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How to compute $\int_0^\infty \frac{x^4}{(x^4+ x^2 +1)^3} dx =\frac{\pi}{48\sqrt{3}}$? $$\int_0^\infty \frac{x^4}{(x^4+ x^2 +1)^3} dx =\frac{\pi}{48\sqrt{3}}$$
I have difficulty to evaluating above integrals.
First I try the substitution $x^4 =t$ or $x^4 +x^2+1 =t$ but it makes integral worse.
Using Mathematica I found the result $\dfrac{\pi}{48\sqrt{3}}$ I want to know the procedure of evaluating this integral.
| A yet another approach:
\begin{align}
I&=\int_0^\infty \frac{x^4}{(x^4+ x^2 +1)^3} dx\\
&=\int_0^1 \frac{x^4}{(x^4+ x^2 +1)^3} dx+\int_1^\infty \frac{x^4}{(x^4+ x^2 +1)^3} dx\\
&=\int_0^1\frac{x^4}{(x^4+ x^2 +1)^3} dx+\int_0^1\frac{x^6}{(x^4+ x^2 +1)^3} dx\\
\end{align}
where I used $x \to 1/x$.
Now use $x\to \tan x$ to obtain
\begin{align}
I&=\int_0^{\pi/4} \frac{32\sin^4 2x}{(7+\cos 4x)^3} dx
\end{align}
which is manageable.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1290669",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "14",
"answer_count": 3,
"answer_id": 1
} |
Normal convergence: $\sum_{n = 1}^{\infty} \frac{x}{(1 + (n-1)x)(1 + nx)}$ I want to prove that:
$$\sum_{n = 1}^{\infty} \frac{x}{(1 + (n-1)x)(1 + nx)}$$
is not normally convergent on $[a, \infty)$ for fixed $a>0$.
Let $U_n(x)$ denote the general term.
We have:
$$||U_n||_{\infty} = \sup_{x \ge a} \left( \frac{x}{(1 + (n-1)x)(1 + nx)} \right) \ge \sup_{x \ge a} \left( \frac{x}{(1 + (n-1)a)(1 + na)} \right) \ge \sup_{x \ge a} \left( \frac{x}{(1 + na)^2} \right) \ge \frac{1}{(1 + a)^2} \sup_{x \ge a} \left( \frac{x}{n^2} \right) \ge \frac{1}{(1+ a)^2} \times \frac{n}{n^2} = \frac{1}{(1+a)^2} \frac{1}{n} = a_n$$
The series of general term $a_n$ is divergent, so by comparison test the series of norms is divergent.
Can someone check/criticize my work? Thanks.
Bonus question: what are the sets on which we have normal convergence?
| Your first inequality is invalid, if you make the (positive) denominator smaller, since the numerator is also positive, you make the fraction larger.
Actually, the series is normally convergent on $[a,+\infty)$ for all $a > 0$, since for $n \geqslant 2$
$$0 < U_n(x) = \frac{x}{(1+(n-1)x)(1+nx)} < \frac{x}{(n-1)xnx} = \frac{1}{n(n-1)x} \leqslant \frac{1}{n(n-1)a},$$
and
$$0 < U_1(x) = \frac{x}{1+x} < 1.$$
The series $\sum_{n=2}^\infty \frac{1}{n(n-1)}$ is convergent, so
$$\sum_{n=1}^\infty \lVert U_n\rVert_\infty \leqslant 1 + \frac{1}{a}\sum_{n=2}^\infty \frac{1}{n(n-1)} = 1+\frac{1}{a} < +\infty.$$
If there is a problem, it occurs at $0$. For $n \geqslant 2$, we compute
$$U_n'(x) = \frac{(1+(n-1)x)(1+nx) - x[(n-1)(1+nx) + n(1+(n-1)x)]}{[(1+(n-1)x)(1+nx)]^2}$$
and to locate the maximum solve
\begin{align}
&&(1+(n-1)x)(1+nx) &= x[2n-1 + 2n(n-1)x]\\
&\iff& 1 + (2n-1)x + n(n-1)x^2 &= (2n-1)x + 2n(n-1)x^2\\
&\iff& 1 &= n(n-1)x^2\\
&\iff& \frac{1}{\sqrt{n(n-1)}} &= x,
\end{align}
where the last is an equivalence since we only consider $x \geqslant 0$. Thus
\begin{align}
\lVert U_n\rVert_\infty &= U_n \biggl(\frac{1}{\sqrt{n(n-1)}}\biggr) \\
&= \frac{1}{\Bigl(1 + \sqrt{\frac{n-1}{n}}\Bigr)\Bigl(1 + \sqrt{\frac{n}{n-1}}\Bigr)\sqrt{n(n-1)}}\\
&= \frac{1}{(\sqrt{n} + \sqrt{n-1})(\sqrt{n-1}+\sqrt{n})}\\
&= \frac{1}{(\sqrt{n}+\sqrt{n-1})^2}\\
&= \frac{1}{2n-1 + 2\sqrt{n(n-1)}}.
\end{align}
Since $\lVert U_n\rVert_\infty > \frac{1}{4n}$, we see that the series is not normally convergent on all of $[0,+\infty)$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1292844",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Derivative of $f(x) = \frac{\cos{(x^2 - 1)}}{2x}$ Find the derivative of the function $$f(x) = \frac{\cos{(x^2 - 1)}}{2x}$$
This is my step-by-step solution: $$f'(x) = \frac{-\sin{(x^2 - 1)}2x - 2\cos{(x^2 -1)}}{4x^2} = \frac{2x\sin{(1 - x^2)} - 2\cos{(1 - x^2)}}{4x^2} = \frac{x\sin{(1 - x^2)} - \cos{(1 - x^2)}}{2x^2} = \frac{\sin{(1 - x^2)}}{2x} - \frac{\cos{(1-x^2)}}{2x^2}$$
and this is the output of WolphramAlpha: http://www.wolframalpha.com/input/?i=derivative+cos(x^2+-+1)%2F(2x)
Where is the mistakes?
| $$f(x) = \frac{\cos{(x^2 - 1)}}{2x}\Longrightarrow$$
$$\frac{df(x)}{dx}\left(\frac{\cos{(x^2 - 1)}}{2x}\right)=$$
$$\frac{1}{2}\left(\frac{df(x)}{dx}\left(\frac{\cos{(x^2 - 1)}}{x}\right)\right)=$$
$$\frac{1}{2}\left(\frac{x\frac{df(x)}{x}(\cos(1-x^2))-\cos(1-x^2)\frac{df(x)}{dx}x}{x^2}\right)=$$
$$\frac{x\frac{df(x)}{x}(\cos(1-x^2))-1\cos(1-x^2)}{2x^2}=$$
$$\frac{-\cos(1-x^2)+2x(x\sin(1-x^2))}{2x^2}=$$
$$\frac{-\cos(1-x^2)+x^2\sin(1-x^2)}{2x^2}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1293272",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
"answer_id": 3
} |
Do the spaces spanned by the columns of the given matrices coincide? Reviewing linear algebra here.
Let $$A = \begin{pmatrix}
1 & 1 & 0 & 0 \\
1 & 1 & 0 & 0 \\
0 & 0 & 1 & 0 \\
0 & 0 & 1 & 1
\end{pmatrix} \qquad \text{and} \qquad
B = \begin{pmatrix}
1 & 0 & 0 \\
1 & 0 & 0 \\
0 & 1 & 0 \\
0 & 0 & 1
\end{pmatrix}.$$
Is the space spanned by the columns of $A$ the same as the space spanned by the columns of $B$?
My answer: Yes. For $A$, we have
$$C(A) =
\left\{\left.a_1\begin{pmatrix}
1 \\
1 \\
0 \\
0
\end{pmatrix}+ a_2
\begin{pmatrix}
1 \\
1 \\
0 \\
0
\end{pmatrix}+ a_3
\begin{pmatrix}
0 \\
0 \\
1 \\
1
\end{pmatrix}+ a_4
\begin{pmatrix}
0 \\
0 \\
0 \\
1
\end{pmatrix} =
\begin{pmatrix}
a_1 + a_2 \\
a_1 + a_2 \\
a_3 \\
a_3 + a_4
\end{pmatrix} \,
\right\vert a_i \in \mathbf{R}\right\}\text{.}
$$
Denote $a_1 + a_2 = c_1$, $a_3 = c_2$, and $a_3 + a_4 = c_3$. Then since $a_4$ is arbitrary, any vector in $C(A)$ can be described as $(c_1, c_1, c_2, c_3)^{\prime}$.
For $B$, we have
$$
C(B) =
\left\{\left.b_1\begin{pmatrix}
1 \\
1 \\
0 \\
0
\end{pmatrix}+ b_2
\begin{pmatrix}
0 \\
0 \\
1 \\
0
\end{pmatrix}+ b_3
\begin{pmatrix}
0 \\
0 \\
0 \\
1
\end{pmatrix} =
\begin{pmatrix}
b_1 \\
b_1 \\
b_2 \\
b_3
\end{pmatrix} \,
\right\vert b_i \in \mathbf{R}\right\}\text{.}
$$
Thus, any vector in $C(B)$ can be described as $(b_1, b_1, b_2, b_3)^{\prime}$, the same form as a vector in $C(A)$.
| There is a very elementary solution. By applying the column transformations $C_1\rightarrow C_1-C_2$ and $C_3 \rightarrow C_3-C_4$ to $A$ you get $$A^*=
\begin{bmatrix}
0 & 1 & 0 & 0 \\
0 & 1 & 0 & 0 \\
0 & 0 & 1 & 0 \\
0 & 0 & 0 & 1
\end{bmatrix}$$
If you look at the colums of $A^*$ and $B$ they are the same. (except of the zero vector which does not contribute anything to the span). And $A^*$ and $A$ are column equivalent. So they have the same span.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1293755",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Evaluate the double sum $\sum_{m=1}^{\infty}\sum_{n=1}^{m-1}\frac{ 1}{m n\left(m^2-n^2\right)^2}$ As a follow up of this nice question I am interested in
$$
S_1=\sum_{m=1}^{\infty}\sum_{n=1}^{m-1}\frac{ 1}{m n\left(m^2-n^2\right)^2}
$$
Furthermore, I would be also very grateful for a solution to
$$
S_2=\sum_{m=1}^{\infty}\sum_{n=m+1}^{\infty}\frac{ 1}{m n\left(m^2-n^2\right)^2}
$$
Following my answer in the question mentioned above and the numerical experiments of @Vladimir Reshetnikov it's very likely that at least
$$
S_1+S_2 = \frac{a}{b}\pi^6
$$
I think both sums may be evaluated by using partial fraction decomposition and the integral representation of the Polygamma function but I don't know how exactly and I guess there could be a much more efficient route.
| Clearly, $S_1$=$S_2$ (this can be shown by reversing the order of summation, as was noted above).
Using
$$
\frac{ 1}{m n\left(m^2-n^2\right)^2}=\frac{ (m+n)^2-(m-n)^2}{4 m^2 n^2\left(m^2-n^2\right)^2}
$$
we get
$$
S_1=\sum_{m=1}^{\infty}\sum_{n=1}^{m-1}\frac{ 1}{m n\left(m^2-n^2\right)^2}=\frac{1}{4}\sum_{m=1}^{\infty}\sum_{n=1}^{m-1}\frac{ 1}{m^2 n^2\left(m-n\right)^2}-\frac{1}{4}\sum_{m=1}^{\infty}\sum_{n=1}^{m-1}\frac{ 1}{m^2 n^2\left(m+n\right)^2},
$$
and after reversing the order of summation in the first sum
$$
S_1=\frac{1}{4}\sum_{n=1}^{\infty}\sum_{m=n+1}^{\infty}\frac{ 1}{m^2 n^2\left(m-n\right)^2}-\frac{1}{4}\sum_{m=1}^{\infty}\sum_{n=1}^{m-1}\frac{ 1}{m^2 n^2\left(m+n\right)^2}=\\
\frac{1}{4}\sum_{n=1}^{\infty}\sum_{m=1}^{\infty}\frac{ 1}{m^2 n^2\left(m+n\right)^2}-\frac{1}{4}\sum_{m=1}^{\infty}\sum_{n=1}^{m-1}\frac{ 1}{m^2 n^2\left(m+n\right)^2}. \qquad\qquad (1)
$$
Let's introduce a third sum
$$
S_3=\sum_{m=1}^{\infty}\sum_{n=1}^{m-1}\frac{ 1}{m^2 n^2\left(m+n\right)^2}
=\sum_{n=1}^{\infty}\sum_{m=n+1}^{\infty}\frac{ 1}{m^2 n^2\left(m+n\right)^2}=\\
\frac{1}{2}\sum_{m=1}^{\infty}\sum_{n=1}^{\infty}\frac{ 1}{m^2 n^2\left(m+n\right)^2}-\frac{1}{8}\zeta(6).
$$
Using An Infinite Double Summation $\sum_{k=1}^{\infty} \sum_{n=1}^{\infty} \frac{1}{n^2k^2(n+k)^2}$? we get
$$
S_3=\frac{1}{2}\cdot\frac{1}{3}\zeta(6)-\frac{1}{8}\zeta(6)=\frac{1}{24}\zeta(6).\qquad\qquad\qquad (2)
$$
From (1) and (2) we get
$$
S_1=\frac{1}{4}\cdot\frac{1}{3}\zeta(6)-\frac{1}{4}\cdot\frac{1}{24}\zeta(6)=\frac{7}{96}\zeta(6)=\frac{7}{96}\frac{\pi^6}{945}=\frac{\pi^6}{12960}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1294185",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "22",
"answer_count": 2,
"answer_id": 0
} |
Calculating $\int_0^\pi \frac{1}{a+b\sin^2(x)} dx $ How do I calculate this integral? $a \gt b$ is given.
$$\int_0^\pi \frac{1}{a+b\sin^2(x)} dx $$
I am confused since WolframAlpha says one the one hand, that $F(\pi) = F(0) = 0 $ , but with some random values it isn't 0.
What am I missing?
Note that I am not really interested in a complete antiderivative, I am more interested in a $G(a,b) = F(\pi) - F(0) = \, ... $
| Suppose we seek to evaluate
$$\frac{1}{2} \int_0^{2\pi} \frac{1}{a+b\sin^2 x} dx.$$
Put $z = \exp(ix)$ so that $dz = i\exp(ix) \; dx$ and hence
$\frac{dz}{iz} = dx$ to obtain
$$\frac{1}{2} \int_{|z|=1}
\frac{1}{a+b(z-1/z)^2/4/(-1)}\frac{dz}{iz}
\\ = \frac{1}{2} \int_{|z|=1}
\frac{4}{4a-b(z-1/z)^2}\frac{dz}{iz}
\\ = \frac{2}{i} \int_{|z|=1}
\frac{z}{4a-b(z-1/z)^2}\frac{dz}{z^2}
\\ = \frac{2}{i} \int_{|z|=1}
\frac{z}{4az^2-b(z^2-1)^2} dz
\\ = \frac{2}{i} \int_{|z|=1}
\frac{z}{-bz^4+(2b+4a)z^2-b} dz.$$
The poles here are all simple and located at
$$\rho_{1,2,3,4}
= \pm\sqrt{\frac{2a+b}{b} \pm
\frac{2\sqrt{a^2+ab}}{b}}.$$
Re-write this as
$$\rho_{1,2,3,4}
= \pm\sqrt{1+\frac{2a}{b} \pm
\frac{2\sqrt{a^2+ab}}{b}}.$$
With $a$ and $b$ positive the first two poles are clearly not inside
the contour (modulus larger than one). That leaves
$$\rho_{3,4}
= \pm\sqrt{1+\frac{2a}{b}
- \frac{2\sqrt{a^2+ab}}{b}}.$$
Now we have
$$1+\frac{2a}{b}
- \frac{2\sqrt{a^2+ab}}{b} < 1$$
and also
$$1+\frac{2a}{b}
- \frac{2\sqrt{a^2+ab}}{b} > 0$$
since
$$(b+2a)^2 = b^2+4ab+4a^2 > 4(a^2+ab)$$
and therefore these poles are indeed inside the contour.
The residues are given by
$$\left. \frac{z}{-4bz^3+2(2b+4a)z}
\right|_{z=\rho_{3,4}}
= \left. \frac{1}{-4bz^2+2(2b+4a)}
\right|_{z=\rho_{3,4}}.$$
This is
$$\frac{1}{-4b-8a+8\sqrt{a^2+ab}
+2(2b+4a)}
= \frac{1}{8\sqrt{a^2+ab}}.$$
It follows that the desired value is
$$\frac{2}{i}\times 2\pi i
\times \frac{2}{8\sqrt{a^2+ab}}
= \frac{\pi}{\sqrt{a}\sqrt{a+b}}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1296351",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
} |
Show that $\frac {\sin x} {\cos 3x} + \frac {\sin 3x} {\cos 9x} + \frac {\sin 9x} {\cos 27x} = \frac 12 (\tan 27x - \tan x)$ The question asks to prove that -
$$\frac {\sin x} {\cos 3x} + \frac {\sin 3x} {\cos 9x} + \frac {\sin 9x} {\cos 27x} = \frac 12 (\tan 27x - \tan x) $$
I tried combining the first two or the last two fractions on the L.H.S to allow me to use the double angle formula and get $\sin 6x$ or $\sin 18x$ but that did not help at all.
I'm pretty sure that if I express everything in terms of $x$, the answer will ultimately appear but I'm also certain that there must be another simpler way.
| In general, we have
$$\sum_{k=1}^n \dfrac{\sin\left(3^{k-1}x \right)}{\cos\left(3^kx\right)} = \dfrac{\tan\left(3^nx\right)-\tan(x)}2$$
We can prove this using induction. $n=1$ is trivial. All we need to make use of is that
$$\tan(3t) = \dfrac{3\tan(t)-\tan^3(t)}{1-3\tan^2(t)}$$
For the inductive step, we have
$$\sum_{k=1}^n \dfrac{\sin\left(3^{k-1}x \right)}{\cos\left(3^kx\right)} = \dfrac{\tan\left(3^nx\right)-\tan(x)}2$$
Adding the last term, we obtain
$$\sum_{k=1}^{n+1} \dfrac{\sin\left(3^{k-1}x \right)}{\cos\left(3^kx\right)} = \dfrac{\tan\left(3^nx\right)-\tan(x)}2 + \dfrac{\sin\left(3^{n}x \right)}{\cos\left(3^{n+1}x\right)} = \dfrac{\tan\left(3^{n+1}x\right)-\tan(x)}2$$
where we again rely on $(\spadesuit)$ to express $\tan\left(3^{n+1}x\right)$ in terms of $\tan\left(3^{n}x\right)$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1299957",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 0
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Proving a differential equation is a circle So, I have solved the differential equation, to find the general solution of:
$$\frac{y^2}{2} = 2x - \frac{x^2}{2} + c$$
I am told that is passes through the point $(4,2)$. Using this information, I found $C$ to be $2$. Then I am asked to prove that the equation forms a circle, and find the radius and centre point. So I know to try and get it into the form:
$$x^2 + y^2 = r^2$$
So I can get:
$$\frac{y^2}{2} + \frac{x^2}{2} = 2x + 2$$
Multiply both sides by $2$:
$$y^2 + x^2 = 4x + 4$$
But from here I am not sure how to group the $x$'s to leave only a constant on the RHS. Any guidance would be much appreciated.
| You have,
$$ y^2 +x^2 = 4x+4,$$
what we need to do is group the $x$'s together and complete the square,
$$ y^2 + \left( x^2-4x \right) = 4.$$
So we need to complete the square for the expression $x^2-4x$. The idea is to write this polynomial as a perfect square plus a constant. To do this we first consider the expression $(x+a)^2 = x^2 + 2ax + a^2$. This motivates us to rewrite our polynomial as,
$$ x^2-4x = x^2-2(2)x = (x-2)^2-2^2 = (x-2)^2-4$$
we can then substitute this back into the original equation to get,
$$ y^2 + \left( x-2\right)^2 = 8.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1300121",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 1
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How to prove that the ring of all algebraic integers is a Bézout domain? I was told that the ring of all algebraic integers (that is, the complex numbers which are roots of a monic polynomials with integral coefficients) is a Bézout domain. But I have no idea how to prove it. Can anyone help me with this? Thanks in advance.
| First, an example:
Consider the algebraic number field $\mathbb{Q}(\sqrt{-5})$. Its class number is two. That means there exist ideals of the ring of integers $\mathbb{Z}[\sqrt{-5}]$ of this field that are not principal, but the square of any ideal is principal. For instance, we have the decomposition
$$6 = 2 \cdot 3 = ( 1 + \sqrt{-5})\cdot (1-\sqrt{-5})$$
The ideals $\mathfrak{p}\colon =(3, 1+\sqrt{-5})$ is prime and of order $2$. That means that its square $\mathfrak{p}^2$ is principal. Indeed, we have from first principles
\begin{gather*}
\mathfrak{p}^2 = (\, 9,\, 3( 1 +\sqrt{-5}),\, (1+\sqrt{-5})^2)=\\
= (\, (2+\sqrt{-5})(2-\sqrt{-5}),\, (-1 + \sqrt{-5})(2 -\sqrt{-5}),\, -2(2 - \sqrt{-5})\, )=\\
(2 - \sqrt{-5})
\end{gather*}
Let's consider the quadratic extension $\mathbb{Q}\left(\sqrt{2 - \sqrt{-5}}\right) $ of the field $\mathbb{Q}(\sqrt{5})$. In the ring of the integers of this number field we have the equality of ideals (extend the ideals)
$$(3, 1+\sqrt{-5})^2 =(2 - \sqrt{-5}) = \left( \sqrt{2 - \sqrt{-5}}\right)^2$$
Because of the unique decomposition of ideals of any ring of integers of an algebraic number field, the above equality implies
$$(3, 1+\sqrt{-5})= \left( \sqrt{2 - \sqrt{-5}}\right)$$
Let's check this equality. For comfort, fix an embedding of $\mathbb{Q}\left(\sqrt{2 - \sqrt{-5}}\right) $ into $\mathbb{C}$, $\sqrt{2 - \sqrt{-5}} \mapsto \frac{ \sqrt{5} - i }{\sqrt{2}}$ ( so $\sqrt{-5} \mapsto \sqrt{5}i$)
We have $3^2 = 9 = (2+ \sqrt{5}i) \cdot (2- \sqrt{5}i)$ so
$$
3= \frac{ \sqrt{5} + i }{\sqrt{2}}\cdot \frac{ \sqrt{5} - i }{\sqrt{2}}
$$
Note that $\frac{ \sqrt{5} + i }{\sqrt{2}}$ is in the ring of integers of $\mathbb{Q}\left(\frac{ \sqrt{5} - i }{\sqrt{2}}\right)$. Indeed, the minimal polynomial of $\frac{ \sqrt{5} + i }{\sqrt{2}}$ is $x^4 - 4 x^2 + 9$ and moreover
$$\frac{ \sqrt{5} + i }{\sqrt{2}}= 4/3 \cdot \frac{ \sqrt{5} - i }{\sqrt{2}}-1/3 \cdot \left(\frac{ \sqrt{5} - i }{\sqrt{2}}\right)^3 $$
Also, from $(1+ \sqrt{5}i)^2 = -2 ( 2 - \sqrt{5}i)$ we get
$$1+ \sqrt{5}i = \sqrt{2} i \cdot \frac{ \sqrt{5} - i }{\sqrt{2}}$$
Again, $\sqrt{2}i$ is an algebraic integers and $\mathbb{Q}\left(\frac{ \sqrt{5} - i }{\sqrt{2}}\right)$, since
$$\sqrt{2} i =1
/3 \cdot \frac{ \sqrt{5} - i }{\sqrt{2}}-1/3 \cdot \left(\frac{ \sqrt{5} - i }{\sqrt{2}}\right)^3 $$
So far, both $3$ and $1+ \sqrt{5}i$ are divisible by $\frac{ \sqrt{5} - i }{\sqrt{2}}$. Now we'll see that $\frac{ \sqrt{5} - i }{\sqrt{2}}$ is an (algebraic) integer combination of $3$ and $1+ \sqrt{5}i$. Indeed, we have
$$\frac{ \sqrt{5} - 3 i}{\sqrt{2}} \cdot (1 + \sqrt{5} i) + \frac{ \sqrt{5} - i }{\sqrt{2}}\cdot 3 = \frac{ \sqrt{5} - i }{\sqrt{2}}$$
And so it goes in general. Consider finitely many algebraic integers $(\alpha_s)$ . Let $K$ the algebraic number field generated by them, $\mathcal{O_K}$ its ring of integers. The ideal class of $K$ is finite. Therefore, the $m$ power of ideal of $\mathcal{O}_K$ generated by the $\alpha_s$ is principal for some integer $m$ ( for instance, if $m$ is the class number of $K$). There exists $\beta$ in $\mathcal{O}_K$ so that $(\alpha_s)^m_s = (\beta)$ ($\beta$ uniquely determined by this $m$ up to a unit element ( in $\mathcal{O}_K
^{\times}$). Extend $K$ by adding $\beta^{1/m}$ to $L = K(\beta^{1/m})$. We have the equality of $\mathcal{O}_L$ ideals
$$(\alpha_s)_s^m = (\beta)$$ and so
$$(\alpha_s)_s = (\beta^{1/m})$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1302481",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
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"answer_id": 0
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Trigonometric relationship in a triangle If in a triangle $ABC$, $$3\sin^2B+4\sin A\sin B+16\sin^2A-8\sin B-20\sin A+9=0$$ find the angles of the triangle. I am unable to manipulate the given expression. Thanks.
| Let $X = \sin A$ and $Y = \sin B$. Then we have
$$16X^2 + 4XY + 3Y^2 - 20X - 8Y + 9 = 0$$
Rearranging the terms to solve for Y, we get
$$3Y^2 + (4X-8)Y + (16X^2 - 20X + 9) = 0$$
The discriminant for this quadratic equation becomes
$$(4X-8)^2 - 4*3*(16X^2-20X+9) = 16X^2 - 64X + 64 - 192X^2 + 240X - 108 = -176X^2 + 176X - 44 = -44(4X^2 - 4X + 1) = -44 (2X-1)^2$$
This must equal $0$ (or $Y$ will not be a real number), which means that $$X = \frac{1}{2}$$.
