Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
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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... | 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",
"question_score": "2",
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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 + 3... | 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 unt... | {
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"timestamp": "2023-03-29T00:00:00",
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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+... | 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(\dfra... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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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... | {
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "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$ imp... | 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$
$... | {
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "2",
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"answer_id": 2
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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... | 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}}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4086523",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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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... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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"answer_id": 4
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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 $$x_{1,2}=\dfrac{a^2+b^2\pm\sqrt{(a^2... | The polynomial $ \ Cx^2 + Bx + A \ $ is referred to as the reciprocal polynomial to $ \ Ax^2 + Bx + C \ \ , $ as its zeroes are the reciprocals of the latter polynomial. In the same vein, a polynomial $ \ Ax^2 + Bx + A \ $ is called palindromic and has the property that if $ \ r \ $ is a zero, then $ \ \frac{1}{r} \ ... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Conditions on coefficients obtained when solving two equations A question defines $$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$$ You're required to find a and b.
Here, they've solved it by taking the $e^x$ in the second equation to the left side and diffe... | Going by your method, we get,
\begin{align}
&f'(x) = e^x(ax+b+a)\\
\implies &f(x)=\int_0^xe^{x-y}e^y(ay+b+a)dy-(x^2-x+1)e^x\\
\implies &f(x) = e^x\int_0^x(ay+b+a)dy-(x^2-x+1)e^x\\
\implies &f(x) = e^x\left(\dfrac{ax^2}{2}+(b+a)x\right)-(x^2-x+1)e^x\\
\implies &(ax+b)e^x = e^x\left(\left(\dfrac{a}{2}-1\right)x^2+(b+a+1)... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4093973",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Coefficient of $x^{12}$ in $(1+x^2+x^4+x^6)^n$ I need to find the coefficient of $x^{12}$ in the polynomial $(1+x^2+x^4+x^6)^n$.
I have reduced the polynomial to $\left(\frac{1-x^8}{1-x^2}\right)^ n$ and tried binomial expansion and Taylor series, yet it seems too complicated to be worked out by hand.
What should I do?... | Here is just a standard trick from generating functions using
$$\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}$$
Hence, for $n\geq 1$ you get
\begin{eqnarray*}[y^6]\frac{(1-y^4)^n}{(1-y)^n}
& = & [y^6]\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4095701",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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What are the possible real values of $\frac{1}{x} + \frac{1}{y}$ given $x^3 +y^3 +3x^2y^2 = x^3y^3$? Let $x^3 +y^3 +3x^2y^2 = x^3y^3$ for $x$ and $y$ real numbers different from $0$.
Then determine all possible values of $\frac{1}{x} + \frac{1}{y}$
I tried to factor this polynomial but there's no a clear factors
| The equation is equivalent to
$$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}$$
If $(2)$ holds tr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4096184",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 1
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Multiplication and division by an integer in systems of linear congruences I have the following system of linear congruences:
\begin{align}
x&\equiv -2 \pmod {6}\\
x&\equiv 2 \pmod {10}\\
x&\equiv 7 \pmod {15}
\end{align}
From the first equation we have: $ x= 6 k-2, k\in \mathbb{Z}$. Then we insert it into 2nd equation... | Another approach:
$\begin{cases} x\equiv -2 \bmod 6\\x=2 \bmod 10\end{cases}$
$\rightarrow 6t_1-2=10t_2 +2\rightarrow 6t_1-10t_2=4 \rightarrow t_1=-10 s-1, t_2=6s -1$
Which gives:
$t_1=-10 s-1\rightarrow x=6(-10s-1)-2=-60 s -8$
$t_2=6s-1 \rightarrow x=10(6s-1)+2=60s -8$
$\begin{cases} x\equiv 7 \bmod 15\\x=2 \bmod 10\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4096768",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Proving global minimum by lower bound of 2-variable function $f(x,y)=x^4+2x^2y+y^2-4x^2-8x-8y$ I would like to prove that the following function $f :\mathbb{R}^2\to\mathbb{R}$ has a global minimum:
$f(x,y)=x^4+2x^2y+y^2-4x^2-8x-8y=(x^2+y)^2-4(x^2+2x+2y)$
$f$ has strict local minimum at
$f(1,3)=-20$
I think that what ... | My favorite way,
$$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(... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4097017",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 0
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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.
The... | 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 coordi... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 2,
"answer_id": 0
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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|. ... | 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 \t... | {
"language": "en",
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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:
$$\begin{align}
&\>\>\>\>\>\frac{3x^4+2x^2+4x+2}{(1-2x)^3}\\&=\frac{A}{(1-2x)}+\frac{B}{(1-2x)^2}+\frac{C}{(1-2x)^3}\\[1ex]... | If you want to use partial fractions for a solution to something, then you first have to check the numerator and denominator's degrees (the degree of a polynomial is the highest power in it, eg the degree of $x^3+5x-4$ is $3$).
