Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
|---|---|---|
counting Number of matrices We have a $2 \times 2$ matrix.
We are given the trace of the matrix as $N$.
Also, all elements of the matrix are greater than or equal to $1$.
And, the determinant of matrix is $\geq 1$.
QUESTIONS:
If given $N$, what are number of $2 \times 2$ matrices that can be formed in consideration to the given constraints?
for example we have $N=3$ then there are $2$ matrices which satisfy the given constraints i.e
1 1
1 2
and
2 1
1 1
| $$A = \left[\begin{array}{cc}x & \frac{x(N-x)-D}{y} \\ y & N - x\end{array}\right]$$
This is the prototype of your matrix where I'm assuming that $N, x, y$ and $D$ are all natural number.
It's easy to show that $\det(A) = D$.
A first constraint is: $x \in \{1, \ldots , N-1\}$.
The second constraint is: $D < x(N-x)$.
Having fixed $x$, then $D \in \{1, \ldots, x(N-x) - 1\}$
The third constraint is: $y ~\text{ divides } ~x(N-x) - D $
Let's indicate with $\alpha(n)$ the number of divisors of $n$.
Then, the number of such matrices having fixed $N, x$ and $D$ is:
$$\alpha(x(N-x)-D)$$
The number of such matrices having fixed $N$ and $x$ is:
$$\sum_{D=1}^{[x(N-x)-1]}\alpha(x(N-x)-D)$$
Finally, the number of such matrices having fixed only $N$ is:
$$\sum_{x=1}^{N-1}\sum_{D=1}^{[x(N-x)-1]}\alpha(x(N-x)-D)$$
The function $\alpha$ is well explained here (denoted with $\sigma_0$)
Using $N=3$, you get:
$$\sum_{x=1}^{2}\sum_{D=1}^{[x(3-x)-1]}\alpha(x(3-x)-D) = $$
$$=\sum_{D=1}^{[1(3-1)-1]}\alpha(1(3-1)-D) + \sum_{D=1}^{[2(3-2)-1]}\alpha(2(3-2)-D)=$$
$$=\sum_{D=1}^{1}\alpha(2-D) + \sum_{D=1}^{1}\alpha(2-D)=$$
$$=\alpha(1) + \alpha(1) = 1 + 1= 2$$
Case $N=4$
$$\sum_{x=1}^{3}\sum_{D=1}^{[x(4-x)-1]}\alpha(x(4-x)-D) = $$
$$=\sum_{D=1}^{2}\alpha(3-D) + \sum_{D=1}^{3}\alpha(4-D) + \sum_{D=1}^{2}\alpha(3-D) = $$
$$=2(\alpha(3-1) + \alpha(3-2)) + \alpha(4-1) + \alpha(4-2) + \alpha(4-3) = $$
$$=2(\alpha(2) + \alpha(1)) + \alpha(3) + \alpha(2) + \alpha(1) = $$
$$=2(2 + 1) + 2 + 2 + 1 = 11$$
$$\left[\begin{array}{cc}1 & 1\\ 1 & 3\end{array}\right], \left[\begin{array}{cc}3 & 1\\ 1 & 1\end{array}\right],
\left[\begin{array}{cc}1 & 2\\ 1 & 3\end{array}\right],
\left[\begin{array}{cc}3 & 2\\ 1 & 1\end{array}\right],
\left[\begin{array}{cc}1 & 1\\ 2 & 3\end{array}\right],
\left[\begin{array}{cc}3 & 1\\ 2 & 1\end{array}\right],
\left[\begin{array}{cc}2 & 1\\ 1 & 2\end{array}\right],
\left[\begin{array}{cc}2 & 2\\ 1 & 2\end{array}\right],
\left[\begin{array}{cc}2 & 1\\ 2 & 2\end{array}\right],
\left[\begin{array}{cc}2 & 3\\ 1 & 2\end{array}\right],
\left[\begin{array}{cc}2 & 1\\ 3 & 2\end{array}\right].$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Is $(A+B)^2 = A^2 + B^2$ if $A$ and $B$ are matrices If $A$ and $B$ are matrices is $(A+B)^2 = A^2 + B^2$?
I thought because, $AB + BA = AB - AB = 0$.
| $$(A+B)^2 = (A+B)(A+B) = A^2 + AB + BA + B^2 \neq A^2 + B^2$$
$(A+B)^2 = A^2 + B^2$ if and only if $AB + BA = 0 \iff AB = -BA$, but this is not true in general.
| {
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What particular solution should I guess for this relation? Just trying to solve a non-homogeneous recurrence relation:
$$f(n) = 2f(n-1) + n2^n$$
$$f(0) = 3$$
Characteristic equation is:
$$f(n) - 2f(n-1) = 0$$
$$a-2 = 0$$
$$a = 2$$
Homogeneous solution is:
$$f_H(n) = b_0\cdot (2)^n$$
Right-hand side is:
$$n2^n$$
What is the particular guess I should be taking based on this?
| Just solve the bugger by using generating functions. Define $F(z) = \sum_{n \ge 0} f(n) z^n$, adjust indices so there aren't subtractions:
$$
f(n + 1) = 2 f(n) + (n + 1) 2^{n + 1}
$$
Multiply by $z^n$, sum over $n \ge 0$, recognize:
\begin{align}
\sum_{n \ge 0} f(n + 1) z^n &= \frac{F(z) - f(0)}{z} \\
\sum_{n \ge 0} (2 z)^n &= \frac{1}{1 - 2 z} \\
\sum_{n \ge 0} n 2^n z^{n - 1} &= \frac{\mathrm{d}}{\mathrm{d} z} \frac{1}{1 - 2 z} \\
\sum_{n \ge 0} (n + 1) 2^{n + 1} z^n
&= \frac{2}{(1 - 2 z)^2}
\end{align}
This gives:
$$
\frac{F(z) - 3}{z} = 2 F(z) + \frac{2}{(1 - 2 z)^2}
$$
Solving for $F(z)$, as partial fractions:
$$
F(z) = \frac{3}{1 - 2 z} - \frac{1}{(1 - 2 z)^2} + \frac{1}{(1 - 2 z)^3}
$$
Using the generalized binomial theorem:
\begin{align}
f(n) &= 3 \binom{-1}{n} (-2)^n - \binom{-2}{n} (-2)^n + \binom{-3}{n} (-2)^n \\
&= 3 \cdot 2^n - (n + 1) 2^n + \frac{(n + 2) (n + 1)}{2} 2^n \\
&= (n^2 + n + 6) \cdot 2^{n - 1}
\end{align}
At first glance, this checks, as $f(0) = 3$.
| {
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"timestamp": "2023-03-29T00:00:00",
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Fair die rolled twice, random variables A fair die is tossed twice. Let $d_1=\text{value of die on roll 1}$ and $d_2=\text{value of die on roll 2}$ Let $X=d_1+d_2$, the sum of the faces; $Y=\max\left\{d_1,d_2\right\}$, the maximum of the two faces; and $Z = |d_1-d_2|$.
What is the value of $X, Y, Z,$ and $ W = XZ$ for each outcome $\omega\in\Omega$?
What is $\mathrm{Range}(X)$, $\mathrm{Range}(Y)$, $\mathrm{Range}(Z)$ and $\mathrm{Range}(W)$?
Describe the partitions $A_X$ and $A_Z$ induced by these random variables.
From my understanding $\left(d_1,d_2\right)=\omega\in\Omega=\left\{1,2,3,4,5,6\right\}^2=\left\{\left(1,1\right),\left(1,2\right),\dotsc,\left(6,5\right),\left(6,6\right)\right\}$
| With the sample space $\Omega=\left\{1,2,3,4,5,6\right\}^2=\left\{\left(1,1\right),\left(1,2\right),\dotsc,\left(6,5\right),\left(6,6\right)\right\}$, we can easily find what is being asked.
The range of a Random Variable is the set of values it can assume, so:
$\mathrm{Range}\left(X\right)=\left\{2,3,\dotsc,11,12\right\}$ as these are the possible sums of the sides of a fair die rolled twice.
$\mathrm{Range}\left(Y\right)=\left\{1,\dotsc,6\right\}$ as whatever is rolled, any of these values could be the maximum.
$\mathrm{Range}\left(Z\right)=\left\{0,1,\dotsc,5\right\}$ as these are the possible differences between the values of the two rolls, note that we have $0$ when $d_1=d_2$ and 6 isn't possible as the largest difference will occur when we get $\left(6,1\right)$ or $\left(1,6\right)$.
$\mathrm{Range}\left(W\right)$ is a little more difficult as we need to calculate $W\left(\omega\right)$, $\forall\omega\in\Omega$. When done, we find that $\mathrm{Range}\left(W\right)=\left\{0,3,5,7,8,9,11,12,15,16,20,21,24,27,32,35\right\}$
A partition of a random variable is the set of points $\omega\in\Omega$ that give rise to each possible value that the random variable can take.
The partitions $A_X$ and $A_Z$ can now be found
\begin{align*}
A_{X_1=2} &= \left\{\left(1,1\right)\right\} \\
A_{X_2=3} &= \left\{\left(1,2\right),\left(2,1\right)\right\} \\
A_{X_2=4} &= \left\{\left(1,3\right),\left(2,2\right),\left(3,1\right)\right\} \\
&\vdots \\
A_{X_{10}=11} &= \left\{\left(5,6\right),\left(6,5\right)\right\} \\
A_{X_{11}=12} &= \left\{\left(6,6\right)\right\} \\
\end{align*}
and
\begin{align*}
A_{Z_1=0} &= \left\{\left(1,1\right),\left(2,2\right),\dotsc,\left(5,5\right),\left(6,6\right)\right\} \\
A_{Z_2=1} &= \left\{\left(1,2\right),\left(2,3\right),\dotsc,\left(3,2\right),\left(2,1\right)\right\} \\
&\vdots \\
A_{Z_6=5} &= \left\{\left(1,6\right),\left(6,1\right)\right\} \\
\end{align*}
In response to the comment about why the value $5\times12\not\in\mathrm{Range}\left(W\right)$, we see that $W\left(\omega\right)=X\left(\omega\right)\times Z\left(\omega\right)$, so therefore it isn't possible to obtain $60=12\times5$ because the only value $\omega\in\Omega$ such that $X\left(\omega\right)=12$ is $\omega=\left(6,6\right)$ but clearly $Z\left(\omega\right)=0$ if $\omega=\left(6,6\right)$. Similarly for $Z\left(\omega\right)=5$, we need $\omega=\left(1,6\right)$ or $\omega=\left(6,1\right)$ and then clearly $X\left(\omega\right)=7$ for both of these $\omega$.
In order to compute $\mathrm{Range}\left(W\right)$, it is necessary to check the value of $W\left(\omega\right)=X\left(\omega\right)\times Z\left(\omega\right)$ for all $\omega\in\Omega$ where $\Omega$ is defined as above.
| {
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Finding circumcentre Tangents are draw from $P(2,3)$ to $x^2+y^2=4$ meeting at $Q,R$ on circle. Parallelogram $PQSR$ is completed. Find the circumcentre of triangle $QSR$.
My attempt: Clearly, the parallelogram is a rhombus. $S$ will be mirror image of $P$ in chord of contact. But, I want to know a good way of finding the circumcentre other than it lies on $PS$ and dropping perpendiculars and other boring stuff. Maybe a nice geometrical solution? Or combining it with coordinate?
| Here's an overview ...
Note that the circumcenter (call it $C$) of $\triangle PQR$ is easy to find: It's the intersection of $\overline{PS}$ (a line through the origin) and the (horizontal) perpendicular bisector of (vertical) segment $\overline{PQ}$ with known endpoints. Now, the circumcenter you want (call it $D$) is clearly the reflection of $C$ in $\overline{QR}$; this implies that $\overline{CQ} \cong \overline{DQ}$. Thus, the circle with center $Q$ through $C$ meets $\overline{PS}$ at $D$.
Details ...
$$\overline{PS}:\;\; y = \frac{3}{2}x,\qquad\text{perp bis of } \overline{PQ}:\;\; y = \frac{3}{2}\qquad \implies \qquad C = \left( 1, \frac{3}{2} \right)$$
Then, writing $D = \left(d, \frac{3}{2}d\right)$, we have
$$\begin{align}
|\overline{DQ}|^2 = |\overline{CQ}|^2 \quad &\implies \quad\left(d-2\right)^2 + \left(\frac{3}{2}d - 0\right)^2 = \left( 1 - 2 \right)^2 + \left( \frac{3}{2} - 0 \right)^2 \\[6pt]
&\implies 13 d^2 - 16 d + 3 = 0 \\[6pt]
&\implies d = \frac{3}{13} \qquad (\text{or } d = 1, \text{ but that gives } C ) \\[6pt]
&\implies D = \left(\frac{3}{13}, \frac{9}{26}\right) \qquad \square
\end{align}$$
| {
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Are there contradictions in math? Someone told me that math has a lot of contradictions.
He said that a lot of things are not well defined.
He told me two things that I do not know.
*
*$1+2+3+4+...=-1/12$
*what is infinity $\infty$?
Since I am not a math specialist and little. How to disprove the previous two?
How can I convince him?
| There is already great general answer by Ittay Weiss, so I will try a different approach. In fact, I will try to explain a bit the infinite sum you stated. As for the infinity, one could write a lot about it (mainly because there are multiple infinities, each with different properties), and I couldn't even hope to fit it here. Unfortunately, I don't know any sources or text about infinity that I would like to recommend.
Consider the following infinite sum
$$G(x) = x^0 + x^1 + x^2 + x^3 + \ldots = \sum_{i = 0}^{\infty}x^i.$$
We have no idea what such a sum might mean (without any proper definition, which wasn't stated). However, suppose that we could manipulate it in a way similar to numbers. Let's start with multiplying by $x$
$$x\cdot G(x) = x^1 + x^2 + x^3 + x^4 + \ldots = \sum_{i = 0}^{\infty}x\cdot x^{i},$$
and then add $x^0 = 1$ (we assume that $x \neq 0$)
$$1 + x\cdot G(x) = x^0 + x^1 + x^2 + x^3 + x^4 + \ldots = x^0 + \sum_{i = 1}^{\infty}x^{i} = \sum_{i = 0}x^i.$$
We got the same expression! In other words, we don't know what $G(x)$ means yet, but if it does mean anything similar to a number, then this number has to satisfy
$$G(x) = 1+x\cdot G(x),\quad \text{ that is, }\quad G(x) = \frac{1}{1-x}.$$
Indeed, there is a way to specify what $G(x)$ means for $|x| < 1$ and then, e.g. $$G\left(\frac{1}{2}\right) = 1+\frac{1}{2} + \frac{1}{4} + \frac{1}{8} + \ldots = \frac{1}{1-\frac{1}{2}} = 2.$$
On the other hand, for other values of $x$ we get peculiar results, like
$$G(2) = 1 + 2 + 4 + 8 + \ldots = \frac{1}{1-2} = -1,$$
which would mean that sum of positive numbers got negative. However, this is because we didn't defined what $G(2)$ would mean, we only said we could do it for $|x| < 1$. Nevertheless, there is a way of dealing with this (e.g. see $p$-adic numbers), but for $x=1$ we get even stranger result, that is,
$$G(1) = 1 + 1 + 1 + \ldots = \frac{1}{1-1} = \frac{1}{0} =\ ?!$$
Similar situation happens with your sequence. Consider
\begin{align}
S(x) &= 0\cdot x^0 + 1\cdot x^1 + 2\cdot x^2 + 3\cdot x^3 + \ldots = \sum_{i = 0}^{\infty} i \cdot x^i \\
x\cdot S(x) &= \color{white}{0 \cdot x^{?}} + 0 \cdot x^1 + 1 \cdot x^2 + 2 \cdot x^3 + \ldots = \sum_{i=0}^{\infty} i\cdot x^{i+1} \\
x\cdot G(x) + x\cdot S(x) &= 0\cdot x^0 + 1\cdot x^1 + 2\cdot x^2 + 3\cdot x^3 + \ldots = \sum_{i = 0}^{\infty} i \cdot x^i \\
S(x) &= x\cdot G(x) + x \cdot S(x) \\
S(x) &= \frac{x \cdot G(x)}{1-x} = \frac{x}{(1-x)^2}
\end{align}
which is true for $|x| < 1$ (i.e. we can define the infinite sum in a way that would make the above equality true with an appropriate $x$), but with $x = 1$ we get
$$ S(1) = \frac{1}{(1-1)^2} = \frac{1}{0} = \ ?! $$
On the other hand, observe that
\begin{align}
S(1) &= 1 + 2 + 3 + 4 + 5 + 6 + \ldots \\
4S(1) &= \color{white}{0} + 4 + \color{white}{0} + 8 + \color{white}{0} + 12 + \ldots \\
-3S(1) &= 1 - 2 + 3 - 4 + 5 - 6 + \ldots \\
-S(-1) &= 1 - 2 + 3 - 4 + 5 - 6 + \ldots = -\frac{-1}{\big(1-(-1)\big)^2} = \frac{1}{4} \\
-3S(1) &= -S(-1) = \frac{1}{4} \\
S(1) &= -\frac{1}{12}
\end{align}
which means that, if there is any number-like interpretation of $S(1)$ (and $S(-1)$ which we have used), then it has to be equal $-\frac{1}{12}$. However, note that we didn't argue whether there actually is such an interpretation. Nevertheless, those interpretations of $S(x)$ which do allow $S(1)$ also explain why the sum of positive values might be considered negative (i.e. given particular interpretation it stops being strange). If you would like to know more about it, you can start here.
I hope this helps $\ddot\smile$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Definite Integral $\int_{-\infty}^{\infty}\frac{x\sin x}{(x^2+a^2)(x^2+b^2)}\,\mathrm{d}x$ $$\int_{-\infty}^{\infty}\frac{x\sin x}{(x^2+a^2)(x^2+b^2)}\,\mathrm{d}x$$
This is easy to evaluate with complex analysis but is there an elementary way (substitution, partial fractions, integration by parts)?
| Note
$$\int_0^\infty e^{-xt}\cos(at)dt=\frac{x}{x^2+a^2}$$
and hence
$$\int_0^\infty e^{-xt}(\cos(at)-\cos(bt))dt=-\frac1{a^2-b^2}\frac{x}{(x^2+a^2)(x^2+b^2)}.$$
Thus
\begin{eqnarray}
\int_{-\infty}^\infty \frac{x\sin x}{(x^2+a^2)(x^2+b^2)}dx&=&2\int_{0}^\infty \frac{x\sin x}{(x^2+a^2)(x^2+b^2)}dx\\
&=&-\frac2{a^2-b^2}\int_{0}^\infty\left(\int_{0}^\infty e^{-xt}(\cos(at)-\cos(bt))dt\right)\sin xdx\\
&=&-\frac1{a^2-b^2}\int_{0}^\infty(\cos(at)-\cos(bt))\left(\int_{0}^\infty e^{-xt}\sin x dx\right)dt\\
&=&-\frac2{a^2-b^2}\int_{0}^\infty\frac{\cos(at)-\cos(bt)}{t^2+1}dt\\
&=&-\frac2{a^2-b^2}\left(\frac{\pi}{2e^a}-\frac{\pi}{2e^b}\right)\\
&=&-\frac1{a^2-b^2}\left(\frac{\pi}{e^a}-\frac{\pi}{e^b}\right).
\end{eqnarray}
Here we used the following result
$$ \int_0^\infty\frac{\cos(a t)}{t^2+1}dt=\frac{\pi}{2e^a}. $$
| {
"language": "en",
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How to integrate $(x^2 - 1)/(x^2 + 1)$? I have gone until separating $(x^2 - 1)/(x^2 + 1)$ into $x^2/(x^2 + 1)$ - $1/(x^2 + 1)$. The latter fraction I can substitute by $\tan u$, but how to solve the first fraction and how does it all come together?
| $$ \int \frac{x^2 -1}{x^2 + 1} dx = \int \frac{x^2 +1 - 1 - 1}{x^2+1}dx = \int dx - 2 \int\frac{dx}{x^2+1} = x - 2 \arctan x + C $$
| {
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Solving integral $ \int \frac{x+\sqrt{1+x+x^2}}{1+x+\sqrt{1+x+x^2}}\:\mathrm{d}x $ there is integral $$ \int \frac{x+\sqrt{1+x+x^2}}{1+x+\sqrt{1+x+x^2}}\:\mathrm{d}x$$
i am trying to separate this :
$$=\int \mathrm{d}x -\int \frac{\mathrm{d}x}{1+x+\sqrt{1+x+x^2}} $$ but have no idea about second
| Decompose the integrand as follows
\begin{align}
&\int \frac{x+\sqrt{1+x+x^2}}{1+x+\sqrt{1+x+x^2}}dx\\
=& \int \frac{2x+1}{2\sqrt{x^2+x+1}}+\frac1{2\sqrt{x^2+x+1}}+\frac1x\bigg(\frac1{\sqrt{x^2+x+1}}
-1\bigg)\ dx\\
= &\ \sqrt{x^2+x+1}+\frac12\sinh^{-1}\frac{2x+1}{\sqrt3}
-\ln \bigg(\frac x2+1+\sqrt{x^2+x+1}\bigg)+C
\end{align}
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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$\int_0^\frac{\pi}{2}\cos ^2x\log(\tan x)dx.$ Evaluate $\int_0^\frac{\pi}{2}\cos ^2x\log(\tan x)dx.$
Sidenote:Via mathalpha I know that answer is $-\pi/4$ but do not know how to derive that.
| Note that
$$
\begin{align}
I(a)
&=\int_0^{\pi/2} \cos^2 x \tan^a xdx&\\
&=\int_0^{\pi/2} \sin^a x \cos^{2-a}xdx&(\mbox{definition of beta function})\\
&=\frac{1}{2}B\left(\frac{1+a}{2},\frac{3-a}{2}\right)& \left(B(x,y)=\frac{\Gamma(x)\Gamma(y)}{\Gamma(x+y)}\right)\\
&=\frac{\Gamma\left(\frac{1+a}{2}\right)\Gamma\left(1+\frac{1-a}{2}\right)}{2\Gamma(2)}&(\Gamma(1+t)=t\Gamma(t))\\
&=\frac{\Gamma\left(\frac{1+a}{2}\right)\frac{1-a}{2}\Gamma\left(\frac{1-a}{2}\right)}{2}\\
&=\frac{1-a}{4}\Gamma\left(1-\frac{1+a}{2}\right)\Gamma\left(1+\frac{1+a}{2}\right)&\left(\Gamma(1-z)\Gamma(z)=\frac{\pi}{\sin(\pi z)}\right)\\
&=\frac{1-a}{2}\frac{\pi}{\sin\left(\pi\frac{1+a}{2}\right)}\\
&=\frac{\pi(1-a)}{4}\sec\left(\frac{\pi a}{2}\right)
\end{align}
$$
So
$$
\begin{align}
\int_0^{\pi/2} \cos^2 x\log(\tan x)dx
&=\frac{d}{da}I(a)\Biggl|_{a=0}\\
&=-\frac{1}{8}\pi\left(\pi(a-1)\tan\left(\frac{\pi a}{2}\right)+2\right)\sec \left(\frac{\pi a}{2}\right)\Biggl|_{a=0}\\
&=-\frac{\pi}{4}
\end{align}
$$
| {
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Proving that $(a+b+c)^n=a^n + b^n + c^n$ Suppose that $(a+b+c)^3=a^3 + b^3 + c^3$.
For what positive integer values of n is it true that $(a+b+c)^n=a^n + b^n + c^n$. Any hint will be much appreciated
| For all odd n it is true.