Plugging this back into the original equation, we get
$$3Y^2 - 6Y + 3 = 3 (Y-1)^2 = 0$$
which means that $$Y = 1$$
Since $Y = \sin B$, we now know that $B$ must be $90$ degrees. Because $X = \sin A$, we know that $A$ is either $30$ degrees or $150$ degrees, but only the former makes sense because the sum of a triangle's internal angles cannot exceed $180$ degrees.
As a final answer, we get that $ABC$ is a $30-60-90$ triangle, with $B$ being the right angle and $A$ being $30$ degrees.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1303454",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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} |
Matrices $\begin{pmatrix} a &b \\ c &d \end{pmatrix}\begin{pmatrix} x\\y \end{pmatrix}=k\begin{pmatrix}x\\y \end{pmatrix}$ If $\begin{pmatrix}
a &b \\
c &d
\end{pmatrix}\begin{pmatrix}
x\\y \end{pmatrix}=k\begin{pmatrix}x\\y \end{pmatrix}$, prove that $k$ satisfies the equation $k^2-(a+d)k+(ad-bc)=0$.
If the roots of this quadractic equation are $\alpha$ and $\beta$, find the value of $\alpha+\beta$ and $\alpha \beta$ in terms of $a,b,c$ and $d$. Hence, or otherwise, prove that $\begin{pmatrix}
a &b \\
c &d
\end{pmatrix}\begin{pmatrix}
b &b \\
\alpha+a &\beta-a
\end{pmatrix}=\begin{pmatrix}
b &b \\
\alpha+a &\beta-a
\end{pmatrix}\begin{pmatrix}
\alpha &0 \\
0&\beta
\end{pmatrix}$
Can anyone give me some hints for solving this question? Thanks
| $Ax=kx$
Roots of the given equations are the eigen-values of $A$.
So sum of roots = trace (A) = a+b
Product of roots = det (A)= ad-bc
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1305179",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Showing that $x^3+2y^3+4z^3=2xyz$ has no integer solutions except $(0,0,0)$. Let $x,y,z\in \mathbb{Z}$ satisfy the equation:
$$
x^3+2y^3+4z^3=2xyz
$$
How do I prove that the only solution is $x=y=z=0$?
| We have $x^3 +2y^3+4z^3=2xyz$ so x is even and $8x_1^3+2y^3+4z^3=4x_1yz$ $\implies$
$4x_1^3+y^3+2z^3=2x_1yz$ so y is even and $4x_1^3+8y_1^3+2z^3=4x_1y_1z$
$\implies$ $2x_1^3+4y_1^3+z^3=2x_1y_1z$ so z is even and
$2x_1^3+4y_1^3+8z_1^3=4x_1y_1z_1$ $\implies$ $x_1^3+2y_1^3+4z_1^3=2x_1y_1z_1$.
Consequently, the procedure can be repeated indefinitely and, by infinite descent (Fermat), there are no solution distinct of (0,0,0).Obviously no matter if we consider negative integers.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1306029",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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$\mathbb{Q}\left ( \sqrt{2},\sqrt{3},\sqrt{5} \right )=\mathbb{Q}\left ( \sqrt{2}+\sqrt{3}+\sqrt{5} \right )$ Prove that
$$\mathbb{Q}\left ( \sqrt{2},\sqrt{3},\sqrt{5} \right )=\mathbb{Q}\left ( \sqrt{2}+\sqrt{3}+\sqrt{5} \right )$$
I proved for two elements, ex, $\mathbb{Q}\left ( \sqrt{2},\sqrt{3}\right )=\mathbb{Q}\left ( \sqrt{2}+\sqrt{3} \right )$, but I can't do it by the similar method.
| This is a boring but purely algebraic derivation.
For any three distinct positive integers $a,b,c$, let $x = \sqrt{a}+\sqrt{b}+\sqrt{c}$, we have
$$\begin{align}
& (x-\sqrt{a})^2 = (\sqrt{b}+\sqrt{c})^2 = b + c + 2\sqrt{bc}\\
\implies & (x^2 + a - b - c) - 2\sqrt{a}x = 2\sqrt{bc}\\
\implies & (x^2 + a - b - c)^2 - 4\sqrt{a}x(x^2 + a - b - c) + 4ax^2 = 4bc\\
\end{align}
$$
It is easy to check $x^2 + a - b - c \ne 0$. This leads to
$$\sqrt{a} = \frac{(x^2+a-b-c)^2 + 4(ax^2-bc)}{4x(x^2+a-b-c)} \in \mathbb{Q}(x)$$
By a similar argument, we have $\sqrt{b}, \sqrt{c} \in \mathbb{Q}(x)$ and hence
$$\mathbb{Q}(\sqrt{a}, \sqrt{b}, \sqrt{c}) \subset \mathbb{Q}(x) = \mathbb{Q}(\sqrt{a}+\sqrt{b}+\sqrt{c})$$
Since $\sqrt{a} + \sqrt{b} + \sqrt{c} \in \mathbb{Q}(\sqrt{a},\sqrt{b},\sqrt{c})$, we have
$$\mathbb{Q}(\sqrt{a}+\sqrt{b}+\sqrt{c}) \subset \mathbb{Q}(\sqrt{a}, \sqrt{b}, \sqrt{c})$$
Combine these two result, we get
$$\mathbb{Q}(\sqrt{a}+\sqrt{b}+\sqrt{c}) = \mathbb{Q}(\sqrt{a}, \sqrt{b}, \sqrt{c})$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1306082",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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find the minimum value of $\sqrt{x^2+4} + \sqrt{y^2+9}$ Question:
Given $x + y = 12$, find the minimum value of $\sqrt{x^2+4} + \sqrt{y^2+9}$?
Key:
I use $y = 12 - x$ and substitute into the equation, and derivative it.
which I got this
$$\frac{x}{\sqrt{x^2+4}} + \frac{x-12}{\sqrt{x^2-24x+153}} = f'(x).$$
However, after that. I don't know how to do next in order to find the minimum value. Please help!
| If you are aware of the rarely seen in action "Minkowski" inequality..then here you go: $\sqrt{x^2+2^2}+\sqrt{y^2+3^2} \geq \sqrt{(x+y)^2+(2+3)^2}=\sqrt{12^2+5^2} = \sqrt{169}=13$, and rest assured that this extrema is achievable.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Need a help to find the series formula for A005836 I need a help in finding the series formula for the Sequence $A005836$
$a(1)=0$
$a(2)=1$
$a(3)=3$
$a(4)=4$
$a(5)=9$
so series of above sequence would be
$A(1)=0$
$A(2)=1$
$A(3)=4$
$A(4)=8$
$A(5)=17$
where $$A(n)=a(1)+a(2)+a(3)+........+a(n)$$
the value of $n$ can be up to $10^{18}$.
So I am interested in finding the nth term of above series or the algorithm to compute $A(n)$ in an efficient way. Can someone help me with this.
Thanks
| I suspect there is a relatively simple recursion with $O(\log n)$ complexity:
$$A(2^n+b) = (3^n-1)2^{n-2}+3^n b+A(b)$$ for $0 \le b \lt 2^b$ and with $A(0)=0$, where the $(3^n-1)2^{n-2}$ represents $A(2^n)$, the $3^n b$ represents the leading ternary digits of the final $b$ terms, and the $A(b)$ representing the other ternary digits of the final $b$ terms
So to take Hyrum Hammon's example
*
*$A(1000)=A(2^9+488)= (3^9-1)\times 2^7 +3^9 \times 488+A(488)$
*$A(488)=A(2^8+232)= (3^8-1)\times 2^6 +3^8 \times 232+A(232)$
*$A(232)=A(2^7+104)= (3^7-1)\times 2^5 +3^7 \times 104+A(104)$
*$A(104)=A(2^6+40)= (3^6-1)\times 2^4 +3^6 \times 40+A(40)$
*$A(40)=A(2^5+8)= (3^5-1)\times 2^3 +3^5 \times 8+A(8)$
*$A(8)=A(2^3+0)= (3^3-1)\times 2^1 +3^3 \times 0+A(0)$
*$A(0)=0$
Adding these up gives $A(1000)=14408732$ as Hyrum Hammon also found.
In general $A(n)$ seems to be between $0.24$ and $0.29$ times $n^{\log_2 6}$ for $n \gt 4$ and $A(10^{18})$ is likely to be of the order of $9 \times 10^{45}$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Finding $\lim\limits_{n\to\infty }\frac{1+\frac12+\frac13+\cdots+\frac1n}{1+\frac13+\frac15+\cdots+\frac1{2n+1}}$ I need to compute: $\displaystyle\lim_{n\rightarrow \infty }\frac{1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\cdots+\frac{1}{n}}{1+\frac{1}{3}+\frac{1}{5}+\frac{1}{7}+\cdots+\frac{1}{2n+1}}$.
My Attempt: $\displaystyle\lim_{n\rightarrow \infty }\frac{1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\cdots+\frac{1}{n}}{1+\frac{1}{3}+\frac{1}{5}+\frac{1}{7}+\cdots+\frac{1}{2n+1}}=\lim_{n\rightarrow \infty }\frac{2s}{s}=2$.
Is that ok?
Thanks.
| By the Stolz-Cesaro theorem from https://en.wikipedia.org/wiki/Stolz%E2%80%93Ces%C3%A0ro_theorem,
$$\lim_{n\rightarrow \infty }\frac{1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\cdots+\frac{1}{n}}{1+\frac{1}{3}+\frac{1}{5}+\frac{1}{7}+\cdots+\frac{1}{2n+1}}=\lim_{n\to\infty}\frac{\frac{1}{n+1}}{\frac{1}{2n+3}}=2.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1309606",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
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How do I rewrite this rational expression? How do I rewrite the rational expression:
$$\frac{x^3+5x^2+3x-10}{x+4}$$
But in the form of:
$$q(x) + \frac{r(x)}{b(x)}$$
| Use Polynomial long division. The only way I know how to format that here would be something like:
$$\begin{split}
x^3 + 5x^2 + 3x - 10 &= (x+4)(x^2) + x^2 + 3x - 10 \\
&= (x+4)(x^2) + (x+4)(x) - x - 10 \\
&= (x+4)(x^2) + (x+4)(x) - (x+4)(1) - 6 \\
&= (x+4)(x^2 + x - 1) - 6
\end{split}$$
So:
$$\frac{x^3 + 5x^2 + 3x - 10}{x+4} = (x^2 + x - 1) + \frac{-6}{x+4}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1310780",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Simplify $\prod_{k=1}^5\tan\frac{k\pi}{11}$ and $\sum_{k=1}^5\tan^2\frac{k\pi}{11}$ My question is:
If $\tan\frac{\pi}{11}\cdot \tan\frac{2\pi}{11}\cdot \tan\frac{3\pi}{11}\cdot \tan\frac{4\pi}{11}\cdot \tan\frac{5\pi}{11} = X$ and $\tan^2\frac{\pi}{11}+\tan^2\frac{2\pi}{11}+\tan^2\frac{3\pi}{11}+\tan^2\frac{4\pi}{11}+\tan^2\frac{5\pi}{11}=Y$ then find $5X^2-Y$.
I couldn't find any way to simplify it. Please help. Thanks.
| Like Roots of a polynomial whose coefficients are ratios of binomial coefficients,
$$\tan(2n+1)x=\dfrac{\binom{2n+1}1\tan x-\binom{2n+1}3\tan^3x+\cdots}{1-\binom{2n+1}2\tan^2x+\cdots}$$
If $\tan(2n+1)x=0,(2n+1)x=r\pi$ where $r$ is ay integer
$\implies x=\dfrac{r\pi}{2n+1}$ where $r\equiv-n,-(n-1),\cdots,0,1,\cdots,n\pmod{2n+1}$
So, the roots of $\displaystyle\binom{2n+1}1\tan x-\binom{2n+1}3\tan^3x+\cdots+(-1)^{n-1}\binom{2n+1}{2n-1}\tan^{2n-1}x+(-1)^n\tan^{2n+1}x=0$
$\displaystyle\iff\tan^{2n+1}x-\binom{2n+1}2\tan^{2n-1}x+\cdots+(-1)^n(2n+1)\tan x=0$
are $\tan x$ where $x=\dfrac{r\pi}{2n+1}$ where $r\equiv-n,-(n-1),\cdots,0,1,\cdots,n\pmod{2n+1}$
So, the roots of $\displaystyle\tan^{2n}x-\binom{2n+1}2\tan^{2n-2}x+\cdots+(-1)^n(2n+1)=0$
are $\tan x$ where $x=\dfrac{r\pi}{2n+1}$ where $r\equiv\pm1,\pm2,\cdots,\pm n\pmod{2n+1}$
As $\tan(-A)=-\tan A,$
the roots of $\displaystyle t^nx-\binom{2n+1}2t^{n-1}x+\cdots+(-1)^n(2n+1)=0\ \ \ \ (1)$
are $\tan^2x$ where $x=\dfrac{r\pi}{2n+1}$ where $r\equiv1,2,\cdots,n\pmod{2n+1}$
Using Vieta's formula on $(1),$ $$\sum_{r=1}^n\tan^2\dfrac{r\pi}{2n+1}=\dfrac{\binom{2n+1}2}1=\cdots$$
and $$(-1)^n\prod_{r=1}^n\tan^2\dfrac{r\pi}{2n+1}=(-1)^n(2n+1)\iff\prod_{r=1}^n\tan^2\dfrac{r\pi}{2n+1}=(2n+1)$$
Now, $\tan\dfrac{r\pi}{2n+1}>0$ for $\dfrac\pi2>\dfrac{r\pi}{2n+1}>0\iff2n+1>2r>0$
$$\implies\prod_{r=1}^n\tan\dfrac{r\pi}{2n+1}=\sqrt{2n+1}$$
| {
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"url": "https://math.stackexchange.com/questions/1311717",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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Determinate set $A\subseteq\mathbb{R}$ so that for every $a\in A$ and every $x\in\mathbb{R}$ the condition $ax^2+x+3\ge0$ is valid Quadratic function is always greater than $0$ if
$$a>0$$ and $$D=0$$
Solving for $D$ we have
$$1-12a=0\Rightarrow a=\frac{1}{12}$$
So, $$a\in[\frac{1}{12},+\infty)$$
Is this the only condition to check?
For found $a\in \mathbb{A}$ how to evaluate
$$\lim\limits_{x\to\infty}(x+1-\sqrt{ax^2+x+3})$$
| $ax^2+x+3\ge 0$ for every $x\in\mathbb R$ if and only if $a\gt 0$ and $D=1-12a\color{red}{\le} 0$, i.e. $a\color{red}{\ge} \frac{1}{12}$.
For the limit, you can have
$$\begin{align}\lim_{x\to\infty}\left(x+1-\sqrt{ax^2+x+3}\right)&=\lim_{x\to\infty}\frac{(x+1-\sqrt{ax^2+x+3})(x+1+\sqrt{ax^2+x+3})}{x+1+\sqrt{ax^2+x+3}}\\&=\lim_{x\to\infty}\frac{(x+1)^2-(ax^2+x+3)}{x+1+\sqrt{ax^2+x+3}}\\&=\lim_{x\to\infty}\frac{(1-a)x^2+x-2}{x+1+\sqrt{ax^2+x+3}}\\&=\lim_{x\to\infty}\frac{(1-a)x+1-\frac 2x}{1+\frac 1x+\sqrt{a+\frac 1x+\frac{3}{x^2}}}\end{align}$$
*
*If $1-a\gt 0$, i.e. $a\lt 1$, then the limit is $+\infty$.
*If $1-a=0$, i.e. $a=1$, then the limit is $\frac 12$.
*If $1-a\lt 0$, i.e. $a\gt 1$, then the limit is $-\infty$.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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Find the maximum value of $a^2+b^2+c^2+d^2+e^2$ Find the maximum value of $a^2+b^2+c^2+d^2+e^2$ with the following condition:
i) $a\ge b\ge c\ge d\ge e\ge 0$
ii) $a+b\le m$
iii) $c+d+e \le n$
Where $m,n\ge 0$.
Playing with numbers I found that for $m=n$ we should get the maximum value is $m^2$.
| It seems the following.
Since a set $C$ of all sequences $(a,b,c,d,e)$ satisfying Conditions i-iii is compact, a continuous function $$f(a,b,c,d,e)= a^2+b^2+c^2+d^2+e^2$$ attains its maximum $M$ on the set $C$ at some point $p=(a, b, c, d, e)$. If $b>c$ then a point $p’=(m-c, c, c, d, e)$ also belongs to the set $M$ and $f(p’)>f(p)$, which contradicts to the maximality of $f(p)$. So $b=c$ and $a=m-c$. Similarly we can show that $d=c$ and $e=n-2c$ or $e=0$ and $d=n-c$. So we have two following cases.
1) $p=(m-c, c, c, c, n-2c)$. Then $m-c\ge c$ and $0\le n-2c\le c$, so $$\min\{m/2, n/2\}\ge c\ge n/3.$$
The function $$g(c)=(m-c)^2+3c^2+(n-2c)^2=8c^2-c(2m+4n)+m^2+n^2$$ is convex (its graph is a parabola), so it attains its maximum at an end of the segment $[n/3, \min\{m/2, n/2\}]$.
If $c=n/3$ then $g(c)=\frac 19(9m^2-6mn+5n^2)$.
If $c=m/2$ then $g(c)=2m^2-2mn+n^2$.
If $c=n/2$ then $g(c)=m^2-mn+n^2$.
Compare these possible value for $f(p)=g(c)$:
$2m^2-2mn+n^2-\frac 19(9m^2-6mn+5n^2)=\frac 19(3m-2n)^2\ge 0.$
$2m^2-2mn+n^2-(m^2-mn+n^2)=m^2-mn\ge 0$ iff $m\ge n.$ But in this case $\min\{m/2, n/2\}=n/2$, and $c=m/2$ is allowed only if $n=m$, so we may skip this case.
$\frac 19(9m^2-6mn+5n^2)-(m^2-mn+n^2)=\frac 19(3mn-4n^2)\ge 0$ iff $m\ge (4/3)n.$
2) $p=(m-c, c, c, n-c, 0)$. Then $m-c\ge c$ and $n-c\le c$ so $$m/2\ge c\ge n/2.$$
The function $$g(c)=(m-c)^2+2c^2+(n-c)^2=4c^2-c(2m+2n)+m^2+n^2$$ is convex (its graph is a parabola), so it attains its maximum at an end of the segment $[n/2, m/2]$. If $c=n/2$ or $c=m/2$ then $g(c)=m^2-mn+n^2$.
Now we can write
The final answer. The maximum is
$\frac 19(9m^2-6mn+5n^2)$, if $m\ge (4/3)n$,
$m^2-mn+n^2$, if $(4/3)n\ge m\ge n$,
$2m^2-2mn+n^2$, if $n\ge m\ge (2/3)n$.
The case $(2/3)n>m$ is impossible, because $m/2=(a+b)/2\ge (c+d+e)/3=n/3$.
| {
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"timestamp": "2023-03-29T00:00:00",
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Generating functions (Counting?) Determine how many ways Brian, Katie, and Charlie can split a $\$50$ dinner bill such that Brian and Katie each pay an odd number of dollars and Charlie pays at least $\$5$.
How can I use generating functions to solve this problem? Or should I use a counting approach?
| The generating function for the number of ways that the three can pay a bill of $n$ dollars is
$$
\begin{align}
&(x+x^3+x^5+\dots)(x+x^3+x^5+\dots)(x^5+x^6+x^7+\dots)\\[3pt]
&=x^7\frac1{(1-x^2)^2}\frac1{1-x}\\
&=x^7\frac1{(1-x)^3}\frac1{(1+x)^2}\\
&=x^7\left[\frac14\frac1{(1-x)^3}+\frac14\frac1{(1-x)^2}+\frac3{16}\frac1{1-x}+\frac18\frac1{(1+x)^2}+\frac3{16}\frac1{1+x}\right]\\
&=\sum_{k=0}^\infty\small\left[\frac14(-1)^k\binom{-3}{k}+\frac14(-1)^k\binom{-2}{k}+\frac3{16}(-1)^k\binom{-1}{k}+\frac18\binom{-2}{k}+\frac3{16}\binom{-1}{k}\right]x^{k+7}\\
&=\sum_{k=0}^\infty\small\left[\frac14\binom{k+2}{2}+\frac14\binom{k+1}{1}+\frac3{16}\binom{k}{0}+\frac18(-1)^k\binom{k+1}{1}+\frac3{16}(-1)^k\binom{k}{0}\right]x^{k+7}\\
&=\sum_{k=0}^\infty\frac1{16}\left[(2k^2+10k+11)+(-1)^k(2k+5)\vphantom{\frac1{16}}\right]x^{k+7}\\
&=\sum_{k=0}^\infty\left[\binom{\lfloor k/2\rfloor}{2}+2\binom{\lfloor k/2\rfloor}{1}+\binom{\lfloor k/2\rfloor}{0}\right]x^{k+7}\\
&=\sum_{k=0}^\infty\binom{\lfloor k/2\rfloor+2}{2}x^{k+7}\tag{1}
\end{align}
$$
Plugging $k=43$ into $(1)$ yields the coefficient of $x^{50}$ to be $253$.
| {
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"timestamp": "2023-03-29T00:00:00",
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solving difficult complex number proving if $z= x+iy$ where $y \neq 0$ and $1+z^2 \neq 0$, show that the number $w= z/(1+z^2)$ is real only if $|z|=1$
solution :
$$1+z^2 = 1+ x^2 - y^2 +2xyi$$
$$(1+ x^2 - y^2 +2xyi)(1+ x^2 - y^2 -2xyi)=(1+ x^2 - y^2)^2 - (2xyi)^2$$
real component
$$(1+ x^2 - y^2)x - yi(2xyi) = x + x^3 + xy^2$$
imaginary component
$$-2yx^2 i +yi + x^2 yi - y^3 i=0i$$
$$-2yx^2 +y + x^2 y - y^3 =0$$
...
can't solve this question
| I tried to solve this with geometry and how the arithmetic operations on complex numbers can be interpreted geometrically.
*
*By dividing two complex numbers, their arguments (angles, $\varphi$)
are subtracted.
*If $w$ should be real, its argument has to be zero.
That means $\varphi(z) = \varphi(1+z^2)$.
Here's a little sketch of both those complex numbers drawn as arrows in the complex plane:
You can see $z$, $z^2$, $1$ and $1+z^2$ in green. I also added the arguments of $z$ and $z^2$ in red, for no real reason (pun not intended) other than to remember that they have a constant ratio of $\varphi(z) = 2\varphi(z^2)$. So you can't do much about that, but $\varphi(1+z^2)$ is a whole 'nother story, because of the $+1$
The true goal is still to line up the green arrow (i.e. $1+z^2$) with $z$.
In order to be lined up, one has to be derivable from the other by multiplying with a real number $a$. I guess this is called "being linear dependent", but am not sure. Anyway, the formula for that looks like that:
$$
\begin{align}
1+z^2 &= az\\
z^2 -az +1 &= 0\\
z_{1/2} &= \frac{a}{2} \pm \sqrt{\left(\frac{a}{2}\right) ^2-1}\\
\end{align}
$$
This is kind of a good thing, because there's $\pm$, which means this is a two for one. And circles aren't functions because they need two lines. So maybe we are on to something here. Let's see if the above formula can be squeezed into being a circle.
You said $z= x+iy$ where $y \neq 0$ and that means there has to be an imaginary part, so whatever $a$ is (cannot be bothered to calculate it), the term under the square will be negative and by that forces $z_{1/2}$ to have an imaginary part.
$$
\begin{align}
z_{1/2} &= \frac{a}{2} \pm \sqrt{(-1)\left(-\left((\frac{a}{2}\right) ^2+1\right)}\\
&= \frac{a}{2} \pm \sqrt{-1} \sqrt{-\left(\frac{a}{2}\right)^2+1}\\
&= \frac{a}{2} \pm i \sqrt{-\left(\frac{a}{2}\right)^2+1}\\
&= x+iy\\
\end{align}
$$
Finally, what's $|z|$?
$$\begin{align}
|z| = |z_{1/2}| &= \sqrt{x^2 + y^2}\\
&= \sqrt{\left(\frac{a}{2}\right)^2 + \left(\sqrt{-\left(\frac{a}{2}\right)^2+1}\right)^2}\\
&= \sqrt{\left(\frac{a}{2}\right)^2 -\left(\frac{a}{2}\right)^2+1}\\
&=1
\end{align}$$
I'm not sure if this is rigorous enough to be a proof, but you just asked to show it, so I hope this is ok and offers a bit of a different approach.
| {
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"timestamp": "2023-03-29T00:00:00",
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Solving for n in the equation $\left ( \frac{1}{2} \right )^{n}+\left ( \frac{1}{4} \right )^{n}+\left ( \frac{3}{4} \right )^{n}=1$
Solving for $n$ in the equation $$\left ( \frac{1}{2} \right )^{n}+\left ( \frac{1}{4} \right )^{n}+\left ( \frac{3}{4} \right )^{n}=1$$
Can anyone show me a numerical method step-by-step to solve this?
Thanks
| Let $f(x)=\left ( \frac{1}{2} \right )^{x}+\left ( \frac{1}{4} \right )^{x}+\left ( \frac{3}{4} \right )^{x}-1$. Now note that $f(1)=\frac12$ and $f(2)=-\frac18$. Hence given that $$f'(x)=-4^{-x}(3^x \log\frac43+2^x\log 2+\log4)<0$$ there is one unique root in $(1,2)$. The rest is numerics, this could be approximated to $1.73051$.
| {
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Express 99 2/3% as a fraction? No calculator My 9-year-old daughter is stuck on this question and normally I can help her, but I am also stuck on this! I have looked everywhere to find out how to do this but to no avail so any help/guidance is appreciated:
The possible answers are:
$1 \frac{29}{300}$
$\frac{269}{300}$
$9 \frac{29}{30}$
$\frac{299}{300}$
$1 \frac{299}{300}$
Whenever I try this I am doing $\frac{99.66} \times 100$ is $\frac{9966}{10000}$ and then simplify down, but I am not getting the answer. I get $\frac{4983}{5000}$? Then I can't simplify more — unless I am doing this totally wrong?
