in the question. I have been unable to find my error. So, I need some help to identify the mistake to solve the problem properly.
EDIT: My fault is making equation (5) from the past.
|
\begin{align}
\operatorname{perimeter}(APQD) &= \operatorname{perimeter}(ABCD) \\
(3+\sqrt 3)x + 2y + CQ+DQ &= (6+2\sqrt 3)x + 2y \\
CQ +DQ &= (3+\sqrt 3)x
\end{align}
\begin{align}
\operatorname{area}(APQD) &= \operatorname{area}(ABCD) \\
\frac 12(3x+2y+CQ)(\sqrt 3x) &= \frac 12(4x+2y)((\sqrt 3x)) \\
CQ &= x \\
\hline
DQ &= (2+\sqrt 3)x \\
(DQ)^2 &= (7+4\sqrt 3)x^2
\end{align}
But, then, $(7+4\sqrt 3)x^2=(DQ)^2 = (\sqrt 3x)^2+(4x)^2=19x^2$
So, by contradiction, there is no solution.
|
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|
If $\cos^6 (x) + \sin^4 (x)=1$, find $x$ if $x\in [0, \dfrac {\pi}{2}]$ If $\cos^6 (x) + \sin^4 (x)=1$, find $x$ if $x\in [0, \dfrac {\pi}{2}]$
My attempt:
$$\cos^6 (x) + \sin^4 (x)=1$$
$$\cos^6 (x) + (1-\cos^2 (x))^{2}=1$$
$$\cos^6 (x) + 1 - 2\cos^2 (x) + \cos^4 (x) = 1$$
$$\cos^6 (x) + \cos^4 (x) - 2\cos^2 (x)=0$$
|
Hint If we rewrite $u = \cos^2 x$ and factor, the equation becomes $$(u + 2) u (u - 1) = 0 .$$
Alternative hint We have $\cos^6 x \leq \cos^2 x$ and $\sin^4 x \leq \sin^2 x$, and in both cases equality holds (for $x \in [0, \frac{\pi}{2}]$) only for $x = 0, \frac{\pi}{2}$.
|
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|
Find all integer solutions to $ x^3 - y^3 = 3(x^2 - y^2) $ The objective is to find all solutions to
$$ x^3 - y^3 = 3(x^2 - y^2) $$
where $x,y \in \mathbb{Z}$.
So far I've got one pair of solution. Try $(x, y)=(0,0)$:
$$ 0^3-0^3=3(0^2-0^2) \\
0=0 \qquad \text{equation satisfied}$$
Another try $ (x, y) = (x, x)$ then
$$ x^3 - x^3 = 3(x^2 -x^2) \\
0 = 0 \qquad \text{equation satisfied}$$
But the above are just particular solutions. To find all $(x, y)$ pairs, I tried:
$$\begin{aligned}
(x-y)(x^2+xy+y^2) &= 3(x+y)(x-y) \\
x^2+xy+y^2 &= 3(x+y) \end{aligned}\\
\begin{aligned}
x^2+xy+y^2-3x-3y &= 0 \\
x^2+x(y-3)+y(y-3) &=0 \\
x^2 + (y-3)(x+y) &= 0 \end{aligned}$$
What now? Am I on the right track?
|
By symmetry, we can say that whenever $(x,y)$ is a solution, $(y,x)$ is also a solution. Also, one of them must be positive integer, as $x+y=\frac{x^2+xy+y^2}{3}>0$
Apply quadratic formula in $3$ specific cases, when $y=1,2,3.$ We get $(2,2), (-1,2), (0,3), (3,0), (2,-1)$ as the integer solutions in these cases. There is no other case, as otherwise $y>3$ or $x>3$ and Left Hand Side of the last equation you wrote will be strictly positive.
Hope it is helpful:)
|
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|
Evaluate $\int_{0}^{1}\frac{1+x+x^2}{1+x+x^2+x^3+x^4}dx$ Evaluate $$I=\int_{0}^{1}\frac{(1+x+x^2)}{1+x+x^2+x^3+x^4}dx$$
My try:
We have:
$$1+x+x^2=\frac{1-x^3}{1-x}$$
$$1+x+x^2+x^3+x^4=\frac{1-x^5}{1-x}$$
So we get:
$$I=\int_{0}^{1}\frac{1-x^3}{1-x^5}dx$$
$$I=1+\int_{0}^{1}\frac{x^3(x^2-1)}{x^5-1}dx$$
Any idea from here?
|
The reduced form you gave in the end is a good starting point. Since we are within the interval $[0,1]$ we may expand the denominator as its corresponding geometric series, followed by changing the order of summation and integration and finally by termwise integration. Overall this leads us to
$$I=\int_0^1\frac{1-x^3}{1-x^5}\mathrm dx=\int_0^1\sum_{n=0}^\infty x^{5n}(1-x^3)\mathrm dx=\sum_{n=0}^\infty\int_0^1x^{5n}-x^{5n+3}\mathrm dx\\=\sum_{n=0}^\infty\left[\frac{x^{5n+1}}{5n+1}-\frac{x^{5n+4}}{5n+4}\right]_0^1
=\sum_{n=0}^\infty\left[\frac{1}{5n+1}-\frac{1}{5n+4}\right]$$
The difficult task is to find a closed-form for this sum. Of course, one could invoke the Digamma Function as Zacky quoted, but we can also note the following due the absolute convergence of the series
$$I=\sum_{n=0}^\infty\left[\frac{1}{5n+1}-\frac{1}{5n+4}\right]=1-\frac14+\frac16-\frac19+\cdots=1+\sum_{n=1}^\infty\left[\frac{1}{5n+1}-\frac{1}{5n-1}\right]$$
The latter sum falls down quite easy by considering a well-known series expansion of the cotangent function
$$\pi\cot(\pi z)~=~\frac1z+\sum_{n=1}^{\infty}\frac{2z}{z^2-n^2} $$
Thus, we rewrite the original sum as
\begin{align}
1+\sum_{n=1}^\infty\left[\frac{1}{5n+1}-\frac{1}{5n-1}\right]&=1-\sum_{n=1}^\infty\frac2{25n^2-1}\\
&=1+\frac15\left[\sum_{n=1}^\infty\frac{2\frac15}{\left(\frac15\right)^2-n^2}\right]\\
&=1+\frac15\left[\pi\cot\left(\frac\pi5\right)-5\right]\\
&=\frac\pi5\cot\left(\frac\pi5\right)
\end{align}
$$\therefore~I=\int_0^1\frac{1+x+x^2}{1+x+x^2+x^3+x^4}\mathrm dx=\sum_{n=0}^\infty\left[\frac{1}{5n+1}-\frac{1}{5n+4}\right]=\frac\pi5\sqrt{1+\frac2{\sqrt 5}}$$
|
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|
Evaluate ${\lim_{x \rightarrow 0} \frac{1-\cos x\cos 2x\cdots \cos nx}{x^2}}$
Evaluate $${\lim_{x \rightarrow 0} \frac{1-\cos x\cos 2x\cdots \cos nx}{x^2}}$$
It should be $$\frac{1}{12}n(n+1)(2n+1)$$ but I don't know how to prove that. I am also aware that $\lim\limits_{\theta \to 0} \dfrac{1-\cos \theta}{\theta ^2}=\dfrac{1}{2}$, but I don't know how to use that.
|
Set $f_n(x)=\cos x\cos2x\cdots\cos nx$ for simplicity. We want to prove by induction that
$$
\lim_{x\to0}\frac{1-f_n(x)}{x^2}=\frac{1}{12}n(n+1)(2n+1)
$$
The statement is true for $n=1$. Assume it is true for $n$. Then
$$
f_{n+1}(x)=f_n(x)(\cos nx\cos x-\sin nx\sin x)
$$
Thus
\begin{align}
\frac{1-f_{n+1}(x)}{x^2}
&=\frac{1-f_n(x)\cos nx\cos x}{x^2}+\frac{\sin nx\sin x}{x^2}\\[6px]
&=\frac{1-f_n(x)}{x^2}\cos nx\cos x+\frac{1-\cos nx\cos x}{x^2}+\frac{\sin nx\sin x}{x^2}
\end{align}
The limit of the second summand is, by standard computations, $(n^2+1)/2$; the limit of the third summand is $n$.
Thus, by the induction hypothesis, the limit is
\begin{align}
\frac{1}{12}n(n+1)(2n+1)+\frac{n^2+1}{2}+n
&=\frac{1}{12}n(n+1)(2n+1)+\frac{(n+1)^2}{2}\\[6px]
&=\frac{n+1}{12}(2n^2+n+6n+6)\\[6px]
&=\frac{1}{12}(n+1)(n+2)(2n+3)
\end{align}
as required.
|
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|
Find all possible values of $x$ if $\frac{\sin\frac{x}{2}+\cos\frac{x}{2}-i\tan x}{1+2i\sin\frac{x}{2}}$ is real
If the expression $\dfrac{\sin\frac{x}{2}+\cos\frac{x}{2}-i\tan x}{1+2i\sin\frac{x}{2}}$ is real, find the set of all possible values of $x$.
My Attempt
$$
-\tan x-2\sin^2\frac{x}{2}-2\sin\frac{x}{2}\cos\frac{x}{2}=0\\
\frac{2\sin\frac{x}{2}\cos\frac{x}{2}}{\cos x}+2\sin^2\frac{x}{2}+2\sin\frac{x}{2}\cos\frac{x}{2}=0\\
\sin\frac{x}{2}=0\text{ or }\frac{\cos\frac{x}{2}}{\cos x}+\sin\frac{x}{2}+\cos\frac{x}{2}=0\\
\frac{x}{2}=n\pi\text{ or }\cos\frac{x}{2}+\sin\frac{x}{2}.\cos x+\cos\frac{x}{2}.\cos x=0\\
\boxed{x=2n\pi}\text{ or _____________}
$$
My reference gives the solution $x=2n\pi$, but what about the other remaining expression ?
|
You are right: there are more solutions. Actually, there is exactly one more solution in $(0,2\pi)$. Note that\begin{align}-\tan x-2\sin^2\frac{x}{2}-2\sin\frac{x}{2}\cos\frac{x}{2}=0&\iff-\tan x-1+1-2\sin^2\left(\frac x2\right)-2\sin x=0\\&\iff-\tan x-1+\cos x-\sin x.\end{align}Now, let $c=\cos x$ and let $s=\sin x$. Then$$\left\{\begin{array}{l}c^2+s^2=1\\-\dfrac sc-1+c-s=0\end{array}\right.$$Solving this system leads you to a quartic. It has a trivial solution ($(c,s)=(1,0)$) and a single real non-trivial solution: $(c,s)\simeq(-0.396608,0.917988)$. This corresponds to taking $x\simeq1.97861$.
|
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|
Prove that $\lim_{(x,y)\to (0,0)}\frac{x^{2}+xy+y^{2}}{\sqrt{x^{2}+y^{2}}}=0$.
Prove that $\displaystyle \lim_{(x,y)\to (0,0)}\frac{x^{2}+xy+y^{2}}{\sqrt{x^{2}+y^{2}}}=0$.
This has been my rough work so far, but I am not sure how to go further or if I am doing it the wrong way...
