Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
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Proof of rational numbers I am recently starting in the subject of pure mathematics and this problem has eluded me.
Any rational number $\frac{p}{q}$ can be expressed in a simple series form:
$$\frac{1}{1},\frac{2}{1},\frac{1}{2},\frac{3}{1},\frac{2}{2},\frac{1}{3},\frac{4}{1},\frac{3}{2},\frac{2}{3},\frac{1}{4},...$$... | Hint. Divide the numbers in rows:
$$\frac{1}{1}$$
$$\frac{2}{1},\frac{1}{2}$$
$$\frac{3}{1},\frac{2}{2},\frac{1}{3}$$
$$\frac{4}{1},\frac{3}{2},\frac{2}{3},\frac{1}{4}.$$
In the first $n$ rows there are $1+2+\dots+n=n(n+1)/2$ fractions.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1935563",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Prove that $\sin^4\theta+\cos^4\theta=\frac{3+\cos4\theta}{4}$ I have a question that goes exactly like this:
By considering $(\sin^2\theta+\cos^2\theta)^2$ and $(\sin^2\theta+\cos^2\theta)^3$ prove that
$$
\text{a) }\sin^4\theta+\cos^4\theta=\frac{3+\cos4\theta}{4},\qquad\text{b) }\sin^6\theta+\cos^6\theta=\frac{5+3\c... | From Euler's Identity and subsequent use of the binomial theorem, we have
$$\begin{align}
\cos(4\theta)&=\text{Re}(e^{i4\theta})\\\\
&=\text{Re}\left(\cos(\theta)+i\sin(\theta)\right)^4\\\\
&=\text{Re}\left(\sum_{k=0}^4\binom{4}{k}\cos^{4-k}(\theta)i^k\sin^k(\theta)\right)\\\\
&=\cos^4(\theta)-6\cos^2(\theta)\sin^2(\th... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1936121",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 8,
"answer_id": 6
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Compute $\int^{\pi/2}_0 \frac{dx}{(a^2\cos^2 x + b^2 \sin ^2 x)^2}$ I have tried solving this for about an hour and will probably resort to head banging in some time:
$$\int ^{\frac{\pi}{2}}_{0} \dfrac{dx}{(a^2\cos^2 x + b^2 \sin ^2 x)^2}$$
I first divided by $\cos^4 x$ and then subsequently put $\tan x = t$, to get:
$... | Say that $ a, b >0 .$ From the substitution $t \mapsto \dfrac{a}{b}t$ followed by $ t \mapsto \dfrac{1}{t}$ we get
$$
I = \int_{0}^{\infty}\dfrac{1+t^{2}}{(a^{2}+b^{2}t^{2})^{2}}\, dt = \dfrac{1}{a^{3}b^{3}} \int_{0}^{\infty}\dfrac{b^2+a^2t^{2}}{(1+ t^{2})^{2}}\, dt
$$
and
$$
I = \dfrac{1}{a^{3}b^{3}} \int_{0}^{\infty}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1939026",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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"answer_id": 0
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How to show this conjecture：$\frac{1}{\sqrt{a_{1}}}+\frac{1}{\sqrt{a_{2}}}+\cdots+\frac{1}{\sqrt{a_{n}}}\ge\frac{2n}{\sqrt{a_{1}}+\sqrt{a_{n}}}$ I have prove this inequality
$$\dfrac{1}{\sqrt{1}}+\dfrac{1}{\sqrt{2}}+\cdots+\dfrac{1}{\sqrt{n}}>2\sqrt{n+1}-2$$
because
$$\sum_{k=1}^{n}\dfrac{1}{\sqrt{k}}>\sum_{k=1}^{n}\d... | Using the AM-HM inequality states: if $x_1\leq\ldots\leq x_n$ are nonnegative numbers, then
\begin{align}
\frac{x_1+\ldots +x_n}{n} \geq \frac{n}{\frac{1}{x_1}+\ldots+\frac{1}{x_n}}
\end{align}
which implies
\begin{align}
\frac{1}{x_1} + \ldots + \frac{1}{x_n} \geq \frac{n^2}{x_1+ \ldots + x_n} \geq \frac{n}{x_1 + x_n}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1939123",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Evaluation of Trigonometric Integration $\int \sin x\sqrt{\sin 2x}dx$.
Evaluation of $$\int \sin x\sqrt{\sin 2x}dx$$
$\bf{My\; Try::}$ Let $$I = \int \sin x\sqrt{\sin 2x}dx = \sqrt{2}\int \frac{\tan x\sqrt{\tan x}}{\sec x}dx = \sqrt{2}\int \frac{(\tan x)^{\frac{3}{2}}}{\sec^3 x}\cdot \sec^2 xdx$$
So $$I = \sqrt{2}\in... | Note that $\sin{2x} \equiv 1-(\sin{x}-\cos{x})^2 \equiv (\sin{x}+\cos{x})^2 -1$
Also note that $\text{d}(\sin{x}-\cos{x}) - \text{d}(\sin{x}+\cos{x}) = 2\sin{x}\,\text{d}x$
Substitute and separate the integral into two different integrals which can be evaluated independently.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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In the expansion of $(1+x+x^2)^n$, find the required value If $n \in $ odd integer, such that:
$$(1+x+x^2)^n=a_0+a_1x+a_2x^2+.....+a_{2n}x^{2n}$$ and
$$(a_1)^2-(2a_2)^2+(3a_3)^2-(4a_4)^2+......-(2na_{an})^2=- \lambda n^2a_{n-1}$$ then find the value of $\lambda$.
Is there any better approach to this question rather tha... | $$y(x):=(1+x+x^2)^n$$
Suppose ' represents differentiation w.r.t $x$.
$$
y'(x)=n(1+x+x^2)^{n-1}(1+2x)=\sum_{k=1}^{2n}ka_kx^{k-1}
$$
Now the constant term in $y'(-x)y'(x^{-1})$ is
$$
a:=a_1-(2a_2)^2+(3a_3)^2-\cdots-(2na_{2n})^2
$$
Now
$$
y'(-x)y'(x^{-1})={n^2(1+x^2+x^4)^{n-1}(2-3x-2x^2)\over x^{2n-1}}
$$
Let $g(x)=(1+x+... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1939679",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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$120^{1492}\pmod{100}$ How can I solve it?
I can solve it using the fact: if $b\mid a$ then $b^{n}\mid a^{n}$ (I have proved it using induction)
$\frac{120}{10}=12$ ,then $\frac{120^{2}}{100}$
this mean $120^{2}\equiv0\pmod{10}$
so, $120^{1492}\equiv (120^{2})^{746}\equiv 0^{746}\equiv 0\pmod{100}$
I'm asking for ano... | Use this teorem: If $a \equiv b \pmod{m} \implies a^n \equiv b^n \pmod{m}$, then: $$120 \equiv 20 \pmod{100} \implies 120^{1492} \equiv 20^{1492} \pmod{100}$$
If you look the values of $20^n$ you gonna see this:
$$\begin{align}
20^1 \equiv 20 \pmod {100} \\
20^2 \equiv 0 \pmod {100} \\
20^3 \equiv 0 \pmod{100} \\
... \... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1939929",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Solve the equation $\cos^n(x) - \sin^n(x)=1$
Solve the equation $\cos^n(x) - \sin^n(x)=1,n \in \mathbb{N}-\{0\}$
If $n$ is even then $\cos^n(x) = \sin^n(x)+1$ is only possible if $\sin(x)=0$ therefore the solution is $x=k\pi, k \in \mathbb{Z}$.
I'm having problems with $n$ odd case.
UPDATE
For $n=1$ we have $\cos(x)... | I found a simple proof. For $n \ge 3$ odd we have $\cos^n(x) + (-\sin(x))^n=1 \tag1$ First, we know $\sin^2(x) + \cos^2(x)=1$
Also $|\cos^n(x)|\lt \cos^2(x)$ and $|\sin^n(x)|\lt \sin^2(x)$ for $x \not = k\pi$ and $x \not = \pm \frac \pi 2 + k\pi$ therefore the equality (1) cannot hold for $x \not = k\pi$ and $x \not ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1940444",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Simplifcation of a sum I’m unsure if I’m allowed to simplify like this:
$$\begin{align*}
\sum_{n=2}^\infty \frac{2^{n+3}+7^{n-1} \cdot(n-1)!}{10^n(n-1)!} &= \sum_{n=2}^\infty \frac{2^{n+3}}{10^n(n-1)!} + \sum_{n=2}^\infty \frac{7^{n-1} \cdot(n-1)!}{10^n(n-1)!}\\
&= \sum_{n=2}^\infty \frac{2^{n+3}}{10^n(n-1)!} + \frac{... | Your first two steps are correct, but then you went astray.
$$\sum_{n=2}^\infty\frac{7^n(n-1)!}{10^n(n-1)!}=\sum_{n=2}^\infty\frac{7^n}{10^n}=\sum_{n=2}^\infty\left(\frac7{10}\right)^n\;,$$
not $\sum_{n=2}^\infty\frac7{10}$. Now
$$\frac17\sum_{n=2}^\infty\left(\frac7{10}\right)^n=\frac17\cdot\frac{\left(\frac7{10}\righ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1942281",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Find $\int_0^{2\pi}\frac1{5-4\cos x}\ dx$ $$\int_0^{2\pi}\frac1{5-4\cos x}\ dx$$
How do I compute this integral? An online integral calculator gives an antiderivative as
$$\frac{2\arctan(3\tan\frac x2)}3$$
but then gives the definite integral as $\frac{2\pi}3$. Obviously this doesn't make sense as the antiderivative va... | Using complex analysis
your integral can be expressed as
$$ \int \limits^{2\pi }_{0}\frac{1}{5-4\cos \left( x\right) } dx=\oint \frac{1}{5-2\left( z+\frac{1}{z} \right) iz} dz$$
Now we have two singularites at $z=2$and at $z=\frac {1}{2} $
Now apply Residue theorem
$$\oint \frac{1}{5-2\left( z+\frac{1}{z} \right) i... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1942983",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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"answer_id": 2
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Properties of the rotation matrix $$R(\theta) = \begin{pmatrix} \cos \theta & -\sin \theta \\ \sin \theta & \cos \theta \end{pmatrix}$$
Consider the above rotation matrix.
$a)$ Show that $R(\theta)$ is non-singular with: $R^{-1}(\theta) = R(-\theta)$
$b)$ Show that $R(\theta)^T = R(-\theta)$
$c)$ For what angles $\thet... | a
A quick singularity check:
$$
\det \mathbf{R} \left( \theta \right)
= \cos^{2} \left( \theta \right) + \sin^{2} \left( \theta \right) = 1
$$
Because the determinant is not $0$, the rotation matrix is not singular.
