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Positive integer $x$ such that $\sqrt{144+x^2}$is an integer How many positive integers $x$ are there so that $\sqrt{144+x^2}$ is an integer?
Attempt a solution:
I used trial and error and only got $x=5$
|
Well then since $144 = 12^2$ then you want to find values of $x$ to satisfy the equation, $$12^2 + x^2 = y^2.$$ Now since $(a + b)^2 = a^2 + b^2 + 2ab$ and $(a - b)^2 = a^2 + b^2 - 2ab$ then when we multiply them together, we get that $$(a^2 - b^2)^2 + (2ab)^2 = (a^2 + b^2)^2.\tag1$$ What you have done to solve for $x$ is by letting $12 = 2ab$ and thus $a = 2$ and $b = 3$. We now want $12 = a^2 - b^2 = (a + b)(a - b)$. Now what divides into $12$? Well the answers are $1, 2, 3, 4, 6, 12$.
Let $a + b = 12$, then $a - b = 1$, therefore $a = b + 1$ and $2b + 1 = 12$. That is a contradiction because $a$ and $b$ are integers if $2ab$ could equal $12$ in the first place, and $12$ is even, so we have to rule that out.
Let $a + b = 6$ then $a - b = 2$, Therefore $a = b + 2$ and $2b + 2 = 2(b + 1) = 6$. Therefore, $b = 2$ and thus $a = 4$.
We have now found another solution for $x$, such that by letting $x = 2ab$ instead (since we let $12 = a^2 - b^2$), we have that $x = 2\times 2\times 4 = 16$.
Now let $a + b = 4$ then $a - b = 3$, Therefore $a = b + 3$ and $2b + 3 = 4$. That is a contradiction, so we rule that out. Now we can’t let $a + b < 4$ because otherwise $a + b < a - b$ which is also a contradiction, so we stop there.
$$***$$
Now in Eq. $(1)$, if we let $b = 1$ then $$(a^2 - 1)^2 + (2a)^2 = (a^2 + 1)^2.$$ By letting $12 = 2a$ since $12 + 1 = 13 \neq$ a squared number, then we get that $a = 6$. By letting $x = a^2 - 1$ then we obtain another solution for $x$, namely $x = 6^2 - 1 = 36 - 1 = 35.$
$$***$$
Now it is also provable that for any integer $n$, $$(2n^2 + 2n)^2 + (2n + 1)^2 = (2n^2 + 2n + 1)^2.$$ Since $12$ is even, then $12\neq 2n + 1$ and therefore $12 = 2n^2 + 2n$. We now have the following quadratic trinomial to solve: $$2n^2 + 2n - 12 = 0,$$ and using the quadratic formula, we get that $$n = \frac{-2\pm \sqrt{4 + 96}}{2} = \frac{-2}{2} \pm \frac{10}{2} = -1\pm 5.$$ Therefore, $n = 4$ or $n = -6$. Now, we let $x = 2n + 1$ for those values of $n$ and we get two more solutions for $x$, namely $x = 9$ and $x = -11$.
However, $x > 0$ because we established earlier that if $x = 2ab$ then $a + b\not< a - b$, and if $x < 0$ then either $a$ or $b$ is negative which arouses a contradiction in the strict inequality, so we get only one new solution for $x$ such that $x = 9$. (This also proves why $x$ must be a positive integer.)
$$***$$
Overall, our solutions for $x$ so far are $x = 5, 9, 16, 35$.
Checking the equation $u^2 + v^2 = w^2$ for $w \leqslant 10,000$, there does not seem to be another solution where either $u$ or $v$ equals $12$ with the other remaining variable on the LHS not equal to any of our listed values of $x$. From this, I conjecture that there only exists $4$ such solutions of $x$.
Main Answer:
According to @Piquito’s answer below $\downarrow\downarrow\downarrow$ there exist only $4$ solutions, and thus the conjecture is valid. I apologise for constantly editing my answer, but I did not have a big calculator with me. I was able to do the calculation concerning $u^2 + v^2 = w^2$ from a link in one of the comments. This meant that I had to find the values of $x$ with pure maths, and I was not going to sit there and exhaustively find solutions for $(x, y)$ in the equation, $144 = (y + x)(y - x)$.
Consequently, I decided to be creative.
|
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"url": "https://math.stackexchange.com/questions/2658837",
"timestamp": "2023-03-29T00:00:00",
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|
How to extract $(x+y+z)$ or $xyz$ from the determinant Prove
$$\color{blue}{
\Delta=\begin{vmatrix}
(y+z)^2&xy&zx\\
xy&(x+z)^2&yz\\
xz&yz&(x+y)^2
\end{vmatrix}=2xyz(x+y+z)^3}
$$
using elementary operations and the properties of the determinants without expanding.
My Attempt
$$
\Delta\stackrel{C_1\rightarrow C_1+C_2+C_3}{=}\begin{vmatrix}
xy+y^2+yz+zx+yz+z^2&xy&zx\\
x^2+xy+xz+xz+yz+z^2&(x+z)^2&yz\\
x^2+xy+xz+xy+y^2+yz&yz&(x+y)^2\\
\end{vmatrix}\\
=\begin{vmatrix}
y(x+y+z)+z(x+y+z)&xy&zx\\
x(x+y+z)+z(x+y+z)&(x+z)^2&yz\\
x(x+y+z)+y(x+y+z)&yz&(x+y)^2\\
\end{vmatrix}\\
=(x+y+z)\begin{vmatrix}
y+z&xy&zx\\
x+z&(x+z)^2&yz\\
x+y&yz&(x+y)^2\\
\end{vmatrix}\\
\stackrel{R_1\rightarrow R_1+R_2+R_3}{=} \\(x+y+z)\begin{vmatrix}
2(x+y+z)&x^2+xy+xz+xz+yz+z^2&x^2+xy+zx+xy+y^2+yz\\
x+z&(x+z)^2&yz\\
x+y&yz&(x+y)^2\\
\end{vmatrix}\\
=(x+y+z)^2\begin{vmatrix}
2&x+z&x+y\\
x+z&(x+z)^2&yz\\
x+y&yz&(x+y)^2\\
\end{vmatrix}\\
=2(x+y+z)^2\begin{vmatrix}
1&\frac{x+z}{2}&\frac{x+y}{2}\\
x+z&(x+z)^2&yz\\
x+y&yz&(x+y)^2
\end{vmatrix}
\stackrel{R_2\rightarrow R_2-(x+z)R_1|R_2\rightarrow R_3-(x+y)R_1}{=}
2(x+y+z)^2\begin{vmatrix}
1&\frac{x+z}{2}&\frac{x+y}{2}\\
0&\frac{(x+z)^2}{2}&\frac{yz-x^2-xy-xz}{2}\\
0&\frac{yz-x^2-xy-xz}{2}&\frac{(x+y)^2}{2}
\end{vmatrix}
$$
How do I extract $x+y+z$ or $xyz$ from the determinant to find the solution ?
Note: In a similar problem How to solve this determinant there seems to be solutions talking about factor theorem and polynomials. I am basically looking for extracting the terms $(x+y+z)$ or $xyz$ from the given determinant to solve it only using the basic properties of determinants.
|
It is easy to calculate
\begin{vmatrix}
1&\frac{x+z}{2}&\frac{x+y}{2}\\
0&\frac{(x+z)^2}{2}&\frac{yz-x^2-xy-xz}{2}\\
0&\frac{yz-x^2-xy-xz}{2}&\frac{(x+y)^2}{2}
\end{vmatrix}
If you still want to use extract $x+y+z$, you can use
\begin{eqnarray}
\begin{vmatrix}
\frac{(x+z)^2}{2}&\frac{yz-x^2-xy-xz}{2}\\
\frac{yz-x^2-xy-xz}{2}&\frac{(x+y)^2}{2}
\end{vmatrix}&=&\frac{(x+z)^2}{2}\begin{vmatrix}
1&\frac{yz-x^2-xy-xz}{(x+z)^2}\\
\frac{yz-x^2-xy-xz}{2}&\frac{(x+y)^2}{2}
\end{vmatrix}\\
&=&\frac{(x+z)^2}{2}\begin{vmatrix}
1&\frac{yz-x^2-xy-xz}{(x+z)^2}\\
0&\frac{(x+y)^2}{2}-\frac{(yz-x^2-xy-xz)^2}{2(x+z)^2}
\end{vmatrix}\\
&=&\frac{(x+z)^2}{2}\begin{vmatrix}
1&\frac{yz-x^2-xy-xz}{(x+z)^2}\\
0&\frac{(x+y)^2(x+z)^2-(yz-x^2-xy-xz)^2}{2(x+z)^2}
\end{vmatrix}\\
&=&\frac{(x+z)^2}{2}\begin{vmatrix}
1&\frac{yz-x^2-xy-xz}{(x+z)^2}\\
0&\frac{xyz(x+y+z)}{2(x+z)^2}
\end{vmatrix}\\
&=&xyz(x+y+z)\begin{vmatrix}
1&\frac{yz-x^2-xy-xz}{(x+z)^2}\\
0&1
\end{vmatrix}.
\end{eqnarray}
|
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"timestamp": "2023-03-29T00:00:00",
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|
Laurent series expansion about z=0 I am to obtain the first 3 terms of the Laurent series about $z = 0$ for:
$$f(z)=1/(e^z-1-z)$$
I know that there is 1 singularity at $z=0$. My denominator is not in a polynomial form so I can convert it into such by taking a Taylor series of the function around z=0 which yields the denominator as:
$$(e^z-1-z) = x^2/2 + x^3/6 + x^4/24 + ...$$
This is the part where I get stuck. I'm supposed to bring that back in and perhaps factor something out of the denominator so the left side is analytic at z=0 and the singularities are shifted to the term on the right side. Any tips on what my next step should be? Many thanks!
|
Borrowing from that answer I commented, we have
\begin{align}
\frac{1}{e^z - 1 - z} &= \frac{1}{(1 + z + z^2/2! + \dots) - 1 - z}\\
&= \frac{1}{z^2/2! + z^3/3! + z^4/4! + \dots}\\
&= \frac{2}{z^2} \cdot \frac{1}{1 + 2z/3! + 2z^2/4! + \dots}
\end{align}
Now letting $P(z) = 2z/3! + 2z^2/4! + \dots$, we have
\begin{align}
\frac{1}{e^z - 1 - z} &= \frac{2}{z^2} \cdot (1 - P(z) + P(z)^2 - P(z)^3 + \dots)\\
&= \frac{2}{z^2} \cdot (1 - 2z/3! + z^2(-2/4! + (2/3!)^2) + \dots)\\
&= 2z^{-2} - \frac{2z^{-1}}{3} + \frac{1}{18} - \dots
\end{align}
Getting higher order terms starts to get a bit complicated, so I'd suggest using Wolfram Alpha to fill in the rest: http://www.wolframalpha.com/input/?i=laurent+series+of+1%2F(e%5Ez+-+1+-+z)
|
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"language": "en",
"url": "https://math.stackexchange.com/questions/2660585",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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|
Find the kernel of a 4x4 matrix $$
\begin{pmatrix}
1 & 2 & 3 & 4\\
5 & 6 & 7 & 8\\
9 & 10 & 11 & 12\\
13 & 14 & 15 & 16\\
\end{pmatrix}
$$
I am asked to find the kernel of the matrix $M$. After doing some row operation I get to
$$
\begin{pmatrix}
1 & 2 & 3 & 4\\
0 & -4 & -8 & -12\\
0 & 0 & 0 & 0\\
0 & 0 & 0 & 0\\
\end{pmatrix}
$$
and for $x$ I find $x = \alpha + 2\beta$, whereas $y = -2\alpha -3\beta$
Therefore,
$$
\begin{pmatrix}
\alpha + 2\beta\\
-2\alpha - 3\beta\\
\alpha\\
\beta\\
\end{pmatrix}
$$
When we take outside alpha and beta:
we get two vectors:
$$
\begin{pmatrix}
1\\
-2\\
1\\
0\\
\end{pmatrix}
$$
and
$$
\begin{pmatrix}
2\\
-3\\
0\\
1\\
\end{pmatrix}
$$
which are linearly independent and form a basis of this $ker(M)$
Could you please confirm with me whether you get the same result? Thank you.
|
Your computations are correct and so is your answer. But you should really improve your MathJax skills.
|
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"url": "https://math.stackexchange.com/questions/2661571",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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|
Stuck on two phase simplex where RHS $= 0$ I'm stuck on the following linear program, it's a minimal example of the problem I'm having:
Minimize $Z = y$ where $x \ge 1$ and $-2x+y \ge 0$
We start by rewriting this as a system of linear equalities, introducing slack and artificial variables:
$\left\{
\begin{array}{c}
x - s_1 + a_1=1\\
-2x +y - s_2=0
\end{array}
\right.$
We shouldn't need an artificial variable for the second one: $a_1 = 1,s_2=0$ is already feasible.
For phase 1 maximize $p = -a_1 = x-s_1-1$. This gives us the starting tableau
$$
\begin{array}{c|ccccc|c}
&x&y&s_1&s_2&a_1&A\\
\hline
a_1&1&0&-1&0&1&1\\
s_2&-2&1&0&-1&0&0\\
\hline
-p&1&0&-1&0&0&1
\end{array}
$$
With solution $a_1=1/1=1,s_1=0/(-1)=0$
The only positive element in in column 1, so $x$ is the entering variable. The only positive element in this column is in row 1, so $a_1$ is the leaving variable. Pivoting around (1,1) results in the tableau:
$$
\begin{array}{c|ccccc|c}
&x&y&s_1&s_2&a_1&A\\
\hline
x&1&0&-1&0&1&1\\
s_2&0&1&-2&-1&2&2\\
\hline
-p&0&0&0&0&-1&0
\end{array}
$$
With solution $x=1/1=1,s_2=2/(-1) = -2$.
Wait what? How did $s_2$ suddenly become negative? What did I do wrong?
|
How did $s_2$ suddenly become negative? What did I do wrong?
You've chosen the wrong entering variable. This give rise to an unfeasible solution.
To set up the initial tableau, multiply the second equation of the system
$$
\left\{
\begin{array}{c}
x - s_1 + a_1=1\\
-2x +y - s_2=0,
\end{array}
\right.
$$
by $-1$, so that the coefficient of the current basic variable $s_2$ becomes $1$. This makes the tableau proper.
\begin{array}{c|ccccc|cc}
&x&y&s_1&s_2&a_1&A&\text{ratio}\\
\hline
a_1&1&0&-1&0&1&1 & 1/1 = 1\\
s_2&\fbox{2}&-1&0&1&0&0 & \fbox{0/2 = 0}\\
\hline
p&-1&0&1&0&0&-1
\end{array}
For feasibility reason, one has to pivot around $(1,2)$.
\begin{array}{c|ccccc|c}
&x&y&s_1&s_2&a_1&A\\
\hline
a_1&0&\fbox{1/2}&-1&-1/2&1&\fbox1\\
x&1&-1/2&0&1/2&0&0\\
\hline
p&0&-1/2&1&1/2&0&-1
\end{array}
\begin{array}{c|ccccc|c}
&x&y&s_1&s_2&a_1&A\\
\hline
y&0&1&-2&-1&2&2\\
x&1&0&-1&0&1&1\\
\hline
p&0&0&0&0&1&0
\end{array}
This gives us a feasible nondegenerate solution $(x,y) = (1,2)$.
To start phase II, we discard the column representing artificial variable $a_1$, calculate the objective function row, and reuse the rest of the above tableau.
$$\min Z = y \iff \max z = -y$$
\begin{array}{c|cccc|c}
&x&y&s_1&s_2&A\\
\hline
y&0&1&-2&-1&2\\
x&1&0&-1&0&1\\
\hline
z&0&1&0&0&0
\end{array}
To make it a simplex tableau, make the entries representing the current basis zero at the objective function row.
\begin{array}{c|cccc|c}
&x&y&s_1&s_2&A\\
\hline
y&0&1&-2&-1&2\\
x&1&0&-1&0&1\\
\hline
z&0&0&2&1&-2
\end{array}
Therefore, the solution is $(x,y) = (1,2)$ with $Z = -(-2) = 2$. Note that it is nondegenerate and unique because we have strictly positive entries on the RHS and entries representing nonbasic variables at the objective function row.
A quicker solution using dual simplex method.
Observe that that $\max z = -y$ has nonpositve coefficients. This gives rise to a nonnegative objective function row, so optimality is satisfied. This allows us to use dual simplex method.
Rewrite the LPP as
$\max z = -y$ where $-x \le -1$ and $2x-y \le 0, x,y \ge 0$.
Note that $y \ge 2x \ge 2(1) \ge 1 > 0$, so adding $x,y \ge 0$ doesn't change the problem.
\begin{array}{c|cccc|c}
&x&y&s_1&s_2&A\\
\hline
s_1&\fbox{-1}&0&1&0&\fbox{-1}\\
s_2&2&-1&0&1&0\\
\hline
z&0&1&0&0&0 \\
\text{ratio} & \fbox0 & -
\end{array}
\begin{array}{c|cccc|c}
&x&y&s_1&s_2&A\\
\hline
x&1&0&-1&0&1\\
s_2&0 & -1 & 2 & 1 & \fbox{-2}\\
\hline
z&0&1&0&0&0 \\
\text{ratio} & & \fbox{-1} & -
\end{array}
\begin{array}{c|cccc|c}
&x&y&s_1&s_2&A\\
\hline
x&1&0&-1&0&1\\
y&0 & 1 & -2 & -1 & 2 \\
\hline
z&0&0&2&1&-2
\end{array}
Observe that we arrived at the same tableau much quicker than the usual two phase method.
|
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"timestamp": "2023-03-29T00:00:00",
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|
real value of $k$ inirrational equation Find value of $k$ for which the equation $\sqrt{3z+3}-\sqrt{3z-9}=\sqrt{2z+k}$ has no solution.
solution i try $\sqrt{3z+3}-\sqrt{3z-9}=\sqrt{2z+k}......(1)$
$\displaystyle \sqrt{3z+3}+\sqrt{3z-9}=\frac{12}{\sqrt{2z+k}}........(2)$
$\displaystyle 2\sqrt{3z+3}=\frac{12}{\sqrt{2z+k}}+\sqrt{2z+k}=\frac{12+2z+k}{\sqrt{2z+k}}$
$\displaystyle 2\sqrt{6z^2+6z+3kz+3k}=\sqrt{12+2z+k}$
$\displaystyle 4(6z^2+6z+3kz+3k)=144+4z^2+k^2+48z+4kz+24k$
Help me how i solve it after that point
|
Hint : From the start : $\sqrt{3z+3}-\sqrt{3z-9}=\sqrt{2z+k}$ leads to
$$ 3z+3 +3z-9-2\sqrt{3z+3}\sqrt{3z-9}=2z+k $$
so
$$2\sqrt{3z+3}\sqrt{3z-9}=4z-6-k $$
and squaring again to
$$ 4(3z+3)(3z-9)=(4z-6-k)^2 $$
so
$$20z^2 +4(48+2k)z-27\times 4 -(6+k)^2=0 \tag{3} $$
You won't have solutions if
$$ \Delta_k= (4(48+2k))^2-4\times20\times -(27\times 4 +(6+k)^2 < 0 $$
With gives you a domain for $k$, by studying this last second degree equation in $k$.
But, since we took the power 2 twice, a solution of (3) may not be a solution of (1) : so it remains to verify if $k$ is such (3) has a solution ($\Delta_k\geq 0$), the quantities before squaring remain positive ; if not you can add those values of $k$ in the set of case where (1) has no solution.
|
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"language": "en",
"url": "https://math.stackexchange.com/questions/2665723",
"timestamp": "2023-03-29T00:00:00",
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|
Roots of $f(x)=x^3-x^2-2x+1$ We can prove using a monotony study that the function $f(x)=x^3-x^2-2x+1$ has three real roots. However, when I solve the equation $f(x)=0$ using Mathematica, I get
$$x_1=\frac{1}{3}+\frac{7^{2/3}}{3 \left(\frac{1}{2} \left(-1+3 i \sqrt{3}\right)\right)^{1/3}}+\frac{1}{3} \left(\frac{7}{2} \left(-1+3 i \sqrt{3}\right)\right)^{1/3}$$
$$x_2=\frac{1}{3}-\frac{\left(\frac{7}{2}\right)^{2/3} \left(1+i \sqrt{3}\right)}{3 \left(-1+3 i \sqrt{3}\right)^{1/3}}-\frac{1}{6} \left(1-i \sqrt{3}\right) \left(\frac{7}{2} \left(-1+3 i \sqrt{3}\right)\right)^{1/3}$$
$$x_3=\frac{1}{3}-\frac{\left(\frac{7}{2}\right)^{2/3} \left(1-i \sqrt{3}\right)}{3 \left(-1+3 i \sqrt{3}\right)^{1/3}}-\frac{1}{6} \left(1+i \sqrt{3}\right) \left(\frac{7}{2} \left(-1+3 i \sqrt{3}\right)\right)^{1/3}$$
Since those roots are real, then maybe there's a way to write them without using the complex number $i$?
|
You need the complex numbers even when there are three real roots. That was a real puzzle when the cubic was first solved.
Read about the casus irreducibilis:
https://en.wikipedia.org/wiki/Casus_irreducibilis
|
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|
$Pr(A+B+C=X+Y+Z)$ with $A, B, C, X, Y, Z$ being iid random variables $A, B, C, X, Y, Z$ are iid random variables and they are uniformly distributed on the set of integers $0$ to $9$ (inclusive). What is $\Pr(A+B+C=X+Y+Z)$?
I am kind of lost and my attempt is to use generating functions such that $$G_{A+B+C}(t)\cdot G_{X+Y+Z}(1/t)$$ because i can transform $\Pr(A+B+C=X+Y+Z)$ into $\Pr((A+B+C)-(X+Y+Z)=k)$ and $k=0.$
So I need to find the coefficient of $t^0$ in the function $$G_{A+B+C}(t) \cdot G_{X+Y+Z}(1/t).$$ But how do i find $G_{A+B+C}(t)$?
|
Let $A,B,C$ and $X,Y,Z$ are iid uniformly distributed on $\{0,1,\dots,9\}$, split the below probability by the value of the sums,
$\mathbb{P}(A+B+C=X+Y+Z)=\displaystyle \sum_{k=0}^{27}\mathbb{P}(A+B+C=k,\ X+Y+Z=k).$
Since $A+B+C$ and $X+Y+Z$ are independents, for all $k\in \mathbb{Z}$:
$\mathbb{P}(A+B+C=k,X+Y+Z=k)=\mathbb{P}(A+B+C=k)\cdot \mathbb{P}(X+Y+Z=k)$
and $\mathbb{P}(A+B+C=k)\cdot \mathbb{P}(X+Y+Z=k)=\mathbb{P}(A+B+C=k)^2$, because $A+B+C$ and $X+Y+Z$ has the same distribution. It means
$\displaystyle \sum_{k=0}^{27}\mathbb{P}(A+B+C=k,\ X+Y+Z=k)=\displaystyle \sum_{k=0}^{27}\mathbb{P}(A+B+C=k)^2$.
There are 28 possible value for $A+B+C$ from $0$ to $27$, please note that distribution of the sum is symmetric, so $\mathbb{P}(A+B+C=k)=\mathbb{P}(A+B+C=27-k)$, which means if one calculates their probabilities, the sum of squares will be
$\mathbb{P}(A+B+C=X+Y+Z)=\displaystyle 2\cdot \sum_{k=0}^{13}\mathbb{P}(A+B+C=k)^2=\\ \displaystyle = 2 \cdot \frac{1^2+3^2+6^2+10^2+15^2+21^2+28^2+36^2+45^2+55^2+63^2+69^2+73^2+75^2}{1000^2}= \\ =5.5252\%.$
|
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"timestamp": "2023-03-29T00:00:00",
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}
|
Show that $\frac{1}{a^n+b^n+c^n} = \frac{1}{a^n} + \frac{1}{b^n} + \frac{1}{c^n}$ Let $n \in N, n=2k+1, and \text{ } \frac{1}{a+b+c} = \frac{1}{a} + \frac{1}{b} + \frac{1}{c}$.
Show that $$\frac{1}{a^n+b^n+c^n} = \frac{1}{a^n} + \frac{1}{b^n} + \frac{1}{c^n}$$
I have tried, but I don't get anything. Can you please give me a hint?
|
The condition gives $$(a+b+c)(ab+ac+bc)=abc$$ or
$$\sum_{cyc}(a^2b+a^2c+abc)=abc$$ or
$$\sum_{cyc}\left(a^2b+a^2c+\frac{2}{3}abc\right)=0$$ or
$$\prod_{cyc}(a+b)=0.$$
We need to prove that
$$\prod_{cyc}(a^n+b^n)=0$$ or $$\prod_{cyc}(a+b)\prod_{cyc}\left(a^{2k}-a^{2k-1}b+...+b^{2k}\right)=0,$$ which is obvious.
We can use also the following reasoning.
