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how to integrate $\int\frac{1}{x^2-12x+35}dx$? How to integrate following
$$\int\frac{1}{x^2-12x+35}dx?$$
What I did is here:
$$\int\frac{dx}{x^2-12x+35}=\int\frac{dx}{(x-6)^2-1}$$
substitute $x-6=t$, $dx=dt$
$$=\int\frac{dt}{t^2-1}$$
partial fraction decomposition,
$$=\int{1\over 2}\left(\frac{1}{t-1}-\frac{1}{t+1}\right)dt$$
$$=\frac12(\ln|1-t|-\ln|1+t|)+c$$
$$=\frac12\left(\ln\left|\frac{1-t}{1+t}\right|\right)+c$$
substitute back to $t$
$$=\frac12\ln\left|\frac{7-x}{x-5}\right|+c$$
I am not sure if my answer correct. Can I integrate this without substitution? Thank you
| Can you skip the substitution? Certainly!
$\frac {1}{x^2 - 12 x + 35} = \frac {1}{(x-7)(x-5)} = \frac {A}{x-7} + \frac {B}{x-5}$
$A+B = 0\\
-5A - 7B = 1\\
A = \frac {1}{2}, B = -\frac {1}{2}$
$12\int \frac {1}{x-7} - \frac {1}{x-5} \ dx\\
\frac 12 (\ln |x-7| - \ln |x-5|)\\
\frac 12 \ln \left|\frac {x-7}{x-5}\right|$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3742215",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Evaluate $\lim\limits_{x \to 1} \left( \frac{1}{x-1}-\frac{3}{1-x^3} \right)$ Evaluate $$\lim_{x \to 1} \left( \frac{1}{x-1}-\frac{3}{1-x^3} \right)$$
My attempt: $$\lim_{x \to 1} \left( \frac{1}{x-1}-\frac{3}{1-x^3} \right) = \lim_{x \to 1} \frac{x+2}{x^2+x+1}=1$$
According to the answer key, this limit does not exist. I turned that into one fraction, then I factored the polynomial on the numerator as $-(x-1)(x+2)$ and the one on the denominator as $(x+1)(-x^2-x-1)$. What did I do wrong?
| Using the identity
$A-B=\dfrac{1}{\dfrac{1}{A}}-\dfrac{1}{\dfrac{1}{B}}$ we get $$\dfrac{1}{x-1}-\dfrac{3}{1-x^3}=\dfrac{2-x^3-x}{-x^4+x^3+x-1}$$ so you go from the form $\infty-\infty$ to the form $\dfrac00$ and you can apply L'Hôspital so you have $$\lim_{x \to 1} \frac{1}{x-1}-\frac{3}{1-x^3}=\lim\dfrac{-3x^2-1}{-4x^3+3x^2+1}$$ and you can verify that the limits to the right and to the left are not equal, the first is $-\infty$ and the second is $\infty$.
Thus the limit does not exist.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3745531",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Let f(x) be a polynomial satisfying $\lim_{x \to\infty} \frac{x^4f(x)}{x^8+1}=3,$ $f(2)=5 $ and $f(3)=10$, $f(-1)=2$ and $f(-6)=37$ Let f(x) be a polynomial satisfying $\lim_{x \to\infty} \frac{x^4f(x)}{x^8+1}=3,$ $f(2)=5 $ and $f(3)=10$, $f(-1)=2$ and $f(-6)=37$ .
And then there are some question related to this.
In the solution it is given that according to the given information, $f(x)$ is a polynomial of degree 4 with leading coefficient 3.
$f(x)=3(x-2)(x-3)(x+1)(x+6)+(x^2+1)$
Now what i think is that he got the coefficient 3 using the limits and the rest of the terms$[(x-2)(x-3)(x+1)(x+6)]$ from the respective $f(x)$ values from what was given and also the $(x^2+1)$ was added for the remainders, as they satisfy all the values that have been given to us in the question. I just wanted to know if my approach is right? or is there another elegant way to do it?.
| the correct and precise reason for writing f(x)=3(x−2)(x−3)(x+1)(x+6)+(x^2+1) is as follows (I am assuming you have understood why f(x) will be a 4th deg polynomial and how its leading coeff shall be 3 from other answers):
consider g(x)= f(x)-(x^2+1)
clearly, deg(g(x))= 4
now, g(2)= f(2)-(2^2+1)= 0
similarly, g(3)=g(-1)=g(-6)= 0
hence, we have 4 roots of g(x) as 2, 3, -1, -6.
hence, by factor theorem g(x)= A(x−2)(x−3)(x+1)(x+6) where A is a constant.
or f(x)-(x^2+1)= A(x−2)(x−3)(x+1)(x+6)
or f(x)= A(x−2)(x−3)(x+1)(x+6)+x^2+1
but A=3 (already seen).
hence f(x)=3(x−2)(x−3)(x+1)(x+6)+(x^2+1)
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3748237",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
There are only two six-digit integers $N$, each greater than $100,000$. for which $N^2$ has $N$ as its final six digits There are only two six-digit integers $N$, each greater than 100,000 for which $N^2$ has $N$ as its final six digits (or $N^2-N$ is divisible by $10^6$). What are these two numbers?
Is the problem solvable by the Chinese Remainder Theorem? If so, how?
| Yes, we can solve $N^2-N\equiv0\bmod10^6$ with the Chinese remainder theorem
to get $N\equiv0 $ or $1\bmod 2^6=64$ and $N\equiv0$ or $ 1\bmod5^6=15625$.
We don't want the solutions $N\equiv0\bmod2^6$ and $5^6$ or $N\equiv1\bmod2^6 $ and $5^6$.
We want the solutions $N\equiv1\bmod 2^6$ and $N\equiv0\bmod5^6$ and vice versa.
The extended Euclidean algorithm gives $15625=244\times64+9$ and $64=7\times9+1$,
so $1=1709\times64-7\times15625$.
Thus, one solution is $1709\times64$, and the other is $(64-7)\times15625$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3748716",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Another way to solve $\int \frac{\sin^4(x)}{1+\cos^2(x)}\ dx$ without the substitution $y=\tan\left(\frac{x}{2}\right)$? Is there another way to solve an integral $$\int \frac{\sin^4(x)}{1+\cos^2(x)}\ dx$$ without the substitution $y=\tan\left(\frac{x}{2}\right)$?
$\large \int \frac{\sin^3(x)}{1+\cos^2(x)}\ dx$ is easily solved using the substitution $y=\cos(x)$.
What if the power of sine is even?
| HINT:
$$\int \frac{\sin^4(x)}{1+\cos^2(x)}\ dx=\int \frac{(1-\cos^2(x))^2}{1+\cos^2(x)}\ dx$$
$$=\int \frac{(1+\cos^2(x))^2-4(1+\cos^2(x))+4}{1+\cos^2(x)}\ dx$$
$$=\int \left(1+\cos^2(x)-4+\frac{4}{1+\cos^2(x)}\right)\ dx$$
$$=\int \left(\frac{1+\cos(2x)}{2}-3+\frac{4\sec^2x}{\tan^2(x)+2}\right)\ dx$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3751405",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Solving a multiple variable system of congruences using the Chinese Remainder theorem I've tried to solve the following system:
$2x+7y\equiv17\mod35\\3x+8y\equiv18\mod35$
My idea was using the Chinese Remainder theorem, so firstly, I've found that
$3\cdot 5 - 2\cdot 7 = 1$
And that $15$ is $1\mod7$, $\space -14\equiv 1\mod5$.
I'm not sure how to proceed from this point, and I'd like to receive a little help.
Thank you!
| Subtracting the first congruence from the second gives
$$ x+y \equiv 1\pmod{35}. $$
Using this in the first congruence gives
$$ 17 \equiv 2(x+y) + 5y \equiv 2+5y \pmod{35}, $$
so that $35 \mid 5(y-3)$, or $7 \mid (y-3)$. From $y \equiv 3\pmod{7}$ and $x+y \equiv 1\pmod{7}$ we get $x \equiv 5\pmod{7}$.
The two congruences are identical modulo $5$; they both give $x+y \equiv 1\pmod{5}$. Thus, there are five pairs $(x\bmod{5},y\bmod{5})$ from the two congruences, as opposed to the unique $(x\bmod{7},y\bmod{7})=(5,3)$ for congruences modulo $7$. Therefore, we have following five solutions $(x\bmod{35},y\bmod{35})$.
Set $x=7a+5$, $y=7b+3$, $a,b \in \{0,1,2,3,4\}$. Then $5 \mid (x+y-1)$ reduces to $5 \mid 7(a+b+1)$, and hence to $5 \mid (a+b+1)$ since $\gcd(5,7)=1$. This yields the following pairs $(a,b)$:
$$ (0,4), \quad (1,3), \quad (2,2), \quad (3,1), \quad (4,0). $$
The corresponding pairs $(x\bmod{35},y\bmod{35})$ are
$$ (5,31), \quad (12,24), \quad (19,17), \quad (26,10), \quad (33,3). $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3751862",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
If $\alpha,\beta,\gamma$ are the roots of $x^3+x+1=0$, then find the equation whose roots are: $(\alpha-\beta)^2,(\beta-\gamma)^2,(\gamma-\alpha)^2$ Question:
If $\alpha,\beta,\gamma$ are the roots of the equation, $x^3+x+1=0$, then find the equation whose roots are: $({\alpha}-{\beta})^2,({\beta}-{\gamma})^2,({\gamma}-{\alpha})^2$
Now, the normal way to solve this question would be to use the theory of equations and find the sum of roots taken one at a time, two at a time and three at a time. Using this approach, we get the answer as $(x+1)^3+3(x+1)^2+27=0$. However, I feel that this is a very lengthy approach to this problem. Is there an easier way of doing it?
| The standard method:
$$a+b+c=0; ab+bc+ca=1; abc=-1;\\
a^2+b^2+c^2=-2;a^2b^2+b^2c^2+c^2a^2=1;a^4+b^4+c^4=2;\\
a^3=-a-1.$$
First coefficient:
$$(a-b)^2+(b-c)^2+(c-a)^2=\\2(a^2+b^2+c^2)-2(ab+bc+ca)=-6$$
Second coefficient:
$$(a-b)^2(b-c)^2+(b-c)^2(c-a)^2+(c-a)^2(a-b)^2=\\
a^4+b^4+c^4+3(a^2b^2+b^2c^2+c^2a^2)-\\
2(a^3(\underbrace{b+c}_{-a})+b^3(\underbrace{c+a}_{-b})+c^3(\underbrace{a+b}_{-c}))=\\
3(a^2b^2+b^2c^2+c^2a^2+a^4+b^4+c^4)=9$$
Third coefficient:
$$(a-b)^2(b-c)^2(c-a)^2=\\
\small{(a^2+b^2-2ab)(b^2+c^2-2bc)(c^2+a^2-2ac)=\\
(c^2-4ab)(a^2-4bc)(b^2-4ac)=\\
16 a^4 b c - 4 a^3 b^3 - 4 a^3 c^3 - 63 a^2 b^2 c^2 + 16 a b^4 c + 16 a b c^4 - 4 b^3 c^3=\\
-16(a^3+b^3+c^3)-4(a^3b^3+b^3c^3+c^3a^3)-63=\\
-16(-a-b-c-3)-4((-a-1)(-b-1)+\\
(-b-1)(-c-1)+(-c-1)(-a-1))-63=}\\
48-4(1+3)-63=-31.$$
Hence, the equation is: $x^3+6x^2+9x+31=0$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3753474",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
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"answer_id": 3
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Given $\cos(a) +\cos(b) = 1$, prove that $1 - s^2 - t^2 - 3s^2t^2 = 0$, where $s = \tan(a/2)$ and $t = \tan(b/2)$ Given $\cos(a) + \cos(b) = 1$, prove that $1 - s^2 - t^2 - 3s^2t^2 = 0$, where $s = \tan(a/2)$ and $t = \tan(b/2)$.
I have tried using the identity $\cos(a) = \frac{1-t^2}{1+t^2}$. but manipulating this seems to have got me nowhere.
| Alternatively:
$$\cos a=2\cos^2 \frac a2-1; \\
\cos^2 \frac a2=\frac 1{1+\tan^2 \frac a2}=\frac1{1+s^2}\\
\cos a+\cos b=2\cos ^2 \frac a2+2\cos^2 \frac b2-2=\\
\frac2{1+s^2}+\frac2{1+t^2}-2=1.$$
I leave to you to simplify it.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3754249",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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How can I show that $A=\begin{pmatrix}a&b\\b&d\end{pmatrix}$ with $b\neq 0$ is diagonalizable?
How can I show that $A=\begin{pmatrix}a&b\\b&d\end{pmatrix}$ with
$b\neq 0$ is diagonalizable?
Sei $b\neq 0$, dann gilt:$$\det(A-\lambda E)=\det \begin{pmatrix} a-\lambda & b \\ b & d-\lambda \end{pmatrix}=(a-\lambda)(d-\lambda)-b^2=\lambda ^2-\lambda (a+d)+ad-b^2 $$
$$\lambda_{1,2}=\frac{a+d}{2}\pm\sqrt{\frac{(a+d)^2}{4}-ad+b^2}=\frac{a+d}{2}\pm \sqrt{\frac{1}{4}(d-a)^2+b^2}$$
Since $\frac{1}{4}(d-a)^2\geq 0$ for all $d,a\in \mathbb{R}$ and $b\neq 0$ (and thus $b^2>0$), we have $\sqrt{\frac{1}{4}(d-a)^2+b^2}>0$. Hence, we have to different Eigenvalues.
Do I have to brute force this problem? Or can I show it without going through all these tedious computations?
| *
*The characteristic equation is $(a-\lambda)(d-\lambda)-b^2=\lambda^2-(a+d)\lambda+ad-b^2$;
*The roots a distinct because the discriminant $(a+d)^2-4(ad-b^2)=(a-d)^2+4b^2$ is certainly non-zero.
*$v=(b,d-\lambda)$ solves the system $Av=\lambda v$, for both $\lambda$s (check by direct subtitution).
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3754866",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Find range of $f(x)=\frac{5}{\sin^2x-6\sin x\cos x+3\cos^2x}$ Find range of $f(x)=\frac{5}{\sin^2x-6\sin x\cos x+3\cos^2x}$
My attempt :
\begin{align*}
f(x)&=\dfrac{5}{9\cos^2x-6\sin x\cos x+\sin^2x-6\cos^2x}\\
&= \dfrac{5}{(3\cos x+\sin x)^2-6\cos^2x}
\end{align*}
The problem is if I'm going to use
$$-1\leqslant\sin x\leqslant1\;\text{and}-1\leqslant\cos x\leqslant1$$
I think I need to have only one term.
Edit : I have made some more progress
$$-3\leqslant 3\cos x\leqslant 3$$
$$\therefore -4\leqslant 3\cos x+\sin x\leqslant 4$$
$$ 0\leqslant (3\cos x+\sin x)^2\leqslant 16$$
| $$f(x)=\frac{5}{\sin^2x-6\sin x\cos x+3\cos^2 x}=\frac{10}{2\sin^2 x-12 \sin x \cos x+6 \cos^2 x}.$$
$$f(x)=\frac{10}{1-\cos 2x-6 \sin 2x +3(1+\cos 2x)}=\frac{10}{4+2\cos 2x-6 \sin 2x}$$
$$\implies f(x)=\frac{5}{2+\cos 2x-3 \sin 2x}=\frac{5}{2+\sqrt{10}(\cos 2x-3\sin 2x)/\sqrt{10}}$$
$$\implies f(x)=\frac{5}{2+\sqrt{10}\cos(2x+a)}~~~~\color{red}{(1)}$$
$$\implies f_{\min}=\frac{5}{2+\sqrt{10}},\quad f_{\max}=\frac{5}{2-\sqrt{10}}.$$
when $\cos (\cdots)=\mp 1$.
These are only local maximum.and minimum. But $f(x)$ $\color{red}{(1)}$ can take values close to $\pm \infty$ real value so the Range it's range is $(-\infty,\frac{5}{2-\sqrt{10}}] \cup [\frac{5}{2+\sqrt{10}}, \infty )$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3755235",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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A locus problem related to circumcenters and conic sections Given a point $A$, a circle $O$ and conic section $e$, if $BC$ is a moving chord of the circle $O$ tangent to $e$, then prove that
the locus of △$ABC$'s circumcenters $T$ is a conic section.
The question was posted in 纯几何吧 by TelvCohl and remained unsolved for many years but regrettably I cannot provide the link because the post was deleted by Baidu accidentally.
It seems that the locus related to circumcenter is often a conic section.Another example:
The directions of two sides of a triangle is fixed and the third side passes through a fixed point, then the locus of the circumcenter is a conic section.(The elementary geometry of conics.1883)
| Here's something of a brute-force vector proof.
In the figure, $P$ is a point on our given conic, $\bigcirc K$ of radius $r$ is our given circle, and $A$ is our given point. The tangent line at $P$ meets $\bigcirc K$ at $R$ and $R'$, and the circumcenter of $\triangle ARR'$ is $Q$. Points $A'$ and $K'$ are the respective projections of $A$ and $K$ onto the tangent line. Finally, $a:=|PA|$, $k := |PK|$, and $\alpha$ and $\kappa$ are the (signed) angles made by the tangent line and the respective vectors $\overrightarrow{PA}$ and $\overrightarrow{PK}$.
First, a bit of geometry. Ignoring the conic, and concentrating how $\bigcirc K$ meets the tangent line and determines circumcenter $Q$, one can show that $|K'Q|$ is given by
$$\begin{align}
|K'Q| &= \frac{|AK'|^2 -|RK'|^2}{2|AA'|} \tag{1}\\[4pt]
&= \frac{\left(a^2+|PK'|^2-2a|PK'|\cos\alpha\right)-\left(r^2-|KK'|^2\right)}{2a\sin\alpha} \tag{2}\\[4pt]
&= \frac{a^2+k^2\cos^2\kappa-2ak\cos\alpha\cos\kappa-r^2+k^2\sin^2\kappa}{2a\sin\alpha} \tag{3}\\[4pt]
&= \frac{a^2+k^2-2ak\cos\alpha\cos\kappa-r^2}{2a\sin\alpha} \tag{4} \\[4pt]
&= \frac{a^2+k^2-2ak\cos(\alpha+\kappa)-2ak\sin\alpha\sin\kappa-r^2}{2a\sin\alpha} \tag{5}\\[4pt]
&= \frac{|AK|^2-r^2-2ak\sin\alpha\sin\kappa}{2a\sin\alpha} \tag{6}\\[4pt]
&= \frac{|AK|^2-r^2}{2a\sin\alpha}-k\sin\kappa \tag{7}\\[4pt]
&= \frac{|AK|^2-r^2}{2a\sin\alpha}-|KK'| \tag{8} \\[4pt]
\end{align}$$
Therefore, we can write, defining $s := |AK|$,
$$Q = K + (P')^\perp (|KK'|+|K'Q|) = K + \frac{s^2-r^2}{2a\sin\alpha}\,(P')^\perp \tag{9}$$
where $(P')^\perp := (P'_y,-P'_x)$ is perpendicular to $P'$;that is, it's a unit normal to the conic. (We could take either normal. The calculation of $a\sin\alpha$ in $(12)$ ensures that $Q$ is properly offset from $K$.)
Now, to bring the conic back into consideration ... Let us suppose that $P$ lies on an origin-focused conic, with corresponding vertex on the positive $x$-axis, latus rectum $p$, and eccentricity $e$. Then $P$ is parameterized by
$$P = \frac{p}{1+e\cos\theta}\;(\cos\theta,\sin\theta) \tag{10}$$
The unit tangent vector is then
$$P' = \frac1{\sqrt{1+e^2+2e\cos\theta}} (\sin\theta, -(e + \cos\theta)) \tag{11}$$
We also have
$$
a \sin\alpha = (P')^\perp\cdot(A-P) =
\frac{p - A_x(e+\cos\theta) - A_y \sin\theta}{\sqrt{1 + e^2 + 2 e \cos\theta}} \tag{12}
$$
Substituting into $(9)$ we find the locus point $(x,y)=Q$ parameterized as
$$(x,y) = K + \frac{s^2-r^2}{2(p - A_x(e+\cos\theta) - A_y \sin\theta)}(e+\cos\theta,\sin\theta) \tag{13}$$
Separating the components and clearing denominators gives equations that happen to constitute a linear system in $\cos\theta$ and $\sin\theta$; solving and substituting into $\cos^2\theta+\sin^2\theta=1$ gives a second-degree polynomial equation in $x$ and $y$ which necessarily represents a conic.
For completeness, the equation is as follows:
$$\begin{align}
0 &= 4 x^2 (p + A_x (1-e)) (p - A_x(1+e))\\
&+ 4 y^2 ( p^2 - A_y^2(1-e^2)) \\
&-8 x y\, A_y (ep+A_x(1-e^2)) \\
&+4 x(A_x (s^2-r^2+2A_x K_x+2A_y K_y) (1-e^2)
+ ep( s^2 - r^2 + 4 A_x K_x + 2 A_y K_y ) - 2 K_x p^2 ) \\
&+4y ( A_y ( s^2 - r^2 + 2 A_x K_x + 2 A_y K_y ) (1-e^2) + 2 ep A_y K_x - 2 p^2 K_y ) \\
&-(s^2 - r^2 + 2 A_x K_x + 2 A_y K_y)^2 (1 - e^2)
- 4 e p K_x (s^2 - r^2 + 2 A_x K_x + 2 A_y K_y)
+ 4 p^2 (K_x^2 + K_y^2)
\end{align} \tag{$\star$}$$
Defining $m^2=s^2-r^2+2(A_xK_x+A_yK_y)=|OA|^2+|OK|^2-r^2$, we can write this as
$$\begin{align}
0 &= 4 x^2 (p + A_x (1-e)) (p - A_x(1+e)) \\
&+ 4 y^2 ( p^2 - A_y^2(1-e^2)) \\
&-8 x y\, A_y (ep+A_x(1-e^2)) \\
&+4 x(A_x m^2 (1-e^2)
+ ep( m^2 +2A_xK_x) - 2 K_x p^2 ) \\
&+4y ( A_ym^2 (1-e^2) + 2 ep A_y K_x - 2 p^2 K_y ) \\
&-m^4 (1 - e^2)- 4 e p K_x m^2+ 4 p^2 |OK|^2
\end{align} \tag{$\star'$}$$
which isn't much of an improvement, but I don't think the equation is the important thing here.
The discriminant (ignoring a factor of $64p^2$) reduces to $|OA|^2 - (p - e A_x)^2$, which indicates that the nature of the resulting conic (ellipse, parabola, hyperbola) depends only upon $A$'s position relative to the focus.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
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Optimizing a quadratic in one variable with parameterized coefficients Recently I was doing a physics problem and I ended up with this quadratic in the middle of the steps:
$$ 0= X \tan \theta - \frac{g}{2} \frac{ X^2 \sec^2 \theta }{ (110)^2 } - 105$$
I want to find $0 < \theta < \frac{\pi}2$ for which I can later take the largest $X$ value that solves this equation, i.e. optimize the implicit curve to maximize $X$.
I tried solving this by implicit differentiation (assuming $X$ can be written as a function of $\theta$) with respect to $\theta$ and then by setting $\frac{dX}{d\theta} = 0$:
\begin{align}
0 &= X \sec^2 \theta + \frac{ d X}{ d \theta} \tan \theta - \frac{g}{2} \frac{ 2 \left( X \sec \theta \right) \left[ \frac{dX}{d \theta} \sec \theta + X \sec \theta \tan \theta \right]}{ (110)^2 } - 105 \\
0 &= X \sec^2 \theta - \frac{g}{2} \frac{ 2( X \sec \theta) \left[ X \sec \theta \tan \theta\right] }{ (110)^2 } \\
0 &= 1 - \frac{ Xg \tan \theta}{(110)^2} \\
\frac{ (110)^2}{ g \tan \theta} &= X
\end{align}
This is still not an easy equation to solve. However, one of my friends told we could just take the discriminant of the quadratic in terms of $X$, and solve for $\theta$ such that $D=0$.
Taking discriminant and equating to 0, I get
$$ \sin \theta = \frac{ \sqrt{2 \cdot 10 \cdot 105} }{110}$$
and, the angle from it is, 24.45 degrees
I tried the discriminant method, but it gave me a different answer from the implicit differentiation method. I ended up with two solutions with the same maximum value of $X$ but different angles: $\theta =24.45^\text{o}$ and $X=1123.54$ (from discriminant method), and
$\theta = 47^\text{o}$ and $X=1123.54$ (from implicit differentiation).
I later realized the original quadratic can only have solutions if $D(\theta) > 0$, where $D$ is the discriminant. Using the discriminant, I can find a lower bound on the angle. Once I have the lower bound, if I can prove that $X$ decreases monotonically as a function of $\theta$, then I can use the lower bound for further calculations of $\theta$.
So then I used the implicit function theorem and got
$$ \frac{dX}{ d \theta} =- \frac{X \sec^2 \theta -\frac{g X^2}{2 (110)^2} 2 \sec \theta ( \sec \theta \tan \theta) } {\tan \theta - \frac{g \sec^2 \theta}{2 (110)^2} 2X }$$
Now the problem here is that I can't prove this function is in monotonic in terms of $\theta$ as the implicit derivative is a function of both $\theta$ and $X$.
| The trick is to write a quadratic in terms of $ \tan \theta $ and not in terms of $X$
$$ 0 = X \tan \theta -\frac{g}{2} \frac{ X^2 (1 +\tan^2 \theta)}{(110)^2} - 105$$
applying the condition that tangent $ D>0$ for the quadratic in terms of $ \tan \theta$(taking g as 10),
I get, this
$$ 1> \frac{20}{(110)^2} ( 105 + \frac{5X^2}{(110)^2})$$
$$ -1100 < X<1100$$
Taking the upper bound, it becomes $X=1100$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3756436",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Does a closed solution of this integral exist I saw a problem on Facebook with this integral except it was a definite integral from -2 to +2 and the answer was 0 since the function was odd. I am wondering if a closed solution exists or if this can even be integrated, its not an obvious integral as trig. Substitution would not work due to the $x$ term being present in the cosine function and outside. Is there perhaps a way to transform this to something a bit manageable?
$$ \int x^3 \cos \bigg( \frac{x}{2} \bigg) \sqrt{4-x^2}\, dx$$
| Let $x=2\sin u$ ,
Then $dx=2\cos u~du$
$\therefore\int x^3\cos\dfrac{x}{2}\sqrt{4-x^2}~dx$
$=16\int\sin^3u\cos^2u\cos\sin u~du$
$=16\int\sin^3u(1-\sin^2u)\cos\sin u~du$
$=16\int\sin^3u\cos\sin u~du-16\int\sin^5u\cos\sin u~du$
$=16\int\sum\limits_{n=0}^\infty\dfrac{(-1)^n\sin^{2n+3}u}{(2n)!}~du-16\int\sum\limits_{n=0}^\infty\dfrac{(-1)^n\sin^{2n+5}u}{(2n)!}~du$
$=16\int\sum\limits_{n=0}^\infty\dfrac{(-1)^n\sin^{2n+4}u}{(2n)!}~d(\cos u)-16\int\sum\limits_{n=0}^\infty\dfrac{(-1)^n\sin^{2n+2}u}{(2n)!}~d(\cos u)$
$=16\int\sum\limits_{n=0}^\infty\dfrac{(-1)^n(1-\cos^2u)^{n+2}}{(2n)!}~d(\cos u)-16\int\sum\limits_{n=0}^\infty\dfrac{(-1)^n(1-\cos^2u)^{n+1}}{(2n)!}~d(\cos u)$
$=16\int\sum\limits_{n=0}^\infty\sum\limits_{k=0}^{n+2}\dfrac{(-1)^kC_k^{n+2}\cos^{2k}u}{(2n)!}~d(\cos u)+16\int\sum\limits_{n=0}^\infty\sum\limits_{k=0}^{n+1}\dfrac{(-1)^kC_k^{n+1}\cos^{2k}u}{(2n)!}~d(\cos u)$
$=16\sum\limits_{n=0}^\infty\sum\limits_{k=0}^{n+2}\dfrac{(-1)^k(n+2)!\cos^{2k+1}u}{(2n)!k!(n-k+2)!(2k+1)}+16\sum\limits_{n=0}^\infty\sum\limits_{k=0}^{n+1}\dfrac{(-1)^k(n+1)!\cos^{2k+1}u}{(2n)!k!(n-k+1)!(2k+1)}+C$
$=16\sum\limits_{n=0}^\infty\sum\limits_{k=0}^{n+2}\dfrac{(-1)^k(n+2)!\left(1-\dfrac{x^2}{4}\right)^k\sqrt{1-\dfrac{x^2}{4}}}{(2n)!k!(n-k+2)!(2k+1)}+16\sum\limits_{n=0}^\infty\sum\limits_{k=0}^{n+1}\dfrac{(-1)^k(n+1)!\left(1-\dfrac{x^2}{4}\right)^k\sqrt{1-\dfrac{x^2}{4}}}{(2n)!k!(n-k+1)!(2k+1)}+C$
$=8\sum\limits_{n=0}^\infty\sum\limits_{k=0}^{n+2}\dfrac{(-1)^k(n+2)!(4-x^2)^k\sqrt{4-x^2}}{4^k(2n)!k!(n-k+2)!(2k+1)}+8\sum\limits_{n=0}^\infty\sum\limits_{k=0}^{n+1}\dfrac{(-1)^k(n+1)!(4-x^2)^k\sqrt{4-x^2}}{4^k(2n)!k!(n-k+1)!(2k+1)}+C$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3756666",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
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Prime dividing $4n^2+1$ congruent to $1\pmod{4}$ Prove that for any integer $n$, any prime $p$ such that $p\mid 4n^2+1$ is congruent to $1\pmod{4}$.
