Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
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A caboodle of Pell's equation in one? $x^2+y^2-5xy+5=0$ I saw this twitter post that reads:
Find all the pairs of positive integers $(x,y)$ satisfying $$ x^2 + y^2 - 5xy + 5 = 0 . $$
I don't know how to tackle this and I ended up summoning WolframAlpha which shows that there are infinitely many solutions. What's interesting is that there are (at least?) 12 general forms that looks like they all resemble the solutions to Pell's equation.
For example, the first general solution presented reads:
$$ \begin{array} { r c l }
x &=&
\dfrac1{42} \Big [ 21 \left(55 - 12\sqrt{21} \right)^n
- \sqrt{21} \left(55 - 12\sqrt{21} \right)^n
+ 21 \left(55 + 12\sqrt{21} \right)^n
+ \sqrt{21} \left(55 + 12\sqrt{21} \right)^n \Big ] \\ \phantom0 \\
y &=&
\dfrac1{42} \Big [ 63 \left(55 - 12\sqrt{21} \right)^n
- 13 \sqrt{21} \left(55 - 12\sqrt{21} \right)^n
+ 63 \left(55 + 12\sqrt{21} \right)^n
+ 13 \sqrt{21} \left(55 + 12\sqrt{21} \right)^n \Big] \end{array} $$ for $n = 0,1,2,3\ldots $.
Plugging $n=0,1,2,3$ gives the first few solutions
$$(x,y) = (1,3), (67,321), (7369,35307), (810523,3883449)
$$
I tried converting these solutions to the form of $X^2 - DY^2 = A$ or something similar to Pell's equation but I got nothing.
Is it a coincidence that this innocuous-looking quadratic Diophantine equation is actually a caboodle of Pell's equation in disguise? If so, how can we derive them all?
| It's $(5y-2x)^2-21y^2=-20$, which is a Pell type equation.
I got that by completing the square:
$x^2-5xy+y^2=-5\implies \left(x-\frac52y\right)^2-\frac{21}4y^2=-5\implies (5y-2x)^2-21y^2=-20$.
So it's $X^2-21Y^2=-20$, with $X=5y-2x$ and $Y=y$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3991684",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 0
} |
weird finding function from the graph
The graph is on the interval [-3,3], defined on R
This function maintain the pattern on all of R
How to find the $f(x)$?
See how the graph goes parallel at certain intervals. I am thinking adding one more variables to indicate one of the direction. However, I cannot merge the two together.
Any thoughts?
| Using the absolute value function together with the floor function, you can write
$$
f(x)=1-\left|x-\left(2\left\lfloor\frac{x+1}{2}\right\rfloor\right)\right|
$$
Verification:
With $f$ as defined above, we get
\begin{align*}
f(x+2)
&=
1-\left|(x+2)-\left(2\left\lfloor\frac{(x+2)+1}{2}\right\rfloor\right)\right|
\\[4pt]
&=
1-\left|x+2-\left(2\left\lfloor\frac{x+1}{2}+1\right\rfloor\right)\right|
\\[4pt]
&=
1-\left|x+2-\left(2\left(\left\lfloor\frac{x+1}{2}\right\rfloor+1\right)\right)\right|
\\[4pt]
&=
1-\left|x+2-\left(2\left(\left\lfloor\frac{x+1}{2}\right\rfloor+1\right)\right)\right|
\\[4pt]
&=
1-\left|x+2-\left(2\left(\left\lfloor\frac{x+1}{2}\right\rfloor+1\right)\right)\right|
\\[4pt]
&=
1-\left|x+2-\left(2\left\lfloor\frac{x+1}{2}\right\rfloor\right)-2\right|
\\[4pt]
&=
1-\left|x-\left(2\left\lfloor\frac{x+1}{2}\right\rfloor\right)\right|
\\[4pt]
&=
f(x)
\\[4pt]
\end{align*}
so $f$ is periodic with period $2$.
It remains to verify that $f$ works correctly on the interval $[-1,1)$.
Thus suppose $f$ is restricted to the interval $[-1,1)$.
\begin{align*}
\text{Then}\;\;&
-1\le x < 1
\\[4pt]
\implies\;&
0\le x+1 < 2
\\[4pt]
\implies\;&
0\le \frac{x+1}{2} < 1
\\[4pt]
\implies\;&
\left\lfloor\frac{x+1}{2}\right\rfloor=0
\\[4pt]
\end{align*}
hence for $-1\le x < 1$ we get
$$
f(x)
=
1-\left|x-\left(2\left\lfloor\frac{x+1}{2}\right\rfloor\right)\right|
=
1-|x|
$$
which matches the graph for the interval $[-1,1)$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3993657",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
Strong inductive proof for recursive sequence given Explicit formula I am stuck on this
Show that the recursively defined sequence $a_n=2·a_{n-1}+3·a_{n-2}, a_1=-1, a_2=13$ has the explicit formula $〖a_n=3^n+4(-1)^n〗$
I can prove the base cases no problem but I keep getting stuck on the transformation
so I have $a_{k+1}=2·a_{k} + 3·a_{k-1}$
=$2(3^k+4(-1)^k)+3( 3^{k+1})+4(-1)^{k-1})$
=$2*3^k+8*(-1)^k+3*3^{k-1}+12*(-1)^{k-1}$
=$$2*3^k+8*(-1)^k+3^k- 12(-1)^k$$
=$$3*3^k-4(-1)^k$$
How do I get that -4(-1)^k to turn into +$4(-1)^{k+1}$?
Did I goof the alegbra somewhere?
| You want to show:
$$3^{n}+4(-1)^{n} = 2[3^{n-1}+4(-1)^{n-1}] + 3[3^{n-2}+4(-1)^{n-2}]$$
Simplifying from the left we have:
$$2[3^{n-1}+4(-1)^{n-1}] + 3[3^{n-2}+4(-1)^{n-2}] = 2[3^{n-1}] + 8[(-1)^{n-1}] + [3^{n-1}] + 12[(-1)^{n-2}] =$$
$$=3[3^{n-1}] - 8[(-1)^{n-2}] + 12[(-1)^{n-2}] = 3^{n} + 4(-1)^{n-2} = 3^{n} + 4(-1)^{n-2}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4003311",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
If $x^5=2$, find $\frac{x}{x^2+1}+\frac{x^2}{x^4+1}+\frac{x^3}{x+1}+\frac{x^4}{x^3+1}$
If $x^5=2$, find
$$\frac{x}{x^2+1}+\frac{x^2}{x^4+1}+\frac{x^3}{x+1}+\frac{x^4}{x^3+1}$$
My attempt-
Since $x^5=2,x^6=2x,x^7=2x^2..$and so on
The equation is equivalent to
$$x^5\left(\frac{1}{2x+x^4}+\frac{1}{2x^2+x^3}+\frac{1}{x^3+x^2}+\frac{1}{x^4+x}\right)$$
Which simplifies to
$$2\left[\frac{3x^4+7x^3+5x^2+8x+12}{2x^4+6x^3+4x^2+6x+12}\right]$$
Now I am struck here.
Any help will be appreciated.
Note: Please don't use root tables and complex numbers too!
| $$
y=\frac{x}{x^2+1}+\frac{x^2}{x^4+1}+\frac{x^3}{x+1}+\frac{x^4}{x^3+1}$$
$$
=\frac{x (x^4 + x^3 + x^2 + 1) (x^6 - x^5 + 2 x^4 + x + 1)}{(x + 1) (x^2 + 1) (x^2 - x + 1) (x^4 + 1)}$$
$$
=\frac{8 + 5 2^{1/5} + 7 2^{2/5} + 3 2^{3/5} + 6 2^{4/5}}{3 + 2 2^{1/5} + 3 2^{2/5} + 2^{3/5} + 3 2^{4/5}}$$
$$x=\sqrt[\large 5]{2}\approx 1.1487 \implies y\approx 2.37411
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4008550",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
} |
Moment of inertia of sphere of changing density I need to calculate moment of inertia of a sphere. The density of sphere changing $\rho=cz^2$ All the three components $I_x,~I_y,~I_z~.$
I did the assumption $dm=\rho dV$ then I found $c$ as $=\frac{15m}{4\pi r^5}~.$ Then called
$$I=\int\rho dv z^2=c\int r^4 \cos^4\theta r^2 \sin\theta dr d\theta d\phi=c\int r^6 dr \int \cos^4\theta \sin\theta d\theta \int d\phi $$
Finally I found the $I=\frac{3}{7}\frac{R^7}{r^5}$
But I guess it's for $I_z$. How can I calculate $I_x=I_y$
| $I_x = \int dm \ (y^2+z^2), I_y = \int dm \ (x^2+z^2), I_z = \int dm \ (x^2+y^2)$
As $\rho = cz^2, dm = \rho dV = cz^2 dV$
Using cylindrical coordinates,
$x^2+y^2 = r^2, r^2 + z^2 = R^2 \ $ where $R$ is the radius of the sphere.
$dm = \rho dV = c z^2 \ r \ dr \ dz \ d\theta \ $
$\displaystyle I_z = c \int_0^{2\pi} \int_{-R}^{R} \int_0^{\sqrt{R^2-z^2}} z^2 r^3 \ dr \ dz \ d\theta = \frac{8 \pi c R^7}{105}$
Please note that density is only a function of $z$, so $I_x$ and $I_y$ will be equal due to symmetry of sphere.
That leads to $\displaystyle I_x = I_y = \frac{I_x + I_y}{2} = \frac{c}{2} \int_0^{2\pi} \int_{-R}^{R} \int_0^{\sqrt{R^2-z^2}} z^2 (r^2+2z^2) \ r \ dr \ dz \ d\theta$
$\displaystyle = \frac{I_z}{2} + c \int_0^{2\pi} \int_{-R}^{R} \int_0^{\sqrt{R^2-z^2}} z^4 \ r \ dr \ dz \ d\theta$
I will leave the integral for you to complete.
Lastly,
$m = c \displaystyle \int_0^{2\pi} \int_{-R}^{R} \int_0^{\sqrt{R^2-z^2}} z^2 \ r \ dr \ dz \ d\theta$
Once you find $m$, you can also write moment of inertia in terms of $m, R$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4009470",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
$\det( J_f^T J_f)$ $f:R^n \rightarrow R^{n+1}, f(x) =\frac{(x_1,\dots,x_n,1) }{(1+x_1^2+\dots+x_n^2)^{\frac{1}{2}}} $
$J_f$ is the Jacobi matrix of f
I want to calculate $\det( J_f^T J_f)$ but:
$J_f^t J_f = \left( \begin{array}{cc} |(x,1)|^2 - x_1^2 & -x_1 x_2 & \dots & -x_1 x_n \\ \vdots & & & \vdots \\- x_1 x^n & -x_2 x^n & \dots & |(x,1)|^2 -x_n^2 \end{array} \right) \cdot \frac{1}{|(x,1)|^{4}}$
I don't know how it's possible to find a general solution for this determinant.
It seems impossible to me.
| Let me use $n = 3$ to make this easier to write and picture
The trick here is that the matrix
$$\begin{pmatrix} x^2 & xy & xz \\ xy & y^2 & yz \\ xz & yz & z^2 \end{pmatrix} = \begin{pmatrix} x \\ y \\ z \end{pmatrix} \begin{pmatrix} x & y & z \end{pmatrix}$$
is rank 1.
So we have
$$J_f^\top J_f = \frac{1}{c^2}\left( cI - u u^\top \right)$$
where $c = (1 + x^2 + y^2 + z^2)$ and $u^\top = \begin{pmatrix} x & y & z \end{pmatrix}$.
And now we use a common formula for the determinant of a rank-1 update:
$$\det(I + uv^\top) = 1 + u^\top v.$$
See e.g. Determinant of rank-one perturbations of (invertible) matrices or https://en.wikipedia.org/wiki/Matrix_determinant_lemma
So we obtain
\begin{align}
\det(J_f^\top J_f) &= \frac{1}{c^{2n}}\det(cI - u u^\top) \\
&= \frac{1}{c^n} \det\left(I - \frac{1}{c} u u^\top \right) \\
&= \frac{1}{c^n} \left( 1 - \frac{1}{c} u^\top u \right) \\
&= \frac{1}{c^{n+1}} \left[ (1 + x_1 + \dots + x_n) - (x_1 + \dots + x_n) \right] \\
&= \frac{1}{c^{n+1}}.
\end{align}
So the final answer is
$$\det(J_f^\top J_f) = \frac{1}{(1 + x_1 + \dots + x_n)^{n+1}}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4011844",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Inequality : $ab^2+bc^2+cd^2+da^2 \leq 4$ I have a excercise like that: let $a;b;c \geq 0$ and $a+b+c=3$, prove: $ab^2+bc^2+ca^2 \leq 4$
We can assume $b=mid\{a;b;c\}$, then $(a-b)(b-c) \geq 0$, by that way, the problem is solved.
But I have a question: what should I do if the problem is for 4 variables, eg:
Let $a;b;c;d \geq 0$ and $a+b+c+d=3$, prove: $ab^2+bc^2+cd^2+da^2 \leq 4$
Thank you.
| We have
If $x,\,y,\,z$ are non-negative real numers satisfy $x+y+z=3,$ then
$$xy^2+yz^2+zx^2\leqslant 4.$$
Now, suppose $b = \max \{a,\,b,\,c,\,d\}.$ From this inequality we put $x=a,\,y=b,\,z=c+d,$ we get
$$a^2(c+d)+ab^2+b(c+d)^2 \leqslant 4.$$
But
$$a^2(c+d)+ab^2+b(c+d)^2$$
$$= ab^2+bc^2+cd^2+da^2 + c(a^2+2bd)+d^2(b-c)$$
$$ \geqslant ab^2+bc^2+cd^2+da^2.$$
Therefore
$$ab^2+bc^2+cd^2+da^2 \leqslant 4.$$
The proof is completed.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4014930",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
How to find the roots of a cubic polynomial? All of the examples on the internet I could find are made so that you can somehow make the cubic equation into a first degree polynomial multiplied by a second degree polynomial. But what if you can't do that?
For example, how would you find the roots of the following equation:
$x^3+7x^2+16x+12=0$
I know the roots are $-2$, $-2$ and $-3$ but don't know how to get them.
If I somehow guess them, is there a way to tell which one is the double root?
| While there is a rather nice test for whether a cubic polynomial has a triple zero by inspecting its coefficients (it will even tell us directly the value of that root), there is not anywhere near as simple a method for testing for a double zero by using the coefficients. (It can be managed, but the calculations do not just require basic arithmetic.)
We can produce a straightforward method of testing with the coefficients if we suspect that a double zero is present. We would write for the polynomial $ \ ax^3 + bx^2 + cx + d \ = \ a · (x - r)^2 · (x - s) \ \ , $ with $ \ r \ $ being the double zero and $ \ s \ $ being the other (real) zero. This alone is not especially helpful, but we could express the "singleton" zero as $ \ s = \sigma · r \ \ . $ This then gives us
$$ a · (x - r)^2 · (x \ - \ \sigma · r) \ \ = \ \ a · ( x^3 \ - \ [ \ 2r \ + \ \sigma · r \ ]· x^2 \ + \ [ \ r^2 \ + \ 2·\sigma·r^2 \ ]·x \ - \ \sigma·r^3 \ \ . $$
The relation among the coefficients will thus be
$$ \frac{b}{a} \ = \ -r · (\sigma \ + \ 2) \ \ , \ \ \frac{c}{a} \ = \ r^2 · (1 \ + \ 2·\sigma) \ \ , \ \ \frac{d}{a} \ = \ -\sigma · r^3 \ \ . $$
This can be helpful for integer and "reasonably simple" rational coefficients. For your example polynomial $ \ x^3 + 7x^2 + 16x + 12 \ \ , $ the fact that the constant coefficient can be factored as $ \ 12 = 2^2 · 3 \ $ might lead us to check for a double zero. We could have either $ \ \frac{d}{a} \ = \ 12 \ = \ - \ 2^2 · (-3) \ \ $ or $ \ -(-2)^2 · (-3) \ \ . $ The Rule of Signs will tell us which of the two factorizations would be the one to try, but we will see that this test also leaves no sign ambiguity:
$$ \mathbf{r = 2 \ , \ \sigma = -\frac32 \ \ :} \quad \frac{b}{a} \ = \ -2 · \left(-\frac32 \ + \ 2 \right) \ = \ -1 \ \ , \ \ \frac{c}{a} \ = \ 2^2 · \left(1 \ + \ 2·\left[-\frac32 \right] \right) \ = \ -8 \ \ ; $$
$$ \mathbf{r = -2 \ , \ \sigma = \frac32 \ \ :} \quad \frac{b}{a} \ = \ 2 · \left(\frac32 \ + \ 2 \right) \ = \ 7 \ \ , \ \ \frac{c}{a} \ = \ (-2)^2 · \left(1 \ + \ 2·\frac32 \right) \ = \ 16 \ \ . $$
Hence, the zeroes are $ \ x \ = \ -2 \ [\text{multiplicity} \ 2] \ , \ -3 \ \ $ or $ \ x^3 + 7x^2 + 16x + 12 \ = \ (x + 2)^2 · (x \ + \ \frac32 · 2) \ = \ (x + 2)^2 · (x + 3) \ \ . $
As an additional demonstration, we could attempt to solve in this manner $ \ 9x^3 + 3x^2 - 56x + 48 \ = \ 0 \ : $
$$ \frac{d}{a} \ = \ \frac{48}{9} \ = \ \frac{16}{3} \ = \ - \left(\frac43 \right)^2 · (-3) \ \ \ [?] $$
(the Rule of Signs tells us there would be one negative zero and two positive real zeroes) so we will want to check $ \ r = \frac43 \ , \ \sigma = -\frac94 \ \ : $
$$ \rightarrow \ \ \frac{b}{a} \ = \ -\frac43 · \left(-\frac94 \ + \ 2 \right) \ = \ \frac13 \ \ , \ \ \frac{c}{a} \ = \ \left(\frac43 \right)^2 · \left(1 \ + \ 2·\left[-\frac94 \right] \right) \ = \ -\frac{56}{9} \ \ . $$
We do have $ \ b = \frac93 = 3 \ $ and $ \ c = -56 \ \ , $ which permits us to conclude that
$$ 9x^3 + 3x^2 - 56x + 48 \ \ = \ \ 9 · \left( x - \frac43 \right)^2 · \ (x + 3 ) \ \ . $$
Naturally, if this coefficient test indicates that we do not have a double zero, we will need to "fall back" to more elaborate methods to determine the zeroes.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4015633",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Smallest possible area for triangle I'm solving the following question:
The two legs of a right triangle lie along the positive x and y axes.
The hypotenuse is tangent to the ellipse $2x^2 + y^2 = 1$. What is the
smallest possible area for such a triangle ?
This is my attempt at solving the problem:
Let $a$ be the length of the x axis and $b$ be the length in it's y axis if the triangle. So it's area is $1/2 a.b$
Let's find the tangent of the ellipse:
$2x^2 + y^2 = 1$
$4x + 2y \dfrac{dy}{dx} = 0$
$\dfrac{dy}{dx} = \dfrac{-2x}{y}$
Since the two points on the triangle are $(0,b)$ and $(a,0)$, we can
find the slope of the line:
$m = \dfrac{b-0}{0-a} = \dfrac{-a}{b}$
So the equation of hypotonuse is $y = \dfrac{-a}{b}x + c$
Since $(0,b)$ is one of the point in the line, we can substitute it in
the above line to find that $b = c$. Solving it with the other
co-ordinate, we can find the following fact $b = a$.
So now area is $1/2 a.b = 1/2 a^2$
Now I used the first derivative test to find that the function attains
it's local minima at $0$. So the smallest possible area for such a
triangle is zero.
I have missed lots of steps in my above solution for brevity,
but I can expand if needed. My question: is zero
the right answer ? Unfortunately, the book I'm using doesn't provide
the answer to this question.
| You have made a small error. You wrote that
$$m = \dfrac{b-0}{0-a}$$ but then said that this is equal to
$$ \dfrac{-a}{b}$$
This is clearly wrong. In fact,
$$m=-\frac{b}{a}$$
The Solution
We have $y=-\frac{b}{a}x+b=\frac{ab-bx}{a}$. Substituting this into $2x^2+y^2=1$ yields
$$\begin{align}
&2x^2+\left(\frac{ab-bx}{a}\right)^2=1\\
\implies& 2a^2x^2+b^2x^2-2ab^2x+a^2b^2=a^2\\
\implies& x^2(2a^2+b^2)+x(-2ab^2)+(a^2b^2-a^2)=0\\
\end{align}$$
We could then go on to try to solve this quadratic, but in fact we can do better. Note that the tangent line will intersect the ellipse once, and only once. Hence, the discriminant of this quadratic in $x$ is equal to $0$:
$$4a^2b^4-4(2a^2+b^2)(a^2b^2-a^2)=0$$
We can divide through by $a^2$, as we don't have a degenerate triangle with $a=0$, simplifying to
$$\begin{align}&4b^4-4(2a^2+b^2)(b^2-1)=0\\
\implies&4b^4-4(2a^2b^2-2a^2+b^4-b^2)=0\\
\implies & a^2(8b^2-8)=4b^2\\
\implies & a^2=\frac{b^2}{2(b^2-1)}
\end{align}$$
Hence, we have, assuming without loss of generality that $a,b\geqslant 0$, that
$$a=\frac{b}{\sqrt{2(b^2-1)}}$$
So we find that the area of the triangle, which we know is equal to $\frac{1}{2}ab$, is in fact equal to
$$\frac{b^2}{2\sqrt{2(b^2-1)}}$$
So we now have an expression for the area, $A$, in terms of one variable only! This means that to find the minimum value of $A$, we can just differentiate with respect to $b$, solve for stationary points, check it is a minimum, and we are basically done!
I hope that was helpful. Thank you for giving me the opportunity to attack this interesting question; I'd never come across a similar one before. What I find particularly pleasing about this solution is that we have no need to look at the derivative of the ellipse, which would've made things rather complicated. If you have any questions please don't hesitate to ask!
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4016294",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 1
} |
$x^3 + y^3 +3x^2 y^2 =x^3y^3$ Find all possible values to $\frac{x+y}{xy}$ $x,y \in \mathbb{R}\setminus\{0\}$
Indeed the original question said :Find all possible values to: $$\frac{1}{x} + \frac{1}{y}$$
But it’s the same thing.
My Attempt:
$$x^3 + y^3 +3x^2 y^2 =x^3y^3 \iff (x+y)(x^2-xy+y^2)=x^3y^3 -3x^2y^2$$
$$\frac{x+y}{xy}=\frac{x^2y^2-3xy}{x^2-xy+y^2}$$
But i don’t know what to do now.
Please if you know the key to this type of problems ‘Find all possible values’ post it.
| Because $x,y \ne 0$, from $x^3 + y^3 +3x^2 y^2 =x^3y^3$, we have
$$\frac{1}{x^3}+\frac{1}{y^3} +\frac{3}{xy} =1 \iff\frac{1}{x^3}+\frac{1}{y^3} =1-\frac{3}{xy} \tag{1}$$
But we have also
$$(\frac{1}{x}+\frac{1}{y})^3 = (\frac{1}{x^3}+\frac{1}{y^3}) + \frac{3}{xy}(\frac{1}{x}+\frac{1}{y}) \tag{2}$$
From (1) and (2), we deduce
\begin{align}
&(\frac{1}{x}+\frac{1}{y})^3 = (1-\frac{3}{xy}) + \frac{3}{xy}(\frac{1}{x}+\frac{1}{y}) \\
&\iff (\frac{1}{x}+\frac{1}{y})^3- 1=\frac{3}{xy}\left( \frac{1}{x}+\frac{1}{y} - 1 \right) \\
&\iff \left( \frac{1}{x}+\frac{1}{y} - 1 \right)\left( (\frac{1}{x}+\frac{1}{y})^2 +(\frac{1}{x}+\frac{1}{y})+1 \right)=\frac{3}{xy}\left( \frac{1}{x}+\frac{1}{y} - 1 \right) \\
&\iff \left( \frac{1}{x}+\frac{1}{y} - 1 \right)\left( (\frac{1}{x}+\frac{1}{y})^2 +(\frac{1}{x}+\frac{1}{y})+1 -\frac{3}{xy} \right)=0\\
&\iff \left( \frac{1}{x}+\frac{1}{y} - 1 \right)\left( (\frac{1}{x}+\frac{1}{y})^2 -\frac{4}{xy} +(\frac{1}{x}+\frac{1}{y})+1 +\frac{1}{xy} \right)=0\\
&\iff \left( \frac{1}{x}+\frac{1}{y} - 1 \right)\left( (\frac{1}{x}-\frac{1}{y})^2 +\left( 1+\frac{1}{x} \right)\left( 1+\frac{1}{y} \right) \right) =0\\
&\iff \left( \frac{1}{x}+\frac{1}{y} - 1 \right)\left( ((1+\frac{1}{x})-(1+\frac{1}{y}))^2 +\left( 1+\frac{1}{x} \right)\left( 1+\frac{1}{y} \right) \right) =0\\
&\iff \left( \frac{1}{x}+\frac{1}{y} - 1 \right)\left( \left(1+\frac{1}{x} \right)^2+\left(1+\frac{1}{y} \right)^2 -\left( 1+\frac{1}{x} \right)\left( 1+\frac{1}{y} \right) \right) =0\\
\end{align}
The last equation happens if and only if $\frac{1}{x}+\frac{1}{y} =1$ or $\left( 1+\frac{1}{x}, 1+\frac{1}{y} \right) = (0,0) $. Hence, the two possible values of $\frac{1}{x}+\frac{1}{y}$ is $1$ and $-2$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4017341",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
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} |
Finding the value of $ax^4+by^4$
If $\quad a+b=23 , \quad ax+by=79,\quad ax^2+by^2=217,\quad
ax^3+by^3=691\quad$ find the value of $ax^4+by^4$.
Here is my attempt:
$$(a+b)(x+y)=ax+by+ay+bx\rightarrow 23(x+y)=79+(ay+bx)$$
$$(ax+by)(x+y)=ax^2+by^2+axy+bxy\rightarrow79(x+y)=217+23 xy$$
In each equation I have two unknowns it seems that doesn't work.
| Your method is not in vain. We have:
$$(ax+by)(x+y) = ax^2 + by^2 +axy + bxy \to 79(x+y) = 217 + 23xy$$
$$(ax^2 + by^2)(x+y) = ax^3 + by^3 + ax^2y + bxy^2 \to 217(x+y) = 691 + 79xy$$
hence $x + y = 1, xy = -6$. Now:
$$(ax^3 + by^3)(x+y) = ax^4 + by^4 + ax^3y + bxy^3 = ax^4 + by^4 + xy(ax^2 + by^2)$$
$$691\times 1 = ax^4 + by^4 -6 \times 217$$
thus $ax^4 + by^4 = 1993$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4018900",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 0
} |
Maximum and minimum value of $\frac{8x - 3y}{\sqrt{4x^2+y^2}}$? For real numbers $\,x, y\neq 0\,$ consider
$$\frac{8x - 3y}{\sqrt{4x^2+y^2}}\,.$$
How to find the maximum and minimum value?
I've already got the maximum by using the Cauchy–Schwarz inequality
$$\big[(2x)^2 + y^2\big]\big[4^2 + (-3)^2\big] \geq (8x - 3y)^2\\[3ex]
25 \geq \frac{(8x - 3y)^2}{(2x)^2 + y^2}\\[4ex]
5 \geq \frac{8x - 3y}{\sqrt{(2x)^2 + y^2}}$$
But I cannot get the minimum value.
| Let $r > 0$ and $0 \le \theta < 2\pi$, and $x = r \cos \theta$, $y = 2r \sin \theta$, so that $$4x^2 + y^2 = 4r^2,$$ hence $$\begin{align}f(x,y) &= \frac{8x-3y}{\sqrt{4x^2 + y^2}} \\ &= \frac{8r \cos \theta - 6r \sin \theta}{2r} \\ &= 4 \cos \theta - 3 \sin \theta \\
&= 5 \left(\frac{4}{5} \cos \theta - \frac{3}{5} \sin \theta \right) \\
&= 5 \left( \sin \psi \cos \theta - \cos \psi \sin \theta \right) \quad \quad \psi = \arcsin \frac{4}{5} \\
&= 5 \sin(\theta - \psi).
\end{align}$$ Now we can see this function depends only on the choice of angle $\theta$. This is obviously maximized when $\theta = \psi + \pi/2$ and the maximum attained is $5$.
This also happens to get us the minimum for free, which occurs when $\theta = \psi + \frac{3\pi}{2},$ and the minimum is $-5$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4020195",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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} |
Show that if $r$ and $s$ are positive roots to the equation $x^3-cx+d=0$, then $r+s$ is a root of $x^3-cx-d=0$ If $r$ and $s$ are roots of $x^3-cx+d=0$, then
$$r^3-cr+d=s^3-cs+d , $$
for $r \neq s$. Solving the equation for $c$ gives me
$$ c=r^2+rs+s^2$$
Now to find an expression for $d$
$$ r^3-cr+d=0 $$
$$ d=r(c-r^2) $$
Using the expression for $c$ gives me
$$d=sr^2+rs^2$$
Assume $t$ is a root of the second equation
$$t^3-ct+d=0$$
We can represent $c$ and $d$ in terms of $r$ and $s$ which gives us
$$s*r^2+r*s^2=t(t^2-r^2-rs-s^2)$$
Plugging this into wolfram alpha gives us $t=r+s$, which is what we wanted to show. But how can I solve for $t$ algebraically? Or perhaps there is another way to solve this question?
| Let $t$ be the $3$rd root of the equation $x^3 - cx + d = 0$
So $$t^3 - ct + d = 0 \tag{1}$$
Also $$r + s + t = 0 \implies t = -(r+s) \tag{2}$$
Substitute the value of $t$ from $(2)$ into $(1)$ to get
$$-(r+s)^3 + c(r+s) + d = 0 \equiv (r+s)^3 - c(r+s) - d = 0$$
Hence the proof.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4020920",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Intersection of perpendicular chords: is this true for a sphere? It is true that in a circle with radius $R$, if the intersection of any two perpendicular chords divides one chord into lengths $a$ and $b$ and divides the other chord into lengths $c$ and $d$, then
$$a^2 + b^2 + c^2 + d^2 = 4R^2.$$
Does this equation hold for a sphere with radius $R$, and three mutually perpendicular chords:
$$a^2 + b^2 + c^2 + d^2 + e^2 + f^2 = 6R^2?$$
| For circles:
From the circle center draw perpendicular lines to the chords. Let the intersection of the lines with the chords be $X$ and $Y$, respectively. Let $|OX|=x$, $|OY|=y$.
We have (see figure):
$$\begin{align}
a&=\sqrt{R^2-x^2}+y,\\
b&=\sqrt{R^2-x^2}-y,\\
c&=\sqrt{R^2-y^2}-x,\\
d&=\sqrt{R^2-y^2}+x,
\end{align}$$
so that
$$
a^2+b^2+c^2+d^2=4R^2,
$$
as claimed.
Try to apply the same construction for the sphere case.
Drawing from the center of the sphere the planes perpendicular to the chords and denoting the intersection points $X,Y,Z$ you will obtain: $$a^2+b^2+c^2+d^2+e^2+f^2=6R^2-2(x^2+y^2+z^2),$$ so that the claim does not hold for a sphere.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4021931",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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} |
With or without squeeze theorem I am doing a proof involving finding the sum of a series $\sum_{x\geq1} \frac{x}{2^x} - \frac{x+1}{2^{x+1}}$ and I need to show that $\lim_{x\to\infty} \frac{x+1}{2^x}$ converges to $0$ without using L'Hopital's theorem.
My attempt:
$\lim_{x\to\infty} \frac{x+1}{2^x} = \lim_{x\to\infty} \frac{x}{2^x} + \lim_{x\to\infty} \frac{1}{2^x}$
We then get that $2^x > x^2 \implies \frac{x}{2^x} < \frac{x}{x^2} = \frac{1}{x}$ and also, $2^x > x \implies \frac{1}{2^x} < \frac{1}{x}$ for sufficiently large $x$. We have that
$$0 \leq \lim_{x\to\infty} \frac{x}{2^x} \leq \lim_{x\to\infty} \frac{1}{x} = 0$$
$$0 \leq \lim_{x\to\infty} \frac{1}{2^x} \leq \lim_{x\to\infty} \frac{1}{x} = 0$$
(Notice the change from $<$ to $\leq$ above. Is that allowed? Also, the series starts at $1$, not $0$.)
Using squeeze theorem, $\lim_{x\to\infty} \frac{x+1}{2^x} = 0$.
Is this correct? Notice my bracketed comment. Any assistance much appreciated.
| Since you are in the context of a series, $x$ is natural.
So, you can apply the binomial formula to $2^x = (1+1)^x=\sum_{n=0}^x\binom xn$:
$$0\leq \frac{x}{2^x} = \frac{x}{(1+1)^x} \stackrel{x\geq 2}{\leq} \frac{x}{\binom x2} = \frac 2{x-1}\stackrel{x\to \infty}{\longrightarrow}0$$
Similarly for the case $\frac 1{2^x}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4022654",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Find the general solution to the ODE: I was asked to find the general solution to these ODEs.