Since the condition is true for $a=-b$ and it's symmetric and degree $3$, we obtain that the condition is equal to $$(a+b)(a+c)(b+c)=0$$ and we need to prove that $$\left(a^{2k+1}+b^{2k+1}\right)\left(a^{2k+1}+c^{2k+1}\right)\left(b^{2k+1}+c^{2k+1}\right)=0.$$
|
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"timestamp": "2023-03-29T00:00:00",
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|
Simplify $\frac14\log(3x+1)+\frac12\log(x-3)-\frac12\log(x^4-x^2-10)$ $$\frac14\log(3x+1)+\frac12\log(x-3)-\frac12\log(x^4-x^2-10)$$
I was able to find
$$\log(3x+1)^{1/4}(x-3)^{1/2}-\log(x^4-x^2-10)^{1/2}$$
through distributing the $^{1/2}$ but this answer is incorrect.
|
I do not know how you would get a "simplified answer", but:
$$\dfrac 14\log(3x+1)+\dfrac 12 \log(x-3)-\dfrac 12\log(x^4-x^2-10)$$
Using the property $\log a^b=b\log a$:
$$\log{\sqrt[4]{3x+1}}+ \log{\sqrt{x-3}}-\log\sqrt{{x^4-x^2-10}}$$
Using the property $\log a+\log b=\log ab$:
$$\log{\sqrt[4]{3x+1}}+ \log{\sqrt{x-3}}-\log\sqrt{{x^4-x^2-10}}=\log\left(\dfrac{{{\sqrt[4]{3x+1}\sqrt{x-3}}}}{\sqrt{x^4-x^2-10}}\right)$$
Rationalizing:
$$\log\left(\dfrac{{{\sqrt[4]{3x+1}\sqrt{x-3}}}}{\sqrt{x^4-x^2-10}}\cdot \dfrac {\sqrt{x^4-x^2-10}}{\sqrt{x^4-x^2-10}}\right)=\log\left(\dfrac{{{\sqrt[4]{3x+1}\sqrt{(x-3)(x^4-x^2-10)}}}}{x^4-x^2-10}\right)$$
The answer does not seem to "simplify" further than the following:
$$\log\left(\dfrac{{{\sqrt[4]{3x+1}\sqrt{(x-3)(x^4-x^2-10)}}}}{x^4-x^2-10}\right)$$
|
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|
Summation of binomial coefficients multiplied by their index
$\sum_{r=0}^{300}a_rx^r = (1+x+x^2+x^3)^{100}.$ If $a= \sum^{300}_{r=0}a_r$ then $\sum_{r=0}^{300}ra_r =? $
a) $300a$
b) $100a$
c) $150a$
d) $75 a$
Attempt:
First of all if we substitute $x=1$, we get $a= 4^{100}$
Then, using formula of summation of geometric progression:
$\sum_{r=0}^{300}a_rx^r= (1-x^4)^{100}(1-x)^{-100}$
Using generalised binomial theorem, coefficient of $x^r$ in $(1-x)^{-100}$ is $\dbinom{99+r}{r}$ and that of $x^{4r}$ in $(1-x^4)^{100}$ is $(-1)^{100-r}\dbinom{n}{r}$
What do I do after this?
|
You have been given
$$\sum_{r=0}^{300}a_rx^r = (1+x+x^2+x^3)^{100}$$
substitute $x = 1 $ to get $a = \sum a_r = 4^{100}$
Differentiate once w.r.t $x$ to get
$$\sum_{r=0}^{300}ra_rx^{r-1} = 100(1+x+x^2+x^3)^{99} (1+2x+3x^2)$$
again substitute $x = 1$ to get $\sum ra_r = 4^{99} \cdot 100 \cdot 6 = 150a$
|
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|
How to calculate $ \int \frac{\sin^{6}(x)}{\sin^{6}(x) + \cos^{6}(x)} dx? $ How to calculate
$$ \int \frac{\sin^{6}(x)}{\sin^{6}(x) + \cos^{6}(x)} dx? $$
I already know one possible way, that is by :
$$ \int \frac{\sin^{6}(x)}{\sin^{6}(x) + \cos^{6}(x)} dx = \int 1 - \frac{\cos^{6}(x)}{\sin^{6}(x) + \cos^{6}(x)} dx $$
$$= x-
\int \frac{1}{1+\tan^{6}(x)} dx $$
Then letting $u=\tan(x)$, we must solve
$$\int \frac{1}{(1+u^{6})(1+u^{2})} du $$
We can reduce the denominator and solve it using Partial Fraction technique. This is quite tedious, I wonder if there is a better approach.
Using same approach, for simpler problem, I get
$$\int \frac{\sin^{3}(x)}{\sin^{3}(x)+\cos^{3}(x)} dx = \frac{x}{2} - \frac{\ln(1+\tan(x))}{6} + \frac{\ln(\tan^{2}(x)- \tan(x)+1)}{3} - \frac{\ln(\sec(x))}{2} + C$$
|
Let us take:
$$I=\int \frac{\sin^{6}(x)}{\sin^{6}(x) + \cos^{6}(x)} dx$$
then
$$I=\int \frac{-\cos^{6}(x)}{\sin^{6}(x) + \cos^{6}(x)} dx+x$$
giving$$2I=\int \frac{\sin^{6}(x)-\cos^{6}(x)}{\sin^{6}(x) + \cos^{6}(x)} dx+x$$
This can be written as (using identities like $a^3-b^3$ and $a^3+b^3$)
$$2I= \int\frac{(\sin^2(x)-\cos^2(x)(1-\sin^2(x)\cos^2(x))}{(1-\sqrt3\sin(x)\cos(x)(1+\sqrt3\sin(x)\cos(x))}dx+x$$
$$
2I=\frac{1}{2}\left(\int\frac{(\sin^2(x)-\cos^2(x)(1-\sin^2(x)\cos^2(x))}{(1+\sqrt3\sin(x)\cos(x))}
+\int\frac{(\sin^2(x)-\cos^2(x)(1-\sin^2(x)\cos^2(x))}{(1-\sqrt3\sin(x)\cos(x))}\right)+x
$$
Evaluating the integrals separately by using $u=1+\sqrt3\sin(x)\cos(x)$
for first one gives
$$\int\frac{(\sin^2(x)-\cos^2(x)(1-\sin^2(x)\cos^2(x))}{(1+\sqrt3\sin(x)\cos(x))}=\frac{1}{\sqrt3}\int\frac{(\sin(x)\cos(x)-1)(\sin(x)\cos(x)+1)}{u}du$$
Now use $\sin(x)\cos(x)=\frac{u-1}{\sqrt3}$
which will evaluate the integral as $\frac{u^2}{6\sqrt3}-\frac{2u}{3\sqrt3}-\frac{2\ln(u)}{3\sqrt3}$...Similar approach for other with $v=1-\sqrt3\sin(x)\cos(x)$
The final value is
$$I=\frac{x}{2}-\frac{\sin(x)\cos(x)}{6}+\frac{\ln(1-\sqrt3\sin(x)\cos(x))}{6\sqrt3}-\frac{\ln(1+\sqrt3\sin(x)\cos(x))}{6\sqrt3}+C$$
|
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|
Greatest Common Divisor of two Polynomials. Find the $\gcd(x^3-6x^2+14x-15, x^3-8x^2+21x-18)$ over $\mathbb{Q}[x]$. Then find two polynomials $a(x),b(x) \in \mathbb{Q}[x]$ such that, $$a(x)(x^3-6x^2+14x-15) + b(x)(x^3-8x^2+21x-18)=\gcd(x^3-6x^2+14x-15, x^3-8x^2+21x-18)$$
I have managed to find,
$$x^3-6x^2+14x-15=(x-3)(x^2-3x+5)$$
$$x^3-8x^2+21x-18=(x-3)(x-3)(x-2)$$
Now since $x^2-3x+5$ is irreducible over $\mathbb{Q}[x]$ and so the greatest common divisor is $(x-3)$. Now to find $a(x)$ and $b(x)$ I have no clue how to do that. I have looked online and it seems there is extended euclidean algorithm for polynomials but I haven't formally learned it in my class yet, so I was wondering if there is another efficient way to find these polynomials. Any help is appreciated, thanks!
|
If I interpreted your question correctly and you want to do something like this, where in:
$a(x)(p)+b(x)(q)=gcd(p,q)$, $a=1$, and $b=1$
You can restate your expression as:
$$a(x)(x-3)(x^2-3x+5)+b(x)(x-3)(x-3)(x-2)$$
$$(x-3)\left(a(x)(x^2-3x+5)+b(x)(x-3)(x-2)\right)$$
Since this has to equal the $gcd$, which is $x-3$:
$$(x-3)\left(a(x)(x^2-3x+5)+b(x)(x-3)(x-2)\right)=x-3$$
$$\left(a(x)(x^2-3x+5)+b(x)(x-3)(x-2)\right)=1$$
Note: There are infinitely many solutions for $a(x)$ and $b(x)$ (but might not satisfy the definition of polynomials), since $0\le a(x)\le 1$ and $0\le b(x)\le 1$
Here, $a(x)(x^2-3x+5)$ and $b(x)(x-3)(x-2)$ sum up to $1$, so one instance is that they can individually sum up to $\dfrac 12$, since $\dfrac 12+\dfrac 12=1$.
So:
$$a(x)(x^2-3x+5)=\dfrac 12$$
$$b(x)(x-3)(x-2)=\dfrac 12$$
Can you figure this out?
|
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|
Rattles with beads or necklace with beads? I came across this problem in a book called limits, sequences combinations great book for intro to combinatorics .
A rattle consists of a ring with $3$ white beads and $7$ red ones strung on it. Some rattles seemingly different can be made identical by arranging the rings and moving the beads in a suitable manner (rotation or flipping). Find the number of different rattles .
I of course thought polya enumeration on this one but was thinking it can be done case by case without being too messy . Can anyone help? Also this is essentially the same problem as a necklace with $n$ beads, $k$ colors is it not ?
|
Let me just observe that with $(3,10) = 1$ and $(7,10) = 1$ Polya will
be very simple in this case. The OEIS uses necklace for cyclic
symmetry and bracelet for dihedral, so we have a bracelet here. The
cycle index is from first principles
$$Z(D_{10}) = \frac{1}{20} \sum_{d|10} \varphi(d) a_d^{10/d}
+ \frac{1}{4} a_2^5 + \frac{1}{4} a_1^2 a_2^4.$$
With the two colors our answer is given by
$$[W^3 R^7] \frac{1}{20} \sum_{d|10} \varphi(d) (W^d+R^d)^{10/d}
\\ + [W^3 R^7] \frac{1}{4} (W^2+R^2)^5
\\ + [W^3 R^7] \frac{1}{4} (W+R)^2 (W^2+R^2)^4.$$
From the first term only $d=1$ contributes, the second does not
contribute at all and from the third the first term must contribute
$WR.$ We get
$$[W^3 R^7] \frac{1}{20} (W+R)^{10}
+ [W^2 R^6] \frac{1}{2} (W^2+R^2)^4
\\ = [W^3 R^7] \frac{1}{20} (W+R)^{10}
+ [W^1 R^3] \frac{1}{2} (W+R)^4.$$
The desired count is thus
$$\frac{1}{20} {10\choose 3} + \frac{1}{2} {4\choose 1}
= 8.$$
|
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|
Area of intersection using integration
Suppose, for the sake of illustration, that $AF=2$, where $F\mathbb{'s}$ coordinates are $(2,0)$ and $A$ is the point of origin; $E=(0,1)$ and circle $A$ has $R=1.5$
How do I solve the area of the intersection of the circle and rectangle by integration?
If we let $f(x)=1$, then the area of the rectangle is defined by:
$$\int_0^Ff(x)dx$$
If we let $g(x)=\sqrt{R^2-x^2}$, then the area of the circle is defined by:
$$4\int_0^Rg(x)dx$$
Now my best attempt to find the area is through:
$$\int_0^Rg(x)dx-\int_0^{\sqrt{R^2-1}}(g(x)-1)dx$$
The first integral gives me a quarter of the area circle $R$, the second gives me the area of the circle that is outside the rectangle, because the circle and the top line intersects at $x=\sqrt{R^2-1}$
Is my solution correct?
|
Consider $x=f(y)$ instead of $y=f(x)$.
Then the area of intersection is just
\begin{align}
\int_0^1 \sqrt{R^2-y^2}\,dy
&=\left.\left(
\tfrac12\,y\sqrt{R^2-y^2}
+\tfrac12\,R^2\arctan\frac{y}{\sqrt{R^2-y^2}}
\right)\right|_0^1
\\
&=
\left(
\tfrac12\,\sqrt{R^2-1}
+\tfrac12\,R^2\arctan\frac{1}{\sqrt{R^2-1}}
\right)
\\
&=
\left(
\tfrac12\,\sqrt{\tfrac{9}{4}-1}
+\tfrac12\,\tfrac{9}{4}\arctan\frac{1}{\sqrt{\tfrac{9}{4}-1}}
\right)
\\
&=\tfrac{\sqrt5}4+\tfrac98\,\arctan(\tfrac{2\sqrt5}5)
\approx 1.38
.
\end{align}
|
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|
Find $\int_0^{\sqrt{R^2-1}}g(x)-f(x)\,\mathrm dx$
How do you evaluate this integral?
$$\int_0^{\sqrt{R^2-1}}g(x)-f(x)\,\mathrm dx$$
where $f(x)=1$ and $g(x)=\sqrt{R^2-x^2}$.
Wolfram tells me I exceeded my computational limit. Mathematica gives me a long answer which is very difficult to read (for me).
EDIT: I was told to consider instead $x=f(y)$, so:
$$\int_0^{1}\sqrt{R^2-y^2}\,\mathrm dy$$
|
Trick: Split the integral into three parts.
$$\begin{align}\int_0^{\sqrt{R^2-1}}g(x)-f(x)\,\mathrm dx&=\int_0^{\sqrt{R^2-1}}\sqrt{R^2-x^2}-1\,\mathrm dx\\&=\left(\int_0^{\sqrt{R^2-1}}\frac{\sqrt{R^2-x^2}}2-\frac{x^2}{\sqrt{R^2-x^2}}+\frac{a^2}{\sqrt{R^2-x^2}}\,\mathrm dx\right)-\sqrt{R^2-1}\\&=\left(\int_0^{\sqrt{R^2-1}}\left(\frac x2\right)'\sqrt{R^2-x^2}+\frac x2\left(\sqrt{R^2-x^2}\right)'+\frac{R^2}2\left(\sin^{-1}\frac xR\right)'\,\mathrm dx\right)-\sqrt{R^2-1}\\&=\left[\frac x2\sqrt{R^2-x^2}+\frac{R^2}2\sin^{-1}\frac xR\right]_0^{\sqrt{R^2-1}}-\sqrt{R^2-1}\\&=\frac{\sqrt{R^2-1}}2+\frac{R^2}2\sin^{-1}\left(\frac{\sqrt{R^2-1}}R\right)-\sqrt{R^2-1}\\&=\boxed{\frac{R^2}2\cos^{-1}\left(\frac1R\right)-\frac{\sqrt{R^2-1}}2}\end{align}$$
|
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|
Area of largest inscribed rectangle in an ellipse. Can I take the square of the area to simplify calculations? So say I have an ellipse defined like this:
$$\frac{x^2}{9} + \frac{y^2}{4} = 1$$
I have to find the largest possible area of an inscribed rectangle.
So the area ($A$) of a rectangle is $2x2y=4xy$. Also we can redefine $y$ in terms of $x$:
$$\frac{y^2}{4} = 1 - \frac{x^2}{9}$$
$$y^2 = 4 - \frac{4x^2}{9}$$
$$y = \sqrt{4 - \frac{4x^2}{9}}$$
So the area function is now:
$$A=4x \cdot \sqrt{4 - \frac{4x^2}{9}}$$
$$A' = \frac{4x}{2 \cdot \frac{-8x}{9}} + \sqrt{4 - \frac{4x^2}{9}} \cdot 4$$
So this track seems too difficult, so I'd like to find another approach. Can I square the area first, find the derivative of that to solve for $x$?
So the $\text{Area} = 4x \cdot \sqrt{4 - \frac{4x^2}{9}}$
Is this valid?
$$\text{Area}^2 = 16x^2 \cdot \left(4 - \frac{4x^2}{9}\right)$$
$$= 64x^2 - \frac{64x^4}{9}$$
Derivative:
$$ \frac{d}{dx} \text{Area}^2 = 128x - \frac{256x^3}{9}$$
$$128x\left(1-\frac{2x^2}{9}\right)$$
So critical values: $x = 0, \frac{3}{\sqrt{2}}$
because the derivative equals $0$ when:
$$2x^2 = 9$$
$$x = \frac{3}{\sqrt{2}}$$
Plugging this value of $x$ into $y$ we get that $y = \sqrt{2}$ so the $\text{Area}$ is $3$.
Is this valid? If so why? Does squaring not cause any problems?
|
$$
\dfrac{x^2}{9} + \dfrac{y^2}{4} = 1 \quad \Rightarrow \quad \dfrac{2x}{9}dx + \dfrac{y}{2}dy = 0 \quad (1)
$$
$$
A = 4xy \quad \Rightarrow \quad 0 = dA = 4ydx + 4xdy \quad (2)
$$
Therefore, from (1) and (2)
$$
\dfrac{dy}{dx} = -\dfrac{4x}{9y} \quad \text{and} \quad \dfrac{dy}{dx} = -\dfrac{y}{x} \quad \Rightarrow \quad 4x^2 = 36\dfrac{y^2}{4} = 36\biggl(1 - \dfrac{x^2}{9}\biggr) \quad \Rightarrow \quad x = \pm \dfrac{3}{\sqrt{2}}
$$
The maximum occurs for $x = 3/\sqrt{2}$.
|
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|
Why does $\int_0^1\binom{2n}{n}x^n(1-x)^ndF(x)=\frac{1}{2n+1}$? Why does $$\int_0^1\binom{2n}{n}x^n(1-x)^ndF(x)=\frac{1}{2n+1},$$
Where $F(x)$ is uniform over the interval $[0,1]$?
It's not clear from simple factoring:
$$\binom{2n}{n}\int_0^1x^n(1-x)^ndF(x)$$
$$=\frac{(2n)!}{n!^2}\int_0^1x^n(1-x)^ndF(x).$$
|
By the binomial theorem,
\begin{align*}
\sum_{k=0}^{n} \left( \int_{0}^{1} \binom{n}{k} x^k(1-x)^{n-k} \, dx \right) z^k
&= \int_{0}^{1} (1 + (z-1)x)^n \, dx \\
&= \frac{1}{n+1} \cdot \frac{z^{n+1} - 1}{z-1} \\
&= \sum_{k=0}^{n} \frac{z^k}{n+1}.
\end{align*}
So we have
$$ \int_{0}^{1} \binom{n}{k} x^k(1-x)^{n-k} \, dx = \frac{1}{n+1} $$
for all $0 \leq k \leq n$.
|
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|
If $\sin(x)=3\cos(x)$, compute $\sin(x)*\cos(x)$ I drew a triangle and was instructed to use a property unknown to me. Apparently the answer is $\frac{3}{\sqrt{10}}*\frac{1}{\sqrt{10}}=\frac{3}{10}$. Is this answer correct, and what theorem/formula is this?
|
$$\begin{align} \\
\sin (2x)&=2\sin (x)\cos (x) \\
&=\frac {2\tan (x)}{1+\tan^2 (x)} \\
&=\frac {2\cdot3}{1+(3)^2} \\
&=\frac {6}{10} \\
\end{align}$$
thus
$$\sin (x)\cos (x)=\frac {3}{10}$$
|
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|
On the series $\sum \limits_{n=1}^{\infty} \frac{1}{n^2-3n+3}$ and $\sum\limits_{n=-\infty}^{\infty} \frac{1}{n^2-3n+3}$ Wolfram Alpha says that
$$\sum_{n=1}^{\infty} \frac{1}{n^2-3n+3} = 1 + \frac{\pi \tanh \left ( \frac{\sqrt{3}\pi}{2} \right )}{\sqrt{3}}$$
However I am unable to get it. It is fairly routine to prove that
$$\sum_{n=-\infty}^{\infty} \frac{1}{n^2-3n+3} = \frac{2\pi \tanh \left ( \frac{\sqrt{3}\pi}{2} \right )}{\sqrt{3}}$$
by using complex analysis ( contour integration ) but honestly I am stuck how to retrieve the original sum. Split up , the last sum gives:
\begin{align*}
\sum_{n=-\infty}^{\infty} \frac{1}{n^2-3n+3} &= \sum_{n=-\infty}^{-1} \frac{1}{n^2-3n+3} + \frac{1}{3} + \sum_{n=1}^{\infty} \frac{1}{n^2-3n+3} \\
&=\frac{1}{3} +\sum_{n=1}^{\infty} \frac{1}{n^2+3n+3} + \sum_{n=1}^{\infty} \frac{1}{n^2-3n+3} \\
&=\frac{1}{3}+ \sum_{n=1}^{\infty} \left [ \frac{1}{n^2-3n+3} + \frac{1}{n^2+3n+3} \right ]
\end{align*}
Am I overlooking something here?
P.S: Working with digamma on the other hand I am not getting the constant. I'm getting $\frac{1}{3}$ instead.
|
The First Sum
$$
\begin{align}
\sum_{n=1}^\infty\frac1{n^2-3n+3}
&=1+\sum_{n=2}^\infty\frac1{n^2-3n+3}\tag1\\
&=1+\sum_{n=2}^\infty\frac1{\left(n-\frac32-i\frac{\sqrt3}2\right)\left(n-\frac32+i\frac{\sqrt3}2\right)}\tag2\\
&=1+\frac1{i\sqrt3}\sum_{n=2}^\infty\left(\frac1{n-\frac32-i\frac{\sqrt3}2}-\frac1{n-\frac32+i\frac{\sqrt3}2}\right)\tag3\\
&=1+\frac1{i\sqrt3}\sum_{n=2}^\infty\left(\frac1{n-\frac32-i\frac{\sqrt3}2}+\frac1{-n+\frac32-i\frac{\sqrt3}2}\right)\tag4\\
&=1+\frac1{i\sqrt3}\sum_{n=-\infty}^\infty\frac1{n+\frac12-i\frac{\sqrt3}2}\tag5\\
&=1+\frac\pi{i\sqrt3}\cot\left(\frac\pi2-i\frac{\pi\sqrt3}2\right)\tag6\\[3pt]
&=1+\frac\pi{i\sqrt3}\tan\left(i\frac{\pi\sqrt3}2\right)\tag7\\[3pt]
&=1+\frac\pi{\sqrt3}\tanh\left(\frac{\pi\sqrt3}2\right)\tag8
\end{align}
$$
Explanation:
$(1)$: separate the $n=1$ term
$(2)$: factor the denominator
$(3)$: apply partial fractions
$(4)$: rewrite the right hand summand
$(5)$: combine the summands into a sum over $\mathbb{Z}$
$(6)$: apply $(7)$ from this answer
$(7)$: $\cot\left(\frac\pi2-x\right)=\tan(x)$
$(8)$: $\tan(ix)=i\tanh(x)$
The Second Sum
$$
\begin{align}
\sum_{n=-\infty}^\infty\frac1{n^2-3n+3}
&=\frac1{i\sqrt3}\sum_{n=-\infty}^\infty\left(\frac1{n-\frac32-i\frac{\sqrt3}2}+\frac1{-n+\frac32-i\frac{\sqrt3}2}\right)\tag9\\
&=\frac2{i\sqrt3}\sum_{n=-\infty}^\infty\frac1{n+\frac12-i\frac{\sqrt3}2}\tag{10}\\
&=\frac2{\sqrt3}\tanh\left(\frac{\pi\sqrt3}2\right)\tag{11}
\end{align}
$$
Explanation:
$\phantom{1}(9)$: partial fractions á la $(3)$
$(10)$: combine two series over $\mathbb{Z}$
$(11)$: apply $(5)$-$(8)$
|
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|
Simplifying $\frac{y}{y^2+b^2}$ where $y=b \cot\theta$ Could appreciate some help with this question.
I want to simplify the following trigonometric equation.
$$\frac{y}{y^2+b^2}$$
where $y=b \cot\theta$.
The solution I got was
$$\frac{1}b \cos\theta$$
Can someone verify and try and guide me through the solution?
|
$$1+\cot^2\theta=1+\frac{\cos^2\theta}{\sin^2\theta}=\frac1{\sin^2\theta}=\csc^2\theta$$ or
$$1+\cot^2\theta=\frac1{\sin^2\theta}$$
Now take reciprocal of both side:
$$\frac{1}{1+\cot^2\theta}=\sin^2\theta$$
If $y=b\cot\theta$, then:
$$\frac{y}{y^2+b^2}=\frac{b\cot\theta}{b^2\cot^2\theta+b^2}=\frac{\cot\theta}{b\left(1+\cot^2\theta\right)}=\frac{1}{b}\cot\theta\sin^2\theta=\frac{1}{b}\frac{\cos\theta}{\sin\theta}\sin^2\theta=\frac{1}{2b}2\cos\theta\sin\theta=\frac{1}{2b}\sin{2\theta}$$
|
{
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"timestamp": "2023-03-29T00:00:00",
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|
$a+b+c=?$ by having $a^2+160=b^2+5$ and $a^2+320=c^2+5$ My question:
$a+b+c=?$ by having $a^2+160=b^2+5$ and $a^2+320=c^2+5$.
My work so far:
$a^2+160=b^2+5\Rightarrow (b-a)(a+b)=155=31\times 5$
$a^2+320=c^2+5\Rightarrow (c-a)(c+a)=315=5\times3^2\times 7$
And now, I'm stuck.
($a,b,c$ are a members of $\mathbb Z$ and are positive)
|
We know that: $$ (b-a)(a+b)=31\times 5=155$$ as both $31$ and $5$ are prime numbers we can say either $b-a=5, a+b=31$ OR $b-a=1, a+b=155$. (becasue $a+b>b-a$)
*
*$b-a=5, a+b=31$, thus: $$a=13 , b=18$$
On the other hand, we have:
$$a^2+320=c^2+5$$
$$\to c^2=13^2+320-5 \to c=22$$
*$b-a=1, a+b=155$, thus: $$a=77 , b=78$$
On the other hand, we have:
$$a^2+320=c^2+5$$
$$\to c^2=77^2+320-5 \to c\approx 79.02 \not\in \Bbb Z$$
Hence the only correct answer is $a=13$ and $b=18$.
|
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|
Find the rational solution of the equation Find the rational solution of the equation:
$\frac{2x - 1}{2016} + \frac{2x - 3}{2014} + \frac{2x - 5}{2012}+ ...+ \frac{2x - 2011}{6} +\frac{2x - 2013}{4} + \frac{2x - 2015}{2} =\\ \frac{2x - 2016}{1} + \frac{2x - 2014}{3} + \frac{2x - 2012}{5}+ ...+ \frac{2x - 6}{2011} + \frac{2x - 4}{2013} + \frac{2x - 2}{2015}$
The problem is from a competition for seven graders. I tried various algebraic manipulations for no avail. Any hint will be highly appreciated.
|
Every term is $\dfrac{2x-2k}{2015-2k+2}-\dfrac{2x-(2k-1)}{2016-2k+2}=\dfrac{2x-2017}{(2016-2k+2)(2015-2k+2)}$
|
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|
Show that $\frac{1}{3}<\int_0^1\frac{1}{1+x+x^2} \, dx <\frac{\pi}{4}$ Show that $$\frac{1}{3}<\int_0^1\frac{1}{1+x+x^2}\,dx <\frac{\pi}{4}$$
I want to use if $f<g<h$ then $\int f<\int g<\int h$ formula for Riemann integration.