I am using this as a step to prove that there are infinitely many primes that are congruent to $1\pmod{4}$, but I want to prove this using this method.
| Suppose that $p$ is a prime dividing $4n^2+1$.
Then, if we define $x=2n$:
$x^2 \equiv -1 \pmod{p}$
$\left(x^2\right)^{\frac{p-1}{2}} \equiv (-1)^{\frac{p-1}{2}} \pmod{p} $
$x^{p-1} \equiv (-1)^{\frac{p-1}{2}} \pmod{p}$
$(-1)^{\frac{p-1}{2}} \equiv 1 \pmod{p}\ $ by Fermat's theorem
$(-1)^{\frac{p-1}{2}} = 1$
And, so: $\ p\equiv 1 \pmod{4}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3756927",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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} |
Prove that $\tan^{-1}\frac{\sqrt{1+x^2}+\sqrt{1-x^2}}{\sqrt{1+x^2}-\sqrt{1-x^2}}=\frac{\pi}{4}+\frac 12 \cos^{-1}x^2$ Let the above expression be equal to $\phi$
$$\frac{\tan \phi +1}{\tan \phi-1}=\sqrt{\frac{1+x^2}{1-x^2}}$$
$$\frac{1+\tan^2\phi +2\tan \phi}{1+\tan^2 \phi-2\tan \phi}=\frac{1+x^2}{1-x^2}$$
$$\frac{1+\tan^2\phi}{2\tan \phi }=\frac{1}{x^2}$$
$$\sin 2\phi=x^2$$
$$\phi=\frac{\pi}{4}-\frac 12 \cos^{-1}x^2$$
Where am I going wrong?
| Your mistake
$\sin y=a\stackrel{\text{to}}{\longrightarrow}\sin^{-1}(\sin y)=\begin{cases}2n\pi+y&y\in\text{I, IV quadrant}\\(2n-1)\pi-y&y\in\text{II, III quadrant}\end{cases}=\sin^{-1}a$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3761071",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "12",
"answer_count": 4,
"answer_id": 3
} |
Exercises 6 and 7 in 1.1 (page 4) of Topology and Groupoids, by Ronald Brown I am not sure which keywords to use to search for previously asked questions dealing with these problems. I did search for "Topology and Groupoids" but didn't find anything relevant.
Problems:
*Let $C$ be a neighborhood of $c \in \mathbb{R}$, and let $a + b = c$. Prove that there are neighborhoods $A$ of $a$ and $B$ of $b$ such that $x \in A$ and $y \in B$ imply $x + y \in C$.
*Write down and prove a similar result to that of Exercise 6, but with $c = ab$.
Other information:
The definition of a neighborhood used in the book is (from page 1) that $N$ is a neighborhood of $a$ if there is a $\delta > 0$ such that $(a - \delta, a + \delta) \subseteq N$.
Attempt for Exercise 6:
By assumption there exists a $\delta > 0$ such that $(c - \delta, c + \delta) \subseteq C$. We know that $a + b = c$, so we see that
\begin{align*}
\Big( a - \frac{\delta}{2} \Big) + \Big( b - \frac{\delta}{2} \Big) &= c - \delta\\
\Big( a + \frac{\delta}{2} \Big) + \Big( b + \frac{\delta}{2} \Big) &= c + \delta.
\end{align*}
Therefore if $x \in \big( a - \frac{\delta}{2}, a + \frac{\delta}{2} \big)$ and $y \in \big( b - \frac{\delta}{2}, b + \frac{\delta}{2} \big)$, we see that $x + y \in (a + b - \delta, a + b + \delta) = (c - \delta, c + \delta) \subseteq C$. Let $A = \big( a - \frac{\delta}{2}, a + \frac{\delta}{2} \big)$ and $B = \big( b - \frac{\delta}{2}, b + \frac{\delta}{2} \big)$ and the proof is complete.
Attempt for Exercise 7:
We know there exists a $\delta > 0$ such that $(c - \delta, c + \delta) \subseteq C$. Since $ab = c$, we see that
\begin{align*}
(a - \sqrt{\delta})(b + \sqrt{\delta}) &= ab - \delta,
\end{align*}
so for $0 < \sqrt{\epsilon} < \sqrt{\delta}$, we have
\begin{align*}
ab - \delta < (a - \sqrt{\epsilon})(b + \sqrt{\epsilon}) < ab.
\end{align*}
This suggests that $x \in (a - \sqrt{\delta}, a)$ and $y \in (b, b + \sqrt{\delta})$ will work.
The problem is that I'm suspicious my proposed solution will not work if I let $x$ and $y$ vary independently. For example, it's not clear to me if
\begin{align*}
ab - \delta < \big( a - \frac{\sqrt{\delta}}{2} \big) \big( b + \frac{\sqrt{\delta}}{4} \big) < ab + \delta
\end{align*}
is true or not.
Question:
Can anyone check my proof for Exercise 6, and tell me what's wrong with my attempted proof of Exercise 7? Thank you.
| Let $ n$ enough great such that
$$0<a-\frac 1n<x<a+\frac 1n$$
$$0<b-\frac 1n<y<b+\frac 1n$$
then
$$0<ab -\frac 1n(a+b)+\frac{1}{n^2}<xy<ab+\frac 1n(a+b)+\frac{1}{n^2}$$
and
$$|ab-xy|<\frac{|a+b|}{n} +\frac{1}{n^2}$$
Since $$\lim_{n\to+\infty}\frac{|a+b|}{n}+\frac{1}{n^2}=0$$
we can choose $ n$ such that
$$\frac{|a+b|}{n}+\frac{1}{n^2}<\delta$$
| {
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"url": "https://math.stackexchange.com/questions/3761635",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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} |
How to evaluate $\int x^5 (1+x^2)^{\frac{2}{3}}\ dx$? I am trying to evaluate $\int x^5 (1+x^2)^{\frac{2}{3}} dx$
This is apparently a binomial integral of the form $\int x^m (a+bx^k)^ndx$. Therefore, we can use Euler's substitutions in order to evaluate it. Since $\dfrac{m+1}{k} = \dfrac{5+1}{2} = 3 \in \mathbb{Z}$ we will use the substitution: $$ a+bx^k = u^{\frac{1}{n}}$$
Therefore,
$$ u^3 = 1 + x^2 \iff x = \sqrt{u^3 +1} \text{ (Mistake Here. Check the comments) } $$ $$\iff dx = \frac{3u^2}{2\sqrt{u^3}+1}$$
So the new integral is,
$$ \int x^5 (1+x^2)^{\frac{2}{3}} dx = \frac{3}{2} \int u^4 (u^3+1)^{\frac{7}{6}} du$$
Instead of simplifying the integral, the substitution did nothing by keeping it at the same form, with different values on the variables $m,k,n$.
I tried to substitute once again and it doesn't seem to lead in any known paths, anytime soon.
Any ideas on how this could be evaluated?
| Alternatively, denote $x^2=t\Rightarrow 2xdx=dt$:
$$\int x^5 (1+x^2)^{\frac{2}{3}}dx=\frac12\int t^2 (1+t)^{\frac{2}{3}}dt=\\
\frac12\int (t^2-1+1)(1+t)^{\frac23}dt=\int(t-1)(1+t)^\frac53dt+\int(1+t)^\frac23dt=\\
\frac12\int(t+1-2)(1+t)^{\frac53}dt+\frac12\int(1+t)^{\frac23}dt=\\
\frac12\int(1+t)^{\frac83}dt-\int (1+t)^{\frac53}dt+\frac12\int(1+t)^{\frac23}dt=\\
\frac3{22}(1+t)^{\frac{11}3}-\frac3{8}(1+t)^{\frac{8}3}+\frac3{10}(1+t)^{\frac{5}3}+C$$
Now, plug $t=x^2$ and you are done.
Wolfram alpha answer and comparison.
| {
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"url": "https://math.stackexchange.com/questions/3762071",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Check and comment my proof of $a+b \geq 2 \sqrt{ab}$ I want to prove
$$
a+b \geq 2 \sqrt{ab}, \quad \text{for} \quad a, b>0 \tag 1
$$
First question:
Isn't this wrong? Shouldn't it be
$$
a+b \geq 2 \sqrt{ab}, \quad \text{for} \quad a, b\geq0 \tag 2
$$
Or
$$
a+b > 2 \sqrt{ab}, \quad \text{for} \quad a, b>0 \tag 3
$$
And the proof:
I start backwards.
\begin{align}
a+b-2&\sqrt{ab}\geq 0 \tag 4 \\
(\sqrt a)^2+(\sqrt b)^2-2&\sqrt{ab}\geq 0 \tag 5 \\
(\sqrt a)^2+(\sqrt b)^2-2&\sqrt{a}\cdot \sqrt{b}\geq 0 \tag 6
\end{align}
Let $x= \sqrt a$ and $y=\sqrt b$, so we see that
\begin{align}
x^2+y^2-2xy&\geq 0 \tag 7\\
(x-y)^2\geq 0 \tag 8
\end{align}
So finally we have
$$
(\sqrt a- \sqrt b)^2 \geq 0 \tag 9
$$
If $a=b=0$ we have $0\geq 0$ which is true for $0=0$ ($0>0$ is always false).
For $a>0$ and $b>0$ the square $(\sqrt a- \sqrt b)^2$ is always strictly positive.
And also, $a<0$ and $b<0$ is not valid because $\sqrt a$ and $\sqrt b$ are only defined for $a\geq 0$ and $b\geq 0$.
| No, it's not wrong. It is simply not as strong as it could be. Your (2) is also right, but that doesn't make (1) wrong.
However, your (3) is wrong: if $a=b$, then $a+b=2\sqrt{ab}$.
| {
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"timestamp": "2023-03-29T00:00:00",
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How to evaluate $\int \frac{\sqrt{1+x}+\sqrt{1-x}}{\sqrt{1+x}-\sqrt{1-x}} dx$? I am trying to evaluate
$$\int \frac{\sqrt{1+x}+\sqrt{1-x}}{\sqrt{1+x}-\sqrt{1-x}} dx \quad (1)$$
The typical way to confront this kind of integrals are the conjugates i.e:
$$\int \frac{\sqrt{1+x}+\sqrt{1-x}}{\sqrt{1+x}-\sqrt{1-x}} dx = $$
$$ \int \left(\frac{\sqrt{1+x}+\sqrt{1-x}}{\sqrt{1+x}-\sqrt{1-x}}\right)\left(\frac{\sqrt{1+x}+\sqrt{1-x}}{\sqrt{1+x}+\sqrt{1-x}}\right)\left(\frac{\sqrt{1+x}-\sqrt{1-x}}{\sqrt{1+x}-\sqrt{1-x}}\right) dx = $$
$$\int \left(\frac{(\sqrt{1+x})^2-(\sqrt{1-x})^2}{(\sqrt{1+x})^2-(\sqrt{1-x})^2}\right)\left(\frac{\sqrt{1+x}-\sqrt{1-x}}{\sqrt{1+x}+\sqrt{1-x}}\right)dx = $$
$$\int 1*\left(\frac{\sqrt{1+x}-\sqrt{1-x}}{\sqrt{1+x}+\sqrt{1-x}}\right)dx $$
That's a dead end.
I also tried other conjugate approaches (only the numerator, only the denominator etc) with no better luck.
Any ideas?
| Substitute $x$ with $\cos{2\theta}$.
The above integral will simplify to:
$-2\int \frac{\cos\theta + \sin\theta}{\cos\theta - \sin\theta} \sin{2\theta} \mathrm d\theta $.
Then you can use integration by parts.
Note that: $-\mathrm d(\cos\theta - \sin\theta) = (\cos\theta + \sin\theta) \mathrm d\theta$.
| {
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"timestamp": "2023-03-29T00:00:00",
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Show that $\lim\limits_{N\rightarrow\infty}\sum\limits_{n=1}^N\frac{1}{N+n}=\int\limits_1^2 \frac{dx}{x}=\ln(2)$ Show that:
$$\lim\limits_{N\rightarrow\infty}\sum\limits_{n=1}^N\frac{1}{N+n}=\int\limits_1^2 \frac{dx}{x}=\ln(2)$$
My attempt:
We build a Riemann sum with:
$1=x_0<x_1<...<x_{N-1}<x_N=2$
$x_n:=\frac{n}{N}+1,\,\,\,n\in\mathbb{N}_0$
That gives us:
$$\sum\limits_{n=1}^N(x_n-x_{n-1})\frac{1}{x_n}=\sum\limits_{n=1}^N \left(\frac{n}{N}+1-\left(\frac{n-1}{N}+1\right)\right)\frac{1}{\frac{n}{N}+1}=\sum\limits_{n=1}^N \frac{1}{N}\frac{N}{N+n}=\sum\limits_{n=1}^N\frac{1}{N+n}$$
We know from the definition, that:
$$\lim\limits_{N\rightarrow\infty}\sum\limits_{n=1}^N\frac{1}{N+n}=\lim\limits_{N\rightarrow\infty}\sum\limits_{n=1}^N(x_n-x_{n-1})\frac{1}{x_n}=\int\limits_1^2 \frac{dx}{x}$$
Now we show that,
$$\int\limits_1^2 \frac{dx}{x}=\ln(2)$$
First we choose another Rieman sum with:
$1=x_0<x_1<...<x_{N-1}<x_N=2$
$x_n:=2^{\frac{n}{N}},\,\,\,n\in\mathbb{N}_0$
We get:
$$\sum\limits_{n=1}^N(x_n-x_{n-1})\frac{1}{x_n}=\sum\limits_{n=1}^N\left(2^{\frac{n}{N}}-2^{\frac{n-1}{N}}\right)\frac{1}{2^{\frac{n-1}{N}}}=\sum\limits_{n=1}^N 2^{\frac{1}{N}}-1=N\left(2^{\frac{1}{N}}-1\right)$$
Since we know that (with $x \in \mathbb{R})$:
$$\lim\limits_{x\rightarrow0}\frac{2^x-1}{x}=\ln(2)\Longrightarrow \lim\limits_{x\rightarrow \infty}x(2^{\frac{1}{x}}-1)=\ln(2)\Longrightarrow \lim\limits_{N\rightarrow \infty}N(2^{\frac{1}{N}}-1)=\ln(2)$$
We get:
$$\ln(2)=\lim\limits_{N\rightarrow \infty}N(2^{\frac{1}{N}}-1)=\lim\limits_{N\rightarrow \infty}\sum\limits_{n=1}^N\left(2^{\frac{n}{N}}-2^{\frac{n-1}{N}}\right)\frac{1}{2^{\frac{n-1}{N}}}=\int\limits_1^2 \frac{dx}{x}=\lim\limits_{N\rightarrow\infty}\sum\limits_{n=1}^N\frac{1}{N+n}$$
$\Box$
Hey it would be great, if someone could check my reasoning (if its correct) and give me feedback and tips :)
| Without Rieman sums.
$$S_N=\sum\limits_{n=1}^N\frac{1}{N+n}=H_{2 N}-H_N$$ Using the asymptotics of harmonic numbers
$$S_N=\log (2)-\frac{1}{4 N}+\frac{1}{16 N^2}+O\left(\frac{1}{N^4}\right)$$
| {
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"url": "https://math.stackexchange.com/questions/3766146",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
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Find the area between $f(x) = x^2+3x+7 $ and $g(x) = xe^{x^3+4}$ for $x \in [3,5]$.
Calculate the area between the two functions, $f(x)$, $g(x)$, for $x \in [3,5]$.
$$f(x)=x^2+3x+7$$
$$g(x)=xe^{x^3+4}$$
To determine the area between the functions I used the formula $A= \int_a^b|f(x)-g(x)|dx$. Therefore, I have:
\begin{align}
A&=\int_3^5|x^2+3x+7-xe^{x^3+4}|dx \\
&= \int_3^5|x^2|dx+\int_3^5|3x|dx+\int_3^5|7|dx-\int_3^5|e^{x^3+4}|dx\\
&= \left(\frac{x^3}{3}+\frac{3x^2}{2}+7x\right)_3^5 - \int_3^5|xe^{x^3+4}|dx
\end{align}
Here I can use $u$-subsitution and the Incomplete Gamma Function to find the integral of $xe^{x^3+4}$. All in all I get:
\begin{align}
A&=\left(\frac{x^3}{3}+\frac{3x^2}{2}+7x-e^4\frac{1}{2}\left(-\frac{x^2Γ\left(\frac{1}{\frac{3}{2}},\:-\left(x^2\right)^{\frac{3}{2}}\right)}{\frac{3}{2}\sqrt[\frac{3}{2}]{-\left(x^2\right)^{\frac{3}{2}}}}\right)\right)_3^5\\
&=-\frac{98}{3}-24-14+e^4\frac{1}{2}\left(-\frac{25Γ\left(\frac{1}{\frac{3}{2}},\:-25^{\frac{3}{2}}\right)}{\frac{3}{2}\sqrt[\frac{3}{2}]{-25^{\frac{3}{2}}}}-\left(-\frac{9Γ\left(\frac{1}{\frac{3}{2}},\:-9^{\frac{3}{2}}\right)}{\frac{3}{2}\sqrt[\frac{3}{2}]{-9^{\frac{3}{2}}}}\right)\right)
\end{align}
Can this be simplified further or is the given solution enough?
| $$\int{x\,e^{x^3+4}}\,dx=-\frac{e^4 x^2 \Gamma \left(\frac{2}{3},-x^3\right)}{3
\left(-x^3\right)^{2/3}}=-\frac{1}{3} e^4 x^2 E_{\frac{1}{3}}\left(-x^3\right)$$
$$\int_3 ^5 {x\,e^{x^3+4}}\,dx=\frac{e^4}{3} \left(9 E_{\frac{1}{3}}(-27)-25 E_{\frac{1}{3}}(-125)\right)$$ Since the arguments ar quite large, for an evaluation, you could use the expansion
$$E_{\frac{1}{3}}\left(-x^3\right)=-e^{x^3} \left(\frac{1}{x^3}+\frac{1}{3 x^6}+\frac{4}{9
x^9}+O\left(\frac{1}{x^{12}}\right)\right)$$
| {
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"url": "https://math.stackexchange.com/questions/3767518",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Finding the Last Eigenvalue for a Matrix $K$ is a $3 \times 3$ real symmetric matrix such that $K = K^3$. Furthermore, we are given that:
\begin{align*}
K(1, 1, 1) \ \ & = \ \ (0, 0, 0) \\
K(1, 2, -3) \ \ & = \ \ (1, 2, -3)
\end{align*}
So we know that $0, 1$ are two of the eigenvalues of $K$. What can I do to ascertain that the last eigenvalue is $0, 1$ or something else?
\begin{align*}
\det(K - \lambda I ) \ \ & = \ \ -\lambda(1 - \lambda)(c - \lambda) \\
& = \ \ -\lambda(\lambda - 1)(\lambda - c) \\
& = \ \ -\lambda^3 +(c + 1)\lambda^2 - c \lambda \\ \\
K^3 - K \ \ & = \ \ K(K - 1)(K + 1)
\end{align*}
If I know for sure that $K$ is not idempotent, then I know that $c = -1$:
\begin{align*}
\det (K - \lambda I) - K^3 + K \ \ = \ \ (c+1)K^2 - (c+1)K \ \ = \ \ 0
\end{align*}
But since it was not given explicitly, I don't think this will work.
| Let $w = (1,1,1)\times (1,2,-3) = (-5,4,1)$ and consider
$$Kw = w, \quad Kw = 0, \quad Kw = -w.$$
In all three cases $K$ is symmetric, $K = K^3$ and we get that the last eigenvalue is $1,0$ and $-1$ respectively. The respective matrices of $K$ in basis $\{(1,1,1), (1,2,-3),(-5,4,1)\}$ are
$$\begin{bmatrix} 0 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{bmatrix}, \quad\begin{bmatrix} 0 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 0\end{bmatrix}, \quad\begin{bmatrix} 0 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & -1\end{bmatrix}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3768994",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
} |
Show that if $a$ and $b$ have the same sign then $|a + b| = |a| + |b| $, and if $a$ and $b$ have opposite signs then $|a+b| < |a| + |b|$ I'm considering the different cases for $a$ and $b$
Case 1) $ a\geq 0$ and $ b\geq 0$
Given both terms are positive, $ a + b \geq 0 $
$$
|a+b| = a + b = |a| + |b|\\
$$
Case 2) $ a< 0$ and $ b< 0$
Given both terms are negative, $ a + b < 0 $
$$
|a+b| = -(a + b) = -a -b = |a| + |b|\\
$$
Case 3) $ a> 0$ and $ b< 0$
Case 3a) $ a + b \geq 0 $
Knowing that $ b < -b $ because $ b < 0 < -b $:
$$
|a+b| = a + b < a - b = |a| + |b|
$$
Case 3b) $ a + b \leq 0 $
Knowing that $ -a < a $ because $ -a < 0 < a $:
$$
|a+b| = -(a + b) = -a - b < a - b = |a| + |b|
$$
Case 4) $ a< 0$ and $ b> 0$
Case 4a) $ a + b \geq 0 $
Knowing that $ a < -a $ because $ a < 0 < -a $:
$$
|a+b| = a + b < -a + b = |a| + |b|
$$
Case 4b) $ a + b \leq 0 $
Knowing that $ -b < b $ because $ -b < 0 < b $:
$$
|a+b| = -(a + b) = -a -b < -a + b = |a| + |b|
$$
Is my proof correct?
I feel like it's somewhat incomplete as I'm not considering when either $a$ or $b$ are equal to zero in the cases 3 and 4.
| Dealing with absolute values, the general strategy is: if you can avoid the cases method, do it.
Here , we have a typical example of this strategy: as both expressions are non-negative numbers, we compare them by comparing their squares, using that $|x|^2=x^2\,$:
$$|a+b|^2=a^2+2ab+b^2,\enspace \text{ and }\enspace (|a|+|b|)^2=a^2+2|a||b|+b^2,$$
so that comparing $|a+b|$ and $|a|+|b|$ amounts to comparing $ab$ and $|a||b|=|ab|$, namely
$$|a+b| <|a|+|b|\iff ab<|ab| \iff ab<0,$$
i.e., in ordinary language $a$ and $b$ have different signs.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3769795",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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$f\left( x \right) = {x^3} + x$, then $\int\limits_1^2 {f\left( x \right)dx} + 2\int\limits_1^5 {{f^{ - 1}}\left( {2x} \right)dx} $
If $f\left( x \right) = {x^3} + x$, then
$$\int\limits_1^2 {f\left( x \right)dx} + 2\int\limits_1^5 {{f^{ - 1}}\left( {2x} \right)dx} $$
is________.
My approach is as follows:
$$g = {f^{ - 1}} \Rightarrow g\left( x \right) = {f^{ - 1}}\left( x \right)$$
$$g\left( {2x} \right) = {f^{ - 1}}\left( {2x} \right)$$
$$2y = {8x^3} + 2x$$
$${f^{ - 1}}\left( {{x^3} + x} \right) = x$$
$$\int\limits_1^2 {f\left( x \right)dx} + 2\int\limits_1^5 {{f^{ - 1}}\left( {2x} \right)dx} $$
I am not able to proceed further.
| $$f(x)=x^3+x$$
Let $$K=\int_{1}^{2} f(x) dx+2\int_{1}^{5} f^{-1}(2x) dx=I+J$$
$$ I=\int_{1}^{2} (x^3+x) dx=\frac{21}{4}$$
$$J=2\int_{1}^{5} f^{-1}(2x)dx=\int_{2}^{10} f^{-1}(z) dz$$
Let $f^{-1}(z)=t \implies z=f(t) \implies dz=f'(t) dt$, then
$$J=\int_{1}^{2} t f'(t) dt=\int_{1}^{2}t(3t^2+1)dt=\frac{51}{4}.$$
Finally $$K=21/4+51/4=18.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3771968",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 2
} |
Given ellipse of axes $a$ and $b$, find axes of tangential and concentric ellipse at angle $t$ Let’s say I have an ellipse with horizontal axis $a$ and vertical axis $b$, centered at $(0,0)$.
I want to compute $a’$ and $b’$ of a smaller ellipse centered at $(0,0)$, with the axes rotated by some angle $t$, tangent to the bigger ellipse and $\frac{a’}{b’}=\frac{a}{b}$.
| Let the ratio $r= \frac{a’}a=\frac{b’}b$. Then, the inscribed ellipse with the tilt angle $t$ is
$$\frac{(x\cos t+y\sin t)^2}{r^2a^2}+ \frac{(-x\sin t+y\cos t)^2}{r^2b^2}=1\tag 1
$$
Also, the inscribed ellipse can be expressed as
$$
\frac{x^2}{a^2}+\frac{y^2}{b^2}-k\left(
\frac{x\sin \theta}{a}-\frac{y\cos \theta}{b}\right)^2=1\tag 2$$
as pointed out by @Ng Chung Tak. Match the terms of $x^2$, $y^2$ and $xy$ between (1) and (2) to establish the system of equations below
\begin{align}k\sin 2\theta &=\frac{b^2-a^2}{r^2ab}\sin 2t\\
1-k\sin^2 \theta &= \frac1{r^2b^2}(b^2\cos^2 t+a^2\sin^2t)\\
1-k\cos^2 \theta &= \frac1{r^2a^2}(a^2\cos^2 t+b^2\sin^2t)\\
\end{align}
Plug the 1st equation into the others to eliminate $k$
$$r^2- \frac{b^2\cos^2 t+a^2\sin^2t}{b^2}= \frac{b^2-a^2}{2ab}\sin 2t\tan\theta
$$
$$r^2- \frac{a^2\cos^2 t+b^2\sin^2t}{a^2}= \frac{b^2-a^2}{2ab}\sin 2t\cot\theta
$$
Then, multiply the two equations to eliminate $\theta$. After simplification
$$r^4 -( 2+\Delta^2)r^2+1=0$$
where $\Delta = \left( \frac a b - \frac b a\right)\sin t$. Solve to obtain the axis ratio
$$r= \sqrt{1+\frac{\Delta^2}4}-\frac{\Delta}2$$
Thus, the axes of the smaller ellipse are
$$a’ = ra, \>\>\>\>\>b’= rb$$
As expected, $r=1$ in the special case of $t=0$, and $r=\frac b a$ of $t=\frac\pi2$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3773593",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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EGMO 2014/P3 : Prove that there exist infinitely many positive integers $n$ such that $\omega(n) = k$ and $d(n)$ does not divide $d(a^2+b^2)$ We denote the number of positive divisors of a positive integer $m$ by $d(m)$ and the number of distinct prime divisors of $m$ by $\omega(m)$. Let $k$ be a positive integer. Prove that there exist infinitely many positive integers $n$ such that $\omega(n) = k$ and $d(n)$ does not divide $d(a^2+b^2)$ for any positive integers $a,b$ satisfying $a+b=n$.