$(a+b+c)^3=a^3+b^3+c^3+3(a+b)(b+c)(c+a)$. Now according to the hypothesis $a+b=0 $ or $c+b=0 $ or $a+c=0 $. Say $a+b=0 $ then substituting it is true for all odd n. Similarly we can check for all other cases.
| {
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} |
Minimum value of $\sqrt{(1+1/y)(1+1/z)}$ If $y,z > 0$ and $y + z = c$ where $c$ is a constant, then what's the minimum value of
$$\sqrt{\left(1+\frac1y\right)\left(1+\frac1z\right)}$$
I am having a hard time solving this.
| This can also be taken on using single-variable calculus by using the constraint to insert $ \ z \ = \ c \ - \ y \ $ into
$$ \left[ \ \frac{y \ (c-y) \ + \ (c-y) \ + \ y \ + \ 1}{y \ (c-y)} \ \right]^{1/2} \ = \ \left[ \ \frac{1 \ + \ c \ + \ cy \ - \ y^2}{y \ (c-y)} \ \right]^{1/2} \ \ . $$
Differentiating with respect to $ \ y \ $ , simplifying the resulting expression, and setting that equal to zero yields
$$ \frac{1}{2} \ \frac{\sqrt{y \ (c-y)} \ \cdot \ (1+c) \ (c - 2y)}{[1+c+cy-y^2]^{1/2} \ \cdot \ y^{3/2} \ (c-y)^{3/2}} \ = \ 0 \ \ . $$
Since the domain for the function is $ \ 0 \ < \ y \ < \ c \ $ , the only value that makes the numerator zero is $ \ y \ = \ \frac{c}{2} \ $ , from which it follows that $ \ z \ = \ \frac{c}{2} \ $ and that the minimal value for the function is $ \ \ \ \frac{c + 2}{c} \ $ , as user2345215 calculates. (It is not difficult to show that this is a local minimum by varying $ \ y \ $ slightly.)
[Just to demonstrate this, if we let $ \ y \ = \ \frac{c}{2} \ + \ \delta \ $ and $ \ z \ = \ \frac{c}{2} \ - \ \delta \ $ , the value of the function becomes
$$ \sqrt{\frac{ \ [ \ \frac{c}{2} \ + \ 1 \ ]^2 \ - \ \delta^2}{ \ ( \frac{c}{2} )^2 \ - \ \delta^2}} \ \ ; $$
since the first term in the numerator is larger than that of the denominator, subtracting $ \ \delta^2 \ $ from it reduces it proportionally less. Thus, the ratio under the square-root becomes larger.]
$$ \ \ $$
That was a bit unpleasant -- Lagrange-multipliers are somewhat easier to work with. Since $ \ y \ $ and $ \ z \ $ are positive, as are the square-roots in this function, we can just extremize $ f(y,z) $ $ = \left(1 + \frac{1}{y} \right) \ \left(1 + \frac{1}{z} \right) \ $ , subject to the constraint $ \ g(y,z) \ = \ y \ + \ z \ - \ c \ $. For the Lagrange equations $ \ \nabla f \ = \ \lambda \ \nabla g \ $ , we have
$$ -\frac{1}{y^2} \ - \ \frac{1}{y^2z} \ = \ \lambda \ \cdot \ 1 \ \ , \ \ -\frac{1}{z^2} \ - \ \frac{1}{yz^2} \ = \ \lambda \ \cdot \ 1 \ \ $$
$$ \Rightarrow \ \ \frac{z \ + \ 1}{y^2z} \ = \ \frac{y \ + \ 1}{yz^2} \ \ \Rightarrow \ \ z^2 \ + \ z \ = \ y^2 \ + \ y \ \ . $$
Now inserting $ \ z \ = \ c \ - \ y \ $ , we obtain
$$ (c \ - \ y)^2 \ + \ (c \ - \ y ) \ = \ y^2 \ - \ y \ \ \Rightarrow \ \ c \ (c + 1) \ = \ 2y \ (c + 1) \ \ , $$
from which we again find $ \ y \ = \ z \ = \ \frac{c}{2} \ $ .
$$ \ \ $$
We can also really just exploit the symmetry of the function itself: since it is unchanged by "exchanging" the variables $ \ y \ $ and $ \ z \ $ , the function has "diagonal symmetry" about the line $ \ \ y \ = \ z \ $ . This line intersects the constraint line $ \ y \ + \ z \ = \ c \ $ at $ \ y \ = \ z \ = \ \frac{c}{2} \ $ , so an extremum for the function lies there (found to be the minimum).
| {
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Find the length of $y=x^2$ from $x=0$ to $x=1$ $$L=\int_0^1 \sqrt{1+4x^2} dx = \frac{1}{4}\left(\theta+2\sqrt{5}\right)$$
It tells you to use $2x=\sinh\theta$.
I've tried the best I could, but I can never get the right side of the equation.
Can you help me?
| Using $2x=\sinh\theta$ yield $x=\frac{1}{2}\sinh\theta$ and $dx=\frac{1}{2}\cosh\theta\,d\theta$. The limit of integral for $0<x<1$ equals $0<\theta<\sinh^{-1}2$. Hence
\begin{align}
\int_0^1 \sqrt{1+4x^2}\,dx&=\int_0^{\sinh^{-1}2} \sqrt{1+4\left(\frac{1}{2}\sinh\theta\right)^2}\frac{1}{2}\cosh\theta\,d\theta\\
&=\frac{1}{2}\int_0^{\sinh^{-1}2} \sqrt{1+\sinh^2\theta}\cosh\theta\,d\theta\\
&=\frac{1}{2}\int_0^{\sinh^{-1}2} \cosh^2\theta\,d\theta\\
&=\frac{1}{2}\int_0^{\sinh^{-1}2} \cosh^2\theta\,d\theta\\
&=\frac{1}{2}\int_0^{\sinh^{-1}2} \frac{1}{2}(1+\cosh2\theta)\,d\theta\\
&=\frac{1}{4}\left[\theta+\frac{1}{2}\sinh2\theta\right]_0^{\sinh^{-1}2}\\
&=\frac{1}{4}(\sinh^{-1}2+2\sqrt{5})
\end{align}
The last step can be calculated as follows, let $y=\sinh^{-1}2$ then $\sinh y=2$ and
$$
\cosh y=\sqrt{1+\sinh^2 y}=\sqrt{5}\qquad\rightarrow\qquad\text{take the principal root}
$$
then
$$
\sinh2\theta=2\sinh\theta\cosh\theta=2\times2\times\sqrt{5}=4\sqrt{5}
$$
| {
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Number of integer solutions by generating functions method I'm stuck in the middle of a problem and not sure where to go next. The original problem is:
Find the number of integer solutions to the equation
$$2x + 3y + 4z + w + s + t = n$$ with $$0 \le w \le 2$$ $$2 \le s \le 5$$ $$0 \le t \le 3$$
Now I was able to create my generating functions equations to get my overall equation to this:
$$ G(x) = (\frac{1}{1-x^2})(\frac{1}{1-x^3}) (\frac{1}{1-x^4})(1+x+x^2)(x^2+x^3+x^4+x^5)(1+x+x^2+x^3) $$
Which with the help of the finite geometric series and some cancelling I was able to simplify down to:
$$ G(x) = (\frac{1}{1-x^2})(\frac{x^2}{1}) (\frac{1}{(1-x)^3})(\frac{1-x^4}{1}) $$
But now I'm stuck. Any help would be appreciated. Thanks.
Edit: With the extra step of simplification
$$ G(x) = (\frac{x^2}{1}) (\frac{1}{(1-x)^3})(\frac{1+x^2}{1}) $$
| Hint: you can factor $1-x^2$ and $1-x^4$.
| {
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$A^3 = I$ ($A$ is real Symmetric matrix). Does it imply that $A = I$? Question of our assignment
$A$ is a $3×3$ real symmetric matrix such that $A^3 = I$ (Identity matrix). Does it imply that $A = I$?
If so, why? If not, give an example.
Any help will be appreciated.
| For any vector $x$, let $y=(A-I)x$. Since $A$ is symmetric and $A^3=I$, we have
\begin{align*}
\|Ay\|^2 + \|A^2y\|^2 + \|y\|^2
&=y^T\left[A^TA+(A^2)^TA^2+I\right]y\\
&=y^T\left(A^2+A^4+I\right)y\\
&=y^T(A^2+A+I)y\\
&=y^T(A^2+A+I)(A-I)x\\
&=y^T(A^3-I)x\\
&=0.
\end{align*}
Therefore $\|y\|$ must be zero, i.e. $y=0$ or $Ax=x$. Since $x$ is arbitrary, we conclude that $A=I$.
Edit. In general, suppose $A$ is real symmetric and $A^m=I$ for some odd positive integer $m$. For any vector $x$, let $y=(A-I)x$. Since $A$ is symmetric, we have $\|A^ry\|^2=(A^ry)^T(A^ry)=y^TA^{2r}y$ for every nonnegative integer $r$. It follows that
\begin{align*}
\|y\|^2
&\le\sum_{\substack{0\le k\le m-1\\ k\text{ is even}}}\|A^{k/2}y\|^2+\sum_{\substack{0\le k\le m-1\\ k\text{ is odd}}}\|A^{(k+m)/2}y\|^2\\
&=y^T\left(\sum_{\substack{0\le k\le m-1\\ k\text{ is even}}}A^k+\sum_{\substack{0\le k\le m-1\\ k\text{ is odd}}}A^{k+m}\right)y\\
&=y^T\left(\sum_{k=0}^{m-1}A^k\right)y\quad\text{(because $A^m=I$)}\\
&=y^T\left(\sum_{k=0}^{m-1}A^k\right)(A-I)x\\
&=y^T(A^m-I)x\\
&=0.\\
\end{align*}
Therefore $\|y\|$ must be zero, i.e., $y=0$ or $Ax=x$. Since $x$ is arbitrary, we conclude that $A=I$.
Similarly, if $A$ is real symmetric and $A^m=I$ for some even positive integer $m$, we may show that $A^2=I$. Indeed, if we put $y=(A^2-I)x$, then
\begin{align*}
\|y\|^2
&\le\|A^{(m-2)/2}y\|^2+\|A^{(m-4)/2}y\|^2+\cdots+\|Ay\|^2+\|y\|^2\\
&=y^T\left(A^{m-2}+A^{m-4}+\cdots+A^2+I\right)y\\
&=y^T\left(A^{m-2}+A^{m-4}+\cdots+A^2+I\right)(A^2-I)x\\
&=y^T(A^m-I)x\\
&=0\\
\end{align*}
and hence $y=(A^2-I)x$ must be zero for any vector $x$.
| {
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How many different systems are there? A Football team has 10 players besides the Keeper, who are allocated into 3 positions, the defense, the center and the attack. A system is the allocation of the players into these positions.
For example the system 4 4 2 means that there are 4 players at the defense, 4 players at the center and 2 players at the attack.
There are 5 players available for the defense, 6 players for the center and 4 players for the attack.
How many different systems can be formed, when we know that at each position must be at least one player??
I found the following sentence:
$$ \text{ The number of integer solutions of the equation} x_1+x_2+ \dots + x_n=r , x_1>a_1 , x_2>a_2 , \dots, x_n>a_n \text{ can be found with the formula } \displaystyle{\frac{r-a_1-a_2- \dots -a_n-1}{n-1}} $$
Can I do it as followed?
$$x_1+x_2+x_3=10 , x_1 \leq 5, x_2 \leq 6, x_3 \leq 4, x_1,x_2,x_3 \geq 1 $$
$$y_1=x_1-1, 0 \leq y_1 \leq 4 $$
$$y_2=x_2-1, 0 \leq y_2 \leq 5 $$
$$y_3=x_3-1, 0 \leq y_3 \leq 3 $$
$$ y_1+1+y_2+1+y_3+1=10 \Rightarrow y_1+y_2+y_3=7 , y_1,y_2,y_3 \geq 0$$
There are $\binom{3+7-1}{7}=36$ integer solutions of the equation $ y_1+y_2+y_3=7 , y_1,y_2,y_3 \geq 0$
We have to substract the cases:
*
*$x_1>5: \binom{10-5-1}{2}=6 $
*$x_2>6: \binom{10-6-1}{2}=3$
*$x_3>4: \binom{10-4-1}{2}=10 $
So,there are $36-6-3-10=17$ different systems that can be formed.
Or couldn't we do it this way?
| How about enumerating all possibilities?
Starting with 1 defender (the minimum) and 4 attackers (the maximum), we get:
*
*1-5-4
*1-6-3
*2-4-4
*2-5-3
*2-6-2
*3-3-4
*3-4-3
*3-5-2
*3-6-1
*4-2-4
*4-3-3
*4-4-2
*4-5-1
*5-1-4
*5-2-3
*5-3-2
*5-4-1
Hmm. That is not what you have. ;)
| {
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Let $f(x)=ax^2+bx+c$ where $a,b,c$ are real numbers. Suppose $f(-1),f(0),f(1) \in [-1,1]$. Prove that $|f(x)|\le \frac{3}{2}$ for all $x \in [-1,1]$. Let $f(x)=ax^2+bx+c$ where $a,b,c$ are real numbers. Suppose $f(-1),f(0),f(1) \in [-1,1]$. Prove that $|f(x)|\le \frac{3}{2}$ for all $x \in [-1,1]$.
I made quite a few attempts but could not explicitly derive the required condition. I can understand that it happens, but cannot derive that. Please help.
| WLOG, let us suppose $f(x)=a(x-u)^2-v$ for $a>0$ and $0< u\le 1$, $v\ge 0$.
Then the premise says $1\ge au^2-v\ge -1$, $1\ge a(1-u)^2-v\ge -1$, $1 \ge a(1+u)^2-v\ge -1$. We need to show that $v\le \frac{3}{2}$.
The above inequalities implies that $1\ge au$ and $2\ge a(1+2u)$.
If $u\in (0,1/2]$, then
$$
v\le 1+au^2\le 1+2u^2/(1+2u)\le\frac{5}{4}.
$$
If $u\in [1/2, 1]$, then
$$
v\le 1+a(1-u)^2\le 1+(1-u)^2/u\le \frac{5}{4}.
$$
Let $u=\frac{1}{2}$, $a=2$, $v=\frac{5}{4}$, then $f=2(x-\frac{1}{2})^2-\frac{5}{4}$ attains the extreme value.
I suppose there should be more effective proof.
| {
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For any integer n greater than 1, $4^n+n^4$ is never a prime number. For any integer n greater than 1, show that $4^n+n^4$ is never a prime number.
My approach: I tried to use mathematical induction. But somehow couldn't manage to prove it. Is it to be done by induction or can it be done by another method?
| This is a nice identity. Try to complete the square: $4^n+n^4=(2^n)^2+(n^2)^2=(2^n+n^2)^2-2\cdot2^n\cdot n^2$.
Now, clearly we may assume that $n$ is odd, else the sum is even and larger than $2$. But if $n$ is odd, then $2\cdot 2^n$ is itself a square, and we can continue: Say $n=2k-1$, then $(2^n+n^2)^2-2\cdot2^n\cdot n^2=(2^n+n^2-2^kn)(2^n+n^2+2^kn)$.
Finally, one just needs to check that neither term in this product is $1$. This is clear for the second term. As for the first, note that $2^n+n^2-2^kn=2^k(2^{k-1}-2k+1)+n^2$, and $2^{k-1}-2k+1\ge0$ if $2^{k-1}+1\ge 2k$, which holds for $k\ge 4$ (as one easily verifies by induction). The remaining cases $k=2,3$ can be checked directly by hand. (Note that $k=1$ corresponds to $n=1$, and $4^1+1^4=5$ is prime.)
| {
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Calculate the inequality $a,b,c$ are the sides of a triangle.Then show that
$$(a+b+c)^3 > 27(a+b-c)(b+c-a)(c+a-b)$$ Also give the case where equality holds i.e. $$(a+b+c)^3=27(a+b-c)(b+c-a)(c+a-b)$$
I tried triangle inequality, AM-GM inequality in many way but cannot help my cause.
| By AM-GM
$$(a+b+c)^3=\left(\sum_{cyc}a\right)^3=\left(\sum_{cyc}(a+b-c)\right)^3\geq$$
$$\geq\left(3\sqrt[3]{\prod_{cyc}(a+b-c)}\right)^3=27(a+b-c)(a+c-b)(b+c-a).$$
Done!
| {
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Calculate $\lim_{n \to \infty} \frac{1^4 + 3^4 + \ldots + (2n-1)^4}{n^5}$ Is my solution correct?
By Faulhaber's formula, $1^4 + 2^4 + \ldots + n^4 = \frac{6n^5+15n^4+10n^3-n}{30} $.
$$\frac{1^4 + 3^4 + \ldots + (2n-1)^4}{n^5} = \frac{1}{n^5} \left[\sum_{k=1}^{2n} k^4 - \sum_{k=1}^{n} (2k)^4 \right] = \frac{1}{n^5} \left[\frac{6\cdot(2n)^5 + o(n^5) - 2^4 \cdot 6 \cdot n^5 - o(n^5)}{30} \right] = \frac{6\cdot2^5 - 2^4 \cdot 6}{30} + o(1) = \frac{16}{5} + o(1)$$
So, $\lim_{n \to \infty} \frac{1^4 + 3^4 + \ldots + (2n-1)^4}{n^5} = 3.2$
| It is correct, of course, but it would've perhaps been much simpler if one were to just recognize the expression as the Riemann sum of $\displaystyle\int_0^1(2x)^4dx$.
| {
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Working out the value of $a^4+b^4$ If $ab = 2$ and $a+b = 5$ then calculate the value of $a^4+b^4$
My approach:
$$a^4+b^4 = (a+b)^4-4a^3b-6a^2b^2-4ab^3$$
$$=(5)^4-6(ab)^2-4ab.a^2-4ab.b^2$$
$$=(5)^4-6(24)-4ab(a^2-b^2)$$
$$=(5)^4-6(24)-8(a+b)(a-b)$$
$$=(5)^4-6(24)-8(5)(a-b)$$
I am a little stuck now and any help will be appreciated.
| You could start with $a+b$ and $ab$,
*
*Note first $(a+b)^2 - 2ab = a^2+ b^2 $
*Note that $(a^2+b^2)^2 - 2(ab)^2 = a^4 + b^4$
As an asside, i use this set of numbers to factorise $b^n-a^n$. Part of the process involves creating a sequence $T_n = a^n+b^n$. You find that $T_{n+1}=(a+b)T_n - ab T_{n-1}$
Applying $T_0=2$ and $T_1$=5 we get this series for the sum of $a^n+b^n$ for n=0 t0 6. The iteration is t(n+1)=5t(n)-2t(n-1).
0 1 2 3 4 5 6
2 5 21 95 433 2070 9484
Doing it this way, allows one to evaluate symmetric equations (ie $f(a,b)=f(b,a)), very quickly. You start at the middle, and work outwards, multiplying the result by ab, before adding the next term.
| {
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Finding rational points on a non-trivial algebraic curve The curve in question is defined as the set of all points $(x,y)$ where both $x$ and $y$ are in $(0,1)$, and that satisfy the following:
$$ \left( x^4 - 6 x^2 + 1 \right) \left( y^4-6y^2+1
\right) -64 x^2 y^2 = 0$$
I do not know how I could identify even a single rational point on this curve, let alone others.
Any assistance would be appreciated.
| The equation
$$
\left( x^4-6x^2+1 \right) \left( y^4-6y^2+1 \right) - 64 x^2 y^2 = 0
$$
has no rational solutions; indeed the only solutions for which both
$X := x^2$ and $Y := y^2$ are rational are those with $x^2 = y^2 = -1$.
Bruno Joyal already noted that $X$ and $Y$ satisfy
$$
\left( X^2 - 6X + 1 \right) \left( Y^2 - 6Y + 1 \right) - 64 X Y = 0,
$$
and noted that this equation still has degree $4$ and that
in general a degree-$4$ curve has genus $3$. The good news is that
in fact this curve has genus $0$. The bad news is that it is not
a rational curve over $\bf Q$, and the only rational solution
is the singularity at $(X,Y) = (-1,-1)$.
For the "good news", note that the curve has degree $2$ in each variable
separately, which makes its genus at most $1$; the node at $(-1,-1)$
then drops the genus to zero.
For the "bad news", compute the discriminant with respect to $Y$,
finding that $8 (X+1)^2 (X^2+10X+1)$ must be a square. If $X=-1$ then
the equation $(X^2 - 6X + 1) (Y^2 - 6Y + 1) - 64 X Y = 0$ becomes
$8(Y+1)^2 = 0$, so we recover $(X,Y) = (-1,-1)$.
If $X \neq -1$ then $2(X^2 + 10 X + 1)$ must be a square. But
$2(X^2 + 10 X + 1) = 2 (X+5)^2 - 3 \cdot 4^2$, and
$2a^2 - 3b^2$ cannot be a square unless $a=b=0$,
because $2$ is not a square mod $3$.
[We could also show this using congruences modulo $8$.]
Hence $(-1,-1)$ is the unique rational solution of
$(X^2 - 6X + 1) (Y^2 - 6Y + 1) = 64 X Y$, and we're done.
| {
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Calculation of sum of infinite products $\sum_{n=1}^\infty n*\frac{r^n}{n!}$ Is there any analytical solution to this sum of infinite products??
$$\sum_{n=1}^\infty n\cdot \frac{r^n}{n!}$$
and
$$\sum_{n=1}^\infty n^2\cdot \frac{r^n}{n!}$$
| Hint for $1$:
$$\sum_{n=1}^\infty n\cdot \frac{r^n}{n!} = r \sum_{n=1}^\infty \frac{r^{n-1}}{(n-1)!} = r \sum_{n=0}^\infty \frac{r^n}{n!}$$
Hint for $2$:
$$\sum_{n=1}^\infty n^2\cdot\frac{r^n}{n!} = \sum_{n=1}^\infty (n(n-1) + n)\cdot\frac{r^n}{n!} = \sum_{n=1}^\infty n(n-1)\cdot\frac{r^n}{n!}+\sum_{n=1}^\infty n\cdot\frac{r^n}{n!}$$
And
$$\sum_{n=1}^\infty n(n-1)\cdot\frac{r^n}{n!}=r^2\sum_{n=1}^\infty n(n-1)\cdot\frac{r^{n-2}}
{n!} \tag{1}$$
You get $(1)$ by differentiating $2$ times w.r.t. $r$ and then multiplying by $r^2$.
$$\sum_{n=0}^\infty \frac{r^n}{n!}$$
| {
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Prove that $\lim_{n\to\infty} (\sqrt{n^2+n}-n) = \frac{1}{2}$ Here's the question: Prove that $\lim_{n \to \infty} (\sqrt{n^2+n}-n) = \frac{1}{2}.$
Here's my attempt at a solution, but for some reason, the $N$ that I arrive at is incorrect (I ran a computer program to test my solution against some test cases, and it spits out an error). Can anyone spot the error for me?
