Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
|---|---|---|
The range of the function $f:\mathbb{R}\to \mathbb{R}$ given by $f(x)=\frac{3+2 \sin x}{\sqrt{1+ \cos x}+\sqrt{1- \cos x}}$ The range of the function $f:\mathbb{R}\to \mathbb{R}$ given by $f(x)=\frac{3+2 \sin x}{\sqrt{1+ \cos x}+\sqrt{1- \cos x}}$ contains $N$ integers. Find the value of $10N$.
I tried to find the mini... | Hint:
The minimum value of the function is $1/2$ and the maximum is $2.5$. The function is clearly continuous. So it takes every value between these numbers, specifically 1 and 2. So $N=2$ which gives $10N$. Can you show that these are indeed the minimum and maximums. I have outlined the general approach
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1386474",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Prove identity: $\frac{1+\sin\alpha-\cos\alpha}{1+\sin\alpha+\cos\alpha}=\tan\frac{\alpha}{2}$ Prove identity: $$\frac{1+\sin\alpha-\cos\alpha}{1+\sin\alpha+\cos\alpha}=\tan\frac{\alpha}{2}.$$
My work this far:
we take the left side
$$\dfrac{1+\sqrt{\frac{1-\cos2\alpha}{2}}-\sqrt{\frac{1+\cos2\alpha}{2}}}{1+\sqrt{\frac... | Express both $\sin\alpha$ and $\cos\alpha$ in terms of $t=\tan\frac{\alpha}{2}$:
\begin{align*}
\sin\alpha&=\frac{2t}{1+t^2},\\
\cos\alpha&=\frac{1-t^2}{1+t^2}.
\end{align*}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1387753",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 3
} |
Evaluate the limit $\lim_{x \to 0} \left(\frac{1}{x^{2}}-\frac{1}{\tan^{2}x}\right)$
Evaluate the limit $$\lim_{x \to 0}\left( \frac{1}{x^{2}}-\frac{1}{\tan^{2}x}\right)$$
My attempt
So we have $$\frac{1}{x^{2}}-\frac{\cos^{2}x}{\sin^{2}x}$$
$$=\frac{\sin^2 x-x^2\cos^2 x}{x^2\sin^2 x}$$
$$=\frac{x^2}{\sin^2 x}\cdot\... | For your final problem I'd use L'Hospital's Rule to obtain:
$$\lim_{x \to 0} \frac{x \sin(x)}{3x^2} \implies \lim_{x \to 0} \frac{\sin(x)}{3x} \\ \hspace{.1cm} \text{using L'Hospital's again}, \hspace{.1cm} \\ \lim_{x \to 0}\frac{\cos(x)}{3} = \frac{1}{3}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1392209",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 1
} |
Perpendicular Bisector of Made from Two Points For a National Board Exam Review:
Find the equation of the perpendicular bisector of the line joining
(4,0) and (-6, -3)
Answer is 20x + 6y + 29 = 0
I dont know where I went wrong. This is supposed to be very easy:
Find slope between two points:
$${ m=\frac{y^2 - y^1}... | Just for variety, a different approach:
If a point $(x,y)$ is on the perpendicular bisector, then it is equidistant from $(4,0)$ and $(-6,-3)$, so
$$
(x-4)^2 + (y-0)^2 = (x+6)^2 + (y+3)^2
$$
Multiplying out, we get
$$
x^2 -8x +16 + y^2 = x^2 +12x +36 \; + \; y^2 +6y + 9
$$
So
$$
20x + 6y + 29 = 0
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1392661",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 4
} |
real values of $x$ in $\sqrt{5-x} = 5-x^2$.
Calculate the real solutions $x\in\mathbb{R}$ to
$$
\sqrt{5-x} = 5-x^2
$$
My Attempt:
We know that $5-x\geq 0$ and thus $x\leq 5$ and
$$
\begin{align}
5-x^2&\geq 0\\
x^2-\left(\sqrt{5}\right)^2&\leq 0
\end{align}
$$
which implies that $-\sqrt{5}\leq x \leq \sqrt{5}$. Now le... | $$
\begin{align}
\sqrt{5-x}&=5-x^2\\
5-x &= \left(5-x^2\right)^2\\
5-x &= x^4-10x^2+25\\
x^4-10x^2+25-5+x &= 0\\
x^4-10x^2+x+20 &= 0\\
(x^2-x-4)(x^2+x-5) &= 0
\end{align}
$$
$$
\begin{align}
x^2-x-4=0 &\vee x^2+x-5=0\\
x=\frac{-(-1)\pm\sqrt{(-1)^2-4\cdot 1 \cdot (-4)}}{2\cdot 1} &\vee x=\frac{-1\pm\sqrt{1^2-4\cdot 1 \c... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1393046",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 2,
"answer_id": 0
} |
$\lim _{ x\to 1 }{ \frac { \sqrt { x } +\sqrt [ 3 ]{ x } -2 }{ x-1 } } $ I have been trying to algebraically solve this limit problem without using L'Hospital's rule but was whatsoever unsuccessful:
$$\lim _{ x\to 1 }{ \frac { \sqrt { x } +\sqrt [ 3 ]{ x } -2 }{ x-1 } } $$
Thanks in advance!
| Using de l'hospital rule you get
$$
\lim_{x \to 1} \frac{x^{\frac{1}{2}}+x^{\frac{1}{3}}-2}{x-1} = \left[\frac{0}{0}\right]=\lim_{x \to 1} \frac{\frac{1}{2}x^{-\frac{1}{2}}+\frac{1}{3}x^{-\frac{2}{3}}}{1} = \frac{5}{6}.
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1393768",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 2
} |
Integral of $\frac{1}{x^2+x+1}$ and$\frac{1}{x^2-x+1}$ How to integrate two very similar integrals. I am looking for the simplest approach to that, it cannot be sophisticated too much as level of the textbook this was taken from is not very high. $$\int \frac{1}{x^2+x+1} dx$$ and$$\int \frac{1}{x^2-x+1} dx$$
| First recall that if $a,b,c$ are real numbers then $ax^2+bx+c$ can be factored using real numbers if and only if $b^2-4ac\ge 0$. For your first polynomial above you have $a=b=c=1$ so $b^2-4ac=-3$, so you would need complex numbers to factor it.
Then recall that there is a standard technique in algebra for reducing a p... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1394731",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 6,
"answer_id": 4
} |
$\int\limits_{0}^{\pi/2}\frac{1+2\cos x}{(2+\cos x)^2}dx$ $\int\limits_{0}^{\pi/2}\frac{1+2\cos x}{(2+\cos x)^2}dx$
I tried to solve it.
$\int\limits_{0}^{\pi/2}\frac{1+2\cos x}{(2+\cos x)^2}dx=\int\limits_{0}^{\pi/2}\frac{4+2\cos x}{(2+\cos x)^2}-\frac{3}{(2+\cos x)^2}dx=\int\limits_{0}^{\pi/2}\frac{2}{2+\cos x}-\frac... | Given $\displaystyle \int_{0}^{\frac{\pi}{2}}\frac{1+2\cos x}{(2+\cos x)^2}dx.\;,$ Now Divide both $\bf{N_{r}}$ and $\bf{D_{r}}$ by $\sin^2 x\;,$ we get
$\displaystyle I = \int_{0}^{\frac{\pi}{2}}\frac{\csc^2 x+2\cot x \cdot \csc x}{\left(2\csc x+\cot x\right)^2}dx\;,$ Now Put $\displaystyle \left(2\csc x+\cot x \right... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1395338",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
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Area of shaded region circle help
Find the area of the shaded region
Area of the sector is $240^\circ$ or $\frac{4\pi}{3}$
Next find $\frac{b\cdot h}{2}$ which is $\frac{2\cdot2}{2}$ which is $2$.
Then subtract the former from the latter: $\frac{4\pi}{3} - 2$
Therefore the answer is $~2.189$?
Is this correct?
| No, the area that corresponds to 120 degrees or $\frac{2\pi}{3}$ is one third of the total area of the circle $4\pi$ the area of the triangle is $\frac{bh}{2}$ but $h$ is not $2$ you need to project it, since $\cos (\, 30 \text{ degrees}) = \frac{\sqrt{3}}{2}$ degrees to find $h = \sqrt{3}$ Therefore
$$\text{area}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1395808",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
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Angle between segments resting against a circle Motivation:
A couple of days ago, when I was solving this question, I had to consider a configuration like this
Now, I didn't intentionally make those two yellow bars stand at what appears to be a $90^{\circ}$-angle but it struck me as an interesting situation, so much s... | I have a solution of the form $\tan\frac{\theta}{2} = f(a,b,c,r)$.
First, note that $\angle QOM = \angle POM$. This comes from the triangles $OPM$ and $OQM$ being similar. They're both rectangular (at $P$ and $Q$), they share the same hypotenuse ($\overline{OM}$), and $\overline{OP} = \overline{OQ} = r$. This also impl... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1396522",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
If $t=\tanh\frac{x}{2}$, prove that $\sinh x = \frac{2t}{1-t^2}$ and $\cosh x = \frac{1+t^2}{1-t^2}$.
If $t=\tanh\frac{x}{2}$, prove that $\sinh x = \frac{2t}{1-t^2}$ and $\cosh x = \frac{1+t^2}{1-t^2}$. Hence solve the equation $7\sinh x + 20 \cosh x = 24$.
I have tried starting by writing out $\tanh\frac{x}{2}$ in ... | Let $t$=$tanh$$\frac{x}{2}$.
Using Identity $sech^2$$\frac{x}{2}$=1-$t^2$
$cosh^2$$\frac{x}{2}$=$\frac{1}{1-t^2}$
$cosh$ $x$=2$cosh^2$$\frac{x}{2}$-1
$cosh$ $x$ =2 ($\frac{1}{1-t^2}$)-1
$cosh$ $x$=$\frac{1+t^2}{1-t^2}$ is obtained by simplifying the above.
Using another identity $cosh^2$$\frac{x}{2}$-$sinh^2$$\frac{x}{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1396994",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
$\sum_{n=1}^\infty \frac{1}{(n^2-1)!} - \sum_{n=1}^\infty \frac{1}{(7n+1)!}$ is almost $1+1/6$ I've recognized, that
$$\mathcal{S} = \sum_{n=1}^\infty \frac{1}{(n^2-1)!} - \sum_{n=1}^\infty \frac{1}{(7n+1)!} \approx 1.1666666666666666666657785992648796$$
which is almost $1+1/6$.
