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How find $\sum_{k \in A} \frac{1}{k-1} $ for $ A = \{ m^n| \text{ } m, n \in Z \text { and } m, n \ge 2 \} $ If $ A = \{ m^n| \text{ } m, n \in Z \text { and } m, n \ge 2 \} $, then how find $\sum_{k \in A} \frac{1}{k-1} $?
| The answer is 1.
The result is known as Goldbach-Euler theorem.
See Wikipedia entry for "proof".
For rigorous proof, you could consider sum of reciprocals of all perfect powers, $S$. Note that sum equals
$$
S = \sum_{x \in B}\sum_{n = 2}^{\infty}\frac{1}{x^{n}} = \frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\ldots+\frac{1}{9}+\frac{1}{27}+\frac{1}{81}+\ldots+\frac{1}{25}+\frac{1}{125}+\ldots
$$
where $B$ is set of all integers not perfect power. Why? All reciprocals of perfect powers are clearly in the sum, and terms in sum are distinct: if
$$
\frac{1}{x^{n}} = \frac{1}{y^{m}}
$$
and $x$ and $y$ are distinct, also $m$ and $n$ are. But since $x$ and $y$ aren't perfect powers, $x^{n}$ and $y^{m}$ are exactly $n$:th and $m$:th powers, respectively, for distinct $m$ and $n$ which is impossible.
But now
$$
S = \sum_{x \in B}\sum_{n = 2}^{\infty}\frac{1}{x^{n}} = \sum_{x = 2}^{\infty}\sum_{n = 2}^{\infty}\frac{1}{x^{n}}-\sum_{x \in A}\sum_{n = 2}^{\infty}\frac{1}{x^{n}} \\
= \sum_{x = 2}^{\infty}\frac{1}{x(x-1)}-\sum_{x \in A}\frac{1}{x(x-1)} \\
= 1- \sum_{x \in A}\left(\frac{1}{x-1}-\frac{1}{x}\right) = 1- \sum_{x \in A}\frac{1}{x-1}+\sum_{x \in A}\frac{1}{x} \\
= 1- \sum_{x \in A}\frac{1}{x-1}+S
$$
so
$$
\sum_{x \in A}\frac{1}{x-1} = 1
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/900884",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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Finding the locus - middle point of a line segment Question:
Find the locus of the middle point of the portion of the line $x\cos \alpha + y\sin \alpha = p$ which is intercepted between the axes, given that $p$ remains constant.
No idea how to even approach this problem. Please help!
| Locus is the set of points that satisfy the given condition.
Here, let $M(h,k)$ is the general point and the given condition is that it must be the midpoint of the distance between the axes of the line $x\cos{\alpha}+y\sin{\alpha}=p$.
We need to find the equation of the locus in terms of $x-$ and $y-$ axis.
$M(h,k)=(\frac{p}{2\cos\alpha},\frac{p}{2\sin\alpha})$
So,
$$
x^2+y^2=(\frac{p}{2\cos\alpha})^2+(\frac{p}{2\sin\alpha})^2=\frac{p^2}{4.\sin^2\alpha\cos^2\alpha}
$$
For $M(h,k)$ we have $x=\frac{p}{2\cos\alpha}$ and $y=\frac{p}{2\sin\alpha}$$\implies \cos\alpha=\frac{p}{2.x}$ and $\sin\alpha=\frac{p}{2y}$
Substituting,
$$
x^2+y^2=\frac{p^2}{4}.\frac{4.y^2.4.x^2}{p^4}=\frac{4.x^2y^2}{p^2}\implies\frac{1}{x^2}+\frac{1}{y^2}=\frac{4}{p^2}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/901752",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Trigonometry Identity: $\tan \theta\sin \theta + \cos \theta = \sec \theta$ Sorry if my question seems too simple. I cannot find a proof and my text book does not provide one either. I am supposed to prove:
$$\tan \theta \times \sin \theta + \cos \theta = \sec \theta$$
I know that $\sec = \frac{1}{\cos\theta}$. But I do not know how to prove that $\tan \theta \times \sin \theta + \cos \theta = \frac{1}{\cos \theta}$.
I appreciate if someone point me to the right direction.
| $$\begin{align*}
\tan \theta \sin \theta + \cos \theta & \stackrel{\text{def.}}= \frac{\sin^2 \theta}{\cos \theta} + \cos \theta \\
& \stackrel{\text{Pythagoras}}= \frac{1-\cos^2 \theta}{\cos \theta} + \cos \theta \\
& = \frac1{\cos\theta} - \cos \theta + \cos \theta \\
& \stackrel{\text{def.}}= \sec\theta
\end{align*}$$
Where we use the definitions of $\tan \theta$ and $\sec\theta$ plus Pythagoras' theorem $\sin^2 \theta + \cos^2 \theta = 1$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/901835",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 0
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Stuck simplifying a fractional expression $$ \frac { \frac { 1 }{ 1+x+h } -\frac { 1 }{ 1+x } }{ h } $$
$$ \frac { 1(1+x) }{ 1+x+h(1+x) } -\frac { 1(1+x+h) }{ 1+x(1+x+h) } $$
$$ \frac { -h }{ (1+x+h)(1+x) } \quad *\quad \frac { 1 }{ h } $$
This is what I have so far. I have no idea what my next step is. I get to:
$$\frac { -h }{ (1+x+h)(1+x)(h) } $$
and don't know where to go from here..
| $$\frac { \frac { 1 }{ 1+x+h } -\frac { 1 }{ 1+x } }{ h }\cdot \frac{1/h}{1/h} = \frac{1}{h(1+x+h)} - \frac{1}{h(1+x)} = \frac{(1+x) - (1+x+h)}{h(1+x+h)(1+x)}=\frac{-h}{h(1+x+h)(1+x)}$$
The $h$ cancels, leaving us with $$\frac{-\require{cancel}\cancel h}{\cancel{h}(1+x+h)(1+x)} = \left(-\frac {1}{(1+x+h)(1+x)}\right)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/901924",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Evaluate $\int{\sin^3(x)\cos^2(x)}dx$ I'm trying to solve $\int{\sin^3(x)\cos^2(x)}dx$.
I got $-\frac{1}{2}\cos(x)+C$, but the memo says $\frac{1}{5}\cos^5(x)-\frac{1}{3}\cos^3(x)+C$
This is my working:
Your help is appreciated!
| I was challenged to do this question without using either substitution or even $\sin^2 x + \cos^2x = 1$,
$\sin^3 x \cdot\cos^2 x \\
= \sin x \cdot \sin^2 x \cdot \cos^2 x\\
= \sin x \cdot (\frac{1}{2}\cdot 2\sin x\cdot\cos x)^2\\
= \sin x \cdot (\frac{1}{2} \sin 2x)^2\\
= \frac{1}{4}\cdot (\sin x \cdot \sin 2x) \cdot \sin 2x\\
= \frac{1}{4} \cdot \frac{1}{2} (\cos x - \cos 3x)\cdot \sin 2x\\
= \frac{1}{8}\cdot (\sin2x\cdot \cos x - \sin2x\cdot \cos3x)\\
=\frac{1}{8}\cdot[\frac{1}{2}(\sin x + \sin 3x) - \frac{1}{2}(\sin(-x) + \sin5x) ]\\
=\frac{1}{8}\cdot\frac{1}{2} (\sin x + \sin 3x + \sin x - \sin 5x)\\
=\frac{1}{16}(2\sin x + \sin3x - \sin 5x)\\$
$$\therefore \int{(\sin^3x\cdot\cos^2x)}\cdot dx \\
= \frac{1}{16}\int{(2\sin x + \sin3x - \sin 5x)}\cdot dx\\
= \frac{(-\cos x)}{8} + \frac{(-\cos3x)}{3\times 16} - \frac{(-\cos5x)}{5\times 16} + C\\
= \frac{\cos5x}{80} - \frac{\cos3x}{48} - \frac{\cos x}{8} + C$$
$\dots$which isn't the answer in the memo but that doesn't mean it isn't right.
Now, if there's any math teacher out there who won't accept my answer on a written examination, please speak or forever hold your peace.
Edit: This video could useful for general questions of this type
| {
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"timestamp": "2023-03-29T00:00:00",
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How prove this limit $\left(\frac{1}{2}+\sum_{k=1}^{n-1}(-1)^{\lfloor\frac{mk}{n}\rfloor}\{\frac{mk}{n}\} \right)^n=\frac{1}{\sqrt{e}}$ let $m$ is even number,and $n$ is odd number,and such $(m,n)=1$,
show this limit:
$$\lim_{n\to\infty}\left(\dfrac{1}{2}+\sum_{k=1}^{n-1}(-1)^{\left\lfloor\dfrac{mk}{n}\right\rfloor}\left\{\dfrac{mk}{n}\right\} \right)^n=\dfrac{1}{\sqrt{e}}$$
where $\{x\}=x-\lfloor x\rfloor$
I think we can find this sum $$\sum_{k=1}^{n-1}(-1)^{\left\lfloor\dfrac{mk}{n}\right\rfloor}\left\{\dfrac{mk}{n}\right\}$$
But I can't
| Oh, I think I am late, but decided not to erase it.
Let $m = 2l$. We claim that the quantity
$$\lfloor mk / n \rfloor \equiv \lfloor 2lk / n \rfloor \pmod 2$$
depends only on $r_{k} = lk \text{ mod } n$. Indeed, write $lk = nq + r_{k}$. Then
$$ \lfloor 2lk / n \rfloor
= \lfloor 2(nq + r) / n \rfloor
= 2q + \lfloor 2r_{k} / n \rfloor. $$
So it follows that
$$ S_{n} := \sum_{k=1}^{n-1} (-1)^{\lfloor mk/n \rfloor} \left\{ \frac{mk}{n} \right\} = \sum_{k=1}^{n-1} (-1)^{\lfloor 2r_{k}/n \rfloor} \left\{ \frac{2r_{k}}{n} \right\} $$
But since $(l, n) = 1$, $r_{k}$ is just a rearrangement of $1, \cdots, n-1$ and we have
$$ S_{n} := \sum_{k=1}^{n-1} (-1)^{\lfloor 2k/n \rfloor} \left\{ \frac{2k}{n} \right\}. $$
Dividing the sum into two parts, with one running over $k = 1, \cdots, \frac{n-1}{2}$ and the other running over $k = \frac{n+1}{2}, \cdots, n-1$, it follows that
$$ S_{n} = \sum_{k=1}^{(n-1)/2} \frac{2k}{n} - \sum_{k=(n+1)/2}^{n-1} \left( \frac{2k}{n} - 1 \right) = \frac{n-1}{2n}. $$
This proves the observation as desired.
A slight generalization: Let $f$ be a $C^{2}$-function on $[0, 1]$. Then utilizing the Taylor's Theorem, it is not hard to show that
\begin{align*}
S_{f,n} := \sum_{k=1}^{n-1} (-1)^{\left\lfloor \frac{mk}{n} \right\rfloor} f \left( \left\{ \tfrac{mk}{n} \right\} \right)
&= \sum_{k=1}^{(n-1)/2} \left( f \left( \tfrac{2k}{n} \right) - f \left( \tfrac{2k}{n} - \tfrac{1}{n} \right) \right) \\
&= \frac{1}{2}(f(1) - f(0)) - \frac{1}{4n}(f'(1) + f'(0) + o(1)).
\end{align*}
So we have
$$ \left( 1 - \frac{f(1) - f(0)}{2} + S_{f,n} \right)^{n} \to \exp\left(-\frac{f'(0)+f'(1)}{4} \right). $$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Find the equation of the curve I've been given the following math question; however, I don't understand the wording of what it's asking.
if $ \dfrac{d^2y}{dx^2} = y''= \dfrac{dy'}{dx} = 3x^2 + x $ for any point on a curve and the tangent line at $ (1,-2) $ is $ y = 2 - 4x $, find the equation of the curve.
| We integrate $y''$ to get $y'$. So, $y'=x^3+\frac{1}{2}x^2+c_{1}$, where $c_{1}$ is the constant of integration (to be found).
Recall that $y'$ is the "slope function", $m$, of the tangent line. In this case, $m=-4$. That is, $y'(x=1)=-4$. We use this condition to find the value of $c_{1}$: $$y'(1)=1+\frac{1}{2}+c_{1}=-4\quad\Rightarrow\quad c_{1}=-\frac{11}{2}$$
So, $y'=x^3+\frac{1}{2}x^2-\frac{11}{2}$. Now integrate $y'$ to get $y$ (which is the equation of the curve). So, $y=\frac{1}{4}x^4+\frac{1}{6}x^3-\frac{11}{2}x+c_{2}$, where $c_{2}$ is the constant of integration (to be found).
Recall that we have a point that lies on the curve, namely, $(x,y)=(1,-2)$. This means that $y(x=1)=-2$. So, $$y(1)=\frac{1}{4}+\frac{1}{6}-\frac{11}{2}+c_{2}=-2\quad\Rightarrow\quad c_{2}=\frac{37}{12}$$
So, $y=\frac{1}{4}x^4+\frac{1}{6}x^3-\frac{11}{2}x+\frac{37}{12}$.
| {
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"timestamp": "2023-03-29T00:00:00",
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prime division problem $a,b,c \in$ {0,1,2,...,9} with at least one of $a,b,c$ nonzero. Prove that the six-digit integer $abcabc$ is divisible by at least 3 distinct primes.
My thinking is not to use induction as there is a definite size to the integer.
So $abcabc$ comes to $c1001 + b10010 + a100100$ with at least one of them nonzero. This factors to $7^{1}11^{1}13^{1}c + 7^{1}11^{1}13^{1}2^{1}5^{1}b + 7^{1}11^{1}13^{1}2^{2}5^{2}a$ which factors to $7^{1}11^{1}13^{1}(c + 2^{1}5^{1}b + 2^{2}5^{2}a$) so with one of $a,b,c$ nonzero $abcabc$ is divisible by at least 3 primes $7,11,13$.
| $$\overline{abcabc}=\overline{abc}(1001)$$
as:
$$\overline{abcabc}\\=10^5a+10^4b+10^3c+10^2a+10^1b+10^0c\\=10^2a(10^3+1)+10^1b(10^3+1)+10^0c(10^3+1)\\=(10^2a+10b+c)(10^3+1)$$
Since $1001$ is divisible by $7,11$ and $13$; you're done.
| {
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"timestamp": "2023-03-29T00:00:00",
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Looking for help with this elementary method of finding integer solutions on an elliptic curve. In the post Finding all solutions to $y^3 = x^2 + x + 1$ with $x,y$ integers larger than $1$, the single positive integer solution $(x,y)=(18,7)$ is found using algebraic integers.
In one of the comments (Finding all solutions to $y^3 = x^2 + x + 1$ with $x,y$ integers larger than $1$, the OP indicated the method suggested (i.e., the “theory of elliptic curves”, torsion groups, etc.) was “far too advanced for [his] level of understanding”. This inspired me to try finding a totally elementary approach. I've made it to a certain stage, and wanted some advice on how to proceed.
Beginning with the original equation
$$y^3 = x^2+x+1,\tag{1}$$
I added $x^3$ to both sides and factored, obtaining
\begin{align}
x^3+y^3 &= x^3+x^2+x+1 \\
(x+y)(x^2-xy+y^2) &= (x^2+1)(x+1).\tag{2}
\end{align}
Since we're looking for $x,y > 1$, and both factors on the right-hand side of (2) are positive, so are both on the left-hand side. Hence there exist positive integers $a,b,c,d$ such that
\begin{align}
x^2+1 &= ab, \\
x+1 &= cd, \\
x+y &= ac, \\
x^2-xy+y^2 &= bd.
\end{align}
Knowing the solution a priori, I'm now faced with trying to show that $(a,b,c,d)=(25,13,1,19)$. Using a few simple congruence and divisibility arguments, it can fairly easily be shown that $c=1$, and hence we have
\begin{align}
x^2+1 &= ab, \tag{3.1} \\
x+1 &= d, \tag{3.2} \\
x+y &= a, \tag{3.3} \\
x^2-xy+y^2 &= bd. \tag{3.4}
\end{align}
But now I'm running in circles. Evidently $(x,y)=(d-1,a-d+1)$, and I can prove other results like $(y-1)\mid x$ and $y \mid (b+1)$, but I can't seem to take it across the goal line. Any help would be greatly appreciated.
EDIT: From $a^2-bd=3xy$ and $d^2-ab=2x$, we have
$$
(a-d)(a+d+b)=x(3y-2). \tag{4}
$$
By the form of the left-hand side of (3.4) and the fact that $b,d$ are odd and relatively prime [because $x^2+1$ and $x+1$ are], we deduce $b\equiv d\equiv 1\!\pmod{6}$. Then (3.1) implies $a \equiv 1\!\pmod{6}$; in fact, by the form of (3.1), we have also $a \equiv b \equiv 1\!\pmod{12}$. In any case, $a+b+d \equiv 3\!\pmod{6}$, and (4) now implies $18 \mid x$.
EDIT: Another thread on the same question is https://mathoverflow.net/questions/56338/is-n-m-18-7-the-only-positive-solution-to-n2-n-1-m3.
| So I've found an elementary and fairly easy method of proof that the only two positive integer solutions are $(x,y)=(0,1)$ and $(x,y)=(18,7)$, with the only two negative integer solutions being $(x,y)=(-1,1)$ and $(x,y)=(-19,7)$.
Using the logic as shown in the question, we quickly deduce that $(y-1) \mid x$, say $x = (y-1)w$ for integer $w$. Substituting into the original equation and solving for $y$ yields
$$
y = \frac{w^2-1 \pm \sqrt{w^4-6w^2+4w-3}}{2}. \tag{$\star$}
$$
Evidently, $w^4-6w^2+4w-3$ must be an integer square. Now observe that
$$
(w^4-6w^2+4w-3)-(w^2-3)^2 = 4w-12 \tag{1}
$$
and
$$
(w^2-2)^2-(w^4-6w^2+4w-3) = 2w^2-4w+7. \tag{2}
$$
The right-hand side of (2) is positive for all integer $w$, and the right-hand side of (1) is positive for $w>3$. So for $w > 3$
$$
(w^2-3)^2 < w^4-6w^2+4w-3 < (w^2-2)^2,
$$
and the expression cannot be a square. For $w < 0$, we determine that
$$
(w^2-4)^2 < w^4-6w^2+4w-3 < (w^2-3)^2,
$$
and again it cannot be a square. Hence we need only consider $0 \le w \le 3$, and it is quickly found that the only solution is $w=3$. By substitution into $(\star)$, we find $y=1$ or $y=7$, as claimed.
p.s. Thanks to Will Jagy for the elegant bounds-based solution to the last step!
| {
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Make $n$ cents with $1$-cent, $2$-cent, and $3$-cent coins I encountered the following problem in Herber Wilf's book Generatingfunctionology:
Prove that, in country that has $1$-cent, $2$-cent, and $3$-cent coins
only, the number of ways of changing $n$ cents is exactly the integer
nearest to $ \frac{(n+3)^2}{12} $.
Of course, the solution provided was one that makes use of a generating function. However, it is unbelievably tedious. Is there a way to do this without making use of a generation function?
For reference, here is the solution given in the book.
If $f(n)$ is that number, then
$$ \begin {gather*}
\sum_{n\ge 0} f(n) x^n =
\frac {1}{(1-x)(1-x^2)(1-x^3)}\qquad\qquad\qquad\qquad \\
= \frac{1}{6(1-x)^3} + \frac{1 }{4(1-x)^2} +
\frac {17}{72(1-x)} + \frac {1}{8(1+x)} + \frac {1}{9(1-\omega x)} +
\frac {1}{9(1-\overline{\omega}x)}.
\end {gather*} $$
If we expand each of
the fractions on the right, we find the formula $$ f(n) = \frac {1}{6}
\dbinom {n+2}{2} + \frac {1}{4} (n+1) + \frac {17}{72} + \frac
{(-1)^n}{8} + \frac {2}{9} \cos \left( \frac {2n \pi}{3} \right), $$
which can be rewritten as
$$ f(n) = \frac {(n+3)^2}{12} + \frac {-7 + 9 (-1)^n
+ 16 \cos \left( \frac {2n\pi}{3} \right)}{72}. $$
The second
fraction cannot exceed $ \frac {32}{72} < \frac {1}{2} $ in absolute
value, so $f(n)$ is the unique integer who distance from
$ \frac {(n+3)^2}{12} $ is less than $\frac{1}{2}$, as required. $\Box$
| This is a solution form the book ''CHALLENGING MATHEMATICAL PROOLEMS WITH ELEMENTARY SOLUTIONS VOL 1'' by A. M. Yaglom and I. M. Yaglom :
In making up $n$ cents out of 1-, 2-, and 3-cent stamps, we can use either no 3-cent stamps at all, or one 3-cent stamp, or 2 of them, etc., up to a maximum of $q = \left[n/3\right]$ of them.
In the first case the $n$ cents would have to be made up entirely out of 1- and 2-cent stamps, which can be
done in $\left[n/2\right] + 1$ ways (Please see the notes). In the second case we must form $n-3$ cents with 1- and 2-cent stamps,
which can be done in $\left[(n-3)/2\right] + l$ ways. In the third case we must form $n-6$ cents, which can be done in $\left[(n-6)/2\right] + 1$ ways. In general, in the $(k+1)^{th}$ case, we must form $n-3k$ cents with 1- and 2-cent stamps, which can be done in $\left[(n-3k)/2\right] + l$ ways.
Let $n = 3q + r$, where $q = \left[n/3\right]$ is the quotient obtained by dividing 3 into $n$, and $r$ is the remainder (thus
$r = 0, l$, or $2$). We then see that in the final case (where $q$ 3-cent stamps are used), the remaining $r$ cents must be made
up out of 1- and 2-cent stamps. This can be done in $[r/2] + 1$ ways, so we get a total of
$$
S = \left(\left[\frac{n}{2}+1 \right] \right) + \left(\left[\frac{n-3}{2}+1 \right] \right) +
\left(\left[\frac{n-6}{2}+1 \right] \right) + \ldots +\left(\left[\frac{r}{2}+1 \right] \right)
$$
Now, we note that for any integer $m$, $\left[\frac{m}{2} \right] = \frac{m}{2} - \frac{1}{4} + \frac{(-1)^m}{4}$. We use this fact to simply the expression for $S$.
$$ \begin{aligned}
S = & \left(\frac{n}{2}+\frac{3}{4}+\frac{(-1)^{n}}{4} \right) + \left(\frac{n-3}{2}+\frac{3}{4}+\frac{(-1)^{n-3}}{} \right) \\
+ & \left(\frac{n-6}{2}+\frac{3}{4}+\frac{(-1)^{n-6}}{4} \right) + \ldots + \left(\frac{r}{2}+\frac{3}{4}+\frac{(-1)^{r}}{4} \right)
\end{aligned}
$$
Since there are $q + 1$ parentheses each containing a $\frac{3}{4}$, we have
$$ \begin{aligned}
S = & \frac{3(q+1)}{4} + \left( \frac{(-1)^n}{2}+\frac{(-1)^{n-3}}{2}+\frac{(-1)^{n-6}}{2}+\ldots+\frac{(-1)^r}{2} \right) \\
+ & \left(\frac{n}{2}+\frac{n-3}{2}+\frac{n-6}{2}+\ldots+\frac{r}{2} \right)
\end{aligned} $$
The terms in the first parenthesis alternate in sign. Therefore this parenthesis is equal to $\frac{1}{4}$ if both $n$ and $r$ are even, $\frac{-1}{4}$ if both are odd, and $0$ otherwise. We will denote its value by $\epsilon$, and note for later purposes that $\epsilon \geq 0$ when $r$ is even. The terms of the second parenthesis are in arithmetic progression. So,
$$ \begin{aligned}
S & = \frac{3(q+1)}{4} + \epsilon + \frac{q+1}{2}\left(\frac{n}{2}+\frac{r}{2} \right) \\
& = \frac{q+1}{4} \left( 3+n+r \right) + \epsilon \\
& = \frac{3q+3}{12} \left( 3+n+r \right) + \epsilon \\
& = \frac{(n+3+r)(n+3-r)}{12} + \epsilon \\
& = \frac{\left(n+3) \right)^2}{12} - \frac{r^2}{12} + \epsilon
\end{aligned} $$
The rest is to show that $\left| - \frac{r^2}{12} + \epsilon \right| \leq \frac{1}{2}$ for $r=0, 1, 2$. Thus $S$, which is an integer, differs from $(n +3)^2/12$ by a quantity whose absolute values is less than $\frac{1}{2}$· Therefore $S$ is the integer nearest to $(n+3)^2/12$.
Note : Please see the problem 22 of the same book about an argument that to make $n$ cents with only pennies and another coin having the value of $k$ cents, then the number of ways would be $\left[\frac{n}{k}\right]+1$ .
