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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 minimum and maximum value of the function.First i simplified the function.
$f(x)=\frac{3+2 \sin x}{\sqrt{1+ \cos x}+\sqrt{1- \cos x}}=\frac{1+4\sin^2\left(\frac{x}{2}+\frac{\pi}{4}\right)}{2\sin \left(\frac{x}{2}+\frac{\pi}{4}\right)}$
Then i differentiated the function and equate it to zero to get the critical points.
Critical point equations are $\cos\left(\frac{x}{2}+\frac{\pi}{4}\right)=0$
$\sin\left(\frac{x}{2}+\frac{\pi}{4}\right)=\frac{1}{2},\sin\left(\frac{x}{2}+\frac{\pi}{4}\right)=\frac{-1}{2}$
When i checked plotted the function on desmos.com graphing calculator,i found minimum value to be $0.5$ and maximum value to be $2.5$.
Which i cannot get by my critical points.Where have i gone wrong?Please help me.
| 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
| {
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"url": "https://math.stackexchange.com/questions/1386474",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 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
$$\dfrac{1+\sqrt{\frac{1-\cos2\alpha}{2}}-\sqrt{\frac{1+\cos2\alpha}{2}}}{1+\sqrt{\frac{1-\cos2\alpha}{2}}+\sqrt{\frac{1+\cos2\alpha}{2}}}=$$
then after some calulations I come to this
$$=\frac{\sqrt{2} +\sqrt{1-\cos2\alpha}-\sqrt{1+\cos2\alpha}}{\sqrt{2} +\sqrt{1-\cos2\alpha}+\sqrt{1+\cos2\alpha}}$$ but now I'm stuck...
| 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,
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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 $$\frac{1}{x^{2}}-\frac{\cos^{2}x}{\sin^{2}x}$$
$$=\frac{\sin^2 x-x^2\cos^2 x}{x^2\sin^2 x}$$
$$=\frac{x^2}{\sin^2 x}\cdot\frac{\sin x+x\cos x}{x}\cdot\frac{\sin x-x\cos x}{x^3}$$
Then I have $3$ limits to evaluate
$$\lim_{x \to 0}\frac{x^2}{\sin^2 x}=\left(\lim_{x \to o}\frac{x}{\sin x}\right)^2=1^2=1$$
$$\lim_{x \to 0}\frac{\sin x+x\cos x}{x}=\lim_{x \to 0}\left(\frac{\sin x}{x} + \cos x\right)=1+1=2$$
Now I'm having trouble with the last one which is
$$\lim_{x \to 0}\frac{\sin x-x\cos x}{x^3}=?$$
Thanks for any help.
| 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",
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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:
$${ m=\frac{y^2 - y^1}{ x^2 - x^1 } = \frac{-3-0}{-6-4} = \frac{3}{10}}$$
Obtain Negative Reciprocal:
$${ m'=\frac{-10}{3}}$$
Get Midpoint fox X
$${ \frac{-6-4}{2} = -5 }$$
Get Midpoint for Y
$${ \frac{-0--3}{2} = \frac{3}{2} }$$
Make Point Slope Form:
$${ y = m'x +b = \frac{-10}{3}x + b}$$
Plugin Midpoints in Point Slope Form
$${ \frac{3}{2} = \frac{-10}{3}(-5) + b}$$
Evaluate b
$${ b = \frac{109}{6}}$$
Get Equation and Simplify
$${ y = \frac{-10}{3}x + \frac{109}{6}}$$
$${ 6y + 20x - 109 = 0 }$$
Is the problem set wrong? What am I doing wrong?
| 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 let $y=\sqrt{5-x}$. Then
$$
\tag1 y^2=5-x
$$
and the equation converts into
$$
\begin{align}
y &= 5-x^2\\
x^2 &= 5-y\\
y^2-x^2 &= 5-x-(5-y)\\
y^2-x^2 &= y-x\\
(y-x)(y+x)-(y-x) &= 0\\
(y-x)(y+x-1) &= 0
\end{align}
$$
So either $y=x$ or $x+y=1$.
Case 1 ($y=x$):
We can plug this into $(1)$ to get
$$
\begin{align}
y^2 &= 5-x\\
x^2 &= 5-x\\
x^2+x-5 &= 0\\
x &= \frac{-1\pm \sqrt{1+20}}{2}
\end{align}
$$
Since $-\sqrt{5}\leq x\leq \sqrt{5}$, the only solution is
$$
x = \frac{-1+\sqrt{21}}{2}
$$
Case 2 ($y=1-x$):
We can plug this into $(1)$ to get
$$
\begin{align}
y^2 &= 5-x\\
(1-x)^2 &= 5-x\\
1+x^2-2x &= 5-x\\
x^2-x-4 &= 0\\
x &= \frac{1\pm\sqrt{17}}{2}
\end{align}
$$
Since $-\sqrt{5}\leq x\leq \sqrt{5}$, the only solution is
$$
x = \frac{1-\sqrt{17}}{2}
$$
So final solution is
$$
x \in \left\{\displaystyle \frac{1-\sqrt{17}}{2}, \frac{-1+\sqrt{21}}{2} \right\}
$$
Is it possible to solve this problem using geometry? For example, could we use the properties of a half-circle and a parabola?
| $$
\begin{align}
\sqrt{5-x}&=5-x^2\\
5-x &= \left(5-x^2\right)^2\\
5-x &= x^4-10x^2+25\\
x^4-10x^2+25-5+x &= 0\\
x^4-10x^2+x+20 &= 0\\
(x^2-x-4)(x^2+x-5) &= 0
\end{align}
$$
$$
\begin{align}
x^2-x-4=0 &\vee x^2+x-5=0\\
x=\frac{-(-1)\pm\sqrt{(-1)^2-4\cdot 1 \cdot (-4)}}{2\cdot 1} &\vee x=\frac{-1\pm\sqrt{1^2-4\cdot 1 \cdot (-5)}}{2\cdot 1}\\
x=\frac{1\pm\sqrt{17}}{2} &\vee x=\frac{-1\pm\sqrt{21}}{2}
\end{align}
$$
Two of the $4$ solutions are good:
$$
\begin{align}
x_1 &= \frac{1-\sqrt{17}}{2}\\
x_2 &= \frac{-1+\sqrt{21}}{2}
\end{align}
$$
I don't see the positive thing in using geometry! This is the fastest way.
| {
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"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 problem involving a quadratic polynomial with a first-degree term to a quadratic polynomial with no first degree term, namely completing the square. You get
$$
x^2+x+1 = \left( x^2 + x + \frac 1 4 \right) + \frac 3 4 = \left( x + \frac 1 2 \right)^2 + \frac 3 4.
$$
Then you would like $\displaystyle \int \frac 1 {(\text{square})+1} \, dx$ so that you get an arctangent. So write
$$
\left( x + \frac 1 2 \right)^2 + \frac 3 4 = \frac 3 4 \left( \left( \frac{2x+1}{\sqrt 3} \right)^2 + 1 \right) = \frac 3 4 (u^2 + 1) \quad\text{and}\quad dx = \frac{\sqrt 3} 2\, du.
$$
| {
"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.
$\int\limits_{0}^{\pi/2}\frac{1+2\cos x}{(2+\cos x)^2}dx=\int\limits_{0}^{\pi/2}\frac{4+2\cos x}{(2+\cos x)^2}-\frac{3}{(2+\cos x)^2}dx=\int\limits_{0}^{\pi/2}\frac{2}{2+\cos x}-\frac{3}{(2+\cos x)^2}dx$
But i could not solve further.Please help me in completing.
| 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 $\displaystyle \left(2\csc x+\cot x \right) = t\;,$
Then $\left(\csc^2 x+2\cot x \cdot \csc x \right)dx = -dt.$
So Integral Convert into $\displaystyle I = -\int\frac{1}{t^2}dt = \frac{1}{t} = \left[\frac{\sin x}{2+\cos x }\right]_{0}^{\frac{\pi}{2}} = \frac{1}{2}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1395338",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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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
$$\text{area} = \frac{4\pi}{3} - \sqrt{3} = 2.4567 \cdots $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1395808",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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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 so that I thought the following question might be an interesting one to solve.
The Question:
Given a circle of radius $r$, a horizontal line a distance $c>r$ from the circle's centre, and two points $A$ and $B$ on that line located as indicated in the picture below, find the angle $\Theta$ as a function of the parameters given ($a,b,c,r$). The blue and red lines passing through $M$ are tangents to the circle, at $P$ and $Q$ respectively.
| 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 implies that $\overline{PM} = \overline{QM}$. Let $\xi \equiv \overline{PM}$.
Next, note that $\angle POM = \theta/2$. This results from the above and $\theta + (\pi/2 - \angle POM) + (\pi/2 - \angle QOM) = \pi$.
Next, note that $\tan\left(\angle POM\right) = \overline{PM}/\overline{OP} = \xi/r$, so
$$
\xi = r\tan\frac{\theta}{2}
$$
We'll now seek an equation for $\xi$ in terms of the various parameters of the problem. Note the following relations:
$$
\overline{AM} = \overline{AQ} - \overline{QM} = \overline{AQ} - \xi \qquad (1)
$$
$$
\overline{OA}^{\,2} = \overline{OQ}^{\,2} + \overline{AQ}^{\,2} = r^2 + \overline{AQ}^{\,2}
$$
$$
\overline{OA}^{\,2} = (a-b)^2 + c^2
$$
Similarly,
$$
\overline{BM} = \overline{BP} + \overline{PM} = \overline{BP} + \xi \qquad (2)
$$
$$
\overline{OB}^{\,2} = \overline{OP}^{\,2} + \overline{BP}^{\,2} = r^2 + \overline{BP}^{\,2}
$$
$$
\overline{OB}^{\,2} = b^2 + c^2
$$
Then
$$
\overline{AQ}^{\,2} = (a-b)^2 + c^2 - r^2 \qquad (3)
$$
$$
\overline{BP}^{\,2} = b^2 + c^2 - r^2 \qquad (4)
$$
Now let
$$
u \equiv \overline{AQ} = \sqrt{(a-b)^2 + c^2 - r^2}
\qquad\mbox{and}\qquad
v \equiv \overline{BP} = \sqrt{b^2 + c^2 - r^2}
$$
From (1)-(4) and the law of cosines,
$$
\overline{AB}^{\,2} = \overline{AM}^{\,2} + \overline{BM}^{\,2} - 2\,\overline{AM}\,\overline{BM}\,\cos\theta
\qquad (5)
$$
we see that we'll need $\cos\theta$ in terms of $\tan\frac{\theta}{2}$. That's easy:
$$
\cos\theta = \frac{1 - \tan^2\frac{\theta}{2}}{1 + \tan^2\frac{\theta}{2}} = \frac{r^2 - \xi^2}{r^2 + \xi^2}
\qquad (6)
$$
Combining all of the above, we can get a quadratic equation for $\xi$, after some tedius but not particularly difficult algebra:
$$
\left[ (u+v)^2 + 4r^2 - a^2 \right] \xi^2 - 4(u-v)r^2\,\xi + \left[ (u-v)^2 - a^2 \right] r^2 = 0
$$
whose solution is
$$
\frac{\xi}{r} =
\frac{2(u-v)r \pm \sqrt{
4(u-v)^2r^2 - \left[ (u+v)^2 + 4r^2 - a^2 \right] \left[ (u-v)^2 - a^2 \right]
}}{\left[ (u+v)^2 + 4r^2 - a^2 \right]}
$$
The second term inside the radical, sans the negative sign, simplifies to
$$
\left[ (u+v)^2 + 4r^2 - a^2 \right] \left[ (u-v)^2 - a^2 \right] =
4r^2(u-v)^2-4a^2c^2
$$
and we find
$$
\frac{\xi}{2r} =
\frac{(u-v)r \pm ac}{\left[ (u+v)^2 + 4r^2 - a^2 \right]}
$$
Thus,
$$
\tan\frac{\theta}{2} =
\frac{2\left[ (u-v)r \pm ac \right]}{\left[ (u+v)^2 + 4r^2 - a^2 \right]}
$$
where
$$
u \equiv \overline{AQ} = \sqrt{(a-b)^2 + c^2 - r^2}
\qquad\mbox{and}\qquad
v \equiv \overline{BP} = \sqrt{b^2 + c^2 - r^2}
$$
It's not surprising that there should be two valid solutions, since $Q$ could also be on the side closer to $B$.
| {
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "2",
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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 exponential form and then squaring it but I can't make any progress from this.
| 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}{2}$=1
$sinh^2$$\frac{x}{2}$=$cosh^2$$\frac{x}{2}$-1
$sinh^2$$\frac{x}{2}$=($\frac{1}{1-t^2}$)-1
$sinh $$\frac{x}{2}$=$\frac{t}{\sqrt(1-t^2)}$
$sinh$ $x$=2$cosh$$\frac{x}{2}sinh$$\frac{x}{2}$
$sinh$ $x$=$\frac{2t}{1-t^2}$ is obtained by simplifying the above.
Moving to the 2nd part after replacing with $t$ in the equation, you will get
12$t^2$+17$t$-2=0.From here I guess you should solve for $t$. You will get something in terms of $\sqrt385$. What you can do is replace it in $cosh$ $x$ and simplify by $x$ = $\underline+$$ln$($x+\sqrt{1+x^2}$). I have not tried it but still I hope it helps.
| {
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"timestamp": "2023-03-29T00:00:00",
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$\sum_{n=1}^\infty \frac{1}{(n^2-1)!} - \sum_{n=1}^\infty \frac{1}{(7n+1)!}$ is almost $1+1/6$ I've recognized, that
$$\mathcal{S} = \sum_{n=1}^\infty \frac{1}{(n^2-1)!} - \sum_{n=1}^\infty \frac{1}{(7n+1)!} \approx 1.1666666666666666666657785992648796$$
which is almost $1+1/6$.
I think it is not a mathematical coincidence, but I'm not sure what are behind the scences.
Why is this value almost $1+1/6$?
In the first terms of the sequences $n^2-1$ [$A005563$] and $7n+1$ [$A016993$] the terms $8$ and $15$ coincide. Furthermore
$$\sum_{n=1}^\infty \frac{1}{(n^2-1)!} \approx 1.166691468254732970341437561639 $$
and
$$\sum_{n=1}^\infty \frac{1}{(7n+1)!} \approx 0.000024801588066303675658962374$$
Are there a family of this type of identities? Also would be nice to see the exact value of $1+1/6-\mathcal{S}$ which is approximately
$$
8.880674017870608390962304909410175868736976 \cdot 10^{-22}
$$
| It follows because the first few terms are the same, and the terms in the series decrease to zero extremely quickly. Since$$\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}{8!}+\frac{1}{15!}\right)\right|\leq\frac{2}{24!}$$ and since
$$\sum_{n=1}^{\infty}\frac{1}{(7n+1)!}=\frac{1}{8!}+\frac{1}{15!}+\frac{1}{22!}+\cdots$$ we have that
$$\left|\sum_{n=1}^{\infty}\frac{1}{(7n+1)!}-\left(\frac{1}{8!}+\frac{1}{15!}\right)\right|\leq\frac{2}{22!},$$ and so the difference of the sums is $$\left|\sum_{n=1}^{\infty}\frac{1}{(n^{2}-1)!}-\sum_{n=1}^{\infty}\frac{1}{(7n+1)!}-\left(1+\frac{1}{6}\right)\right|\leq\frac{3}{22!} .$$ Now, if we were a bit more careful we could replace $3$ with $1+\epsilon$ where $\epsilon>0$ is a small positive constant around $1/100$. Notice that the error is almost exactly $1/22!$ since $$\frac{1}{22!}=8.8967913924505732867\dots \times 10^{-22}.$$
| {
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"timestamp": "2023-03-29T00:00:00",
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$\int_{0}^{\infty}\frac{\ln x dx}{x^2+2x+2}$ $$\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$$ and then lost track,answer is $\frac{\pi \ln 2}{8}$.
Any hint will solve my problem.
| Integrate $$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 \log 2 + 3/4\pi i)^2}{2i}
- \frac{(1/2 \log 2 + 5/4\pi i)^2}{2i}
\\ = \frac{1}{2i} (-1/2 (\log 2) \pi i + \pi^2)
= - \frac{1}{4} (\log 2) \pi
- \frac{1}{2} \pi^2 i$$
for a contribution of
$$2\pi i \times
\left(- \frac{1}{4} (\log 2) \pi
- \frac{1}{2} \pi^2 i\right)
= -\frac{1}{2} (\log 2) \pi^2 i
+ \pi^3.$$
The contribution from the circular arc vanishes and on the lower part
of the key slot we obtain after cancellation of the square of the
logarithm
$$-\int_0^\infty \frac{2\times 2\pi i \log x - 4\pi^2}{x^2+2x+2} dx.$$
Equating real and complex parts we obtain the two integrals
$$-4\pi \int_0^\infty \frac{\log x}{x^2+2x+2} dx
= -\frac{1}{2} (\log 2) \pi^2$$
or
$$\int_0^\infty \frac{\log x}{x^2+2x+2} dx
= \frac{1}{8} (\log 2) \pi$$
and
$$4\pi^2 \int_0^\infty \frac{1}{x^2+2x+2} dx
= \pi^3$$
or
$$\int_0^\infty \frac{1}{x^2+2x+2} dx = \frac{1}{4} \pi.$$
We may verify by ML that on the large circle say of radius $R$ we get
the estimate of the modulus of the integral
$$2\pi R \times \frac{\log^2 R}{R^2}
\rightarrow 0$$
as $R\rightarrow\infty.$
We get for the small circle of radius $\epsilon$ encircling the origin
$$2\pi \epsilon \times \frac{\log^2 \epsilon}{2}
\sim \pi \epsilon \times \log^2 \epsilon
\rightarrow 0$$
as $\epsilon\rightarrow 0.$
| {
"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 would suffice, because my own attempts to generate examples seem to end in failure. I tried starting with $10\equiv 2 \bmod 8$ and $c=4$, but $10\equiv 26\mod 32$ is false ($26 = 2 + 3\cdot8$ and $32 = 4\cdot8$).
