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How to prove Vector Angles?
Any vector $\mathbf{v}$ is a unit vector if $\|\mathbf{v}\| = 1$.
Let $\mathbf{x}$, $\mathbf{y}$, and $\mathbf{z}$ be unit vectors, such
that $\mathbf{x} + \mathbf{y} + \mathbf{z} = \mathbf{0}$. Show that
the angle between any two of these vectors is $120^\circ$.
I know how to prove this using geometry, but the problem instructs me not to. How should I start to/prove this?
Thanks in advance!
| We have that $$(\mathbf{x} + \mathbf {y} + \mathbf {z})^2 =0$$ $$\Rightarrow \mathbf {x}^2 + \mathbf {y}^2 + \mathbf {z}^2 +2 (\mathbf {x}\cdot \mathbf {y} + \mathbf {y}\cdot \mathbf {z} + \mathbf {z}\cdot \mathbf {x}) =0$$ $$\Rightarrow (\mathbf {x}\cdot \mathbf {y} + \mathbf {y}\cdot \mathbf {z} + \mathbf {z}\cdot \mathbf {x}) = -\frac {3}{2} $$ $$(|\mathbf {x}||\mathbf {y}|\cos \theta_{xy} + |\mathbf {y}||\mathbf {z}|\cos \theta_{yz} + |\mathbf {z}||\mathbf {x}|\cos \theta_{zx}) = -\frac {3}{2} $$ $$=\cos \theta_{xy} + \cos \theta_{yz} + \cos \theta_{zx} = -\frac {3}{2}$$
Can you conclude using this? Hope it helps.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2132975",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Showing matrices in $SU(2)$ are of form $\begin{pmatrix} a & -b^* \\ b & a^*\end{pmatrix}$ Matrices $A$ in the special unitary group $SU(2)$ have determinant $\operatorname{det}(A) = 1$ and satisfy $AA^\dagger = I$.
I want to show that $A$ is of the form $\begin{pmatrix} a & -b^* \\ b & a^*\end{pmatrix}$ with complex numbers $a,b$ such that $|a|^2+|b|^2 = 1$.
To this end, we put $A:= \begin{pmatrix} r & s \\ t & u\end{pmatrix}$ and impose the two properties.
This yields \begin{align}\operatorname{det}(A) &= ru-st \\ &= 1 \ ,\end{align}
and
\begin{align}
AA^\dagger &= \begin{pmatrix} r & s \\ t & u\end{pmatrix} \begin{pmatrix} r^* & t^* \\ s^* & u^* \end{pmatrix} \\&= \begin{pmatrix} |r|^2+|s|^2 & rt^* +su^* \\ tr^*+us^* & |t|^2 + |u|^2\end{pmatrix} \\
&= \begin{pmatrix} 1 & 0 \\ 0 & 1\end{pmatrix} \ .\\
\end{align}
The latter gives rise to
\begin{align}
|r|^2+|s|^2 &= 1 \\
&= |t|^2+|u|^2 \ ,
\end{align}
and
\begin{align}
tr^*+us^* &= 0 \\
&= rt^*+su^* \ .
\end{align}
At this point, I don't know how to proceed. Any hints would be appreciated.
@Omnomnomnom's remark
\begin{align}
A A^\dagger &= \begin{pmatrix} |r|^2+|s|^2 & rt^* +su^* \\ tr^*+us^* & |t|^2 + |u|^2\end{pmatrix} \\
&= \begin{pmatrix} |r|^2+|t|^2 & sr^* +ut^* \\ rs^*+tu^* & |s|^2 + |u|^2\end{pmatrix} = A^\dagger A \ ,
\end{align}
gives rise to
$$
|t|^2 = |s|^2 \\
|r|^2 = |u|^2
$$
and
$$
AA^\dagger :\begin{pmatrix}
rt^* +su^* = sr^* +ut^* \\
tr^*+us^* = rs^*+tu^*
\end{pmatrix}: A^\dagger A $$
At this point, I'm looking in to find a relation between $t,s$ and $r,u$ respectively.
| Using @Omnomnomnom's suggestion $AA^\dagger =A^\dagger A$, we first obtain the relations
\begin{align}
AA^\dagger: r &= -\frac{su^*}{t^*}\ , \ u= -\frac{tr^*}{s^*} \\
A^\dagger A: r &= -\frac{tu^*}{s^*}\ , \ u= -\frac{sr^*}{t^*} \ .
\end{align}
Noticing the common factor $\frac{-t}{s^*}$ for $r_{A^\dagger A}$ and $u_{AA^\dagger}$, we put $x:=\frac{-t}{s^*}$.
This allows us to write $u = xr^*$.
Similarly, we have
\begin{align}
AA^\dagger: s &= -\frac{rt^*}{t^*}\ , \ t= -\frac{us^*}{s^*} \\
A^\dagger A: s &= -\frac{ut^*}{s^*}\ , \ t= -\frac{rs^*}{t^*} \ ,
\end{align}
and $y:= \frac{-u}{s^*}$. Which yields $s = yt^*$.
Hence, so far, we have
$$
A = \begin{pmatrix}r & yt^* \\ t & xr^*\end{pmatrix} \ .
$$
We now notice that, in fact, we have
$$
y = -\frac{u}{r^*} = -\frac{(xr^*)}{r^*} = -x \ .
$$
Our matrix now looks like
$$
A = \begin{pmatrix}r & -xt^* \\ t & xr^*\end{pmatrix} \ .
$$
Now, finally, at last, we use $\operatorname{det}(A) = 1$ to show that $x=1$:
\begin{align}
\operatorname{det}(A) &= 1 \\
&= x(|r|^2+|t|^2) \\
&= x \cdot 1 \ .
\end{align}
We now conclude with
$$
A = \begin{pmatrix}r & -t^* \\ t & r^*\end{pmatrix} \ .
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2133790",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
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} |
Evaluate the integral $\int \frac{x^2(x-2)}{(x-1)^2}dx$
Find $$\int \frac{x^2(x-2)}{(x-1)^2} dx .$$
My attempt:
$$\int \frac{x^2(x-2)}{(x-1)^2}dx = \int \frac{x^3-2x^2}{x^2-2x+1}dx $$
By applying polynomial division, it follows that
$$\frac{x^3-2x^2}{x^2-2x+1} = x + \frac{-x}{x^2-2x+1}$$
Hence $$\int \frac{x^3-2x^2}{x^2-2x+1}dx = \int \left(x + \frac{-x}{x^2-2x+1}\right) dx =\int x \,dx + \int \frac{-x}{x^2-2x+1} dx \\ = \frac{x^2}{2} + C + \int \frac{-x}{x^2-2x+1} dx $$ Now using substitution $u:= x^2-2x+1$ and $du = (2x-2)\,dx $ we get $dx= \frac{du}{2x+2}$.
Substituting dx in the integral:
$$\frac{x^2}{2} + C + \int \frac{-x}{u} \frac{1}{2x-2} du =\frac{x^2}{2} + C + \int \frac{-x}{u(2x-2)} du $$
I am stuck here. I do not see how using substitution has, or could have helped solve the problem. I am aware that there are other techniques for solving an integral, but I have been only taught substitution and would like to solve the problem accordingly.
Thanks
| HINT: write your Integrand in the form $$x- \left( x-1 \right) ^{-1}- \left( x-1 \right) ^{-2}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2135003",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Find the value of $\binom{2000}{2} + \binom{2000}{5} + \binom{2000}{8} + \cdots \binom{2000}{2000}$
Find the value of $\binom{2000}{2} + \binom{2000}{5} + \binom{2000}{8} + \cdots \binom{2000}{2000}$
I've seen many complex proofs. I am looking for an elementary proof. I know the fact that $\binom{2000}{0} + \binom{2000}{1} + \binom{2000}{2} + \cdots \binom{2000}{2000} = 2^{2000}$. This may help here.
| For $n\ge0$ let
$$a_n=\binom n0+\binom n3+\binom n6+\cdots=\sum_{k=0}^\infty\binom n{3k},$$
$$b_n=\binom n1+\binom n4+\binom n7+\cdots=\sum_{k=0}^\infty\binom n{3k+1},$$
$$c_n=\binom n2+\binom n5+\binom n8+\cdots=\sum_{k=0}^\infty\binom n{3k+2};$$
we seek the value of $c_{2000}.$ Observe that
$$a_n+b_n+c_n=2^n$$
and, for $n\ge1,$ from Pascal's rule we get the recurrences
$$a_n=a_{n-1}+c_{n-1},$$
$$b_n=a_{n-1}+b_{n-1},$$
$$c_n=b_{n-1}+c_{n-1}.$$
Hence, for $n\ge3,$ we have
$$c_n=b_{n-1}+c_{n-1}=a_{n-2}+2b_{n-2}+c_{n-2}=3a_{n-3}+3b_{n-3}+2c_{n-3}$$
$$=3(a_{n-3}+b_{n-3}+c_{n-3})-c_{n-3}=3\cdot2^{n-3}-c_{n-3}$$
and, for $n\ge6,$
$$c_n=3\cdot2^{n-3}-c_{n-3}=3\cdot2^{n-3}-(3\cdot2^{n-6}-c_{n-6})=c_{n-6}+21\cdot2^{n-6},$$
that is:
$$\boxed{c_n=c_{n-6}+21\cdot2^{n-6}}$$
Since $2000\equiv2\pmod6,$ we establish a closed formula for this case, namely
$$\boxed{c_n=\frac{2^n-1}3\text{ when }n\equiv2\pmod6}\ ,$$
by induction.
$c_2=\binom22=1=\frac{2^2-1}3.$
If $c_n=\frac{2^n-1}3,$ then
$$c_{n+6}=c_n+21\cdot2^n=\frac{2^n-1}3+21\cdot2^n=\frac{2^{n+6}-1}3.$$
In particular, when $n=2000,$ we have:
$$\boxed{\sum_{k=0}^\infty\binom{2000}{3k+2}=\sum_{k=0}^{666}\binom{2000}{3k+2}=c_{2000}=\frac{2^{2000}-1}3}$$
By the way, since $c_0=0=\frac{2^0-1}3,$ the identity $c_n=\frac{2^n-1}3$ also holds when $n\equiv0\pmod6.$
The general formula is
$$\boxed{\sum_{k=0}^\infty\binom n{3k+2}=\sum_{k=0}^{\left\lfloor\frac{n-2}3\right\rfloor}\binom n{3k+2}=c_n=\frac{2^n+2\cos\frac{(n+2)\pi}3}3}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2136738",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
"answer_id": 2
} |
If $m$th term and $n$th term of arithmetic sequence are $1/n$ and $1/m$ then the sum of the first $mn$ terms of the sequence is $(mn+1)/2$
If $m$th term and $n$th term of arithmetic sequence are $1/n$ and $1/m$ respectively then prove that the sum of the first $mn$ terms of the sequence is $(mn+1)/2$.
My Attempt ;
$$\textrm t_{m}=\dfrac {\textrm 1}{\textrm n}$$
$$\textrm a + \textrm (m-1)d =\dfrac {1}{n}$$
And,
$$\textrm t_{n}=\dfrac {1}{m}$$
$$\textrm a+\textrm (n-1)d=\dfrac {1}{m}$$
What do I do further?
| I am assuming equations created by you as equation (1) and (2) respectively.
From equation (1),
$a = \frac 1n - (m - 1)d$
Put value of a in equation (2),
$\frac 1n - (m - 1)d + (n - 1)d = \frac 1m$
$\implies (-m + 1 + n - 1)d = \frac 1m - \frac 1n$
$$\implies (n - m)d = \frac {n - m}{mn}$$
$$\implies d = \frac 1{mn}$$
Then $$a = \frac 1n - (m - 1) \cdot \frac{1}{mn}$$
$$a = \frac 1n - \frac 1n + \frac 1{mn}$$
$$a = \frac 1{mn}$$
Using sum formula -
$S_{mn}=\frac{mn}2 \left[ 2 \cdot \frac {1}{mn} + (mn - 1) \cdot \frac {1}{mn} \right]$
Taking $\frac {1}{mn}$ common,
$= \frac{mn}2 × \frac {1}{mn}\left[ 2 + (mn - 1) \right]$
$= \frac{1}2 × \left( mn + 1) \right)$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2138109",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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In right triangle $ABC$ ($\angle A=90$), $E$ is a point on $AC$.Find $AE$ given that...
In right triangle $ABC$ ($\angle A=90$), $AD$ is a height and $E$ is a point on $AC$ so that $BE=EC$ and $CD=CE=EB=1$. $\color {red} {Without}$ using trigonometric relations find $AE$.
I do have a solution USING trigonometric relations ( $AE=\sqrt[3]{2}-1$ ),but it seems other solutions are troublesome.My attempt led to a complicated polynomial equation...
| Here is a proof with only Pythagoreans:
(as I see, Michael Rozenberg used also the altitude theorem of right triangles)
We have 5 eqns in 5 variables AE, AC, BA, BD, AD:
Obviously
$AE = AC - CE = AC - 1;$
and 4 Pythagoreans:
$$
AC^2 = - AB^2 + (BD+DC)^2 = - AB^2 + (BD+1)^2;\\
AC^2 = AD^2 + DC^2 = AD^2 + 1;\\
AD^2 + BD^2 = AB^2;\\
1 = BE^2 = AB^2 + AE^2;
$$
Using the third one to eliminate AB everywhere:
$$
AE = AC - 1;\\
AC^2 = - AD^2 - BD^2 + (BD+1)^2;\\
AC^2 = AD^2 + 1;\\
AD^2 + BD^2 = 1 - AE^2;
$$
Using the third one of this block to eliminate AD everywhere:
$$AE = AC - 1;\\
2 AC^2 = 1 - BD^2 + (BD+1)^2;\\
AC^2 + BD^2 = 2 - AE^2;
$$
Using the first one of this block to eliminate AC everywhere:
$$2 (AE +1)^2 = 1 - BD^2 + (BD+1)^2 = 2 + 2 BD;\\
BD^2 = 2 - AE^2 - (AE + 1)^2;$$
Using the second one of this block to eliminate $BD^2$:
$$((AE +1)^2 -1 )^2 = 2 - AE^2 - (AE + 1)^2;$$
or, with $AE = x-1$,
$$ 0 = -(x^2 -1 )^2 + 2 - (x-1)^2 - x^2 = x (2 - x^3)$$
So we have the only positive real solution $AE = \sqrt[3] 2 -1$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2138416",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Prove, by contradiction, that, if $cx^2 + bx + a$ has no rational root, then $ax^2 + bx + c$ has no rational root. Proposition: Suppose that $a$, $b$, and $c$ are real numbers with $c \not = 0$. Prove, by contradiction, that, if $cx^2 + bx + a$ has no rational root, then $ax^2 + bx + c$ has no rational root.
Hypothesis: $cx^2 + bx + a$ has no rational root where $a$, $b$, and $c$ are real numbers with $c \not = 0$.
Conclusion: $ax^2 + bx + c$ has no rational root
To form a proof by contradiction, we take the negation of the conclusion:
$\neg B$: $ax^2 + bx + c$ has a rational root.
We now have a suitable hypothesis and conclusion for proof by contradiction:
A (Hypothesis): $cx^2 + bx + a$ has no rational root where $a$, $b$, and $c$ are real numbers with $c \not = 0$.
A1: $ax^2 + bx + c$ has a rational root.
Given that this is a proof by contradiction, we can work forward from both the hypothesis and conclusion, as shown above.
My Workings
A2: Let $x = \dfrac{p}{q}$ where $p$ and $q \not = 0$ are integers. This is the definition of a rational number (in this case, $x$): A rational number is any number that can be expressed as the quotient/fraction of two integers.
A3: $a\left(\dfrac{p}{q}\right)^2 + b\left(\dfrac{p}{q}\right) + c = 0$
$\implies \dfrac{ap^2}{q^2} + \dfrac{bp}{q} + c = 0$ where $q \not = 0$.
$\implies ap^2 + bpq + cq^2 = 0$
A4: $ap^2 + bpq + cq^2 = 0$ where $c \not = 0$
$\implies ap^2 + bpq = -cq^2$ where $-cq \not = 0$ since $c \not = 0$ and $q \not = 0$.
A5: $ap^2 + bpq + cq^2 = 0$ where $ap^2 + bpq \not = 0$ and $cq^2 \not = 0$.
But $ap^2 + bpq + cq^2 = 0$? Contradiction. $Q.E.D.$
I would greatly appreciate it if people could please take the time to review my proof and provide feedback on its correctness.
| Since you have proved that $ap^2+bpq \neq 0$ and $cq^2 \neq 0$, you are very close to the answer.
Notice that $p(ap+bq)=-cq^2$. But $p(ap+bq) \neq 0 \Rightarrow p \neq 0$.
Divide both sides by $p^2$. Then you get $a+b\frac qp=-c\frac {q^2}{p^2}.$
This shows that $cx^2+bx+a=0$ has a rational root $\frac qp$. WHICH is contradiction.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2139056",
"timestamp": "2023-03-29T00:00:00",
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"question_score": "1",
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Solve an integral $\int\frac{\cos^3 x}{\sin^3 x+\cos^3 x}dx$ Solve an integral $$\int\frac{\cos^3 x}{\sin^3 x+\cos^3 x}dx$$
I tried to divide the numerator and denominator by $\cos^4 x$ to get $\sec x$ function but the term ${\sin^3 x}/{\cos^4 x}$ gives $\tan^2 x\sec^2 x\sin x$. How to get rid of $\sin x$ term?
| Let $$I = \int \frac{\cos^3x}{\sin^3x+\cos^3x} dx$$ $$I_1 = \int\frac{\sin^3x+\cos^3x}{\sin^3x+cos^3x}dx = x + C$$ and $$I_2 = \int\frac{\cos^3x-\sin^3x}{\sin^3x+\cos^3x}dx$$Then $$I = \frac{I_1 + I_2}{2}$$
$$I_2 = \int\frac{(\cos x-\sin x)(1+\frac{\sin2x}{2})}{(\sin x+\cos x)(1-\frac{\sin 2x}{2})}dx$$
Now substitute $$t = \sin x+\cos x$$
and you get $$dt = (\cos x- \sin x)dx$$ and $$\frac{\sin2x}{2} = \frac{t^2-1}{2}$$
Then $$I_2 = \int\frac{t^2+1}{t(3-t^2)}dt = \frac{\ln t - 2\ln(3-t^2)}{3} + C$$
Finally, when putting this back together :
$$I = \frac{x}{2} + \frac{\ln(\sin x+\cos x)-2\ln(2-\sin2x)}{6} + C$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2139198",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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A series involve combination I want find another Idea to find sum of $\left(\begin{array}{c}n+3\\ 3\end{array}\right)$ from $n=1 ,to,n=47$
or $$\sum_{n=1}^{47}\left(\begin{array}{c}n+3\\ 3\end{array}\right)=?$$ I do it first by turn $\left(\begin{array}{c}n+3\\ 3\end{array}\right)$ to $\dfrac{(n+3)(n+2)(n+1)}{3!}=\dfrac16 (n^3+6n^2+11n+6)$ and find sum of them by separation $$\sum i=\dfrac{n(n+1)}{2}\\\sum i^2=\dfrac{n(n+1)(2n+1)}{6}\\\sum i^3=(\dfrac{n(n+1)}{2})^2$$
then I think more and do like below ...
I think there is more Idea to find this summation .
please hint, thanks in advanced
| By the well known hockey stick identity
$$ \sum_{n=0}^{47}\binom{n+3}{3} = \binom{47+3+1}{3+1} $$
and the problem is trivial from there.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2139408",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Minimize $\big(3+2a^2\big)\big(3+2b^2\big)\big(3+2c^2\big)$ if $a+b+c=3$ and $(a,b,c) > 0$
Minimize $\big(3+2a^2\big)\big(3+2b^2\big)\big(3+2c^2\big)$ if $a+b+c=3$ and $(a,b,c) > 0$.
I expanded the brackets and applied AM-GM on all of the eight terms to get :
$$\big(3+2a^2\big)\big(3+2b^2\big)\big(3+2c^2\big) \geq 3\sqrt{3}abc$$
, which is horribly weak !
I can not use equality constraint whichever method I use. Thanks to Wolfram|Alpha, I know the answer is $125$ when $(a,b,c) \equiv (1,1,1).$
Any help would be appreciated. :)
| More way.
Let $a+b+c=3u$, $ab+ac+bc=3v^2$ and $abc=w^3$.
Hence, we need to prove that
$$(3u^2+2a^2)(3+2b^2)(3+2c^2)\geq125u^6$$ or
$$27u^6+18u^4(9u^2-6v^2)+12u^2(9v^4-6uw^3)+8w^6\geq125u^6$$ or $f(w^3)\geq0$, where
$$f(w^3)=2w^6-18u^3w^3+16u^6-27u^4v^2+27u^2v^4.$$
But $f'(w^3)=2w^3-18u^3<0$, which says that $f$ is a decreasing function.
Thus, it remains to prove our inequality for a maximal value of $w^3$,
which happens for equality case of two variables.
Let $b=a$. Hence, $c=3-2a$ and we get
$$(a-1)^2(8a^4-8a^3+21a^2-22a+16)\geq0$$
again. Done!
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2141009",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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$\tan^{-1}(x^2-\frac{x^4}{2}+\frac{x^6}{4}-\cdots)+ \cot^{-1}(x^4-\frac{x^8}{2}+\frac{x^{12}}{4}-\cdots)=\pi/ 2$
Find $x$ with $0<|x|<2$ such that
$$\tan^{-1}(x^2-\frac{x^4}{2}+\frac{x^6}{4}-\cdots)+ \cot^{-1}(x^4-\frac{x^8}{2}+\frac{x^{12}}{4}-\cdots)=\pi/ 2$$
My try:
$(x^2-\frac{x^4}{2}+\frac{x^6}{4}-\cdots)=\alpha$
$(x^4-\frac{x^8}{2}+\frac{x^{12}}{4}-\cdots)=\beta$
$\alpha=\frac{\pi }{2}-\beta \to \tan \alpha=\tan(\frac{\pi }{2}-\beta)=\cot \beta$
Now?
| What I can come up with is as below
$\tan^{-1}(\alpha) =A \implies \tan(A)=\alpha$
$\cot^{-1}(\beta) =B \implies \cot(B)=\beta$
Also, we have
$A+B=\frac{\pi}{2}$
Add $\frac{\pi}{2}$ to both sides
$A+B+\frac{\pi}{2}=\pi$
Take $\tan$ from both sides and use the famous formula of $\tan(x+y)=\frac{\tan(x)+\tan(y)}{1-\tan(x)\tan(y)}$
$0=\tan(\pi)=\frac{\tan(A)+\tan(B+\frac{\pi}{2})}{1-\tan(A)\tan(B+\frac{\pi}{2})}$
Knowing $\tan(x+\frac{\pi}{2})=\cot(x)$, we get
$0=\tan(\pi)=\frac{\tan(A)+\cot(B)}{1-\tan(A)\cot(B)}=\frac{\alpha+\beta}{1-\alpha \beta}$
So, we need to have $\alpha+\beta=0$. Of course, we cannot have $1-\alpha \beta=0$, at the same time. Because, the n we have $\alpha=\frac{1}{\beta}$ and as we want to have $\alpha = -\beta$, by multiplying them we get the contradiction $\alpha^2=-1$.
Then, we need to find a closed formula for $(x^2-\frac{x^4}{2}+\frac{x^6}{4}-\cdots)=\alpha$ and $(x^4-\frac{x^8}{2}+\frac{x^{12}}{4}-\cdots)=\beta$, which can be found by separating each of them into positive and negative terms and using geometric series summation formula. Note that the limit on $0<|x|<2$ comes useful here, as we need the geometric summations to be convergent.
Finally, if I have not made any mistakes, you need to solve a polynomial of order $10$ to find $x$.
| {
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"url": "https://math.stackexchange.com/questions/2141349",
"timestamp": "2023-03-29T00:00:00",
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Compute $\lim\limits_{x \to 0} \frac{ \sqrt[3]{1+\sin x} - \sqrt[3]{1- \sin x }}{x}$
Compute $\lim\limits_{x \to 0} \dfrac{ \sqrt[3]{1+\sin x} - \sqrt[3]{1- \sin x }}{x}$
Original question was to solve $\lim\limits_{x \to 0} \dfrac{ \sqrt[3]{1+ x } - \sqrt[3]{1-x }}{x}$ and it was solved by adding and subtracting 1 in denominator. Making it in the form $\lim\limits_{x \to a}\dfrac {x^n - a^n}{x-a} = ax^{n-1}$
How to solve for above limit without using lhopitals rule?
| $$\lim _{ x\to 0 }{ \frac { { (1+\sin x) }^{ \frac { 1 }{ 3 } }-{ (1-\sin x) }^{ \frac { 1 }{ 3 } } }{ x } } \frac { \left( { (1+\sin x) }^{ \frac { 2 }{ 3 } }+{ (1+\sin x) }^{ \frac { 1 }{ 3 } }{ (1-\sin x) }^{ \frac { 1 }{ 3 } }+{ (1-\sin x) }^{ \frac { 2 }{ 3 } } \right) }{ \left( { (1+\sin x) }^{ \frac { 2 }{ 3 } }+{ (1+\sin x) }^{ \frac { 1 }{ 3 } }{ (1-\sin x) }^{ \frac { 1 }{ 3 } }+{ (1-\sin x) }^{ \frac { 2 }{ 3 } } \right) } =\\ =\lim _{ x\to 0 }{ \frac { 1+\sin { x-1+\sin { x } } }{ x } \frac { 1 }{ \left( { (1+\sin x) }^{ \frac { 2 }{ 3 } }+{ (1+\sin x) }^{ \frac { 1 }{ 3 } }{ (1-\sin x) }^{ \frac { 1 }{ 3 } }+{ (1-\sin x) }^{ \frac { 2 }{ 3 } } \right) } } =\\ =\lim _{ x\to 0 }{ \frac { 2\sin { x } }{ x } \frac { 1 }{ \left( { (1+\sin x) }^{ \frac { 2 }{ 3 } }+{ (1+\sin x) }^{ \frac { 1 }{ 3 } }{ (1-\sin x) }^{ \frac { 1 }{ 3 } }+{ (1-\sin x) }^{ \frac { 2 }{ 3 } } \right) } } =\frac { 2 }{ 3 } \\ \\ $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2143650",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 3
} |
concurrence of three lines in a quadrilateral Prove that the lines joining the midpoints of opposite sides of a quadrilateral and the line joining the midpoints of the diagonals of the quadrilateral are concurrent.
| Let $E=\frac{A+B}{2}$ and $F=\frac{C+D}{2}$ be the midpoints of two opposite sides.