My progress: Really beautiful but hard problem!
For $k=1$,we can take $n=2^{p-1}$, where p is an odd prime. Let's say for some $a+b=n$ and write $a=2^ke$ and $b=2^kf$ with $e, f$ odd and $0\le k<p-1$. If $d(n)|d(a^2+b^2)$, then$$p|d \left ( 2^{2k+1}\cdot \dfrac{e^2+f^2}{2} \right )=2^{2k+2}\cdot d\left ( \dfrac{e^2+f^2}{2} \right )$$.
So $p|d\left (\dfrac{e^2+f^2}{2}\right) $ .Now, for $p$ to divide $d\left (\dfrac{e^2+f^2}{2}\right) $, we should have $\left (\dfrac{e^2+f^2}{2}\right)=l^{p-1}\cdot x, l $ is a prime and $gcd(l,x)=1$. But note that both 2 and 3 does not divide $\left (\dfrac{e^2+f^2}{2}\right)$. But Max$(a^2+b^2)=4^{p-1}<5^{p-1}$ .
So we are done for $k=1$ .
I thought that this would be almost same for $k>1$ , but I am not able to prove.
I have conjectured that for any $k$ we can take $n = 2^{p-1}j$ such that $j$ has only $k-1$ primes.
But no progress!Please, if possible post hints rather than solution.
Thanks in advance.
| I couldn't do this solution without @Raheel 's hint. It was all about $5 \pmod 6$ ! Also, I will be really grateful if someone proof reads it?Thanks in advance.
Case 1 : For $k=1$,we can take $n=2^{p-1}$, where p is an odd prime. Let's say for contradiction, there exist some $a$ and $b$ such that $a+b=n$ and $d(n)|d(a^2+b^2)$ . Let $a=2^ke$ and $b=2^kf$ where $e, f$ odd and $0\le k<p-1$. $d(n)|d(a^2+b^2)$, then$$p|d \left ( 2^{2k+1}\cdot \dfrac{e^2+f^2}{2} \right )=2^{2k+2}\cdot d\left ( \dfrac{e^2+f^2}{2} \right )$$.
Since $e,f$ are odd we have $e^2+f^2 \equiv 2\pmod 4$ .
Now, since $0\le k < {p-1} \implies 0\le 2k <2(p-1) \implies 0 \le 2k+2 < 2p$ and also $2k+2 \ne p$ ( as $p$ is odd) , we have, $$p|d\left (\dfrac{e^2+f^2}{2}\right) $$ .Now, for $p$ to divide $d\left (\dfrac{e^2+f^2}{2}\right) $, we should have $\left (\dfrac{e^2+f^2}{2}\right)=l^{p-1}\cdot x, l $ is a prime and $gcd(l,x)=1$.
Now since $3\nmid e$ and $3\nmid f$, by modulo $3$ , we get that $3 \nmid \left (\dfrac{e^2+f^2}{2}\right)$.
But note that both 2 and 3 does not divide $\left (\dfrac{e^2+f^2}{2}\right)$. So we should $\left (\dfrac{e^2+f^2}{2}\right)\ge 5^{p-1}$
But Max$(a^2+b^2)=4^{p-1}<5^{p-1}$ . A contradiction!
So we are done for $k=1$ .
Case 2 : For $k>1$.
Consider $n=2^{p-1}\cdot s$ , where $s \equiv 5 \pmod 6$ and $w(s)=k-1$ .
Now, note that $w(n)=k$ and $d(n)=p\cdot d(s)$.
Let's say for contradiction, there exist some $a$ and $b$ such that $a+b=n=2^{p-1}\cdot s$ and $d(n)|d(a^2+b^2)$.
Using the same reasoning like we did for $k=1$ case , let $a=2^ke$ and $b=2^kf$ where $e, f$ odd and $0\le k<p-1 $ $\implies p|d\left (\dfrac{e^2+f^2}{2}\right) $
Hence, we should have $\left (\dfrac{e^2+f^2}{2}\right)=l^{p-1}\cdot x,l $ is a prime and $gcd(l,x)=1$.
Now, here comes the $5 \pmod 6$ part! Since, both $2$ and $3$ does not divide $\left(\dfrac{e^2+f^2}{2}\right)$ ,and so we should have $\left (\dfrac{e^2+f^2}{2}\right)\ge 5^{p-1}$
But Max$(a^2+b^2)=4^{p-1}<5^{p-1}.$ A contradiction!
And we are done!
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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If $a_n=100a_{n-1}+134$, find least value of n for which $a_n$ is divisible by $99$
Let $a_{1}=24$ and form the sequence $a_{n}, n \geq 2$ by $a_{n}=100 a_{n-1}+134 .$ The first few terms are
$$
24,2534,253534,25353534, \ldots
$$
What is the least value of $n$ for which $a_{n}$ is divisible by $99 ?$
We have to find. $a_n \equiv 0\pmod{99}$
$$ a_n=100a_{n-1}+134 \\ \implies a_n-a_{n-1}\equiv 35 \pmod{99}$$
Now how do I proceed from here?
I Concluded some results, verified for smaller values, which proved to be wrong. How does doing this get me the $n$? Or am I even correct to proceed like this?
The (unofficial) solution isn't very good either:
$a_{3}=253534 \\
a_{4}=25353534 \\
\therefore a_{n}=2 \underbrace{535353 \ldots 53}_{(n-1) \text { Times } 53}4$
Now, $a_n \rightarrow$ divisible by $99 \Rightarrow$ by $\ 9 \ \& \ 11$ both.
Sum of digits $=6+8(n-1)$
To be divisible by 9
$\mathrm{n}=7,16,25,34,43,52,61,70,79,88, \ldots$
$a_{7}=2\underbrace{535353535353 }_{6 \text { Times } 53} 4$
But $a_{7} \rightarrow$ Not divisible by 11 .
$a_{16}=2\underbrace{5353535353 \ldots \ldots 53 }_{15 \text { Times } 53}4$
Similarly, $a_{16} \rightarrow$ Not divisible by 11 .
Now, $n=88$
$a_{88}=2 \underbrace{5353 \ldots \ldots 53}_{87 \text { Times } 53}4$
Divisibility by $11 \rightarrow|(2+3+3 \ldots \ldots)-(5+5+\ldots+ 4)|$
$$
\begin{array}{l}
=|263-439| \\
=176
\end{array}
$$
$\therefore$ Least $n=88$
Hints are more appreciated than the solution.
| Claim 1: a number is divisible by $9$ if and only if the sum of its digits (written in base $10$) is $0\pmod 9$.
Proof: if $n=\sum_{k=0}^md_k10^k$, then since $10^k\equiv 1\pmod 9$, we have $n\equiv \sum_{k=0}^m d_k\pmod 9$. Thus $9$ divides $n$ if and only if $9$ divides $\sum_{k=0}^md_k$.
Claim 2: a number is divisible by $11$ if and only if the alternating sum of its digits is $0\pmod {11}$.
Proof: the same thing, but now $10^k\equiv (-1)^k\pmod{11}$.
Now $a_n$ is divisible by $99$ if and only if it is divisible by $9$ and $11$, and the sequence is designed so we can easily tell what the digits are. Writing $b_n$ for the sum of the digits of $a_n$, we have
$$b_2=7+7, b_3=7+8+7,b_4=7+8+8+7,$$
etc. Mod $9$, each successive term subtracts one. Inductively, one sees that $a_n$ is divisible by nine if and only if $n=9k+7$ for some $k$.
Now let $c_n$ be the alternating sum of the digits. We have
$$c_2=2-5+3-4,c_3=2-5+3-5+3-4,$$
etc. In other words,
$$c_2=-4,c_3=-6,c_4=-8,$$
etc. Inductively, $c_n=-2n$, and $a_n$ is divisible by $11$ if and only if $n=11k$.
All together, $a_n$ is divisible by $99$ if and only if $n\equiv 7\pmod 9$ and $n\equiv 0\pmod {11}$. It is then straightforward to use the Euclidean algorithm to find the set of solutions $\{88+99k:k\in\Bbb Z\}$.
| {
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"timestamp": "2023-03-29T00:00:00",
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Proving a solution of a Bernoulli type equation Prove that
\begin{equation}
y(x) = \sqrt{\dfrac{3x}{2x + 3c}}
\end{equation}
is a solution of
\begin{equation}
\dfrac{dy}{dx} + \dfrac{y}{2x} = -\frac{y^3}{3x}
\end{equation}
All the math to resolve this differential equation is already done. The exercise simply asks to prove the solution.
I start by pointing out that it has the form
\begin{equation}
\dfrac{dy}{dx} + P(x)y = Q(x)y^3
\end{equation}
where
\begin{equation}
P(x) = \dfrac{1}{2x}, \qquad Q(x) = -\frac{1}{3x}
\end{equation}
Rewriting y(x) as
\begin{equation}
y(x) = (3x)^{\frac{1}{2}} (2x + 3c)^{-\frac{1}{2}}
\end{equation}
Getting rid of that square root, I'll need it later on to simplify things
\begin{equation}
[y(x)]^2 = 3x(2x + 3c)^{-1}
\end{equation}
Calculating dy/dx
\begin{align}
\dfrac{dy}{dx} &= \frac{1}{2}(3x)^{-\frac{1}{2}}(3)(2x + 3c)^{-\frac{1}{2}} + \left(-\dfrac{1}{2}\right)(2x + 3c)^{-\frac{3}{2}}(2)(3x)^{\frac{1}{2}} \\
&= \frac{3}{2}(3x)^{-\frac{1}{2}}(2x + 3c)^{-\frac{1}{2}} - (3x)^{\frac{1}{2}}(2x + 3c)^{-\frac{3}{2}} \\
&= (3x)^{\frac{1}{2}}(2x + 3c)^{-\frac{1}{2}} \left[\dfrac{3}{2}(3x)^{-1} - (2x + 3c)^{-1}\right] \\
&= y \left[\dfrac{3}{2}(3x)^{-1} - (2x + 3c)^{-1}\right] \\
&= \dfrac{y}{2x} - y(2x + 3c)^{-1} \\
&= \dfrac{y}{2x} - y\left(\dfrac{y^2}{3x}\right) \\
&= \dfrac{y}{2x} - \dfrac{y^3}{3x}
\end{align}
Finally
\begin{align}
\dfrac{dy}{dx} + P(x)y &= \dfrac{y}{2x} - \dfrac{y^3}{3x} + \dfrac{y}{2x} \\
&= \dfrac{y}{x} - \dfrac{y^3}{3x}
\end{align}
which obviously isn't the same as equation 2. I don't know where I screwed up.
| Your work seems fine and we obtain
$$\dfrac{dy}{dx} =\dfrac{y}{2x} - \dfrac{y^3}{3x}$$
which should be the correct differential equation.
I don't understand why you have added the term $P(x)y$ in the last step.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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If $a^2 + b^2 + c^2 = 1$, what is the the minimum value of $\frac {ab}{c} + \frac {bc}{a} + \frac {ca}{b}$?
Suppose that $a^2 + b^2 + c^2 = 1$ for real positive numbers $a$, $b$, $c$. Find the minimum possible value of $\frac {ab}{c} + \frac {bc}{a} + \frac {ca}{b}$.
So far I've got a minimum of $\sqrt {3}$. Can anyone confirm this? However, I've been having trouble actually proofing that this is the lower bound. Typically, I've solved problems where I need to prove an inequality as true, but this problem is a bit different asking for the minimum of an inequality instead, and I'm not sure how to show that $\sqrt {3}$ is the lower bound of it. Any ideas?
| We have
$$\left(\frac {ab}{c} + \frac {bc}{a} + \frac {ca}{b}\right)^2 = \frac {a^2b^2}{c^2} + \frac {b^2c^2}{a^2} + \frac {c^2a^2}{b^2}+2(a^2+b^2+c^2).$$
Using the AM-GM inequality, we get
$$\frac {a^2b^2}{c^2} + \frac {b^2c^2}{a^2} + \frac {c^2a^2}{b^2} = \frac{1}{2} \sum \left(\frac {a^2b^2}{c^2} + \frac {b^2c^2}{a^2}\right) \geqslant \sum \sqrt{\frac {a^2b^2}{c^2} \cdot \frac {b^2c^2}{a^2}}=a^2+b^2+c^2.$$
Therefore
$$\left(\frac {ab}{c} + \frac {bc}{a} + \frac {ca}{b}\right)^2 \geqslant 3(a^2+b^2+c^2) = 3,$$
or
$$\frac {ab}{c} + \frac {bc}{a} + \frac {ca}{b} \geqslant \sqrt 3.$$
Equality occur when $a=b=c=\frac{1}{\sqrt{3}}.$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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Orthocenter, Circumcenter, and Circumradius In triangle $ABC,$ let $a = BC,$ $b = AC,$ and $c = AB$ be the sides of the triangle. Let $H$ be the orthocenter, and let $O$ and $R$ denote the circumcenter and circumradius, respectively. Express $HO^2$ in terms of $a,$ $b,$ $c,$ and $R.$
I know what orthocenter, circumcenter, and circumradius are, however, I am having trouble expressing it in terms of a, b, c, and R.
| In the standard notation we obtain: $$OH^2=\vec{OH}\cdot\vec{OH}=\left(\vec{OA}+\vec{OB}+\vec{OC}\right)^2=$$
$$=3R^2+2\sum_{cyc}\vec{OA}\cdot\vec{OB}=3R^2+2R^2\sum_{cyc}\cos\measuredangle AOB=$$
$$=3R^2+2R^2\sum_{cyc}\cos2\gamma=3R^2+2R^2\sum_{cyc}\left(2\left(\frac{a^2+b^2-c^2}{2ab}\right)^2-1\right)=$$
$$=R^2\sum_{cyc}\left(\left(\frac{a^2+b^2-c^2}{ab}\right)^2-1\right)=$$
$$=\frac{R^2}{a^2b^2c^2}\sum_{cyc}c^2(a^4+b^4+c^4+a^2b^2-2a^2c^2-2b^2c^2)=$$
$$=\frac{R^2}{a^2b^2c^2}\sum_{cyc}(a^6-a^4b^2-a^4c^2+a^2b^2c^2).$$
Now, we can prove that $$\frac{R^2}{a^2b^2c^2}\sum_{cyc}(a^6-a^4b^2-a^4c^2+a^2b^2c^2)=9R^2-a^2-b^2-c^2.$$
Indeed, we need to probe that:
$$\frac{1}{16S^2}\sum_{cyc}(a^6-a^4b^2-a^4c^2+a^2b^2c^2)=\frac{9a^2b^2c^2}{16S^2}-a^2-b^2-c^2$$ or
$$\sum_{cyc}(a^6-a^4b^2-a^4c^2-2a^2b^2c^2)+\sum_{cyc}(2a^2b^2-a^4)\sum_{cyc}a^2=0$$ or
$$\sum_{cyc}(a^6-a^4b^2-a^4c^2-2a^2b^2c^2)+$$
$$+\sum_{cyc}(2a^4b^2+2a^4c^2+2a^2b^2c^2-a^6-a^4b^2-a^4c^2)=0,$$
which is obvious.
Id est, we got also the following known nice formula:
$$OH^2=9R^2-a^2-b^2-c^2.$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Proving that $\int_{-1}^{1} \frac{\{x^3\}(x^4+1)}{(x^6+1)} dx=\frac{\pi}{3}$, where $\{.\}$ is positive fractional part Here, $\{-3.4\}=0.6$.
The said integral can be solved using $\{z\}+\{-z\}=1$, if $z$ is a non-zero real number;
after using the property that $$\int_{-a}^{a} f(x) dx= \int_{0}^{a} [ f(x)+f(-x)] dx$$
So here $$I=\int_{-1}^{1} \frac{\{x^3\}(x^4+1)}{(x^6+1)} dx=\int_{0}^{1} \frac{[\{x^3\}+\{-x^3\}](x^4+1)}{(x^6+1)} dx =\int_{0}^{1} \frac{(x^4+1)}{(x^6+1)} dx.$$
$$\implies I= \int_{0}^{1} \frac{(1+x^2)^2-2x^2}{(x^6+1)}dx=\int_{0}^{1}\frac{(1+x^2) dx}{x^4-x^2+1}-\int_{0}^{1} \frac{2x^2 dx}{x^6+1}=I_1-I_2.$$
In $I_1$, divide up and down by $x^2$ and use $x-1/x=u$, then
$$I_1=\int_{-\infty}^{0} \frac{du}{1+u^2}=\frac{\pi}{2}$$
Next, use $x^3=v$ $$I_2=\int_{0}^{1} \frac{2x^2 dx}{x^6+1}=\frac{2}{3} \int_{0}^{1} \frac{dv}{1+v^2}=\frac{\pi}{6}$$
Finally $$I=\frac{\pi}{2}-\frac{\pi}{6}=\frac{\pi}{3}$$
It will be interesting to see other approaches/methods of proving this integral.
| $$I=\int\dfrac{x^4+1}{x^6+1}\,dx=\int\dfrac{x^4+1}{\left(x^2+1\right)\left(x^4-x^2+1\right)}\,dx=\int\left(\dfrac{x^2+1}{3\left(x^4-x^2+1\right)}+\dfrac{2}{3\left(x^2+1\right)}\right)dx$$
$$\int\dfrac{x^2+1}{x^4-x^2+1}\,dx=\int\dfrac{x^2+1}{\left(x^2-\sqrt{3}x+1\right)\left(x^2+\sqrt{3}x+1\right)}\,dx=\int\left(\dfrac{1}{2\left(x^2+\sqrt{3}x+1\right)}+\dfrac{1}{2\left(x^2-\sqrt{3}x+1\right)}\right)dx$$ Complete the squares and finish to get
$$I=\dfrac{\arctan\left(2x+\sqrt{3}\right)+\arctan\left(2x-\sqrt{3}\right)+2\arctan\left(x\right)}{3}$$ Combine the arctangents to end with
$$I=\frac{1}{3} \left(\tan ^{-1}\left(\frac{x}{1-x^2}\right)+2 \tan ^{-1}(x)\right)$$
| {
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"source": "stackexchange",
"question_score": "3",
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How to prove that $\frac{\cos(x)-\cos(2x)}{\sin(x)+\sin(2x)} = \frac{1-\cos(x)}{\sin(x)}$ in a simpler way. EDIT: Preferably a LHS = RHS proof, where you work on one side only then yield the other side.
My way is as follows:
Prove: $\frac{\cos(x)-\cos(2x)}{\sin(x)+\sin(2x)} = \frac{1-\cos(x)}{\sin(x)}$
I use the fact that $\cos(2x)=2\cos^2(x)-1, \sin(2x)=2\sin(x)\cos(x)$
(1) LHS = $\frac{\cos(x)-2\cos^2(x)+1}{\sin(x)(1+2\cos(x))}$
(2) Thus it would suffice to simply prove that $\frac{\cos(x)-2\cos^2(x)+1}{1+2\cos(x)}=1-\cos(x)$
(3) Then I just used simple algebra by letting $u = \cos(x)$ then factorising and simplifying.
(4) Since that equals $1-\cos(x)$ then the LHS = $\frac{1-\cos(x)}{\sin(x)} = $ RHS.
Firstly, on the practice exam, we pretty much only had maximum 2-2.5 minutes to prove this, and this took me some trial and error figuring out which double angle formula to use for cos(2x).
This probably took me 5 minutes just experimenting, and on the final exam there is no way I can spend that long.
What is the better way to do this?
EDIT: I also proved it by multiplying the numerator and denominator by $1-\cos(x)$, since I saw it on the RHS. This worked a lot better, but is that a legitimate proof?
| Using compound angle formula, we have
$$
\begin{aligned}
\frac{\cos x-\cos 2 x}{\sin x+\sin 2 x} & =\frac{2 \sin \frac{3 x}{2} \sin \frac{x}{2}}{2 \sin \frac{3 x}{2} \cos \frac{x}{2}} \\
& =\frac{2 \sin ^2 \frac{x}{2}}{2 \sin \frac{x}{2} \cos \frac{x}{2}} \\
& =\frac{1-\cos x}{\sin x}
\end{aligned}
$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Square equal to sum of three squares For which integers $n$ there exists integers $0\le a,b,c < n$ such that $n^2=a^2+b^2+c^2$?
I made the following observations:
*
*For $n=1$ and $n=0$ those integers doesn't exist.
*If $n$ is a power of 2 those integers doesn't exist. Let $n=2^m$ with $m>0$ the smallest power of 2 for which there exists $a,b,c$ such that $\left (2^m\right )^2=4^m=a^2+b^2+c^2$. Since $4^m$ is divisible by 4, $a^2+b^2+c^2$ has to be divisible by 4 too. This is only possible if $a^2\equiv b^2\equiv c^2\equiv 0\pmod 4$, so we can write $a=2a',b=2b',c=2c'$ with $a',b',c'\in \mathbb{N}$. But then we get $\left (2^{m-1}\right )^2=4^{m-1}=a'^2+b'^2+c'^2$, so $m=1$, otherwise $2^m$ wouldn't be the smallest power of two with this property. It is easy to check that $n=2$ doesn't work, so for $n=2^m$ the statement doesn't hold.
*I suspect (but can't prove) that for all other values the statement holds. It would be enough to prove that for all odd primes $p$ there exists $a,b,c$ such that $p^2=a^2+b^2+c^2$, since for all other values of $n$ there exist some $p,m$ such that $n=pm$. Then we get $n^2=(pm)^2=(ma)^2+(mb)^2+(mc)^2$.
| If we want $a^2+b^2+c^2=n^2$, then $a^2+b^2=n^2-c^2=(n-c)(n+c)$.
So a way to generate solutions is to choose $a$ and $b$ and then try to find $n$ and $c$ that work.
Example: $a=10$, $b=11$. So $a^2+b^2=100+121=221$.
Now $221=13*17=(15-2)(15+2)$. So a solution should be $10^2+11^2+2^2=15^2$.
You could also write $221=221=1*221=(111-110)(111+110)=111^2-110^2$. So another solution would be $10^2+11^2+110^2=111^2$.
I will leave it to you to explore this idea further if you wish.
| {
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How to find $a$, $b$, $c$ such that $P(x)=ax^3+bx^2+cx$ and $P\left(x\right)-P\left(x-1\right)=x^2$ I'm trying to find $a$, $b$ and $c$ such that $P(x)=ax^3+bx^2+cx$ and $P\left(x\right)-P\left(x-1\right)=x^2$.
After expanding the binomial in $P(x-1)$, I end up getting
$3ax^2-3ax+2bx+a-b=x^2$. What next? Using $3a = 1$ doesn't work.
| Alternatively. We see that $p(0)=0$.
If $x=1$ we get $$1 = p(1)-p(0) = a+b+c \implies \boxed{a+b+c=1}$$
If $x=2$ we get $$4 = p(2)-p(1) = 7a+3b+c \implies \boxed{7a+3b+c=4}$$
If $x=-1$ we get $$1 = p(0)-p(-1) = a-b+c \implies \boxed{a-b+c=1}$$
Now solve this system...
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Finding the determinant of a $5\times 5$ matrix Let
$$A = \left[\begin{array}{rrrrr}6 & 2 & 2 & 2 & 2 \\ 2 & 6 & 2 & 2 & 2 \\ 2 & 2 & 6 & 2 & 2 \\2 & 2 & 2 & 6 & 2 \\ 2 & 2 & 2 & 2 & 6\end{array}\right] \in {M}_{5}(\mathbb{R})$$
Which of following options is $\det(A)$ ?
*
*$4^4 \times 14$
*$4^3 \times 14$
*$4^2 \times 14$
*$4 \times 14$
I think we have
$$\det \left[\begin{array}{ll}6 & 2 \\ 2 & 6\end{array}\right] = 4 \times 8$$
$$\det \left[\begin{array}{lll}6 & 2 & 2 \\ 2 & 6 & 2 \\ 2 & 2 & 6\end{array}\right] = 4^{2} \times10$$
and for any $n$ we have $\det(A_n)= 4^{n-1} \times (6+2×(n-1))$ so "1" is true.
| $$\begin{vmatrix}6 & 2 & 2 & 2 & 2 \\ 2 & 6 & 2 & 2 & 2 \\ 2 & 2 & 6 & 2 & 2 \\2 & 2 & 2 & 6 & 2 \\ 2 & 2 & 2 & 2 & 6\end{vmatrix}=\begin{vmatrix}4 & 0 & 0 & 0 & -4 \\ 0 & 4 & 0 & 0 & -4 \\ 0 & 0 & 4 & 0 & -4 \\0 & 0 & 0 & 4 & -4 \\ 2 & 2 & 2 & 2 &\ \ 6\end{vmatrix}=\begin{vmatrix}4 & 0 & 0 & 0 & -4 \\ 0 & 4 & 0 & 0 & -4 \\ 0 & 0 & 4 & 0 & -4 \\0 & 0 & 0 & 4 & -4 \\ 0 & 0 & 0 & 0 & 14\end{vmatrix}$$
| {
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"timestamp": "2023-03-29T00:00:00",
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A Problem on Inverse Trigonometric Functions and Differential Equation.
$f(x)=\left(\sin \left(\tan ^{-1} x\right)+\sin \left(\cot ^{-1} x\right)\right)^{2}-1,\ |x|>1$
If $\displaystyle\frac{\mathrm{d} y}{\mathrm{d} x}=\frac{1}{2} \frac{\mathrm{d}}{\mathrm{d} x}\left(\sin ^{-1}(f(x))\right)$ and
$y(\sqrt{3})=\frac{\pi}{6}$, then $y(-\sqrt{3})$ is equal to :
Options:
$1. \quad-\frac{\pi}{6}\\
2. \qquad \frac{2 \pi}{3}\\
3. \qquad \frac{5 \pi}{6}\\
4. \qquad\frac{\pi}{3}$
It can be easily shown that $f(x) = [\sin(\tan^{-1}x) + \sin(\cot^{-1}x )]^2 -1 = \frac{2x}{1+x^2}$.
$\frac{d}{dx} (\sin^{-1}\frac{2x}{1+x^2}) = \frac{-2}{1+x^2}$
Now $\frac{dy}{dx} = \frac{1}{2} \frac{ d(\sin^{-1}f(x))}{ dx} = \frac{-1}{1+x^2}$ from which we get $y = - \tan^{-1}x + C. $
Now $y (\sqrt 3) = \frac{\pi}{6}$. So $C= \frac{\pi}{2}$.
So $y = \frac{\pi}{2} - \tan^{-1}x $.