*
*$$\frac{dy}{dx}=\frac{(-8x+3y-31)}{(-3x+y-11)}$$
I tried by rearranging to the form $M(x,y)+N(x,y)\frac{dy}{dx}=0$ and let $y=vx$, but what I got was $$(v^2x-6vx-11v+8x+31)dx+x(-3x+vx-11)dv=0$$
and I could not separate $x$ and $v$ for integration.
*$$\frac{dy}{dx}=\frac{y}{x}+\frac{9}{2}xexp(-\frac{2y}{x})+\frac{9}{2}xexp(\frac{2y}{x})$$
I have totally no idea which method to be used in Question 2.
May I know if I'm on the wrong track, and how should I solve them? Thanks.
| We have
$$
\frac{dy}{dx} = \frac{-8x+3y-31}{-3x+y-11}.
$$
One way to solve it is to introduce time variable $t$, obtaining a system
$$
\left\{
\begin{align*}
&\dot{y} = -8x+3y-31 \\
&\dot{x} = -3x+y-11
\end{align*}
\right. ,
$$
which is linear. A matrix of this system
$$
A =
\begin{pmatrix}
-8 & 3\\
-3 & 1
\end{pmatrix}
$$
is nondegenerate: $\det A = 1$.
Thus we can find a particular solution almost immediately substituting
$$
\left\{
\begin{align*}
&y = Y \\
&x = X
\end{align*}
\right. .
$$
That gives us a system of linear equations:
$$
\left\{
\begin{align*}
&-8X+3Y = 31\\
&-3X+Y = 11
\end{align*}
\right. ,
$$
which has a unique solution $Y = 5$, $X = -2$. Now everything what we we have to do is to find a general solution of the homogeneous system
$$
\left\{
\begin{align*}
&\dot{y} = -8x+3y\\
&\dot{x} = -3x+y
\end{align*}
\right. .
$$
So we consider a characteristic equation $\det (A - \lambda I) = \lambda^2 + 7\lambda + 1 = 0$. We have two distinct roots:
$$
\left\{
\begin{align*}
&\lambda_1 = \frac{3\sqrt{5}-7}{2}\\
&\lambda_2 = \frac{-3\sqrt{5}-7}{2}
\end{align*}
\right. .
$$
Corresponding eigenvectors are
$$
e_1 = \begin{pmatrix} \frac{-\sqrt{5}+3}{2} \\ 1 \end{pmatrix}, \quad e_2 = \begin{pmatrix} \frac{\sqrt{5}+3}{2} \\ 1 \end{pmatrix}.
$$
Hence the general solution is
$$
\begin{pmatrix} y \\ x \end{pmatrix} = C_1\begin{pmatrix} \frac{-\sqrt{5}+3}{2} \\ 1 \end{pmatrix} \exp{\left(\frac{3\sqrt{5}-7}{2}t\right)} + C_2\begin{pmatrix} \frac{\sqrt{5}+3}{2} \\ 1 \end{pmatrix} \exp{\left(\frac{-3\sqrt{5}-7}{2}t\right)},
$$
$C_1, C_2 = \mathrm{const}$. Finally, we combine two parts together:
$$
\begin{pmatrix} y \\ x \end{pmatrix} = C_1\begin{pmatrix} \frac{-\sqrt{5}+3}{2} \\ 1 \end{pmatrix} \exp{\left(\frac{3\sqrt{5}-7}{2}t\right)} + C_2\begin{pmatrix} \frac{\sqrt{5}+3}{2} \\ 1 \end{pmatrix} \exp{\left(\frac{-3\sqrt{5}-7}{2}t\right)} + \begin{pmatrix} 5 \\ -2 \end{pmatrix}.
$$
To proceed to the answer we have to say a few words about singularities where the right hand side of our original equation doesn't exist. A condition is $-3x + y - 11 = 0$ (we look for singularities). It gives us
$$
C_1\left(\frac{-\sqrt{5}+3}{2} - 3 \right)\exp{\left(\frac{3\sqrt{5}-7}{2}t\right)} + C_2\left(\frac{\sqrt{5}+3}{2} - 3 \right)\exp{\left(\frac{-3\sqrt{5}-7}{2}t\right)} = 0,
$$
$$
C_1\frac{\sqrt{5}+3}{2}\exp{\left(\frac{3\sqrt{5}-7}{2}t\right)} = C_2\frac{\sqrt{5}-3}{2}\exp{\left(\frac{-3\sqrt{5}-7}{2}t\right)},
$$
$$
C_1\frac{\sqrt{5}+3}{\sqrt{5}-3}\exp{\left(3\sqrt{5}t\right)} = C_2.
$$
But those points are not a real problem: in those points a tangent line to our integral curves is vertical. Not a big deal. So the answer is above in the form of a family of curves parametrised by time.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4022824",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Given a recursion $a_{n+ 1}= \sqrt{na_{n}+ 2n+ 1}$ with $a_{1}\geq 1.$ Prove that $a_{n}\sim n\,{\rm as}\,n\rightarrow\infty$
Given a recursion $a_{n+ 1}= \sqrt{na_{n}+ 2n+ 1}$ with $a_{1}\geq 1.$ Prove that
$$a_{n}\sim n\,{\rm as}\,n\rightarrow\infty$$
Let $b_{n}= \dfrac{a_{n}}{n},$ we need to prove $\lim b_{n}= 1,$ then we find a value $\beta\in\left ( 0, 1 \right )$ satisfying $\left | b_{n+ 1}- 1 \right |\leq\beta\left | b_{n}- 1 \right |$ so that
$$\beta\rightarrow 0\,{\rm aut}\,n\rightarrow\infty$$
I can't predict the relationship among them, I need the the help to go to the induction, thank you.
| You don't need $\beta \rightarrow 0$, you just need that for sufficiently large $n$ $$\beta < 1$$ Then $$|b_{n+M}-1|<\beta^n|b_M-1| \rightarrow 0$$
You can check that
$$ b_{n+1} = \sqrt{\frac{b_nn^2 +2n+1}{(n+1)^2}}{} = \sqrt{(b_n-1)\frac{n^2}{(n+1)^2}+1}$$
that is
$$ b_{n+1}-1 = \sqrt{(b_n-1)\frac{n^2}{(n+1)^2}+1} - 1= \frac{(b_n-1)\frac{n^2}{(n+1)^2}}{\sqrt{(b_n-1)\frac{n^2}{(n+1)^2}+1} +1} $$
Since $a_1 \ge 1$, then $b_1 -1 \ge 0$ and by induction it's easy to show that $b_n -1 \ge 0$. We have then
$$ \frac{\frac{n^2}{(n+1)^2}}{\sqrt{(b_n-1)\frac{n^2}{(n+1)^2}+1} +1} \le \frac{\frac{n^2}{(n+1)^2}}{\sqrt{1}+1} = \frac{n^2}{2(n+1)^2} \le \frac12 $$
Therefore
$$b_{n+1} -1 \le \frac12(b_n-1)$$
As showed before, it is enough to prove that $b_{n+1}-1 \rightarrow 0$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4024146",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 2
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If $x+2y=8$, find the minumum of $x+y+\frac{3}{x}+\frac{9}{2y}$ (x, y ∈ R+) Question : If $x+2y=8$, find the minumum of $x+y+\dfrac{3}{x}+\dfrac{9}{2y}$ $(x, y \in \mathbb{R}^+)$
I tried to use $x=8-2y$, and I thought I can plug in. But I think I'm not doing it well. I thought about AM-GM, and how can I expand this using $x+2y=8$?
| $$x+2y=8 \implies x = 8 - 2y\tag{1}$$
Let $$f(x, y) = x+y+\frac{3}{x}+\frac{9}{2y} \tag{2}$$
Using $(1)$ in $(2)$,
$$f(x, y) = 8 - 2y +y+\frac{3}{8 - 2y}+\frac{9}{2y} = 8 - y +\frac{3}{8 - 2y}+\frac{9}{2y} \tag{3}$$
Differentiate $(3)$ wrt $y$, set it equal to $0$ and solve for $y$
$$\frac{-9}{2y^2} + \frac{3}{2(4-y)^2} - 1 = 0$$
The real root is $y = 3$ and the second derivative $$\left[\frac{9}{y^3} + \frac{3}{(4-y)^3} \right]_{y = 3} = \frac{10}{3} > 0$$
From $(1)$, $x = 2$ and the minimum value is $f(2, 3) = 8$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4024960",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Show that $\triangle ABC$ satisfying $B^2=AC$ and $2b=a+c$ is equilateral
A $\triangle ABC$ satisfies the conditions below:
$$B^2=AC \qquad 2b=a+c$$
Show that it's an equilateral one, where $a=|BC|$, $b=|AC|$, $c=|AB|$ and $B=\angle ABC$, $A=\angle BAC$, $C=\angle ACB$.
What I have done:
When trying to solve these question, I want to find a solution in all triangles which satisfy $2b=a+c$. It means that point $B$ can be regarded as a point in an ellipse.
Then I try to prove that for all points $B$ in that ellipse, we have $B^2\leq AC$. When proving it I think I can get the equal conditions and then show that $AC=B^2$ if and only if $A=B=C$ or $AC=0$.
Then I meet a terrible problem, meaning that I need to show ($x=A$):
$$
\frac{4-5\cos x}{5-4\cos x}\leq\cos\frac{(2\pi-x)-\sqrt{x(4\pi-3x)}}{2}=-\cos \frac{x+\sqrt{x(4\pi -3x)}}{2}，x\in[0,\frac{\pi}{3}]\\
$$
I have never seen such a hard problem before. I try to prove the inequality by segment amplification and minification, derivative, and many other ways but all failed.
Can you help me or give me some hints on this problems?
| I am assuming $A\ge C$ without loss of generality. Therefore $A=Br$ and $C=\frac Br$ for some $r\ge1$, by the first equation. From the second equation, note that
\begin{align}
&2b=a+c\\
\implies &\dfrac{2b}{2R} = \dfrac{a}{2R}+\dfrac{c}{2R}\\
\implies &2\sin B = \sin A+\sin C&&(\text{By sine rule})\\
\implies &4\cos\dfrac B2\sin\dfrac B2 = 2\sin\left(\dfrac{A+C}{2}\right)\cos\left(\dfrac{A-C}{2}\right)\\
\implies &2\sin\dfrac B2=\cos\left(\dfrac{A-C}{2}\right)&&\left(\because\cos\frac B2 = \sin\left(\dfrac{A+C}{2}\right)\right)\\
\implies &2=\dfrac{\cos\left(\dfrac{A-C}{2}\right)}{\cos\left(\dfrac{A+C}{2}\right)}\\
\implies &3=\cot\left(\dfrac{A}{2}\right)\cot\left(\dfrac{C}{2}\right)&&(\text{By componendo and dividendo})\\
\implies &\tan\left(\frac A2\right)\tan\left(\frac C2\right)=\frac13\\
\implies &\tan\left(\dfrac{Br}{2}\right)\tan\left(\dfrac{B}{2r}\right)=\frac13\tag1\label{1}
\end{align}
Also note that
\begin{align}
&A+B+C=\pi\\
\implies &Br+B+\dfrac Br=\pi\\
\implies &B = \dfrac{\pi r}{r^2+r+1}\tag2\label{2}
\end{align}
Using \eqref{2} in \eqref{1}, we get,
\begin{equation}
\tan\left(\dfrac{\frac\pi2\times r^2}{r^2+r+1}\right)\tan\left(\dfrac{\frac\pi2}{r^2+r+1}\right)=\frac13
\end{equation}
Now, since LHS is a decreasing function of $r$ for all $r\ge 1$ and equality holds at $r=1$, hence $r=1$ and we're done.
| {
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"url": "https://math.stackexchange.com/questions/4027351",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
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Show that $G\left(n,p\right)e^{-\frac{1}{12np\left(1-p\right)}}<{n\choose pn} < G\left(n, p\right)$ The problem:
Using the inequalities
$\sqrt{2\pi n}\left(\frac{n}{e}\right)^{n}e^{\left(\frac{1}{12n}-\frac{1}{360n^{3}}\right)} < n! < \sqrt{2\pi n}\left(\frac{n}{e}\right)^ne^{\frac{1}{12n}}$
show that, for a constant $p$ such that $0<p<1$, where $pn$ is an integer,
$G\left(n, p\right)=\frac{2^{nH\left(p\right)}}{\sqrt{2\pi n p\left(1-p\right)}}$, and $H\left(p\right)$ is the binary entropy function, show that
$G\left(n,p\right)e^{-\frac{1}{12np\left(1-p\right)}}<{n\choose pn} < G\left(n, p\right)$
My work so far:
First I have noted that
${n\choose pn} = \frac{n!}{\left(pn\right)!\left(\left(1-p\right)n\right)!}$
$ \approx \frac{\sqrt{2\pi n} \left( \frac{n}{e} \right)^{n}}{\left(pn\right)!\left(\left(1-p\right)n\right)!}$
$\approx \frac{ \sqrt{2\pi \left(1-p\right)n}\left( \frac{\left(1-p\right)n}{e}\right)^{\left(1-p\right)n}}{2\pi \sqrt{\left(1-2p\right)}\left(\frac{pn}{e}\right)^{pn} \left(\frac{\left(1-2p\right)n}{e}\right)^{\left(1-2p\right)n}} e^\left({\frac{1}{12\left(1-p\right)n}}-\frac{1}{12pn+1}-\frac{1}{12\left(1-2p\right)n+1}\right)$
$\leq 2^{nH\left(p\right)} \frac{e^{\frac{1}{12n}}}{\sqrt{2\pi np\left(1-p\right)}}$
and from here I am not sure where to go. Any hints or help would be greatly appreciated.
| I put $q=1-p$. First, I shall use the inequalities
$$
\sqrt {2\pi n} \left( {\frac{n}{e}} \right)^n < n! < \sqrt {2\pi n} \left( {\frac{n}{e}} \right)^n e^{\frac{1}{{12n}}}
$$
which hold for all positive integer $n$. We obtain
$$
\binom{n}{pn} > \frac{{\sqrt {2\pi n} \left( {\frac{n}{e}} \right)^n }}{{\sqrt {2\pi pn} \left( {\frac{{pn}}{e}} \right)^{pn} e^{\frac{1}{{12pn}}} \sqrt {2\pi qn} \left( {\frac{{qn}}{e}} \right)^{qn} e^{\frac{1}{{12qn}}} }} = \frac{{2^{nH(p)} }}{{\sqrt {2\pi pqn} }}e^{ - \frac{1}{{12npq}}} .
$$
To prove the upper bound, I shall use the inequalities
$$
\sqrt {2\pi n} \left( {\frac{n}{e}} \right)^n e^{\frac{1}{{12n+1}}} < n! < \sqrt {2\pi n} \left( {\frac{n}{e}} \right)^n e^{\frac{1}{{12n}}}
$$
which hold for all positive integer $n$ (cf. H. Robbins, A remark on Stirling's formula, Amer. Math. Monthly 62 (1955), pp. 26–29.). We deduce
$$
\binom{n}{pn} < \frac{{\sqrt {2\pi n} \left( {\frac{n}{e}} \right)^n e^{\frac{1}{{12n}}} }}{{\sqrt {2\pi pn} \left( {\frac{{pn}}{e}} \right)^{pn} e^{\frac{1}{{12np + 1}}}\sqrt {2\pi qn} \left( {\frac{{qn}}{e}} \right)^{qn} e^{\frac{1}{{12nq + 1}}}}} = \frac{{2^{nH(p)} }}{{\sqrt {2\pi pqn} }}e^{\frac{1}{{12n}} - \frac{1}{{12np + 1}} - \frac{1}{{12nq + 1}}} .
$$
Finally, note that
$$
\frac{1}{{12n}} - \frac{1}{{12pn + 1}} - \frac{1}{{12qn + 1}} = - \frac{{144n^2 (pq + 1) + 12n - 1}}{{12n(12np + 1)(12nq + 1)}} < 0.
$$
| {
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"url": "https://math.stackexchange.com/questions/4030148",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Limit of $(x^3 - 3x^2 + 2x)$ as $x \to 1$ using epsilon delta definition I need to prove the following limit using epsilon delta method. I have come up this on my own just to practice skills.
$$\lim \limits_{x \to 1} \,(x^3 - 3x^2 + 2x) = 0 $$
So, I need to come up with some $\delta >0$ such that given $0 < \lvert x - 1 \rvert <\delta$, I need to prove that
$$ |(x^3 - 3x^2 + 2x) - 0| < \varepsilon \tag{1}$$
I can simplify eq $1$ as follows.
$$ |x| |x-1||x-2| < \varepsilon $$
When we have a non linear term, then we restrict $x$ to some distance from $1$. So, suppose we have
$ 0 < |x-1| < 10$. This leads to $-10 < x-1 < 10$. Doing some algebra, we can get inequalities
$$ 0 \leqslant |x| < 11 \tag{2}$$
$$ 0 \leqslant |x-2| < 11 \tag{3}$$
From eq 2 and 3, it follows that
$$ 0 \leqslant |x||x-2| < 121 $$
And since $0 < |x-1| $, it follows that
$$ |x||x-1||x-2| < 121 |x-1| \tag{4}$$
To show that $|x||x-1||x-2| < \varepsilon$ , its sufficient to show that $ 121 |x-1| < \varepsilon$. And this will happen if $ |x-1| < \frac{\varepsilon}{121} $. So, this is another restriction on $|x-1|$. So, we can let
$$\delta = \text{min}\left(10, \frac{\varepsilon}{121}\right) $$
So, we can let this be our choice of $\delta$. Its clear that $\delta > 0$. So, now the official proof will follow.
Let $\varepsilon > 0$ be some arbitrary real. Suppose $0 < |x-1| < \delta$. With the choice of $\delta$ we have done, this means that $ 0 < |x-1| < 10$. As demonstrated above, equation 4 follows from this inequality
$$ |x||x-1||x-2| < 121 |x-1| $$
But now, we also have $ |x-1| < \frac{\varepsilon}{121} $. It means that $ 121 |x-1| < \varepsilon $. Using this, we get that
$$ |x||x-1||x-2| < \varepsilon $$
which can be rewritten as
$$ | (x^3 - 3x^2 + 2x) - 0 | < \varepsilon $$
Since, $\varepsilon > 0$ was arbitrary to begin with, its proven that
$$\lim \limits_{x \to 1} \,(x^3 - 3x^2 + 2x) = 0 $$
Is the proof correct ?
| Other methode which easier
$f(x) =x^3 - 3x^2 +2x$
Suppose $|f(x)-l|<\epsilon $
Let's show $|x-1|<\alpha$
$|f(x)-l|=|f(x)-0|=|x^3 - 3x^2 +2x|<\epsilon $
$|x^3 - 3x^2 +2x|=|x^2-2x|×|x-1|<\epsilon $
Suppose $x\in [\frac{1}{2}, \frac{3} {2}]$
$\Rightarrow $$ \frac{1} {4}\leq x^2 \leq \frac{9}{4}$
And
$-3\leq-2x\leq - 1$
So :
$\frac{-11} {4}\leq x^2-2x \leq \frac{5}{4}<\frac{11}{4}$$\Rightarrow $$|x^2 - 2x|\leq\frac{11}{4}$
$\Rightarrow $$|x^2 - 2x||x-1|\leq\frac{11}{4}|x-1|$
We know
$|x^2-2x|×|x-1|<\epsilon$
So $\frac{11}{4}|x-1|<\epsilon $$\Rightarrow $$|x-1|<\frac{4\epsilon }{1 1} $
Pose
$\alpha =\frac{4\epsilon }{1 1} $
So we can see $|x-1|<\alpha$
Finally:
After the definition of limite we showed $lim_{x\to 1} x^3 - 3x^2 +2x=0$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4031616",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Entropy of extractions A box contains $3$ white and $6$ black balls. We draw $2$ balls consequentially without replacement. Find the entropy of first and second extractions and the entropy for both of them.
| For the first case, when $2$ balls are drawn sequentially, the corresponding pmf is
$$P(B_1B_2)=\frac{6}{9}.\frac{5}{8}=\frac{5}{12}$$
$$P(W_1B_2)=\frac{3}{9}.\frac{6}{8}=\frac{1}{4}$$
$$P(B_1W_2)=\frac{6}{9}.\frac{3}{8}=\frac{1}{4}$$
$$P(W_1W_2)=\frac{3}{9}.\frac{2}{8}=\frac{1}{12}$$
Hence, the entropy is $$H(P) = -\sum\limits_{i}p_{i}\log_2{p_i}=1.825011$$
For the second case, when $2$ balls are drawn sequentially, the corresponding pmf is
$$P(2B) = \frac{{6 \choose 2}}{{9 \choose 2}} = \frac{5}{12}$$
$$P(2W) = \frac{{3 \choose 2}}{{9 \choose 2}} = \frac{1}{12}$$
$$P(1B1W) = \frac{{3 \choose 1}{6 \choose 1}}{{9 \choose 2}} = \frac{1}{2}$$
Hence, the entropy is $$H(P) = -\sum\limits_{i}p_{i}\log_2{p_i}=1.325011$$
Which is as per our intuition that in the second case entropy is lower because the uncertainty is lower (the order of drawing need not be considered).
| {
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"timestamp": "2023-03-29T00:00:00",
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Inequality proof $\frac{n}{a_1a_{2n+1}}<\frac{1}{a_1a_2}+\frac{1}{a_3a_4}+...+\frac{1}{a_{2n-1}a_{2n}}<\frac{n}{a_0a_{2n}}$ Can someone give me an idea about how to prove the inequality?
Mention that $a_0, a_1, a_2, ..., a_{2n}, a_{2n+1}$ is an AP and $0<a_0<a_1<...<a_{2n}<a_{2n+1}$
$$\frac{n}{a_1a_{2n+1}}<\frac{1}{a_1a_2}+\frac{1}{a_3a_4}+...+\frac{1}{a_{2n-1}a_{2n}}<\frac{n}{a_0a_{2n}}$$
Mention that I tried ideas like: $$\frac{1}{a_{2n+1}}<\frac{1}{a_1}<\frac{1}{a_{0}};$$
and then add, but with no result.
A possible solution is:
$$\frac{n}{a_1a_{2n+1}}<\frac{1}{r}(\frac{1}{a_1}-\frac{1}{a_2}+\frac{1}{a_3}-\frac{1}{a_4}+...+\frac{1}{a_{2n-1}}-\frac{1}{a_{2n}})<\frac{n}{a_0a_{2n}}, r=a_n-a_{n-1}
$$
$$\frac{rn}{a_1a_{2n+1}}<\frac{1}{a_1}+\frac{1}{a_3}+...+\frac{1}{a_{2n-1}}-(\frac{1}{a_2}+\frac{1}{a_4}+...+\frac{1}{a_{2n}})<\frac{rn}{a_0a_{2n}}
$$
Let be $S_1>I_1$:
$$\frac{1}{a_1}+\frac{1}{a_1}+...+\frac{1}{a_1}>\frac{1}{a_1}+\frac{1}{a_3}+...+\frac{1}{a_{2n-1}}
$$
Let be $S_2>I_2$:
$$\frac{1}{a_2}+\frac{1}{a_2}+...+\frac{1}{a_2}>\frac{1}{a_2}+\frac{1}{a_4}+...+\frac{1}{a_{2n}}
$$
The following difference can be taken into account in this case?
$$S_1-S_2>I_1-I_2 \rightarrow S_1-S_2=\frac{nr}{a_1a_2}
$$
Then
$$
\frac{nr}{a_1a_{2n+1}}<\frac{nr}{a_1a_2}<\frac{nr}{a_0a_{2n}}
$$
Inequality comes out immediately if this difference occurs. Can give me some advice to solve it or if this approach is wrong, why is it wrong?
Thank you!
| For simplicity let us assume that the common difference of the arithmetic progression is equal to one (the same calculation works in the general case with small modifications).
You correctly noticed that the middle expression can be written as
$$
\frac{1}{a_1a_2}+\frac{1}{a_3a_4}+\cdots+\frac{1}{a_{2n-1}a_{2n}} \\
= \frac{1}{a_1}-\frac{1}{a_2} + \frac{1}{a_3}-\frac{1}{a_4} + \cdots + \frac{1}{a_{2n-1}} - \frac{1}{a_{2n}} \, .
$$
Now the “trick“ is to group this sum as
$$
= \frac{1}{2a_0} - \left( \frac{1}{2a_0}-\frac{1}{a_1}+\frac{1}{2a_2}\right) - \cdots - \left( \frac{1}{2a_{2n-2}}-\frac{1}{a_{2n-1}}+\frac{1}{2a_{2n}}\right) - \frac{1}{2a_{2n}} \, .
$$
and show that all terms in the parentheses are positive. It follows that the whole expression is
$$
< \frac{1}{2a_0}- \frac{1}{2a_{2n}} = \frac{n}{a_0 a_{2n}}
$$
which proves the right inequality.
The proof of the left inequality is similar.
| {
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"timestamp": "2023-03-29T00:00:00",
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Quaternion sandwich simplification gives me a term too many Context:
I'm reading Foundations of Game Engine Development by Eric Lengyel. On the topic of rotations with quaternions it is explained that a vector v can be rotated by being "sandwiched" by quaternions: v' = $qvq^{-1}$. When the quaternion is a unit-quaternion, the conjugate of q (denoted q*) is equal to $q^{-1})$ , such that v' = qvq*
To demonstrate how the sandwich product shown in this equation performs a rotation, the quaternion is split into a vector and a scalar part: q = b + c.
v is treated like a pure quaternion so doesn't have a scalar part.
The book gives the following formula:
$$\mathbf{q}\mathbf{v}\mathbf{q^*} = (\mathbf{b} \times \mathbf{v} + c\mathbf{v} - \mathbf{b} \cdot \mathbf{v})(-\mathbf{b} + c)$$
Problem:
According to the author, the right side can be simplified to:
$$=(c^2 - b^2)\mathbf{v} + 2(\mathbf{v} \cdot \mathbf{b})\mathbf{b} + 2c(\mathbf{b} \times \mathbf{v})$$
But when I try to work this out myself, I have a term more, and I don't know how he got rid of it.
For me it looks like this:
$$ = (\color{red}{\mathbf{b} \times \mathbf{v} \times \mathbf{-b})} -c(\mathbf{v} \times \mathbf{b}) + \mathbf{b}(\mathbf{b} \cdot \mathbf{v}) + c(\mathbf{b} \times \mathbf{v}) + c^{2}\mathbf{v} + c(\mathbf{b} \cdot \mathbf{v}) $$
$$ = \color{red}{\mathbf{v}(\mathbf{b} \cdot \mathbf{-b}) + \mathbf{b}(\mathbf{b} \cdot \mathbf{v})}\color{blue}{ -c(\mathbf{v} \times \mathbf{b})} + \mathbf{b}(\mathbf{b} \cdot \mathbf{v})\color{blue}{ + c(\mathbf{b} \times \mathbf{v})} + c^{2}\mathbf{v} + c(\mathbf{b} \cdot \mathbf{v}) $$
$$ = \color{red}{-b^2\mathbf{v} + \mathbf{b}(\mathbf{b} \cdot \mathbf{v})} + \color{blue}{2c(\mathbf{b} \times \mathbf{v})} + \mathbf{b}(\mathbf{b} \cdot \mathbf{v}) + c^{2}\mathbf{v} + c(\mathbf{b} \cdot \mathbf{v}) $$
$$ = \color{green}{-b^2\mathbf{v}} \color{brown}{+ \mathbf{b}(\mathbf{b} \cdot \mathbf{v})} + 2c(\mathbf{b} \times \mathbf{v}) \color{brown}{+ \mathbf{b}(\mathbf{b} \cdot \mathbf{v})} \color{green}{+ c^{2}\mathbf{v}} + c(\mathbf{b} \cdot \mathbf{v}) $$
$$ = \color{green}{(c^2-b^2)\mathbf{v}} \color{brown}{+ 2\mathbf{b}(\mathbf{b} \cdot \mathbf{v})} + 2c(\mathbf{b} \times \mathbf{v})+ c(\mathbf{b} \cdot \mathbf{v}) $$
I hope the colours make it clear how I simplified. My simplification looks the same except that I end up with an extra term $c(\mathbf{b} \cdot \mathbf{v})$.
Did I make a mistake? Or does the last term equate to zero somehow in this context?
Some help would be greatly appreciated.
| Your mistake comes at the first step. Remember quaternion multiplication is $(a+\mathbf{b})(c+\mathbf{d})=(ac-\mathbf{b}\cdot\mathbf{d})+(c\mathbf{b}+a\mathbf{d}+\mathbf{b}\times\mathbf{d})$ so you should get
\begin{align*}\require{cancel}
&(\mathbf{b} \times \mathbf{v} + c\mathbf{v} - \mathbf{b} \cdot \mathbf{v})(-\mathbf{b} + c)\\
&=[(- \mathbf{b} \cdot \mathbf{v})(c)-((\mathbf{b} \times \mathbf{v} + {\color{blue}{c\mathbf{v}}})\cdot(-\mathbf{b})]\\
&\quad+[(- \mathbf{b} \cdot \mathbf{v})(-\mathbf{b})+c(\mathbf{b}\times\mathbf{v}+c\mathbf{v})+(\mathbf{b}\times\mathbf{v}+c\mathbf{v})\times(- \mathbf{b} \cdot \mathbf{v})(-\mathbf{b})]\\
\end{align*}
you missed my blue term there.
Also, since $\times$ is not associative, $\mathbf{b}\times\mathbf{v}\times(-\mathbf{b})$ is slightly ambiguous.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Solve in the set of integers $2^x +5^x = 3^x + 4^x$
Find the number of integer solutions(both positive and negative) of the equation:
$$2^x +5^x = 3^x + 4^x$$
With induction we see that for $x\geq 2$ we have $$5^x\geq 3^x+4^x$$
It is trivially true for $x=2$ and $x=3$. Now say it is true for $x$ and prove it for $x+2$: $$5^{x+2} \geq 5^2(3^x+4^x) = (3^2+4^2)(3^x+4^x) > 3^{x+2}+4^{x+2}$$
So equation has no solution for $x\geq 2$. Clearly $x=0$ and $x= 1$ are solutions.
If $x<0$ you can write $x=-n$ where $n$ is positive integer and we get $$3^n\cdot 4^n(2^n+5^n) = (3^n+4^n)2^n\cdot 5^n$$
so $ 3^n\mid 40^n$ which is clearly nonsense and thus no solution (for negative $x$).
Any easier solution?
| You can show $5^n\ge 3^n+4^n$ for $n\ge2$ faster: $5^2\ge 3^2+4^2$ and in the induction step
$$ 5^{n+1}=5\cdot 5^n\ge 5\cdot3^n+5\cdot4^n>3^{n+1}+4^{n+1}.$$
An alternative idea for negative $x$ s also by size comparison with $2^{-n}$ being quite big
| {
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"url": "https://math.stackexchange.com/questions/4037810",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Supremum of $x^2y^2(x^2+y^2)$ Let $S$ be the set of all tuples $(x, y) $ with $x, y$ non-negative real numbers satisfying $x+y=2n$,for a fixed $n\in \Bbb{N} $. Then the supremum value of $x^2y^2(x^2+y^2)$ on the set $S$ is
*
*$3n^2$
*$2n^6$
*$4n^6$
*$n^6$
My thinking : take $f(x, y) =x^2y^2(x^2+y^2)$ then express $f(x, y) $ as a function of $x$ by putting $y=2n-x$. Then applying calculus we can find the maximum for some value of $x$.
But calculation will be cumbersome. Is there any smart way?
Thanks in advance.
| By the AM-GM inequality, we have
$$xy \leqslant \left(\frac{x+y}{2}\right) ^2 =n^2,$$
and
$$xy(x^2+y^2) \leqslant \frac{1}{2} \cdot \left(\frac{2xy+(x^2+y^2)}{2}\right)^2=\frac{1}{2} \cdot \left(\frac{(x+y)^2}{2}\right)^2 = 2n^4.$$
Therefore
$$x^2y^2(x^2+y^2) \leqslant 2n^6.$$
| {
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"url": "https://math.stackexchange.com/questions/4040417",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Finding the general solution of a differential equation using two methods. Can somebody please help me in finding the general solution of $2x^3dy=y(y^2+3x^2)dx$ using another method?
