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Let x and y be real numbers such that $6x^2 + 2xy + 6y^2 = 9 $.Find the maximum value of $x^2+y^2$ I tried to re-arrange the terms
$6x^2 + 2xy + 6y^2 = 9 $
$6x^2 + 6y^2 = 9 - 2xy $
$6 (x^2 + y^2) = 9 - 2xy $
$x^2 + y^2 = \frac{9 - 2xy}{6} $
Using A.M $\geq$ G.M
$\frac{x^2 + y^2}{2} \geq xy $
Can ayone help me from here? Am I going correct?
| Let $x= r\cos\theta , y= r\sin\theta$ , put it back into the given relation and get a function of $r^2 $in terms of $\theta$ we need to maximize this $r^ 2 $wrt to theta ,use calculus or just bounds of trigo function
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Find the minimal polynomial of $\sqrt[3]{3} + \sqrt{5}$ over $\mathbb{Q}$. Let $x = \sqrt[3]{3} + \sqrt{5}$
Notice that $(x+y)^2 = x^2 + 2xy + y^2$
Then, $x^2 = (\sqrt[3]{3} + \sqrt{5})^2 = \sqrt[3]{3}^2 + 2\sqrt[3]{3}\sqrt{5} + 5$
Then, $x^3 - 5 = \sqrt[3]{3}^2 + 2\sqrt[3]{3}\sqrt{5}$
Or notice that $(x+y)^3 = x^3 + 3x^2y + 3xy^2 + y^3$
Then, $x^3 = (\sqrt[3]{3} + \sqrt{5})^3 = 3 + 2\sqrt[3]{3}^2\sqrt{5} + 10\sqrt[3]{3} + 5\sqrt{5}$
Then, $x^3 - 3 = 2\sqrt[3]{3}^2\sqrt{5} + 10\sqrt[3]{3} + 5\sqrt{5}$
Which I think the degree of the minial polynomial of $x$ over $\mathbb{Q}$ should have a degree of $6$.
Which means the minimal polynomial can be written as $x^6 + a_5 x^5 + \cdots + a_0$ for some $a_5, ..., a_0 \in \mathbb{Q}$.
But I just don't how get it from both the ways I provide above.
Or that $\sqrt[3]{3} + \sqrt{5}$ is not algebraic. So, I can not write such minimal polynomial out algebraic ?
| Set of algebraic numbers form a field, so the sum of two algebraic numbers $\sqrt[3]{3} + \sqrt{5}$ is again an algebraic number.
To find the minimal polynomial, you want to continue in either of the two ways that you have started.
If it is the first way,
you have to take the suitable powers of $x^2-5$ and continue until getting a rational polynomial.
| {
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Prove if $x > 0$, $1+\frac{1}{2}x-\frac{1}{8}x^2 \le \sqrt{1+x}$ Prove if $x > 0$, $1+\frac{1}{2}x-\frac{1}{8}x^2 \le \sqrt{1+x}$.
I find online, one person suggested using Taylor Theorem to expand the right-hand side, and apply Bernoulli's inequality.
So, if $x_0 = 0$, $\sqrt{1+x} = 1+\frac{1}{2}x-\frac{1}{4(2!)}x^2+R$, where R is larger than $0$, this makes sense to me, but I'm trying to find another way to prove the equality. Like Mean Value Theorem for inequality $\sqrt{1+x} \le 1+\frac{1}{2}x$.
I see the part where $1+\frac{1}{2}x$ is the same, but get trouble to put $\frac{1}{8}x^2$ into the equation.
| The inequality can be proved using elementary algebra. If $\dfrac{1}{8}x^2 < 1+\dfrac{1}{2}x\implies 2-2\sqrt{3} <x < 2+2\sqrt{3} < 8$, then both sides are positive, and square both of them: $\left(1+\dfrac{1}{2}x-\dfrac{1}{8}x^2\right)^2 < 1+x\iff \left(1+\dfrac{1}{2}x\right)^2-2\left(1+\dfrac{1}{2}x\right)\left(\dfrac{1}{8}x^2\right)+\left(\dfrac{1}{8}x^2\right)^2< 1+x\iff -x+\dfrac{1}{8}x^2 < 0\iff \dfrac{1}{8}x^2 < x\iff x^2 - 8x < 0\iff x(x-8) < 0\iff 0 < x < 8$, and this is true since we've shown that $8 > x > 0$, and if $\dfrac{1}{8}x^2 \ge 1+\dfrac{1}{2}x$, then it is obviously true since the left side is non-positive and the right side is non-negative. You're done !
| {
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Find polynomials $M_1(x)$ and $M_2(x)$ such that $(x+1)^2M_1(x) + (x^2 + x + 1)M_2(x) = 1$ I have been trying to solve this with no success. Could you suggest me a solution?
| Back to the basics! Remember how you do it for integers? For example, how would you find $x$, $y$ such that $23x+47y=1$? That's right, Euclid's algorithm.
Let's try to calculate the GCD of $(x+1)^2$ and $x^2+x+1$.
$$
(x+1)^2 - (x^2+x+1) = x\\
(x^2+x+1) - x(x+1) = 1
$$
Hence,
$$(x^2+x+1) - ((x+1)^2 - (x^2+x+1))(x+1) = 1$$
Therefore,
$$- (x+1)(x+1)^2 + (x+2)(x^2+x+1) = 1$$
| {
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Congruency application I want to prove whether the following is true using congruencies: If $n$ is odd and $3\not | n$, then $n^2\equiv 1\pmod{24}$.
I tried a direct proof.
Let $n=2k+1$ for an integer $k$ such that $3\not| n$, or $3c=n+r$ such that $r=1$ or $r=2$ for an integer $c$. Then in the first case, $3c=n+1$ implies $n\equiv 1\pmod 3$. Or in the second case, $3c=n+2$ implies $n\equiv 2 \pmod 3$. Then $$n^2\equiv 1 \pmod 3 \text{ or } n^2\equiv 4 \pmod 3$$
iff $$3|{n^2-1} \text{ or } 3|n^2-4$$
or $$8n^2\equiv8, 32 \pmod{24}$$
This was the farthest I got. I'm having a hard time going from mod 3 to mod 24 just given the conjunction that $n$ is odd and that $3\not| n$
| Let $n$ $=$ $12k + r$ r is not equal to $2,4,6,8$ and $10$ since n is odd and r is not equal to $3$ since r is not divisible by $3$.
Therefore $r={1,5,7,11}$
$Case 1:$ $r = 1$
$n²= 144k²+24k+1$
→$n =24p + 1$ ,$∃p∈ℤ$
$Case 2:$ $r = 5$
$n²= 144k²+24⋅5k+25$
→$n²= 144k²+24k+24+1$
→$n =24p + 1$ ,$∃p∈ℤ$
$Case 3:$ $r = 7$
$n²= 144k²+24⋅7k+49$
→$n²= 144k²+24k+48+1$
→$n =24p + 1$ ,$∃p∈ℤ$
$Case 4:$ $r = 11$
$n²= 144k²+24k+121$
→$n²= 144k²+24⋅11k+120+1$
→$n =24p + 1$ ,$∃p∈ℤ$
Therefore $n^2 \equiv 1 \pmod{24}$.
| {
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How to prove that $\sqrt{2-\sqrt{2}} \in \mathbb{Q}(\sqrt{2+\sqrt{2}})$ I am trying to prove a statement about the decomposition field of a polynomial that has both $\sqrt{2-\sqrt{2}}$ and $\sqrt{2+\sqrt{2}}$ as roots. I cannot find a way to prove that $\sqrt{2-\sqrt{2}} \in \mathbb{Q}(\sqrt{2+\sqrt{2}})$. I have tried writing it in the basis $1,\sqrt{2+\sqrt{2}}(\sqrt{2+\sqrt{2}})^2,(\sqrt{2+\sqrt{2}})^3$ but nothing works and without this I cannot prove that the decomposition field of $t^4-4t^2+2$ is $\mathbb{Q}(\sqrt{2+\sqrt{2}})$ over $\mathbb{Q}$
| Hint: what is $\sqrt{2 - \sqrt 2} \cdot \sqrt{2 + \sqrt 2}$ ?
It's $\sqrt 2$ ! Therefore $\sqrt{2 - \sqrt 2} = \frac{\sqrt 2}{\sqrt{2 + \sqrt 2}} = \frac{\left(\sqrt{2 + \sqrt 2}\right)^2 - 2}{\sqrt{2 + \sqrt 2}}$
| {
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"timestamp": "2023-03-29T00:00:00",
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If $\tan A, \tan B, \tan C$ are roots of $x^3-ax^2+b=0$, find $(1+\tan^2 A)(1+\tan^2B)(1+\tan^2C)$ Here are all the results I got
$$\tan(A+B+C)=a-b$$
And
$$(1+\tan^2A)(1+\tan^2B)(1+\tan^2C)=(\frac{1}{\cos A\cos B\cos C})^2$$
And
$$\cot A+\cot B + \cot C=0$$
How should I use these results?
| It is basically an algebraic problem:
If $p,q,r$ are the roots of $$x^3-ax^2+b=0$$
it is sufficient to find the cubic equation whose roots are $w=p^2+1,y=q^2+1,z=r^2+1$
From $ap^2-b=p^3,$ we have
$$(ap^2-b)^2=(p^3)^2=(p^2)^3$$
Replace $p^2$ with $w-1$ to find $$(a(w-1)-b)^2=(w-1)^3$$
$$\iff w^3+(\cdots)w^2+(\cdots)w-1-(a+b)^2=0$$
Obviously, we shall reach at the same equation if start with $y=q^2+1$ or $z=r^2+1$
Using Vieta's formula, $$wyz=-\dfrac{-1-(a+b)^2}1$$
See also :
*
*Finding $\sum_{k=0}^{n-1}\frac{\alpha_k}{2-\alpha_k}$, where $\alpha_k$ are the $n$-th roots of unity
*If $\alpha$, $\beta$, $\gamma$ and $\delta$ be the roots of the polynomial $x^4+px^3+qx^2+rx+s$, prove the following.
| {
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"timestamp": "2023-03-29T00:00:00",
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Factorize $abx^2-(a^2+b^2)x+ab$ Factorize the quadratic trinomial $$abx^2-(a^2+b^2)x+ab.$$
The discriminant of the trinomial is $$D=(a^2+b^2)^2-(2ab)^2=\\=(a^2+b^2-2ab)(a^2+b^2+2ab)=(a-b)^2(a+b)^2=\\=\left[(a-b)(a+b)\right]^2=(a^2-b^2)^2\ge0 \text{ } \forall a,b.$$
So the roots are $$x_{1,2}=\dfrac{a^2+b^2\pm\sqrt{(a^2-b^2)^2}}{2ab}=\dfrac{a^2+b^2\pm\left|a^2-b^2\right|}{2ab}.$$ How can I expand the modulus here? Have I calculated the discriminant in a reasonable way? Can we talk about "the discriminant of a quadratic trinomial" or only the corresponding quadratic equation (the trinomial=0) has a discriminant? What about "the roots of a trinomial" (or of the corresponding quadratic equation)?
| The polynomial $ \ Cx^2 + Bx + A \ $ is referred to as the reciprocal polynomial to $ \ Ax^2 + Bx + C \ \ , $ as its zeroes are the reciprocals of the latter polynomial. In the same vein, a polynomial $ \ Ax^2 + Bx + A \ $ is called palindromic and has the property that if $ \ r \ $ is a zero, then $ \ \frac{1}{r} \ $ is as well. (This applies to polynomials of higher degree as well as quadratics.)
This permits us to write
$$ ab·x^2 \ - \ (a^2+b^2)·x \ + \ ab \ \ = \ \ ab · \left[ \ x^2 \ - \ \frac{a^2+b^2}{ab}·x \ + \ 1 \ \right] $$ $$ = \ \ ab · \left[ \ x^2 \ - \ \left(\frac{b}{a } + \frac{a}{b}\right)·x \ + \ 1 \ \right] $$ $$ = \ \ ab·(x - r)· \left(x - \frac{1}{r} \right) \ \ = \ \ ab·[x^2 - \left(r + \frac{1}{r} \right)·x \ + 1] \ \ . $$
The reciprocal nature of the two zeroes is thus confirmed and the factorization proceeds as in Taladris's answer.
| {
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"timestamp": "2023-03-29T00:00:00",
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Conditions on coefficients obtained when solving two equations A question defines $$f(x)= (ax+b)e^x$$ and states that it satisfies the equation $$f(x)=\int _0 ^x e^{x-y}f'(y)dy - (x^2-x+1)e^x$$ You're required to find a and b.
Here, they've solved it by taking the $e^x$ in the second equation to the left side and differentiating using the Leibniz rule. You end up with $(a-2)x+(b-1)=0$, which gives you a and b as 2 and 1, since this equation has to hold for all x, I assume. This makes sense.
But I did it by differentiating f to get $f'(x)$, plugged that into the integral and integrated it, substituted f(x) for the expression in the first equation, and simplified it to get $$x^2(\frac{a}{2}-1)+x(b+1)-(1+b)=0$$ Using the same thing about this having to satisfy all x, I tried the values 0 and 1 for x in the equation to get $$b=-1, a=0$$
What did I get wrong here?
| Going by your method, we get,
\begin{align}
&f'(x) = e^x(ax+b+a)\\
\implies &f(x)=\int_0^xe^{x-y}e^y(ay+b+a)dy-(x^2-x+1)e^x\\
\implies &f(x) = e^x\int_0^x(ay+b+a)dy-(x^2-x+1)e^x\\
\implies &f(x) = e^x\left(\dfrac{ax^2}{2}+(b+a)x\right)-(x^2-x+1)e^x\\
\implies &(ax+b)e^x = e^x\left(\left(\dfrac{a}{2}-1\right)x^2+(b+a+1)x-1\right)
\end{align}
Cancelling $e^x$ and equating coefficient both sides, you get $a=2$ and $b=-1$.
Going by their method, we get,
\begin{align}
&e^{-x}f(x) = \int_0^xe^{-y}f'(y)dy-(x^2-x+1)\\
\implies &\dfrac{d}{dx}(ax+b) = e^{-x}f'(x)-(2x-1)\\
\implies &a = (ax+b+a)-(2x-1)\\
\implies &0=(a-2)x+(b+1)=0
\end{align}
which leads to the same solution. Note that the given answer that $b=1$ is wrong.
| {
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Coefficient of $x^{12}$ in $(1+x^2+x^4+x^6)^n$ I need to find the coefficient of $x^{12}$ in the polynomial $(1+x^2+x^4+x^6)^n$.
I have reduced the polynomial to $\left(\frac{1-x^8}{1-x^2}\right)^ n$ and tried binomial expansion and Taylor series, yet it seems too complicated to be worked out by hand.
What should I do?
| Here is just a standard trick from generating functions using
$$\frac 1{(1-y)^n} = \sum_{k=0}^{\infty}\binom{k+n-1}{n-1}y^k$$.
To simplify the expressions set
$$y=x^2\Rightarrow \text{ we look for }[y^6]\frac{(1-y^4)^n}{(1-y)^n}$$
Hence, for $n\geq 1$ you get
\begin{eqnarray*}[y^6]\frac{(1-y^4)^n}{(1-y)^n}
& = & [y^6]\left((1-y^4)^n\sum_{k=0}^{\infty}\binom{k+n-1}{n-1}y^k\right) \\
& = & [y^6]\left((1-ny^4)\sum_{k=0}^{\infty}\binom{k+n-1}{n-1}y^k\right) \\
& = & \boxed{\binom{6+n-1}{n-1} - n\binom{2+n-1}{n-1}}
\end{eqnarray*}
| {
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"timestamp": "2023-03-29T00:00:00",
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What are the possible real values of $\frac{1}{x} + \frac{1}{y}$ given $x^3 +y^3 +3x^2y^2 = x^3y^3$? Let $x^3 +y^3 +3x^2y^2 = x^3y^3$ for $x$ and $y$ real numbers different from $0$.
Then determine all possible values of $\frac{1}{x} + \frac{1}{y}$
I tried to factor this polynomial but there's no a clear factors
| The equation is equivalent to
$$x^3+y^3+(-xy)^3 -3xy(-xy)=0 \tag{1}$$
Applying this formula
$$a^3+b^3+c^3 -3abc= \frac{1}{2}(a+b+c)((a-b)^2+(b-c)^2+(c-a)^2)$$
for $(a,b,c) = (x,y,-xy)$
then $(1)$ holds true if and only if either of these 2 equations holds true $$x+y-xy = 0\tag{2}$$ $$x=y=-xy \tag{3}$$
If $(2)$ holds true then $\frac{1}{x}+\frac{1}{y} = 1$
if $(3)$ holds true then $x = y = -1 \implies \frac{1}{x}+\frac{1}{y} = -2$
Conclusion:
$$\frac{1}{x}+\frac{1}{y} = 1 \text{ or } \frac{1}{x}+\frac{1}{y} =-2$$
| {
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Multiplication and division by an integer in systems of linear congruences I have the following system of linear congruences:
\begin{align}
x&\equiv -2 \pmod {6}\\
x&\equiv 2 \pmod {10}\\
x&\equiv 7 \pmod {15}
\end{align}
From the first equation we have: $ x= 6 k-2, k\in \mathbb{Z}$. Then we insert it into 2nd equation:
$ 6 k -2 \equiv 2 (\mod 10) $ i.e. $ 6 k \equiv 4 (\mod 10) $
from which we obtain
$ 3 k \equiv 2 (\mod 5) $
Now at this point, is it allowed to multiply the last congruence by 2 to get:
$ k \equiv -1 (\mod 5) $
and conclude $k = 5 l -1, l\in \mathbb{Z}$ and $ x = 30 n -8 , n\in \mathbb{Z}$ ?
I wonder if this is correct, is the obtained congruence equivalent to the first one? Thanks in advance.
| Another approach:
$\begin{cases} x\equiv -2 \bmod 6\\x=2 \bmod 10\end{cases}$
$\rightarrow 6t_1-2=10t_2 +2\rightarrow 6t_1-10t_2=4 \rightarrow t_1=-10 s-1, t_2=6s -1$
Which gives:
$t_1=-10 s-1\rightarrow x=6(-10s-1)-2=-60 s -8$
$t_2=6s-1 \rightarrow x=10(6s-1)+2=60s -8$
$\begin{cases} x\equiv 7 \bmod 15\\x=2 \bmod 10\end{cases}$
$10t_3+2=15t_4+7$
which gives:
$t_3=15s_1-1\rightarrow x=10(15 s_1-1)+2=150 s_1-8$
$t_4=-10 s_1-1\rightarrow x=15(-10s_1-1)+7=-150 s_1-8$
$ lcm(6, 10, 15)=30|60, 150$, therefore:
$$x\equiv -8 \bmod 30\rightarrow x=30n -8$$
| {
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Proving global minimum by lower bound of 2-variable function $f(x,y)=x^4+2x^2y+y^2-4x^2-8x-8y$ I would like to prove that the following function $f :\mathbb{R}^2\to\mathbb{R}$ has a global minimum:
$f(x,y)=x^4+2x^2y+y^2-4x^2-8x-8y=(x^2+y)^2-4(x^2+2x+2y)$
$f$ has strict local minimum at
$f(1,3)=-20$
I think that what I need to show is that $-20$ is a lower bound of this function, and then conclude that's a global minimum, but I didn't manage to do so.
Please advise.
Thank you.
| My favorite way,
$$f(x,y)+20=y^2+y(2x^2-8)+x^4-4x^2-8x+20$$
$$\begin{align}\Delta_{\text{half}}&=(x^2-4)^2-(x^4-4x^2-8x+20)\\
&=-4(x-1)^2≤0.\end{align}$$
This means, $f(x,y)+20≥0.$
Hence, for minimum of $f(x,y)+20$, we need to take $x=1$ and $y=4-x^2$, which gives $f(x,y)+20=0.$
Finally, we deduce that
$$\min\left\{f(x,y)+20\right\}=0~ \\
\text {at}~ (x,y)=(1,3)$$
$$\min\left\{f(x,y)\right\}=-20~ \\ \text {at}~ (x,y)=(1,3)$$
where $f(x,y)=x^4+2x^2y+y^2-4x^2-8x-8y.$
Small Supplement:
Using the formula
$$ay^2+by+c=a(y-m)^2+n$$
where $m=-\dfrac{b}{2a}, n=-\dfrac{\Delta}{4a}$
$$\begin{align}y^2+y(2x^2-8)+x^4-4x^2-8x+20=(y+x^2-4)^2+4(x-1)^2≥0.\end{align}$$
| {
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Cubic equation with circle intersection to form a square
A cubic equation and circle (unit radius) has intersection at A,B,C,D. ABCD is a square. Find the angle $\theta$.
I tried:
*
*$(0,0)$ is a solution so constant term is $0$
*Substituting A(x,y) and C(-x,-y) and adding them gives coefficient of $x^2$ is 0.
Then the cubic becomes f(x) = $ax^3+bx$.
3.Substituting A and B and added the two equations.
I found it interesting-for n points given we can find a unique n+1 degree polynomial
Also - Can complex number be used here?
Please note: I am not sure whether we can find the angle(integer) without knowing the coefficients of the cubic.
EDIT: From the answers
1.putting A $(cos\theta,sin\theta)$ in f(x) :
$acos^3\theta + b cos\theta = sin\theta$
2.putting B $(-sin\theta,cos\theta)$ in f'(x) :
$asin^2\theta + b = tan\theta$ [ as circle has $tan\theta$ slope at B]
$1,2 $ eqn gives $3asin^2\theta = acos^2\theta$
So, $sin^2\theta = \frac{1}{4}$
But I getting the value of $\theta$ but a answer shows plot of many cubics
-> because in my case $ABCD$ is a square.
| Let
$$f(x)=ax^3+bx. \quad \quad (1)$$
Let's find out the values of $a$ and $b$ for a specific value of $R$.
Let the points $ A=(R \cos \theta, R \sin \theta)$ and $B=(-R \sin \theta, R \cos \theta)$ two intersection points of the circle $\Lambda (O, R)$ with $f(x)$, so that B is a tangent point.
Substituting the coordinates of $A$ in $f(x)$, we get:
$$\tan \theta =aR^2\cos^2\theta+b. \quad \quad (2)$$
Substituting the coordinates of $A$ and $B$ in $f(x)$, we get with some algebra:
$$a =\frac{4}{R^2 \sin4\theta}. \quad \quad (3)$$
Substituting the $x$ of $B$ in $f'(x)$, which is equal to $\tan \theta$, we get:
$$\tan \theta =3aR^2 \sin^2\theta +b. \quad \quad (4)$$
Choosing $R =2$ and substituting in $(2)$ and $(4)$, we get:
$$\theta = 30°.$$
The values of $a$ and $b$ are:
$$a =\frac{2}{\sqrt 3}$$
and
$$b =-\frac{5\sqrt 3}{3}.$$
Plot:
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
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Angle between two vectors given magnitudes and difference in magnitude
Given: |a| = 31 , |b| = 23 and |a – b| = 42 , what is the angle between a and b and the magnitude of |a + b|
How would you solve this, and what kind of diagrams can I use to solve this question?
Edit: I got 78.92 as the angle between |a| and |b|. I got 34.87 as the magnitude of |a + b|
| Your answer for $|\textbf{a} + \textbf{b}|$ is correct. For angle between $\textbf{a}$ and $\textbf{b}$, it should be $\gt 90^0$ as $\cos\theta$ is negative. I think you missed the negative sign in your working.
$|\textbf{a} - \textbf{b}|^2 = |\textbf{a}|^2 + |\textbf{b}|^2 - 2 \ \textbf{a} \cdot \textbf{b} \implies \textbf{a} \cdot \textbf{b} = \frac{1}{2} (31^2 + 23^2 - 42^2) = -137$
So $|\textbf{a} + \textbf{b}|^2 = |\textbf{a}|^2 + |\textbf{b}|^2 + 2 \ \textbf{a} \cdot \textbf{b} = 31^2 + 23^2 - 274 = 1216$
$\implies |\textbf{a} + \textbf{b}| = 8 \sqrt{19}$
For angle between $\textbf{a}$ and $\textbf{b}$,
$\textbf{a} \cdot \textbf{b} = |a| \ |b| \cos\theta \implies \cos\theta = \displaystyle \small - \frac{137}{713}$
$\implies \theta \approx 101.1^0 $
| {
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"timestamp": "2023-03-29T00:00:00",
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Integrate $\int\frac{6x^4+4x^2+8x+4}{(1-2x)^3}dx$ Integrate $$\int\frac{6x^4+4x^2+8x+4}{(1-2x)^3}dx$$
I tried $2\cdot\int\frac{3x^4+2x^2+4x+2}{(1-2x)^3}dx\:\:$ now after partial fractions i have:
$$\begin{align}
&\>\>\>\>\>\frac{3x^4+2x^2+4x+2}{(1-2x)^3}\\&=\frac{A}{(1-2x)}+\frac{B}{(1-2x)^2}+\frac{C}{(1-2x)^3}\\[1ex]
&=A(1-2x)^2(1-2x)^3+B(1-2x)(1-2x)^3+C(1-2x)(1-2x)^2\\[1ex]
&=(1-2x)^3\cdot[A(1-2x)^2+B(1-2x)+C]\\[1ex]
&=(1-2x)^3\cdot[A-4Ax+4Ax^2+B-2Bx+C]
\end{align}$$
Need a bit help to find out $A,B,C$ and evaluation of problem.
Thank you in advance:)
| If you want to use partial fractions for a solution to something, then you first have to check the numerator and denominator's degrees (the degree of a polynomial is the highest power in it, eg the degree of $x^3+5x-4$ is $3$).
*
*If the degree of the numerator is $\ge$ the degree of the denominator, then before using partial fractions we must use long division or some other algebraic technique to transform the fraction we're dealing with into a fraction with the degree of the numerator that is less than the degree of the denominator, and a remainder term. For example,
$$\frac{x^2}{x^2-1}=\frac{x^2-1+1}{x^2-1}=1+\frac{1}{(x+1)(x-1)}$$
and now we're in a position to use partial fractions.
*If the degree of the denominator is $>$ than the degree of the numerator than we can immediately proceed with partial fractions.
Your integral fits into the first category, so you have a bit more work to do before you can use partial fractions.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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If $\left(1+x^{2}+x^{4}\right)^{8}={^{16} C_{0}}+{^{16}C_{1}} x^{2}+{^{16} C_{2}} x^{4}+\cdots +{^{16} C_{16}} x^{32}$, how to prove... If I am given that $$\left(1+x^{2}+x^{4}\right)^{8}={^{16} C_{0}}+{^{16}C_{1}} x^{2}+{^{16} C_{2}} x^{4}+\cdots +{^{16} C_{16}} x^{32},$$
how can I prove that $${^{16} {C}_{1}}+{^{16} {C}_{4}}+{^{16} {C}_{7}}+{^{16} {C}_{10}}+{^{16} {C}_{13}}+{^{16}{C}_{16}}=3^{7}$$ is true?
My approach: I usually try to evaluate these by putting $x =\sqrt{-1}, \sqrt[3]{-1}$ however I'm not sure how to approach this. Please suggest any hints or methods as to how to solve this particular and in general how to approach these types of expressions.
| The difference between the powers of $x^6$ between subsequent coefficients, so I suggest dividing this polynomial by $x^6-1$ and checking the remainder. If $$ (1+x^2+x^4)^8 = {}^{16}C_0 +{}^{16}C_1 x^2 + \dots + {}^{16}C_{16} x^{32} = w(x) (x^6-1) + r(x)$$
we have
$$r(x) = a + b x^2 + c x^4$$
where
$$ a = {}^{16}C_0 +{}^{16}C_3 + \dots + {}^{16}C_{30} $$
$$ b = {}^{16}C_1 +{}^{16}C_4+ \dots + {}^{16}C_{31} $$
$$ c = {}^{16}C_2 +{}^{16}C_5 + \dots + {}^{16}C_{32} $$
If you put $x^2 \in\{1, (-1)^\frac23, (-1)^\frac43\}$ you have $x^6-1 =0 $, so
$$a + b+ c = (1+1+1)^8 = 3^8$$
$$a + b (-1)^\frac23 + c (-1)^\frac43 = (1+(-1)^\frac23+(-1)^\frac43)^8 = 0$$
$$a + b (-1)^\frac43 + c (-1)^\frac83 = (1+(-1)^\frac43+(-1)^\frac83)^8 = 0$$
Solving these equations gives you
$$ a = b= c= 3^7$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Uniform convergence of $\sum_{n = 1}^\infty x \sin \frac{1}{(x n)^2}$ on $(0, 1)$. Functional sequence $x \sin \frac{1}{(x n)^2} {}^\longrightarrow_\longrightarrow 0$ because
$$
\left| x \sin \frac{1}{(x n)^2} \right| \le |x| \sqrt{\left|\sin \frac{1}{(x n)^2}\right|} \le |x| \sqrt{\frac{1}{(x n)^2}} = \frac1n \to 0.
$$
It is easy to see that the series $\sum_{n = 1}^\infty x \sin \frac{1}{(x n)^2}$ converges pointwise on $(0, 1)$:
$$
\left| \sum_{n = 1}^\infty x \sin \frac{1}{(x n)^2} \right| \le \sum_{n = 1}^\infty \left| x \sin \frac{1}{(x n)^2} \right| \le |x| \sum_{n = 1}^\infty \frac{1}{n^2 x^2} = \frac{1}{x} \sum_{n = 1}^\infty \frac{1}{n^2} = \frac{\pi^2}{6 x}.
$$
But whether the series converges uniformly on $(0, 1)$?
| A necessary and sufficient condition for uniform convergence of a series $\sum f_n(x)$ for $x \in (a,b)$ is the Cauchy criterion. That is, for all $\epsilon > 0$ there exists $N \in \mathbb{N}$, independent of $x$, such that for all $m > n > N$ and all $x \in (a,b)$ we have $\left|\sum_{k=n+1}^m f_k(x) \right| < \epsilon$.
