Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
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Given that $f(x)$ is a polynomial of degree $3$, and its remainders are $2x - 5$ and $-3x + 4$ when divided by $x^2 - 1$ and $x^2 - 4$ respectively. So here is the Question :-
Given that $f(x)$ is a polynomial of degree $3$, and its remainders are $2x - 5$ and $-3x + 4$ when divided by $x^2 - 1$ and $x^2 - 4$ respectiv... | Hint:
Let $$\dfrac{f(x)}{(x-1)(x+1)(x-2)(x+2)}=\dfrac A{x-1}+\dfrac B{x+1}+\dfrac C{x-2}+\dfrac D{x+2}$$
$$\implies f(x)=?$$
where $A,B,C,D$ which are arbitrary constants which can be found by putting $x=1,-1,2,-2$ one by one.
For example, put $x=1,$ $$2\cdot1-5=A(1+1)(1-2)(1+2)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3770874",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 5,
"answer_id": 1
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Correct way to integrate $\int x(x^2-16)dx$
Evaluate:
$$\int x(x^2-16)dx$$
I have noticed that this integral can be solved using two different methods, but I am not sure which one is the correct one.
Way 1: Using $u$-subtitution
Let $u=x^2-16, du = 2xdx$
Then, we have $$\int x(x^2-16)dx$$ $$= \frac{1}{2}\int udu$$ $$... | Alternatively, use hyperbolic trigonometric substitution.
Let $x = 4\cosh t\Rightarrow x^2 - 16 = 16\sinh^2 t$ and $\frac{dx}{dt} = 4\sinh t$.
The integral becomes
$$\begin{array} {r c l }
\displaystyle \int 4\cosh t \cdot (16\sinh^2 t)(4\sinh t) \, dt
&=& 4^4 \displaystyle \int (\cosh t)(\sinh t)^3 \, dt = 4^3 \sinh^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3771677",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Find $ED$ in the triangle $ABC$
$AB = 6$, $AB \| EG$, $BC = 8$, $AC = 10$
I only have this:
From here, we know $BF =FG, AF=DA, GC=DC, OG=OF=OD$, also $$\frac{6}{GE}=\frac{8}{GC}=\frac{10}{EC}$$
And, we can see $$\frac{OG}{OC} = \frac{GC}{EC} \implies OG = \frac{4}{5}OE$$
From the same $$\frac{4}{5}EC=CG=DC=EC - ED \im... |
Let
$BC=8=a$,
$AC=10=b$,
$|AB|=6=c$,
$\angle BAC=\angle GEC=\alpha$.
Since $b^2=a^2+c^2=100$,
$\angle CBA=\angle CGE=90^\circ$.
Inradius of the right triangle
$r=\tfrac12(a+c-b)=2$.
\begin{align}
|ED|&=
r\,\cot\alpha
=r\cdot\frac ca
=\frac32
.
\end{align}
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 2
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Write the polynomial of degree $4$ with $x$ intercepts of $(\frac{1}{2},0), (6,0)$ and $(-2,0)$ and $y$ intercept of $(0,18)$. Write the polynomial of degree $4$ with $x$ intercepts of $(\frac{1}{2},0), (6,0) $ and $ (-2,0)$ and $y$ intercept of $(0,18)$.
The root ($\frac{1}{2},0)$ has multiplicity $2$.
I am to write t... | $$\Big(x-\dfrac{1}{2}\Big)^2 = \Big(\color{red}{\frac{1}{2}}\Big(2x-1\Big)\Big)^2=\color{red}{\frac{1}{4}}\Big(2x-1\Big)^2$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3772371",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"answer_id": 1
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Does $-6(x-\frac{1}{2})^2$ = $-\frac{3}{2}(2x-1)^2$? I'm confused about a question I posted this morning.
I am trying to understand if $-6(x-\frac{1}{2})^2$ can be rewritten as $-\frac{3}{2}(2x-1)^2$?
I tried multiplying out the expression $-6(x-\frac{1}{2})^2$ to a polynomial form $36x^2-36x+9$ but that didn't take me... | $$-6\left(x-\frac{1}{2} \right)^2 = -\frac{3}{2}\left(2x-1 \right)^2$$
$$-6\left( x^2 -x +\frac{1}{4}\right) = -\frac{3}{2} \left(4x^2-4x+1 \right)$$
$$-6x^2+6x - \frac{6}{4} = -6x^2+6x -\frac{3}{2} $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3772654",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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"answer_id": 2
} |
For which values of $\eta\in\mathbf{N}$, $\frac{10^2\cdot\eta}{\varphi\cdot\eta-1}$ is an integer ($\varphi\in\mathbf{N}_0$)? Recently, I have found this problem:
Determine for which values of $\eta\in\mathbf{N}$ and $\varphi\in\mathbf{N}_0$, the quantity:
$$\gamma=\frac{10^2\cdot\eta}{\varphi\cdot\eta-1}$$
is a posit... | $\eta$ must be a factor of $\gamma$ (because $\phi \eta - 1$ and $\eta$ are relatively prime). Then $\gamma$ is a positive integer iff $\eta \phi - 1$ is a positive factor of 100. Clearly, there are only finitely many such combinations.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3774283",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Set of prime integers Let $S$ be a set of primes such that $a,b\in S$ ($a$ and $b$ need not be distinct) implies $ab+4\in S$. Show that $S$ must be empty. (Hint use modulo 7)
I don't have an idea how to use the hint, any further hint will be appreacited
| Suppose $p \in S$.
Suppose $p\equiv 1 \pmod 7$ and $p\in S$. Then $p_2=p^2 +4 \equiv 5 \pmod 7$ is prime and in $S$. so $p_3 =p_1p_2 + 4\equiv 9\equiv 2\pmod 7$. So $p_4 = p_3p_2 +4\equiv 2*5+4 \equiv 0 \pmod 7$ is in $S$. But $7|p_4$ and $p_4$ is prime so $p_4 = p_3p_2 + 4 =7$ so $p_3p_2=3$ but $3$ is primes so th... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3775960",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Determine the number of ways to color a 1-by-n chessboard with the colors red, blue, green and orange. Determine the number of ways to color a 1-by-n chessboard, using the colors red, blue, green and orange if an even number of squares is to be colored red and an even number is to be colored green.
Current result:
I wi... | You can simply read off the coefficients $h_n$ from the sum
$$\sum_{n=0}^\infty\left(\frac{4^n+2\cdot 2^n}4\right)\frac{x^n}{n!}+\frac14\;:$$
$h_0=\frac{1+2}4+\frac14=1$, and
$$h_n=\frac{4^n+2\cdot 2^n}4=4^{n-1}+2^{n-1}=2^{n-1}\left(2^{n-1}-1\right)$$
for $n\ge 1$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3777193",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Coordinate Geometry Question using matrices
Let on the x-y plane, the distance between the points $A(x_1,y_1)$ and $B(x_2,y_2)$ be $d$. Another point $P(a,b)$ satisfies the equations $x_1a+y_1b=1$ and $x_2a+y_2b=1$ and the distance between point $P$ from the origin,$O(0,0)$ is $p$. Find the area of triangle $\Delta OA... | Points $A$ and $B$ lie on the line $\ell \ldots ax+by-1=0$. The distance of $\ell$ from the origin is
$$d(\ell,0) = \frac{|a\cdot 0+b\cdot 0-1|}{\sqrt{a^2+b^2}}=\frac1p$$
since $\sqrt{a^2+b^2} = d(P,O) = p$. The area of your triangle is
$$\text{area} = \frac12 \,\text{base}\cdot \text{height} = \frac12 d(A,B)\cdot d(\e... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3777536",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
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Proof that $2^{ab} - 1$ is divisible by $2^a - 1$ Could you please comment on the correctness and quality of the following proof?
I'm a bit suspicious of the line $n = ab, a = 1, r = 2^a$. Should it be $r = 2$? And are there any "off by one errors? E.g. $n = ab$ vs $n = ab -1$?
To prove:
For $a, b > 1$
$2^{ab} - 1$ is ... | If we pick the common ratio to be $2^a$,
\begin{align}
\sum_{i=0}^{ab-1}2^i = \sum_{t=0}^{b-1}\sum_{i=0}^{a-1}2^{at+i} =\sum_{t=0}^{b-1}(2^a)^t \left(\sum_{i=0}^{a-1}2^i\right)
\end{align}
the first term is $\left(\sum_{i=0}^{a-1}2^i\right)$ and we have $b$ terms in total.
That is $$2^{ab}-1=\sum_{t=0}^{b-1}(2^a)^t \le... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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List all the pairs $(x,y)$ s.t. $x^2 - y^2 = 2020$ List all of the pairs $(x,y) \in \mathbb{Z}^2$ s.t. $^2 - ^2 = 2020$.
The prime factorization of $2020$ is $2^2 \cdot 5 \cdot 101$. I used the fact that there exists a solution $(x,y) \in \mathbb{Z}^2$ to the Diophantine equation $x^2 - y^2 = n$ if and only if $n$ is o... | Hint
$x+y,x-y$ have the same parity as $x+y+x-y$ or $x+y-(x-y)$ is even
Now as $2020$ is even, both must be even
$$\implies\dfrac{x+y}2\cdot\dfrac{x-y}2=\dfrac{2020}4$$
Now if $x,y>0$ $$x+y>x-y$$
As $505=1\cdot505=5\cdot101,$
$\dfrac{x-y}2=1$ or $5$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3778174",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 1
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Estimating Error due to replacing the sum $\sum\limits_{n=1}^{\infty} \frac{1}{n!} (\frac{1}{2})^n$ by the first $n$ terms Question: Estimating Error due to replacing the sum $\sum\limits_{n=1}^{\infty} \frac{1}{n!} (\frac{1}{2})^n$ by the first $n$ terms
All I can really say at this point is that the remainder $R_n$ w... | Note that we have
$$\begin{align}
\sum_{m=n+1}^\infty \frac{1}{m!2^m}&=\frac{1}{n!2^n}\sum_{m=n+1}^\infty \frac{n!2^n}{m!2^m}\\\\
&=\frac{1}{n!2^n}\sum_{m=n+1}^\infty \frac{n!}{m!2^{m-n}}\\\\
&=\frac{1}{n!2^n}\sum_{m=1}^\infty \frac{n!}{(m+n)!2^{m}}\\\\
&\le \frac1{n!2^n}\sum_{m=1}^\infty \frac1{(n+1)^m2^m}\\\\
&=\frac... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3778724",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Proving $(1+a^2)(1+b^2)(1+c^2)\geq8 $ I tried this question in two ways-
Suppose a, b, c are three positive real numbers verifying $ab+bc+ca = 3$. Prove that $$ (1+a^2)(1+b^2)(1+c^2)\geq8 $$
Approach 1:
$$\prod_{cyc} {(1+a^2)}= \left({a^2\over2}+1+{a^2\over2}\right)\left({b^2\over2}+{b^2\over2}+1\right)\left(1+{c^2\... | Another way.
Since $$(x^2+y^2)(z^2+t^2)=(xz+yt)^2+(xt-yz)^2,$$ we obtain:
$$(1+a^2)(1+b^2)(1+c^2)=((1-ab)^2+(a+b)^2)(1+c^2)=$$
$$=(1-ab-(a+b)c)^2+((1-ab)c+a+b)^2=$$
$$=(1-ab-ac-bc)^2+(a+b+c-abc)^2=4+(a+b+c-abc)^2.$$
Id est, it's enough to prove that
$$a+b+c-abc\geq2,$$ which is true by Maclaurin.