My attempt for the first method:
$2x^3y^{-3}dy-3x^2y^{-2}dx=dx$
Let $v=y^{-2}$; $dv=-2y^{-3}dy$
The equation now becomes:
$-x^3dv-3x^2vdx=dx$
$\frac{dv}{dx}+\frac{3v}{x}=-\frac{1}{x^3}$
$P(x)=\frac{3}{x}$, I.F.$=e^{3\int x^{-1}dx}=x^3$
$vx^3=-\int \frac{x^3}{x^3}dx$
$vx^3=-x+C$
$x^3=y^2(C-x)$
| \begin{gather*}
\frac{dy}{dx} =\frac{y^{3}}{2x^{3}} +\frac{3y}{2x}\\
\\
Let\ y=tx\\
\frac{dy}{dx} =t+x\frac{dt}{dx}\\
t+x\frac{dt}{dx} =\frac{t^{3}}{2} +\frac{3t}{2}\\
x\frac{dt}{dx} =\frac{t^{3} +t}{2}\\
\\
\int \frac{2dt}{t^{3} +t} =\int \frac{dx}{x}\\
\\
Now,\ \int \frac{2dt}{t^{3} +t} =\int \frac{2dt}{t\left( t^{2} +1\right)} =\int \frac{2tdt}{t^{2}\left( t^{2} +1\right)}\\
\\
Let\ t^{2} =u\\
2tdt=du\\
\\
\int \frac{2dt}{t^{3} +t} =\int \frac{du}{u( u+1)} =\ln\frac{u}{u+1} =\ln\frac{t^{2}}{t^{2} +1} +C\\
\\
So,\ form\ the\ original\ differential\ equation,\\
\ln\frac{t^{2}}{t^{2} +1} +C_{1} =\ln x+C_{2}\\
\ln\frac{y^{2}}{y^{2} +x^{2}} -\ln x=C\\
\\
So,\ the\ general\ solution\ is\ \\
y^{2} =e^{c} x\left( x^{2} +y^{2}\right)
\end{gather*}
Hope this answers your question!
| {
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Solution of a trigonometric linear equation with the method of the added angle This morning I have done for my students of an high school an exercise with the method of the added angle. I not write all the steps but the principals.
The equation is:
$$\bbox[5px,border:3px solid #FF7F50]{-2\sin x+\cos x = 1} \tag 1$$
Surely $x=2\Bbb Z \pi$ it is solution of the $(1)$. In fact it is an identity. Being $$\tan \varphi=-\frac 12 \implies \varphi =-\arctan \frac 12, \quad A=\sqrt{5}$$
Now the $(1)$ becomes
$$\bbox[5px,border:3px solid #AA7F50]{\sqrt 5 \sin \left(x-\arctan \frac 12\right)=1} \tag 2$$
Hence I will have the other two solutions:
$$x=\arctan \frac 12+\arcsin\left( \frac{\sqrt{5}}5\right)+2\Bbb Z\pi$$ and $$ \quad x=\pi +\arctan \frac 12-\arcsin\left( \frac{\sqrt{5}}5\right)+2\Bbb Z\pi \tag 3$$
The solution of my textbook are $x=k\pi$ and $x=\alpha+2k\pi$ where $\cos \alpha =-3/5$ and $\sin \alpha =-4/5$.
Please, do can be that the $(3)$ are equivalent to $x=2k\pi$ and $x=\alpha+2k\pi$ where $\cos \alpha =-3/5$ and $\sin \alpha =-4/5$?
| Putting together the two comments we have that the initial equation is equivalent to
$$\sqrt 5 \sin\left(x - \arctan \frac12\right) = -1$$
Leading to the solutions
$$ x = \arctan \frac12 - \arcsin \frac{\sqrt 5}5 + 2k \pi = 2k \pi$$
and
$$x = \pi + \arcsin \frac{\sqrt 5}5 + \arctan \frac12 + 2k \pi=\underbrace{\pi + 2 \arcsin \frac{\sqrt 5}5}_{\alpha}+ 2 k \pi.\tag{1} \label{1}$$
Now let $\alpha = \pi + 2\arcsin \frac{\sqrt 5}5$. We get
$$\sin \alpha = -\sin\left( 2\arcsin \frac{\sqrt 5}5\right) = -2 \sin\left( \arcsin \frac{\sqrt 5}5\right)\cos\left(\arcsin\frac{\sqrt 5}5\right)=-\frac45,$$
and similarly for the cosine of $\alpha$.
Note that $k \pi$ is not a solution for odd $k$.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Radius of Convergence of $(1-x)^{1/4}$ This is my first post so I am very new to MathJax formatting - i apologize in advance for the messy formatting
This equation becomes $$\sum_{n=2}^\infty \frac{(4n-5) x^n}{(4^n)n!}$$
The textbook says the ratio test would be:
$$\lim_{n\to \infty} \left\lvert\frac{(4n-1)(4n-5)(x^{n+1})(4^n)n!}{(4n-5)(x^n)(4^{n+1})(n+1)n!}\right\rvert$$
But I thought it would be:
$$\lim_{n\to \infty} \left\lvert\frac{(4n-1)(x^{n+1})(4^n)n!}{(4n-5)(x^n)(4^{n+1})(n+1)!}\right\rvert$$
So no extra $(4n-5)$ - but that would result in an addition $(n+1)$ in denominator (because of the expanded factorial) and the radius of convergence would now be infinite (and not $|x| < 1$ as above fraction suggests).
Thanks
| The generalized binomial series is
$$ (1 \ + \ x)^p \ \ = \ \ \sum_{n=0}^\infty \ \left( \begin{array}{c} p \\ n \end{array} \right) \ x^n \ \ , $$
which becomes $$ (1 \ - \ x)^p \ \ = \ \ \sum_{n=0}^\infty \ (-1)^n \ \left( \begin{array}{c} p \\ n \end{array} \right) \ x^n \ \ $$
for the base $ \ (1 \ - \ x) \ $ . Writing out the first few terms of the series with the fractional binomial coefficients for $ \ p \ = \ \frac14 \ $ gives us
$$ (1 \ - \ x)^{1/4} \ \ = \ \ 1 \ · \left( \begin{array}{c} \frac14 \\ 0 \end{array} \right) \ x^0 \ \ + \ \ -1 \ · \left( \begin{array}{c} \frac14 \\ 1 \end{array} \right) \ x^1 \ \ + \ \ 1 \ · \left( \begin{array}{c} \frac14 \\ 2 \end{array} \right) \ x^2 \ \ + \ \ -1 \ · \left( \begin{array}{c} \frac14 \\ 3 \end{array} \right) \ x^3 \ \ + \ \ldots $$
$$ = \ \ 1 \ · \frac{ 1}{0!} \ x^0 \ \ - \ \ 1 \ · \frac{ \frac14 }{1!} \ x^1 \ \ + \ \ 1 \ · \frac{ \frac14 · \left( \frac{-3}{4} \right)}{2!} \ x^2 \ \ - \ \ 1 \ ·\frac{ \frac14 · \left( \frac{-3}{4} \right) · \left( \frac{-7}{4} \right) }{3!} \ x^3 \ \ + \ \ldots $$
$$ = \ \ 1 \ \ - \ \ \frac{ 1 }{4 \ · \ 1!} \ x^1 \ \ + \ \ \frac{ 1 · ( -3)}{4^2 \ · \ 2!} \ x^2 \ \ - \ \ \frac{ 1 · (-3) · (-7) }{4^3 \ · \ 3!} \ x^3 \ \ + \ \ldots $$
There are various ways that people choose for writing the general term; my guess is that this author(s) for your textbook decided to leave the first two terms outside of the general summation to start it at $ \ n \ = \ 2 \ $ (this has no effect on the interval of convergence):
$$ (1 \ - \ x)^{1/4} \ \ = \ \ 1 \ \ - \ \ \frac14 \ x \ \ - \ \
\sum_{n=2}^\infty \ \frac{3 \ · \ 7 \ · \ · \ldots \ · \ (4n-5) }{4^n \ · \ n!} \ x^n \ \ . $$
This is a product of factors in arithmetic progression in the numerator of the general term; that is what must be used in the Ratio Test. Almost all of them will cancel in the calculation, but what will remain is
$$ \lim_{n\to \infty} \ \left\lvert \ \frac{(4[n+1]-5)(4n-5)(x^{n+1})}{4^{n+1} \ · \ (n+1)!} \ · \ \frac{4^n \ · \ n!}{(4n-5)(x^n) } \ \right\rvert \ \ = \ \ \lim_{n\to \infty} \ \left\lvert \ \frac{(4n-1)(x)}{4 \ · \ (n+1)} \ \right\rvert \ \ , $$
which will yield the radius of convergence $ \ R \ = \ 1 \ $ that obtains for the series representing $ \ (1 \ + \ x)^p \ $ . [The generalized binomial series is exactly the same as you would get by deriving a Maclaurin series for these functions.]
| {
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$(x^2-9)^{(3x+5)}=(x-3)^{(x-1)}(x+3)^{(x-1)}$ What is $x$? $(x^2-9)^{(3x+5)}=(x-3)^{(x-1)}(x+3)^{(x-1)}$
$(x^2-9)^{(3x+5)}=(x^2-9)^{(x-1)}$
$3x+5=x-1$
$x=-3$
But when I try to use WolfarmAlpha the integer solution is $3$ instead of $-3$. The numerical solution is $x\approx 3.1622776600...$
So I tried to do other things, such as making sure the base is larger than $0$.
$x^2-9>0$
$x<-3$ or $x>3$
But I still didn't obtain the $x$.
So what is $x$?
| The domain is $x\ge3$ due to the power functions involved, and it may easier to examine the equation with $x=t+3$, which leads to
$$(t^2+6t)^{t+2}[( t^2+6t)^{2(t+6)} -1]=0$$
The first factor produces the solution $t=0$, or $x=3$, and the second factor reduces to $t^2+6t=1$,
which yields $t=\sqrt{10}-3$, or $x=\sqrt{10}$.
| {
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Looking for general approach to find number of $(x,y,z)$ so that $x+y=z$ How many $(x,y,z)$ exist so that $x+y=z$ holds for $x<y<z\le10$ ($x,y,z\in \mathbb{Z}^+$ ) ?
$1)15\quad\quad\quad\quad\quad\quad2)20\quad\quad\quad\quad\quad\quad3)40\quad\quad\quad\quad\quad\quad4)45\quad\quad\quad\quad\quad\quad5)50\quad\quad\quad\quad\quad\quad$
I solved this problem with counting all the possibilities: $z$ should be at least $3$ and for $z=3$ the only possibility is $3=1+2$ I continued to count for other values of $z$:
$$z=3\to1\quad z=4\to1\quad z=5\to2\quad z=6\to2\quad z=7\to3\quad z=8\to3\quad z=9\to4\quad z=10\to4$$
So the answer is $1+1+2+2+3+3+4+4=20$.
But I wonder is it possible to solve this problem without counting all the possibilities like I did? for example I don' know what to do if it was $x<y<z\le100$.
| One of the ways would be to first look at the number of positive integer solutions of $x+y \leq 10$. Taking $x = a + 1, y = b +1$, it is equivalent to,
Number of non-negative solutions of $a + b + c = 8$ and that is $\displaystyle {10 \choose 2} = 45$. We discard solutions where $a = b = 0$ (as $x+y \gt 2)$ and we discard in general all solutions where $a = b $ (i.e $x = y$). That gives us $40$ solutions where $x \ne y$. This includes solutions where $x \gt y$ and $y \gt x$ and due to symmetry, the solutions that we are interested in is $\frac{40}{2} = 20$.
| {
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Simple method to prove $a^3+b^3=1$ has no integer solutions if $ab\neq 0$ Any simple method to prove $a^3+b^3=1$ has no integer solutions if $ab\neq 0$ that does not involve Fermat's last theorem?
| Well $a^3 + b^3 = (a+b)(a^2 -ab + b^2)=1$
The only factors of $1$ are either $1\cdot 1$ or $-1 \cdot (-1)$.
So we must have $a+b = a^2 - ab + b^2 =\pm 1$.
And that just won't work unless $ab=0$.
It would follow that $b= \pm 1 -a$
And therefore $a^2 - a(\pm 1 - a) + (\pm 1 - a) = \pm 1$ so
$a^2+a^2 \mp a -a \pm 1 = \pm 1$
SO either $2a^2 -2a+1=1$ or $2a^2 -1 = -1$ so
$2a(a-1) = 0;b= 1-a$ or $2a^2 = 0;b = -1-a$.
So we have either $a=0$ and $b=1$. Of $a =1$ and $b =0$. Or $a=0$ and $b = -1$.
In all three cases we must have $ab = 0$.
| {
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Show that $a^2 - 2021b^2 = 13$ has no integer solutions
Show that $a^2 - 2021b^2 = 13$ has no integer solutions.
I'm no number theory expert, but here's what I've thought of so far. If there exist $a,b$ that satisfy the above equation, then there must also exist $a,b$ that satisfy $$a^2 \bmod 4 = b^2\bmod 4 + 1$$
A square is $0$ or $1 \bmod 4$, so we get $a^2\equiv 1\bmod 4$ and $b^2 \equiv 0 \bmod 4$. Thus, $2|b$ and $a^2 = 4k+1$. I do not see a contradiction yet. What else should I try?
Also tried to use the fact that squares have units digit only $0,1,4,5,6,9$, so if $a,b$ exist that satisfy the equation, we must have $a^2$ ending in $\{0,1,4,5,6,9\} \cap \{3,4,7,8,9,2\} = \{4,9\}$. Moreover, we must have $b^2$ ending in $\{0,1,4,5,6,9\} \cap \{7, 8, 1,2,3,6\} = \{1,6\}$.
If $b^2$ ends in $1$, then $a^2$ ends in $4$ - but we just saw that $a^2 = 4k+1$, so this is not a possibility.
If $b^2$ ends in $6$, then $a^2$ ends in $9$. How do I contradict this? Thank you!
| A slightly different alternative that does not require separate consideration of zero residues:
Because any odd number $n$ is a product of only odd primes, quadratic reciprocity implies $(p|n)=(n|p)$ for any $4m+1$ prime $p$. Thus $(13|2021)=(2021|13)=(6|13)=-1$, and then we can't have $a^2\equiv13\bmod 2021$.
| {
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Determine all $f:\Bbb Z \to \Bbb Z$ st $f(x^3+y^3+z^3)=f(x)^3+f(y)^3+f(z)^3$. (Vietnamese TST 2005) Determine all $f:\Bbb Z \to \Bbb Z$ st $$f(x^3+y^3+z^3)=f(x)^3+f(y)^3+f(z)^3$$ for all $x,y,z\in\Bbb Z$.
Source : Vietnamese TST 2005
I'm sure that the only solutions are the zero function, the identity, and the negative of the identity. First, when $x=y=z=0$ we have $f(0)=3f(0)^3$ so $f(0)=0$ is the only integer solution. Next if $y=-x$ and $z=0$ then $f(-x)=-f(x)$ so $f$ is odd. Then $x=y=1$ and $z=-1$ give $f(1)^3=f(1)$ so $f(1)=0,1,-1$.
Since $f$ is a soln iff $-f$ is, we can take $f(1)=0,1$. Then $f(2)=f(1^3+1^3+0^3)=0,2$. Then $f(3)=f(1^3+1^3+1^3)=0,3$. Also $f(4^3-3^3-3^3)=f(10)=f(1^3+1^3+2^3)$, so $f(4)=0,4$. Then $f(5^3-4^3-4^3)=f(-3)$ so $f(5)=0,5$. Then $f(6)=f(2^3-1^3-1^3)=0,6$. Then, $f(7)=f(2^3-1^3+0^3)=0,7$. Then $f(8)=f(2^3+0^3+0^3)=0,8$. Then $f(9)=f(2^3+1^3+0^3)=0,9$ and $f(10)=f(2^3+1^3+1^3)=0,10$. Etc...
There should be a smart way to make induction, but I haven't found that. May be a nice identity. This is the closest to the claim that $n^3$ is the sum of five cubes of integers with absolute values $<|n|$:
$$n^3=6^3+(n-2)^3+(n-3)^3-(n-6)^3-(n-11)^3+(n-12)^3.$$
At least the identity above gives me $f(11)=0,11$ and $f(12)=0,12$.
| Using hgmath's comment, here is a way to write $n^3$ as a sum of five cubes of integers with absolute values $<|n|$ for $n=2k$ and $k\ge 8$. We write
$$(2k)^3=(2k-4)^3+(k+7)^3-(k-9)^3-10^3-2^3.$$
For this to work, we need to have $f(0)$, $f(1)$, $f(2)$, ..., $f(15)$. I already got $f(0)$ until $f(12)$.
For $f(13)$ and $f(15)$ we use hgmath's comment (note that hgmath's comment is good for $k\geq 4$, but I already found until $f(12)$). For $f(14)$, we can use
$14^3=12^3+10^3+2^3+2^3$.
| {
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Big O for error terms The following link from wikipedia explains the Big O notation really good. I have only one problem, which is to formalize the usage of Big O notation for error terms in polynomials. In the example give here we have
$$
e^x=1+x+\frac{x^2}{2!}+\frac{x^3}{3!}+\frac{x^4}{4!}+...=1+x+\frac{x^2}{2!}+\mathcal{O}(x^3)=1+x+\mathcal{O}(x^2)
$$
as $x\rightarrow 0$. Now we find a similar notation also for the error terms in taylor polynomials. I would like to understand why this is right just formaly.
1.) Why is $\frac{x^3}{3!}+\frac{x^4}{4!}+... = \mathcal{O}(x^3)$
Is this because
$$
\frac{x^3}{3!}\frac{1}{x^3}+\frac{x^4}{4!}\frac{1}{x^3}+\frac{x^5}{5!}\frac{1}{x^3}+...=\frac{1}{3!}+\frac{x}{4!}+\frac{x^2}{5!}+...\leq M, \quad x\rightarrow 0
$$
for some $M$ that has to be bigger than $\frac{1}{3!}$ or can one show this in a different way formaly?
2.) Why is $\frac{x^2}{2!}+\mathcal{O}(x^3) = \mathcal{O}(x^2)$?
I appreciate your help! :)
| The intuitive answer to why
$$\frac{x^3}{3!}+ \frac{x^4}{4!} + \cdots = \mathcal{O}(x^3)$$
as $x \to 0$ is that $x^3$ is the term that goes to $0$ at the lowest rate. All higher-order terms go to $0$ faster, so the expression will "at worst" approach $0$ like $x^3$. More precisely, the reason in that
$$|x^3| \geq C|x^4 + x^5 + \cdots |$$
for $x$ small enough. To see this, observe that for $|x|<\frac{1}{2}$, we have
$$\left| x^4 + x^5 + \cdots \right| = \left| \frac{x^4}{1-x} \right| = \left| \frac{x}{1-x} \right| |x^3| \leq C |x^3|$$
since $\frac{x}{1-x}$ is bounded for $|x| < \frac{1}{2}$.
| {
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how to prove this trig inequality: $\cos x \cos y + z \sin x \sin y \geq \cos \left( \sqrt{x^2 + y^2 - 2 x y z} \right)$ The following inequality has come up in my work. I'm trying to prove that it is true:
$$
\begin{equation}
\cos x \cos y + z \sin x \sin y \geq \cos \left( \sqrt{x^2 + y^2 - 2 x y z} \right)
\end{equation}
$$
where
$$
\begin{align}
-\pi \leq{} &x \leq \pi \\
-\pi \leq{} &y \leq \pi \\
-1 \leq{} &z \leq 1
\end{align}
$$
The LHS of the inequality is derived from the spherical law of cosines, the RHS is derived from the law of cosines, and $z$ represents the cosine of another variable. Note that the inequality holds with equality when $z=1$.
I have also put the inequality in the following form using the product-to-sum trig identity, and rearranging the RHS:
$$
\begin{equation}
(1+z) \cos(x-y) + (1-z) \cos(x+y) \geq
2 \cos \left( \sqrt{(x-y)^2 + 2 x y (1-z)} \right)
\end{equation}
$$
I have also tried other things like applying Cauchy-Schwarz to the LHS, and applying arccosine to both sides, but I can't seem to prove that the inequality is true.
I am quite certain that the inequality holds because I have performed a Monte Carlo simulation (code shown below) which shows that the inequality holds for randomly sampled values.
python Monte Carlo random sampling code
import numpy as np
n = 10**6
x = np.random.uniform(-np.pi, np.pi, n)
y = np.random.uniform(-np.pi, np.pi, n)
z = np.random.uniform(-1, 1, n)
lhs = np.cos(x)*np.cos(y) + z*np.sin(x)*np.sin(y)
rhs = np.cos(np.sqrt(x**2 + y**2 - 2*x*y*z))
check = lhs>=rhs
print('inequality is true', np.sum(check), '/', n, 'times')
| Replacing $(x, z)$ by $(-x, -z)$ or $(y, z)$ by $(-y, -z)$ does not change the inequality, therefore we can assume that
$$
0 \le x, y \le \pi \, , \, -1 \le z \le 1 \, .
$$
Case 1: $x+y \le \pi$. The solution for this case is inspired by Proving that $\cosh^{-1} (\cosh(x)\cosh(y) ) \geq \sqrt{x^2 + y^2}$ and the CAT(0) inequality.
The function $f(u) = \cos(\sqrt u)$ is convex on the interval $0 \le u \le \pi^2$, this is straightforward to verify, e.g. by calculating the second derivative.
We can assume that $y \le x$. The convexity condition with $\lambda = (1+z)/2$ and $1-\lambda = (1-z)/2$ then gives
$$
\begin{align}
\cos x \cos y + z \sin x \sin y &=
\frac{1+z}2 \cos(x-y) + \frac{1-z}2 \cos(x+y) \\
&= \frac{1+z}2 f((x-y)^2) + \frac{1-z}2 f((x+y)^2) \\
&\ge f \left( \frac{1+z}2 (x-y)^2 + \frac{1-z}2 (x+y)^2\right) \\
&= f( x^2+y^2-2xyz) \\
&= \cos \left( \sqrt{x^2 + y^2 - 2 x y z} \right) \, .
\end{align}
$$
$f$ is in fact strictly convex, so that equality holds exactly in the following cases:
*
*$\lambda = 0$ or $1$, that is if $z = \pm 1$.
*$(x-y)^2 = (x+y)^2$, that is if $x=0$ or $y=0$.
Case 2: $\pi < x+y \le 2\pi$. This case can be reduced to the first case, credit goes to ipst. I have only simplified the solution a bit.
Setting $x' = \pi - x$ and $y' = \pi - y$ we can apply the first case to $(x', y', z)$:
$$
\begin{align}
\cos x \cos y + z \sin x \sin y &= \cos x' \cos y' + z \sin x' \sin y' \\
&\ge \cos \left( \sqrt{x'^2 + y'^2 - 2 x' y' z} \right) \\
&\overset{(*)}{\ge} \cos \left( \sqrt{x^2 + y^2 - 2 x y z} \right) \, .
\end{align}
$$
The last inequality $(*)$ needs to be proved. For $\alpha, \beta \in [0, 2 \pi]$ we have
$$
\cos \alpha \ge \cos \beta \iff \alpha \le \beta \le 2\pi -\alpha \, .
$$
Therefore we have to show that
$$
\sqrt{x'^2 + y'^2 - 2 x' y' z} \overset{(1)}{\le} \sqrt{x^2 + y^2 - 2 x y z}\overset{(2)}{\le} 2 \pi - \sqrt{x'^2 + y'^2 - 2 x' y' z} \, .
$$
The left part $(1)$ follows from
$$
(x^2 + y^2 - 2 x y z) - (x'^2 + y'^2 - 2 x' y' z) = 2 \pi(x+y-\pi)(1-z) \ge 0
$$
and the right part $(2)$ follows from
$$
\sqrt{x^2+y^2-2xyz} + \sqrt{x'^2+y'^2-2x'y'z}
\le (x+y) + (x'+y') = 2 \pi \, .
$$
Equality holds in the following cases:
*
*$z = \pm 1$.
*$x'y' = 0$ and $x+y= \pi$, that is if $(x, y) = (\pi, 0)$ or $(x, y) = (0, \pi)$.
| {
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How to evaluate $ \frac{1}{3} + \frac{2}{3^{2} } + \frac{1}{3^{3}} + \frac{2}{3^{4}} + \frac{1}{3^{5}} + ...$? I was given the series:
$$ \frac{1}{3} + \frac{2}{3^{2} } + \frac{1}{3^{3}} + \frac{2}{3^{4}} + \frac{1}{3^{5}} + ...$$
Making some observations I realized that the $ a_{n} $ term would be the following:
$$ a_{n} = \frac{1}{ 3^{n} } \left(\frac{3}{2}+ \frac{(-1)^{n}}{2} \right)$$
What I wanted to do is to find the result of the series, so the answer would be:
$$\sum_{n=1}^{ \infty } \frac{1}{ 3^{n} } \left(\frac{3}{2}+ \frac{(-1)^{n}}{2} \right) = \frac{3}{2} \sum_{n=1}^{ \infty } \left[ \frac{1}{3^{n} } \right]+ \sum_{n=1}^{ \infty } \left[\frac{1}{3^{n} } \frac{(-1)^{n}}{2}\right]$$
I can tell that the first term is convergent because it is a geometric series, in fact, the result is $\frac{3}{4}$. However, I have no clue in how to solve the second term series. I should say that the series given in the beginning is convert and its result is 5/8. How to arrive to it is a mystery to me.
| There is a much simpler way to solve this.
You know how to compute the geometric series
$$\frac{1}{3} + \frac{1}{3^2}+ \frac{1}{3^3}+ \frac{1}{3^4}+ \cdots = \frac{1}{2}$$
you also know how to compute the geometric series
$$\frac{1}{9} + \frac{1}{9^2}+ \frac{1}{9^3}+ \frac{1}{9^4}+ \cdots= \frac{1}{8}$$
But this can be rewritten in terms of $3^2$, so that you get
$$\frac{1}{3^2} + \frac{1}{3^4}+ \frac{1}{3^6}+ \frac{1}{3^8}+ \cdots$$
Now the answer to your question is simply $\frac{1}{2} +\frac{1}{8}$.
| {
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Computing $\int_0^\infty \frac{\ln x}{(x^2+1)^2}dx$ I'm trying to compute
$$I=\int_0^\infty \frac{\ln x}{(x^2+1)^2}dx$$
The following is my effort,
$$I(a)=\int_0^\infty\frac{\ln x}{x^2+a^2}dx$$
Let $x=a^2/y$ so that $dx=-(a^2/y^2)dy$ which leads to
$$I(a)=\int_0^\infty \frac{\ln(a/y)}{a^2+y^2}dy=\int_0^\infty\frac{\ln a}{y^2+a^2}dy-I(a)$$
$$I(a)=\frac{1}{2}\int_0^\infty\frac{\ln a}{y^2+a^2}dy=\frac{1}{2}\frac{\ln a}{a}\arctan\left( \frac{y}{a}\right)_0^\infty=\frac{\ln a}{a}\frac{\pi }{4}$$
Differentiating with respect to $a$ then
$$\frac{dI(a)}{a}=-2aI'(a)=\frac{\pi}{4}\left( \frac{1}{a^2}-\frac{\ln a}{a^2}\right)$$
where $$I'(a)=\int_0^\infty \frac{\ln y}{(y^2+a^2)^2}dx$$
$$I'(a)=\frac{\pi}{-8a}\left( \frac{1}{a^2}-\frac{\ln a}{a^2}\right)$$
$$I'(a=1)=-\frac{\pi}{8}$$
But the correct answer is $-\pi/4$.
Can you help me figure where I mistake? Please give some method if there is which is much better than what I have done?
| Contour Integration Approach
Consider the keyhole contour, $\gamma$, that goes from $0$ to $\infty$ just above the positive real axis, circles the complex plane counterclockwise, then returns to $0$ just below the positive real axis.
Now compute the integral
$$\newcommand{\Res}{\operatorname*{Res}}
\begin{align}
\int_\gamma\frac{\log(z)^2}{\left(1+z^2\right)^2}\,\mathrm{d}z
&=\int_0^\infty\frac{\log(x)^2}{\left(1+x^2\right)^2}\,\mathrm{d}x
-\int_0^\infty\frac{(\log(x)+2\pi i)^2}{\left(1+x^2\right)^2}\,\mathrm{d}x\tag1\\
&=-4\pi i\int_0^\infty\frac{\log(x)}{\left(1+x^2\right)^2}\,\mathrm{d}x
+4\pi^2\int_0^\infty\frac1{\left(1+x^2\right)^2}\,\mathrm{d}x\tag2\\
&=2\pi i\left(\color{#C00}{\Res_{z=i}\left(\frac{\log(z)^2}{\left(1+z^2\right)^2}\right)}+\color{#090}{\Res_{z=-i}\left(\frac{\log(z)^2}{\left(1+z^2\right)^2}\right)}\right)\tag3\\
&=2\pi i\left(\color{#C00}{-\frac\pi4+i\frac{\pi^2}{16}}+\color{#090}{\frac{3\pi}4-i\frac{9\pi^2}{16}}\right)\tag4\\[9pt]
&=i\pi^2+\pi^3\tag5
\end{align}
$$
Explanation:
$(1)$: integrate along the contour above and below the positive real axis
$\phantom{\text{(1):}}$ the integral along the huge circle vanishes
$\phantom{\text{(1):}}$ the integral near zero also vanishes
$(2)$: expand $(\log(x)+2\pi i)^2$ and combine
$(3)$: the original integral is $2\pi i$ times the sum of the residues of the integrand
$(4)$: compute the residues
$(5)$: simplify
Thus, we get not only that
$$
\int_0^\infty\frac{\log(x)}{\left(1+x^2\right)^2}\,\mathrm{d}x=-\frac\pi4\tag6
$$
but also that
$$
\int_0^\infty\frac1{\left(1+x^2\right)^2}\,\mathrm{d}x=\frac\pi4\tag7
$$
A Real Approach
This is similar to what is attempted in the question.
$$
\begin{align}
\int_0^\infty\frac{\log(x)}{x^2+a^2}\,\mathrm{d}x
&=\frac1a\int_0^\infty\frac{\log(x)+\log(a)}{x^2+1}\,\mathrm{d}x\tag8\\
&=\frac{\log(a)}a\int_0^\infty\frac1{x^2+1}\,\mathrm{d}x\tag9\\
&=\frac{\pi\log(a)}{2a}\tag{10}
\end{align}
$$
Explanation:
$\phantom{1}\text{(8)}$: substitute $x\mapsto ax$
$\phantom{1}\text{(9)}$: substituting $x\mapsto1/x$ shows that $\int_0^\infty\frac{\log(x)}{x^2+1}\,\mathrm{d}x=0$
$(10)$: evaluate the integral
Taking the derivative of $(10)$ in $a$ and dividing by $-2a$ gives
$$
\int_0^\infty\frac{\log(x)}{\left(x^2+a^2\right)^2}\,\mathrm{d}x=\frac\pi4\frac{\log(a)-1}{a^3}\tag{11}
$$
Plug in $a=1$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4069094",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 8,
"answer_id": 0
} |
Existence of a particular function Given positive integers $n,d,\ell\ge1$ such that $n\ge 1+(d-1)\ell$, I must find a a function
$$
a=a(n,d,\ell),
$$
such that
\begin{align*}
0< a(n,d,\ell)&<1,\\
\frac{n}{n-(d-1)(\ell-1)}a(n-1,d-1,\ell)&\le a(n,d,\ell),
\end{align*}
for all $n,d,\ell\ge1,\ n\ge1+(d-1)\ell.$
I've been able to solve only the case $\ell=1$. In such a case, the function
$$
a(n,d,1)=\frac{n-d}{n-d+1}
$$
satisfy the requests.