I think it is not a mathematical coincid... | It follows because the first few terms are the same, and the terms in the series decrease to zero extremely quickly. Since$$\sum_{n=1}^{\infty}\frac{1}{(n^{2}-1)!}=1+\frac{1}{3!}+\frac{1}{8!}+\frac{1}{15!}+\frac{1}{24!}+\cdots,$$ we have that $$\left|\sum_{n=1}^{\infty}\frac{1}{(n^{2}-1)!}-\left(1+\frac{1}{3!}+\frac{1}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1398698",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 1,
"answer_id": 0
} |
$\int_{0}^{\infty}\frac{\ln x dx}{x^2+2x+2}$ $$\int_{0}^{\infty}\frac{\ln x .dx}{x^2+2x+2}$$
$$\int_{0}^{\infty}\frac{\ln x .dx}{x^2+2x+2}=\int_{0}^{\infty}\frac{\ln x .dx}{(x+1)^2+1}\\
=\ln x\int_{0}^{\infty}\frac{1}{(x+1)^2+1}-\int_{0}^{\infty}\frac{1}{x}\frac{1}{(x+1)^2+1}dx$$ and then lost track,answer is $\frac{\p... | Integrate $$f(z) = \frac{\log^2 z}{(z+1)^2+1}$$ along a keyhole
contour with the branch cut of the logarithm on the positive real axis and
its argument between $0$ and $2\pi.$
The poles are at $$\rho_{1,2} = -1 \pm i$$ with residues
$$\frac{\log^2 \rho_{1,2}}{2\rho_{1,2}+2}$$
and these sum to
$$\frac{(1/2 \lo... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1398875",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 3
} |
Unfamiliar Property of Modular Arithmetic I saw this property listed in Princeton Review's Math GRE book:
"For any positive integer $c$, the statement $a\equiv b\mod n$ is equivalent to the congruences $a\equiv b,b+n,b+2n,\ldots,b+(c-1)n\mod cn$."
Now, my problem is that I have no idea what it's telling me. An example ... | Example: $a = 10, b = 2, n = 8, c = 4$.
\begin{align}
10 &\equiv 2 \pmod{8}.
\end{align}
The equivalence guarantees
\begin{align}
a &\equiv b + dn \pmod{32}
\end{align}
for some $0 \leq d < 4$. Indeed,
\begin{align}
10 &\equiv 2 + 1 \cdot 8 \\
&\equiv 10\pmod{32}.
\end{align}
Justification: Here is a proof for the fir... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1399985",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Prove that, if $A, B$ are matrices from $M_4(R)$ so that $AB=BA$ Prove that, if $A, B$ are matrices from $M_4(\Bbb R)$ so that $AB=BA$ and $\det(A^2 −AB + B^2) = 0$ then:
$$
\det(A + B) + 3\det(A − B) = 6 (\det(A) + \det(B)) \tag 1
$$
What I tried:
Because of $AB=BA$ we can use, let's say, the Newton's binomial expan... | Let $\omega$ be a third root of unity. Since $A$ and $B$ commute, the condition that $\det(A^2-AB+B^2)=0$ becomes
$$\det(A+\omega B)\det(A+\omega^2 B)=0$$
and so either $\det(A+\omega B)=0$ or $\det(A+\omega^2 B)=0$.
Now consider the function $p(x)=\det(A+xB)$. This is a polynomial of degree at most $4$ with real coeff... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1403900",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 1,
"answer_id": 0
} |
$\int\frac{1+x^2}{x^4+3x^3+3x^2-3x+1}dx$ $$\int\frac{1+x^2}{x^4+3x^3+3x^2-3x+1}dx$$
I tried to solve it.
$$\int\frac{1+x^2}{x^4+3x^3+3x^2-3x+1}dx=\int\frac{1+x^2}{(x^2+1)^2+3x^3+x^2-3x+1}dx=\int\frac{1+x^2}{(x^2+1)^2+3x(x^2-1)+x^2+1}dx$$
But this does not seem to be solving.Please help.
| First rewrite the integral as
$$\mathcal{I}\stackrel{def}{=}\int \frac{x^2+1}{x^4 + 3x^3 + 3x^2 - 3x + 1} dx
= \int \frac{x+x^{-1}}{(x^2 + x^{-2}) + 3(x - x^{-1}) + 3}\frac{dx}{x}
$$
Using the identity: $\displaystyle\;\frac{dx}{x} = \frac{d(x-x^{-1})}{x+x^{-1}}\;$,
we can change variable to $u = x-x^{-1}$ and get
$$
\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1403985",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 6,
"answer_id": 3
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$\int_{0}^{\infty}f\left(\frac{a}{x}+\frac{x}{a}\right)\frac{\ln x}{x}dx=\ln a\int_{0}^{\infty}f\left(\frac{a}{x}+\frac{x}{a}\right)\frac{dx}{x}$ Show that $\displaystyle \int_{0}^{\infty}f\left(\frac{a}{x}+\frac{x}{a}\right)\frac{\ln x}{x}dx=\ln a\int_{0}^{\infty}f\left(\frac{a}{x}+\frac{x}{a}\right)\frac{dx}{x}$
What... | No, but split the integral up into 2 pieces. Consider the difference between the LHS and the RHS:
$$\int_0^\infty \frac{dx}{x} f \left ( \frac{a}{x} + \frac{x}{a} \right ) \log{\frac{x}{a}} = \\ \int_0^a \frac{dx}{x} f \left ( \frac{a}{x} + \frac{x}{a} \right ) \log{\frac{x}{a}} + \int_a^\infty \frac{dx}{x} f \left (... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1404050",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 0
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Upper and lower bounds for $S(n) = \sum_{i=1}^{2^{n}-1} \frac{1}{i} = 1+\frac{1}{2}+ \cdots +\frac{1}{2^n-1}.$
For a positive integer $n$ let $S(n) = \sum_{i=1}^{2^{n}-1} \frac 1i = 1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+ \cdots +\frac{1}{2^n-1}.$
Then which of the following are true.
*
*(a) $S(100)\leq 100$.
... | Yes, there is another method that is easy to implement. Recall that we have
$$\int_1^N \frac{1}{x}\,dx<\sum_{k=1}^N\frac1k <1+\int_1^N \frac{1}{x}\,dx$$
For $N=2^n-1$ this gives
$$\log (2^n-1)<\sum_{k=1}^{2^n-1}\frac1k <1+\log (2^n-1)$$
Then, $\log (2^n-1)=n\log 2+\log (1-2^{-n})$ and therefore, we can write
$$n\log(2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1404602",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
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Prove that the length of the common chord is $\frac{2ab\sin \theta}{\sqrt{a^2+b^2+2ab\cos \theta}}$ Two circles ,of radii $a$ and $b$,cut each other at an angle $\theta.$Prove that the length of the common chord is $\frac{2ab\sin \theta}{\sqrt{a^2+b^2+2ab\cos \theta}}$
Let the center of two circles be $O$ and $O'$ and ... | The above method is too long, a faster method would be
Let A & B be the centers of the circles with radii a & b respectively which intersect each other at an angle θ such that they have a common chord PQ.
Angle between AP and BP is $180-θ$
Using cosine rule in $\triangle APB$,
$$AB=\sqrt {AM^2+BM^2-2(AM)(BM)\cos (\angl... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1404695",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Where did I go wrong in my evaluation of the integral of cosine squared? $$\int{\cos^2(x)}dx$$
Where did I go wrong in my evaluation of this integral?
$$=x\cos^2x - \int-2x\sin(x)\cos(x)\,dx$$
$$=x\cos^2x + \int x\sin(2x)\,dx$$
$$=x\cos^2x + \left(\frac {-x\cos(2x)}2 -\int \frac{-\cos(2x)}2\,dx\right)$$
$$=x\cos^2x + \... | Notice,
$$\int \cos^2 x dx=\int \cos x \cos x dx$$
$$=\int \cos x\sqrt{1-\sin ^2 x}dx$$
Let $\sin x=t\implies \cos x dx=dt$
$$\int \sqrt {1-t^2}dt$$
$$=\frac{1}{2}\left[t\sqrt {1-t^2}+\sin^{-1}(t)\right]$$
$$=\frac{1}{2}\left[\sin x\sqrt {1-sin^2x}+\sin^{-1}(\sin x)\right]$$
$$=\frac{1}{2}\left[\sin x\cos x+x\right]... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1408634",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "12",
"answer_count": 6,
"answer_id": 4
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Prove that $\frac{2^{x+1}+(x+1)^2}{2^x+x^2}\rightarrow 2$ as $x \rightarrow \infty$ Could someone please show me the proof that $\frac{2^{x+1}+(x+1)^2}{2^x+x^2}\rightarrow 2$ as $x \rightarrow \infty$
I have no idea where to begin with this one.
Thanks.
| $$\lim_{x\to \infty}\frac{2\cdot (2^{x}+x^2)+(x+1)^2-2x^2}{2^x+x^2}$$
$$=2+\lim_{x\to \infty}\frac{(x+1)^2-2x^2}{2^x+x^2}$$
($\frac{\infty}{\infty}$)form so using L-Hospital's rule twice:
$$=2+\lim_{x\to \infty}\frac{-2}{2^x (ln2)^2+2}$$
$$=2$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1408900",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
The number of ordered pairs of positive integers $(a,b)$ such that LCM of a and b is $2^{3}5^{7}11^{13}$ I started by taking two numbers such as $2^{2}5^{7}11^{13}$ and $2^{3}5^{7}11^{13}$. The LCM of those two numbers is $2^{3}5^{7}11^{13}$. Similarly, If I take two numbers like $2^{3-x}5^{7-y}11^{13-z}$ and $2^{3}5^{... | Hint:
*
*One of $a$ and $b$ must have 3 factors two, the other less.
*One of $a$ and $b$ must have 7 factors five, the other less.
*One of $a$ and $b$ must have 13 factors eleven, the other less.
*No other prime factors.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1409388",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
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If $f\in C^2(\mathbb R)$ then $M_1^2 \le 2M_0 M_2$, where $M_k = \text {sup}_x |(d/dx)^k f(x)|$ for $k=0,1,2.$ I wanna prove this problem. I tried it with Mean Value Theorem but cannot proceed to any plausible result. So could I have some hints?
| Mean value theorem generally doesn't work on problems involving $C^2$ functions, but the generalization does. From Taylor's theorem, for any $x,h \in \mathbb R$,
\begin{align*}
f(x + h) &= f(x) + h\cdot f'(x) + h^2 \cdot \frac{f''(\zeta_1)}{2} \\
f(x - h) &= f(x) - h\cdot f'(x) + h^2 \cdot \frac{f''(\zeta_2)}{2}
\end{a... | {
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"answer_id": 0
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bend measurement and calculating $\int_4^8 \sqrt{1+{\left(\frac{{x^2-4}}{4x}\right)^2}} $ How can i get the measure of this bend : $y=\left(\frac{x^2}{8}\right)-\ln(x)$ between $4\le x \le 8$. i solved that a bit according to the formula $\int_a^b \sqrt{1+{{f'}^2}} $:$$\int_4^8 \sqrt{1+{\left(\frac{x^2-4}{4x}\right)^2}... | HINT:
$$\sqrt{1+\left(\frac{x^2-4}{4x}\right)^2}=\frac{\sqrt{16x^2+(x^4-8x^2+16)}}{4x}=\frac{x^2+4}{4x}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1410466",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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calculating the characteristic polynomial I have the following matrix: $$A=\begin{pmatrix}
-9 & 7 & 4 \\
-9 & 7 & 5\\
-8 & 6 & 2
\end{pmatrix}$$
And I need to find the characteristic polynomial so I use det(xI-A) which is $$\begin{vmatrix}
x+9 & -7 & -4 \\
9 & x-7 & -5\\
8 & -6 & x-2
\end{vmatri... | I am not sure if this method is "faster", but it does seem to involve less numbers in the actual calculation:
We know from a different theorem that two similar matrices share the same characteristic polynomial, see this post Elegant proofs that similar matrices have the same characteristic polynomial?.