Any help is appreciated!
| 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 first direction. Let $c$ be a positive integer. Suppose
\begin{align}
a \equiv b \pmod{n}.
\end{align}
Then $a = b + d n$ for some integer $d$. If we reduce modulo $cn$ we have
\begin{align}
a \equiv b + dn \pmod{cn}.
\end{align}
for some integer $d$. The set $\{0,n,2n,\dots,(c-1)n\}$ are all possible reductions of $dn$ modulo $cn$. Hence
\begin{align}
a \equiv b + dn \pmod{cn}.
\end{align}
for some integer $0 \leq d < c$. Moreover only one integer $d$ in this range will satisfy this congruence, since $jn \not\equiv kn \pmod{cn}$ for integers $j,k$ such that $0 \leq j < k < c$.
The reverse direction is much easier, just reduce modulo $n$. That is, suppose
\begin{align}
a \equiv b + dn \pmod{cn}.
\end{align}
for some integer $0 \leq d < c$. Then $a = b + dn + kcn$ for some integer $k$. Thus,
\begin{align}
a &\equiv b + dn + kcn \\
&\equiv b \pmod{n}.
\end{align}
| {
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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
$$
What I tried:
Because of $AB=BA$ we can use, let's say, the Newton's binomial expansion for $A$ and $B$, but it didn't take me to the solution.
Also it's easy to show that $\det (A^3 + B^3) = 0$.
| Let $\omega$ be a third root of unity. Since $A$ and $B$ commute, the condition that $\det(A^2-AB+B^2)=0$ becomes
$$\det(A+\omega B)\det(A+\omega^2 B)=0$$
and so either $\det(A+\omega B)=0$ or $\det(A+\omega^2 B)=0$.
Now consider the function $p(x)=\det(A+xB)$. This is a polynomial of degree at most $4$ with real coefficients, and from the above we see that either $\omega$ or $\omega^2$ is a root of $p$. Since $p$ has real coefficients, we then see that in fact both $\omega$ and $\omega^2$ are roots of $p$, and that $x^2+x+1$ is a factor of $p$.
Let $p(x)=(x^2+x+1)q(x)$ where $q$ is a polynomial of degree at most $2$, and let $q(x)=ax^2+bx+c$ where $a,b$ and $c$ are some real numbers.
Now consider the polynomial $r(x)=\det(xA+B)$.
For any $x\neq 0$, we have that
$$r(x)=\det\left(xA+B\right)=\det\left(x\left(A + \frac{1}{x}B\right)\right) = x^4\det\left(A + \frac{1}{x}B\right) = x^4p\left(\frac{1}{x}\right)$$
Then we have that
$$r(x)=x^4\left(\frac{1}{x^2}+\frac{1}{x}+1\right)\left(\frac{a}{x^2}+\frac{b}{x}+c\right)=(x^2+x+1)(a+bx+cx^2)$$
for any $x\neq 0$. The left and right hand sides of this expression are polynomials which agree at every point except possibly $x=0$, and so they must in fact be equal for all real $x$, including $x=0$. We see that
$$\det(B)=r(0)=a$$
Now we note that
$$\det(A+B)+3\det(A-B)=p(1)+3p(-1)=3q(1)+3q(-1)$$
which is equal to
$$6(a+c) = 6(\det(B)+q(0))=6(\det(B)+p(0))=6(\det(B)+\det(A))$$
as required.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1403900",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
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$\int\frac{1+x^2}{x^4+3x^3+3x^2-3x+1}dx$ $$\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}
$$
Using the identity: $\displaystyle\;\frac{dx}{x} = \frac{d(x-x^{-1})}{x+x^{-1}}\;$,
we can change variable to $u = x-x^{-1}$ and get
$$
\mathcal{I}
= \int\frac{du}{u^2 + 3u + 5} = \int\frac{du}{(u+\frac32)^2 + \frac{11}{4}}
= \sqrt{\frac{4}{11}}
\tan^{-1}\left(\sqrt{\frac{4}{11}}\left(u+\frac32\right)\right) + C\\
= \frac{2}{\sqrt{11}}
\tan^{-1}\left(\frac{2}{\sqrt{11}}\left(x - \frac1x +\frac32\right)\right) + C
$$
for some integration constant $C$.
Update
As pointed out by @Andrei, the expression above is discontinuous at $x = 0$. If one want to use it to compute definite integral, one need to use a different $C$
for the region $x > 0$ and $x < 0$. Let $C_{+}$ and $C_{-}$ be the integration constant for this two regions. Since the integrand is well behaved at $x = 0$,
as a function of $x$, the indefinite integral $\mathcal{I}$ is continuous at $x = 0$. This impose a constraint
$$C_{+} - C_{-} = \frac{2\pi}{\sqrt{11}}$$
and leads to
$$\mathcal{I} = \frac{2}{\sqrt{11}}
\left[
\tan^{-1}\left(\frac{2}{\sqrt{11}}\left(x - \frac1x +\frac32\right)\right) +
\Delta(x)
\right] + C'
$$
where
$\displaystyle\;C' = \frac{C_{+} + C_{-}}{2}\;$
and $\displaystyle\;
\Delta(x) = \begin{cases}+\frac{\pi}{2}, & x > 0\\-\frac{\pi}{2},& x < 0\end{cases}$.
To further simplify this, we use following representation of $\Delta(x)$:
$$\Delta(x) = \tan^{-1}x + \tan^{-1}\frac1x,\quad\text{ for } x \ne 0$$
This leads to
$$\begin{align}
\mathcal{I}
&= \frac{2}{\sqrt{11}}
\left[
\tan^{-1}\left(\frac{2}{\sqrt{11}}\left(x - \frac1x +\frac32\right)\right) +
\tan^{-1}\left(\frac{2}{\sqrt{11}x}\right)
+ \tan^{-1}\left(\frac{\sqrt{11}x}{2}\right)
\right] + C'\\
&= \frac{2}{\sqrt{11}}
\left[
\tan^{-1}\left(\frac{\sqrt{11}x^2(2x+3)}{7x^2-6x+4}\right)
+ \tan^{-1}\left(\frac{\sqrt{11}x}{2}\right)
\right] + C'
\end{align}
$$
An expression of $\mathcal{I}$ which is continuous for all $x$ and one can use to compute definite integral.
| {
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$\int_{0}^{\infty}f\left(\frac{a}{x}+\frac{x}{a}\right)\frac{\ln x}{x}dx=\ln a\int_{0}^{\infty}f\left(\frac{a}{x}+\frac{x}{a}\right)\frac{dx}{x}$ Show that $\displaystyle \int_{0}^{\infty}f\left(\frac{a}{x}+\frac{x}{a}\right)\frac{\ln x}{x}dx=\ln a\int_{0}^{\infty}f\left(\frac{a}{x}+\frac{x}{a}\right)\frac{dx}{x}$
What substitution should i make for this.Both sides are looking alike,how to transform one
into another.Putting $\displaystyle \frac{a}{x}+\frac{x}{a}=t$ will not work,i think.
| No, but split the integral up into 2 pieces. Consider the difference between the LHS and the RHS:
$$\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 ( \frac{a}{x} + \frac{x}{a} \right ) \log{\frac{x}{a}} $$
In the latter integral, sub $x=a^2/u$ and you will find that the integral is the negative of the former integral. viz.
$$\int_a^\infty \frac{dx}{x} f \left ( \frac{a}{x} + \frac{x}{a} \right ) \log{\frac{x}{a}} = \int_0^a \frac{du}{u} f \left ( \frac{a}{u} + \frac{u}{a} \right ) \log{\frac{a}{u}}$$
The integral over the infinite interval is thus zero, and the sought-after equality is true.
ADDENDUM
Not for nothing that
$$\int_0^\infty \frac{dx}{x} f \left ( \frac{a}{x} + \frac{x}{a} \right ) = 2 \int_2^{\infty} du \frac{f(u)}{\sqrt{u^2-4}} $$
| {
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"source": "stackexchange",
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Upper and lower bounds for $S(n) = \sum_{i=1}^{2^{n}-1} \frac{1}{i} = 1+\frac{1}{2}+ \cdots +\frac{1}{2^n-1}.$
For a positive integer $n$ let $S(n) = \sum_{i=1}^{2^{n}-1} \frac 1i = 1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+ \cdots +\frac{1}{2^n-1}.$
Then which of the following are true.
*
*(a) $S(100)\leq 100$.
*(b) $S(100)>100$.
*(c) $S(200)\leq 100$.
*(d) $S(200)>100$.
My attempt
*
*For the upper bound
$$\begin{align}
S(n) &= 1 + \left( \frac 12 + \frac 13 \right) + \left( \frac 14 + \frac 15 + \frac 16 + \frac 17 \right) + \cdots + \left( \frac 1{2^{n-1}} + \frac 1{2^{n-1}+1} + \cdots + \frac 1{2^n-1} \right)
\\ &< 1 + \left( \frac 12 + \frac 12 \right) + \left( \frac 14 + \frac 14 + \frac 14 + \frac 14 \right) + \cdots + \left( \frac 1{2^{n-1}} + \frac 1{2^{n-1}} + \cdots + \frac 1{2^{n-1}} \right) \\
&= \underbrace{1 + 1 + \cdots + 1}_{n\text{ times}} \\
&= n.
\end{align}$$
So we get $S(n) < n$ (for $n > 1$), and in particular $S(100) < 100$.
Now I did not understand how to calculate a lower bound, or if there is any other method by which we can solve this.
| Yes, there is another method that is easy to implement. Recall that we have
$$\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)-\frac{1}{2^n-1}<\log (2^n-1)<1+n\log 2-2^{-n}$$
For purposes of approximating for $n=100$, we have
$$69<100\,\log (2)-\frac{1}{2^{100}-1}<\sum_{k=1}^{2^{100}-1}\frac1k <1+100\,\log (2)-2^{-100}<71$$
so that
$$\bbox[5px,border:2px solid #C0A000]{69<\sum_{k=1}^{2^{100}-1}\frac1k <71}$$
For $n=200$, we have
$$138<200\,\log (2)-\frac{1}{2^{200}-1}<\sum_{k=1}^{2^{200}-1}\frac1k <1+100\,\log (2)-2^{-200}<140$$
so that
$$\bbox[5px,border:2px solid #C0A000]{138<\sum_{k=1}^{2^{200}-1}\frac1k <140}$$
| {
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"url": "https://math.stackexchange.com/questions/1404602",
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"source": "stackexchange",
"question_score": "3",
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Prove that the length of the common chord is $\frac{2ab\sin \theta}{\sqrt{a^2+b^2+2ab\cos \theta}}$ Two circles ,of radii $a$ and $b$,cut each other at an angle $\theta.$Prove that the length of the common chord is $\frac{2ab\sin \theta}{\sqrt{a^2+b^2+2ab\cos \theta}}$
Let the center of two circles be $O$ and $O'$ and the points where they intersect be $P$ and $Q$.Then angle $OPO'=\theta$
$\cos \theta=\frac{a^2+b^2-OO'^2}{2ab}$
$OO'^2=a^2+b^2-2ab\cos\theta$
In triangle $PO'Q$,angle $PO'Q=\pi-\theta$
$\cos(\pi-\theta)=\frac{b^2+b^2-l^2}{2b^2}$ Then i am stuck.Please help me to reach upto proof.
| The above method is too long, a faster method would be
Let A & B be the centers of the circles with radii a & b respectively which intersect each other at an angle θ such that they have a common chord PQ.
Angle between AP and BP is $180-θ$
Using cosine rule in $\triangle APB$,
$$AB=\sqrt {AM^2+BM^2-2(AM)(BM)\cos (\angle APB)}$$
$$AB=\sqrt {a^2+b^2-2(a)(b)\cos (180-θ)}$$
We have,
Area of quadilateral $APBQ = \frac{(PQ*AB)}{2}$
Also, area of quadilateral $APBQ = AP*BP*sin(\angle APB)$
So,$$\frac{(PQ*AB)}{2} = AP*BP*sin(\angle APB)$$
$$\frac{(PQ*\sqrt {a^2+b^2-2(a)(b)\cos (180-θ)})}{2} = a*b*sin(180-θ)$$
$$ PQ=\frac{2ab\sin\theta}{\sqrt{a^2+b^2+2ab\cos\theta}}$$
Sorry for the late answer.
| {
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Where did I go wrong in my evaluation of the integral of cosine squared? $$\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 + \left(\frac {-x\cos(2x)}2 + \frac 12\cdot\frac{\sin(2x)}2\right)$$
$$x\cos^2x-\frac{x\cos(2x)}2+\frac{\sin(2x)}4 + C$$
And this is clearly wrong, but I don't know where I messed up in my calculations. Would anyone mind correcting me somewhere?
| Notice,
$$\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]+C$$
$$=\frac{1}{2}\sin x\cos x+\frac{x}{2}+C$$
$$=\frac{x}{2}+\frac{1}{4}\sin2x+C$$
| {
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"question_score": "12",
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"answer_id": 4
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Prove that $\frac{2^{x+1}+(x+1)^2}{2^x+x^2}\rightarrow 2$ as $x \rightarrow \infty$ Could someone please show me the proof that $\frac{2^{x+1}+(x+1)^2}{2^x+x^2}\rightarrow 2$ as $x \rightarrow \infty$
I have no idea where to begin with this one.
Thanks.
| $$\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$$
| {
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"timestamp": "2023-03-29T00:00:00",
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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^{7}11^{13}$, the LCM is $2^{3}5^{7}11^{13}$. Here $0\le x\le3$, $0\le y\le7$, $0\le z\le13$
The number of ways of choosing 3 numbers $x,y,z$ is $^4C_1\cdot^8C_1\cdot^{14}C_1$
The above value is only for $b=2^{3}5^{7}11^{13}$.But now I have to consider another value of $b$ and the the solution becomes lengthy. Is this the correct approach?
| 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.
| {
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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{align*}
for some $\zeta_1$ between $x$ and $x + h$ and $\zeta_2$ between $x$ and $x - h$. Subtracting,
\begin{align*}
f(x+h) - f(x-h) &= 2h \cdot f'(x) - \frac{h^2}{2} \left(\,f''(\zeta_1) - f''(\zeta_2) \right)
\end{align*}
Solving for $f'(x)$ and bounding,
\begin{align*}
|f'(x)| &= \left| \frac{1}{2h} \left( f(x+h) - f(x-h) \right) + \frac{h}{4}\left(\,f''(\zeta_1) - f''(\zeta_2) \right) \right| \\
&\leq \frac{M_0}{h} + \frac{hM_2}{2}
\end{align*}
Taking the sup over all $x$ gives
$$M_1 \leq \frac{M_0}{h} + \frac{hM_2}{2}.$$
Rearranging and simplifying,
$$M_2h^2 - 2M_1h + 2M_0 \geq 0.$$
I'll let you take it from here.
| {
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bend measurement and calculating $\int_4^8 \sqrt{1+{\left(\frac{{x^2-4}}{4x}\right)^2}} $ How can i get the measure of this bend : $y=\left(\frac{x^2}{8}\right)-\ln(x)$ between $4\le x \le 8$. i solved that a bit according to the formula $\int_a^b \sqrt{1+{{f'}^2}} $:$$\int_4^8 \sqrt{1+{\left(\frac{x^2-4}{4x}\right)^2}} $$ $$= \cdots$$I don't know how I calculate this integration.the answer is: 6+ln2
| HINT:
$$\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}$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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calculating the characteristic polynomial I have the following matrix: $$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{vmatrix}$$
Is there a way to calculate the determinate faster or is way is:
$$(x+9)\cdot\begin{vmatrix}
x-7 & -5 \\
-6 & x-2 \\
\end{vmatrix}+7\cdot\begin{vmatrix}
9 & -5 \\
8 & x-2 \\
\end{vmatrix} -4\begin{vmatrix}
9 & x-7 \\
8 & -6 \\
\end{vmatrix}=$$
$$=(x+9)[(x-7)(x-2)-30]+7[9x-18+40]-4[54-8x+56]=(x+9)[x^2-9x-16]+7[9x+22]-4[-8x+2]=x^3-2x+2$$
| I am not sure if this method is "faster", but it does seem to involve less numbers in the actual calculation:
We know from a different theorem that two similar matrices share the same characteristic polynomial, see this post Elegant proofs that similar matrices have the same characteristic polynomial?.
From this fact, we can put $A$ in row echelon form. That is $A = \begin{bmatrix}
{-9} & {7} & {4} \\
{-9} & {7} & {5} \\
{-8} & {6} & {2}
\end{bmatrix}$ and for some invertable matrix $E, F = E*A*E^{-1} = \begin{bmatrix}
{1} & {-7/9} & {-4/9} \\
{0} & {-2/9} & {-14/9} \\
{0} & {0} & {1}
\end{bmatrix}$, that is $A$ is similar to $F$. From the previous theorem, it is clear that $det(F-I*x)=det(A-I*x)=(1-x)^2(\frac{-2}{9}-x)$.
But what you did is correct.
| {
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"url": "https://math.stackexchange.com/questions/1411724",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Inequality problem: Application of Cauchy-Schwarz inequality Let $a,b,c \in (1, \infty)$ such that $ \frac{1}{a} + \frac{1}{b} + \frac{1}{c}=2$. Prove that:
$$
\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$$ or
$$\sum_{cyc}\frac{x}{x+1}=1.$$
Thus, by C-S
$$1=\sum_{cyc}\frac{x}{x+1}\geq\frac{\left(\sqrt{x}+\sqrt{y}+\sqrt{z}\right)^2}{x+y+z+3},$$
which gives $\sqrt{xy}+\sqrt{xz}+\sqrt{yz}\leq\frac{3}{2}$.