*
*If the degree of the numerator is $\ge$ the degree of the denominator, then before usin... | {
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"timestamp": "2023-03-29T00:00:00",
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If $\left(1+x^{2}+x^{4}\right)^{8}={^{16} C_{0}}+{^{16}C_{1}} x^{2}+{^{16} C_{2}} x^{4}+\cdots +{^{16} C_{16}} x^{32}$, how to prove... If I am given that $$\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} ... | 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
$$ a = {}^{16}C_0 +{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4104489",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Uniform convergence of $\sum_{n = 1}^\infty x \sin \frac{1}{(x n)^2}$ on $(0, 1)$. Functional sequence $x \sin \frac{1}{(x n)^2} {}^\longrightarrow_\longrightarrow 0$ because
$$
\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.
$... | 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| < \eps... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4104912",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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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 ... | 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... | {
"language": "en",
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"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?
$$\sqrt{2020}(\sqrt{a} + \sqrt{b} )= \sqrt{ (c+ 2020)(d+ 2020)}$$
$$\sqrt{2020}(\sqrt{b} + \sqrt{c} )=... | Applying the Cauchy–Schwarz inequality to $(u_1,v_1,u_2,v_2) = (\sqrt{c},\sqrt{2020},\sqrt{2020},\sqrt{d})$ to the right hand side of the first equation
$$\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}$$
Do the ... | {
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"url": "https://math.stackexchange.com/questions/4107950",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Calculus 2 test question that even the professor could not solve (Parametric Arc length) everyone got this question wrong from my calculus 3 midterm.
" find the arc length of the curve on the given interval: $x=\sqrt(t), \space y=8t-6,\space on\space 0 \leq t \leq 3$"
I set the problem up just fine, however at that po... | There is a way to perform the evaluation without resorting to trigonometric substitution, although it is hardly ideal. Note that with the substitution $$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 ... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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"answer_id": 0
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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}... | {
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"url": "https://math.stackexchange.com/questions/4113433",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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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
$... | 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
$$x_1^{(n)}=2 \pi ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4119198",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Let $\int_0^2 f\left(x\right) dx = a+\frac{b}{\log 2}$. Find $a,b$ Let $f$ be a real-valued continuous function on $\mathbb{R}$ such that $2^{f\left(x\right)}+f\left(x\right)=x+1$ for all $x\in \mathbb{R}$. Assume that $\int_0^2 f\left(x\right) dx = a+\dfrac{b}{\log 2}$ with $a,b$ are rational numbers . Find $a,b$.
I ... | Let $\alpha = \log 2$, differenital $x + 1 = 2^f + f$ on both sides, we get
$$\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{alig... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4127503",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
} |
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
$$\mathcal{O}((x - \... | Expanding only the terms to the third order, $$\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).$$
Check:
Using the c... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4129449",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
How can I prove that $\sum_{n=1}^\infty \dfrac{n^4}{2^n} = 150 $? I can easily prove that this series converges, but I can't imagine a way to prove the statement above. I tried with the techniques for finding the sum of $\sum_{n=1}^\infty \frac{n}{2^n}$, but I didn't find an useful connection with this case. Could some... | We have that
$$\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... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4132076",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 1
} |
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 :
$$\begin{align}\mathcal{I}& =\displaystyle\int_{-a}^a \dfrac{\cos(\pi n x/a)}{x^2+a^2}\mathrm dx \\ & = 2\displaystyle\int_0^a \dfrac{\cos(\pi n x/a)}{x^2+... | I do not know if it helps, but...
$$\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 (... | {
"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 $\li... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4134718",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 6,
"answer_id": 1
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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 th... | 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... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4135773",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
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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 f... | 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 ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4137316",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
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 $$\frac{a^2}{ab+ac^2}+\frac{b^2}{bc+ba^2}+\frac{c^2}{ca+cb... | By C-S and by the Vasc's inequality we obtain:$$\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\s... | {
"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... | 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... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4152868",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
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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... | 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, \ta... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4153846",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
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",
"answer_count": 6,
"answer_id": 1
} |
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:
$$A =\exp\left(\lim_{x\to\infty}(2x-5)\ln\frac{x^2-3x+1}{x^2+x+2}\rig... | One of favorite way to see this so quickly as follows
$$\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)} $$
So, since $\lim_{x\to\inf... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4162919",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 1
} |
Function related polynomial problem
My approach is as follow $f\left( x \right) = 2{x^4} + 3{x^3} - 3{x^2} - 6x + a$
$f'\left( x \right) = 8{x^3} + 9{x^2} - 6x - 6$
$f''\left( x \right) = 24{x^2} + 18x - 6 = 6\left( {4{x^2} + 3x - 1} \right)$
The roots of $f''(x)$ are $-1$ and $\frac{1}{4}$, how do I arrive at the pro... | It's easy to show that there are no other roots on the $(1,2)$ besides $y$. Indeed, from the positive sign of the second derivative, we see that $f'$ is increasing on the interval $(1,2)$. Since $f'(1) > 0$, we obtain that $f$ is an increasing function that's why it can have only one root on the $(1,2)$, namely $y$.
