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generalized binomial coefficient I know that the coefficient $$-\frac{1}{2} \choose k$$ can be simplified by multiplying both the nominator and the denominator by $$2^k$$ and then represented as $$ (-\frac{1}{4})^k {2k\choose k}$$
But what if the upper index is $$\frac{1}{3}$$it seems that our trick do not work anymore, is there anything we can do about it?
|
There is no nice generalisation from $\binom{-\frac{1}{2}}{k}$ to $\binom{\pm \frac{1}{n}}{k}, 0\leq k \leq n$ available.
We obtain for $n=2$:
\begin{align*}
\binom{-\frac{1}{2}}{k}&=\frac{1}{k!}
\left(-\frac{1}{2}\right)\left(-\frac{3}{2}\right)\cdots\left(-\frac{1}{2}-(k-1)\right)\\
&=\frac{1}{k!}\frac{(-1)^k}{2^k}(2k-1)!!\\
&=\frac{1}{k!}\frac{(-1)^k}{2^k}\frac{(2k)!}{(2k)!!}\tag{1}\\
&=\frac{(-1)^k}{2^{2k}}\frac{(2k)!}{k!k!}\\
&=\frac{(-1)^k}{2^{2k}}\binom{2k}{k}\\
\end{align*}
and for $n=3$:
\begin{align*}
\binom{-\frac{1}{3}}{k}&=\frac{1}{k!}
\left(-\frac{1}{3}\right)\left(-\frac{4}{3}\right)\cdots\left(-\frac{1}{3}-(k-1)\right)\\
&=\frac{1}{k!}\frac{(-1)^k}{3^k}(3k-2)!!!\\
&=\frac{1}{k!}\frac{(-1)^k}{3^k}\frac{(3k)!}{(3k)!!!(3k-1)!!!}\tag{2}\\
&=\frac{(-1)^k}{3^{2k}}\frac{(3k)!}{k!k!}\frac{1}{(3k-1)!!!}\\
&=\frac{(-1)^k}{3^{2k}}\binom{3k}{2k}\binom{2k}{k}\frac{1}{(3k-1)!!!}\\
\end{align*}
Comment:
*
*In (1) we use the double factorials
\begin{align*}
(2k)!&=(2k)!!(2k-1)!!\qquad \text{and} \qquad (2k)!!=2^kk!\\
\end{align*}
*
*The formula with triple factorials in (2) is somewhat more complicated due to additional factors $(3k-1)!!!$
\begin{align*}
(3k)!&=(3k)!!!(3k-1)!!!(3k-2)!!!\qquad \text{and} \qquad (3k)!!!=3^kk!\\
\end{align*}
Note: You might find the generalisation to $\binom{\alpha}{k}$ with $\alpha \in\mathbb{C}, k\geq 0$ used in binomial series or the even more generalized binomial coefficients based upon Gamma functions useful.
|
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"timestamp": "2023-03-29T00:00:00",
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|
Which of the following is an irreducible factor of $x^{12}-1$ over $\mathbb Q$? Which of the following is an irreducible factor of $x^{12}-1$ over $\mathbb Q$?
*
*$x^8+x^4+1$
*$x^4+1$
*$x^4-x^2+1$
*$x^5-x^4+x^3-x^2+x-1$
Is the answer $(1)$ correct? I am not sure which one is.
|
By repeatedly using the simple factorizations:
$$x^{2n}-1=(x^n+1)(x^n-1)$$
and
$$x^{3n}-1=(x^{2n}+x^n+1)(x^n-1)$$
and
$$x^{3n}+1=(x^{2n}-x^n+1)(x^n+1)$$
We get:
$$x^{12}-1=(x^4-x^2+1)(x^2+1)(x^2-x+1)(x+1)(x^2+x+1)(x-1)$$
Ruling out every choice except for number 3:
$$x^4-x^2+1$$
Call this $p(x)$. There are many ways to show that $p(x)$ is irreducible. For example, we can note that it has no real roots, and hence it must be a product of two 2nd degree polynomials, if it is not irreducible. But $p(2)=p(-2)=13$ and $p(3)=p(-3)$ and $p(4)=p(-4)=241$, so it is a prime number for at least 6 different integer values of $x$. However, a product of two second degree polynomials with integer coefficients can only be prime for $4$ integer values of $x$, since one of the factors must be either $1$ or $-1$ for $p(x)$ to be prime.
|
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|
Laurent series in various regions I have the question: "Find the Laurent series which represents the function
$$
f(z) = (z^2 - 1)/(z + 2)(z + 3)\
$$
in the regions
(i) $\mid z\mid < 2\ $
(ii) $ 2 < \mid z\mid < 3\ $
(iii) $\mid z\mid > 3\ $"
I can rewrite this function into partial fractions as:
$$
f(z) = 5/(z+2) - 10/(z+3)
$$
I understand that for part (ii) this is the region within the annulus.
I also know that I can write:
$$
5/(z+2) = 5/2 \sum (-1)^n (z/2)^n $$ for $ \mid z\mid < 2\ $
$$
10/(z+3) = 10/3 \sum (-1)^n (z/3)^n $$ for $ \mid z\mid < 3\ $
I also know how to find the respective sums for $ \mid z\mid > 2\ $ and $ \mid z\mid > 3\ $
But I'm not at all sure how to combine these to get the answers for the whole function $f$, for parts (i), (ii) and (iii).
Any help would be much appreciated
|
(i) $|z| < 2$
$$\frac5{z+2}-\frac{10}{z+3} = \sum_{n=0}^\infty (-1)^n \left(\frac52\left(\frac{z}2\right)^n-\frac{10}3\left(\frac{z}3\right)^n\right)$$
(ii) $2<|z|<3$
\begin{align*}
& \frac{5}{z+2} - \frac{10}{z+3} \\
=& \frac{5}{z} \frac{1}{1+\frac{2}{z}} - \frac{10}{3} \frac{1}{1+\frac{z}{3}} \\
=& -\frac{10}{3} \sum_{n = 0}^\infty (-1)^n \left( \frac{z}{3} \right)^n + \frac{5}{z} \sum_{n = 0}^\infty (-1)^n \left( \frac{2}{z} \right)^n \\
=& \sum_{n = 0}^\infty (-1)^{n + 1} \frac{10}{3} \left( \frac{z}{3} \right)^n + \sum_{n = 1}^\infty (-1)^{n - 1} 5 \left( \frac{2}{z} \right)^n
\end{align*}
(iii) $|z|>3$
\begin{align*}
& \frac{5}{z+2} - \frac{10}{z+3} \\
=& \frac{5}{z} \frac{1}{1+\frac{2}{z}} - \frac{10}{z} \frac{1}{1+\frac{3}{z}} \\
=& \frac{1}{z} \sum_{n = 0}^\infty (-1)^n \left( 5 \left( \frac{2}{z} \right)^n - 10 \left( \frac{3}{z} \right)^n \right) \\
=& \sum_{n = 1}^\infty (-1)^{n - 1} \left( 5 \left( \frac{2}{z} \right)^n - 10 \left( \frac{3}{z} \right)^n \right)
\end{align*}
|
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|
Computing an infinite trigonometric sum $\sum \frac{2}{n^2 \pi^2} \cos(n \pi x)\cos(n \pi y)$ Let $G(x,y) = \sum_{n=1}^\infty \frac{2}{n^2 \pi^2} \cos(n \pi x)\cos(n \pi y)$
I'm trying to compute this sum by understanding it as an integral kernel. This question comes from Dym and Mckean Fourier Series and Integrals Ex 1.7.14:
Consider the cosine basis on $L^2[0,1]$ defined by $f_n(x) = \sqrt 2 \cos(n \pi x)$ for $n \ge 1$ and $f_0 = 1$. Define the operator $F$ by $F f_n = \frac{1}{n^2\pi^2} f_n$ acting only on the subspace $n \ge 1$.
It turns out that $F f(x) = \int_0^1 G(x, y) f(y) dy$ for $f \in L^2[0,1]$.
Now let $u = f_n''$ for some unspecified $n \ge 1$ .
$$F (f_n'') = -f_n $$
$$(F(f_n''))'' = -f_n''$$
$$(F u)'' = -u$$
$$(Fu)'(x) =-\int_0^x u(y)dy + K_1$$
$K_1 = 0$ because $(F u)'(0) = C \sin(0) = 0$
$$(Fu)(x) = -\int_0^x \int_0^y u(z)dz\,dy + K_2$$
$K_2 = \int_0^1\int_0^x \int_0^y u(z)dz\,dy\,dx$ because $\int_0^1 (Fu)(x) dx = \int_0^1 C \cos (2 \pi n x) dx = 0$
Now changing order of integration
$$ (Fu)(x) = - \int_0^x (x - z)u(z) dy + K_2$$
$$ = - \int_0^x (x - z)u(z) dy + \int_0^1\int_0^x (x - z)u(z) dz\,dx$$
$$ =... + \int_0^1\int_z^1(x -z)u(z)dx\,dz$$
$$ =... + \int_0^1u(z) [(1 - z^2)/2 - z(1-z)] dz $$
$$ = ... + \int_0^1 u(z) (z - 1)^2/2 dz$$
$$ = - \int_0^x (x - z)u(z) dz + \int_0^1 u(z) (z - 1)^2/2 dz$$
$$ = - \int_0^x (x - z)u(z) dz + \int_0^x u(z) (z - 1)^2/2 dz+ \int_x^1 u(z) (z - 1)^2/2 dz$$
$$ = \int_0^1 T(x,z) u(z) dz$$
where $T(x,y) = (y^2 + 1)/2 - x $ for $y < x$ and $(y-1)^2/2$ for $y > x$
and by necessity $T(x,y) = G(x,y)$
Unfortunately, I don't think this is the right function and it doesn't even look symmetric in $x,y$. I am looking for help for the correct derivation which should lead to $G(x,y) = \frac{x^2 - x + y^2 -y - |x-y|}{2} + 1/3$
Edit:
Even though $T(x,y)$ gives rise to the same operator $F$, it may not equal to $G(x,y)$ since they can differ by any function $\Delta(x)$ so that $T(x,y) + \Delta(x) = G(x,y)$, since $\int_0^1\Delta(x) f_n(y) dy = 0$. Since $\int G(x,y) dy = 0$, we should require $\Delta(x) = -\int_0^1T(x,y)dy = -1/6 + x^2/2$ so that $G(x,y) = (y^2 + 1)/2 - x - x^2/2 + 1/6 $ for $y < x$ and $(y-1)^2/2 + x^2/2 + 1/6$ for $y > x$ and this is the book's answer.
|
Marty Cohen is correct in his approach to this $G(x,y)$ but there is the sign error in the trig expressions.
Defined on the interval $0 < t < P$, it is fairly easy to demonstrate that the function:
\begin{equation}
h(t)=\pi^2 \left[ \left( \frac{t}{P} \right)^2 -\left( \frac{t}{P} \right) + \frac{1}{6} \right]
\end{equation}
has a Fourier series expansion of:
\begin{equation}
h(t) = \sum\limits_{n = 1}^{\infty}
\frac{
\cos\left( 2 \pi \frac{n}{P} t \right)
}{n^2}
\end{equation}
The polynomial, $h(t)$, is Bernoulli Polynomial, $B_2(\frac{t}{P})$ which has been scaled appropriately.
Using $h(t)$ above and setting $P=2$ the closed form of $g(z)$ above is given by:
\begin{equation}
g(z) = \left[ \left(\frac{z}{2}\right)^2 - \left(\frac{z}{2}\right) + \frac{1}{6} \right]
\end{equation}
This is exactly the same value of $g(z)$ found above, so his exposition on the approch is correct.
Proceeding with the reasoning of Marty Cohen but with the sign correction for $cos(a)cos(b)$:
\begin{align}
G(x,y) &= g(x-y) + g(x+y)
\\
&= \left[ \left( \frac{x-y}{2} \right)^2 - \left( \frac{x-y}{2} \right) + \frac{1}{6} \right]
\\
&+ \left[ \left( \frac{x+y}{2} \right)^2 - \left( \frac{x+y}{2} \right) + \frac{1}{6} \right]
\\
&= \left[ \frac{x^2}{4} - xy + \frac{y^2}{4} - \frac{x-y}{2} + \frac{1}{6} \right]
\\
&+ \left[ \frac{x^2}{4} + xy + \frac{y^2}{4} - \frac{x+y}{2} + \frac{1}{6} \right]
\\
&= \frac{x^2 + y^2}{2} -x +\frac{1}{3}
\end{align}
where the domain for $x$ and $y$ are:
*
*$0 \le (x + y) \le 2$
*$0 \le (x - y) \le 2$
This is different way to answer to the question: "What is G(x,y) in closed form?" It is not in keeping with the the essence of the question posed by the OP so it is not a proper answer.
None the less, this confirms Marty Cohen's approach (less the sign error) is correct.
|
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|
How many rectangles are there on an $8 \times 8$ checkerboard?
How many rectangles are there on an $8 \times 8$ checkerboard?
\begin{array}{|r|r|r|r|r|r|r|r|}
\hline
& & & & & & & \\ \hline
& & & & & & & \\ \hline
& & & & & & & \\ \hline
& & & & & & & \\ \hline
& & & & & & & \\ \hline
& & & & & & & \\ \hline
& & & & & & & \\ \hline
& & & & & & & \\ \hline
\end{array}
Attempt
I just counted them up via casework:
a
$1 \times 1: 64 $
$1 \times 2: 56 $
$\vdots$
$1 \times 8: 8$
Then,
$2 \times 2: 49$
$2 \times 3: 42 $
$\vdots$
The pattern continues like it looks like it should.
Thus, we can sum up all of these solutions as $8(1+\cdots+8)+7(1+\cdots+7)+\cdots+2(1+2)+1(1) = 750$, but the correct answer is $1296$. Where did I go wrong?
|
First, the easy way to count them is to notice that each rectangle is completely determined by its top and bottom edges and its left and right edges. Pick any two of the nine horizontal lines and any two of the nine vertical lines, and you’ve picked out a rectangle. Conversely, each rectangle determines two horizontal and two vertical lines, those on which its edges lie. Thus, there must be
$$\binom92^2=36^2=1296$$
rectangles.
Your calculation is off because you forgot that a rectangle can be wider than it is tall, so you missed half of the non-square rectangles.
|
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|
Positive Integer solutions to $y = \frac{x z}{-x - z + x z}$ I'm trying to find positive integer solutions to the following diophantine equation:
$$y = \frac{x z}{-x - z + x z}$$
The first thing I did was split the fraction as follows:
$$ \frac{x z}{-x - z + x z} = 1 + \frac{x+z}{x z-x-z}$$
This doesn't seem to help any though. The only thing I can pin down is that the numerator of the fraction must be larger than the denominator to get a positive value for y, placing a constraint on $x$ and $z$ as follows:
$$z>2\land x<\frac{2 z}{z-2}$$.
Now what though?
|
If $-x-z+xz=0$, then $(x-1)(z-1)=1$, so $(x,z)=(2,2)$, which doesn't give a solution.
If $-x-z+xz\neq 0$, then the equation is equivalent to $xy+yz+zx=xyz$, i.e. $\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=1$.
Let wlog $x\ge y\ge z\ge 1$. Then $\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=1\le \frac{3}{z}$, so $z\le 3$.
If $z=1$, then $\frac{1}{x}+\frac{1}{y}=0$; no solutions.
If $z=2$, then $\frac{1}{x}+\frac{1}{y}=\frac{1}{2}$, i.e. $2y+2x=xy$, i.e. $(x-2)(y-2)=4$.
$x-2\ge y-2\ge -1$, so either $(x-2,y-2)=(4,1)$ or $(2,2)$. We get the solutions $(x,y,z)=(6,3,2),(4,4,2)$.
If $z=3$, then $\frac{1}{x}+\frac{1}{y}=\frac{2}{3}$, so $3y+3x=2xy$, so $(2x-3)(2y-3)=9$.
$2x-3\ge 2y-3\ge -1$, so either $(2x-3,2y-3)=(9,1)$ or $(3,3)$. In the first case $y=2<z=3$. In the second case we get the solution $(x,y,z)=(3,3,3)$.
All the solutions are $$(x,y,z)=(6,3,2),(6,2,3),(3,6,2),(3,2,6),(2,6,3),\\(2,3,6),(4,4,2),(4,2,4),(2,4,4),(3,3,3)$$
|
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|
Problem with a residue calculation I have a problem with calculating a residue. I want to calculate Res($f,\frac{1}{2}$), where
$$ f(z) = \frac{z^6 +1}{z^3(2z-1)(z-2)} = \frac{z^6 +1}{2z^5 -5z^4+2z^3}.$$
$$ $$
One method is to view $f(z)$ as $\frac{p(z)}{q(z)}$, where both $p$ and $q$ are analytic; here the residue of $\frac{1}{2}$ is then:
$$ Res(f,\frac{1}{2})= \frac{p(\frac{1}{2})}{q'(\frac{1}{2})} = \frac{(\frac{1}{2})^6 +1}{10(\frac{1}{2})^4 -20(\frac{1}{2})^3+6(\frac{1}{2})^2} = -\frac{65}{24}.$$
$$ $$
Another method we learnt is this: We can view $f(z)$ as $f(z) = \frac{\phi(z)}{(z-a)^n}$, where $\phi(z)$ is analytic. The residue here is then given by: $Res(f,a) = \frac{\phi^{(n-1)}(a)}{(n-1)!} $.
Applying this to our example, where $\phi(z) = \frac{z^6 +1}{z^3(z-2)}$, gives:
$$Res(f,\frac{1}{2}) = \frac{z^6 +1}{z^3(z-2)}|_\frac{1}{2} = \frac{(\frac{1}{2})^6 +1}{(\frac{1}{2})^3(\frac{1}{2}-2)} =-\frac{65}{12}. $$
$$ $$
Why are the two residues different? The first residue is the correct one. Why can't I here apply the second rule? In other examples both methods give the same residue.
|
$\text{Res}(f,\frac 12)=\lim_{z \to \frac 12}(z-\frac 12)f(z)=\lim_{z\to \frac 12} \frac {(z-\frac 12)(z^6+1)}{(z^3)(z-2)(2z-1)} =\lim_{z\to \frac 12}\frac {(z-\frac 12)(z^6+1)}{2(z^3)(z-2)(z-\frac 12)}=\lim_{z\to \frac 12} \frac {z^6+1}{2(z^3)(z-2)}=-\frac {65}{24}$.
|
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|
minimum $x^2 + y^2$ on $\frac{(x-12)^2}{16} + \frac{(y+5)^2}{25} = 1 $ ellipse Given $\frac{(x-12)^2}{16} + \frac{(y+5)^2}{25} = 1$.
Then minimum value of $x^2 + y^2 = ?$
P.S. My solution: Suppose that $x = 4\cos{\theta}+12$and $y = 5\sin{\theta}-5$
and expand $x^2 + y^2$ to find minimum value, but stuck in the end.
Thank you for every comment.
|
Another, maybe not as elegant, way is to solve for $y$, getting $$y=-5\pm\frac54\sqrt{16-\left( x -12\right)^2}$$ and then, since $$y_+(x):=\left(-5+\frac54\sqrt{16-\left( x -12\right)^2}\right)^2\le\left(-5-\frac54\sqrt{16-\left( x -12\right)^2}\right)^2=:y_-(x), $$ find the zero $x_0$ of the derivative of $x^2+(y_+(x))^2$ , which is $$-\frac98 x +\frac{25}{2}\left(3+\frac{x-12}{\sqrt{16-(x-12)^2}}\right) \tag{$\star$}$$ and compute $x_0^2+(y_+(x_0))^2$.
Note that $x_0$ is unique and is approximately $8.345$. The other solution mentioned by Mirko is produced by squaring, which is necessary to solve exactly $(\star)$, ending up with $$\frac{(x-12)^2}{16-(x-12)^2}=\left(\frac{9}{100}x-3\right)^2,$$ whose solutions are indeed the roots of the quartic found by Mirko.
|
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|
Arithmetic: Prove that is multiple of 30 Prove that $n^{19}-n^7$ is multiple of $30$
I've seen $6$ can divide it because
$$n^{19}-n^7=n^7(n^{12}-1) = n^7(n^6+1)(n^6-1)=n^4(n^6+1)(n^3-1)n^3(n^3+1)$$
And there are three consecutive numbers, so, at least one is multiple of $3$ and up to two even numbers.
But, how to prove that is multiple of $5$?
|
Say $$n \equiv 0,\pm1,\pm2 \pmod 5$$ or,
$$n^2 \equiv 0,1,4 \pmod 5$$ or,
$$n^6 \equiv 0,1,64 \pmod 5$$ or,
$$n^6 \equiv 0,1,-1 \pmod 5$$
Therefore, $5$ divides at least one of $n^6,n^6-1$ or $n^6+1$, that is, $5$ divides $n^6(n^6-1)(n^6+1)$ .
And
$n^{19}-n^7=n\cdot n^6(n^{12}-1)=n\cdot n^6(n^6-1)(n^6+1)$
Hence given expression is divisible by $5$ and thus by $30$.
Hope this helps.
|
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|
How does the Schwarz inequality prove the following? I don't quite see how the Schwarz inequality proves that, for
$$f_n(x) = \frac{x}{1+nx^2}$$
we have
$$|f_n(x)| \leq \frac{|x|}{2\sqrt{n}|x|} = \frac{1}{2\sqrt{n}}.$$
|
Consider the vectors $a = (1,x\sqrt{n})$ and $b = (x\sqrt{n},1)$. By the Cauchy-Schwarz Inequality:
$$|a \cdot b| \le \|a\| \cdot \|b\|$$
$$|1 \cdot x\sqrt{n} + x\sqrt{n} \cdot 1| \le \sqrt{(1)^1+(x\sqrt{n})^2} \cdot \sqrt{(1)^1+(x\sqrt{n})^2}$$
$$|2x\sqrt{n}| \le 1+(x\sqrt{n})^2$$
$$2\sqrt{n}|x| \le 1+nx^2$$
Therefore, $|f_n(x)| = \dfrac{|x|}{1+nx^2} \le \dfrac{|x|}{2\sqrt{n}|x|} = \dfrac{1}{2\sqrt{n}}$, as desired.
It might be easier to use the AM-GM inequality, i.e. $\dfrac{1+nx^2}{2} \ge \sqrt{1 \cdot nx^2}$, i.e. $1+nx^2 \ge 2\sqrt{n}|x|$.
|
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|
Asymptotic expansion at order 2 of $\int_0^1 \frac{x^n}{1+x} \, dx$ I'd like to get an asymptotic expansion of $\int_0^1 \frac{x^n}{1+x} \, dx$ at order two in $\frac{1}{n}$.
I'm able to prove that $$\lim\limits_{n \to \infty} n \int_0^1 \frac{x^n}{1+x} \, dx = \frac{1}{2}$$ which provides an asymptotic expansion at order $1$. How can I go one step further? Even better, is there a way to get an asymptotic expansion at any order $m$?
|
Method 1.
An elementary approach.
You may just integrate by parts twice,
$$
\begin{align}
I_n&=\int_0^1\frac{x^n}{1+x}\:dx
\\&=\left. \frac{x^{n+1}}{(n+1)}\frac{1}{1+x}\right|_0^1+\frac{1}{(n+1)}\int_0^1\frac{x^{n+1}}{(x+1)^2}\:dx\\
&=\frac1{2(n+1)}+\frac{1}{n+1}\int_0^1\frac{x^{n+1}}{(x+1)^2}\:dx\\
&=\frac1{2(n+1)}+\frac{1}{n+1}\left(\left. \frac{x^{n+2}}{(n+2)}\frac{1}{(1+x)^2}\right|_0^1+\frac{2}{(n+1)}\int_0^1\frac{x^{n+1}}{(x+1)^3}\:dx \right)\\
&=\frac1{2(n+1)}+\frac1{4(n+1)(n+2)}+\frac{2}{(n+1)^2}\int_0^1\frac{x^{n+1}}{(x+1)^3}\:dx
\end{align}
$$ but
$$
0\leq \int_0^1\frac{x^{n+1}}{(x+1)^3}\:dx\leq\int_0^1x^{n+1}dx=\frac{1}{(n+2)}
$$ thus
$$
\frac{2}{(n+1)^2}\int_0^1\frac{x^{n+1}}{(x+1)^3}\:dx=\mathcal{O}\left(\frac{1}{n^3} \right)
$$
Finally, as $n \to \infty$,
$$ I_n=\int_0^1\frac{x^n}{1+x}dx=\frac1{2n}-\frac{1}{4n^2}+\mathcal{O}\left(\frac{1}{n^3} \right).$$
$$
$$
Method 2.
One may use the standard integral representation of the digamma function and its asymptotics, as $n \to \infty$,
$$
\begin{align}
I_n=\int_0^1\frac{x^n}{1+x}\:dx&=\int_0^1\frac{x^n-x^{n+1}}{1-x^2}\:dx\\
&=\frac12\int_0^1\frac{(1-t^{n/2})-(1-t^{(n-1)/2})}{1-t}\:dt\\
&=\frac12\psi\left(\frac{n}{2}+1\right)-\frac12\psi\left(\frac{n}{2}+\frac12\right)\\
&=\frac1{2n}-\frac{1}{4n^2}+\mathcal{O}\left(\frac{1}{n^3} \right).
\end{align}
$$
|
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"timestamp": "2023-03-29T00:00:00",
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|
Prove that for all real numbers $x,y$ and $z$ that $x^2+y^2+z^2 \geq x\sqrt{y^2+z^2}+y\sqrt{x^2+z^2}$
Prove that for all real numbers $x,y$ and $z$ that $x^2+y^2+z^2 \geq x\sqrt{y^2+z^2}+y\sqrt{x^2+z^2}$.
It looks like AM-GM should be used here but the square roots make it difficult. So maybe Cauchy-Schwarz works?
|
Note that (by AM-GM),
$\frac{x^2+(y^2+z^2)}{2}\geq\sqrt{x^2(y^2+z^2)}=x\sqrt{y^2+z^2}$.
and similarly,
$\frac{y^2+(x^2+z^2)}{2}\geq\sqrt{y^2(x^2+z^2)}=y\sqrt{x^2+z^2}$.