Is
$$
\mathbf{R} \left( \theta \right)^{-1} = \mathbf{R} \left( -\theta \right)?
$$
Check via multipl... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1943175",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
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Double Integral Math Olympiad Problem I was taking a Math Olympiad test and one of the questions was to calculate the following double integral:
$$\int_0^\infty\int_0^\infty\frac{\log|(x+y)(1-xy)|}{(1+x^2)(1+y^2)}\ \mathrm{d}x\ \mathrm{d}y$$
Here, as usual, $\log a$ and $|a|$ are the natural logarithm and absolute valu... | Here is a solution following @Sirzh's hint. Let
$$ I = \int_{0}^{\infty}\int_{0}^{\infty} \frac{\log|(x+y)(1-xy)|}{(1+x^2)(1+y^2)} \, dxdy $$
and
$$ J = \int_{0}^{\frac{\pi}{2}} \log \cos \theta \, d\theta = -\frac{\pi}{2}\log 2. \tag{1}$$
Applying the substitution $x = \tan u$ and $y = \tan v$ to $I$, we get
\begin{a... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1945637",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "13",
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Find one side of a triangle using trigonometry impossible question In the isosceles triangle $ABC$ where $a=|BC|,\ b=|CA|,\ c=|AB|$ and angle at corner $A = x$
and angle at corner $B = y$ and angle at corner $C = z$ ($z>90$ degrees)
$a=b=4$ ; $\sin z = 3/8$
Find value of $c$ (which is side $|AB|$ as stated above)
My f... | @grixor's answer, using the law of cosines is (currently, almost) correct, but perhaps uses a tool that is not yet in the student's toolbox. We can derive this result from the problem as given.
Extend $\overline{AC}$ until it intersects the perpendicular to $\overline{AC}$ through $B$ and call this intersection $D$. ... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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Solving $\lim\limits_{x \to 0} \frac{\ln(1+x)}{\ln(1+4x+x^2)}$ without L'Hospital or Sandwiching Using the L'Hospital I got: $\lim\limits_{x \to 0} \frac{\ln(x+1)}{\ln(1+4x+x^2)} = \lim\limits_{x \to 0}\frac{\frac{1}{1+x}}{\frac{4+2x}{1+4x+x^2}}= \frac{1}{4}$,
I then wondered, if I could sandwich it andobserved that ... | Note that we can write $\log(1+4x+x^2)=\log(1+4x)+\log\left(1+\frac{x^2}{1+4x}\right)$.
Therefore, we have
$$\begin{align}
\frac{\log(1+x)}{\log(1+4x+x^2)}&=\frac{\log(1+x)}{\log(1+4x)+\log\left(1+\frac{x^2}{1+4x}\right)}\\\\&=\frac{\log(1+x)}{\log(1+4x)}\left(1+\frac{\log\left(1+\frac{x^2}{1+4x}\right)}{\log(1+4x)}\ri... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1946768",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
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Is my proof that $x^2+y^2+z^2 ≥ xy+yz+xz$ correct? The question:
Prove that $x^2+y^2+z^2 ≥ xy+yz+xz$ for all real numbers $x$, $y$ and $z$.
This problem has been posed before, but my question is whether my proof below is correct, since it seems the other answers to this problem are different.
If $x$, $y$ and $z$ are re... | another way $$xy+yz+zx\le \sqrt { \left( x^{ 2 }+{ y }^{ 2 }+{ z }^{ 2 } \right) \left( x^{ 2 }+{ y }^{ 2 }+{ z }^{ 2 } \right) } =x^{ 2 }+{ y }^{ 2 }+{ z }^{ 2 }\\ \\ \\ $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1948633",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Find the equation of the circle passing ... Find the equation of the circle passing through the points $P(5,7)$, $Q(6,6)$ and $R(2,-2)$.
My Attempt:
Let the equation of the circle be:
$$x^2+y^2+2gx+2fy+c=0$$
The point $P(5,7)$ lies on the circle then,
$$5^2+7^2+10g+14f+c=0$$
$$10g+14f+c=-74$$-----(1)
The point $Q(6,6)$... | Another way to do this is to use the geometric fact that the perpendicular bisector of a chord passes through the center of a circle. The line through P(5,7) and Q(6,6)I has slope (7- 6)/(5- 6)= -1 and midpoint (11/2, 13/2). The perpendicular bisector is y= (x- 11/2)+ 13/2)= x+ 1. The line through Q(6, 6) and R(2,−2... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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Find $a$ such that $(x+a)(x+1991)+1=(x+b)(x+c)$ with $a,b,c\in\Bbb Z$
Find all integer values of $a$ such that the quadratic expression $(x+a)(x+1991)+1$ can be factored as a product $(x+b)(x+c)$ where b and c are integers.
I tried to do it by comparing the two expressions but I can't proceed.
| We have that $(x+a)(x+1991) + 1 = x^2 + (1991 + a)x + (1991a + 1)$. So if such factorisation exists we must have that the discriminant of the quadratic equation is a square. Now:
$$D = (1991 + a)^2 - 4(1991a + 1) = 1991^2 + 2 \cdot 1991a + a^2 - 4 \cdot 1991a - 4 = (1991 - a)^2 - 4$$
But then $\sqrt{D}, 2, (1991 - a)$ ... | {
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"timestamp": "2023-03-29T00:00:00",
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How to find the inverse of $f(x)=\dfrac{-x|x|}{1+x^2}$? I can see that
$f(x)= \dfrac{-(x^2)}{1+x^2}$ for $x \geq 0$
and $f(x)= \dfrac{(x^2)}{1+x^2}$ for $x<0$.
Help me proceed to find the inverse.
| You can usted the fact that f(x) is a Odd function, so if $x>0$ then $y=f(x)=\frac{-x^2}{1+x^2}$, so $y+yx^2=-x^2$ and $x^2=\frac{-y}{1+y}$ finally $f(x)^{-1}=\sqrt{\frac{-x}{1+x}}$ for $x>0$, and $f(x)^{-1}=\sqrt{\frac{x}{1-x}}$ for $x\leq0$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1953547",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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What will be the value of the following determinant without expanding it? $$\begin{vmatrix}a^2 & (a+1)^2 & (a+2)^2 & (a+3)^2 \\ b^2 & (b+1)^2 & (b+2)^2 & (b+3)^2 \\ c^2 & (c+1)^2 & (c+2)^2 & (c+3)^2 \\ d^2 & (d+1)^2 & (d+2)^2 & (d+3)^2\end{vmatrix} $$
I tried many column operations, mainly subtractions without any succ... | If you expand the binomials and you subtract the $1^{\rm st}$ column from the 2nd, 3rd, and 4th you get
$$
\begin{vmatrix}a^2 & 2a+1 & 4a+4 & 6a+9 \\ b^2 & 2b+1 & 4b+4 & 6b+9 \\ c^2 & 2c+1 & 4c+4 & 6c+9 \\ d^2 & 2d+1 & 4d+4 & 6d+9\end{vmatrix}.
$$
Next subtract 2 times the 2$^{\rm nd}$ column from the 3$^{\rm rd}$ one... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1953843",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "27",
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"answer_id": 3
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If $x*y$ is the operation $x*y = xy/(x+y+1)$, then how does $x*y*z = xyz(x+y+1)/(x+y+z+xy+xz+1)$? $$x*y = \frac{xy}{x+y+1}$$
What I got was,
$$x*(y*z) = x* \frac{yz}{y+z+1} =x\frac{\frac{yz}{y+z+1}}{x+\frac{yz}{y+z+1}+1}$$
The book's answer is
$$x*(y*z) = x*\frac{yz}{y+z+1} = \frac{xyz(y+z+1)}{x+y+z+xy+xz+1}$$
| $x*y=\frac{xy}{x+y+1}$
We put $ b=x*y=\frac{xy}{x+y+1}. $
$b*z=\frac{bz}{b+z+1}=\frac{\frac{xy}{x+y+1}z}{\frac{xy}{x+y+1}+z+1}$
$x*y*z=\frac{zxy}{xy+xz+yz+z+x+y+1}$
So: the book is wrong
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1954170",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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If the value of $xy$ , If $ x^2 + y^2 = 34$ and $x + y =8$ If $ x^2 + y^2 = 34$ and $x + y =8$ find the value of xy
So what should I be considering here the sum of two squares or..
I'm struggling here & any help to figure this out will be kindly appreciated
| $x^2$ + $y^2$ = $34$ $-->$ Equation 1
$x + y = 8$ $-->$ Equation 2
Now, we can rewrite Equation 2 with respect to x. This value will be substituted into Equation 1.
$x = 8 - y$
$x^2 + y^2 = 34$
$(8 - y)^2 + y^2 = 34$
$64 - 16y + y^2 + y^2 = 34$
$64 - 16y + 2y^2 = 34$
$2y^2 - 16y + 30 = 0$
Solving the quadratic above, t... | {
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"url": "https://math.stackexchange.com/questions/1962121",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 2
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Upper bound on $\displaystyle\sum_{\text{cyc}}\dfrac{a}{a^3+b^2+c}$ Let $a,b,c$ be positive reals such that $a+b+c=3$. Determine the largest possible value of $$\dfrac{a}{a^3+b^2+c}+\dfrac{b}{b^3+c^2+a}+\dfrac{c}{c^3+a^2+b}.$$
Experimenting for some values of $(a,b,c)$ I conjectured that the maximum value attained is ... | Hint:
Using Holder: $(a^3+b^2+c)(1+b+c)(1+1+c) \ge (a+b+c)^3$ to yield $$\frac{a}{a^3+b^2+c} \le \frac{a(1+b+c)(2+c)}{(a+b+c)^3}$$ and reduce the inequality to $5\sum ab + \sum a^2b + 3abc \le 21$ which is easy.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1962291",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 1,
"answer_id": 0
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Expressing $(\alpha-1)^{-1}$ where $\alpha^3-\alpha^2+\alpha+2=0$ Let $E=\Bbb{Q}[\alpha]$ where $\alpha^3-\alpha^2+\alpha+2=0$
I would like to express $(\alpha^2+\alpha+1)(\alpha^2-1)$ and $(\alpha-1)^{-1}$ on the form $a+b\alpha+c\alpha^2$ where $(a,b,c)\in\Bbb{Q}^3.$
For the first one I get $-4\alpha-2$, but for the ... | Let's try to represent $(\alpha - 1)^{-1}$ as $a + b\alpha + c\alpha^2$. We must have
$$ (\alpha - 1)(a + b\alpha + c\alpha^2) = a\alpha + b\alpha^2 + c\alpha^3 - a - b\alpha - c\alpha^2 \\
= c\alpha^3 + (b - c)\alpha^2 + (a-b)\alpha -a = \\
c(\alpha^3 - \alpha^2+\alpha +2) +b\alpha^2+(a-b-c)\alpha - (2c + a) = 1. $$
T... | {
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Prove the inequality $\frac{1}{1+a+b}+\frac{1}{1+c+b}+\frac{1}{1+a+c}+a+b+c \le 3 + \frac 13 (ab+bc+ac)$ Let $a,b,c \in [0;1]$. Prove that
$$\frac{1}{1+a+b}+\frac{1}{1+c+b}+\frac{1}{1+a+c}+a+b+c \le 3 + \frac 13 (ab+bc+ac).$$
I used Buffalo way. Can anyone offer an easier way
| Let $f(a,b,c)=\sum\limits_{cyc}\left(\frac{1}{1+a+b}+a-1-\frac{ab}{3}\right)$.