Hello All - just an update - I managed to teach this to her right now, and she cracked it pretty much the first time! Thanks sooooooo much for the answers and here is some of the working she did:
| There are multiple roads that lead to Rome:
First, take out the $\%$. We find $\frac{299}{3}$
$99 \frac{2}{3} = 99 + \frac{2}{3} = \frac{1}{3} \cdot ( 297 + 2) = \frac{1}{3} \cdot 299 = \frac{299}{3}$
or
$99 \frac{2}{3} = 99 + \frac{2}{3} = 3 \cdot \frac{99}{3} + \frac{2}{3} = \frac{297}{3} + \frac{2}{3} = \frac{299}{3}$
Lastly, add the $\%$ back: $\frac{299}{3} \cdot \frac{1}{100} = \frac{299}{300}$
| {
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How to calculate $\sum_{n=1}^\infty \frac{2}{(2n-1)(2n+1)}$ How can I calculate the sum of the following infinite series: $$\sum_{n=1}^\infty \frac{2}{(2n-1)(2n+1)}$$
| This is how you do it formally:
$$\sum_{i=1}^N\frac{2}{(2n-1)(2n+1)} = \sum_{i=1}^N \left (\frac{1}{2n-1}-\frac{1}{2n+1} \right)= \sum_{i=1}^N \left (\frac{1}{2n-1} \right) - \sum_{i=1}^N \left (\frac{1}{2n+1} \right) = 1 + \sum_{i=2}^{N} \left (\frac{1}{2n-1} \right) - \sum_{i=1}^{N-1}\left (\frac{1}{2n+1} \right) - \frac{1}{2N+1} =1 + \sum_{i=1}^{N-1} \left (\frac{1}{2n+1} \right) - \sum_{i=1}^{N-1} \left (\frac{1}{2n+1} \right) - \frac{1}{2N+1} =1 - \frac{1}{2N+1} $$
Now take the limit $$\lim_{N\to\infty} \left(1 - \frac{1}{2N+1}\right) =1$$
So the series converges towards 1.
| {
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"timestamp": "2023-03-29T00:00:00",
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why the convolution of two functions of moderate decrease is again function of moderate decrease? I need to prove that a convolution of 2 functions of moderate decrease is a function of moderate decrease.
I tried to split the integral into two integrals but I couldn't manage to bound any one of them by $\frac{A}{x^2+1}$.
Thanks
| Take $z = x - y/2$.
$$ \begin{align*}\int_{-\infty}^{\infty} \frac{1}{1 + x^2} \cdot \frac{1}{1 + (y-x)^2} \mathrm{d}x &= 2 \int_0^\infty \frac{1}{1 + (y/2 + z)^2} \cdot \frac{1}{1 + (y/2 - z)^2} \mathrm{d}z \\
& \leq \frac{2}{1 + (y/2)^2} \int_0^\infty \frac{1}{1 + (y/2 - z)^2} \mathrm{d}z \\
& \leq \frac{2}{1 + (y/2)^2} \int_{-\infty}^\infty \frac{1}{1 + (y/2 - z)^2} \mathrm{d}z \\
& = \frac{2}{1 + (y/2)^2} \int_{-\infty}^\infty \frac{1}{1 + w^2} \mathrm{d}w \end{align*}$$
Lastly observe
$$ \frac{2}{1 + (y/2)^2} \leq \frac{2}{1/4 + (y/2)^2} \leq \frac{8}{1 + y^2} $$
The power $2$ in $(1 + x^2)^{-1}$ is not special. For any power $1+\epsilon$ where $\epsilon > 0$ you get a similar result.
| {
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Bounding $\sum_{k=1}^N \frac{1}{1-\frac{1}{2^k}}$ I'm looking for a bound depending on $N$ of $\displaystyle \sum_{k=1}^N \frac{1}{1-\frac{1}{2^k}}$.
The following holds $\displaystyle \sum_{k=1}^N \frac{1}{1-\frac{1}{2^k}} = \sum_{k=1}^N \sum_{i=0}^\infty \frac{1}{2^{ki}}=\sum_{n=0}^\infty \sum_{\substack{q\geq 0 \\ 1\leq p \leq N\\pq=n}}\frac{1}{2^{pq}}\leq 2N$ which does not seem accurate.
Numerical trials suggest that $\displaystyle N-\sum_{k=1}^N \frac{1}{1-\frac{1}{2^k}}$ is a convergent sequence (it's easy to prove it is decreasing).
| Note that
$$\frac{1}{1-\frac{1}{2^k}} = 1+ \frac{1}{2^k-1}$$
so
$$N \le \sum_{k=1}^N\frac{1}{1-\frac{1}{2^k}} \le N+ \sum_{k=1}^{+\infty} \frac{1}{2^k-1} \le N+2$$
since $1\le \sum_{k=1}^{+\infty} \frac{1}{2^k-1} \le 2$
| {
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Finding a polynomial by divisibility
Let $f(x)$ be a polynomial with integer coefficients. If $f(x)$ is divisible by $x^2+1$ and $f(x)+1$ by $x^3+x^2+1$, what is $f(x)$?
My guess is that the only answer is $f(x)=-x^4-x^3-x-1$, but how can I prove it?
| This is a straightforward application of CRT = Chinese Remainder Theorem.
By hypothesis $\ (x^2\!+1)g = f = -1 + (x^3\!+x^2\!+1)h\ $ for some polynomials $\,g,h,\,$ so
${\rm mod}\ x^2\!+\!1\!:\ x^2\equiv -1\,\Rightarrow\,1\equiv (x^3\!+x^2\!+1)h\equiv -xh\,\Rightarrow\, h \equiv -1/x\equiv x^2/x\equiv \color{#c00}x$
Therefore $\ h = \color{#c00}x + (x^2\!+1)h',\ $ for some polynomial $\,h'$
$\! \begin{align}{\rm hence}\ \ f\, =&\, -1 + (x^3\!+x^2\!+1)(\color{#c00}x+(x^2\!+1)h')\\ =&\ \ x^4\!+x^3\!+x-1 + (x^3\!+x^2\!+1)(x^2\!+1) h'\end{align}$
| {
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How to calculate the gcd of two polynomials $\mod 7$ I need to find gcd of $x^4-3x^3-2x+6$ and $x^3-5x^2+6x+7$ in $\mathbb Z/7 \mathbb Z[x]$, the integer polynomials mod $7$. Please any help will be appreciated.
| In general, you use the same method as gcd of numbers: the Euclidean algorithm (detailed below). In this specific (i.e. non-general) case you can take a short-cut: since you see that the constant term in your second polynomial is $0$ mod $7$, we know that $0$ is a root mod $7$. So the second polynomial factors as $(ax^2+bx+c)x$, and we might hope there's another root. Testing, we see that $x=2,3$ are also roots, so the factorization is
$$x^3-5x^2+6x+7=x(x+4)(x+5)\mod 7.$$
Checking the first polynomial for common roots shows only $x=3$ is a common root, so $(x+4)$ is the gcd.
The more general way:
Write
$$x^4-3x^3-2x+6=x(x^3-5x^2+6x+7)+(2x^3+x^2-2x-1)\mod 7$$
$$x^3-5x^2+6x+7=4(2x^3+x^2-2x-1)+(-2x^2+4)\mod 7$$
$$2x^3+x^2-2x-1=(-x-4)(-2x^2+4)+(2x+1)\mod 7$$
$$-2x^2+4=(-x)(2x+1)+(x+4)\mod 7$$
$$2x+1=2(x+4)+0\mod 7$$
so the gcd is $x+4$ since $2$ is a unit, though you can write $2(x+4)$ if you want, as well.
| {
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Is $2|x|/(1+x^{2})<1$ true if $|x|<1$? [Solved] I was asked to show that the series
$$\sum_{k\geq 0}a_{k}\left ( \frac{2x}{1+x^{2}} \right )^{k}$$
is convergent for $x\in (-1,1)$, where $\{ a_{k}\}_{k\geq 0}$ is a real bounded sequence.
My main question is, how do I know that $|2x/(1+x^{2})|<1$, that is, $2|x|/(1+x^{2})<1$ is true for $|x|<1$?
EDIT: Thanks for your answers. It took time to find it out on my own. I hope this too is correct. Let $y(x)=2x/(1+x^{2})$. Since $y'(x)>0$ for all $x\in (-1,1)$ then $y(x)$ is strictly increasing for all $x\in (-1,1)$. Hence $y(-1)<y(x)<y(1)$ for all $x\in (-1,1)$. Since $y(-1)=-1$ and $y(1)=1$, then $| y(x)|<1$ is true for $|x|<1$.
| Looking at positive values only, $\frac{2x}{1 + x^2} <1$ leads to $2x < 1 + x^2$ which leads to $0 < x^2 - 2x + 1 = (x - 1)^2$--this is true for any value of $x \neq 1$--that is all values of $0 \leq x < 1 \cup 1 < x < \infty$. If we consider negative values then we have that $\frac{-2x}{1 + x^2} < 1$ which leads to $0 < x^2 + 2x + 1 = (x + 1)^2$. In this case we see the inequality is true for $-\infty < x < -1 \cup -1 < x \leq 0$.
By combining these two intervals we get that:
$$
\frac{2|x|}{1 + x^2} < 1 \text{ when } x \in (-\infty, -1)\cup(-1, 1)\cup (1, +\infty)
$$
Otherwise, when $x = \pm 1$, $\frac{2|x|}{1 + x^2} = \frac{2}{2} = 1 \nless 1$.
| {
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Integral of Absolute Value of $\sin(x)$ For the Integral: $\int |\sin (ax)|$, it is fairly simple to take the Laplace transform of the absolute value of sine, treating it as a periodic function.
$$\mathcal L(|\sin (ax)|) = \frac{\int_0^\frac{\pi} {a} e^{-st} \sin (ax)} {1 - e^{-\frac{ \pi s} {a}}} = \frac{a (1 + e^{-\frac{ \pi s} {a}})} {(a^2 +s^2)(1 - e^{-\frac{ \pi s} {a}})} $$
Then the integral of the absolute value would simply be $ 1/s $ times the Laplace transform. Splitting into partial fractions we have:
$$\mathcal L(\int |\sin (ax)|) = \frac{a (1 + e^{-\frac{ \pi s} {a}})} {s(a^2 +s^2)(1 - e^{-\frac{ \pi s} {a}})} = \frac{(1 + e^{-\frac{ \pi s} {a}})} {as(1 - e^{-\frac{ \pi s} {a}})} - \frac{s (1 + e^{-\frac{ \pi s} {a}})} {a(a^2 +s^2)(1 - e^{-\frac{ \pi s} {a}})} $$
The first term is close to the floor function, when separated:
$$ \frac{(1 + e^{-\frac{ \pi s} {a}})} {as(1 - e^{-\frac{ \pi s} {a}})} = \frac{(1 - e^{-\frac{ \pi s} {a}}) + 2e^{-\frac{ \pi s} {a}}} {as(1 - e^{-\frac{ \pi s} {a}})} = \frac{1} {as} + \frac{2e^{-\frac{ \pi s} {a}}} {as(1 - e^{-\frac{ \pi s} {a}})}$$
This leads to $$ \frac{1} {a} + \frac {2\cdot\text{floor}(\frac{ax}{\pi})} {a}$$
Since the second term looks to be $ s/a^2 $ multiplied by the original function, this points to the derivative, or $$-1\cdot\text{sign}(\sin(ax))\cos (ax)/a$$, for a total integral is:
$$ \frac{1} {a} + \frac {2\cdot \text{floor}(\frac{ax}{\pi})} {a} -\text{sign}(\sin(ax))\cos(ax)/a$$
The Laplace transform method to evaluate an integral seems bulky - is there a better method than this? And what is the cosine version?
| Suppose $F(x)$ is an antiderivative of $|\sin(x)|$.
On the interval $[n\pi, (n+1)\pi]$, $|\sin(x)| = (-1)^n \sin(x)$, so we must have $F(x) = (-1)^{n+1}\cos(x) + c_n$ there, where $c_n$ is some constant.
Comparing the values at $n\pi$ from both sides, since $\cos(n\pi) = (-1)^n$,
we get
$ 1 + c_{n-1} = -1 + c_n$, i.e. $c_n = c_{n-1} + 2$. Thus we could take
$c_n = 2 n$, i.e. since $n = \lfloor x/\pi \rfloor$,
$$F(x) = 2 \lfloor x/\pi \rfloor - (-1)^{\lfloor x/\pi \rfloor} \cos(x)$$
| {
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High School Trigonometry ( Law of cosine and sine) I am preparing for faculty entrance exam and this was the question for which I couldn't find the way to solve (answer is 0). I guess they ask me to solve this by using the rule of sine and cosine:
Let $\alpha$, $\beta$ and $\gamma$ be the angles of arbitrary triangle with sides a, b and c respectively. Then $${b - 2a\cos\gamma \over a\sin\gamma} + {c-2b\cos\alpha \over b\sin\alpha} + {a - 2c\cos\beta \over c\sin\beta}$$ is equal to (answer is zero but I need steps).
| From Sine Rule in the triangle, we have $$\frac{\sin\alpha}{a}=\frac{\sin\beta}{b}=\frac{\sin\gamma}{c}=k \space (\text{let any arbitrary constant})$$ Thus by substituting the values of $a=k\sin \alpha$, $b=k\sin \beta$ & $c=k\sin \gamma$ in the given expression, we get $$\frac{b-2a\cos\gamma}{a\sin\gamma}+\frac{c-2b\cos\alpha}{b\sin\alpha}+\frac{a-2c\cos\beta}{c\sin\beta}$$ $$=\frac{k\sin\beta-2k\sin\alpha\cos\gamma}{k\sin\alpha\sin\gamma}+\frac{k\sin\gamma-2k\sin\beta\cos\alpha}{k\sin\beta\sin\alpha}+\frac{k\sin\alpha-2k\sin\gamma\cos\beta}{k\sin\gamma\sin\beta}$$ $$=\frac{\sin\beta-2\sin\alpha\cos\gamma}{\sin\alpha\sin\gamma}+\frac{\sin\gamma-2\sin\beta\cos\alpha}{\sin\beta\sin\alpha}+\frac{\sin\alpha-2\sin\gamma\cos\beta}{\sin\gamma\sin\beta}$$ $$=\frac{\sin(\pi-(\alpha+\gamma))-2\sin\alpha\cos\gamma}{\sin\alpha\sin\gamma}+\frac{\sin(\pi-(\alpha+\beta))-2\sin\beta\cos\alpha}{\sin\beta\sin\alpha}+\frac{\sin(\pi-(\beta+\gamma))-2\sin\gamma\cos\beta}{\sin\gamma\sin\beta}$$ $$=\frac{\sin\alpha\cos\gamma+\cos\alpha\sin\gamma-2\sin\alpha\cos\gamma}{\sin\alpha\sin\gamma}+\frac{\sin\alpha\cos\beta+\cos\alpha\sin\beta-2\sin\beta\cos\alpha}{\sin\beta\sin\alpha}+\frac{\sin \beta\cos\gamma+\cos\beta\sin\gamma-2\sin\gamma\cos\beta}{\sin\gamma\sin\beta}$$ $$=\frac{\sin\gamma\cos\alpha-\cos\gamma\sin\alpha}{\sin\alpha\sin\gamma}+\frac{\sin\alpha\cos\beta-\cos\alpha\sin\beta}{\sin\beta\sin\alpha}+\frac{\sin \beta\cos\gamma-\cos\beta\sin\gamma}{\sin\gamma\sin\beta}$$
$$=(\cot\alpha-\cot\gamma)+(\cot\beta-\cot\alpha)+(\cot\gamma-\cot\beta)$$ $$=0$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Determining irreducible components of $Z(y^4 - x^6, y^3 - x y^2 - y x^3 + x^4) \subset \mathbb A^2$ I'm trying to determine the irreducible components of the zero set $Z(I)$ for the ideal $I = (y^4 - x^6,\, y^3 - x y^2 - y x^3 + x^4)$, in the affine space $\mathbb A^2$ over an algebraically closed field $k$.
I can factor $I = (y^2 - x^3)(y^2 +x^3,\, y - x)$, hence $Z(I) = Z(y^2 - x^3) \cup Z(y^2 + x^3,\, y - x)$, but I'm stuck trying to prove these ideals are prime (and is the second one even prime? If not, how do I reduce it further?).
| The key fact that we will use to find the irreducible components is that an algebraic subset of a variety is irreducible iff its coordinate ring is an integral domain. Therefore, consider the coordinate ring of the algebraic set $Z(y^2 - x^3)$: it is $\frac{k[x,y]}{(y^2 - x^3)}$, which is an integral domain, hence $Z(y^2 - x^3)$ is irreducible.
In order to apply this fact to the second algebraic set, it must be simplified further:
$$
Z(y^2 + x^3, y-x) = Z(x^2 + x^3, y-x) = Z(x^2(x+1), y-x) = Z(x^2, y-x) \cup Z(x+1, y-x).
$$
Now, $\frac{k[x,y]}{(x+1,y-x)} \simeq \frac{k[x]}{(x+1)} \simeq k$ is an integral domain, so $Z(x+1,y-x)$ is irreducible. Finally, the coordinate ring of $Z(x^2, y-x)$ is $\frac{k[x,y]}{(x,y-x)} \simeq k$, which is an integral domain, so $Z(x^2, y-x)$ is irreducible, as well (it's important here that, when constructing the coordinate ring of $Z(x^2,y-x)$, one takes the radical of the ideal $(x^2,y-x)$ rather than the ideal itself).
Therefore, we conclude that the irreducible components of $Z(y^4-x^6, y^3-xy^2-yx^3+x^4)$ are
$$
Z(y^2 - x^3) \textrm{ and } Z(x^2, y-x) \textrm{ and } Z(x+1, y-x).
$$
| {
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Sums $a_k=[\frac{2+(-1)^k}{3^k}, (\frac{1}{n}-\frac{1}{n+2}), \frac{1}{4k^2-1},\sum_{l=0}^k{k \choose l}\frac{1}{2^{k+l}}]$
Determien the sums of the following series'.
1:$\sum_{k=0}^\infty \frac{2+(-1)^k}{3^k}$
2:$\sum_{k=0}^\infty (\frac{1}{n}-\frac{1}{n+2})$
3:$\sum_{k=0}^\infty \frac{1}{4k^2-1}$
4:$\sum_{k=0}^\infty \sum_{l=0}^k{k \choose l}\frac{1}{2^{k+l}}$
I have always been bad with infinite series', so I decided to practice them strongly for the next couple of days. I found an exercise in a textbook and kinda troubling with some of them.
Here's what I was thinking of doing.
1: I thought about dividing this into two sums, $\sum_{k=0}^\infty \frac{2}{3^k}+\sum_{k=0}^\infty \frac{(-1)^k}{3^k}$, which should be okay since they converge, right? Anyway, then I can see the geometric series there, giving me $3+\frac{3}{4}=3,75$. Is that correct?
2: I think that's what's called a telescope series? So I tried writing down some of the first numbers to see where it starts canceling each other out:
$\frac{1}{1}-\frac{1}{3}+\frac{1}{2}-\frac{1 }{ 4}+\frac{ 1}{ 3}-\frac{ 1}{ 5}+\frac{1 }{ 4}-\frac{1 }{6 } + ...$. Since $\frac{1}{n}$ converges to zero for $n\to \infty$, the sum of this infinite series is $\frac{3}{2}$?
About 3 and 4. I'm pretty much lost there. Can't seem to find an approach to find the sum.
Any tips?
| $1-$
$$\sum_{k=0}^{\infty }\frac{2+(-1)^k}{3^k}=\sum_{k=0}^{\infty }\frac{2}{3^k}+\frac{(-1)^k}{3^n}=2\frac{1}{1-1/3}+\frac{1}{1+1/3}$$
$3-$
$$\sum_{k=0}^{\infty }\frac{1}{4k^2-1}=\sum_{k=0}^{\infty }0.5[\frac{1}{2k-1}-\frac{1}{2k+1}]=0.5[1/1-1/3+1/3-1/5+1/5+....]=\frac{1}{2}$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
Calculating fluxintegral out of the surface $1-x^2-y^2$ I am trying to calculate the flux integral of the vector field
$$
\vec{F} = (x,y,1+z)
$$
Out of the surface $z=1-x^2-y^2$, $z\geq 0$
Answer : $\frac{5\pi}{2}$
I begin by defining a vector that traces out the surface and calculate the cross product of its derivate to get normal vector.
$$
\vec{r} = (x,y,1-x^2-y^2)\\
\vec{r}_x\times \vec{r}_y = (2x,2y,1)
$$
Next, I calculate the corresponding double integral using polar coordinates:
$$
\iint \vec{F} \cdot \hat{n}dS = \iint_D\vec{F}\cdot\vec{n}dxdy = \iint_D2+x^2+y^2dxdy\\
=\int_0^{2\pi}\int_0^{1}(2+r^2)*rdrd\theta \neq \frac{5\pi}{2}
$$
Where am I going wrong?
| You should consider that $ds$=$\frac{dxdy}{\sigma}$ and thus projecting the whole surface on the XOY plane where $\sigma$=$|\vec n . \vec k|$ and $\vec n$ is the unit vector that is in this case $\vec n$ = $\frac{\vec grad(S)}{||\vec grad(S)||}$ = $\frac{2x\vec i + 2y\vec j + 2z\vec k}{\sqrt{4x^2 +4y^2 +4z^2}}$=$\frac{x\vec i + y\vec j + z\vec k}{\sqrt{x^2 +y^2 +z^2}}$=$\frac{x\vec i + y\vec j + z\vec k}{1}$ where $x^2 +y^2 +z^2=1$ and so you can evaluate your integral.
$$
\iint \vec{F} \cdot \hat{n}dS = \iint_D\vec{F}\cdot\vec{n}\frac{dxdy}{\sigma} = \iint_Dx^2+y^2+z(1-z)\frac{dxdy}{z}=\iint_D \frac{x^2+y^2-(1-x^2-y^2)+\sqrt{1-x^2-y^2}dxdy}{\sqrt{1-x^2-y^2}}=2 \iint_D \frac{x^2+y^2}{\sqrt{1-x^2-y^2}}dxdy -\iint_D \frac{1}{\sqrt{1-x^2-y^2}}dxdy + \iint_D dxdy$$. With D= {${{ (x,y) \in R^2 \ , x^2+y^2=1}}$} Now considering $x=rcos(\theta)$, $y=rsin(\theta)$ with the Jaacobian $|J|$ = $r$ you can solve the integral that will surely lead to $\frac{5\pi}{2}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1327807",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Computing $\sum_{n\geq 0}n\frac{1}{4^n}$ Can I compute the sum
$$
\sum_{n\geq 0}n\frac{1}{4^n}
$$
by use of some trick?
First I thought of a geometrical series?
| The sum can be expanded as follows:
\begin{align*}
\begin{array}{l|cccc}
\text{once}&1/4^1&&&\\
\text{twice}&1/4^2&1/4^2&&\\
\text{3 times}&1/4^3&1/4^3&1/4^3&&\\
\vdots\\
\text{$n$ times}&1/4^n&1/4^n&\cdots&1/4^n\\
\vdots
\end{array}
\end{align*}
Now, instead of summing row-by-row, try summing column-by-column:
\begin{align*}
\left(\frac{1}{4^1}+\frac{1}{4^2}+\frac{1}{4^3}+\ldots\right)+\left(\frac{1}{4^2}+\frac{1}{4^3}+\ldots\right)+\left(\frac{1}{4^3}+\ldots\right)+\ldots
\end{align*}
Each summand is of the form $$\sum_{j=k}^{\infty}\frac{1}{4^j}=\frac{1}{4^k}\sum_{j=0}^{\infty}\frac{1}{4^j}=\frac{1}{4^k}\times\frac{1}{1-1/4}=\frac{1}{4^k}\times\frac{4}{3}$$
for $k\in\{1,2,\ldots\}$. Now sum over $k$ to get
$$\sum_{k=1}^{\infty}\frac{1}{4^k}\times \frac{4}{3}=\frac{1/4}{1-1/4}\times\frac{4}{3}=\frac{1}{3}\times\frac{4}{3}=\frac{4}{9}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1328050",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 5,
"answer_id": 2
} |
evaluate $\int_0^{2\pi} \frac{1}{\cos x + \sin x +2}\, dx $ This is supposed to be a very easy integral, however I cannot get around.
Evaluate:
$$\int_0^{2\pi} \frac{1}{\cos x + \sin x +2}\, dx$$
What I did is:
$$\int_{0}^{2\pi}\frac{dx}{\cos x + \sin x +2} = \int_{0}^{2\pi} \frac{dx}{\left ( \frac{e^{ix}-e^{-ix}}{2i}+ \frac{e^{ix}+e^{-ix}}{2} +2\right )}= \oint_{|z|=1} \frac{dz}{iz \left ( \frac{z-z^{-1}}{2i} + \frac{z+z^{-1}}{2}+2 \right )}$$
Although the transformation is correct I cannot go on from the last equation. After calculations in the denominator there still appears $i$ and other equations of that. How can I proceed?
| Note that: $$\frac{1}{iz \left ( \frac{z-z^{-1}}{2i} + \frac{z+z^{-1}}{2}+2 \right )}$$
is a rational function. Write it as a quotient of polynomials, then factor the denominator and apply partial fractions.
If you change the integral to:
$$\int_{-\pi}^{\pi} \frac{1}{\cos x + \sin x +2}\, dx$$ you can apply the Weierstrass substitution and get:
$$\int_{-\infty}^{\infty} \frac{\frac{2}{1+t^2}}{\frac{1-t^2}{1+t^2}+\frac{2t}{1+t^2}+2}\,dt=\int_{-\infty}^{\infty}\frac{2\,dt}{2+(t+1)^2}\\=\int_{-\infty}^{\infty}\frac{dt}{1+\left(\frac{1+t}{\sqrt 2}\right)^2}$$
Substituting $u=\frac{1+t}{\sqrt{2}}$ gives:
$$=\sqrt{2}\int_{-\infty}^{\infty} \frac{du}{1+u^2}=\sqrt{2}\pi$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1328371",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Centroid of a Triangle on a inscribed circle $AB$ is the hypotenuse of the right $\Delta ABC$ and $AB = 1$.