Summing them up gives:
$x^2+y^2+z^2\geq x\sqrt{y^2+z^2}+y\sqrt{x^2+z^2}$
|
{
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|
Prove that the sum $1^k+2^k+\cdots+n^k$ where $n$ is an arbitrary integer and $k$ is odd, is divisible by $1+2+\cdots+n$.
Prove that the sum $$1^k+2^k+\cdots+n^k$$
where $n$ is an arbitrary integer and $k$ is odd, is divisible by $1+2+\cdots+n$.
Question
In the solution to this problem it splits it up into two cases: ($1$) $n$ is an even integer ($2$) and $n$ is an odd integer. In the case where $n$ is an odd integer it says the following: $$1^k+n^k,2^k+(n-1)^k,3^k+(n-2)^k,\ldots, \left (\dfrac{n-1}{2} \right )^k + \left(\dfrac{n+3}{2} \right )^k \left (\dfrac{n+1}{2} \right )^k$$ are all divisible by $\dfrac{n+1}{2}$.
I get how the beginning terms are all divisible by $\dfrac{n+1}{2}$, but did they make a typo when they said $\left(\dfrac{n+3}{2} \right )^k \left (\dfrac{n+1}{2} \right )^k$? If not, then how is $\left (\dfrac{n-1}{2} \right )^k + \left(\dfrac{n+3}{2} \right )^k \left (\dfrac{n+1}{2} \right )^k$ divisible by $\dfrac{n+1}{2}$?
|
Using Proof of $a^n+b^n$ divisible by a+b when n is odd,
$$r^k+(n-r)^k$$ is divisible by $r+n-r=n$ as $k$ is odd
$$\implies\sum_{r=1}^n(r^k+(n-r)^k)$$ will be divisible by $n$
Similarly,
$$\sum_{r=1}^n(r^k+(n+1-r)^k)$$ will be divisible by $r+n+1-r=n+1$
$$\implies\sum_{r=1}^n(r^k+(n+1-r)^k)=2\sum_{r=1}^n r^k$$ will be divisible by lcm$(n+1,n)$
|
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"timestamp": "2023-03-29T00:00:00",
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|
Prove that $\frac{1}{a}+\frac{1}{b}+\frac{1}{c} \geq 2 \left (\frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{c+a} \right) \geq \frac{9}{a+b+c}$
Let $a,b,$ and $c$ be positive real numbers, prove that $\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c} \geq 2 \left (\dfrac{1}{a+b}+\dfrac{1}{b+c}+\dfrac{1}{c+a} \right) \geq \dfrac{9}{a+b+c}$.
Should I use AM-GM for the expression in the middle of the inequality? We have $a+b \geq 2\sqrt{ab}$ etc.?
|
To prove:
$2(\frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{c+a}) \geq \frac{9}{a+b+c}$,
let $x=a+b, y=b+c, z=c+a$, then we have:
$\frac{1}{x}+\frac{1}{y}+\frac{1}{z} \geq \frac{9}{x+y+z} \Rightarrow 3+(\frac{y}{x}+\frac{x}{y}) +(\frac{y}{z}+\frac{z}{y})+ (\frac{z}{x}+\frac{x}{z}) \geq 3+2+2+2=9$
|
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|
Algebra: Rational roots of a polynomial of degree $4$ Consider the following polynomial with real coefficients: $x^4(t^5 - 1) - x^3(1+4t^5) + x^2(6t^5 -1) - x(1+4t^5) + (t^5 -1 ) = 0$, where $x$ and $t$ are both rational and $t$ is fixed.
By simple algebra, we find that the sum of the roots is $\dfrac{1+4t^5}{1-t^5}$ and their product is $1$. Prove that any rational solution $x$ to this equation is in fact an integer.
From my initial attempts, the ideas of the sum and product of the roots seem not to be offering much help.
|
Let $x$ be a rational solution to the equation. We can easily check that $x \not = 1$ and $t \not = 0$. Then
$$\begin{array}{c c c c}
&x^4 (t^5-1) - x^3 (4t^5+1) + x^2 (6t^5-1) - x (4t^5+1) + (t^5-1)&=&0 \\
\Leftrightarrow&t^5 (x^4 - 4x^3 + 6x^2 - 4x + 1) - (x^4 + x^3 + x^2 +x + 1) &=&0\\
\Leftrightarrow&t^5 (x-1)^4&=&x^4 + x^3 + x^2 +x + 1\\
\Leftrightarrow&t^5 (x-1)^5&=&(x-1)(x^4 + x^3 + x^2 +x + 1)\\
\Leftrightarrow&t^5 (x-1)^5&=&x^5-1
\end{array}$$
Now let's have $t = \dfrac{p}{q}$ and $x = \dfrac{a}{b}$. This gives us
$$\begin{array}{c c c c}
&\dfrac{p^5}{q^5} (\dfrac{a}{b}-1)^5&=&\dfrac{a^5}{b^5}-1\\
\Leftrightarrow&p^5 \dfrac{(a-b)^5}{b^5}&=&\dfrac{a^5-b^5}{b^5} q^5\\
\Leftrightarrow&p^5 (a-b)^5&=&(a^5-b^5) q^5\\
\end{array}$$
The last equality implies that there exists $c \in \mathbb{Z}$ such that $a^5-b^5=c^5$.
By Fermat's Last Theorem, we either have $a = 0$ or $b = 0$ or $c = 0$. The last two cases are impossible since $b \not = 0$ and $x \not = 1$. Hence we must have $a = 0$.
$$\boxed{\text{So the only possible rational solution to the equation is }x = 0}$$
And we can easily see it is indeed a solution if and only if $t = 1$.
|
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|
Help with $\tan(x)\sin^2(y) + \cos^2(x) \cot(y) y’=0$ I need help with differential equation $$\tan(x)\sin^2(y) + \cos^2(x) \cot(y) y’=0$$
I know that $y’=\frac{dy}{dx}$, but i have no idea what to do next.
|
$$\tan(x)\sin^2(y) + \cos^2(x) \cot(y) y’=0$$
We know that $$y’=\frac{dy}{dx}$$
$$\tan(x)\sin^2(y) + \cos^2(x) \cot(y) \frac{dy}{dx}=0
-\tan(x)\sin^2(y)=\cos^2(x)\cot(y)\frac{dy}{dx}
-\frac{\tan(x) dx}{\cos^2(x)}=\frac{\cot(y) dy}{\sin^2(y)} $$
We also know that $$ \tan(x)= \frac{\sin(x)}{\cos(x)} $$ and $$ \cot(x)=\frac{\cos(x)}{\sin(x)}$$.
$$ -\frac{\sin(x) dx}{\cos^2(x)}=\frac {\cos(y)}{\sin^3(y)} $$
$$-\int{\frac{\sin(x) dx}{\cos^2(x)}}=\int {\frac {\cos(y)}{\sin^3(y)}} $$
And we know that :
$$\int{\frac{\sin(x)}{\cos^2(x)}}=$$
Setting : $$t=\sin(x)$$ we have got: $$dx=\frac{dt}{\cos(x)}$$
$$\int{\frac{t}{1-t^2}dt}=$$
Setting $$ u= t^2$$ we have got: $${\frac{du}{2t}}=dt$$
And finally we get: $$\int{\frac{\sin(x)}{\cos^2(x)}}= \frac {1}{2\cos^2(x)}$$
Next:
When you setting $$t=cos(y)$$ and $$u=t^2 $$
$$ \int{\frac{\cos(y) dy}{\sin^3(y)}}= \frac{1}{2\sin^2(y)} $$
Finally we have:
$$-\frac{1}{2\cos^2(x)}=\frac{1}{2\sin^2(y)} $$
$$-\sin^2(y)=\cos^2(x)$$
$$0=\cos^2(x)+\sin^2(y)$$
And the solutions for the variable y are:
$$ y= 2\pi C – i\sinh^-1 (\cos(x))$$
$$y= 2\pi C – i\sinh^-1 (\cos(x)) +\pi$$
$$y= 2\pi C + i\sinh^-1 (\cos(x))$$
$$y= 2\pi C + i\sinh^-1 (\cos(x))+\pi $$
|
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|
Prove that $5^n + 2\cdot3^{n-1} + 1$ is multiple of $8$ Prove that $5^n + 2\cdot3^{n-1}+ 1$ is multiple of $8$.
I've tried using induction (it isn't):
For $n=1$:
$$5^1 + 2\cdot3^{n-1} + 1 = 8$$
If it is true for $n$, then $n+1$?
\begin{align}
5^{n+1} + 2\cdot3^n + 1
=
&(4+1)^n\cdot(4+1)+ 2\cdot(2+1)^n + 1
\\
=& (4^n + n4^{n-1} + 1)\cdot(4+1) + 2\cdot(2^n + n2^{n-1} + 1) + 1
\\
=
&
(4k+1)\cdot(4+1) + 2(2r+1) + 1
\\
= &16k+4k+4 +1+4r+2+1
\\
=
&20k + 4r + 8 = 4(5+r+2)
\end{align}
But i've only proved it is multiple of $4$.
|
Suppose it's true for $n\ge1$: then
$$
5^n+2\cdot3^{n-1}+1=8k
$$
for some integer $k$; in particular, $5^n=8k-2\cdot3^{n-1}-1$. Then
\begin{align}
5^{n+1}+2\cdot3^{n}+1
&=5(8k-2\cdot3^{n-1}-1)+2\cdot3^{n}+1 \\[3px]
&=40k-10\cdot 3^{n-1}-5+6\cdot 3^{n-1}+1\\[3px]
&=40k-4\cdot 3^{n-1}-4
\end{align}
Thus you just need to show $4\cdot 3^{n-1}+4$ is divisible by $8$, that is, $3^{n-1}+1$ is divisible by $2$, which is obvious for $n\ge1$, because $3$ is odd and so is any of its power (including $3^0=1$).
|
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|
Prove that $n(1+n)^{\frac{1}{n}} < n+H_n$ for every $n \geq 2$.
For every positive integer $n$ set $H_n = \dfrac{1}{1}+\dfrac{1}{2}+\cdots+\dfrac{1}{n}$. Prove that $n(1+n)^{\frac{1}{n}} < n+H_n$ for every $n \geq 2$.
Attempt
I will prove this result by induction. The base case holds since for $n = 2$ we have $(1+2)^{\frac{1}{2}} < 2 + H_2 = 3.5$. Now assume that $k(1+k)^{\frac{1}{k}} < k+H_k$ holds for some $k$. We need to show that $(k+1)(2+k)^{\frac{1}{k+1}} < 1+k+H_{k+1}$. How do I show this? Also, is induction the best way to prove this?
|
Notice that $$n+1 = \frac{2}{1}\frac{3}{2} \cdots \frac{n}{n-1}\frac{n+1}{n} $$
Hence by AM-GM we get
$$ \left (n+1\right )^{\frac{1}{n}} < \frac{ \frac{2}{1}+\frac{3}{2}+ \cdots \frac{n}{n-1}+\frac{n+1}{n} }{n} $$
But observe that
$$\frac{ \frac{2}{1}+\frac{3}{2}+ \cdots \frac{n}{n-1}+\frac{n+1}{n} }{n}= \frac{n + H_n}{n}.$$
And we are done.
|
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|
Multiplying the denominator $$\frac{2}{3} \gt -4y - \frac{25}{3}$$
My question is to solve this problem we need to multiply both sides by $3$. The result will be
$$2 \gt -12y - 25.$$
Why doesn't the numerator of both $\frac{2}{3}$ and $\frac{25}{3}$ get multiplied?
|
$$\begin{align}
\frac{2}{3}
&>
-4y-\frac{25}{3} \\
3\left(\frac{2}{3}\right)
&>
3\left(-4y-\frac{25}{3}\right) \\
3·\frac{2}{3}
&>3·\left(-4y\right)-3·\frac{25}{3} \\
\require{cancel} \cancel{3}·\frac{2}{\require{cancel} \cancel{3}}
&>
3·\left(-4y\right) -\require{cancel} \cancel{3} ·\frac{25}{\require{cancel} \cancel{3}} \\
2
&>
-12y -25 \\
\end{align}$$
|
{
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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|
Compute $\int\frac{x}{2x^2+x+3}dx$ $$\int\frac{x}{2x^2+x+3}\,dx$$
Well, to approach this kind of exercises I know that I need to check the derivative of the denominator. which is $4x + 1$.
Then, I can re-write the integral: $\int\frac{0.25(4x+1) - 0.25}{2x^2+x+3}\,dx$.
Then, I get:
$$0.25\int\frac{(4x+1)}{2x^2+x+3}\,dx-0.25\int\frac{1}{2x^2+x+3}\,dx.$$
The left one is pretty straight forward with $\ln|\cdot|$,
Problem: does anyone have some "technique" to solve the right integral? hints would be appreciated too.
Edit: maybe somehow: $$0.25\int\frac{1}{2(2x^2/2+x/2+3/2)}\,dx = 0.25\int\frac{1}{(x+0.25)^2 + \frac{23}{16}}\,dx$$
|
$$\int \frac{1}{2x^2+x+3}\, dx=\frac{1}{\sqrt{2}}\int \frac{d\left(\sqrt{2}x+\frac{1}{2\sqrt{2}}\right)}{\left(\sqrt{2}x+\frac{1}{2\sqrt{2}}\right)^2+\frac{23}{8}}\, $$
Let $\sqrt{2}x+\frac{1}{2\sqrt{2}}=\sqrt{\frac{23}{8}}\tan u$.
$$=\frac{1}{\sqrt{2}}\int \frac{\sqrt{\frac{23}{8}}\frac{1}{\cos^2 u}}{\frac{23}{8}(\tan^2 u+1)}\, du=\frac{2}{\sqrt{23}}\int du=\frac{2}{\sqrt{23}}u+C$$
|
{
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"timestamp": "2023-03-29T00:00:00",
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|
find the tangent to the sphere obtain the equations of tangent to sphere
$$x^2+ y^2+z^2+6x-2z+1 = 0$$
which pass through the line
$$3 (16-x) = 3z=2y+30$$
Now I know if the plane is $$lx +my+n z=p$$
then $$-I/3 +m/2+n/3=0$$
also $(16,-15,0)$ is a point on the plane
I know that there is $2$ answers , but how to proceed
|
The sphere has for equation $$S \equiv (x+3)^2-9+y^2+(z-1)^2-1+1=(x+3)^2+y^2+(z-1)^2-9=0$$ which means that its center is $C= (-3,0,1)$ and its radius is equal to $r=3$. The line has for equations $$L \equiv \begin{cases}
x+z=16\\
2y-3z=-30
\end{cases}$$ It goes through the point $P_L=(16,-15,0)$ and has $v_L=(-2,3,2)$ as a direction vector.
The general equation of a plane $\mathscr P$ is $a x+by+cz+d=0$ where $a^2+b^2+c^2=1$. $\mathscr P$ is passing through $P_L$ so you must have $16a-15b+d=0$. And $\mathscr P$ contains the direction of the line, so $-2a+3b+2c=0$. Finally as $\mathscr P$ is supposed to be tangent to the sphere, the distance of $\mathscr P$ to $C$ is equal to the radius. Hence $\vert -3a+c+d \vert = 3$.
Therefore, you have to solve following system of equations $$\begin{cases}
a^2+b^2+c^2=1\\
16a-15b+d=0\\
-2a+3b+2c=0\\
\vert -3a+c+d \vert = 3
\end{cases}$$
You'll get two different solutions depending on the sign of $\vert -3a+c+d \vert$.
|
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"timestamp": "2023-03-29T00:00:00",
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|
Integrate $\int\frac{1}{x}\sqrt{\frac{1-x^2}{1+x^2}}\,\mathrm{d}x$ How do I go about integrating:
$$\int\frac{1}{x}\sqrt{\frac{1-x^2}{1+x^2}}\,\mathrm{d}x$$
The common trigonometric substitutions don't seem to work here.
I think it requires to take some power of $x$ outside the square root but I am not able to solve further.
|
The answer is (after a request from OP) updated with more details
I suggest you to set
$$
u=\sqrt{\frac{1-x^2}{1+x^2}}.
$$
Then
$$
x^2=\frac{1-u^2}{1+u^2}
$$
and so
$$
2x\,dx=-\frac{4u}{(1+u^2)^2}\,du.
$$
Thus
$$
\begin{aligned}
\int \frac{1}{x}\sqrt{\frac{1-x^2}{1+x^2}}\,dx&=\int\frac{1}{2x^2}\sqrt{\frac{1-x^2}{1+x^2}}\,2x\,dx\\
&=\int\frac{1}{2}\frac{1+u^2}{1-u^2}u\cdot\Bigl(-\frac{4u}{(1+u^2)^2}\Bigr)\,du\\
&=-\int\frac{2u^2}{(1-u^2)(1+u^2)}\,du.
\end{aligned}
$$
Next, do a partial fraction decomposition, and you will end up with
$$
\int\frac{1}{1+u^2}\,du-\int\frac{1}{1-u^2}\,du.
$$
I guess you can take it from here? If not, ask for more details.
|
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"timestamp": "2023-03-29T00:00:00",
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|
To solve differential equation ($xy^{3} + x^{2}y^{7}) \frac{dy}{dx} = 1$ The ODE is
($xy^{3} + x^{2}y^{7}) \frac{dy}{dx} = 1$
I have tried everything like integrating factor,it is not homogenous and not linear differential equation..What should be done now?
|
Re-arranging your differential equation, we have
$$\frac{dx}{dy}=xy^3+x^2y^7$$
$$\frac{dx}{dy}-xy^3=x^2y^7$$
$$\frac{1}{x^2}\cdot \frac{dx}{dy}-\frac{1}{x}\cdot y^3=y^7$$
$$-\frac{d}{dy}\left(\frac{1}{x}\right)-y^3\cdot \left(\frac{1}{x}\right) =y^7$$
Put $u=\frac{1}{x}$
You get $$\frac{du}{dy}+uy^3=y^7$$
The integrating factor comes out to be $e^\frac{y^4}{4}$.
Then we have $$\frac{d}{dy}\left(ue^\frac{y^4}{4}\right)=y^7e^\frac{y^4}{4}$$
Now $$\int y^7e^\frac{y^4}{4} dy = \int y^4 e^\frac{y^4}{4} \cdot y^3 dy$$
$$=\int 4ze^z dz$$
where $z=\frac{y^4}{4}$.
Can you complete the integration now?
|
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"timestamp": "2023-03-29T00:00:00",
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|
Given matrix A. To find $A^{2010}$
Let $\theta = 2\pi/67$. Now consider the matrix
$$
A =
\begin{pmatrix}
\phantom{-}\cos \theta & \sin \theta \\
-\sin \theta & \cos \theta
\end{pmatrix}.
$$
Then the matrix $A^{2010}$ is
\begin{align*}
&\text{(A)}\;
\begin{pmatrix}
\phantom{-}\cos \theta & \sin \theta \\
-\sin \theta & \cos \theta
\end{pmatrix},
&
&\text{(B)}\;
\begin{pmatrix}
1 & 0 \\
0 & 1
\end{pmatrix}, \\
&\text{(C)}\;
\begin{pmatrix}
\phantom{-}\cos^{30} \theta & \sin^{30} \theta \\
-\sin^{30} \theta & \cos^{30} \theta
\end{pmatrix},
&
&\text{(D)}\;
\begin{pmatrix}
\phantom{-}0 & 1 \\
-1 & 0
\end{pmatrix}.
\end{align*}
I think answer is B. But I am not sure.
|
$A$ can be thought as $e^{-i\theta}$. So that for the natural homeomorphism $\phi:\mathbb{C}\rightarrow \mathbb{R}^2:x+iy \mapsto (x,y)$, $(\phi^{-1}\circ A\circ \phi )(z)=e^{-i\theta } z$. Hence, $(\phi^{-1} \circ A^{2010} \circ \phi)(z)= e^{- 2010 i \theta} z = e^{-60\pi i}z=z$. Hence, the answer is (B).
|
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|
How do I evaluate the sum $\sum_{k=1}^\infty\left(\ln\big(1+\frac{1}{k+a}\right)-\ln\left(1+\frac{1}{k+b}\big)\right)$ How do I evaluate the sum
$$\sum_{k=1}^\infty\left(\ln\Big(1+\frac{1}{k+a}
\Big)-\ln\Big(1+\frac{1}{k+b}\Big)\right)$$
where $0 <a<b<1$?
Hints will be appreciated
Thanks
|
Note that, for each fixed $k$, $\ln\left(1+\frac{1}{k+a}\right)-\ln\left(1+\frac{1}{k+b}\right)=\ln\left(\dfrac{1+\frac{1}{k+a}}{1+\frac{1}{k+b}}\right)$. Then:
\begin{eqnarray*}
\sum_{k=1}^n\ln\left(1+\frac{1}{k+a}\right)-\ln\left(1+\frac{1}{k+b}\right)&=&\sum_{k=1}^n\ln\left(\dfrac{1+\frac{1}{k+a}}{1+\frac{1}{k+b}}\right)\\
&=&\ln\left(\prod_{k=1}^n\dfrac{1+\frac{1}{k+a}}{1+\frac{1}{k+b}}\right)\\
&=&\ln\left(\dfrac{\frac{n+a+1}{a+1}}{\frac{n+b+1}{b+1}}\right)\\
&=&\ln\left(\dfrac{1+\frac{n}{a+1}}{1+\frac{n}{b+1}}\right)\\
&=&\ln\left(\dfrac{\frac{1}{n}+\frac{1}{a+1}}{\frac{1}{n}+\frac{1}{b+1}}\right)
\end{eqnarray*}
Thus, taking $n\to\infty$, the limit is $\ln(b+1)-\ln(a+1)$
|
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|
probability of coloring a $3\times 3$ table with two colors such that no $2\times 2$ square exists imagine we have a $3\times3$ table ( or $3\times3$ square ) and we want to color each place of that $9$ places with two colors ( red and blue ).
find the probability that no $2\times2$ square exists after coloring places .
|
There are $2^9$ ways to color the square. We now want to count how many ways there are to make a $2\times 2$ square of the same color.
Pick any $2\times 2$ square. There are $2^5$ ways to color the squares not inside the square we chose. The square we chose can be colored in two ways, so we get $2^6$ colors that result in this square being all the same color. There are 4 different $2\times 2$ squares in the $3\times 3$ square, giving us $2^8$.
This double counts some of the boards, for example, if the bottom and middle rows are all blue. Thus we need to exclude the double countings that occur due to $3\times 2$ solid color boards. There are $4\cdot 2^3=2^5$ of these.
The above paragraph also double counts something! Here, we removed the $3\times 3$ solid color board multiple times! We need to add back in the $2$ ways to color the entire board, giving us our final answer of $2^8-2^5+2$. Giving us $$p=\frac{2^8-2^5+2}{2^9}$$This process is called the Inclusion/Exclusion principle.
|
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|
How to find the Maclaurin series for $(\cos x)^6$ using the Maclaurin series for $\cos x$?
Find the Maclaurin series for $(\cos x)^6$ using the Maclaurin series for $\cos x$ for the terms up till $x^4$.
Here is what I've worked out:
Let $f(x) = \cos x,\ g(x) = (\cos x)^6$.
$$g(x) = (f(x))^6$$
$$\cos x = 1 - \frac{1}{2}x^2 + \frac{1}{24}x^4+\cdots$$
So, $$g(x)=\left(1 - \frac{1}{2}x^2 + \frac{1}{24}x^4+\cdots\right)^6$$
However I'm stucked from here on. Thank you in advance!
|
If you want to use the Maclaurin series for $\cos$ in order to obtain the first few terms of the Maclaurin series for $\cos^6$ proceed as follows: Start with
$$\cos x=1-{x^2\over2}+{x^4\over24}+?x^6=1-{x^2\over2}\left(1-{x^2\over12}+?x^4\right)\ .$$Using the binomial theorem you then obtain
$$\cos^6 x=1-{6\choose 1}{x^2\over2}\left(1-{x^2\over12}+?x^4\right)+{6\choose2}{x^4\over4}(1+?x^2)+?x^6\ .$$
Now collect terms:
$$\cos^6 x=1-3x^2+4x^4+?x^6\ .$$
|
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|
Combinatorial proof that $\frac{({10!})!}{{10!}^{9!}}$ is an integer I need help to prove that the quantity of this division :
$\dfrac{({10!})!}{{10!}^{9!}}$
is an integer number, using combinatorial proof
|
Here's a simple non-combinatorial proof. Still eager to see a truly combinatorial one. The denominator contains only powers of $2$, $3$, $5$, and $7$:
$$
(10!)^{9!}=(10\cdot 9 \cdot 8 \cdot 7 \cdot 6 \cdot 5 \cdot 4 \cdot 3 \cdot 2)^{9!}=\left(7\cdot 5^2 \cdot 3^4 \cdot 2^7\right)^{9!}=7^{9!}5^{2\cdot 9!}3^{4\cdot 9!}2^{8\cdot 9!}.
$$
The numerator, on the other hand, can be written as a product of $10!$ factors, of which $1/7$ are divisible by $7$, $1/5$ are divisible by $5$ (and $1/25$ by $25$), $1/3$ are divisible by $3$ (and $1/9$ by $9$), and $1/2$ are divisible by $2$ (and $1/4$ by $4$, and $1/8$ by $8$). So the numerator is divisible by at least $$10!/7 = (10/7) \cdot 9! > 9!$$ factors of $7$, at least $$10!/5 + 10!/25 = (10/5 + 10/25)\cdot 9! > 2\cdot 9!$$ factors of $5$, at least $$10!/3 + 10!/9 = (10/3 + 10/9)\cdot 9! > 4\cdot 9!$$ factors of $3$, and at least $$10!/2 + 10!/4 + 10!/8 = (10/2 + 10/4 + 10/8)\cdot 9! > 8\cdot 9!$$ factors of $2$. Hence the ratio is an integer.