Indeed, let $a+b+c=3u$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3779630",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 5
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Solve $\sqrt[4]{x}+\sqrt[4]{x+1}=\sqrt[4]{2x+1}$ Solve $\sqrt[4]{x}+\sqrt[4]{x+1}=\sqrt[4]{2x+1}$
My attempt:
Square both sides three times
$$\begin{align*}
36(x^2+x)&=4(\sqrt{x^2+x})(2x+1+\sqrt{x^2+x})\\
(\sqrt{x^2+x})(35\sqrt{x^2+x}-4(2x+1))&=0
\end{align*}$$
This means $0,-1$ are solutions but I can't make sure that... | It is trivial that for any $a>0$ and $b>0$ (just raise both side to the $n^{th}$ power)
$$a^\frac1n +b^\frac1n >(a+b)^\frac1n$$
therefore $\forall x>0$
$$\sqrt[4]{x}+\sqrt[4]{x+1}> \sqrt[4]{2x+1}$$
and for $x=0$
$$\sqrt[4]{x}+\sqrt[4]{x+1}= \sqrt[4]{2x+1}=1$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3779710",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 4,
"answer_id": 3
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Simplifying $\frac{(\ln a)^3+(\ln b)^3+(\ln c)^3+(\ln d)^3}{\ln ab\;(\ln c\ln d-\ln a\ln b)}$ for positive reals with $abcd=1$ The problem is as follows:
Given the condition for $a$, $b$, $c$, and $d$ to be real positive numbers whose product equals to $1$, find the value of $B$.
$$B=\frac{(\ln a)^3+(\ln b)^3+(\ln c)^... | Following the idea given in the comments, let $a=e^A$, $b=e^B$, $c=e^C$, $d=e^D$ then
$$\frac{A^3+B^3+C^3+D^3}{(A+B)\cdot\left(CD-AB\right)}=\frac{(A+B)(A^2-BC+B^2)+(C+D)(C^2-CD+D^2)}{(A+B)\cdot\left(CD-AB\right)}=$$
$$=\frac{(A+B)(A^2-AB+B^2)-(A+B)(C^2-CD+D^2)}{(A+B)\cdot\left(CD-AB\right)}=$$
$$=\frac{(A^2-AB+B^2)-(C... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3780764",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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How do I solve $(\cos2x+1)^2=1/2$? $x$ belongs to $$[0,\pi/2[$$
$$(\cos(2x+1))^2 = \frac{1}{2}$$
I tried to find $x$ using
$$2x+1=\frac{\pi}{4} +\frac{n\pi}{2}
\qquad\text{or}\qquad
2x+1=\frac{3\pi}{4} + n\pi$$
but I didn't find my answer in MCQ which is
$$\frac{3\pi-4}{8}
\qquad\text{and}\qquad
\frac{5\pi-4}{8}.$$
I ... | $$\implies\cos2(2x+1)=2\cos^2(2x+1)-1=0$$
$2(2x+1)= \dfrac{(2n+1)\pi}2 $ where $n$ is any integer
Now $1\le2x+1\le1+\pi$
If $1\le\dfrac{(2n+1)\pi}4\le1+\pi\iff\dfrac4\pi-2\le n\le\dfrac4\pi\implies n=0,1$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3783070",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 0
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Diophantine Equation with Square Root I want to resolve the diophantine equation:
$\sqrt{x^2+5x+12} ≡ x-2\pmod 5$
I have thought 2 ways:
1.
$(\sqrt{x^2+5x+12})^2 ≡ (x-2)^2\pmod 5$
$ x^2+5x+12 ≡ x^2 -4x+4\pmod 5$
$ 9x+8 ≡ 0\pmod 5$
$ 4x+3 ≡ 0\pmod 5$
$ 4x+3 = 5y$
$ 4x-5y = -3$
I resolve:
$x=5n+3$,
n
∈
Z
$y=4n+3$,
n... | Writing the congruence as an equation in the field $\Bbb F_5$ we obtain
$$
x^2+5x+12=(x-2)^2=x^2-4x+4,
$$
which gives $4x=2$. In $\Bbb F_5$ this has the unique solution $2^{-1}=3$, because $2\cdot 3=6=1$, so that $3$ is the inverse of $2$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3787527",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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If $f(x)=\begin{cases} \cos (x^3 )& x<0 \\ \sin (x^3 )- |x^3-1| & x\ge 0 \end {cases}$, find the number of points where $f(|x|)$ is non differentiable Since we are dealing with $f(|x|)$, the $\cos (x^3)$ part can be ignored. (And $\cos$ is differentiable everywhere anyway)
For $\sin |x|^3$, the graph would break its sm... | $$f(x)=\begin{cases}\cos(x^3 )& x<0 \\ \sin (x^3 )- |x^3-1| & x\geq 0 \end {cases}=\begin{cases}\cos(x^3 )& x<0 \\ \sin (x^3 )- 1+x^3 & 0\leq x<1\\ \sin(x^3)-x^3+1&x\geq1\end {cases}$$
$$f(|x|)=\begin{cases}\sin(x^3)-x^3+1&x\geq1\\\sin(x^3)+x^3-1&0\leq x<1\\-\sin(x^3)-x^3-1&-1\leq x<0\\-\sin(x^3)+x^3+1&x<-1\end {cases}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3788623",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
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Find the minimum value of $x_1^2+x_2^2+x_3^2+x_4^2$ subject to $x_1+x_2+x_3+x_4=a$ and $x_1-x_2+x_3-x_4=b$. Question: Find the minimum value of $x_1^2+x_2^2+x_3^2+x_4^2$ subject to $x_1+x_2+x_3+x_4=a$ and $x_1-x_2+x_3-x_4=b$.
My attempt: It can be easily seen that $x_1+x_3=\frac{a+b}{2}$ and $x_2+x_4=\frac{a-b}{2}$. Fu... | Using algebra.
Use the two equality constraints to get $x_3$ and $x_4$ as linear functions of $x_1$ and $x_2$.
This makes
$$x_1^2+x_2^2+x_3^2+x_4^2=x_1^2+x_2^2+\frac{1}{4} (a+b-2 x_1)^2+\frac{1}{4} (-a+b+2 x_2)^2$$
Compute the partial derivatives wrt $x_1$ and $x_2$ and set them equal to $0$. This would give $x_1=\frac... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3790537",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 2
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One matrix identity implies the other I have matrices $A\in\mathbb R^{n\times n}$, $C\in\mathbb R^{m\times m}$, and $B\in\mathbb R^{n\times m}$. I know that $A$ and $C$ are symmetric and positive semidefinite and that $K_A = A^2+BB^T$ and $K_C = C^2+B^TB$ are positive definite. Now assume that $AB = BC$. I can prove in... | I have it! If $AB=BC$, then also $B^TA = CB^T$ and thus $CB^TB = B^TAB = B^TBC$. Also,
$$
BK_C = BC^2 + BB^TB = A^2B + BB^TB = K_AB.
$$
Hence, $BK_C^{-1} = K_A^{-1}B$. Therefore,
$$
AK_A^{-1}B = ABK_C^{-1} = BCK_C^{-1} = BK_C^{-1}C,
$$
where the last identity follows from $CB^TB = B^TBC$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3790894",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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How many four digits numbers are there not containing zero and multiplication of its digits divisible by 7? I saw a question in my math book, it seems very trivial, it says that:
How many four digits numbers are there not containing zero and
multiplication of its digits divisible by 7?
I thought of:
(all four digits ... | Your first answer is right in regard of the statement you gave, indeed:
Let $(a,b,c,d) \in \{1,2,3,4,5,6,7,8,9\}^4$. So $A = 1000a+100b+10c+d $ is a four-digit number.
Morevover, $a\cdot b \cdot c \cdot d $ is divisible by $7$, if and only if, its prime factorization contains at least one time $7$ so, if and only if, a... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3791577",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 1,
"answer_id": 0
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How to integrate $ \int\frac{2x-\sqrt{4x^{2}-x+1}}{x-1}dx $ Im trying to integrate $ \int\frac{2x-\sqrt{4x^{2}-x+1}}{x-1}dx $.
I've tried integration by parts, and any resonable substitution that came to my mind. Neither seem to work. I tried to use integral online calculator, the result was too complicated, and I'm pr... | Take,$$t=\sqrt{4x^2-x+1}-2x$$
Then,$$\sqrt{4x^2-x+1}=2x+t \;\;\Longrightarrow\;\;4x^2-x+1=(2x+t)^2 \;\;\Longrightarrow\;\; x=\frac{1-t^2}{4t+1}\,.$$
And $$dx = -\frac{4t^2+2t+4}{(4t+1)^2}dt\,.$$
So,
$$I=\int\frac{2x-\sqrt{4x^2-x+1}}{x-1}dx=\int\frac{t}{(1-t^2)/(4t+1)-1}\cdot\frac{4t^2+2t+4}{(4t+1)^2}dt\,,$$
$$\begin{al... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3791702",
"timestamp": "2023-03-29T00:00:00",
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"answer_id": 0
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Why does $-8^{\frac{1}{3}}$ have $2$, $e^{\frac{\pi}{3}}$ and $e^{\frac{5\pi}{3}}$? Use DeMoivre’s theorem to find $-8^{\frac{1}{3}}$. Express your answer in complex form.
Select one:
a. –2
b. – 2, 2 cis ($\pi$/3)
c. – 2, 2 cis ($\pi$/3), 2 cis (5$\pi$/3)
d. 2, 2 cis ($\pi$/3), 2 cis (5$\pi$/3)
e. None of these
The c... | $-8 = 8(-1) = 8\operatorname{cis} (\pi) = 8\operatorname{cis}( (2k+1)\pi)$
So $(-8)^{\frac 13} = \sqrt[3]{8}\operatorname{cis}(\frac {2k+1}3\pi)=$.
$2\operatorname{cis}(\frac \pi 3 + \frac {2k\pi}3); k=0,1,2$
$2\operatorname{cis}(\frac \pi 3), 2\operatorname{cis}(\pi)= 2\cdot (-1)=-2, 2\operatorname{cis}(\frac {5\pi} 3... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3791784",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Let $L=\lim_{x\to 0} \frac{ a-\sqrt {a^2-x^2} -\frac{x^2}{2}}{x^4}$, $a>0$. If $L$ is finite, find $a$ and $L$ I need a hint to start this question, because I have no idea how to do it. It’s a $\frac 00$ form, so L’Hospital can be applied, but that would be extremely tedious. Expansion can’t be used because there is no... | You have
$$\frac{a - \sqrt{a^2-x^2}- \frac{x^2}{2}}{x^4} = \frac{a-a\sqrt{1-\frac{x^2}{a^2}} - \frac{x^2}{2}}{x^4} = \frac{a-a\left( 1- \frac{x^2}{2a^2}-\frac{x^4}{8a^4} + o(x^5)\right) - \frac{x^2}{2}}{x^4}$$
$$= \frac{\frac{x^2}{2} \left( \frac{1}{a}-1\right)+\frac{x^4}{8a^3} + o(x^5) }{x^4}$$
You see that if $a \neq... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3792963",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 4
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Show for any odd prime $p\geq 5,$ $(-3/p)=1$ or $ -1$ Show for any odd prime $$p\geq 5,$$
$$\left ( \frac{-3}{p} \right ) =\begin{cases}
1 & \text{ if } p\equiv 1,-5\pmod{12} \\
-1& \text{ if } p\equiv -1,5\pmod{12}
\end{cases}$$
So far I have that
(1) Let $$p\equiv 1\pmod{4}$$ then $$p\equiv 1\pmod{3}$$ to get
$$\lef... | You messed up some calculations. When $p\equiv1\pmod4$, $\left(\dfrac{-3}p\right)=\left(\dfrac{-1}p\right)\left(\dfrac{3}p\right)=1\left(\dfrac p3\right)$,
so in those cases $\left(\dfrac{-3}p\right)=1$ when $p\equiv1\pmod3$ and $\left(\dfrac{-3}p\right)=-1$ when $p\equiv2\pmod3$ .