Fr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4165829",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
If $f(x) = \frac{x^3}{3} -\frac{x^2}{2} + x + \frac{1}{12}$, then $\int_{\frac{1}{7}}^{\frac{6}{7}}f(f(x))\,dx =\,$? This is a question from a practice workbook for a college entrance exam.
Let
$$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.$$
While I kn... | We know,
$ \displaystyle \int_{a}^{1-a}f(f(x))\,dx = \int_{a}^{1-a}f(f(1-x))\,dx$.
So,
$\displaystyle \int_{a}^{1-a}f(f(x))\,dx = \frac{1}{2} \int_a^{1-a}\left[f(f(x))+f(f(1-x)) \right] \ dx$
Now for the given function, observe that $f(x) + f(1-x) = 1 \implies f(1-x) = 1 - f(x)$
So, $f(f(x)) + f(f(1-x)) = f(f(x)) + f(1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4168939",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "20",
"answer_count": 3,
"answer_id": 0
} |
Can I do this rearrangement with an infinite series to arrive at divergent parts? $$\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} - \f... | 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\righ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4170727",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
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 $$\frac{a^3+a^2b+ab^2+b^3+2a-... | Your approach works (It seems that you have an error in your approach).
Letting $z=a+bi$ where $a,b$ are real numbers, we have
$$\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
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4172279",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
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 gettin... | 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 ro... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4172715",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
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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, ... | 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(... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4175239",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
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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 ei... | 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... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4175558",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
} |
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*}
\... | 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$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4177770",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 0
} |
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... | $$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}$$
Let $a=\frac{p}{q}$ with p,q are integers, hence ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4180314",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 5,
"answer_id": 3
} |
Comparing numbers of the form $c+\sqrt{b}$ (eg, $3+3\sqrt{3}$ and $4+2\sqrt{5}$) without a calculator It is easy to compare to numbers of the form $a\sqrt{b}$, simply by comparing their squares, for example $3\sqrt{3}$ and $2\sqrt{5}$.
But what if we have $a=3+3\sqrt{3}$ and $b=4+2\sqrt{5}$ for example?
How to compare ... | You can use binomial theorem result
$(1\pm x)^n\approx (1\pm nx)$ provided that $x<<1$
Therefore you have $3+3\sqrt3=3+3(4-1)^{\frac{1}{2}}\approx3+6(1-0.25)^{\frac{1}{2}}= 3+6(1-0.125)=9-0.75=8.25$
Similarly $4+2\sqrt5=4+2(4+1)^{\frac{1}{2}}\approx4+4(1+0.25)^{\frac{1}{2}}= 4+4(1+0.125)=8+0.5=8.5$
This would work bet... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4180606",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 2
} |
Let $n$ and $m$ be integers such that $5$ divides $1+2n^2+3m^2.$ Then show that $5$ divides $n^2-1.$
Let $n$ and $m$ be integers such that $5$ divides $1+2n^2+3m^2.$ Then show that $5$ divides $n^2-1.$
$\textbf{My attempts} :$
From the condition, we can write $$1+2n^2+3m^2\equiv 0(\mod 5)\tag 1$$
Now, since $5$ is pr... | If $1+2n^2+3m^2\equiv 0\pmod 5$, then $n^2\equiv m^2+2$.
Since the only squares modulo $5$ are $0,\pm 1$, the only values for $n^2$ and $m^2$ are $n^2\equiv 1\pmod 5$ and $m^2\equiv -1\pmod 5$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4183071",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
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Finding the sum of the cubes of the first n odd natural numbers. I know that the sum of the cubes of the first $n$ natural numbers is $({\frac{n(n+1)}{2}})^2$, which in expanded form is $(1)^3+(2)^3+(3)^3+...+(n)^3$. Multiplying by $2^3$ on both sides gives $2(n(n+1))^2$, which I can see is the sum of cube of first $n$... | Use a specific example.
Let $A = 1^3 + 3^3 + 5^3 + 7^3 + 9^3.$
Let $B = 2^3 + 4^3 + 6^3 + 8^3.$
Let $C = 1^3 + 2^3 + 3^3 + 4^3.$
Assume that you want to evaluate $A$.
You can use the formula to evaluate $(A + B)$, so the problem is reduced to evaluating $B$.
This can be done by using the formula to evaluate $C$, and th... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4184291",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Find constant in system of function equations
Question: Let $a \in \mathbb{R}$. Find all possible values of $a$ such that there exists a function $f: \mathbb{R} \to \mathbb{R}$ satisfying $f(x + 2) = -f(x)$ and $f(x + 3) = f(x) + a$ for any real number $x \in \mathbb{R}$.