Per the higher-ranked answer, note that generally ${a!}/{(b!)^{a/b}}$, where $b$ divides $a$, is the number of ways to partition $a$ items into groups of size $b$:
$$
N_{a,b}={{a}\choose{b}}\cdot{{a-b}\choose{b}}\cdots{{2b}\choose{b}}\cdot{{b}\choose{b}}=\frac{a!}{(a-b)!b!}\frac{(a-b)!}{(a-2b)!b!}\cdots\frac{(2b!)}{b!}\frac{b!}{b!}=\frac{a!}{(b!)^{a/b}}.
$$
This is clearly an integer; and here $a=10!$ and $b=10$.
|
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|
Incorrect cancellation of fraction with correct answer If you cancel (incorrectly) the 6 in $\frac{26}{65}$, you get the (correct) equation $\frac{26}{65} = \frac{2}{5}$. What other fraction exhibits this property?
I tried considering only the simple case where the fraction is of the form $\frac{10a+n}{10n+b}$ (as in the example), then attempting to solve $\frac{a}{b} = \frac{10a+n}{10n+b}$ to get $n(10a-b) = 9ab$. I was thinking of using divisibility tricks, but am not getting anywhere.
The case of the forms $\frac{10n+a}{10n+b}$ and $\frac{10a+n}{10b+n}$ are easy; you just get $a = b$. I haven't attempted higher than two digits or non-rationals. It would be interesting to see a complicated formula not necessarily involving a ratio of integers which works out to a correct answer with incorrect cancellation.
|
We'll solve $$\frac{10 a + n}{10 n + b} = \frac{a}{b},$$ or equivalently $n (10 a - b) = 9 a b$; to interpret the l.h.s. as a ratio of $2$-digit numbers, we restrict to solutions for which $1 \leq a, b, n \leq 9$.
Reducing modulo $9$ reduces $n(10 a - b) = 9 ab$ to $$n (a - b) \equiv 0 \pmod 9.$$ In particular $n$ and $(a - b)$ must have two factors of $3$ between them. This gives three cases:
CASE 1 $(3^2) \mid n$ : This forces $n = 9$, and canceling gives $10 a - b = ab$. Rearranging gives $(a + 1)(10 - b) = 10$, so since $1 \leq a, b \leq 10$ we must have either $a + 1 = 10$, or one of $a + 1$ and $10 - b$ is $2$ and the other is $5$. These cases respectively give $$\frac{99}{99}, \quad \frac{19}{95}, \quad \frac{49}{98}.$$
CASE 2 $3 \mid n$ and $3 \mid (a - b)$: We may as well assume $9 \nmid n$, as this is the content of case $1$, so, $n = 3$ or $n = 6$. If $n = 3$, simplifying gives $$10 a - b = 3 a b ,$$ and rearranging gives $(3 a + 1)(10 - 3 b) = 10$, which forces $a = b = 3$, giving $$\frac{33}{33} .$$ If $n = 6$, simplifying and rearranging gives $(3 a + 2)(20 - 3 b) = 40$. The factors of $40$ of the form $3 a + 2$ for $1 \leq a \leq 9$ are $5, 8, 20$, and these lead to $a = 1, b = 4$ and $a = 2, b = 5$, $a = 6, b = 6$, which respectively give
$$\frac{16}{64}, \quad \frac{26}{65}, \quad \frac{66}{66}.$$
CASE 3 $9 \mid a - b$: Since $1 \leq a, b \leq 9$, we have $a = b$, and so $n (10 a - b) = 9 a b$ simplifies to $n = a$, giving the trivial cases
$$\frac{11}{11}, \ldots, \frac{99}{99} .$$
In summary, the nontrivial solutions are
$$\color{#bf0000}{\boxed{\frac{16}{64}, \qquad \frac{19}{95}, \qquad \frac{26}{65}, \qquad \frac{49}{98}}}. $$
Of course, we can also look for "false cancellations" of the form $$\frac{10 n + a}{10 b + n} = \frac{a}{b} ,$$ but this is exactly the equation formed by taking the reciprocals of both sides of the above equation and interchanging the roles of $a, b$, so this equation leads precisely to the four reciprocals, $\frac{64}{16}$, etc., of the above solutions.
(There are other two-digit "false cancellation" fractions, corresponding to other equations, but there are all trivial in some sense; they have the forms $\frac{d0}{e0}$ and $\frac{dd}{ee}$ for digits $d, e$.)
|
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|
geometric meaning of complex cubic polynomial coefficients A complex cubic polynomial arranged as $f(z)=z^3-3az^2+3b^2z-c^3$ with coefficients $a,b,c\in\mathbb{C}$ can be represented as the product of (unknown) factors $f(z)=(z-p)(z-q)(z-r)=z^3-3\left(\frac{p+q+r}{3}\right)z^2+3\left(\sqrt \frac{pq+qr+rp}{3}\right)^2z-\left(\sqrt[3]{pqr}\right)^3$ with roots $p,q,r\in\mathbb{C}$.
Thus we can read the centroid $a=\frac{p+q+r}{3}$ of the Steiner ellipse of the triangle $p,q,r\in\mathbb{C}$ right from the coefficients (e.g. reasoning with Marden’s theorem).
Do the coefficients $b=\sqrt \frac{pq+qr+rp}{3}$ and $c=\sqrt[3]{pqr}$ have similar nice intuitive geometric interpretation, e.g. describe characteristic points or features of the triangle or its Steiner ellipse?
|
Vieta's Formulae
\begin{align}
f(z) &= z^3-3az^2+3b^2z-c^3 \\
&= (z-p)(z-q)(z-r) \\
a &= \frac{p+q+r}{3} \\
b^2 &= \frac{pq+qr+rp}{3} \\
c^3 &= pqr
\end{align}
Foci of Steiner ellipse
\begin{align}
f'(\lambda) &=0 \\
\lambda &= a\pm \sqrt{a^2-b^2} \\
&= \frac{p+q+r \pm \sqrt{(p+q+r)^2-3(pq+qr+rp)}}{3}
\end{align}
Resolvents
\begin{align}
u &=
\sqrt[3]{\frac{c^3+3ab^2-2a^3}{2}+
\sqrt{(b^2-a^2)^3+
\left( \frac{c^3+3ab^2-2a^3}{2} \right)^2}} \\
v &=
\sqrt[3]{\frac{c^3+3ab^2-2a^3}{2}-
\sqrt{(b^2-a^2)^3+
\left( \frac{c^3+3ab^2-2a^3}{2} \right)^2}} \\
0 &= f(a+u\, \omega^n+v\, \omega^{2n}) \tag{$\omega=e^{2\pi i/3}$}
\end{align}
*
*Centroid
$$a=\frac{p+q+r}{3}$$
*Linear eccentricity of Steiner ellipse
$$\sqrt{|a^2-b^2|}=\sqrt{|uv|}$$
*Semi-major axis of Steiner ellipse
$$\frac{|u|+|v|}{2}$$
*Semi-minor axis of Steiner ellipse
$$\frac{||u|-|v||}{2}$$
|
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|
$\lim_{x \to 1} \frac{\ln(x^3)}{x^2-1}$ My first idea was to use the definition of derivative (but it's giving me the wrong answer).
$\displaystyle \lim_{x \to 1}\frac{\ln(x^3)}{x^2-1} =\lim_{t \to 1}\frac{\ln(t^{3/2})}{t-1} = \frac{3}{2}\lim_{t\to 0}\frac{\ln(t+1)}{t} =\frac{3}{2}\lim_{t\to 0}\frac{\ln(t+1)-\ln(1)}{t} = \frac{d}{dx}\ln(x)\bigg|_{x=1} = 0.$
But using L'hopital's rule I get $\displaystyle \lim_{x \to 1}\frac{\ln(x^3)}{x^2-1} =\frac{3\frac{d}{dx}\ln(x)}{\frac{d}{dx}(x^2-1)} = \frac{3}{2}\lim_{x \to 1}\frac{1}{x^2} = \frac{3}{2}.$
So where have I messed up in the first approach?
|
Your error is that
$$\frac{d}{dx} \ln x \big|_{x = 1} = \frac 1 x \big|_{x = 1} = 1$$
Here's a slightly different way that doesn't involve making a substitution before applying the definition of the derivative:
\begin{align*}
\lim_{x \to 1} \frac{\ln(x^3)}{x^2 - 1} &= \lim_{x \to 1} \frac{3}{x + 1} \frac{\ln x}{x - 1} \\
&= \frac{3}{1 + 1} \frac{d}{dx} \ln x \big|_{x = 1} \\
&= \frac 3 2
\end{align*}
as desired.
|
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|
Euler's theorem for congruences Using Euler's theorem, how to compute:
a) $3^{1000} \pmod{2500}$
b) $7^{1001} \pmod{2500}$
c) $101^{21600} \pmod{81000}$
|
First note that $(3,2500)=1$ and then compute $\phi(2500)=5^3 \cdot 2 \cdot 4=1000$ .
Now Euler's theorem tells us that :
$$3^{\phi(2500)} \equiv 1 \pmod{2500}$$ or :
$$3^{1000} \equiv 1 \pmod{2500}$$
In the same way $7^{1000} \equiv 1 \pmod{2500}$ so :
$$7^{1001} \equiv 7 \pmod{2500}$$
Finally : $$\phi(81000)=3^3 \cdot 5^2 \cdot 2^2 \cdot 4 \cdot 2=21600$$ and also $(101,81000)=1$ so :
$$101^{21600} \equiv 1 \pmod{81000}$$
|
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If $P$ be permutation matrix then $({P^T}AP)x\mathop = \limits^? \left( {\begin{array}{*{20}{c}} * \\ 0 \\ \end{array}} \right)$
*
*Let $A\in M_n$ is nonnegative(all $a_{ij}\ge0$).
*$x\in C^n$ be eigenvector of $A$ with $r ≥ 1$ positive entries and $n − r$ zero entries .
*There is $P\in M_n$(permutation matrix) such that $Px = \left( {\begin{array}{*{20}{c}}
{x'} \\
0 \\
\end{array}} \right)$ where $x' > 0$.
Can we say that $({P^T}AP)x\mathop = \limits^? \left( {\begin{array}{*{20}{c}}
* \\
0 \\
\end{array}} \right)$
.
|
No, this won't hold for general matrices $A$.
For instance, you can take $n=2$, and $x = \left( {\begin{array}{*{20}{c}}
0 \\
1 \\
\end{array}} \right)$. Then the permutation matrix you are looking for is $P = \left( {\begin{array}{*{c}{c}}
0 & 1 \\
1 & 0\\
\end{array}} \right)$, and $Px = \left( {\begin{array}{*{20}{c}}
1 \\
0 \\
\end{array}} \right).$
Now take $A = \left( {\begin{array}{*{c}{c}}
1 & 1 \\
1 & 1\\
\end{array}} \right)$. Then $APx = A(Px) = \left( {\begin{array}{*{20}{c}}
1 \\
1 \\
\end{array}} \right)$, and so $(P^TAP)x = P^T(APx) = \left( {\begin{array}{*{20}{c}}
1 \\
1 \\
\end{array}} \right)$.
|
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|
Prove that the equations of common tangents to the two hyperbolas $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$ and $\frac{y^2}{a^2}-\frac{x^2}{b^2}=1$ are Prove that the equations of common tangents to the two hyperbolas $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$ and $\frac{y^2}{a^2}-\frac{x^2}{b^2}=1$ are $y=x+\sqrt{a^2-b^2},y=x-\sqrt{a^2-b^2},y=-x+\sqrt{a^2-b^2},y=-x-\sqrt{a^2-b^2}$.
I tried to solve the two equations and find the points of intersections of the hyperbolas but these hyperbolas do not intersect each other.
Let the point of tangency be $(x_1,y_1)$ on the $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$ and $(x_2,y_2)$ on the $\frac{y^2}{a^2}-\frac{x^2}{b^2}=1$For $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$
$\frac{dy}{dx}=\frac{b^2x}{a^2y}$
For $\frac{y^2}{a^2}-\frac{x^2}{b^2}=1$
$\frac{dy}{dx}=\frac{a^2x}{b^2y}$
$\frac{b^2x_1}{a^2y_1}=\frac{a^2x_2}{b^2y_2}$
$b^4x_1y_2=a^4x_2y_1$
I do not know how to take it further.
|
Intersecting the dual conics is one way to go.
The dual conics are $$a^2X^2-b^2Y^2=1\quad (I)$$ and $$-b^2X^2+a^2Y^2=1\quad(II).$$ $b^2(I)+a^2(II)$ gives $(-b^4+a^4)Y^2=b^2+a^2$ or $(a^2-b^2)Y^2=1$. Substituting back to get the corresponding $X$s, we get the four intersection points to be $(X,Y)=(\frac1{\sqrt{a^2-b^2}},\pm\frac1{\sqrt{a^2-b^2}})$ and $(X,Y)=(-\frac1{\sqrt{a^2-b^2}},\pm\frac1{\sqrt{a^2-b^2}})$.
The four common tangents are $$Xx+Yy+1=0.$$
|
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|
An application of Bernoulli's Inequality - relevant to an expression for $e^{x}$ I saw on another post the following proposition.
\begin{equation*}
1 + x + \frac{x^{2}}{2!} + ... + \frac{x^{n}}{n!}
≤ \left( 1 − \frac{x}{n} \right)^{−n}
\end{equation*}
for every real number $0 < x < n$. The member posting this proposition said that it can be deduced from Bernoulli's Inequality. How? Is the inequality valid if a negative value is substituted for $x$?
|
Using the inequality $\ln(1+y) \leqslant y$, we have for $0 \leqslant y < 1$,
$$1+y \leqslant e^y = \sum_{k=0}^{\infty} \frac{y^k}{k!} < \sum_{k=0}^{\infty} y^k = \frac1{1-y},$$
Take $y = x/n$ with $x > 0$. It follows that for $n$ sufficiently large
$$1 + \frac{x}{n} \leqslant e^{x/n} < \left(1 - \frac{x}{n}\right)^{-1},$$
and
$$\left(1 + \frac{x}{n}\right)^n \leqslant e^x < \left(1 - \frac{x}{n}\right)^{-n}.$$
Hence, for $0 < x < n$
$$1 + x + \frac{x^{2}}{2!} + ... + \frac{x^{n}}{n!}
\leqslant e^x \leqslant \left( 1 − \frac{x}{n} \right)^{−n}.$$
|
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|
Limit $\lim_{x \rightarrow \infty} (\sin {\sqrt{1+x}} - \sin {\sqrt{x}})$ I need to calculate limit:
$\lim_{x \rightarrow \infty} (\sin {\sqrt{1+x}} - \sin {\sqrt{x}})$
I was thinking of using the formula for $\sin \alpha - \sin \beta$, but what can be the next step?
|
$$
\sin\sqrt{x+1}-\sin\sqrt{x}=
2\cos\frac{\sqrt{x+1}+\sqrt{x}}{2}
\sin\frac{\sqrt{x+1}-\sqrt{x}}{2}
$$
Now
$$
\lim_{x\to\infty}\frac{\sqrt{x+1}-\sqrt{x}}{2}=0
$$
and
$$
\cos\frac{\sqrt{x+1}+\sqrt{x}}{2}
$$
is bounded.
|
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|
Best Fit Line with 3d Points Okay, I need to develop an alorithm to take a collection of 3d points with x,y,and z components and find a line of best fit. I found a commonly referenced item from Geometric Tools but there doesn't seem to be a lot of information to get someone not already familiar with the method going. Does anyone know of an alorithm to accomplish this? Or can someone help me work the Geometric Tools approach out through an example so that I can figure it out from there? Any help would be greatly appreciated.
|
We start with a sequence of $m$ measurements $\left\{ x_{k}, y_{k}, z_{k}, f_{k} \right\}_{k=1}^{m}$, and the trial function
$$
f(x,y,z) = a_{0} + a_{1} x + a_{2} y + a_{3} z.
$$
The linear system is
$$
\begin{align}
\mathbf{A} a &= f \\
%
\left[ \begin{array}{cccc}
1 & x_{1} & y_{1} & z_{1} \\
1 & x_{2} & y_{2} & z_{2} \\
\vdots & \vdots & \vdots & \vdots \\
1 & x_{m} & y_{m} & z_{m}
\end{array} \right]
%
\left[ \begin{array}{cccc}
a_{0} \\
a_{1} \\
a_{2} \\
a_{3}
\end{array} \right]
%
&=
%
\left[ \begin{array}{cccc}
f_{1} \\
f_{2} \\
\vdots \\
f_{m}
\end{array} \right].
%
\end{align}
$$
The normal equations are
$$
\begin{align}
\mathbf{A}^{*} \mathbf{A} a &= \mathbf{A}^{*} y \\
%
\left[ \begin{array}{cccc}
\mathbf{1} \cdot \mathbf{1} & \mathbf{1} \cdot x & \mathbf{1} \cdot y & \mathbf{1} \cdot z \\
x \cdot \mathbf{1} & x \cdot x & x \cdot y & x \cdot z \\
y \cdot \mathbf{1} & y \cdot x & y \cdot y & y \cdot z \\
z \cdot \mathbf{1} & z \cdot x & z \cdot y & z \cdot z \\
\end{array} \right]
%
\left[ \begin{array}{cccc}
a_{0} \\
a_{1} \\
a_{2} \\
a_{3}
\end{array} \right]
%
&=
%
\left[ \begin{array}{cccc}
\mathbf{1} \cdot f \\
x \cdot f \\
y \cdot f \\
z \cdot f
\end{array} \right]
%
\end{align}
$$
The solution is
$$
\begin{align}
a_{LS} &= \left( \mathbf{A}^{*} \mathbf{A} \right)^{-1} \mathbf{A}^{*} y \\
%
\left[ \begin{array}{cccc}
a_{0} \\
a_{1} \\
a_{2} \\
a_{3}
\end{array} \right]
%
&=
%
\left[ \begin{array}{cccc}
\mathbf{1} \cdot \mathbf{1} & \mathbf{1} \cdot x & \mathbf{1} \cdot y & \mathbf{1} \cdot z \\
x \cdot \mathbf{1} & x \cdot x & x \cdot y & x \cdot z \\
y \cdot \mathbf{1} & y \cdot x & y \cdot y & y \cdot z \\
z \cdot \mathbf{1} & z \cdot x & z \cdot y & z \cdot z \\
\end{array} \right]^{-1}
%
\left[ \begin{array}{cccc}
\mathbf{1} \cdot f \\
x \cdot f \\
y \cdot f \\
z \cdot f
\end{array} \right]
%
\end{align}
$$
|
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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|
If the $81$ digit number $111\cdots 1$ is divided by $729$, the remainder is?
If the $81$ digit number $111\cdots 1$ is divided by $729$, the remainder is?
$729=9^3$
For any number to be divisible by $9$, the sum of the digits have to be divisible by $9$. The given number is divisible by $9$.
Then I tried dividing the given number by $9$. The quotient was like $123456789012\cdots$
So the quotients sum is $(45\times 8)+1=361$
Is this the way to proceed? Is there a shorter way?
|
Put $$a=10^9=729\cdot1371742+82\equiv 1+9^2\ (mod\space 9^3)$$
$$N=111111111=729\cdot 152415+576\equiv 576 \ (mod\space 9^3)$$
Let $M$ be the number so $$M=N(a^8+a^7+a^6+a^5+a^4+a^3+a^2+a+1)$$
Furthermore $$a\equiv 1+9^2\ (mod\space 9^3)$$ $$a^2\equiv 1+2\cdot9^2\ (mod\space 9^3)$$ $$a^3\equiv 1+3\cdot9^2\ (mod\space 9^3)$$ $$......$$ $$a^8\equiv 1+8\cdot9^2\ (mod\space 9^3)$$
It follows $$M\equiv N(9+9^2\cdot 9\cdot 4)\equiv 576\cdot 9\equiv \color{red} {81}\ (mod\space 9^3)$$ because $576\cdot 9=5184=7\cdot 729+81$
|
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"timestamp": "2023-03-29T00:00:00",
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|
If $\frac{\sec^8 \theta}{a}+\frac{\tan^8 \theta}{b} = \frac{1}{a+b}\;,$ Then prove that $ab\leq 0$
If $\displaystyle \frac{\sec^8 \theta}{a}+\frac{\tan^8 \theta}{b} = \frac{1}{a+b}\;,$ Prove that $ab\leq 0$
$\bf{My\; Try::}$ I am Trying To solve it Using Inequality.
Using $\bf{Cauchy\; Schwartz\; Inequality::}$
$$\frac{(\sec^4 \theta)^2}{a}+\frac{(\tan^4 \theta)^2}{b}\geq \frac{(\sec^4 \theta+\tan^4\theta)^2}{a+b}$$
and equality hold when $$\frac{\sec^4 \theta}{a}=\frac{\tan^4 \theta}{b}$$
Now I did not understand How can I solve after that
Help me
Thanks
|
Trivially, $\sec^8 \theta \ge 1$ and $\tan^8 \theta \ge 0$ for all $\theta \in \mathbb{R}$ for which $\sec \theta$ and $\tan \theta$ are defined.
If $a > 0$ and $b > 0$. Then, $\dfrac{\sec^8 \theta}{a}+\dfrac{\tan^8 \theta}{b} \ge \dfrac{1}{a}+\dfrac{0}{b} = \dfrac{1}{a} > \dfrac{1}{a+b}$, a contradiction.
If $a < 0$ and $b < 0$. Then, $\dfrac{\sec^8 \theta}{a}+\dfrac{\tan^8 \theta}{b} \le \dfrac{1}{a}+\dfrac{0}{b} = \dfrac{1}{a} < \dfrac{1}{a+b}$, a contradiction.
Thus, either $a \ge 0$ and $b \le 0$ or $a \le 0$ and $b \ge 0$. Therefore, $ab \le 0$.
|
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"timestamp": "2023-03-29T00:00:00",
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|
Calculate complex determinant $$\left| {\begin{array}{*{20}{c}}{{a^2}}&{{{(a + 1)}^2}}&{{{(a + 2)}^2}}&{{{(a + 3)}^2}}\\{{b^2}}&{{{(b + 1)}^2}}&{{{(b + 2)}^2}}&{{{(b + 3)}^2}}\\{{c^2}}&{{{(c + 1)}^2}}&{{{(c + 2)}^2}}&{{{(c + 3)}^2}}\\{{d^2}}&{{{(d + 1)}^2}}&{{{(d + 2)}^2}}&{{{(d + 3)}^2}}\end{array}} \right|
$$
It's very stupid desicion for me to expand it by row or column.
Any suggestions?
|
By the fourth column (resp. the third column, the second column) subtract the third column (resp. the second column, the first column), you get
$\left|\begin{array}{cccc}
a^2 & 2a+1 & 2a+3 & 2a+5 \\
b^2 & 2b+1 & 2b+3 & 2b+5 \\
c^2 & 2c+1 & 2c+3 & 2c+5 \\
d^2 & 2d+1 & 2d+3 & 2d+5 \\
\end{array} \right|=0$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1613093",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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|
Evaluating $\int_{-\pi}^{\pi} (e^{ix} + e^{-ix})^n dx $ In an exercise following identity is used:
$$ \int_{-\pi}^{\pi} (e^{ix} + e^{-ix})^n dx = \begin{cases} 0, \hspace{2.1cm} n = 2m+1 \\
2\pi {2m \choose m}, \hspace{1cm} n=2m. \end{cases}, $$
Does anybody know how to prove this result or has some ideas to do so?