Done!
| {
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"url": "https://math.stackexchange.com/questions/1412095",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Trying to solve $\sqrt{7-4\sqrt2 \sin x}=2\cos(x)-\sqrt2 \tan(x)$ The equation is
$$\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^2(x)+cos^2(x))-4\sqrt 2 \sin(x)=4\cos^2(x)-2\sqrt2\sin(x)+2\tan^2(x)$$
$$7\sin^2(x)+3cos^2(x)-2\sqrt 2 \sin(x)-2\frac{\sin^2(x)}{1-sin^2(x)}=0$$
I multiply the whole equation by $(1-sin^2(x))$ and then substitute sin(x) with t:
$$4t^4-2\sqrt2 t^3+t^2+2\sqrt2 t - 3 =0$$
And here I'm stuck. The polynomial is seemingly non-factorizable.
A hint would be welcome. (0:
P.S. The problem as it is presented in the texbook:
| Hint:
The solutions of $$\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)$$
Let $t=\cos^2 x$, so $(1)$ can be seen as
$$4t+2\left(\frac{1}{t}-1\right)-7=0\iff 4t^2-9t+2=0$$
which can be solved by the quadratic formula giving us $t\in\{2,\,\frac{1}{4}\}$. Since no real $x$ satisfies $\cos^2 x =2$ we take $\cos^2 x =\frac{1}{4}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1412884",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Need help with tangents to a quadratic The quadratic $y=kx^2+(3k-1)x-1$ and the straight line $y=(k+1)x-11$ meet.
Find the range of value(s) of $k$ such that the line is a tangent to the curve.
Got this question for school. Seems really simple and it's a non-calculator question but I'm not sure how to go about it.
| Given curve $y=kx^2+(3k-1)x- 1$ and curve $y=(k+1)x-11$ are intersect each
other exactly at one ponit
(Means only one value of $x$ and Corrosponding value of $y$)
which is also called Condition of tangency.
So Equating $y\;,$ We get $$kx^2+(3k-1)x-1 = (k+1)x-11$$
So $$\displaystyle kx^2+2(k-1)x+10=0\Rightarrow x=\frac{-2(k-1)x\pm\sqrt{4(k-1)^2-4\cdot k\cdot 10}}{2k}$$
Now for only one value of $x\;,$ We have $$\displaystyle 4(k-1)^2-4\cdot k\cdot 10=0$$
So $$\displaystyle(k-1)^2-10k=0\Rightarrow k^2-12k+1=0\Rightarrow k = \frac{12\pm\sqrt{140}}{2} = 6\pm \sqrt{35}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1414343",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Find the solution set of the equation $5.(\frac{1}{25})^{\sin^2x}+4.5^{\cos2x}=25^{\frac{\sin2x}{2}}$ Problem :
Find the solution set of the equation $5.(\frac{1}{25})^{\sin^2x}+4.5^{\cos2x}=25^{\frac{\sin2x}{2}}$ where $x \in [0,2\pi]$
My approach :
$5.(\frac{1}{25})^{\sin^2x}+4.5^{1-2\sin^2x}=25^{\frac{\sin2x}{2}}$
Unable to understand how to use $\sin2x$ in R.H.S. to solve further , can you please guide further thanks
| Hint:
Observe
\begin{align*}
5^{1-2\sin^2 x}+4\cdot 5^{1-2\sin^2 x}&=5^{2\sin x\cos x}\\
5\cdot 5^{1-2\sin^2 x}&=5^{2\sin x\cos x}\\
\end{align*}
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Prove that $1^2 + 2^2 + ..... + (n-1)^2 < \frac {n^3} { 3} < 1^2 + 2^2 + ...... + n^2$ I'm having trouble on starting this induction problem.
The question simply reads : prove the following using induction:
$$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,$$
and it's quite obvious that current integral is exactly $\frac{n^3}{3}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1416472",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Calculate simple expression: $\sqrt[3]{2 + \sqrt{5}} + \sqrt[3]{2 - \sqrt{5}}$ Tell me please, how calculate this expression:
$$
\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 + \sqrt{5}\right)^2} - \sqrt[3]{\left(2 + \sqrt{5}\right)\left(2 - \sqrt{5}\right)} + \sqrt[3]{\left(2 - \sqrt{5}\right)^2}\right)}{\left(\sqrt[3]{\left(2 + \sqrt{5}\right)^2} - \sqrt[3]{\left(2 + \sqrt{5}\right)\left(2 - \sqrt{5}\right)} + \sqrt[3]{\left(2 - \sqrt{5}\right)^2}\right)} =
$$
$$
= \frac{2 + \sqrt{5} + 2 - \sqrt{5}}{\sqrt[3]{\left(2 + \sqrt{5}\right)^2} + 1 + \sqrt[3]{\left(2 - \sqrt{5}\right)^2}}
$$
what next?
| , Let $$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-\sqrt{5}\right)}\right]\cdot (-x)$$
So $$4-x^3 = -3x\Rightarrow x^3+3x-4=0\Rightarrow (x-1)\cdot (x^2+x+4)=0$$
So we get $$x=1\Rightarrow \sqrt[3]{2+\sqrt{5}}+\sqrt[3]{2-\sqrt{5}} = 1$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1416720",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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"answer_id": 3
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Another messy integral: $I=\int \frac{\sqrt{2-x-x^2}}{x^2}\ dx$ I found the following question in a practice book of integration:-
$Q.$ Evaluate $$I=\int \frac{\sqrt{2-x-x^2}}{x^2}\ dx$$ For this I substituted $t^2=\frac {2-x-x^2}{x^2}\implies x^2=\frac{2-x}{1+t^2}\implies 2t\ dt=\left(-\frac4{x^3}+\frac 1{x^2}\right)\ dx$. Therefore $$\begin{align}I&=\int\frac {\sqrt{2-x-x^2}}{x^2}\ dx\\&=\int \left(\frac tx\right)\left(\frac{2t\ dt}{-\frac4{x^3}+\frac 1{x^2}}\right)\\&=\int \frac{2t^2\ dt}{\frac{x-4}{x^2}}\\&=\int \frac{2t^2(1+t^2)\ dt}{{x-4}\over{2-x}}\\&=\int \frac{2t^2(1+t^2)(5+4t^2-\sqrt{8t^2+9})\ dt}{\sqrt{8t^2+9}-(8t^2+9)}\end{align}$$ Now I substituted $8t^2+9=z^2 \implies t^2=\frac {z^2-9}8 \implies 2t\ dt=z/4\ dz$. So, after some simplification, you get $$\begin{align}I&=-\frac1{512}\int (z^2-9)(z+1)(z-1)^2\ dz\end{align}$$ I didn't have the patience to solve this integration after all these substitutions knowing that it can be done (I think I have made a mistake somewhere but I can't find it. There has to be an $ln(...)$ term, I believe). Is there an easier way to do this integral, something that would also strike the mind quickly? I have already tried the Euler substitutions but that is also messy.
| use so so-called Euler substitution and set $$\sqrt{2-x-x^2}=xt+\sqrt{2}$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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How to solve the integration to get the desired answer How to prove the following:
If
$$\int \frac{adx}{(ax-2)\sqrt{(ax-1)}}=\frac{y}{\sqrt{5}}$$
then show that $\frac{1}{x}=\frac{a}{2}\left(1+sech({\frac{y}{\sqrt{5}}})\right)$
Given, at $y=0$, $x=1/a$.
Approach
I have assumed $ax-1= \cosh^2z$ then $ax-2=\sinh^2z$, $adx=2\cosh{z}\sinh{z} dz$
Then we get,
$$2\int \frac{dz}{\sinh {z}}=\frac{y}{\sqrt{5}}$$
But I failed to show the desired.
| Let $u=\sqrt{ax-1}$, then $x=\frac{u^2+1}{a}\,\,\,$ and $\,\,\,dx=\frac{2u}{a}du$, so
\begin{align}
\int\frac{adx}{(ax-2)\sqrt{ax-1}}&=\int\frac{a\left(\frac{2u}{a}\right)du}{(u^2-1)u}\\
&=\int\frac{2du}{u^2-1}\\
&=\int\left(\frac{1}{u-1}-\frac{1}{u+1}\right)du
\end{align}
You can do the remaining part.
| {
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Finding the $n$th derivative of trigonometric function.. My maths teacher has asked me to find the $n$th derivative of $\cos^9(x)$. He gave us a hint which are as follows:
if $t=\cos x + i\sin x$,
$1/t=\cos x - i\sin x$,
then $2\cos x=(t+1/t)$.
How am I supposed to solve this? Please help me with explanations because I am not good at this. And yes he's taught us Leibniz Theorem. Thanks.
| Using his hint, you get
\begin{align*}
[\cos x]^9
&= \left[ \frac12 \left( t + \frac1t \right) \right]^9 \\
&= \frac{1}{2^9} \sum_{k=0}^9 {9 \choose k} (t)^k \left( \frac1t \right)^{9-k} \\
&= \frac{1}{2^9} \sum_{k=0}^9 {9 \choose k} t^{2k-9}. \tag{1}
\end{align*}
Also,
$$
\frac{dt}{dx} = -\sin x + i \cos x = i(\cos x + i \sin x) = it;
$$
Therefore,
$$
\frac{d}{dx} t^n = n t^{n-1} (it) = in t^n;
$$
it follows that
$$
\frac{d^N}{dx^N} t^n = (in)^N t^n.
$$
Applying this to (1),
\begin{align*}
\frac{d^N}{dx^N} [\cos x]^9
&= \frac{d^N}{dx^N} \frac{1}{2^9} \sum_{k=0}^9 {9 \choose k} t^{2k-9} \\
&= \frac{1}{2^9} \sum_{k=0}^9 {9 \choose k} \frac{d^N}{dx^N} t^{2k-9} \\
&= \frac{1}{2^9} \sum_{k=0}^9 {9 \choose k} (2k-9)^N i^N t^{2k-9} \\
&= \frac{1}{2^{10}}
\left[ \sum_{k=0}^9 {9 \choose k} (2k-9)^N i^N t^{2k-9}
+ \sum_{k=0}^9 {9 \choose {9-k}} (2(9-k)-9)^N i^N t^{2(9-k)-9} \right] \\
&= \frac{1}{2^{10}}
\left[ \sum_{k=0}^9 {9 \choose k} \left( (2k-9)^N i^N t^{2k-9}
+ (9-2k)^N i^N t^{9-2k} \right) \right] \\
&= \frac{1}{2^{10}}
\left[ \sum_{k=0}^9 {9 \choose k} (2k-9)^N i^N \left( t^{2k-9}
+ (-1)^N t^{9-2k} \right) \right] \\
&= \begin{cases}
\frac{1}{2^9}
\sum_{k=0}^9 {9 \choose k} (2k-9)^N i^N \cos((2k-9)x)
&N \text{ even} \\
\frac{1}{2^9}
\sum_{k=0}^9 {9 \choose k} (2k-9)^N i^{N+1} \sin((2k-9)x)
&N \text{ odd} \\
\end{cases} \\
&= \begin{cases}
\frac{1}{2^9}
\sum_{k=0}^9 {9 \choose k} (2k-9)^N (-1)^{N/2} \cos((2k-9)x)
&N \text{ even} \\
\frac{1}{2^9}
\sum_{k=0}^9 {9 \choose k} (2k-9)^N (-1)^{(N+1)/2} \sin((2k-9)x)
&N \text{ odd}. \\
\end{cases}
\end{align*}
| {
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"timestamp": "2023-03-29T00:00:00",
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How to compute $\int{\ln(\sqrt{x}+\sqrt{x+1})dx}$? The integral I am trying to work out is
$$\int{\ln(\sqrt{x}+\sqrt{x+1})dx}$$
So first step is substitute $u=x+1$:
$$\int{\ln(\sqrt{u-1}+\sqrt{u})du}.$$
Now substitute $u=\cos^2(t)$.
$\int{-\ln(\sqrt{-\sin^2(t)}+\sqrt{\cos^2(t)})2\sin(t)\cos(t)dt}$. This works out to $\int{-\ln(\cos(t)+i\sin(t))\sin(2t)dt}$. Remarkably, we obtain $\int{-\ln(e^{it})\sin(2t)dt}$ using Euler's identity.
Now we have $\int{-it\sin(2t)dt}$ and filling in reverse substitution $t=\arccos(\sqrt{x+1})$ after performing partial integration eventually yields me $0.5i\arccos(\sqrt{x+1})(2x+1)+0.5\sqrt{x^2+x}$.
I tried it on wolfram alpha and it gave a different result with hyperbolic functions. So that was not too useful to check this answer with. I feel this is not the right answer due to the imaginary inverse cosine factor. Where did it go wrong?
| Here is a solution not working with hyperbolic functions. Now updated with some more details.
Integrating by parts, and using that (by the chain rule, and writing the expression inside square brackets on common denominator)
$$
\begin{aligned}
D\log(\sqrt{x}+\sqrt{x+1})&=\frac{1}{\sqrt{x}+\sqrt{x+1}}D\bigl(\sqrt{x}+\sqrt{x+1}\bigr)\\
&= \frac{1}{\sqrt{x}+\sqrt{x+1}}\Bigl[\frac{1}{2\sqrt{x}}+\frac{1}{2\sqrt{x+1}}\Bigr]\\
&=\frac{1}{\sqrt{x}+\sqrt{x+1}}\frac{\sqrt{x+1}+\sqrt{x}}{2\sqrt{x}\sqrt{x+1}}\\
&=\frac{1}{2\sqrt{x}\sqrt{x+1}},
\end{aligned}
$$
we find that
$$
\int \log(\sqrt{x}+\sqrt{x+1})\,dx=x\log(\sqrt{x}+\sqrt{x+1})-\int \frac{\sqrt{x}}{2\sqrt{x+1}}\,dx
$$
Then, integrating by parts again, rewriting, splitting, and using the derivative on top,
$$
\begin{aligned}
\int \frac{\sqrt{x}}{2\sqrt{x+1}}\,dx&=\sqrt{x}\sqrt{x+1}-\int \frac{\sqrt{x+1}}{2\sqrt{x}}\,dx\\
&=\sqrt{x}\sqrt{x+1}-\int\frac{x+1}{2\sqrt{x}\sqrt{x+1}}\,dx\\
&=\sqrt{x}\sqrt{x+1}-\int\frac{\sqrt{x}}{2\sqrt{x+1}}\,dx-\log(\sqrt{x}+\sqrt{x+1}).
\end{aligned}
$$
Moving the integral to the left-hand side, we find that
$$
\int \frac{\sqrt{x}}{2\sqrt{x+1}}\,dx=\frac{1}{2}\sqrt{x}\sqrt{x+1}-\frac{1}{2}\log(\sqrt{x}+\sqrt{x+1}),
$$
and so, finally, one primitive is given by
$$
\int\log(\sqrt{x}+\sqrt{x+1})\,dx = \Bigl(x+\frac{1}{2}\Bigr)\log(\sqrt{x}+\sqrt{x+1})-\frac{1}{2}\sqrt{x}\sqrt{x+1}.
$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Limit involving trigonometric series summation How should I go about this one?
$\large{L = \lim_{n \to \infty} \sum_{k=1}^n \dfrac{1}{2^k} \tan\left(\dfrac{\pi}{3\cdot2^{k+1}}\right)}$
Hints please!
| Let $\displaystyle x= \frac{\pi}{6}\;,$ Then $\displaystyle L = \lim_{n\rightarrow \infty}\sum^{n}_{k=1}\frac{1}{2^k}\tan\left(\frac{x}{2^k}\right)$
Now using $$\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}\right)-2\cot x$$
So $$\displaystyle \bullet \; \frac{1}{2}\tan\left(\frac{x}{2}\right) = \frac{1}{2}\cot \left(\frac{x}{2}\right)-\cot x$$
So $$\displaystyle \lim_{n\rightarrow \infty}\sum^{n}_{k=1}\frac{1}{2^k}\tan\left(\frac{x}{2^k}\right) = \lim_{n\rightarrow \infty}\left[\frac{1}{2^n}\cot\left(\frac{x}{2^n}\right)-\cot x\right] = \frac{1}{x}-\cot x$$
So we get $$\displaystyle \lim_{n\rightarrow \infty}\sum^{n}_{k=1}\frac{1}{2^k}\tan\left(\frac{x}{2^k}\right) = \frac{6}{\pi}-\sqrt{3}$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Proving that if $n \in \mathbb{Z}$ and $n^2 − 6n + 5$ is even, then $n$ must be odd. Prove that if $n \in \mathbb{Z}$ and $n^2 − 6n + 5$ is even, then $n$ must be odd.
$p= n^2 - 6n + 55$ is even, $Q= n$ is odd
Proof: Assume on contrary $n$ is even. Then $n= 2k$ for some $k \in \mathbb{Z}$. Then, $$n^2 -6n + 5= 2k^2-6(2k)+5=2k^2-12k + 5$$
Unsure of where to go from here.
| $n^2 − 6n + 5$ even $\implies$ $n^2+5$ even because $6n$ is even.
$n^2+5$ even implies $n^2$ odd because $5$ is odd.
$n^2$ odd implies $n$ odd.
Or use the contrapositive: $n$ even $\implies$ $n^2 − 6n + 5$ odd:
$n$ even $\implies$ $n=2k$ $\implies$ $n^2 − 6n + 5=4k^2-12k+5=2(2k^2-6k+2)+1$, which is odd.
| {
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Prove that any integer that is both square and cube is congruent modulo 36 to 0,1,9,28 This is from Burton Revised Edition, 4.2.10(e) - I found a copy of this old edition for 50 cents.
