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How do I write a vector as a linear combination of other vectors. Write $\begin{pmatrix} 5 \\ 3 \\15 \end{pmatrix}$ as a linerar combination of the following vectors: $u=\begin{pmatrix} 1 \\ 2 \\5 \end{pmatrix}$, $v=\begin{pmatrix} 3 \\ -4 \\-1 \end{pmatrix}$, $w=\begin{pmatrix} -1 \\ 1 \\1 \end{pmatrix}$.
My attempt:
$$\begin{bmatrix}
1 & 3& -1 & 5\\
2 & -4 & 1& 3\\
5&-1&1&15\\
\end{bmatrix}\sim\to\begin{bmatrix}
1 & 0& 0 & 3\\
0 & 1 & 0 & 1\\
0&0&1&1\\
\end{bmatrix}$$
Obviously I skipped a lot of reduction steps, because it's a pain to type matrices here, but I was wondering if I even did the right thing.
|
So...you've done the hard part, now finish it off:
$$c_1\begin{pmatrix} 1 \\ 2 \\5 \end{pmatrix} + c_2 \begin{pmatrix} 3 \\ -4 \\-1 \end{pmatrix}+ c_3\begin{pmatrix} -1 \\ 1 \\1 \end{pmatrix} = \begin{pmatrix} 5 \\ 3 \\15 \end{pmatrix}$$
So what have you found to be $c_1, c_2, c_3$?
Fill in the correct values for each constant, and you're done!
|
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|
Prove by Mathematical Induction: $1(1!) + 2(2!) + \cdot \cdot \cdot +n(n!) = (n+1)!-1$ Prove by Mathematical Induction . . .
$1(1!) + 2(2!) + \cdot \cdot \cdot +n(n!) = (n+1)!-1$
I tried solving it, but I got stuck near the end . . .
a. Basis Step:
$(1)(1!) = (1+1)!-1$
$1 = (2\cdot1)-1$
$1 = 1 \checkmark$
b. Inductive Hypothesis
$1(1!) + 2(2!) + \cdot \cdot \cdot +k(k!) = (k+1)!-1$
Prove k+1 is true.
$1(1!) + 2(2!) + \cdot \cdot \cdot +(k+1)(k+1)! = (k+2)!-1$
$\big[RHS\big]$
$(k+2)!-1 = (k+2)(k+1)k!-1$
$\big[LHS\big]$
$=\underbrace{1(1!) + 2(2!) + \cdot \cdot \cdot + k(k+1)!} + (k+1)(k+1)!$ (Explicit Last Step)
$= \underbrace{(k+1)!-1}+(k+1)(k+1)!$ (Inductive Hypothesis Substitution)
$= (k+1)!-1 + (k+1)(k+1)k!$
$= (k+1)k!-1 + (k+1)^{2}k!$
My [LHS] looks nothing like my [RHS] did I do something wrong?
EDIT:
$ = (k+1)k! + (k+1)^2k! -1 $
$ = (k+1)(k!)(1 + (k+1))-1$
$ = (k+1)(k!)(k+2)-1 = (k+2)(k+1)k!-1$
|
$$1(1!) + 2(2!) + \cdot \cdot \cdot +n(n!) = (n+1)!-1$$
then
$$1(1!) + 2(2!) + \cdot \cdot \cdot +n(n!)+(n+1)(n+1)! =(n+1)!-1+(n+1)(n+1)!=$$
$$(n+1)!(n+1+1)-1=(n+2)!-1=((n+1)+1)!-1$$
|
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|
If $a$ and $b$ are two roots of $x^4 + x^3 - 1 = 0$ prove that $ab$ is a root of $x^6 + x^4 + x^3 - x^2 - 1$. If $a$ and $b$ are two roots of $x^4 + x^3 - 1 = 0$ prove that $ab$ is a root of $x^6 + x^4 + x^3 - x^2 - 1$.