Hence, $f$ is a convex function of $a$, of $b$ and of $c$.
Thus, $\max\limits_{\{a,b,c\}\subset[0,1]}f=\max\limits_{\{a,b,c\}\subset\{0,1\}}f=f(0,0,0)=0$.
Done!
| {
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Trying to prove, by induction, that $2^{4n} + 5 $ is divible by $21.$ I want show by induction
$$ 21 \mid (2^{4n}+5) $$
So I assume:
$ 2^{4k}+5= 21p$
to prove that $ 21 \mid 2^{4(k+1)}+5 $
So I get it:
$2^{4(k+1)}+5 = 2^{4k+4}+5 = 2^{4k}2^{4}+
2^{4}2^{4k}+5 = 2^{4k} 16 +5 $ =
$16(2^{4k} +5 -5 )+5 = 16(21p-5)+5 = 1... | for $n=2$
$2^{4n}+5=256+5=261$
and
$261=21\times12+9$
so your identity is not true for $n=2$.
| {
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conditional probability that a freshman will graduate
A university finds that 75% of its graduating seniors scored above 80 on the entrance
exam, while only 50% of those who fail to graduate score above 80. They also find that half
of the entering freshmen graduate. What is the conditional probability that a fresh... | We introduce the following notation.
*
*$\Omega$ - the set of admitted students, which we assume to be finite and non-empty.
*$A$ - the set of graduating students.
*$B$ - the set of students who scored above $80$ on the entrance exam.
*$P$ - the uniform probability on $\Omega$.
It is given that
$$
\begin{align}... | {
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Is this the shortest proof of Fermat's last theorem for exponent 3? To begin with, here we only consider the case of $n^3 + (n+1)^3 = (n+2)^3$, that is the case of consecutive integers $(x,y,z)$ in the well known equation. The general case $x^3 + y^3 = z^3$ is much more difficult to prove (and the work is yet to be don... | No.
The presumably shortest proof for the case you consider is this
Assume $$n^3+(n+1)^3=(n+2)^3. $$
Then
$$ 0=n^3+(n+1)^3-(n+2)^3= n^3-3n^2-9n-7=(n^2-3n-9)\cdot n-7.$$
We conclude that $n\mid 7$, but neither $1^3+2^3=3^3$, nor $7^3+8^3=9^3$. $\square$
| {
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Prove $30|(a^3b-ab^3) $ Prove that if three distinct integers are chosen at random then there will exists two among them, say $a$ and $b$ such that $30|(a^3 b-ab^3)$
| By Fermat's little theorem (or simply by direct calculation), $x^2\equiv x$ mod $2$ and $x^3\equiv x$ mod $3$ for any integer $x$. It follows that
$$a^3b\equiv ab\equiv ab^3\mod 6$$
for any pair of integers $a,b$, hence $6$ divides $a^3b-ab^3$ for any pair. It remains to show that $5$ divides $a^3b-ab^3=ab(a^2-b^2)$ ... | {
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Develop $(\frac{2}{x}-x)^8$ $(\frac{2}{x}-x)^8$
How do I develop this?
My book gives me four options to what that one of the terms of the expression is supposed to look like after the expression is developed. The right option is:$$^8C_3*(\frac{2}{x})^3*(-x)^5$$
I even tried graphing both expressions on my graphical cal... | The function is:
$$ f(x) = \left( \frac{2}{x} - x \right)^8 = \frac{(2-x^2)^8}{x^8} = x^8 \, \left(\frac{2}{x^2} - 1 \right)^8. $$
These have the expansions:
\begin{align}
f(x) &= \sum_{k=0}^{8} \binom{8}{k} \, \left(- \frac{2}{x}\right)^{8-k} \, (-x)^k \\
&= \frac{1}{x^8} \, \sum_{k=0}^{8} \binom{8}{k} \, 2^{8-k} \, (... | {
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Prove that any odd number can be a leg in a Pythagorean Triple $(a,b,c)$ where $a^2+b^2=c^2$ is a Pythagorean triple.
My first thought was to do a proof by cases. I have three cases :
*
*a is odd, b is odd.
*a is odd, b is even.
*a is even, b is odd.
I'll just show case one :
$(2k+1)^2 + (2k)^2 = c^2$
I am kind ... | Hint:
Pythagorean Triples comes in the form: $(x^2-y^2,2xy,x^2+y^2)$ where $x$ and $y$ are any positive integers with $x>y$.
Can you use this to show that any odd number is possible?
Edit:
$(x^2-y^2,2xy,x^2+y^2)$ is equivalent to $(a,b,c)$ because:
$$\big(x^2-y^2\big)^2\ +\ \big(2xy\big)^2\ =\ \big(x^2+y^2\big)^2$$
Ans... | {
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If $n > 1$ is an integer not of the form $6k + 3$, prove that $n^{2} + 2^{n}$ is composite. I try to divide in to $5$ case $6k, 6k+1, 6k+2, 6k+4, 6k+5$
and put them in $n^{2}$ and consider the possible last digit 0f $2^n$
and sum with those form
I can only prove $6k, 6k+2, 6k+4$ case .the last digit is even number and ... | $n = 6k + i$. If $i$ is even then $n$ is even and $n^2 + 2^n$ is even and as $n^2 + 2^n > 2$ it must be composite.
So it remains to check for $n = 6k \pm 1$.
$(6k \pm 1)^2 + 2^n = 36k^2 \pm 12k + (1 + 2^n)$.
$3|36k^2 \pm 12k$ so if we can show $3|1+2^n$ we are done as $3|n^2 + 2^n$ and as $n^2 + 2^n > 3$ if $n > 1$.
W... | {
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Looking for $\tan\frac{1}{2}\theta$ Given that $$\frac{\cos \theta}{1-\sin\theta}=a,$$
how can I find the value of $\tan\frac{1}{2}\theta$?
I went in this way,
but the matter is in the box as shown (the fallacies).
| $$\frac{\cos\theta}{1-\sin\theta}=a$$
$$\frac{\frac{1-\tan^2\frac{\theta}{2}}{1+\tan^2\frac{\theta}{2}}}{1-\frac{2\tan\frac{\theta}{2}}{1+\tan^2\frac{\theta}{2}}}=a$$
$$\frac{1-\tan^2\frac{\theta}{2}}{1+\tan^2\frac{\theta}{2}-2\tan\frac{\theta}{2}}=a$$
$$(a+1)\tan^2\frac{\theta}{2}-2a\tan\frac{\theta}{2}+(a-1)=0$$
now ... | {
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A curious case of $1729$ Ramanujan's Taxicab number 1729 is famous for being the smallest positive integer which can be written as the sum of two positive cubes in two different ways. On a different note, I observed that
$12^3 + 1^3 = 1729$
$12^2 + 1^2 = 1\cdot7\cdot2\cdot9 + 1 + 7 + 2 + 9$
$12^1 + 1^1 = -1+7-2+9$
Ques... | If I understood correctly the problem is
Find numbers $n$ such that there exist a pair $(a,b)$ with the property that
$a^3+b^3=n$
$a^2+b^2$=products of the digits of n + Sum of the digits of n
Then, the number $1547$ is a solution, for the pair $(a,b)=(11,6)$:
$11^3+6^3=1547\\11^2+6^2=1\cdot 5 \cdot 4 \cdot 7 + 1 + 5... | {
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prove that if x is a real number, then $\lfloor2x\rfloor=\lfloor x\rfloor+\lfloor x + \frac{1}{2}\rfloor$ prove that if $x$ is a real number, then $\lfloor2x\rfloor=\lfloor x\rfloor + \lfloor x + \frac{1}{2}\rfloor$
I've seen the proof in my discrete math book, but it was very confusing. I don't understand why they con... | Let $x=n+\varepsilon$ where $n$ is an integer and $0\le\varepsilon\lt1.$ Then $$\lfloor2x\rfloor=\lfloor2n+2\varepsilon\rfloor=2n+\lfloor2\varepsilon\rfloor,$$
and
$$\lfloor x\rfloor+\lfloor x+\frac12\rfloor=\lfloor n+\varepsilon\rfloor+\lfloor n+\varepsilon+\frac12\rfloor=n+\lfloor\varepsilon\rfloor+n+\lfloor\varepsil... | {
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Find matrix $A$. How many solutions are there? The matrices below is matrix multiplcation of $AB=C$ but someone has changed the order of the coloumns in $C$ and erased what was in $A$. What is matrix $A$? Is it the only one? I have no idea how to start. I appreciate all the help I can get.
$$\begin{bmatrix} & & & \\ ... | No solution (In case you don't shuffle $C$)
P.S: Thanks @JeanMarie
Denote the $(i,j)^{th}$ entry of $A$ by $a_{ij}$.
$$\begin{bmatrix} a_{11} &a_{12} &a_{13} &a_{14} \\ a_{21} &a_{22} &a_{23} &a_{24} \\ a_{31} &a_{33} &a_{33} &a_{34} \\a_{41} &a_{42} &a_{43} &a_{44} \end{bmatrix}
\cdot\begin{bmatrix}2 & 0 &-1 & -... | {
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Find the sum of $\sum\limits_{n=1}^{\infty} \frac{x^{n}}{n(n+1)}$
Find the sum of $\sum\limits_{n=1}^{\infty} \frac{x^{n}}{n(n+1)}$ on its domain of convergence.
This is my idea.
We have the radius of convergence is $R=1$. And
$\sum_{n=1}^{\infty} \dfrac{x^{n}}{n(n+1)}=\dfrac{1}{x}\cdot \sum_{n=1}^{\infty} \dfrac{x^{... | If you know (or can assume) the logarithm series for $|x|<1$:
$$-\log(1-x) = \sum_{n=1}^{\infty} \frac{x^n}{n}$$
then you can forego the derivation/integration steps by noticing that:
$$
\begin{align}
\sum_{n=1}^{\infty} \frac{x^{n}}{n(n+1)} & = \sum_{n=1}^{\infty} x^n \left(\frac{1}{n} - \frac{1}{n+1} \right) \\
& = \... | {
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find $b$ if function $f(x)$ is strictly decreasing function if $\displaystyle f(x) = \frac{2}{\sqrt{3}}\tan^{-1}\left(\frac{2x+1}{\sqrt{3}}\right)-\ln(x^2+x+1)+(b^2-5b+3)x+c$ is strictly decreasing
function forall real values of $x.$ then $b$
$\displaystyle f'(x) = 2\sqrt{3} \; \frac{1}{3+(2x+1)^2}-\frac{2x+1}{x^2+x+1}... | I think that you made a mistake in the derivative $$f(x)=\frac{2}{\sqrt{3}}\tan^{-1}\left(\frac{2x+1}{\sqrt{3}}\right)-\ln(x^2+x+1)+(b^2-5b+3)x+c$$ should lead, after basic simplifications, to $$f'(x)=(b^2-5b+3)-\frac{2 x}{x^2+x+1}$$ Since we are in the real domain, $(x^2+x+1)$ does not show any real root, then $f'x)<... | {
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When does $\int_0^{\infty}\frac{7x}{x^2+1}-\frac{7C}{3x+1}dx$ converge? I am required to find the values of $C$ for which the integral
$$\int_0^{\infty}\frac{7x}{x^2+1}-\frac{7C}{3x+1}dx$$ converges.