In this case, we can show that the series is not uniformly convergent on $(0,1)$ by violation of the Cauchy criterion. For any $N$, take $n = \lceil \sqrt{2}N\rceil$, $m = 2N$, and $x_N = \frac{1}{N \sqrt{\pi}}$. For $n < k \leqslant m$, we have
$$\frac{\pi}{4}= \frac{1}{(2n)^2 x_N^2} \leqslant \frac{1}{(x_Nk)^2} < \frac{1}{\lceil \sqrt{2}N\rceil^2 x_N^2}< \frac{\pi}{2}, $$
and
$$\frac{1}{\sqrt{2}} < \sin \frac{1}{(x_N k)^2} < 1$$
Hence,
$$\left|\sum_{k=n+1}^{m}x_N\sin \frac{1}{(x_Nk)^2} \right| = \sum_{k=n+1}^{m}x_N\sin \frac{1}{(x_Nk)^2}> (2N - \lceil{\sqrt{2}N\rceil})\cdot \frac{1}{N \sqrt{\pi}}\cdot \frac{1}{\sqrt{2}} \\ > \frac{2 - \sqrt{2}}{\sqrt{2 \pi}},$$
which violates the Cauchy criterion for uniform convergence (when $\epsilon$ less than the RHS is chosen).
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Integrate $\int\frac{x}{x^3-8}dx$ Integrate
$$\int\frac{x}{x^3-8}dx$$
I solved this integral by dividing it in partial fractions, than i came in two integrals $I_1,I_2$. Then, partioning $I_2$ into $I_3,I_4$ and partioning $I_4$ into $I_5,I_6$. But it took me so much work even though i got correct answer.
Is there any simple way to solve it?
Thank you in advance :)
| Partial fractions will indeed work without the need for too many integrals. We see that
\begin{align*}
\int \frac{x}{x^3 -8} \ dx &= \int \frac{1}{6}\frac{1}{x-2} - \frac{1}{6} \frac{x-2}{x^2 +2x+4}\ dx \\
&= \frac{1}{6} \ln(x-2) - \frac{1}{6} \int \frac{\frac{1}{2}(2x+2)-3}{x^2 +2x+4}\ dx\\
&\overset{\color{blue}{u=x^2 +2x+4}}= \frac{1}{6} \ln(x-2) - \frac{1}{12} \int \frac{1}{u}\ du + \frac{1}{2} \int \frac{1}{(x+1)^2 +3} \ dx\\
&\overset{\color{blue}{s=x+1}}= \frac{1}{6} \ln(x-2) - \frac{1}{12} \ln(x^2 +2x+4)+ \frac{1}{2} \int \frac{1}{s^2 +(\sqrt{3})^2} \ ds\\
&= \frac{1}{6} \ln(x-2) - \frac{1}{12} \ln(x^2 +2x+4)+ \frac{1}{2} \frac{\arctan\left(\frac{x+1}{\sqrt{3}} \right)}{\sqrt{3}} + C
\end{align*}
| {
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"url": "https://math.stackexchange.com/questions/4106334",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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HKMO-2020 Math contest Question from math contest (already concluded in 2020) which I have tried but not yet found any elegant solution.
How many positive integer solutions does the following system of equations have?
$$\sqrt{2020}(\sqrt{a} + \sqrt{b} )= \sqrt{ (c+ 2020)(d+ 2020)}$$
$$\sqrt{2020}(\sqrt{b} + \sqrt{c} )= \sqrt{ (d+ 2020)(a+ 2020})$$
$$\sqrt{2020}(\sqrt{c} + \sqrt{d} )= \sqrt{ (a+ 2020)(b+ 2020)}$$
$$\sqrt{2020}(\sqrt{d} + \sqrt{a} )= \sqrt{ (b+ 2020)(c+ 2020)}$$
My attempt:
After few manipulations with an aim to remove the radicals, I end up with:
$$4040 \left(\sqrt{ab}+\sqrt{bc}+\sqrt{cd}+\sqrt{da}\right)=ab+bc+cd+da+4\cdot 2020^2$$
Before I go further, I feel that it is only going to become more unwieldy. Help is appreciated.
| Applying the Cauchy–Schwarz inequality to $(u_1,v_1,u_2,v_2) = (\sqrt{c},\sqrt{2020},\sqrt{2020},\sqrt{d})$ to the right hand side of the first equation
$$\sqrt{(c+2020)(2020+d)} \ge \sqrt{c}\sqrt{2020}+\sqrt{2020}\sqrt{d}= \sqrt{2020}(\sqrt{c}+\sqrt{d})$$
Then
$$\sqrt{a}+\sqrt{b}\ge \sqrt{c}+\sqrt{d} \tag{1}$$
Do the same method for the next 3 equation, we deduce
$$\sqrt{b}+\sqrt{c}\ge \sqrt{d}+\sqrt{a} \tag{2}$$
$$\sqrt{c}+\sqrt{d}\ge \sqrt{a}+\sqrt{b}\tag{3}$$
$$\sqrt{d}+\sqrt{a}\ge \sqrt{b}+\sqrt{c}\tag{4}$$
From $(1)(3)$ we deduce $$\sqrt{a}+\sqrt{b}= \sqrt{c}+\sqrt{d}$$
and the equality occurs when
$$\frac{c}{2020} = \frac{2020}{d}$$
$$\frac{a}{2020} = \frac{2020}{b}$$
Do the same thing, from $(2)(4)$, we must have
$$\frac{b}{2020} = \frac{2020}{c}$$
$$\frac{d}{2020} = \frac{2020}{a}$$
Hence, we have
$$\frac{a}{2020}=\frac{2020}{b}=\frac{c}{2020}=\frac{2020}{d} $$
or
$$(a,b,c,d)=\left(t,\frac{2020^2}{t},t,\frac{2020^2}{t} \right) \tag{5}$$
From $(5)$, we have $t| 2020^2 = (2^4 5^2 101^2)$ So, there are $(4+1)(2+1)(2+1)=45$ different values of $t$.
Conclusion: there are $45$ solutions
$$(a,b,c,d)=\left(t,\frac{2020^2}{t},t,\frac{2020^2}{t} \right) \qquad \forall t| 2020^2$$
| {
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"url": "https://math.stackexchange.com/questions/4107950",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Calculus 2 test question that even the professor could not solve (Parametric Arc length) everyone got this question wrong from my calculus 3 midterm.
" find the arc length of the curve on the given interval: $x=\sqrt(t), \space y=8t-6,\space on\space 0 \leq t \leq 3$"
I set the problem up just fine, however at that point I got stuck.
$$ \int_{0}^{3}\sqrt{\left ( \frac{1}{2\sqrt{t}} \right )^{2}+64} dt$$
we asked the teacher for a review, he got the problem set up then quit and said he would not count it against us. but it's killing me I need to know how to do it!
thanks
| There is a way to perform the evaluation without resorting to trigonometric substitution, although it is hardly ideal. Note that with the substitution $$u^2 = \frac{1}{4t} + 64, \quad t = \frac{1}{4(u^2 - 64)}, \quad dt = - \frac{u}{2(u^2 - 64)^2} \, du,$$ and letting $c = \sqrt{769/12}$,
$$\begin{align}
\int_{t=0}^3 \sqrt{\frac{1}{4t} + 64} \, dt &= \int_{u=\infty}^c -\frac{u^2}{2(u^2 - 64)^2} \, du \\
&= \frac{1}{64} \int_{u=c}^\infty \frac{1}{u-8} - \frac{1}{u+8} + \frac{8}{(u-8)^2} + \frac{8}{(u-8)^2} \, du \\
&= \frac{1}{64} \left[\log \frac{u-8}{u+8} - \frac{8}{u-8} - \frac{8}{u+8} \right]_{u=c}^\infty \\
&= -\frac{1}{64} \log \frac{c-8}{c+8} + \frac{c}{4(c^2-64)} \\
&= \frac{\sqrt{2307}}{2} - \frac{1}{64} \log (1537 - 32 \sqrt{2307}).
\end{align}$$
While tractable, I would not expect such a computation to be performed by hand without a calculator within a short amount of time on a test. The partial fraction decomposition is not particularly difficult, but the evaluation of the antiderivative at $c$ involves a lot of tedious arithmetic that is not relevant for testing proficiency at calculus.
| {
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Counting question - binomial coefficients Why is there a difference in choosing $2$ out of $n$ elements and after that $2$ out of the $n-2$ remaining elements compared to choosing $4$ elements. So why is, intuitively (not algebraically), ${n \choose 2} \cdot {n-2 \choose 2} \cdot \dfrac{1}{2}\neq {n \choose 4}$?
| Consider $n=6$, and assume that set $S = \{1,2,3,4,5,6\}.$
Consider the specific collection of 4 items represented by the subset $T = \{1,2,3,4\}.$
There are 3 distinct ways of partitioning set $T$ into two pairs: $\{(1,2),(3,4)\}$, $\{(1,3),(2,4)\}$, and $\{(1,4),(2,3)\}$.
This means that since there are $\binom{6}{4} = 15$ ways of forming a collection of four items (similar to subset $T$) from set $S$, there are $(3 \times 15) = 45$ ways of partitioning set $S$ into two distinct pairs, with two items left over.
Superficially, you would expect that the # of ways of forming the two distinct pairs from set $S$ would be
$\displaystyle \binom{6}{4} \times \binom{4}{2}$, (which is wrong)
rather than $\displaystyle \binom{6}{4} \times \binom{4}{2} \times \frac{1}{2}$, (which is right). The last factor, $\frac{1}{2}$, which is an overcounting adjustment scalar is needed, because otherwise each of the partitions into two pairs would be double counted.
As an illustration of this, note that the partitioning of set $T$ above, into $2$ pairs could only be done in $3$ ways, even though $\binom{4}{2} = 6.$
In general, there will be $\binom{n}{4}$ ways of forming collections of $4$ items from a set of $n$ items, and there will be
$\displaystyle \binom{n}{4} \times \binom{4}{2} \times \frac{1}{2}$ ways of forming two distinct pairs of two items each, from a set of $n$ items.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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show this $x_{n+1}-x_{n}\ge 2\pi$ let $f(x)=e^x\cos{x}-\sin{x}-1$,and $n$ be postive integer,such $x_{n}$ be a root of $f(x)=0$ ,and
$\dfrac{\pi}{3}+2n\pi<x_{n}<\dfrac{\pi}{2}+2n\pi$,show that
$$x_{n+1}-x_{n}\ge 2\pi,\forall n\in ^{+}\tag{1}$$
My try: since
$$\dfrac{7\pi}{3}+2n\pi<x_{n+1}<\dfrac{5\pi}{2}+2n\pi$$
and
$$-(\dfrac{\pi}{2}+2n\pi)<-x_{n}<-(\dfrac{\pi}{3}+2n\pi)$$
so
$$x_{n+1}-x_{n}\ge \left(\dfrac{7\pi}{3}+2n\pi\right)-\left(-(\dfrac{\pi}{2}+2n\pi)\right)=\dfrac{11\pi}{6}$$
But $2\pi>\dfrac{11\pi}{6}$,so How to prove this inequality $(1)$?
| You are looking for the zero's of function$$f(x)=e^x\cos(x)-\sin(x)-1$$ Because of the $\cos(x)$, they are very close to the right bound $x_0^{(n)}=\left(2 \pi n+\frac{\pi }{2}\right)$.
A first estimate of the solution can be obtained performing one single iteration of Newton method. This would give
$$x_1^{(n)}=2 \pi n+\frac{\pi }{2}-2 e^{-2 \pi n-\frac{\pi }{2}}$$
$$x_1^{(n+1)}-x_1^{(n)}=2 \pi+4 \sinh (\pi )\,e^{-\frac{4n+3}{2} \pi } > 2\pi$$
We could obtain better results using one single iteration of Halley method. This would give
$$x_1^{(n)}=2 \pi n+\frac{\pi }{2}-\frac{1}{2} \text{csch}^2\left( \left(n+\frac{1}{4}\right)\pi\right)$$
$$x_1^{(n+1)}-x_1^{(n)}=2\pi+\frac{1}{2} \left(\text{csch}^2\left( \left(n+\frac{1}{4}\right)\pi\right)-\text{csch}^2\left(
\left(n+\frac{5}{4}\right)\pi\right)\right) > 2\pi$$ For sure, we could improve the value of the difference using iterative methods of higher order.
For illustration purposes, consider the second and third roots; computed rigorously, they give
$$x_3-x_2=2\pi +\color{red}{1.447189010436905}\times 10^{-6}$$
Using the first formula, we should have
$$x_3-x_2=2\pi+\color{red}{1.44718690}82458479\times 10^{-6}$$
Using the second formula, we should have
$$x_3-x_2=2\pi+\color{red}{1.44718901043}48732\times 10^{-6}$$
| {
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "1",
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Let $\int_0^2 f\left(x\right) dx = a+\frac{b}{\log 2}$. Find $a,b$ Let $f$ be a real-valued continuous function on $\mathbb{R}$ such that $2^{f\left(x\right)}+f\left(x\right)=x+1$ for all $x\in \mathbb{R}$. Assume that $\int_0^2 f\left(x\right) dx = a+\dfrac{b}{\log 2}$ with $a,b$ are rational numbers . Find $a,b$.
I have no idea how to use the assumption $2^{f\left(x\right)}+f\left(x\right)=x+1$ for all $x\in \mathbb{R}$ except taking integral of the both sides:
$$\int_0^2 2^{f\left(x\right)}+f\left(x\right) dx=4 $$
Please help me an idea. Thank you.
| Let $\alpha = \log 2$, differenital $x + 1 = 2^f + f$ on both sides, we get
$$\begin{align}
1 &= (\alpha 2^f + 1)f'
= (\alpha(x + 1 - f) + 1)f'\\
&= \left[\alpha\left((x + 1)f - \frac{f^2}{2}\right) + f\right]' - \alpha f\\
\implies \alpha f &= \left[\alpha\left((x + 1)f - \frac{f^2}{2}\right) + f - x\right]'
\end{align}
$$
Integrate both sides for $x$ over $[0,2]$ and notice $f(0) = 0$ and $f(2) = 1$, we get
$$\begin{align}
\alpha\int_0^2 f dx
&= \left[\alpha\left((2 + 1)1 - \frac{1^2}{2}\right) + 1 - 2\right]
- \left[\alpha\left((0 + 1)0 - \frac{0^2}{2}\right) + 0 - 0\right]\\
&= \alpha\frac52 - 1
\end{align}
$$
This means $\displaystyle\;\int_0^2 fdx = \frac52 - \frac1{\log 2}$
| {
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"timestamp": "2023-03-29T00:00:00",
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How does one treat big O when Taylor expanding $\sin(\sin x)$? I am working on finding the Taylor (Maclaurin expansion) of $\sin(\sin(x))$ to the third order.
We have $$\sin{t} = t - \frac{t^3}{6} + \mathcal{O}(t^5)$$
If I then set $t = \sin(x)$, and expand once more, I end up with a big O which is
$$\mathcal{O}((x - \frac{x^3}{6} + \mathcal{O}(x^5))^5)$$
How does one treat this mathematically? How do I know which order the error will be in my final expansion?
| Expanding only the terms to the third order, $$\sin(\sin x)\\
= x - \frac{x^3}{6} + \mathcal{O}(x^5) - \frac{x^3\left(1 - \dfrac{x^2}{6} + \mathcal{O}(x^4)\right)^3}{6} + \mathcal{O}\left(x^5\left(1- \frac{x^2}{6} + \mathcal{O}(x^4)\right)^5\right)\\
=x-\frac{x^3}{6}-\frac{x^3}6+\mathcal O(x^5).$$
Check:
Using the chain rule,
$$\frac{df}{dx}=\cos t\cos x\to1$$
$$\frac{d^2f}{dx^2}=-\sin t\cos^2x-\cos t\sin x\to0$$
$$\frac{d^3f}{dx^3}=-\cos t\cos^3x+2\sin t\cos x\sin x+\sin t\sin^2 x-\cos t\cos x\to-2.$$
| {
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"timestamp": "2023-03-29T00:00:00",
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How can I prove that $\sum_{n=1}^\infty \dfrac{n^4}{2^n} = 150 $? I can easily prove that this series converges, but I can't imagine a way to prove the statement above. I tried with the techniques for finding the sum of $\sum_{n=1}^\infty \frac{n}{2^n}$, but I didn't find an useful connection with this case. Could someone help me?
| We have that
$$\sum_{n\ge 1} \frac{1}{2^n} n^4
= \sum_{n\ge 1} \frac{1}{2^n}
\sum_{m=1}^4 n^{\underline{m}} {4\brace m}
= \sum_{m=1}^4 {4\brace m} m!
\sum_{n\ge 1} \frac{1}{2^n} {n\choose m}.$$
The inner sum is
$$\sum_{n\ge m} \frac{1}{2^n} {n\choose m}
= \frac{1}{2^m} \sum_{n\ge 0} \frac{1}{2^n} {n+m\choose m}
= \frac{1}{2^m} \frac{1}{(1-1/2)^{m+1}} = 2.$$
We thus obtain evaluating the Stirling numbers combinatorially (e.g.
${4\brace 2} = {4\choose 1} + \frac{1}{2} {4\choose 2} = 7$)
$$2\sum_{m=1}^4 {4\brace m} m!
\\ = 2 \times 1 \times 1 + 2 \times 7 \times 2
+ 2 \times 6 \times 6 + 2 \times 1 \times 24
= 150$$
as claimed.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4132076",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
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} |
An integral : $\int_{-a}^a \frac{\cos(\pi n x/a)}{x^2+a^2}\mathrm dx$
How can I evaluate
$$\int_{-a}^a \frac{\cos(\pi n x/a)}{x^2+a^2}\,\mathrm dx$$?
My attempt :
$$\begin{align}\mathcal{I}& =\displaystyle\int_{-a}^a \dfrac{\cos(\pi n x/a)}{x^2+a^2}\mathrm dx \\ & = 2\displaystyle\int_0^a \dfrac{\cos(\pi n x/a)}{x^2+a^2}\, \text{ ,even integrand}\\ &=2\displaystyle\int_0^{\pi n} \dfrac{\cos u}{\frac{a^2u^2}{n^2\pi^2}+1}\left(\dfrac{a\mathrm du}{n\pi}\right)\, \text{ ,via substituting $u=\dfrac{\pi nx}{a}$}\\ &=\dfrac{2\pi n}{a}\displaystyle\int_0^{\pi n} \dfrac{\cos u}{u^2+n^2\pi^2}\, \mathrm du\\ & =\dfrac{2\pi n}{a}\displaystyle\int_0^{n\pi} \dfrac{\cos u}{(u-i\pi n)(u+i\pi n)}\,\mathrm du\\ & =\dfrac{i}{a}\underbrace{\displaystyle\int_0^{n\pi} \dfrac{\cos u}{u+i\pi n}\mathrm du}_{\mathcal{I_1}}-\dfrac{i}{a}\overbrace{\displaystyle\int_0^{n\pi}\dfrac{\cos u}{u-in\pi}\mathrm du}^{\mathcal{I_2}}\, \text{ ,via partial fraction decomposition}\end{align}$$
$$\begin{align}\mathcal{I_1}&=\displaystyle\int_0^{n\pi}\dfrac{\cos u}{u+i\pi n}\mathrm du \\&=\displaystyle\int_0^{(i+1)n\pi}\dfrac{\cos(v-i\pi n)}{v}\mathrm dv\, \text{ ,via substituting $v=u+i\pi n$}\\ & = \displaystyle\sum_{k=0}^\infty \dfrac{(-1)^k}{(2k)!}\displaystyle\int_0^{(i+1)n\pi} \dfrac{(v-i\pi n)^{2k}}{v}\mathrm dv\, \text{ ,via Maclaurin series}\end{align}$$
from where I can't figure out further.
Any help is appreciated.
| I do not know if it helps, but...
$$\begin{align}
A = \frac{1}{a^2}\int_{-a}^a \frac{\cos(\frac{x}{a}\pi n)}{1+(\frac xa)^2}dx
&= \frac{2}{a} \int_0^1 \frac{\cos(\pi n x)}{1+x^2}dx \\
&= \frac{2}{a} \int_0^1 \cos(\pi n x) \sum_{k = 0}^\infty (-x^2)^k dx \\
& = \frac{2}{a} \sum_{k=1}^\infty (-1)^k\int_0^1 x^{2k} \cos (\pi n x)dx \\
&= \frac{2}{a} \sum_{k=1}^\infty (-1)^k I_{2k} \end{align}$$
For $I_{2k}$, we have:
$$\begin{align}
I_{2k}
&= \int_0^1 x^{2k} \cos(\pi n x) dx \\
&= \frac{1}{\pi n}x^{2k} \sin (\pi n x)\bigg|_0^1 - \frac{2k}{\pi n}\int_0^1 x^{2k-1} \sin(\pi n x) dx \\
&= - \frac{2k}{\pi n} \int_0^1 x^{2k-1}\sin(\pi n x) dx \\
&= -\frac{2k}{\pi n} \left(-\frac{1}{\pi n} x^{2k-1} \cos (\pi n x)\bigg|_0^1 + \frac{2k-1}{\pi n} \int_0^1 x^{2k-2} \cos(\pi n x)dx \right) \\
&= -\frac{2k}{\pi n}\left(\frac{1}{\pi n} + \frac{2k - 1}{\pi n}\int_0^1 x^{2k-2} \cos(\pi n x)dx \right) \\
&= \frac{1}{(\pi n)^2}\left((2k)(2k-1)\int_0^1 x^{2k-2} \cos(\pi n x)dx - 1\right) \end{align}$$
Therefore,
$$\begin{align}
I_{2k} = \frac{2k(2k-1)}{(\pi n)^2}I_{2k-2} - \frac{1}{(\pi n)^2}\end{align}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4133637",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Determine $\lim_{(x,y)\to(0,0)}\frac{\sin(xy)}{(x^4+y^4)}$ Determine $$\lim_{(x,y)\to(0,0)}\frac{\sin(xy)}{(x^4+y^4)}$$ | $$\lim_{(x,y)\to(0,0)}\frac{x^2 y^2\sin(xy^2)}{(x^4+y^4)\sqrt{x^2+y^2}}=\lim_{r\to0}\frac{r^4\cos^2(\theta)\sin^2(\theta)\sin\left(r^3\cos(\theta)\sin^2(\theta)\right)}{r^5\left(\cos^4(\theta)+\sin^4(\theta)\right)}$$If $\sin\theta\cdot\cos\theta=0$, the numerator is zero. If $\sin\theta\cdot\cos\theta\ne0$, using $\lim_{x\to0}\sin x/x=1$, we get$$\begin{align*}&=\lim_{r\to0}\frac{r^4\cos^2(\theta)\sin^2(\theta)\cdot r^3\cos(\theta)\sin^2(\theta)}{r^5\left(\cos^4(\theta)+\sin^4(\theta)\right)}\\&=\lim_{r\to0}\frac{r^2\cos^3(\theta)\sin^4(\theta)}{\cos^4(\theta)+\sin^4(\theta)}\end{align*}$$
Now $\cos^4(\theta)+\sin^4(\theta)=(\cos^2(\theta)+\sin^2(\theta))^2-2\cos^2(\theta)\sin^2(\theta)=1-\frac12\sin^2(2\theta)\ge1/2$ and thus $\frac1{\cos^4(\theta)+\sin^4(\theta)}\le2$ which means the factor $\frac{\cos^3(\theta)\sin^4(\theta)}{\cos^4(\theta)+\sin^4(\theta)}$ is bounded. Thus the limit is $0$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4134718",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Intersection of sphere of radius c and a general ellipsoid Question: Given the ellipsoid $$E:\ \ \frac{x^2}{a^2}+\frac{y^2}{b^2}+\frac{z^2}{c^2}=1, \ \ 0 < a < c < b,$$
let $\pmb{\alpha}$ denotes the intersection of $E\ $ with the sphere $$S^2(c): \ \ x^2 + y^2 +z^2 =c^2.$$
Show that $\pmb{\alpha}$ is the union [of the traces] of two regular curves $\pmb{\alpha_1}$ and $\pmb{\alpha_2}$. Parametrise either curve and compute its torsion, $\tau$.
Attempts: I've not gotten past the first part.
I've found that a local parametrisation for $E$ is $\mathbf{x}(u,v)=(a\cos u \sin v,\ b \sin u \sin v,\ c \ \cos v)$ which hasn't really made a difference in my confusion.
In both the books I have and on related questions found on this site, the intersection seems to be given specifically for the unit sphere and I'm not sure how to extend the theory to a general sphere.
Further, any computation I do seems to return the solution $a=b=c=1$ which clearly contradicts the fact that $0 < a < c < b$.
| you asked this question 10 months ago, so perhaps this answer is of no use to you now, but I present my solution here:
The points $(x,y,z)$ on the ellipsoid satisfy : $\frac{x^2}{a^2}+\frac{y^2}{b^2}+\frac{z^2}{c^2} = 1$ and similarly The points $(x,y,z)$ on the sphere satisfy : $\frac{x^2}{c^2}+\frac{y^2}{c^2}+\frac{z^2}{c^2} = 1$.
Since $1=1$, we equate the two equations to get:
$\frac{x^2}{a^2}+\frac{y^2}{b^2}+\frac{z^2}{c^2} = \frac{x^2}{c^2}+\frac{y^2}{c^2}+\frac{z^2}{c^2}$
Cancelling the z term we get:
$\frac{x^2}{a^2}+\frac{y^2}{b^2} = \frac{x^2}{c^2}+\frac{y^2}{c^2}$
which can be re-arranged for a relation of $y$ and $x$:
$y^2 = \frac{c^2-a^2}{b^2-c^2}x^2$
The co-efficient of $x^2$ here is positive since $a<c<b$. Then taking square roots, we get 2 solutions:
$y = \pm(\frac{c^2-a^2}{b^2-c^2})^{1/2}x$
These are 2 distinct curves. Parameterising one of them shouldn't be too difficult, hope that helps :)
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Prove that $8(5)^n + 6(3)^{2n} - 80n - 14$ is divisible by $512$ for all $n \in \mathbb{N}$ by induction. Note: question was updated to correct the constant term $-14$ (vs. $-40$).
We need to prove that $8(5)^n + 6(3)^{2n} - 80n - 14$ is divisible by $512$ for all $n \in \mathbb{N}$.
I started by taking $n = 1$, then for $n = k$ and then for $n = k+1$.
I am stuck here. Not able to solve further but still trying. What to do next?
Thanks for your time.
| I assume that you are trying to prove that the difference is divisible by $512$, as the expression is not.
For simplicity, let $f(n) = 8(5)^{n} + 6(3)^{2n} - 80n - 40$. We can simplify this as $$f(n) = 8(5)^{n} + 2(3)^{2n + 1} - 40(2n + 1).$$
We are trying to prove that $512$ divides $f(n) - f(n - 1)$. We see that it is
$$\left[8(5)^{n} + 2(3)^{2n + 1} - 40(2n + 1)\right] - \left[8(5)^{n - 1} + 2(3)^{2n - 1} - 40(2n - 1)\right] \\ 8(5)^{n} - 8(5)^{n - 1} + 2(3)^{2n + 1} - 2(3)^{2n - 1} - 40(2n + 1) + 40(2n - 1) \\ 8(5)^{n - 1}(5 - 1) + 2(3)^{2n - 1}(3^{2} - 1) - 80 \\ 32(5)^{n - 1} + 16(3)^{2n - 1} - 80 \\ 16(2(5)^{n - 1} + (3)^{2n - 1} - 5)$$
Because we factored $16$, we see that $512/16 = 32$. Hence, we need to prove that $32$ divides $2(5)^{n - 1} + (3)^{2n - 1} - 5$. Let this expression be $g(n)$.
The condition $n \in \mathbb{N}$ tells us that the base case should be $n = 1$. We see that $g(1) = 0$ which is trivial.
Can you do the inducting step?
| {
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"url": "https://math.stackexchange.com/questions/4137316",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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} |
Prove that $\frac{a}{b+c^2}+\frac{b}{c+a^2}+\frac{c}{a+b^2}\geq \frac{3}{2}$
My question:
Let $a,b,c$ be positive real numbers satisfy $a+b+c=3.$
Prove that $$\frac{a}{b+c^2}+\frac{b}{c+a^2}+\frac{c}{a+b^2}\geq \frac{3}{2}.$$
I have tried to change the LHS to $$\frac{a^2}{ab+ac^2}+\frac{b^2}{bc+ba^2}+\frac{c^2}{ca+cb^2}$$
And using Cauchy–Schwarz inequality for it
$$\frac{a^2}{ab+ac^2}+\frac{b^2}{bc+ba^2}+\frac{c^2}{ca+cb^2}\geq \frac{(a+b+c)^2}{ab+bc+ca+ac^2+ba^2+cb^2}$$
Then because $$ab+ca+ca\leq \frac{(a+b+c)^2}{3}=\frac{3^2}{3}=3,$$
$$\frac{(a+b+c)^2}{ab+bc+ca+ac^2+ba^2+cb^2}\geq \frac{9}{3+ac^2+ba^2+cb^2}$$
Finally, I can't prove $ac^2+ba^2+cb^2\leq 3$ $$ $$
I look forward to your help, thank you!
| By C-S and by the Vasc's inequality we obtain:$$\sum_{cyc}\frac{a}{b+c^2}=\sum_{cyc}\frac{a^3}{a^2b+a^2c^2}\geq\frac{\left(\sum\limits_{cyc}\sqrt{a^3}\right)^2}{\sum\limits_{cyc}(a^2b+a^2b^2)}=$$
$$=\frac{3\sum\limits_{cyc}(a^3+2\sqrt{a^3b^3})}{\sum\limits_{cyc}a^2b\sum\limits_{cyc}a+3\sum\limits_{cyc}a^2b^2}=\frac{3\sum\limits_{cyc}(a^3+2\sqrt{a^3b^3})}{\sum\limits_{cyc}(4a^2b^2+a^3b+a^2bc)}\geq$$
$$\geq\frac{3\sum\limits_{cyc}(a^3+2\sqrt{a^3b^3})}{\sum\limits_{cyc}(4a^2b^2+a^2bc)+\frac{1}{3}(a^2+b^2+c^2)^2}=\frac{(a+b+c)\sum\limits_{cyc}(a^3+2\sqrt{a^3b^3})}{\sum\limits_{cyc}(4a^2b^2+a^2bc)+\frac{1}{3}(a^2+b^2+c^2)^2}$$ and it's enough to prove that:
$$\frac{(a+b+c)\sum\limits_{cyc}(a^3+2\sqrt{a^3b^3})}{\sum\limits_{cyc}(4a^2b^2+a^2bc)+\frac{1}{3}(a^2+b^2+c^2)^2}\geq\frac{3}{2}$$ or
$$\sum_{cyc}\left(a^{4}+2a^{3}b+2a^{3}c+4\sqrt{a^5b^3}+4\sqrt{a^5c^3}+4\sqrt{a^3b^3c^2}-14a^{2}b^{2}-3a^{2}bc\right)\geq0,$$ which is smooth.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4149973",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 3,
"answer_id": 0
} |
If bisector of angle $C$ of $\triangle ABC$ meet $AB$ in $D$ and circumcircle in $E$ prove that $\frac{CE}{DE}=\frac{(a+b)^2}{c^2}$. If bisector of angle $C$ of $\triangle ABC$ meets $AB$ in point $D$ and the circumcircle in point $E$ then prove that $$\frac{CE}{DE}=\frac{(a+b)^2}{c^2}$$.