$\displaystyle|f(x,y)-L|=\left |\frac{x^{2}+xy+y^{2}}{\sqrt{x^{2}+y^{2}}} \right | \leq \frac{|x^{2}+xy+y^{2}|}{\big|\sqrt{x^{2}+y^{2}}\big|}\cdot \frac{\sqrt{x^{2}+y^{2}}}{\sqrt{x^{2}+y^{2}}} = \frac{\big(\sqrt{x^{2}+y^{2}}\big)|x^{2}+xy+y^{2}|}{|x^{2}+y^{2}|}$
I wanted the square root on top to help determine what I should set my $\delta$ as. Any ideas on how to finish this or do it in a better way?
|
To begin with, notice that
\begin{align*}
\begin{cases}
|x| = \sqrt{x^{2}} \leq \sqrt{x^{2}+y^{2}}\\\\
|y| = \sqrt{y^{2}} \leq \sqrt{x^{2}+y^{2}}
\end{cases}\Longrightarrow
\begin{cases}
\displaystyle\frac{|x|}{\sqrt{x^{2}+y^{2}}} \leq 1\\\\
\displaystyle\frac{|y|}{\sqrt{x^{2}+y^{2}}} \leq 1
\end{cases}
\end{align*}
Furthermore, the given expression can be split as the following sum
\begin{align*}
E(x,y) = \frac{x^{2}+2xy+y^{2}}{\sqrt{x^{2}+y^{2}}} = \frac{x^{2}}{\sqrt{x^{2}+y^{2}}} + \frac{2xy}{\sqrt{x^{2}+y^{2}}} + \frac{y^{2}}{\sqrt{x^{2} + y^{2}}}
\end{align*}
As a consequence of the squeeze theorem applied to each summand, the given limit tends to zero.
|
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|
Integrate using trigonometric substitution. Am I on the right path? I have been trying to solve:
$$\int \frac{\sqrt{x^2-9}}{x^3} dx$$
I am letting $ x = 3\sec \theta$ and so $dx = 3 \sec \theta \tan \theta$
So then I have:
$$\int \frac{\sqrt{9\sec^2 \theta - 9}}{27 \sec^3 \theta} dx$$
$$\int \frac{\sqrt{9(\sec^2 \theta - 1)}}{27 \sec^3 \theta} 3 \sec \theta \tan \theta\, d \theta$$
$$ \int \frac{3\tan \theta}{27 \sec^3 \theta} 3 \sec \theta \tan \theta\, d \theta$$
$$ = \int \frac{9 \tan ^2 \theta}{27 \sec ^2 \theta} d \theta$$
$$ = \int \frac{\tan ^2 \theta}{3 \sec ^2 \theta} d \theta$$
$$ \int \frac{\sin^2 \theta}{\cos^2 \theta} \cdot \frac{\cos^2 \theta}{3} d \theta$$
$$ \int \frac{\sin^2 \theta}{3} d \theta$$
$$\frac{1}{3} \int \sin^2 \theta d \theta$$
Am I on the right track?
|
Yes, your solution so far is correct. Now, to integrate $\int \sin^2{\theta}\,d\theta$, use the half-angle formula for the sine function:
$$
\sin^2{\theta}=\frac{1-\cos(2\theta)}{2}.
$$
Also, a bit later, you're going to need this formula:
$$
\sin{(2\theta)}=2\sin{\theta}\cos{\theta}.
$$
|
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|
Why $\frac{1}{2} \arctan(x) = \arctan(x-\sqrt{x^2+1}) + \frac{\pi}{4}$? Since
$$\dfrac{d}{dx} \left( \dfrac{1}{2} \arctan(x) \right) = \dfrac{d}{dx} \left( \arctan(x-\sqrt{x^2+1}) \right) $$
then the format of their graphs are the same, but separated by a constant, which is $\dfrac{\pi}{4}$. That is,
$$\dfrac{1}{2} \arctan(x) = \arctan(x-\sqrt{x^2+1}) + \dfrac{\pi}{4} $$
My question is: why is this constant and what procedure is done to discover it.
|
It's even nicer if you use simple trigonometry.
Consider the following Figure, for positive $x$ (for negative $x$ see edit, thanks to Steven for pointing this out in comment).
$\hskip1.in$
The triangle $OAB$ is right-angled, and such that $\overline{OA} = 1$, and $\overline{AB} = x$. We have
\begin{equation}\angle BOA = 2\alpha = \arctan x\tag{1}\label{eq:0}\end{equation}
and
$$\overline{OB} = \sqrt{x^2+1}.$$
Extend $BA$ to a segment $BC$ such that
$$ BC \cong OB.$$
Then
$$\overline{AC} = \sqrt{1+x^2}-x,$$
and so
\begin{equation}\beta = \angle AOC = \arctan\left(\sqrt{1+x^2}-x\right).\tag{2}\label{eq:0p}\end{equation}
By considering the right-angled triangle $OAB$, you
obtain
\begin{equation}\angle OBA = \frac{\pi}{2}-2\alpha,\tag{3}\label{eq:1}\end{equation}
and, by considering the isosceles triangle $OBC$, you get
\begin{equation}\angle OBA = \pi - 2(2\alpha+\beta).\tag{4}\label{eq:2}\end{equation}
Equating \eqref{eq:1} and \eqref{eq:2} yields
$$\alpha = -\beta + \frac{\pi}{4}.$$
Plugging in \eqref{eq:0} and \eqref{eq:0p} and using the odd symmetry of the $\arctan(\cdot )$ function leads to
$$ \frac{1}{2}\cdot \arctan x = \arctan\left(x-\sqrt{1+x^2}\right) + \frac{\pi}{4},$$
the desired result. $\blacksquare$
EDIT
The same result applies if $x<0$. Use then the Figure below.
$\hskip1.in$
Now $\overline{AB} = -x$, $\overline{OB} = \overline{BC} = \sqrt{1+x^2}$, and $\overline{CA} = \sqrt{1+x^2}-x$. Then, again using right-angled triangle $OAC$ and isosceles triangle $OBC$, we get
$$\angle OCA = \frac{\pi}{2}-\beta = \beta - 2\alpha,$$
where $2\alpha = \angle BOA$, and $\beta = \angle COA$. The equality of OP follows.
|
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|
$\sum_{n \leq x} \frac {\log n}{n^3} = C - \frac {\log x}{2x^2} -\frac {1}{4x^2} +O \bigg (\frac {\log x}{x^3} \bigg)$ I try to prove the following equation:
for $x \geq 2$
$$\sum_{n \leq x} \frac {\log n}{n^3} = C - \frac {\log x}{2x^2} -\frac {1}{4x^2} +O \bigg (\frac {\log x}{x^3} \bigg)$$
To my opinion I have to use Euler's formula, but I can't get to the right equation.
Any help appreciated
|
From Euler's summation formula (also page 50 here)
$$\sum\limits_{y< n\leq x}f(n)= \int\limits_{y}^{x}f(t)dt + \int\limits_{y}^{x}\{t\}f'(t)dt-\{x\}f(x)+\{y\}f(y) \tag{1}$$
and
$$\int \frac{\ln{x}}{x^3} dx = -\frac{\ln{x}}{2x^2}- \frac{1}{4 x^2} + C \tag{2}$$
we have
$$\sum\limits_{2\leq n \leq x}\frac{\ln{n}}{n^3}=
\int\limits_{2}^{x}\frac{\ln{t}}{t^3}dt + \int\limits_{2}^{x}\{t\}\frac{1-3\ln{t}}{t^4}dt-\{x\}\frac{\ln{x}}{x^3}+\{2\}\frac{\ln{2}}{2^3}=\\
-\frac{\ln{x}}{2x^2}-\frac{1}{4x^2}-\left(-\frac{\ln{2}}{2\cdot2^2}-\frac{1}{4\cdot2^2}\right)+\int\limits_{2}^{x}\{t\}\frac{1-3\ln{t}}{t^4}dt-\{x\}\frac{\ln{x}}{x^3}=\\
\frac{1+\ln{4}}{16}-\frac{\ln{x}}{2x^2}-\frac{1}{4x^2}+\int\limits_{2}^{x}\{t\}\frac{1-3\ln{t}}{t^4}dt-\{x\}\frac{\ln{x}}{x^3}=... \tag{3}$$
Now, $\forall x\geq2$
$$\left|\{x\}\frac{\ln{x}}{x^3}\right|\leq \frac{\ln{x}}{x^3}$$
and
$$\left|\int\limits_{2}^{x}\{t\}\frac{1-3\ln{t}}{t^4}dt\right| \leq \int\limits_{2}^{x}\left|\{t\}\frac{1-3\ln{t}}{t^4}\right|dt \leq
\int\limits_{2}^{x}\left|\frac{1-3\ln{t}}{t^4}\right|dt=\\
\int\limits_{2}^{x}\frac{3\ln{t}-1}{t^4}dt=-\frac{\ln{x}}{x^3}+\frac{\ln{2}}{2^3}$$
this shows that
$\int\limits_{2}^{\infty}\{t\}\frac{1-3\ln{t}}{t^4}dt=I$ exists (by taking $\lim\limits_{x\rightarrow\infty}$) and
$$\int\limits_{2}^{x}\{t\}\frac{1-3\ln{t}}{t^4}dt=I-\int\limits_{x}^{\infty}\{t\}\frac{1-3\ln{t}}{t^4}dt=
I+\int\limits_{x}^{\infty}\{t\}\frac{3\ln{t}-1}{t^4}dt\leq \\
I+\int\limits_{x}^{\infty}\frac{3\ln{t}-1}{t^4}dt=I+\frac{\ln{x}}{x^3}$$
Summarising all these together (plus definitions) and continuing from $(3)$
$$...=\frac{1+\ln{4}}{16}-\frac{\ln{x}}{2x^2}-\frac{1}{4x^2}+I+O\left(\frac{\ln{x}}{x^3}\right)+O\left(\frac{\ln{x}}{x^3}\right)=\\
\left(\frac{1+\ln{4}}{16}+I\right)-\frac{\ln{x}}{2x^2}-\frac{1}{4x^2}+O\left(\frac{\ln{x}}{x^3}\right)$$
where $C=\frac{1+\ln{4}}{16}+I$
|
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|
Conflicting situations arising while solving for $x$ Given $$x^{1/4}+\frac{1}{x^{1/4}}=1$$
find $$x^{1024}+\frac{1}{x^{1024}}=?$$
My approach let $a=x^{1/4}$
$$a+\frac{1}{a}=1$$
$$a^2+1-a=0 $$
multiplying by $(a+1)$
$$(a+1)(a^2+1-a)=0 $$
$$a^3+1=0, a^3=-1$$
by solving we get value of $x$
$$x^3=1$$
Now the answer to the should be 2.
$$x^{1024}+\frac{1}{x^{1024}}=2$$
But if we square the equation $$x^{1/4}+\frac{1}{x^{1/4}}=1$$
we will get $-1$ as the recurring digit on the RHS as we square on both the sides. Thus it will result in $-1$ as the answer.
Can anyone help me to figure out where I am wrong
|
Since $a^3=-1$, one has $a^6=1$ and
$$ x=(x^\frac14)^4=a^4=-a. $$
So
$$ x^{1024}+\frac{1}{x^{1024}}=a^{1024}+\frac{1}{a^{1024}}=a^{6\cdot17+4}+\frac{1}{a^{6\cdot17+4}}=a^4+\frac{1}{a^4}=-a-\frac{1}{a}=-1$$
|
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|
Find $21^{1234}\pmod{100}\equiv \ ?$ The I'm having trouble to do this only by hand (no software or calculator). I tried the following:
\begin{align}21^{1234}(\text{mod} \ 100) &= 21^{1000}21^{200}21^{20}21^4(\text{mod} \ 100)\equiv41^{500}41^{100}41^{15}41^2(\text{mod} \ 100)\\
\end{align}
It's not reasonable to continue taking powers of 21, takes too long with pen and paper. Is there a more efficient way?