$1+x^2<1+x+x^2$ and it will give RHS as $$\frac{1}{1+x+x^2}<\frac{1}{1+x^2}$$
How to choose function $f$ and $h.$
|
You want to show $$\frac{1}{3}<\int_0^1\frac{1}{1+x+x^2}dx <\frac{\pi}{4}$$
Note that for $0<x<1$ we have $$ 1+x^2 < 1+x+x^2 <3 $$ Thus $$ \frac {1}{3} < \frac {1}{1+x+x^2}< \frac {1}{1+x^2}$$
Upon integration we get $$\frac{1}{3}=\int^{1}_0\frac{1}{3}\,dx<\int^1_0\frac{1}{1+x+x^2}\,dx<\int^1_0\frac{1}{1+x^2}\,dx=\arctan(x)\Big|^1_0=\frac{\pi}{4}$$
|
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|
What kind of matrix is this and why does this happen? So I was studying Markov chains and I came across this matrix \begin{align*}P=\left( \begin{array}{ccccc}
0 & \frac{1}{4} & \frac{3}{4} & 0 & 0\\
\frac{1}{4} & 0 & 0 & \frac{1}{4} & \frac{1}{2}\\
\frac{1}{2} & 0 & 0 & \frac{1}{4}& \frac{1}{4}\\
0 & \frac{1}{4} & \frac{3}{4} & 0& 0\\
0 & \frac{1}{4} & \frac{3}{4} & 0& 0\\
\end{array} \right).\end{align*}
I noticed (by brute force) that \begin{align*}P^2=\left( \begin{array}{ccccc}
\frac{7}{16} & 0 & 0 & \frac{1}{4} & \frac{5}{16}\\
0 & \frac{1}{4} & \frac{3}{4} & 0& 0\\
0 & \frac{1}{4} & \frac{3}{4} & 0& 0\\
\frac{3}{8} & 0 & 0 & \frac{1}{4}& \frac{1}{2}\\
\frac{7}{16} & 0 & 0 & \frac{1}{4} & \frac{5}{16}\\
\end{array} \right),\end{align*}
and \begin{align*}P^3=\left( \begin{array}{ccccc}
0 & \frac{1}{4} & \frac{3}{4} & 0& 0\\
\frac{7}{16} & 0 & 0 & \frac{1}{4} & \frac{5}{16}\\
\frac{7}{16} & 0 & 0 & \frac{1}{4} & \frac{5}{16}\\
0 & \frac{1}{4} & \frac{3}{4} & 0& 0\\
0 & \frac{1}{4} & \frac{3}{4} & 0& 0\\
\end{array} \right).\end{align*}
In fact; using a computer I found that every even power takes the form of the $P^2$ matrix and every odd power takes the form of the $P^3$ matrix.
I just wanted to know why that oscillation occurs? Is there a special name for the kind of matrix that $P$ is for it to exhibit that kind of behaviour?
|
This is a periodic Markov chain (with period $2$). Otherwise, there's not much that's terribly unusual about it.
User "Iwillnotexist Idonotexist" raised an important point in the comments:
Well, something that can be noted for periodic Markov chains is that
by definition they cannot be ergodic, another very important property
of MCs that you may encounter soon. Roughly speaking, an ergodic MC
that runs long enough "forgets" everything about its initial state. If
the MC is periodic, then clearly you must remember some information
about the contents of the initial state, because you're stuck in a
loop of states that you keep on coming back to and aren't forgetting.
|
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|
Prove that the area of $\triangle DEF$ is twice the area of $\triangle ABC$ Let $\triangle ABC$ be an equilateral triangle and let $P$ be a point on its circumcircle. Let lines $PA$ and $BC$ intersect at $D$ , let lines $PB$ and $CA$ intersect at $E$, and let lines $PC$ and $AB$ intersect at $F$. Prove that the area of $\triangle DEF $ is twice the area of $\triangle ABC$.
I had used my method and it required that $\triangle DEF $ have to be isosceles triangle which means $ P $ should be the mid point of $\widehat {BC}$ or $\widehat{AC}$ or $\widehat {BC}$. But the problem does not said so. So, how to solve it correctly?
|
It is very interesting to solve this problem by only using our greek friends Pythagoras and Menelaus, together with some simple algebra.
Initial Observations
Without loss of generality assume that $P$ lies on the half-plane determined by $BC$ and not containing $A$. Note that there is no need to use the circle, once we fix $\angle BPC = 120°$. In the figure above I added point $G$ given by the intersection of line $FD$ with $AC$. Suppose also $\overline{AB}=1$.
Once $\overline{BF} = x$ is chosen, the entire Figure is defined. We aim therefore at showing that, independently of $x$,
$$[DEF] = 2[ABC].$$
Characterization of $\triangle BFC$
Below the triangle in question has been isolated, to better help you in the demontrations.
Recall that $\overline{BC} = 1$ and $\overline{BF} = x$.
Use Pythagorean Theorem on $\triangle CKF$ in oder to determine
$$\overline{CF} = \sqrt{x^2+x+1}.$$
From the similarity
$$\triangle BPC \sim \triangle BCF,$$
show that
$$\overline{CF}\cdot\overline{CP} = 1,$$
so that, in the end, you have
$$
\begin{cases}
\overline{CF} = \sqrt{x^2+x+1},\\
\overline{CP} = \frac{1}{\sqrt{x^2+x+1}},\\
\overline{FP} = \frac{x(x+1)}{\sqrt{x^2+x+1}}.
\end{cases}
$$
Four Applications of Menelaus's Theorem (MT)
MT on $\triangle ACF$ with the line $BE$, gives
$$\frac{\overline{CE}}{\overline{AE}} = \frac{1}{x+1}.$$
Together with $\overline{AE}-\overline{CE} = 1$ this leads to
$$\overline{CE} = \frac{1}{x}$$
and
$$\overline{AE} = \frac{x+1}{x}.$$
Similarly, MT on $\triangle BFC$ and line $AP$, with the known fact that $\overline{BD} + \overline{CD} = 1$ yields
$$\overline{CD} = \frac{1}{x+1}$$
and
$$\overline{DB} = \frac{x}{x+1}.$$
MT on $\triangle ABC$ and line $FG$, with $\overline{AG} + \overline{CG} = 1$, gives you
$$ \overline{AG} = \frac{x+1}{x+2}$$
and
$$ \overline{CG} = \frac{1}{x+2}.$$
Finally, MT on $\triangle CGF$ with line $AP$ yields
$$ \frac{\overline{GD}}{\overline{DF}} = \frac{1}{x(x+2)}.$$
Areas Computation
Now we only need to compute areas of triangles with fixed altitude and a given ratio between bases.
Firstly we have
$$[ACF] = [ABC](1+x).$$
Then
$$[AFE] = [ACF] \frac{\overline{AE}}{\overline{AC}},$$
that is
$$[AFE] = [ABC]\frac{(x+1)^2}{x}.$$
We also have
$$[GFE] = [AFE]\frac{\overline{GE}}{\overline{AE}},$$
yielding
$$ [GFE] =2[ABC]\frac{(x+1)^2}{x(x+2)}.$$
Finally observe that
$$\frac{[DFE]}{[GDE]} = \frac{\overline{GD}}{\overline{DF}} = \frac{1}{x(x+2)}.$$
Thus we have the system of equations
$$
\begin{cases}
\frac{[DFE]}{[GDE]} = \frac{1}{x(x+2)}\\
[DFE] + [GDE] =2[ABC]\frac{(x+1)^2}{x(x+2)}.
\end{cases}
$$
leading, once solved, to the desired result, i.e.
$$\boxed{[DFE] = 2[ABC]} $$
$\blacksquare$
|
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|
Find an upper bound for the modulus of a complex function For real $x$ let $r(x)$ be defined (implicitly) through
$$
e^{i\,x} = \left(1+i\,x\right)\,e^{\left(-(1/2)\,x^2+r(x)\right)},
$$
where $i$ is the complex unit. I need to prove that
$$
\left|r(x)\right|\leq \left|x\right|^3,\quad \text{for} ~~\left|x\right|\leq 1.
$$
Unfortunately I do not have any clue on this.
===Some computations====
Using expansion I get
$$
r(x) = i\,x+\frac{1}{2}\,x^2-\ln(1+i\,x)=i\,\frac{x^3}{3}+\frac{x^4}{4}-i\,\frac{x^5}{5}-\frac{x^6}{6}+i\,\frac{x^7}{7}+\frac{x^8}{8}+...
$$
that is
$$
r(x)=\sum_{k=2}^{\infty}(-1)^k\frac{x^{2\,k}}{2\,k}+i\,\sum_{k=1}^{\infty}(-1)^{k+1}\frac{x^{2\,k+1}}{2\,k+1} = \frac{1}{2} \left(x^2-\log \left(x^2+1\right)\right)+i\,\left(x-\tan ^{-1}(x)\right)
$$
so that
$$
|r(x)| = \sqrt{\frac{1}{4} \left(x^2-\log \left(x^2+1\right)\right)^2+\left(x-\tan ^{-1}(x)\right)^2}
$$
|
I think that the inequality does not hold. Let's find $r(x)$ explicitly. Let $p(x)$ and $q(x)$ be real functions such that $r(x)=p(x)+iq(x)$. To find $p(x)$, we take modulus on given equation:
$$
1=\sqrt{1+x^2}e^{-\frac{1}{2}x^2+p(x)}
$$
Then we get
$$
p(x)=\frac{1}{2}x^2 - \frac{1}{2}\ln(1+x^2).
$$
Substitute $p(x)$ to the original equation, then
$$
e^{ix}=(1+ix)e^{-\frac{1}{2}\ln(1+x^2)+iq(x)}=\frac{1+ix}{\sqrt{1+x^2}}e^{iq(x)}=e^{i(\theta(x)+q(x))},
$$
where $\theta(x)$ is a real number such that $\cos\theta(x)=\dfrac{1}{\sqrt{1+x^2}}$ and $\sin\theta(x)=\dfrac{x}{\sqrt{1+x^2}}$. Thus
$$
q(x)=x-\theta(x)+2n\pi
$$
for some $n\in \mathbb{Z}$. Define a complex-valued function $f:\mathbb{R}\to\mathbb{C}$ by
$$
f(x)=\frac{1}{2}x^2-\frac{1}{2}\ln(1+x^2)+i\left(x-\arccos\frac{1}{\sqrt{1+x^2}}+2018\pi\right),
$$
then $f$ satisfies the original equation. However, $f(0)\ne 0$.
|
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|
Solving a linear congruence system How can this linear congruence system be solved by using Chinese remainder theorem?
\begin{align}
12x&\equiv -7 \pmod {13}\\
4x&\equiv 7 \pmod {9}\\
2x&\equiv -3 \pmod {11}
\end{align}
As far as I understand, we have to create some M numbers as it follows:
\begin{align}
M = 13 * 9 * 11 = 1287\\
M_1 = 9 * 11 = 99\\
M_2 = 13 * 11 = 143\\
M_3 = 13 * 9 = 117
\end{align}
But from here forward it becomes confusing for me. What should be done next?
Thank you in advance.
|
Note that CRT guarantees that an unique solution exist mod $9\cdot 11 \cdot 13$ since those are relatively primes but doesn’t give any particular method to solve the problem others that those used for the proof that you can find here.
To find the solution here we can proceed as follow
*
*$12x \equiv -7 \iff x\equiv 7 \pmod {13}$
*$4x\equiv 7 \iff x\equiv 4 \pmod {9}$
*$2x\equiv -3 \iff x\equiv 4 \pmod {11}$
this is the standard form for the system , then
*
*$x\equiv 7 \pmod {13}\implies x=7+13k$
*$x\equiv 4\implies 7+4k\equiv 4\implies 4k\equiv 6 \implies k\equiv 6 \pmod {9}\implies x=7+13(6+9h)=85+9\cdot13h$
*$x\equiv 4 \implies 8+7h\equiv4 \implies h\equiv 1 \pmod {11} \implies x=85+9\cdot13(1+11s)=202+9\cdot11\cdot 13s$
that is $$x=202 \pmod{9\cdot11\cdot 13}$$
|
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|
Limit simplification I've been practicing limits today, and I've came across this exercise which confused me a bit:
$\lim_{x\to -3} \frac{(x^2-9)^2}{(x+3)^2}$
My approach was to do this:
$\lim_{x\to -3} \frac{(x^2-9)^2}{x^2+3x+9}$
$\lim_{x\to -3} \frac{0}{9}$
Can anyone explain why this is not a valid approach? I realised that as long as the denominator is not 0, I can start inserting the x.
The valid solution is 36, and I know how to get there, I just don't get it why my approach is wrong.
Thanks!
|
$$\begin{align} x^2 - 9 &= x^2 - 3^2 \\ \\ &= (x+3)(x-3) \\ \Rightarrow \dfrac{(x^2-9)^2}{(x+3)^2} &= \dfrac{\require{cancel}\cancel{(x+3)^2}(x-3)^2}{\cancel{(x+3)^2}} \\ &= (x-3)^2. \\ \\ \therefore \lim_{x\to -3}\frac{(x^2-9)^2}{(x+3)^2} &= \lim_{x\to -3} (x-3)^2 \\ \\ &= (-3 - 3)^2 \\ \\ &= 36.\end{align}$$
|
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|
Let C be the unit square with diagonal corners at $−1 − i$ and $1 + i$. Evaluate $\oint_C\frac{1}{2z+1}\mathrm{d}z$ I'm having trouble trying to understand what I'm doing wrong in solving this complex integral.
I partitioned the square into 4 different parameterized curves (each curve being 1 of the sides of the square) and added the 4 integrals. The final answer I got was zero, but the book says the answer is $πi$. I cannot for the life of me understand how they got such an answer. Any tip will be greatly appreciated. Thanks!
EDIT:
I parameterized the square as such
$c_1(t)=t-i$
$c_2(t)=1+it$
$c_3(t)=-t+i$
$c_4(t)=-1-it$, with $-1\leq t\leq 1$ for all the $c$ curves.
I obtained the following complex integrals via
$\int_{-1}^{1} f(c(t))c'(t)dt$:
For $c_1$, $\int_{-1}^{1}\frac{1}{2(t-i)+1}dt$ = $\frac{1}{2}ln(3-2i) - \frac{1}{2}ln(-1-2i)$.
For $c_2$,
$\int_{-1}^{1}\frac{i}{2(1+it)+1}dt$ = $\frac{1}{2}ln(3+2i) - \frac{1}{2}ln(3-2i)$.
For $c_3$, $\int_{-1}^{1}\frac{-1}{2(-t+i)+1}dt$ = $\frac{1}{2}ln(-1+2i) - \frac{1}{2}ln(3+2i)$.
For $c_4$, $\int_{-1}^{1}\frac{-i}{2(-1-it)+1}dt$ = $\frac{1}{2}ln(-1-2i) - \frac{1}{2}ln(-1+2i)$.
I added the 4 integrals up and they all cancelled out to zero. But I don't see how $πi$ was obtained.
|
$$ I:=\oint_C\frac{1}{2z+1}\mathrm{d}z $$
The idea of your integration path is correct. Only in $c_3$ and $c_4$ you fail to implement it correctly, when you use $-t+i$ you actually have to go from $-1$ to $+1$ in positive direction $dt$ not in negative $-dt$ to get the right orientation. You do not have to transform anything, but just construct the right path.
(I use the name $ln$ for the complex logarithm like OP did. Conventions on that are not the same in all countries.).
$$\begin{align}I & = \int_{-1}^{1}\frac{1}{2(t-i)+1}dt + \int_{-1}^{1}\frac{i}{2(1+it)+1}dt + \int_{-1}^{1}\frac{1}{2(-t+i)+1}dt + \int_{-1}^{1}\frac{i}{2(-1-it)+1}dt \\ & = \frac{1}{2}\bigg[ \ln(-1-2 i + 2t) + \ln(-i+2t) + \ln(-3i + 2t) + \ln(1-2i+2t) \bigg]^{1}_{-1}\\ &= \frac{1}{2}[\ln(3-2i)-\ln(-3-2i)+\ln(2-i)-\ln(-2-i)+\ln(2-3i)-\ln(-2-3i)+\ln(1-2i)-\ln(-1-2i)] \\ &= \frac{i}{2}\big[(\arg(3-2i)-\arg(-3-2i))+(\arg(2-i)-\arg(-2-i))+(\arg(2-3i)-\arg(-2-3i))+(\arg(1-2i)-\arg(-1-2i))\big]\\ &=\frac{i}{2}\big[\arctan(-\frac{2}{3})-(\arctan(\frac{2}{3})-\pi)+\arctan(-\frac{3}{2})-(\arctan(\frac{3}{2})-\pi)+\\ &\;\;\;+ \arctan(-\frac{1}{2})-(\arctan(\frac{1}{2})-\pi)+\arctan(-\frac{2}{1})-(\arctan(\frac{2}{1})-\pi)\big] \\ &= \frac{i}{2}\big[\pi - 2 \arctan(\frac{2}{3}) +\pi - 2 \arctan(\frac{3}{2}) + \pi - 2 \arctan(\frac{1}{2}) +\pi - 2 \arctan(\frac{2}{1})\big] \\ &= \frac{i}{2}\big[4 \pi -2[\arctan(\frac{2}{3})+\arctan(\frac{3}{2})+\arctan(\frac{1}{2})+\arctan(2) ] \big]\\ &= \frac{i}{2}(4\pi-2(\frac{\pi}{2}+\frac{\pi}{2}))\\ &= i \pi.
\end{align}$$
Using $\arctan|\frac{a}{b}|+\arctan|\frac{b}{a}|=\frac{\pi}{2}.$
Edit: Since the paramterisations of the curves (lines) of the integrals are continuously differentiable we can evaluate the integrals in the way like integrals of a real variable. The residue theorem was not used here, since the choice of the loop suggests that it should be done on elementary level.
|
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|
Complex numbers $\left(\frac{1+i}{1-i}\right)^k = 1$ what is $k$? The smallest possible integer $k$ for which $\left(\frac{1+i}{1-i}\right)^k = 1$ is?
I tried solving this, but my answer doesn't match the given answer. Correct me if I'm wrong at some place
My solution:
\begin{align}
\left(\frac{1+i}{1-i}.\frac{1+i}{1+i}\right)^k&= 1\\
\left(\frac{1+2i+i^2}{1-i^2}\right)^k&= 1\\
\left(\frac{1+2i-1}{1-(-1)}\right)^k&= 1\\
\left(\frac{2i}{2}\right)^k&= 1\\
i^k&= 1\\
i^4&= 1\\
\end{align}
EDIT: The question is part of the multiple choice section and the answer is 2.
Other options include: 4, 8, 16
|
$(\frac {1+i}{1-i})^k=1$
$(1+i)^k = (1-i)^k$
$\sum_{j=0}^k {k\choose j}i^j = \sum_{j=0}^k {k\choose j}i^j*(-1)^j$
As $(-1)^{even} = 1$ and $(-1)^{odd} = -1$
$\sum_{j=0;j odd}{k\choose j}i^j =0$
As $i^{4k + 1} = i$ and $i^{4k - 1} = -i$
we need to find the smallest $k$ where $\sum_{h=0}^{4h+1 \le k}{k\choose 4h+1} =\sum_{h=1}^{4h-1 \le k}{k\choose 4h-1}$.
${k \choose 4h -1} = {k\choose k -4h + 1}$ so if $k = 4m$ and $g = m-h$ we will have ${k\choose 4h -1} = {k \choose k - 4h + 1} = {k\choose 4(m-h) + 1} = {k\choose 4g + 1}$ so any $k = 4m$ will be a solution so $k =4$ will be a solution.
It's easy to show directly that $k = 1,2,3$ are not:
$1 + i \ne 1 -i$
$(1+i)^2 =1 + 2i -1 \ne 1 - 2i -1 =(1-i)^2$
$(1+i)^3 = 1 + 3i + 3i^2 + i^3 = -2 + 2i \ne 2-2i = 1 - 3i + 3i^2 -i^3 = (1+i)^3$
However $(1 + i)^4 = 1 + 4i + 6i^2 + 4i^3 + i^4 = 1 - 4i^3 + 6i^2 - 4i + i^4 = 1 - 4i + 6i^2 - 4i^3 + i^4 = (1 - i)^4$.
|
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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|
solving $x\,dx + xy^2\,dx + y\,dy + yx^2\,dy=0$ I have to solve $x\,dx + xy^2\,dx + y\,dy + yx^2\,dy=0$
Dividing by $dx$ we have
$x + xy^2 + yy' + yy'x^2=0$
From where,
$$\frac{yy'}{1+y^2}+\frac{x}{1+x^2}=\frac{y\,dy}{1+y^2}+\frac{x\,dx}{1+x^2}=\\ =\frac{d(y^2+1)}{1+y^2}+\frac{d(x^2+1)}{1+x^2}= \frac{1}{2}d\ln(1+y^2)+\frac{1}{2} d\ln(1+x^2)=\frac{1}{2}d\ln(1+y^2)(1+x^2)=0$$
Let $c=(1+y^2)(1+x^2)$, so our equation becomes:
$$
d\ln c=0
$$
So what should I do here, should I integrate, or should I divide by $dx$?
If I divide by dx I get the expression $2x+2yy'+2xy^2+2x^2yy'=0$ which has $x$, $y$ and $y'$ and doesn't help me get anywhere.
Thanks in advance.
|
$$d( \text{something})=0 \implies \text{something = constant}$$
So you get the solution $$\ln(1+y^2)(1+x^2) = C$$ (Where $C$ is arbitrary constant)
|
{
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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|
Vector spaces and basis I came across this problem on vector space basis:
Verify that $2-x^2, x^3-x, 2-3x^2$ and $3-x^3$ form a basis for $P^4$ and express each of the polynomial $x^2$ as a linear combination of this basis.
I have tried solving the problem with my knowledge of vector space but still don't know how to go about it.
Any helpful solution to this?
|
1) Let's verify that the system of vectors $(2−x^2,x^3−x,2−3x^2, 3−x^3)$(1) is a basis for $P^4$. Write the canonical basis for $P^4$: $(1 ,x ,x^2 ,x^3)$(2) and decompose vectors of the system (1) by basis (2):
\begin{pmatrix}
2 & 0 & 2 & 3\\
0 & -1 & 0 & 0\\
-1 & 0 & -3 & 0\\
0 & 1 & 0 & -1\\
\end{pmatrix}
Now we'll proof linear independence of (1):
$
A=\begin{pmatrix}
2 & 0 & 2 & 3\\
0 & -1 & 0 & 0\\
-1 & 0 & -3 & 0\\
0 & 1 & 0 & -1\\
\end{pmatrix}
\sim
\begin{pmatrix}
1 & 0 & -1 & 3\\
0 & -1 & 0 & 0\\
-1 & 0 & -3 & 0\\
0 & 0 & 0 & -1\\
\end{pmatrix}
\sim
\begin{pmatrix}
1 & 0 & -1 & 3\\
0 & -1 & 0 & 0\\
0 & 0 & -4 & 3\\
0 & 0 & 0 & -1\\
\end{pmatrix}
\sim
\begin{pmatrix}
1 & 0 & 0 & 0\\
0 & 1 & 0 & 0\\
0 & 0 & 1 & 0\\
0 & 0 & 0 & 1\\
\end{pmatrix}$
Proofed,(1) is linear independent, $rankA=rank(1)=4$. But it is maximal linear independent for $P^4$( if we add any vector to (1), it becomes linear dependent (definition of max. linear independent system)($dimP^4=rank(1)=4$)), so we can state that (1) is a basis for $P^4$.
2)Since (1) is a basis for $P^4$, we can decompose any vectors of the $P^4$ by (1)(definition of basis), including each of the polynomial $x^2$.
|
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"timestamp": "2023-03-29T00:00:00",
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|
Evaluating $\lim _{t\to \infty}\frac{1-\frac{\sqrt{t}}{\sqrt{t+1}}}{2-\frac{\sqrt{4t\:+\:1}}{\sqrt{t+2}}}$ I am having difficulties evaluating this limit:
$$\lim _{t\to \infty \:}\left(\frac{1-\frac{\sqrt{t}}{\sqrt{t+1}}}{2-\frac{\sqrt{4t\:+\:1}}{\sqrt{t+2}}}\right)$$
I have tried to divide out by $\frac{\sqrt{t}}{\sqrt{t+1}}$ in the numerator and denominator but I run into problems and I also tried to divide through with $\sqrt{t}$ but I still get 0/0.
I've been stumped for hours and need a heads up on this.
|
The Hospital Rule :
We separate the numerator and the denominator :
$$ f(x) = \frac{\sqrt{x}}{\sqrt{x+1}}$$
$$ g(x) = \frac{\sqrt{4x+1}}{\sqrt{x+2}}$$
we have that :
$$ f'(x) = \frac{1}{2 \sqrt{x} \cdot (x+1)^{3/2}} $$
$$g'(x) = \frac{2}{\sqrt {x+2} \sqrt{4x+1} } - \frac{4x+1}{2(x+2)^{3/2}}$$
then you do the quotient, you find :
$$ \frac{(x+2)^{3/2} \sqrt{4x+1}}{7\sqrt{x}(x+1)^{3/2}} $$
taking the limit you find :
$$\lim _{t\to \infty \:}\left(\frac{1-\frac{\sqrt{t}}{\sqrt{t+1}}}{2-\frac{\sqrt{4t\:+\:1}}{\sqrt{t+2}}}\right)
=
\lim_{x\to \infty \:} \left( \frac{(x+2)^{3/2} \sqrt{4x+1}}{7\sqrt{x}(x+1)^{3/2}}\right) = \frac{2}{7} $$
|
{
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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|
Help with a trigonometry problem. I have encountered some problems to solve the left side to the right.
$$ \cos^ 2x \sin x = \frac{\sin 3x + \sin x}{4}$$
I trying to solve a differential equation on the form $$ y'' + y = \cos^ 2x * \sin x$$ and need to rewrite it to $$ y'' + y = \frac{\sin 3x + \sin x}{4}$$
Have tried with different combinations but not really got it to $$\frac{\sin 3x + \sin x}{4}$$
I have also tried to start backwards, but it seems not to be the method I would have chosen if I started from the left.
\begin{align}
\frac{\sin 3x+\sin x}{4}
& =\frac{\sin x+\sin 3x}{4} \\
& =\frac{\sin x+\sin \left( 2x+x \right)}{4} \\
& =\frac{\sin x+\left( \sin 2x\cos x+\sin x\cos 2x \right)}{4} \\
& =\frac{\sin x+\left( \left( 2\sin x\cos x \right)\cos x+\sin x\left( {{\cos }^{2}}x-{{\sin }^{2}}x \right) \right)}{4} \\
& =\frac{\sin x+\left( 2\sin x{{\cos }^{2}}x+\sin x\left( {{\cos }^{2}}x-{{\sin }^{2}}x \right) \right)}{4} \\
& =\frac{\sin x+\left( 2\sin x\left( 1-{{\sin }^{2}}x \right)+\sin x\left( 1-2{{\sin }^{2}}x \right) \right)}{4} \\
& =\frac{\sin x+\left( (2\sin x-2{{\sin }^{3}}x)+(\sin x-2{{\sin }^{3}}x) \right)}{4} \\
& =\frac{\sin x+\left( 3\sin x-4{{\sin }^{3}}x \right)}{4} \\
& =\frac{4\sin x-4{{\sin }^{3}}x}{4} \\
& =\frac{4\sin x(1-{{\sin }^{2}}x)}{4} \\
& =\sin x{{\cos }^{2}}x \\
\end{align}
Would be grateful if you could give out some kind of guidance.
|
Note that by product to sum formula
$$2\cos \theta \sin \varphi = {{\sin(\theta + \varphi) - \sin(\theta - \varphi)} }\implies \cos 2x \cdot \sin x = \frac{\sin 3x - \sin x}{2}
$$
|
{
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"timestamp": "2023-03-29T00:00:00",
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|
If $a_k = c^k-1$, where $c > 1$, what can be said about $s(n) =\sum_{k=1}^n \frac{a_k}{a_{k+1}} $? If
$a_k = c^k-1$,
where $c > 1$,
what can be said about
$s(n)
=\sum_{k=1}^n \dfrac{a_k}{a_{k+1}}
$?