Let $G=\frac{A+D}{2}$ and $H=\frac{C+B}{2}$ be the midpoints of another two opposite sides.
Let $J=\frac{A+C}{2}$ and $K=\frac{B+D}{2}$ be the midpoints of diagonals.
The midpoint of section $EF$ is a point $\frac{E+F}{2}=\frac{A+B+C+D}{2}$. Also the midpoint of $GH$ and $JK$ is $\frac{G+H}{2}=\frac{J+K}{2}=\frac{A+B+C+D}{2}$. Thuse each section $EF$, $GH$ and $JK$ contains point $L=\frac{A+B+C+D}{2}$
Therefore the lines connecting the midpoints of opposite sides of a quadrilateral and the line joining the midpoints of the diagonals of the quadrilateral are concurrent. QED
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2143928",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
If $\gcd(x,y)=1$, and $x^2 + y^2$ is a perfect sixth power, then $xy$ is a multiple of $11$ This is a problem that I don't know how to solve:
Let $x, y, z$ integer numbers such that $x$ and $y$ are relatively primes and $x^2+y^2=z^6$ . Show that $x\cdot y$ is a multiple of $11$.
| It's not complete answer
By computer search, it seems to be valid for co-prime $x$ and $y$.
Now, I'm trying to give the general solution of $(x,y,z)$.
\begin{align*}
(a+bi)^3 &= a(a^2-3b^2)+b(3a^2-b^2)i \\[7pt]
(a^2+b^2)^3 &=
\underbrace{a^2(a^2-3b^2)^2}_{\Large{m^2}}+
\underbrace{b^2(3a^2-b^2)^2}_{\Large{n^2}} \\[7pt]
\begin{pmatrix} x \\ y \\ z \end{pmatrix} &=
\begin{pmatrix}
m^2-n^2 \\
2mn \\
\sqrt[3]{m^2+n^2}
\end{pmatrix} \\[7pt] &=
\begin{pmatrix}
(a^2-b^2)(a^2+4ab+b^2)(a^2-4ab+b^2) \\
2ab(a^2-3b^2)(3a^2-b^2) \\
a^2+b^2 \end{pmatrix}
\end{align*}
Take absolute value if the solution is limited to positive.
The lowest non-trivial solution is $(117,44,5)$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2144094",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 3,
"answer_id": 2
} |
Sum of a series using squeeze theorem How do I find using the Squeeze theorem
$$\lim_{n\to \infty}\sum_{k=1}^n \frac{1}{\sqrt{n^2+k}} \;,$$
using the fact that
$$ \lim_{n\to \infty}\frac{n}{\sqrt{n^2+n}}=1.$$
Thank you very much for your help,
C.G
| Note that
$$\frac{1}{\sqrt{n^2+1}}+\frac{1}{\sqrt{n^2+2}}+\cdots+\frac{1}{\sqrt{n^2+n}}\leq \frac{n}{\sqrt{n^2+1}}$$
because $\sqrt{n^2+1}\leq\sqrt{n^2+k}$ for $k\geq1$.
On the other hand, by a similar reasoning, using that $\sqrt{n^2+n}\geq\sqrt{n^2+k}$ for $k\geq1$:
$$\frac{n}{\sqrt{n^2+n}}\leq\frac{1}{\sqrt{n^2+1}}+\frac{1}{\sqrt{n^2+2}}+\cdots+\frac{1}{\sqrt{n^2+n}}$$
so in the end,
$$\frac{n}{\sqrt{n^2+n}}\leq \sum_{k=1}^n \frac{1}{\sqrt{n^2+k}}\leq \frac{n}{\sqrt{n^2+1}}$$
and use that $$\lim_{n} \frac{n}{\sqrt{n^2+n}}=1=\lim_{n} \frac{n}{\sqrt{n^2+1}}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2144308",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Let $\alpha = \omega + \omega^2 + \omega^4$ and $\beta = \omega^3 + \omega^5 + \omega^6$. Let $\omega$ be a complex number such that $\omega^7 = 1$ and $\omega \neq 1$. Let $\alpha = \omega + \omega^2 + \omega^4$ and $\beta = \omega^3 + \omega^5 + \omega^6$. Then $\alpha$ and $\beta$ are roots of the quadratic
[x^2 + px + q = 0]
for some integers $p$ and $q$. Find the ordered pair $(p,q)$.
I got that $p=-1$ but does not know how to go on to finding $q$. All help is appreciated!
| method is due to Gauss. This book 1875. $x^2 + x + 2.$
gp-pari to check; note a^8 = a
? x = a + a^2 + a^4
%1 = a^4 + a^2 + a
? q = x^2 + x + 2
%2 = a^8 + 2*a^6 + 2*a^5 + 2*a^4 + 2*a^3 + 2*a^2 + a + 2
?
If we switched to one of the real numbers $$ t = \omega + \omega^6, $$ we would have a root of
$$ t^3 + t^2 - 2 t - 1 $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2144834",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
Factor $9(a-1)^2 +3(a-1) - 2$ I got the equation $9(a-1)^2 +3(a-1) - 2$ on my homework sheet. I tried to factor it by making $(a-1)=a$ and then factoring as a messy trinomial. But even so, I couldn't seem to get the correct answer; they all seemed incorrect.
Any help would be greatly appreciated.
Thank you so much in advance!
| $x=a-1\\
9x^2+3x-2=0\\
\Delta=9+72=81\\
\sqrt{\Delta}=9\\
x=\frac{-3 \pm 9}{18} = \pm \frac{1}{2}-\frac{1}{6}\\
a=x+1=\pm \frac{1}{2}-\frac{1}{6}+1 = \pm \frac{1}{2}+\frac{5}{6}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2147120",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 6,
"answer_id": 3
} |
Inequality of arithmetic and geometric mean I did a proof for inequality below, anyone has a other proof?
Let $a$ and $b$ be positive real numbers, and $t$ the parameter. Prove that:
$$a+b\geq 2\sqrt{1-t^2}\sqrt{ab}+(a-b)t$$
| Let $f (t)=Rhs $ of inequality . So differentiating and setting it equal to $0$ gives $\frac {2t}{\sqrt {1-t^2}}=\frac {a-b}{\sqrt {ab}} $ squaring both sides and solving we have $t=\frac {a-b}{a+b} $ thus putting the value of $t $ in original equation and simplifying we get it as $a^2+b^2$ now both $a,b $ are positive so $a^2+b^2\leq (a+b)^2$. Thus its proved. $$\text {another way} $$ put $t=\sin (x) $ thus $\sqrt {1-t^2}=\cos (x) $ we can do this as for the equality to give real values t has to be in $[0,1] $ thus we have $f (x)=2\sqrt {ab}\cos (x)+(a-b)\sin (x)\leq \sqrt {(2\sqrt {ab})^2+(a-b)^2}$ which is same as $\sqrt {(a+b)^2}=(a+b) $ as both $a,b $ are positive.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2148700",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
} |
How to find the maximum and minimum value of $\left|(z_1-z_2)^2 + (z_2-z_3)^2 + (z_3-z_1)^2\right|$ (where $|z_1|=|z_2|=|z_3|=1$)? How to find the maximum and minimum value of $\left|(z_1-z_2)^2 + (z_2-z_3)^2 + (z_3-z_1)^2\right|$ (where $|z_1|=|z_2|=|z_3|=1$ are complex numbers.) ?
My try:
$$\begin{align}\left|(z_1-z_2)^2 + (z_2-z_3)^2 + (z_3-z_1)^2\right| &\leq |z_1-z_2|^2 + |z_2-z_3|^2 + |z_3-z_1|^2 \\
&\leq (|z_1|+|z_2|)^2 + (|z_2|+|z_3|)^2 + (|z_3|+|z_1|)^2 \\
&\leq 2^2+2^2+2^2 \leq 12\end{align}$$
However the answer given is $8$. Where am I going wrong and how to do it correctly?
| Given
$$(z_1-z_2)+(z_2-z_3)+(z_3-z_1)=0$$
and
$$\left|(z_1-z_2)^2+(z_2-z_3)^2+(z_3-z_1)^2\right|=\\
\left|z_1^2-2z_1z_2+z_2^2+z_2^2-2z_2z_3+z_3^2+z_3^2-2z_3z_1+z_1^2\right|=\\
2\left|z_1^2+z_2^2+z_3^2-z_1z_2-z_2z_3-z_3z_1\right|=\\
2\left|z_1(z_1-z_2)+z_2(z_2-z_3)+z_3(z_3-z_1)\right|=\\
2\left|z_1(z_1-z_2)+z_2(z_2-z_3)+z_3(-(z_1-z_2)-(z_2-z_3))\right|=\\
2\left|(z_1-z_2)(z_1-z_3)+(z_2-z_3)^2)\right|=...$$
replacing $z_1=1$
$$...=2\left|(1-z_2)(1-z_3)+(z_2-z_3)^2)\right|=2\left|(1-z_2)(1-z_3)+(z_2-1+1-z_3)^2\right|=\\
2\left|(1-z_2)(1-z_3)+(z_2-1)^2+(1-z_3)^2+2(z_2-1)(1-z_3)\right|=\\
2\left|(1-z_2)(1-z_3)+(z_2-1)^2+(1-z_3)^2-2(1-z_2)(1-z_3)\right|=\\
2\left|(1-z_2)^2+(1-z_3)^2-(1-z_2)(1-z_3)\right|=...$$
which is
$$...=2\left|\frac{(1-z_2)^3+(1-z_3)^3}{1-z_2+1-z_3}\right|=...$$
using law of sines ...
$$...=2\left|\frac{2^3\sin^3{\alpha}+2^3\sin^3{\beta}}{2\sin{\alpha}+2\sin{\beta}}\right|=8\left|\frac{\sin^3{\alpha}+\sin^3{\beta}}{\sin{\alpha}+\sin{\beta}}\right|\leq ...\tag{1}$$
both $\alpha, \beta \in (0,\pi)$ (corner cases can be treated individually), which means $$0<\sin{\alpha}\leq 1,0<\sin{\beta}\leq 1$$
or
$$0<\sin^3{\alpha}\leq \sin{\alpha}<1,0<\sin^3{\beta}\leq \sin{\beta}<1$$
thus $$0<\sin^3{\alpha} + \sin^3{\beta} \leq \sin{\alpha} + \sin{\beta}$$
and, continuing (1)
$$...\leq 8$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2149243",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "12",
"answer_count": 2,
"answer_id": 1
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How to show $(a + b)^n \leq a^n + b^n$, where $a, b \geq 0$ and $n \in (0, 1]$? Does anyone happen to know a nice way to show that $(a+b)^n \le a^n+b^n$, where $a,b\geq 0$ and $n \in (0,1]$? I figured integrating might help, but I've been unable to pull my argument full circle. Any suggestions are appreciated :)
| Assume that
$a \ge b$.
If $a = b = 0$,
the result is immediate.
If $a > 0$,
divide
$(a+b)^n \le a^n+b^n
$
by
$a^n$
to get
$(1+b/a)^n \le 1+(b/a)^n
$.
Since
$b \le a$,
$0 \le b/a \le 1$,
so this becomes
$(1+x)^n \le 1+x^n
$
where
$x = b/a$.
Let $f(x)
=1+x^n-(1+x)^n
$.
$f(0) = 0$
and
$f(1)
=2-2^n
\ge 0
$
since
$0 < n \le 1$.
$f'(x)
=nx^{n-1}-n(1+x)^{n-1}
=n(x^{n-1}-(1+x)^{n-1})
=n(\frac1{x^{1-n}}-\frac1{(1+x)^{1-n}})
$.
Since
$0 < n \le 1$,
$1-n \ge 0$
so that
$x^{1-n}
\le (1+x)^{1-n}
$
so that
$\frac1{x^{1-n}}\ge\frac1{(1+x)^{1-n}}
$
so that
$f'(x) \ge 0$.
Since
$f(0) = 0$
and
$f'(x) \ge 0$
for $0 < x\le 1$,
$f(x)
\ge 0
$
for
$0 \le x \le 1$.
Note that
the inequality goes the other way
if $n > 1$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2150372",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
"answer_id": 3
} |
A formula for $\sin(\pi/2^n)$ May be this a duplicate, but I did not find any question related.
I found the following formula, but there was no proof of it:
$$2\sin\left(\frac{\pi}{2^{n+1}}\right)=\sqrt{2_1-\sqrt{2_2+\sqrt{2_3+\sqrt{2_4+\cdots\sqrt{2_n}}}}}$$
where
$$2_k=\underbrace{222\cdots222}_{k\text { times}}.$$
(The number $22$ is twenty-two for instance, and not $2\times 2=4$.)
Do you know a proof of this result? Do you know any references?
I think one way to prove it would be to deal with regular polygons inside a circle and play the angles and trigonometry.
Do you think it would work?
Is there a different way to proceed?
| Here's my repeated half-angle approach (I know, this is definitely not a great way to deal with it, but still am posting it here. This is my first answer, here in this website, so please bear with me..):
We know
$2\cos^2 \theta =1+\cos 2\theta\implies \cos \theta =\sqrt{\frac{1+\cos 2\theta}{2}}.$
Taking positive sign because I am going to take $\theta=\frac{π}{2^n}, n\ge 2.$
So $2\cos \theta =\sqrt{2+2\cos 2\theta}.$
Let $\theta=\frac{π}{2^n}, n\ge 2.$ Then
\begin{align}
2\cos \left(\frac{π}{2^n}\right)& =\sqrt{2+2\cos \left(\frac{π}{2^{n-1}}\right)} \;\;(1 \text{ radical}) \\\\
&=\sqrt{2+\sqrt{2+2\cos \left(\frac{π}{2^{n-2}}\right) } }\;\;(2\text{ radicals})\\\\
&\vdots\\\\
&=\sqrt{2+\sqrt{2+\cdots+\sqrt{2+\cos \frac{π}{2}}}}\;\;(n-1\text{ radicals}) \\\\
&=\sqrt{2+\sqrt{2+\cdots+\sqrt{2}}}\;\;(n-1 \text{ radicals})\\\\
&=A_{n-1},\text{ say}.
\end{align}
Therefore,
$2\cos \left(\frac{2π}{2^{n+1}}\right) =A_{n-1}$
$\implies 2\left[1-2\sin^2 \left(\frac π{2^{n+1}}\right) \right]=A_{n-1}$
$\implies 4\sin^2 \left(\frac π{2^{n+1}}\right) =2-A_{n-1}$
$\implies 2\sin \left(\frac{π}{2^{n+1}}\right) =\sqrt{2-A_{n-1}}$
Thus,
$\sin \left(\frac{π}{2^{n+1}}\right) =\frac 12 \sqrt{2-\sqrt{2+\sqrt{2+\cdots+\sqrt{2}}}}\;\;(n\text{ radicals}), \forall n\ge 2.$
As for example,
$\sin \frac{π}{8}=\sin \left(\frac{π}{2^{2+1}}\right)=\frac 12 \sqrt{2-\sqrt{2}}.$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2151405",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
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Solve the equation : $\tan \theta + \tan 2\theta + \tan 3\theta = \tan \theta \tan 2\theta \tan 3\theta $ I've been having some trouble solving this equation. (The solution in my book is given as $ \frac {n \pi}{3}, n \in Z $)
Here is what I've done
$$\frac {\sin \theta}{\cos \theta} + \frac {\sin 2\theta} {\cos 2\theta} + \frac{\sin 3\theta}{\cos 3\theta}= \frac {\sin \theta}{\cos \theta} \frac {\sin 2\theta} {\cos 2\theta} \frac{\sin 3\theta}{\cos 3\theta}$$
$$ \frac {\sin \theta \cos 2\theta \cos 3\theta + \cos \theta \sin 2\theta \cos 3\theta + \cos \theta \cos 2\theta \sin 3\theta - \sin \theta \sin 2\theta \sin 3\theta }{\cos\theta \cos 2\theta \cos 3\theta} = 0 $$
$$\cos 2\theta \{\sin\theta \cos 3\theta + cos \theta \sin 3\theta \} + \sin 2\theta \{\cos \theta \cos 3\theta - \sin \theta \sin 3\theta \} = 0 $$
$$\cos 2\theta \sin(3\theta + \theta) +\sin2\theta \cos(3\theta + \theta) = 0 $$
$$ \cos 2\theta \sin 4\theta + sin 2\theta cos 4\theta = 0$$
$$ \sin (2\theta + 4\theta) = 0$$
$$\sin 6\theta = 0 $$
$$ \theta = \frac {n\pi}{6}, n \in Z$$
I understand from this question that whatever mistake I am making is in the third step, where I remove $\cos \theta \cos 2\theta \cos 3\theta $ from the denominator. However, despite reading through the aforementioned post, I couldn't really get the intuition behind why this is wrong.
I'd like :
*
*To understand the intuition behind why removing $\cos \theta \cos 2\theta cos 3\theta $ is a mistake.
*To know how to solve this question correctly
*How do I avoid making these types of mistakes when solving trigonometric equations
| We have
$$\tan (A+B+C) = \frac{\tan A + \tan B + \tan C - \tan A \tan B \tan C}{1-(\tan A \tan B + \tan B \tan C + \tan C \tan A)}$$
and the given condition implies that $\tan (6\theta) = 0$. Thus $\theta = \frac{n\pi}{6}$, $n \in \mathbb{Z}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2154221",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 3
} |
How to take the integral? $\int \frac{x^2-3x+2}{x^2+2x+1}dx$ $$\int \frac{x^2-3x+2}{x^2+2x+1}dx$$
So after all I had
$$ \frac{-5x+1}{(x+1)^2} = \frac{A}{(x+1)} + \frac{B}{(x+1)^2}$$
and of course $$ \int xdx $$
but it is easy to solve, I do not know how to act with devided things, probably solve the system, or is there easier way to find A and B?
After all steps I finally got:
$$-5x + 1 = Ax + A + B$$
| \begin{align*}
\frac{x^2 - 3x + 2}{x^2 + 2x + 1} &= 1 + \frac{1-5x}{(x+1)^2} \\
&= 1 + \frac{A}{x+1} + \frac{B}{(x+1)^2}
\end{align*}
Then
\begin{align*}
\frac{1-5x}{(x+1)^2} &= \frac{A}{x+1} + \frac{B}{(x+1)^2} \\
&= \frac{A(x+1)+B}{(x+1)^2} \\
1-5x &= Ax+(A+B)
\end{align*}
So $A = -5$ and $B=6$.
Therefore
\begin{align*}
\frac{x^2 - 3x + 2}{x^2 + 2x + 1} &= 1 - \frac{5}{x+1} + \frac{6}{(x+1)^2} \\
\int\frac{x^2 - 3x + 2}{x^2 + 2x + 1}\,dx &= \int \left( 1 - \frac{5}{x+1} + \frac{6}{(x+1)^2} \right) dx \\
&= x - 5 \ln \lvert x+1 \rvert - \frac{6}{x+1} + C
\end{align*}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2154334",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 0
} |
Factorise an ideal into a product of prime ideals So for the ideal $I = (20+\sqrt{-5})_{R}$ when $K=\mathbb{Q}(\sqrt{-5})$, how do I factorise this into a product of prime ideals. Do you start by taking the norm of $I$ and decomposing it into a product of primes? So
$$N(I) = 405 = 3^{4} \times 5.$$
If so, where do I go next?
| Recall that the primes $ 3, 5 $ split as $ 3R = \mathfrak p \mathfrak p' $ and $ 5R = (\sqrt{-5})^2 $ in $ R = \mathcal O_K = \mathbf Z[\sqrt{-5}] $. The norm suggests that the ideal $ I = (20 + \sqrt{-5}) $ factors as $ \mathfrak p^i \mathfrak p'^j (\sqrt{-5}) $, where $ i + j = 4 $. It follows upon division that
$$ \mathfrak p^i \mathfrak p'^j = (1 - 4\sqrt{-5}) $$
This is not divisible by $ 3 $, therefore either $ i = 0 $ or $ j = 0 $. The primes lying over $ 3 $ are $ (3, 1 \pm \sqrt{-5}) $; by calculation we have
$$ (3, 1 - \sqrt{-5})^2 = (9, 3 - 3 \sqrt{-5}, -4 - 2\sqrt{-5}) = (9, 7 - \sqrt{-5}, -4 - 2\sqrt{-5}) = (9, 7 - \sqrt{-5}) = (2 + \sqrt{-5}) $$
$$ (2 + \sqrt{-5})^2 = (1 - 4 \sqrt{-5}) $$
It follows that
$$ (20 + \sqrt{-5}) = (3, 1 - \sqrt{-5})^4 (\sqrt{-5}) $$
is the desired factorization.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2154454",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Find $\frac{\sum_\limits{k=0}^{6}\csc^2\left(x+\frac{k\pi}{7}\right)}{7\csc^2(7x)}$ Find the value of
$\dfrac{\sum_\limits{k=0}^{6}\csc^2\left(x+\dfrac{k\pi}{7}\right)}{7\csc^2(7x)}$
when $x=\dfrac{\pi}{8}$.
The Hint given is: $n\cot nx=\sum_\limits{k=0}^{n-1}\cot\left(x+\dfrac{k\pi}{n}\right)$
I dont know how it comes nor how to use it
| For future reference with this problem being tagged complex-numbers we
show how to evaluate the sum using residues. Suppose we are interested
in
$$S(n) = \sum_{k=0}^{n-1} \csc^2\left(x+\frac{k\pi}{n}\right)
= \sum_{k=0}^{n-1} \frac{2}{1-\cos\left(2x+\frac{2k\pi}{n}\right)}.$$
where we take $x$ to be a real number.
With $$f(z) = \frac{4}{2-\exp(2ix)z-1/\exp(2ix)/z}
\frac{nz^{n-1}}{z^n-1}$$
or alternatively
$$f(z) = \frac{4}{2-\exp(2ix)z-1/\exp(2ix)/z} \frac{1}{z}
\frac{n}{z^n-1}
\\ = \frac{4}{2z-\exp(2ix)z^2-1/\exp(2ix)}
\frac{n}{z^n-1}
\\ = -\frac{4\exp(-2ix)}{z^2 - 2z\exp(-2ix) + \exp(-4ix)}
\frac{n}{z^n-1}
\\ = -\frac{4\exp(-2ix)}{(z-\exp(-2ix))^2}
\frac{n}{z^n-1}$$
we get for the sum with $\zeta_k = \exp(2\pi i k/n)$
$$\sum_{k=0}^{n-1} \mathrm{Res}_{z=\zeta_k} f(z)$$
which means we can evaluate the sum using the negative of the residues
at $z=\exp(-2ix)$ and at infinity. Note however that with $R$ the
radius of a circle going to infinity we get that $f(z)$ is
$\theta(1/R^{n+2})$ and $2\pi R \times 1/R^{n+2} = 2\pi \times
1/R^{n+1}$ vanishes so the residue at infinity is zero. That leaves
for the other residue
$$\left.
\left(-\frac{4n\exp(-2ix)}{z^n-1}\right)'\right|_{z=\exp(-2ix)}
= \left.\frac{4n\exp(-2ix)}{(z^n-1)^2} \times n z^{n-1}
\right|_{z=\exp(-2ix)}
\\ = \frac{4n^2\exp(-2inx)}{(\exp(-2inx)-1)^2}
= -\frac{(2i)^2 n^2}{(\exp(-inx)-\exp(inx))^2}.$$
We obtain
$$S(n) -\frac{(2i)^2 n^2}{(\exp(inx)-\exp(-inx))^2} = 0$$
or
$$\bbox[5px,border:2px solid #00A000]{
S(n) = n^2 \csc^2(nx).}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2154639",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Proving Saalschutz Theorem I saw this in a pdf, and I'm wondering
Questions:
*
*How do you prove Saalschutz Theorem:
$$_3F_2\left[\begin{array}{c,c}-x,-y,-z\\n+1,-x-y-z-n\end{array}\right]=\dfrac {\Gamma(n+1)\Gamma(x+y+n+1)\Gamma(y+z+n+1)\Gamma(z+x+n+1)}{\Gamma(x+n+1)\Gamma(y+n+1)\Gamma(z+n+1)\Gamma(x+y+z+n+1)}\tag{1}$$
I'm somewhat relatively new to Hypergeometrical Series. I understand that the general Hypergeometrical series takes the form$$_pF_q\left[\begin{array}{c,c}\alpha_1,\alpha_2,\ldots,\alpha_p\\\beta_1,\beta_2,\ldots,\beta_q\end{array};x\right]=\sum\limits_{k=0}^{\infty}\dfrac {(\alpha_1)_k(\alpha_2)_k\ldots(\alpha_p)_k}{(\beta_1)_k(\beta_2)_k\ldots(\beta_q)_k}\dfrac {x^k}{k!}\tag{2}$$
So therefore, by $(2)$, we should have$$_3F_2\left[\begin{array}{c,c}-x,-y,-z\\n+1,-x-y-z-n\end{array}\right]=\sum\limits_{k=0}^{\infty}\dfrac {(-x)_k(-y)_k(-z)_k}{(n+1)_k(-x-y-z-n)_k}\tag{3}$$
However, I'm not sure how to manipulate the RHS of $(3)$ to get the RHS of $(1)$.
EDIT: Since $(a)_k=\Gamma(a+k)/\Gamma(a)$, the RHS of $(3)$ becomes$$\dfrac {(-x)_k(-y)_k(-z)_k}{(n+1)_k(-x-y-z-n)_k}=\dfrac {\Gamma(k-y)\Gamma(n+1)\Gamma(k-x)\Gamma(-x-y-z-n)\Gamma)k-z)}{\Gamma(n+k+1)\Gamma(-x)\Gamma(-y)\Gamma(-z)\Gamma(-x-y-z-n+k)}$$Now, I need to figure out how$$\Gamma(k-y)\Gamma(k-x)\Gamma(k-z)\Gamma(-x-y-z-n)=\Gamma(x+y+n+1)\Gamma(y+z+n+1)\Gamma(x+z+n+1)$$$$\Gamma(n+k+1)\Gamma(-x)\Gamma(-y)\Gamma(-z)\Gamma(-x-y-z-n+k)=\Gamma(x+n+1)\Gamma(y+n+1)\Gamma(z+n+1)\Gamma(x+y+z+n+1)$$
Extra: I also believe that using the same general approach, we can prove$$\begin{align*} & _7F_6\left[\begin{array}{c,c}n,\frac 12n+1,-x,-y,-z,-u,x+y+z+u+2n+1\\\frac 12n,x+n+1,y+n+1,z+n+1,u+n+1,-x-y-z-u-n\end{array}\right]\\ & =\dfrac {\Gamma(x+n+1)\Gamma(y+n+1)\Gamma(z+n+1)\Gamma(u+n+1)\Gamma(x+y+z+n+1)}{\Gamma(n+1)\Gamma(x+y+n+1)\Gamma(y+z+n+1)\Gamma(x+u+n+1)\Gamma(z+u+n+1)}\\ & \times\dfrac {\Gamma(y+z+u+n+1)\Gamma(x+u+z+n+1)\Gamma(x+y+u+n+1)}{\Gamma(x+z+n+1)\Gamma(y+u+n+1)\Gamma(x+y+z+u+n+1)}\end{align*}\tag{4}$$
| The development of the Saalschütz's identity that I know proceeds along the following path.