PLEASE NOTE: It seems that the above identity was not quite correct, instead of $(e^{ix} - e^{-ix})^n $ it should be $(e^{ix} + e^{-ix})^n$. I'm sorry for that.
|
Notice, $e^{ix}+e^{-ix}=2\cos x$, hence one should have
$$\int_{-\pi}^{\pi}(e^{ix}+e^{-ix})^n\ dx=\int_{-\pi}^{\pi}(2\cos x)^n\ dx$$
$$=2^n\int_{-\pi}^{\pi}\cos^n x\ dx$$
since, $\cos(-x)=\cos x$,
$$=2\cdot 2^{n}\int_{0}^{\pi}\cos^n x\ dx$$
$$=2^{n+1}\int_{0}^{\pi}\cos^n (\pi-x)\ dx$$
Case-1: if $n=2m+1$ then $\cos^{2m+1}(\pi-x)=-\cos^{2m+1}(x)$ hence
$$2^{n+1}\int_{0}^{\pi}\cos^n (\pi-x)\ dx=2^{2m+2}\int_{0}^{\pi}\cos^{2m+1} (\pi-x)\ dx=2^{2m+2}(0)=\color{red}{0}$$
Case-2: if $n=2m$ then $\cos^{2m}(\pi-x)=\cos^{2m}(x)$, hence
$$2^{n+1}\int_{0}^{\pi}\cos^n (\pi-x)\ dx=2^{2m+1}\int_{0}^{\pi}\cos^{2m} (\pi-x)\ dx$$$$=2\cdot 2^{2m+1}\int_{0}^{\pi/2}\cos^{2m} (x)\ dx$$$$=2^{2m+2}\int_{0}^{\pi/2}\cos^{2m} (x)\ dx$$$$=2^{2m+2}\int_{0}^{\pi/2}\sin^{0}(x)\cos^{2m} (x)\ dx$$
Using formula: $\color{blue}{\int_{0}^{\pi/2}\sin^mx\cos^n x\ dx=\frac{\Gamma\left(\frac{m+1}{2}\right)\Gamma\left(\frac{n+1}{2}\right)}{2\Gamma\left(\frac{m+n+2}{2}\right)}}$,
$$=2^{2m+2}\frac{\Gamma\left(\frac{0+1}{2}\right)\Gamma\left(\frac{2m+1}{2}\right)}{2\Gamma\left(\frac{0+2m+2}{2}\right)}$$
$$=2^{2m+1}\frac{\Gamma\left(\frac{1}{2}\right)\Gamma\left(\frac{2m+1}{2}\right)}{\Gamma\left(m+1\right)}$$
$$=2^{2m+1}\frac{\sqrt \pi\left(\underbrace{\frac{2m-1}{2}\cdot \frac{2m-3}{2}\cdot \frac{2m-5}{2}\cdots \frac{3}{2}\cdot \frac{1}{2}}_{\text{m times }}\Gamma\left(\frac{1}{2}\right)\right)}{m!}$$
$$=2^{2m+1}\frac{\sqrt \pi\left(\frac{(2m-1)(2m-3)(2m-5)\cdots 3\cdot 1 }{2^{m}}\sqrt \pi\right)}{m!}$$
$$=2^{m+1}\pi \frac{\frac{(2m)(2m-1)(2m-2)(2m-3)(2m-4)(2m-5)\cdots 4\cdot 3\cdot 2\cdot 1 }{(2m)(2m-2)(2m-4)\cdots 4\cdot 2}}{m!}$$
$$=2^{m+1}\pi \frac{\frac{(2m)!}{2^m\cdot m!}}{m!}$$
$$=2\pi\frac{(2m)!}{(m!)^2}$$
$$=\color{red}{2\pi \binom{2m}{m}}$$
hence, one should have
$$\int_{-\pi}^{\pi}(e^{ix}+e^{-ix})^n\ dx=\begin{cases}0, \ \ \ \ \ \ \ \ \ \ \ \ \ n=2m+1\\ 2\pi \binom{2m}{m}, \ \ \ n=2m
\end{cases}$$
|
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|
Prove that $0 \leq ab + ac + bc - abc \leq 2.$
Let $a,b,$ and $c$ be nonnegative real numbers such that $a^2+b^2+c^2+abc = 4$. Prove that $$0 \leq ab + ac + bc - abc \leq 2.$$
I tried using rearrangement to get $a^2+b^2+c^2+abc = 4 \geq ab+bc+ac+abc$. Then I just need to show that $0\leq ab + ac + bc - abc$ and $4-2abc \leq 2$. I am not sure if this method will work, though.
|
I have a diferent solution for this question using a trigonometric substitution.Let:
$\\ \\ \displaystyle u=a\sqrt{\frac{2a+bc}{(2b+ac)(2c+ab)}};v=b\sqrt{\frac{2b+ac}{(2a+bc)(2c+ab)}};w=c\sqrt{\frac{2c+ab}{(2a+bc)(2b+ac)}} \\ \\$
This implies that:
$\\ \displaystyle a=2\sqrt{\left(\frac{1}{1-uv}-1\right)\left(\frac{1}{1-uw}-1\right)}$
$\\ \displaystyle b=2\sqrt{\left(\frac{1}{1-uv}-1\right)\left(\frac{1}{1-vw}-1\right)};$
$\\ \displaystyle c=2\sqrt{\left(\frac{1}{1-uw}-1\right)\left(\frac{1}{1-vw}-1\right)};\\ $
Observe that:
$\\ \displaystyle a^{2}+b^{2}+c^{2}+abc=4\Rightarrow \left(\frac{1}{1-vw}-1\right)\left(\frac{1}{1-uw}-1\right)+ \left(\frac{1}{1-vw}-1\right)\left(\frac{1}{1-uv}-1\right)+\left(\frac{1}{1-uw}-1\right)\left(\frac{1}{1-uv}-1\right)+
2\left(\frac{1}{1-uv}-1\right)\left(\frac{1}{1-uw}-1\right)\left(\frac{1}{1-vw}-1\right)=1\Rightarrow uv+vw+uw=1 $
Using that $\displaystyle uv+uw+vw=1$, our inequality is equivalent to:
.
$\\ \\ \displaystyle \frac{2uv\sqrt{(u+w)(v+w)}}{(u+w)(v+w)(v+u)}+ \frac{2uw\sqrt{(v+w)(u+v)}}{(u+w)(v+w)(v+u)}+\frac{2vw\sqrt{(u+v)(u+w)}}{(u+v)(v+w)(u+w)} -
\frac{4uvw}{(u+v)(v+w)(u+w)}\leq1\\ \\$
Let $\displaystyle a'=v+w,b'=u+w, c'=u+v$, observe that $\displaystyle a',b',c'$ will be sides of an arbitrary triangle.Let $\displaystyle S=\frac{a'+b'+c'}{2}$, we get $ \displaystyle u=S-a',v=S-b', w=S-c' $, and our inequality is equivalent to:
.
\begin{equation}
\frac{2(S-a')(S-b')\sqrt{a'b'}}{a'b'c'}+\frac{2(S-a')(S-c')\sqrt{a'c'}}{a'b'c'}+\frac{2(S-b')(S-c')\sqrt{b'c'}}{a'b'c'} -\frac{4(S-a')(S-b')(S-c')}{a'b'c'}\leq 1 \tag{1}
\end{equation}
We have to prove the inequality (1).Let's prove it, for that, consider that the square of every real number is positive, so we will have:
$ \\ \\ \displaystyle 0 \leq(\sqrt{a'}-\sqrt{b'})^2 \Rightarrow 2\sqrt{a'b'} \leq a'+b' \Rightarrow \frac{2(S-a')(S-b')\sqrt{a'b'}}{a'b'c'} \leq \frac{(a'+b')(S-a')(S-b')}{a'b'c'} \\ \\ $
$\displaystyle 0 \leq(\sqrt{a'}-\sqrt{c'})^2 \Rightarrow 2\sqrt{a'c'} \leq a'+c' \Rightarrow \frac{2(S-a')(S-c')\sqrt{a'c'}}{a'b'c'} \leq \frac{(a'+c')(S-a')(S-c')}{a'b'c'} \\ \\ $
$\displaystyle 0 \leq(\sqrt{b'}-\sqrt{c'})^2 \Rightarrow 2\sqrt{b'c'} \leq b'+c' \Rightarrow \frac{2(S-b')(S-c')\sqrt{b'c'}}{a'b'c'} \leq \frac{(b'+c')(S-b')(S-c')}{a'b'c'} \\ \\ $
Adding all the inequalities above and subtracting $\displaystyle \frac{4(S-a')(S-b')(S-c')}{a'b'c'} $ we get:
\begin{equation*}
\frac{2(S-a')(S-b')\sqrt{a'b'}}{a'b'c'}+\frac{2(S-a')(S-c')\sqrt{a'c'}}{a'b'c'}+\frac{2(S-b')(S-c')\sqrt{b'c'}}{a'b'c'}-\frac{4(S-a')(S-b')(S-c')}{a'b'c'} \leq
\end{equation*}
\begin{equation}
\frac{(a'+b')(S-a')(S-b')}{a'b'c'}+\frac{(a'+c')(S-a')(S-c')}{a'b'c'}+\frac{(b'+c')(S-b')(S-c')}{a'b'c'} -\frac{4(S-a')(S-b')(S-c')}{a'b'c'} \tag{2}
\end{equation}
Observe that:
$\\ \\ \displaystyle a'+b'=2\left(S-c'+\frac{c'}{2}\right); \displaystyle a'+c'=2\left(S-b'+\frac{b'}{2}\right);\displaystyle b'+c'=2\left(S-a'+\frac{a'}{2}\right)\\ \\$
Take the RHS of (2):
$\\ \\ \displaystyle \frac{(a'+b')(S-a')(S-b')}{a'b'c'}+\frac{(a'+c')(S-a')(S-c')}{a'b'c'}+\frac{(b'+c')(S-b')(S-c')}{a'b'c'} -\frac{4(S-a')(S-b')(S-c')}{a'b'c'}=$
$\\ \\ \displaystyle \frac{2\left(S-c'+\frac{c'}{2}\right)(S-a')(S-b')}{a'b'c'}+\frac{2\left(S-b'+\frac{b'}{2}\right)(S-a')(S-c')}{a'b'c'}+\frac{2\left(S-a'+\frac{a'}{2}\right)(S-b')(S-c')}{a'b'c'} -\frac{4(S-a')(S-b')(S-c')}{a'b'c'}=$
$\\ \\ \displaystyle \frac{(S-a')(S-b')}{a'b'}+\frac{(S-a')(S-c')}{a'c'}+\frac{(S-b')(S-c')}{b'c'} +\frac{2(S-a')(S-b')(S-c')}{a'b'c'} =\sin^{2}\frac{\alpha}{2}+ \sin^{2}\frac{\beta}{2}+ \sin^{2}\frac{\gamma}{2}+2\sin\frac{\alpha}{2} \sin\frac{\beta}{2} \sin\frac{\gamma}{2}=1$
By the law of cosines.Replacing this result in RHS of (2).Thus (1)
yields.
|
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|
Find all integer solutions to $a+b+c|ab+bc+ca|abc$ As you can see from the title, I am trying to find all integer solutions $(a,b,c)$ to $$(a+b+c) \ \lvert\ (ab+bc+ca) \ \lvert\ abc$$ (that is, $a+b+c$ divides $ab+bc+ca$, and $ab+bc+ca$ divides $abc$). Unfortunately, I could not find anything on this problem (although I find it hard to believe that nobody though of this before).
What I've found so far
I have looked at the simpler case: $(a+b) \ \lvert\ ab$. I was able to solve this, and all solutions are $$(a,b)=(\alpha(\alpha+\beta)\gamma,\beta(\alpha+\beta)\gamma)$$ with $\alpha,\beta,\gamma\in\mathbb{Z}$.
I was also able to reduce the given problem to only one division. If we are able to solve $$(a_0b_0+b_0c_0+c_0a_0) \ \lvert\ (a_0+b_0+c_0)a_0b_0c_0 $$ with $\gcd(a_0,b_0,c_0)=1$, then we know that
\begin{align}
a&=a_0(a_0+b_0+c_0)\cdot k\\
b&=b_0(a_0+b_0+c_0)\cdot k\\
c&=c_0(a_0+b_0+c_0)\cdot k\\
\end{align}
For $k\in\mathbb{Z}$ are all solutions to the original problem. However, I was not able to solve this. I have computed a few solutions to the last (and the corresponding solutions for the original problem) but was not able to find a pattern. Any progress on the problem is welcome!
|
$a,b,c$ are roots of $X^3-(a+b+c)X^2+(ab+ac+bc)X=abc$.
Putting $$\begin{cases}a+b+c=A\\ab+ac+bc=mA\\abc=n(ab+ac+bc)\end{cases}\qquad (*)$$
one gets $$X^3-AX^2+mAX=mnA$$ it follows
$$\begin{cases}a^3-Aa^2+mAa=mnA\\b^3-Ab^2+mAb=mnA\\c^3-Ac^2+mAc=mnA\end{cases}$$
Hence $a^3\equiv b^3\equiv c^3\pmod A$ so that
$a^3+b^3+c^3=kA$.
We take a look to the equation $$X^3+Y^3+Z^3=r$$ for cases in which there are solutions and we try to condition these to our problem.
Example.-For $r=s^3$. $$X^3+Y^3+Z^3= s^3 \Rightarrow (X,Y,Z)=(t,-t,s)$$ To fit with (*) we choose $t=2s$ which gives $(a,b,c)=(2s,-2s,s)$ so one has $$\begin{cases}a+b+c=2s-2s+s=s\\ab+ac+bc=-4s^2=(-4s)s\\ abc=-4s^3=(-4s^2)s\end{cases}$$ Thus $$a+b+c=s$$ $$\frac{ab+ac+bc}{a+b+c}=-4s$$ $$\frac{abc}{ab+ac+bc}=s$$
I will return to this problem…., if I may.
|
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"timestamp": "2023-03-29T00:00:00",
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|
Finding the infinite series: $3 \cdot \frac{9}{11} + 4 \cdot \left(\frac{9}{11}\right)^2 + 5 \cdot \left(\frac{9}{11}\right)^3 + \cdots$? I'm trying to find the infinite sum that is defined by:
$$
3 \cdot \frac{9}{11} + 4 \cdot \left(\frac{9}{11}\right)^2 + 5 \cdot \left(\frac{9}{11}\right)^3 + \cdots
$$
However, I do not know of any known formula to do this. Am I missing something really simple? Thanks!
|
For $|r|<1$ we have $$\sum_{k=0}^{\infty}r^k=\frac{1}{1-r}$$
Then, by taking derivatives respect to $r$
$$\sum_{k=0}^{\infty}kr^{k-1}=\frac{1}{(1-r)^2}\quad\implies\quad \frac{1}{r}+2+\sum_{k=3}^{\infty}kr^{k-2}=\frac{1}{r(1-r)^2}$$
Then
$$\sum_{k=3}^{\infty}kr^{k-2}=\frac{1}{r(1-r)^2}-\frac{1}{r}-2$$
Thus
$$\sum_{k=3}^{\infty}k\left(\frac{9}{11}\right)^{k-2}=\frac{1}{\frac{9}{11}\left(\frac{2}{11}\right)^2}-\frac{1}{\frac{9}{11}}-2=\frac{1331-44-72}{36}=\boxed{\color{blue}{\frac{135}{4}}}$$
|
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"timestamp": "2023-03-29T00:00:00",
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|
Integers represented by $x^2 + 3 y^2$ vs. integers represented by $x^2 + x y + y^2$. How does one show that the quadratic forms $x^2 + 3 y^2$ and $x^2 + x y + y^2$ represent the same set of integers?
I think it relates to a classical result of Euler about primes of form $6k+1$. In fact a positive integer $n$ is of the form $x^2 + 3 y^2$ if and only if the the primes $\equiv -1 \mod 3$ have an even exponent in $n$. Similarly for $x^2 + x y + y^2$.
However somebody told me that this can be shown without so much number theoretical background. Any ideas would be appreciated!
|
Hint
$$x^2+xy+y^2=(x+\frac{1}{2}y)^2+3 (\frac{y}{2})^2=(\frac{1}{2}x+y)^2+3 (\frac{x}{2})^2= (\frac{x-y}{2})^2+3(\frac{x+y}{2})^2$$
Split it in three cases by parity.
Backwards you need to do the back substitutions, i.e. denote the brackets on the RHS by $x',y'$ and solve for $x, y$. Then you get relations of the form
$$x'^2+3y'^2= x^2+3y^2=( ... )^2+3 (..)^2$$
|
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|
Let a,b be rationals and x irrational. Show that if $\frac{x+a}{x+b}$ is rational, then $a=b$. I'm trying to solve the following problems:
*
*Let $a$,$b$ be rationals and $x$ irrational. Show that if $\frac{x+a}{x+b}$ is rational, then $a=b$
*Let $x$,$y$ be rationals such that $\frac{x^2+x+\sqrt{2}}{y^2+y+\sqrt{2}}$ is also rational. Prove that either $x=y$, or $x+y=-1$.
Only thing I can think of is that if $\frac{x+a}{x+b}$ and $\frac{x^2+x+\sqrt{2}}{y^2+y+\sqrt{2}}$ are rationals, then there's a number $\frac{m}{n}$ where integers $m$ and $n$ are co-prime.
I would appreciate any form of help.
Thank you.
|
1. There are $p,q\in\mathbb{Z}$, with $gcd(p,q)=1$ such that $\frac{x+a}{x+b}=\frac{p}{q}$. Then $$(x+a)q=(x+b)p\Leftrightarrow x(q-p)=bp-aq\Leftrightarrow x=\frac{bp-aq}{q-p}\in\mathbb{Q}\Leftrightarrow x=0.$$ This is absurd. Therefore $p=q$ and $x+a=x+b$, so $a=b$.
2. Let $r\in\mathbb{Q}$ such that $\frac{x²+x+\sqrt{2}}{y²+y+\sqrt{2}}=r$ and suppose that $\frac{x}{y}$ is a irreducible fraction. Then $$\frac{x²+x+\sqrt{2}}{y²+y+\sqrt{2}}=r\Leftrightarrow x²+x+\sqrt{2}=(y²+y+\sqrt{2})r\Leftrightarrow x²+x-y²r-yr=\sqrt{2}(r-1).$$ If $r\neq 1$ we have $\sqrt{2}=\frac{x²+x-y²r-yr}{r-1}\in\mathbb{Q}$, but this is absurd. It follows that $r=1$ and $$x²+x+\sqrt{2}=y²+y+\sqrt{2}\Rightarrow y(y+1)=x(x+1).$$ If $x=-1$, we have $y(y+1)=0$ and so $y=0=x+1$ or $y=x=-1$. If $x\neq -1$ we have $$x=y(\frac{y+1}{x+1}).$$ If $y=-1$, it is analogous. We can suppose that $y\neq -1$. If $y=0$, $x=0$. Suppose $-1\neq y\neq 0$. We have $$\frac{x}{y}=\frac{y+1}{x+1}.$$ Since $\frac{x}{y}$ is irreducible, follows the last equality that $x=y$.
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"url": "https://math.stackexchange.com/questions/1620177",
"timestamp": "2023-03-29T00:00:00",
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|
Prove that $3 \le a+b+c \le 2\sqrt{3}$ in a triangle
Let $a,b,$ and $c$ be the lengths of the sides of a triangle satisfying $ab+bc+ca = 3.$ Prove that $3 \le a+b+c \le 2\sqrt{3}$.
The idea I had was $(a+b+c)^2 = a^2+b^2+c^2+2(ab+bc+ca) = a^2+b^2+c^2+6 \geq 9$ by rearrangement. That takes care of the first inequality. How do I show the other inequality?
|
For the other side, we note that
\begin{align*}
(a+b+c)^2 &\ge 3 (ab + bc + ca)\\
&= 9.
\end{align*}
|
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"timestamp": "2023-03-29T00:00:00",
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|
If $a^4 + 4^b$ is prime, then $a$ is odd and $b$ is even. We say an integer $p>1$ is prime when its only positive divisors are $1$ and $p$. Let $a$ and $b$ be natural number not both $1$. Prove that if $a^4+4^b$ is prime, then $a$ is odd and $b$ is even.
I'm also given a hint:
Consider the expression $(x^2-2xy+2y^2)(x^2+2xy+2y^2)$.
What I have so far is:
Let $x=a^2, y=2^b$ and substitute it into the expression.
But I am not sure how to proceed. Can anyone help please?
|
I think that hint is terrible and only the geniuses in the class could get it right away. I put $(x^2 - 2xy + 2y^2)(x^2 + 2xy + 2y^2)$ into Wolfram Alpha and it gave me $x^4 + 4y^4$ as an "alternate form."
But the problem says "$4^b$", not "$4y^4$." If you're a precocious genius you can right away see that if $b$ is odd then $$4^b = 4 \left(2^{\frac{2b - 2}{4}} \right)^4.$$ I'm not a genius, so it took me a while (three days, to be precise) to arrive at this. It was only then that the problem finally unraveled. Now we have $$a^4 + 4 \left(2^{\frac{2b - 2}{4}} \right)^4 = (a^2 - 2^{\frac{2b - 2}{2}} a + 2^b)(a^2 + 2^{\frac{2b - 2}{2}} a + 2^b).$$
This means that with $b$ odd, the only way for the resulting number to be prime is for $(a^2 - 2^{\frac{2b - 2}{2}} a + 2^b)$ to equal $1$ or $-1$ somehow. There are four possible solutions to that equation, and they're all ruled out with the definition of the problem (because $a$ and $b$ are natural numbers not both equal to $1$, and $0$ is or isn't a "natural number" -- but that's a whole other can of worms). Therefore, under the constraints of the problem, odd $b$ gives a composite number, so $b$ must be even in order for the expression to give a prime (this is a necessary but not sufficient condition, of course).
Let's have a concrete example: $a = 1$, $b = 3$. Then $a^4 + 4^b = 65$, which is obviously composite. But still take the time to run it through Sophie Germain's identity: $1 + 4^3 = 1 + 4 (2^1)^4 = (1 - 2 \times 1 \times 2 + 2 \times 2^2)(1 + 2 \times 1 \times 2 + 2 \times 2^2) = 5 \times 13$.
As for the parity of $a$, that, as Bob Happ said in a comment, "is the easy part here."
|
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|
Finding $(a+\sqrt b)^n+(a-\sqrt b)^n$ where $n$ is natural For the expression $\left(a+\sqrt{b}\right)^n+\left(a-\sqrt{b}\right)^n$ where $n \in \mathbb{N}$, and $a,b, \in \mathbb{Q}$, the radical is always ends up cancelled, and the result is always in $\mathbb{Q}$. Is there any way that this could be reexpressed with the assumption that n is always a positive integer without the use of a square root operation, such as an alternative closed form for even $n$, and another closed form for odd $n$, or am I always stuck having to calculate a square root?
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Since
$$\begin{bmatrix}
A & B \\
1 & A
\end{bmatrix} = \begin{bmatrix}
1 & 1 \\
\frac{-1}{\sqrt B} & \frac{1}{\sqrt B}
\end{bmatrix} \begin{bmatrix}
A - \sqrt B & 0 \\
0 & A + \sqrt B
\end{bmatrix} \begin{bmatrix}
1 & 1 \\
\frac{-1}{\sqrt B} & \frac{1}{\sqrt B}
\end{bmatrix}^{-1}
$$
It follows that
$$\begin{bmatrix}
A & B \\
1 & A
\end{bmatrix}^n = \begin{bmatrix}
1 & 1 \\
\frac{-1}{\sqrt B} & \frac{1}{\sqrt B}
\end{bmatrix} \begin{bmatrix}
A - \sqrt B & 0 \\
0 & A + \sqrt B
\end{bmatrix}^n \begin{bmatrix}
1 & 1 \\
\frac{-1}{\sqrt B} & \frac{1}{\sqrt B}
\end{bmatrix}^{-1}
$$
$$Q = \begin{bmatrix}
1 & 1 \\
\frac{-1}{\sqrt B} & \frac{1}{\sqrt B}
\end{bmatrix}^{-1} \begin{bmatrix}
A & B \\
1 & A
\end{bmatrix}^n \begin{bmatrix}
1 & 1 \\
\frac{-1}{\sqrt B} & \frac{1}{\sqrt B}
\end{bmatrix} = \begin{bmatrix}
(A - \sqrt B)^n & 0 \\
0 & (A + \sqrt B)^n
\end{bmatrix}
$$
If you assume that $\begin{bmatrix}
A & B \\
1 & A
\end{bmatrix}^n = \begin{bmatrix} H & I \\ J & K \end{bmatrix}$ then multiply out the above it follows that $(A - \sqrt B)^n + (A + \sqrt B)^n = Q_{1,1} + Q_{2,2} = H + K$.
In summary, $$(A - \sqrt B)^n + (A + \sqrt B)^n = \text{trace}\left(\begin{bmatrix}
A & B \\
1 & A
\end{bmatrix}^n\right)$$
|
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"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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|
Laurent series of f(z) Can you check my calculations?
Find Laurent series for: $f(z)=\frac{6z+8}{(2z+3)(4z+5)}=\frac{1}{4z+5}+\frac{1}{2z+3}$
when a)$\frac{5}{4}<|z|<\frac{3}{2}$
b)$|z|>\frac{3}{2}$
a)$\sum_{n=0}^\infty \frac{(-2)^nz^n}{3^{n+1}} + \sum_{n=1}^\infty \frac{(-4)^nz^{-n}}{5^{n+1}}$
b)$\sum_{n=0}^\infty \frac{(-2)^nz^{-n}}{3^{n+1}} + \sum_{n=1}^\infty \frac{(-4)^nz^{-n}}{5^{n+1}}$
Is it important which series I will take since n=1? Thanks in advance
|
Note: Regarding your question, the series we take is important. Here I provide a complete Laurent expansion around $z=0$. From this you should be able to check our results (which slightly differ).
The function
\begin{align*}
f(z)&=\frac{1}{4}\frac{1}{z+\frac{5}{4}}+\frac{1}{2}\frac{1}{z+\frac{3}{2}}\\
\end{align*}
has two simple poles at $-\frac{5}{4}$ and $-\frac{3}{2}$.
Since we want to find a Laurent expansion with center $0$, we look at the poles $-\frac{5}{4}$ and $-\frac{3}{2}$ and see they determine three regions.
\begin{align*}
|z|<\frac{5}{4},\qquad\quad
\frac{5}{4}<|z|<\frac{3}{2},\qquad\quad
\frac{3}{2}<|z|
\end{align*}
*
*The first region $ |z|<\frac{5}{4}$ is a disc with center $0$, radius $\frac{5}{4}$ and the pole $-\frac{5}{4}$ at the boundary of the disc. In the interior of this disc all two fractions with poles $-\frac{5}{4}$ and $-\frac{3}{2}$ admit a representation as power series at $z=0$.
*The second region $\frac{5}{4}<|z|<\frac{3}{2}$ is the annulus with center $0$, inner radius $\frac{5}{4}$ and outer radius $\frac{3}{2}$. Here we have a representation of the fraction with pole $\frac{5}{4}$ as principal part of a Laurent series at $z=0$, while the fraction with pole at $\frac{3}{2}$ admits a representation as power series.
*The third region $|z|>\frac{3}{2}$ containing all points outside the disc with center $0$ and radius $\frac{3}{2}$ admits for all fractions a representation as principal part of a Laurent series at $z=0$.