My (maybe wrong) solution:
Start listing down... | This makes sense. Here is a different approach leading to the same place. Note that using the first relationship,
$$
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,
$$
so we have $f(x)+2a = -f(x) \iff f(x) = -a$, so $f$ is constant.
Now from the first relationsh... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4186151",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
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$a \geq b \geq c \geq 0$ and $a + b + c \leq 1$. Prove that $a^2 + 3b^2 + 5c^2 \leq 1$. Positive numbers $a,b,c$ satisfy $a \geq b \geq c$ and $a + b + c \leq 1$. Let $f(a, b, c) = a^2 + 3b^2 + 5c^2$. Prove that $f(a, b, c) \leq 1$.
One observation is that the bound is met: $f(1, 0, 0) = f\left(\frac{1}{2}, \frac{1}{2}... | Let $ c = x$, $b = x+y$, $a = x+y+z$ where $ x, y, z \geq 0$.
Then, the question becomes:
Given non-negative $x, y,z$ such that $ 3x+2y+z \leq 1$, show that $(x+y+z)^2 + 3 (x+y)^2 + 5 x^2 \leq 1$.
This is true because
$$(x+y+z)^2 + 3 (x+y)^2 + 5 x^2 = (3x+2y+z)^2 - (4xy+4xz+2yz) \leq 1^2 $$
We have equality iff $3x... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4186722",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 3,
"answer_id": 1
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What is the minimum value of $8 \cos^2 x + 18 \sec^2 x$? As per me the answer should be $26$.
But when we apply AM-GM inequality it gives $24$ as the least value but as per the graph 24 can never come.
What I think is that in AM-GM, it gives $8 \cos^2 x = 18 \sec^2 x$ which gives $\cos x > 1$ which is not possible and ... | $$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$$ at $x=0$
And min of $10\sec^2(x)$ is $10$ at $x=0$. So, the minimum of the net function is $26$ at $x=... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4186848",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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$2^\sqrt{10}$ vs $3^2$ Is there a neat way to show that $2^\sqrt{10} < 3^2$?
I have tried raising to larger powers, like $(2^\sqrt{10})^{100}$ vs $3^{200}$ but the problem is the two functions $2^{x\sqrt{10}}$ and $3^{2x}$ are almost equivalent, and there is no point (that I can find) where one function is "obviously" ... | It follows from the maclaurian series of the function $\sqrt{k^2+h}$ for $|h|\leq k$ that ,
$$\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\c... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4188357",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 4,
"answer_id": 1
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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(... | 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 ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4188690",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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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:
$$ \int^{x=2}_{x=0} \int^{y=3x-x^2}_{y=x} \left({x^2-xy}\right)dy\ dx$$
$$\int^2_0 \left[x^2y-... | $y=3x-x^2 \implies y = - (x-\frac{3}{2})^2 + \frac{9}{4}$
So equation of parabola is $ \ 9 - 4y = (2x-3)^2 \ $ and its vertex is $\left(\frac{3}{2}, \frac{9}{4}\right)$.
Intersection of parabola and the line $y = x$ is $(2, 2)$. So you can see that it is below the vertex and to the right of the axis of symmetry.
So if ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4189572",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Locating roots of a special class of polynomials in $\mathbb{Z}[X]$ I am reading Prof. Ram Murty's Prime number and Irreducible polynomials. I am having problem in understanding a part of the following lemma:
Statement:
Suppose that $\alpha$ is a complex root of a polynomial
$$
f(x)=x^m+a_{m-1}x^{m-1}+\cdots+a_1x+a_0
$... | Here is a proof of the first inequality.
$$f(z) = z^m + a_{m-1}z^{m-1} + a_{m-2}z^{m-2} + a_{m-3}z^{m-3}+ \dots+a_0\\
\frac{f(z)}{z^m} = \left(1 + \frac{a_{m-1}}z + \frac{a_{m-2}}{z^2}\right) + \left(\frac{a_{m-3}}{z^3}+ \dots+\frac{a_0}{z^m}\right)\\
\frac{f(z)}{z^m} - \left(\frac{a_{m-3}}{z^3}+ \dots+\frac{a_0}{z^m}\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4189906",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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True or false? Any triangle we can $h_A^2+h_B^2 \geq 2h_C^2$, where $h_A, h_B, h_C$ are lengths of the heights from vertices A, B and C Let $\triangle ABC$ any triangle and the lengths of the heights from vertices A, B and
C are, respectively, $h_A$, $h_B$ , $h_C$. Then how should we prove that,
$$h_A^2+h_B^2 \geq 2h_... | 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:
$$t=\frac 94 p^2-3p\cdot h_C-2h_Ah_B\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4190563",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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How to evaluate $\lim\limits_{x\to \infty}\frac {2x^4-3x^2+1}{6x^4+x^3-3x}$?
