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Let $S = \{1, 2, 3, 4, 5, 6, 7, 8, 9, 10\}$. Find the number of subsets $A$ of $S$ such that $x \in A$ and $2x \in S \implies 2x \in A$. Let $S=\{1, 2, 3, 4, 5, 6, 7, 8, 9, 10\}$. Find the number of subsets $A\subseteq S$ such that $x\in A$ and $2x\in S$ $\implies 2x\in A$.
My Attempt
I broke the problem into cases. I made pairs $(1,2),(2,4),(3,6),(4,8),(5,10)$.
If considering these pairs only there are
$$\binom{5}{1}+\binom{5}{2}+\binom{5}{3}+\binom{5}{4}+\binom{5}{5}=2^5-1=31$$
Considering solely elements of $\{6,7,8,9,10 \}$ there are similarly $31$ subsets.
Now there may be cases where both occur.
$$2^4\binom{5}{1}+2^3\binom{5}{2}+2^2\binom{5}{3}+2^1\binom{5}{4}+2^0\binom{5}{5}=211$$
Clearly answer after summing all and including empty set is $31+31+211+1=274$.
It is not likely to be correct please give a systematic proceeding of solution.
Edit: On making an important comment by @Desparado
I would change tuples to $(\pmb1,2,4,8),(\pmb2,4,8),(\pmb3,6),(\pmb4,8),(\pmb5,10),(6),(7),(8),(9),(10)$.
Boldface indicates the entry which must be accompanied by the following in that particular tuple in the subset.
Correct answer is 180 given by @lulu and @Desperado in the comments
|
A subset $A$ of $S$ satisfying your property is exactly the same as a set that satisfies the following properties :
*
*If $A$ contains 1, then $A$ contains $1$,$2$,$4$ and $8$ ;
*If $A$ contains 2, then $A$ contains $2$,$4$ and $8$ ;
*If $A$ contains 3, then $A$ contains $3$, $6$ ;
*If $A$ contains 4, then $A$ contains $4$ and $8$ ;
*If $A$ contains 5, then $A$ contains $5$ and $10$ ;
Five families of numbers appear : $(1,2,4,8)$, $(3,6)$, $(5,10)$, $(7)$ and $(9)$ (two numbers which can or cannot be in $A$ without any consequence on the property).
$A$ is therefore given by exactly 5 independant informations, which are the four smallest numbers that $A$ contains in these families (or if it contains none of the numbers of a family).
We have $5$ choices for the first family ($A$ contains no one / $A$ contains only $8$ / $A$ contains $4$ and $8$ / $A$ contains $2,4,8$ / $A$ contains $1,2,4,8$) ; $3$ choices for the 2nd family ; $3$ choices for the 3rd family and 2 choices for each of the last two families ($A$ contains 7 or $A$ does not contain 7). Which gives a total of
$$5 \times 3 \times 3 \times 2 \times 2 = 180 \text{ possibilities}.$$
|
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"url": "https://math.stackexchange.com/questions/4584150",
"timestamp": "2023-03-29T00:00:00",
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|
Reduction of the Degree of a Curve by a Substitution Let $y^2=P_{2n}(x)$ be an (hyper)elliptic curve, where $P_{2n}$ is a polynomial of degree $2n.$
It is said that the substitution $$x=x_1^{-1}+\alpha\qquad \text{and} \qquad y=y_1x_1^{-n}$$ reduces the degree of the curve to some $y_1^2=P_{2n-1}(x_1).$ Here $\alpha$ is a zero of the polynomial $P_{2n}.$
I do not understand the rationale behind this substitution. Can somebody explain to me what is happening here? Is there a geometric meaning for this substitution?
|
TL;DR: For an even model hyperelliptic curve, the line at infinity $Z=0$ intersects the curve in two distinct points. For an odd model, the line at infinity intersects the curve in just one point, and is tangent to the curve at this point. The substitution you've given maps a Weierstrass point $(\alpha, 0)$, at which the vertical line $x=\alpha$ is tangent to the curve, to infinity, and maps the vertical line $x = \alpha$ to the line at infinity. This ensures that we obtain an odd model.
First, the complete model of a hyperelliptic curve $C$ given by $y^2 = P_{2n}(x)$ or $y^2 = P_{2n-1}(x)$ naturally lives in the weighted projective space $\mathbb{P}(1,n,1) = \mathbb{P}(1,g+1,1)$, where $g$ is the genus of $C$. (If one simply tries to take the closure in $\mathbb{P}^2$, there is a singularity at infinity; see this chapter (Waybackup) by Galbraith for more on weighted projective models.) This already explains to some extent the factor of $1/x_1^n$ in the expression for $y$, as in $\mathbb{P}(1,n,1)$ we have $[X:Y:Z] = [\lambda X : \lambda^n Y : \lambda Z]$ for all $\lambda \neq 0$.
Given a curve with affine equation $y^2 = P_{2n}(x)$, where
$$
P_{2n}(x) = a_{2n} x^{2n} + a_{2n-1} x^{2n-1} + \cdots + a_2 x^2 + a_1 x + a_0 \, ,
$$
we can compute its projective closure in $\mathbb{P}(1,n,1)$ as follows. Writing $X,Y,Z$ for the weighted projective coordinates, then $x = X/Z$ and $y = Y/Z^n$ on the affine open where $Z \neq 0$, so the affine equation above becomes
\begin{align*}
\frac{Y^2}{Z^{2n}} = \left(\frac{Y}{Z^n}\right)^2 = a_{2n} \frac{X^{2n}}{Z^{2n}} + \cdots + a_2 \frac{X^2}{Z^2} + a_1 \frac{X}{Z} + a_0 \, .
\end{align*}
Multiplying both sides by $Z^{2n}$, we find that the closure of $C$ in $\mathbb{P}(1,n,1)$ is given by
$$
Y^2 = a_{2n} X^{2n} + a_{2n-1} X^{2n-1} Z + \cdots + a_2 X^2 Z^{2n-2} + a_1 X Z^{2n-1} + a_0 Z^{2n} \, .
$$
To compute the curve's intersection with the line at infinity $Z = 0$, we set $Z = 0$, which yields $Y^2 = a_{2n} X^{2n}$. Thus $Y = \pm \sqrt{a_{2n}} X^n$, so
$$
[X:Y:Z] = [X : \pm \sqrt{a_{2n}} X^n : 0] = [1 : \pm \sqrt{a_{2n}} : 0]
$$
and we find two points at infinity.
If instead we have an odd model given by $y^2 = P_{2n-1}(x)$, then the projective closure is given by
$$
Y^2 = a_{2n-1} X^{2n-1} Z + a_{2n-2} X^{2n-2} Z^2 + \cdots + a_2 X^2 Z^{2n-2} + a_1 X Z^{2n-1} + a_0 Z^{2n} \, .
$$
Since every term on the right has a factor of $Z$, setting $Z = 0$ yields $Y^2 = 0$, so $[X:Y:Z] = [1:0:0]$ is the only point at infinity.
Suppose that $C$ is given by $y^2 = f(x)$, where $f$ has degree $2n$. Replacing $x$ by $x+\alpha$, we may assume that $f(0) = 0$, so $(0,0)$ is a Weierstrass point of the curve. Then $f$ has the form
$$
f(x) = a_{2n} x^{2n} + a_{2n-1} x^{2n-1} + \cdots + a_2 x^2 + a_1 x \, .
$$
Note the lack of constant term!
As before, we find that the closure of $C$ in $\mathbb{P}(1,n,1)$ is given by
$$
Y^2 = a_{2n} X^{2n} + \cdots + a_2 X^2 Z^{2n-2} + a_1 X Z^{2n-1} \, .
$$
We now dehomogenize with respect to $X$, letting $u = Z/X$ and $v = Y/X^n$ on the affine open where $X \neq 0$. This amounts to applying the transformation $[X:Y:Z] \mapsto [Z:Y:X]$, which interchanges the line $X = 0$ and the line at infinity $Z = 0$ as described in the first paragraph. Dividing both sides of the above equation by $X^{2n}$ yields
\begin{align*}
v^2 &= \left(\frac{Y}{X^n}\right)^2 = a_{2n} + \cdots + a_2 \frac{Z^{2n-2}}{X^{2n-2}} + a_1 \frac{Z^{2n-1}}{X^{2n-1}} = a_{2n} + \cdots + a_2 u^{2n-2} + a_1 u^{2n-1}
\end{align*}
which is an odd, model, as desired. Note that
\begin{align*}
u &= \frac{Z}{X} = \frac{1}{x}\\
v &= \frac{Y}{X^n} = \frac{Y}{Z^n} \, \frac{Z^n}{X^n} = y \frac{1}{x^n}
\end{align*}
which is the substitution you gave in the question.
Here's an illustration for the curve $y^2 = x^6 - x$, for which the corresponding odd model is $v^2 = 1 - u^5$. The point $(0,0) = [0:0:1]$ is mapped to the point at infinity $[1:0:0]$ and red vertical line $X = 0$ is mapped to $Z = 0$.
$\hspace{5cm}$
|
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"timestamp": "2023-03-29T00:00:00",
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|
For which integer values of $m \geq 0$ and $n \geq 0$ is $A_m(n) = (\frac 2 3) ^ m (n + 2) - 2$ a positive integer? For which integer values of $m \geq 0$ and $n \geq 0$ is $A_m(n) = (\frac 2 3) ^ m (n + 2) - 2$ a positive integer?
I made a table of the first few expressions for $m \in [0, 5]$
$$
\begin{align}
m && A_m(n) = (\frac 2 3) ^ m (n+2) - 2 && \text{integer iff}\\
0 && n && \text{always}\\
1 && \frac 2 3 (n - 1) && n \equiv 1 \pmod 3\\
2 && \frac 1 9 (4n - 10) && 4n \equiv 10 \equiv 1 \pmod 9\\
3 && \frac 1 {27} (8n - 38) && 8n \equiv 38 \equiv 11 \pmod {27}\\
4 && \frac 1 {81} (16n - 130) && 16n \equiv 130 \equiv 49 \pmod {81}\\
5 && \frac 1 {243} (32n - 422) && 32n \equiv 422 \equiv 179 \pmod {243}
\end{align}
$$
In general,
$$
A_m(n) = (\frac 2 3) ^ m (n + 2) - 2 = \frac 1 {3^m} (2^m n - 2(3^m - 2^m))
$$
so $A_m(n)$ is an integer when
$$
\begin{align}
2^m n &\equiv 2(3^m - 2^m) &\pmod {3^m}\\
2^{m-1} n &\equiv 3^m - 2^m &\pmod {3^m}\\
2^{m-1} n &\equiv -(2^m) &\pmod {3^m}\\
2^{m-1} n &\equiv -2(2^{m-1}) &\pmod {3^m}\\
n &\equiv -2 &\pmod {3^m}\\
n &\equiv 3^m - 2 &\pmod {3^m}
\end{align}
$$
Is this correct? I'm not $100%$ on the rules of manipulating modular equivalencies. Thanks for any help!
Background
Let $A(n) = \frac 2 3 (n - 1) = \frac 2 3 n - \frac 2 3$.
Let $A_m(n) = \underbrace{A(A(A(...A}_{m \text{ times}}(n)..)))$. For example, $A_3(n) = A(A(A(n)))$. Then,
$$
\begin{align}
A_0(n) &= n\\
A_1(n) &= \frac 2 3 n - \frac 2 3\\
A_2(n) &= \frac 2 3 (\frac 2 3 n - \frac 2 3) - \frac 2 3 = (\frac 2 3)^2 n - (\frac 2 3)^2 - \frac 2 3\\
A_3(n) &= (\frac 2 3)^3 n - (\frac 2 3)^3 - (\frac 2 3)^2 - \frac 2 3
\end{align}
$$
In general,
$$
\begin{align}
A_m(n) &= (\frac 2 3)^m n - \sum_{k=1}^m (\frac 2 3)^k\\
&= r^m n - \frac {r - r^{m+1}} {1 - r} &\text{Geometric series with } r = \frac 2 3\\
&= r^m n - r\frac {1 - r^m} {\frac 1 3}\\
&= r^m n - 3r(1 - r^m)\\
&= r^m n - 2(1 - r^m)\\
&= r^m n - 2 + 2r^m\\
&= r^m (n + 2) - 2\\
&= (\frac 2 3)^m (n + 2) - 2\\
&= \frac 1 {3^m} (2^m n - 2(3^m - 2^m))
\end{align}
$$
|
$\left(\dfrac 2 3\right) ^ m (n + 2) - 2=k\iff 2^m(n+2)=3^m(k+2)$ so if you take
$$n+2=3^m\\k+2=2^m$$ and you get infinitely many solutions..
|
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"url": "https://math.stackexchange.com/questions/4587257",
"timestamp": "2023-03-29T00:00:00",
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|
A series sum involving Catalan numbers I was trying to compute $$\sum_{k=0}^{n} \left(-\frac{1}{2}\right)^k \, \binom{2k}{k} \, \frac{k}{k+1} = \sum_{k=0}^n \left(-\frac12\right)^k k C_k$$
(where $C_k$ is the $k^{\rm th}$ Catalan number) but could not come up with a good idea.
I found out it is equal to $$\sum_{k=0}^{n} \left(-\frac{1}{2}\right)^k \, \binom{2k}{k+1}.$$ Doing some computation I see it is also equal to $$\sum_{k=0}^{n} \left(\frac{1}{2}\right)^k \, \binom{-k-1}{k}.$$ If someone can help, I would be very greatful. Thanks... My attempts might be wrong.
|
One can view the Oeis sequences to find A014138 which is a finite sum of Catalan numbers given as $$ a_{n} = \sum_{k=0}^{n} C_{k}. $$ So there is a base case of the series. Now, consider the series
$$ a_{n}(x) = \sum_{k=0}^{n} C_{k} \, x^k $$
which leads to
$$ x \, \frac{d \, a_{n}(x)}{d x} = b_{n}(x) = \sum_{k=0}^{n} k \, C_{k} \, x^k $$
which have the generating functions
$$ \sum_{n=0}^{\infty} a_{n}(x) \, t^n = \frac{1 - \sqrt{1 - 4 x t}}{2 \, x \, t \, (1-t)} $$
and
$$ \sum_{n=0}^{\infty} x \, \frac{d \, a_{n}(x)}{dx} \, t^n = \frac{1 - 2 x t - \sqrt{1 - 4 x t}}{2 \, x^2 \, t \, (1-t) \, \sqrt{1 - 4 x t}}, $$
respectively. Setting $x = - 1/2$ gives the desired generating function, which has been given in another answer.
Another way to consider the series is the use the integral representation
$$ C_{n} = \frac{2^{2 n +1}}{\pi} \, \int_{-1}^{1} t^{2 n } \, \sqrt{1- t^2} \, dt $$
which leads to
\begin{align}
b_{n}\left(- \frac{1}{2}\right) &= \frac{2}{\pi} \, \int_{-1}^{1} \sqrt{1 - t^2} \, \left( \sum_{k=0}^{n} k \, (- 2 t^2)^k \right) \, dt \\
&= - \frac{4}{\pi} \, \left( I_{1} - (n+1) \, (-2)^n \, I_{2} + n \, (-2)^{n+1} \, I_{3} \right),
\end{align}
where
\begin{align}
I_{1} &= \int_{-1}^{1} \frac{t^2 \, \sqrt{1 - t^2}}{(1 + 2 t^2)^2} \, dt = \frac{2 \sqrt{3} - 3}{12} \\
I_{2} &= \int_{-1}^{1} \frac{t^{2n +2} \, \sqrt{1 - t^2}}{(1 + 2 t^2)^2} \, dt = - \frac{\sqrt{\pi} \, \Gamma\left(n + \frac{3}{2}\right)}{6 \, (n+2)!} \, \left((3 n +2) F_{1} - n - 2 \right) \\
I_{3} &= \int_{-1}^{1} \frac{t^{2 n + 4} \, \sqrt{1 - t^2}}{(1 + 2 t^2)^2} \, dt = - \frac{\sqrt{\pi} \, \Gamma\left(n + \frac{5}{2}\right)}{6 \, (n+3)!} \, \left((3 n +5) F_{2} - n - 3 \right) \\
F_{1} &= {}_{2}F_{1}\left(1, \, n + \frac{3}{2}; \, n+3; \, -2 \right) \\
F_{2} &= {}_{2}F_{1}\left(1, \, n + \frac{5}{2}; \, n+4; \, -2 \right).
\end{align}
This leads to
\begin{align}
b_{n}\left(- \frac{1}{2}\right) &= \frac{3 - 2 \, \sqrt{3}}{3 \, \pi} + \frac{(-1)^n}{3 \cdot 2^{n+2}} \, \left(2 (n+1) \, \binom{2n +2}{n+1} + n \, \binom{2n+4}{n+2} \right) \\
& \hspace{10mm} - \frac{(-2)^{n+2}}{\pi} \, \left((n+1) \, F_{1} - 2 \, n \, F_{2} \right).
\end{align}
|
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"timestamp": "2023-03-29T00:00:00",
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|
Minimum square sum of inscripted triangle Problem: In a right triangle $ABC$, the hypotenuse is long $4$ and the angle in $B$ is $30°$. Calling $N$ the midpoint on the side $AB$ (the hypotenuse), and $M$ the middle point on the side $CB$, consider a random point on the side $AC$, lets call it $P$ and $AP = x$. Find the value of $x$ such that is minimum the sum of the squares of the sides of triangle $PNM$.
My Solution so far:
I have to minimise $NM^2+NP^2 + PM^2$ and
*
*$NM$ is fixed and equal to $1$, so I should only care about $NP$ and $MP$
*$ABC$ is a right triangle with two angles of one angle of $30°$ then the other angle will be $60°$ and therefore the side $AC$ is half the hypotenuse, therefore $AC=2$, and from this follow that $CB = 2 \sqrt{3}$
*$PM$ is the hypothenuse of the right triangle $CPM$ and therefore its equation is $PM^2 = CM^2 + PC^2 = (\sqrt{3})^2 + (2-x)^2 = x^2 -4x + 7$
But now I'm stuck since I don't know how to correctly parametrise $PN$ as a function of $AP$.
Once found a way to write $PN$ as a function of $AP$ I'm done, since I can write them inside my, equation, take the derivative and find the minimum of the function of $AP$
Final Solution:
Thanks to @mathlove, the triangle $APN$, can be solved by using the Law of Cosines, in our case used on the side $PN$ and with angle $\beta$ it results in:
\begin{equation}\begin{aligned}
PN^2 &= AP^2 + AN^2 - 2\cdot AP\cdot AN\cdot cos(\beta) = \\
&= x^2 + 2^2 - 4xcos(60) =\\
&= x^2 -2x + 4
\end{aligned}\end{equation}
Now I finally have an equation of both $PN^2$ and $PM^2$ both depending on $x$ and therefore I can find the minimum:
\begin{equation}\begin{aligned}
PN^2+PM^2 &= x^2 -2x + 4 + x^2 -4x +7 = \\
& = 2x^2 + -6x + 11
\end{aligned}\end{equation}
Which by deriving in $x$ we obtain $4x -6 = 0$ which implies $x = 3/2$
|
I like mindless approaches.
Using Analytical Geometry, you can assign $(x,y)$ coordinates of
*
*C : $~\displaystyle (0,0).$
*B : $~\displaystyle \left(2\sqrt{3},0\right).$
*A : $~\displaystyle \left(0,2\right).$
*M : $~\displaystyle \left(\sqrt{3},0\right).$
*N : $~\displaystyle \left(\sqrt{3},1\right).$
First, consider what happens if you set $P = (0,1)$.
This must be superior to setting $P = (0,r) ~: 1 < r \leq 2$ because as $r$ moves from $1$ to $2$ both line segments $\overline{MP}$ and $\overline{NP}$ are increasing.
So, you can assume that $0 \leq r \leq 1$, assign $P = (0,r)$ calculate the sum of the squares of the line segments as a function of $r$, and then minimize the sum.
$$\left[\overline{MP}\right]^2 =
\left[\overline{ ~\left(\sqrt{3},0\right) ~(0,r) ~}\right]^2 = r^2 + 3. \tag1 $$
$$\left[\overline{NP}\right]^2 =
\left[\overline{ ~\left(\sqrt{3},1\right) ~(0,r) ~}\right]^2 = \left(1 - r\right)^2 + 3. \tag2 $$
Examination of (1) and (2) above indicates that you want to minimize the sum
$$r^2 + (1 - r)^2 ~: ~0 \leq r \leq 1.$$
This simplifies to choosing $r$ to minimize $r(r-1)$.
In Calculus (AKA Real Analysis) this would be simple. In Pre-Calculus, you can employ the following trick:
Set $r = \dfrac{1}{2} + s ~: ~-\dfrac{1}{2} \leq s \leq \dfrac{1}{2}.$
Then, you are trying to minimize
$$r(r-1) = \left[ ~s + \frac{1}{2} ~\right] \times \left[ ~s - \frac{1}{2} ~\right] = s^2 - \frac{1}{4}.$$
Immediate that $r(r-1)$ is minimized by $s = 0 \implies r = \dfrac{1}{2}.$
|
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|
Proof that $\binom{n}{k}\frac{1}{n^k} \le \frac{1}{k!}$ In a Calculus textbook, I was faced with the following
Problem: If $n$ is a natural number with $n \ge 1$, proof that $\binom{n}{k}\frac{1}{n^k} \le \frac{1}{k!}$ for all $k \in \mathbb{N}$.
The book presented a different solution than the one, I came up with. So maybe mine is wrong. My question now is, if my solution is correct and - more importantly - if not, where I went wrong. Thanks in advance for any comments and advice.
Solution: I proof the proposition by induction on $k$. For $k=0$ we have
$$\binom{n}{0}\frac{1}{n^0} = 1 \le \frac{1}{0!},$$
a true statement. Now notice, that if $k \gt n$ we have $\binom{n}{k} = 0$, so the proposition is plainly true. It thus suffices to proof it for $0 \le k \le n$. In this case we have as inductive step
$$\begin{align} \binom{n}{k+1} \frac{1}{n^{k+1}} & = \binom{n}{k} \frac{1}{n^{k}} \cdot \frac{1}{n(k+1)(n-k+1)} \\ & \le \frac{1}{k!} \cdot \frac{1}{n(k+1)(n-k+1)} && \text{by inductive hypothesis} \\ & \le \frac{1}{k!}\frac{1}{k+1} = \frac{1}{(k+1)!} && \text{since $n \ge 1$ and $n-k+1 \ge 1$}\end{align}$$
as required. $\blacksquare$
As additional comment, I have used the following steps to manipulate the binomial coefficient:
$\binom{n}{k+1} = \frac{n!}{(k+1)!(n-k+1)!} = \frac{n!}{(k+1)k!(n-k+1)(n-k)!} = \frac{n!}{k!(n-k)!} \cdot \frac{1}{(k+1)(n-k+1)} = \binom{n}{k} \frac{1}{(k+1)(n-k+1)}$
|
Here is an alternative method:
We want to show $\frac{n!}{(n-k)!}\le n^k$. The LHS is the number of ordered subsets of size $k$ of a set of $n$ elements, the RHS is the number of ordered lists of size $k$ taken from a set of $n$ elements, where its ok if two elements are equal. Clearly the inequality is satisfied with this interpretation.
|
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|
To prove $1^1\cdot2^2\cdot 3^3...\cdot n^n<(\frac{2n+1}{3})^{\frac{n(n+1)}{2}} $ So we have to prove the following for $n\in N $ $$1^1\cdot 2^2\cdot 3^3...\cdot n^n<\left(\frac{2n+1}{3}\right)^{\frac{n(n+1)}{2}} $$
So I used concept of weighted means (arithmetic and geometric) used AM GM inequality.
$$AM=\frac{a_1w_1+a_2w_2+...+a_nw_n}{w_1+w_2+...+w_n}$$
$$GM=(a_1^{w_1}\cdot a_2^{w_2}\cdot...\cdot a_n^{w_n})^{\frac{1}{w_1+w_2+...+w_n}}$$
So here I let $w_1=1, w_2=2^1,w_3=3^1..$ and of course $a_1=1,a_2=2^1,a_3=3^2...$
So we get:
$$\frac{1^1+ 2^2+ 3^3...+ n^n}{\frac{n(n+1)}{2}}>(1^1\cdot 2^2\cdot 3^3...\cdot n^n)^{\frac{1}{\frac{n(n+1)}{2}}}$$
However on lhs, I cant deal with numerator, and I feel that if it can be simplified, I would get the answer. So please help or if possible suggest new method.
|
This is not a proof since working for large values of $n$
$$\prod_{k=1}^n k^k=H(n)$$ where $H(n)$ is the hyperfactorial function.
Expanding its logarithm
$$\log (H(n))=-\frac 14 n^2+\frac 1{12} \left(6 n^2+6 n+1\right)\log(n)+\log (A)+\frac{1}{720 n^2}\left(1-\frac{1}{7 n^2}+\frac{1}{14
n^4}+O\left(\frac{1}{n^6}\right) \right)$$
Doing the same for the logarithm of the rhs
$$\log\left(\frac{\text{rhs}}{\text{lhs}}\right)=\log \left(\frac{4 e}{9}\right)\,\frac{n(n+1)}4-\frac 1{12}\log(n)+\left(\frac{3}{16}-\log (A)\right)+O\left(\frac{1}{n}\right)$$
|
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"timestamp": "2023-03-29T00:00:00",
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|
Rational approximation of $\pi$ by recursion I wonder if the following result is already known and may be considered as interesting : let $\mathcal{C}$ be the real algebra of continuous functions $f:\left[0,1\right]\to\mathbf{R}$, $T:\mathcal{C}\to \mathcal{C}$ the map defined by
$$
T\left(f\right)\left(x\right)=f\left(x\right)\left(2-\left(1+x^2\right)f\left(x\right)\right)
$$
$\left(f_n\right)$ the sequence of $\mathcal{C}$ defined by
$$
f_n=
\begin{cases}
T\left(f_{n-1}\right) & \text{if } n>0\\
1/2 & \text{if } n=0
\end{cases}
$$
and $\left(I_n\right)$ the real sequence defined by $I_n=\int_0^1 f_n\left(x\right) \, dx$. One has
$$
\forall n\in\mathbf{N} \qquad 0 \leq \frac{\pi}{4} - I_n \leq 2^{-2^n}
$$
and
$$
\forall n\in\mathbf{N} \qquad I_n = \frac{1}{2} \sum_{k=0}^{2^n-1} \frac{2^k}{\left(2k+1\right)\binom{2k}{k}}
$$
|
Too long for a comment.
Using the Gaussian hypergeometric function
$$I_n = \frac{1}{2} \sum_{k=0}^{2^n-1} \frac{2^k}{\left(2k+1\right)\binom{2k}{k}}=\frac \pi 4-J_n$$
$$J_n=\frac{2^{2^n-1}}{\left(2^{n+1}+1\right) \binom{2^{n+1}}{2^n}}\,\,\, _2F_1\left(1,2^n+1;2^n+\frac{3}{2};\frac{1}{2}\right)$$
$$J_n < \frac{2^{2^n}}{\left(2^{n+1}+1\right) \binom{2^{n+1}}{2^n}}$$
$$\log \big[-\log (J_n)\big]=n \log (2)-\frac{2}{5}+\frac{3}{5
n}+O\left(\frac{1}{n^2}\right)$$
$$\log \left(-\log \left(\frac{J_{n+1}}{J_n}\right)\right) \sim n\log(2)-\frac 13$$
Edit
The first $I_n$ form the sequence
$$\left\{\frac{1}{2},\frac{2}{3},\frac{16}{21},\frac{11776}{15015},
\frac{208481288192}{265447707525},\frac{3374092439600157908008
96}{429602792188525533252675}\right\}$$
|
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|
Find a line that passes through a point A and cuts two other lines. Let $L_1$ be the straight line in $R^3$ given by (x, y, z) = (2, 2, 0) + t(3, 0, 2). A plane containing the line $L_1$ and the point A = (8, 2, 3)is given by (x,y,z) = (2,2,0) + s(6,0,3) + t(3,0,2)
The line $L_2$ is given by (x, y, z) = (5, 1, 0) + $\tau$(2, 1, 1). Determine an equation for a line that passes through the point A = (8, 2, 3) and meets both $L_1$ and $L_2$.
What I've understood is that the $L_3$ has to be on the plane since it goes thru two points on it. So far I have tried solving the problem by making $L_3 = L_2$ and $L_3 = L_1$ with gaussian elimination but it didn't lead me nowhere. At this point I am clueless.
|
$L_1: (2, 2, 0) + t (3,0,2) $
$L_2: (5,1,0) + s(2,1,1) $
$ A = (8,2,3) $
The vector from $A$ to a point on $L_1$ must be a multiple of the vector from $A$ to a point on $L_2$. Thus define the first vector as follows
$ V_1 = (2, 2, 0) + t (3, 0, 2) - (8, 2, 3) = (-6, 0, -3) + t (3, 0, 2) $
and the second vector
$ V_2 = (5,1,0) + s(2,1,1) - (8,2,3) = (-3, -1, -3) + s (2,1,1) $
If $V_1 $ is along $V_2$ then $V_1 \times V_2 = 0 $
Now,
$ V_1 \times V_2 = \begin{vmatrix} \mathbf{i} && \mathbf{j} && \mathbf{k} \\
-6 + 3 t && 0 && -3 + 2 t \\
-3 + 2 s && -1 + s && -3 + s \end{vmatrix} \\= (-(-1+s)(-3+2t) , (-3+2t)(-3+2s) - (-3+s)(-6+3t) , (-6+3t)(-1+s)) $
Simplifying, this becomes
$ V_1 \times V_2 = ( -3 + 3 s + 2 t - 2 t s , -9 +3 t + t s , 6 - 3 t - 6 s + 3 t s ) $
Now we want to find $t, s$ such that all the above components are $0$, i.e.