Students and I are unsure how to go about this problem. Also will this be a problem I can solve and prove in front of a class in 20 minutes?
|
There is probably a shorter way, but I think this way sheds some light on why it's true:
Let $a,b,c,d$ be the four roots of $x^4+x^3-1$, so that $x^4+x^3-1=(x-a)(x-b)(x-c)(x-d)$. Now set
$$
g(x) = (x-ab)(x-ac)(x-ad)(x-bc)(x-bd)(x-cd).
$$
This polynomial is symmetric in the roots $a,b,c,d$, and so its coefficients will be rational numbers (since the coefficients of $x^4+x^3-1$ are rational). This explains why there is such a sextic; to work out its coefficients, we need to mess around with symmetric polynomials.
The coefficients of the original polynomial tell us that $a+b+c+d=-1$, $abcd=-1$, and $ab+ac+ad+bc+bd+cd=abc+abd+acd+bcd=0$. Therefore the coefficient of $x^5$ in $g(x)$ equals $0$; the coefficient of $x^4$ equals
$$
abac+abad+abbc+abbd+abcd+acad+acbc+acbd+accd+adbc+adbd+adcd+bcbd+bccd+bdcd = (a + b + c + d) (a b c + a b d + a c d + b c d) - abcd = 1;
$$
and so on - the coefficient of $x^0$ equals $(abcd)^3=-1$.
(One can play the same game with $f(x)=(x-a^2)(x-b^2)(x-c^2)(x-d^2)$ to obtain the polynomial $x^4−x^3−2x^2+1$.)
|
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|
Solve the congruence$x^3+4x+8\equiv{0}\pmod{15}$ Solve (if possible)the congruence involving polynomial
$x^3+4x+8\equiv{0}\pmod{15}$
My work:
Since $15=3\cdot5$, we have
$x^3+4x+8\equiv{0}\pmod{3}$ and $x^3+4x+8\equiv{0}\pmod{5}$
In $\mathbb{Z}_3$,
We have $[0],[1],[2]$
They all dont work
In $\mathbb{Z}_5$,
We have $[0],[1],[2],[3],[4]$
They all dont work
So does it mean I have NO solution?
Thank you!!
|
Modulo $3$, there is a solution: $x\equiv 2\pmod{3}$ does work.
But modulo $5$, there is no solution.
So there is no solution modulo $15$. For if $x^3+4x+8\equiv 0\pmod{15}$, then $x^3+4x+8\equiv 0\pmod{5}$.
|
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|
The remainder of $1^2+3^2+5^2+7^2+\cdots+1013^2$ divided by $8$ How to find the remainder of $1^2+3^2+5^2+7^2+\cdots+1013^2$ divided by $8$
|
The sum is of the form
$$\sum_{k=0}^{125}((8k+1)^2+(8k+3)^2+(8k+5)^2+(8k+7)^2)+((8.126+1)^2+(8.126+3)^2+(8.126+5)^2)$$
This above one is congruent to(mod 8),
$$(\sum_{k=0}^{125}((1)^2+(3)^2+(5)^2+(7)^2))+(1+3^2+5^2)$$
This is again congruent to,
$$(\sum_{k=0}^{125}(1+1+1+1))+(1+1+1)=126\times 4+3$$
This is congruent to ,
$3(\mod 8)$
This means it will leave a remainder of 3 when divided by 8.
|
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|
Proving $4^{47}\equiv 4\pmod{12}$ I know this is a simple exercise, but I was wondering if I can make the following logical jump in my proof:
We see that $4\equiv 4\pmod{12}$ and $4^2\equiv 4\pmod{12}$. Then we can recursively multiply by $4$ to get $4^{47}\equiv 4\pmod{12}$.
|
$4^{47}$ and $4$ are divisible by $4$ thus $4^{47}-4$ is a multiple of $4$.