From this fact... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Inequality problem: Application of Cauchy-Schwarz inequality Let $a,b,c \in (1, \infty)$ such that $ \frac{1}{a} + \frac{1}{b} + \frac{1}{c}=2$. Prove that:
$$
\sqrt {a-1} + \sqrt {b-1} + \sqrt {c-1} \leq \sqrt {a+b+c}.
$$
This is supposed to be solved using the Cauchy inequality; that is, the scalar product inequality... | Let $a=x+1$, $b=y+1$ and $c=z+1$.
Hence, $x$, $y$ and $z$ are positives and we need to prove that
$$\sqrt{x}+\sqrt{y}+\sqrt{z}\leq\sqrt{x+y+z+3}$$ or
$\sqrt{xy}+\sqrt{xz}+\sqrt{yz}\leq\frac{3}{2}$.
But in another hand, the condition geves
$$\sum_{cyc}\frac{1}{x+1}=2$$ or
$$\sum_{cyc}\left(\frac{1}{x+1}-1\right)=-1$$ or... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1412095",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
} |
Trying to solve $\sqrt{7-4\sqrt2 \sin x}=2\cos(x)-\sqrt2 \tan(x)$ The equation is
$$\sqrt{7-4\sqrt2 \sin x}=2\cos(x)-\sqrt2 \tan(x)$$
We get the system
$$
\begin{cases}
7-4\sqrt 2 \sin(x)=4\cos^2(x)-2\sqrt2\cos(x)\tan(x)+2\tan^2(x) \\
2\cos(x)-\sqrt2 \tan(x)\ge 0
\end{cases}
$$
I transformed the equation thus:
$$7(\sin... | Hint:
The solutions of $$\sqrt{7-4\sqrt2 \sin x}=2\cos(x)-\sqrt2 \tan(x)$$
are also solutions of
\begin{align*}
7-4\sqrt{2}\sin x&=4\cos^2 x-4\sqrt{2}\cos x\tan x+2\tan^2 x\\
7-4\sqrt{2}\sin x&=4\cos^2 x-4\sqrt{2}\sin x+2\tan^2 x
\end{align*}
Last equation is equivalent to
$$4\cos^2 x+2\tan^2 x -7=0...(1)$$
Let $t=\co... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Need help with tangents to a quadratic The quadratic $y=kx^2+(3k-1)x-1$ and the straight line $y=(k+1)x-11$ meet.
Find the range of value(s) of $k$ such that the line is a tangent to the curve.
Got this question for school. Seems really simple and it's a non-calculator question but I'm not sure how to go about it.
| Given curve $y=kx^2+(3k-1)x- 1$ and curve $y=(k+1)x-11$ are intersect each
other exactly at one ponit
(Means only one value of $x$ and Corrosponding value of $y$)
which is also called Condition of tangency.
So Equating $y\;,$ We get $$kx^2+(3k-1)x-1 = (k+1)x-11$$
So $$\displaystyle kx^2+2(k-1)x+10=0\Rightarrow x=\fra... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1414343",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Find the solution set of the equation $5.(\frac{1}{25})^{\sin^2x}+4.5^{\cos2x}=25^{\frac{\sin2x}{2}}$ Problem :
Find the solution set of the equation $5.(\frac{1}{25})^{\sin^2x}+4.5^{\cos2x}=25^{\frac{\sin2x}{2}}$ where $x \in [0,2\pi]$
My approach :
$5.(\frac{1}{25})^{\sin^2x}+4.5^{1-2\sin^2x}=25^{\frac{\sin2x}{2}}$... | Hint:
Observe
\begin{align*}
5^{1-2\sin^2 x}+4\cdot 5^{1-2\sin^2 x}&=5^{2\sin x\cos x}\\
5\cdot 5^{1-2\sin^2 x}&=5^{2\sin x\cos x}\\
\end{align*}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1414941",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Prove that $1^2 + 2^2 + ..... + (n-1)^2 < \frac {n^3} { 3} < 1^2 + 2^2 + ...... + n^2$ I'm having trouble on starting this induction problem.
The question simply reads : prove the following using induction:
$$1^{2} + 2^{2} + ...... + (n-1)^{2} < \frac{n^3}{3} < 1^{2} + 2^{2} + ...... + n^{2}$$
| $$0^2 + 1^2 + \ldots + (n - 1)^2 \leqslant \int_0^nx^2\,dx \leqslant 1^2 + 2^2 + \ldots + n^2,$$
and it's quite obvious that current integral is exactly $\frac{n^3}{3}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1416472",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 2
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Calculate simple expression: $\sqrt[3]{2 + \sqrt{5}} + \sqrt[3]{2 - \sqrt{5}}$ Tell me please, how calculate this expression:
$$
\sqrt[3]{2 + \sqrt{5}} + \sqrt[3]{2 - \sqrt{5}}
$$
The result should be a number.
I try this:
$$
\frac{\left(\sqrt[3]{2 + \sqrt{5}} + \sqrt[3]{2 - \sqrt{5}}\right)\left(\sqrt[3]{\left(2 + \sq... | , Let $$x = \sqrt[3]{2+\sqrt{5}}+\sqrt[3]{2-\sqrt{5}}\;,$$ Then we can write as $$\sqrt[3]{2+\sqrt{5}}+\sqrt[3]{2+\sqrt{5}}+(-x) = 0$$
Now Using If $$\bullet \; a+b+c = 0\;,$$ Then $$a^3+b^3+c^3 = 3abc$$
So $$\left(2+\sqrt{5}\right)+\left(2-\sqrt{5}\right)-x^3 = 3\left[\sqrt[3]{\left(2+\sqrt{5}\right)\cdot \left(2-\sqr... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Another messy integral: $I=\int \frac{\sqrt{2-x-x^2}}{x^2}\ dx$ I found the following question in a practice book of integration:-
$Q.$ Evaluate $$I=\int \frac{\sqrt{2-x-x^2}}{x^2}\ dx$$ For this I substituted $t^2=\frac {2-x-x^2}{x^2}\implies x^2=\frac{2-x}{1+t^2}\implies 2t\ dt=\left(-\frac4{x^3}+\frac 1{x^2}\right)\... | use so so-called Euler substitution and set $$\sqrt{2-x-x^2}=xt+\sqrt{2}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1423103",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
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How to solve the integration to get the desired answer How to prove the following:
If
$$\int \frac{adx}{(ax-2)\sqrt{(ax-1)}}=\frac{y}{\sqrt{5}}$$
then show that $\frac{1}{x}=\frac{a}{2}\left(1+sech({\frac{y}{\sqrt{5}}})\right)$
Given, at $y=0$, $x=1/a$.
Approach
I have assumed $ax-1= \cosh^2z$ then $ax-2=\sinh^2z$, $ad... | Let $u=\sqrt{ax-1}$, then $x=\frac{u^2+1}{a}\,\,\,$ and $\,\,\,dx=\frac{2u}{a}du$, so
\begin{align}
\int\frac{adx}{(ax-2)\sqrt{ax-1}}&=\int\frac{a\left(\frac{2u}{a}\right)du}{(u^2-1)u}\\
&=\int\frac{2du}{u^2-1}\\
&=\int\left(\frac{1}{u-1}-\frac{1}{u+1}\right)du
\end{align}
You can do the remaining part.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Finding the $n$th derivative of trigonometric function.. My maths teacher has asked me to find the $n$th derivative of $\cos^9(x)$. He gave us a hint which are as follows:
if $t=\cos x + i\sin x$,
$1/t=\cos x - i\sin x$,
then $2\cos x=(t+1/t)$.
How am I supposed to solve this? Please help me with explanations bec... | Using his hint, you get
\begin{align*}
[\cos x]^9
&= \left[ \frac12 \left( t + \frac1t \right) \right]^9 \\
&= \frac{1}{2^9} \sum_{k=0}^9 {9 \choose k} (t)^k \left( \frac1t \right)^{9-k} \\
&= \frac{1}{2^9} \sum_{k=0}^9 {9 \choose k} t^{2k-9}. \tag{1}
\end{align*}
Also,
$$
\frac{dt}{dx} = -\sin x + i \cos x = i(\cos x ... | {
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"url": "https://math.stackexchange.com/questions/1425755",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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How to compute $\int{\ln(\sqrt{x}+\sqrt{x+1})dx}$? The integral I am trying to work out is
$$\int{\ln(\sqrt{x}+\sqrt{x+1})dx}$$
So first step is substitute $u=x+1$:
$$\int{\ln(\sqrt{u-1}+\sqrt{u})du}.$$
Now substitute $u=\cos^2(t)$.
$\int{-\ln(\sqrt{-\sin^2(t)}+\sqrt{\cos^2(t)})2\sin(t)\cos(t)dt}$. This works out to $\... | Here is a solution not working with hyperbolic functions. Now updated with some more details.
Integrating by parts, and using that (by the chain rule, and writing the expression inside square brackets on common denominator)
$$
\begin{aligned}
D\log(\sqrt{x}+\sqrt{x+1})&=\frac{1}{\sqrt{x}+\sqrt{x+1}}D\bigl(\sqrt{x}+\sqr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1428166",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 0
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Limit involving trigonometric series summation How should I go about this one?
$\large{L = \lim_{n \to \infty} \sum_{k=1}^n \dfrac{1}{2^k} \tan\left(\dfrac{\pi}{3\cdot2^{k+1}}\right)}$
Hints please!
| Let $\displaystyle x= \frac{\pi}{6}\;,$ Then $\displaystyle L = \lim_{n\rightarrow \infty}\sum^{n}_{k=1}\frac{1}{2^k}\tan\left(\frac{x}{2^k}\right)$
Now using $$\displaystyle \bullet \; \cot \left(\frac{x}{2}\right)-\tan \left(\frac{x}{2}\right) = 2\tan (x)\Rightarrow \tan \left(\frac{x}{2}\right)=\cot \left(\frac{x}{2... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Proving that if $n \in \mathbb{Z}$ and $n^2 − 6n + 5$ is even, then $n$ must be odd. Prove that if $n \in \mathbb{Z}$ and $n^2 − 6n + 5$ is even, then $n$ must be odd.
$p= n^2 - 6n + 55$ is even, $Q= n$ is odd
Proof: Assume on contrary $n$ is even. Then $n= 2k$ for some $k \in \mathbb{Z}$. Then, $$n^2 -6n + 5= 2k^2-6(2... | $n^2 − 6n + 5$ even $\implies$ $n^2+5$ even because $6n$ is even.
$n^2+5$ even implies $n^2$ odd because $5$ is odd.
$n^2$ odd implies $n$ odd.
Or use the contrapositive: $n$ even $\implies$ $n^2 − 6n + 5$ odd:
$n$ even $\implies$ $n=2k$ $\implies$ $n^2 − 6n + 5=4k^2-12k+5=2(2k^2-6k+2)+1$, which is odd.
| {
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"timestamp": "2023-03-29T00:00:00",
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Prove that any integer that is both square and cube is congruent modulo 36 to 0,1,9,28 This is from Burton Revised Edition, 4.2.10(e) - I found a copy of this old edition for 50 cents.