This is a generalization of
$a_n=3^n-1$, prove that $\frac{a_1}{a_2}+\frac{a_2}{a_3}+\dots+\frac{a_n}{a_{n+1}}>\frac{n}{3}-\frac{1}{8}.$
which is the case
$c=3$.
I can show that
$\dfrac{n}{c}-\dfrac{1}{c^2-1}
\lt s(n)
\lt \dfrac{n}{c}-\dfrac1{c^2}
$
and I wondered
if better inequalities
can be found.
Here is my method.
$\begin{array}\\
s(n)
&=\sum_{k=1}^n \dfrac{a_k}{a_{k+1}}\\
&=\sum_{k=1}^n \dfrac{c^k-1}{c^{k+1}-1}\\
&=\sum_{k=1}^n \dfrac{c^k-1/c+1/c-1}{c^{k+1}-1}\\
&=\sum_{k=1}^n (\dfrac1{c}-\dfrac{1-1/c}{c^{k+1}-1})\\
&=\dfrac{n}{c}-\sum_{k=1}^n \dfrac{1-1/c}{c^{k+1}-1}\\
&=\dfrac{n}{c}-(1-\frac1{c})\sum_{k=1}^n \dfrac{1}{c^{k+1}-1}\\
&=\dfrac{n}{c}-\dfrac{c-1}{c}r(n)
\qquad\text{where }r(n)=\sum_{k=1}^n \dfrac{1}{c^{k+1}-1}\\
r(n)
&=\sum_{k=1}^n \dfrac{1}{c^{k+1}-1}\\
&=\dfrac1{c^2-1}+\sum_{k=2}^n \dfrac{1}{c^{k+1}-1}\\
&>\dfrac1{c^2-1}-\dfrac1{c^2}+\dfrac1{c^2}+\sum_{k=2}^n \dfrac{1}{c^{k+1}}\\
&=\dfrac1{c^2(c^2-1)}+\sum_{k=1}^n \dfrac{1}{c^{k+1}}\\
&=\dfrac1{c^2(c^2-1)}+\dfrac{1-1/c^{n}}{c^2(1-1/c)}\\
&=\dfrac1{c^2(c^2-1)}+\dfrac{1}{c(c-1)}-\dfrac{1}{c^{n+1}(c-1)}\\
&=\dfrac{1}{c(c-1)}+\dfrac1{c^2(c^2-1)}(1-\dfrac1{c^{n-1}})\\
&\ge\dfrac{1}{c(c-1)}\\
\text{so}\\
s(n)
&<\dfrac{n}{c}-\dfrac1{c^2}\\
\text{and}\\
r(n)
&=\sum_{k=1}^n \dfrac{1}{c^{k+1}-1}\\
&<\sum_{k=1}^n \dfrac{1}{c^{k+1}-c^{k-1}}\\
&<\sum_{k=1}^n \dfrac{1}{c^{k-1}(c^2-1)}\\
&=\dfrac{1}{c^2-1}\sum_{k=0}^{n-1} \dfrac{1}{c^{k}}\\
&<\dfrac{1}{c^2-1}\sum_{k=0}^{\infty} \dfrac{1}{c^{k}}\\
&=\dfrac{1}{(c^2-1)(1-1/c)}\\
&=\dfrac{c}{(c^2-1)(c-1)}\\
\text{so}\\
s(n)
&>\dfrac{n}{c}-\dfrac{c-1}{c}\dfrac{c}{(c^2-1)(c-1)}\\
&>\dfrac{n}{c}-\dfrac{1}{c^2-1}\\
\end{array}
$
|
Given the identity
$$ s(n)=\frac{n}{c}-\frac{c-1}{c}\left[\frac{1}{c^2-1}+\sum_{k=2}^{n}\frac{1}{c^{k+1}-1}\right] $$
we have $s(n)\leq \frac{n}{c}-\frac{1}{c(c+1)}$ for any $n\geq 2$. Under the same assumption
$$ \sum_{k=2}^{n}\frac{1}{c^{k+1}-1}\leq \sum_{k=2}^{+\infty}\frac{1}{c^{k+1}-1}\leq \sum_{k=2}^{+\infty}\frac{1}{c^{k+1}-c^{k-2}}=\frac{1}{c-\frac{1}{c^2}}\cdot\frac{1}{c(c-1)} $$
leads to $s(n)\geq \frac{n}{c}-\frac{1}{c(c+1)}-\frac{1}{(c-1)(c^2+c+1)}$.
|
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "3",
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|
Variation on Harmonic Series I'm trying to establish the convergence or divergence of the following variant of the harmonic series:
$$\frac{1}{1}+\frac{1}{2}-\frac{1}{3}-\frac{1}{4}-\frac{1}{5}+\frac{1}{6}+\frac{1}{7}-\frac{1}{8}-\frac{1}{9}-\frac{1}{10}+\frac{1}{11}\cdots$$
Where the sign pattern has period 5, ie, it looks like this: ++---++---++---....
My thought has been to find a regrouping that diverges, since I only need to find one in order to show the series is divergent. I tried to bound this series below by increasing the denominator on the positive terms, and decreasing it on the negative terms to yield a series like $$\left(\frac{1}{2}+\frac{1}{2}\right)-\left(\frac{1}{3}+\frac{1}{3}+\frac{1}{3}\right)+\left(\frac{1}{7}+\frac{1}{7}\right)+ \cdots$$
but I'm pretty sure this will converge. I don't really know what approach to take next. A hint would be greatly appreciated! Thanks!
|
The sum of the first two is
$\dfrac1{5n+1}+\dfrac1{5n+2}
\lt \dfrac{2}{5n}
$
and the sum of the last 3 is
$\dfrac1{5n+3}+\dfrac1{5n+4}+\dfrac1{5n+5}
\gt \dfrac{3}{5n+5}
$.
Therefore
$\begin{array}\\
\dfrac1{5n+1}+\dfrac1{5n+2}
-(\dfrac1{5n+3}+\dfrac1{5n+4}+\dfrac1{5n+5})
&\lt \dfrac{2}{5n}-\dfrac{3}{5n+5}\\
&=\dfrac{2 - n}{5 n (n + 1)}\\
&=\dfrac{2}{5 n (n + 1)}-\dfrac{1}{5 (n + 1)}\\
\end{array}
$
and the sum of this diverges
to $-\infty$.
|
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"timestamp": "2023-03-29T00:00:00",
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|
Finding $\int\frac{1}{(1+\sin^2\theta)^{\frac{3}{2}}}d\theta$
Finding $$\int\frac{1}{(1+\sin^2\theta)^{\frac{3}{2}}}d\theta$$
Try: $$I=\int\frac{1}{(1+\sin^2\theta)^{\frac{3}{2}}}d\theta =2\sqrt{2}\int \frac{1}{(3-\cos 2\theta)^{\frac{3}{2}}}d\theta$$
How can I write this in terms of the elliptic integral? Could someone help me to explain it? Thanks
|
As stated by SmarthBansal in the comments, this can be described as an Incomplete Elliptic Integral of the Third Kind, defined as
$$\Pi(n;\phi,m)=\int_0^\phi \frac{1}{1-n\sin^2(\theta)}\frac{1}{\sqrt{1-m\sin^2(\theta)}} d\theta$$
So we can write this integral as $\Pi(-1,x|-1)$. However, this is not a very satisfying answer, which I believe is why SmarthBansal mentioned this as a comment. Jack D'Aurizio mentions another solution in terms of the Incomplete Elliptic Integral of the Second Kind, defined as
$$E(\phi, m)=\int_0^\phi \sqrt{1-m\sin^2(\theta)} d\theta$$
These two solutions can be reconciled by an elegant relation, namely,
$$E(\phi, m) = (1-m) \Pi(m;\phi ,m) + \frac{m \sin(2\phi)}{2\sqrt{1-m\sin^2(\phi)}}$$
Now let's derive that equation! If we start with the left side of the equation we get,
$$
\int_0^\phi \sqrt{1-m\sin^2(\theta)}d\theta
= \int_0^\phi \frac{(1-m\sin^2(\theta))^2}{(1-m\sin^2(\theta))^{3/2}}d\theta $$$$
=\int_0^\phi \frac{1- 2 m \sin^2(\theta)+m^2 \sin^4 (\theta)}{(1-m\sin^2(\theta))^{3/2}}d\theta $$$$
=\int_0^\phi \frac{1-m}{((1-m\sin^2(\theta))^{3/2})}d \theta + m\int_0^\phi \frac{\cos^2(\theta)-\sin^2(\theta)+m\sin^4(\theta)}{(1-m\sin^2(\theta))^{3/2}}d\theta$$$$
=(1-m)\Pi(m; \phi, m) + m\int_0^\phi \frac{(\cos^2(\theta)-\sin^2(\theta))(1-m\sin^2(\theta)) + m \sin^2(\theta)\cos^2(\theta)}{(1-m\sin^2(\theta))^{3/2}}d\theta$$$$
=(1-m)\Pi(m; \phi, m) + m\int_0^\phi \frac{(\cos(2\theta))(1-m\sin^2(\theta)) + m\sin(2\theta)\sin(\theta)\cos(\theta)}{(1-m\sin^2(\theta))^{3/2}}d\theta$$
Now dividing each side by $(1-m\sin^2(\theta))^{1/2}$, we can use the "reverse quotient rule" to get
$$=(1-m)\Pi(m; \phi, m) + \frac{m}{2}\int_0^\phi \frac{2\cos(2\theta)\sqrt{1-m\sin^2(\theta)} - \sin(2\theta)\frac{-2m\sin(\theta)\cos(\theta)}{\sqrt{1-m\sin^2(\theta)}}}{(1-m\sin^2(\theta))}d\theta$$
$$=(1-m) \Pi(m;\phi ,m) + \frac{m}{2} \frac{\sin(2\phi)}{\sqrt{1-m\sin^2(\phi)}}$$
Thus we have our relationship, and solving for $\Pi(m; x, m)$, we get
$$\Pi(m;\phi ,m)= \frac{E(\phi,m)}{1-m} - \frac{m\sin(2\phi)}{2(1-m)\sqrt{1-m\sin^2(\phi)}}$$
And plugging in $m=-1$, we get
$$\int \frac{1}{(1+\sin^2(\theta))^{3/2}}d\theta= \Pi(-1;\phi ,-1)= \frac{E(\phi,-1)}{2} + \frac{\sin(2\phi)}{4\sqrt{1+m\sin^2(\phi)}}$$
|
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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|
Evaluating trigonometric limit $\csc^2(2x) - \frac{1}{4x^2}$
$$\lim_{x\rightarrow 0} \left[ \csc^2(2x) - \frac{1}{4x^2} \right]$$
I've tried to use l'Hôpital's rules but still can't find the answer. Here's my approach:
$$
\begin{aligned}
&\lim_{x\rightarrow 0} \left[ \csc^2(2x) - \frac{1}{4x^2} \right] \\
=& \lim_{x\rightarrow 0} \left[ \frac{1}{\sin^2(2x)} - \frac{1}{4x^2} \right] \\
=& \lim_{x\rightarrow 0} \left[ \frac{4x^2-\sin^2(2x)}{4x^2(\sin^2(2x))} \right] \\
=& \lim_{x\rightarrow 0} \left[ \frac{8x-4\sin(2x)\cos(2x)}{8x\sin^2(2x)+16x^2\sin(2x)\cos(2x)} \right] \\
=& \lim_{x\rightarrow 0} \left[ \frac{8x-2\sin(4x)}{8x\sin^2(2x)+8x^2\sin(4x)} \right] \\
=& \lim_{x\rightarrow 0} \left[ \frac{4x-\sin(4x)}{4x\sin^2(2x)+4x^2\sin(4x)} \right] \\
\end{aligned}
$$
Am I using the correct way? How to solve it correctly?
P. S. I tried to use calculator and it outputs one third $1/3$.
|
Almost.
I'll edit your answer.
$\begin{aligned}
&\lim_{x\rightarrow 0} \left[ \csc^2(2x) - \frac{1}{4x^2} \right] \\
=& \lim_{x\rightarrow 0} \left[ \frac{1}{\sin^2(2x)} - \frac{1}{4x^2} \right] \\
=& \lim_{x\rightarrow 0} \left[ \frac{1}{\sin(2x)} - \frac{1}{2x} \right]\left[ \frac{1}{\sin(2x)} + \frac{1}{2x} \right] \\
=& \lim_{x\rightarrow 0} \left[ \frac{1}{2x-(2x)^3/6+O(x^5))} - \frac{1}{2x} \right]\left[ \frac1{2x} + \frac{1}{2x} \right] \\
=& \lim_{x\rightarrow 0} \left[ \frac{2x-(2x-(2x)^3/6+O(x^5))}{2x(2x-(2x)^3/6+O(x^5)))} \right]\left[ \frac1{x}\right] \\
=& \lim_{x\rightarrow 0} \left[ \frac{(2x)/6+O(x^3))}{(1-(2x)^2/6+O(x^4)))} \right]\left[ \frac1{x}\right] \\
=& \lim_{x\rightarrow 0} \left[ \frac{1/3+O(x^2))}{1-(2x)^2/6+O(x^4))} \right] \\
=&\frac13\\
\end{aligned}
$
|
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|
Find all integer solutions of: $\;\frac{1}{m}+\frac{1}{n}-\frac{1}{mn^2}=\frac{3}{4}$ I found the following problem from the 10th Iranian Mathematical Olympiad in Crux Magazine.
Find all integer solutions of $$\frac{1}{m}+\frac{1}{n}-\frac{1}{mn^2}=\frac{3}{4}$$
Initially it looked like a typical quadratic problem, however I hit a dead end each time I solve it.
My methodology is as follows,
$$\frac{n^2+mn-1}{mn^2}=\frac{3}{4}$$
$$\implies 4n^2+4mn-4 = 3mn^2$$
$$\implies (4-3m)n^2+4m \cdot n-4=0$$
I used the quadratic formula, and got,
$$n = \frac{-4m \pm \sqrt{(4m)^2-4\cdot(4-3m)\cdot(-4)}}{2(4-3m)}$$
I do the usual algebraic manipulations and drop at,
$$n = \frac{-2m \pm 2\sqrt{m^2-3m+4}}{4-3m}$$
I am unsure how I go ahead from this.
Some help would be much appreciated.
Cheers!
|
As an alternative to Dr. Sonnhard Graubner's answer on how to find the complete set of solutions from
$m=\frac{4(n^{2}−1)}{3n^{2}−4n}$
consider the following:
We can factor that expression further to
$m=\frac{4(n+1)(n-1)}{n(3n−4)}$
That tells us a lot about $n$. First off, if $n$ is 1, then $m$ must be 0, but that is not a valid solution because the original expression involves a division by $m$. So $n$ cannot be 1. For the same reason, $n$ cannot be -1.
Since $n$ is an integer, both the numerator and denominator of that expression are integers. In order for their quotient $m$ to also be an integer, the numerator must be divisible by the denominator. Since $n\neq\pm1$, both $(n+1)$ and $(n-1)$ must be relatively prime to $n$, so neither $n$ nor any of its prime factors can divide either of these terms. Therefore, if $m$ is an integer, 4 must be divisible by $n$. The only candidates are $\pm1$, $\pm2$, and $\pm4$, and we already excluded $\pm1$.
Substituting $n=2$ gives us $m=3$, so $(m, n) = (3, 2)$ is a valid solution.
Substituting $n=4$ gives us $m=\frac{15}{8}$, which is not an integer and therefore not a valid solution. Likewise, $n=-2$ yields $m=\frac{3}{5}$ and $n=-4$ yields $m=\frac{15}{64}$.
|
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Is this proof of the irrationality of the square root of $3$ valid? Assume that $\sqrt{3}=\frac{a}{b}$ where $\frac{a}{b}$ is in its simplest possible form. So $3=\frac{a^2}{b^2}$, hence $3b^2=a^2$. If $b$ is even then $a^2$ and $a$ are also even. That means they have a common factor of $2$. But this is impossible since $\frac{a}{b}$ is in its simplest form. So if $\frac{a}{b}$ does exist, $a$ and $b$ must both be odd. If $a=2k+1$ and $b=2m+1$ then $3=\frac{4k^2+4k+1}{4m^2+4m+1}$. After a few steps we get to $4(3m^2+3m)−4(k^2+k)=2$. If $3m^2+3m=c$ and $k^2+k=d$ then $4c−4d=2$ and so $4(c−d)=2$. This has no whole number solutions and that means that $a$ and $b$ are not even and not odd and that cannot be. So $\frac{a}{b}$ does not exist.
|
It looks fine to me, except for a tiny detail: after saying that if $b$ is even, then $a$ must be even too, you should have added that the reverse is true: if $a$ is even, then $b$ must be even too.
And there's no need no introduce the numbers $c$ and $d$, though. You could just say that we can't have $4(3m^2+3m-k^2-k)=2$, since $4$ doesn't divide $2$.
|
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|
Find all rational points on $x^2 + y^2 = 17$ So yes, this was asked before but I'm stuck on a specific step.
So to solve this, I tried to find the the intersection between the line $y=m(x-1)+4$ (we have $(1, 4)$ as a rational point on the circle) and $x^2 + y^2 = 17$. So simply substitution, $x^2 + (m(x-1)+ 4)^2 = 17 \implies m^2(x-1)^2 + 8m(x-1) + x^2 - 1 = 0$.
Now I want to find the roots to get all the rational points but I'm not sure how one would get the roots of this equation. Stuck on the algebra. Would appreciate help.
|
It comes out better if you do
$$ (x,y) = (1,4) + t(p,q) $$
with integers $p,q$ and $\gcd(p,q) = 1.$
$$ x = 1 + tp, \; \; y = 4 + t q \; . $$
$$ 17 = x^2 + y^2 = 1 + 2 p t + p^2 t^2 + 16 + 8 q t + q^2 t^2 \; , $$
$$ 17 = 17 + (2p + 8 q) t + (p^2 + q^2 ) t^2 \; , $$
$$ 0 = (2p + 8 q) t + (p^2 + q^2 ) t^2 \; . $$
Next, we see that $t=0$ is a waste, and divide through by $t \neq 0$
$$ 0 = (2p + 8 q) + (p^2 + q^2 ) t \; , $$
$$ (p^2 + q^2 ) t = - (2p + 8 q) \; , $$
$$ t = - \frac{2p + 8 q}{p^2 + q^2} \; \; .$$
Then $$ x = 1 + tp, \; \; y = 4 + t q \; $$
gives
$$ x = \frac{p^2 + q^2 -2p^2 - 8 pq}{p^2 + q^2} = \frac{-p^2 - 8pq + q^2 }{p^2 + q^2} \; \; ,$$
$$ y = \frac{4p^2 + 4q^2 -2pq - 8 q^2}{p^2 + q^2} = \frac{4p^2 - 2pq -4 q^2 }{p^2 + q^2} \; \; .$$
Note that both binary quadratic forms $-p^2 - 8pq + q^2 \; , \; \; 4p^2 - 2pq -4 q^2 \; \;$ have discriminant $68 = 4 \cdot 17.$
|
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If $\sqrt{1-x^2}+\sqrt{1-y^2}=a(x-y)$, prove that $\frac{dy}{dx}=\sqrt{\frac{1-y^2}{1-x^2}}$
If $\sqrt{1-x^2}+\sqrt{1-y^2}=a(x-y)$, prove that $\frac{dy}{dx}=\sqrt{\frac{1-y^2}{1-x^2}}$
My Attempt
$$
\frac{-2x}{2\sqrt{1-x^2}}-\frac{2y}{2\sqrt{1-y^2}}.\frac{dy}{dx}=a-a\frac{dy}{dx}\\
\implies \frac{dy}{dx}\bigg[a-\frac{y}{\sqrt{1-y^2}}\bigg]=a+\frac{x}{\sqrt{1-x^2}}\\
\frac{dy}{dx}=\frac{a\sqrt{1-x^2}+x}{\sqrt{1-x^2}}.\frac{\sqrt{1-y^2}}{a\sqrt{1-y^2}+x}=\sqrt{\frac{1-y^2}{1-x^2}}.\frac{a\sqrt{1-x^2}+x}{a\sqrt{1-y^2}-y}
$$
How do I poceed further and find the derivative ?
|
Let us remark that the looked for differential equation can be written under the form
$$\frac{dx}{\sqrt{1-x^2}}=\frac{dy}{\sqrt{1-y^2}}\tag{1}$$
involving solutions of the form :
$$-\arccos(x)=-\arccos(y)+a\tag{2}$$
otherwise said with cartesian equation :
$$y=\cos(\arccos(x)-a)\tag{3}$$
where $a$ is any real.
Setting $\alpha=\arccos(x)$ in (3), we get the equivalent parametric equations :
$$\begin{cases}x&=&\cos(\alpha)\\y&=&\cos(\alpha-a)\end{cases}\tag{4}$$
in which we recognize that we are working with elliptical arcs, as shown on the graphical representation of the family of curves $C_a$ displayed below (we have to pay attention to the domains of variables $x$ and $y$ : in general we will not have the whole ellipses as solutions but arcs of them).
From (4), it is easy to establish a connection with relationship:
$$\sqrt{1-x^2}+\sqrt{1-y^2}=a(x-y)$$
(in the spirit of the solution given by @CY Aries).
Fig. 1:Curves $C_a$ for $a=-\pi$ to $a=\pi$ with step $\pi/8$ (progressively changing from blue to red). The curves are elliptical arcs with two degenerate cases (straight lines).
|
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|
Elementary geometry problem about two squares.
Consider the picture above, where $ABCD$ and $EFGC$ are squares with areas respectively $A$ and $B$. Find the shaded area.
Well, I observerd the following: If the intersection between $BG$ and $EF$ is the midpoint of $EF$, then the area will be $\dfrac{A}{2} + \dfrac{3B}{4}$, since the square $EFGC$ will be divided into a rectangle plus a triangle with half the area of the rectangle, leaving out an identical triangle. But then when I change the square $EFGC$, the intersection will change, and the expression will change. Is there a more general formula that includes this case? I can't see how that's possible.
A colleague of mine did the following:
"Area of triangle $DBC$ is $\dfrac{A}{2}$. Area of triangle $GCB$ is $\dfrac{\sqrt{A}\sqrt{B}}{2}$. If $H$ is $BG \cap EF$, triangle $BEH$ has height $\sqrt{A}-\sqrt{B}$ and is similar to $GFH$, with ratio $\dfrac{\sqrt{A}-\sqrt{B}}{\sqrt{B}} = \dfrac{\sqrt{A}}{\sqrt{B}}-1$, then base of $BEH$ is $\dfrac{A}{\sqrt{B}} - \sqrt{A}$. Then the trapezoid $EHGC$ has area $\dfrac{\Bigg(\dfrac{A}{\sqrt{B}} - \sqrt{A}+\sqrt{B}\Bigg)\sqrt{B}}{2} = \dfrac{A - \sqrt{AB} + B}{2}$, then the shaded area is equal to $A + \dfrac{B-\sqrt{AB}}{2} $."
But I don't understand this part: "then base of $BEH$ is $\dfrac{A}{\sqrt{B}} - \sqrt{A}$".
I think his formula is not correct because I tried an example with $A=2$,$B=1$ and $H$ being midpoint of $EF$, and it didn't work, but maybe I'm doing something wrong.
Can someone please explain what's happening? Thanks.
|
Let $BC=p$ and $EC=q.$ The area of square $ABCD$ is $p^2,$ which we will also call $a.$ The area of $EFGC$ is $q^2,$ which we will also call $b.$ (The letters $A,B$ are already in use, for vertices.)
The area of $\triangle BCD$ is $\frac {p^2}{2}.$ The area of $\triangle BCG$ is $\frac{ pq}{2}.$
Let $H$ be the point of intersection of $EF$ with $BG.$ By similar triangles the area of $\triangle BEH$ is $$ \text {Area}(BCG)\cdot \left(\frac {BE}{BC}\right)^2=\frac {pq}{2}\cdot \left(\frac {BE}{BC}\right)^2=\frac {pq}{2}\cdot \left(\frac {p-q}{p}\right)^2.$$
So the area of $EHCG$ is therefore Area$(BCG)$-Area$(BEH)$, which is $$\frac {pq}{2}-\frac {pq}{2}\left(\frac {p-q}{p}\right)^2$$ which we will call $X.$
Therefore the shaded area is Area$(BCD)+X=\frac {p^2}{2}+X,$ which, using the above value for $X$, simplifies to $$\frac {p^2}{2}+q^2-\frac {q^2}{2}\cdot \frac {q}{p}=\frac {a}{2}+b-\frac {b}{2}\sqrt {\frac {b}{a}}\;.$$
|
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How to factor this polynomial over $\mathbb{F}_2$ On page 587 in Dummit and Foote, the authors say the polynomial $\frac{x^{16}-x}{x(x-1)(x^2+x+1)}$ can be factored into quartics over $\mathbb{F}_2$ as $(x^4+x^3+x^2+x+1)(x^4+x^3+1)(x^4+x+1)$.
I am having trouble seeing this. When I divide the polynomial $x^{16}-x$ by $x(x-1)(x^2+x+1)$ using long division, I get $x^{12}+x^9+x^6+x^3+1$. However, I am not sure how to factor this polynomial into quartics. How do I do this?
|
The Berlekamp algorithm gives
$$
x^{16}-x=(x^4 + x^3 + x^2 + x + 1)(x^4 + x^3 + 1)(x^4 + x + 1)(x^2 + x + 1)(x + 1)x.
$$
So the statement in Dummit and Foote is correct.
|
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Find the horizontal asymptotes of $f(x) = \frac{x^3-2}{\lvert x\rvert^3+1}$ Taken from Thomas' Calculus 12e
Find the horizontal asymptote of the graph of:
$$f(x) = \frac{x^3-2}{\lvert x\rvert^3+1}$$
Solution: We calculate the limits as ${x \to \pm \infty}$
For $x\ge0$: $$\lim\limits_{x\to\infty}\frac{x^3-2}{\lvert x\rvert ^3+1} =\lim\limits_{x\to\infty}\frac{x^3-2}{x^3+1}=\lim\limits_{x\to\infty}\frac{1-(2/x^3)}{1+(1/x^3)}=1$$
I understand all except why the answer is one. I actually cannot find out how to get the answer.
|
As you find out, the limit as $x\to \infty$ is $1$ then $y=1$ is an horizontal asymptote.
For $x\to -\infty$ we have
$$\lim\limits_{x\to-\infty}\frac{x^3-2}{\lvert x\rvert ^3+1} =\lim\limits_{x\to-\infty}\frac{x^3-2}{-x^3+1}=\lim\limits_{x\to-\infty}\frac{1-(2/x^3)}{-1+(1/x^3)}=-1$$
then $y=-1$ is an horizontal asymptote for $x\to -\infty$.
|
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In any triangle, if $\frac {\cos A+2\cos C}{\cos A+2\cos B}=\frac {\sin B}{\sin C}$, then the triangle is either isosceles or right-angled In any triangle, if $\dfrac {\cos A+2\cos C}{\cos A+2\cos B}=\dfrac {\sin B}{\sin C}$, prove that the triangle is either isosceles or right angled.