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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Why $\cos^2 x-\sin^2 x = \cos 2x\;?$ I was hoping someone could explain how $\cos^2 x-\sin^2 x = \cos 2x$
After using the product rule to differentiate $\sin x \cdot \cos x$ I get the answer $\cos^2 x - \sin ^2 x$ I've come across this problem twice now and each time I've gotten the same wrong so I'm hoping someone can point out what I'm missing.
| Algebraic proof. $$\cos(2x)=\cos(x+x)=\cos x \cos x - \sin x \sin x = \cos^2 x - \sin^2 x$$
Using de Moivre's formula. $$\cos(2x) + i \sin(2x) = (\cos x + i \sin x)^2 = \\ \cos^2 x + i^2 \sin^2 x + 2i \cos x \sin x = \cos^2 x - \sin^2 x + 2i \cos x \sin x .$$
From here we get $\cos(2x) = \cos^2 x - \sin^2 x.$
You can find more hints at ProofWiki.
| {
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"timestamp": "2023-03-29T00:00:00",
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} |
Factoring $(x+1)(x+2)(x+3)(x+4)+1$ I have to factor this polynomial:
$$(x+1)(x+2)(x+3)(x+4)+1$$
WolframAlpha gives
$$(x^2+5x+5)^2$$
How can I prove it without expanding the result?
Thanks!
| $(x^2+5x+5)^2-1^2=(x^2+5x+4)(x^2+5x+6)=(x+1)(x+4)(x+2)(x+3)$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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} |
summing the powers of a complex number Let $z=e^{\frac{2\pi i}{5}}$, then $1+z+z^2+z^3+5z^4+4z^5+4z^6+4z^7+4z^8+5z^9=?$
I am kind of confused since by drawing a graph, $1+z+z^2+z^3+z^4$ should be zero, but using computational softwares the result is different, and hence I do not know how to solve this problem.
Thanks for helping!
| We have $$1+z+z^2+z^3+z^4+5z^4+4z^5+4z^6+4z^7+4z^8+4z^8+5z^9=$$
$$=(1+z+z^2+z^3+z^4)+4z^4(1+z+z^2+z^3+z^4)+5z^9.$$
Using the fact that the $\left(e^{2\pi i/5}\right)^5=1$, we see that $z=e^{2\pi i/5}$ is a root of $z^5-1=(z-1)(1+z+z^2+z^3+z^4)$, in particular, it must be a root of $1+z+z^2+z^3+z^4$, so that we are left with $$5z^9=5e^{(2\pi i/5)\cdot 9}=5e^{18\pi i/5}=5e^{8\pi i/5},$$where we have used the fact that $e^{2\pi i}=1$.
| {
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"timestamp": "2023-03-29T00:00:00",
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Complete the squares to find the center and radius of the circle I've had trouble on this one problem for a couple days.
Complete the square on the X and Y terms to find the center and radius of the circle.
$x^2+2x+y^2-4y=-4\:\:$
| $$
\begin{align*}
x^2+2x+y^2-4y &= -4 \\
\Rightarrow x^2+2x+\left( \frac{2}{2} \right)^2+y^2-4y+\left( -\frac{4}{2} \right)^2 &=-4+\left( \frac{2}{2} \right)^2+\left( -\frac{4}{2} \right)^2 \\
\Rightarrow \,\,\,\,\, \,\left(x+\frac{2}{2}\right)^2\quad \,\,\, +\quad \,\,\,\,\left( y-\frac{4}{2} \right)^2\qquad &= -4+\left( \frac{2}{2} \right)^2+\left( -\frac{4}{2} \right)^2 \ldots
\end{align*}
$$
Here I have successfully completed the square. I will leave you to clean up the result. Two days is a long time to work with no help, best of luck.
| {
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Construction of a polynomial of degree 4 with some conditions Exercise Let $P(x)$ be a polynomial of degree $4$, the question is :
Find this $P$ such that :
*
*The coefficient oh highest degree is $1$
*P is divisible by $x^2+x+1$
*The rest of the division of $P$ by $x^2-1$ is $-3x+9$
The first condition implies that $P=a_0+a_1x+a_2x^2+a_3x^3+x^4$. The second one can be written as
$$
P(x)=\bigl(x(x+1)+1\bigr)Q(X).
$$
So $Q(x)=\frac{a_0+a_1x+a_2x^2+a_3x^3+x^4}{x(x+1)+1}$ with $P(0)=Q(0)$ and $P(-1)=Q(-1)$.
The last condition I wrote
$$
P(x)=-3X+9+R(x)\bigl((x-1)(x+1)\bigr).
$$
Then we have a system,
$$
\left\{
\begin{array}{ll}
a_0+a_1+a_2+a_3=5 \\
a_0-a_1+a_2-a_3=13&
\end{array}
\right.
$$
because $P(1)=6$ and $P(-1)=12$. Which is equivalent to
$$
\left\{
\begin{array}{ll}
a_0+a_2=9 \\
a_1+a_3=-4&
\end{array}
\right.
$$
How can I continue?
| Since there is a $(a,b)$ such that
$$P(x)=(x^2+x+1)(x^2+ax+b),$$
you can find $(a,b)$ from
$$P(1)=6\iff (1+1+1)(1+a+b)=6,$$$$P(-1)=12\iff (1-1+1)(1-a+b)=12.$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Find $\int_0^\pi \sin(x)\,dx$ explicitly A book asks me to prove that:
$$\int_0^{\pi}\sin(x)\,dx = 2$$
Using the identity:
$$\sin\left(\frac{\pi}{n}\right) + \sin\left(\frac{2\pi}{n}\right) + \cdots + \sin\left(\frac{n\pi}{n}\right) = \frac{\cos\left(\frac{\pi}{2n}\right)-\cos\left(\frac{(2n+1)\pi}{2n}\right)}{2\sin\left(\frac{\pi}{2n}\right)}$$
And the famous $\lim_{x\to0}\frac{\sin(x)}{x} = 1$
What I tried:
Using the Right Riemann Sum Method:
$$\int_a^{b}\sin(x)\,dx \approx \Delta x\left[f(a + \Delta x) + f(a + 2\,\Delta x) + \cdots + f(b)\right]$$
By taking $\Delta x = \frac{\pi}{n}$, $a = 0$ and $b = \pi$ we have:
$$\int_0^{\pi}\sin(x)\,dx \approx \Delta x\left[f(\Delta x) + f(2\,\Delta x) + \cdots + f\left(\frac{n\pi}{n}\right)\right] = \frac{\pi}{n}\left[\sin\left(\frac{\pi}{n}\right) + \sin\left(\frac{2\pi}{n}\right) + \cdots + \sin\left(\frac{n\pi}{n}\right)\right] = \frac{\pi}{n}\left[\frac{\cos\left(\frac{\pi}{2n}\right)-\cos\left(\frac{(2n+1)\pi}{2n}\right)}{2\sin\left(\frac{\pi}{2n}\right)}\right]$$
So
$$\int_0^\pi \sin(x)\,dx = \lim_{n\to\infty} \frac{\pi}{n}\frac{\cos\left(\frac{\pi}{2n}\right)-\cos\left(\frac{(2n+1)\pi}{2n}\right)}{2\sin\left(\frac{\pi}{2n}\right)}$$
I can't see, however, how to prove this limit to be $=2$
| Without simplifying your last expression (thr numerator is twice the same term of it), we could use Taylor series and, for large values of $n$, you could write $$\cos\left(\frac{\pi}{2n}\right)=1-\frac{\pi ^2}{8 n^2}+\frac{\pi ^4}{384
n^4}+O\left(\left(\frac{1}{n}\right)^5\right)$$ $$\cos\left(\frac{(2n+1)\pi}{2n}\right)=-1+\frac{\pi ^2}{8 n^2}-\frac{\pi ^4}{384
n^4}+O\left(\left(\frac{1}{n}\right)^5\right)$$ $$\sin\left(\frac{\pi}{2n}\right)=\frac{\pi }{2 n}-\frac{\pi ^3}{48 n^3}+O\left(\left(\frac{1}{n}\right)^5\right)$$ So $$\frac{\pi}{n}\frac{\cos\left(\frac{\pi}{2n}\right)-\cos\left(\frac{(2n+1)\pi}{2n}\right)}{2\sin\left(\frac{\pi}{2n}\right)}=2-\frac{\pi ^2}{6 n^2}+O\left(\left(\frac{1}{n}\right)^4\right)$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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Find the side of an equilateral triangle given only the distance of an arbitrary point to its vertices Triangle $ABC$ is an equilateral triangle and $P$ is an arbitrary point inside it. The distance from $P$ to $A$ is $4$ and the distance from $P$ to $B$ is $6$ and the distance from $P$ to $C$ is $5$. How to find the side of an equilateral triangle from this information?
| You may be knowing this result for an equilateral triangle of side a and distance from P to vertices as x,y and z.
$$\frac{a^2+x^2-y^2}{2xa} = \frac{\sqrt{3}}{2}\frac{a^2+x^2-z^2}{2xa}+\frac{1}{2} \sqrt{1- \Big(\frac{a^2+x^2-z^2}{2xa}\Big)^2}$$
After putting values $(x,y,z)=(4,5,6)$ it becomes:
$$\frac{a^2+4^2-5^2}{8a} = \frac{\sqrt{3}}{2}\frac{a^2+4^2-6^2}{8a}+\frac{1}{2} \sqrt{1- \Big(\frac{a^2+4^2-6^2}{8a}\Big)^2}$$
Solving we get:
$$a=1/2\sqrt{154\pm30\sqrt{21}}\sim 8,2$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Evaluation of Integral $ \int\frac{1}{(2+3\sin x)^2}dx$ Evaluation of Integral $\displaystyle \int\frac{1}{(2+3\sin x)^2}dx$
$\bf{My\; Try::}$ Using Integration by parts,
Let $$\displaystyle I=\int\frac{1}{(2+3\sin x)^2}dx = \int \frac{1}{\cos x}\cdot \frac{\cos x}{(2+3\sin x)^2}dx$$
$$\displaystyle I = -\frac{1}{3 \cos x}\cdot \frac{1}{(2+3\sin x)}+\frac{1}{3}\int \frac{\sin x}{\cos^2 x}\cdot \frac{1}{(2+3\sin x)}dx$$
Now Let $$\displaystyle J=\int \cdot \frac{\sin x}{(2+3\sin x)}\cdot \frac{1}{\cos^2 x} = \int \frac{\sin x}{(2+3\sin x)}\cdot \sec^2 xdx$$
again Using Integration parts
$$\displaystyle J = \frac{\sin x}{(2+3\sin x)}\cdot \tan x - \int \frac{d}{dx}\left\{\frac{\sin x}{(2+3\sin x)}\right\}\cdot \tan xdx$$
So $$\displaystyle J = \frac{\sin x}{(2+3\sin x)}\cdot \tan x-2\int\frac{\sin x}{(2+3\sin x)^2}dx$$
$$\displaystyle J = \frac{\sin x}{(2+3\sin x)}\cdot \tan x-\frac{2}{3}\int \frac{[(2\sin x+3)-3]}{(2\sin x+3)^2}dx$$
So $$\displaystyle J = \frac{\sin x}{(2+3\sin x)}\cdot \tan x- \frac{2}{3}\int\frac{1}{(2\sin x+3)}dx+4I$$
Now Let $$\displaystyle K = \int\frac{1}{2\sin x+3}dx = \int\frac{2\sin x-3}{4\sin^2 x-9}dx = -2\int\frac{\sin x}{5+4\cos^2 x}dx+3\int\frac{\sec^2 x}{5\tan^2 x+9}dx$$
My Question is can we solve it Without Using Integration by parts
If Yes, The plz explain here
thanks
| If you set $x=2\arctan\theta$, you end with:
$$\frac{1}{2}\int\frac{1+\theta^2}{(1+3\theta+\theta^2)^2}\,d\theta$$
that can be computed through simple fractions decomposition.
| {
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"timestamp": "2023-03-29T00:00:00",
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How prove this sum $\sum\limits_{n=k}^{\infty}\dbinom{n}{k}\left(\dfrac{-z}{1-z}\right)^n$ prove or disprove
$$\sum_{n=k}^{\infty}\binom{n}{k}\left(\dfrac{-z}{1-z}\right)^n=
(1-z)(-z)^k$$
my try: since
$$\binom{k}{k}\left(\dfrac{-z}{1-z}\right)^k+\binom{k+1}{k}\left(\dfrac{-z}{1-z}\right)^{k+1}+\binom{k+2}{k}\left(\dfrac{-z}{1-z}\right)^{k+2}+\cdots=\left(\dfrac{-z}{1-z}\right)^{k}\left[\binom{k}{k}+\binom{k+1}{k}\dfrac{-z}{1-z}+\binom{k+2}{k}\left(\dfrac{-z}{1-z}\right)^2+\cdots\right]
=\left(\dfrac{-z}{1-z}\right)^k\left[\binom{k}{0}\left(\dfrac{-z}{1-z}\right)^0+\binom{k+1}{1}\dfrac{-z}{1-z}+\binom{k+2}{2}\left(\dfrac{-z}{1-z}\right)^2+\cdots\right]
$$
| $$\sum_{n=k}^{\infty}\binom{n}{k}\left(\dfrac{-z}{1-z}\right)^n=
(1-z)(-z)^k$$
\begin{align*}
&\binom{k}{k}\left(\dfrac{-z}{1-z}\right)^k+\binom{k+1}{k}\left(\dfrac{-z}{1-z}\right)^{k+1}+\binom{k+2}{k}\left(\dfrac{-z}{1-z}\right)^{k+2}+\cdots\\
&=\left(\dfrac{-z}{1-z}\right)^{k}\left[\binom{k}{k}+\binom{k+1}{k}\dfrac{-z}{1-z}+\binom{k+2}{k}\left(\dfrac{-z}{1-z}\right)^2+\cdots\right]\\
&=\left(\dfrac{-z}{1-z}\right)^k\left[\binom{k}{0}\left(\dfrac{-z}{1-z}\right)^0+\binom{k+1}{1}\dfrac{-z}{1-z}+\binom{k+2}{2}\left(\dfrac{-z}{1-z}\right)^2+\cdots\right]
\end{align*}
since
$$\dfrac{1}{(1-x)^{n+1}}=\binom{n}{n}+\binom{n+1}{n}x+\binom{n+2}{n}x^2+\binom{n+k}{n}x^k+\cdots$$
let$x=\dfrac{-z}{1-z}$
| {
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"timestamp": "2023-03-29T00:00:00",
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A infinite sum with harmonic serie Proof or disproof the folowing statement:
$$\sum_{n=1}^{+\infty}\frac{2n+1}{(n^2+n)^2}H_n=\sum_{n=1}^{+\infty}\frac{1}{n^3}$$
where $\displaystyle H_n=\sum_{k=1}^{n}\frac{1}{k}$.
| Method 1a.
Since
$$
\begin{align}
\frac{2n+1}{(n^2+n)^2} H_n & = \frac{(n+1)^2 - n^2}{n^2(n+1)^2} H_n \\
&= \frac{H_n}{n^2} - \frac{H_n}{(n+1)^2},\\
&= \frac{H_n}{n^2} - \left(\frac{H_{n+1}}{(n+1)^2}-\frac{1}{(n+1)^3}\right),
\end{align}
$$
then, summing from $n=1$ to $+\infty$, by telescoping you get
$$\sum_{n=1}^{\infty}\frac{2n+1}{(n^2+n)^2}H_n = \frac{H_1}{1^2}+\sum_{n=1}^{\infty}\frac{1}{(n+1)^3}=\zeta(3).$$
Method 1b.
By using Abel's transformation, you get
$$\sum_{n=1}^{\infty}\frac{2n+1}{(n^2+n)^2}H_n = \sum_{n=1}^{\infty}\frac{H_n}{(n+1)^2}=\zeta(3),$$
you have diverse proofs of the last equality, the Euler sum identity $\zeta(2,1)=\zeta(3)$, there : Thirty-Two Goldbach Variations.
Method 2.
You have
$$\frac{2n+1}{(n^2+n)^2} x^n= \frac{(n+1)^2 - n^2}{n^2(n+1)^2} x^n= \frac{x^n}{n^2} - \frac1x\frac{x^{n+1}}{(n+1)^2},$$ then, using the standard integral representation for the harmonic numbers and the standard expansion for the dilogarithm function:
$$
H_n=\int_0^1 \frac{1-x^n}{1-x} {\rm d} x, \qquad {\rm L}_2(x)=\sum_{n=1}^{\infty}\frac{x^n}{n^2},
$$ we deduce that
$$\sum_{n=1}^{\infty}\frac{2n+1}{(n^2+n)^2}H_n =\int_0^1\frac{{\rm L}_2(x)}{x} {\rm d} x= [{\rm L}_3(x)]_0^1=\zeta(3).$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Proving a second derivative Given that $$y = \sin^3 x + \cos^3 x$$
prove that $$\frac{d^2 y}{dx^2} = \frac{3}{2} (\cos x + \sin x)(3 \sin 2x - 2)$$
I began with differentiating the equation as it is and it took me around twelve steps to reach the answer, and I'm guessing that factoring the sum of the cubes before differentiating wouldn't make it any easier. Is there a simpler approach to this problem?
Thanks in advance.
| $y'=3\sin^2 x \cos x - 3\cos^2 x \sin x \\
= 3\sin x \cos x(\sin x - \cos x) \\
=\frac{3}{2}\sin(2x)(\sin x - \cos x)\\
y''=\frac{3}{2}\cos(2x)(2)(\sin x - \cos x)+\frac{3}{2}\sin(2x)(\cos x+\sin x)\\
=\frac{3}{2}(2(\cos^2 x-\sin^2 x)(\sin x - \cos x)+\sin(2x)(\cos x+\sin x))\\
=\frac{3}{2}(-2(\sin x - \cos x)(\sin x + \cos x)(\sin x - \cos x))+\sin(2x)(\sin x + \cos x))\\
=\frac{3}{2}(\sin x + \cos x)(-2(\sin x - \cos x)^2+\sin(2x))\\
=\frac{3}{2}(\sin x + \cos x)(-2(\sin^2 x -2\sin x \cos x + \cos^2 x)+\sin(2x))\\
=\frac{3}{2}(\sin x + \cos x)(-2(1 -\sin(2x))+\sin(2x))\\
=\frac{3}{2}(\sin x + \cos x)(3\sin(2x)-2)\\$
As far as I can tell, the result given is wrong, so watch out :)
| {
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Finding a specific term in an expansion $(a+b)^n$ without expanding How can I find a term within an expansion without actually expanding or using Pascal's Triangle?
For example: 5th term of
$$
\left(\dfrac{x}{y}-\dfrac{y}{x}\right)^8
$$
| $$(a+b)^n=\binom{n}{0}a^nb^0 +\binom{n}{1}a^{n-1}b^1+\binom{n}{2}a^{n-2}b^2+\binom{n}{3}a^{n-3}b^3+....\\so\\5th\\term\\is\\\binom{n}{4}a^{n-4}b^4\\(\frac{x}{y}-\frac{y}{x})^8\\5th-term\\
\binom{8}{4}(\frac{x}{y})^{8-4}(-\frac{y}{x})^4=\\\frac{8*7*6*5}{4*3*2*1}(\frac{x}{y})^{4}(\frac{y}{x})^4=\\\frac{8*7*6*5}{4*3*2*1} $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/917103",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Prove $a^2+b^2+c^2=\frac{6}{5}$ if $a+b+c=0$ and $a^3+b^3+c^3=a^5+b^5+c^5$
if $a,b,c$ are real numbers that $a\neq0,b\neq0,c\neq0$ and $a+b+c=0$ and $$a^3+b^3+c^3=a^5+b^5+c^5$$
Prove that $a^2+b^2+c^2=\frac{6}{5}$.
Things I have done: $a+b+c=0$ So $$a^3+b^3+c^3=a^5+b^5+c^5=3abc$$
also $$a^2+b^2+c^2=-2ab-2bc-2ca$$
I tried multiplying $(a^3+b^3+c^3)$ and $a^2+b^2+c^2$ but I was not able to reach a useful result from that.
| You should be able to get something by looking at the product $a^3+b^3+c^3$ and $a^2+b^2+c^2$. Note that
$$a^2b^3+a^3b^2=a^2b^2(a+b)=-(ab)(abc).$$
| {
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"url": "https://math.stackexchange.com/questions/917847",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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$M=\{a+b\sqrt{2}: a,b \in \mathbb{Q} \}$ and $N=\{c+d\sqrt{3}: c,d \in \mathbb{Q}\}$. $M \cap N \subseteq \mathbb{Q}$. Let $M=\{a+b\sqrt{2}: a,b \in \mathbb{Q} \}$ and $N=\{c+d\sqrt{3}: c,d \in \mathbb{Q}\}$.
Prove $M \cap N \subseteq \mathbb{Q}$.
| What we have to prove is this:
If $a+b\sqrt 2 = c+d\sqrt 3$ with $a,b,c,d \in \mathbb Q$, then $b = d = 0$.
$$\begin{align}
a+b\sqrt 2 & = c+d\sqrt 3 \\
(a-c)^2 & = (d\sqrt 3 - b\sqrt 2)^2 \\
& = 3d^2-2bd\sqrt 6 + 2b^2
\end{align}$$
So
$$bd \sqrt 6 = \frac12(3d^2+2b^2-(a-c)^2)$$
is rational. We know $\sqrt 6$ is irrational, so we must have $bd = 0$. Suppose $b = 0$. Then $a = c + d\sqrt 3$, so $d\sqrt 3$ is rational, hence $d = 0$. Similarly, $d = 0 \implies b = 0$.
Hence $b = d = 0$.
| {
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Prove |cos(x−1)|+|cos(x)|+|cos(x+1)|≥3/2 I'm working on an induction proof, but I keep coming up against a brick wall.
While working through the induction proof process I keep ending up with $$|\cos(m)|\ge\frac12$$ ,but clearly this isn't true for all m. From another post someone gave me the idea to look at triples and to consider $$|\cos(x−1)|+|\cos(x)|+|\cos(x+1)|$$ to try and find a lower bound. If you plug the above equation into Wolfram Alpha it tells you that $$|\cos(x−1)|+|\cos(x)|+|\cos(x+1)|\ge\frac32$$ Which would prove my induction proof. I just don't know how to go about proving that it does truly exceed $\frac32$
If it helps the original statement that I was trying to prove is $$\sum_{k=0}^n|\cos(k)|\ge\frac n2$$
| Well I have a proof for you but its not the most elegant.
Assume first that $\cos(x+1)$ and $\cos(x-1)$ have opposite signs, then
$$|\cos(x+1)|+|\cos(x-1)|=|\cos(x+1)-\cos(x-1)|=|2\sin (1) \sin x|$$
Now $$|2\sin 1 \sin x+\cos x |\leq |2\sin 1 \sin x|+|\cos x|$$
Now a bit of calculus, the minimum of $a\sin x+\cos x$ occurs when $\tan x =a$, so its minimum value is
$$a\sin x+\cos x=\cos x(a\tan x+1)=\cos x (\tan^2 x+1) =\sec x=\sqrt{a^2+1}$$
Now $\frac{3}{2}\leq \sqrt{a^2+1}$ holds if $\frac{\sqrt{5}}{2} \leq a$ and since $a=2\sin 1$ we have to see that $\frac{\sqrt{5}}{4} \leq \sin 1$. But $\frac{\pi}{4} < 1$ so we are reduced to showing $\frac{\sqrt{5}}{4} \leq \frac{1}{\sqrt{2}}$ which is easy.
Now consider what happens if $\cos(x+1)$ and $\cos(x-1)$ have equal signs. Just to fix ideas lets assume they are positive. Then for an appropriate $n$,
$$-(\frac{\pi}{2}-1) < x+2\pi n < \frac{\pi}{2}-1$$ so in particular
$$\sin 1=\cos (\frac{\pi}{2}-1)\leq \cos x$$.
Now in this case we have
$$|\cos(x)|+|\cos(x+1)|+|\cos(x-1)|=|\cos(x)+\cos(x+1)+\cos(x-1)|=|\cos(x)(2\cos 1 +1)|$$
Thus with the above inequality we have to show
$$\frac{3}{2}\leq |\sin 1(2\cos 1 +1)|.$$
Now
$\sin 1(2\cos 1 +1)=\sin 2+\sin 1$
using $2 < \frac{2\pi}{3}$ and $\frac{\pi}{4}<1$ we have
$$\frac{\sqrt{3}}{2}\leq \sin 2$$
$$\frac{\sqrt{2}}{2}\leq \sin 1$$
So we have $$\frac{3}{2}\leq \frac{\sqrt{3}}{2}+\frac{\sqrt{2}}{2}$$ which is easily checked.
| {
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"url": "https://math.stackexchange.com/questions/918877",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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} |
Solve $ x^2+y^2=4, z^2+t^2=9, xt+yz=6 $ in integers
find answers of this system of equations in integers$$
\left\{
\begin{array}{c}
x^2+y^2=4 \\
z^2+t^2=9 \\
xt+yz=6
\end{array}
\right.
$$
things I have done: we can observe that $$(x^2+y^2)(z^2+t^2)=(xt+yz)^2+(xz-yt)^2 \rightarrow xz-yt=0$$
summing up this with first and second equality $$(x+z)^2+(y-t)^2=13$$
at this stage I used guessing answers. putting like $x= 0,z=3,y=2,t=0$.is there a better way to doing this without guessing and making sure that all answer found?
| Perhaps you are working too hard. The solutions of the first equation are $x=\pm 2$, $y=0$, or the opposite. The solutions of $z^2+t^2=9$ are equally simple.