Prove that if an integer $a$ is both a square and a cube then $a \equiv 0,1,9, \textrm{ or } 28 (\textrm{ mod}\ 36)$
An outline of ... | First establish that $a$ must be a sixth power. We have $a=b^2=c^3$ so that $a^3=b^6$ and $a^2=c^6$ whence $$a=\cfrac {a^3}{a^2}=\cfrac {b^6}{c^6}=\left(\cfrac bc\right)^6$$
And if $q$ is a rational number whose sixth power is an integer, it must be an integer itself. [see below]
Now, let's have a look at the sixth pow... | {
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"url": "https://math.stackexchange.com/questions/1431396",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 1
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Find the general integral of $ px(z-2y^2)=(z-qy)(z-y^2-2x^3).$ $ p=\frac{\partial z}{\partial x} $ and $ q=\frac{\partial z}{\partial y} $
Find the general integral of the linear PDE $ px(z-2y^2)=(z-qy)(z-y^2-2x^3). $
My attempt to solve this is as follows:
$ p=\frac{\partial z}{\partial x} $ and $ q=\frac{\partial z}{... | i think you made error in question - in place of $2x^3$ it should be $2x^2$
$$ \frac{dx}{x(z-2y^2)}=\frac{dy}{y(z-y^2-2x^2)}=\frac{dz}{z(z-y^2-2x^2)}\Rightarrow1)$$
then take $0,-2y,1$ as multipliers
$$ \frac{-2ydy+dz}{-2y^2(z-y^2-2x^2)+z(z-y^2-2x^2)}=\frac{d(z-y^2)}{(z-2y^2)(z-y^2-2x^2)}\Rightarrow2)$$
Combining fract... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1432235",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 3
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Prove that $g$ is continuous at $x=0$ Given, $g(x) = \frac{1}{1-x} + 1$. I want to prove that $g$ is continuous at $x=0$. I specifically want to do an $\epsilon-\delta$ proof. Related to this Is $g(x)\equiv f(x,1) = \frac{1}{1-x}+1$ increasing or decreasing? differentiable $x=1$?.
My work: Let $\epsilon >0$ be given.... | $$\begin{align}\left|\frac{1}{1-x}\right|<\frac{1}{2}\\
\Longleftrightarrow -\frac{1}{2}<\frac{1}{1-x}<\frac{1}{2}\\
\Longleftrightarrow \frac{x-1}{2}<1< \frac{1-x}{2}\\
\Longleftrightarrow x-1<2<1-x\\
\Longleftrightarrow x<3<2-x\end{align}$$
Which is clearly not true for an $x$ arbitrarily near zero. I think there cou... | {
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"timestamp": "2023-03-29T00:00:00",
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Assume this equation has distinct roots. Prove $k = -1/2$ without using Vieta's formulas. Given $(1-2k)x^2 - (3k+4)x + 2 = 0$ for some $k \in \mathbb{R}\setminus\{1/2\}$, suppose $x_1$ and $x_2$ are distinct roots of the equation such that $x_1 x_2 = 1$.
Without using Vieta's formulas, how can we show $k = -1/2$ ?
Here... | You don't need Vieta's formulas to conclude that a quadratic of the form $ax^2+bx+c=0$ has distinct roots that are reciprocals if and only if $a=c$. If $r$ and $1/r$, with $r\not=1/r$, are roots of the quadratic, then
$$ar^2+br+c=0\quad\text{and}\quad {a\over r^2}+{b\over r}+c=0$$
Multiplying the second equation throu... | {
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"url": "https://math.stackexchange.com/questions/1438522",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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If $x>0$we have $(1+x^2)f'(x)+(1+x)f(x)=1$ and $g'(x)=f(x), f(0)=g(0)=0$Prove: If $x>0$ we have $(1+x^2)f'(x)+(1+x)f(x)=1$. And $g'(x)=f(x), f(0)=g(0)=0$
Prove that:$\displaystyle \frac14<\sum_{n=1}^{\infty}g(\frac1n)<1$
I tried solving the ODE,But it seems very complex.and I still have no idea about it.Could someone ... | Using integrating factor $m$ we first solve given ODE:
\begin{align*}
f'+\frac{1+x}{1+x^2}f&=\frac{1}{1+x^2}\\
m'&=m\frac{1+x}{1+x^2}\\
(\ln(m))'&=\frac{1+x}{1+x^2}\\
\ln(m)&=\int_0^x\frac{1+y}{1+y^2}dy=arctg(x)+\frac{1}{2}\int_{1}^{1+x^2}\frac{1}{z}dz=arctg(x)+\frac{1}{2}\ln(1+x^2)\\
m&=\exp(arctg(x))+\sqrt{1+x^2}\\
f... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1440660",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
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Solving $\lim _{x\to 1}\left(\frac{1-\sqrt[3]{4-3x}}{x-1}\right)$ $$\lim _{x\to 1}\left(\frac{1-\sqrt[3]{4-3x}}{x-1}\right)$$
So
$$\frac{1-\sqrt[3]{4-3x}}{x-1} \cdot \frac{1+\sqrt[3]{4-3x}}{1+\sqrt[3]{4-3x}}$$
Then
$$\frac{1-(4-3x)}{(x-1)(1+\sqrt[3]{4-3x})}$$
That's
$$\frac{3\cdot \color{red}{(x-1)}}{\color{red}{(x-1)}... | $ (1-(4-3x)^(1/3))(1+(4-3x)^(1/3)) = 1 - (4-3x)^(2/3) $
I couldn't re-format it right but others beat me to it
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 6,
"answer_id": 4
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Evaluate limit as x approaches infinity of $\lim_{x\to\infty}\frac{\sqrt{x^3 +7x}}{\sqrt{4x^3+5}}$ I am having trouble figuring out how to answer this question by determining the degree of the numerator and/or denominator:
$$\lim_{x\to\infty}\frac{\sqrt{x^3 +7x}}{\sqrt{4x^3+5}}$$
I have tried deriving the first coeffic... | Notice, $$\lim_{x\to \infty}\frac{\sqrt{x^3+7x}}{\sqrt{4x^3+5}}$$
$$=\lim_{x\to \infty}\sqrt{\frac{x^3+7x}{4x^3+5}}$$
$$=\lim_{x\to \infty}\sqrt{\frac{x^3\left(1+\frac{7}{x^2}\right)}{x^3\left(4+\frac{5}{x^3}\right)}}$$
$$=\lim_{x\to \infty}\sqrt{\frac{1+\frac{7}{x^2}}{4+\frac{5}{x^3}}}$$
$$=\sqrt{\frac{1+0}{4+0}}=\col... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1448697",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
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$\lim_{x\to a}\frac{1}{(x^2-a^2)^2}\left(\frac{a^2+x^2}{ax}-2\sin\left(\frac{a\pi}{2}\right)\sin\left(\frac{\pi x}{2}\right)\right)$ Prove that $\lim_{x\to a}\frac{1}{(x^2-a^2)^2}\left(\frac{a^2+x^2}{ax}-2\sin\left(\frac{a\pi}{2}\right)\sin\left(\frac{\pi x}{2}\right)\right)=\frac{\pi^2 a^2+4}{16a^4}$ where $a$ is an o... | We have $$\lim_{x\rightarrow a}\frac{1}{\left(x^{2}-a^{2}\right)^{2}}\left(\frac{x^{2}+a^{2}}{ax}-2\sin\left(\frac{a\pi}{2}\right)\sin\left(\frac{\pi x}{2}\right)\right)=
$$ $$=\frac{1}{4a^{2}}\lim_{x\rightarrow a}\frac{1}{\left(x-a\right)^{2}}\left(\frac{x^{2}+a^{2}}{ax}-2\sin\left(\frac{a\pi}{2}\right)\sin\left(\fra... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Solving $\lim_{x\to0}\frac{1-\sqrt{\cos x}}{x^2}$ I've been trying to solve this over and over without L'Hopital but keep on failing:
$$\lim_{x\to0}\frac{1-\sqrt{\cos x}}{x^2}$$
My first attempt involved rationalizing:
$$\frac{1-\sqrt{\cos x}}{x^2} \cdot \frac{1+\sqrt{\cos x}}{1+\sqrt{\cos x}} = \frac{1-\cos x}{x\cdot... | You are supposed to use the limit
$$
\lim_{x \to 0} \frac{1-\cos x}{x^2}=\frac12.
$$
This should be known to you as soon as you know that $$\lim_{x \to 0} \frac{\sin x}{x}=1.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1453288",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Why can there be an infinite difference between two functions as x grows large, but a ratio of 1? I learned in grade school that the closer $a$ and $b$ are to one another, the closer $\frac{a}{b}$ is going to be to $1$. For example, $\frac{3}{\pi}$ is pretty close to 1, and $\frac{10^{100}}{42}$ isn't even close to 1.
... | $\lim_{x\to\infty} \frac{x^{2}}{x^{2}+x} = 1$
As stated, this is because $x$ grows much more slowly than $x^2$. So the ratio goes to 1.
$1^2\over 1^2+1$$={1\over 1+1}$$={1\over 2}$$=0.5$
$2^2\over 2^2+2$$={2\over 4+2}$$={4\over 6}$$=0.66$
$3^2\over 3^2+3$$={9\over 9+3}$$={9\over 12}$$=0.75$
$10^2\over 10^2+10$$={100\ov... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1455826",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "40",
"answer_count": 5,
"answer_id": 1
} |
Stuck solving a logarithmic equation $$\log _{ 2 }{ 2x } =\log _{ 4 }{ 4x^{ 6 } } -\log _{ 2 }{ 2x } $$
Steps I took:
$$\frac { \log _{ 4 }{ 2x } }{ \log _{ 4 }{ 2 } } =\log _{ 4 }{ 4x^{ 6 } } -\frac { \log _{ 4 }{ 2x } }{ \log _{ 4 }{ 2 } } $$
$$2\log _{ 4 }{ 2x } +2\log _{ 4 }{ 2x } =\log _{ 4 }{ 4x^{ 6 } } $$
$$... | Using: $\log_a b=\frac{\log a}{\log b}, \log ab=\log a + \log b, \log_a a=1$
$$\implies\log _{ 2 }2+\log _{ 2 }{ x } =\log _{ 4 }{ 4}+\log _{ 4 }{ x^{ 6 } } -\log _{ 2 }{ 2}-\log _{ 2 }{ x } $$
$$\implies1+\log _{ 2 }{ x } =6\frac{\log _{ 2 }{ x }}{\log_2 4} -\log _{ 2 }{ x } +1-1$$
$$\implies1+\log _{ 2 }{ x } =6\frac... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1462022",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Is there an easier way to solve this logarithmic equation? $$2\log _{ 8 }{ x } =\log _{ 2 }{ x-1 } $$
Steps I took:
$$\frac { \log _{ 2 }{ x^{ 2 } } }{ \log _{ 2 }{ 8 } } =\log _{ 2 }{ x-1 } $$
$$\frac { \log _{ 2 }{ x^{ 2 } } }{ 3 } =\log _{ 2 }{ x-1 } $$
$$\log _{ 2 }{ x^{ 2 } } =3\log _{ 2 }{ x-1 } $$
$$2\log _{ ... | $2\log_{8}x=\log_{2}x-1$
$2\log_{2}x=3\log_{2}x-3$ based on $\log_{8}x=\frac{\log_{2}x}{\log_{2}8}=\frac{\log_{2}x}{3}$
$3=\log_{2}x$
$8=2^{3}=x$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1462075",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 3
} |
An example of a series $\sum a_n$ that converges conditionally but $\sum a_n^3$ does not converge Give an example of a series $\sum a_n$ that converges conditionally but $\sum a_n^3$ does not converge conditionally.