$ -3 + 3 s + 2 t - 2 t s = 0 $
$ -9 + 3 t + t s = 0 $
$ 6 - 3 t - 6 s + 3 t s = 0 $
Solving, gives the following values for $t$ and $s$:
$ t = 2.25 , s = 1 $
Therefore, the points of intersection of $L_3$ with $L_1$ and $L_2$ are
$ P_1 = (2, 2, 0) + 2.25 (3, 0, 2) = (8.75, 2, 4.50) $
$ P_2 = (5, 1, 0) + (2, 1, 1) = (7, 2, 1 ) $
Constructing $L_3$ from these two points, gives us
$ L3: (7, 2, 1) + r (1.75, 0, 3.50) = (7, 2, 1) + r' (1, 0, 2) $
with $r' = 1$ we get our point $A = (8,2,3) $ on $L_3$.
|
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|
Why is $2\sqrt{x+1}-2\geq\frac{\sqrt{x}}{2}$ for $x \in \mathbb{N}$? I have found following formula stands:
$2\sqrt{x+1}-2\geq\frac{\sqrt{x}}{2}$
($x \in \mathbb{N}$)
I have wondered why this is true. I have checked its rightness on the graph drawing tools, but I want to know the mathematical proof of this.
Although I can prove it for $x \in \mathbb{N}, x\geq2$, as following:
$2\sqrt{x+1} - \sqrt{x}/2 \geq 2\sqrt{x} - \sqrt{x}/2 = \frac{3}{2}\sqrt{x}\geq\frac{3}{2}\cdot \sqrt{2} \geq2$
and prove the rightness in case $x=1$ by direct calculation,
I want to know whether it is possible to prove this formula without dividing the case of $x=1$ and others.
Thank you.
|
Using the fact that $\sqrt{a}-\sqrt{b}=\dfrac{a-b}{\sqrt{a}+\sqrt{b}}$ , we have
$$\begin{aligned}2\sqrt{x+1}-\frac{\sqrt{x}}{2}&=\frac12\left(4\sqrt{x+1}-\sqrt{x}\right)\\
&=\frac12\cdot\frac{16(x+1)-x}{4\sqrt{x+1}+\sqrt{x}}\\
&=\frac{15x+16}{2(4\sqrt{x+1}+\sqrt{x})}\\
&\geqslant\dfrac{15(x+1)}{10\sqrt{x+1}}\\
&=\dfrac32\sqrt{x+1}\\
&\geqslant\frac32\sqrt{2}\\
&=\sqrt{\frac92}>2 \end{aligned}$$
for $x\in\mathbb{N}$, so in fact the inequality is strict.
|
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|
Assuming x is small, expand $\frac{\sqrt{1-x}}{\sqrt{1+2x}}$ up to and including the term in $x^{2}$ I have tried this many times but can't quite land on the correct answer.
The correct answer:
$1-\frac{3x}{2}+\frac{15x^{2}}{8}$
These are the steps I took:
*
*Re wrote it as: $\left ( 1-x \right )^{\frac{1}{2}}\left ( \left ( 1+2x \right )^{\frac{1}{2}} \right )^{-1}$
*Using bionmal expansion I expanded $\left ( 1-x \right )^{\frac{1}{2}}$ up to $x^{2}$
*
*For this I get : $\left (1-\frac{x}{2}-\frac{x^{2}}{8} \right)$
*Then to solve the second bracket written in step 1 I did $\left (1+2x \right )^{\frac{1}{2}}$
*
*For this I got $\left (1-x+\frac{x^{2}}{4} \right)$
*Then finally I raised this to the power one so $\left (1-x+\frac{x^{2}}{4} \right)^{-1}$
*For this I got $\left (1+x+\frac{3x^{2}}{4} \right)$
*Then I just expaned those two brakcets so: $\left (1-\frac{x}{2}-\frac{x^{2}}{8} \right)$$\left (1+x+\frac{3x^{2}}{4} \right)$
Once expanded, I then neglected powers bigger than $x^{2}$ (mentioned in question). Then collected like terms. However either my method or the algebra is going wrong, and I just need some help with this.
|
$\frac{1}{1+2x}=1-2x+4x^2-8x^3+O(x^4)$
so
$\frac{1-x}{1+2x}=1-2x+4x^2-8x^3-x+2x^2-4x^3+8x^4+O(x^5)=$
$1-3x+6x^2-12x^3+8x^4+O(x^5).$
Therefore, when $x$ is small,
$\sqrt{\frac{1-x}{1+2x}}\sim 1+\frac{1}{2}\left(-3x+6x^2-12x^3+8x^4\right)-\frac{1}{8}\left(-3x+6x^2-12x^3+8x^4\right)^2\sim$
$1-\frac{3}{2}x+3x^2-\frac{9}{8}x^2=1-\frac{3}{2}x+\frac{15}{8}x^2$
|
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|
Evaluating a double integral I was trying to evaluate the following integral
$$\int_{x=0}^{\infty}\int_{y=0}^{\infty}\frac{y \ln y \ln x}{(x^2+ y^2)( 1+y^2)} dy dx.$$
I have a guess that the value of this integral is $\frac{\pi^4}{8}$. But I am unable to prove it.
Could someone please help me in evaluating this integral?
Or, can we show the following identity holds without much calculation?
$$\int_{x=0}^{\infty}\int_{y=0}^{\infty}\frac{2y \ln y \ln x}{(x^2+ y^2)( 1+y^2)} dy dx= \int_{x=0}^{\infty}\int_{y=0}^{\infty}\frac{y (\ln y )^2}{(x^2+ y^2)( 1+y^2)} dy dx.$$
Any help or hint would be appreciated. Thanks in advance.
|
To show your identity holds, make the one-dimensional change of variables $x=ty$ to get
\begin{align*}
\int_{0}^\infty \int_0^\infty \frac{y \ln(y) \ln (x)}{(y^2+1)(x^2+y^2)} \ dy \ dx &= \int_{0}^\infty \int_0^\infty \frac{y^2 \ln(y) \ln (ty)}{(y^2+1)(t^2+1) y^2} \ dy \ dt \\
&=\int_{0}^\infty \int_0^\infty \frac{ \ln(y) \ln (t)}{(y^2+1)(t^2+1)} \ dy \ dt \\
&\qquad + \int_{0}^\infty \int_0^\infty \frac{ \ln^2(y)}{(y^2+1)(t^2+1)} \ dt \ dy \\
&= \left(\int_0^\infty \frac{ \ln (t)}{t^2+1} \ dt \right)^2 + \frac{\pi}{2} \int_0^\infty \frac{\ln^2(y)}{y^2+1} \ dy.
\end{align*}
Note upon a change of variables $t=1/u,$ we have
\begin{align*}
\int_0^1\frac{ \ln (t)}{t^2+1} \ dt &= -\int_1^\infty \frac{ \ln (t)}{t^2+1} \ dt,
\end{align*}
which implies $\int_0^\infty\frac{ \ln (t)}{t^2+1} \ dt =0.$ Hence,
\begin{align*}
\int_{0}^\infty \int_0^\infty \frac{y \ln(y) \ln (x)}{(y^2+1)(x^2+y^2)} \ dy \ dx &= \frac{\pi}{2} \int_0^\infty \frac{\ln^2(y)}{y^2+1} \ dy.
\end{align*}
In order to evaluate the right hand side, consider the triple integral
$$I=\int_0^\infty \int_0^\infty \int_0^\infty \frac{xz}{(x^2+1)(z^2+1)(x^2y^2z^2+1)} \ dy \ dx \ dz.$$ By carrying out the integration directly, we can find $I=\frac{\pi^3}{8}.$ On the other hand, using Fubini's Theorem to integrate with respect to $z$ first, we get by partial fractions (or Mathematica)
\begin{align*}
I &=\int_0^\infty \int_0^\infty \int_0^\infty \frac{xz}{(x^2+1)(z^2+1)(x^2y^2z^2+1)} \ dz \ dx \ dy \\
&= \int_0^\infty \int_0^\infty \frac{x \ln \left(xy\right)}{\left(x^2+1\right) \left( x^2 y^2-1\right)} \ dx \ dy \\
&= \int_0^\infty \int_0^\infty \frac{x \ln \left(x\right)}{\left(x^2+1\right) \left( x^2 y^2-1\right)} \ dy \ dx + \int_0^\infty \int_0^\infty \frac{x \ln \left(y\right)}{\left(x^2+1\right) \left( x^2 y^2-1\right)} \ dx \ dy \\
&= \int_0^\infty \int_0^\infty \frac{x \ln \left(y\right)}{\left(x^2+1\right) \left( x^2 y^2-1\right)} \ dx \ dy \\
&= \int_0^\infty \frac{\ln^2(y)}{y^2+1} \ dy,
\end{align*}
in which the first double integral in the third to last equality has Cauchy Principal Value $0$ and thus vanishes, and we carried out the double integral with respect to $x$ in the second to last equality using partial fractions once again.
Hence, $$I=\int_0^\infty \frac{\ln^2(y)}{y^2+1} \ dy=\frac{\pi^3}{8},$$ and so the desired integral is equal to $$\frac{\pi}{2} I=\frac{\pi^4}{16}.$$
Remark: These type of integral calculations are prominent in my joint AMS paper with Daniele Ritelli. See here: https://www.ams.org/journals/qam/2018-76-03/S0033-569X-2018-01499-3/
|
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|
A line $4x+y-1=0$ through $A(2,-7)$ meets the line $BC$ whose equation is $3x-4y-1=0$ at point $B$. Find Equation of Line $AC$, such that $AB=AC$. A line $4x+y-1=0$ through $A(2,-7)$ meets the line $BC$ whose equation is $3x-4y-1=0$ at point $B$. Find Equation of Line $AC$, such that $AB=AC$.
My Approach:
I took slope of line $AC$ as $m$.
And since $AB=AC$ that means Line $BC$ is equally inclined to $AC, AB$.
Now I used the formula for angle between two lines $\tan \theta = \bigg|\dfrac{(m_{1}-m_{2})}{1+m_{1}m_{2}}\bigg|$.
Slope of $AB$ is $-4$, slope of $BC$ is $\dfrac{3}{4}$ and Slope of $AC$ is $m$.
So, $\bigg| \dfrac{4m-3}{3m+4}\bigg|=\bigg |\dfrac {19}{8}\bigg |$.
Note: $\dfrac{19}{8}$ is $\tan$ of angle between $AB$ and $BC$ and $\bigg| \dfrac{4m-3}{3m+4}\bigg|$ is $\tan$ of angle between $AB$ and $AC$.
My Doubt: After solving above relation I am obtaining two values of $m$. And those values are $\dfrac{-110}{25}$ and $\dfrac{-52}{89}$.
According to me Both answer must be correct but my professor said only $\dfrac{-52}{89}$ is correct.
Similar Question Straight lines - equation of line
|
Solve the system of equations corresponding to the two given lines to find the coordinates of their point of intersection, which is $B.$ Find the distance between $A$ and $B.$ We'll call this distance $r.$ Solve the system of equations $$3x-4y=1,(x-2)^2+(y+7)^2=r^2$$. There are two solutions. One solution is the point $B$; the other solution is the point $C$. Since you now have the coordinates of both $A$ and $C$, you can find the equation of the line $AC$.
|
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|
How do we prove $x^6+x^5+4x^4-12x^3+4x^2+x+1\geq 0$? Question
How do we prove the following for all $x \in \mathbb{R}$ :
$$x^6+x^5+4x^4-12x^3+4x^2+x+1\geq 0 $$
My Progress
We can factorise the left hand side of the desired inequality as follows:
$$x^6+x^5+4x^4-12x^3+4x^2+x+1=(x-1)^2(x^4+3x^3+9x^2+3x+1)$$
However, after this I was unable to make any further progress in deducing the desired inequality.
I appreciate your help
|
Your quartic polynomial is called self-reciprocal or palindromic, in mathematics.
Let,
$$
\begin{align}P(x):=&x^4+3x^3+9x^2+3x+1\end{align}
$$
The case $x=0$ is trivial. Therefore, we can divide all terms of the polynomial $P(x)$ by $x^2\thinspace : (x\neq 0\thinspace)$
$$
\begin{align}\frac {P(x)}{x^2}=&\left(x^2+\frac {1}{x^2}\right)+3\left(x+\frac 1x\right)+9\end{align}
$$
Then, using the standard substitution $x+\frac 1x=u$, you have:
$$
\begin{align}&\frac {P(x)}{x^2}=\underbrace{ u^2+3u+7}_{\Delta_u<0}\thinspace >0\\
\implies &P(x)>0\thinspace , \forall x\neq 0\end{align}
$$
This means: $\thinspace P(x)>0,\thinspace \forall x\in\mathbb {R}\thinspace.$
|
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|
How do I integrate this rational function? Evaluate this integral
$$I=∫\dfrac{1-x^2}{x^4+x^2+1}dx$$ (with $x$ is different from 0).
I tried divided both the numerator and denominator by $x^2$, and I got $$∫\dfrac{\frac{1}{x^2}-1}{x^2+\frac{1}{x^2}+1}dx.$$
I then put $t=x+1/x$ and got this $$I=∫-\dfrac{dt}{t^2-1}.$$
Finally I got the answer
$$\dfrac12{\ln|-x^2-\dfrac{1}{x^2}-1|}+C$$
or the alternative form is
$$\dfrac{\ln\left(\left|x^2+x+1\right|\right)+\ln\left(\left|x^2-x+1\right|\right)-2\ln\left(\left|x\right|\right)}{2}+C. $$ Where did I make mistake? Because when I compute the function, the result is different $$\dfrac{\ln\left(x^2+x+1\right)-\ln\left(x^2-x+1\right)}{2}+C.$$
|
Since xpaul answered your question. Notice that another way is using partial fraction and we get
$$\frac{1-x^2}{x^4+x^2+1}=\frac{1}2\underbrace{\left(\frac{1+2x}{1+x+x^2}\right)}_{(*)}+\frac{1}{2}\underbrace{\left(\frac{1-2x}{1-x+x^2}\right)}_{(**)}$$
For $(*)$ substitution $u(x)=1+x+x^2$ and for $(**)$ substitution $v(x)=1-x+x^2$ in the integration.
Edit:
*
*$x^4+x^2+1=(x^2+1)^2-x^2=(x^2+x+1)(x^2-x+1)$.
*In partial fraction we want find constant $A,B,C$ and $D$ such that
$$\frac{1-x^2}{x^4+x^2+1}=\frac{A+Bx}{x^2+x+1}+\frac{C+Dx}{x^2-x+1}$$ That implies logically that
$$1-x^2=(A+Bx)(x^2+x+1)+(C+Dx)(x^2-x+1)$$
Then, a way to find the constants is using roots of the polynomial (in this case it is not useful) and other way it is solving the linear system with the coefficients (expanding and grouping). The second method work here.
|
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|
If $α,β,γ$ are the roots of the equation $f(x)=x^3+qx+r=0$ then find the equation whose roots are, If $α,β,γ$ are the roots of the equation $f(x)=x^3+qx+r=0$ then find the equation whose roots are, $\frac{\beta^2+\gamma^2}{\alpha^2}$,$\frac{\alpha^2+\gamma^2}{\beta^2}$,$\frac{\beta^2+\alpha^2}{\gamma^2}$.
My solution goes like this:
We consider, $a=\frac{\alpha^2+\beta^2}{\gamma^2}$,$b=\frac{\beta^2+\gamma^2}{\alpha^2}$,$c=\frac{\gamma^2+\alpha^2}{\beta^2}$. Now, $$\alpha+\beta+\gamma=0,\alpha\beta+\beta\gamma+\gamma\alpha=q,\gamma\alpha\beta=-r$$ and hence,$\alpha^2+\beta^2+\gamma^2=-2q$. Also, $a=\frac{\alpha^2+\beta^2}{\gamma^2}=\frac{-2q-\gamma^2}{\gamma^2}=\frac{-2q}{\gamma^2}-1$ or $\gamma^2=\frac{-2q}{a+1}$. Also, $$a=\frac{\alpha^2+\beta^2}{\gamma^2}=\frac{\gamma^2-2\alpha\beta}{\gamma^2}=1-\frac{2\alpha\beta\gamma}{\gamma^2\gamma}=1+\frac{2r}{\frac{-2q}{a+1}\gamma}=1-\frac{r(a+1)}{q\gamma}$$ and hence,$\gamma=\frac{r(a+1)}{q(1-a)}$. Also, $a=\frac{\alpha^2+\beta^2}{\gamma^2}=\frac{\gamma^2-2\alpha\beta}{\gamma^2}=1-2\frac{\alpha\beta\gamma}{\gamma^3}=1+\frac{2r}{\gamma^3}$. Thus, $\gamma^3=\frac{2r}{a-1}$. Now, we have, $\gamma^3+q\gamma+r=0$. Thus, $\frac{2r}{a-1}+\frac{r(a+1)}{(1-a)}+r=0$, which implies $a^2-a-2=0$. Thus, the required equation is $x^2-x-2=0$.
Is the above solution correct? If not, then where is it going wrong? I dont get where is the mistake occuring?
|
To answer your question, rather than suggest an alternative method: the error is in "Thus, $\frac{2r}{a-1}+\frac{r(a+1)}{(1-a)}+r=0$, which implies $a^2-a-2=0$". In fact (since $a \ne 1$) it implies that $2r -r(a+1) +r(a-1) = 0$, which is identically true. Instead, use your (correct) expression for $\gamma^3$ and substitute it into $(\gamma^3+r)^3=-q^3\gamma^3$.
|
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|
Tricky integral in a complex plane How to integrate using residues:
$$\oint\limits_{|z|=2}\frac{1}{(z^6-1)(z-3)} dz $$
if the idea probably requires to change the sum of 6 residues to aditive inverse of residue at infinity.
I believe that the sum of 6 residues is:
$$2 \pi i\sum_{i=1}^6 Res_i \frac{1}{(z^6-1)(z-3)}=-2\pi i(Res_\inf \frac {1}{(z^6-1)(z-3)}+Res_3 \frac {1}{(z^6-1)(z-3)})$$
because $${3>|z|=2, \infty>|z|=2.}$$
Since $$\frac {1}{(z^6-1)(z-3)} = \frac {1}{(z^7)(1-\frac{1}{z^6})(1-\frac {3}{z})} = \frac {1}{z^7}+ \frac {3}{z^8}+ ...$$
$$=> -2\pi iRes_\inf \frac {1}{(z^6-1)(z-3)}=0$$
|
Thank you for adding your work. Expanding on my comment: for $|z|\ge R\gt3$, we have
$$
\begin{align}
\left|\frac1{\left(z^6-1\right)(z-3)}\right|
&\le\frac1{|z|^7}\frac1{\left(1-\frac1{|z|^6}\right)\left(1-\frac3{|z|}\right)}\tag{1a}\\
&\le\frac{\frac{R^6}{R^6-1}\frac{R}{R-3}}{|z|^7}\tag{1b}
\end{align}
$$
For $R=7$, $\frac{R^6}{R^6-1}\frac{R}{R-3}\lt2$. Therefore, for $|z|\ge7$, we have
$$
\left|\frac1{\left(z^6-1\right)(z-3)}\right|\le\frac2{|z|^7}\tag2
$$
Thus, for $R\ge7$, the absolute value of the sum of the residues at all $7$ poles is
$$
\left|\frac1{2\pi i}\int_{|z|=R}\frac{\mathrm{d}z}{\left(z^6-1\right)(z-3)}\right|\le\frac2{R^6}\tag3
$$
That is, the sum of the residues at all $7$ poles is $0$.
The residue at $z=3$ is $\frac1{3^6-1}$. The integral around $|z|=2$ contains all the residues except at $z=3$, thus the sum of the residues inside $|z|=2$ is $-\frac1{728}$.
Therefore,
$$
\int_{|z|=2}\frac{\mathrm{d}z}{\left(z^6-1\right)(z-3)}=-\frac{\pi i}{364}\tag4
$$
|
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"timestamp": "2023-03-29T00:00:00",
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|
Find the maximum value of $xy+yz+xz-2xyz$
If $x+y+z=1$ and $0\le x,y,z\le1$ then find the maximum value of expression $$xy+yz+xz-2xyz$$
Solution that I have
$$(1-2x)(1-2y)(1-2z)=1-2\sum x+4\sum xy-8xyz=4\sum xy-8xyz-1$$
$\implies$
$$\sum xy-2xyz=\frac{1+(1-2x)(1-2y)(1-2z)}{4}$$
Now
$$\frac{\sum(1-2x)}{3}\ge\{(1-2x)(1-2y)(1-2z)\}^{\frac13}$$
$$\implies (1-2x)(1-2y)(1-2z)\le\frac{1}{27}$$
$$\implies xy+yz+xz-2xyz\le\frac{1+\frac{1}{27}}{4}=\frac{7}{27}$$
My own solution
Consider $f(x,y,z)=xy+yz+zx-2xyz$
Since $f(x,y,z)$ is symmetric, it will achieve it's maximum value when $x=y=z=\frac13$ thereby giving $\frac{7}{27}$ as the answer.
What I want to know, is whether there is any other practical method to solve this question as considering $(1-2x)(1-2y)(1-2z)$ seems somewhat impractical and unimpressive to me. Like how will one think of such expression.
Any help is greatly appreciated.
|
Another way for proving that $f(x,y,z)=xy+yz+zx-2xyz\leq \frac{7}{27}$ if $x,y,z\in(0,1)$ and $x+y+z=1.$ We denote $x=X+1/3,y=Y+1/3, z=Z+1/3$ with $X,Y,Z\in [-1/3.2/3]$ and $X+Y+Z=0.$ We denote
$$A=-2(XY+YZ+ZX)=X^2+Y^2+Z^2$$ leading to
$$f(x,y,z)=\frac{7}{27}-\frac{1}{6}(A+12XYZ).$$ If 2 of the three numbers $X, Y,Z$ are negative we have $A+12XYZ\geq 0.$ If not, without loss of generality we assume $Y$ and $Z$ non negative. As a consequence $0\leq Y+Z=-X\stackrel{(*)}
{\leq} 1/3$ and we get
$$A+12XYZ=(Y+Z)^2+Y^2+Z^2-12 (Y+Z)YZ\stackrel{(*)}
{\geq }2(Y^2+Z^2+YZ) -4YZ=2((Y-Z)^2+YZ)\geq 0.$$
|
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|
Finding the value of given integral. It was asked to find the correct option(s) for the given integral: $$I_n = \displaystyle\int_{\frac{n}{2}}^{\frac{n+1}{2}}\dfrac{\sin{(\pi(\sin^2{\pi x}}))}{(\sqrt2)^x} \, dx$$
(a)$\dfrac{I_n}{I_{n+4}}=2$
(b)$\dfrac{I_n}{I_{n+4}}=\dfrac{1}{\sqrt2}$
(c)$\dfrac{\displaystyle\sum_{n=0}^\infty I_{8n}}{I_0}=\dfrac{4}{3}$
(d)$ \dfrac{\displaystyle\sum_{n=0}^\infty I_{n}}{I_0}=2$
I tried using King's property to solve this one, but it is not working here, instead it is making it more complicated.
Any help is appreciated :)
|
First, notice that for any even integer $2k$
$$I_{n+2k} = \int_{\frac{n}{2}+k}^{\frac{n+1}{2}+k}\frac{\sin\left[\pi \sin^2(\pi x)\right]}{\left(\sqrt{2}\right)^x}dx = \int_{\frac{n}{2}}^{\frac{n+1}{2}}\frac{\sin\left[\pi \sin^2(\pi x + \pi k)\right]}{\left(\sqrt{2}\right)^{x+k}}dx = \frac{1}{\left(\sqrt{2}\right)^k}I_n$$
Selecting $k=2$ gives us
$$\frac{I_n}{I_{n+4}} = \frac{I_n}{\frac{1}{\left(\sqrt{2}\right)^2}I_n} = 2$$
Selecting $k=4n$ gives us
$$\frac{\sum\limits_{n=0}^{\infty}I_{8n}}{I_0} = \frac{\sum\limits_{n=0}^\infty \frac{1}{\left(\sqrt{2}\right)^{4n}}I_0}{I_0} = \sum_{n=0}^\infty \frac{1}{4^n} = \frac{4}{3}$$
Selecting $k=2n$ gives us
$$\frac{\sum\limits_{n=0}^\infty I_n}{I_0} > \frac{\sum\limits_{n=0}^\infty I_{4n}}{I_0} = \frac{\sum\limits_{n=0}^\infty \frac{1}{\left(\sqrt{2}\right)^{2n}}I_0}{I_0} = \sum_{n=0}^\infty \frac{1}{2^n} = 2$$
because
$$\pi \sin^2(\pi x) \in [0,\pi] \implies I_n > 0$$
Therefore, the only correct choices are options $(a)$ and $(c)$
|
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|
$A=\begin{pmatrix}a&c&-1\\ 1-c&-a&0\\ 5&3&b\end{pmatrix}$ has an eigenvector $(-1,1,-1)'$, and $|A|=0$, can we find $a,b,c$?
$A=\begin{pmatrix}a&c&-1\\ 1-c&-a&0\\ 5&3&b\end{pmatrix}$ has an eigenvector $(-1,1,-1)'$, and $|A|=0$, can we find $a,b,c$?
By $Ax=tx$ for some $t$, we see $c=1+t-a, b=t-2$ (with the other equality superfluous), and then
$0=|A|=t^3-(2a+1)t^2+(3a+1)t-6a$. Oh my god, we can not do anymore.
|
Since $\det(A)=0$, then we have
$$-ba^{2}-5a+(c-1)(bc+3)=0,\quad (1)$$
Since $(-1,1-1)$ is an eigenvector for some eigenvalue $k$ of $A$ we have
$$-a+c+k=-1,\quad (2)$$
$$-a+c-k=1,\quad (3)$$
$$-b+k=2,\quad (4)$$
Solving the non-linear system with equations $(1),(2),(3)$ and $(4)$ with Mathematica we have one solution
$$(a,b,c,k)=(3,-3,3,-1).$$
It works, since
*
*$\det(A)=\det\begin{pmatrix}3&3&-1\\ 1-3&-3&0\\ 5&3&-3\end{pmatrix}=0$.
*$Av=\begin{pmatrix}3&3&-1\\ 1-3&-3&0\\ 5&3&-3\end{pmatrix}\begin{pmatrix}-1\\1\\-1\end{pmatrix}=\begin{pmatrix}1\\-1\\1\end{pmatrix}=-1\begin{pmatrix} -1\\1\\-1\end{pmatrix}=kv$.
So a priori with that information we can find the values $a,b,c$. Of course, it would be interesting to be able to avoid using the software.
|
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|
Show that $|\tan x-x|\leq 8x^2$ if $|x|\leq \pi/3$ Show that $|\tan x-x|\leq 8x^2$ if $|x|\leq \pi/3$. I think this is supposed be solved using the maclaurin series.
Let $f(x)=\tan x$ then $f'=1/\cos ^2x$ and $f''=\frac{-2\cos x \sin x}{\cos^4 x}=\frac{-\sin 2x}{\cos^4 x}$. Since $f(0)=0$ and $f'(0)=1$ we have that
$|x +\frac{-\sin 2\theta x}{\cos^4 x\theta}-x|=|\frac{-\sin 2\theta x}{\cos^4 x\theta}|$,where $\theta$ is between $0$ and $1$. Now what is left is to show that $|\frac{-\sin 2\theta x}{\cos^4 x\theta}|\leq 8x^2$ if $|x|\leq \pi/3$.
|
Let
$$ f(x)=\tan x-x. $$
Then there is $\theta\in(0,1)$ such that
$$ f(x)-f(0)=f'(\theta x)x. $$
Note
$$ f'(\theta x)=\sec^2(\theta x)-1=\tan^2(\theta x)=\frac{\sin^2(\theta x)}{1-\sin^2(\theta x)}. $$
Using
$$ \frac{2}{\pi}\theta\le\sin\theta\le\theta, \theta\in[0,\frac\pi2]$$
one has
$$ \sin^2(\theta x)\le \theta^2x^2\le x^2 $$
and
$$ 1-\sin^2(\theta x)\ge 1-\frac{4}{\pi^2}\theta^2x^2\ge 1-\frac{4}{\pi^2}x^2. $$
So
$$ f(x)\le \frac{x^3}{1-\frac{4}{\pi^2}x^2}=\frac{x}{1-\frac{4}{\pi^2}x^2}\cdot x^2\le 8x^2. $$
It is easy to show
$$ \frac{x}{1-\frac{4}{\pi^2}x^2}\le 8.$$
|
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|
How to order $x = \sqrt{3}-1, y = \sqrt{5}-\sqrt{2}, z = 1+\sqrt{2}$ ascendingly? How would I order $x = \sqrt{3}-1, y = \sqrt{5}-\sqrt{2}, z = 1+\sqrt{2} \ $ without approximating the irrational numbers? In fact, I would be interested in knowing a general way to solve such questions if there is one.