Now look at the third equation.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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} |
Prove that the series$\sum a_n$, with $a_n$ defined recursively, converges to 1 Let $a_n$ , with $n\geq1$ be the sequence of real numbers defined by $a_1=\frac12$ and the following relation
$$a_{n+1}=\frac{a_n^2}{a_n^2-a_n+1}$$Prove that $\sum a_n$ converges to $1$.
Now, I have only been able to study the sequence defined by the recurrence relation. I know that it converges to zero by the monotone convergence theorem. I think I should prove that this is a geometric series like this $\sum_{n=0}^\infty \frac12\left(\frac12\right)^n$. What is the strategy I should follow in these cases? How would you approach this little problem?
Thank you for your help.
| Let $S_n = \sum_{k=1}^n a_k$, then we observe the following pattern:
$$S_1 = \frac{1}{2}, a_2 = \frac{1}{3}$$
$$S_2 = \frac{5}{6}, a_3 = \frac{1}{7}$$
$$S_3 = \frac{41}{42}, a_4 = \frac{1}{43}$$
Actually, if $S_n = \dfrac{k-1}{k} = 1 -\dfrac{1}{k}$ and $a_{n+1} = \dfrac{1}{k+1}$ then $S_{n+1} = \dfrac{k^2 + k -1}{k^2 +k} = 1- \dfrac{1}{k^2 + k}$ and $a_{n+2} = \dfrac{1}{k(1 +k) + 1}$.
ALl these suggest us to prove by induction $S_n = 1- \dfrac{1}{b_{n+1} - 1}$ and $a_{n+1} = \dfrac{1}{b_{n + 1}}$ with $b_{n+1} = b_n^2 - b_n + 1 $ and $b_1 =2$. It's easy to see $b_n \to +\infty$
This is obvious when $n = 1$.
If this is true when $n = k$, i.e. if $S_k = 1- \dfrac{1}{b_{k+1} - 1}$ and $a_{k+1} = \dfrac{1}{b_{k + 1}}$ is true. then we can see $$a_{k+2} = \dfrac{a_k^2}{a_k^2 - a_k + 1} = \dfrac{1}{b_{k+1}^2 - b_{k+1} + 1} = \dfrac{1}{b_{k+2} }$$ and $$S_{k+1} = S_k + a_{k+1} =1 - \dfrac{1}{b_{k+1}^2 - b_{k+1}}= 1 - \dfrac{1}{b_{k+2} - 1}$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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} |
Prove $\frac{a}{(b-c)^2}+\frac{b}{(c-a)^2}+\frac{c}{(a-b)^2}=0$ if $\frac{a}{b-c}+\frac{b}{c-a}+\frac{c}{a-b}=0$
if $a,b,c$ are real numbers and $$\frac{a}{b-c}+\frac{b}{c-a}+\frac{c}{a-b}=0$$
Prove $$\frac{a}{(b-c)^2}+\frac{b}{(c-a)^2}+\frac{c}{(a-b)^2}=0$$
things i have done: using the assumption i deduced that $$\frac{a}{b-c}=-\frac{(b-c)(a-b-c)}{(c-a)(a-b)}\rightarrow\frac{a}{(b-c)^2}=\frac{(c+b-a)}{(c-a)(a-b)}$$
So some rewriting $$\frac{(c+b-a)}{(c-a)(a-b)}+\frac{b}{(c-a)^2}+\frac{c}{(a-b)^2} =\frac{(c+b-a)(c-a)(a-b)+b(a-b)^2+a(c-a)^2}{(c-a)^2(a-b)^2}$$
I tried to write numerator in expanded form but it was to large and it seemed like there was not going to be something useful after factoring. so is there any better way than my brute force ? can my solution get continued?
| Put
$$P(x)=(x-b+c)(x-c+a)(x-a+b)=x^{3}+\alpha x^2+\beta x+\gamma$$
We immediately have $\alpha=0$, and $\gamma\not =0$ (as $a,b,c$ are distincts).
Put $w_n=a(b-c)^n+b(c-a)^n+c(a-b)^n$ for $n\in \mathbb{Z}$. We see $w_1=0$ (easy computation) and $w_{-1}=0$ (by hypothesis). Now we have $w_{n+3}+\beta w_{n+1}+\gamma w_n=0$ for all $n\in \mathbb{Z}$. Take $n=-2$ and we get $w_{-2}=0$.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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"answer_id": 1
} |
Limit of $\frac{1}{x^2}-\frac{1}{\sin^2(x)}$ as $x$ approaches $0$
Evaluate
$$\lim \limits_{x \rightarrow 0}\left(\frac{1}{x^2}-\frac{1}{\sin^2(x)}\right)$$
I tried to combine the fractions
$$\frac{1}{x^2}-\frac{1}{\sin^2(x)} = \frac{\sin^2(x)-x^2}{x^2\sin^2(x)}$$
and apply L'Hopitals which only made a mess.
I feel like there is a simpler way of doing this but I'm not quite sure what to do
| $$\begin{aligned}L &= \lim_{x \to 0}\frac{1}{x^{2}} - \frac{1}{\sin^{2}x}\\
&= \lim_{x \to 0}\frac{\sin^{2} x - x^{2}}{x^{2}\sin^{2}x}\\
&= \lim_{x \to 0}\frac{(\sin x - x)(\sin x + x)}{x^{3}\cdot x}\cdot\frac{x^{2}}{\sin^{2}x}\\
&= \lim_{x \to 0}\frac{(\sin x - x)(\sin x + x)}{x^{3}\cdot x}\\
&= \lim_{x \to 0}\frac{\sin x - x}{x^{3}}\cdot\lim_{x \to 0}\frac{\sin x + x}{x}\\
&= \lim_{x \to 0}\frac{\sin x - x}{x^{3}}\cdot\left(\lim_{x \to 0}\frac{\sin x}{x} + 1\right)\\
&= 2\lim_{x \to 0}\frac{\sin x - x}{x^{3}}\\
&= 2\lim_{x \to 0}\frac{\cos x - 1}{3x^{2}}\text{ (applying L'Hospital's Rule)}\\
&= -\frac{2}{3}\lim_{x \to 0}\frac{1 - \cos x}{x^{2}}\cdot\frac{1 + \cos x}{1 + \cos x}\\
&= -\frac{2}{3}\lim_{x \to 0}\frac{1 - \cos^{2} x}{x^{2}(1 + \cos x)}\\
&= -\frac{2}{3}\lim_{x \to 0}\frac{\sin^{2} x}{x^{2}}\cdot\frac{1}{1 + \cos x}\\
&= -\frac{2}{3}\cdot 1\cdot\frac{1}{2} = -\frac{1}{3}\end{aligned}$$
Thus as mentioned by OP in comments, it is doable without Taylor series. In fact the limit of $(\sin x - x)/x^{3}$ is also doable without L'Hospital Rule, but it requires more work as shown by user robjohn in a beautiful answer.
| {
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"timestamp": "2023-03-29T00:00:00",
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Summation of $2^n$ where $n$ is even or odd I need the summation formula for the following:
$2^0+2^1+2^3+2^5+\cdots+2^n$, $n$ is even i.e.
$2^0+2^2+2^4+2^6+\cdots+2^n$, $n$ is odd
I am aware that the formula of
$2^0+2^1+2^2+2^3+\cdots+2^n = 2^{n+1}-1$. But can anyone help me with a approach so solve such kind of problems.
| [EDIT] Thanks for accepting. This is an even more straightforward approach:
$$\begin{align}
S&=2^0+2^2+2^4+ \cdots +2^{2m}\\
&=4^0+4^1+4^2+\cdots +4^m\\
&=\frac 13 (4^{m+1}-1)\end{align}$$
$$\begin{align}
S'&=2^1+2^3+2^5+ \cdots +2^{2m+1}\\
&=2(4^0+4^1+4^2+\cdots +4^m)\\
&=\frac 23 (4^{m+1}-1)\end{align}$$
[ORIGINAL ANSWER] Consider the case where $n$ is even, i.e. $n=2m$.
Let
$$\begin{align}
S&=2^0+2^2+2^4+\cdots +2^{2m}\\
2S&=2^1+2^3+2^5+\cdots+2^{2m+1}\end{align}$$
Adding:
$$\begin{align}3S&=2^0+2^1+2^2+2^3+2^4+\cdots+2^{2m+1}\\
&=\sum_{r=0}^{2m+1}2^r\\
&=2^{2m+2}-1\\
S&=\frac 13 \left(2^{2m+2}-1\right)\\
&=\frac 13 \left(4^{m+1}-1\right)\end{align}$$
Similarly, for the case where $n$ is odd, i.e. $n=2m+1$,
$$\begin{align}S'&=2^1+2^3+2^5+\cdots+2^{2m+1}\\
&=2(2^0+2^2+2^4+\cdots +2^{2m})\\
&=2S\\
&=\frac 23 \left(4^{m+1}-1\right)\end{align}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/923762",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 2
} |
Indefinite integral of a rational function: $\int\frac{6x+4}{x^2+4}\,dx$ Find $\displaystyle\int\frac{6x+4}{x^2+4}\,dx$
The question asks to find integral of the expression so I divided them into two parts:
$$
\int\frac{6x}{x^2+4}\,dx
$$
and
$$\int\frac{4}{x^2+4}.
$$
So, for the first integral, I set $u=x^2+4$ and got $\int\frac{3}{u}\,du$ which is $3\ln|x^2+4|+c$. But I don't know how to integrate the second part.
| Since $\displaystyle \frac{d}{dx} \tan^{-1} x = \frac{1}{x^2+1}$,
$$
\frac{1}{2}\frac{d}{dx} \tan^{-1} (x/2) = \frac{1}{2} \cdot \frac{1}{1+(x/2)^2} \cdot \frac{1}{2} = \frac{1}{4+x^2}
$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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How to prove the equivalence of the two functions? $$f_1(k,n):=\sum_{0\leqslant v\leqslant n}\dfrac{\left(2(k+v)\right)!}{(k+v)!v!(2k+v)!(n-v)!2^v}, \quad k,n \in \mathbb{N}
$$
$$
f_2(k,n):=\sum_{0\leqslant m\leqslant \lfloor{\frac{n}{2}}\rfloor}\dfrac{1}{(k+m)!m!(n-2m)!2^{4m-n}},\quad k,n \in \mathbb{N}
$$
I checked many small natural number (up to several hundred) $k$'s and $n$'s, and all indicate the two functions are equivalent.
How to prove or disprove it?
| Suppose we seek to verify that $f_1(n,k) = f_2(n,k)$
where
$$f_1(n,k) = \sum_{v=0}^n
\frac{(2k+2v)!} {(k+v)!\times v!\times (2k+v)!\times (n-v)!} 2^{-v}$$
and
$$f_2(n,k) = \sum_{m=0}^{\lfloor n/2\rfloor}
\frac{1}{(k+m)!\times m! \times (n-2m)!} 2^{n-4m}.$$
Multiplying by $(n+k)!$ we obtain
$$g_1(n,k) = \sum_{v=0}^n {n+k\choose n-v} {2k+2v\choose v} 2^{-v}$$
and
$$g_2(n,k) = 2^n \sum_{m=0}^{\lfloor n/2\rfloor}
{n+k\choose m} {n+k-m\choose n-2m} 2^{-4m}.$$
We will work with the latter two.
Re-write the first sum as follows:
$$2^{-n} \sum_{v=0}^n {n+k\choose v}
{2k+2n-2v\choose n-v} 2^{v}$$
Introduce
$${2k+2n-2v\choose n-v} =
\frac{1}{2\pi i}
\int_{|z|=\epsilon}
\frac{1}{z^{n-v+1}} (1+z)^{2k+2n-2v} \; dz.$$
This integral is zero when $v\gt n$ so we may extend $v$ to infinity.
We get for $g_1(n,k)$
$$2^{-n} \frac{1}{2\pi i}
\int_{|z|=\epsilon}
\frac{1}{z^{n+1}} (1+z)^{2k+2n}
\sum_{v\ge 0} {n+k\choose v} \frac{z^v}{(1+z)^{2v}} 2^v
\; dz
\\ = 2^{-n} \frac{1}{2\pi i}
\int_{|z|=\epsilon}
\frac{1}{z^{n+1}} (1+z)^{2k+2n}
\left(1+2\frac{z}{(1+z)^2}\right)^{n+k}
\; dz
\\ = 2^{-n} \frac{1}{2\pi i}
\int_{|z|=\epsilon}
\frac{1}{z^{n+1}}
\left(1+4z+z^2\right)^{n+k}
\; dz.$$
For the second sum introduce
$${n+k-m\choose n-2m} =
\frac{1}{2\pi i}
\int_{|z|=\epsilon}
\frac{1}{z^{n-2m+1}} (1+z)^{n+k-m} \; dz.$$
This is zero when $2m\gt n$ so we may extend $m$ to infinity.
We get for $g_2(n,k)$
$$2^n \frac{1}{2\pi i}
\int_{|z|=\epsilon}
\frac{1}{z^{n+1}} (1+z)^{n+k}
\sum_{m\ge 0} {n+k\choose m} \frac{z^{2m}}{(1+z)^m} 2^{-4m}
\; dz
\\ = 2^n \frac{1}{2\pi i}
\int_{|z|=\epsilon}
\frac{1}{z^{n+1}} (1+z)^{n+k}
\left(1+\frac{1}{16}\frac{z^2}{1+z}\right)^{n+k}
\; dz
\\ = 2^n \frac{1}{2\pi i}
\int_{|z|=\epsilon}
\frac{1}{z^{n+1}}
\left(1+z+\frac{1}{16}z^2\right)^{n+k}
\; dz.$$
Finally put $z=4w$ in this integral to get
$$2^n \frac{1}{2\pi i}
\int_{|w|=\epsilon}
\frac{1}{4^{n+1} w^{n+1}}
\left(1+4w+w^2\right)^{n+k}
\; 4 dw
\\ = 2^{-n} \frac{1}{2\pi i}
\int_{|w|=\epsilon}
\frac{1}{w^{n+1}}
\left(1+4w+w^2\right)^{n+k}
\; dw.$$
This concludes the argument.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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derive limit to make a function continuous Here's the problem:
Choose the value of k that makes the following
function continuous at $x = 1$:
$f(x)=\begin{cases}
\frac{-8x^2 + 48x - 40}{x - 1} & x < 1\\
-2x + k &x \geq 1
\end{cases}$
My steps:
$\lim_{x\uparrow 1} f(x) = \lim_{x\downarrow 1} f(x)$
$\frac{-8x^2 + 48x - 40}{x - 1} = -2x + k$
$\frac{-8(x^2 - 6x + 5)}{x - 1} = -2x + k$
$\frac{-8(x - 5)(x - 1)}{x - 1} = -2x + k$
$-8(x - 5) = -2x + k$
$-8x - (-8)\cdot5 = -2x + k$
$-8x + 40 = -2x + k$
$-8x + 2x + 40 = k$ (See note at bottom)
$-6x + 40 = k$
$-6\cdot1 + 40 = k$
$-6 + 40 = k$
$36 = k$
The solution steps:
For f to be continuous at $x = 1$ we need $\lim_{x\uparrow 1} f(x) = \lim_{x\downarrow1} f(x) = f(1)$
First lets evaluate lim x→1− f(x). Since x < 1 as x approaches 1 from the left,
f(x) = (−8x^2 + 48x − 40)/(x − 1) as x→1−
So lim x→1− f(x) = lim x→1− (−8x^2 + 48x − 40)/(x − 1)
Start by factoring the numerator.
In this case we find:
lim x→1− (−8x^2 + 48x − 40)/(−3x^2 − 15x + 18) = lim x→1− (−8x + 40)(x − 1)/(x − 1)
Aside: can anyone explain (−3x^2 − 15x + 18) in the denominator?
Since we are taking the limit as x→1−, we may assume that x ≠ 1.
Canceling factors we see:
\lim x→1− (−8x + 40)(x − 1)/(−3x − 18)(x − 1) = \lim x→1− (−8x + 40)
We can evaluate this linear function at x = 1.
Hence,
lim x→1− −8x^2 + 48x − 40)/(−3x^2 − 15x + 18) = −8(1) + 40.
Thus, lim x→1− f(x) = 32.
Thus lim x→1− f(x) = 32
lim x→1+ f(x) = lim x→ (−2x + k), because x > 1 as x approaches 1 from the right.
lim x→ (−2x + k) = −2⋅1 + k = −2 + k = f(1) because linear functions are continuous.
For f to be continuous, these limits have to be equal. Thus 32 = −2 + k
When k = 34, f is continuous.
NOTE: This is where I diverged from the solution by moving -2x to the left. My question is, assuming the given solution is correct, why is this nominally valid algebraic step not permitted?
| You solution is correct, but it's an overkill. After you have correctly found limit from the above, set limit from the below $f(1) = -2 \cdot 1 +k = 34$ and solve for $k$. Continuity is ensured by the fact that $1$ is included in the definition of the limit from below.
| {
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"timestamp": "2023-03-29T00:00:00",
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Turning $2\le x$ into $\sqrt{1+\frac{4}{x^6}}\le \sqrt{5}$? I am supposed to turn $2\le x$ into $\sqrt{1+\frac{4}{x^6}}\le \sqrt{5}$, and I have no idea on how to approach this. I'll post my steps, even though I don't think they'll be of much help.
$$2\le x \rightarrow \frac{1}{x} \ge \frac{1}{2} \rightarrow\frac{4}{x}\ge2$$
and that is pretty much all I got. Thanks for all the help in advance!
| Your first implication $2\leq x\implies \frac{1}{x}\geq\frac{1}{2}$ is actually false: you have to flip the inequality direction, roughly because taking the reciprocal of a large number will give you a small number. Here is how you can do it:
\begin{align*}
2 &\leq x \\
\implies \frac{1}{x} &\leq \frac{1}{2} \\
\implies \frac{1}{x^6} &\leq \frac{1}{64} \\
\implies \frac{4}{x^6} &\leq \frac{1}{16} \\
\implies 1+\frac{4}{x^6} &\leq \frac{17}{16}<5 \\
\implies \sqrt{1+\frac{4}{x^6}} &< \sqrt{5}
\end{align*}
where taking the square root accross the inequality was justified because everything is positive.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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} |
Prove the sum to n terms of the series Prove that the sum to $n$ terms of a sequence:
$$\frac{1^2}{1\times 3}+\frac{2^2}{3\times 5}+\frac{3^2}{5\times 7}+\cdots $$
is
$$
\frac{n(n+1)}{2(2n+1)}
$$
| Assuming that $n^2, (2n-1), (2n+1)$ are the general terms for $\{1^2, 2^2, 3^3, \cdots, \}$, $\{1, 3, 5, \cdots, \}$, $\{3, 5, 7 \cdots, \}$ we see that $t_n=\frac{n^2}{(2n-1)(2n+1)}$ which partial fraction decomposition gives as $\frac 18[\frac{1}{2n-1}-\frac{1}{2n+1}]$. take the sum over $n$ from 1 to $n$ and get desired answer.
| {
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"timestamp": "2023-03-29T00:00:00",
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Solving algebraic equations for x So I was able to find the least common denominator which is $12$ but I'm struggling to solving the equation: $$\frac{4(x - 2)}{6} - \frac{2(x + 4)}{4} = -\frac{2}{3}.$$
| Multiplying both sides by $12$ gives you
$$12\times\frac{4(x - 2)}{6} - 12\times \frac{2(x + 4)}{4} =12\times\frac{-2}{3}\iff 8(x-2)-6(x+4)=-8.$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Find the Taylor expansion of $\frac{1}{x^2+2x-3}$ around $x=-1$. Find the Taylor expansion of $\frac{1}{x^2+2x-3}$ around $x=-1$. What is its radius of convergence?
So I write the fraction as $\frac{1}{(x-1)(x+3)}$ and what should I do now?
| You may write
$$
x^2+2x-3=-4+(x+1)^2
$$ then
$$
\frac{1}{x^2+2x-3}=\frac{1}{-4+(x+1)^2}=-\frac{1}{4}\frac{1}{1-\frac{(x+1)^2}{4}}
$$ and then use the expansion
$$
\frac{1}{1-u}=1+u+u^2+u^3+ \ldots
$$ as $|u|<1.$
Hence
$$
\frac{1}{x^2+2x-3}=-\sum_{n=0}^{\infty}\frac{1}{4^{n+1}}(x+1)^{2n}, \quad |x+1|<2.
$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Find $f$ such that $f'(x)=ax^2+bx$, given the values of $f'(1)$, $f''(1)$, and $\int_0^2 f(x)\,dx$ The question is : Find the solution $f(x)$ if $f'(x)=ax^2+bx$, and
(i) $f'(1)=6$,
(ii) $f''(1)=18$,
(iii) $\int_0^2 f(x)dx=18$.
My solution is:
According to $(i)$, we know $6=a+b$, and $a=6-b$.
Since $f''(x)=2ax+b$, and according to $(ii)$, we know $18=2a+b$.
Therefore, $b=-6, a=12$. And, $f'(x)=12x^2-6x$.
Now, we have
$$
f(x)=\int f'(x)dx=\int (12x^2-6x)dx=4x^3-3x^2+C
$$
According to $(iii)$,
$$
\bigl |_0^2 f(x) = 4*8-12-0=20
$$
It means $(\bigr|_0^2 f(x) )+ C=18$, and $C=-2$. The solution of $f(x)$ is $4x^3-3x^2-2$.
But the correct answer should be $4x^3-3x^2+5$. Where is my mistake?
Thank you!!
| You used $f'$ instead of $f$ in (iii), which should read $$18=\int_0^2f(x)\,\mathrm dx=\int_0^2(4x^3-3x^2+C)\,\mathrm dx=\left.x^4-x^3+Cx\,\right|_0^2=8+2C.$$
| {
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Prove that $a+b+1 = 0$ The polynomials $x^2+ax+b$ and $x^2+bx+a$ have common factors.prove that $a+b+1=0$.
My attempt- I could do nothing other than dividing the polynomials to get $x^2+bx+a$=$x^2+ax+b+bx-ax+a-b$.Please help me what to do.
| If $x-c$ is the common factor,
$x-c$ will divide $(x^2+bx+a)-(x^2+ax+b)=(b-a)(x-1)\implies c=1$
If $x^2+bx+a=(x-1)(x-d),$
$\iff x^2+bx+a=x^2-x(d+1)+d$
Comparing the constants & the coefficients of $x,$
$ d=a,b=-(d+1)=-(a+1)$
| {
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Evaluating $\lim_{x \to 0} \frac{(1 + \sin x + \sin^2 x)^{1/x} - (1 + \sin x)^{1/x}}{x}$ I did this:
$$\begin{align}
\lim_{x \to 0} \frac{(1 + \sin x + \sin^2 x)^{1/x} - (1 + \sin x)^{1/x}}{x} &\sim \lim_{x \to 0} \frac{(1 + x + x^2)^{1/x} - (1 + x)^{1/x}}{x} = \\
&= \lim_{x \to 0} \left [ (1+x)^{1/x} \frac{\left ( \frac{1 + x + x^2}{1 + x} \right )^{1/x} - 1}{x} \right ] =\\
&= \lim_{x \to 0} (1+x)^{1/x} \cdot \lim_{x \to 0} \frac{\left ( 1 + \frac{x^2}{1 + x} \right )^{1/x} - 1}{x} =\\
&= e \cdot \lim_{x \to 0} \frac{e^{\frac{1}{x} \cdot \ln \left ( 1 + \frac{x^2}{1+x} \right )} - 1}{x} \sim \\
&\sim e \cdot \lim_{x \to 0} \frac{e^{\frac{x}{1 + x}} - 1}{x} \sim \\
&\sim e \cdot \lim_{x \to 0} \frac{1}{1 + x} = e
\end{align}$$
Is it right? If it is, how to evaluate the limit faster? It was pretty long the way I did it.
| We have $$\begin{aligned}L &= \lim_{x \to 0}\frac{(1 + \sin x + \sin^{2}x)^{1/x} - (1 + \sin x)^{1/x}}{x}\\
&= \lim_{x \to 0}(1 + \sin x)^{1/x}\dfrac{\left(1 + \dfrac{\sin^{2}x}{1 + \sin x}\right)^{1/x} - 1}{x}\\
&= \lim_{x \to 0}e\cdot\dfrac{\left(1 + \dfrac{\sin^{2}x}{1 + \sin x}\right)^{1/x} - 1}{x}\\
&= e\lim_{x \to 0}\dfrac{\left(1 + \dfrac{\sin^{2}x}{1 + \sin x}\right)^{1/x} - 1}{x}\\
&= e\lim_{x \to 0}g(x) = eA\end{aligned}$$ The limit $A = \lim_{x \to 0}g(x)$ can be handled with elementary techniques. Let $$f(x) = \frac{\sin^{2}x}{1 + \sin x}$$ so that $f(x) \to 0$ as $x \to 0$, then we can see that $$\begin{aligned}g(x) &= \frac{(1 + f(x))^{1/x} - 1}{x}\\
&= \dfrac{\exp\left(\dfrac{\log(1 + f(x))}{x}\right) - 1}{x}\\
&= \dfrac{\exp(t) - 1}{x}\end{aligned}$$ Note that $$\begin{aligned}t &= \frac{\log(1 + f(x))}{x}\\
&= \frac{\log(1 + f(x))}{f(x)}\cdot\frac{f(x)}{x}\\
&= \frac{\log(1 + f(x))}{f(x)}\cdot\frac{\sin x}{x}\cdot\frac{\sin x}{1 + \sin x}\\
&\to 1\cdot 1\cdot 0 = 0\end{aligned}$$ Thus we can write $$\begin{aligned}A &= \lim_{x \to 0}g(x)\\
&= \lim_{x \to 0}\frac{e^{t} - 1}{t}\cdot\frac{t}{x}\\
&= \lim_{x \to 0}1\cdot\frac{\log(1 + f(x))}{x^{2}}\\
&= \lim_{x \to 0}\frac{\log (1 + f(x))}{f(x)}\cdot\frac{f(x)}{x^{2}}\\
&= \lim_{x \to 0}1\cdot\frac{\sin^{2}x}{x^{2}(1 + \sin x)}\\
&= \lim_{x \to 0}\frac{\sin^{2}x}{x^{2}}\cdot\frac{1}{1 + \sin x} = 1\cdot 1 = 1\end{aligned}$$ Thus $L = eA = e\cdot 1 = e$ and the final answer is $e$.