My Attempt
Using the fact that $$CD=\frac{2ab}{a+b}\cos\frac{C}{2}$$
and $$AD.DB=CD.DE$$
we get$$\left(\frac{bc}{a+b}\right)\left(\frac{ac}{a+b}\right)=\left(\frac{2ab}{a+b}\cos\frac{C}{2}\right)DE\Rightarrow DE=\frac{c^2}{2(a+b)\cos\frac{C}{2}}$$ and then I did lot of calculation to obtain $$CE=CD+DE=\frac{a+b}{\cos\frac{C}{2}}$$ to get the final ratio.
But I believe there would be a more generic way to do it perhaps a geometrical solution rather than relying upon the formula of angle bisector $CD=\frac{2ab}{a+b}\cos\frac{C}{2}$
| Using formula for length of angle bisector, $ \ \displaystyle CD^2 = \frac{ab}{(a+b)^2} ((a+b)^2 - c^2)$
By Intersecting Chords theorem,
$ \displaystyle CD.DE = AD \cdot DB = (\frac{a}{a+b} \cdot c) \cdot (\frac{b}{a+b} \cdot c) = \frac{ab c^2}{(a+b)^2}$
$ \displaystyle \frac{CE}{DE} = 1 + \frac{CD^2}{ CD \cdot DE} = 1 + \frac{(a+b)^2}{abc^2} \cdot CD^2 = 1 + \frac{(a+b)^2 - c^2}{c^2} = \frac{(a+b)^2}{c^2}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4152868",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Solution Verification: Find the Global Max and Min of $z=x+y$, in $D=\{x\ge 0, y\ge 0, y\le 3x, x^2+y^2\le 25 \}$
$z=x+y$
$D=\{x\ge 0, y\ge 0, y\le 3x, x^2+y^2\le 25 \}$
My attempt:
Plan:
Find inner critical points.
Find maxmin of border curves.
Find intersecting points of border curves.
Find values of all points and decide global maximum and minimum.
Inner critical points:
First of all, I checked the inner points of the domain by taking $z_x,z_y=0$. And obviously there's no critical points there since $z_x=z_y=1$.
Borders or domain curves (not sure what it's called):
Next step, is finding the maximum and minimum on the domain curves (Or the borders),
$x^2+y^2=25 \Longrightarrow x = \sqrt{25-y^2}$, Substituting that into my function:
$z = \sqrt{25-y^2}+y \Longrightarrow z'=\frac{-2y}{2\sqrt{25-y^2}}+1=0$ (To find critical points).
$-y+\sqrt{25-y^2}=0 \Longrightarrow y^2=25-y^2 \Longrightarrow y^2 =\frac{25}{2} \Longrightarrow y=\frac{5}{\sqrt{2}}$ (I took positive value only because $y \ge 0$).
The point $(x,\frac{5}{\sqrt{2}})$ must be on the circle, so $x^2+\frac{25}{2}=25 \Longrightarrow x=\frac{5}{\sqrt{2}}$ ($x \ge 0$).
So I've found a point $(\frac{5}{\sqrt{2}},\frac{5}{\sqrt{2}})$
Next, $y=0 \Longrightarrow z=x \Longrightarrow z_x=1$ There's no critical points. (same for $x=0$).
Next, $y=3x\Longrightarrow z=4x \Longrightarrow z_x = 4$ (no critical points too).
Intersecting points of border curves:
Now my next step is to check points where $x^2+y^2=25$ and $y=3x$ and $y=0$ and $x=0$ meet,
I can see points $(0,0)$, $(0,5)$, now I need to find the intersection of the circle and $y=3x$.
$(x^2 + 9x^2)=25 \Longrightarrow 10x^2=25 \Longrightarrow x=\frac{5}{\sqrt{10}}$
Substituting in the $y=3x \Longrightarrow y=\frac{15}{\sqrt{10}}$ And so my last point is $(\frac{5}{\sqrt{10}}, \frac{15}{\sqrt{10}})$
Calculating the value of the function in the points I've found:
My next and last step, is to calculate the value of $z=f(x,y)$ in all the points I have found:
$f(\frac{5}{\sqrt{2}},\frac{5}{\sqrt{2}})= \frac{10}{\sqrt{2}} = 7.071...$
$f(0,0)=0$
$f(0,5)=5$
$f(\frac{5}{\sqrt{10}}, \frac{15}{\sqrt{10}})= \frac{20}{\sqrt{10}}=6.324...$
Final Answer:
And so, my global maximum is $(\frac{5}{\sqrt{2}},\frac{5}{\sqrt{2}})$, Global minimum is $(0,0)$.
I would love to hear feedback about my solution and to know if I missed some points, checked some unnecessary points or some of my work flow doesn't seem okay. Thanks in advance!
| Geometrically, we are bound by circle $x^2 + y^2 = 25 \ $ between $x$-axis and line $y = 3x$ in the first quadrant.
In polar coordinates, $x = r \cos\theta, y = r \sin\theta$.
Constraint is given as $x^2 + y^2 \leq 25, x,y \geq 0 \implies r \leq 5, 0 \leq \theta \leq \frac{\pi}{2}$
But we are also bound by $y = 3x, \tan \theta = 3 \implies 0 \leq \theta \leq \arctan(3)$
So we need to find minimum and maximum of $z = x + y = r ( \cos\theta + \sin\theta) \ $ in region $0 \leq r \leq 5, 0 \leq \theta \leq \arctan(3)$
We know $ \ 0 \lt \cos\theta + \sin\theta \leq \sqrt2 \ $ for $0 \leq \theta \leq \frac{\pi}{2}$.
We also know (or can quickly take derivative to find it), maximum of $\sqrt2 \ $ occurs at $ \theta = \frac{\pi}{4} \lt \arctan (3)$, which is in our region.
So it is clear that maximum of function $z$ occurs at $r = 5$ and $\theta = \frac{\pi}{4}$ which is point $(\frac{5}{\sqrt2}, \frac{5}{\sqrt2})$.
As $\cos\theta + \sin\theta \gt 0$ and $r \geq 0$, it is easy to see that $z = r (\cos\theta + \sin\theta)$ will have minimum at $r = 0$, which is point $(0, 0)$.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
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How to show that $ 1 + x + x^2 +x^3 +...= \frac{1}{1-x} $? By long division, it is easy to show that $$ \frac{1}{1-x} = 1 + x + x^2 +x^3 +... $$
But how to show that
$$ 1 + x + x^2 +x^3 +...= \frac{1}{1-x} $$
| You have following expression-
$ 1 + x + x^2 +x^3 +...$
Now multiplying numerator and denominator by $x-1$, provided $x\neq1$, we have
$\frac{(1-x)( 1 + x + x^2 +x^3 +...)}{1-x}=\frac{(1+x+x^2...)-(x+x^2+x^3)}{1-x}=\frac{1}{1-x}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4157510",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 1
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Limit of Function raised to a power of another function I'm trying to evaluate the following limit:
$$A =\lim_{x\to\infty}\left(\frac{x^2-3x+1}{x^2+x+2}\right)^{2x-5}$$
So far, I have exponentiated the limit and whatnot and now I am at this stage:
$$A =\exp\left(\lim_{x\to\infty}(2x-5)\ln\frac{x^2-3x+1}{x^2+x+2}\right)$$
Now, I don't know what to do here, I know via simple algebra that:
$$\lim_{x\to\infty}\ln\frac{x^2-3x+1}{x^2+x+2}=0$$
But $2x-5$ has no limit, and thus I cannot separate $A$ into the product of two limits. Perhaps I am missing something? WolframAlpha says
$$\lim_{x\to\infty}(2x-5)\ln\frac{x^2-3x+1}{x^2+x+2}=-8$$
And that therefore $A = e^{-8}$, but gives no insight as to how this is the case.
| One of favorite way to see this so quickly as follows
$$\lim_{x\to\infty}\left(\frac{x^2-3x+1}{x^2+x+2}\right)^{2x-5}= \lim_{x\to\infty}\left(1-\frac{4x+2}{x^2+x+2}\right)^{2x-5}=\lim_{x\to\infty}\left(1-\frac{1}{\frac{x^2+x+2}{4x+2}}\right)^{\frac{x^2+x+2}{4x+2}\frac{4x+2}{x^2+x+2} (2x-5)} $$
So, since $\lim_{x\to\infty}\frac{(4x+2)(2x-5)}{x^2+x+2}=8.$ Then,
$$\lim_{x\to\infty}\left(\frac{x^2-3x+1}{x^2+x+2}\right)^{2x-5}= e^{-8}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4162919",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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Function related polynomial problem
My approach is as follow $f\left( x \right) = 2{x^4} + 3{x^3} - 3{x^2} - 6x + a$
$f'\left( x \right) = 8{x^3} + 9{x^2} - 6x - 6$
$f''\left( x \right) = 24{x^2} + 18x - 6 = 6\left( {4{x^2} + 3x - 1} \right)$
The roots of $f''(x)$ are $-1$ and $\frac{1}{4}$, how do I arrive at the proper conclusion.
$f\left( 1 \right) = a - 4$
$f\left( 2 \right) = 32 + 24 - 12 - 12 + a = a + 32$
$f\left( 1 \right).f\left( 2 \right) < 0$
$\left( {a - 4} \right)\left( {a + 32} \right) < 0;a \in \left( { - 32,4} \right)$
I found this question on the ineternet.
| It's easy to show that there are no other roots on the $(1,2)$ besides $y$. Indeed, from the positive sign of the second derivative, we see that $f'$ is increasing on the interval $(1,2)$. Since $f'(1) > 0$, we obtain that $f$ is an increasing function that's why it can have only one root on the $(1,2)$, namely $y$.
From the mean value theorem we have
$$
\left|f(y) - f\left( \frac{p}{q} \right) \right| = |f'(\xi)| \cdot \left| y - \frac{p}{q} \right|.
$$
$\xi$ is a point on the interval between $y$ and $p/q$. This gives
$$
\left| y - \frac{p}{q} \right| = \frac{\left| f\left( \frac{p}{q} \right) \right|}{|f'(\xi)|}.
$$
Now we need to estimate the right-hand side.
Denote
$$
M = \max_{x \in [1,2]} |f'(x)|
$$
and let's see what happens when we plug $\frac{p}{q}$ in our polynomial
$$
f\left( \frac{p}{q} \right) = \frac{2p^4}{q^4} + \frac{3p^3}{q^3} - \frac{3p^2}{q^2} - \frac{6p}{q} + a = \frac{2p^4 + 3p^3q - 3p^2q^2 - 6pq^3 + aq^4}{q^4}.
$$
The numerator is a non-zero integer (it's not equal to $0$ since $p/q$ is not a root). That's why its absolute value is at least one and
$$
\left| f\left( \frac{p}{q} \right) \right| \ge \frac{1}{q^4}.
$$
This gives us the needed estimate
$$
\left| y - \frac{p}{q} \right| \ge \frac{1}{Mq^4}.
$$
It should be noted that this result can be easily generalised to the polynomial of any degree $n$. It's actually a key argument in the Liouville theorem. A precise statement of this theorem can be found for example here.
| {
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If $f(x) = \frac{x^3}{3} -\frac{x^2}{2} + x + \frac{1}{12}$, then $\int_{\frac{1}{7}}^{\frac{6}{7}}f(f(x))\,dx =\,$? This is a question from a practice workbook for a college entrance exam.
Let
$$f(x) = \frac{x^3}{3} -\frac{x^2}{2} + x + \frac{1}{12}.$$
Find
$$\int_{\frac{1}{7}}^{\frac{6}{7}}f(f(x))\,dx.$$
While I know that computing $f(f(x))$ is an option, it is very time consuming and wouldn't be practical considering the time limit of the exam. I believe there must be a more elegant solution.
Looking at the limits, I tried to find useful things about $f(\frac{1}{7}+\frac{6}{7}-x)$
The relation I obtained was that $f(x) + f(1-x) = 12/12 = 1$. I don't know how to use this for the direct integral of $f(f(x)).$
| We know,
$ \displaystyle \int_{a}^{1-a}f(f(x))\,dx = \int_{a}^{1-a}f(f(1-x))\,dx$.
So,
$\displaystyle \int_{a}^{1-a}f(f(x))\,dx = \frac{1}{2} \int_a^{1-a}\left[f(f(x))+f(f(1-x)) \right] \ dx$
Now for the given function, observe that $f(x) + f(1-x) = 1 \implies f(1-x) = 1 - f(x)$
So, $f(f(x)) + f(f(1-x)) = f(f(x)) + f(1-f(x)) = 1$
So we have,
$\displaystyle \int_{a}^{1-a}f(f(x))\,dx = \frac{1}{2} (1-2a)$
Here $a = \dfrac{1}{7}$ and that leads to $\dfrac{5}{14}$.
| {
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Can I do this rearrangement with an infinite series to arrive at divergent parts? $$\sum_{i=3}^{\infty}\frac{i}{\left(i-2\right)\left(i-1\right)\left(i+1\right)} = \sum_{i=3}^\infty \frac{2}{3\left(i-2\right)} - \frac{1}{2\left(i-1\right)} - \frac{1}{6\left(i+1\right)} \\**= \frac{2}{3}\sum_{i=1}^\infty\frac{1}{i} - \frac{1}{2}\left(\left(\sum_{i=1}^\infty\frac{1}{i}\right) - 1\right) - \frac{1}{6}\left(\left(\sum_{i=1}^\infty\frac{1}{i}\right) - 1 - \frac{1}{2} - \frac{1}{3}\right)** \\= \left(\frac{2}{3} - \frac{1}{2} - \frac{1}{6}\right) \left(\sum_{i=1}^\infty\frac{1}{i}\right) + \frac{1}{2} + \frac{1}{6} + \frac{1}{12} + \frac{1}{18} \\= 0\cdot\left(\sum_{i=1}^\infty\frac{1}{i}\right) + \frac{29}{36} \\= \frac{29}{36} \space or \space \infty?$$
When summing up the series above I came by the harmonic series multiplied by $0$. I simply equated it to be $0$ while my friend keeps arguing that the harmonic series is an infinity and infinity multiplied by $0$ is undefined. Now I find it difficult to argue with it. Does this mean that the series above diverge?
Or maybe I was thinking, in the starred line, the $\frac{2}{3}\left(\sum_{i=1}^\infty\frac{1}{i}\right) - \frac{1}{2}\left(\sum_{i=1}^\infty\frac{1}{i}\right) - \frac{1}{6}\left(\sum_{i=1}^\infty\frac{1}{i}\right)$ is cancelling out anyway, right? I mean I could just right it term by term and show every term cancels. If they are cancelling out then this rearrangement should not be problematic, right? So, am I right to say that $0\cdot\left(\sum_{i=1}^\infty\frac{1}{i}\right) = 0$ to show it's convergence?
| Well you can use this method for verification
$\displaystyle\sum_{i=3}^{\infty}\frac{i}{\left(i-2\right)\left(i-1\right)\left(i+1\right)}$
$\displaystyle\sum_{i=3}^{\infty}\frac{i^2}{\left(i-2\right)\left(i-1\right)(i)\left(i+1\right)}$
$\displaystyle\sum_{i=3}^{\infty}\frac{(i-1)(i+1)+1}{\left(i-2\right)\left(i-1\right)(i)\left(i+1\right)}$
$\displaystyle\sum_{i=3}^{\infty}\frac{(i+1)(i-1)}{\left(i-2\right)\left(i-1\right)(i)\left(i+1\right)}+\sum_{i=3}^{\infty}\frac{1}{\left(i-2\right)\left(i-1\right)(i)\left(i+1\right)}$
$\displaystyle\sum_{i=3}^{\infty}\frac{1}{\left(i-2\right)(i)}+\sum_{i=3}^{\infty}\frac{1}{\left(i-2\right)\left(i-1\right)(i)\left(i+1\right)}$
$\displaystyle\frac{1}{2}\sum_{i=3}^{\infty}\frac{i-(i-2)}{\left(i-2\right)(i)}+\frac{1}{3}\sum_{i=3}^{\infty}\frac{(i+1)-(i-2)}{\left(i-2\right)\left(i-1\right)(i)\left(i+1\right)}$
$\displaystyle\frac{1}{2}\sum_{i=3}^{\infty}(\frac{1}{\left(i-2\right)}-\frac{1}{i})+\frac{1}{3}\sum_{i=3}^{\infty}(\frac{1}{\left(i-2\right)\left(i-1\right)(i)}-\frac{1}{\left(i-1\right)\left(i\right)(i+1)})$
$\displaystyle\frac{1}{2}(1+\frac{1}{2})+\frac{1}{3}(\frac{1}{6})$
$\displaystyle\frac{3}{4}+\frac{1}{18}={\frac{29}{36}}$
| {
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Maximizing magnitude given a complex valued function I am presented with a problem, which asks you to maximize $|z|$ with the constraint of $$\left|z+\frac 2z \right| = 2.$$
I tried to approach this with components, however this degraded into a messy cubic, letting $z=a+bi$ only to receive $$\frac{a^3+a^2b+ab^2+b^3+2a-2b}{a^2+b^2},$$ trying to maximize $\sqrt{a^2+b^2}.$
How does one go through solving this, without calculus?
| Your approach works (It seems that you have an error in your approach).
Letting $z=a+bi$ where $a,b$ are real numbers, we have
$$\begin{align}&\bigg|a+bi+\frac 2{a+bi}\bigg|=2
\\\\&\iff \bigg|a+bi+\frac{2(a-bi)}{a^2+b^2}\bigg|=2
\\\\&\iff \bigg|\bigg(a+\frac{2a}{a^2+b^2}\bigg)+\bigg(b-\frac{2b}{a^2+b^2}\bigg)i\bigg|=2
\\\\&\iff \sqrt{\bigg(a+\frac{2a}{a^2+b^2}\bigg)^2+\bigg(b-\frac{2b}{a^2+b^2}\bigg)^2}=2
\\\\&\iff \frac{(a(a^2+b^2)+2a)^2+(b(a^2+b^2)-2b)^2}{(a^2+b^2)^2}=4
\\\\&\iff \frac{(a^2 + b^2) (a^4 + 2 a^2 b^2 + 4 a^2 + b^4 - 4 b^2 + 4)}{(a^2+b^2)^2}=4
\\\\&\iff \frac{a^4 + 2 a^2 b^2 + 4 a^2 + b^4 - 4 b^2 + 4}{a^2 + b^2}=4
\\\\&\iff a^4 + 2 a^2 b^2 + 4 a^2 + b^4 - 4 b^2 + 4-4(a^2+b^2)=0
\\\\&\iff b^4 +(2 a^2 - 8) b^2 +a^4 + 4=0
\\\\&\iff b^2=4-a^2\pm\sqrt{12-8a^2}\end{align}$$
So, we can write
$$|z|=\sqrt{a^2+b^2}=\sqrt{4\pm\sqrt{12-8a^2}}$$
So, the maximum value of $|z|$ is $$\sqrt{4+\sqrt{12-8\times 0^2}}=\sqrt{4+2\sqrt 3}=\sqrt{(1+\sqrt 3)^2}=\color{red}{1+\sqrt 3}$$ which is attained when $z=\pm(1+\sqrt 3)i$.
Another approach (there was an answer using this approach, but the answer was deleted) :
For any two complex numbers $z_1$ and $z_2$,
$$|z_1+z_2|\le |z_1|+|z_2|$$
(see here for a proof)
Using this inequality, we have
$$|z|=\bigg|z+\frac 2z-\frac 2z\bigg|\le \bigg|z+\frac 2z\bigg|+\bigg|-\frac 2z\bigg|=2+\frac{2}{|z|}$$
$$\implies |z|\le 2+\frac{2}{|z|}\implies |z|^2-2|z|-2\le 0\implies |z|\le 1+\sqrt{3}$$
If $z=(1+\sqrt 3)i$, then $|z|=1+\sqrt 3$ and $\bigg|z+\dfrac 2z\bigg|=2$ hold.
Therefore, the maximum value of $|z|$ is $\color{red}{1+\sqrt 3}$.
| {
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Find $x$ and $y$ such that $(x+3i)(x+iy)=1-i$ I tried expanding the brackets, equating Real and Imaginary parts and finally substituting $x$ in terms of $y$. I ended up with the equation:
$$3y^3 + 19y^2 + 33y +8=0$$
Using a calculator, I got a root at $y\simeq-0.288$ which is approx. $\frac{-7}{25}$.
I should be getting $y=\frac{-7}{25}$ and $x=\frac{1}{25}$. How should I go about finding $x$ and $y$ without a calculator?
| Expanding the left hand side and comparing real and imaginary parts yields
\begin{eqnarray*}
x^2-3y&=&\hphantom{-}1\tag{1}\\
xy+3x&=&-1\tag{2}
\end{eqnarray*}
The first equation shows that $y=\tfrac{x^2-1}{3}$ and substituting into the second yields
$$-1=x\frac{x^2-1}{3}+3x=\frac13x^3+\frac{8}{3}x,$$
and so $x$ is a root of the cubic equation
$$X^3+8X+3=0.\tag{3}$$
By the rational root test, this cubic has no rational roots. From the second equation we find that $x\neq0$ and $y=-\frac1x-3$, which shows that also $y$ cannot be rational. In particular, the solutions you give cannot actually be solutions.
The roots of $(3)$ can be expressed in terms of radicals by means of the cubic formula. You may find that the cubic has precisely one real root because its discriminant is negative. It does not simplify to anything 'nice'.
The only real solution of the equation $(3)$ is
$x=\sqrt[3]{-\dfrac32+\dfrac1{18}\sqrt{6873}}-\sqrt[3]{\dfrac32+\dfrac1{18}\sqrt{6873}}$
Moreover ,
$y=\dfrac13\sqrt[3]{\dfrac{1267}{54}+\dfrac16\sqrt{6873}}+ \dfrac13\sqrt[3]{\dfrac{1267}{54}-\dfrac16\sqrt{6873}}-\dfrac{19}9$
| {
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How to find $f'(e^2)$ given that $f(x) + \ln (x^2 + f(x)^2) = 4$? So, as the title says, I'm looking to find $f'(e^2)$, being How can I find $f'(e^2)$ being $y = f(x)$ in $y + \ln (x^2 + y^2) = 4$. I found this interesting exercise in a peruan book, and I think that it'd be a great way to share its mechanics.
Firstly, I did the $f(x)$ equation:
\begin{align*}
y + \ln(x^2 + y^2) &= 4\\
f(x) + \ln[x + f(x)^2] &= 4\\
f(e²) + \ln[e^2 + f(e^2)^2] &= 4
\end{align*}
Then, the $f'(x)$ part:
\begin{align*}
[y + \ln(x^2 + y^2)]' &= [4]'\\
y' + \displaystyle\frac{1}{x^2 + y^2} \cdot (2x + 2yy') &= 0\\
y' &= -\displaystyle\frac{2x + 2yy'}{x^2 + y^2}\\
y' \cdot (x^2 + y^2) &= –(2x + 2yy')\\
y' \cdot (x^2 + y^2) &= –2x – 2yy'\\
y' \cdot (x^2 + y^2) + 2yy' &= –2x \\
y' \cdot (x^2 + 2y + y²) &= –2x \\
y' &= –\displaystyle\frac{2x}{x^2 + 2y + y^2}\\
f'(x) &= –\displaystyle\frac{2x}{x^2 + 2f(x) + f(x)^2}\\
f '(e^2) &= –\displaystyle\frac{2e^2}{e^4 + 2f(e^2) + f(e^2)^2²}
\end{align*}
Couldn't go on from here.
| So firstly we start differentiating and then look for corresponding value of the function.
$$y=4-\ln (x^2+y^2)$$
$$y'=-\frac{1}{x^2+y^2} ( 2x+2yy')$$
Simplifying gives
$$y'=-\frac{2x}{x^2+y^2+2y}$$
Evaluating at $x=e^2$ you'll be needing $f(e^2)$.
So try to substitute this in what is given and find y that is
$$y=4-\ln(e^4+y^2)$$
Now we substitute for few values and observe how the function behaves and substituting $y=0$ perfectly fits the condition as then you get $4-4=0$ on rhs.
Put this back in our calculated derivative to get
$$y'(e^2)=-\frac{2e^2}{e^4}=-\frac{2}{e^2}$$
| {
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Finding the range of $a$ for which line $y=2x+a$ lies between circles $(x-1)^2+(y-1)^2=1$ and $(x-8)^2+(y-1)^2=4$ without intersecting either
Find the range of parameter $a$ for which the variable line $y = 2x + a$ lies between the circles $(x-1)^2+(y-1)^2=1$ and $(x-8)^2+(y-1)^2=4$ without intersecting or touching either circle.
Now, how I solved it was realising the line has a positive slope of $+2$, and thus if it's a tangent to circle #1 then its intercept should be negative. And so, using the condition of tangency,
$$a^2= m^2(r_1)^2+(r_1)^2= 4+1 $$
And thus, as $a$ can only be negative (otherwise a positive value of $a$ will make the line intersect the circle). Thus, making the lower bound of $a> - \sqrt 5$.
Similarly for the bigger circle $a<-\sqrt{20}=-\sqrt{(2^2)(2^2)+(2^2)}$
Hence I find that the solution should be $(-\sqrt5,-\sqrt{20})$, but the actual solution is $\left(2\sqrt 5-15,-\sqrt 5-1\right)$.
| The line passing through the center of the first circle is $y=2x-1$. Its perpendicular line passing through its center is $y=-\frac12x+\frac12$. It crosses the first circle at $(x,y)=\left(1+\frac2{\sqrt5},1-\frac1{\sqrt5}\right)$. Hence, $y=2x-\sqrt5-1$ touches the first circle from the right side.
Similarly, the line passing through the center of the second circle is $y=2x-15$. Its perpendicular line passing through its center is $y=-\frac12x+5$. It crosses the first circle at $(x,y)=\left(8-\frac4{\sqrt5},1+\frac2{\sqrt5}\right)$. Hence, $y=2x+2\sqrt5-15$ touches the second circle from the left side.
Hence, the value of $a$ ranges between $2\sqrt5-15$ and $-\sqrt5-1$.
See the Desmos graph for reference.
| {
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Geometric reason why this determinant can be factored to (x-y)(y-z)(z-x)? The determinant $\begin{vmatrix}
1 & 1 &1 \\
x & y & z \\
x^2 & y^2 &z^2 \\
\end{vmatrix}$ can be factored to the form $(x-y)(y-z)(z-x)$
Proof:
Subtracting column 1 from column 2, and putting that in column 2,
\begin{equation*}
\begin{vmatrix}
1 & 1 &1 \\
x & y & z \\
x^2 & y^2 &z^2 \\
\end{vmatrix}
=
\begin{vmatrix}
1 & 0 &1 \\
x & y-x & z \\
x^2 & y^2-x^2 &z^2 \\
\end{vmatrix}
\end{equation*}
$
= z^2(y-x)-z(y^2-x^2)+x(y^2-x^2)-x^2(y-x)
$
rearranging the terms,
$
=z^2(y-x)-x^2(y-x)+x(y^2-x^2)-z(y^2-x^2)
$
taking out the common terms $(y-x)$ and $(y^2-x^2)$,
$
=(y-x)(z^2-x^2)+(y^2-x^2)(x-z)
$
expanding the terms $(z^2-x^2)$ and $(y^2-x^2)$
$
=(y-x)(z-x)(z+x)+(y-x)(y+x)(x-z)
$
$
=(y-x)(z-x)(z+x)-(y-x)(z-x)(y+x)
$
taking out the common term (y-x)(z-x)
$
=(y-x)(z-x) [z+x-y-x]
$
$
=(y-x)(z-x)(z-y)
$
$
=(x-y)(y-z)(z-x)
$
Is there a geometric reason for this?
The determinant of this matrix is the volume of a parallelopiped with sides as vectors whose tail is at the origin and head at x,y,z coordinates being equal to the columns(or rows) of the matrix.$^{[1]}$
So is the volume of this parallelopiped equals $(x-y)(y-z)(z-x)$ in any obvious geometric way?
References
[1] Nykamp DQ, “The relationship between determinants and area or volume.” From Math Insight. http://mathinsight.org/relationship_determinants_area_volume
| Since $x=y, y=z, z=x$ give thre determinant $D$ as the determinant as zero, and it being homogeneous cubic (see the product of diagonal element), D needs to be
$D=A(x-y)(y-z)(z-x)$. Further, set $z=0, x=1,y=2$ to get $A=1$.
| {
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Number of integer solutions of $a^2+b^2=10c^2$ Find the number of integer solutions of the equation $a^2+b^2=10c^2$.
I can only get by inspection that $a=3m, b=m,c=m$ satisfies for any $m \in Z$.
Is there a formal logic to find all possible solutions? Any hint?
Also i tried taking $a=p^2-q^2$, $b=2pq$ and $10c^2=(p^2+q^2)^2$
which gives $$\frac{p^2+q^2}{c}=\sqrt{10}$$ which is invalid, since a rational can never be an irrational.
| $$x^2+y^2=nz^2\tag{1}$$
In general, if equation $(1)$ has one integer solution $(x,y,z)=(x_0,y_0,z_0)$ then there exists an infinitely many integer solutions.