What I tried to so far, because they are all positive numbers, is to square $x,y,z$ but, obviously, the rational parts will not be equal so I cannot compare the radicals. Proving that $x<z$ is easy and so is $y<z$, but I'm stuck at $x < y \text{ or } x>y$.
|
Suppose $ x \geqslant y $, then
$$ \begin{array} { r c l }
\sqrt3 + \sqrt2 &\geqslant& \sqrt5 + 1 \\
(\sqrt3 + \sqrt2)^2 &\geqslant& (\sqrt5 + 1)^2 \\
5 + 2 \sqrt6 &\geqslant& 6 + 2 \sqrt5 \\
2( \sqrt6 - \sqrt5) &\geqslant& 1 \\
2( \sqrt6 - \sqrt5)( \sqrt6 + \sqrt5) &\geqslant& ( \sqrt6 + \sqrt5) \\
2(6 - 5) &\geqslant& \sqrt6 + \sqrt5 \\
2 &\geqslant& \sqrt6 + \sqrt5 > \sqrt4 + \sqrt4 = 4 \\
\end{array}
$$
A contradiction.
|
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|
Prove that $\lfloor {(\frac{\sqrt{5}+1}{2})}^{4n-2}\rfloor-1$ is a square number where $n$ is a natural number. I found this problem in a junior high school math competition.
Prove that $\lfloor {(\frac{\sqrt{5}+1}{2})}^{4n-2}\rfloor-1$ is a
square number where $n$ is a natural number.
Here's what I think
Suppose $x$ is ${(\frac{\sqrt{5}+1}{2})}^2$
Then we get
$\lfloor {(\frac{\sqrt{5}+1}{2})}^{4n-2}\rfloor-1$
$=(x^{2n-1}+\frac{1}{x^{2n-1}}-1)-1$
$=(x^{2n-1}-1)(1-\frac{1}{x^{2n-1}})$
$=(x-1)(1-\frac{1}{x})(x^{2n-2}+x^{2n-3}+x^{2n-4}+\ldots+1)(1+\frac{1}{x}+\frac{1}{x^2}+\ldots+\frac{1}{x^{2n-2}})$
$=(x^{2n-2}+x^{2n-3}+x^{2n-4}+\ldots+1)(1+\frac{1}{x}+\frac{1}{x^2}+\ldots+\frac{1}{x^{2n-2}})$
$=(x^{n-1}+x^{n-2}+x^{n-3}+\ldots+\frac{1}{x^{n-1}})(x^{n-1}+x^{n-2}+x^{n-3}+\ldots+\frac{1}{x^{n-1}})$
$=(x^{n-1}+x^{n-2}+x^{n-3}+\ldots+\frac{1}{x^{n-1}})^2$
Now I hope to prove that $(x^{n-1}+x^{n-2}+x^{n-3}+\ldots+\frac{1}{x^{n-1}})\in\mathbb{N}$
But I don't know how to continue, but I think there is something with the golden ratio, can someone help me, or provide a better solution for the original problem? Thanks a lot!
NEW: I found the same question on AoPS, I see that the original formula is equal to $((\frac{\sqrt5+1}{2})^{2n-1}-(\frac{\sqrt5-1}{2})^{2n-1})^2$, but it just said that is actually the general form for the recurrence relation $a_n=3a_{n-1}+a_{n-2}$ with $a_0=0$ and $a_1=1$, which can be shown by the characteristic polynomials, but I still don't know why...
|
Not to be confused with this exceedingly similar question.
I'll reproduce the solution found here (with a few added details).
Because $\frac{\sqrt{5} - 1}{2} < 1$, we also know that $\left(\frac{\sqrt{5} - 1}{2}\right)^{4n - 2} < 1$. Then
$$\begin{align}
\left\lfloor\left(\frac{\sqrt{5} + 1}{2}\right)^{4n - 2}\right\rfloor -1 &= \left(\frac{\sqrt{5} + 1}{2}\right)^{4n - 2} + \left(\frac{\sqrt{5} - 1}{2}\right)^{4n - 2} - 2\\
&= \left(\left(\frac{\sqrt{5} + 1}{2}\right)^{2n - 1}-\left(\frac{\sqrt{5} - 1}{2}\right)^{2n - 1}\right)^2
\end{align}$$
Thus, we only need to show that
$$\left(\frac{\sqrt{5} + 1}{2}\right)^{2n - 1}-\left(\frac{\sqrt{5} - 1}{2}\right)^{2n - 1} \in \mathbb{N}.$$
Using the Newton binomial formula we have
\begin{align}
\left(\frac{\sqrt{5} + 1}{2}\right)^{2n - 1} &= A_n \sqrt{5} +B_n\\
\left(\frac{\sqrt{5} - 1}{2}\right)^{2n - 1} &= A_n \sqrt{5} -
B_n
\end{align}
so
$$\left(\frac{\sqrt{5}+1}{2}\right)^{2 n-1}-\left(\frac{\sqrt{5}-1}{2}\right)^{2 n-1}=2 B_n$$
where $A_n, B_n$ there are positive integers
Every term in this recurrence is an integer so
$$\left(\frac{\sqrt{5} + 1}{2}\right)^{2n - 1}-\left(\frac{\sqrt{5} - 1}{2}\right)^{2n - 1} \in \mathbb{N}$$
and we are done.
|
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|
Derivative of $(2x^2+3)^2(x^3-2x)^4$ I want to know if there's another method in shorter way. I came up to this problem
Find the $f'(x)$
$$
f(x) = (2x^2+3)^2\cdot(x^3-2x)^4
$$
Applying product rules
$$
\frac{d}{dx}f(x)g(x) = f(x)\frac{d}{dx}g'(x) + g'(x)\frac{d}{dx}f(x)
$$
Solve:
$$ f = (2x^2+3)^2 $$
$$ g = (x^3-2x)^4$$
$$
\frac{d}{dx}(2x^2+3)^2(x^3+2x)^4 + \frac{d}{dx}(x^3-2x)^4(2x^2+3)^2
$$
$$
2(2x^2+3) 4x(x^3-2x)^4 + 4(x^3-2x)^3(3x^2-2)(2x^2+3)^2
$$
$$
8x(2x^2+3)(x^3-2x)^4 + 4(x^3-2x)^3(3x^2-2)(2x^2+3)^2
$$
$$
8x(2x^2+3)(x^3-2x)^4 + (x^3-2x)^3(12x^2-8)(2x^2+3)^2
$$
Answer:
$$
f'(x) =8x(2x+3)(x^3-2x)^4 + (x^3-2x)^3(12x^2-8)(2x^2+3)^2
$$
Is this correct way or are there other methods?
Thanks in advance.
|
Your method is correct, and since the question states that you should make use of the product rule, it is the right solution.
If you're new to derivatives, you may not yet have learned about derivatives of logs or about implicit differentiation; however you're clearly already familiar with the chain rule, so you should be learning about these other topics very soon.
Once you do, there is a more efficient approach which takes advantage of the properties of logs:
$$\begin{align}
f(x) &= (2x^2+3)^2\cdot(x^3-2x)^4 \\[12pt]
\ln f(x) &= \ln \left[ (2x^2+3)^2\cdot(x^3-2x)^4 \right] \\
&= 2 \ln (2x^2+3) + 4 \ln (x^3-2x) \\[12pt]
\frac{f'(x)}{f(x)} &= 2 \left( \frac{4x}{2x^2+3} \right) + 4 \left( \frac{3x^2-2}{x^3-2x} \right) \\[12pt]
f'(x) &= f(x) \left[ \frac{8x}{2x^2+3} + \frac{12x^2-8}{x^3-2x} \right] \\
&= 8x (2x^2+3) (x^3-2x)^4 + (12x^2-8) (2x^2+3)^2 (x^3-2x)^3
\end{align}$$
As @RyszardSzwarc notes in the comments below, technically the above method would not work when $x = 0, \pm \sqrt{2}$ as the function value would be zero, but that ultimately does not affect the solution.
|
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|
BMO2 1994 Q1 - Need Help Find the first number such that the average of the sum of the squares from $1$ to $n$ (where $n > 1$) equals $k^2$.
Here is what I have done so far:
The sentence is equivalent to saying that $\frac{\sum_{r=1}^{n}r^2}{n} = k^2$. Knowing the sum of the squares formula, the average of the sum of the squares is $\frac{(n+1)(2n+1)}{6}$. Equating it to $k^2$ and simplifying leads to $2n^2 + 3n + 1 = 6k^2$. This is where I am stuck at. I have tried everything in regards to algebraic tricks and manipulation. Does anybody know any methods or techniques from https://archive.ukmt.org.uk/docs/BMO%20Preparation%20Sheet.pdf that would help?
|
Even tough the other answer provided is far more elegant, I would like to show another possible approach:
First of all, we need $\frac{(n+1)(2n+1)}{6}$ to be an integer, and checking the possibilities modulo $2$ and $3$ we get that $n$ must be of the form $6a\pm 1$. However, if $n$ were of the form $6a-1$, plugging it in gets us $k^2=a(12a-1)$. Now since $a$ and $12a-1$ don't share any common factors, in order for their product to be a perfect square they both need to be perfect squares, but $12a-1=c^2$ doesn't have solutions modulo $3$, so $n=6a+1$.
Plugging in we have $(3a+1)(4a+1)=k^2$, and again since $3a+1$ and $4a+1$ don't share any common divisors, they both need to be perfect squares, so set $x^2=4a+1$ and $y^2=3a+1$. Substituting gets you to $3x^2=4y^2-1=(2y-1)(2y+1)$. Once again, we notice that $2y-1$ and $2y+1$ don't have common factors,so this means that there exist integers $p,q$ such that $x=pq$, $gcd(p,q)=1$, and $p^2|(2y+1)$, $q^2|(2y-1)$. So now we have two possibilities:
Case 1: $2y+1=p^2, 2y-1=3q^2$. Subtracting the two equations gives $p^2-3q=2$, which has no solution modulo $3$.
Case 2: $2y+1=3p^2$, $2y-1=q^2$, subtracting we get $3p^2-q^2=2$, and you can easily see that the smallest possible solution is $(p,q)=(3,5)$, and substituting back gives as finally $x=15$, so $a=56$ and $n=337$.
|
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|
Evaluation of $\lim_{n\to\infty}\sum_{k=1}^{n}kA^{k-1}$ Let
$$\begin{align}
A:&=\begin{pmatrix}{11\over 6}&2\\-1&-1\end{pmatrix}\\
B:&=\lim_{n\to\infty}\sum_{k=1}^{n}kA^{k-1}\\
A^{0}&=I
\end{align}$$
I want to evaluate $B$.
My tries:
$$\begin{align}
A&=\begin{pmatrix}{11\over 6}&2\\-1&-1\end{pmatrix}\\
&={6\over 6}\begin{pmatrix}{11\over 6}&2\\-1&-1\end{pmatrix}\\
&={1\over 6}\begin{pmatrix}11&12\\-6&-6\end{pmatrix}\\
\therefore A^{k-1}&=\left({1\over 6}\right)^{k-1}\begin{pmatrix}11&12\\-6&-6\end{pmatrix}^{k-1}\\
A^{0}&=I\\
A^{1}&={1\over 6}\begin{pmatrix}11&12\\-6&-6\end{pmatrix}\\
A^2&={1\over 36}\begin{pmatrix}49&60\\-30&-36\end{pmatrix}\\
A^3&={1\over 216}\begin{pmatrix}179&228\\-114&-144\end{pmatrix}
\end{align}$$
I coudn't find a regurarity for a power of $A$.
I also tried to use $P^{-1}AP$ but it seems that this kind of approach may not work.
$$
\lambda=\frac{1}{2},~\frac{1}{3}\leftarrow\text{Eigenvalues for}~A
$$
I got $\begin{pmatrix}2&0\\0&1\end{pmatrix}$ after doing elementary roe operations for $\lambda={1\over 2}$ and definitely this set of matrices can't compose a non-zero vector.
I need your help.
|
We have
$$
(I-2A+A^2)\sum_{k=1}^{n} kA^{k-1}
= \sum_{k=1}^{n} kA^{k-1}
-2\sum_{k=1}^{n} kA^k
+\sum_{k=1}^{n} kA^{k+1} \\
=I+2A+\sum_{k=3}^{n} kA^{k-1} \\
-2\left(A+\sum_{k=2}^{n-1} kA^k+nA^n\right) \\
+\sum_{k=1}^{n-2} kA^{k+1}+(n-1)A^n+nA^{n+1} \\
=I+2A-2A-2nA^n+(n-1)A^n+nA^{n+1} \\
+\sum_{k=2}^{n-1} (k+1)A^k
-2\sum_{k=2}^{n-1} kA^k
+\sum_{k=2}^{n-1} (k-1)A^k \\
=I-(n+1)A^n+nA^{n+1} +\sum_{k=2}^{n-1} (k+1-2k+k-1)A^k \\
=I-(n+1)A^n+nA^{n+1}
$$
Therefore, if $I-2A+A^2 = (I-A)^2$ is invertible, we have
$$
\sum_{k=1}^{n} kA^{k-1} = (I-A)^{-2}\left(I-(n+1)A^n+nA^{n+1}\right)
$$
As the absolute values of the eigenvalues of $A$ are smaller than $1$ (the eigenvalues are $\frac12$ and $\frac13$), we know $\lim_{n\rightarrow\infty}\left(p(n)A^n\right) = 0$ for any polynomial $p$, see below.
Therefore
$$ \lim_{n\rightarrow\infty}\sum_{k=1}^{n} kA^{k-1}
=(I-A)^{-2}\left(I-0+0\right) = (I-A)^{-2}
$$
The details of $\lim_{n\rightarrow\infty}\left(p(n)A^n\right) =0$
We know that $A$ is diagonalizable and that the eigenvalues are $\frac12$ and $\frac13$. Therefore, we know that there is an invertible matrix $P$ such that
$$ P^{-1}AP = \begin{pmatrix} \frac12 & 0 \\ 0 & \frac13\end{pmatrix}
$$
which means
$$ A = P \begin{pmatrix} \frac12 & 0 \\ 0 & \frac13\end{pmatrix} P^{-1}
$$
Therefore, for a given polynomial $p$:
$$
p(n)A^n = P \begin{pmatrix} \frac{p(n)}{2^n} & 0 \\ 0 & \frac{p(n)}{3^n}\end{pmatrix} P^{-1}
$$
As matrix multiplication is a continuous operation, we can swap the evaluation of the limits and the matrix multiplication:
$$
\lim_{n\rightarrow\infty}\left(p(n)A^n\right)
= P\begin{pmatrix} \lim\limits_{n\rightarrow\infty}\frac{p(n)}{2^n} & 0 \\ 0 & \lim\limits_{n\rightarrow\infty}\frac{p(n)}{3^n}\end{pmatrix} P^{-1}
$$
It is a well known fact that exponential growth is "stronger" than polynomial growth in the sense that the latter divided by the former approaches $0$ as $n$ approaches infinity. Therefore,
$$
\lim\limits_{n\rightarrow\infty}\frac{p(n)}{2^n}
=\lim\limits_{n\rightarrow\infty}\frac{p(n)}{3^n} = 0
$$
for any given polynomial and
$$
\lim\limits_{n\rightarrow\infty}\left(p(n)A^n\right) = 0
$$
|
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"url": "https://math.stackexchange.com/questions/4622717",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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|
Determine whether this series converges: Determine whether this series converges:
$ \sum_{n=1}^{\infty}\frac{\sqrt n\cos(n^2)}{n^{4/3}+\cos(n^2) }$
I tried
$$-1\leq \cos(n^2) \leq 1 \Rightarrow \sum_{n=1}^{\infty}\frac{- \sqrt n}{n^{4/3}-1 } \leq \sum_{n=1}^{\infty}\frac{\cos(n^2)\cdot \sqrt n}{n^{4/3}+\cos(n^2) } \leq \sum_{n=1}^{\infty}\frac{ \sqrt n}{n^{4/3}-1 }$$
Applying limit comparison test, we get
$- \infty \leq \sum_{n=1}^{\infty}\frac{\sqrt n\cos(n^2)}{n^{4/3}+\cos(n^2) } \leq \infty$
Comparison test doesn't work. What test should I use?
|
Denote
\begin{align}
a_n = \frac{\sqrt{n}}{n^{4/3} +\cos(n^2)}, \quad b_n = \cos(n^2), \quad \text{ and } \quad B_N = \sum^N_{n=0} \cos(n^2).
\end{align}
Then the given series can be expressed as
\begin{align}
\sum^\infty_{n=0} a_nb_n.
\end{align}
Now, using Abel's summation formula (or the so called summation-by-parts formula), we have that
\begin{align}
S_N:=\sum^N_{n=0} a_nb_n =&\, a_N B_N -\sum^{N-1}_{n=0} B_n\left(a_{n+1}-a_n\right)\\
=&\, \frac{\sqrt{N}}{N^{4/3} +\cos(N^2)}\sum^N_{n=0}\cos(n^2)+\sum^{N-1}_{n=0} \left(\frac{\sqrt{n}}{n^{4/3} +\cos(n^2)}-\frac{\sqrt{n+1}}{(n+1)^{4/3} +\cos((n+1)^2)}\right)\sum^n_{k=0}\cos(k^2).
\end{align}
First, notice that
\begin{align}
\frac{\sqrt{n}}{n^{4/3} +\cos(n^2)}-\frac{\sqrt{n+1}}{(n+1)^{4/3} +\cos((n+1)^2)} =&\, \frac{1}{n^{4/3}}\left(\frac{\sqrt{n}}{1 +\frac{\cos(n^2)}{n^{4/3}}}-\frac{\sqrt{n+1}}{(1+\frac{1}{n})^{4/3} +\frac{\cos((n+1)^2)}{n^{4/3}}}\right)\\
\simeq&\, \frac{1}{n^{4/3}\left(\sqrt{n+1}+\sqrt{n}\right)}\left(1+\mathcal{O}(\frac{1}{n^\alpha})\right)
\end{align}
provided $n$ is sufficiently large.
If we show that
\begin{align}
|B_N|=\left|\sum^N_{n=0} \cos(n^2)\right| \le C\sqrt{N} \log N
\end{align}
where $C$ is some constant independent of $N$ then we have that
\begin{align}
|S_N-S_M| \le&\, C\frac{N \log N}{N^{4/3} +\cos(N^2)}+C\frac{M \log M}{M^{4/3} +\cos(M^2)}\\
& +C\sum^{N-1}_{n=M} \left|\frac{\sqrt{n}}{n^{4/3} +\cos(n^2)}-\frac{\sqrt{n+1}}{(n+1)^{4/3} +\cos((n+1)^2)}\right|\left|\sum^n_{k=0}\cos(k^2)\right|\\
\le &\, C\frac{N \log N}{N^{4/3} +\cos(N^2)}+C\frac{M \log M}{M^{4/3} +\cos(M^2)}\\
&\, + C \sum^{N-1}_{n=M} \frac{\sqrt{n}\log n}{n^{4/3}\left(\sqrt{n+1}+\sqrt{n}\right)}.
\end{align}
Fix $\varepsilon>0$. Hence if we choose $K=K(\varepsilon)$ sufficiently large, then for any $M, N>K$ we have that
\begin{align}
|S_N-S_M|<\varepsilon.
\end{align}
This shows that the series converges.
The proof of the bound on $B_N$ can be found here.
|
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|
Find the coefficient of $f(x)=e^x \sin x$ for the 5th derivative $f^{(5)}(0)$ using maclaurin series
Find the coefficient of $f(x)=e^x \sin x$ for the 5th derivative $f^{(5)}(0)$ using maclaurin series
the maclaurin polynomial is supposed to be $M_5(0)=f(0)+ \frac{f'(0)}{1!}+...+ \frac{f^{(5)}(0)}{5!}$
We know that $e^x=1+x+ \frac{x^2}{2!}+...$
we also know $\sin x = x-\frac{x^3}{3!}+\frac{x^5}{5!}+...$
the answer is supposed to be $-4$ and the solution it shows is just $(\frac{1}{5!}- \frac{1}{12}+\frac{1}{4!})x^5$ and then $(\frac{1}{5!}- \frac{1}{12}+\frac{1}{4!})=\frac{f^{(5)}(0)}{5!}$
but I do not understand how they got to this? I assumed they took the expansion of each $e^x$ and $sinx$ till the fifth order and multiplied but that didn't work for me
there has to be a simple way and I am unable to figure it
Thanks for any tips and help!
|
You have that
$$e^x=1+x+\frac{1}{2!}x^2+\frac{1}{3!}x^3+\frac{1}{4!}x^4+\text{higher order terms}$$
and
$$\sin x=x-\frac{1}{3!}x^3+\frac{1}{5!}x^5+\text{higher order terms}.$$
If we multiply these together and just look at the resulting $x^5$-term, we get
$$e^x\sin x=\text{lower order terms}+\left(\frac{1}{4!}\cdot1-\frac{1}{2!}\cdot\frac{1}{3!}+1\cdot\frac{1}{5!}\right)x^5+\text{higher order terms}.$$
Now since the factor in front of the $5$'th order term in the expansion corresponds to $\frac{1}{5!}f^{(5)}(0)$ (with $f(x)=e^x\sin x$), we get that
$$f^{(5)}(0)=5!\left(\frac{1}{4!}\cdot1-\frac{1}{2!}\cdot\frac{1}{3!}+1\cdot\frac{1}{5!}\right)=-4.$$
|
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"timestamp": "2023-03-29T00:00:00",
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|
Range of Trigonometric function having square root
Finding range of function
$\displaystyle f(x)=\cos(x)\sin(x)+\cos(x)\sqrt{\sin^2(x)+\sin^2(\alpha)}$
I have use Algebric inequality
$\displaystyle -(a^2+b^2)\leq 2ab\leq (a^2+b^2)$
$\displaystyle (\cos^2(x)+\sin^2(x))\leq 2\cos(x)\sin(x)\leq \cos^2(x)+\sin^2(x)\cdots (1)$
And
$\displaystyle -[\cos^2(x)+\sin^2(x)+\sin^2(\alpha)]\leq 2\cos(x)\sqrt{\sin^2(x)+\sin^2(\alpha)}\leq [\cos^2(x)+\sin^2(x)+\sin^2(\alpha)]\cdots (2)$
Adding $(1)$ and $(2)$
$\displaystyle -(1+1+\sin^2(\alpha))\leq 2\cos(x)[\sin(x)+\sqrt{\sin^2(x)+\sin^2(\alpha)}]\leq (1+1+\sin^2(\alpha))$
$\displaystyle -\bigg(1+\frac{\sin^2(\alpha)}{2}\bigg)\leq f(x)\leq \bigg(1+\frac{\sin^2(\alpha)}{2}\bigg)$
I did not know where my try is wrong.
Please have a look on it
But actual answer is $\displaystyle -\sqrt{1+\sin^2(\alpha)}\leq f(x)\leq \sqrt{1+\sin^2(\alpha)}$
|
Let $a=|\sin\alpha|.$
First notice that (since $|\sin x|\le\sqrt{a^2+\sin^2x}$) when the maximum value $f_{\max}$ is attained the $\sin$ and $\cos$ of $x$ are positive, and that the minimum value is $f_{\min}=-f_{\max},$ corresponding to the same $\sin$ and the opposite $\cos.$
$$f'(x)=\left(1+\frac{\sin x}{\sqrt{a^2+\sin^2x}}\right)g(x)$$
where
$$g(x):=\cos^2x-\sqrt{a^2+\sin^2x}\,\sin x$$
decreases continuously from $g(0)=1-0=1$ to $g(\pi/2)=0-a=-a$ hence vanishes for a unique $x_0\in[0,\pi/2].$ One easily checks that
$$\sin^2x_0=\frac1{2+a^2},\quad\cos^2x_0=\frac{1+a^2}{2+a^2}.$$
Therefore, $f_\max=f(x_0)=\sqrt{1+a^2}$ and
$$\operatorname{range}(f)=\left[-\sqrt{1+a^2},\sqrt{1+a^2}\right].$$
|
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|
Is it true that if $a, b, x, y>0$, $a+b \ge x+y$ and $ab \le xy$ then $a^n+b^n≥x^n+y^n$ for $n>1$? Let $a, b, x, y>0$, $a+b \ge x+y$ and $ab \le xy$ then $a^n+b^n \ge x^n+y^n$ where $n=2$.
Is it true that if $a, b, x, y>0$, $a+b \ge x+y$ and $ab \le xy$ then $a^n+b^n \ge x^n+y^n$ for $n \ge 1$?
I have just been checked it is true with $1< a, b, x, y \le 1000$ with $n=3,4$ by my computer. I am looking for a proof if this is true inequality.
|
Let
$d_n
= a^n+b^n
$
and let $a+b=d$.
$\begin{array}\\
d_n(a+b)
&=(a^n+b^n)(a+b)\\
&=a^{n+1}+ba^n+ab^n+b^{n+1}\\
&=a^{n+1}+b^{n+1}+ab(a^{n-1}+b^{n-1})\\
&=d_{n+1}+abd_{n-1}\\
\text{so}\\
d_{n+1}
&=d_n(a+b)-abd_{n-1}\\
&=dd_n-abd_{n-1}\\
\end{array}
$
Similarly,
if
$e_n=x^n+y^n
$
and
$e=x+y$,
$e_{n+1}
=ee_n-xye_{n-1}
$.
$d_1 \ge e_1$
and
$\begin{array}\\
d_2
&=a^2+b^2\\
&=a^2+2ab+b^2-2ab\\
&=(a+b)^2-2ab\\
&\ge(x+y)^2-2xy\\
&=e_2\\
\end{array}
$
If
$d_n \ge e_n$
and
$d_{n-1} \ge e_{n-1}
$,
then
$\begin{array}\\
d_{n+1}
&=dd_n-abd_{n-1}\\
&\ge ee_n-xyd_{n-1}
\qquad d\ge e, xy\ge ab, d_n\ge e_n\\
&= ee_n-xy(e_{n-1}-e_{n-1}+d_{n-1})\\
&= ee_n-xye_{n-1}-xy(e_{n-1}-d_{n-1})\\
&= e_{n+1}+xy(d_{n-1}-e_{n-1})\\
&\ge e_{n+1}
\qquad d_{n-1}\ge e_{n-1}\\
\end{array}
$
Therefore
$d_n \ge e_n$
for all $n \ge 1$.
|
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"timestamp": "2023-03-29T00:00:00",
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|
Quadratic polynomials $f(x)$ with rational coefficients such that $f(p)=q, f(q)=r, f(r)=p$ where $p,q,r$ are the roots of $x^3-7x+7$. Let $p,q,r$ be the three roots of $x^3-7x+7$. Find all quadratic polynomials $f(x)$ with rational coefficients such that $f(p)=q, f(q)=r, f(r)=p$.
So far, I let $f(x)=ax^2+bx+c$, and I have
$$ap^2+bp+c=q,$$
$$aq^2+bq+c=r,$$
$$ar^2+br+c=p.$$
Adding the three equations and apply Vieta's formulas, I get $$a(p^2+q^2+r^2)+b(p+q+r)+3c=p+q+r\implies 14a+3c=0.$$
This is because $p^2+q^2+r^2=(p+q+r)^2-2(pq+qr+pr)=0^2-2(-7)=14,$ and $p+q+r=0$. However, I am unable to continue from here. Can I have some help?
Thanks!
|
Multiply each of your three equations by the root on its left hand side
$$ap^3+bp^2+cp=pq,$$
$$aq^3+bq^2+cq=qr,$$
$$ar^3+br^2+cr=rp.$$
Add the three equations again and apply Vieta's formulas, but also using $p^3 = 7p-7$ etc (they are roots!), so $$p^3+q^3+r^3 =-21$$ gives $-21a+14b=-7$, so $3a=2b+1$, which now gives all three coefficients in terms of one of them.
It is also true that if $f(x)=x^3-7x+7$ and $g(x)=ax^2+bx+c$, then $f(g(x))$ is zero at all the roots of $p$ (as $q$ just permutes these roots), so f(x) has to be a factor of the sixth order polynomial $h(x)=f(g(x))$. We can also find the other cubic factor of $h$: apart from $p$, $q$ and $r$, its other roots are the other values of $x$ that make $g(x)=p$ or $q$ or $r$. That is, $-\dfrac{b}{a}-p$, $-\dfrac{b}{a}-q$,$-\dfrac{b}{a}-r$. The cubic with these roots is (using $a^3$ as the coefficient of $x^3$) $$a^3x^3+3ba^2x^2+(3b^2a-7a^3)x+(b^3-7a^2b-7a^3).$$
This means that the constant coefficient of $h$ is $7(b^3-7a^2b-7a^3)$, from the two factors we have; but it is also, from the definition of $h$, $f(g(0))=c^3-7c+7.$ We know $b$ and $c$ in terms of $a$, from earlier calculations, so we have a cubic satisfied by $a$. This cubic turns out to be $$(a^2-9)(85a+27)=0,$$ so there are 3 possible values of $a$ and corresponding values of $b$, $c$.
|
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|
Bounds on $\max_i p_i$ in terms of $\sum_i p_i^2$ Suppose $(p_1,p_2,\ldots,p_d)$ is a discrete probability distribution over $d$ outcomes with $p_i\ge p_{i+1}$
Given the value of $\rho=\sum_{i=1}^d p_i^2$ (aka purity), what are the possible values of $p_1$?