I know by experimentation that it converges when $C=3$. I am, however, unable to show this in a rigorous way.
I get stuc... | Hints:
$$\frac{7x}{x^2+1}-\frac{7C}{3x+1}=7\cdot\,\frac{(3-C)x^2+x-C}{(x^2+1)(3x+1)}$$
Now, if $\;C\neq3\;$ then comparison with some multiple of $\;\frac1x\;$ will give you divergence...
| {
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How to factorise quadratic equation in the form of $x^2+6xy+y^2$? My Text book has some questions of quadratic equations in the form of
$x^2+xy+y^2$
1)Is there any strategy or proper way to factorise quadratic equations in such way?
2)Is that a type of quadratic trinomials?if yes why?
| Hint: let $z = \frac{x}{y}$ then $a x^2+b xy+c y^2=y^2(a z^2+b z+c)$. The latter is a quadratic in $z$. Factoring it, then substituting $\frac{x}{y}$ back gives the factorization in terms of $x,y$.
[ EDIT ] Here is the fully worked out example for factoring $8x^2+6xy+y^2$.
*
*Let $z=\frac{x}{y}$ then write $8x^2+6x... | {
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Geometry with Trigonometry From the given figure, prove that
$$\cot\ \theta=\cot\ A+\cot\ B+\cot\ C$$
I could not even start to the first point. Please help me to solve this.
Thanks in advance
| If $O$ is the intersection of the inner lines,
$\angle ABO=B-\theta\implies\angle AOB=\pi-B,$
$\angle BCO=C-\theta\implies\angle COB=\pi-C$
Writing $AB=c$, $BC=a$, $CA=b$ (conventionally)
Using Sine Law for $\triangle AOB,$ $$\dfrac{AB}{\sin(\pi-B)}=\dfrac{OB}{\sin\theta}\implies\dfrac c{\sin B}=\dfrac{OB}{\sin\theta}\... | {
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The sum of two integers is a square. The sum of their squares is also a square. The sum of two integers is a square. The sum of their squares is also a square. E.g.
$$
-3 + 4 = 1^2, \ (-3)^2 + 4^2 = 5^2
$$
$$
9 + 40 = 7^2, \ 9^2 + 40^2 = 41^2
$$
Question: General solution/form of all such pairs.
Clarification: By gener... | WLOG for $d^2=a^2+b^2, a=2mn, b=m^2-n^2$
$a+b=m^2-n^2+2mn$
Let $m^2-n^2+2mn=(m+pn)^2\iff n(p^2+1)=2m(1-p)$
$\dfrac n{2(1-p)}=\dfrac m{1+p^2}=r$(say)
Now $(1+p^2,1-p)|(1+p^2,1-p^2)$ which must divide $1+p^2+1-p^2=2$
Case $\#1:$ $(1+p^2,2(1-p))=1$ if $p$ is even
$n=2r(1-p^2),m=r(1+p^2)$
Case $\#2:$
For odd $p, p^2\equiv... | {
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Divisibility of $8^{2000}+12^{2000}-5^{2000}-1$ Choose the correct option:
$8^{2000}+12^{2000}-5^{2000}-1$ is divisible by:
$(1)$ $323$
$(2)$ $221$
$(3)$ $299$
$(4)$ $237$
Could someone give me slight hint to solve this problem?
| The expression $E := 8^{2000}+12^{2000}-5^{2000}-1$ will obviously be divisible by $K$ if each term $ \equiv 1 \bmod K$, but may also be divisible by $K$ even if this is not true.
The first thing to do is factor the four numbers given : $323= 17\times 19$, $221 = 13 \times 17$, $299 = 13 \times 23$, $237=3 \times 79$.
... | {
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Show that $\{a_n\}$ is an increasing sequence and bounded from above Question
Let $a_1$ = 1 and $a_{n+1} = \sqrt{1+2a_n}$. Show that $\{a_n\}$ is increasing and bounded from above.
Attempted solution
showing that the sequence is increasing:
\begin{align*}
a_n &\geq 1 \quad \forall n = {1,2,3,...} \\
\Rightarrow 1+2a_n ... | You have taken $a_1 = 1$. For this value $ { a_{n+1}} $ keeps on increasing. But had you taken $ a_1$ greater than $ 1\, + \, √2, a_{n+1}$ decreases. Now, for all values of $ a_1$ less than $1\, +√2, \, a_{n+1} $ increases. And as long as $a_n$ is less than $ 1+√2,$ the $a_{n+1}$ is also less than $1+√2.$ Thus for $... | {
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Integrate $I=\int_0^1\frac{\arcsin{(x)}\arcsin{(x\sqrt\frac{1}{2})}}{\sqrt{2-x^2}}dx$ How to prove
\begin{align}
I &= \int_0^1\frac{\arcsin{(x)}\arcsin{(x\sqrt\frac{1}{2})}}{\sqrt{2-x^2}}dx \\
&= \frac{\pi}{256}\left[ \frac{11\pi^4}{120}+2{\pi^2}\ln^2{2}-2\ln^4{2}-12\zeta{(3)}\ln{2} \right]
\end{align}
By asking ... | This integral can be done by recognizing that
$$\frac{\arcsin{\frac{x}{\sqrt{2}}}}{\sqrt{2-x^2}} = \sum_{n=0}^{\infty} \frac{2^n x^{2 n+1}}{(2 n+1) \binom{2 n}{n}}$$
and that
$$\int_0^1 dx \, x^{2 n+1} \arcsin{x} = \frac{\pi}{4 (n+1)} \left [1- \frac1{2^{2 n+2}} \binom{2 n+2}{n+1}\right ] $$
To see this, integrate by p... | {
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Derivative of sum over sum? Consider the following expression:
$$f(r)=\sum_{t=1}^T\frac{c}{(1+r)^t\sum_{k=1}^T\frac{c}{(1+r)^k}}\cdot t$$
with $c,r,T>0$.
Is it true, that $f(r)$ is decreasing in $r$?
I.e. $f(0)\geq f(r),\quad \forall r>0$?
| Hint: The function can be considerably simplified, which should ease the analysis of monotonicity.
In the following I use for convenience only $f_T$ instead of $f$ to indicate the dependency of the parameter $T>0$.
Observe that the inner sum does not depend on the index $t$ of the outer sum and can so be factored out.... | {
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Show that $a^{bc}+b^{ca}+c^{ab}>2$ for $a,b,c\in(0,1)$
Show that $a^{bc}+b^{ca}+c^{ab}>2$ for $a,b,c\in(0,1)$.
I have no idea how to attempt this exponential inequality and couldn't find the condition for equality. Any hints will be appreciated.
| Let $\{x,y\}\subset(0,1)$.
Hence, by Bernoulli $\left(\frac{1}{x}\right)^y=\left(1+\frac{1}{x}-1\right)^y\leq1+y\left(\frac{1}{x}-1\right)=\frac{x+y-xy}{x}$.
Id est, $\sum\limits_{cyc}a^{bc}\geq\sum\limits_{cyc}\frac{a}{a+bc-abc}$.
Thus, it remains to prove that $\sum\limits_{cyc}\frac{a}{a+bc-abc}\geq2$, which is
$2a^... | {
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Find $w+\frac{1}{w} + \frac{1}{w^2}$ where $w^4=1$, $w\neq 1$ Let $w$ be a root of the equation $z^4 = 1$ and $w$ is not equal to $1$.
I would like to find the value of $w+\frac{1}{w} + \frac{1}{w^2}$.
So for this question since w is a root,
I first substituted w and got $w^4=1$
Which is $w * w^3 = 1$.
Since $w$ is ... | Solution one:
The four roots are $1,-1,i,-i$. Simply plugging in the last three we get $-1$.
Solution two:
$z^4-1=0$ implies $(z^2-1)(z^2+1)=0$ implies $(z-1)(z+1)(z^2+1)=0$. Since we don't consider $1$, we may divide by $(z-1)$ to get: $(z+1)(z^2+1)=0$ or $z^3+z^2+z+1=0$.
Note $$\omega+\frac{1}{\omega}+\frac{1}{\omega... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2000113",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
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functional equation of type $f(x+f(y)+xf(y)) = y+f(x)+yf(x)$
If $f:\mathbb{R}-\{-1\}\rightarrow \mathbb{R}$ and $f$ is a differentiable function that satisfies $$f(x+f(y)+xf(y)) = y+f(x)+yf(x)\forall x,y \in \mathbb{R}-\{-1\}\;,$$ Then value of $\displaystyle 2016(1+f(2015)) = $
$\bf{My\; Try::}$ Using partial Differ... | Your last equation tells us that
$$
\frac{(1+x)f'(x)}{1+f(x)}
$$
has to be constant, and that this constant has to be equal to its reciprocal, i.e., it is $\pm1$.
Integration gives
$$
\ln|1+f(x)|=\pm \ln|1+x|+C
$$
and thus
$$
f(x)=C(1+x)-1\text{ or } f(x)=\frac{C}{1+x}-1
$$
This now needs to be tested against the origi... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2001644",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
"answer_id": 0
} |
Find the inverse of 17 mod 41 Questions
(1) Find the inverse of $17 \mod 41$.
(2) Find the smallest positive number n so that $17n \equiv 14 \pmod{41}$
For the first question, my attempt is as follows:
$$41-17\cdot2=7$$
$$17-7\cdot2=3$$
$$7-3\cdot2=1$$
$$7-2(17-7\cdot2)=1$$
$$7-2\cdot17=1$$
$$41-17\cdot2-2\cdot17=1$$
... | \begin{align*}
41 &= 2 \cdot 17 + 7 & 7 &= 1 \cdot 41 - 2 \cdot 17 \\
17 &= 2 \cdot 7 + 3 & 3 &= 1 \cdot 17 - 2 \cdot (1 \cdot 41 - 2 \cdot 17) \\
&& &= 5 \cdot 17 - 2 \cdot 41 \\
7 &= 2 \cdot 3 + 1 & 1 &= 1 \cdot 7 - 2 \cdot 3 \\
&& &= 1\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2003116",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 8,
"answer_id": 2
} |
Limit of product with cubes $\lim\limits_{n \to \infty}\frac{2^3-1}{2^3+1}\frac{3^3-1}{3^3+1}\dots\frac{n^3-1}{n^3+1}$ I am trying to evaluate
$$\lim\limits_{n \to \infty}{2^3-1 \over 2^3+1}{3^3-1 \over 3^3+1}\dots{n^3-1 \over n^3+1}$$
It seems to be the perfect candidate for a factoring formula, however, I get stuck r... | $$
\begin{align}
\prod_{k=2}^\infty\frac{k^3-1}{k^3+1}
&=\lim_{n\to\infty}\prod_{k=2}^n\frac{(k-1)\left(k^2+k+1\right)}{(k+1)\left(k^2-k+1\right)}\\
&=\lim_{n\to\infty}\underbrace{\prod_{k=2}^n\frac{k-1}{k+1}}\underbrace{\prod_{k=2}^n\frac{k(k+1)+1}{(k-1)k+1}}\\
&=\lim_{n\to\infty}\,\frac{1\cdot2}{n(n+1)}\quad\frac{n(n... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2004071",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
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How to symbolically solve a system of linear equations A fellow user had an interesting question:
Less-tedious way of solving this system of linear equations?