Given that the centroid of the triangle $G$ lies on the incircle of $\Delta ABC$, what is the perimeter of the triangle?
| Assume that $C=(0,0),B=(a,0),A=(0,b)$ and $c=\sqrt{a^2+b^2}=AB$. We have:
$$ G = \left(\frac{a}{3},\frac{b}{3}\right) \tag{1}$$
and the inradius is given by
$$ r = \frac{2\Delta}{a+b+c} = \frac{ab}{a+b+c},\tag{2}$$
so the equation of the incircle is given by:
$$ \left(x-\frac{ab}{a+b+c}\right)^2 + \left(y-\frac{ab}{a+b+c}\right)^2 = \left(\frac{ab}{a+b+c}\right)^2\tag{3} $$
and since $G$ lies on the incircle, by setting $a=\sin\theta,b=\cos\theta$ we get:
$$ 5-3\sin\theta-3\cos\theta-3\cos\theta\sin\theta\tag{4} $$
or:
$$ (1+\sin\theta)(1+\cos\theta)=\frac{8}{3} \tag{5}$$
hence:
$$ a+b+c = \sin\theta+\cos\theta+1 = \color{red}{\frac{4}{\sqrt{3}}}\tag{6}$$
since $(4)$ can be solved as a quadratic equation in $\sin\theta+\cos\theta$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1329263",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
} |
Minimum value of $\sqrt{\frac{a}{b+c}}+\sqrt{\frac{b}{c+a}}+\frac{24}{5\sqrt{5a+5b}}$ Let $a\ge b\ge c\ge 0$ such that $a+b+c=1$
Find the minimum value of $P=\sqrt{\dfrac{a}{b+c}}+\sqrt{\dfrac{b}{c+a}}+\dfrac{24}{5\sqrt{5a+5b}}$
I found that the minimum value of $P$ is $\dfrac{78}{5\sqrt{15}}$ when $a=b=\dfrac{3}{8};c=\dfrac{1}{4}$
And this is my try
Applying AM-GM inequality, we get: $\dfrac{b+c}{a}+\dfrac{5}{3}\ge2\sqrt{\dfrac{5}{3}}.\sqrt{\dfrac{b+c}{a}}$
This implies $\sqrt{\dfrac{a}{b+c}}\ge2\sqrt{\dfrac{5}{3}}.\dfrac{3a}{3+2a}$
Similarly, $\sqrt{\dfrac{b}{a+c}}\ge2\sqrt{\dfrac{5}{3}}.\dfrac{3b}{3+2b}$
We need to prove that: $2\sqrt{\dfrac{5}{3}}\left(\dfrac{3a}{3+2a}+\dfrac{3b}{3+2b}\right)+\dfrac{24}{5\sqrt{5a+5b}}\ge\dfrac{78}{5\sqrt{15}}$
But I have no idea how to continue. Who can help me or have any other idea?
| $P=\sqrt{\dfrac{a}{1-a}}+\sqrt{\dfrac{b}{1-b}}+\dfrac{24}{5\sqrt{5a+5b}}$
$m=\sqrt{\dfrac{a}{1-a}},n=\sqrt{\dfrac{b}{1-b}} \implies a=\dfrac{m^2}{m^2+1},b=\dfrac{n^2}{n^2+1},\\a\ge b\ge c \implies a\ge \dfrac{1}{3} \implies m^2\ge \dfrac{1}{2} ,a+b=1-c \ge 1-b \implies n\ge \sqrt{\dfrac{1}{2m^2+1}}\implies mn\ge \sqrt{\dfrac{m^2}{2m^2+1}}=\sqrt{\dfrac{1}{2+\dfrac{1}{m^2}}}\ge \dfrac{1}{2} \implies 4mn \ge 2 $
$\dfrac{1}{a+b}=\dfrac{m^2n^2+m^2+n^2+1}{2m^2n^2+m^2+n^2}\ge \dfrac{(\dfrac{m+n}{2})^2+1}{2(\dfrac{m+n}{2})^2} \iff (n-m)^2(n^2+m^2+4mn-2) \ge 0$
$(\dfrac{m+n}{2}) =\dfrac{1}{t} \implies P \ge \dfrac{2}{t}+\dfrac{24}{5\sqrt{10}}\sqrt{1+t^2}=2f(t) \\ f'(t)=-\dfrac{1}{t^2}+\dfrac{12}{5\sqrt{10}}\sqrt{\dfrac{t^2}{1+t^2}}=0 \implies t^2=\dfrac{5}{3}$
it is easy to verify this is the min point.
$f_{min}=\dfrac{39}{5\sqrt{15}}, P_{min}=\dfrac{78}{5\sqrt{15}}$ when $m=n$
so $m=n=\dfrac{1}{t} \implies a=b=\dfrac{1}{1+t^2}=\dfrac{3}{8}>\dfrac{1}{3}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1330429",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 0
} |
Quick way to solve the system $\displaystyle \left( \frac{3}{2} \right)^{x-y} - \left( \frac{2}{3} \right)^{x-y} = \frac{65}{36}$, $xy-x+y=118$. Consider the system
$$\begin{aligned} \left( \frac{3}{2} \right)^{x-y} - \left( \frac{2}{3} \right)^{x-y} & = \frac{65}{36}, \\ xy -x +y & = 118. \end{aligned}$$
I have solved it by performing the substitutions $x-y=u$ and $xy=v$. Then I multiplied the first equation by $6^u$ and used $a^2-b^2=(a+b)(a-b)$ to find
$$(3^u+2^u)(3^u-2^u) = 65 \cdot 6^{u-2}.$$
By inspection I found $u=2$ and $v=120$. I solved the original system in $x,y$ and got the answers. Is there another quicker way to solve this without resorting to this sort of ninja inspection? I have found a second solution by solving $a^u +1/a^u = 65/36$, which assures $u=2$ but takes much more time.
Could there be a third way faster than these?
| Consider $$ \left( \frac{3}{2} \right)^{x-y} - \left( \frac{2}{3} \right)^{x-y} = \frac{65}{36}$$ and set $a=\left( \frac{3}{2} \right)^{x-y}$. So, you have $a-\frac 1a=\frac{65}{36}$, that is to say $a=\frac{9}{4}$. So $$\left( \frac{3}{2} \right)^{x-y}=\frac{9}{4}=\left( \frac{3}{2} \right)^2$$ so $x-y=2$.
For the remaining, the same as in lab bhattacharjee'answer.
No ninja inspection required.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1331134",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 1
} |
Find the cubic equation of $x=\sqrt[3]{2-\sqrt{3}}+\sqrt[3]{2+\sqrt{3}}$ Find the cubic equation which has a root $$x=\sqrt[3]{2-\sqrt{3}}+\sqrt[3]{2+\sqrt{3}}$$
My attempt is
$$x^3=2-\sqrt{3}+3\left(\sqrt[3]{(2-\sqrt{3})^2}\right)\left(\sqrt[3]{(2+\sqrt{3})}\right)+3\left(\sqrt[3]{(2-\sqrt{3})}\right)\left(\sqrt[3]{(2+\sqrt{3})^2}\right)+2-\sqrt{3}$$
$$x^3=4+\left(\sqrt[3]{(2-\sqrt{3})^2}\right)\left(\sqrt[3]{(2+\sqrt{3})}\right)+3\left(\sqrt[3]{(2-\sqrt{3})}\right)\left(\sqrt[3]{(2+\sqrt{3})^2}\right)$$
then what I will do??
| Let $a^3=2-\sqrt3,b^3=2+\sqrt3$, we have
$$a^3+b^3=4$$
$$a+b=x$$
$$ab=1$$
Consider
\begin{align}
x^3=(a+b)^3&=3ab(a+b)+a^3+b^3\\&=3x+4
\end{align}
Therefore
$$x^3-3x-4=0$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1331417",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 8,
"answer_id": 5
} |
Finding $ \lim_{x\to 2}\frac{\sqrt{x+2}-\sqrt{2x}}{x^2-2x} $ I'm kind of stuck on this problem, I could use a hint.
$$
\lim_{x\to 2}\frac{\sqrt{x+2}-\sqrt{2x}}{x^2-2x}
$$
After some algebra, I get
$$
{\lim_{x\to 2}\frac{x+2 - 2x}{x(x-2)-\sqrt{x+2}+\sqrt{2x}}}
$$
EDIT above should be:
$$
\lim_{x\to 2}\frac{x+2 - 2x}{x(x-2)(\sqrt{x+2}+\sqrt{2x)}}
$$
I'm stuck at this point, any help would be greatly appreciated.
| \begin{align}
\lim_{x\to 2}\frac{\sqrt{x+2}-\sqrt{2x}}{x^2-2x}&=
\lim_{x\to 2}\frac{\sqrt{x+2}-\sqrt{2x}}{x(x-2)}\frac{\sqrt{x+2}+\sqrt{2x}}{\sqrt{x+2}+\sqrt{2x}} \\
&=\lim_{x \to 2} \frac{(x+2)-(2x)}{x(x-2)(\sqrt{x+2}+\sqrt{2x})} \\
&=\lim_{x \to 2} \frac{-(x-2)}{x(x-2)(\sqrt{x+2}+\sqrt{2x})} \\
&=\lim_{x \to 2} \frac{-1}{x(\sqrt{x+2}+\sqrt{2x})} \\
&=\frac {-1}{2(\sqrt{2+2}+\sqrt{2(2)})} \\
&=\boxed{-\frac 18}
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1333110",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
Value of $x$ in $\sin^{-1}(x)+\sin^{-1}(1-x)=\cos^{-1}(x)$ How can we find the value of $x$ in $\sin^{-1}(x)+\sin^{-1}(1-x)=\cos^{-1}(x)$?
Note that $\sin^{-1}$ is the inverse sine function.
I'm asking for the solution $x$ for this equation.
Please work out the solution.
| $\bf{My\; Solution::}$ Given $$\sin^{-1}(x)+\sin^{-1}(1-x) = \cos^{-1}(x)$$
$$\displaystyle \sin^{-1}(1-x) = \cos^{-1}(x)-\sin^{-1}(x)\Rightarrow (1-x) = \sin \left[\cos^{-1}(x)-\sin^{-1}(x)\right]$$
Using $$\sin^{-1}(x)+\cos^{-1}(x) = \frac{\pi}{2}$$
So $$\displaystyle (1-x) = \sin\left[\frac{\pi}{2}-2\sin^{-1}(x)\right]=\cos(2\sin^{-1}(x)) = \cos(\cos^{-1}(1-2x^2))$$
Using $$2\sin^{-1}(x)=\cos^{-1}(1-2x^2)$$
So $$\displaystyle (1-x) = (1-2x^2)\Rightarrow 2x^2-x = 0$$
So $$\displaystyle 2x^2-x = 0\Rightarrow x = 0\;\;,x = \frac{1}{2}$$
Now Put into Original Equation we get $\displaystyle x = \frac{1}{2}$ and
$x=0$ satisfy the Given equation.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1333609",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 1
} |
How to find sum of the infinite series $\sum_{n=1}^{\infty} \frac{1}{ n(2n+1)}$ $$\frac{1}{1 \times3} + \frac{1}{2\times5}+\frac{1}{3\times7} + \frac{1}{4\times9}+\cdots $$
How to find sum of this series?
I tried this: its $n$th term will be = $\frac{1}{n}-\frac{2}{2n+1}$; after that I am not able to solve this.
| $$\frac{1}{n(2n+1)}= \frac 1n-\frac{2}{2n+1}=2\left(\frac{1}{2n}-\frac{1}{2n+1}\right)$$
Hence
$$\sum_{n=1}^{\infty} \frac{1}{ n(2n+1)} =2 \sum_{n=1}^{\infty} \left(\frac{1}{2n}-\frac{1}{2n+1}\right)= 2\left(1-\sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{n}\right)=\color{red}{ 2(1-\ln 2)} $$
since $${\displaystyle \ln(1+x)=\sum _{n=1}^{\infty }{\frac {(-1)^{n+1}}{n}}x^{n}=x-{\frac {x^{2}}{2}}+{\frac {x^{3}}{3}}-\cdots ,}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1334870",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 6,
"answer_id": 2
} |
Product of Matrices I Given the matrix
\begin{align}
A = \left( \begin{matrix} 1 & 2 \\ 3 & 2 \end{matrix} \right)
\end{align}
consider the first few powers of $A^{n}$ for which
\begin{align}
A = \left( \begin{matrix} 1 & 2 \\ 3 & 2 \end{matrix} \right)
\hspace{15mm} A^{2} = \left( \begin{matrix} 7 & 6 \\ 9 & 10 \end{matrix} \right) \hspace{15mm}
A^{3} = \left( \begin{matrix} 25 & 26 \\ 39 & 38 \end{matrix} \right).
\end{align}
Notice that the first rows have the values $(1,2)$, $(7,6)$, $(25,26)$ of which the first and second elements are rise and fall in a cyclical pattern.
The same applies to the bottom rows.
*
*Is there an explanation as to why the numbers rise and fall in order of compared columns?
*What is the general form of the $A^{n}$ ?
| It is often possible to diagonalize a square matrix, that is find a diagonal matrix $D$ and another matrix $P$ such that $$A=PDP^{-1}.$$ This is a nice property because $$A^n=PDP^{-1}PDP^{-1}...PDP^{-1}=PD^nP^{-1}.$$
The $P$ and $P^{-1}$ matrices cancel except on the ends, so we only have to take a power of the diagonal matrix, which is easy. If
$$D=\left(\begin{array}{cc}
a & 0\\
0 & b\\ \end{array}\right),$$
then
$$D^n=\left(\begin{array}{cc}
a^n & 0\\
0 & b^n\\ \end{array}\right).$$
In fact, it is possible to diagonalize your matrix as
$$A=\left(\begin{array}{cc}
2 & -1\\
3 & 1\\ \end{array}\right)
\left(\begin{array}{cc}
4 & 0\\
0 & -1\\ \end{array}\right)
\left(\begin{array}{cc}
1/5 & 1/5\\
-3/5 & 2/5\\ \end{array}\right).$$
Thus,
$$A^n=\left(\begin{array}{cc}
2 & -1\\
3 & 1\\ \end{array}\right)
\left(\begin{array}{cc}
4^n & 0\\
0 & (-1)^n\\ \end{array}\right)
\left(\begin{array}{cc}
1/5 & 1/5\\
-3/5 & 2/5\\ \end{array}\right)=\left(
\begin{array}{cc}
\frac{3 (-1)^n}{5}+\frac{1}{5} 2^{2 n+1} & \frac{1}{5} (-2) (-1)^n+\frac{1}{5} 2^{2 n+1} \\
\frac{1}{5} (-3) (-1)^n+\frac{3\ 4^n}{5} & \frac{2 (-1)^n}{5}+\frac{3\ 4^n}{5} \\
\end{array}
\right).$$
As for the top row business, you can see that each term in the top row has two terms. The second of each is the same. In $n$ is odd, the first terms are $-3/5$ and $2/5$ respectively. If $n$ is even, the two terms are $3/5 $ and $-2/5$. Can you see why this relationship causes the behavior you noticed?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1336283",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
The partial fraction decomposition of $\dfrac{x}{x^3-1}$ I was trying to decompose $\dfrac{x}{x^3-1}$ into Partial Fractions. I tried the following:
$$\dfrac{x}{(x-1)(x^2+x+1)}=\dfrac{A}{(x-1)}+\dfrac{B}{(x^2+x+1)}$$
$$\Longrightarrow A(x^2+x+1)+B(x-1)=x$$
Putting $x=1,$
$$A(1+1+1)+B(1-1)=1\Rightarrow A=\dfrac{1}{3}$$
$$\Rightarrow\dfrac{1}{3}(x^2+x+1) +B(x-1)=x$$
$$\Rightarrow (x^2+x+1)+3Bx-3B=x$$
$$x^2+(3B+1)x +(1-3B)=x$$
At this point I gave up thinking I must have done something wrong since there is no $x^2$ on the RHS while it is there in the LHS.
I would be truly grateful if somebody would be so kind to help me understand my mistake. Many, many thanks in advance!
| $\frac{x}{x^3-1}=\frac{x}{(x-1)(x^2+x+1)}$
$\frac{x}{(x-1)(x^2+x+1)}=\frac{A}{x-1}+\frac{Bx+C}{x^2+x+1}$
$x=A(x^2+x+1)+(Bx+C)(x-1)$
Let $x=1$
$1=A(1^2+1+1)$
$1=3A$
$A=\frac{1}{3}$
So, $x=\frac{1}{3}(x^2+x+1)+(Bx+C)(x-1)$
$x=\frac{1}{3}x^2+\frac{1}{3}x+\frac{1}{3}+Bx^2-Bx+Cx-C$
$x=(\frac{1}{3}+B)x^2+(\frac{1}{3}-B+C)x+\frac{1}{3}-C$
Compare the coefficient,
$0x^2=(\frac{1}{3}+B)x^2$
$\frac{1}{3}+B=0$, $B=-\frac{1}{3}$
$x=(\frac{1}{3}-B+C)x$
$\frac{1}{3}-B+C=1$
$\frac{1}{3}-(-\frac{1}{3})+C=1$
$C=\frac{1}{3}$
Therefore,
$$\frac{x}{(x-1)(x^2+x+1)}=\frac{1}{3(x-1)}+\frac{-\frac{1}{3}x+\frac{1}{3}}
{x^2+x+1}$$
Simplify it and get $\frac{1}{3(x-1)}+\frac{1-x}{3(x^2+x+1)}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1337329",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 2,
"answer_id": 1
} |
How do I prove this combinatorial identity using inclusion and exclusion principle? $$\binom{n}{m}-\binom{n}{m+1}+\binom{n}{m+2}-\cdots+(-1)^{n-m}\binom{n}{n}=\binom{n-1}{m-1}$$
Note that we can show this with out using inclusion and exclusion principle by using Pascal's Identity i.e. $C(n,k)=C(n-1,k)+C(n-1,k-1)$.
| Note: This answer is inspired by the deleted answer of @diracpaul.
We analyze lattice paths generated from $(1,0)$ steps in $x$-direction and $(0,1)$ steps in $y$-direction. The number of paths from $(0,0)$ to $(m,n-m)$ is $\binom{n}{m}$. This is valid since for paths of length $n$ we can select $m$ steps in $x$-direction leaving $n-m$ steps in $y$-direction.
Let's write $\binom{n}{m}$ in the more symmetrical version using multinomial coefficients $\binom{n}{m,n-m}$. Using this notation OPs equation becomes
\begin{align*}
\binom{n-1}{m-1,n-m}&=\binom{n}{m,n-m}-\binom{n}{m+1,n-m-1}\\
&\qquad+\binom{n}{m+2,n-m-2}-\cdots+(-1)^{n-m}\binom{n}{n,0}\tag{1}
\end{align*}
For convenience only we change the representation by setting $x=m$ and $y=n-m$. Then OPs equation (1) becomes
\begin{align*}
\binom{x+y-1}{x-1,y}&=\binom{x+y}{x,y}-\binom{x+y}{x+1,y-1}\\
&\qquad+\binom{x+y}{x+2,y-2}-\cdots+(-1)^{y}\binom{x+y}{x+y,0}\tag{2}
\end{align*}
Before we interpret the identity (2) one additional aspect. We can use only $(1,0)$ steps or $(0,1)$ steps. So, when we go from $(0,0)$ to a point $(x,y)$ with $x,y>0$, we have to pass either $(x-1,y)$ or $(x,y-1)$. This gives the well known identity
\begin{align*}
\binom{x+y}{x,y}=\binom{x+y-1}{x-1,y}+\binom{x+y-1}{x,y-1}\tag{3}
\end{align*}
Now we claim the following:
*
*The LHS of (2) gives the number of paths of length $x+y-1$ from $(0,0)$ to $(x-1,y)$.
*The RHS of (2) gives the number of paths of length $x+y$ from $(0,0)$ to $(x,y)$ which pass $(x-1,y)$.
Since there is only one possibility to go from $(0,0)$ to $(x,y)$ via $(x-1,y)$, namely by adding a $(1,0)$ step from $(x-1,y)$ to $(x,y)$ it's obvious that both parts give the same number. According to (3) we conclude
\begin{align*}
\binom{x+y-1}{x-1,y}&=\binom{x+y}{x,y}-\binom{x+y-1}{x,y-1}\tag{4}
\end{align*}
The RHS of expression (4) is correct, but not the same as the RHS of (3).
The idea is to approach the representation (3) step by step. We note that $\binom{x+y-1}{x,y-1}$ is the number of paths of length $x+y-1$ from $(0,0)$ to $(x,y-1)$. We argue now similar as above and say that this is the same as the number of paths of length $x+y$ from $(0,0)$ to $(x+1,y-1)$ passing through $(x,y-1)$.
This is realised according to (3) via
\begin{align*}
\binom{x+y-1}{x,y-1}&=\binom{x+y}{x+1,y-1}-\binom{x+y-1}{x+1,y-2}\tag{5}
\end{align*}
Combining this with (4) gives
\begin{align*}
\binom{x+y-1}{x-1,y}&=\binom{x+y}{x,y}-\binom{x+y-1}{x,y-1}\\
&=\binom{x+y}{x,y}-\binom{x+y}{x+1,y-1}+\binom{x+y-1}{x+1,y-2}
\end{align*}
And again we substitute the surplus $\binom{x+y-1}{x+1,y-2}$ by the number of pathes of length $x+y$ from $(0,0)$ to $(x+2,y-2)$ passing through $(x+1,y-2)$. This results in
\begin{align*}
\binom{x+y-1}{x-1,y}&=\binom{x+y}{x,y}-\binom{x+y-1}{x,y-1}\\
&=\binom{x+y}{x,y}-\binom{x+y}{x+1,y-1}+\binom{x+y}{x+2,y-2}-\binom{x+y-1}{x+2,y-3}
\end{align*}
Proceeding in this way we finally reach step by step the RHS of (2). Observe that the amounts of surplus are compensated according to the inclusion-exclusion principle.
We conclude: The binomial identity
\begin{align*}
\binom{x+y-1}{x-1,y}&=\binom{x+y}{x,y}-\binom{x+y}{x+1,y-1}\\
&\qquad+\binom{x+y}{x+2,y-2}-\cdots+(-1)^{y}\binom{x+y}{x+y,0}=\tag{2}
\end{align*}
shows that the number of paths of length $x+y-1$ from $(0,0)$ to $(x-1,y)$ is the same as the number of paths of length $x+y$ from $(0,0)$ to $(x,y)$ passing through $(x-1,y)$, whereby the surplus can be compensated by successively adding and subtracting paths of length $x+y$ according to the inclusion-exclusion principle.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1337914",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "12",
"answer_count": 3,
"answer_id": 2
} |
Conditions of a differential equation Consider the differential equation
\begin{align}
2 x^2 y'' + x(x^2 - 1) y' + (2 x^2 - x +1)y = 0 \hspace{5mm} y(0) = 0, y'(0)=1.
\end{align}
A solution readily found is
\begin{align}
y(x) &= B_{0} \, \sqrt{x} (x+1) \, e^{- \frac{x^{2}}{4}} \\
& \hspace{5mm} + B_{1} \, e^{- \frac{3 \pi i}{4}} \, \sqrt{x} \, e^{- \frac{x^{2}}{4}} \, \left[ 4 \, e^{\frac{3 \pi i}{4}} \, x \, e^{- \frac{x^{2}}{4}}
+ \sqrt{2} (x+1) \left( 2 \Gamma\left(\frac{3}{4}, - \frac{x^{2}}{4}\right) - i \, \Gamma\left(\frac{1}{4}, - \frac{x^{2}}{4}\right) \right) \right]
\end{align}
How are the boundary conditions to be aplied such that $B_{0}$ and $B_{1}$ can be obtained?
| We develop $y(x)$ in the vicinity of $x=0$
$$2\Gamma\left(\frac{3}{4},-\frac{x^2}{4}\right)+i\Gamma\left(\frac{1}{4},-\frac{x^2}{4}\right)=2\Gamma\left(\frac{3}{4}\right)+i\Gamma\left(\frac{1}{4}\right)-(2i)^{3/2}\sqrt{x}+O(x)$$
$$4e^{\frac{3\pi i}{4}}e^{-\frac{x^2}{4}}\sqrt{2}(x+1)\left(2\Gamma\left(\frac{3}{4},-\frac{x^2}{4}\right)+i\Gamma\left(\frac{1}{4},-\frac{x^2}{4}\right)\right)=$$ $$=2\Gamma\left(\frac{3}{4}\right)+i\Gamma\left(\frac{1}{4}\right)-(2i)^{3/2}\sqrt{x}+O(x)$$
$$e^{-\frac{3\pi i}{4}}\sqrt{x}e^{-\frac{x^2}{4}} \left[4e^{\frac{3\pi i}{4}}e^{-\frac{x^2}{4}}\sqrt{2}(x+1)\left(2\Gamma\left(\frac{3}{4},-\frac{x^2}{4}\right)+i\Gamma\left(\frac{1}{4},-\frac{x^2}{4}\right)\right)\right]=$$ $$= e^{-\frac{3\pi i}{4}}\left(2\Gamma\left(\frac{3}{4}\right)+i\Gamma\left(\frac{1}{4}\right)\right)\sqrt{x} -2^{3/2}x +O(x^{3/2})$$
$$\sqrt{x}(x+1)e^{-\frac{x^2}{4}}=\sqrt{x}+O(x^{3/2})$$
$$y(x)=B_0\sqrt{x}+B_1\left(e^{-\frac{3\pi i}{4}}\left(2\Gamma\left(\frac{3}{4}\right)+i\Gamma\left(\frac{1}{4}\right)\right)\sqrt{x}-2^{3/2}x\right) +O(x^{3/2}) $$
The condition $y(0)=0$ is fulfilled.