Prove that if an integer $a$ is both a square and a cube then $a \equiv 0,1,9, \textrm{ or } 28 (\textrm{ mod}\ 36)$
An outline of the proof I have is
Any such integer $a$ has $a = x^2$ and $a = y^3$ for some integers $x,y$
Then by the Division Algorithm, $x = 36s + b$ for some integers $s,b$ with $0 \le b \lt 36$ and $y = 36t + c$ for some integers $t,c$ with $0 \le c \lt 36$
Using binomial theorem, it is easy to show that $x^2 \equiv b^2$ and $y^3 \equiv c^3$
Then $a \equiv b^2$ and $a \equiv c^3$
By computer computation (simple script), the intersection of the possible residuals for any value of $b$ and $c$ in the specified interval is 0,1,9,28
These residuals are possible but not actual without inspection which shows $0^2 = 0^3 \equiv 0$ , $1^2 = 1^3 \equiv 1$ , $27^2 = 9^3 \equiv 9$, and $8^2 = 4^3 \equiv 28$ $\Box$
There is surely a more elegant method, can anyone hint me in the right direction.
| First establish that $a$ must be a sixth power. We have $a=b^2=c^3$ so that $a^3=b^6$ and $a^2=c^6$ whence $$a=\cfrac {a^3}{a^2}=\cfrac {b^6}{c^6}=\left(\cfrac bc\right)^6$$
And if $q$ is a rational number whose sixth power is an integer, it must be an integer itself. [see below]
Now, let's have a look at the sixth powers modulo $36$. Every integer is congruent to a number of the form $6a+b$ where $-2\le a,b \le 3$. Then a simple application of the binomial theorem gives that:
$$(6a+b)^6\equiv b^6 \bmod 36$$
Finally, checking all the possibilities for $b$ we see $$(-2)^6=2^6=64\equiv 28; (-1)^6=1^6=1; 0^6=0; 3^6=81^2\equiv 9^2=81\equiv 9$$
Suppose $a,m,n \in \mathbb N$ with $a=\left(\frac mn\right)^6$ with $\frac mn$ in lowest terms and suppose $p$ is a prime factor of $n$ so that $n=pd$ with $d\in \mathbb N$. Then we have $an^6=m^6=ap^6d^6$ whence $p|m^6$ and because $p$ is prime $p|m$. But this is a contradiction since $m$ and $n$ were constructed to have no common factor. Hence $n$ has no prime factors and $n=1$.
| {
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"timestamp": "2023-03-29T00:00:00",
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Find the general integral of $ px(z-2y^2)=(z-qy)(z-y^2-2x^3).$ $ p=\frac{\partial z}{\partial x} $ and $ q=\frac{\partial z}{\partial y} $
Find the general integral of the linear PDE $ px(z-2y^2)=(z-qy)(z-y^2-2x^3). $
My attempt to solve this is as follows:
$ p=\frac{\partial z}{\partial x} $ and $ q=\frac{\partial z}{\partial y} $
$$px(z-2y^2)+qy(z-y^2-2x^3)=z(z-y^2-2x^3)$$
\begin{align*}
\text{The Lagrange's auxiliary equation is:} \frac{dx}{x(z-2y^2)}=\frac{dy}{y(z-y^2-2x^3)}=\frac{dz}{z(z-y^2-2x^3)}
\end{align*}
Now consider the 2nd and 3rd ratios,
\begin{align*}
\frac{dy}{y(z-y^2-2x^3)} & =\frac{dz}{z(z-y^2-2x^3)}\\
\implies \frac{dy}{y} & =\frac{dz}{z}\\
\implies \ln(y) & =\ln(z)+\ln(c_1)\\
\implies \frac{y}{z} & =c_1.
\end{align*}
But I am unable to get the 2nd integral surface. Kindly, help me.
Thanks in advance.
| i think you made error in question - in place of $2x^3$ it should be $2x^2$
$$ \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)$$
Combining fraction $\Rightarrow2)$ with $1 st $ fraction of $\Rightarrow 1) $
$$\frac{dx}{x(z-2y^2)}=\frac{d(z-y^2)}{(z-2y^2)(z-y^2-2x^2)}$$
$$\frac{d(z-y^2)}{dx}=\frac{z-y^2-2x^2}{x} $$
take $z-y^2=u$ we get $$\frac{du}{dx}=\frac{u-2x^2}{x} $$
$$ \frac{du}{dx}-\frac{u}{x}=-2x $$
IF $$ e^{\int\frac{-dx}{x}}=\frac{1}{x} $$
its solution =$$\frac{z-y^2}{x}=-2x+c_{1}$$ or
$$ \frac{z-y^2+2x^2}{x}=c_1$$
final solution =$$\phi(\frac{y}{z},\frac{z-y^2-2x^2}{x})=0$$
| {
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Prove that $g$ is continuous at $x=0$ Given, $g(x) = \frac{1}{1-x} + 1$. I want to prove that $g$ is continuous at $x=0$. I specifically want to do an $\epsilon-\delta$ proof. Related to this Is $g(x)\equiv f(x,1) = \frac{1}{1-x}+1$ increasing or decreasing? differentiable $x=1$?.
My work: Let $\epsilon >0$ be given. The challenging part for me is to pick the $\delta$.
Let $\delta = 2 \epsilon$ and suppose that $|x-0| = |x| < \delta$ and $|\frac{1}{1-x}| < \frac{1}{2}$.
So $$|g(x) - g(0)|$$
$$=|\frac{2-x}{1-x} - 2|$$
$$=|\frac{x}{1-x}|$$
$$=|\frac{1}{1-x}||x|$$
$$<\frac{1}{2} 2 \epsilon = \epsilon$$.
So $g$ is continuous at $x=0$. Is my proof correct?
EDIT: Rough work on how I picked $\delta$. Suppose $|x-0|<\delta$ and since $\delta \leq 1$, we have $|x|<1$, so $-1<x<1$ Then this implies $0<1-x<2$. The next step I'm not too sure of, I have $0<\frac{1}{1-x}<\frac{1}{2} \implies |\frac{1}{1-x}|<\frac{1}{2}$.
| $$\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}$$
Which is clearly not true for an $x$ arbitrarily near zero. I think there could be a problem with your choice of $\delta$.
| {
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Assume this equation has distinct roots. Prove $k = -1/2$ without using Vieta's formulas. Given $(1-2k)x^2 - (3k+4)x + 2 = 0$ for some $k \in \mathbb{R}\setminus\{1/2\}$, suppose $x_1$ and $x_2$ are distinct roots of the equation such that $x_1 x_2 = 1$.
Without using Vieta's formulas, how can we show $k = -1/2$ ?
Here is what I have done so far:
$(1-2k)x_1^2 - (3k+4)x_1 + 2 = 0$
$(1-2k)x_2^2 - (3k+4)x_2 + 2 = 0$
$\to (1-2k)x_1^2 - (3k+4)x_1 + 2 = (1-2k)x_2^2 - (3k+4)x_2 + 2$
$\to (1-2k)x_1^2 - (3k+4)x_1 = (1-2k)x_2^2 - (3k+4)x_2$
$\to (1-2k)x_1^2 - (3k+4)x_1 = (1-2k)(1/x_1)^2 - (3k+4)(1/x_1)$
$\to (1-2k)[x_1^2 - (1/x_1)^2] - (3k+4)[x_1 - (1/x_1)] = 0$
$\to (1-2k)[x_1 - (1/x_1)][x_1 + (1/x_1)] - (3k+4)[x_1 - (1/x_1)] = 0$
$\to (1-2k)[x_1 + (1/x_1)] - (3k+4) = 0$ or $[x_1 - (1/x_1)] = 0$
$\to (1-2k)[x_1 + (1/x_1)] - (3k+4) = 0$, $x_1 = 1$ or $x_1 = -1$
Since the later two cases violate the distinct roots assumption, we have:
$(1-2k)[x_1 + (1/x_1)] - (3k+4) = 0$
This gives:
The answer is $x_1 = \frac{5-i\sqrt{39}}{8}$ or $x_1 = \frac{5+i\sqrt{39}}{8}$. How do I get that (without Vieta's)?
| You don't need Vieta's formulas to conclude that a quadratic of the form $ax^2+bx+c=0$ has distinct roots that are reciprocals if and only if $a=c$. If $r$ and $1/r$, with $r\not=1/r$, are roots of the quadratic, then
$$ar^2+br+c=0\quad\text{and}\quad {a\over r^2}+{b\over r}+c=0$$
Multiplying the second equation through by $r^2$ and writing the terms in reverse order gives
$$cr^2+br+a=0$$
Subtracting this from the first equation gives $(a-c)r^2+(c-a)=0$, better written as
$$(a-c)(r^2-1)=0$$
Since $r\not=1/r$ by assumption, we have $r^2-1\not=0$, which implies $a-c=0$.
Applying this to the case at hand, $a=1-2k$ and $c=2$, we get $1-2k=2$, hence $k=-1/2$, as desired.
It's maybe worth noting that the middle term, $b=-(3k+4)$, is irrelevant to the proof.
| {
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If $x>0$we have $(1+x^2)f'(x)+(1+x)f(x)=1$ and $g'(x)=f(x), f(0)=g(0)=0$Prove: If $x>0$ we have $(1+x^2)f'(x)+(1+x)f(x)=1$. And $g'(x)=f(x), f(0)=g(0)=0$
Prove that:$\displaystyle \frac14<\sum_{n=1}^{\infty}g(\frac1n)<1$
I tried solving the ODE,But it seems very complex.and I still have no idea about it.Could someone help me?
Thanks!
| Using integrating factor $m$ we first solve given ODE:
\begin{align*}
f'+\frac{1+x}{1+x^2}f&=\frac{1}{1+x^2}\\
m'&=m\frac{1+x}{1+x^2}\\
(\ln(m))'&=\frac{1+x}{1+x^2}\\
\ln(m)&=\int_0^x\frac{1+y}{1+y^2}dy=arctg(x)+\frac{1}{2}\int_{1}^{1+x^2}\frac{1}{z}dz=arctg(x)+\frac{1}{2}\ln(1+x^2)\\
m&=\exp(arctg(x))+\sqrt{1+x^2}\\
f&=\frac{1}{m}\int_0^x\frac{m}{1+y^2}dy=\frac{1}{m}\Big(\int_0^xarctg'(y)\exp(arctg(y))dy+\int_0^x\frac{1}{\sqrt{1+y^2}}dy\Big)=\frac{1}{m}\Big(\exp(arctg(x))-1+\ln(x+\sqrt{1+x^2})\Big)
\end{align*}
Thus
\begin{align*}
g(x)&=\int_0^xf(y)dy=\int_0^x\frac{\exp(arctg(y))-1+\ln(y+\sqrt{1+y^2})}{\exp(arctg(y))+\sqrt{1+y^2}}dy
\end{align*}
Using monotonicity of $f$ we get that
\begin{align*}
g(1)&\leq\int_0^1\frac{\exp(arctg(1))-1+\ln(1+\sqrt{1+1^2})}{\exp(arctg(1))+\sqrt{1+1^2}}dy=\frac{\exp(\pi/4)-1+\ln(1+\sqrt{2})}{\exp(\pi/4)+\sqrt{2}}<0,575
\end{align*}
| {
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Solving $\lim _{x\to 1}\left(\frac{1-\sqrt[3]{4-3x}}{x-1}\right)$ $$\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+\sqrt[3]{4-3x})}$$
Finally
$$\frac{3}{(1+\sqrt[3]{4-3x})}$$
But this evaluates to
$$\frac{3}{2}$$
When the answer should be
$$1$$
Where did I fail?
| $ (1-(4-3x)^(1/3))(1+(4-3x)^(1/3)) = 1 - (4-3x)^(2/3) $
I couldn't re-format it right but others beat me to it
| {
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Evaluate limit as x approaches infinity of $\lim_{x\to\infty}\frac{\sqrt{x^3 +7x}}{\sqrt{4x^3+5}}$ I am having trouble figuring out how to answer this question by determining the degree of the numerator and/or denominator:
$$\lim_{x\to\infty}\frac{\sqrt{x^3 +7x}}{\sqrt{4x^3+5}}$$
I have tried deriving the first coefficient of the numerator and denominator, but not sure how to proceed to find the limit as $x \to \infty$.
| Notice, $$\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}}=\color{red}{\frac{1}{2}}$$
| {
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"timestamp": "2023-03-29T00:00:00",
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$\lim_{x\to a}\frac{1}{(x^2-a^2)^2}\left(\frac{a^2+x^2}{ax}-2\sin\left(\frac{a\pi}{2}\right)\sin\left(\frac{\pi x}{2}\right)\right)$ Prove that $\lim_{x\to a}\frac{1}{(x^2-a^2)^2}\left(\frac{a^2+x^2}{ax}-2\sin\left(\frac{a\pi}{2}\right)\sin\left(\frac{\pi x}{2}\right)\right)=\frac{\pi^2 a^2+4}{16a^4}$ where $a$ is an odd integer.
I tried to apply L Hospital rule in this question but it is not coming in $\frac{0}{0}$ form,neither series expansion seems to be helpful.What should i do to prove this limit.
| We have $$\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(\frac{\pi x}{2}\right)\right)=$$ $$=\frac{1}{4a^{4}}\lim_{x\rightarrow a}\frac{x^{2}+a^{2}-2ax\sin\left(\frac{a\pi}{2}\right)\sin\left(\frac{\pi x}{2}\right)}{\left(x-a\right)^{2}}
$$ so apply De L'Hopital twice $$=\frac{1}{4a^{4}}\lim_{x\rightarrow a}\frac{2-2a\pi\cos\left(\frac{\pi x}{2}\right)\sin\left(\frac{a\pi}{2}\right)+2^{-1}\pi^{2}ax\sin\left(\frac{\pi x}{2}\right)\sin\left(\frac{a\pi}{2}\right)}{2}=\frac{4+\pi^{2}a^{2}}{16a^{4}}.
$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Solving $\lim_{x\to0}\frac{1-\sqrt{\cos x}}{x^2}$ I've been trying to solve this over and over without L'Hopital but keep on failing:
$$\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 x \cdot (1+\sqrt{\cos x})}$$
Using the rule $\frac{1-\cos x}{x} = 0$ for $x\to0$ is useless because we would end up with
$$\frac{0}{0\cdot x \cdot \sqrt{\cos x}} = \frac{0}{0}$$
But hey, perhaps we can rationalize again?
$$\frac{1-\cos x}{(x^2+x^2\sqrt{\cos x})}\cdot \frac{(x^2-x^2\sqrt{\cos x})}{(x^2-x^2\sqrt{\cos x})}$$
Resulting in
$$\frac{(1-\cos x)(x^2-x^2\sqrt{\cos x})}{x^4 - x^4\cdot\cos x} = \frac{(1-\cos x)(x^2-x^2\sqrt{\cos x})}{x^4 \cdot (1-\cos x)}$$
Cancelling
$$\frac{(x^2-x^2\sqrt{\cos x})}{x^4} = \frac{1-\sqrt{\cos x}}{x^2}$$
Well that was hilarious. I ended up at the beginning! Dammit.
My second attempt was to use the definition $\cos x = 1 - 2\sin \frac{x}{2}$:
$$\lim_{x\to0}\frac{1-\sqrt{\cos x}}{x^2} = \frac{1-\sqrt{1 - 2\sin \frac{x}{2}}}{x^2}$$
And then rationalize
$$\frac{1-\sqrt{1 - 2\sin \frac{x}{2}}}{x^2} \cdot \frac{1+\sqrt{1 - 2\sin \frac{x}{2}}}{1+\sqrt{1 - 2\sin \frac{x}{2}}}$$
$$\frac{1-(1 - 2\sin \frac{x}{2})}{x^2+x^2\sqrt{1 - \sin \frac{x}{2}}} = \frac{- 2\sin \frac{x}{2}}{x^2\left(1+\sqrt{1 - \sin \frac{x}{2}}\right)}$$
I want to make use of the fact that $\frac{\sin x}{x} = 1$ for $x\to0$, so I will multiply both the numerator and denominator with $\frac{1}{2}$:
$$\frac{-\sin \frac{x}{2}}{\frac{x}{2}\cdot x\left(1+\sqrt{1 - \sin \frac{x}{2}}\right)}$$
Then
$$\frac{-1}{x\left(1+\sqrt{1 - \sin \frac{x}{2}}\right)}$$
Well clearly that's not gonna work either. I will still get $0$ in the denominator.
The correct answer is $\frac{1}{4}$. I can kind of see why is the numerator $1$, but no idea where is that $4$ going to come out of.
I don't know how am I supposed to solve this without L'Hopital.
| You are supposed to use the limit
$$
\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.$$
| {
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Why can there be an infinite difference between two functions as x grows large, but a ratio of 1? I learned in grade school that the closer $a$ and $b$ are to one another, the closer $\frac{a}{b}$ is going to be to $1$. For example, $\frac{3}{\pi}$ is pretty close to 1, and $\frac{10^{100}}{42}$ isn't even close to 1.
So, why is:
$$\lim_{x\to\infty} \frac{x^{2}}{x^{2}+x} = 1$$
But:
$$\lim_{x\to\infty}[(x^2+x)-(x^2)] = \infty$$ ?
Seems pretty counterintuitive. What's going on here?
| $\lim_{x\to\infty} \frac{x^{2}}{x^{2}+x} = 1$
As stated, this is because $x$ grows much more slowly than $x^2$. So the ratio goes to 1.
$1^2\over 1^2+1$$={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$$={100\over 100+10}$$={100\over 110}$$=0.91$
$100^2\over 100^2+100$$={10000\over 10000+100}$$={10000\over 10100}$$=0.99$
$\lim_{x\to\infty}[(x^2+x)-(x^2)] = \infty$
This isn't a ratio, so the non-ratio can't go to 1. In fact, it's exactly equal to $x$ for any values:
$\lim_{x\to\infty}[(x^2+x)-(x^2)]$
$=\lim_{x\to\infty}[x^2+x-x^2]$
$=\lim_{x\to\infty}[x^2-x^2+x]$
$=\lim_{x\to\infty}[(x^2-x^2)+x]$
$=\lim_{x\to\infty}[(0)+x]$
$=\lim_{x\to\infty}[x]$
$=\infty$
Obviously, as $x$ goes to infinity, the limit diverges to infinity.