Substitute $x=t+x_0, y=t+y_0, z=at+z_0$ to equation $(1)$, then we get
$$t = \frac{2(x_0+y_0-nz_0a)}{-2+na^2}$$
Let $a=\frac{p}{q}$ with p,q are integers, hence we get a parametric solution $(x,y,z)=(x_0np^2-2qnz_0p+2q^2y_0, y_0np^2-2qnz_0p+2q^2x_0, nz_0p^2-2qx_0p-2qy_0p+2q^2z_0).$
Example of $x^2+y^2=10z^2.$
Let $n=10, (x_0,y_0,z_0)=(3,1,1).$
$(x,y,z)=(15p^2-10qp+q^2, 5p^2-10qp+3q^2, 5p^2-4qp+q^2).$
| {
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"timestamp": "2023-03-29T00:00:00",
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Comparing numbers of the form $c+\sqrt{b}$ (eg, $3+3\sqrt{3}$ and $4+2\sqrt{5}$) without a calculator It is easy to compare to numbers of the form $a\sqrt{b}$, simply by comparing their squares, for example $3\sqrt{3}$ and $2\sqrt{5}$.
But what if we have $a=3+3\sqrt{3}$ and $b=4+2\sqrt{5}$ for example?
How to compare them without using a calculator? Is there a method that works for all numbers of the form $c+\sqrt{b}$?
I found a similar question to this on the site, but that one is a bit different and I am looking for an answer suitable for elementary students to understand. (but I would love to know any kind of answer even if it does not fit the elementary level).
I was thinking about subtracting $b-a$ and checking wether it is negative or positive,
I got $b-a=1+2\sqrt{5}-3\sqrt{3}$ but $2\sqrt{5}-3\sqrt{3}<0$
(I don't want to continue from here by approximating $\sqrt{5}=2.something$, because we will be somehow using the calculator in our mind; so I am stuck here)
| You can use binomial theorem result
$(1\pm x)^n\approx (1\pm nx)$ provided that $x<<1$
Therefore you have $3+3\sqrt3=3+3(4-1)^{\frac{1}{2}}\approx3+6(1-0.25)^{\frac{1}{2}}= 3+6(1-0.125)=9-0.75=8.25$
Similarly $4+2\sqrt5=4+2(4+1)^{\frac{1}{2}}\approx4+4(1+0.25)^{\frac{1}{2}}= 4+4(1+0.125)=8+0.5=8.5$
This would work better when you have large numbers involved inside radical
| {
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Let $n$ and $m$ be integers such that $5$ divides $1+2n^2+3m^2.$ Then show that $5$ divides $n^2-1.$
Let $n$ and $m$ be integers such that $5$ divides $1+2n^2+3m^2.$ Then show that $5$ divides $n^2-1.$
$\textbf{My attempts} :$
From the condition, we can write $$1+2n^2+3m^2\equiv 0(\mod 5)\tag 1$$
Now, since $5$ is prime so by Fermat's little theorem, we can write $$n^4\equiv 1(\mod 5)\quad\text{and} \quad m^4\equiv 1(\mod 5).$$
So, we get $n^4-m^4\equiv 0(\mod 5)$.
Since $5$ prime so, either $5|(m^2+n^2)$ or $5|(m^2-n^2).$
Now if $5|(m^2+n^2)$ then from $(1)$ we get $$1-n^2+3n^2+3m^2\equiv 0(\mod 5)$$ So, we are done.
Now, if $5|(m^2-n^2)$ then from $(1)$ we get $$1+5n^2-3n^2+3m^2\equiv 0(\mod 5).$$ So, we shall arrive at a contradiction that $1\equiv 0(\mod 5).$
In this way, I have tried to solve this problem. I will be highly obliged if you kindly check this or correct me.
Thanks in advance.
| If $1+2n^2+3m^2\equiv 0\pmod 5$, then $n^2\equiv m^2+2$.
Since the only squares modulo $5$ are $0,\pm 1$, the only values for $n^2$ and $m^2$ are $n^2\equiv 1\pmod 5$ and $m^2\equiv -1\pmod 5$.
| {
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"timestamp": "2023-03-29T00:00:00",
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Finding the sum of the cubes of the first n odd natural numbers. I know that the sum of the cubes of the first $n$ natural numbers is $({\frac{n(n+1)}{2}})^2$, which in expanded form is $(1)^3+(2)^3+(3)^3+...+(n)^3$. Multiplying by $2^3$ on both sides gives $2(n(n+1))^2$, which I can see is the sum of cube of first $n$ even natural numbers. But my question is why do we take the difference between the sum of cubes of all $2n$ (i.e. $({\frac{2n(2n+1)}{2}})^2$) numbers and sum of cubes of first $n$ even numbers (i.e. $2n^2(n+1)^2$) to obtain the odd sum? How would you distinguish between the two?
| Use a specific example.
Let $A = 1^3 + 3^3 + 5^3 + 7^3 + 9^3.$
Let $B = 2^3 + 4^3 + 6^3 + 8^3.$
Let $C = 1^3 + 2^3 + 3^3 + 4^3.$
Assume that you want to evaluate $A$.
You can use the formula to evaluate $(A + B)$, so the problem is reduced to evaluating $B$.
This can be done by using the formula to evaluate $C$, and then reasoning that $8C = B$.
| {
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"url": "https://math.stackexchange.com/questions/4184291",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Find constant in system of function equations
Question: Let $a \in \mathbb{R}$. Find all possible values of $a$ such that there exists a function $f: \mathbb{R} \to \mathbb{R}$ satisfying $f(x + 2) = -f(x)$ and $f(x + 3) = f(x) + a$ for any real number $x \in \mathbb{R}$.
My (maybe wrong) solution:
Start listing down some values of $f(x)$ with $f(2) = -f(0)$ and $f(3) = f(0) + a$.
$f(2) + f(3) = - f(0) + f(0) + a = a$.
Similarly, you can write $f(1) = -(f(-1))$ and $f(2) = f(-1) + a$.
Hence, $f(1) + f(2) = -(f(-1)) + f(-1) + a = a$.
So, $f(1) + f(2) = f(2) + f(3) \Longrightarrow f(1) = f(3).$
But, according to the problem, $f(x + 2) = -f(x) \Longrightarrow f(3) = -f(1)$.
Since $f(1) = f(3)$ and $f(1) = -f(3) \Longrightarrow f(1) = f(3) = 0.$
Doing similar steps, we find out that $f(0) = f(2) = 0$. Hence, $a = f(2) + f(3) = 0 + 0 = 0$.
The only possible value of $a$ is $\boxed{0}$.
$\\$
If I'm doing something wrong, can someone please explain? Thanks in advance!
(Thanks to the answers, I now know I only need to prove that $a = 0$ can satisfy a valid function :))
| This makes sense. Here is a different approach leading to the same place. Note that using the first relationship,
$$
f(x+6) = -f(x+4) = f(x+2) = -f(x)
$$
and using the second relationship,
$$
f(x+6) = f(x+3)+a = f(x) + 2a,
$$
so we have $f(x)+2a = -f(x) \iff f(x) = -a$, so $f$ is constant.
Now from the first relationship,
$$
-f(x) = f(x+2) = f(x) \iff f(x) = 0 \ \forall x \in \mathbb{R}.
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4186151",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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$a \geq b \geq c \geq 0$ and $a + b + c \leq 1$. Prove that $a^2 + 3b^2 + 5c^2 \leq 1$. Positive numbers $a,b,c$ satisfy $a \geq b \geq c$ and $a + b + c \leq 1$. Let $f(a, b, c) = a^2 + 3b^2 + 5c^2$. Prove that $f(a, b, c) \leq 1$.
One observation is that the bound is met: $f(1, 0, 0) = f\left(\frac{1}{2}, \frac{1}{2}, 0\right) = f\left(\frac{1}{3}, \frac{1}{3}, \frac{1}{3}\right) = 1$.
Another observation is that clearly $a + b + c = 1$ at a maximum of $f$ since increasing any of $a,b,c$ increases $f$. So I'm just going to assume that $a + b + c = 1$ from now on.
My progress so far is that using Lagrange multipliers, you can see that $f$ is minimised subject to the constraint $a + b + c = 1$ at the point $\left(\frac{15}{23}, \frac{5}{23}, \frac{3}{23}\right)$. And since there are no other minimums it must be increasing as you choose $(a, b, c)$ away from this point. Setting $a = 1 - b - c$, we just need to optimise in $b, c$. The optimisation is bounded by the triangle formed from the lines $$c = 0, b = \frac{1}{2} - \frac{1}{2}c, \textrm{ and } c = b.$$
As $f$ is increasing as you move away from the minimum, the optimisation occurs at one of the corners of this triangle, and these give the three solutions that I gave at the start.
I think my
it must be increasing as you choose $(a, b, c)$ away from this point.
claim is a bit vague and I'm not sure how to formulate it properly. Hopefully there is a clearer method of proving this.
I found the puzzle here in case it's of interest.
| Let $ c = x$, $b = x+y$, $a = x+y+z$ where $ x, y, z \geq 0$.
Then, the question becomes:
Given non-negative $x, y,z$ such that $ 3x+2y+z \leq 1$, show that $(x+y+z)^2 + 3 (x+y)^2 + 5 x^2 \leq 1$.
This is true because
$$(x+y+z)^2 + 3 (x+y)^2 + 5 x^2 = (3x+2y+z)^2 - (4xy+4xz+2yz) \leq 1^2 $$
We have equality iff $3x+2y+z = 1$ and $4xy + 2xz + 2yz = 0 $.
The second condition holds when at least 2 of these non-negative terms are equal to 0, so we have the following equality cases.
*
*$ x = 0, y = 0$, we get $c = 0, b = 0 , a = 1$.
*$ x = 0, z = 0 $, we get $ c = 0, b = 1/2, a = 1/2$.
*$y = 0, z = 0 $, we get $ c = 1/3, b = 1/3, a = 1/3$.
*$x=0, y = 0, z = 0$ has no solution.
Notes
*
*I was bounding with $(3x+2y+z)^2$ because the square terms on the LHS were $ 9x^2, 4y^2, z^2$, and that expression also nicely matched with our condition.
*In hindsight, the $4xy+4xz+2yz = 2z(2x+y) + 4xy = 2(a-b)(b+c) + 4c(b-c) $ is what Paresseux had. However, IMO their factorization was hard to guess at, whereas this approach was immediate.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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What is the minimum value of $8 \cos^2 x + 18 \sec^2 x$? As per me the answer should be $26$.
But when we apply AM-GM inequality it gives $24$ as the least value but as per the graph 24 can never come.
What I think is that in AM-GM, it gives $8 \cos^2 x = 18 \sec^2 x$ which gives $\cos x > 1$ which is not possible and because of this, AM-GM is giving a wrong minimum value.
If we had $18 \cos^2 x + 8 \sec^2 x$, then AM-GM would have worked and $24$ would be a right answer since $18 \cos^2 x = 8 \sec^2 x$, which gives $\cos x < 1$ which is true.
Is this reason correct?
| $$8\cos^2(x)+18\sec^2(x)=8\cos^2(x)+8\sec^2(x)+10\sec^2(x)$$
Now, apply $AM-GM$ on the first two terms.
$$\frac{8\cos^2(x)+8\sec^2(x)}{2}\ge\sqrt{8\cos^2(x)\cdot8\sec^2(x)}$$
$$\implies {8\cos^2(x)+8\sec^2(x)}\ge 16$$ at $x=0$
And min of $10\sec^2(x)$ is $10$ at $x=0$. So, the minimum of the net function is $26$ at $x=0$
| {
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"timestamp": "2023-03-29T00:00:00",
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$2^\sqrt{10}$ vs $3^2$ Is there a neat way to show that $2^\sqrt{10} < 3^2$?
I have tried raising to larger powers, like $(2^\sqrt{10})^{100}$ vs $3^{200}$ but the problem is the two functions $2^{x\sqrt{10}}$ and $3^{2x}$ are almost equivalent, and there is no point (that I can find) where one function is "obviously" larger than the other.
Looking at $2^{2\sqrt{10}}$ vs $3^4$ I tried to find a way of showing that $2^{2\sqrt{10}} < 2^6+2^4 = 2^4(2^2+1)$ but couldn't see any neat solution.
Any help or hints are appreciated.
edit: I should have specified when I say "neat solution" I was looking for a method readily done by hand. I realise this might be an unrealistic limitation, but it was why I'm interested.
Final thoughts before going to bed: I was looking at the functions $f(x)=2^{\sqrt{1+x^2}}$ and $g(x)=x^2$. At $x=0$, clearly $f>g$. They are equal at $x=2\sqrt{2}$. At $x=4$, again $f>g$. This mean that somewhere in the interval $2<x<4$, $g>f$ (as they are both convex). The task then is to try to find a point such that $x>3$ and $g>f$. But then another fun inequality pops out... $2^{\sqrt{11}}$ vs $10$...
| It follows from the maclaurian series of the function $\sqrt{k^2+h}$ for $|h|\leq k$ that ,
$$\sqrt{k^2+h}<k+\frac{h}{2k}$$
Put $k=3$ and $h=1$ we get ,
$$\sqrt{10}<3+\frac{1}{2\cdot 3}=3+\frac{1}{6}$$
Hence we must have
$$2^{\sqrt{10}}<2^3\cdot \sqrt[6]{2}=8\cdot \sqrt[6]{2}$$
Therefore we only need to show that
$$8\cdot \sqrt[6]{2}<3^2$$
Or in other words we need to show that
$$\sqrt[6]{2}<\frac{9}{8}=1+\frac{1}{8}$$
or in other words if we raise both sides to $6th$ power ,we need to show that
$$2<\left( \frac{9}{8}\right)^6=\left(1+\frac{1}{8}\right)^6$$
Expanding right hand side using binomial theorem we're left to prove that
$$2<\sum \limits_{k=0}^6 {}^6C_k \cdot \frac{1}{8^k}$$
As the first term in the expansion on right hand side is obviously 1 therefore subtracting 1 both sides leaves us to prove that
$$1<\sum \limits_{k=1}^6 {}^6C_k \cdot \frac{1}{8^k}$$
Now it's working time we don't need to know the full expansion. We'll just leave where the rhs exceeds 1.
Now ,$${}^6C_1\cdot \frac{1}{8}=\frac{6}{8}=\frac{3}{4}$$
$${}^6C_2\cdot \frac{1}{8^2}=\frac{30}{128}=\frac{15}{64}$$
$${}^6C_3 \cdot \frac{1}{8^3}=\frac{20}{512}=\frac{5}{128}$$
let's check now whether this exceeds 1 or not. Adding we get
$$\frac{3}{4}+\frac{15}{64}+\frac{5}{128}=\frac{63}{64}+\frac{5}{128}=\frac{131}{128}=\frac{128+3}{128}=1+\frac{3}{128}>1$$
(Yay!! ;) ) Therefore the inequality is proven. Hence we have
$$2^{\sqrt{10}}<3^2$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Two functions of two random variables v= sqrt(x^2+y^2) w = y/x This is a home work problem
$v = \sqrt{x^{2} + y^{2}} $
$w = \frac{y}{x} $
x and y are random variables with the following joint density function:
$f(x,y) = \frac{1}{2 \pi \sigma^{2}}exp[-\frac{1}{2 \sigma^{2}}(x^{2}+y^{2})]$
find f(v,w) as a function of f(x,y)
The following is my solution. Could someone please tell me if I am right and if not what I am doing wrong?
Change to polar coordinates:
$v = r \hspace{1cm}w = tan(\theta) $
inverse functions:
$v = r \hspace{1cm} $
$ arctan(w) = \theta_{1} \hspace{3mm}arctan(w) + \pi = \theta_{2} $
$$J(r, \theta) = det\begin{bmatrix}\frac{\partial v}{\partial r} & \frac{\partial v}{\partial \theta} \\\frac{\partial w}{\partial r} & \frac{\partial w}{\partial \theta} \end{bmatrix} = \begin{bmatrix}1 & 0 \\0 & sec^{2}(\theta) \end{bmatrix} = sec^{2}(\theta)$$
$\frac{\partial v}{\partial r} = 1 \hspace{5mm} \frac{\partial v}{\partial \theta} = 0 $
$\frac{\partial w}{\partial r} = 0 \hspace{5mm} \frac{\partial w}{\partial \theta} = sec^{2}(\theta) $
express the joint density function in polar coordinates
$f_{r \theta}(r, \theta) = \frac{1}{2 \pi \sigma^{2}}exp[-\frac{1}{2 \sigma^2} r^{2}]$
$$f_{vw}(v,w) = (f_{r \theta}(r, \theta_{1}) + f_{r \theta}(r, \theta_{2}))\frac{1}{|J(r, \theta)|}$$
there for
$$f_{vw}(v,w)= \frac{1}{\pi \sigma^2} exp[ -\frac{1}{2 \sigma^{2}}r^{2}]|cos^2(\theta)| = \frac{1}{\pi \sigma^2} exp[ -\frac{1}{2 \sigma^{2}}v^{2}] \frac{1}{w^{2}+1}$$
| You map $(r, \theta) \to (v, w)$ using Jacobian but you forgot about the Jacobian to map $(x, y) \to (r, \theta)$, which is $r$. That should give you $v$ in the numerator.
Alternatively, without converting to polar coordinates,
$v = \sqrt{x^2+y^2}, w = \dfrac{y}{x}$
$J^{-1} = \begin{vmatrix} \frac{\partial v}{\partial x} & \frac{\partial v}{\partial y} \\ \frac{\partial w}{\partial x} & \frac{\partial w}{\partial y}\end {vmatrix} = \begin{vmatrix} \frac{x}{\sqrt{x^2+y^2}} & \frac{y}{\sqrt{x^2+y^2}} \\ -\frac{y}{x^2} & \frac{1}{x}\end {vmatrix}$
$ = \dfrac{\sqrt{x^2+y^2}}{x^2}$
Now note that $(x,y) \to (v,w)$ is not one-to-one mapping -
$y = wx, v^2 = x^2 + y^2 = x^2 (1+w^2)$
$x = \pm \dfrac{v}{\sqrt{1+w^2}}$
Using symmetry, we can multiple $|J|$ by $2$.
$f(v,w) = \dfrac{2v^2}{v (1+w^2)} \dfrac{1}{2 \pi \sigma^{2}} \ e^{- \left(\dfrac{v^2}{2 \sigma^{2}}\right)}$
$ \displaystyle = \dfrac{v}{\pi \sigma^2 (1+w^2)} e^{- \left( \dfrac{v^2}{2 \sigma^{2}} \right)}$
| {
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Given a general region, find the double integral bounded between $y = x$ and $y=3x-x^2 $ $$ J = \iint_R (x^2-xy)\,dx \,dy, $$
Suppose region R is bounded between $y = x$ and $y=3x-x^2 $
My attempt using vertical integration:
$$ \int^{x=2}_{x=0} \int^{y=3x-x^2}_{y=x} \left({x^2-xy}\right)dy\ dx$$
$$\int^2_0 \left[x^2y-x\frac{y^2}{2}\right]^{3x-x^2}_{x}\, dx$$
$$\int^2_0 \frac{-x^5+4x^4-4x^3}{2} \,dx $$
$$\boxed{J = -\frac{8}{15}}$$
My attempt using horizontal integration :
$$ \int^{y=2}_{y=0} \int^{x=y}_{x=3\,\pm \sqrt{9-y}} \left({x^2-xy}\right)dx\ dy$$
For $ x = 3+\sqrt{9-y}$
$$ \int^2_0 \left[\frac{x^3}{3}-\frac{x^2}{2}y\right]^y_{3+\sqrt{9-y} }\,dy$$
For $ x = 3-\sqrt{9-y}$
$$ \int^2_0 \left[\frac{x^3}{3}-\frac{x^2}{2}y\right]^y_{3-\sqrt{9-y} }\,dy$$
My doubts :
1.) How do I set my limit of integration for horizontal integration, if there is $\pm$ to be considered ?
2.) the answer as negative what does that imply in questions related to double integrals?
Could you guys please help
| $y=3x-x^2 \implies y = - (x-\frac{3}{2})^2 + \frac{9}{4}$
So equation of parabola is $ \ 9 - 4y = (2x-3)^2 \ $ and its vertex is $\left(\frac{3}{2}, \frac{9}{4}\right)$.
Intersection of parabola and the line $y = x$ is $(2, 2)$. So you can see that it is below the vertex and to the right of the axis of symmetry.
So if you are doing horizontal integration,
for $0 \leq y \leq 2, $ $x$ is bound between parabola and on the right by the line $y = x$. But for $2 \leq y \leq \frac{9}{4}$, $x$ is bound between parabola at both ends.
Left of the axis of parabola is given by $2x - 3 = - \sqrt{9-4y}$ and right of it is given by $2x - 3 = \sqrt{9-4y}$
So the integral should be
$\displaystyle \int_0^2 \int_{(3 - \sqrt{9-4y})/2}^y (x^2 - xy) \ dx \ dy \ \ $ +
$\displaystyle \int_2^{9/4} \int_{(3 - \sqrt{9-4y})/2}^{(3 + \sqrt{9-4y})/2} (x^2 - xy) \ dx \ dy \ \ $
$ = - \frac{47}{120} - \frac{17}{120} = - \frac{8}{15}$
which matches your answer using vertical integration.
| {
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"timestamp": "2023-03-29T00:00:00",
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Locating roots of a special class of polynomials in $\mathbb{Z}[X]$ I am reading Prof. Ram Murty's Prime number and Irreducible polynomials. I am having problem in understanding a part of the following lemma:
Statement:
Suppose that $\alpha$ is a complex root of a polynomial
$$
f(x)=x^m+a_{m-1}x^{m-1}+\cdots+a_1x+a_0
$$
with coefficients $a_i$ equal to $0$ or $1$. If $|arg \ \alpha| \le \frac{\pi}{4}$ then $|\alpha| < \frac{3}{2}$. Otherwise $\mathfrak{R}(\alpha) < \frac{1+\sqrt{5}}{2\sqrt{2}}$.
Proof: The cases $m=1$ and $m=2$ can be verified directly. Assuming that $m \ge 3$, we compute for $z \ne 0$:
$$
\left| \frac{f(z)}{z^m}\right| \ge \left| 1+\frac{a_{m-1}}{z}+\frac{a_{m-2}}{z^2}\right| - \left(\frac{1}{|z|^3}+\cdots+\frac{1}{|z|^m}\right).
$$
For $z$ satisfying $|arg \ z| \le \frac{\pi}{4}$ it is true that $\mathfrak{R}(\frac{1}{z^2}) \ge 0$, so for such $z$ we have
$$
\left|\frac{f(z)}{z^m}\right| > 1 - \frac{1}{|z|^2(|z|-1)}
$$
*
*How to obtain the cases when $m=1,2$?
*How are the above two inequalities obtained?
*Where are we using the fact: For $z$ satisfying $|arg \ z| \le \frac{\pi}{4}$ it is true that $\mathfrak{R}(\frac{1}{z^2}) \ge 0$?
| Here is a proof of the first inequality.
$$f(z) = z^m + a_{m-1}z^{m-1} + a_{m-2}z^{m-2} + a_{m-3}z^{m-3}+ \dots+a_0\\
\frac{f(z)}{z^m} = \left(1 + \frac{a_{m-1}}z + \frac{a_{m-2}}{z^2}\right) + \left(\frac{a_{m-3}}{z^3}+ \dots+\frac{a_0}{z^m}\right)\\
\frac{f(z)}{z^m} - \left(\frac{a_{m-3}}{z^3}+ \dots+\frac{a_0}{z^m}\right)= \left(1 + \frac{a_{m-1}}z + \frac{a_{m-2}}{z^2}\right)$$
$$\left|\frac{f(z)}{z^m}\right| + \left(\frac{|a_{m-3}|}{|z|^3}+ \dots+\frac{|a_0|}{|z|^m}\right)\ge \left|1 + \frac{a_{m-1}}z + \frac{a_{m-2}}{z^2}\right|\tag{1}$$
$$\left|\frac{f(z)}{z^m}\right| + \left(\frac{1}{|z|^3}+ \dots+\frac{1}{|z|^m}\right)\ge \left|1 + \frac{a_{m-1}}z + \frac{a_{m-2}}{z^2}\right|\tag{2}\\
\left|\frac{f(z)}{z^m}\right|\ge \left|1 + \frac{a_{m-1}}z + \frac{a_{m-2}}{z^2}\right| - \left(\frac{1}{|z|^3}+ \dots+\frac{1}{|z|^m}\right)$$
Where $(1)$ is the triangle inequality, and $(2)$ is because $\forall i, |a_i| \le 1$.
[this only works for $|z| > 1$]For the second inequality, note that if $|z| < 1$, then
$$\begin{align}\frac{1}{|z|^3}+ \dots+\frac{1}{|z|^m} &= \frac{1}{|z|^3}\left(1 + |z|^{-1} + \dots+|z|^{-(m-3)}\right)\\
&=\frac{1}{|z|^3}\left(\dfrac{1-|z|^{-(m-2)}}{1-|z|^{-1}}\right)\\
&\le\frac{1}{|z|^2(|z|-1)}\end{align}$$
and if $|z| > 1$, then
$$\begin{align}\frac{1}{|z|^3}+ \dots+\frac{1}{|z|^m} &\le \frac{1}{|z|^3}\left(1 + |z|^{-1} + \dots\right)\\
&=\frac{1}{|z|^3}\left(\dfrac1{1-|z|^{-1}}\right)\\
&=\frac{1}{|z|^2(|z|-1)}\end{align}$$
So if we can show $\left|1 + \frac{a_{m-1}}z + \frac{a_{m-2}}{z^2}\right| \ge 1$, the inequality you are after follows. This is where $\mathfrak R\left(\frac1{z^2}\right) \ge 0$ comes in. Though not mentioned, the argument restriction also implies $\mathfrak R\left(\frac1{z}\right) \ge 0$. Hence $$\mathfrak R\left(1 + \frac{a_{m-1}}z + \frac{a_{m-2}}{z^2}\right) \ge 1 + 0 + 0 = 1$$
But the modulus of any number with real part $\ge 1$ must be $\ge 1$ as well. Thus $$\left|1 + \frac{a_{m-1}}z + \frac{a_{m-2}}{z^2}\right| \ge 1$$ as desired.
| {
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True or false? Any triangle we can $h_A^2+h_B^2 \geq 2h_C^2$, where $h_A, h_B, h_C$ are lengths of the heights from vertices A, B and C Let $\triangle ABC$ any triangle and the lengths of the heights from vertices A, B and
C are, respectively, $h_A$, $h_B$ , $h_C$. Then how should we prove that,
$$h_A^2+h_B^2 \geq 2h_C^2, \quad h_C^2+h_A^2\geq 2 h_B^2 \quad and \quad h_B^2+h_C^2 \geq 2 h_A^2$$
I think that is true but only I can bounded like this:
$$BC^2 \geq h_B^2, \quad AB^2 \geq h_B^2 \quad \text{By Exterior Angle Theorem}$$
So, $BC^2+AB^2\geq h_B^2$ and I can't prove that $h_A^2+ h_C^2 \geq BC^2+AB^2$
| We may use following inequality:
$$h_A+h_b+h_c< a+b+c=2p\space\space\space\space\space (1)$$
Also:
$$h_A+h_B+h_C> p$$
Let's suppose:
$$h_A+h_B+h_C\approx p+\frac p2=\frac{3p}2$$
Squaring both sides we get:
$$h_A^2+h_B^2=h_C^2+\frac 94 p^2-3p\cdot h_C-2h_Ah_B$$
So we must show that:
$$t=\frac 94 p^2-3p\cdot h_C-2h_Ah_B\geq h_C^2$$
If $h_A=h_B=\epsilon$ then $p\approx h_C$ and we have:
$$t=\frac94 \times 4h_C^2-3 h_C^2-\epsilon=6h_C^2-\epsilon$$
So there is possibility to have a triangle in which:
$h_A^2+h_B^2\geq h_C^2$
But there is no guarantee for this relation to be symmetric. for example Consider Right angle triangle with sides 3, 4, 5 where :
$h_B=4$, $h_C=3$ and $h_A=2.4$ we have:
$h_A^2+h_B^2=2.4^2+4^2=21.76>2\times 3^2=18$
$h_B^2+h_C^2=9+16=25> 2\times 2.4^2=11.54$
But:
$h_A^2+h_C^2=14.76< 2\times 4^2=32$
| {
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How to evaluate $\lim\limits_{x\to \infty}\frac {2x^4-3x^2+1}{6x^4+x^3-3x}$?
Evaluate: $\lim\limits_{x\to \infty}\frac {2x^4-3x^2+1}{6x^4+x^3-3x}$.
I've just started learning limits and calculus and this is an exercise problem from my textbook.
To solve the problem, I tried factorizing the numerator and denominator of the fraction. The numerator can be factorized as $(x-1)(x+1)(2x^2-1)$ and the numerator can be factorized as $x(6x^3+x^2-3x)$. So, we can rewrite the problem as follows: $$\lim\limits_{x\to\infty}\frac {2x^4-3x^2+1}{6x^4+x^3-3x}=\lim\limits_{x\to\infty}\frac {(x-1)(x+1)(2x^2-1)}{x(6x^3+x^2-3)}$$ But this doesn't help as there is no common factor in the numerator and denominator. I've also tried the following: $$\lim\limits_{x\to\infty}\frac {(x-1)(x+1)(2x^2-1)}{x(6x^3+x^2-3)}=\lim\limits_{x\to\infty}\frac{x-1}{x}\cdot \lim\limits_{x\to\infty}\frac {(x+1)(2x^2-1)}{6x^3+x^2-3}=1\cdot \lim\limits_{x\to\infty}\frac {(x+1)(2x^2-1)}{6x^3+x^2-3}$$ Here I used $\frac{x-1} x=1$ as $x$ approaches infinity. Yet this does not help.
The answer is $\frac 1 3$ in the book but the book does not include the solution.