Is there a shortcut to solving for $w_1, w_2, w_3$ in $$
\left[ \begin{array}{ccc} 1 & 1 & 1\\ \frac{3}{4}a+\frac{1}{4}b &
\frac{1}{2}a+\frac{1}{2}b & \fra... | One way is to use the free Maxima computer algebra system.
You can enter your matrix like this: | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2004873",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Periodic Continued Fraction Formula for $\sqrt{a^2+4}$ for any non-zero integer $a$ If $$x = \sqrt{ a^2 + 4 } $$ then I have conjectured that:
$$ x = [ a ; \overline{ \frac{a}{2} , 2a }] \ \ \ \text{ if a is even}$$
and $$ x = [ a ; \overline{ \frac{a}{2}, 1, 1, \frac{a}{2}, 2a } ]\ \ \ \text{ if a is odd} ... | Follow the usual development of a continued fraction. It is important to express the first term as $\frac {a-1}2$ as that can be used in algebra instead of $\frac a2$ rounded down.
$$\frac 1{\sqrt{a^2+4}-a}=\frac {a-1}2+\frac {\sqrt{a^2+4}-a+2}4\\
\frac 4{\sqrt{a^2+4}-a+2}=1+\frac {\sqrt{a^2+4}-2}a\\
\frac a{\sqrt{a^2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2008944",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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How to find the coefficient of $x^{98}$ in $(x-1)(x-2)(x-3).....(x-100)$ Anyone please teach me how to solve such questions... I am thinking of solving it using binomial theory but still no idea how to do it..
Plz help
| $$(x-1)(x-2)(x-3)…(x-100)=x^{100} (1-\frac{1}{x})(1-\frac{2}{x})...(1-\frac{100}{x})$$
Denote $y=\frac{1}{x}$. Then
$$x^{100} (1-\frac{1}{x})(1-\frac{2}{x})...(1-\frac{100}{x}) =x^{100} (1-y)(1-2y)...(1-100y)$$
and the problem becomes: Find the coefficient of $y^2$ in
$$P(y)=(1-y)(1-2y)...(1-100y)$$
This coefficient ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2010183",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Inequality $9^x-k\cdot 3^x-k+3\leq 0$ is satisfied for at least one real $x$
If the Inequality $9^x-k\cdot 3^x-k+3\leq 0$ is satisfied for at least one real $x$ in $(0,1)\;,$ Then $k\in $
$\bf{My\; Try::}$ Let $3^x=y>0\;,$ Then inequality is
$$y^2-ky-k+3\leq 0\Rightarrow y^2-ky-k+3\leq 0\forall y\in (1,3)$$
So $$y... | Given: the inequality $9^x-k3^x -k + 3 \leq 0$ has at least one real x in $(0,1)$.
Transform the above inequality by replacing $3^x$ by $y$ to get
$y^2 - ky -k + 3 \leq 0$ which has atleast one real $y$ in $(3^0, 3^1) = (1,3)$. Now consider the equality $y^2 - ky -k + 3 = 0$. Note that the coefficient of $y^2$ is great... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2015281",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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To show: $\det\left[\begin{smallmatrix} -bc & b^2+bc & c^2+bc\\ a^2+ac & -ac & c^2+ac \\ a^2+ab & b^2+ab & -ab \end{smallmatrix}\right]=(ab+bc+ca)^3$ I've been having quite some trouble with this question.
I'm required to show that the below equation holds, by using properties of determinants (i.e. not allowed to direc... | If $a=b=c=0$ then the equality holds.
WLG we assume $a\neq 0$ then
$$\begin{vmatrix}
-bc & b^2+bc & c^2+bc\\
a^2+ac & -ac & c^2+ac \\
a^2+ab & b^2+ab & -ab
\end{vmatrix}\\=\frac{1}{a}\begin{vmatrix}
a(-bc)+b(a^2+ac)+c(a^2+ab) & b(ac+bc+ca) & c(ac+bc+ca)\\
a^2+ac & -ac & c^2+ac \\
a^2+ab & b^2+ab & -ab
\end{vmatrix}\\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2016733",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 4,
"answer_id": 0
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Prove the inequality $xyz \geq xy+yz+xz \implies \sqrt{x y z } \geq \sqrt{x}+\sqrt{y}+\sqrt{z}$ Let $x,y,x>0$ and $xyz \geq xy+yz+xz.$ Prove that
$$\sqrt{x y z } \geq \sqrt{x}+\sqrt{y}+\sqrt{z}. $$
Solution.
Using AM GM inequality we have
\begin{gather*}
x y+xz \geq 2 \sqrt{x y x z}=2 x \sqrt{yz},\\
xy+yz \geq 2 y ... | You could use this famous inequality: $$xy+yz+xz\geq \sqrt{3xyz(x+y+z)}$$ (which can be easily proved by AM-GM).
Using that inequality we have $$xyz\geq xy+yz+zx\geq \sqrt{3xyz(x+y+z)},$$
from where is $$\sqrt{xyz}\geq \sqrt{3(x+y+z)}=\sqrt{(1+1+1)(x+y+z)}$$
By Cauchy-Schwartz inequality, we have
$$\sqrt{(1+1+1)(x+y+z)... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2017063",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 1
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Show the coefficient of $x^3$ in the expansion of $(1+x+x^2+x^3)(1+x^2)(1+x^3)$ is the same as the number of partitions of $3$ (let a partition of $n \in \Bbb N, n>0 $ be a sum of positive integers)
How would you show that the number of different partitions of $3$ is the coefficient of
$x^3$
in the
expansion of:
$(1+x... | Here are two variations, both being equally valid with respect to the stated problem. Maybe the second variant is preferable since it provides some additional insight to the representation of the polynomial.
Algebraic variant:
We calculate the coefficient of $x^3$ of the polynomial and compare it with $p(3)$, the numb... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2019205",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
} |
Showing an infimum of an expression involving alternating series is positive in a given variable as another variable tends to infinity Let $\Bbb D$ be the unit disc in the complex plane. For the last month or so, I have been trying to prove that, if $\lambda \in \Bbb D$, and if $n \geq 4$ is an integer,
$$\inf_{\lambda... | The following partial answer only shows that the infimum above is positive, and does so in a bit of a hack-y manner. An elegant answer would be interesting.
Let $\alpha_n$ and $\beta_n$ be the expressions multiplied respectively by $x$ and $y$ in the question. We first show that $\alpha_n^2 + \beta_n^2 < 1$ for $ n \g... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2019397",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
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Factorise $f(x) = x^3+4x^2 + 3x$ Not sure if this belongs here, but I'm slowly trudging through my studies for Math 1 and wondered if y'all could give feedback and/or corrections on the following factorisation question:
$$ \text{Factorise}: f(x) = x^3+4x^2+3x $$
Firstly, the GCD of the above is $x$:
$$x(x^2+4x+3)$$
Now... | Everything is more or less correct about the way you approach the problem. You could have also opted for the middle term factorisation method or the method of vanishing method. However you have to correct one thing ...
Now factorise the rows and columns:
$$ x^2 + 1x = x(x+1)\\
x^2 + 3x = x(x+3)\\
1x + 3=1(x+3)\\
3x ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2020321",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
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Find all positive three - digit integers $\overline{abc}$ such that $\overline{abc}=5abc$ Number is divisible by $5$ if the last digit is $0$ or $5$. Since $c\neq 0$, $c=5$.
Using brute force (checking each combination with three digits), I get that the only solution is $175$ : $a=1,b=7,c=5$.
How to solve a problem ef... | Once you have $c=5$, it becomes clear that $100a+10b+5=25ab$. Manipulating this, we have:
$$
0=5ab-20a-2b-1=5a(b-4)-2(b-4)-8-1\implies \color{blue}{\boxed{9=(5a-2)(b-4)}}.
$$
Note that $5a-2$ can only be $-2$, $3$, $8$ or bigger than $9$. We conclude that $5a-2=3=b-4$, yielding the answer you have found.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2020594",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
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Solve for $x$ in $3\cdot 5^{x^2}+3^{x^2+1}-8\cdot 3^{x^2}=0$ $3\cdot 5^{x^2}+3^{x^2+1}-8\cdot 3^{x^2}=0$
Note that I didn't forget any parenthesis. I copied it as it is in my book.
I tried:
$$3\cdot 5^{x^2}+3^{x^2+1}-8\cdot 3^{x^2}=0 \Leftrightarrow 3\cdot 5^{x^2} - 5 \cdot 3^{2x} = 0 \Leftrightarrow 5 \cdot 3^{x^2} = ... | $3^{x^2+1}$ indeed equals $3\cdot 3^{x^2}$. Proceeding from your last equality:
$$
5\cdot 3^{x^2}=3\cdot 5^{x^2}\iff 3^{x^2-1}=5^{x^2-1}\iff(x^2-1)\log(3/5)=0.
$$
The last $\iff$ above is obtained by taking $\log$ and simplifying. Can you continue from here?
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Prove that $\triangle ABC$ is equilateral, If $(a+b-c)^{a}\cdot (b+c-a)^{b}\cdot (c+a-b)^{c}\geq a^a\cdot b^b\cdot c^c$
If $a,b,c$ are the sides of a $\triangle ABC\;,$ such that $$\left(1+\frac{b-c}{a}\right)^{a}\left(1+\frac{c-a}{b}\right)^{b}\left(1+\frac{a-b}{c}\right)^{c}\geq 1,$$
then proving $\triangle$ is equi... | The title's inequality is wrong. I'll prove a reversed inequality.
Let $a=y+z$, $b=x+z$ and $c=x+y$.
Since our inequality does not changed after substitution $a\rightarrow ka$...,
we can assume that $x+y+z=3$.
And we need to prove that $\sum\limits_{cyc}(3-x)\ln\frac{2x}{3-x}\leq0$ or
$\sum\limits_{cyc}\left(3(x-1)-(3-... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2021321",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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For 1/A = 1/B + 1/C, why is A $\frac{1}{a}$ =$\frac{1}{b}$ +$\frac{1}{c}$
I read that $a$ is always < the smaller of $b$ and $c$
In the case of $0$ < $B$ $<1$ and $0$ < $C$ $<1$,
I can understand the rule as:
Firstly,
$$\frac{1}{A}=\frac{1}{B} +\frac{1}{C}=\frac{C+B}{B*C}$$
Now looking at 3 cases:
Case 1
$$\fr... | Since everything is strictly positive, it's simply the fact that $$\frac1A=\frac1B+\frac1C>\frac1B$$ And $$\frac1A>\frac1B\implies B>A$$ Same for $C$. Since $A$ is smaller than boh, it is smaller than the minimum.