$1^2+1-1^2$$=1+1-1$$=1$
$2^2+2-2^2$$=4+2-4$$=2$
$3^2+3-3^2$$=9+3-9$$=3$
$10^2+10-10^2$$=100+10-100$$=10$
$100^2+100-100^2$$=10000+100-10000$$=100$
| {
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Stuck solving a logarithmic equation $$\log _{ 2 }{ 2x } =\log _{ 4 }{ 4x^{ 6 } } -\log _{ 2 }{ 2x } $$
Steps I took:
$$\frac { \log _{ 4 }{ 2x } }{ \log _{ 4 }{ 2 } } =\log _{ 4 }{ 4x^{ 6 } } -\frac { \log _{ 4 }{ 2x } }{ \log _{ 4 }{ 2 } } $$
$$2\log _{ 4 }{ 2x } +2\log _{ 4 }{ 2x } =\log _{ 4 }{ 4x^{ 6 } } $$
$$4\log _{ 4 }{ 2x } =\log _{ 4 }{ 4x^{ 6 } } $$
At this point I get stuck I don't think turning this into $\log _{ 4 }{ (2x)^{ 4 } } =\log _{ 4 }{ 4x^{ 6 } } $ is the right answer. It leads to: $16x^{ 4 }=4x^{ 6 }$ and this has what seem to be extraneous solutions.
| Using: $\log_a b=\frac{\log a}{\log b}, \log ab=\log a + \log b, \log_a a=1$
$$\implies\log _{ 2 }2+\log _{ 2 }{ x } =\log _{ 4 }{ 4}+\log _{ 4 }{ x^{ 6 } } -\log _{ 2 }{ 2}-\log _{ 2 }{ x } $$
$$\implies1+\log _{ 2 }{ x } =6\frac{\log _{ 2 }{ x }}{\log_2 4} -\log _{ 2 }{ x } +1-1$$
$$\implies1+\log _{ 2 }{ x } =6\frac{\log _{ 2 }{ x }}{2}-\log _{ 2 }{ x } $$
$$\implies1+\log _{ 2 }{ x } =3{\log _{ 2 }{ x }}-\log _{ 2 }{ x } $$
$$\implies{\log _{ 2 }{ x }}=1 $$
$$\implies x=2^1$$
| {
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Is there an easier way to solve this logarithmic equation? $$2\log _{ 8 }{ x } =\log _{ 2 }{ x-1 } $$
Steps I took:
$$\frac { \log _{ 2 }{ x^{ 2 } } }{ \log _{ 2 }{ 8 } } =\log _{ 2 }{ x-1 } $$
$$\frac { \log _{ 2 }{ x^{ 2 } } }{ 3 } =\log _{ 2 }{ x-1 } $$
$$\log _{ 2 }{ x^{ 2 } } =3\log _{ 2 }{ x-1 } $$
$$2\log _{ 2 }{ x } =3\log _{ 2 }{ x-1 } $$
$$\log _{ 2 }{ x } =\frac { 3 }{ 2 } \log _{ 2 }{ x-1 } $$
$$\log _{ 2 }{ x } =\log _{ 2 }{ (x-1)^{ \frac { 3 }{ 2 } } } $$
This method seems to be very inefficient and I don't know how I would go from here. Can someone please point me in the right direction. Hints only please. No actual solution.
| $2\log_{8}x=\log_{2}x-1$
$2\log_{2}x=3\log_{2}x-3$ 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.
$\frac{1}{\sqrt[3]2}-\frac{1}{2\sqrt[3]2}-\frac{1}{2\sqrt[3]2}+\frac{1}{\sqrt[3]3}-\frac{1}{3\sqrt[3]3}-\frac{1}{3\sqrt[3]3}-\frac{1}{3\sqrt[3]3}+\cdots$.
While the sum of the cubes is
$\frac{1}{2}-\frac{1}{8\cdot 2}-\frac{1}{8\cdot 2}+\frac{1}{3}-\frac{1}{27\cdot 3}-\frac{1}{27\cdot 3}-\frac{1}{27\cdot 3}+\cdots$
Now the series seems to converge to 0, however, I cannot show using an epsilon argument that it does. Also, the sum of the cubes looks like $\frac{1}{4}\cdot \frac{1}{2}+\frac{8}{9}\cdot \frac{1}{3}+ \frac{15}{16}\cdot \frac{1}{4}+\cdots$, so I can see that it diverges, but likewise, cannot supply this with a rigorous argument.
I would greatly appreciate it if anyone can help me with this part.
| 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 harmonic series in one subsequence of the sequence of partial sums and thus not convergent.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1462893",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "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 $$\lim_{x \to 0}\frac{2\sin x\log \cos x + x^{3}}{x^{5}} = 0\tag{2}$$ From the fact that the numerator in the above limit expression is an odd function, I guessed that the limit in $(1)$ would also be $0$ (it would be great if this guess can be supported by a proof). However I was not able to do this via simple algebraic manipulation. Also note that evaluating $(1)$ is equivalent to solving the linked question (without the assumption of existence of limit).
I think it is better to go one step ahead and establish that $$\lim_{x \to 0}\frac{2\sin x \log \cos x + x^{3}}{x^{7}} = -\frac{1}{40}\tag{3}$$ It is possible to evaluate the above limit via Taylor's series very easily, but I would prefer to have a solution of either $(1)$ or $(3)$ without using Taylor's series.
Update: I provide an evaluation of limit $(2)$ as an illustration of the kind of answer I would prefer. We have
\begin{align}
L &= \lim_{x \to 0}\frac{2\sin x\log \cos x + x^{3}}{x^{5}}\notag\\
&= \lim_{x \to 0}\frac{\sin x\log (1 - \sin^{2}x) + x^{3}}{x^{5}}\notag\\
&= \lim_{x \to 0}\frac{\sin x\log (1 - \sin^{2}x) + \sin^{3}x + x^{3} - \sin^{3}x}{x^{5}}\notag\\
&= \lim_{x \to 0}\frac{\sin x\log (1 - \sin^{2}x) + \sin^{3}x}{x^{5}} + \frac{x^{3} - \sin^{3}x}{x^{5}}\notag\\
&= \lim_{x \to 0}\frac{\sin x\log (1 - \sin^{2}x) + \sin^{3}x}{\sin^{5}x}\cdot\frac{\sin^{5}x}{x^{5}} + \lim_{x \to 0}\frac{x - \sin x}{x^{3}}\cdot\frac{x^{2} + x\sin x + \sin^{2}x}{x^{2}}\notag\\
&= \lim_{x \to 0}\frac{\log (1 - \sin^{2}x) + \sin^{2}x}{\sin^{4}x}\cdot 1 + \lim_{x \to 0}\frac{1 - \cos x}{3x^{2}}\cdot (1 + 1 + 1)\text{ (via LHR)}\notag\\
&= \lim_{t \to 0}\frac{\log (1 - t) + t}{t^{2}} + \frac{1}{2}\notag\\
&= \lim_{t \to 0}\dfrac{-\dfrac{1}{1 - t} + 1}{2t} + \frac{1}{2}\text{ (via LHR)}\notag\\
&= -\frac{1}{2} + \frac{1}{2} = 0\notag
\end{align}
Further Update: I have finally found a solution which uses algebraic manipulation and L'Hospital's Rule. The rule has been applied 4 times in total and resulting expressions are simple. See the details in my answer.
| 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 $$-2 \left(720 \sec ^7(x)-864 \sec ^5(x)+204 \sec ^3(x)-4 \sec (x)+\cos (x) (\log (\cos
(x))+7)\right)$$ the value of which being $-126$ for $x=0$. Then, your result.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1465203",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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} |
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 $$\lim_{n \to \infty} \frac {a_n}{2^{n-1}}=\frac 4{\pi}$$ I tried many things none of which seemed fruitful. First thing I did was to define $a_n=2^{n-1}b_n$. Substituting this into the condition, we get $$2^{n}b_n=2^{n-1}b_n+\sqrt{1+2^{2n-2}b_n^2}\implies b_{n+1}=\frac {b_n}2+\sqrt{\frac1{2^{2n}}+\left(\frac {b_n}2\right)^2}$$ Simplifying this gives $$b_{n+1}^2-b_{n+1}b_n=\frac1{2^{2n}}\implies b_{n+1}(b_{n+1}-b_n)=\frac1{2^{2n}}$$ This doesn't lead anywhere. One lead that I got was by substituting $a_n=\tan \theta_n$. This gives $$\begin{align}a_{n+1}&=\tan \theta_n+\sec \theta_n\\&=\frac{\sin\theta_n+1}{\cos\theta_n}\\&=\frac{1+\tan\frac{\theta_n}2}{1-\tan\frac{\theta_n}2}\\&=\tan\left(\frac {\theta_n}2+\frac{\pi}4\right)\end{align}$$ Hence $$\tan\theta_{n+1}=\tan\left(\frac {\theta_n}2+\frac{\pi}4\right)\implies \theta_{n+1}=\frac {\theta_n}2+\frac{\pi}4\implies \theta_n=\frac {\pi}2+c\left(\frac 12\right)^n$$From initial conditions we get $\theta_0=0\implies c=-\frac{\pi}2$. Therefore $$\lim_{n \to \infty} \frac{a_n}{2^{n-1}}=\lim_{n\to\infty}\frac{\tan\left(\frac {\pi}2-\frac{\pi}2 \left( \frac 12 \right)^n\right)}{2^{n-1}}=\lim_{n\to\infty}\frac{2\left(\frac12\right)^n}{\tan\left(\frac {\pi}2\left(\frac12\right)^n\right)}=\lim_{n\to\infty}\frac{2\left(\frac12\right)^n}{\frac{\pi}2\left(\frac12\right)^n}=\frac4{\pi}$$ Is this proof valid and are there any other ways of doing it?
| 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",
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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
\end{array}
\begin{array}{c l}
\text{'cross' computation} &\text{ partial result}\\
2\cdot 5=10 & 10\\
2\cdot 4+3\cdot 5=23 & 123\\
2\cdot 3+3\cdot 4 +7\cdot5=53 &1283\\
3\cdot 3+7\cdot 4=37 & 12867\\
7\cdot 3=21 & 128691\\
\end{array}
To compure the square of a number use $a^2=(a+b)(a-b)+b^2$.
For example $$78^2=80\cdot 76+2^2$$
Other methods and examples in this MSE thread.
| {
"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
} |
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 find the integral values of $a,b$ for which this equation is satisfied,how should I procced from here??
Also is there any shorter much elegant way to do this question.
(Note-A hint would suffice)
| 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 $$k^2-k=l^2\Rightarrow 4k^2-4k = 4l^2\Rightarrow (4k^2-4k+1) = (2l)^2+1$$
So we get $$(2k-1)^2-(2l)^2=1\Rightarrow (2k+2l-1)\cdot (2k-2l-1) = 1\times 1 = -1\times 1$$
$\bullet\; $ If $2k+2l-1 = 1$ and $2k-2l-1 = 1\;,$ after solving we get $(k,l) = (1,0)$
$\bullet\; $ If $2k+2l-1 = -1$ and $2k-2l-1 = -1\;,$ after solving we get $(k,l) = (0,0)$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1467285",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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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...
$$\lim_{x\to 0}\frac{\sqrt[5]{1+\sin(x)}-1}{\ln(1+\tan(x))}$$
$$\sqrt[5]{1+x} = 1 + \dfrac x5 - \dfrac{2}{25}x^2 + O(x^3)$$
$$\sqrt[5]{1+\sin(x)}-1 \approx \dfrac 15 \sin x - \dfrac{2}{25}\sin{x^2}$$
$$\ln(1+x) = x - \dfrac 12x^2 + O(x^3)$$
$$\ln(1+ \tan x) \approx \tan x - \dfrac 12 \tan^2 x $$
\begin{align}
\frac{\sqrt[5]{1+\sin(x)}-1}{\ln(1+\tan(x))}
& \approx
\frac{\dfrac 15 \sin x-\dfrac{2}{25}\sin{x^2}}{\tan x-\dfrac 12 \tan^2 x} \\
& \approx
\frac{\dfrac 15 \sin x-\dfrac{2}{25}\sin{x^2}}{\tan x-\dfrac 12 \tan^2 x} \\
& \approx
\dfrac{\sin x}{\tan x}
\frac{\dfrac 15 - \dfrac{2}{25}\sin x}{1 - \dfrac 12 \tan x} \\
& \approx
\cos x \,
\frac{\dfrac 15 - \dfrac{2}{25}\sin x}{1 - \dfrac 12 \tan x} \\
& \to \dfrac 15 \quad \text{as} \quad x \to 0
\end{align}
| {
"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 multiplicity = algebric multiplicity = 1.
The eigenvectors of 1 are $(1, 0, 0, -1),(1, 0, -1, 0),(1, -1, 0, 0)$ and of 7 is $(0, 1, 1, 1)$.
I know that $A$ is diagonalizable and is similar to, for example:
$$D=
\begin{pmatrix}
1 & 0 & 0 & 0\\
0 & 1 & 0 & 0\\
0 & 0 & 1 & 0\\
0 & 0 & 0 & 7\\
\end{pmatrix}
$$
but how can I show that $A$ is similar to
$$D_1=
\begin{pmatrix}
1 & 1 & 0 & 0\\
0 & 7 & 0 & 0\\
0 & 0 & 1 & 0\\
0 & 0 & 0 & 1\\
\end{pmatrix}
$$
but not to:
$$D_2=
\begin{pmatrix}
1 & 1 & 0 & 0\\
0 & 1 & 0 & 0\\
0 & 0 & 1 & 0\\
0 & 0 & 0 & 7\\
\end{pmatrix}
$$
I see that the difference is in the order we choose the basis that the eigenvectores span, but how does it make difference? Suddenly $A$ is similar to non-diagonal matrix? so what the difference between $D_1$ and $D_2$?
| 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",
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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 that you attempt eight questions and from each part you attempt at least three questions?
Please don't answer with the three case method. What I want is the error in my method?
My method: For three questions from each part, we do: $${5 \choose 3}*{7\choose3} $$. But 2 questions are lett, so for that, we multiply the expression by $${6\choose2} $$. Because 6 questions are left to choose from. But this does not give the answer, why?
| 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 questions from part A, three questions from part B, and two additional questions counts the same selection more than once.
If we choose three questions from part A and five questions from part B, the second method counts the same selection of questions ten times since there are $\binom{5}{3}$ ways to choose three of the five selected questions from part B.
To make this concrete, suppose the selected questions are $A_1, A_2, A_3, B_1, B_2, B_3, B_4, B_5$. We can make this particular selection once. However, the second method counts this particular selection ten times.
\begin{align*}
& \color{red}{A_1}, \color{red}{A_2}, \color{red}{A_3}, \color{blue}{B_1}, \color{blue}{B_2}, \color{blue}{B_3}, B_4, B_5\\
& \color{red}{A_1}, \color{red}{A_2}, \color{red}{A_3}, \color{blue}{B_1}, \color{blue}{B_2}, B_3, \color{blue}{B_4}, B_5\\
& \color{red}{A_1}, \color{red}{A_2}, \color{red}{A_3}, \color{blue}{B_1}, \color{blue}{B_2}, B_3, B_4, \color{blue}{B_5}\\
& \color{red}{A_1}, \color{red}{A_2}, \color{red}{A_3}, \color{blue}{B_1}, B_2, \color{blue}{B_3}, \color{blue}{B_4}, B_5\\
& \color{red}{A_1}, \color{red}{A_2}, \color{red}{A_3}, \color{blue}{B_1}, B_2, \color{blue}{B_3}, B_4, \color{blue}{B_5}\\
& \color{red}{A_1}, \color{red}{A_2}, \color{red}{A_3}, \color{blue}{B_1}, B_2, B_3, \color{blue}{B_4}, \color{blue}{B_5}\\
& \color{red}{A_1}, \color{red}{A_2}, \color{red}{A_3}, B_1, \color{blue}{B_2}, \color{blue}{B_3}, \color{blue}{B_4}, B_5\\
& \color{red}{A_1}, \color{red}{A_2}, \color{red}{A_3}, B_1, \color{blue}{B_2}, \color{blue}{B_3}, B_4, \color{blue}{B_5}\\
& \color{red}{A_1}, \color{red}{A_2}, \color{red}{A_3}, B_1, \color{blue}{B_2}, B_3, \color{blue}{B_4}, \color{blue}{B_5}\\
& \color{red}{A_1}, \color{red}{A_2}, \color{red}{A_3}, B_1, B_2, \color{blue}{B_3}, \color{blue}{B_4}, \color{blue}{B_5}\\
\end{align*}
By similar reasoning, if we choose five questions from part A and three questions from part B, the second method counts the same selection of questions ten times since there are $\binom{5}{3}$ ways to choose three of the five selected questions from part A.
If we choose four questions from part A and four questions from part B, the second method counts the same selection sixteen times since we count the same set of four questions from part A four times for each of the $\binom{4}{3}$ ways we choose three of those four selected questions and the same set of four questions from part B four times for each of the $\binom{4}{3}$ ways we choose three of those four selected questions.