I've come up with an example.
$\frac{1}{\sqrt[3]2}-\frac{1}{2\sqrt[3]2}-\frac{1}{2\sqrt[3]2}+\frac{1}{\... | Use the simplified example
$$
\frac1{\sqrt[3]2}-\frac1{2\sqrt[3]2}-\frac1{2\sqrt[3]2}+\frac1{\sqrt[3]3}-\frac1{2\sqrt[3]3}-\frac1{2\sqrt[3]3}+\frac1{\sqrt[3]4}-\frac1{2\sqrt[3]4}-\frac1{2\sqrt[3]4}+…
$$
Then it is easy to see that this series is conditionally convergent, however the third power series is $3/4$ of the h... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1462893",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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Evaluate limit of $(2\sin x\log \cos x + x^{3})/x^{7}$ as $x \to 0$ While trying to solve this question, I came across the following limit $$\lim_{x \to 0}\frac{2\sin x\log \cos x + x^{3}}{x^{6}}\tag{1}$$ Using some algebraic manipulation (and L'Hospital's Rule) I was able to show that $$\lim_{x \to 0}\frac{2\sin x\log... | If you use L'Hôpital rule, if there is a limit, because of the $x^7$, you would need to differentiate seven times the numerator. After these seven differentiations (have fun !), the denominator will be $7!=5040$ and, after a long series of successive simplifications, the seventh derivative of numerator would write $$-... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1465203",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 1,
"answer_id": 0
} |
Limit $\lim_{n \to \infty} \frac {a_n}{2^{n-1}}=\frac 4{\pi}$ for $a_{n+1}=a_n+\sqrt{1+a_n^2}$ and $a_0=0$ I found the following question in a book:-
$Q:$Let $a_1, a_2, ... , a_n$ be a sequence of real numbers with $a_{n+1}=a_n+\sqrt{1+a_n^2}$ and $a_0=0$. Prove that $$\lim_{n \to \infty} \frac {a_n}{2^{n-1}}=\frac 4{\... | Make use of :
$$ \cot\frac{x}{2}=\cot x+\sqrt{1+\cot^2 x} $$
Since $a_0=\cot\frac{\pi}{2}$, one can notice ( use an inductive argument):
$$ a_n = \cot\frac{\pi}{2^{n+1}} $$
So the limit obviously becomes:
$$\frac{4}{\pi} $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1465556",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 1,
"answer_id": 0
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Solving large multiplications in my head What would be the best approach to solve 73 x 42 in my head?
I started with 70 x 40 and then 3 x 40 and combined, but at this point I forgot what I had done and ended up getting lost and not figuring it out.
Is there a good method for solving multiplication as such in my head?
| Many mental calculators do mutliplications from left to right this way :
\begin{array}{r}
73\\
\times\; 42
\end{array}
\begin{array}{c l}
\text{'cross' computation} &\text{ partial result}\\
7\cdot 4=28 & 28\\
7\cdot 2+3\cdot 4=26 & 306\\
3\cdot 2=6 &3066
\end{array}
Other example :
\begin{array}{r}
237\\
\times\;543
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1465689",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 5,
"answer_id": 0
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Finding condition for integral roots of a quadratic equation. I need to find the values of k(possible) for which the quadratic equation $$x^2+2kx+k =0$$ will have integral roots.
So I assumed roots to be $a,b$
Then I got the condition $a+b=-2k$and $a\cdot b=k$; so combining these I get $a+b+2ab=0$;
And now I need to fi... | Given $$x^2+2kx+k=0\Rightarrow x^2+2kx+k^2 = k^2-k\Rightarrow (x+k)^2=k^2-k$$
So we get $$\displaystyle (x+k)^2= \left(\sqrt{k^2-k}\right)^2\Rightarrow x+k =\pm \sqrt{k^2-k} $$
So we get $$\displaystyle x= \pm \sqrt{k^2-k}-k\;,$$ Now here $x\in \mathbb{Z}$
So let $$k^2-k=l^2\Rightarrow 4k^2-4k = 4l^2\Rightarrow (4k^2-4... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1467285",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Evaluate the limit without using the L'Hôpital's rule $$\lim_{x\to 0}\frac{\sqrt[5]{1+\sin(x)}-1}{\ln(1+\tan(x))}$$
How to evaluate the limit of this function without using L'Hôpital's rule?
| I'm not sure if this can be made formal...
$$\lim_{x\to 0}\frac{\sqrt[5]{1+\sin(x)}-1}{\ln(1+\tan(x))}$$
$$\sqrt[5]{1+x} = 1 + \dfrac x5 - \dfrac{2}{25}x^2 + O(x^3)$$
$$\sqrt[5]{1+\sin(x)}-1 \approx \dfrac 15 \sin x - \dfrac{2}{25}\sin{x^2}$$
$$\ln(1+x) = x - \dfrac 12x^2 + O(x^3)$$
$$\ln(1+ \tan x) \approx \tan x - ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1468993",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
} |
Diagonalizable matrix is similar to non-diagonal matrix??? $$A=
\begin{pmatrix}
1 & 0 & 0 & 0\\
2 & 3 & 2 & 2\\
2 & 2 & 3 & 2\\
2 & 2 & 2 & 3\\
\end{pmatrix}
$$
I know that the eigenvalues are 1 of geometric multiplicity = algebric multiplicity = 3, and 7 of geometric mul... | Well, I figured it out...
$D_1$ and $D_2$ have the same eigenvalues as $A$ with the same algebric multiplicity but $D_2$ isn't diagonalizable because geometric multiplicity $\ne$ algebric multiplicity for every eigenvalue.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1469130",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Combinations problem: Choosing ways to select $8$ questions out of $12$. In an exam, there are $12$ questions in total. He has to attempt $8$ questions in all. There are two parts: Part A and Part B of the question paper containing $5$ and $7$ questions respectively. How many ways are there to attempt the exam, such th... | As you are aware, the actual number of ways to select eight questions from the examination given the restrictions that at least three questions must be selected from each part is
$$\binom{5}{3}\binom{7}{5} + \binom{5}{4}\binom{7}{4} + \binom{5}{5}\binom{7}{3}$$
The alternative method you proposed of selecting three qu... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1470160",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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$\frac{x^3+y^3+z^3}{x+y+z}\in \mathbb N$ has infinitely many non-trivial solutions Trying to solve this I find out the following problem in which it is not necessary the condition $x^3=y^3=z^3$ in some $\mathbb F_p$:
Prove there are infinitely many pairwise coprime triples of distinct natural numbers, $(x,y,z)$, such t... | Let $x=y=z$ then:
$$
\frac {x^3+y^3+z^2}{x+y+z}=\frac {3x^3}{3x}=x^2\in \Bbb N, \forall x\in\Bbb Z-\{0\}
$$
Furthermore, let $z=0,y=1$, then:
$$
\frac {x^3+y^3+z^2}{x+y+z}=\frac {x^3+1}{x+1}=x^2-x+1\in\Bbb N, \forall x\in\Bbb N-\{1\}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1470655",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 0
} |
find the solution set of the following inequality with so many radical I get lost
$\sqrt[4]{\frac{\sqrt{x^{2}-3x-4}}{\sqrt{21}-\sqrt{x^{2}-4}}}\geqslant x-5$
Edit
I get online with wolfram
-5 < x <= -2 || x == -1 || 4 <= x < 5
| Clearly $\sqrt[4]{\dfrac{\sqrt{x^{2}-3x-4}}{\sqrt{21}-\sqrt{x^{2}-4}}}\ge0$
So, one immediate solution is $x-5<0\iff x<5$
Otherwise i.e., if $x\ge5$
$$\dfrac{\sqrt{x^2-3x-4}}{\sqrt{21}-\sqrt{x^2-4}}=\dfrac{\sqrt{(x-4)(x+1)(\sqrt{21}+\sqrt{x^2-4})}}{25-x^2}$$ which will be $<0$ if $x>5$
and what if $x=5$?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1471835",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
Maximizing $\sin \beta \cos \beta + \sin \alpha \cos \alpha - \sin \alpha \sin \beta$ I need to maximize
$$ \sin \beta \cos \beta + \sin \alpha \cos \alpha - \sin \alpha \sin \beta \tag{1}$$
where $\alpha, \beta \in [0, \frac{\pi}{2}]$.
With numerical methods I have found that
$$ \sin \beta \cos \beta + \sin \alpha \co... | A nice solution with elementary methods was provided by user arqady in AoPS (see here). It is as follows:
It is easy to see, using the trigonometric addition formulas that
$$ \sin \beta \cos \beta + \sin \alpha \cos \alpha - \sin \alpha \sin \beta=\sin(\alpha+\beta)\cos(\alpha-\beta)-\frac{1}{2}\cos(\alpha-\beta)+\frac... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1475601",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 2
} |
How many roots are rational?
If $P(x) = x^3 + x^2 + x + \frac{1}{3}$, how many roots are rational?
EDIT:
$3x^3 + 3x^2 + 3x + 1 = 0$, if any rat roots then,
$x = \pm \frac{1}{1, 3} = \frac{-1}{3}, \frac{1}{3}$, and none of these work. Complete?
| There's none, the proof is basically the same as for $\sqrt 2$ being irrational. Assume that $x=p/q$ where $p/q$ is can't be reduced further. Now
$$ x^3 + x^2 + x + 1/3 = p^3/q^3 + p^2/q^2 + p/q + 1/3 = 0$$
Multiply both sides with $3q^3$:
$$ 3p^3 + 3p^2q + 3pq^2 + q^3 = 0$$
Now we see that $3|q$ so we rewrite $q = 3r$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1476281",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
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How do you factorize quadratics when the coefficient of $x^2 \gt 1$? So I've figured out how to factor quadratics with just $x^2$, but now I'm kind of stuck again at this problem:
$2x^2-x-3$
Can anyone help me?
| We can factor $2x^2 - x - 3$ with respect to the rationals if we can find two numbers with product $2 \cdot -3 = -6$ and sum $-1$. The factors of $-6$ are
\begin{align*}
-6 & = 1 \cdot -6 & -6 & = -1 \cdot 6\\
& = \color{blue}{2 \cdot -3} & & = -2 \cdot 3
\end{align*}
Of these four pairs of factors, only $2$ and $... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1477307",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Source and/or combinatorial interpretation for $F_{n+k} = \sum_{i=0}^{k} \binom{k}{i}F_{n-i}$ Through some fussing with Taylor's Theorem in the discrete calculus described here (among other places), I found what I believe to be an identity:
$$F_{n+k} = \sum_{i=0}^{k} \binom{k}{i}F_{n-i}$$
For example, with $n = 3$ and ... | The explicit formula for Fibonacci numbers gives:
$$ F_n = \frac{1}{\sqrt{5}}\left(\sigma^n-\overline{\sigma}^n\right) $$
where $\sigma,\overline{\sigma}$ are solutions of $x^2=1+x$. By the binomial theorem it follows that:
$$ \sum_{i=0}^{k}\binom{k}{i}F_{n-i} = F_{n+k}. \tag{1}$$
On the other hand, the LHS of $(1)$ is... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1478155",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Find the inverse of the cubic function What is the resulting equation when $y=x^3 + 2x^2$ is reflected in the line $y=x$ ?