My Attempt:
Given:
$$\dfrac {\cos A+2\cos C}{\cos A+2\cos B}=\dfrac {\sin B}{\sin C}$$
$$\dfrac {\dfrac {b^2+c^2-a^2}{2bc}+2\dfrac {a^2+b^2-c^2}{2ab}}{\dfrac {b^2+c^2-a^2}{2bc}+2\dfrac {a^2+c^2-b^2}{2ac}}=\dfrac {b}{c}$$
On simplification,
$$\dfrac {ab^2+ac^2-a^3+2a^2c+2b^2c-2c^3}{ab^2+ac^2-a^3+2a^2b+2bc^2-2b^3}=\dfrac {b}{c}$$
|
In your work try to factor out a term like $(b-c)$ from the expression:
$$c(ab^2+ac^2-a^3+2a^2c+2b^2c-2c^3) = b(ab^2+ac^2-a^3+2a^2b+2bc^2-2b^3)\\
ab^2c+ac^3-a^3c + 2a^2c^2 +2b^2c^2 - 2 c^4 = ab^3 + abc^2 -a^3b + 2a^2 b^2 + 2b^2 c^2 -2b^4\\
abc(b-c) + a(c^3 - b^3 ) +a^3 (b-c)+2a^2 (c^2 - b^2) +2(b^4-c^4)$$
From here either $b=c$, or
$$abc - a(c^2+cb+b^2)+a^3 - 2a^2(c+b) + 2(b^3 + b^2c+bc^2 +c^3)=0\\
- ac^2 -ab^2 +a^3 -2a^2c - 2a^2 b + 2b^3 + 2b^2c+2bc^2 + 2c^3 = 0 \\
-2c(a^2-b^2 -c^2)+a(a^2-b^2-c^2)-2b(a^2-b^2-c^2) = 0\\
(a-2b-2c)(a^2-b^2-c^2)=0$$
This is only possible if $a^2 = b^2 +c^2 $ as $a-2b-2c$ is always negative due to triangle inequality.
|
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Integrate $\int\frac{1}{x^3+1}dx$ The problem is, as stated:
$$\int\frac{1}{x^3+1}dx$$
I tried using substitution: $t^3 = x^3 + 1$ but didn't get far with that. I also tried setting: $t = x^3 + 1$, with no luck again.
I tried partial decomposition but I didn't know how to integrate $$\int\frac{1}{x^2-x+1}$$ and I kept getting that term when expanding $x^3 + 1$
Any help would be much appreciated.
|
$$\int\frac {1}{x^2-x+1} dx=\int\frac {1}{(x-\frac{1}{2})^2 +\frac{3}{4}}dx$$
And that take $x-\frac{1}{2}=t$.
Always use $ax^2+bx+c=a(x+\frac{b}{2a}) ^2 - \frac{b^2-4ac}{4a}$
|
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In any $\triangle ABC$, prove that: $\frac {\cos B-\cos C}{\cos A +1}=\frac {c-b}{a}$ In any $\triangle ABC$, prove that: $\dfrac {\cos B-\cos C}{\cos A +1}=\dfrac {c-b}{a}$
My Attempt:
$$\begin{align}
\text{R.H.S.}&=\dfrac {c-b}{a} \\[4pt]
&=\frac {a\cos B+b\cos A-a\cos C-c\cos A}{b\cos C+c\cos B} \\[4pt]
&=\dfrac {a(\cos B-\cos C)+(b-c)\cos A}{b\cos C+c\cos B}
\end{align}$$
|
We have$$a\cos B+b\cos A=c,$$$$a\cos C+c\cos A=b.$$Subtracting gives$$a(\cos B-\cos C)-(c-b)\cos A=c-b,$$which on rearranging yields the required result.
|
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|
proof verification $\frac{3+2\sqrt{6}}{1-\sqrt{6}}$ is an algebraic integer
Is $$\frac{3+2\sqrt{6}}{1-\sqrt{6}}$$
an algebraic integer?
An algebraic integer means an algebraic number in some algebraic number field $K\supset \Bbb Q$ that is the root of a monic polynomial $f\in \Bbb Z[x]$. Here I guess we are in $\Bbb Q({\sqrt{6}})=\Bbb{Q}(\alpha)/(\alpha^2-6)$, and with that:
$$\beta =\frac{3+2\sqrt{6}}{1-\sqrt{6}}\implies \beta= \frac{(3+2\sqrt{6})(1+\sqrt6)}{-5}\implies -5\beta =15+5\sqrt{6}\implies (-5\beta-15)^2=25(6)\implies 5^2\beta^2+2(15)(5)\beta+15^2-5^2(6)=0,$$
and thus $5^2x^2+6(5)^2x+5^2(3)\in\Bbb{Z}[x]$ is the desired polynomial.
Is this correct?
|
Your approach - finding the minimal polynomial and observing that it is a monic polynomial over $\Bbb Z$ - is correct, but the following is probably easier:
$$
\frac{3 + \sqrt6}{1 - \sqrt6} =
\frac{3 + \sqrt6}{1 - \sqrt6}
\frac{1 + \sqrt6}{1 + \sqrt6} =
\frac{15 + 5 \sqrt6}{-5} =
-3 + \sqrt6.
$$
Since $\sqrt 6$ is an algebraic integerer, so is $-3 + \sqrt6$ and its minimal polynomial is $(x + 3)^2 - 6 = x^2 + 6x + 3$.
|
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Determine the value of $ax^5 + by^5$ given that $a$, $b$, $x$, and $y$ satisfy the following system: Given
$\begin{cases} ax + by = 4 \\ ax^2 + by^2 = 8 \\ ax^3 + by^3 = 17 \\ ax^4 + by^4 = 42 \end{cases} $
Determine the value of $ax^5 + by^5$
Do you have any brilliant idea to solve this problem? By looking at the right-hand side, it is actually a sequence of the sum of quadratic numbers. But, of course it is so irrational to say that $ax^5 + by^5 = 42 + 36 = 78$ just by looking at that point.
|
You may exploit the idea behind the Berlekamp-Massey algorithm. By denoting as $U_n = a x^n +b y^n $ we have that $\{U_n\}_{n\geq 0}$ is a linear recurrent sequence with a quadratic characteristic polynomial and such that
$$ U_0=a+b,\quad U_1=4,\quad U_2=8,\quad U_3=17,\quad U_4=42 $$
These informations are enough to reconstruct the characteristic polynomial, hence the next term of the sequence. Indeed
$$ \begin{pmatrix}17 & 8 \\ 8 & 4 \end{pmatrix}\begin{pmatrix}u\\ v\end{pmatrix} = \begin{pmatrix}42\\ 17\end{pmatrix} $$
implies that the characteristic polynomial $x^2-ux-v$ is $x^2-8x+\frac{47}{4}$ and
$$ U_5 = 8U_4-\frac{47}{4}U_3 =\color{red}{\frac{545}{4}}. $$
In compact form:
$$ U_5 = (U_4\,U_3)\begin{pmatrix}U_3 & U_2 \\ U_2 & U_1 \end{pmatrix}^{-1}\begin{pmatrix} U_4\\ U_3\end{pmatrix}. $$
|
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Real Analysis problem concerning recursive sequence limit. Define $$a_1=1$$, and define
$$
a_{n+1} = \left\{\begin{aligned}
&a_n+1/n &&: a_n^2\leq 2\\
&a_n-1/n &&: a_n^2>2
\end{aligned}
\right.$$
I am then asked to show that $$|a_n-\sqrt{2}|<2/n$$ for all indices n in order to show convergence by comparison.
My initial thought was to use induction but my algebra during the inductive step doesn't go very far. Any thoughts would be greatly appreciated.
|
Induction works. Suppose $|a_n - \sqrt 2| < 2/n$, i.e.
\begin{equation*}
(1) \quad \sqrt 2 - 2/n < a_n < \sqrt 2 + 2/n.
\end{equation*}
There are two cases:
Case 1. If $a_n^2 \le 2$, i.e. $a_n \le \sqrt 2$, we can refine (1) to
\begin{equation*}
\sqrt 2 - 2/n < a_n \le \sqrt 2.
\end{equation*}
And we have $a_{n + 1} = a_n + 1/n$, so by adding $1/n$ to this inequality, we get
\begin{equation*}
\sqrt 2 - 1/n < a_{n + 1} \le \sqrt 2 + 1/n,
\end{equation*}
i.e. $|a_{n + 1} - \sqrt 2| < 1/n$.
Case 2. If $a_n^2 \ge 2$, i.e. $a_n \ge \sqrt 2$, we can refine (1) to
\begin{equation*}
\sqrt 2 < a_n \le \sqrt 2 + 2/n.
\end{equation*}
And we have $a_{n + 1} = a_n - 1/n$, so by subtracting $1/n$ from this inequality, we get
\begin{equation*}
\sqrt 2 - 1/n < a_n \le \sqrt 2 + 1/n,
\end{equation*}
i.e. $|a_{n + 1} - \sqrt 2| < 1/n$.
Either way, we have $|a_{n + 1} - \sqrt 2| < 1/n$. And for all $n \ge 1$, we have $1/n \le 2/(n + 1)$.
(Proof of the last inequality:
\begin{equation*}
2/(n + 1) - 1/n = (2n - n - 1)/(n(n + 1)) = (n - 1)/(n^2 + n),
\end{equation*}
which is nonnegative).
|
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|
Calculate the following limit: $\lim_{n\to \infty} a_n = \frac{1}{n^5}(1^4 +2^4+....+n^4)$ Calculate the following sequence: $\lim_{n\to \infty} a_n = \frac{1}{n^5}(1^4 +2^4+....+n^4)$.
I know how to solve this using Riemann sums of the function $x^4$.
I was wondering if there's another way using other methods. Any help would be appreciated.
|
The Binomial Theorem gives
$$
(k+1)^5-k^5=5k^4+10k^3+10k^2+5k+1\tag1
$$
Summing both sides yields
$$
\begin{align}
(n+1)^5-1^5
&=\sum_{k=1}^n\left(5k^4+10k^3+10k^2+5k+1\right)\\
&=5\sum_{k=1}^nk^4+O\!\left(n^4\right)\tag2
\end{align}
$$
since
$$
\begin{align}
\sum_{k=1}^n\left(10k^3+10k^2+5k+1\right)
&\le\sum_{k=1}^n\left(10n^3+10n^2+5n+1\right)\\
&=10n^4+10n^3+5n^2+n\\[6pt]
&=O\!\left(n^4\right)\tag3
\end{align}
$$
Dividing $(2)$ by $5n^5$ and taking the limit as $n\to\infty$ leads to
$$
\frac15=\lim_{n\to\infty}\frac1{n^5}\sum_{k=1}^nk^4\tag4
$$
|
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|
Solving differential equation using Frobenius Method I've been given the problem to solve the following differential equation
\begin{equation}
x^2y''+(2x+3x^2)y'-2y=0
\end{equation}
using Frobenius Method around the regular singular point $x=0$. From the indicialequation I get $r=-2$ or $r=1$, which differ by an integer, giving logaritmic terms in Frobenius method. However, the problem wants me to give two independent power series solutions.
|
Let $y=\sum_{n=r}^{\infty}{a_nx^n}$. Then,
$x^2y''+(2x+3x^2)y'-2y=0$ gives us
$\sum_{n=r}^{\infty}{a_nx^n(n(n-1)+2n-2}+a_nx^{n+1}(3n)=0$
$\sum_{n=r}^{\infty}a_nx^n{(n^2+n-2)}=\sum_{n=r}^{\infty}{a_nx^{n+1}(3n)}$
Looking at the $x^r$ term gives us that $r^2+r-2=0$ and $r=-2,1$ as you mentioned.
Comparing the subsequent terms, we get the recursive formula,
$a_n(n^2+n-2)=a_{n-1}3(n-1)$
$a_n=\frac{3}{n+2} a_{n-1}$
So, for $r=1$, we get $a_n=\frac{2\cdot 3^{n}}{(n+2)!}a_1$
And $y=a_1\sum_{n=1}^{\infty}{\frac{2\cdot 3^{n}}{(n+2)!}x^n}$
Do something similar for $r=-2$
|
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|
$y^2+5xy+6x^2-9x-4y=0$ where $x$ and $y$ are integers Need some help solving this for integers $x$ and $y$:
$$
y^2+5xy+6x^2-9x-4y=0
$$
I managed to make something like this:
$$
(y+3x-4)(y+2x)=x\\
(y+3x)(y+2x-3)=y
$$
Find integers for $x$ and $y$ that satisfy the equations above.
But, what do I do next, or is this a bad approach?
|
$$ (y+3x -3)(y+2x - 1)= y^2 + 5 yx + 6 x^2 -9x-4y +3 $$
$$ (y+3x -3)(y+2x - 1)= 3 $$
|
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|
Find $\frac{\tan 54-\tan 2}{\frac{\sin 2}{\cos 6}+\frac{\sin 6}{\cos 18}+\frac{\sin 18}{\cos 54}}$ Find $$S=\frac{\tan 54-\tan 2}{\frac{\sin 2}{\cos 6}+\frac{\sin 6}{\cos 18}+\frac{\sin 18}{\cos 54}}$$
All the angles are in degrees.
My Try: I tried converting everything to $\sin$ and $\cos$ we get
$$S=\frac{\sin 52}{\cos 2 \cos 54} \times \frac{\cos 6 \cos 18 \cos 54}{\sin 2 \cos 18 \cos 54+\sin 6 \cos 6 \cos 54+\sin 18 \cos 18 \cos 6}$$
any clue here?
|
Have a look at @labbhattacharjee's comment, we will use that identity in the answer.
\begin{align}
\frac{\sin x}{\cos3x} & = \frac{\sin(3x-x)}{2 \cos x \cos 3x} \\ \tag{Angle Difference}
& = \frac{\sin 3x \cos x - \cos 3x \sin x}{2\cos x \cos 3x} \\
& = \frac{\cos x \cos 3x(\tan 3x - \tan x)}{2\cos x \cos 3x} \\
&= \frac{\tan 3x - \tan x}{2}
\end{align}
We have
\begin{align}
S & = \frac{\tan 54-\tan 2}{\frac{\sin 2}{\cos 6}+\frac{\sin 6}{\cos 18}+\frac{\sin 18}{\cos 54}}\\
& = \frac{\tan 54-\tan 2}{{\frac{\tan6- \tan2}{2}}+{\frac{\tan18- \tan6}{2}}+{\frac{\tan54- \tan18}{2}}} \\
& = \frac{ \tan54 - \tan 2}{\frac{\tan 54 - \tan 2}{2}}\\
& = 2
\end{align}
|
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|
Probability two withdrawn balls are the same color Suppose we have $n$ white and $m$ black balls in a urn. First, randomly withdraw two balls, what is the probability (Call it $P_1$) that they are the same color? Now, suppose a ball is randomly withdrawn and then replaced before second one is drawn, what is the probability (Call it $P_2$) that withdrawn balls are same color? Finally prove that $P_2 > P_1$.
try
For the first situation sample space size is ${m + n \choose 2 }$. Now, in how many ways can we withdraw balls the same color? If both are white, then can do this in ${n \choose 2}$ ways and if both are black can do in ${m \choose 2}$. Thus
$$ P_1 = \frac{ {m \choose 2 } + {n \choose 2} }{ {m+n \choose 2 } } $$
Now, for second situation, two cases. If the first ball drawn is white, then the probability this happens is ${n \choose 1 } / {m+n \choose 1 } = \frac{n}{m+n} $ . For the seecond ball we want it to be white so this can be done in ${n-1 \choose 1 } / {m+n-1 \choose 1 } = \frac{n-1}{m+n-1} $ so for this case we have $\frac{n(n-1) }{(m+n)(m+n-1)}$. Similarly if the first ball drawn is black we obtain probability $ \frac{m(m-1) }{(m+n)(m+n-1)}$.Thus,
$$ P_2 = \frac{ m(m-1) + n(n-1) }{(m+n)(m+n-1) } $$
But, Im stuck in trying to prove $P_2 > P_1$. Is my approach correct?
|
When working without replacement, we can either select $2$ white balls with probability
$$\frac{n}{n+m}\cdot\frac{n-1}{n+m-1}$$
or $2$ black balls with probability
$$\frac{m}{n+m}\cdot\frac{m-1}{n+m-1}$$
giving $$P_1=\frac{n}{n+m}\cdot\frac{n-1}{n+m-1}+\frac{m}{n+m}\cdot\frac{m-1}{n+m-1}$$
which is equivalent to what you have done.
When working with replacement, the probability does not change after the first draw giving
$$P_2=\frac{n}{n+m}\cdot\frac{n}{n+m}+\frac{m}{n+m}\cdot\frac{m}{n+m}$$
It suffices to show that for two positive integers $$\frac{x}{x+y}\gt\frac{x-1}{x+y-1}$$
Can you go from here?
|
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|
How to solve this equation for c? I want to solve the equation
$$
-\frac{a}{2}\left(c+\sqrt{c^2+4}\right)=-\frac{a-1}{2}\left(c-\sqrt{c^2+4}\right)
$$
for $c$, where $a$ is just a constant.
What I get is
$$
\frac{c-\sqrt{c^2+4}}{c+\sqrt{c^2+4}}=\frac{a}{a-1}.
$$
I think there now is some "trick" to solve this for $c$.
|
Rationalise the denominator to get $$\frac{2c^2+4-2c\sqrt{c^2+4}}{2c^2+4}=\frac a{a-1}.$$ Then after simplifying, you get $$\frac 1{1-a}=\frac c{c^2+2}\sqrt{c^2+4}.$$ If you then square both sides and make the substitution $B=1/(1-a)^2$, you should get a quadratic equation in $c^2$, which ultimately leads where you're going.
Better still, use a computer algebra system.
|
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|
Using the $N - \varepsilon$ definition to find the limit of a sequence Yesterday I had a post about this, and it cleared a lot up; however, even though I feel like I understand how to go about solving problems like this, I don't seem to get the right answers. For example:
$u_n = \frac{2n+3}{2n+1}$. We know that this sequence converges to 1, but I want to prove it using the definition.
So I want to show that , $\forall \varepsilon > 0$, $\exists N$ where,$ n > N \implies u_n - 1 < \varepsilon$.
We can manipulate the expression such that $u_n = \frac{2n+3}{2n+1} =\frac{n}n \frac{2+\frac{3}n}{2+\frac{1}n} = \frac{2+\frac{3}n}{2+\frac{1}n}$
Thus, we can show that $\forall \varepsilon > 0$, $\exists N$ where,$ n > N \implies \frac{2+\frac{3}n}{2+\frac{1}n} - 1 < \varepsilon$
Now, we can say that 0 < $\frac{2+\frac{3}n}{2+\frac{1}n}$ < $\frac{2+\frac{3}n}{2} = 1 + \frac{3}{2n}$.
We can use this logic to say that
$\frac{2+\frac{3}n}{2+\frac{1}n} - 1 < 1 + \frac{3}{2n} - 1 < \varepsilon$ (since $1 + \frac{3}{2n}$ converges to 1).
So, finding an $N$ for $\frac{3}{2n} - 1$ would be the same as finding an $N$ for $u_n$, so we do:
$1 + \frac{3}{2n} - 1 < \varepsilon$ is the same as $\frac{3}{2n} < \varepsilon$.
Thus, if $ N > \frac{3}{2\varepsilon}$, then the initial implication follows. However, I'm not sure how to complete things from here; am I close, or did I fall off the rails at some point? Thanks.
|
It might be useful to rewrite the expression in question before doing "epsilontic":
$$u_n = \frac{2n+3}{2n+1} = \frac{2n+1+2}{2n+1}= 1+\frac{2}{2n+1}$$
Now,
$$u_n - 1 = \frac{2}{2n+1} \lt \frac{2}{2n}= \frac{1}{n}$$
From here. you can surely find your $N_{\varepsilon}$ easily.
|
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|
If $\frac{b+c}{2k-1}=\frac{c+a}{2k}=\frac{a+b}{2k+1}$, then $\frac{\sin A}{k+1}=\frac{\sin B}{k}=\frac{\sin C}{k-1}$
If $$\frac{b+c}{2k-1}=\frac{c+a}{2k}=\frac{a+b}{2k+1}$$
show that
$$\frac{\sin A}{k+1}=\frac{\sin B}{k}=\frac{\sin C}{k-1}$$
where $k$ is an integer such that $2 < k \neq 4$, and where $a$, $b$, $c$ are sides of $\triangle ABC$.
Actually, I don't have any idea. Please someone help. It would be great if someone who answers include how he/she found where to begin.
|
Hint:
$$\frac{b+c}{2k-1}=\frac{c+a}{2k}=\frac{a+b}{2k+1}=\frac{(b+c)+(c+a)+(a+b)}{(2k-1)+(2k)+(2k+1)}=\frac{a+b+c}{3k}$$
$$\frac{b+c}{2k-1}=\frac{a+b+c}{3k}=\frac{(a+b+c)-(b+c)}{(3k)-(2k-1)}$$
You can easily find $a:b:c$ and apply the sine formula.
|
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|
Factoring and expanding $1-x^8$ It's been a while since I've studied factoring, and I need it for a question. How did this go from $1-x^8$ to $(1-x^4)(1+x^4)$ and then to $(1-x)(1+x+...+x^7)$? I remember studying this a few years go but unfortunately I don't remember.
$$1-x^8=(1-x^4)(1+x^4)=(1-x)(1+x+\dots + x^7)$$
|
Taking your question, “How did this go from $1-x^8$ to $(1-x^4)(1+x^4)$ and then to...” literally, it involves repeated applications of the identity $1-z^2=(1-z)(1+z)$: $$\begin{align} 1+x^8 &= (1-x^4)(1+x^4) \\ &= (1-x^2)(1+x^2)(1+x^4) \\ &= (1-x)(1+x)(1+x^2)(1+x^4) \\ &= (1-x)(1+x^2+x^3+x^4+x^5+x^6+x^7). \end{align}$$
|
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|
Find $\lim_{x\rightarrow\frac{\pi}{6}}\frac{1-2\sin x}{\cos3x}$
Find $$\lim_{x\rightarrow\frac{\pi}{6}}\frac{1-2\sin x}{\cos3x}$$
without L'Hôpital's rule.
My work:
1) I know that $$\lim_{x\rightarrow0}\frac{\sin x}{x}=1$$
2) Let $x=t-\frac{\pi}{6}$. Then
$$\lim_{x\rightarrow\frac{\pi}{6}}\frac{1-2\sin x}{\cos3x}=\lim_{t\rightarrow 0}\frac{3t}{\sin3t}\cdot\frac{1-2\sin \left(t-\frac{\pi}6\right)}{3t}$$
|
Hint: Multiply numerator and denominator by $$1+2\sin(x)$$
you can write your term in the form $$\frac{\csc \left(\frac{\pi }{4}-\frac{x}{2}\right) \csc
\left(\frac{x}{2}+\frac{\pi }{4}\right)}{2 (2 \sin
(x)+1)}$$
|
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|
Kummer solution to second order ODE I need to solve a second order linear ODE with non-constant coefficients of the form
$$
\frac{d^2Z}{dt^2}+(a+be^{-ct})\frac{dZ}{dt}+dZ=0
$$
where $a,b,c,f$ are positive real constants.
Mathematica provides me with this solution
$$
i^{(a - \sqrt{a^2 - 4 d})/c}\, b^{(a - \sqrt{a^2 - 4 d})/2 c}\,
c^{-(a - \sqrt{a^2 - 4 d})/2 c} \,{e^{-c t}}^{(
a - \sqrt{a^2 - 4 d})/2 c}\, C[1]\, \text{Hypergeometric1F1}\left[\frac{a}{2c} - \frac{\sqrt{a^2 - 4 d}}{2c}, 1 - \frac{\sqrt{a^2 - 4 d}}{c}, \frac{be^{-ct}}{c}\right] +\\
i^{(a + \sqrt{a^2 - 4 d})/c}\, b^{(a + \sqrt{a^2 - 4 d})/2 c}\,
c^{-(a + \sqrt{a^2 - 4 d})/2 c} \,{e^{-c t}}^{(
a + \sqrt{a^2 - 4 d})/2 c}\, C[2]\, \text{Hypergeometric1F1}\left[\frac{a}{2c} + \frac{\sqrt{a^2 - 4 d}}{2c}, 1 + \frac{\sqrt{a^2 - 4 d}}{c}, \frac{be^{-ct}}{c}\right]
$$
but I don't quite get how to derive such an expression. Any hint?
|
Making
$$
\xi = a+b e^{-ct}\Rightarrow \dot \xi = -c(\xi-a)
$$
Now changing variables
$$
\ddot Z + (a+b e^{-ct})\dot Z + d Z \equiv c^2\ddot Z(t-a)^2+c(c-1)\dot Z (t-a)+dZ = 0
$$
This last equation has an easy solution as
$$
Z(\xi) = C_1 (a-\xi)^{-\frac{\left(-\frac{\sqrt{1-4 d}}{\sqrt{d}}-\frac{1}{\sqrt{d}}\right) \sqrt{d}}{2 c}}+C_2 (a-\xi)^{-\frac{\left(\frac{\sqrt{1-4
d}}{\sqrt{d}}-\frac{1}{\sqrt{d}}\right) \sqrt{d}}{2 c}}
$$
|
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|
attempt to solve a Bernoulli equation I tried solving the Bernoulli equation $y'-y\tan x = y^4 \cos x $ by equating the left hand side to 0 and finding the homogeneous solution, however the equation turned out to be too complex and without the ability to isolate x. Any hints on how to solve this?
|
Another trick
$$y'-y\tan x = y^4 \cos x$$
$$\cos(x)y'-y\sin x = y^4 \cos^2 x$$
$$(\cos(x)y)' = y^4 \cos^2 x$$
Substitute $z=\cos(x)y$
$$z' = \frac {z^4} {\cos^2 x}$$
It's separable
$$\frac 1{z^3} =-3\int \frac {dx} {\cos^2 x}$$
$$\frac 1{z^3} =-3\tan x +K$$
$$z^3 =\frac 1 {-3\tan x +K}$$
$$\boxed{y^3(x) =\frac 1{\cos^2(x)(K\cos(x)-3\sin x)}}$$
$$......$$
|
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|
Rational function in integral form
If $f$ is quadratic function such that $f(0)=1$ and $\int\frac{f(x)}{x^2(x+1)^3}dx $ is a rational function, find the value of $f'(0)$.
I already tried solving this question by using general quadratic equation $ax^2+bx+c$ and then using partial fraction method but it became very complicated.
|
We have that $b=f'(0)$ and $c=f(0)=1$. Moreover, by using partial fraction method we get
$$\frac{ax^2+bx+1}{x^2(x+1)^3}=\frac{A}{x}+\frac{B}{x^2}+\frac{C}{x+1}+\frac{D}{(x+1)^2}+\frac{E}{(x+1)^3}.$$
If the integral is a rational function, i.e. a ratio of polynomials, then $A=0$ and $C=0$ because the integrals of those terms yield logarithms. Hence
$$ax^2+bx+1=B(x+1)^3+Dx^2(x+1)+Ex^2,$$
that is
$$(B+D)x^3+(3B+D+E-a)x^2+(3B-b)x+ (B-1)=0.$$
Are you able to find $b$?
|
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|
Isolating $a_n$ in a recursive formula I have three equations with three sequences, $a_n, b_n, c_n$.