I don't really understand why in you decided to consider only three numbers.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Simplify $\frac {2\sqrt{x^2+x+1}-2-x} {x^2} $ In order to find the limit of the function at $0$ I need to simplify the fraction $$\frac {2\sqrt{x^2+x+1}-(2+x)} {x^2} $$
so that at $0$ it didn't look like $\frac 0 0$. I'm almost certain that it can be solved with a clever substitution of the variable but I didn't manage... | Hint. By Taylor expansion, it is easy to see that for $t\to 0$,
$$\sqrt{1+t}=1+\frac{t}{2}-\frac{t^2}{8}+o(t^2).$$
Hence
$$2\sqrt{x^2+x+1}=2\sqrt{1+(x+x^2)}=2+x+x^2-\frac{(x+x^2)^2}{4}+o((x+x^2)^2)\\=2+x+\frac{3}{4}x^2+o(x^2).
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2022345",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
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Let $\frac{{{x^5}}}{{1 - {x^5} + {x^{10}}}} = k$. What is $\frac{{{x^{15}}}}{{1 - {x^{15}} + {x^{30}}}}$? Let $\frac{{{x^5}}}{{1 - {x^5} + {x^{10}}}} = k$ and $k\in \mathbb{R}$.
What is $\frac{{{x^{15}}}}{{1 - {x^{15}} + {x^{30}}}}$?
| Note that
$$
k=\frac{x^5}{1-x^5+x^{10}}=\frac{1}{x^{-5}-1+x^{5}}\implies x^{-5}-1+x^{-5}=k^{-1}.
$$
It follows that
\begin{aligned}
k^{-2}&=1+(x^{-10}+x^{10})-2(x^{-5}-1+x^5)\implies(x^{-10}+x^{10})=k^{-2}-1+2k^{-1},\\
k^{-3}&=(x^{-5}-1+x^{-5})^3=(x^{-15}-1+x^{15})+6(x^{-5}-1+x^5)-3(x^{-10}+x^{10}).
\end{aligned}
We co... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Is $\frac{a^4}{(b-a)(c-a)}+\frac{b^4}{(c-b)(a-b)}+\frac{c^4}{(a-c)(b-c)} $ always an integer? In a textbook I found the rather strange identity:
$$ \frac{2^4}{(5-2)(3-2)}+\frac{3^4}{(5-3)(3-2)}+\frac{5^4}{(5-3)(5-2)}= \frac{414}{6}=69 $$
just kind if out of nowhere and I wonder if it generalizes and why. Perhaps it is ... | The statement is true even when you replace the numerators by any polynomial with integer coefficients or increase the number of variables. More precisely,
For any $f(x) \in \mathbb{Z}[x]$ and distinct $a_1, \ldots, a_n \in \mathbb{Z}$,
$\displaystyle\;\sum\limits_{k=1}^n \frac{f(a_k)}{\prod\limits_{j=1,\ne k}^n (a_... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2023628",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 3,
"answer_id": 1
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Solving $a^2-b^2-a=0$ ∧ $-2ab+b={1\over 2}$ I want to solve
$$a^2-b^2-a=0 ∧ -2ab+b={1\over 2}$$
I'm a bit stuck in the first part. I thought of the solutions, $a={1±\sqrt {1+4b^2} \over 2}$. But can't figure out solutions to the second part. Am I approaching it the right way?
| I would solve it as $b=\pm \sqrt{a^2-a}$ and $b=\frac{1}{2(1-2a)}$.
So $$a^2-a = \frac{1}{4(1-2a)^2}$$
Or:
$$\left(2a-1\right)^2 - 1 = \frac{1}{(1-2a)^2}$$
Letting $u=(2a-1)^2$, you have $u-1=\frac{1}{u}$ or $u^2-u-1=0$ or $u=\frac{1\pm\sqrt{5}}{2}$. Then:
$$a=\frac{1\pm\sqrt{\frac{1\pm\sqrt{5}}{2}}}{2}$$
Not that $a$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2030374",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
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Finding the length of the shortest ladder
A wall 8 feet high is 1 foot from a house. Find the length $L$ of the
shortest ladder over the wall to the house. Draw a triangle with
height $y$, base $1 + x$, and hypotenuse $L$.
The shortest ladder is placed straight up close to the wall. It has length $8+8=16$ for goi... |
The triangles $ABC$ and $AB'C'$ satisfy the relation
$$
\frac{y}{1 + x} = \frac{8}{x}
$$
So that
$$
y = \frac{8}{x}(1 +x) \tag{1}
$$
The length of the ladder (red line) follows from
$$
L^2 = (1 + x)^2 + y^2 \stackrel{(1)}{=} (1 + x)^2 + \frac{64}{x^2}(1 + x)^2 = (1 + x)^2\frac{x^2 + 64}{x^2}
$$
Call $f(x) = L^2$, a... | {
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"url": "https://math.stackexchange.com/questions/2030973",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Evaluation of $\sum_{n=1}^{\infty} (\frac{1}{4n-3}-\frac{1}{4n-1}) $ Prove that
$$\sum_{n=1}^{\infty} (\frac{1}{4n-3}-\frac{1}{4n-1})=\frac{\pi}{4}$$
Could someone give me slight hint as how to convert above expression to $\int_{0}^{1} \tan^{-1}x.dx$
| By IBP,
\begin{eqnarray*}\int \tan^{-1}x.dx&=&x\times tan^{-1}x|-\int x\times \frac{1}{1+x^2}.dx \\
&=&x\times tan^{-1}x|-\int \frac{d(x^2+1)}{2dx}\times \frac{1}{1+x^2}.dx\\
&=&x\times tan^{-1}x|-\ln|x^2+1| |
\end{eqnarray*}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2032479",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 1
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Finding and proving linear independence of solutions of second order differential equation Show that the solutions of the differential equation $2x^2y'' + 10xy' + 8y = 0$ are linearly
independent in two ways and find its particular solution satisfying the initial conditions
$y(1) = 3$ and $y'(1) = 5$
I have tried this ... | Your equation is a Cauchy-Euler equation which can be solved by assuming that the solution is of the form $y(x) = x^k$ for some $k \in\mathbb{R}$.
Plugging this in gives you:
$$2x^2(k-1)kx^{k-2} + 10xkx^{k-1} + 8x^k = 0$$
Factoring out $x^k$, we obtain:
$$x^k(2k^2 - 2k + 10k + 8) = x^k(2k^2+8k+8) = 0$$
However, this mu... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2033118",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Convergence and evaluation of $\sum_{n=1}^{\infty}\ln {(2n+1)n \over (n+1)(2n-1)}$ I am trying to evaluate
$$\sum_{n=1}^{\infty}\ln{(2n+1)n \over (n+1)(2n-1)}$$
First, we notice
$${(2n+1)n \over (n+1)(2n-1)} = {2n^2+n -1 + 1 \over 2n^2+n-1} = 1 + {1 \over 2n^2+n-1}= 1 + {1 \over (n+1)(2n-1)}$$
Then, we use ${x \over... | Hint. This may be seen a limit of two telescoping sums:
$$
\sum_{n=1}^N\ln{(2n+1)n \over (n+1)(2n-1)}=\sum_{n=1}^N\left(\ln(2n+1)-\ln(2n-1)\right)-\sum_{n=1}^N\left(\ln(n+1)-\ln(n)\right).
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2034917",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
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Calculating sign of a permutation of unknown size, but with a pattern I'm trying to calculate the determinant of the following matrix, using the Leibniz formula:
$$
\begin{pmatrix}
0 & 1 & 0 & \cdots & 0 & 1 \\
1 & 0 & 1 & \ddots & \ddots & 0\\
0 & 1 & \ddots & \ddots & \ddots & \vdots\\
\vdots & \ddots & \ddots & \ddo... | $\mathrm{sgn}(\sigma)=(-1)^{\mathrm{inv}(\sigma)}$, where $\mathrm{inv}(\sigma)$ is the number of inversions of $\sigma$, that is to say, the number of pairs of positions $(i,j)$ where $i<j$ but $\sigma(i)>\sigma(j)$ (a larger value is to the left of a smaller value).
For the permutations you listed, the numbers of in... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2040819",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
An interesting definite integral $\int_0^1(1+x+x^2+x^3+\cdot\cdot\cdot+x^{n-1})^2 (1+4x+7x^2+\cdot\cdot\cdot+(3n-2)x^{n-1})~dx=n^3$ How to prove
$~~ \forall n\in\mathbb{N}^+$,
\begin{align}I_n=\int_0^1(1+x+x^2+x^3+\cdot\cdot\cdot+x^{n-1})^2 (1+4x+7x^2+\cdot\cdot\cdot+(3n-2)x^{n-1})~dx=n^3.\end{align}
My Try:
Define ... | First apply the substitution $x = t^3$. Then
\begin{align*}
I_n
&= \int_{0}^{1} (1 + t^3 + \cdots + t^{3n-3})^2 (1 + 4t^3 + \cdots + (3n-2)t^{3n-3}) \cdot 3t^2 \, dt \\
&= \int_{0}^{1} 3 (t + t^4 + \cdots + t^{3n-2})^2 (1 + 4t^3 + \cdots + (3n-2)t^{3n-3}) \, dt.
\end{align*}
Now let $u = u(t) = t + t^4 + \cdots + t^{3n... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2042986",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "21",
"answer_count": 1,
"answer_id": 0
} |
Prove that $\vert\sin(x)\sin(2x)\sin(2^2x)\cdots\sin(2^nx)\vert < \left(\frac{\sqrt{3}}{2}\right)^n$ For $x$ in $\mathbb R^*$ and $n$ in $\mathbb N$, define
$$U_n = \sin(x)\sin(2x)\sin(2^2x)\cdots\sin(2^nx) = \prod_{k=0}^n\sin(2^k x)$$
Prove that $$\vert U_n\vert \leq \left(\frac{\sqrt{3}}{2}\right)^n$$
Can anyone help... | Easy to see that $|U_1(x)|\leq\frac{\sqrt3}{2}$ and $|U_2(x)|\leq\frac{3}{4}$ for every $x$.
Assume that $|U_k(x)|\leq\left(\frac{\sqrt3}{2}\right)^k$ for every real $x$ and every $k\leq n$, for some $n\ge2$.
*
*If $|\sin{x}|\leq\frac{\sqrt3}{2}$, then $$|U_{n+1}(x)|=|\sin x|\cdot|U_n(2x)|\leq\frac{\sqrt3}{2}\cdot\l... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2043127",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 2,
"answer_id": 0
} |
Rate of convergence of ${x_n}=\arctan ({x_{n-1}})$ Find the rate of convergence of :
${x_n}=\arctan ({x_{n-1}})$. Well,
$$\lim\limits_{n\to\infty}\frac{{x_{n+1}}}{{x_n}}=\lim\limits_{n\to\infty}\frac{\arctan({x_n})}{\arctan({x_{n-1}})}=l,$$
where $l$ is the point where the series converges to. I can't calculate this ... | As pointed out in comments, both the limits $\lim_{n\to\infty} \frac{x_{n+1}}{x_n}$ and $\lim_{n\to\infty} \frac{x_{n+1}}{x_n^p}$ are easy to compute given the observation that $x_n \to 0$. So let me demonstrate a method of finding an asymptotic expansion of $(x_n)$ instead.