Also, since $4 \equiv 1 \pmod{3}$ we have $4^{46}\equiv 1^{46}\pmod{3}$ and then
$$4^{47} \equiv 4 \pmod 3 \,.$$
Thus $4^{47}-4$ is a multiple of $3$ and $4$, and hence of 12...
|
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|
The area of the triangle with vertices $(3, 2), (3, 8)$, and $(x, y)$ is $24$. What is $x$? The area of the triangle with vertices $(3, 2), (3, 8)$, and $(x, y)$ is $24$. A possible value for $x$ is:
*
*a) $7$
*b) $9$
*c) $11$
*d) $13$
*e) $15$
Please show your work and explain.
|
We know the area of a triangle (Article#25) having vertices $(x_i,y_i)$ for $i=1,2,3$ is
$$\frac12\det\begin{pmatrix} x_1 & y_1 & 1\\x_2&y_2&1\\ x_3 & y_3 &1\end{pmatrix}$$
Now, $$\det\begin{pmatrix}x & y &1\\ 3 & 2 & 1 \\ 3 & 8 & 1 \end{pmatrix}$$
$$=\det\begin{pmatrix}x-3 & y-2 &0\\ 3 & 2 & 1 \\ 0 & 6 & 0 \end{pmatrix}\text { (Applying } R_3'=R_3-R_2,R_1'=R_1-R_2)$$
$$=6(x-3)$$
As we take the area in absolute value,the are here will be $\frac12\cdot6|x-3|=3|x-3|$
If $x\ge 3, 3(x-3)=24\implies x=11$
If $x<3, 3(3-x)=24\implies -5$
|
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|
Prove that $\frac{a}{b^2+5}+ \frac{b}{c^2+5} + \frac{c}{a^2+5} \le \frac 12$ Let $a,b,c>0$ and $a^3+b^3+c^3=3$. Prove that $$\dfrac{a}{b^2+5}+ \dfrac{b}{c^2+5} + \dfrac{c}{a^2+5} \le \dfrac 12$$
I have an ugly solution for this solution.
|
Lohwater's "Introduction to inequalities" is a tour de force in proving all sorts of inequalities using mostly elementary means.
|
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|
Complex Integration : $\int_1^{1+i}\frac{1}{1+z^2}dz$
Integrate alonf the line segment from $z=1$ to $z=1+i$ :
$$\int_1^{1+i}\frac{1}{1+z^2}dz$$
If I integrate, it is just the identity $tan^{-1}z$, but the answer to this question is $$\frac{\pi}{4}-\frac{1}{2}\arctan2+\frac{i}{4}\log5$$ which I don't understand how they got?
|
Set $z=1+i t$, $dz = i dt$. Then the integral is
$$\begin{align}i \int_0^1 \frac{dt}{1+(1+i t)^2} &= i \int_0^1 \frac{dt}{2+i 2 t -t^2} \\ &= -i \int_0^1 \frac{dt}{(t-(1+i))(t+(1-i))}\\ &= -\frac{i}{2} \int_0^1 dt \left (\frac{1}{t-(1+i)} - \frac{1}{t+(1-i)} \right ) \\ &= -\frac{i}{2} \left [ \log{ \left ( \frac{t-(1+i)}{t+(1-i)} \right )} \right ]_0^1 \\ &= -\frac{i}{2} \left [ \log{ \left ( \frac{-i}{2-i} \right )} - \log{ \left ( \frac{-(1+i)}{1-i} \right )} \right ]\\ &= -\frac{i}{2}\log{ \left ( \frac{i (1-i)}{(2-i)(1+i)} \right )}\\ &= \frac{i}{2}\log{(2-i)} \end{align}$$
Now write $2-i$ as $\sqrt{5} e^{i (\pi - \arctan{(1/2)})} = \sqrt{5} e^{i ((\pi/2) - \arctan{2})}$ Then
$$i \int_0^1 \frac{dt}{1+(1+i t)^2} = \frac{\pi}{4} - \frac{1}{2} \arctan{2} + i \frac{\log{5}}{4}$$
|
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|
Silly question: Why is $\sqrt{(9x^2)} $ not $3x$? I had to find the derivative of $f(x) = \sqrt{(9x^2)}$. I applied chain rule with the following steps.
Let $f(x)$ be $\sqrt{x}$ and $g(x)$ be $9x^2$
$$ \begin{align} &f'(g(x)) \times g'(x) \\
& = \frac{1}{2\sqrt{(9x^2)}} \times 18x \\
& = \frac{18x}{2\sqrt{9x^2}} \\
& = \frac{9x}{3\sqrt{x^2}}\\
& = \frac{9x}{3\sqrt{x^2}} \\
& = \frac{3x}{\sqrt{x^2}} \end{align}$$
I got the answer but I don't understand why the last bit doesn't simplify to $3$ because $\sqrt{x^2}$ is $x$ and if it does then why does the back of my textbook and W|A say that it is not?