$\left|\sqrt{n^2+n}-n-\frac{1}{2}\right| < \epsilon$
$\Rightarrow \left|\frac{n}{\sqrt{n^2+n}+n} - \frac{1}{2}\right| < \epsilon$
$\Rightarrow \frac{1}{2}-\frac{1}{\sqrt{1+\frac{1}{n}}+1} < \epsilon$
$\Rightarrow \frac{1}{\sqrt{1+\frac{1}{n}}+1} > \frac{1}{2} - \epsilon = \frac{1-2 \epsilon}{2}$
$\Rightarrow \frac{1}{\sqrt{1+\frac{1}{n}}} > \frac{1-2 \epsilon}{2}$
$\Rightarrow \frac{1}{\sqrt{\frac{1}{n}}} > \frac{1-2 \epsilon}{2}$
$\Rightarrow \sqrt{n} > \frac{1-2 \epsilon}{2}$
$\Rightarrow n > \frac{4 {\epsilon}^2-4 \epsilon +1}{4}$
| (Yet another solution - sorry!)
$$
\left(1 + \frac1{2n}\right)^2 = 1 + \frac1n + \frac1{4n^2} > 1 + \frac1n,
$$
therefore
$$
\sqrt{1 + \frac1n} < 1 + \frac1{2n},
$$
therefore
$$
\sqrt{n^2 + n} - n = n\left(\sqrt{1 + \frac1n} - 1\right) < \frac12,
$$
but also
$$
\sqrt{n^2 + n} - n = \frac1{\sqrt{1 + \frac1n} + 1} > \frac1{2\left(1 + \frac1{4n}\right)}
> \frac{1 - \frac1{4n}}{2} = \frac12 - \frac1{8n},
$$
therefore
$$
\frac12 - \epsilon < \sqrt{n^2 + n} - n < \frac12
\quad \left(\epsilon > 0, \ n \geqslant \frac1{8\epsilon}\right).
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/783536",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "13",
"answer_count": 10,
"answer_id": 5
} |
Point on $z = \frac{1}{xy}$ closest to origin Where $x>0$ and $y>0$.
I want to work with the square of the distance formula from the origin, so I went with $f(x,y) = x^2 + y^2 + \frac{1}{(xy)^2}$.
Then I found the first partial derivatives:
$f_x = 2x - \frac{2}{x^3y^2}$ and $f_y = 2y - \frac{2}{y^3x^2}$.
Set them equal to $0$ to find the critical points, and multiply both sides by respective $x$ and $y$ to get a common term to set equal to each other:
$2xy = \frac{2}{x^3y}$ and $2xy = \frac{2}{y^3x}$ so $\frac{2}{x^3y} = \frac{2}{y^3x}$. Simplfying, we get $y^2 = x^2$, or $y = x$ since we are given that $x>0$ and $y>0$.
I'm not sure what to do from here.
| The surface you are describing can be described as $xyz = 1$. Since this is symmetric in $x$ versus $y$ versus $z$, any point where $|x| = |y| = |z|$ would be an extremum in distance from the origin. Visually, we can see that that such points are minima.
Therefore the solutions must be $x,y,z = (1,1,1), (1,-1,-1), (-1,1,-1),$ or $(-1,-1,1)$. Only one of these has $x,y > 0$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/783591",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Value of $ \cos 52^{\circ} + \cos 68^{\circ} + \cos 172^{\circ} $? I am a little weak in trigonometry. I have two questions:
*
*Find the value of $\cos 52^{\circ} + \cos 68^{\circ} + \cos 172^{\circ} $
*Find the value of $\sin 28^{\circ}+ \cos 17^{\circ} + \cos 28^{\circ} + \sin 17^{\circ} $
I am asking these questions because:
1. I am weak and unable to solve these.
2. I want to know the difference in these questions.
| Note that,
$$
\cos(\alpha\pm \beta)=\cos \alpha\cos \beta\mp\sin \alpha\sin \beta
$$
and
$$
\cos(180^\circ-\alpha)=-\cos\alpha.
$$
Hence
$$
\begin{align}
\cos52^\circ+\cos68^\circ+\cos172^\circ&=\color{blue}{\cos(60^\circ-8^\circ)+\cos(60^\circ+8^\circ)}+\color{red}{\cos(180^\circ-8^\circ)}\\
&=\color{blue}{2\cos60^\circ\cos8^\circ}+(\color{red}{-\cos8^\circ})\\
&=\color{blue}{2\cdot\frac12\cdot\cos8^\circ}-\color{red}{\cos8^\circ}\\
&=0.
\end{align}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/783695",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
Telescoping trigonometric series: $\sin(x) \sec(3x) + \cdots+ \sin(3^{n}x) \sec(3^{n+1} x)$ How would you evaluate the sum of this series:
$$\sin(x) \sec(3x) + \sin(3x) \sec(3^{2}x) + \sin(3^{2}x) \sec(3^{3} x) + \cdots+ \sin(3^{n}x) \sec(3^{n+1} x)$$
I can tell it is a telescoping series but I am not able to get that one step that will cancel all terms
| Lab Bhattacharjee has provided the necessary components to evaluate the series.
The series is seen as:
\begin{align}
\sum_{k=0}^{n} \frac{ \sin(3^{k} x) }{ \cos(3^{k+1} x)} &= \frac{1}{2} \sum_{k=0}^{n} \left( \tan(3^{k+1} x) - \tan(3^{k} x) \right) \\
&= \frac{1}{2} \left( \tan( 3^{n+1} x) - \tan(x) \right).
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/787457",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
how can we get from $x \equiv 4 \pmod 9$ to $x \equiv 1 \pmod 3$ I want to solve the system:
$$ x \equiv 1 \pmod 3 , x \equiv 2 \pmod 5, x \equiv 3 \pmod 7, x \equiv 4 \pmod 9, x \equiv 5 \pmod {11}$$
The numbers $3,5,7,9,11$ are not pairwise coprime,especially the numbers $3,9$ are not coprime.
According to my notes, $x \equiv 4 \pmod 9 \Rightarrow x \equiv 4 \pmod 3 \Rightarrow x \equiv 1 \pmod 3$
But..how can we get from the relation $x \equiv 4 \pmod 9$ to this one: $x \equiv 1 \pmod 3$ ?
| Since 3 is a factor of 9, $x \equiv_9 4 \Rightarrow x \equiv_3 4 \Rightarrow x \equiv_3 1$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/788956",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Calculate $\lim_{x \to \infty} x - \sqrt{x^2 + 2x}$ without derivations How can I calculate $\displaystyle \lim_{x \to \infty} x - \sqrt{x^2 + 2x}$?
Here is what I've done so far:
Multiplying by $\displaystyle \frac{x + \sqrt{x^2 + 2x}}{x + \sqrt{x^2 + 2x}}$
I got $\displaystyle \frac {-2x}{x+\sqrt{x^2 + 2x}}$
Multiplying this by $\displaystyle \frac{\frac{1}{x}}{\frac{1}{x}}$
I got $\displaystyle \frac{-2}{1+\frac{\sqrt{x^2 + 2x}}{x}}$
I know I'm very close to the answer wich is $-1$, but I have no idea what to do next. I can´t just say that $\displaystyle \lim_{x \to \infty} \frac{\sqrt{x^2 + 2x}}{x} = 1$, as far as I know...
| Note that $x - \sqrt{x^2 + 2x} = x - \sqrt{(x+1)^2 - 1}$. For large $x$, the second term is approximately $x+1$. That is to say, for $x > -1$, $$x - \sqrt{(x+1)^2 - 1} = x - (x+1)\sqrt{1 - \frac{1}{(x+1)^2}}$$ and the limit of the term in the square root is clearly $1$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/789772",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 2
} |
Interesting Base summation contest math problem The problem is as follows:
Let $N_b=1_b+2_b+\cdots+100_b$ where $b$ is an integer greater than $2$. Compute the number of values of $b$ for which the sum of the squares of the digits of $N_b$ is at most $512$.
I originally tried to find the maximum case for which this held true, by going until it no longer worked, but I realized that being less than or equal to $512$ was an intermittent pattern that might not be easy to tell when it stopped. We were given $20$ minutes, no calculators. If I remember correctly, I think the answer happened to be $30$. Can anyone tell me why it is this and how to prove it quickly?
| Observe that $$N_b= 101_b\left(\frac{100_b}{2}\right)=\frac{b^4+b^2}{2}=b^2\left(\dfrac{b^2+1}{2}\right).$$
If $b$ is odd,
$$b^2\left(\frac{b^2+1}{2}\right)=b^2\left(\frac{b-1}{2}b+\frac{b+1}{2}\right)=\frac{b-1}{2}b^3+\frac{b+1}{2}b^2+0b+0,$$
so $N_b = \left(d_3d_2d_1d_0\right)_b$ where $d_3=\dfrac{b-1}{2}d_2=\dfrac{b+1}{2}$ and $d_1=d_0=0$, and the sum of the squares of the digits of $N_b$ is $(\dfrac{b-1}{2})^2+(\dfrac{b+1}{2})^2$.
If $b$ is even,
$$b^2\left(\frac{b^2+1}{2}\right)=(b^2+1)\left(\frac{b^2}{2}\right)=\frac{b}{2}b^3+0b^2+\frac{b}{2}b+0,$$
so $N_b = \left(d_3d_2d_1d_0\right)_b$ where $d_3=\dfrac{b}{2},d_2=0,d_1=\dfrac{b}{2},$ and $d_0=0$, and the sum of the squares of the digits of $N_b$ is $2(\dfrac{b}{2})^2$.
It’s not hard to see that $N_b$ is an increasing sequence in $b$, and that the sum of the digits in $N_{32}=(g0g0)_{32}$ is 512.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/789862",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Find a basis of a nullspace of a matrix
Find a basis of nullspace of S.
I have already found a parameterized solution of $S*x=0$. But I dont Know how to get a basis out of that.
| Note that
$$
S\begin{bmatrix}x_1\\x_2\\x_3\\x_4\\x_5\\x_6\\x_7\end{bmatrix}=\begin{bmatrix}0\\0\\0\\0\\0\end{bmatrix}
$$
if and only if
\begin{align*}
x_1 &= 8x_4+2x_5-x_6+x_7\\
x_2 &= x_4-x_5+2x_6+x_7\\
x_3 &= -x_4+2x_5+3x_6-x_7
\end{align*}
It follows that the vectors in $\DeclareMathOperator{Null}{Null}\Null(S)$ are exactly the vectors of the form
$$
\begin{bmatrix}x_1\\x_2\\x_3\\x_4\\x_5\\x_6\\x_7\end{bmatrix}
=\begin{bmatrix} 8x_4+2x_5-x_6+x_7\\x_4-x_5+2x_6+x_7\\-x_4+2x_5+3x_6-x_7\\x_4\\x_5\\x_6\\x_7\end{bmatrix}=
x_4
\begin{bmatrix}
8\\1\\-1\\1\\0\\0\\0
\end{bmatrix}
+
x_5
\begin{bmatrix}
2\\-1\\2\\0\\1\\0\\0
\end{bmatrix}
+
x_6
\begin{bmatrix}
-1\\2\\3\\0\\0\\1\\0
\end{bmatrix}
+
x_7
\begin{bmatrix}
1\\1\\-1\\0\\0\\0\\1
\end{bmatrix}
$$
Hence
$$
\Null(S)=
\DeclareMathOperator{Span}{Span}\Span\left\{
\begin{bmatrix}
8\\1\\-1\\1\\0\\0\\0
\end{bmatrix},
\begin{bmatrix}
2\\-1\\2\\0\\1\\0\\0
\end{bmatrix},\begin{bmatrix}
-1\\2\\3\\0\\0\\1\\0
\end{bmatrix},\begin{bmatrix}
1\\1\\-1\\0\\0\\0\\1
\end{bmatrix}
\right\}
$$
So, to provide a basis for $\Null(S)$ we need only show that these four vectors are linearly independent. Can you show this?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/791397",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
How to derive the least squares solution for linear regression? We have $n$ dots: $(x_1,y_1)\cdots (x_n,y_n)$.
We know that if we use the Least squares method we will get a line $y=mx+b$ that giving the minimal value for the function $w=\sum_{i=1}^n (mx_i+b-y_i)^2$.
I need to prove that $w$ maintains:
$$b=\frac{\sum_{i=1}^n x_i^2\sum_{i=1}^ny_i -\sum_{i=1}^nx_i\sum_{i=1}^nx_iy_i}{n(\sum_{i=1}^nx_i^2)-(\sum_{i=1}^nx_i)^2 } \\ and: \\m=\frac{n(\sum_{i=1}^nx_iy_i-\sum_{i=1}^nx_i \sum_{i=1}^ny_i}{n\sum_{i=1}^nx_i^2-(\sum_{i=1}^nx_i)^2}$$
I'm really don't know how to begin, and I'd like to get any help.
Thank you!
| This post is an adjunct to the solution of @SA-255525, who solved the problem via the calculus. Another option is to use linear algebra.
Start with a crisp definition. We have a sequence of measurements of the form $\left\{ x_{k}, y_{k} \right\}_{k=1}^{m}$ and the trial function is $y(x) = c_{0} + c_{1}x$.
This implies the linear system
$$
\begin{align}
\mathbf{A} c &= y \\
\left[
\begin{array}{ccc}
1 & x_{1} \\
1 & x_{2} \\
\vdots & \vdots \\
1 & x_{m}
\end{array}
\right]
\left[
\begin{array}{c}
c_{0} \\
c_{1}
\end{array}
\right]
&=
\left[
\begin{array}{}
y_{1} \\
y_{2} \\
\vdots \\
y_{m}
\end{array}
\right].
\end{align}
$$
By definition the least squares solution $c$ minimizes the sum of the squares of the residuals given by
$$
\left[
\begin{array}{c}
c_{0} \\
c_{1}
\end{array}
\right]
_{LS}
=
\left\{
\left[
\begin{array}{c}
c_{0} \\
c_{1}
\end{array}
\right]
\in
\mathbb{R}^{2}
\colon
\lVert
\mathbf{A} c - y
\rVert_{2}^{2}
\text{ is minimized}
\right\}
$$
One solution path uses column vectors. The column structure of the system matrix is
$$
\mathbf{A} =
\left[
\begin{array}{cc}
\mathbf{1} & x
\end{array}
\right].
$$
Using the normal equations $\mathbf{A}^{\mathrm{T}}\mathbf{A}c = \mathbf{A}^{\mathrm{T}} y$, the least squares solution is
$$
c_{LS} = \left( \mathbf{A}^{\mathrm{T}}\mathbf{A} \right)^{-1} \mathbf{A}^{\mathrm{T}} y.
$$
This is the same solution found earlier. Note that
$$
\mathbf{A}^{\mathrm{T}} y =
\left[
\begin{array}{c}
\mathbf{1}\cdot y\\
x \cdot y
\end{array}
\right],
\qquad
\mathbf{A}^{\mathrm{T}}\mathbf{A} =
\left[
\begin{array}{cc}
\mathbf{1}\cdot\mathbf{1} & \mathbf{1}\cdot x \\
x \cdot \mathbf{1} & x \cdot x
\end{array}
\right],
\qquad
\left( \mathbf{A}^{\mathrm{T}}\mathbf{A} \right)^{-1} =
\left( \det \mathbf{A}^{\mathrm{T}}\mathbf{A} \right)^{-1}
\left[
\begin{array}{rr}
x \cdot x & -\mathbf{1}\cdot x \\
-x \cdot \mathbf{1} & \mathbf{1}\cdot\mathbf{1}
\end{array}
\right].
$$
To connect the two notations use
$$
\det \mathbf{A}^{\mathrm{T}}\mathbf{A} =
\left(
\mathbf{1}\cdot\mathbf{1}
\right) \left(
x \cdot x
\right)
-
\left(
\mathbf{1}\cdot x
\right)^{2},
$$
and
$$
\mathbf{1}\cdot\mathbf{1} = \sum_{k=1}^{m}(1) = m,
\quad
\mathbf{1}\cdot x = x\cdot\mathbf{1} = \sum_{k=1}^{m}x_{k},
\quad
\mathbf{1}\cdot y = \sum_{k=1}^{m} y_{k},
\quad
x \cdot y = \sum_{k=1}^{m} x_{k} y_{k}.
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/792863",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Find the points in the graph (my solution) - high school. my math problem is on Swedish so i'll try my best to translate it so you can understand. I'd appreciate it if someone could point out if I did anything wrong and if there is anything that I should add that's essential to these kind of problems.
(I'm a high school student in my second year)
The function $f(x) = x^3 - 6x^2 + 9x + 2$ gives us the points A and B
on the graph and the second function, $g(x) = x^2 - 7x + 14$ gives C
and D, where C is a mutual point for these two functions. C has a
gradient that is -3.
To start off with, I set the x for $f(x)$ to 0 in order to see where on the y-axis the point A crosses.
$f(0) = 0^3 - 6*0^2 + 9*0 + 2 = 2$
$A(0,2)$
The next step is to figure out B. To be able to get the coordinates for this point, I have to derivate $f(x)$, and then set $f'(x)=0$.
$f'(x) = 3x^2 - 12x + 9 = 0$
$3x^2 - 12x + 9 = 0$
Dividing everything with 3 to use the pq-formula.
$x^2 - 4x + 3 = 0$
$x = 2 \pm{\sqrt {2^2-3}}$
$x_1 = 2+1 = 3$
$x_2 = 2-1 = 1$
With these 2 x-values I have to choose one that gives the biggest value for $f(x)$ point B is a Maximum point on the graph that has been provided. I can find this out with the help of Second Derivative.
$f''(x) = 6x - 12$
$f''(3) = 6*3 - 12 = 6$ This is a minimum point because of $f'(3) = 0$ and $f''(3) > 0$.
$f''(1) = 6*1 - 12 = -6$ This is the x-value that gives the maximum point that I am looking for. $f'(1) = 0$ and $f''(1) < 0$
$f(1) = 1^3 - 6*1^2 + 9*1 + 2 = 6$
$B(1,6)$
As stated earlier, C is a mutual point with a gradient -3. Derivating both functions and setting them equal to -3 and solving these will give us which x-value they have in common.
$f'(x) = 3x^2 - 12x + 9 = -3$
$3x^2 - 12x + 9 = -3$
+3 on both sides to get everything on one side then dividing everything with 3 to get x^2 alone so I can use pq-formula.
$x^2 - 4x + 4 = 0$
$x = 2 \pm {\sqrt {2^2-4}}$
$x = 2$
On to the second function: $g(x) = x^2 - 7x + 14$ I set $g'(x) = -3$
By derivating this function and moving -7 to the right side will give us:
$2x = 4$
Dividing both sides with 2
$x = 2$
$f(2) = 2^3 - 6*2^2 + 9*2 + 2 = 4$
$g(2) = 2^2 - 7*2 + 14 = 4$
$f(2)=g(2)=4$
$C(2,4)$
To get the last point (D) I do the exact same way as I did for point B. I derivate the function and set g'(x) = 0 to solve x. To confirm that D is a Min.Point I derivate the first derivative and this gives me $g''(x) = 2$. We can see that $g''(x) > 0$.
$g'(x) = 2x - 7 = 0$
$2x = 7$
$x = 3.5$
$g(3.5) = 3.5^2 - 7*3.5 + 14 = 1.75$
$D(3.5;1.75)$
Answer:
$A(0.2)$
$B(1.6)$
$C(2.4)$
$D(3.5;1.75)$
$${}$$
| My solution above is correct.
Answer:
$A(0.2)$
$B(1.6)$
$C(2.4)$
$D(3.5;1.75)$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/793638",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 1,
"answer_id": 0
} |
matrix inverse and elementary row operations I'm trying to see if I can find a relationship between the inverse of 2 matrices.
Matrix A is a $n$ x $n$ matrix.
Matrix B is a $(n-1)$ x $(n-1)$ matrix
The transformation from A to B can be described in the following 2 steps:
1) Take the last 2 rows of A and add them together.
2) take the last 2 columns from 1) and add them together.
The resulting top-left $(n-1)$ x $(n-1)$ submatrix is B. I know that both A and B are invertible but is there any way to relate the inverses to each other?
thanks
| I don't think there would be an easy way to relate the two. In going from $A$ to $B$ we lose information about the structure of $A$ which is important in determining the inverse of $A$.
In the $2 \times 2$ case, the $B$ matrix is just the sum of the entries. This just tells us that
$$A=\begin{bmatrix}a_{11} & a_{12} \\ a_{21} & b_{11}-a_{11}-a_{12}-a_{21}\end{bmatrix}$$
so inverse of $A$, if it exists, is $$\tfrac{1}{a_{11}(b_{11}-a_{11}-a_{12}-a_{21})-a_{12}a_{21}} \begin{bmatrix} b_{11}-a_{11}-a_{12}-a_{21} & -a_{12} \\ -a_{21} & a_{11} \end{bmatrix}.$$
So there are infinitely many $A$ matrices with the same $B$ matrix, but with distinct inverses.
If e.g.
$$
A \in \left\{ \begin{bmatrix}
1 & 0 & 0 \\
0 & 1 & 1 \\
0 & 0 & 1 \\
\end{bmatrix},
\begin{bmatrix}
1 & 0 & 0 \\
0 & 0 & 1 \\
0 & 1 & 1 \\
\end{bmatrix},
\begin{bmatrix}
1 & 0 & 0 \\
0 & 0 & 2 \\
0 & 1 & 0 \\
\end{bmatrix}
\right\}
$$
then
$$
B=
\begin{bmatrix}
1 & 0 \\
0 & 3 \\
\end{bmatrix}.
$$
But the possible $A$ matrices have distinct inverses:
$$\begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & -1 \\ 0 & 0 & 1 \\ \end{bmatrix},
\begin{bmatrix} 1 & 0 & 0 \\ 0 & -1 & 1 \\ 0 & 1 & 0 \\ \end{bmatrix},
\begin{bmatrix} 1 & 0 & 0 \\ 0 & 0 & 1 \\ 0 & 1/2 & 0 \\ \end{bmatrix}.$$
The third of these even has a distinct determinant.
And even the non-invertible matrices
$$
\begin{bmatrix}
1 & 0 & 0 \\
0 & 3 & 0 \\
0 & 0 & 0 \\
\end{bmatrix}, \text{ and }
\begin{bmatrix}
1 & 0 & 0 \\
0 & 2 & 0 \\
0 & 1 & 0 \\
\end{bmatrix}$$
have the same $B$ matrix.
We can't even tell if $A$ or $B$ is invertible from its counterpart:
*
*$\begin{bmatrix}
1 & 0 & 0 \\
0 & 1 & -2 \\
0 & 0 & 1 \\
\end{bmatrix}$ is invertible and
$\begin{bmatrix}
1 & 0 & 0 \\
0 & 0 & 0 \\
0 & 0 & 0 \\
\end{bmatrix}$ is not invertible but
$\begin{bmatrix}
1 & 0 \\
0 & 0 \\
\end{bmatrix}$
is not invertible.
*$\begin{bmatrix}
1 & 0 & 0 \\
0 & 1 & 0 \\
0 & 0 & 1 \\
\end{bmatrix}$ is invertible and
$\begin{bmatrix}
1 & 0 & 0 \\
0 & 1 & 0 \\
0 & 1 & 0 \\
\end{bmatrix}$ is not invertible but
$\begin{bmatrix}
1 & 0 \\
0 & 2 \\
\end{bmatrix}$
is invertible.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/793851",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Expected value of number of draws We have $5$ number in a bag: $(1,3,5,7,9)$. We draw one from the bag and then put it back. We do this until the sum of the numbers can be divided by $3$. Whats the expected value of the number of draws?
My idea was to solve it with Markov-chains:
States: $0,1,2$. So numbers$\mod 3$.
The matrix will be:
$\begin{pmatrix}
2/5 & 2/5 & 1/5 \\
1/5 & 2/5 & 2/5 \\
2/5 & 1/5 & 2/5
\end{pmatrix}$
Then we have an equation system:
$k_1=1+2/5k_1+2/5k_2$, $\ k_2=1+1/5k_1+2/5k_2$ and $k_3=0$.
Even if I solve this, I'm not sure how to continue. Thanks for help.
| We use your idea, but partly for ease of typing we use different notation. Say that we are in State $1$ if the sum so far is congruent to $1$ modulo $3$, and that we are in State $2$ if the sum is congruent to $2$. Let $m$ be the mean waiting time. Let $a_1$ be the additional mean waiting time, given we are in State $1$, and $a_2$ the additional mean waiting time given that we are in State $2$.