Evaluate: $\lim\limits_{x\to \infty}\frac {2x^4-3x^2+1}{6x^4+x^3-3x}$.
I've just started learning limits and calculus and this is an exercise problem from my textbook.
To solve the problem, I tried factorizing the numerator and denominator o... | Essentially, divide the numerator and denominator by the largest power of numerator:$$\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{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4190851",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
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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 wou... | *
*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 $$A(p,k)=\dfrac{1+(8 m ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4191369",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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If $\alpha$ be a multiple root of the order 3 of the equation $x⁴+bx²+cx+d=0$ then $α= 8d/3c$ I found this statement online here and I think this statement is false.I would first prove a general statement and then I would use it to find $\alpha$ in order to discard the statement given in online link.
Theorem:If the e... | Verification of the Theorem
Expanding
$$
(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+6... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4191533",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 2
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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 $\fra... | 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=\p... | {
"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 $$\lim_{ n \to \infty} \left(\frac{1^2+1}{1-n^3}+\frac{2^2+2}{2-n^3}+\frac{3^2+3}{3-n^3}+..... | Hint: $k-n^{3}$ lies between $1-n^{3}$ and $n-n^{3}$ for $1 \leq k \leq n$. This allows you to show that $\lim \sup$ and $\lim \inf$ are both equal to $-\frac 1 3$ using the formulas for $\sum\limits_{k=1}^{n} k$ and $\sum\limits_{k=1}^{n} k^{2}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4193228",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Probability two blocks have exactly 2 out of 4 properties the same?
I have 120 blocks. Each block is one of 2 different materials, 3 different colors, 4 different sizes, and 5 different shapes. No two blocks have exactly the same of all four properties. I take two blocks at random. What is the probability the two bloc... | No. The answer given on the linked website is correct.
After picking the first block there are 119 left. Assume that the second block has material and color the same, whereas size and shape are different. Of the $4*5 = 20$ blocks that meet the first criterium (note: minus 1 because the first block can not be picked aga... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4196967",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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If $f(x) = x^2 - 2x + 5$ what is $f^{-1}(x)$?
If $f(x) = x^2 - 2x + 5$ what is $f^{-1}(x)$?
with the condition : $x > 1$
I solved this problem in this way:
$f(x) = x^2 - 2x + 5 -1 +1 \longrightarrow (x-2)^2 + 1 = f(x) $
$f^{-1}(x) = \sqrt{x-1} + 2$
But I saw someone else solved it in this way:
$f(x) = x^2 - 2x +... | $$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$$
for $x>1$ inverse of f, $f^{-1}(x)=y=\sqrt{x-4}+1$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4198095",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Calculating $\lim_{x \to 0} \frac{\sin(x^2)-\sin^2(x)}{x^2\ln(\cos x)}$ without L'Hospital's rule this is my first post here so excuse the lack of knowledge about how things usually go.
My question revolves around calculating the limit as $x$ approaches $0$ of the following function:
$$\lim_{x \to 0} \frac{\sin(x^2)-\s... | More elementary solution (without Taylor series expansion or L'Hospital's rule) can be:
$$
\begin{align*}
\lim_{x \to 0} \frac{\sin(x^2)-\sin^2(x)}{x^2\ln(\cos x)}&=\lim_{x \to 0} \frac{\sin(x^2)-\sin^2(x)}{x^4}\cdot\frac{x^2}{1-\cos x}\cdot\frac{-(\cos x-1)}{\ln(\cos x)-0}\\
&=\lim_{x \to 0} \frac{\sin(x^2)-\sin^2(x)}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4198444",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Prove by limit definition $\lim _{x\to 3}\left(\frac{x^2-9}{x^3-3x^2-2x+6}\right)=\frac{6}{7}$. Prove, using the limit definition, that
$\lim _{x\to 3}\left(\frac{x^2-9}{x^3-3x^2-2x+6}\right)=\frac{6}{7}$.
I tried to do it this way, but I can’t move forward. Can anyone help me how to continue?
By definition we have
$\f... | If you want to avoid complications, just replace $x\to 3$ by $x=u+3$ with $u\to 0$.
Then $\left|f(x)-\dfrac 67\right|=\underbrace{\left|\dfrac{6u+29}{7(u^2+6u+7)}\right|}_\text{bounded}\times\underbrace{|u|}_{<\varepsilon}\to 0$
It is bounded because $|u|<1\implies \begin{cases}|6u+29|<35\\|u^2+6u+7|\ge |6u+7|\ge\big|7... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4200177",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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How do i approach ahead in this question
Let $a,b \in\Bbb N$, $a$ is not equal to $b$ and the quadratic equations $(a-1)x^2 -(a^2 + 2)x +a^2 +2a=0$ and $(b-1)x^2 -(b^2 + 2)x +b^2 +2b=0$ have a common root, then the value of $ab/5$ is
So what I did was,
I subtracted the two equations and got
$x=(a+b+2)/(a+b)$
I tried ... | Let $q_a(x)=(a-1)x^2-(a^2+2)x+a^2+2a$. Any root $r$ of $q_a(x)$ and of $q_b(x)$ is also a root of$$(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$$But you can't have $r=1$, because\begin{align}1\text{ is a root of }q_a(x)&\iff q_a(1)=0\\&\i... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4201360",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Proving $x^2 + x + 1$ is a factor of $P_n (x)=(x+1)^{2n+1} + x^{n+2}$. Further says to consider $P_n(\omega)$ and $P_n(\omega^2)$ wherein $\omega$ is a cube root of unity. $\omega \not=1$.