I have tried and tried and am unable to come up with the answer.
The furthest I was able to get without making any mistakes or getting confused was $x= y^3 + 2y^2$. What am I supposed to do after... | Given $$ y=x^3+2*x^2 $$ solve for x:
We can use the cubic formula:
for $$0 = a x^3 + b x^2 + c x + f$$
x equals:
$$-\frac{\sqrt[3]{\sqrt{\left(27 a^2 f-27 a^2 y-9 a b c+2 b^3\right)^2+4 \left(3 a
c-b^2\right)^3}+27 a^2 f-27 a^2 y-9 a b c+2 b^3}}{3 \sqrt[3]{2}
a}+\frac{\sqrt[3]{2} \left(3 a c-b^2\right)}{3 a \sqrt... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1483839",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 3
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Find solutions of linear congruences: $x\equiv 0 \pmod 2$ , $x\equiv 0 \pmod 3$, $x\equiv 1 \pmod5$, $x\equiv 6 \pmod7$
$x\equiv 0 \pmod 2$
$x\equiv 0 \pmod 3$
$x\equiv 1 \pmod5$
$x\equiv 6 \pmod7$
Find all the solutions of each of the following systems of linear congruences.
I know how to find solutions of t... | The big idea is that the mapping
$$f:\mathbb Z_{210} \to
\mathbb Z_2 \times \mathbb Z_3 \times\mathbb Z_5 \times\mathbb Z_7$$
(where $210 = 2\cdot3\cdot5\cdot7$) defined by
$f(\bar n)= (\bar n,\bar n,\bar n,\bar n)$ is an isomorphism between additive groups.
You need to find an integer, $n$, such that
$f(\bar n)... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1485643",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 0
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If $x^2+4y^2=4.$ Then find range of $ x^2+y^2-xy$
If $x^2+4y^2=4.$ Then find range of $ x^2+y^2-xy$
$\bf{My\; Try::}$ Given $$x^2+4y^2 = 4\Rightarrow \frac{x^2}{4}+\frac{y^2}{1} = 1$$, So parametric Coordinate for Ellipse are
$x = 2\cos \phi$ and $y = \sin \phi$.
Now Let $$f\left(x,y\right) = x^2+y^2-xy = 4-4y^2+y^2-... | Perhaps I do not understand what is your question, but maybe this can be of some help: there exists $\alpha\in \mathbb{R}$ such that $\displaystyle \cos(\alpha)=\frac{3}{\sqrt{13}}$, and $\displaystyle\sin(\alpha)=\frac{2}{\sqrt{13}}$, and so $\displaystyle f(\phi)=\frac{5}{2}+\frac{\sqrt{13}}{2}\cos(2\phi+\alpha)$, an... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1487085",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 2,
"answer_id": 0
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Find the area of the triangle There are two points $N$ and $M$ on the sides $AB$ and $BC$ of the triangle $ABC$ respectively. The lines $AM$ and $CN$ intersect at point $P$. Find the area of the triangle $ABC$, if areas of triangles $ANP, CMP, CPA$ are $6,8,7$ respectively.
| Let
$$
\begin{cases}
a &= \verb/Area/(ANP) = 6\\
b &= \verb/Area/(CPA) = 7\\
c &= \verb/Area/(CMP) = 8
\end{cases}$$
and $(\alpha,\beta,\gamma)$ be the barycentric coordinate of $P$ with respect to $\triangle ABC$. i.e
the triplet of numbers such that
$$\vec{P} = \alpha \vec{A} + \beta \vec{B} + \gamma \vec{C}\quad\te... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1490393",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
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Proof by induction $\frac{1}{1^2} + \frac{1}{2^2} + \cdots + \frac{1}{n^2} < 2$ Proof by induction $\frac{1}{1^2} + \frac{1}{2^2} + \cdots + \frac{1}{n^2} < 2 \ \ \ n \in \mathbb{N}$
So for $n=1$
$$ 1 < 2$$
For $n > 1$
Assumption: $$\frac{1}{1^2} + \frac{1}{2^2} + \cdots + \frac{1}{n^2} < 2$$
Hypothesis (inductive ste... | HINT: It is easier to prove by induction this: $\frac{1}{1^2} + \frac{1}{2^2} + \cdots + \frac{1}{n^2} < 2 - \frac{1}{n}$ for $n > 1$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1491412",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
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Finding the coordinate C. A triangle $A$, $B$, $C$ has the coordinates:
$A = (-1, 3)$
$B = (3, 1)$
$C = (x, y)$
$BC$ is perpendicular to $AB$. Find the coordinates of $C$
My attempt:
Grad of $AB$ =
$$\frac{3-1}{-1-3} = -0.5$$
Grad of $BC = 2$ ($-0.5 \times 2 = -1$ because AB and BC are perpendicular).
Equation of $BC$... | HINT:
Length of $AC =$
$\sqrt{((x+1)^2 +(y-3)^2)} = \sqrt{(\mathrm{Length} \space \mathrm{of} \space (AB)^2+\mathrm{Length} \space \mathrm{of} \space (BC)^2)}\tag{1}$
because you have a right-angled triangle as $BC$ is perpendicular to $AB$
From $(1)$ $$(x+1)^2 +(y-3)^2=(3+1)^2+(1-3)^2+(x-3)^2+(y-1)^2$$
$$\implies x^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1493877",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
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Is there a way to parametrise general quadrics? A general quadric is a surface of the form:
$$ Ax^2 + By^2 + Cz^2 + 2Dxy + 2Eyz + 2Fxz + 2Gx + 2Hy + 2Iz + J = 0$$
It can be written as a matrix expression
$$ [x, y, z, 1]\begin{bmatrix}
A && D && F && G \\
D && B && E && H \\
F && E && C && I \\
G && H && I && J
\end{bma... | Yes. Since $M$ is symmetric, for an appropriate choice $P$ we can factor $M$ as
$$M = P^T D P$$
where $D$ is diagonal and has entries in $\{-1, 0, 1\}$, and in fact, we can choose $P$ so that its columns are orthogonal, which is convenient for some purposes. Then, in the new coordinates defined by transforming the orig... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1495645",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
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Why does $\lim\limits_{n \to \infty } ((1 + x)(1 + {x^2})(1 + {x^4})\ldots(1 + {x^{{2^n}}})) = \frac{1}{{1 - x}}$? Let $\left| x \right| < 1$.
Why does $$\lim\limits_{n \to \infty } ((1 + x)\cdot(1 + {x^2})\cdot(1 + {x^4})\cdot\ldots\cdot(1 + {x^{{2^n}}})) = \frac{1}{{1 - x}}$$
| We have
$$
(1+x)(1+x^2)(1+x^4) \cdots (1+x^{2^n}) = \frac{x^{2^{n+1}}-1}{x-1} \stackrel{n \to \infty}{\longrightarrow} \frac{1}{1-x}
$$
for $n \to \infty$. We used that
$$
x^{2^{n+1}}-1 = (x^{2^n}+1)(x^{2^n}-1) = (x^{2^n}+1)(x^{2^{n-1}}+1)(x^{2^{n-1}}-1) = \cdots = (x^{2^{n}} + 1) (x^{2^{n-1}} + 1) \cdots (x^{4}+1)(x^2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1495750",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 0
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What is the amplitude of this function? What is the amplitude of the function f(x) = 2sin(x) + 2cos(x) ?
| Notice $$f(x)=2\sin x+2\cos x$$
$$=2(\sin x+\cos x)$$
$$=\frac{2\sqrt 2}{\sqrt 2}(\sin x+\cos x)$$
$$=2\sqrt{2}\left(\sin x\frac{1}{\sqrt 2}+\cos x \frac{1}{\sqrt 2}\right)$$
$$=2\sqrt{2}\left(\sin x\cos \frac{\pi}{4}+\cos x \sin\frac{\pi}{4}\right)$$
$$f(x)=2\sqrt{2}\sin \left(x+\frac{\pi}{4}\right)$$
The above functi... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1495947",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Prove: If $|z|<\frac{1}{2}\Rightarrow |(1+i)z^3+iz|<\frac{3}{4},z\in\mathbb{C}$ Prove: If $|z|<\frac{1}{2}\Rightarrow |(1+i)z^3+iz|<\frac{3}{4},z\in\mathbb{C}$
$|z|=\sqrt{x^2+y^2}<\frac{1}{2}\Rightarrow x^2+y^2 <\frac{\sqrt{2}}{2}$
$$|(1+i)z^3+iz|=|(x^3-3xy-3x^2y-y-y^3)+i(x^3-3xy+3x^2y+x-y^3)|=\sqrt{(x^3-3xy-3x^2y-y-y^... | Hint
$$|(1+i)z^3+iz|\le |(1+i)||z|^3+|iz|<\dfrac{\sqrt{2}}{8}+\dfrac{1}{2}<\dfrac{1}{4}+\dfrac{1}{2}=\dfrac{3}{4}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1500203",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Proof that given equation(quartic) doesn't have real roots $$
(x^2-9)(x-2)(x+4)+(x^2-36)(x-4)(x+8)+153=0
$$
I need to prove that the above equation doesn't have a real solution. I tried breaking it up into an $(\alpha)(\beta)\cdots=0$ expression, but no luck. Wolfram alpha tells me that the equation doesn't have real r... | Expand $x^2-9=(x-3)(x+3)$ and $x^2-36=(x-6)(x+6)$. Let $t=x+\frac 1 2$, $u=x+1$. Then
$$LHS=(t^2-(2.5)^2)(t^2-(3.5)^2)+(u^2-25)(u^2-49)+153\geq -(\frac {3.5^2-2.5^2} 2)^2-(\frac {49-25} 2)^2+153=0$$
with equality iff $t^2=\frac {2.5^2+3.5^2} 2$ and $u^2=\frac {25+49} 2$ which is incompatible. Hence $LHS>0$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1500705",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Calculating two specific limits with Euler's number I got stuck, when I were proving that
$$\lim_{n \to \infty} \frac {\sqrt[2]{(n^2+5)}-n}{\sqrt[2]{(n^2+2)}-n} = \frac {5}{2}$$
$$\lim_{n \to \infty}n(\sqrt[3]{(n^3+n)}-n) = \frac {1}{3}$$
First one I tried to solve like
$$\lim_{n \to \infty} \frac {\sqrt[2]{(n^2+5)}-n... | When you arrive at
$$\lim_{n \to \infty} \frac {\sqrt[2]{(1+\frac {5}{n^2})}-1}{\sqrt[2]{(1+\frac{2}{n^2})}-1},$$
you can continue by
$$\lim_{n \to \infty} \frac {\sqrt[2]{(1+\frac {5}{n^2})}-1}{\sqrt[2]{(1+\frac{2}{n^2})}-1}=\lim_{n \to \infty} \frac {\sqrt[2]{(1+\frac {5}{n^2})}-1}{\sqrt[2]{(1+\frac{2}{n^2})}-1}\cdo... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1502198",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
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In the real number system,the equation $\sqrt{x+3-4\sqrt{x-1}}+\sqrt{x+8-6\sqrt{x-1}}=1$ has how many solutions? In the real number system,the equation $\sqrt{x+3-4\sqrt{x-1}}+\sqrt{x+8-6\sqrt{x-1}}=1$ has how many solutions?