EDIT:
Okay, so from what I understand, it should be actually $3|x|$. If for example, I had $\sqrt{4x^2}$, I will have $2|x|$, if my understanding is correct.
|
Remember that $\sqrt{x^2}$ is the positive number whose square is $x^2$ (this is the definition of the square root sign). If $x$ is positive, then that number is $x$, otherwise, it is $-x$. For example, $\sqrt{(-3)^2} = \sqrt{9} = 3$, not $-3$. You can always say that $\sqrt{x^2} = \left| x \right|$.
|
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|
To check the solution via radicals Given equations are
$$x^5+y^5=a \tag 1$$
$$5xy(x^2+xy+y^2)=b \tag 2$$
Is it possible to find $x$ or $y$ via using radicals?
My attempt
$$x^2+y^2=\frac{b}{5xy}-xy$$
$$x^4+y^4=\frac{b^2}{5^2x^2y^2}-x^2y^2-\frac{2b}{5}$$
$$(x^2+y^2)(x^4+y^4)=\left(\frac{b}{5xy}-xy\right)\left(\frac{b^2}{5^2x^2y^2}-x^2y^2-\frac{2b}{5}\right)$$
$$x^6+y^6+x^2y^2(x^2+y^2)=\left(\frac{b}{5xy}-xy\right)\left(\frac{b^2}{5^2x^2y^2}-x^2y^2-\frac{2b}{5}\right)$$
$$x^6+y^6+x^2y^2\left(\frac{b}{5xy}-xy\right)=\left(\frac{b}{5xy}-xy\right)\left(\frac{b^2}{5^2x^2y^2}-x^2y^2-\frac{2b}{5}\right)$$
$$x^6+y^6=\left(\frac{b}{5xy}-xy\right)\left(\frac{b^2}{5^2x^2y^2}-2x^2y^2-\frac{2b}{5}\right)$$
I haven't got $x^5+y^5$ in my way,it looks impossible to solve it via radicals but I need to proof.
Thanks for answers
|
No. First, notice that
$$ (x+y)^5 = (x^5+y^5) + 5 x y (x^2+xy+y^2) (x+y)$$
So your equations imply
$$(x+y)^5 = a + b(x+y)$$
Putting $s = x+y$, if $x$ and $y$ could be expressed in terms of $a$ and $b$ using radicals, then so could the root of $s^5 -bs -a$. This quintic cannot be solved in radicals. (In fact, this quintic is in what is known as Bring-Jerrard normal form. Every quintic can, by a change of variables be put into this form so, if we could solve equations in this form by radicals, we could also solve general equations.)
So, how did I find this? I always try to use symmetries of the equations to reduce their degree before starting. In this case, the only symmetry I noticed was switching $x$ and $y$. So I set $s=x+y$ and $t=xy$, and used the fundamental theorem of symmetric functions to write:
$$s^5 - 5 s^3 t + 5 s t^2 = a \quad 5 t (s^2-t) = b $$
The next thing I wanted to do was eliminate one of the variables. I wasn't smart enough to do it by hand, so I asked Mathematica:
(* Find an equation satisfied by t, containing no copies of s. *)
GroebnerBasis[{s^5 - 5 s^3 t + 5 s t^2 - a, 5 t (s^2 - t) - b}, {t}, {s}]
Mathematica output:
{b^5 - 25 b^4 t^2 + 125 b^3 t^4 - 3125 a^2 t^5 + 625 b^2 t^6 - 3125 b t^8 + 3125 t^10}
A degree $10$ equation. It probably isn't solvable by radicals, but it seems hard to prove. Let's try again:
(* This time, keep s and eliminate t *)
GroebnerBasis[{s^5 - 5 s^3 t + 5 s t^2 - a, 5 t (s^2 - t) - b}, {s}, {t}]
Output:
{-a - b s + s^5}
Ah. much simpler. And clearly not solvable by radicals, so neither is the original problem.