Note: We have used the the limit $(1 + \sin x)^{1/x} \to e$ as $x \to 0$. This is easily handled because $$\log(1 + \sin x)^{1/x} = \frac{\log(1 + \sin x)}{x} = \frac{\log(1 + \sin x)}{\sin x}\cdot\frac{\sin x}{x}\to 1\cdot 1 = 1$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Inequality for a rational function of three variables $x,y,z$ are positive real numbers such that $$x^2+y^2+z^2=1$$ Prove that $\dfrac{x^2}{1+2yz}+\dfrac{y^2}{1+2xz}+\dfrac{z^2}{1+2xy} \geqslant \dfrac{3}{5}$.Again, I try with Engel form of Cauchy inequality...
| Hint: By AM-GM:
$$\frac{x^2}{1+2yz} \ge \frac{x^2}{1+y^2+z^2} = \frac{x^2}{2-x^2}$$
Now apply Jensen to the convex function $t \mapsto \dfrac{t^2}{2-t^2}$ to conclude...
| {
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Value of constants when a limit is finite?
If $\displaystyle \lim_{x\to0}\frac{a\cos x+bx\sin x-5}{x^4}$ is finite. Find the value of 'a' and 'b'.
$$\lim_{x\to0}\frac{a\left[1-\frac{x^2}{2!}+\frac{x^4}{4!}....\right]+bx\left[x-\frac{x^3}{3!}+\frac{x^5}{5!}....\right]-5}{x^4}$$
$$\implies\lim_{x\to0}\frac{a-5+a\left[-\frac{x^2}{2!}+\frac{x^4}{4!}....\right]+bx\left[x-\frac{x^3}{3!}+\frac{x^5}{5!}....\right]}{x^4}$$
Here we take $a-5=0$ or $a=5$
And then we proceed further, equating constant terms (terms which do not hold 'x' of any degree) to zero and finding the values.
But if we take the case of:
$$\lim_{x\to0}\frac{6+x^2-6e^x}{x}=\lim_{x\to0}2x-6e^x=-6$$
(using L'Hopital Rule) is a finite limit, but in this case the independent value '6' is not equal to zero. So why do we assume in the above case that the independent values 'a','b' etc. when summed up gives zero?
Hope the question is clear.
| The main problem with your thinking is that the coefficient of limit must not always be zero. Only the coefficient of terms that give undefined limit such as $x^{-k},k>0,k\in\mathbb Z$ must be zero, rest all can be anything.
"equating constant terms (terms which do not hold 'x' of any degree) to zero"
is wrong, actually
"equating constant terms (terms which hold 'x' of degree less than denominator) to zero"
$$\lim_{x\to0}\frac{a\cos x+bx\sin x-5}{x^4}=\lim_{x\to0}\frac{a\color{fuchsia}{(1-x^2/2+x^4/24+O(x^6))}+bx\color{blue}{(x-x^3/6+O(x^5))}-5}{x^4}\\=\lim_{x\to0}\frac{\color{red}{x^0}(a-5)+\color{red}{x^2}(-a/2+b)+\color{green}{x^4}(a/24-b/6)+\color{green}{O(x^6)}}{x^4}$$
So $a=5$ and $b=a/2=5/2$ since $\displaystyle \lim_{x\to0}\frac{\color{red}{x^0}}{x^4},\frac{\color{red}{x^2}}{x^4}$ are undefined, so we rid of these terms by making their coefficient zero.
$$\lim_{x\to0}\frac{6+x^2-6e^x}{x}=\lim_{x\to0}\frac{6+x^2-6\color{red}{(1+x+x^2/2+O(x^3))}}{x}=\lim_{x\to0}\frac{-6x+O(x^2)}{x}=-6$$
$$\lim_{x\to0}\frac{a+x^2-ae^{x}}{x}=\lim_{x\to0}\frac{a+x^2-a\color{red}{(1+x+x^2/2+O(x^3))}}{x}=\lim_{x\to0}\frac{-ax+3x^2/2+O(x^3)}{x}=-a$$
| {
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"timestamp": "2023-03-29T00:00:00",
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How find this minimum of the $q$, if such $\frac{95}{36}>\frac{p}{q}>\frac{96}{37}$ let $p,q$ is postive integer,and such
$$\dfrac{95}{36}>\dfrac{p}{q}>\dfrac{96}{37}$$
Find the minimum of the $q$
maybe can use
$$95q>36p$$
and $$37p>96q$$
and then find this minimum of the value?
before I find a
$$2.638\approx \dfrac{95}{36}>\dfrac{49}{18}\approx 2.722>\dfrac{96}{37}\approx 2.59 $$ is not such condition
idea 2: since
$$\dfrac{95}{36}=\dfrac{95\cdot 37}{36\cdot 37}=\dfrac{3515}{1332}$$
$$\dfrac{96}{37}=\dfrac{96\cdot 36}{36\cdot 37}=\dfrac{3456}{1332}$$
so
$$\dfrac{3515}{1332}>\dfrac{p}{q}>\dfrac{3456}{1332}$$
so
$$p\in(3456,3515),q=1332$$
| We have
$$2.64\gt a=\frac{17575}{5\cdot 36\cdot 37}=\frac{95}{36}\gt \color{red}{\frac{13}{5}}=2.6=\frac{17316}{5\cdot 36\cdot 37}\gt\frac{96}{37}=\frac{17280}{5\cdot 36\cdot 37}=b\gt 2.59.$$
Note that
$$\frac{11}{4}\gt a\gt b\gt\frac{10}{4}$$
$$\frac{8}{3}\gt a\gt b\gt\frac{7}{3}$$
$$\frac{6}{2}\gt a\gt b\gt\frac{5}{2}$$
$$\frac{3}{1}\gt a\gt b\gt\frac{2}{1}$$
Hence, the minimum of $q$ is $5$.
| {
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"timestamp": "2023-03-29T00:00:00",
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$P(X^2+Y^2<1)$ of two independent n(0,1) random variables Suppose that X and Y are independent n(0,1) random variables.
a) Find $P(X^2+Y^2<1)$
Attempt:
a) Let $U = X^2 + Y^2$, $V = Y$.
Then $X = \sqrt{V^2 -U}$, $Y = V$.
$J = \left| \begin{array}{ccc}
\frac{-1}{\sqrt{V^2-U}} & \frac{V}{V^2-U} \\
0 & 1\\
\end{array} \right| $
Then the joint distribution of $f_{u,v}(u,v)$ is:
$$f_{u,v}(u,v)= \frac{1}{2\pi}e^{\frac{-\sqrt{v^2-u}}{2}}e^{\frac{-u^2}{2}}\frac{1}{\sqrt{v^2-u}}$$
Then $P(X^2 +Y^2 <1)$ is:
$$\int_0^\infty \int_0^{v^2-u} \frac{1}{2\pi}e^{\frac{-\sqrt{v^2-u}}{2}}e^{\frac{-u^2}{2}}\frac{1}{\sqrt{v^2-u}}dudv$$
However, at this point I simply do not know how any tricks to complete this integration.
| Box-Muller transform suggests that if $X, Y$ are two independent standard normal random variables, then if we let $X = R cos\Theta, Y = R sin\Theta, R \in (0, \infty), \Theta \in [0, 2\pi)$ (polar coordinate), then this bivariate transform is 1-to-1 and onto. Further, $R, \Theta$ are also independent. And $\frac{1}{2}R^2 \sim Exponential(1), \Theta \sim Uniform(0, 2\pi)$, i.e., $\frac{1}{2}R^2$ follows an Exponential distribution with $\lambda = 1$ and $\Theta$ follows a Uniform distribution on $(0, 2\pi)$.
Hence, $P(X^2 + Y^2 < 1) = P(R^2cos^2\Theta + R^2sin^2\Theta < 1) = P(R^2 < 1) = P(\frac{1}{2}R^2 < \frac{1}{2})$. And since $\frac{1}{2}R^2 \sim Exponential(1)$, we can denote $Z = \frac{1}{2}R^2$. Using the pdf of an $Exponential(1)$ distribution, $P(\frac{1}{2}R^2 < \frac{1}{2}) = P(Z < \frac{1}{2}) = \int_{-\infty}^{\frac{1}{2}} e^{-z} \, dz = 1 - e^{-\frac{1}{2}}$.
| {
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summation of series (odd and even case) Can anyone please answer this for me,it involves alternate signs which is different from normal summation formula.
Q:Find the sum to n terms of the series $ 1^3 - 2^3 + 3^3 - 4^3 + \ldots -(n-2)^3 + (n-1)^3 - n^3 $.
| The situation is a little different in the case $n$ even than in the case $n$ odd. We deal with even $n$, say $n=2k$, and leave the case $n$ odd to you.
Our sum is equal to
$$\left(1^3+2^3+3^3+\cdots +(2k)^3\right) -2\left(2^3+4^3+6^3+\cdots+(2k)^3\right).$$
This is equal to
$$\left(1^3+2^3+3^3+\cdots +(2k)^3\right) -16\left(1^3+2^3+3^3+\cdots+k^3\right).$$
Now use the perhaps familiar fact that
$$1^3+2^3+3^3+\cdots +w^3=\left(\frac{w(w+1)}{2}\right)^2.$$
| {
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Finding limits using $\epsilon$ - $\delta$ method How do I prove the following function is continuous at $(0,0)$ using epsilon-delta approach?
$$\frac{x^5-y^5}{x^2+y^2}$$
| I suppose your function $$f(x,y) = \frac{x^5-y^5}{x^2+y^2}$$ is defined as $0$ for $(x,y) = (0,0)$. Then it is continuous if for every $\epsilon > 0$ a $\delta > 0$ exists, so that for $||(x,y)|| < \delta \implies ||f(x,y)|| < \epsilon$. Choose $$\delta := \frac{\sqrt[3]{\epsilon}}{\sqrt[3]{2}},$$
because of:
$$0 \leq |\frac{x^5-y^5}{x^2+y^2}| \leq |\frac{|x^5|+|y^5|}{x^2+y^2}| = |\frac{|x^5|}{x^2+y^2} + \frac{|y^5|}{x^2+y^2}| \leq |\frac{|x^5|}{x^2} + \frac{|y^5|}{y^2}| = ||x^3| +|y^3||, $$ and
$$||x^3| +|y^3|| < |\delta^3 + \delta^3| = |(\frac{\sqrt[3]{\epsilon}}{\sqrt[3]{2}})^3 + (\frac{\sqrt[3]{\epsilon}}{\sqrt[3]{2}})^3| = |\frac{\epsilon}{2} + \frac{\epsilon}{2}| = \epsilon. $$
| {
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"timestamp": "2023-03-29T00:00:00",
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Area of similar triangle Suppose that we are given a triangle whose area is known. put a circle C of radius r inside that triangle. How can we find the area of a triangle similar to the first one and whose inscribed circle is C?
| The area $a$ of the smaller triangle whose inscribed circle has radius $r$ cannot be defined in an univocal manner only knowing the area $A$ of the larger triangle. For a given value of $A$, there are infinite possible values of $a$, since the area of the smaller triangle depends on the "shape" of the larger triangle.
Let us make an example. Imagine that the larger triangle is an equilateral one with side $\displaystyle 2 \sqrt{\frac{A}{\sqrt {3}} } $. Its height is therefore $\displaystyle \sqrt{A \sqrt {3}} $, and its area is $\displaystyle \sqrt{\frac{A}{\sqrt {3}} } \cdot \sqrt{A \sqrt {3}} =A $. In this case, the smaller triangle is also equilateral. It is not difficult to show that an equilateral triangle whose inscribed circle has radius $r$ has side equal to $2 \sqrt{3} \,r$ and height equal to $3r$, so that its area is $a=r^2 \cdot 3\sqrt{3} \approx r^2 \cdot 5.196...$.
Now let us consider another initial larger triangle, with unchanged area and different shape. For instance, we could choose an isosceles right triangle where the length of each leg is $\sqrt{2A}$, so that the area is still equal to $A$. In this case, the smaller triangle is also an isosceles right triangle. We can find its leg length $s$ by noting that the area is $\displaystyle \frac{s^2}{2}$ and the perimeter is $\displaystyle (2+\sqrt{2})s$. Applying the standard formula for the calculation of the radius of the inscribed circle (area/semiperimeter) we get $\displaystyle \frac{s^2}{ s(2+ \sqrt{2})}=r$, and then $s=r(2+\sqrt{2})$. The area of the smaller triangle is therefore given by $\displaystyle a=\frac{r^2}{2}(2+\sqrt{2})^2\approx r^2 \cdot 5.828...$.
In conclusion, to determine the area $a$ of the smaller triangle, some other information about the larger triangle - in addition to the area $A$ - is necessary. In particular, the shape of the larger triangle has to be unambiguously defined. The simplest information that could allow to identify the shape of the larger triangle is its semiperimeter $P$. Knowing this information, we could directly calculate the radius of the circle inscribed in the larger triangle as $\displaystyle R=\frac{A}{P}$, and obtain the scale factor $\displaystyle \frac{r}{R}= \frac{Pr}{A}$. Since the ratio of the areas of two similar triangles is equal to the square of the scale factor, we can conclude that $\displaystyle a=A (\frac{Pr}{A})^2=\frac{(Pr)^2}{A}$.
As a confirmation, we can apply this formula to the triangles of the two mentioned examples. In the first case, that of the equilateral larger triangle, we have $P=\displaystyle 3 \sqrt{\frac{A}{\sqrt {3}} } $, and the formula yields the correct value of $\displaystyle (3 \sqrt{\frac{A}{\sqrt {3}} }r )^2/A= r^2 \cdot 3\sqrt{3}$. In the second case, that of the isosceles right triangle, we have $P=\frac{1}{2}(2+\sqrt{2})\sqrt{2A}$, and again the formula yields the correct value of $\displaystyle a= \frac
{[\frac{1}{2}(2+\sqrt{2})\sqrt{2A}]^2r^2}{A}= \frac{r^2}{2}(2+\sqrt{2})^2$.
| {
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What algebraic manipulation is used to express the solution to this integral? According to WolframAlpha, $$
\int \frac{2}{\sqrt{x^2 + 4}} dx = 2 \sinh^{-1}\left(\frac{x}{2}\right) + c.$$
I'm wondering how this was obtained, as I got the following:
Let $x = 2\tan\theta$. Then $dx = 2\sec^2\theta$, so \begin{align*}\int\frac{2}{\sqrt{x^2 + 4}}dx &= \int\frac{4\sec^2\theta}{\sqrt{4\tan^2\theta + 4}}d\theta = \int\frac{2\sec^2\theta}{\sqrt{\tan^2\theta + 1}}d\theta = 2\int\frac{\sec^2\theta}{\sec\theta}d\theta\\ &= 2\int\sec\theta{}d\theta = 2\ln\mid{\tan\theta + \sec\theta}\mid + c.\end{align*}
The substitution $x = 2\tan\theta$ implies that $\sec\theta = \frac{\sqrt{x^2 + 4}}{2}$ and $\tan\theta = \frac{x}{2}$. Hence $$
\int \frac{2}{\sqrt{x^2 + 4}} dx = 2\ln\left|{\frac{x}{2} + \frac{\sqrt{x^2 + 4}}{2}}\right| + c.$$
Either I made a mistake or I don't know the right algebraic manipulation.
| your integral can transformed into
$\int\frac{1}{\sqrt{\left(\frac{x}{2}\right)^2+1}}dx=\int\frac{2dt}{\sqrt{t^2+1}}$
and this is $2arsinh(t)$
| {
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Recurrence relation $T(n) = T(n/2) + n\log(n)$ So I've been working on this recurrence equation and I'm stumped at the end. $T(n) = T(n/2) + n\log(n);\: T(1) = 1;\: n = 2^k$ and $\log $ is base $2$.
$T(2^k) = T(2^{k-1}) + 2^k \log(2^k)$
$T(2^k) = T(2^{k-1}) + (2^k) k$
$T(2^{k-1}) = T(2^{k-2}) + 2^{k-1} \log (2^{k-1})$
$T(2) = T(1) + 2^{1}\cdot(1)$
$T(1) = 1$
$T(2^k) = 1 + \displaystyle \sum_{i=0}^{ k} (2^i) (k-i)$
I'm not sure if I did everything correct, but I'm confused about tackling the summation.
| To assure you about the summation I separated the terms of the final sums in a likely better patternizable form:
$ \qquad \qquad \displaystyle \small \begin{array} {}
T(1)&=&1 &=&1&=&3-1\cdot2\\
T(2^1)&=&1+1 \cdot2^1 &=&3&=&3+0\cdot4 \\
T(2^2)&=&1+1 \cdot2^1+2 \cdot2^2 &=&11&=&3+1\cdot8\\
T(2^3)&=&1+1 \cdot2^1+2 \cdot2^2 +3 \cdot2^3 &=& 35&=&3+2\cdot16\\
T(2^4)&=&1+ ...+4 \cdot2^4 &=& 99&=&3+3\cdot32\\
T(2^5)&=&1+ ...+5 \cdot2^5 &=& 259&=&3+4\cdot64\\
\end{array}$
The pattern $1x+2x^2+3x^3+4x^4+...$ (where here $x=2$) can also be represented by a generating function of the style $ a/ (1-x)^2$ and this can be adapted to match the sum-expressions above as truncation of the occuring series to make the final formula below rigorous. This is the generating function:
$ \qquad \qquad \displaystyle G(m,x)= 1+{x \cdot (1-x^{m+1})-(m+1) \cdot x^{m+1} \cdot (1-x) \over (1-x)^2}$
From this, evaluated at $x=2$ we get the expression for $T(n$)
$ \displaystyle \qquad \qquad T(n)=G(\log_2(n),2) \\
\qquad \qquad \qquad = 1+2 \cdot { (1-2^m)+m \cdot 2^m \over (-1)^2}
\qquad \qquad \qquad \text{ ( where } m=\log_2(n) \text{ )}\\
\qquad \qquad \qquad = 1+2 \cdot (1-n+\log_2(n) \cdot n) $
and finally
$ \qquad \qquad \displaystyle { T(n) = 3 + (\log_2(n)-1)\cdot2 \cdot n }$
(which is the result as was previously given in the answer of pjhuxford)
| {
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"url": "https://math.stackexchange.com/questions/943879",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Proving the summation formula using induction: $\sum_{k=1}^n \frac{1}{k(k+1)} = 1-\frac{1}{n+1}$ I am trying to prove the summation formula using induction:
$$\sum_{k=1}^n \frac{1}{k(k+1)} = 1-\frac{1}{n+1}$$
So far I have...
Base case:
*
*Let n=1 and test
$\frac{1}{k(k+1)} = 1-\frac{1}{n+1}$
$\frac{1}{1(1+1)} = 1-\frac{1}{1+1}$
$\frac{1}{2} = \frac{1}{2}$
*
*True for n=1
Induction Hypothesis:
*
*Assume the statement is true for the n-th case
$\sum_{k=1}^n \frac{1}{k(k+1)} = 1-\frac{1}{n+1}$
Inductive Step:
*
*Prove, using the Inductive Hypothesis as a premise, that
$$\sum_{k=1}^{n+1} \frac{1}{k(k+1)} = \sum_{k=1}^{n} \frac{1}{k(k+1)} + \frac{1}{(n+1)(n+2)} = 1-\frac1{n+1} + \frac{1}{(n+1)(n+2)} = \frac{(n+1)(n+2)}{(n+1)(n+2)}+\frac{-2-n}{(n+1)(n+2)}+\frac{1}{(n+1)(n+2)} = \frac{(n+1)(n+2)-2-n+1)}{(n+1)(n+2)} = \frac{(n+1)(n+2)-n-1}{(n+1)(n+2)} = \frac{n^2+2n+1}{(n+1)(n+2)} = \frac{(n+1)(n+1)}{(n+1)(n+2)} = \frac{n+1}{n+2}$$
To prove
$$ 1-\frac{1}{n+2} = \frac{n+1}{n+2} $$
Multiply both sides by $n+2$ to get an equivalent expression.
$$ (1-\frac{1}{n+2}) * (n+2) = (\frac{n+1}{n+2}) * (n+2) $$
$$ n+1=n+2−1 $$
Does this all make sense? How can this be improved upon?
| What do you know about induction proof?
You assume that statement is valid for $P(n)$, and show that is then valid for $P(n+1)$.
(basically, you prove $P(n) \implies P(n+1)$.
Then, if the statement is valid for $P(0)$, is valid for $P(1)$, then is valid for $P(2)$ and so on.
This way you proved your statement for every $n \in \mathbb N$.
Back to your problem.
Assume $$\sum_{k=1}^n \frac{1}{k(k+1)} = 1-\frac{1}{n+1}$$
Your goal is to show that
$$\sum_{k=1}^{n+1} \frac{1}{k(k+1)} = 1-\frac{1}{n+2}$$
which should not be too difficult given the previous assumption.
Ask if you have any troubles!
EDIT
How do you manipulate that expression? The goal is to make the premises appear! So just do
$$\sum_{k=1}^{n+1} \frac{1}{k(k+1)} = \sum_{k=1}^{n} \frac{1}{k(k+1)} + \frac{1}{(n+1)(n+2)} = $$
thanks to the inductive step
$$= 1-\frac1{n+1} + \frac{1}{(n+1)(n+2)}$$
You just have to prove that this equals $\displaystyle 1- \frac{1}{n+2}$ and you are done
EDIT 2
How do you prove that $$\frac{n+1}{n+2} = 1 - \frac{1}{n+2}$$?
You can multiply both sides by $n+2$ to get an equivalent expression.
$$n+1 = n+2 - 1$$ which is true, and so $\displaystyle \frac{n+1}{n+2} = 1 - \frac{1}{n+2}$ is also true
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/946712",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 0
} |
If $R$ is the circumradius of $\triangle ABC$, and $\cos A=\frac1{2R}$, $\cos B=\frac1{R}$ and $\cos C= \frac3{2R}$, then is it unique and its area? Given that $R$ is the circumradius of $\triangle ABC$, and $\cos A=\frac1{2R}$, $\cos B=\frac1{R}$ and $\cos C= \frac3{2R}$. Then would the $\triangle ABC$ be unique? If so how easily we may find its area or otherwise the maximum possible area?
Any help is appreciated :)
| Since
$$
\cos(B)=2\cos(A)\tag{1}
$$
and
$$
\cos(C)=3\cos(A)\tag{2}
$$
and furthermore, since $A+B+C=\pi$,
$$
\cos(C)=-\cos(A+B)\tag{3}
$$
Using $(1)$ and $(2)$ in $(3)$ says
$$
3\cos(A)=\sin(A)\sin(B)-2\cos^2(A)\tag{4}
$$
Moving $2\cos^2(A)$ to the left and squaring yields
$$
\begin{align}
4\cos^4(A)+12\cos^3(A)+9\cos^2(A)&=(1-\cos^2(A))(1-4\cos^2(A))\\
12\cos^3(A)+14\cos^2(A)-1&=0\\
\cos(A)&=0.24312617957188074493\tag{5}
\end{align}
$$
Since $R=\dfrac1{2\cos(A)}$ and the area is $2R^2\sin(A)\sin(B)\sin(C)$, we get
$$
\begin{align}
\text{Area}
&=\frac{\sqrt{1-\cos^2(A)}\sqrt{1-4\cos^2(A)}\sqrt{1-9\cos^2(A)}}{2\cos^2(A)}\\
&=4.90482198456130394784\tag{6}
\end{align}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/947307",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
Indefinite integral with partial fractions $$\int{ -5x^3-2x^2+32\over x^4-4x^3 } dx $$
How should I solve this indefinite integral using partial fractions?
I have already checked the online calculators but the answer they give me is incorrect whenever I check it.
I have used partial fractions of type ${A\over x}+{B\over x^2} + {C\over x^3} + {D\over (x-4)}.$
The answer I get when I solve the problem is ${2(x-2) \over x^2 }-5\ln(x-4) + C.$
However, the answer shows as incorrect when I input it.
| Using partial fraction decomposition on the integrand, we can write
\begin{equation*}
\int\frac{-5x^3-2x^2+32}{(x-4)x^3}dx=\int (-\frac{8}{x^3}-\frac{2}{x^2}-\frac{5}{x-4})dx=\frac{4}{x^2}+\frac{2}{x}-5\int\frac{1}{x-4}dx.