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3794662",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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limit of $\lim\limits_{x \to \infty} (1+ \frac{\pi}{2} - \arctan(x))^x$ Question: Limit of $\lim\limits_{n \to \infty} (1+ \frac{\pi}{2} - \arctan(x))^x$
The first things I notice are : $\lim\limits_{x \to \infty} \arctan(x) = \frac{\pi}{2}$ and the limit looks something of the form $(1 + \frac{1}{x})^x$. Unfortunetly,... | $$A=\left(1+ \frac{\pi}{2} - \tan^{-1}(x)\right)^x=\left(1+\tan ^{-1}\left(\frac{1}{x}\right)\right)^x$$
$$\log(A)=x \log\left(1+\tan ^{-1}\left(\frac{1}{x}\right)\right)$$
By Taylor
$$\tan ^{-1}\left(\frac{1}{x}\right)=\frac{1}{x}-\frac{1}{3 x^3}+O\left(\frac{1}{x^5}\right)$$
$$\log\left(1+\tan ^{-1}\left(\frac{1}{x}\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3796775",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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Solve $\lim_{x\rightarrow +\infty}\frac{1}{x}\log\left(\frac{x+1}{1+x^2}\right) = 0$ without L'Hôpital's rule How would you solve the limit
$$\lim_{x\rightarrow +\infty}\frac{1}{x}\log\left(\frac{x+1}{1+x^2}\right) = 0$$
without using L'Hôpital's rule?
| \begin{align}
\lim_{x\to+\infty}\frac{1}{x}\log\left(\frac{x+1}{1+x^2}\right)
&= \lim_{x\to+\infty}\left(\frac{1}{x}\cdot\frac{1+x^2}{x+1}\right)\left[\frac{x+1}{1+x^2}\log\left(\frac{x+1}{1+x^2}\right)\right]=\\
&= \lim_{x\to+\infty}\frac{1+x^2}{x^2+x}\cdot\lim_{y\to0}y\log y=1\cdot0=0\\
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3798628",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 4
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Find all the $\bf{x}\in\mathbb{R}^3$, which has a property that $\displaystyle \lim_{ n \to \infty } A^n\bf{x} = \bf{x}$ $A$ is a matrix with elements
$$\begin{pmatrix}
\frac{2}{3} & -\frac{2}{3}& -\frac{1}{3} \\
-\frac{2}{3}& 1 & \frac{2}{3} \\
\frac{4}{3} & 0 & -\frac{1}{3}
\end{pmatrix}$$
Find all the $\bf{x}\in\mat... | Let $e_1, e_2, e_3$ be three linearly independent eigenvectors which are of unit norm. (You have shown that these exist.)
Suppose that they are corresponding to the eigenvalues $1, -1/3, 2/3$, respectively.
Let $\mathbf{x} \in \Bbb R^3$ be any such with the property you stated. Since $\{e_1, e_2, e_3\}$ form a basis, w... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3800965",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
If $a = \frac{1+\sqrt5}2$, then what is $a^{18} + \frac{323}{a^6}$? My question is this:
If $a = \frac{1+\sqrt5}2,\frac{1-\sqrt5}2$, then what is $a^{18} + \frac{323}{a^6}$?
It is an AMC style question and is timed, so I will not be able to use solutions with a lot of case work.
I defined $a$ from part a of a 2 part ... |
Lemma. Define $$L_n = a^n + (-a)^{-n}.$$ Then $L_n$ is a Lucas number satisfying the recurrence $$L_{n+1} = L_n + L_{n-1}, \\ L_0 = 2, \quad L_1 = 1.$$
Proof. Since $a^2 = a + 1$, or equivalently, $$a - a^{-1} = 1,$$ we have $$(a^n + (-a)^{-n})(a - a^{-1}) = a^{n+1} + (-a)^{-(n+1)} - \left(a^{n-1} + (-a)^{-(n+1)}\r... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3802260",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 3
} |
How to prove $\frac{a}{b}=\frac{c}{d} \implies \frac{a+c}{b+d}=\frac{a-c}{b-d}$? My question:
How do I prove the following property of ratio?
$$\frac{a}{b}=\frac{c}{d} \implies \frac{a+c}{b+d}=\frac{a-c}{b-d} \tag{1}$$
$a,b,c,d \in \mathbb{R} \backslash \{ 0 \}$
I want to use the result in argument below.
Use the s... | But why not use this simple solution:
$$
\frac{a}{b}=\frac{c}{d}=x \implies a=bx , c = dx
$$
$$
\begin{align}
\frac{a+c}{b+d}&=\frac{bx+dx}{b+d}\\
\\
&=x\\
\\
\frac{a-c}{b-d}&=\frac{bx-dx}{b-d}\\
\\
&=x
\end{align}
$$
For the desired result, use sine rule
$$
\begin{align}
a&=2R\sin{(A)}\\
b&=2R\sin{(B)}\\
c&=2R\sin{(C)... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3802366",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Find $r$ such that the equation $x^4+x^2(1-2r)-2x+1=0$ has only one real solution I'm looking for $r$ such that the equation $$x^4+x^2(1-2r)-2x+1=0$$
has only one real solution. I've made some attempts to this problem, but it seems that I even didn't get close to the solution. The approximation for $r$ is 0.3347498
Is... | We can write the function in following form where $c>0$.
$(x-a)^2((x-b)^2+c)=0$
$(x^2-2ax+a^2)(x^2-2bx+b^2+c)=0$
$x^4-(2a+2b)x^3+(a^2+b^2+4ab+c)x^2-(2ab^2+2ac+2a^2b)x+a^2b^2+a^2c=0$
Therefore,
$b=-a, c={1-a^4\over a^2} > 0$
$2ab^2+2ac+2a^2b=2a^3+{2-2a^4\over a}-2a^3=-2,a^4-a-1=0$
$1-2r=a^2+b^2+4ab+c=2a^2-4a^2+{1-a^4\ov... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3803011",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
Find the minimum value of $f(x) = x + \frac{x}{x^2 + 1} + \frac{x(x + 4)}{x^2 + 2} + \frac{2(x + 2)}{x(x^2 + 2)}$ Find the minimum value of
$$f(x) = x + \frac{x}{x^2 + 1} + \frac{x(x + 4)}{x^2 + 2} + \frac{2(x + 2)}{x(x^2 + 2)}$$ for $x > 0.$
I'm not sure how to start this problem. I think AM-GM is the best way, but I'... | Here is your solution by A.M-G.M
$$f(x)=x+\frac{x}{x^2+1}+\frac{x(x+4)}{x^2+2}+\frac{2(x+2)}{x(x^2+2)}$$
$$f(x)=x+\frac{x}{x^2+1}+\frac{x^2+4x}{x^2+2}+\frac{2}{x^2+2}+\frac{4}{x(x^2+2)}$$
$$f(x)=\frac{x(x^2+2)}{x^2+1}+\frac{x^2}{x^2+2}+\frac{4x}{x^2+2}+\frac{2}{x^2+2}+\frac{4}{x(x^2+2)}$$
$$f(x)=\frac{x(x^2+2)}{x^2+1}+... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3804362",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
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Does the series $\frac{2}{4-1}+\frac{4}{16-1}+\dots+\frac{2k}{4k^2-1}$ have a sum up to $\infty$? If there's one, then how do I find the sum?
I tried to rewrite the common term into partial fractions to see if it's a telescoping series but got a dead end there. How can I proceed now?
| To expand on @TheSilverDoe's answer...
The sum is approximately $$\sum_{k=1}^\infty \frac{1}{2k}.$$
Note that this is $$\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+\frac{1}{8}+\frac{1}{10}+\frac{1}{12}+\frac{1}{14}+\frac{1}{16}+\dots$$
Group the sum as follows:
$$\left(\frac{1}{2}\right)+\left(\frac{1}{4}\right)+\left(\frac{1}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3805505",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 2
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Prove that$(1-\omega+\omega^2)(1-\omega^2+\omega^4)(1-\omega^4+\omega^8)…$ to 2n factors$=2^{2n}$ Prove that $(1-\omega+\omega^2)(1-\omega^2+\omega^4)(1-\omega^4+\omega^8)…$ to 2n factors$=2^{2n}$ where $\omega$ is the cube root of unity
My attempt:
$(1+\omega^n+\omega^{2n})=0$
$\Rightarrow (1-\omega^n+\omega^{2n})=-2\... | Hint : Evaluate first few factors and see.
$$
1st : (1-\omega+\omega^2)=-2\omega \\
2nd : (1-\omega^2+\omega^4)=-2\omega^2 \\
1st \times 2nd = \ldots \\
3rd : (1-\omega^4+\omega^8)=-2\omega \\
4th : (1-\omega^8+\omega^{16})= \ldots
$$
Can you complete?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3805820",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
} |
Is this alternative proof of the inequality $\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}\geq\frac{3}{2}$ correct? Prove that for all positive real numbers:
$$\dfrac{a}{b+c}+\dfrac{b}{c+a}+\dfrac{c}{a+b}\geq\dfrac{3}{2}$$
This is same as this question but a different approach is used there whereas I want to verify my appr... | Your solution is right.
Also, SOS helps:
$$\sum_{cyc}\frac{a}{b+c}-\frac{3}{2}=\sum_{cyc}\left(\frac{a}{b+c}-\frac{1}{2}\right)=\sum_{cyc}\frac{2a-b-c}{2(b+c)}=$$
$$=\sum_{cyc}\frac{a-b-(c-a)}{2(b+c)}=\sum_{cyc}\left(\frac{a-b}{2(b+c)}-\frac{c-a}{2(b+c)}\right)=$$
$$=\sum_{cyc}\left(\frac{a-b}{2(b+c)}-\frac{a-b}{2(c+a)... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3810467",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 0
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SOS proof for $\sum_{cyc}\frac{a^3}{bc}\ge a+b+c$ I need an SOS(sum of squares) proof for $$\sum_{cyc}\frac{a^3}{bc}\ge a+b+c$$
if $a,b,c>0$
I already have a am-gm proof but is there a way to use SOS.
Am-gm proof :
$\frac{a^3}{bc}+b+c\ge 3a$ .....by(AM-GM ineq.)
thus $$\sum \frac{a^3}{bc}+2\sum a... | The idea of the SOS's proof it's the following.
Let $P$ be a symmetric function of three variables $a$, $b$ and $c$ and let we can get:
$$P(a,b,c)=\sum_{cyc}((a-b)Q(a,b,c)-(c-a)Q(a,c,b)).$$
Thus, $$P(a,b,c)=\sum_{cyc}((a-b)Q(a,b,c)-(c-a)Q(a,c,b))=$$
$$=\sum_{cyc}((a-b)Q(a,b,c)-(a-b)Q(b,a,c))=\sum_{cyc}(a-b)(Q(a,b,c)-Q(... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3810623",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 0
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A hint on finding all solutions of $S = \frac{a}{a+b+d} + \frac{b}{a+b+c} + \frac{c}{b+c+d} + \frac{d}{a+c+d}$? I've been working at this for awhile but I haven't been able to figure out the right approach. The question is to find all values of $S = \frac{a}{a+b+d} + \frac{b}{a+b+c} + \frac{c}{b+c+d} + \frac{d}{a+c+d}$... | For $c=d\rightarrow0^+$ we see that $$S\rightarrow\frac{a}{a+b}+\frac{b}{a+b}=1$$ and
$$\sum_{cyc}\frac{a}{a+b+d}>\sum_{cyc}\frac{a}{a+b+c+d}=1.$$
For $b=d\rightarrow0^+$ we see that $S$ close to $2$.