Any help is appreciated.
| Consider the function $a(n, d, l) = \frac {1} {b(n-d, \ell)} \cdot \frac {\Gamma(n+1)} {\Gamma(d+1)}$ where $b(c, l) = \Gamma(3+ \frac {c \cdot \ell} {\ell-1})$. This function satisifies the constraint since:
\begin{align*}
\frac{n}{n-(d-1)(\ell-1)}a(n-1,d-1,\ell)&=\frac{n}{n-(d-1)(\ell-1)} \cdot \frac {1} {b(n-1-(d-1), \ell)} \cdot \frac {\Gamma(n)} {\Gamma(d)} \\
&\le \frac{n}{1 + (d-1)\ell-(d-1)(\ell-1)} \cdot \frac {1} {b(n-d, \ell)} \cdot \frac {\Gamma(n)} {\Gamma(d)} \ \textit{(Since $n\ge 1+(d-1)\ell$)}\\
&\le \frac{n}{1 + (d-1)} \cdot \frac {1} {b(n-d, \ell)} \cdot \frac {\Gamma(n)} {\Gamma(d)} \\
&\le \frac{n}{d} \cdot \frac {1} {b(n-d, \ell)} \cdot \frac {\Gamma(n)} {\Gamma(d)} \\
&\le \frac {1} {b(n-d, \ell)} \cdot \frac {\Gamma(n+1)} {\Gamma(d+1)} \\
&\le a(n, d, l)
\end{align*}
Clearly $a(n,d,l)>0$ since gamma function is always positive for positive argument. To prove $a(n,d,l)<1$, let us prove the following claim:
\begin{equation*}
n < 2 + \frac {(n-d) \cdot \ell} {\ell-1}
\end{equation*}
Proof of claim:
\begin{align*}
n &\ge 1 + (d-1)\ell \\
n-n\ell &\ge 1 + d\ell -\ell -n\ell \\
n(1-\ell) &\ge 1-\ell - (n-d)\ell \\
n &\le 1- \frac {(n-d)\ell} {1-\ell} \\
n &\le 1+ \frac {(n-d)\ell} {\ell-1} \\
\implies n &< 2 + \frac {(n-d)\ell} {\ell-1}
\end{align*}
From the above claim,we get $n +1 < 3 + \frac {(n-d)\ell} {\ell-1}$ which implies that $\frac {\Gamma(n+1)} {b(n-d, l)} = \frac {\Gamma(n+1)} {\Gamma(3 + \frac {(n-d) \cdot \ell} {\ell-1})} < 1$. Hence $a(n, d, l) = \frac {1} {b(n-d, \ell)} \cdot \frac {\Gamma(n+1)} {\Gamma(d+1)} < 1$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Find the value of $T=\mathop {\lim }\limits_{n \to \infty } {\left( {1+ \frac{{1+\frac{1}{2}+ \frac{1}{3}+ . +\frac{1}{n}}}{{{n^2}}}} \right)^n}$ I am trying to evaluate
$$T = \mathop {\lim }\limits_{n \to \infty } {\left( {1 + \frac{{1 + \frac{1}{2} + \frac{1}{3} + \cdots + \frac{1}{n}}}{{{n^2}}}} \right)^n}.$$
My solution is as follow
$$T = \mathop {\lim }\limits_{n \to \infty } {\left( {1 + \frac{{1 + \frac{1}{2} + \frac{1}{3} + \cdots + \frac{1}{n}}}{{{n^2}}}} \right)^n} \Rightarrow T = {e^{\mathop {\lim }\limits_{n \to \infty } n\left( {1 + \frac{{1 + \frac{1}{2} + \frac{1}{3} + \cdots + \frac{1}{n}}}{{{n^2}}} - 1} \right)}} = {e^{\mathop {\lim }\limits_{n \to \infty } \left( {\frac{{1 + \frac{1}{2} + \frac{1}{3} + \cdots + \frac{1}{n}}}{n}} \right)}}$$
$$T = {e^{\mathop {\lim }\limits_{n \to \infty } \left( {\frac{1}{n} + \frac{1}{{2n}} + \frac{1}{{3n}} + \cdots + \frac{1}{{{n^2}}}} \right)}} = {e^{\left( {0 + 0 + \cdots + 0} \right)}} = {e^0} = 1$$
The solution is correct but I presume my approach $\mathop {\lim }\limits_{n \to \infty } \left( {\frac{{1 + \frac{1}{2} + \frac{1}{3} + \cdot + \frac{1}{n}}}{n}} \right) \Rightarrow \mathop {\lim }\limits_{n \to \infty } \left( {\frac{1}{n} + \frac{1}{{2n}} + \frac{1}{{3n}} + \cdot + \frac{1}{{{n^2}}}} \right) = 0$ is wrong.
Is there any generalized method
| If
$$T = \mathop {\lim }\limits_{n \to \infty } {\left( {1 + \frac{{1 + \frac{1}{2} + \frac{1}{3} + . + \frac{1}{n}}}{{{n^2}}}} \right)^n}$$
then
$$\log T = \mathop {\lim }\limits_{n \to \infty } n\log{\left( {1 + \frac{H_n}{{{n^2}}}} \right)}\le \mathop {\lim }\limits_{n \to \infty } \frac{H_n}{n}.$$
It can be easily shown that this limit tends to $0$ (for example by Cesaro-Stolz.) So
$$\log T \le 0 \Rightarrow \log T = 0 \Rightarrow T=1.$$
Addendum: Showing that $\mathop {\lim }\limits_{n \to \infty } \frac{H_n}{n}\to0$.
Cesaro-Stolz:
We may set $a_n=H_n$ and $b_n={n}$. (We can apply Cesaro-Stolz since $(n)_{n\ge 1}$ is strictly monotone and divergent. The limit
$$\lim_{n\to\infty}\frac{H_{n+1}-H_n}{n+1-n}=\lim_{n\to\infty}\frac{1}{n+1},$$
exists so
$$\lim_{n\to\infty}\frac{H_n}{n}=\lim_{n\to\infty}\frac{1}{n}=0.$$
Bounding $H_n$
In this question a user proved $H_n<\log(n)+1.$ That means that
$$\lim_{n\to\infty}\frac{H_n}{n}\le\lim_{n\to\infty}\frac{\log(n)+1}{n}=0.$$
Since our oringal limit is non-negative we can conclude $\lim_{n\to\infty}\frac{H_n}{n}=0$.
l'hopital
One may also stick to the classic l'hopital. Differentiating numerator and denominator yields
$$\mathop {\lim }\limits_{n \to \infty } \frac{H_n}{n}=\mathop {\lim }\limits_{n \to \infty } \frac{\pi^2/6-H_n^{(2)}}{1}.$$
Since $H_n^{(2)}\to\pi^2/6$ for $n\to\infty$, the overall limit equals zero too.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4071409",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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} |
Prove that the sum $2^0+2+2^2+\cdots+2^{p-1}(2^p-1)$ is $2^{p-1}(2^p-1)$ Attempt: In an attempt to find the sum of the series given that $2^{p-1}$ is prime,
$$1+2+2^2+\cdots+2^{p-2}+2^{p-1}+(2^p-1)+2(2^p-1)+2^2(2^p-1)+\cdots+2^{p-1}(2^p-1)$$
A teacher grouped elements as follows,
$$(1+2^p-1)+2(1+2^p-1)+2^2(1+2^p-1)+\cdots+2^{p-2}(1+2^p-1)+2^{p-1}$$
Question: I am confused a little about the last two terms in the summation above as what I think it should be is as follows,
$$(1+2^p-1)+2(1+2^p-1)+2^2(1+2^p-1)+\cdots+2^{p-2}(1+2^p-1)+2^{p-1}(1+2^p-1)$$
Edit:
Question: I should get the sum $2^{p−1}(2^p − 1)$, but with the way I wrote it I did not get the sum wanted, but I got $2^p(2^p-1)$ for,
$$(1+2^p-1)+2(1+2^p-1)+2^2(1+2^p-1)+\cdots+2^{p-2}(1+2^p-1)+2^{p-1}(1+2^p-1)$$
Edit 2: The only solution to this is that the sum is $2^p(2^p-1)$.
| If I read well you are evaluating the sum
$$ \sum_{k=0}^{p-1}2^k + \sum_{k=0}^{p-1}2^k(2^p-1) = 2^p\sum_{k=0}^{p-1}2^k = 2^p\frac{1-2^p}{1-2} = 2^p(2^p-1)$$
| {
"language": "en",
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"source": "stackexchange",
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} |
$Q(x)=f(x^3)+xg(x^3)$ and $Q(x)$ is divisible by $x^2+x+1$ Let $Q\left( x \right) = f\left( {{x^3}} \right) + xg\left( {{x^3}} \right)$ and $Q(x)$ is divisible by $x^2+x+1$, then find the value of $Q(1)$
My approach is as follow
$Q\left( x \right) = f\left( {{x^3}} \right) + xg\left( {{x^3}} \right)$
$Q\left( x \right) = \phi \left( x \right)\left( {{x^2} + x + 1} \right)$
$Q\left( \omega \right) = \phi \left( \omega \right)\left( {{\omega ^2} + \omega + 1} \right) = 0$
$Q\left( {{\omega ^2}} \right) = \phi \left( \omega \right)\left( {{\omega ^4} + {\omega ^2} + 1} \right) = 0$
$Q\left( \omega \right) = f\left( {{\omega ^3}} \right) + \omega g\left( {{\omega ^3}} \right) = f\left( 1 \right) + \omega g\left( 1 \right) = 0$
$Q\left( {{\omega ^2}} \right) = f\left( {{\omega ^6}} \right) + {\omega ^2}g\left( {{\omega ^6}} \right) = f\left( 1 \right) + {\omega ^2}g\left( 1 \right) = 0$
Adding we get $2f(1)-g(1)=0$
$Q\left( 1 \right) = f\left( 1 \right) + g\left( 1 \right)$
Not able to proceed from here
| Hint: if $x^2+x+1|Q(x)$, then
$$x^3-1\ \big| (x-1)Q(x)=(x^2-x)g(x^3)+(x-1)f(x^3).$$
See if you can use this, combined with the property that
$$f(x)-g(x)\ \big|\ h(f(x))-h(g(x))$$
for any polynomials $f,g,h$, to find $f(1)$ and $g(1)$.
| {
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"source": "stackexchange",
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"answer_count": 1,
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} |
Find the matrix of the linear transformation $T$ Let $A=\begin{pmatrix}0\\0\end{pmatrix},B=\begin{pmatrix}1\\0\end{pmatrix},C=\begin{pmatrix}0\\1\end{pmatrix}$ and $D=\begin{pmatrix}1\\1\end{pmatrix}$. The images of these vectors under the linear transformation is $A^*=\begin{pmatrix}0\\0\end{pmatrix},B^*=\begin{pmatrix}2\\2\end{pmatrix},C^*=\begin{pmatrix}0\\1\end{pmatrix}$ and $D^*=\begin{pmatrix}2\\3\end{pmatrix}$ respectively. Find the matrix of the linear transformation $T$.
Let $T=\begin{pmatrix}a & b\\c & d\end{pmatrix}$
$$
T(A)=A^*\implies\begin{pmatrix}a & b\\c & d\end{pmatrix}
\begin{pmatrix}0\\0\end{pmatrix}=\begin{pmatrix}0\\0\end{pmatrix}
$$
$$
T(B)=B^*\implies\begin{pmatrix}a & b\\c & d\end{pmatrix}
\begin{pmatrix}1\\0\end{pmatrix}=\begin{pmatrix}a\\c\end{pmatrix}=
\begin{pmatrix}2\\2\end{pmatrix}
$$
$$
T(C)=C^*\implies\begin{pmatrix}a & b\\c & d\end{pmatrix}
\begin{pmatrix}0\\1\end{pmatrix}=\begin{pmatrix}b\\d\end{pmatrix}=
\begin{pmatrix}0\\1\end{pmatrix}
$$
$$
T(D)=D^*\implies\begin{pmatrix}a & b\\c & d\end{pmatrix}
\begin{pmatrix}1\\1\end{pmatrix}=\begin{pmatrix}a+b\\c+d\end{pmatrix}=
\begin{pmatrix}1\\1\end{pmatrix}
$$
$(2)$ and $(3)$ implies $T=\begin{pmatrix}2&0\\2&1\end{pmatrix}$ but it's not satisfy $(4)$. Am I doing it in wrong way?
| The last equation should be $$\begin{pmatrix}a+b\\c+d\end{pmatrix}=
\begin{pmatrix}2\\3\end{pmatrix}.$$
The right hand side of the equation is $D^*$, not $D$.
| {
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"source": "stackexchange",
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Given the number $N=3^{12}-1$. Find how many even and odd divisors of $N$ exist, which divide $N$ but don't divide $3^k-1$, for $1\le k\le 11$ Given the number $N=3^{12}-1$. Find how many even and odd divisors of $N$ exist, which divide $N$ but don't divide $3^k-1$, for $1\le k\le 11$
Initially I tried working out how many odds and evens divide $N$, without the extra restriction.
$3^{12}-1=(3-1)(3^2+3+1)(3+1)(3^2-3+1)(3^6+1)=2*13*4*7*730$
$=2^4*5*7*13*73$
Hence there are $16$ odd divisors and $64$ even divisors.
I don't know how to finish off the question with the extra condition, without exmining all $80$ possibilities individually. Could you please explain to me how to solve this question?
Also a solution to the question is the following, despite the fact that I have tried to understand it, but it is too complex for me:
If we have an integer $m$, $3^{12}\equiv 1\mod{m}$, then the smallest integer for which $d<12$ for which it holds true that $3^d\equiv 1\mod{m}$, must divide $12$, since if this is not true, then we can write $12=dp+r$, with $0<r<d$ and then we have $3^r\equiv 1\mod{m}$, which is not true, from the hypothesis that $d$ is the smallest number for which $3^d\equiv 1\mod{m}$.
Hence to be sure that $m$ does not divide $3^k-1$, for $1\le k\le 11$, we examine all values of $k$ which divide $12$, in other words $k=1,2,3,4,6, 3^1-1=2, 3^2-1=2^3,3^3-1=2*13, 3^4-1=2^4*5 and 3^6-1=2^3*7*13$.
Odd divisors of $N$ divide $N$ but do not divide $3^k-1$, for $1\le k \le 11$ is: $16-5=11$.
Even divisors of $N$ which divide $N$, but do not divide $3^k-1$ for $1\le k\le 11$, are: $64-17=47$
| I think the "too-complicated" solution you cite would be more understandable if it were phrased constructively, for example:
For a given integer $m$ with $3\nmid m$, there is a smallest $k>0$ for which $3^{k}\equiv 1\bmod{m}$ - the multiplicative order of $3 \bmod m$. Then for any $j>0$, $3^j\equiv 1 \bmod m$ $\iff$ $k\mid j$. Thus any number $m$ which divides $3^{12}{-}1$ can only divide a smaller $3^i{-}1$ if $i\mid 12$.
Now we can assign factors of $3^k-1$ as being "primary" if they do not belong to a lower $k$ - we find these simply by subtracting off the primary factors of lower dividing powers. Then considering $k=1,2,3,4,6,12$:
$$\begin{array}{c} \hline
k & 3^k-1 & \text{# factors}& \text{dividing } k & \text{# primary}& \\
& & \text{ odd + even} & & \text{ odd + even} \\ \hline
1 & 2 & 1+ 1& - & 1+ 1 \\
2 & 2^3 & 1+ 3 & 1 & 0+ 2 \\
3 & 2\cdot 13 & 2+ 2 & 1 & 1+ 1 \\
4 & 2^4\cdot 5 & 2+ 8 & 1,2 & 1+ 5 \\
6 & 2^3\cdot 7\cdot 13 & \;4+12 & 1,2,3 & 2+ 8 \\
12 &\quad 2^4\cdot 5\cdot 7\cdot 13\cdot 73 & 16+ 64 & 1,2,3,4,6 & 11+ 47 \\
\end{array}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4075126",
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"source": "stackexchange",
"question_score": "3",
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How to separate real and imaginary parts of $\arccos(x+iy)$? I know how to separate the real and imaginary parts of $\ln z$ and $\arctan z$:
$$\begin{align}\ln(x+iy)&=\frac{1}{2}\ln(x^2+y^2)+i\arctan\frac{y}{x}\\
\arctan(x+iy)&=\frac{1}{2}\arctan\frac{2x}{1-x^2-y^2}+\frac{i}{2}\operatorname{artanh}\frac{2y}{1+x^2+y^2}\end{align}$$
and I know how to prove both of these; in particular for the second one we can make use of the fact that $\arctan z=-i\operatorname{artanh}(iz)$. I would now like to know how to separate the real and imaginary parts of $\arccos(x+iy)$; I've thought of using the identity $\arccos(z)=\operatorname{arcosh}(iz)$, but I don't know how to use this, as the argument of the $\ln$ becomes very complicated.
Thank you very much for your help.
| Let
$$
\arccos(x+iy)=u+iv.\tag1
$$
Then
$$\begin{align}
x+iy&=\cos(u+iv)=\cos u\cosh v-i\sin u\sinh v\\
\\
&\implies
\begin{cases}
x=\hphantom-\! \cos u\,\!\cosh v\\
y=-\sin u\,\sinh v
\end{cases}.
\end{align}$$
Eliminating $u$ one obtains:
$$\begin{align}
&\left(\frac x{\cosh v}\right)^2+\left(\frac y{\sinh v}\right)^2=1\\
&\implies\cosh^4v-(x^2+y^2+1)\cosh^2v+x^2=0\\
&\implies\cosh^2v=\frac12\left[x^2+y^2+1+\sqrt{(x^2+y^2+1)^2-4x^2}\right]\tag{$*$}\\
&\implies\cosh^2v=\frac14\left[\sqrt{(x+1)^2+y^2}+\sqrt{(x-1)^2+y^2}\right]^2\\
&\implies\cosh v=\frac12\left[\sqrt{(x+1)^2+y^2}+\sqrt{(x-1)^2+y^2}\right],\tag2
\end{align}$$
and from this:
$$
\cos u=\frac{x}{\cosh v}=\frac12\left[\sqrt{(x+1)^2+y^2}-\sqrt{(x-1)^2+y^2}\right].\tag3
$$
Resolving $(2)$ and $(3)$ and plugging the obtained values of $u$ and $v$ into $(1)$ gives rise to the final expression:
$$\begin{align}
\arccos(x+iy)&=\arccos\frac{\sqrt{(x+1)^2+y^2}-\sqrt{(x-1)^2+y^2}}2\\
&+i\operatorname{arccosh}\frac{\sqrt{(x+1)^2+y^2}+\sqrt{(x-1)^2+y^2}}2.\tag4
\end{align}$$
Note added:
In implication $(*)$ the possibility
$$
\cosh^2v=\frac12\left[x^2+y^2+1-\sqrt{(x^2+y^2+1)^2-4x^2}\right]
$$
was disregarded since for $y\ne0$ it would imply $\cosh^2v<1$, which is impossible for real $v$.
| {
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"source": "stackexchange",
"question_score": "1",
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Showing that $\frac{x^{2x}}{(x+1)^{x+1}}\rightarrow +\infty$ as $x\rightarrow +\infty$ I am trying to show that $$\frac{x^{2x}}{(x+1)^{x+1}}\rightarrow +\infty \ \ \text{as} \ \ x\rightarrow +\infty.$$
My attempt is as follows:
\begin{align}
\frac{x^{2x}}{(x+1)^{x+1}}&=\frac{x^{x}}{x+1}\left(\frac{x^x}{(x+1)^x}\right) \\
&=\frac{x^{x}}{x+1}\left(\frac{x}{x+1}\right)^x \\
&=\frac{x^{x}}{x+1}\left(\frac{1}{(1+1/x)^x}\right).
\end{align}
I can see that the second fraction will converge to $1/e$, but I am unsure of how to approach the first fraction.
| $$y=\frac{x^{2x}}{(x+1)^{x+1}}\implies \log(y)=2x \log(x)-(x+1)\log(x+1)$$
$$\log(y)=2x \log(x)-(x+1)\left(\log(x)+\log \left(1+\frac{1}{x}\right)\right)$$
$$\log(y)=(x-1) \log (x)-(x+1)\log \left(1+\frac{1}{x}\right)$$
Now, by Taylor series
$$(x+1)\log \left(1+\frac{1}{x}\right)=(x+1)\Bigg[\frac{1}{x}-\frac{1}{2 x^2}+O\left(\frac{1}{x^3}\right) \Bigg]$$
$$(x+1)\log \left(1+\frac{1}{x}\right)=1+\frac{1}{2 x}+O\left(\frac{1}{x^2}\right)$$ Putting all together
$$\log(y)=(x-1) \log (x)-1-\frac{1}{2 x}+O\left(\frac{1}{x^2}\right)$$ which already tends to $\infty$. If you want to continue, use
$y=e^{\log(x)}$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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find the roots $(2z+3)^3=\frac{1}{64}$ There are 3 roots 1 real and 2 imaginary i found one z by doing $\frac{\frac{1}{4}-3}{2}$ so $z=\frac{-11}{8}$ however there are two more complex roots which are $z=\frac{-25+i√3}{16}$ and $z=\frac{-25-i√3}{16}$ but i dont know how to get to it any help is much appricated
. thank you
| Hope you got it now, it's a property of the polynomial, the equation $z^2-1$ is satisfied by $z =1, z= -1$, because a quadratic has two roots
While $z^3-1$ is satisfied by $z=1$ as the obvious real solution, but a cubic has three roots, so the other two are unreal, but if we can factor it or substitute $z=a+bi$, to get the remaining roots
$$ z^2+z+1$$
$$ z_1 = -\frac{1}{2} + \frac{\sqrt{ -3}}{2}$$
$$ z_2 = -\frac{1}{2} - \frac{\sqrt{ -3}}{2}$$
They are conjugate of eachother, If you cube it using the property of complex number the result is $z^3 = 1$, so $z_0 , z_1, z_2$ all from the cube root of unity
Anytime you take the cube root of a number $n$, the solutions are $\sqrt[3]{n} , \sqrt[3]{n} \cdot z_1, \sqrt[3]{n} \cdot z_2$
| {
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Show that $O_2(\mathbb{R})$ contains only rotational and reflective symmetries. Show that $O_2(\mathbb{R})$ contains only rotational and reflective matrices.
I know that rotational and reflective symmetries are part of $O_2(\mathbb{R})$. I want to show that there is no other matrix that satisfies $O_2(\mathbb{R})$. This is what I have and I don't know what to do after this or if I'm going about this in the right way at all.
Suppose there exists a matrix $A \in O_2(\mathbb{R})$ such that is not a rotational or reflexive symmetry.
$A^T = A^{-1}$ by definition
Therefore we get the following simultaneous equations:
$$a = \frac{d}{ad - bc}; d = \frac{a}{ad - bc}; b = \frac{-c}{ad - bc}; c = \frac{-b}{ad - bc}$$
using the first two equations, you get $ad - bc =1$ or $a = -d$ and from the second two $ad - bc = -1$ or $b = -c$.
I don't know where to go from here to show that $a,b,c,d$ form either a rotational or reflective matrix.
| By definition $A \in O_2(\mathbb{R})$ iff $AA^T = A^TA = I$.
Let
\begin{align}
A =
\begin{pmatrix}
a & b\\
c & d
\end{pmatrix}
\end{align}
then we have that
\begin{align}
\begin{pmatrix}
a & b\\
c & d
\end{pmatrix}
\begin{pmatrix}
a & c\\
b & d
\end{pmatrix}
=
\begin{pmatrix}
a^2+b^2 & ac+bd\\
ac+bd & c^2+d^2
\end{pmatrix}
=
\begin{pmatrix}
1 & 0\\
0 & 1
\end{pmatrix}.
\end{align}
Hence, $(a, b)$ and $(c, d)$ are unit vectors. Moreover, $(a, b)$ is orthogonal $(c, d)$ since $ac+bd=0$. Then we can rewrite $A$ as follows
\begin{align}
A =
\begin{pmatrix}
a & -b\\
b & a
\end{pmatrix} \ \ \text{ or } \ \ A =
\begin{pmatrix}
a & b\\
b & -a
\end{pmatrix}
=
\begin{pmatrix}
0 & 1\\
1 & 0
\end{pmatrix}
\begin{pmatrix}
b & -a\\
a & b
\end{pmatrix}
\end{align}
Since $a^2+b^2=1$, we can reparametrize the matrix by noting that $\cos^2\theta+\sin^2\theta = 1$. Hence, we have that either
\begin{align}
A =
\begin{pmatrix}
\cos\theta & -\sin\theta\\
\sin\theta & \cos\theta
\end{pmatrix}
\ \ \text{ or } \ \
A =
\begin{pmatrix}
\cos\theta & \sin\theta\\
\sin\theta & -\cos\theta
\end{pmatrix}
=
\begin{pmatrix}
0 & 1\\
1 & 0
\end{pmatrix}
\begin{pmatrix}
\sin\theta & -\cos\theta\\
\cos\theta & \sin\theta
\end{pmatrix}.
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4082619",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
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Something maybe related to stochastic process We have $5$ floors and we are on the ground(treat the ground as floor$0$).
The probability of jumping from the ground to floor$1$ is $1$.
For $i\in\{1,2,3,4\}$,
the probability of jumping from the $i$-th floor to the $i+1$-th floor is $\frac{1}{3^i}$,
the probability of staying at the $i$-th floor is $\frac{3^i-1}{2\times 3^i}$,
the probability of jumping from the $i$-th floor to the $i-1$-th floor is $\frac{3^i-1}{2\times 3^i}$,
What is the expectation of the number of turns that you jump from the ground and reach floor$5$?
My thought:
I first came up with the probability matrix $P = \begin{pmatrix}
0& \frac{1}{3}& 0& 0& 0& 0\\
1& \frac{1}{3}& \frac{4}{9}& 0& 0& 0\\
0& \frac{1}{3}& \frac{4}{9}& \frac{13}{27}& 0& 0\\
0& 0& \frac{1}{9}& \frac{13}{27}& \frac{40}{81}& 0\\
0& 0& 0& \frac{1}{27}& \frac{40}{81}& 0\\
0& 0& 0& 0& \frac{1}{81}& 0
\end{pmatrix}$
and the initial state is $e_1 = \begin{pmatrix}
1\\
0\\
0\\
0\\
0\\
0
\end{pmatrix}$
so the expectation should be
$$\sum_{x=0}^\infty x[(P^xe_1)\cdot e_6]$$ where $\cdot$ is the inner product and $e_6 = \begin{pmatrix}
0\\
0\\
0\\
0\\
0\\
1
\end{pmatrix}$
Is my thought correct so far? And I can't go through it further. I used programming and got an approximation of the series that the expectation should be $14543$
| Labelling floors $0 - 5$ as $a - f$, and using first-step-analysis, we have
$\displaylines{a = 1+b\\
b= 1+c/3 + (1/3)(b+a)\\
c= 1+d/9 + (4/9)(c+b)\\
d=1+e/27 +(13/27)(d+c)\\
e = 1 +(40/81)(e+d)}$
and wolfram confirms your answer of $\;\fbox{14543}$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Solving $\cos4\theta=\cos3\theta$ to find the roots of $8x^3+4x^2-4x-1=0$. Find all the solution to the equation
$\cos 4\theta=\cos 3\theta$. Hence or otherwise, show that the roots of the equation
$$8x^3+4x^2-4x-1=0$$ are $$\cos \frac{2\pi}{7},\cos \frac{4\pi}{7},\cos \frac{6\pi}{7}$$
Therefore prove that
$$\sec \frac{2\pi}{7}+\sec \frac{4\pi}{7}+\sec \frac{6\pi}{7}=-4$$
I am slightly stuck! I have solved the first equation using a graphical calculator and found that it is $(2n\pi)/7$ where n is any positive integer. I have then algebraically rewritten $cos(4\theta) = cos(3\theta)$ using double angle identities.
Now I am stuck... Is there are mistake in the question? What would be your model answer?
Thank you so very much.
| $\cos(x) = \cos(y) \Leftrightarrow x = 2n\pi \pm y, n\in \Bbb Z$
So,
$4\theta = 2n\pi \pm3\theta \Rightarrow \begin{cases}\theta = 2n\pi\\\text{ or} \\\theta = \dfrac{2n\pi}{7}\end{cases}$
Now using compound angle formula,
$\begin{align}&\cos4\theta = \cos3\theta \\\Rightarrow &2\cos^2(2\theta )-1 = 4\cos^3\theta-3\cos\theta \\\Rightarrow &2(2\cos^2\theta-1)^2 -1 = 4\cos^3\theta-3\cos\theta \\
\Rightarrow &8\cos^4\theta-8\cos^2\theta+1 = 4\cos^3\theta-3\cos\theta \\
\Rightarrow&8\cos^4\theta-4\cos^3\theta- 8\cos^2\theta+3\cos\theta+1 = 0\end{align}$
We don't need the case where $\theta = 2n\pi$, that'll lead to $\cos\theta =1$.
So, we can factor out $\cos\theta -1 $ from the 4th order equation in $\cos$ (Perform long division or any other suitable method)
$\Rightarrow (\cos\theta-1)(8\cos^3\theta+4\cos^2\theta-4\cos\theta-1) = 0$
$\Rightarrow 8\cos^3\theta+4\cos^2\theta-4\cos\theta-1 =0$ or $\boxed{8x^3+4x^2-4x-1 = 0}$.
As $\cos$ has a periodicity of $2\pi$, we only consider for $n = \pm1,\pm2,\pm3$ or $\cos\frac{\pm2\pi}{7} = \cos\frac{2\pi}{7},\cos\frac{\pm4\pi}{7}=\cos\frac{4\pi}{7},\cos\frac{\pm6\pi}{7}=\cos\frac{6\pi}{7} $.
So the roots are $\cos\frac{2\pi}{7},\cos\frac{4\pi}{7},\cos\frac{6\pi}{7}$
If $a,b,c$ are roots of $Ax^3 +Bx^2+Cx+D = 0$, then
$ab +bc + ca = \frac{C}{A}$ and $abc = -\frac{D}{A} \Rightarrow \frac{ab+bc+ca}{abc}= \frac1a+\frac1b+\frac 1c = -\frac CD$
As $\frac1{\cos x} = \sec x$, here,
$$\frac1a+\frac 1b +\frac 1c =\sec\frac{2\pi}7 +\sec\frac{4\pi}7+\sec\frac{6\pi}7- \frac{-4}{-1} = 4$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4084498",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Maximum value of a symmetric function on a square In a paper of Frits Beukers "A note on the irrationality of $\zeta(2)$ and $\zeta(3)$", he says
It is a matter of straightforward computation to show that
$$\frac{y(1-y)x(1-x)}{1-xy} \le \left\{\frac{\sqrt{5}-1}{2}\right\}^5$$ for all $0\le x\le1, 0\le y \le 1$.
I noticed (using calculus) that if we let $x = y$,
$$
\frac{x^2(1-x)^2}{1-x^2} = \frac{x^2(1-x)}{1+x}
$$
has a maximum value of $\left\{\frac{\sqrt{5}-1}{2}\right\}^5$ on the interval $0\le x\le1$.
My question is why is the maximum of $\frac{y(1-y)x(1-x)}{1-xy}$ attained when $x=y$? Is it because of the symmetry in $x$ and $y$?.
Is it true in general that a symmetric function $f(x_1,x_2,\dots,x_n)$ attains its maximum on an $n$-dimensional cube when $x_1=x_2=\dots=x_n$?
| $$\frac{y(1-y)x(1-x)}{1-xy} = \frac{xy(1-(x+y)+xy)}{1-xy} \le \frac{xy(1-2\sqrt{xy}+xy)}{1-xy}= \frac{xy(1-\sqrt{xy})}{1+\sqrt{xy}}$$
Denote $t = \sqrt{xy}$ then $0 \le t\le 1$.
The function $f(t) = \frac{t^2(1-t)}{1+t}$ reaches its maximum in $t \in [0,1]$ when $t = \frac{\sqrt{5}-1}{2}$ (it suffices to calculate $f'(t)$)
The equality occurs when $x = y$ (so that $x+y = 2\sqrt{xy}$) and $\sqrt{xy} = \frac{\sqrt{5}-1}{2}$, or $x = y = \frac{\sqrt{5}-1}{2}$
PS: about your last question concerning the relationship between the symmetry of $f(x_1,...,x_n)$ and the maximum of $f$, this relationship doesn't exists in general. And in this particular problem, the equality attains because $x+y = \sqrt{xy} \iff x = y$.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Proving that odd partitions and distinct partitions are equal I am working through The Theory of Partitions by George Andrews (I have the first paperback edition, published in 1998).
Corollary 1.2 is a standard result that shows that the number of partitions of $n$ into odd parts is the same as the number of partitions of $n$ into distinct parts.
Andrews's proof uses generating functions, but it contains this step that has me confounded:
$$
\prod_{n = 1}^\infty \frac{1 - q^{2 n }}{1 - q^n} = \prod_{n = 1}^\infty \frac{1}{1 - q^{2 n - 1}}.
$$
Can anybody point me in the right direction?
I tried turning $1 - q^n$ into an infinite sum and then expanding, to get
$$
\frac{1 - q^{2 n }}{1 - q^n} = 1 + q + q^2 + \dots + q^{2 n -1},
$$
but I don't see how this manipulation will help me.
| See if this helps?
\begin{align*}
\prod_{n = 1}^\infty \frac{1 - q^{2 n }}{1 - q^n} &= \frac{1 - q^{2}}{1 - q}\cdot\frac{1 - q^{4 }}{1 - q^2}\cdot\frac{1 - q^{6 }}{1 - q^3}\cdot\dots\\
&= \left[\frac{1}{1-q}\cdot\frac{1}{1-q^3}\cdot\frac{1}{1-q^5}\dots\right]\left[\frac{1 - q^{2}}{1 - q^2}\cdot\frac{1 - q^{4 }}{1 - q^4}\cdot\frac{1 - q^{6 }}{1 - q^6}\cdot\dots\right]\\
&= \left[\frac{1}{1-q}\cdot\frac{1}{1-q^3}\cdot\frac{1}{1-q^5}\dots\right]\left[1\right]\\
&= \prod_{n = 1}^\infty \frac{1}{1 - q^{2 n - 1}}\\
\end{align*}
Where the second step indicates a rearrangement of the terms. Now one must be careful in infinite rearrangements - you should check for the absolute convergence of an infinite sum. see here
You should check this for validity of the second step.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4092376",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Show that $\langle 2,1+\sqrt{-5}\rangle \langle3,1+\sqrt{-5} \rangle = \langle1-\sqrt{-5} \rangle$ In $\mathbb Z[\sqrt{-5}]$ I compute
$\langle 2,1+\sqrt{-5} \rangle \langle 3,1+\sqrt{-5} \rangle = \langle 6,2+2\sqrt{-5},3+3\sqrt{-5},-4+2\sqrt{-5} \rangle $
So there must be $a+b\sqrt{-5}\in\mathbb Z[\sqrt{-5}]$ that $(a+b\sqrt{-5})(1-\sqrt{-5})=2+2\sqrt{-5}$ but then we have $b=\frac{2}{3}$ that it's not possible. Am I doing it wrong? How can I prove this?
| As you've showed, \begin{align} \langle 2,1+\sqrt{-5} \rangle \langle 3,1+\sqrt{-5} \rangle &= \langle 6,\color{blue}{2+2\sqrt{-5}},\color{blue}{3+3\sqrt{-5}},-4+2\sqrt{-5} \rangle \\ \tag{1}&= \langle \color{blue}{6},1+\sqrt{-5}, \color{blue}{-4+2\sqrt{-5}} \rangle \\ \tag{2}&=\langle 6, \color{blue}{1+\sqrt{-5}},\color{blue}{2+2\sqrt{-5}} \rangle \\ \tag{3}& =\langle 6,1+\sqrt{-5} \rangle \\& \tag{4}= \langle \color{red}{1+ \sqrt{-5}} \rangle \end{align}
*
*Both $2 + 2 \sqrt{-5}$ and $3 + 3\sqrt{-5}$ are multiples of $1 + \sqrt{-5}$, hence belong to $\langle 6, 1+ \sqrt{-5}, -4+2\sqrt{-5} \rangle$. On the other hand, $1+\sqrt{-5}$ = $(3 + 3\sqrt{-5}) - (2 + 2\sqrt{-5}) \in \langle 6, 2+ 2\sqrt{-5}, 3+ 3\sqrt{-5}, -4+2\sqrt{-5} \rangle$.