Note that $$10\binom{5}{3}\binom{7}{5} + 16\binom{5}{4}\binom{7}{4} + 10\binom{5}{5}\binom{7}{3} = \binom{5}{3}\binom{7}{3}\binom{6}{2}$$
| {
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"url": "https://math.stackexchange.com/questions/1470160",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
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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 that:
$$\frac{x^3+y^3+z^3}{x+y+z}\in \mathbb N$$
| 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
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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
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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
$$ \sin \beta \cos \beta + \sin \alpha \cos \alpha - \sin \alpha \sin \beta \leq 2 \sin \frac{\alpha + \beta}{2} \cos \frac{\alpha + \beta}{2} - \sin^2 \frac{\alpha + \beta}{2}. \tag{2} $$
If $(2)$ is true then I can denote $x = \frac{\alpha + \beta}{2}$ and prove (using Cauchy inequality) that
$$ 2 \sin x \cos x - \sin^2 x \leq \frac{\sqrt{5}-1}{2}. \tag{3}$$
But is $(2)$ true? How do I prove it? Maybe I need to use a different idea to maximize $(1)$?
| 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
$$ \sin \beta \cos \beta + \sin \alpha \cos \alpha - \sin \alpha \sin \beta=\sin(\alpha+\beta)\cos(\alpha-\beta)-\frac{1}{2}\cos(\alpha-\beta)+\frac{1}{2}\cos(\alpha+\beta) $$
Using Cauchy's inequality,
$$ \sin(\alpha+\beta)\cos(\alpha-\beta)-\frac{1}{2}\cos(\alpha-\beta)+\frac{1}{2}\cos(\alpha+\beta) \leq \sqrt{\cos(\alpha - \beta)^2+\frac{1}{4}}-\frac{1}{2} \cos (\alpha - \beta) $$
and, denoting $\cos(\alpha - \beta) = x$
$$ \sqrt{x^2+\frac{1}{4}}-\frac{1}{2} x \leq\frac{\sqrt5-1}{2} $$
because, after squaring,
$$3x^2 +2x - 2\sqrt{5}x + 2\sqrt{5} - 5 \leq 0 \\
(x-1)(3x+5-2\sqrt{5}) \leq 0$$
which holds because $\alpha, \beta \in [0, \frac{\pi}{2}]$ so $x \in [0, 1]$. (The first term is nonnegative, the second one is positive.)
Equality holds if $\frac{\sin(\alpha + \beta)}{\cos(\alpha - \beta)} = \frac{\cos(\alpha + \beta)}{\frac{1}{2}}$ and $x = 1$. Therefore $\alpha = \beta$ and $\tan 2\alpha = 2$ so
$$ \sin \beta \cos \beta + \sin \alpha \cos \alpha - \sin \alpha \sin \beta \leq \frac{\sqrt5-1}{2}$$
with equality at $\alpha = \beta = \frac{1}{2} \arctan 2$.
| {
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"timestamp": "2023-03-29T00:00:00",
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"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 $q = 3r$:
$$ 3p^3 + 3p^23r + 3p9r^2 + 27r^3 = 3p^3 + 9p^2r + 27pr^2 + 27r^3 = 0$$
Now divide by $3$:
$$ p^3 + 3p^2r+9pr^2 + 9r^3 = 0$$
So also $3|p$ wich contradicts the assumtion.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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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 $-3$ have sum $-1$. Hence,
\begin{align*}
2x^2 - x - 3 & = 2x^2 + 2x - 3x - 3 && \text{split the linear term}\\
& = 2x(x + 1) - 3(x + 1) && \text{factor by grouping}\\
& = (2x - 3)(x + 1) && \text{extract the common factor}
\end{align*}
Note that if we multiply $2x - 3$ and $x + 1$, we carry out the steps of the factorization in reverse order.
Why does this work? Suppose we have the factorization with respect to the rationals
\begin{align*}
ax^2 + bx + c & = (rx + s)(tx + u)\\
& = rx(tx + u) + s(tx + u)\\
& = \color{blue}{rt}x^2 + \color{green}{ru}x + \color{green}{st}x + \color{blue}{su}\\
& = \color{blue}{rt}x^2 + (\color{green}{ru + st})x + \color{blue}{su}
\end{align*}
Observe that the product of the coefficients of the quadratic and constant terms is equal to the product of the two coefficients that sum to the coefficient of the linear term, that is
$$(\color{blue}{rt})(\color{blue}{su}) = (\color{green}{ru})(\color{green}{st}) = rstu$$
Matching coefficients gives $a = rt$, $b = ru + st$, and $c = su$. Thus, we can factor a quadratic with respect to the rationals if we can find two numbers with product $ac$ and sum $b$.
| {
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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 $k = 3$:
$$
\begin{aligned}
F_6 &= 8\\
&= \binom{3}{0}F_3 + \binom{3}{1}F_2 + \binom{3}{2}F_1 + \binom{3}{3}F_0\\
&= F_3 + 3F_2 + 3F_1 + F_0\\
&= 2 + 3(1) + 3(1) + 0\\
\end{aligned}
$$
and with $n = 4$ and $k = 2$:
$$
\begin{aligned}
F_6 &= 8\\
&= \binom{2}{0}F_4 + \binom{2}{1}F_3 + \binom{2}{2}F_2\\
&= F_4 + 2F_3 + F_2\\
&= 3 + 2(2) + 1\\
\end{aligned}
$$
I'd like to find a source/alternate proof, but I don't know how to search for this kind of thing, beyond googling "Fibonacci binomial identity" which hasn't so far been successful.
I know there are an insane number of Fibonacci identities out there; is this a special case of something else?
Also, is there a nice combinatorial interpretation?
| 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 the coefficient of $x^n$ in the product between $(1+x)^k$ and the generating function of Fibonacci numbers:
$$\begin{eqnarray*} \sum_{i=0}^{k}\binom{k}{i}F_{n-i} = [x^n]\frac{x(1+x)^k}{1-x-x^2}&=&[x^{n+k}]\frac{x(x+x^2)^k}{1-x-x^2}\\&=&[x^{n+k}]\frac{x\left(1-(1-x-x^2)\right)^k}{1-x-x^2}\\&=&[x^{n+k}]\frac{x}{1-x-x^2}=F_{n+k}.\tag{2}\end{eqnarray*}$$
| {
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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 that step?
| 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:
$$-\frac{\sqrt[3]{\sqrt{\left(27 a^2 f-27 a^2 y-9 a b c+2 b^3\right)^2+4 \left(3 a
c-b^2\right)^3}+27 a^2 f-27 a^2 y-9 a b c+2 b^3}}{3 \sqrt[3]{2}
a}+\frac{\sqrt[3]{2} \left(3 a c-b^2\right)}{3 a \sqrt[3]{\sqrt{\left(27 a^2 f-27
a^2 y-9 a b c+2 b^3\right)^2+4 \left(3 a c-b^2\right)^3}+27 a^2 f-27 a^2 y-9 a b
c+2 b^3}}-\frac{b}{3 a}
$$
This gives us:
$$y(x) = \frac{1}{3} (-(3 \sqrt{3} \, \sqrt{27 x^2-32 x}-27 x+16)^{1/3}/2^{1/3}-(4 2^{1/3})/(3 \sqrt{3} \, \sqrt{27 x^2-32 x}-27 x+16)^{1/3}-2)$$
| {
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "1",
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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 three congruence equations, but
I don't know how to solve the 4 equations system...
I can't find it in my text book but it is in exercise sample... help me pls.
| 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
$f(\bar n) = (\bar 0, \bar 0, \bar 1, \bar 6)$.
Because $f$ is an isomorphism, there exists
$e_1, e_2, e_3, e_4 \in \mathbb Z_{210}$ such that
\begin{align}
f(e_1) &= (\bar 1, \bar 0, \bar 0, \bar 0)\\
f(e_2) &= (\bar 0, \bar 1, \bar 0, \bar 0)\\
f(e_3) &= (\bar 0, \bar 0, \bar 1, \bar 0)\\
f(e_4) &= (\bar 0, \bar 0, \bar 0, \bar 1)
\end{align}
It follows that
$$n \equiv 0 e_1 + 0 e_2 + 1 e_3 + 6 e_4 \equiv 1 e_3 + 6 e_4 \pmod{210}.$$
We compute $e_3$
Because $e_3 \equiv 0$ modulo $2$, modulo $3$, and modulo $7$, then $e_3$ is a multiple of $42 = 2\cdot3\cdot 7$.
So $e_3 \equiv 42x \pmod{210}$ for some integer, $x$.
Because $e_3 \equiv 1 \pmod 5,$
\begin{align}
42x &\equiv 1 \pmod 5 \\
2x &\equiv 1 \pmod 5 \\
x &\equiv 3 \pmod 5
\end{align}
So $e_3 \equiv 42\cdot3 \equiv 126 \pmod{210}$
We compute $e_4$
Because $e_4 \equiv 0$ modulo $2$, modulo $3$, and modulo $5$, then $e_4$ is a multiple of $30 = 2\cdot3\cdot 5$.
So $e_4 \equiv 30x \pmod{210}$ for some integer, $x$.
Because $e_4 \equiv 1 \pmod 7,$
\begin{align}
30x &\equiv 1 \pmod 7 \\
2x &\equiv 1 \pmod 7 \\
x &\equiv 4 \pmod 7
\end{align}
So $e_4 \equiv 30\cdot4 \equiv 120 \pmod{210}$
We compute $n$
\begin{align}
n
&\equiv 1 e_3 + 6 e_4 \pmod{210}\\
&\equiv 1 \cdot 126 + 6 \cdot 120 \pmod{210}\\
&\equiv 126 + 90 \pmod{210} \\
&\equiv 6 \pmod{210}
\end{align}
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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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 $$f\left(x,y\right) = x^2+y^2-xy = 4-4y^2+y^2-xy = 4-3y^2-xy = 4 - 3\sin^2 \phi - 2\sin \phi \cdot \cos \phi$$
So $$f\left(\phi \right) = 4-\frac{3}{2}\left(1-\cos 2\phi \right) - \sin 2\phi = \frac{5}{2} + \frac{1}{2}\left(3\cos 2 \phi -2 \sin 2\phi \right)$$
Now Range of $$-\sqrt{13}\leq \left(3\cos 2 \phi - 2\sin 2\phi \right)\leq \sqrt{13}$$
So $$\frac{1}{2}\cdot \left(5-\sqrt{13} \right) \leq f\left(\phi \right)\leq \frac{1}{2}\cdot \left(5+\sqrt{13} \right)$$
My question is can we solve it using Inequality or any other method, If yes Then plz explain here
Thanks
| 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 $\displaystyle f(\phi)=\frac{5}{2}+\frac{\sqrt{13}}{2}\cos(2\phi+\alpha)$, and as the range of $\displaystyle \cos(2\phi+\alpha)$ is $[-1,+1]$, you are done.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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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
$$\vec{P} = \alpha \vec{A} + \beta \vec{B} + \gamma \vec{C}\quad\text{ subject to }\quad \alpha + \beta + \gamma = 1$$
Following is a picture illustrating the arrangement of points and labeling
(not drawn to scale because the actual triangle is highly acute and
hard to display)
$\hspace1in$
Since
$$
\begin{align}
\alpha &= \frac{\verb/Area/(CMP)}{\verb/Area/(CMA)} = \frac{c}{c+b}\\
\gamma &= \frac{\verb/Area/(ANP)}{\verb/Area/(ANC)} = \frac{a}{a+b}\\
\end{align}
\quad\text{ and }\quad
\frac{\verb/Area/(CPA)}{\verb/Area/(ABC)} = \beta = 1 - \alpha - \gamma
$$
This leads to $$\verb/Area/(ABC) = \frac{\verb/Area/(CPA)}{1 - \frac{c}{c+b} - \frac{a}{a+b}}
= \frac{b}{1 -\frac{c}{c+b} - \frac{a}{a+b}}
= \frac{b(a+b)(c+b)}{b^2-ac}\\ = \frac{7(6+7)(7+8)}{7^2-6\cdot 8} = 1365$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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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 step):
$$\frac{1}{1^2} + \frac{1}{2^2} + \cdots + \frac{1}{n^2} + \frac{1}{(n+1)^2} < 2$$
So using assumption and hypothesis I have:
$$\frac{1}{1^2} + \frac{1}{2^2} + \cdots + \frac{1}{n^2} < 2 - \frac{1}{(n+1)^2} < 2 $$
So then: $$ \frac{1}{(n+1)^2} > 0 $$ which is always true
I was told it's relatively "hard" one. Thus I think I made sth stupid here.
| 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 $n > 1$
| {
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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 $BC$
$(y-1) = 2(x-3)$
$y = 2x - 5$
Equation of $AC$
$(y-3) = m(x--1)$
$y = mx+m+3$
I do not know how to proceed further. Please help me out.
| 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 $(1)$ $$(x+1)^2 +(y-3)^2=(3+1)^2+(1-3)^2+(x-3)^2+(y-1)^2$$
$$\implies x^2+2x+1 +y^2-6y+9=20+x^2-6x+9+y^2-2y+1$$
| {
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"timestamp": "2023-03-29T00:00:00",
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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
$$ [x, y, z, 1]\begin{bmatrix}
A && D && F && G \\
D && B && E && H \\
F && E && C && I \\
G && H && I && J
\end{bmatrix}
\begin{bmatrix}
x \\ y \\ z \\ 1
\end{bmatrix}
= \mathbf{p}^\intercal \mathbf{Q} \mathbf{p} = 0
$$
Is it possible to represent this quadric as a parametric surface $\mathbf{p}(u, v): \mathbb{R}^2 \to \mathbb{R}^3$?
$$ \forall u, v, \mathbf{p}(u, v)^\intercal \mathbf{Q}\mathbf{p}(u, v) = 0
$$
| 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 original ones via $P$, the quadric has equation
$$a x^2 + b y^2 + c z^2 = d,$$ where $a, b, c \in \{-1, 0, 1\}$, and by changing signs if necessary, we may assume that the first nonzero coefficient among $d, a, b, c$ (if any) is $1$. This leaves, up to relabeling of the variable names, just a few possibilities for nonempty quadrics (these thus serve as normal forms for the various types of quadrics, including degenerate cases):
\begin{align}
x^2 + y^2 + z^2 = 1 & \qquad \textrm{sphere} \\
x^2 + y^2 - z^2 = 1 & \qquad \textrm{hyperboloid of one sheet} \\
x^2 - y^2 - z^2 = 1 & \qquad \textrm{hyperboloid of two sheets} \\
x^2 + y^2 + z^2 = 0 & \qquad \textrm{point} \\
x^2 + y^2 - z^2 = 0 & \qquad \textrm{cone} \\
x^2 + y^2 = 1 & \qquad \textrm{cylinder} \\
x^2 - y^2 = 1 & \qquad \textrm{hyperbolic cylinder} \\
x^2 + y^2 = 0 & \qquad \textrm{line} \\
x^2 - y^2 = 0 & \qquad \textrm{intersection union of two distinct planes} \\
x^2 = 1 & \qquad \textrm{disjoint union of two distinct planes} \\
x^2 = 0 & \qquad \textrm{plane}
\end{align}
All of these can be readily parameterized. Hence, we can produce an explicit parameterization in the original coordinates by changing variables in our parameterization using $P^{-1}$
| {
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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
$$
x^{2^{n+1}}-1 = (x^{2^n}+1)(x^{2^n}-1) = (x^{2^n}+1)(x^{2^{n-1}}+1)(x^{2^{n-1}}-1) = \cdots = (x^{2^{n}} + 1) (x^{2^{n-1}} + 1) \cdots (x^{4}+1)(x^2+1)(x+1)(x-1),
$$
and that $x^{2^{n+1}} \to 0$ for $n \to \infty$.
| {
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "1",
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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 function is of form $f(x)=x_0\sin (x+\phi)$ with amplitude $x_0$
hence, the amplitude of the given function $f(x)$ is $\color{red}{2\sqrt 2}$
| {
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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 $|z|<\frac{1}{2}\Rightarrow |(1+i)z^3+iz|<\frac{3}{4},z\in\mathbb{C}$
$|z|=\sqrt{x^2+y^2}<\frac{1}{2}\Rightarrow x^2+y^2 <\frac{\sqrt{2}}{2}$
$$|(1+i)z^3+iz|=|(x^3-3xy-3x^2y-y-y^3)+i(x^3-3xy+3x^2y+x-y^3)|=\sqrt{(x^3-3xy-3x^2y-y-y^3)^2+(x^3-3xy+3x^2y+x-y^3)^2}$$
Is there another approach rather than expanding these expressions?
| Hint
$$|(1+i)z^3+iz|\le |(1+i)||z|^3+|iz|<\dfrac{\sqrt{2}}{8}+\dfrac{1}{2}<\dfrac{1}{4}+\dfrac{1}{2}=\dfrac{3}{4}$$
| {
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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 roots, but I'm sure there's simpler way to solve this than working trough the quartic this gives.
| 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
$$\lim_{n \to \infty} \frac {\sqrt[2]{(n^2+5)}-n}{\sqrt[2]{(n^2+2)}-n} = \lim_{n \to \infty} \frac {n\sqrt[2]{(1+\frac {5}{n^2})}-n}{n\sqrt[2]{(1+\frac{2}{n^2})}-n}= \lim_{n \to \infty} \frac {\sqrt[2]{(1+\frac {5}{n^2})}-1}{\sqrt[2]{(1+\frac{2}{n^2})}-1}$$
and now I think, that this one sholud go like $$\lim_{n \to \infty} \frac{\frac{5}{n^2}}{\frac{2}{n^2}}=\frac{5}{2} $$
but I have no idea how to prove this.
In the second one I made
$$\lim_{n \to \infty}n(\sqrt[3]{(n^3+n)}-n) = \lim_{n \to \infty}n(n\sqrt[3]{(1+\frac{1}{n^2}}-n)= \lim_{n \to \infty}n^2(\sqrt[3]{(1+\frac{1}{n^2}}-1)= \lim_{n \to \infty}n^2(e^{\frac{1}{3n^2}}-1) $$
And now I do not know what to do next...