We'll prove that $$\sum_{cyc}\frac{a}{a+b+d}<2.$$
Indeed, we need to prove that
$$\sum_{cyc}\left(\frac{a}{a+b+d}-1\ri... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3813717",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"answer_id": 0
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Finding the difference of square root of conjugate complex number Find the imaginary part of $\left( {{{\left( {3 + 2\sqrt { - 54} } \right)}^{\frac{1}{2}}} - {{\left( {3 - 2\sqrt { - 54} } \right)}^{\frac{1}{2}}}} \right)$
(1) $-\sqrt 6$
(2) $-2\sqrt 6$
(3) $\sqrt 6$
(4) $6$
My Approach is as follow and none of the an... | Let $$z=\sqrt{3+6\sqrt{6}i)}=x+iy ~~~(1)$$
Squaring we get $3+6\sqrt{6}i=x^2-y^2+2ixy$
$$\implies x^2-y^2=3, xy=3\sqrt{6}~~~~(2)$$
$$\bar z=\sqrt{3-6\sqrt{6}i}=x-iy~~~~(3),$$
multiplying the two (1) and (3)we get $$x^2+y^2=\sqrt{9+36.6}=15~~~(4)$$
Using (4) in (2) we get $x=\pm 3, y=\pm \sqrt{6}.$
So $z-\bar z=2y=\pm 2... | {
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"url": "https://math.stackexchange.com/questions/3813966",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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prove $a^3+b^3+c^3+3abc\ge \sum_{cyc}ab(a+b)$
prove $$a^3+b^3+c^3+3abc\ge \sum_{cyc}ab(a+b),$$$a,b,c>0$
Obviously this is a direct consequence of the third degree schur's inequality.
I was wondering if this could be proved without this theorem ,or uvw but through basic methods like AM-GM,C-S etc.
| Suppose $a \geqslant b \geqslant c,$ we get
$$a^3+b^3+c^3+3abc - \sum ab(a+b)$$
$$=(c^3+b^2c+ca^2-abc-ac^2-bc^2)+(a^3+b^3+4abc-a^2b-ab^2-2ca^2-2b^2c)$$
$$=(a^2+b^2+c^2-ab-bc-ca)c+(a+b-2c)(a-b)^2 \geqslant 0.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3815419",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 6,
"answer_id": 4
} |
Solving 2-degree equations in 3 variables. We are given 3 equations:
$x^2+\sqrt3 xy + y^2 = 25$
$y^2 + z^2 = 9$
$x^2 +xz+ z^2 = 16$.
$x,y,z$ are positive real numbers.
Then we have to find value of $xy + 2yz + \sqrt3 xz$.
| Based on system of equations being sides of a right angled triangle and a point P
inside the triangle such that -
$\angle BPC = 90^0, \angle APC = 150^0, \angle APB = 120^0 $
and, $AP = x, CP = y, BP = z$
We know area of a triangle is $\frac{1}{2} \times$ length of side 1 $\times$ length of side 2 $\times \sin \theta$
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3815573",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 0
} |
Finding $k$ such that $\frac{3x^2+kx-44}{x^2-121}$ simplifies to $\frac{3x-4}{x-11}$
If the rational expression $\frac{3x^2+kx-44}{x^2-121}$, where $k$ is an element of $\mathbb{W}$, simplifies to $\frac{3x-4}{x-11}$, then the value of $k$ must be?
my work:
$$\frac{3x^2+kx-44}{x^2-121}= \frac{3x-4}{x-11}$$
so the ans... | Notice $x^2-121=(x+11)(x-11)$. Then we have
$$
3x^2+kx-44=(3x-4)(x+11)=3x^2+29x-44.
$$
So $k=29$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3820408",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
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Inequality manipulation: $\frac{1}{\sqrt{4n + 1}} \cdot \frac{2n + 1}{2n + 2} > \frac{1}{\sqrt{4n + 5}}$ I'm reading Analytic Inequalities by Nicholas D. Kazarinoff. On page 5, we are trying to use induction to prove the inequality
$$ \frac{1}{\sqrt{4n + 1}} < \frac{1}{2} \cdot \frac{3}{4} \cdots \frac{2n - 3}{2n - 2} ... | although an elementary solution is aesthetically preferable (and a good algebraic solution was posted and then for some reason deleted by Soumyadwip Chanda), it may be of interest to see the number-theoretic result posted by OP as a corollary of a more general theorem of real analysis:
suppose that for $a, b \in (0,1)$... | {
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"url": "https://math.stackexchange.com/questions/3823022",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 0
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How does this division work? $\frac{\;\frac{6^6}{1}\;}{2^{-3}}\cdot2^{-10}$ I came across this division and can't wrap my head around how it is solved:
$$\frac{\;\;\frac{6^6}{1}\;\;}{2^{-3}}\cdot2^{-10}$$
They subtract the exponent of $2^{-10}$ from the denominator's exponent $2^{-3}$:
$$2^{-3-(-10)}$$
Which gives us:
... | First off, the way in which the problem is typeset is almost intended to cause confusion. Adding some parentheses and not using \dfrac everywhere should make it easier to understand the expression:
$$\frac{\left(\frac{6^6}{1}\right)}{2^{-3}} \cdot 2^{-10}. $$
Division is the inverse operation of multiplication and so,... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3827352",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 7,
"answer_id": 6
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Epsilon Delta proof for rational function containing radicals I am having trouble constructing the delta-epsilon proof for this limit. I am currently trying to find the delta in terms of epsilon:
$$\lim_{x \to 1} \frac{x^3-1}{\sqrt{x}-1}$$
So far I have gotten to
$$6-\epsilon < (x-1)*\frac{x^2+x+1}{\sqrt{x}-1} < 6+\eps... | We have that for $x\neq 1$
$$\frac{x^3-1}{\sqrt{x}-1}=\frac{(x-1)(x^2+x+1)}{\sqrt{x}-1}=$$
$$=\frac{(\sqrt{x}-1)(\sqrt{x}+1)(x^2+x+1)}{\sqrt{x}-1}=(\sqrt{x}+1)(x^2+x+1)$$
and since for $|x-1|<1 \iff 0<x<2$, using triangle inequality and that $\forall a,b \:a\ge b\ge 0 \implies |x^a-1|\le |x^b-1|$, we obtain
$$\left|\fr... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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"answer_id": 1
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Convergence of the series $ \frac{2x}{1+x^2}+\frac{4x^3}{1+x^4}+\frac{8x^7}{1+x^8}+...$ The series $ \frac{2x}{1+x^2}+\frac{4x^3}{1+x^4}+\frac{8x^7}{1+x^8}+...$
(A) is uniformly convergent for all x
(B) is convergent for all x but the convergence is not uniform
(C) is convergent only for $|x|\le \frac{1}{2}$ but the co... | We have
$$f(x)=\sum_{k=1}^\infty \frac{2^kx^{2^k-1}}{1+x^{2^k}}$$
and by root test
$$\sqrt[k]{\frac{2^kx^{2^k-1}}{1+x^{2^k}}}=2\sqrt[k]{\frac{x^{2^k-1}}{1+x^{2^k}}}$$
and for $|x|<1$
$$2\sqrt[k]{\frac{x^{2^k-1}}{1+x^{2^k}}} \to 0$$
$|x|\ge1$
$$2\sqrt[k]{\frac{x^{2^k-1}}{1+x^{2^k}}} \to 2$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3837139",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Alternative approach for proving that for any $x\in\mathbb R^+$, $x^2+3x+\frac{1}{x} \ge \frac{15}{4}$. Let $x \in \mathbb{R^+}$. Prove that:
$$x^2+3x+\frac{1}{x} \ge \frac{15}{4}.$$
While this is indeed easily proven using derivatives, where the minimum is obtained when $x=\frac{1}{2}$, is it possible to prove it thro... | If we know equality when $x=\frac{1}{2}$ then find AM-GM solution very easy.
For $x=\frac 12,$ then
$$x^2 = \color{blue}{\frac{1}{4}}, \quad 3x = \frac{3}{2} = \color{red}{6} \cdot \color{blue}{\frac{1}{4}}, \quad \frac 1x = 2 = \color{red}{8} \cdot \color{blue}{\frac{1}{4}}.$$
Now, using the AM-GM inequality, we have
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3838183",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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How to prove $5^{2n+1}-3^{2n+1}-2$ is divisible by $30$ for even $n$? How to prove $5^{2n+1}-3^{2n+1}-2$ is divisible by $30$ for all positive even integer $n$?
Here are my efforts:
Let $f(n)=5^{2n+1}-3^{2n+1}-2$.
For $n=2$,
$$f(1)=5^{2\cdot 2+1}-3^{2\cdot2+1}-2=2880=30\times 96$$
So, the statement is true for $n=2$.
... | Since $$5^{2n+1}-3^{2n+1}-2=5^{2n+1}-5-3\cdot3^{2n}+3=5^{2n+1}-5-3\left(81^{\frac{n}{2}}-1\right)=$$$$=5^{2n+1}-5-3\cdot80\left(81^{\frac{n}{2}-1}+...+1\right),$$
we see that $5^{2n+1}-3^{2n+1}-2$ is divisible by $5$.
Also, $$5^{2n+1}-3^{2n+1}-2=5^{2n+1}+1-\left(3^{2n+1}+3\right)=(5+1)(5^{2n}-...+1)-\left(3^{2n+1}+3\ri... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3838423",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 3
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Prove that $\lim_{(x,y)\to(1,2)}\frac{x}{x+y}=\frac{1}{3}$ How to prove that $$\lim_{(x,y)\to(1,2)}\frac{x}{x+y}=\frac{1}{3}$$ with $\varepsilon-\delta$ limit definition?
I know so far that $\forall \varepsilon > 0 \ \exists \delta > 0: 0 < d((x, y), (1, 2)) < \delta \Rightarrow \Big|\frac{x}{x+y} - \frac{1}{3}\Big| <... | By $|x-1|<\delta$ and $|y-2|<\delta$ assuming wlog $\delta\le\frac12$ we have
$$\left|\frac{x}{x+y} - \frac{1}{3}\right|=\left|\frac{2x-y}{x+y}\right|=\left|\frac{2(x-1)-(y-2)}{x+y}\right| \le 2\left|\frac{x-1}{x+y}\right|+\left|\frac{y-2}{x+y}\right|\le$$
$$\le 2\left|\frac{x-1}{x+y}\right|+\left|\frac{y-2}{x+y}\right... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3838615",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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} |
how to show that $\csc x - \csc\left(\frac{\pi}{3} + x \right) + \csc\left(\frac{\pi}{3} - x\right) = 3 \csc 3x$?
How to show that $\csc x - \csc\left(\frac{\pi}{3} + x \right) + \csc\left(\frac{\pi}{3} - x\right) = 3 \csc 3x$?