I'm particularly interested in "large-d" regime.
Below is the graph of purity vs $p_1$ for distributions of the form $p_i\propto i^{-c}$. Does the set of feasible points $(\rho,p_1)$ concentrate around this curve when $d\to \infty$?
notebook
Animating the bounds provided in answer below:
|
The range of $p_1$ is given by $[\alpha, \beta]$ where
$$\beta = \frac{1}{d} + \sqrt{\rho - \frac{\rho}{d} - \frac{1}{d} + \frac{1}{d^2}}$$
and
$$\alpha = \frac{1}{m} + \frac{1}{m}\sqrt{\frac{m\rho - 1}{m-1}}$$
where $m = \lfloor 1/\rho\rfloor + 1$.
Moreover, $p_1 = \beta$ if $p_2 = p_3 = \cdots = p_d = \frac{1}{d} - \frac{1}{d}\sqrt{\frac{d\rho - 1}{d-1}}$;
$p_1 = \alpha$ if
$p_1 = p_2 = \cdots = p_{m-1} = \alpha$
and $p_m = \frac{1}{m} - \sqrt{\rho - \frac{\rho}{m} - \frac{1}{m} + \frac{1}{m^2}}$ and $p_{m+1} = \cdots = p_d = 0$.
Proof:
(1) Prove that $p_1 \le \beta$
We have
$$1 - p_1 = p_2 + p_3 + \cdots + p_d
\le \sqrt{(d-1)(p_2^2 + p_3^2 + \cdots + p_d^2)} = \sqrt{(d-1)(\rho - p_1^2)}$$
or
$$-dp_1^2 + 2p_1 + d\rho - \rho - 1 \ge 0$$
which results in
$$p_1 \le \beta.$$
(2) Prove that $p_1 \ge \alpha$
Let $y_1 = y_2 = \cdots = y_{m-1} = \alpha$ and $y_m = \frac{1}{m} - \sqrt{\rho - \frac{\rho}{m} - \frac{1}{m} + \frac{1}{m^2}}$ and $y_{m+1} = \cdots = y_d = 0$.
If $p_1 < \alpha$, then
$(p_1, p_2, \cdots, p_d)$ is majorized by
$(y_1, y_2, \cdots, y_d)$.
By Karamata's inequality,
we have
$$p_1^2 + p_2^2 + \cdots + p_d^2
< y_1^2 + y_2^2 + \cdots + y_d^2 = \rho.$$
This is impossible.
We are done.
|
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|
Determining whether $\frac{1}{2^{2}} + \frac{2}{3{^2}} + ... +\frac{n}{(n+1)^{2}}$ is a Cauchy sequence. My attempt so far: If $n \leq m$, then
$a_{m} - a_{n} = \frac{n+1}{(n+2)^2}+\frac{n+2}{(n+3)^2}+...+\frac{m}{(m+1)^2} < \frac{n+1}{(n+2)(n+1)}+\frac{n+2}{(n+3)(n+2)}+...+\frac{m}{(m+1)m} = \frac{1}{(n+2)}+\frac{1}{(n+3)}+...+\frac{1}{(m+1)} < \frac{1}{n}+\frac{1}{n}+...+\frac{1}{m} \leq \frac{1}{n}+\frac{1}{n}+...+\frac{1}{n}$.
A bit stuck here though. The hint in the book said it was not Cauchy.
Edit: If it's not Cauchy, then let me take a new approach..
$a_{m} - a_{n} = \frac{n+1}{(n+2)^2}+\frac{n+2}{(n+3)^2}+...+\frac{m}{(m+1)^2} \geq \frac{n+1}{(n+2)^2} + \frac{n+2}{(n+3)^2}+...+\frac{m}{(m+1)^2} \geq \frac{1}{(m+1)^2} + \frac{1}{(m+1)^2}+...+\frac{1}{(m+1)^2}$.
The $\frac{1}{(m+1)^2}$ occurs $m-n$ times here.
|
Try working with inequalities in the other directions:
$$a_{m} - a_{n} =\sum_{k=n+1}^m \frac{k}{(k+1)^2}>\sum_{k=n+1}^m \frac{k}{(k+k)^2}=\frac{1}{4}\sum_{k=n+1}^m \frac{1}{k}>\frac{1}{4}\sum_{k=1}^m \frac{1}{k}-\frac{n}{4}$$
Now what do you know about the sum $\sum_{k=1}^\infty \frac{1}{k}$?
|
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"timestamp": "2023-03-29T00:00:00",
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|
Inequality $\frac{a^3}{3ab^2+2c^3} +\frac{b^3}{3bc^2+2a^3} +\frac{c^3}{3ca^2+2b^3} \geq \frac{3}{5} $ I have trouble with solving this inequality:
Prove $\frac{a^3}{3ab^2+2c^3} +\frac{b^3}{3bc^2+2a^3} +\frac{c^3}{3ca^2+2b^3} \geq \frac{3}{5}$ for a,b,c>0.
Using Cauchy-Schwartz I got this: $\frac{a^3}{3b^3+c^3} +\frac{b^3}{3c^3+2a^3} +\frac{c^3}{3a^3+2b^3} \geq \frac{3}{5} $
Now I have to prove $\frac{a^3}{3ab^2+2c^3} +\frac{b^3}{3bc^2+2a^3} +\frac{c^3}{3ca^2+2b^3} \geq \frac{a^3}{3b^3+c^3} +\frac{b^3}{3c^3+2a^3} +\frac{c^3}{3a^3+2b^3}$
Any help is appreciated!
|
By the Cauchy-Schwartz inequality, we have:
$$(\frac{a^3}{3ab^2+2c^3}+\frac{b^3}{3bc^2+2a^3}+\frac{c^3}{3ca^2+2b^3})(a(3ab^2+2c^3)+b(3bc^2+2a^3)+c(3ca^2+2b^3)) \\ \geq (a^2+b^2+c^2)^2 \\ \implies \frac{a^3}{3ab^2+2c^3}+\frac{b^3}{3bc^2+2a^3}+\frac{c^3}{3ca^2+2b^3} \geq \frac{(a^2+b^2+c^2)^2}{3(a^2b^2+a^2c^2+b^2c^2)+2(ac^3+ba^3+cb^3)}.$$
Therefore, we are done if we show that:
$$\frac{(a^2+b^2+c^2)^2}{3(a^2b^2+a^2c^2+b^2c^2)+2(ac^3+ba^3+cb^3)} \geq \frac{3}{5},$$
or equivalently:
$$5(a^4+b^4+c^4)+(a^2b^2+a^2c^2+b^2c^2)\geq 6(ac^3+ba^3+cb^3).$$
However, we already have: $$a^4+a^2b^2 \geq 2ba^3 \\b^4+b^2c^2 \geq 2cb^3 \\c^4+a^2c^2 \geq 2ac^3 .$$
Thus, it is enough to show that:
$$a^4+b^4+c^4 \geq ac^3+ba^3+cb^3,$$
which is a consequence of the rearrangement inequality (note that $a,b,c$ and $a^3,b^3,c^3$ are sorted in the same way).
|
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|
Prove that there is no triple $(a,b,c)$ Prove that there is no triple $(a,b,c)$ such that
$$a,b,c \geq 0,ab+bc+ca=2+abc \textrm{ and } abc \in (1;10+6\sqrt{3}).$$
I transform the expression as follows
$$ab = \frac{m}{c},a + b = \frac{{2 + m - \frac{m}{c}}}{c}.$$
I expect the following inequality to be false and thus conclude that there is no satisfying set of numbers
$${\left( {a + b} \right)^2} \ge 4ab \Leftrightarrow {\left( {\frac{{2 + m - \frac{m}{c}}}{c}} \right)^2} - 4\frac{m}{c} \ge 0.$$
But it's quite complicated
|
$a,b,c \geq 0, \; ab+bc+ca=2+abc, \; abc \in (1,10+6\sqrt{3})$
By AM-GM inequality, $\frac{ab+bc+ca}{3} \geq (abc)^{\frac{2}{3}} \implies ab+bc+ca \geq 3(abc)^{\frac{2}{3}} \implies 2+abc \geq 3(abc)^{\frac{2}{3}}$
Consider $abc = x^3$, then $x^3-3x^2+2 \geq 0 \implies (x-1)(x-1+\sqrt{3})(x-1-\sqrt{3}) \geq 0$
It can be easily checked that this is only possible when $x \in [1-\sqrt{3},1] \cup [1+\sqrt{3}, \infty)$
So, $abc = x^3 \in [10-6\sqrt{3},1] \cup [10+6\sqrt{3}, \infty)$
Therefore, there is no such triple (a,b,c) such that $abc \in (1, 10+6\sqrt{3})$
|
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|
Hints on solving $x^{2x}-(x^2+x)x^x+x^3=0$
Solve this equation over $\mathbb{R}^+$: $x^{2x}-(x^2+x)x^x+x^3=0$
I’ve been trying to solve this exponential equation but can’t get the answer because normal substitution ($y=x^x$) isn’t working. Any tips/hints that don’t use logs?
(the section of the book I got this from is before the introduction of logarithms). Thanks.
|
There are only three obvious solutions
$x=-1,x=1$ and $x=2$
The other intervals are inconsistent with the sign of the simplified equation:
$x^{2x}-(x^2+x)x^x+x^3$ = 0 { expand }
$x^3+x^{2x}-x^{1+x}-x^{2+x}=0$ {a}
$1+x^{-3+2x}-x^{-2+x}-x^{-1+x}=0$ Divide {a} by $x^3$
Simplified to
$-x^{x-2}-x^{x-1}+x^{x^2-3}=-1$ (1)
for example if x>2
$-x^{x-2}-x^{x-1}+x^{x^2-3}$ is always positive sign where is inconsistent with the equation simplified (1):
Cause: when x>2
$-x^{x-2}-x^{x-1} < x^{x^2-3}$ So
$-x^{x-2}-x^{x-1}+x^{x^2-3}$ is always positive.
So No solution for x>2.
You can continued the demonstration with the intervals : 1<x<2 , 0<x<1, x=0 , -1<x<0 , x<-1 . All this intervals gives no solutions in Reals.
...
|
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|
I need to solve $yy'' +y^2 = C$ (constant) Is there an analytic solution? - if so please show me how to derive it.
If not, then boundary conditions for a numerical solution would be $y(0) = 35$, $y(400) = 8$.
I am pretty new at this . . . so apologies if further information/conditions/explanation required. I have access to Mathematica (though I am far from expert) but could not obtain either an analytic or a numerical solution.
|
I don't think there is a closed form solution that is not a constant, but you can still find a solution.
It is an ODE of the form $f\left( y\left( x \right),\, \frac{\operatorname{d}y\left( x \right)}{\operatorname{d}x},\, \frac{\operatorname{d}^{2}y\left( x \right)}{\operatorname{d}x^{2}} \right) = 0$ where you can reduce the order from $1$ to $2$ using the substitution $z\left( y\left( x \right) \right) := \frac{\operatorname{d}y\left( x \right)}{\operatorname{d}x}$:
$$
\begin{align*}
z\left( y\left( x \right) \right) &= \frac{\operatorname{d}y\left( x \right)}{\operatorname{d}x}\\
z\left( y\left( x \right) \right) \cdot \frac{\operatorname{d}z\left( y\left( x \right) \right)}{\operatorname{d}y\left( x \right)} &= \frac{\operatorname{d}^{2}y\left( x \right)}{\operatorname{d}^{2}x}\\
\end{align*}
$$
$$
\begin{align*}
y\left( x \right) \cdot \frac{\operatorname{d}^{2}y\left( x \right)}{\operatorname{d}x^{2}} + \left( y\left( x \right) \right)^{2} - c &= 0\\
y\left( x \right) \cdot z\left( y\left( x \right) \right) \cdot \frac{\operatorname{d}z\left( y\left( x \right) \right)}{\operatorname{d}y\left( x \right)} + \left( y\left( x \right) \right)^{2} - c &= 0\\
z\left( y\left( x \right) \right) \cdot \frac{\operatorname{d}z\left( y\left( x \right) \right)}{\operatorname{d}y\left( x \right)} &= \frac{c}{y\left( x \right)} - y\left( x \right)\\
\left( z\left( y\left( x \right) \right) \cdot \frac{\operatorname{d}z\left( y\left( x \right) \right)}{\operatorname{d}y\left( x \right)} \right)\, \operatorname{d}y\left( x \right) &= \left( \frac{c}{y\left( x \right)} - y\left( x \right) \right)\, \operatorname{d}y\left( x \right)\\
z\left( y\left( x \right) \right) \cdot \operatorname{d}z\left( y\left( x \right) \right) &= \left( \frac{c}{y\left( x \right)} - y\left( x \right) \right)\, \operatorname{d}y\left( x \right)\\
\int z\left( y\left( x \right) \right)\, \operatorname{d}z\left( y\left( x \right) \right) &= \int \frac{c}{y\left( x \right)} - y\left( x \right)\, \operatorname{d}y\left( x \right)\\
\frac{1}{2} \cdot \left( z\left( y\left( x \right) \right) \right)^{2} &= c \cdot \ln\left( y\left( x \right) \right) - \frac{1}{2} \cdot \left( y\left( x \right) \right)^{2} + k_{1}\\
\frac{1}{2} \cdot \left( \frac{\operatorname{d}y\left( x \right)}{\operatorname{d}x} \right)^{2} &= c \cdot \ln\left( y\left( x \right) \right) - \frac{1}{2} \cdot \left( y\left( x \right) \right)^{2} + k_{1}\\
\end{align*}
$$
You can further reduce the order of the ODE from $1$ to $0$ by solving for $\operatorname{d}x$:
$$
\begin{align*}
\frac{1}{2} \cdot \left( \frac{\operatorname{d}y\left( x \right)}{\operatorname{d}x} \right)^{2} &= c \cdot \ln\left( y\left( x \right) \right) - \frac{1}{2} \cdot \left( y\left( x \right) \right)^{2} + k_{1}\\
\frac{\operatorname{d}y\left( x \right)}{\operatorname{d}x} &= \pm\sqrt{2 \cdot c \cdot \ln\left( y\left( x \right) \right) - \left( y\left( x \right) \right)^{2} + 2 \cdot k_{1}}\\
1\, \operatorname{d}x &= \pm\frac{1}{\sqrt{2 \cdot c \cdot \ln\left( y\left( x \right) \right) - \left( y\left( x \right) \right)^{2} + 2 \cdot k_{1}}}\, \operatorname{d}y\left( x \right)\\
\int 1\, \operatorname{d}x &= \int \pm\frac{1}{\sqrt{2 \cdot c \cdot \ln\left( y\left( x \right) \right) - \left( y\left( x \right) \right)^{2} + 2 \cdot k_{1}}}\, \operatorname{d}y\left( x \right)\\
x + k_{2} &= \pm\int_{a}^{y\left( x \right)} \frac{1}{\sqrt{2 \cdot c \cdot \ln\left( x \right) - x^{2} + 2 \cdot k_{1}}}\, \operatorname{d}x\\
\end{align*}
$$
If we now define this integral as a special function $I_{c,\, k_{1},\, a}\left( y\left( x \right) \right) := \pm\int_{a}^{y\left( x \right)} \frac{1}{\sqrt{2 \cdot c \cdot \ln\left( x \right) - x^{2} + 2 \cdot k_{1}}}\, \operatorname{d}x$, then we can also use the equation for $y(x)$:
$$
\begin{align*}
x + k_{2} &= \pm\int_{a}^{y\left( x \right)} \frac{1}{\sqrt{2 \cdot c \cdot \ln\left( x \right) - x^{2} + 2 \cdot k_{1}}}\, \operatorname{d}x\\
x + k_{2} &= I_{c,\, k_{1},\, a}\left( y\left( x \right) \right)\\
y\left( x \right) &= I_{c,\, k_{1},\, a}^{-1}\left( x + k_{2} \right)\\
\end{align*}
$$
where $I_{c,\, k_{1},\, a}^{-1}\left( z \right)$ is the inverse if $I_{c,\, k_{1},\, a}\left( z \right)$ aka $I_{c,\, k_{1},\, a}^{-1}\left( I_{c,\, k_{1},\, a}\left( z \right) \right) = z$ or
$$
y\left( x \right) = \sqrt{c}\\
$$
Wolfram|Alpha finds the same solution as you can see here.
|
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|
Proof that $n^3+2n$ is divisible by $3$ I'm trying to freshen up for school in another month, and I'm struggling with the simplest of proofs!
Problem:
For any natural number $n , n^3 + 2n$ is divisible by $3.$
This makes sense
Proof:
Basis Step: If $n = 0,$ then $n^3 + 2n = 0^3 +$
$2 \times 0 = 0.$ So it is divisible by $3.$
Induction: Assume that for an arbitrary natural number $n$,
$n^3+ 2n$ is divisible by $3.$
Induction Hypothesis: To prove this for $n+1,$ first try to express $( n + 1 )^3 + 2( n + 1 )$ in terms of $n^3 + 2n$ and use
the induction hypothesis. Got it
$$( n + 1 )^3+ 2( n + 1 ) = ( n^3 + 3n^2+ 3n + 1 ) + ( 2n + 2 ) \{\text{Just some simplifying}\}$$
$$ = ( n^3 + 2n ) + ( 3n^2+ 3n + 3 ) \{\text{simplifying
and regrouping}\}$$
$$ = ( n^3 + 2n ) + 3( n^2 + n + 1 ) \{\text{factored out
the 3}\}$$
which is divisible by $3$, because $(n^3 + 2n )$ is divisible by $3$
by the induction hypothesis. What?
Can someone explain that last part? I don't see how you can claim $(n^3+ 2n ) + 3( n^2 + n + 1 )$ is divisible by $3.$
|
RTP : $a^3+3a$ is in the form of $3k$
Proof : $a^3+2a=
a^3-a+3a=
a(a^2-1)+3a=
a(a+1)(a-1)+3a$
We know that sum of three consecutive numbers is divisible by 3 $$3k+3a=3(k+a)=3k$$
|
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|
Modular exponentiation using Euler’s theorem How can I calculate $27^{41}\ \mathrm{mod}\ 77$ as simple as possible?
I already know that $27^{60}\ \mathrm{mod}\ 77 = 1$ because of Euler’s theorem:
$$ a^{\phi(n)}\ \mathrm{mod}\ n = 1 $$
and
$$ \phi(77) = \phi(7 \cdot 11) = (7-1) \cdot (11-1) = 60 $$
I also know from using modular exponentiation that $27^{10} \mathrm{mod}\ 77 = 1$ and thus
$$ 27^{41}\ \mathrm{mod}\ 77 = 27^{10} \cdot 27^{10} \cdot 27^{10} \cdot 27^{10} \cdot 27^{1}\ \mathrm{mod}\ 77 = 1 \cdot 1 \cdot 1 \cdot 1 \cdot 27 = 27 $$
But can I derive the result of $27^{41}\ \mathrm{mod}\ 77$ using $27^{60}\ \mathrm{mod}\ 77 = 1$ somehow?
|
You can use exponentiation by squaring (all operations are modulo 77):
$27^{41} = 27^{32+8+1} = 27 \cdot 27^8 \cdot (27^8)^4 = (*)$
$\big[ 27^8 = ((27^2)^2)^2 = (36^2)^2 = 64^2 = 15 \big]$
$(*) = 27 \cdot 15 \cdot 15^4 = 27 \cdot 15 \cdot (15^2)^2 = 27 \cdot 15 \cdot 71^2 = 27 \cdot 15 \cdot 36 = 27$
This uses only 7 multiplications instead of 41.
|
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|
Proof that $1+2+3+4+\cdots+n = \frac{n\times(n+1)}2$ Why is $1+2+3+4+\ldots+n = \dfrac{n\times(n+1)}2$ $\space$ ?
|
If you knew of the geometric series, you would know that
$$\frac{1-r^{n+1}}{1-r}=1+r+r^2+r^3+\dots+r^n$$
If we differentiate both sides, we have
$$\frac{nr^{n+2}-(n+1)r^{n+1}+r}{(1-r)^2}=1+2r+3r^2+\dots+nr^{n-1}$$
Letting $r\to1$ and applying L'Hospital's rule on the fraction, we end up with
$$\frac{n(n+1)}2=1+2+3+\dots+n$$
|
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Help in getting the Quadratic Equation I'm starting a chapter on Functions and they had the steps shown to reach the p-q equation.
$$ x_{1,2} = -\frac{p}{2} \pm\sqrt{\left(\frac{p}{2}\right)^2 - q}$$
So I wanted to do the same with the Quadratic Equation. I'm using the base linear equation
$$ax+by+c = 0.$$
The solution I have so far is as follows:
$$x^2 + \frac{b}{a}x + \frac{c}{a}= 0$$
$$x^2 + \frac{b}{a}x = -\frac{c}{a}$$
$$x^2 + \frac{b}{a}x + \left(\frac{b}{2a}\right)^2 = -\frac{c}{a} + \left(\frac{b}{2a}\right)^2$$
$$\left(x + \frac{b}{2a}\right)^2 = \left(\frac{b}{2a}\right)^2 - \frac{c}{a}$$
$$\left(x + \frac{b}{2a}\right) = \pm\sqrt{\left(\frac{b}{2a}\right)^2 - \frac{c}{a}}$$
$$x = -\frac{b}{2a} \pm\sqrt{\left(\frac{b}{2a}\right)^2 - \frac{c}{a}}$$
My problem comes from trying to solve the insides of the square root:
$$\sqrt{(\frac{b}{2a})^2 - \frac{c}{a}} = \sqrt{\frac{b^2}{4a2} - \frac{c}{a}}$$
$$= \sqrt{\frac{b^2}{4a^2} - \frac{c}{a} \left(\frac{4a}{4a}\right)} = \sqrt{\frac{b^2 - 4ac}{4a^2}}$$
$$= \sqrt{\frac{b^2 - 4ac}{\left(2a\right)^2}}$$
Then:
$$x_{1,2} = \frac{-\left(\frac{b}{2a}\right) \pm\sqrt{b^2 -4ac}}{2a}$$
but there is still the problem of the -(b/2a) outside of the sqrt.
What am I doing wrong? Also, Tex is awesome; is there a better way to do the 1,2 subscripts than _1,_2?
Answer:
Instead of
$$x_{1,2} = \frac{-\left(\frac{b}{2a}\right) \pm\sqrt{b^2 -4ac}}{2a}.$$
The solution goes
$$\frac{-b}{2a}\pm \sqrt{\frac{b^{2}-4ac}{4a^{2}}}=\frac{-b}{2a}\pm \frac{\sqrt{b^{2}-4ac}}{2a}$$
$$= \frac{-b \pm\sqrt{b^2 - 4ac}}{2a}$$
|
Sorry I don't know how to do tex on websites, but I'm trying to learn.
You just made a small mistake on the final step. In the second to last step, we actually have our full equation as:
$$\frac{-b}{2a}\pm \sqrt{\frac{b^{2}-4ac}{4a^{2}}}=\frac{-b}{2a}\pm \frac{\sqrt{b^{2}-4ac}}{2a}$$
Now we can collect the common factor of $\frac1{2a}$ and get:
$$\frac{-b\pm \sqrt{b^{2}-4ac}}{2a}$$
Does that make sense?
|
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|
Finding the $N$-th derivative of $f(x)=\frac {x} {x^2-1}$ I'm practicing some problems from past exams and found this one:
Find the n-th derivative of this function:
$$f(x)=\frac {x} {x^2-1}$$
I have no idea how to start solving this problems. Is there any theorem for finding nth derivative?
|
related problem: (I). I am referring you to this book where you will find the complete answer to the problem of finding the nth derivative of rational polynomials and other classes of functions.
Added Some people suggested the post should be self-contained. Here is the example I am referring to
\begin{equation}
f(x) = \frac{ x^2 + x - 1 } { x^4 + x^3 - x - 1 }\,.
\end{equation}
The roots of the denominator are
\begin{equation}\nonumber
x = -1, 1, -\frac{1}{2} + \frac{ i \sqrt{3}}{2}, -\frac{1}{2} -
\frac{ i \sqrt{3}}{2}\,.
\end{equation}
Expressing the function in the partial fraction form
\begin{equation}\nonumber
f(x) = \frac{ x^2 + x - 1 } { x^4 + x^3 - x - 1 } =
\frac{a_1}{x-1} + \frac{a_2}{x+1} + \frac{a_3}{ x + \frac{1 -
i\sqrt{3}}{2}} + \frac{a_4}{ x + \frac{1 + i\sqrt{3}}{2}}\,.
\end{equation}
To find $a_i$, one can use the formula
$$ a_{ i_{\ell}} = \mathrm{Res}( ( x - \alpha_{\ell} )^{{i_\ell} - 1
} f(x)\,, x = \alpha_\ell) \,,\quad i_\ell = 1,2,...,m_\ell, $$
where $ \alpha_{\ell} $ are the roots of the denominator and the formula takes care of the multiplicities $m_{\ell}$ of the roots in case there is any. For instance,
\begin{align}\nonumber
a_1 & = \mathrm{Res}\left( \frac{ x^2 + x - 1 } { x^4 + x^3 - x -
1 }, x = 1 \right)&
\\ \nonumber
\\ \nonumber
& = \lim_{x = 1} (x-1)\frac{ x^2 + x - 1 } { x^4 + x^3 - x - 1 }&
\\ \nonumber
\\ \nonumber
& = \frac{1}{6}\,,&
\end{align}
Once we have the function in the partial fraction form, the $nth$ derivative
can be found directly by using the formula
\begin{equation}
D^n ( x - \alpha )^{(-k)} =
{\frac { \left( -1 \right) ^{n}\Gamma \left( n+k \right) }{\Gamma
\left( k \right) }}(x-\alpha)^{-k-n}.
\end{equation}
Following the above techniques, the final answer is
$$ \begin{align}\nonumber
f^{(n)}(x) = (-1)^n \Gamma(n+1) \, &\left(
\frac{1}{2\,(x+1)^{n+1}} + \frac{1}{6\,(x-1)^{n+1}} \right. &
\\ \nonumber
\\ \nonumber
& \left. + \frac{\frac{ - 1 - i\sqrt{3}}{3}}{ \left(x + \frac{1 -
i\sqrt{3}}{2}\right)^{n+1} } + \frac{\frac{ - 1 + i\sqrt{3}}{3}}{
\left(x + \frac{1 + i\sqrt{3}}{2}\right)^{n+1}} \right) \,. &
\end{align}.$$
|
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|
Find polynomials such that $(x-16)p(2x)=16(x-1)p(x)$ Find all polynomials $p(x)$ such that for all $x$, we have $$(x-16)p(2x)=16(x-1)p(x)$$
I tried working out with replacing $x$ by $\frac{x}{2},\frac{x}{4},\cdots$, to have $p(2x) \to p(0)$ but then factor terms seems to create a problem.
Link: http://web.mit.edu/rwbarton/Public/func-eq.pdf
|
Putting $x=1$ and $x=16$ we see that $p(2)=p(16) = 0$.
Let $p(x) = (x-2)(x-16)g(x)$.
Thus we see that
$$(x-16)(2x-2)(2x-16)g(2x) = 16(x-1)(x-2)(x-16)g(x)$$
Thus $$ (x-8)g(2x) = 4(x-2)g(x)$$
We now see that $g(4) = g(8) = 0$
Thus $g(x) = (x-4)(x-8)h(x)$
Thus
$$ (x-8)(2x-4)(2x-8)h(2x) = 4(x-2)(x-4)(x-8)h(x)$$
i.e
$$ h(2x) = h(x) $$
This implies that $h(x)$ is a constant (as otherwise the non-identically zero polynomial $h(2x) - h(x)$ will have an infinite number of roots.)
Thus
$$p(x) = C(x-2)(x-4)(x-8)(x-16)$$
|
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|
Finding a Function That Approaches Another Function One of my math professors gave me the following challenge. It isn't graded, it's just for fun.
Consider the function:
\begin{equation*}
f_n(x)=x+3^3x^3+5^3x^5+...+(2n-1)^3x^{2n-1},~x \in (0, 1).
\end{equation*}
I want to find which of the following functions $f_n$ is getting close to as $n$ gets larger:
$ \displaystyle a)\frac{x(x+1)(x^4+22x^2+1)}{(x-1)^4(x+1)^4}$ $\displaystyle b) \frac{x(x^2+1)(x^4+22x^2+1)}{(x-1)^4(x+1)^4}$
$\displaystyle c) \frac{x^2(x+1)(x^4+22x^2+1)}{(x-1)^4(x+1)^4}$ $\displaystyle d) \frac{x^2(x^2+1)(x^4+22x^2+1)}{(x-1)^4(x+1)^4}$
Based on some tests i ran in mathematica by giving $n$ and $x$ values, it looks like $b)$ is the answer, but I am not sure. Can anyone confirm or deny this, and show how one might find the right answer, either with pen and paper or by using mathematica or maple or some other software?
|
Here's yet another way to verify that it's (b). Let Mathematica do partial fractions decomposition:
Apart[x(x^2+1)(x^4+22x^2+1)/(x^2-1)^4]
This gives $\frac{1}{2 (x+1)}-\frac{7}{2(x+1)^2}+\frac{6}{(x+1)^3}-\frac{3}{(x+1)^4}+\frac{1}{2 (x-1)}+\frac{7}{2(x-1)^2}+\frac{6}{(x-1)^3}+\frac{3}{(x-1)^4}$, which can be expanded into power series using the geometric series and its derivatives.