Yes I know about the Euler theorem and his totient function but please I don't want to use it, only elementary methods.
|
$3^4< 100 < 3^5 = 243 \equiv 43 \pmod{100}$
So $3^{5k}\equiv 43^{k}\pmod {100}$.
$43^2 = (40 + 3)^2 = 1600 + 2*3*40 + 9 \equiv 49 \pmod {100}$
So $3^{10k} \equiv 49^k \pmod {100}$.
$49^2 = (50-1)^2 = 2500 - 100 + 1 \equiv 1 \pmod {100}$.
So $3^{20k}\equiv 1\pmod{100}$ so
$3^{1234} \equiv 3^{14} \equiv 49*3^4 = (50-1)(80+1) \equiv 4000-80 + 50 -1 \equiv -31\equiv 69\pmod{100}$.
Do similar crap for $7$.
$7^2 = 49$ and so $7^4 = (50-1)^2 = 2500- 100 + 1\equiv 1 \pmod {100}$
So $7^{4k}\equiv 1 \pmod {100}$ and $7^{1234} \equiv 7^2 \equiv 49\pmod {100}$.
So $21^{1234} = 3^{12347}7^{1234} \equiv 69*49 \equiv (70 -1)(50-1) \equiv 3500 -70-50 + 1\equiv 81\pmod{100}$.
|
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|
Find $f(x)$ where $f(x)(A-\frac{B}{x+B/A})+Cf(x+\frac{B}{A})=0$. $A, B, C > 0$, $x$ is complex and $Re(x)>0$. My guess is that $f(x)=0$ but I don't know how to prove it.
|
For $f(x)\left(A-\dfrac{B}{x+C}\right)+Df(x+C)=0$ ,
$Df(x+C)=\dfrac{B-A(x+C)}{x+C}f(x)$
$f(Cx+C)=\dfrac{B-A(Cx+C)}{D(Cx+C)}f(Cx)$
$f(C(x+1))=\dfrac{B-AC(x+1)}{CD(x+1)}f(Cx)$
$f(C(x+1))=\dfrac{-AC\left(x-\dfrac{B}{AC}+1\right)}{CD(x+1)}f(Cx)$
$f(C(x+1))=-\dfrac{A}{D}\dfrac{x-\dfrac{B}{AC}+1}{x+1}f(Cx)$
With reference to http://eqworld.ipmnet.ru/en/solutions/fe/fe1105.pdf,
The general solution is $f(Cx)=\Theta_1(x)\dfrac{A^x\Gamma\left(x-\dfrac{B}{AC}+1\right)}{D^x\Gamma(x+1)}$, where $\Theta_1(x)$ is an arbitrary unit antiperiodic function
$f(x)=\Theta(x)\dfrac{A^\frac{x}{C}\Gamma\left(\dfrac{x}{C}-\dfrac{B}{AC}+1\right)}{D^\frac{x}{C}\Gamma\left(\dfrac{x}{C}+1\right)}$, where $\Theta(x)$ is an arbitrary antiperiodic function with period $C$
Similarly, for $f(x)\left(A-\dfrac{B}{x+\dfrac{B}{A}}\right)+Cf\left(x+\dfrac{B}{A}\right)=0$ ,
The general solution is $f(x)=\Theta(x)\dfrac{A^\frac{Ax}{B}\Gamma\left(\dfrac{Ax}{B}\right)}{C^\frac{Ax}{B}\Gamma\left(\dfrac{Ax}{B}+1\right)}=\Theta(x)\dfrac{A^{\frac{Ax}{B}-1}B}{C^\frac{Ax}{B}x}$, where $\Theta(x)$ is an arbitrary antiperiodic function with period $\dfrac{B}{A}$
|
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probability of $3$ girls sitting together in back row of adjacent seat in vans In how many ways $9$ boys and $3$ girls can be selected in two vans,each having numbered seats,$3$ in the front and $4$ at the back? How many sitting arrangements are possible $3$ girls should sit together in a back row of adjacent seat? Now if all the sitting arrangements are equal likely, what is the probability of $3$ girls sitting together in back row of adjacent seat
what i try
we have $14$ seats in $2$ vans and there are $9$ boys and $3$ girls. The number of ways of arranging $12$ people on $14$ seats without ristriction is $\displaystyle \binom{14}{12}\times 12=7\times 13!$
how do i arrange $3$ girls sit together in back seat help me please
|
We have $9$ boys $+$ $3$ girls $=12$ people
Total seats in one van $=3+4=7$
Total seats in $2$ vans $=7\times2=14$
So, we have to arrange $12$ people in $14$ seats.
Number of ways $= \ ^{14}P_{12}=\dfrac{14!}{(14-12)!}=7(13!)$
Now the number of ways to choose back seats is $2$ and we can arrange girls in adjacent seats in $2(3!)$ ways.
Number of ways of arranging $9$ bos in the remaining $11$ seats is $^{11}P_9$ ways.
Therefore, the total number of ways are $2\times2\times3!\times^{11}P_9=12!$
Therefore, the required probability is $\dfrac{12!}{7\times13!}=\dfrac{1}{91}$
|
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|
How many solutions has the following equation? Determine how many real solutions has the following equation:
$\sqrt[5]{2x+1+2\sqrt{{x^2+1}}}+\sqrt[5]{2x+1-2\sqrt{{x^2+1}}} = 2$
My first idea was to use the graph for the square root function: f:C->D, f(x)= $\sqrt[a]{b}$, if a is odd then f(x) is increasing on R so it should have only one solution but it doesn't 'feel' right.
Thanks in advance.
|
Let $f(x)$ be a function defined by the expression.
Notice that the second term has a value of $0$ when $x=\frac{3}{4}$ and the first term has a value of $\sqrt[5]{5}$ which is less than $2$.
The derivative of $f(x)$ is
$$ \frac{1}{5}(2x+1+2\sqrt{x^2+1})^{-4/5}\left(2+\frac{2x}{\sqrt{x^2+1}}\right)+ \frac{1}{5}(2x+1-2\sqrt{x^2+1})^{-4/5}\left(2-\frac{2x}{\sqrt{x^2+1}}\right)$$
This $f^\prime(x)$ is positive for all $x>\frac{3}{4}$ so the function is increasing on the interval $\left(\frac{3}{4},\infty\right)$ and has a value smaller than $2$ at $x=\frac{3}{4}$.
Notice that $f(n^5)>n$, so the expression has no upper bound.
Thus, since the function is continuous and is increasing for $x>\frac{3}{4}$ and has a value less than $2$ at $x=\frac{3}{4}$, and has no upper bound, there is only one solution to the equation.
|
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|
Prove that if $a < b$ and $c < d$ then $a + c < b + d$ I'm having a tough time with this questions. Could someone perhaps give me a hint? I'm only allowed to use the axioms of real numbers:
A1. $a + b = b + a$
A2. $a + (b + c) = (a+b) + c$
A3. $a + 0 = a$
A4. $a + (-a) = 0$
M1. $a \cdot b = b \cdot a$
M2. $a \cdot (b \cdot c) = (a \cdot b) \cdot c$
M3. $a \cdot 1 = a$
M4. if $a \neq 0$ then $a \cdot \frac{1}{a} = 1$
D. $a \cdot (b+c) = a \cdot b + a \cdot c$
|
Use $x < y\iff x + d < y+d$, to prove that $a < b \iff a+c < b+ c$ and to prove that $c< d \iff b + c < b + d$.
and finish it off with $x < y; y< z \implies x < z$.
to prove $a + c < b+c$ and $b+c < b+d\implies a+c < b+d$.
|
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|
How to compute remainder of division of $P(x)$ by $x^2 -3x+2$?
The remainder of division of $P(x)$ by $x^2−1$ is $2x+1$, and the remainder of division of the same polynomial by $x^2−4$ is $x+4$. Compute the remainder of division of $P(x)$ by $x^2−3x+2$.
I will translate these into math equations
$$P(x) = (x^2-1)Q(x)+ 2x+1$$
$$P(x) = (x^2 -4)R(x)+x+4$$
And let
$$f(x) = P(x) $$
We're asked to find the remainder when this polynomial is divided by $x^2 -3x+2$. So, there are two equations, which is why I'm confused with what to use in the equation $f(x) = P(x)$. What am I missing here?
Regards
|
$P(x)=(x^2-1)Q(x)+2x+1;$
$P(x)=(x^2-4)R(x)+x+4;$
$P(x)=(x-1)(x-2)S(x)+ax+b.$
1)$P(1)=2(1)+1=a+b;$
2)$P(2)=2+4=2a+b;$
$3=a+b$; and $6=2a+b;$
$a=3$; $b=0;$
Remainder: $ax+b=3x.$
|
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Decide all solutions to the following trigonometric equation: $\sin \left( 4\,x \right) =\cos \left( 3\,x \right).$ in the interval $[0, 2π[ $. Decide all solutions to the following trigonometric equation:
$$
\sin \left( 4\,x \right) =\cos \left( 3\,x \right).
$$
in the interval $[0, 2π[$.
I start by expanding $\sin \left( 4\,x \right)$ and $\cos \left( 3\,x \right)$ to get to get that
$\sin \left( 4\,x \right) =4\,\sin \left( x \right) \cos \left( x
\right) \left( 1-2\, \left( \sin \left( x \right) \right) ^{2}
\right)$
$\cos \left( 3\,x \right) =\cos \left( x \right) \left( 1-4\, \left(
\sin \left( x \right) \right) ^{2} \right)$.
I cancel $\cos \left( x \right)$ in both equations above and get that
$4\,\sin \left( x \right) \left( 1-2\, \left( \sin \left( x \right)
\right) ^{2} \right) =1-4\, \left( \sin \left( x \right) \right) ^{2}$
which is equivalent to
$8\, \left( \sin \left( x \right) \right) ^{3}-4\, \left( \sin \left(
x \right) \right) ^{2}-4\,\sin \left( x \right) +1=0$.
Now I use the substitution $u=\sin \left( x \right)$ to get to
$8\,{u}^{3}-4\,{u}^{2}-4\,u+1=0$ ⟺ $u \left( u-1 \right) \left( 2\,u+1 \right) =-1/4$.
Now, I'm not so sure I'm on the right track. To solve the cubic equation for $u$, seem to lead to a real nasty piece of a solution. Appreciate any help!
|
Hint: Note that
$$
\sin(4x)=\cos(3x) = \sin\left(\frac{\pi}2-3x\right),
$$ and $\sin x = \sin y$ if and only if $x-y = 2n\pi$ or $x+y= (2n+1)\pi$ for some $n\in \Bbb Z$.
|
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In a geometric sequence, the sum of the first six terms is 9 times the sum of the first three terms. If the first term is f5, what is the third term? I solved this question and my answer was 20, but apparently, it's supposed to be -25/2 and I don't understand how that works out.