A power series expansion of $\frac{1}{z+a}$ at $z=0$ is
\begin{align*}
\frac{1}{z+a}&=\frac{1}{a}\cdot\frac{1}{1+\frac{z}{a}}\\
&=\sum_{n=0}^{\infty}\frac{1}{a^{n+1}}(-z)^n
\end{align*}
The principal part of $\frac{1}{z+a}$ at $z=0$ is
\begin{align*}
\frac{1}{z+a}&=\frac{1}{z}\cdot\frac{1}{1+\frac{a}{z}}=\frac{1}{z}\sum_{n=0}^{\infty}\frac{a^n}{(-z)^n}
=-\sum_{n=0}^{\infty}\frac{a^n}{(-z)^{n+1}}\\
&=-\sum_{n=1}^{\infty}\frac{a^{n-1}}{(-z)^n}
\end{align*}
We can now obtain the Laurent expansion of $f(x)$ at $z=0$ for all three regions
*
*Region 1: $|z|<\frac{5}{4}$
\begin{align*}
f(z)&=\frac{1}{4}\sum_{n=0}^{\infty}\left(\frac{4}{5}\right)^{n+1}(-z)^n
+\frac{1}{2}\sum_{n=0}^{\infty}\left(\frac{2}{3}\right)^{n+1}(-z)^n\\
&=\sum_{n=0}^{\infty}\left(\frac{4^n}{5^{n+1}}+\frac{2^{n}}{3^{n+1}}\right)(-z)^n
\end{align*}
*
*Region 2: $\frac{5}{4}<|z|<\frac{3}{2}$
\begin{align*}
f(z)&=-\frac{1}{4}\sum_{n=1}^{\infty}\left(\frac{5}{4}\right)^{n-1}\frac{1}{(-z)^n}
+\frac{1}{2}\sum_{n=0}^{\infty}\left(\frac{2}{3}\right)^{n+1}(-z)^n\\
&=-\sum_{n=1}^{\infty}\frac{5^{n-1}}{4^n}\frac{1}{(-z)^n}+\sum_{n=0}^{\infty}\frac{2^n}{3^{n+1}}(-z)^n
\end{align*}
*
*Region 3: $\frac{3}{2}<|z|$
\begin{align*}
f(z)&=-\frac{1}{4}\sum_{n=1}^{\infty}\left(\frac{5}{4}\right)^{n-1}\frac{1}{(-z)^n}
-\frac{1}{2}\sum_{n=0}^{\infty}\left(\frac{3}{2}\right)^{n-1}\frac{1}{(-z)^n}\\
&=-\sum_{n=1}^{\infty}\left(\frac{5^{n-1}}{4^n}+\frac{3^{n-1}}{2^n}\right)\frac{1}{(-z)^n}
\end{align*}
|
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"timestamp": "2023-03-29T00:00:00",
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|
Is $x^2+x+1$ divisible by $101$, if $x\in\mathbb Z$? Prove $x^2+x+1$ isn't divisible by $101$, for any $x\in\mathbb Z$?
I think the way of solving the problem it by using "Fermat's Little Theorem".
|
Oh, well. If $$ x^2 + x + 1 \equiv 0 \pmod {101}, $$ then multiplying by $4$ gives
$$ (2x+1)^2 + 3 \equiv 0 \pmod {101}, $$ and
$$ (2x+1)^2 \equiv -3 \pmod {101}. $$
However, Legendre
$$ (-3|101) = (-1|101) (3|101) = (3|101) = (101|3) = -1 $$
NOTE: for any odd prime $p,$ we always get $(-3|p) = (p|3),$ all that matters is that $3 \equiv 3 \pmod 4.$ IF $p \equiv 1 \pmod 4,$
$$ (-3|p) = (-1|p) (3|p) = (3|p) = (p|3)$$
IF $q \equiv 3 \pmod 4,$
$$ (-3|q) = (-1|q) (3|q) = - (3|q) = (q|3)$$
SIMILAR PROBLEM FOR ILLUSTRATION: If $p$ is an odd prime and $$ x^2 + x + 2 \equiv 0 \pmod {p}, $$ then multiplying by $4$ gives
$$ (2x+1)^2 + 7 \equiv 0 \pmod {p}, $$ and
$$ (2x+1)^2 \equiv -7 \pmod {p}. $$
However, Legendre
$$ (-7|p) = (p|7). $$
So, if $(p|7) = -1,$ it is impossible to have $ x^2 + x + 2$ divisible by $p.$ These primes, with $(p|7) = -1,$ are
$$ p \equiv 3,5,6 \pmod 7$$ and $101 = 98 +3$ is one of them.
ANOTHER SIMILAR PROBLEM FOR ILLUSTRATION: If $p$ is an odd prime and $$ x^2 + x + 3 \equiv 0 \pmod {p}, $$ then multiplying by $4$ gives
$$ (2x+1)^2 + 11 \equiv 0 \pmod {p}, $$ and
$$ (2x+1)^2 \equiv -11 \pmod {p}. $$
However, Legendre
$$ (-11|p) = (p|11). $$
So, if $(p|11) = -1,$ it is impossible to have $ x^2 + x + 3$ divisible by $p.$ These primes, with $(p|11) = -1,$ are
$$ p \equiv 2,6,7,8,10 \pmod {11}$$ and $101 = 99 +2$ is one of them.
|
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"timestamp": "2023-03-29T00:00:00",
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|
Solving $\lim_{n\to\infty}\sum_{k=1}^{n}\frac{2015^{k}}{\sum_{i=0}^{k-1}2015^i\sum_{l=0}^k2015^l}=$ $$\lim_{n\to\infty}\sum_{k=1}^{n}\frac{2015^{k}}{\sum_{i=0}^{k-1}2015^i\sum_{l=0}^k2015^l}=$$
$$\lim_{n\to\infty}\sum_{k=1}^{n}\frac{2015^k}{\frac{1-2015^k}{1-2015}\cdot\frac{1-2015^{k+1}}{1-2015}}=$$
$$\lim_{n\to\infty}\sum_{k=1}^{n}\frac{2015^k}{\frac{1-2015^{k+1}-2015^k+2015^{2k+1}}{2014^2}}$$
I'm not sure what to do next, I'm stuck there. Any tips?
Edit: Just to save for later.
$$\lim_{n\to\infty}\sum_{k=1}^{n}2014^2\bigg(\frac{1}{2014}\cdot\bigg(\frac{1}{2015^k-1}-\frac{1}{2015^{k+1}-1}\bigg)\bigg)=$$
$$\lim_{n\to\infty}2014\frac{2015(2015^n-1)}{2014(2015^{n+1}-1)}=$$
$$\lim_{n\to\infty}\frac{2015(2015^n-1)}{2015^{n+1}-1}=$$
$$\lim_{n\to\infty}\frac{2015^{n+1}-2015}{2015^{n+1}-1}\bigg/\frac{:2015^{n+1}}{:2015^{n+1}}=$$
$$\lim_{n\to\infty}\frac{1-\frac{1}{2015^n}}{1-\frac{1}{2015^{n+1}}}=1$$
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More general, $\frac{x^k}{(1-x^{k})(1-x^{k+1})}=\frac{1}{x-1}\left(\frac{1}{x^k-1}-\frac{1}{x^{k+1}-1}\right)$.
Thus
$\begin{eqnarray}
\sum_{k=1}^n\frac{x^k}{(1-x^{k})(1-x^{k+1})}&=&\frac{1}{x-1}\sum_{k=1}^n\left(\frac{1}{x^k-1}-\frac{1}{x^{k+1}-1}\right)=\frac{x(x^n-1)}{(x-1)^2(x^{n+1}-1)}
\end{eqnarray}$
Therefore
$$\sum_{k=1}^{n}\frac{x^k}{\frac{1-x^k}{1-x}\cdot\frac{1-x^{k+1}}{1-x}}=(1-x)^2\sum_{k=1}^n\frac{x^k}{(1-x^k)\cdot(1-x^{k+1})}=\frac{x(x^n-1)}{x^{n+1}-1}$$
Can you continue?
|
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|
Prove that $\int_{0}^{\pi}\frac{r}{1-2r\cos x+r^2}dx=\int_{0}^{\pi}\frac{\cos x}{1-2r\cos x+r^2}dx$ Prove for every $r\in(-1,1)$: $$\int_{0}^{\pi}\frac{r}{1-2r\cos x+r^2}\,dx=\int_{0}^{\pi}\frac{\cos x}{1-2r\cos x+r^2}\,dx$$
I tried proving that $$\int_{0}^{\pi}\frac{\cos x-r}{1-2r\cos x+r^2}dx=0$$ using variable substitution with $t=\tan(x/2)$ but it didn't get me anywhere.
|
By symmetry this is the same as proving that
$$\int_0^{2\pi} \frac{r}{1-2r \cos x + r^2} \; dx
= \int_0^{2\pi} \frac{\cos x}{1-2r \cos x + r^2} \; dx$$
for $r\in (-1,1).$
Put $z = \exp(ix)$ so that $dz = i\exp(ix) \; dx$ and hence
$\frac{dz}{iz} = dx$ to obtain for the first integral
$$\int_{|z|=1}
\frac{r}{1-r(z+1/z) + r^2} \; \frac{dz}{iz}
\\ = \frac{1}{i}
\int_{|z|=1}
\frac{r}{z-r(z^2+1) + zr^2} \; dz
\\ = \frac{1}{i}
\int_{|z|=1}
\frac{r}{-rz^2 + z(r^2+1) -r} \; dz.$$
This has poles at $z=r$ and $z=1/r$ and with $r\in(-1,1)$ only $z=r$
is inside the contour. We thus get for the first integral
$$2\pi i \times \frac{1}{i} \times
\left.\frac{r}{-2rz + r^2+1}\right|_{z=r}
= \frac{2\pi r}{1-r^2}.$$
We get for the second integral
$$\frac{1}{i}
\int_{|z|=1}
\frac{1/2(z+1/z)}{-rz^2 + z(r^2+1) -r} \; dz.$$
We thus obtain from the pole at $z=r$ the residue
$$\left.\frac{1/2(r+1/r)}{-2rz + r^2+1}\right|_{z=r}
= \frac{1}{2}(r+1/r) \times \frac{1}{1-r^2}.$$
and from the pole at $z=0$
$$-\frac{1}{2r}
= -\frac{1-r^2}{2r} \frac{1}{1-r^2} .$$
This yields for the second integral
$$2\pi i \times \frac{1}{i} \times
\left(\frac{1}{2}(r+1/r) - \frac{1}{2}(1/r-r)\right)
\frac{1}{1-r^2}
\\ = 2\pi \times
\left(\frac{1}{2}\times 2r\right)
\frac{1}{1-r^2}
= \frac{2\pi r}{1-r^2}.$$
We have equality of the two integrals as claimed.
|
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|
Integration by parts: $\int{\frac{dx}{(x^2 + a^2)^n}}$. I need to show that the following holds using integration by parts:
\begin{equation}
\int{\frac{dx}{(x^2 + a^2)^n}} = \frac{x}{2a^2(n-1)(x^2 + a^2)^{n-1}} + \frac{2n - 3}{2a^2(n-1)} \int{\frac{dx}{(x^2 + a^2)^{n-1}}}
\end{equation}
I really just don’t know where to start. It’s trivial to construct some solution of the form $\int u'v dx = uv - \int uv' dx$ to the integral on the left, but I can’t see how to get at this exact one.
EDIT:
I have tried to solve it by splitting it up,
\begin{equation}
\int{\frac{dx}{(x^2 + a^2)^n}} = \int{\Big( \frac{1}{(x^2 + a^2)^{n-1}} \cdot \frac{1}{(x^2 + a^2)} \Big)}dx
\end{equation}
but as far as I can tell this results in something rather different from where I am supposed to end up:
\begin{equation}
\int{\Big( \frac{1}{(x^2 + a^2)^{n-1}} \cdot \frac{1}{(x^2 + a^2)} \Big)}dx = \frac{1}{a}arctan \Big(\frac{x}{a}\Big) \cdot \frac{1}{(x^2 + a^2)^{n-1}} + \frac{2(n-1)}{a} \int{\frac{x^2 arctan \big(\frac{x}{a}\big)}{(x^2 + a^2)^{n}}}dx
\end{equation}
It is quite possible that I have made a very obvious mistake, so apologies in advance.
I have also tried this:
\begin{equation}
\int{(x^2 + a^2)^{-n}dx} = \int{\Big(1 \cdot (x^2 + a^2)^{-n}\Big) dx}
= x(x^2 + a^2)^{-n} + n\int{\frac{2x^2}{(x^2 + a^2)^{n+1}} dx}
\end{equation}
Again, it doesn’t seem to lead me nearer the specific solution I need.
|
Standart way of such:
$$J = \int\dfrac{dx}{(x^2+a^2)^n} = \dfrac1{a^2}\int\dfrac{(a^2+x^2)-x^2}{(x^2+a^2)^n}dx$$
$$J = \dfrac1{a^2}\int\dfrac{dx}{(x^2+a^2)^{n-1}} - \dfrac1{a^2}\int\dfrac{x^2}{(x^2+a^2)^n}dx$$
$$J = \dfrac1{a^2}\int\dfrac{dx}{(x^2+a^2)^{n-1}} + \dfrac{1}{2a^2(n-1)}\int x\,d\, \dfrac1{(x^2+a^2)^{n-1}}.$$
By parts:
$$J = \dfrac1{a^2}\int\dfrac{dx}{(x^2+a^2)^{n-1}} + \dfrac{1}{2a^2(n-1)} \dfrac {x}{(x^2+a^2)^{n-1}} - \dfrac{1}{2a^2(n-1)}\int \dfrac{dx}{(x^2+a^2)^{n-1}}, $$
$$\boxed{J = \dfrac{1}{2a^2(n-1)} \dfrac {x}{(x^2+a^2)^{n-1}} + \dfrac{2n-3}{2a^2(n-1)}\int \dfrac{dx}{(x^2+a^2)^{n-1}}}$$
|
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|
Prove $\frac{2ab}{a+b}\leq\sqrt {ab}$ $a$ and $b$ are both positive real numbers. I'm supposed to work backwards (i.e. start with what I'm trying to prove and change it until something is absolutely true, then start from what is absolutely true in my proof).
Here's my attempt:
$\frac{2ab}{a+b}\leq\sqrt {ab}$
$\frac{2a^2b^2}{(a+b)^2}\leq{ab}$
$\frac{2a^2b^2}{a^2+2ab+b^2} - ab \leq 0$
$\frac{2a^2b^2-ab(a^2+2ab+b^2)}{a^2+2ab+b^2}\leq 0$
$\frac{2a^2b^2-a^3b-2a^2b^2-ab^3}{a^2+2ab+b^2}\leq 0$
$\frac{-a^3b-ab^3}{a^2+2ab+b^2}\leq0$
Because $a$ and $b$ are positive, this guarantees the left side is negative, making the inequality true.
|
This is just a rearrangement of the famous AM-GM inequality!
It states that $\frac{a+b}{2} \geq \sqrt{ab}$
for $a,b \geq 0$.
The proof is clear from the picture!
Now multiply by $\sqrt{ab}$ on both sides and rearrange with ease to get the desired: $\frac{2ab}{a+b}\leq\sqrt {ab}$.
|
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|
Say for what values of $a \in \mathbb {R} $ this matrix system has solutions
Let $a \in \mathbb{R}$ and
$$
A_a =
\begin{pmatrix}
1 & a & 1 \\
a & 2 & 3 \\
2 & 3 & 4
\end{pmatrix}.
$$
*
*Say for each values of $a \in \mathbb{R}$ the system:
$$
A_a
\begin{pmatrix}
x \\ y \\ z
\end{pmatrix}
=
\begin{pmatrix}
1 \\ 0 \\ 1
\end{pmatrix}
$$
accepts one solution only and for what values have no solutions.
*Resolve the system for $a = 1$.
I got this problem from this image.
I think I know how to resolve system of equations using inverse matrices $3 \times 3$, $2 \times 2$ and more by finding the determinant. Then, I can reflect the matrix about the diagonal and multiply it with the inverse to find the solution. However, here, I get a little lost, because it seems like I should work backwards. What should I do?
|
Writing $A_a$ short as $A$, and the inhomogenous system $A x = b$ as augmented matrix $[A|b]$ we get these transformations:
$$
\left[
\begin{array}{rrr|r}
1 & a & 1 & 1 \\
a & 2 & 3 & 0 \\
2 & 3 & 4 & 1
\end{array}
\right]
\to
\left[
\begin{array}{rrr|r}
1 & a & 1 & 1 \\
a-3 & 2-3a & 0 & -3 \\
-2 & 3-4a & 0 & -3
\end{array}
\right]
\to
\left[
\begin{array}{rrr|r}
1 & a & 1 & 1 \\
a-1 & a-1 & 0 & 0 \\
-2 & 3-4a & 0 & -3
\end{array}
\right]
$$
The second row would benefit from division by $a-1$, so it is important if $a-1$ is zero or not.
Case 1:
If $a = 1$ this augmented matrix turns into
$$
\left[
\begin{array}{rrr|r}
1 & 1 & 1 & 1 \\
0 & 0 & 0 & 0 \\
-2 & 3 & 0 & -3
\end{array}
\right]
\to
\left[
\begin{array}{rrr|r}
1 & 1 & 1 & 1 \\
-2 & 3 & 0 & -3 \\
0 & 0 & 0 & 0 \\
\end{array}
\right]
\to
\left[
\begin{array}{rrr|r}
1 & 1 & 1 & 1 \\
0 & 5 & 2 & -1 \\
0 & 0 & 0 & 0 \\
\end{array}
\right]
\to
\left[
\begin{array}{rrr|r}
1 & 0 & 3/5 & 6/5 \\
0 & 1 & 2/5 & -1/5 \\
0 & 0 & 0 & 0 \\
\end{array}
\right]
$$
Only two equations for three variables, one variable stays free and in this case we have infinite many solutions
$$
(x, y, z) = (6/5-3/5 z, -1/5 - 2/5z, z) \quad (z \in \mathbb{R})
$$
where we used $z$ as free variable.
Other Cases:
If $a \ne 1$ we are led to
$$
\left[
\begin{array}{rrr|r}
1 & a & 1 & 1 \\
1 & 1 & 0 & 0 \\
-2 & 3-4a & 0 & -3
\end{array}
\right]
\to
\left[
\begin{array}{rrr|r}
1-a & 0 & 1 & 1 \\
1 & 1 & 0 & 0 \\
-2+4a & 3 & 0 & -3
\end{array}
\right]
\to
\left[
\begin{array}{rrr|r}
1-a & 0 & 1 & 1 \\
1 & 1 & 0 & 0 \\
-5+4a & 0 & 0 & -3
\end{array}
\right]
$$
Here it makes a difference if $-5 + 4a$ is zero or not.
Case 2:
If $a = 5/4$ we have
$$
\left[
\begin{array}{rrr|r}
-1/4 & 0 & 1 & 1 \\
1 & 1 & 0 & 0 \\
0 & 0 & 0 & -3
\end{array}
\right]
$$
and the equation $0 x + 0 y + 0 z = -3$ corresponding to the last row can not be fulfilled, this means no solution in this case.
Case 3:
If $a \ne 1$ and $a \ne 5/4$ we have:
$$
\left[
\begin{array}{rrr|r}
1-a & 0 & 1 & 1 \\
1 & 1 & 0 & 0 \\
1 & 0 & 0 & -3/(-5+4a)
\end{array}
\right]
$$
and the unique solution
$$
(x, y, z) = \left(
-\frac{3}{-5+4a},
\frac{3}{-5+4a},
1+\frac{3}{(1-a)(-5+4a)}
\right)
$$
|
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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|
I need some help with Geometry. Is this a correct answer to this problem? Good day,
I have a question regarding geometry. I don't know whether my answer is correct because the answer in my book uses a totally different method for solving this particular problem.
Here's the problem:
Given is a triangle, ABC, in which the middle of AB is also the middle of the circle drawn outside of the triangle. Finally, we draw a line from C to M where M signifies the point in the middle of AB. Now you have angle C sub 1 and angle C sub 2 ( together: angle C sub 12.)
We have created 2 equal-sided triangles, within the triangle ABC, within the circle, namely:
triangle ACM and triangle BCM.
Proof that angle C sub 12 is equal to 90 degrees.
My answer:
Given:
AM = MC = MB (2 equal-sided triangles)
To Proof:
$$\angle C_{12} = 90^{\circ}$$
Proof:
$$
\left.\begin{matrix}
& & & & & \\
& & & & & \\
& & & & & \\
\angle A + \angle M_{1} + \angle C_{1} = 180^{\circ}\\
\angle B + \angle M_{2} + \angle C_{2} = 180^{\circ}\\
\angle M_{2} + \angle 2B = 180^{\circ} \\
\angle M_{1} + \angle 2A = 180^{\circ} \\
\angle M_{1} + \angle M_{2} = 180^{\circ}\\
\angle A + \angle B + \angle C = 180^{\circ} (= Q)\\
\\
\end{matrix}\right\}general
\\
$$
Calculate for angle A and angle B:
$$ \angle M_{2} + \angle 2B = 180^{\circ}\\
\angle 2B = 180^{\circ} - \angle M_{2}\\
\angle B = 90^{\circ} - (\angle M_{2} / 2) $$
$$and$$
$$ \angle M_{1} + \angle 2A = 180^{\circ}\\
\angle 2A = 180^{\circ} - \angle M_{1}\\
\angle A = 90^{\circ} - (\angle M_{1} / 2) $$
Also:
$$
\angle M_{2} = 180^{\circ} - \angle M_{1}\\
\angle M_{1} = 180^{\circ} - \angle M_{2} (= U)\\
$$
Now we can put angle A and angle B in the sum of angles (indicated by Q in general section.):
$$(90^{\circ} - (\angle M_{1} / 2)) + (90^{\circ} - (\angle M_{2} / 2)) + \angle C = 180^{\circ}$$
Now we substitute $$ \angle M_{1}$$ inside $$((\angle M_{1} / 2)) $$ with U:
$$(90^{\circ} - (\frac{180^{\circ} - \angle M_{2}}{2}) + ((90^{\circ} - (\angle M_{2} / 2)) + \angle C = 180^{\circ}\\
(90^{\circ} - (\frac{180^{\circ}}{2}-\frac{\angle M_{2}}{2})) + (90^{\circ} - (\angle M_{2} / 2)) + \angle C = 180^{\circ}\\
90^{\circ} - \frac{180^{\circ}}{2}+\frac{\angle M_{2}}{2} + 90^{\circ} -\frac{\angle M_{2}}{2} + \angle C = 180^{\circ}\\
\Rightarrow 90^{\circ} + \angle C = 180^{\circ}\\
Conclusion: \angle C_{12} = 90^{\circ}
$$
(not the prettiest proof but hopefully you get the idea.)
Here's the more concise ( and beautiful ) proof from the answers in my book:
$$\angle A + \angle B + \angle C = 180^{\circ} \Rightarrow \angle C_{1} + \angle C_{2} + \angle C_{12} = 180^{\circ} \Rightarrow \angle 2C_{12} = 180^{\circ} \Rightarrow \angle C_{12} = 90^{\circ} $$
If you need more information please ask and I'll provide the necessities.
Thanks!
Picture:
|
Note: My figure is misleading. ABC in figure is not isosceles.
$$2x+2y = 180^\circ$$
$$x+y = ?$$
|
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|
Why is the solution to $\sqrt{6-5x}=x$ only $x=1$ and not $x=-6$? I solved the equation $\sqrt{6-5x}=x$ as follows:
$$(\sqrt{6-5x})^2=x^2$$
$$6-5x=x^2$$
$$0=x^2+5x-6=(x+6)(x-1)$$
$$x=-6 \quad \text{or} \quad x=1$$
If I plug in $x=-6$ into the original equation, I get $\sqrt{6+30}=\sqrt{36}=\pm 6$ and if I plug in $x=1$, I get $\sqrt{6-5}=\sqrt{1}=\pm 1$.
To me it seems that both values satisfy the original equation. I am using an online education system for my class called MyMathLab and the solution is only $x=1$. Why is that?
Thank you in advance.
|
The confusion you are having comes from the concept of finding the solutions to $z^2 = 36$. Definitely $z = \pm 6$ are solutions because $(-6)^2 = 36 = 6^2$, but this is not the same thing as $y = \sqrt{36}$. Otherwise, we get nonsense like $-6 = 6$ which isn't true.
If we plug the answer $x = -6$ back into the original equation, we have $$\sqrt{6 - 5(-6)} = \sqrt{36} = 6 = -6.$$ Again, we get nonsense of $-6 = 6$. Even it were positive and negative $6$, you'd have $-6 = \pm 6$. This is true for only one of the values, which shows it's not valid to assume that $\sqrt{36} = \pm 6$.
|
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"timestamp": "2023-03-29T00:00:00",
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|
How can I solve $\int \frac{3x+2}{x^2+x+1}dx$ I want to compute this primitive $$I=\int \frac{3x+2}{x^2+x+1}dx.$$
I split this integral into two part:
$$\int \frac{3x+2}{x^2+x+1}dx=\int \frac{2x+1}{x^2+x+1}dx+\int \frac{x+1}{x^2+x+1}dx,$$
For the first part:
$$\int \frac{2x+1}{x^2+x+1}dx= \ln(|x^2+x+1|)+c_1$$
For the second part: due a change of variable $u=x+1$, I find
$$\int \frac{x+1}{x^2+x+1}dx= \int \frac{x+1}{x(x+1)+1}dx=\int \frac{u}{u(u-1)+1}dx=\dots$$
|
Hint For the second:
$$\int \frac{x+1}{x^2+x+1}dx$$
Rewrite as $$\int\bigg( \frac{2x+1}{2(x^2+x+1)}+\frac{1}{2(x^2+x+1)} \bigg)dx$$
$$=\underbrace{\frac 1 2 \int\frac{2x+1}{x^2+x+1}dx}_{:=I}+\underbrace{\frac 1 2 \int \frac{1}{x^2+x+1}dx}_{:=J}$$
For $I$ substitute $t=x^2+x+1$ and $dt=(2x+1)dx$
For $J$ complete the square : $\frac 1 2\displaystyle\int \frac{1}{(x+1/2)^2+3/4}$ and now substitute $\varphi=x+1/2$ and $d\varphi=dx\Longrightarrow \frac 1 2\displaystyle\int \frac{1}{\varphi ^2 +3/4}d \varphi$
factor out $3/4$ from the denominator $\frac 2 3\displaystyle\int \frac{1}{(4\varphi ^2)/(3)+1}d \varphi$ and then substitute $z=\frac{2 \varphi}{\sqrt 3}$ and $dz=\frac{2}{\sqrt 3}d \varphi$ so you will get $\frac {1}{\sqrt 3}\displaystyle\int \frac{1}{z^2+1}dz$ and from here it is easy
|
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"timestamp": "2023-03-29T00:00:00",
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|
How to prove that $k^3+3k^2+2k$ is always divisible by $3$? How can I prove that the following polynomial expression is divisible by 3 for all integers $k$?
$$k^3 + 3k^2 + 2k$$
|
Just to be different. If $k \in \mathbb Z$ then $k = 3m + i$ where $i = $ either $0, 1,$ or $-1$ and $m \in \mathbb Z$.