$y(-\sqrt 3) = \frac{5\pi}{6}$
Have I gone wrong any where? Can anyone please check my solution?
| $$\dfrac{d}{dx}(\sin^{-1}f)= \dfrac{1}{\sqrt{1-f^2}}\cdot \dfrac{df}{dx}=P\cdot Q \;;$$
$$P=\dfrac{1}{\sqrt{1-(2x/(1+x^2))^2}} = \dfrac{1+x^2}{1-x^2} $$
Differentiating by Quotient rule
$$Q=\dfrac{2}{(1+x)^2}; \quad PQ=\dfrac{2}{1-x^2};\; $$
EDIT1:
Stopped too early by mistake. Continuing,
$$\dfrac{dy}{dx}= \dfrac{1}{1-x^2}$$
Integrates to
$$y=\log_e\sqrt {\dfrac{1+x}{1-x}} +c$$
given
$$ y(\sqrt3)=\dfrac{\pi}{6}$$
$y$ is not real, constant of integration cannot be evaluated. For this result no option available.
| {
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"timestamp": "2023-03-29T00:00:00",
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On proving $a^3+b^3+c^3-3abc \geq 2\left({b+c\over 2}-a\right)^3$. This problem was a "warm-up" problem by the author. Note: $a, b, c$ are non-negative numbers.
$$a^3+b^3+c^3-3abc \geq 2\left({b+c\over 2}-a\right)^3$$
I tried to remove the $2$ from ${b+c\over 2}$ and got this-
$$ 4(a^3+b^3+c^3-3abc) \geq (b+c-2a)^3 $$
$$ \Rightarrow 2(a+b+c)((a-b)^2+(b-c)^2+(c-a)^2) \geq (b+c-2a)^3 $$
But couldn't take it any further, but it looks as if Hölder's Inequality may help. And also I don't think that it is a "warm-up" problem.
Any help will be appreciated.
| Because $$a^3+b^3+c^3-3abc-2\left(\frac{b+c}{2}-a\right)^3$$
$$=\frac{3}{4}\Big[4a^3-4(b+c)a^2+2(b^2+c^2)a+b^3-b^2c-bc^2+c^3\Big]$$
$$=\frac{3}{4}\Big[a(2a-b-c)^2+a(b-c)^2+(b+c)(b-c)^2\Big]\geqslant 0.$$
The equality occurs for $$a(2a-b-c)^2=a(b-c)^2=(b+c)(b-c)^2=0.$$
*
*$a=0$.
Thus, $(b+c)(b-c)^2=0,$ which gives $b=c$.
*$a\neq0$.
Thus, since $a(b-c)^2=0,$ we obtain $b=c$ and since $a(2a-b-c)^2=0,$
we obtain $a=b=c$.
| {
"language": "en",
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"source": "stackexchange",
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Given $\triangle ABC$ can we construct point $O$ such that $AO\times BC=BO\times AC=CO\times AB$? Given a $\triangle ABC$, is it possible to construct, with compass and straightedge, a point $O$ such that
$$AO\cdot BC=BO\cdot AC=CO\cdot AB$$
Does that point exist?
| These points are known
in ETC
as 1st and 2nd isodynamic points,
the triangle centers $X_{15}$ and $X_{16}$.
Isodynamic point:
In Euclidean geometry, the isodynamic points of a triangle are points
associated with the triangle, with the properties that ... the
distances from the isodynamic point to the triangle vertices are
inversely proportional to the opposite side lengths of the triangle.
The barycentric coordinates of these points are
\begin{align}
X_{15}:\quad&
a\sin(\alpha + \tfrac\pi3) &: b\sin(\beta + \tfrac\pi3) : c \sin(\gamma + \tfrac\pi3)
,\\
X_{16}:\quad&
a\sin(\alpha - \tfrac\pi3) &: b\sin(\beta - \tfrac\pi3) : c \sin(\gamma - \tfrac\pi3)
.
\end{align}
As a linear combination of the vertices,
\begin{align}
X_{15}&=\frac{u\cdot A+v\cdot B+w\cdot C}{u+v+w}
,\\
u&=a\sin(\alpha + \tfrac\pi3)
,\\
v&=b\sin(\beta + \tfrac\pi3)
,\\
w&=c \sin(\gamma + \tfrac\pi3)
,
\end{align}
and the invariants are
\begin{align}
a\cdot |AX_{15}|&=b\cdot |BX_{15}|=c\cdot |CX_{15}|
\\
&=
\frac{\sqrt2\,abc}{\sqrt{a^2+b^2+c^2+4\sqrt3 S}}
,\\
a\cdot |AX_{16}|&=b\cdot |BX_{16}|=c\cdot |CX_{16}|
\\
&=
\frac{\sqrt2\,abc}{\sqrt{a^2+b^2+c^2-4\sqrt3 S}}
,
\end{align}
where $S$ is the area of $\triangle ABC$.
Example: for the the nominal $6-9-13$ triangle,
\begin{align}
a&=6,\quad b=9,\quad c=13,\quad S=4\sqrt{35}
,\\
a\cdot |AX_{15}|&=b\cdot |BX_{15}|=c\cdot |CX_{15}|
=\frac{702 \sqrt2}{\sqrt{286+16\sqrt{105}}}
\approx 46.80
,\\
a\cdot |AX_{16}|&=b\cdot |BX_{16}|=c\cdot |CX_{16}|
=\frac{702 \sqrt2}{\sqrt{286-16\sqrt{105}}}
\approx 89.86
.
\end{align}
Construction.
Points $A_b.A_e$ and $B_b,B_e$
are the feet of the internal and external bisectors
of the angles $CAB=\alpha$
and $ABC=\beta$, respectively.
Points
$O_a=\tfrac12(A_b+A_e)$,
$O_b=\tfrac12(B_b+B_e)$
are the centers of the circles
$\mathcal{C_a}$ and $\mathcal{C_b}$
through points
$A,A_b,A_e$
and $B,B_b,B_e$, respectively.
Intersection of the circles
$\mathcal{C_a}$ and $\mathcal{C_b}$
gives the pair of isodynamic points,
1st, $X_{15}$ inside $\triangle ABC$
and 2nd, $X_{16}$, outside of $\triangle ABC$.
| {
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "5",
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Express the value $z$ below in polar form, and the value $w$ in the form $a+bi$. I have been having a lot of issues on determining how to work through problems of the sort and I would be very grateful if somebody could provide me with a guided/ explained answer to enable me to understand how to work through these.
Given:
Picture of Question
Express the value $z$ below in polar form, and the value $w$ in the form $a+bi$. Use the square root symbol $\sqrt{\ \ }$ where needed to give an exact value for your answer. Be sure to include parentheses where necessary, e.g. to distinguish $\frac{1}{2k}$ from $\frac{1}{2}k$. .
| Just do it. If $z = a + bi$ than $|z| =\sqrt{a^2 + b^2}$ and
$z = \sqrt{a^2+ b^2}(\frac a{a^2 + b^2} + \frac b{a^2 + b^2} i)$. Let $r= \sqrt{a^2 + b^2} = |z|$.
Now $\frac a{a^2 + b^2}=x$ and $\frac b{a^2 + b^2}=y$ are two numbers such that $x^2 + y^2 = 1$. That means that there must be some angle $\theta$ so that $x = \cos \theta$ and $y = \sin \theta$. What angle can that be?
Well, $\tan \theta = \frac {\sin \theta}{\cos \theta} = \frac yx = \frac {\frac b{a^2 + b^2}}{\frac a{a^2 + b^2}}= \frac ba$. So $\theta = \arctan \frac ba$.
so $z =a+bi = \sqrt{a^2 + b^2}(\cos \arctan \frac ba + \sin \arctan \frac ba i)$.
Now by definition $e^{\psi i} = \cos \psi + i\sin \psi$ so
$z = \sqrt{a^2 + b^2}e^{\arctan \frac ba i}$.
So for $z = a+bi; z\ne 0$ we can always convert it to $z= re^{\theta i}$ form by letting $r = |z| =\sqrt{a^2 + b^2}$ and letting $\theta = \arctan \frac ba$. Always.
....
And converting $z= re^{\theta i}$ to $a + bi$ form is simply noting:
$z = re^{\theta i} = r(\cos \theta + i \sin\theta) =r\cos\theta + r\sin \theta i$ .
So $a = r\cos \theta$ and $b = r\sin \theta$.
That's all there is to it.
so just do them.
If $z = \frac 52 - \frac {5\sqrt 3}2i$ then $r = \sqrt {(\frac 52)^2 + (\frac {5\sqrt 3}2)^2}=???$ and $\theta = \arctan \frac {\frac {5\sqrt 3}2}{\frac 52}=\arctan \frac {5\sqrt 3}{5}=\arctan \sqrt 3 = ????$ and $z = re^{\theta i}$. That's all.
And if $w = 2e^{\frac {i5\pi}4}$ then $w = 2\cos (\frac {5\pi}4) +2\sin(\frac {5\pi}4)i$.
That's all...
..... but.... wait a minute. The problem you posted has them both equal to $0$????? That makes utterly zero sense.
| {
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Let $x, y, z$ be positive distinct integers. Prove that $(x+y+z)(xy+yz+zx-2)\ge9xyz$ I just came across the following question, in a book, which has as its topic contest-math:
Let $x, y, z$ be positive distinct integers. Prove that $(x+y+z)(xy+yz+zx-2)\ge9xyz$
I solved the question, in the following way:
Without loss of generality, from symmetry, we have that $x<y<z$.
However, $xy+yz+xz\ge\sqrt[3]{x^2y^2z^2}$ (AM-GM)
The equality holds if $x=y=z$. Impossible.
So, the following holds:
$xy+yz+xz\ge3*\sqrt[3]{x^2y^2z^2}+2$, since $y\ge x+1$ and $z\ge x+2$, so $y-1\ge x$ and $z-2\ge x$ and hence only this way can we have an $\ge$ in the inequality. (Which is false as proven in the comments)
We also have that $x+y+z\ge3*\sqrt[3]{xyz}$
So the inequality $(x+y+z)(xy+yz+xz-2)\ge3*\sqrt[3]{xyz}*3\sqrt[3]{x^2y^2z^2}=9xyz$, holds true.
I am not to certain about my proof and I also believe that there must exist far simpler solutions. Could you please verify that what I've written is indeed correct and show me some alternative methods?
| Let $x<y<z$, $y=x+1+a$ and $z=x+2+a+b$,
where $a$ and $b$ are non-negative integers.
Thus, we need to prove that
$$(x+y+z)(xy+xz+yz)-9xyz\geq2(x+y+z)$$ or $$\sum_{cyc}z(x-y)^2\geq2(x+y+z)$$ or
$$(x+2+a+b)(1+a)^2+(x+1+a)(2+a+b)^2+x(b+1)^2\geq2(3x+3+2a+b).$$
Can you end it now?
| {
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How to prove $\frac{a^{n+1}+b^{n+1}+c^{n+1}}{a^n+b^n+c^n} \ge \sqrt[3]{abc}$? Give $a,b,c>0$. Prove that: $$\dfrac{a^{n+1}+b^{n+1}+c^{n+1}}{a^n+b^n+c^n} \ge \sqrt[3]{abc}.$$
My direction: (we have the equation if and only if $a=b=c$)
$a^{n+1}+a^nb+a^nc \ge 3a^n\sqrt[3]{abc}$
$b^{n+1}+b^na+b^nc \ge 3b^n\sqrt[3]{abc}$
$c^{n+1}+c^na+c^nb \ge 3c^n\sqrt[3]{abc}$
But from these things, i can't prove the problem.
| Because for natural $n$ by AM-GM we obtain: $$\begin{aligned} \sum_{\text{cyc}}a^{n+1}&=\frac{1}{3(n+1)}\sum_{\text{cyc}}\left((3n+1)a^{n+1}+b^{n+1}+c^{n+1}\right) \\
&\geq \sum_{\text{cyc}}\sqrt[3n+3]{a^{(3n+1)(n+1)}b^{n+1}c^{b+1}} \\
&=\sum_{\text{cyc}}a^{n+\frac{1}{3}}b^{\frac{1}{3}}c^{\frac{1}{3}} \\
&=\sqrt[3]{abc}\sum_{\text{cyc}}a^n \end{aligned}$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Partial reciprocal sum How can I show that $$\sum_{k=1}^{n}\frac{1}{n+k}\leq\frac{3}{4}$$ for every integer $n \geq 1$?
I tried induction, estimates with logarithms and trying to bound the sum focusing on the larger terms or things like $\frac{1}{n+1}+\frac{1}{n+2}\leq\frac{2}{n+1}$ but nothing seems to work. Do you have any suggestion? Thanks
| Here is an elementary proof. Let $$ u_n = \sum_{k=1}^n \frac{1}{n+k}$$
For all $n \geq 1$, we have
$$u_{n+1}-u_n = \sum_{k=1}^{n+1} \frac{1}{n+1+k} - \sum_{k=1}^n \frac{1}{n+k}$$ $$= \sum_{k=2}^{n+2} \frac{1}{n+k} - \sum_{k=1}^n \frac{1}{n+k} = \frac{1}{2n+2} + \frac{1}{2n+1} - \frac{1}{n+1}$$ $$ = \frac{1}{2n+1}-\frac{1}{2n+2} = \frac{1}{(2n+1)(2n+2)} > 0.$$
So the sequence $(u_n)_{n \geq 1}$ is strictly increasing.
Moreover, you have
$$u_n = \frac{1}{n} \sum_{k=1}^n \frac{1}{1 + \frac{k}{n}} \longrightarrow \int_0^1 \frac{1}{1+x} \mathrm{dx} = \ln(2)$$
So the sequence is increasing and converges to $\ln(2)$, so you have $u_n \leq \ln(2)$ for all $n \geq 1$. You deduce
$$u_n \leq \frac{3}{4}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3788424",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 5,
"answer_id": 1
} |
For $a>1$, show that $\frac{1}{1+x}-\frac{1}{1+ax} \leq \frac{\sqrt{a}-1}{\sqrt{a}+1}$, $x \geq 1$ I'm self-learning the analysis I "by Herbert Amann" and got stuck in this problem. It's in Chapter IV Taylor's theorem.
For $a>1$ and $x\geq 1$ show that $$\frac{1}{1+x}-\frac{1}{1+ax} \leq \frac{\sqrt{a}-1}{\sqrt{a}+1}$$
This is what I've tried:
Let $f(t)=\frac{1}{1+tx}$ which is convex and,
$f(a) \geq f(1)+f'(1)(a-1)$
so $\frac{1}{1+x}-\frac{1}{1+ax} \leq \frac{x}{(1+x)^2}(a-1) \leq \frac{a-1}{4}$
| We need to prove that $$\frac{ax-x}{(1+ax)(1+x)}\leq\frac{\sqrt{a}-1}{\sqrt{a}+1}$$ or
$$\frac{x(\sqrt{a}+1)}{(1+ax)(x+1)}\leq\frac{1}{\sqrt{a}+1},$$ which is true by C-S:
$$\frac{x(\sqrt{a}+1)}{(1+ax)(x+1)}\leq\frac{x(\sqrt{a}+1)}{(\sqrt{x}+\sqrt{ax})^2}=\frac{1}{\sqrt{a}+1}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3789967",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
If $x$, $y$ and $z$ satisfy $xy=1$, $yz=2$ and $xz=3$, what is the value of $x^2+y^2+z^2$? Express your answer as a common fraction.
If $x$, $y$ and $z$ satisfy $xy=1$, $yz=2$ and $xz=3$, what is the value of $x^2+y^2+z^2$? Express your answer as a common fraction.
I tried going along the path of computing $(x+y+z)^2$, which expands to $(x^2+y^2+z^2) + 2\cdot (xy+yz+xz)$, but I couldn't go anywhere from there (other than substituting xy, yz, xz, but that's not enough information.
| $$\text{Compilation of all *Hints*}$$
We know that $$(x+y+z)=xyz(\frac{1}{xy}+\frac{1}{yz}+\frac{1}{zx})$$
From here we see that $$(x+y+z)=xyz(\frac{11}{6})$$
Also, $(xyz)^2=6$ , so it would be nice to say that
$$(x+y+z)^2=6\big(\frac{11}{6}\big)^2$$
Then write the expansion for $$(x+y+z)^2$$ and keep substituting till you get the answer.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3790065",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 3
} |
Why $8^{\frac{1}{3}}$ is $1$, $\frac{2\pi}{3}$, and $\frac{4\pi}{3}$ The question is:
Use DeMoivre’s theorem to find $8^{\frac{1}{3}}$. Express your answer in complex form.
Select one:
a. 2
b. 2, 2 cis (2$\pi$/3), 2 cis (4$\pi$/3)
c. 2, 2 cis ($\pi$/3)
d. 2 cis ($\pi$/3), 2 cis ($\pi$/3)
e. None of these
I think that $8^{\frac{1}{3}}$ is $(8+i0)^{\frac{1}{3}}$
And, $r = 8$
And, $8\cos \theta = 8$ and $\theta = 0$.
So, $8^{\frac{1}{3}}\operatorname{cis} 0^\circ = 2\times (1+0)=2$
I just got only $2$. Where and how others $\frac{2\pi}{3}$, and $\frac{4\pi}{3}$ come from?
| We could look at it like this:
$$8^{\frac13}=2.1^{\frac13}=2\cdot \text{CiS}\left(\frac{2k\pi}{n}\right)$$
Now for different values of $k$, we have different answers: (here $n$ is $3$)
$$k=1\implies 8^{\frac13}=2\cdot\text{CiS} \left(\frac{2\pi}{3}\right)$$
$$k=2\implies8^{\frac13}=2\cdot\text{CiS}\left(\frac{4\pi}{3}\right)$$
$$k=3\implies8^{\frac13}=2\cdot\text{CiS}(2\pi)=2$$
You could read up on $n^{\text{th}}$ roots of unity on Wikipedia to get a better picture
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3791438",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 0
} |
Finding $\lim_{x \to \infty} (x + \frac{2x^{3}}{3} - \frac{2(x^2+1)^{\frac{3}{2}}}{3})$ $\lim_{x \to \infty} (x + \frac{2x^{3}}{3} - \frac{2(x^2+1)^{\frac{3}{2}}}{3})$ this limit according to wolframalpha is equal to $0$.
So this is my work thus far
$\lim_{x \to \infty} (x + \frac{2x^{3}}{3} - \frac{2(x^2+1)^{\frac{3}{2}}}{3})$ output is $\infty - \infty$ which is indeterminate form.
So next I basically but it on the same denominator: $\frac{1}{3}$ $((3x + 2x^3 - 2(x^2+1)^{\frac{3}{2}})$ and turned $2(x^2+1)^{\frac{3}{2}}$ into something easier to work with $2\sqrt{x^2+1}+2x^{2}\sqrt{x^2+1}$
now the limit is $\frac{1}{3} \lim_{x \to \infty} ((3x + 2x^3-2\sqrt{x^2+1} -2x^{2}\sqrt{x^2+1})$ and this is where I am stuck to do next and lost.
| Hint:
WLOG $x=\tan y\implies y\to\dfrac\pi2$
$$\dfrac{3\tan y+2\tan^3y-2\sec^3y}3$$
$$=\dfrac{3\sin y\cos^2y+2\sin^3y-2}{3\cos^3y}$$
The numerator $$=3(1-\sin^2y)\sin y+2\sin^3y-2=\cdots=(1-\sin y)^2(2\sin y+1)$$
Finally use $$\dfrac{1-\sin y}{\cos y}=\dfrac{\cos y}{1+\sin y}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3794325",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 6,
"answer_id": 3
} |
Directional derivative and gradient of a differentiable function Let $u = f(x,y,z)$ be a differentiable function in $\mathbb{R^3}$
Given the function satisfies: $f(x,y,x^2 + y^2) = 2x+y$ for all $x,y$
And the directional derivative of the point $(0,2,4)$ in the direction: $(-2,1,2)$ is equal to $-\frac{5}{3}$.
Calculate: $\nabla f(0,2,4)$
My try so far:
I first normalized the vector: $\frac{(-2,1,2)}{||(-2,1,2)||} = (-\frac{2}{3}, \frac{1}{3}, \frac{2}{3})$
we know that $f(0,2,4) = f(x,y,x^2 + y^2)$ because $x=0, y=2 , x^2+y^2 = 4$ and so:
$f(0,2,4) = 2 \cdot 0 + 2 = 2$.
By the definition: $\frac{\partial f}{\partial (-2,1,2)}(0,2,4) = -\frac{5}{3} \Rightarrow \nabla f(0,2,4) \cdot (-\frac{2}{3}, \frac{1}{3}, \frac{2}{3}) = -\frac{5}{3}$
Even by the definition I am stuck:
$\nabla f(0,2,4) = \lim_{h \rightarrow 0} \frac{f( (0,2,4) + h(-2,1,2)) - f(0,2,4)}{h} = \lim_{h \rightarrow 0} \frac{f(-2h,2+h, 4+2h) - 2}{h}$
Because the point $(-2h,2+h, 4+2h)$ does not satisfy that $x^2 + y^2 = z$ (coordinates) ...
The problem is that I don't know how to find the gradient of that point, because the function is not given in its explicit form.. I would appreciate your help, thank you!
| For simplicity, I will denote $\nabla f(0,2,4) = (f_x(0,2,4), f_y(0,2,4) , f_z(0,2,4)) = (a,b,c)$
You need to use the chain rule on this known fact $f(x,y,x^2+y^2) = 2x+y$:
$$\frac{ \partial }{\partial x}f(x,y,x^2+y^2) = f_x(x,y,x^2+y^2) \cdot \frac{\partial}{\partial x}x + f_y(x,y,x^2+y^2) \cdot \frac{\partial}{\partial x}y + f_z(x,y,x^2+y^2) \cdot \frac{\partial}{\partial x}(x^2 + y^2) = \overbrace{f_x(x,y,x^2+y^2) +2xf_z(x,y,x^2+y^2)}^{\text{left side}} = \overbrace{\frac{ \partial }{ \partial x}(2x+y)}^{\text{right side}} = 2$$
You want to find $\nabla f(0,2,4)$ and thus we can substitute:
$$f_x(0,2,4) + 0 = a = 2$$
$2x \cdot f_z(0,2,4) = 0$ because it is being multiplied by the $x$ value - which is $0$.
Same thing for $\frac{\partial}{\partial y}$:
$$\frac{ \partial }{\partial y}f(x,y,x^2+y^2) = f_x(x,y,x^2+y^2) \cdot \frac{\partial}{\partial y}x + f_y(x,y,x^2+y^2) \cdot \frac{\partial}{\partial y}y + f_z(x,y,x^2+y^2) \cdot \frac{\partial}{\partial y}(x^2 + y^2) = f_y(x,y,x^2+y^2) +2yf_z(x,y,x^2+y^2) = \frac{ \partial }{ \partial y}(2x+y) = 1$$
And so keep in mind this equation (1): $$ \fbox{b+4c =1}$$
Now we can use the fact that the directional derivative at $(-2,1,2)$ at the point $(0,2,4)$ is $-\frac{5}{3}$ by definition, the directional derivative for this differentiable function at the given point and direction is:
$$ \nabla(0,2,4) \cdot ( -\frac{2}{3}, \frac{1}{3}, \frac{2}{3}) = - \frac{5}{3}$$
Recall that $\nabla f(0,2,4) = (a,b,c)$:
$$ - \frac{2a}{3} + \frac{b}{3} + \frac{2c}{3} = - \frac{5}{3}$$
$$ -2a +b +2c = -5$$
We found that $a=2$:
$$ \fbox{ b + 2c = -1}$$
Now we have a set of two equations with two unknowns, this should be fast forward:
$$ \left\{\begin{matrix}
b+4c =1\\ b + 2c = -1
\end{matrix}\right.$$
We get that $$a=2, ~~~~ b = -3 , ~~~~ c = 1$$
And thus the answer is:
$$ \nabla f(0,2,4) = (2,-3,1)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3800263",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
} |
Solving $\sqrt[3]{x+1} - \sqrt[3]{x-1} = \sqrt[3]{x^2-1}$ for real $x$
Solve the equation in the Real number system:
$$\sqrt[3]{x+1} - \sqrt[3]{x-1} = \sqrt[3]{x^2-1}$$
I have attempted using $(A-B)^3 = A^3 - B^3 - 3.A.B.(A-B)$ with $A = \sqrt[3]{x+1}$ , $B = \sqrt[3]{x-1}$ and $(A-B) = \sqrt[3]{x^2-1}$, however I end up getting $-3.(\sqrt[3]{x^2-1})^2 = x^2-3$ which is not equivalent to the original one since 0 is solution only of $-3.(\sqrt[3]{x^2-1})^2 = x^2-3$. Could someone explain to me why does it not work here?
By the way, the answer according to the book is $\left\lbrace \frac{\sqrt 5}{2},-\frac{\sqrt 5}{2} \right\rbrace$ but I could not get there.
| Put $a=\sqrt[3]{x+1}$ and $b= - \sqrt[3]{x-1}$.
Then $a+b=-ab$, hence $a$ satisfy $a^2+pa+p=0$ where $p=ab$ from which
$$a=\frac{-p\pm\sqrt{p^2-4p}}2$$
Consequently,
\begin{align}
x
&=a^3-1\\
&=-pa(a+1)-1\\
&=-p(-pa-p+a)-1\\
&=(p-1)\frac{-(p^2-2p-2)\pm p\sqrt{p^2-4p}}2
\end{align}
On the other hand, we have $a^3+b^3=2$.
Cubing $a+b=-ab$ gives $a^3+b^3+3ab(a+b)=-a^3b^3$ from which $2-3a^2b^2=-a^3b^3$.
If $p=ab$, then
$$0=p^3-3p^2+2=(p-1)(p^2-2p-2)$$
from which $p=1$ and $p=1\pm\sqrt 3$.
Excluding $p=1$ (which gives $x=0$) we get
$$x=\pm\frac{(p+2)\sqrt 2\sqrt{1-p}}2$$
hence $p=1-\sqrt 3$ and
$$x=\pm\frac{(3-\sqrt 3)\sqrt 2\sqrt[4]3}2$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3800380",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Find the argument of $z = {\left( {2 + i} \right)^{3i}}$ $z = {\left( {2 + i} \right)^{3i}}$
My approach is as follow
$z = {\left( {2 + i} \right)^{3i}} = {\left( {{{\left( {2 + i} \right)}^3}} \right)^i} = {\left( {8 + {i^3} + 12i - 6} \right)^i} = {\left( {2 + 11i} \right)^i}$
$\ln z = i\ln \left( {2 + 11i} \right)$
$2 + 11i = r\cos \theta + ir\sin \theta \Rightarrow r = 5\sqrt 5 ;\theta = {\tan ^{ - 1}}\frac{{11}}{2}$
$ \Rightarrow \ln z = i\ln \left( {2 + 11i} \right) = i\ln \left( {r{e^{i\theta }}} \right) = i\left( {\ln \left( r \right) + \ln \left( {{e^{i\theta }}} \right)} \right) = i\left( {\ln 5\sqrt 5 + i\theta } \right) = - \theta + i\left( {\ln 5\sqrt 5 } \right)$
$ \Rightarrow z = {e^{\left( { - \theta + i\left( {\ln 5\sqrt 5 } \right)} \right)}} = {e^{ - \theta }}{e^{i\left( {\ln 5\sqrt 5 } \right)}} \Rightarrow r{e^{i\phi }}$
The modulus is $r = {e^{ - {{\tan }^{ - 1}}\frac{{11}}{2}}}$, Argument $= \phi = \ln 5\sqrt 5 $
How do we convert this argument into angle
| $$Z=(2+i)^{3i} \implies \ln Z=3i \ln (2+i)=3 i[\ln \sqrt{5}+i \tan^{-1}\frac{1}{2}]$$
$$\implies \ln Z= -3\tan^{-1}(1/2)+3i\ln \sqrt{5} \implies Z=e^{-3\cot^{-1}2}~~ e^{3i \ln \sqrt{5}}$$
$$\implies \arg(Z)=3\ln\sqrt{5}=2.414<\pi$$
AS this lies in $(-\pi, \pi]$, it is also the principal value of the argument.
Here we have used $\ln z=\ln\sqrt{x^2+y^2}+i\tan^{-1}(y/x), \arg (r e^{i\theta})=\theta$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3800640",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
$x+y+z=xyz$ and $x,y,z>0$. Proof that $\frac{x}{1+x^2}+\frac{2y}{1+y^2}+\frac{3z}{1+z^2}=\frac{xyz(5x+4y+3z)}{(x+y)(y+z)(z+x)}$ $x+y+z=xyz$ and $x,y,z>0$.