|
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|
Formula for $\sum _{i=1}^n (n+1-i) (n-i)$ It is easy to show that
$$\sum _{i=1}^n (n+1-i) (n-i) = n (n-1)+(n-1) (n-2)+...+1 (1-1)=\frac{1}{3} \left(n^3-n\right)$$
using induction. But how do I derive this formula? I couldn't find any substitution to do this.
|
You must already know that
$$
\begin{align}
\sum_{i = 1}^n 1 &= n \\
\sum_{i = 1}^n i &= \frac{n(n + 1)}{2} \\
\sum_{i = 1}^n n &= n \cdot \sum_{i = 1}^n 1 \\
& = n^2 \\
\sum_{i = 1}^n i^2 &= \frac{n(n + 1)(2n + 1)}{6} \\
\end{align}
$$
Now, from the given expression;
$$
\begin{align}
\sum_{i=1}^n (n+1-i) (n-i) &= \sum_{i = 1}^n\left( n^2 + i^2 + n - i - 2 n i\right) \\
&= n^2 \cdot \sum_{i=1}^n 1 + \sum_{i=1}^n i^2 + n \cdot \sum_{i=1}^n1 - \sum_{i=1}^ni -2n \cdot \sum_{i=1}^ni \\
&= n^3 + \frac{n(n + 1)(2n + 1)}{6} + n^2 - \frac{n(n + 1)}{2} - n^2(n + 1) \\
&= \frac{n(n + 1)}{6} \cdot \left( 2n + 1 - 3\right) \\
&= \frac{n (n + 1)}{6} \cdot 2(n - 1) = \frac{n (n + 1) (n - 1)}{3} \\
&= \frac{n^3 - n}{3}
\end{align}
$$
|
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|
Prove $\int_0^\infty \frac{\sin^4x}{x^4}dx = \frac{\pi}{3}$ I need to show that
$$
\int_0^\infty \frac{\sin^4x}{x^4}dx = \frac{\pi}{3}
$$
I have already derived the result $\int_0^\infty \frac{\sin^2x}{x^2} = \frac{\pi}{2}$ using complex analysis, a result which I am supposed to start from. Using a change of variable $ x \mapsto 2x $ :
$$
\int_0^\infty \frac{\sin^2(2x)}{x^2}dx = \pi
$$
Now using the identity $\sin^2(2x) = 4\sin^2x - 4\sin^4x $, we obtain
$$
\int_0^\infty \frac{\sin^2x - \sin^4x}{x^2}dx = \frac{\pi}{4}
$$
$$
\frac{\pi}{2} - \int_0^\infty \frac{\sin^4x}{x^2}dx = \frac{\pi}{4}
$$
$$
\int_0^\infty \frac{\sin^4x}{x^2}dx = \frac{\pi}{4}
$$
But I am now at a loss as to how to make $x^4$ appear at the denominator. Any ideas appreciated.
Important: I must start from $ \int_0^\infty \frac{\sin^2x}{x^2}dx $, and use the change of variable and identity mentioned above
|
Integrating the integral by parts thrice yields
$$
\begin{aligned}
& \int_{0}^{\infty} \frac{\sin ^{4} x}{x^{4}} d x \\
=& \int_{0}^{\infty} \frac{\left(\sin ^{4} x\right)^{(3)}}{3 \times 2 \cdot x} d x \\
=& \int_{0}^{0} \frac{4 \sin 2 x-8 \sin 4 x}{6 x} d x \\
=& \frac{4}{3}\left(-\int_{0}^{\infty} \frac{\sin 2 x}{x} d x+2 \int_{0}^{\infty} \frac{\sin 4 x}{x}\right) \\
=& \frac{2}{3}\left[-\frac{\pi}{2}+2\left(\frac{\pi}{2}\right)\right] \\
=& \frac{\pi}{3}
\end{aligned}
$$
|
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|
Connecting midpoints of sides of a triangle In triangle $\triangle ABC$, $AB=8$, $BC=14$ and $CA=10$. Points $M$, $N$, and $P $ are the midpoints of sides $AB$, $BC$, and $CA$, respectively. If $M$, $N$, and $P$ are connected to form a triangle, what is the perimeter of triangle $\triangle MNP$?