*$-4+2\sqrt{-5} = (2 + 2\sqrt{-5}) - 6 \in \langle 6, 1+\sqrt{-5}, 2+2\sqrt{-5} \rangle $. Conversely, $2+2\sqrt{-5} = (-4 + 2\sqrt{-5}) + 6 \in \langle 6, 1+\sqrt{-5}, -4+2\sqrt{-5} \rangle$.
*$2+2\sqrt{-5}$ is an element of $\langle 6, 1+\sqrt{-5} \rangle$ as it is a multiple of $1+\sqrt{-5}$.
*$ 6 = (1 + \sqrt{-5})(1-\sqrt{-5}) \in \langle 1 + \sqrt{-5} \rangle $.
The result $\langle 1+ \sqrt{-5} \rangle$ is not equal to $\langle 1 - \sqrt{-5} \rangle$, so your title presents wrong equality. To see this, suppose $\langle 1 + \sqrt{-5} \rangle = \langle 1 - \sqrt{-5} \rangle$. Then there exists $a + b\sqrt{-5} \in \mathbb{Z}[\sqrt{-5}]$ such that $1 + \sqrt{-5} = (a+b\sqrt{-5})(1-\sqrt{-5})$. Taking the norm of both sides yields $N(a+b\sqrt{-5}) = a^2 + 5b^2 =1$, whence $a=\pm 1$ and $b=0$. Equivalently, $1+\sqrt{-5} = \pm ( 1 - \sqrt{-5})$; contradiction.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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On bounds for the quantity $n^2 / D(n^2)$ when $q^k n^2$ is an odd perfect number with special prime $q$ and $k > 1$ Let $N = q^k n^2$ be an odd perfect number given in Eulerian form (i.e. $q$ is the special/Euler prime satisfying $q \equiv k \equiv 1 \pmod 4$ and $\gcd(q,n)=1$).
In this post, I would like to compute the general expression for the quantity
$$\frac{n^2}{D(n^2)}$$
where $D(x) = 2x - \sigma(x)$ is the deficiency of $x$. (Here, $\sigma(x)=\sigma_1(x)$ is the classical sum of divisors of $x$.) I would also like to derive bounds for the said quantity, which are hopefully sharp.
It is known that, for an odd perfect number $N = q^k n^2$ given in Eulerian form, we have
$$\frac{\sigma(n^2)}{q^k}=\frac{2n^2}{\sigma(q^k)}=\frac{D(n^2)}{s(q^k)}=\frac{2s(n^2)}{D(q^k)},$$
where $s(x)=\sigma(x)-x$ is the aliquot sum of $x$.
We obtain, immediately from the middle equation, the equality
$$\frac{n^2}{D(n^2)}=\frac{\sigma(q^k)}{2s(q^k)}.$$
This quantity is an integer if and only if $k=1$ i.e. when the RHS is $(q+1)/2$. Here I am interested in deriving bounds for this quantity when $k>1$.
We rewrite the RHS as follows:
$$\dfrac{\sigma(q^k)/q^k}{2(s(q^k)/q^k)} = \frac{I(q^k)}{2(I(q^k) - 1)},$$
where $I(x)=\sigma(x)/x$ is the abundancy index of $x$.
Now, use the following bounds (which hold since $k>1$ and $k \equiv 1 \pmod 4$ imply that $k \geq 5$):
$$I(q^5) \leq I(q^k) < \frac{q}{q-1}.$$
This gives
$$\dfrac{I(q^5)}{2\bigg(\dfrac{q}{q-1} - 1\bigg)} < \frac{n^2}{D(n^2)}=\frac{I(q^k)}{2(I(q^k) - 1)} < \dfrac{\dfrac{q}{q-1}}{2(I(q^5) - 1)}.$$
The LHS and RHS of the last inequality then simplifies to
$$\text{LHS} = \dfrac{I(q^5)}{2\bigg(\dfrac{q}{q-1} - 1\bigg)} = \frac{q}{2} - \frac{1}{2q^5}$$
while
$$\text{RHS} = \dfrac{\dfrac{q}{q-1}}{2(I(q^5) - 1)} = \frac{q^6}{2(q^5 - 1)}.$$
The function
$$f_1(q) = \frac{q}{2} - \frac{1}{2q^5}$$
is increasing, since its derivative is positive. Hence
$$f_1(q) \geq f_1(5) = \frac{7812}{3125}$$ since $q$ is a prime satisfying $q \equiv 1 \pmod 4$ implies that $q \geq 5$.
On the other hand, the function
$$f_2(q) = \frac{q^6}{2(q^5 - 1)}$$
is also increasing, since its derivative is also positive. Thus, we may rewrite $f_2(q)$ as follows:
$$f_2(q) = \frac{q\cdot{q^5}}{2(q^5 - 1)} = \frac{q(q^5 - 1)}{2(q^5 - 1)} + \frac{q}{2(q^5 - 1)} = \frac{q}{2} + \frac{q}{2(q^5 - 1)}.$$
Consequently, we have the following proposition:
PROPOSITION: If $N = q^k n^2$ is an odd perfect number with special/Euler prime $q$ and $k > 1$ holds, then we have
$$\frac{7812}{3125} \leq \frac{q}{2} - \frac{1}{2q^5} < \frac{n^2}{D(n^2)} < \frac{q}{2} + \frac{q}{2(q^5 - 1)}.$$
In particular:
$$k > 1 \implies \Bigg(\bigg(\frac{n^2}{D(n^2)} > \frac{7812}{3125}\bigg) \iff \bigg(\frac{D(n^2)}{n^2} < \frac{3125}{7812}\bigg) \iff \bigg(I(n^2) > 2 - \frac{3125}{7182} = \frac{12499}{7182} \approx 1.59997439836\bigg)\Bigg)$$
which does not improve on the currently known
$$I(n^2) > \frac{8}{5}.$$
Here are my:
QUESTIONS: Can you improve on the bounds for $n^2 / D(n^2)$ (in terms of $q$) when $k>1$? If it is no longer possible to refine the bounds given here, can you explain/show why?
| Yes.
We have
$$\frac{n^2}{D(n^2)}=\frac{\sigma(q^k)}{2s(q^k)}=\frac{q^{k+1}-1}{2(q-1)\bigg(\dfrac{q^{k+1}-1}{q-1}-q^k\bigg)}=\frac{q^{k+1}-1}{2q^k-2}$$
Let $f(k):=\dfrac{q^{k+1}-1}{2q^k-2}$. Then, we get $f'(k)=-\dfrac{(q-1)q^k\log q}{2(q^k-1)^2}$ which is negative.
Since $f(k)$ is decreasing, we obtain
$$\lim_{k\to\infty}f(k)\lt f(k)\le f(5),$$
i.e.
$$\frac q2\lt \frac{n^2}{D(n^2)}\le \frac q2+\dfrac{q-1}{2(q^5-1)}$$
| {
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"url": "https://math.stackexchange.com/questions/4097665",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Bounds on $\sum\limits_{k=1}^n \frac{\sin(k)}{k}$ $\sum\limits_{k=1}^n \frac{\sin(k)}{k}$ converges as $n$ increases, to a limit of $\frac12(\pi-1) \approx 1.0708$
Empirically, it seems to be bounded by about $\frac12(\pi-1) \pm \frac{1.043}{n}$, as shown in the chart below.
What is the precise value of this $1.043$ term?
Added: I think the three answers from Oliver Diaz, Gary, and Gabriel Romon, plus the comments, have demonstrated that the bounds are $$\frac12(\pi-1) \pm \frac1{2\sin(1/2)}\frac{1}{n}$$ not only asymptotically but also as tight actual bounds; $\frac1{2\sin(1/2)}\approx 1.0429148214667441$
created in R with
n <- 1:11000
plot(n, cumsum(sin(n)/n), ylim=c(1.07,1.0715), xlim=c(1,10000), pch=46)
abline(h=(pi-1)/2, col="red")
curve((pi-1)/2 + 1.043/x, from=1, to=max(n), add=TRUE, col="red")
curve((pi-1)/2 - 1.043/x, from=1, to=max(n), add=TRUE, col="red")
| A slightly worse estimate may be obtained as follows. If $|\arg z|<\pi$ then
$$
\log (1 + z) = \sum\limits_{k = 1}^N {( - 1)^{k + 1} \frac{{z^k }}{k}} + ( - 1)^N z^{N + 1} \int_0^1 {\frac{{t^N }}{{1 + zt}}dt}
$$
for any $N\geq 0$. Substituting $z = - e^i$, taking the imaginary part of each side and rearranging, we find
$$
\sum\limits_{k = 1}^N {\frac{{\sin (k)}}{k}} - \frac{{\pi - 1}}{2} = \Im \left[ { - e^{i(N + 1)} \int_0^1 {\frac{{t^N }}{{1 - e^i t}}dt} } \right].
$$
Here
\begin{align*}
\left| {\Im \left[ { - e^{i(N + 1)} \int_0^1 {\frac{{t^N }}{{1 - e^i t}}dt} } \right]} \right| & \le \int_0^1 {\frac{{t^N }}{{\left| {1 - e^i t} \right|}}dt} \le \frac{1}{{\sin 1}}\int_0^1 {t^N dt} \\ & = \frac{1}{{(N + 1)\sin 1}} < \frac{{1.1884}}{{N + 1}}.
\end{align*}
This is larger than the expected result by a factor of $\frac{1}{{\cos (1/2)}}=1.13949\ldots$. An exact formula for the remainder can be obtained by computing the imaginary part explicitly, but it results in a complicated formula which does not seem to be useful for better estimates.
Addendum. From the link you posted, the remainder term is
$$
{\int_0^1 {\left( {\frac{1}{{2\sin \frac{t}{2}}} - \frac{1}{t}} \right)\sin \left( {\left( {n + \tfrac{1}{2}} \right)t} \right)dt} - \int_{n + 1/2}^\infty {\frac{{\sin t}}{t}dt} }.
$$
Integration by parts gives
$$
\int_0^1 {\left( {\frac{1}{{2\sin \frac{t}{2}}} - \frac{1}{t}} \right)e^{i\left( {n + \frac{1}{2}} \right)t} dt} = ie^{i\left( {n + \frac{1}{2}} \right)} \left( {1 - \frac{1}{{2\sin \left( {\frac{1}{2}} \right)}}} \right)\frac{1}{n} + \mathcal{O}\!\left( {\frac{1}{{n^2 }}} \right).
$$
Taking imaginary parts
$$
\int_0^1 {\left( {\frac{1}{{2\sin \frac{t}{2}}} - \frac{1}{t}} \right)\sin \left( {\left( {n + \tfrac{1}{2}} \right)t} \right)dt} = \left( {1 - \frac{1}{{2\sin \left( {\frac{1}{2}} \right)}}} \right)\frac{{\cos \left( {n + \frac{1}{2}} \right)}}{n} + \mathcal{O}\!\left( {\frac{1}{{n^2 }}} \right).
$$
By another integration by parts
$$
\int_{n + 1/2}^\infty {\frac{{\sin t}}{t}dt} = \frac{1}{n}\cos \left( {n + \tfrac{1}{2}} \right) + \mathcal{O}\!\left( {\frac{1}{{n^2 }}} \right).
$$
Thus, your remainder term is
$$
-\frac{1}{{2\sin \left( {\frac{1}{2}} \right)}}\frac{1}{n}\cos \left( {n + \tfrac{1}{2}} \right) + \mathcal{O}\!\left( {\frac{1}{{n^2 }}} \right),
$$
which for large $n$ is at most
$$
\approx \frac{1}{{2\sin \left( {\frac{1}{2}} \right)}}\frac{1}{n}
$$
in absolute value. Because of the big-$\mathcal{O}$, in principle, it could be slightly larger, so this is not a complete proof.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4102159",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "14",
"answer_count": 4,
"answer_id": 2
} |
$ \int \frac{x^3}{\sqrt{x^2+x}}\, dx$ I'm trying to solve this irrational integral $$ \int \frac{x^3}{\sqrt{x^2+x}}\, dx$$
doing the substitution
$$ x= \frac{t^2}{1-2 t}$$
according to the rule.
So the integral becomes:
$$ \int \frac{-2t^6}{(1-2t)^4}\, dt= \int (-\frac{1}{8}t^2-\frac{1}{4}t-\frac{5}{16}+\frac{1}{16}\frac{-80t^3+90t^2-36t+5}{(1-2t)^4})\, dt=\int (-\frac{1}{8}t^2-\frac{1}{4}t-\frac{5}{16}+\frac{1}{16}(\frac{10}{1-2t}-\frac{15}{2} \frac{1}{(1-2t)^2}+\frac{3}{(1-2t)^3}-\frac{1}{2} \frac{1}{(1-2t)^4}))\, dt=-\frac{1}{24}t^3-\frac{1}{8}t^2-\frac{5}{16}t-\frac{5}{16}\cdot \ln|1-2t|
-\frac{15}{64}\frac{1}{1-2t}+\frac{3}{64} \frac{1}{(1-2t)^2}-\frac{1}{16 \cdot 12} \frac{1}{(1-2t)^3}+cost $$
with $t=-x+ \sqrt{x^2+x}$.
The final result according to my book is instead $(\frac{1}{3}x^2-\frac{5}{12}x+\frac{15}{24})\sqrt{x^2+x}-\frac{5}{16}\ln( x+\frac{1}{2}+ \sqrt{x^2+x})$
And trying to obtain the same solution putting t in the formulas I'm definitely lost in the calculation...
I don't understant why this difference in the complexity of the solution...
Can someone show me where I'm making mistakes?
| $$x=\frac{t^2}{1-2t}$$
$$dx=\frac{2t(1-2t)+2t^2}{(1-2t)^2}dt$$
$$=\frac{2(t-t^2)}{(1-2t)^2}dt$$
$$\sqrt{x^2+x}=x+t=\frac{t-t^2}{1-2t}$$
$$\frac{x^3dx}{\sqrt{x^2+x}}=$$
$$\frac{t^6}{(1-2t)^3}\frac{1-2t}{t-t^2}\frac{2(t-t^2)dt}{(1-2t)^2}$$
$$=\frac{2t^6dt}{(1-2t)^4}$$
To make the integration easier, put $$u=2t$$
to get
$$\frac{1}{64}\int \frac{u^6du}{(1-u)^4}$$
$$=\frac{1}{64}\int \frac{(u^6-u^5+u^5-u^4+u^4-u^3+u^3)du}{(1-u)^4}$$
Here is an other way
$$x^2+x=(x+\frac 12)^2-\frac 14$$
put
$$x+\frac 12=\frac 12\cosh(t)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4107707",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 1
} |
Repeated eigenvalue = 0 I have this quite straightforward system:
$$ \displaystyle
\begin{array}{l}
\frac{d x}{d t}=\left[\begin{array}{cc}
3 & 9 \\
-1 & -3
\end{array}\right] x \\
\text { with } x(0)=\left[\begin{array}{c}
2 \\
4
\end{array}\right] .
\end{array}
$$
My reasoning:
Using the traditional eyeball-method, one can see that this system is not linearly independant. Taking $det(A-\lambda I)$ yield $-(3+\lambda)(3-\lambda)+9 = 0 \longrightarrow \boxed{\lambda^2 = 0} $.
Since it is a repeated root, I have to get a solution in the form:
$$
\mathbf{u}(t)=c_{1} e^{\lambda t}\left(\begin{array}{l}
a_1 \\
a_2
\end{array}\right)+c_{2}\left(t e^{\lambda t}\left(\begin{array}{l}
a_1 \\
a_2
\end{array}\right)+e^{\lambda t}\left(\begin{array}{l}
b_1 \\
b_2
\end{array}\right)\right)
=e^{\lambda t}\left(c_{1}\left(\begin{array}{l}
a_1 \\
a_2
\end{array}\right)+c_{2}\left(\begin{array}{c}
b_1+a_1t \\
b_2+a_2t
\end{array}\right)\right).
$$
With eigenvectors $$
\mathbf{v_1}=\left[\begin{array}{l}
a_1 \\
a_2
\end{array}\right],
\mathbf{v_2}=\left[\begin{array}{l}
b_1 \\
b_2
\end{array}\right],
$$
*
*$v_1$ is then found with the null space of $A-\lambda I$ =
$$\begin{pmatrix} 3 & 9 \\ -1 & -3 \end{pmatrix} \thicksim \text{gaussa-gaussa}\thicksim \begin{pmatrix} 1 & 3 \\ 0 & 0 \end{pmatrix} \longrightarrow \left[\begin{array}{l}
a_1 \\
a_2
\end{array}\right] = \left[\begin{array}{l}
\hfill 3 \\
-1
\end{array}\right]$$
*$v2$ must fullfil the condition:
$$
\left(A-\lambda I\right) \mathbf{v}_{2}=\mathbf{v}_{1}
$$
$$\begin{pmatrix} 3 & 9 \\ -1 & -3 \end{pmatrix} \cdot\left[\begin{array}{l}
b_1 \\
b_2
\end{array}\right] = \left[\begin{array}{l}
\hfill 3 \\
-1
\end{array}\right]$$
I find that $v_2 = \left[\begin{array}{l}
\frac{1}{2} \\
\frac{1}{6}
\end{array}\right]$ does the job.
My solution should be:
$$e^{ \lambda t}\left(c_{1}\left(\begin{array}{l}
3 \\
-1
\end{array}\right)+c_{2}\left(\begin{array}{c}
1/2+3t \\
1/6-t
\end{array}\right)\right)$$, with $e^\lambda$ disappearing since $\lambda = 0$
I find $$c_1 = -5/3 \text{ and } c_2 = 14$$
What is wrong in my reasoning?
The problem: When entering solutions, I get wrong answer:
$$
x_1 = 21t - \frac{3}{2} \\
x_2 = -7t - \frac{1}{2}
$$
| Your solution is correct up until you plug the constants into your general solution:
$$-\frac53\begin{bmatrix}3 \\ -1\end{bmatrix} + 14\begin{bmatrix}\frac12 + 3t \\ \frac16 - t\end{bmatrix} = \begin{bmatrix}-5 \\ \frac53\end{bmatrix} + \begin{bmatrix}7+42t \\ \frac73 - 14t\end{bmatrix} = \begin{bmatrix}2 + 42t \\ 4 - 14t\end{bmatrix}$$
Verifying the solution:
$$x(0) = \begin{bmatrix}2 + 42(0) \\ 4 - 14(0)\end{bmatrix} = \begin{bmatrix}2 \\ 4\end{bmatrix}$$
$$x' = \begin{bmatrix}42 \\ -14\end{bmatrix}$$
$$Ax = \begin{bmatrix}3 & 9 \\ -1 & -3\end{bmatrix}\begin{bmatrix}2 + 42t \\ 4 - 14t\end{bmatrix} = \begin{bmatrix}(6 + 126t) + (36 - 126t) \\ (-2-42t) + (-12 + 42t)\end{bmatrix} = \begin{bmatrix}42 \\ -14\end{bmatrix}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4110811",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Showing an equality of complex numbers Let $e_k \in \mathbb{C}$ such that
$$e_1+e_2+e_3 =0$$ and
$$g =4(e_1 e_2 + e_2 e_3 + e_1 e_3)$$
Prove that $$
g^2=16(e_1^2e_2 ^2 + e_2^2 e_3^2+e_1^2 e_3^2)
$$
What I got :
$$
g^2 =16(e_1^2e_2 ^2 + e_2^2 e_3^2+e_1^2 e_3^2+2e_1^2e_2e_3+2e_1e_2^2e_3+2e_1e_2e_3^2)
$$
However, I fail to see how $2e_1^2e_2e_3+2e_1e_2^2e_3+2e_1e_2e_3^2 = 0$
Would appreciate any help
| You can factor out $2e_{1}e_{2}e_{3}$ from $$2e_{1}^{2}e_{2}e_{3} + 2e_{1}e_{2}^{2}e_{3} + e_{1}e_{2}e_{3}^{2},$$ which gives you $$2e_{1}e_{2}e_{3}(e_{1} + e_{2} + e_{3}).$$ Because $e_{1} + e_{2} + e_{3} = 0$, then $2e_{1}^{2}e_{2}e_{3} + 2e_{1}e_{2}^{2}e_{3} + e_{1}e_{2}e_{3}^{2} = 0$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4119918",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 2,
"answer_id": 1
} |
Calculate the value of the expression for polynomial roots The roots of the polynomial equation: $ x ^ 3 - 3x -1 = 0 $ are: $ a, b $ and $ c $. Based on this, calculate the value of:$$ L = \frac {1} {a ^ 2-2a + 2} + \frac {1} {b ^ 2-2b + 2} + \frac {1} {c ^ 2-2c + 2} $$
Attemp:
$L=\frac{1}{(a-1)^2+1}+\frac{1}{(b-1)^2+1}+\frac{1}{(c-1)^2+1}$.
First, we want to find a polynomial whose roots are $a-1$, $b-1$ and $c-1$. This polynomial is just $(x+1)^3-3(x+1)-1=x^3+3x^2-3$. If we say the roots of this are $a'$, $b'$ and $c'$, we want to find $\frac{1}{a'^2+1}+\frac{1}{b'^2+1}+\frac{1}{c'^2+1}$.
Now we want to find what polynomial has roots $a'^2$, $b'^2$ and $c'^2$. In order to find what this polynomial is, we need to know $a'^2+b'^2+c'^2$, $a'^2b'^2+b'^2c'^2+c'^2a'^2$ and $a'^2b'^2c'^2$. $a'^2+b'^2+c'^2=(a'+b'+c')^2-2(a'b'+b'c'+c'a')=3^2-2(0)=9$. $a'^2b'^2+b'^2c'^2+c'^2a'^2=\frac{(a'^2+b'^2+c'^2)^2-a'^4-b'^4-c'^4}{2}$. To find $a'^4+b'^4+c'^4$, we can realize that $a'^3+b'^3+c'^3=(a'^2+b'^2+c'^2)(a'+b'+c')-(a'+b'+c')(a'b'+b'c'+c'a')+$ $a'b'c'(a'^0+b'^0+c'^0)=-3(a'^2+b'^2+c'^2)+3(a^0+b^0+c^0)=-18$, and similarly $a'^4+b'^4+c'^4=-3(-18)+3(-3)=45$. Our expression is therefore $\frac{(a'^2+b'^2+c'^2)^2-a'^4-b'^4-c'^4}{2}=\frac{81-45}{2}=18$. Also, it's easy to see $(a'b'c')^2=9$. Our polynomial is $x^3-9x^2+18x-9$
Our sum is the sum of the reciprocals of the $(x-1)^3-9(x-1)^2+18(x-1)-9=x^3-12x^2+39x-37$ which equals $\frac{39}{37}$
attemp2: I also think about using the logarithmic derivative you can think that
$\frac{1}{(1 - a) ^2} + \frac{1}{(1 - b)^2} + \frac{1 }{(1 - c)^2} = \frac{(p (1) p '' (1) - p '(1)^2) } {p (1 )^2} = 2$, then there must be some magical way to make that 1 appear by adding to the denominator, without having to appeal to polynomial transforms
Am I right? Is there an easier way to solve this problem?
| This page gives a nice summary of common polynomial root transforms, which we can employ here.
First, let's find a polynomial that has roots that are $a-1, b-1, c-1$. This is just $$Q(x) = P(x+1) = (x+1)^3 -3(x+1) -1 = x^3 +3x^2 -3.$$
Next, we find the polynomial that has roots $(a-1)^2, (b-1)^2, (c-1)^2.$ This is
$$R(x) = -Q(\sqrt{x}) Q(-\sqrt{x}) = x^3 - 9x^2 + 18x - 9$$
Then, the polynomial having roots $(a-1)^2 + 1, (b-1)^2 + 1, (c-1)^2 +1$ is
$$S(x) = R(x-1) = x^3 - 12x^2 + 39x - 37.$$
We could find the polynomial that has roots that are the reciprocals of the roots of $S(x)$, however, we can apply Vieta to obtain that the sum is indeed $\frac{39}{17}.$
What you did is of course correct, but you were more explicit with these computations. These tricks are nice to know so you don't have to do all the work every time!
As for your second approach, it's not clear to me how to proceed. Maybe someone else could shed some light?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4121424",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
} |
Solve the inequality $\log_3(2^{x+1}-8)\le x$ Find x
$$\log_3(2^{x+1}-8)≤x$$
I found that $x>2$
$$\log_3(2^{x+1}-8)\le x\log_33$$
$$(2^{x+1}-8)\le 3^x$$
$$(2^{x+1}-8)> 0$$
$$(2^{x+1})>8$$
$$2^{x+1}>2^3$$
$${x+1}>3$$
$${x}>2$$
| We have
$$\log_3(2^{x+1}-8)≤x$$
$$\log_3(2^{x+1}-8)≤\log_33^x$$
$$\begin{align}&\begin{cases} 2^{x+1}-3^x-8≤0 \\x+1>3 \end{cases} \\\\ \iff &\begin{cases}3^x-2^{x+1}≥-8 \\ x>2\end{cases}\end{align}$$
Then, note that $f(x)$ is a strictly increasing function, $\forall x>2.$
This means,
$$f(x)=3^x-2^{x+1}>1≥-8, \forall x>2$$
Finally, you get the right answer
$$x\in(2, +\infty)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4123192",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 1,
"answer_id": 0
} |
Use generating functions to calculate the sum $\sum_{k=0}^n (k+1)(n-k+1)$ Use generating functions to calculate the sum $\sum_{k=0}^n (k+1)(n-k+1)$
And Prove it by induction.
Attempt:
$$A(x)= \sum_{k=0}^n (k+1)(n-k+1)$$
$$\sum_{n=0}^\infty(\sum_{k=0}^n (k+1)(n-k+1))x^n$$
The series of differences is :$\{(n+1),2(n),3(n-1),4(n-2)…,1,…\}$
The generating functions of this series is : $(1-x)A(x)$
The series of differences is :$\{(n+1),(n-1),(n-3),(n-5),…,2-n,…\}$
The generating functions of this series is : $(1-x)^2A(x)$
The series of differences is :$\{(n+1),-2,-2,-2,…,-2,…\}$
The generating functions of this series is : $(1-x)A^3(x)$
The series of differences is :$\{(n+1),-(n+3),0,0,…,0,…\}$
The generating functions of this series is : $(1-x)^4A(x)$
On the other hand, the generating functions $(n+1)+(-n-3)x$,
Therefore: $(1-x)^4A(x)=(n+1)+(-n-3)x$
$$A( x) =\frac{( n+1) +( -n-3) x}{( 1-x)^{4}} =(( n+1) +( -n-3) x) \cdot \sum _{n=0}^{\infty }\binom{n+4-1}{4-1} x^{n}$$
$$\displaystyle \sum _{k=0}^{n}( k+1)( n-k+1) =\sum _{n=0}^{\infty }( n+1) \cdot \binom{n+4-1}{4-1} x^{n} +\sum _{n=0}^{\infty }( -n-3) \cdot \binom{n+4-1}{4-1} x^{n+1}$$
Prove in with induction :
\begin{aligned}
( n+1) \cdot \binom{n+3}{3} +( -n-3)\binom{n+2}{3} & =\frac{( n+3)( n+2)( n+1)^{2} n}{3!} +\frac{( n+2)( n+1)( n-1) n( -n-3)}{3!}\\
& =\frac{( n+3)( n+2)( n+1)^{2} -( n+2)( n+1)( n+3) n}{6}\\
& =\frac{( n+3)( n+2)( n+1)}{6}\\
& =\frac{1}{6}\left( n^{3} +6n^{2} +11n+6\right)
\end{aligned}
$$\displaystyle \sum _{k=0}^{n}( k+1)( n-k+1) =\frac{1}{6}\left( n^{3} +6n^{2} +11n+6\right)$$
induction basis: $n=0$
$$\displaystyle \sum _{k=0}^{n}( 0+1)( 0-0+1) =1=\frac{1}{6}\left( 0^{3} +6\cdotp 0^{2} +11\cdotp 0+6\right)$$
For $n+1$:
\begin{aligned}
\sum _{k=0}^{n+1}( k+1)( n-k+1) & =\sum _{k=0}^{n}( k+1)( n-k+1) +( n+2)(( n+1) -( n+1) +1)\\
& =\frac{1}{6}\left( n^{3} +6n^{2} +11n+6\right) +( n+2)\\
\end{aligned}
If anyone can enlighten me, and find my mistakes I appreciate that.
Unfortunately, I did not succeed.
| I have no idea what you are doing, but the generating function of $(k+1)(n-k+1)$ is the derivative of the generating function of $(n-k+1).$ In other words,
if $$A_n(x) = \sum_{i=1}^{n+1} (n-k+1) x^k,$$ then
$$A^\prime_n = \sum_{i=0}^n (k+1)(n-k+1),$$ and your sum is just equal to $A_n^\prime(1).$
Now, $$A_n(x) = n \sum_{i=1}^{n+1} x^k - \sum_{i=1}^{n+1}(k+1)x^k.$$
The geometric series you (hopefully) know how to sum, while the second term is the derivative of a geometric series.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4124684",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
} |
Use laws of logarithms to expand $\ln\left(2b\sqrt{\frac{b+1}{b-1}}\right)$ I am to expand $\ln\left(2b\sqrt{\frac{b+1}{b-1}}\right)$ using the laws of logarithms. The solution is provided as $\ln(2)+\ln(b)+\frac{\ln(b+1)-\ln(b-1)}{2}$ whereas I got $\ln(2)+\ln(b)+\frac{1}{2}\ln(b+1)-\frac{1}{2}\ln(b-1)$
My working:
$$\ln\left(2b\sqrt{\frac{b+1}{b-1}}\right)$$
$$=\ln(2)+\ln(b)+\ln\left(\sqrt{\frac{b+1}{b-1}}\right)$$
$$=\ln(2)+\ln(b)+\frac{1}{2}\ln\left(\frac{b+1}{b-1}\right)$$
$$=\ln(2)+\ln(b)+\frac{1}{2}\ln(b+1)-\frac{1}{2}\ln(b-1).$$
I suspect that I might have made a mistake using the power rule to use $1/2$ for the root of the expression $\frac{b+1}{b-1}$, but I'm unsure.
How can I arrive at the correct solution?
| $$\frac{1}{2}A-\frac{1}{2}B=\frac{A-B}{2}$$
where $A=\ln(b+1)$, $B=\ln(b-1)$.
In short, you arrived at the given answer.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4125471",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
An immediate alternative to a trigonometry problem (high school) I have a right triangle, $AH\perp BC$, where $\cos \beta=\sqrt 5/5$ and $\overline{AH}+\overline{CH}+\overline{HB}=7$.
I have to found the area.
My steps (or solution):
$$\mathcal A(\triangle ABC)=\frac 12\overline{AB}\cdot \overline{BC} \sin \beta$$
with $0<\beta<\pi$. Hence $\sin \beta=\sqrt{1-5/25}=2\sqrt5/5$. But $\overline{AB}=\overline{BC}\cos \beta$ and the area
$$\mathcal A(\triangle ABC)=\frac12(\overline{BC}\cos \beta)\cdot (\overline{BC}\sin \beta)=\frac12\overline{BC}^2 \tag 1$$
We know that:
$$\overline{AH}+\overline{CH}+\overline{HB}=7, \quad \text{with}\quad \overline{CH}+\overline{HB}=\overline{BC}$$
Hence
$$\overline{AH}+\overline{BC}=7 \iff \overline{BC}=7-\overline{AH}$$
and $\overline{AH}=\overline{AB}\cos\gamma=\overline{BC}\cos\gamma\cos\beta$.