Of course this is ugly, and requires you to know the answer in advance. Moron's answer is much better. ;-)
|
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|
How to prove $\cos \frac{2\pi }{5}=\frac{-1+\sqrt{5}}{4}$? I would like to find the apothem of a regular pentagon. It follows from
$$\cos \dfrac{2\pi }{5}=\dfrac{-1+\sqrt{5}}{4}.$$
But how can this be proved (geometrically or trigonometrically)?
|
By my post, $$
\cos \frac{\pi}{5}-\cos \frac{2 \pi}{5}=\frac{1}{2}
$$
By $\cos (\pi-\dfrac{\pi}{5} )=-\cos \dfrac{\pi}{5}, $ we have
$$
-\cos \frac{4 \pi}{5}-\cos \frac{2 \pi}{5}=\frac{1}{2}
$$
By double-angle formula, $$
-\left(2 \cos ^{2} \frac{2 \pi}{5}-1\right)-\cos \frac{2 \pi}{5}=\frac{1}{2} \Leftrightarrow
4 \cos ^{2} \frac{2 \pi}{5}+2 \cos \frac{2 \pi}{5}-1=0
$$
Using quadratic formula gives $$
\boxed{\cos \frac{2\pi}{5}=\frac{-1+\sqrt{5}}{4}}
$$
By the way, $$\cos \frac{\pi}{5}= \cos \frac{2\pi}{5}+\frac{1}{2}= \frac{1+\sqrt{5}}{4}
$$
|
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|
Prove that $\prod_{k=1}^{n-1}\sin\frac{k \pi}{n} = \frac{n}{2^{n-1}}$ Using $\text{n}^{\text{th}}$ root of unity
$$\large\left(e^{\frac{2ki\pi}{n}}\right)^{n} = 1$$
Prove that
$$\prod_{k=1}^{n-1}\sin\frac{k \pi}{n} = \frac{n}{2^{n-1}}$$
|
$$ \begin{align*}
P & = \prod_{k=1}^{n-1}\sin(k\pi/n) \\
& = (2i)^{1-n}\prod_{k=1}^{n-1}(e^{ik\pi/n}-e^{-ik\pi/n}) \\
& = (2i)^{1-n} e^{-i \frac{n(n-1)}{2}\frac{\pi}{n}} \prod_{k=1}^{n-1}(e^{2ik\pi/n}-1) \\
& = (-2)^{1-n}\prod_{k=1}^{n-1}(\xi^k-1) \\
& = 2^{1-n}\prod_{k=1}^{n-1}(1-\xi^k) \\
\end{align*}
$$
where $\xi=e^{2i\pi/n}$.
Now note that $x^n-1=(x-1)\sum_{k=0}^{n-1}x^k$ and $x^n-1=\prod_{k=0}^{n-1} (x-\xi^k)$.
Cancelling $(x-1)$ we have $\prod_{k=1}^{n-1} (x-\xi^k) =\sum_{k=0}^{n-1}x^k$.
Substituting $x=1$ we have $\prod_{k=1}^{n-1} (1-\xi^k)=n$. $$ \therefore \boxed{P=n2^{1-n}}$$
Edit:
In order to note that $x^n-1=\prod_{k=0}^{n-1} (x-\xi^k)$, note that $1,\xi,\dots,\xi^{n-1}$ are roots of $x^n-1$. Therefore by polynomial reminder theorem we have $x^n-1=Q(x) \prod_{k=0}^{n-1} (x-\xi^k)$. Comparing degrees we find $Q(x)$ has degree $0$. Comparing highest coefficients we conclude $Q(x)=1$.
Edit:
We may instead use the identity $\left\lvert 1 - e^{2ik\pi/n} \right\rvert = 2\sin(k\pi/n), k = 1, ..., n - 1,$ to establish immediately that $P \equiv \prod_{k=1}^{n-1}\sin(k\pi/n)= 2^{1-n}\prod_{k=1}^{n-1}\left\lvert 1 - e^{2ik\pi/n} \right\rvert = 2^{1 - n}\left\lvert \prod_{k=1}^{n-1}(1 - e^{2ik\pi/n}) \right\rvert$, and continue by applying the foregoing logic to the product to obtain $P=n2^{1-n}$.
|
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How to solve this equation for $x$? How do I solve for $x$ from this equation?
$$ -\frac{1}{x^2} + \frac{9}{(4-x-y)^2} = 0.$$
I need to get this into $x=$"blah"?
|
If you add $\frac{1}{x^{2}}$ to both sides of your equation
$$-\frac{1}{x^{2}}+\frac{9}{\left( 4-x-y\right) ^{2}}=0\qquad (1)$$
you get this equivalent one (provided that $\frac{1}{x^{2}}$ is finite,
i.e $x\neq 0$)
$$\frac{1}{x^{2}}=\frac{9}{\left( 4-x-y\right) ^{2}}.\qquad (2)$$
It is satisfied if the square root of one side is equal or
symmetric to the square root of the other side:
$$\frac{1}{x}=\pm \frac{3}{4-x-y}.\qquad (3)$$
Equation $(3)$ is equivalent to
$$3x=\pm \left( 4-x-y\right) \qquad (4)$$
provided that $x\neq 0$ and $4-x-y\neq 0$.
The equation $(4)$ represents two equations. One is
$$3x=4-x-y,\qquad (5)$$
which is equivalent to
$$4x=4-y\Leftrightarrow x=1-\frac{1}{4}y\qquad (6)$$
and the other
$$3x=-\left( 4-x-y\right), \qquad (7)$$
is equivalent to
$$3x=-4+x+y\Leftrightarrow x=-2+\frac{1}{2}y.\qquad (9)$$
Thus $(1)$ is equivalent to
$$x=1-\frac{1}{4}y\qquad (10)$$
or
$$x=-2+\frac{1}{2}y,\qquad (11)$$
provided that $y\neq 4$ because the conditions $x\neq 0$ and $4-x-y\neq 0$
correspond to
$$x\neq 0\Leftrightarrow 1-\frac{1}{4}y\neq 0\Leftrightarrow y\neq 4$$
$$x\neq 0\Leftrightarrow -2+\frac{1}{2}y\neq 0\Leftrightarrow y\neq 4$$
and
$$4-x-y\neq 0\Leftrightarrow 4-\left( 1-\frac{1}{4}y\right) -y\neq
0\Leftrightarrow y\neq 4$$
$$4-x-y\neq 0\Leftrightarrow 4-\left( -2+\frac{1}{2}y\right) -y\neq
0\Leftrightarrow y\neq 4.$$
|
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|
Expressions of $\sin \frac{A}{2}$ and $\cos \frac{A}{2}$ in terms of $\sin A$ I am trying to understand the interpretation of $\sin \frac{A}{2}$ and $\cos \frac{A}{2}$ in terms of $\sin A$ from my book, here is how it is given :
We have $ \bigl( \sin \frac{A}{2} + \cos \frac{A}{2} \bigr)^{2} = 1 + \sin A $ and $ \bigl( \sin \frac{A}{2} - \cos \frac{A}{2} \bigr)^{2} = 1 - \sin A $
By adding and subtracting we have $ 2 \cdot \sin \frac{A}{2} = \pm \sqrt{ 1 + \sin A } \pm \sqrt{ 1 - \sin A } $ ---- (1) and $ 2 \cdot \cos \frac{A}{2} = \mp \sqrt{ 1 + \sin A } \mp \sqrt{ 1 - \sin A } $ ---(2)
I have understood upto this far well,
Now they have broke the them into quadrants :
In 1st quadrant :
$$ 2 \cdot \sin \frac{A}{2} = \sqrt{ 1 + \sin A } - \sqrt{ 1 - \sin A } $$
$$ 2 \cdot \cos \frac{A}{2} = \sqrt{ 1 + \sin A } + \sqrt{ 1 - \sin A } $$
In 2nd quadrant :
$$ 2 \cdot \sin \frac{A}{2} = \sqrt{ 1 + \sin A } + \sqrt{ 1 - \sin A } $$
$$ 2 \cdot \cos \frac{A}{2} = \sqrt{ 1 + \sin A } - \sqrt{ 1 - \sin A } $$
In 3rd quadrant :
$$ 2 \cdot \sin \frac{A}{2} = \sqrt{ 1 + \sin A } - \sqrt{ 1 - \sin A } $$
$$ 2 \cdot \cos \frac{A}{2} = \sqrt{ 1 + \sin A } + \sqrt{ 1 - \sin A } $$
In 4th quadrant :
$$ 2 \cdot \sin \frac{A}{2} = - \sqrt{ 1 + \sin A } - \sqrt{ 1 - \sin A } $$
$$ 2 \cdot \cos \frac{A}{2} = - \sqrt{ 1 + \sin A } + \sqrt{ 1 - \sin A } $$
Now, In knew the ALL-SINE-TAN-COSINE rule but still I am not able to figure out how the respective signs are computed in these (above) cases.
|
The easiest way of computing the signs is to make them match; we know that sin x > 0 if 0 < x < π and that cos x > 0 if -π/2 < x < π/2. Knowing whether sin A is greater than 0 or less than zero tells you whether $\sqrt{1-\mathrm{sin} A}$ is greater or less than $\sqrt{1+\mathrm{sin} A}$; that in turn lets you figure out what the overall sign on all of the right-hand terms is, and each quadrant corresponds to one of the four positive/negative pairs on the right-hand terms.
|
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|
Evaluate $\int \frac{1}{\sin x\cos x} dx $ Question: How to evaluate $\displaystyle \int \frac{1}{\sin x\cos x} dx $
I know that the correct answer can be obtained by doing:
$\displaystyle\frac{1}{\sin x\cos x} = \frac{\sin^2(x)}{\sin x\cos x}+\frac{\cos^2(x)}{\sin x\cos x} = \tan(x) + \cot(x)$ and integrating.
However, doing the following gets a completely different answer:
\begin{eqnarray*}
\int \frac{1}{\sin x\cos x} dx
&=&\int \frac{\sin x}{\sin^2(x)\cos x} dx\\
&=&\int \frac{\sin x}{(1-\cos^2(x))\cos x} dx.
\end{eqnarray*}
let $u=\cos x, du=-\sin x dx$; then
\begin{eqnarray*}
\int \frac{1}{\sin x\cos x} dx
&=&\int \frac{-1}{(1-u^2)u} du\\
&=&\int \frac{-1}{(1+u)(1-u)u}du\\
&=&\int \left(\frac{-1}{u} - \frac{1}{2(1-u)} + \frac{1}{2(1+u)}\right) du\\
&=&-\ln|\cos x|+\frac{1}{2}\ln|1-\cos x|+\frac{1}{2}\ln|1+\cos x|+C
\end{eqnarray*}
I tested both results in Mathematica, and the first method gets the correct answer, but the second method doesn't. Is there any reason why this second method doesn't work?
|
The second method gives the same answer as the first. By the first method, the answer you get is $-\log(\cos x) + \log(\sin x)$. The first term is the same as what you get by the second method.
What you need to show is that $\log(\sin x) = \frac{1}{2}\log(1-\cos x) + \frac{1}{2}\log(1+\cos x)$.
\begin{equation}
\begin{split}
\frac{1}{2}\log(1-\cos x) + \frac{1}{2}\log(1+\cos x) &= \frac{1}{2}\left( \log(2 \sin^2 \frac{x}{2}) + \log(2 \cos^2 \frac{x}{2})\right)\\ & = \log(2 \sin \frac{x}{2} \cos \frac{x}{2})
\end{split}
\end{equation}
|
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|
Finding roots of the fourth degree polynomial: $2x^4 + 3x^3 - 11x^2 - 9x + 15 = 0$. My son is taking algebra and I'm a little rusty. Not using a calculator or the internet, how would you find the roots of $2x^4 + 3x^3 - 11x^2 - 9x + 15 = 0$. Please list step by step. Thanks, Brian
|
Guessing one root sometimes opens up the whole equation for you.
First notice that $\displaystyle x=1$ gives 0. So $\displaystyle x-1$ is a factor.
Next, rewrite as
$\displaystyle 2x^4 - 2x^3 + 5x^3 - 5x^2 -6x^2 + 6x - 15x + 15$
This is to try and get $x-1$ as a factor.
This gives us
$\displaystyle 2x^3(x-1) + 5x^2(x-1) - 6x(x-1) - 15(x-1) = (x-1)(2x^3 + 5x^2 - 6x - 15)$
Now notice that $\displaystyle 2x^3 - 6x = 2x(x^2-3)$ and $5x^2 - 15 = 5(x^2 - 3)$
Thus
$\displaystyle (x-1)(2x^3 + 5x^2 - 6x - 15) = (x-1)(2x(x^2 - 3) + 5(x^2 - 3)) = (x-1)(x^2-3)(2x+5)$
and so the roots are $\displaystyle 1, \pm\sqrt{3}, -\frac{5}{2}$
|
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|
Factorize polynomial I am trying to factorize $-6x^5+15x^4-30x^2+30x-13$ for hours:( Could someone help me? I tried making a system of equations from $(Ax^3 + Bx^2 + Cx + D) (Ex^2 + Fx + G)$ but it is a nightmare:(
In case you are interested, the system is:
$AE = -6$
$AF + BE = 15$
$AG + BF + CE = 0$
$BG + CF + DE = -30$
$CG + DF = 30$
$DG = -13$
Thanks in advance!
Edit: The original task is to draw the following function: $\ln \dfrac{x^2 - 3x + 2}{x^2 + 1}$. The polynomial above is the numerator of the second derivative of the function.
Best regards,
Petar
|
First note that the domain over which the function makes sense in real variables is when $x^2 - 3x + 2 > 0$ i.e. when $(x-1)(x-2) > 0$ i.e. when $x > 2$ or $x < 1$. Now the way out is to rewrite
$\log \frac{x^2 - 3x + 2}{x^2 + 1}$ as $\log (x-1) + \log (x-2) - \log (x^2+1)$ if $x > 2$
and
$\log \frac{x^2 - 3x + 2}{x^2 + 1}$ as $\log (1-x) + \log (2-x) - \log (x^2+1)$ if $x < 1$
The individual plots for the three terms can be drawn trivially. All you need to do now is to superpose these three together.
You could do further analysis which will help you with your plotting.
For instance, when $\frac{x^2-3x+2}{x^2+1} \geq 1$, the function will be non-negative and when $\frac{x^2-3x+2}{x^2+1} < 1$, the function will be negative.
So the function will be non-negative when $\frac{x^2-3x+2}{x^2+1} \geq 1 \Rightarrow -3x+2 \geq 1 \Rightarrow x \leq \frac{1}{3}$ and will be negative when $x > \frac{1}{3}$. The zero crossing is at $x= \frac{1}{3}$.
As mentioned previously the function is not-defined for $1 \leq x \leq 2$. And the function tends to $-\infty$ as $x \rightarrow 2^+$ or as $x \rightarrow 1^{-}$.
Further as $x \rightarrow \pm \infty$, the function tends to $0$.
The derivative when $x>2$ is $-\frac{2x}{x^2+1} + \frac{1}{x-2} + \frac{1}{x-1} > - \frac{2x}{x^2} + \frac{1}{x} + \frac{1}{x} = 0$ when $x>2$. So in the domain $(2,\infty)$ we have the function to be increasing and $f(2^+) = - \infty$ and $\displaystyle \lim_{x \rightarrow \infty}f(x) = 0$. Hence, $f(x) < 0$, $\forall x \in (2, \infty)$.
So we have $f(x)$ negative and it increases from $-\infty$ to $0$ in the domain $(2,\infty)$.
In the domain $(-\infty,1)$, we know that $\displaystyle \lim_{x \rightarrow -\infty}f(x) = 0$ and $f(1^-) = -\infty$ and we know that there is only one zero crossing, which means there has to be at least one maximum.
Setting the derivative to zero, we get a quadratic in $x$ which gives $x = \frac{1 \pm \sqrt{10}}{3}$. The positive root falls in $[1,2]$ and hence can be ruled out. The negative root is where the maximum occurs. And there is only one maximum.
Further, $x=0$ gives $f(0) = \log (2) > 0$ as expected since $f(\frac{1}{3}) = 0$.
So the summary is,
The function increases from $0$ to $f(\frac{1 - \sqrt{10}}{3})$ in the domain $(-\infty,\frac{1 - \sqrt{10}}{3})$.
The function decreases from $f(\frac{1 - \sqrt{10}}{3})$ to $-\infty$ in the domain $[f(\frac{1 - \sqrt{10}}{3}),1)$ with zero crossing at $x = \frac{1}{3}$.
The function is not defined in the domain $[1,2]$.
The function $f(x)$ negative and it increases from $-\infty$ to $0$ in the domain $(2,\infty)$.
This information should be enough to help you make a sketch of the plot.
|
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Proving $(2n+1) + (2n+3) + \cdots + (4n-1) = 3n^{2}$ for all positive integers $n$
Prove $(2n+1) + (2n+3) + (2n+5) +
\cdots + (4n-1) = 3n^{2}$ for all
positive integers $n$.
So the provided solution avoids induction and makes use of the fact that $1 + 3 + 5 + \cdots + (2n-1) = n^{2}$ however I cannot understand the first step: $(2n+1) + (2n+3) + (2n+5) + \cdots + (4n-1) = (1 + 3 + 5 + \cdots + (4n-1)) -(1 + 3 + 5 + \cdots + (2n-1))$. Once that has been established I can follow the rest, but I was hoping someone could help me understand why $(1 + 3 + 5 + \cdots + (4n-1))$ which already looks like less than the LHS can be made equal to the LHS by subtracting a positive number.
Additionally, I wanted to prove the equality using induction, but had trouble with that as well. I think I am thrown off by the last term $4n-1$. Even the initial case of $n=1$ is not totally clear to me: does it hold because $2(1)+1 = 3(1)^{2}$ or is it because $4(1)-1 = 3(1)^{2}$?
Either way, my approach was to replace every $n$ on the LHS with $n+1$ which resulted in: $$(2(n+1)+1) + (2(n+1)+3)+ \cdots + (4(n+1)-1)$$ and I am not sure what the second to last term in the sequence would be... I tried simplifying anyway $$(2n+3) + (2n+5) + \cdots + (4n+3)$$ and I this point I thought I could subtract $(2n+1)$ from both sides of the induction assumption resulting in $(2n+3) + (2n+5) + \cdots ? = 3n^{2} - (2n+1) - (4n-1)$ substituting this yields: $$3n^{2} - 2n - 1 - 4n + 1 + 4n + 3 = 3n^{2} + 6n + 3 - 8n + 1 = 3(n+1)^{2} - 8n + 1$$ but I guess I don't want the $-8n + 1$... I also tried substituting $3n^{2} - (2n+1)$ without that last term being subtracted, but that did not work out either. If anyone can help me understand how to do this properly I would really appreciate it. Thanks!
|
Your initial problem seems to be seeing
$$1 + 3 + 5 + \cdots + (4n-1) = 1 + 3 + 5 + \cdots + (2n-1) + (2n+1) + (2n+3) + \cdots + (4n-1)$$
so the LHS is $(2n)^2 = 4n^2$ and the left half of the RHS is $n^2$, leaving $3n^2$ for the right half of the RHS.
For induction, you need to start with one term where $(2 \times 1 +1) = 3 \times 1^2$, and then given $(2n+1) + (2n+3) + \cdots + (4n-1) = 3n^2$ that you can prove $(2n+3) + \cdots + (4n-1) + (4n+1) + (4n+3) = 3(n+1)^2$, i.e. that $3n^2 - (2n+1) + (4n+1) + (4n+3) = 3(n+1)^2$, which is not difficult.
|
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How to solve $x^3 + 2x + 2 \equiv 0 \pmod{25}$? My attempt was:
$x^3 + 2x + 2 \equiv 0 \pmod{25}$
By inspection, we see that $x \equiv 1 \pmod{5}$. is a solution of $x^3 + 2x + 2 \equiv 0 \pmod{5}$. Let $x = 1 + 5k$, then we have:
$$(1 + 5k)^3 + 2(1 + 5k) + 2 \equiv 0 \pmod{25}$$
$$\Leftrightarrow 125k^3 + 75k^2 + 25k + 5 \equiv 0 \pmod{25}$$
$$\Leftrightarrow 5 \equiv 0 \pmod{25}$$
And I was stuck here :( because k was completely cancelled out, how can we find solution for this equation?
Thanks,
|
This shows, that there is no modular zero for $x^3+2x+2\equiv 0 \; (\text{mod}\; 25)$ of the form $x = 1 + 5k$, since clearly $ 5 \not\equiv 0 \; (\text{mod}\; 25)$.
Now go ahead and check $x = 3 + 5k$, because $x \equiv 3 \; (\text{mod}\; 5)$ is the other solution to $x^3+2x+2\equiv 0 \; (\text{mod}\; 5)$.
|
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|
Factorization of sum of two square The difference is $a^2 - b^2 = (a - b).(a + b)$
But what about when I have $a^{25} + 1$ ? According to wolfram alpha, the alternate form is:
*
*$(a+1) (a^4 -a^3 + a^2 -a + 1)( a^{20} - a^{15} + a^{10} -a^5 +1)$
However, the square root of 25 is a rational number 5.
But If I had 50, where the square root of 50 is irrational ?
*
*$(a^2 + 1) (a^8 -a^6 +a^4 -a^2 +1) (a^{40} -a^{30} +a^{20} -a^{10} +1 )$
In fact, I'm just wondering and trying to find out patterns, I'm very curious about that, since I haven't found anything related, only the alternate forms generated by wolfram. My question is how and what mathematical algorithm they used to find $k$ forms for $a^n + 1$ ?
Thanks in advance.
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If $n$ is odd then the polynomial $p(x)=x^n+1$ has a zero at $x=-1$, $p(-1)=(-1)^n+1=-1+1=0$. By the factor theorem, $x+1$ is therefore a factor of $p(x)$. Using polynomial long division (or synthetic division) you can show that $x^n+1=(x+1)(x^{n-1}-x^{n-2}+x^{n-3}-\cdots-x+1)$.
For example, consider $p(x)=x^5+1$. By the above, since $5$ is odd, $p(x)=(x+1)(x^4-x^3+x^2-x+1)$. Next notice that
$\begin{align*}x^{25}+1&=(x^5)^5+1=p(x^5)\\
&=(x^5+1)((x^5)^4-(x^5)^3+(x^5)^2-x^5+1)\\
&=(x+1)(x^4-x^3+x^2-x+1)(x^{20}-x^{15}+x^{10}-x^5+1).\end{align*}$
Finally, notice that $x^{50}+1=(x^2)^{25}+1=p((x^2)^5)$, so you can just substitute $x^2$ for $x$ in the factorization of $x^{25}+1$.
|
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exponential equation $$\sqrt{(5+2\sqrt6)^x}+\sqrt{(5-2\sqrt6)^x}=10$$
So I have squared both sides and got:
$$(5-2\sqrt6)^x+(5+2\sqrt6)^x+2\sqrt{1^x}=100$$
$$(5-2\sqrt6)^x+(5+2\sqrt6)^x+2=100$$
I don't know what to do now
|
$(\sqrt{(5+2\sqrt6)^x}+(\sqrt{(5-2\sqrt6)^x}=10$
We have, $5+2\sqrt6 = (\sqrt{2} + \sqrt{3})^2$ and $5 - 2\sqrt6 = (\sqrt3-\sqrt2)^2$
The equation $\Leftrightarrow \left(\sqrt{2} + \sqrt{3}\right)^x + \left(\sqrt3-\sqrt2\right)^x = 10$
Let $$t = \left(\sqrt3+\sqrt2\right)^x = \frac{1}{\left(\sqrt3-\sqrt2\right)^x}$$ with $t\neq 0$
The equation $\Leftrightarrow t + \frac1t - 10 = 0 \Leftrightarrow t^2 - 10t + 1 = 0 $
$\Leftrightarrow t = 5 \pm 2\sqrt6$
With case $t = 5 + 2\sqrt6 \Rightarrow \left(\sqrt3+\sqrt2\right)^x = 5 + 2\sqrt6 \Rightarrow x = \log_{\sqrt3+\sqrt2}(5+2\sqrt6) = 2$
With case $t = 5 - 2\sqrt6 \Rightarrow \left(\sqrt3+\sqrt2\right)^x = 5 - 2\sqrt6 \Rightarrow x = \log_{\sqrt3+\sqrt2}(5-2\sqrt6) = -2$.
Done!
|
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Sum of cubed roots I need to calculate the sums
$$x_1^3 + x_2^3 + x_3^3$$
and
$$x_1^4 + x_2^4 + x_3^4$$
where $x_1, x_2, x_3$ are the roots of
$$x^3+2x^2+3x+4=0$$
using Viete's formulas.
I know that $x_1^2+x_2^2+x_3^2 = -2$, as I already calculated that, but I can't seem to get the cube of the roots. I've tried
$$(x_1^2+x_2^2+x_3^2)(x_1+x_2+x_3)$$
but that did work.
|
If $x_1,x_2,x_3$ are the roots of $x^3+2x^2+3x+4=0$ then $$x^3+2x^2+3x+4 = (x-x_1)(x-x_2)(x-x_3) $$ $$= x^3 - (x_1 + x_2 + x_3)x^2 + (x_1 x_2 + x_1 x_3 + x_2 x_3)x - x_1 x_2 x_3 = x^3 - e_1 x^2 + e_2 x - e_3.$$ So $e_1 = -2$, $e_2 = 3$ and $e_3 = -4$.
Now the trick is to express the power sums $x_1^3 + x_2^3 = x_3^3$ and $x_1^4 + x_2^4 = x_3^4$ in terms of the elementary symmetric polynomials $\{x_1 + x_2 + x_3,x_1 x_2 + x_1 x_3 + x_2 x_3,x_1 x_2 x_3\}$.
See my answer to the question here for details on how to do that Three-variable system of simultaneous equations
In the case of the fourth power sums you should get $x_1^4 + x_2^4 + x_3^4 = e_1^4 - 4 e_1^2 e_2 + 4 e_1 e_3 + 2 e_2^2 = 18$.
|
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How to find $x$ for $1 + \sin(x/2) = \cos x$? How to find $x$ for $1 + \sin(x/2) = \cos x$ ?
From the equation, I can figure out that it is satisfied at $x = 0$ by looking. How do I find the other solutions to this equation?
|
Re-writing this as $\cos(2y) = 1 + \sin(y)$ for $y = \frac{x}{2}$ and using the fact that $\sin^2(y) = \frac{1 - \cos(2y)}{2}$, we can simplify the equation to $1 - 2\sin^2(y) = 1 + \sin(y)$ and thus $2\sin^2(y) + \sin(y) = (2\sin(y) + 1)\sin(y) = 0$. This means either $\sin(y) = 0$ or $\sin(y) = \frac{-1}{2}$, which has solutions of the form $y = n\pi$ or $y = \frac{12n\pi - \pi}{6}$ or $\frac{12n\pi + 7\pi}{6}$ for $n$ an integer. Clearly $x = 2y$.
|
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|
Integral of $\frac{1}{(1+x^2)^2}$ I am in the middle of a problem and having trouble integrating the following integral:
$$\int_{-1}^1\frac1{(1+x^2)^2}\mathrm dx$$
I tried doing partial fractions and got:
$$1=A(1+x^2)+B(1+x^2)$$
I have no clue how to solve this since it is obvious there is no way to cancel out either $A$ or $B$ to get the other variable. Please guide me.
Thank you.
|
$$\frac{1}{(1+x^2)^2} = \left(\frac{1}{1+x^2}-\frac{x^2}{(1+x^2)^2}\right)= \left(\frac{1}{1+x^2}+\frac x2\left(\frac{1}{ 1+x^2 }\right)'\right)$$$$=\left(\frac{1}{1+x^2}+\left(\frac 12 \frac {x}{1+x^2}\right)'-\frac 12 \frac {1}{1+x^2}\right) = \frac 12 \left(\arctan x+\frac{x}{1+x^2}\right)'.$$
|
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|
How to prove an alternating series is convergent? How to prove that the sequence $\sum_{n=1}^{\infty} (-1)^{n-1} \frac{1}{\sqrt{n}}$ is convergent? I was trying to find the upper bound and lower bound of the partial sum $s_k$ and use Squeeze Theorem to figure out the limit, but I couldn't find the lower bound for $s_k$. Any suggestions? Thanks!
|
This is an alternating series. Letting $a_n = \frac{1}{\sqrt{n}}$, for $n\geq 1$, then the series is $\sum (-1)^{n-1}a_n$. Notice that the sequence $a_n$ is strictly decreasing, as $a_1\gt a_2\gt a_3\gt\cdots$.