Found this from a examination paper with no solutions. I understand it relates to roots of unity, but I'm unsure how I can bring th... | Let $\omega$ be any root of $x^2+x+1$. We will show that $\omega$ is a root of $(x+1)^{2n+1} + x^{n+2}$.
First note that we can rearrange $x^2+x+1$ to $x+1 = -x^2$ so we know that $(\omega+1)^{2n+1} = (-\omega^2)^{2n+1} = -\omega^{4n+2}$.
Thus $(\omega+1)^{2n+1} + \omega^{n+2} = -\omega^{4n+2} + \omega^{n+2} = -\omega^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4202061",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Facing issue in rectifying the solution. Gauss Divergence Theorem problem Evaluate
$$
\iint_S (y^2z^2 \textbf{i} \, +z^2x^2\textbf{j}+z^2y^2\textbf{k}).\textbf{n} ~\mathrm{d}S
$$
where S is the part of sphere $x^2+y^2+z^2=1$ above the $xy$ plane and bounded by this plane.
I've tried solving this using Gauss Divergence... | $ \displaystyle \iiint_V \nabla \cdot (y^2z^2, z^2x^2, z^2y^2) \ dV = \iiint_V 2zy^2 \ dV$
Your bounds are correct so you may have made some mistake in evaluating it.
$ \displaystyle \iiint_V 2 z y^2 \ dV = 2 \int_{-1}^1 \int_{-\sqrt{1-x^2}}^{\sqrt{1-x^2}} \int_{0}^{\sqrt{1-x^2-y^2}} z y^2 \ dz \ dy \ dx$
$ \display... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4204013",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Using Rearrangement Inequality .
Let $a,b,c\in\mathbf R^+$, such that $a+b+c=3$. Prove that $$\frac{a^2}{b+c}+\frac{b^2}{c+a}+\frac{c^2}{a+b}\ge\frac{a^2+b^2+c^2}{2}$$
$Hint$ : Use Rearrangement Inequality
My Work :-$\\$
Without Loss of Generality let's assume that $0\le a\le b\le c$ then we can infer that $c^2\ge b^... | Another way:
We need to prove that:
$$\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,$$ which is true because $(... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4204601",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
"answer_id": 2
} |
Prove $a^2 + b^2 + c^2 + ab + bc +ca \ge 6$ given $a+b+c = 3$ for $a,b,c$ non-negative real. I want to solve this problem using only the AM-GM inequality. Can someone give me the softest possible hint? Thanks.
Useless fact: from equality we can conclude $abc \le 1$.
Attempt 1:
Adding $(ab + bc + ca)$ to both sides of i... | By AM-GM $$\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.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4208010",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 6,
"answer_id": 0
} |
Determinant of a matrix of squares equals $4(x-y)(x-z)(y-z)$ I need to find the value of the following determinant.
$$\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... | 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^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4212425",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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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=\fr... | 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
$$
\log \Gamma \!\left( {t... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4212570",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
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Denoting $2\le\frac{6-4x}5\le3$ by $|mx-n|\le5$. What is the value of $|n-m|$?
If we denote solution set of the inequality $2\le\frac{6-4x}5\le3$ by
$|mx-n|\le5$, what is the value of $|n-m|$?
$1)7\qquad\qquad2)5\qquad\qquad3)21\qquad\qquad4)23$
I solved this problem with following approach:
$$2\le\frac{6-4x}5\le3\qu... | $$2 \le \frac{6-4x}{5} \le 3$$
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 $|n-m|=|13-(-8)|=21.$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4213018",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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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}{... | 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 \ta... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4213775",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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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... | 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 ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4214040",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
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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 ... | 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.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4218465",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
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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)(... | $\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}\righ... | {
"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
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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 ... | 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... | {
"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}
10... | 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 latte... | {
"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 $u=x^2+1$:
$$
\int \sqrt{x^2+1}\;x^4\c... | As suggested in the comments we can use the substitution $x=\mbox{sinh}(u)$ and some other well known trig identities:
$1)$ $dx=(\mbox{sinh}(u))'du=\mbox{cosh}(u)du$
$2)$ $\mbox{cosh}^{2}(u)-\mbox{sinh}^{2}(u)=1$
$3)$ $\mbox{cosh}(u)=\frac{e^{u}+e^{-u}}{2}$
$4)$ $\mbox{arcsinh}(u)=\mbox{log}(u+\sqrt{1+u^{2}})$
$5)$ $\i... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4223596",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 4,
"answer_id": 2
} |
Show that for a, b, c > 0 with a + b + c = 1, $(a + \frac 1a)^2 + (b + \frac 1b)^2 + (c + \frac 1c)^2 ≥ \frac {100}{3}$ For this question I took the function as $x^2$ and using the second derivative I found it to be 2. This is greater than 0 for all x values meaning that the function is convex. Then I used Jensen's Ine... | Putting it all together from the comments.