So, how to solve the problem?
| Essentially, divide the numerator and denominator by the largest power of numerator:$$\lim\limits_{x\to \infty}\frac {2x^4-3x^2+1}{6x^4+x^3-3x} = \lim\limits_{x\to \infty}\frac {\frac{2x^4-3x^2+1}{x^4}}{\frac{6x^4+x^3-3x}{x^4}} = \lim\limits_{x\to \infty}\frac {2-\frac{3}{x^2} + \frac{1}{x^4}}{6 + \frac{1}{x} - \frac{3}{x^3}} = \frac{2 - 0 + 0}{6 + 0 - 0}= \frac13$$
Factorization usually works when you have $0/0$ expressions. For instance,
$$\lim_{x\to2} \frac{x^2-5x + 6}{x^2 - 3x + 2}$$
Note that both numerator and denominator are equal to zero when $x = 2$. From this, we can suspect that one of the roots of those expressions is $2$. Then, let's factorize and cancel that out (note that we can cancel since $x \to 2$, not $x = 2$)
$$\lim_{x\to2} \frac{x^2-5x + 6}{x^2 - 3x + 2} = \lim_{x\to2}\frac{(x-2)(x-3)}{(x-2)(x-1)} = \lim_{x\to2}\frac{x-3}{x-1}$$
which is fine at $x = 2$ so we can just go ahead and plug this in to get
$$\lim_{x\to2}\frac{x-3}{x-1} = \frac{2-3}{2-1} = -1$$
Note: We can plug in $x=2$ because rational functions are continuous.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4190851",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
For what primes $p$ and positive integers $k$ is this algebraic expression divisible by $3$? My initial question is as is in the title:
For what prime $p$ and positive integers $k$ is this algebraic expression divisible by $3$?
$$A(p,k):=\dfrac{p^{2k+2} - 4p^{2k+1} + 6p^{2k} + 2p^{k+1} - 8p^k + 3}{2(p - 1)^2}$$
I would like to qualify that I am specifically interested in those values for $p$ and $k$ satisfying the congruence $p \equiv k \equiv 1 \pmod 4$.
MY ATTEMPT
Here, I will evaluate my expression for
$$p_1 = 5, k_1 = 1,$$
which gives $A(p_1,k_1)=9$ (which is divisible by $3$),
$$p_2 = 13, k_2 = 1,$$
which gives $A(p_2,k_2)=73$ (which is not divisible by $3$), and
$$p_3 = 17, k_3 = 1,$$
which gives $A(p_3,k_3)=129$ (which is divisible by $3$).
Here is my final question:
Will it be possible to consider the cases $p \equiv 1 \pmod 3$ and $p \equiv 2 \pmod 3$ separately, then use the Chinese Remainder Theorem afterwards? (I know the concept, but I would have forgotten how to do that.)
CONJECTURE: If $p \equiv 2 \pmod 3$, then $3 \mid A(p,k)$.
Alas, this is where I get stuck.
| *
*If $p\equiv 2\pmod 3$, then the numerator of $$A(p,k)=\dfrac{p^{2k+2} - 4p^{2k+1} + 6p^{2k} + 2p^{k+1} - 8p^k + 3}{2(p - 1)^2}$$ is divisible by $3$ while the denominator is not divisible by $3$, so $3\mid A(p,k)$.
*If $p\equiv 1\pmod 3$, then since $p$ can be written as $p=12m+1$, we have $$A(p,k)=\dfrac{1+(8 m - 2) (12 m + 1)^k + (48m^2 - 8 m + 1) (12 m + 1)^{2 k} }{96m^2}$$Now, we have, by the binomial theorem,$$\begin{align}&1+(8 m - 2) (12 m + 1)^k + (48m^2 - 8 m + 1) (12 m + 1)^{2 k}
\\\\&=1+(8 m - 2)\sum_{i=0}^{k}\binom{k}{i}(12m)^i + (48m^2 - 8 m + 1)\sum_{i=0}^{2k}\binom{2k}{i}(12m)^{i}
\\\\&=1+(8 m - 2)\bigg(1+12km+\underbrace{\sum_{i=2}^{k}\binom{k}{i}(12m)^i}_{\text{divisible by $288m^2$}}\bigg)
\\&\qquad\quad + (48m^2 - 8 m + 1)\bigg(1+24km+\binom{2k}{2}(12m)^{2}+\underbrace{\sum_{i=3}^{2k}\binom{2k}{i}(12m)^{i}}_{\text{divisible by $288m^2$}}\bigg)
\\\\&\equiv 1+(8 m - 2)(1+12km)
\\&\qquad\quad+ (48m^2 - 8 m + 1)\bigg(1+24km+\binom{2k}{2}(12m)^{2}\bigg)\pmod{288m^2}
\\\\&\equiv 48(k+1)m^2\pmod{288m^2}
\end{align}$$So, there is an integer $N$ such that $$A(p,k)=\frac{48(k+1)m^2+288m^2N}{96m^2}=3\bigg(\frac{k+1}{6}+N\bigg)$$Since $k+1$ is even, we see that $3\mid A(p,k)$ if and only if $k+1\equiv 0\pmod 3$, i.e. $k\equiv 2\pmod 3$.
In conclusion, considering $p\equiv k\equiv 1\pmod 4$, we can say that
*
*If $p\equiv 5\pmod{12}$, then $3\mid A(p,k)$.
*If $p\equiv 1\pmod{12}$ and $k\equiv 5\pmod{12}$, then $3\mid A(p,k)$.
*If $p\equiv 1\pmod{12}$ and $k\not\equiv 5\pmod{12}$, then $3\not\mid A(p,k)$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4191369",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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If $\alpha$ be a multiple root of the order 3 of the equation $x⁴+bx²+cx+d=0$ then $α= 8d/3c$ I found this statement online here and I think this statement is false.I would first prove a general statement and then I would use it to find $\alpha$ in order to discard the statement given in online link.
Theorem:If the equation $x^4+ax^3+bx^2+cx+d=0$ has three equal roots, show that each of them is equal to $\frac{6c-ab}{3a^2-8b}$
Proof;Three roots of the given equation are same hence, the roots can be assumed as $\alpha,\alpha,\alpha,\beta$
Here, $S_1=3α+β=−a;S_2
=3α(α+β)=b;S_3=α^2(α+3β)=−c;S_4=α^3β=d$
We need to evaluate the value of $\frac{6c-ab}{3a^2-8b}$.
$6c−ab=α(3α^2−6αβ+3β^2)$
$3a^2−8b=3α^2−6αβ+3β^2$
Therefore,$\frac{6c-ab}{3a^2-8b}=\frac{α(3α^2−6αβ+3β^2)}{(3α^2−6αβ+3β^2)}=\alpha$
Now according to this ,putting $a=0$ we get $\alpha=\frac{-3c}{4b}$ which is not same as $\frac{8d}{3c}$ as claimed in online site.
My question is: Is the claim in online site wrong?or I have gone wrong somewhere in my proof.
Any help is much appreciated!
| Verification of the Theorem
Expanding
$$
(x-u)^3(x-v)=x^4\ \overbrace{-(3u+v)}^{\large a}x^3\ \overbrace{+3u(u+v)}^{\large b}x^2\ \overbrace{-u^2(u+3v)}^{\large c}x\overbrace{\ +u^3v\ \ }^{\large d}\tag1
$$
The first two derivatives are $0$ at $u$
$$
u^4+au^3+bu^2+cu+d=0\tag2
$$
$$
4u^3+3au^2+2bu+c=0\tag3
$$
$$
12u^2+6au+2b=0\tag4
$$
Substituting $v=d/u^3$ into $c+u^2(u+3v)=0$ gives
$$
u^4+cu+3d=0\tag5
$$
Writing $(2)$-$(5)$ as a matrix equation yields
$$
\begin{align}
0
&=\begin{bmatrix}
1&-\frac a4&\frac{3a^2-4b}{48}&-1
\end{bmatrix}
\begin{bmatrix}
1&a&b&c&d\\
0&4&3a&2b&c\\
0&0&12&6a&2b\\
1&0&0&c&3d
\end{bmatrix}
\begin{bmatrix}
u^4\\u^3\\u^2\\u\\1
\end{bmatrix}\tag{6a}
\\
&=\begin{bmatrix}
0&0&0&\frac{3a^3-8ab}8&\frac{3a^2b-4b^2-6ac-48d}{24}
\end{bmatrix}
\begin{bmatrix}
u^4\\u^3\\u^2\\u\\1
\end{bmatrix}\tag{6b}
\end{align}
$$
Since $b^2+12d=3ac$, $\text{(6b)}$ gives
$$
\begin{align}
u
&=\frac{3a^2b-4b^2-6ac-48d}{24ab-9a^3}\tag{7a}\\
&=\frac{3a^2b-18ac}{24ab-9a^3}\tag{7b}\\
&=\frac{ab-6c}{8b-3a^2}\tag{7c}
\end{align}
$$
When $\boldsymbol{a=0}$
Plugging $a=0$ into $\text{(7c)}$ gives
$$
u=-\frac{3c}{4b}\tag8
$$
If we use $a=0$, that is, $v=-3u$, we get $b=-6u^2$, $c=8u^3$, and $d=-3u^4$, we get not only that $u=-\frac{3c}{4b}$, as given in $(8)$, but also that
$$
u=-\frac{8d}{3c}\tag9
$$
which says that the problem in question is a factor of $-1$ off.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4191533",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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"answer_id": 2
} |
How to make use of angle sum and difference identities to find the value of sine and cosine?
Calculate: $\cos\left({5\pi\over12}\right)$ and $\cos\left({\pi\over12}\right)$
What is the easiest way to find $\cos\left({5\pi\over12}\right)$ and $\cos\left({\pi\over12}\right)$ (without a calculator)? If I know that $\frac{5\pi }{12}=\frac{\pi }{4}+\frac{\pi }{6}$ and $\frac{\pi }{12}=\frac{\pi }{3}-\frac{\pi }{4}$, then I can apply angle sum and difference identities. But how do I know $\frac{5\pi }{12}= \frac{\pi }{4}+\frac{\pi }{6}$ and $\frac{\pi }{12}= \frac{\pi }{3}-\frac{\pi }{4}$ in the first place. I know $ \frac{\pi }{4}+\frac{\pi }{6} = \frac{5\pi }{12}$ and $ \frac{\pi }{3}-\frac{\pi }{4}=\frac{\pi }{12}$ but I can't go the other way round.
I gave $\frac{5\pi}{12}$ and $\frac{\pi}{12}$ as an example, I want the general solution for any value in pi rational form $\frac{\pi p}{q}$.
| We want to find the values of $\displaystyle\cos\frac{5\pi}{12}$ and $\displaystyle\cos\frac{\pi}{12}$.
Recall the sum to product formulae below:
$$\begin{align}\cos A+\cos B&=2\cos\frac{A+B}{2}\cos\frac{A-B}{2}\\
\cos A-\cos B&=-2\sin\frac{A+B}{2}\sin\frac{A-B}{2}\\\end{align}.$$
In our case, let $A=5\pi/12$ and $B=\pi/12$. Then we find that
$$\begin{align}\cos\frac{5\pi}{12}+\cos\frac{\pi}{12}&=2\cos\frac{\frac{5\pi}{12}+\frac{\pi}{12}}{2}\cos\frac{\frac{5\pi}{12}-\frac{\pi}{12}}{2}=2\cos\frac{\pi}{4}\cos\frac{\pi}{6}\\
&=\frac{\sqrt6}{2}\tag{1}\\
\cos\frac{5\pi}{12}-\cos\frac{\pi}{12}&=-2\sin\frac{\frac{5\pi}{12}+\frac{\pi}{12}}{2}\sin\frac{\frac{5\pi}{12}-\frac{\pi}{12}}{2}=-2\sin\frac{\pi}{4}\sin\frac{\pi}{6}\\
&=-\frac{\sqrt2}{2}\tag{2}.\end{align}$$
Adding equations $(1)$ and $(2)$ together and dividing by $2$, we find that
$$\cos\frac{5\pi}{12}=\frac{\sqrt6-\sqrt2}{4}$$
and subtracting equation $(2)$ from equation $(1)$ and dividing by $2$ gives
$$\cos\frac{\pi}{12}=\frac{\sqrt6+\sqrt2}{4}.$$
However, this method is not always guaranteed to work, for the simple reason that $\cos x\pi$ does not always have a closed form for rational $x$.
I hope that helps. If you have any questions please don't hesitate to ask :)
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4191686",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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} |
Find the value of $\lim_{ n \to \infty} \left(\frac{1^2+1}{1-n^3}+\frac{2^2+2}{2-n^3}+\frac{3^2+3}{3-n^3}+...+\frac{n^2+n}{n-n^3}\right)$ The following question is taken from the practice set of JEE Main exam.
Find the value of $$\lim_{ n \to \infty} \left(\frac{1^2+1}{1-n^3}+\frac{2^2+2}{2-n^3}+\frac{3^2+3}{3-n^3}+...+\frac{n^2+n}{n-n^3}\right)$$
I wrote it as$$\lim_{ n \to \infty} \sum_{r=1}^n\frac{r^2+r}{r-n^3}\\=\lim_{ n \to \infty} \sum_{r=1}^n\frac{r(r+1)}{r-n^3}\\=\lim_{ n \to \infty} \sum_{r=1}^n\frac{r(\dfrac rn+\dfrac1n)}{\dfrac rn-n^2}\\=\lim_{ n \to \infty} \sum_{r=1}^n\frac{\dfrac rn(\dfrac rn+\dfrac1n)n}{\dfrac rn-n^2}$$
To convert it into into integration, $\dfrac rn$ can be written as $x$. $\dfrac1n$ is written as $dx$, but instead we have $n$. How to tackle that? Or any other approach for the question? Just a hint would suffice. Thanks.
| Hint: $k-n^{3}$ lies between $1-n^{3}$ and $n-n^{3}$ for $1 \leq k \leq n$. This allows you to show that $\lim \sup$ and $\lim \inf$ are both equal to $-\frac 1 3$ using the formulas for $\sum\limits_{k=1}^{n} k$ and $\sum\limits_{k=1}^{n} k^{2}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4193228",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
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Probability two blocks have exactly 2 out of 4 properties the same?
I have 120 blocks. Each block is one of 2 different materials, 3 different colors, 4 different sizes, and 5 different shapes. No two blocks have exactly the same of all four properties. I take two blocks at random. What is the probability the two blocks have exactly two of these four properties the same?
I ended up getting:
$${{\binom{2}{1}\binom{3}{1}\binom{4}{2}\binom{5}{2} + \binom{2}{1}\binom{3}{2}\binom{4}{1}\binom{5}{2} + \binom{2}{1}\binom{3}{2}\binom{4}{2}\binom{5}{1} + \binom{2}{2}\binom{3}{1}\binom{4}{1}\binom{5}{2} + \binom{2}{2}\binom{3}{1}\binom{4}{2}\binom{5}{1} + \binom{2}{2}\binom{3}{2}\binom{4}{1}\binom{5}{1}}\over{\binom{120}{2}}} = {{41}\over{238}}$$Is this correct?
EDIT: The answer given here is ${{35}\over{119}} = {5\over{17}}$:
https://web2.0calc.com/questions/probability-question-blocks
Who is correct?
| No. The answer given on the linked website is correct.
After picking the first block there are 119 left. Assume that the second block has material and color the same, whereas size and shape are different. Of the $4*5 = 20$ blocks that meet the first criterium (note: minus 1 because the first block can not be picked again!), it is easy to see that there are $(4-1)*(5-1) = 12$ that satisfy the second criterium. Analyzing the other combinations in the same way we get:
$$P = \frac {3*4 + 2*4 + 2*3 +1*4 + 1*3+1*2} {119} = \frac {12+8+6+4+3+2}{119} = \frac {35}{119}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4196967",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
If $f(x) = x^2 - 2x + 5$ what is $f^{-1}(x)$?
If $f(x) = x^2 - 2x + 5$ what is $f^{-1}(x)$?
with the condition : $x > 1$
I solved this problem in this way:
$f(x) = x^2 - 2x + 5 -1 +1 \longrightarrow (x-2)^2 + 1 = f(x) $
$f^{-1}(x) = \sqrt{x-1} + 2$
But I saw someone else solved it in this way:
$f(x) = x^2 - 2x + 1 + 4 \longrightarrow (x+1)^2 + 4 = f(x) $
$f^{-1}(x) = \sqrt{x-4} + 1$
Which one is correct? If the second one is correct why mine is wrong?
| $$f(x)=y= x^2-2x+5=x^2-2x+4+1=(x-1)^2+4$$
$$(x-1)^2=y-4$$
$$x-1=\pm\sqrt{y-4} $$
$$x=\pm\sqrt{y-4}+1$$
Interchange $x$ and $y$
$$f^{-1}(x)=y=\pm\sqrt{x-4}+1$$
for $x>1$ inverse of f, $f^{-1}(x)=y=\sqrt{x-4}+1$
| {
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"timestamp": "2023-03-29T00:00:00",
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Calculating $\lim_{x \to 0} \frac{\sin(x^2)-\sin^2(x)}{x^2\ln(\cos x)}$ without L'Hospital's rule this is my first post here so excuse the lack of knowledge about how things usually go.
My question revolves around calculating the limit as $x$ approaches $0$ of the following function:
$$\lim_{x \to 0} \frac{\sin(x^2)-\sin^2(x)}{x^2\ln(\cos x)}$$
The question came up in a test about a month ago and while I couldn't solve it in the test I've been working on it since then but I can't seem to get it. I know the limit is supposed to be $\frac{-2}{3}$ from some online calculators which abused l'hopital rule over and over again. I've tried playing around with it in so many ways but I always seem to get 0 over 0 or the so called indeterminate form. I've even tried calculating it by substituting in the Taylor series for the functions given but no luck. If anyone could show me a method of calculating this without using l'hopital rule or better yet, give me a hint as to how I should proceed I would be grateful.
| More elementary solution (without Taylor series expansion or L'Hospital's rule) can be:
$$
\begin{align*}
\lim_{x \to 0} \frac{\sin(x^2)-\sin^2(x)}{x^2\ln(\cos x)}&=\lim_{x \to 0} \frac{\sin(x^2)-\sin^2(x)}{x^4}\cdot\frac{x^2}{1-\cos x}\cdot\frac{-(\cos x-1)}{\ln(\cos x)-0}\\
&=\lim_{x \to 0} \frac{\sin(x^2)-\sin^2(x)}{x^4}\cdot\frac{x^2(1+\cos x)}{1-\cos^2x}\cdot\frac{-1}{\ln'(1)}\\
&=\lim_{x \to 0} \frac{\sin(x^2)-\sin^2(x)}{x^4}\cdot\frac{2x^2}{\sin^2x}\cdot(-1)\\
&=-2\cdot\lim_{x \to 0} \frac{\sin(x^2)-\sin^2(x)}{x^4}=L
\end{align*}
$$$$
\begin{align*}
L_1&=\lim_{x \to 0} \frac{\sin(x^2)-\sin^2(x)}{x^4}\\
&=\lim_{2x \to 0} \frac{\sin(4x^2)-\sin^2(2x)}{16x^4}=\lim_{x \to 0} \frac{2\sin(2x^2)\cos(2x^2)-4\sin^2(x)\cos^2(x)}{16x^4}\\
&=\lim_{x \to 0} \frac{4\sin(x^2)\cos(x^2)(1-2\sin^2(x^2))-4\sin^2(x)(1-\sin^2(x))}{16x^4}\\
&=\lim_{x \to 0} \frac{4\sin(x^2)(1-2\sin^2(x^2))-4\sin^2(x)(1-\sin^2(x))}{16x^4}\\
&=\lim_{x \to 0} \frac{4\sin(x^2)-4\sin^2(x)}{16x^4}-\lim_{x \to 0}\frac{8\sin^3(x^2)}{16x^4} +\lim_{x \to 0}\frac{4\sin^4(x)}{16x^4} \\
&=\frac{L_1}{4}-\lim_{x^2 \to 0}\frac{8}{16}\left(\frac{\sin x}{x}\right)^2\cdot\sin(x)+\lim_{x \to 0}\frac{4}{16}\left(\frac{\sin x}{x} \right)^4\\
&=\frac{L_1}{4}+\frac14\\
\therefore L_1&=\frac13
\end{align*}
$$
$$\therefore L=-2L_1=-\frac23$$
| {
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"url": "https://math.stackexchange.com/questions/4198444",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Prove by limit definition $\lim _{x\to 3}\left(\frac{x^2-9}{x^3-3x^2-2x+6}\right)=\frac{6}{7}$. Prove, using the limit definition, that
$\lim _{x\to 3}\left(\frac{x^2-9}{x^3-3x^2-2x+6}\right)=\frac{6}{7}$.
I tried to do it this way, but I can’t move forward. Can anyone help me how to continue?
By definition we have
$\forall \epsilon >0,\exists \delta >0\::\:0<\left|x-a\right|<\delta \:\Rightarrow \left|f\left(x\right)-L\right|<\epsilon \Leftrightarrow a-\delta<x<a+\delta \Rightarrow L-\epsilon<f(x)<L+\epsilon.$
Proof:
\begin{eqnarray*}
3-\delta <x<3+\delta \:\Rightarrow \:\frac{6}{7}-\epsilon <\:f\left(x\right)<\frac{6}{7}+\epsilon &\:\Leftrightarrow& \:\frac{6}{7}-\epsilon <\frac{\left(x-3\right)\left(x+3\right)}{x^2\left(x-3\right)+2\left(x-3\right)}<\frac{6}{7}+\epsilon\\ &\:\Leftrightarrow& \:\frac{6}{7}-\epsilon <\frac{\left(x-3\right)\left(x+3\right)}{\left(x-3\right)\left(x^2-2\right)}<\frac{6}{7}+\epsilon\\ &\:\Leftrightarrow& \:\frac{6}{7}-\epsilon <\frac{\left(x+3\right)}{\left(x^2-2\right)}<\frac{6}{7}+\epsilon \\
&\:\Leftrightarrow& \:\:?
\end{eqnarray*}
| If you want to avoid complications, just replace $x\to 3$ by $x=u+3$ with $u\to 0$.
Then $\left|f(x)-\dfrac 67\right|=\underbrace{\left|\dfrac{6u+29}{7(u^2+6u+7)}\right|}_\text{bounded}\times\underbrace{|u|}_{<\varepsilon}\to 0$
It is bounded because $|u|<1\implies \begin{cases}|6u+29|<35\\|u^2+6u+7|\ge |6u+7|\ge\big|7-6|u|\big|>1\end{cases}$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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How do i approach ahead in this question
Let $a,b \in\Bbb N$, $a$ is not equal to $b$ and the quadratic equations $(a-1)x^2 -(a^2 + 2)x +a^2 +2a=0$ and $(b-1)x^2 -(b^2 + 2)x +b^2 +2b=0$ have a common root, then the value of $ab/5$ is
So what I did was,
I subtracted the two equations and got
$x=(a+b+2)/(a+b)$
I tried putting it in equation,it didn’t work,then i tried adding the equations and then put the value,still didn’t work.
I cant seem to figure out how to approach this problem.Can anybody help out?
| Let $q_a(x)=(a-1)x^2-(a^2+2)x+a^2+2a$. Any root $r$ of $q_a(x)$ and of $q_b(x)$ is also a root of$$(b-1)q_a(x)-(a-1)q_b(x)=\left(-a^2b+a^2+a b^2+2 a-b^2-2b\right)(x-1).$$Therefore, $r=1$ or$$-a^2b+a^2+a b^2+2 a-b^2-2b=0.\tag1$$But you can't have $r=1$, because\begin{align}1\text{ is a root of }q_a(x)&\iff q_a(1)=0\\&\iff3a-3=0\\&\iff a=1.\end{align}So, if $1$ was a root of both $q_a(x)$ and $q_b(x)$, you would have $a=b=1$, but you are assuming that $a\ne b$.
If, on the other hand, you have $(1)$, then $a=b$ or $a=\frac{b+2}{b-1}=1+\frac3{b-1}$. Since $a,b\in\Bbb N$, this can only occur in two case: when $b=2$ (in which case $a=4$) and when $b=4$ (in which case $a=2$). In both cases, you have $\frac{ab}5=\frac85$.
| {
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"timestamp": "2023-03-29T00:00:00",
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Proving $x^2 + x + 1$ is a factor of $P_n (x)=(x+1)^{2n+1} + x^{n+2}$. Further says to consider $P_n(\omega)$ and $P_n(\omega^2)$ wherein $\omega$ is a cube root of unity. $\omega \not=1$.
Found this from a examination paper with no solutions. I understand it relates to roots of unity, but I'm unsure how I can bring that into play. Any explanation on how to approach these types of questions is appreciated thank you.
| Let $\omega$ be any root of $x^2+x+1$. We will show that $\omega$ is a root of $(x+1)^{2n+1} + x^{n+2}$.
First note that we can rearrange $x^2+x+1$ to $x+1 = -x^2$ so we know that $(\omega+1)^{2n+1} = (-\omega^2)^{2n+1} = -\omega^{4n+2}$.
Thus $(\omega+1)^{2n+1} + \omega^{n+2} = -\omega^{4n+2} + \omega^{n+2} = -\omega^{n+2}(\omega^{3n} - 1).$
So we just need to show that $\omega^{3n} - 1 = 0$.
This can been shown via the identity $u^3-1 = (u-1)(u^2+u+1)$ with $u = \omega^n$.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Facing issue in rectifying the solution. Gauss Divergence Theorem problem Evaluate
$$
\iint_S (y^2z^2 \textbf{i} \, +z^2x^2\textbf{j}+z^2y^2\textbf{k}).\textbf{n} ~\mathrm{d}S
$$
where S is the part of sphere $x^2+y^2+z^2=1$ above the $xy$ plane and bounded by this plane.
I've tried solving this using Gauss Divergence Theorem, as follows,
$$
\iiint_V \text{div} (y^2z^2\textbf{i} \, +z^2x^2\textbf{j}+z^2y^2\textbf{k})~\mathrm{d}V = \iiint_V (2zy^2)~\mathrm{d}V
$$
For the limits of the volume V,
$-1\le x\le 1$
$-\sqrt{1-x^2}\le y \le \sqrt{1-x^2}$
$0\le z\le \sqrt{1-x^2-y^2}$
Upon integrating with these limits, I'm getting the asnwer as $\frac{\pi}{10}$, which is not the correct answer. And I'm unable to make out where I'm going wrong.
Kindly guide me to understand my error and rectify the solution. Thank you.
| $ \displaystyle \iiint_V \nabla \cdot (y^2z^2, z^2x^2, z^2y^2) \ dV = \iiint_V 2zy^2 \ dV$
Your bounds are correct so you may have made some mistake in evaluating it.
$ \displaystyle \iiint_V 2 z y^2 \ dV = 2 \int_{-1}^1 \int_{-\sqrt{1-x^2}}^{\sqrt{1-x^2}} \int_{0}^{\sqrt{1-x^2-y^2}} z y^2 \ dz \ dy \ dx$
$ \displaystyle = \int_{-1}^1 \int_{-\sqrt{1-x^2}}^{\sqrt{1-x^2}} y^2 (1 - x^2 - y^2) \ dy \ dx$
Converting to polar coordinates,
$x = r \cos\theta, y = r \sin\theta$
$ \displaystyle = \int_0^{2\pi} \int_0^1 r^2 \sin^2\theta (1 - r^2) \ r \ dr \ d\theta$
$ \displaystyle = \int_0^{2\pi} \int_0^1 \sin^2\theta (r^3 - r^5) \ dr \ d\theta$
$ \displaystyle = \int_0^{2\pi} \cfrac{\sin^2\theta}{12} \ d\theta = \cfrac{\pi}{12}$
Alternatively use spherical coordinates, as the other answer shows.
| {
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Using Rearrangement Inequality .
Let $a,b,c\in\mathbf R^+$, such that $a+b+c=3$. Prove that $$\frac{a^2}{b+c}+\frac{b^2}{c+a}+\frac{c^2}{a+b}\ge\frac{a^2+b^2+c^2}{2}$$
$Hint$ : Use Rearrangement Inequality
My Work :-$\\$
Without Loss of Generality let's assume that $0\le a\le b\le c$ then we can infer that $c^2\ge b^2\ge a^2$ also $b+c\ge c+a \ge a+b$. Thus $\frac{1}{b+c}\le \frac{1}{c+a}\le \frac{1}{a+b}$. Hence by Rearrangement Inequality $\frac{a^2}{b+c}+\frac{b^2}{c+a}+\frac{c^2}{a+b}$ is the greatest permutation however I am not able to express $\frac{a^2+b^2+c^2}{2}$ as another permutation of the same, hence I require assistance.
Thank You
| Another way:
We need to prove that:
$$\sum_{cyc}\frac{a^2}{b+c}\geq\frac{3(a^2+b^2+c^2)}{2(a+b+c)}$$ or
$$\sum_{cyc}\left(\frac{a^2}{b+c}-\frac{3a^2}{2(a+b+c)}\right)\geq0$$ or
$$\sum_{cyc}\frac{a^2(a-b-(c-a))}{b+c}\geq0$$ or
$$\sum_{cyc}(a-b)\left(\frac{a^2}{b+c}-\frac{b^2}{a+c}\right)\geq0,$$ which is true because $(a,b)$ and $\left(\frac{a^2}{b+c},\frac{b^2}{a+c}\right)$ have the same ordering for any positive $c$.
| {
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"url": "https://math.stackexchange.com/questions/4204601",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Prove $a^2 + b^2 + c^2 + ab + bc +ca \ge 6$ given $a+b+c = 3$ for $a,b,c$ non-negative real. I want to solve this problem using only the AM-GM inequality. Can someone give me the softest possible hint? Thanks.
Useless fact: from equality we can conclude $abc \le 1$.
Attempt 1:
Adding $(ab + bc + ca)$ to both sides of inequality and using the equality leaves me to prove: $ab + bc + ca \le 3$.
Final edit: I found a easy way to prove above.
$18 = 2(a+b+c)^2 = (a^2 + b^2) + (b^2 + c^2) + (c^2 + a^2) + 4ab + 4bc + 4ca \ge 6(ab + bc + ca) \implies ab + bc + ca \le 3$
(please let me know if there is a mistake in above).
Attempt 2: multiplying both sides of inequality by $2$, we get:
$(a+b)^2 + (b+c)^2 + (c+a)^2 \ge 12$. By substituting $x = a+b, y = b+c, z = c+a$ and using $x+y+z = 6$ we will need to show:
$x^2 + y^2 + z^2 \ge 12$. This doesnt seem trivial either based on am-gm.
Edit: This becomes trivial by C-S.
$(a+b).1 + (b+c).1 + (c+a).1 = 6 \Rightarrow \sqrt{((a+b)^2 + (b+c)^2 + (c+a)^2)(1 + 1 + 1)} \ge 6 \implies (a+b)^2 + (b+c)^2 + (c+a)^2 \ge 12$
Attempt 3:
$x = 1-t-u$, $y = 1+t-u$, $z = 1 + 2u$
$(1-u-t)^2 + (1-u+t)^2 + (1+2u)^2 + (1-u-t)(1-u+t) + (1+t-u)(1+2u) + (1-t-u)(1+2u)$
$ = 2(1-u)^2 + 2t^2 + (1 + 2u)^2 + (1-u)^2 - t^2 + 2(1+2u)$
expanding we get:
$ = 3(1 + u^2 -2u) + t^2 + 1 + 4u^2 + 4u + 2 + 4u = 6 + 7u^2 + t^2 + 2u\ge 6$.