My attempt:
\begin{align}
LHS &= \csc x - \csc\left(\frac{\pi}{3} + x\right) + \csc\left(\... | If $\csc3x=\csc3y\iff \sin3y=\sin3x$
$$3y=3x+(-1)^nn\pi$$ where $n$ is any integer
$y=x+\dfrac{2n\pi}3; n=-1,0,1$
Now as $\sin3y=3\sin y-4\sin^3y,$
the roots of $$4\sin^3y-3\sin y+\sin3x=0$$ are
$\sin\left(x-\dfrac{2\pi}3\right)=\sin\left(-\pi+\left(x+\dfrac\pi3\right)\right)=-\sin\left(x+\dfrac\pi3\right)=x_1,$
$\sin ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3840253",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 3
} |
$\lim_{n\to\infty}\left( \frac1{4\cdot 7}+\frac1{7\cdot 10}+\ldots+\frac1{(3n+1)(3n+4)} \right) $ I am having trouble finding the infinite sum
$$
\lim_{n\to\infty}\left( \frac1{4\cdot 7}+\frac1{7\cdot 10}+\ldots+\frac1{(3n+1)(3n+4)} \right).
$$
I know that
$$
\lim_{n\to\infty}\frac1{(3n+1)(3n+4)} =0,
$$
but I have no i... | You want to compute $$\lim_{n\rightarrow \infty}\sum_{i=1}^{n}\frac{1}{(3i+1)(3i+4)}.$$
Using partial fractions we want to find constants $A$ and $B$ with $$\frac{1}{(3n+1)(3n+4)}=\frac{A}{3n+1}+\frac{B}{3n+4}=\frac{A(3n+4)+B(3n+1)}{(3n+1)(3n+4)}$$
So comparing numerators we want $1=A(3n+4)+B(3n+1)$. Then choosing $n=-... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3843359",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
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Evaluate $\lim_{x\to \infty} \frac{e^{\frac{1}{x^2}}-1}{2\tan^{-1} (x^2)-\pi}$ $$\lim_{x\to \infty} \frac{ e^{\frac{1}{x^2}} -1}{\frac{1}{x^2}} \frac{1}{(2\tan^{-1}(x^2)-\pi)x^2}$$
$$=\lim_{x\to \infty} \frac{1}{x^2(2\tan^{-1} x^2 -\pi)}$$
How do I proceed?
| Apply l'Hôpital's rule:
$$L=\lim\limits_ {x \to \infty} \dfrac {x^{-2}} {2\arctan x^2 -\pi}=\lim\limits_ {x \to \infty} \dfrac {-2x^{-3}} {2\dfrac {2x}{x^4+1}}$$
$$L==\lim\limits_ {x \to \infty} -{\dfrac {x^4}{2x^4}}=-\dfrac 12$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3843885",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
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Find the minimum of $P = (a - b)(b - c)(c - a)$
Given $a, b, c$ are real numbers such that $a^2 + b^2 + c^2 = ab + bc + ca + 6$. Find the minimum of
$$P = (a - b)(b - c)(c - a)$$
My solution:
*
*We have:
$$a^2 + b^2 + c^2 = ab + bc + ca + 6$$
$$\implies 2a^2 + 2b^2 + 2c^2 = 2ab + 2bc + 2ca + 12$$
$$\implies (a - b... | Using $x=a-b,y=b-c$, then $-(x+y)=c-a$, so $P=-xy(x+y)=-(x^2y+xy^2)$.
The condition $(a−b)^2+(b−c)^2+(c−a)^2=12$ becomes $g=x^2+y^2+xy=6$.
Now using Lagrange multipliers:
\begin{align}
\nabla P&=-\langle2xy+y^2,x^2+2xy\rangle\\
&=-\langle y(2x+y),x(x+2y)\rangle\\
\\
\nabla g&=\langle2x+y,x+2y\rangle
\end{align}
Now let... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3844805",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 6,
"answer_id": 2
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For any real positive numbers $a, b, c$, prove that $3(a^2b+b^2c+c^2a)(ab^2+bc^2+ca^2) \geq abc(a+b+c)^3$ My progress is that I applied Hölder’s for this,
$3(a^2b+b^2c+c^2a) \frac{(ab^2+bc^2+ca^2)}{abc} \geq (a+b+c)^3$
whereas $3(a^2b+b^2c+c^2a) \frac{(ab^2+bc^2+ca^2)}{abc} = (1+1+1)(a^2b+b^2c+c^2a)(\frac{b}{c} + ... | Now, you can use Holder so:
$$(1+1+1)\left(a^2b+b^2c+c^2a\right)\left(\frac{b}{c} + \frac{c}{a} + \frac{a}{b}\right)=\sum_{cyc}1\sum_{cyc}a^2b\sum_{cyc}\frac{a}{b}\geq$$
$$\geq\left(\sum_{cyc}\sqrt[3]{1\cdot a^2b\cdot\frac{a}{b}}\right)^3=(a+b+c)^3.$$
Another way.
We need to prove that:
$$\sum_{cyc}(3a^3b^3+3a^4bc+3a^2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3846173",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
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Evaluating a definite integral as the limit of a Riemann sum I'm having trouble evaluating the following problem using the limit of a Riemann sum:
$\int_1^4x^2-4x+2dx$
Using $\lim_{n->\infty}\sum_{i=1}^n f(x_i)\Delta x$ where $a=1$ and $b=4$, $\Delta x = \frac{b-a}{n}$, and $x_i = a + i\Delta x$,
I come up with:
$\lim_... | Wow this confused me for a bit also until I realized we were both making the same mistake. In your case, you wrote
$$\sum_{i=1}^n\frac3n=\frac3n\sum_{i=1}^n1=\frac3n,$$
but $\sum_{i=1}^n1=n$ and this should have been
$$\sum_{i=1}^n\frac3n=\frac3n\sum_{i=1}^n1=3.$$
Since $x_i=1+\frac{3i}{n}=\frac{n+3i}{n}$, the sum is
... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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If $x,y,z\in(1,3)$ and $xy+yz+zx=26$ then prove that $x+y+z>\frac{35}{4}$(Sweden 1971) If $x,y,z\in(1,3)$ and $xy+yz+zx=26$ then prove that $x+y+z>\frac{35}{4}$
I was trying to do the question above using homogenization. I was trying to do it as follows:
$\sqrt{xy+yz+xz}=\sqrt{26}$
Hence $x+y+z>\frac{35}{4*\sqrt{26}}*\... | Maybe the following is better.
Let $x=\sqrt{\frac{26}{3}}a$, $y=\sqrt{\frac{26}{3}}b$ and $z=\sqrt{\frac{26}{3}}c$.
Thus, $$ab+ac+bc=3$$ and $$x+y+z=\sqrt{\frac{26}{3}}(a+b+c)\geq\sqrt{\frac{26}{3}}\sqrt{3(ab+ac+bc)}=\sqrt{78}.$$
Thus, it's enough to prove that $$\sqrt{78}>\frac{35}{4}$$ or
$$1248>1225$$ and we are don... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3847360",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Venn Diagram Conditional Probability Questions Solution I was having a go at this question, I got an answer of $\frac{2}{29}$, not sure if this is correct, I would appreciate if someone could explain the solution to it.
| You are correct, please include your working next time.
From the definition of conditional probability we have $$P(B\cap A'|C)=\frac{P(B\cap A'\cap C)}{P(C)}$$
$$=\frac{\frac{x}{5}}{x+\frac{x}{5}+\frac{x}{2}+2x^2+x}=\frac{\frac{x}{5}}{\frac{27}{10}x+2x^2}=\frac{\frac{1}{5}}{\frac{27}{10}+2x}.$$
But the sum of the proba... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3851301",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
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Use the $\varepsilon - \delta$ definition of the limit to verify that $\lim_{(x,y)\to(2,5)} xy = 10$.
Question:
Use the $\varepsilon - \delta$ definition of the limit to verify that
$\lim\limits_{(x,y)\to(2,5)}xy = 10$.
Hint given:
$xy − 10 = (x − 2)(y − 5) + 5(x − 2) + 2(y − 5).$
My solution
$\forall \space\varepsil... | First observe that:
$$
|xy-10|=|xy-2y+2y+10|\le |y||x-2|+2|y-5|
$$
Then set the first restriction: $\delta_1=1$
$$
|y-5|\le \big((x-2)^2+(y-5)^2)\big)^{½}<1=\delta_1,
$$
which implies, that
$$
|y|\le |y-5|+5<6,
$$
and hence, if $\big((x-2)^2+(y-5)^2)\big)^{½}<1$, then
$$
|xy-10|=|xy-2y+2y+10|\le |y||x-2|+2|y-5|\le 6|x... | {
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"url": "https://math.stackexchange.com/questions/3853806",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Please give an algebraic proof of the following inequality: $\frac{a}{a+b} \gt \frac{a-1}{a+b-1}$ for $a,b \gt 0$. For $ a,b > 0 $ and $\left(\frac{a}{a+b}\right)\cdot\left(\frac{a-1}{a+b-1}\right)=\frac{1}{3}$, show that $$\frac{a}{a+b} > \frac{a-1}{a+b-1}$$
I am not getting how to do this. Please enlighten me. What i... | You have
$$
\frac{a}{a+b}-\frac{a-1}{a+b-1}=\frac{b}{(a+b)(a+b-1)}
$$
Given that $a,b>0$, your statement is equivalent to $a+b-1>0$.
If $a$ and $b$ are integer, as you seem to hint in the comments, then the statement is completely obvious.
For general real $a,b$, we'd like to see whether this follows from
$$
\frac{a}{a... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3853921",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Prove that if x,y,z are positive integers such that $x^3+y^3+z^3=3(x+y+z+xyz)$ then they must be consecutive numbers I was thikning going for contradiction, assuming they're not consecutive but I found it very hard to find a statement that contradicts that condition. Maybe stating that there exist some $m, n \in \Bbb{Z... | First, subtract $3xyz$ from both sides of the equation:
$$x^3+y^3+z^3=3(x+y+z+xyz) \\
x^3+y^3+z^3-3xyz = 3(x+y+z)$$
Now, using this factorisation, observe:
$$
(x+y+z)(x^2+y^2+z^2-xy-yz-zx)=3(x+y+z)$$
Since $x,y$, and $z$ are positive integers which implies $x+y+z\neq 0$, you can divide the equation by $x+y+z$ to get
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3856174",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 1
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If $a+b+c=3$, then show $\sqrt{a^2 + ab +b^2}+ \sqrt{b^2 + bc +c^2}+\sqrt{c^2 + ac +a^2} \geq \sqrt{3}$ If $a+b+c=3$, and $a,b,c$ are positive real numbers, then show $\sqrt{a^2 + ab +b^2} + \sqrt{b^2 + bc +c^2} +\sqrt{c^2 + ac +a^2}\geq \sqrt{3}$
Normally when I do inequalities I try to first find where equality would... | $$\left(\frac{a+b}{2}\right)^2+\frac{(a-b)^2}{12}=\frac{a^2+ab+b^2}{3}$$
$$\therefore\ \sqrt{\frac{a^2+ab+b^2}{3}}\ge\frac{a+b}{2}$$
$$\therefore\ \sum_{cyc}\sqrt{\frac{a^2+ab+b^2}{3}}\ge\sum_{cyc}\frac{a+b}{2}=a+b+c=3$$
So in fact a stronger identity is possible $\sum_{cyc}\sqrt{a^2+ab+b^2}\ge 3\sqrt{3}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3861614",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 2
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Closed form $\int_{0}^{\frac{\pi}{2}}\int_{0}^{\frac{\pi}{2}}\int_{0}^{\frac{\pi}{2}}\sin(xyz)\,dx\,dy\,dz$ This is part of an assignment from a multivariable calculus course. We've only just defined triple integrals and we're not using special functions or anything like that, so I'm pretty sure this was a mistake. Sti... | Incomplete (?) solution.