I tried shifting the second term to the rhs and squaring.Even after that i'm left with square ... | $$x+3-4\sqrt{x-1}=x-1+4-4\sqrt{x-1}=(x-1)-4\sqrt{x-1}+4=(\sqrt{x-1}-2)^2$$ Similarly $$x+8-6\sqrt{x-1}=(\sqrt{x-1}-3)^2$$ from which the answer.
It is not an equation but an identity in its domain of definition.This is the reason there are infinitely many solutions and not the expected finite number of them of a true e... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1503958",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Can a pre-calculus student prove this?
a and b are rational numbers satisfying the equation $a^3 + 4a^2b = 4a^2 + b^4$
Prove $\sqrt a - 1$ is a rational square
So I saw this posted online somewhere, and I kind of understand what the question is saying. I'm interesting in doing higher order mathematics but don't quite... | My solution doesn't differ much from the other ones but I want to explain a bit how I did the factorization:
We are mainly interested in $\sqrt{a}-1$ and not so much in $a$, so let's define $c:=\sqrt{a}-1$. Our goal is to show that $\sqrt{c}$ is a rational number, given that $a$ and $b$ are rational, too. We rewrite th... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1505661",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "33",
"answer_count": 8,
"answer_id": 7
} |
Show that $a^3+a+1$ is in the subfield $L$ with $4$ elements of the field $K=F_2[X]/(X^4+X^3+1)$ The field $K$ is constructed in the following way: $K=F_2[X]/(X^4+X^3+1)$, where $F_2$ is short for $\mathbb{Z}$/$2$$\mathbb{Z}$.
Let $a$ be the class of $X$ in $K$ (so $a=X+(X^4+X^3+1))$.
The field $K$ contains a unique s... | We just need to perform a little of linear algebra to check that $a^3+a+1$ is an element of order $3$. In the given field, $a^4=a^3+1$, so:
$$ (a^3+a+1)^2 = a^6+a^2+1 = a^2(a^3+1)+a^2+1 = a+a^3,$$
$$ (a^3+a+1)^3 = (a^3+a)^2+(a^3+a) = a^6+a^2+a^3+a = 1 $$
and the set $\{0,a^3+a+1,a^3+a,1\}$ is a subfield with four elem... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1506828",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
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In a triangle prove that $\sin^2({\frac{A}{2}})+\sin^2(\frac{B}{2})+\sin^2(\frac{C}{2})+2\sin(\frac{A}{2})\sin(\frac{B}{2})\sin(\frac{C}{2})= 1$ Let ABC be a triangle. Thus prove that
$$\sin^2\left({\frac{A}{2}}\right)+\sin^2\left(\frac{B}{2}\right)+\sin^2\left(\frac{C}{2}\right)+2\sin\left(\frac{A}{2}\right)\sin\left... | $$F=\sin^2\dfrac A2+\sin^2\dfrac B2+\sin^2\dfrac C2 =1-\left(\cos^2\dfrac A2-\sin^2\dfrac B2\right)+\sin^2\dfrac C2$$
Using Prove that $\cos (A + B)\cos (A - B) = {\cos ^2}A - {\sin ^2}B$,
$$F=1-\cos\left(\dfrac{A-B}2\right)\cos\left(\dfrac{A+B}2\right)+\sin^2\dfrac C2$$
Using $\cos\left(\dfrac{A+B}2\right)=\cos\dfrac{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1507437",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
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Integrating $\sqrt{x^2+a^2}$ I'm trying to integrate this function wrt $x$, substituting $x = a \tan \theta$
$$ \int \sqrt{x^2+a^2} dx = a^2 \int \frac {d\theta}{\cos^3\theta} = $$
$$= a^2 \cdot \frac 12 \left( \tan\theta \sec\theta + \ln\lvert \tan\theta + \sec\theta \rvert \right) = \frac 12 \left( x \sqrt{a^2 + x^2... | Notice, your result is correct because adding or subtracting a constant will make no difference in the result. This can also be obtained using integration by parts
let $$I=\int \sqrt{x^2+a^2}\ dx\tag 1$$
$$I=\int \underbrace{\sqrt{x^2+a^2}}_{I}\cdot \underbrace{1}_{II}\ dx$$
using integration by parts
$$I=\sqrt{x^2+a... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1508157",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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How many sets contains 6 or its multiple given the following conditions?
MyApproach
I created
@Edit
S1={1,2,3,4,5} ...B)
S2={2,3,4,5,6}
S3={3,4,5,6,7}
S4={4,5,6,7,8}
S5={5,6,7,8,9}
S6={6,7,8,9,10}
S7={7,8,9,10,11} ....A)
S8={8,9,10,11,12}
From this information I analyzed that these $8$ sets have $6$ sets that have 6... | Note that if $x \equiv 0$ mod$(6)$, the $S_x$ contains a multiple of $6$.
If $x \equiv 2$ mod$(6)$, then $x+4 \equiv 0 $ mod$(6)$, and so $S_x$ contains a multiple of $6$.
Similarly you can show that $S_x$ contains a multiple of $6$, whenever $x \equiv 3,4,5$ mod$(6)$.
It remains the case $x \equiv 1$ mod$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1511174",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 1
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Epsilon and Delta proof for limit $\lim_{x\to-1} \frac{x+5}{2x+3}=4$ I have a function $f(x)=\frac{x+5}{2x+3}$ and want to show that the limit
$$\lim_{x\to -1} f(x) = 4$$
is true using $\epsilon$-$\delta$. I have a trouble finding $\delta$ so that $f(x)-4$ is less than every $\epsilon$.
Can anyone help?
| Hint: We want to find a bound of the form
$$\left| \frac{x + 5}{2x + 3} - 4 \right| \leq M |x+1|$$
Start with analyzing
$$\left|\frac{x + 5}{2x + 3} - 4\right| = \left| \frac{7(x+1)}{2x + 3}\right|$$
Now put a bound on $\displaystyle \left|\frac{7}{2x + 3}\right|$ for $x$ in some interval around $-1$. There's no canon... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1511991",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
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An identity for the factorial function A friend of mine was doodling with numbers arranged somewhat reminiscent of Pascal's Triangle, where the first row was $ 1^{n-1} \ \ 2^{n-1} \ \cdots \ n^{n-1} $ and subsequent rows were computed by taking the difference of adjacent terms. He conjectured that the number we get at... | Let's say we want to prove that iterating $n^3$ three times gives us $3!$. (Assume we already know that your thing works for $n^2$, $n$, and $1$.)
Well, if you iterate $n^3$ once, you get $(n+1)^3-n^3=3n^2+3n+1$. We also know that iterating $3n^2$ two times gives us $3\times2!=3!$, and we know that iterating $3n$ and $... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1512743",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "30",
"answer_count": 8,
"answer_id": 7
} |
Solving $\sqrt\frac{x^2+2x-2}{x^2+3x+2} + \sqrt\frac{x^2+3x+2}{x^2+2x-2}=\frac52 $ Got these two problems on an exam recently and was unsure if I managed to answer them correctly:
$$\sin^4x + \cos^4x \ge \frac58, \ x \in \left\{-\frac\pi4,\frac\pi4\right\}$$
Second:
$$\sqrt\frac{x^2+2x-2}{x^2+3x+2} + \sqrt\frac{x^2+3x+... | Hint...for the second one, set $u$ as the first square root expression and solve the quadratic $$u+\frac 1u=\frac 52$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1513586",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
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Finding the derivative of $(\frac{a+x}{a-x})^{\frac{3}{2}}$ This is a very simple problem, but I am stuck on one step:
Differentiate $(\frac{a+x}{a-x})^{\frac{3}{2}}$
Now, this is what I have done:
$$
(\frac{a+x}{a-x})^{\frac{3}{2}} \\
\implies \frac{\delta}{\delta y}\frac{f}{g} \\
\implies gf' = (a-x)^{\frac{3}{2}} \t... | So, there was something flawed in my calculation:
$$
y = f(g(x)) \\
y' = \frac{dy}{du} \times \frac{du}{dx} \\
$$
Let's see it again:
$$
y = (\frac{a+x}{a-x})^{\frac{3}{2}} \\
y = u^{\frac{3}{2}} \hspace{0.5cm} ; \hspace{0.5cm} u = \frac{a+x}{a-x} \\
\frac{dy}{du} = \frac{3}{2} \times u^{\frac{1}{2}} \implies \frac{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1514719",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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Solve $ \left|\frac{x}{x+2}\right|\leq 2 $ I am experiencing a little confusion in answering a problem on Absolute Value inequalities which I just started learning. This is the problem: Solve:
$$
\left|\frac{x}{x+2}\right|\leq 2
$$
The answer is given to be $x\leq-4$ or $x \geq-1$
This is my attempt to solve the probl... | Given that, $$\left|\frac{x}{x+2}\right|\le 2$$$$\iff |x|\le 2|x+2| $$
Notice, $x=-2$ & $x=0$ are two critical points on the number line. Now, let's consider the following cases,
Case 1: If $\color{blue}{x\le -2}$ $$-x\le -2(x+2)\iff x\le -4\ \ \ ({\text{True}})$$
$$\implies \color{red}{x\in(-\infty, -4]}$$
Case 2: I... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1515986",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 2
} |
Show that $\sqrt{x-1}+\sqrt{y-1}\leq \sqrt{xy}$ How can one show that $\sqrt{x-1}+\sqrt{y-1}\leq \sqrt{xy}$
Assuming that :
$\sqrt{x-1}+\sqrt{y-1}\leq \sqrt{xy}$
So
$(\sqrt{x-1}+\sqrt{y-1})^2\leq xy$
$\sqrt{(x-1)(y-1)} \leq xy-x-y+2$
$ (y-1)(x-1)+3 \leq \sqrt{(x-1)(y-1)}$
Here I'm stuck !
| Starting with “Assuming that $X$” when $X$ is the thing to be proved is not the right thing to do. You can make your computations easier if you set
$$
t=\sqrt{x-1},\quad u=\sqrt{y-1}
$$
so $x=t^2+1$ and $y=u^2+1$. The inequality to be proved becomes
$$
t+u\le\sqrt{(t^2+1)(u^2+1)}
$$
that has a single radical. Since $t+... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1519682",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
Prove $ \forall x >0, \quad \sqrt{x +2} - \sqrt{x +1} \neq \sqrt{x +1}-\sqrt{x}$
I would like to prove
$$ \forall x >0, \quad \sqrt{x +2} - \sqrt{x +1} \neq \sqrt{x +1}-\sqrt{x}$$
*
*I'm interested in more ways of proving it
My thoughts:
\begin{align}
\sqrt{x+2}-\sqrt{x+1}\neq \sqrt{x+1}-\sqrt{x}\\
\fr... | I will show that
if
$\sqrt{x+a}-\sqrt{x+b}
=\sqrt{x+b}-\sqrt{x+c}
$
for two different values of $x$,
then
$a=c$
and
$b = a+c
=2a
$.