I tried to isolate $a_n$ with no luck.
$$a_n = 2b_{n-1}+c_{n-1}$$
$$b_n=2a_{n-1}+2b_{n-1}+c_{n-1}$$
$$c_n = 4a_{n-1}+4b_{n-1}$$
Is it even possible to get an expression based only on $a_n$ terms here?
|
Your coefficient matrix is
$$
M =
\left(
\begin{array}{ccc}
0 & 2 & 1 \\
2 & 2 & 1 \\
4 & 4 & 0 \\
\end{array}
\right)
$$
which satisfies (Cayley-Hamilton)
$$ M^3 - 2 M^2 - 12 M - 8I = 0 $$
$$ a_{n+3} = 2 a_{n+2} + 12 a_{n+1} + 8 a_n \; . $$
You also get
$$ b_{n+3} = 2 b_{n+2} + 12 b_{n+1} + 8 b_n \; , $$
$$ c_{n+3} = 2 c_{n+2} + 12 c_{n+1} + 8 c_n \; . $$
If we make the column vector
$$
x_n =
\left(
\begin{array}{c}
a_n \\
b_n \\
c_n \\
\end{array}
\right) \; ,
$$
we find $x_{n+1} = M x_n \; ,$ then $x_{n+2} = M x_{n+1} = M^2 x_n \; ,$ finally $x_{n+3}= M x_{n+2} = M^3 x_n.$ Cayley Hamilton says
$$ x_{n+3} = M^3 x_n = \left( 2M^2 + 12 M + 8 I \right)x_n = 2M^2 x_n + 12 M x_n + 8 I x_n = 2 x_{n+2} + 12 x_{n+1} + 8 x_n $$
|
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|
Map From the Cantor Set to the Unit Interval Let $f: C \rightarrow I$ map each point of the middle third Cantor set $C$, expressed as a ternary number which contains only digits $0$ and $2$, to the set of real numbers in $I=[0,1]$ expressed in base $2$ according the the rule:
$0.a_1a_2a_3...\rightarrow 0.b_1b_2b_3...$
where $b_i = \frac{a_i}{2}$.
How do I prove this map is continuous?
|
Fix a ternary expansion
$$x = \frac{a_1}{3} + \frac{a_2}{3^2} + \frac{a_3}{3^3} + \ldots,$$
where $a_n = 0, 2$ for all $n$. Further, fix $\varepsilon > 0$. Choose an $m \in \mathbb{N}$ such that $2^{-m} < \varepsilon$. Note that two binary expansions are within $2^{-m}$ if they agree up to their $m$th bit. That is, necessarily,
$$\left|\left(\frac{b_1}{2} + \ldots + \frac{b_m}{2^m} + \frac{b_{m+1}}{2^{m+1}}+\ldots\right) - \left(\frac{b_1}{2} + \ldots + \frac{b_m}{2^m} + \frac{b'_{m+1}}{2^{m+1}}+\ldots\right)\right| < 2^{-m} < \varepsilon.$$
Consider another ternary expansion
$$y = \frac{b_1}{3} + \frac{b_2}{3^2} + \frac{b_3}{3^3} + \ldots.$$
If $x \neq y$, then the ternary expansions must differ (note: this is not true in general, but true in our case since we are taking only ternary expansions with $0$ and $2$ trits). Let $k$ be the first digit that differs, that is, $b_k \neq a_k$, but $b_n = a_n$ for $n < k$. Then $|a_k - b_k| = 2$, and
\begin{align*}
|x - y| &\ge \frac{|a_k - b_k|}{3^k} - \frac{|a_2 - b_2|}{3^{k+1}} - \frac{|a_3 - b_3|}{3^{k+2}} - \ldots \\
&\ge \frac{2}{3^k} - \frac{2}{3^{k+1}} - \frac{2}{3^{k+2}} - \ldots = \frac{1}{3^k}.
\end{align*}
Therefore, if we set $\delta = 3^{-m}$, then $|y - x| < \delta$ implies that the ternary expansion agrees at least to $m$ trits. Thus, $f(y) - f(x)$ agree to $m$ bits, hence $|f(y) - f(x)| < 2^{-m} < \varepsilon$. This proves continuity.
|
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|
Prove the following determinant Prove the following:
$$\left|
\begin{matrix}
(b+c)&a&a \\
b&(c+a)&b \\
c&c&(a+b) \\
\end{matrix}\right|=4abc$$
My Attempt:
$$\left|
\begin{matrix}
(b+c)&a&a \\
b&(c+a)&b \\
c&c&(a+b) \\
\end{matrix}\right|$$
Using $R_1\to R_1+R_2+R_3$
$$\left |
\begin{matrix}
2(b+c)&2(a+c)&2(a+b) \\
b&(c+a)&b \\
c&c&(a+b) \\
\end{matrix}\right|$$
Taking common $2$ from $R_1$
$$2\left|
\begin{matrix}
(b+c)&(a+c)&(a+b) \\
b&(c+a)&b \\
c&c&(a+b) \\
\end{matrix}\right|$$
How do I proceed further?
|
You can use the rule of Sarrus in this case:
$$\left|
\begin{matrix}
(b+c)&a&a \\
b&(c+a)&b \\
c&c&(a+b) \\
\end{matrix}\right|$$
\begin{align}
&=(b+c)(c+a)(a+b) +abc +abc
- c(c+a)a - cb(b+c) -ab(a+b)\\
&=a^2 b + a^2 c + a b^2 + 2 a b c + a c^2 + b^2 c + b c^2 +2abc -c^2a-a^2c-c^2b-b^2c -a^2b-b^2a\\
&=4abc.
\end{align}
|
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|
Find $f_{10}$ of the sequence defined by $f_{n+1}=\frac{8f_n}{5}+\frac{6 \sqrt{4^n-f_n^2}}{5}$ Find $f_{10}$ of the sequence defined by $$f_{n+1}=\frac{8f_n}{5}+\frac{6 \sqrt{4^n-f_n^2}}{5}$$ given $f_0=0$
My Approach:
Letting $f_n=2^n b_n$ we get
$$b_{n+1}=\frac{4b_n}{5}+\frac{3 \sqrt{1-b_n^2}}{5}$$
Now letting $b_n=\cos(x_n)$ we get
$$\cos(x_{n+1})=\cos(x_n-\theta)$$ where $\cos(\theta)=\frac{4}{5}$
Now Since $f_0=0$ we have $b_0=0$ and $x_0=\frac{\pi}{2}$
Now we have
$$x_{n+1}=x_n-\theta$$
Putting $n=0,1,2,3 \cdots 10$ and adding all we get
$$x_{10}=\frac{\pi}{2}-10\theta$$
Hence
$$b_{10}=\cos\left(\frac{\pi}{2}-10\theta\right)=\sin(10\theta)=\sin\left(10\arcsin\left(\frac{3}{5}\right)\right)$$
How to proceed further?
|
$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
\newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace}
\newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
\newcommand{\dd}{\mathrm{d}}
\newcommand{\ds}[1]{\displaystyle{#1}}
\newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
\newcommand{\ic}{\mathrm{i}}
\newcommand{\mc}[1]{\mathcal{#1}}
\newcommand{\mrm}[1]{\mathrm{#1}}
\newcommand{\pars}[1]{\left(\,{#1}\,\right)}
\newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
\newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
\newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
\newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$
$\ds{f_{n} =
{8f_{n - 1} \over 5} + {6\root{4^{n - 1} - f_{n - 1}^{2}} \over 5}:\ {\large ?}\,,\qquad f_{0} = 0}$.
Lets $\ds{f_{n} = 2^{n}\cos\pars{x_{n}}}$:
\begin{align}
2^{n}\cos\pars{x_{n}} & =
{8 \over 5}\,2^{n - 1}\cos\pars{x_{n - 1}} +
{6\root{2^{2n - 2} - 2^{2n - 2}\cos^{2}\pars{x_{n - 1}}} \over 5}
\\[5mm] & =
{8 \over 5}\,2^{n - 1}\bracks{\cos\pars{x_{n - 1}} +
{3 \over 4}\verts{\sin\pars{x_{n - 1}}}}\quad
\pars{\begin{array}{l}
\mbox{Note that the correct term is}
\\
\ds{\color{red}{\verts{\sin\pars{x_{n - 1}}}}}\,\,\, \mbox{instead of}
\\
\ds{\sin\pars{x_{n - 1}}}\ \mbox{since}
\\
\ds{\left.\root{a^{2}}\right\vert_{\ a\ \in\ \mathbb{R}} = \verts{a}}.
\end{array}}
\\[5mm] & =
{8 \over 5}\,2^{n - 1}\bracks{\cos\pars{x_{n - 1}} +
\tan\pars{\theta}\verts{\sin\pars{x_{n - 1}}}}
\end{align}
where $\ds{\theta = \arctan\pars{3 \over 4}}$ and $\ds{x_{0} = \pi/2}$.
\begin{align}
\cos\pars{x_{n}} & =
{4 \over 5}\,\sec\pars{\theta}\cos\pars{x_{n - 1} -
\mrm{sign}\pars{\sin\pars{x_{n - 1}}}\theta}
\\[5mm] & =
{4 \over 5}\,\root{\pars{3/4}^{2} + 1}
\cos\pars{x_{n - 1} - \mrm{sign}\pars{\sin\pars{x_{n - 1}}}\theta}
\\[5mm] \implies &
\bbx{\cos\pars{x_{n}} = \cos\pars{x_{n - 1} -
\mrm{sign}\pars{\sin\pars{x_{n - 1}}}\theta}}
\end{align}
Moreover,
\begin{align}
x_{n} & = x_{n - 1} -
\mrm{sign}\pars{\sin\pars{x_{n - 1}}}\theta
\\[5mm]
\sum_{k = 1}^{n}x_{k} & = \sum_{k = 1}^{n}x_{k - 1} -
\theta\sum_{k = 1}^{n}\mrm{sign}\pars{\sin\pars{x_{k - 1}}}
\\[5mm]
-\,{\pi \over 2} + \sum_{k = 0}^{n}x_{k} & = \sum_{k = 0}^{n - 1}x_{k} -
\theta\sum_{k = 1}^{n}\mrm{sign}\pars{\sin\pars{x_{k - 1}}} =
\sum_{k = 0}^{n - 1}x_{k} -
\theta\sum_{k = 0}^{n - 1}\mrm{sign}\pars{\sin\pars{x_{k}}}
\\[5mm]
\implies &
\bbx{x_{n} =
{\pi \over 2} -
\bracks{\sum_{k = 0}^{n - 1}\mrm{sign}\pars{\sin\pars{x_{k}}}}\theta\,,
\qquad x_{0} = {\pi \over 2}\,,\quad\theta = \arctan\pars{3 \over 4}}
\end{align}
$$
\mbox{Then,}\quad x_{0} = {\pi \over 2}\,,\ x_{1} = {\pi \over 2} - \theta\quad \mbox{and}\quad
\left\{\begin{array}{rcl}
\ds{x_{2}} & \ds{=} & \ds{{\pi \over 2} - 2\theta}
\\[1mm]
\ds{x_{3}} & \ds{=} & \ds{{\pi \over 2} - 3\theta}
\\[1mm]
\ds{x_{4}} & \ds{=} & \ds{{\pi \over 2} - 2\theta}
\\[1mm]
\ds{x_{5}} & \ds{=} & \ds{{\pi \over 2} - 3\theta}
\\[1mm]
\ds{\vdots} & \ds{\vdots} & \ds{\phantom{AA}\vdots}
\\[1mm]
\ds{x_{10}} & \ds{=} & \ds{{\pi \over 2} - 2\theta}
\end{array}\right.
$$
$$
f_{10} = 2^{10}\
\underbrace{\cos\pars{{\pi \over 2} - 2\arctan\pars{3 \over 4}}}
_{\ds{=\ {24 \over 25}}} = \bbx{24576 \over 25} = 983.04
$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2753141",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 6,
"answer_id": 1
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|
Three factor pairs summing to consecutive numbers The number $144$ can be factored in three different ways such that $$12\times12=9\times16=8\times18$$ with$$12+12=24;\quad 9+16=25; \quad8+18=26$$ the sum of the factor pairs being consecutive numbers, Similarly $180$ also has three such factor pairs, $(12,15), (10,18), (9,20)$.
Is there any algorithm or way to find such numbers which have three-factor pairs? Any help is appreciated.
|
I have a partial algorithm to give you a start.
You are looking for numbers with factors $a\cdot b = k$ such that $(1-a-b)^2-4ab = a^2+b^2-2(a+ab+b)+1$ is a perfect square. Additionally, you need $a+b-1+\sqrt{a^2+b^2-2(a+ab+b)+1}$ be divisible by 2. Finally, you want $k$ to be divisible by $\dfrac{a+b-1+\sqrt{a^2+b^2-2(a+ab+b)+1}}{2}$. Then, you will have found a pair factors. Extending this to a triple could follow a similar pattern. The pair would be:
Let $c = \dfrac{2k}{a+b-1+\sqrt{a^2+b^2-2(a+ab+b)+1}}$
Then you have:
$c+\dfrac{a+b-1+\sqrt{a^2+b^2-2(a+ab+b)+1}}{2}+1 = a+b$
and
$c\cdot \dfrac{a+b-1+\sqrt{a^2+b^2-2(a+ab+b)+1}}{2}=ab=k$
|
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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|
Where have I gone wrong in evaluating $\lim_{x \to \infty}\sqrt x(\sqrt{x+c}- \sqrt x )$?
Evaluate
$$\lim_{x \to \infty}\sqrt x(\sqrt{x+c}- \sqrt x )$$
Attempt:
$$\begin{align}
\lim_{x \to \infty}\sqrt x(\sqrt{x+c}- \sqrt x ) &= \lim_{x\to \infty }(\sqrt{x^2+cx}- x) \\
&= \lim _{x\to \infty}x\left(\sqrt{\left(1+\dfrac{c}{x}\right)}-1\right)\\
&= \lim _{x \to \infty} x \times 0 \\
&= 0 \times \infty \\
&=0
\end{align}$$
But the answer given is :
$$\frac c 2$$
|
Write like this (using $a^2-b^2=(a-b)(a+b)$)
$$ \sqrt{x^2+cx} - x = \frac{ x^2+cx - x^2}{\sqrt{x^2+cx}+x} = \frac{cx}{\sqrt{x^2+cx}+x}$$
Now, factor $x$ numerator and denominator cancel and we obtain
$$ \frac{ c }{ \sqrt{ 1 + \frac{c}{x} } + 1 } $$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2755909",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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|
$\sqrt[5]{2^4\sqrt[3]{16}} = ? $ $$\sqrt[5]{2^4\sqrt[3]{16}} = ? $$
Rewriting this and we have
$$\sqrt[5\cdot 3]{2^{4\cdot 3}4}$$
$$\sqrt[15]{2^{12}2^2}$$
Finally we get
$$\sqrt[15]{2^{12}2^2} = \sqrt[15]{2^{14}} = 2^{\frac{15}{14}}$$
Am I right?
|
$$\sqrt[5]{2^4\sqrt[3]{16}} = (2^4)^{1/5}(16^{1/3})^{1/5}=2^{4/5}(2^4)^{1/15}=2^{4/5}2^{4/15}=2^{16/15}$$
|
{
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"url": "https://math.stackexchange.com/questions/2756282",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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|
Show that $\int_0^{\pi/2}\frac{\sin x\cdot \cos x}{x+1}dx=\frac12(\frac12+\frac1{\pi +2}-A)$
Show that $\int_{0}^{\pi/2}\frac{\sin x\cdot \cos x}{x+1}dx=\frac{1}{2}(\frac{1}{2}+\frac{1}{\pi +2}-A)$
where $A=\int_0^\pi\frac{\cos x}{(x+2)^2}dx$.
I tried using partial integration on the integral $\int_0^{\pi/2}\frac{\sin x\cdot \cos x}{x+1}dx$ but I am in no luck. Also tried using $\frac12A=\int_0^{\pi/2}\frac{\cos x}{(x+2)^2}dx$. But still something is going wrong . Please suggest.
The best I could think of is :
$\int {\dfrac{sin2x}{2x+2}dx}=-\dfrac{1}{2x+2}cos2x+\dfrac{1}{4}\int {\dfrac{cos2x}{(x+1)^2}dx}$
Notice carefully that $\int_0^\pi\frac{\cos x}{(x+2)^2}dx$ has the same structure as $\int {\dfrac{cos2x}{(x+1)^2}dx}$
P.S. See I purposefully avoided the long details of my efforts as that would hamper the understandibility of the problem and may mislead the answerer as well.
|
Just a couple of tiny missteps by the OP.
\begin{align*}
\mathcal{I} \equiv\int_0^{\pi/2} \frac{ \sin x \cos x}{x+1}\,\mathrm{d}x &= \int_0^{\pi/2} \frac{ \sin 2x }{2x+2}\,\mathrm{d}x \tag*{, then denote $u = 2x$} \\
&= \frac12\int_0^{\pi} \frac{ \color{blue}{\sin u} }{u+2}\,\color{blue}{\mathrm{d}u} \tag*{, then by-part ...}
\end{align*}
... let $\color{blue}{\sin u \,\mathrm{d}u}$ go together
\begin{align*}
\mathcal{I} &= \frac12 \left[ \frac{ -\cos u }{u+2}\Bigg|_0^{\pi} - \int_0^{\pi} (-\cos u) \cdot \frac{ -1 }{ (u+2)^2 }\,\mathrm{d}u\right ] \\
&= \frac12 \left[ \frac1{ \pi + 2} - \frac{-1}{2} - \int_0^{\pi} \frac{ \cos u}{ (u+2)^2 }\,\mathrm{d}u\right]
\end{align*}
which is the desired result.
|
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 3,
"answer_id": 0
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|
Function that sends $\sum_{i=0}^n x^i$ to $\sum_{i=0}^n x^{-i}$? Is there a known function $f(S,x)$ that takes
$$S=1+x+x^2+x^3+x^4+\dots + x^n$$
and $x$ as its input and returns
$$S'=1+\frac{1}{x}+\frac{1}{x^2}+\frac{1}{x^3}+\frac{1}{x^4}+\dots+\frac{1}{x^n} \ $$
for $x \in$ R, $x>0$ and $n>0$?
Edit: Simply evaluating $S(\frac{1}{x})$ is disallowed, as mentioned in the comments. I am looking for a more elegant formula, which does not require knowledge of $n$.
|
The following formula inverts the terms of the geometric series $\displaystyle{S=\sum_{i=0}^nx^i}$ independent of n:
$$S'=\frac{x}{\frac{1}{S}+x-1}.$$
As should be clear,
$$S'=\sum_{i=0}^nx^{-i}=\frac{S}{x^n}.$$
Since $S$ is a geometric series, $S=\frac{1-x^n}{1-x}$ and
$$
\begin{align}
1-x^n &= (1-x)S \\
x^n &= 1-S+xS \\
n &= \log_x(1-S+xS).
\end{align}
$$
Therefore,
$$
\begin{align}
S' &=\frac{S}{x^n} \\
&=\frac{S}{x^{\log_x(1-S+xS)}} \\
&= \frac{S}{1-S+xS} \\
&= \frac{x}{\frac{1}{S}+x-1}
\end{align}
$$
for $x \in$ R and $x>0, n\geq 0$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2756730",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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|
For what interval of $k$ does the equation have one postitive and one negative root? I was given an equation that looked like this
$(k-x)(1-x) + 4$
and was told to find the interval of k when the equation would have one positive root and one negative root.
So far I have found the solutions in terms of k
$x_1 = \frac{1}{2}(k+1-\sqrt{k^2-2k-15})$
$x_2 = \frac{1}{2}(k+1+\sqrt{k^2-2k-15})$
I am a bit stuck of where to go from here, do we have to figure it out using
$\sqrt{b^2-4ac}$?
|
I made some silly errors
in my original answer.
The corrections now seem to show that
k < -4 is the condition.
Following dxiv's advice,
the roots are
$x_1 = \frac{1}{2}(-\sqrt(k^2-2k-15)+k+1)
$
and
$x_2 = \frac{1}{2}(\sqrt(k^2-2k-15)+k+1)
$.
For there to be
two real roots,
we must have
$k^2-2k-15 > 0$
or
$k^2-2k+1 > 16$
or
$(k-1)^2 > 16
$
or
$k-1 > 4$
or
$k-1 < -4$
or
$k > 5$
or
$k < -3$.
(Error - had -5 here)
To make the roots
of different signs,
consider the cases separately.
If $k > 5$
then
$\sqrt(k^2-2k-15)+k+1 > 0$
so we want
$-\sqrt(k^2-2k-15)+k+1
\lt 0$
or
$\sqrt(k^2-2k-15)
\gt k+1
$
or
$k^2-2k-15
\gt k^2+2k+1
$
or
$0 > 4k+16$
which never happens.
(I messed up the inequality below
also)
If $k < -3$
then
$\sqrt(k^2-2k-15)+k+1 > 0$
so we want
$\sqrt(k^2-2k-15)
\gt -k-1
$
or
$k^2-2k-15
\gt k^2+2k+1
$
or
$0 > 4k+16$
or
$k <-4$,
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2761257",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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|
Find the integral $\int\frac{dx}{\sqrt{x^2-a^2}}$
Evaluate the integral of $\frac{1}{\sqrt{x^2-a^2}}$
Put $x=a\sec\theta\implies dx=a\sec\theta\tan\theta d\theta$
$$
\begin{align}
& \ \ \ \int \frac{dx}{\sqrt{x^2-a^2}} \\
&=\int \frac{a\sec\theta\tan\theta d\theta}{\sqrt{a^2\sec^2\theta-a^2}} \\
&=\int \frac{a\sec\theta\tan\theta d\theta}{a\sqrt{\tan^2\theta}}\\
&= \int\frac{a\sec\theta\tan\theta d\theta}{a\color{red}{\tan\theta}}\\
&=\int\sec\theta d\theta \\
&=\log \lvert \sec\theta+\tan\theta \rvert|+C_0 \\
& \mbox{where $C_0$ is an arbitrary constant of integration } \\
&=\log \left\lvert \frac{x}{a}+\sqrt{\frac{x^2}{a^2}-1} \right\rvert + C_0 \\
&= \log \left\lvert \frac{x+\sqrt{x^2-a^2}}{a} \right\rvert +C_0 \\
&= \log \left\lvert x+\sqrt{x^2-a^2} \right\rvert -\log|a|+C_0 \\
&= \log|x+\sqrt{x^2-a^2}|+ C,
& \mbox{ where $C = \log \lvert a \rvert + C_0$}.
\end{align}
$$
This is how it is solve in my reference. But, $\sqrt{\tan^2\theta}=|\tan\theta|$ right ?
Then, does that imply
$$
\int\frac{dx}{\sqrt{x^2-a^2}}=\int\frac{a\sec\theta\tan\theta d\theta}{a\color{red}{|\tan\theta|}}=\color{red}{\pm}\int\sec\theta d\theta
$$
Why am I getting this confusion and is the first solution complete ?
|
Suppose that $a>0$.
The work is just for the case when $x>a$. The case for $x<-a$ is different, but the finals result is the same.
Let $x=a\sec\theta$, where $\theta\in[0,\frac{\pi}{2})\cup(\frac{\pi}{2},\pi]$. This is the domain of $\textrm{arcsec}$.
For $x< -a$, $\theta\in(\frac{\pi}{2},\pi]$ and so $\tan\theta\le0$.
$$\sqrt{x^2-a^2}=\sqrt{a^2\tan^2\theta}=-a\tan\theta$$
\begin{align*}
\int\frac{dx}{\sqrt{x^2-a^2}}&=\int\frac{a\sec\theta\tan\theta}{-a\tan\theta}d \theta\\
&=-\int\sec\theta d\theta\\
&=-\ln|\sec\theta+\tan\theta|+C\\
&=\ln|\sec\theta-\tan\theta|+C\\
&=\ln\left|\frac{x}{a}-\frac{-\sqrt{x^2-a^2}}{a}\right|+C\\
&=\ln\left|x+\sqrt{x^2-a^2}\right|-\ln|a|+C
\end{align*}
There are two minus signs and they cancel each other to reach the final result.
|
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"timestamp": "2023-03-29T00:00:00",
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|
Find all integral ordered pairs $(n,k)$ such that $\left\lfloor\frac{n^2+18n+10}{2}\right\rfloor = k^2$. I had two problems that I want to solve. The first one was easy, but the second one... not so much:
First Problem:
Find all values of $n$ such that$$\frac{(n+1)(n+9)+8n+1}{2} = n^2.$$
My Attempt:
$$(n+1)(n+9) = n^2 + 10n + 9.$$ Therefore, the problem is asking me to find all values of $n$ such that $$\begin{align} \frac{n^2 + 18n + 10}{2} &= n^2 \\ \\ \Leftrightarrow n^2 + 18n + 10 &= 2n^2 \\ \\ \Leftrightarrow 18n + 10 &= n^2 \\ \\ \Leftrightarrow -n^2 + 18n + 10 &= 0.\end{align}$$ By the Fundamental Theorem of Algebra (FTOA), I know that there must exist only two distinct solutions for $n$. Then, using the quadratic formula, I obtain that
$$\begin{align} n &=\frac{-18\pm\sqrt{18^2 - (-4\cdot 10)}}{2} \\ \\ &= \frac{18}{2}\pm\frac{\sqrt{324 + 40}}{2} \\ \\ &= 9\pm\sqrt{364}.\end{align}$$ These are the only solutions, and since $364$ is not a square number, then these solutions are also irrational.
Second Problem:
Find all ordered pairs $(n,k)\in\mathbb{Z}^2$ such that $$\left\lfloor\frac{(n+1)(n+9)+8n+1}{2}\right\rfloor=k^2$$ for which $n\neq k$ and we denote by $\left\lfloor x\right\rfloor$ rounding $x$ to the lowest integer.
How must I approach this problem? Because we have the $\left\lfloor\ldots\right\rfloor$ function, I assume that in at least one case, $(n+1)(n+9)+8n+1$ is odd. Therefore, in at least one case, $n$ must be odd. By trial and error, it seems like there exist only three solutions: $(2, 5), (3, 6), (4, 7)$.
My Attempt:
Executed Trial and Error and expanded the numerator, but it did not help.
Thank you in advance.
Edit (Another Attempt):
If $n$ is even, then $\exists a \in\mathbb{Z}$, the set of all integers, such that $n = 2a$. Simplifying results in $2a^2 + 24a$ $+ \, 5 = k^2$. Since the quadratic is congruent to $1$ modulo $4$ (or $a$ is a quadratic residue modulo $4$), then it definitely can be a square.