Start with the known identity about the sum of the product of three binomials:
$$ \bbox[lightyellow] {
\begin{gathered}
F(m,n,r,s)\quad \left| {\;0 \leqslant \text{integers}\,m,n} \right.\quad = \hfill \\
= \sum\limits_{\left( {0\, \leqslant } \right)\;k\,\left( { \leqslant \,n} \right)} {\left( \begin{gathered}
m - r + s \\ k \\
\end{gathered} \right)\left( \begin{gathered}
n + r - s \\ n - k \\
\end{gathered} \right)\left( \begin{gathered}
r + k \\ m + n \\
\end{gathered} \right)} = \hfill \\
= \sum\limits_{\begin{subarray}{l}
\left( {0\, \leqslant } \right)\;k\,\left( { \leqslant \,n} \right) \\
\left( {0\, \leqslant } \right)\;j\,\left( { \leqslant \,m + n} \right)
\end{subarray}} {\left( \begin{gathered}
m - r + s \\ k \\
\end{gathered} \right)\left( \begin{gathered}
n + r - s \\ n - k \\
\end{gathered} \right)\left( \begin{gathered}
r \\ m + n - j \\
\end{gathered} \right)\left( \begin{gathered}
k \\ j \\
\end{gathered} \right)} = \hfill \\
= \sum\limits_{\begin{subarray}{l}
\left( {0\, \leqslant } \right)\;k\,\left( { \leqslant \,n} \right) \\
\left( {0\, \leqslant } \right)\;j\,\left( { \leqslant \,m + n} \right)
\end{subarray}} {\left( \begin{gathered}
m - r + s \\ j \\
\end{gathered} \right)\left( \begin{gathered}
m - r + s - j \\ k - j \\
\end{gathered} \right)\left( \begin{gathered}
n + r - s \\ n - k \\
\end{gathered} \right)\left( \begin{gathered}
r \\ m + n - j \\
\end{gathered} \right)} = \hfill \\
= \sum\limits_{\left( {0\, \leqslant } \right)\;j\,\left( { \leqslant \,m + n} \right)} {\left( \begin{gathered}
m - r + s \\ j \\
\end{gathered} \right)\left( \begin{gathered}
n + m - j \\ n - j \\
\end{gathered} \right)\left( \begin{gathered}
r \\ m + n - j \\
\end{gathered} \right)} = \hfill \\
= \sum\limits_{\left( {0\, \leqslant } \right)\;j\,\left( { \leqslant \,m + n} \right)} {\left( \begin{gathered}
m - r + s \\ j \\
\end{gathered} \right)\left( \begin{gathered}
n + m - j \\ m \\
\end{gathered} \right)\left( \begin{gathered}
r \\ m + n - j \\
\end{gathered} \right)} = \hfill \\
= \sum\limits_{\left( {0\, \leqslant } \right)\;j\,\left( { \leqslant \,m + n} \right)} {\left( \begin{gathered}
m - r + s \\ j \\
\end{gathered} \right)\left( \begin{gathered}
r \\ m \\
\end{gathered} \right)\left( \begin{gathered}
r - m \\ n - j \\
\end{gathered} \right)} = \hfill \\
= \left( \begin{gathered}
r \\ m \\
\end{gathered} \right)\left( \begin{gathered}
s \\ n \\
\end{gathered} \right) \hfill \\
\end{gathered} \tag{1}
} $$
where the steps are:
- inverse convolution on last b.;
- trinomial revision on 1st and 4th b.;
- convolution in $k$ on 2nd and 3rd b.;
- symmetry on 2nd b.;
- trinomial revision on 2nd and 3rd b.;
- convolution in $j$ on 1st and 3rd b.
Then consider that the addenda in the sum are
$$
\begin{gathered}
t_{\,k} = \left( \begin{gathered}
m - r + s \\
k \\
\end{gathered} \right)\left( \begin{gathered}
n + r - s \\
n - k \\
\end{gathered} \right)\left( \begin{gathered}
r + k \\
m + n \\
\end{gathered} \right) = \hfill \\
= \frac{{\left( {m - r + s} \right)^{\,\underline {\,k\,} } }}
{{k!}}\frac{{\left( {n + r - s} \right)^{\,\underline {\,n - k\,} } }}
{{\left( {n - k} \right)!}}\frac{{\left( {r + k} \right)^{\,\underline {\,m + n\,} } }}
{{\left( {m + n} \right)!}} \hfill \\
\end{gathered}
$$
where $x^{\,\underline {\,k\,} }$ indicates the falling factorial.
The value of the initial term is
$$
t_{\,0} = \frac{{\left( {n + r - s} \right)^{\,\underline {\,n\,} } }}
{{n!}}\frac{{r^{\,\underline {\,m + n\,} } }}
{{\left( {m + n} \right)!}} = \left( \begin{gathered}
n + r - s \\
n \\
\end{gathered} \right)\left( \begin{gathered}
r \\
m + n \\
\end{gathered} \right)
$$
and the ratio of consecutive terms is a rational function in $k$
$$
\begin{gathered}
\frac{{t_{\,k + 1} }}
{{t_{\,k} }} = \frac{{\left( {m - r + s} \right)^{\,\underline {\,k + 1\,} } \left( {n + r - s} \right)^{\,\underline {\,n - k - 1\,} } \left( {r + k + 1} \right)^{\,\underline {\,m + n\,} } }}
{{\left( {m - r + s} \right)^{\,\underline {\,k\,} } \left( {n + r - s} \right)^{\,\underline {\,n - k\,} } \left( {r + k} \right)^{\,\underline {\,m + n\,} } }}\frac{{k!\left( {n - k} \right)!\left( {m + n} \right)!}}
{{\left( {k + 1} \right)!\left( {n - k - 1} \right)!\left( {m + n} \right)!}} = \hfill \\
= \frac{{\left( {m - r + s - k} \right)\left( {r + k + 1} \right)}}
{{\left( {r - s + 1 + k} \right)\left( {r + k - m - n + 1} \right)}}\frac{{\left( {n - k} \right)}}
{{\left( {k + 1} \right)}} = \hfill \\
= \frac{{\left( {r - m - s + k} \right)\left( {r + 1 + k} \right)\left( { - n + k} \right)}}
{{\left( {r - s + 1 + k} \right)\left( {r - m - n + 1 + k} \right)\left( {1 + k} \right)}} \hfill \\
\end{gathered}
$$
so that we can write
$$
\begin{gathered}
F(m,n,r,s) = t_0 \;{}_3F_2 \left[ {\left. \begin{gathered}
r + 1,\;r - m - s,\; - n \hfill \\
r - s + 1,\;r - m - n + 1\; \hfill \\
\end{gathered} \right|\;1} \right] = \hfill \\
= \left( \begin{gathered}
n + r - s \\
n \\
\end{gathered} \right)\left( \begin{gathered}
r \\
m + n \\
\end{gathered} \right)\;{}_3F_2 \left[ {\left. \begin{gathered}
r + 1,\;r - m - s,\; - n \hfill \\
r - s + 1,\;r - m - n + 1\; \hfill \\
\end{gathered} \right|\;1} \right] \hfill \\
\end{gathered}
$$
that is, changing $n$ to $q$ so not to get confused with your notation:
$$ \bbox[lightyellow] {
\begin{gathered}
{}_3F_2 \left[ {\left. \begin{gathered}
r + 1,\;r - m - s,\; - q \hfill \\
r - s + 1,\;r - m - q + 1\; \hfill \\
\end{gathered} \right|\;1} \right]\quad \left| {\;0 \leqslant \text{integers}\,m,q} \right.\quad = \hfill \\
= \left( \begin{gathered}
s \\
q \\
\end{gathered} \right)\left( \begin{gathered}
r \\
m \\
\end{gathered} \right)\;\mathop /\limits_{} \;\left( {\left( \begin{gathered}
q + r - s \\
q \\
\end{gathered} \right)\left( \begin{gathered}
r \\
m + q \\
\end{gathered} \right)} \right) = \hfill \\
= \frac{{\Gamma \left( {s + 1} \right)\Gamma \left( {r + 1} \right)\Gamma \left( {r - s + 1} \right)\Gamma \left( {q + 1} \right)\Gamma \left( {r - m - q + 1} \right)\Gamma \left( {m + q + 1} \right)}}
{{\Gamma \left( {s - q + 1} \right)\Gamma \left( {q + 1} \right)\Gamma \left( {r - m + 1} \right)\Gamma \left( {m + 1} \right)\Gamma \left( {q + r - s + 1} \right)\Gamma \left( {r + 1} \right)}} = \hfill \\
= \frac{{\Gamma \left( {s + 1} \right)\Gamma \left( {r - s + 1} \right)\Gamma \left( {r - m - q + 1} \right)\Gamma \left( {m + q + 1} \right)}}
{{\Gamma \left( {s - q + 1} \right)\Gamma \left( {r - s + q + 1} \right)\Gamma \left( {r - m + 1} \right)\Gamma \left( {m + 1} \right)}} = \hfill \\
= \frac{{s^{\,\underline {\,q\,} } \left( {r - s} \right)^{\,\underline {\, - \,q\,} } }}
{{\left( {r - m} \right)^{\,\underline {\,q\,} } m^{\,\underline {\, - q\,} } }} \hfill \\
\end{gathered} \tag{2}
} $$
which you can compare with your notation ( not forgetting the final $1/k!$ term) to get
$$
\left\{ \begin{gathered}
q = z \hfill \\
m = y + n \hfill \\
r = - x - 1 \hfill \\
s = - x - 1 - n \hfill \\
\end{gathered} \right.
$$
and finally
$$ \bbox[lightyellow] {
\begin{gathered}
{}_3F_2 \left[ {\left. \begin{gathered}
- x,\; - y,\; - z \hfill \\
n + 1,\; - x - y - z - n\; \hfill \\
\end{gathered} \right|\;1} \right]\quad \left| {\;0 \leqslant \text{integers}\,z,\left( {y + n} \right)} \right.\quad = \hfill \\
= \left( \begin{gathered}
- x - 1 - n \\
z \\
\end{gathered} \right)\left( \begin{gathered}
- x - 1 \\
y + n \\
\end{gathered} \right)\;\mathop /\limits_{} \;\left( {\left( \begin{gathered}
z + n \\
z \\
\end{gathered} \right)\left( \begin{gathered}
- x - 1 \\
y + n + z \\
\end{gathered} \right)} \right) = \hfill \\
= \frac{{\Gamma \left( { - x - n} \right)\Gamma \left( {n + 1} \right)\Gamma \left( { - x - y - n - z} \right)\Gamma \left( {y + n + z + 1} \right)}}
{{\Gamma \left( { - x - n - z} \right)\Gamma \left( {n + 1 + z} \right)\Gamma \left( { - x - y - n} \right)\Gamma \left( {y + n + 1} \right)}} = \hfill \\
= \frac{{\left( { - x - 1 - n} \right)^{\,\underline {\,z\,} } \,n^{\,\underline {\, - \,\,z\,} } }}
{{\left( { - x - 1 - n - y} \right)^{\,\underline {\,z\,} } \;\left( {n + y} \right)^{\,\underline {\, - \,z\,} } }} = \hfill \\
= \frac{{\Gamma \left( { - x - n} \right)\Gamma \left( {n + 1} \right)\Gamma \left( { - x - y - n - z} \right)\Gamma \left( {y + n + z + 1} \right)}}
{{\Gamma \left( { - x - y - n} \right)\Gamma \left( {y + n + 1} \right)\Gamma \left( { - x - n - z} \right)\Gamma \left( {n + 1 + z} \right)}} = \hfill \\
= \frac{{\left( { - x - 1 - n} \right)^{\,\underline {\,y\,} } \,n^{\,\underline {\, - \,\,y\,} } }}
{{\left( { - x - 1 - n - z} \right)^{\,\underline {\,y\,} } \;\left( {n + z} \right)^{\,\underline {\, - \,y\,} } }} \hfill \\
\end{gathered} \tag{3}
} $$
---- Addendum -------
Identity (1) is famous because it is one of the very few involving the product of more than two binomials,
and can be found in good collections of binomial identities .
A hint on how it could be related to binomial expansion can be got from considering
how it works for the last step, i.e.
$$
\begin{array}{l}
\left( {1 + x} \right)^{\,r} \left( {1 + y} \right)^{\,s} = \sum\limits_{\left( {0\, \le } \right)\,m} {\sum\limits_{\left( {0\, \le } \right)\,n} {\left( \begin{array}{c}
r \\
m \\
\end{array} \right)\left( \begin{array}{c}
s \\
n \\
\end{array} \right)x^{\,m} y^{\,n} } } = \\
= \sum\limits_{\left( {0\, \le } \right)\,m} {\;\sum\limits_{\left( {0\, \le } \right)\,n} {\sum\limits_{\left( {0\, \le } \right)\;j\,\left( { \le \,n} \right)} {\left( \begin{array}{c}
m - r + s \\
j \\
\end{array} \right)\left( \begin{array}{c}
r - m \\
n - j \\
\end{array} \right)z^{\,j} y^{\,n} \left( \begin{array}{c}
r \\
m \\
\end{array} \right)x^{\,m} } } } = \\
= \sum\limits_{\left( {0\, \le } \right)\,m} {\;\left( {1 + yz} \right)^{\,m - r + s} \left( {1 + y} \right)^{\,r - m} \left( \begin{array}{c}
r \\
m \\
\end{array} \right)x^{\,m} } = \\
= \left( {1 + yz} \right)^s \sum\limits_{\left( {0\, \le } \right)\,m} {\left( {\frac{{1 + y}}{{1 + yz}}} \right)^{\,r - m} \left( \begin{array}{c}
r \\
m \\
\end{array} \right)x^{\,m} } = \\
= \left( {1 + yz} \right)^s \left( {\frac{{1 + y}}{{1 + yz}} + x} \right)^r \\
\end{array}
$$
and then put $z=1$.
Note: the bounds put in brackets are to indicate that, from an algebraic point of view, they are superfluous because inherent to the binomial coeff. (which is null outside them); this is important to carry on in order to correctly apply the convolution.
A combinatorial proof can be found in this interesting paper
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2155025",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 0
} |
What is the sum of the solutions to $6x^3+7x^2-x-2=0$ What is the easy way to solve the problem?
The sum of the solutions to $6x^3+7x^2-x-2=0$ is:
$$A) \ \frac{1}{6}$$ $$B) \ \frac{1}{3}$$ $$C) \ \frac{-7}{6}$$ $$D) -2$$ $$E) \text{ none of above}$$
| For a cubic equation of the form $ax^3 + bx^2 + cx +d$, the sum of all the roots is given by $\frac{-b}{a}$. Here, $b=7,a=6$ so sum of the roots $= \frac{-7}{6}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2155713",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
If $a^2+b^2+c^2+d^2+e^2=5$ so $\sum\limits_{cyc}\frac{1}{7-2a}\leq1$.
Let $a$, $b$, $c$, $d$ and $d$ be non-negative numbers such that $a^2+b^2+c^2+d^2+e^2=5$. Prove that:
$$\frac{1}{7-2a}+\frac{1}{7-2b}+\frac{1}{7-2c}+\frac{1}{7-2d}+\frac{1}{7-2e}\leq1$$
The equality occurs also for $a=2$ and $b=c=d=e=\frac{1}{2}$.
I tried TL:
$$1-\sum_{cyc}\frac{1}{7-2a}=\sum_{cyc}\left(\frac{1}{5}-\frac{1}{7-2a}\right)=\frac{2}{5}\sum_{cyc}\frac{1-a}{7-2a}=$$
$$=\frac{2}{5}\sum_{cyc}\left(\frac{1-a}{7-2a}+\frac{a^2-1}{10}\right)=\frac{1}{25}\sum_{cyc}\frac{(a-1)^2(3-2a)}{7-2a}$$
and I don't see what is the rest.
| We must to prove that $f(a_1)+f(a_2)+f(a_3)+f(a_4)+f(a_5) \le 1$ ,
for non-negative $a_1+a_2+a_3+a_4+a_5=5$
$$f(x)=\dfrac{1}{7-2\sqrt{x}}$$
Since $f''(x)=\dfrac{(7-6\sqrt{x})}{2(2x-7\sqrt{x})^3}$ , we only need to consider the inequality in case $0< a_1=a_2=a_3=a_4=t^2 \le 1 \ , \ a_5=5-4t^2$
Clearly
$g(t)=\dfrac{4}{7-2t}+\dfrac{1}{7-2\sqrt{5-4t^2}} $
$ g'(t)=\dfrac{8}{(7-2t)^2}-\dfrac{8t}{\sqrt{5-4t^2}(7-2\sqrt{5-4t^2})^2}\ge 0 \Leftrightarrow $
$\sqrt{5-4t^2}\cdot(69-16t^2) \ge 4t^3-140t^2+49t+140>0 \Leftrightarrow$
$(t-1)(2t-1)(t+1)(104t^3-60t^2+1062t-841) \le 0 \ ,\ t\in [0,1]$
So maximum of $g(t)$ is attained at $t_1=\dfrac{1}{2}$ and $t_2=1$.
$$g(t_1)=g(t_2)=1$$
Equality holdes for : $(a=b=c=d=e=1) $ and $\left( a=b=c=d=\dfrac{1}{2}, e=2\right)$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2157579",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Which of the following numbers is greater? Which of the following numbers is greater?
Without using a calculator and logarithm.
$$7^{55} ,5^{72}$$
My try
$$A=\frac{7^{55} }{5^{55}×5^{17}}=\frac{ 7^{55}}{5^{55}}×\frac{1}{5^{17}}= \left(\frac{7}{5}\right)^{55} \left(\frac{1}{5}\right)^{17}$$
What now?
| Note: $7^2<2\cdot 5^2$ and $5>2^2$
$7^{55}<7\cdot 5^{54}\cdot 2^{27}<5^{55}\cdot 2^{28}<5^{69}<5^{72}$ as required
With an extra jink into factors of $3$, we can show $7^{55}<5^{67}$
Extra notes: $3^3>5^2$ and $5^5>3\cdot2^{10}$
$7^{55}<7\cdot 5^{54}\cdot 2^{27}<5^{54}\cdot 2^{30}<5^{52}\cdot 2^{30}\cdot 3^{3}<5^{67}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2158143",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
"answer_count": 6,
"answer_id": 1
} |
What will happen to the roots of $ax^2 + bx + c = 0$ if the $a \to 0$? Exercise:
What will happen to the roots of the quadratic equation
$$ax^2 + bx + c = 0$$
if the coefficient $a$ approaches zero while the coefficients $b$ and $c$ are constant, and $b \neq 0$?
Attempt:
$\lim\limits_{a \to 0}{(ax^2 + bx + c)} = bx + c = 0 \longrightarrow x = -\frac{c}{b}$
However, I don't think my solution is complete; shouldn't I end up with $2$ roots? (I've only found $1$.)
Request:
Is there indeed another root to find? If so, how to I find it?
| If $b=0$, then the roots are (if real), $\pm\sqrt{-c/a}$. If $c\ne0$, both roots tend to infinity (positive and negative).
If $b\ne0$, it's not restrictive to assume $b>0$ (otherwise multiply by $-1$).
If $c=0$, the roots are $0$ and $-b/a$, the latter tending to infinity ($\infty$ if $a$ approaches $0$ from the negative side, $-\infty$ if $a$ approaches $0$ from the positive side).
Assume $c\ne0$. The discriminant will be positive when $a$ belongs to a suitable (punctured) neighborhood of $0$, namely for $0<|a|<b^2/|4c|$, so we have two roots.
Now we can rationalize the expression for the roots.
First root:
$$
\frac{-b+\sqrt{b^2-4ac}}{2a}=
\frac{b^2-b^2+4ac}{2a(-b-\sqrt{b^2-4ac})}=
-\frac{2c}{\sqrt{b^2-4ac}+b}
$$
For $a\to0$ this has limit $-c/b$. This is to be expected, because the polynomial will become $bx+c$ for $a=0$.
Second root:
$$
\frac{-b-\sqrt{b^2-4ac}}{2a}=
\frac{b^2-b^2+4ac}{2a(-b+\sqrt{b^2-4ac})}=
\frac{2c}{\sqrt{b^2-4ac}-b}
$$
If $a\to0$, then this has limit $\pm\infty$ (according to the sign of $c$).
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Inequality for positive real numbers less than $1$: $8(abcd+1)>(a+1)(b+1)(c+1)(d+1)$ If $a,b,c,d$ are positive real numbers, each less than 1, prove that the following inequality holds:
$$8(abcd+1)>(a+1)(b+1)(c+1)(d+1).$$
I tried using $\text{AM} > \text{GM}$, but I could not prove it.
| First, note that $2^0(a+1) \ge a + 1$.
Then we prove by induction that $2^{k-1}(x_1\dots x_k+1) > \prod_{i=1}^k(x_i + 1)$. For the ease of notation, there I will focus on the case $k = 4$.
$$(d+1)(a+1)(b+1)(c+1)\le 4(d+1)(abc+1)< 4(abcd+abc+1+d)< 8(abcd+1)$$
The last inequality is provided by
$$abc+d< abcd+1 \Leftarrow Q+d<Qd+1 \Leftarrow (Q-1)(d-1)>0,$$
where $0 < Q = abc < 1.$
| {
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Let $a,b,c\in \Bbb R^+$ such that $(1+a+b+c)(1+\frac{1}{a}+\frac{1}{b}+\frac{1}{c})=16$. Find $(a+b+c)$ Let $a,b,c\in \Bbb R^+$ such that $(1+a+b+c)(1+\frac{1}{a}+\frac{1}{b}+\frac{1}{c})=16$.
Find $(a+b+c)$.
I computed the whole product ;If $(a+b+c)=x\implies (1+x)(1+\frac{bc+ca+ab}{abc})=16$. Unable to view how to proceed further.
Please help.
| By $AM \ge GM$ inequality,$$(1+a+b+c)\left(1+\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\ge \left(4\sqrt[4]{abc}\right)\left(4\sqrt[4]{\frac{1}{abc}}\right)=16$$and equality holds when $1=a=b=c$.
| {
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equation $x^4+ax^3-6x^2+ax+1 = 0$ has two distinct positive roots Finding parameter of $a$ for which equation $x^4+ax^3-6x^2+ax+1 = 0$
has two distinct positive roots
Attempt: writing equation $\displaystyle \bigg(x^2+\frac{1}{x^2}\bigg)+a\bigg(x+\frac{1}{x}\bigg)-6=0\;,$ where $x\neq 0$
So $\displaystyle \bigg(x+\frac{1}{x}\bigg)^2+a\bigg(x+\frac{1}{x}\bigg)-8=0$
put $\displaystyle \bigg(x+\frac{1}{x}\bigg)=t\;,$ where $|t|\geq 2$
so $t^2+at-8=0.$ So for real roots $D\geq 0.$ So $a^2+32\geq 0$
So $\displaystyle t = \frac{-a\pm \sqrt{a^2+32}}{2}.$
could some help me how to solve it, thanks
| With $\displaystyle \bigg(x+\frac{1}{x}\bigg)=t$ the roots occur when $t^2+at-8=0$.
$f(x)=x^4+ax^3-6x^2+ax+1 = 0$ shows $f(0)=1>0$ and $\displaystyle \lim_{x\to+\infty} f=+\infty$, which show $f$ has two distinct positive roots if there is $x_0>0$ for which $f(x_0)<0$. Let $\displaystyle \bigg(x_0+\frac{1}{x_0}\bigg)=t_0\geq2$ and $f(x_0)=t_0^2+at_0-8<0$ or
$$4+2a-8\leq t_0^2+at_0-8<0$$
Shows $a<2$.
| {
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Calculate the following limit I need to evaluate the following limit, however, in doing so, I let $\sqrt{1+c}=1$ which I came to undertand that it's not valid. My procedure was the following:
$$\lim_{c\to0}\left(-\ln(c)\sqrt{1+c}-\ln\left(\frac{1+\sqrt{1+c}}{1-\sqrt{1+c}}\right)+\ln\left(\frac{1+\sqrt2}{1-\sqrt2}\right)\right)$$ $$=\lim_{c\to0}\ln\left(\frac{\left(1-\sqrt{1+c}\right)\left(1+\sqrt2\right)}{c\left(1+\sqrt{1+c}\right)\left(1-\sqrt2\right)}\right)=\ln\left(\lim_{c\to0}\frac{\left(1-\sqrt{1+c}\right)\left(1+\sqrt2\right)}{c\left(1+\sqrt{1+c}\right)\left(1-\sqrt2\right)}\right)$$
$$$$We can use L'Hôspital's rule on the fraction to obtain the simplified version:
$$\ln\left(\lim_{c\to0}\left(\frac{3+2\sqrt2}{2+3c+2\sqrt{1+c}}\right)\right)=\boxed{\ln\left(\frac{3+2\sqrt2}4\right)}$$
The answer that I obtained matches with the answer of Wolfram Alpha even though there is a mistake. If I don't let $\sqrt{1+c}=1$ the limit becomes really difficult and I don't know how to solve it. Does someone have any ideas on how to solve it? And why is my answer correct even though the procedure is not?
| You can write this as
$$-\ln(c)\sqrt{1+c}-\ln\left(\frac{1+\sqrt{1+c}}{1-\sqrt{1+c}}\right)+\ln\left(\frac{1+\sqrt2}{1-\sqrt2}\right) \\
= -\ln(c)\sqrt{1+c} + \ln c- \ln c -\ln\left(\frac{1+\sqrt{1+c}}{1-\sqrt{1+c}}\right)+\ln\left(\frac{1+\sqrt2}{1-\sqrt2}\right) \\ = \ln c ( 1 - \sqrt{1 + c}) + A(c) $$
where
$$A(c) = \ln\left(\frac{\left(1-\sqrt{1+c}\right)\left(1+\sqrt2\right)}{c\left(1+\sqrt{1+c}\right)\left(1-\sqrt2\right)}\right).$$
As you have shown
$$\lim_{c \to 0} A(c) = \ln\left(\frac{3+2\sqrt2}4\right)$$
You will get the correct limit for the original expression now because $\lim_{c \to 0} \ln c ( 1 - \sqrt{1 + c}) = 0$
| {
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Let X~geometric (1/3), and let Y=|X-5|. Find the range n and PMF of Y. Let X~geometric (1/3), and let Y=|X-5|. Find the range n and PMF of Y.