\end{equation*}
Using the substitution $u=x-4,~du=dx$ for the final integral, we get
\begin{equation*}
-5\ln(u)+\frac{4}{x^2}+\frac{2}{x}+C.
\end{equation*}
Finish by substituting back for $u=x-4$ to get
$$
\int\frac{-5x^3-2x^2+32}{x^4-4x^3}dx=-5\ln(x-4)+\frac{4}{x^2}+\frac{2}{x}+C.~_{\square}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/950282",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Solve the differential equation $p=\sin(10x+6y)$
$$p=\sin(10x+6y),p:=dy/dx$$
What I did:
$z=10x+6y\implies 6y=z-10x$
So:
$$dz/dx-10=6\sin z\\
dz/dx=2(5+3\sin z)\\
\int dz/(5+3\sin z)=2\int dx\\
\int 2dt/(5+5t^2+6t)=\int2dx,t:=\tan z/2\\
\int(1/5)dt/((t+3/5)^2-4/5)=2x+a\\
(1/5)(1/(2(\sqrt{4/5})))\ln((t+3/5-2/\sqrt5)/(t+3/5+2/\sqrt5))=2x+b\\
\frac{1}{4\sqrt5}\ln\left(\frac{10x+6y+3/5-2/\sqrt5}{10x+6y+3/5+2/\sqrt5}\right)=2x+c$$
Given answer:
$$y=\frac13\arctan\left(\frac{5\tan 4x}{4-3\tan 4x}\right)-\frac{5x}3$$
| I believe you made a slight mistake in your integration:
$$\int \frac{2}{5t^2+6t+5}\, dt = \frac{2}{5}\int \frac{dt}{\left(t+\frac{3}{5}\right)^2 + \left(\frac{4}{5}\right)^2},$$
from which you get
$$2x = \frac{1}{2}\arctan\left(\frac{5t+3}{4}\right)+C.$$
Continuing from here, you should arrive at the answer.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/953554",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 0
} |
Proving $9$ divides $n^3 + (n+1)^3 + (n+2)^3$ I'm trying to prove by MI. I have already distributed n+1, but now I'm stuck on how I can show 9 divides the RHS since $42n$ and $3n^3$ does not divide evenly.
$$(n+1)^3+(n+2)^3+(n+3)^3=3n^3+18n^2+42n+36$$
| $$(n-1)^3+n^3+(n+1)^3=3n^3+6n=3(n^3-n)+9n$$
Now $n^3-n=(n-1)n(n+1)$ is product of three consecutive integers, hence divisible by $3!$
Also, $$(n+1)^3+(n+2)^3+(n+3)^3=3n^3+18n^2+42n+36\equiv3(n^3-n)\pmod9$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/953974",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 1
} |
Integration without substitution of $\frac{x^2+3}{x^6\left(x^2+1\right)}$ This is a repost of a question i had written incorrectly earlier.
How do I integrate this without substitutions ?
$$
\frac{x^2+3}{x^6\left(x^2+1\right)}
$$
I got:
$$
\frac{1}{x^6}+\frac{2}{x^6\left(x^2+1\right)},
$$
but wasn't able to eliminate the 2.
| $\bf{My\; Solution}$ Given $\displaystyle \int\frac{x^2+3}{x^6(x^2+1)}dx = \int\frac{(x^2+1)+2}{x^6(x^2+1)}dx = \int x^{-6}dx+\int\frac{2}{x^6(x^2+1)}dx$
$\displaystyle =-\frac{1}{5}x^{-5}+I,$ where $\displaystyle I = \int\frac{2}{x^6(x^2+1)}dx$
Now Using $\displaystyle x= \frac{1}{t}$ and $\displaystyle dx = -\frac{1}{t^2}$. So Let $\displaystyle I = -\int \frac{2t^6}{1+t^2}dt = -\int \frac{2(t^6+1)-2}{1+t^2}dt$
So $\displaystyle I = -2\int \frac{t^6+1}{t^2+1}dt+\int \frac{2}{1+t^2}dt = -2\int (t^4-t^2+1)dt+2\tan^{-1}(t)$
$\displaystyle = -\frac{2t^5}{5}+\frac{2t^3}{3}-\frac{2t^2}{2}+2\tan^{-1}(t)+\mathcal{C} = -\frac{2}{5x^5}+\frac{2}{3x^3}-\frac{2}{2x^2}+2\tan^{-1}\left(\frac{1}{x}\right)+\mathcal{C}$
So $\displaystyle \displaystyle \int\frac{x^2+3}{x^6(x^2+1)}dx = -\frac{1}{5x^5}-\frac{2}{5x^5}+\frac{2}{3x^3}-\frac{2}{2x^2}+2\tan^{-1}\left(\frac{1}{x}\right)+\mathcal{C}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/954101",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 3
} |
Find range of $\sin x+\cos^{2}x$ By differentiating and equating to $0$ I know that the maximum must be $5/4$. The minimum is where I am confused. $-1$ would be logical but I'm not sure if this function can ever be equal to $0$. Answer is apparently $1$ to $5/4$ but I think that is wrong.
| $\sin(x) + \cos^2(x) = - \sin^2(x) + \sin(x) + 1$ since $\cos^2(x) = 1 - \sin^2(x)$. Let $t=\sin(x)$.
$\sin(x) + \cos^2(x) = -t^2 + t + 1$ and $t \in [-1,1]$ because $\sin(x)\in [-1,1]$. Draw the curve of the quadratic $-t^2 + t + 1$ for $t \in [-1,1]$. The quadratic $-t^2 + t + 1$ has a root $\frac{1-\sqrt{5}}{2}$ between $-1$ and $1$. The function $\sin(x) + \cos^2(x)$ therefore has a root at $\sin(x) = \frac{1-\sqrt{5}}{2}$ i.e., the function $\sin(x) + \cos^2(x)$ attains the value $0$ for $\sin(x)=\frac{1-\sqrt{5}}{2}$. The quadratic attains its minimum at $t = -1$ and has the value of $-1$. Hence, the original function too has the minimum value of $-1$ when $t = \sin(x) = -1$ or $x = -\frac{\pi}{2}$. The quadratic attains a maximum at $t=\frac{1}{2}$ and has the value $\frac{5}{4}$. The original function thus has a maximum value of $\frac{5}{4}$ at $t=\sin(x)=\frac{1}{2}$ or $x = \frac{\pi}{6}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/956550",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 3
} |
Integrate $\int \cos^4 x\,\mathrm dx.$ $$\int \cos^4 x\,\mathrm dx = \int (1 - \sin^2x)^2\,\mathrm dx.$$
I tried using $\cos^2(x) + \sin^2(x)$ = 1. This was on the integration by parts section of my textbook.
The integral I came out with is given me a hard time. I double checked my work and I feel I am on the right track but I feel stuck. I tried u substitution but there is no $\cos x$ to leverage with.
| In general$\newcommand{\dx}{\,\mathbb d x}$\begin{align*} I_n = \int \cos^n x \dx &= \int (1-\sin^2 x)\cos^{n-2} x \dx \\ &= I_{n-2} - \int \underbrace{\sin x \cos^{n-2}x}_{v^\prime} \underbrace{\sin x}_u \dx \\ &= I_{n-2} + \frac{1}{n-1}\cos^{n-1} x \sin x - \frac{1}{n-1}\int \cos^{n-1} x \cos x \dx \\ &= I_{n-2} + \frac{1}{n-1}\cos^{n-1} x \sin x - \frac{1}{n-1}I_n \\ \frac{n}{n-1} I_n &= I_{n-2} + \frac{1}{n-1}\cos^{n-1} x \sin x \\[12pt] I_n &= \frac{n-1}{n}I_{n-2} + \frac{1}{n}\cos^{n-1} x \sin x \end{align*}
Valid for $n \gt 1$
So \begin{align*}\int \cos^4 x \dx &= \frac34\int\cos^2 x \dx + \frac13\cos^3 x \sin x \\ &= \frac34\left( \frac12\int\dx+\frac12\cos x \sin x \right) + \frac13\cos^3 x \sin x \\ &= \frac38x + \frac38 \cos x \sin x + \frac13 \cos^3 x \sin x +c\end{align*}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/958733",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 5,
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Solve the equation $4^{7 x - 10} = 10^{7 x - 6}$ for $x$ Solve for $x$ of the following equation.
$$4^{7 x - 10} = 10^{7 x - 6}$$
I tried to make $4$ and $10$ have a common base but I could not find one so I don't know where to go from here.
| $$4^{7x - 10} = 10^{7x - 6}$$
$$(7x - 10)\log4 = (7x - 6)\log10$$
$$7x = \frac{6\log10 - 10\log4}{\log10 - \log4}$$
$$ = \frac{6\log5 + 6\log2 - 20\log2}{\log5 + \log2 - 2\log2}$$
$$ = \frac{6\log5 - 14\log2}{\log5 - \log2}$$
$$ x = \frac{2(3\log5 - 7\log2)}{7(\log5 - \log2)}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/959918",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 3
} |
Checking some work on finding roots OK, I have the following response function:
$$H(\omega) = \frac{1-\omega^2 LC}{1+\omega^2 LC - i \omega RC}$$
I want to find where it becomes $\frac{1}{\sqrt{2}}$.
This should be simple enough. First I multiply the whole thing by its complex conjugate, which gives me the absolute value squared, or 1/2:
$$\frac{1-\omega^2 LC}{1+\omega^2 LC - i \omega RC}=\frac{(1-\omega^2 LC)^2}{(1+\omega^2 LC)^2 + \omega^2 R^2C^2} = \frac{1}{2}$$
Then I want to solve for omega. Since this is kind of ugly as it is I multiply both sides by $2((1+\omega^2 LC)^2 + \omega^2 R^2C^2)$ and end up with
$$2(1-\omega^2 LC)^2=(1+\omega^2 LC)^2 + \omega^2 R^2C^2$$
which yields
$$(1-\omega^2 LC)^2= \omega^2 R^2C^2$$
which I can take the square roots of both sides and turn it into a quadratic
$$(1-\omega^2 LC)= \omega RC \rightarrow 1-\omega RC - \omega^2LC = 0$$
I pull out the old quadratic formula. $\omega = \frac{RC \pm \sqrt{R^2C^2+4LC}}{2LC}$
On the plus side we end up with $\frac{2R^2C^2 + 2RC\sqrt{R^2C^2+4LC}+4LC}{2LC}=\frac{R^2C}{L}+\frac{R}{L}\sqrt{R^2C^2+4LC}+2$
and on the minus side $\frac{2R^2C^2 - 2RC\sqrt{R^2C^2+4LC}+4LC}{2LC}=\frac{R^2C}{L}-\frac{R}{L}\sqrt{R^2C^2+4LC}+2$
Anther method I thought of was to try and break up the quadratic "by hand" -- that is, try to come up with some square root of the coefficient of omega squared and half of RC, but down that path lieth madness.
OK, seems fine. But I am told that the answer you ought to get is $\omega = \frac{1}{RC}$. So either a) I messed up badly or b) I am being told wrong.
Did I miss something here? This isn't even calculus.
EDIT: looking at where I messed up:
OK Seeing as I messed up here:
$$2(1-\omega^2LC)^2 = (1+\omega^2LC)^2 + \omega^2 R^2C^2$$
That really ought to be
$$(2 - 4\omega^2LC + 2\omega^4L^2C^2) = 1 + 2\omega^2LC + \omega^4L^2C^2+ \omega^2 R^2C^2$$
Which then becomes
$$(1 - 6\omega^2LC + \omega^4L^2C^2) = \omega^2 R^2C^2$$
turning that into something like a quadratic
$$1+(6LC-R^2C^2)\omega^2+ \omega^4L^2C^2=0$$
(and substituting $u$ for $\omega^2$ means the LHS reduces to $ (1+(6LC-R^2C^2)u + L^2C^2u^2)$
We hit it with the quadratic formula. $$\frac{R^2C^2-6LC \pm \sqrt{36L^2C^2-12R^2LC^3+R^4C^4-4L^2C^2}}{2L^2C^2}=\frac{R^2C^2-6LC \pm \sqrt{32L^2C^2-12R^2LC^3+R^4C^4}}{2L^2C^2}=\frac{R^2}{2L^2}-\frac{3}{L}\pm \frac{\sqrt{32L^2-12R^2LC+R^4C^2}}{2L^2C}$$
Still getting crazy numbers.... hmmm.
| You messed up at least once, in this step:
$$2(1-\omega^2 LC)^2=(1+\omega^2 LC)^2 + \omega^2 R^2C^2$$
which yields
$$(1-\omega^2 LC)^2= \omega^2 R^2C^2$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/960187",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Prove ${_2F_1}\left({{\tfrac16,\tfrac23}\atop{\tfrac56}}\middle|\,\frac{80}{81}\right)=\frac 35 \cdot 5^{1/6} \cdot 3^{2/3}$ I've found the following hypergeometric function value by numerical observation. The identity matches at least for $100$ digits.
$${_2F_1}\left(\begin{array}c\tfrac16,\tfrac23\\\tfrac56\end{array}\middle|\,\frac{80}{81}\right) \stackrel{?}{=} \frac 35 \cdot 5^{1/6} \cdot 3^{2/3}$$
Or using a Pfaff transformation in an equivalent form
$$81^{1/6} \cdot {_2F_1}\left(\begin{array}c\tfrac16,\tfrac16\\\tfrac56\end{array}\middle|\,-80\right) \stackrel{?}{=} \frac 35 \cdot 5^{1/6} \cdot 3^{2/3}$$
How could we prove it?
Other related problem: How could we prove that
$${_2F_1}\left(\begin{array}c\tfrac16,\tfrac23\\\tfrac56\end{array}\middle|\,\frac{80}{81}\right) \stackrel{?}{=} {_2F_1}\left(\begin{array}c\tfrac12,\tfrac56\\\tfrac12\end{array}\middle|\,\frac{4}{9}\right) = {_1F_0}\left(\begin{array}c\tfrac56\\\ - \,\end{array}\middle|\,\frac{4}{9}\right)$$
| (This may not be complete, but generalizes the observation for context.) The OP's answer mentioned Reshetnikov's integral which can be expressed as,
$$\frac{1}{48^{1/4}\,K(k_3)}\,\int_{-1}^1\frac{dx}{\sqrt[3]{9+4\sqrt{5}\,x}\,\left(1-x^2\right)^{\small2/3} } =\,_2F_1\big(\tfrac{1}{3},\tfrac{1}{3};\tfrac{5}{6};-4\big)= \frac{3}{5^{5/6}} $$
so it will be fruitful to find a transformation between Reshetnikov's use of $\,_2F_1\big(\tfrac{1}{3},\tfrac{1}{3};\tfrac{5}{6};z_1\big)$ and the OP's use of $\,_2F_1\big(\tfrac{1}{6},\tfrac{2}{3};\tfrac{5}{6};z_2\big)$. Employing well-known transformations, we get,
$$G(y)=\,_2F_1\big(\tfrac{1}{3}, \tfrac{1}{3};\tfrac{5}{6}; -y\big) = \big(\tfrac1{1+2y}\big)^{1/3}\,_2F_1\Big(\tfrac{1}{6}, \tfrac{2}{3};\tfrac{5}{6}; \tfrac{(1+2y)^2-1}{(1+2y)^2}\Big)$$
It is conjectured that if $y$ are certain algebraic numbers, than $G(y)$ is also an algebraic number. Specializing the RHS, define as in this post,
$$H(\tau)=\,_2F_1\big(\tfrac16,\tfrac23;\tfrac56;(1-2\delta_4)^2\big),\quad\text{where}\;\frac1{\delta_4}-1=\frac1{27}\left(\tfrac{\eta\big((\tau+1)/3\big)}{\eta(\tau)}\right)^{12}$$
with Dedekind eta function $\eta(\tau)$. If $\tau = \frac{1+N\sqrt{-3}}2$ for some integer $N>1$, then the conjecture implies both $\delta_4$ and $H(\delta_4)$ are algebraic numbers. For example, if we use $\tau = \frac{1+5\sqrt{-3}}2$ and $\tau = \frac{1+7\sqrt{-3}}2$, respectively, then we recover your,
$$H\big(\tfrac{1+\color{blue}5\sqrt{-3}}2\big)=\,_2F_1\big(\tfrac16,\tfrac23;\tfrac56;\tfrac{80}{81}\big) = \tfrac3{\color{blue}5}\,(9\sqrt5)^{1/3}$$
as well as,
$$H\big(\tfrac{1+\color{blue}7\sqrt{-3}}2\big)=\,_2F_1\big(\tfrac16,\tfrac23;\tfrac56;\tfrac{3024}{3025}\big) = \tfrac4{\color{blue}7}\,55^{1/3}$$
and so on.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/961515",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "12",
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What is the flaw in both of my approaches to this limit? I have been solving a 3D limit problem: $$\lim_{(x,y) \to (0,0)} \frac{x^2+y^2}{\sqrt{x^2+y^2+1}-1}$$which apparently, is $2$. However, I cannot find the flaw in my approach. Here it goes:
$$\mathrm{Let \space y=mx} \\\lim_{(x,y) \to (0,0)} \frac{x^2+m^2x^2}{\sqrt{x^2+m^2x^2+1}-1}= \lim_{(x,y) \to (0,0)} \frac{x^2(1+m^2)}{\sqrt{2x^2+1}-1}=\frac{0}{0} \\ \mathrm{Apply \space L'Hôpital's \space rule}\\ \lim_{(x,y) \to (0,0)}\frac{2x(1+m^2)}{(\frac{2x}{\sqrt{2x^2+1}})}$$
I've solved this two different ways and I keep getting the answer as $(1+m^2)$. The Student Solutions Manual, alas, says the answer it $2$. Just in case my mistake lies somewhere after this step, even though I got the same answer solving it two different ways after this point I'll give you one of my methods, the shorter of the two:
$$...\space=\lim_{(x,y) \to (0,0)} \frac{2x(1+m^2)\sqrt{2x^2+1}}{2x}=\lim_{(x,y) \to (0,0)}\frac{(1+m^2)\sqrt{2x^2+1}}{1}=(1+m^2)$$
No idea how to get to $2$ here. Also, am I abusing notation by leaving the bounds of the limits in terms of $(x,y)$ rather than reducing it to $x$ when I have let $y$ be equal to some terms of $x$?
| Rationalize the denominator:
\begin{equation}
\frac{x^2+y^2}{\sqrt{x^2+y^2+1}-1}=\frac{x^2+y^2}{\sqrt{x^2+y^2+1}-1}\cdot \frac{\sqrt{x^2+y^2+1}+1}{\sqrt{x^2+y^2+1}+1}
\end{equation}
\begin{equation}=\frac{(x^2+y^2)(\sqrt{x^2+y^2+1}+1)}{(x^2+y^2+\sqrt{x^2+y^2+1}-\sqrt{x^2+y^2+1}+1)-1}
\end{equation}
\begin{equation}=\frac{(x^2+y^2)(\sqrt{x^2+y^2+1}+1)}{x^2+y^2}
\end{equation}
\begin{equation}=\sqrt{x^2+y^2+1}+1 \, .
\end{equation}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/962507",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 1
} |
Find all positive integers $(x,y,z)$ such that $\frac{1}{x}+\frac{1}{y}+\frac{1}{z}$ is integer As stated in title, I would like to find solution to this problem:
Find all positive integers $(x,y,z)$ such that $\frac{1}{x}+\frac{1}{y}+\frac{1}{z}$ is also integer.
I need idea how to solve this?
| Let $1\leq x\leq y\leq z$ and $N= \frac{1}{x} + \frac{1}{y} +
\frac{1}{z}$
If $N=3$ then $$x=y=z=1$$
If $N=2$ then $$ x=1,\ y=z=2$$
If $N=1$ then $$ 2\leq x\leq 3 $$
If $x=3$ then $$ (x,y,z)=(3,3,3) $$
If $x=2$ then $$ 3\leq y \leq 4$$
So $$ (x,y,z)=(2,3,6),\ (2,4,4)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/965117",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Solving $2\arcsin\frac{x}{2}+\arcsin(x\sqrt{2})=\frac{\pi}{2}$ How do I solve this equation:
$$
2\arcsin\frac{x}{2}+\arcsin(x\sqrt{2})=\frac{\pi}{2}
$$
We know that:
$$\sin(\alpha+\beta)=\sin\alpha\cos\beta+\cos\alpha\sin\beta$$
So letting $\alpha = 2\arcsin\frac{x}{2}$ and $\beta=\arcsin(x\sqrt{2})$ leads to: $\sin\frac{\alpha}{2} = \frac{x}{2}$ and $\sin\beta = x\sqrt{2}$. Finding $\sin\alpha$ and $\cos\alpha$ first:
$$\begin{align}
\cos \frac{\alpha}{2} & = \frac{\sqrt{(4-x^2)}}{2} \\
\sin\alpha & = \sin\left(\frac{\alpha}{2} + \frac{\alpha}{2}\right) = 2\cos\frac{\alpha}{2}\sin\frac{\alpha}{2}\\
& = \frac{x\sqrt{(4-x^2)}}{2} \\
\cos \alpha & = \frac{\sqrt{(4 - (x\sqrt{(4-x^2)})^2)}}{2} = \frac{\sqrt{(4 - x^2(4-x^2))}}{2}
\end{align}$$
And now $\cos\beta$:
$$\begin{align}
\sin \beta & = x\sqrt2 \\
\cos \beta & = \sqrt{1 - 2x^2}
\end{align}$$
Plugging everything together:
$$
1 = \frac{x\sqrt{(4-x^2)} \times \sqrt{1 - 2x^2}}{2} + \frac{\sqrt{(4 - x^2(4-x^2))} \times x\sqrt2}{2} \\
2 = x\sqrt{4-9x^2+2x^4} + x\sqrt{8-8x^2+2x^4} \\
4 = x^2(4-9x^2+2x^4) + x^2(8-8x^2+2x^4) \\
0 = 4x^4 -17x^3 + 12x^2 - 4
$$
Which is incorrect - the correct answer is $\sqrt{6-4\sqrt2}$. Where did I go wrong?
| Let $\arcsin\dfrac x2=y\implies x=2\sin y$
So we have,
$$\arcsin(\sqrt2\cdot2\sin y)=\dfrac\pi2-2y$$
Applying sine on both sides,
$$\sin\left[\arcsin(\sqrt2\cdot2\sin y)\right]=\sin\left(\dfrac\pi2-2y\right)=\cos2y$$
$$\implies\sqrt2\cdot2\sin y=1-2\sin^2y$$
Rearrange to form a Quadratic Equation in $\sin y$
Check if the values of $x(=2\sin y)$ satisfy the given equation
Observe that $x>0$ as $-\dfrac\pi2<\arcsin(u)<0$ for $u<0$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/967769",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
} |
Evaluate $\int \frac 1{x^{12}+1} \, dx$
Evaluate $\displaystyle \int \frac 1{x^{12}+1} \, dx$
I tried writing this in partial fractions.
$$\int \frac 1{x^{12}+1} \, dx=\int \frac{1}{[(x^6+1)+\sqrt{2}x^3][(x^6+1)-\sqrt{2}x^3]} \, dx$$
So I did:
\begin{align}
\frac{1}{[(x^6+1)+\sqrt{2}x^3][(x^6+1)-\sqrt{2}x^3]} &= \frac{Ax^5+Bx^4+Cx^3+Dx^2+Ex+F}{[(x^6+1)-\sqrt{2}x^3]}\\&{}\qquad+\frac{A_1x^5+B_1x^4+C_1x^3+D_1x^2+E_1x+F_1}{[(x^6+1)+\sqrt{2}x^3]}\end{align}
After messy work, I found $A=A_1=B=B_1=D=D_1=E=E_1=0$ and
$$C=-\frac 1{2\sqrt{2}},\qquad C_1=\frac 1{2\sqrt{2}},\qquad F=F_1=\frac 12$$
I now get the integral
$$\int \frac{-\frac 1{2\sqrt{2}}x^3+\frac 12}{(x^6+1)-\sqrt{2}x^3}\, dx + \int \frac{\frac 1{2\sqrt{2}}x^3+\frac 12}{(x^6+1)+\sqrt{2}x^3} \, dx$$
But how can I go from here? I am stuck.
| Decompose and arrange the integrand as follows
\begin{align}\frac{6}{1+x^{12}}
& =\frac2{x^4+1}-\frac{\sqrt3x^2-2}{x^4-\sqrt3x^2+1}
+\frac{\sqrt3x^2+2}{x^4+\sqrt3x^2+1} \\
& =\frac{(1+x^2)+(1-x^2)}{x^4+1}+\frac{2+\sqrt3}2\left(\frac{1+x^2}{x^4+\sqrt3x^2+1}+\frac{1-x^2}{x^4-\sqrt3x^2+1} \right)\\
& \hspace{30mm}+\frac{2-\sqrt3}2\left(\frac{1+x^2}{x^4-\sqrt3x^2+1}+\frac{1-x^2}{x^4+\sqrt3x^2+1} \right)\\
\end{align}
Then
\begin{align}
\int \frac{dx}{1+x^{12}}
& =\frac16J(0)
+\frac{2+\sqrt3}{12}J(\sqrt3) +\frac{2-\sqrt3}{12}J(-\sqrt3) +C\\
\end{align}
where
\begin{align}
J(a)&= \int \left(\frac{1+x^2}{x^4+ax^2+1}+ \frac{1-x^2}{x^4-ax^2+1}\right)dx \\
&=\int \frac{d(x-\frac1x)}{(x-\frac1x)^2+(2+a)}+\int \frac{d(x+\frac1x)}{(2+a)-(x+\frac1x)^2} \\
&=\frac1{\sqrt{2+a}}\left( \tan^{-1}\frac{x^2-1}{x\sqrt{2+a}} +\coth^{-1}\frac{x^2+1}{x\sqrt{2+a}} \right)\\
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/971973",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
"answer_count": 5,
"answer_id": 1
} |
Field extensions and irreducibility I'm having trouble trying to show that the function f=x^3 + x + 3 is irreducible in the rationals. I tried using Eisensteins criterion but it didn't work as it doesnt satisfy all conditions.
the second part of the question is to show that f has one real root which i showed using Rolle's Theorem from Analysis
the third part is showing that denoting the real root as x
Q(x)=Q(x^2) where Q(a) [represents the field extension over rationals]
My attempt at this question is to show that every member of Q(x) is contained in Q(x^2), by using closure under field operations, but I'm struggling to show that every member of Q(x^2) is inside Q(x).