|
$\displaystyle \frac{8}{14}=\frac{4}{MP}\Rightarrow MP=7$
$\displaystyle \frac{14}{8}=\frac{7}{NP}\Rightarrow NP=4$
$\displaystyle\frac{14}{10}=\frac{7}{MN}\Rightarrow MN=5$
Perimeter$=MP+NP+MN=7+4+5=16$
|
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|
Evaluating $\int_{0}^{\infty}\sin^3(x)\cos[a\tan(x)]\frac{dx}{x}$ $$I(a)=\int_{0}^{\infty}\sin^3(x)\cos[a\tan(x)]\frac{dx}{x}$$ I'd like to evaluate the integral by differentiating with respect to parameter $a$ but no success yet. Seems impossible. What would the other options?
Edit: A hypothetical closed form solution:
$$I(a)=\frac{\pi}{4}(1-a)e^{-a}$$
|
Let us assume that $f(z)$ has a Maclaurin series expansion with real coefficients that converges on the unit circle on the complex plane.
Then
$$ \text{Re} \int_{0}^{\infty} \frac{\sin^{3} x}{x} f(e^{2ix})\, dx = \int_{0}^{\infty} \frac{\sin^3 x}{x} \sum_{n=0}^{\infty} \frac{f^{(n)}(0)}{n!} \cos(2nx) \, dx .$$
Now assuming we can switch the order of summation and integration (which would be more difficult to justify if the series doesn't converge absolutely),
$$ \begin{align} &\text{Re} \int_{0}^{\infty} \frac{\sin^{3} x}{x} f(e^{2ix})\, dx \\ &= \sum_{n=0}^{\infty} \frac{f^{(n)}(0)}{n!} \int_{0}^{\infty} \frac{\sin^{3} x}{x} \cos(2nx) \, dx \\ &= f(0) \int_{0}^{\infty} \frac{\sin^{3} x}{x} \, dx + f'(0) \int_{0}^{\infty} \frac{\sin^{3} x}{x} \cos(2x) \, dx + \sum_{n=2}^{\infty}\frac{f^{(n)}(0)}{n!} \int_{0}^{\infty} \frac{\sin^{3} x}{x} \cos(2nx) \, dx. \end{align}$$
I will now show that all the integrals vanish except for the first two.
Notice that for $a \ge 0$,
$$ \begin{align} &\int_{0}^{\infty} \frac{\sin^{3} x}{x} \cos(ax) \, dx \\ &= \frac{1}{4} \int_{0}^{\infty} \frac{3 \sin(x) - \sin (3x)}{x} \cos(ax) \, dx \\ &= \frac{1}{8} \int_{0}^{\infty} \frac{3 \sin [(1+a)x] + 3 \sin [(1-a)x] - \sin [(3+a)x] - \sin [(3-a)x]}{x} \, dx \\ &= \frac{\pi}{16} \Big(3 \, \text{sgn}(1+a) + 3 \, \text{sgn}(1-a) - \text{sgn}(3+a) - \text{sgn}(3-a) \Big) \tag{1} \\ &= \begin{cases} 0 & \text{if} \ a >3 \\ - \frac{\pi}{16} & \text{if} \ a = 3 \\ - \frac{\pi}{8} & \text{if} \ 1<a <3 \\ \frac{\pi}{16} & \text{if} \ a = 1 \\ \frac{\pi}{4} & \text{if} \ a <1 \end{cases} \end{align}$$
Therefore,
$$ \begin{align} \text{Re} \int_{0}^{\infty} \frac{\sin^{3} x}{x} f(e^{2ix})\, dx &= f(0) \left(\frac{\pi}{4} \right) + f'(0) \left(-\frac{\pi}{8} \right) + \sum_{n=2}^{\infty}\frac{f^{n}(0)}{n!} (0) \\ &= \frac{\pi}{8} \Big(2 f(0) - f'(0) \Big). \end{align}$$
$(1)$ $\int_{0}^{\infty} \frac{\sin (ax)}{x} \, dx = \frac{\pi}{2} \, \text{sgn}(a)$
As a simple example, let us consider the entire function $f(z) = e^{z}$.