Putting this last identity to the condition $\overline{BC}=7-\overline{AH}$ I will have:
$$\overline{BC}=7-\overline{BC}\cos\gamma\cos\beta \iff \overline{BC}(1+\cos\gamma\cos\beta)=7$$
After,
$$\overline{BC}=\frac{7}{1+\cos\gamma\cos\beta}$$
with $\cos \gamma=\cos(\pi/2-\cos\beta)=\sin\beta=2\sqrt5/5$.
Definitively
$$\overline{BC}^2=\frac{49}{\left(1+\frac 25\right)^2}=\frac{49}{\frac{49}{25}}$$
and
$$\mathcal{A}(\triangle ABC)=\frac15\overline{BC}^2=5 \tag 2$$
I'm very tired and often don't find immediate solutions to problems. I had thought of using Euclid's second theorem by placing
$\overline{AH}=x$ and $\overline{BC}=y$, and $x+y=7$. After I will have $x^2=\overline{BH}\cdot \overline{CH}$. But I have left this path.
Can this possible alternative be used or is there another more immediate way?
| $$\cos \beta = \frac{|AB|}{|BC|} = \frac{\sqrt 5}{5},\quad \sin \beta = \frac{|AH|}{|AB|} = \frac{2\sqrt 5}{5}$$
Given $|AB|: |BC| = \sqrt5:5$, we assume $$|AB| = \sqrt5k, |BC| = 5k $$ where $k$ is a constant. Now using $$\frac{|AH|}{|AB|} = \frac{2\sqrt 5}{5}, \quad |AH| = 2k$$
$$|AH| + |BH| + |HC| = |AH| + |BC| = 7k = 7\implies k = 1.$$
So, $$|AH| = 2, |BC| = 5 \implies \mathcal{A}(\triangle ABC) = \frac{1}{2} \cdot |AH| \cdot |BC| = 5.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4127013",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 5,
"answer_id": 1
} |
For positive integer $n$, why is $\lfloor \log_{10}(2^n)\rfloor + \lfloor \log_{10}(5^n)\rfloor + 2 = n+1$?
For positive integer $n$, why is $\lfloor \log_{10}(2^n)\rfloor + \lfloor \log_{10}(5^n)\rfloor + 2 = n+1$?
This question comes from counting the number of digits of $10^n$ in terms of the number of digits of $2^n$ and $5^n$. Number of digits of $10^n$ is $n+1$ which equals to the sum of the number of digits of $2^n$ and $5^n$. I know that the number of digits of a positive integer $x$ is $\lfloor \log_{10}(x)\rfloor + 1$.
Using programming, I've checked that this is true for $n$ less than $100$.
| There is a slightly more general result:
Claim 1: Let $a$ and $b$ be 2 positive nonintegral reals satisfying $a+b=n$ for some natural number $n$. Then $$\lfloor a \rfloor + \lfloor b\rfloor =n-1.$$
To see this, note that as $a+b$ is integral, it follows that $(a-\lfloor a \rfloor) + (b-\lfloor b \rfloor)$ must also be integral, and as $a$ and $b$ are each not integral, it follows that each of $(a-\lfloor a \rfloor)$, $(b -\lfloor b \rfloor)$ must be strictly between $0$ and $1$, and so the sum must be strictly between $0$ and $2$. So as this sum still must be integral, it follows that this sum must be $1$, which gives
$$(a-\lfloor a \rfloor) + (b-\lfloor b \rfloor) =1,$$ and therefore
$$\lfloor a \rfloor + \lfloor b \rfloor = (a+b)-1=n-1.$$ $\surd$
So let $a \doteq \log_{10}(2^n)$ and $b\doteq \log_{10}(5^n)$; $n$ integral, note that $a$ and $b$ are each nonintegral [why is that] and $$a+b = \log_{10}(5^n)+\log_{10}(2^n) =\log_{10}10^n = n.$$
Can you finish then using Claim 1?
| {
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"timestamp": "2023-03-29T00:00:00",
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Finding all tangent lines to circle $x^2+y^2=9$ having exactly one common point with parabola $y=x^2+3$
I have to find all tangent lines to the circle $x^2+y^2=9$, which have exactly one common point with the parabola $y=x^2+3$.
I drew it and I saw that $(0,3)$ is a common point circle and parabola. I found that three tangent lines are: $y=3$, $x=-3$, $x=3$. And now the question is whether these are all? I don't know how to explain it or find others lines.
Thanks for your help!
| At any given point on the circle, there is a unique tangent line. At the point $(0,3)$ you have identified 3. Only one is tangent.
The other lines:
Since this is tagged pre-calculus, I will not use calculus in this answer.
Suppose $x_1,y_1$ lies on the circle.
Slope of the tangent line though that point is negative reciprocal of the slope of the radius.
$m = -\frac {x_1}{y_1}$
Equations of the tangent line in point slope and y-intercept form:
$y-y_1 = -\frac {x_1}{y_1} (x-x_1)\\
y = -\frac {x_1}{y_1} x + \frac {x_1^2 +y_1^2}{y_1}\\
y = -\frac {x_1}{y_1} x + \frac {9}{y_1}\\
$
Remember $x_1,y_1$ lie on the circle so $x_1^2 + y_1^2 = 9$ we will be using this fact a couple more times.
The line is also tangent to $y = x^2 + 3$
Substituting:
$-\frac {x_1}{y_1} x + \frac {9}{y_1} = x^2 + 3\\
y_1x^2 + x_1x + 3y_1 - 9 = 0$
Since our line is tangent, when we plug this into the quadratic formula, we must get exactly one root. i.e. $b^2 - 4ac = 0$.
$x_1^2 - 4y_1(3y_1-9) = 0\\
x_1^2 + 36y_1 - 12y_1^2 = 0$
We can use $x_1^2+ y_1^2 = 9$ again. Substitute $x_1^2 = 9 - y_1^2$
$(9-y_1^2) + 36y_1 - 12y_1^2 = 0\\
13y_1^2 - 36y_1 - 9 = 0\\
(y_1 - 3)(13y_1 + 3) = 0$
We have two values for $y_1.$ Solve for $x_1.$
$x_1^2 + 9 = 9\\
x_1^2 + \frac {9}{169} = 9$
$(0,3),(\frac {3\sqrt {168}}{13},-\frac {3}{13}), (-\frac {3\sqrt {168}}{13},-\frac {3}{13})$ are our 3 points of tangency.
And, plug into our earlier equations of the tangent lines.
$y = 3\\
y = \sqrt {168} x - 39\\
y = -\sqrt {168} x - 39$
| {
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Evaluating limits using Taylor expansions The limits are
$$\lim_{x\to 0}(\frac{\cos{x}-e^{-x^2/2}}{x^4})$$
$$\lim_{x\to 0}\frac{e^x\cdot\!\sin{x}-x(1+x)}{x^3}$$
Probably wrong things that I've tried
$\lim_{x\to 0}\frac{\cos{x}-e^{-x^2/2}}{x^4}=\lim_{x\to 0}(\frac{1-\frac{x^2}{2!}+\frac{x^4}{4!}+o(x^5)-1+\frac{x^2}{2}+\frac{x^4}{4}+o(x^4)}{x^4})$
$\lim_{x\to 0}\frac{e^x\cdot\!\sin{x}-x(1+x)}{x^3}=\lim_{x\to 0}\frac{(1+x+\frac{x^2}{2}+o(x^2))(x+o(x^2))-x-x^2)}{x^3}=\lim_{x\to 0}\frac{\frac{x^3}{2}+o(x^2)}{x^3}=1/2$
Could you please help me understand how these kinds of limits can be computed I keep getting all of them wrong and if this goes on for a little bit longer I may have a panic attack.
| The first rule is that, when you use Taylor's expansion, you use it at the same order for all terms. The second rule is that, when you have a product of expansions, or a composition, you expand it with the usual rules for polynomials, and truncate the result at the chosen order.
I'll start it for the first example, for which you had an error in the expansion of the exponential. We'll expand at order $4$:
\begin{align}
\cos x- \mathrm e^{-x^2/2}
&=1-\frac{x^2}2+\frac{x^4}{24}+o\bigl(x^4\bigr)-\Bigl(1-\frac{x^2}2+\frac{x^4}8+o\bigl(x^4\bigr)\Bigr)\\
&=\frac{x^4}{24}-\frac{x^4}8+o\bigl(x^4\bigr)=-\frac{x^4}{12}+o\bigl(x^4\bigr)
\end{align}
| {
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Showing $\tan70° = \tan20° + 2\tan50°$ Q. Prove that $\tan 70° = \tan 20° + 2\tan 50°$.
My approach:
$ LHS = \tan 70° = \dfrac1{\cot 70°} = \dfrac1{\tan 20°}$
$ \begin{align}
RHS
&= \tan 20° + 2\tan 50° \\
&= \tan 20° + 2\tan (20+30)° \\
&= \tan 20° + \dfrac{2(\tan 20° + 1/√3)}{1 - \tan 20°/√3} \\
&=\dfrac{2 + 3√3 \tan 20° - \tan^2 20°)}{√3 - \tan 20°}
\end{align}
$
Why are the two sides not equal despite being expressed in the same terms? Can someone offer some help?
Much to my surprise, my friend just expanded tan70° and cross-multiplied the terms to prove it.
| You can do the following: $\tan (50) = \tan (70 - 20)= \frac{\tan (70) - \tan (20)} {1+ \tan( 70). \tan (20)}.$ The bottom is $2,$ so you are done.
In general, if $ A +B = \pi/2,$ then $\tan A = \tan B + 2\tan(A-B)$, where $A$ is the bigger one.
Edit: If you can also start from the right/left side. Essentially do the same trick to get equality. For example,
RHS= $\tan 20° + 2\tan 50° \\
=\tan 20° + 2\tan (70-20)° \\
= \tan 20° + \frac{2(\tan 70° - \tan 20° )}{1 + \tan 70° \tan 20°}
= \tan 20° + \frac{2(\tan 70° - \tan 20° )}{2}\\
=\tan 20° + \tan 70° - \tan 20° .\\
$
To show this equality, you split the angle differently, and the expression has become a bit complicated-looking and Tavish explained why they have to be equal.
| {
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Proof of Least upper bound of $\frac{x}{x+1}$ The questions states:
Prove that there is a unique positive real number $m$ that has the following two properties.
*
*For every positive real number $x$, $\frac{x}{x+1}<m$.
*If $y$ is any positive real number with the property that for every positive real number $x$ , $\frac{x}{x+1}<y$ then $m\leq y$.
My concern is my proof of the second part as I am not sure if its valid or there is a clearer way of phrasing it.
For the first part I used $m=1$. My doubt is with my second portion
My proof:
Suppose for the sake of contradiction, that there exists a $y<m$ such that $\frac{x}{x+1}<y$. Then $0<\frac{x}{x+1}<y<1$. Thus $y$ can be written as $\frac{p}{q}$ where $p$ and $q$ are positive integers and $q>p$. As $p$ and $q$ are positive integers, then $q-p\geq1$. Thus $q+1\geq p+2$. So we have $\frac{x}{x+1} < \frac{p}{q} < \frac{p+1}{q+1} \leq \frac{p+1}{p+2}$. However if we choose $x=p+1$ then $\frac{x}{x+1} = \frac{p+1}{p+2} \geq y$ , we arrive at a contradiction. So it follows that if $y$ is any positive real number with the property that for every positive real number $x$ , $\frac{x}{x+1}<y$ then $m\leq y$.
My specific questions are :
*
*Is it valid to express $\frac{p+1}{p+2}$ as $\frac{x}{x+1}$, or do I need to choose another variable $y$ to show that its the same thing.
*Is it needed to show $\frac{p}{q} < \frac{p+1}{q+1}$ ? Or does it make the proof too long winded?
*Is there a more succinct way of proving this?
| $$ \dfrac{x}{x+1} < \dfrac{x+1}{x+1} \implies \dfrac{x}{x+1} < 1 $$
So 1 is an upper bound. Let $\epsilon < 1$ We will show that there is an $x$ for which
$\dfrac{x}{x+1} = 1-\epsilon$
\begin{align}
\dfrac{x}{x+1} &= 1-\epsilon \\
x(1-\epsilon) + 1(1-\epsilon) &= x \\
-\epsilon x &= -(1-\epsilon) \\
x &= \dfrac{1-\epsilon}{\epsilon}
\end{align}
So, for all positive \epsilon close to $0$, $x = \dfrac{1-\epsilon}{\epsilon}$ will give us $\dfrac{x}{x+1} = 1-\epsilon$.
It follows that $1$ is the LUB of $\dfrac{x}{x+1}$.
| {
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Denesting Cardano's Formula For a depressed cubic equation $x^3 + px + q =0$ having exactly one real root, Cardano's formula gives the real root as $$\sqrt[3]{-\frac{q}{2} +\sqrt{\frac{q^2}{4} + \frac{p^3}{27}}} + \sqrt[3]{-\frac{q}{2} - \sqrt{\frac{q^2}{4} + \frac{p^3}{27}}}.$$ Suppose we know that the cubic equation has exactly one real root which is an integer or rational number. For example, consider $f = x^3 +x -2$ which has exactly one real root 1. Cardano's formula gives the root as $$\sqrt[3]{1 +\sqrt{\frac{28}{27}}} + \sqrt[3]{1 -\sqrt{\frac{28}{27}}}.$$ I referred to this post and used the technique mentioned there to observe that $\sqrt[3]{1 +\sqrt{\frac{28}{27}}}$ can be denested as $\frac{1}{2}+\frac{\sqrt{21}}{6}$ and $$\sqrt[3]{1 -\sqrt{\frac{28}{27}}} = \frac{1}{2}-\frac{\sqrt{21}}{6}.$$ This recovers our integral root $1,$ as desired. I tried few more examples and I was able to denest the outer cubic radical in those cases also.
This brings us to my question. Suppose we have a cubic polynomial $f \in \mathbb{Q}[x]$ which is promised to have exactly one integral/rational root and other two to be complex conjugates. Then, can we always denest the outer cubic radical as shown in the above example? Is there any counterexample and if not, what is the proof/proof idea for denesting.
I refered to this paper in search of a solution but it is heavily loaded with Galois theory which is a bit overwhelming for me at this stage. However, I did start reading the basics of Galois theory - field extensions, Galois correspondence etc. and I hope that I would be able to understand the layman terms of any answer to this question which involves Galois theory.
| If $n$ is your one real integer solution, then $$x^3+px+q=(x-n)(x^2+nx+b)$$ where $-nb=q, b-n^2=p$ and $n^2<4b,$ that last because there are real roots to $x^2+nx+b$ otherwise.
Substitute $p=b-n^2, q=nb$ into your formula, then see if you can eliminate the cube roots in the general case. You should be able to.
In particular:
$$\begin{align}D=\frac{q^2}{4}+\frac{p^3}{27}
&=\frac{n^2b^2}{4} +\frac{(b-n^2)^3}{27}\\
&=\frac{27n^2b^2+4b^3-12n^2b^2+12n^4b-4n^6}{108}\\
&=\frac{4b^3+15n^2b^2+12n^4b-4n^6}{108}\\
&=\frac{(4b-n^2)(b+2n^2)^2}{108}
\end{align}$$
So $$-\frac q2+\sqrt{D}=\frac{bn}2+\frac{ b+2n^2}{6}\sqrt{\frac{4b-n^2}{3}}$$
It turns out this has a simple cube root:
$$\begin{align}\frac{bn}2+\frac{ b+2n^2}{6}\sqrt{\frac{4b-n^2}{3}}&=\left(\frac{n}{2}+\frac 12\sqrt{\frac{4b-n^2}3}\right)^3
\end{align}$$
So you can always reduce the cube root.
In your case $n=1,b=2,$ and this can be written:
$$1+\frac{2}{3}\sqrt{\frac 7 3}=\left(\frac12+\frac12\sqrt{\frac73}\right)^3$$
Nothing in this result requires $n$ to be an integer. The equalities are all true regardless of what $n$ is. It even holds when there are other real roots, or if $n$ is a complex root.
For example if $p=-7,q=6,$ then $x^3-7x+6=0$ has roots $1,2,-3.$ Cardano requires us to s find the cube root of:
$$S:=\frac72 +\frac{10}{9}\sqrt{-3}$$
When $n=1, b=-6$ and you get:
$$S=\left(\frac12+\frac{1}{2}\sqrt{\frac{-25}{3}}\right)^3$$
When $n=2, b=-3,$ and
$$S=\left(1+\frac{1}{2}\sqrt{\frac{-16}{3}}\right)^3$$
When $n=-3, b=2$ and
$$S=\left(-\frac32+\frac{1}{2}\sqrt{\frac{-1}{3}}\right)^3$$
| {
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Indefinite integration of $\int\frac{1}{\cos(x-1)\cos(x-2)\cos(x-3)}\,\textrm dx$
Integrate $$\int\dfrac{1}{\cos(x-1)\cos(x-2)\cos(x-3)}\,\textrm dx$$
My Attempt:
Using, $$\tan A-\tan B=\dfrac{\sin(A-B)}{\cos A\cdot \cos B}$$
The given integral can be transformed as
$$\int\dfrac{\tan(x-1)}{\cos(x-2)}\,\textrm dx - \int\dfrac{\tan(x-3)}{\cos(x-3)}\,\textrm dx$$
The right most integral can be calculated easily by writing $\tan(x-3)$ as $\frac{\sin(x-3)}{\cos(x-3)}$ and then by a substituiton $\cos(x-3)$ as $t$. But I have no clue for the left most integral. How to evaluate that?
| Note $\cos(x-n) = \cos x\cos n + \sin x \sin n$ and rewrite the integrand as
\begin{align}
\frac{1}{{\cos(x-1)\cos(x-2)\cos(x-3)}}
={}\frac{\csc1\csc2\csc3\sec^3 x}{(\tan x+\cot 1) (\tan x+\cot 2) (\tan x+\cot 3)}
\end{align}
Substitute $t=\tan x $ and perform the partial fractionization
\begin{align}
&\int \frac{1}{\cos(x-1)\cos(x-2)\cos(x-3)}\,dx \\
={}& \frac{1}{\sin 2}\int\frac{\sqrt{1+t^2}}{t+\cot1}\,dt
-\frac{1}{\sin1}\int\frac{\sqrt{1+t^2}}{t+\cot2}\,dt
+ \frac{\sin 3}{\sin1\sin2}\int\frac{\sqrt{1+t^2}}{t+\cot3} \, dt\\
\end{align}
The three integrals are of the same form and can be readily carried out.
| {
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Prove that $\sum_{cyc} \sqrt[n]{\frac{a}{b+c}} \ge \frac{3}{\sqrt[n]{2}}$ Given a,b,c are positive real numbers and n is positive integer number.
Prove that
$$\sum_{cyc} \sqrt[n]{\frac{a}{b+c}} \ge \frac{3}{\sqrt[n]{2}}$$
I tried expanding it, using equivalence transformation, using inequalities AM-GM, Holder.., but no success. I can only prove the above inequality for n=1
$$\sum_{cyc} \frac{a}{b+c}=\sum_{cyc} \frac{a^2}{ab+ac}\geq \frac{(a+b+c)^2}{2(ab+bc+ca)}\geq \frac{3(ab+bc+ca)}{2(ab+bc+ca)}=\frac{3}{2}$$
I hope your help, thanks
| Your conjecture does not hold for $n > 1$. Let $a = b > 0$ and $c \to 0$ then, for $n > 1$,
$$
\sum_{cyc} \sqrt[n]{\frac{a}{b+c}} \to 2 \le \frac{3}{\sqrt[n]{2}}
$$
Indeed, $2$ is a tight lower limit for $n > 1$.
| {
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Finding $\sum_{0 \le r We know that $$\sum_{0 \le r <s \le n} A_r A_s=\frac{1}{2}\left [(\sum_{t=0}^{n} A_t)^2-\sum_{t=0}^{n} A^2_t\right]$$
Nest using $A_r={n \choose r}x^r$, we can get
$$S(x)=\sum_{0 \le r <s \le n} {n \choose r}{n \choose s}x^{r+s}=\frac{1}{2}\left[(\sum_{t=0}^{n} {n \choose r} x^r)^2=\sum_{t=0}^{n} {n \choose t}^2 x^{2t}\right]=\frac{1}{2}\left[(1+x)^{2n}~-~_2F_1(-n,-n,1,x^2)\right]$$ Differentiating w.r.t. $x$ noting that $$\frac{d}{dz}~_2F_1(a,b,c;z) =\frac{ab}{c}~_2F_1(1+a,1+b,1+c;z),$$ we get
$$S'(x)=\sum_{0 \le r <s \le n} (r+s){n \choose r}{n \choose s}x^{r+s-1}=n(1+x)^{2n-1}-n^2x~_2F_1(1-n,1-n,2,x^2)]$$
Let $x=1$, then $$\sum_{0 \le r <s \le n} (r+s) {n \choose r}{n \choose s}=n2^{2n-1}-\frac{n}{2} {2n \choose n}.~~~~~(*)$$
The question is how else one can get this summation (*)?
| $$ \begin{align} \tag{1} S &\ = \sum_{0 \le r <s \le n}~ (r+s){n \choose r}{n \choose s} \\
&\ = \sum_{0 \le r <s \le n}~ (r+s){n \choose n-r}{n \choose n-s} \\
&\ \tag{2} = \sum_{0 \le r <s \le n}~ (n-r+n-s){n \choose n-r}{n \choose n-s} \\
\end{align}$$
Add $(1)$ and $(2)$
$$ 2S= \sum_{0 \le r <s \le n}~ 2n{n \choose r}{n \choose s}$$
$$\begin{align} S &\ = n\sum_{0 \le r <s \le n}~ {n \choose r}{n \choose s} \\ &\ =n \left(2^{2n-1}-\dfrac12 \binom{2n}{n}\right) \\ \end{align} $$
NOTE: Finding $ \sum_{0 \le r <s \le n}~ {n \choose r}{n \choose s}$
$$ (1+x)^{2n}=\sum_{0 \le r,s \le n}~ {n \choose r}{n \choose s}x^{r+s}$$
Put $x=1$
$$ \begin{align} 2^{2n} &\ =\sum_{0 \le r,s \le n}~ {n \choose r}{n \choose s} \\ &\ = \sum_{0 \le r<s \le n}~ {n \choose r}{n \choose s} + \sum_{0 \le s<r \le n}~ {n \choose r}{n \choose s}+ \sum_{0 \le r=s \le n}~ {n \choose r}{n \choose s} \\
&\ = S_1 + S_1 + \sum_{k=0}^{n} \binom{n}{k}^2 \\
&\ = 2S_1 + \binom{2n}{n} \\
\end{align}$$
$$\therefore S_1 = 2^{2n-1} - \dfrac12 \binom{2n}{n}$$
| {
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Proving $\left(a^2+b^2\right)^2\geqslant(a+b+c)(a+b-c)(b+c-a)(c+a-b)$ for positive reals $a$, $b$, $c$ Question $5$ of BMO1 $2008$:
For positive real numbers $\;a,\;b,\;c,\;$ prove that $$\left(a^2+b^2\right)^2\geqslant(a+b+c)(a+b-c)(b+c-a)(c+a-b)$$
I noticed that the right side can be grouped, but did not get further.
| If the right hand side is negative then the inequality trivially holds. Otherwise $a,b,c$ are the sides of a triangle, and by Heron's formula for the area $S$ of a triangle the inequality reduces to:
$$
a^2+b^2 \ge 4 S
$$
This follows from $S = \frac{1}{2} ab \sin C \le \frac{1}{2} ab$ and the means inequalities $ab \le \left(\frac{a+b}{2}\right)^2\le \frac{a^2+b^2}{2}\,$.
[ EDIT ] It is not possible for two factors on the right hand side of the original expression to be negative, since for example $a+b-c\lt 0$ and $b+c-a \lt 0$ would imply $b \lt 0$. Therefore, when the right hand side is non-negative each factor must be non-negative, so $a,b,c$ observe the triangle inequalities and must be the sides of a (possibly degenerate) triangle.
| {
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Sequence $ a_0=1,a_1=3$ and $a_{n+2}=4a_{n+1}-a_n $. Prove that $2017|a^2_{2017}-4a_{2017}+3$ I need a solution verification for next problem.
Let sequence $a_0,a_1,...$ be such that:
$ a_0=1,a_1=3$ and $a_{n+2}=4a_{n+1}-a_n $. Prove that $$2017\mid a^2_{2017}-4a_{2017}+3$$
By solving this recurrence we get $$a_n = {3+\sqrt{3}\over 6}(2+\sqrt{3})^n+{3-\sqrt{3}\over 6}(2-\sqrt{3})^n$$
Notice that $p:=2017$ is prime. By Gauss reciprocity law we have $\big({3\over p}\big) = 1$ so there exists $a\in \mathbb{Z}_p$ such that $a^2\equiv 3 \pmod p$. Consider now this sequence in the field $\mathbb{Z}_p$. So we have $$6a_n = (3+a)(2+a)^n+(3-a)(2-a)^n$$
By Fermat little theorem we have $$6a_p\equiv _p (3+a)(2+a)+(3-a)(2-a) \equiv_p 18$$
So, since $p\nmid 6$ we have $a_p \equiv_p 3$ and thus a conclusion.
| Your proof works.
I wondered why they chose $$f(m)=m^2-4m+3=(m-1)(m-3),$$ I see when $p=5,7$ that $a_p\equiv 1\pmod p.$
I wonder if this is true for all primes. To prove that, you’d want to to show that if $\left(\frac 3p\right)=-1$ then $a_p\equiv 1\pmod p.$
Proof as outlined by commenter Derive Foiler.
If $\left(\frac 3p\right)=-1$ then we have $F_{p^2}=\mathbb F_{p}[\sqrt{3}].$ And the map $f:F_{p^2}\to F_{p^2}$ sending $u\to u^p$ is a non-trivial automorphism. So $f(a+b\sqrt{3})=a-b\sqrt 3.$
Then:
$$6a_n = (3+\sqrt{3})(2-\sqrt{3})+(3-\sqrt{3})(2+\sqrt 3)=6$$
So $a_n\equiv 1\pmod p.$
This shows more generally, if $T_{n+1}=bT_n+cT_{n-1}$ with $b,c$ integers and $T_0,T_1$ integers, and $b^2+4c$ is not a perfect square, then when $p\not\mid b^2+4c$ is an odd prime, then $T_p$ is congruent to one of two values, depending on the value of $\left(\frac{b^2+4c}{p}\right).$
When $\left(\frac{b^2+4c}p\right)=1,$ the specific values is $T_p\equiv T_1\pmod p.$
You also get when $\left(\frac{b^2+4c}{p}\right)=-1,$ that $T_p \equiv -cT_{-1}\pmod{p}$ where $T_1=bT_0+cT_{-1},$ or $-cT_{-1}=bT_0-T_1.$ For example, in our current question, $c=-1,b=4$ we get $-cT_{-1}=1.$
So you also get only one value if $bT_0-T_1=-cT_{-1}=T_0,$ or if $(b-1)T_0=T_1.$
A closed form when $T_0=0$:
$$T_p\equiv \left(\frac{b^2+4c}{p}\right)T_1$$
which becomes much easier if $T_0=0.$
For another example, for Fibonacci numbers, $b^2+4c=5,$ then if $p\equiv\pm 1\pmod{5}$ then $F_p\equiv F_1=1\pmod p$ and when $p\equiv \pm 2\pmod 5$ then $F_p\equiv F_0-F_1=-1\pmod{p}.$
(If $b^2+4c$ is a perfect square, including $0,$ then there is one value $d$ such that $T_p\equiv d\pmod p$ for all odd primes $p\not\mid b^2+4c.$ (For $b^2+4c=0,$ it is one value for any $p\not\mid b.$)
| {
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In the following question find the ratio of $a:b:c$
If the equations $x^2+2x+3=0$ and $ax^2+bx+c=0$; $\enspace$ $a,b,c \in \mathbb{R}$, have a common root, then $a:b:c$ is ?
This is my approach:
let the common root be $\alpha$. Then
$$\alpha^2+2\alpha+3=0------(1)$$
$$a\alpha^2+b\alpha+c=0------(2)$$.
Now the above two eqations can be solved in $\alpha \enspace and\enspace \alpha^2$.
$$\frac{\alpha^2}{2c-3b}=\frac{\alpha}{3a-c}=\frac{1}{b-2a}$$
$$\alpha^2=\frac{2c-3b}{b-2a};\alpha=\frac{3a-c}{b-2a}$$
$$\implies \frac{2c-3b}{b-2a}=\left(\frac{3a-c}{b-2a}\right)^2$$
On further simplifying we get:
$$9a^2+3b^2+c^2-6ab-2ac-2bc=0$$.
After this I am unable to proceed further. I thought a lot on how to obtain their ratios, but I am stuck.
The ratio given is $a:b:c=1:2:3$
| Here is a direct answer to your question. You should have continued with a Gauss decomposition of the quadratic expression:
$$9a^2+3b^2+c^2-6ab-2ac-2bc=0\tag{1}$$
into:
$$(3a-b-c/3)^2+2(b-2c/3)^2=0 \ \ \iff \ \ \begin{cases}3a-b-c/3=0 & Eq. (2)\\b-2c/3=0& Eq. (3)\end{cases}$$
From (3), one deduces that $b=2c/3$ ; plugging this expression of $b$ into (2) gives $3a=c$. Therefore we have established the desired proportionality.
Remark 1: expression (1) can be directly obtained by calculating the so-called resultant
$$\det\begin{pmatrix}1&2&3&0\\
0&1&2&3\\
a&b&c&0\\
0&a&b&c \end{pmatrix}$$
Remark 2: There was a completely different way to solve this question.
As the roots of the first polynomial are $-1+i\varepsilon \sqrt{3}$, plugging (one of) them into the second equation gives:
$$(-2a-b+c)+i\varepsilon \sqrt{3}(-2a+b)=0$$
Identifiying real part and imaginary part with zero gives at once the result.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4153973",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Why the PIs using different methods are different? What am I doing wrong? I have a PDE
$$
\text { Solve }\left(D^{2}+3 D D^{\prime}+2 D^{\prime 2}\right) z=x+y .
$$
where $D = \frac{\partial }{\partial x}$, $D^{\prime} = \frac{\partial }{ \partial y}, D^2 = \frac{\partial^2}{\partial x^2}, DD^{\prime}= \frac{\partial^2 }{\partial x \partial y} $.
I solved for Particular Integral as follows:
$$
\begin{array}{l}
P.I. &=\frac{1}{D^{2}\left(1+\frac{3 D^{\prime}}{D}+\frac{2 {D^{\prime}}^{2}}{D^{2}}\right)}[x+y] \\
&=\frac{1}{D^{2}}\left[1+\frac{3 D^{\prime}}{D}+\frac{2 D^{\prime 2}}{D^{2}}\right]^{-1}(x+y) \\
&=\frac{1}{D^{2}}\left[1-\frac{3 D^{\prime}}{D}-\frac{2 {D^{\prime}}^{2}}{D^{2}}+\left(\frac{3 D^{\prime}}{D}+\frac{2 {D^{\prime}}^{2}}{D^{2}}\right)^{2}-. .\right](x+y) \\
&=\frac{1}{D^{2}}\left[x+y-\frac{3}{D}(0+1)\right] \\
&=\frac{1}{D^{2}}[x+y-3 x] \\
&=\frac{1}{D^{2}}[y-2 x] \\
&=\frac{1}{D}\left(yx-\frac{2 x^{2}}{2}\right) \\
&=\frac{yx^2}{2}-\frac{x^{3}}{3}
\end{array}
$$
But the textbook uses a different method, I have seen the derivation that looks correct.
$$
\begin{array}{l}
\text { When } F(a, b) \neq 0 \text { and } F\left(D, D^{\prime}\right) \text { is a homogeneous function of degree } n \text { , then }\\
\text { P.I. }=\frac{1}{F\left(D, D^{\prime}\right)} \phi(a x+b y)=\frac{1}{F(a, b)} \iint \ldots \int f(v) d v d v \ldots d v\\
\text { where } \quad v=a x+b y
\end{array}
$$
Now, P.I. $=\frac{1}{D^{2}+3 D D^{\prime}+2 D^{\prime 2}}(x+y)$
$=\frac{1}{1^{2}+3 \times 1 \times 1+2 \times 1^{2}} \iint v d v d v$, where $v=x+y$,
$$
=(1/6) \times \int\left(v^{2} / 2\right) d v=(1 / 6) \times\left(v^{3} / 6\right)=(1 / 36) \times(x+y)^{3} .
$$
Hence the required general solution is $\quad z=$ C.F. + P.I., i.e.,
$$
z=\phi_{1}(y-x)+\phi_{2}(y-2 x)+(1 / 36) \times(x+y)^{3} .
$$
Can someone point out what am I doing wrong?
| $$
\frac{(x+y)^3}{36} = \frac{(y-x)^3}{12} - \frac{(y-2x)^3}{18} + \frac{x^2 y}{2} -\frac{x^3}{3}
$$
So the two answers are equivalent.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4158154",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Harmonic-analysis Fourier series proof - general fourier series. Let $f(x) \in L^2[-\pi,\pi]$ such that $f(x)$ suffice: $f(x) \sim \frac{a_0}{\sqrt{2}} + \sum_{n=1}^{\infty}a_n\cos(nx)+b_n\cos(nx) $
$f(x)$ is also even function. Prove:
$
\int_{-\pi}^{\pi} f^2(x)\cos^2(x)\, dx = \frac{1}{2}
\int_{-\pi}^{\pi} f^2(x)\ \iff \frac{1}{2}a_1^2 +\frac{a_0}{\sqrt{2}}a_2+\sum_{n=2}^{\infty}a_{n-1}a_{n+1} =0
$
I've proved that $f(x)\cos(x)$ has a Fourier coefficients (such that Fourier sum is exists). But I don't have any idea how to solve the proof? May anyone help me/ guide me how should I prove it?