Yes, this works.. (not using am-gm or any such thing).
| By AM-GM $$\sum_{cyc}(a^2+ab)-6=\sum_{cyc}(a^2+ab)-\frac{2}{3}(a+b+c)^2=$$
$$=\sum_{cyc}\left(a^2+ab-\frac{2}{3}(a^2+2ab)\right)=\frac{1}{3}\sum_{cyc}(a^2-ab)=$$
$$=\frac{1}{6}\sum_{cyc}(2a^2-2ab)=\frac{1}{6}\sum_{cyc}(a^2+b^2-2ab)\geq\frac{1}{6}\sum_{cyc}(2ab-2ab)=0.$$
| {
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "8",
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Determinant of a matrix of squares equals $4(x-y)(x-z)(y-z)$ I need to find the value of the following determinant.
$$\det\begin{pmatrix}x^2&(x+1)^2&(x+2)^2\\ \:y^2&(y+1)^2&(y+2)^2\\ \:z^2&(z+1)^2&(z+2)^2\end{pmatrix}$$
By long calculations (by minors and properties), I found that the value is
$$4x^2y-4x^2z-4xy^2+4xz^2-4yz^2+4y^2z=4(x-y)(x-z)(y-z)$$
I'm wondering if there is an easier way to calculate this. Any help is welcome.
| Here's a possibility:\begin{align}\begin{vmatrix}x^2&(x+1)^2&(x+2)^2\\y^2&(y+1)^2&(y+2)^2\\ z^2&(z+1)^2&(z+2)^2\end{vmatrix}&=\begin{vmatrix}x^2&x^2+2x+1&x^2+4x+4\\y^2&y^2+2y+1&y^2+4y+4\\ z^2&z^2+2z+1&z^2+4z+4\end{vmatrix}\\&=\begin{vmatrix}x^2&2x+1&4x+4\\y^2&2y+1&4y+4\\ z^2&2z+1&4z+4\end{vmatrix}\\&=2\begin{vmatrix}x^2&2x+1&2x+2\\y^2&2y+1&2y+2\\ z^2&2z+1&2z+2\end{vmatrix}\\&=2\begin{vmatrix}x^2&2x+1&1\\y^2&2y+1&1\\ z^2&2z+1&1\end{vmatrix}\\&=2\begin{vmatrix}x^2&2x&1\\y^2&2y&1\\z^2&2z&1\end{vmatrix}\\&=4\begin{vmatrix}x^2&x&1\\y^2&y&1\\z^2&z&1\end{vmatrix}\\&=4(x^2 y-x^2 z-x y^2+x z^2+y^2 z-y z^2)\\&=4(x-y)(x y-x z-y z+z^2)\\&=4(x-y)(x-z)(y-z).\end{align}
| {
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"url": "https://math.stackexchange.com/questions/4212425",
"timestamp": "2023-03-29T00:00:00",
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Show an inequality about digamma function Let $\psi$ be the digamma function, i.e. $\psi(x)=\frac{\Gamma'(x)}{\Gamma(x)}$. Show that
for $x>0$,
\begin{equation}
\label{1}
\psi\left(1+\frac{1}{2x}\right)-\psi\left(\frac{1}{2}+\frac{1}{2x}\right)>x-\frac{1}{2}x^2. \tag{1}
\end{equation}
I have tried the using that $t=\frac{1}{2x}$, then the inequality can be rewritten as
$$ \psi\left(1+t\right)-\psi\left(\frac{1}{2}+t\right)>\frac{1}{2t}-\frac{1}{8t^2}$$
holds for all $t>0$. Using the wolframalpha I get the asymptotic expansion
$$ \psi\left(1+t\right)-\psi\left(\frac{1}{2}+t\right)=\frac{1}{2t}-\frac{1}{8t^2}+\frac{1}{64t^{4}}+O\left(t^{-5} \right)$$
for $t\to +\infty$.
How can I prove inequality (1)? Can someone help me? Thank you very much!
| From equation (25) in https://doi.org/10.2298/AADM130124002N, we have
$$
\log \Gamma (t + 1) = \left( {t + \frac{1}{2}} \right)\log t - t + \frac{1}{2}\log (2\pi ) + \frac{1}{{12t}} + \frac{1}{{\pi t^3 }}\int_0^{ + \infty } {\frac{{t^2 }}{{t^2 + s^2 }}s^2 \log (1 - e^{ - 2\pi s} )ds}
$$
and
$$
\log \Gamma \!\left( {t + \frac{1}{2}} \right) = t\log t - t + \frac{1}{2}\log (2\pi ) - \frac{1}{{24t}} + \frac{1}{{\pi t^3 }}\int_0^{ + \infty } {\frac{{t^2 }}{{t^2 + s^2 }}s^2 \log (1 + e^{ - 2\pi s} )ds}
$$
for all $t>0$. Accordingly,
$$
\log \frac{{\Gamma (t + 1)}}{{\Gamma \!\left( {t + \frac{1}{2}} \right)}} = \frac{1}{2}\log t + \frac{1}{{8t}} - \frac{2}{{\pi t^3 }}\int_0^{ + \infty } {\frac{{t^2 }}{{t^2 + s^2 }}s^2 \tanh ^{ - 1} (e^{ - 2\pi s} )ds}
$$
for all $t>0$. Differentiating both sides gives
$$
\psi (t + 1) - \psi \left( {t + \frac{1}{2}} \right) = \frac{1}{{2t}} - \frac{1}{{8t^2 }} + \frac{2}{{\pi t^2 }}\int_0^{ + \infty } {\frac{{3t^2 + s^2 }}{{(t^2 + s^2 )^2 }}s^2 \tanh ^{ - 1} (e^{ - 2\pi s} )ds} > \frac{1}{{2t}} - \frac{1}{{8t^2 }}
$$
for all $t>0$.
| {
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Denoting $2\le\frac{6-4x}5\le3$ by $|mx-n|\le5$. What is the value of $|n-m|$?
If we denote solution set of the inequality $2\le\frac{6-4x}5\le3$ by
$|mx-n|\le5$, what is the value of $|n-m|$?
$1)7\qquad\qquad2)5\qquad\qquad3)21\qquad\qquad4)23$
I solved this problem with following approach:
$$2\le\frac{6-4x}5\le3\quad\Rightarrow-\frac94\le x\le-1$$
And $|mx-n|\le5$ is equivalent to $\frac{-5+n}m\le x\le\frac{5+n}{m}$. Hence we have $\frac{-5+n}m=-\frac94$ and $\frac{5+n}m=-1$ and by solving system of equations I got $m=8$ and $n=-13$. So the final answer is $21$.
I wonder, can we solve this problem with other approahces?
| $$2 \le \frac{6-4x}{5} \le 3$$
I first centralize it,
$$2-\frac52 \le \frac{6-4x}{5}-\frac52\le 3 - \frac52$$
$$-\frac12 \le \frac{12-8x-25}{10}\le \frac12$$
$$-1 \le \frac{-8x -13}{5}\le 1$$
$$|-8x-(13)|\le 5$$
Hence $|n-m|=|13-(-8)|=21.$
| {
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"timestamp": "2023-03-29T00:00:00",
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For which value of $t \in \mathbb R$ the equation has exactly one solution : $x^2 + \frac{1}{\sqrt{\cos t}}2x + \frac{1}{\sin t} = 2\sqrt{2}$ For which value of $t \in R $ the equation has exactly one solution : $x^2 + \frac{1}{\sqrt{\cos t}}2x + \frac{1}{\sin t} = 2\sqrt{2}$
Here $t \neq n\pi, t \neq (2n+1)\frac{\pi}{2}$
Therefore , for the given equation to have exactly one solution we should have :
$(\frac{2}{\sqrt{\cos t}})^2 -4.(\frac{1}{\sin t} - 2\sqrt{2}) = 0 $
$\Rightarrow \frac{4}{\cos t} - 4 (\frac{1}{\sin t} - 2\sqrt{2}) = 0 $
$\Rightarrow \sin t -\cos t +2 \sqrt{2}\sin t\cos t = 0 $
$\Rightarrow \sqrt{2}( \frac{1}{\sqrt{2}}\sin t - \frac{1}{\sqrt{2}}\cos t) = -2\sqrt{2} \sin t\cos t$
$\Rightarrow \sqrt{2}(\cos(\pi/4)\sin t -\sin(\pi/4)cos t = -\sqrt{2}\sin2t $ [Using $\sin x\cos y -\cos x\sin y = \sin(x-y)$]
$\Rightarrow \sqrt{2}\sin(\frac{\pi}{4}-t) =-\sqrt{2}\sin2t$
$\Rightarrow \sin(\frac{\pi}{4}-t) =-\sin2t $ [ Using $-\sin x = \sin(-x)$ and comparing R.H.S. with L.H.S. ]
$\Rightarrow \frac{\pi}{4}-t = -2t $
$\Rightarrow t = - \frac{\pi}{4}$
Is it correct answer, please suggest.. thanks
| You face a quadratic equation in $x$
$$x^2+a x+b=0 \qquad \text{with} \qquad a=\frac{2}{\sqrt{\cos (t)}}\quad \text{and} \quad b=\csc (t)-2 \sqrt{2}$$ The discriminant is
$$\Delta=a^2-4 b=4 \left(\sec (t)-\csc (t)+2 \sqrt{2}\right)$$ must be zero to have a double root.
Using the tangent half-angle substitution $t=2 \tan ^{-1}(x)$
$$\sec (t)-\csc (t)+2 \sqrt{2}=-\frac{x^2+1}{2 x}-\frac{2}{x^2-1}+2 \sqrt{2}-1$$ So, what is left is
$$x^4+\left(2-4 \sqrt{2}\right) x^3+\left(2+4 \sqrt{2}\right) x-1=0$$ whih has a double root
$$x_{1,2}=1+\sqrt{2}$$ and what is left is
$$x^2+\left(4-2 \sqrt{2}\right) x+2 \sqrt{2}-3=0$$ which shows the ugly roots
$$x_3=-2+\sqrt{2}-\sqrt{9-6 \sqrt{2}}\qquad \text{and} \qquad x_4=-2+\sqrt{2}-\sqrt{9-6 \sqrt{2}}$$
Use your pocket (or Google) calculator; you will find whole numbers in degrees. Convert to radians and you will obtain the results given by Wolfram Alpha (do not forget the modulo $2\pi$).
| {
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Understanding the proof that $A_1+A_2\mathop{\longrightarrow}\limits^{\tiny\begin{pmatrix}1&0\\0&1\end{pmatrix}}A_1\times A_2$ is an isomorphism Fix an abelian category. Let $A_1$ and $A_2$ be objects and let $A_1\times A_2$ be their product and $A_1+A_2$ their coproduct (i.e., sum). I am trying to understand the proof of the following theorem in Peter Freyd's "Abelian Categories."
Theorem 2.35 for abelian categories $A_1+A_2\mathop{\longrightarrow}\limits^{\begin{pmatrix}1&0\\0&1\end{pmatrix}}A_1\times A_2$ is an isomorphism.
Proof: Let $K\to A_1+A_2$ be the kernel of $\begin{pmatrix}1&0\\0&1\end{pmatrix}$. Then $K\longrightarrow A_1+A_2\mathop{\longrightarrow}\limits^{\begin{pmatrix}1&0\\0&1\end{pmatrix}}A_1\times A_2\mathop{\longrightarrow}\limits^{p_2}A_2\ =\ K\longrightarrow A_1+A_2\mathop{\longrightarrow}\limits^{\begin{pmatrix}0\\1\end{pmatrix}} A_2$ and $K\longrightarrow A_1+A_2$ is contained in $A_1\mathop{\longrightarrow}\limits^{u_1}A_1+A_2$. Similarly it is contained in $A_2\mathop{\longrightarrow}\limits^{u_2}A_1+A_2$, and hence it is contained in their intersection, which is zero. Thus $K=O$ and $\begin{pmatrix}1&0\\0&1\end{pmatrix}$ is monomorphic. Dually it is epimorphic and hence an isomorphism.
I don't understand the second sentence of the proof. Why are $K\longrightarrow A_1+A_2\longrightarrow A_1\times A_2\longrightarrow A_2$ and $K\longrightarrow A_1+A_2 \longrightarrow A_2$ the same, and why is $K\longrightarrow A_1+A_2$ contained in $A_1\longrightarrow A_1+A_2$?
| By definition, $p_2\circ\begin{pmatrix}1&0\\0&1\end{pmatrix} = \begin{pmatrix}0\\1\end{pmatrix}$, so the composites $K\to A_1 + A_2 \to A_1\times A_2 \to A_2 = K\to A_1 + A_2\to A_2$ are the same. But by definition of $K$, $K\to A_1+A_2\to A_1\times A_2$ is zero, so we conclude that $K\to A_1 + A_2\to A_2$ is zero.
By Proposition 2.22 and Theorem 2.31 for abelian categories, $u_1$ is a kernel of $\begin{pmatrix}0\\1\end{pmatrix}$, so the map $K\to A_1\times A_2$ factors though $u_1$, and hence $K\to A_1+A_2$ is contained in $A_1\stackrel{u_1}{\longrightarrow} A_1 + A_2$.
| {
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$\sqrt{a+b-c}+\sqrt{b+c-a}+\sqrt{c+a-b}\leq \sqrt a+\sqrt b+ \sqrt c.$ I was practicing some inequality problems when I saw the following problem:
Problem: If $a,b,c$ are the three sides of a triangle, prove that $$\sqrt{a+b-c}+\sqrt{b+c-a}+\sqrt{c+a-b}\leq \sqrt a+\sqrt b+ \sqrt c.$$
Solution: We assume, without loss of generality, that $a\geq b\geq c$ and notice that $(a+b-c,b+c-a,c+a-b)\succ (a,b,c)$. Then the rest of the solution is just Karamata.
But my question is on the statement: $(a+b-c,b+c-a,c+a-b)\succ (a,b,c)$. According to the definition of majorization, $\{a_i\}_{i=1}^n\succ \{b_i\}_{i=1}^n$ if and only if for all $1\leq k\leq n$, $\sum_{i=1}^ka_i\geq \sum_{i=1}^kb_i$ ( see here as reference). Of course, the inequality holds for $k=1$ and $k=3$ in our original case. But for $k=2$, we have $a+b-c+b+c-a=2b\leq a+b$. Then how come $(a+b-c,b+c-a,c+a-b)\succ (a,b,c)$? Is the solution above wrong or am I missing something?
| For $a\geq b\geq c$ it should be $$(a+b-c,a+c-b,b+c-a)\succ(a,b,c).$$
Because $$a+b-c\geq a+c-b\geq b+c-a,$$ $$a+b-c\geq a,$$ $$a+b-c+a+c-b\geq a+b$$ and
$$a+b-c+a+c-b+b+c-a=a+b+c.$$
Now, we can use Karamata.
| {
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Increasing the number of sides of a polygon by $k$ increases the number of diagonals by $6k$ For reference:
In a certain polygon when increasing the number of sides in $k$, the number of diagonals increases by $6k$. How many polygons meet this condition? (Answer: $5$)
My progress:
$\frac{n(n-3)}{2}+ 6k = \frac{(n+k)(n+k-3)}{2}\rightarrow\\
{\frac{n^2}{2}}-{\frac{3n}{2}}+6k = \frac{k^2}{2}+kn-\frac{3k}{2}+{\frac{n^2}{2}}-{\frac{3n}{2}}\rightarrow\\
\frac{k^2}{2}+kn-\frac{15k}{2}\rightarrow{k = 0} ~or ~k =15 -2n \implies n = \frac{15-k}{2}\\
n \ is \ integer \geq3 \therefore n={7, 6, 5, 4, 3}\rightarrow 5~polygons$
Is my resolution correct?
| $\frac{n(n-3)}{2}+ 6k = \frac{(n+k)(n+k-3)}{2}\rightarrow\\
{\frac{n^2}{2}}-{\frac{3n}{2}}+6k = \frac{k^2}{2}+kn-\frac{3k}{2}+{\frac{n^2}{2}}-{\frac{3n}{2}}\rightarrow\\
\frac{k^2}{2}+kn-\frac{15k}{2}\rightarrow{k = 0} ~or ~k =15 -2n \implies n = \frac{15-k}{2}\\
n \ is \ integer \geq3 \therefore n={7, 6, 5, 4, 3}\rightarrow 5~polygons$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4220954",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Integration using inverse function If $3f\left( x \right) = 3 {x^4} + {x^3} + 3{x^2}$ and $\mathop {\lim }\limits_{a \to \infty } \int\limits_{2a}^{8a} {\frac{1}{{{{\left( {{f^{ - 1}}\left( x \right)} \right)}^2} + {{\left( {{f^{ - 1}}\left( x \right)} \right)}^4}}}dx} = \ln(n)$, find the value of $n$.
My approach is as follow , I tried to get the value of $f^{-1}(x)$ by putting a separate question in Maths-type Inverse of a quartic function but it has been closed so I am putting the original question. I cross checked it. Iam not able to approach it.
| Like already mentioned $f$ is not injective. For every $y>0$ there are exactly two $x\in \mathbb R$ with $y=f(x)$, one of the $x$ is greater than zero and the other one is less than zero. Let's take only the positive one.
For every $x>0$ holds $x^4<x^4+\frac 13 x^3+x^2$. For $y:=x^4+\frac 13 x^3+x^2$ we get $\sqrt [4]y<x$. By inserting $f^{-1}(y)>\sqrt [4]y$ we have
$$\int_{2a}^{8a}\frac{1}{(f^{-1}(y))^2+(f^{-1}(y))^4}\,\mathrm dy\leq \int_{2a}^{8a} \frac{1}{\sqrt y+y}\,\mathrm dy=\left[2 \ln(1 + \sqrt y)\right]_{2a}^{8a}=2\ln\left(\frac{1+2\sqrt{2a}}{1+\sqrt{2a}}\right)\xrightarrow{a\to \infty}2\ln(2)$$
On the other hand for every $C\in (0,1)$ and every sufficently large $x$, we have $y>Cx^4$. Hence,
$$\int_{2a}^{8a}\frac{1}{(f^{-1}(y))^2+(f^{-1}(y))^4}\,\mathrm dy\geq \int_{2a}^{8a} \frac{1}{C^2\sqrt y+C^4y}\,\mathrm dy= \left[\frac{2}{C^4} \ln(1 + C^2 \sqrt{y})\right]_{2a}^{8a}=\frac{2}{C^4}\ln\left(\frac{1+2C^2\sqrt{2a}}{1+C^²\sqrt{2a}}\right)\xrightarrow{a\to \infty}\frac{2}{C^4}\ln(2).$$
Since this holds for all $C<1$, we get $n=2^2=4$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4221106",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Finding $a+b+c+d$, where $ab+c+d=15$, $bc+d+a=24$, $cd+a+b=42$, $da+b+c=13$ Let $a,b,c,d \in \mathbb{R}$. Consider the following constraints:
\begin{cases} ab+c+d=15 \\ bc+d+a=24 \\ cd+a+b=42 \\da+b+c=13 \end{cases}
Calculate the value of $a+b+c+d$.
It is easy to use the Gröbner basis to get the value:
\begin{cases}
10849-4501 d+380d^2,-39409+2320c+3420d,-20+29b-9d,1801+2320 a-380 d\}
\end{cases}
so the value of $a+b+c+d$ is $\frac{169}{10}$.
What I am curious about is how to use high schools mathematics to get an answer without too much complicated mathematical calculations ?
| Let $\,x=a+b+c+d\,$ then:
$$ab+c+d=ab\color{red}{-a-b+1-1+a+b}+c+d=(a-1)(b-1)-1+x\,$$
The system can then be written as:
$$
\begin{cases}
(a-1)(b-1)=16-x
\\ (b-1)(c-1)=25-x
\\ (c-1)(d-1)=43-x
\\ (d-1)(a-1)=14-x
\end{cases}
$$
It follows that:
$$
(a-1)(b-1)(c-1)(d-1) \;=\; (16-x)(43-x) \;=\; (25-x)(14-x)
$$
The latter equality gives $\,2\left(169 - 10 x\right) = 0\,$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4222409",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 2,
"answer_id": 1
} |
Integrate via substitution: $x^2\sqrt{x^2+1}\;dx$ I need to integrate the following using substitution:
$$
\int x^2\sqrt{x^2+1}\;dx
$$
My textbook has a similar example:
$$
\int \sqrt{x^2+1}\;x^5\;dx
$$
They integrate by splitting $x^5$ into $x^4\cdot x$ and then substituting with $u=x^2+1$:
$$
\int \sqrt{x^2+1}\;x^4\cdot x\;dx\\
=\frac{1}{2}\int \sqrt{u}\;(u-1)^2\;du\\
=\frac{1}{2}\int u^{\frac{1}{2}}(u^2-2u+1)\;du\\
=\frac{1}{2}\int u^{\frac{5}{2}}-2u^{\frac{3}{2}}+u^{\frac{1}{2}}\;du\\
=\frac{1}{7}u^{\frac{7}{2}}-\frac{2}{5}u^{\frac{5}{2}}+\frac{1}{3}u^{\frac{3}{2}}+C
$$
So far so good. But when I try this method on the given integral, I get the following:
$$
\int x^2\sqrt{x^2+1}\;dx\\
=\frac{1}{2}\int \sqrt{x^2+1}\;x\cdot x\;dx\\
=\frac{1}{2}\int \sqrt{u}\;\sqrt{u-1}\;du\;(u=x^2+1)\\
=\frac{1}{2}\int u^{\frac{1}{2}}(u-1)^\frac{1}{2}\;du
$$
Here is where it falls down. I can't expand the $(u-1)^\frac{1}{2}$ factor like the $(u-1)^2$ factor above was, because it results in an infinite series. I couldn't prove, but I think any even exponent for the $x$ factor outside the square root will cause an infinite series to result. Odd exponents for $x$ will work, since it will cause the $(u-1)$ term to have a positive integer exponent.
How should I proceed? I don't necessarily want an answer. I just want to know if I'm missing something obvious or if it is indeed above first year calculus level and probably a typo on the question.
| As suggested in the comments we can use the substitution $x=\mbox{sinh}(u)$ and some other well known trig identities:
$1)$ $dx=(\mbox{sinh}(u))'du=\mbox{cosh}(u)du$
$2)$ $\mbox{cosh}^{2}(u)-\mbox{sinh}^{2}(u)=1$
$3)$ $\mbox{cosh}(u)=\frac{e^{u}+e^{-u}}{2}$
$4)$ $\mbox{arcsinh}(u)=\mbox{log}(u+\sqrt{1+u^{2}})$
$5)$ $\int \mbox{sinh}^{n}(u)du=\frac{\mbox{cosh}(u)\mbox{sinh}^{n-1}(u)}{n}-\frac{n-1}{n}\int \mbox{sinh}^{n-2}(u)du$
You can prove the last identity using integration by parts. So your integral becomes:
$\begin{align} \int x^{2}\sqrt{x^{2}+1}\;dx &=\int\mbox{sinh}^2(u)\sqrt{\mbox{sinh}^2(u)+1}\:\mbox{cosh}(u)\;du\\
&=\int \mbox{sinh}^{2}(u)\mbox{cosh}^{2}(u)\;du\\
&=\int \mbox{sinh}^{2}(u)(1+\mbox{sinh}^{2}(u))\;du\\
&=\int \mbox{sinh}^{2}(u)\;du\:+\:\int \mbox{sinh}^{4}(u)\;du\\
&=\int \mbox{sinh}^{2}(u)\;du\:+\:\begin{bmatrix}\frac{\mbox{cosh}(u)\mbox{sinh}^{3}(u)}{4}\;-\;\frac{3}{4}\int \mbox{sinh}^{2}(u)\;du\end{bmatrix}\\
&=\frac{1}{4}\int \mbox{sinh}^{2}(u)\;du\:+\:\frac{\mbox{cosh}(u)\mbox{sinh}^{3}(u)}{4}\\
&=\frac{1}{4}\begin{bmatrix} \frac{\mbox{cosh}(u)\mbox{sinh}(u)}{2}-\frac{1}{2} \int du \end{bmatrix}\:+\:\frac{\mbox{cosh}(u)\mbox{sinh}^{3}(u)}{4}\\
&=\bigstar
\end{align}$
Let's compute something useful: $u=\mbox{arcsinh}(x)=\mbox{log}(x+\sqrt{1+x^{2}})$.
So we have that:
$\mbox{sinh}(u)=\mbox{sinh}(\mbox{arcsinh}(x))=x$
$\begin{align} \mbox{cosh}(u)
&=\mbox{cosh}(\mbox{arcsinh}(x))\\
&=\frac{x+\sqrt{1+x^{2}}+\frac{1}{x+\sqrt{1+x^{2}}}}{2}\\
&=\frac{x^{2}+x\sqrt{1+x^{2}}+1}{x+\sqrt{1+x^{2}}}\\
&=\frac{x^{2}+x\sqrt{1+x^{2}}+1}{x+\sqrt{1+x^{2}}} \cdot \frac{\sqrt{1+x^{2}}-x}{\sqrt{1+x^{2}}-x}\\
&=...=\sqrt{1+x^{2}}
\end{align}$
So the integral becomes:
$\begin{align} \bigstar
&=\frac{\mbox{cosh}(u)\mbox{sinh}(u)-\mbox{arcsinh}(u)}{8}\:+\:\frac{\mbox{cosh}(u)\mbox{sinh}^{3}(u)}{4}+C\\
&=\frac{x\sqrt{1+x^{2}}\:-\:\mbox{log}(x+\sqrt{1+x^{2}})}{8}\:+\:\frac{x^{3}\sqrt{1+x^{2}}}{4}+C\\
&=\frac{(2x^{3}+x)\sqrt{1+x^{2}}\:-\:\mbox{log}(x+\sqrt{1+x^{2}})}{8}+C
\end{align}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4223596",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 4,
"answer_id": 2
} |
Show that for a, b, c > 0 with a + b + c = 1, $(a + \frac 1a)^2 + (b + \frac 1b)^2 + (c + \frac 1c)^2 ≥ \frac {100}{3}$ For this question I took the function as $x^2$ and using the second derivative I found it to be 2. This is greater than 0 for all x values meaning that the function is convex. Then I used Jensen's Inequality which got me $$\frac {(a + \frac 1a)^2 + (b + \frac 1b)^2 + (c + \frac 1c)^2}{3} ≥ \frac {(a + b + c + \frac 1a + \frac 1b + \frac 1c)^2}{9}$$ I'm not sure how to find the value of $ \frac 1a + \frac 1b + \frac 1c$ or if this even is the correct approach. Can I have some hints for the next step or direction of attack?
| Putting it all together from the comments.
$$\left(a + \frac 1a\right)^2 + \left(b + \frac 1b\right)^2 + \left(c + \frac 1c\right)^2 ≥ \frac {(a + b + c + \frac 1a + \frac 1b + \frac 1c)^2}{3}$$
$$≥ \frac {\left(a + b + c + \frac{9}{a+b+c} \right)^2}{3} = \frac{100}{9}$$
Note that applying Cauchy-Schwarz or QM-AM yields the first inequality, so no need to use Jensen's here.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4226251",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
If $A$ and $B$ are solutions to $7\cos\theta+4\sin\theta+5=0$, then $\cot\frac{A}{2}+\cot\frac{B}{2}=-\frac{2}{3}$ If $A$ and $B$ are the solutions to ${\displaystyle 7\cos\theta+4\sin\theta+5=0\mbox{ where }A>0,0<B<2\pi;}$
Without finding the solutions to the trig equation, show that;
$$\cot\left(\frac{A}{2}\right)+\cot\left(\frac{B}{2}\right)=-\frac{2}{3}$$
This is my effort so far:
Since A and B are solutions then:
\begin{align*}
7\cos A+4\sin A+5 & =0\\
7\left( 2\cos^{2}\frac{A}{2}-1\right) +4\times2\sin\frac{A}{2}\cos\frac{A}{2}+5 & =0\\
14\cos^{2}\frac{A}{2}+8\sin\frac{A}{2}\cos\frac{A}{2}-2 & =0\\
\mbox{Now divide by }\sin\frac{A}{2}\cos\frac{A}{2} & \mbox{ gives:}\\
14\cot\frac{A}{2}+8-\frac{2}{\sin\frac{A}{2}\cos\frac{A}{2}} & =0\\
\mbox{similarly for B;}\\
14\cot\frac{B}{2}+8-\frac{2}{\sin\frac{B}{2}\cos\frac{B}{2}} & =0
\end{align*}
When I add these two equations I have $\cot\left(\frac{A}{2}\right)+\cot\left(\frac{B}{2}\right)$,
but am unsure how to progress.
| $\displaystyle 7\cos\theta+4\sin\theta+5=0$
$14\cos^{2}\frac{\theta}{2}+8\sin\frac{\theta}{2}\cos\frac{\theta}{2}-2 =0$
Dividing by $\sin ^2 \frac{\theta}{2} \ $,
$14 \cot^2\frac{\theta}{2} + 8 \cot \frac{\theta}{2} - 2 \csc^2\frac{\theta}{2} = 0$
Using $\csc^2\frac{\theta}{2} = 1 + \cot^2\frac{\theta}{2}$,
$6 \cot^2\frac{\theta}{2} + 4 \cot \frac{\theta}{2} - 1 = 0$
As we know, sum of roots of quadratic $ax^2 + bx + c = 0$ is $ - \frac{b}{a}$.
So, $\cot \frac{A}{2} + \cot \frac{B}{2} = - \frac{2}{3}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4226367",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
Given the function $f(x,y)=\frac{5xy^2}{x^2+y^2}$, argue analytically, if the discontinuity of $f$ at $(0,0)$ is avoidable or unavoidable. It is clear that $f(0,0)$ does not exist, and therefore the function is discontinuous. Now to know if the discontinuity is avoidable or inevitable, we must see that the limit at that point exists or not. To see the above we will use polars.
\begin{align*}
\lim_{(x,y)\rightarrow(0,0)}\frac{5xy^2}{x^2+y^2}&=\lim_{r\rightarrow0}\frac{5(r\cos(\theta))(r\sin(\theta))^2}{(r\cos(\theta))^2+(r\sin(\theta))^2}\\
& =\lim_{r\rightarrow0}\frac{5r^3\cos(\theta)\sin^2(\theta)}{r^2(\cos^2(\theta)+\sin^2(\theta))}\\
& =\lim_{r\rightarrow0}\frac{5r^3\cos(\theta)\sin^2(\theta)}{r^2}\\
& =\lim_{r\rightarrow0}5r\cos(\theta)\sin^2(\theta)=0.