Proof that $$\frac{x}{1+x^2}+\frac{2y}{1+y^2}+\frac{3z}{1+z^2}=\frac{xyz(5x+4y+3z)}{(x+y)(y+z)(z+x)}$$
So far, I have managed to reduce $(x+y)(y+z)(z+x)$:
$$(x+y)(y+z)(z+x)=xyz+x^2y+xz^2+x^2z+y^2z+y^2x+yz^2+xyz$$
$$=2xyz+x^2y+xy^2+yz^2+y^2z+x^2z+xz^2$$
$$=2xyz+(x+y)xy+(y+z)yz+(x+z)xz$$
$$=2xyz+(xy-1)xyz+(yz-1)xyz+(xz-1)xyz$$
$$=xyz(xy+yz+zx-1)$$
So the original equation equals to:
$$\frac{x}{1+x^2}+\frac{2y}{1+y^2}+\frac{3z}{1+z^2}=\frac{5x+4y+3z}{xy+yz+zx-1}$$
So I was stuck here: How to simplified the left part of the equation? Also, I have noticed $5x+4y+3z=6(x+y+z)-(x+2y+3z)$ but they seemed to be unhelpful.
Any help is much appreciated, thanks!
| Though tedious, this is a purely mechanical problem:
Expanding the denominators, the problem becomes equivalent to the polynomial
$$
(x + y) (y + z) (z + x)(x(1 + z^2)(1 + y^2) + 2y(1 + z^2)(1 + x^2) + 3z(1 + y^2)(1 + x^2)) \\ -(1 + z^2)(1 + y^2)(1 + x^2)(x y z (5 x + 4 y + 3 z))
$$
being 0 given $xyz=x+y+z$.
Expand the products and each time you reach a product $xyz$ replace it by $x+y+z$ and keep expanding and cancelling terms. In the end, all terms will cancel.
This is really the job for a computer algebra system of your choice.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3801704",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"answer_id": 0
} |
Finding triangle of maximal area given two vertices and condition on internal bi-sector
$A=(2,5), B=(5,11)$ and a point $P$ moves such that internal bi-sector of $\angle APB$ passes through $(4,9).$ The maximum area of $ \triangle APB\;$ is __?
My attempt:
I found that $(4,9)$ lies on the line segment of $AB.$ Then I thought that due to the internal bi-sector condition it must have something to do with ellipse. Because, the two triangles formed by connecting focii of ellipse of perimeter are a congruent. So, if $P$ is a point on perimeter and $Q= (4,9),$ then $\angle APQ= \angle BPQ.$
I initially thought the optimum would happen when $ \angle ABP = \angle BAP =45^{\circ},$ but seems like that leads to a situation where the constraint is not obeyed.
So, how exactly do I get the optimum area while holding the constraint true?
|
Since the point $D=(4,9)$ belongs to $AB$,
it is a foot of the bisector of $\angle ACB$,
and a known expression for it is
\begin{align}
D&=\frac{aA+bB}{a+b}
\tag{1}\label{1}
.
\end{align}
With known coordinates we have
\begin{align}
(4,9)&=
\left(\frac{2a+5b}{a+b},\, \frac{5a+11b}{a+b} \right)
\tag{2}\label{2}
\end{align}
or a system
\begin{align}
\frac{2a+5b}{a+b}&=4
\tag{3}\label{3}
,\\
\frac{5a+11b}{a+b}&=9
\tag{4}\label{4}
\end{align}
with solution $b=2a$.
Expression for the area of $\triangle ABC$ squared is
\begin{align}
S_2(a,b,c)&=
\tfrac1{16}\,(4a^2b^2-(a^2+b^2-c^2)^2)
\tag{5}\label{5}
,
\end{align}
so for $b=2a$ we must have
\begin{align}
S_2(a,2a,c)&=
-\tfrac9{16}\,(a^2)^2+\tfrac58\,c^2\,a^2-\tfrac1{16}\,c^4
\tag{6}\label{6}
.
\end{align}
It's easy to find that
\begin{align}
\max_a S_2(a,2a,c)&=
S_2(a,2a,c)\Big|_{a=5}
=225
,
\end{align}
hence
the maximal area of such triangle is $15$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3804895",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 1
} |
If $a$, $b$, $c$ are the roots of $x^3-6x^2+3x+1=0$, find all possible values of $a^2b+b^2c+c^2a$
Let $a$, $b$, $c$ be the roots of
$$x^3 - 6x^2 + 3x + 1 = 0$$ Find all possible values of $a^2 b + b^2 c + c^2 a$. Express all the possible values, in commas.
I've already tried to bash out all the roots, Vieta's Formula, and try to manipulate the equation to become easier to work with. However, Vieta's didn't get me anywhere, and I couldn't find a way to make the equation simpler or anything. Any hints to start this problem?
| Let $p = a + b + c$, $q = ab + bc + ca$ and $r = abc$.
As pointed out by @Donald Splutterwit, we may let $A = a^2b + b^2c + c^2a$ and $B = ab^2 + bc^2 + ca^2$,
then $A + B$ and $AB$ are both symmetric which can both be expressed in terms of polynomials of $p, q, r$.
Indeed, we have
\begin{align}
A + B &= a^2b + b^2c + c^2a + ab^2 + bc^2 + ca^2\\
&= a^2(b+c) + b^2(c+a) + c^2(a+b)\\
&= (a^2+b^2+c^2)(a+b+c) - (a^3 + b^3 + c^3)\\
&= (p^2 - 2q)p - [3r + p(p^2 - 2q - q)]\\
&= pq - 3r
\end{align}
where we have used the known identity
$$a^3 + b^3 + c^3 - 3abc = (a+b+c)(a^2+b^2+c^2 - ab - bc - ca).$$
Also, we have
\begin{align}
AB &= (a^3+b^3+c^3)abc + (a^3b^3 + b^3c^3 + c^3a^3) + 3(abc)^2\\
&= [3r + p(p^2 - 2q - q)]r + [q^3 - 6r^2 - 3r(pq - 3r)] + 3r^2\\
&= p^3r-6pqr+q^3+9r^2
\end{align}
where we have used
\begin{align}
(ab+bc+ca)^3 &= a^3b^3 + b^3c^3 + c^3a^3 + 6(abc)^2\\
&\quad + 3abc(a^2b + b^2c + c^2a + ab^2 + bc^2 + ca^2).
\end{align}
Now, by Vieta's formula, we have $p = 6, q = 3, r = -1$.
Thus, $A + B = 21$ and $AB = -72$ which results in $A = 24, B = -3$ or $A = -3, B = 24$.
Thus, all the possible values of $A$ are $24, -3$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3805574",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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For $\alpha\in(0^\circ;90^\circ)$ simplify $\sin^2\alpha+\tan^2\alpha+\sin^2\alpha.\cos^2\alpha+\cos^4\alpha$ For $\alpha\in(0^\circ;90^\circ)$ simplify $\sin^2\alpha+\tan^2\alpha+\sin^2\alpha\cdot\cos^2\alpha+\cos^4\alpha.$
My try: $\sin^2\alpha+\tan^2\alpha+\sin^2\alpha\cdot\cos^2\alpha+\cos^4\alpha=\sin^2\alpha+\dfrac{\sin^2\alpha}{\cos^2\alpha}+\sin^2\alpha\cdot\cos^2\alpha+\cos^4\alpha=\dfrac{\sin^2\alpha\cdot\cos^2\alpha+\sin^2\alpha+\sin^2\alpha\cdot\cos^4\alpha+\cos^6\alpha}{\cos^2\alpha}.$
This doesn't seem to help much.
| Express everything in terms of $ \cos \alpha=c $
$$ (1-c^2) + (1-c^2)/c^2 + (1-c^2) c^2 + c^4$$
$$= \dfrac{c^2-c^4+1-c^2+c^4-c^6+c^6}{c^2} = \dfrac{1}{c^2}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3807240",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
} |
Is it possible to solve this identity by "inspection"? I was asked to prove the following identity (starting from the left-hand side):
$$(a+b)³(a⁵+b⁵)+5ab(a+b)²(a⁴+b⁴)+15a²b²(a+b)(a³+b³)+35a³b³(a²+b²)+70a⁴b⁴=(a+b)^8.$$
I'm trying to solve it by a sort of "inspection", but I haven't made it yet. Of course I could try to expand the left-hand polynomial and come to a more recognizable form of $(a+b)^8$, but of course that would be the hard way (assuming that there is an easy one).
As an example of why I am talking of "inspection" I can state a similar problem:
Show that $$(x+\frac{5}{2}a)⁴-10a(x+\frac{5}{2}a)³+35a²(x+\frac{5}{2}a)²-50a³(x+\frac{5}{2}a)+24a⁴=(x²-\frac{1}{4}a²)(x²-\frac{9}{4}a²).$$
Here by "inspection" we can deduce that the left-hand side of the identity is equivalent to $$[(x+\frac{5}{2}a)-a][(x+\frac{5}{2}a)-2a][(x+\frac{5}{2}a)-3a][(x+\frac{5}{2}a)-4a]$$ and then after a few steps come to the the desire result.
I would appreciate any help you could give me.
| $$(a+b)^3(a^5+b^5)+5ab(a+b)^2(a^4+b^4)+15a^2b^2(a+b)(a^3+b^3)+35a^3b^3(a^2+b^2)+70a^4b^4=(a+b)^8$$
Note that $a+b|(a+b)^3$, so except the last two terms in the LHS, every term is divisible by $(a+b)^2$, in fact you can take $35a^3b^3$ common from the last two terms to have $$35a³b³(a²+b²)+70a⁴b⁴=35a^3b^3(a+b)^2$$ so that the LHS becomes
$$(a+b)^2\times((a+b)(a^5+b^5)+5ab(a^4+b^4)+15a^2b^2(a^2-ab+b^2)+35a^3b^3)$$
and if you leave out $(a+b)^2$ from this now, the multiplying and expanding becomes easier, you can verify that it is indeed $(a+b)^6$
| {
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"source": "stackexchange",
"question_score": "4",
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Inverse of a matrix and matrix relation Let $A$ be the matrix $$ A= \begin{pmatrix}
2 & -1 & -1\\
0 & -2 & -1 \\
0 & 3 & 2 \\
\end{pmatrix} $$ I am trying to find $A^{-1}$ as a relation of $I_{3}, A$ and $A^{2}$ and also to prove that $A^{2006}-2A^{2005}=A^{2}-2A$. For the first one I noticed that $$A^{n}= \begin{pmatrix}
2^{n} & -(2^{n}-1) & -(2^{n}-1)\\
0 & -2 & -1 \\
0 & 3 & 2 \\ \end{pmatrix} , A^{k}= \begin{pmatrix}
2^{k} & -(2^{k}-1) & -(2^{k}-1)\\
0 & 1 & 0 \\
0 & 0 & 1\\ \end{pmatrix}$$ for $n$ odd and $k$ even but I don't know how to proceed from here. Also the the equality $A^{2006}-2A^{2005}=A^{2}-2A$ leads to $A^{2004}(A-2I_{3})=A-2I_{3}$ but again I don't know what to do next since $(A-2I_{3})$ is not invertible. Any help?
| Since the characteristic polynomial of $A$ is $-\lambda ^3+2\lambda^2+\lambda-2$, you know, by the Cayley-Hamilton theorem, that $-A^3+2A^2+A-2\operatorname{Id}_3=0$. Therefore,$$A^{-1}=-\frac12A^2+A+\frac12\operatorname{Id}_3.$$On the other hand,\begin{align}A^{2006}-2A^{2005}&=A^{2003}(A^3-2A^2)\\&=A^{2003}(A-2\operatorname{Id}_3)\\&=A^{2001}(A^3-2A^2)\\&=A^{2001}(A-2\operatorname{Id}_3)\\&=\cdots\end{align}
| {
"language": "en",
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Verify my proof that for any $n>1$, if $n^n+1$ is prime, then $n=2^{2^k}$ for some integer $k$. I am solving a problem and I am respectfully asking someone to critique my work and offer suggestions on formatting or point out any glaring logical errors. Here is the problem:
Prove that for any $n>1$, if $n^n+1$ is prime, then $n=2^{2^k}$ for some integer $k$. Use this to prove that $2020^{2020}+1$ is not prime.
Proof. Let $n>1$ be an arbitrary positive integer such that $n^n+1$ is prime. We need to prove that $n$ must be of the form $2^{2^k}$ for some integer $k$. We see that if $n$ is odd, then $n^n$ is odd and thus $n^n+1$ is even, so $n$ must be even. Since $n$ is even, $n=2^pq$ for some integers $p$ and $q$, where $q$ is odd. We then see that $n^n+1=(2^pq)^{2^pq}+1=[(2^pq)^{2^p}]^q+1^q$. However, the only way that $n^n+1$ cannot be factored is if $q=1$, so $n=2^p$ for some integer $p$. If $p=1$, then we just notice that $p=2^0$, so $n=2^{2^0}$. Otherwise, if $p$ is odd and $p\neq 1$, then we see that $n^n+1=(2^p)^{2^p}+1=(2^{2^p})^p+1^p$, and since $p\neq 1$ and $p$ is odd, this expression can be factored, so $p$ must be even. This means that $p=2^kb$ for some integer $k$ and some odd integer $b$. We see that $n^n+1=(2^{2^kb})^{2^{2^kb}}+1=(2^{2^k2^{2^kb}})^b+1^b$, which can be factored if $b\neq 1$, so we see that $b=1$ and $p=2^k$. This means that $n=2^p=2^{2^k}$. Now, we see that $n^n+1=(2^{2^k})^{2^{2^k}}+1$, and we see that $2^k\cdot 2^{2^k}$ is even for all $k$, so $n^n+1$ cannot be factored and thus it is not necessary to go further. This means that if $n^n+1$ is prime, than $n=2^{2^k}$ for some integer $k$, completing the proof. $\blacksquare$
Now assume for the sake of contradiction that $2020^{2020}+1$ is prime. By our theorem, this means that $2020=2^{2^k}$ for some $k$. However, $2020$ is not a power of 2, contradiction. This means our assumption is false and $2020^{2020}+1$ is not prime.
| Your proof is correct in its ideas, but contains a few unclear and even false statements. I'll list a few points of improvement:
...if $n$ is odd, then $n^n$ is odd and thus $n^n+1$ is even, so $n$ must be even.
How does it follow that $n$ must be even? If you are concluding this because you have reached a contradiction from the assumption that $n$ is odd, I would recommend to state this explicitly.
We then see that $n^n+1=(2^pq)^{2^pq}+1=[(2^pq)^{2^p}]^q+1^q$. However, the only way that $n^n+1$ cannot be factored is if $q=1$...
Again I would recommend to be more explicit here; show that you can factor $n^n+1$ if $q>1$, and conclude that if $n^n+1$ is prime then $q=1$. In fact it is not true that $n^n+1$ cannot be factored if $q=1$; for $n=2^3$ you have
$$n^n+1=(2^3)^{2^3}+1=97\times257\times673.$$
Moreover, both arguments up to this point can be combined into a single argument. Instead of first proving that $n$ is even, simply write $n=2^pq$ with $p$ and $q$ nonnegative integers, where $q$ is odd.
...if $p$ is odd and $p\neq 1$, then we see that $n^n+1=(2^p)^{2^p}+1=(2^{2^p})^p+1^p$, and since $p\neq 1$ and $p$ is odd, this expression can be factored...
This is again the same argument as before, which can be bundled with the previous arguments as well.
We see that $n^n+1=(2^{2^kb})^{2^{2^kb}}+1=(2^{2^k2^{2^kb}})^b+1^b$, which can be factored if $b\neq 1$...
This is again the same argument as before, which can be bundled with the previous arguments as well.
Now, we see that $n^n+1=(2^{2^k})^{2^{2^k}}+1$, and we see that $2^k\cdot 2^{2^k}$ is even for all $k$, so $n^n+1$ cannot be factored and thus it is not necessary to go further.
It is not true that $n^n+1$ cannot be factored if $n=2^{2^k}$: For $k=4$ you have
$$n^n+1=(2^{2^4})^{2^{2^4}}+1=274177\times67280421310721.$$
In fact the sentence I quote above is redundant, no such thing asked in the question. Leaving it out would improve the proof. Here's a version of your proof with the suggested improvements:
Write $n=2^pq$ with $p$ and $q$ positive integers, and $q$ odd. Similarly write $p=2^kb$ with $k$ and $b$ positive integers, and $b$ odd, so that $n=2^{2^kb}q$ and
$$n^n+1=(2^{2^kb}q)^{2^{2^kb}q}+1.\tag{1}$$
Lemma. If $x$ and $y$ are positive integers and $y$ is odd, then $x^y+1$ is divisible by $x+1$.
Proof. Exercise.$\quad\square$
Taking $x=(2^{2^kb}q)^{2^{2^kb}}$ and $y=q$ shows that $n^n+1$ is divisible by $x+1>1$, so if $n^n+1$ is prime then $x+1=x^y+1$ and hence $y=q=1$. Then $(1)$ becomes
$$n^n+1=(2^{2^kb})^{2^{2^kb}}+1=2^{2^k2^{2^kb}b}+1.\tag{2}$$
Taking $x=2^{2^k2^{2^kb}}$ and $y=b$ shows that $n^n+1$ is divisible by $x+1>1$, so if $n^n+1$ is prime then $x+1=x^y+1$ and hence $y=b=1$, and so $n=2^{2^k}$.$\quad\square$
| {
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"url": "https://math.stackexchange.com/questions/3809033",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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Distance between $\left\{(x,y,z):\frac{x^2}{a^2}+\frac{y^2}{b^2}+\frac{z^2}{c^2}=1\right\}$ and $\big\{(x,y,z): x+y+z=h\big\}$ Calculate the distance between $$M=\left\{(x,y,z) \in \mathbb{R}^{3}: \frac{x^2}{a^2}+ \frac{y^2}{b^2}+\frac{z^2}{c^2}=1\right\}$$ and $$H=\big\{(x,y,z) \in \mathbb{R}^{3}: x+y+z=h\big\}$$ where $a,b,c,h \in \mathbb{R}^+$.
| I'm assuming when you say the distance between $M$ and $H$ you mean $Dist(M, H) = \inf\{||x-y|| \ |\ x\in M, y\in H\}$. Additionally, I'll assume that $M$ and $H$ don't intersect.
Let $\tilde{M} = \{(x,y,z) \in \mathbb{R}^{3}: \frac{x^2}{a^2}+ \frac{y^2}{b^2}+\frac{z^2}{c^2} \leq 1\}$ and notice that this is convex set. Given a point $p$ the closest point in $X$ to $p$ is the projection of $p$ onto $X$, i.e., $proj_{X}(p)$. Therefore, to find the minimum distance between two sets we need to find two points $v\in M$ and $w\in H$ such that $w = proj_M(v)$ and $v = proj_H(w)$. To find these points we are going to use the fact that the vector $n = p-proj_{X}(p)$ is normal to the set $X$ at $proj_{X}(p)$. Notice that if $w = proj_M(v)$ and $v = proj_H(w)$ then $n_v = v-proj_{X}(v) = v-w$ and $n_w = w-proj_{X}(w) = w-v$ so that $n_v = -n_w$.
The vector $\begin{pmatrix}
\frac{2s}{a^2}x \\
\frac{2s}{b^2}y \\
\frac{2s}{c^2}z \\
\end{pmatrix}
$ where $s\in (0,\infty)$ is normal to $\tilde{M}$ at $(x,y,z)\in M$ and the vector $
\begin{pmatrix}
-t\\
-t\\
-t
\end{pmatrix}
$ where $t\in (0,\infty)$ is normal to $H$ towards $\tilde{M}$ at $(x,y,z)$.
In order for $n_v = -n_w$ we need the normal of $\tilde{M}$ at $(x,y,z)\in M$ to be a positive scalar multiple of $(1,1,1)$. This occurs when $\frac{2}{a^2}x = \frac{2}{b^2}y$ and $\frac{2}{a^2}x = \frac{2}{c^2}z$. Therefore, $y = \frac{b^2}{a^2}x$ and $z = \frac{c^2}{a^2}x$. To determine the point $v$ we see that,
\begin{align}
1 &= \frac{x^2}{a^2}+ \frac{y^2}{b^2}+\frac{z^2}{c^2} \\
&=\frac{x^2}{a^2}+ \frac{b^2}{a^4}x^2+\frac{c^2}{a^4}x^2 \\
&=\frac{a^2+b^2+c^2}{a^4}x^2 \\
x &= \frac{a^{2}}{\sqrt{a^2+b^2+c^2}}
\end{align}
Similarly, you can derive that $y = \frac{b^{2}}{\sqrt{a^2+b^2+c^2}}$ and $z = \frac{c^{2}}{\sqrt{a^2+b^2+c^2}}$. Therefore, the point on $M$ that is closest to $H$ is,
\begin{align}
v =\begin{pmatrix}
\frac{a^{2}}{\sqrt{a^2+b^2+c^2}}\\
\frac{b^{2}}{\sqrt{a^2+b^2+c^2}}\\
\frac{c^{2}}{\sqrt{a^2+b^2+c^2}}
\end{pmatrix}
\end{align}
Next, we want to determine the value of $k$ such that
\begin{align}
w &= v + k\begin{pmatrix}
1\\
1\\
1
\end{pmatrix}
\end{align}
Note that the vector (1,1,1) is the normal vector of $M$ at $v$. The point $w = (w_1,w_2,w_3)$ is in $H$ if $w_1+w_2+w_3 = h$ therefore,
\begin{align}
h &= w_1+w_2+w_3 \\
&= v_1+v_2+v_3 + 3k \\
&= \frac{a^{2}}{\sqrt{a^2+b^2+c^2}} + \frac{b^{2}}{\sqrt{a^2+b^2+c^2}} + \frac{c^{2}}{\sqrt{a^2+b^2+c^2}} + 3k \\
&= \sqrt{a^2+b^2+c^2} + 3k \\
k &= \frac{h-\sqrt{a^2+b^2+c^2}}{3}
\end{align}
Lastly, $Dist(M, H) = || w- v|| = ||k\begin{pmatrix}
1\\
1\\
1
\end{pmatrix}|| = \sqrt{3}k = \frac{h-\sqrt{a^2+b^2+c^2}}{\sqrt{3}}$.
Sorry, if the explanation is not clear. If you have any questions add a comment below.
| {
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"url": "https://math.stackexchange.com/questions/3809873",
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Distribution of $Z=\left\{\begin{matrix} X+Y & \operatorname{if} & X+Y<1\\ X+Y-1 & \operatorname{if} & X+Y>1 \end{matrix}\right.$ Let $X\perp Y$ be two random variables with uniform distribution on $[0,1]$. How is it possible that $F_Z(z)=z$?
Initially I wrote $F_Z(z)=\mathbb{P}(Z\leq z)=\mathbb{P}(X+Y\leq z,X+Y<1)+\mathbb{P}(X+Y-1\leq z,X+Y>1)$, and I started to study the first probability. For the triangle with vertices $(0,0),(0,1),(1,0)$ and longest side bounded by the line $y=1-x$, I plotted the line $y=z-x$ and I tried to study the different cases ($0<z<\frac{1}{2}$ and $\frac{1}{2}<z<1$). I thought the my approach was correct, but when I saw $F_Z(z)=z$ my entire argument was lacking. Where was I wrong? Could you please give me any hints on how solve the problem?
Thanks in advance.
| For $z\in\left(0,1\right)$ we find:
$$\begin{aligned}P\left(Z\leq z\right) & =P\left(Z\leq z,X+Y<1\right)+P\left(Z\leq z,X+Y>1\right)\\
& =P\left(X+Y\leq z,X+Y<1\right)+P\left(X+Y-1\leq z,X+Y>1\right)\\
& =P\left(X+Y<z\right)+P\left(1<X+Y<1+z\right)\\
& =\frac{1}{2}z^{2}+\left[\frac{1}{2}-\frac{1}{2}\left(1-z\right)^{2}\right]=z
\end{aligned}
$$
Addendum:
To find the two probabilities it is handsome to make a picture and to find the areas that are involved.
Integration in order to find $P(1<X+Y<1+z)$ works like this:
$$\begin{aligned}P\left(1<X+Y<1+z\right) & =\int_{0}^{z}\int_{1-x}^{1}dydx+\int_{z}^{1}\int_{1-x}^{1-x+z}dydx\\
& =\int_{0}^{z}xdx+\int_{z}^{1}zdx\\
& =\frac{1}{2}z^{2}+z\left(1-z\right)\\
& =z-\frac{1}{2}z^{2}
\end{aligned}
$$
Or - making use of the fact that $1-X\stackrel{d}{=}X$ and $1-Y\stackrel{d}{=}Y$
$$\begin{aligned}P\left(1<X+Y<1+z\right) & =P\left(1<X+Y\right)-P\left(1+z<X+Y\right)\\
& =\frac{1}{2}-P\left(1-X+1-Y<1-z\right)\\
& =\frac{1}{2}-\frac{1}{2}\left(1-z\right)^{2}\\
& =z-\frac{1}{2}z^{2}
\end{aligned}
$$
| {
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How to show that $ C := A^3-3A^2+2A = 0$?
Let $$A = \begin{pmatrix} -1 & 1 & 2 \\ 0 & 2 & 0 \\ -1 & 1 & 2 \end{pmatrix}$$
Let $C:= A^3-3A^2+2A $.
Show that $C=0$.
I know that $A$ is diagonalisable with $\operatorname{spec}(A)=\{0,1,2\}$.
I have no clue how to approach that problem. Any advice?
| Method 1: you can solve it by direct computation:
$$A^2 = \begin{pmatrix} -1 & 3 & 2 \\ 0 & 4 & 0 \\ -1 & 3 & 2 \end{pmatrix}$$
and
$$A^3 = \begin{pmatrix} -1 & 7 & 2 \\ 0 & 8 & 0 \\ -1 & 7 & 2 \end{pmatrix}.$$
You can actually argue that
$$A^n = \begin{pmatrix} -1 & \sum_{k=0}^{n-1} 2^k & 2 \\ 0 & 2^n & 0 \\ -1 & \sum_{k=0}^{n-1} 2^k & 2 \end{pmatrix}.$$
Thus if you sum up you easily see that $C\neq 0$.
Method 2: you can use Cayley-Hamilton theorem for these kinds of problems
https://en.wikipedia.org/wiki/Cayley%E2%80%93Hamilton_theorem.
By this theorem you would get that
$$A^3-3A^2+2A=0,$$
which is actually verifiable by direct computation. Conversely, always by Cayley-Hamilton theorem, if you knew the eigenvalues you could have just plugged them in $\lambda^3+3\lambda^2+2\lambda=0$ and verified that this equation is not satisfied.
| {
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Is my Integral Solution Correct? Compute $\int \frac{n^x}{n^{2x} + 8n^x + 12} \, dx$, where $n$ is a positive real number.