Now, consider the sequences of even and odd terms of the partial sum sequence:
$$\begin{align*}
s_1 &= 1\\
s_3 &= 1 - \left(\frac{1}{\sqrt{2}} - \frac{1}{\sqrt{3}}\right)\\
s_5 &= 1 - \left(\frac{1}{\sqrt{2}} - \frac{1}{\sqrt{3}}\right) - \left(\frac{1}{\sqrt{4}} - \frac{1}{\sqrt{5}}\right)\\
&\vdots\\
s_{2n+1} &= 1 - \left(\frac{1}{\sqrt{2}} - \frac{1}{\sqrt{3}}\right) -\cdots -\left(\frac{1}{\sqrt{2n}} - \frac{1}{\sqrt{2n+1}}\right)\\
&\vdots
\end{align*}$$
and
$$\begin{align*}
s_2 &= \left( 1- \frac{1}{\sqrt{2}}\right)\\
s_4 &= \left(1 - \frac{1}{\sqrt{2}}\right) + \left(\frac{1}{\sqrt{3}} - \frac{1}{\sqrt{4}}\right)\\
s_6 &= \left(1 - \frac{1}{\sqrt{2}}\right) + \left(\frac{1}{\sqrt{3}}-\frac{1}{\sqrt{4}}\right) + \left(\frac{1}{\sqrt{5}} - \frac{1}{\sqrt{6}}\right)\\
&\vdots\\
s_{2n} &= \left(1 - \frac{1}{\sqrt{2}}\right) + \left(\frac{1}{\sqrt{3}}-\frac{1}{\sqrt{4}}\right) + \cdots + \left(\frac{1}{\sqrt{2n-1}} - \frac{1}{\sqrt{2n}}\right)\\
&\vdots
\end{align*}$$
We have:
$$s_2 \lt s_4 \lt s_6 \lt\cdots \lt s_{2n} \lt \cdots \lt s_{2k+1} \lt \cdots \lt s_3 \lt s_1.$$
The sequence of odd terms of the partial sum sequence is strictly decreasing, and bounded below by each of the even terms, so it converges to some $L$. The sequence of even terms of the partial sum sequence is strictly increasing and bounded above by each of the odd terms, so it converges to some $M$; we have $M\leq L$. But notice that $\lim\limits_{n\to\infty}a_n = 0$, so
$$L-M = \lim_{n\to\infty}(s_{2n+1} - s_{2n}) = \lim_{n\to\infty}\left(\sum_{k=1}^{2n+1}(-1)^{k-1}a_k - \sum_{k=1}^{2n}(-1)^{k-1}a_k\right) = \lim_{n\to\infty}(-1)^{2n+1}a_n = 0,$$
so $L=M$. It is now straightforward to verify that the sequence of partial sums $s_n$ converges to this common limit $L$, so the series converges.
(This is just the usual proof that the Alternating Series Test "works).
|
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|
Linear Algebra Question My question is; How can I find an equation relating $a,b$, and $c$ so that the linear system
$$\begin{cases}2x+y-z=a\\ x-2y-3z=b\\ -3x-y+2z=c\end{cases}$$
is consistent for any values of $a,b$, and $c$ that satisfy that equation.
Thanks,
|
I don't know what tools you have available, but here is how I would consider this problem. If it uses terminology you do not know yet, you might come back to my answer later. This is intentionally not a direct path to the solution, rather the path I went though to solve the problem.
Write this system as a matrix equation:
$$
\left( \begin{array}{ccc} 2 & 1 & -1 \\ 1 & -2 & -3 \\ -3 & -1 & 2 \end{array} \right)
\left( \begin{array}{c} x \\ y \\ z \end{array} \right)
= \left( \begin{array}{c} a \\ b \\ c \end{array} \right)
$$
If the matrix were invertible, we could multiply both sides by the inverse and get relations between $x,y,z$ and $a,b,c$.
$$
\left( \begin{array}{c} x \\ y \\ z \end{array} \right) =
\left( \begin{array}{ccc} 2 & 1 & -1 \\ 1 & -2 & -3 \\ -3 & -1 & 2 \end{array} \right)^{-1}
\left( \begin{array}{c} a \\ b \\ c \end{array} \right)
$$
But that matrix is not invertible: it has determinant 0, or you can see that the first column plus the third column equals the second column, so that the columns are not linearly independent. There has to be a dependence among the rows as well, but it is not as obvious. The "row reduced echelon form" is a good tool for finding column dependencies, so we use it on the transpose of the coefficient matrix (making the rows we want to relate into columns).
$$\text{rref} \left( \begin{array}{ccc} 2 & 1 & -3 \\ 1 & -2 & -1 \\ -1 & -3 & 2 \end{array} \right) = \left( \begin{array}{ccc} 1 & 0 & -7/5 \\ 0 & 1 & -1/5 \\ 0 & 0 & 0 \end{array} \right) $$
For solving $\alpha (2, 1, -1) + \beta(1,-2,-3) + \gamma(-3,-1,2) = 0$, this result means $\alpha - (7/5)\gamma = 0$ and $\beta - (1/5)\gamma = 0$. Since $\gamma$ in independent, make things easy by choosing $\gamma = 5$, which gives $\alpha = 7$ and $\beta = 1$.
So on the left-hand side of the original system, $$7(2x+y-z) + (x-2y-3z) + 5(-3x-y+2z) = 0.$$ Keeping track of the right-hand side gives the dependency you want, $7a + b + 5c = 0$. (It's interesting how row reduction and inner products as used by Luboš lead to the same result.)
|
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|
Explaining an algebra step in $ \frac{n^2(n+1)^2}{4} + (n+1)^3 = \frac{(n+1)^2}{4}(n^2+4n+4)$ I have encountered this step in my textbook and I do not understand it, could someone please list the intermediate steps?
$$ \frac{n^2(n+1)^2}{4} + (n+1)^3 = \frac{(n+1)^2}{4}(n^2+4n+4). $$
Thanks,
|
Add the fractions, factor out common terms in the numerator and then rearrange:
$$
\displaystyle\; \; \frac{n^2(n^2 + 1)^2}{4} + (n + 1)^3
$$
$$
= \displaystyle\frac{n^2(n^2 + 1)^2}{4} + \frac{4(n + 1)^3}{4}
$$
$$
= \displaystyle\frac{n^2(n^2 + 1)^2 + 4(n+1)^3}{4}
$$
$$
= \displaystyle\frac{(n^2 + 1)^2(n^2 \cdot 1 + 4(n + 1))}{4}
$$
$$
= \displaystyle\frac{(n^2 + 1)^2(n^2 + 4n + 4)}{4}
$$
$$
= \displaystyle\frac{(n^2 + 1)^2}{4}(n^2 + 4n + 4)
$$
|
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|
How do I get the square root of a complex number? If I'm given a complex number (say $9 + 4i$), how do I calculate its square root?
|
You can also do following (technique often advised at school) :
*
*Let's write $z² = 9 + 4i$ with $z = a + bi$. The goal is to find $z$
Thus we have $(a + bi)² = 9 + 4i$ and if you expend we get $a²+ 2abi - b² = 9 + 4i$
If you identify the real and imaginary parts, you obtain :
$a²-b² = 9$ (1)
and
$2ab= 4$ (2)
*Now, as $z² = 9 + 4i$, the modulus of $z²$ and $9 + 4i$ are equal so we can write :
$a²+b² = \sqrt{9²+4²}$
$a²+b² = \sqrt{97}$ (3)
*Now find $a$ and $b$ with the the equations (1) , (2) and (3) :
(1) + (3) $\Leftrightarrow 2a² = 9+\sqrt{97} $
so $a = \sqrt{\frac{1}{2}(9+\sqrt{97})} $ or $a = - \sqrt{\frac{1}{2}(9+\sqrt{97})} $
With equation (2) and the previous result you can now find $b$ :
$2ab= 4$
$b= 2/a$
so $b = 2\sqrt{\frac{2}{9+\sqrt{97}}} $ or $b = - 2\sqrt{\frac{2}{9+\sqrt{97}}} $
The answer is : $z = \sqrt{\frac{1}{2}(9+\sqrt{97})} + 2i\sqrt{\frac{2}{9+\sqrt{97}}} $ or $z = - \sqrt{\frac{1}{2}(9+\sqrt{97})} - 2i\sqrt{\frac{2}{9+\sqrt{97}}} $
|
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|
Sum of First $n$ Squares Equals $\frac{n(n+1)(2n+1)}{6}$ I am just starting into calculus and I have a question about the following statement I encountered while learning about definite integrals:
$$\sum_{k=1}^n k^2 = \frac{n(n+1)(2n+1)}{6}$$
I really have no idea why this statement is true. Can someone please explain why this is true and if possible show how to arrive at one given the other?
|
Another take, similar to Euler's (?) proof given by @leonbloy. We know that if:
$\begin{align}
A(z)
&= \sum_{n \ge 0} a_n z^n
\end{align}$
then (writing $\mathtt{D}$ for the derivative):
$\begin{align}
z \mathtt{D} A(z)
&= \sum_{n \ge 0} n a_n z^n
\end{align}$
Also:
$\begin{align}
\frac{A(z)}{1 - z}
&= \sum_{n \ge 0} \left( \sum_{0 \le k \le n} a_n \right) z^n \\
\frac{1}{1 - z}
&= \sum_{n \ge 0} z^n
\end{align}$
This you can repeat and combine. In our case, we get that:
$\begin{align}
\sum_{n \ge 0} n^2
&= (z \mathtt{D})^2 \frac{1}{1 - z} \\
&= \frac{z + z^2}{(1 - z)^3} \\
\sum_{n \ge 0} \left( \sum_{0 \le k \le n} k^2 \right) z^n
&= \frac{z + z^2}{(1 - z)^4}
\end{align}$
We are interested in the coefficient of $z^n$:
$\begin{align}
[z^n] \frac{z + z^2}{(1 - z)^4}
&= [z^n] \frac{z}{(1 - z)^4} + [z^n] \frac{z^2}{(1 - z)^4} \\
&= [z^{n - 1}] (1 - z)^{-4} + [z^{n - 2}] (1 - z)^{-4} \\
&= (-1)^{n - 1} \binom{-4}{n - 1} + (-1)^{n - 2} \binom{-4}{n - 2} \\
&= \binom{n - 1 + 4 - 1}{4 - 1} + \binom{n - 2 + 4 - 1}{4 - 1} \\
&= \binom{n + 2}{3} + \binom{n + 1}{3} \\
&= \frac{(n + 2) (n + 1) n}{3!} + \frac{(n + 1) n (n - 1)}{3!} \\
&= \frac{(2 n + 1) (n + 1) n}{6}
\end{align}$
This approach is less messy than the cited one (no horrible derivatives and then l'Hôpital thrice).
|
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|
Derive $\frac{d}{dx} \left[\sin^{-1} x\right] = \frac{1}{\sqrt{1-x^2}}$ Derive $\frac{d}{dx} \left[\sin^{-1} x\right] = \frac{1}{\sqrt{1-x^2}}$ (Hint: set $x = \sin y$ and use implicit differentiation)
So, I tried to use the hint and I got:
$x = \sin y$
$\frac{d}{dx}\left[x\right] = \sin y\frac{d}{dx}$
$\frac{dx}{dx} = \cos y \frac{dy}{dx}$
$\frac{dy}{dx} = \frac{1}{\cos y}$
$\frac{dy}{dx} = \sec y$
From here I need a little help.
*
*Did I do the implicit
differentiation correctly?
*How do I use this to help with
the original question?
|
First, note that $\displaystyle \sin^{-1}: [-1,1] \to [-\frac{\pi}{2}, \frac{\pi}{2}]$
The range is important, as for this range, you have that if $y = \sin^{-1} x$ then $\cos y = \sqrt{1 - x^2}$, as we have $\sin y = x$ and $\cos y \ge 0$ whenever $y \in [-\frac{\pi}{2}, \frac{\pi}{2}]$.
i.e. since $\sin^2 y + \cos^2 y = 1$, we get $\cos y = \pm \sqrt{1 - x^2}$ and since $\cos y \ge 0$, we can say $\cos y = \sqrt{1- x^2}$.
Exercise
Suppose we defined $\sin^{-1}x$ as the unique angle $\theta$ in $[\frac{\pi}{2}, \frac{3 \pi}{2}]$ such that $\sin \theta = x$, what is the derivative of $\sin^{-1} x$?
|
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|
What is going on in this step? (from arc length problem) I am confused as to what is occurring in this step in an arc length problem:
Could anyone take a stab at trying to explain it to me? thanks
|
First, the square root of the quotient is the quotient of the square roots:
$$\sqrt{\frac{x^{2/3}+1}{x^{2/3}}} = \frac{\sqrt{x^{2/3}+1}}{\sqrt{x^{2/3}}}.$$
Next, the square root in the denominator simplifies with the exponent, since:
$$\sqrt{x^{2/3}} = \left(x^{2/3}\right)^{1/2} = x^{1/3}.$$
Next, introduce a factor of $1$, "disguised" as $\frac{3}{2}\times\frac{2}{3}$; finally, pull one of the two factors out of the integral, since it is constant. Thus:
$$\begin{align*}
\int_1^8\sqrt{\frac{x^{2/3}+1}{x^{2/3}}}\,dx &= \int_1^8\frac{\sqrt{x^{2/3}+1}}{\sqrt{x^{2/3}}}\,dx\\
&= \int_1^8\frac{\sqrt{x^{2/3}+1}}{x^{1/3}}\,dx\\
&= \int_1^8\left(\sqrt{x^{2/3}+1}\right)\times\frac{3}{2}\times\frac{2}{3}\times\frac{1}{x^{1/3}}\,dx\\
&= \frac{3}{2}\int_1^8\sqrt{x^{2/3}+1}\left(\frac{2}{3x^{1/3}}\right)\,dx
\end{align*}$$
|
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|
Help with integrating 101: $\int y \ln{y}\,\mathbb dy$ Would appreciate it if someone would please help me solve this
$$\int y\;\ln y\, \mathbb dy$$
taking time to explain reason for each step taken.
Thanks in advance!
|
Here is a method of doing this by substitution. Put $y=e^{x}$. Then you have $dy = e^{x} \ dx$. Substituting we have
\begin{align*}
\int e^{2x} \cdot x \ dx &= \frac{1}{4}\int e^{t} \cdot t\ dt \qquad \Bigl[ \text{substituting} \ t = 2x \Bigr] \\ &= \frac{1}{4}\int e^{t} \cdot \bigl( t + 1 \bigr) \ dt - \frac{1}{4}\int e^{t} \ dt \\ &= \frac{1}{4}e^{t} \cdot t -\frac{1}{4} e^{t} + C \qquad\qquad\qquad \Bigl[ \because \small \int e^{x} \cdot \Bigl( f(x) + f'(x) \Bigr) \ dx = e^{x} \cdot f(x) + C \ \Bigr] \\ &=\frac{1}{4}\cdot e^{2x} \cdot 2x - \frac{1}{4}\cdot e^{2x} +C \\ &=\log{y} \cdot y^{2} \times \frac{1}{2} - \frac{1}{4} \cdot y^{2} +C
\end{align*}
|
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|
Find $\cos(x+y)$ if $\sin(x)+\sin(y)= a$ and $\cos(x)+\cos(y)= b$ Find $\cos(x+y)$ if $\sin(x)+\sin(y)= a$ and $\cos(x)+\cos(y)= b$.
|
An approach using complex numbers/geometry:
If $c = b + ia$, and if $z = \cos x + i \sin x$ and $z_1 = c - z = \cos y + i \sin y$, you are looking at the real part of $w = \cos(x+y) + i \sin (x+y) = zz_1 = z(c-z)$ with the restriction that $|z| = 1$ and $|z_1| = |c - z| = 1$.
Now the points satisfying $|z| = 1$ and $|c-z| = 1$ are given by the intersection of two unit circles: one centered at the origin and the other at $c$.
Thus $w = z(c-z)$ is a constant dependent only on $c$: it is the product of the two intersection points and can be computed easily as follows:
Each point of intersection can be written as $\dfrac{c}{2} \pm d$, where $d$ is perpendicular to $c$ (Why?) i.e. $d = kic$ for some real constant $k$. (Why?) (It might help to draw a figure here).
Thus the product of points of intersection is $w = (\dfrac{c}{2} + d)(\dfrac{c}{2} - d) =\dfrac{c^2}{4} + k^2 c^2$.
This implies that $w$ is a multiple of $c^2$ and the argument of $w$ is twice that of $c$. Since $|w| = 1$, $w$ can be computed easily without having to worry about $k$.
We get $w = \cos 2 \alpha + i \sin 2 \alpha$ where $\alpha = \tan^{-1}(\frac{a}{b})$.
PS:
We find that $|c|^2(1/4 + k^2) = 1$ and hence $d = (\sqrt{\frac{1}{|c|} - \frac{1}{4}})\ i c$ and so we can easily solve the given system of equations: i.e. find $\cos x, \cos y, \sin x, \sin y$
|
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|
Two problems on number theory I need some ideas (preferable some tricks) for solving these two problems:
Find the largest number $n$ such that $(2004!)!$ is divisible by
$((n!)!)!$
For which integer $n$ is $2^8 + 2^{11} + 2^n$ a perfect square?
For the second one the suggested solution is like this : $ 2^8 + 2^{11} + 2^n = ((2^4)^2 + 2\times2^4\times2^6 + (2^ \frac{n}{2})^2 ) \Rightarrow n=12$
But I can't understand the approach,any ideas?
|
A way to do the second problem is the following. Check small values of $n<8$ by hand (nothing there). Then assume that $n\ge8$. Now $2^8+2^{11}+2^n=2^8(1+8+2^{n-8})$ is a perfect square, iff the latter factor $9+2^{n-8}$ is a perfect square also. But
$$
9+2^{n-8}=m^2\Leftrightarrow 2^{n-8}=m^2-9=(m-3)(m+3).
$$
So by the unique factorization both $m-3$ and $m+3$ must be powers of two. The difference between these two factors is 6, and the differences between powers of 2 are larger than 6 unless both powers are at most 8. The only solution is thus $m=5$, $n=12$.
|
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|
Optimizing $a+b+c$ subject to $a^2 + b^2 + c^2 = 27$
If $a,b,c \gt 0$ and $a^2+b^2+c^2=27$, find the maximum and minimum values of $a+b+c$.
How to solve this one?
(Here's the source of inspiration for the problem.)
|
You can also use the method of Lagrange multipliers to maximize/minimize $f(a,b,c)=a+b+c$ given $g(a,b,c)=a^2+b^2+c^2=27$. We need the gradient of $f$ to be a multiple of the gradient of $g$, i.e.,
$$1=\lambda 2a,\quad 1=\lambda 2b,\quad 1=\lambda 2c,$$
where $\lambda$ is some real number. Hence:
$$\lambda = \frac{1}{2a}=\frac{1}{2b}=\frac{1}{2c}$$
and we must have $a=b=c$. This yields $g(a,a,a)=3a^2=27$, so that $a=b=c=3$ or $a=b=c=-3$. We clearly have a maximum at $a=b=c=3$. Since you assumed $a,b,c>0$, the minimum at $a=b=c=-3$ is not an allowed solution. Thus, the minimum must occur in the boundary of your domain (outside of your domain, so there is an infimum, but no minimum), i.e., when one at least on of $a,b,c$ is zero. However, say $c=0$, then we are trying to minimize $a+b$ given $a^2+b^2=27$. The min occurs when $a=b=-\sqrt{27/2}$ which is again outside our domain $a,b>0$, so the infimum occurs at the boundary when one of $a$ or $b$ is zero. And this leads to an infimum for $f(a,b,c)=a+b+c$ at $(\sqrt{27},0,0)$, $(0,\sqrt{27},0)$ and $(0,0,\sqrt{27})$, and the infimum value is $\sqrt{27}=3\sqrt{3}$.
|
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|
Obtaining Differential Equations from Functions I can now recognise the order and the type of differential equations.Let's say
$$\frac{dy}{dx} = x^2 - 1$$
is a first order ODE,
$$\frac{d^2y}{dx^2} + 2\left(\frac{dy}{dx}\right)^2 + y = 0$$
is a second order ODE and so on. I am having trouble to obtain a differential equation from a given function. I could find the differential equation for $y = e^x(A \cos x + B \sin x)$ and the steps that I followed are as follows.
$$
\begin{align*}
\frac{dy}{dx} &= e^x(A \cos x + B \sin x) + e^x(-A \sin x + B \cos x)
\\ &= y +e^x(-A \sin x + B \cos x) \tag{1}
\\
\frac{d^2 y}{dx^2} &= \frac{dy}{dx} + e^x(-A \sin x + B \cos x) + e^x(-A \cos x - B \sin x)
\\ &=
\frac{dy}{dx}+\left(\frac{dy}{dx} - y\right) - y
\end{align*}
$$
using the orginal function and $(1)$. Finally,
$$
\frac{d^2y}{dx^2} - 2 \frac{dy}{dx} + 2y = 0 ,$$
which is the required differential equation.
Similarly, if the function is $y=(A\cos2t + B\sin2t)$, the differential equation that I get is
$$
\frac{d^2y}{dx^2} + 4y = 0
$$
following similar steps as above.
My question is how do I obtain the differential equations for the following functions using similer procedures
$$y = Ae^{3x} + Be^{2x}$$
$$xy = Ae^{x} + Be^{-x} + x^{2}$$
I am not looking for a solution in determinant form using vector spaces or any other linear algebra/matrices.Please provide step by step solution for the function in order to obtain a particular differential equation.
Thank you in advance.
|
Eliminate $a$ and $b$ from $$y = ae^{2x} +be^{3x}\tag{1}$$
$$\frac{dy}{dx} = 2ae^{2x} + 3be^{3x}\tag{2}$$
$$d^2y/dx^2 = 4ae^{2x} + 9be^{3x}\tag{3}$$
$(1)\cdot 2 -(2)$ we get
$$2y - \frac{dy}{dx} = -be^3x$$
$$b = e^{-3x} \left(\frac{dy}{dx} - 2y\right)\tag{4}$$
$(2)*2 - (3)$
$$2 \frac{dy}{dx} - \frac{d^2y}{dx^2} = - 3 b e^{3x}$$
$$b = 1/ 3e ^{3x} \left(\frac{d^2y}{dx^2} - 2 \frac{dy}{dx}\right)\tag{5}$$
compare $(4)$ and $(5)$
$$(dy/dx - 2y) = 1/3 \left(\frac{d^2y}{dx^2} - 2 \frac{dy}{dx}\right)$$
$$3\frac{dy}{dx} - 6y = \frac{d^2y}{dx^2} -2 \frac{dy}{dx}$$
$$\frac{d^2y}{dx^2} -5 \frac{dy}{dx} + 6y = 0$$
|
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|
Olympiad Inequality Problem Consider three positive reals $x,y,z$ such that $xyz=1$.
How would one go about proving:
$$\frac{x^5y^5}{x^2+y^2}+\frac{y^5z^5}{y^2+z^2}+\frac{x^5z^5}{x^2+z^2}\ge \frac{3}{2}$$
I really dont know even where to begin! It looks a BIT like Nesbitts? Maybe?
|
By Holder $$\sum_{cyc}\frac{x^5y^5}{x^2+y^2}\geq\frac{(xy+xz+yz)^5}{(1+1+1)^3\sum\limits_{cyc}(x^2+y^2)}=\frac{(xy+xz+yz)^5}{54(x^2+y^2+z^2)}.$$
Hence, it remains to prove that $$(xy+xz+yz)^5\geq81(x^2+y^2+z^2).$$
Let $x+y+z=3u$, $xy+xz+yz=3v^2$ and $xyz=w^3$.
Thus, since $xy+xz+yz$ and $x^2+y^2+z^2$ does not depend on $w^3$,
it's enough to prove the last inequality for a maximal value of $w^3$,
which happens for equality case of two variables.
Let $y=x$. Hence, $z=\frac{1}{x^2}$ and we need to prove that
$$(x^3+2)^5\geq81x(2x^6+1).$$
Let $x^3=t$.
Hence, we need to prove that $f(t)\geq0$, where
$$f(t)=5\ln(t+2)-\ln(2t^2+1)-\frac{1}{3}\ln{t}-4\ln3$$
But $$f'(t)=\frac{5}{t+2}-\frac{4t}{2t^2+1}-\frac{1}{3t}=\frac{(t-1)(2t-1)(4t-1)}{3t(t+2)(2t^2+1)}$$
and since $f\left(\frac{1}{4}\right)>0$, we are done!
|
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|
Solve equation: $\frac{7}{\sqrt{343^{5x-1}}} = 2401^{-6.7}$ I need help to solve this equation, thanks in advance.
$$\frac{7}{\sqrt{343^{5x-1}}} = 2401^{-6.7}$$
|
$$\frac{7}{\sqrt{343^{5x-1}}} = 2401^{-6.7}$$
Since $343=7^3$ and $2401 = 7^4$, we can write
$$
\frac{7}{\sqrt{7^{3(5x-1)}}} = 7^{4\cdot(-6.7)}
$$
and then
$$
\frac{7}{\left(7^{3(5x-1)}\right)^{1/2}} = 7^{4\cdot(-6.7)}
$$
So
$$
7^{1 - (1/2)(3)(5x-1)} = 7^{4\cdot(-6.7)}.
$$
Hence
$$
1 - \frac12 \cdot3(5x-1) = 4\cdot(-6.7).
$$
etc.
|
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|
Solving $8=x(2(1-\sqrt{5}))+(1-x)(2(1+\sqrt{5}))$ I came up with this equation during my homework :
$8=x(2(1-\sqrt{5}))+(1-x)(2(1+\sqrt{5}))$
My algebra is weak and I can't seem to find a way to solve for x nicely
Could someone please show me a decent way of doing this?
Thanks alot, Jason
|
$$8 = x\left [ 2-2\sqrt{5} \right ] + \left ( 1 - x \right )\left [ 2 + 2\sqrt{5} \right ]$$
use foil
$$8 = 2x - 2x\sqrt{5} + 2 + 2\sqrt{5} - 2x - 2x\sqrt{5}$$
$$8 = -4x\sqrt{5} + 2\sqrt{5} + 2$$
subtract 2 and square both sides
$$36 = 16x^25 + 20$$
subtract 20 from both sides to obtain
$$80x^2 = 16$$
Now divide both sides by 80 and you get $$x^2 = 16/80$$
therefore $$x = \pm 1 / \sqrt{5}$$
|
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|
How to solve this equation: $\frac{\sqrt[5]{x^3 \sqrt{x\sqrt[3]{x^{-2}}}}}{\sqrt[4]{x\sqrt[3]{x}}}=3$ Please, help me to solve this equation:
$$\frac{\sqrt[5]{x^3\sqrt{x\sqrt[3]{x^{-2}}}}}{\sqrt[4]{x\sqrt[3]{x}}}=3$$
I tried to shorten fraction, but I get very weird numbers like
$$\frac{\sqrt[30]{x^{19}}}{\sqrt[3]{x}}=3,$$ and I'm stuck there :(
|
You're almost done. Note that $\frac{1}{x} = x^{-1}$, so
$$\frac{\sqrt[30]{x^{19}}}{\sqrt[3]{x}} = \sqrt[30]{x^{19}}(\sqrt[3]{x})^{-1} = x^{19/30}x^{-1/3} = x^{9/30}$$
So then
$$x^{9/30} = 3 \quad \Longleftrightarrow \quad x = 3^{30/9} = 3^{10/3} = \sqrt[3]{3^{10}}$$
|
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|
Inverse Laplace Transform -s domain How can I find the inverse Laplace transforms of the following function?
$$ G\left(s\right)=\frac{2(s+1)}{s(s^2+s+2)} $$
I solved so far. After that, how do I do?
$$ \frac{1}{s}+\frac{1}{s^2+s+2}+\frac{s}{s^2+s+2}=G\left( s \right)$$
|
To find the inverse Laplace transforms of the function $\ G\left(s\right)=\dfrac{2\left(s+1\right)}{s\left(s^2+s+2\right)} $
You have solved up to partial fraction form of $G\left(s\right)$ i.e
$$G\left(s\right)=\frac{1}{s}+\frac{1}{s^2+s+2}+\frac{s}{s^2+s+2}$$
Now taking the Laplace inverse $$\begin{align}\mathcal{L^{-1}}\left\{G\left(s\right)\right\}&=\mathcal{L^{-1}}\left\{\frac{1}{s}+\frac{1}{s^2+s+2}+\frac{s}{s^2+s+2}\right\}\\&=\mathcal{L^{-1}}\left\{\frac{1}{s}\right\}+\mathcal{L^{-1}}\left\{\frac{1}{s^2+s+2}\right\}+\mathcal{L^{-1}}\left\{\frac{s}{s^2+s+2}\right\}\\
\end{align}$$
Now the first term $$\mathcal{L^{-1}}\left\{\frac{1}{s}\right\}=1 \\ \qquad$$
Second term is
$$\begin{align}\mathcal{L^{-1}}\left\{\frac{1}{s^2+s+2}\right\}&=\mathcal{L^{-1}}\left\{\frac{1}{s^2+2\times s\times\frac{1}{2}+\left(\frac{1}{2}\right)^2+2-\left(\frac{1}{2}\right)^2}\right\}\\
&=\mathcal{L^{-1}}\left\{\frac{1}{\left(s+\frac{1}{2}\right)^2+\left(\frac{\sqrt{7}}{2}\right)^2}\right\}\\
&=\mathcal{L^{-1}}\left\{\frac{2}{\sqrt{7}}\frac{\frac{\sqrt{7}}{2}}{\left(s+\frac{1}{2}\right)^2+\left(\frac{\sqrt{7}}{2}\right)^2}\right\}\\
&=\frac{2}{\sqrt{7}}\mathcal{L^{-1}}\left\{\frac{\frac{\sqrt{7}}{2}}{\left(s+\frac{1}{2}\right)^2+\left(\frac{\sqrt{7}}{2}\right)^2}\right\}\\
&=\frac{2}{\sqrt{7}}e^{\frac{-1}{2}t}\sin\left(\frac{\sqrt{7}}{2}t\right)\\
\left[\text{since}\quad \mathcal{L^{-1}}\left\{\frac{b}{\left(s-a\right)^2+b^2}\right\}=e^{at}\sin\left(bt\right)\right]
\end{align}$$
Third term is
$$\begin{align}\mathcal{L^{-1}}\left\{\frac{s}{s^2+s+2}\right\}&=\mathcal{L^{-1}}\left\{\frac{s}{s^2+2.s.\frac{1}{2}+\left(\frac{1}{2}\right)^2+2-\left(\frac{1}{2}\right)^2}\right\}\\
&=\mathcal{L^{-1}}\left\{\frac{s+\frac{1}{2}-\frac{1}{2}}{\left(s+\frac{1}{2}\right)^2+\left(\frac{\sqrt{7}}{2}\right)^2}\right\}\\
&=\mathcal{L^{-1}}\left\{\frac{\left(s+\frac{1}{2}\right)}{\left(s+\frac{1}{2}\right)^2+\left(\frac{\sqrt{7}}{2}\right)^2}\right\}-\mathcal{L^{-1}}\left\{\frac{\frac{1}{2}}{\left(s+\frac{1}{2}\right)^2+\left(\frac{\sqrt{7}}{2}\right)^2}\right\}\\
&=e^{\frac{-1}{2}t}\cos\left(\frac{\sqrt{7}}{2}t\right)-\frac{1}{2}\mathcal{L^{-1}}\left\{\frac{1}{\left(s+\frac{1}{2}\right)^2+\left(\frac{\sqrt{7}}{2}\right)^2}\right\}\\
&=e^{\frac{-1}{2}t}\cos\left(\frac{\sqrt{7}}{2}t\right)-\frac{1}{2}\frac{2}{\sqrt{7}}e^{\frac{-1}{2}t}\sin\left(\frac{\sqrt{7}}{2}t\right) \quad \left(\text{By Second term}\right)\\
&=e^{\frac{-1}{2}t}\cos\left(\frac{\sqrt{7}}{2}t\right)-\frac{1}{\sqrt{7}}e^{\frac{-1}{2}t}\sin\left(\frac{\sqrt{7}}{2}t\right)\\
\qquad \qquad \qquad \left[\text{since}\quad \mathcal{L^{-1}}\left\{\frac{\left(s-a\right)}{\left(s-a\right)^2+b^2}\right\}=e^{at}\cos\left(bt\right)\right].