On the first pick, we either get something divisible by $3$, in which case we have spent $1$ pick, and our additional waiting time is $0$. Or else we get a $1$ or a $7$, and our mean additional waiting time is $a_1$. Or else we get a $5$, and our mean additional waiting time is $a_2$. Thus
$$m=1+\frac{2}{5}a_1+\frac{1}{5}a_2.\tag{1}$$
Suppose we are in State $1$, and do another trial. If we get a $3$ or a $9$, we stay in State $1$, and our additional expected time remains at $a_1$. If we get a $1$ or a $7$, our additional expected waiting time becomes $a_2$. And if we get $5$, we are finished. Thus
$$a_1=1+\frac{2}{5}a_1+\frac{2}{5}a_2.\tag{2}$$
Similarly,
$$a_2=1+\frac{1}{5}a_1+\frac{2}{5}a_2.\tag{3}$$
Solve the last two equations for $a_1$ and $a_2$, and substitute in Equation (1).
| {
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How to find the sum: $1^{\frac{1}{3}}+2^{\frac{1}{3}}+3^{\frac{1}{3}}+ . . . +(50)^{\frac{1}{3}}$ Can some one help me to find the sum of the following expression?
$$1^{\frac{1}{3}}+2^{\frac{1}{3}}+3^{\frac{1}{3}}+ . . . +(50)^{\frac{1}{3}}$$
| You can bound the sum for any upper limit b.
$\frac{3 b^{4/3}}{4}<\sum _{n=1}^b \sqrt[3]{n}< \frac{3}{4} \
\left(-1+\sqrt[3]{1+b}+b \sqrt[3]{1+b}\right)$
Plugging in b = 50 you would get
$ 138.151<\sum _{n=1}^{50} \sqrt[3]{n}<141.097 $
| {
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If $ -3\left(x-\lfloor x \rfloor \right)^2+2(x-\lfloor x \rfloor )+a^2=0$ has no integral solution, then $a$ is if $a$ is real number and $\displaystyle -3\left(x-\lfloor x \rfloor \right)^2+2(x-\lfloor x \rfloor )+a^2=0$.has no real integral solution,
Then all possible values of $a$ lie in the interval..
(where $\lfloor x \rfloor $ means floor function of $x$)
$\bf{My\; Solution::}$ Let $x-\lfloor x \rfloor = \{x\}= y$, where $ 0 \leq \{x\}<1\Rightarrow 0\leq y<1$. Then equation..
$$\Rightarrow -3y^2+2y+a^2 = 0\Rightarrow 3y^2-2y-a^2 = 0$$
$$\displaystyle \Rightarrow y = \frac{2\pm \sqrt{4+12a^2}}{6} = \frac{1\pm \sqrt{1+3a^2}}{3}$$
Now If equation has no real solution , Then $y<0 \;\; \cup \;\; y\geq 1$.
So $$\Rightarrow \frac{1\pm \sqrt{1+3a^2}}{3}<0\;\; \cup \;\; \frac{1\pm \sqrt{1+3a^2}}{3}\geq 1$$
Now how can I calculate value of $a,$ after that
Help me
Thanks
| Since $$\frac{1- \sqrt{1+3a^2}}{3}<0,\quad \forall a\neq 0$$
then you only need to find $a$ such that
$$\frac{1+\sqrt{1+3a^2}}{3}>1\iff \sqrt{1+3a^2}>2\iff a^2\ge 1\iff a\le -1 \vee a\ge 1.$$
So, your equation has no solutions for $a\le -1 \vee a\ge 1$.
| {
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Prove that $\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\le\frac{3}{\sqrt{7}}$
Let $a,b,c>0$ such that
$$\dfrac{1}{a^2+2}+\dfrac{1}{b^2+2}+\dfrac{1}{c^2+2}=\dfrac{1}{3}.$$
Show that
$$\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\le\dfrac{3}{\sqrt{7}}.$$
My try: since
$$\dfrac{1}{2+a^2}=\dfrac{1}{2}\left(1-\dfrac{a^2}{a^2+2}\right)$$
so
$$\dfrac{a^2}{a^2+2}+\dfrac{b^2}{b^2+2}+\dfrac{c^2}{c^2+2}=\dfrac{7}{3}$$
we only prove
$\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\le\dfrac{3}{\sqrt{7}}$
and I want use Cauchy-Schwarz inequality to prove it,But I can't works,such
$$\left(\dfrac{a^2}{a^2+2}+\dfrac{b^2}{b^2+2}+\dfrac{c^2}{c^2+2}\right)(a^2+2+b^2+2+c^2+2)\ge (a+b+c)^2$$
$$(a^2+b^2+c^2+6)\ge \dfrac{3}{7}(a^2+b^2+c^2+2ab+2bc+2ac)$$
$$\Longrightarrow 4(a^2+b^2+c^2)+42\ge 6(ab+bc+ac)$$
and let $$p=a+b+c,q=ab+bc+ac,r=abc$$
so
$$2p^2+21\ge 7q$$
and we only prove
$$\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}=\dfrac{ab+bc+ac}{abc}=\dfrac{q}{r}\le\dfrac{3}{\sqrt{7}}$$
maybe this is not true.
But this not usefull to solve this problem .
Thank you
| An analytical proof is proposed below :
| {
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Partial fraction decomposition,how? I need to decompose this fraction:
$${x^2+1\over (x-1)^3\cdot(x+3)}$$
I tried to write it up like this: $${A\over (x-1)}+{B\over (x-1)^2}+{C\over (x-1)^3}+{D\over (x+3)}$$
But now i get $$A\cdot(x-1)^5\cdot (x+3) +B\cdot (x-1)^4\cdot (x+3) +C\cdot(x-1)^3\cdot (x+3) +D\cdot (x-1)^6 =x^2+1$$ which is not correct i think.
What should i do?
| Your first step is correct, in fact to find the partial fraction decomposition the first thing you need to do is:
$$\dfrac{x^2+1}{(x-1)^3 (x+3)} = \dfrac{A}{x-1} + \dfrac{B}{(x-1)^2} + \dfrac{C}{(x-1)^3} + \dfrac{D}{x+3}$$
Now what you need to do is to sum the fractions on the left side in this way:
$$\dfrac{x^2+1}{(x-1)^3 (x+3)} = \dfrac{A}{x-1} + \dfrac{B}{(x-1)^2} + \dfrac{C}{(x-1)^3} + \dfrac{D}{x+3} $$
$$\iff \dfrac{x^2+1}{(x-1)^3 (x+3)}= \dfrac{A(x-1)^2(x+3) + B(x-1)(x+3) + C(x+3) + D(x-1)^3}{(x-1)^3(x+3)}$$
$$\iff x^2+1 = A(x-1)^2(x+3) + B(x-1)(x+3) + C(x+3) + D(x-1)^3$$
Now you can expand the polynomial $(x-1)^2 (x+3)$, $(x-1)(x+3)$, $(x-1)^3$
and you should get that the left hand side is equal to
$$Ax^3 +Ax^2-5Ax+3A + Bx^2+2Bx-3B+Cx+3C+Dx^3 -3Dx^2 +3Dx -D$$
now you can compare the coefficient, for example you see that $x^2+1$ has degree $2$ and hence you can conclude that $Ax^3 + Dx^3 = 0$. If you do this reasoning for $x^2, x, \text{and the constants}$ you should get a linear system like this
$$\begin{pmatrix}1 & 0 & 0 & 1 \\ 1 & 1 & 0 & -3 \\ -5 & 2 & 1 & 3 \\ 3 & -3 & 3 & -1 \end{pmatrix} \begin{pmatrix} A \\ B \\ C \\ D\end{pmatrix} = \begin{pmatrix} 0 \\ 1 \\ 0 \\ 1 \end{pmatrix}$$
The solutions are given by
$$\begin{pmatrix}A \\B\\C\\D\end{pmatrix} = \begin{pmatrix} \frac{5}{32} \\ \frac{3}{8} \\ \frac{1}{2} \\ \frac{-5}{32}\end{pmatrix}$$
and therefore you partial fraction decomposition is
$$\dfrac{x^2+1}{(x-1)^3 (x+3)} = \dfrac{5}{32(x-1)} + \dfrac{3}{8(x-1)^2} + \dfrac{1}{2(x-1)^3} - \dfrac{5}{32(x+3)}$$
| {
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"question_score": "1",
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log and poisson-like integral Here is a fun looking one some may enjoy.
Show that:
$$\int_{0}^{1}\log\left(\frac{x^{2}+2x\cos(a)+1}{x^{2}-2x\cos(a)+1}\right)\cdot \frac{1}{x}dx=\frac{\pi^{2}}{2}-\pi a$$
| I understand that this an old question but I would like to share my 2 cents. First of all we recall the following Fourier identities:
Lemma 1: Let $x \in [-\pi, \pi]$ then
$$\sum_{n=1}^{\infty} \frac{\sin nx}{n} = \frac{\pi-x}{2} \Rightarrow \sum_{n=1}^{\infty} \frac{\cos n x}{n^2} = \frac{\pi^2}{6} -\frac{\pi x}{2} + \frac{x^2}{4} \tag{1}$$
Lemma 2: Let $x \in [-\pi, \pi]$ then
$$\sum_{n=1}^{\infty} \frac{(-1)^n \sin nx}{n} = - \frac{x}{2} \Rightarrow \sum_{n=1}^{\infty} \frac{(-1)^n \cos nx}{n^2} = \frac{x^2}{4} - \frac{\pi^2}{12} \tag{2}$$
Lemma 3: It holds that:
$$\ln \left ( 1 - 2x \cos a + x^2 \right ) = -2\sum_{n=1}^{\infty} \frac{x^n \cos na}{n} \tag{3}$$
Hence for $a \in [0, \pi]$
\begin{align*}
\ln \left ( \frac{1+2x \cos a + x^2}{1-2x \cos a + x^2} \right ) &= \ln \left ( 1 + 2x \cos a + x^2 \right ) - \ln \left ( 1 - 2x \cos a + x^2 \right ) \\
&=-2 \sum_{n=1}^{\infty} \frac{(-x)^n \cos na}{n} + 2 \sum_{n=1}^{\infty} \frac{x^n \cos na}{n} \\
&= -2 \sum_{n=1}^{\infty} \frac{\left ( (-x)^n - x^n \right ) \cos na}{n} \\
&= 4 \sum_{n=1}^{\infty} \frac{x^{2n+1} \cos (2n+1)a}{2n+1}
\end{align*}
since $(-x)^n - x^n$ is $0$ whenever $n$ is even. Hence,
\begin{align*}
\int_0^1 \frac{1}{x} \ln \left( \frac{1+2x\cos a+x^2}{ 1-2x \cos a+x^2 } \right) \, \mathrm{d}x &= 4\int_{0}^{1} \frac{1}{x} \sum_{n=0}^{\infty} \frac{x^{2n+1} \cos (2n+1)a}{2n+1} \, \mathrm{d}x \\
&=4 \sum_{n=0}^{\infty} \frac{\cos(2n+1)a}{2n+1} \int_{0}^{1} x^{2n} \, \mathrm{d}x \\
&=4 \sum_{n=0}^{\infty} \frac{\cos (2n+1)a}{(2n+1)^2}
\end{align*}
This is nothing else than $\chi_2$ Legendre function directly associated with the series in $(1)$, $(2)$. Hence,
\begin{align*}
\sum_{n=0}^{\infty} \frac{\cos (2n+1)a}{(2n+1)^2} & = \mathfrak{Re} \left ( \sum_{n=0}^{\infty} \frac{e^{i(2n+1)a}}{(2n+1)^2} \right ) \\
&=\frac{1}{2}\mathfrak{Re} \left ( \mathrm{Li}_2 \left ( e^{ia} \right ) - \mathrm{Li}_2 \left ( - e^{ia} \right ) \right )
\end{align*}
The real part of $\mathrm{Li}_2(e^{ix})$ is equation $(1)$ whereas the real part of $\mathrm{Li}_2(-e^{ix})$ is equation $(2)$. Thus,
\begin{align*}
4\sum_{n=0}^{\infty} \frac{\cos (2n+1)a}{\left ( 2n+1 \right )^2} &= 2 \left ( \frac{\pi^2}{6} - \frac{\pi a}{2} + \frac{a^2}{4} - \frac{a^2}{4} + \frac{\pi^2}{12} \right ) \\
&=\frac{\pi^2}{2} - \pi a
\end{align*}
Because the LHS is even so must be the RHS. So, the result can be extended for $a \in [-\pi, 0]$. Therefore,
$$\int_0^1 \frac{1}{x} \ln \left( \frac{1+2x\cos a+x^2}{ 1-2x \cos a+x^2 } \right) \, \mathrm{d}x = \frac{\pi^2}{2} - \pi \left| a \right|$$
| {
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Finding the function of x in a differential equation $y=f(x)$ If the graph of $y=f(x)$ contains the point $(0,2)$, $\frac{dy}{dx}=\frac{e^{2x}}{y}$ and $f(x) > 0$ for all $x$, then $f(x) = ?$
$dy/dx = \frac{e^{2x}}y$
$y\cdot dy = e^{2x}dx$
(The following step confuses me, because of the $dy$).
$\int y\cdot dy = \int e^{2x}dx$
$\frac{1}2y^2 = \frac{1}2e^{2x}+C$
$y=\sqrt{e^{2x}+2C}$
or
$y=\sqrt{e^{2x}+C}$
from $y(0) = 2$ I get $C = 2^2-e^{2x}$
I can't get to the answer for $f(x) = ?$
| If we have $$\frac{1}{2}y^{2} = \frac{1}{2}e^{2x} + C$$ then $$y = \sqrt{e^{2x}+C'}$$ where $C'=2C$.
For the initial value problem:
$y(0) = 2 = \sqrt{e^{2(0)}+C'} \iff C' = 4 - e^{2(0)} = 4 - 1 = 3$
So $$y(x) = \sqrt{e^{2x} + 3}$$
| {
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Double Integral $\iint_D\ (x+2y)\ dxdy$ $$\iint_D (x+2y)\ dxdy $$
If the area is range by $x=2,\ x=3,\ y=x,\ y=2x$,
how to include the lines? How limits for integral will looks like?
You mean something like this? ( I made mess)
$$\iint_D (x+2y)\ dxdy $$
If the area is range by $x=2,\ x=3,\ y=x,\ y=2x$,
how to include the lines? How limits for integral will looks like?
So
$$ \int_{y=x}^{2x}\ (x+2y)\ dx = \int_{y=x}^{2x} x dx + \int_{y=x}^{2x}\ 2y dx = \frac{1}2 [x^{2}] + 2y \int_{y=x}^{2x}\ 1 dx = \frac{1}{2}(2x)^{2}-y^{2} + 2y(2x-y) $$
and
$$ \int_{x=2}^3 [ \frac{1}{2}(2x)^{2}-y^{2} + 2y(2x-y) ]dy = \frac{1}{2}\int_{x=2}^3 (2x)^{2} dy - \int_{x=2}^3 y^{2} dy + \int_{x=2}^3 4xy dy - \int_{x=2}^3 2y^{2} dy = \frac{1}{2} (2x)^{2} \int_{x=2}^3 1 dy - [\ \frac{1}{3}y^{3}] + 4x[\frac{1}{2}y^{2}] - 2[\frac{1}{3}y^{3}]= \frac{1}{2} (2x)^{2}... $$ and I do not know.
| Hint :
$$
\iint_D\ (x+2y)\ dx\,dy=\int_{x=2}^3\int_{y=x}^{2x}\ (x+2y)\ dx\,dy.
$$
| {
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find p,q to the expression A does not depend on x? Find $p,q$ to the expression $A = p (\cos^{8}x-\sin^{8}x) + 4(\cos^{6}x-2\sin^{6}x) + q\sin^{4}x$ does not depend upon $x$
| $$A=p((1-\sin^2x)^4- \sin^8x) + 4((1-\sin^2x)^3 - 2\sin^6x) + q\sin^4x=\\
=p(\sin^8x - 4\sin^6x + 6\sin^4x - 4\sin^2 x + 1 - \sin^8x) + 4(-\sin^6x + 3\sin^4x -3\sin^2x+1-2\sin^6x) + q\sin^4x = \\
=(-4p-12)\sin^6x + (6p+12+q)\sin^4x + (-4p-12)\sin^2x + p+4.$$
| {
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A definite integral with tanh and sin In an old textbook, Thomas John I'Anson Bromwich asks the reader to use the expansion
$$ \tanh \left(\frac{\pi x}{2} \right) = 1 + 2 \sum_{n=1}^{\infty} (-1)^{n} e^{-n \pi x}$$
to show $$\lim_{t \to 0^{+}} \int_{0}^{\infty} e^{-tx} \tanh \left(\frac{\pi x}{2} \right) \sin (ax) \, dx = \text{csch}(a).$$
The reader is then supposed to assign this value to the divergent integral $$\int_{0}^{\infty} \tanh \left(\frac{\pi x}{2} \right) \sin (ax) \, dx $$
and deduce
$$\int_{0}^{\infty}\frac{\tanh(\frac{\pi x}{2})\sin(ax)}{x^{2}+1} \, dx= \frac{1}{2} \left[e^{-a}\log(e^{2a}-1)-e^{a}\log(1-e^{-2a})\right] \tag{1}.$$
This approach seems quite unusual and not easy to justify.
Can we prove $(1)$ in another way?
| Consider the complex function $ \displaystyle f(z) = \tanh \left( \frac{\pi z}{2} \right) \frac{e^{iaz}}{1+z^{2}}$ where $a > 0$.
On the right, top, and left sides of a rectangular contour with vertices at $\pm 2N, \pm 2N + i2N$ (where N is a positive integer), $\tanh \left(\frac{\pi z}{2} \right)$ is uniformly bounded.
So using a slight modification of Jordan's lemma, we can conclude
$$ \int_{-\infty}^{\infty} \tanh \left( \frac{\pi z}{2} \right) \frac{e^{iaz}}{1+z^{2}} \, dx = 2 \pi i \left(\text{Res}[f(z),i] + \sum_{n=1}^{\infty} \text{Res}[f(z),i(2n+1)]\right) .$$
Now since
$$ \begin{align} f(z+i) &= e^{-a} \coth \left(\frac{\pi z}{2} \right) e^{iaz}\frac{1}{z} \frac{1}{z+2i} \\ &= e^{-a} \left(\frac{2 }{\pi z} + \mathcal{O}(z) \right) \left(1+iaz + \mathcal{O}(z^{2}) \right) \frac{1}{z} \left(- \frac{i}{2}+ \frac{z}{4} + \mathcal{O}(z^{2}) \right) \\ &= e^{-a} \left(-\frac{i}{\pi z^{2}} + \frac{2a+1}{2 \pi z} + \mathcal{O}(1) \right) , \end{align}$$
we get
$$ \text{Res}[f(z),i] = \text{Res}[f(z+i), 0] = \frac{e^{-a}(2a+1)}{2 \pi} .$$
And for $n \ge 1$,
$$ \begin{align}\text{Res}[f(z),i(2n+1)] &= \lim_{n \to i(2n+1)} \frac{\sinh (\frac{\pi z}{2}) e^{iaz}}{\frac{d}{dz} \, \cosh (\frac{\pi z}{2}) (1+z^{2}) } \\ &= - \frac{1}{2 \pi} \frac{e^{-(2n+1)a}}{n(n+1)} . \end{align}$$
Therefore,
$$ \begin{align} \int_{-\infty}^{\infty} \tanh \left(\frac{\pi z}{2} \right) \frac{e^{iax}}{1+x^{2}} \ dx &= i e^{-a} (2a+1) - i \sum_{n=1}^{\infty} \frac{e^{-(2n+1)a}}{n(n+1)} \\ &= i e^{-a} (2a+1) + ie^{-a} \sum_{n=1}^{\infty} \left(\frac{1}{n+1} -\frac{1}{n} \right) e^{-2na} \\ &= 2i a e^{-a} + i e^{-a} + ie^{-a} \sum_{n=1}^{\infty} \frac{e^{-2na}}{n+1} - ie^{-a}\sum_{n=1}^{\infty} \frac{e^{-2na}}{n} \\ &= 2i a e^{-a} + ie^{-a} + i e^{-a} \left(- \frac{\log(1-e^{-2a})}{e^{-2a}} -1\right) + i e^{-a} \log (1-e^{-2a}) \\ &= 2ia e^{-a} - ie^{a} \log(1-e^{-2a})+ie^{-a} \log(1-e^{-2a}) . \end{align}$$
And equating the imaginary parts on both sides of the equation, we get
$$ \begin{align} \int_{0}^{\infty} \tanh \left(\frac{\pi x}{2} \right) \frac{\sin ax}{1+x^{2}} \ dx &= \frac{1}{2} \left( 2a e^{-a} + e^{-a} \log(1-e^{-2a}) - e^{a} \log(1-e^{-2a}) \right) \\ &= \frac{1}{2} \left(e^{-a} \log(e^{2a} (1-e^{-2a})) - e^{a} \log(1-e^{-2a})\right) \\ &= \frac{1}{2} \left(e^{-a}\log(e^{2a}-1)-e^{a}\log(1-e^{-2a})\right) . \end{align}$$
| {
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Value of $\sin (2^\circ)\cdot \sin (4^\circ)\cdot \sin (6^\circ)\cdots \sin (90^\circ) $ How can I calculate the value of
*
*$\sin (1^\circ)\cdot \sin (3^\circ)\cdot \sin (5^\circ)\cdots \sin (89^\circ)$
*$\sin (2^\circ)\cdot \sin (4^\circ)\cdot \sin (6^\circ)\cdots \sin (90^\circ)$
My solution: Let $$A = \sin (1^\circ)\cdot \sin (3^\circ)\cdot \sin (5^\circ)\cdots \sin (89^\circ)$$
$$B = \sin (2^\circ)\cdot \sin (4^\circ)\cdot \sin (6^\circ)\cdots \sin (90^\circ)$$
Now $$
\begin{eqnarray}
A\cdot B &=& (\sin 1^\circ \cdot \sin 89^\circ)\cdot (\sin2^\circ\cdot \sin 88^\circ)\cdots(\sin 44^\circ\cdot \sin 46^\circ) \cdot \sin 45^\circ \\
&=& (\sin 1^\circ\cdot \cos1^\circ)\cdot (\sin2^\circ\cdot \cos 2^\circ)\cdots(\sin 44^\circ\cdot \cos 44^\circ)\cdot \sin 45^\circ \\
&=& \frac{1}{2^{44}}\left[\sin(2^\circ)\cdot \sin (4^\circ)\cdots \sin(88^\circ) \cdot \sin (90^\circ)\right]\cdot \frac{1}{\sqrt{2}} \\
&=& \frac{1}{2^{\frac{89}{2}}}\cdot B
\end{eqnarray}
$$
Which implies that $$B\cdot \left(A-2^{-\frac{89}{2}}\right) = 0$$
So $A = 2^{-\frac{89}{2}}$ because $B\neq 0$.
But I do not understand how can I calculate the value of $B$. Can we calculate these values using complex numbers?
| Using the identity
$$\prod_{k=1}^{n-1}\sin \left( \frac{k\pi}{n}\right) = \frac{n}{2^{n-1}} \tag{1}$$
Putting $n=180$, it gives
$$ \left(\sin (1^\circ)\sin (2^\circ)\dots \sin (89^\circ)\right)^2= \frac{180}{2^{179}} \tag{2}$$
The value of
$$\boxed{ \sin(2^{\circ})\sin(4^{\circ}) \dots \sin(90^{\circ}) = \sqrt{\frac{180}{2^{179}}}\div \sqrt{\frac 1 {2^{89}}} = \sqrt{\frac{180}{2^{90}}}} \tag{3}$$
The required value seems to be in agreement with calculated value.