$S_6=9(S_3)$, $t_1=5$
$\frac{1-r^6}{1-r} = 9t_1(\frac{1-r^3}{1-r})$
$(1-r^3)(1-r^3)= 9(1-r^3)$
$(1+r^3)=9$
$r^3=8,\ 2^3$
$r=2$
therefore: $t_3= 5(2)^{3-1}$
$t_3=20$
|
Your answer is correct. Here it is worked out again in a different manner:
Remembering a geometric sequence is in the form $a,ax,ax^2,ax^3,\cdots$ we are told the sum of the first six terms is equal to 9 times the sum of the first three terms.
$a+ax+ax^2+ax^3+ax^4+ax^5 = 9(a+ax+ax^2)$
Letting $a=5$ and moving to one side, this gives
$5x^5+5x^4+5x^3-40x^2-40x-40=0$
Factoring a bit:
$5x^3(x^2+x+1) - 40(x^2+x+1)=0$
$(5x^3-40)(x^2+x+1)=0$
Now... one of the solutions can be seen from the first factor as being when $5x^3=40$ and so $x=\sqrt[3]{8}=2$
The remaining four solutions can all be found with the quadratic formula and can be seen to all be complex.
So, we have found that for the sequence to be comprised only of real entries, the ratio between terms must be $2$.
The first term being $5$, that leaves the second term being $10$ and the third term being $20$.
|
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|
a,b,c are three real numbers where $a + \frac{1}{b} = b + \frac{1}{c} = c + \frac{1}{a}$. Now $abc$ = ? Here will the answer be a number? I want to know whether it is possible to get a real number (not an algebraic expression) as the product of $a$, $b$ and $c$.
I tried for a long time and this is what I got.
$$3abc = a^2 b + b^2 c + c^2 a$$
But it seems like there is a way to determine the value.
|
If $a=b$ then $b=c$ and $a=b=c$, which says that $abc$ is not defined.
Let $(a-b)(a-c)(b-c)\neq0.$
Thus, since
$$a-b=\frac{1}{c}-\frac{1}{b}=\frac{b-c}{bc},$$
$$b-c=\frac{c-a}{ac}$$ and
$$c-a=\frac{a-b}{ab},$$ we obtain
$$(a-b)(b-c)(c-a)=\frac{(a-b)(b-c)(c-a)}{a^2b^2c^2},$$ which gives
$$abc=1$$ or $$abc=-1.$$
|
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|
Stuck at proving whether the sequence is convergent or not I have been trying to determine whether the following sequence is convergent or not. This is what I got:
Exercise 1: Find the $\min,\max,\sup,\inf, \liminf,\limsup$ and determine whether the sequence is convergent or not:
$X_n=\sin\frac{n\pi}{3}-4\cos\frac{n\pi}{3}$
I wrote down a few cases:
$X_1 = \sin\frac{\pi}{3}-4\cos\frac{\pi}{3}= \frac{\sqrt3-4}{2} $
$X_2 = \sin\frac{2\pi}{3}-4\cos\frac{2\pi}{3}= \frac{\sqrt3+4}{2} $
$X_3 = \sin\frac{3\pi}{3}-4\cos\frac{3\pi}{3}=4 $
$X_4 = \sin\frac{4\pi}{3}-4\cos\frac{4\pi}{3}= \frac{4-\sqrt3}{2} $
$X_5 = \sin\frac{5\pi}{3}-4\cos\frac{5\pi}{3}= \frac{4-\sqrt3}{2} $
$X_6 = \sin\frac{6\pi}{3}-4\cos\frac{6\pi}{3}= -4 $
So as found above, $\min = -4$, $\max = 4$, $\inf = -4$, $\sup = 4$, $\liminf = -4$, $\limsup = 4$.
Let's check whether its convergent or not:
The sequence is bounded as stated above so lets check if its decreasing or increasing.
$X_n \geq X_{n+1}$
$\sin\frac{n\pi}{3}-4\cos\frac{n\pi}{3} \geq\sin\frac{(n+1)\pi}{3}-4\cos\frac{(n+1)\pi}{3}$
$\sin\frac{n\pi}{3} - \sin\frac{(n+1)\pi}{3} \geq -4\cos\frac{(n+1)\pi}{3} + 4\cos\frac{n\pi}{3}$
I used trigonometrical identity for $\sin\alpha+\sin\beta$ and $\cos\alpha-\cos\beta$ :
$-\cos\frac{\pi(2n+1)}{6} \geq 4\sin\frac{\pi(2n+1)}{6}$
What should I do next? I am stuck here.
Thanks, and sorry if I made mistakes.
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Hint: Note that this is a periodic sequence, meaning: $x_{n+6}=x_n$.
Therefore, this sequence diverges.
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|
Radical equation - can I square both sides with more than 1 radical on one side? I'm familiar with equations like:
$\sqrt{x+1} - \sqrt{x+2} = 0 $
Has no solutions, it's just an example off the top of my head
Just move the negative square root to the other side, square both sides and solve.
$\sqrt{x+1} = \sqrt{x+2}$
$x+1 = x+2$
0 = 1
My question is, if there are two square roots on one side, then can I still square both sides in this way:
$\sqrt{x+1} - \sqrt{x+2} = \sqrt{x+3}$
$x+1 - (x+2) = x+3$
$x+1 - (x-2) = x+3$
$x = -4$
Or does squaring both sides cause something strange to happen on the left hand side?
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Yes, you can square both sides. But you do actually have to square both sides.
$\sqrt{x+1} - \sqrt{x+2} = \sqrt{x+3}$
$(\sqrt{x+1} - \sqrt{x+2})^2 = (\sqrt{x+3})^2$
$x+1 - 2\sqrt{x+1}\sqrt{x+2} + x + 2 = x + 3$.
$-2\sqrt{x+1}\sqrt{x+2} = -x$
$(-2\sqrt{x+1}\sqrt{x+2})^2 = (-x)^2$
$4(x+1)(x+2) = x^2$
$4x^2 +12x + 8 = x^2$
$3x^2 + 12x + 8 = 0$
$x = \frac {-12\pm{144-4*3*8}}{6} =4
$-2 \pm \frac{\sqrt{48}}6 = -2 \pm \frac 2{\sqrt 3}$.
BUt note: Squaring both sides gives extraneous solutions because it ignores whether terms given are positive or negative. These answers all don't actually work if we plug them back in.
.....
Obviously you can't just square parts of each side and expect the result to make any sense.
$\sqrt{x+1} - \sqrt{x+2} = \sqrt{x+3}$ obviously does not mean
$(\sqrt{x+1})^2 - (\sqrt{x+2})^2 = (\sqrt{x+3})^2$
Because $(\sqrt{x+1} - \sqrt{x+2})^2 \ne (\sqrt{x+1})^2 - (\sqrt{x+2})^2$.
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|
Show that the pseudoinverse of $T$ is $1/4T$. If $T$ is a selfadjoint matrix then, by wiki, the pseudo inverse of $T$ is given by
$$T^{+}= \lim_{t \to 0}(T^2+tI)^{-1} T. \tag{1} $$
*
*I don't understand under which norms we have
$$\lim_{t \to 0}\|T^{+}-(T^2+tI)^{-1} T \|=0?$$
*Also if $T=\left(\begin{array}{cc}1&1\\1&1\end{array}\right)$. I want to use the formula $(1)$ in order to show that
$$T^+=\frac{1}{4}T.$$
If I don't made mistakes, I find
$$T^2+tI=\left(\begin{array}{cc}2+t&2\\2&2+t\end{array}\right)$$
and so
$$(T^2+tI)^{-1}=\frac{1}{(2+t)^2-t}\left(\begin{array}{cc}2+t&-2\\-2&2+t\end{array}\right).$$
Also
$$(T^2+tI)^{-1} T=\frac{1}{(2+t)^2-t}\left(\begin{array}{cc}t&t\\t&t\end{array}\right).$$
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The norm is irrelevant, as all norms on $M_2(\mathbb{C})$ are equivalent, being a finite dimensional vector space.
The determinant of $T^2+tI$ is $(2+t)^2\color{red}{-4}=t^2+4t$. So you need to compute the limit of
$$
\frac{1}{t^2+4t}\begin{pmatrix}t & t \\ t & t \end{pmatrix}=
\frac{1}{t+4}\begin{pmatrix}1 & 1 \\ 1 & 1 \end{pmatrix}
$$
With a different technique: the LU decomposition of $T$ is
$$
T=
\begin{pmatrix} 1 & 0 \\ 1 & 1 \end{pmatrix}
\begin{pmatrix} 1 & 1 \\ 0 & 0 \end{pmatrix}
$$
so a full rank decomposition $T=AB$ is
$$
T=
\begin{pmatrix} 1 \\ 1 \end{pmatrix}
\begin{pmatrix} 1 & 1 \end{pmatrix}
$$
Then
$$
A^+=A(A^HA)^{-1}=\begin{pmatrix} 1/2 \\ 1/2 \end{pmatrix}
\qquad
B^+=(BB^H)^{-1}B=\begin{pmatrix} 1/2 & 1/2 \end{pmatrix}
$$
and
$$
T^+=B^+A^+=
\begin{pmatrix} 1/2 & 1/2 \end{pmatrix}
\begin{pmatrix} 1/2 \\ 1/2 \end{pmatrix}
=
\begin{pmatrix} 1/4 & 1/4 \\ 1/4 & 1/4 \end{pmatrix}
$$
|
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|
Evaluate : $\lim_{x\to -\infty} \frac{\sqrt{4x^2-2x}}{\sqrt[3]{x^3+1}}$
Evaluate: $\lim_{x\to -\infty} \frac{\sqrt{4x^2-2x}}{\sqrt[3]{x^3+1}}$
My solution:
\begin{align}
\lim_{x\to -\infty} \frac{\sqrt{4x^2-2x}}{\sqrt[3]{x^3+1}} & = \lim_{x\to -\infty} \frac{\sqrt{4-\frac{2}{x}}}{\sqrt[3]{1+\frac{1}{x^3}}}\\
& = \frac{\sqrt{4-0}}{\sqrt[3]{1+0}} \\
& = 2
\end{align}
Despite the steps I've taken seems plausible to me, the answer is given as $-2$.