Here is my trial
If $x=0$, $P(Y=|0-5|)=P(Y=5)=\left(\frac{2}{3}\right)^5 \frac{1}{3}$
If $x=1$, $P(Y=|1-5|)=P(Y=4)=\left(\frac{2}{3}\right)^4 \frac{1}{3}$
If $x=2$, $P(Y=|2-5|)=P(Y=3)=\left(\frac{2}{3}\right)^3 \frac{1}{3}$
If $x=3$, $P(Y=|3-5|)=P(Y=2)=\left(\frac{2}{3}\right)^2 \frac{1}{3}$
If $x=4$, $P(Y=|4-5|)=P(Y=1)=\left(\frac{2}{3}\right)^1 \frac{1}{3}$
If $x=5$, $P(Y=|5-5|)=P(Y=0)= \frac{1}{3}$
Range of $Y={0,1,2,3,4,5}$
and PMF is I $P(Y=y)=\left(\frac{2}{3}\right)^y \frac{1}{3}$ for $y=0,1,2,3,4,5$
Am I correct. If not please correct me.
| Range of $Y$ is $(0,\infty)$.
$$P(Y=k) \;=\; (2/3)^{k-1} * (1/3).$$
| {
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Calculate the image and a basis of the image (matrix)
What's the image of the matrix? What's the basis of the image?
$M=\begin{pmatrix}
-1 & 1 & 1\\
-2 & -3 & 6\\ 0 & -1 & 1 \end{pmatrix}$
First transposed the matrix:
$M^{T}=\begin{pmatrix}
-1 & -2 & 0\\
1 & -3 & -1\\
1 & 6 & 1
\end{pmatrix}$
Now we use Gauss and get zero lines. Take the first line and add it to the third:
$M^{T}=\begin{pmatrix}
-1 & -2 & 0\\
0 & -5 & -1\\
1 & 6 & 1
\end{pmatrix}$
Take the first line and add it to the third:
$M^{T}=\begin{pmatrix}
-1 & -2 & 0\\
0 & -5 & -1\\
0 & 4 & 1
\end{pmatrix}$
Multiply the second line with $4$, multiply the third line with $5$, then add second line to third:
$M^{T}=\begin{pmatrix}
-1 & -2 & 0\\
0 & -20 & -4\\
0 & 0 & 1
\end{pmatrix}$
Transpose back:
$M=\begin{pmatrix}
-1 & 0 & 0\\
-2 & -20 & 0\\
0 & -4 & 1
\end{pmatrix}$
The image of the matrix is $\text{Im(M)}= \text{span} \left ( \left\{ \begin{pmatrix}
-1\\
-2\\
0
\end{pmatrix}, \begin{pmatrix}
0\\
-20\\
4
\end{pmatrix},\begin{pmatrix}
0\\
0\\
1
\end{pmatrix} \right\} \right)$
The basis of the image is $\left\{ \begin{pmatrix}
-1\\
-2\\
0
\end{pmatrix}, \begin{pmatrix}
0\\
-20\\
4
\end{pmatrix},\begin{pmatrix}
0\\
0\\
1
\end{pmatrix} \right\}$
Please tell me if I did everything correctly? It's very important for me to know as I would do it like that in the exam :)
I hope it's correct and please also tell me if the notation is.
| The image of a matrix is the same as its column space. To find column space, you first find the row echelon form of the given matrix (do not transpose it). The definition of row-echelon form is:
*
*Rows with all zero's are below any nonzero rows
*The leading entry in each nonzero row is a one
*All entries below each leading "1" are zero
With the matrix in row-echelon form, the image (and column space) basis of the matrix comprises of the columns that contain a leading 1.
It is also useful to note that the dimensions (dim) of im(M) = dim(colM) = rank of M
| {
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Prove that $\cos 20^{\circ} + \cos 100^{\circ} + \cos {140^{\circ}} = 0$ Assume $A = \cos 20^{\circ} + \cos 100^{\circ} + \cos 140^{\circ}$ . Prove that value of $A$ is zero.
My try : $A = 2\cos 60^{\circ} \cos 40^{\circ} + \cos 140^{\circ}$ and I'm stuck here
| $\cos 20^{\circ} + \cos 100^{\circ} = \cos (60^{\circ}-40^{\circ}) + \cos (60^{\circ}+40^{\circ}) = \cos 40^{\circ}$ by using the addition formulae for $\cos$.
Then $\cos 140^{\circ} = \cos (180^{\circ} - 40^{\circ}) = -\cos (40^{\circ})$. So $$\cos 20^{\circ} + \cos 100^{\circ} + \cos 140^{\circ} = \cos 40^{\circ} - \cos 40^{\circ} = 0$$
| {
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"timestamp": "2023-03-29T00:00:00",
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$\lim_{x\to0}\,(a^x+b^x-c^x)^{\frac1x}$
Given $a>b>c>0$, calculate$\displaystyle\,\,\lim_{x\to0}\,(a^x+b^x-c^x)^{\frac1x}\,$
I tried doing some algebraic manipulations and squeeze it, but couldn't get much further.
| $$\lim_{x\to0}(a^x+b^x-c^x)^\frac{1}{x}=[1^\infty]=\exp\lim_{x\to 0}(a^x+b^x-c^x-1)\frac{1}{x}\boxed=\\(a^x+b^x-c^x-1)\frac{1}{x}=(a^x-c^x+b^x-1)\frac{1}{x}=c^x\cdot\frac{\left(\frac{a}{c}\right)^x-1}{x}+\frac{b^x-1}{x}\\ \boxed =\exp \lim_{x\to 0}\left(c^x\cdot\frac{\left(\frac{a}{c}\right)^x-1}{x}+\frac{b^x-1}{x} \right)=\exp\left(\ln\frac{a}{c}+\ln b\right)=\exp\left( \ln \frac{ab}{c} \right)=e^\frac{ab}{c}$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Two different expansions of $\frac{z}{1-z}$ This is exercise 21 of Chapter 1 from Stein and Shakarchi's Complex Analysis.
Show that for $|z|<1$ one has $$\frac{z}{1-z^2}+\frac{z^2}{1-z^4}+\cdots +\frac{z^{2^n}}{1-z^{2^{n+1}}}+\cdots =\frac{z}{1-z}$$and
$$\frac{z}{1+z}+\frac{2z^2}{1+z^2}+\cdots \frac{2^k z^{2^k}}{1+z^{2^k}}+\cdots =\frac{z}{1-z}.$$
Justify any change in the order of summation.
[Hint: Use the dyadic expansion of an integer and the fact that $2^{k+1}-1=1+2+2^2+\cdots +2^k$.]
I don't really know how to work this through. I know that $\frac{z}{1-z}=\sum_{n=1}^\infty z^n$ and each $n$ can be represented as a dyadic expansion, but I don't know how to progress from here. Any hints solutions or suggestions would be appreciated.
| Since minimalrho has explained how to proceed with the given hint, I'll give an alternative method. The $k$th summand of the first series can be written
$$\frac{z^{2^k}}{1 - z^{2^{k}}} - \frac{z^{2^{k+1}}}{1-z^{2^{k+1}}}$$
and the $k$th summand of the second series can be written
$$\frac{2^kz^{2^k}}{1 - z^{2^k}} - \frac{2^{k+1}z^{2^{k+1}}}{1-z^{2^{k+1}}}$$
Hence, the $N$th partial sums of the two series telescope to
$$\frac{z}{1 - z} - \frac{z^{2^{N+1}}}{1 - z^{2^{N+1}}}\quad \text{and}\quad \frac{z}{1 - z} - \frac{2^{N+1}z^{2^{N+1}}}{1 - z^{2^{N+1}}}$$
respectively. Using the condition $\lvert z\rvert < 1$, argue that $z^{2^{N+1}}/(1 - z^{2^{N+1}})$ and $2^{N+1}z^{2^{N+1}}/(1 - z^{2^{N+1}})$ tend to $0$ as $N\to \infty$. Then the results follow.
| {
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Find all possible positive integers $n$ such that $3^{n-1} + 5^{n-1} \mid 3^n + 5^n $. Proof explanation Question: Find all possible positive integers $n$ such that $3^{n-1} + 5^{n-1} \mid 3^n + 5^n $.
Solution: Let's suppose that $3^{n-1}+5^{n-1}\mid 3^n+5^n$, so there is some positive integer $k$ such that $3^n+5^n=k(3^{n-1}+5^{n-1})$. Now, if $k\ge 5$ we have $$k(3^{n-1}+5^{n-1})\ge 5(3^{n-1}+5^{n-1})=5\cdot 3^{n-1}+5\cdot 5^{n-1}>3^n+5^n.$$
This means that $k\le 4$. On the other hand, $3^n+5^n=3\cdot 3^{n-1}+5\cdot 5^{n-1}>3(3^{n-1}+5^{n-1})$, then $k\ge 4$ and thus we deduce that $k=4$. In this case we have $3^n+5^n=4(3^{n-1}+5^{n-1})$, which gives us the equation $5^{n-1}=3^{n-1}$, but if $n>1$ this equation is impossible.
Hence $n=1$
The above solution is from this answer. Could someone please explain a few things:
*
*why was the value of $k\le4$ and $k\ge4$ picked? where did they get 4?
*why does $3\cdot 3^{n-1}+5\cdot 5^{n-1}>3(3^{n-1}+5^{n-1})$ mean that $k\ge4$ ?
*Why does $5\cdot 3^{n-1}+5\cdot 5^{n-1}>3^n+5^n.$ mean that $k\le4$ ?
| Perhaps an easier way to look at it: the equation
$$3^n+5^n=k(3^{n-1}+5^{n-1})$$
can be rewritten
$$5^{n-1}(5-k)=3^{n-1}(k-3)\ .$$
Therefore $5-k$ and $k-3$ must have the same sign. They can't both be negative as then $k$ would be simultaneously less than $3$ and greater than $5$, and they clearly can't both be zero, so they must both be positive. Hence
$$k<5\quad\hbox{and}\quad k>3\ ,$$
so $k=4$. This makes our second equation
$$5^{n-1}=3^{n-1}\ ,$$
and this is true if and only if $n-1=0$.
| {
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Solution of a trigonometric equation involving double sines,cosines What is the sum of all the solutions of $$\sin^2 (2\sin (x-\frac {\pi}{6}))+\sec^2 (x-\frac {\pi}{2}\tan^2 (x))=1$$ in $[0-4\pi]$ writing $\sec (..)=\frac {1}{\cos (..)} $ .. and rearranging we have $\cos (a).\cos (b)=1$ now as $\cos $ is $-1\leq \cos^2 (..)\leq 1$ . So both a and b have to be $n\pi $ thus for bracket of $\sin (..) $ the solution will be $\frac {\pi}6,\frac {7\pi}{6},2\pi+\frac {\pi}{6},2\pi+\frac {7\pi}{6} $ but I dont know what to do of the next bracket involving $x,\tan (x) $ . Thanks
| We have $$\sin^2\left(2\sin\left(x-\dfrac\pi6\right)\right)+\tan^2\left(x-\dfrac\pi2\tan^2x\right)=0$$
For real $x,$
$\sin\left(2\sin\left(x-\dfrac\pi6\right)\right)=\tan\left(x-\dfrac\pi2\tan^2x\right)=0$
$\implies 2\sin\left(x-\dfrac\pi6\right)=m\pi\ \ \ \ (1)$ where $m$ is any integer
Now as $-1\le\sin y\le1$
$-1\le\dfrac{m\pi}2\le1\implies m=0$
$\implies\sin\left(x-\dfrac\pi6\right)=0\implies x-\dfrac\pi6=r\pi$ where $r$ is any integer
and $x-\dfrac\pi2\tan^2x=n\pi\ \ \ \ (2)$ where $n$ is any integer
$\implies \dfrac\pi6+r\pi-n\pi=\dfrac\pi2\tan^2\left(\dfrac\pi6+r\pi\right)$
$\iff\dfrac16+r-n=\dfrac16\implies r=n$
| {
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A hard integral Looking for a solution for an integral:
$$I(k)=\int_0^{\infty } \frac{e^{-\frac{(\log (u)-k)^2}{2 s^2}}}{\sqrt{2 \pi } s \left(1+u\right)} \, du .$$
So far I tried substitutions and by parts to no avail.
| Here is a start: $I(0) = \frac{1}{2}$
Proof:
$$I(0) = \int\limits_0^\infty \frac{\exp\left[-\frac{(\log u)^2}{2s^2}\right]}{\sqrt{2\pi} s (1+u)} \rm{d}u$$
Put $\log u = x$
\begin{align}
I(0) &= \int\limits_{-\infty}^\infty \frac{\exp\left[-\frac{x^2}{2s^2}\right]}{\sqrt{2\pi} s} \frac{e^x}{1+e^x} \rm{d}x \\
&= \int\limits_{-\infty}^\infty \frac{\exp\left[-\frac{x^2}{2s^2}\right]}{\sqrt{2\pi} s}\rm{d}x - \int\limits_{-\infty}^\infty \frac{\exp\left[-\frac{x^2}{2s^2}\right]}{\sqrt{2\pi} s} \frac{1}{1+e^x} \rm{d}x
\end{align}
The first integral is $1$. Call the second integral $K$.
$$K=\int\limits_{-\infty}^\infty \frac{\exp\left[-\frac{x^2}{2s^2}\right]}{\sqrt{2\pi} s} \frac{1}{1+e^x} \rm{d}x$$
Flipping the range around $0$,
$$K=\int\limits_{-\infty}^\infty \frac{\exp\left[-\frac{x^2}{2s^2}\right]}{\sqrt{2\pi} s} \frac{1}{1+e^{-x}} \rm{d}x$$
Now take the average of the two expressions,
\begin{align}
K &=\frac{1}{2}\int\limits_{-\infty}^\infty \frac{\exp\left[-\frac{x^2}{2s^2}\right]}{\sqrt{2\pi} s} \left[\frac{1}{1+e^x}+\frac{1}{1+e^{-x}}\right] \rm{d}x\\
&=\frac{1}{2}\int\limits_{-\infty}^\infty \frac{\exp\left[-\frac{x^2}{2s^2}\right]}{\sqrt{2\pi} s}\rm{d}x\\
&=\frac{1}{2}\\
I(0) &= 1 - K = \frac{1}{2}
\end{align}
| {
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Substitution or comparison? For the question,
$X^4 + \frac{9}{X^4} = (X^2 - \frac{a}{X^2})^2 + b$ I have to find the values of $a$ and $b$. I tried two solutions both included the expansion of brackets.
By comparing terms I obtained $3$ for $a$ and $6$ for $b$.
However when I tried to substitute using $b$, I obtained a value of $0$ for $b$.
Why is it not possible to solve by substituting and on a more general tone when is it possible to solve by substituting?
| $X^4 + \frac{9}{X^4} = (X^2 - \frac{a}{X^2})^2 + b$
$x^4 + \frac {9}{x^4} = x^4 + \frac {a^2}{x^4} + b - 2a$
So by comparing $a^2 = 9$ and $b-2a = 0$.
So $a = 3$ or $a = -3$ and $b = 6$ or $b = -6$.
I'm not sure what you mean by substituting? Do you mean picking an arbitrary value for $x$ and getting two equations for $a$ and $b$.
Let $x = 1$. then $10 = (1-a)^2 + b = 1 -2a + a^2 +b$ or $b= 9 + 2a -a^2$
if $x = \sqrt{2}$ then $4 +\frac 94 = (2- \frac a2)^2 +b = 4-2a + \frac{a^2}4 +b$ or $b=\frac 94 + 2a -\frac {a^2}4$
So $9-a^2 = \frac 94 - \frac {a^2}4 = \frac 14( 9 - a^2)$ so $9-a^2 = 0$ so $a = 3$ or $-3$.... and $b = 6$ or $-6$.
So how did you get $b= 0$? Perhaps if we knew how you got that we can figure out what went wrong. Did you divide by zero? Did you get an redundant equation $0 = 0$ and mistake that for $b = 0$?
If $x = 0$ you get $0 + \frac 90 = (0 + \frac a0)^2 +b$ but this doesn't help us as dividing by zero is an absolute forbidden no-no that makes no sense. If you make the error and assume $\frac 90 = 0$ (It does !!!!!!!!NOT!!!!!!!) then you get $0 = 0 + b$ (which is !!!!!!!!WRONG!!!!!!)
Of if you did $x = k$ and $x = -k$ you will end up with the redundant two equations: $k^4 + \frac 9{k^4} = (k^2 - \frac a{k^2})^2 + b$ repeated twice which gets us no where.
=====
from the comments it sounds like what you did was
$X^4 + \frac{9}{X^4} = (X^2 - \frac{a}{X^2})^2 + b$
$b = X^4 + \frac{9}{X^4}-(X^2 - \frac{a}{X^2})^2$ and then plug that back into the original
$X^4 + \frac{9}{X^4} = (X^2 - \frac{a}{X^2})^2+ X^4 + \frac{9}{X^4}-(X^2 - \frac{a}{X^2})^2$
$X^4 + \frac{9}{X^4}=X^4 + \frac{9}{X^4}$
$0 = 0$
Which is true. Zero does equal 0. But that doesn't help us in the least as it says nothing about $a$ or $ b$ or $x$.
If this is what you mean by "substitution" it will NEVER work. You are simply defining something in terms of an expression and trying to solve the same expression with no new information and you simply eliminate everything in circular reasoning.
It's a bit like trying to do "Snoopy = Charlie Brown's dog. Solve for Charlie Brown"
Charlie Brown = Snoopy's owner
Snoopy = Snoopy's owner's dog
Snoopy = Snoopy
Snoopy - Snoopy = Snoopy - Snoopy
0 = 0
Therefore Charlie Brown is nobody!
Hopefully you can see why that is wrong and how it will never work.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2181952",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Does $\int _0^{\infty }\:\frac{1}{1+x^2\left(\sin x\right)^2}\ \operatorname dx$ converge? I have been trying to prove the following integral:
$$\int _0^{\infty }\:\frac{1}{1+x^2\left(\sin x\right)^2}\ dx$$
diverges (please correct me if I am mistaken).
I have tried to use different comparison tests (as this is an integral of a positive function) with no success.
Any ideas?
| The idea is to bound the integral below on intervals where $\displaystyle \frac{1}{1+x^2\left(\sin x\right)^2}$ has spikes, that is to say, it suffices to find some $\varepsilon_k$ such that $$\sum_{k\geq1}\int_{k\pi -\varepsilon_k}^{k\pi +\varepsilon_k}\frac{1}{1+x^2\left(\sin x\right)^2}dx$$ diverges.
On each of these intervals, since $\sin^2(x)$ is $\pi$-periodic, $1+x^2\sin^2(x)\leq 1+(k\pi + \varepsilon_k)^2\sin^2(\varepsilon_k)$, hence $$\sum_{k=1}^N\int_{k\pi -\varepsilon_k}^{k\pi +\varepsilon_k}\frac{1}{1+x^2\left(\sin x\right)^2}dx \geq \sum_{k=1}^N \frac{2 \varepsilon_k}{1+(k\pi + \varepsilon_k)^2\sin^2(\varepsilon_k)}$$
Some rough asymptotics suggest $$\frac{2 \varepsilon_k}{1+(k\pi + \varepsilon_k)^2\sin^2(\varepsilon_k)}\sim \frac{2 \varepsilon_k}{\pi^2k^2\epsilon^2_k}=\frac 2\pi \frac{1}{k^2\varepsilon_k} $$
Setting $\varepsilon_k=\frac 1k$ seems therefore like a sound idea, since we would get something like the harmonic series, which diverges.
Indeed, $$\begin{align}\frac{ \frac 2k}{1+(k\pi + \frac 1k)^2\sin^2(\frac 1k)}&=\frac 2k \frac{1}{1+(\pi^2k^2+2\pi +\frac{1}{k^2})(\frac{1}{k^2}+o(\frac{1}{k^2}))} \\
&=\frac 2k \frac{1}{1+\pi^2 + o(1)}\\
&\sim \frac{2}{1+\pi^2}\frac{1}k
\end{align}$$
With this choice of $\varepsilon_k$, $\displaystyle \sum_{k\geq 1}^\infty \frac{2 \varepsilon_k}{1+(k\pi + \varepsilon_k)^2\sin^2(\varepsilon_k)}$ diverges, which concludes the proof.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Minimize $A=\frac{1+2^{x+y}}{1+4^x}+\frac{1+2^{x+y}}{1+4^y}$ For $a,b>0$. Minimize $$A=\frac{1+2^{x+y}}{1+4^x}+\frac{1+2^{x+y}}{1+4^y}$$
i think we let $2^x=a;2^y=b$
Hence $A=\frac{1+ab}{1+a^2}+\frac{1+ab}{1+b^2}$
We need pro $A\geq 2$(Wolfram Alpha) but $x,y$ is a very odd number and i can't find how to prove it $\geq2$
| we have to prove that $$(1+ab)\left(\frac{1}{1+a^2}+\frac{1}{1+b^2}\right)\geq 2$$ and this is equivalent to $${\frac { \left( ab-1 \right) \left( a-b \right) ^{2}}{ \left( {a}^{2}
+1 \right) \left( {b}^{2}+1 \right) }}
\geq 0$$ this is right if $$ab\geq 1$$
| {
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"source": "stackexchange",
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maximum number of number of roots of $p(x) = 0$ is
Let $p(x)=x^6+ax^5+bx^4+x^3+bx^2+ax+1.$ Given that $x=1$ is a root of $p$ but $x=-1$ is not, find the maximum number of number of roots of $p$.
My attempt:
$x=0$ in not a root of $p(x)=0.$ So
$$\left(x^3+\frac{1}{x^3}\right)+a\left(x^2+\frac{1}{x^2}\right)+b\left(x+\frac{1}{x}\right)+1=0$$
So
$$\displaystyle \left(x+\frac{1}{x}\right)^3-3\left(x+\frac{1}{x}\right)+a\left(x+\frac{1}{x}\right)^2-2+b\left(x+\frac{1}{x}\right)+1=0$$
So
$$ t^3+at^2+(b-3)t+1=0,$$
where
$$\left(x+\frac{1}{x}\right) = t \quad\text{ and }\quad |t|\geq 2.$$
Given $x=1$ is a root of $p$, $1+a+b+1+a+b+1=0$, so
$$\displaystyle a+b = -\frac{3}{2}$$
so
$$t^3+at^2+\left(-\frac{9}{2}-a\right)t+(1-2a) = 0.$$
$t=2$ is a root, so
$$(t-2)\bigg[t^2+(a+2)t+\left(\frac{2a-1}{2}\right)\bigg]=0,$$
So discriminant of above quadratic equation is $\displaystyle D = a^2+6>0$. So above equation has $2$ distinct real roots.
But I did not know how I use $x=-1$ is not a root. Could some help me to solve it? Thanks.
| This answer assumes that you want to find the maximum number of the real roots of $p(x)$.
You already have
$$(t-2)\left(t^2+(a+2)t+a-\frac 12\right)=0$$
where $t=x+\frac 1x$ (which is correct though you have a typo in the part $a(x^2+\frac{1}{x^2})=a(x+\frac 1x)^2-2\color{red}{a}$).
Let $t_{\pm}$ where $t_-\lt t_+$ be the roots of $t^2+(a+2)t+a-\frac 12$ whose discriminant is $a^2+6\gt 0$.
Then, we have that
$$\text{$p(x)$ has six real roots if and only if $|t_-|\ge 2$ and $|t_+|\ge 2$}$$
Here, we have
$$\begin{align}|t_-|\ge 2&\iff \left|\frac{-a-2-\sqrt{a^2+6}}{2}\right|\ge 2\\\\&\iff \left|-a-2-\sqrt{a^2+6}\right|^2\ge 4^2\\\\&\iff (-a-2)^2+2(a+2)\sqrt{a^2+6}+a^2+6\ge 16\\\\&\iff (a+2)\sqrt{a^2+6}\ge -(a+3)(a-1)\tag2\end{align}$$
*
*For $a\gt 1$, $(2)$ holds since the LHS is non-negative and the RHS is negative.
*For $-3\le a\lt -2$, $(2)$ does not hold since the LHS is negative and the RHS is non-negative.
*For $-2\le a\le 1$, since the both sides are non-negative,$$(2)\iff (a+2)^2(a^2+6)\ge (-(a+3)(a-1))^2\iff \left(a+\frac 12\right)\left(a+\frac{5}{2}\right)\ge 0$$
*For $a\lt -3$, since the both side are negative,$$(2)\iff (-a-2)^2(a^2+6)\le (a+3)^2(a-1)^2\iff \left(a+\frac 12\right)\left(a+\frac{5}{2}\right)\le 0$$
So, we get
$$|t_-|\ge 2\iff (2)\iff a\ge -\frac 12\tag3$$
Similarly, we get
$$|t_+|\ge 2\iff a\le -\frac{5}{2}\tag4$$
Since there are no $a$ satisfying $(3)$ and $(4)$, we have that the number of the real roots of $p(x)$ is equal to or less than $4$. (Note that the number of the real roots of $p(x)$ is even since the number of the real solutions of $t=x+\frac 1x$ is even counted with multiplicity.)
By the way, for $a=-\frac 52$, we have
$$p(x)=(x-1)^4\left(\left(x+\frac 34\right)^2+\frac{7}{16}\right)$$
of which $x=-1$ is not a root.
The number of real roots of $p(x)$ for $a=-\frac 52$ is $4$.
Therefore, the maximum number of real roots of $p(x)$ is $\color{red}{4}$.
| {
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"source": "stackexchange",
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Faster way to solve a equation. Solve the equation:
$\sqrt[3]{x-2} + \sqrt[3]{x} + \sqrt[3]{x+2} = 0$
$f(x) = \sqrt[3]{x-2} + \sqrt[3]{x} + \sqrt[3]{x+2}$
Firstly I check the amount of solutions.