The proof of identity $(1)$ can be found at the end of this pdf.
| {
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"timestamp": "2023-03-29T00:00:00",
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Prove that $(4/5)^{\frac{4}{5}}$ is irrational Prove that $(4/5)^{\frac{4}{5}}$ is irrational.
My proof so far:
Suppose for contradiction that $(4/5)^{\frac{4}{5}}$ is rational.
Then $(4/5)^{\frac{4}{5}}$=$\dfrac{p}{q}$, where $p$,$q$ are integers.
Then $\dfrac{4^4}{5^4}=\dfrac{p^5}{q^5}$
$\therefore$ $4^4q^5=5^4p^5$
I've got to this point and now I don't know where to go from here.
| This is essentially the same as the standard proof that $\sqrt 2$ is irrational.
Let $z = (4/5)^{(4/5)}$. Now to calculate $z^5$:
$$
z^5 = \left(\left({4\over 5}\right)^{4\over 5}\right)^5 = \left({4\over 5}\right)^{4}
= {4^4\over 5^4}
$$
Clearly $z \neq 0$ as $0^{(5/4)} = 0 \neq 4/5$.
Let a rational number $r = p/q$ where $p$ and $q$ are positive integers.
Every positive integer is known to have a unique prime factorization. Specifically, we know there exist integers $i \geq 0$ and $t \geq 1$ such that $p = t \times 5^i$ and $t \not \equiv 0 \mod 5$.
Likewise, $q = u \times 5^j$. Thus, we have
$$
r = \frac pq = \frac tu \times 5^{(i-j)}
$$
and neither $t$ nor $u$ is divisible by $5$.
Now, we calculate $r^5$:
$$
r^5 = \frac{u^5}{v^5} \times 5^{5 (i-j)}
$$
If $r^5 = z^5$, then we must have $5(i-j) = -4$. There are clearly no integers $i,j$ which satisfy that. Thus we have shown that for any rational number $r$ it cannot be the case that $r = z = (4/5)^{(4/5)}$.
QED
| {
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Simplify expression, skanavi 2.001 How can I simplify this expression?
$$
\frac {\sqrt{x}+1}{x\sqrt{x} + x +\sqrt{x}} \colon \frac {1}{x^2 - \sqrt{x}}
$$
Solving:
$$
a = \sqrt x
$$
$$
\frac {a+1}{a^3 + a^2 + a} \colon \frac{1}{a^4-a} = \frac{ a(a + 1)(a-1)(a^2 + a + 1) }{a(a^2 + a + 1)} = (a +1)(a -1)=a^2-1
$$
Anser is: $x - 1$
| Notice that $x^2 - \sqrt{x} = \sqrt{x}\Big( (\sqrt{x})^3 - 1 \Big)$ and use the formula $a^3 - 1 = (a -1)(a^2+a+1)$
| {
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Inequality $\left(a-1+\frac{1}b\right)\left(b-1+\frac{1}c\right)\left(c-1+\frac{1}a\right)\leq1$ Let $a,b,c\in\mathbb{R}^*_+$, $abc=1$.
How can i show that $\left(a-1+\frac{1}b\right)\left(b-1+\frac{1}c\right)\left(c-1+\frac{1}a\right)\leq1$ ?
I got $\left(ab-b+1\right)\left(bc-c+1\right)\left(ca-a+1\right)\leq abc=1$, but i can't go any further ...
If i expand it :
$2-ab-\frac{b}a-c-ac-\frac{a}c-b+2a+\frac{2}a-bc-\frac{c}b-a+2b+\frac{2}b+2c+\frac{2}c-4$
Which gives using $c=\frac{1}{ab}$ :
$5ab-\frac{b}a-\frac{1}{ab}-a^2b+\frac{1}a+b+\frac{1}b-2$
| Make the substitution $a=x/y$, $b=y/z$, $c=z/x$ where $x$, $y$ and $z$ are positive reals. The inequality becomes
$$
\left(\frac xy - 1 + \frac zy\right) \left( \frac yz - 1 + \frac xz \right) \left( \frac zx - 1 + \frac yx \right) \stackrel{?}{\leq} 1
$$
or
$$ (x-y+z)(x+y-z)(-x+y+z) \stackrel{?}{\leq} xyz. $$
At most one of the factors on the left hand side can be nonpositive and in such a case the inequality is obvious. If all factors are positive, the result follows by multiplying
$$ (x-y+z)(x+y-z) \leq \left(\frac{(x-y+z)+(x+y-z)}{2}\right)^2 = x^2 $$
and similar inequalities. The last inequality follows by AM-GM on $x-y+z$ and $x+y-z$.
| {
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Trigonometry equation solution I have to solve this equation for all solutions
$$\sin(2x) = -\cos(2x)$$
Here are my steps
$$\sin(2x) + \cos(2x) = 0$$
$$\cos(2x)(\tan(2x) + 1) = 0$$
Upon solving these two equations, I find:
For cosine, $x = \frac{\pi}{4} + \pi n$
For both, $x = \frac{3 \pi}{8} + \pi n$
For both, $x = \frac{7 \pi}{8} + \pi n$
Where n is an integer.
The problem is that the first solution is incorrect, because
$$\sin\left(2\times\frac{\pi}{4}\right) + \cos\left(2\times\frac{\pi}{4}\right) = 1 + 0 = 1$$
But,
$$\cos\left(2 \times \frac{\pi}{4}\right) = 0$$
I know this solution is invalid, but why? Why, upon factorization, did one of the solutions give me an incorrect answer? Shouldn't it also be correct, mathematically speaking, because when expanding the equation I get $\sin(2x) = -\cos(2x)$
| Note that $\;\tan 2x = \frac{\sin 2x}{\cos 2x}\;$ is undefined when $\cos 2x = 0$, i.e., when $x = \pi/4 + k\pi$, so it cannot be a solution to either the original or factored equation.
| {
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Proving $\cos x < 1 - \frac{x^2}{2} +\frac{x^4}{24}$ I wish to prove the following inequality for $x\ne 0$:
$$\cos x < 1 - \frac{x^2}{2} +\frac{x^4}{24}$$
Using the fact that I already prove:
$$\cos x > 1 - \frac{x^2}{2}$$
My try:
$\cos x = 1 - \frac{x^2}{2} +\frac{x^4}{24} + R_2(x)$ where $R_2(x)$ is the remainder with the order of $n=2$.
By Lagrange's form of the remainder:
$$R_2(x) = \frac{cos^{(3)}(c)}{(3)!}x^3 = \frac{\sin c}{6}x^3$$
WLOG we assume $x>0$ and $0<c<x$
Now, what's left to prove is that: $$R_2(x) < \frac{x^4}{24}$$
But, I got the strange result: $x<4$.
What did I do wrong / what should I do instead?
I wish to rely on this inequality: $\cos x > 1 - \frac{x^2}{2}$
| By the symmetry of $\cos$ about $x=0$, we only need to prove it for positive $x$. Since $\cos t\gt 1-\frac{t^2}{2!}$, we have for positive $x$ that
$$\sin x=\int_0^x \cos t\,dt\gt \int_0^x \left(1-\frac{t^2}{2!}\right)\,dt=x-\frac{x^3}{3!}.$$
Do it again. We have
$$1-\cos x=\int_0^x \sin t\,dt\gt \int_0^x \left(t-\frac{t^3}{3!}\right)\,dt,$$
which gives us what we want. And if we want to continue, there is nothing to stop us.
Remark: One can do it with differentiation. We first prove that for positive $x$ we have $\sin x\gt x-\frac{x^3}{3!}$.
Let $f(x)=\sin x-\left(x-\frac{x^3}{3!}\right)$. We have $f(0)=0$. To show that $f(x)\gt 0$ for $x\gt 0$, it is enough to show that $f(x)$ is increasing.
Note that $f'(x)=\cos x-\left(1-\frac{x^2}{2!}\right)$. By the result you want to use, we have $f'(x)\gt 0$ if $x\gt 0$. It follows that $f(x)$ is increasing.
For your question about $1-\frac{x^2}{2!}+\frac{x^4}{4!}$, do it again. Let $g(x)=1-\frac{x^2}{2!}+\frac{x^4}{4!}-\cos x$. We have $g(0)=0$, and want to prove $g(x)$ is increasing for $x\gt 0$.
For this, we will use the fact that $g'(x)\gt 0$. Note that $g'(x)=\sin x-\left(x-\frac{x^3}{3!}\right)$, and we have just shown that this is positive.
| {
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How to use Legendre symbol to find a prime which divides $ax^2+b$? I'm trying to prove that $\dfrac{x^2-2}{2y^2+3}$is never an integer if $x,y\in\mathbb{Z}$.
It can be proven if $\forall p\in\mathbb{P}\:$doesn't suffice both of the following congruences:
$$\: x^2\equiv 2 \pmod{p}$$
$$2y^2\equiv -3\pmod{p}$$
If I could know that which prime $p$ suffices the congruence $ax^2+b\equiv 0\pmod{p}$ where $a\not\equiv 0\pmod{p}$ (in this case, $2y^2\equiv -3\pmod{p}$), I can prove the claim.
Could you give me any idea?
| There are odd primes $p$ such that the congruences $x^2\equiv 2\pmod{p}$ and $2y^2\equiv -3\pmod{p}$ have solutions. For example, $p=7$ works ($x=y=3$).
OP has asked for which (odd) primes $p$ does the congruence $2y^2\equiv -3\pmod{p}$ have a solution. It is not difficult to show that this congruence has a solution if and only if the congruence $z^2\equiv -6\pmod{p}$ has a solution.
Let $p\gt 3$. We want to find the primes such that the Legendre symbol $(-6/p)$ is equal to $1$. Since $(-6/p)=(2/p)(-3/p)$, there are two cases to consider, $(2/p)=1$ and $(2/p)=-1$.
In the first case, we want $(-3/p)=1$, and in the second case we want $(-3/p)=-1$.
We deal with the first case only. The analysis of the second case is very similar.
Because $(2/p)=1$, we want $p\equiv \pm 1\pmod{8}$. We have $(-3/p)=1$ if and only if $p\equiv 1\pmod{6}$, that is, $p\equiv 1$ or $7$ or $13$ or $19$ modulo $24$. The condition $p\equiv \pm 1\pmod{8}$ rule out the last two. So the first case holds if $p\equiv 1\pmod{24}$ or $p\equiv 7\pmod{24}$.
Remark: Finding for which $p$ we have $(-3/p)=1$ is straightforward. We divide into $p\equiv 1\pmod{4}$, which makes $(-1/p)=1$, and $p\equiv 3\pmod{4}$, which makes $(-1/p)=-1$.
For the case $p\equiv 1\pmod{4}$, we get $(3/p)=(p/3)$, so we want $p\equiv 1\pmod{3}$. In the case $p\equiv 3\pmod{4}$, we have $(3/p)=-(p/3)$, so again we end up wanting $p\equiv 1\pmod{3}$.
| {
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Solving a separable differential equation Solve the differential equation:
$$y'=\frac{1-y^2}{1-x^2}$$
My book says the solution is: $$y=\frac{x+c}{cx+1},$$ where $c$ is a constant.
It's been ten minutes I tried to verify if it was correct but I'm pretty sure the book is wrong. Can someone confirm it?
| \begin{align*}
\text{Separate:} && \frac{dy}{1-y^2} & =\frac{dx}{1-x^2} \\
\text{Partial fractions:} && \frac{1}{2}\left(\int\frac{1}{1+y}dy+\int\frac{1}{1-y}dy\right) & =\frac{1}{2}\left(\int\frac{1}{1+x}dx+\int\frac{1}{1-x}dx\right) \\
\text{Integrate:} && \log\frac{1+y}{1-y} & = \log c+\log\left(\frac{1+x}{1-x}\right) \\
\text{Solve for $y$:} && y & =\frac{c+cx-1+x}{c+cx+1-x} \\
\text{Factor:} && y & =\frac{x(c+1)+c-1}{x(c-1)+c+1} \\
\text{Divide by $c+1$:} && y & =\frac{x+\frac{c-1}{c+1}}{\frac{c-1}{c+1}x+1} \\
\text{Let $k=\frac{c-1}{c+1}$:} && y & =\frac{x+k}{kx+1}.
\end{align*}
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Help with finding eigenvectors and polynomial factorization I have a real and invertible matrix
$$ A = \left( \begin{matrix} 5 & 4 \\
4&5 \end{matrix} \right)$$
and I need to find the eigenvectors.
The characteristic polynomial is $(5-\lambda)^2 - 16$. According to Wolfram Alpha this factors to $(1-\lambda)(9-\lambda)$ (How does one get there, are there any good tutorials on polynomial factorization?), so the eigenvalues are $1$ and $9$.
Calculating the eigenvector to the eigenvalue $1$ given by $(A-I)\vec x = \vec 0$.
I get that $\vec x$ is the null vector $=\left(
\begin{array}{c}
0\\
0\\
\end{array}
\right)$(can that be right?).
For the second eigenvector $(A-9I)\vec x = 0$ I get
$$x_{1} = 4x_{2}$$
$$-x_{2} = 5x_{1}$$
Is this right, and how can I write the second eigenvector with this information??
| Well $(5-\lambda)^2-4^2$ is a difference of squares, but you really need to learn factoring.
$$\begin{pmatrix}
4&4\\4&4\\\end{pmatrix}\begin{pmatrix}x\\y\\\end{pmatrix}=0$$
gives $x=-y$ so $\begin{pmatrix}1\\-1\\\end{pmatrix}$ is one eigenvector,
and $$\begin{pmatrix}
-4&4\\4&-4\\\end{pmatrix}\begin{pmatrix}x\\y\\\end{pmatrix}=0$$
gives $x=y$ so $\begin{pmatrix}1\\1\\\end{pmatrix}$ is the other eigenvector,
| {
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"timestamp": "2023-03-29T00:00:00",
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$ a \cos A + b \cos B + c \cos C = \dfrac{a+b+c}2 $ $\implies$ the triangle is equilateral? If in a triangle $ a \cos A + b \cos B + c \cos C = \dfrac{a+b+c}2 $ , then is the triangle equilateral ?
| $\displaystyle a=2R\sin A\implies\cos A=R(\sin2A)$
$\displaystyle\implies\sum a\cos A=R(\sum\sin2A)$
Using Prove that $\sin(2A)+\sin(2B)+\sin(2C)=4\sin(A)\sin(B)\sin(C)$ when $A,B,C$ are angles of a triangle, $\displaystyle\sum a\cos A=4R\prod\sin A=4R\prod2\sin\frac A2\cos\frac A2$
Again, $\displaystyle\frac{a+b+c}2=R(\sum \sin A)$
Using this, $\displaystyle\frac{a+b+c}2=4R\prod\cos\frac A2$
As $\displaystyle0<A,B,C<\pi\implies\cos\frac A2$ (etc.) $\ne0,$ the problem reduces to $\displaystyle\prod2\sin\frac A2=1$
$\implies4\sin\frac A2\left(\cos\frac{B-C}2-\cos\frac{B+C}2\right)=1 $
As $\displaystyle B+C=\pi-A\implies \frac{B+C}2=\frac\pi2-\frac A2$
$\displaystyle\iff4\sin^2\frac A2-4\sin\frac A2\cos\frac{B-C}2+1=0 $ which is a Quadratic Equation in $\sin\frac A2$
As $A$ is real the discriminant $\displaystyle\left(4\cos\frac{B-C}2\right)^2-4\cdot4=-16\sin^2\frac{B-C}2$ must be $\ge0$
$\displaystyle\implies\sin^2\frac{B-C}2=0\iff\cos(B-C)=1-2\sin^2\frac{B-C}2=1\implies B-C=2n\pi $
As $\displaystyle0<B,C<\pi\implies B=C\implies\sin\frac A2=\frac12$
Can you prove $A=60^\circ?$
| {
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"timestamp": "2023-03-29T00:00:00",
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how to solve this Complex Equation? I have a doubt about the resolution of this equation:
$z^2 +|z^2 -1|=\frac{1}{2}(z+\bar{z})$
I tried to replace $z$ by $x+iy$ and $\bar{z}$ by $x-iy$.
From theory, I know that $|z|^2=z \cdot \bar{z}$, but I do not think this is the case ..
My question is: how do I evaluate $|z^2 -1|$ ?
After that, can I continue with the substitutions to solve the equation?
Thanks in advance: D
| $\left| z^2-1\right|\equiv \left| (x+yi)^2-1 \right|\equiv\left| x^2+2xyi-y^2-1\right|\equiv \left|[\color{green}{x^2-y^2-1}]+[\color{red}{2xy}]i\right|\equiv\sqrt{(\color{green}{x^2-y^2-1})^2+(\color{red}{2xy})^2}\equiv\sqrt{x^4+2x^2y^2-2x^2+y^4+2y^2+1}\equiv\sqrt{(x^2-2x+y^2+1)(x^2+2x+y^2+1)}$.
| {
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"timestamp": "2023-03-29T00:00:00",
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Proving trigonometric equation $\cos(36^\circ) - \cos(72^\circ) = 1/2$ Please help me to prove this trigonometric equation.
$\cos \left( 36^\circ \right)-\cos \left( 72^\circ \right) = \frac{1}{2}$
Thank you.
| From a recent post which relates to the above trig functions at the given angles, we have:
$\cos36^{\circ} = \sin54^{\circ} = \dfrac{\sqrt{5}+1}{4}$, and:
$\cos72^{\circ} = \sin18^{\circ} = \dfrac{\sqrt{5}-1}{4}$. Thus:
$\cos36^{\circ} - \cos72^{\circ} = \dfrac{\sqrt{5}+1}{4} - \dfrac{\sqrt{5}-1}{4} = \dfrac{1}{2}$
| {
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"timestamp": "2023-03-29T00:00:00",
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$\frac{1}{\sin^2(x)}+\frac{1}{\cos^2(x)} = \cos^2(x)+\sin^2(x) ?$ This is a simple question. Since $\cos(\theta)^2 + \sin(\theta)^2 = 1$
Can I take the inverse of this $\frac{1}{\cos^2(\theta)}+\frac{1}{\sin^2(\theta)} = \frac{1}{1}$?
Finally getting $\frac{1}{\cos^2(\theta)}+\frac{1}{\sin^2(\theta)} = \cos(\theta)^2 + \sin(\theta)^2$
If this is incorrect thinking can someone please put me on the right path?
Thanks
| No, since: $$\dfrac{1}{\cos^2\theta}+\dfrac{1}{\sin^2\theta}\neq \dfrac1{\cos^2\theta+\sin^2\theta}.$$
$$\dfrac{1}{\sin^2\theta}+\dfrac{1}{\cos^2\theta}=\dfrac{\cos^2\theta+\sin^2\theta}{\sin^2\theta\cos^2\theta}=\dfrac{1}{\sin^2\theta\cos^2\theta}\neq1.$$
In general, $\tfrac1a+\tfrac1b=\tfrac1{a+b}$ is not true, that's like saying that one half plus one half equals one quarter.
| {
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Apollonius’ Identity inner product space $||z-x||^2+||z-y||^2=\frac{1}{2}||x-y||^2+2||z-\frac{x+y}{2}||^2$
I proved it by expanding both sides and i found both sides are equal. Are there any easy way to prove it?
| In a Hilbert space you can use the Parallelogram law :
$$\|a+b\|^2 + \|a-b\|^2 = 2(\|a\|^2 + \|b\|^2)$$
So :
$$\begin{align}
&\|z - \frac{x+y}{2} \|^2\\
&= \|\frac{z}{2} - \frac{x}{2} + \frac{z}{2} - \frac{y}{2}\|^2\\
&= 2(\|\frac{z}{2} - \frac{x}{2}\|^2 + \|\frac{z}{2} - \frac{y}{2}\|^2) - \|\frac{y}{2} + \frac{x}{2}\|^2\\
&= \frac12(\|z - x\|^2 + \|z - y\|^2 - \frac{1}{4}\|y - x\|^2\\
&\square\end{align}$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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Evaluate the integrals $\int \sin{x} \cot^2{x} \,dx$ and $\int \cos{x} \cot^2{x} \,dx$. Can you please show how to evaluate the integrals $$\int \sin{x} \cot^2{x} \,dx$$ and $$\int \cos{x} \cot^2{x} \,dx.$$
I know that $\cot x=\dfrac{\cos x}{\sin x}$, which can simplify the integrands a bit.
$$\sin x\cot^2x = \frac{\cos^2x}{\sin x}\\
\cos x\cot^2x = \frac{\cos^3x}{\sin^2 x}$$
But I still do not know where to continue from there.
| HINT:
$$\sin x\cot^2x=\frac{\cos^2x}{\sin x}=\frac{1-\sin^2x}{\sin x}=\csc x-\sin x$$
$$\cos x\cot^2x=\frac{\cos^3x}{\sin^2x}=\frac{\cos^3x-\cos x+\cos x}{1-\cos^2x}=-\cos x+\frac{\cos x}{\sin^2x}$$
| {
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Bernoulli trials are run until 3 successful trials are completed. If probability of a success is 0.3 what is the expected number of failures? Bernoulli trials are run until 3 successful trials are completed. If probability of a success is 0.3 what is the expected number of failures?
My attempt:
Average number of successes in a n trials is $E[X] = np$. We know that the last trial must be a success, so perhaps we could find how many trails were likely made before the last trial in this way $n*0.3=2$, we get $n=6.7\approx 7$, then the answer is 7-2=5. I have no faith in this solution and I would like to see what you think is the correct way to solve this.
| Let $p$ be the probability of success (here $p = 0.3$), and $q = 1 - p$, the probability of failure (here $q = 0.7$). $X = k$ if there are $k - 1$ trials with exactly $2$ successes and $k - 3$ failures, and the $k$th trial is success.
The probability of exactly $2$ successes in a sequence of $k - 1$ trials is $^{k-1}\text{C}_2\, p^2 q^{k - 3}$ (by using a binomial distribution with $n = k - 1$). Multiplying this with $p$, the probability of success in the $k$th trial, we obtain
$$P(X = k) = ^{k-1}\text{C}_2\, p^3 q^{k - 3},\ k = 3, 4, \ldots$$
$\begin{align}
E[X] & = \sum_{k = 3}^{\infty} kP(X = k)\\
\mu & = \sum_k \dfrac{k(k - 1)(k - 2)}{2} p^3 q^{k - 3}\\
& = \dfrac{p^3}{2}\sum_k k(k - 1)(k - 2) q^{k - 3}\\
\int \mu\, dq & = \dfrac{p^3}{2}\sum_k k(k - 1) q^{k - 2}\\
\iint \mu\, dq\,dq & = \dfrac{p^3}{2}\sum_k k q^{k-1}\\
\iiint \mu\, dq\,dq\,dq & = \dfrac{p^3}{2}\sum_k q^k\\
& = \dfrac{p^3}{2} \left(\dfrac{1}{1 - q}\right)\\
\mu & = \dfrac{p^3}{2}\dfrac{d^3}{dq^3}\left( \dfrac{1}{1 - q} \right)\\
& = \dfrac{p^3}{2}\left( \dfrac{6}{(1 - q)^4} \right)\\
& = \dfrac{p^3}{2}\left( \dfrac{6}{p^4} \right)\\
& = \boxed{\dfrac{3}{p}}
\end{align}$
Thus, the expected number of trials is $\dfrac{3}{p} = \dfrac{3}{0.3} = 10$. Then the expected number of failures is $10 - 3 = 7$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/831697",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Systems of equations with three variables Consider:
$a + b = 5$
$2a + b + c = 4$
$a - b - c = 5$
I like to use substituiton for solving systems of equations, so I firstly look at equation 1 and solve for $a$
$a + b = 5$
$a = 5 - b$
I substitute this into equation 2
$2(5-b) + b + c = 4$
$10 -2b + b + c = 4$
$-b = 4 - c - 10$
$b = -4 + c + 10$
What next? Am I on the right track? Can I use substitution method for any simultaneous equation?
| You have the right concept, and we can see from three equations and three unknowns that these variables are solvable (taking into account equations are linearly independent)
equation 1: $\ a+b=5$
equation 2:$\ 2a+b+c= 4$
equation 3:$\ a-b-c =5$
lets solve for b first:
from equation 1 and 3, we say
$a=5-b$ and $\ c = a-b-5$
Now we plug these two 'new' equations into equation 2:
$\ 2(5-b) + b + ((5-b) - b -5) = 4 $
simplifying for b gives that $ b=2$.