Best regards,
Sababoni.
| Since $f$ is even, we have
$$f(x) = \frac{a_0}{\sqrt2}+\sum_{n=1}^\infty a_n \cos (nx)$$
\begin{align}
f(x)\cos x &= \frac{a_0}{\sqrt2}\cos x + \sum_{n=1}^\infty a_n \cos (nx) \cos (x)\\
&=\frac{a_0}{\sqrt2}\cos x + \sum_{n=1}^\infty \frac{a_n}{2}\left[ \cos(n+1)x + \cos (n-1) x \right] \\
&= \frac{a_0}{\sqrt2}\cos x + \sum_{n=2}^\infty \frac{a_{n-1}}{2}\cos (nx) + \sum_{n=0}^\infty \frac{a_{n+1}}{2}\cos (nx)\\
&=\frac{a_1}{2} + \left( \frac{a_0}{\sqrt2}+\frac{a_2}{2}\right)\cos x + \sum_{n=2}^\infty\left(\frac{a_{n-1}+a_{n+1}}2\right) \cos (nx)
\end{align}
Also,
$$\frac{f(x)}{\sqrt2}=\frac{a_0}{2}+\sum_{n=1}^\infty \left(\frac{a_n}{\sqrt2} \right) \cos nx$$
Hence, by Parserval theorem,
\begin{align}
&\int_{-\pi}^\pi f^2(x) \cos^2(x) \, dx= \frac12 \int_{-\pi}^{\pi}f^2(x) \, dx = \int_{-\pi}^\pi \left( \frac{f(x)}{\sqrt2}\right)^2 \, dx \\
& \iff \frac{a_1^2}2 + \left( \frac{a_0}{\sqrt2} + \frac{a_2}2\right)^2 +\frac14 \sum_{n=2}^\infty \left( a_{n-1}+a_{n+1}\right)^2 = \frac{a_0^2}2 + \sum_{n=1}^\infty \frac{a_n^2}{2}\\
&\iff \frac{a_2^2}{4}+\frac{a_0a_2}{\sqrt2} + \frac{a_1^2}{4} + \frac14 \sum_{n=2}^\infty a_n^2 + \frac14 \sum_{n=3}^\infty a_n^2 + \frac12 \sum_{n=2}^\infty a_{n-1}a_{n+1}= \sum_{n=2}^\infty \frac{a_n^2}{2}\\
&\iff \frac{1}{4}a_1^2 + \frac{a_0a_2}{\sqrt2} + \frac12 \sum_{n=2}^\infty a_{n-1}a_{n+1}=0\\
&\iff\frac12 a_1^2 + \color{red}{\sqrt2} a_0a_2 + \sum_{n=2}^\infty
a_{n-1}a_{n+1}=0\end{align}
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "1",
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Compute $\iiint x+y+z$ over the region inside $x^2+y^2+z^2 \le 1$ in the fist octant I feel this should be an easy question, but I seem to be struggling with it. So, I started by finding my bounds of integration. In this case, I get $0 \le x \le 1$, $0 \le y \le \sqrt{1-x^2}$ and $0 \le z \le \sqrt{1-x^2-y^2}$. Then, upon integrating, I get:
\begin{align}
= {} & \int_0^1 \int_0^{\sqrt{1-x^2}}\int_0^{\sqrt{1-x^2-y^2}}(x+y+z) \, dz \, dy \, dx \\[8pt]
= {} & \int_0^1 \int_0^{\sqrt{1-x^2}}(\sqrt{1-x^2-y^2})(x+y) - \frac{1-x^2-y^2}{2} \, dy \, dx
\end{align}
From here the integration got pretty hairy, and using an online calculator the next inner integral with respect to $y$ resulted in imaginary numbers, which seems way too complex for a final answer of $\frac{3\pi}{16}$. My guess is my bounds of integration are wrong, but I'm not sure why or what the right ones should be.
Thanks!
| If $E$ is the region $x^2 + y^2 + z^2 \leq 1$ in first octant. Please note that due to symmetry,
$$I = \iiint_E (x+y+z) \ dV = 3 \iiint_E z \ dV$$
In spherical coordinates,
$x = \rho \cos\theta \sin\phi, y = \rho \sin\theta \sin\phi, z = \rho \cos\phi$
So $x^2 + y^2 + z^2 \leq 1 \implies \rho \leq 1$
As we are in first octant, $0 \leq \phi \leq \pi/2, 0 \leq \theta \leq \pi/2$
So, $ \displaystyle I = 3 \int_0^{\pi/2} \int_0^{\pi/2} \int_0^1 \rho^3 \cos\phi \sin\phi \ d\rho \ d\phi \ d\theta = \frac{3\pi}{16}$
Edit:
In polar coordinates,
$$x = r \cos\theta, y = r \sin\theta$$
$x^2 + y^2 + z^2 \leq 1 \implies z \leq \sqrt{1-r^2}$ in first octant.
Also, $0 \leq r \leq 1$ and $0 \leq \theta \leq \pi/2$.
So, integral is $I = 3 \displaystyle \int_0^{\pi/2} \int_0^1 \int_0^{\sqrt{1-r^2}} r \ z \ dz \ dr \ d\theta$
which is a straightforward integral.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4164263",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Reversing the divisibility of 2 out of 3 formula I was looking into a couple of exercises:
First exercise:
Prove that if $x \mid (x^2 + 1)$ the $x = 1$ or $x = -1$
My solution was that if $x \mid (x^2 + 1)$ this means that $x | x^2$ and $x | 1$. The only number that is a divisor of $1$ is 1/-1. Hence $x = 1$ or $x = -1$
Second exercise:
Prove that if $6 \mid (x^3 - x)$ for every integer $x$ (hint among $3$ consecutive integers, one must be a multiple of $3$
Now for the second exercise I started with the same approach i.e. that if $6 \mid (x^3 - x)$ then $6 \mid x^3$ and $6 \mid x$ and then I realized that the question is about to prove that it holds for all integers which is not obvious to me that it does. So I plugged in some value e.g. for $x = 7$ we have $7^3 - 7 = 343 - 7 = 336$ which is a multiple of $6$. Also for $x = 5$ we have $5^3 - 5 = 125 - 5 = 120$ again is multiple of $6$.
So the $6 \mid (x^3 - x)$ seems to hold but I notice that for $x = 7$ we indeed have $6 \mid (7^3 - 7)$ but it is not the case that $6 \mid 7$.
So I have two questions here:
a) Is it wrong to consider that if $x | (a + b)$ then $x | a$ and $x | b$? Or are there specific conditions that this formula holds? Does that mean that my solution to the first exercise is wrong?
b) how can I use the hint to solve the exercise? It is not clear to me
| As mentioned by shoteyes in the comments, if $x \mid (a + b)$ and either $x \mid a$ or $x \mid b$, then $x$ divides both $a$ and $b$.
In the first exercise, since it is known that $x \mid x^2$ and it is given that $x \mid (x^2 + 1)$, $x \mid 1$. Therefore, the only values of $x$ for $x \mid 1$ to be true are $-1$ and $1$. Your solution to the first exercise is correct, though the way it was shown was not made entirely clear. It would be best to mention the specific conditions for $x \mid a$ and $x \mid b$ to be true.
In the second exercise, even though it is known that $6 \mid (x^3 - x)$, it does not hold true for $6 \mid x^3$ nor $6 \mid x$ for every integer of $x$.
However, it might be helpful to factorise $x^3 - x$, as follows:
$$x^3 - x = x(x^2 - 1) = x(x - 1)(x + 1) = (x - 1) \cdot x \cdot (x + 1)$$
Since $x$ is an integer, $(x - 1)$, $x$ and $(x + 1)$ are consecutive integers. Therefore, by using the hint given in the question, one of the integers must be divisible by $3$ (take any $3$ consecutive integers and you will find that one will be divisible by $3$). Moreover, since $(x - 1)$, $x$ and $(x + 1)$ are consecutive integers, at least one of the integers must be divisible by $2$. Thus, since there is at least one integer which is divisible by $2$ and one integer which is divisible by $3$, $x(x - 1)(x + 1)$ must be divisible by $6$.
I hope that helps!
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Find the number of 5-digit number divisible by 6 which can be formed using $0,1,2,3,4,5$ if repetition of digits is not allowed.
Find the number of $5$-digit number divisible by $6$ which can be formed using $0,1,2,3,4,5$ if repetition of digits is not allowed.
I started by considering the following cases:
*
*Unit digit = $0$
I can fill four places with digits ${1,2,4,5}$ , so number of 5-digit numbers $= P(4,4) = 24$
*Unit digit = $2$
I can fill four places with digits ${0,1,3,4,5}$ such that (a) $0$ does not come in the first place (b) either of $1,4$ is used (c) $0,3,5$ are always used. I have to fill 4 places with $0,(1,4),3,5$. So, total number of arrangements $=P(4,4)$. Number of arrangements when $0$ comes as first digit $=P(3,3)$. Number of arrangements of $(1,4) = 2$. Therefore , number of 5-digit numbers $=2(P(4,4)-P(3,3)) = 36$
*Unit digit = $4$
Similarly, number of 5-digit numbers $= 36$
Therefore, the total number of 5-digit numbers $= 24+36+36 = 96$.
But the correct answer is $108$. Where did I make a mistake?
| The number has to be divisible by $3$, so the sum of the digits must also have that property. Since the sum of the six digits is $15$, the missing digit has to be either $0$ or $3$. (This was your mistake -- you thought that either 1 or 4 needed to be unused).
*
*If the missing digit is $0$, then the units digit can be chosen in two ways and the remaining four digits can be arranged in $4!=24$ ways. This leads to $2\cdot24=48$ cases.
*If the missing digit is $3$ and $0$ is the units digit, then the remaining four digits can be arranged in $4!=24$ ways,
*If the missing digit is $3$ and $0$ is not the units digit, then the units digit can be chosen in two ways. From the remaking four digits, the ten-thousands digit can be chosen in three ways (to avoid a leading zero), and the remaining three digits can be arranged in $3!=6$ ways. This leads to $2\cdot3\cdot6=36$ cases.
The total is $48+24+36=108$ possible cases.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Solve $y' = (1+\frac{y-1}{2x})^2$ Solve $y' = (1+\frac{y-1}{2x})^2$
My first thought was to expand to see if I can get a linear form:
$$y' = 1 + \frac{y-1}{x} + (\frac{y-1}{2x})^2 = 1 + \frac{y-1}{x} + \frac{y^2-2y+1}{4x^2}$$
$$y' = 1 + \frac{y}{x} -\frac{1}{x} + \frac{y^2}{4x^2} + \frac{-2y}{4x^2} + \frac{1}{4x^2} \rightarrow y'+y(\frac{2}{4x^2}-\frac{1}{x}) = \frac{(2x-1)^2+y^2}{4x^2}$$
However, this is not even a bernoulli equation. How should I approach this problem?
| The first thing I would try is $v = y - 1$
$$
v' = \left(1+\frac{v}{2x}\right)^2
$$
Let $v = wx$
$$
w + xw' = \left(1+\frac{w}{2}\right)^2
$$
which is basically
$$
xw' = g(w)
$$
now see if you can solve this ODE.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Finding Dual Basis
In $\mathbb{R^{3}}$, let be $v_{1}=\begin{bmatrix} 1 \\2 \\0 \end{bmatrix}$i $v_{2}=\begin{bmatrix} 0 \\3 \\1 \end{bmatrix}$ and $v_{3}=\begin{bmatrix} 0 \\0 \\1 \end{bmatrix}$. If f is a linear functional on $\mathbb{R}^{3}$ such that $f(v_1)=-1$,$f(v_2)=2$ and $f(v_3)=4$ and if $v=\begin{bmatrix} x \\y \\z \end{bmatrix}$, find $f(v)$.
In addition to that question I want to find Dual Basis. But now, I will write my solution for $f(v)$.
Solution:
$$f\left(1 \cdot \begin{bmatrix} 1 \\0 \\0 \end{bmatrix}+2 \cdot \begin{bmatrix} 0 \\1 \\0 \end{bmatrix} + 0 \cdot \begin{bmatrix} 0 \\0 \\1 \end{bmatrix} \right) = f(e_1)+2f(e_2)=-1$$
$$f\left(0 \cdot \begin{bmatrix} 1 \\0 \\0 \end{bmatrix}+3 \cdot \begin{bmatrix} 0 \\1 \\0 \end{bmatrix} + 1 \cdot \begin{bmatrix} 0 \\0 \\1 \end{bmatrix} \right) = 3f(e_2)+f(e_3)=2$$
$$f\left(0 \cdot \begin{bmatrix} 1 \\0 \\0 \end{bmatrix}+0 \cdot \begin{bmatrix} 0 \\1 \\0 \end{bmatrix} + 1 \cdot \begin{bmatrix} 0 \\0 \\1 \end{bmatrix} \right) = f(e_3)=4$$
And we get;
$$f(v)=\dfrac{1}{3}x-\dfrac{2}{3}y+4z$$
And I want to find Dual Basis, any help? Thanks
| Let $f=(p,q,r)_{1\times 3}$, then $f(v_1)=-1 \implies p+2q=-1, f(v_2)=3q+r=2, f(v_3)=4 \implies r=4 \implies p=-2/3, q=1/3,r=4$. Finally
$$f(v)=(-2/3,1/3,4)\begin{pmatrix} x\\ y\\ z \end{pmatrix}=-2x/3+y/3+4z.$$
We can write $v_1=(1,2,0), v_2=(0,3,1), v_3=(0,0,3)$, then the dual basis is given by cross products:
$$u_1=v_1\times v_2=(2,-1,3), u_2=v_2\times v_3=(3,0,0), u_3=v_3 \times v_1=(-2,1,0).$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Find value of $|(\overrightarrow{a} \times\overrightarrow{b}) \times\overrightarrow{c}|$
let $\overrightarrow{a} = 2 \hat {i}+ \hat {j}-2 \hat {k}$ and $\overrightarrow{b}= \hat {i} + \hat{j}$. $\:$ if $\overrightarrow{c}$ is a vector such that $\overrightarrow{a} \cdot \overrightarrow{c}+2|\overrightarrow{c}|=0$ and $|\overrightarrow{a}-\overrightarrow{c}|= \sqrt{14}$ and angle between $\overrightarrow{a} \times \overrightarrow{b}$ and $\overrightarrow{c}$ is $30 ^\circ$. Then value of $|(\overrightarrow{a} \times\overrightarrow{b}) \times\overrightarrow{c}|$ is ?
My first approach:
$|\overrightarrow{a}-\overrightarrow{c}|=\sqrt{|\overrightarrow{a}|^2+|\overrightarrow{c}|^2-2\overrightarrow{a} \cdot\overrightarrow{c}}= \sqrt{14}$
after substituting value of $\overrightarrow{a} \cdot \overrightarrow{c}=-2|\overrightarrow{c}|$ and $|\overrightarrow{a}|=3$ in above equation. I obtained $9+|\overrightarrow{c}|^2+4|\overrightarrow{c}|=14$
$\implies$ $|\overrightarrow{c}|^2+4|\overrightarrow{c}|-5=0$
$\implies$ $|\overrightarrow{c}|=1$
$\overrightarrow{a} \times \overrightarrow{b}=2 \hat{i}-2\hat{j}+ \hat{k}$
$\implies$ $|\overrightarrow{a} \times \overrightarrow{b}|=3$
So from above two result
$|(\overrightarrow{a} \times \overrightarrow{b}) \times \overrightarrow{c}|= |\overrightarrow{a} \times \overrightarrow{b}||\overrightarrow{c}|sin(30^\circ)=(3)(1) \frac{1}{2}= \frac{3}{2}$.
My second Approach:
$|(\overrightarrow{a} \times\overrightarrow{b}) \times\overrightarrow{c}|=|(\overrightarrow{a} \cdot \overrightarrow{c})\overrightarrow{b}-(\overrightarrow{b} \cdot \overrightarrow{c})\overrightarrow{a}|$
To calculate $\overrightarrow{b} \cdot \overrightarrow{c}$ $\;$I took help of
$[\overrightarrow{a} \; \overrightarrow{b} \; \overrightarrow{c}]^2=\begin{vmatrix} \overrightarrow{a}\cdot \overrightarrow{a} & \overrightarrow{a}\cdot \overrightarrow{b} &\overrightarrow{a}\cdot \overrightarrow{c} \\ \overrightarrow{b}\cdot \overrightarrow{a} & \overrightarrow{b}\cdot \overrightarrow{b} &\overrightarrow{b}\cdot \overrightarrow{c} \\ \overrightarrow{c}\cdot \overrightarrow{a} & \overrightarrow{c}\cdot \overrightarrow{b} &\overrightarrow{c}\cdot \overrightarrow{c} \end{vmatrix}$
$\overrightarrow{a} \cdot \overrightarrow{a}=9$ & $\overrightarrow{b} \cdot \overrightarrow{b}=2$ $\overrightarrow{c} \cdot \overrightarrow{c}=1$ & $\overrightarrow{a} \cdot \overrightarrow{b}=3$ & $\overrightarrow{a} \cdot \overrightarrow{c}=-2$ and let $\overrightarrow{b} \cdot \overrightarrow{c}=x$
$[\overrightarrow{a} \; \overrightarrow{b} \; \overrightarrow{c}]=(\overrightarrow{a} \times \overrightarrow{b}) \cdot \overrightarrow{c} = |\overrightarrow{a} \times \overrightarrow{b}| |\overrightarrow{c}| cos(\theta)=\frac{3 \sqrt3}{2}$
here $\theta$ = angle between $\overrightarrow{a} \times \overrightarrow{b}$ and $\overrightarrow{c} $
$\frac {27}{4}=\begin{vmatrix} 9 & 3 & -2 \\ 3 & 2 & x \\ -2 & x & 1\end{vmatrix}$
After solving above determinant I obtained a equation which is $36 x^2+48x+23=0$
$\implies$ no real value of $x$ that is no real $\overrightarrow{b} \cdot \overrightarrow{c}$
Doubt: Why am I not getting answer using second method?
| The first approach gives you a result for $(\vec a \times \vec b) \times \vec c \ $ as it does not test the compatibility of the given data.
Just simply applying,
$|(\overrightarrow{a} \times\overrightarrow{b}) \times\overrightarrow{c}|=|(\overrightarrow{a} \cdot \overrightarrow{c})\overrightarrow{b}-(\overrightarrow{b} \cdot \overrightarrow{c})\overrightarrow{a}|$
shows that there cannot be vectors $\vec a, \vec b$ and $\vec c$ with given conditions.
We know $|a \times b| = 3, |a| = 3, |b| = \sqrt2$
Say the angle between $\vec b$ and $\vec c$ is $\alpha$.
$\displaystyle \frac{3}{2} \ |c| = | \ (-2 \ |c|) \ (\hat i + \hat j) - \sqrt2 \ |c| \cos\alpha \ (2\hat i + \hat j - 2 \hat k) \ | $
Dividing by $|c|$ on both sides,
$\displaystyle \frac{3}{2} = |-2 \hat i - 2 \hat j - \sqrt2 \cos\alpha \ (2 \hat i + \hat j - 2 \hat k) \ |$
$\dfrac{9}{4} = (2 + 2 \sqrt2 \cos\alpha)^2 + (2+ \sqrt2 \cos\alpha)^2 + (2\sqrt2\cos\alpha)^2$
Simplifying, $ \displaystyle \cos^2\alpha + \frac{2\sqrt2}{3} \cos\alpha + \frac{4}{9} = \frac{1}{8}$
$\displaystyle \left(\cos\alpha + \frac{\sqrt2}{3}\right)^2 = - \frac{7}{72}$
which has no real solution.
| {
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How to solve this integral: $\int_0^\infty{\frac{ln x}{x^2+2x+4}}\cdot dx$ I have been trying to solve this integral:- $$\int_0^\infty{\frac{\ln x}{x^2+2x+4}}\cdot dx$$ But I have been unsuccessful so far. I have tried integration by parts ( taking $1/(x^2+2x+4)$ as the second function does not work because it ends of as tan-inverse function which is more compicated to integrate.)
taking $\ln x$ as the second function was interesting, I ended up in this dead end:-
$$\begin{align} I&=\int_0^\infty{\frac{\ln x}{x^2+2x+4}}\cdot dx \\ &= \left[ {\frac{x\ln x -x}{x^2+2x+4}}\right]_0^\infty + \int_0^\infty \frac{2(x+1)(x\ln x-x)}{(x^2+2x+4)^2}\cdot dx \\&= 2\int_0^\infty \frac{(x^2+2x+4-(x+4))(\ln x-1)}{(x^2+2x+4)^2}\cdot dx \\ &= 2\int_0^\infty\frac{\ln x}{x^2+2x+4}\cdot dx - 2\int_0^\infty\frac{1}{x^2+2x+4}\cdot dx -2\int_0^\infty \frac{(x+4)(\ln x-1)}{(x^2+2x+4)^2}\cdot dx \\&= 2I - \left[\frac{1}{\sqrt{3}} \tan^{-1}{\frac{x+1}{\sqrt{3}}} \right]_0^\infty - 2\int_0^\infty \frac{(x+4)(\ln x-1)}{(x^2+2x+4)^2}\cdot dx \end{align}$$
the function in the second step limits to $0$ so I didn't mention it in the third step. Now I have quartic polynomial at the bottom and don't know how to proceed. I have also tried trigonometric substitution ($ x=\tan\theta$)
If possible please try to end this process, I am a high school student so please keep that in mind if you show another method :)
| Hint
For $$\int_0^\infty\dfrac{\ln x}{x^2+bx+b^2}dx$$
set $x=by$
then we can easily solve $$\int_0^\infty\dfrac{dy}{y^2+y+1}$$ using trigonometric substitution
Set $\dfrac1y=u$ in $$J=\int_0^\infty\dfrac{\ln y}{y^2+y+1}dy$$ to find $$J=-J$$
| {
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"source": "stackexchange",
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How can I evaluate $\int _0^1\frac{\operatorname{Li}_2\left(-x^2\right)}{\sqrt{1-x^2}}\:\mathrm{d}x$ I've been trying to find and prove that:
$$\int _0^1\frac{\operatorname{Li}_2\left(-x^2\right)}{\sqrt{1-x^2}}\:\mathrm{d}x=\pi \operatorname{Li}_2\left(\frac{1-\sqrt{2}}{2}\right)-\frac{\pi }{2}\left(\ln \left(\frac{1+\sqrt{2}}{2}\right)\right)^2$$
result obtained via WolframAlpha.
Here $\operatorname{Li}_2\left(z\right)$ denotes the dilogarithm function.
However it seems very difficult to prove it, the integral equals:
$$-\frac{\pi ^3}{24}+2\int _0^1\frac{\arcsin \left(x\right)\ln \left(1+x^2\right)}{x}\:\mathrm{d}x$$
Using the series representation of the dilogarithm function yields:
$$\sum _{n=1}^{\infty }\frac{\left(-1\right)^n}{n^2}\int _0^1\frac{x^{2n}}{\sqrt{1-x^2}}\:\mathrm{d}x$$
Perhaps using the beta function can allow us to proceed further? any kind of hint or solution is well regarded.
| $$\int _0^1\frac{\operatorname{Li}_2\left(-x^2\right)}{\sqrt{1-x^2}}\:\mathrm{d}x=\int _0^1\frac{1}{\sqrt{1-x^2}}\left(\int_0^1\frac{x^2\ln(y)}{1+x^2 y}\mathrm{d}y\right)\mathrm{d}x$$
$$=\int _0^1\ln(y)\left(\int_0^1\frac{x^2}{(1+x^2y)\sqrt{1-x^2} }\mathrm{d}x\right)\mathrm{d}y$$
$$=\int _0^1\ln(y)\left(\frac{\pi }{2y}\left(1-\frac{1}{\sqrt{1+y}}\right)\right)\mathrm{d}y$$
$$=\int _0^1\ln(y)\left(\frac{\pi }{2y}\left(-\sum_{n=1}^\infty\frac{2n\choose n}{4^n}(-y)^n\right)\right)\mathrm{d}y$$
$$=-\frac{\pi}{2}\sum_{n=1}^\infty(-1)^n\frac{2n\choose n}{4^n}\int_0^1 y^{n-1}\ln(y)\mathrm{d}y$$
$$=\frac{\pi}{2}\sum_{n=1}^\infty(-1)^n\frac{2n\choose n}{4^n n^2}.\qquad (*)$$
To get this sum, first replace $x$ by $-x$ in
$$\sum_{n=1}^\infty\frac{{2n\choose n}}{4^n}\frac{x^n}{n}=-2\ln\left(\frac{1+\sqrt{1-x}}{2}\right)$$
then divide both sides by $x$ and integrate from $0$ to $1$,
$$\sum_{n=1}^\infty(-1)^n\frac{2n\choose n}{4^n n^2}=-2\int_0^1\frac{\ln\left(\frac{1+\sqrt{1+x}}{2}\right)}{x}\mathrm{d}x$$
$$\overset{\frac{1+\sqrt{1+x}}{2}=t}{=}2\int_1^{\frac{1+\sqrt{2}}{2}}\frac{\ln(t)}{1-t}\mathrm{d}t-2\int_1^{\frac{1+\sqrt{2}}{2}}\frac{\ln(t)}{t}\mathrm{d}t$$
$$=2\operatorname{Li}_2(1-t)-\ln^2(t)\bigg|_1^{\frac{1+\sqrt{2}}{2}}$$
$$=2 \operatorname{Li}_2\left(\frac{1-\sqrt{2}}{2}\right)-\ln^2 \left(\frac{1+\sqrt{2}}{2}\right).$$
Addendum:
You can get the summation representation in $(*)$ by expanding $\operatorname{Li}_2(-x^2)$ in series as you mentioned in your question body, then using $\int_0^1\frac{x^{2n}}{\sqrt{1-x^2}}\mathrm{d}x=\frac{\sqrt{\pi}}{2}\frac{\Gamma\left(\frac12+n\right)}{n\Gamma(n)}=\frac{\pi}{2}\frac{{2n\choose n}}{4^n}$ which follows from using the beta function and the Legendre duplication formula:
$$\Gamma\left(\frac12+n\right)=\frac{2\sqrt{\pi}\,\Gamma(2n)}{4^n\Gamma(n)}=\sqrt{\pi}n\Gamma(n)\frac{2n\choose n}{4^n}.$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Showing that that x^2 ≡ (−1)^k (mod p). I'm stumped on using my observations to properly solve this problem. The problem is as follows:
Suppose p is an odd prime expressible as $p = 2k − 1$ and let $x$ be the product of all positive odd integers less than p. Show that $x^2 ≡ (−1)^k $(mod p).
Here are my thoughts on the problem thus far:
• First, since x is the product of all positive odd integers less than p, so we can express x as $x = (p-2)(p-4)(p-6)…(3)(1)$
• Thus, it follows that $x^2 = (p-2)^2 * (p-4)^2 * (p-6)^2…(3)^2 * (1)^2$.
• Also, if $p = 2k – 1$, then $k = (p + 1) / 2$. In this case, dividing is permissible, since $(p + 1)$ is always an even number, so dividing that by 2 always results in an integer.
• I also noticed that whenever k is even, then $(-1)^k$ is always 1, and whenever k is odd, then $(-1)^k$ is odd.
• We want to show, somehow, that $(p-2)^2 * (p-4)^2 * (p-6)^2…(3)^2 * (1)^2 ≡ (−1)^k$ (mod p). By plugging in some examples, this statement was demonstrably true, but I’m stuck about how to leverage all of these observations to get to what I want to show. Also, my apologies for not using the MathJax as I'm still getting used to all this.
| Recall that $(p-1)! \equiv -1 \pmod{p}.$ Note that
$$(p-1)! = x \cdot (2 \cdot 4 \cdots (p-1)) = x \cdot 2^{\frac{p-1}{2}} \cdot \left(1 \cdot 2 \cdots \frac{p-1}{2}\right).$$
So,
$$-1 \equiv x \cdot 2^{\frac{p-1}{2}} \cdot \left(\frac{p-1}{2}\right)! \pmod{p}.$$
Squaring this, we get that
$$1 \equiv x^2 \cdot 2^{p-1} \cdot \left(\left(\frac{p-1}{2}\right)!\right)^2 \pmod{p}.$$
$2^{p-1} \equiv 1\pmod{p}$ by Fermat's little theorem. So, now we just show that the factorial squared is $(-1)^k.$
Note that we can rewrite
$$\left(\left(\frac{p-1}{2}\right)!\right)^2 \equiv (-1)^{\frac{p-1}{2}} \cdot (p-1)! \pmod{p},$$
since
$$(-1)^{\frac{p-1}{2}} \left(\frac{p-1}{2}\right)! \equiv (p-1)(p-2)\cdots \left(p - \frac{p-1}{2}\right),$$
so that we get the latter half of the factorial. Using that $k-1 = \frac{p-1}{2},$ we can conclude
$$\left(\frac{p-1}{2}\right)!^2 \equiv (-1)^{k-1} \cdot (-1) \equiv (-1)^k \pmod{p},$$
which we showed above was equivalent to your desired identity.
| {
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How can one evaluate $\int_0^1 \frac{3x^3-x^2+2x-4}{\sqrt{x^2-3x+2}}\mathrm dx$ using only elementary means? How can one evaluate $$\int_0^1 \frac{3x^3-x^2+2x-4}{\sqrt{x^2-3x+2}}\mathrm dx$$ using only elementary means?
By using only elementary means I mean methods not exceeding the basic methods taught in a first course in the integral calculus. This is because I want an explanation accessible to secondary school students, as it was asked of me from a student, and I haven't so far come up with a simple way to evaluate this.
Thank you.
| One way:
Let $3x^2-x^2+x-4=A(x^2-3x-2)(2x-3)+B(x^2-3x+4)+C(2x-3)+D.$
Find $A,B,C,D$ by comparing the coefficients of various powers of $x$. then
$$\int_0^1 \frac{3x^3-x^2+2x-4}{\sqrt{x^2-3x+2}}dx=A\int_{0}^{1}\sqrt{x^2-3x+2} ~(2x-3)~dx+B\int_{0}^{1} \sqrt{x^2-3x+2} ~dx+ C\int_{0}^{1}\frac{(2x-3)dx}{\sqrt{x^2-3x+2}}+D\int_{0}^{1} \frac{dx}{\sqrt{x^2-3x+2}}.$$
Solve first and third integrals by letting $2x-3=t$, second and fourth integra;ls are standard.
| {
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"timestamp": "2023-03-29T00:00:00",
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mathematical induction proof aren't solving i am trying to solve for a mathematical induction for a while now but i seem to be getting the incorrect answer
$$
\frac{1}{10}+\frac{1}{40}\:+\frac{1}{88}+_...+\frac{1}{\left(3n-1\right)\left(3n+2\right)}=\frac{n}{6n+4}
$$
Equation
i am unable to solve the left hand side to the right hand side i have tried multiple ways but I do not seem to be able to get it to be equal can some one please help
i am able to go up till this step
$$
\frac{k}{6k+4}+\frac{1}{\left(3\left(k+1\right)-1\right)\left(3\left(k+1\right)+2\right)}=\frac{k+1}{6\left(k+1\right)+4}
$$
i think am doing somthing incorrect after this
i tried following this youtube video
https://www.youtube.com/watch?v=twA6vZgX_U4
but it somehow doesnt work for this question
| HINT: Write the $n^{\text{th}}$ term.
$$T_n = \frac{1}{(3n-1)(3n+2)} = \frac{1}{3}\frac{(3n+2) - (3n-1)}{(3n-1)(3n+2)} = \frac{1}{3}\frac{1}{3n-1} - \frac{1}{3}\frac{1}{3n+2}$$
Can you identify what happens on adding successive terms?
Using Mathematical Induction, first of all, verify the base case of $n=1$ which you can do. After that, assume that the given formula for some is true for $n$ terms i.e.
$$S_n = \frac{1}{10}+ \frac{1}{40} + \cdots + \frac{1}{(3n-1)(3n+2)} = \frac{n}{6n+4}$$
Now, for the sum till $n+1$ terms i.e.
$$S_{n+1} = \frac{1}{10}+ \frac{1}{40} + \cdots + \frac{1}{(3n-1)(3n+2)} + \frac{1}{(3n+2)(3n+5)}$$
you have
$$S_{n+1} = S_n + \frac{1}{(3n+2)(3n+5)} = \frac{n}{6n+4} + \frac{1}{(3n+2)(3n+5)}$$
This gives you
$$S_{n+1} = \frac{3n^2+5n+2}{(6n+4)(3n+5)} = \frac{(n+1)(3n+2)}{(6n+4)(3n+5)} = \frac{n+1}{6n+10} = \frac{n+1}{6(n+1) + 4}$$
Here, the RHS is the exactly what you get when you put $n+1$ in the proposed sum expression.
| {
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Prove that $(n+1)a\leq a^{n+1}+n, \forall a,n\in\mathbb{N}$. We can start from the fact that:
\begin{align*}
0\leq a^n + a^{n-1} +\ldots + a^2 + a-n,\forall a,n\in\mathbb{N}.