I would be really grateful, for any help, or prompt, how to solve these ones (or information, where is the mistake).
| 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
$$\lim_{n \to \infty} \frac {\sqrt[2]{(1+\frac {5}{n^2})}-1}{\sqrt[2]{(1+\frac{2}{n^2})}-1}=\lim_{n \to \infty} \frac {\sqrt[2]{(1+\frac {5}{n^2})}-1}{\sqrt[2]{(1+\frac{2}{n^2})}-1}\cdot
\frac {\sqrt[2]{(1+\frac {5}{n^2})}+1}{\sqrt[2]{(1+\frac{2}{n^2})}+1}\cdot
\frac {\sqrt[2]{(1+\frac{2}{n^2})}+1}{\sqrt[2]{(1+\frac {5}{n^2})}+1}$$
$$=\lim_{n \to \infty} \frac { 1+\frac {5}{n^2} -1}{ { 1+\frac{2}{n^2}}-1}
\cdot
\frac {\sqrt[2]{(1+\frac{2}{n^2})}+1}{\sqrt[2]{(1+\frac {5}{n^2})}+1}=\frac 52\lim_{n \to \infty}\frac {\sqrt[2]{(1+\frac{2}{n^2})}+1}{\sqrt[2]{(1+\frac {5}{n^2})}+1},$$
and the last limit should be easy to find (both numerator and denominator converge to a finite non-zero number).
The same trick works for the second limit. Just observe that
$$n(\sqrt[3]{(n^3+n)}-n) = n\frac{\big((n^3+n)^{1/3}-n\big)\big((n^3+n)^{2/3}+(n^3+n)^{1/3}n+n^2\big)}{(n^3+n)^{2/3}+(n^3+n)^{1/3}n+n^2} = n\frac{n^3+n-n^3}{(n^3+n)^{2/3}+(n^3+n)^{1/3}n+n^2} = \frac{1}{(1+1/n^2)^{2/3}+(1+1/n^2)^{1/3}+1}.$$
| {
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"timestamp": "2023-03-29T00:00:00",
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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 roots.No idea how to proceed.Help!
| $$x+3-4\sqrt{x-1}=x-1+4-4\sqrt{x-1}=(x-1)-4\sqrt{x-1}+4=(\sqrt{x-1}-2)^2$$ 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 equation with coefficients in a field. For example the "equation"
$$\sqrt[3]\frac{x^3-3x+(x^2-1)\sqrt{x^2-4}}{2}+\sqrt[3]\frac{x^3-3x-(x^2-1)\sqrt{x^2-4}}{2}=x$$ has all the reals as solutions.
| {
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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 have the mathematical skills yet to grasp them (hence, precalculus) but is this problem way more complex than it seems or are the tools within the reach of a precalculus student?
| 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 the equation as
$$(c+1)^6+4\cdot(c+1)^4\cdot b = 4\cdot(c+1)^4+b^4.$$
In a first step, we want to find all the possible $b$ for a given $c$. Since the highest order in $b$ is $4$, we know that there are at most $4$ solutions. In the following, we make some assumptions about the shape of $b$ as an expression of $c$ and hope that we find some solutions:
First, we notice that $(c+1)$ occurs multiple times in our equation. Wouldn't it be cool if $b$ had those $(c+1)$-brackets, too? Well, let's assume that $b$ has the following shape:
$$b=(c+1)\cdot f(c).$$
Then the equation becomes
$$(c+1)^6+4\cdot(c+1)^5\cdot f(c) = 4\cdot(c+1)^4+(c+1)^4\cdot f^4(c)$$
$$(c+1)^2+4\cdot(c+1)\cdot f(c) = 4+ f^4(c).$$
Let's assume that $f(c)$ can be written as
$$f(c)=\alpha\cdot c^\beta+g(c)$$
where $g(c)$ has an order of less than $\beta$. Surely, the highest order on the right side of the equation is $4\cdot\beta$, but what is the highest order on the left side? It is either $2$, coming from the first term which contains $c^2$, or it is $\beta+1$, coming from the second term which contains $4\alpha c \cdot c^\beta$.
*
*If the highest order on the left side was $\beta+1$, then we'd have $2\lt\beta+1$ and $4\beta=\beta+1$ which is impossible.
*If the highest order on the left side is $2$, then we have $\beta+1\lt2$ and $4\beta=2$. This works and we get $\beta=1/2$.
If we compare the coefficients of these two leading terms, we get that $\alpha=\pm1$. Let's plug in everything we have:
$$c^2+2c+1\pm 4\cdot(c+1)\cdot c^{1/2}+4\cdot(c+1)\cdot g(c)$$
$$=4+c^2\pm 4c^{3/2}\cdot g(c)+6c\cdot g^2(c)\pm4c^{1/2}\cdot g^3(c)+g^4(c).$$
If we stare at this a little bit, we see that $g(c)=1$ works. So we have found two solutions for $b$:
$$b_1=(c+1)\cdot(1+c^{1/2})=c^{3/2}+c+c^{1/2}+1$$
$$b_2=(c+1)\cdot(1-c^{1/2})=-c^{3/2}+c-c^{1/2}+1$$
Now we wonder if there are even more real solutions for $b$. Having found two solutions, we can apply polynomial long division on $b^4-4\cdot(c+1)^4\cdot b +4\cdot(c+1)^4-(c+1)^6$ and get:
$$b^4-4\cdot(c+1)^4\cdot b +4\cdot(c+1)^4-(c+1)^6$$
$$=\left(b-(1+c)(1+c^{1/2})\right)\left(b-(1+c)(1-c^{1/2})\right)\left(b^2+2b(c+1)+(c+1)^2(3+c)\right)$$
If we calculate the discriminant of the last term, we'll find that it will be negative, except for $c=-1$, i.e. $a=0$. So, there are no further real expressions, unless $a=0$. (But then we have $c=-1$ which isn't a squared rational number).
In the following we assume that $a\neq0$. Then $b_1$ and $b_2$ are the only solutions to our equation. As Matt Samuel showed, $c=\sqrt a -1$ is a rational number. But we want to show that $\sqrt{c}$ is a rational number, too! Take
$$b=b_1=(c+1)(1+c^{1/2})$$
Then we have
$$c^{1/2}=\frac{b}{c+1}-1$$
which is rational (because $b$ and $c$ are rational). Finally, take
$$b=b_2=(c+1)(1-c^{1/2})$$
Then we have
$$c^{1/2}=-\frac{b}{c+1}+1$$
which is rational, too. We have reached our goal.
| {
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"timestamp": "2023-03-29T00:00:00",
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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$$\mathbb{Z}$.
Let $a$ be the class of $X$ in $K$ (so $a=X+(X^4+X^3+1))$.
The field $K$ contains a unique subfield $L$ with $4$ elements.
I need to show that $a^3+a+1$ is in that subfield $L$.
I know that a subfield of field $K$ is a subset containing $0,1$ and that it is closed with respect to addition, multiplication and inverses. But I don't know how to show that the element $a^3+a+1$ is in $L$. How can I do this?
| We just need to perform a little of linear algebra to check that $a^3+a+1$ is an element of order $3$. In the given field, $a^4=a^3+1$, so:
$$ (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 $$
and the set $\{0,a^3+a+1,a^3+a,1\}$ is a subfield with four elements.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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In a triangle prove that $\sin^2({\frac{A}{2}})+\sin^2(\frac{B}{2})+\sin^2(\frac{C}{2})+2\sin(\frac{A}{2})\sin(\frac{B}{2})\sin(\frac{C}{2})= 1$ Let ABC be a triangle. Thus prove that
$$\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(\frac{B}{2}\right)\sin\left(\frac{C}{2}\right)= 1$$
How do i go about solving this?
| $$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$$
Using $\cos\left(\dfrac{A+B}2\right)=\cos\dfrac{\pi-C}2=\sin\dfrac C2,$
$$F=1-\sin\dfrac C2\left(\cos\left(\dfrac{A-B}2\right)-\cos\left(\dfrac{A+B}2\right)\right)$$
Now $\cos\left(\dfrac{A-B}2\right)-\cos\left(\dfrac{A+B}2\right)=2\sin\dfrac A2\sin\dfrac B2$
Hope you can take it from here!
| {
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"url": "https://math.stackexchange.com/questions/1507437",
"timestamp": "2023-03-29T00:00:00",
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Integrating $\sqrt{x^2+a^2}$ I'm trying to integrate this function wrt $x$, substituting $x = a \tan \theta$
$$ \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} + a^2 \ln \left| \frac x a + \frac{\sqrt{a^2+x^2}}{a} \right| \right)$$
But WolframAlpha says it should be
$$= \frac 12 \left( x \sqrt{a^2 + x^2} + a^2 \ln \left| x + \sqrt{a^2+x^2} \right| \right)$$
What am I doing wrong?
| Notice, your result is correct because adding or subtracting a constant will make no difference in the result. This can also be obtained using integration by parts
let $$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^2}\int \ dx-\int\left(\frac{d}{dx}(\sqrt{x^2+a^2})\cdot \int 1\ dx\right)\ dx$$
$$I=x\sqrt{x^2+a^2}-\int\left(\frac{2x}{2\sqrt{x^2+a^2}}\cdot x\right)\ dx$$
$$I=x\sqrt{x^2+a^2}-\int\left(\frac{x^2}{\sqrt{x^2+a^2}}\right)\ dx$$
$$I=x\sqrt{x^2+a^2}-\int\frac{(x^2+a^2)-a^2}{\sqrt{x^2+a^2}}\ dx$$
$$I=x\sqrt{x^2+a^2}-\int \sqrt{x^2+a^2}\ dx+a^2\int\frac{1}{\sqrt{x^2+a^2}}\ dx$$
from (1)
$$I=x\sqrt{x^2+a^2}-I+a^2\int\frac{1}{\sqrt{x^2+a^2}}\ dx$$
$$I=\frac{1}{2}\left(x\sqrt{x^2+a^2}+a^2\int\frac{1}{\sqrt{x^2+a^2}}\ dx\right)\tag 2$$
Now, let $x=a\tan \theta\implies dx=a\sec^2\theta\ d\theta$ $$ \int\frac{1}{\sqrt{x^2+a^2}}\ dx= \int\frac{1}{\sqrt{a^2\tan^2\theta+a^2}}(a\sec^2\theta \ d\theta)$$
$$=\int \sec\theta\ d\theta=\ln|\sec\theta +\tan\theta|$$
$$=\ln\left|\tan\theta+\sqrt{\tan^2\theta+1}\right|=\ln\left|x+\sqrt{x^2+a^2}\right|+c$$
setting the value in (2), we get
$$I=\frac{1}{2}\left(x\sqrt{x^2+a^2}+a^2\ln\left|x+\sqrt{x^2+a^2}\right|\right)$$
| {
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How many sets contains 6 or its multiple given the following conditions?
MyApproach
I created
@Edit
S1={1,2,3,4,5} ...B)
S2={2,3,4,5,6}
S3={3,4,5,6,7}
S4={4,5,6,7,8}
S5={5,6,7,8,9}
S6={6,7,8,9,10}
S7={7,8,9,10,11} ....A)
S8={8,9,10,11,12}
From this information I analyzed that these $8$ sets have $6$ sets that have 6 or its multiple.
Thus $80$ sets will have $60$ elements which contain $6$ or its multiple.
Is my Ans right?Please correct me if I am wrong?
| Note that if $x \equiv 0$ mod$(6)$, the $S_x$ contains a multiple of $6$.
If $x \equiv 2$ mod$(6)$, then $x+4 \equiv 0 $ mod$(6)$, and so $S_x$ contains a multiple of $6$.
Similarly you can show that $S_x$ contains a multiple of $6$, whenever $x \equiv 3,4,5$ mod$(6)$.
It remains the case $x \equiv 1$ mod$(6)$. Indeed, if $x \equiv 1$ mod$(6)$, then $x+1\equiv 2$ mod$(6)$, $x+2\equiv 3$ mod$(6)$ , $x+3\equiv 4$ mod$(6)$ , and $x+4\equiv 5$ mod$(6)$ . Hence $S_x$ doesnt contain any of the multiples of $6$.
Thus the number of $S_x$s having no multiple of $6$, is the number of integers between $1$ and $80$ that have remainder one when divided by $6$. These integers have the form $6k+1$ with $k \in \mathbb{N}$.
Now find $K$, such tha $6K+1 \leq 80 $, so that $K \leq 13.1$, so $K=13$. Hence the number of such sets is 14. So $80-14= 66$. $66$ is the answer.
| {
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"timestamp": "2023-03-29T00:00:00",
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Epsilon and Delta proof for limit $\lim_{x\to-1} \frac{x+5}{2x+3}=4$ I have a function $f(x)=\frac{x+5}{2x+3}$ and want to show that the limit
$$\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|$$
Now put a bound on $\displaystyle \left|\frac{7}{2x + 3}\right|$ for $x$ in some interval around $-1$. There's no canonical choice for such an interval and the bound can be any finite, positive number.
Here's one option: for $|x+1| < 1/4$, that is $-5/4 < x < -3/4$, we have $$-5/2 + 3 < 2x + 3 < -3/2 + 3$$ and thus
$$|x+1| < \frac 14 \quad \Longrightarrow \quad \frac 23 < \frac{1}{2x + 3} < 2 \quad \Longrightarrow \quad \left|\frac{7}{2x + 3}\right| < 14$$
In other words,
$$|x+1| < \frac 14 \quad \Longrightarrow \quad \left| \frac{x + 5}{2x + 3} - 4 \right| < 14 \ |x+1| \quad - \color{red}{(*)}$$
Can you finish from here?
Added: The final choice of $\delta$ is not $1/4$.
The basic proof strategy is this:
If we can establish a bound
$$\left| \frac{x + 5}{2x + 3} - 4 \right| \leq M |x+1| \quad\text{ for some } M > 0$$
then given an arbitrary $\epsilon > 0$ choosing $\delta = \epsilon/M$ enables us to write the target deduction of
$$0 < |x + 1| < \delta \quad \Longrightarrow \quad \left| \frac{x + 5}{2x + 3} - 4 \right| \leq M |x+1| < M \cdot\frac\epsilon M = \epsilon $$
The wrinkle in this plan is that such a bound does not exist for all $x$; the expression blows up in any neighborhood of $-3/2$. However what we have now shown with $\color{red}{(*)}$ that there does exist such a bound if we start with $|x+1| < 1/4$. (The idea was to constrain $x$ away from such 'blow up' points and to center it around the value of $x = -1$. This is a standard trick in $\epsilon$-$\delta$ proofs; look again at the proof that $\lim_{x\to a} 1/x = 1/a$.)
Hence we modify the strategy and can finish the proof this way: given an arbitrary $\epsilon > 0$, choose $\displaystyle \delta = \min\left( \frac 14, \frac \epsilon{14}\right)$. With that choice of $\delta$ and by $ \color{red}{(*)}$,
$$0 < |x+1| < \delta \ \Rightarrow \ |x+1| < \frac 14 \ \Rightarrow \ \left| \frac{x + 5}{2x + 3} - 4 \right| < 14 \ |x+1|$$
We also have
$$0 < |x+1| < \delta \ \Rightarrow \ |x+1| < \frac\epsilon{14}$$
These last two deductions together imply that
$$0 < |x+1| < \delta \ \Rightarrow \ \left| \frac{x + 5}{2x + 3} - 4 \right| < 14 |x+1| < 14 \cdot \frac\epsilon{14} = \epsilon$$
as desired.
| {
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An identity for the factorial function A friend of mine was doodling with numbers arranged somewhat reminiscent of Pascal's Triangle, where the first row was $ 1^{n-1} \ \ 2^{n-1} \ \cdots \ n^{n-1} $ and subsequent rows were computed by taking the difference of adjacent terms. He conjectured that the number we get at the end is $ n! $ but I've not been able to prove or disprove this. The first few computations are given below:
$$
\begin{pmatrix}
1 \\
\end{pmatrix}
$$
$$
\begin{pmatrix}
1 & & 2 \\
& 1 & \\
\end{pmatrix}
$$
$$
\begin{pmatrix}
1 & & 4 & & 9 \\
& 3 & & 5 & \\
& & 2 & & \\
\end{pmatrix}
$$
$$
\begin{pmatrix}
1 & & 8 & & 27 & & 64 \\
& 7 & & 19 & & 37 & \\
& & 12 & & 18 & & \\
& & & 6 & & & \\
\end{pmatrix}
$$
$$
\newcommand\pad[1]{\rlap{#1}\phantom{625}}
\begin{pmatrix}
1 & & 16 & & 81 & & 256 & & 625 \\
& 15 & & 65 & & 175 & & 369 & \\
& & 50 & & 110 & & 194 & & \\
& & & 60 & & 84 & & & \\
& & & & 24 & & & & \\
\end{pmatrix}
$$
I attempted to write down the general term and tried to reduce that to the required form. The general term worked out as
$$
\sum_{i=0}^n (-1)^{n-i} \binom{n}{i} (i+1)^{n}.
$$
I tried applying various identities of the binomial coefficients but I'm barely making any progress. Any help would be appreciated.
Small note: If I instead start with the first row as $ 0^{n} \ \ 1^{n} \ \cdots \ n^{n} $ then I still get $n!$ at the end of the computation, and the general formula in this case works out as
$$
\sum_{i=0}^n (-1)^{n-i} \binom{n}{i} i^{n}.
$$
In fact, we can start with any $n$ consecutive natural numbers, each raised to the $(n-1)$th power, and we still get $n!$ at the end of the computation.
| Let's say we want to prove that iterating $n^3$ three times gives us $3!$. (Assume we already know that your thing works for $n^2$, $n$, and $1$.)
Well, if you iterate $n^3$ once, you get $(n+1)^3-n^3=3n^2+3n+1$. We also know that iterating $3n^2$ two times gives us $3\times2!=3!$, and we know that iterating $3n$ and $1$ two times both give us $0$ (since they become constant after one or zero iterations respectively).