Now taking that b=2, we can sub that back into equation 1 and solve for a and once we have a we can use a and b to solve for c. Which I'm sure you'll be able to solve for it.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Mass of ellipsoid
We are given the elipsoid $$\frac{x^2}{a^2}+\frac{y^2}{b^2}+\frac{z^2}{c^2} \leq 1$$ with density function $$p(x,y,z)=x^2+y^2+z^2$$ Find the mass of the elipsoid.
I used the transformation $x=ar\sin\theta \cos\phi$, $y=br\sin \theta \sin \phi$ and $z=cr\cos\theta$. Spherical coordinates but I multiplied $x,y,z$ by $a,b,c$ respectively.
I calculate the jacobian of this transformation, it is $abcr^2\sin\theta$.
So the mass is $$\int_{0}^{2\pi} \int_{0}^{\pi}\int_{0}^{1}(a^2r^2\sin^2\theta \cos^2\phi+b^2r^2\sin^2 \theta \sin ^2\phi+c^2r^2\cos^2 \theta)abcr^2\sin\theta drd\theta d\phi$$
But I'm having trouble calculating this integral. I thought maybe we can use that $$\sin^2x+\cos^2x=1$$ but because the coefficients are different it's difficult to use that here. How would I calculate this integral?
| You may integrate $\theta$ first, then $\phi$, and finally $r$.
$$I(r,\cos\theta,\phi):=(a^2r^2(1-\cos^2\theta) \cos^2\phi+b^2r^2(1-\cos^2 \theta) \sin ^2\phi+c^2r^2\cos^2 \theta)$$
$$J(r,\phi)=\int_{0}^{\pi}I(r,\cos\theta,\phi)\sin\theta d\theta=
\int_{-1}^{1}I(r,x,\phi)dx=\frac{2}{3}r^4(a^2 + b^2 + c^2 + (a^2 - b^2)\cos(2\phi))$$
$$K(r)=\int_0^{2\pi}J(r,\phi)d\phi=\frac{4}{3}(a^2 + b^2 + c^2)\pi r^4$$
$$L=\int_0^1 K(r)r^2 dr=\frac{4}{15}(a^2 + b^2 + c^2)\pi$$
Final result is
$$\frac{4}{15}abc(a^2 + b^2 + c^2)\pi$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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} |
Find the orthogonal trajectories of the family of curves given by $x^2 + y^2 + 2Cy =1$. Find the orthogonal trajectories of the family of curves given by $$x^2 + y^2 + 2Cy =1.$$
The ordinary differential equation for the family of curves is given by $y'=\frac{2xy}{x^2-y^2-1}$.Therefore, the differential equation for the orthogonal curves is given by $y'=\frac{1-x^2-y^2}{2xy}$.
This is an exact differential equation. So, solving by the standard method for exact differential equation gives $x-x^3/3+xy^2=C$. But this is not the correct answer according to the answers given at back of the book.
The answer given at the back of the book is $x^2 - y^2 - Cx +1 = 0$.
Can someone please find out at which step I am maing a mistake or provide a solution that leads to the correct answer?
| The answer at the end of your book does not seem correct. The correct one is $y^2 + x^2 - Cx + 1 = 0$.
The equation is solved most easily with the substitution (all derivatives in Lagrange notation are with respect to $x$):
$$ (y^2 + x^2 + 2Cy)' = (1)' \quad \Rightarrow \quad 2yy' + 2x + 2Cy' = 0 $$
Express $2C$ with the help of the initial equation:
$$y^2 + x^2 + 2Cy = 1 \quad \Rightarrow \quad 2C = \frac{1 - x^2 - y^2}{y}$$
Substitute this into the derivative equation:
$$2yy' + 2x + \frac{1 - x^2 - y^2}{y}y' = 0 \quad \Rightarrow \quad 2y^2y' + 2xy + y' -x^2y' - y^2y' = 0, y \neq 0$$
where we removed the $y$ in the denominator, by restricting the sought function $y$ to be nonzero. Now grouping the terms:
$$2xy + (y^2 - x^2 + 1)y' = 0 \quad \Rightarrow \quad y' = \frac{-2xy}{y^2 - x^2 + 1}$$
The family of orthogonal curves would have their direction field as a reciprocal of the direction field we found above^. Thus it can be expressed as $y_0$:
$$y_0' = \frac{y_0^2 - x^2 + 1}{2xy_0} \quad \Rightarrow \quad 2xy_0y_0' = y_0^2 - x^2 + 1$$
It is easy to notice now, that if we take $v = y_0^2$. then this expression is equivalent to:
$$xv' = v - x^2 + 1 \quad \Rightarrow \quad v' - \frac{1}{x}v =
\frac{1}{x} - x$$
But this is the first order differential equation for which we have an existence uniqueness theorem, and a formula, where $P(x) = -\frac{1}{x}$, and $Q(x) = \frac{1}{x} - x$, and $A(x) = \int_1^x P(t)dt = -\log(x)$. The point $x = 1$ is selected as an initial value point for simplicity, since the logarithm which appears in the integration becomes 0 there. Moreover, the fact that the term $1/x$ is integrated from 1 means that the integral is not defined at $x = 0$, so if the initial value is taken positive, the solution will be only defined for $x > 0$. This allows to remove the absolute value from the logarithm when after integration. The solution is then given by:
$$v = v(1)e^{-A(x)} + e^{-A(x)} \int_{1}^x e^{A(t)}Q(t)dt$$
or
$$v = v(1)e^{\log(x)} + e^{\log(x)}\int_1^x (\frac{1}{t^2} - 1)dt = v(1)x + x(-\frac{1}{x} + 1 - x + 1) = v(1)x - 1 + 2x - x^2$$
Now, $v(1)$ is the initial condition of the equation, which we assume to be some constant $C$. We group it with other similar terms:
$$v = (v(1) + 2)x - x^2 - 1 = -x^2 + Cx - 1$$
However, we remember that we substituted $v = y_0^2$. Thus:
$$y_0^2 = -x^2 + Cx - 1 \quad \Rightarrow \quad y_0^2 + x^2 - Cx + 1 = 0$$
which is the required answer.
PS. I do not know if this linear ODE solution formula is common knowledge, but it directly follows from the proof of uniqueness/existence theorem and sure is defined as such in Apostol calculus 1 page 310.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Evaluate $\int\frac{1}{\sec x\tan^2 x}dx$ $$\int\frac{1}{\sec x\tan^2 x}dx$$
I first thought of using the property $\int\sec^2 x = \tan x$ but figured it's no use.
How would I go by solving this question?
| $\int\dfrac{(1 - \sin^2 x)\cos x}{\sin^2 x}dx = \int\dfrac{(1 - \sin^2 x}{\sin^2 x}d(\sin x) = \int\biggl(\dfrac{1}{u^2} - 1\biggr)du = -\dfrac{1}{u} - u \quad \Rightarrow$
$\int \dfrac{\cos^3x}{\sin^2x}dx = -\dfrac{1}{\sin x} - \sin x + C$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/835881",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Limit of $\lim_{n\rightarrow \infty} \sum^{n}_{i=1} \frac{1}{\sqrt {n^2 +i} }$ Direct way leads to nowhere.
$$\lim_{n\rightarrow \infty} \sum^{n}_{i=1} \frac{1}{\sqrt {n^2 +i} } = \lim_{n\rightarrow \infty} \sum^{n}_{i=1} \frac{1}{n \sqrt {1 +\frac{i}{n^2}} } = $$
$$ = \lim_{n\rightarrow \infty} \frac{1}{n} \cdot \lim_{n\rightarrow \infty} \sum^{n}_{i=1} \frac{1}{\sqrt {1 +\frac{i}{n^2}} } = 0 \cdot \sum^{n}_{i=1} 1 = 0 \cdot n $$
So, I tried to apply squeeze theorem:
$$\lim_{n\rightarrow \infty} \sum^{n}_{i=1} \frac{1}{\sqrt {n^2 +i} } \leq \lim_{n\rightarrow \infty} \sum^{n}_{i=1} \frac{1}{n} = 1$$
Could you help me to figure out the lower bound?
| Notice that $$\frac{n}{\sqrt{n^2+n}} =\sum_{i=1}^n \frac{1}{\sqrt{n^2+n}}\leq \sum_{i=1}^n\frac{1}{\sqrt{n^2+i}}\leq \sum_{i=1}^n \frac{1}{n} =1$$ Since $\frac{n}{\sqrt{n^2+n}}=\frac{1}{\sqrt{1+\frac{1}{n}}}\to 1$, you get that $1$ as your limit by squeeze theorem.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/836006",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Calculation of $ \lim_{x\rightarrow 1}\frac{(1-x)\cdot(1-x^2)\cdot(1-x^3)\cdots(1-x^{2n})}{\{(1-x)\cdot(1-x^2)\cdot (1-x^3)\cdots(1-x^n)\}^2} = $ Calculation of $\displaystyle \lim_{x\rightarrow 1}\frac{(1-x)\cdot(1-x^2)\cdot(1-x^3)\cdots (1-x^{2n})}{\{(1-x)\cdot(1-x^2)\cdot (1-x^3)\cdots(1-x^n)\}^2} = $
My Trial After simplification, we get $$\displaystyle \lim_{x\rightarrow 1}\frac{(1-x^{n+1})\cdot(1-x^{n+2})\cdot(1-x^{n+3})\cdots(1-x^{2n})}{(1-x)\cdot(1-x^2)\cdot (1-x^3)\cdots(1-x^n)}$$
Now I did not understand How can I solve after that, Help me
Thanks
| Try the factorization $(1-x^n) = (1-x)(1+x+\dots+x^{n-1})$.
| {
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"timestamp": "2023-03-29T00:00:00",
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How to solve this equation in general $ax^3+bx^2+cx+d=0 ?$ For example if you have $ax^2+bx+c=0$ then solution given by formula $$x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}.$$
Is there formula like this for $ax^3+bx^2+cx+d=0 ?$
| Yes, such formula exists, and it is called the cubic formula, however it isn't nice looking: $$\begin{align}x &= \sqrt[\displaystyle3\,]{\left(\dfrac{-b^3}{27a^3} + \dfrac{bc}{6a^2} - \dfrac{d}{2a}\right) + \sqrt{\left(\dfrac{-b^3}{27a^3} + \dfrac{bc}{6a^2}-\dfrac{d}{2a}\right)^2 + \left(\dfrac{c}{3a}-\dfrac{b^2}{9a^2}\right)^3}}\\
& + \sqrt[\displaystyle3\,]{\left(\dfrac{-b^3}{27a^3}+\dfrac{bc}{6a^2}-\dfrac{d}{2a}\right) - \sqrt{\left(\dfrac{-b^3}{27a^3}+\dfrac{bc}{6a^2}-\dfrac{d}{2a}\right)^2 + \left(\dfrac{c}{3a}-\dfrac{b^2}{9a^2}\right)^3}} - \dfrac{b}{3a}.\end{align}$$
It can be made more simple, as suggested by @SenexÆgyptiParvi, if you let $P=3ac-b^2$ and $Q=27a^2d-9abc+2b^3$, then $$x={{{\sqrt[3]{4\left(-Q+\sqrt{4P^3+Q^2}\right)}}+{\sqrt[3]{4\left(-Q-\sqrt{4P^3+Q^2}\right)}}-{2ab}}\over{6a}}.$$
| {
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"timestamp": "2023-03-29T00:00:00",
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$\lim_{n \rightarrow \infty} \sqrt{n+1}(\sqrt{n^3 + n} - \sqrt{n^3 + 1})$ I wish to compute the following limit
$$\lim_{n \rightarrow \infty} \sqrt{n+1}(\sqrt{n^3 + n} - \sqrt{n^3 + 1})$$
Wolfram Alpha gives the answer as $1/2$. I've tried L'Hopital's rule, but I can't seem to get it to work.
| $$ \sqrt{n+1}(\sqrt{n^3 + n} - \sqrt{n^3 + 1}) \cdot \frac{\sqrt{n^3 + n}+ \sqrt{n^3 + 1}}{\sqrt{n^3 + n}+ \sqrt{n^3 + 1}} = \frac{\sqrt{n+1}(n-1)}{\sqrt{n^3 + n}+ \sqrt{n^3 + 1}}$$ Now divide numerator and denominator by $\sqrt{n^3}.$
| {
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"timestamp": "2023-03-29T00:00:00",
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Limit of a sequence $a_1=1;a_{n+1}=(n+1)(1+a_n)$ Let a sequence be $a_1=1;a_{n+1}=(n+1)(1+a_n)$
If $P_n=\prod_1^n(1+a_i^{-1})$
then $$\lim_{n\to\infty}P_n $$is ?
I did:
$$P_n=\prod_1^n\frac{(1+a_i)}{a_i}
=\frac{a_{n+1}}{a_1}\prod1^{n-1}\frac{(1+a_i)}{a_{i+1}}
=\frac{a_{n+1}}{a_1}\prod_1^{n-1}\frac1{i+1}=\frac{a_{n+1}}{n!}$$
Now how do I find the limit?
| Let a sequence be $a_1=1;a_{n+1}=(n+1)(1+a_n)$
We have $$P_n=\prod_{i=1}^n(1+a_i^{-1})$$
then
$$
P_n
=\prod_{i=1}^n (1 + \frac{1}{a_i})
=\left (1 + \frac{1}{n(1+a_{n-1})}\right )\prod_{i=1}^{n-1} (1+a_i^{-1} )
=\left (\frac{n(1+a_{n-1}) + 1}{n(1+a_{n-1})}\right )\prod_{i=1}^{n-1} (1+a_i^{-1} )
=\left (\frac{n(1+a_{n-1}) + 1}{n(1+a_{n-1})}\right )
\left (\frac{(n-1)(1+a_{n-2}) + 1}{(n-1)(1+a_{n-2})}\right )\prod_{i=1}^{n-2} (1+a_i^{-1} )
= \ldots
\approx \frac{n! + \prod_{i=1}^n a_i}{n! + \prod_{i=1}^n a_i} = 1
$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Local minimum of $f(x) = 4x + \frac{9\pi^2}{x} + \sin x$ What's the minimum value of the function
$$f(x) = 4x + \frac{9\pi^2}{x} + \sin x$$
for $0 < x < +\infty$? The answer should be $12\pi - 1$, but I get stuck with the expression involving both $\cos x$ and $x^2$ in the derivative. Taking the derivative, we have:
$$f'(x) = 4 - \frac{9\pi^2}{x^2} + \cos x.$$
In order to find the local extrema of the function, we set $f'(x) = 0$. Therefore,
\begin{align}
4 - \frac{9\pi^2}{x^2} + \cos x &= 0 \\
4x^2 - 9\pi^2 + x^2 \cos x &= 0 \\
x^2 (4 + \cos x) &= 9\pi^2.
\end{align}
However, I'm not sure what to do from here or if, indeed, I'm doing it right at all. Any help would be appreciated.
| Important fact: the minimum of the function is NOT the same as where $f'(x) = 0$. The latter is a local minimum - a point that it is minimal compared to it's local points, but taking as a simple example the function $f(x) = x^3$ on $[-1,1]$ the function is minimised at $-1$ despite this not being a turning point.
Despite this, in your example the local minimum happens to also be the global minimum. To solve the equation $x^2(4 + \cos(x ))= 9\pi^2$, the simplest way is to spot that $x = \frac{3\pi}{2}$ is a solution. As far as I'm aware, this kind of equation is not normally soluble by standard non-numerical methods, but the logic behind the guess is that if $\cos(x) = 0$, we're left with the equation $4x^2 = 9\pi^2$, and the solution to this gives $\cos(x) = 0$.
| {
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"url": "https://math.stackexchange.com/questions/843107",
"timestamp": "2023-03-29T00:00:00",
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Providing solutions for the intersection of two trigonometric functions I'm trying to find a general solution for the intersection of two trigonometric functions:
$$a(x)=500\sin \left( \frac{\pi x}{2} \right)+150$$
$$a(x)=200\cos \left( \frac{\pi }{2}\left( x+\frac{\pi }{2} \right) \right)+250$$
When graphing the equations together, it can be seen that they periodically intersect. However, I am trying to algebraically prove their intersection point by obtaining a general solution for the x-coordinates of their intersection, without using a graph.
Here's how I'm trying to solve it:
$$500\sin \left( \frac{\pi x}{2} \right)+150=200\cos \left( \frac{\pi }{2}\left( x+\frac{\pi }{2} \right) \right)+250$$
$$\frac{500\sin \left( \frac{\pi x}{2} \right)}{200}-\frac{100}{200}=\cos \left( \frac{\pi }{2}\left( x+\frac{\pi }{2} \right) \right)$$
$$\left( \frac{500\sin \left( \frac{\pi x}{2} \right)}{200}-\frac{100}{200} \right)^{2}=1-\sin ^{2}\left( \frac{\pi }{2}\left( x+\frac{\pi }{2} \right) \right)$$
so that
$$6.25\sin ^{2}\left( \frac{\pi x}{2} \right)-2.5\sin \left( \frac{\pi x}{2} \right)+\sin ^{2}\left( \frac{\pi x}{2}+\frac{\pi ^{2}}{4} \right)-0.75=0$$
But here's where I'm stuck. How can I eliminate the $\frac{\pi ^{2}}{4}$ term from the $\sin ^{2}\left( \frac{\pi x}{2}+\frac{\pi ^{2}}{4} \right)$ term so that I can have 2 like terms and be able to add the coefficients and solve the quadratic? Then I'd be able to find the reference angle for the general solution and provide an algebraic solution.
Any help appreciated.
| Hint:
let $a=\pi/4$.
Step (1) expand $\sin(2a(x+2a))$:
$$\sin(2a(x+2a))=\sin(2ax)\cos(4a^2)+\cos(2ax)\sin(4a^2)......(1)$$
Step (2) Change variable:
$$\sin(2ax)=\frac{2t}{1+t^2}, \cos(2ax)=\frac{1-t^2}{1+t^2},t=\tan(ax)$$
Step (3) solve the resulting equation for $t$.
Step (4) solve $t=\tan(ax)$ for $x$.
| {
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"timestamp": "2023-03-29T00:00:00",
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Show that $\lim_{x \rightarrow 1} \frac{x^4-2x+1}{x-1} + \sqrt{x} =3$
Show that $\lim_{x \rightarrow 1} \frac{x^4-2x+1}{x-1} + \sqrt{x} =3$ from the definition (using $\epsilon-\delta$)
Why can't I do something like this?
We want:
$|\frac{x^4-2x+1}{x-1} + \sqrt{x}-3| = |\frac{x^4-2x+1}{x-1} + \frac{x-9}{\sqrt{x}+3}| \le |\frac{x^4-2x+1}{x-1}| + |\frac{x-9}{\sqrt{x}+3}| \le |x^3+x^2+x-1| + |x-9| \le |x|^3 + |x|^2+|x-1| + |x-9| < \epsilon$
Now suppose $|x-1| < 1$, then $|x|\le 2$ and $|x-9| \le 9$, so
$$|x|^3 + |x|^2+|x-1| + |x-9| < 2^3+2^2+9+|x-1| < \epsilon$$
Now take $\delta = \min\{1, \epsilon -21\}$.
But this clearly is wrong because $\delta$ needs to be positive for every $\epsilon>0$, what part of my working is incorrect? (I know how to get the answer properly, just wondering why this way is incorrect, I'm quite new to $\epsilon-\delta$ proofs)
| Note: This answer was posted before the condition of using the definition with $(\epsilon-\delta)$ was specified.
Factorize $x^4-2x+1$, to remove the singularity in the denominator:
\begin{aligned}
\lim_{x\to1}\frac{x^4-2x+1}{x-1} + \sqrt{x} & = \lim_{x\to1}\frac{(x-1)(x^3+x^2+x-1)}{x-1} + \sqrt{x} = \\
& = \lim_{x\to1}(x^3+x^2+x-1) + \sqrt{x} = \\
& = 3
\end{aligned}
| {
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"timestamp": "2023-03-29T00:00:00",
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$A$ is not similar to a diagonal matrix over the reals Let $A = \begin{bmatrix} 6 & -3 & -2 \\ 4 & -1 & -2 \\ 10 & -5 & -3 \end{bmatrix} $ then $A$ is not similar to a diagonal matrix over the reals and it is not similar to a diagonal matrix over the complexes.
We know that $A$ is similar to a diagonal matrix over the reals(complexes) if there exist $D$ diagonal matriz and $P$ invertible matrix both $n \times n$ with real entries (complex entries) such that $A = PDP^{-1}$
I find that the inverse of $A$ is $A^{-1} = \frac{1}{2}\begin{bmatrix} -7 & 1 & 4 \\ -8 & 2 & 4 \\ -10 & 0 & 6 \end{bmatrix}$ and i diagonalize the matrix $A$ and got $A = PBP^{-1}$ with $ P = \begin{bmatrix} 1 & \frac{3}{5}-\frac{i}{2} & \frac{3}{5}+\frac{i}{2} \\ 0 & \frac{3}{5}-\frac{i}{2} & \frac{3}{5}+\frac{i}{2} \\ 2 & 1 & 1 \end{bmatrix} $ , $ P^{-1} = \begin{bmatrix} 1 & -1 & 0 \\ -1 + 3i & 1-\frac{i}{2} & \frac{1}{2}-\frac{3i}{2} \\ -1 - 3i & 1+\frac{i}{2} & \frac{1}{2}+\frac{3i}{2} \end{bmatrix} $ and finally i got $D = \begin{bmatrix} 2 & 0 & 0 \\ 0 & -i & 0 \\ 0 & 0 & i \end{bmatrix}$ then A is diagonalizable but my question is how can i conclude that $A$ is not similar to a diagonalize matrix over the reals? , please some help for this.
| *
*$A$ is diagonalisable to a matrix over $\mathbb{C}$.
*The diagonal components contain complex numbers.
*Matrix diagonals are unique (up to a permutation).
Therefore it is not diagonalisable over $\mathbb{R}$.
| {
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system of two quadratic equations with two variables Is there a general way to solve exactly a system of this shape (the $a_i$ are constants):
$$\begin{array}{cc}a_1x^2+a_2x+a_3y^2+a_4y+a_5=0\\
a_6xy+a_7x+a_8y+a_9=0
\end{array}
$$
It comes from a geometrical problem: the first equation states two vectors have same length, and the second ones state that they are perpendicular.