Any help would be greatly appreciated.
| Here is what I think for the first question:
Suppose $f(x) = x^{3} + x + 3$ is not irreducible over $\mathbb{Q}$. Then we can write $f(x)$ as the product $(x + a)(x^{2} + bx + c)$ for some rational numbers $a, b, c$.
But then, multiplying this out gives the expression $x^{3} + bx^{2} + cx + ax^{2} + abx + ac = x^{3} + (b + a)x^{2} + (ab + c)x + ac$
And we know this must equal $x^{3} + x + 1$, so we know that:
*
*$b + a = 0$
*$ab + c = 1$
*$ac = 3$
But if a polynomial is irreducible over $\mathbb{Z}$, then it is irreducible over $\mathbb{Q}$, so we really only need to find a contradiction with the above equations if $a,b,c$ are integers.
Equation 3 means either $a = 1$, $c = 3$ or $a = 3$, $c = 1$, or $a = -1$, $c = -3$, or $a = -3, c = -1$. But equation 2 implies $b = -a$, so:
If $a = 1$, $c = 3$, then $b = -1$, so from equation 2 we get $ab + c = 1(-1) + 3 = -1 + 3 = 2 \neq 1$. So equation 2 is not satisfied.
If $a = 3$, $c = 1$, then $b = -3$, and by equation 2 we get $3(-3) + 1 = -9 + 1 = -8 \neq 1$, so again equation 2 is not satisfied.
If $a = -1$, $c = -3$, then $b = 1$ and so $(-1)1 - 3 = -4 \neq 1$, which means equation 2 is yet again not satisfied.
Finally, if $a = -3$, $c = -1$, then $b = 3$, and so $(-3)3 - 1 = -10 \neq 1$, which means equation 2 is yet again not satisfied.
So in every case, we can't find a solution of integers that satisfies the three equations, and thus the polynomial is not reducible over $\mathbb{Z}$. Consequently, it is not reducible over $\mathbb{Q}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/973471",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Proving these are equal We have the following equality (in physics) $$c^2t^2-x^2 = c^2t'^2-x'^2$$
where:
$t' = \gamma (t- \dfrac{vx}{c^2})$
$x' = \gamma(x-vt)$
$\gamma = \dfrac{1}{\sqrt{1-\dfrac{v^2}{c^2}}}$
My textbook gives no proof of this, and I thought it would be a nice algebraic exercise to perform, but I have failed to find it. I get stuck at the following expression:
$$c^2t'^2-x'^2 = \gamma ^2 (c^2t^2 + \dfrac{v^2x^2}{c^2} - x^2 - v^2t^2)$$
This very slightly resembles $c^2t^2-x^2$, but I still can't figure out how to get to that expression. Can anybody help?
| \begin{align*}
c^2t'^2 - x'^2
&= c^2\gamma^2\left(t - \frac{v x}{c^2}\right)^2 - \gamma^2(x-vt)^2 \\
&= \frac{1}{1-\frac{v^2}{c^2}}\left(c^2t^2 - 2tvx + \frac{v^2x^2}{c^2} - x^2 + 2xvt - v^2t^2\right) \\
&= \frac{c^2}{c^2-v^2}\left(c^2t^2 + \frac{v^2x^2}{c^2} - x^2 - v^2t^2\right) \\
&= \frac{c^2}{c^2-v^2}(c^2-v^2)t^2 + \frac{c^2}{c^2-v^2}\left(\frac{v^2x^2}{c^2} - x^2\right) \\
&= c^2t^2 + \frac{c^2}{c^2-v^2}\cdot\frac{(v^2-c^2)x^2}{c^2} \\
&= c^2t^2 - x^2.
\end{align*}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/974336",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
How to find $\left\lfloor\sum_{r=1}^{80}\int_0^1x^{\sqrt r-1}dx\right\rfloor$
$$\left\lfloor\sum_{r=1}^{80}\int_0^1x^{\sqrt r-1}dx\right\rfloor$$
My try:
$$K=\left\lfloor\sum_{r=1}^{80}\int_0^1x^{\sqrt r-1}dx\right\rfloor=\left\lfloor\sum_{r=1}^{80}\frac1{\sqrt r}\right\rfloor=\left\lfloor\frac1{\sqrt 1}+\frac1{\sqrt 2}+\ldots+\frac1{\sqrt{80}}\right\rfloor$$
I can only say:
$$\left\lfloor\sum_{r=1}^{80}\frac1{\sqrt {80}}\right\rfloor<K<\left\lfloor\sum_{r=1}^{80}\frac1{\sqrt 1}\right\rfloor\\
\lfloor\sqrt{80}\rfloor<K<\lfloor80\rfloor\\
8<K<80$$
So I don't know how to find K?
| Since $\frac{1}{\sqrt{x}}$ is decreasing, we have
$$\int_2^{81} \frac{1}{\sqrt{x}} \,dx < \sum_{k = 1}^{80} \frac{1}{\sqrt{k}} < \int_1^{81} \frac{1}{\sqrt{x}} \,dx,$$
and evaluating gives
$$2 (9 - \sqrt{2}) < \sum_{k = 1}^{80} \frac{1}{\sqrt{k}} < 16.$$
Now, $8 < 9$ implies that $\sqrt{2} < \frac{3}{2}$, so we may weaken the first inequality to $$15 = 2 \left(9 - \frac{3}{2}\right) < \sum_{k = 1}^{80} \frac{1}{\sqrt{k}},$$ and so $$\left\lfloor \sum_{k = 1}^{80} \frac{1}{\sqrt{k}} \right\rfloor = 15.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/976237",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
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How to Solve : $ A =\frac{1}{6}\left((\log_2(3))^3-(\log_2(6))^3-(\log_2(12))^3+(\log_2(24))^3\right) $ $ A =\frac{1}{6}\left((\log_2(3))^3-(\log_2(6))^3-(\log_2(12))^3+(\log_2(24))^3\right).$
Solve for $2^A.$
(no calculators or graphs are permitted)
The way I went about solving this problem was using the properties of base $2$. For example, $2^x = 8$ , $x$ is obviously $3$, and $2^y = 16$, $y = 4 = x+1$, so adding $1$ to the exponent will give twice the answer. Since the arguments are twice each other, I used this pattern:
$ A =\frac{1}{6}\left((x)^3-(x+1)^3-(x+2)^3+(x+3)^3\right)$
$= \frac{1}{6}\left((x^3-(x^3 + 3x^2 + 3x + 1)-(x^3 + 6x^2 + 12x + 8)+(x^3 + 9x^2 + 27x + 27)\right)$
Simplifying this we get $A = (12x + 18)/6 = 2x + 3.$
So now I solved for $2^{2x+3}$ using $2^{2x} =2^x * 2^x = 3^2 = 9.$ Now, since we add $3$ to the exponent, we have $2^{2x+3} = (2^3)(9) = 72.$
Would this be the correct answer, and if not, what would be a good way of going about it?
| A good choice of substitution might help cut through the clutter of logs and nested parentheses, giving a visually pleasing and symmetrical solution.
Let $\text{L}^mn=(\log_2 n)^m$ and put $u=L3=\log_23$.
Hence
$$\begin{align}
\text{L}^3 3&=(L3)^3&=u^3\\
\text{L}^3 6&=(L3+L2)^3&=(u+1)^3\\
\text{L}^3 12&=(L2+L4)^3&=(u+2)^3\\
\text{L}^3 24&=(L2+L8)^3&=(u+3)^3\end{align}$$
$$\begin{align}
A&=\frac16 ((\log_2(3))^3-(\log_2(6))^3-(\log_2(12))^3+(\log_2 (24))^3)\\
&=\frac16 (L^33-L^36-L^312+L^324)\\
&=\frac 16 \left[u^3-(u+1)^3-(u+2)^3+(u+3)^3\right]\\
&=\frac 16 \left[ 12u+18\right]\\
&=2u+3\\
&=2L3+L8\\
&=L72=\log_2 72\\
2^A&=72\qquad \blacksquare
\end{align}$$
Previous answer below:
$$\begin{align}\text{L}^3 3&=(\log_2 3)^3&&=u^3\\
\text{L}^3 6&=(\log_2 6)^3&=(\log_2 3+\log_2 2)^3&=(u+1)^3\\
\text{L}^3 12&=(\log_2 12)^3&=(\log_2 3+\log_2 4)^3&=(u+2)^3\\
\text{L}^3 24&=(\log_2 24)^3&=(\log_2 3+\log_2 8)^3&=(u+3)^3\end{align}$$
$$\begin{align}
A&=\frac16 ((\log_23)^3-(\log_2 6)^3-(\log_2 12)^3+(\log_2 24)^3)\\
&=\frac 16 \left[u^3-(u+1)^3-(u+2)^3+(u+3)^3\right]\\
&=\frac 16 \left[ 12u+18\right]\\
&=2u+3\\
&=2\log_2 3+\log_28\\
&=\log_2 72\\
2^A&=72\qquad \blacksquare
\end{align}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/978930",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 2,
"answer_id": 1
} |
Integer solutions of $x^3-x+9=5y^2$ What are the solutions in integers of $x^3-x+9=5y^2$?
[Source: Hungarian competition problem]
| If $x$ were odd, then we would have $x^3 \equiv x \pmod {8}$. But $5y^2 \equiv 1\pmod{8}$ has no solutions, so $x$ is even.
We have $(x-1)x(x+1) = 5y^2 - 9 \equiv 1\pmod{5}$. We have $2\cdot 3\cdot 4 \equiv -1\pmod{5}$, so we must have $x\equiv 2\pmod{5}$.
Finally, we observe that $x^3-x$ is divisible by $3$, so $y$ is divisible by $3$, so $x^3-x$ is divisible by $9$. However, $5y^2 \equiv 9 \pmod{27}$ has no solutions, so $x^3-x$ is divisible by $9$, but not $27$. This implies that either $x+1$ is not divisible by $3$, or it is divisible by $9$, but not $27$.
In the former case, we have $x+1 \equiv 3\pmod{5}$. In the latter, $\frac{x+1}{9} \equiv 2\pmod{5}$. Since $x+1$ is odd, in both cases there is a factor of $x+1$ that is in $\{2,3\}\pmod{5}$, but is not divisible by $2$ or $3$. That implies that there exists a prime factor $p\neq 2,3$ of $x+1$ such that $p\in \{2,3\}\pmod{5}$.
So $p$ is not a square$\pmod{5}$. By quadratic reciprocity, $5$ is not a square$\pmod{p}$. But we have $5y^2 - 9 \equiv 0 \pmod{p}$, so $5\equiv (3/y)^2\pmod{p}$, contradiction.
We conclude that there are no solutions.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/979044",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "13",
"answer_count": 2,
"answer_id": 0
} |
Calculate $1+(1+2)+(1+2+3)+\cdots+(1+2+3+\cdots+n)$ How can I calculate $1+(1+2)+(1+2+3)+\cdots+(1+2+3+\cdots+n)$? I know that $1+2+\cdots+n=\dfrac{n+1}{2}\dot\ n$. But what should I do next?
| Sums we know:
$\sum^n_{i=1} i = 1+2+\cdots+n=\frac{n^2+n}{2}$
$\sum^n_{i=1} i^2 = 1^2 + 2^2 + \dots + n^2 = \frac{n(n+1)(2n+1)}6$
Your sum is $$(1+2+3+ \cdots + n) + (1 + 2 + \cdots + (n-1)) + (1 + 2 + \cdots + (n-2)) + \cdots + (1)$$ $$= \sum^n_{k=1} \sum^k_{i=1} i$$ $$= \sum^n_{k=1} \frac{k^2+k}{2} = \frac 12 (\sum^n_{k=1} k^2 + \sum^n_{k=1} k)$$
NOTE: You can reorder the terms if the are a finite number of them. So if you're going to be taking a limit as $n \to \infty$ don't do it this way.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/980941",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 8,
"answer_id": 6
} |
Solving Yoshida equations I want to solve $a$, $b$ and $c$ out of the following set of equations
\begin{cases}
a + b + c = 1 \\
a^{p+1} + b^{p+1} + c^{p+1} = 0 \\
a = c \\
\end{cases}
where $p$ is even. But I absolutely have no idea how to even start solving it...
Edit: based on the hint given below, the solution goes as follows
\begin{cases}
2a + b = 1 \\
2a^{p+1} + b^{p+1} = 0 \\
\end{cases}
because $a=c$. By writing $b = 1-2a$ and substituting this into the second equation, we get $2a^{p+1} + (1-2a)^{p+1} = 0$. Because $p$ is even, this becomes $2a^{p+1} = (2a-1)^{p+1}$. Now taking the $(p+1)th$ root of both sides we obtain
$$
a = \frac{1}{2- \sqrt[\leftroot{-2}\uproot{2}p+1]{2} }
$$
| Hint
So, you have (since $a=c$)
$$\begin{cases}
2a + b= 1, \\
2a^{p+1} + b^{p+1} = 0 ,
\end{cases}$$
from where $b=1-2a$ and then $2a^{p+1}=-(1-2a)^{p+1}.$ Since $p$ is even, we have $\sqrt[p+1]{2}a=2a-1.$
Can you continue from here?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/981366",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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How to find nonnegative solutions of a linear system? I have the following system of $M$ linear equations in $N$ unknowns.
$$
\begin{bmatrix}
3 & 0 & 1 & 0 & -1 & -3 & 2\\
1 & 2 & 0 & 4 & 0 & 0 & -1\\
1 & 1 & 0 & 0 & -1 & -1 & -2\\
0 & 0 & 1 & 0 & -3 & -1 & 1 \\
\end{bmatrix}
\begin{bmatrix}
x_{1}\\
x_{2}\\
x_{3}\\
x_{4} \\
x_{5} \\
x_{6} \\
x_{7} \\
\end{bmatrix} =
\begin{bmatrix}
1\\
0\\
0\\
-1\\
\end{bmatrix}$$
Is there any algorithm for finding answers of this equations that ${x_{i} \ge 0}$?
Comment: I just want that $x_i \ge 0$.
It can change to
$$
\begin{bmatrix}
1 & 0 & 0 & 0 & 2/3 & -2/3 & 1/3 & 2/3\\
0 & 1 & 0 & 0 & -5/3 & -1/3 & -7/3 & -2/3 \\
0 & 0 & 1 & 0 & -3 & -1 & 1 & -1 \\
0 & 0 & 0 & 1 & 2/3 & 1/3 & 5/6 & 1/6 \\
\end{bmatrix}
$$
| You can solve the linear programming problem given below by the simplex method:
$\max z=0$ subject to the constraints given by the equations.
Of course, you would have to add artificial variables for all the constraints which would make it a far too large problem to be solved by hand. Probably some software would be of assistance.
Wolfe's method in quadratic programming relies on the same idea for finding non negative solutions of the Kuhn Tucker conditions.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/982780",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
Improper integral : $\int_0^{+\infty}\frac{x\sin x}{x^2+1}$ How to evaluate the following improper integral : $$\int_0^{+\infty}\frac{x\sin x}{x^2+1}\,dx$$
I have tried integration by parts and variable substitution, but it didn't work.
| According to the residue theory,
$$
\int_0^{+\infty}\frac{1}{s^2+x^2}\mathrm{d}x=\frac{\pi}{2s} ~ , ~
I(\alpha)=\int_0^{+\infty}\frac{x\sin \alpha x}{1+x^2}\mathrm{d}x
$$
Laplace transform:
\begin{align}
\mathcal{L}\left[ I(\alpha)\right] &= \int_0^{+\infty}\frac{x}{1+x^2}\cdot \frac{x}{s^2+x^2}\mathrm{d}x \\
&= \int_0^{+\infty}\frac{x^2+1-1}{1+x^2} \cdot \frac{1}{s^2+x^2} \mathrm{d}x \\
&= \int_0^{+\infty}\frac{1}{s^2+x^2} \mathrm{d}x - \int_0^{+\infty}\frac{1}{1+x^2} \cdot \frac{1}{s^2+x^2} \mathrm{d}x \\
&= \int_0^{+\infty}\frac{1}{s^2+x^2} \mathrm{d}x - \frac{1}{s^2-1}\int_0^{+\infty}\left( \frac{1}{1+x^2} - \frac{1}{s^2+x^2} \right) \mathrm{d}x \\
&= \frac{\pi}{2} \cdot \frac{1}{s+1}
\end{align}
Inverse transform:
$$\mathcal{L}^{-1}\left[ I(\alpha)\right] = \frac{\pi}{2}e^{-\alpha} \Longrightarrow I(1)=\int_0^{+\infty}\frac{x\sin x}{1+x^2} \mathrm{d}x=\frac{\pi}{2e}$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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"answer_count": 3,
"answer_id": 2
} |
Evaluating $\int_0^{\frac{\pi}{2}}\ln\left(\frac{\ln^2\sin\theta}{\pi^2+\ln^2\sin\theta}\right)\,\frac{\ln\cos\theta}{\tan\theta}\,d\theta$ Prove
$$\int_0^{\frac{\pi}{2}}\ln\left(\frac{\ln^2\sin\theta}{\pi^2+\ln^2\sin\theta}\right)\,\frac{\ln\cos\theta}{\tan\theta}\,d\theta = \frac{\pi^2}{4}$$
| Note: I will be making this post Community Wiki as it bears resemblance to Chris's sis's answer.
I will use the following result:
$$\lim_{N\to\infty}\left[\sum^\infty_{k=1}\frac{(-1)^{k+1}}{2k(2k)!}N^{2k}-\ln{N}\right]=\gamma$$
It's derivation can be found here.
Letting $\ln(\sin{\theta})=-x$ and using $\mathcal{I}$ to denote the integral in question,
\begin{align}
\mathcal{I}
=&\frac{1}{2}\int^\infty_0\ln(1-e^{-2x})\ln\left(\frac{x^2}{\pi^2+x^2}\right)\ {\rm d}x\\
=&-\frac{\partial}{\partial a}\Bigg{|}_{a=0}\sum^\infty_{n=1}\frac{1}{n}\int^\infty_0x^ae^{-2nx}\ {\rm d}x+\frac{1}{2}\sum^\infty_{n=1}\frac{1}{n}\int^\infty_0e^{-2nx}\ln(\pi^2+x^2)\ {\rm d}x\\
=&-\frac{\partial}{\partial a}\Bigg{|}_{a=0}\frac{\Gamma(a+1)\zeta(a+2)}{2^{a+1}}+\frac{1}{2}\sum^\infty_{n=1}\frac{\ln{\pi}}{n^2}+\frac12\sum^\infty_{n=1}\frac{1}{n^2}\int^\infty_0\frac{xe^{-2nx}}{\pi^2+x^2}{\rm d}x\\
=&-\frac{1}{2}\Gamma'(1)\zeta(2)-\frac{1}{2}\Gamma(1)\zeta'(2)+\frac{1}{2}\Gamma(1)\zeta(2)\ln{2}+\frac{1}{2}\zeta(2)\ln{\pi}\\
&+\frac{1}{2}\sum^\infty_{n=1}\frac{1}{n}\int^\infty_0\int^\infty_0e^{-2nx}e^{-xy}\cos{\pi y}\ {\rm d}x\ {\rm d}y\\
=&\frac{\pi^2}{12}\left(\gamma+\ln{2\pi}\right)-\frac{1}{2}\zeta'(2)+\frac12\sum^\infty_{n=1}\frac{1}{n^2}\int^\infty_0\frac{\cos{\pi y}}{y+2n}{\rm d}y\\
=&\frac{\pi^2}{12}\left(\gamma+\ln{2\pi}\right)-\frac{1}{2}\zeta'(2)+\frac12\sum^\infty_{n=1}\frac{1}{n^2}\int^\infty_0\frac{\cos{(y+2n\pi)}}{y+2n\pi}{\rm d}y\\
=&\frac{\pi^2}{12}\left(\gamma+\ln{2\pi}\right)-\frac{1}{2}\zeta'(2)+\frac12\sum^\infty_{n=1}\frac{1}{n^2}\int^\infty_{2n\pi}\frac{\cos{y}}{y}{\rm d}y\\
=&\frac{\pi^2}{12}\left(\gamma+\ln{2\pi}\right)-\frac{1}{2}\zeta'(2)+\frac12\sum^\infty_{n=1}\frac{1}{n^2}\left[\left(\int^\infty_0-\int^{2n\pi}_0\right)\frac{\cos{y}-1}{y}{\rm d}y+\ln{y}\Bigg{|}^\infty_{2n\pi}\right]\\
=&\color{grey}{\frac{\pi^2}{12}\left(\gamma+\ln{2\pi}\right)-\frac{1}{2}\zeta'(2)}+\frac12\sum^\infty_{n=1}\frac{1}{n^2}\int^{2n\pi}_0\frac{1-\cos{y}}{y}{\rm d}y\color{grey}{-\frac{1}{2}\sum^\infty_{n=1}\frac{\ln(2n\pi)}{n^2}}\\
&+\color{grey}{\frac{1}{2}\lim_{N\to\infty}\sum^\infty_{n=1}\frac{1}{n^2}\left[\ln{N}-\sum^\infty_{k=1}\frac{(-1)^{k+1}}{2k(2k)!}N^{2k}\right]}\\
=&\frac{1}{2}\sum^\infty_{n=1}\frac{1}{n^2}\int^1_0\frac{1-\cos{2n\pi y}}{y}{\rm d}y\\
=&\frac{1}{2}\int^1_0\left(\frac{\pi^2}{6y}-\pi^2y+\pi^2-\frac{\pi^2}{6y}\right)\ {\rm d}y=\frac{\pi^2}{2}\int^1_0(1-y)\ {\rm d}y=\Large{\frac{\pi^2}{4}}
\end{align}
as was to be shown.
| {
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "47",
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$f\left(x + \frac1x\right)= x^3+x^{-3},$ find $f(x)$ $$f\left(x + \frac1x\right)= x^3+x^{-3},$$ find $f(x)$.
What i do know at this state is that..
express x as a function of y :
$y= x + 1/x$
$x^2−xy+1=0$
Quad formula:
$x= (y ± \sqrt {y^2-4}) / 2$
When i substitute this into the original equation, i can't solve it.
| $x^3+x^{-3}=(x + \frac{1}{x})^3-3(x + \frac{1}{x})$
set $t=x + \frac{1}{x}, f(t)= t^3-3t$.
| {
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"source": "stackexchange",
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"answer_count": 7,
"answer_id": 1
} |
Prove that $2\sqrt{(a^2-ab+b^2)(x^2-xy+y^2)}\ge 2ax+2by-ay-bx$? Let $a,b,x,y$ be real numbers , then is it true that
$$2\sqrt{(a^2-ab+b^2)(x^2-xy+y^2)}\ge 2ax+2by-ay-bx?$$ Actually I am trying to prove the triangle inequality for the norm of numbers in $K(\rho)$ , where $\rho$ is the imaginary cube-root of unity , $|a+b\rho|:=\sqrt{(a+b\rho)(a+b \rho^2)}=a^2-ab+b^2$
| $4\sqrt{(a^2-ab+b^2)(x^2-xy+y^2)} = \sqrt{(3(a-b)^2+(a+b)^2)(3(x-y)^2+(x+y)^2)} $
$\ge_{c.s.} (3|a-b||x-y|+|a+b||x+y|) \ge 3(a-b)(x-y)+(a+b)(x+y)$
$ = 4ax+4by - 2ay-2bx$
| {
"language": "en",
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"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
} |
Proving a triangle equilateral given condition $al_a^2+bl_b^2+cl_c^2=9R\Delta$ $ABC$ is a triangle, with $l_a$, $l_b$, $l_c$ as angle bisectors, $R$ as circumradius and $\Delta$ as area, such that:
$$al_a^2+bl_b^2+cl_c^2=9R\Delta$$
Is it true that $ABC$ is equilateral?