\end{align}$$ Collecting all three terms,
$$\begin{align}\mathcal{L^{-1}}\left\{G\left(s\right)\right\}&=1+\frac{2}{\sqrt{7}}e^{\frac{-1}{2}t}\sin\left(\frac{\sqrt{7}}{2}t\right)+e^{\frac{-1}{2}t}\cos\left(\frac{\sqrt{7}}{2}t\right)-\frac{1}{\sqrt{7}}e^{\frac{-1}{2}t}\sin\left(\frac{\sqrt{7}}{2}t\right)\\
&=1+\frac{1}{\sqrt{7}}e^{\frac{-1}{2}t}\sin\left(\frac{\sqrt{7}}{2}t\right)+e^{\frac{-1}{2}t}\cos\left(\frac{\sqrt{7}}{2}t\right)\\
\end{align}$$
|
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|
Parametric form of an ellipse given by $ax^2 + by^2 + cxy = d$ If $c = 0$, the parametric form is obviously $x = \sqrt{\frac{d}{a}} \cos(t), y = \sqrt{\frac{d}{b}} \sin(t)$.
When $c \neq 0$ the sine and cosine should be phase shifted from each other. How do I find the angular shift and from there how do I adjust the factors multiplying the sine and cosine?
|
You would complete squares: $\left(a x + \frac{1}{2} c y\right)^2 + \left(a b - \frac{c^2}{4} \right) y^2 = a d$.
From there: $a x + \frac{c}{2} y = \sqrt{a d} \sin(t)$ and $\sqrt{a b - \frac{c^2}{4}} y = \sqrt{a d} \cos(t)$, assuming $c^2 < 4 a b$, and $a d > 0$.
Solving for $x$ and $y$ and denoting $\mathcal{D} = 4 a b - c^2$
$$
x(t) = \sqrt{\frac{d}{a}} \left( \sin(t) - \frac{c}{\sqrt{\mathcal{D}} } \cos(t) \right) \qquad
y(t) = \frac{2 \sqrt{a d}}{\sqrt{\mathcal{D}}} \cos(t)
$$
|
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|
Explicit solution to $100(y + \frac{100}{10-y}) = x + 1000$ $$
100\left(y + \frac{100}{10-y}\right) = x + 1000
$$
I'm looking to have the equation in the form $y(x) = ...$ Seems pretty simple, but I'm just not seeing it.
The answer is:
$$
y(x) = \frac{1}{200}\sqrt{x^2 + 4000x} − x
$$
Wolfram|Alpha returns a slightly different answer:
$$
y(x) = \frac{1}{200} (\sqrt{x^2+4000000}+x+2000)
$$
There must be something I'm missing. Thoughts?
|
Wolfram Alpha is correct. Starting with
$$100\left(y + \frac{100}{10-y}\right) = x + 1000$$
and dividing both sides by $100$:
$$y+\frac{100}{10-y}=\frac{x}{100}+10$$
multiplying both sides by $(10-y)$:
$$-y^2+10y+100=\left(\frac{x}{100}+10\right)(10-y)$$
collecting everything on one side:
$$y^2-\left(20+\frac{x}{100}\right)y+\frac{x}{10}=0$$
and then applying the quadratic formula gives
$$y=\frac{\left(20+\frac{x}{100}\right)\pm\sqrt{\left(20+\frac{x}{100}\right)^2-\frac{2x}{5}}}{2}=\frac{1}{200}\left(2000+x\pm\sqrt{(x+2000)^2-4000x}\right)=$$
$$\frac{1}{200}\left(2000+x\pm\sqrt{x^2+4000000}\right)$$
|
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|
Complex numbers, solutions of $1-z+z^2=0$ $$z_1 \text{ and } z_2 \text{ are the solutions of } 1-z+z^2=0$$
$$E=(z_1^4-z_1^3+2z_1^2-2z_1+1)^{2005}+(z_2^4-z_2^3+2z_2^2-2z_2+1)^{2005}$$
Which is the value of $E$ ?
I have solved the equation:
\begin{align*}\Delta = 1-4=-3=3i^2&\Rightarrow
z_{1,2}=\frac{1\pm i\sqrt{3}}{2}
\end{align*}
One solution would be to write these numbers in trigonometric form. But I am sure there is an easier way if I write E differently, but I can't find it.
|
If you know that $z_i^2-z_i+1 = 0$, then since
$$x^4 - x^3 + 2x^2 - 2x + 1 = (x^2-x+1)(x^2+1) - x,$$
it follows that:
$$z_1^4 - z_1^3 + 2z_1^2 - 2z_1 + 1 = (z_1^2 - z_1 + 1)(z_1^2 + 1) - z_1 = 0(z_1^2+1) - z_1 = -z_1.$$
Much simpler than trying to work directly with the roots.
|
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|
Maximum Value Question I have a question about this following question:
Let $a>0$. Show that the maximum value of $f(x):=\frac{1}{1+|x|}+\frac{1}{1+|x-a|}$ is $\frac{2+a}{1+a}$
I am wondering if I am headed in the right direction with the following process. I first adress the issue of the absolute values by rewriting the function:
$$f(x)=:\frac{1}{1+\sqrt{x^{2}}}
+\frac{1}{1+\sqrt{(x-a)^{2}}}$$
I then take the derivative of $f(x)$, which yields:
$$f'(x)=\frac{a-x}{(1+\sqrt{(a-x)^{2}})^{2}\sqrt{(a-x)^{2}}}+\frac{x}{\sqrt{x^{2}}(1+\sqrt{x^{2}})^{2}}$$
The derivative I've taken from Wolfram Alpha for the moment. I am a bit confused about my next step, though I have an idea. Would I input $\frac{2+a}{1+a}$ into the first derivative, which should give me back zero? This would show that there is a maximum or minimum at that point. The next step being to check the value of the second derivative to show that it is indeed a maximum? I suppose this still wont show that $\frac{2+a}{1+a}$ is the global maximum.
|
This was a response to the question $f(x):=\frac{1}{1+|x|}-\frac{1}{1+|x-a|}$
Perhaps I am misunderstanding something, but doesn't the graph look something like this?
If $x \le 0 $ then $f(x)=\frac{1}{1+|x|}-\frac{1}{1+|x-a|}=\frac{1}{1-x}-\frac{1}{1+a-x} =\frac{a}{(1-x)(1+a-x)}$ which is an positive increasing function for that $x$.
If $0 \le x \le a $ then $f(x)=\frac{1}{1+|x|}-\frac{1}{1+|x-a|}=\frac{1}{1+x}-\frac{1}{1+a-x} =\frac{a-2x}{(1+x)(1+a-x)}$ which is a decreasing function for that $x$, positive for $x\lt a/2$ and negative for $a/2 \lt x$.
If $a \le x $ then $f(x)=\frac{1}{1+|x|}-\frac{1}{1+|x-a|}=\frac{1}{1+x}-\frac{1}{1+x-a} =\frac{-a}{(1+x)(1+x-a)}$ which is a negative but increasing function for that $x$.
So the maximum must occur when $x=0$ and takes the value $f(0)=\frac{a}{1+a}$.
With the change in the question this becomes
and the maximum occurs when $x=0$ or $x=a$ and takes the value $f(0)=\frac{1}{1}+\frac{1}{1+a} = \frac{2+a}{1+a}$.
|
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|
Showing whether two numbers are equal or not $\dfrac{\sin (2x+y)}{\sin (2x)} =\dfrac{\sin (x+2y)}{\sin (2y)}$,where $0<x,y\le\dfrac{\pi}{4}$ .
Can I show that $x=y $ or find two numbers $x,y$ such that $x\not=y$?
|
Edited: I think I got it...there was typo in the first try.
After cross-multiplication, we get
$[2\sin y\sin(2x+y)]\cos y-[2\sin x\sin(x+2y)]\cos x=0$
$\Rightarrow[\cos(2x)-\cos2(x+y)]\cos y-[\cos(2y)-\cos2(x+y)]\cos x=0$
$\Rightarrow\cos2(x+y)[\cos x-\cos y]+[(2\cos^2x-1)\cos y-(2\cos^2y-1)\cos x]=0$
$\Rightarrow[\cos x-\cos y][\cos2(x+y)+2\cos x\cos y+1]=0$
$\Rightarrow[\cos x-\cos y][\cos^2(x+y)+\cos x\cos y]=0$
Note that in the specified range, the second factor is strictly positive. Hence we must have $\cos x=\cos y\Rightarrow x=y$
|
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|
Modular exponentiation by hand ($a^b\bmod c$) How do I efficiently compute $a^b\bmod c$:
*
*When $b$ is huge, for instance $5^{844325}\bmod 21$?
*When $b$ is less than $c$ but it would still be a lot of work to multiply $a$ by itself $b$ times, for instance $5^{69}\bmod 101$?
*When $(a,c)\ne1$, for instance $6^{103}\bmod 14$?
Are there any other tricks for evaluating exponents in modular arithmetic?
|
Wikipage on modular arithmetic is not bad.
*
*When $b$ is huge, and $a$ and $c$ are coprime, Euler's theorem applies:
$$
a^b \equiv a^{b \, \bmod \, \phi(c)} \, \bmod c
$$
For the example at hand, $\phi(21) = \phi(3) \times \phi(7) = 2 \times 6 = 12$.
$$
\Rightarrow 844325 \bmod 12 = 5,\ \text{so}\ 5^5 = 5 \times 25^2 \equiv 5 \times 4^2 = 80 \equiv 17 \mod 21
$$.
*When $a$ and $c$ are coprime, but $0<b<\phi(c)$, repeated squaring (or using other compositions of powers) is the fastest way to go (manually):
$$
\begin{eqnarray}
5^4 \equiv 5 \times 5^3 \equiv 5 \times 24 \equiv 19 &\pmod{101}\\
19^4 \equiv (19^2)^2 \equiv 58^2 \equiv (-43)^2 \equiv 1849 \equiv 31 &\pmod{101} \\
31^4 \equiv (31^2)^2 \equiv (961)^2 \equiv 52^2 \equiv 2704 \equiv 78 &\pmod{101} \\
5^{69} \equiv 5 \times 5^4 \times ((5^4)^4)^4 \equiv 5 \times 19 \times 78 \equiv 5 \times 19 \times (-23)\\
\equiv 19 \times (-14) \equiv -266 \equiv 37 & \pmod{101}
\end{eqnarray}
$$
*When $a$ and $c$ are not coprime, let $g = \gcd(a,c)$. Let $a = g \times d$ and $c = g \times f$, then, assuming $b > 1$:
$$
a^b \bmod c = g^b \times d^b \bmod (g \times f) = ( g \times (g^{b-1} d^b \bmod f) ) \bmod c
$$
In the example given, $\gcd(6,14) = 2$. So $2^{102} \times 3^{103} \mod 7$, using Euler'r theorem, with $\phi(7) = 6$, and $102 \equiv 0 \mod 6$, $2^{102} \times 3^{103} \equiv 3 \mod 7$, so $6^{103} \equiv (2 \times 3) \equiv 6 \mod 14 $.
|
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|
Proving inequality $x^{10}-x^6+x^2-x+1>0$ How can the inequality $x^{10}-x^6+x^2-x+1>0$ be proved
a) using elementary mathematical methods？
b) using higher mathematical methods？
|
Another elementary method:
We want to show that $x^{10}+x^2+1 >x^6+x$. If we can show it for $x \ge 0$, it will be true always. This is because if $x$ is positive, replacing $x$ by $-x$ does not change $x^{10}+x^2+1$, but turns $x^6+x$ into the smaller $x^6-x$.
We now take care of $x \ge 0$. If $x\ge 1$, then $x^{10}\ge x^6$ and $x^2 \ge x$, so
$x^{10}+x^2+1 >x^6+x$.
If $0 \le x < 1$, then $x^2 \ge x^6$ and $1 >x$, so $x^{10}+x^2+1 >x^6+x$.
|
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|
A Recurrence Relation Problem
In a standard elimination tournament, a player wins $\$100k$ when she/he wins a match in the $k$th round. Develop and solve a recurrence relation for $a_n$, the total amount of money given away in a tournament with $n$ entrants, where $n$ is assumed to be a power of $2$.
I seem always to fail build recurrence relations... I hope someone could explain it to me in details.
|
I think Ross's answer is the beginning of things but that he had it wrong. Let me explain.
You have $2^n$ players to begin with, hence $2^{n-1}$ matches for the first round. This gives us $2^{n-1}$ winners for the first round, hence $1 \times 100 \times 2^{n-1}$ dollars given away. Then $2^{n-1}$ players go to the second round, we have $2^{n-2}$ winners and they win $2 \times 100 \times 2^{n-2}$. We are left with $2^{n-2}$ players for the third round, $2^{n-3}$ winners, and they win $3 \times 100 \times 2^{n-3}$ dollars. It is easy to see that this keeps going on by induction on $k$, the round, and that the total amount of winnings at the $k^{\text{\th}}$ round will be $k \times 100 \times 2^{n-k}$ for $k = 1, \dots, n$. Thus
$$
\text{Total}(n) = \sum_{k=1}^n \, (k \times 100 \times 2^{n-k}) = 2^n \times 100 \times \left( \sum_{k=1}^n \frac{k}{2^k} \right).
$$
If you compute $\text{Total}(n+1)$ a little, you obtain
$$
\begin{align}
\text{Total}(n+1)= 2^{n+1} \times 100 \times \left( \sum_{k=1}^{n+1} \frac k{2^k} \right) \\
= 2^{n+1} \times 100 \times \left( \sum_{k=1}^{n} \frac k{2^k} + \frac{n+1}{2^{n+1}} \right) \\
= (n+1) \times 100 + 2 \times 2^n \times 100 \times \left( \sum_{k=1}^n \frac k{2^k} \right) \\
= (n+1) \times 100 + 2 \,\, \text{Total}(n) \\
\end{align}
$$
Thus a recurrence formula for $T(n)$ would be $T(n+1) = (n+1)100 + 2T(n)$.
Hope that helps,
EDIT : You also have a more explicit formula for $T(n)$ by the following. Consider the geometric sum
$$
\sum_{k=1}^n x^k = \frac{x^{n+1} - 1}{x-1}.
$$
Thus
$$
\begin{align}
\sum_{k=1}^n kx^k &= x \left( \sum_{k=1}^n kx^{k-1} \right) \\
&= x \frac{d}{dx} \left( \sum_{k=1}^n x^k \right) \\
&= x \frac {d}{dx} \left( \frac{x^{n+1} - 1}{x-1} \right) \\
&= x \left( \frac{(n+1)x^n}{x-1} - \frac{x^{n+1} - 1}{(x-1)^2} \right) \\
&= \frac x{(x-1)^2} \left( (n+1)x^n(x-1) - x^{n+1} + 1 \right) \\
&= \frac x{(x-1)^2} \left( nx^{n+1} - (n+1) x^n + 1 \right).
\end{align}
$$
Let $x=1/2$ and you get
$$
\sum_{k=1}^n \frac{k}{2^k} = 2 \left( \frac n{2^{n+1}} - \frac{(n+1)}{2^n} + 1 \right) = \frac{2^{n+1} - n -2}{2^n}.
$$
This means that
$$
T(n) = 2^n \times 100 \times \left( \frac{2^{n+1} - n - 2}{2^n} \right) = 100 (2^{n+1} - n - 2).
$$
|
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"timestamp": "2023-03-29T00:00:00",
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|
Show that there are infinitely many reducible polynomials of the form $x^n+x+1$ in $\mathbf{F}_2[x]$ Here is a question from an old exam:
Show that there are infinite $n\in \mathbf{N}, A= x^{n}+x+1 $ which are reducible over $\mathbf{F}_{2}[x]$.
Using André Nicolas' and Qiaochu Yuan's hint: $x^{2}+x+1$ as dividing polynomial. $x^{2}+x+1$ is irreducible over $\mathbf{F_{2}}$. If an irreducible polynomial divides another polynomial which is not itself, that means that polynomial must be reducible. We want to show that $x^{2}+x+1$ divides all polynomials of the form $x^{3n+5}+x+1$. I can't figure the induction steps, but in $\mathbf{F_{2}}$ the polynomial belongs to the residue class $\tilde{1}$, therefore there must be an infnite amount of them.
Concerning Gerry Myerson's hint, how can I use cubic roots in $\mathbf{F_{2}}$, wouldn't I need $\mathbf{R}[i]$ for that?
Help is greatly appreciated.
|
Proof by induction.
Base case ($n=0$):
$$\begin{align}
(x^2+x+1)\left(x^3+\sum\limits_{i=0}^0 (x^{3i}+x^{3i+2})\right)&=(x^2+x+1)(x^3+x^2+1)\\&=x^5+2x^4+2x^3+2x^2+x+1\\&=x^5+x+1=x^{3(0)+5}+x+1
\end{align}$$ (working in $\mathbb{F}_2[x]$).
Inductive hypothesis ($n \geq 0$):
Suppose that $$(x^2+x+1)\left(x^{3(n+1)}+\sum\limits_{i=0}^n (x^{3i}+x^{3i+2})\right)=x^{3n+5}+x+1$$
Then:
$$\begin{align}
&(x^2+x+1)\left(x^{3(n+2)}+\sum\limits_{i=0}^{n+1} (x^{3i}+x^{3i+2})\right)=\\
&(x^2+x+1)\left(x^{3(n+2)}+x^{3(n+1)}+x^{3(n+1)+2}+\sum\limits_{i=0}^{n} (x^{3i}+x^{3i+2})\right)=\\
&(x^2+x+1)(x^{3(n+2)}+x^{3(n+1)+2}) +
(x^2+x+1)\left(x^{3(n+1)}+\sum\limits_{i=0}^{n} (x^{3i}+x^{3i+2})\right)
\end{align}$$
using our inductive hypothesis we get
$$\begin{align}
&=(x^2+x+1)(x^{3(n+2)}+x^{3(n+1)+2}) + x^{3n+5}+x+1\\
&=(x^2+x+1)(x^{3n+6}+x^{3n+5}) + x^{3n+5}+x+1\\
&=x^{3n+8}+2x^{3n+7}+2x^{3n+6}+2x^{3n+5}+x+1\\
&=x^{3(n+1)+5}+x+1
\end{align}$$
Therefore, $x^{3(n+1)+5}+x+1$ is reducible in $\mathbb{F}_2[x]$ (or in any polynomial ring with coefficients in a field of characteristic 2) for all non-negative integers $n$.
|
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"timestamp": "2023-03-29T00:00:00",
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|
How do you divide a polynomial by a binomial of the form $ax^2+b$, where $a$ and $b$ are greater than one? I came across a question that asked me to divide $-2x^3+4x^2-3x+5$ by $4x^2+5$. Can anyone help me?
|
The answer in no way depends on whether $a$ and $b$ are more than $1$ or less; the only fact about those that you need is that the polynomial you're dividing by is not zero. You have this:
$$
\begin{array}{ccccccccccccccccc}
\\
4x^2+5 & \big) & -2x^3 & + & 4x^2 & - & 3x & + & 5 \\
& &
\end{array}
$$
So ask what you need to multiply $4$ by to get $-2$. In other words, divide $-2$ by $4$. You get $-2/4=-1/2$. So write
$$
\begin{array}{ccccccccccccccccc}
& & \frac{-1}{2} x \\
\hline 4x^2+5 & \big) & -2x^3 & + & 4x^2 & - & 3x & + & 5 \\
& &
\end{array}
$$
Then multiply:
$$
\begin{array}{ccccccccccccccccc}
& & \frac{-1}{2} x \\
\hline 4x^2+5 & \big) & -2x^3 & + & 4x^2 & - & 3x & + & 5 \\
& & -2x^3 & & & -& \frac 52 x
\end{array}
$$
Then subtract:
$$
\begin{array}{ccccccccccccccccc}
& & \frac{-1}{2} x \\
\hline 4x^2+5 & \big) & -2x^3 & + & 4x^2 & - & 3x & + & 5 \\
& & -2x^3 & & & -& \frac 52 x \\ \hline
& & & & 4x^2 & - & \frac 12 x & + 5 \\ \hline
\end{array}
$$
The ask what you have to multiply $4x^2$ by to get $4x^2$, then multiply, then subtract.....
|
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"url": "https://math.stackexchange.com/questions/86190",
"timestamp": "2023-03-29T00:00:00",
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|
Solve $a, b$ in the improper integral Find $a, b \in\mathbb{R}$ such that $$\int^\infty_1\left(\dfrac{2x^2+bx+a}{x(2x+a)}-1\right) \mathrm{d}x=1$$
|
$$I = \int_1^{\infty} \left(\frac{bx+a-ax}{x(2x+a)} \right) dx$$ $$ \left(\frac{bx+a-ax}{x(2x+a)} \right) = \frac{1}{x} + \frac{b-a-2}{2x+a}$$ $$I = \int_1^{\infty} \left(\frac{1}{x} + \frac{b-a-2}{2x+a} \right) dx = \left[ \log(x) + \frac{b-a-2}{2} \log \left(x + \frac{a}{2} \right)\right]_1^{\infty}$$ Since, we get a definite answer for this, we need $b-a-2 = -2 \implies b = a$. This is so since if $b > a$, then the integral blows up to $+\infty$ and if $a > b$, the integral blows up to $- \infty$. Hence, $$I = \int_1^{\infty} \left(\frac{1}{x} + \frac{b-a-2}{2x+a} \right) dx = \left[ \log \left(\frac{x}{x+\frac{a}{2}} \right)\right]_1^{\infty} = - \log \left( \frac{1}{1+\frac{a}{2}}\right) = \log \left( 1 + \frac{a}{2} \right) = 1$$
Hence, $1 + \frac{a}{2} = e \implies a = 2 (e-1) = b$.
|
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|
How to get closed form from generating function? I have this generating function:
$$\frac{1}{2}\, \left( {\frac {1}{\sqrt {1-4\,z}}}-1 \right) \left( \,{
\frac {1-\sqrt {1-4\,z}}{2z}}-1 \right)$$
and I know that $\frac {1}{\sqrt {1-4\,z}}$ is the generating function for the sequence $\binom {2n} {n}$, and $\frac {1-\sqrt {1-4\,z}}{2z}$ is the generating function for the sequence $\frac {1}{n+1}\binom{2n} {n}$.
Now, I thought that I could substitute those in there, and where they multiply I'll use a summation like this:
$$\frac{1}{2}\left( 1-\frac{1}{n+1}\binom{2n} {n}-\binom{2n} {n} +
\sum_{k=0}^n \frac{1}{k+1} \binom{2k}{k}\binom{2(n-k)}{n-k}
\right)$$
Could this be right? It doesn't seem to work when I try in Maple. What else could I do?
I already know that the end sequence will be $\binom{2n-1}{n-2}$ if this can help...
|
It’s not necessary to use the generalized binomial theorem and the gamma function. Let $$g(x)=\frac12\left(\frac1{\sqrt{1-4x}}-1\right) \left(\frac{1-\sqrt{1-4x}}{2x}-1\right)$$ and $u=\sqrt{1-4x}$. Then
$$\begin{align*}
g(x)&= \frac12\left(\frac1u-1\right)\left(\frac{1-u}{2x}-1\right)\\
&=\frac12\left(\frac{1-u}{2xu}-\frac{1-u}{2x}-\frac1u+1\right)\\
&=\frac12\left(\frac1{2xu}-\frac1x+\frac{u}{2x}-\frac1u+1\right)\\
&=\frac12\left(\frac{1+u^2}{2xu}-\frac1x-\frac1u+1\right)\\
&=\frac12\left(\frac{1-2x}{xu}-\frac1x-\frac1u+1\right)\\
&=\frac12\left(\frac1{xu}-\frac1x-\frac3u+1\right)\\
&=\frac12\left(\frac1x\left(\frac1u-1\right)+1-\frac3u\right)\\
&=\frac12\left(\sum_{k\ge 1}\binom{2k}k x^{k-1}+1-3\sum_{k\ge 0}\binom{2k}k x^k\right)\\
&=\frac12\left(1+\sum_{k\ge 0}\left(\binom{2k+2}{k+1}-3\binom{2k}k\right)x^k\right)\\
&=\frac12\sum_{k\ge 2}\left(\binom{2k+2}{k+1}-3\binom{2k}k\right)x^k.
\end{align*}$$
Now
$$\begin{align*}
\binom{2k+2}{k+1}-3\binom{2k}k&=\binom{2k+1}k+\binom{2k+1}{k+1}-3\binom{2k}k\\
&=\binom{2k+1}k+\binom{2k}k+\binom{2k}{k-1}-3\binom{2k}k\\
&=\binom{2k+1}k+\binom{2k}{k+1}-2\binom{2k}k\\
&=\binom{2k}{k-1}+\binom{2k}k+\binom{2k}{k+1}-2\binom{2k}k\\
&=\binom{2k}{k-1}-\binom{2k}k+\binom{2k}{k+1}\\
&=\binom{2k}{k-1}-\binom{2k}k+\binom{2k-1}k+\binom{2k-1}{k+1}\\
&=\binom{2k}{k-1}-\binom{2k-1}{k-1}-\binom{2k-1}k+\binom{2k-1}k+\binom{2k-1}{k+1}\\
&=\binom{2k}{k-1}-\binom{2k-1}{k-1}+\binom{2k-1}{k+1}\\
&=\binom{2k-1}{k-2}+\binom{2k-1}{k+1}\\
&=2\binom{2k-1}{k-2},
\end{align*}$$
so $\displaystyle[x^k]g(x)=\binom{2k-1}{k-2}$, as desired.
|
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|
How to prove :If $p$ is prime greater than $3$ and $\gcd(a,24\cdot p)=1$ then $a^{p-1} \equiv 1 \pmod {24\cdot p}$? I want to prove following statement :
If $p$ is a prime number greater than $3$ and $\gcd(a,24\cdot p)=1$ then :
$a^{p-1} \equiv 1 \pmod {24\cdot p}$
Here is my attempt :
The Euler's totient function can be written in the form :
$n=p_1^{k_1}\cdot p_2^{k_2} \ldots \cdot p_r^{k_r} \Rightarrow \phi(n)=p_1^{k_1}\cdot\left(1-\frac{1}{p_1}\right)\cdot p_2^{k_2}\cdot\left(1-\frac{1}{p_2}\right)\ldots p_r^{k_r}\cdot \left(1-\frac{1}{p_r}\right)$
So,
$\phi(24 \cdot p)=2^3\cdot \left(1-\frac{1}{2}\right)\cdot3^1\cdot\left(1-\frac{1}{3}\right)\cdot p\cdot\left(1-\frac{1}{p}\right)=8\cdot(p-1)$
Euler's totient theorem states that :
if $\gcd(a,n)=1$ then $a^{\phi(n)} \equiv 1 \pmod n$
Therefore we may write :
$a^{\phi(n)}-1 \equiv 0 \pmod n \Rightarrow a^{\phi(24\cdot p)}-1=a^{8\cdot(p-1)}-1 \equiv 0 \pmod{24\cdot p} \Rightarrow$
$\Rightarrow \left(a^{p-1}\right)^8-1=(a^{p-1}-1)\cdot \displaystyle \sum_{i=0}^7 a^{(p-1)\cdot i} \equiv 0\pmod{24\cdot p}$
So we may conclude :
$(a^{p-1}-1) \equiv 0 \pmod {24\cdot p}$ , or $\displaystyle \sum_{i=0}^7 a^{(p-1)\cdot i} \equiv 0\pmod{24\cdot p}$
How can I prove that $\displaystyle \sum_{i=0}^7 a^{(p-1)\cdot i} \not\equiv 0\pmod{24\cdot p}$ ?
|
Since $(a,24\cdot p)=1$, it also follows that $(a,p)=(a,3)=(a,8)=1$.