UPDATED SOLUTION:
First, set $n^x = u.$ We have $du = n^x\log(n)\, dx.$
Subbing in $u$ and $\frac{du}{u\log(n)},$ we have $\frac{1}{\log n}\int \frac{1}{u^{2} + 8u + 12} \, du.$
Using partial fractions, we have $\frac1{u^2+8u+12}=\frac1{(u+6)(u+2)}=\frac A{u+6}+\frac B{u+2}.$
Solving for $A$ and $B,$ $Au+Bu=0$ and $2A+6B=1.$
Since $A=-B$ and $B=\frac{1}{4}, A=-\frac{1}{4}.$
Therefore, $\frac{1}{\log n}\int \frac{1}{u^{2} + 8u + 12} \, du =\frac{1}{4\log n}\int \frac{1}{u+2} - \frac {1}{u+6}\, du.$
Taking the integral of each fraction and simplifying, we get $\frac{1}{4\log n}\log\frac{|u+2|}{|u+6|} + C.$
Subbing in $n^x$ for $u,$ we get $\frac{1}{4\log n}\log\frac{|n^x+2|}{|n^x+6|} + C.$
Since $n^x$ is always positive, our answer is $\boxed{\frac{1}{4\log n}\log\frac{n^x + 2}{n^x + 6} + C},$ where $C$ is a constant and $n$ doesn't equal $1.$
OLD SOLUTION:
First, set $n^x = u.$ We have $du = n^x\log(n) dx.$
Subbing in $u$ and $\frac{du}{u\log(n)},$ we have $\frac{1}{\log n}\int \frac{1}{u^{2} + 8u + 12} \, du.$
Completing the square, $\frac{1}{\log n}\int \frac{1}{(u+4)^2 - 4} \, du.$
Next, we set $u + 4 = v,$ so $dv = du.$
Subbing in $v$ and $dv,$ we have $\frac{1}{\log n}\int \frac{1}{v^2 - 4} \, dv.$
Factoring out -1/4, we have $-\frac{1}{4\log n}\int \frac{1}{1-\frac{v^2}{4}} \, dv.$
Again, we set $\frac{v}{2} = w,$ so $dw = \frac{dv}{2}.$
Subbing in $w$ and $2dw$, we have $-\frac{1}{2\log n}\int \frac{1}{1-w^2} \, dw.$
Making partial fractions, we get $-\frac{1}{2\log n}\int \frac{1}{1-w^2} \, dw = -\frac{1}{2\log n}\int \frac{1}{(1+w)(1-w)} \, dw = -\frac{1}{4\log n}\int \frac{1}{1+w} + \frac{1}{1-w} \, dw.$
Taking the integral of each fraction and simplyfying, we get $-\frac{1}{4\log n}\log\frac{|1+w|}{|1-w|}.$
Subbing in $ \frac{n^x}{2} + 2$ for $w,$ we get $-\frac{1}{4\log n}\log\frac{|1+ \frac{n^x}{2} + 2|}{|1-(\frac{n^x}{2} + 2)|} = -\frac{1}{4\log n}\log\frac{|n^x + 6|}{|-n^x - 2|}.$
Taking out the negative from $-n^x - 2$ because of the absolute value and taking the reciprocal of the logarithm because of the negative in front of $\frac{1}{4\log n}$, we get $\frac{1}{4\log n}\log\frac{|n^x + 2|}{|n^x + 6|},$ and since $n^x$ is always positive, our answer is $\boxed{\frac{1}{4\log n}\log\frac{n^x + 2}{n^x + 6}}.$
| Alternatively, you could have used the roots of the quadratic to write
$$\frac1{u^2+8u+12}=\frac1{(u+6)(u+2)}=\frac A{u+6}+\frac B{u+2}$$
You determine $A$ and $B$ using $$(A+B)u=0$$ and $$2A+6B=1$$
So $A=-B$ and $4B=1$. Therefore $$\frac1{u^2+8u+12}=\frac 14\left(\frac 1{u+2}-\frac 1{u+6}\right)$$
You will get the same result, but less chances to make mistakes.
| {
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Complex Roots in a Complex Equation
Problem: The equation $$\frac{x}{x+1} + \frac{x}{x+2} = kx$$ has exactly two complex roots. Find all possible complex values for $k.$
Progress: I was online today, and I saw this problem. Heres what I tried: $$(x+1)x+(x+2)x=kx(x+1)(x+2),$$ and I immediately got stuck :(. I'm quite new at complex numbers. Help? Thanks.
| Good!
Yes, multiplying both sides by $(x+ 1)(x+ 2)$ "reduces" the equation to $(x+ 1)x+ (x+ 2)x= kx(x+ 1)(x+ 2)$. Now, I would do the indicated multiplications to get $x^2+ x+ x^2+ 2x= kX*(x^2+ 3x+ 2)= kx^3+ 3kx^2+ 2kx$ which can be written as $kx^3+ (3k- 1)x^2+ (2k- 1)x= 0$. The left side can be factored as $x(kx^2+ (3k- 1)x+ (2k- 1))= 0$. One root is $x= 0$ and, if $x$ is not $0$, $kx^2+ (3k- 1)x+ (2k-1)= 0$ can be solved by the "quadratic formula".
| {
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"timestamp": "2023-03-29T00:00:00",
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Relationship between constants so that the center of curvature of the helix is contained in the cylinder I am studying the topic of torsion and curvature and I ran into the following problem:
Consider the helix $\alpha(t)=(a\sin t,a\cos t,bt)$. Get the relationship between constants
so that the center of curvature of the helix is contained in the cylinder
where the helix is contained as well.
First I decided to parameterize the curve with the arc length to get the curvature:
$s(t)=\int_{0}^{t}|\alpha'(t)|dt=\int_{0}^t\sqrt{a^2(\sin^2t+\cos^2t)+b^2}dt=\int_0^t\sqrt{a^2+b^2}dt=\sqrt{a^2+b^2}t$
Hence, $t(s)=\cfrac{s}{\sqrt{a^2+b^2}}$, so
$\beta(s)=\alpha\left(t(s)\right)=\left(a\cos\cfrac{s}{\sqrt{a^2+b^2}},a\sin\cfrac{s}{\sqrt{a^2+b^2}},\cfrac{bs}{\sqrt{a^2+b^2}}\right)$
is a reparametrization of the curve by arc length. Differentiating, we see
that:
$\alpha'(s)=\left(\cfrac{-a\sin\cfrac{s}{\sqrt{a^2+b^2}}}{\sqrt{a^2+b^2}},\cfrac{a\cos\cfrac{s}{\sqrt{a^2+b^2}}}{\sqrt{a^2+b^2}},\cfrac{b}{\sqrt{a^2+b^2}}\right)$
Therefore,
$\alpha''(s)=\left(\cfrac{-a\cos\cfrac{s}{\sqrt{a^2+b^2}}}{a^2+b^2},\cfrac{-a\sin\cfrac{s}{\sqrt{a^2+b^2}}}{a^2+b^2},0\right)$
and so the curvature of the helix is given by:
$k(s)=|\alpha''(s)|=\sqrt{\cfrac{a^2\cos^2\left(\cfrac{s}{\sqrt{a^2+b^2}}\right)}{(a^2+b^2)^2}+\cfrac{a^2\sin^2\left(\cfrac{s}{\sqrt{a^2+b^2}}\right)}{(a^2+b^2)^2}}=\sqrt{\cfrac{a^2}{a^2+b^2}}=\cfrac{|a|}{a^2+b^2}$
So I checked that it has a curvature that depends on a and b and that it is always constant.
Does this help me to get the relationship between constants
so that the center of curvature of the propeller is contained in the cylinder
where in turn is said helix contained?
I appreciate your time and help.
| Since the principal normal is horizontal, you will need $1/\kappa<2|a|$, so that seems to require $|b|<|a|$.
| {
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Proving $6(x^3+y^3+z^3)^2 \leq (x^2+y^2+z^2)^3$, where $x+y+z=0$ Question :
Let $x,y,z$ be reals satisfying $x+y+z=0$. Prove the following inequality:$$6(x^3+y^3+z^3)^2 \leq (x^2+y^2+z^2)^3$$
My Attempts :
It’s obvious that $x,y,z$ are either one negative and two positive numbers or one positive and two negative numbers.
In addition, putting $(x,y,z)$ and $(-x,-y,-z)$ into the inequality has the same outcome.
Therefore, without loss of generality, I suppose $x,y \geq 0$ and $z\leq0$.
$x+y=-z\\ \Longrightarrow (x+y)^2=z^2 \\ \Longrightarrow (x^2+y^2+z^2)^3=8(x^2+xy+y^2)^3$
is what I have got so far, and from here I can’t continue.
Am I on the right direction? Any suggestions or hints will be much appreciated.
| The difference
$$
(x^2+y^2+z^2)^3-6(x^3+y^3+z^3)^2
$$
for $x+y+z=0$ can be written
$$
2 (x-y)^2 (y-z)^2 (z-x)^2
$$
| {
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If $a, b, c, d>0$ and $abcd=1$ prove that an inequality holds true If $a, b, c, d>0$ and $abcd=1$ prove that:
$$\frac{a+b+c+d}{4}\ge\frac{1}{a^3+b+c+d}+\frac{1}{a+b^3+c+d}+\frac{1}{a+b+c^3+d}+\frac{1}{a+b+c+d^3}$$
I attempted to solve it in the following way:
$$\begin{equation}\frac{1}{a^3+b+c+d}+\frac{1}{a+b^3+c+d}+\frac{1}{a+b+c^3+d}+\frac{1}{a+b+c+d^3}\leq\\
\frac{1}{4\sqrt[4]{a^3bcd}}+\frac{1}{4\sqrt[4]{ab^3cd}}+\frac{1}{4\sqrt[4]{abc^3d}}+\frac{1}{4\sqrt[4]{abcd^3}} = \\
\frac{1}{4\sqrt{a}}+\frac{1}{4\sqrt{b}}+\frac{1}{4\sqrt{c}}+\frac{1}{4\sqrt{d}} = \\
\frac{\sqrt{bcd}+\sqrt{acd}+\sqrt{abd}+\sqrt{abc}}{4\sqrt{abcd}} = \\
\frac{\sqrt{bcd}+\sqrt{acd}+\sqrt{abd}+\sqrt{abc}}{4}
\end{equation}$$
This is as far as I got. Could you please help me finish off my thought pattern and finish the question the way I was attempting to solve it?
| This is just a thought about the possibility to continue the proof. The inequality:
$$ \frac{a+b+c+d}{4} \ge \frac{\sqrt{bcd}+\sqrt{acd}+\sqrt{abd}+\sqrt{abc}}{4}$$
cannot hold for all $a,b,c,d>0$ with $abcd=1$. In fact, for $N>1$ set $a=b=c=N$, $d=1/N^3$. Then $LHS < N$ and $ RHS > N^{3/2}/4$. Moreover, if $N>16$ then $N^{3/2}/4>N$. Thus, when $N>16$ holds, $LHS<RHS$.
| {
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If $a^2+b^2-ab=c^2$ for positive $a$, $b$, $c$, then show that $(a-c)(b-c)\leq0$
Let $a$, $b$, $c$ be positive numbers. If $a^2+b^2-ab=c^2$. Show that
$$(a-c)(b-c)\leq0$$
I have managed to get the equation to $(a-b)^2=c^2-ab$, but I haven't been able to make any progress.
Can someone help me?
| WLOG, we assume that $a \le b$ and we want to show that
$$a \le \sqrt{a^2+b^2-ab}\le b$$
which is equivalent to
$$a^2\le a^2+b^2-ab \le b^2$$
The first inequality reduces to $0 \le b(b-a)$ which is clearly true.
The second inequality is $a^2-ab=a(a-b) \le 0$ which is true again.
| {
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Prove $\lim\limits_{x^2 + y^2 \to +\infty} x^2 -2xy + 2y^2 = +\infty$
Prove that $$\lim_{x^2 + y^2 \to +\infty} x^2 -2xy + 2y^2 = +\infty$$
My attempt:
$$x^2 + 2y^2 = x^2+y^2 + y^2 \implies \lim_{x^2 + y^2 \to +\infty}x^2 +2y^2 = +\infty$$
Then, from Cauchy-Schwarz:
$$x^2 + 2y^2 \geq 2\sqrt2xy \geq 2xy $$
Thus,
$$x^2+2y^2 -2xy \geq 0$$
I think I am on the correct path, but I don't know how to proceed.
| Note: If $xy<0$, then $2\sqrt{2}xy \leq 2xy$. So the Cauchy-Schwartz argument may not hold.
However, $x^2-2xy+y^2 =(x-y)^2 \geq 0$. So as $x^2+y^2\rightarrow\infty$, $x^2-2xy+y^2+y^2\rightarrow\infty$.
| {
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Prove that $a^3 + b^3 + 3abc > c^3$ Suppose $a,b,c$ are the sides of a non-degenerate triangle. Prove that $a^3 + b^3 + 3abc > c^3.$
I was thinking that this inequality looked suspiciously like $a^3 + b^3 + c^3 - 3abc,$ which factors as $(a + b + c) (a^2 - a b + b^2 - a c - b c + c^2).$ However, I have little to no idea how to get that. Can someone give me a hint?
| By your hint we need to prove that: $$(a+b-c)(a^2+b^2+c^2-ab+ac+bc)>0.$$
Can you end it now?
$$a^2+b^2+c^2-ab+ac+bc=b(b+c-a)+a^2+c^2+ac>0.$$
| {
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Minimize $(x+y)(y+z)(z+x)$ given $xyz(x+y+z) = 1$ $x,y,z$ are positive reals and I am given $xyz(x+y+z) = 1$. Need to minimize $(x+y)(y+z)(z+x)$. Here is my approach.
Using AM-GM inequality
$$ (x+y) \geqslant 2 \sqrt{xy} $$
$$ (y+z) \geqslant 2 \sqrt{yz} $$
$$ (z+x) \geqslant 2 \sqrt{zx} $$
So, we have
$$ (x+y)(y+z)(z+x) \geqslant 8xyz $$
Also, I got
$$ \frac{x+y+z+(x+y+z)}{4} \geqslant \bigg[ xyz(x+y+z) \bigg] ^{1/4} $$
$$ \therefore x+y+z \geqslant 2 $$
But, I am stuck here. Any hints ?
| $(x+y)(y+z)(z+x)=(z+x)(y(x+y+z)+xz)=(\frac{1}{zx}+zx)(x+z)$
now we can use $$\frac{1}{zx}+zx\ge 4{(\frac{1}{27{(xz)}^2})}^{1/4}$$
(HINT:$\frac{1}{zx}=\frac{1}{3zx}+\frac{1}{3zx}+\frac{1}{3zx}$)
also we can use $$x+z\ge 2\sqrt{xz}$$
Multiplying we get $$(x+y)(y+z)(z+x)\ge \frac{8}{3^{3/4}}$$
| {
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Probability that atleast $3$ from $5$ dice show the same face I think I have solved this problem correctly, but my solution does not seem to match the one in the textbook - Problem IV.2 from Feller's introduction to probability theory & applications page 101. Could you help with where I've possible gone wrong?
Note. The author expects the student to use the result for the realisation of at least $m$ among $N$ events.
$$P_m = S_m - {m \choose m-1}S_{m+1} + {m + 1 \choose m-1}S_{m+2}-{m+2 \choose m-1}S_{m+3} + \ldots \pm {N-1 \choose m-1}S_N$$
Five dice are thrown. Find the probability that at least three of them show the same face. (Verify by the methods of chapter II, section 5).
Solution.
Define $A_i := \text{the event that dice i shows the common face}$.
$P(\text{3 selected dice show the same face})=P(A_i A_j A_k)=\frac{6}{6^3}$
$P(\text{4 selected dice show the same face})=P(A_i A_j A_k A_l)=\frac{6}{6^4}$
$P(\text{5 selected dice show the same face})=P(A_i A_j A_k A_l A_m)=\frac{6}{6^5}$
$S_3 = \sum_{1 \le i < j < k \le 5} p_{ijk}={5 \choose 3}\frac{6}{6^3}$
$S_4 = \sum_{1 \le i < j < k < l \le 5} p_{ijkl}={5 \choose 4}\frac{6}{6^4}$
$S_5 = \sum_{1 \le i < j < k < l < m \le 5} p_{ijklm}={5 \choose 5}\frac{6}{6^5}$
\begin{align*}
P_3 &= S_3 - {3 \choose 2}S_4 + {4 \choose 2}S_5\\
&= \frac{15}{36}-\frac{15}{216} + \frac{1}{216}\\
&= \frac{76}{216} \\
&= \frac{19}{54} \approx 0.35
\end{align*}
However, the answer in the book is $0.21$.
| In case you are interested in alternative approaches, here is a slightly different solution, also using the formula for the realization of at least $m$ among $N$ events.
Let's see if we can solve a simpler problem: What is the probability that at least $3$ of $5$ dice roll a $1$? Here $A_i$ is the event that die $i$ rolls a $1$, and
$$S_k = \binom{5}{k} \left( \frac{1}{6} \right)^k$$
for $1 \le k \le 5$.
Applying the formula for the probability of at least $m$ of $N$ events with $m=3$ and $N=5$,
$$P_3 = S_3 - \binom{3}{2} S_4 + \binom{4}{2} S_5 = \frac{23}{648}$$
So $P_3$ is the probability that at least $3$ of $5$ dice will roll a $1$. The same probability holds for $3$ of $5$ dice rolling a $2,3,4,5$ or $6$. So the probability that at least $3$ of $5$ dice roll the same number is
$$6 \cdot P_3 = \frac{23}{108}$$
| {
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How to solve system of equations using inverse matrix? System of equations is the following:
$$x + 4y + 2z = 10$$
$$4x - 3y+0z = 6$$
$$2x + 2y + 2z = 14$$
Here is my solution:
$$det(A) = 1 *(-3 * 2 - 0 * 2) -4 * (4 * 2 - 0 * 2) + 2 * (4 * 2 - (-3) * 2)$$
$$= -6 -32 + 28$$
$$= -10$$
$$
+\begin{pmatrix}
-3 & 0 \\
2 & 2 \\
\end{pmatrix}
$$
$$
-\begin{pmatrix}
4 & 0 \\
2 & 2 \\
\end{pmatrix}
$$
$$
+\begin{pmatrix}
4 & -3 \\
2 & 2 \\
\end{pmatrix}
$$
$$
-\begin{pmatrix}
4 & 2 \\
2 & 2 \\
\end{pmatrix}
$$
$$
+\begin{pmatrix}
1 & 2 \\
2 & 2 \\
\end{pmatrix}
$$
$$
-\begin{pmatrix}
1 & 4 \\
2 & 2 \\
\end{pmatrix}
$$
$$
+\begin{pmatrix}
4 & 2 \\
-3 & 0 \\
\end{pmatrix}
$$
$$
-\begin{pmatrix}
1 & 2 \\
4 & 0 \\
\end{pmatrix}
$$
$$
+\begin{pmatrix}
1 & 4 \\
4 & -3 \\
\end{pmatrix}
$$
Above equals to:
$$
\begin{pmatrix}
-6 & -8 & 14 \\
-4 & -2 & 6 \\
-6 & -8 & -15\\
\end{pmatrix}
$$
Then I multiply it by 1/-10 and the result is:
$$
\begin{pmatrix}
0,6 & 0,4 & 0,6 \\
0,8 & 0,2 & -0,8 \\
-1,4 & -0,6 & 1,5\\
\end{pmatrix}
$$
Then I multiply it by:
$$
\begin{pmatrix}
10 \\
4 \\
16\\
\end{pmatrix}
$$
Result is:
$$
\begin{pmatrix}
6 & 2,4 & 8,4 \\
8 & 1,2 & -11,2 \\
-14 & -3,6 & 21\\
\end{pmatrix}
$$
Which results in:
$$
\begin{pmatrix}
16,8 \\
-2\\
3,4\\
\end{pmatrix}
$$
So according to this logic:
$$x = 16,8$$
$$y = -2$$
$$z = 3,4$$
However when I test this solution it is incorrect, can anyone tell me what I'm doing wrong? Thanks.
| There are mistakes in the last 3 determinants.
$+\begin{vmatrix}4 & 2 \\-3 & 0\end{vmatrix} = 0 + 6 = 6$
$-\begin{vmatrix}1 & 2 \\4 & 0\end{vmatrix} = -(0-8)=+8$
$+\begin{vmatrix}1 & 4 \\4 & -3\end{vmatrix} = -3 - 16 = -19$
Try using these values.
| {
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If complex number $a, b, c, d,$ and $|a|=|b|=|c|=|d|=1$, why $|a(c+d)|+|b(c-d)|\leq 2\sqrt{2}$? If we have 4 complex number $a, b, c, d,$ and $|a|=|b|=|c|=|d|=1$, So, how to prove that $|a(c+d)|+|b(c-d)|\leq 2\sqrt{2}$？
I try to separate $|a(c+d)|+|b(c-d)|$ to $|a||(c+d)|+|b||(c-d)|$ than I get $|(c+d)|+|(c-d)|$. SO, if $c=d, c+d=2c=2b, c-d=0$
Is my idea good?
| Let $c=\operatorname{cis}\alpha$ and $d=\operatorname{cis}\beta$.
Thus, by C-S we obtain: $$|a(c+d)|+|b(c-d)|=|c+d|+|c-d|=\sqrt{2+2\cos(\alpha-\beta)}+\sqrt{2-2\cos(\alpha-\beta)}=$$
$$=2|\cos\frac{\alpha-\beta}{2}|+2|\sin\frac{\alpha-\beta}{2}|\leq2\sqrt{(1+1)\left(\cos^2\frac{\alpha-\beta}{2}+\sin^2\frac{\alpha-\beta}{2}\right)}=2\sqrt2.$$
| {
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Getting a cubic equation from 3 equations Here is the problem:
Three numbers have a sum of $5$ and the sum of their squares is $29$.
If the product of the three numbers is $−10$, what are the three
numbers? Express your answer in simplest radical form.
I used Vieta's formulas to get $x^3-5x^2+12x+10=0$
What should I do after this?
| As @WhatsUp noted in the comments, the equation should have been $x^3-5x^2-2x+10=0$, and so you can find by inspection or by the Rational Root Theorem that $x=5$ is a root. Divide and get $x^2-2=0$, and so your answers should be $x=5, \sqrt{2}, -\sqrt{2}$.
Edit: Another (essentially equivalent, but more detailed) method to get the coefficient of $-2$ for the linear term:
Assume the roots are $a, b, c$.
Square the first equation:
$(a+b+c=5)^2\Rightarrow (a^2+b^2+c^2+2(ab+ac+bc))=25$
We know that $a^2+b^2+c^2=29$ by the second equation, and so we get $2(ab+ac+bc)=-4\Rightarrow ab+ac+bc=-2$
| {
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Determine convergence of the sequence $x_0=1 , x_{n+1}=x_n (1+ 2^{-(n+1)})$ I want to check if the following sequence converges:
$$x_0=1 , x_{n+1}=x_n \left(1+ \frac{1}{2^{n+1}}\right)$$
I proved the sequence is increasing :
$\cfrac{x_{n+1}}{x_n}=1+ \cfrac{1}{2^{n+1}} \gt 1$
Now I should prove it is bounded above. let's write some terms of the equation:
\begin{align}
x_0&=1 \\[2ex]
x_1&=1\cdot\left(1+\dfrac{1}{2^1}\right)\\[2ex]
x_2&=\left(1+\dfrac{1}{2^1}\right)\cdot\left(1+ \dfrac{1}{2^2}\right)\\[2ex]
x_3&=\left(1+\dfrac{1}{2^1}\right)\cdot\left(1+ \dfrac{1}{2^2}\right)\cdot\left(1+ \dfrac{1}{2^3}\right)\\[2ex]
\end{align}
So, we can write:
$$x_{n+1}=\left(1+\frac{1}{2^1}\right)\cdot\left(1+ \frac{1}{2^2}\right)\cdots\left(1+\frac{1}{2^{n}}\right)\cdot\left(1+\frac{1}{2^{n+1}}\right)$$
Here, I'm not sure how to prove it is bounded above.
| First show that $x_n \leq \sqrt{2}^{n+1}$ through mathematical induction. Then,
$0 \leq x_{n+1} - x_n = x_n \frac{1}{2^{n+1}} \leq \frac{1}{\sqrt{2}^{n+1}}$. You can easily check that given sequence is Cauchy sequence.
| {
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How to integrate this area? I need to find the following area between the shapes made by $r=3+2\sin{t}$ and $r=2$:
My problem is if I need to integrate with any function or not. First, to check the angle I just did $2=r=3+2\sin{t} \Rightarrow \sin{t} = -\frac{1}{2} \Rightarrow$ the angle goes from $\frac{7\pi}{6}$ and $\frac{11\pi}{6}$. For the radius, it's from $2$ to $3+2 \sin{t}$. My problem is in the integrals.
It it $\int^{\frac{11\pi}{6}}_{\frac{7\pi}{6}} \int^{3+2 \sin{\theta}}_{2} (1) r drd\theta$? Or $\int^{\frac{11\pi}{6}}_{\frac{7\pi}{6}} \int^{3+2 \sin{\theta}}_{2} (3+2 \sin{\theta} )r drd\theta$?
| Consider the following diagram. Here $dθ$ is the infinitesimally small angle such that we the variation of $r$ with $θ$ can be neglected.
Clearly, we need to find the summation (integral) of the orange areas :
Now consider the following diagrams:
Our Orange area = Blue Area - Green Area
Also, the figures can be approximated as sectors of a circle as the variation of $r$ with $θ$ is negligible.
So, the Orange area,
$$dA=\frac{1}{2}\left(r_{blue}\right)^{2}d\theta\ -\ \frac{1}{2}\left(r_{green}\right)^{2}d\theta$$
$$\int_{-\frac{\pi}{6}}^{\frac{7\pi}{6}}dA=\int_{-\frac{\pi}{6}}^{\frac{7\pi}{6}}\frac{1}{2}\left(3+2\sin\theta\right)^{2}d\theta\ -\ \int_{-\frac{\pi}{6}}^{\frac{7\pi}{6}}\frac{1}{2}\left(2\right)^{2}d\theta$$
Which comes out to be approximately as $24.1870451584$
| {
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In the Euclidean Ring ${\Bbb Z}_3[x]$, what is the remainder of the euclidean division of $x^4+2x^3+x^2+2x+2$ by $x^2+2$? So far I've $(x^2+2)(x^2+2x-1)+4-2x$ but the options for the remainder are:
$2x+1$,
$2x$,
$x+1$, or
$x$.
| Hint:
$$-2=1\pmod 3\;,\;\;4=1\pmod 3\;$$
| {
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How to solve system of modular equivalence with parameter Given system of modular equivalences. Find the smallest natural parameter $a$ that system has solutions
$$\begin{cases}
x \equiv a \mod 140\\
x \equiv 3 \mod 91\\
x \equiv 2a \mod 39
\end{cases}$$
Of course, solution will be appreciated but if you don't want to solve this then please give me some hint, I don't even know how to approach this problem
| If the system is consistent then it has solutions.
Since $140=4\cdot 5\cdot 7$ then the first system is equivalent to $$x \equiv a \pmod {4, 5, 7}$$
and similarly for the other congruences we have $x \equiv 3 \pmod{7, 13}$ and $x\equiv 2a \pmod{3,13}$.
From the first we have $x\equiv a \pmod{7}$ and from the second $x\equiv 3 \pmod{7}$. So for the system to be consistent we must have $a\equiv 3 \pmod{7}$.
Similarly from the second and third congruences we see that $a$ must also satisfy $2a\equiv 3\pmod {13}.$ Then since $2^{-1} \equiv 7 \pmod{7}$, we multiply the congruence by $7$ and obtain that $$14a\equiv a \equiv21\equiv8 \pmod{13}$$
Thus for the system to be consistent it has to satisfy $$a\equiv 3 \pmod{7}\space(1)$$ $$a \equiv8 \pmod{13}\space(2)$$
Since $7$ and $13$ are co-prime, there exists solutions modulo $7\times13=91$ by the CRT.
From $(2)$ we know that $a=8+13k_1$ for $k_1\in\mathbb Z$ and substituting into the first gives $$8+13k_1\equiv 3 \pmod{7}$$
$$13k_1\equiv 2 \pmod{7}$$
So $$13\times 6 k_1\equiv k_1\equiv 12\equiv 5 \pmod{7}$$
Thus $k_1=5+7k_2$ for $k_2\in\mathbb Z$ and hence we have $a=8+13k_1=8+13(5+7k_2)=73+91k_{2}.$
Hence $a\equiv 73 \pmod {91}$ and the smallest natural parameter of $a$ is $73$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3840093",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
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How to prove $\sum_{k=0}^{n}2^{2k}\binom{2n}{2k}=\frac{9^{n}+1}{2}$ by using mathematical induction? My question is how to prove $\sum_{k=0}^{n}2^{2k}\binom{2n}{2k}=\frac{9^{n}+1}{2}$ by using mathematical induction?
I can finish this question by considering the binomial $(1-x)^{2n}$. First, prove that
$$\binom{2n}{0}+\binom{2n}{2}(2)^{2}+\binom{2n}{4}(2)^{4}+\dots+\binom{2n}{2n}(2)^{2n}\\=\binom{2n}{0}+\binom{2n}{1}(2)+\binom{2n}{3}(2)^{3}+\dots+\binom{2n}{2n-1}(2)^{2n-1}$$ by taking $x=2$.