\begin{align}J&=\int_0^{\frac{\pi}{2}}\int_0^{\frac{\pi}{2}}\int_0^{\frac{\pi}{2}}\sin(xyz)dxdydz\\
&\overset{u=\frac{2x}{\pi},v=\frac{2y}{\pi},w=\frac{2z}{\pi}}=\frac{\pi^3}{8}\int_0^1\int_0^1\int_0^1\sin\left(\frac{\pi^3uvw}{8}\right)dudvdw\\
&=-\frac{\pi^3}{8}\int_0^1\int_0^1 \ln x\sin\left(... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3863508",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 1
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Have I done my homogenization correctly for this question and if so how do I finish it off? We have $a,b,c>0$ with $a^2+b^2+c^2=1$ prove that:
$\frac{a}{1+bc}+\frac{b}{1+ac}+\frac{c}{1+ab}>1$
As soon as I saw this question, I immediately thought of using homogenization in the following manner:
$\frac{a}{1+bc}+\frac{b}{... | we will prove $$\sum_{cyc}\frac{a}{1+bc}=3-\sum_{cyc}\frac{abc}{1+bc}>1$$
or $$\sum\frac{abc}{1+bc}<2$$
we know will make denominator homogenous in degree 2
$1+bc=a^2+b^2+c^2+bc\ge 4a^{1/2}b^{3/4}c^{3/4}$
thus $$\sum\frac{abc}{1+bc}\le \frac{1}{4}a^{1/2}b^{1/4}c^{1/4}\le \frac{3}{4}abc\le \frac{1}{4\sqrt{3}}<2$$\
here... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3865582",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 1
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Triple integration with $x^2+y^2-z^2$ I'm having this problem.
$$\iiint_{K}(x^2+y^2-z^2)dxdydz,\quad K: x^2+y^2+z^2\leq 1$$
My solution.
$$x=\rho \sin\phi \cos\Theta,\ y=\rho \sin\phi \sin\Theta, z=\rho \cos \phi\\
dxdydz=\rho^2 \sin\phi d\rho d\phi d\theta\\
0\leq \rho \leq 1\\
0 \leq \phi \leq \pi\\
0 \leq \theta \le... | $\int \int \int _{K}(x^2+y^2-z^2)dxdydz $
$= \displaystyle \int _0^{2\pi }\int _0^{\pi} \int_0^1 ((\rho \sin \phi cos \theta)^2 + (\rho \sin \phi \sin\theta)^2 - (\rho \cos \phi)^2) \, \rho^2 \, sin\phi \, d\rho \, d\phi \, d\theta$
$= \displaystyle \int _0^{2\pi} \int_0^{\pi} \int_0^1 (\rho^2 \sin^2 \phi - \rho^2 \c... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3866733",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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System of equations from roots of polynomial I'm given the equation $3072x^4-2880x^3+840x^2-90x+3=0$ and told that its roots are $\alpha, \alpha r, \alpha r^2, \alpha r^3,$ for some $r\in \mathbb{R}$.
By considering the sum of the roots, the product, etc. I've found that \begin{gather}\alpha(1+r+r^2+r^3)=\frac{15}{16} ... | Going with your equations... Dividing third equation by first, we get $\alpha^2 r^3=\frac{1}{32}$. This implies $r > 0$ because $\alpha$ must be real as well (follows from the first equation). Plugging this into the second equation we get
$$
\frac{35}{128}=\frac{1}{32}\left(\frac{1}{r^2}+\frac{1}{r}+2+r+r^2\right)=\fra... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3869455",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
"answer_id": 1
} |
Generating primitive Pythagorean triples Given this equation $a^2 + b^2 = c^2$ , in order to generate all primitive Pythagorean triples all we nedd to do is : write (a,b,c) as :
$a = 2mn $
$b = m^2-n^2$
$c= m^2+n^2$
with conditions : $\gcd(m,n)=1$ and $(m+n)$%$2= 1 $
But what about if we add a parameter $K$ to the eq... | We need only show how side $C$ of one triple is the same as side $A$ of another triple. For example, if we have $(3,4,5)$ and $(5,12,13)$, then
$$A_1^2+B_1^2=C_2^2-K_2^2\quad\text{ is simply }\quad 3^2+4^2=13^2-12^2$$
In this "easiest" case, $GCD(A_1,B_1,C_2)=1$.
There is a formula to identify these matches, if they ex... | {
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"url": "https://math.stackexchange.com/questions/3870745",
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"source": "stackexchange",
"question_score": "3",
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Find all $n,x \in \Bbb N$ such that $3\cdot2^x+4=n^2$
Find all $n,x \in \Bbb N$ such that $3\cdot2^x+4=n^2$
Arranging the equation a bit one has that $$(n-2)(n+2)=3\cdot2^x.$$
Now considering the cases $n-2=3, n+2=2^x$ from $n-2=3 \Rightarrow n=5$. Substituting this to $n+2=2^x \Rightarrow 7=2^x$ which isn't an integ... | You're off to a good start; writing
$$(n-2)(n+2)=3\cdot2^x,$$
shows, by unique factorization of integers, that
$$n-2=2^a3^b,\qquad\text{ and }\qquad n+2=2^c3^d,$$
for nonnegative integers $a$, $b$, $c$ and $d$ with $a+c=x$ and $b+d=1$. Of course $b+d=1$ leaves very few options for $b$ and $d$. Can you continue from her... | {
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Prove by epsilon delta $\lim\limits_{(x,y) \to (0,0)} \frac{(x-1)^2\sin(x^2)y}{x^2+y^4} = 0$ I need to prove this using $\epsilon$ $\delta$
$$\lim\limits_{(x,y) \to (0,0)} \frac{(x-1)^2\sin(x^2)y}{x^2+y^4} = 0$$
I know that $|x| < \delta $ and $|y| < \delta $, so then
$|\frac{(x-1)^2sin(x^2)y}{x^2+y^4}-0|$ $\lt$ $\frac... | $|\frac {(x-1)^{2} \sin (x^{2}) y} {x^{2}+y^{4}}| \leq (x-1)^{2} \frac {x^{2}|y|} {x^{2}}$ since $|\sin (x^{2})| \leq x^{2}$. Hence $|\frac {(x-1)^{2} \sin (x^{2}) y} {x^{2}+y^{4}}| <\epsilon$ if $|x|<1$ and $|y|<\frac {\epsilon } 4 $
| {
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Sum of digits of square number raised to itself From testing a few different square numbers, it seems to be the case that when raising a square number to the power of itself, the sum of the digits of the result satisfy the property that the sum of their digits is the square number itself.
I realise the above sentence i... | Your original claim is poorly worded in that it does not adequately distinguish between sums of digits and sums of sums of digits and what numbers should be compared against one another. As written the claim is false, as pointed out by other users, since you have the sum of digits or the sum of the sum of the digits n... | {
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Find the Limit.? What is the limit of
$Y_{n} := \frac{1}{3}Y_{n-1} + \frac{2}{3}Y_{n-2} $
for $n > 2$ and $Y_{1}<Y_{2}$.
I have tried solving, but the limit tends to exist between $y_{1}$ and $y_{2}$ and hence I am not able to evaluate what would be the ratio between which they would lie?
| Let me show you the general method for dealing with these linear recurrences. The idea is to convert the linear recurrence into a matrix recurrence: $v_{n} = Av_{n-1}$ so that $v_n = A^nv_0$.
Here, you have
$$ \begin{pmatrix} Y_n \\ Y_{n - 1} \end{pmatrix} = \begin{pmatrix} \frac13 & \frac23 \\ 1 & 0 \end{pmatrix} \beg... | {
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minimum value of $\frac{xy}{z}+\frac{yz}{x}+\frac{xz}{y}$ if $x^2+y^2+z^2=1$ If $x^2+y^2+z^2=1$ what is the minimum value of $\frac{xy}{z}+\frac{yz}{x}+\frac{xz}{y}$ for $x,y,z \gt 0$ ?
I would like to know if the minimum could be found using simpler ways.(like $AM \ge GM \ge HM$).
knowing $xy+yz+xz \geq \frac{-1}{2}$ ... | First, notice that $\frac{xy}{z}+\frac{yz}{x}+\frac{xz}{y} = \frac{xyz}{z^2}+\frac{xyz}{x^2}+\frac{xyz}{y^2}=xyz({\frac{1}{z^2}+\frac{1}{x^2}+\frac{1}{y^2}})$
Using the inequality of arithmetic and geometric means, we have:
$$ \frac{1}{x^2}+\frac{1}{y^2}\ge2\sqrt{\frac{1}{x^2}\frac{1}{y^2}}\iff \frac{1}{x^2}+\frac{1}{y... | {
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How do I solve $(3x^2 \tan y- 2y^3/x^3 )dx + (4y^3 + x^3 \cos^2 y+ 3y^2/x^2 ) dy=0$ Somebody please help me to solve this Differential equation.
$\left(3x^2\tan (y)- \frac{2y^3}{x^3}\right)dx+ \left(4y^3 + \frac{x^3}{\cos^2y} + \frac{3y^2}{x^2}\right) dy=0 $
| It is an exact DE, solution is $x^3\tan{y}+y^4+\frac{y^3}{x^2}+C=0$
| {
"language": "en",
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"source": "stackexchange",
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Prove a limit in two variables is $0$ I have the limit
\begin{align}
\lim_{(x,y) \rightarrow(0,0)} \frac{1-\cos(xy)}{\sqrt{x^2+y^2}(x^2+y^2)}
\end{align}
First thing I did is to use Taylor, so $$1-\cos(xy)=\frac{(xy)^2}{2} -\frac{(xy)^4}{4} + o((xy)^4)$$
Therefore the limit is $$\lim_{(x,y) \rightarrow (0,0)} \frac{1}{... | It seems correct, to avoid Taylor we can use that
$$\frac{1-\cos(xy)}{\sqrt{x^2+y^2}(x^2+y^2)}=\frac{1-\cos(xy)}{(xy)^2}\frac{(xy)^2}{\sqrt{x^2+y^2}(x^2+y^2)}$$
with $\frac{1-\cos(xy)}{(xy)^2} \to \frac12$ by standard limits.
For the conclusion by squeeze theorem, we should also consider a part the case $x=0$.
| {
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Factorizing the determinant Here, in this question, we have to factorize this determinant. I understood the (a-b), (b-c), (c-a) part but I am not able to understand
*
*how to find if determinant is symmetrical in a, b, c.
*I know that rest factor will be of degree 2 as we have got (a-b), (b-c), (c-a), but as here (a... | (1) If you put $a=b,b=c$ or $c=a$ then, $\Delta=0$.
Being symmetric about $a,b,c$ means that if $(a_0,b_0,c_0)$ is a solution to $\Delta=0$, then $(b_0,a_0,c_0)$ is also a solution and so are all other permutations of $(a_0,b_0,c_0)$. This can be seen easily. For now, we verify it for the $3$ permutations $(a_0,b_0,c_0... | {
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Combinatorics problem - $5$-digit code with $3,4,5$ present I have a combinatorics problem that I'm struggling with. Here it is: "How many 5-digit codes have among their digits each of $3,4,5$?"
I do understand the general strategy, where we first see how we can permute $3,4,5$ with the other empty slots ($\frac{5!}{2}... | There are a couple of strategies we could employ here.
Method 1: We consider cases depending on how often each digit appears.
Case 1: The digits $3$, $4$, $5$ each appear exactly once.