$$\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}$$
Note that applying Cauchy-Schwarz or QM-AM yield... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4226251",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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If $A$ and $B$ are solutions to $7\cos\theta+4\sin\theta+5=0$, then $\cot\frac{A}{2}+\cot\frac{B}{2}=-\frac{2}{3}$ If $A$ and $B$ are the solutions to ${\displaystyle 7\cos\theta+4\sin\theta+5=0\mbox{ where }A>0,0<B<2\pi;}$
Without finding the solutions to the trig equation, show that;
$$\cot\left(\frac{A}{2}\right)+\c... | $\displaystyle 7\cos\theta+4\sin\theta+5=0$
$14\cos^{2}\frac{\theta}{2}+8\sin\frac{\theta}{2}\cos\frac{\theta}{2}-2 =0$
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\f... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4226367",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"answer_id": 0
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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 tha... | 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, ... | 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 $F(x)$ w... | {
"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
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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... | 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
$$
\begin{pmatrix} a & b \\ c & d \end{pmatrix}^{-1}
=
\frac{1}{ad-bc... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4230194",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
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Complex number related problem Let $z_1,z_2,z_3$ be complex numbers such that $|z_1|=|z_2|=|z_3|=|z_1+z_2+z_3|=2$ and $|z_1–z_2| =|z_1–z_3|$,$(z_2 \ne z_3)$, then the value of $|z_1+z_2||z_1+z_3|$ is_______
My solution is as follow
${z_1} = 2{e^{i{\theta _1}}};{z_2} = 2{e^{i{\theta _2}}};{z_3} = 2{e^{i{\theta _3}}}$ &... | We have three points on the circle of radius $2$ centered at the origin. We know that the triangle formed by these points is isosceles. We can rotate the triangle by some angle so that point $z_1$ is on either the real axis or the imaginary axis. Note that the modulus will not change because rotation = multiplying by $... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4230819",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 5,
"answer_id": 1
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Calculating $\int\frac{a^4-x^4}{x^4+a^2x^2+a^4}dx$
$$\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$$ So, beside the linear term (and the integ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4231586",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Expressing the $n-$th derivative of $f(x)=\frac{3x^2-6x+5}{x^3-5x^2+9x-5}$ In an attempt to express the $n-$th derivative of the rational function $f(x)=\frac{3x^2-6x+5}{x^3-5x^2+9x-5}$ ,
I split it into $\left( \frac{1+2i}{(x-2-i)} + \frac{1-2i}{(x-2+i)} + \frac{1}{(x-1)} \right)$ ,
then using $y= \frac{1}{ax+b} \Righ... | From your well-asked question, I assume that you are only interested in a trick for the part with complex coefficients
$$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{al... | {
"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
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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: $$\begin{align} x^2-x-2 &=(x+1)(x-2) \\ 3x^2-x-2 &=(x-1)(3x-2) \\ 6x^2-x-2 &= (2x+1)(3x-2) \\ 10x^2-x-2 &= (2x-1)(5x+2) \\ 15x^2-x-2 &= (3x+1)(5x-2) \\ 21x^2-x-2 &= (3x-1)(7x+2) ... | Hint
To get factors, write $$\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)$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4232401",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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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
$$
g(x,y) = x^2 + \frac{y^2}{4} - c, ~0 \leq c \leq 1.... | Clearly, $\nabla f(x,y)=0\iff(x,y)=(0,0)$, but $(0,0)$ is a saddle-point. So, the maximum and the minimum can be attained only at the boundary of $D$.
Let $h(x,y)=x^2+\frac{y^2}4$. Then apply the method of Lagrange multipliers:$$\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... | {
"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
$$\int\frac{dx}{\sqrt{\alpha+\beta x+\gamma x^2}}=\frac{1}{\sqrt{-\gamm... | Assuming that $q$ is equal to $\beta ^2-4 \alpha \gamma$ otherwise the integral won't make any sense. In the integral $\int\frac{dx}{\sqrt{\alpha+\beta x+\gamma x^2}}$, the coefficient of $x^2$ plays a major role.