When $k=0$ then $a=-6\pm\sqrt{536}\notin\mathbb{Z}$ so $k\neq 0$.
When $k = 1$ then $a=-6\pm\sqrt{544}\notin\mathbb{Z}$ so $k\neq 1$.
When $k=2$ then $a=-6\pm\sqrt{568}\notin\mathbb{Z}$ so $k\neq 2$.
Now I go back to trial and error....
|
A different approach to obtaining the Pell equations:
Note that $(n+1)(n+9)+8n+1=n^2+18n+10=(n+9)^2-71$ Let's write $N=n+9$. Then
$$\begin{align}
\left\lfloor(n+1)(n+9)+8n+1\over2 \right\lfloor=k^2
&\iff\left\lfloor N^2-71\over2\right\rfloor=k^2\\
&\iff k^2\le{N^2-71\over2}\lt k^2+1\\
&\iff71\le N^2-2k^2\lt73\\
&\iff 71=N^2-2k^2\quad\text{or}\quad 72=N^2-2k^2
\end{align}$$
Note that $71=11^2-2\cdot5^2$ and $72=12^2-2\cdot6^2$, so you do get solutions for both.
|
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "2",
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|
how many positive dividers that aren't multiple of 2 are there in the number 52920? i need to know how many positive dividers that aren't multiple of 2 are there in the number 52920.
How do i eliminate multiples of 2?
|
Just keep dividing by $2$ until it is no longer even:
$52920=2 \cdot 26460$
$2646- = 2 \cdot 13230$
$13230 = 2 \cdot 6615$
OK, so find out the dividers of $6615$.
Now, since it ends in $5$ it's divisible by $5$, so divide by $5$:
$6615 = 5 \cdot 1323$
Also, any number is divisible by $3$ iff the sum of its digits is divisible by $3$, so this number is divisible by $3$ since $1+3+2+3=9$:
$1323 = 3 \cdot 441$
$4+4+1=9$, so still divisible by $3$:
$441=3 \cdot 147$
$1+4+7=12$ so still divisible by $3$:
$147=3 \cdot 49$
OK, and we know $49 = 7 \cdot 7$
OK, so we know $6615=3^3 \cdot 5 \cdot 7^2$
So, any positive divisor will be of the form $3^i \cdot 5^j \cdot 7^k$ with $0 \le i \le 3$, $0\le j \le 1$, and $0 \le k \le 2$
So, final question left for you: in how many ways can I pick $i,j,k$?
|
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"timestamp": "2023-03-29T00:00:00",
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|
Find a corresponding eigenvector for each eigenvalue Let A =$\begin{pmatrix} 6 & 1 \\ 2 & 7 \end{pmatrix}$
(a) Find the eigenvalues of A.
(b) Find a corresponding eigenvector for each eigenvalue in part (a).
My attempt
a) Eigenvalues:
$$\begin{vmatrix} 6-\lambda& 1 \\ 2 & 7-\lambda \end{vmatrix}=0 \Rightarrow \lambda^2-13\lambda +40=0 \Rightarrow \lambda_1=5, \lambda_2=8.$$
b) If $\lambda=5$, then
$\begin{pmatrix} 6-5 & 1 \\ 2 & 7-5 \end{pmatrix}$ = $\begin{pmatrix} 1 & 1 \\ 2 & 2 \end{pmatrix}$ $\Rightarrow $ Assuming this as B.
Then $ B\bar { x } =\bar { 0 } $,
$\begin{pmatrix} 1 & 1 \\ 2 & 2 \end{pmatrix}$ $\begin{pmatrix}X_1\\X_2\end{pmatrix}=0$
$\begin{pmatrix} 1 & 1 & 0 \\ 2 & 2 & 0 \end{pmatrix}$ By doing row reduction $=>$ $R_2->R_2-2R_1$
$\begin{pmatrix} 1 & 1 & 0 \\ 0 & 0 & 0 \end{pmatrix}$ By doing row reduction $
$X_1+X_2=0$
$X_1=-X_2$
Let $X_2=1$, then $X_1=-1$
$\begin{pmatrix}X_1\\X_2\end{pmatrix}=$ $\begin{pmatrix}-1\\1\end{pmatrix}$
But, the answer is $\begin{pmatrix}X_1\\X_2\end{pmatrix}=$ $\begin{pmatrix}1\\-1\end{pmatrix}$..
I verified many times but I ended up getting the answer as $\begin{pmatrix}X_1\\X_2\end{pmatrix}=$ $\begin{pmatrix}-1\\1\end{pmatrix}$
Can anyone please verify which is the correct answer.
|
Both answers are correct. Eigenvectors that are multiples of each other share the same eigenvalue for a particular matrix.
$$Ae=\lambda e\\
A(ke)=k(Ae)=k(\lambda e)=\lambda (ke)$$
|
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|
Solve differential equation $(1+x^2) \frac{dy}{dx} - 2xy = x$ Solve differential equation $$(1+x^2) \frac{dy}{dx} - 2xy = x$$
I simplified it to $$\frac{1}{1+2y} dy = \frac{x}{1+x^2} dx$$
$$ \int \frac{1}{1+2y} dy =\int \frac{x}{1+x^2} dx $$
$$\frac{1}{2} \int \frac{2}{1+2y} dy = \frac{1}{2} \int \frac{2x}{1+x^2} dx$$
$$\frac{1}{2} \ln | 1 + 2y | = \frac{1}{2} \ln | 1+x^2 | + C $$
From here, I got stuck. I have to remove y from here to solve it.
The answer in the textbook gave $ y= (k(1+x^2) - 1)/{2}$ I believe $k$ is the integration constant. How do I remove the $\ln$ from both sides?
|
solving the equation $$\ln(1+2y)=ln(1+x^2)+C$$ for $y$ we get
$$y=\frac{e^c}{2}(x^2+1)-\frac{1}{2}$$ now substitute $$k=\frac{e^C}{2}$$
|
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"timestamp": "2023-03-29T00:00:00",
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|
Show that $\cos(x) \le e^{-x^2/2}$ for $0 \le x \lt \pi/2$. Show that $\cos(x) \le e^{-x^2/2}$ for $0 \le x \lt \pi/2$.
This inequality came up
in my solution to
Show that the sequence $\sum\limits_{k=1}^n\cos\left(\frac kn\right)^{2n^2/k}$ converges.
This is in the category of
"There should be
a number of ways
to prove this."
Here is one way
I came up with.
I am interested in seeing
how many others
there are.
$(\ln(\cos(x))'
=-\tan(x)
\le -x$
for $0 \le x \lt \pi/2$.
Integrating from $0$ to $x$,
since the two sides are $0$
at $x=0$,
$\ln(\cos(x))
\le -x^2/2$
so
$\cos(x)
\le e^{-x^2/2}
$.
|
The function $\tan x$ has a Maclaurin series with positive coefficients, convergent for $|x|<\pi/2$, see series . Its integral $\log(\sec x)$ also has a series with positive coefficients
$$\log(\sec x)=\frac{x^2}{2} + \frac{x^4}{12} + \frac{x^6}{45} + \frac{17 x^8}{2520} + \frac{31 x^{10}}{14175}+\mathcal{O}(x^{12})$$
We conclude
$$\cos x = \exp(-\frac{x^2}{2} - \frac{x^4}{12} - \frac{x^6}{45}-\cdots )$$
for $|x|<\pi/2$, so truncating at any point will give an inequality, for instance
$$\cos x \le \exp(-\frac{x^2}{2} - \frac{x^4}{12} - \frac{x^6}{45})$$ for $|x|\le |3\pi/2|$. The inequality holds in fact on $[-\delta, \delta]$, where $\delta$ is the smallest positive root of $\cos(x)=\exp(-\frac{x^2}{2} - \frac{x^4}{12} - \frac{x^6}{45})$, a number $>3\pi/2$ but very close to it.
|
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|
What order of operation should I use calculating $2^3 \times 4 - 6 \times 3 ÷ (4 + 3) - 16 ÷ 4$? I am struggling if I should use PEMDAS or BODMAS in this equation. What is the right method to get around these type of equations?
$$2^3 \times 4 - 6 \times 3 ÷ (4 + 3) - 16 ÷ 4 = \ ?$$
|
In this case both PEMDAS and BODMAS give the same result:
$2^3 \times 4 - 6 \times 3 ÷ (4 + 3) - 16 ÷ 4 =$
$8\times 4 - 6\times 3\div 7 - 16\div 4 = $
$32 - 18\div 7 - 16\div 4 = $(PEMDAS) or $8\times 4 - 6\times \frac 37 - 4$ (BODMAS)
$32 - \frac {18}7 - 4=$
$\frac {206}7 - 4 = \frac {178}7$.
The ambiguity lies when you have something like $16\div 4 \times 2$.
Whether that is supposed to be $(16\div 4)\times 2$ or $16\div (4\times 2)$.
I would say treat Multiplication and Division with equal weight and go from left to right and do $16\div 4 \times 2 = 4\times 2 = 8$, just as you'd do $16 - 4 + 2 = 12 +2 = 14$ and you'd NEVER do $16 - 4 + 2 = 16 - 6 = 10$. EVER!
Except.... many text do give multiplication when written like this heavier weight. So SOME text will do $16\div 4\times 2 = 16\div (4\times 2) = 16\div 8 = 2$.
But those texts are rare and it's usually easy to tell from context what to do.
|
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"timestamp": "2023-03-29T00:00:00",
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|
When does $\frac{a+b}{2}$ and $\sqrt{ab}$ have inversed tens digits and ones digits? Let $a$ and $b$ be natural numbers, and
$$A = \frac{a+b}{2}$$
$$B = \sqrt{ab}$$
It's given that $A$ and $B$ are two-digit numbers such that the tens digit of $A$ is the same as the ones digit of $B$, and the tens digit of $B$ is the same as the ones digit of $A$.
So $A = 10x + y\;\,$and$\;B = 10y + x$.
Also given is $A\ne B$.
What is $a$ and $b$?
My teacher gave us the answer without explaining it as: $a = 98$ and $b = 32$, which makes $A = 65$ and $B=56$.
My question is: How do you prove this? I know $98 = 2\cdot 7^2$ and $32 = 2^5$, but I don't understand how to find this specific answer.
|
Note that $$A+B = 11(x+y)$$
and $$A-B = 9(x-y)$$
where $$1\le y \le x \le 9$$
Considering all our options, we see $x+y=9$ and $x-y=1$ provides the given solution of A=54 and B=45.
|
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|
Prove $\sin(\alpha -\beta)+\sin(\alpha-\gamma)+\sin(\beta-\gamma)=4\cos\frac{\alpha-\beta}{2}\sin\frac{\alpha-\gamma}{2}\cos\frac{\beta-\gamma}{2}$ Here is a problem from Gelfand's Trigonometry:
Let $\alpha, \beta, \gamma$ be any angle, show that $$\sin(\alpha -\beta)+\sin(\alpha-\gamma)+\sin(\beta-\gamma)=4\cos\left(\frac{\alpha-\beta}{2}\right)\sin\left(\frac{\alpha-\gamma}{2}\right)\cos\left(\frac{\beta-\gamma}{2}\right).$$
I have tried to worked through this problem but cannot complete it. If I let $A= \alpha -\beta$, $B=\beta-\gamma$ and $C= \beta-\gamma$, and $A+B+C=\pi$ (now $A$, $B$ and $C$ are angles of a triangle), then I could prove the equality. But without this condition, I am stuck.
Could you show me how to complete this exercise?
|
Use
\begin{eqnarray*}
\sin A + \sin B = 2 \sin \left( \frac{A+B}{2} \right) \cos \left( \frac{A-B}{2} \right)
\end{eqnarray*}
to give
\begin{eqnarray*}
\sin (\alpha-\beta) + \sin (\alpha-\gamma) = 2 \sin \left( \frac{2 \alpha-\beta-\gamma}{2} \right) \cos \left( \frac{\beta-\gamma}{2} \right).
\end{eqnarray*}
Now use the double angle formula
\begin{eqnarray*}
\sin(\beta-\gamma)=2 \sin \left(\frac{\beta-\gamma}{2} \right) \cos \left(\frac{\beta-\gamma}{2}\right).
\end{eqnarray*}
Use the first formula again & the result follows.
|
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|
Exact value of a convergent series: $\sum_{n=1}^\infty \frac{1}{n^3+6n^2+8n}$ I have a series
$$\sum_{n=1}^\infty \frac{1}{n^3+6n^2+8n}.$$
I know the series converges because
$$\frac{1}{n^3+6n^2+8n}\le \frac{1}{n^3}, $$
since $p=3>1$, I know that $\sum 1/n^3$ converges. But I am not sure how to figure out what it converges to.
|
Note
$$ \frac{1}{n^3+6n^2+8n}=\frac1{n(n+2)(n+4)}=\frac14\bigg[\frac{1}{n(n+2)}-\frac{1}{(n+2)(n+4)}\bigg]. $$
Let
$$ f(x)=\sum_{n=1}^\infty\frac{1}{n(n+2)}x^n, g(x)=\sum_{n=1}^\infty\frac{1}{(n+2)(n+4)}x^{n+2}. $$
Then
\begin{eqnarray}
f(1)&=&\int_0^1\frac1{x^3}\int_0^x\frac{t^2}{1-t}dtdx\\
&=&\int_0^1\int_0^x\frac{t^4}{x^2(1-t)}dtdx\\
&=&\int_0^1\int_t^1\frac{t^4}{x^2(1-t)}dxdt\\
&=&\frac12\int_0^1\frac{t^2}{1-t}\bigg(\frac1{t^2}-1\bigg)dt\\
&=&\frac12\int_0^1(1+t)dt\\
&=&\frac{3}{4},\\
g(1)&=&\int_0^1\frac1{x^3}\int_0^x\frac{t^4}{1-t}dtdx\\
&=&\int_0^1\int_0^x\frac{t^4}{x^3(1-t)}dtdx\\
&=&\int_0^1\int_t^1\frac{t^4}{x^3(1-t)}dxdt\\
&=&\frac12\int_0^1\frac{t^4}{1-t}\bigg(\frac1{t^2}-1\bigg)dt\\
&=&\frac12\int_0^1t^2(1+t)dt\\
&=&\frac{7}{24}.
\end{eqnarray}
So
$$ \sum_{n=1}^\infty\frac1{n(n+2)(n+4)}=\frac14\bigg(\frac23-\frac{7}{24}\bigg)=\frac{11}{96}.$$
|
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|
Simplifying $\operatorname{tanh}(\operatorname{arsinh}(x))$ So I am trying to simplify $\tanh(\operatorname{arsinh}(x))$ to $\frac{x}{\sqrt{1+x^2}}$
In general,
$$\tanh(x) = \frac{e^x-e^{-x}}{e^x+e^{-x}}$$
and $$\operatorname{arsinh}(x)= \ln(x+\sqrt{x^2-1})$$
therefore
\begin{align}\tanh(\operatorname{arsinh}(x)) & =\frac{e^{\ln(x+\sqrt{x^2-1})}-e^{\ln(\frac{1}{x+\sqrt{x^2-1}})}}{e^{\ln(x+\sqrt{x^2-1})}+e^{\ln(\frac{1}{x+\sqrt{x^2-1}})}}\\
&=\frac{x+\sqrt{x^2-1}-\frac{1}{x+\sqrt{x^2-1}}}{x+\sqrt{x^2-1}+\frac{1}{x+\sqrt{x^2-1}}} \\
&=\frac{(x+\sqrt{x^2-1})^2-1}{(x+\sqrt{x^2-1})^2+1} \\
&=\frac{x^2+2x\sqrt{x^2-1}+x^2-1-1}{x^2+2x\sqrt{x^2-1}+x^2-1+1} \\
&=\frac{x^2+x\sqrt{x^2-1}-1}{x^2+x\sqrt{x^2-1}} \\
&=\frac{x^2+x\sqrt{x^2-1}}{x^2+x\sqrt{x^2-1}}-\frac{1}{x^2+x\sqrt{x^2-1}} \\
&=1-\frac{1}{x(x+\sqrt{x^2-1})} \\
&=1-\frac{x-\sqrt{x^2-1}}{x(x+\sqrt{x^2-1})(x-\sqrt{x^2-1})} \\
&=1-\frac{x-\sqrt{x^2-1}}{x(x^2-x^2+1)} \\
&=1-\frac{x-\sqrt{x^2-1}}{x} \\
&=\frac{\sqrt{x^2-1}}{x} \\
&\ne \frac{x}{\sqrt{1+x^2}} \end{align}
I'm struggling to get my answer into the required form.
|
Let $$ arsinh(x)=y$$
$$ x= sinh (y)$$
$$ \sqrt {1+x^2} = cosh (y)$$
$$tanh ^2 (y) = 1- sech ^2 (y) = 1-\frac {1}{1+x^2}=\frac {x^2}{1+x^2} $$
$$tanh (y)=\frac {x}{\sqrt {1+x^2}}$$
|
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|
Finding the residue of $\frac{1}{z^3 \sin z}$ at z = 0 Given the function $f(z) = \frac{1}{z^3 \sin{(z)}}$, what is the residue of $f(z)$ at $z = 0$?
I want to find $b_1$ from the Laurent expansion. So I did the following:
\begin{align*}
\frac{1}{z^3 \sin{(z)}}
&= ( \dots \frac{b_5}{z^5} + \frac{b_4}{z^4} + \frac{b_3}{z^3} + \frac{b_2}{z^2} + \frac{b_1}{z} + a_0 + a_1z + \dots )\\
1& = \Big ( \dots \frac{b_5}{z^5} + \frac{b_4}{z^4} + \frac{b_3}{z^3} + \frac{b_2}{z^2} + \frac{b_1}{z} + a_0 + a_1z + \dots \Big ) \cdot \Big ( z^3 \sin{(z)} \Big )\\
&= \Big ( \dots \frac{b_5}{z^2} + \frac{b_4}{z} + b_3 + b_2z + b_1z^2 + a_0z^3 + a_1z^4 + \dots \Big ) \cdot \Big ( \sin{(z)} \Big )\\
&= \Big ( \dots \frac{b_5}{z^2} + \frac{b_4}{z} + b_3 + b_2z + b_1z^2 + a_0z^3 + a_1z^4 + \dots \Big ) \cdot \Big ( z - \frac{z^3}{3!} + \frac{z^5}{5!} - \dots \Big )\\
\end{align*}
After some more thought...
Is it true to say that because f(z) has a pole of order 4 at $z=0$ that our $b_n$'s only go out to the 4th term? Meaning there are no $b_5$, $b_6$, etc like how wrote previously. That is,
$\frac{1}{z^3 \sin{(z)}}
= \Big ( \frac{b_4}{z^4} + \frac{b_3}{z^3} + \frac{b_2}{z^2} + \frac{b_1}{z} + a_0 + a_1z + \dots \Big )\\$
followed by
\begin{align*}
1
&= \Big ( \frac{b_4}{z} + b_3 + b_2z + b_1z^2 + a_0 + \dots \Big ) \cdot \Big ( z - \frac{z^3}{3!} + \frac{z^5}{5!} - \dots\Big )\\
\end{align*}
Which then when multiplying out $b_1z^2$ with each term from sin(z)'s Laurent expansion will never yield a $\frac{1}{z}$ term, concluding that the coefficient $b_1 = 0$?
|
$\displaystyle z \mapsto -z$ does not change
$\displaystyle {1 \over z^{3}\sin\left(z\right)}.\quad$ So,
$\displaystyle {1 \over z^{3}\sin\left(z\right)}
\,\,\,\stackrel{\mathrm{as}\ z\ \to\ 0}{\sim}\,\,\, {b_{4} \over z^{4}}
+ {b_{2} \over z^{2}} + {\color{red}{\large 0} \over z} + a_{0}$
|
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|
Prove that $x^{3m}+x^{3n+1}+x^{3p+2}$ is divisible by $x^2+x+1$ Prove that $x^{3m}+x^{3n+1}+x^{3p+2}$ is divisible by $x^2+x+1$ in ring $ {\displaystyle \mathbb {R}}$$[x]$ where $m, n, p \in {\displaystyle \mathbb {N}}$
I have tried to use factor theroem to write down $x^{3m}+x^{3n+1}+x^{3p+2}$ as a product of $x^2+x+1$ and something. But I cannot find the second factor.
Thanks in Advance.
|
Note:
$$\frac{x^{3m}+x^{3n+1}+x^{3p+2}}{x^2+x+1}=\frac{x^{3m}-1+x^{3n+1}-x+x^{3p+2}-x^2+(x^2+x+1)}{x^2+x+1}=\\
\frac{(x^{3m}-1)+x(x^{3n}-1)+x^2(x^{3p}-1)+(x^2+x+1)}{x^2+x+1}.$$
And:
$$x^{3k}-1 \equiv 0 \pmod{x^2+x+1}, k\in \mathbb N.$$
|
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"url": "https://math.stackexchange.com/questions/2779155",
"timestamp": "2023-03-29T00:00:00",
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|
Power series of $f(x)=\ln (x^2+4)$ I am supposed to find a power series representation of
$$f(x)=\ln\left(x^{2}+4\right).$$
Then, I am to graph it and observe what happens as $n$ increases. My attempt at a solution:
$$\ln\left(x^2+4\right) = \int \frac{1}{x^2+4}\,dx = \frac{1}{4}\int \frac{1}{\frac{x^2}{4}+1}\,dx = \frac{1}{4}\int \frac{1}{1-\left(-\frac{x^2}{4}\right)}\,dx.$$
Now that it is in the $\frac{1}{1-x}$ format, the power series representation is $\sum_{n=0}^{\infty} \int \frac{1}{4}\left(-\frac{x^2}{4}\right)^n$. The first terms are
$$\frac{1}{4} \left(x-\frac{x^5}{80}+\frac{x^7}{448}-\frac{x^9}{2304}+\cdots\right).$$
But as I graph these, they look nothing like the graph of $f(x)=\ln(x^2 +4)$. I am not sure if I turned the revised formula into a sum correctly.
|
At first: the cubic member missed in OP, and the signs of the other ones inverted. So the series is wrong at all.
Corrrect realizaion of the OP idea is:
$$\ln(4+x^2) = \dfrac14x - \dfrac1{48}x^3+ \dfrac1{320}x^5- \dfrac1{1792}x^7+\dots.$$
Then, there are known the series
$$\ln(1+t) = t-\dfrac12t^2+\dfrac13t^3+\dots,\tag1$$
which converges only if $|t|<1.$
Then
$$\ln(4+x^2) = \ln 4 + \ln\left(1+\dfrac{x^2}4\right)\approx \ln4 + \dfrac14 x^2 - \dfrac1{32}x^4+\dfrac1{192}x^6-\dots.$$
|
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"timestamp": "2023-03-29T00:00:00",
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|
Integrate $\sin^{-1}\frac{2x}{1+x^2}$
Integrate $\sin^{-1}\frac{2x}{1+x^2}$
The solution is given in my reference as: $2x\tan^{-1}x-\log(1+x^2)+C$.
But, is it a complete solution ?
My Attempt
$$
\int 2\tan^{-1}x \, dx=\int \tan^{-1}x \cdot 2\,dx=\tan^{-1}x\int2\,dx-\int\frac{1}{1+x^2}\int2\,dx\cdot dx\\
=\tan^{-1}x \cdot2x-\int\frac{2x}{1+x^2}\,dx=2x\tan^{-1}x-\log(1+x^2)+C
$$
$$
2\tan^{-1}x=\begin{cases}\sin^{-1}\frac{2x}{1+x^2}\text{ if }|x|\leq{1}\\
\pi-\sin^{-1}\frac{2x}{1+x^2}\text{ if }|x|>{1}\text{ and }x>0\\
-\pi-\sin^{-1}\frac{2x}{1+x^2}\text{ if }|x|>{1}\text{ and }x<0
\end{cases}\\
\sin^{-1}\frac{2x}{1+x^2}=\begin{cases}2\tan^{-1}x\text{ if }|x|\leq{1}\\
\pi-2\tan^{-1}x\text{ if }|x|>{1}\text{ and }x>0\\
-\pi-2\tan^{-1}x\text{ if }|x|>{1}\text{ and }x<0
\end{cases}
$$
$$
\int\sin^{-1}\frac{2x}{1+x^2}\,dx=\begin{cases}\int2\tan^{-1}x\,dx&\text{ if } |x|\leq{1}\\\int\pi\, dx-\int2\tan^{-1}x\,dx&\text{ if }|x|>{1}&\text{ and } x>0\\-\int\pi \,dx-\int2\tan^{-1}x\,dx&\text{ if }|x|>{1}&\text{ and } x<0\end{cases}=\begin{cases}\color{red}{2x\tan^{-1}x-\log(1+x^2)+C\text{ if } |x|\leq{1}}\\\pi x-2x\tan^{-1}x+\log(1+x^2)+C&\text{ if }|x|>{1}&\text{ and } x>0\\-\pi x-2x\tan^{-1}x+\log(1+x^2)+C&\text{ if }|x|>{1}&\text{ and } x<0\end{cases}$$
So don't we have two more cases for our solution rather than that is given in my reference, right ?
|
According to the standard tangent half-angle substitution
\begin{align}
& x=\tan\frac\theta 2 \\[8pt]
& 2\arctan x = \theta \\[8pt]
& \frac {2\,dx}{1+x^2} = d\theta \\[8pt]
& dx = \sec^2\frac\theta 2 \,\,\frac{d\theta} 2 \\[8pt]
& \frac{2x}{1+x^2} = \sin\theta \\[8pt]
& \frac{1-x^2}{1+x^2} = \cos\theta
\end{align}
Thus we have $\theta = \arcsin\dfrac{2x}{1+x^2}.$
So
\begin{align}
& \int \arcsin\frac{2x}{1+x^2} \, dx = \int \theta \left( \sec^2\frac\theta 2 \,\, \frac{d\theta} 2 \right) = \int\theta \, dv = \theta v - \int v\,d\theta \\[10pt]
= {} & \theta\tan\frac\theta 2 - \int \tan\frac\theta 2 \, d\theta = \theta\tan\frac\theta 2 +2\log\left|\cos\frac\theta 2\right| + C \\[10pt]
= {} & \theta \tan \frac \theta 2 + \log\left|\frac 1 2 + \frac 1 2 \cos\theta \right| + C = \theta\tan\frac \theta 2 + \log\left| \frac 1 2 + \frac{1-x^2}{2(1+x^2)} \right| + C \\[10pt]
= {} & 2x\arctan x - \log(1+x^2) + C
\end{align}
|
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Prove the polynomial $x^2+2x+1$ divides the polynomial $x^{2(2n+1)}+2x^{2n+1}+1$ for every $n\in \mathbb{N}$ I've been trying to prove that the polynomial $x^2+2x+1$ divides the polynomial $x^{2(2n+1)}+2x^{2n+1}+1$ for all $n\in \mathbb{N}$ as follows and reached a "roadblock":
I wrote $x^2+2x+1$ as $(x+1)^2$ and showed that $x=-1$ is a root of the 2nd polynomial, hence by Bezout's Little Theorem the polynomial $x+1$ divides $x^{2(2n+1)}+2x^{2n+1}+1$, but how do I continue from here?