It can be reduced to one equation of degree four with one variable, but is it the best we can do?
| Assume your $a_1\neq0,\,a_6\neq0$. Dividing by these, you can have the slightly simpler system with two fewer constants,
$$x^2 + b_1 x + b_2 y^2 + b_3 y + b_4 = 0$$
$$x y + b_5 x + b_6 y + b_7 = 0$$
Eliminating $y$ results in a monic quartic of form,
$$x^4+ax^3+bx^2+cx+d=0\tag1$$
The solution of the general quartic is not that bad. It is just,
$$x_{1,2} = -\tfrac{1}{4}a+\tfrac{1}{2}\sqrt{u}\pm\tfrac{1}{4}\sqrt{3a^2-8b-4u+\frac{-a^3+4ab-8c}{\sqrt{u}}}\tag2$$
$$x_{3,4} = -\tfrac{1}{4}a-\tfrac{1}{2}\sqrt{u}\pm\tfrac{1}{4}\sqrt{3a^2-8b-4u-\frac{-a^3+4ab-8c}{\sqrt{u}}}\tag3$$
where,
$$u = \frac{a^2}{4} +\frac{-2b+v_1^{1/3}+v_2^{1/3}}{3}$$
and the $v_i$ are the two roots of the quadratic,
$$v^2 + (-2 b^3 + 9 a b c - 27 c^2 - 27 a^2 d + 72 b d)v + (b^2 - 3 a c + 12 d)^3 = 0$$
P.S. However, one has to be take note that some older CAS have $(-n)^{1/3} = \text{complex}$, even for real $n$.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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How to establish this inequality: $(1-a)(1-b)(1-c) \geq 8abc$ for $a+b+c=1$? Let $a$, $b$, and $c$ be positive real numbers such that $a+b+c = 1$. Then how to establish the following inequality? $$ (1-a)(1-b)(1-c) \geq 8abc.$$
My effort:
Since $a+b+c =1$, we can write $$ (1-a)(1-b)(1-c) = 1 - (a+b+c) +(ab+bc+ca) - abc = 1 - 1 +(ab+bc+ca) - abc = (ab+bc+ca) - abc = abc(a^{-1} + b^{-1} + c^{-1} ) - abc = 3abc (\frac{a^{-1} + b^{-1} + c^{-1}}{3}) - abc.$$ What next?
| Because $x+y \ge 2\sqrt{xy}$ for any $x>0,y>0$, and $1-a = (a+b+c) - a = b+c$ (similarly for $1-b$ and $1-c$) we have
$$(1-a)(1-b)(1-c) = (b+c)(c+a)(a+b) \ge 2\sqrt{bc}\cdot 2\sqrt{ca}\cdot 2\sqrt{ab} = 8abc.$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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For the matrix A as given below, which of them satisfy $A^6=I$? For the matrix A as given below, which of them satisfy $A^6=I$?
*
*$A=\left(\begin{array}{ccc}
\cos\frac{\pi}{4}&\sin\frac{\pi}{4}&0\\
-\sin\frac{\pi}{4}&\cos\frac{\pi}{4}&0\\
0&0&1
\end{array}\right)$
*$A=\left(\begin{array}{ccc}
1&0&0\\
0&cos\frac{\pi}{3}&\sin\frac{\pi}{3}\\
0&-\sin\frac{\pi}{3}&\cos\frac{\pi}{3}
\end{array}\right)$
*$A=\left(\begin{array}{ccc}
\cos\frac{\pi}{6}&0&\sin\frac{\pi}{6}\\
0&1&0\\
-\sin\frac{\pi}{6}&0&\cos\frac{\pi}{6}
\end{array}\right)$
*$A=\left(\begin{array}{ccc}
\cos\frac{\pi}{2}&\sin\frac{\pi}{2}&0\\
-\sin\frac{\pi}{2}&\cos\frac{\pi}{2}&0\\
0&0&1
\end{array}\right)$
I tried and conclude that the option 3 is correct by the technique of solving the equation$A^6=I$ and getting the sixth roots of unity. but i don't know whether it is correct or not. Kindly help me
| Let us take the option (3). the matrix is a rotation matrix. The rotation is about the Y-axis. The X and Z axes are rotated by $\frac{\pi}{6}$. So $A^6$ is same as rotating X and Z axes by $\frac{\pi}{6}$ about the Y Axis, sizx times. This is same as rotating X and Z axes by $\pi$ about the Y Axis, the rotation matrix for which has -I.
Can you do the others now?
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Finding Integral of $y = e^{ax}\sin (bx)$ I was solving questions like $ \int e^{2x} \sin x dx. $ I decided to find the general term for $$\int e^{ax} \sin (bx) $$ which can be directly used in questions like (above).
I hope it helps others :) (Just wanted to share my work. I will appreciate feedback/suggestions or any alternate method available)
| $\qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \large{\textbf{Method - 2}}$
Let : $I = \int e^{ax} \sin (bx) $
And say :
$$y_1 = e^{ax} \sin (bx) \\
y_2 = e^{ax} \cos (bx) $$
And : $$y'_1 = ae^{ax}\sin (bx) + e^{ax}b\cos (bx) \\
y'_2 = -e^{ax} b\sin (bx) + ae^{ax} \cos (bx) $$
Or we can also write $y'_1$ and $y'_2$ in the form of $y_1$ and $y_2$ as: $$y'_1 = ay_1 + by_2 \\ y'_2 = -by_1 + ay_2 $$
In a matrix, this can be written as : $$\bar y ' = \left[ \begin{matrix} a & -b \\ b & a \end{matrix} \right] \bar y$$
Inverting the Derivative Matrix :
$$\begin{align} D^{-1} = \cfrac{1}{a^2 + b^2} \left[ \begin{matrix} a & b \\ -b & a \end{matrix} \right] \\
= \left[ \begin{matrix} \cfrac{a}{a^2+b^2} & \cfrac{b}{a^2 + b^2} \\ \cfrac{-b}{a^2 + b^2} & \cfrac{a}{a^2 + b^2} \end{matrix} \right] \times \left( \begin{matrix} 1 \\ 0 \end{matrix} \right) \\
= \cfrac{1}{a^2 + b^2} \left[ \begin{matrix} a \\ -b \end{matrix} \right] \\
I = \cfrac{e^{ax}}{a^2+b^2} (a \sin bx - b \cos bx ) + C \end{align}$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Producing a CDF from a given PDF So I have this PDF:
$$
f(x)=
\begin{cases}
x + 3 & \text{ for } -3 \leq x < -2\\
3 - x & \text{ for } 2 \leq x < 3\\
0 & \text{ otherwise}
\end{cases}
$$
To make this a CDF, I have integrated the PDF from $-\infty$ to some value, $x$.
$$
F(x)=
\begin{cases}
\frac{x^2}{2} + 3x + \frac{9}{2} & \text{ for } -3 \leq x <-2\\
\frac{1}{2} & \text{ for } -2 \leq x<2\\
\frac{-x^2}{2} + 3x + \frac{7}{2} & \text{ for } 2 \leq x<3
\end{cases}
$$
My friend argues that the first term in this CDF which is $(x^2/2 + 3x + 9/2)$ should actually be $(x^2/2 + 3x)$. But isn't this impossible? At $x = -3$, the CDF must be $0$, am I correct?. This is only true in the case where the first term is $(x^2/2 + 3x + 9/2)$.
If someone could shed light on this topic, that would be much appreciated.
| Note first of all that the cdf is defined for all $x$, so a complete answer should take care of all $x$.
If $x\lt -3$, everything is easy. We have $\int_{-\infty}^x f_X(t)\,dt=0$. So $F_X(x)=0$ if $x\lt -3$.
Now for $-3\le x\lt -2$, in effect we are integrating $t+3$ from $-3$ to $x$, since there is no "mass" before $-3$. An antiderivative is $t^2/2+3t$. Plug in. We get $\frac{x^2}{2}+3x+\frac{9}{2}$.
Note that this is $\frac{1}{2}$ at $x=-2$. The density function is then $0$ for a while, so exactly as you had it, if $-2\le x\lt 2$ then $F_X(x)=\frac{1}{2}$.
Now for $2\le x\lt 3$, we want the integral up to $2$, plus the integral from $2$ to $x$ of $3-t$. We get
$$\frac{1}{2}+(3x-\frac{x^2}{2})- 4.$$
Finally, for $x\ge 3$, the cdf is $1$.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Discrete math induction problem. I am stuck at this step in the inductive process and I was wondering if someone can help me out from where I am stuck.
Question:
if $n$ is a positive integer, prove that,
$$1\cdot2+2\cdot3+3\cdot4+\cdots+n\cdot(n+1) = \frac{n(n+1)(n+2)}3$$
Basic step is $1$, then assume $k$ is true and we get,
$$1\cdot2+2\cdot3+3\cdot4+\cdots+k\cdot(k+1) = \frac{k(k+1)(k+2)}3$$
After that I am stuck because the answer key is telling me that the next step is to add $(k+1)(k+2)$, but I don't know why we do that. Aren't we supposed to just plugin $k+1$ into all $k$ variables. Why are we adding $(k+1)(k+2)$ to both sides?
| Finishing up the induction proof,\begin{align}
1 \cdot 2 + 2 \cdot 3 + 3 \cdot 4 + \cdots + (k+1)(k+2) &= [1 \cdot 2 + 2 \cdot 3 + 3 \cdot 4 + \cdots + k(k+1)]+(k+1)(k+2) \\
&= \frac{k(k+1)(k+2)}{3}+(k+1)(k+2) \\
&= \frac{k(k+1)(k+2)}{3}+\frac{3(k+1)(k+2)}{3} \\
&= \frac{k(k+1)(k+2)+3(k+1)(k+2)}{3} \\
&= \frac{(k+1)(k+2)(k+3)}{3}
\end{align}
where the last equality is by hypothesis. Now simplify the last expression accordingly and you will get your result.
| {
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"timestamp": "2023-03-29T00:00:00",
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confirmation of method: Write the vector $(2,-3)$ as a linear combination of vectors $(-1,4)$ and $(1,2)$ Write the vector $(2,-3)$ as a linear combination of vectors $(-1,4)$ and $(1,2)$
this was on an assignment rather hw so I want to be sure I'm using the right method and have the right answer.
I confirm $w = k_1u + k_2v$ to prove I have the right linear combination of vectors.
I solve for $k_1$ and $k_2$
$(-1,4)k_1 + (1,2)k_2 = w$
guassian elimination
$|-1,4 = 2|$
$| 1,2 = -3|$
$=$
$|-1,4 = 2|$
$| 0,6 = -1|$
back substitution
$k_2 = -1/6$
$-1k_1 = 6/3+2/3$
$k_1 = -8/3$
plug $k$'s into vector equation
$(-1,4)(-1/6) + (1,2)(-8/3) = w$
$(1/6,2/3) + (-8/3,-16/3) = w$
$-15/6,-14/3 = w$ .....nope... where did I do it wrong.. and am I using the right method?
| It shouldn't be this matrix
$$
\left(
\begin{array}{cc|c}
-1 & 4 & 2 \\
1 & 2 & -3
\end{array}
\right)
$$
since this corresponds to the system of equations
\begin{align*}
-k_1+4k_2 &= 2 \\
k_1+2k_2 &= -3 \\
\end{align*}
whereas you want the system of equations
\begin{align*}
-k_1+k_2 &= 2 \\
4k_1+2k_2 &= -3 \\
\end{align*}
to solve $k_1(-1,4)+k_2(1,2)=(2,-3)$.
So it should instead be
$$
\left(
\begin{array}{cc|c}
-1 & 1 & 2 \\
4 & 2 & -3
\end{array}
\right)
$$
Aside from that, the method should work fine.
| {
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"timestamp": "2023-03-29T00:00:00",
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How does $n+2\mid(n^2+n+1)(n^2+n+2)$ imply $n+2\mid12$? I'm trying to show that if $n+2\mid(n^2+n+1)(n^2+n+2)$, then $n+2 \mid 12$ for all $n\in \mathbb{N}$.
| Observe that $\displaystyle n^2+n=n(n+2)-(n+2)+2=(n+2)(n-1)+2$
$\displaystyle\implies (n^2+n+1)(n^2+n+2)=\{\underbrace{(n+2)(n-1)+2}+1\}\{\underbrace{(n+2)(n-1)+2}+2\}\equiv(2+1)(2+2)\pmod{(n+2)(n-1)}\equiv12\pmod{(n+2)} $
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Solving a quadratic trigonometric equation? The equation is $6 \cos^2x+\cos x=1$,
My work:
$6x^2+x-1=0$
$(3x-1)(2x+1)$
$3x-1=0 ∨ 2x+1=0$
$x=\frac{1}{3} ∨ x= \frac{-1}{2}$
But I do not know how to progress further.
| It is preferable to change the variable so let $t=\cos x$ and we solve the quadratic equation
$$6t^2+t=1$$
and we find $t=\frac13$ and $t=-\frac12$. Now
$$\cos x=\frac 13\iff \left(x=\arccos\left(\frac13\right)+2k\pi\right)\lor\left( x=2k\pi-\arccos\left(\frac13\right)\right)$$
and
$$\cos x=-\frac12\iff \left(x=\frac{2\pi}3+2k\pi\right)\lor\left( x=2k\pi-\frac{2\pi}3\right)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/858415",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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If $0\leq a \leq 3; 0\leq b \leq 3$ and the equation $x^2 +4+3 cos(ax+b)=2x$ has at least one solution , then find the value of a+b Problem : If $0\leq a \leq 3; 0\leq b \leq 3$ and the equation $x^2 +4+3 cos(ax+b)=2x$ has at least one solution , then find the value of a+b.
Solution :
We can write the given equation : $x^2 +4+3 cos(ax+b)=2x$ as $x^2-2x+4 =-3cos(ax+b)$
Since the L.H.S. of this problem is always $+ve$ $\forall x \in \mathbb{R}$ , and we know that $ -1 \leq cosx \leq 1$
But I am not getting any clue how to proceed further please help.
Thanks
| $$x^2-2x+4 =-3\cos(ax+b)$$
$$(x-1)^2 +3 =-3\cos(ax+b)$$
$$\frac{(x-1)^2 +3}{-3} =\cos(ax+b)$$
how $-1\leq\cos{(ax+b)}\leq 1$, then $x=1$
$$\cos(a+b)=-1$$
$$\cos(a+b)=-1$$
$$a+b=\pi$$
because $a,b\leq3$
| {
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"url": "https://math.stackexchange.com/questions/858547",
"timestamp": "2023-03-29T00:00:00",
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Trigonometry calculation problem Problem: Calculate $\cos(\alpha+60)$, if $\sin(\alpha)=\frac{2}{3}$ and $\alpha\in[90^\circ,180^\circ]$.
I have tried following:
$$\cos(60^\circ)\cos\alpha-\sin(60^\circ)\sin\alpha=\cos(\alpha+60^\circ)\\\frac{\cos\alpha}{2}-\frac{\sqrt{3}}{3}=\cos(\alpha+60^\circ)$$
I would appreciate any hints.
| Using the identity $$\sin^2 \alpha + \cos^2 \alpha = 1$$
$$\frac{4}{9} + \cos^2 \alpha = 1$$
$$\cos^2 \alpha = \frac{5}{9}$$
$$\cos \alpha = \frac{\sqrt{5}}{3}$$
Using another identity:
$$\cos(60^o)\cos\alpha-\sin(60^o)\sin\alpha=\cos(\alpha+60^o)\\\frac{\cos\alpha}{2}-\frac{\sqrt{3}}{3}=\cos(\alpha+60^o)$$
$$\cos(\alpha+60^o) = \frac{\frac{\sqrt{5}}{3}}{2} - \frac{\sqrt{3}}{3}$$
$$\cos(\alpha+60^o) = \frac{\sqrt{5}-2\sqrt{3}}{6}$$
| {
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "3",
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Easier way to solve $\lim \limits_{x \to 0} \space \cot{x}-{1 \over x}$ using L'Hospital's Rule? This is what I did for:
$$\lim \limits_{x \to 0} \space \cot{x}-{1 \over x}$$
*
*Check form: $\infty - \infty$.
*Rearrange it to be a quotient:
$$\begin{align} \\
& =\lim \limits_{x \to 0} \space {\cos{x} \over \sin{x}}-{1 \over x} \\
&\\
& =\lim \limits_{x \to 0} \space {x\cos{x} - \sin{x} \over x\sin{x}} \\
\end{align}$$
*Check form: $0 \over 0$
*Apply L'Hospital's Rule:
$$ = \lim \limits_{x \to 0} \space {(-x\sin{x} + \cos{x}) - \cos{x} \over (x\cos{x} + \sin{x})} $$
*Check form: $0 \over 0$.
*Apply L'Hospital's Rule Again:
$$ = \lim \limits_{x \to 0} \space {((-x\cos{x} -\sin{x}) +\sin{x}) -\sin{x} \over (-x\sin{x} + \cos{x}) + \cos{x}} $$
*Check form: $0\over2$.
Therefore:
$$\lim \limits_{x \to 0} \space \cot{x}-{1 \over x} = 0$$
This works, but seems that apply the rule twice made things really messy. Was there a simplification step that I missed along that way that would have made this easier to deal with (other than removing the parenthesis that I put in while using the power rule)?
| Speaking of simplicity, often by L'Hospital rule can be applied in combination with so-called remarkable limits. If this exercise can do so:
$$\lim \limits_{x \to 0}( \cot{x}-{1 \over x})=\lim \limits_{x \to 0} \frac{x\cos x-\sin x}{x\sin x} = \lim \limits_{x \to 0} \frac{x\cos x-\sin x}{x^2}\cdot\lim \limits_{x \to 0}\frac{x}{\sin x}=$$$$=\lim \limits_{x \to 0} \frac{x\cos x-\sin x}{x^2} = \lim \limits_{x \to 0}\frac{\cos x-x\sin x- \cos x}{2x}= \lim \limits_{x \to 0}\frac{-x\sin x}{2x}=0.$$
Simplification is more obvious in the case of the more complicated example:
$$\lim \limits_{x \to 0}( \cot^2 x-{1 \over x^2}) = \lim \limits_{x \to 0} \frac{x^2\cos^2 x-\sin^2 x}{x^2\sin^2 x} = \lim \limits_{x \to 0} \frac{x\cos x+\sin x}{x}\cdot\lim \limits_{x \to 0} \frac{x\cos x-\sin x}{x^3} \cdot\lim \limits_{x \to 0}\frac{x^2}{\sin^2 x}=\lim \limits_{x \to 0} (\cos x +\frac{\sin x}{x})\cdot\lim \limits_{x \to 0} \frac{x\cos x-\sin x}{x^3} = 2 \cdot \lim \limits_{x \to 0}\frac{\cos x-x\sin x- \cos x}{3x^2}= $$$$=-\frac{2}{3} $$
| {
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"timestamp": "2023-03-29T00:00:00",
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Find the number of integers solutions How many solutions are there to the equation $$x_1+x_2+x_3+x_4=39,$$
I) where $x_1,x_2,x_3,x_4$ are nonnegative integers,
II) where $x_1,x_2,x_3,x_4$ are nonnegative integers such that $3 \leq x_1,x_2,x_3 \leq 12$ and $0 \leq x_4$.
I) There is a 1-1 correspondence between the solutions and reordering of 39 ones and 3 zeros. Hence the answer is $\binom{42}{3}$
I did not understand very well this type of solution. What about the second case?
| Assume $a$ as a real number, such that $0<a <1$
I.
The number of integral solutions should be the same as:
The coefficient of $a ^ {39}$ in
$ \left(1+a + a^{2} + a^{3} + ... \right)^4$
Which is coeffient of $a^{39}$ in the series for
$ \left(1-a\right)^{-4}$
which is $\binom{39+4}{3}$
II.
Answer is same as the coefficient of $a ^ {39}$ in
$ \left(a^{3} + a^{4} + a^{5} + a^{6} + ... + a^{12} \right)^3\left(1+a + a^{2} + a^{3} + ... \right)$
= coeffient of $a^{30}$ in
$ \left(1+a+a^2+a^3+ ... + a^{9} \right)^3\left(1+a + a^{2} + a^{3} + ... \right)$
= coeffient of $a^{30}$ in
$ \left(1-a^{10}\right)^3\left(1-a\right)^{-4}$
= coeffient of $a^{30}$ in
$ \left(1-3a^{10} + 3a^{20} -a^{30}\right)\left(1-a\right)^{-4}$
Since $\left(1-a\right)^{-4}$ is
$1 +\binom{3+1}{3}a+\binom{3+2}{3}a^3+\binom{3+3}{3}a^3+ ...$
Ignore all powers more than 30
Answer
$= \binom{33}{3}-3\binom{23}{3}+3\binom{13}{3}-\binom{3}{3}$
| {
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Global Max and Min Hi can someone please help me figure out how to start this question?
Temperature within a circular region $R = \{(x, y)| x^2 + y^2 \le 49\}$ is given by $T(x,y) = 4x^2 -4xy + y^2$
Find the global minimum and maximum temperatures in the region and the points where these extreme temperatures occur.
I'm supposed to use Lagrange multipliers and parametrization which I don't really understand. If someone can please help me get started I would really appreciate it.
| Let $G(x,y) = x^2 + y^2 - 49$. At the interior points using critical points approach we have:
$T_x = 8x - 4y = 0$, and $T_y = 2y - 4x = 0$ both gives $y = 2x$. Thus. $T(x,2x) = 4x^2 - 4x(2x) + (2x)^2 = 0$.
At the boundary, using Lagrange Multiplier we have:$T_x = \lambda G_x$, and $T_y = \lambda G_y$ gives:
$8x - 4y = \lambda 2x$, and
$2y - 4x = \lambda 2y$.
Thus: $0 = 8x - 4y + 2(2y - 4x) = 2\lambda (x + 2y)$
So either $\lambda = 0$ or $x = -2y$.
Case 1: $\lambda = 0$,and $x \neq -2y$, then $y = 2x$. Thus: $x^2 + (2x)^2 = 49 \to 5x^2 = 49 \to x = \pm \dfrac{7}{\sqrt{5}}$. So $y = \pm \dfrac{14}{\sqrt{5}}$. Thus: $T\left(\pm \dfrac{7}{\sqrt{5}},\pm \dfrac{14}{\sqrt{5}}\right) = 0$
Case 2: $x = -2y$, and $\lambda \neq 0$, then $(-2y)^2 +y^2 = 49 \to 5y^2 = 49 \to y = \pm \dfrac{7}{\sqrt{5}}$, and $x = \mp \dfrac{14}{\sqrt{5}}$. Thus: $T\left(\dfrac{7}{\sqrt{5}}, -\dfrac{14}{\sqrt{5}}\right) = \dfrac{784}{5} = T\left(-\dfrac{7}{\sqrt{5}}, \dfrac{14}{\sqrt{5}}\right)$
Case 3: $\lambda = 0$, and $x = -2y$. So: $y = 2x = -4y$, and this gives: $y = 0 = x$. Thus: $T(0,0) = 0$.
In summary: $T_{min} = T(x,2x) = 0$, and $T_{max} = \dfrac{784}{5}$ achieved at $(x,y) = \left(\pm \dfrac{7}{\sqrt{5}}, \mp \dfrac{14}{\sqrt{5}}\right)$
Note: $T(x,y) = 4x^2 - 4xy + y^2 = (2x - y)^2 \geq 0$. This implies that $T_{min} = 0$ when $y = 2x$, and this is justified by the critical points method.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/866434",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
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} |
Solve this with CBS: minimum value of $ 1/x + 4/y + 9/z $ with $x+y+z=1$ How can you see the minimum value of $ 1/x + 4/y + 9/z $ with $x+y+z=1$ using the CBS inequality?
I have seen a proof of that that use trigonometric substitutions, but I don't see as one-step the solution using the CBS inequality.
| Consider positive values $x,y,z$.