*
*Graph of the function starts at the bottom and ends at the top.
*The derivative is always greater than 0, so the function is always growing.
*After all of that we know that there is just 1 solution.
Now I try to get the solution:
*
*$\sqrt[3]{x-2} + \sqrt[3]{x+2} = -\sqrt[3]{x} $
*$(\sqrt[3]{x-2} + \sqrt[3]{x+2})^3 = -x $
*$x-2 + x + 2 + 3\sqrt[3]{(x-2)^2(x+2)} + 3\sqrt[3]{(x-2)(x+2)^2} = -x$
*$3x + 3\sqrt[3]{(x-2)(x+2)}(\sqrt[3]{x-2} + \sqrt[3]{x+2})= 0$
*$3x + 3\sqrt[3]{(x-2)(x+2)}(-\sqrt[3]{x})= 0$
*$3(x - \sqrt[3]{x^2-4} * \sqrt[3]{x}) = 0$
*$\sqrt[3]{x^2} * \sqrt[3]{x} - \sqrt[3]{x^2-4} * \sqrt[3]{x} = 0$
*$\sqrt[3]{x}(\sqrt[3]{x^2} - \sqrt[3]{x^2-4}) = 0 $
So the solution is $x = 0$.
I am wondering if there is a faster way to do this, without checking the amount of solutions.
| Observe that for the given function, $f(a)=-f(-a)$ for any value of $a$.
Now put $a=0$.
Thus $f(0)=-f(0)$.
=> $2f(0)=0$
=> $f(0)=0$
=> $x=0$ is a solution.
Also, $f'(x)>0$ for all real numbers $x$.
So, the function is always increasing and thus $x=0$ is the only solution.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2192212",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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show that for $n=1,2,...,$ the number $1+1/2+1/3+...+1/n-\ln(n)$ is positive show that for $n=1,2,...,$ the number $1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{n}-\ln(n)$ is positive, that it decreases as $n$ increases, and hence that the sequence of
these numbers converges to a limit between $0$ and $1$ (Euler's constant).
I'm trying to prove this by induction on $n$ and I made the base step, I could not with the inductive step because to do so suppose that for $n=1,2,\dots,$ it is true that $1+\frac{1}{2}+\frac{1}{3}+\dots+\frac{1}{n}-\ln(n)$ is positive and let's see that $1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{n}+\frac{1}{n+1}-\ln(n+1)$ is positive,
We see that
\begin{align}
&1+\frac{1}{2}+\frac{1}{3}+\dots+\frac{1}{n}+\frac{1}{n+1})-\ln(n+1)\\
=&1+\frac{1}{2}+\frac{1}{3}+\dots+\frac{1}{n}-\ln(n)+\frac{1}{n+1}-\ln(n+1)+\ln(n)\\
>&\frac{1}{n+1}-\ln(n+1)+\ln(n)
\end{align}
But I do not know how to prove that $\frac{1}{n+1}-\ln(n+1)+\ln(n)>0$ what do you say? Can you do what I did?
| Note that the sequence in question, let's call it $\alpha_n$, can be written as
$$\alpha_n = \Big(\sum^n_{k=1} 1/k\Big) - ln(n).$$
Note that
\begin{align}
ln(n) &:= \int^n_1 \frac {1} {t} dt \\\ &= \int_1^2 \frac {1}{t} dt + \int_2^3 \frac {1}{t} dt + \dots + \int^n_{n-1} \frac {1}{t} dt \\\
& \le (2-1) \cdot \frac {1}{1} + (3-2) \cdot \frac {1}{2} + \dots + (n-(n-1)) \cdot \frac {1}{n-1} \\\
&= 1 + \frac {1}{2} + \dots + \frac {1}{n-1}.
\end{align}
Therefore, for a fixed $n$, $\alpha_n \ge \frac {1} {n} > 0$, so the sequence is positive.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Junior olympiad question: Minimum value of 3 digit number divided by sum of its digits I recently had a maths competition where we were given this problem. I solved the question, but I narrowed down the possibilities then did more of a guess and check method. I was hoping someone else could help me get the answer to this question using a more efficient, less time consuming way. The answer that I got was 189/18 which was 10.5. I know it seems significantly easy but I just can't find the correct method for it.
Let A be a number consisting of three different nonzero digits and let B be the sum of all the three digits. Find the minimum value of A/B.
| Let the number be $\overline{abc}=100a+10b+c\,$ with digits $1 \le a,b,c \le 9\,$. Then:
$$
\begin{align}
\frac{100a+10b+c}{a+b+c} & = 1 + 9\cdot\frac{11a+b}{a+b+c} \\[3px]
& \ge 1 + 9\cdot\frac{11a+b}{a+b+\color{red}{9}} \quad\quad\quad\quad\text{(*)}\\[3px]
& = 1 + 9 + 9 \cdot \frac{10a-9}{a+b+9} \\[3px]
& \ge 1 + 9 + 9 \cdot \frac{10a-9}{a+\color{red}{8}+9} \\[3px]
& = 10 + 9 \cdot \frac{10a+170-179}{a+17} \\[3px]
& = 10 + 9 \cdot 10 - 9 \cdot \frac{179}{a+17} \\[3px]
& \ge 100 - 9 \cdot \frac{179}{\color{red}{1}+17} \\[3px]
& = \frac{189}{18}
\end{align}
$$
The minimum is attained when all the inequalities above are equalities i.e. $\overline{abc}=189\,$.
$(*)\,$ Given the condition that the digits must be different, and given that the next digit $b$ will also compete for the highest value, the alternative is to assign $c=8$ at this step, and save the digit $9$ to be assigned to $b=9$ at the next step. However, that leads to the solution $\overline{abc}=198\,$, which gives a higher ratio $198 / 18 \gt 189 / 18\,$, therefore the unique minimum is indeed attained for $\overline{abc}=189\,$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2192758",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Find the generating function or closed form for the recurrence relation $a_n = a_{n-1} + 4a_{n-2} + 2a_{n-3}$ I was trying to solve this recurrence relation using generating function
$a_n = a_{n-1} + 4a_{n-2} + 2a_{n-3} \qquad : \quad a_0 =1,a_1 =1,a_2 =5, $
I did in the following way
$
\begin{align*}
&G(x) = \sum_{n=0}^{\infty}a_n.x^n \\
&G(x) = a_0x^0 +a_1x^1+a_2x^2+\sum_{n=3}^{\infty}a_n.x^n \\
&G(x) = 1.x^0 +1.x^1+5.x^2+\sum_{n=3}^{\infty}\left ( a_{n-1} +4.a_{n-2} + 2.a_{n-3} \right ).x^n \\
&G(x) = 1.x^0 +1.x^1+5.x^2+ x.\sum_{n=2}^{\infty}a_nx^n + 4x^2.\sum_{n=1}^{\infty}a_nx^n + 2x^3.\sum_{n=0}^{\infty}a_nx^n\\
&G(x) = 1.x^0 +1.x^1+5.x^2+ x.\left [ G(x) - 1 - x \right ] + 4x^2.\left [ G(x) - 1 \right ] + 2x^3.G(x)\\
&G(x) = \frac{1}{1-x-4x^2-2x^3} \\
\end{align*}$
Now how to get the closed form in terms of $n$ after this?
If any other methods available to find the closed form please mention.
Thanks !
| Here are two variants to derive $a_n$. The first one gives a closed form, the other one an explicit expression, which results in a nice binomial identity.
First variant: Partial fractions
In case it's easy to derive the zeros of the denominator of
\begin{align*}
G(x) = \frac{1}{1-x-4x^2-2x^3}
\end{align*}
the partial fraction decomposition is a convenient method. As @J.G. indicated is $x=-1$ a zero.
Omitting some intermediary calculations we obtain
\begin{align*}
G(x)&=\frac{1}{1-x-4x^2-2x^3}\\
&=\frac{1}{1+x}-\frac{2x}{2x^2+2x-1}\\
&=\frac{1}{1+x}-\frac{x}{(x+\frac{1}{2}(1+\sqrt{3}))(x+\frac{1}{2}(1-\sqrt{3}))}\\
&=\frac{1}{1+x}-\frac{1}{2\sqrt{3}}\cdot\frac{1+\sqrt{3}}{\left(x+\frac{1}{2}+\frac{\sqrt{3}}{2}\right)}
+\frac{1}{2\sqrt{3}}\cdot\frac{1-\sqrt{3}}{\left(x+\frac{1}{2}-\frac{\sqrt{3}}{2}\right)}\\
&=\frac{1}{1+x}-\frac{1}{\sqrt{3}}\cdot\frac{1}{1-(1-\sqrt{3})x}+\frac{1}{\sqrt{3}}\cdot\frac{1}{1-(1+\sqrt{3})x}\\
&=\sum_{n=0}^\infty\left[(-1)^n-\frac{1}{\sqrt{3}}(1-\sqrt{3})^n+\frac{1}{\sqrt{3}}(1+\sqrt{3})^n\right]x^n\tag{1}
\end{align*}
Second variant: Geometric series
We can also directly apply a geometric series expansion. It is convenient to use the coefficient of operator $[x^n]$ to denote the coefficient of $x^n$ in a series and obtain
\begin{align*}
[x^n]G(x)&=[x^n]\frac{1}{1-x-4x^2-2x^3}\\
&=[x^n]\sum_{j=0}^\infty x^j(1+4x+2x^2)^j\\
&=\sum_{j=0}^n[x^{n-j}]\sum_{k=0}^j\binom{j}{k}(2x)^k(2+x)^k\tag{2}\\
&=\sum_{j=0}^n\sum_{k=0}^{\min\{j,n-j\}}\binom{j}{k}2^k[x^{n-j-k}](2+x)^k\\
&=\sum_{j=0}^n\sum_{k=0}^{\min\{j,n-j\}}\binom{j}{k}\binom{k}{n-j-k}2^{3k-n+j}\tag{3}
\end{align*}
Comment:
*
*In (2) we use the linearity of the coefficient of operator and apply the rule
\begin{align*}
[x^{p-q}]A(x)=[x^p]x^qA(x)
\end{align*}
We also set the upper limit of the outer sum to $n$ since the exponent of $x^{n-j}$ is non-negative.
*In (3) we select the coefficient of $x^{n-j-k}$.
Binomial identity: We derive from (1) and (3) the following binomial identity by changing the order of summation in the outer sum of (3), i.e. $j\rightarrow n-j$.
\begin{align*}
\sum_{j=0}^n&\sum_{k=0}^{\min\{j,n-j\}}\binom{n-j}{k}\binom{k}{j-k}2^{3k-j}\\
&=(-1)^n-\frac{1}{\sqrt{3}}(1-\sqrt{3})^n+\frac{1}{\sqrt{3}}(1+\sqrt{3})^n\qquad\qquad n\geq 0
\end{align*}
| {
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "4",
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Prove that for all $n\ge9$, there exist natural numbers $x,y$ such that $n=2x+5y$. How would you use induction to prove this?
| If $n=9$ then $n = 2\cdot 2 + 5\cdot 1$; taking $x:= 2$ and $y:=1$ suffices.
If $n \geq 9$ is an integer such that
$n-1= 2x + 5y$ for some integers $x,y > 0$, then
$n = n-1 + 1 = 2x+5y + 1 = 2x' + 5y'$.
Note that $2x+5y+1 = 2x+5y+(5-4) = 2(x-2) + 5(y+1)$.
So the preceding equalities are equivalent to
$$
n = 2x' + 5y' = 2(x-2) + 5(y+1).
$$
Taking $x' := x-2$ and $y' := y+1$ suffices,
which are still integers.
Note that for $n=10$ we have $x' = 2-2 = 0$, which is not a natural number if you don't count $0$ as one such.
| {
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Algebra Precalculus
$a> 0 < b$ For all the integer value of $a$ and $b$
$X= (a^2 + ab)-(ab^2-b)/(2a^2+b^2 -ab)$
Quantity I: $x $
Quantity II: $1.5$
(a) Quantity I $\lt$ Quantity II
(b) Quantity I $\gt$ Quantity II
(c) Quantity I $\ge$ Quantity II
(d) Quantity I $=$ Quantity II
(e) No relation
$(x^a)^c = x^c$
$x^{2b}/x^a = (x^{5a}) * (x^d)*(x^b)$
Quantity I = $b$
Quantity II = $d$
(a) Quantity I $\gt$ Quantity II
(b) Quantity I $\lt$ Quantity II
(c) Quantity I $\ge$ Quantity II
(d) Quantity I $=$ Quantity II
(e) No relation
Generally How to solve this sum and how to approach this Question,please guide me the steps with the answer
| I assume $a = 1$
Then,
$$x^{2b}/x^a = (x^{5a}) * (x^d)*(x^b) \implies x^{2b}/x = (x^{5}) * (x^d)*(x^b) \implies x^{2b} = (x^{6}) * (x^{d+1})*(x^{b+1}) \implies x^{2b - b - 1} = (x^{6}) * (x^{d+1})\implies x^{b - 1} = (x^{d+7}) \implies x^{b - 1 - d - 7} = 1 \implies x^{b- d - 8} = 1 \implies b = d+8$$
Therefore $b > d$ option a is correct.
$x= (a^2 + ab)-(ab^2-b)/(2a^2+b^2 -ab)$
Since $b <0$ let $-c = b, c \in \mathbb{R^+}$
$$x= (a^2 - ac)-{(ac^2+c)\over(2a^2+c^2 +ac)}$$
$$x= {(a^2 - ac)(2a^2+c^2 +ac)-(ac^2+c)\over(2a^2+c^2 +ac)} = {2a^4 - a^3c-c^3a-ac^2-c\over(2a^2+c^2 +ac)}$$
By AM-GM,
$$x\ge {2a^4 - 2a^2c^2-ac^2-c\over(2a^2+c^2 +ac)} \ge {4a^4 - 4a^2c^2-2ac^2-2c\over(5a^2+3c^2)}$$
Let $c = 1$
Therefore,
$$x\ge {4a^4 - 4a^2-2a-2\over(5a^2+3)}$$
Now let $a = 10$ , then $x \ge 78$
and as you said $x = 0$ for $a = 1, b= -1$.
Therefore option e is correct.
| {
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"timestamp": "2023-03-29T00:00:00",
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How find this maximum of the value $\sum_{i=1}^{6}x_{i}x_{i+1}x_{i+2}x_{i+3}$? Let
$$x_{1},x_{2},x_{3},x_{5},x_{6}\ge 0$$ such that
$$x_{1}+x_{2}+x_{3}+x_{4}+x_{5}+x_{6}=1$$
Find the maximum of the value of
$$\sum_{i=1}^{6}x_{i}\;x_{i+1}\;x_{i+2}\;x_{i+3}$$
where
$$x_{7}=x_{1},\quad x_{8}=x_{2},\quad x_{9}=x_{3}\,.$$
| For $x_i=\frac{1}{6}$ we get $\frac{1}{216}$.
We'll prove that it's a maximal value.
Indeed, let $x_1=\min\{x_i\}$, $x_2=x_1+a$, $x_3=x_1+b$, $x_4=x_1+c$, $x_5=x_1+d$ and $x_6=x_1+e$.
Hence, $a$, $b$, $c$, $d$ and $e$ are non-negatives and we need to prove that:
$$216\sum_{i=1}^6x_ix_{i+1}x_{i+2}x_{i+3}\leq\left(\sum_{i=1}^6x_i\right)^4$$ or
$$216(a^2+b^2+c^2+d^2+e^2-ab-bc-cd-de)x_1^2+$$
$$24((a+b+c+d+e)^3-9(2abc+abd+abe+acd+ade+2bcd+bce+bde+2cde))x_1+$$
$$+(a+b+c+d+e)^4-216(abcd+bcde),$$
which is true because
$$a^2+b^2+c^2+d^2+e^2-ab-bc-cd-de\geq$$
$$\geq a^2+b^2+c^2+d^2+e^2-ab-bc-cd-de-ea=\frac{1}{2}\sum_{cyc}(a-b)^2\geq0,$$
$$216(abcd+bcde)=216bcd(a+e)\leq216\left(\frac{a+b+c+d+e}{4}\right)^4=$$
$$=\frac{216}{256}(a+b+c+d+e)^4\leq(a+b+c+d+e)^4$$ and
$$(a+b+c+d+e)^3\geq9(2abc+abd+abe+acd+ade+2bcd+bce+bde+2cde),$$
but my proof of this statement is very ugly.
| {
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Finding $ \int \frac{5x^2-x-4}{x^5+x^4+1}dx$ Finding $\displaystyle \int \frac{5x^2-x-4}{x^5+x^4+1}dx$
Attempt : $\displaystyle I = \int\frac{5x^2-x-4}{x^5+x^4+1}dx = \int\frac{5x^2-x-4}{(x^2+x+1)(x^3-x+1)}dx$
because $\omega,\omega$ are the roots of $x^5+x^4+1 = 0$
so one factor is $(x-\omega)(x-\omega^2) = (x^2+x+1)$
could some help me how to solve it, thanks
| HINT:Using partial fraction decomposition
$$\frac{5x^2-x-4}{(x^2+x+1)(x^3-x+1)}=\frac{Ax+B}{x^2+x+1}+\frac{Cx^2+Dx+E}{x^3-x+1}$$
$${5x^2-x-4}=(Ax+B)(x^3-x+1)+(Cx^2+Dx+E)(x^2+x+1)$$
Solving gives..
$$\frac{5x^2-x-4}{(x^2+x+1)(x^3-x+1)}=\frac{-3x-3}{x^2+x+1}+\frac{3x^2-1}{x^3-x+1}$$
Another hint:
Maybe at some point you might require:$$\int\frac{1}{x^2+x+1}dx$$
Complete the square and use $$\int\frac{1}{a^2+x^2}dx=\frac{\arctan(\frac xa)}{a}+C$$$$$$
Can you do it now?
| {
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How to find the domain and range of $f(x) = \sqrt {\frac{x+1}{x+2}}$
Find the domain and range of $$f(x) = \sqrt {\frac{x+1}{x+2}}$$
I got the domain $[-1, \infty)$ but the answer contains $(-\infty, -2)$ along with it. And how to calculate range?
| We must have $$\frac{x+1}{x+2}\geq 0.$$ We consider the following cases:
Case 1. Suppose that $x+1\geq 0$ and $x+2>0$. Then $x\geq -1$ and $x>-2$. Thus,
$$SS_1=[-1,\infty).$$
Case 2. Suppose that $x+1\leq 0$ and $x+2<0$. Then $x\leq -1$ and $x<-2$. Thus,
$$SS_2=[-\infty,-2).$$
Hence, domain$=SS_1\cup SS_2$
Let $y=f(x)=\sqrt{\frac{x+1}{x+2}}$. Note that $y\geq 0$. Now,
$$y^2=\frac{x+1}{x+2}=1-\frac{1}{x+2}$$ and we get
$$x=\frac{1}{1-y^2}-2\qquad;y\neq\pm 1$$
Because $y\geq 0$, it follows that the range is $\Bbb R\smallsetminus\{1\}$. For completeness the graph is given below:
| {
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Proof by induction, dont know how to represent range The question asks for me to prove the following through induction:
$1 + \frac{1}{2} + \frac{1}{3} + \cdots + \frac{1}{2^n} \geq 1 + \frac{n}{2}$
This is my proof thus far:
Proving true for $n = 1$
\begin{align*}
1 + \frac{1}{2} &\geq 1 + \frac{1}{2}\\
\end{align*}
Assuming true for $n = k$
\begin{align*}
1 + \frac{1}{2} + \frac{1}{3} + \cdots + \frac{1}{2^k} \geq 1 + \frac{k}{2}
\end{align*}
Proving true for $n = k + 1$
\begin{align*}
1 + \frac{1}{2} + \frac{1}{3} + \cdots + \frac{1}{2^k} + \frac{1}{2^{k+1}} &\geq 1 + \frac{k}{2} + \frac{1}{2}\\
(1 + \frac{1}{2} + \frac{1}{3} + \cdots + \frac{1}{2^k}) + \frac{1}{2^{k+1}} &\geq (1 + \frac{1}{2} + \frac{1}{3} + \cdots + \frac{1}{2^k}) + \frac{1}{2}\\
\frac{1}{2^{k+1}} &\geq \frac{1}{2}\\
\end{align*}
I realized at the last statement that something was off, because the last statement contradicts the thing I'm trying to prove. I then realized that it was because the difference between $\frac{1}{2^k}$ and $\frac{1}{2^{k + 1}}$ sets was not simply the addition of $\frac{1}{2^{k + 1}}$, but rather, all numbers between $\frac{1}{2^k}$ and $\frac{1}{2^{k + 1}}$.
For example n = 1 is $1 + \frac{1}{2}$, but $n = 2$ is $1 + \frac{1}{2} + \frac{1}{3} + \frac{1}{4}$. (note the addition of $\frac{1}{3}$.)
So how can I represent this range?
I believe my proof works except for the fact that $\frac{1}{2^{k + 1}}$ needs to be replaced with something else, I just don't know what that is.
| $$
\underbrace{ 1 + \frac 1 2 + \frac 1 3 + \cdots + \frac 1 {2^k} + \frac 1 {2^{k+1}} }_{\Large\text{This is wrong.}}
$$
$$
\overbrace{ \underbrace{ \underbrace{1 + \frac 1 2 + \frac 1 3 + \cdots + \frac 1 {2^k}}_{\Large\text{The is the case }n=k.} + \frac 1 {2^k+1} + \frac 1 {2^k+2} + \frac 1 {2^k+3} + \cdots + \frac 1 {2^{k+1}} } }_{\Large\text{This is the case } n = k+1.}^{\Large\text{This is what you need instead.}}
$$
For example, suppose $n=k=3.$ Then
$$
1+ \frac 1 2 + \frac 1 3 + \cdots + \frac 1 {2^k} = 1 + \frac 1 2 + \frac 1 3 + \frac 1 4 + \frac 1 5 + \frac 1 6 + \frac 1 7 + \frac 1 8.
$$
And the next case, $n=k+1 = 3+1=4$ is this
$$
\underbrace{ \underbrace{1 + \frac 1 2 + \frac 1 3 + \frac 1 4 + \frac 1 5 + \frac 1 6 + \frac 1 7 + \frac 1 8}_{\Large\text{This is the case } n=3.} + \frac 1 9 + \frac 1{10} + \frac 1 {11} + \frac 1 {12} + \frac 1 {13} + \frac 1 {14} + \frac 1 {15} + \frac 1 {16}.}_{\Large\text{This is the case } n = 4.}
$$
You don't just add one more term when you increment $n$ from $k$ to $k+1;$ you add $2^k$ terms to get up to $2^{k+1}.$
| {
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"timestamp": "2023-03-29T00:00:00",
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Find the limit or prove that the limit does not exist $\lim_{x \to c}x^2 + x + 1,$ for any $c \in R$
This is what I tried.
For $\epsilon > 0$ there exists $\delta >0$ such that $0< \mid x - c \mid < \delta (x \in R)$ => $\mid x^2 + x + 1 - c^2 - c - 1 \mid = \mid x^2 + x - c^2 - c \mid < \epsilon$
After that I tried to use triangle inequlity $\mid x^2 - c^2 + x - c \mid \leq \mid x^2 - c^2 \mid + \mid x - c \mid$
if $\mid x^2 - c^2 \mid + \mid x - c \mid$ is less then $ \mid x^2 - c^2 + x - c \mid$, I was going to make $\mid x^2 - c^2 \mid + \mid x - c \mid$ to $\mid x + c \mid \mid x- c \mid +\mid x - c\mid$ and then $\mid x - c \mid \mid x+c\mid +1 \mid$
and then use $\mid x - c \mid \mid x+c\mid +1 \mid < \epsilon $
However, I cannot
I am stuck now.
I need help.
| Here is a trick:
Start by choosing $\delta \le |c|+1$, then $|x| \le |c|+|x-c| \le 2(|c|+1)$.
Then $|x^2-c^2| = |x+c||x-c| \le 3 (|c|+1) |x-c|$.
Then $|x^2+x+1 - (c^2+c+1)| \le (3|c|+2)|x-c|$.