Is dividing both the numerator and the denominator by $x$ allowed here? Where am I making a mistake?
|
It is faster using equivalents. Recall that a polynomial is asymptotically equivalent to its leading term, so
$$\frac{\sqrt{4x^2-2x}}{\sqrt[3]{x^3+1}}\sim_{-\infty}\frac{\sqrt{4x^2}}{\sqrt[3]{x^3}}=\frac{2|x|}{x}=-2.$$
|
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|
Having trouble computing $\int_3^5\frac{t}{1+0.1t} dt $ $$\int_3^5\frac{t}{1+0.1t} dt $$
For some reason this is equal to:
1/0.1 (2 - (1/0.1 (ln1.5 - ln1.3)))
I have no idea how to reduce to that.
|
$$
\frac{x}{1+0.1x}=\frac{x}{1+0.1x}\cdot\frac{10}{10}=
\frac{10x}{10+x}=10\left(\frac{x}{10+x}\right)=\\
10\left(\frac{-10+10+x}{10+x}\right)=
10\left(\frac{-10}{10+x}+\frac{10+x}{10+x}\right)=
10\left(-\frac{10}{10+x}+1\right)=\\
10\left(1-\frac{10}{10+x}\right)=10-\frac{100}{10+x}.
$$
$$
\int\left(10-\frac{100}{10+x}\right)\,dx=
10\int\,dx-100\int\frac{1}{10+x}\frac{d}{dx}(10+x)\,dx=\\
10x-100\int\frac{1}{10+x}\,d(10+x)=
10x-100\ln{|10+x|}+C.
$$
$$
\int_3^5\frac{t}{1+0.1t}\,dt=
\bigg[10t-100\ln{|10+t|}\bigg]_3^5=\\
50-100\ln{15}-(30-100\ln{13})=
20-100\ln{15}+100\ln{13}=\\
20-100(\ln{15}-\ln{13})=20-100\ln{\frac{15}{13}}.
$$
The answer you gave is equivalent to what I got:
$$
\frac{1}{0.1}\left(2-\frac{1}{0.1}\left[\ln{1.5}-\ln{1.3}\right]\right)=
10\left(2-10\left[\ln{\frac{15}{10}}-\ln{\frac{13}{10}}\right]\right)=\\
20-100\ln{\left(\frac{15}{10}\div\frac{13}{10}\right)}=
20-100\ln{\frac{15}{13}}.
$$
|
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|
If $a$, $b$ and $c$ are sides of a triangle, then prove that $a^\text{2}(b+c-a) + b^\text{2}(c+a-b) + c^\text{2}(a+b-c)$ $\leqslant$ $3abc$
Let $a$, $b$ and $c$ be the sides of a triangle. Prove that $$a^\text{2}(b+c-a) + b^\text{2}(c+a-b) + c^\text{2}(a+b-c) \leqslant 3abc$$
SOURCE: BANGLADESH MATH OLYMPIAD (Preparatory Question.)
I am very new and novice at this problem. I did a little try but couldn't succeed because I was unable to substitute the left term of the inequality into formula. I know a formula that
$a^\text{3} + b^\text{3} + c^\text{3} - 3abc$ = $(a+b+c)(a^\text{2} + b^\text{2} + c^\text{2} - ab -bc - ca)$. But how to use this formula in that case isn't known to me. And how to show the relation of the both side and when they will become equal?
A small help will be enough for me. Thanks in advance.
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Using the Ravi substitution $$a=y+z,b=x+z,c=x+y$$ we have to prove that
$$x^2y+x^2z+xy^2+xz^2+y^2z+yz^2-6xyz\geq 0$$ But this is AM-GM:
$$x^2y+x^2z+xy^2+xz^2+y^2z+yz^2\geq 6\sqrt[6]{x^6y^6z^6}=6xyz$$
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|
How to express $z +\frac{1}{z}$ in polar form If we write $z=re^{i\theta}$ then we can express $z + \frac{1}{z}= \frac{r^{2}e^{i2\theta} +1}{r}. e^{-i\theta}$ Can we simplify it again and express it as $Re^{i \phi}$ ?
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$$z+\frac{1}{z}=\left(r+\frac{1}{r}\right)\cos\theta+i\left(r-\frac{1}{r}\right)\sin\theta=$$
$$=\sqrt{\left(r+\frac{1}{r}\right)^2\cos^2\theta+\left(r-\frac{1}{r}\right)^2\sin^2\theta}e^{\left(\arctan\left(\frac{r^2-1}{r^2+1}\tan\theta\right)\right)i}=$$
$$=\sqrt{r^2+\frac{1}{r^2}+2\cos2\theta}e^{\left(\arctan\left(\frac{r^2-1}{r^2+1}\tan\theta\right)\right)i}.$$
|
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|
Confusion with Summation notation I need to compute the value of this:
$$\frac{1}{3}\left(\sum_{n=1}^{3} \frac{z_{n}}{f} + N(2,2)\right)$$
the $N$ is a gaussian noise with mean=2 and standard deviation=2.
The question:
is this equivalent to:
$$
\frac{1}{3}\left(\frac{z_{1}}{f} + N(2,2)+\frac{z_{2}}{f} + N(2,2)+\frac{z_{3}}{f} + N(2,2)\right)
$$
or
$$
\frac{1}{3}\left(\left(\frac{z_{1}}{f} +\frac{z_{2}}{f} +\frac{z_{3}}{f}\right) + N(2,2)\right)
$$
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The scope of the sigma operator $\Sigma$ is solely defined via arithmetic precedence rules. The scope is given by the expression that follows immediately the $\Sigma$ and is valid respecting the arithmetic precedence rules up to an operator with precedence level equal to '$+$' or up to the end if no such operator follows.
This implies that
\begin{align*}
\color{blue}{\sum_{n=1}^{3} \frac{z_{n}}{f} + N(2,2)}&=\sum_{n=1}^{3}\left( \frac{z_{n}}{f} \right)+ N(2,2)\\
&=\left(\sum_{n=1}^{3} \frac{z_{n}}{f}\right) + N(2,2)\\
&\,\,\color{blue}{=\frac{z_{1}}{f} + \frac{z_{2}}{f} + \frac{z_{3}}{f} + N(2,2)}\\
\end{align*}
Hint: You might find chapter 2: Sums in Concrete Mathematics by R.L. Graham, D.E. Knuth and O. Patashnik helpful. It provides a thorough introduction in the usage of sums.
|
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|
Does this pattern of summing polygonal numbers to get a square repeat indefinitely? I am using the table of polygonal numbers on this site:
http://oeis.org/wiki/Polygonal_numbers
The first row of the table gives the value of $n=0,1,2,3,...$. The first column gives the polygonal numbers $P_{N}(n)$ starting with $N=3$.
Here's the pattern, the sum is done by adding elements of the same column starting with the triangular numbers.
$$1+1+1+1=4=2^2$$
$$3+4+5+6+7=25=5^2$$
$$6+9+12+15+18+21=81=9^2$$
$$10+16+22+28+34+40+46=196=14^2$$
...
The squares on the right hand side are given by $n+T_{n}$. The sum of index $(n+1)$ has one more element than that of $n$.
1-Does this pattern repeat indefinitely? (I suspect the answer is yes but I can't prove it)
2-Do we know why we have such a pattern?
Edit 03-05-2019
Following the suggestion of Eleven-Eleven, I looked for other patterns similar to the one above. I found one that is even simpler. This time we skip the triangular numbers when we calculate the sum. We start with the squares and sum up enough terms to get another square.
$$1=1^2=T_{1}^2$$
$$4+5=3^2=9=T_{2}^2$$
$$9+12+15=6^2=36=T_{3}^2$$
$$16+22+28+34=10^2=100=T_{4}^2$$
$$25+35+45+55+65=15^2=225=T_{5}^2$$
We see the same pattern as above. The square with index $(n+1)$ requires the addition of one more term than the square with index $n$. The number of elements to sum up to get the square $T_{n}^2$ is simply $n$.
Can this be stated as the following theorem?
The square of a triangular number $T_{n}$ can be expressed as the sum of $n$ polygonal numbers excluding the triangular number itself.
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This is a much more drawn out approach using summations. It doesn't have the elegance of Oscar's solutions, but it still works. You have that with $T_k$ being the $k$-th Triangular number (here $T_0=0$)
$$\sum_{k=1}^{4+0}{[T_1+(k-1)T_0]}=(T_1+1)^2$$
$$\sum_{k=1}^{4+1}{[T_2+(k-1)T_1]}=(T_2+2)^2$$
$$\sum_{k=0}^{4+2}{[T_3+(k-1)T_2]}=(T_3+3)^2$$
This suggests that
$$\sum_{k=1}^{m+4}{[T_{m+1}+(k-1)T_{m}]}=(T_{m+1}+(m+1))^2$$
Since summation is linear, we have
\begin{eqnarray*}\sum_{k=1}^{m+4}{[T_m+(k-1)kT_{m-1}]}&=&T_{m+1}\sum_{k=1}^{m+4}1+T_m\sum_{k=1}^{m+4}{(k-1)}\\&=&T_{m+1}(m+4)+T_m\sum_{k=1}^{m+3}{k}
\end{eqnarray*}
Using the formula for the $m$-th Triangular number, we have
$$=\left[\frac{(m+1)(m+2)}{2}\right](m+4)+\left[\frac{m(m+1)}{2}\right]\left[\frac{(m+3)(m+4)}{2}\right]$$
Factoring and simplifying gives
$$\frac{(m+1)(m+4)}{2}\left[(m+2)+\frac{m(m+3)}{2}\right]=\frac{(m+1)^2(m+4)^2}{4}$$
Now, the right hand side
\begin{eqnarray*}[T_{m+1}+(m+1)]^2&=&T_{m+1}^2+2(m+1)T_{m+1}+(m+1)^2\\&=&\frac{(m+1)^2(m+2)^2}{4}+2(m+1)^2\frac{(m+2)}{2}+(m+1)^2
\end{eqnarray*}
Factoring out an $(m+1)^2$,
\begin{eqnarray*}\frac{(m+1)^2(m+2)^2}{4}+2(m+1)^2\frac{(m+2)}{2}+(m+1)^2&=&(m+1)^2\left[\frac{(m+2)^2}{4}+\frac{4(m+2)}{4}+\frac{4}{4}\right]\\&=&\frac{(m+1)^2}{4}\left[(m^2+4m+4)+(4m+8)+4\right]\\&=&\frac{(m+1)^2}{4}(m^2+8m+16)\\&=&\frac{(m+1)^2(m+4)^2}{4}
\end{eqnarray*}
Since both expressions equal $\frac{(m+1)^2(m+4)^2}{4}$, we are done.
EDIT:
For the new problem, note the first term in each line is a successive square and as above, the common difference is a triangular number. Thus we have to prove, in general,
$$\sum_{k=1}^n[n^2+(k-1)T_{n-1}]=T_n^2$$
The left hand side then can be written as
\begin{eqnarray*}\sum_{k=1}^n[n^2+(k-1)T_{n-1}]&=&\sum_{k=1}^n n^2+T_{n-1}\sum_{k=1}^n(k-1)\\&=&n^3+T_{n-1}\sum_{k=1}^{n-1}k\\&=&n^3+\frac{(n-1)n}{2}\cdot\frac{(n-1)n}{2}\\&=&n^2\left(n+\frac{(n-1)^2}{2^2}\right)\\&=&\frac{n^2}{2^2}(n+1)^2\\&=&\left[\frac{n(n+1)}{2}\right]^2\\&=&T_n^2
\end{eqnarray*}
And we're done.
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|
evaluate the surface integral $f(x, y, z) = x^2 + y^2$ and $S$ is part of the cone $z = \sqrt{x^2 + y^2}$ with $z \leq 2$ Evaluate the surface integral of scalar function $\int_{S} f dS$
$f(x, y, z) = x^2 + y^2$ and $S$ is part of the cone $z = \sqrt{x^2 + y^2}$ with $z \leq 2$
my attempt
$f(x,y,z) = x^2 + y^2$ and $z = \sqrt{x^2 + y^2}, z \leq 2$
$\frac{dz}{dx} = \frac{x}{\sqrt{x^2 + y^2}}$ and $\frac{dz}{dy} = \frac{y}{\sqrt{x^2 + y^2}}$
$ds = \sqrt{\left(\frac{dz}{dx} \right)^2 + \left(\frac{dz}{dy} \right)^2} dx dy= \sqrt{\frac{x^2}{x^2 + y^2} + \frac{y^2}{x^2 + y^2}}dx dy=dx dy$
$\int_S f dS = \iint(x^2 + y^2)dx dy$, such that $x^2 + y^2 = 4, z \leq 2$
$\int_{0}^{2\pi} \int_{0}^{2} r^2 r dr d \theta = 8 \pi$
would this be right?
|
Your approach is almost correct but for one erroneous formula:
$\mathrm{d}S = \sqrt{z^2_x+z^2_y + 1} \ \mathrm{d}A $
$$
\begin{align*}
\mathrm{d}\vec{S} = \langle -z_x,-z_y, \ 1 \rangle \ \mathrm{d}A
\end{align*}
$$
therefore, $$|\mathrm{d}\vec{S}| = \sqrt{z^2_x+z^2_y + 1} \ \mathrm{d}A $$
$z_x$ and $z_y$ values what you got are correct.