\end{align*}
Now let's test for the definition $\epsilon-\delta$, to see that the limit exists and is zero.
\begin{align*}
\left|\frac{5xy^2}{x^2+y^2}-0\right| & =\left|\frac{5xy^2}{x^2+y^2}\right|\\
& =\frac{5|x|y^2}{x^2+y^2}\\
& \leq 5|x| \text{ since $y^2\leq x^2+y^2$}\\
& \leq 5\sqrt{x^2+y^2} \text{ since $|x|\leq\sqrt{x^2+y^2}$}\\
& <5\delta = \epsilon.
\end{align*}
Therefore we have:
\begin{align*}
\forall\epsilon>0, \exists\delta>0 \text{ such that, if } 0<\sqrt{x^2+y^2}<\delta \Rightarrow |f(x,y)-0| &\leq5 \sqrt{x^2+y^2}\\
&<5 \delta = 5\left(\frac{\epsilon}{5}\right)=\epsilon.
\end{align*}
Thus the limit exists and is zero, therefore the discontinuity of $f$ is avoidable.
I think this is the correct solution, I await your comments. If anyone has a different solution or correction of my work I will be grateful.
| Both your proofs (the analytic proof, and the proof using polar coordinates) that $\lim_{(x,y) f(x,y)\rightarrow (0,0)}=0$ are correct.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4227337",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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Recursive formula for a combinatorial problem and define the generating function Question:
Let $\sigma=\{a,b,c\}$. How many words can we assemble without the substrings $'ab'$ and $'bc'$?
*
*define the recursive formula.
*define the generating function for this formula.
$Solution.A.$
For the first sub-question, we consider 3 disjoint cases: $$a_{n}^{c}=\text{the number of valid words with length of $n$, starting with 'c' }$$
$$a_{n}^{b}=\text{the number of valid words with length of $n$, starting with 'b' }$$
$$a_{n}^{a}=\text{the number of valid words with length of $n$, starting with 'a' }$$
For the first case, we can add $'c'$ at the beginning of the word and then take a valid word with a length of $n-1$. Therefore, we have $a_{n-1}$ options for this case.
For the second case, we use the complement method. We can take all valid words starting with $'b'$ and then get rid of all the options for words with a length of $n-2$ starting with $'c'$. So we have: $a^{b}_{n}=a_{n-1}-a_{n-2}$ options.
For the third case, we use the complement method. We can take all valid words starting with $'a'$ and then get rid of all the options for words with a length of $n-2$ starting with $'b'$. So we have: $$a_{n}^{a} =a_{n-1} -a_{n-1}^{b} =a_{n-1} -( a_{n-2} -a_{n-3}) =a_{n-1} -a_{n-2} +a_{n-3}$$
options.
Now, since all three cases are disjoint then it is clear that: $a_{n} =a_{n}^{a} +a_{n}^{b} +a_{n}^{c}$, so therefore,$$a_{n} =a_{n}^{a} +a_{n}^{b} +a_{n}^{c} =a_{n-1} +a_{n-1} -a_{n-2} +a_{n-1} -a_{n-2} +a_{n-3}\\
=3a_{n-1} -2a_{n-2} +a_{n-3}$$
$Solution.B.$
Let $$F( x) =\sum _{n=0}^{\infty } a_{n} x^{n}$$
Thus,$$ \begin{array}{l}
F( x) =\sum _{n=0}^{\infty } a_{n} x^{n} =a_{0} +a_{1} x+a_{2} x^{2} +\sum _{n=3}^{\infty }( 3a_{n-1} -2a_{n-2} +a_{n-3}) x^{n}\\
\\
=1+3x+7x^{2} +\sum _{n=3}^{\infty }( 3a_{n-1} -2a_{n-2} +a_{n-3}) x^{n}\\
\\
=1+3x+7x^{2} +3\sum _{n=3}^{\infty } a_{n-1} x^{n} -2\sum _{n=3}^{\infty } a_{n-2} x^{n} +\sum _{n=3}^{\infty } a_{n-3} x^{n}\\
\\
=1+3x+7x^{2} +3x\underbrace{\sum _{n=3}^{\infty } a_{n-1} x^{n-1}}_{F( x) \ -\ a_{1} x\ -\ a_{0}} -2x^{2}\underbrace{\sum _{n=3}^{\infty } a_{n-2} x^{n-2}}_{F( x) \ -\ a_{0}} +x^{3}\underbrace{\sum _{n=3}^{\infty } a_{n-3} x^{n-3}}_{F( x)}\\
\\
=1+3x+7x^{2} +3x( F( x) -a_{1} x-a_{0}) -2x^{2}( F( x) -a_{0}) +x^{3} F( x)\\
\\
=1+3x+7x^{2} +3x( F( x) -3x-1) -2x^{2}( F( x) -1) +x^{3} F( x)\\
\\
=1+3x+7x^{2} +3xF( x) -9x^{2} -3x-2x^{2} F( x) +2x^{2} +x^{3} F( x)\\
\\
=1+\left( 3x-2x^{2} +x^{3}\right) F( x)
\end{array}$$
therefore, we have that: $$ \begin{array}{l}
F( x) =1+\left( 3x-2x^{2} +x^{3}\right) F( x) /-\left( 3x-2x^{2} +x^{3}\right) F( x)\\
\\
F( x) -\left( 3x-2x^{2} +x^{3}\right) F( x) =1\\
\\
\left( 1-\left( 3x-2x^{2} +x^{3}\right)\right) F( x) =1\\
\\
\left( 1-3x+2x^{2} -x^{3}\right) F( x) =1/:\left( 1-3x+2x^{2} -x^{3}\right)\\
\\
\boxed{F( x) =\frac{1}{1-3x+2x^{2} -x^{3}}}
\end{array}$$
Now, is it correct?
| The following answer is based upon the Goulden-Jackson Cluster Method. We consider the set of words of length $n\geq 0$ built from an alphabet $$\mathcal{\sigma}=\{a,b,c\}$$ and the set $B=\{ab,bc\}$ of bad words, which are not allowed to be part of the words we are looking for. We derive a generating function $F(x)$ with the coefficient of $x^n$ being the number of wanted words of length $n$.
According to the paper (p.7) the generating function $F(x)$ is
\begin{align*}
F(x)=\frac{1}{1-dx-\text{weight}(\mathcal{C})}\tag{1}
\end{align*}
with $d=|\mathcal{V}|=3$, the size of the alphabet and $\mathcal{C}$ is the weight-numerator of bad words with
\begin{align*}
\text{weight}(\mathcal{C})=\text{weight}(\mathcal{C}[ab])+\text{weight}(\mathcal{C}[bc])\tag{2}
\end{align*}
We calculate according to the paper
\begin{align*}
\text{weight}(\mathcal{C}[bc])&=-x^2\\
\text{weight}(\mathcal{C}[ab])&=-x^2-x\cdot\text{weight}(\mathcal{C}[bc])\tag{3}\\
\end{align*}
so that
\begin{align*}
\text{weight}(\mathcal{C})=-x^2+\left(-x^2-x\cdot\left(-x^2\right)\right)=-2x^2+x^3
\end{align*}
The additional term on the right-hand side of (3) takes account of the overlapping of $\color{blue}{b}c$ with $a\color{blue}{b}$.
We obtain according to (1) and (3)
\begin{align*}
\color{blue}{F(x)}&=\frac{1}{1-dx-\text{weight}(\mathcal{C})}\\
&\; \color{blue}{=\frac{1}{1-3x+2x^2-x^3}}\\
&=1+3x+7x^2+16x^3+\color{blue}{37}x^4+86x^5\\
&\qquad+200x^6+465x^7+1\,081x^8+\cdots
\end{align*}
in accordance with OPs result.
The last line was calculated with some help of Wolfram Alpha. The blue marked coefficient of $x^4$ tells us that out of $3^4=81$ words of length $4$ there are $37$ valid words which do not contain $ab$ or $bc$. These $37$ words are listed below.
\begin{align*}
\begin{array}{cccccc}
\text{aaaa}&\text{aaac}&\text{aaca}&\text{aacb}&\text{aacc}&\text{acaa}\\
\text{acac}&\text{acba}&\text{acbb}&\text{acca}&\text{accb}&\text{accc}\\
\text{baaa}&\text{baac}&\text{baca}&\text{bacb}&\text{bacc}&\text{bbaa}\\
\text{bbac}&\text{bbba}&\text{bbbb}&\text{caaa}&\text{caac}&\text{caca}\\
\text{cacb}&\text{cacc}&\text{cbaa}&\text{cbac}&\text{cbba}&\text{cbbb}\\
\text{ccaa}&\text{ccac}&\text{ccba}&\text{ccbb}&\text{ccca}&\text{cccb}\\
\text{cccc}
\end{array}
\end{align*}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4228313",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 1,
"answer_id": 0
} |
Solving a second order ODE of the form $y'' + P(x)y' + Q(x)y = f(x)$, given one part of the solution I have the following ODE
$$x(x-1)y'' - (2x-1)y' + 2y = 2x^3-3x^2$$ where I'm given
$$y_1 = x^2$$
Now, my workbook states that if we encounter such an ODE, where one part of the solution is given, the other part $y_2$ is $$y_2 = y_1 \int \frac{e^{-\int P(x)dx}}{y_1^2}dx$$
After I divide my equation by $x(x-1)$ and substituting into the formula for $y_2$, I get
$$y_2 = x^2\int\frac{(x-1)x}{x^4}$$
$$y_2 = \frac{1-2x}{2}$$
Now, $$y=c_1x^2 + c_2\frac{1-2x}{2}$$
Which is the solution of the homogenous equation. In order to find the particular solution, I used variation of constants, where I got the system:
$$
\left\{
\begin{array}{c}
c_1'x^2 + c_2'\frac{1-2x}{2} = 0 \\
c_1' 2x - c_2' = \frac{2x^2 - 3x}{x-1}
\end{array}
\right.
$$
The solution for which are
$c_1' = \frac{(2x-3)(1-2x)}{x-1}$
and
$c_2' = \frac{x^3(2x-3)}{x(x-1)^2}$
Which after integrating end up being
$$c_1 = \ln|x-1| - 2(x-1)^2 + C_1$$
and $$c_2 = \frac{x^3}{x-1} + C_2$$
After plugging in into the solution, I get
$$y = x^2 ( \ln|x-1| - 2(x-1)^2 + C_1) + \frac{1-2x}{2}(\frac{x^3}{x-1}+C_2)$$
However, this is not the same as the solution in my workbook which is
$$y=C_1x^2 + C_2(-x+\frac{1}{2}) + x^3 - \frac{x^2}{2} + x-\frac{1}{2}$$
Where did I go wrong?
| Your equations for $c_1'$ and $c_2'$ are
$$
\begin{pmatrix}
x^2 & \frac12 (1-2x) \\
2x & -1
\end{pmatrix}
\begin{pmatrix}
c_1' \\
c_2'
\end{pmatrix}
=
\begin{pmatrix}
0 \\
\frac{2x^2-3x}{x-1}
\end{pmatrix}
,
$$
and with the help of the useful formula
$$
\begin{pmatrix} a & b \\ c & d \end{pmatrix}^{-1}
=
\frac{1}{ad-bc}
\begin{pmatrix} d & -b \\ -c & a \end{pmatrix}
$$
we get
$$
\begin{split}
\begin{pmatrix}
c_1' \\
c_2'
\end{pmatrix}
&=
\frac{1}{x(x-1)}
\begin{pmatrix}
-1 & -\frac12 (1-2x) \\
-2x & x^2
\end{pmatrix}
\begin{pmatrix}
0 \\
\frac{2x^2-3x}{x-1}
\end{pmatrix}
\\ &
=
\frac{2x-3}{(x-1)^2}
\begin{pmatrix}
\frac12 (2x-1) \\
x^2
\end{pmatrix}
.
\end{split}
$$
So you're missing a factor of $1/2$ and the square in the denominator in your formula for $c_1'$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4230194",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Complex number related problem Let $z_1,z_2,z_3$ be complex numbers such that $|z_1|=|z_2|=|z_3|=|z_1+z_2+z_3|=2$ and $|z_1–z_2| =|z_1–z_3|$,$(z_2 \ne z_3)$, then the value of $|z_1+z_2||z_1+z_3|$ is_______
My solution is as follow
${z_1} = 2{e^{i{\theta _1}}};{z_2} = 2{e^{i{\theta _2}}};{z_3} = 2{e^{i{\theta _3}}}$ & $Z = {z_1} + {z_2} + {z_3} = 2\left( {{e^{i{\theta _1}}} + {e^{i{\theta _2}}} + {e^{i{\theta _3}}}} \right)$
$\left| {{z_1} - {z_2}} \right| = \left| {{z_1} - {z_3}} \right| \Rightarrow \left| {{e^{i{\theta _1}}} - {e^{i{\theta _2}}}} \right| = \left| {{e^{i{\theta _1}}} - {e^{i{\theta _3}}}} \right|$
Let ${\theta _1} = 0$
$\left| {{z_1} - {z_2}} \right| = \left| {{z_1} - {z_3}} \right| \Rightarrow \left| {1 - \left( {\cos {\theta _2} + i\sin {\theta _2}} \right)} \right| = \left| {1 - \left( {\cos {\theta _3} + i\sin {\theta _3}} \right)} \right|$
$ \Rightarrow \left| {1 - \cos {\theta _2} - i\sin {\theta _2}} \right| = \left| {1 - \cos {\theta _3} - i\sin {\theta _3}} \right| \Rightarrow \left| {2{{\sin }^2}\frac{{{\theta _2}}}{2} - 2i\sin \frac{{{\theta _2}}}{2}\cos \frac{{{\theta _2}}}{2}} \right| = \left| {2{{\sin }^2}\frac{{{\theta _3}}}{2} - 2i\sin \frac{{{\theta _3}}}{2}\cos \frac{{{\theta _3}}}{2}} \right|$
$\Rightarrow \left| { - 2{i^2}{{\sin }^2}\frac{{{\theta _2}}}{2} - 2i\sin \frac{{{\theta _2}}}{2}\cos \frac{{{\theta _2}}}{2}} \right| = \left| { - 2{i^2}{{\sin }^2}\frac{{{\theta _3}}}{2} - 2i\sin \frac{{{\theta _3}}}{2}\cos \frac{{{\theta _3}}}{2}} \right| \Rightarrow \left| { - 2i\sin \frac{{{\theta _2}}}{2}\left( {\cos \frac{{{\theta _2}}}{2} + i\sin \frac{{{\theta _2}}}{2}} \right)} \right| = \left| { - 2i\sin \frac{{{\theta _3}}}{2}\left( {\cos \frac{{{\theta _3}}}{2} + i\sin \frac{{{\theta _3}}}{2}} \right)} \right|$
$ \Rightarrow \left| { - 2i\sin \frac{{{\theta _2}}}{2}\left( {{e^{i\frac{{{\theta _2}}}{2}}}} \right)} \right| = \left| { - 2i\sin \frac{{{\theta _3}}}{2}\left( {{e^{i\frac{{{\theta _3}}}{2}}}} \right)} \right|$
$ \Rightarrow \left| {2\sin \frac{{{\theta _2}}}{2}\left( {{e^{ - i\frac{\pi }{2}}}} \right)\left( {{e^{i\frac{{{\theta _2}}}{2}}}} \right)} \right| = \left| {2\sin \frac{{{\theta _3}}}{2}\left( {{e^{ - i\frac{\pi }{2}}}} \right)\left( {{e^{i\frac{{{\theta _3}}}{2}}}} \right)} \right| \Rightarrow \left| {2\sin \frac{{{\theta _2}}}{2}\left( {{e^{i\left( {\frac{{{\theta _2}}}{2} - \frac{\pi }{2}} \right)}}} \right)} \right| = \left| {2\sin \frac{{{\theta _3}}}{2}\left( {{e^{i\left( {\frac{{{\theta _3}}}{2} - \frac{\pi }{2}} \right)}}} \right)} \right|$
$\Rightarrow \left| {2\sin \frac{{{\theta _2}}}{2}} \right|\left| {\left( {{e^{i\left( {\frac{{{\theta _2}}}{2} - \frac{\pi }{2}} \right)}}} \right)} \right| = \left| {2\sin \frac{{{\theta _3}}}{2}} \right|\left| {\left( {{e^{i\left( {\frac{{{\theta _3}}}{2} - \frac{\pi }{2}} \right)}}} \right)} \right| \Rightarrow \left| {2\sin \frac{{{\theta _2}}}{2}} \right| = \left| {2\sin \frac{{{\theta _3}}}{2}} \right|$
${\theta _2} \ne {\theta _3}$
How do I proceed further?
| We have three points on the circle of radius $2$ centered at the origin. We know that the triangle formed by these points is isosceles. We can rotate the triangle by some angle so that point $z_1$ is on either the real axis or the imaginary axis. Note that the modulus will not change because rotation = multiplying by $e^{iθ}$ for some $θ$. Now the point $z_2,z_3$ are symmetric about the axis you chose; what can you deduce about their sum?
Spoilers
After the rotation $z_1$ is either $\pm 2$ or $\pm 2i$. WLOG assume it's $2$. Then the rotated $z_2$, $z_3$ are $x+iy$ and $x-iy$. We have $|2+2x|=2$ meaning $x=0$. Thus, $z_2=\pm 2i$ and $z_3=\mp 2i$. Thus, $|z_1+z_2||z_1+z_3|=8$.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 5,
"answer_id": 1
} |
Calculating $\int\frac{a^4-x^4}{x^4+a^2x^2+a^4}dx$
$$\int\frac{a^4-x^4}{x^4+a^2x^2+a^4}dx$$
I tried factoring as $${(a^2-x^2)(a^2+x^2)}$$ $$\text{and}$$ $$(x^2+a^2+ax)(x^2+a^2-ax)$$ but didn't get much after this. I sew that the expression is symmetric in a,x but I don't know how to use that.
| $$I=\int\frac{a^4-x^4}{x^4+a^2x^2+a^4}\,dx=a\int\frac{1-y^4}{y^4+y^2+1}\,dy=a\int\frac{1-y^4}{\left(y^2-y+1\right) \left(y^2+y+1\right) }\,dy$$
$$I=\frac a2\int\Bigg[\frac{\left(y+\frac 12\right)+\frac 32}{y^2+y+1}-\frac{\left(y-\frac 12\right)-\frac 32}{y^2-y+1}-1 \Bigg]\,dy$$ So, beside the linear term (and the integration constant), you should end with a logarithm and an arctangent.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4231586",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Expressing the $n-$th derivative of $f(x)=\frac{3x^2-6x+5}{x^3-5x^2+9x-5}$ In an attempt to express the $n-$th derivative of the rational function $f(x)=\frac{3x^2-6x+5}{x^3-5x^2+9x-5}$ ,
I split it into $\left( \frac{1+2i}{(x-2-i)} + \frac{1-2i}{(x-2+i)} + \frac{1}{(x-1)} \right)$ ,
then using $y= \frac{1}{ax+b} \Rightarrow y_{n}=\frac{(-1)^n n! a^n}{(ax+b)^{n+1}}$ I ended up with:
$$(-1)^n n! [(1+2i)(x-2-i)^{-n-1}+(1-2i)(x-2+i)^{-n-1} +(x-1)^{-n-1}]$$
Now I don’t know how to get back to $\mathbb{\mathbb{R}}$ from $\mathbb{\mathbb{C}}$.
What I'd like to get is a form that doesn't make use of trigonometric functions but uses the binomial theorem.
Can anyone help me achieve this? Thank you so much in advance.
| From your well-asked question, I assume that you are only interested in a trick for the part with complex coefficients
$$A(x)=:(1+2i)(x-2-i)^{-n-1}+(1-2i)(x-2+i)^{-n-1}$$ or equivalently
$$A(x)=\frac{1+2i}{(x-2-i)^{n+1}}+\frac{1-2i}{(x-2+i)^{n+1}}$$
We will profit from the fact that the terms are conjugate:
$$\begin{aligned}A(x)&=\frac{(1+2i)\big((x-2)+i\big)^{n+1}+(1-2i)\big((x-2)-i\big)^{n+1}}{\big((x-2)^2+1\big)^{n+1}}\end{aligned}$$
Let us split the numerator into
$$\left[\big((x-2)+i\big)^{n+1}+\big((x-2)-i\big)^{n+1}\right]+2i\left[\big((x-2)+i\big)^{n+1}-\big((x-2)-i\big)^{n+1}\right]$$
and use the binomial theorem
$$\left[\sum_{k=0}^{n+1}\binom{n+1}{k}(x-2)^{n+1-k}\cdot i^k + \sum_{k=0}^{n+1}\binom{n+1}{k}(x-2)^{n+1-k}\cdot (-i)^k \right]\\+2i\cdot \left[\sum_{k=0}^{n+1}\binom{n+1}{k}(x-2)^{n+1-k}\cdot i^k - \sum_{k=0}^{n+1}\binom{n+1}{k}(x-2)^{n+1-k}\cdot (-i)^k\right]$$
In the first part all terms with odd $k$ fall out, in the second part with even $k.$ We obtain (please check it)
$$A(x)=\frac{2\cdot \sum\limits_{k=0}^{\lfloor\frac{n+1}{2}\rfloor}\binom{n+1}{2k}(x-2)^{n+1-2k}(-1)^k+4\cdot \sum\limits_{k=0}^{\lfloor\frac{n}{2}\rfloor}\binom{n+1}{2k+1}(x-2)^{n-2k}(-1)^{k+1}}{\big((x-2)^2+1\big)^{n+1}}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4232019",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 1,
"answer_id": 0
} |
Factoring $\frac{n(n+1)}2x^2-x-2$ for $n\in\mathbb Z$ I was factoring quadratic polynomials for high-school practice and I noticed a pattern: $$\begin{align} x^2-x-2 &=(x+1)(x-2) \\ 3x^2-x-2 &=(x-1)(3x-2) \\ 6x^2-x-2 &= (2x+1)(3x-2) \\ 10x^2-x-2 &= (2x-1)(5x+2) \\ 15x^2-x-2 &= (3x+1)(5x-2) \\ 21x^2-x-2 &= (3x-1)(7x+2) \\ &\vdots\end{align}$$ So it seemed that, for any integer $n$, we have:
$$\frac{n(n+1)}2x^2-x-2=\Big(\left\lceil{\frac n2}\right\rceil x+(-1)^{n+1}\Big)\Big(\left\lceil{n+\frac{(-1)^n}2}\right\rceil x + 2(-1)^n\Big).$$
where the pattern I showed above begins with $n=1$ and ends with $n=6$.
I am not sure how to prove this (assuming it is true). I know that $-2=(-1)^{n+1}\cdot 2(-1)^n$ but I don't know how to prove that: $$\left\lceil{\frac n2}\right\rceil\cdot \left\lceil{n+\frac{(-1)^n}2}\right\rceil=\frac{n(n+1)}2\tag{$\star$}$$ and $$2(-1)^n\left\lceil{\frac n2}\right\rceil+(-1)^{n+1}\left\lceil{n+\frac{(-1)^n}2}\right\rceil=-1$$ which I believe is necessary to prove this conjecture. Since $n$ is an integer, I was thinking of letting $n=\left\lceil\frac k2\right\rceil$ for any real $k$ or something, and I'd assume that $$\left\lceil{\frac n2}\right\rceil=\left\lceil{\frac{\left\lceil{\frac k2}\right\rceil}2}\right\rceil=\left\lceil{\frac k4}\right\rceil$$ but I'm not sure how "round-off arithmetic" works (informally speaking). Any help is appreciated.
Edit:
Thanks to @lone_student's comment, I have shown $(\star)$ to be true for all $n\in\mathbb Z$ by considering $n$ even and odd.
Lemma:$$\left\lceil {\frac nm}\right\rceil=\left\lfloor{\frac{n-1}m+1}\right\rfloor\tag1$$
Here, $m=2$.
Also: $$n-\left\lfloor\frac n2\right\rfloor=\left\lceil\frac n2\right\rceil\tag2$$
Using these, we can show that:
$$\frac{n(n+1)}2=\left\lceil\frac n2\right\rceil\cdot\left\lceil{n+\frac{(-1)^n}2}\right\rceil=\underbrace{\Big(n-\left\lfloor\frac n2\right\rfloor\Big)}_{\text{By } (2)}\cdot\underbrace{\left\lfloor\frac{2n+(-1)^n-1}2+1\right\rfloor}_{\text{By (1)}}$$
When $n=2k\in\mathbb Z$ we have $$k(2k+1)=\big(2k-\left\lfloor k\right\rfloor\big)\lfloor2k+1\rfloor=k(2k+1)$$ since $k$ and $2k+1$ are integers, and, by definition, $\lceil \alpha\rceil =\lfloor \alpha\rfloor = \alpha$ iff $\alpha\in\mathbb Z$.
Similarly, when $n=2k-1$, we have: $$k(2k-1)=\Big(2k-1-\left\lfloor k-\frac 12\right\rfloor\Big)\lfloor 2k-1\rfloor = k(2k-1)$$ since obviously $\left\lfloor k-\frac 12\right\rfloor = k-1$ for $k\in\mathbb Z$.
I believe there was some confusion towards my question: did I mean to factorise the quadratic in terms of $n$ or did I mean to prove specifically the ceiling-function product identity? I did intend to ask a question on the latter subject, but I had falsely assumed that the case-by-case polynomial pattern I showed above could only be represented through the ceiling functions. This was wrong.
| Hint
To get factors, write $$\frac{n (n+1)}{2}x^2 - x- 2 = \frac{n (n+1)}{2} \left(x^2 -\frac{2}{n(n+1)}x - \frac{2}{n} \cdot \frac{2}{n+1} \right) = \frac{n (n+1)}{2} \left(x^2 -(\frac{2}{n} -\frac{2}{n+1})x - \frac{2}{n} \cdot \frac{2}{n+1} \right) = \frac{n (n+1)}{2} \left( (x- \frac2n)(x+ \frac{2}{n+1}) \right)$$
Can you proceed?
| {
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"url": "https://math.stackexchange.com/questions/4232401",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Calculate maximum and minimum when second partial derivative test fail
Calculate the maximum and minimum of function $z = f(x,y) = x^2 - y^2 +2$ subjected to the inequality constraint $D=\{(x,y)| x^2 + \frac{y^2}{4} \leq 1\}$.
My solution:
First form the function
$$
g(x,y) = x^2 + \frac{y^2}{4} - c, ~0 \leq c \leq 1.
$$
Then form the Lagrangian function
$$
L(x,y,\lambda) = x^2 - y^2 + 2 + \lambda\left(x^2 + \frac{y^2}{4} - c\right).
$$
Therefore we have
$$
\left \{
\begin{array}{ll}
L_x' = 2x +2\lambda x = 0\\
L_y' = -2y + \frac{\lambda}{2}y = 0 \\
L_\lambda' = x^2 + \frac{y^2}{4} - c
\end{array} \right.
$$
After solving above equations, so we can got saddle point like $(\varphi(c),\psi(c))$.
*
*when $c = 0$, we have $x = y = 0$.
*when $c \neq 0$, there are two kind of solutions: (1 $x = 0$, we have $y = \pm 2\sqrt{c}$; (2 $y=0$, we have $ x = \pm \sqrt{c}$.
My problem is that i cant use second partial derivative test for judging $(0,\pm 2\sqrt{c})$ and $(\pm \sqrt{c},0)$ are maximum or minimum, obviously $AC-B^2 = 0$. How can i do next? Thanks in advance!
| Clearly, $\nabla f(x,y)=0\iff(x,y)=(0,0)$, but $(0,0)$ is a saddle-point. So, the maximum and the minimum can be attained only at the boundary of $D$.
Let $h(x,y)=x^2+\frac{y^2}4$. Then apply the method of Lagrange multipliers:$$\left\{\begin{array}{l}f_x(x,y)=\lambda h_x(x,y)\\ f_y(x,y)=\lambda h_y(x,y)\\h(x,y)=1,\end{array}\right.$$or$$\left\{\begin{array}{l}2x=2\lambda x\\-2y=\frac12\lambda y\\x^2+\frac{y^2}4=1.\end{array}\right.$$It's solutions are $(x,y,\lambda)=(0,-2,-4)$, $(x,y,\lambda)=(0,2,-4)$, $(x,y,\lambda)=(-1,0,1)$ and $(x,y,\lambda)=(1,0,1)$. Since $f(0,\pm2)=-2$, and $f(\pm1,0)=3$, the maximum is $3$ and the minimum is $-2$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4232806",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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$\int\frac{dx}{\sqrt{\alpha+\beta x+\gamma x^2}}=\frac{1}{\sqrt{-\gamma}}\arccos\bigl(-\frac{\beta+2\gamma x}{\sqrt{q}}\bigr)$? I'm studying Kepler's laws from Classical Mechanics, 2nd ed. Goldstein. In page 95 there is given an indefinite integral
$$\int\frac{dx}{\sqrt{\alpha+\beta x+\gamma x^2}}=\frac{1}{\sqrt{-\gamma}}\arccos\biggl(-\frac{\beta+2\gamma x}{\sqrt{q}}\biggr).$$
However, when I took a look the source given in a book (A Short Table of Integrals), there is the result
$$\int\frac{dx}{\sqrt{\alpha+\beta x+\gamma x^2}}=-\frac{1}{\sqrt{-\gamma}}\arcsin\biggl(\frac{\beta+2\gamma x}{\sqrt{q}}\biggr).$$
Then, I tried relations of $\arcsin$ and $\arccos$, so $\arcsin(x)=\frac{\pi}{2}-\arccos(x)$ and also negative argument $-\arcsin(x)=\arcsin(-x)$, but just ended up to result
$$-\frac{1}{\sqrt{-\gamma}}\arcsin\biggl(\frac{\beta+2\gamma x}{\sqrt{q}}\biggr)=-\frac{1}{\sqrt{-\gamma}}\arccos\biggl(-\frac{\beta+2\gamma x}{\sqrt{q}}\biggr)+\frac{\pi}{2\sqrt{-\gamma}}.$$
So is there something I don't see or understand, or is there just a misprint in the book?