I am not sure how to approach this problem. Ideas, anyone? :)
| Use the fact that $l_a$ divides the side $BC = a$ (at $A'$) in the ratio $BA':CA' = b:c$, and from Stewart's Theorem we have $\displaystyle l_a^2 = bc - \frac{a^2bc}{(b+c)^2}$.
Thus, $\displaystyle \sum\limits_{cyc} al_a^2 = 3abc - abc\sum\limits_{cyc}\frac{a^2}{(b+c)^2} = 9R\Delta = \frac{9}{4}abc \implies \sum\limits_{cyc}\frac{a^2}{(b+c)^2} = \frac{3}{4}$
Now by Cauchy-Schwarz: $\displaystyle \sum\limits_{cyc}\frac{a^2}{(b+c)^2} \ge \frac{1}{3}\left(\sum\limits_{cyc}\frac{a}{b+c}\right)^2$ and $\displaystyle R.H.S. \ge \frac{3}{4}$ from Nesbitt's Inequality.
Equality holds iff $a=b=c$. Hence, the triangle must be equilateral.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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A finding (?) about some primitive Pythagorean triples I have just stumbled on the fact that the sum of the three absolute differences between each pair of a primitive Pythagorean triple [absolute values of (a-b), (b-c) and (c-a), where a,b,c constitute the triple ] add up to the square of the integer that is one less than the smallest of the integers in the triple.
Example 1: starting with the {3,4,5} triple, 4-3=1, 5-4=1, and 5-3=2; sum of the positive differences =1+1+2 = 4 = 2^2.
Example 2: {7,24,25} triple; 24-7 +25-24 + 25-7 = 17+1+18 = 36 = 6^2
Moreover, I find that this result applies only to primitive triples where the least of the three integers is odd.
For primitive triples where the lowest integer is even (actually a multiple of 4), as in [8,15,17}, {12,35,37}, [16,63,65}, etc. the sum of the three positive differences turns out to be twice a square number.
I have a strong feeling that this can be proven. That is my next step.
| Let the triple have smallest term odd. Call that term $a$. Then there exist integers $s$ and $t$ of opposite parity such that $a=s^2-t^2$, $b=2st$, and $c=s^2+t^2$. Our sum of absolute values is
$$(2st-s^2+t^2)+(s^2+t^2-2st)+(s^2+t^2-s^2+t^2).$$
This is $4t^2$, That is certainly a perfect square, but it is not necessarily the square of $1$ less than the smallest element in the triple.
For example, let $s=7$ and $t=4$. Then we get the triple $(33, 56,65)$. Our sum of absolute values of differences is $64$, which is not the square of $32$.
In the case where the even side is smallest, our sum of absolute values is
$$(s^2-t^2-2st)+(s^2+t^2-2st)+(s^2+t^2-s^2+t^2).$$
This is $2s^2-4st+2t^2$, which is indeed twice a perfect square.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/989799",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
What is the most elementary way of proving a sequence is free of non-trivial squares? Given the sequence A001921
$$
0, 7, 104, 1455, 20272, 282359, 3932760, 54776287, 762935264, 10626317415, 148005508552, 2061450802319, 28712305723920, 399910829332567, \dots
$$
which obeys the recurrence relation
$$
a_0 := 0,\ a_1 := 7, \quad a_{n+2} = 14a_{n+1} - a_n + 6.
$$
How can I prove (or, I suppose, disprove) my conjecture that $a_0 = 0$ is the only square?
In case it helps, the conjecture is equivalent to saying that $x$ and $3x^2+3x+1$ cannot both be [positive integer] squares, though I am particularly interested in any general method of proof that attacks the recurrence relation directly.
EDIT: Here's an application.
| Theorem. If $a$ and $b$ are integers such that $3a^4+3a^2+1=b^2\!$, then $a=0$.
Proof. Assume, contrary to the claim, that $a \ge 1$ and $b>1$ are integers satisfying the equation. Evidently $b$ is odd, say $b=2c+1$ for an integer $c \ge 1$. Hence
\begin{align*}
3a^4 + 3a^2+1 &= (2c+1)^2 \\
3a^2(a^2+1) &= 4c(c+1).
\end{align*}
Since $4 \nmid 3(a^2+1)$, regardless of the parity of $a$, we must have $2 \mid a^2$; hence $a$ is even, say $a=2d$ for an integer $d \ge 1$. Now
\begin{align*}
3d^2(4d^2+1) &= c(c+1).
\end{align*}
We now show that this equation has no solutions with $c,d \ge 1$. Assume to the contrary that there exist integers $p,q,r,s \ge 1$ such that
\begin{align*}
3d^2 &= pq, &&&
c &= pr, \\
4d^2+1 &= rs, &&&
c+1 &= qs.
\end{align*}
Adding and factoring gives
\begin{align*}
3d^2+c+1 &= pq+qs = q (p+s), \\
4d^2+c+1 &= rs+pr = r(p+s),
\end{align*}
and then by subtraction we obtain $(p+s)(r-q) = d^2\!$. Since $3d^2=pq$, we have either $p \mid 3$ or $\gcd(p,d)>1$. The latter together with $(p+s) \mid d^2$ would contradict $\gcd(p,s) \mid \gcd(pr,qs) = \gcd(c,c+1) = 1$. Hence $p \mid 3$, so $p=1$ or $p=3$.
Case 1: $p=1$. Then $q=3d^2$, and $c=r$. Now $c+1 = qs$, so by substitution $r+1 = 3d^2s$. On the other hand, $rs-1 = 4d^2$, and adding these two relations yields
\begin{align*}
(rs-1)+(r+1) &= 4d^2 + 3d^2s \\
r(s+1) &= d^2(3s+4).
\end{align*}
Since $3s+4=3(s+1)+1$ implies $\gcd(s+1,3s+4)=1$, and $rs=4d^2+1$ implies $\gcd(r,d)=1$, we conclude $r=3s+4$ and $s+1=d^2$. Hence
\begin{align*}
4d^2+1 &= rs \\
&= \bigl(3(d^2-1)+4\bigr)(d^2-1) \\
2 &= 3d^2(d^2-2),
\end{align*}
which is clearly impossible.
Case 2: $p=3$. Then $q=d^2$ and $c=3r$. Now $c+1=qs$ becomes $3r+1=d^2s$. Since it is still true that $rs-1=4d^2$, we add the two relations to obtain
\begin{align*}
(rs-1) + (3r+1) &= 4d^2 + d^2s \\
r(s+3) &= d^2(4+s).
\end{align*}
Again, by considering common factors, we conclude $r=s+4$ and $s+3=d^2$. Hence
\begin{align*}
4d^2+1 &= rs \\
&= (d^2+1)(d^2-3) \\
d^2(d^2-6) &= 4.
\end{align*}
This is also impossible, completing the proof.
| {
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"url": "https://math.stackexchange.com/questions/996949",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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"answer_id": 1
} |
$\frac{1}{1+2}+\frac{1}{1+2+3}+\dots+\frac{1}{1+2+3+\dots+x}=\frac{2011}{2013}$ I want to see OTHER approaches than this one. Make sure they are significantly different and not a direct restatement.
$$\frac{1}{1+2}+\frac{1}{1+2+3}+\dots+\frac{1}{1+2+3+\dots+x}=\frac{2011}{2013}\tag{1}$$
$$\sum_{n=1}^x n=\frac{x(x+1)}{2} \; \forall x >0\tag{2}$$
$$\begin{align*}
(1)&\stackrel{(2)}{\iff} \frac{2}{2\cdot 3}+\frac{2}{3\cdot 4}+\frac{2}{4\cdot 5}+\dots+\frac{2}{x(x+1)}=\frac{2011}{2013}\\\\
&\iff 2\left (\frac12 -\frac13+\frac13-\frac14+\frac14-\dots+\frac{1}{x}-\frac{1}{x+1}\right )=\frac{2011}{2013}\\\\
&\iff 1-\frac{2}{x+1}=\frac{2011}{2013}\\\\
&\iff x=2012
\end{align*}$$
| Let's prove by induction that
$$\sum_{m=2}^n \frac1{\sum_{k=1}^m k} = \frac{n-1}{n+1}$$
The base case $n=2$ is true. Now the induction step:
$$\sum_{m=2}^{n+1} \frac1{\sum_{k=1}^m k}=\sum_{m=2}^n \frac1{\sum_{k=1}^m k}+\frac1{\sum_{k=1}^{n+1}k}=\frac{n-1}{n+1}+\frac2{(n+1)(n+2)}=\frac{n^2+n}{(n+1)(n+2)}=\frac{n}{n+2}$$
Therefore the answer is $n=2012$.
(I know this is kind of lame, but the formula is easy to guess)
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
"answer_id": 1
} |
Maximum of $\sin A\sin B\cos C+\sin B\sin C\cos A+\sin C\sin A\cos B$ in triangle What is the maximum value of $$\sin A\sin B\cos C+\sin B\sin C\cos A+\sin C\sin A\cos B,$$ where $A,B,C$ are angles in a triangle?
We can rewrite as $$-\sin A\sin B\sin(A+B)+\sin B\sin(A+B)\cos A+\sin(A+B)\sin A\cos B$$
Expanding, this becomes
$$-\sin^2 A\sin B\cos B-\sin^2B\sin A\cos A+2\sin B\sin A\cos B\cos A+\cos^2A\sin^2B+\sin^2A\cos^2B$$
| I will need the fact that, if $A,B,C$ are the angles of a triangle, then
$$ \cos A + \cos B + \cos C \le \tfrac32 \tag{$\ast$} $$
with equality iff $A=B=C=\frac\pi3$. With this result in hand, your problem can be solved thus:
\begin{align*}
&\cos A\sin B\sin C + \sin A\cos B\sin C + \sin A\sin B\cos C \\
&= \cos A\cos B\cos C - \cos(A+B+C) \\
&= \cos A\cos B\cos C + 1 \\
&\le \left(\frac{\cos A+\cos B+\cos C}{3}\right)^3 + 1
&&\text{(AM/GM)} \\
&\le \tfrac98
&&\text{(by ($\ast$))}
\end{align*}
with, again, equality iff $A=B=C=\frac\pi3$.
To prove ($\ast$), first note that since $A+B+C=\pi$ and all three are positive, at most one of them is greater than $\frac\pi2$. Wlog, $A,B\in[0,\frac\pi2]$. On this interval, $\cos$ is concave, so by Jensen's inequality,
\begin{align*}
\cos A + \cos B + \cos C
&\le 2\cos(\tfrac{A+B}{2}) + \cos C \\
&= 2\cos(\tfrac{A+B}{2}) - \cos(A+B) \\
&= 2\cos(\tfrac{A+B}{2}) - 2\cos^2(\tfrac{A+B}{2}) + 1 \\
&= \tfrac32 - 2\left(\cos(\tfrac{A+B}{2})-\tfrac12\right)^2 \\
&\le \tfrac32
\end{align*}
with equality iff $A=B=C=\frac\pi3$.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Conceptual problem in solving quadratic equation The sum of all real roots of the equation $$|x-2|^2 + |x-2| - 2 = 0$$
is?
I tried this problem by taking two cases $x<2$ and $x>2$ and solving the corresponding equations and I got $8$ as the answer, but in my book the author solves it by taking this equation as a quadratic in $x-2$ and gets $4$ as the answer. What's wrong with my way of solving this problem?
| When $x < 2$, the equation becomes
$$
x^2 - 4x + 4 + 2 - x - 2 = 0,
$$
or $x^2-5x+4=0$. This gets solutions $x = 4$ or $x = 1$, but since $x < 2$ we only take $x = 1$.
When $x > 2$, we have
$$
x^2 - 4x + 4 + x - 2 - 2 = 0,
$$
or $x^2 - 3x = 0$. This gets solutions $x = 0$ or $x = 3$, but since $x > 2$ we only take $x = 3$.
EDIT: the book's way is probably nicer though. Let $y = |x-2|$, then $y^2 - y + 2 = 0$ and $y = -2$ or $y = 1$. Since $y \geq 0$, we only take $|x-2| = y = 1$, which again gets $x = 3$ or $x = 1$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/998369",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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} |
Proving this formula $1+\sum_{n=0}^{\infty }\frac{1}{\pi \left(2n+\frac{3}{4}\right)\left(2n+\frac{5}{4}\right)}=\sqrt2$ I tried to prove this formula but I couldn't do.
$$1+\sum_{n=0}^{\infty }\frac{1}{\pi \left(2n+\frac{3}{4}\right)\left(2n+\frac{5}{4}\right)}=\sqrt{2}$$
| $$\begin{align}
1+\sum_{n=0}^{\infty }\frac{1}{\pi \left(2n+\frac{3}{4}\right)\left(2n+\frac{5}{4}\right)}
&=1+\frac1{4\pi}\sum_{n=0}^\infty\frac1{\left(n+\frac38\right)\left(n+\frac58\right)}\tag{1}\\
&=1+\frac1\pi\sum_{n=0}^\infty\left(\frac1{n+\frac38}-\frac1{n+\frac58}\right)\tag{2}\\
&=1+\frac1\pi\sum_{n=0}^\infty\int_0^1\left(x^{n-5/8}-x^{n-3/8}\right)\,\mathrm dx\tag{3}\\
&=1+\frac1\pi\int_0^1\left(\frac{x^{-5/8}-x^{-3/8}}{1-x}\right)\,\mathrm dx\tag{4}\\
&=1+\frac{1}{\pi} \Big[\pi\Big(\sqrt2-1\Big)\Big]\tag{5}\\
&=1+\sqrt2-1\tag{6}\\
&=\sqrt2\tag{7}\\
\end{align}$$
$$\large1+\sum_{n=0}^{\infty }\frac{1}{\pi \left(2n+\frac{3}{4}\right)\left(2n+\frac{5}{4}\right)}
=\sqrt2
$$
I can add explanations if needed
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
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How to evaluate the integral $\int\frac{\sqrt{x^2-9}}{x^3}\;\mathrm d x$? $$\int\frac{\sqrt{x^2-9}}{x^3}\;\mathrm d x$$
The question is ask me to evaluate the integral but I have no idea how to start?
If there are any formulas required for this question, can you please list them ?
Thank you for any help!
| If you know the formula
\begin{gather*}\tag{$\star$}
\int \sqrt{1-t^2}\,\rm{d}t=\frac{1}{2}\left(t\sqrt{1-t^2}+\arcsin(t)+C\right),
\end{gather*}
you can evaluate the integral as follows.
Since
\begin{align*}
&\quad \frac{\sqrt{x^2-9}}{x^3}\rm{d} x=\frac{\sqrt{x^2\big(1-(3/x)^2\big)}}{x^3}\rm{d} x\\
&=\sgn(x)\cdot\frac{x\sqrt{1-(3/x)^2}}{x^3}\rm{d} x=\sgn(x)\sqrt{1-(3/x)^2}\cdot\frac{1}{x^2}\rm{d} x\\
&=-\frac{\sgn(x)}{3}\sqrt{1-\left(\frac{3}{x}\right)^2}\rm{d}\left(\frac{3}{x}\right),
\end{align*}
together with $(\star)$ we have
\begin{align*}
\int\frac{\sqrt{x^2-9}}{x^3}\rm{d} x&=-\frac{\sgn(x)}{6}\left(\frac{3}{x}\sqrt{1-\left(\frac{3}{x}\right)^2}+\arcsin\left(\frac{3}{x}\right)\right)+C\\
&=-\frac{\sgn(x)}{2}\left(\frac{1}{x}\cdot\sqrt{1-\frac{9}{x^2}}+\frac{1}{3}\arcsin\left(\frac{3}{x}\right)\right)+C.
\end{align*}
Here $\sgn(x)$ is the signum function.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1005362",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 4
} |
Unable to find the sum of a series I am trying to find the sum of the following series:
$$\sum_{n=1}^{\infty} {\frac{1+7^n}{9^n}}$$
which I rewrote as
$$\sum_{n=1}^{\infty} \left(\frac{1}{9^n}+
\left(\frac{7}{9}\right)^n\right)$$
I am assuming that it is a geometric series and the initial value is
$$a_1=\frac{1}{9} + \frac{7}{9}$$
I also see that
$$a_2 = \frac{1}{9^2} + \frac{7^2}{9^2}$$
I know that in a geometric series the first term is $a$ and the second term is $ar$.
This allows me to see that
$$\left(\frac{1}{9}+\frac{7}{9}\right)r=\frac{1}{9^2}+\frac{7^2}{9^2}$$
which when solved for $r$ gives the value $\frac{25}{36}$.
Using the formula to find the sum of a geometric series $\frac{a}{1-r}$, I find that the sum is equal to $\frac{32}{11}$.
But this value is incorrect and the sum is actually $\frac{29}{8}$. How does one find that value?
| \begin{align*}\sum_{n=1}^\infty\frac{1+7^n}{9^n} &= \sum_{n=1}^\infty\bigg[ \frac{1}{9^n} + \frac{7^n}{9^n}\bigg] \\ &= \sum_{n=1}^\infty\frac{1}{9}\bigg(\frac{1}{9}\bigg)^{n-1} + \sum_{n=1}^\infty \frac{7}{9}\bigg(\frac{7}{9}\bigg)^{n-1} \\ &=\frac{1}{9}\bigg(\frac{1}{1-\frac{1}{9}}\bigg)+\frac{7}{9}\bigg(\frac{1}{1-\frac{7}{9}}\bigg) \\ &= \frac{1}{8}+\frac{7}{2} \\ &= \frac{29}{8} \end{align*}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1006700",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 3
} |
Evaluate $\int\frac{1}{1+x^6} \,dx$ I came across following problem
Evaluate $$\int\frac{1}{1+x^6} \,dx$$
When I asked my teacher for hint he said first evaluate
$$\int\frac{1}{1+x^4} \,dx$$
I've tried to factorize $1+x^6$ as
$$1+x^6=(x^2 + 1)(x^4 - x^2 + 1)$$
and then writing
$$I=\int\frac{1}{1+x^6} \,dx=\int\frac{1}{(x^2 + 1)(x^4 - x^2 + 1)} \,dx=\int\frac{1+x^2-x^2}{(x^2 + 1)(x^4 - x^2 + 1)} \,dx$$
$$I=\int\frac{1}{x^4 - x^2 + 1} \,dx-\int\frac{x^2}{(x^2 + 1)(x^4 - x^2 + 1)} \,dx$$
However $$x^4-x^2+1=\left(x^2-\frac12\right)^2+\frac{3}{4}$$
But I can't see how it helps
I've also tried to reverse engineer the solution given by Wolfram Alpha
And I need to have terms similar to
$$\frac{x^2-1}{x^4-x^2+1} \quad , \quad \frac{1}{1+x^2} \quad , \quad \frac{1}{(x+c)^2+1}\quad , \quad \frac{1}{(x+c)^2+1}$$ in integrand, How can I transform my cute looking integrand into these huge terms?
Since in exams I will neither have access to WA nor time to reverse engineer the solution moreover it does not seem intuitive,is there any way to solve this problem with some nice tricks or maybe substitutions?
| Here's a nice "trick" my former professor taught me
$$ \int\frac{dx}{1+x^6} = \frac{1}{2} \int \frac{(1-x^2+x^4)+x^2+(1-x^4)}{(1+x^2)(1-x^2+x^4)} dx \\
= \frac{1}{2}\int \frac{dx}{1+x^2} + \frac{1}{2} \int \frac{x^2}{1+x^6} dx + \frac{1}{2} \int \frac{1-x^2}{1-x^2+x^4} dx \\
= \frac{1}{2}\int \frac{dx}{1+x^2} + \frac{1}{2} \int \frac{x^2}{1+x^6} dx - \frac{1}{2} \int \frac{1-\frac{1}{x^2}}{x^2-1+\frac{1}{x^2}} dx $$
The first integral is simply the arctangent of $x$. The second can be solved by substituting $u = x^3$. The third can be solved by substituting $t = x + \frac{1}{x}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1007399",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 5,
"answer_id": 1
} |
Proving $\sqrt{a+\sqrt{a+\sqrt{a+\sqrt{a.....}}}}-\sqrt{a-\sqrt{a-\sqrt{a-\sqrt{a....}}}}=1$ I checked many values of this inequality, but I don't have the complete proving.
$$\sqrt{a+\sqrt{a+\sqrt{a+\sqrt{a.....}}}}-\sqrt{a-\sqrt{a-\sqrt{a-\sqrt{a....}}}}=1$$ if $$a>1$$
| Define $x = \sqrt{a + \sqrt{a + \sqrt{a +\cdots}}}$ and $y = \sqrt{a - \sqrt{a - \sqrt{a -\cdots}}}.$ Then we have $x^2 = a+ x$ and $y^2 = a-y$.
Now compute $(x+y)(x-y) = x^2 - y^2 = x+y$. Cancelling $x+y$ from both sides gives $x-y = 1$, which is the result you're after.
(Of course, I've ignored convergence issues here and just worked formally).
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1011146",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 1
} |
Converting rational equations into polynomial equations. A question on my practice exam asks to solve the equation:
$$
\frac{1}{x^4} - \frac{1}{x^2} = 12
$$
In the answers section it says to first convert the equation to the equivalent equation:$$
12x^4 + x^2 - 1 = 0
$$
However, we have not yet learned how to do this and I am unable to contact my teacher before the exam next week.
So, my question is how does one convert an equation like this into a polynomial like has been achieved in this example?
Any help would be much appreciated!
Jack.
| To eliminate all the fractions in a single step, just multiply each term in the equation by the LCM of all the denominators. In this case, $\text{lcm}(x^4, x^2) = x^4$, so we obtain:
\begin{align*}
\frac{1}{x^4} - \frac{1}{x^2} &= 12 \\
x^4 \cdot \frac{1}{x^4} - x^4 \cdot \frac{1}{x^2} &= x^4 \cdot 12 \\
1 - x^2 &= 12x^4 \\
0 &= 12x^4 + x^2 - 1
\end{align*}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1011258",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Complex numbers and geometric series a) Use the formula for the sum of a geometric series to show that
$$\sum _{k=1}^n\:\left(z+z^2+\cdots+z^k\right)=\frac{nz}{1-z}-\frac{z^2}{\left(1-z\right)^2}\left(1-z^n\right),\:z\ne 1$$
I thought the formula for geometric series is $$\frac{a\left(1-r^n\right)}{1-r}=\frac{z\left(1-z^n\right)}{1-z}$$
How do I appraoch this?
b) Let $$z=\cos\left(\theta \right)+i\sin\left(\theta \right),\text{ where }0<\theta <2\pi.$$
By considering the imaginary part of the left-hand side of the equation of $a$, deduce that
$$\sum _{k=1}^n (\sin(\theta)+\sin(2\theta)+\cdots+\sin(k\theta ))=\frac{(n+1)\sin(\theta ) -\sin(n+1)\theta }{4\sin^2\left(\frac{\theta }{2}\right)}$$
assuming
$$\frac{z}{1-z}=\frac{i}{2\sin\left(\frac{\theta }{2}\right)} \left(\cos\left( \frac{\theta }{2} \right) +i\sin\left(\frac{\theta }{2}\right)\right)$$
| $$\sum_{k=1}^n (z + z^2 + \cdots + z^k) = \sum_{k=1}^n z \frac{1 - z^k}{1-z} = \frac{z}{1-z} \sum_{k=1}^n ( 1 - z^k )$$
See where to go now?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1012704",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
} |
Determine the existence of a function Does there exist a function $f:\mathbb{R}^n \to \mathbb{R}$ such that for all $x = (x_1,...,x_n) \in \mathbb{R}^n$:
*
*$f(A) = f(x)$ for every permutation A of $\{x_1,...,x_n\}$.
*$f(x + (a,...,a)) = f(x) + a$ for every $a \in \mathbb{R}$.
*$f(ax) = af(x)$ for every $a \in \mathbb{R}$.
*There exist $y = (y_1,...,y_n) \in \mathbb{R}^n$ such that $f(y) > max\{y_1,...,y_n\}$
?
I have managed to prove that for $n = 1$ and $n = 2$ such a function doesn't exist, but I don't know to prove it for $n>2$.
| Let
\begin{align*}
m(x) &:= \min\{x_k \colon\ k=1,\dots,n\}, \\
M(x) &:= \min\{x_k \colon\ k=1,\dots,n\}, \\
h(x) &:= -\frac{n}{2}(m(x) + M(x)) + \sum_{k=1}^n x_k, \\
f(x) &:= \frac{m(x) + M(x)}{2} + h(x).
\end{align*}
We have already determined that $\frac{m(x) + M(x)}{2}$ satisfies $1-3$. Let us now check the behaviour of $h(x)$.
Notice that $h(x) = h(p(x))$ for any permutation $p$ of $x$ because neither $n$, $m(x)$, $M(x)$, nor sum of the elements of $x$ depend on the ordering of the elements.
As for what happens when we add $(a,\dots,a)$,
\begin{align*}
h(x+(a,\dots,a)) &= -\frac{n}{2}(m(x) + a + M(x) + a) + \sum_{k=1}^n (x_k + a), \\
&= -na - \frac{n}{2}(m(x) + M(x)) + na + \sum_{k=1}^n x_k = h(x).