Choose one of the positions for the $3$, one of the remaining four positions for the $4$, and one of the remaining three positions fo... | {
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Find every point whose distance from each of the two coordinate axes equals its distance from the point $(4, 2)$ Here is my attempt:
Let $(x, y)$ be any such point.
Then the distance of $(x, y)$ from the $x$-axis is $\lvert y \rvert$, whereas the distance of that point from the $y$-axis is $\lvert x \rvert$. Thus we... | If you had actually checked each candidate point for equality of distance from the coordinate axes and $(4,2)$, you would have seen that only $(2,2)$ and $(10,10)$ satisfy the conditions of the original question.
| {
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Determine convergence and limit I want to determine whether each of the sequences converges, and if so, to what limit.
$$\begin{align} 1. \quad \quad & a_n=\frac{n^{2}}{n+1}-\frac{n^{2}+1}{n}\\
2. \quad \quad & a_n=\sqrt{n+1}-\sqrt{n}
\end{align}$$
For the first one, I tried to use ratio test, but it's not working. For... | For the first sequence, note that
$$\begin{align}
a_n & = \frac{n^2}{n+1} - \frac{n^2+1}{n}\\
& = \frac{n^3-n^3-n-n^2-1}{n^2+n}\\
& = \frac{-n^2-n-1}{n^2+n}.
\end{align}$$
And taking the limit you obtain
$$\lim_{n \to \infty} a_n = -1 \quad (\in \mathbb{R})$$
so it converges.
For the second sequence, as you said, multi... | {
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Showing that a sequence is decreasing I need to show that the sequence defined by $a_1 = \frac{5}{2} $ and $a_{n+1} = \frac{1}{5}(a_n^2 + 6) $
is decreasing. The fact that $2 < a_n < 3 $ has been proven already.
I now just need to show that $a_n$ is decreasing.
I have tried the usual ways of showing $a_{n+1} - a_n \le ... | Using the definition of a decreasing sequence, we starter by computing $a_{n+1}-a_n.$
$$\begin{align}
a_{n+1}-a_n & \implies \frac{1}{5}(a_n^2+6)-a_n\\
& \implies \frac{a_n^2-5a_n+6}{5}\\
& \implies \frac{(a_n - \frac{5}{2})^2-\frac{1}{4}}{5}\\
& \implies \frac{(\frac{2a_n-5}{2})^2-\frac{1}{4}}{5}\\
& \implies \frac{(2... | {
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Find the residue of $1/(z^2+1)^{n+1}$ at $z=i$. I'm trying to find the residue of $\frac{1}{(z^2+1)^{n+1}}$ at $z=i$. I believe this should be approached combinatorially but I am getting stuck in the combinatorics. Here's what I have so far.
$$ \frac{1}{(z^2+1)^{n+1}} = \frac{1}{(z-i)^{n+1}} \cdot \frac{1}{(z-i+2i)^{... | There is agreement between the residue and the integral of $f(x)$ along the real axis. We show that $$\int_{-\infty}^\infty \frac{dx}{(x^2+a^2)^{n+1}} = \frac{\pi}{2^{2n}a^{2n+1}} \frac{(2n)!}{n!n!}.$$
This is true for $n=0$:
$$\int_{-\infty}^\infty \frac{dx}{x^2+a^2} = \frac{1}{a}\int_{-\infty}^\infty \frac{\frac{dx}... | {
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A bag contains 7 blue balls and 3 green balls. What is the probability of the following events? A bag contains 7 blue balls and 3 green balls.
I am not sure about my answer but please allow me to share.
*
*If 5 balls are drawn in succession at random without replacement, what is the probability of dra... | Problem 1.
Your answer to problem 1. is wrong, because you have computed the probability
of getting G-B-G-B in the first four drawings. The problem specifically
asked for the probability of the satisfying sequence occuring anywhere
within the first five drawings.
The actual problem is much more complicated, and is bes... | {
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"question_score": "1",
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On the nearest-square function - Part 2 and the quantity $m^2 - p^k$ where $p^k m^2$ is an odd perfect number (Note: This question has been cross-posted to MO.)
This question is an offshoot of this earlier one and this other question.
Let $n = p^k m^2$ be an odd perfect number with special prime $p$ satisfying $p \equ... | If you don't have a proof that the smallest square larger than $m^2-p^k$ is not $m^2$, then your method does not work.
Otherwise, your method works.
Using your idea, one can prove that if $\lfloor\sqrt{4z}+1\rfloor\lt m$, then $m\lt p^k$.
Proof :
Subtracting $\lfloor\sqrt{4z}+1\rfloor^2$ which is the smallest square la... | {
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If $2^{2k}-x^2\bigm|2^{2k}-1$ then $x=1$ This is the $y=2^k$ case of this question.
Suppose that $k\geq1$ and $0<x<2^k$ and $2^{2k}-x^2\bigm|2^{2k}-1$. Is it necessarily the case that $x=1$?
Equivalently: Suppose that there are two positive divisors of $2^{2k}-1$ which average to $2^k$. Is it necessarily the case that ... | Following from the OP's self-answer, the quadratic Diophantine equation $$nx^2-4(n-1)y^2=1\tag1$$ means that $nx^2\equiv1\pmod4$ which forces $n=4N+1$ as $x^2\equiv0,1\pmod4$. Thus $$(4N+1)x^2-NY^2=1$$ where $Y=4y$ so $Y^2\equiv4\pmod{4N+1}$ and $x^2\equiv1\pmod N$.
Suppose that $Y\equiv\pm2\pmod{4N+1}$. Then $y=(rn\pm... | {
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Find the probability that a 4 will appear on the second dice. Two dice are rolled. If the sum of the two dice is 7 or less than 7. Find the probability that a 4 will appear on the second dice.
I am trying to do this exercise. But my result is different from the one provided by the book. In the book it appears that the ... | Your working is correct.
With one of the dice being $1$, you can have max sum of $7$ with total of $6 \times 2 - 1 = 11$ possibilities. (Numbers can interchange places on dice and subtract one as we did count $(1, 1)$ twice.)
Now with $2$, there are $10$ possibilities (but subtract 2 possibilities for $(1,2)$ which is ... | {
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Changing bounds of integration question; spot the mistake. Suppose we wanted to compute $$\int \frac{1}{x^2 + x + 1} ~ dx.$$ One such way is to fix $c$ and let $x$ vary as bounds, thus we have that $$\int \frac{1}{x^2 + x + 1} ~ dx = \int_c^x \frac{1}{t^2 + t + 1} ~ dt = \int_c^x \frac{1}{(t + 1/2)^2 + 3/4} ~ dt.$$ We ... | $$\int_{a}^b f(x + k) ~ dx = \frac{1}{c} \int_{ac + k}^{bc + k} f(x/c) ~ dx.$$ is wrong. It should be:
$$\int_{a}^b f(x + k) ~ dx = \frac{1}{c} \int_{a(c + k)}^{b(c + k)} f(x/c) ~ dx.$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Is $h(n)$ independent of $n$? Let $n$ be a positive integer.
Let $f(n) = \lfloor\frac{8n+13}{25}\rfloor$ and $g(n) = \lfloor\frac{n-17}{25}\rfloor$.
Now consider $$h(n) = f(n) - \lfloor\frac{n - 12 - g(n)}{3}\rfloor$$
Then I calculated values of $h(n)$ for various values of $n$ and I'm getting $h(n)=4 $ everytime. Infa... | Let $n=25k+r$ for some non-negative integers $k,r$ with $r<25$. We then have
$$\left\lfloor\frac{n-12-\left\lfloor\frac{n-17}{25}\right\rfloor}{3}\right\rfloor = \left\lfloor\frac{24k+r-12-\left\lfloor\frac{r-17}{25}\right\rfloor}{3}\right\rfloor = 8k-4+\left\lfloor\frac{r-\left\lfloor\frac{r-17}{25}\right\rfloor}{3}\r... | {
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Find the smallest possible number of different numbers between $a + b, b + c, c + a, ab + 1, bc + 1, ca + 1, abc$
Find the smallest possible number of different real positive number
between $a + b, b + c, c + a, ab + 1, bc + 1, ca + 1, abc$
$a$, $b$, $c$ are real positive numbers and they're all
different. From that,... | Consider the inequality chain:
$$ a+b =ab+1 < a+c = ac+1 = abc < b+c < bc+1,$$
which has general solutions $ a = 1, 1 < b < \frac{1+\sqrt{5}}{2} , c = \frac{1}{b-1}$.
As an explicit example, with $a = 1, b = 1.5, c = 2$, we get 4 distinct values: $2.5, 3, 3.5, 4.$
We will show that 4 distinct values is the best we can... | {
"language": "en",
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"source": "stackexchange",
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Bessel function, characteristic function, semicircle distribution I am trying to prove that $\dfrac{1}{x} J_1(2x)$, where $J_n$ is the Bessel function of order n, is the characteristic function of the semicircle distribution, i.e. $σ(x)=\dfrac{1}{2π}\sqrt{4-x^2}
\textbf{1}_{|x|<2} $. Basically, I would like to calcul... | $$J_1(z) = \dfrac{1}{iπ} \int_{0}^π e^{iz\cos θ}\cos θ dθ $$
Now, we want to calculate the integral shown above to find the density function corresponding to the characteristic function $ \dfrac{1}{x} J_1(2x) $.
\begin{align*}
f(x) &= \dfrac{1}{2π} \int_{\mathbb{R}} φ(t) e^{-itx} dt = \dfrac{1}{2π} \int_{\mathbb{R}} ... | {
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$a_n = u^n+v^n+(-1)^n$ Anyone has solutions for this problem?
Let $u, v$ are distinct roots of equation:
$$x^2-tx+1=0 (t \in \mathbb{N}, t>2)$$
and sequence $$(a_n): a_n = u^n+v^n+(-1)^n$$
Prove that: $$a_{3^m}\mid a_{3^mn} \forall m,n \in \mathbb{N} , gcd(n,3)=1$$
Notes
$S_n=\frac{a_{3^mn}}{a_{3^m}}\\$
$a=u^{3^m}$... | By either using Vieta's formulas or just expanding $(x - u)(x - v) = x^2 - tx + 1$ and comparing coefficients, we get
$$\begin{equation}\begin{aligned}
u + v & = t \\
uv & = 1
\end{aligned}\end{equation}\tag{1}\label{eq1A}$$
For non-negative integers $i$, define $f_i(t)$ to be $u^i + v^i$ expressed as a function of $t$... | {
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Moment generating function of Laplace distribution step by step There are two similar posts but none of them helped me to get through the full derivation of the very simple MGF that should be according to Wikipedia:
$$
\frac {\exp(\mu t)}{1-b^{2}t^{2}}
$$
Here's my attempt:
Definition of MGF
$$
M_X(t) = \mathbb{E}\left... | Alright, here's the full correct solution if anyone is interested.