Let's proceed:
$$
\begin{align*}
\iint_S f\mathrm{d}S &= \iint_S x^2+y^2 \ \sqrt{z^2_x+z^2_y + 1} \ \mathrm{d}A \\
\\
&= \iint_S x^2+y^2 \sqrt{\frac{x^2}{x^2+y^2}+\frac{y^2}{x^2+y^2}+1} \ \mathrm{d}x\mathrm{d}y \\
\\
\end{align*}
$$
Now switch to polar coordinates:
$$
\begin{align*}
\iint_S f\mathrm{d}S &= \iint_S r^2\sqrt{2} \cdot r\mathrm{d}r\mathrm{d}r\theta \\
\\
&= \sqrt{2}\int_{0}^{2\pi}\int_{0}^{2}r^3 \ \mathrm{d}r\mathrm{d}\theta \\
\\
&= 8\sqrt{2}\pi
\end{align*}
$$
|
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|
Solving $\int_0^\frac{\pi}{2} \frac{\sqrt[3]{\sin x}}{4-\sin^2 x}dx$
$$\int_0^\frac{\pi}{2} \frac{\sqrt[3]{\sin x}}{4-\sin^2 x}dx$$
I tried much of elementary methods to solve above integral but is not advancing.
Any methods from elementary to advanced are appreciated.
|
Hint. One may write
$$
\frac{\sqrt[3]{\sin x}}{4-\sin^2 x}=\frac14\sum_{n=0}^\infty\frac1{4^n}\left(\sin x\right)^{2n+\frac13}
$$ then one is allowed to perform a termwise integration
$$
\int_0^{\Large \frac{\pi}2}\frac{\sqrt[3]{\sin x}}{4-\sin^2 x}\,dx=\sum_{n=0}^\infty\frac1{4^{n+1}}\int_0^{\Large \frac{\pi}2}\left(\sin x\right)^{2n+\frac13}\,dx
$$using the classic Euler evaluation
$$
\int_0^{\Large \frac{\pi}2}\left(\sin x\right)^s\,dx=\frac{\sqrt{\pi} \,\Gamma \left(\frac{s+1}{2}\right)}{2\, \Gamma \left(\frac{s}{2}+1\right)},\qquad s>0,
$$ obtaining
$$
\int_0^{\Large \frac{\pi}2}\frac{\sqrt[3]{\sin x}}{4-\sin^2 x}\,dx=\frac{3 \sqrt{\pi} \,\, _2F_1\left(\frac{2}{3},1;\frac{7}{6};\frac{1}{4}\right)\, \Gamma \left(\frac{2}{3}\right)}{4\, \Gamma \left(\frac{1}{6}\right)}.
$$
A path to the simplification $\dfrac{\pi}{2^{2/3} 3^{3/2}}$ would be interesting.
|
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|
Calculate the sum $S_n = \sum\limits_{k=1}^{\infty}\left\lfloor \frac{n}{2^k} + \frac{1}{2}\right\rfloor $ I am doing tasks from Concrete Mathematics by Knuth, Graham, Patashnik for trainning, but there are a lot of really tricky sums like that:
Calculate sum $$S_n = \sum_{k=1}^{\infty} \left\lfloor \frac{n}{2^k} + \frac{1}{2} \right\rfloor $$
My idea
I had the idea to check when $$\frac{n}{2^k} < \frac{1}{2}$$
because then $$ \forall_{k_0 \le k} \left\lfloor \frac{n}{2^k} + \frac{1}{2} \right\rfloor=0$$
It should be $$ k_0 = \log_2(2n) $$ but I don't know how it helps me with this task (because I need not only "stop moment" but also sum of considered elements
Book idea
Let $$S_n = \sum_{k=1}^{\infty} \left\lfloor \frac{n}{2^k} + \frac{1}{2} \right\rfloor $$
then $$ S_n-S_{n-1} = 1$$
and then solve this recursion. But I write $S_n - S_{n-1}$ and I don't see how it can be $1$ , especially that is an infinite sum.
|
From $\left\lfloor x+\frac{1}{2}\right\rfloor=\lfloor 2x\rfloor-\lfloor x\rfloor$ we have
$$\left\lfloor \frac{n}{2^k}+\frac{1}{2}\right\rfloor=\left\lfloor \frac{n}{2^{k-1}}\right\rfloor-\left\lfloor \frac{n}{2^k}\right\rfloor$$
therefore
$$\sum_{k=1}^\infty \left\lfloor \frac{n}{2^k}+\frac{1}{2}\right\rfloor=\sum_{k=1}^\infty
\left\lfloor \frac{n}{2^{k-1}}\right\rfloor-\left\lfloor \frac{n}{2^k}\right\rfloor\\
=\lfloor n\rfloor -\lim_{k\to \infty}\left\lfloor \frac{n}{2^k}\right\rfloor\\=n$$
|
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|
Find range of $x$ satisfying $\left \lfloor \frac{3}{x} \right \rfloor+\left \lfloor \frac{4}{x} \right \rfloor=5$ Find range of $x$ satisfying $$\left \lfloor \frac{3}{x} \right \rfloor +\left \lfloor \frac{4}{x} \right \rfloor=5$$ Where $\lfloor\cdot\rfloor$ is the floor function
My try:
As far as domain of LHS is concerned we have $x \ne 0$ and since RHS is positive, we have $x \gt 0$
Now since LHS is sum of two positive integers, let us suppose:
$$\left \lfloor \frac{3}{x} \right \rfloor=m$$ and
$$\left \lfloor \frac{4}{x} \right \rfloor=5-m$$
Thus we have:
$$ m \le \frac{3}{x} \lt m+1$$
$$5-m \le \frac{4}{x} \lt 6-m$$
Adding both we get:
$$5 \le \frac{7}{x} \lt 7$$
$\implies$
$$1 \lt x \le \frac{7}{5}$$
Hence $$x \in (1, 1.4]$$
But answer in book is given as $$x \in (1,\frac{4}{3})$$
What went wrong?
|
Since $\frac{4}{x}>\frac{3}{x}$, we have three cases:
*
*$\left [ \frac{3}{x} \right ]=0$ and $\left [ \frac{4}{x} \right ]=5.$ Easy to show that it's impossible.
*$\left [ \frac{3}{x} \right ]=1$ and $\left [ \frac{4}{x} \right ]=4,$ which is impossible again and
*$\left [ \frac{3}{x} \right ]=2$ and $\left [ \frac{4}{x} \right ]=3,$ which gives the answer: $\left(1,\frac{4}{3}\right]$.
Can you end it now?
|
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|
Solve a system equation in $\mathbb{R}$ - $\sqrt{x+y}+\sqrt{x+3}=\frac{1}{x}\left(y-3\right)$ how to solve a system equation with radical
$$\sqrt{x+y}+\sqrt{x+3}=\frac{1}{x}\left(y-3\right)$$
And $$\sqrt{x+y}+\sqrt{x}=x+3$$
This system has $1$ root is $x=1;y=8$,but i have no idea which is more clearly to solve it. I tried substituting and squaring to find the factor but failed.
|
Remember that $$ \sqrt{a}+\sqrt{b}= {a-b\over \sqrt{a}-\sqrt{b}}$$
so $$\sqrt{x+y}+\sqrt{x+3}={y-3\over \sqrt{x+y}-\sqrt{x+3}} $$
and now we have $${y-3\over \sqrt{x+y}-\sqrt{x+3}}=\frac{1}{x}\left(y-3\right)$$
Obviuosly $y\neq 3$ so we have $$\sqrt{x+y}-\sqrt{x+3}=x$$
Since $$\sqrt{x+y}+\sqrt{x}=x+3$$
we have now $$\underbrace{\sqrt{x}+\sqrt{x+3}}_{f(x)} = 3\implies x= 1$$
($f$ is strictly increasing function so this equation has at most $1$ solution.) And then we get $y=8$.
|
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|
Evaluate $\int \frac{x+4}{x^2 + 2x + 5}$ I am having issues with this integral. I am not sure if it is irreducible or not. I can't use the quadratic formula, but I can rearrange the integral to be $\int \frac{x+4} {(x^2 + 2x + 1) + 4}$, but I don't know how to deal with the $+4$.
Here is my work treating the quadratic equation as irreducible with no repeating factors.
$\int \frac{x+4}{x^2 + 2x + 5}$
= $\frac {Ax + B} {x^2 + 2x + 5} $
= $Ax+B(x^2 + 2x + 5)$
= $Ax^3 + 2Ax^2 + 5Ax + Bx^2 + 2Bx + 5B$
= $Ax^3 + (2A + 2B)x^2 + (5A + 2B)x + 5B$
However I get stuck trying to solve my system of equations. This leads me to believe that I did the partial fractions improperly.
$\begin{bmatrix}
1 & 0 & 0 \\
2 & 2 & 0 \\
5 & 2 & 1 \\
0 & 5 & 4 \\
\end{bmatrix}$
|
Given $$\int\dfrac{x+4}{x^2+2x+5}dx=\int\dfrac{x+4}{(x+1)^2+4}dx$$
Use u-substitution: $u=x+1$ and we get
\begin{align}
\int\dfrac{u+3}{u^2+4}du &= \int\dfrac{u}{u^2+4}du+\int\dfrac{3}{u^2+4}du \\
&= \dfrac12\ln|u^2+4|+\dfrac32\arctan\left(\dfrac u2\right) +c \\
&= \dfrac12\ln|(x+1)^2+4|+\dfrac32\arctan\left(\dfrac{x+1}{2}\right) +c
\end{align}
Therefore, $$\int\dfrac{x+4}{x^2+2x+5}dx=\dfrac12\ln|(x+1)^2+4|+\dfrac32\arctan\left(\dfrac{x+1}{2}\right) +c$$
|
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|
Evaluating $\lim\limits_{x \to 0} \frac{(1+x)^{1/x} - e + \frac{1}{2}ex}{x^2}$ without expansions in limits
Evaluate $\lim\limits_{x \to 0} \frac{(1+x)^{1/x} - e + \frac{1}{2}ex}{x^2}$
One way that I can immediately think of is expanding each of the terms and solving like,
$$(1+x)^{1/x} = e^{\log_e (1+x)^{1/x}} = e^{\frac{1}{x} (x-\frac{x^2}{2} -\frac{x^3}{3}+...)}$$
and then after complete expansion of each and every and substuting into back to limit and solving I get $\frac{11e}{24}$ as an answer.