By the generalization of Fermat's little theorem, $a^{p-1}\equiv 1\pmod{p}$, $a^2\equiv 1\pmod{3}$, and $a^4\equiv 1\pmod{8}$. But $a^4\equiv 1\pmod{8}$, implies $a\equiv 1,3,5,7\pmod{8}$, and in all cases $a^2\equiv 1\pmod{8}$.
Since $p-1$ is even, $a^{p-1}\equiv 1$ in all cases, since $a^{p-1}\equiv (a^2)^{(p-1)/2}\equiv 1^{(p-1)/2}\pmod{3,8}$.
Then $a^{p-1}-1$ is divisible by $3$, $8$, and $p$, and thus as a common multiple, is divisible by $\mathrm{lcm}(3,8,p)=24\cdot p$, so $a^{p-1}\equiv 1\pmod{24\cdot p}$.
|
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|
How to find the equation of the tangent line to $y=x^2+2x-4$ at $x=2$?
I'm given a curve $$y=x^2+2x-4$$
How do I find the tangent line to this curve at $x = 2$?
|
Here's an algebraic approach that avoids the explicit use of derivatives.
We are given a quadratic function $f(x) = x^2 + 2x -4$, and we want to find the equation of the tangent to the parabola $y = f(x)$ at the point $(2, 4)$. (Note that $f(2) = 2^2 + 2 \cdot 2 - 4 = 4$.) Assume that it is given by the equation
$$
y = m(x-2) + 4, \tag{$\ast$}
$$
where $m$ is its slope.
Let's consider the intersection of the parabola with the tangent; this is given by the system of equations
$$
\begin{cases}
y &=& x^2 + 2x - 4,
\\ y &=& m(x-2)+4.
\end{cases}
$$
In other words, to find the intersection, we should solve the quadratic equation
$ x^2 + 2x - 4 = m(x-2)+4$, or
$$
x^2 + (2-m)x+(2m-8) = 0. \tag{$\ast\ast$}
$$
using the quadratic formula like so
$$
\frac{-(2-m)\pm\sqrt{(2-m)^2-4.1.(2m-8)}}{2.1}
$$
Pictorially it is clear that the tangent meets the parabola meets in exactly one point, so we want $(\ast \ast)$ to have a unique solution. This implies that the discriminant of $(\ast \ast)$ vanishes:
$$
\begin{array}{crcl}
&(2-m)^2 - 4 \cdot (2m-8) &=& 0
\\ \implies \qquad & m^2 - 12m + 36 &=& 0.
\end{array}
$$
Conveniently (although this is not a numerical coincidence), this equation has a unique solution $m=6$: this is the solution we are after. Plugging $m=6$ in $(\ast)$, we get the equation of the tangent at $(2, 4)$ to be $y = 6x-8$.
|
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|
Sangaku: Show line segment is perpendicular to diameter of container circle "From a 1803 Sangaku found in Gumma Prefecture. The base of an isosceles triangle sits on a diameter of the large circle. This diameter also bisects the circle on the left, which is inscribed so that it just touches the inside of the container circle and one vertex of the triangle. The top circle is inscribed so that it touches the outsides of both the left circle and the triangle, as well as the inside of the container circle. A line segment connects the center of the top circle and the intersection point between the left circle and the triangle. Show that this line segment is perpendicular to the drawn diameter of the container circle. (T. Rothman)"
Source: http://hermay.org/jconstant/wasan/sangaku/index.html
Enjoy!
EDIT
Note 1: The radius of the smaller circle touching the left vertex of the triangle is not necessarily half the radius of the enclosing circle. There IS variability in the base of the triangle.
Note 2: Two of the vertices of triangle necessarily are on the enclosing circle.
|
For those who do not like inversion... =P
Let $A$ be the big circle with centre $O$ and diameter $PQ$
Let $B$ be the circle internally tangent to $A$ at $P$ and intersecting $PQ$ again at $M$ and having centre $J$
Let $R$ be on $A$ such that $\overline{MR} = \overline{QR}$
Let $C$ be the circle that is tangent to $A$, $B$, $MR$ and inside $A$ and outside both $B$ and completely on the same side of $PQ$ as $R$
Let $K$ be the point on $MQ$ such that $RK \perp MQ$
Let $D$ be the circle tangent to $B$ and $MR$ with centre $N$ such that $PQ \perp MN$ and $N$,$R$ are on the same side of $PQ$
Let $Z$ be the point where $D$ is tangent to $MR$
Let $\overrightarrow{MO} = r \overrightarrow{OQ}$ and WLOG $\overline{OQ} = 1$
Let $x = \overline{MN}$
Let $y$ be the radius of $D$
Then $\overline{MR}^2 = \overline{RK}^2 + \overline{MK}^2 = \overline{OR}^2 - \overline{OK}^2 + \overline{MK}^2 = 1 - (\frac{1-r}{2})^2 + (\frac{1+r}{2})^2 = 1+r$
Since $\triangle MNZ \sim \triangle RMK$, $\frac{x}{y} = \overline{MN} / \overline{NZ} = \overline{RM} / \overline{MK} = \sqrt{1+r} / \frac{1+r}{2} = \frac{2}{\sqrt{1+r}}$
Thus $(1+r)x^2 = 4y^2$
Also $(y+\frac{1-r}{2})^2 = \overline{NJ}^2 = \overline{MN}^2 + \overline{MJ}^2 = (\frac{1-r}{2})^2 + x^2$
Thus $(1+r)y^2 + (1+r)(1-r)y = (1+r)x^2 = 4y^2$
Since $y\not=0$, $(3-r)y = 1-r^2$
$\overline{OQ} = y + \overline{ON}$
$\Leftrightarrow (1-y)^2 = \overline{ON}^2 = \overline{MN}^2 + \overline{MO}^2 = x^2 + r^2$
$\Leftrightarrow (1+r)(1-2y+y^2) = 4y^2 + (1+r)r^2$
$\Leftrightarrow (3-r)y^2 + (2+2r)y - (1+r)(1-r^2) = 0$
$\Leftrightarrow (1-r^2)y + (2+2r)y = (1+r)(1-r^2)$
$\Leftrightarrow (1+r)(3-r)y = (1+r)(1-r^2)$ which is clearly true
Thus $D$ is tangent to $A$
Since $C$ is uniquely defined and $D$ is an identically defined circle, $N$ is the centre of $C$
Therefore the line through both $M$ and the centre of $C$ is perpendicular to $PQ$
(QED)
|
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|
Very curious properties of ordered partitions relating to Fibonacci numbers I came across some interesting propositions in some calculations I did and I was wondering if someone would be so kind as to provide some explanations of these phenomenon.
We call an ordered Partition of a positive integer $n$ as the way of writing $n$ as a sum of one or more positive integers, where the order of the sum DOES matter. For example, there are $4$ ordered partitions of $3$, namely $1+1+1, 1+2,2+1,3$
Now suppose we replace the last term of each above partition with a $1$, multiply the terms of each individual partition, then add the results all together. In this manner we get:
$(1\cdot 1 \cdot 1 )+ (1 \cdot 1 )+ (2 \cdot 1) +(1)$ respectively, which equals $5$, which curiously is a Fibonacci number! Can someone please explain why this result is always a Fibonacci number? (Do some more calculations if you are not yet convinced.)
Now here is a more challenging relationship I found. Given any partition sum, keep only the first, third, fifth,... term. Replace each such term $x$ by $2^{x-1}$ and multiply the results. So as an example, the sum $2+4+1+3+5$ would become $2^12^02^4=32$. Now do this to every ordered partition of $n$ and add the result. Curiously, the sum is always $F_{2n}$. Can someone please explain this as well?
Thank you so much for your time!
Edit: this is problematic since I can't write comments as I am not a user, but yes In the first example replacing a 1 or not will still yield a fibonacci sum-product, but I have already proven the case where you do nt replace with a 1
|
This problem has a solution using ordinary generating functions.
First question.
Observe that
$$\sum_{q\ge 0} q z^q = \frac{z}{(1-z)^2}.$$
Therefore the generating function of the contribution from partitions
with $k$ terms is given by
$$\left(\frac{z}{(1-z)^2}\right)^{k-1} \frac{z}{1-z}$$
and the contribution from all partitions is
$$p(z) =
\sum_{k\ge 1} \left(\frac{z}{(1-z)^2}\right)^{k-1} \frac{z}{1-z}
= \frac{z}{1-z} \frac{1}{1-z/(1-z)^2}
\\= \frac{z}{1-z} \frac{(1-z)^2}{(1-z)^2-z}
= \frac{z(1-z)}{1 - 3z + z^2}.$$
Recall that the generating function of the Fibonacci numbers
is $$\sum_{q\ge 0} F_q z^q = \frac{z}{1-z-z^2}.$$
It follows that
$$\sum_{q\ge 0} F_{2q+1} z^{2q+1}
= \frac{1}{2}\frac{z}{1-z-z^2}
+ \frac{1}{2}\frac{z}{1+z-z^2}
\\= z \frac{1-z^2}{(z^2+z-1)(z^2-z-1)}
= z \frac{1-z^2}{(z^2-1)^2-z^2}$$
so that
$$\sum_{q\ge 0} F_{2q+1} z^{q+1}
= z \frac{1-z}{1-3z+z^2}$$
which finally yields
$$[z^n] p(z) = F_{2n-1}.$$
Second question.
To start observe that
$$\frac{1}{2} \sum_{q\ge 1} 2^q z^q = \frac{z}{1-2z}.$$
Here the generating function is for $k=2q$ even
($k$ -- number of elements in the partition)
$$\left(\frac{z}{1-2z}\right)^q \left(\frac{z}{1-z}\right)^q$$
and for $k=2q+1$ odd
$$\left(\frac{z}{1-2z}\right)^{q+1} \left(\frac{z}{1-z}\right)^q.$$
Collecting both contributions we obtain
$$q(z) = \sum_{q\ge 1}
\left(\frac{z}{1-2z}\right)^q \left(\frac{z}{1-z}\right)^q +
\sum_{q\ge 0}
\left(\frac{z}{1-2z}\right)^{q+1} \left(\frac{z}{1-z}\right)^q.$$
Simplify this to get
$$\frac{z^2/(1-2z)/(1-z)}{1-z^2/(1-2z)/(1-z)}
+ \frac{z}{1-2z} \frac{1}{1-z^2/(1-2z)/(1-z)}$$
or
$$\frac{z^2}{(1-2z)(1-z)-z^2}
+ \frac{z(1-z)}{(1-2z)(1-z)-z^2}$$
which is
$$\frac{z}{(1-2z)(1-z)-z^2}
= \frac{z}{1-3z+z^2}.$$
Recall the generating function of the Fibonacci numbers to
get
$$\sum_{q\ge 0} F_{2q} z^{2q}
= \frac{1}{2}\frac{z}{1-z-z^2}
- \frac{1}{2}\frac{z}{1+z-z^2}
= \frac{z^2}{(1-z^2)^2-z^2}$$
so that
$$\sum_{q\ge 0} F_{2q} z^q
= \frac{z}{(1-z)^2-z}
= \frac{z}{1-3z+z^2}.$$
This finally yields
$$[z^n] q(z) = F_{2n}.$$
Addendum.
The following Maple code can be used to verify the above results and further investigate these numbers. Maple has a Fibonacci routine with the name fibonacci.
with(combinat);
q := proc(n)
option remember;
local s, p, q, x, k;
s := 0;
for p in partition(n) do
for q in permute(p) do
x := [seq(q[k], k = 1 .. nops(q) - 1), 1];
s := s + mul(x[k], k = 1 .. nops(x))
od;
od;
s
end;
q2 := proc(n)
option remember;
local s, p, q, x, k;
s := 0;
for p in partition(n) do
for q in permute(p) do
x := [seq(q[2*k+1], k=0..iquo(nops(q)-1, 2))];
s := s + mul(2^(x[k]-1), k = 1 .. nops(x))
od;
od;
s
end;
|
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|
How to use modulo to find the last character of an exponentiation? I already found interesting answers in other questions to this topic. Yet, I still don't get it well enough to use it. Can someone please show it with an example? I won't tell you mine, because it's a homework and I really want to understand it myself.
|
For $a^N\equiv x\pmod{10}$ (with $0\leq x<10$), a general strategy is to start finding a small exponent $n$ such that $a^n\equiv 0,1,-1\pmod{10}$.
Example 1:
$3^{1000}\equiv x\pmod{10}$. We have $3^2\equiv -1\pmod{10}$ thus $3^4=(3^2)^2\equiv (-1)^2\equiv 1\pmod{10}$, then $3^{1000}=(3^4)^{250}\equiv 1^{250}\equiv 1\pmod{10}$.
Example $1+\epsilon$:
$3^{1001}\equiv x\pmod{10}$. Then $3^{1001}=3.3^{1000}\equiv 3.1\equiv3\pmod{10}$.
Example 2: sometimes, we cannot follow this strategy as in this example.
$2^{1000}\equiv x\pmod{10}.$
$$
2^3\equiv -2\pmod{10}\quad\textrm{thus}\quad 2^9=(2^3)^3\equiv (-2)^3=-2^3\equiv 2 \pmod{10}.
$$
Then
$$
2^{1000}=2.2^{999}=2.(2^9)^{111}\equiv2.2^{111}=2^{112}=2^4.(2^9)^{12}\equiv 2^4.2^12=2^{16}\pmod{10},
$$
finally
$$
2^{1000}\equiv 2^7.2^9\equiv 128.2\equiv -4\equiv 6\pmod{10}.
$$
|
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|
Factoring polynomial with complex coefficients
Given the equation $z^2+4iz-13=0$, solve for $z$ without the quadratic formula.
In real numbers set, when I find this kind of equations I usually complete the perfect square trinomial.In this case:
$(z^2+4iz-4)-13+4=0$
$(z+2i)^2-9=0$
I chosen $-4$ because the number whose double is $4i$, is $2i$. And the square of $2i$ is $-4$.
$z+2i= \pm \sqrt{9}$
$z=3-2i \vee z=-3-2i$
Is this correct?Thanks
|
What you did is correct.
You can check by substitution: If $z=3-2i$ then
$$(z+2i)^2-9= ((3-2i)+2i)^2-9 = 3^2-9=0.$$
If $z=-3-2i$ then
$$
(z+2i)^2 - 9 = ((-3-2i) + 2i)^2 - 9 = (-3)^2 - 9 = 0.
$$
|
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|
Given that $[n(n+1)(n+2)]^2 = 303916253\square96$, find the value of $\square$. Given that $[n(n+1)(n+2)]^2 = 303916253\square96$, find the value of $\square$.
Given that $[n(n+1)(n+2)]^2 = 30391625\square796$, find the value of $\square$.
Problem
Given that $[n(n+1)(n+2)]^2 = 3039162537\square6$, find the value of $\square$.
Solution
In any three consecutive integers, $n, n+1, n+2$, at least one of the numbers will be
even, and one of them will be a multiple of 3. Hence the product, $n(n+1)(n+2)$, will
be even and divisible by 3. Furthermore, the square of an even number will be
divisible by 4, and the square of a multiple of 3 will be divisible by 9.
If a number is divisible by 9, the sum of the digits will also be divisible by $9: 3 + 3 + 9+ 1 + 6 + 2 + 5 + 3 + 7 + 6 = 45$, so the value of $\square$ must be 0 or 9.
However, if the number is divisible by 4, the last two digits (either 06 or 96) must be
divisible by 4. Hence the value of $\square$ is 9.
Find the value of n.
How would you solve the equation, $[n(n+1)(n+2)] ^2 = k$, in general?
|
Then, there's brute force:
303,916,253,?96 ~= 30.3 e+10 ~= (5.5 e+5) squared ~= (n cubed) squared.
So, 80 < n < 90 since 80 cubed = 512,000 and 90 cubed = 729,000.
Since 303,916,253,?96 ends with 6, the number that is squared must end with 4 or 6. Hence, none of n, n+1, and n+2 can have 0 or 5 in the one's digit.
This limits possible choices to 81*82*83 or 82*83*84 or 86*87*88 or 87*88*89.
As it turns out, with n=81, 81*82*83 = 551,286, which when squared = 303,916,253,796.
The missing digit (?) is 7.
|
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|
Cumulative distribution function question
Possible Duplicate:
Normal random variable $X$ and the cdf of $Y=aX+b$
I'm given a standard random variable $X$, and $Y = aX + b$:
How can I find the cumulative distribution function for Y as an integral of $f(x)=(\frac{1}{\sqrt{2}\pi\sigma}e^{-\frac{(x-u)^2}{\sigma^2}}$?
I know $F_y(y)=F_x(\frac{y-b}{a})$, but cant figure out where to go from there.
|
If $a>0$ then
$$
\begin{align}
& f_Y(y) = \frac{d}{dy} F_Y(y) = \frac{d}{dy}\Pr(Y\le y) = \frac{d}{dy}\Pr(aX+b\le y) = \frac{d}{dy}\Pr\left(X \le \frac{y-b}{a}\right) \\ \\
& = \frac{d}{dy}F_X\left( \frac{y-b}{a} \right) = f_X\left(\frac{y-b}{a}\right) \cdot \frac{d}{dy} \frac{y-b}{a} = \frac 1 a f_X\left(\frac{y-b}{a}\right) \\ \\ \\
& = \frac 1 a \cdot \frac{1}{\sqrt{2\pi}}\exp\left(\frac{-1}{2} \cdot \left(\frac{y-b}{a}\right)^2\right).
\end{align}
$$
Therefore
$$
F_Y(y) = \frac 1 a \int_{-\infty}^y \frac{1}{\sqrt{2\pi}} \exp\left(\frac{-1}{2} \left(\frac{u-a}{a}\right)^2\right)\; du.
$$
The equivalence between $aX+b\le y$ and $X \le \frac{y-b}{a}$ holds only if $a>0$.
|
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|
What is this formula? There is some formula that I can't precisely remember for polynomials, which goes something like $x^n-1 = (x-1)(\text{a lot of stuff})$. It could be more general, like $x^n - k$, or maybe it is just for the same powers, so $x^m - k^m$, but I think it's not just for the same powers. Does anyone know what I am talking about? I realize this is vague.
|
What that formula is stating is that, when $x^n-1$ is dividided through $x-1$ one gets $$1+x+x^2+x^3+\cdots+x^{n-1}$$
This is derived in several ways. One is
Let $$S = 1+x+x^2+x^3+\cdots+x^{n-1}$$
Then one has that
$$Sx= x+x^2+x^3+\cdots+x^{n}$$
Thus substracting this two equation one gets
$$Sx-S = x^n-1$$
or
$$S(x-1) =x^n-1$$
Which is what we wanted to prove.
Similarily, let $x=\dfrac{b}{a}$, then we have that.
$$x^n-1 = (x-1)(1+x+x^2+\cdots+x^{n-1})$$
$$\eqalign{
& \frac{{{b^n}}}{{{a^n}}} - 1 = \left( {\frac{b}{a} - 1} \right)\left( {1 + \frac{b}{a} + \frac{{{b^2}}}{{{a^2}}} + \cdots +\frac{{{b^{n - 2}}}}{{{a^{n - 2}}}}+ \frac{{{b^{n - 1}}}}{{{a^{n - 1}}}}} \right) \cr
& \frac{{{b^n} - {a^n}}}{{{a^n}}} = \left( {\frac{{b - a}}{a}} \right)\left( {1 + \frac{b}{a} + \frac{{{b^2}}}{{{a^2}}} + \cdots +\frac{{{b^{n - 2}}}}{{{a^{n - 2}}}}+ \frac{{{b^{n - 1}}}}{{{a^{n - 1}}}}} \right) \cr
& {b^n} - {a^n} = a\left( {\frac{{b - a}}{a}} \right){a^{n - 1}}\left( {1 + \frac{b}{a} + \frac{{{b^2}}}{{{a^2}}} + \cdots + \frac{{{b^{n - 1}}}}{{{a^{n - 1}}}}} \right) \cr
& {b^n} - {a^n} = \left( {b - a} \right)\left( {{a^{n - 1}} + b{a^{n - 2}} + {b^2}{a^{n - 3}} + \cdots + ab^{n-2}+ {b^{n - 1}}} \right) \cr} $$
or succintly and with a change of exponent, for notation's sake:
$$\frac{{{b^{n + 1}} - {a^{n + 1}}}}{{b - a}} = \sum\limits_{k = 0}^n {{a^{n - k}}{b^k}} $$
|
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|
the cube of integer can be written as the difference of two square This Exercise $4$, page 7, from Burton's book Elementary Number Theory.
Prove that the cube of any integer can be written as the difference of two squares. [Hint: Notice that $n^{3}=(1^{3}+2^{3}+\cdots+n^{3})-(1^{3}+2^{3}+\cdots+(n-1)^{3}).$]
Is there a way to prove that in a more natural way?
I would appreciate your help.
|
A more natural approach is to work out exactly which integers can be written as the difference of two squares, and then notice that all cubes are such.
So suppose $n=a^2-b^2$, with $n$, $a$ and $b$ all integers. Then we have $n=(a+b)(a-b)$. Setting $x=a+b$, $y=a-b$ we have $n=xy$ with $a=\frac{x+y}{2}$, $b=\frac{x-y}{2}$. Since $a$ and $b$ are integers, $x$ and $y$ must be of the same parity.
Thus if $n$ is a different of two squares, then it can be factorized into two integers of the same parity. Either the factors are both odd, in which case $n$ is odd, or the factors are both even, in which case $n$ is divisible by $4$. Conversely, if $n$ is odd then $n=1\cdot n=\left (\frac{n+1}{2}\right)^2-\left(\frac{n-1}{2}\right)^2$, while if $n$ is divisible by $4$ then $n=2\cdot \frac{n}{2}=\left (\frac{n+4}{4}\right)^2-\left (\frac{n-4}{4}\right)^2$.
So the integers which are a difference of two squares are precisely those which are either odd or a multiple of $4$ (in other words, those not congruent to $2$ mod $4$). Since the cube of an odd number is odd and the cube of an even number is divisible by $2^3=8$ and hence by $4$, every cube is a difference of two squares.
|
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|
Fibonacci's final digits cycle every 60 numbers How would you go about to prove that the final digits of the Fibonacci numbers recur after a cycle of 60?
References:
The sequence of final digits in Fibonacci numbers repeats in cycles of 60. The last two digits repeat in 300, the last three in 1500, the last four in , etc. The number of Fibonacci numbers between and is either 1 or 2 (Wells 1986, p. 65).
http://mathworld.wolfram.com/FibonacciNumber.html
|
Note that
$$
\begin{pmatrix} F_{n+1}\\ F_{n+2} \end{pmatrix} = \begin{pmatrix} 0 & 1 \\ 1 & 1 \end{pmatrix} \begin{pmatrix} F_n \\ F_{n+1} \end{pmatrix}
$$
and
$$
\begin{pmatrix} 0 & 1 \\ 1 & 1 \end{pmatrix}^{60} \equiv \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} \mod 10.
$$
One can verify that $60$ is the smallest power for which this holds, so it is the order of the matrix mod 10. In particular
$$
\begin{pmatrix} F_{n+60}\\ F_{n+61} \end{pmatrix} \equiv \begin{pmatrix} F_n \\ F_{n+1} \end{pmatrix} \mod 10
$$
so the final digits repeat from that point onwards.
|
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|
Intuitive explanation for a polynomial expansion? Is there an ituitive explanation for the formula:
$$
\frac{1}{\left(1-x\right)^{k+1}}=\sum_{n=0}^{\infty}\left(\begin{array}{c}
n+k\\
n
\end{array}\right)x^{n}
$$
?
Taylor expansion around x=0
:
$$
\frac{1}{1-x}=1+x+x^{2}+x^{3}+...
$$
differentiate this k
times will prove this formula. but is there an easy explanation for this? Any thing similar to the binomial law to show that the coefficient of $x^{n}$
is $\left(\begin{array}{c}
n+k\\
n
\end{array}\right)$
.
Thanks in advance.
|
If by the binomial law you mean
$$
(1+x)^n=\sum_k\binom{n}{k}x^k\tag{1}
$$
then yes. Note that
$$
\binom{n}{k}=\frac{n(n-1)(n-2)\dots(n-k+1)}{k!}\tag{2}
$$
Consider what $(2)$ looks like for a negative exponent, $-n$:
$$
\begin{align}
\binom{-n}{k}
&=\frac{-n(-n-1)(-n-2)\dots(-n-k+1)}{k!}\\
&=(-1)^k\frac{(n+k-1)(n+k-2)(n+k-3)\dots n}{k!}\\
&=(-1)^k\binom{n+k-1}{k}\tag{3}
\end{align}
$$
Plug $(3)$ into $(1)$ and we get
$$
\begin{align}
\frac{1}{(1-x)^{k+1}}
&=(1-x)^{-(k+1)}\\
&=\sum_n\binom{-(k+1)}{n}(-x)^n\\
&=\sum_n(-1)^k\binom{n+k}{n}(-x)^n\\
&=\sum_n\binom{n+k}{n}x^n\tag{4}
\end{align}
$$
|
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|
Let $x$ and $y$ be positive integers such that $xy \mid x^2+y^2+1$. Let $x$ and $y$ be positive integers such that $xy \mid x^2+y^2+1$.
Show that $$ \frac{x^2+y^2+1}{xy}= 3 \;.$$
I have been solving this for a week and I do not know how to prove the statement. I saw this in a book and I am greatly challenged. Can anyone give me a hint on how to attack the problem? thanks
|
Suppose $xy\mid x^2+y^2+1$ and let $t=\displaystyle\frac{x^2+y^2+1}{xy}$ such that $t\in\mathbb{N}$.
Construct the set,
$$S=\left\{(x,y)\in\mathbb{N}\times\mathbb{N} : \frac{x^2+y^2+1}{xy}=t\in\mathbb{N}\right\}$$
We deduce that $\displaystyle\frac{x^2+y^2+1}{xy} \ge 3$ because $\displaystyle\frac{x^2+y^2+1}{xy}<3$ implies $x^2+y^2+1 \le 2xy \le x^2+y^2$ which is clearly a contradiction. Now fix $t$ and suppose that $t>3$. Since $S\neq \varnothing$, we can choose $(a,b)\in S$ such that $a+b$ is minimal and $t>3$. WLOG assume $a\ge b>0$.
Let us consider the quadratic
$$p(w)=w^2-tbw+b^2+1=0$$
It follows that $a$ is a solution since $(a,b)\in S$ and hence satisfies the quadratic equation, that is $a$ is a root. By applying Vieta's formulas we obtain the other root $c$. Hence $a+c=tb$ and $ac=b^2+1$. Since $c=tb-a$ we have $c\in\mathbb{Z}$. Now it remains to prove that $(a,c)\in S$.
To this end suppose $c<0$. It follows that
$$\displaystyle 0<a^2+ac+1-3ab=a^2+ac-\frac{3c}{t}(a+c)<0$$
which is clearly a contradiction. It also immediately follows that $c\neq 0$ as this implies $b^2+1=0$. Therefore $c\in\mathbb{N}$ and $(a,c)\in S$.
Now we show that this $c$ contradicts the minimality of $a$, that is $c<a$. Suppose $c>a$ so it follows that $a+1\le c$. But from Vieta's equations we obtain $\displaystyle a+1\le c=\frac{b^2+1}{a}\le a+\frac{1}{a}$ which is impossible since this inequality holds in $\mathbb{N}$ if and only if $a=1$ and hence $a=b=1$ implying $t=3$ which contradicts our assumption that $t>3$. Therefore $c\le a$. But if $c=a$ then this implies that $\displaystyle a^2=b^2+1>\frac{9}{4}b^2$ which again is a contradiction. Hence we conclude that $c<a$ and as a result $c+b$ contradicts the minimality of $a+b$. Hence $t=3$
|
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Subsets and Splits
Fractions in Questions and Answers
The query retrieves a sample of questions and answers containing the LaTeX fraction symbol, which provides basic filtering of mathematical content but limited analytical insight.