\end{align*}
Now, for $a \ge 0$, we have:
\begin{align*}
h(ax) &= -\frac{n}{2}(m(ax) + M(ax)) + \sum_{k=1}^n ax_k, \\
&= -\frac{an}{2}(m(x) + M(x)) + a\sum_{k=1}^n x_k = ah(x).
\end{align*}
If we prove the same for $a = -1$, this will work in general. Notice that
$$m(-x) = -M(x), \quad M(-x) = -m(x).$$
So,
\begin{align*}
h(-x) &= -\frac{n}{2}(m(-x) + M(-x)) + \sum_{k=1}^n (-x_k), \\
&= -\frac{n}{2}(-M(x) - m(x)) - \sum_{k=1}^n x_k \\
&= - \left(-\frac{n}{2}(m(x) + M(x)) + \sum_{k=1}^n x_k\right) = -h(x).
\end{align*}
Hence, $h(ax) = ah(x)$ for any $a \in \mathbb{R}$.
Using all of this, it is easy to verify $1-3$ for $f$:
*
*Neither $m(x)$, $M(x)$, nor $h(x)$ change with a permutation.
*$f(x+(a,\dots,a)) = \frac{m(x)+a + M(x)+a}{2} + h(x+(a,\dots,a)) = a + \frac{m(x) + M(x)}{2} + h(x) = f(x) + a.$
*$f(ax) = \frac{m(ax) + M(ax)}{2} + h(ax) = a\frac{m(x) + M(x)}{2} + ah(x) = af(x)$.
Now, notice that for $n = 1$ and $n = 2$, $h(x) = 0$, so we did nothing to those cases. But, for $n > 2$, this is no longer true. Let $n = 5$ and let $r > 2$ be any real number bigger than $2$. We construct $y$:
$$y := (1, r-1, r-1, r-1, r-1).$$
Now,
\begin{align*}
f(y) &= \frac{m(y) + M(y)}{2} + h(y) = \frac{m(y) + M(y)}{2} - \frac{n}{2}(m(y) + M(y)) + \sum_{k=1}^n y_k \\
&= \frac{1 + (r-1)}{2} - \frac{5}{2}(1 + (r-1)) + (1+(r-1)+(r-1)+(r-1)+(r-1)) \\
&= \frac{r}{2} - \frac{5r}{2} + 4r - 3 = 2r - 3 = r + (r-3) > r - 1 = M(x).
\end{align*}
My guess is that we could probably construct an appropriate $y$ for $n=3,4$, but the above example with $n=5$ seemed neater. Obviously, $n>5$ can be constructed the same way as the above example.
Also, notice that
$$f(x) := \frac{m(x) + M(x)}{2} + rh(x)$$
also satisfies $1-3$ for any $r \in \mathbb{R}$, which gives us pretty much freedom in constructing examples for $n=3,4$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1013474",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 3
} |
Linear Transformation using Trig Identities
Let $T:\mathbb{R}^2\longrightarrow \mathbb{R}^2$ and $ S:\mathbb{R}^2\longrightarrow \mathbb{R}^2$ be linear transformation defined by
$$
T(x) =Rx \;\text{and} \;S(x)=Qx
$$
Where $$R = \begin{bmatrix}
\cos\theta & -\sin\theta \\
\sin\theta & \cos\theta
\end{bmatrix}$$
and $$Q= \begin{bmatrix}
\cos\theta & \sin\theta \\
\sin\theta & -\cos\theta
\end{bmatrix}$$
Show that $T$ represents a rotation of $\theta$ counterclockwise around the origin and S represents a reflection in the line $y=mx$ with $m=\tan(\theta/2)$
I have absolutely no clue how to do this question and my textbook does not have any similar example. Please help
| Take an arbitrary vector $v=\begin{pmatrix} x\\ y\end{pmatrix}$, and find the solutiuon $T(v)$ and $S(v)$.
For the first one we get $T(v)=\begin{pmatrix}\cos\theta & -\sin\theta\\ \sin\theta & \cos\theta \end{pmatrix} \begin{pmatrix} x\\ y\end{pmatrix}=\begin{pmatrix}x\cos\theta-y\sin\theta\\ x\sin\theta + y\cos\theta\end{pmatrix}$.
We can show that this is a rotation of $\theta$ degrees about the origin. Let $\phi$ be an angle such that $(x^2+y^2)\cos\phi=x$ and $(x^2+y^2)\sin\phi=y$. Now we wish to move this by a degree $\theta+\phi$. The new coordinates are then $(x^2+y^2)\cos(\theta+\phi)=(x^2+y^2)(\cos\theta\cos\phi-\sin\theta\sin\phi)$. But this is simply $(x^2+y^2)\cos\theta\cos\phi -(x^2+y^2)\sin\theta\sin\phi= x\cos\theta - y\sin\theta$.
We can do a similar explanation for the $y$ component (I leave this to you).
As for $S(v)=\begin{pmatrix}\cos\theta & \sin\theta\\ \sin\theta & -\cos\theta \end{pmatrix} \begin{pmatrix} x\\ y\end{pmatrix}=\begin{pmatrix}x\cos\theta+y\sin\theta\\ x\sin\theta - y\cos\theta\end{pmatrix}$.
Now the angle from the $x$ axis to the line $y=\tan\frac{\theta}{2}x$ is simply $\frac{\theta}{2}$. If we wish to reflect a point $(x,y)$ across this line, we need to take the angle of a line passing through the origin and this point to the line $y=\tan\frac{\theta}{2}x$, and then add this to the angle from the $x$ axis to the line $y=\tan\frac{\theta}{2}x$. Sorry if that sounds wordy, I had to explain it to myslef!
So let $\phi$ be the angle such that for an arbitrary point $p=(x,y)$ we have $(x^2+y^2)\cos\phi=x$ and $(x^2+y^2)\sin\phi=y$. Now we wish to reflect this so we need to let $\psi=\frac{\theta}{2}-\phi$. Then a reflection is the equivalent of to $\frac{\theta}{2}+\psi=\theta-\phi$. Then $(x^2+y^2)\cos(\theta-\phi)=(x^2+y^2)(\cos\theta\cos\phi+\sin\theta\sin\phi)=x\cos\theta+y\sin\theta$.
Again, I leave the $y$-component to you.
| {
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"url": "https://math.stackexchange.com/questions/1013762",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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Understanding $ \frac {e^x}{x^2} $ as $ x \to -\infty $ Given the limit of this function
$$
\lim_{x\to -\infty} \frac{x^2 + e^x + 1}{x^2 + x \sin x} =
\begin{bmatrix}
\frac {\infty}{\infty}
\end{bmatrix} = \lim_{x\to -\infty} \frac {e^x (1 + \frac{x^2}{e^x} + \frac{1}{e^x})} {x^2(1 + \frac{1}{x} \sin x)}
$$
Since $ \lim_{x\to -\infty} e^x = 0$ and $ \lim_{x\to -\infty} x^2 = +\infty$ , I'd give the result $ \frac{0}{+\infty} = 0$ but it is not correct.
What am I mistaking? Have I made any wrong assumption?
| Why not:
$$
\lim_{x\to -\infty} \frac{x^2 + e^x + 1}{x^2 + x \sin x} =
\begin{bmatrix}
\frac {\infty}{\infty}
\end{bmatrix} = \lim_{x\to -\infty} \frac {x^2 (1 + \frac{e^x}{x^2} + \frac{1}{x^2})} {x^2(1 + \frac{1}{x} \sin x)}=1?
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1015602",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
Lesser known derivations of well-known formulas and theorems What are some lesser known derivations of well-known formulas and theorems?
I ask because I recently found a new way to derive the quadratic formula which didn't involve completing the square as is commonly taught. Doing so I was wondering what other proofs and derivations for other formulas that have remained unknown to most people?
Whether it be because the proof is too complex or less pretty, I still find it insightful to see different ways to solve a problem. To me it makes me understand proofs better, and thus also giving a better comprehension of it.
$$
\begin{align}
&\text{Given a quadratic function } f:\\[0.1em]
f &=ax^2+bx+c = a(x-r_1)(x-r_2) = ax^2-a(r_1+r_2)+ar_1r_2\\[0.1em]
a &= a,\enspace \frac{b}{a} = -(r_1+r_2),\enspace \frac{c}{a} = r_1r_2\\[1em]
f' &= 2ax+b, \enspace f'(x) = 0 \Rightarrow x = -\frac{b}{2a}\\[0.2em]
\text{This is an}& \text{ extremum of } f \text{, and is equidistant from each root } r_1, \enspace r_2 \text{ as shown:}\\[0.4em]
\frac{b}{a} &= -(r_1+r_2) \iff -\frac{b}{2a} = \frac{r_1+r_2}{2} \\[1em]
\Rightarrow \enspace &\text{The roots are of the form } r= -\frac{b}{2a}\pm d\\[1em]
\frac{c}{a} = r_1r_2 &= (-\frac{b}{2a}+d)(-\frac{b}{2a}-d) = \frac{b^2}{4a^2}-d^2\\[0.2em]
\Rightarrow\enspace& d^2 = \frac{b^2}{4a^2}-\frac{c}{a} = \frac{b^2}{4a^2}-\frac{4ac}{4a^2} = \frac{b^2-4ac}{4a^2}\\[0.2em]
\Rightarrow\enspace&d = \pm\frac{\sqrt{b^2-4ac}}{2a}\\[1em]
\text{Which yields }& r = -\frac{b}{2a}\pm d = \frac{-b\pm\sqrt{b^2-4ac}}{2a}\enspace\square
\end{align}
$$
| Brahmagupta's Formula which finds the area for a cyclic quadrilateral based on side lengths.
The formula is shown below where K is the area and $a, b, c$, and $d$ are the side lengths.
$$K=\frac{1}{4} \sqrt{(-a+b+c+d)(a-b+c+d)(a+b-c+d)(a+b+c-d)}$$
Here is a proof of the formula.
| {
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"url": "https://math.stackexchange.com/questions/1017602",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "16",
"answer_count": 9,
"answer_id": 8
} |
Intuition behind sum of multiplication arithmetic sequence Maybe this is a stupid question but please guide and enlighten me patiently. I have just known something fact that quite shocking me. Let start from this simple fact
$$\sum_{k=1}^n k=\frac{n(n+1)}{2}\tag{1}$$
The summation above is a sum of arithmetic progression with common difference of 1 and I have already known it. Then, it turns out (I realized these when playing with Wolfram|Alpha)
$$\begin{align}\sum_{k=1}^n k(k+1)&=\frac{n(n+1)(n+2)}{3}\tag{2}\\\sum_{k=1}^n k(k+1)(k+2)&=\frac{n(n+1)(n+2)(n+3)}{4}\tag{3}\\\sum_{k=1}^n k(k+1)(k+2)(k+3)&=\frac{n(n+1)(n+2)(n+3)(n+4)}{5}\tag{4}\\\end{align}$$
and it seems (I haven't proved it yet)
$$\sum_{k=1}^n k(k+1)(k+2)\cdots(k+r)=\frac{n(n+1)(n+2)(n+3)(n+4)\cdots(n+r+1)}{r+2}\tag{5}$$
We have an obvious pattern here. I know the intuition of $(1)$, but I am wondering what are the intuitutions for the other sums: $(2),\,(3),\,(4),\,(5)$?
I can derive $(2)$ using well-known formulas for arithmetic series and square pyramidal number, but how do the other formulas, $(3),\,(4),\,(5)$, derive? Does it use Faulhaber's formula?
| The derivation is by straightforward induction on $n$. Suppose
$\sum_{k=1}^n k(k+1)...(k+r)=\frac{n(n+1)(n+2)...(n+r+1)}{r+2}$
Then
$$\begin{align}
\sum_{k=1}^{n+1} k(k+1)...(k+r) & =\frac{n(n+1)(n+2)...(n+r+1)}{r+2} + (n+1)(n+2)...(n+r+1) \\
& = (n+1)(n+2)...(n+r+1)\left(\frac{n}{r+2} + 1\right) \\
& = (n+1)(n+2)...(n+r+1)\left(\frac{n + r + 2}{r+2}\right) \\
& =\frac{(n+1)(n+2)...(n+r+1)(n+r+2)}{r+2}
\end{align}$$
Now you just have to check the base case $n=1$ to establish the formula for all $n$.
| {
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"url": "https://math.stackexchange.com/questions/1021942",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 0
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Find all polynomials with real coefficients that satisfy $(x^2-6x+8)P(x)=(x^2+2x)P(x-2)$ Find all polynomials with real coefficients that satisfy $$(x^2-6x+8)P(x)=(x^2+2x)P(x-2)\forall x\in\Bbb R$$
My work;
$$\frac{P(x)}{P(x-2)}=-\frac{4}{x-2}+\frac{12}{x-4}+1\tag{1}$$
$$\frac{P(x-2)}{P(x)}=-\frac{12}{x+2}+\frac{4}{x}+1\tag{2}$$
I also factorised the two known polynomial which didn't give anything useful. What should I extract from these two ratios?
| $$(x^2-6x+8)P(x)=(x^2+2x)P(x-2)\iff (x-2)(x-4)P(x)=(x+2)xP(x-2)$$
When $x=2,4$ LHS=0, so $RHS=0$, hence $x=0,2$ is a root for $P(x)$.
When $x=0,-2$ RHS=0, so $LHS=0$, hence $x=0,-2$ is a root for $P(x)$.
Write $P(x)=a(x)\cdot(x-2)x(x+2)$, plug in the assumption, we have
$$a(x)\cdot(x-4)(x-2)^2x(x+2)=a(x-2)\cdot(x-4)(x-2)x^2(x+2)$$
$$\implies a(x)\cdot(x-2)=a(x-2)\cdot x$$
$$\implies a(x)=ax$$
For the last step, we first argue $a(x)$ can't have constant term, since $x=0$ is a root. Hence $a(x)=xb(x)$, which implies $b(x)=b(x−2)\forall x\in \mathbb{R}$.
Then we use the fact if polynomial $p(x)$ takes value $p_0$ infinitely many times (which means $p(x)-p_0=0$ has infinite zeros), then $p(x)$ is constant.
Hence $P(x)=ax^2(x-2)(x+2)$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1023114",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 4,
"answer_id": 0
} |
Infinite Series with Pi I have this homework problem, that I'm stuck on.
We know that:$$\sum_{n=1}^{\infty}\frac{1}{n^2}=\frac{\pi^2}{6}$$
I have to find the sum of: $$\frac{1}{1^2}+\frac{1}{3^2}+\frac{1}{5^2}+\frac{1}{7^2}+\cdots$$
I came up with this equation:
$$\sum_{n=1}^{\infty}\frac{1}{(2n-1)^2}=\sum_{n=1}^{\infty}\left(\frac{1}{2n-1}\right)^2$$
I know that the answer is $\frac{\pi^2}{8}$ I found $a=1$, but can't seem to figure it out...
| Of more historical than mathematical interest, Euler's original approach:
$\cos x = 1 - \frac{x^2}{2!}+ \frac{x^4}{4!}- \frac{x^6}{6!}+...$
So $\cos x$ may be regarded as a polynomial with roots $\pm \pi/2, \pm 3\pi/2, \pm 5\pi/2,...$
That is,
$$\cos x = (x - \frac{\pi}{2})(x + \frac{\pi}{2})(x - \frac{3\pi}{2})(x + \frac{3\pi}{2})...$$
$$= (x^2-\frac{\pi^2}{4})(x^2-\frac{3^2\pi^2}{4})(x^2-\frac{5^2\pi^2}{4})... $$
which we can write as
$$(*) \hspace{10mm}1 - \frac{x^2}{2!}+ \frac{x^4}{4!}- \frac{x^6}{6!}+... = A(1-\frac{4x^2}{\pi^2})(1-\frac{4x^2}{3^2\pi^2})(1-\frac{4x^2}{5^2\pi^2})...$$
for some constant A; and since $\cos x \to 1 ~~\text{as}~~ x\to 0,$ A must equal $1$.
Then from (*) we can equate coefficients of $x^2:$
$$-\frac{1}{2!} = -\frac{4}{\pi^2}- \frac{4}{3^2\pi^2}- \frac{4}{5^2\pi^2}-... $$
so finally
$$\frac{\pi^2}{8} = 1 + \frac{1}{3^2} + \frac{1}{5^2}+\frac{1}{7^2}+... $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1023195",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 2,
"answer_id": 0
} |
Algebraic - Issue with simple equation problem I have this equation: $x+7-(\frac{5x}8 + 10) = 3 $
I've used step-by-step calculators online but I simply don't understand it. Here is how I've tried to solve the problem:
$$x+7-\left(\frac{5x}8+10\right) = x + 7 - \frac{5x}8 - 10 = 3$$
$$x + 7 - \frac{5x}8 - 10 + 10 = 3 + 10$$
$$x + 7 - 7 - \frac{5x}8 = 13 - 7$$
$$x - \frac{5x}8 = 6$$
$$x - 8\times\frac{5x}8 = 6\times8$$
$$x - 5x = 48$$
$$\frac{-4x}{-4} = \frac{48}4$$
$$x = -12$$
Now obviously, it's wrong. The right answer is $16$, but I don't know how to get to that answer. Therefore, I'm extremely thankful if someone truly can show what I need to do, and why I need to do it, because I'm completely lost right now. Thanks.
| $x+7-(\frac{5x}{8}+10)=3$
$\Rightarrow x+7-(\frac{5x+80}{8})=3$
$\Rightarrow x+7+\frac{-5x-80}{8}=3$
$\Rightarrow \frac{8x+7*8-5x-80}{8}=3$
$\Rightarrow \frac{8x+7*8-5x-80}{8}=3$
$\Rightarrow 8x+56-5x-80=3*8$
$\Rightarrow 8x-5x=24+80-56$
$\Rightarrow 3x=48$
$\Rightarrow x=16$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1024794",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 4
} |
Least squares problem: find the line through the origin in $\mathbb{R}^{3}$ The problem is as follows:
"Please set up (but do not solve) the normal equations for the following least squares
approximation problem: Find $(a, b, c, d)$ such that the plane H described by $ax + by + cz = d$
minimizes $\sum |ax_i + by_i + cz_i − d|^2$ where the $(x1, y1, z1), \cdots ,(x6, y6, z6)$ are the following points: $(2,3,4)$, $(99,−85,0)$, $(0,1,8)$, $(5,2,2)$, $(3,3,3)$, $(1,2,4)$."
I solved a similar problem with points representing ordered pairs, for example $(2,1)$, $(3,5)$ and so on, but the question was to get a line that approximates such ordered pairs. In that case I represented the points as $y = mx + b$ and got $m$ and $b$ for the new line. I wonder if for the new problem which is a 3D space I also have to represent the ordered triplets as $z = ax + by + c$ or something like that.
I will very much appreciate any clue.
| We have our data points $\left\{ x_{k}, y_{k}, z_{k}, d_{k} \right\}_{k=1}^{m}$, and our trial function, a line in $\mathbb{R}^{3}$ through the origin:
$$
d\left( x, y, z \right) = a x + b y + c z.
$$
The linear system is
$$
\begin{align}
\mathbf{A} a &= d \\
%
\left[ \begin{array}{ccc}
x & y & z
\end{array} \right]
%
\left[ \begin{array}{c}
a
\end{array} \right]
%
& =
%
\left[ \begin{array}{c}
d
\end{array} \right]\\[3pt]
%
\left[ \begin{array}{ccc}
x_{1} & y_{1} & z_{1} \\
x_{2} & y_{2} & z_{2} \\
\vdots & \vdots & \vdots \\
x_{m} & y_{m} & z_{m} \\
\end{array} \right]
%
\left[ \begin{array}{c}
a \\ b \\ c
\end{array} \right]
%
&=
%
\left[ \begin{array}{c}
d_{1} \\ d_{2} \\ \vdots \\ d_{m}
\end{array} \right].
%
\end{align}
$$
The normal equations are
$$
\begin{align}
\mathbf{A}^{*} \mathbf{A} a &= \mathbf{A}^{*} d \\
\left[
\begin{array}{ccc}
x \cdot x & x \cdot y & z \cdot z \\
y \cdot x & y \cdot y & y \cdot z \\
x \cdot z & x \cdot z & z \cdot z \\
\end{array}
\right]
%
\left[
\begin{array}{c}
a \\
b \\
c
\end{array}
\right]
%
&=
%
\left[
\begin{array}{c}
x \cdot d \\
y \cdot d \\
z \cdot d
\end{array}
\right].
%
\end{align}
%
$$
The solution is
$$
a_{LS} =
%
\left[
\begin{array}{c}
a \\
b \\
c
\end{array}
\right]_{LS}
%
= \left( \mathbf{A}^{*}\mathbf{A} \right)^{-1} \mathbf{A}^{*} d.
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1026451",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Integral $\int_0^{2\pi}\frac{dx}{2+\cos{x}}$ How do I integrate this?
$$\int_0^{2\pi}\frac{dx}{2+\cos{x}}, x\in\mathbb{R}$$
I know the substitution method from real analysis, $t=\tan{\frac{x}{2}}$, but since this problem is in a set of problems about complex integration, I thought there must be another (easier?) way.
I tried computing the poles in the complex plane and got
$$\text{Re}(z_0)=\pi+2\pi k, k\in\mathbb{Z}; \text{Im}(z_0)=-\log (2\pm\sqrt{3})$$
but what contour of integration should I choose?
| Here is an elementary treatment:
First note that $\displaystyle2+\cos x=\frac{3+\tan ^2 \frac{x}{2}}{1+\tan^2 \frac{x}{2}}$. Also note that for $\displaystyle f(x)=\frac{3+\tan ^2 \frac{x}{2}}{1+\tan^2 \frac{x}{2}}$, it holds that $\displaystyle f(x+\pi)=f(x)$ for $0<x<\pi$. Therefore
$$\begin{align}\int_0^{2\pi}\frac{1}{2+\cos x}dx&=\int_0^{2\pi}\frac{1+\tan^2 \frac{x}{2}}{3+\tan ^2 \frac{x}{2}}dx\\
&=2\int_0^{\pi}\frac{1+\tan^2 \frac{x}{2}}{3+\tan ^2 \frac{x}{2}}dx \\
&=2\pi-4\int_0^{\pi}\frac{1}{3+\tan ^2 \frac{x}{2}}dx \\
&=2\pi-8\int_0^{\infty}\frac{1}{(3+u^2)(1+u^2)}du\\
&=2\pi-4\int_0^{\infty}\frac{1}{1+u^2}du+4\int_0^{\infty}\frac{1}{3+u^2}du\\
&=\frac{2\pi}{\sqrt{3}}
\end{align}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1027486",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
Number of ways to distribute 100 identical chairs among 4 different rooms In how many ways can 100 identical chairs be divided among 4 different rooms so that each room will have 10,20,30,40 or 50 chairs?
I'm having problems coming up with the generating function for this question.
The answer given is 68.
| The problem can also be done without generating functions.
After dividing by $10$ we need to find the number of ordered quadruples of numbers in $\{1,2,3,4,5\}$ that add to $10.$ This is equivalent to the number of ordered quadruples of numbers in $\{0,1,2,3,4\}$ that add to $6.$ If we drop the restriction that the numbers be no greater than $4,$ we can use stars-and-bars to find that the number of quadruples is $\binom{6+3}{3}=84.$
Now we need to subtract the number of quadruples that contain a number greater than or equal to $5.$ Such a quadruple contains exactly one such number, given that the numbers add to $6.$ We represent a quadruple in which the first number is greater than or equal to $5$ by $(5,0,0,0)+q,$ where $q$ is an ordered quadruple of non-negative integers that add to $1.$ We represent quadruples where the number greater than or equal to $5$ is in positions $2,$ $3,$ or $4$ similarly. Since there are four choices of position, and four choices of $q,$ there are $16$ quadruples containing a number greater than or equal to $5.$ This gives the desired result.
Added: I should add that the calculation above can be seen in the generating function approach. Using the expression in Thijs' answer, we get
$$
\begin{aligned}
\left[x^{10}\right]\left(x+x^2+x^3+x^4+x^5\right)^4&=\left[x^6\right]\left(1+x+x^2+x^3+x^4\right)^4\\
&=\left[x^6\right]\left(\frac{1-x^5}{1-x}\right)^4\\
&=\left[x^6\right]\frac{1-4x^5}{(1-x)^4}\\
&=\left[x^6\right]\left(1-4x^5\right)\sum_{j=0}^\infty\binom{-4}{j}(-1)^jx^j\\
&=1\cdot\binom{-4}{6}(-1)^6-4\cdot\binom{-4}{1}(-1)^1\\
&=\binom{9}{6}-4\binom{4}{1}.
\end{aligned}
$$
The notation $[\text{blah}]$ means "coefficient of blah". In the third line we discarded numerator terms of order $x^{10}$ or higher. In line four, we used the generalized binomial theorem to expand the denominator. In line six, we rewrote the binomial coefficients $\binom{n}{r}$ with negative $n$ in terms of positive $n.$
If one traces through the steps of this calculation, one sees that they correspond with the steps of the previous calculation.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1027707",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Subsets and Splits
Fractions in Questions and Answers
The query retrieves a sample of questions and answers containing the LaTeX fraction symbol, which provides basic filtering of mathematical content but limited analytical insight.