In order to have a finite value of $y'(0)$ it is necessay that there is no term $\sqrt{x}$ because, if not the derivative $\frac{1}{2\sqrt{x}}$ would be infinite. Hense :
$$B_0+B_1e^{-\frac{3\pi i}{4}}\left(2\Gamma\left(\frac{3}{4}\right)+i\Gamma\left(\frac{1}{4}\right)\right) =0$$
$$y(x)=-B_1 2^{3/2}x +O(x^{3/2}) $$
and the condition $y'(0)=1$ leads to :
$$B_1=-2^{-3/2}$$
$$B_0=2^{-3/2}e^{-\frac{3\pi i}{4}}\left(2\Gamma\left(\frac{3}{4}\right)+i\Gamma\left(\frac{1}{4}\right)\right)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1341549",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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} |
Minimization on compact region I need to solve the minimization problem
$$\begin{matrix} \min & x^2 + 2y^2 + 3z^2 \\ subject\;to & x^2 + y^2 + z^2 =1\\ \; & x+y+z=0
\end{matrix}$$
I was trying to verify the first order conditions using lagrange multipliers
$$\left\{\begin{matrix}
2x = 2\lambda_1x + \lambda_2 \\
4y = 2 \lambda_1 y + \lambda_2 \\
6z = 2\lambda_1 z + \lambda_2
\end{matrix}\right.$$
Summing the 3 equations we get
$$ 2y + 4z = 3 \lambda_2$$
Multiplying the first one for x, the second for $y$ and third for $z$ and summing:
$$ 2 + 2y^2 + 4z^2 = 2 \lambda_1$$
I just got stuck there and I have no idea how to proceed. We can check the answer here but I couldn't find none of the variables. Any idea?
Thanks in advance!
| $x^2+2y^2+3z^2=x^2+y^2+z^2+y^2+2z^2=1+y^2+2z^2, x=-(y+z),(y+z)^2+y^2+z^2=1 \implies y^2+yz+z^2=\dfrac{1}{2}$, now the problem become : find min of $y^2+2z^2$ with $y^2+yz+z^2=\dfrac{1}{2}$,it shoud be easier now.
or we can go further :$y=\dfrac{-z \pm \sqrt{2-3z^2}}{2},f(z)=\left(\dfrac{-z \pm \sqrt{2-3z^2}}{2}\right)^2+2z^2$ which is single variable function.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1342012",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 3
} |
How can I prove irreducibility of polynomial over a finite field?
I want to prove what $x^{10} +x^3+1$ is irreducible over a field $\mathbb F_{2}$ and $x^5$ + $x^4 +x^3 + x^2 +x -1$ is reducible over $\mathbb F_{3}$.
As far as I know Eisenstein criteria won't help here.
I have heard about a criteria of irreducibility which sounds like
"If $f(x)$ divides $x^{p^n-1}-1$ (where $n=\deg(f))$, $p$ is from $\mathbb F_{p}$ then $f(x)$ is irreducible."
But I don't know how to show if $f(x)$ can be divided by this or not.
| Neither polynomial has zeros in the respective prime field, so they do not have linear factors either.
Let's check the first polynomial $p(x)=x^{10}+x^3+1\in\Bbb{F}_2[x]$. The claim is that it is irreducible. A more efficient way of using the cited result is to observe that if $p(x)$ were a product of two or more irreducible factors at least one of those factors would have degree $\le5$. We eliminate these possibilities one-by-one. Testing for the presence of higher degree factors becomes progressively a little bit harder as the degree goes up, but the principle is always the same, and this result gives us a tool.
Could it have an irreducible quadratic factor? By the cited result all the irreducible quadratic polynomials are factors of $p_2(x):=x^{2^2-1}-1=x^3+1$. To eliminate the possibility of $p(x)$ having a quadratic factor it thus suffices to show that $\gcd(p(x),p_2(x))=1$. Because $p(x)-p_2(x)=x^{10}$ any common factor they have is also a factor of $x^{10}$, but the only irreducible factor of $x^{10}$ is $x$, so we have cleared this obstacle.
To eliminate the possibility of $p(x)$ having a cubic factor it similarly suffices to check that $\gcd(p(x),p_3(x))=1$, where $p_3(x)=x^{2^3-1}-1=x^7+1$. Here we see (the first step in calculating this gcd by Euclid's algorithm) that
$$
p(x)-x^3p_3(x)=1,
$$
so no common factors there either.
To eliminate the possibility of a quartic factor of $p(x)$, we need to calculate $\gcd(p(x),p_4(x))$ with $p_4(x)=x^{2^4-1}-1=x^{15}+1.$ Let's run Euclid:
$$
\begin{aligned}
p_4(x)-x^5p(x)&=x^8+x^5+1=: r_1(x)\\
p(x)-x^2r_1(x)&=x^7+x^3+x^2+1=:r_2(x).
\end{aligned}
$$
Here we could continue with Euclid, but we can also save a step or two by factoring the remainder whenever we can do so easily. Here
$$
r_2(x)=x^3(x^4+1)+(x^2+1)=x^3(x^2+1)^2+(x^2+1)=(x^2+1)(x^5+x^3+1).
$$
We know from earlier steps that $p(x)$ is not divisible by $(x+1)$. Because
$(x^2+1)=(x+1)^2$, we can discard that factor as a possibility, and continue Euclid with $\gcd(r_1(x), r_3(x))$ where $r_3(x)=x^5+x^3+1$.
We see that
$$
\begin{aligned}
r_1(x)-x^3r_3(x)&=x^6+x^5+x^3+1=(x+1)(x^5+x^2+x+1)\\
&=(x+1)^2(x^4+x^3+x^2+1)=(x+1)^3(x^3+x+1).
\end{aligned}
$$
Here I used extensively the trick (based on geometric sums) that whenever $0<a<b$ we have
$$
\frac{x^a+x^b}{x+1}=x^a+x^{a+1}+x^{a+2}+\cdots+x^{b-1}.
$$
Anyway, the above calculation shows that any eventual common factor of $p(x)$ and $p_4(x)$ is also a factor of the cubic $x^3+x+1$, but we already showed that $p(x)$ has no cubic factors. Therefore we are done with this case, too.
The last step of proving that $p(x)$ as no quintic factors needs another run of Euclid. This time checking that $\gcd(p(x),p_5(x))=1$, where $p_5(x)=x^{31}+1$. To get started we use square-and-multiply. We already calculated in the previous step that
$$
x^{15}\equiv x^8+x^5\pmod{p(x)}.\qquad(*)
$$
Consequently $x^{16}\equiv x^9+x^6\pmod{p(x)}$. Squaring $(*)$ (Freshman's dream simplifies squaring in characteristic two) gives
$$
x^{30}\equiv x^{16}+x^{10}\equiv (x^9+x^6)+x^{10}\pmod{p(x)}.
$$
Here
$$
x^{10}+x^9+x^6-p(x)=x^9+x^6+x^3+1.
$$
Therefore (multiply by $x$)
$$
x^{31}\equiv x^{10}+x^7+x^4+x\equiv x^7+x^4+x^3+x+1\pmod{p(x)}.
$$
So we know the remainder of $p_5(x)$ modulo $p(x)$. The remaining task is thus to check that
$$
\gcd(p(x),x^7+x^4+x^3+x)=1.
$$
Leaving that to you :-)
Your other polynomial
$$
q(x)=x^5+x^4+x^3+x^2+x-1\in\Bbb{F}_3(x)
$$
is quintic. So if it is reducible, it must have one factor that is either linear or a quadratic. We eliminated the possibility of linear factors by checking that it has no zeros. Using the above technique we find eventual quadratic factors by calculating
$$
\gcd(q(x),q_2(x))\in\Bbb{F}_3[x].
$$
Here $q_2(x)=x^{3^2-1}-1=x^8-1$. Leaving it to you to run this instance of Euclid, and find the quadratic factor.
| {
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"url": "https://math.stackexchange.com/questions/1343450",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "23",
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How to get the results of this logarithmic equation? How to solve this for $x$:
$$\log_x(x^3+1)\cdot\log_{x+1}(x)>2$$
I have tried to get the same exponent by getting the second
multiplier to reciprocal and tried to simplify $(x^3+1)$.
| Keep in mind $(\log_a b)(\log_c a) = \log_c b$.
Setting $a = x, b = x^3 + 1$, and $c = x + 1$, we have
$\log_x (x^3 + 1)\cdot \log_{x+1}x = \log_{x + 1}(x^3 + 1)$.
So your inequality is equivalent to
$\log_{x + 1}(x^3 + 1) > 2
\\ x^3 + 1 > (x + 1)^2
\\ x^3 + 1 > x^2 + 2x + 1
\\ x^3 - x^2 - 2x > 0
\\ x(x + 1)(x - 2) > 0$
Then $-1 < x < 0$ or $x > 2$. One case is ruled out.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1343570",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 1
} |
Proving a trigonometric identity with tangents Prove that: $$\tan^227^\circ +2 \tan27^\circ \tan36^\circ=1$$
any help, I appreciate it.
| The solution is nothing more than some computation and observing $2\cdot27 = 90 - 36.$
\begin{align*}
\tan^2(27^\circ) + 2 \tan(27 ^ \circ) \tan (36^\circ) &= \tan^2(27^\circ) + 2 \tan(27 ^ \circ) \cot (54^\circ) \\[1ex]
&= \tan^2(27^\circ) + 2 \, \frac{\tan(27 ^ \circ)}{ \tan(54^\circ)} \\[1ex]
&= \tan^2(27^\circ) + 2 \, \frac{\tan(27 ^ \circ)}{\frac{2 \tan{27^\circ}}{1 - \tan^2(27^\circ)}} \\[1ex]
&= \tan^2(27^\circ) + 2 \, \frac{\tan(27 ^ \circ) (1 - \tan^2(27^\circ))}{2 \tan{27^\circ}} \\[1ex]
&= \tan^2(27^\circ) + (1 - \tan^2(27^\circ)) \\
&= 1,
\end{align*}
where we have used the following identities:
$$\tan(\theta) = \cot(90^\circ - \theta),$$
$$\tan(2\theta) = \frac{2 \tan{\theta}}{1 - \tan^2(\theta)}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1344074",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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} |
Prove equality of Fibonacci sequence
Let $u(n)$ — the Fibonacci sequence. Prove that
$$u(1)^3+...+u(n)^3=\frac{ 1 }{ 10 } \left[ u(3n+2)+(-1)^{n+1}6u(n-1)+5 \right].$$
I suppose we need prove that equality by iduction in $n$. It's obviously holds at $n=1$. So let $P(k)$ holds, then also must $P(k + 1)$ holds. What do we do next?
| Consider the series
\begin{align}
\sum_{r=1}^{n} t^{r} = \frac{t \, (1-t^{n})}{1-t}
\end{align}
and
\begin{align}
F_{n}^{3} &= \frac{1}{5 \sqrt{5}} \, (\alpha^{n} - \beta^{n})^{3} \\
&= \frac{1}{5} \left( F_{3n} - 3 (-1)^{n} \, F_{n} \right),
\end{align}
since $\sqrt{5} \, F_{n} = \alpha^{n} - \beta^{n}$ and $2 \alpha = 1 + \sqrt{5}$, $2 \beta = 1 - \sqrt{5}$.
Now,
\begin{align}
\sum_{r=1}^{\infty} F_{n}^{3} &= \frac{1}{5 \sqrt{5}} \, \left[ \frac{\alpha^{3} (1 - \alpha^{3n})}{1 - \alpha^{3}} - \frac{\beta^{3} (1 - \beta^{3n})}{1 - \beta^{3}} \right] - \frac{3}{5 \sqrt{5}} \, \left[ \frac{-\alpha (1 - (-1)^{n} \alpha^{n})}{1 + \alpha} - \frac{- \beta (1 - (-1)^{n} \beta^{n})}{1 + \beta} \right] \\
&= \frac{1}{10} \, \left( F_{3n+2} - 6 \, (-1)^{n} \, F_{n-1} + 6 - F_{2} \right) \\
&= \frac{1}{10} \, \left( F_{3n+2} - 6 \, (-1)^{n} \, F_{n-1} + 5 \right)
\end{align}
| {
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"url": "https://math.stackexchange.com/questions/1345634",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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solve $|x-6|>|x^2-5x+9|$
solve $|x-6|>|x^2-5x+9|,\ \ x\in \mathbb{R}$
I have done $4$ cases.
$1.)\ x-6>x^2-5x+9\ \ ,\implies x\in \emptyset \\
2.)\ x-6<x^2-5x+9\ \ ,\implies x\in \mathbb{R} \\
3.)\ -(x-6)>x^2-5x+9\ ,\implies 1<x<3\\
4.)\ (x-6)>-(x^2-5x+9),\ \implies x>3\cup x<1 $
I am confused on how I proceed.
Or if their is any other short way than making $4$ cases than I would like to know.
I have studied maths up to $12$th grade. Thanks.
| HINT:
As $x^2-5x+9=\dfrac{(2x-5)^2+11}4\ge\dfrac{11}4>0\forall $ real $x$
So, $|x^2-5x+9|=+(x^2-5x+9)\forall $ real $x$
So, the problem reduces to two cases for $|x-6|$
Also, $|x-6|>\dfrac{11}4\implies \pm(x-6)>\dfrac{11}4$
| {
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"url": "https://math.stackexchange.com/questions/1346024",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Quadratic Absolute Value Equation Problem:
Find all $x$ such that $|x^2+6x+6|=|x^2+4x+9|+|2x-3|$
I can't understand how to get started with this. I thought of squaring both sides of the equation to get rid of the modulus sign, but then couldn't understand how to deal with the following term:$$2|(x^2+4+9)(2x+3)|$$Any help or tips in general would be greatly appreciated. Many thanks!
| You-know-me has already provided a good answer.
But I decided to undelete my answer in the hope that this answer might be of some help.
Since we have $x^2+4x+9=(x+2)^2+5\gt 0$, we have
$$|x^2+4x+9|=x^2+4x+9.$$
Then, since we have $x^2+6x+6=0\iff x=-3\pm\sqrt{3}$ and $2x-3=0\iff x=\frac 32$, separate it into three cases :
*
*For $x\lt -3-\sqrt 3$ or $-3+\sqrt 3\lt x\lt\frac 32$, we have $x^2+6x+6\gt 0$ and $2x-3\lt 0$. So we have $x^2+6x+6=x^2+4x+9-(2x-3)\iff x=\frac 32.$ So, in this case, there exists no such $x$.
*For $-3-\sqrt 3\le x\le -3+\sqrt 3$, we have $x^2+6x+6\le 0$ and $2x-3\le 0$. So we have $-(x^2+6x+6)=x^2+4x+9-(2x-3)\iff x^2+4x+9=0.$ So, in this case, there exists no such $x$.
*For $x\ge \frac 32$, we have $x^2+6x+6\ge 0$ and $2x-3\ge 0$, and we know that $x^2+6x+6=x^2+4x+9+2x-3$ always holds. So, in this case, we have $x\ge \frac 32$.
Hence, the answer is $x\ge\frac 32$.
| {
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"timestamp": "2023-03-29T00:00:00",
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One-One Correspondences Adam the ant starts at $(0,0)$. Each minute, he flips a fair coin. If he flips heads, he moves $1$ unit up; if he flips tails, he moves $1$ unit right.
Betty the beetle starts at $(2,4)$. Each minute, she flips a fair coin. If she flips heads, she moves $1$ unit down; if she flips tails, she moves $1$ unit left.
If the two start at the same time, what is the probability that they meet while walking on the grid?
How do I go about solving this problem, and what's the answer?
| Adam and Betty may meet only after three steps from the start, in the points $(2,1),(1,2)$ or $(0,3)$.
After three steps, Adam is in $(2,1)$ with probability $\frac{3}{8}$, in $(1,2)$ with probability $\frac{3}{8}$ and in $(0,3)$ with probability $\frac{1}{8}$.
After three steps, Betty is in $(2,1)$ with probability $\frac{1}{8}$, in $(1,2)$ with probability $\frac{3}{8}$ and in $(0,3)$ with probability $\frac{3}{8}$.
So the probability that Adam and Betty meet is:
$$ \frac{3}{8}\cdot\frac{1}{8}+\frac{3}{8}\cdot\frac{3}{8}+\frac{3}{8}\cdot\frac{1}{8}=\color{red}{\frac{15}{64}}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1348533",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Does the equation $2\cos^2 (x/2) \sin^2 (x/2) = x^2+\frac{1}{x^2}$ have real solution?
Do the equation
$$2\cos^2 (x/2) \sin^2 (x/2) = x^2+\frac{1}{x^2}$$
have any real solutions?
Please help. This is an IITJEE question.
Here $x$ is an acute angle.
I cannot even start to attempt this question. I cannot understand.
| the right hand of the equation, you have $$x^2 + \frac1{x^2} = \left(x-\frac 1x\right)^2 + 2 \ge 2 \tag 1$$and equality occurs for $$x = \pm 1.$$
on the left hand side, we have $$ \frac12 \sin^2 x = 2\cos^2 (x/2) \sin^2 (x/2) \le \frac12 \tag 2$$ equality for $$\sin x = \pm 1.$$
but $(1)$ and $(2)$ are inconsistent, therefore $$2\cos^2 (x/2) \sin^2 (x/2) = x^2+\frac{1}{x^2} $$ has no real solutions.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Differentiate the Function $ h(z)=\ln\sqrt{\frac{a^2-z^2}{a^2+z^2}}$
Differentiate the function $$h(z)=\ln\sqrt{\frac{a^2-z^2}{a^2+z^2}}$$
My try:
$$h(z) = \frac{1}{2}\ln\left(a^2-z^2\right)-\frac{1}{2}\ln\left(a^2+z^2\right)$$
so
$$h'(z) = \frac{1}{2}\cdot\frac{2a-2z}{a^2-z^2}-\frac{1}{2}\cdot\frac{2a+2z}{a^2+z^2}$$
My answer is therefore $$h'(z) = \frac{-2a^2z+2z^2a}{(a^2-z^2)(a^2+z^2)}$$
Is it correct?
| Albeit you used logarithmic differentiation, you must never forget the power of substitutions.
Let $$u=\sqrt{\frac{a^2-z^2}{a^2+z^2}}.$$ And, for more simplicity, let $$v=\frac{a^2-z^2}{a^2+z^2}. $$ Then $u=\sqrt{v} $ with $h(z)=\ln(u).$
We find: $${d}h(z)=\frac{du}{u}$$ $$du=\frac{1}{2}v^{-1/2}dv $$ $$dv=\frac{-2z}{a^2+z^2}-\frac{2z(a^2-z^2)}{(a^2+z^2)^2}dz. $$
Before we undo our substitutions, we mentally note that $$\frac{dh(z)}{du}\times\frac{du}{dv}\times\frac{dv}{dz}=\frac{d}{dz}h(z)=h'(z). $$
Substituting back in yields the final result of $$h'(z)=\frac{-z}{a^2-z^2}-\frac{z}{a^2+z^2}. $$ To match other results on this page, a little algebraic manipulation yields the result $$h'(z)=\frac{-2a^2z}{(a^2-z^2)(a^2+z^2)}. $$
| {
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"timestamp": "2023-03-29T00:00:00",
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How to calculate $\int \frac{\sin x}{\tan x+\cos x} \, dx$ How to calculate $$\int \frac{\sin x}{\tan x+\cos x} \, dx\text{ ?}$$ I got to $$\int \frac{-u}{u^2-u-1} \, du$$ while $u=\sin x$ but can I continue from here?
| By replacing $\sin x$ with $u$ we have:
$$I=\int\frac{\sin x}{\tan x+\cos x}\,dx = \int \frac{\sin x\cos x}{1+\sin x-\sin^2 x}\,dx = \int \frac{u}{1+u-u^2}\,du\tag{1}$$
but the roots of $1+u-u^2$ occur at $u=\frac{1\pm\sqrt{5}}{2}$, so that:
$$ \frac{u}{1+u-u^2} = \frac{-\frac{5+\sqrt{5}}{10}}{u-\left(\frac{1+\sqrt{5}}{2}\right)}+\frac{-\frac{5-\sqrt{5}}{10}}{u-\left(\frac{1-\sqrt{5}}{2}\right)}\tag{2}$$
and:
$$ I = -\frac{1}{10}\left((5+\sqrt{5})\log(1+\sqrt{5}-2u)+(5-\sqrt{5})\log(-1+\sqrt{5}-2u)\right).\tag{3}$$
| {
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"timestamp": "2023-03-29T00:00:00",
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The number of numbers lying between 1 and 200 which are divisible by either of 2 , 3 or 5? The number of numbers lying between 1 and 200 which are divisible by either of two , three or five?
| A)numbers divisible by 2: $\frac{200}{2} = 100$
B)numbers divisible by 3: $\frac{200}{3} = 66$
C)numbers divisible by 5: $\frac{200}{5} = 40$
counting twice
AB)numbers divisible by 6: $\frac{200}{6} = 33$
AC)numbers divisible by 10: $\frac{200}{10} = 20$
BC)numbers divisible by 15: $\frac{200}{15} = 13$
counting 3 times
ABC)numbers divisible by 30: $\frac{200}{30} = 6$
Total of numbers = A + B + C - AB - AC - BC + ABC = 100 + 66 + 40 - 33 - 20 - 13 + 6 = 146
| {
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Product of cosines: $ \prod_{r=1}^{7} \cos \left(\frac{r\pi}{15}\right) $
Evaluate
$$ \prod_{r=1}^{7} \cos \left({\dfrac{r\pi}{15}}\right) $$
I tried trigonometric identities of product of cosines, i.e, $$\cos\text{A}\cdot\cos\text{B} = \dfrac{1}{2}[ \cos(A+B)+\cos(A-B)] $$
but I couldn't find the product.
Any help will be appreciated.
Thanks.
| I think it's worth noting the product is also evaluable just remembering, besides the well known $\displaystyle \cos\frac{\pi}{3}=\frac{1}{2},$ the somewhat nice $$\displaystyle\cos\frac{\pi}{5}=\frac{1+\sqrt{5}}{4}=\frac{\phi}{2}$$ (where $\phi$ is the golden section) and iterating the sum/difference formula for the cosine and the product formula you mention. Your product is, once we rearrange factors and simplify fractions, equal to $${\color\red{\cos\frac{\pi}{3}\cdot\cos\frac{\pi}{5}}}\cdot\color\orange{\cos\frac{2\pi}{5}} \cdot\color\navy{\cos\frac{\pi}{15}\cdot\cos\frac{4\pi}{15}}\cdot\color\green{\cos\frac{2\pi}{15}\cdot\cos\frac{7\pi}{15}} \\ ={\color\red{\frac{\phi}{4}}}\color\orange{\left(2\cos^2\frac{\pi}{5}-1\right)}\color\navy{\frac{1}{2}\left(\cos\frac{\pi}{3}+\cos\left(-\frac{\pi}{5}\right)\right)}\color\green{\frac{1}{2}\left(\cos\left(-\frac{\pi}{3}\right)+\cos\frac{3\pi}{5}\right)} \\ = \frac{\phi}{16}\left(\frac{\phi^2}{2}-1\right)\frac{\phi+1}{2}\left(\frac{1}{2}+\cos\frac{\pi}{5}\cdot\left(2\cos^2\frac{\pi}{5}-1\right)-2\cos\frac{\pi}{5}\left(1-\cos^2\frac{\pi}{5}\right)\right)\\=\frac{\phi(\phi^2-1)}{64}\left(\frac{1}{2}+\frac{\phi(\phi^2-2)}{4}-\frac{\phi(4-\phi^2)}{4}\right)\\=\frac{\phi^2}{64}\left(\frac{1}{2}+\frac{\phi(\phi-1)-\phi(3-\phi)}{4}\right)\\=\frac{\phi+1}{64}\left(\frac{1}{2}+\frac{2-2\phi}{4}\right)=\frac{\phi+1}{128}-\frac{\phi^2-1}{128}=\frac{\phi+1}{128}-\frac{\phi}{128}=\frac{1}{128}.$$
| {
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"timestamp": "2023-03-29T00:00:00",
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this inequality $\prod_{cyc} (x^2+x+1)\ge 9\sum_{cyc} xy$ Let $x,y,z\in R$,and $x+y+z=3$
show that:
$$(x^2+x+1)(y^2+y+1)(z^2+z+1)\ge 9(xy+yz+xz)$$
Things I have tried so far:$$9(xy+yz+xz)\le 3(x+y+z)^2=27$$
so it suffices to prove that
$$(x^2+x+1)(y^2+y+1)(z^2+z+1)\ge 27$$
then the problem is solved. I stuck in here
| Assume that $x,y,z$ are roots of the polynomial
$$ q(w) = w^3-3w^2+s w-p \tag{1}$$
where $s=xy+xz+yz$ and $p=xyz$.
In order that $x,y,z\in\mathbb{R}$, the discriminant of $q$ must be non-negative, hence:
$$ 54ps + 9s^2 \geq 4s^3+27p^2+108 p.\tag{2}$$
On the other hand, $2x+1,2y+1,2z+1$ are roots of the monic polynomial $8\cdot q\left(\frac{w-1}{2}\right)$, hence $(2x+1)^2,(2y+1)^2,(2z+1)^2$ are roots of the monic polynomial:
$$ w^3 + (8s-51) w^2 + (99-144 p+48 s+16 s^2) w + (-49 - 112 p - 64 p^2 - 56 s - 64 p s - 16 s^2) $$
and $4(x^2+x+1),4(y^2+y+1),4(z^2+z+1)$ are roots of the previous polynomial evaluated at $w-3$. Viete's theorem hence gives:
$$\prod_{cyc}(x^2+x+1) = 13-5 p+p^2+2 s+p s+s^2\tag{3}$$
and our problem boils down to proving that:
$$ 13+p^2+p s+s^2 \geq 7s+5p.\tag{4} $$
However, that is trivial since the quadratic form $p^2+ps+s^2-7s-5p+13$ is positive definite and achieve its minimum, zero, at $(p,s)=(1,3)$.
| {
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"timestamp": "2023-03-29T00:00:00",
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How find $I= \int_{x=0}^{ \frac{1}{2} } \int_{y=x}^{1-x} ( \frac{x-y}{x+y})^{2}\, dy\,dx$ In $$I= \int_{x=0}^{ \frac{1}{2} } \int_{y=x}^{1-x} \left( \frac{x-y}{x+y}\right)^{2} \,dy\,dx$$
follow the change of variables on $x= \frac{1}{2} (r-s),y= \frac{1}{2} (r+s)$ and find$I$
My try
$J=\begin{vmatrix} x'_{r} & x'_{s}\\ y'_{r} & y'_{s} \end{vmatrix}=\begin{vmatrix} \frac{1}{2} & -\frac{1}{2}\\ \frac{1}{2} & \frac{1}{2} \end{vmatrix}=\frac{1}{2}$
$x-y=\frac{1}{2}r-\frac{1}{2}s-\frac{1}{2}r-\frac{1}{2}s=-s$
$x+y=\frac{1}{2}r-\frac{1}{2}s+\frac{1}{2}r+\frac{1}{2}s=r$
and what next?
| Off to a good start. Draw the picture! The hard part is determining the integration limits. Really, you just need to understand what the lines of constant $r$ and $s$ mean. Basically, they are lines at 45-degree angles to the axes. So, really, imagine that if $r \in [0,1]$, then what are the limits of $s$? Well, in fact, $s$ ranges from $0$ at the origin, to its maximum value at the top of the triangle on the $y$ axis, or $r$. Thus, the integral is equal to
$$\frac12 \int_0^1 dr \, \int_0^r ds \, \left (\frac{s}{r} \right )^2 $$
This is easily doable. I get $1/12$.
| {
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"timestamp": "2023-03-29T00:00:00",
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Why does the equation of a circle have to have the same $x^2,y^2$ coefficients? In one of my geometry texts, it tells me they should be the same but not why. I am unsatisfied with this.