Now choose $\delta < \min (|c|, {\epsilon \over 3|c|+2})$, then you will
obain the required bound.
| {
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"timestamp": "2023-03-29T00:00:00",
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Show that $\sum_{n=1}^\infty \frac{n^2}{(n+1)!}=e-1$ Show that:
$$\sum^\infty_{n=1} \frac{n^2}{(n+1)!}=e-1$$
First I will re-define the sum:
$$\sum^\infty_{n=1} \frac{n^2}{(n+1)!} = \sum^\infty_{n=1} \frac{n^2-1+1}{(n+1)!} - \sum^\infty_{n=1}\frac{n-1}{n!} + \sum^\infty_{n=1} \frac{1}{(nm)!}$$
Bow I will define e:
$$e^2 = 1+ \frac{2}{1!} + \frac{x^2}{2!} + ... + \infty$$
$$e' = 1 + \frac{1}{1!} + \frac{1}{2!} + ... + \infty$$
$$(e'-2) = \sum^\infty_{n=1} \frac{1}{(n+1)!}$$
Now I need help.
| $$\frac{n^2}{(n+1)!} = \frac{(n+1)(n-1) + 1}{(n+1)!} = \frac{(n-1)}{n!} + \frac{1}{(n+1)!}$$
Remembering that we're summing to infinity, evaluating the first terms and paying careful attention to the indices,
$$
\begin{align}
\sum_{n=1}^\infty \left( \frac{(n-1)}{n!} + \frac{1}{(n+1)!} \right) &= \sum_{n=2}^\infty \frac{(n-1)}{n!} + \sum_{n=2}^\infty \frac{1}{n!}\\ &= \sum_{n=2}^\infty \frac{n}{n!}\\ &=\left( \sum_{n=1}^\infty \frac{n}{n!} \right) - 1\\
&= e - 1
\end{align} $$
| {
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How prove this inequality $\sum\limits_{cyc}\sqrt{\frac{yz}{x^2+2016}}\le\frac{3}{2}$ Given $x,y,z$ are positive real number satisfy $xy+yz+xz=2016$. Prove that $$\sqrt{\dfrac{yz}{x^2+2016}}+\sqrt{\dfrac{xy}{z^2+2016}}+\sqrt{\dfrac{xz}{y^2+2016}}\le\dfrac{3}{2}$$
I tried
$\sqrt{\frac{yz}{x^2+2016}}=\sqrt{\frac{yz}{x^2+xy+yz+xz}}=\sqrt{\frac{yz}{\left(x+y\right)\left(x+z\right)}}$
And by C-S $\sqrt{\left(x+y\right)\left(x+z\right)}\ge x+\sqrt{yz}$
i can't continues. Help me !
| I believe there is something called the Purkiss Principle which would imply that in this case the maximum of $f$ is achieved when $x=y=z=\sqrt{2016/3}$. Thus, $$f(\sqrt{2016/3},\sqrt{2016/3},\sqrt{2016/3}) = 3/2.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2216107",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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$\int_{|z-2i|=1}\frac{(z-2i+\frac{1}{2})^2 \sin(2i\pi z)}{\overline{z}^{2} (z-2i)^4} \ dz$ Let $C$ be the circle $|z -2i|=1$
How Can I Compute this Integral :
$$\int_{C}\frac{(z-2i+\frac{1}{2})^2 \sin(2i\pi z)}{\overline{z}^{2} (z-2i)^4} \ dz$$
Thank you
| I have tried solving this using cauchy integral formula as follows :
$|z-2i|= 1\Rightarrow (z-2i)(\overline{z}+2i)=1 \Rightarrow (z-2i+\frac{i}
{2}-\frac{i}{2})(\overline{z}+2i)=1$
Now we have : $$(z-2i+\frac{i}{2}-\frac{i}{2})(\overline{z}+2i)=1 \Rightarrow(z-2i+\frac{i}{2})(\overline{z}+2i)=\frac{i \overline{z}}{2}$$
Which implies that :$$\frac{(z-2i+\frac{i}{2})^2}{\overline{z}^2}=\frac{-1}{4(\overline{z}+2i)^2}$$
hence the integral become :
$$\begin{align}\int_{|z-2i|=1}\frac{(z-2i+\frac{1}{2})^2 \sin(2i\pi z)}{\overline{z}^{2} (z-2i)^4} \ dz&=\frac{-1}{4}\int_{|z-2i|=1}\frac{\sin(2i\pi z)}{(\overline{z}+2i)^2 (z-2i)^2(z-2i)^2} \ dz \\ \\
&=\frac{-1}{4}\int_{|z-2i|=1}\frac{\sin(2i\pi z)}{(z-2i)^2} \ dz \\ \\
&=\frac{-1}{4} 2\pi i \cos(4\pi) \\ \\
&=0
\end{align}
$$
| {
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Convergence of $1+\frac{1^2\cdot2^2}{1\cdot3\cdot5}+ \frac{1^2\cdot2^2\cdot3^2}{1\cdot3\cdot5\cdot7\cdot9}+...$ I am trying to use the ratio test, for that, I need the general formula for the series.
The general formula for the numerator is $(n!)^2$
The denominator is a sequence of odd numbers that grows by two terms every time but how do I represent it?
Also, any tips for how I can guess the series from a sequence would be greatly appreciated.
| Lets try writing the general term
$$a_n=\frac{(n!)^2}{1\cdot 3\cdot 5\cdots (4n-5)(4n-3)}\\a_{n+1}=\frac{((n+1)!)^2}{1\cdot 3\cdot 5\cdots (4n-1)(4n+1)}\\\frac{a_{n+1}}{a_n}=\frac{((n+1)!)^2}{1\cdot 3\cdot 5\cdots(4n-1)(4n+1)}\cdot\frac{1\cdot 3\cdot 5\cdots(4n-5)(4n-3)}{(n!)^2}\\\frac{a_{n+1}}{a_n}=\frac{(n+1)^2}{(4n-1)(4n+1)}$$
also you can notice that
$$1\cdot 3\cdot 5\cdots (2k-1)(2k+1)=\frac{(2k+1)!}{2^kk!}$$
So $$a_n=\frac{(n!)^2(2k-2)!2^{k-2}}{(4k-3)!}$$
Doing the ratio test should give the same result as above.
| {
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Tangent Points for Common Tangent to Two Ellipses This is somewhat similar to my other question here.
Consider the two ellipses given by the equations
\begin{equation}
\frac{x^2}{2^2} + \frac{(y-1)^2}{1^2} = 1
\end{equation}
and
\begin{equation}
\frac{x^2}{1^2} + \frac{(y-4)^2}{(1/2)^2} = 1.
\end{equation}
How do I find the coordinates for the two tangent points of their common tangent at the left side of the ellipses? (I hope the question makes sense.)
| Let's introduce the following new variables:
$$2u=x\ \text{ and }\ y-1=v.$$
With these new variables, we have
$$u^2+v^2=1\ \text{ and }\ u^2+(v-3)^2=\frac14$$
that is, we have two circles as shown in the figure below.
We have similar triangles and we can see that $OD=6$. Also, by the Pythagorean theorem $DC=\frac{\sqrt{35}}2$ and by the similarity of $OBD$ and $O'CD$: $OB=\frac6{\sqrt{35}}$. So, the slope of the red straight lien is $-\sqrt{35}$. Don't forget a about the other tangent, not shown, whose slope is $\sqrt{35}$.
The two tangent lines in the $u,v$ system are
$$v=-\sqrt{35}u+6\ \text{ and } \ v=\sqrt{35}u+6.$$
Returning to the $x,y$ coordinate system, we get
$$y=-\frac{\sqrt{35}}2x+7\ \text{ and } \ y=\frac{\sqrt{35}}2x+7.$$
EDIT
Unforgivable! I forgave the other pair of tangents:
After similar calculations we get the equations of the other pair of tangent lines.
$$y=-\frac{\sqrt3}{2}x+3\ \text{ and }y=\frac{\sqrt3}{2}x+3 \ $$
| {
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Proof: if $a$ and $b$ are integers, then $a^2-4b-3\neq 0$. I was wondering if someone could take the time to look over this proof and make sure it is correct. I greatly appreciate the help.
Proposition: If $a$ and $b$ are integers, then $a^2-4b-3\neq 0$.
Proof: Assume $a,b\in\mathbb{Z}$ and, for contradiction's sake, $a^2-4b-3=0$. Solving for $a^2$, we find $a^2=4b+3$. Clearly, $a^2 \equiv 3($mod $4)$.
Now, we can factor 2 out of the left-hand side of $a^2=4b+3$ yielding $a^2=2(2b+1)+1$. Thus, by the definition of odd, $a^2$ is odd. Since $a^2$ is odd, $a$ must be odd. By the definition of odd, we can write $a=2c+1$ where $c\in\mathbb{Z}$.
Now we can substitute for $a$ in $a^2$ to find $a^2=(2c+1)^2=4c^2+4c+1$. Factoring 4 out from the first two terms, we discover $a^2=4(c^2+c)+1$. Clearly, $a^2\equiv 1($mod $4)$. Earlier, however, we found that $a^2 \equiv 3($mod $4)$. Since $a$ can not be congruent to both 1 and 3 modulo 4, we have a contradiction. Therefore, if $a,b\in\mathbb{Z}$, then $a^2-4b-4\neq0$.
| Yes, your proof is correct! Also, when you discorver that a has a remainder of 3 when divided by 4, you can go straight to the fact that a is odd.
| {
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Prove $\prod_1^\infty (1+p_n)$ converges Let $p_{2n-1} = \frac{-1}{\sqrt{n}}$, and $p_{2n} = \frac{1}{n}+\frac{1}{\sqrt{n}}$.
Prove $\prod_1^\infty (1+p_n)$ converges.
By numerical simulations, it appears to converge (to something around $0.759$). However, I'm not sure how to prove this. I know we can skip the first term since it's $0$. Then we can write it in the following form.
\begin{align*}
\prod_1^\infty (1+p_n) &= \prod_1^\infty \left(1+\frac{1}{2n}+\frac{1}{\sqrt{2n}}\right)\left(1-\frac{1}{\sqrt{2n+1}}\right) \\
&= \left(1+\frac{1}{2}+\frac{1}{\sqrt{2}}\right)\left(1-\frac{1}{\sqrt{3}}\right)\left(1+\frac{1}{4}+\frac{1}{\sqrt{4}}\right)\left(1-\frac{1}{\sqrt{5}}\right)...
\end{align*}
Any thoughts?
| Note that
$$\prod\limits_{k=2}^{2n}(1+p_k) = \prod\limits_{k=2}^{n}(1+p_{2k-1})(1+p_{2k}) = \prod\limits_{k=2}^{n}\left(1-\dfrac{1}{\sqrt{k}}\right)\left(1+\dfrac{1}{k} +\dfrac{1}{\sqrt{k}}\right) = \prod\limits_{k=2}^{n} \left(1- \dfrac{1}{k\sqrt{k}}\right)$$
It is also known that for any sequence $\{a_k\}$ such that $\forall k \geqslant 2: 0 \leqslant a_k < 1$:
$$\prod\limits_{k=2}^{n} \left(1- a_k\right) \leqslant e^{-\sum\limits_{k=2}^n a_k}$$
The inequality above is true because $1 -x \leqslant e^{x}$ for $0 \leqslant x < 1$.
Hence, for $a_k = \dfrac{1}{k\sqrt{k}}$:
$$\prod\limits_{k=2}^{n} \left(1- \dfrac{1}{k\sqrt{k}}\right) \leqslant \exp\left(-\sum\limits_{k=2}^n \dfrac{1}{k\sqrt{k}}\right)$$
Thus, to finish the proof it is enough to note that $\sum\limits_{k=2}^n \dfrac{1}{k\sqrt{k}}$ converges.
| {
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Find the solution of the $x^2+ 2x +3=0$ mod 198 Find the solution of the $x^2+2x+3 \equiv0\mod{198}$
i have no idea for this problem i have small hint to we going consider $x^2+2x+3 \equiv0\mod{12}$
| Here's a more-or-less generalizable, manual way of finding all of the solutions:
First, as P. Vanchinathan does, change variable to $a := x + 1$, which transforms the equation into one with zero linear term:
$$a^2 + 2 \equiv 0 \pmod {198} .$$
(This step is option, but reduces the amount of later work.)
Now, we exploit the prime factorization $198 = 2 \cdot 3^2 \cdot 11$. Reducing modulo $2$ gives $a^2 \equiv 0 \pmod {198}$, so any solution $a$ to the above display equation has the form $a = 2b$. Substituting into the previous display equation gives
$$4 b^2 + 2 \equiv 0 \pmod {198},$$
which is equivalent to $$2 b^2 + 1 \equiv 0 \pmod {99}.$$
Now reducing modulo $11$ (and multiplying by $6$ to produce a monic polynomial on the l.h.s.) leaves
$$b^2 + 6 \equiv 0 \pmod {11} .$$
The l.h.s. factors as $(b + 4)(b - 4)$, so any solution $b$ to the equation modulo $99$ has the form $$b = 11 c \pm 4 .$$
Substituting in the above equation modulo $99$ and proceeding as before (in particular, multiplying both sides of the resulting equation by $7$, which is coprime to $9$ and hence produces an equivalent equation) gives
$$c^2 \pm 4 c + 3 \equiv 0 \pmod 9 .$$
We may factor the left-hand side as $(c \pm 1)(c \pm 3)$. Since the prime factorization of $9$ is $9 = 3^2$, the equation in $c$ has a solution iff either factor is $0$ or both of the above factors are congruent to $0$ modulo $3$. The latter is impossible since the difference of those factors modulo $3$ is $\pm 1$, so the solutions are exactly
$c = \mp 1, \mp 3$. Substituting these four solutions successively into our equations for $b, a, x$ gives all of the solutions to the original equation: $$x \equiv 13, 57, 139, 183 ,$$
which agrees with the solution lioness99a produced using W.A.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2227709",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
Evaluate $\int_0^{2\pi}\frac{\cos(\theta)}{1 + A\sin(\theta) + B\cos(\theta)}\, \mathrm{d}\theta \ \text{ for }\ (A^2+B^2) <<1$ I need to evaluate the definite integral $\int_0^{2\pi}\frac{\cos(\theta)}{1 + A\sin(\theta) + B\cos(\theta)}\, \mathrm{d}\theta \ \text{ for }\ A<<1, B<<1, (A^2+B^2) <<1,$
For unresricted (but real) A&B, Wolfram Alpha provides the following indefinite general solution:-
$$\int \frac{\cos(\theta)}{1 + A\sin(\theta) + B\cos(\theta)}\, \mathrm{d}\theta = \left( \frac{1}{A^2+B^2}\right) \left[ \frac{2B}{K} \tanh^{-1} \left( \frac{A-(B-1)\tan\left(\frac{\theta}{2}\right)}{K}\right) + F(\theta)\right]$$
where $F(\theta) = A \ln(1 + A\sin\theta+B\cos\theta)+B\theta$,
and $K = \sqrt{A^2 + B^2 -1}$, therefore K is complex for the range of A,B I am interested in.
In a previous question seeking a solution for the similar, but slightly simpler, definite integral (with numerator $1$ rather than $\cos\theta$) user Dr. MV found a solution given by:
$$\int_0^{2\pi}\frac{1}{1 + A\sin(\theta) + B\cos(\theta)}\, \mathrm{d}\theta = \frac{2\pi}{\sqrt{1-A^2-B^2}}\text{ } (\text { for} \sqrt{A^2+B^2}<1) $$.
My question: Can similar solutions be found for these two definite integrals $$\int_0^{2\pi}\frac{\cos\theta}{1 + A\sin(\theta) + B\cos(\theta)}\, \mathrm{d}\theta \tag 1$$
and
$$\int_0^{2\pi}\frac{\sin\theta}{1 + A\sin(\theta) + B\cos(\theta)}\, \mathrm{d}\theta \tag 2$$?
EDIT
I have taken the solution proposed by user Chappers.
By simultaneous equations in A,B,I,J it turns out that
$$I=\int_0^{2\pi}\frac{\cos\theta}{1 + A\sin(\theta) + B\cos(\theta)}\, \mathrm{d}\theta = \frac{B}{A^2+B^2} 2\pi (1-\frac{1}{\sqrt{1-A^2-B^2}})$$ and
$$J=\int_0^{2\pi}\frac{\sin\theta}{1 + A\sin(\theta) + B\cos(\theta)}\, \mathrm{d}\theta = \frac{A}{A^2+B^2} 2\pi (1-\frac{1}{\sqrt{1-A^2-B^2}})$$.
These were confirmed in a numerical model.
| HINT: set $$t=\tan(x/2)$$, $$\sin(x)=\frac{2t}{1+t^2}$$, $$\cos(t)=\frac{1-t^2}{1+t^2}$$ and $$dx=\frac{2dt}{1+t^2}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2230930",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
How to prove $\sin(x) = x + O(x^3)$? I was doing an exercise in which I have to find the rate of convergence of
$$\lim\limits_{h\to 0}\dfrac{\sin h}{h} = 1$$
and the answer is $O(h^2)$. I don't understand why. The only thing I have found is that $$\sin(x) = x + O(x^3)$$ when $x$ tends to zero, and with that the exercise is solved, but how I demonstrate that fact?
| Start with
$\sin' = \cos
$,
$\cos' = -\sin
$,
$\sin(0) = 0$,
$\cos(0) = 1$,
and
$\sin^2+\cos^2 = 1$.
For small $t$,
$1 \ge \cos(t)
\ge 0
$
so
$\sin(x)
=\int_0^x \cos(t)dt
\le x
$.
Therefore
$1-\cos(x)
=\int_0^x \sin(t) dt
\le \int_0^x t dt
= \frac{x^2}{2}
$
so
$\cos(x)
\ge 1-\frac{x^2}{2}
$.
Therefore
$\sin(x)
=\int_0^x \cos(t)dt
\ge\int_0^x (1-\frac{t^2}{2})dt
=x-\frac{x^3}{6}
$.
So we already have
$x-\frac{x^3}{6}
\le \sin(x)
\le x
$.
This is actually
enough for what you want.
Doing this again,
$1-\cos(x)
=\int_0^x \sin(t) dt
\ge \int_0^x (t-\frac{t^3}{6}) dt
= \frac{x^2}{2}-\frac{x^4}{24}
$
so
$\cos(x)
\ge 1-\frac{x^2}{2}+\frac{x^4}{24}
$.
Doing this one more time,
$\sin(x)
=\int_0^x \cos(t)dt
\ge\int_0^x (1-\frac{t^2}{2}+\frac{t^4}{24})dt
=x-\frac{x^3}{6}+\frac{x^5}{120}
$.
By induction,
we can get the power series
for $\sin$ and $\cos$.
Note:
This is not original.
I first saw this in
"100 Great Problems of Elementary Mathematics"
by Heinrich Dorrie
(less than $15 from Dover).
Get it.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2233089",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 1
} |
$\int ^{\infty}_0 \frac{\ln x}{x^2 + 2x+ 4}dx$ We have to find the integration of
$$\int ^{\infty}_0 \frac{\ln x}{x^2 + 2x+ 4}dx$$
In this I tried to do substitution of $x=e^t$
After that got stuck .
| Let $I$ be defined by the integral
$$I=\int_0^\infty \frac{\log(x)}{x^2+2x+4}\,dx \tag1$$
and let $J$ be the contour integral
$$J=\oint_{C}\frac{\log^2(z)}{z^2+2z+4}\,dz \tag2$$
where the contour $C$ is the classical "key-hole" contour for which the keyhole coincides with the branch cut along the positive real axis.
Then, we can write $(2)$ as
$$\begin{align}
J&=\int_0^R \frac{\log^2(x)}{x^2+2x+4}\,dx-\int_0^R \frac{(\log(x)+i2\pi)^2}{x^2+2x+4}\,dx+\int_{C_R}\frac{\log^2(z)}{z^2+2z+4}\,dz\\\\
&=-i4\pi \int_0^R \frac{\log(x)}{x^2+2x+4}\,dx+4\pi^2\int_0^R\frac{1}{x^2+2x+4}\,dx+\int_{C_R}\frac{\log^2(z)}{z^2+2z+4}\,dz\tag 3
\end{align}$$
As $R\to \infty$ the integral over $C_R$ vanishes and we have from $(1)$
$$\lim_{R\to \infty}J=-i4\pi I +4\pi^2\int_0^\infty\frac{1}{x^2+2x+4}\,dx\tag 4$$
Next, we apply the residue theorem to evaluate the left-hand side of $(4)$. Proceeding we find that
$$\begin{align}
\lim_{R\to \infty}J&=2\pi i \text{Res}\left(\frac{\log^2(z)}{z^2+2z+4}\,dz, z=-1\pm i\sqrt{3}\right)\\\\
&=2\pi i \left(\frac{(\log(2)+i2\pi/3)^2}{i2\sqrt{3}}+\frac{(\log(2)+i4\pi/3)^2}{-i2\sqrt{3}}\right)\\\\
&=-\frac{i4\pi^2\log(2)}{3\sqrt{3}}+\frac{4\pi^3}{3\sqrt{3}}\tag 5
\end{align}$$
Equating real and imaginary parts of $(3)$ and $(5)$ reveals
$$\int_0^\infty \frac{\log(x)}{x^2+2x+4}\,dx =\frac{\pi \log(2)}{3\sqrt{3}}$$
and as a bonus
$$\int_0^\infty\frac{1}{x^2+2x+4}\,dx=\frac{\pi}{3\sqrt{3}}$$
And we are done!
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2233396",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "12",
"answer_count": 4,
"answer_id": 0
} |
Proving the inequality $0\leq \frac{\sqrt{xy}}{1-p}\frac{x^{\frac{1}{p}-1}-y^{\frac{1}{p}-1}}{x^{\frac{1}{p}}-y^{\frac{1}{p}}} \leq 1$ Suppose $p\in(0,1)$. How might one show that
\begin{equation}\tag{1}
0\leq \frac{\sqrt{xy}}{1-p}\frac{x^{\frac{1}{p}-1}-y^{\frac{1}{p}-1}}{x^{\frac{1}{p}}-y^{\frac{1}{p}}} \leq 1
\end{equation}
for all $x,y\in[0,1]$?
It is clearly non-negative, so the hard part is to show that it is never greater than 1.
I was hoping to use a technique similar to the one to prove that
$$
0\leq \sqrt{xy}\frac{\log x - \log y}{x-y}\leq 1
$$
for all $x,y\in[0,1]$. We can use an integral representation and see that
\begin{align*}
\sqrt{xy}\frac{\log x - \log y}{x-y}
&= \int_{0}^{\infty} \frac{\sqrt{xy}}{(x+t)(y+t)}dt\\
&\leq \int_{0}^{\infty} \frac{\sqrt{xy}}{(\sqrt{xy}+t)^2}dt\\
& = 1.
\end{align*}
Is there a suitable integral representation that can prove (1)?
| I've figured out the correct integral representation to use here. For $a\in(-1,1)$, consider the following integral representations:
\begin{align*}
\frac{x^a-y^a}{x-y} &= \frac{\sin(a\pi)}{\pi}\int_{0}^{\infty}\frac{t^a}{(x+t)(y+t)}dt\\
\text{and}\qquad ax^{a-1} &= \frac{\sin(a\pi)}{\pi}\int_{0}^{\infty}\frac{t^a}{(x+t)^2}dt.
\end{align*}
Similar to the example in the original post, we have
\begin{align*}
\frac{1}{a}\frac{x^a-y^a}{x-y}
&\leq \frac{\sin(a\pi)}{a\pi}\int_{0}^{\infty}\frac{t^a}{(\sqrt{xy}+t)^2}dt\\
& = (\sqrt{xy})^{a-1}.
\end{align*}
Thus, if we let $a=1-p$, we have
\begin{align*}
\frac{1}{1-p}\frac{x^{\frac{1}{p}-1}-y^{\frac{1}{p}-1}}{x^{\frac{1}{p}}-y^{\frac{1}{p}}}
= \frac{1}{1-p}\frac{x^{\frac{1-p}{p}}-y^{\frac{1-p}{p}}}{x^{\frac{1}{p}}-y^{\frac{1}{p}}}
&= \frac{1}{a}\frac{x^{\frac{a}{p}}-y^{\frac{a}{p}}}{x^{\frac{1}{p}}-y^{\frac{1}{p}}}\\
&\leq \left(\sqrt{x^{\frac{1}{p}}y^{\frac{1}{p}}}\right)^{a-1} \\
& = \left(\sqrt{x^{\frac{1}{p}}y^{\frac{1}{p}}}\right)^{-p}\\
&=\frac{1}{\sqrt{xy}}
\end{align*}
which proves the desired result.
Hence, even though I only originally conjectured it for $p\in(0,1)$, the claim holds for $p\in(1,2)$ as well!
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2234190",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Solve$\int\frac{x^4}{1-x^4}dx$ Question: Solve $\int\frac{x^4}{1-x^4}dx.$
My attempt:
$$\int\frac{x^4}{1-x^4}dx = \int\frac{-(1-x^4)+1}{1-x^4}dx = \int 1 + \frac{1}{1-x^4}dx$$
To integrate $\int\frac{1}{1-x^4}dx,$ I apply substitution $x^2=\sin\theta.$ Then we have $2x \frac{dx}{d\theta} = \cos \theta.$ which implies that $\frac{dx}{d\theta}=\frac{\cos \theta}{2\sqrt{\sin \theta}}.$
So we have $\int \frac{1}{1-x^4}dx=\int\frac{1}{\cos^2\theta} \cdot \frac{\cos \theta}{2\sqrt{\sin \theta}} d\theta = \int\frac{1}{2\cos\theta \sqrt{\sin\theta}}d\theta.$ Then I stuck here.
Any hint would be appreciated.
| By doing long division, you'll get $$-\int \left(-\frac{1}{2(x^2+1)}-\frac{1}{4(x+1)}+\frac{1}{4(x-1)}+1\right) dx$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2235305",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 2
} |
Find the constant term in the expansion of $(x^2+1)(x+\frac{1}{x})^{10}$ I can't solve this problem. How to solve it?
The Problem is
"Find the constant term in the expansion of $
\left({{x}^{2}\mathrm{{+}}{1}}\right){\left({{x}\mathrm{{+}}\frac{1}{x}}\right)}^{\mathrm{10}}
$"
| $f(x)=(x^2+1)(x+\dfrac{1}{x})^{10}$
We can rewrite $f(x)$ like below:
$f(x) = x^2(x+\dfrac{1}{x})^{10} + (x+\dfrac{1}{x})^{10}$
In the first term, the power of $x$, must be $-2$ in the parenthesis, so the when it's multiplyed by $x^2$ the power of $x$ becomes $0$. And we know that:
$(x+\dfrac{1}{x})^{10}=\sum_{k=0}^{10}\binom{10}{k}x^k(\dfrac{1}{x})^{10-k}$
So $k+k-10=-2=>k=4$. So the coefficient of that term is $\binom{10}{4}$.
On the other hand, the coefficient of the constant term of the second term of $f(x)$ which is just $(x+\dfrac{1}{x})^{10}$, is when $k=5$. So the coefficient would be $\binom{10}{5}$
So the answer is:
$\binom{10}{4}+\binom{10}{5}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2235394",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 1
} |
Triangle - length of the sides - proof a, b and c are the lengths of the sides of a triangle. Prove that
$$a^2+b^2 \ge \frac{1}{2}c^2$$
Let $\gamma$ be the angle between sides a and b. then:
$$a^2 + b^2 - 2ab\cos(\gamma) = c^2$$
Hence we need to prove that
$$a^2+b^2 \ge \frac{1}{2}c^2$$
$$2a^2+2b^2 \ge a^2+b^2 -2ab\cos(\gamma) $$
$$a^2+b^2+2ab\cos(\gamma) \ge0$$
Knowing that $\cos(\gamma) \ge-1$, we get $a^2+b^2+2ab\cos(\gamma) \ge a^2 +b^2 -2ab =(a-b)^2$
I have carried this problem so far but now I am stuck. How should I proceed?
| You're already there. Your last line is
$$a^2+b^2+2ab\cos(\gamma) \geq (a-b)^2$$
And so you only have to note $(a-b)^2\geq 0$ to see that
$$a^2+b^2+2ab\cos(\gamma) \geq 0$$
Note that we can make this inequality strict ($>$) since $\cos(\gamma)=-1$ can't happen, for one angle is then stretched, so we could've taken $\cos(\gamma)>-1$ rather than $\cos(\gamma)\ge-1$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2236586",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
Question on infinite geometric series. Problem- The sum of the first two terms of an infinite geometric series is 18. Also, each term of the series is seven times the sum of all the terms that follow. Find the first term and the common ratio of the series respectively.