Thus, iterating $3n^2+3n+1$ two times gives us $3!$, and therefore iterating $n^3$ three times gives us $3!$.
Can you generalize this argument?
| {
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Solving $\sqrt\frac{x^2+2x-2}{x^2+3x+2} + \sqrt\frac{x^2+3x+2}{x^2+2x-2}=\frac52 $ Got these two problems on an exam recently and was unsure if I managed to answer them correctly:
$$\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+2}{x^2+2x-2}=\frac52 $$
Anyone able to provide the correct answer or method of solving these?
| Hint...for the second one, set $u$ as the first square root expression and solve the quadratic $$u+\frac 1u=\frac 52$$
| {
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"answer_id": 0
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Finding the derivative of $(\frac{a+x}{a-x})^{\frac{3}{2}}$ This is a very simple problem, but I am stuck on one step:
Differentiate $(\frac{a+x}{a-x})^{\frac{3}{2}}$
Now, this is what I have done:
$$
(\frac{a+x}{a-x})^{\frac{3}{2}} \\
\implies \frac{\delta}{\delta y}\frac{f}{g} \\
\implies gf' = (a-x)^{\frac{3}{2}} \times \frac{3}{2} (a+x)^{\frac{1}{2}} \times 2 \\
\implies fg' \implies (a+x)^{\frac{3}{2}} \times \frac{3}{2} (a-x)^{\frac{1}{2}} \times 0 = 0 \\
\implies \frac{(a-x)^{\frac{3}{2}} \times 3 (a+x)^{\frac{1}{2}}}{(a-x)^3}\\
\implies \frac{(a-x)^{\frac{3}{2}} - 3\sqrt{a+x}}{(a-x)^3}
$$
But the answer is:
$$
\frac{3\times a (a+x)^{\frac{1}{3}}}{(a-x)^{\frac{5}{2}}}
$$
WolframAlpha shows:
$$
\frac{3a \sqrt{\frac{a+x}{a-x}}}{(a-x)^2}
$$
Another Answer (Somehow I got this):
$$
\frac{3 \sqrt{\frac{a+x}{a-x}}}{2(a-x)}
$$
==================
EDIT 1:
What about:
$$
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}\\
\implies \frac{3}{2}u^{\frac{1}{2}} \hspace{0.5cm} ; \hspace{0.5cm} \frac{(0+1)\times (a-x) - [ -1 (a+x) ]}{(a-x)^2} \\
\implies \frac{2a}{(a-x)^2} \\
\implies \frac{3}{2}\sqrt{\frac{2a}{(a-x)^2}} = \frac{3}{2} \times \frac{\sqrt{2a}}{a-x}
$$
| So, there was something flawed in my calculation:
$$
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{3}{2} \times (\frac{a+x}{a-x})^{\frac{1}{2}} \\
\frac{du}{dx} = \frac{2a}{(a-x)^2} \\
y' = \frac{dy}{du} \times \frac{du}{dx} \\
...\\
\frac{3a(a+x)^{\frac{1}{2}}}{(a-x)^{\frac{5}{2}}}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1514719",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Solve $ \left|\frac{x}{x+2}\right|\leq 2 $ I am experiencing a little confusion in answering a problem on Absolute Value inequalities which I just started learning. This is the problem: Solve:
$$
\left|\frac{x}{x+2}\right|\leq 2
$$
The answer is given to be $x\leq-4$ or $x \geq-1$
This is my attempt to solve the problem:
By dividing, $\left|\frac{x}{x+2}\right|\leq 2$ is equivalent to $\left |1-\frac{2}{x+2}\right|\leq2$ which is also equivalent to $\left |\frac{2}{x+2}-1\right|\leq2$
So, $-2\leq\frac{2}{x+2}-1\leq2$ which is equivalent to $-\frac{1}{2}\leq\frac{1}{x+2}\leq\frac{3}{2}$
Case 1: $x+2>0$. Solving $-\frac{1}{2}\leq\frac{1}{x+2}\leq\frac{3}{2}$, I get $x\geq-4$ and $x\geq-\frac{4}{3}$ which is essentially $x\geq-\frac{4}{3}$.
Case 2: $x+2<0$. Solving $-\frac{1}{2}\times(x+2)\geq{1}\geq\frac{3}{2}\times(x+2)$, I get $x\leq-4$ and $x\leq-\frac{4}{3}$ which is essentially $x\leq-4$.
So, the solutions are: $x\leq-4$ or $x\geq-\frac{4}{3}$.
I couldn't get $x \geq-1$ as a solution. Did I do anything wrong? The book I am using is Schaum's Outlines-Calculus. Another question I would like to ask is that am I using 'and' and 'or' correctly in the above attempt to solve the problem? I have had this problem many times.
| Given that, $$\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: If $\color{blue}{-2<x<0}$ $$-x\le 2(x+2)\iff x\ge -\frac{4}{3}\ \ \ $$
but, $x<0$ hence, $$-\frac{4}{3}<x<0\iff \color{red}{x\in\left[-\frac{4}{3}, 0\right)}$$
Case 3: If $\color{blue}{x\ge 0}$ $$x\le 2(x+2)\iff x\ge -4\ \ \ $$
but, $x\ge 0$ hence, $$x\ge 0\iff \color{red}{x\in\left[0, \infty\right)}$$
Hence, the combining the above cases, the complete solution is given as
$$\color{red}{x\in (-\infty, -4]\cup \left[-\frac{4}{3}, \infty\right) }$$
| {
"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 $t+u\ge0$, the inequality is equivalent to
$$
(t+u)^2\le(t^2+1)(u^2+1)
$$
that is,
$$
t^2+2tu+u^2\le t^2u^2+t^2+u^2+1
$$
and so equivalent to
$$
0\le t^2u^2-2tu+1
$$
(by transporting terms to the right-hand side) which becomes
$$
0\le(tu-1)^2
$$
which is true.
Don't forget the $2$ when you square! You basically did $(a+b)^2=a^2+ab+b^2$ and this is where you got stuck.
| {
"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}\\
\frac{x+2-x-1}{\sqrt{x+2}+\sqrt{x+1}}&\neq \frac{x+1-x}{\sqrt{x +1}+\sqrt{x}}\\
\frac{1}{\sqrt{x+2}+\sqrt{x+1}}&\neq \frac{1}{\sqrt{x +1}+\sqrt{x}}\\
\sqrt{x +1}+\sqrt{x} &\neq \sqrt{x+2}+\sqrt{x+1}\\
\sqrt{x} &\neq \sqrt{x+2}\\
\end{align}
*
*Is my proof correct?
*I'm interested in more ways of proving it.
| 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 one $x$,
since
$a+c-2b = 0$
and
$(a-4b+c)^2-4ac
=2^2-0
=4
$,
this does not hold for
any $x$.
then
Suppose
$\sqrt{x+a}-\sqrt{x+b}
=\sqrt{x+b}-\sqrt{x+c}
$,
or
$\sqrt{x+a}+\sqrt{x+c}
=2\sqrt{x+b}
$.
Squaring,
$x+a+x+c+2\sqrt{(x+a)(x+c)}
=4(x+b)
$
or
$2x-a+4b-c
=2\sqrt{(x+a)(x+c)}
$.
Squaring again,
$4x^2-4x(a-4b+c)+(a-4b+c)^2
=4(x+a)(x+c)
=4(x^2+x(a+c)+ac)
=4x^2+4x(a+c)+4ac
$
or
$4x(a+c+a-4b+c)
=(a-4b+c)^2-4ac
$
or
$4x(2a+2c-4b)
=(a-4b+c)^2-4ac
$
or
$8x(a+c-2b)
=(a-4b+c)^2-4ac
$.
If this is true for
more than one $x$,
then
$a+c-2b = 0$
and
$(a-4b+c)^2-4ac
= 0
$.
From the first equation,
$2b = a+c$.
Substituting this
in the second equation,
$4ac
=(a-4b+c)^2
=(a+c-2(a+c))^2
=(a+c)^2
=a^2+2ac+c^2
$
or
$0
=a^2-2ac+c^2
=(a-c)^2
$,
so that
$a=c$.
Therefore,
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
$.
| {
"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
$$3^{3n+4}+2^{n+2}=3^{3n+1}3^3+2^{n+1}2$$
I tried to prove that that the difference is a multiple of $5$.
$$3^{3n+1}3^3+2^{n+1}2-3^{3n+1}+2^{n+1}=2(3^{3n+1}\cdot 13+2^n)$$
Therefore it's enough to prove that $3^{3n+1}\cdot 13+2^n$ is a multiple of $5$. But if I do again this method applied to this "new problem" is get something similar. I think that there exist a different method to do this using induction.
| 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
} |
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.
$\lim _{x\to 1}\left(\frac{\left(\sqrt[3]{x+7}-2\right)\cdot \left(\sqrt[3]{\left(x+7\right)^2}+2\sqrt[3]{x+7}+4\right)}{\left(2x^2+3x-5\right)\:\cdot \left(\sqrt[3]{\left(x+7\right)^2}+2\sqrt[3]{x+7}+4\right)}\right)=\lim _{x\to 1}\left(\frac{x+7-8}{\left(2x^2+3x-5\right)\:\cdot \left(\sqrt[3]{\left(x+7\right)^2}+2\sqrt[3]{x+7}+4\right)}\right)=\lim _{x\to 1}\left(\frac{x-1}{\left(2x^2+3x-5\right)\:\cdot \left(\sqrt[3]{\left(x+7\right)^2}+2\sqrt[3]{x+7}+4\right)}\right)$
Here's what to do with the denominator?
and whether I chose the idea of a solution?
| 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,
"answer_id": 0
} |
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 and exponential limits?
Any kind of help would be appreciated. I gave the problem simply as an example to reflect the above confusion.
EDIT: Even though tag says so, you are still allowed to use l hospital's rule to help evaluate the limit.
The question is edited now. It is e^(2x^4) in denominator.
Again, thanks a lot for taking your time to reply to my doubt.
| 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:
$$\lim_{x\to 0}\frac{\sin x^4-x^4\cos x^4+x^{20}}{x^4(e^{2x}-1-2x^4)} = \lim_{x\to 0}\frac{\left(x^4-\frac{(x^4)^3}{3!} + \frac{(x^4)^5}{5!} - \ldots\right)-x^4\left(1-\frac{(x^4)^2}{2!} + \frac{(x^4)^4}{4!} - \ldots\right)+x^{20}}{x^4\left(\left(1 + 2x + \frac{(2x)^2}{2} + \frac{(2x)^3}{6} + \ldots\right) +-1-2x^4\right)}= \frac{\frac{x^{12}}{12} + \ldots}{2x^5 + \ldots + \frac{256x^{12}}{40320} + \ldots} = 0$$
If you notice, our leading coefficient for the numerator is $x^{12}$ and for the denominator is $x^5$.
I don't believe the answer is $1$. Therefore, the answer is $0$. Comment if you have questions.
| {
"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
} |
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 $\displaystyle \frac{1}{2}(u-v)^2+ \left(\sqrt{2-u^2}-\frac{9}{v}\right)^2 \geq \frac{1}{4}\left(u-v+\sqrt{2-u^2}-\frac{9}{v}\right)^2$ by Chebychev inequality. Is anyone is able to give me a hint how to finish it?
I think I have to use the Arithmetic and Geometric Means Inequalities.
| 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 $\sqrt{\frac{2v^4}{81+v^4}}\lt u\lt \sqrt{2}$. So, $F$ is minimized when $u=\sqrt{\frac{2v^4}{81+v^4}}$.
Thus, we can have
$$F(u,v)\ge F\left(\sqrt{\frac{2v^4}{81+v^4}},v\right)=\frac{81+v^4}{v^2}-2\sqrt 2\sqrt{\frac{81+v^4}{v^2}}+2$$
By the way, letting $\frac{81+v^4}{v^2}=k$ gives
$$(v^2)^2-kv^2+81=0.$$
Considering the discriminant gives that $(-k)^2-4\cdot 81\ge 0\Rightarrow k\ge 18$.
Since $x-2\sqrt 2\sqrt x$ is increasing for $x\gt 2$, we can see that $F$ is minimized when $\frac{81+v^4}{v^2}=18$, i.e. $v=3$.
Thus, the minimum of the expression is $\color{red}{8}$ for $(u,v)=(1,3)$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1530179",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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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:
$a_k = (-1)^k\sigma _k(x_1,x_2,x_3,x_4), k\in \{1,2,3,4\}$, where $\sigma _k$ is the $k$-th elementary symmetrical polynomial. Therefore:
$x_1+x_2+x_3+x_4 = 2$
$x_1x_2+x_1x_3+x_1x_4+x_2x_3+x_2x_4+x_3x_4 = 0\ (*)$
$x_1x_2x_3 +x_1x_2x_4+x_1x_3x_4+x_2x_3x_4 = 1$
$x_1x_2x_3x_4 = 2/7$
Now how to determine the sum of the cubed roots?
$2^3 = 8= (x_1+x_2+x_3+x_4)(x_1+x_2+x_3+x_4)^2 = (x_1+x_2+x_3+x_4)(x_1^2+x_2^2+x_3^2+x_4^2 + 2(*))$
Here's where things go out of hand:
$(x_1+x_2+x_3+x_4)(x_1^2+x_2^2+x_3^2+x_4^2) = (x_1^3 + x_2^3 + x_3^3+x_4^3) + x_1^2(x_2+x_3+x_4)+x_2^2(x_1+x_3+x_4)+x_3^2(x_1+x_2+x_4)+x_4^2(x_1+x_2+x_3) = 8$
What should I do here?
| 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
$$AB=E+3C.$$
So,
$$x_1^3+x_2^3+x_3^3+x_4^3=A^3-3(AB-3C)-6C=\color{red}{11}.$$
| {
"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
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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 class of the sum/product", however I am unsure how to apply this properly, any suggestions?
| $$1^6+2^6+3^6+4^6+5^6+\dots+100^6\equiv20(1^6+2^6+3^6+4^6+5^6)\\\equiv0\text{ (mod }5)$$
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, $2^6\equiv7^6\equiv12^6\equiv\dots\equiv97^6\text{ (mod }5)$, and the same can be done with the numbers which are congruent to 3, 4 and 0 modulo 5.
Hence, we can replace all these terms with $1^6+2^6+3^6+4^6+5^6$ repeated 20 times and therefore the sum is a multiple of 20 and is divisible by 5 exactly. (Comment made by David K.)
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1535216",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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"answer_id": 0
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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
$$\lim_{x \to 0} \bigg(\frac{\sin x}{x} \bigg)^{\frac{1}{x^2}} \approx \lim_{x \to 0} \bigg(\frac{x - \frac{x^3}{6}}{x} \bigg)^{\frac{1}{x^2}} \approx \lim_{x \to 0} \bigg(1 - \frac{x^2}{6} \bigg)^{\frac{1}{x^2}}$$ How can we say answer that is $e$ in the power of $-\frac{1}{6}$. I want some proving of that fact.
Thank you.
| 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",
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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}-bx+\frac{bx^{3}}{6}}{x^{3}} \\
&=&\frac{(1+a-b)x+x^{3}(\frac{b-3a}{6})}{x^{3}} \\
&=&\frac{(1+a-b)}{x^{2}}+\frac{b-3a}{6}
\end{eqnarray*}
It suffices to choose $a$ and $b$ such that
\begin{equation*}
(1+a-b)=0\ \ \ \ \ and\ \ \ \ \ \ b-3a=6
\end{equation*}
that is
\begin{equation*}
a=-\frac{5}{2},\ \ \ and\ \ b=-\frac{3}{2}
\end{equation*}
| {
"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:
$$\begin{align}
\frac{\sqrt{x+7}-3}{\sqrt{x^{2}+5}-x-1}&=\frac{(\sqrt{x+7}-3)(\sqrt{x+7}+3)(\sqrt{x^{2}+5}+x+1)}{(\sqrt{x^{2}+5}-x-1)(\sqrt{x^{2}+5}+x+1)(\sqrt{x+7}+3)}\\
&=\frac{(x-2)(\sqrt{x^{2}+5}+x+1)}{(x^2+5-(x+1)^2)(\sqrt{x+7}+3)}\\
&=-\frac{\sqrt{x^{2}+5}+x+1}{2(\sqrt{x+7}+3)}.
\end{align}$$
| {
"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 $k=6^{n-3}$ is a good choice.
| {
"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[2 Blue] = Pr[2 White] = $ (\frac{5}{20})^2 $
Pr[Event] = $ (\frac{10}{20})^2\cdot(\frac{5}{20})^4 \approx 0.001$
| 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 of the remaining four positions can be filled with a white marble is $\binom{4}{2}$. The number of ways the remaining two positions can be filled with two blue marbles is $\binom{2}{2}$.
As you determined,
\begin{align*}
P(\text{two red}) & = \left(\frac{10}{20}\right)^2 = \left(\frac{1}{2}\right)^2\\
P(\text{two white}) & = \left(\frac{5}{20}\right)^2 = \left(\frac{1}{4}\right)^2\\
P(\text{two blue}) & = \left(\frac{5}{20}\right)^2 = \left(\frac{1}{4}\right)^2
\end{align*}
Hence, the probability that two red marbles, two white marbles, and two marbles are selected when six marbles are selected from ten marbles, five white marbles, and five blue marbles with replacement is
$$\binom{6}{2}\binom{4}{2}\binom{2}{2}\left(\frac{1}{2}\right)^2\left(\frac{1}{4}\right)^2\left(\frac{1}{4}\right)^2$$
Note. The number
$$\binom{6}{2}\binom{4}{2}\binom{2}{2} = \frac{6!}{2!4!} \cdot \frac{4!}{2!2!} \cdot \frac{2!}{2!0!} = \frac{6!}{2!2!2!}$$
is the multinomial coefficient of the term $r^2b^2g^2$ in the multinomial expansion $(r + b + g)^6$.
| {
"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
} |
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