Suppose that:
$$ax^2+by^2 + cx + dy + f = 0 \text{ such that } a \neq b$$
is the equation of some circle. Upon completing the square and rearranging, I obtain
$$a\left(x+\frac{c}{2a}\right)^2 + b\left(y+\frac{d}{2b}\right)^2 = \frac{c^2}{4a} + \frac{d^2}{4b} - f$$
I know that a circle is defined as the set of points a fixed distance from a fixed point.
How can i arrive at a satisfying contradiction? At the moment I just can't see it.
| If $a\ne b$ and both of them are non-zero, then your equation above
$$\begin{align*}
a\left(x+\frac{c}{2a}\right)^2 + b\left(y+\frac{d}{2b}\right)^2 &= \frac{c^2}{4a} + \frac{d^2}{4b} - f\\
\left(x+\frac{c}{2a}\right)^2 + \frac ba\left(y+\frac{d}{2b}\right)^2 &= \frac{c^2}{4a^2} + \frac{d^2}{4ab} - \frac fa\\
\left(x+\frac{c}{2a}\right)^2 + \left(y\sqrt{\frac ba}+\frac{d}{2b}\sqrt\frac ba\right)^2 &= \frac{c^2}{4a^2} + \frac{d^2}{4ab} - \frac fa\\
\end{align*}$$
If $\frac ba > 0$, this means the equation only represents a circle in the scaled $\left(x,y\sqrt{\frac ba}\right)$ plane. In other words, the equation represents an ellipse.
If $\frac ba < 0$, this is not a circle either.
| {
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"timestamp": "2023-03-29T00:00:00",
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Probability that team $A$ has more points than team $B$
Seven teams play a soccer tournament in which each team plays every other team exactly once. No ties occur, each team has a $50\%$ chance of winning each game it plays, and the outcomes of the games are independent. In each game, the winner is awarded a point and the loser gets 0 points. The total points are accumulated to decide the ranks of the teams. In the first game of the tournament, team $A$ beats team $B.$ The probability that team $A$ finishes with more points than team $B$ is $m/n,$ where $m$ and $n$ are relatively prime positive integers. Find $m+n.$
I got that since, Team $B$ already has one loss, it doesnt matter how many games team $B$ wins.
We must find the probability that team $A$ wins the rest $5$ games.
Since it says: "outcome of games is independent, I am confused."
My first approach was:
$$P(\text{A Wins 5}) = \frac{1}{32} \implies m + n = 33$$
This was wrong.
Second approach.
Suppose $A$ has a match with Team $C$.
$$P(\text{Team A wins, Team C loses}) = \frac{1}{2}\frac{1}{2} = \frac{1}{4}$$
But then overall: $\frac{1}{4^5} > 1000$ too big of an answer ($m + n < 1000$ requirement).
HINTS ONLY PLEASE!!
EDIT:
I did some casework and the work is very messy and I don't think I got the right answer anyway. I have to find:
$$\binom{5}{1} = 5, \binom{5}{2} = 10, \binom{5}{3} = 10, \binom{5}{4} = 5, \binom{5}{5} = 1.$$
Let $A = x$
$$P(B=1, x \ge 1) + P(B = 2, x\ge 2) + P(B=3, x\ge 3) + ... + P(B=5, x = 5)$$
$$P(B=1, x \ge 1) = \binom{5}{1}(0.5)^{5} \cdot \bigg(\binom{5}{1} (0.5)^5 + \binom{5}{2} (0.5)^5 + ... + (0.5)^5\binom{5}{1} \bigg) = \frac{5}{1024} \cdot \bigg(31\bigg) = \frac{155}{1024} $$
$$P_2 = \frac{10}{1024} \bigg(\binom{5}{2} + ... + \binom{5}{5}\bigg) = \frac{260}{1024}$$
$$P_3 = \frac{10}{1024} \bigg(\binom{5}{3} + ... + \binom{5}{5} \bigg) = \frac{160}{1024}$$
$$P_4 = \frac{5}{1024} \bigg( \binom{5}{4} + \binom{5}{5}\bigg) = \frac{60}{1024}$$
$$P_5 = \frac{1}{1024} \bigg( 1\bigg) = \frac{1}{1024}$$
$$P(\text{Total}) = \frac{636}{1024} = \frac{318}{512} = \text{wrong}$$
What is wrong with this method?
| Hints:
*
*Team A has $5$ more games to play, and so does team B.
*The outcome of games is independent. Thus, for team A or B, the probability that the team wins exactly $k$ more games (after their game with each other) has binomial distribution, namely
$$P(\text{exactly $k$ more games won})={5 \choose k}\left(\frac 12\right)^5$$
*There are really four main ways that B can finish with more points than A: (a) A wins $0$ more games and B wins $2$ through $5$ more games; (b) A wins $1$ more game and B wins $3$ through $5$ more games; (c) A wins $2$ more games and B wins $4$ through $5$ more games; (d) A wins $3$ more games and B wins $5$ more games.
*The probability of (a) happening is
$$P(\text{A wins $0$ more games}) \times [P(\text{B wins $2$ more games}) + P(\text{B wins $3$ more games}) + P(\text{B wins $4$ more games}) + P(\text{B wins $5$ more games})]$$
*Possibilities (a) through (d) are mutually exclusive, so you can just add their probabilities.
I'm sure you get the idea. The final calculation is made easier by the fact that $\left(\frac 12\right)^5$ factors out nicely. Calculating everything else has $6$ additions, $4$ multiplications, $3$ more additions, and $1$ more multiplication. This may be more calculating than you would like, but it does get the job done in a reasonable amount of time. Some calculators, like the TI-84 and TI-Nspire, can add ranges of binomial probabilities together all at once, which makes the calculations easier.
Of course, when you find the final probability, finding $m+n$ is easy.
| {
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Differentiate the Function: $y=\frac{ae^x+b}{ce^x+d}$ $y=\frac{ae^x+b}{ce^x+d}$
$\frac{(ce^x+d)\cdot [ae^x+b]'-[(ae^x+b)\cdot[ce^x+d]'}{(ce^x+d)^2}$
numerator only shown (') indicates find the derivative
$(ce^x+d)\cdot(a[e^x]'+(e^x)[a]')+1)-[(ae^x+b)\cdot (c[e^x]'+e^x[c]')+1]$
$(ce^x+d)\cdot(ae^x+(e^x))+1)-[(ae^x+b)\cdot (ce^x+e^x)+1]$
$\frac{(ace^x)^2+de^x+dae^x+de^x+1-[(ace^x)^2(ae^x)^2(bce^x)(be^x)+1]}{(ce^x+d)^2}$
Am I doing this problem correctly? I got to this point and was confused as to how to simplify. Can someone help me?
| Another method would be to use logarithmic differentiation. Your function $$y=\frac{ae^x+b}{ce^x+d} $$ is identical to $$\ln(y)=\ln\bigg(\frac{ae^x+b}{ce^x+d}\bigg)=\ln(ae^x+b)-\ln(ce^x+d). $$ We find the differentials for both variables $x$ and $y$. $$\frac{1}{y}dy=\bigg(\frac{ae^x}{ae^x+b}-\frac{ce^x}{ce^x+d}\bigg)dx.$$ We solve for $\frac{dy}{dx}$ and find $$\frac{dy}{dx}=y\bigg(\frac{ae^x}{ae^x+b}-\frac{ce^x}{ce^x+d}\bigg). $$ Substituting the initial equation for $y$ yields a solution of $$\frac{dy}{dx}=\frac{ae^x+b}{ce^x+d}\bigg(\frac{ae^x}{ae^x+b}-\frac{ce^x}{ce^x+d}\bigg). $$ We simplify then distribute through $$\begin{align}\frac{dy}{dx}&=\frac{ae^x+b}{ce^x+d}\bigg(\frac{ae^x(ce^x+d)-ce^x(ae^x+b)}{(ae^x+b)(ce^x+d)}\bigg)\\ &= \frac{e^x[ace^x+ad-(ace^x+bc)]}{(ce^x+d)^2}\\ &=\frac{e^x(ad-bc)}{(ce^x+d)^2}.\end{align}$$ This is your answer.
| {
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Find the roots of this 6th degree polynomial Hey guys I'm reviewing for a test and I'm getting stuck on one part, I can't remember what to do next.
$x^6+16x^3+64$
$(x^3)^2+16x^3+64$
let $x^3=w$
$w^2+16w+64$
$(w+8)^2$ now substitute again
$(x^3+8)^2$
Now what do I do?
| Then you have to solve $x^3+8=0$:
\begin{align*}
x^3+8=0\Longleftrightarrow
x^3=-8
\end{align*}
But
$$
-8=8(-1)=8e^{i\pi}
$$
Thus
$$
x=2e^{ki\pi/3}\;\;\;\;k=0,1,2.
$$
Hence
$$
x^6+16x^3+64=(x^3+8)^3=(x-2)^3(x-2e^{i\pi/3})^3(x-2e^{2i\pi/3})^3
$$
| {
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Given that $2\cos(x + 50) = \sin(x + 40)$ show that $\tan x = \frac{1}{3}\tan 40$ Given that:
$$
2\cos(x + 50) = \sin(x + 40)
$$
Show, without using a calculator, that:
$$
\tan x = \frac{1}{3}\tan 40
$$
I've got the majority of it:
$$
2\cos x\cos50-2\sin x\sin50=\sin x\cos40+\cos x\sin40\\
$$
$$
\frac{2\cos50 - \sin40}{2\sin50 + \cos40}=\tan x
$$
But then, checking the notes, it says to use $\cos50 = \sin40$ and $\cos40 =\sin50$; which I don't understand. Could somebody explain this final step?
| Using the well-known identity $$\cos (90^{\circ} - x) = \sin x$$ since the cosine function is just a $90^{\circ}$ horizontal translation of the sine function.
Taking $x = 40^{\circ}$ gives us $\cos 50^{\circ} = \sin 40^{\circ}$ and letting $x=50^{\circ}$ establishes the second result. Then $$\begin{align}2\cos x\cos50-2\sin x\sin50=\sin x\cos40+\cos x\sin40 \\ \iff 2 \cos x \sin 40 - 2\sin x\cos 40 = \sin x \cos 40 +\cos x \sin 40 \\ \iff 2 \cos x\sin 40 - \cos x\sin 40 = \sin x\cos 40 + 2\sin x\cos 40 \\ \iff \tan 40 \cos x = 3 \sin x \end{align}$$
so that $$\tan x = \frac{\tan 40^{\circ}}{3}$$
| {
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How to prove $\cos{\frac{\pi}{11}}$ is a root of I want to show that $x=\cos{\frac{\pi}{11}}$ is a solution of equation :
$$8x^2-4x+\frac{1}{x}-4=4\sqrt{\frac{1-x}{2}}$$
Thanks in advance.
| We have to prove that for $x=\cos\frac{\pi}{11}$ we have:
$$ 4x(2x^2-1)-(4x^2-1)=4x\sqrt{\frac{1-x}{2}}\tag{1}$$
or:
$$ 4\cos\frac{\pi}{11}\cos\frac{2\pi}{11}-\frac{\sin\frac{3\pi}{11}}{\sin\frac{\pi}{11}}=4\cos\frac{\pi}{11}\sin\frac{\pi}{22}\tag{2}$$
that is equivalent to:
$$ \sin\frac{4\pi}{11}-\sin\frac{3\pi}{11}=2\sin\frac{2\pi}{11}\sin\frac{\pi}{22}\tag{3}$$
or to:
$$ 2\cos\frac{7\pi}{22}\sin\frac{\pi}{22}=2\sin\frac{2\pi}{11}\sin\frac{\pi}{22}\tag{4}$$
that is trivial since $\cos\frac{7\pi}{22}=\sin\frac{4\pi}{22}=\sin\frac{2\pi}{11}.$
| {
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Finding all groups of order $7$ up to isomorphism? I'm learning group theory but I didn't learn any concepts of building groups.
I know that there exists the identity group $\{e\}$, and the group with 2 elements: $\{e,a\}$.
If I try to create a group with 3 elements, let's say: $\{e,a,b\}$ then we would have:
$ea = a, eb = b, aa = ?$, and what about $ab$?
Am I supposed to try this for the $7$ elements?
What does the statement ``up to isomorphism" mean? In particular, how many groups of order $7$ are there up to isomorphism?
I'm really confused, and my book says nothing about it.
| Groups of order $1$, $2$, $3$, $5$ and $7$ are unique up to isomorphism: for $1$ it's obvious, the others are prime numbers.
Let's tackle the $4$ case; one of the elements is the identity, so the group is $\{1,a,b,c\}$. Every group of even order has an element of order $2$ (a very easy case of Cauchy's theorem). Let's say it's $a$, so $a^2=1$.
Make the Cayley table for the group, where we fill in the information we already have.
$$
\begin{array}{c|cccc}
& 1 & a & b & c \\
\hline
1 & 1 & a & b & c \\
a & a & 1 \\
b & b\\
c & c\\
\end{array}
$$
For the product $ab$ we have only one choice, namely $c$, because an element cannot appear twice in a row or column. Hence $ac=b$. Similarly, $ba=c$ and $ca=b$. So we have
$$
\begin{array}{c|cccc}
& 1 & a & b & c \\
\hline
1 & 1 & a & b & c \\
a & a & 1 & c & b \\
b & b & c \\
c & c & b \\
\end{array}
$$
Now we have two choices for $c^2$: either $c^2=1$ or $c^2=a$. Let's look at what we have, because once we decide for $c^2$ the other slots can be filled in uniquely:
$$
\begin{array}{c|cccc}
& 1 & a & b & c \\
\hline
1 & 1 & a & b & c \\
a & a & 1 & c & b \\
b & b & c & 1 & a \\
c & c & b & a & 1
\end{array}
\qquad\qquad
\begin{array}{c|cccc}
& 1 & a & b & c \\
\hline
1 & 1 & a & b & c \\
a & a & 1 & c & b \\
b & b & c & a & 1 \\
c & c & b & 1 & a
\end{array}
$$
The structure on the left is Klein's group, the one on the right is the cyclic group. So we have two groups of order $4$.
For the groups with six elements we need something more advanced, because the Cayley table method is too long if started from nothing. Another application of Cauchy's theorem says that the group has a cyclic subgroup of order $3$. So there are three elements $1,a,b$ such that $a^2=b$, $ab=ba=1$ and $b^2=a$. Let's fill in this information for the set $\{1,a,b,c,d,e\}$. Again we can also assume $c^2=1$.
$$
\begin{array}{c|cccccc}
& 1 & a & b & c & d & e \\
\hline
1 & 1 & a & b & c & d & e \\
a & a & b & 1 & & & \\
b & b & 1 & a & & & \\
c & c & & & 1 & & \\
d & d \\
e & e
\end{array}
$$
For $ac$ we can choose either $ac=d$ or $ac=e$; it doesn't matter which one, because at the end exchanging $d$ with $e$ will give an isomorphism; so let's choose $ac=d$ that forces $ad=e$ and $ae=c$
$$
\begin{array}{c|cccccc}
& 1 & a & b & c & d & e \\
\hline
1 & 1 & a & b & c & d & e \\
a & a & b & 1 & d & e & c \\
b & b & 1 & a & & & \\
c & c & & & 1 & & \\
d & d \\
e & e
\end{array}
$$
Let's now examine $ca$: again we have either $ca=d$ or $ca=e$, but this time we can't choose one and we need to go with two alternatives:
$$
\begin{array}{c|cccccc}
& 1 & a & b & c & d & e \\
\hline
1 & 1 & a & b & c & d & e \\
a & a & b & 1 & d & e & c \\
b & b & 1 & a & & & \\
c & c & d & & 1 & & \\
d & d & e \\
e & e & c
\end{array}
\qquad\qquad
\begin{array}{c|cccccc}
& 1 & a & b & c & d & e \\
\hline
1 & 1 & a & b & c & d & e \\
a & a & b & 1 & d & e & c \\
b & b & 1 & a & & & \\
c & c & e & & 1 & & \\
d & d & c \\
e & e & d
\end{array}
$$
We now have no choice for $bc$: in both cases it must be $e$; similarly, $bd=c$ and also $cb$ has a determined result (and we can also fill in some other slots):
$$
\begin{array}{c|cccccc}
& 1 & a & b & c & d & e \\
\hline
1 & 1 & a & b & c & d & e \\
a & a & b & 1 & d & e & c \\
b & b & 1 & a & e & c & d \\
c & c & d & e & 1 & & \\
d & d & e & c & & & \\
e & e & c & d & & &
\end{array}
\qquad\qquad
\begin{array}{c|cccccc}
& 1 & a & b & c & d & e \\
\hline
1 & 1 & a & b & c & d & e \\
a & a & b & 1 & d & e & c \\
b & b & 1 & a & e & c & d \\
c & c & e & d & 1 & & \\
d & d & c & e & & & \\
e & e & d & c
\end{array}
$$
We're almost done. Observe that interchanging $a$ and $b$ is irrelevant, because both are generators of the (normal) subgroup with three elements. So we can arbitrarily choose $cd=a$ in both cases and we can complete the tables. Indeed, if in the table on the left we choose $dc=b$, we'd have $a(dc)=ab=1$, but $(ad)c=ec=a$ against associativity. Similarly for the table on the right.
$$
\begin{array}{c|cccccc}
& 1 & a & b & c & d & e \\
\hline
1 & 1 & a & b & c & d & e \\
a & a & b & 1 & d & e & c \\
b & b & 1 & a & e & c & d \\
c & c & d & e & 1 & a & b \\
d & d & e & c & a & b & 1 \\
e & e & c & d & b & 1 & a
\end{array}
\qquad\qquad
\begin{array}{c|cccccc}
& 1 & a & b & c & d & e \\
\hline
1 & 1 & a & b & c & d & e \\
a & a & b & 1 & d & e & c \\
b & b & 1 & a & e & c & d \\
c & c & e & d & 1 & a & b \\
d & d & c & e & a & b & 1 \\
e & e & d & c & b & 1 & a
\end{array}
$$
So we have a maximum of two groups of order six. But we know two non isomorphic one: the cyclic group and $S_3$. So we're done.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1361137",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 4,
"answer_id": 2
} |
is there a general formula for cases like $\sqrt{2}$ = $\frac {2}{\sqrt{2}}$? I just noticed that $\sqrt{2}$ is equal to $\frac {2}{\sqrt{2}}$:
$\sqrt{2} = 1.414213562$
$\frac {2}{\sqrt{2}} = 1.414213562$
It is confirmed by a hand-calculator.
I tried to proof this as follows:
$\sqrt{2}$ = $\frac {2}{\sqrt{2}}$
$2^{\frac 12} = \frac {2}{2^{\frac 12}}$
$2 = 2^{\frac 12} \cdot 2^{\frac 12}$
$2 = 2^{\frac 12 + \frac 12} $
$ 2 = 2^1 $
$ 2 = 2$
It is also true for:
$-\frac {2}{\sqrt5} = \frac {-2}{5}\sqrt 5$
I didn't know this relationship beforehand and it was new for me, my question is: Is there a general rule for this?
| It will be true for any positive real $x$ that
$$\frac{x}{\sqrt{x}} = \frac{x}{x^{1/2}} = x^1\cdot x^{-1/2} = x^{1-1/2} = x^{1/2} = \sqrt{x}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1362633",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 5,
"answer_id": 4
} |
How many $5$ element sets can be made?
Let $m$ be the number of five-element subsets that can be chosen from the set of the first $14$ natural numbers so that at least two of the five numbers are consecutive. Find the remainder when $m$ is divided by $1000$.
I think the opposite would be to find, the number of sets with no consecutive numbers.
There are: $\binom{14}{5} = 2002$ total sets.
$M = \{a, b, c, d, e\}$
Now I am confused, how do I find sets with no consecutive numbers?
| We wish to calculate the number of subsets of five elements of the set $\{1, 2, 3, \ldots, 14\}$ in which no two of the numbers are consecutive.
Suppose the five non-consecutive numbers selected from the set $\{1, 2, 3, \ldots, 14\}$ are $x_1, x_2, x_3, x_4, x_5$, where $x_1 < x_2 < x_3 < x_4 < x_5$. Let
\begin{align*}
y_1 & = x_1\\
y_2 & = x_2 - x_1\\
y_3 & = x_3 - x_2\\
y_4 & = x_4 - x_3\\
y_5 & = x_5 - x_4\\
y_6 & = 14 - x_5
\end{align*}
Since $y_1$ is the smallest number in the subset, $y_1 \geq 1$. Since no two elements of the subset $\{x_1, x_2, x_3, x_4, x_5\}$ are consecutive, $y_2, y_3, y_4, y_5 \geq 2$. The number $y_6$ represents the difference between $14$ and the largest number in the subset, so $y_6 \geq 0$. Moreover,
$$y_1 + y_2 + y_3 + y_4 + y_5 + y_6 = 14 \tag{1}$$
Let
\begin{align*}
z_1 & = y_1 - 1\\
z_2 & = y_2 - 2\\
z_3 & = y_3 - 2\\
z_4 & = y_4 - 2\\
z_5 & = y_5 - 2\\
z_6 & = y_6
\end{align*}
Substitution into equation 1 yields
\begin{align*}
z_1 + 1 + z_2 + 2 + z_3 + 2 + z_4 + 2 + z_5 + 2 + z_6 & = 14\\
z_1 + z_2 + z_3 + z_4 + z_5 + z_6 & = 5 \tag{2}
\end{align*}
which is an equation in the nonnegative integers. The number of solutions of equation 2 is the number of ways five addition signs can be inserted into a row of five ones, which is
$$\binom{5 + 5}{5} = \binom{10}{5}$$
Hence, the number of subsets in which no two of the numbers are consecutive is $\binom{10}{5}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1363424",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 4
} |
How find $\lim\limits_{n \to \infty } \sqrt{n} \cdot \sqrt[n]{\left( \lim\limits_{n \to \infty } a_n\right)-a_n}$? Let $a_n= \sqrt{1+ \sqrt{2+\cdots+ \sqrt{n} } }$
How find
$\lim\limits_{n \to \infty } \sqrt{n} \cdot \sqrt[n]{\left( \lim\limits_{n \to \infty } a_n\right)-a_n}$ ?
| Call the limit $\lim\limits_{n\to \infty} a_n = a$.
Let us denote: $$a_{k,n} = \sqrt{k+\sqrt{k+1+\sqrt{\cdots + \sqrt{n}}}}$$
Since, $\displaystyle (a_{k,n+1} - a_{k,n}) = \frac{a_{k+1,n+1} - a_{k+1,n}}{a_{k,n+1}+a_{k,n}}$ we have: $
\displaystyle (a_{n+1} - a_n) = \frac{\sqrt{n+1}}{\prod\limits_{k=1}^{n}(a_{k,n+1}+a_{k,n})}$
Now, $\displaystyle a_{k,n} = \sqrt{k+a_{k+1,n}}$, where, $1 \le k \le n-1$.