Now, this is a relatively long and complicated way to solve as you can see. I want to know if there is an easier way to solve this problem. Please help. Thank you!
|
If I may suggest, the problem of
$$y=\frac{(1+x)^{\frac1 x} - e + \frac{1}{2}ex}{x^2}$$ is not so difficult if you use another way.
$$a=(1+x)^{\frac1 x}\implies \log(a)= {\frac1 x}\log(1+x)$$
$$ \log(a)={\frac1 x}\left(x-\frac{x^2}{2}+\frac{x^3}{3}-\frac{x^4}{4}+O\left(x^5\right) \right)=1-\frac{x}{2}+\frac{x^2}{3}-\frac{x^3}{4}+O\left(x^4\right)$$ Now, continuing with Taylor
$$a=e^{\log(a)}=e-\frac{e x}{2}+\frac{11 e x^2}{24}-\frac{7 e x^3}{16}+O\left(x^4\right)$$
$$y=\frac{\frac{11 e x^2}{24}-\frac{7 e x^3}{16}+O\left(x^4\right) }{x^2}=\frac{11 e}{24}-\frac{7 e x}{16}+O\left(x^2\right)$$ which gives not only the limit but also how it is approached.
|
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|
Prove $\sum \sqrt{\frac{a^2}{6a^2+5ab+b^2}}\le \frac{\sqrt{3}}{2}$ Let $a,b,c\in R^+$ prove that the inequality $$\sqrt{\frac{a^2}{6a^2+5ab+b^2}}+\sqrt{\frac{b^2}{6b^2+5bc+c^2}}+\sqrt{\frac{c^2}{6c^2+5ca+a^2}}\le \frac{\sqrt{3}}{2}$$
My try:$$\sum\limits_{cyc} \sqrt{\frac{a^2}{6a^2+5ab+b^2}}=\sum\limits_{cyc} \sqrt{\frac{a^2}{\left(3a+b\right)\left(2a+b\right)}}=\sum\limits_{cyc} \sqrt{\frac{1}{\left(3+\frac{b}{a}\right)\left(2+\frac{b}{a}\right)}}=\sum\limits_{cyc} \frac{1}{\sqrt{\left(x+3\right)\left(x+2\right)}}\le \sum\limits_{cyc} \frac{1}{4\sqrt{3}}\left(\frac{4}{x+3}+\frac{3}{x+2}\right),$$ where $\frac{b}{a}=x;\frac{c}{b}=y;\frac{a}{c}=z\left(xyz=1\right)$
It is easy now but i also solve it by SOS but stuck, here is my solution by SOS and Cauchy-Schwarz
$$\left(\sum_{cyc}\sqrt{\frac{a^2}{6a^2+5ab+b^2}}\right)^2\le 3\left(\sum_{cyc} \frac{a^2}{6a^2+5ab+b^2}\right)\le \frac{3}{4}$$
Or $$\sum_{cyc} \left(\frac{a^2}{6a^2+5ab+b^2}-\frac{1}{12}\right)\le 0\Leftrightarrow \sum_{cyc} \frac{\left(a-b\right)\left(6a+b\right)}{12\left(2a+b\right)\left(3a+b\right)}\le 0$$
And i tried to taking $6a^2-5ab-b^2$ into $(a-b)\cdot Q(a,b,c)-(c-a)\cdot P(a,b,c)$ but unsuccessful
I want to solve it by SOS, help.
|
SOS works for symmetric inequalities. For cyclic inequalities SOS helps sometimes,
but for your inequality it does not help or at least, does not help immediately.
Also, your second way gives a wrong inequality.
Try $c=0.0001$, $b=0.01$ and $a=1$.
Your first idea gives a proof.
Indeed, by AM-GM
$$\sum_{cyc}\frac{a}{\sqrt{6a^2+5ab+b^2}}=\sum_{cyc}\left(\frac{a}{2\sqrt3}\cdot\sqrt{\frac{1}{\frac{3a+b}{4}}\cdot\frac{1}{\frac{2a+b}{3}}}\right)\leq\sum_{cyc}\frac{a}{4\sqrt3}\left(\frac{4}{3a+b}+\frac{3}{2a+b}\right).$$
Thus, it's enough to prove that
$$\sum_{cyc}\left(\frac{4a}{3a+b}+\frac{3a}{2a+b}\right)\leq6$$ or
$$\sum_{cyc}(12a^4b^2+19a^4c^2+53a^3b^3+53a^4bc-112a^3b^2c+78a^3c^2b-103a^2b^2c^2)\geq0,$$
which is true by SOS, Rearrangement and AM-GM.
For example,
$$\sum_{cyc}(a^4c^2-a^3b^2c)=a^2b^2c^2\sum_{cyc}\left(\frac{a^2}{b^2}-\frac{a}{c}\right)=$$
$$=\frac{1}{2}a^2b^2c^2\sum_{cyc}\left(\frac{a^2}{b^2}+\frac{b^2}{c^2}-\frac{2a}{c}\right)=\frac{1}{2}a^2b^2c^2\sum_{cyc}\left(\frac{a}{b}-\frac{b}{c}\right)^2\geq0.$$
Can you end it now?
Also, the Tangent Line method helps.
Let $\frac{b}{a}=x$, $\frac{c}{b}=y$ and $\frac{a}{c}=z$.
Thus, $xyz=1$ and
$$\frac{\sqrt3}{2}-\sum_{cyc}\sqrt{\frac{a^2}{6a^2+5ab+b^2}}=\sum_{cyc}\left(\frac{1}{2\sqrt3}-\frac{1}{\sqrt{x^2+5x+6}}\right)=$$
$$=\sum_{cyc}\left(\frac{1}{2\sqrt3}-\frac{1}{\sqrt{x^2+5x+6}}-\frac{7}{48\sqrt3}\ln{x}\right)\geq0$$ because easy to show that
$$\frac{1}{2\sqrt3}-\frac{1}{\sqrt{x^2+5x+6}}-\frac{7}{48\sqrt3}\ln{x}\geq0$$ for all $x>0$.
|
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Solve the recurrence $b_1 = 2$ for $b_n = 3b_{n-1} + 5$
Solve the recurrence $b_1 = 2$ for $$b_n = 3b_{n-1} + 5$$
I've tried solving this problem using iteration, but the formula I get in the end is wrong. It is not a closed formula since there's still recurrence. I think my error is starts from the last line below. I'm not sure how to fix it. The formula I get in the end is:
$2(3^{n-1})+ 5((3^n - 1) / 3)$
which I got by setting $k = n -1$ and using $\sum_{k=0}^{n-1}3^k\ = (3^n-1)/2$
\begin{align*}
b_n&=3b_{n-1}+5\\
&=3(3b_{n-2}+5)+5\\
&=3^2b_{n-2}+3\cdot5+5\\
&=3^2(3b_{n-3}+5)+3\cdot5+5\\
&=3^3b_{n-3}+3^2\cdot5-3\cdot5+5\\
&\;\vdots\\
&=3^kb_{n-k}+3^{k-1}\cdot5+3^{k-2}\cdot5+\ldots+3\cdot5+5\\
\end{align*}
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Let $a_{n} = b_n+c$ for some $c$ so that $a_{n+1}=3a_n$ (i.e. $a_n$ is geometric sequence). Then $$ a_{n+1}-c= 3a_n-3c+5\implies c = 5/2$$
and $$a_n= a_03^n$$ so $$ b_n = a_03^n-5/2$$
Since $b_1 = 2$ we get $3a_0=9/2$ so $a_0 =3/2$ so we finaly have $$ b_n = {3^{n+1}-5\over 2}$$
|
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|
All the roots of $5\cos x - \sin x = 4$ in the interval $0^{\circ} \leq x \leq 360^{\circ}$? This is a problem that I stumbled upon in one of my books.
Representing $5\cos x - \sin x$ in the form $R\cos(x + \alpha)$ (as demanded by the question):
$
\rightarrow R = \sqrt{5^2 + ({-}1)^2} = \sqrt{26}\\
\rightarrow R\cos x \cos \alpha - R\sin x \sin \alpha = 5\cos x - \sin x \\
\rightarrow ➊\hspace{0.25cm}5 = \sqrt{26}\cos \alpha \\
\rightarrow ➋\hspace{0.25cm}{-}1 = \sqrt{26}\sin \alpha \\
$
So here are my question(s): $\\$
• Why is only ➊ working out?
• When I find one solution in the interval, which is 27°, why is another solution like $(360 - x)$ not working out (in this case 333°)?
• I see 310.4° as a solution to this equation in my book. How do you get to 310.4° from 27° (which is the solution I got)? Which $\cos$ identity am I missing? And surprisingly, my book doesn't include 27° as a solution!
|
\begin{eqnarray}
5\cos x - \sin x &=& 4\\
\frac{5}{\sqrt{26}}\cos x-\frac{1}{\sqrt{26}}\sin x&=&\frac{4}{\sqrt{26}}\\
\cos\alpha\cos x-\sin\alpha\sin x&=&\frac{4}{\sqrt{26}}\\
\end{eqnarray}
Giving
$$\cos(x+\alpha)=\frac{4}{\sqrt{26}}$$
So either
$$ x+\alpha=\arccos\left(\frac{4}{\sqrt{26}}\right)=38.33^\circ $$
or
$$ x+\alpha=360^\circ-\arccos\left(\frac{4}{\sqrt{26}}\right)=321.67^\circ $$
with
$$\alpha=\arccos\left(\frac{5}{\sqrt{26}}\right)=11.31$$
So you get that either $x=27.02^\circ$ or $x=310.36^\circ$.
|
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|
$e^{\frac{1}{x}} < 1 + \frac{1}{x-1} $ I want to prove that $e^{1/x} < 1 + \frac{1}{x-1}$ for $x > 1$.
The first thing I tried is differentiating $f(x) = e^{1/x} - 1 - \frac{1}{x-1}$: this gives
$$ \frac{1}{x^2} \left( \left(1 + \frac{1}{x-1}\right)^2 - e^{\frac{1}{x}} \right) $$
If I could show that $\left(1 + \frac{1}{x-1} \right)^2 > e^\frac{1}{x} $ for $x>1$, then $f(x)$ would be increassing, and since $ \lim_{x \to \infty} f(x) = 0$ this would mean that $f(x) < 0$ for $x>1$. However, proving that inequality is very similar to the first one, and still involves bounding above $e^\frac{1}{x}$.
The other thing I tried is considering $f(x) = e^{x-1} - (x-1) - 1$ which is increasing for $x > 1$. Then $f(\frac{1}{x})$ is decreasing, so $e^{\frac{1}{x}-1} - \frac{1}{x}$ is decreasing; However this also does not seem to help too much.
Any ideas?
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For $x>1$, $\frac{1}{x-1}=\frac{1}{x}\frac{1}{1-1/x}$ is a convergent geometric series. Let's return to your title inequality. The left-hand side is $\sum_{n\ge 0}\frac{1}{n!x^n}$; the right-hand side is $\sum_{n\ge 0}\frac{1}{x^n}$.
|
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|
How to find the integration factor and solve $(x^3+xy^2-y) dx$ $+$ $(y^3+x^2y+x) dy$ $=$ $0$? $(x^3+xy^2-y)dx+(y^3+x^2y+x)dy=0$
I tried to find on Wolfram Alpha but it showed only the solution not in step-by-step, I know that the factor is $\frac{1}{x^2+y^2}$ but how to find it?
|
$$(x^3+xy^2-y)dx+(y^3+x^2y+x)dy=0 \tag 1$$
If we a-priori know that the integrating factor $\mu$ is a function of $(x^2+y^2)$ it is easy to find that $\mu=\frac{1}{x^2+y^2}$ : For example see the Aleksas Domarkas's answer.