So
\begin{align*}
k^3 + 3k^2 + 2k
&=(3m + i)^3 + 2(3m + i)+ 3k^2 \\
&= 3^3m^3 + 3 \cdot 3^2m^2 \cdot i + 3 \cdot 3m \cdot i^2 + i^3 + 2 \cdot 3m + 2i + 3k^2 \\
&= i^3 + 2i + 3\big[3^2m^3 + 3^2m^2i + 3m \cdot i^2 + k^2\big]
\end{align*}
Now $i^3 = i$ for $i = 0,1,-1$ so $i^3 + 2i = 3i$.
So
$$
k^3 + 3k^2 + 2k = 3\big[i + 3^2m^3 + 3^2m^2i + 3m \cdot i^2 + k^2\big].
$$
...
But seriously, factoring is the better way to do it.
|
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|
Calculate the area enclosed by the curve and line Calculate the are enclosed by ${y = 2x - 1}$ and ${y= x^2 + 6x + 2}$
First of all I combine the equations into:
${x^2 + 4x + 3 = 0}$
${(x + 3)(x + 1), x = -3, x = -1}$
They intersect at ${(-3 -7) (-1, -3)}$
I would say that ${y = 2x -1}$ is the top equation so to work out the area, I would use:
${\int_{-3}^{-1} (2x -1) - (x^2 + 6x + 2) dx}$
=> ${-{x^3\over 3} - 2x^2 - 3x}$
=> ${(-{1\over 3} - 2 + 3) - ({-27 \over 3} -18 + 9)}$
=> ${{2\over 3} - (-18)}$
=> ${18{2\over 3}}$
This is not the correct answer, can anyone point out where I have gone wrong?
|
$(\frac13-2+3)-(9-18+9)=\frac43$ when there are two equation, there are two roots. You should use
$$\int_{α}^{β}a(x-α)(x-β)dx=-\frac{a}6(β-α)^3$$
for example
$$\frac16(-1-(-3)^2)=\frac43$$
|
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"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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|
Prove that $\frac 12\leq \frac{1}{n+1}+\frac {1}{n+2}+\frac{1}{n+3}...+\frac{1}{n+n}$ How do I prove $$\frac 12\leq \frac{1}{n+1}+\frac {1}{n+2}+\frac{1}{n+3}...+\frac{1}{n+n}$$
without using induction? Note that clearly $n\neq 0$
Thanks for any help!!
|
This inequality does not hold since for any $n\in \mathbb N$ you have
$$\frac{1}{2}\leq \frac{1}{n+1}+\frac {1}{n+2}+\frac{1}{n+3}+...+\frac{1}{n+n}.$$
To see this, note first that any term above is greater than $\frac{1}{n+n}$, so that your sum is greater than $n\cdot \frac{1}{2n}=\frac{1}{2}.$
|
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|
Solving $x^2 \equiv -x\pmod{2015}$
Problem: Find all integer solutions of $x^2 \equiv -x \pmod{2015}$.
I proceeded this way: first, I realized that $2015 = 5 \times 13 \times 31$. I rewrote $x^2 \equiv -x$ as $x^2 + x \equiv 0$.
Then, for each factor $d$, I have that $d\ |\ x(x+1)$, therefore $d\ |\ x$ or $d\ |\ x + 1$.
This gives eight cases:
*
*$3,13,31\ |\ x$
*$3,13\ |\ x\quad{\rm and}\quad 31\ |\ (x + 1)$
*etc.
Solving each case separately and taking union of the results gives correct results (the conditions are both necessary and sufficient), but it's really tedious.
I am sure someone can come up with a better approach.
|
There's this general trick for solving quadratic congruences: if $\gcd(4a,n)=1$, then:
$$ax^2+bx+c\equiv 0\pmod{n}\stackrel{\cdot 4a}\iff (2ax+b)^2\equiv b^2-4ac\pmod{n}$$
In your case, $2015=5\cdot 13\cdot 31$ and:
$$x^2\equiv -x\pmod{2015}\iff (2x+1)^2\equiv 1\pmod{2015}$$
$$\iff \begin{cases}(2x+1)^2\equiv 1\pmod{5}\\(2x+1)^2\equiv 1\pmod{13}\\(2x+1)^2\equiv 1\pmod{31}\end{cases}$$
$$(2x+1)^2-1=((2x+1)+1)((2x+1)-1),$$ so by Euclid's Lemma:
$$\iff \begin{cases}2x+1\equiv \pm 1\pmod{5}\\2x+1\equiv \pm 1\pmod{13}\\2x+1\equiv \pm 1\pmod{31}\end{cases}$$
$$\iff \begin{cases}x\equiv \{-1,0\}\pmod{5}\\x\equiv \{-1,0\}\pmod{13}\\x\equiv \{-1,0\}\pmod{31}\end{cases}$$
By Chinese Remainder Theorem (CRT), this gives $8$ solutions mod $2015$.
E.g., if $x\equiv -1\pmod{5}$, $x\equiv -1\pmod{13}$, $x\equiv 0\pmod{31}$, then $x\equiv 1364\pmod{2015}$.
Etc. You should already know how to apply CRT. You'll get: $$x\equiv \{-1,0,155,650,805,1209,1364,1859\}\pmod{2015}$$
|
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|
What to do if the modulus is not coprime in the Chinese remainder theorem? Chinese remainder theorem dictates that there is a unique solution if the congruence have coprime modulus.
However, what if they are not coprime, and you can't simplify further?
E.g. If I have to solve the following 5 congruence equations
$x=1 \pmod 2$
$x=1 \pmod 3$
$x=1 \pmod 4$
$x=1 \pmod 5$
$x=1\pmod 6$
as gcd (2,3,4,5,6) is not coprime, how would you do it. I have heard that you can't use the lcm of the numbers, but how does it work?
Sorry for the relatively trivial question and thank you in advance.
|
\begin{align}
x &\equiv 1 \pmod 2 \\
x &\equiv 1 \pmod 3 \\
x &\equiv 1 \pmod 4 \\
x &\equiv 1 \pmod 5 \\
x &\equiv 1 \pmod 6 \\
\end{align}
Note that
$x \equiv 1 \pmod 6 \implies
\left\{
\begin{array}{l}
x\equiv 1 \pmod 2 \\
x \equiv 1 \pmod 3
\end{array}
\right.$
Replace $x \equiv 1 \pmod 6$ in your original list with those two equivalences and sort the list by prime number bases:
\begin{align}
x &\equiv 1 \pmod 2 \\
x &\equiv 1 \pmod 2 \\
x &\equiv 1 \pmod 4 \\\hline
x &\equiv 1 \pmod 3 \\
x &\equiv 1 \pmod 3 \\ \hline
x &\equiv 1 \pmod 5 \\
\end{align}
Note first that $x \equiv 1 \pmod 4 \implies x \equiv 1 \pmod 2$. This means that the equivalence $x \equiv 1 \pmod 2$ is included in the equivalence $x \equiv 1 \pmod 4$ and is therefore superfluous - it can be removed.
So we can simplifiy the list to
\begin{align}
x &\equiv 1 \pmod 4 \\
x &\equiv 1 \pmod 3 \\
x &\equiv 1 \pmod 5 \\
\end{align}
The solution is $x \equiv 1 \pmod{30}$.
|
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|
Find the projection of the line $x+y+z-3=0=2x+3y+4z-6$ on the plane $z=0$
Find the projection of the line $x+y+z-3=0=2x+3y+4z-6$ on the plane $z=0$
The equation represents the line of intersection of two planes. Using augmented matrix
$$
\begin{bmatrix}
1 & 1 & 1 & 3 \\
2 & 3 & 4 & 6
\end{bmatrix}$$
$R_2\rightarrow R_2-2R_1$
$$
\begin{bmatrix}
1 & 1 & 1 & 3 \\
0 & 1 & 2 & 0
\end{bmatrix}$$
$$y+2z=0$$
$$x+y+z=3$$
Equation of line is
$$\frac{x-0}{1}=\frac{y-0}{-2}=\frac{z-0}{1}$$
Angle made by line and normal is $$\cos(90-\theta)=\frac{1}{\sqrt{6}}$$
where $\theta$ is the angle made by line and plane.
How should I proceed?
|
A line passes through two points.
You can get the first point intersecting your line with the plane $z=0$. $p=(3,0,0)$.
Simply project another random point of your line, say $(0,6,-3)$, in the plane $z=0$, you get $q=(0,6,0)$.
The line you are looking for is thus $z=0$, $y=-2x+6$.
|
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|
If $ \frac{x}{b-c} = \frac{y}{c-a} = \frac{z}{a-b}$, Prove that $x+y+z$ $=0$
Question If $$ \frac{x}{b-c} = \frac{y}{c-a} = \frac{z}{a-b} $$ Prove that $x+y+z$ $=0$
I've attmempted this question by cross multiplying so that
$$ x(c-a)(a-b) = y(b-c)(a-b) = (b-c)(c-a)z $$
but that did not work
I also tried splitting the equation so
$ \frac{x}{b-c} = \frac{y}{c-a}$ and $\frac{y}{c-a} = \frac{z}{a-b}$
and tried forming systems of equation but that didn't lead me to the proof as well...
Could someone please help!
|
Using $k$-method, let $\displaystyle \frac{x}{b-c}=\frac{y}{c-a}=\frac{z}{a-b}=k$, then $\displaystyle \sum_{xyz} x=\sum_{abc} (b-c)k=0$.
|
{
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"timestamp": "2023-03-29T00:00:00",
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|
integrate $\int \sin^{4}x\cos^{2}x$
$$\int \sin^{4}x\cos^{2}xdx$$
$$\int \sin^{4}x\cos^{2}xdx=\int (\sin x \cos x)^{2}\sin^2xdx=\int \left(\frac{\sin^{2}2x}{2}\right)\left(\frac{1}{2}-\frac{\cos2x}{2}\right)dx=\int \left(\frac{\sin^{2}2x}{4}-\frac{\sin^{2}2x\cos2x}{4}\right)dx=\frac{1}{4}\int ({\sin^{2}2x}-{\sin^{2}2x\cos2x})dx$$
I still have not mange to find $u$ substitution
|
$$\int\sin^4(x)\cos^2(x)\space\text{d}x=$$
$$\int\sin^4(x)\left(1-\sin^2(x)\right)\space\text{d}x=$$
$$\int\left(\sin^4(x)-\sin^6(x)\right)\space\text{d}x=$$
$$\int\sin^4(x)\space\text{d}x-\int\sin^6(x)\space\text{d}x=$$
You've to use twice the reduction formula:
$$\int\sin^m(x)\space\text{d}x=-\frac{\cos(x)\sin^{m-1}(x)}{m}+\frac{m-1}{m}\int\sin^{m-2}(x)\space\text{d}x$$
$$\frac{\sin^5(x)\cos(x)}{6}+\frac{1}{6}\int\sin^4(x)\space\text{d}x=$$
$$\frac{\cos(x)\sin^3(x)\left(4\sin^2(x)-1\right)}{24}+\frac{1}{8}\int\sin^2(x)\space\text{d}x=$$
$$\frac{\cos(x)\sin^3(x)\left(4\sin^2(x)-1\right)}{24}+\frac{1}{8}\int\left[\frac{1}{2}-\frac{\cos(2x)}{2}\right]\space\text{d}x=$$
$$\frac{\cos(x)\sin^3(x)\left(4\sin^2(x)-1\right)}{24}+\frac{1}{8}\left[\frac{1}{2}\int1\space\text{d}x-\frac{1}{2}\int\cos(2x)\space\text{d}x\right]=$$
$$\frac{\cos(x)\sin^3(x)\left(4\sin^2(x)-1\right)}{24}+\frac{1}{8}\left[\frac{x}{2}-\frac{1}{2}\int\cos(2x)\space\text{d}x\right]=$$
Substitute $u=2x$ and $\text{d}u=2\space\text{d}x$:
$$\frac{\cos(x)\sin^3(x)\left(4\sin^2(x)-1\right)}{24}+\frac{1}{8}\left[\frac{x}{2}-\frac{1}{2}\int\cos(u)\space\text{d}u\right]=$$
$$\frac{\cos(x)\sin^3(x)\left(4\sin^2(x)-1\right)}{24}+\frac{1}{8}\left[\frac{x}{2}-\frac{1}{2}\int\cos(u)\space\text{d}u\right]=$$
$$\frac{\cos(x)\sin^3(x)\left(4\sin^2(x)-1\right)}{24}+\frac{1}{8}\left[\frac{x}{2}-\frac{\sin(u)}{4}\right]+\text{C}=$$
$$\frac{\cos(x)\sin^3(x)\left(4\sin^2(x)-1\right)}{24}+\frac{1}{8}\left[\frac{x}{2}-\frac{\sin(2x)}{4}\right]+\text{C}$$
|
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|
Integrate $\int \frac{x\cos x}{\sin^2x}dx$
$$\int \frac{x\cos x}{\sin^2x}dx$$
$$\int \frac{x\cos x}{\sin^2x}dx=\int \frac{x\cos x}{1-\cos^2x}dx=\int \frac{x\cos x}{(1-\cos x)(1+\cos x)}dx$$
How can I find the two fractions? if there are at all?
|
Since you wanted to solve using partial fractions,
$$\int\frac{x\cos x}{(1+\cos x)(1-\cos x)}dx=\frac{1}{2}\int\frac{x(1+\cos x)-x(1-\cos x)}{(1+\cos x)(1-\cos x)}dx$$
$$=\frac{1}{2}\int\frac{x}{1-\cos x}dx- \frac{1}{2}\int{\frac{x}{1+\cos x}}dx$$
$$= \frac{1}{2}\int \frac{x}{2\sin^2\frac{x}{2}}dx- \frac{1}{2}\int \frac{x}{2\cos^2\frac{x}{2}}dx$$
$$= \frac{1}{2}\int x\sec^2\frac{x}{2}dx-\frac{1}{2}\int x\csc^2\frac{x}{2}dx$$
Now use integration by parts.
I will solve one term
$$\int x\sec^2\frac{x}{2}dx=2x\tan\frac{x}{2}-2\int\tan\frac{x}{2}dx$$
|
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|
Integrate $\int \frac{x^7}{(1-x^4)^2}dx$
$$\int \frac{x^7}{(1-x^4)^2}dx$$
I have tried to simplify the expression, to use U substation, any idea where to start from?
|
$$\int\frac{x^7}{\left(1-x^4\right)^2}\space\text{d}x=$$
$$\int\frac{x^7}{\left(x^4-1\right)^2}\space\text{d}x=$$
Substitute $u=x^4$ and $\text{d}u=4x^3\space\text{d}x$:
$$\frac{1}{4}\int\frac{u}{\left(u-1\right)^2}\space\text{d}u=$$
$$\frac{1}{4}\int\left[\frac{1}{u-1}+\frac{1}{(u-1)^2}\right]\space\text{d}u=$$
$$\frac{1}{4}\left[\int\frac{1}{u-1}\space\text{d}u+\int\frac{1}{(u-1)^2}\space\text{d}u\right]=$$
Substitute $s=u-1$ and $\text{d}s=\text{d}u$:
$$\frac{1}{4}\left[\int\frac{1}{s}\space\text{d}s+\int\frac{1}{s^2}\space\text{d}s\right]=$$
$$\frac{1}{4}\left[\ln\left|s\right|-\frac{1}{s}\right]+\text{C}=$$
$$\frac{1}{4}\left[\ln\left|u-1\right|-\frac{1}{u-1}\right]+\text{C}=$$
$$\frac{1}{4}\left[\ln\left|x^4-1\right|-\frac{1}{x^4-1}\right]+\text{C}=$$
$$\frac{1}{4}\left[\ln\left|x^4-1\right|+\frac{1}{1-x^4}\right]+\text{C}$$
|
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|
Where did my simplification go wrong? Sum and difference formula simplification I'm struggling with the following: We are to use the sum and difference formulas to find the exact value of the expression. The problem is simplification has been tough. As a last resort I decided to use Symbolab to find the answer and steps but the steps were not to be found. Despite lack of steps, the answer is $(\sqrt {2+\sqrt{3}})/2$
Here are my steps so far:
\begin{align*}
\sin(135^\circ - 30^\circ) & = (\sin 135^\circ\cos 30^\circ)-(\cos 135^\circ \sin 30^\circ)\\
& = \left(\frac{\sqrt{2}}{2}\right)\left(\frac{\sqrt{3}}{2}\right)-\left(-\frac{\sqrt{2}}{2}\right)\left(\frac{1}{2}\right)\\
& = \left(\frac{\sqrt{6}}{4}\right)-\left(-\frac{\sqrt{2}}{4}\right)
\end{align*}
[insert final simplification step]
$(\sqrt {2+\sqrt{3}})/2$
What's the missing step here?
|
\begin{align*}
\frac{\sqrt{6}}{4}- \left(-\frac{\sqrt{2}}{4}\right) & = \frac{1}{4}(\sqrt6+\sqrt2)\\
& = \frac{\sqrt{2}}{4}(\sqrt{3}+1)\\
& = \frac{\sqrt{2}}{4}\sqrt{(\sqrt3+1)^2}\\
& = \frac{\sqrt{2}}{4}\sqrt{4+2\sqrt3}\\
& = \frac{\sqrt{2}}{4}\sqrt{2}\sqrt{2+\sqrt{3}}\\
& = \frac{1}{2}\sqrt{2+\sqrt{3}}
\end{align*}
|
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|
Is $\lim_{x\to -3}\frac{x^2+9}{\sqrt{x^2+16}-5} = \infty$? It was asked in our test, and below is what I did:
$$\lim_{x\to -3}\frac{x^2+9}{\sqrt{x^2+16}-5} $$
$$=\lim_{x\to -3}\frac{x^2+9}{\sqrt{x^2+16}-5}\times\frac{\sqrt{x^2+16}+5}{\sqrt{x^2+16}+5} $$
$$=\lim_{x\to -3}\frac{x^2+9}{x^2-9}\times \left(\sqrt{x^2+16}+5\right) $$
Now no terms cancel. We get 0 in numerator and denominator too.
Ans: My teacher told me that the limit is $+\infty$, but didn't tell how.
|
$$\lim_{x\to -3}\frac{x^2+9}{\sqrt{x^2+16}-5} $$
As $x\rightarrow-3^+$, we have the numerator, $x^2+9\rightarrow18$, and the denominator, $\sqrt{x^2+16}-5\rightarrow0$ from the left side on the number line.
As $x\rightarrow-3^-$, we have the numerator, $x^2+9\rightarrow18$, and the denominator, $\sqrt{x^2+16}-5\rightarrow0$ from the right side on the number line.
Thus, the fraction, $\lim_{x\to-3^+}\frac{x^2+9}{\sqrt{x^2+16}-5}\rightarrow-\infty$ and $\lim_{x\to-3^-}\frac{x^2+9}{\sqrt{x^2+16}-5}\rightarrow+\infty$.
|
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|
Schurs Lemma for Upper Triangular matrix Schurs Lemma Says : For any $A\in M_n(\mathbb{C})$ with eigen values $\lambda_1,\dots,\lambda_n$ There exists an unitary matrix $U$ such that $$UAU^*=T$$ where $T$ is an upper triangular matrix with diagonal $\lambda_1,\dots,\lambda_n$
If In particular $A$ is itself an upper triangular matrix with diagonals $(x_1,\dots,x_n)$ then $T$ and $A$ will be same matrix?
what if $B$ is another upper triangular matrix with diagonals $(y_1,\dots,y_n)$ which is a permutation of $(x_1,\dots,x_n)$ , then how can I conclude $A,B$ are equivalent?
Thanks for helping.
|
To answer your first question: no, $A$ and $T$ are not equal in general (since Schur decomposition is not unique--even in the case where each $\lambda_1,\dots,\lambda_n$ are distinct). Take
$$A = \begin{pmatrix} 5 & 5 &5 \\ 0 & 6 & 5 \\ 0 & 0 & 7 \end{pmatrix},$$
we can Schur factor $A$ as follows:
$$\begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{pmatrix} \begin{pmatrix} 5 & 5 &5 \\ 0 & 6 & 5 \\ 0 & 0 & 7 \end{pmatrix} \begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{pmatrix}^{-1}=\begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{pmatrix} \begin{pmatrix} 5 & 5 &5 \\ 0 & 6 & 5 \\ 0 & 0 & 7 \end{pmatrix} \begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{pmatrix}$$
$$=\begin{pmatrix} 5 & 5 &5 \\ 0 & 6 & 5 \\ 0 & 0 & 7 \end{pmatrix}$$
$$=A$$
And the matrix $\begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{pmatrix} $ is definitely unitary. So we found a Schur decomposition that was our original matrix $A$.
But we can also Schur factor $A$ as follows:
$$\begin{pmatrix} -1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & -1 \end{pmatrix} \begin{pmatrix} 5 & 5 &5 \\ 0 & 6 & 5 \\ 0 & 0 & 7 \end{pmatrix} \begin{pmatrix} -1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & -1 \end{pmatrix}^{-1}=\begin{pmatrix} -1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & -1 \end{pmatrix} \begin{pmatrix} 5 & 5 &5 \\ 0 & 6 & 5 \\ 0 & 0 & 7 \end{pmatrix} \begin{pmatrix} -1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & -1 \end{pmatrix}$$
$$=\begin{pmatrix} 5 & 5 &-5 \\ 0 & 6 & -5 \\ 0 & 0 & 7 \end{pmatrix}$$
$$ \neq A$$
and it is easily checked that $\begin{pmatrix} -1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & -1 \end{pmatrix}$ is unitary, and so we found another Schur decomposition for $A$ that was not equal $A$.
|
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|
Determinant equality issue Hi I am studying for an exam tomorrow and I have a question, How do I prove that the two determinants are equal ? is there a short way ?
$2abc\left|\begin{array}{ccc}
1 & 1 & 1\\
a & b & c\\
a^{2} & b^{2} & c^{2}
\end{array}\right|$
$
\left|\begin{array}{ccc}
b+c & c+a & a+b\\
b^{2}+c^{2} & c^{2}+a^{2} & a^2+b^2\\
b^{3}+c^{3} & c^{3}+a^{3} & a^{3}+b^{3}
\end{array}\right|
$
Thanks !!!!
|
Use
$$abc\left|\begin{array}{ccc}
1 & 1 & 1\\
a & b & c\\
a^{2} & b^{2} & c^{2}
\end{array}\right| = \left|\begin{array}{ccc}
a & b & c\\
a^2 & b^2 & c^2\\
a^{3} & b^{3} & c^{3}
\end{array}\right| \quad\text{and}\quad
\left|\begin{array}{ccc}
0 & 1 & 1\\
1 & 0 & 1\\
1 & 1 & 0
\end{array}\right| = 2.$$
|
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|
My answer don't match with the answer of the book Let $x^{2}-4x+6=0$. What can be the result of $1-\frac{4}{3x}+\frac{2}{x^{2}}$ ?
A) $-\frac{2}{3}~~$ B) $-\frac{1}{3}~~$ C) $\frac{1}{3}~~$ D) $\frac{2}{5}~~$ E) $2$
My answer is: $1-\frac{4}{3x}+\frac{2}{x^{2}}=1-\frac{1}{x^{2}}(\frac{4x}{3}-2)=1-\frac{1}{x^{2}}(\frac{4x-6}{3})=1-\frac{1}{x^{2}}(\frac{x^{2}}{3})=\frac{2}{3}$.
But according to the book the right answer is option E, i.e, $2$.
Is the book right or me ?
|
Divide both sides by $x^2$ to get $$1-\dfrac4x+\dfrac6{x^2}=0\iff\dfrac4x-\dfrac6{x^2}=1$$
$$1-\dfrac4{3x}+\dfrac2{x^2}=1-\dfrac13\left(\dfrac4x-\dfrac6{x^2}\right)=?$$
|
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|
integrate $\int_{2}^{4} \frac{\sqrt{16-x^2}}{x}$
$$\int_{2}^{4} \frac{\sqrt{16-x^2}}{x}$$
$x=4\sin(u)$
$dx=4\cos(u)du$
$$\int_{2}^{4} \frac{\sqrt{16-x^2}}{x}=\int_{2}^{4} \frac{\sqrt{16-16\sin^2u}}{4\sin u}\cos u\,du=\int_{2}^{4} \frac{4\sqrt{1-\sin^2u}}{4\sin u}\cos u \,du=\int_{2}^{4} \frac{\cos^2u}{\sin u}du=\int_{2}^{4} \frac{1-\sin^2u}{\sin u}du=\int_{2}^{4} \frac{1}{\sin u}du-\int_{2}^{4} {\sin u}du$$
$$=\ln\left(\tan\left(\frac{u}{2}\right)\right)+\cos u$$
$\frac{x}{4}=\sin u$
$\tan u=\frac{\sqrt{x}}{x^2-16}$
$\ln\left(\frac{{x}}{8(\sqrt{x^2-16})}\right)+\left(\frac{\sqrt{16-x^2}}{4}\right)$ form $2$ to $4$
but I get a $\frac{4}{0}$ the end result according to Wolfram is $1.80$
|
If we write $$\frac{\sqrt{16-x^2}}{x} = \frac{x \sqrt{16-x^2}}{x^2},$$ then the substitution $u^2 = 16-x^2$, $x^2 = 16-u^2$, $x \, dx = -u \, du$ immediately yields $$\int_{x=2}^4 \frac{\sqrt{16-x^2}}{x} \, dx = \int_{u=\sqrt{12}}^0 \frac{-u^2}{16-u^2} \, du = \int_{u=0}^{\sqrt{12}} \frac{16}{16-u^2} - 1 \, dy.$$ Then we proceed by partial fraction decomposition, giving $$2 \int_{u=0}^{\sqrt{12}} \left(\frac{1}{4-u} + \frac{1}{4+u}\right) \, du - \sqrt{12} = 2 \log \frac{4+\sqrt{12}}{4-\sqrt{12}} - \sqrt{12},$$ which simplifies to $$4 \log (2 + \sqrt{3}) - 2 \sqrt{3}.$$
Since people seem to think that the above calculation is incorrect, here it is via trigonometric substitution along the same lines as attempted by the original question:
$$\begin{align*} \int_{x=2}^4 \frac{\sqrt{16-x^2}}{x} \, dx &= \int_{u = \pi/6}^{\pi/2} \frac{4 \sqrt{1 - \sin^2 u}}{4 \sin u} \cdot 4 \cos u \, du \\ &= 4 \int_{u=\pi/6}^{\pi/2} \frac{\cos^2 u}{\sin u} \, du \\ &= 4 \int_{u=\pi/6}^{\pi/2} \csc u - \sin u \, du \\ &= 4 \int_{u=\pi/6}^{\pi/2} \frac{\csc u (\csc u + \cot u)}{\csc u + \cot u} \, du - \left[ - 4\cos u \right]_{u=\pi/6}^{\pi/2} \\ &= 4\left[-\log|\csc u + \cot u| \right]_{u=\pi/6}^{\pi/2} - 2 \sqrt{3} \\ &= 4\log (2 + \sqrt{3}) - 2 \sqrt{3}.\end{align*}$$
And here is the WolframAlpha link.
|
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|
Let $x_1 = 2 $, and define $x_{n+1} = \frac{1} {2} (x_n + \frac {2} {x_n})$
Show that $x^2_n $ is always greater than or equal to $2$. And then use this to prove that $$x_n - x_{n+1} \geqslant 0 \quad \text{for any } n \in \mathbb{N}$$ Hence conclude that $\lim\limits_{n \to \infty} x_n = \sqrt{2}$.