If the equality
holds for one value of $x$,
then
$8x(a+c-2b)
=(a-4b+c)^2-4ac
$.
Since,
in the original problem,
$a=2, b=1, c=0$,
this can not hold for
two values of $x$.
If it holds for on... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1523179",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 7,
"answer_id": 6
} |
Prove that $5$ divides $3^{3n+1}+2^{n+1}$
Prove that $5$ divides $3^{3n+1}+2^{n+1}$
I tried to prove the result by induction but I couldn't.
The result is true for $n=1$.
Suppose that the result is true for $n$ i.e $3^{3n+1}+2^{n+1}=5k$ for some $k\in \mathbb{N}$. We study the term
$$3^{3n+4}+2^{n+2}=3^{3n+1}3^3+2^{... | Your approach works if you just subtract once more:
$$
3^{3n+4} + 2^{n+2} - 2\cdot 5k\\
= 27\cdot 3^{3n+1} + 2\cdot 2^{n+1} - 2(3^{3n+1} + 2^{n+1})\\
= 25\cdot 3^{3n+1}
$$
and you're done.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1526409",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 7,
"answer_id": 0
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Solve the limit without using L'Hopitals $\lim _{x\to 1}\left(\frac{\sqrt[3]{x+7}-2}{2x^2+3x-5\:}\right)$ $$\lim _{x\to 1}\left(\frac{\sqrt[3]{x+7}-2}{2x^2+3x-5\:}\right)$$
I think I need to do in the numerator by multiplying the difference of cubes.
$\lim _{x\to 1}\left(\frac{\left(\sqrt[3]{x+7}-2\right)\cdot \left(\s... | Hint: Just note that $$2x^2+3x-5=(x-1)(2x+5)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1526961",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
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Finding the value of this limit $$\lim_{x\to 0}\frac{\sin x^4-x^4\cos x^4+x^{20}}{x^4(e^{2x^4}-1-2x^4)}$$
The answer is given to be $1$. Can someone give any hint/s related to this problem?
More importantly, why am I not able to get the answer by simply using L hopital's rule and by basic substitution of trigonometric... | You can use taylor series (a lot easier than L'Hopital's):
We have the following limit:
$$\lim_{x\to 0}\frac{\sin x^4-x^4\cos x^4+x^{20}}{x^4(e^{2x}-1-2x^4)}$$
Using taylor series:
$$\lim_{x\to 0}\frac{\sin x^4-x^4\cos x^4+x^{20}}{x^4(e^{2x}-1-2x^4)} = \lim_{x\to 0}\frac{\left(x^4-\frac{(x^4)^3}{3!} + \frac{(x^4)^5}{5!... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1530115",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
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Find the minimum of $(u-v)^2+ (\sqrt{2-u^2}-\frac{9}{v})^2$ for $00$.
Find the minimum of $\displaystyle (u-v)^2+ \left(\sqrt{2-u^2}-\frac{9}{v}\right)^2$ for
$0<u<\sqrt{2}$ and $v>0$.
I think I have to use the Arithmetic and Geometric Means Inequalities. Or $\displaystyle \frac{1}{2}(u-v)^2+ \left(\sqrt{2-u^2}-\fr... | We want to minimize
$$F=-2uv-\frac{18}{v}\sqrt{2-u^2}+v^2+\frac{81}{v^2}+2$$
Then, we have
$$\frac{\partial F}{\partial u}=\frac{18u-2v^2\sqrt{2-u^2}}{v\sqrt{2-u^2}}$$
So, if we see $v\gt 0$ as a constant, then we know that $F$ is decreasing for $0\lt u\lt\sqrt{\frac{2v^4}{81+v^4}}$ and is increasing for $\sqrt{\frac{2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1530179",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
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Homework: Sum of the cubed roots of polynomial Given $7X^4-14X^3-7X+2 = f\in R[X]$, find the sum of the cubed roots.
Let $x_1, x_2, x_3, x_4\in R$ be the roots. Then the polynomial $X^4-2X^3-X+ 2/7$ would have the same roots. If we write the polynomial as $X^4 + a_1X^3 + a_2X^2 +a_3X + a_4$ then per Viete's theorem:
$a... | Let
$$A=x_1+x_2+x_3+x_4=2$$
$$B=x_1x_2+x_1x_3+x_1x_4+x_2x_3+x_2x_4+x_3x_4=0$$
$$C=x_1x_2x_3+x_1x_2x_4+x_1x_3x_4+x_2x_3x_4=1$$
$$D=x_1x_2x_3x_4=\frac 27.$$
$$E=x_1^2x_2+x_1x_2^2+x_1^2x_3+x_1x_3^2+x_1^2x_4+x_1x_4^2+x_2^2x_3+x_2x_3^2+x_2^2x_4+x_2x_4^2+x_3^2x_4+x_3x_4^2$$
We have
$$A^3=x_1^3+x_2^3+x_3^3+x_4^3+3E+6C$$
and
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1533363",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 1
} |
What is the remainder when $1^6 + 2^6 + 3^6 + ... + 99^6 + 100^6$ is divided by 5? What is the remainder when $1^6 + 2^6 + 3^6 + ... + 99^6 + 100^6$ is divided by 5?
I think that the only way to solve this would be to applying to the proposition that “the sum/product of congruence classes is equal to the congruence cla... | $$1^6+2^6+3^6+4^6+5^6+\dots+100^6\equiv20(1^6+2^6+3^6+4^6+5^6)\\\equiv0\text{ (mod }5)$$
The first conguruence can be applied since $1\equiv6\equiv11\equiv\dots\equiv96\text{ (mod }5)$, $1^6\equiv6^6\equiv11^6\equiv\dots\equiv96^6\text{ (mod }5)$
Likewise, $2^6\equiv7^6\equiv12^6\equiv\dots\equiv97^6\text{ (mod }5)$, a... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1535216",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 0
} |
Evaluate $\lim_{x \to 0} \left(\frac{ \sin x }{x} \right)^{\frac{1}{x^2}}$ $$\lim_{x \to 0} \bigg(\frac{\sin x}{x} \bigg)^{\frac{1}{x^2}}$$
The task should be solved by using Maclaurin series so I did some kind of asymptotic simplification
$$\lim_{x \to 0} \bigg(\frac{\sin x}{x} \bigg)^{\frac{1}{x^2}} \approx \lim_{x \... | When $x\to 0$
$$
\bigg(\frac{\sin x}{x} \bigg)^{\frac{1}{x^2}}
= \exp\left(\frac{\log\frac{x - x^3/6 + o(x^6)}{x}}{x^2}\right)
= \exp\left(\frac{-x^2/6 + o(x^3)}{x^2}\right)
= \exp(-1/6 + o(1)) \to \exp(-1/6)
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1535827",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
Limits with L'Hôpital's rule Find the values of $a$ and $b$ if $$ \lim_{x\to0} \dfrac{x(1+a \cos(x))-b \sin(x)}{x^3} = 1 $$
I think i should use L'Hôpital's rule but it did not work.
| use the equivalent, near $0$
\begin{eqnarray*}
\cos x &\approx &1-\frac{x^{2}}{2} \\
\sin x &\approx &x-\frac{x^{3}}{6}
\end{eqnarray*}
\begin{eqnarray*}
\frac{x(1+a\cos x)-b\sin x}{x^{3}} &\approx &\frac{x(1+a\left( 1-\frac{x^{2}%
}{2}\right) )-b\left( x-\frac{x^{3}}{6}\right) }{x^{3}} \\
&=&\frac{x+ax-a\frac{x^{3}}{2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1537881",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
Solving $\lim_{x\to2^{+}}\frac{\sqrt{x+7}-3}{\sqrt{x^{2}+5}-x-1}$ I need to solve $$\lim_{x\to2^{+}}\frac{\sqrt{x+7}-3}{\sqrt{x^{2}+5}-x-1}$$ without using L'hopital or taylor. tried Conjugate multiplication to no end. Any ideas?
| You should give conjugate multiplication a second chance:
$$\begin{align}
\frac{\sqrt{x+7}-3}{\sqrt{x^{2}+5}-x-1}&=\frac{(\sqrt{x+7}-3)(\sqrt{x+7}+3)(\sqrt{x^{2}+5}+x+1)}{(\sqrt{x^{2}+5}-x-1)(\sqrt{x^{2}+5}+x+1)(\sqrt{x+7}+3)}\\
&=\frac{(x-2)(\sqrt{x^{2}+5}+x+1)}{(x^2+5-(x+1)^2)(\sqrt{x+7}+3)}\\
&=-\frac{\sqrt{x^{2}+5}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1540837",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
number theory for finding value of $k$ How do I find what is the smallest positive integer $k$ such that $(3^3 + 4^3 + 5^3)\cdot k = a^n$ for some positive integers $a$ and $n$, with $n > 1$?
| Since $3^3+4^3+5^3=6^3$, we have
$$
6^3k=a^n
$$
Thus $2|a$ and $3|a$, so $6|a$.
Now for $n=1$, $k=1$ suffices, and it is easy to see that for $n=2$, $k=6$ is good, and no smaller number suffices. If $n=3$, then $k=1$ is good as well. Otherwise, $a\geq 6\Rightarrow$ $a^n\geq 6^n$ and thus $k\geq 6^{n-3}$. But $k=6^{n-3... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1541174",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
How Can I Find the Probability of Drawing This Sample? A random sample of size $6$ is selected with replacement from an urn that contains $10$ red, $5$ white and $5$ blue marbles. What is the probability that the sample contains $2$ marbles of each color?
This is what I got so far:
Pr[2 Red]= $ (\frac{10}{20})^2 $
Pr[... | As indicated in the comments, you have to take into account the number of sequences in which two red marbles, two white marbles, and two blue marbles are selected. Since the sequence has length $6$, the number of ways two of the six positions can be filled with a red marble is $\binom{6}{2}$. The number of ways two o... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1541692",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
What is $\frac{x^{10} + x^8 + x^2 + 1}{x^{10} + x^6 + x^4 + 1}$ given $x^2 + x - 1 = 0$?
Given that $x^2 + x - 1 = 0$, what is $$V \equiv \frac{x^{10} + x^8 + x^2 + 1}{x^{10} + x^6 + x^4 + 1} = \; ?$$
I have reduced $V$ to $\dfrac{x^8 + 1}{(x^4 + 1) (x^4 - x^2 + 1)}$, if you would like to know.
| Another approach is to note the following
$$\begin{align} x^2&=1-x\\
x^4&=(1-x)^2=1-2x+1-x&=2-3x\\
x^6&=(2-3x)(1-x)=2-5x+3(1-x)&=5-8x\\
x^8&=(5-8x)(1-x)=5-13x+8(1-x)&=13-21x\\
x^{10}&=(13-21x)(1-x)=13-34x+21(1-x)&=34-55x
\end{align}$$
Our fraction becomes
$${49-77x\over 42-66x}={7\over 6}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1542378",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 5,
"answer_id": 4
} |
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.