Definition of MGF
$$
M_X(t) = \mathbb{E}\left[\exp(t X)\right]
$$
by LOTUS
$$
= \int \exp(t x) \cdot p(x) dx
$$
Plugging-in PDF
$$
= \int \exp(t x) \cdot \frac{1}{2b}\exp \left(- \frac{\mid x - \mu \mid }{b} \right) dx
$$
$$
= \frac{1}{2b} \int \exp(t x... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Eigenvalues of an almost diagonal matrix I know that the eigenvalue of a diagonal matrix is simply the values in the diagonal. However, if I have a matrix of the following form:
$$
\begin{bmatrix}
a & b & 0 & 0 \\
b & c & 0 & 0 \\
0 & 0 & d & e \\
0 & 0 & e & f
\end{bmatrix}.
$$
Is there a closed form way to express t... | The eigenvalues of a block diagonal matrix are the eigenvalues of each block. The corresponding eigenvectors are the eigenvectors of each block padded with zeros. For example:
The eigenvalues of the matrix $$A = \begin{bmatrix}4 & 3 \\ 3 & 4 \end{bmatrix}$$ are $7$ and $1$, and the corresponding eigenvectors are respec... | {
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Evaluating $\left.\int_0^{\pi/2}\sqrt{1+\frac1{\sqrt{1+\tan^nx}}}\text dx \middle/\int_0^{\pi/2}\sqrt{1-\frac1{\sqrt{1+\tan^nx}}}\text dx\right.$
Evaluate the integral ratio$$\dfrac{I_1}{I_2}=\dfrac{\displaystyle \int_{0}^{\frac{\pi}{2}} \sqrt{1+\frac{1}{\sqrt{1+\tan^nx}}} \mathrm{d}x}{\displaystyle \int_{0}^{\frac{\p... | Note
\begin{align}
\frac{I_1}{I_2}&=1+ \frac{I_1-I_2}{I_2} \\
&=1+\frac1{I_2}\int_0^{\frac\pi2} \left( \sqrt{1+\frac{1}{\sqrt{1+\tan^nx}}}- \sqrt{1-\frac{1}{\sqrt{1+\tan^nx}}} \right)dx \\
&=1+\frac1{I_2}\int_0^{\frac\pi2}
\frac{\sqrt{\sqrt{\sin^nx+\cos^nx}+\sqrt{\cos^nx}}
- \sqrt{\sqrt{\sin^nx+\cos^nx}-\sqrt{\cos^nx... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3939493",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "30",
"answer_count": 1,
"answer_id": 0
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Range of quadratic function using discriminant
Let $x^2-2xy-3y^2=4$. Then find the range of $2x^2-2xy+y^2$.
Let $2x^2-2xy+y^2=a$.
Then $ax^2-2axy-3ay^2=4a=8x^2-8xy+4y^2\implies (a-8)x^2-(2a-8)xy-(3a+4)y^2=0$.
We divide both side by $y^2$ and let $t=\frac{x}{y}$.
Then it implies $(a-8)t^2-(2a-8)t-(3a+4)=0$.
Since its ... | Consider the curves $x^2-2xy-3y^2=4 \ (C)$ and $2x^2-2xy+y^2=k^2+4 \ (E)$ At the points where they intersect, we can substitute the value of $-2xy$ from the first into the second to get $$x^2+4y^2=k^2 \ (E)$$ This represents an ellipse centered at the origin. Now, either with a bit of graph sketching, or by considering... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3942812",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 2
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A polynomial has the same remainder when divided by $x+k$ or $x-k$; what is $k$? Question
Given that $y = 3x^3 + 7x^2 - 48x + 49$ and that $y$ has the same remainder when it is divided by $x + k$ or $x - k$, find the possible values of $k$.
My attempt
Let $f(x) = 3x^3 + 7x^2 - 48x + 49$
$\text{Using Remainder Theorem... | \begin{align}
f(-k) &= f(k) \\
3(-k)^3 + 7(-k)^2 - 48(-k) + 49 &= 3(k)^3 + 7(k)^2 - 48(k) + 49 \\
\bf-3k^3 - 7k^2 + 48k + 49 &= 3k^3 + 7k^2 - 48k + 49 \\
-3k^3 --3k^3 - 7k^2 - 7k^2 + 48k + 48k + 49 - 49 &= 0 \\
-6k^3 + 14k^2 + 96k &= 0 \\
\frac{-6k^3}{k} + \frac{14k^2}{k} + \frac{96k}{k} &= \frac{0}{k} \\
6k^2 + 14k -... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3943318",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 0
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$\lim_{x \to 0} \frac{\ln (1+x \arctan x)-e^{x^2}+1}{\sqrt{1+2x^4}-1}$ I've tried to solve this limit:
$$\lim_{x \to 0} \frac{\ln (1+x \arctan x)-e^{x^2}+1}{\sqrt{1+2x^4}-1}$$
Here,
$$\frac{\ln (1+x \arctan x)-e^{x^2}+1}{\sqrt{1+2x^4}-1}$$ $$\sim \frac{(x \arctan x-x^2)(\sqrt{1+2x^4}+1)}{2x^4}$$ $$\sim \frac{(x \arctan... | The handling of denominator is easy. Multiplying the expression under limit with $$\frac{\sqrt{1+2x^4}+1}{\sqrt{1+2x^4}+1}$$ we can see that the limit in question is equal to the limit of the expression $$\frac{\log(1+x\arctan x) - e^{x^2}+1}{x^4}$$ Split this expression into three parts as $$\frac{\log(1+x\arctan x) -... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3944331",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
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$4q=a^2+27b^2$ elementary proof of uniqueness I’m trying to prove that if $q$ is prime then the representation $$4q=a^2+27b^2$$ is unique (up to sign). Any help would be appreciated.
Edit: By elementary I mean only using divisibility rules, modular arithmetic, etc.
| Suppose that $4q=a^2+27b^2=c^2+27d^2$ and $a,b,c,d> 0$ (equals 0 is not possible).
We have $a^2d^2-b^2c^2\equiv 0 $ mod $q$.
Then $(ad-bc)(ad+bc)\equiv 0$ mod $q$.
If $ad+bc\equiv 0$ mod $q$, then by $16q^2=(a^2+27b^2)(c^2+27d^2)=(ac-27bd)^2+27(ad+bc)^2$, we must have $ad+bc=0$. This is impossible, so we have $ad-bc \e... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3945743",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 1
} |
Generalize a sum given first elements Given $\rho\in(0,1)$ and $f\colon\mathbb N\to\mathbb R$ defined as
$$\begin{aligned}
f(0)&\mapsto1\\
f(1)&\mapsto2+\rho\\
f(2)&\mapsto3+2\rho+\rho^2\\
f(3)&\mapsto4+3\rho+2\rho^2+\rho^3
\end{aligned}$$
generalize $f$ for an arbitrary $n$ and simplify the summator (if possible)
$$\s... | I would like to add one more approach.
Consider
$$
f(n) = (n+1) + n \rho + (n-1) \rho^2 + \cdots + 1 \rho^n = \sum_{k=1}^{n+1} k \rho^{n+1-k}.
$$
Obviously $f(0)= 1$, $f(1) = 2 + \rho$ and so on.
Now make this function a bit more complicated
$$
f(n,x) = (n+1) x^n + n x^{n-1} \rho + (n-1) x^{n-2} \rho^2 + \cdots + 1 \rh... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3946005",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 6,
"answer_id": 4
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Find a polynomial $P$ of the smallest degree such that $\int_{-1}^{1}\left|x^{2 / 3}-P(x)\right|^{2} d x<0.01$ Find a polynomial P of the smallest degree such that $$\int_{-1}^{1}\left|x^{2 / 3}-P(x)\right|^{2} d x<0.01$$
I am ususally solve this kind of "best approximation" problems by taking using $GS$ but I think th... | First, let us try $\mathrm{deg} P(x) \le 2$.
Let $P(x) = ax^2 + bx + c$.
We have $$\int_{-1}^1 |\sqrt[3]{x^2} - P(x)|^2 \mathrm{d} x =
\frac{2}{5}a^2 + \frac{4}{3}ac + \frac{2}{3}b^2 + 2c^2 - \frac{12}{11}a - \frac{12}{5} c + \frac{6}{7}.$$
It is easy to minimize a quadratic form. The best $P(x)$
is given by $P(x) = \... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3951685",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
} |
Best way to integrate this? $$ \int_{\frac{\pi}{3}}^\frac{\pi}{2} \frac{\sin^2(x)}{\sqrt{1+\cos(x)}}dx $$
What would be the best way to go about integrating and evaluating this integral?
| Substitute $\cos x =t \implies -\sin x dx =dt $: $$\int_{0}^{0.5} \frac{\sqrt{1-t^2}}{\sqrt{1+t}} dt =\int_0^{0.5} \sqrt{1-t}dt = \left[ -\frac 23 (1-t)^{3/2} \right]_0^{0.5} = \frac 23-\frac 23 \cdot \frac{1}{2\sqrt 2} = \frac 23 -\frac{1}{3\sqrt 2}$$
Second way: $$\frac{\sin^2x}{\sqrt{1+\cos x}}= \frac{4\sin^2(x/2)\c... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3953946",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
} |
Evaluating $\int \sqrt{16-x^2} \,dx$ Are the steps below correct? Can you tell me another way to solve it, just hint?
$ \int \sqrt{16-x^2} \,dx $
$ x = 4 \sin u \rightarrow u=\arcsin(\frac{x}{4}) \rightarrow dx = 4\cos u \,dx$
$ \int \sqrt{16-x^2} \,dx = 16 \int \cos^2 u \,du = 16 \int \frac{\cos 2u +1}{2}... | Let $I = \int\sqrt{16-x^2} \ dx$. Then applying integration by parts with $u = \sqrt{16-x^2}, du = -\frac{x}{\sqrt{16-x^2}}$ and $dv = 1, v = x$:
$$I = x \sqrt{16-x^2} + \int \frac{x}{\sqrt{16-x^2}} \ dx$$
and now let $w = 16-x^2$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3954842",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Prove by definition that $\frac{1-2n^3}{4n^3-2n^2-1} \to -\frac12$ $a_n = \dfrac{1-2n^3}{4n^3-2n^2-1}$
My attempt (of the scratch):
$\left| \dfrac{1-2n^3}{4n^3-2n^2-1} + \dfrac{1}{2}\right|=\left| \dfrac{2-4n^3+4n^3-2n^2-1}{8n^3-4n^2-2}\right|=\left|\dfrac{-2n^2+1}{8n^3-4n^2-2} \right|=\dfrac{2n^2-1}{8n^3-4n^2-2}<\dfra... | Let $N \in \mathbb{N}$ s.t. $4N-2>n'+\delta$ for $n' \in \mathbb{N}$ and $\delta >0$, then $n \geq N$ implies $4n-2 \geq 4N-2 > n'+\delta$.
Let $N' \in \mathbb{N}$ s.t. $\frac{1}{N'^2}< \delta$ for $\delta >0$, then $n\geq N'$ implies $\frac{1}{n^2} \leq \frac{1}{N'^2} < \delta$.
If $n \geq \text{max}\{N, N'\}$ then $4... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3955581",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Solve $3x^3=({x^2+\sqrt{18}x+\sqrt{32}})(x^2-\sqrt{18}x-\sqrt{32})-4x^2$ Solve for $x$ in
$$3x^3=({x^2+\sqrt{18}x+\sqrt{32}})(x^2-\sqrt{18}x-\sqrt{32})-4x^2$$
I'm really not good with 4th degree equations but since the 1st term in the RHS looked a simple (a+b)(a-b) application, I tried solving that but I'm really not a... | Note that the given equation
$$3x^3=({x^2+\sqrt{18}x+\sqrt{32}})(x^2-\sqrt{18}x-\sqrt{32})-4x^2$$
is purposefully structured to allow convenient factorization as follows
$$3x^3=x^4-(\sqrt{18}x+\sqrt{32})^2 -4x^2$$
$$x^4-3x^3-4x^2-(\sqrt{18}x+\sqrt{32})^2 =0$$
$$x^4-(3x+4)x^2-2(3x+4)^2 =0$$
$$(x^2+(3x+4))(x^2-2(3x+4))=0... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3957453",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
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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.