For the clarification, the result is used to solve this equation:
$$\varphi=\varphi_{0}-\int\frac{du}{ \sqrt{\frac{2mE}{l^2}+\frac{2mku}{l^2}-u^2}}$$
and the book ends up to result
$$\varphi=\varphi'-\arccos\left(\frac{\frac{l^2u}{mk}-1}{\sqrt{1+\frac{2El^2}{mk^2}}}\right)$$
and by solving the $u=1/r$ we got final result:
$$\frac{1}{r}=\frac{mk}{l^2}\left(1+\sqrt{1+\frac{2El^2}{mk^2}}\cdot \cos(\varphi-\varphi')\right)$$
| Assuming that $q$ is equal to $\beta ^2-4 \alpha \gamma$ otherwise the integral won't make any sense. In the integral $\int\frac{dx}{\sqrt{\alpha+\beta x+\gamma x^2}}$, the coefficient of $x^2$ plays a major role.
Assumption: $q=\beta^2-4 \alpha \gamma>0$
When $\gamma <0$
$\begin{align} \int\frac{dx}{\sqrt{\alpha+\beta x+\gamma x^2}}&=\int\frac{dx}{\sqrt{- \gamma (-x^2-\frac{\beta}{\gamma}x-\frac{\alpha}{\gamma})}} \\
&=\frac1{\sqrt{-\gamma}} \int \frac{dx}{\sqrt{\frac{\beta^2}{4 \gamma^2}-\frac{\alpha}{\gamma}-\left(x+\frac{\beta}{2\gamma}\right)^2}}
\quad \;(\because -\gamma >0) \\
&=\frac1{\sqrt{-\gamma}} \int \frac{dx}{\sqrt{\frac{\beta^2-4 \alpha \gamma}{4 \gamma^2}-\left(x+\frac{\beta}{2\gamma}\right)^2}}\\
&=\frac1{\sqrt{-\gamma}} \int \frac{dx}{\sqrt{\left(\frac{\sqrt{\beta^2-4 \alpha \gamma}}{2 \gamma}\right)^2-\left(x+\frac{\beta}{2\gamma}\right)^2}} \text{ $\quad \;$ since ${\beta^2-4 \alpha \gamma}>0$} \\
&=\frac1{\sqrt{-\gamma}} \arcsin{\left( \frac{ \frac{2 \gamma x+\beta}{2\gamma}}{\frac{\sqrt{\beta^2-4 \alpha \gamma}}{2 |\gamma|}}\right)}+c \;\quad \left(\because \int \frac{dx}{\sqrt{a^2-x^2}}=\arcsin\left(\frac{x}{\left|a\right|}\right)+C\right)\\
&=\frac1{\sqrt{-\gamma}} \arcsin{\left( -\frac{2 \gamma x+\beta}{\sqrt{\beta^2-4 \alpha \gamma}}\right)}+c \; \quad\; (\because |\gamma|= -\gamma)\\
&=-\frac1{\sqrt{-\gamma}} \arcsin{\left( \frac{2 \gamma x+\beta}{\sqrt{\beta^2-4 \alpha \gamma}}\right)}+c \;\quad(\because \arcsin({-x}) = -\arcsin(x)) \\
&=-\frac{1}{\sqrt{-\gamma}}\arcsin\left(\frac{\beta+2\gamma x}{\sqrt{q}}\right)+c \tag{1}\end{align}$
When $\gamma>0$
$\begin{align} \int\frac{dx}{\sqrt{\alpha+\beta x+\gamma x^2}}&=\int\frac{dx}{\sqrt{\gamma (x^2+\frac{\beta}{\gamma}x+\frac{\alpha}{\gamma})}} \\
&=\frac1{\sqrt{\gamma}} \int \frac{dx}{\sqrt{\left(x+\frac{\beta}{2\gamma}\right)^2-\left(\frac{\sqrt{\beta^2-4 \alpha \gamma}}{2 \gamma}\right)^2}} \text{ $\;\quad$ since ${\beta^2-4 \alpha \gamma}>0$} \\
&= \frac1{\sqrt{\gamma}} \cosh^{-1}{\left( \frac{ \frac{2 \gamma x+\beta}{2\gamma}}{\frac{\sqrt{\beta^2-4 \alpha \gamma}}{2 \gamma}}\right)}+c \; \quad \left(\because \int \frac{dx}{\sqrt{x^2-a^2}}=\cosh^{-1}\left(\frac{x}{a}\right)+C\right)\\
&= \frac1{\sqrt{\gamma}} \cosh^{-1}{\left( \frac{2 \gamma x+\beta}{\sqrt{\beta^2-4 \alpha \gamma}}\right)}+c\\
&=\frac1{\sqrt{\gamma}} \cosh^{-1}{\left( \frac{2 \gamma x+\beta}{\sqrt{ q}}\right)} \tag{2}\end{align}$
Since, in your case, $\gamma<0, \;$ $$\int\frac{dx}{\sqrt{\alpha+\beta x+\gamma x^2}}=\frac{1}{\sqrt{-\gamma}}\arcsin\left(-\frac{\beta+2\gamma x}{\sqrt{q}}\right)+c$$ And this result can be rewritten in terms of $\arccos()$ as $$-\frac{1}{\sqrt{-\gamma}}\arccos\left(-\frac{\beta+2\gamma x}{\sqrt{q}}\right)+c'\tag{3}$$ where $c'=c+{ \pi\over 2 \sqrt{-\gamma}}$
Clearly, $(3)$ and $$\frac{1}{\sqrt{-\gamma}}\arccos\left(-\frac{\beta+2\gamma x}{\sqrt{q}}\right) \tag4$$ don't just differ by a constant.
So, you are right, that is a misprint.
Edit: Proceeding with $(3)$ we get,
$\begin{align} &\int\frac{du}{ \sqrt{\frac{2mE}{l^2}+\frac{2mku}{l^2}-u^2}}=-\arccos\left(\frac{\frac{l^2u}{mk}-1}{\sqrt{1+\frac{2El^2}{mk^2}}}\right)\\
\therefore & \; \:\varphi=\varphi'+\arccos\left(\frac{\frac{l^2u}{mk}-1}{\sqrt{1+\frac{2El^2}{mk^2}}}\right)\\
\implies & \cos(\varphi-\varphi')=\frac{\frac{l^2u}{mk}-1}{\sqrt{1+\frac{2El^2}{mk^2}}}\\
\implies &\frac{1}{r}=u=\frac{mk}{l^2}\left(1+\sqrt{1+\frac{2El^2}{mk^2}}\cdot \cos(\varphi-\varphi')\right)\end{align}$
which is same as the final result.
It can be noticed that it doesn't matter if $\varphi=\varphi'+\arccos\left(\frac{\frac{l^2u}{mk}-1}{\sqrt{1+\frac{2El^2}{mk^2}}}\right)$ or $\varphi=\varphi'-\arccos\left(\frac{\frac{l^2u}{mk}-1}{\sqrt{1+\frac{2El^2}{mk^2}}}\right)$, the final result would be same since $\cos(\varphi-\varphi')=\cos(\varphi'-\varphi)$
| {
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"timestamp": "2023-03-29T00:00:00",
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Prove $\left(\frac{u}{a}\right)^a.\left(\frac{v}{b}\right)^b.\left(\frac{w}{c}\right)^c \le \left(\frac{u+v+w}{a+b+c}\right)^{(a+b+c)} $ Let $u,v,w>0$ and $a,b,c$ are positive constant. Prove that $\left(\frac{u}{a}\right)^a.\left(\frac{v}{b}\right)^b.\left(\frac{w}{c}\right)^c \le \left(\frac{u+v+w}{a+b+c}\right)^{(a+b+c)} $
First, I prove with $x+y+z=1$ so $x^ay^bz^c\le\left(\frac{a}{a+b+c}\right)^a\left(\frac{b}{a+b+c}\right)^b\left(\frac{c}{a+b+c}\right)^c$ by Largrange theorem
And it become $\left(\frac{x}{a}\right)^a\left(\frac{y}{b}\right)^b\left(\frac{z}{c}\right)^c\le\left(\frac{1}{a+b+c}\right)^{a+b+c}=\left(\frac{x+y+z}{a+b+c}\right)^{a+b+c}$
So it true with $x+y+z=1$ but i can't prove it true with $x,y,z>0$. Please help me! Thank you.
| The inequality is homogenenous in $(u, v, w)$: If you have proven
$$
\left(\frac{x}{a}\right)^a\left(\frac{y}{b}\right)^b\left(\frac{z}{c}\right)^c\le\left(\frac{1}{a+b+c}\right)^{a+b+c}
$$
under the condition $x+y+z=1$ then you can use that to prove the general case: With
$$
x = \frac{u}{u+v+w}, y = \frac{v}{u+v+w},z = \frac{w}{u+v+w}
$$
one has $x+y+z=1$ and
$$
\left(\frac{u}{a}\right)^a\left(\frac{v}{b}\right)^b\left(\frac{w}{c}\right)^c = \left(\frac{x}{a}\right)^a\left(\frac{y}{b}\right)^b\left(\frac{z}{c}\right)^c (u+v+w)^{a+b+c} \\
\le \left(\frac{1}{a+b+c}\right)^{a+b+c}(u+v+w)^{a+b+c}
= \left(\frac{u+v+w}{a+b+c}\right)^{a+b+c} \, .
$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Staver's identity relating $\sum_{k=1}^{n}\binom{2k}{k}\frac{1}{k}$ and $\sum_{k=1}^{n}\left(k\binom{n}{k}\right)^{-2}$ I am looking for a proof of the following identity, very relevant for evaluating certain binomial sums $\pmod{p}$:
$$ \sum_{k=1}^{n}\binom{2k}{k}\frac{1}{k} = \frac{2n+1}{3}\binom{2n}{n}\sum_{k=1}^{n}\frac{1}{k^2\binom{n}{k}^2} \tag{1}$$
It is due to Tor B. Staver, Om summasjon av potenser av binomialkoeffisienten, Norsk Mat.
Tidsskrift 29 (1947), but I did not manage to get access to the original article. I have attempted a few things, so far unsuccessfully:
A) Exploit
$$ \sum_{k=0}^{n}\frac{\lambda^k}{\binom{n}{k}}=(n+1)\sum_{r=0}^{n}\frac{\lambda^{n+1}+\lambda^{n-r}}{(r+1)(1+\lambda)^{n-r+1}}\tag{A}$$
which holds for any $\lambda > -1$, as shown by Bala Sury through Euler's Beta function.
B) Exploit a partial fraction decomposition of the reciprocal squared binomial coefficient, namely
$$ \frac{1}{k^2\binom{n}{k}^2} = \sum_{r=0}^{k-1}\frac{\binom{k-1}{r}^2}{(n-r)^2}+2\sum_{r=0}^{k-1}\frac{\binom{k-1}{r}^2}{(n-r)}(H_{k-1-r}-H_{r-1}) \tag{B}$$
C) Exploit the Beta function and symmetry in order to get
$$ \sum_{k=1}^{n}\frac{1}{k^2\binom{n}{k}^2}=\frac{1}{4^n n!^2}\iint_{(-1,1)^2}\frac{(1+x+y+xy)^{n+1}-(1+xy)^{n+1}}{x+y}\,dx\,dy \tag{C}$$
D) Apply Wilf's snake oil method, proving that the LHS and the RHS of (1) have the same OGF. Quite clearly
$$ \sum_{k=1}^{n}\binom{2k}{k}\frac{1}{k} = 2[z^n]\frac{\log(2)-\log(1+\sqrt{1-4z})}{1-z}.\tag{D}$$
E) I am also able to show through the Beta function that
$$\begin{eqnarray*} \sum_{r=0}^{n}\frac{1}{\binom{n}{r}^2}&=&2\frac{(n+1)^2}{(n+2)}\sum_{r=0}^{n}\frac{1}{(n-r+1)\binom{n+r+2}{r}}\\&=&2\frac{(n+1)^2}{(n+2)}[z^{n+1}]\left(-\log(1-z)\cdot\phantom{}_2 F_1\left(1,1;n+3;x\right)\right)\\&=&
2(n+1)^2 [z^{2n+3}](1-z)^{n+1}\left(\log(1-z)-\log^2(1-z)\right)\tag{E}\end{eqnarray*}$$
| Denote $$F(k,n) = \frac{1}{k^2 \binom{n}{k}^2} \qquad G(k,n) = \binom{2k}{k}\frac{1}{k} \frac{3}{2n+1}\binom{2n}{n}^{-1}$$
$$S(n) = \sum_{k=1}^n F(n,k) \qquad T(n) = \sum_{k=1}^n G(n,k)$$
let's show $S=T$ by showing they satisfy same recurrence (plus some trivial checks on initial conditions).
Recurrence for $T$ is easy. Note that
$$(-n-1) G(k,n)+2 (2 n+3) G(k,1+n) = 0$$
summing this over $k$ from $1$ to $n$ gives $$(-n-1) T(n)+2 (2n+3) (T(1+n)-G(n+1,n+1)) = 0$$
that is $$(-n-1)T(n) + 2(2n+3) T(n+1) = \frac{6}{n+1}$$
On the other hand, one checks, with $R(k,n) = \frac{(2 k-3 n-5) (-k+n+1)^2}{(n+1)^2}$,
$$\tag{*}(-n-1) F(k,n)+2 (2 n+3) F(k,1+n)=F(k+1,n) R(k+1,n) -F(k,n) R(k,n)$$
note that RHS telescopes in $k$. Summing over $k$ from $1$ to $n-1$,
$$(-n-1)(S(n)-\frac{1}{n^2}) + 2(2n+3) (S(n+1)-\frac{1}{(n+1)^2} - \frac{1}{n^2(n+1)^2}) = F(n,n)R(n,n)-F(1,n)R(1,n)$$
which rearranges into
$$(-n-1)S(n) + 2(2n+3) S(n+1) = \frac{6}{n+1}$$
QED.
| {
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"timestamp": "2023-03-29T00:00:00",
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Let $x^3+\frac{1}{x^3}$ and $x^4+\frac{1}{x^4}$ are rational numbers. Show that $x+\frac{1}{x}$ is rational.
$x^3+\dfrac{1}{x^3}$ and $x^4+\dfrac{1}{x^4}$ are rational numbers. Prove that $x+\dfrac{1}{x}$ is rational number.
My solution:
$x^3+\dfrac{1}{x^3}=\left(x+\dfrac{1}{x}\right)\left(x^2+\dfrac{1}{x^2}-1\right)$
$x^4+\dfrac{1}{x^4}=x^4+2+\dfrac{1}{x^4}-2=\left(x^2+\dfrac{1}{x^2}\right)^2-1-1=\left(x^2+\dfrac{1}{x^2}-1\right)\left(x^2+\dfrac{1}{x^2}+1\right)-1$
Because $x^4+\dfrac{1}{x^4}$ is rational number so
$\left(x^2+\dfrac{1}{x^2}-1\right)$ is rational number too and
$x^3+\dfrac{1}{x^3}$ is rational number so
$\left(x+\dfrac{1}{x}\right)$ is rational number.
Am I wrong? Please check my solution, thank you.
| With $\alpha= x+\frac{1}{x}$ we get
$$x^3 + \frac{1}{x^3} =\alpha^3 - 3 \alpha = u\\
x^4 + \frac{1}{x^4} = \alpha^4 - 4 \alpha^2 + 2 = v$$
The polynomials $t^3 - 3 t - u$ and $t^4 - 4 t^2 + 2 - v$ have a common root $\alpha$. If that were the only common root, then their gcd $t-\alpha$ would have rational coefficients. So assume that they have another common root $\beta \ne \alpha$. Now we get the system
$$(\alpha^3 - 3 \alpha) -(\beta^3 - 3 \beta) = 0\\
(\alpha^4 - 4 \alpha^2) -(\beta^4 - 4 \beta^2) = 0$$
that is
$$(\alpha - \beta) (\alpha^2 + \alpha \beta + \beta^2 - 3)=0 \\
(\alpha-\beta)(\alpha+ \beta) (\alpha^2 + \beta^2 - 4)=0$$
and since $\alpha- \beta \ne 0$ we get finitely many solutions
$\pm (\sqrt{3}, -\sqrt{3}), \pm(\sqrt{2-\sqrt{3}}, -\sqrt{2-\sqrt{3}}(2+\sqrt{3}) )$ and its conjugate. For neither of these $\alpha$'s do we get both rational values for $x^3 + 1/x^3$, $x^4 + 1/x^4$.
| {
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Proving divisibility by $9$ for $10^n + 3 \times 4^{n+2} + 5$. I am trying to prove that for all $n$, $9$ divides $10^n + 3 \times 4^{n+2} + 5$. I first tried induction on $n$, but couldn't get anywhere with the induction step. I then tried to use modular arithmetic. The furthest I could get was:
As $10 \equiv 1 \mod 9$, $10^n \equiv 1^n = 1$ for all $n$, so modulo $9$, we have
\begin{align*}
10^n + 3 \cdot 4^{n+2} + 5 & = 3 \cdot 4^{n+2} + 6 \\
& = 3\left(4^{n+2} + 2\right) \\
& = 3\left(\left(2^2\right)^{n+2} + 2 \right) \\
& = 3\left(2^{2(n+2)} + 2 \right) \\
& = 3\left(2\left(2^{2(n+2) - 1} + 1\right) \right)
\end{align*}
I need to something get a factor of $3$ to show that the entire expression is divisible by $9$ and hence equal to $0$, mod $9$. But, with a sum of even terms, this does not appear possible.
Any hints on how to proceed would be appreciated. Is induction the standard way to prove something like this?
| $$10^n\equiv 1^n\equiv 1\pmod 9$$
$$(3m+1)^t\equiv 1\pmod 3\implies 3(3m+1)^t\equiv 3\pmod 9$$ which then lowers everything to: $$1+3+5\equiv 0\pmod 9$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Why does Solving system of quadratic equations gives extra roots? Consider these system of Equations
\begin{align*}
\begin{cases}
x^2+4x+4=0\\\\
x^2+5x+6=0
\end{cases}
\end{align*}
For solving them
We have
Method 1-
Subtract both equations
So
$-x-2=0$
Hence,
$x=-2$
Method-2
Add both equations
$2x^2+9x+10=0$
After applying quadratic formula,we get
$x=-2$ or $x=-5/2$. But only $x=-2$ satisfies the system of equation.
Why is the $-5/2$ not satisying the system of equations,what is intuition behind the error in method 2?
| We could also describe the difference in your two approaches in this way. You began with the system $ \ p(x) \ = \ x^2 + 4x + 4 \ = \ 0 \ , \ q(x) \ = \ x^2 + 5x + 6 \ = \ 0 \ \ . $ When you subtract one equation from the other, you have $ \ p(x) - q(x) \ = \ 0 \ \ , $ which is equivalent to the equation we would set up for finding intersections of the curves represented by these functions, $ \ p(x) \ = \ q(x) \ \ . $ You found the single solution $ \ x \ = \ -2 \ \ $ from $ \ -x - 2 \ = \ 0 \ \ , $ which is correct. It so happens for this system that this also locates a common factor of $ \ p(x) \ $ and $ \ q(x) \ $ because both functions are equal to zero at $ \ x \ = \ -2 \ \ , $ but this method would be correct to use in any case (as we'll show shortly for a different system).
When you add the two equations, as you did in your second calculation, you are now solving $ \ p(x) + q(x) \ = \ 0 \ \rightarrow \ p(x) \ = \ -q(x) \ \ , $ which is no longer the original problem. Here, it really is only because both $ \ p(x) \ $ and $ \ q(x) \ $ are equal to zero at $ \ x \ = \ -2 \ $ that this appears as a solution to $ \ 2x^2 + 9x + 10 \ = \ (2x + 5 )·(x + 2) \ = \ 0 \ \ , $ since $ \ (x + 2)·(x + 2) \ + \ (x + 2)·(x + 3) $
$ = \ (x + 2)·[ \ (x + 2) + (x + 3) \ ] \ \ . $ Other systems may not produce any solutions at all by adding the equations.
If we take, for example, the system $ \ p(x) \ = \ x^2 - 5x + 6 \ = \ ( x - 2)·(x - 3) \ = \ 0 \ , $
$ q(x) \ = \ x^2 + 7x + 10 \ = \ (x + 2)·(x + 5) \ = 0 \ \ , $ the two polynomials have no factors in common, but surely must intersect since they both "open upward". We find $ \ p(x) \ = \ q(x) $ $ \Rightarrow \ p(x) - q(x) \ = \ -12x - 4 \ = \ 0 \ \Rightarrow \ x \ = \ -\frac13 \ \ , $ a solution which is not suggested immediately by any of the polynomial factors. [Indeed, we could have chosen two "upward-opening" parabolas represented by polynomials which cannot be factored using real numbers and would still be able to find the intersection(s).]
On the other hand, $ \ p(x) + q(x) \ = \ 2x^2 + 2x + 16 \ \ $ has no real zeroes (or $ \ p(x) \ = \ -q(x) \ $ has no real-number solutions); we see that the function curves do not intersect. So adding the equations together in this system provides no information about the solutions of the original system of equations.
| {
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "3",
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Simple way to compute the finite sum $\sum\limits_{k=1}^{n-1}k\cdot x^k$ I'm looking for an elementary method for computing a finite geometric-like sum,
$$\sum_{k=1}^{n-1} k\cdot3^k$$
I have a calculus-based solution: As a more general result, I replace $3$ with $x$ and denote the sum by $f(x)$. Then
$$f(x) = \sum_{k=1}^{n-1} k\cdot x^k = x\sum_{k=1}^{n-1}k\cdot x^{k-1} = x\frac{\mathrm d}{\mathrm dx}\left[\sum_{k=1}^{n-1}x^k + C\right]$$
for some constant $C$. I know that
$$\sum_{k=1}^{n-1}x^k = \frac{x-x^n}{1-x}$$
so it follows that
$$\begin{align}
f(x) &= x\frac{\mathrm d}{\mathrm dx}\left[\frac{x-x^n}{1-x}+C\right] \\[1ex]
&= x\cdot\frac{(1-x)\left(1-nx^{n-1}\right) + x-x^n}{(1-x)^2} \\[1ex]
&= \frac{x-nx^n(1-x)-x^{n+1}}{(1-x)^2}
\end{align}$$
which means the sum I started with is
$$\sum_{k=1}^{n-1}k\cdot3^k = \frac{3+2n\cdot3^n-3^{n+1}}4$$
I am aware of but not particularly adept with summation by parts, and I was wondering if there was another perhaps simpler method I can employ to get the same result?
| Hint:
Take $S = 3 + 2 \cdot 3^2 + \ldots + (n-1)3^{n-1}$ then $3S = 3^2 + 2\cdot 3^3 + \ldots + (n-2)3^{n-1} + (n-1)3^n$
Then $3S - S = -3 + ( 3^2 + 3^3 + \ldots + 3^{n-1}) + (n-1) 3^n$
The expression in bracket is geometric series.
| {
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Number of solutions to $x+y+z=x^2+y^2+z^2=x^3+y^3+z^3=3$
Determine all the roots, real or complex, of the system of simultaneous equations (USAMO 1973/4)
$$x+y+z=3$$
$$x^2+y^2+z^2=3$$
$$x^3+y^3+z^3=3$$
Using exactly the same approach as this answer here:
$$x+y+z=3$$
$$xy+yz+zx=3$$
$$xyz=1$$
And we then conclude that these are the roots of a cubic polynomial. This restricts the maximum number of solutions to $3$.
How do we know or show that the number of solutions is maximum $3$?
| The fundamental theorem of algebra states that any polynomial will have the same number of roots as the degree of the polynomial (counting repeated roots).
Looking at the last set of equations, we can use Viète's formulas to find the cubic which solves the original set of equations.
$x + y + z = \frac{-b}{a}$, $xy + xz + yz = \frac{c}{a}$, $xyz = \frac{-d}{a}$, therefore plugging in the values from the last set of equations gives you: $3a = -b, 3a = c, a = -d$.
Knowing that the general form of a cubic is $ax^3 + bx^2 + cx + d$ and using the relations, the cubic used to solve the last set of equations is $ax^3-3ax^2+3a-a=0$. Since $a \neq 0$, $x^3 - 3x^2 + 3x - 1 = 0$.
Therefore, the cubic is guaranteed to have $3$ roots (including repeated roots) because of the fundamental theorem of algebra and hence, the last set of equations.
| {
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Why is this approach wrong?
If $\alpha$ and $\beta$ are the roots of $x^{2} - 4 x - 3$, then find $$\frac{1}{(\alpha-4)^{2}}+\frac{1}{(\beta-4)^{2}}$$
Solution
This is the approach that gives wrong answer.
\begin{array}{l}
x^{2}-4 x-3=0 \\
(x-4)^{2}=19 \\
So, {(\alpha-4)^{2}} {\&(\beta-4)^{2}}=19 \\
\text { Hence, } \frac{1}{(\alpha-4)^{2}}+\frac{1}{(\beta-4)^{2}}=\frac{2}{19}
\end{array}
This is how it is solved :
\begin{array}{l}
x^{2}-4 x-3=0 \\
\alpha^{2}-4 \alpha-3=0 \\
\alpha(\alpha-4)=3 \\
\alpha-4=\frac{3}{\alpha} \\
\& \beta-4=\frac{3}{\beta}
\end{array}
So , $$
\frac{1}{(\alpha-4)^{2}}+\frac{1}{(\beta-4)^{2}}
$$
\begin{array}{l}
=\frac{\alpha^{2}}{9}+\frac{\beta^{2}}{9} \\
=\frac{1}{9}\left(\alpha^{2}+\beta^{2}\right) \\
=\frac{22}{9}
\end{array}
Please tell me why the first approach is wrong .
| Another approach would be to go ahead and perform the ratio addition (but don't fully multiply out the binomial-squares in the denominator):
$$ \frac{1}{(\alpha-4)^{2}}+\frac{1}{(\beta-4)^{2}} \ \ = \ \ \frac{(\beta-4)^{2} \ + \ (\alpha-4)^{2}}{(\alpha-4)^{2}· (\beta-4)^{2}} \ \ = \ \ \frac{\alpha^2 \ + \ \beta^2 \ - \ 8·(\alpha + \beta) \ + \ 32}{[ \ \alpha · \beta \ - \ 4·(\alpha + \beta) \ + 16 \ ]^2} $$
$ [ \ \alpha^2 + \beta^2 \ = \ (\alpha + \beta)^2 \ - \ 2·\alpha·\beta \ ] $
$$ = \ \ \frac{(\alpha \ + \ \beta)^2 \ - \ 2·\alpha·\beta \ - \ 8·(\alpha + \beta) \ + \ 32}{[ \ \alpha · \beta \ - \ 4·(\alpha + \beta) \ + 16 \ ]^2} $$
[we then apply Viete's relations: $ \ \alpha + \beta \ = \ -(-4) \ = \ 4 \ \ , \ \ \alpha·\beta \ = \ -3 \ ] $
$$ = \ \ \frac{4^2 \ - \ 2·(-3) \ - \ 8·(4) \ + \ 32}{[ \ (-3) \ - \ 4·4 \ + 16 \ ]^2} \ \ = \ \ \frac{16 \ + \ 6 \ - \ 32 \ + \ 32}{[ \ (-3) \ - \ 16 \ + 16 \ ]^2} \ \ = \ \ \frac{16 \ + \ 6 }{(-3)^2} \ \ = \ \ \frac{22}{9} \ \ . $$
This method also allows us to proceed without determining the zeroes of the polynomial themselves.
| {
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inequality for positive real numbers Given $$a,b,c,x,y,z \in R^+$$ how to show the following inequality:
$$\frac{a^3}{x} + \frac{b^3}{y} + \frac{c^3}{z} \geq \frac{(a+b+c)^3}{3(x + y + z)}$$
I rearranged the inequality since all are positive, the inequality would be true iff
$$\frac{a^3}{x(a+b+c)^3}+\frac{b^3}{y(a+b+c)^3}+\frac{c^3}{z(a+b+c)^3}\geq\frac{1}{3(x+y+z)}$$
which further simplifies to
$$\frac{a^3yz}{(a+b+c)^3}+\frac{b^3xz}{(a+b+c)^3}+\frac{c^3xy}{(a+b+c)^3}\geq\frac{xyz}{3(x+y+z)}$$
and now I am struggling how to rearrange further to apply Holding's inequality
| By Holder $$\frac{a^3}{x} + \frac{b^3}{y} + \frac{c^3}{z}=\frac{(1+1+1)\left(\frac{a^3}{x} + \frac{b^3}{y} + \frac{c^3}{z}\right)(x+y+z)}{3(x+y+z)}\geq\frac{(a+b+c)^3}{3(x+y+z)}.$$
| {
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Prove by contradiction that if $n^3$ is a multiple of $3$, then $n$ is a multiple of $3$ Problem statement:
Using proof by contradiction, prove that if $n^3$ is a multiple of $3$ , then $n$ is a multiple of $3.$
Attempt 1:
Assume that there is exist $n$ which is a multiple of $3$ such that $n^3$ is not a multiple of $3.$
Then $n = 3k $ , $n^3 = 27 k^3 $ which a multiple of $3$, which contradicts the assumption that $n^3$ is a multiple of $3.$
Attempt 2:
Assume that there exist $n^3$ which is a multiple of $3$ such that $n$ is not a multiple of $3.$
Then $n = k+ 1 $ or $n= k+2.$
Then $n^3 = k^3 + 3k^2 + 3k + 1$
or $n^3 = k^3 + 6 k^2 + 12 k + 8.$
Both cases contradict the assumption that $n^3 $ is a multiple of $3.$
Which solution is correct ? Or do both work ?
| given statement: “If $n^3$ is a multiple of $3,$ then $n$ is a multiple of $3.$”
*
*Your Attempt 1 is wrong, because it's attempting to prove the
converse of the given statement.
*Attempt 2 (“Assume (to the contrary) that there exists $n^3$ which
is a multiple of $3$ such that $n$ is not a multiple of $3$”) does start correctly. However, the correct translation of <$n^3$ is a multiple of $3$> ought to be <$n^3=3k+1\,$ or $\,3k+2$> instead.
*Alternatively and equivalently, you could also start with “Suppose
that $n^3$ is a multiple of $3,$ and assume (to the contrary) that
that $n$ is not a multiple of $3.$”
| {
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Subsets and Splits