My approach- Let $a+ar+ar^2+\dots$ be the series. Then,
$a+ar=18$ and,
$a=7\frac{1}{1-r}-a$,
solving I get $r = \frac{29}{43}$ but Given answer is $a=16,r=1/8$. where I'm doing wrong?
|
The sum of the first two terms of an infinite geometric series is 18.
$$
S(a, r) = \sum_{k=0}^\infty a r^k = \frac{a}{1-r} \\
a + a r = 18 \quad (*)
$$
Also, each term of the series is seven times the sum of all the terms
that follow.
$$
a r^n = 7 \sum_{k=n+1}^\infty a r^k \quad (**)
$$
Find the first term and the common ratio of the series respectively.
The first term is $a r^0 = a$ and the common ratio is
$$
\frac{a r^{n+1}}{a r^n} = r
$$
From $(*)$ we infer $a \ne 0$.
The instance $n=0$ of $(**)$ is
$$
a = 7 \sum_{k=1}^\infty a r^k
$$
so $a \ne 0$ implies $r \ne 0$.
We rewrite $(**)$ into
$$
\frac{1}{7} r^n
= \frac{1}{1-r} - \sum_{k=0}^n r^k
= \frac{1}{1-r} - \frac{1-r^{n+1}}{1-r}
= \frac{r^{n+1}}{1-r} \iff \\
\frac{1}{7} = \frac{r}{1-r} \iff \\
\frac{1}{7} - \frac{1}{7} r = r \iff \\
\frac{1}{7} = \frac{8}{7} r \iff \\
r = \frac{1}{8}
$$
We rewrite $(*)$ into
$$
18 = a (1+r) = a \frac{9}{8} \iff \\
a = 18 \frac{8}{9} = 16
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2236689",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Finding out the numbers under given conditions Let $M$ be a $2$ digit number $ab$, $N$ is a $3$ digit number $cde$ , and $X=M\times N$ is such that $9(X)=abcde$.The question is to find out the ratio $\frac NM$
I tried to solve it using trial and error and examined a number of cases but couldn't reach the answer so far.What I have got that $d,e$ cannot be simultaneously zero and $a+b+c+d+e$ is a multiple of $9$ I know there should be some logic behind this question instead of dwelling on trial and error.Any help shall be highly appreciated.Thanks.
| Rewrite $9(X) = abcde$ as $9MN = abcde = 1000ab + cde = 1000M + N$, then divide by $M$ to get $9N = 1000 + \frac{N}{M}$ or $\frac{N}{M} = 9N - 1000$. Notice then that $\frac{N}{M}$ must be a whole number, call it $k$.
Replace $N$ with $kM$ to get $9kM = 1000 + k$. Since the left hand side is divisible by $k$, the right hand side must also be divisible by $k$, and $1000 + k \equiv 1000 \equiv 0 \pmod{k}$ implies that $k$ must be a divisor of $1000$. Since we know that $N$ is three-digit and $M$ is two-digit, the possible values of $k$ are $\{2, 4, 5, 8, 10, 20, 25, 40, 50\}$.
Notice that we also know that $1000 + k$ is divisible by $9$, due to $9kM = k + 1000$, so that leaves only one possibility, namely $k = 8$. Plugging this back in, we have $9\cdot 8 M = 72M = 1000 + 8 = 1008$, thus $M = \frac{1008}{72} = \boxed{14}$ and $N = kM = 8\cdot 14 = \boxed{112}$, and this is the only solution.
Indeed, $9(X) = 9(14\cdot 112) = 14112$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2237149",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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A question about Taylor expansion Is this statement true? Statement: Let $n$ be a positive integer. Consider the Taylor expansion of $\sqrt[n]{1+x}$ to the $k$th order, that is,
\begin{gather*}
\sqrt[n]{1+x}=\sum_{j=0}^{k}\binom{\frac{1}{n}}{j}x^j+o(x^k), \qquad \text{as $x\to 0$.}
\end{gather*}
Then the Taylor polynomial of $k$th order satisfies
\begin{gather*}
\left(\sum_{j=0}^{k}\binom{\frac{1}{n}}{j}x^j\right)^n=1+x+\sum_{j=k+1}^{nk}a_jx^j,\tag{1}
\end{gather*}
where $a_j$s are the coefficients which can be determined.
The interesting part of this statement is that the right hand side of (1) just a monic polynomial of first order, plus an infinitesimal of very higher-order.
I have checked for $n=2.$ For instance,
\begin{gather*}
\sqrt{1+x}=1+{\frac{1}{2}}x-{\frac{1}{8}}{x}^{2}+{\frac{1}{16}}{x}^{3}-{\frac{5}
{128}}{x}^{4}+O \left( {x}^{5} \right),
\end{gather*}
and
\begin{align*}
&\left(1+{\frac{1}{2}}x-{\frac{1}{8}}{x}^{2}+{\frac{1}{16}}{x}^{3}-{\frac{5}
{128}}{x}^{4}\right)^2\\
=&1+x-{\frac {5\,{x}^{7}}{1024}}+{\frac {25\,{x}^{8}}{16384}}+{\frac {7\,{
x}^{6}}{512}}-{\frac {7\,{x}^{5}}{128}}.
\end{align*}
Furthermore, if expand it to $6$th order, then
\begin{gather*}
\sqrt{1+x}=1+{\frac{1}{2}}x-{\frac{1}{8}}{x}^{2}+{\frac{1}{16}}{x}^{3}-{\frac{5}
{128}}{x}^{4}+{\frac{7}{256}}{x}^{5}-{\frac{21}{1024}}{x}^{6}+O
\left( {x}^{7} \right),
\end{gather*}
then we have
\begin{align*}
&\left(1+{\frac{1}{2}}x-{\frac{1}{8}}{x}^{2}+{\frac{1}{16}}{x}^{3}-{\frac{5}
{128}}{x}^{4}+{\frac{7}{256}}{x}^{5}-{\frac{21}{1024}}{x}^{6}\right)^2\\
=&1+x-{\frac {147\,{x}^{11}}{131072}}+{\frac {441\,{x}^{12}}{1048576}}-{
\frac {77\,{x}^{9}}{16384}}+{\frac {77\,{x}^{10}}{32768}}-{\frac {33\,
{x}^{7}}{1024}}+{\frac {165\,{x}^{8}}{16384}}.
\end{align*}
Thus, by examples, it seems that the statement above is true. But I do not know how to prove it. Can you help me?
| This is actually a very good exercise in combinatorics and in multiplying polynomials so I will post an answer.
Firstly note that the following identity holds true:
\begin{equation}
1= \sum\limits_{J=0}^{n k} \delta_{j_1+\cdots+j_n,J}
\end{equation}
Now we insert the unity into the left hand side of (1) and we expand the whole thing and we get:
\begin{eqnarray}
lhs(1) = \sum\limits_{J=0}^{n k} x^J \cdot \underbrace{\left( \sum\limits_{j_1=0}^k \sum\limits_{j_2=0}^k \cdots \sum\limits_{j_n=0}^k \left[\prod\limits_{\xi=1}^n \binom{\frac{1}{n}}{j_\xi}\right] \cdot \delta_{j_1+\cdots+j_n,J}\right)}_{{\mathfrak a}_J}
\end{eqnarray}
Now we evaluate the coefficients $\left\{ {\mathfrak a}_J \right\}_{J=0}^{n k}$ for consecutive values of $J=0,1,2,3,4,\cdots,k,k+1,\cdots,n \cdot k$. We have:
\begin{eqnarray}
{\mathfrak a}_0 &=& 1 \\
{\mathfrak a}_1 &=& n \binom{\frac{1}{n}}{1} = 1 \\
{\mathfrak a}_2 &=& \binom{n}{2} \cdot \binom{\frac{1}{n}}{1}^2 + n \binom{\frac{1}{n}}{2} = 0 \\
{\mathfrak a}_3 &=& \binom{n}{3} \cdot \binom{\frac{1}{n}}{1}^3+\binom{n}{2}(2 \binom{\frac{1}{n}}{2} \binom{\frac{1}{n}}{1}) + n \binom{\frac{1}{n}}{3} = 0 \\
{\mathfrak a}_4 &=& \binom{n}{4} \cdot \binom{\frac{1}{n}}{1}^4+ \binom{n}{3}(3 \binom{\frac{1}{n}}{2} \binom{\frac{1}{n}}{1}^2) +\binom{n}{2} (2 \binom{\frac{1}{n}}{3} \binom{\frac{1}{n}}{1} + \binom{\frac{1}{n}}{2}^2)+n \binom{\frac{1}{n}}{4} =0
\end{eqnarray}
Let us analyze the last line above. We can achieve $J=4$ in four different ways(from the left to the right), firstly by taking four distinct $j$-indices equal to unity, secondly by taking one $j$-index equal to two and two other ones equal to one, thirdly by taking one index equal to three and another one to one or two distinct indices being equal to two and fourthly by taking exactly one index equal to four. The corresponding numbers of ways for achieving that are given by the respective binomial factors which stand in front of each term in the right hand side of the last line.
Now, clearly for bigger values of $J$ nothing much will change except that we only have to enumerate more possibilities and the expressions become lengthy. The only interesting thing happens when $J$ hits the value $J=k+1$. Then clearly the very last term on the right hand side is missing and therefore:
\begin{eqnarray}
{\mathfrak a}_{k+1} &=& -n \binom{\frac{1}{n}}{k+1}
\end{eqnarray}
which gives ${\mathfrak a}_5 = - 7/128$ for $(n,k)=(2,4)$ and ${\mathfrak a}_7 = - 33/1024$ for $(n,k)=(2,6)$ as in the question above. The higher order coefficients are more complicated but it is clear how they can be extracted. As $J$ increases there will be more and more terms missing on the very right of the right hand side. One needs to extract those terms and carefuly sum them up to get the coefficient in question.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2239782",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 0
} |
Find $\sin A$ and $\cos A$ if $\tan A+\sec A=4 $ How to find $\sin A$ and $\cos A$ if
$$\tan A+\sec A=4 ?$$
I tried to find it by $\tan A=\dfrac{\sin A}{\cos A}$ and $\sec A=\dfrac{1}{\cos A}$, therefore
$$\tan A+\sec A=\frac{\sin A+1}{\cos A}=4,$$
which implies
$$\sin A+1=4\cos A.$$
Then what to do?
| Together with
$$\sin A+1=4\cos A $$
you can use $\sin^2A+\cos^2A=1$ as
$$(\sin A+1)(\sin A-1)=\cos^2 A\ .$$
Putting the two together you easily get
$$4\cos A(\sin A-1)=\cos^2 A\ ,$$
and hence
$$4\sin A-4=\cos A\ .$$
You now just have to solve the linear system in $\sin A$ and $\cos A$:
$$\begin{cases}
4\sin A-4=\cos A\\
\sin A+1=4\cos A
\end{cases}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2243571",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 5,
"answer_id": 4
} |
The number of solutions of the equation $\tan x +\sec x =2\cos x$ lying in the interval $[0, 2\pi]$ is: The number of solutions of the equation $\tan x +\sec x =2\cos x$ lying in the interval $[0, 2\pi]$ is:
$a$. $0$
$b$. $1$
$c$. $2$
$d$. $3$
My Attempt:
$$\tan x +\sec x=2\cos x$$
$$\dfrac {\sin x}{\cos x}+\dfrac {1}{\cos x}=2\cos x$$
$$\sin x + 1=2\cos^2 x$$
$$\sin x +1=2-2\sin^2 x$$
$$2\sin^2 x +\sin x - 1=0$$
$$2\sin^2 x +2\sin x -\sin x - 1=0$$
$$2\sin x (\sin x +1) -1(\sin x+1)=0$$
$$(\sin x +1)(2\sin x-1)=0$$
So, what's the next?
| The answer must be 2
Since tanx and secx domain would consider every x, rejecting (2n+1)π/2 values in the interval [0,2π] such values would be rejected from the solution.
In the last step sinx=1 the solution would be 3π/2 and this would be rejected due to domain condition. The other part, that is, sinx=1/2 has solution x=5π/6 and x= π/6 in the Interval [0,2π].
Thus the number of solutions are 2
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2247408",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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} |
why$2 (-1 + 2^{1 + n})$ is the answer to the recurrence relation $a_{n}=2a_{n-1}+2$? $a_{0}=2$
$a_{1}=2(2)+2$
$a_{2}=2(2(2)+2)+2$
$a_{3}=2(2(2(2)+2)+2)+2$
$a_{4}=2(2(2(2(2)+2)+2)+2)+2$
$a_{5}=2(2(2(2(2(2)+2)+2)+2)+2)+2$
To simplifiy
$a_{6}=2^{6}+2^{5}...2^{1}$
so my answer is
$a_{n}=2^{n+1}+2^{n}+...2^{1}$
The correct answer is $2 (-1 + 2^{1 + n})$
How do I make this transition?
| The inhomogeneous recurrence relation
$$
a_n = 2 a_{n-1} + 2
$$
can be turned into a homogeneous recurrence
$$
a_n - a_{n-1} = 2 a_{n-1} + 2 - (2 a_{n-2} + 2) = 2 a_{n-1} - 2 a_{n-2} \iff \\
a_n = 3 a_{n-1} - 2 a_{n-2}
$$
and solved by the usual algorithm.
The characteristic polynomial is
$$
p(t) = t^2 - 3 t + 2
$$
with roots $r_1 = 1$ and $r_2 = 2$.
So the solution is
$$
a_n = k_1 r_1^n + k_2 r_2^n = k_1 + k_2 2^n
$$
The initial elements give
$$
a_0 = k_1 + k_2 = 2 \\
a_1 = k_1 + 2 k_2 = 6
$$
This gives $k_2 = 4$ and $k_1 = -2$. The solution is
$$
a_n = -2 + 4 \cdot 2^n = 2^{n+2} - 2
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2249216",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 6,
"answer_id": 1
} |
$2x^2 + 3x +4$is not divisible by $5$ I tried by $x^2 \equiv 0, 1, 4 \pmod 5$ but how can I deal with $3x$?
I feel this method does not work here.
| $$5|\;(2x^2+3x+4)\iff$$ $$ \iff5|\;3(2x^2+3x+4)=(6x^2+9x+12)\iff$$ $$\iff 5|\;((6x^2+9x+12)-(5x^2+5x+10))=$$ $$=(x^2+4x+2)=(x+2)^2-2.$$ But no square is $2$ more than a multiple of $5.$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2250718",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 6,
"answer_id": 5
} |
$\lim_{x \to 0} \frac{\log \left(\cosh\left(x^2-xc\right) \right)}{x^2}=\frac{c^2}{2}$ without L'Hospital's rule How to show that without using L'Hospital's rule
\begin{align}
\lim_{x \to 0} \frac{\log \left(\cosh\left(x^2-xc\right) \right)}{x^2}=\frac{c^2}{2}
\end{align}
I was able to show the upper bound by using the bound $\cosh(x) \le e^{x^2/2}$
\begin{align}
\lim_{x \to 0} \frac{\log \left(\cosh\left(x^2-xc\right) \right)}{x^2} \le \lim_{x \to 0} \frac{\left(x^2-xc\right)^2}{2x^2}=\frac{c^2}{2}
\end{align}
My question: How finish this argument.
| Denote $a=x^2-cx$ for simplicity. Then
\begin{align}
\frac{\ln(\cosh a )}{x^2}&=\frac{\ln(\cosh^2a)}{2x^2}=\color{blue}{\frac{\ln(1+\sinh^2a)}{2x^2}}=\frac12\cdot\frac{\ln(1+\sinh^2a)}{\sinh^2a}\cdot\left(\frac{\sinh a}{x}\right)^2=\\
&=\frac12\cdot\frac{\ln(1+\sinh^2a)}{\sinh^2a}\cdot\left(\frac12\cdot\left[\frac{e^a-1}{a}+\frac{e^{-a}-1}{-a}\right]\cdot\frac{a}{x}\right)^2\to\frac{c^2}{2}.
\end{align}
P.S. If you want a lower bound then using the inequality $e^a-1\ge a$ we can estimate
$$
\sinh^2 a=\left(\frac12\left[\frac{e^a-1}{a}+\frac{e^{-a}-1}{-a}\right]\right)^2a^2\ge a^2
$$
and continue from the blue expression above as
$$
\color{blue}{\frac{\ln(1+\sinh^2a)}{2x^2}}\ge\frac{\ln(1+a^2)}{2x^2}=\frac12\cdot\frac{\ln(1+a^2)}{a^2}\cdot\frac{a^2}{x^2}\to \frac{c^2}{2}.
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2252022",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 5,
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If $ab$ is a square number and $\gcd(a,b) = 1$, then $a$ and $b$ are square numbers.
Let $n, a$ and $b$ be positive integers such that $ab = n^2$. If $\gcd(a, b) = 1$, prove that there exist positive integers $c$ and $d$ such that $a = c^2$ and $b = d^2$
So far I have tried this:
Since $n^2 = ab$ we have that $n = \sqrt{ab}$.
Because $\gcd(a,b) = 1$, there exists integers $k$ and $l$ such that $ak + bl = 1$. This means that $\sqrt{a}(k\sqrt{}) + \sqrt{b}(l\sqrt{b}) = 1$.
Hence $\sqrt{a}$ and $\sqrt{b}$ are both positive integers and we can set $\sqrt{a} = c$ for some arbitrary integer $c$ and $\sqrt{b} = d$ for some arbitrary integer $d$. Therefore, $a = c^2$ and $b = d^2$.
| Let $ab=c^2$ for some $c\in N$ then the result will hold if any one of the integers is 1 as 1^2=1.
So let us take a>1,b>1 and c>1. We can use prime factorization and represent the integers as follows:
a=$p_1^{d_1} * p_2^{d_2}$
b= $q_1^{e_1} * q_2^{e_2}$
and c=$k_1^{l_1} *k_2^{l_2}$
thus $ab=c^2$ becomes
$p_1^{d_1} * p_2^{d_2} * q_1^{e_1} * q_2^{e_2} = k_1^{2l_1} *k_2^{2l_2}$
(a,b)=1 so all the primes $p_1...$ are different from $q_1...$. Hence we can say that p and q are just rearrangement of k and each of $d_i$ and $e_j$ are also similar rearrangement of $2l_1....2_r$. Hence each of $d_i$ and $e_j$ are even.
Using this we can also prove that $ab=c^n$ then $a=x^n$ and $b=y^n$
P.s Extension of Alex's answer
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 4
} |
Find a non-zero integer matrix $X$ such that $XA=0$ where $X,A,0$ are all $4 \times 4$ Let $A$ be the following $4 \times 4$ matrix.
\begin{bmatrix}1&2&1&3\\1&3&2&4\\2&5&3&7\\1&4&1&5\end{bmatrix}
How can we find a non-zero integer matrix $C$ such that $CA = 0_{4 \times 4}$
Note that $0$ is a $4 \times 4$ matrix.
| Hint : The rank of $A$ is $3$ so : $$\exists P,Q \in GL_n(\mathbb R),A=P\begin{bmatrix}1&0&0&0\\0&1&0&0\\0&0&1&0\\0&0&0&0\end{bmatrix}Q$$
$$CA=0 \Rightarrow CP\begin{bmatrix}1&0&0&0\\0&1&0&0\\0&0&1&0\\0&0&0&0\end{bmatrix}Q=0\Rightarrow C\begin{bmatrix}1&0&0&0\\0&1&0&0\\0&0&1&0\\0&0&0&0\end{bmatrix}=Q^{-1}P^{-1}$$
Calculate $P$ and $Q$ to find a necessary condition on $C$ and take a matrix fulfilling this condition and verify it is indeed a solution to your problem.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2255388",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 2
} |
Solving $8yy'^2 - 2xy' + y = 0$ I'm solving the differential equation $8yy'^2 - 2xy' + y = 0$
My attempt:
We divide both sides by $x$, obtaining:
$$8\frac{y}{x}y'^2 - 2y' + \frac{y}{x} = 0$$
Then, we introduce $t = y'$, hence the differential equation becomes:
$$8\frac{y}{x}t^2 - 2t + \frac{y}{x} = 0$$
from which follows:
$$\frac{y}{x} = \frac{2t}{8t^2 +1}$$
Hence, in parametric equation we obtain:
$$\begin{cases} y' = t \\ \frac{y}{x} = \frac{2t}{8t^2 +1}\end{cases}$$
From the second equation, after differentiating with respect to $t$, we obtain:
$$\frac{dy}{dt} = \frac{dx}{dt}\frac{2t}{8t^2 + 1} + \frac{-16t^2 + 2}{(8t^2 +1)^2}$$
Hence:
$$dy = dx\frac{2t}{8t^2 + 1} + \frac{-16t^2 + 2}{(8t^2 +1)^2}dt$$
From the first equation we have $dy = tdx$
Therefore:
$$tdx= dx\frac{2t}{8t^2 + 1} + \frac{-16t^2 + 2}{(8t^2 +1)^2}dt$$
Or after rearranging:
$$dx = \frac{-2dt}{(8t^2+1)dt}$$
And by integrating:
$$x(t) = -2\ln|t| - \frac{1}{2}\ln|8t^2 + 1| + c$$
And by the first equation:
$$y(t) = \frac{2t}{8t^2+1}(-2\ln|t| - \frac{1}{2}\ln|8t^2 + 1| + c)$$
Hence, the solution in parametric equation, is:
$$\begin{cases} x(t) = -2\ln|t| - \frac{1}{2}\ln|8t^2 + 1| + c \\ y(t) = \frac{2t}{8t^2+1}(-2\ln|t| - \frac{1}{2}\ln|8t^2 + 1| + c) \end{cases}$$
Can someone verify whether this is correct? The answer my book gives is $$y^2 - 4cx + 32c^2 = 0$$ with singular integral $$8y^2 - x^2 = 0$$ How would I derive this answer?
Thanks in advance.
| You lost the factor $x$ in the last term while differentiating the parametric equation.
It is even easier to multiply with $y$ and then substitute $u=y^2$ to get
$$
2u'^2-xu'+u=0\iff u=xu'-2u'^2
$$
which is a Clairaut differential equation. This has the lines
$$
u=cx-2c^2
$$
as solutions and their envelope which is the non-linear solution to $0=(x-4u')u''$. Thus inserting $u'=x/4$ gives
$$
u=\frac{x^2}4-\frac{x^2}8=\frac{x^2}8
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2255914",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Find the maximum and minimum value of $P =x+y+z+xy+yz+zx$ Let $x^2+y^2+z^2\leq27$ and $P = x+y+z+xy+yz+zx$. Find the value of $x, y, z$ such that $P$ is the maximum value and minimum value.
My attempt :
$$(x-y)^2 + (y-z)^2 + (z-x)^2 \geq 0$$
$$27 \geq x^2+y^2+z^2 \geq xy+yz+zx\tag{1}$$
$$(x+y+z)^2 \leq 3(x^2+y^2+z^2) \le 3 \cdot 27$$
$$(x+y+z)^2 \leq 81$$
$$x+y+z \leq 9\tag{2}$$
From $(1), (2)$, $ x+y+z+xy+yz+zx \leq 36$, so $P_{\text{max}} = 36$ with equality hold at $x=y=z=3$.
Please suggest how to find $P_{\text{min}}$.
| You may use the same method to find the minimum. First, we obtain a lower bound for $P$:
$$
\begin{align}
P&=x+y+z+xy+yz+zx\\
&=\frac12 [ (x+y+z+1)^2 - (x^2+y^2+z^2) - 1 ]\\
&\ge\frac12 (0 - 27 - 1)\tag{1}\\
&= -14.
\end{align}
$$
Next, note that at $\left(\frac{\sqrt{53}-1}2,-\frac{\sqrt{53}+1}2,0\right)$, we have $x+y+z+1=0$ and $x^2+y^2+z^2=27$. Hence tie can occur in $(1)$ and the lower bound $-14$ is attainable.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2256277",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 0
} |
Find the exact value of $A(\beta)=8\pi-16\sin(2\beta)$ with $\tan(\beta)= \frac{1}{2}$
The picture below represents a semi-circumference of diameter [AB] and
center C. Point D belongs to the semi-circumference and it's one of
the vertices of the triangle $ABC$. Consider that BÂD = $\beta (\beta
\in ]0,\frac{\pi}{2}[)$ and AC = 4.
The area of the pink part of the picture is given by $$A(\beta) =
8\pi-16\sin(2\beta)$$
Find the exact value of the area of the pink part with $\tan(\beta)=
\frac{1}{2}$
I tried:
$$\tan \beta = \frac{\sin(\beta)}{\cos(\beta)} = \frac{1}{2}\\ \Leftrightarrow \sin(\beta) = \frac{\cos(\beta)}{2} \\ \Leftrightarrow \sin(\beta) = \frac{\sqrt{1-\sin^2\beta}}{2}\\ \Leftrightarrow ???$$
What do I do next?
| Given: $\tan \beta = \frac 12$, so $2\sin\beta = \cos \beta$
$\sin 2\beta = 2\sin\beta\cos\beta = \cos^2\beta = \frac 1{\sec^2 \beta} = \frac 1{1 + \tan^2 \beta} = \frac 1{ 1 + \frac 14} = \frac 45$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2260722",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 0
} |
How can we integrate integral(s) of this type? So I was able to free $dx$ from the power. Now only wolfram can solve this Integral. How can I do this on my own?
$$r=\int_0^1\left(\frac{x^{12}}{(1-x^4)^3}+1\right)^{1/4}~dx$$
| Hint:
$\int_0^1\left(\dfrac{x^{12}}{(1-x^4)^3}+1\right)^\frac{1}{4}~dx$
$=\int_0^1\left(\dfrac{x^3}{(1-x)^3}+1\right)^\frac{1}{4}~d\left(x^\frac{1}{4}\right)$
$=\dfrac{1}{4}\int_0^1x^{-\frac{3}{4}}\left(\dfrac{x^3}{(1-x)^3}+1\right)^\frac{1}{4}~dx$
$=\dfrac{1}{4}\int_1^0(1-x)^{-\frac{3}{4}}\left(\dfrac{(1-x)^3}{x^3}+1\right)^\frac{1}{4}~d(1-x)$
$=\dfrac{1}{4}\int_0^1(1-x)^{-\frac{3}{4}}\left(\dfrac{3x^2-3x+1}{x^3}\right)^\frac{1}{4}~dx$
$=\dfrac{1}{4}\int_0^1x^{-\frac{3}{4}}(1-x)^{-\frac{3}{4}}(3x^2-3x+1)^\frac{1}{4}~dx$
Which relates to Appell Hypergeometric Function
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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} |
Parametrization of two surfaces $\frac{x^2}{a^2}-\frac{y^2}{b^2}-\frac{z^2}{c^2}=1$ and $\frac{x^2}{p}+\frac{y^2}{q}=2z$. Can someone please help me to parametrize the following surfaces in terms of hyperbolic(for second it might not be possible but i need some more convenient set of parametric equation than mine ) and trigonometric functions
$$\frac{x^2}{a^2}-\frac{y^2}{b^2}-\frac{z^2}{c^2}=1 $$ and $$\frac{x^2}{p}+\frac{y^2}{q}=2z$$
I have tried to do but the set of parametric equations I got were too complicated as I have to use those in some further calculation which makes the result very ugly.