$\bullet~$ I can show that $x^2_n \geqslant 2$ by induction:
*
*Basis step: $x_1=2, x_1^2 = 4 >2$.
*Inductive step:
Assume we have some $k$ for which $x^2_K \geqslant 2$, then
\begin{align*}
(x_{k+1})^2 =&~ \bigg(\dfrac{1} {2} \bigg(x_k + \dfrac{2} {x_k}\bigg)\bigg)^2\\
=&~ \dfrac{1} {4} x_k^2 + 1 + \dfrac{1} {x_k^2}\\
\geqslant&~ \dfrac {1} {4} x_k^2 + 1 \geqslant 2\\
&~\dfrac {1} {4} x_k^2 \geqslant 1 \quad \forall~ n \geqslant 2
\end{align*}
How do I use this to show that $$x_n - x_{n+1} \geqslant 0 \quad \text{for any } n \in \mathbb{N}$$
and hence conclude that $\lim\limits_{n \to \infty} x_n = \sqrt{2}$?
Thanks!!
|
$$x_n-x_{n+1}=x_n-\frac12\left(x_n+\frac 2{x_n}\right)=\frac{2x_n^2-x_n^2-2}{2x_n^2}=\frac{x_n^2-2}{2x_n^2}.$$
|
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|
Prove the inequality $\frac{a^2+b^2}{c^2+a+b}+\frac{b^2+c^2}{a^2+b+c}+\frac{c^2+a^2}{b^2+c+a} \ge 2$ For every $a,b,c>0$ such that $abc=1$ prove the inequality
$$\frac{a^2+b^2}{c^2+a+b}+\frac{b^2+c^2}{a^2+b+c}+\frac{c^2+a^2}{b^2+c+a} \ge 2$$
My work so far:
$abc=1 \Rightarrow \frac 13 (a+b+c)\ge 1$
Then $$c^2+a+b \le c^2+(a+b)\frac 13(a+b+c)$$
Or more
$$ (\frac{a^2+b^2}{c^2+a+b}+\frac{b^2+c^2}{a^2+b+c}+\frac{c^2+a^2}{b^2+c+a})((c^2+a+b)+(a^2+b+c)+(b^2+c+a)) \ge $$
$$\ge(\sqrt{a^2+b^2}+\sqrt{b^2+c^2}+\sqrt{c^2+a^2})^2$$
|
Let $a=x^3$, $b=y^3$ and $c=z^3$.
Hence, by Muirhead $\sum\limits_{cyc}\frac{a^2+b^2}{c^2+a+b}=\sum\limits_{cyc}\frac{x^6+y^6}{z^6+xyz(x^3+y^3)}\geq\sum\limits_{cyc}\frac{x^6+y^6}{z^6+z(x^5+y^5)}=\frac{\sum\limits_{cyc}(x^7y+x^7z)}{xyz(x^5+y^5+z^5)}\geq2$
|
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|
If $x^2=y+z$, $y^2=x+z$ and $z^2=x+y$, prove If $x^2=y+z$, $y^2=x+z$ and $z^2=x+y$,
Prove that
$$\frac{1}{x+1}+\frac{1}{y+1}+\frac{1}{z+1}=1$$.
My attempt:
$$L.H.S=\frac{1}{x+1}+\frac{1}{y+1}+\frac{1}{z+1}$$
$$=\frac{(y+1)(z+1)+(x+1)(z+1)+(x+1)(y+1)}{(x+1)(y+1)(z+1)}$$
$$=\frac{x^2+y^2+z^2+yz+xz+xy+3}{(x+1)(y+1)(z+1)}$$
What should I do next?
|
Note that
$$x^2 + x = y^2 + y = z^2 + z = x + y + z$$
$$x(x + 1) = y(y + 1) = z(z + 1) = x + y + z$$
so
$$\begin{align}\frac{1}{x + 1} + \frac{1}{y + 1} + \frac{1}{z + 1} &= \frac{x}{x + y + z} + \frac{y}{x + y + z} + \frac{z}{x + y + z}\\
&= \frac{x + y + z}{x + y + z}\\&= 1\end{align}$$
|
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|
Prove that $\frac{2\sin(a)+\sec(a)}{1+\tan(a)}$ = $\frac{1+\tan(a)}{\sec(a)}$
Prove that $\frac{2\sin(a)+\sec(a)}{1+\tan(a)}$ = $\frac{1+\tan(a)}{\sec(a)}$
My attempt using the LHS
$$\frac{2\sin(a)+\sec(a)}{1+\tan(a)}$$
$$ \frac{2\sin(a)+\frac{1}{\cos(a)}}{1+\frac{\sin(a)}{\cos(a)}} $$
$$ \frac{\frac{2\sin(a)+1}{\cos{a}}}{\frac{\cos(a)+\sin(a)}{\cos(a)}} $$
$$ {\frac{2\sin(a)+1}{\cos{a}}} * {\frac{\cos(a)}{\cos(a)+sin(a)}} $$
$$ \frac{2\sin(a)+1}{\cos(a)+sin(a)} $$
Now I am stuck...
|
Let $s=\sin(a)$, $c=\cos(a)$, $k=\sec(a) = 1/c$, $t=\tan(a)=s/c \;$ to eliminate visual clutter.
$$\frac{2\sin(a)+\sec(a)}{1+\tan(a)} = \frac{2s+k}{1+t} = \frac{2s+1/c}{1+s/c}=\frac{2sc+1}{c+s}=\frac{(c+s)^2}{c+s}= c+s \\= \frac{1+s/c}{1/c}=\frac{1+t}k = \frac{1+\tan(a)}{\sec(a)}$$
|
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|
What function does this Taylor Series represent? What is the function $f$ who's Taylor series is $1 - \frac{x}{4} + \frac{x^2}{7} - \frac{x^3}{10} + \cdots$ ?
I need to find the value of the series $ \sum^{\infty}_{n = 0}a_n = 1 - \frac{1}{4} + \frac{1}{7} - \frac{1}{10} + \cdots$ by finding $\lim_{x\rightarrow1^-} \sum^{\infty}_{n = 0}a_n x^n$. (Abel Summability)
I already did this for another problem. I was given the series $\sum^{\infty}_{n = 0}b_n = \frac{1}{2\cdot 1} - \frac{1}{3\cdot 2} + \frac{1}{4\cdot 3} - \frac{1}{5\cdot 4} + \cdots$
I showed that $\sum^{\infty}_{n = 0}b_n x^n = \frac{\ln(1+x)}{x^2}+\frac{\ln(1+x)}{x}-\frac{1}{x}\rightarrow 2\ln(2)-1$ as $x\rightarrow1^- \Rightarrow \sum^{\infty}_{n = 0}b_n = 2\ln(2)-1$. (This checks out numerically, as well)
However, I'm having trouble finding the functional representation of $ \sum^{\infty}_{n = 0}a_n x^n$
|
If
$f(x)
=\sum_{k=0}^{\infty} a_k x^k
$,
then
$f_{n, j}(x)
=\sum_{k=0}^{\infty} a_{j+kn} x^{j+kn}
=\frac1{n}\sum_{i=0}^{n-1} w^{ij} f(w^i x)
$
where
$w = e^{2\pi i /n}
$.
This is known as
multisection of series.
Here is one of many
available discussions:
http://mathworld.wolfram.com/SeriesMultisection.html
Since
$f(x)
=\frac{-\ln(1-x)}{x}
=\sum_{k=0}^{\infty} \frac{x^k}{k+1}
$,
$\sum_{k=0}^{\infty} \frac{x^{3k}}{3k+1}
=f_{0, 3}(x)
=\frac1{3}\sum_{i=0}^{2} f(w^i x)
$
where
$w = e^{2\pi i /3}
$.
Since you want
$g(x)
=\sum_{k=0}^{\infty} \frac{(-1)^kx^k}{3k+1}
$,
$g(x)
=f_{0, 3}((-x)^{1/3})
$.
|
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|
why $(\sum 8(n-1)) +1$ is equals to $(2n-1)^2$ (Forgive me if it is a silly question)
When I was solving a puzzle, I observed a sequence
1, 1+8, 1+8+16, 1+8+16+24, 1+8+16+24+32....
is equals to
1, 9, 25, 49, 81.....
for which I see it as:
$(8 \times 0) +1, (8 \times 0 + 8 \times 1) +1, (8 \times 0 + 8 \times 1 + 8 \times 2) +1, (8 \times 1 + 8 \times 2 + 8 \times 3 + 8 \times 4) +1 ....$
equals to
$1^2, 3^2, 5^2, 7^2 ...$
and my brain stuck here and cannot find out the relationship between two series.
Can someone give me a hint why $(\sum 8(n-1)) +1$ is equals to $(2n-1)^2$
|
Hint summation of$n$ odd integers which are positive is $n^2$ which can be expresed as $\sum (2n-1)=n^2$ where $n\in N$
|
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|
Show that $\binom{n}{0}\binom{n}{1}...\binom{n}{n}\leq ({\frac{2^n-2}{n-1}})^{n-1}$ If $n$ is a positive integer, show that
$\binom{n}{0}\binom{n}{1}...\binom{n}{n}\leq ({\frac{2^n-2}{n-1}})^{n-1}$
This is how I tried to solve it:
the given inequality is equivalent to
$\binom{n}{1}...\binom{n}{n-1}\leq ({\frac{2^n-2}{n-1}})^{n-1}$
$\iff \sqrt[n-1] {\binom{n}{1}...\binom{n}{n-1}}\leq \frac{2^n-2}{n-1}$
Now, we know that
$\sqrt[n-1] {\binom{n}{1}...\binom{n}{n-1}} \leq \frac{\binom{n}{1}+...+\binom{n}{n-1}}{n-1}$
Hence, we are left with proving $\binom{n}{1}+...+\binom{n}{n-1} \leq 2^n-2$
or, $\binom{n}{0}+...+\binom{n}{n} \leq 2^n$
Can we prove the last inequality? Am I on the right track? Thanks for help.
|
As noted by @LeGrandDODOM from AM-GM that $\sqrt[n-1] {\binom{n}{1}...\binom{n}{n-1}} \leq \frac{\binom{n}{1}+...+\binom{n}{n-1}}{n-1}$.
However, note that $\binom{n}{1}+...+\binom{n}{n-1}=2^n-2$ which follows from the binomial theorem.
$\sqrt[n-1] {\binom{n}{1}...\binom{n}{n-1}} \leq \frac{2^n-2}{n-1}$.
This gives us that ${\binom{n}{1}...\binom{n}{n-1}} \leq (\frac{2^n-2}{n-1})^{n-1}$.
From here, $\binom{n}{0}\binom{n}{1}...\binom{n}{n}\leq ({\frac{2^n-2}{n-1}})^{n-1}$.
Thus, you appear to mistaken.
|
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|
Integral Calculus: Plane Areas in Rectangular Coordinates Find the area bounded by the given curves:
$y^2+2x-2y-3=0$ and the $y$-axis
(using horizontal & vertical strip)
|
Finding the points:
$x = 0$ so
$$ y^2+2(0)-2y-3=0 $$
$$ y^2-2y-3=0 $$
$$ (y-3)(y+1)=0 $$
Thus points $P$ are $(0,3),(0,-1)$.
By completing the square
$y^2-2y+1=-2x+3+1$.
$$ (y-1)^2=-2(2x-2) $$
$ y=1, x=2 ,
V = (2,1) $
The parabola opens left.
|
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|
Is $x^3+x^2+x+1$ divisible by $x^3+5x^2+x$ in $\mathbb{Z}_5\left[x\right]$? (for simplicity, denote $a(x)=x^3+x^2+x+1$ and $b(x)=x^3+5x^2+x$)
My assignment asks to show that $b(x)$ does not divide $a(x)$ in $\mathbb{Z}\left[x\right]$, but rather that it does divide it in $\mathbb{Z}_5\left[x\right]$.
Long division shows that the first is obviously true, $\frac {a(x)} {b(x)}$ leaves a remainder of $r(x)=1-4x$.
But so far, I'm drawing a blank on the second. Whenever I try to use long division in this new polynomial field, I get a nonzero remainder of $r(x)=x^2+1$. (which is the remainder of the first division, with coefficients made modulo 5)
Since we're now in $\mathbb{Z}_5\left[x\right]$, $b(x)=x^3+x$ (since $5=0$) So this is true if we're using the division algorithm:
$$x^3+x^2+x+1=(1)(x^3+x)+(x^2+1)=x^3+x^2+x+1$$
Where am I having a memory leak?
|
I'm not sure where you had a leak, but:
$$x^3+x^2+x+1=(x+1)(x^2+1)$$
and this much is true in the integers $\;\Bbb Z\;$ , so it is also true in $\;\Bbb Z_5\;$ . It comes from noticing that $\;-1=4\pmod5\;$ is a root of the leftmost polynomial. Observe further that since $\;5=1\pmod4\;,\;\;-1\;$ is a square in this last field, so in fact
$$x^2+1=(x-2)(x+2)\pmod5= (x+3)(x+2)\pmod5$$
so in the end
$$x^3+x^2+x+1=(x+1)(x^2+1)=(x+1)(x+2)(x+3)\pmod5$$
|
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|
Finding the inverse of a matrix given an equation So I've been given this equation:
$A\begin{bmatrix}
2&3&1&5\\
1&0&3&1\\
0&2&-3&2\\
0&2&3&1
\end{bmatrix} = \begin{bmatrix}
0&1&0&0\\
0&0&0&1\\
0&0&1&0\\
1&0&0&0
\end{bmatrix}$
I'm supposed to find the inverse of A using this but I'm not really sure where to start. Any ideas?
|
Bernard's answer above gives a lot of intuition about what's really going on, but in case you are not yet familiar with some of the terminology here is a more brutish answer.
The inverse of the right hand matrix can be calculated by examination. Notice that there is only one non-zero entry per column and row of the matrix, and these are all $1$. This has the effect that every element in a matrix multiplied with this right hand matrix appears exactly once in the product as such
$~\\ \left( \begin{array}{cccc}
0 & 1 & 0 & 0 \\
0 & 0 & 0 & 1\\
0 & 0 & 1 & 0\\
1 & 0 & 0 & 0 \\
\end{array} \right)
\cdot
\left( \begin{array}{cccc}
a_{11} & a_{12} & a_{13} & a_{14}\\
a_{21} & a_{22} & a_{23} & a_{24}\\
a_{31} & a_{32} & a_{33} & a_{34}\\
a_{41} & a_{42} & a_{43} & a_{44}\\
\end{array} \right)
=
\left( \begin{array}{cccc}
a_{21} & a_{22} & a_{23} & a_{24}\\
a_{41} & a_{42} & a_{43} & a_{44}\\
a_{31} & a_{32} & a_{33} & a_{34}\\
a_{11} & a_{12} & a_{13} & a_{14}\\
\end{array} \right)\\$
So to get the identity matrix we choose $a_{21} = a_{42} = a_{33} = a_{14} = 1$ and all other entries to be zero. Then we have computed the inverse of the right hand side matrix as
$~\\
\left( \begin{array}{cccc}
0 & 0 & 0 & 1 \\
1 & 0 & 0 & 0\\
0 & 0 & 1 & 0\\
0 & 1 & 0 & 0 \\
\end{array} \right)\\$
Hence if we multiply both sides of your equation on the right by the above matrix, on the left side we get $A$ multiplied by the product of two matrices and on the right side we get the identity matrix. Hence we have found the inverse of $A$.
|
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|
integral for range in x axis Calculate the area in the shaded region:
${f(x) = x^3 -2x + 7}$
${{{\int_{-1}^2}} f(x) dx = x^4 - x^2 + 7x}$
$= {[(2^4) -(2^2) + 14] - [(-1)^4 - (-1)^2 + 7(-1)]}$
$= [16 - 4 + 14] - [- 7] = 19$
But the answer in the book is $21{3\over 4}$.
|
As @Sean said ,you forgot to divide $x^4$ by $4$
Also if you want to find the integral of ${f(x) = x^3 -2x + 7}$
So the result will be $\require{cancel} \cancel{\int_{-1}^{2}} \frac{x^4}{\color{blue}4}-x^2+7x+C$
$$\int_{-1}^{2}\left(x^3-2x+7\right)dx=\left(\frac{x^4}{\color{blue}4}-x^2+7x\right)\bigg|_{-1}^{2}=4-4+14-(1/4-1-7)=\color{red}{21\frac{3}{4}}$$
|
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|
Differentiation Calculus: $\tan^{-1} \text{Problem}$ Well, Today at Math Revision exam I have to answer for $\frac{dy}{dx}$
Question:$$ y= \arctan\frac{{2x}}{{1+x^2}}$$
I got the answer $$ \frac{2}{1+(\frac{2x}{1+x})^2}\frac{cos(2\tan^{-1}x)}{1+x^2}$$
i think this not correct.
I have to know the right solution with steps.
|
$$ y= \arctan\frac{{2x}}{{1+x^2}}$$
You need to use the chain rule here,
$$ y'=\frac{\frac{d}{dx}\left(\frac{2x}{1+x^2}\right)}{1+\frac{4x^2}{(1+x^2)^2}}=\frac{(x^2+1)\frac{d}{dx}(x)-x\frac{d}{dx}(1+x^2)}{(x^2+1)^2}\frac{2}{1+\frac{4x^2}{(1+x^2)^2}}=\frac{2(1-x^2)}{(1+x^2)^2\left(1+\frac{4x^2}{(1+x^2)^2}\right)}=\color{red}{\frac{2-2x^2}{x^4+6x^2+1}}$$
|
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|
Show that if a and b have the same sign then |a| + |b| = |a + b| We start with a, b > 0.
We know that $|a| = \sqrt{a^2}$, $|b| = \sqrt{b^2}$ and $|a+b| = \sqrt{(a+b)^2}$
We do the following :
$|a| + |b| = \sqrt{a^2} + \sqrt{b^2}$
$=a + b$
So : $|a| + |b| = a+b$
We take this result :
$|a| + |b| = a+b$
$(|a| + |b|)^2 = (a+b)^2$
$\sqrt{(|a| + |b|)^2} = \sqrt{(a+b)^2}$
$a+b =|a+b|$
Thus : |a+b|= |a| + |b|
For the case where a, b < 0, we use the same technique but with minuses before the constants Opinion ?
|
You can shorten your approach and deal with the negative case at the same time, by writing
$$
(|a|+|b|)^2=a^2+2|a||b| + b^2\stackrel{(1)}=a^2+2ab+b^2=(a+b)^2\tag{*}
$$
where (1) uses the fact that $|a||b|=ab$ if $a$ and $b$ have the same sign. Now take square roots of both ends of (*).
|
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|
Equations of the line which intersects the lines $\frac{x-1}{2}=\frac{y-2}{3}=\frac{z-3}{4}$ and $\frac{x+2}{1}=\frac{y-3}{2}=\frac{z+1}{4}$ Find the equations of the line which intersects the lines $\frac{x-1}{2}=\frac{y-2}{3}=\frac{z-3}{4}$ and $\frac{x+2}{1}=\frac{y-3}{2}=\frac{z+1}{4}$ and passes through the point $(1,1,1)$.
First I thought required line passes through point on intersection of given lines, but as I checked, lines are non-intersecting. So doesn't that mean that there will be infinite lines which satisfy the given conditions?
|
Can't comment:
To get to the result that J. Wu and Dr. Sonnard Graubner gets, you can assume that everything is equal to $t$ or seperately
$$t=\frac{x-1}{2}; t=\frac{y-2}{3};t=\frac{z-3}{4}$$
and solve for $x,y,$ and $z$ in each seperate equations
|
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|
How to find solutions for $a$ and $b$ where $9 \equiv 4a+b \pmod {26} $ and $10 \equiv 19a+b \pmod {26}$? $$9 \equiv 4a+b \pmod {26}$$
$$10 \equiv 19a+b \pmod {26}$$
How can I solve the following system?
|
Notice that substracting the two equations gives us that $15a \equiv 1 \pmod {26}$.
This gives us that $a \equiv 7 \pmod {26}$.
Since $4 \times 7+b \equiv 9 \equiv 35 \equiv 5 \times 7 \pmod {26}$, this implies that $b \equiv 7 \pmod {26}$.
The answer is $a \equiv b \equiv 7 \pmod {26}$.
|
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|
Using the $\epsilon$-$\delta$ definition of the limit, evaluate $ \lim_{x \to a} \frac{x+2}{x^2-5}$
Using the $\epsilon$-$\delta$ definition of the limit, evaluate $$ \lim_{x \to a} f(x)$$ where $$f(x) = \frac{x+2}{x^2-5}$$
Attempt
We need to show that $\forall \epsilon, \exists \delta$ such that $$0 < |x-a| < \delta \quad \implies \quad \left |\dfrac{x+2}{x^2-5}-\dfrac{a+2}{a^2-5} \right| < \epsilon.$$ I don't see an easy way to do this algebraically so I am going to say let $|x-a| < 1$ and then $a - 1 < x < a+1$. We then have $\left | \dfrac{a+1}{(a-1)^2}-\dfrac{a+2}{a^2-5}\right | < \epsilon$. Then I am not sure how to simplify it to prove the statement.
|
To solve these types of problems, you usually start with what you want to show and work backwards to try to figure out a $\delta$ that will work.
So let $\epsilon > 0$ be arbitrary. At the end of the day, we want $\left |\dfrac{x+2}{x^2-5}-\dfrac{a+2}{a^2-5} \right| < \epsilon$. Let's try to manipulate the left hand side in ways we are allowed to do and hope we get something useable (you have to guess and try things and see if they lead you anywhere).
Ok, well one thing we can do is try to add the fractions up, and factor or cancel or do any other algebra that seems obvious to try. So we have
$$\begin{split} \left |\dfrac{x+2}{x^2-5}-\dfrac{a+2}{a^2-5} \right|
&= \left | \dfrac{(x + 2)(a^{2} - 5) - (a + 2)(x^{2} - 5)}{(a^{2} - 5)(x^{2} - 5)} \right | \\
&=\left | \dfrac{xa^{2} - 5x + 2a^{2} - 10 - ax^{2} + 5a - 2x^{2} + 10}{(a^{2} - 5)(x^{2} - 5)} \right | \\
&=\left | \dfrac{xa^{2} -ax^{2} - 5x + 5a + 2a^{2} - 2x^{2} }{(a^{2} - 5)(x^{2} - 5)} \right | \\
&= |x-a| \left | \dfrac{xa - 5 - 2(x + a) }{(a^{2} - 5)(x^{2} - 5)} \right |. \end{split}$$
Ok, well after trying some algebra, we ended up with a factor of $|x - a|$ which is good! Usually, if you end up with something like $M|x-a|$ for some number $M$, then you can let $\delta = \epsilon/M$, because then we would get $M |x -a| \leq M (\epsilon/M) = \epsilon$ (if $|x - a| < \delta$).
The only problem is the other factor multiplied by our $|x - a|$ isn't just a number -- it depends on $x$. We don't want any $x$ presence in the other factor. So let's do a standard trick which you already seem to know. Let's assume without loss of generality that $\delta < $ some number (do you understand why we can do this?). Once you understand why we can do this, let's talk about how we do this: if there are no denominators to worry about, we can usually pick $\delta < 1$. But in this case, we have $x^{2} - 5$ in the denominator. We don't want this to be $0$ since otherwise the expression would be undefined! So we want to pick a number $\delta$ so that $x$ can't possibly be $\pm \sqrt{5}$. Since $a$ is either in $(-\sqrt{5}, \sqrt{5})$ or otherwise is in $(-\infty, -\sqrt{5}) \cup (\sqrt{5},\infty)$ (because $a$ cannot equal $\pm \sqrt{5}$), there is some number $\rho > 0$ so that $(a - \rho, a + \rho)$ doesn't contain $\pm \sqrt{5}$. Pick $\delta$ small enough so that if $|x - a| < \delta$, then $x \in (a - \rho, a + \rho)$. So basically, pick $\delta \leq \rho$, where $\rho$ is the fixed number described above. This will ensure $x$ cannot be $\pm \sqrt{5}$.
So then if $\delta \leq \rho$, we have $|x - a| < \delta$ implies $|x - a| < \rho$ so that $-\rho + a < x < \rho + a$. Let $M = \max\{|-\rho + a|, |\rho + a| \}$. So $-M < x < M$, which means $|x| < M$.
Now, try to convince yourself of why the following is true (based on $|x|<M$):
$\left | \dfrac{xa - 5 - 2(x + a) }{(a^{2} - 5)(x^{2} - 5)} \right | \leq \dfrac{|xa| + 5 + 2|x| + 2|a| }{|(a^{2} - 5)(x^{2} - 5)|} \leq \dfrac{M|a| + 5 + 2M + 2|a| }{|(a^{2} - 5)(0^{2} - 5)|} = \dfrac{(M + 2)|a| + 5 + 2M}{5|(a^{2} - 5)|}$
Let's call the last expression on the right hand side the letter $K$. It's just a constant number that doesn't depend on $x$.