Denote
$$(a,b,c)=\left( \frac{1}{\sqrt{x}}, \frac{2}{\sqrt{y}}, \frac{3}{\sqrt{z}}\right),$$
$$(A,B,C)=\left( \sqrt{x}, \sqrt{y}, \sqrt{z}\right),$$
then (by CBS inequality)
$$(a^2+b^2+c^2)(A^2+B^2+C^2)\ge (aA+bB+cC)^2,$$
so
$$
\left(\frac{1}{x}+\frac{4}{y}+\frac{9}{z}\right) \left(x+y+z\right)\ge (1+2+3)^2=36;
$$
$$
\frac{1}{x}+\frac{4}{y}+\frac{9}{z}\ge 36.
$$ (I have added the square)
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/866858",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
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} |
Find the radius Consider the parabola $y=x^2$ and a circle which is tangent to the parabola at the points $(1,1)$ and $(-1,1)$.Find the radius of circle.
My try:I write the general equation of circle
$(x-h)^2+(y-k)^2=r^2$
and substitute the points $(1,1)$ and $(-1,1)$ in the equation of circle,i find $h=0$.
Now further what i should do to find r?
| Substitute $(1,1),(-1,1)$ in $(x-h)^2+(y-k)^2=r^2$
you will get
$h^2+k^2-2h-2k+2=r^2$ and $h^2+k^2+2h-2k+2=r^2$
From here you get $h=0$, substitute this in one of the above to obtain
$k^2-2k+2=r^2$
$m_1$=slope of tangent line at $(1,1)=2$ (here use the $y=x^2$)
Where $m_2$=the slope of radial line at $(1,1)= 1-k$ (here use the equation of circle)
since $m_1m_2=-1$, so we have $k=\frac{3}{2}$
Substitute $k=\frac{3}{2}$ in
$$k^2-2k+2=r^2$$
Finally you will get the answer $r=\frac{\sqrt{5}}{2}$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Integral $\int_1^{\sqrt{2}}\frac{1}{x}\ln\left(\frac{2-2x^2+x^4}{2x-2x^2+x^3}\right)dx$
Calculate the following integral:
\begin{equation}
\int_1^{\sqrt{2}}\frac{1}{x}\ln\left(\frac{2-2x^2+x^4}{2x-2x^2+x^3}\right)dx
\end{equation}
I am having trouble to calculate the integral. I tried to use by parts method but it didn't help. Wolfram Alpha gives me $0$ as the answer but I don't know how to get it. I also tried to search the similar question here and I got this:
$$I=\int_{-1}^1\frac1x\sqrt{\frac{1+x}{1-x}}\ln\left(\frac{2\,x^2+2\,x+1}{2\,x^2-2\,x+1}\right)\ \mathrm dx,$$
yet it didn't help much. Besides, I don't understand the answers there. Could anyone here please help me to calculate the integral preferably (if possible) with elementary ways (high school methods)? Any help would be greatly appreciated. Thank you.
| Update: Finally, a complete solution. Sorry it took so long.
Split the integral up into 3.
\begin{align}
I
&=-\int^{\sqrt{2}}_1\frac{\log{x}}{x}dx+\int^{\sqrt{2}}_1\frac{\log{((x^2-1)^2+1)}}{x}dx-\int^{\sqrt{2}}_1\frac{\log{((x-1)^2+1)}}{x}dx\\
&=-\frac{1}{8}(\log{2})^2+\frac{1}{2}\int^1_0\frac{\log(1+x^2)}{1+x}dx-\int^{\sqrt{2}-1}_0\frac{\log(1+x^2)}{1+x}dx
\end{align}
The second integral is rather easy to evaluate.
\begin{align}
\frac{1}{2}\int^1_0\frac{\log(1+x^2)}{1+x}dx
&=\frac{1}{2}\int^1_0\int^1_0\frac{x^2}{(1+x)(1+ax^2)}dx \ da\tag1\\
&=\frac{1}{2}\int^1_0\frac{1}{1+a}\int^1_0\frac{1}{1+x}+\frac{x-1}{1+ax^2}dx \ da\\
&=\frac{1}{2}\int^1_0\frac{\log{2}}{1+a}+\frac{\log(1+a)}{2a(1+a)}-\underbrace{\frac{\arctan(\sqrt{a})}{\sqrt{a}(1+a)}}_{\text{Let} \ y=\arctan{\sqrt{a}}}da\\
&=\frac{1}{2}\left[(\log{2})^2+\frac{1}{2}\underbrace{\int^1_0\frac{\log(1+a)}{a}da}_{-\operatorname{Li}_2(-1)=\frac{\pi^2}{12}}-\frac{1}{2}\underbrace{\int^1_0\frac{\log(1+a)}{1+a}da}_{\frac{1}{2}(\log{2})^2}-\frac{\pi^2}{16}\right]\\
&=\frac{3}{8}(\log{2})^2-\frac{\pi^2}{96}
\end{align}
The third integral can be evaluated using dilogarithms.
\begin{align}
\int^{\sqrt{2}-1}_0\frac{\log(1+x^2)}{1+x}dx
&=\sum_{r=\pm i}\int^{\sqrt{2}-1}_0\frac{\log(r+x)}{1+x}dx\tag2\\
&=-\sum_{r=\pm i}\int^{\frac{\lambda}{\sqrt{2}}}_{\lambda}\log\left(r-1+\frac{\lambda}{y}\right)\frac{dy}{y}\tag3\\
&=-\sum_{r=\pm i}\int^{\frac{r-1}{\sqrt{2}}}_{r-1}\frac{\log(1+y)}{y}-\frac{1}{y}\log\left(\frac{y}{r-1}\right)dy\tag4\\
&=\frac{1}{4}(\log{2})^2+\sum_{r=\pm i}\mathrm{Li}_2\left(\frac{1-r}{\sqrt{2}}\right)-\mathrm{Li}_2(1-r)\tag5\\
&=\frac{1}{4}(\log{2})^2+\mathrm{Li}_2(e^{i\pi/4})+\mathrm{Li}_2(e^{-i\pi/4})-\mathrm{Li}_2(\sqrt{2}e^{i\pi/4})-\mathrm{Li}_2(\sqrt{2}e^{-i\pi/4})\\
&=\frac{1}{4}(\log{2})^2-\frac{\pi^2}{96}\tag6\\
\end{align}
It follows that
$$I=-\frac{1}{8}(\log{2})^2+\frac{3}{8}(\log{2})^2-\frac{\pi^2}{96}-\frac{1}{4}(\log{2})^2+\frac{\pi^2}{96}=0$$
Explanation
$(1)$: Differentiate under the integral sign
$(2)$: Factorise $1+x^2$, let $r=\pm i$
$(3)$: Let $\displaystyle y=\frac{\lambda}{1+x}$
$(4)$: Let $\lambda=r-1$
$(5)$: Recognise that $\displaystyle\int\frac{\ln(1+y)}{y}dy=-\mathrm{Li}_2(-y)+C$ and $\displaystyle\int\frac{\ln(ay)}{y}dy=\frac{1}{2}\ln^2(ay)+C$
$(6)$: Use the identities here
| {
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"url": "https://math.stackexchange.com/questions/870751",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "15",
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Linear Algebra - elimination and linear systems By given this matrix:
\begin{pmatrix}1&1&1&0\\2&3&k&1\\3&k&5&1\end{pmatrix}
I need to find, what are the values of k the system has infinity/single/no solution.
So I did elimination and I got:
\begin{pmatrix}1&1&1&0\\0&1&k-2&1\\0&k-4&4-k&0\end{pmatrix}
As one can see, for k=4 there are infinity solutions and for any k which is not 4 there is only a single solution.
But my friend told me that for k=1 there's no solution.
So what I did wrong? I have a mistake at the elimination procces?
I'd like to get some help.
Thanks.
| You have obtained
$$\left(\begin{array}{ccc|c}
1 & 1 & 1 & 0 \\
0 & 1 & k-2 & 1 \\
0 & k-4 & 4-k & 0
\end{array}\right)$$
and correctly explained what happens for $k=4$.
Now let us have a look at the case $k\ne 4$. In this case $k-4\ne0$ and you can divide the last row by $(k-4)$:
$$\left(\begin{array}{ccc|c}
1 & 1 & 1 & 0 \\
0 & 1 & k-2 & 1 \\
0 & k-4 & 4-k & 0
\end{array}\right)\sim
\left(\begin{array}{ccc|c}
1 & 1 & 1 & 0 \\
0 & 1 & k-2 & 1 \\
0 & 1 & -1 & 0
\end{array}\right)\sim
\left(\begin{array}{ccc|c}
1 & 1 & 1 & 0 \\
0 & 0 & k-1 & 1 \\
0 & 1 & -1 & 0
\end{array}\right)
$$
Now for $k=1$, the second row corresponds to the equation $0x_1+0x_2+0x_3=1$; which obviously has no solution.
For $k\ne1$ you can divide the second row by $(k-1)$ and you get unique solution.
You can even compute the solution if you with to do so:
$$
\left(\begin{array}{ccc|c}
1 & 1 & 1 & 0 \\
0 & 0 & k-1 & 1 \\
0 & 1 & -1 & 0
\end{array}\right)\sim
\left(\begin{array}{ccc|c}
1 & 1 & 1 & 0 \\
0 & 0 & 1 & \frac1{k-1} \\
0 & 1 & -1 & 0
\end{array}\right)\sim
\left(\begin{array}{ccc|c}
1 & 1 & 0 & -\frac1{k-1} \\
0 & 0 & 1 & \frac1{k-1} \\
0 & 1 & 0 & \frac1{k-1}
\end{array}\right)\sim
\left(\begin{array}{ccc|c}
1 & 0 & 0 & -\frac2{k-1} \\
0 & 1 & 0 & \frac1{k-1} \\
0 & 0 & 1 & \frac1{k-1} \\
\end{array}\right)
$$
So the unique solution (for $k\ne1,4$) is $x_1=-\frac2{k-1}$, $x_2=\frac1{k-1}$, $x_3=\frac1{k-1}$.
You can plug these values into the original system and check that $x_1+x_2+x_3=0$, $2x_1+3x_2+kx_3=\frac{-4+3+k}{k-1}=1$ and $3x_1+kx_2+5x_3=\frac{-6+k+5}{k-1}=1$; so it is indeed a solution.
If you also want to describe the solution set for the parameter $k=4$, you have to solve this case separately.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Suggestion for Computing an Integral Let $$A=\left\{(x,y,z)\in \mathbb R^3:\dfrac{x^2}{2}+\dfrac{y^4}{4}+\dfrac{z^6}{6}\leq1\right\}.$$
Then I want to compute the following integral:
$$\frac{1}{\operatorname{vol}(A)}\displaystyle\int_{\partial A}^{}\!\frac{1}{\sqrt{x^2+y^6+z^{10}}}\, d(x,y,z)$$
Should I use spherical coordinates or something like that?
I am a beginner to this topic, so any help would be nice.
Edit: I put the factor $\frac{1}{\operatorname{vol}(A)}$ before the integral (that's exactly my task now), but there should be no different concerning our problem...
$$I := \frac{1}{\lambda_3(A)} \int\limits_{\partial A} \frac{1}{\sqrt{x^2 + y^6 + z^{10}}} \, dS_{\partial A}$$
| Notice that the normal vector of $A's$ boundary is $\hat{n}=\frac{(x, y^3, z^5)}{\sqrt{x^2+y^6+z^10}}$. Then we choose $F(x,y,z)=(\frac{x}{2},\frac{y}{4},\frac{z}{6})$, on the boundary, we have $$F\cdot \hat{n}=\frac{\frac{x^2}{2}+\frac{y^4}{4}+\frac{z^6}{6}}{\sqrt{x^2+y^6+z^{10}}}=\frac{1}{\sqrt{x^2+y^6+z^{10}}}$$ So the integral
$$\frac{1}{vol(A)}\int_{\partial A}\frac{1}{\sqrt{x^2+y^6+z^{10}}}ds=\frac{1}{vol(A)}\int_{\partial A} F\cdot \hat{n} ds=\frac{1}{vol(A)}\int_{ A}\nabla\cdot F dxdydz \quad= \frac{1}{vol(A)}\int_{ A} \frac{1}{2}+\frac{1}{4}+\frac{1}{6}=\frac{11}{12}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/871950",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
"answer_count": 3,
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} |
Integration problem $\displaystyle \int \frac{dx}{x(x^3+8)}$ $$\int \frac{dx}{x(x^3+8)}$$
I think I'm supposed to use partial fractions, but I am unsure of how to start the problem. Any help would be appreciated.
| Here's the set up:
$$\int \frac{dx}{x(x^3 + 8)} = \int \frac{dx}{x(x+2)(x^2 -2x + 4)} = \int \left(\frac{A}{x} + \frac{B}{x+2} + \frac{Cx + D}{x^2-2x + 4}\right) \,dx$$
Now you can solve for $A, B, C, D$.
Note: Differences and sums of cubes factor predictably: $$a^3 \pm b^3 = (a \pm b)(a^2 \mp ab + b^2)$$
In this question, $x^3 + 8 = x^3 + 2^3$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/872036",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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$\ \sqrt{x+39}-\sqrt{x+7}=4 $ So I tried to solve this problem for x
$\ \sqrt{x+39}-\sqrt{x+7}=4 $
I multiplied both sides ($\ \sqrt{m}\cdot\sqrt{n}=\sqrt{mn} $)
$\ (\sqrt{x+39}-\sqrt{x+7})^2=16 $
$\ (x+39)-2(x^2+46x+273)-(x+7) $
$\ 0x+32+(-2x^2-92x-546) $
$\ -2x^2-92x-514 $
divide the 2 out
$\ x^2+46x+257=-8 $
$\ x^2+46x+265=0 $
Use the quadratic formula (or scientific calculator) and the answers are -6.752 and -39.248. I know the answer is exactly -3. What went wrong?
| One can square twice. I prefer to invert and obtain
$$\frac{1}{\sqrt{x+39}-\sqrt{x+7}}=\frac{1}{4},$$
and then by rationalizing the denominator get
$$\frac{\sqrt{x+39}+\sqrt{x+7}}{32}=\frac{1}{4},$$
or equivalently
$$\sqrt{x+39}+\sqrt{x+7}=8.$$
Now "add" to the original equation, divide by $2$. We get $\sqrt{x+39}=6$, and now it's over.
Remark: In the original post, the first step used was to square. This to some degree complicates things, the middle term should have been $-2\sqrt{(x+39)(x+7)}$. Rearrangement and squaring now get us to where we want.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Sum of all real solutions for $x$ to the equation $\displaystyle (x^2+4x+6)^{{(x^2+4x+6)}^{\left(x^2+4x+6\right)}}=2014.$ Find the sum of all real solutions for $x$ to the equation $\displaystyle (x^2+4x+6)^{{(x^2+4x+6)}^{\left(x^2+4x+6\right)}}=2014.$
$\bf{My\; Try::}$ Let $y=x^2+4x+6 = (x+2)^2+2\geq 2$.
So our exp. equation convert into $\displaystyle y^{y^{y}} = 2014\;,$ where $y\geq 2$
Now at $y=2\;,$ We Get $\displaystyle 2^{2^{2}} = 16<2014$ and $\displaystyle 3^{3^{3}} = 3^{27}>2014$
So $y$ must be lie between $2$ and $3$.
But I did not Understand How can I calculate it..
Help me
Thanks
| As you have written if one substitutes $y=x^2 + 4x +6 = (x+2)^2+2\ge 2$, then the resulting equation is $f(y)=2014$ where $f(y)=y^{y^y}$. Due the the fact that $2014>2^{2^2}$, it accepts solutions in ${\mathbb R}$. Now note that $f(y)=y^{y^y}$ is an strictly increasing function on ${\mathbb R}_{\ge 2}$. So for $f(y)=2014$ there exist a unique $y_0\in {\mathbb R}_{\ge 2}$ such that $f(y_0)=2014$. So one have that
$$
x^2 + 4x +6-y_0 = 0,
$$
where the sum of solutions is $-4$.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Which one is correct for $\sqrt{-16} \times \sqrt{-1}$? $4$ or $-4$? As we can find in order to evaluate $\sqrt{-16} \times \sqrt{-1}$, we can do it in two ways.
FIRST
\begin{align*}
\sqrt{-16} \times \sqrt{-1} &= \sqrt{(-16) \times (-1)}\\
&= \sqrt{16}\\
&=4
\end{align*}
SECOND
\begin{align*}
\sqrt{-16} \times \sqrt{-1} &= \sqrt{16i^2} \times \sqrt{i^2}\\
&= 4i \times i\\
&=4i^2\\
&=-4
\end{align*}
Incidentally if the above is input in complex mode of Casio scientific calculator, the result comes out as $-4$.
Which of the above solutions is correct?
| $$\sqrt[n]{a b} = \sqrt[n]{a} \, \sqrt[n]{b} \quad (*)$$
if $a$ and $b$ are negative, then $(*)$ works only for $n$-th roots with odd $n$,
alas $n = 2$ is even.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/876084",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 2
} |
Easy algebra question: show equality? Let $c:=(u+d)/2$ and $-1<d<0<u$.
Show:
Can you show the steps to arrive from lhs to rhs?
| $$
\frac{1 + u}{2}(1 + 2c) + \frac{1+d}{2}(1 + c) = \frac{1 + u}{2}(1 + c) + \frac{1 + u}{2}c + \frac{1+d}{2}(1 + c)
$$
$$
=(1 + c)\biggl(\frac{1 + u}{2} + \frac{1 + d}{2} \biggr) + \frac{1 + u}{2}c = (1 + c)\biggl(1 + \frac{u + d}{2}\biggr) + \frac{1 + u}{2}c
$$
$$
= (1 + c)^2 + \frac{1 + u}{2}c
$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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Why $ (1- \sin \alpha + \cos \alpha)^2 = 2 (1 - \sin \alpha)(1+ \cos \alpha)$? Why $ (1- \sin \alpha + \cos \alpha)^2 = 2 (1 - \sin \alpha)(1+ \cos \alpha)$?
I am learning trigonometric identities one identity I have to proof is the next:
$ (1- \sin \alpha + \cos \alpha)^2 = 2 (1 - \sin \alpha)(1+ \cos \alpha)$
so I tried to resolve the identity for the left:
$ 1 + \sin^2 \alpha+ \cos^2\alpha - 2\sin\alpha + 2\cos\alpha - 2\sin\alpha\cos\alpha $
$= 1 + 1 - 2\sin\alpha + 2\cos\alpha - 2\sin\alpha\cos\alpha $
$= 2 (1 - \sin\alpha + \cos\alpha - \sin\alpha\cos\alpha)$
And I got stuck, I did not know what to do, so I went to see the problem's answer and I was going fine, the part I was not able to resolve is the next one:
$ 2 (1 - \sin\alpha + \cos\alpha - \sin\alpha\cos\alpha)$
$= 2 (1 + \cos\alpha - \sin\alpha(1+\cos\alpha))$
$= 2 (1 - \sin \alpha)(1+ \cos \alpha)$
So the question is how did the teacher do the last three steps? I cant figure it out.
| They are just the distributive property
$$a(b+c)=ab+ac$$
in the first one we have $a=-\sin\alpha, b= 1, c=\cos\alpha$ which comes from the terms
$$2(1+\underbrace{(-\sin\alpha)}_{a}\cdot\underbrace{1}_{b}+\cos\alpha+\underbrace{(-\sin\alpha)}_{a}\cdot\underbrace{\cos\alpha}_{c})$$
The next step is the same property with $a=(1+\sin\alpha), b=1, c=\cos\alpha$ as seen
$$2(\underbrace{(1+\cos\alpha)}_{a}\cdot \underbrace{1}_{b}+\underbrace{(1+\cos\alpha)}_{a}\underbrace{(1-\sin\alpha)}_{c}).$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Determining the infinite limit of a Riemann' sum I need to evaluate the following $f(x)=x^2 + 2x - 5$, on $[1,4]$, by using the riemans sum and limiting it to infinity. I have set up everything $\Delta x=\frac{3}{n}$.
$x_i=1+\frac{3i}{n}$, I would like to just check my workings against someone who actually finds this fairly simple, the part which i struggle with the most is the simplifying of the summation.
Pg. 1/2 working out
Pg. 2/2 working out
P.S: Apologies for untidy hand writing
| My suggestion is that when you calculate an integral using Riemann sums it might be easier to find $\int_0^4 f-\int_0^1 f$ than $\int_1^4 f$ in one step -avoiding terms of the form $(h+\frac{k}{n})^c$-.
Let $R_n^{4}$ be the Riemann sum of n-terms from $0$ to $1 $ and $R_n^1$ the Riemann sum from $0$ to $1$. It should be fairly straight forward to compute each one of these separately.
$$R_n^4=\sum_{i=1}^n f\left(\frac{4}{n}\right)\left(\frac{4}{n}\right)\\=\sum_{i=1}^n\left(\frac{16}{n^2}i^2+\frac{8}{n}i-5\right)\left(\frac{4}{n}\right)\\=\left(\frac{64}{n^3}\right)\sum_{i=1}^n i^2+\frac{32}{n^2}\sum_{i=1}^n i-\frac{20}{n}\sum_{i=1}^n 1\\=\left(\frac{64}{n^3}\right)\left(\frac{n^3}{3}+\frac{n^2}{2}+\frac{n}{6}\right)+\left(\frac{32}{n^2}\right)\left(\frac{n(n+1)}{2}\right)-\frac{20}{n}n\\=\frac{64}{3}+\frac{64}{2n}+\frac{64}{6n^2}+16+\frac{16}{n}-20.$$
Take $n\to+\infty$ and you get $\int_0^4=\frac{64}{3}-4$. And
$$R^1_n=\sum_{i=1}^n f\left(\frac{1}{n}\right)\left(\frac{1}{n}\right)\\=\sum_{i=1}^n \left(\frac{1}{n^2}i^2+\frac{2}{n}i-5\right)\left(\frac{1}{n}\right)\\=\frac{1}{n^3}\left(\frac{n^3}{3}+\frac{n^2}{2}+\frac{n}{6}\right)+\frac{2}{n^2}\left(\frac{n(n+1)}{2}\right)-5\\=\frac{1}{3}+\frac{1}{2n}+\frac{1}{6n^2}+1+\frac{1}{n}-5.$$
Taking $n\to +\infty$ we get $\int_0^1=\frac{1}{3}-4$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/881726",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Factoring $x^4+x^2+2x+6$ We have to factor $x^4+x^2+2x+6$.Factoring through factor theorem is not helpful in this question as the question does not follow the integral root theorem i.e. the root of this expression is not any of the factors of 6. I am totally confused.Please help in factoring it!
| The polynomial $x^4+x^2+2x+6$ is $x^4+(x+1)^2+5\geqslant5$ for every $x$ hence it has no real root. Thus, there is no factor of degree $1$ hence one can only have two factors of degree $2$, that is, the factorization to be found is $$x^4+x^2+2x+6=(x^2+ax+b)(x^2+cx+d),$$ for some $(a,b,c,d)$. The coefficients of $x^3$ and $x^0$ in the product on the RHS should be $0$ and $6$ respectively hence $a+c=0$ and $bd=6$, which brings us to $$x^4+x^2+2x+6=(x^2+ax+b)(x^2-ax+6/b).$$ Since the coefficients of $x$ and $x^2$ in this product should be $2$ and $1$ respectively, this yields two equations in $(a,b)$.
Can you write these down and see what happens?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/882670",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Fractions in Questions and Answers
The query retrieves a sample of questions and answers containing the LaTeX fraction symbol, which provides basic filtering of mathematical content but limited analytical insight.