Then, I can prove $\sum_{k=0}^{n}2^{2k}\binom{2n}{2k}=\frac{9^{n}+1}{2}$ by taking $x=-2$.
When I ask to prove this question by using mathematical induction, I can prove that $P(0)$ is true. However, I am stuck in how to use the hypothesis of $P(a)$ is true for some positive integer $a$ to prove $P(a+1)$ is true.
Since I think that it is not correct that using the way which is similar to the above method (considering the binomial $(1-x)^{2a+2}$) to prove the $P(a+1)$ is true (As we need to prove $P(a+1)$ is true if $P(a)$ is true, if we use this way to do then we have ignored the hypothesis of $P(a)$ is true).
May anyone give me some suggestions to finish this question by mathematical induction? Thank you so much.
| Use Pascal’s identity twice to get the identity
$$\binom{n}k=\binom{n-2}{k-2}+2\binom{n-2}{k-1}+\binom{n-2}k\,,$$
and then use that at the start of the induction:
$$\begin{align*}
\sum_{k=0}^{n+1}2^{2k}\binom{2n+2}{2k}&=\sum_{k=0}^{n+1}2^{2k}\left(\binom{2n}{2k-2}+2\binom{2n}{2k-1}+\binom{2n}{2k}\right)\\
&=\sum_{k=1}^{n+1}2^{2k}\binom{2n}{2k-2}+2\sum_{k=1}^n2^{2k}\binom{2n}{2k-1}\\
&\qquad+\sum_{k=0}^n2^{2k}\binom{2n}{2k}\\
&=4\sum_{k=0}^n2^{2k}\binom{2n}{2k}+4\sum_{k=1}^n2^{2k-1}\binom{2n}{2k-1}+\frac{9^n+1}2\\
&=5\left(\frac{9^n+1}2\right)+4\left(\sum_{k=0}^{2n}2^k\binom{2n}k-\frac{9^n+1}2\right)\\
&=5\left(\frac{9^n+1}2\right)+4\left(3^{2n}-\frac{9^n+1}2\right)\\
&=4\cdot9^n+\frac{9^n+1}2\\
&=\frac{9\cdot9^n+1}2\\
&=\frac{9^{n+1}+1}2\,.
\end{align*}$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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By first expanding $(\cos^2x + \sin^2x)^3$, otherwise, show that $ \cos^6x + \sin^6x = 1 - (3/4)\sin^2(2x)$ By first expanding $(\cos^2x + \sin^2x)^3$, otherwise, show that
$$ \cos^6x + \sin^6x = 1 - (3/4)\sin^2(2x)$$
Here's what I've done so far (starting from after expansion):
$\cos^6x + (3\cos^4x\sin^2x) + (3\cos^2x\sin^4x) + \sin^6x$
$\cos^6x + (3\cos^2x\sin^2x)(\cos^2x+\sin^2x) + \sin^6x$
$\cos^6x + (3\cos^2x\sin^2x) + \sin^6x$
$\cos^6x + \sin^6x = -3\cos^2x\sin^2x$
$\cos^6x + \sin^6x = (-3/2)(2\cos^2x\sin^2x)$
$\cos^6x + \sin^6x = (-3/2)(\sin^22x)$
How can I get it into $ 1 - (3/4)\sin^2(2x)$?
| Hint Use $$a^3+b^3=(a+b)\left({(a+b)}^2-3ab \right)$$
with
$a=\cos^2 x, b=\sin^2 x$.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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"answer_id": 2
} |
The identity $\arctan(x) + \arctan(x^3) = \arctan(2x+\sqrt{3}) + \arctan(2x-\sqrt{3})$ I came to this identity while doing some indefinite integrals.
$\arctan(x) + \arctan(x^3) = \arctan(2x+\sqrt{3}) + \arctan(2x-\sqrt{3})$
Seems weird to me, no idea why it's correct but it is.
I wonder if there's some geometric or trigonometric reasoning/insight behind it,
say something which can be useful to high-school students for solving some problems.
Any ideas?
| Hint:
\begin{eqnarray*}
\tan^{-1}(A)+\tan^{-1}(B)=\tan^{-1} \left( \frac{A+B}{1-AB} \right).
\end{eqnarray*}
$$\tan^{-1}(x)+\tan^{-1}(x^3)=\tan^{-1} \left( \frac{x(1+x^2)}{1-x^4}\right)=\tan^{-1} \left( \frac{x}{1-x^2} \right).$$
\begin{eqnarray*}\tan^{-1}(2x+\sqrt{3})+\tan^{-1}(2x-\sqrt{3})&=&\tan^{-1} \left( \frac{4x}{1-(4x^2-3)} \right)\\ &=&\tan^{-1} \left( \frac{x}{1-x^2} \right).\end{eqnarray*}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3843570",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 0
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If $x+y+z=xyz$, prove $\frac{2x}{1-x^2}+\frac{2y}{1-y^2}+\frac{2z}{1-z^2}=\frac{2x}{1-x^2}\times\frac{2y}{1-y^2}\times\frac{2z}{1-z^2}$
If $x+y+z=xyz$, prove
$\frac{2x}{1-x^2}+\frac{2y}{1-y^2}+\frac{2z}{1-z^2}=\frac{2x}{1-x^2}\times\frac{2y}{1-y^2}\times\frac{2z}{1-z^2}$
given that $x^2~,~y^2~,~z^2\ne1$
I came across this question in an ancient ($19$th century) Trigonometry book, and this is the method they use to prove the result (please note: I understand this method fully):
Let $x=\tan A$ ,$~y=\tan B$ and $z=\tan C$ which is acceptable without any loss of generality.
This means that we are saying that
$$\tan A+\tan B+\tan C=\tan A\tan B\tan C$$
Consider $\tan (A+B+C)$:
$$\tan (A+B+C)=\frac{\tan A+\tan B+\tan C-\tan A\tan B\tan C}{1- \tan A \tan B- \tan C \tan A - \tan B \tan C}$$
So if $~\tan A+\tan B+\tan C=\tan A\tan B\tan C~~$ then $~~\tan (A+B+C)=0$. Hence, let $~A+B+C=\pi$.
Now consider $\tan (2A+2B+2C)$:
$$\tan (2A+2B+2C)=\frac{\tan 2A+\tan 2B+\tan 2C-\tan 2A\tan 2B\tan 2C}{1- \tan 2A \tan 2B- \tan 2C \tan 2A - \tan 2B \tan 2C}=0$$
$$\implies \tan 2A+\tan 2B+\tan 2C=\tan 2A\tan 2B\tan 2C$$
$$\implies \frac{2\tan A}{1-\tan^2 A}+\frac{2\tan B}{1-\tan^2 B}+\frac{2\tan C}{1-\tan^2 C}=\frac{2\tan A}{1-\tan^2 A}\times\frac{2\tan B}{1-\tan^2 B}\times\frac{2\tan C}{1-\tan^2 C}$$
$$\therefore\frac{2x}{1-x^2}+\frac{2y}{1-y^2}+\frac{2z}{1-z^2}=\frac{2x}{1-x^2}\times\frac{2y}{1-y^2}\times\frac{2z}{1-z^2}$$
as required.
My question is, is there any other way of proving this without this rather heavy use of trigonometry? I also would prefer not to work through heaps of algabraic manipulation and expansion to obtain the required result, although if that's necessary I will put up with it ;)
Thank you for your help.
| Let $z=0$.
Thus, $x+y=0$ and
$$\sum_{cyc}\frac{2x}{1-x^2}=\frac{2x}{1-x^2}-\frac{2x}{1-x^2}=0=\prod_{cyc}\frac{2x}{1-x^2}.$$
Now, let $xyz\neq0.$
Thus, we need to prove that:
$$\sum_{cyc}\frac{1}{\frac{2x}{1-x^2}\cdot\frac{2y}{1-y^2}}=1.$$
Indeed, $$\sum_{cyc}\frac{1}{\frac{2x}{1-x^2}\cdot\frac{2y}{1-y^2}}=\sum_{cyc}\frac{(1-x^2)(1-y^2)}{4xy}=\sum_{cyc}\frac{(\frac{xyz}{x+y+z}-x^2)(\frac{xyz}{x+y+z}-y^2)}{4xy}=$$
$$=\sum_{cyc}\frac{(yz-x(x+y+z))(xz-y(x+y+z))}{4(x+y+z)^2}=$$
$$=\sum_{cyc}\frac{x^2yz-(x+y+z)(x^2y+x^2z)+xy(x+y+z)^2}{4(x+y+z)^2}=$$
$$=\sum_{cyc}\frac{\frac{1}{3}xyz-x^2y-x^2z+xy(x+y+z)}{4(x+y+z)}=\sum_{cyc}\frac{\frac{1}{3}xyz+xyz}{4xyz}=1.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3844870",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "16",
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$f(m, n) = f(m-1, n) + f(m, n-1) + f(m-1, n-1)$ show that $\sum_{n=0}^{\infty} f(n, n) x^n = \frac{1}{\sqrt{1 - 6x + x^2}}$ Problem Statement: Let $f(m, 0) = f(0, n) = 1$ and $f(m, n) = f(m-1, n) + f(m, n-1) + f(m-1, n-1)$ for $m, n > 0$. Show that
$$\sum_{n=0}^{\infty} f(n, n) x^n = \frac{1}{\sqrt{1 - 6x + x^2}}$$
I was able to figure out the generating function for $f(i, j)$ but not for $f(n, n)$ as shown below.
Let $F(x, y) = \sum_{i\geq 0}\sum_{j \geq 0} f(i, j)x^iy^j$ be the generating function for $f(i,j)$. By the recurrence we have
\begin{equation}
F(x, y) - xF(x, y) - yF(x, y) - xyF(x, y) = 1
\end{equation}
so that $F(x, y) = \frac{1}{1 - x - y - xy}$.
Here I'm thinking that I should plug in appropriate values for $x$, and $y$. Naturally $x = y$ is the first thing that comes to mind but that doesn't give what we want. Any hints would be greatly appreciated!
| Let
$$G(x,t)=F(xt,xt^{-1}).$$
Then
$$G(x,t)=\sum_{i,j}f(i,j)x^{i+j}t^{i-j}
=\sum_{r,s}f((r+s)/2,(r-s)/2)x^rt^s$$
so the terms with $t^0$ therein add to $\sum_r f(r,r)x^{2r}$, essentially what you're
after.
But
$$G(x,t)=\frac{1}{1-x^2-xt-xt^{-1}}$$
and
\begin{align}
1-x^2-xt-xt^{-1}&=
-xt^{-1}\left(t-\frac{1-x^2}{2x}+\frac{\sqrt{1-6x^2+x^4}}{2x}\right)
\left(t-\frac{1-x^2}{2x}-\frac{\sqrt{1-6x^2+x^4}}{2x}\right)\\
&=-xt^{-1}(t-u(x))(t-u(x)^{-1})\\
&=xu(x)^{-1}(1-u(x)t)(1-u(x)t^{-1})
\end{align}
where
$$u(x)=\frac{1-x^2}{2x}-\frac{\sqrt{1-6x^2+x^4}}{2x}$$
is actually a power series in $x$ with zero constant term.
This has a partial fraction expansion
$$G(x,t)=\frac{x^{-1}u(x)}{(1-u(x)t)(1-u(x)t^{-1})}=
\frac{x^{-1}}{u(x)^{-1}-u(x)}\left(\frac{1}{1-u(x)t}
+\frac{u(x)t^{-1}}{1-u(x)t^{-1}}\right).$$
The $t^0$ terms therein, are
$$\frac{x^{-1}}{u(x^{-1}-u(x)}=\frac1{\sqrt{1-6x^2+x^4}}.$$
Then
$$\sum_{n=0}^\infty f(n,n)x^{2n}=\frac1{\sqrt{1-6x^2+x^4}}.$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
} |
Representing $\cos(\frac{π}{11})$ as cyclic infinite nested square roots of $2$ How can we represent $2\cos(\frac{\pi}{11})$ and $2\cos(\frac{\pi}{13})$ as cyclic infinite nested square roots of 2
I have partially answered below for $2\cos(\frac{\pi}{11})$ which I derived it accidentally.
Currently I have figured out for $q$ as denominator with patterns of $2^{n\pm1}$here
Is there any way to figure out pattern of nested radicals for other rational number $p \over q$ in $2\cos(\frac{p}{q})\pi$.
P.S. I am able to figure out that the rational number $p \over q$ must be as follows $1\over4$ < $p\over q$ < $ 1 \over 2$ as $1<$ $\sqrt{2\pm\sqrt{2\pm\sqrt{2\pm...}}}$ $<2$
| For your solution
$ 2 \cos \left(\frac{\pi }{11}\right)=\sqrt{2+\sqrt{2+\sqrt{2-\sqrt{2-\sqrt{2-2 \cos \left(\frac{\pi }{11}\right)}}}}} $
I have add it (with your name) to Page 3 of my Notebook:
http://eslpower.org/Notebook.htm
I found the same result independently on July,2021.
Here is my method:
Since $(2^5+1) \bmod 33=0$, we have
$ 2 \cos \left(\frac{\pi }{33}\right)=\sqrt{2+\sqrt{2+\sqrt{2+\sqrt{2+\sqrt{2-2 \cos \left(\frac{\pi }{33}\right)}}}}} $
and
$ 2 \cos \left(\frac{3\pi }{33}\right)=\sqrt{2+\sqrt{2+\sqrt{2-\sqrt{2-\sqrt{2-2 \cos \left(\frac{3\pi }{33}\right)}}}}} $
then we have
$ 2 \cos \left(\frac{\pi }{11}\right)=\sqrt{2+\sqrt{2+\sqrt{2-\sqrt{2-\sqrt{2-2 \cos \left(\frac{\pi }{11}\right)}}}}} $
For $ 2 \cos \left(\frac{\pi }{13}\right) $, we may use the same method:
Since $(2^6+1) \bmod 13=0$, we have
$ 2 \cos \left(\frac{\pi }{65}\right)=\sqrt{2+\sqrt{2+\sqrt{2+\sqrt{2+\sqrt{2+\sqrt{2-2 \cos \left(\frac{\pi }{65}\right)}}}}}} $
and
$ 2 \cos \left(\frac{5 \pi }{65}\right)=\sqrt{2+\sqrt{2+\sqrt{2-\sqrt{2-\sqrt{2+\sqrt{2-2 \cos \left(\frac{5 \pi }{65}\right)}}}}}} $
then we have
$ 2 \cos \left(\frac{\pi }{13}\right)=\sqrt{2+\sqrt{2+\sqrt{2-\sqrt{2-\sqrt{2+\sqrt{2-2 \cos \left(\frac{\pi }{13}\right)}}}}}} $
For more of my new results, I have post as my answers to your another question:
https://math.stackexchange.com/a/4232525/954936
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Role of coprimality in proof of Fermat's Little Theorem This is the start of the proof for FLT:
I was curious -- I know that all the elements of S are unique because gcd(a,p) = 1, but I was wondering --
What would be an example in which the elements were not unique?
Also, in the next step:
what happened to the a? it just disappeared
|
I was curious -- I know that all the elements of S are unique because gcd(a,p) = 1, but I was wondering -- What would be an example in which the elements were not unique?
Well, one where $p$ isn't prime and if $\gcd(a,p) \ne 1$ then you'd have $\frac p{\gcd(a,p)}$ distinct terms and the repeated $a$ times (minus a final term).
Example $a = 3$ and $p=24$ then $a\cdot 1,a\cdot 2, ......, a\cdot (p-1)\equiv 3,6,9,12,15,18,21,0,3,6,9,12,15,18,21,0,3,6,9,12,15,18,21$
what happened to the a? it just disappeared
Not quite
$a, a\cdot 2, a\cdot 3,...., a\cdot (p-1)$ represents a residue system (without $0$) in no particular order. And $1,2,3,......, p-1$ represent a residue system (without $0$). So each $a\cdot j \equiv k_j \pmod p$ for some $0\le k_j \le p-i$.
So $a\cdot a(2)a(3)......a(p-1)\equiv k_1k_2k_3.....k_{p-1}\pmod p$.
And $\{k_1,k_2,k_3,......k_{p-1}\} = \{1,2,3,....,(p-1)\}$ but not in any particular order.
So $k_1k_2k_3.....k_{p-1} = 1\cdot 2\cdot 3.....(p-1)$ because multiplication is commutative and we can put them in numeric order.
Here's an example:
$3^6 \equiv 1 \pmod 7$ because
$S = \{3,6,9,12,15,18\}$ contains a representative of each class $1....,6$. $1\equiv 15;2\equiv 9,3\equiv 3;4\equiv 18;5\equiv 12; 6\equiv 6$.
$3^6\cdot (1\cdot 2\cdot 3\cdot 4\cdot 5\cdot 6) =$
$3\cdot 6\cdot 9\cdot 12\cdot 15\cdot 18 \equiv $
$3\cdot 6 \cdot 2\cdot 5\cdot 1\cdot 4 \pmod 7\equiv$ (note those are not in any particular order)
$1\cdot 2\cdot 3\cdot 4\cdot 5\cdot 6 \pmod 7$ (but now we just rearranged them into order)
So $3^6(1\cdot ....\cdot 6) \equiv (1\cdot ....\cdot 6)\pmod 7$.
And.... $1\cdot ....\cdot 6$ is relatively prime to $7$ so it is invertible. (Actually be Wilson's Th it is congruent to $-1 \pmod 7$ but ... we don't need to know that.)
$3^6[(1\cdot ....\cdot 6)]((1\cdot ....\cdot 6)^{-1}\equiv (1\cdot ....\cdot 6)(1\cdot ....\cdot 6)^{-1}\pmod 7$ so
So $3^6 \equiv 1\pmod 7$.
| {
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"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 3
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Understand how to evaluate $\lim _{x\to 2}\frac{\sqrt{6-x}-2}{\sqrt{3-x}-1}$ We are given this limit to evaluate:
$$\lim _{x\to 2}\frac{\sqrt{6-x}-2}{\sqrt{3-x}-1}$$
In this example, if we try substitution it will lead to an indeterminate form $\frac{0}{0}$. So, in order to evaluate this limit, we can multiply this expression by the conjugate of the denominator.
$$\lim _{x\to 2}\left(\dfrac{\sqrt{6-x}-2}{\sqrt{3-x}-1} \cdot \dfrac{\sqrt{3-x}+1}{\sqrt{3-x}+1} \right) = \lim _{x\to 2}\left(\dfrac{(\sqrt{6-x}-2)(\sqrt{3-x}+1)}{2-x}\right) $$
But it still gives the indeterminate form $\frac{0}{0}$ .
But multiplying the expression by the conjugate of the demoninator and numerator we get
$$\lim _{x\to 2}\left(\dfrac{\sqrt{6-x}-2}{\sqrt{3-x}-1} \cdot \dfrac{\sqrt{3-x}+1}{\sqrt{3-x}+1} \cdot \dfrac{\sqrt{6-x}+2}{\sqrt{6-x}+2}\right) $$
$$\lim _{x\to 2}\left(\dfrac{6-x-4}{3-x-1} \cdot \dfrac{\sqrt{3-x}+1}{1} \cdot \dfrac{1}{\sqrt{6-x}+2}\right)$$
$$\lim _{x\to 2}\left(\dfrac{6-x-4}{3-x-1} \cdot \dfrac{\sqrt{3-x}+1}{\sqrt{6-x}+2}\right)$$
$$\lim _{x\to 2}\left(\dfrac{2-x}{2-x} \cdot \dfrac{\sqrt{3-x}+1}{\sqrt{6-x}+2}\right)$$
$$\lim _{x\to 2}\left(\dfrac{\sqrt{3-x}+1}{\sqrt{6-x}+2}\right)$$
Now we can evaluate the limit:
$$\lim _{x\to 2}\left(\dfrac{\sqrt{3-2}+1}{\sqrt{6-2}+2}\right) = \dfrac{1}{2}$$
Taking this example, I would like to understand why rationalization was used. What did it change in the expression so the evaluation was possible? Especially, why multiplying by the numerator's and denominator's conjugate?
I am still new to limits and Calculus, so anything concerning concepts I'm missing is appreciated. I still couldn't understand how a limit supposedly tending to $\frac{0}{0}$ went to be $\frac{1}{2}$, I really want to understand it.
Thanks in advance for you answer.
| On the 4th row of the second calculation, you arrive at the following expression:
$$
\frac{2-x}{2-x}
$$
This expression is actually $\frac{0}{0}$ (undefined) at $x=2$. Equating it to $1$ - which is what you did is the same as removing the singularity (the point of where the function is not defined) in the expression and that's why you get the correct limit. It is correct since we have replaced a function that we could not evaluate at $x=2$ with another function that can be evaluated and have the same limit as the one we could not evaluate.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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What is the equation of the circle I've been experiencing a difficulties in answering this. I hope someone will help me in solving this
Find the equation of the circle through the points $(2,8),(7,3)$ and $(-2,0)$.
|
The question can be restated as
Given triangle $ABC$ with coordinates of vertices $A=(-2,0)$,
$B=(7,3)$, $C=(2,8)$, find the equation of its circumscribed circle.
First, find the squares of the side lengths of $\triangle ABC$:
\begin{align}
a^2&=50
,\quad
b^2= 80
,\quad
c^2=90
\tag{1}\label{1}
.
\end{align}
Second, the coordinates of the circumcenter
is known to be found as
\begin{align}
O&=
\frac{a^2(b^2+c^2-a^2)\cdot A+b^2(a^2+c^2-b^2)\cdot B+c^2(b^2+a^2-c^2)\cdot C}
{a^2(b^2+c^2-a^2)+b^2(a^2+c^2-b^2)+c^2(b^2+a^2-c^2)}
\\
&=\tfrac1{12}\cdot(5\,A+4\,B+3\,C)
=\tfrac1{12}\cdot(5\,(-2,0)+4\,(7,3)+3\,(2,8))
=(2,3)
\tag{2}\label{2}
.
\end{align}
Third, find the radius:
\begin{align}
R&=|O-A|=|O-B|=|O-C|=5
\tag{3}\label{3}
.
\end{align}
Hence, the equation of the circle is
\begin{align}
(x-2)^2+(y-3)^2&=25
\tag{4}\label{4}
.
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3859430",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 2
} |
How to arrive at correct parametric form In question 2 of this quiz, the resulting parametric vector form is $x = x_3 \begin{align}
\begin{bmatrix}
1 \\
1 \\
1 \\
\end{bmatrix}
\end{align}$. How do they arrive at $x = x_3 \begin{align}
\begin{bmatrix}
1 \\
1 \\
1 \\
\end{bmatrix}
\end{align}$?
I reduced the augmented matrix given is to $
\left[
\begin{matrix}
1 & 0 & -1 & 0\\
0 & 1 & -1 & 0\\
0 & 0 & 0 & 0\\
\end{matrix}
\right]
$ from that I get $x_1 = x_3$ and $x_2 = x_3$.
I then get the parametric vector form of $x_1\begin{align}
\begin{bmatrix}
1 \\
0 \\
0 \\
\end{bmatrix}
\end{align}$ + $x_2\begin{align}
\begin{bmatrix}
0 \\
1 \\
0 \\
\end{bmatrix}
\end{align}$ + $x_3\begin{align}
\begin{bmatrix}
-1 \\
-1 \\
0 \\
\end{bmatrix}
\end{align}$ = $\begin{align}
\begin{bmatrix}
0 \\
0 \\
0 \\
\end{bmatrix}
\end{align}$
From that please help me understand how to arrive at $x = x_3 \begin{align}
\begin{bmatrix}
1 \\
1 \\
1 \\
\end{bmatrix}
\end{align}$. Thank you.
| If $x=(x_1,x_2,x_3)^T$ is a solution, then you already commented that you know that $x_1=x_2=x_3$. But then
$$
\begin{pmatrix} x_1 \\ x_2 \\ x_3 \end{pmatrix}= \begin{pmatrix} x_3 \\ x_3 \\ x_3\end{pmatrix}= x_3 \begin{pmatrix} 1 \\ 1 \\ 1 \end{pmatrix}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3861413",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
The equation $x^4-x^3-1=0$ has roots $\alpha,\beta,\gamma,\delta$. Find $\alpha^6+\beta^6+\gamma^6+\delta^6$ The equation $x^4-x^3-1=0$ has roots $\alpha,\beta,\gamma,\delta$.
By using the substitution $y=x^3$, or by any other method, find the exact value of
$\alpha^6+\beta^6+\gamma^6+\delta^6$
This is a problem from Further Mathematics(9231) Paper 1, Question 1, 2009. I tried to solve it but was unable to figure it out, especially how to find the value of $\alpha^6+\beta^6+\gamma^6+\delta^6$. Could anyone try to solve this question and explain how they got the value?
| Define the polynomial function $p$ by
$$p(x) = x^4-x^3-1$$
Also define the function $q$ by
$q(x)=p\left(x^{1/3}\right)$. Then,
$$q(t)=t^{4/3}-t-1$$
By definition of $q$, we have that $\alpha^3,\beta^3,\gamma^3,\delta^3$ are roots of the equation $q(t)=0$.
$$q(t)=0\iff t^4 - (t+1)^3=0$$
The polynomial $t^4-(t+1)^3$ is a fourth degree polynomial in $t$, and we know four of its roots -- and since any polynomial of degree $n\geqslant 1$ has exactly $n$ roots, these are the $\textit{only}$ roots.
Now, Vieta's formulae give
$$\alpha^3+\beta^3+\gamma^3+\delta^3 = 1 \:\:\text{ and } \:\:\alpha^3\beta^3+\alpha^3\gamma^3+\alpha^3\delta^3+\beta^3\gamma^3+\beta^3\delta^3+\gamma^3\delta^3=-3$$
Thus, $$\begin{align}\alpha^6+\beta^6+\gamma^6+\delta^6&=\left(\alpha^3+\beta^3+\gamma^3+\delta^3\right)^2-2\left(\alpha^3\beta^3+\alpha^3\gamma^3+\alpha^3\delta^3+\beta^3\gamma^3+\beta^3\delta^3+\gamma^3\delta^3\right)\\ &=7\end{align}$$
The problem is solved. $\square$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3861550",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 1
} |
Range of $f(z)=|1+z|+|1-z+z^2|$ when $ |z|=1$
Find the Range of $f(z)=|1+z|+|1-z+z^2|$ when $z$ is a complex $ |z|=1$
I am able to get some weaker bounds using triangle inequality.
$f(z)< 1+|z|+1+|z|+|z^2|=5$
also $f(z)>|1+z+1-z+z^2|=|z^2+2|>||z^2|-2|=1$,but
these are too weak!.as i have checked with WA
Is there an elgant solution using minimum calculus?. Substituiting $x+iy$ makes things very complex indeed.
Answer:
$[\sqrt{\frac{7}{2}},3\sqrt{\frac{7}{6}}]$
| Note that $1-z+z^2=(1+r)(1+r^2)$ where $r$ is the cube root of unity.
Also, since $|z|=1$, $z\bar z=1$
Now rewrite the given expression as $√ (1+z)(1+\bar z) + √(1-z+z^2)(1-\bar z +\bar z^2)$
$\Rightarrow √ 1+z+\bar z+1 + √(z+r)(z+r^2)(\bar z+r)(\bar z+r^2)$
$\Rightarrow √(2+2\cos\theta) + √(z\bar z+zr+\bar zr+ r^2)(z\bar z+zr^2+\bar zr^2+ r)$
$\Rightarrow √(2+2\cos\theta) + √(2(r^2+r)z+2(r^2+r)\bar z +r+r^2+z^2+\bar z^2+4)$
$\Rightarrow √(2+2\cos\theta) + √(2(r^2+r+1)z+2(r^2+r+1)\bar z -1+(z+\bar z)^2-2z\bar z)$
$\Rightarrow √(2+2\cos\theta) + √((2\cos\theta)^2-3)$
$\Rightarrow (√2)\cos\frac{\theta}{2} + √(4\cos^2\theta-3)$
All that's left is minimising and maximsing this expression
which can be done with simple calculus.