Note that we have $\sqrt{k} \le a_{k,n} \le \sqrt{k} + 1$, so that plugging $a_{k,n} = \sqrt{k}+\mathcal{O}(1)$ in the recursive relation we get
$$\begin{align}a_{k,n} = \sqrt{k}\left(1+\dfrac{a_{k+1,n}}{k}\right)^{1/2} &= \sqrt{k}\left(1+\frac{\sqrt{k+1}+\mathcal{O}(1)}{k}\right)^{1/2}\\ &= \sqrt{k} + \frac{1}{2} + \mathcal{O}\left(\frac{1}{\sqrt{k}}\right)\end{align}$$
Again, $$\begin{align}a_{k,n} = \sqrt{k}\left(1+\frac{a_{k+1,n}}{k}\right)^{1/2} &= \sqrt{k}\exp \left\{\frac{1}{2}\log \left(1 + \frac{a_{k+1,n}}{k}\right)\right\} \\ &= \sqrt{k}\exp \left\{\frac{1}{2} \left(\frac{a_{k+1,n}}{k} - \frac{a_{k+1,n}^2}{2k^2} + \mathcal{O}\left(\frac{a_{k+1,n}^3}{k^3}\right)\right)\right\} \\ &= \sqrt{k}\exp\left\{\frac{1}{2\sqrt{k}} + \mathcal{O}\left(\frac{1}{k^{3/2}}\right)\right\}\end{align}$$
Hence, $$\begin{align}a_{n+1} - a_{n} &= \frac{(1+\mathcal{o}(1))\sqrt{n+1}}{2^{n}(n!)^{1/2}}\exp\left\{-\sum\limits_{k=1}^{n}\left(\dfrac{1}{2\sqrt{k}} + \mathcal{O}\left(\dfrac{1}{k^{3/2}}\right)\right)\right\}\\&= \frac{(1+\mathcal{o}(1))\sqrt{n+1}}{2^{n}(n!)^{1/2}}\exp\left\{-\sqrt{n} + \mathcal{O}(1)\right\}\end{align}$$
Therefore, $$a - a_N = \sum\limits_{n\ge N} (a_{n+1} - a_n) = e^{\mathcal{O}(1)}\sum\limits_{n \ge N}\frac{\sqrt{n+1}}{2^n(n!)^{1/2}}e^{-\sqrt{n}}$$
Now, by Stirling's Approximation we have:
$$\left(\frac{\sqrt{n+1}}{2^n(n!)^{1/2}}e^{-\sqrt{n}}\right)^{1/n} \sim \frac{\sqrt{e}}{2}e^{-\frac{1}{2}\log n}$$
Hence, $\displaystyle \lim\limits_{n \to \infty}\sqrt{n}\cdot \sqrt[n]{a-a_n} = \lim\limits_{n \to \infty} \sqrt{n} \cdot \sqrt[n]{a_{n+1}-a_n} = \frac{\sqrt{e}}{2}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1363520",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "12",
"answer_count": 1,
"answer_id": 0
} |
Find lim$_{x \to 0}\left(\frac{1}{x} - \frac{\cos x}{\sin x}\right).$ $$\lim_{x \to 0} \left(\frac{1}{x} - \frac{\cos x}{\sin x}\right) = \frac{\sin x - x\cos x}{x\sin x}= \frac{0}{0}.$$
L'Hopital's: $$\lim_{x \to 0} \frac{f'(x)}{g'(x)} = -\frac{1}{x^2} + \frac{1}{\sin ^2x} = \frac{0}{0}.$$
Once again, using L'Hopital's: $$\lim_{x \to 0} \frac{f''(x)}{g''(x)} = \frac{2}{x^3}- \frac{2\cos x}{\sin ^3x} = \frac{0}{0}\,\ldots$$ The terms are getting endless here. Any help? Thanks.
| You have
$$
-\frac{1}{x^2} + \frac{1}{\sin ^2x}
$$
You can't apply L'Hopital's rules to either of the two fractions above since the numerators do not approach $0$ or $\pm\infty$. But you can first add the fractions and then use L'Hopital's rule:
$$
-\frac{1}{x^2} + \frac{1}{\sin ^2x} = \frac{-\sin^2 x + x^2}{x^2\sin^2 x}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1363719",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 7,
"answer_id": 4
} |
Bounds for $\frac{x-y}{x+y}$ How can I find upper and lower bounds for $\displaystyle\frac{x-y}{x+y}$? So I do see that
$$\frac{x-y}{x+y} = \frac{1}{x+y}\cdot(x-y) = \frac{x}{x+y} - \frac{y}{x+y} > \frac{1}{x+y} - \frac{1}{x+y} = \frac{0}{x+y} = 0$$
(is it correct?) but I don't get how to find the upper bound.
| Look at pairs $(x,y)$ such that $x = 1-y$. For these pairs, $\frac{x-y}{x+y} = 1-2y$ and we have $\mbox{lim}_{y\rightarrow \pm\infty}(1-2y) = \mp \infty$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1363835",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
Number of combinations of selecting $r$ numbers from first $n$ natural numbers of which exactly $m$ are consective. Number of combinations of selecting $r$ numbers from first $n$ natural numbers of which exactly $m$ are consective.
Say $g(n,r,m)$ is the number of such combinations. The two cases of $m=r$ and $m=1$ are easy:
$g(n,r,r) = n-r+1 $
$g(n,r,1) = \binom{n-r+1}{r}$
Example: $n=5, r=3, m=1$
Here we have to select three numbers from set $\{1,2,3,4,5\}$
So $g(5,3,3) = 5-2+1 = 3$, The subsets being $\{\{1,2,3\},\{2,3,4\},\{3,4,5\}\}$
So $g(5,3,2) = 6$, With subsets $\{\{1,2,4\},\{1,3,4\},\{1,2,5\},\{2,3,5\},\{1,4,5\},\{2,4,5\}\}$
$g(5,3,1) = \binom{5-3+1}{3}=\binom{3}{3}=1$, The subsets $\{1,3,5\}$
I am looking for a general formula to enumerate for values of $m$ other then $m=1$ and $m=r$.
By trial I found that for $n$ even and $r=n/2$ the enumerations are given by A203717 in $n/2^{nd}$ row and the value is $n/2+1$ times the $m^{th}$entry.
| You can do this using generating functions. If we let $x^k$ represent $k$ consecutive numbers and one gap, the number of ways to make them fill $n+1$ slots (plus $1$ for the extra gap) is
$$
[x^r]\left(1+x+\cdots+x^m\right)^{n-r+1}=[x^r]\left(\frac{1-x^{m+1}}{1-x}\right)^{n-r+1}\;,
$$
where $[x^r]$ denotes extracting the coefficient of $x^r$. Since this allows up to $m$ consecutive numbers and we want exactly $m$ consecutive numbers, we need to subtract the number of selections with up to $m-1$ consecutive numbers, so the result is
$$
[x^r]\left(\left(\frac{1-x^{m+1}}{1-x}\right)^{n-r+1}-\;\left(\frac{1-x^m}{1-x}\right)^{n-r+1}\right)=[x^r]\frac{\left(1-x^{m+1}\right)^{n-r+1}-\left(1-x^m\right)^{n-r+1}}{(1-x)^{n-r+1}}\;.
$$
This contains your two results as special cases.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1364168",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Prove that $\cos^2(\theta) + \cos^2(\theta +120^{\circ}) + \cos^2(\theta-120^{\circ})=3/2$
Prove that $$\cos^2(\theta) + \cos^2(\theta +120^{\circ}) + \cos^2(\theta-120^{\circ})=\frac{3}{2}$$
I thought of rewriting $$\cos^2(\theta +120^{\circ}) + \cos^2(\theta-120^{\circ})$$ as $$\cos^2(90^{\circ}+ (\theta +30^{\circ})) + \cos^2(90^{\circ}-(\theta-30^{\circ}))$$ However I don't seem to get anywhere with this.
Unfortunately I don't know how to solve this question. I would be really grateful for any help or suggestions. Many thanks in advance!
| Notice,
$$\color{blue}{\sin (A+B)\sin(A-B)=\sin^2A-\sin^2 B}$$
$$\color{blue}{\sin (A+B)+\sin(A-B)=2\sin A\cos B}$$
Now, I will follow OP's steps
$$LHS=\cos^2 \theta+\cos^2(\theta +120^{\circ}) + \cos^2(\theta-120^{\circ})$$
$$=\cos^2 \theta+\cos^2(90^{\circ}+ (\theta +30^{\circ})) + \cos^2(90^{\circ}-(\theta-30^{\circ}))$$ $$=\cos^2 \theta+\sin^2(\theta +30^{\circ}) + \sin^2(\theta-30^{\circ})$$ $$=\cos^2 \theta+(\sin(\theta+30^\circ)+\sin(\theta-30^\circ))^2 -2\sin(\theta+30^\circ)\sin(\theta-30^\circ)$$ $$=\cos^2 \theta+(2\sin\theta\cos 30^\circ)^2 -2(\sin^2\theta-\sin^2 30^\circ)$$ $$=\cos^2 \theta+\left(2\sin\theta\frac{\sqrt{3}}{2}\right)^2 -2\left(\sin^2\theta-\frac{1}{4}\right)$$ $$=\cos^2 \theta+3\sin^2\theta -2\sin^2\theta+\frac{1}{2}$$
$$=\cos^2 \theta+\sin^2\theta+\frac{1}{2}$$
$$=1+\frac{1}{2}$$$$=\frac{3}{2}=RHS$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1364788",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 8,
"answer_id": 6
} |
Value of an expression with cube root radical What is the value of the following expression?
$$\sqrt[3]{\ 17\sqrt{5}+38} - \sqrt[3]{17\sqrt{5}-38}$$
| The expression strangely reminds of Cardano's formula for the cubic equation
$$x=\sqrt[3]{-\frac q2-\sqrt{\frac{q^2}4+\frac{p^3}{27}}}+\sqrt[3]{-\frac q2+\sqrt{\frac{q^2}4+\frac{p^3}{27}}}.$$
Identifying, we have
$$q=-76,p=3,$$
and $x$ is a root of $$x^3+3x-76=0.$$
By inspection (trying values close to $\sqrt[3]{76}$), $\color{green}4$ is a solution.
As $$x^3+3x-76=(x-4)(x^2+4x+19),$$ the other roots are complex.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1365489",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 7,
"answer_id": 2
} |
Maximize the area of a triangle inscribed in a semicircle. http://i.imgur.com/Q5gjaSG.png
Consider the semicircle with radius 1, the diameter is AB. Let C be a point on the semicircle and D the projection of C onto AB. Maximize the area of the triangle BDC.
My attempt so far, I'm new at these problems and have only done a few so far. Thanks for any hints. I feel I'm not setting up the correct equation.
$x^{2}$+$y^2$=$r^2$
$y=\sqrt{r^2 - x^2}=\sqrt{1-x^2}\;(\text{since}\:r=1).$
The area is A=$\dfrac{1}{2}$($x+1$)($y$)
A'=$\frac{2x^2-x-1}{2\sqrt{1-x^2}}$
A'=$0$ When $x$ = - $\dfrac{1}{2}$
$y$=$\dfrac{\sqrt{3}}{2}$
Plugging everything back into the original area equation I get $\dfrac{ \sqrt{3}}{8}$
| Another approach is to use $\angle CBD=\phi\in(0,\frac\pi2)$:
\begin{align}
|BC|&=|AB|\cos\phi=2\cos\phi
\\
|BD|&=|BC|\cos\phi=2\cos^2\phi
\\
S_{\triangle BCD}&=
\tfrac12|BD|\cdot|BC|\sin\phi
\\
&=
2\sin\phi\cos^3\phi
\\
S'(\phi)
&=
2\cos^2\phi(\cos^2\phi-3\sin^2\phi)
\\
&=
2\cos^2\phi(1-4\sin^2\phi)
\end{align}
The only suitable solution to $S'(\phi)=0$
is $\phi=\arcsin\tfrac12=\tfrac\pi6$,
hence
\begin{align}
S_{\max}
&=
2\cdot\tfrac12\cdot\left(\tfrac{\sqrt3}{2}\right)^3
=\tfrac{3\sqrt3}{8}.
\end{align}
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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} |
How to solve $\sin78^\circ-\sin66^\circ-\sin42^\circ+\sin6^\circ$ Question:
$ \sin78^\circ-\sin66^\circ-\sin42^\circ+\sin6° $
I have partially solved this:-
$$ \sin78^\circ-\sin42^\circ +\sin6^\circ-\sin66^\circ $$
$$ 2\cos\left(\frac{78^\circ+42^\circ}{2}\right) \sin\left(\frac{78^\circ-42^\circ}{2}\right) + 2\cos\left(\frac{6^\circ+66^\circ}{2}\right)\sin\left(\frac{6^\circ-66^\circ}{2}\right) $$
$$ 2\cos(60^\circ)\sin(18^\circ) + 2\cos(36^\circ)\sin(-30^\circ) $$
$$ 2\frac{1}{2}\sin(18^\circ) - 2\cos(36^\circ)\cdot\frac{1}{2} $$
$$ \sin(18^\circ) - \cos(36^\circ) $$
At this point I had to use a calculator. Does anyone know a way to solve it without a calculator.Thanks in advance.
| You have done your question already. Note that $sin18^\circ=\frac{\sqrt{5}-1}{4}$ and $cos36^\circ=\frac{\sqrt{5}+1}{4}$.If you had done mulitiple angles in trigonometry you can easily calculate these.
For $sin18^\circ$ let, $A=18^\circ$. Then $5A=90^\circ$
$2A=90^\circ-3A$
$Sin2A=cos3A$. Use $Sin2A=2sinAcosA$ and $Cos3A=4cos^3A-3CosA.$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1366530",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 5,
"answer_id": 2
} |
Triple Integral with difficult limits. I don't understand how to count integral bounds from the inequality.
$$\iiint\limits_{V}(x-2)dV $$ where
$$V=\left\{(x, y, z):\frac{(x-2)^2}{9}+\frac{(y+3)^2}{25}+\frac{(z+1)^2}{16} < 1\right\}$$
| As @JohnMa hinted, we infer from the symmetry about $x-2=0$ that the volume integral is zero. If one wishes to forgo using the symmetry argument, then we can proceed to evaluate the volume integral directly.
METHOD 1:
We use Cartesian coordinates and take as the inner integral, the integration over $x$ and write
$$\begin{align}
\int_{2-3\sqrt{1-\left(\frac{y+3}{5}\right)^2-\left(\frac{z+1}{4}\right)^2}}^{2-3\sqrt{1-\left(\frac{y+3}{5}\right)^2+\left(\frac{z+1}{4}\right)^2}} (x-2)\,dx&=\int_{-\sqrt{1-\left(\frac{y+3}{5}\right)^2-\left(\frac{z+1}{4}\right)^2}}^{\sqrt{1-\left(\frac{y+3}{5}\right)^2-\left(\frac{z+1}{4}\right)^2}}9x\,dx\\\\
&=\frac92 \left.x^2\right|_{-\sqrt{1-\left(\frac{y+3}{5}\right)^2-\left(\frac{z+1}{4}\right)^2}}^{\sqrt{1-\left(\frac{y+3}{5}\right)^2-\left(\frac{z+1}{4}\right)^2}}\\\\
&=\frac92\left(1-\left(\frac{y+3}{5}\right)^2-\left(\frac{z+1}{4}\right)^2\right)\\\\
&-\frac92\left(1-\left(\frac{y+3}{5}\right)^2-\left(\frac{z+1}{4}\right)^2\right)\\\\
&=0
\end{align}$$
Thus, the volume integral is zero as expected!
METHOD 2:
We first change variables with
$$\frac{x-2}{3}=u$$
$$\frac{y+3}{5}=v$$
$$\frac{z+1}{4}=w$$
The Jacobian for the transformation is trivially $60$. Thus,
$$\int_V (x-2) dx\,dy\,dz=\int_{V'}(3u)\,60du\,dv\,dw$$
where $V'$ is the spherical region defined by $u^2+v^2+w^2\le1$. Now, we change coordinates again, this time using a spherical coordinate system with
$$\begin{align}
u&=R\sin t\,\cos s\\\\
v&=R\sin t\,\sin s\\\\
w&=R\cos t
\end{align}$$
The Jacobian here is $R^2\,\sin t$ and thus
$$\begin{align}
\int_{V'}(3u)\,60du\,dv\,dw&=180\int_{0}^{2\pi}\int_{0}^{\pi}\int_{0}^{1}R \sin t\cos s R^2 \sin t dR\, dt\,ds\\\\
&=\frac{45\pi}{2}\int_{0}^{2\pi}\cos s\,ds\\\\
&=0
\end{align}$$
as expected again!!
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1366717",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Contradiction between integration by partial fractions and substitution Integration by substitution:
$$\int \frac {dx}{x^2-1}$$
Let $x=\sec\theta$ and $dx=\sec\theta\tan\theta \,d\theta$
$$\int \frac {dx}{x^2-1} = \int \frac{\sec\theta\tan\theta \,d\theta}{\sec^2\theta-1} = \int \frac {\sec\theta\tan\theta\, d\theta}{\tan^2\theta} = \int \frac {\sec\theta \,d\theta}{\tan\theta} $$
$$\int \frac {\sec\theta \,d\theta}{\tan\theta} = \int \frac {\cos \theta \,d\theta}{\cos\theta \sin \theta} = \int \csc\theta\, d\theta$$
$$=\ln|\csc\theta-\cot\theta|+C$$
$$=\ln| \frac{x}{\sqrt{x^2-1}}-\frac{1}{\sqrt{x^2-1}}|+C=\ln| \frac{x-1}{\sqrt{x^2-1}}|+C$$
Which is $Undefined$ for $|x|<1$
Integration by partial fractions:
$$\int \frac {dx}{x^2-1}$$
$$\int \frac {dx}{x^2-1}= \frac 12\int\frac{dx}{x-1}- \frac12\int \frac{dx}{x+1} = \frac12 \ln | x-1| - \frac12 \ln|x+1| +C$$
$$ = \frac12 \ln | \frac{x-1}{x+1}|+C$$
Which is $Defined$ for $|x|<1$ and this is right because the integrand is defined for $|x|<1$
What is the problem in the substitution method ?
| $$
\ln\frac{|x-1|}{\sqrt{|x^2-1|}} = \ln\frac{\sqrt{|x-1|}\sqrt{|x-1|}}{\sqrt{|x-1|}\sqrt{|x+1|}} = \ln\sqrt{\frac{|x-1|}{|x+1|}} = \frac 1 2 \ln\left|\frac{x-1}{x+1}\right|
$$
In response to comments I've made this more complete than it was. Notice that
*
*I take no square roots of anything except non-negative numbers; and
*$|AB| = |A||B|$, so $|x^2-1|=|x-1||x+1|$; and
*$\sqrt{AB} =\sqrt A \sqrt B$ if $A\ge0$ and $B\ge0\vphantom{\dfrac 1 1}$, so the separation into two square roots is valid; and
*$\sqrt A/\sqrt B = \sqrt{A/B\,{}}$, if $A\ge0$ and $B\ge0$, so the second equality is valid; and
*$|A|/|B| = |A/B|$, so the last equality is valid.
PS in response to comments: The problem with the trigonometric substitution is only that it is valid only when $|x|>1$, since $\sec\theta\ge 1$ for all values of $\theta$ and those points where $|x|=1$ are not in the domain.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1368006",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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Non-linear system of equations Solve following system of equations over real numbers:
$$
x-y+z-u=2\\
x^2-y^2+z^2-u^2=6\\
x^3-y^3+z^3-u^3=20\\
x^4-y^4+z^4-u^4=66
$$
This does not seem as hard problem. I have tried what is obvious here, to write $x^2-y^2$ as $(x-y)(x+y)$, $x^3-y^3$ as $(x-y)(x^2+xy+y^2)$ etc. Problem is because if I substitute $x-y=u-z+2$ in second equation, I cannot eliminate $x+y$ or if I substitute $x^2-y^2=u^2-z^2+6$ in fourth equation, I cannot eliminate $x^2+y^2$ etc. When I realized that this is not the best way, I made a second attempt.
My second attempt is to find come constraints, because all of these numbers are real. I have proven that if $m+n=a$ for some $m,n\in\mathbb{R}^+$, then $\frac{a^2}2\le m^2+n^2\le a^2$. I have applied this formula to second and fourth equation (because in these equations variables have even exponents and that means they are positive), but this didn't give me anything.
What should be the easiest way to solve it?
| rewrite the equations:
$x+z-(y+u)=2\\
x^2+z^2-(y^2+u^2)=6\\
x^3+z^3-(y^3+u^3)=20\\
x^4+z^4-(y^4+u^4)=66$
$a=x+z,c=xz,b=y+u,d=yu, \\x^2+z^2=(x+z)^2-2xz=a^2-2c,\\x^3+z^3=(x+z)(x^2+z^2-xz)=(x+z)((x+z)^2-3xz)=a(a^2-3c)\\x^4+z^4=(x^2+z^2)^2-2(xz)^2=((x+z)^2-2xz)^2-2(xz)^2=a^4-4a^2c+2c^2$
you can go from here now.
| {
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"url": "https://math.stackexchange.com/questions/1368243",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Notation - matrix calculatoin i am self-studying linear algebra, and came across the following statement about singular value decomposition (X = $USV^T$). I am somewhat confused on how to interpret the (|) and (--) symbols. Should i interpret that X is the sum of r matrices, of which the first one is equal to the (scalar?) $\sigma_1$ multiplied by the row vector $U_1$ multiplied by the column vector $V_1^T$?
| In the equation, $u_1$ is an $n\times 1$ matrix (i.e., a column vector) and $v_1^t$ is a $1 \times n$ matrix, i.e., a row vector. The product shown is an "outer product", i.e., the result is an $n \times n$ matrix. The little bars are indeed meant to indicate the "shape" of the iterm, i.e., that $u_1$ consists of a bunch of things organized vertically.
To give a concrete example, if
$$
X = \begin{bmatrix} -5 & -7 \\ 5 & -1\end{bmatrix},
$$ then the SVD is
\begin{align}
X
&= \begin{bmatrix} -5 & -7 \\ 5 & -1\end{bmatrix}\\
&= \begin{bmatrix} -3\frac{\sqrt{10}}{10} & \frac{\sqrt{10}}{10} \\ \frac{\sqrt{10}}{10} & 3\frac{\sqrt{10}}{10}\end{bmatrix}
\begin{bmatrix} 4\sqrt{5} & 0 \\ 0 & 2\sqrt{5} \end{bmatrix}
\begin{bmatrix} \frac{\sqrt{2}}{2} & \frac{\sqrt{2}}{2} \\ \frac{\sqrt{2}}{2} & -\frac{\sqrt{2}}{2}\end{bmatrix}\\
&= \begin{bmatrix} -3& 1 \\ 1 & 3\end{bmatrix}
\begin{bmatrix} 2& 0 \\ 0 & 1 \end{bmatrix}
\begin{bmatrix} 1 & 1 \\ 1 & -1\end{bmatrix}
\end{align}
The claim is then that $X$ can be rewritten as
\begin{align}
X &=
2\begin{bmatrix} -3 \\ 1 \end{bmatrix}
\begin{bmatrix} 1 & 1 \end{bmatrix} +
1\begin{bmatrix} 1 \\ 3 \end{bmatrix}
\begin{bmatrix} 1 & -1 \end{bmatrix}
\end{align}
which indeed turns out to be true. In this case, $u_1 = \begin{bmatrix} -3 \\ 1 \end{bmatrix}$ and $v_1^t = \begin{bmatrix} 1 & 1 \end{bmatrix}$, and so on.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1368614",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Given that $p$ is an odd prime, is the GCD of any two numbers of the form $2^p + 1$ always equal to $3$? I have checked it for some numbers and it appears to be true. Also I am able to reduce it and get the value $3$ for specific primes $p_1$, $p_2$ by using the Euclidean algorithm but I am not able to find a general argument for all numbers.
| Let $p_1, p_2$ be any two distinct prime numbers and assume that
$$q \mid2^{p_1} + 1 \, \mbox{ and } \, q \mid 2^{p_2} + 1$$
Then
$$2^{p_1} \equiv -1 \equiv 2^{p_2} \pmod{q}$$
In particular, there exists a power of $2$ which is $-1 \pmod{q}$. Let $a$ be the smallest number for which $2^a \equiv -1 \pmod{q}$ and let $b$ be the order of $2 \pmod{q}$. Then $b \mid 2a$, thus $2a = bk$ for some integer $k$.
Now, since $2^{p_1} = 2^a$ we have
$$b \mid p_1 - a$$ and hence
$$b \mid 2p_1 - 2a = 2p_1 - bk$$
This shows that $b \mid 2p_1$. Exactly same way we get that $b \mid 2p_2$. As $p_1, p_2$ are distinct primes we get $b \mid 2$.
This shows that
$$2^2 \equiv 1 \pmod{q}$$
and hence $q = 3$.
This shows that $\gcd(2^{p_1} + 1, 2^{p_2} + 1) = 3^\mathcal{L}$ for some $\mathcal{L}$.
Finally, if $9 \mid 2^p + 1$ then
$$2^p \equiv -1 \pmod{9} \Rightarrow 2^p \equiv 2^3 \pmod{9} \Rightarrow 6 \mid p - 3 $$
This shows that $9 \mid 2^p + 1$ only in the case $p = 3$.
Therefore, for all distinct primes $p_1, p_2$ we must have
$$\gcd(2^{p_1} + 1, 2^{p_2} + 1) \mid 3$$
The claim can be easily completed by observing that $2^p + 1$ is divisible by $3$ for odd primes $p$.
P.S. An alternate way to continue after $2a = bk$ is as follows:
$2 \mid bk$ implies $2 \mid b$ or $2 \mid k$. But $2 \mid k$ is impossible, as it would imply that $b \mid a$ and hence $2^a \equiv 1 \pmod{q}$.
Therefore $b$ is even, and as
$$(2^{\frac{b}{2}})^2=1 \pmod{q}$$
we get
$$2^{\frac{b}{2}}= \pm 1 \pmod{q}$$
Since $b$ is the order, the above cannot be $1$ and thus it is $-1$. Therefore, the minimality of $a$ implies that
$$a \leq \frac{b}{2} \Rightarrow 2a \leq b$$
As $b \mid 2a$ we get $b = 2a$.
Now,
$$a \mid b \mid p_{1, 2} - a$$
implies that $a \mid p_{1,2}$ therefore $a = 1$. This shows that
$$2^{1} \equiv -1 \pmod{q}$$
and hence $q = 3$.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Finding $\frac {a}{b} + \frac {b}{c} + \frac {c}{a}$ where $a, b, c$ are the roots of a cubic equation, without solving the cubic equation itself Suppose that we have a equation of third degree as follows:
$$
x^3-3x+1=0
$$
Let $a, b, c$ be the roots of the above equation, such that $a < b < c$ holds. How can we find the answer of the following expression, without solving the original equation?
$$
\frac {a}{b} + \frac {b}{c} + \frac {c}{a}
$$
| By Vieta's formulas, we have
$$a+b+c=-\frac{0}{1}=0\tag1$$
$$ab+bc+ca=\frac{-3}{1}=-3\tag2$$
$$abc=-\frac{1}{1}=-1\tag3$$
From $(1)(2)(3)$, we have
$$P=a^2b+ab^2+b^2c+bc^2+c^2a+ca^2=(a+b+c)(ab+bc+ca)-3abc=3\tag4$$
Now, set
$$Q=\frac ab+\frac bc+\frac ca,\ \ \ R=\frac ba+\frac cb+\frac ac.$$
Then, we have
$$Q+R=\frac ab+\frac bc+\frac ca+\frac ba+\frac cb+\frac ac=\frac{P}{abc}=-3\tag5$$
and
$$\begin{align}QR&=\left(\frac ab+\frac bc+\frac ca\right)\left(\frac ba+\frac cb+\frac ac\right)\\&=3+\frac{bc}{a^2}+\frac{ca}{b^2}+\frac{ab}{c^2}+\frac{c^2}{ab}+\frac{a^2}{bc}+\frac{b^2}{ca}\\&=3-\frac{1}{a^3}-\frac{1}{b^3}-\frac{1}{c^3}+\frac{a^3+b^3+c^3}{abc}\\&=3-\frac{1}{3a-1}-\frac{1}{3b-1}-\frac{1}{3c-1}+\frac{3abc+(a+b+c)((a+b+c)^2-3(ab+bc+ca))}{abc}\\&\small=3+\frac{-9(ab+bc+ca)+6(a+b+c)-3}{27abc-9(ab+bc+ca)+3(a+b+c)-1}+\frac{3abc+(a+b+c)((a+b+c)^2-3(ab+bc+ca))}{abc}\\&=-18\tag6\end{align}$$
Also, since it is easy to see
$$a\lt 0\lt b\lt c,$$
we have
$$\frac ab\lt 0,\frac bc\lt 1,\frac ca\lt 0\Rightarrow Q\lt 1\tag 7$$
So, as a result, from $(5)(6)(7)$ we have
$$\color{red}{\frac ab+\frac bc+\frac ca=Q=-6}$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Tangent line parallel to another line At what point of the parabola $y=x^2-3x-5$ is the tangent line parallel to $3x-y=2$? Find its equation.