A hint is that you can estimate the growth ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2043709",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Finding the joint Probability distribution of $X$ and $Y$? If the joint probability distribution of X and Y is given by
$$f(x,y)= \frac{(x-y)^2}{7}, \text{for }x=1,2,3;y=1,2 $$
$(1)$ Find the probability distribution of $U = X + Y; $
$(2)$ Find the conditional probability distribution of $X$ given $U =4.$
In order to s... | Ok so the first thing you notice is that so far your attempt has
$$
\begin{align}
P(U = 2) &= 0, \\
P(U = 3) &= \frac{1}{7}, \\
P(U = 4) &= \frac{4}{7}, \\
P(U = 5) &= \frac{1}{7}
\end{align}
$$
and zero elsewhere, but summing over all possible situations only takes us to $\frac{6}{7}$ so something has clearly gone wro... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2043984",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Finding the area inside the plot $x^4+y^4=x^2+y^2$ Find the area inside the plot $x^4+y^4=x^2+y^2$.
| Using polar coordinates, we have $x = r\cos \theta, y = r\sin \theta\implies x^4+y^4 = r^4(\cos ^4 \theta + \sin^4 \theta)= r^4(1 - 2\sin^2\theta\cos^2\theta)= x^2+y^2 = r^2\implies r^2 = \dfrac{1}{1-2\sin^2\theta\cos^2\theta}\implies r = \dfrac{1}{\sqrt{1-2\sin^2\theta\cos^2\theta}}\implies \text{Area} = 4\displaystyl... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2044460",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 5,
"answer_id": 0
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How to find the coordinates under a basis when variables are involved I understand how to use row operations when trying to find coordinates of a vector under a basis when both are involving numbers. Simply create an augmented matrix and row reduce. But how do I go about finding the coordinates of the vector ($2x+1+ \f... | Writing the vector as a combination of the base we have:
$$2x+1+ \frac{x^3}{3} + \frac{x^2}{2}=a(x)+b(x^2+x)+c(3-2x)+d(x^3)$$
$$2x+1+ \frac{x^3}{3} + \frac{x^2}{2}=dx^3+bx^2+(a+b-2c)x+3c$$
and using polynomial identity we get:
$$d=1/3$$
$$b=1/2$$
$$a+b-2c=2$$
$$3c=1$$
so
$$a=13/6$$
$$b=1/2$$
$$c=1/3$$
$$d=1/3$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2045263",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
} |
Solve $\frac{5}{x+3}+\frac{2}{2x+6}=4$ I want to solve $\frac{5}{x+3}+\frac{2}{2x+6}=4$ for $x$
I try this;
$5(2x+6)+2(x+3)=4(x+3)(2x+6)$
$12x+36 = 4(2x^2+12x+18)$
$8x^2+36x+36=0$
Where I would then go to the formula, however the answer says -1.5, what am I doing wrong.
| The quadratic equation you have obtained has two solutions: $-3$ and $-\frac{3}{2}$. The latter one is the correct answer. The former one is incorrect because it would lead to division by $0$ when we plug it into the original equation.
However, the problem need not be solved via quadratic equation. Observe that
$\frac{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2046493",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 7,
"answer_id": 1
} |
Evaluate the limit (if it exists) $\{(\frac{n^3+4n-5}{n^6+2n^2-3},n-7[\frac{n}{7}])\}_{n=1}^{\infty}$
$$\left\{\frac{n^3+4n-5}{n^6+2n^2-3},n-7\left[\frac{n}{7}\right]\right\}_{n=1}^{\infty}$$
where $\left[\frac{n}{7}\right]$ is the floor and ceiling function
The sequence converges if each element convarges so:
$$\lim... | The remainder-like function
$$
n-7\left[\frac n7\right]
$$
cycles among the values $\{0,1,2,3,-3,-2,-1\}$. Since one of the components does not converge, the sequence of pairs does not converge.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2050851",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Prove that $\sum\limits_{cyc}\frac{a}{\sqrt{a+3b}}\geq\sqrt{a+b+c+d}$ Let $a$, $b$, $c$ and $d$ be positive numbers. Prove that:
$$\frac{a}{\sqrt{a+3b}}+\frac{b}{\sqrt{b+3c}}+\frac{c}{\sqrt{c+3d}}+\frac{d}{\sqrt{d+3a}}\geq\sqrt{a+b+c+d}$$
I tried Holder, AM-GM and more, but without success.
| WLOG, assume that $d = \min(a,b,c,d)$.
Squaring both sides and using AM-GM, it suffices to prove that
\begin{align}
&\sum_{\mathrm{cyc}} \frac{a^2}{a+3b}
+ \sum_{\mathrm{cyc}} \frac{4ab}{a+3b + b+3c}
+ \frac{4ac}{a+3b + c+3d} + \frac{4bd}{b+3c + d+3a} \\
\ge\ & a + b + c + d.
\end{align}
After clearing the denominators... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2051278",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "22",
"answer_count": 2,
"answer_id": 1
} |
Evaluate $\displaystyle \int_C \frac{1}{4z^2+4z-3}$ where $C=C^{+}_{1}(0)$.
Evaluate $\displaystyle \int_C \frac{1}{4z^2+4z-3}$ where $C=C^{+}_{1}(0)$.
I am not sure whether I understand how to use Cauchy's theorem which states that:
If $f$ is a analytic in a simply connected domain $D$, and $C$ is a simple closed c... | $\displaystyle\frac{1}{4}\frac{1}{2z-1}=\frac{1}{8}\frac{1}{z-\frac{1}{2}}$, so by Cauchy Integral Formula for discs,
\begin{equation*}
\frac{1}{8}\int_{B_1(0)}\frac{dz}{z-\frac{1}{2}}=\frac{1}{8}2\pi i=\frac{1}{4}\pi i
\end{equation*}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2052220",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Simple partial fraction expansion I would like to ask for help solving a simple Partial fraction example:
(It contains complex roots)
$\frac{1}{(x^3 - 3x^2 + 4x - 2)}$
According to my book we can factor it as a product of the real roots and quadratic factors:
$\frac{1}{(x-1)(x^2-2x+2)}$
Quadratic form should adopt the ... | You need to start from
$$\frac{a}{x-1}+\frac{bx+c}{x^2-2x+2}=\frac{1}{(x-1)(x^2-2x+2)}$$
$$a(x^2-2x+2)+(bx+c)(x+1)=1$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2052705",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Proving one of the Binomial Identities The binomial identity is
$$\sum^{\left \lfloor{n/3}\right \rfloor }_{k=-\left \lfloor{n/3}\right \rfloor} (-1)^k \binom{2n}{n+3k} = 2 \cdot 3^{n-1} $$
and this is valid for all positive integers $n$. What would be some proofs to show that this identity is true?
| This can be proved by induction on $n$. Instead of writing it up as efficiently as possible, I’ll show you how I got there.
First note that there’s no need to specify limits on the index $k$: the sum is taken over all $k$ for which the binomial coefficient is non-zero, with the usual convention that $\binom{n}k=0$ if $... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Four Equations in Four Unknowns
Four Equations in Four Unknowns
Completely solve the following equation!
$$\begin{eqnarray}
x&+&y&+&z&+&w&=&10 \\
x^2&+&y^2&+&z^2&+&w^2&=&30 \\
x^3&+&y^3&+&z^3&+&w^3&=&100\\
&&&&&&xyzw&=&24
\end{eqnarray}$$
This is problem 3 on page 3 of
Mathematical Quickies
270 Stimulating Prob... | I. Yes, you are correct that a univariate polynomial of degree $n$ has $n$ zeros (counting multiplicity). However, your system are just the roots of a univariate in disguise.
The clue is the elementary symmetric polynomials $x+y+z+w$ and $xyzw$. If these unknowns are indeed the roots of the quartic,
$$F(u)=u^4+au^3+bu... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2055106",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 1,
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Inverse laplace transform $1/(s^2+9)^2$
Find inverse laplace transform of $$\frac{1}{(s^2+9)^2}$$
I've tried to decompose the fraction using
$$\frac{As+B}{s^2+9}+\frac{Cs+D}{(s^2+9)^2}$$
$$1=(As+B)(s^2+9)+Cs+D$$
yet D=1, still giving me the same exact equation
$$\frac{1}{(s^2+9)^2}$$
any help?
| One could make an ansatz with "known" expressions from a simple Laplace Transform Table:
$$\frac1{(s^2+a^2)^2}=\frac{Aa}{s^2+a^2}+\frac{Bs}{s^2+a^2}+\frac{2Cs}{(s^2+a^2)^2}+\frac{D(s^2-a^2)}{(s^2+a^2)^2},$$
given the Laplace transforms of $\sin at$, $\cos at$, $t\sin at$, and $t\cos at$, respectively. Hence, we get $A=... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2056955",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 3,
"answer_id": 2
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Is there a "clean" way of parametrizing the intersection of the sphere and saddle? I was playing around, trying to find a parametrization for the intersection of the two surfaces,
$$x^2+y^2+z^2=1$$
$$z=x^2-y^2,$$
but I wasn't able to get anything nice-looking. Any suggestions, or is this a hopeless endeavor?
EDIT: Tha... | $$x^2+y^2=1-z^2\\
x^2-y^2=z$$
So $x^2=\frac{1+z-z^2}{2}$ and $y^2=\frac{1-z-z^2}{2}$.
Now you just need to pick the range of values for $z$ where both $1-z-z^2$ and $1+z-z^2$ are positive. This is when $1-z^2\geq |z|$ or $1-3z^2+z^4\geq 0$ and $|z|\leq 1$. But $1-3z^2+z^4=\left(z^2-\frac{3}{2}\right)^2 -\frac{5}{4}$, s... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2057231",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
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How to work with triangle inequality using rational coordinates and squared distance? Given three points in some Euclidean space, $\mathbf{r}, \mathbf{g}, \mathbf{b} \in \mathbb{R}^d$, I know via the triangle inequality that:
$$\|\mathbf{r} - \mathbf{b}\| \leq \|\mathbf{r} - \mathbf{g}\| +\|\mathbf{g} - \mathbf{b}\|.$$... | The bound for the triangle inequality is, and always is, the sum of the two euclidean distances. However, the question is how to represent this sum, and how to do calculations with it.
Let's say you have $A = |r - g|^2$ and $B = |g - b|^2$, which are rational numbers. Then the bound you want is simply $x = \sqrt{A} + \... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2057364",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
Find the Sum of an Infinite Series Compute
$${1 \cdot \frac {1}{2} + 2 \cdot \frac {1}{4} + 3 \cdot \frac {1}{8} + \dots + n \cdot \frac {1}{2^n} + \dotsb.}$$
I tried creating a partition for this but no such luck.