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Take the equation
$$(x^{2n}-x^{2n-1}+x^{2n-2}-\ ...\ +x^2-x+1)(x+1) = x^{2n+1}+1$$
This holds true, since
$$(x^{2n}-x^{2n-1}+x^{2n-2}-\ ...\ +x^2-x+1)(x+1) = \\ =x^{2n+1}-x^{2n}+x^{2n-1}-\ ...\ +x^3-x^2+x + \\ +\ x^{2n}-x^{2n-1}+x^{2n-2}-\ ...\ +x^2-x+1 = \\ = x^{2n+1} + 1$$
You can see that everything in the center cancels out, only leaving the first and last terms.
Now squaring both sides of the equation we get:
$$(x^{2n}-x^{2n-1}+x^{2n-2}-\ ...\ +x^2-x+1)^2(x+1)^2 = (x^{2n+1}+1)^2$$
$$(x^{2n}-x^{2n-1}+x^{2n-2}-\ ...\ +x^2-x+1)^2(x^2+2x+1) = x^{2(2n+1)}+2x^{2n+1}+1$$
Which means that $x^2+2x+1$ divides $x^{2(2n+1)}+2x^{2n+1}+1$, and the divisor is:
$$(x^{2n}-x^{2n-1}+x^{2n-2}-\ ...\ +x^2-x+1)^2.$$
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Given that $a \gt b\ge 1$,show that $f(b)\gt f(a)$ Given that:
$$f(x)=\frac{x}{x^{2}+1}$$
If $a \gt b \ge 1$ and both are real numbers,show that $f(b)\gt f(a)$.
My attempt:
\begin{equation}\begin{aligned} f(b)-f(a)
&=\dfrac{b(a^2+1)-a(b^2+1)}{(b^2+1)(a^2+1)} \\
&=\dfrac{a^2b-ab^2+b-a}{(b^2+1)(a^2+1)} \\
&=\dfrac{ab(a-b)-(a-b)}{(b^2+1)(a^2+1)} \\
&=\dfrac{(ab-1)(a-b)}{(b^2+1)(a^2+1)}\\
\end{aligned}\end{equation}
Both $b^{2}+1$ and $a^{2}+1$ are $\gt 1$
And $(a-b)\gt 0$
As $ab\gt 1$,then $ab-1\gt 0$
Since all the factors of$f(b)-f(a)$ are positive and not zero,
$f(b)- f(a)\gt 0$
$f(b) \gt f(a)$
Is my attempt correct? If not, is there are any mistakes? If so,please check for me and show it to me. Thanks.
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Your way is completely correct.
Another way to see it is just to calculate the derivative
*
*$f'(x) = \frac{1-x^2}{(x^2+1)^2} < 0$ for $x>1 \Rightarrow f$ is strictly decreasing on $[1,\infty)$
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How to solve $2c x + x\dot x + (c^2-1)t = cd$ After computation on a certain problem, for a specific case I came across the differential equation $$2c x + x \dot x + (c^2-1) t = cd$$
with initial conditions $0 < x(0) = d $ and $1 < - \dot x(0) = c $. Does this have a closed form?
This differential equation doesn't appear to conform to solution by factors, integrating factors, Clairaut's equation, homogeneous methods, differentiating methods, Riccati's equation, or any other special form or method I know of.
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$$2c x + \frac{dx}{dt} x + (c^2-1) t =cD$$
The symbol $D$ is used instead of $d$ to avoid confusion with the differential symbol.
Let $\quad cD=a$
$$2c x + \frac{dx}{dt} x + (c^2-1) t - a=0$$
$y(t)=(c^2-1)t-a \quad;\quad \frac{dx}{dt}= \frac{dx}{dy}\frac{dy}{dt}= (c^2-1)\frac{dx}{dy}$
$$2c x + (c^2-1)x\frac{dx}{dy} + y=0$$
$x=yu(y)\quad;\quad \frac{dx}{dy}=u+y\frac{du}{dy}$
$2c yu + (c^2-1)yu(u+y\frac{du}{dy}) + y=0$
$$2c u + (c^2-1)(u^2+yu\frac{du}{dy}) + 1=0$$
$y=e^z$
$$2c u + (c^2-1)(u^2+u\frac{du}{dz}) + 1=0$$
$$\frac{du}{dz} =-\frac{1+2cu}{(c^2-1)u}-u=-\frac{1+2cu+(c^2-1)u^2}{(c^2-1)u}=-\frac{ ((c+1)u+1)((c-1)u+1) }{(c^2-1)u}$$
$$z=(c^2-1)\int \frac{u}{((c+1)u+1)((c-1)u+1)}du+c_1$$
$$z=\frac{1}{2}(c+1)\ln|(c-1)u+1|-\frac{1}{2}(c-1)\ln|(c+1)u+1|+\text{constant}$$
$$y=e^z=c_1\frac{ ((c-1)u+1)^{(c+1)/2} }{((c+1)u+1)^{(c-1)/2}}$$
$$y=c_1\frac{ ((c-1)\frac{x}{y}+1)^{(c+1)/2} }{((c+1)\frac{x}{y}+1)^{(c-1)/2}}$$
$$y^2=c_1\frac{ ((c-1)x+y)^{(c+1)/2} }{((c+1)x+y)^{(c-1)/2}}$$
$$( (c^2-1)t-a) ^2=c_1\frac{ ((c-1)x+(c^2-1)t-a))^{(c+1)/2} }{((c+1)x+(c^2-1)t-a))^{(c-1)/2}}$$
The general solution of the ODE is expressed on the form of an implicit equation (with $a=cD$):
$$( (c^2-1)t-cD) ^2=c_1\frac{ ((c-1)x+(c^2-1)t-cD))^{(c+1)/2} }{((c+1)x+(c^2-1)t-cD))^{(c-1)/2}}$$
Condition $x(0)=D\quad\to\quad (-cd) ^2=c_1\frac{ ((c-1)D-cD))^{(c+1)/2} }{((c+1)D-cD))^{(c-1)/2}}\quad$ After simplification : $c_1=c^2D$
$$( (c^2-1)t-cD) ^2=c^2D\frac{ ((c-1)x+(c^2-1)t-cD))^{(c+1)/2} }{((c+1)x+(c^2-1)t-cD))^{(c-1)/2}}$$
This is the solution of the problem expressed on the form of implicit equation.
In case of first order ODE, only one boundary condition is sufficient. It is up to you to add an extra condition $\left(\frac{dx}{dt}\right)_{t=0}=-c$. Just test if it is compatible with the solution already found. If not, the problem would have no solution according to two boundary conditions instead of only one.
In the present case, at $t=0$ we have $x=D$ and $\frac{dx}{dt}=-c$. Putting them into the ODE gives : $2cD-cD=cD$ which is correct. So, the extra condition is implicitly satisfied. No need to worry about it.
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Testing $\sum\limits_{k=1}^∞(\frac{k+1}k)^{k^2}3^{-k}$ for convergence and absolute convergence
Test $$\sum_{k=1}^{\infty}\left(\frac{k+1}{k}\right)^{k^2}3^{-k}$$ for convergence and absolute convergence.
We apply the ratio test for $\displaystyle \sum_{k=1}^{\infty}\left|\left(\frac{k+1}{k}\right)^{k^2}3^{-k}\right|$:
$$
\left|\frac{\left(\dfrac{k+2}{k+1}\right)^{(k+1)^2}3^{-(k+1)}}{\left(\dfrac{k+1}{k}\right)^{k^2}3^{-k}}\right|=\left|\frac{\left(\left(\dfrac{k+2}{k+1}\right)^{k}\right)^2\left(\dfrac{k+2}{k+1}\right)^{2k+1}3^{-k}\dfrac{1}{3}}{\left(\dfrac{k+1}{k}\right)^{k^2}3^{-k}}\right|→\frac{e^2\cdot e^2\cdot\dfrac13}{e^2}=\frac{1}{3}e^2.
$$
Since $\dfrac{1}{3}e^2 \geq 1$, the series diverges. Is this correct? I feel like I made a mistake somewhere that I cannot pin down.
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FYI, estimation of upper bound:
$\sum\limits_{k=1}^{\infty}\left(\frac{k+1}{k}\right)^{k^2}3^{-k}=\sum\limits_{k=1}^{\infty}\left(\frac{1}{k}+1\right)^{k^2}3^{-k}$
Using binominal theorem and the following unequality: $\begin{pmatrix}n \\ {k} \end{pmatrix}\lt \frac{n^k}{k!}$ we get:
$\sum\limits_{k=1}^{\infty}\frac{1}{3^k}\sum\limits_{j=0}^{k^2}\begin{pmatrix}{k^2} \\ {j} \end{pmatrix} k^{-j}=\sum\limits_{k=1}^{\infty}\frac{1}{3^k}\sum\limits_{j=0}^{k^2}\frac{k^{2j}}{j!} k^{-j}=\sum\limits_{k=1}^{\infty}\frac{1}{3^k}\sum\limits_{j=0}^{k^2}\frac{k^{j}}{j!}\lt\sum\limits_{k=1}^{\infty}\big(\frac{e}{3}\big)^k=\frac{e}{3-e}$
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Inequality in number theory Prove that:
(i)$5<5^{\frac{1}{2}}+5^{\frac{1}{3}}+5^{\frac{1}{4}}$
(ii)$8>8^{\frac{1}{2}}+8^{\frac{1}{3}}+8^{\frac{1}{4}}$
(iii)$n>n^{\frac{1}{2}}+n^{\frac{1}{3}}+n^{\frac{1}{4}}$ for all integer $n\geq9$
Can raising both sides to exponent $12$ help
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The last one is equivalent to showing
$$1 > \frac{1}{n^{\frac{1}{2}}}+\frac{1}{n^{\frac{2}{3}}}+\frac{1}{n^{\frac{3}{4}}}$$
and
$$n\geq 9 \Rightarrow \frac{1}{3}\geq \frac{1}{n^{\frac{1}{2}}}$$
$$n^2\geq 9^2 > 3^3 \Rightarrow \frac{1}{3}> \frac{1}{n^{\frac{2}{3}}}$$
$$n^3\geq 9^3 > 3^4 \Rightarrow \frac{1}{3}> \frac{1}{n^{\frac{3}{4}}}$$
and the result follows.
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Finding factors of second order complex polynomial. It concerns with finding roots of complex polynomial:
$x^2+(i-1)x+(2+i)$. One way is to find roots by factoring, as:$(x-i)(x -(1-2i))=(x-i)(x-1+2i)=(x^2 -x +ix +i +2)=x^2 +x(i-1)+(i+2)$
But, this is a guess game for me, with no formal process to get in this case.
The logical process is to take roots of the quadratic equation, but I hope the quadratic formula does not work for complex coefficients.
Update -- Based on comments by @WillJagy, the square root is $\pm1\pm3i$, hence the equation is having roots as :$\frac{-(i-1)\pm\sqrt{-6i-8}}2\implies \frac{(1-i)\pm(\pm 1\pm 3i)}2$ can have possible values: (i) for $1+3i$, get: $x_{11} = 1+i, x_{12} = -2i$
(ii) for $1-3i$, get: $x_{21} = 1-2i, x_{22} = i$
(iii) for $-1+3i$, get: $x_{11} = i, x_{21} = 1-2i$
(iv) for $-1-3i$, get: $x_{11} = -2i, x_{21} = 1+i$
So, the $4$ root pairs should be tried one by one:
(i) $1+3i =>(x-1-i)(x+2i) = x^2 +ix -x -2i +2$, mismatch
(ii) $1-3i => (x-i)(x-1+2i) = x^2 -x +ix +i +2$, matches original equation
(iii) same as for (ii),
(iv) same as for (i)
Why only two root pairs match, and the other don't?
Some answers on mse : 1
Update 2 -- To add to the answer by @Skip, where the fourth quadrant angle was transformed to the first one by taking out $\sqrt{-1} = i$ as common factor. For mixed sign discriminant, let $D = -6 +8i$, then angle cannot be taken\manipulated to be in the first quadrant; & can lie in the 2nd or 3rd quadrant.
In 2nd quad., $\sin$ of negative angle is positive, while in 3rd quad. both $\sin, \cos$ are negative.
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$$x^2+(i-1)x+2+i=0$$
So
$$x=\frac{-(i-1)\pm\sqrt{-8-6i}}{2}=\frac{-(i-1)\pm i\sqrt{8+6i}}{2}$$
$$\sqrt{8+6i}=\sqrt{10}\left(\cos\left(\frac{\tan^{-1}\left(\frac{3}{4}\right)+2\pi k}{2}\right)+i\sin\left(\frac{\tan^{-1}\left(\frac{3}{4}\right)+2\pi k}{2}\right)\right)\space\space\space\text{k=0, 1}$$
Take $k=0$ since if $k=1$, we'll just get the negative of when $k=0$ and the $\pm$ in front of the square root takes care of this for us
These are nonstandard trig arguments but
$$\cos\left(\frac{1}{2}\tan^{-1}\left(\frac{3}{4}\right)\right)=\frac{3}{\sqrt{10}}$$
$$i\sin\left(\frac{1}{2}\tan^{-1}\left(\frac{3}{4}\right)\right)=i\frac{1}{\sqrt{10}}$$
Therefore
$$\sqrt{8+6i}=3+i$$
So
$$x=\frac{1-i\pm (-1+3i)}{2}$$
$$\boxed{x=i}$$
$$\boxed{x=1-2i}$$
Quadratic formula works
I'll keep the above because that is what appeared in the original problem, but here's how to evaluate $\sqrt{-6+8i}$.
$-6+8i$ is in the second quadrant
$$\theta=\tan^{-1}\left(\frac{8}{-6}\right)$$
But this angle describes a complex number in the fourth quadrant since $tan^{-1}(x)$ has the range $-\frac{\pi}{2}< x<\frac{\pi}{2}$, so add $\pi$ to swing it to the second quadrant
$$\theta=\tan^{-1}\left(\frac{8}{-6}\right)+\pi$$
Its magnitude is still $10$, so for $k=0,\space 1$
$$\sqrt{-6+8i}=\sqrt{10}\left(\cos\left(\frac{\tan^{-1}\left(\frac{-4}{3}\right)+\pi+2\pi k}{2}\right)+i\sin\left(\frac{\tan^{-1}\left(\frac{-4}{3}\right)+\pi+2\pi k}{2}\right)\right)$$
These arguments are not standard by any means, so I admittedly used a calculator and will again for evaluating this one too, which is why there is no work shown in evaluating them. The point of this post though is that the quadratic formula works for complex numbers.
I like Siong's way for solving the square root; the results will be the same and it's nicer to work with in this case. Sometimes this way is easier in case the polynomial isn't easily factorable, it depends on the problem
Taking $k=0$
$$\cos\left(\frac{\tan^{-1}\left(\frac{-4}{3}\right)+\pi}{2}\right)=\frac{1}{\sqrt{5}}$$
$$i\sin\left(\frac{\tan^{-1}\left(\frac{-4}{3}\right)+\pi}{2}\right)=i\frac{2}{\sqrt{5}}$$
So
$$\sqrt{-8+6i}=\sqrt{2}(1+2i)$$
Taking $k=1$ will result in $\sqrt{-8+6i}=\sqrt{2}(-1-2i)$ but the plus and minus takes care of this for us
$$\pm\sqrt{2}(1+2i)=\mp\sqrt{2}(-1-2i)$$
Say if the radicand were $6-8i$, then this lies in the fourth quadrant, so we would not have added $\pi$ when finding its angle.
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Prove that $\frac{1}{\sqrt {a^2 - b^2}} = \frac{1/a}{\sqrt {1 - \frac{b^2}{a^2}}}$ Recently I have been reading physics book and saw interesting equation, like this:
$$\frac{1}{\sqrt {a^2 - b^2}} = \frac{1/a}{\sqrt {1 - \frac{b^2}{a^2}}}$$
But I still don't understand how to get the right part of the equation from left part, and I ask for the explaining of this. Thanks a lot!
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For nonnegative $u$ and $v$, you can write $\sqrt{uv}=\sqrt{u}\cdot\sqrt{v}$. You also have, for any $u$, $\sqrt{u^2}=|u|$. Use this here:
Assume $a^2\ge b^2$ and $a\ne0$, then
$$\sqrt{a^2-b^2}=\sqrt{a^2\left(1-\frac{b^2}{a^2}\right)}=\sqrt{a^2}\cdot\sqrt{1-\frac{b^2}{a^2}}=|a|\sqrt{1-\frac{b^2}{a^2}}$$
If you take the inverse of the preceding, assuming further that $a\ne b$, then
$$\dfrac{1}{\sqrt{a^2-b^2}}=\dfrac{1}{|a|\sqrt{1-\frac{b^2}{a^2}}}$$
Assume further that $a>0$, and
$$\dfrac{1}{\sqrt{a^2-b^2}}=\dfrac{1}{a\sqrt{1-\frac{b^2}{a^2}}}$$
When you take a factor $a^2$ out of a square root, never forget the absolute value unless you are absolutely certain $a\ge0$.
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Help finding the limit of the following function as x tends to 0 $$
\lim_{x\to0}\left(\sqrt{\frac{1}{x} + \sqrt{\frac{1}{x} + \sqrt{\frac{1}{x}}}} - \sqrt{\frac{1}{x} - \sqrt{\frac{1}{x} + \sqrt{\frac{1}{x}}}}\,\right)
$$
From Demidovich 5000 problems in mathematical analysis
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$\lim \limits_{x\rightarrow0} \sqrt{\frac{1}{x} + \sqrt{\frac{1}{x} + \sqrt{\frac{1}{x}}}} - \sqrt{\frac{1}{x} - \sqrt{\frac{1}{x} + \sqrt{\frac{1}{x}}}}=?$
Let $u=\sqrt{\frac{1}{x} + \sqrt{\frac{1}{x}}}$ then the function is
$\sqrt{\frac{1}{x}+ u} -\sqrt{\frac{1}{x}-u}=\frac{1}{\sqrt{x}}(1+ux)^{\frac{1}{2}}-\frac{1}{\sqrt{x}}(1-ux)^{\frac{1}{2}}$
Using the generalized Binominal theorem we get:
$\frac{1}{\sqrt{x}}\sum \limits_{k=0}^\infty\dbinom{\frac{1}{2}}{k}(xu)^k-\frac{1}{\sqrt{x}}\sum \limits_{k=0}^\infty\dbinom{\frac{1}{2}}{k}(-1)^k(xu)^k$
Dependig on that $k$ odd or even the expression becomes for zero or twice one.
When $k=2i+1$ (k, i are integers) we get:
$\frac{2}{\sqrt{x}}\sum \limits_{i=0}^\infty\dbinom{\frac{1}{2}}{2i+1}(xu)^{2i}(xu)=2\sqrt{x}u\sum \limits_{i=0}^\infty\dbinom{\frac{1}{2}}{2i+1}(xu)^{2i}$
and substitue back the expression of $u$:
$2(1+\sqrt{x})^{\frac{1}{2}}\sum \limits_{i=0}^\infty\dbinom{\frac{1}{2}}{2i+1}(x)^i(1+\sqrt{x})^{i}=2\sum \limits_{i=0}^\infty\dbinom{\frac{1}{2}}{2i+1}(x)^i(1+\sqrt{x})^{i+\frac{1}{2}}$
Separate $i=0$ case:
$2\dbinom{\frac{1}{2}}{1}(x)^0(1+\sqrt{x})^{\frac{1}{2}}+2\sum \limits_{i=1}^\infty\dbinom{\frac{1}{2}}{2i+1}(x)^i(1+\sqrt{x})^{i+\frac{1}{2}}$
As
$\dbinom{\frac{1}{2}}{1}=\frac{\Gamma(\frac{3}{2})}{\Gamma(\frac{1}{2})}=\frac{\frac{1}{2}\Gamma(\frac{1}{2})}{\Gamma(\frac{1}{2})}=\frac{1}{2}$ the function is the following:
$(1+\sqrt{x})^{\frac{1}{2}}+2\sum \limits_{i=1}^\infty\dbinom{\frac{1}{2}}{2i+1}(x)^i(1+\sqrt{x})^{i+\frac{1}{2}}$ and take its limit:
$\lim \limits_{x\rightarrow0}(1+\sqrt{x})^{\frac{1}{2}}\rightarrow1$ and for all other $i$ the sum keeps to $0$
Finally
$\lim \limits_{x\rightarrow0} \sqrt{\frac{1}{x} + \sqrt{\frac{1}{x} + \sqrt{\frac{1}{x}}}} - \sqrt{\frac{1}{x} - \sqrt{\frac{1}{x} + \sqrt{\frac{1}{x}}}}=1$
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Cool way of finding $\cos\left(\frac{\pi}{5}\right)$ while trying to solve a problem, I stumbled upon a way of finding $\cos\left(\frac{\pi}{5}\right)$ using identities and the cubic formula. Is it possible to find other values of sine or cosine in a similar way ?
Consider
$$\cos\left(\frac{\pi}{5}\right) - \cos\left(\frac{2\pi}{5}\right).$$
Using the difference of cosines identity, we have
$$\cos\left(\frac{\pi}{5}\right) - \cos\left(\frac{2\pi}{5}\right) = -2\sin\left(\frac{3\pi}{10}\right)\sin\left(-\frac{\pi}{10}\right).$$
Now we change the RHS using the identity $\sin(x) = \cos\left(\frac{\pi}{2}-x\right)$ and the fact that $\sin(x)$ is odd.
$$-2\sin\left(\frac{3\pi}{10}\right)\sin\left(-\frac{\pi}{10}\right) = 2\cos\left(\frac{\pi}{5}\right) \cos\left(\frac{2\pi}{5}\right)$$
So,
$$\cos\left(\frac{\pi}{5}\right) - \cos\left(\frac{2\pi}{5}\right) = 2\cos\left(\frac{\pi}{5}\right) \cos\left(\frac{2\pi}{5}\right)$$
Now we make use of the identity $\cos(2x)=2\cos^2(x)-1$.
$$\cos\left(\frac{\pi}{5}\right) - 2\cos^2\left(\frac{\pi}{5}\right)+1 = 2\cos\left(\frac{\pi}{5}\right) \left(2\cos^2\left(\frac{\pi}{5}\right)-1\right)$$
Let $y=\cos\left(\frac{\pi}{5}\right)$ and we have
$$y-y^2+1=2y(2y^2-1)$$
$$4y^3+2y^2-3y-1=0$$
which has the correct solution
$$ y=\frac{\sqrt{5}+1}{4} =\cos\left(\dfrac{\pi}{5}\right) $$
One of the roots is also $\sin\left(\dfrac{\pi}{10}\right)$ which I'm guessing is because you end up with the same cubic if you apply the above to sin too.
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There is a slightly shorter way:
$$\sin\frac{\pi}{5}=\sin\frac{4\pi}{5}=2\sin\frac{2\pi}{5}\cos\frac{2\pi}{5}=4\sin\frac{\pi}{5}\cos\frac{\pi}{5}\cos\frac{2\pi}{5}$$
Hence
$$\cos\frac{\pi}{5}\cos\frac{2\pi}{5}=\frac14$$
You can also write this
$$\cos\frac{\pi}{5}\cos\frac{3\pi}{5}=-\frac14$$
Then, using the formula $2\cos a\cos b=\cos(a+b)+\cos(a-b)$, you have:
$$\frac12=2\cos\frac{\pi}{5}\cos\frac{2\pi}{5}=\cos\frac{\pi}{5}+\cos\frac{3\pi}{5}$$
Hence $\cos\frac{\pi}{5}$ and $\cos\frac{3\pi}{5}$ are the roots of $t^2-\frac12t-\frac14$. The rest is easy, and the positive root is $\cos\frac{\pi}5$.
To answer your question: in general, it's not possible to find values of $\cos\frac{\pi}{n}$ only by radicals, even though they are algebraic numbers. Even in the case you obtain an irreducible cubic equation (which can be solved by radicals), it's the "trigonometric case" here, which has no expression with real radicals (and a complex cubic root needs the trigonometric functions anyway). For instance, $\cos1^\circ$ can't be computed with real radicals. But $\cos3^\circ$ can.
Even if in general it's not possible, there are a few tricks you can apply, for instance
$$\cos\frac{\pi}{12}=\cos\left(\frac{\pi}{3}-\frac{\pi}{4}\right)$$
Also, if you can compute $\cos x$ by radicals, then you can compute $\cos\dfrac{x}{2^n}$ too.
So, which angles of the form $\frac{\pi}{n}$ leads to trigonometric functions computable by radicals? The answer is given by the Gauss-Wantzel theorem. For instance, one can compute $\cos\dfrac{\pi}{17}$. However, the computation is not obvious, see http://mathworld.wolfram.com/TrigonometryAnglesPi17.html
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}
|
Splitting up an infinite sum I am playing with the following example trying to determine if $\frac{1}{n}-\frac{1}{\sqrt{n}}$ is in $\ell^2$.
Here is what I have,
\begin{align}
\sum\limits_{n=1}^\infty \left( \frac{1}{n}-\frac{1}{\sqrt{n}}\right)^2 &= \sum\limits_{n=1}^\infty \left(\frac{\sqrt{n}-n}{n\sqrt{n}}\right)^2\\
&=\sum\limits_{n=1}^\infty\frac{n-2n\sqrt{n}+n^2}{n^3}\\
&=\sum\limits_{n=1}^\infty\frac{1}{n^2}-2\sum\limits_{n=1}^\infty\frac{1}{n\sqrt{n}}+\sum\limits_{n=1}^\infty\frac{1}{n}
\end{align}
The first summand is in $\ell^2$, the second summand is in $\ell^2$, but the third summand is not. Thus the sequence $\frac{1}{n}-\frac{1}{\sqrt{n}}$ is not in $\ell^2$. I think I may be making an error when I split the sums. Allowing such splitting would result in things like,
\begin{align}
\sum\limits_{n=1}^\infty\frac{1}{n^2} &= \sum\limits_{n=1}^\infty\frac{1-n^2}{n^2}+\sum\limits_{n=1}^\infty\frac{n^2}{n^2}
\end{align}
What am I missing here?
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You could note that$$\sum_{n=1}^\infty\frac{n-2n\sqrt n+n^2}{n^3}=\sum_{n=1}^\infty\frac{1-2\sqrt n+n}{n^2}$$and that$$\lim_{n\to\infty}\frac{\frac{1-2\sqrt n+n}{n^2}}{\frac1n}=\lim_{n\to\infty}\frac{1-2\sqrt n+n}n=1.$$Therefore, your series diveres, by the comparison test.
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{
"language": "en",
"url": "https://math.stackexchange.com/questions/2812293",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
}
|
Subsets and Splits
Fractions in Questions and Answers
The query retrieves a sample of questions and answers containing the LaTeX fraction symbol, which provides basic filtering of mathematical content but limited analytical insight.