I don't know what the slope of the tangent line will be. Is it the negative reciprocal?
| Edit: since the tangent is parallel to the given line: $3x-y=2$ hence the slope of tangent line to the parabola is $\frac{-3}{-1}=3$
Let the equation of the tangent be $y=3x+c$
Now, solving the equation of the tangent line: $y=3x+c$ & the parabola: $y=x^2-3x-5$ by substituting $y=3x+c$ as follows $$3x+c=x^2-3x-5$$ $$\implies x^2-6x-(c+5)=0\tag 1$$ For tangency we have the following condition $$ \text{determinant},\ B^2-4AC=0$$ $$\implies (-6)^2-4(1)(-(c+5))$$ $$\implies c=\frac{-56}{4}=-14$$ Hence, setting the value of $c=-14$ we get $$x^2-6x-(-14+5)=0$$ $$\implies x^2-6x+9=0$$ $$\implies (x-3)^2=0\implies x=3$$ Now, setting the value of $x=3$ in the equation of parabola as follows $$y=(3)^2-3(3)-5=-5$$ Hence, the point of tangency is $\color{blue}{(3, -5)}$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Find the sum: $\sum_{i=1}^{n}\dfrac{1}{4^i\cdot\cos^2\dfrac{a}{2^i}}$ Find the sum of the following :
$S=\dfrac{1}{4\cos^2\dfrac{a}{2}}+\dfrac{1}{4^2\cos^2\dfrac{a}{2^2}}+...+\dfrac{1}{4^n\cos^2\dfrac{a}{2^n}}$
| $$S=S+\frac{1}{4^n \sin^2\frac{a}{2^n}}-4^{-n}\csc^2 (a/2^n)\\=\sum_{k=1}^{n-1}\frac{1}{4^k\cos^2(a/2^k)}+\frac{1}{4^{n-1}\sin^2{a/2^{n-1}}}-4^{-n}\csc^2 (a/2^n)\\
=\cdots=\frac{1}{\sin^2(a)}-4^{-n}\csc^2 (a/2^n)=\csc^2(a)-4^{-n}\csc^2 (a/2^n)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1371239",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
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$U_n=\int_{n^2+n+1}^{n^2+1}\frac{\tan^{-1}x}{(x)^{0.5}}dx$ . $U_n= \int_{n^2+n+1}^{n^2+1}\frac{\tan^{-1}x}{(x)^{0.5}}dx$ where
Find $\lim_{n\to \infty} U_n$ without finding the integration
I don't know how to start
| Let $f(x)=\frac{\arctan x}{\sqrt x}$ and then
$$ f'(x)=\frac{2x-(x^2+1)\arctan x}{2x\sqrt x\arctan x}<0\text{ for large }x>0. $$
So $f(x)$ is decreasing for large $x>0$. By the Mean Value Theorem, there is $c\in(n^2+1,n^2+n+1)$ such that
\begin{eqnarray}
\int^{n^2+n+1}_{n^2+1}\frac{\arctan x}{\sqrt x}dx&=&n\frac{\arctan c}{\sqrt c}.
\end{eqnarray}
But for large $n$,
$$ \frac{\arctan (n^2+n+1)}{\sqrt{n^2+n+1}}\le \frac{\arctan c}{\sqrt c} \le \frac{\arctan (n^2+1)}{\sqrt{n^2+1}} $$
and thus
$$ \lim_{n\to\infty}n\frac{\arctan c}{\sqrt c}=\frac{\pi}{2}.$$
So
$$\lim_{n\to\infty}\int^{n^2+n+1}_{n^2+1}\frac{\arctan x}{\sqrt x}dx=\frac{\pi}{2}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1371340",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Cauchy-Schwarz inequality problem The problems:
*Prove that $$\frac{\sin^3 a}{\sin b} + \frac{\cos^3 a}{\cos b} \geqslant \sec (a-b),$$ for all $a,b \in \bigl(0,\frac{\pi}{2}\bigr)$.
*Prove that $$\frac{1}{a+b} + \frac{1}{b+c} + \frac{1}{c+a} + \frac{1}{2\sqrt[3]{abc}} \geqslant \frac{(a+b+c+\sqrt[3]{abc})^2}{(a+b)(b+c)(c+a)},$$ for all $a,b,c > 0$.
Can someone give me a hint for the two problems. They are all based on the Cauchy-Schwarz inequality.
Just a hint.
| You already have an answer for (1). For (2), note by Cauchy Schwarz inequality,
$$LHS = \sum_{cyc} \frac{c^2}{c^2(a+b)}+\frac{(\sqrt[3]{abc})^2}{2abc} \ge \frac{(a+b+c+\sqrt[3]{abc})^2}{\sum_{cyc} c^2(a+b) + 2abc}$$
and further the denominator factorises as
$$\sum_{cyc} c^2(a+b) + 2abc = (a+b)(b+c)(c+a)$$
| {
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"timestamp": "2023-03-29T00:00:00",
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How to find: $\int^{2\pi}_0 (1+\cos(x))\cos(x)(-\sin^2(x)+\cos(x)+\cos^2(x))~dx$? How to find: $$\int^{2\pi}_0 (1+\cos(x))\cos(x)(-\sin^2(x)+\cos(x)+\cos^2(x))~dx$$
I tried multiplying it all out but I just ended up in a real mess and I'm wondering if there is something I'm missing.
| $\displaystyle\int_0^{2\pi}(\cos x+\cos^2x)(\cos 2x+\cos x)dx=\int_0^{2\pi}\big(\cos x+\frac{1}{2}+\frac{1}{2}\cos 2x\big)\big(\cos 2x+\cos x\big)dx$
$\displaystyle\int_0^{2\pi}\big(\cos x\cos 2x+\cos^{2}x+\frac{1}{2}\cos 2x+\frac{1}{2}\cos x+\frac{1}{2}\cos^{2} 2x+\frac{1}{2}\cos 2x\cos x\big)dx$
$\displaystyle\int_0^{2\pi}\big(\frac{3}{2}\cos x\cos 2x+\frac{1}{2}+\cos 2x+\frac{1}{2}\cos x+\frac{1}{4}+\frac{1}{4}\cos 4x\big)dx$
$\displaystyle\int_0^{2\pi}\big(\frac{3}{4}(\cos 3x+\cos x)+\frac{3}{4}+\cos 2x+\frac{1}{2}\cos x+\frac{1}{4}\cos4x\big)dx=\frac{3}{4}(2\pi)=\frac{3\pi}{2}$
(since $\sin n\pi=0$ for any integer n)
| {
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"timestamp": "2023-03-29T00:00:00",
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find total integer solutions for $(x-2)(x-10)=3^y$ I found this questions from past year maths competition in my country, I've tried any possible way to find it, but it is just way too hard.
How many integer solutions ($x$, $y$) are there of the equation $(x-2)(x-10)=3^y$?
(A)1 (B)2 (C)3 (D)4 (E)5
If let $y=0$, we had
$$x^2 - 12x + 20 = 1$$
$$x^2 - 12x + 19 = 0$$
no integer solution for x
let $y=1$, we had $x^2 - 12x + 17 = 0$, no integer solution too.
let $y=2$, we had $x^2 - 12x + 11 = 0$, we had $x = 1, 11$.
let $y=3$, we had $x^2 - 12x - 7 = 0$, no integer solution.
let $y=4$, we had $x^2 - 12x - 61 = 0$, no integer solution.
and going on....
is there any other efficient way to find it? "brute-forcing" it will wasting a lot of time.
| By the Fundamental Theorem of Arithmetic, $x-2$ and $x-10$ must be powers of $3$, so we may write $$x-2= \pm 3^a $$ and $$x-10 = \pm 3^b.$$ Substracting the latter from the former we get $$8=\pm 3^a\mp3^b \\ 2^3 =3^a(\mp3^{b-a}\pm1)= 3^b (\pm3^{a-b} \mp1),$$ whence, again by FTA, $(a,b)= (0,2),( 2,0)$ and thus $(x,y)= (1,2),(11,2)$ are the only pairs of integer solutions.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1374534",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
Identities involving binomial coefficients, floors, and ceilings I found the following four apparent identities:
$$
\begin{align}
\sum_{k=0}^n
2^{-\lfloor\frac{n+k}{2}\rfloor}
{\lfloor\frac{n+k}{2}\rfloor\choose k}
&=
\frac{4}{3}-\frac{1}{3}(-2)^{-n},\\
\sum_{k=0}^n
2^{-\lceil \frac{n+k}{2}\rceil }
{\lfloor\frac{n+k}{2}\rfloor\choose k}
&=
1,\\
\sum_{k=0}^n
2^{-\lfloor\frac{n+k}{2}\rfloor}
{\lceil \frac{n+k}{2}\rceil \choose k}
&=
2-2^{-n},\\
\sum_{k=0}^n
2^{-\lceil \frac{n+k}{2}\rceil }
{\lceil \frac{n+k}{2}\rceil\choose k}
&=
\frac{4}{3}-\frac{1}{3}(4)^{-\lfloor\frac{n}{2}\rfloor}.
\end{align}
$$
I want to know how to prove them. I also want to know whether they have (after multiplying both sides by $2^n$) nice combinatorial interpretations.
| Assuming that $n$ is even, $n=2m$, we have:
$$\begin{eqnarray*} \sum_{k=0}^{n} 2^{-\left\lfloor\frac{n+k}{2}\right\rfloor}\binom{\left\lfloor\frac{n+k}{2}\right\rfloor}{k}&=&\sum_{j=0}^{m}2^{-(m+j)}\binom{m+j}{2j}+\sum_{j=0}^{m-1}2^{-(m+j)}\binom{m+j}{2j+1}\\&=&4^{-m}+2^{-m}\sum_{j=0}^{m-1}2^{-j}\binom{m+j+1}{m-j}\\&=&4^{-m}+4^{-m}\sum_{j=1}^{m}2^j\binom{2m+1-j}{j}\end{eqnarray*}$$
so your sums can be computed by using Fibonacci polynomials.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1374609",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 1,
"answer_id": 0
} |
Is this:$\sum_{n=1}^{\infty}{(-1)}^{\frac{n(n-1)}{2}}\frac{1}{n}$ a convergent series? Is there someone who can show me how do I evaluate this sum :$$\sum_{n=1}^{\infty}{(-1)}^{\frac{n(n-1)}{2}}\frac{1}{n}$$
Note : In wolfram alpha show this result and in the same time by ratio test it's not a convince way to judg that is convergent series
Thank you for any help
| After $ \ n = 1 \ $ , the exponents of (-1) are binomial coefficients which are "double-alternating" between even and odd integers. So the series looks like
$$ 1 \ - \frac{1}{2} \ - \ \frac{1}{3} \ + \ \frac{1}{4} \ + \ \frac{1}{5} \ - \ \frac{1}{6} \ - \ \frac{1}{7} \ + \ \ldots \ \ . $$
As lulu I think properly objected to my separation of terms originally, since the absolute series here is the harmonic series, the "Riemann derangement" theorem may have something to say against it. It is perhaps safer then to group the terms as
$$ 1 \ + \ \sum_{k=1}^{\infty} \ (-1)^k\left( \ \frac{1}{2k} \ + \ \frac{1}{2k+1} \ \right) \ = \ 1 \ + \ \sum_{k=1}^{\infty} \ (-1)^k\left[ \ \frac{4k + 1}{2k \ (2k+1)} \ \right] $$
$$ = \ 1 \ + \ \sum_{k=1}^{\infty} \ (-1)^k\left[ \ \frac{4k \ + \ 1}{4k^2 \ + \ 2k} \ \right] \ \ . $$
The term $ \ b_k \ = \ \frac{4k \ + \ 1}{4k^2 \ + \ 2k} \ $ proves to be monotonically decreasing toward zero for $ \ k \ \ge \ 1 \ $ , so the series converges by the alternating-series test. (And now my post looks a lot more like corindo's ...)
EDIT: Having thought about this a bit more, it occurred to me that I could have gathered this up a bit more (and now it comes still closer to what corindo is describing). The alternating series could have simply been written as
$$ 1 \ + \ \sum_{k=1}^{\infty} \ \left( \ \frac{-1}{4k-2} \ + \ \frac{-1}{4k-1} \ + \ \frac{1}{4k} \ + \ \frac{1}{4k+1} \ \right) $$
$$ = \ 1 \ - \ \sum_{k=1}^{\infty} \ \left[ \ \frac{32k^2 \ - \ 8k \ - \ 1}{4 \ x \ (2x-1) \ (4x-1) \ (4k+1)} \ \right] $$
$$ = \ 1 \ - \ \frac{1}{4} \ \sum_{k=1}^{\infty} \ \left( \ \frac{32k^2 \ - \ 8k \ - \ 1}{32k^4 \ - \ 16k^3 \ - \ 2k^2 \ + \ k} \ \right) \ \ $$
(with a little help from WolframAlpha). This removes the "alternating-sign" behavior and allow us to apply the "limit comparison test" against $ \ \sum_{k=1}^{\infty} \ \frac{1}{k^2} \ $ .
We can use this to set bounds on the sum (I've only pursued this a little at this point). It is not hard to show that $ \ \frac{32k^2 \ - \ 8k \ - \ 1}{32k^4 \ - \ 16k^3 \ - \ 2k^2 \ + \ k} \ < \ \frac{2}{k^2} \ $ , from which we obtain
$$ 1 \ - \ \frac{1}{4} \ \sum_{k=1}^{\infty} \ \left( \ \frac{32k^2 \ - \ 8k \ - \ 1}{32k^4 \ - \ 16k^3 \ - \ 2k^2 \ + \ k} \ \right) \ > \ 1 \ - \ \frac{1}{4} \ \sum_{k=1}^{\infty} \ \ \frac{2}{ k^2 } $$
$$ = \ 1 \ - \ \frac{1}{2} \ \zeta(2) \ = \ 1 \ - \ \frac{\pi^2}{12} \ \ . $$
Others here are likely better able to set tighter bounds on the sum than I am at the moment.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1375069",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 2
} |
Proving the Fibonacci sum $\sum_{n=1}^{\infty}\left(\frac{F_{n+2}}{F_{n+1}}-\frac{F_{n+3}}{F_{n+2}}\right) = \frac{1}{\phi^2}$ and its friends In this article, (eq.92) has,
$$\sum_{n=1}^{\infty}\frac{(-1)^{n+1}}{F_{n+1}F_{n+2}} = \frac{1}{\phi^2}\tag1$$
and I wondered if this could be generalized to the tribonacci numbers. It seems it can be. Given the Fibonacci, tribonacci, tetranacci (in general, the Fibonacci k-step numbers) starting with $n=1$,
$$F_n = 1,1,2,3,5,8\dots$$
$$T_n = 1, 1, 2, 4, 7, 13, 24,\dots$$
$$U_n = 1, 1, 2, 4, 8, 15, 29, \dots$$
and their limiting ratios, $x_k$, the root $x_k \to 2$ of,
$$(2-x)x^k = 1$$
with Fibonacci constant $x_2$, tribonacci constant $x_3$, etc, it can be empirically observed that,
$$\sum_{n=1}^{\infty}\left(\frac{F_{n+2}}{F_{n+1}}-\frac{F_{n+3}}{F_{n+2}}\right) = \frac{1}{x_2^2}\tag2$$
$$\sum_{n=1}^{\infty}\left(\frac{T_{n+2}}{T_{n+1}}-\frac{T_{n+3}}{T_{n+2}}\right) = \frac{1}{x_3^3}$$
$$\sum_{n=1}^{\infty}\left(\frac{U_{n+2}}{U_{n+1}}-\frac{U_{n+3}}{U_{n+2}}\right) = \frac{1}{x_4^4}$$
and so on. Q: How do we rigorously prove the observation indeed holds for all integer $k\geq2$?
Edit:
To address a comment that disappeared, to transform $(1)$ to $(2)$, we use a special case of Catalan's identity,
$$F_{n+2}^2-F_{n+1}F_{n+3} = (-1)^{n+1}$$
so,
$$\begin{aligned}
\frac{(-1)^{n+1}}{F_{n+1}F_{n+2}}
&= \frac{F_{n+2}^2-F_{n+1}F_{n+3}}{F_{n+1}F_{n+2}}\\
&= \frac{F_{n+2}}{F_{n+1}} - \frac{F_{n+3}}{F_{n+2}}
\end{aligned}$$
hence the alternating series $(1)$ is equal to $(2)$.
| Let's start off with the case of the Fibonacci sequence. We have
$$\sum_{n = 1}^k \left( \frac{F_{n+2}}{F_{n+1}} -\frac{F_{n+3}}{F_{n+2}} \right) = \left(\frac{F_3}{F_2} - \frac{F_4}{F_3}\right) + \left(\frac{F_4}{F_3} - \frac{F_5}{F_4}\right) + \cdots + \left(\frac{F_{k+2}}{F_{k+1}}- \frac{F_{k+3}}{F_{k+2}}\right)\\
= \frac{F_3}{F_2} - \frac{F_{k+3}}{F_{k+2}} = 2 - \frac{F_{k+3}}{F_{k+2}}.$$
Setting the limit as $k$ approaches infinity, we get
$$2 - \lim_{k\to\infty} \frac{F_{k+3}}{F_{k+2}} = 2- x = \frac{1}{x^2}$$
If you want more rigor, you can easily transform this into an induction argument.
Since $(2 - x)(x^2) = 1 \implies 2 - x = \frac{1}{x^2}$, the result follows. The proof generalizes really easily.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1375202",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
} |
Find expression in terms of x Knowing that $$\frac{dy}{dx}= k\cdot x^{\frac{1}{3}}$$ and given that it passes through points $(1,4)$ and $(8,16)$, find an expression for the path in terms of $x$.
I found out that $$y= \frac34 k x^{\frac43}$$ by integrating $$\frac{dy}{dx}.$$
What do I do next?
| We have, $$\frac{dy}{dx}=kx^{\frac{1}{3}}$$ $$\implies dy=kx^{\frac{1}{3}}dx$$ Integrating both the sides w.r.t. $x$ as follows $$\int dy=\int kx^{\frac{1}{3}}dx$$ $$\implies y=k\int x^{\frac{1}{3}}dx=k\left(\frac{x^{\frac{1}{3}+1}}{\frac{1}{3}+1}\right)+C$$ $$\implies y=\frac{3kx^{\frac{4}{3}}}{4}+C$$ Since, the curve passes through $(1, 4)$ hence satisfying the above equation by this point as follows $$4=\frac{3k(1)^{\frac{4}{3}}}{4}+C\implies 4=\frac{3k}{4}+C$$$$\implies C=\frac{16-3k}{4}\tag 1$$
Similarly, satisfying the above equation by the point $(8, 16)$ as follows $$16=\frac{3k(8)^{\frac{4}{3}}}{4}+C\implies 16=\frac{48k}{4}+C$$$$\implies C=16-12k\tag 2$$ Equating (1) & (2), we get
$$\frac{16-3k}{4}=16-12k$$ $$\implies k=\frac{48}{45}=\frac{16}{15}$$ $$\implies C=16-12\left(\frac{16}{15}\right)=\frac{80-64}{5}=\frac{16}{5}$$ Now, substituting the values of $k$ & $C$, we get the equation of the curve $$y=\frac{3\left(\frac{16}{15}\right)x^{\frac{4}{3}}}{4}+\frac{16}{5}$$ $$\color{blue}{y=\frac{4}{5}x^{\frac{4}{3}}+\frac{16}{5}}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1375447",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
remainder of $a^2+3a+4$ divided by 7
If the remainder of $a$ is divided by $7$ is $6$, find the remainder when $a^2+3a+4$ is divided by 7
(A)$2$ (B)$3$ (C)$4$ (D)$5$ (E)$6$
if $a = 6$, then $6^2 + 3(6) + 4 = 58$, and $a^2+3a+4 \equiv 2 \pmod 7$
if $a = 13$, then $13^2 + 3(13) + 4 = 212$, and $a^2+3a+4 \equiv 2 \pmod 7$
thus, we can say that any number, $a$ that divided by 7 has remainder of 6, the remainder of $a^2 + 3a + 4$ is 2.
is there any other way to calculate it? (Let say it given b as the remainder of a divided by 7, not 6)
| $a = 6 \quad(\mathrm{mod} 7)$
$a^2 = 36 = 1 \quad(\mathrm{mod} 7)$
$3a = 18 = 4\quad (\mathrm{mod} 7)$
$a^2 + 3a + 4 = 1 + 4 + 4 = 9 = 2 \quad(\mathrm{mod} 7)$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1376498",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 0
} |
Geometric problem based on angle bisectors I am not asking a question,i just want to conform,is my method of solving problem correct?
Given a triangle ABC.It is known that AB=4,AC=2,and BC=3.The bisector of angle A intersects the side BC at point K.The straight line passing through point B and being parallel to AC intersects the extension of angle bisector AK at point M.Find $(AM)^2$
My work:
$\angle CAK=\angle KMB$ (AC and BM are parallel)
$\angle ACK=\angle KBM$ (AC and BM are parallel)
$\angle AKC=\angle BKM$ (vertically opposite angles)
so,$\bigtriangleup AKC and \bigtriangleup KBM$ are similar
In triangle ABM,angle BAK= angle KMB
therefore,AB=BM=4(side opposite to equal angles are equal.)
therefore,$\frac{AC}{BM}=\frac{KC}{KB}=\frac{AK}{KM}=\frac{1}{2}$
so,$KM=2 AK$
Also,$\frac{AB}{AC}=\frac{BK}{KC}=\frac{4}{2}=\frac{2}{1}$(angle bisector theorem)
BC=3(given)
therefore,$BK=\frac{2}{2+1}\times 3=2$,$KC=\frac{1}{2+1}\times 3=1$
For finding KM,i need to find AK,which i will find by using Stewart's theorem
$n=2,m=1,b=2,c=4,a=3$
$b^2n+c^2m=a(d^2+mn)$ Plugging values we get $d=AK=2$,
so,KM=2 AK=4,$KM^2=16$
| You have a mistake
$$b^2 n + c^2 m =a(d^2 + mn)$$
$$2^2 (2) + 4^2 (1) = 3(d^2 + 2)$$
$$8 + 16 = 3(d^2 + 2)$$
$$24 = 3(d^2 + 2) \Rightarrow d^2 = 6$$
Hence $KM^2 = (2AK)^2 = 4(AK^2) = 24$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1376660",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Find $x$ in the triangle
the triangle without point F is drawn on scale, while I made the point F is explained below
So, I have used $\sin, \cos, \tan$ to calculate it
Let $\angle ACB = \theta$, $\angle DFC = \angle BAC = 90^\circ$, and $DF$ is perpendicular to $BC$ (the reason for it is to have same $\sin, \cos, \tan$ answer)
$$\sin \angle ACB = \frac {DF}{CD} = \frac{AB}{BC}$$
$$\cos \angle ACB = \frac {CF}{CD} = \frac{AC}{BC}$$
$$\tan \angle ACB = \frac {DF}{CF} = \frac{AB}{AC}$$
putting known data into it
\begin{align}
\frac {DF}{CD} &= \frac {AB}{12} \quad(1) \\
\frac {EF+3}{CD} &= \frac {2CD}{12} \\
\frac {EF+3}{CD} &= \frac {CD}{6} \quad(2)\\
\frac {DF}{EF+3} &= \frac {AB}{2CD} \quad(3)
\end{align}
I've stuck at here, how do I find their length?
| Let, $\angle ACB=\alpha$ then we have $$AC=BC\cos \alpha=(9+3)\cos\alpha=12\cos\alpha$$ $$\implies DC=AD=\frac{AC}{2}=6\cos\alpha$$
in right $\triangle DFC$ we have
$$DF=(DC)\sin\alpha=6\sin\alpha\cos\alpha$$ $$FC=(DC)\cos\alpha=6\cos\alpha\cos\alpha=6\cos^2\alpha$$ $$FE=FC-CE=6\cos^2\alpha-3$$ Now, applying Pythagorean theorem in right $\triangle DFE$ as follows $$(DE)^2=(DF)^2+(FE)^2$$ $$\implies x^2=(6\sin\alpha\cos\alpha)^2+(6\cos^2\alpha-3)^2$$ $$\implies x^2=36\sin^2 \alpha\cos^2\alpha+36\cos^4\alpha+9-36\cos^2\alpha$$ $$=36(1-\cos^2\alpha)\cos^2\alpha+36\cos^4\alpha+9-36\cos^2\alpha$$ $$=36\cos^2\alpha-36\cos^4\alpha+36\cos^4\alpha+9-36\cos^2\alpha$$ $$\implies x^2=9$$ $$\implies \color{blue}{x=3}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1376735",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 3
} |
Finding a triangle ABC if $2\prod (\cos \angle A+1)=\sum \cos(\angle A-\angle B)+\sum \cos \angle A+2$ Find $\triangle ABC$ if $\angle B=2\angle C$ and $$2(\cos\angle A+1)(\cos\angle B+1)(\cos\angle C+1)=\cos(\angle A-\angle B)+\cos(\angle B-\angle C)+\cos(\angle C-\angle A)+\cos\angle A+\cos\angle B+\cos\angle C+2$$
| Hint:$$A+B+C=\pi,B=2C\Rightarrow A=\pi-3C$$
$\cos A=-\cos 3C$
$\cos B=\cos 2C$
$\cos(A-B)=-\cos 5C$
$\cos(B-C)=\cos C$
$\cos(C-A)=-\cos 4C$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1378033",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
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