Appendix:
$1-z+z^2=(z-\frac{1}{2})^2+\frac{3}{4}=0$
$\Rightarrow z-\frac{1}{2}=±i√\frac{3}{2}$
$\Rightarrow z=\frac{1}{2}+(±i√\frac{3}{2})$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3865237",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 3,
"answer_id": 2
} |
Proving that $\sum_{k=0}^{k=n} \binom{2n}{k} \cdot k = 2^{2n -1} \cdot n$ I have to show that $\sum_{k=0}^{k=n} \binom{2n}{k} \cdot k = 2^{2n -1} \cdot n$.
What I know is that $\sum_{k=0}^{k=n} \binom{n}{k} \cdot k = 2^{n -1} \cdot n$.
How do I proceed from there?
| \begin{align}
\sum_{k=0}^n k\binom{2n}{k}
&= \sum_{k=1}^n k\binom{2n}{k} \\
&= \sum_{k=1}^n 2n\binom{2n-1}{k-1} \\
&= n \sum_{k=1}^n \left(\binom{2n-1}{k-1} + \binom{2n-1}{k-1}\right) \\
&= n \sum_{k=1}^n \left(\binom{2n-1}{k-1} + \binom{2n-1}{2n-k}\right) \\
&= n \sum_{j=0}^{2n-1} \binom{2n-1}{j} \\
&= n \cdot 2^{2n-1}
\end{align}
Alternatively, a combinatorial proof is to count the number of committees of size at most $n$ with one chairperson from $2n$ people. The LHS conditions on the size $k$ of the committee. The RHS selects the chairperson (in $2n$ ways) and then any subset of size at most $n-1$ from the remaining $2n-1$ people (to see that there are $2^{2n-1}/2$ of these, consider complementary pairs).
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3866083",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
Find asymptotic behavior of $a_n$ We are given $a_1=1$ and $a_{n+1}=\dfrac 1 {\sqrt n \ a_n}$.
At first I found through continuous analogue that $a_n=2\sqrt{\sqrt n}$.
Then I got the first inequality : $a_{n+1}^2 -a_1 ^2 >2 \left(1+ \dots +\dfrac 1 {\sqrt n} \right)$.
Then I obtained $a_n^2 > \dfrac {2n}{\sqrt n}$.
For the second inequality I obtained $a_{n+1}^2 - a_1^2 < \dfrac 2 {\sqrt n}+\dfrac 1 {2 \sqrt n} \left(1+\dots+\dfrac 1n\right)$,
but I am not sure if I approached the asymptotes of $a_n$ correctly.
| It follows by induction that for any $n\geq 2$,
$$
a_{2n} = \frac{{\sqrt {2n - 2} \cdots \sqrt 2 }}{{\sqrt {2n - 1} \sqrt {2n - 3} \cdots \sqrt 3 }} = \sqrt {\frac{{(2n - 2)!!}}{{(2n - 1)!!}}}
$$
and
$$
a_{2n + 1} = \frac{{\sqrt {2n - 1} \cdots \sqrt 3 }}{{\sqrt {2n} \sqrt {2n - 2} \cdots \sqrt 2 }} = \sqrt {\frac{{(2n - 1)!!}}{{(2n)!!}}} .
$$
Then, by using for example Stirling's formula,
$$
a_{2n} \sim \sqrt {\frac{1}{2}\sqrt {\frac{\pi }{n}} } ,\quad a_{2n + 1} \sim \frac{1}{{\sqrt {\sqrt {\pi n} } }}
$$
as $n\to +\infty$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3869523",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
solving for $x$:$\cos^2 3x+\frac{\cos^2 x}{4}=\cos 3x\cos^4 x$
solve for x:$$\cos^2 3x+\frac{\cos^2 x}{4}=\cos 3x\cos^4 x$$
My attempt:
completing square: $${(\cos 3x-\frac{\cos x}{2})}^2=\cos 3x\cos x (1-\cos^3 x)$$ or
$$\cos x\cos 3x\ge 0$$
also by some basic identities :$$1+\cos 6x+\frac{1+\cos 2x}{4}=(\cos 4x+\cos 2x)\left(\frac{\cos 3x+3\cos x}{4}\right)$$
I cant proceed now.....
| $$\color{blue}{\cos^2{3x}} - (\cos^4{x})\color{blue}{\cos{3x}} + \dfrac{\cos^2{x}}{4}=0$$
$$ \color{blue}{\cos{3x}} = \dfrac{\cos^4{x} \pm \sqrt{\cos^8{x} - \cos^2{x}}}{2}$$
Use $\cos^8{x} - \cos^2{x} \le 0$.
Similarly,
$$\cos{3x}\color{red}{\cos^4{x}} - \dfrac{\color{red}{\cos^2{x}}}{4} - \cos^2{3x}=0$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3871800",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 1,
"answer_id": 0
} |
I have a problem from Karush-Kuhn-Tucker condition maximize: $xy-x$ and s.t. $x^2+y^2 \leq 9$ and $x \ge 0$
Then, I tried to find points using lagrange method. I used 4 cases and they are $(1)$ $\lambda_1 =0$ and $\lambda_2 = 0$
forrect?
| Maximize $xy - x$, given constraints: $x^2 + y^2 \leq 9, x \ge 0$
Taking $xy - x = \lambda (x^2 + y^2 -9)$
Taking derivative wrt $x$ and $y$, we get -
$y-1 = 2 \lambda x$ ...(i)
$x = 2 \lambda y$ ...(ii)
From (ii), $\lambda = \frac{x}{2y}$ and substituting in (i)
we get $y(y-1) = x^2 \implies y(y-1) + y^2 \leq 9$
That gives, $y^2 - \frac{y}{2} \leq \frac{9}{2} \implies (y - \frac{1}{4})^2 \leq \frac{73}{16}$
So $y \leq \frac{\sqrt 73 + 1}{4} \approx 2.386001\, $ (only taking positive values as $x \ge 0$ and we need $xy$ to be positive for max value)
taking max value of $y$, gives $x \approx 1.818516$
And max value of $xy-x \approx 2.520465$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3874809",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
} |
Find the roots of the polynomial $x^3-2$. Find the roots of the polynomial $x^3-2$.
If $\alpha$ be the root of this polynomial i.e., $\alpha^3=2$,
then $(\zeta \alpha)^3=2$, where $\zeta$ is $3^{rd}$ root of unity.
Hence the solutions of this equations are
$$\zeta{2}^\frac{1}{3}$$, $\zeta$ a $p^{th}$ root of unity
And we know that $\zeta_n=e^{\frac{2\pi i}{n}}$
So the roots are $2^\frac{1}{3}, -(2^\frac{1}{3}) ~and~ 2^\frac{1}{3}\frac{1}{2}(\sqrt{3}i-1).$
Is this correct?
Note: I don't like factoring method. I studied in schools time. So don't help with that method.
Thanks!
| Forward
i) Euler's formula: $e^{i\theta}=\cos\theta + i\sin\theta$
ii) De Moivre's formula: $\left(\cos x+i\sin x\right)^n=\cos (nx) + i\sin (nx)$
Theorem
If $z\ne 0$, and if $n$ is a positive integer, then there are exactly $n$ distinct complex numbers $z_0,z_1,\dots,z_{n-1}$ such that $z_k^n=z$, $k=0,\dots,n-1$. These roots are given by the formula:
$$z_k=\lvert{z}\rvert^{\frac1n}\exp\left[ i\left(\frac{\text{arg}(z)}{n}+\frac{2\pi k}{n}\right)\right],\qquad k=0,\dots,n-1$$
(e.g. Apostol, Mathematical Analysis, Theorem 1.51)
Roots of unity
Since $\lvert 1\rvert=1$ and $\text{arg}(1)=0$, the $n$ roots of unity are:
\begin{align*}z_k&=1^{\frac1n}\exp\left[i\left(\frac{\text{arg}(z)}{n}+\frac{2\pi k}{n}\right)\right]=\exp\left(i\frac{2\pi k}{n}\right)\\&=\cos\frac{2\pi k}{n}+i\sin\frac{2\pi k}{n}\end{align*}
When $n=3$, the three roots of unity are:
i) $k=0$: $$z_0=\cos 0+i\sin 0=1$$and $1^3=1$;
ii) $k=1$: $$z_1=\cos \frac{2\pi}{3}+i\sin\frac{2\pi}{3}=\frac{-1+i\sqrt{3}}{2}$$and $\left(\cos \frac{2\pi}{3}+i\sin\frac{2\pi}{3}\right)^3=\cos (2\pi)+i\sin(2\pi)=1$;
iii) $k=2$: $$z_2=\cos \frac{4\pi}{3}+i\sin\frac{4\pi}{3}=\frac{-1-i\sqrt{3}}{2}$$and $\left(\cos \frac{4\pi}{3}+i\sin\frac{4\pi}{3}\right)^3=\cos (4\pi)+i\sin(4\pi)=1$.
Roots of a real positive number
If $x^3=2$, then the three roots are:
$$x_0=2^\frac13,\quad x_1=2^\frac13\left(\frac{-1+i\sqrt{3}}{2}\right),\quad x_2=2^\frac13\left(\frac{-1-i\sqrt{3}}{2}\right)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3876285",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 4
} |
Solving ODE, check my answer. I'm trying to solve this ODE but I'm not sure with my answer. Anyone please check my answer. If my work is not true, tell my mistake. Thanks.
Solve the ODE $$\left(x^2+2xy\right)\dfrac{dy}{dx}=y^2-2xy.$$
Solution.
\begin{alignat}{2}
&& \left(x^2+2xy\right)\dfrac{dy}{dx}&=y^2-2xy\nonumber\\
\Longleftrightarrow\quad
&&\left(y^2-2xy\right)dx-\left(x^2+2xy\right)dy&=0\nonumber\\
\Longleftrightarrow\quad
&&\left(\left(\dfrac{y}{x}\right)^2-2\left(\dfrac{y}{x}\right)\right)dx-\left(1+2\left(\dfrac{y}{x}\right)\right)dy&=0.
\end{alignat}
Let $u=\dfrac{y}{x}$, then $y=ux$. We have
\begin{alignat}{1}
dy&=u dx+xdu\label{p8}
\end{alignat}
and
\begin{alignat}{2}
&&\left(u^2-2u\right)dx-\left(1+2u\right)dy&=0.\label{p9}
\end{alignat}
Now, we have
\begin{alignat}{2}
&&\left(u^2-2u\right)dx-\left(1+2u\right)\left(u dx+xdu\right)&=0\nonumber\\
\Longleftrightarrow\quad
&&\left(u^2-2u\right)dx-\left(u+2u^2\right) dx-\left(1+2u\right)xdu&=0\nonumber\\
\Longleftrightarrow\quad
&&\left(-u^2-3u\right)dx-\left(1+2u\right)xdu&=0\nonumber\\
\Longleftrightarrow\quad
&&\left(u^2+3u\right)dx&=-\left(1+2u\right)xdu\nonumber\\
\Longleftrightarrow\quad
&&\dfrac{dx}{x}&=-\dfrac{1+2u}{u^2+3u}du\label{p10}
\end{alignat}
By Integrating last equation, we have
\begin{alignat}{2}
&&\int \dfrac{dx}{x}&=\int -\dfrac{1+2u}{u^2+3u}du\nonumber\\
\Longleftrightarrow\quad
&&\ln x &= -\left(\int \dfrac{2u+3}{u^2+3u}du
-
\int \dfrac{2}{u^2+3u}du
\right)\label{p11}
\end{alignat}
We find the integral on RHS as below.
\begin{alignat*}{1}
\int \dfrac{2u+3}{u^2+3u}du&=\int \dfrac{1}{u^2+3u}d\left(u^2+3u\right)\\
&= \ln(u^2+3u)\\
&= \ln u +\ln(u+3)
\end{alignat*}
\begin{alignat*}{1}
\int \dfrac{2}{u^2+3u}du&=\int \dfrac{2}{u(u+3)}du\\
&=\int \dfrac{\frac{2}{3}}{u}du + \int \dfrac{-\frac{2}{3}}{u+3} du\\
&=\dfrac{2}{3} \ln u-\dfrac{2}{3}\ln{\left(u+3\right)}\\
\end{alignat*}
Now we have
\begin{alignat*}{2}
&&\ln x &= -\left(\left(\ln u +\ln(u+3)\right)
-
\left(\dfrac{2}{3} \ln u-\dfrac{2}{3}\ln{\left(u+3\right)}\right)+C
\right)\\
\Longleftrightarrow\quad
&&\ln x &= -\left(\dfrac{1}{3}\ln u +\dfrac{5}{3}\ln(u+3)
\right)+C\\
\Longleftrightarrow\quad
&&\ln x &= \ln \left(u^{\frac{1}{3}}(u+3)^{\frac{5}{3}}\right)^{-1}+C\\
\Longleftrightarrow\quad
&&x &= K\left(u^{\frac{1}{3}}(u+3)^{\frac{5}{3}}\right)^{-1}, \text{ } K=e^C\\
\Longleftrightarrow\quad
&&x &= K\left(\left(\dfrac{x}{y}\right)^{\frac{1}{3}}\left(\dfrac{x}{y}+3\right)^{\frac{5}{3}}\right)^{-1}.
\end{alignat*}
We get the general solution of ODE
$$x = K\left(\left(\dfrac{x}{y}\right)^{\frac{1}{3}}\left(\dfrac{x}{y}+3\right)^{\frac{5}{3}}\right)^{-1}.$$
or we can write
\begin{alignat*}{2}
&&x &= K\left(\left(\dfrac{x}{y}\right)^{\frac{1}{3}}\left(\dfrac{x}{y}+3\right)^{\frac{5}{3}}\right)^{-1}\\
\Longleftrightarrow\quad
&&x\left(\dfrac{x}{y}\right)^{\frac{1}{3}}\left(\dfrac{x}{y}+3\right)^{\frac{5}{3}} &= K\\
\Longleftrightarrow\quad
&&\left(\dfrac{x^4}{y}\right)\left(\dfrac{x}{y}+3\right)^{5} &= c, \text{} c=K^3.
\end{alignat*}
| Since that $$-\frac{5}{3}\ln(u+3)-\frac{1}{3}\ln(u)=\ln(x)+c \overbrace{\implies}^{u=\frac{y}{x}}\boxed{-\frac{5}{3}\ln\left(\frac{y}{x}+3 \right)-\frac{1}{3}\ln \left(\frac{y}{x} \right)=\ln(x)+c}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3877643",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Differential equation with integrating factor
Hey I am supposed to solve the following differential equation:
$(1-x^{2}y)dx+x^{2}(y-x)dy=0$
I found integrating factor:
$\varphi (x)=-\frac{2}{x}$
So I multiply my original equation and I got:
$\left ( \frac{1}{x^{2}} -y\right )dx+\left ( y-x \right )dy=0$
But then I try to integrate it and I got stuck. The answer should be:
$y^{2}-2xy-\frac{2}{x}=C$
Can somebody help me?
Thanks
| $(1-x^2y)~dx+x^2(y-x)~dy=0$
$x^2(y-x)~dy=(x^2y-1)~dx$
$(y-x)\dfrac{dy}{dx}=y-\dfrac{1}{x^2}$
Let $u=y-x$ ,
Then $y=u+x$
$\dfrac{dy}{dx}=\dfrac{du}{dx}+1$
$\therefore u\left(\dfrac{du}{dx}+1\right)=u+x-\dfrac{1}{x^2}$
$u\dfrac{du}{dx}+u=u+x-\dfrac{1}{x^2}$
$u\dfrac{du}{dx}=x-\dfrac{1}{x^2}$
$u~du=\left(x-\dfrac{1}{x^2}\right)$
$\int u~du=\int\left(x-\dfrac{1}{x^2}\right)~dx$
$\dfrac{u^2}{2}=\dfrac{x^2}{2}+\dfrac{1}{x}+c$
$u^2=x^2+\dfrac{2}{x}+C$
$(y-x)^2=x^2+\dfrac{2}{x}+C$
$y^2-2xy+x^2=x^2+\dfrac{2}{x}+C$
$y^2-2xy=\dfrac{2}{x}+C$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3880485",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Simplify $\frac{\sin t - \cos t}{\sin t + \cos t}$ $$\frac{\sin t - \cos t}{\sin t + \cos t}$$
I am to simplify this into $$\tan(t - \frac{\pi}{4})$$
I'm not sure how to carry on, though. Multiplying numerator and denominator by denominator (or numerator) doesn't get me anywhere.
| Step-by-step
$$ \dfrac{\sin t - \cos t}{\sin t + \cos t}$$
$$= \dfrac{\dfrac{\sin t}{\cos t} - \dfrac{\cos t}{\cos t}}{\dfrac{\sin t}{\cos t} + \dfrac{\cos t}{\cos t}}$$
$$= \dfrac{\tan t - 1}{\tan t + 1}$$
$$= \dfrac{\tan t - 1}{1 + \tan t \color{blue}{\cdot 1}}$$
Substitute $\color{blue}{\tan \frac{\pi}{4} = 1}$
$$= \dfrac{\tan t - \tan \frac{\pi}{4}}{1 + \tan t \color{blue}{\tan \frac{\pi}{4}}}$$
$$= \tan (t - \frac{\pi}{4})$$
We have used the formula
$$ \tan(A-B) = \dfrac{\tan A - \tan B}{1 + \tan A \tan B} $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3883650",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
Find the max. and min. points of $g(x,y,z) = x^2 + y^2 + z^2$ The problem
Let $X$ a submanifold of $\mathbb{R}^3$ defined by $$ X = \left\{ (x,y,z) \mid (x^2 + y^2 + z^2 - 5)^2 = 16(1-z^2) \right\} $$
Let $\Phi(x,y,z) = (x^2 + y^2 + z^2 - 5)^2 - 16(1-z^2) $. Let $f$ a function defined by
$$ f : \mathbb{R}^3 \rightarrow \mathbb{R} \mbox{, } (x,y,z) \mapsto x^2 + y^2 + z^2 $$
We define $g$ by the restriction of $f$ on $X$ : $ g = \left. f \right|_X $
Give the points in which the differential of $g$ is zero and deduce the max. and min. points.
My works
$X$ is compact and $g$ is continuous on $X$, then $g$ has a global max. and min. point. Thus, $g$ is $\mathscr{C}^{\infty}$ on $X$. Then, we can apply the method of Lagrange multipliers. We have
$$ \frac{ \partial g }{ \partial x } = 2x \mbox{, } \frac{ \partial g }{ \partial y } = 2y \mbox{, } \frac{ \partial g }{ \partial z } = 2z $$
and
$$ \frac{ \partial \Phi }{ \partial x } = 4x(x^2 + y^2 + z^2 - 5) \mbox{, } \frac{ \partial \Phi }{ \partial y } = 4y(x^2 + y^2 + z^2 - 5) \mbox{, } \frac{ \partial \Phi }{ \partial z } = 4z(x^2 + y^2 + z^2 + 3) $$
We get the following system
$$ \left( \begin{array}{c}
2x\\ 2y\\ 2z\\
\end{array} \right) = \lambda \left( \begin{array}{c}
4x(x^2 + y^2 + z^2 - 5) \\
4y(x^2 + y^2 + z^2 - 5) \\
4z(x^2 + y^2 + z^2 + 3) \\
\end{array} \right) $$
with $\lambda$ a real scalar. And I don't know how to resolve this system... Actually, I am not sure if this is the system we have to get. I think I made a mistake somewhere but I don't know where.
Thank you for your help.
| Your working is correct -
$\left( \begin{array}{c}
2x\\ 2y\\ 2z
\end{array} \right) = \lambda \left( \begin{array}{c}
4x(x^2 + y^2 + z^2 - 5) \\
4y(x^2 + y^2 + z^2 - 5) \\
4z(x^2 + y^2 + z^2 + 3) \\
\end{array} \right)$
Also, $(x^2 + y^2 + z^2 - 5)^2 = 16(1-z^2)$.
From above system of equations, please note the possible points to check for max and min.
One is $(x = 0, y = 0, z = 0)$ but it does not meet the fourth equation constraint.
If $x \ne 0$ then from first equation $\lambda = \frac{1}{2(x^2+y^2+z^2-5)}$ but that will lead to $z = 0$ so you have another set of points $(x, y, 0)$.
Similarly if $z \ne 0$ then from the third equation $\lambda = \frac{1}{2(x^2+y^2+z^2+3)}$ but that leads to $x = y = 0$. So the next set of possible points are $(0, 0, z)$.
Taking points $(x, y, 0)$ and plugging into the fourth equation, you have -
$x^2+y^2 - 5 = \pm4 \implies x^2+y^2 = 9$ or $x^2+y^2 = 1$.
Substituting $(0, 0, z)$ in fourth equation shows no real solution.
So max and min of $x^2 + y^2 + z^2$ for the given equality constraint seem to be $9$ and $1$ resp, and for all points on $x^2 + y^2 = 9, z = 0$ and $x^2 + y^2 = 1, z = 0$ resp.
| {
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"url": "https://math.stackexchange.com/questions/3888140",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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} |
prove thatt $\sum_{cyc} \frac{1}{{(x+y)}^2}\ge 9/4$
prove that $$\sum_{cyc} \frac{1}{{(x+y)}^2}\ge 9/4$$ where $x,y,z$ are positives such that $xy+yz+xz=1$
By Holder;$$\left(\sum_{cyc} \frac{1}{{(x+y)}^2} \right){\left(\sum yz+zx \right)}^2\ge {\sum \left(z^{2/3} \right)}^{3}$$.
Hence it suffices to prove $$\sum {\left(z^{2/3} \right)}^{3}\ge 3^{2/3}$$ which is falsse.
I was able to get a weaker bound.since $xy+yz+zx=1$ we have $x^2+y^2\le 2$. By C-S and $x^2+y^2<2$
$$\sum_{cyc} \frac{1}{{(x+y)}^2}\ge \frac{1}{2}\sum \frac{1}{x^2+y^2}>3/4$$
I am interested on a solution using standard inequalities (C-S AM-GM,chebyshov etc) rather than fully expanding and using uvw/pqr/schur.
| This is the solution by Vo Quoc Ba Can.
Need to prove:
$$(xy+yz+zx)\left(\frac{1}{(x+y)^2}+\frac{1}{(y+z)^2}+\frac{1}{(z+x)^2}\right)\geqq \frac{9}{4}$$
Wlog, assume $x\geq y \geq z,$ we have the following lemma:
$$\sum\limits_{cyc} \frac{1}{(x+y)^2}\geqq \frac{1}{4xy}+\frac{2}{(x+z)(y+z)}\quad \quad(1)$$
$$\Leftrightarrow \left(\frac{1}{x+z}-\frac{1}{y+z}\right)^2\geqq \frac{(x-y)^2}{4xy(x+y)^2}$$
$$\Leftrightarrow 4xy(x+y)^2 \geqq (x+z)^2(y+z)^2$$
Since $x\geqq y \geqq z \therefore 4xy\geqq 4y^2\geqq (y+z)^2;(x+y)^2 \geqq (x+z)^2.$
So $(1)$ is true, need to prove:
$$(xy+yz+zx)\left(\frac{1}{4xy}+\frac{2}{(x+z)(y+z)}\right)\geqq\frac{9}{4}$$
But it's $$(x+y)(y+z)(z+x)\geq 8xyz,$$
which is true by AM-GM.
| {
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"url": "https://math.stackexchange.com/questions/3889789",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Is it valid to apply L'hopital rule to evaluate the limit? $$\lim_{x\to 0^+}\frac{2\tan x(1-\cos x)}{\sqrt{x^2+x+1}-1}$$
In the book I am reading, the limit evaluated in this way:
$$\lim_{x\to 0^+}\frac{2\tan x(1-\cos x)}{\sqrt{x^2+x+1}-1}\times \frac{\sqrt{x^2+x+1}+1}{\sqrt{x^2+x+1}+1}=\lim_{x\to 0^+}\frac{2\tan x(1-\cos x)}{x^2+x}\times\left(\sqrt{x^2+x+1}+1\right)$$
Then it used equivalence and wrote:
$$\lim_{x\to 0^+}\frac{2x\left(\frac12x^2\right)}{x^2+x}\times\left(\sqrt{x^2+x+1}\right)=\lim_{x\to 0^+}\frac{x^3}{x(x+1)}.\left(\sqrt{x^2+x+1}\right)=0$$
I wonder why we should do all these calculation? Is it possible to use L'hopital rule and get $\frac0{\tfrac12}=0$ ?
| Yes since the expression is in the form $\frac 0 0$ we are allowed to use l'Hospital to obtain
$$\lim_{x\to 0^+}\frac{2\tan x(1-\cos x)}{\sqrt{x^2+x+1}-1}=\lim_{x\to 0^+}\frac{2 \sec^2 x - 2 \cos x}{\frac{2 x + 1}{2 \sqrt{x^2 + x + 1}}}=\frac{0}{\frac12}=0$$
Note that often use l'Hospital's rule is esplicitely not allowed because when we use it blindly we can't really understand what it is going to determine the limit. For these reason is good to solve limits in more than one way, using standard limits when possible.
In this case as an alternative we can proceed as follows
$$\frac{2\tan x(1-\cos x)}{\sqrt{x^2+x+1}-1}=2\frac{\tan x}{x}\frac{1-\cos x}{x^2}\frac{x^3}{\sqrt{x^2+x+1}-1}\frac{\sqrt{x^2+x+1}+1}{\sqrt{x^2+x+1}+1}$$
which allows to conclude by standard limits.
I also exhort not expert users to do not proceed blindly by asymptotic equivalence since it can leads to wrong results if we use this way without the necessary attention. For example as $x \to 0$ for $\frac{1- \cos x}{x^2}$ we could wrongly state that limit is zero since $\cos x \sim 1$ or with $\frac{x- \sin x}{x^3}$ also since $\sin x \sim x$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3890051",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 1
} |
Stronger than Nesbitt's inequality using convexity and functions Hi it's a refinement of Nesbitt's inequality and for that, we introduce the function :
$$f(x)=\frac{x}{a+b}+\frac{b}{x+a}+\frac{a}{b+x}$$
With $a,b,x>0$
Due to homogeneity we assume $a+b=1$ and we introduce the function :
$$g(a)=\frac{a}{1-a+x}$$
Showing that $g(a)$ is convex on $(0,1)$ is not hard so we have :
$$\frac{b}{x+a}+\frac{a}{b+x}\geq 2\frac{a+b}{2(\frac{a+b}{2}+x)}=h(x)$$
So we have :
$$f(x)\geq h(x)+\frac{x}{a+b}$$
Now we put $u=\frac{x}{a+b}$ and we want to show :
$$h(x)+\frac{x}{a+b}=u+\frac{1}{0.5+u}\geq \frac{3}{2}$$
The last inequality is obvious.
My question :
It is correct?
Do you know other refinements?
Thanks in advance!
Ps: I add the tag reference request for the last question.
| There are very many refinements of the Nesbitt's inequality.
For example. For positives $a$, $b$ and $c$ we have:
1.$$\frac{a}{\sqrt[3]{4(b^3+c^3)}}+\frac{b}{a+c}+\frac{c}{a+b}\geq\frac{3}{2};$$
2.$$\frac{a}{b+c}+\frac{b}{a+c}+\frac{c}{a+b}\geq\frac{(a+b+c)^2}{2(ab+ac+bc)};$$
3.$$\frac{a}{b+c}+\frac{b}{a+c}+\frac{c}{a+b}\geq\frac{3\sqrt{5(a^2+b^2+c^2)-ab-ac-bc}}{4(a+b+c)};$$
4. For any reals $a$, $b$ and $c$ such that $ab+ac+bc>0$ prove that:
$$\frac{a}{b+c}+\frac{b}{a+c}+\frac{c}{a+b}\geq\frac{3}{2}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3890210",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
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