For first equation the set of parametric equations is: $$x=a\sqrt{1+\frac{u^2}{c^2}}\cos v, \ \ y=b\sqrt{1+\frac{u^2}{c^2}}\sin v \ \ z=u$$
and for second: $$x=\sqrt{2pu} \cos v ,\ \ y=\sqrt{2qu} \sin v, \ \ z=u $$
| For the first: $$\frac{x^2}{a^2}\color{red}{-}\left(\frac{y^2}{b^2}+\frac{z^2}{c^2}\right)=1$$
\begin{eqnarray}
&x&=a\cosh\theta,\\
&y&=b\cos\phi\sinh\theta,\\
&z&=c\sin\phi\sinh\theta.\\
\end{eqnarray}
| {
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"url": "https://math.stackexchange.com/questions/2263853",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Computing maximum dimension of a vector subspace given that it's every element is a symmetric matrix and is closed under matrix multiplication
Q. Let $S$ be a subspace of the vector space of all $11 \times 11$ real matrices such that (i) every matrix in $S$ is symmetric and (ii) $S$ is closed under matrix multiplication. What is the maximum possible dimension of $S$?
Attempt : Subspace of symmetric matrices has dimension $\frac {n(n+1)}2$ for the vector space of $n \times n$ matrices.
Here in this case, it is $\frac {11\times 12}2=66$.
Next, using (ii) we have that $(AB)^T=AB$. But $(AB)^T=B^T A^T=BA. (\because A^T=A, B^T=B)$. Hence $AB=BA$. So we conclude that elements in $S$ commute with each other.
In case of vector spaces of $2\times 2$ and $3 \times 3$ symmetric matrices, I found that only diagonal matrices commute with each other.
For e.g. in $2 \times 2$ case, I just multiplied the basis elements of the subspace of symmetric matrices $$
\begin{pmatrix}
1 & 0 \\
0 & 0 \\
\end{pmatrix}
\cdot
\begin{pmatrix}
0 & 0 \\
0 & 1 \\
\end{pmatrix}
= \begin{pmatrix}
0 & 0 \\
0 & 1 \\
\end{pmatrix} \cdot \begin{pmatrix}
1 & 0 \\
0 & 0 \\
\end{pmatrix}$$ but $$ \begin{pmatrix}
1 & 0 \\
0 & 0 \\
\end{pmatrix} \cdot \begin{pmatrix}
0 & 1 \\
1 & 0 \\
\end{pmatrix} \neq \begin{pmatrix}
0 & 1 \\
1 & 0 \\
\end{pmatrix} \cdot \begin{pmatrix}
1 & 0 \\
0 & 0 \\
\end{pmatrix}$$
Similarly checking the commutativity of basis elements of space of $3 \times 3$ symmetric matrices also gives only the space of diagonal matrices as a candidate having commuting elements.
Like, $$
\begin{pmatrix}
0 & 1 & 0 \\
1 & 0 & 0 \\
0 & 0 & 0 \\
\end{pmatrix}
\cdot \begin{pmatrix}
0 & 0 & 0 \\
0 & 0 & 1 \\
0 & 1 & 0 \\
\end{pmatrix} \neq \begin{pmatrix}
0 & 0 & 0 \\
0 & 0 & 1 \\
0 & 1 & 0 \\
\end{pmatrix} \cdot \begin{pmatrix}
0 & 1 & 0 \\
1 & 0 & 0 \\
0 & 0 & 0 \\
\end{pmatrix}$$
Hence the vector space of all diagonal matrices satisfies (i) & (ii). Thus minimum possible 'maximum dimension' of $S$ is $11$. So we have that $11 \le \max \dim S \lt 66$.
Am I in right direction so far? What can be my next step?
| If $A$ and $B$ are symmetric, then $AB$ symmetric means $AB=(AB)^T=B^TA^T=BA$. All the matrices therefore commute. A symmetric
real matrix is diagonalisable, and pairwise commuting real matrices
are simultaneously diagonalisable. So the dimension of such a space is
at most $11$.
| {
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"timestamp": "2023-03-29T00:00:00",
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} |
Maximum of $x^3+y^3+z^3$ with $x+y+z=3$ It is given that, $x+y+z=3\quad 0\le x, y, z \le 2$ and we are to maximise $x^3+y^3+z^3$.
My attempt : if we define $f(x, y, z) =x^3+y^3 +z^3$ with $x+y+z=3$ it can be shown that,
$f(x+z, y, 0)-f(x,y,z)=3xz(x+z)\ge 0$ and thus $f(x, y, z) \le f(x+z, y, 0)$. This implies that $f$ attains it's maximum whenever $z=0$. (Is this conclusion correct? I have doubt here).
So the problem reduces to maximise $f(x, y, 0)$ which again can be shown that $f(x, y, 0)\le f(x, 2x,0)$ and this completes the proof with maximum of $9$ and equality at $(1,2,0)$ and it's permutations.
Is it correct? I strongly believe even it might have faults there must be a similar way and I might have made mistakes. Every help is appreciated
| $$
\begin{eqnarray}
&(x+y+z)^3 &=& x^3 + y^3 + z^3 + 3(x+y)(y+z)(z+x)\\
\implies & x^3 + y^3 + z^3 &=& (x+y+z)^3 - 3(x+y)(y+z)(z+x)\\
&& = & 27 - 3(x+y)(y+z)(z+x)
\end{eqnarray}
$$
Now, $x^3+y^3+z^3$ is maximum when $t = (x+y)(y+z)(z+x)$ is minimum. Now since $x$, $y$ and $z$ are each non-negative, therefore $t$ is non-negative. Also, $x,\,y,\,z \in [0,\,2]$. So, $t$ takes minimum value when the variables take values $0,\,1,\,2$. So, $t_\text{min} = (0+1)(1+2)(2+0) = 6$.
So $\max (x^3+y^3+z^3)=27-3\times6=9$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2268524",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 4
} |
If $\sin x + \sin^2 x =1$ then find the value of $\cos^8 x + 2\cos^6 x + \cos^4 x$ If $\sin x + \sin^2 x =1$ then find the value of $\cos^8 x + 2\cos^6 x + \cos^4 x$
My Attempt:
$$\sin x + \sin^2 x=1$$
$$\sin x = 1-\sin^2 x$$
$$\sin x = \cos^2 x$$
Now,
$$\cos^8 x + 2\cos^6 x + \cos^4 x$$
$$=\sin^4 x + 2\sin^3 x +\sin^2 x$$
$$=\sin^4 x + \sin^3 x + \sin^3 x + \sin^2 x$$
$$=\sin^3 x(\sin x +1) +\sin^2 x(\sin x +1)$$
$$=(\sin x +1) (\sin^3 x +\sin^2 x)$$
How do I proceed further?
| Hint:
$$\cos^8x+2\cos^6x+\cos^4x=(\cos^4x+\cos^2x)^2$$
Now as $\cos^2x=\sin x,\cos^4x=(\cos^2x)^2=?$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2269298",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 6,
"answer_id": 4
} |
$\epsilon - \delta$ proof for $\frac{x^2 - 16}{x + sin x}$ limit I'm having difficulty writing an $\epsilon - \delta$ proof for the following limit:
$\lim_{x\to 4} \frac{x^2-16}{x+\sin x} = 0$
I've factored it to $\frac{(x+4)(x-4)}{x+\sin x} = 0$
and guessed that I need $\delta = \frac{2}{5}\epsilon$ for $|x-4| < \delta \implies |\frac{x^2-16}{x+\sin x}| < \epsilon$
I've also bounded $|x+4|$ by $\delta + 8$ but I don't know how to control $|x + \sin x|$.
| $$|x-4| < \delta$$
$$4-\delta < x < 4+ \delta$$
$$3 - \delta< x+ \sin x < 5 + \delta$$
$$\frac{1}{5+\delta} < \frac{1}{x+\sin x} < \frac{1}{3-\delta}$$
If $\delta < 1$, $-\delta > -1$, $3-\delta > $2, $\frac{1}{3-\delta} < \frac12$
$$\left| \frac{x^2-16}{x+\sin x}\right| \leq \frac12 |x^2-16|$$
Also, if $\delta < 1$, $|x+4|<9,$
$$\frac12 |x^2-16|\leq \frac 92 |x-4|$$
Hence, for example, I can choose
$$\delta = \min ( \frac12, \frac29 \epsilon)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2270291",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Show that $a+b+c+\sqrt {\frac {a^2+b^2+c^2}{3}}\le4$, Let $a,b,c$ be positive real numbers such that $a^2+b^2+c^2+abc=4.$ Show that $$a+b+c+\sqrt {\frac {a^2+b^2+c^2}{3}}\le4.$$
| Let $a=\frac{2x}{\sqrt{(x+y)(x+z)}}$ and $b=\frac{2y}{\sqrt{(x+y)(y+z)}},$ where $x$, $y$ and $z$ are positives.
Hence, $c=\frac{2z}{\sqrt{(x+z)(y+z)}}$ and we need to prove that
$$\sum_{cyc}\frac{2x}{\sqrt{(x+y)(x+z)}}+\sqrt{\frac{4}{3}\sum_{cyc}\frac{x^2}{(x+y)(x+z)}}\leq4$$ or
$$\sum_{cyc}x\sqrt{y+z}+\sqrt{\frac{1}{3}\sum_{cyc}(x^2y+x^2z)}\leq2\sqrt{(x+y)(x+z)(y+z)}.$$
Let $\sum\limits_{cyc}(x^2y+x^2z)=6kxyz$.
Hence, by C-S
$$\sum_{cyc}x\sqrt{y+z}\leq\sqrt{\sum_{cyc}x\sum_{cyc}x(y+z)}=\sqrt{2\sum_{cyc}(x^2y+x^2z+xyz)}.$$
Thus, it's enough to prove that
$$\sqrt{2(6k+3)}+\sqrt{\frac{1}{3}\cdot6k}\leq2\sqrt{6k+2}$$ or
$$\sqrt{6k+3}+\sqrt{k}\leq2\sqrt{3k+1},$$
which is C-S again:
$$\sqrt{6k+3}+\sqrt{k}=3\sqrt{\frac{2k+1}{3}}+\sqrt{k}\leq\sqrt{(3+1)\left(3\cdot\frac{2k+1}{3}+k\right)}=2\sqrt{3k+1}.$$
Done!
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2271863",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 1
} |
Find the sum $\sum_{k=1}^\infty {\frac{6^k}{\left(3^k-2^k\right) \left(3^{k+1}-2^{k+1}\right)}}$ I need to find the sum,
$$\sum_{k=1}^\infty {\frac{6^k}{\left(3^k-2^k\right) \left(3^{k+1}-2^{k+1}\right)}}$$
I have tried to break the terms into partial fractions (method of differences) but am not able to do so. How to proceed?
| First we can try to split things into two pieces:
$$\frac{6^k}{\left(3^k-2^k\right) \left(3^{k+1}-2^{k+1}\right)} = \dfrac{A}{3^k-2^k} + \dfrac{B}{3^{k+1}-2^{k+1}}$$
So we have $A \cdot (3^{k+1}-2^{k+1}) + B \cdot (3^k-2^k) = 6^k$ which can be arranged to $3^k (3 A + B) - 2^k (2 A + B) = 6^k$.
If we make $2A+B=0$ and $3A+B=2^k$, this equality will hold. This suggests that $B=-2A$ and we also see that $3A-2A=A=2^k$. Therefore $B=-2^{k+1}$.
$$S = \sum_{k=1}^{\infty} \left(\dfrac{2^k}{3^k - 2^k} - \dfrac{2^{k + 1}}{3^{k + 1} - 2^{k + 1}}\right)$$
Now it's looking like a telescoping series. This is more easily seen by displaying a few terms:
$$S = \left(\dfrac{2}{1} - \dfrac{4}{5}\right) + \left(\dfrac{4}{5} - \dfrac{8}{19}\right) + \left(\dfrac{8}{19} - \dfrac{16}{65}\right) + ...$$
Most of these fractions cancel each other out.
$$S = \lim_{K \rightarrow \infty} \sum_{k=1}^{K} \left(\dfrac{2^k}{3^k - 2^k} - \dfrac{2^{k + 1}}{3^{k + 1} - 2^{k + 1}}\right)= \lim_{K \rightarrow \infty} \left(2 - \dfrac{2^{K + 1}}{3^{K + 1} - 2^{K + 1}}\right) = 2$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2272522",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Calculate $\sqrt{2i}$ I did:
$\sqrt{2i} = x+yi \Leftrightarrow i = \frac{(x+yi)^2}{2} \Leftrightarrow i = \frac{x^2+2xyi+(yi)^2}{2} \Leftrightarrow i = \frac{x^2-y^2+2xyi}{2} \Leftrightarrow \frac{x^2-y^2}{2} = 0 \land \frac{2xy}{2} = 1$
$$\begin{cases}
\frac{x^2-y^2}{2} = 0 \\
xy = 1\\ \end{cases} \\
=\begin{cases}
x^2-y^2 = 0 \\
x = \frac{1}{y}\\ \end{cases} \\
=\begin{cases}
\frac{1}{y}-y^2 = 0 \\
x = \frac{1}{y}\\ \end{cases} \\=
\begin{cases}
\frac{1-y^3}{y} = 0 \\
-\\ \end{cases}
\\=
\begin{cases}
y^3 = 1 \\
-\\ \end{cases}
\\=
\begin{cases}
y = 1 \\
x =1\\ \end{cases} $$
And so $\sqrt{2i} = 1+i$, but my book states the solution is $\sqrt{2i} = 1+i$ and $\sqrt{2i} = -1-i$.
What did I forget?
| Write $2i=2e^{i \pi/2+2k\pi}$. Then square root to get: $\sqrt{2} e^{i\pi/4+k\pi}$. So your roots are $\sqrt{2}e^{i\pi/4}$ and $\sqrt{2}e^{3i\pi/4}$. Which are $\pm(1+i)$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2273345",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 2
} |
Domain of $f(x)=\sqrt{\lfloor x\rfloor-1+x^2}$ I drew the number line and tested with different values, getting the correct domain $(-\infty,-\sqrt3)\cup[1,\infty)$. However, how do I solve this faster by manipulating the function?
| You must find the values of $x$ such that $\lfloor x\rfloor-1+x^2\geq0$.
The easy part:
*
*It's true for $x\geq1$, since $x^2\geq1$ and $\lfloor x\rfloor>0$.
*For $0\leq x<1$, it's false, because $\lfloor x\rfloor=0$ and $x^2<1$.
Now, the case $x<0$. First notice that for $x\in[n,n+1[$, for integer $n$, you have $\lfloor x\rfloor=n$, hence $\lfloor x\rfloor-1+x^2=x^2+n-1$. For $n<0$, its infimum, on $[n,n+1[$, is found when $x\to n+1$, and this infimum is $(n+1)^2+n-1=n^2+3n=n(n+3)$. Hence, for $n<-2$, it's nonnegative.
There are two intervals left, $[-2,-1[$ and $[-1,0[$.
*
*On $[-1,0[$, $\lfloor x\rfloor-1+x^2=x^2-2<0$ because $x^2\leq1$.
*On $[-2,-1[$, $\lfloor x\rfloor-1+x^2=x^2-3$. This is nonnegative for $x\leq-\sqrt{3}$, hence, for $-2\leq x\leq-\sqrt3$.
All in all, $\lfloor x\rfloor-1+x^2\geq0$ for $x\in]-\infty,-\sqrt3]\cup[1,+\infty[$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2274330",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 0
} |
Completing the square of $x^2 - mx = 1$ is not giving me the right answer. This is my attempt
$$
\begin{align}
x^2 - mx &= 1 \\
x^2 - mx - 1 &= 0 \\
\left(x^2 - mx + \frac{m^2}{4} - \frac{m^2}{4}\right) - 1 &= 0 \\
\left(x^2 - mx + \frac{m^2}{4}\right) - \frac{m^2}{4} - 1 &= 0 \\
\left(x^2 - mx + \frac{m^2}{4}\right) - \frac{m^2}{4} - \frac{4}{4} &= 0 \\
\left(x^2 - mx + \frac{m^2}{4}\right) - \frac{m^2 - 4}{4} &= 0 \\
\left(x^2 - mx + \frac{m^2}{4}\right) &= \frac{m^2 - 4}{4} \\
\left(x - \frac{m}{2}\right)^2 &= \frac{m^2 - 4}{4} \\
\sqrt{\left(x - \frac{m}{2}\right)^2} &= \sqrt{\frac{m^2 - 4}{4}} \\
x - \frac{m}{2} &= \pm \frac{\sqrt{m^2 - 4}}{\sqrt{4}} \\
x &= \frac{m}{2} \pm \frac{\sqrt{m^2 - 4}}{2} \\[20pt]
x_1 &= \frac{m}{2} - \frac{\sqrt{m^2 - 4}}{2} \\
x_1 &= \frac{m - \sqrt{m^2 - 4}}{2} \\[16pt]
x_2 &= \frac{m}{2} + \frac{\sqrt{m^2 - 4}}{2} \\
x_2 &= \frac{m + \sqrt{m^2 - 4}}{2} \\
\end{align}
$$
However, the correct answer according to the text is:
$$
\begin{align}
x_1 &= \frac{m}{2} - \frac{\sqrt{m^2 + 4}}{2} \\
x_2 &= \frac{m + \sqrt{m^2 + 4}}{2} \\
\end{align}
$$
Why $\sqrt{m^2 + 4}$ instead of $\sqrt{m^2 - 4}$ ???
| (Not an answer, just a long comment.)
Your actual question has already been answered, but I want to point out another mistake, namely when you go from
$$\left(x - \frac{m}{2}\right)^2 = \frac{m^2 - 4}{4}$$
to
$$\sqrt{\left(x - \frac{m}{2}\right)^2} = \sqrt{\frac{m^2 - 4}{4}}.$$
At this point, there should be $\pm$ signs before the $\surd$ signs (it's redundant to put $\pm$ on both sides of the equation, but there needs to be a $\pm$ on at least one of them). You've added the $\pm$ signs in the next line, so it's not the end of the world, but conceptually it's important to understand where to put them.
Here's a simpler example: suppose we have $x^2=9$. Taking square roots, we have
$$x = \pm\sqrt9$$
so $x=\pm3$. Note that $\sqrt9$ in itself means only $3$, not $\pm 3$! It is not the $\surd$ sign, but rather the act of taking the square root, that engenders the $\pm$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2275086",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Area of triangle and determinant The area of a $\vartriangle ABC$ with given vertices $(a,a^2),(b,b^2),(c,c^2)$ is $\frac{1}{4}$ $sq. units$ and area of another $\vartriangle PQR$ with given vertices $(p,p^2),(q,q^2),(r,r^2)$ is $3$ $sq. units$.
Then what is the value of
$$
\begin{vmatrix}
(1+ap)^2 & (1+bp)^2 & (1+cp)^2 \\
(1+aq)^2 & (1+bq)^2 & (1+cq)^2 \\
(1+ar)^2 & (1+br)^2 & (1+cr)^2 \\
\end{vmatrix}
$$
I could not even begin attempting it , i don't know where to begin from,someone kindly help.
| Let $A, B, C$, $P, Q, R$ be the $6$ column vectors
$$
\begin{cases}
A^T = (1, \sqrt{2}a, a^2),\\
B^T = (1, \sqrt{2}b, b^2),\\
C^T = (1, \sqrt{2}c, c^2)
\end{cases}
\quad\text{ and }\quad
\begin{cases}
P^T = (1, \sqrt{2}p, p^2),\\
Q^T = (1, \sqrt{2}q, q^2),\\
R^T = (1, \sqrt{2}r, r^2)
\end{cases}
$$
Using identites of the form
$$(1+ap)^2 = 1 + 2ap + a^2p^2 = 1\cdot 1 + \sqrt{2}a\cdot\sqrt{2}p + a^2\cdot
p^2 = A\cdot P$$
We can rewrite the determinant at hand as
$$\Delta \stackrel{def}{=}\begin{vmatrix}
(1+ap)^2 & (1+bp)^2 & (1+cp)^2 \\
(1+aq)^2 & (1+bq)^2 & (1+cq)^2 \\
(1+ar)^2 & (1+br)^2 & (1+cr)^2 \\
\end{vmatrix}
= \begin{vmatrix}
A\cdot P & B\cdot P & C\cdot P \\
A\cdot Q & B\cdot Q & C\cdot Q \\
A\cdot R & B \cdot R & C\cdot R \\
\end{vmatrix}
$$
Notice the matrix for rightmost determinant is a product of two $3 \times 3$ matrices
$$
\begin{bmatrix}
A\cdot P & B\cdot P & C\cdot P \\
A\cdot Q & B\cdot Q & C\cdot Q \\
A\cdot R & B \cdot R & C\cdot R \\
\end{bmatrix}
= \left[ P, Q, R\right]^T \left[A, B, C\right]
$$
This leads to (up to a sign),
$$\Delta = \begin{vmatrix}
1 & \sqrt{2}p & p^2 \\
1 & \sqrt{2}q & q^2 \\
1 & \sqrt{2}r & r^2 \\
\end{vmatrix}
\begin{vmatrix}
1 & 1 & 1\\
\sqrt{2}a & \sqrt{2}b & \sqrt{2}c \\
a^2 & b^2 & c^2 \\
\end{vmatrix}
= 2
\begin{vmatrix}
1 & p & p^2 \\
1 & q & q^2 \\
1 & r & r^2 \\
\end{vmatrix}
\begin{vmatrix}
1 & a & a^2 \\
1 & b & b^2 \\
1 & c & c^2 \\
\end{vmatrix}
= 2(2\times 3)(2\times\frac14) = 6
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2276979",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
If $ x^4+x^3+x^2+x+1=0$ then what is the value of $x^5$ If $$x^4+x^3+x^2+x+1=0$$ then what's the value of $x^5$ ??
I thought it would be $-1$ but it does not satisfy the equation
| Well, we have $x^5-1=(x-1)(x^4+x^3+x^2+x+1)=(x-1)0=0$ so that $x^5=1$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2277392",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
How to find the maximum of the value $\frac{\sin{x}+1}{\sqrt{3+2\cos{x}+\sin{x}}}$ Find the maximun of the value
$$f(x)=\dfrac{\sin{x}+1}{\sqrt{3+2\cos{x}+\sin{x}}}$$
I use wolframpha this found this maximum is $\dfrac{4\sqrt{2}}{5}$,But How to prove and how to find this value?(without derivative)
idea 1
let $\tan{\dfrac{x}{2}}=t$,then we have
$$f=\dfrac{(t+1)^2}{\sqrt{t^4+2t^3+6t^2+2t+5}}$$
Therefore,it suffices to prove that
$$\dfrac{(t+1)^4}{t^4+2t^3+6t^2+2t+5}\le\dfrac{32}{25}$$
or$$32(t^4+2t^3+6t^2+2t+5)-25(t+1)^4\ge 0$$
or
$$ (t-3)^2(7t^2+6t+15)\ge 0$$
But this method if we without derivative,we don't known the maximum is $\dfrac{4\sqrt{2}}{5}$.
idea 2
$$f(x)=\dfrac{\sin{x}+1}{\sqrt{(\sin{x}+1)+2(\cos{x}+1)}}$$
Let $u=\sin{x}+1,v=\cos{x}+1$,then $(u-1)^2+(v-1)^2=1$,find the maximum of the
$$\dfrac{u}{\sqrt{u+2v}}$$
| We need to minimize $$\dfrac{3+2\cos x+\sin x}{(1+\sin x)^2}$$
Now WLOG let $x=\dfrac\pi2-2y$ to get $$\dfrac{3+2\sin2y+\cos2y}{(1+\cos2y)^2}$$
Using Weierstrass substitution, writing $\tan y=t$
we get $$2f(t)=(t^2+1)(t^2+2t+2)$$
Now use Second derivative test, to find the minimum value of $f(t)$ occurs at $-\dfrac12$
i.e., $$f(t)\le\dfrac{25}{32}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2277893",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 3,
"answer_id": 1
} |
$\sqrt{x^2-x+1}+\sqrt{y^2-y+1}=\sqrt{x^2+xy+y^2}$ fine $x,y$ :
$$\sqrt{x^2-x+1}+\sqrt{y^2-y+1}=\sqrt{x^2+xy+y^2} \ \ \ : x ,y \in \mathbb{Z}$$
My Try :
$$\sqrt{x^2-x+1}+\sqrt{y^2-y+1}=\sqrt{x^2+xy+y^2} \\ x^2-x+1+y^2-y+1+2\sqrt{(x^2-x+1)(y^2-y+1)}=x^2+xy+y^2 \\ 2\sqrt{(x^2-x+1)(y^2-y+1)}=xy +x+y-2$$
Now ?
| You can simplify as follows
$$\begin{align}
2\sqrt{(x^2-x+1)(y^2-y+1)}&=xy +x+y-2\\
4(x^2-x+1)(y^2-y+1)&=x^2y^2+x^2+y^2+4+2x^2y+2xy^2-4xy\\&\,\,\,\,\,+2xy-4x-4y\\
4x^2y^2-4x^2y+4x^2-4xy^2+4xy+4y^2&=x^2y^2+2x^2y+x^2+2xy^2-2xy+y^2\\
3x^2y^2-6x^2y+3x^2-6xy^2+6xy+3y^2&=0\\
x^2y^2-2x^2y+x^2-2xy^2+2xy+y^2&=0\\
(xy-x-y)^2&=0
\end{align}$$
From here you can get that either $x(y-1)=y$ or $y(x-1)=x$ - either way gives the same solution of $(x,y)=(2,2)$.
As pointed out in a comment by John Bentin, $(0,0)$ is also a solution to this equation. However this is not a solution to the original equation - we have squared some terms which has created this extra solution. So it is always wise to go back and check if the solution does indeed work - $(2,2)$ does, $(0,0)$ does not.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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