Then, based on all of the work above, if we let $\delta = \min\{\rho, \epsilon/K\}$ (we are taking the minimum with $\rho$ because we assumed $\delta \leq \rho$ above), then we get $|x -a| < \delta \implies$ $$\left |\dfrac{x+2}{x^2-5}-\dfrac{a+2}{a^2-5} \right| = |x-a|\left | \dfrac{xa - 5 - 2(x + a) }{(a^{2} - 5)(x^{2} - 5)} \right | < |x-a| K < (\epsilon/K)K = \epsilon$$
|
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|
How is $\frac{1}{1^4}+\frac{1}{2^4}+\frac{1}{3^4}\cdots=\frac{\pi^4}{90}$
If $\frac{1}{1^4}+\frac{1}{2^4}+\frac{1}{3^4}\cdots=\frac{\pi^4}{90}$, then find the value of $\frac{1}{1^4}+\frac{1}{3^4}+\frac{1}{5^4}\cdots$
Firstly how is $\frac{1}{1^4}+\frac{1}{2^4}+\frac{1}{3^4}\cdots=\frac{\pi^4}{90}$?
Secondly, I thought $$\frac{1}{1^4}+\frac{1}{3^4}+\frac{1}{5^4}\cdots=\frac{1}{2^4}+\frac{1}{4^4}+\frac{1}{6^4}\cdots=\frac{S}{2}$$
But answer given is $\frac{\pi^4}{96}$. Whats the mistake in this?
Edit:
I found a way to get the answer.
$$\frac{1}{1^4}+\frac{1}{2^4}+\frac{1}{3^4}\cdots=\frac{1}{1^4}+\frac{1}{3^4}+\frac{1}{5^4}\cdots+\frac{1}{2^4}\left(\frac{1}{1^4}+\frac{1}{2^4}+\frac{1}{3^4}\cdots\right)$$
$$S=S_1+\frac{1}{2^4}S$$
|
If you already know or take as given the first result, then
$$\frac{\pi^4}{90}=\sum_{n=1}^\infty\frac1{n^4}=\sum_{n=1}^\infty\frac1{(2n)^4}+\sum_{n=1}^\infty\frac1{(2n-1)^4}=\frac1{16}\sum_{n=1}^\infty\frac1{n^4}+\sum_{n=1}^\infty\frac1{(2n-1)^4}\implies$$
$$\sum_{n=1}^\infty\frac1{(2n-1)^4}=\left(1-\frac1{16}\right)\frac{\pi^4}{90}=\frac{\pi^4}{96}$$
The first result though is way over the high school level, at least for me.
|
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|
Determining whether an orthogonal matrix represents a rotation or reflection The exercise asks us to determine whether the given orthogonal matrix represents a rotation or a reflection. If it is a rotation, give the angle of rotation; if it is a reflection, give the line of reflection.
$$
A = \begin{bmatrix}
-\frac{3}{5} & -\frac{4}{5}\\[0.3em]
-\frac{4}{5} & \frac{3}{5}\\[0.3em]
\end{bmatrix}
$$
I know you can check whether it is a reflection or rotation by calculating the determinant. So for example for the matrix above
$$det(A) = -\frac{3}{5}\cdot\frac{3}{5}-(-\frac{4}{5})\cdot(-\frac{4}{5}) = -1$$
And so that means that matrix $A$ corresponds to a reflection in $R^2$, but how do you get the line of reflection from this?
$$
B = \begin{bmatrix}
-\frac{1}{2} & \frac{\sqrt{3}}{2}\\[0.3em]
-\frac{\sqrt{3}}{2} & -\frac{1}{2}\\[0.3em]
\end{bmatrix}
$$
And for a rotation, so for example matrix $B$ given above, can you simply say that it corresponds to the rotation matrix $R$
$$
R = \begin{bmatrix}
cos(\theta) & -sin(\theta)\\[0.3em]
sin(\theta) & cos(\theta)\\[0.3em]
\end{bmatrix}
$$
So this gives
$$cos(\theta) = -\frac{1}{2}, \theta = cos^{-1}(-\frac{1}{2}) = 120^{\circ}\\
sin(\theta) = -\frac{\sqrt{3}}{2}, \theta = sin^{-1}(-\frac{\sqrt{3}}{2}) = -60^{\circ}
$$
Which means that matrix $B$ corresponds to a counterclockwise rotation of $120^{\circ}$, right?
|
for the line of reflection look for a mirror lines that stays fixed under the transformation. for example, in the case of $A = \begin{bmatrix}
-\frac{3}{5} & -\frac{4}{5}\\[0.3em]
-\frac{4}{5} & \frac{3}{5}\\[0.3em]
\end{bmatrix}$ try the line $x = \pmatrix{5\\5a}.$ we want $Ax = x$ that is
$$\pmatrix{-3 -4a\\-4 + 3a } = \pmatrix{5\\5a}$$ which gives you $a= -2.$ therefore the line $y = -2x$ is the mirror.
|
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"url": "https://math.stackexchange.com/questions/1685906",
"timestamp": "2023-03-29T00:00:00",
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|
How to evaluate $\int_{0}^{\infty }\frac{\ln( 1+x^{4} )}{1+x^{2}}{d}x~,~\int_{0}^{1}\frac{\arctan x}{x}\frac{1-x^{4}}{1+x^{4}}{d}x$ How to evaluate these two integrals below
$$\int_{0}^{\infty }\frac{\ln\left ( 1+x^{4} \right )}{1+x^{2}}\mathrm{d}x$$
$$\int_{0}^{1}\frac{\arctan x}{x}\frac{1-x^{4}}{1+x^{4}}\mathrm{d}x$$
For the first one, I tried to use
$$\mathcal{I}'(s)=\int_{0}^{\infty }\frac{x^4}{(1+sx^4)(1+x^{2})}\mathrm{d}x$$
but it seems hard to solve.
|
Using the identity $\ln \left(x^2+y^2\right)=2 \operatorname{Re}(\ln (x+y i))$ to reduce the power $4$ of $x$ in the numerator makes the life easier.
$$
\begin{aligned}
\int_0^{\infty} \frac{\ln \left(1+x^4\right)}{1+x^2} d x
= & \frac{1}{2} \int_{-\infty}^{\infty} \frac{\ln \left(1+x^4\right)}{1+x^2} d x \\
= & \operatorname{Re}\left(\int_{-\infty}^{\infty} \frac{\ln \left(x^2+i\right)}{1+x^2}\right)\\=&\operatorname{Re} \int_{-\infty}^{\infty} \frac{\ln \left(x^2+\left(\frac{1+i}{\sqrt{2}}\right)^2\right)}{1+x^2}
\end{aligned}
$$
Using the particular case $\int_{-\infty}^{\infty} \frac{\ln \left(x^2+a^2\right)}{1+x^2} d x=2 \pi \ln (1+a)$ in the post, we have
$$
\begin{aligned}
\int_0^{\infty} \frac{\ln \left(1+x^4\right)}{1+x^2} d x
& =\operatorname{Re}\left[2 \pi \ln \left(1+\frac{1+i}{\sqrt{2}}\right)\right] \\
& =2 \pi \operatorname{Re}\left[\ln \left(1+\frac{1}{\sqrt{2}}+\frac{i}{\sqrt{2}}\right)\right]\\
& =\pi \ln \left(1+\frac{1}{2}+\sqrt{2}+\frac{1}{2}\right) \\
& =\pi \ln (2+\sqrt{2})
\end{aligned}
$$
|
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"timestamp": "2023-03-29T00:00:00",
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|
Prove: $\sin x \ln \left(\frac{1 + \sin x}{1 - \sin x}\right) \geq 2x^2$
Prove: $$\sin x \ln \left(\frac{1 + \sin x}{1 - \sin x}\right) \geq 2x^2$$
for $\: -\frac{\pi}{2} < x < \frac{\pi}{2}$
I have thought of three things so far that can be useful.
The Cauchy–Schwarz inequality
$$
\int_Efg\,\mathrm{d}x\le\left(\int_Ef^2\,\mathrm{d}x\right)^{1/2}\left(\int_Eg^2\,\mathrm{d}x\right)^{1/2}
$$
Both sides of the inequality are even functions so its enough to consider
$x \in [0, \frac{\pi}{2})$.
Maybe I can use that $\int_0^x \cos t \:dt = \sin x$.
I would appreciate a hint or a few on how to continue.
edit: The inequality is supposed to be proved using integration methods.
|
Without loss of generality, let $0\le x < \frac{\pi}{2}$. Define
$$ f(x)=\sin x \ln \left(\frac{1 + \sin x}{1 - \sin x}\right). $$
Then $f(0)=0,f'(0)=0$ and
\begin{eqnarray}
f''(x)=\frac{1}{8\cos^2x}(4 \cos (2 x)-\sin x \ln (\frac{1+\sin
x}{1-\sin x})-\sin (3 x) \ln(\frac{1+\sin x}{1-\sin
x})+12).
\end{eqnarray}
We want to show that $f''(x)\ge 4$. Note
$$ f''(x)-4=-\frac{1}{2} \tan x \sec x \left(-4 \sin (x)+\ln \left(\frac{1+\sin
x}{1-\sin x}\right)+\cos (2 x) \ln \left(\frac{1+\sin x}{1-\sin
x}\right)\right). $$
Define
$$g(x)= 4 \sin x-\ln \left(\frac{1+\sin x}{1-\sin x}\right)-\cos (2 x) \ln
\left(\frac{1+\sin x}{1-\sin x}\right). $$
Now we show that $g(x)\ge0$ for $0\le x<\pi/2$. In fact, easy calculation shows
$$ g'(x)=2\sin(2x)\ln \left(\frac{1+\sin x}{1-\sin x}\right)\ge0$$
for $0\le x<\pi/2$ and hence $g(x)$ is increasing. So $g(x)\ge 0$ and hence
$$ f(x)=f(0)+f'(0)+\frac{1}{2}f''(c)x^2=\frac{1}{2}f''(c)x^2\ge 2x^2 $$ for
some $c\in(0,\pi/2)$.
|
{
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"timestamp": "2023-03-29T00:00:00",
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|
Coordinates of the center of curvature of $y^2 = 4px$ I'm trying to find the coordinates of the center of curvature at any point of
$$
y^2 = 4px
$$
I start with the first and second derivatives
$$
y' = \sqrt{\frac{p}{x}} \\
y'' = - \frac{1}{2} \sqrt{\frac{p}{x^3}} \\
$$
I then use the curvature formula
$$
K = \frac{|y''|}{(1 + y'^2)^{3/2}} = \frac{\sqrt{p}}{2(x+p)^{3/2}}
$$
I then get the radius of curvature from the reciprocal of the curvature
$$
\rho = \frac{2(x+p)^{3/2}}{\sqrt{p}}
$$
Now, I think I need to find the direction of the radius to get any point. I know that the tangent of the curve is given by the first derivative
$$
\tan \phi = \sqrt{\frac{p}{x}}
$$
I also know that the slope of the radius is perpendicular to the tangent
$$
m = -\sqrt{\frac{x}{p}}
$$
That's where I'm stuck finding my way towards the answer $(3x+2p,\frac{-y^3}{4p^2})$.
|
Given $m = -\sqrt{\frac{x}{p}}$, let $\theta$ be the angle of your radius of curvature
$$
\cos \theta = \frac{\sqrt{p}}{\sqrt{x+p}} \\
\sin \theta = \frac{-\sqrt{x}}{\sqrt{x+p}} \\
$$
So the coordinates of the center of curvature are
$$
x = \rho \cos \theta + x = \frac{2(x+p)^{3/2}}{\sqrt{p}} \frac{\sqrt{p}}{\sqrt{x+p}} + x = 2(x+p) + x = 3x + 2p \\
y = \rho \sin \theta + y = \frac{2(x+p)^{3/2}}{\sqrt{p}} \frac{-\sqrt{x}}{\sqrt{x+p}} + y = \frac{-y(x+p)}{p} + y = \frac{-yx}{p} = \frac{-y^3}{4p^2} \\
$$
Note that the first and second derivatives are simpler with respect to $y$
$$
x' = \frac{y}{2p} \\
x'' = \frac{1}{2p} \\
$$
The curvature is calculated the same way but, be careful, the slope of the tangent is $\frac{1}{x'}$.
|
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"timestamp": "2023-03-29T00:00:00",
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|
Prove that $C^nx = \frac{1}{2^n}au + \frac{1}{3^n}bv$, for every $n \in N$ EDIT: I forgot to mention $C = 0.5uu^T + 0.33vv^T$ and now if I use it, I solve it easily.
Given: $Cx = \frac{1}{2}au + \frac{1}{3}bv$, $x \in R^2$, $u,v$ are orthonormal vectors in $R^2$, $x = au + bv$ and $a,b \in R$, $C$ is a matrix $2x2$. Note: $a,b$ are scalars.
Prove that $$C^nx = \frac{1}{2^n}au + \frac{1}{3^n}bv$$, for every $n \in N$
Well, it's trivial that we need to show that with induction, so for $n = 1$ this works because it is given (actually I proved it and it is really true).
Now I assume that it works for $k = n$, and I want to prove for $k = n + 1 $. I use $k$ because it is more comfortable for me that way, and I got stuck with $a^2$ and $u,v$ are gone in the last equation, instead of $a$ and multiplied with $u,v$ as required.
$C^{k+1}x=\frac{1}{2^{k+1}}au + \frac{1}{3^{k+1}}bv$ I need to prove.
Going from the left side of the equation and I try to prove the right side:
$C^{k+1}x = C^k(Cx) = C^k(\frac{1}{2}au + \frac{1}{3}bv) = \frac{1}{2^{k}}au + \frac{1}{3^{k}}bv(\frac{1}{2}au + \frac{1}{3}bv) =
\frac{1}{2^{k+1}}a^2 + \frac{1}{3^{k+1}}b^2$.
Edit: if I take $au$ out, then I get was it required. can I do that?
I mean: $\frac{1}{2^{k}}au + \frac{1}{3^{k}}bv(\frac{1}{2}au + \frac{1}{3}bv) =$ long equation $= au\frac{1}{2^{k+1}} + bu\frac{1}{3^{k+1}}$
|
Think of $C$ as a linear transformation on $\mathbb R^2$.
Then, the matrix of $C$ with respect to the basis $(u,v)$ is the diagonal matrix $\pmatrix{ 1/2 & 0 \\ 0 &1/3}$.
Therefore, the matrix of $C^n$ with respect to the basis $(u,v)$ is the diagonal matrix $\pmatrix{ 1/2^n & 0 \\ 0 &1/3^n}$.
Concretely, if $Cx=\frac{1}{2}au + \frac{1}{3}bv$ whenever $x = au + bv$, then by taking $a=1, b=0$ we get $Cu=\frac{1}{2}u$, and by taking $a=0, b=1$ we get $Cv=\frac{1}{3}v$, which solves the problem in your induction.
|
{
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"timestamp": "2023-03-29T00:00:00",
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|
Quaternion exponential problem I have problem with Euler´s form of quaternion. My quaternion $q=\frac{1}{\sqrt{2}}i+\frac{1}{\sqrt{2}}j,$ so $q^2=-1$, because $$q^2=(\frac{1}{\sqrt{2}}i+\frac{1}{\sqrt{2}}j)(\frac{1}{\sqrt{2}}i+\frac{1}{\sqrt{2}}j)=-\frac{1}{2}+\frac{1}{2}k-\frac{1}{2}k-\frac{1}{2}=-1.$$ Thus I can write $q$ in Euler´s form as $q=e^{q\frac{\pi}{2}}=\cos\frac{\pi}{2}+q \sin\frac{\pi}{2}=q.$ However, I can also write $$q=e^{q\frac{\pi}{2}}=e^{(\frac{1}{\sqrt{2}}i+\frac{1}{\sqrt{2}}j)\frac{\pi}{2}}=e^{\frac{\pi}{2\sqrt{2}}i+\frac{\pi}{2\sqrt{2}}j}=e^{\frac{\pi}{2\sqrt{2}}i}e^{\frac{\pi}{2\sqrt{2}}j}=\\=(\cos\frac{\pi}{2\sqrt{2}}+i \sin\frac{\pi}{2\sqrt{2}})(\cos\frac{\pi}{2\sqrt{2}}+j \sin\frac{\pi}{2\sqrt{2}})=\\=\cos^2\frac{\pi}{2\sqrt{2}}+(i+j) \sin\frac{\pi}{2\sqrt{2}}\cos\frac{\pi}{2\sqrt{2}}+k\sin^2\frac{\pi}{2\sqrt{2}},$$ but this is not $q$. Please can you somebody say me where is the mistake?
|
For a quaternion $z=a+b\mathbf{i}+c\mathbf{j}+d\mathbf{k} = a+\mathbf{v}$ the exponential is defined as:
$$
e^z = e^{a+\mathbf{v}}=e^a \left( \cos |\mathbf{v}| +\dfrac{\mathbf{v}}{|\mathbf{v}|} \,\sin |\mathbf{v}| \right)
$$
(see :Exponential Function of Quaternion - Derivation)
and in general, since quaternions are not commutative, we have:
$$
e^xe^y \ne e^ye^x \ne e^{x+y}
$$
In your case: $q=\frac{1}{\sqrt{2}}i+\frac{1}{\sqrt{2}}j,$ is a pure imaginary quaternion $\mathbf {v} =q$
and $|q|=1$, so we have:
$$
e^q= e^{\mathbf{v}}=\left( \cos (1) +\mathbf{v}\sin (1) \right)=\cos (1)+ \sin(1)\left(\frac{1}{\sqrt{2}}i+\frac{1}{\sqrt{2}}j \right)
$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1693413",
"timestamp": "2023-03-29T00:00:00",
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|
Point P such that perimeter is least Given two points $A(-2,0)$ and $B(0,4)$ then find coordinate of point $P$ lying on the line $2x-3y=9$ so that perimeter of triangle $APB$ is least.
Doing it by traditional calculus is making calculations very complicated. Is there smart geometrical way to solve this?
|
Well, it was easy to say what might help, but a bit less simple to find a way to avoid unpleasant radicals, which are essentially the same ones you would get using path-minimization in calculus. Here is a geometric argument that dodges them.
The big problem with a direct approach is that the lengths of $ \ AP \ $ and $ \ BP \ $ are going to be unequal, and a number of approaches still force us to determine them. Instead, extend a line parallel to $ \ 2x \ - \ 3y \ = \ 9 \ $ that passes through $\ A \ (-2, \ 0 ) \ $ : this is $ \ y \ = \ \frac{2}{3} x \ + \ \frac{4}{3} \ $ . This line intersects $ \ BP \ $ at a point we'll call $ \ Q \ $ . If we designate $ \ P \ ( X, \ Y) \ = \ (X, \ \frac{2}{3}X \ - \ 3) \ $ , then the line containing $ \ BP \ $ is $ \ y \ - \ 4 \ = \ M x \ $ , which has slope
$$ M \ = \ \frac{4 \ - \ [ \ \frac{2}{3}X \ - \ 3 \ ]}{0 \ - \ X \ } \ \ = \ \ \frac{2}{3} \ - \ \frac{7}{X} \ \ . $$
The coordinates of the intersection point $ \ Q \ $ are then found from equating
$$ \frac{2}{3} x_Q \ + \ \frac{4}{3} \ = \ \left( \ \frac{2}{3} \ - \ \frac{7}{X} \ \right) x_Q \ + \ 4 \ \ \Rightarrow \ \ x_Q \ = \ \frac{8X}{21} \ \ , \ \ y_Q \ = \ \frac{16X \ + \ 84}{63} \ \ . $$
(I am sparing you a certain amount of algebra in this.)
Here is where we apply the "optical law" that we minimize the length of $ \ AP \ $ plus $ \ PB \ $ by equating the angles these segments make to the normal line to $ \ 2x \ - \ 3y \ = \ 9 \ $ . The normal line has slope $ \ -\frac{3}{2} \ $ and is the perpendicular bisector of $ \ AQ \ $ when $ \Delta APQ \ $ is isoceles ; then the segment $ \ PM \ $ bisects $ \angle APQ \ $ , and so also $ \ \angle APB \ $ . The midpoint $ \ M \ $ of segment $ \ AQ \ $ is
$$ \ x_M \ = \ \frac{\frac{8X}{21} \ + \ (-2)}{2} \ = \ \frac{4X}{21} \ - \ 1 \ \ , \ \ y_M \ = \ \frac{8X \ + \ 42}{63} $$
and lies on the normal line through $ \ P \ $ , which is
$$ y \ - \ ( \ \frac{2}{3}X \ - \ 3 \ ) \ = \ -\frac{3}{2} \ (x \ - \ X ) \ \ . $$
Omitting some further algebra, after inserting
$$ y_
M \ - \ ( \ \frac{2}{3}X \ - \ 3 \ ) \ = \ -\frac{3}{2} \ (x_
M \ - \ X ) \ \ , $$
we solve for $ \ X \ $ to obtain
$$ X \ = \ \frac{21}{17} \ \ , \ \ Y \ = \ -\frac{111}{51} \ \ . $$
The graph below shows the geometrical situation.
There is no symmetry in the arrangement, so there's no good reason to expect the coordinates to come out "nicely", or that there is any "easy" method.
EDIT (added 12 March) --
Apart from "verifying" this graphically by plotting the function for the total length $ \ AP \ + \ BP \ $ , we can confirm the exact value for $ \ X \ $ by differentiating implicitly the total length
$$ s \ = \ s_1 \ + \ s_2 \ = \ \sqrt{(X \ + \ 2)^2 + (\frac{2}{3}Y \ - \ 3)^2} \ + \ \sqrt{X^2 \ + \ ( \ [ \ \frac{2}{3}X \ - \ 3 \ ] \ - \ 4 \ )^2} $$
to produce
$$ \frac{ds}{dX} \ = \ 0 \ = \ \frac{ds_1}{dX} \ + \ \frac{ds_2}{dX} \ \ \Rightarrow \ \ \frac{ds_1}{dX} \ = \ -\frac{ds_2}{dX} \ \ ;$$
squaring the individual terms $ \ s_1 \ $ and $ \ s_2 \ $ to remove the square-roots and then differentiating, we have
$$ 2 \ s_1 \ \frac{ds_1}{dX} \ = \ 2 \ (X \ + \ 2) \ + \ 2 \ ( \frac{2}{3}X \ - \ 3) \ \cdot \frac{2}{3} \ \ , $$
$$ 2 \ s_2 \ \frac{ds_2}{dX} \ = \ 2 \ X \ + \ 2 \ ( \frac{2}{3}X \ - \ 7) \ \cdot \frac{2}{3} $$
$$ \Rightarrow \ \ \frac{(X \ + \ 2) \ + \ \frac{4}{9}X \ - \ 2 }{X \ + \ ( \frac{4}{9}X \ - \ \frac{14}{3}) } \ = \ - \frac{s_1}{s_2} $$
$$ \Rightarrow \ \ \left( \ \frac{\frac{13}{9}X }{ \frac{13}{9}X \ - \ \frac{14}{3} } \ \right)^2 \ = \ \frac{(X \ + \ 2)^2 + (\frac{2}{3}Y \ - \ 3)^2}{X^2 \ + \ ( \ \frac{2}{3}X \ - \ 7 \ )^2} \ \ . $$
With a bit of help from a computer-algebra system (WA), we obtain two real solutions, $ \ X \ = \ \frac{21}{17} \ $ and the "spurious" solution $ \ -\frac{21}{4} \ $ . The differentiation itself isn't particularly hard; it's the algebra that requires a lot of patience...
|
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|
Find the minimum value of $\sqrt{a+1}+\sqrt{2b+1}+\sqrt{3c+1}$ if $a+b+c=1$
Let $a,b,c\ge 0$, and $a+b+c=1$. Find the minimum value of
$$\sqrt{a+1}+\sqrt{2b+1}+\sqrt{3c+1}$$
I think the minimum value is $\sqrt{2} + 2$? when $a=1,b=c=0$.
Of course, I can't prove it. Can anyone help?
|
Let
$$
U=\{(x,y,z)\in\mathbb{R}^3:x+y+z=1\text{ and }x,y,z\ge 0\}\\
S=\{(x,y,z)\in\mathbb{R}^3:x+y+z=1\text{ and }x,y,z>0\}.
$$
Define $f,g:U\to \mathbb{R}$ such that
\begin{align}
f(a,b,c)&=\sqrt{a+1}+\sqrt{2b+1}+\sqrt{3c+1}\\
g(a,b,c)&=a+b+c-1,
\end{align}
then
\begin{align}
\nabla f&=\left(\frac{1}{2\sqrt{a+1}},\frac{1}{\sqrt{2b+1}},\frac{3}{2\sqrt{3c+1}}\right)\\
\nabla g&=(1,1,1).
\end{align}
If a local minimum exists in $S$, there is $\lambda$ such that $\nabla f = \lambda \nabla g$, Thus
\begin{align}
a&=\frac{1}{4\lambda^2}-1\\
b&=\frac{1}{\lambda^2}-\frac{1}{2}\\
c&=\frac{9}{4\lambda^2}-\frac{1}{3}\\
\end{align}
Substitute $a,b,c$ in $g(a,b,c)=0$, then we get
$$
\frac{1}{\lambda^2}=\frac{17}{9}.
$$
But in this case, we find $a <0$, so there is no critical point of $f$ restricted to $g(x,y,z)=0$ in $S$.
Now we will set $c$ as $0$, then
$$
f(a)=\sqrt{a+1}+\sqrt{3-2a}
$$
and so
$$
f'(a)=\frac{1}{2\sqrt{a+1}}-\frac{1}{\sqrt{3-2a}}.
$$
Since $a\ge 0$,
$$
\frac{1}{2\sqrt{a+1}}-\frac{1}{\sqrt{3-2a}} > 0
$$
and so $f$ increases. Therefore $f(a,b,c)$ has its minimum $2+\sqrt{2}$ at $(1,0,0)$.
Exercise?: $f$ restricted to $g(x,y,z)=0$ has its maximum $1+\sqrt{\frac{55}{6}}$.
|
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|
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