\end{align*}
The above is true, since if $a>1$, then $a^n>n, \forall n \in\mathbb{N}$. Also if $a = 1$, then we will have $1 + 1 + \ldots + 1 + 1-n = n-n = 0, \forall n \in \mathbb{N}$. Thus we have:
\begin{align*}
0 &\leq (a^{n} + a^{n-1} + \ldots + a^2 + a-n)(a-1)\\
0 &\leq (a^{n} + a^{n-1} + \ldots + a^2 + a)(a-1)-n(a-1)\\
0 &\leq a(a^{n-1} + a^{n-2} + \ldots + a + 1)(a-1) -na + n\\
0 &\leq a(a^{n}-1) -na + n\\
0 &\leq a^{n + 1} -a-na + n\\
0 &\leq a^{n + 1} -(n + 1) a + n\\
(n + 1)a &\leq a^{n + 1} + n\\
\end{align*}
Another form of reason is using the inequality of the arithmetic mean with the geometric mean, as follows:
\begin{align*}
\frac{a^{n + 1} + n}{n + 1} &= \frac{a^{n + 1} + 1 + 1 + \ldots + 1 + 1}{n + 1}\\
&\geq \sqrt[n + 1]{a^{n + 1}(1)(1)\ldots(1)(1)}\\
&= \sqrt[n + 1]{a^{n + 1}}\\
& = a.
\end{align*}
Finally it is solved and we have $a^{n + 1} + n \geq (n + 1)a$.
I think this is the correct solution, I await your comments. If anyone has a different solution or correction of my work I will be grateful.
| I found another way to solve it, by Taylor series for $a^x$, we have:
\begin{align*}
a^x\geq 1+x.
\end{align*}
As long as $a>1$. Now if $x = y-1$, we have:
\begin{align*}
a^{y-1}&\geq 1+(y-1)=y\\
a^{y-1}&\geq y\\
a^{y}&\geq ya
\end{align*}
Thus, if $y = n+1$,
\begin{align*}
n+a^{n+1}\geq a^{n+1}\geq (n+1)a.
\end{align*}
Finally for $ a = 1 $ we have trivially.
| {
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Finding the maximum and minimum values of $f(x)=|x-1|+|x^2-2x|$ for $x \in [0,2]$ Can someone help me solve this problem?
If a function $f(x)$ in the domain $x \in [0,2]$ is
$$f(x)=|x-1|+|x^2-2x|$$
then the minimum value is [A] and the maximum one is [B].
I've tried a lot of things:
As I said earlier, I tried to set everything to zero.
The first and second part:
$f(0)=|0-1|+|0^2-2(0)|=>0-1+0^2-2(0)=>f(0)=-1$
$f(2)=|2-1|+|2^2-2(2)|=>2-1+2^2-2(2)=>f(2)=1$
Only the first part:
$f(0)=|0-1|=>0-1=>f(0)=-1$
$f(0)=|0^2-2(0)|=>0^2-2(0)=>f(0)=0$
Only the second part:
$f(2)=|2-1|=>2-1=>f(2)=1$
$f(2)=|2^2-2(2)|=>2^2-2(2)=>f(2)=0$
In all cases, I don't know what to do next. I also tried to do the same thing this guy in this video did (first problem) https://www.youtube.com/watch?v=3wrXDw5ETh4&t=131s
$\frac{d}{dx}(x-1+x^2-2x)=>-1+2x=0=>x=\frac{1}{2}$
\begin{array} {|r|r|}\hline X & Y \\ \hline 0 & -1 \\ \hline \frac{1}{2} & -\frac{11}{4} \\ \hline 2 & 1 \\ \hline \end{array}
But $-\frac{11}{4}$ and $1$ are not the minimum and maximun respectively.
One way this site (https://www.wikihow.com/Find-the-Maximum-or-Minimum-Value-of-a-Quadratic-Function-Easily#:~:text=Categories%3A%20Algebra-,To%20find%20the%20maximum%20or%20minimum%20value%20of%20a%20quadratic,%5E2%20%2B%205x%20%2B%204.) says you can find the minimum and maximum value of a quadratic equation is by taking the derivative of that equation in its general form.
$f(x)=|x-1|+|x^2-2x|$
$f(x)=x-1+x^2-2x=x^2-x-1$
$f'(x)=2x−1$
$0=2x-1$
$x=\frac{1}{2}$
$f(\frac{1}{2})=|(\frac{1}{2})-1|+|(\frac{1}{2})^2-2(\frac{1}{2})|=\frac{5}{4}$
Which is one of the answers on the answer sheet, $\frac{5}{4}$ is the maximum value, but the minimum value is 1 and not $\frac{1}{2}$.
| I think you are looking for a hint rather than a full solution.So this is how I would go about it.
We start by getting rid of modulus symbol
Case 1: $0\leq x<1$
$f(x)=-(x-1)+x(-(x-2))=-x^2+x+1=\frac{5}{4}-(x-\frac{1}{2})^2$
Case 2: $1\leq x\leq 2$
$f(x)=(x-1)+x(-(x-2))=-x^2+3x-1=\frac{5}{4}-(x-\frac{3}{2})^2$
| {
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"timestamp": "2023-03-29T00:00:00",
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Given $7$ blank spaces and four letters $A,B,C,D$, what is the number of ways each space can be filled up so that all four letters are used up. Given $7$ blank spaces and four letters $A,B,C,D$, what is the number of ways each space can be filled up so that all four letters are used up.
The only way I can think of is: Fill up all the squares in any way possible, then remove the arrangements that use less than $4$ letters. This can be done in $4^7 - 4 \times3^7$.
Is this correct? Is there a way we can use PIE (Principle of Inclusion-Exclusion) to solve this?
| This question is very good place to use exponential generating functions .
It is said that all letter will be used ,so
the generating function for $A$ is $$x + \frac{x^2}{2} +\frac{x^3}{6} + \frac{x^4}{24}$$
the generating function for $B$ is $$x + \frac{x^2}{2} +\frac{x^3}{6} + \frac{x^4}{24}$$
the generating function for $C$ is $$x + \frac{x^2}{2} +\frac{x^3}{6} + \frac{x^4}{24}$$
the generating function for $D$ is $$x + \frac{x^2}{2} +\frac{x^3}{6} + \frac{x^4}{24}$$
We wrote our exponential generating functions starting from $x$ because each letter must be used and if each letter will be used then a letter can be used at most $4$ times.
Now, we should find their product . Moreover , when we find their expansion , we should look for the term $[x^{7}]$ and multiply it by $7!$.
Such that https://www.wolframalpha.com/input/?i=expanded+form+of+%28x+%2B+x%5E2+%2F+2+%2B+x%5E3+%2F+6++%2B+x%5E4+%2F+24%29%5E4
We found that the coefficient of $[x^7]$ is $\frac{5}{3}$ ,so $7! \times \frac{5}{3}=8400$
| {
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"timestamp": "2023-03-29T00:00:00",
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Prove that: $(a+b)(b+c)(c+a)\ge(a+1)(b+1)(c+1)$ Let $a,b,c>0$ satisfy $abc=1 $
Prove that: $$(a+b)(b+c)(c+a)\ge(a+1)(b+1)(c+1)$$
I have tried:
Let $p=a+b+c;q=ab+bc+ca;r=abc$, the problem is: $$pq-r\ge r+p+q+1 \;\;\text{or}\;\; pq-p-q\ge3 $$
I have $pq\ge 9r$, so we need to prove $$9-p-q\ge3$$ or $$p+q\le 6$$ I have $p\le \dfrac{q^2}{3r}$ so the problem is $$\dfrac{q^2}{3}+q\le6$$ or $$(q-3)(q+6)\le 0$$ but I can't prove it and maybe it isn't true, pls help me
| Alternative solution:
Since we are given that $abc=1$, we can assume that $a=\frac x y, b=\frac y z, c=\frac z x$.
We have to prove that $$({x^2} + yz)({y^2} + zx)({z^2} + xy) \ge xyz(x+ y)(y + z)(z+ x)$$ Notice that we have
$ ({x^2} + yz)({y^2} + zx) - xy(z + x)(z + y) = z(x + y){(x - y)^2} \ge 0 $
$ \Rightarrow ({x^2} + yz)({y^2} + zx) \ge xy(z + x)(y+z) $
(Use this to complete the solution.)
| {
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How can I evaluate the following limit $\lim_{x\to+ \infty} x^2\left(e^{\frac{1}{x}} -e^{\frac{1}{x+1}}\right)$? Evaluate this limit :
$$\lim_{x\to+ \infty} x^2\left(e^{\frac{1}{x}} -e^{\frac{1}{x+1}}\right)$$
I tried to simplify the function :
$$x^2e^{\frac{1}{x}}-x^2e^{\frac{1}{x+1}}=\frac{e^{\frac{1}{x}}}{\frac{1}{x^2}}-\frac{e^{\frac{1}{x+1}}}{\frac{1}{x^2}}$$
So by the substitution $X=\frac{1}x$ I'll have :
$$\lim_{X\to 0} \frac{e^X}{X^2} -\frac{e^{\frac{X}{X+1}}}{X^2}$$
Am I in the right path ?
| I think you're already on the right path, try to use the classical equivalents :
$$e^x\overset{x\to 0}\sim x+1$$
Using your results :
When $x$ tends to $0$ you'll have :
\begin{align}
\lim_{x\to 0} \frac{e^x}{x^2} -\frac{e^{\frac{x}{x+1}}}{x^2}&= \lim_{x\to 0} \frac{x+1}{x^2}-\frac{\frac{x}{x+1}+1}{x^2}\\
&=\lim_{x\to 0} \frac{x+1-\frac{x}{x+1}-1}{x^2}\\
&=\lim_{x\to 0} \frac{x^2+x-x}{x^2(x+1)}\\
&=\lim_{x\to 0} \frac{1}{x+1}\\
&=1\end{align}
Therefore your limit is $1$.
You may take this equivalence as a technique to evaluate limits :
$$e^{f(x)} \overset{f(x)\to 0}\sim f(x)+1$$
Make sure that you have :
$$f(x)\to 0$$
| {
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$\left( \frac{r}{\sqrt{r^2-1}}, \frac{r}{\sqrt{r^2+1}} \right)$ are roots of the equation $x^2 - bx + 3 = 0$. What is the value of $b$?
If roots of the equation $x^2 - bx + 3 = 0$ are $\left( \frac{r}{\sqrt{r^2-1}}, \frac{r}{\sqrt{r^2+1}} \right)$, then what is the value of $b$ ?
$1)\pm2\sqrt6\qquad\qquad2)\pm2\sqrt3\qquad\qquad3)2\sqrt6\qquad\qquad4)2\sqrt3$
Here is my approach:
We have $\dfrac{r^2}{\sqrt{r^4-1}}=3$. Hence $\dfrac{r^4}{r^4-1}=9$ and $r^4=\dfrac98\Rightarrow r^2=\dfrac{3}{2\sqrt2}$. And $b$ is equal to sum of the roots:
$$b=\frac{r}{\sqrt{r^2-1}}+\frac{r}{\sqrt{r^2+1}}=\frac{r(\sqrt{r^2+1}+\sqrt{r^2-1})}{\sqrt{r^4-1}}=\frac{\sqrt{r^4+r^2}+\sqrt{r^4-r^2}}{\sqrt{r^4-1}}$$
$$=2\sqrt2\times(\sqrt{\frac98+\frac{3\sqrt8}{8}}+\sqrt{\frac98-\frac{3\sqrt8}{8}})=\sqrt{9+3\sqrt8}+\sqrt{9-3\sqrt8}$$
We have $b^2=24$. So $b=\pm2\sqrt6$.
My question is, can we solve this problem with other approaches?
| Let $\alpha$ and $\beta $ be the roots of the equation with $\beta>\alpha$.
$${1\over\alpha^2}=1-\frac1{r^2}$$
$${1\over\beta^2}=1+\frac1{r^2}$$
$$\implies \frac{1}{\alpha^2}+\frac{1}{\beta^2}=2$$
$$\implies \alpha^2+\beta^2=18$$
$$\implies \alpha^2+\beta^2+2\alpha\beta=24$$
$$\implies b^2=24$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Is $y^2+f(y)b+c$ a quadratic equation? The solution to the question:
Let $x, y, z \in R$ such that $x+y+z=6$ and $x y+y z+z x=7$. Then find the range of values of $x, y$, and $z$.
given in book is as follows:
$x, y, z \in R$
$x+y+z=6$ and $x y+y z+z x=7$
$\Rightarrow y(6-y-z)+y z+z(6-y-z)=7$
$\Rightarrow \quad-y^{2}+(6-z+z-z) y+z(6-z)-7=0$
$\Rightarrow \quad y^{2}+(z-6) y+7+z(z-6)=0$
.
Now, $y$ is real. Therefore,
$(z-6)^{2}-4[7+z(z-6)] \geq 0$...(1)
or $3 z^{2}-12 z-8 \leq 0$
or $\frac{12-\sqrt{144+96}}{6} \leq z \leq \frac{12+\sqrt{144+96}}{6}$
or $\frac{6-2 \sqrt{15}}{3} \leq z \leq \frac{6+2 \sqrt{15}}{3}$.
From symmetry, $x$ and $y$ have same range.
But I doubt if this is correct; since $y$ is function of $z$, hence the author cannot solve the question this way. Let me explain my point clearly.
Quadratic equations are defined as those which can be expressed in the form $ax^2+bx+c=0$, where $a \not=0$ and $a, b, c$ are constants. But clearly in this case $z$ is not constant, it is function of $y$, hence $(1)$ is not a quadratic equation so we cant the do the steps after $(1)$ hence the solution is wrong.
So the question is, is $y^2+f(y)b+c$ a quadratic equation?
| First, the quadratic formula deals with polynomials. If you introduce some unspecified function $f$, you muddy the waters, since $f$ needn't be a polynomial. For example, if $f(x)=\sin x$, then $x^2+f(x)$ is not quadratic in $x$.
I understand that you saw the quadratic formula derived, it was assumed that $a,b,c$ are constants, but the proof works so long as they aren't functions of $x$.
When we say that the roots of $p(x)=ax^2+bx+c$ are $$x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$$ this is true so long as $a\neq0$ and $a,b,$ and $c$ do not depend on $x$. You can verify this by substituting them into the formula for $p$.
It doesn't matter what $a,b,$ and $c$ are, so long as they don't depend on $x$. They could be trigonometric functions of some other variable, for example.
| {
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"source": "stackexchange",
"question_score": "2",
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} |
Prove that $\frac{\ 1}{a^2+b^2+ab}+\frac{\ 1}{b^2+c^2+bc}+\ \frac{\ 1}{c^2+a^2+ca}\ge\ \ \frac{\ 9}{\left(a+b+c\right)^2}$.
Let $a,b,c$ be positive real numbers. Prove that $$\frac{\ 1}{a^2+b^2+ab}+\frac{\ 1}{b^2+c^2+bc}+\ \frac{\ 1}{c^2+a^2+ca}\ge\ \ \frac{\ 9}{\left(a+b+c\right)^2}.$$
I want to prove the inequality with elementary inequalities ( Mean inequalities, Cauchy-Schwarz etc.). I tried to use Cauchy-Schwarz, but I got $$\frac{\ 1}{a^2+b^2+ab}+\frac{\ 1}{b^2+c^2+bc}+\ \frac{\ 1}{c^2+a^2+ca}\ge\frac{9}{2(a^2+b^2+c^2)+ab+bc+ca}.$$
Then we have to show that $$2(a^2+b^2+c^2)+ab+bc+ca\leq (a+b+c)^2 $$
which is not true.
Multiplying the numerators of the fractions in LHS by $a^2,b^2,c^2$ also makes the problem more difficult.
So, how to solve the problem with elementary inequalities?
| Another way.
Let $a\geq b\geq c$.
Thus, $$\sum_{cyc}\frac{1}{a^2+b^2+ab}-\frac{9}{(a+b+c)^2}=\sum_{cyc}\left(\frac{1}{a^2+b^2+ab}-\frac{3}{(a+b+c)^2}\right)=$$
$$=\frac{1}{(a+b+c)^2}\sum_{cyc}\frac{c^2-2a^2-2b^2-ab+2ac+2bc}{a^2+b^2+ab}=$$
$$=\frac{1}{2(a+b+c)^2}\sum_{cyc}\frac{(c-a)(c+b+4a)-(b-c)(a+c+4b)}{a^2+b^2+ab}=$$
$$=\frac{1}{2(a+b+c)^2}\sum_{cyc}(a-b)\left(\frac{a+c+4b}{b^2+c^2+bc}-\frac{c+b+4a}{a^2+c^2+ac}\right)=$$
$$=\frac{\sum\limits_{cyc}(a-b)^2(a^2+b^2-2c^2+5ab+2ac+2bc)(a^2+b^2+ab)}{2(a+b+c)^2\prod\limits_{cyc}(a^2+b^2+ab)}\geq$$
$$\geq\frac{\sum\limits_{cyc}(a-b)^2c(a+b-c)(a^2+b^2+ab)}{(a+b+c)^2\prod\limits_{cyc}(a^2+b^2+ab)}\geq\frac{(a-c)^2b(a+c-b)(a^2+c^2+ac)+(b-c)^2a(b+c-a)(b^2+c^2+bc)}{(a+b+c)^2\prod\limits_{cyc}(a^2+b^2+ab)}\geq$$
$$\geq\frac{(b-c)^2b(a-b)(a^2+c^2+ac)+(b-c)^2a(b-a)(b^2+c^2+bc)}{(a+b+c)^2\prod\limits_{cyc}(a^2+b^2+ab)}=$$
$$=\frac{(b-c)^2(a-b)^2(ab-c^2)}{(a+b+c)^2\prod\limits_{cyc}(a^2+b^2+ab)}\geq0.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4197535",
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"source": "stackexchange",
"question_score": "7",
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} |
Prove that $ \det(A^4 + A^2 B^2 + 2A^2 + I) \geq 0 $ Problem:
Let $ A $ and $ B $ be an $ n \times n $ matrices with real entries. If $ AB = -BA $, prove that
$$ {\det}{\left(A^4 + A^2B^2 + 2A^2 + I\right)} \geq 0. $$
My Approach:
If $ A $ invertible, then
$$ AB = -BA \implies AA^{-1}B = -BA^{-1}A \implies B = -B \implies B = O. $$
So,
$$ {\det}{\left(A^4 + A^2B^2 + 2A^2 + I\right)} = {\det}{\left(A^4 + 2A^2 + I\right)} = {\det}{\left(\left(A^2 + I\right)^{2}\right)} = \left({\det}{\left(A^{2} + I\right)}\right)^2 \geq 0. $$
If $ B $ invertible, then
$$ AB = -BA \implies AB^{-1}B = -BB^{-1}A \implies A = -A \implies A = O. $$
So,
$$ {\det}{\left(A^4 + A^2B^2 + 2A^2 + I\right)} = {\det}{(I)} = 1 \geq 0. $$
My Questions:
*
*Can we 'center' multiply both sides of the equation $ AB = -BA $ with a matrix? For example, in my solutions, $ AB = -BA \implies AA^{-1}B = -BA^{-1}A $.
*How do we prove it when $ A $ and $ B $ not invertible?
Thanks
| This answer is not really different from Anurag A's, but I think it is clearer to present the answer in the way below.
Let $A=\pmatrix{0&A_2\\ 0&A_4}$ where $A_4$ is invertible. The condition $AB=-BA$ implies that $B=\pmatrix{B_1&B_2\\ 0&B_4}$ where $A_4$ and $B_4$ anti-commute. Hence it suffices to prove the inequality for the matrix pairs $(0,B_1)$ and $(A_4,B_4)$, i.e., for the cases where $A$ is either zero or invertible.
When $A$ is zero, the determinant is $1$ and the inequality is true. When $A$ is invertible, we have
$$
\det(A^4+A^2B^2+2A^2+I)
=\det(A)^2\det\left(B^2+(A+A^{-1})^2\right).\tag{1}
$$
Since $A$ anti-commutes with $B$, so do $A^{-1}$ and $A+A^{-1}$. In general, if $BC+CB=0$, then $B^2+C^2=(B+C)^2$. Therefore the RHS of $(1)$ is equal to $\left[\det(A)\det(B+A+A^{-1})\right]^2$, which is nonnegative because $A$ and $B$ are real.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4201287",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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What is the sum of the maximum and minimum values BC can take? $\small AB$, $\small BD$ and $\small BC$ are are integer measurements
If $\small AB + BD = k$, find the maximum and minimum values that the $\small BC$ side can assume and then add the values found. (Answer: $k$)
My progress:
$\small \triangle ADB: |AB - BD |< AD < AB + BD \therefore \boxed{|AB-BD| < AD < k}\\$ $\small
\triangle BCD: \boxed{|BD-DC| < BC < BD+DC}\\$ $\small
\boxed{\measuredangle B = 180^o -4\theta}\\$ $\small
\triangle ABC:\dfrac{AB}{\sin\theta}=\dfrac{BC}{\sin3\theta}\rightarrow \boxed{\dfrac{AB}{\sin\theta}=\dfrac{BC}{3\sin\theta-4\sin^3\theta}}\\\small \triangle BCD: \boxed{\frac{BC}{\sin 2\theta}=\frac{BD}{\sin\measuredangle C}}$
These are the relationships I found but I can't "see" how to get the answer.
| $AB=a, BD=b, BC=c$
$\triangle ABC: a\sin 3\theta=c\sin \theta \Rightarrow$
$a(3\sin\theta-4\sin^3\theta)=c\sin\theta\Rightarrow$
$a(3-4\sin^2\theta)=c\Rightarrow$
$\sin^2\theta=\frac{3a-c}{4a}\Rightarrow$
$c<3a$
$\triangle BCD: c\sin\angle BCD=b\sin 2\theta \Rightarrow$
$\sin\angle BCD=\frac{b}{c}\sin 2\theta$
$0 < \angle CBD < 180^\circ-4\theta$,
$\angle BCD=180^\circ-\angle CBD-2\theta\Rightarrow$
$2\theta < \angle BCD < 180^\circ-2\theta\Rightarrow$
$\sin\angle BCD > \sin 2\theta\Rightarrow c<b$
Any $c<{\rm min}(3a,b)$ is satisfying conditions.
Addition of positive integer requirements gives
$c_{min}=1$, $c_{max}=k-1$, $c_{min}+c_{max}=k$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4202557",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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General Formula for an arbitrary Rotation of Sphere I am currently reading F. Klein's book "Lectures on the Icosahedron and the Solution of Equations of the Fifth Degree" and in Part 1, Chapter 2, we desire to deduce a general formula for an arbitrary rotation of sphere. The idea is as follows: first we use stereographic projection to identify $S^2 = \{(\xi,\eta,\zeta)\in \mathbb{R}^3| \quad \xi^2+\eta^2+\zeta^2=1\}$ with Riemann sphere $\mathbb{C} \cup \{\infty\}$ via the following formula: $z = x + i y = \frac{\xi+i\eta}{1-\zeta}$. If we further identify $\mathbb{C}\cup \{\infty\}$ with $\mathbb{CP}^1$, we can conclude every rotation will be represented by a fractional linear substitution.
Now, given a rotation which fixes antipodal points $(\xi,\eta,\zeta), (-\xi,-\eta,-\zeta)$ , which correspondences to $\frac{\xi+i\eta}{1-\zeta}, -\frac{\xi+i\eta}{1+\zeta}$ in $\mathbb{C}\cup \{\infty\}$ and rotate through an angle $\alpha$ counterclockwise. Above process can be decomposed into two steps: first we move $-\frac{\xi+i\eta}{1+\zeta}$ to $0$ and move $\frac{\xi+i\eta}{1-\zeta}$ to $\infty$, which corresponds to the following fractional linear substitution up to some constant:
$$
C\cdot \frac{z+\frac{\xi+i\eta}{1+\zeta}}{z -\frac{\xi+i\eta}{1-\zeta} }
$$
Then we get our new $0,\infty$ axis and then we rotate our $(\xi',\eta')$-plane (equatorial plane) by an angle $\alpha$ counterclockwise, which correspond to multipyling factor $e^{i\alpha}$.
Suppose $z'$ is the coordinate after the rotation, we must have:
$$
\frac{z'+\frac{\xi+i\eta}{1+\zeta}}{z' -\frac{\xi+i\eta}{1-\zeta} } = e^{i\alpha}\frac{z+\frac{\xi+i\eta}{1+\zeta}}{z -\frac{\xi+i\eta}{1-\zeta} } \quad(*)
$$
The author claimed that if we do the following change of notation:
$$
\xi \sin(\frac{\alpha}{2})=a, \quad \eta \sin(\frac{\alpha}{2})=b, \quad \zeta\sin(\frac{\alpha}{2})=c, \quad \cos(\frac{\alpha}{2})=d,
$$
Then $(*)$ can be rewritten as
$$
z' = \frac{(d+ic)z-(b-ia)}{(b+ia)z+(d-ic)} \quad (**)
$$
That is where I stuck. I think I understand the process of deduction of general rotation formula but I am not sure how to get the simple form $(**)$. Now the problem is purely elementary and I try to verify $(**)$ by brute force but it doesn't seem to be a correct way . I guess it will involve with some trigonometric formulas to simplify the computation. Could you please offer me some suggestions on how to start from $(*)$ to derive $(**)$? Thank you in advance.
| Letting $A:=\frac{\xi+i\eta}{1+\zeta}$ and $B:=\frac{\xi+i\eta}{1-\zeta}$ we obtain from $(*)$:
\begin{align*}
e^{-i\frac{\alpha}{2}}\ \frac{z^{\prime}+A}{z^{\prime}-B}&=e^{i\frac{\alpha}{2}}\ \frac{z+A}{z-B}\\
e^{-i\frac{\alpha}{2}}\left(z^{\prime}+A\right)\left(z-B\right)&=e^{i\frac{\alpha}{2}}(z+A)\left(z^{\prime}-B\right)\\
e^{-i\frac{\alpha}{2}}\left(zz^{\prime}+Az-Bz^{\prime}-AB\right)
&=e^{i\frac{\alpha}{2}}\left(zz^{\prime}-Bz+Az^{\prime}-AB\right)\tag{1}\\
\end{align*}
Extracting $z^{\prime}$ from (1) we obtain
\begin{align*}
\color{blue}{z^{\prime}}&=\frac{\left(Be^{i\frac{\alpha}{2}}+Ae^{-i\frac{\alpha}{2}}\right)z
+AB\left(e^{i\frac{\alpha}{2}}-e^{-i\frac{\alpha}{2}}\right)}
{\left(e^{i\frac{\alpha}{2}}-e^{-i\frac{\alpha}{2}}\right)z+Ae^{i\frac{\alpha}{2}}+Be^{-i\frac{\alpha}{2}}}\\
&\,\,\color{blue}{=\frac{\left(Be^{i\frac{\alpha}{2}}+Ae^{-i\frac{\alpha}{2}}\right)z
+2iAB\sin\left(\frac{\alpha}{2}\right)}
{2i\sin\left(\frac{\alpha}{2}\right)z+Ae^{i\frac{\alpha}{2}}+Be^{-i\frac{\alpha}{2}}}}\tag{2}
\end{align*}
Since we want to derive (**):
\begin{align*}
z' = \frac{(d+ic)z-(b-ia)}{(b+ia)z+(d-ic)}
\end{align*}
and the coefficient of $z$ of the denomintor in (2) is $2i\sin\left(\frac{\alpha}{2}\right)$ instead of $b+ia$, we consequently expand numerator and denominator of (2) with
\begin{align*}
\frac{b+ia}{2i\sin\left(\frac{\alpha}{2}\right)}
\end{align*}
We calculate the constant part of the denominator of (**) from (2) and obtain
\begin{align*}
&\color{blue}{\frac{b+ia}{2i\sin\left(\frac{\alpha}{2}\right)}}
\color{blue}{\left(Ae^{i\frac{\alpha}{2}}+Be^{-i\frac{\alpha}{2}}\right)}\\
&\quad=\frac{b+ia}{2i\sin\left(\frac{\alpha}{2}\right)}
\left(\frac{a+ib}{\sin\left(\frac{\alpha}{2}\right)+c}\,e^{i\frac{\alpha}{2}}
+\frac{a+ib}{\sin\left(\frac{\alpha}{2}\right)-c}\,e^{-i\frac{\alpha}{2}}\right)\tag{3}\\
&\quad=\frac{a^2+b^2}{2\sin\left(\frac{\alpha}{2}\right)}
\left(\frac{e^{i\frac{\alpha}{2}}}{\sin\left(\frac{\alpha}{2}\right)+c}
+\frac{e^{-i\frac{\alpha}{2}}}{\sin\left(\frac{\alpha}{2}\right)-c}\right)\\
&\quad=\frac{1}{2\sin\left(\frac{\alpha}{2}\right)}
\left(e^{i\frac{\alpha}{2}}\left(\sin\left(\frac{\alpha}{2}\right)-c\right)
+e^{-i\frac{\alpha}{2}}\left(\sin\left(\frac{\alpha}{2}\right)+c\right)\right)\tag{4}\\
&\quad=\frac{e^{i\frac{\alpha}{2}}+e^{-i\frac{\alpha}{2}}}{2}
-\frac{c}{\sin\left(\frac{\alpha}{2}\right)}\frac{e^{i\frac{\alpha}{2}}-e^{-i\frac{\alpha}{2}}}{2}\\
&\quad=\cos\left(\frac{\alpha}{2}\right)-ic\\
&\,\,\quad\color{blue}{=d-ic}
\end{align*}
according to the claim. The coefficients of the numerator in (**) can be calculated similarly.
Comment:
*
*In (3) we use
\begin{align*}\xi \sin\left(\frac{\alpha}{2}\right)=a,\
\eta \sin\left(\frac{\alpha}{2}\right)=b,\
\zeta \sin\left(\frac{\alpha}{2}\right)=c,\
\cos\left(\frac{\alpha}{2}\right)=d\text{.}
\end{align*}
*In (4) we use $a^2+b^2+c^2=\sin^2\left(\frac{\alpha}{2}\right)$ since $\xi^2+\eta^2+\zeta^2=1$. This implies \begin{align*}
a^2+b^2&=\sin\left(\frac{\alpha}{2}\right)^2-c^2\\
&=\left(\sin\left(\frac{\alpha}{2}\right)-c\right)\left(\sin\left(\frac{\alpha}{2}\right)+c\right)
\end{align*}
| {
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"url": "https://math.stackexchange.com/questions/4206576",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Prove that $2^n + 5^n + 56$ is divisible by $9$, where $n$ is an odd integer
Prove that $9 \mid2^n + 5^n + 56$ where n is odd
I have proved this by using division into cases based on the value of $n\bmod3$ but it seems a little bit clumsy to me and I wonder if there are other ways to prove it, probably by using modular arithmetic or induction? Below is my proof:
$\text{Case 1, }n\bmod3=0,\text{then $n=3k$ for some odd integer k:}$ $$\begin{align}
2^n+5^n+56 & =2^{3k}+5^{3k}+56
\\ & = 8^k+125^k+56
\\ & \equiv (-1)^k+(-1)^k+2\quad&\left(\bmod9\right)
\\ & \equiv 0\quad&\left(\bmod9\right)
\end{align}$$
$\text{Case 2, }n\bmod3=1,\text{then $n=3k+1$ for some even integer k:}$ $$\begin{align}
2^n+5^n+56 & =2^{3k+1}+5^{3k+1}+56
\\ & = 2\cdot8^k+5\cdot125^k+56
\\ & \equiv 2\cdot(-1)^k+5\cdot(-1)^k+2\quad&\left(\bmod9\right)
\\ & \equiv 9\equiv0\quad&\left(\bmod9\right)
\end{align}$$
$\text{Case 3, }n\bmod3=2,\text{then $n=3k+2$ for some odd integer k:}$ $$\begin{align}
2^n+5^n+56 & =2^{3k+2}+5^{3k+2}+56
\\ & = 4\cdot8^k+25\cdot125^k+56
\\ & \equiv 4\cdot(-1)^k+25\cdot(-1)^k+2\quad&\left(\bmod9\right)
\\ & \equiv -27\equiv0\quad&\left(\bmod9\right)
\end{align}$$
| Here is a one-liner:
$$2^{2k+1}+5^{2k+1}+56\equiv2(1+3)^k+5(1+24)^k-7\equiv2(1+3k)+5(1+24k)-7\equiv126k$$
where we used binomial theorem in middle step.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4206716",
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"source": "stackexchange",
"question_score": "5",
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