But what to do if we don't know that $\mu$ is a function of $(x^2+y^2)$ ?
Suppose that we did try various forms of integrating factor such as $\mu=f(x)$ or $\mu=f(y)$ or $\mu=f(xy)$ or $\mu=f(x/y)$, etc. and that all failed. What to do, that is the question !
In order to make $\quad (x^3+xy^2-y)\mu dx+(y^3+x^2y+x)\mu dy\quad$ an exact differential it is necessary that :
$$\frac{\partial}{\partial y}\left((x^3+xy^2-y)\mu \right) = \frac{\partial}{\partial x}\left((y^3+x^2y+x)\mu \right)$$
$$(x^3+xy^2-y)\mu_y+(2xy-1)\mu = (y^3+x^2y+x)\mu_x+(2xy+1)\mu$$
$$(y^3+x^2y+x)\mu_x -(x^3+xy^2-y)\mu_y = -2\mu$$
This is a PDE. Solving it would provide an infinity of solutions $\mu(x,y)$ since the general solution of a first order linear PDE without boundary condition involves an arbitrary function.
But this would be a vicious circle because solving this PDE for the general solution requires to solve the ODE $(1)$ when we look for a first characteristic equation.
In fact, we don't need the general solution. We only need a characteristic equation with $\mu$ in it. This will be made clear latter.
The Charpit-Lagrange system of ODEs is :
$$\frac{dx}{(y^3+x^2y+x)}=\frac{dy}{-(x^3+xy^2-y)} =\frac{d\mu}{-2\mu}$$
A first characteristic equation would come from $\frac{dx}{(y^3+x^2y+x)}=\frac{dy}{-(x^3+xy^2-y)}$ . But this is a vicious circle as already mentioned.
A second characteristic equation comes from
$\frac{dx}{(y^3+x^2y+x)}=\frac{dy}{- (x^3+xy^2-y) }=\frac{xdx+ydy}{x(y^3+x^2y+x)-y(x^3+xy^2-y)} =\frac{xdx+ydy}{x^2+y^2}=\frac{d\mu}{-2\mu}$
$\frac{d(x^2+y^2)}{x^2+y^2}=\frac{d\mu}{-\mu}$
$$\mu(x^2+y^2)=c_2$$
This is a characteristic equation which contains $\mu$. This is sufficient to have at least one solution of the PDE with $c_2=1$ for example :
$$\mu=\frac{1}{x^2+y^2}$$
So, thanks to this method, we found an integrating factor straightforward, without by trial and error and with no initial clue. The drawback is that one has to be familiar with the method of solving first order linear PDEs.
Once the integrating factor known, it is easy to solve the exact differential equation :
$$(x^3+xy^2-y)\frac{1}{x^2+y^2}dx+(y^3+x^2y+x)\frac{1}{x^2+y^2}dy=0$$
$\begin{cases}
\int \frac{x^3+xy^2-y}{x^2+y^2}dx = \frac{x^2}{2}+\tan^{-1}(\frac{y}{x}) +f(y)\\
\int \frac{y^3+x^2y+x}{x^2+y^2}dy = \frac{y^2}{2}+\tan^{-1}(\frac{y}{x}) +g(x)
\end{cases}
\quad\implies\quad f(y)=\frac{y^2}{2} \text{ and }g(x)=\frac{x^2}{2}$
$$d\left(\frac{x^2+y^2}{2}+\tan^{-1}(\frac{y}{x}) \right)=0$$
$$\frac{x^2+y^2}{2}+\tan^{-1}(\frac{y}{x}) =c$$
|
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|
Find all polynomials such that $p(x^2-2x)=p(x-2)^2$
Find all polynomials $p\in \mathbb{C}[x]$ such that $$p(x^2-2x)=p(x-2)^2$$
We can not say anything specific about the degree since both sides are of the degree $2n$.
Also by copering the coeficients we see that the leading coefficent must be $1$.
Setting $$x^2-2x = x-2\implies x\in\{1,2\}$$
If $x= 2$ we get $p(0)=p(0)^2$ so $p(0)\in\{0,1\}$.
If $x= 1$ we get $p(-1)=p(-1)^2$ so $p(-1)\in\{0,1\}$.
Now it sems there is a lot of options. Also if $p$ is linear we have $$a(x^2-2x)+b= (ax-2a+b)^2 =a^2x^2+2a(b-2a)x+(b-2a)^2$$
so $a=a^2$ and $b=(b-2a)^2$ and $-2a=2a(b-2a)$ so $a=1$ and $b=1$ thus $p(x)=x+1$. But how to finish in general?
|
Substitute $y = x-1$, we see that $p(y^2-1) = (p(y-1))^2$. Take $q(x) = p(x-1)$ we get $q(x^2) = (q(x))^2$. Take $r(x) = q(e^x))$. we get $r(2x) = r(x)^2$. Take $s(x) = \log(r(x))$ we get $s(2x) = 2s(x)$. As $s(x)$ is defined at $(0,\infty)$ and continuous we know that the only possibility is $s(x) = cx$ for some constant $c$. Plugging all the way back you see $p(x) = (x + 1)^n$ for some $n$.
|
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"language": "en",
"url": "https://math.stackexchange.com/questions/3149891",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
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|
integral $C(a,b)=\int_0^{2\pi}\frac{xdx}{a+b\cos^2 x}$ I am looking for other methods to find the general integral
$$C(a,b)=\int_0^{2\pi}\frac{xdx}{a+b\cos^2x}$$
To do so, I first preformed $u=x-\pi$:
$$C(a,b)=\int_{-\pi}^{\pi}\frac{xdx}{a+b\cos^2x}+\pi\int_{-\pi}^{\pi}\frac{dx}{a+b\cos^2x}$$
The sub $x\mapsto -x$ provides
$$\int_{-\pi}^\pi \frac{xdx}{a+b\cos^2x}=0$$
and with symmetry,
$$C(a,b)=2\pi\int_0^\pi \frac{dx}{a+b\cos^2x}$$
Then we use $t=\tan(x/2)$:
$$C(a,b)=4\pi\int_0^\infty \frac1{a+b\left[\frac{t^2-1}{t^2+1}\right]^2}\frac{dt}{t^2+1}$$
$$C(a,b)=\frac{4\pi}{a+b}\int_0^\infty \frac{x^2+1}{x^4+2\frac{a-b}{a+b}x^2+1}dx$$
Then we consider
$$N_s(k)=\int_0^\infty\frac{x^{2s}}{x^4+2kx^2+1}dx$$
Then $x\mapsto 1/x$ gives
$$N_s(k)=N_{1-s}(k)$$
and with $s=0$:
$$C(a,b)=\frac{8\pi}{a+b}\int_0^\infty \frac{dx}{x^4+2kx^2+1}\qquad k=\frac{a-b}{a+b}$$
Then with
$$x^4+(2-c^2)x^2+1=(x^2+cx+1)(x^2-cx+1)$$
we have that
$$\begin{align}
u(c)=N_0\left(\frac{2-c^2}2\right)=&\frac1{4c}\int_0^\infty\frac{2x+c}{x^2+cx+1}dx-\frac1{4c}\int_0^\infty\frac{2x-c}{x^2-cx+1}dx\\
&+\frac14\int_0^\infty\frac{dx}{x^2+cx+1}+\frac14\int_0^\infty\frac{dx}{x^2-cx+1}
\end{align}$$
The first two integrals vanish and we have
$$u(c)=\frac14I(1,c,1)+\frac14I(1,-c,1)$$
where
$$I(a,b,c)=\int_0^\infty\frac{dx}{ax^2+bx+c}=\frac2{\sqrt{4ac-b^2}}\left[\frac\pi2-\arctan\frac{b}{\sqrt{4ac-b^2}}\right]$$
So
$$4u(c)=\frac2{\sqrt{4-c^2}}\left[\frac\pi2-\arctan\frac{c}{\sqrt{4-c^2}}+\frac\pi2-\arctan\frac{-c}{\sqrt{4-c^2}}\right]$$
$$u(c)=\frac{\pi}{2\sqrt{4-c^2}}$$
$$N_0(a)=\frac\pi{2\sqrt{2+2a}}$$
And then
$$C(a,b)=\frac{8\pi}{a+b}N_0\left(\frac{a-b}{a+b}\right)$$
$$C(a,b)=\frac{4\pi^2}{\sqrt{a^2+ab}}$$
How else can you prove this? Have fun ;)
As it turns out, we may be missing a factor of $1/2$, so we may actually have
$$C(a,b)=\frac{2\pi^2}{\sqrt{a^2+ab}}$$
an issue discussed in the comment section of @Song's answer.
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By making change of variable $u=2\pi -x$, we have
$$\begin{align*}
C(a,b)&=\int_0^{2\pi}\frac{2\pi -u}{a+b\cos^2 u}du=\int_0^{2\pi}\frac{2\pi}{a+b\cos^2 u}du-C(a,b),
\end{align*}$$ hence $\displaystyle C(a,b)=\pi\int_0^{2\pi}\frac{1}{a+b\cos^2 u}du$. On the other hand, we find that
$$\begin{align*}
\int_0^{2\pi}\frac{1}{a+b\cos^2 u}du&=4\int_0^{\frac\pi 2}\frac{1}{a\sin^2 u+(b+1)\cos^2 u}du
\\&=4\int_0^{\frac\pi 2}\frac{1}{a\tan^2 u+b+1}\sec^2 u\ du
\\&=4\int_0^\infty \frac{dv}{av^2+(b+1)}\tag{$v=\tan u$}
\\&=\frac{4}{\sqrt{a^2+ab}}\left[\arctan\left(\frac{\sqrt{a}v}{\sqrt{b+1}}\right)\right]^\infty_0\\&=\frac{2\pi}{\sqrt{a^2+ab}}.
\end{align*}$$ This gives $$
C(a,b)=\frac{2\pi^2}{\sqrt{a^2+ab}}.
$$ (I guess $2\pi^2$ is the right constant since this gives the right result $2\pi^2$ when $a=1,b=0$.)
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"language": "en",
"url": "https://math.stackexchange.com/questions/3154937",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 0
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Equality of $2^x + 2^{4-x} \geq 8$
Want to find the value(s) of $x$ for which equality holds in $$2^x + 2^{4-x} \geq 8$$
I've found it by solving $2^x + 2^{4-x} = 8$:
$$2^x + 2^{4-x} = 8 \Rightarrow 2^{2x} - 8 \cdot 2^x + 16 = 0 \Rightarrow (2^x - 4)^2 = 0$$
so clearly $x = 2$.
However, the notes I've been reading goes straight to simply say that equality occurs when $2^x = 2^{4-x}$, i.e. when $x = 4 - x$ and again $x=2$.
I'm probably missing something really simple here, but why does equality hold simply when $$2^x = 2^{4-x}?$$
Edit: to clarify, I want to know why is this is an 'obvious' condition for equality without writing the equals sign and solving the quadratic?
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$(2^{x/2}-2^{(2-x/2)})^2 \ge 0;$
$2^x +2^{4-x} -2(2^{x/2}2^{(2-x/2)}) \ge 0;$
$2^x +2^{4-x} \ge 8;$
Equality for $2^{x/2}= 2^{(2-x/2)}$ (Why?);
$2^x=2^2$, or $x=2$.
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{
"language": "en",
"url": "https://math.stackexchange.com/questions/3155460",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
}
|
Subsets and Splits
Fractions in Questions and Answers
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