What would the the equation to generate the sum for the nth term
| Let $S=\sum_{n\geq 1}\frac{n}{2^n}.$ Then
$$2S=\sum_{n\geq 1}\frac{2n}{2^n}=\sum_{n\geq 1}\frac{n}{2^{n-1}}=\sum_{n\geq 1}\frac{(n-1)+1}{2^{n-1}} = 2+\sum_{n\geq 1}\frac{n-1}{2^{n-1}} = 2+S $$
hence $S=2$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2059172",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 2
} |
How many principal minors a 3x3 matrix can have? I'm reading the book Polarimetric Radar Imaging: From basics to applications
On page 75, there's this claim:
$$T_3 = \begin{bmatrix}
2A_0 & C-jD & H+jG\\
C+jD & B_0+B & E+jF\\
H-jG & E-jF & B_0-B
\end{bmatrix}$$
As the coherency $T_3$ matrix in such a case is a ra... | For a general $3\times 3$ matrix,
$$A =\begin{bmatrix}
a_{11} & a_{12} & a_{13}\\
a_{21} & a_{22} & a_{23}\\
a_{31} & a_{32} & a_{33}
\end{bmatrix}
$$ there is one third order principal minor namely $|A|$.
There are three second order principal minors:
$\begin{vmatrix}
a_{11} & a_{12}\\
a_{21} & a_{22}
\end{vmatrix}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2059590",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 1,
"answer_id": 0
} |
Prove that $\sum\limits_{cyc}\frac{a^2}{a^3+2}\leq\frac{4}{3}$ if $a, b, c, d > 0$ and $abcd=1$
Let $a$, $b$, $c$ and $d$ be positive numbers such that $abcd=1$. Prove that:
$$\frac{a^2}{a^3+2}+\frac{b^2}{b^3+2}+\frac{c^2}{c^3+2}+\frac{d^2}{d^3+2}\leq\frac{4}{3}.$$
Vasc's LCF Theorem does not help here. Also I tried ... | Let : $a= e^{x_1}\ , \ b=e^{x_2}\ , \ c=e^{x_3} \ , \ d =e^{x_4}$
So we must to prove:
$$f(x_1)+f(x_2)+f(x_3)+f(x_4) \le \dfrac{4}{3}$$
for $x_1+x_2+x_3+x_4=0 $
$$f(x)= \dfrac{e^{2x}}{e^{3x}+2}$$
Since : $f'(x)= \dfrac{e^{2x}(4-e^{3x})}{(2+e^{3x})^2}$ and $f''(x)=\dfrac{e^{2x}(e^{6x}-26e^{3x}+16)}{(2+e^{3x})^3}$
We on... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2061972",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 3,
"answer_id": 2
} |
Find the value of csc$\frac{16\pi}{3}$. Q. Find the value of csc$\frac{16\pi}{3}$.
A. This angle is equal to two full revolutions plus $\frac{4\pi}{3}$. I got this by subtracting, $\frac{16\pi}{3} - 4\pi = \frac{16\pi}{3} - \frac{12\pi}{3} = \frac{4\pi}{3}$. The terminal side is in Quadrant III. The reference angle i... | Just observe that $16=5\cdot 3+1$.
The cosecant of $16\pi/3$ is the reciprocal of the sine; from $16=5\cdot3+1$, we get
$$
\frac{16\pi}{3}=5\pi+\frac{\pi}{3}=\pi+\frac{\pi}{3}+4\pi
$$
you have
$$
\sin\frac{16\pi}{3}=\sin\left(\pi+\frac{\pi}{3}\right)=
-\sin\frac{\pi}{3}=-\frac{\sqrt{3}}{2}
$$
Therefore
$$
\csc\frac{16\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2063947",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 0
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Logic: Proving tautological consequence I'm having trouble proving this tautological consequence. I'd hope that you guys can maybe oversee my process and identify errors, because I went over this couple of times and I arrive at the same conclusion.
The question goes like this:
$A \Rightarrow B$
$C \Rightarrow B$
Ther... | The problem is really:
A implies B
C implies B is
(A implies B) and (C implies B)
And then it should be easily provable
edit:
\begin{array}{ccc|c@{}ccc@{}ccccc@{}ccc@{}ccc@{}ccc@{}c}
a&b&c&(&a&\lor&c&)&\rightarrow&b&\leftrightarrow&(&a&\rightarrow&b&)&\land&(&c&\rightarrow&b&)\\\hline
1&1&1&&1&1&1&&1&1&\mathbf{1}&&1&1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2066844",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Number of positive integers $n$ The question is to find out the number of positive integers $n$ such that $nx^4+4x+3 \leq 0 $ for some real $x$ (without using a graphic calculator).
My attempt at the solution:
We have $nx^4+4x+3 \leq 0 $ which is $nx^4 \leq -3-4x$ or $n \leq -\frac{(3+4x)}{x^4}$. It can easily be obser... | Solution WITHOUT calculus.
Note that
$$nx^4+4x+3=(n-1)x^4+x^4+4x+3=(n-1)x^4+((x-1)^2+2)(x+1)^2.$$
Hence for $n=1$, we have that the above polynomial is $0$ at $x=-1$.
If $n\geq 2$ then, for any $x\in\mathbb{R}$,
$$nx^4+4x+3\geq x^4+2(x+1)^2>0$$
because the squares $x^4$ and $(x+1)^2$ are zero for different values of $... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2067147",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 0
} |
prove $\frac{a}{b}+\frac{b}{a}+\frac{c}{d}+\frac{d}{c}\le\frac{1}{64abcd}$ Let $a,b,c,d$ be positive real no such that $a+b+c+d\le1$
prove that$$\frac{a}{b}+\frac{b}{a}+\frac{c}{d}+\frac{d}{c}\le \frac{1}{64abcd}$$
I did this way$$\frac{a^2cd+b^2cd+c^2ab+d^2ab}{abcd}\le\frac{1}{64abcd}$$
$\implies$ $a^2cd+b^2cd+c^... | By AM-GM $(ac+bd)(ad+bc)\leq\left(\frac{ac+bd+ad+bc}{2}\right)^2=\left(\frac{(a+b)(c+d)}{2}\right)^2\leq\left(\frac{\left(\frac{a+b+c+d}{2}\right)^2}{2}\right)^2\leq\frac{1}{64}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2067312",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
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Evaluation of limit $\lim_{n\rightarrow \infty}\sum^{n}_{k=1}\left(\sqrt{1+\frac{k}{n^2}}-1\right)$
Evaluation of the limit $$\lim_{n\rightarrow \infty}\sum^{n}_{k=1}\left(\sqrt{1+\frac{k}{n^2}}-1\right).$$
I don't understand how can I start the above problem because Riemann sums and integrals are not applicable here... | Note that $\sqrt{1+k/n^2} -1 = \frac{k}{2n^2} + O\left(\left( \frac{k}{n^2} \right)^2\right)$ by the Maclaurin series of $\sqrt{1+x}$. We use the notation $O(f(n))$ for some term thats bounded for sufficiently large $n$ by $C |f(n)|$ for some constant $C>0$ .
So, $\sum_{k=1}^n \left( \sqrt{1+k/n^2} -1 \right) = \sum_... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2069244",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Show that $\arctan\left({1\over A}\right)=\arctan\left({1\over B}\right)+\arctan\left({1\over C}\right)$ $1,1,2,3,5,8,...$ for $n=1,2,3,4,5,...$ ;It is the n-th Fibonacci numbers.
Show that,
$\arctan\left({1\over xF_{2n-1}+F_{2n}}\right)=\arctan\left({1\over xF_{2n-1}+F_{2n+1}}\right)+\arctan\left({1\over x^2F_{2n-1}+x... | We make use of the identity:$$F_{2n-1}^2 -F_{2n}^2 + F_{2n}F_{2n-1} = 1...(1)$$ We can rewrite this as: $$2F_{2n}F_{2n-1} + F_{2n-1}^2 = F_{2n}^2 + F_{2n}F_{2n-1} + 1$$ $$\implies F_{2n-1}[2F_{2n} + F_{2n-1}] = F_{2n}[F_{2n} + F_{2n-1}] + 1...(2)$$ Now, we use the identity: $$F_{n+m} = F_nF_{m+1} + F_{n-1}F_m$$ to chan... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2069332",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
} |
How to solve $x+x^2+x^3+x^4+x^5=62$? $$x+x^2+x^3+x^4+x^5=62$$
The roots in the set of Real numbers is $x=2$, but how solve it?
| Apply the rational roots theorem:
$$62\implies1,2,31,62$$
Thus, it is easy enough to check if $x=\pm1,\pm2,\pm31,\pm62$ are roots. Indeed, if we check, we find that $x=2$ is the only rational root. The remaining roots can be deduced to be complex since after factoring out, we find it has only one real relative minimu... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2069590",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Proving $\sqrt{x+y-x^2}+\sqrt{y+z-y^2}+\sqrt{z+w-z^2}+\sqrt{w+x-w^2}\le4\sqrt2-3$ I found this inequality using unusual calculations in maths Olympics
and I wonder if some clever teenager could prove it using their elementary knowledge of mathematics.
Let $x,y,z,w$ be non-negative numbers such that $$x+y+z+w=1$$ Pro... | It depends on what you consider to be elementary.
By cauchy-schwarz:
$$ \sum\limits_{cyc} \sqrt{x+y-x^2} \leq2\sqrt{\sum\limits_{cyc}x+y-x^2}=2\sqrt{\frac{7}{4}+\frac{1}{4}(x+y+w+z)^2-(x^2+y^2+w^2+z^2)}.$$
Now, by jenson we have that $$(x+y+w+z)^2\leq 4(x^2+y^2+z^2+w^2). $$
Hence we conclude that $$ \sum\limits_{cyc} ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2069671",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
$t^3$ is never equal to
Let $t$ be a real number such that $t^2=at+b$ for some positive integers $a$ and $b$. Then for any choice of positive integers $a$ and $b$, $t^3$ is never equal to:
(A). $4t+3$
(B). $8t+5$
(C). $10t+3$
(D). $6t+5$
We have $t^3 = at^2+bt$ and so for (A) we have $t^3 = 4t+3 = 4(at^2+bt)+3$. I do... | $$t^2=at+b\\ t^3=at^2+bt \to t^3=a(t^2)+bt\\t^3=a(at+b)+bt \\t^3=(a^2+b)t+ab$$
$$t^3=(a^2+b)t+ab=4t+3 \to ab=3 \to \\a=1 ,b=3 \to t^3=(1^2+3)t+1.3 \\$$
$$t^3=(a^2+b)t+ab=10t+3 \to ab=3 \to \\a=3 ,b=1 \to t^3=(3^2+1)t+1.3 \\$$
$$t^3=(a^2+b)t+ab=6t+5 \to ab=5 \to \\a=1 ,b=5 \to t^3=(1^2+5)t+1.5 \\$$
so answer is :"B"
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2071932",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
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