Assumption: $q=\beta^2-4 \alpha \gamma>0$
When $\gamma <0$
$\begin{align} \int\frac{dx}{\sqrt{\alpha+\bet... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4233492",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Prove $\left(\frac{u}{a}\right)^a.\left(\frac{v}{b}\right)^b.\left(\frac{w}{c}\right)^c \le \left(\frac{u+v+w}{a+b+c}\right)^{(a+b+c)} $ Let $u,v,w>0$ and $a,b,c$ are positive constant. Prove that $\left(\frac{u}{a}\right)^a.\left(\frac{v}{b}\right)^b.\left(\frac{w}{c}\right)^c \le \left(\frac{u+v+w}{a+b+c}\right)^{(a+... | The inequality is homogenenous in $(u, v, w)$: If you have proven
$$
\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... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4234995",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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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 $\pmod{p}$:
$$ \sum_{k=1}^{n}\binom{2k}{k}\frac{1}{k} = \frac{2n+1}{3}\binom{2n}{n}\sum_{k=1}^{n... | Denote $$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)$$
let's show $S=T$ by showing they satisfy same recurrence (plus some trivial checks on initial conditions).
Recurrence for $T$ is ea... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4235768",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "12",
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Let $x^3+\frac{1}{x^3}$ and $x^4+\frac{1}{x^4}$ are rational numbers. Show that $x+\frac{1}{x}$ is rational.
$x^3+\dfrac{1}{x^3}$ and $x^4+\dfrac{1}{x^4}$ are rational numbers. Prove that $x+\dfrac{1}{x}$ is rational number.
My solution:
$x^3+\dfrac{1}{x^3}=\left(x+\dfrac{1}{x}\right)\left(x^2+\dfrac{1}{x^2}-1\right... | With $\alpha= x+\frac{1}{x}$ we get
$$x^3 + \frac{1}{x^3} =\alpha^3 - 3 \alpha = u\\
x^4 + \frac{1}{x^4} = \alpha^4 - 4 \alpha^2 + 2 = v$$
The polynomials $t^3 - 3 t - u$ and $t^4 - 4 t^2 + 2 - v$ have a common root $\alpha$. If that were the only common root, then their gcd $t-\alpha$ would have rational coefficients.... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4237191",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "14",
"answer_count": 5,
"answer_id": 3
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Proving divisibility by $9$ for $10^n + 3 \times 4^{n+2} + 5$. I am trying to prove that for all $n$, $9$ divides $10^n + 3 \times 4^{n+2} + 5$. I first tried induction on $n$, but couldn't get anywhere with the induction step. I then tried to use modular arithmetic. The furthest I could get was:
As $10 \equiv 1 \mod ... | $$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$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4237927",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 6,
"answer_id": 4
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Why does Solving system of quadratic equations gives extra roots? Consider these system of Equations
\begin{align*}
\begin{cases}
x^2+4x+4=0\\\\
x^2+5x+6=0
\end{cases}
\end{align*}
For solving them
We have
Method 1-
Subtract both equations
So
$-x-2=0$
Hence,
$x=-2$
Method-2
Add both equations
$2x^2+9x+10=0$
After apply... | We could also describe the difference in your two approaches in this way. You began with the system $ \ p(x) \ = \ x^2 + 4x + 4 \ = \ 0 \ , \ q(x) \ = \ x^2 + 5x + 6 \ = \ 0 \ \ . $ When you subtract one equation from the other, you have $ \ p(x) - q(x) \ = \ 0 \ \ , $ which is equivalent to the equation we would... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4238314",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 3
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Simple way to compute the finite sum $\sum\limits_{k=1}^{n-1}k\cdot x^k$ I'm looking for an elementary method for computing a finite geometric-like sum,
$$\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... | 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.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4238437",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
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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 ... | 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... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4239778",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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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, }... | Another approach would be to go ahead and perform the ratio addition (but don't fully multiply out the binomial-squares in the denominator):
$$ \frac{1}{(\alpha-4)^{2}}+\frac{1}{(\beta-4)^{2}} \ \ = \ \ \frac{(\beta-4)^{2} \ + \ (\alpha-4)^{2}}{(\alpha-4)^{2}· (\beta-4)^{2}} \ \ = \ \ \frac{\alpha^2 \ + \ \beta^2 \ - \... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4239918",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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inequality for positive real numbers Given $$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)... | By Holder $$\frac{a^3}{x} + \frac{b^3}{y} + \frac{c^3}{z}=\frac{(1+1+1)\left(\frac{a^3}{x} + \frac{b^3}{y} + \frac{c^3}{z}\right)(x+y+z)}{3(x+y+z)}\geq\frac{(a+b+c)^3}{3(x+y+z)}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4240064",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
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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 $... | 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 sta... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4241609",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Subsets and Splits
Fractions in Questions and Answers
The query retrieves a sample of questions and answers containing the LaTeX fraction symbol, which provides basic filtering of mathematical content but limited analytical insight.