Then $$\text{Re} \int_{0}^{\infty} \frac{\sin^{3} x}{x} \, e^{e^{2ix}} \, dx = \int_{0}^{\infty} \frac{\sin^{3} x}{x} \, e^{\cos (2x)} \cos (\sin 2x) \, dx = \frac{\pi}{8} \Big( 2(1)-1 \Big) = \frac{\pi}{8}.$$
Now let us consider the function $$ f(z) = \exp \left(a \, \frac{z-1}{z+1} \right)$$ for $a\ge 0$.
Since $\exp \left(a \, \frac{z-1}{z+1} \right)$ has an essential singularity at $z=-1$, we know that its Maclaurin series has a radius of convergence of $1$.
What we don't immediately know, and what is proving hard to ascertain, is if the series converges on the unit circle (excluding the point $z=-1$).
But assuming that it does we have
$$ \begin{align} \text{Re} \int_{0}^{\infty} \frac{\sin^{3}(x)}{x} \, \exp \left(a \, \frac{e^{2ix}-1}{e^{2ix}+1} \right) \, dx &= \text{Re} \int_{0}^{\infty} \frac{\sin^{3}(x)}{x} \, \exp (ia \tan x) \, dx \\ &= \int_{0}^{\infty} \frac{\sin^{3}(x)}{x} \cos(a \tan x) \, dx \\ &= \frac{\pi}{8} \Big(2 (e^{-a}) -2ae^{-a} \Big) \\ &= \frac{\pi}{4} (1-a) e^{-a}. \end{align}$$
|
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"timestamp": "2023-03-29T00:00:00",
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|
Integrate $\int_{0}^{\pi}{\frac{x\cos{x}}{1+\sin^{2}{x}}dx}$ Integrate
$$\displaystyle \int_{0}^{\pi}{\frac{x\cos{x}}{1+\sin^{2}{x}}dx}$$
|
Here's a more elementary way:$$J=\int_0^\pi\frac{x \cos(x) dx}{1+\sin^2(x)}= \int_0^\frac{\pi}{2} \frac{x \cos(x)dx}{1+\sin^2(x)}+\int_0^\frac{\pi}{2} \frac{-x \sin(x)dx}{1+\cos^2(x)}-\frac{\pi}{2} \int_\frac{\pi}{2}^0 \frac{- \sin (x) dx}{1+\cos^2(x)}$$
$$J=I_1-I_2-\frac{\pi^2}{8}$$
$$I_1=\int_0^\frac{\pi}{2} \frac{ x\cos(x) dx}{1+\sin^2(x)}= \sum_{k \ge 0}(-1)^k \int_0^\frac{\pi}{2} x \cos(x) \sin^{2k}(x)dx$$
$$= \sum_{k \ge 0}(-1)^k \left( \frac{\pi}{2}\frac{1}{2k+1}- \frac{1}{2k+1}\int_0^\frac{\pi}{2} \sin^{2k+1}(x) dx \right)$$
, using the beta function,
$$= \sum_{k \ge 0}(-1)^k \left( \frac{\pi}{2}\frac{1}{2k+1}- \frac{1}{2k+1}\frac{\Gamma(k+1)\sqrt{\pi}}{\Gamma(k+1+\frac{1}{2})} \right)$$
, and the Legendre duplication formula,
$$= \sum_{k \ge 0}(-1)^k \left( \frac{\pi}{2}\frac{1}{2k+1}- \frac{1}{2k+1}\frac{\Gamma(k+1)^22^{2k+1}}{\Gamma(2k+2)} \right)$$
$$= \sum_{k \ge 0}(-1)^k \left( \frac{\pi}{2}\frac{1}{2k+1}- \frac{1}{(2k+1)^2}\frac{2^{2k+1}}{\binom{2k}{k}} \right)$$
$$= \frac{\pi}{2}\arctan(1)-\sum_{k \ge 0}(-1)^k \frac{1}{(2k+1)^2}\frac{2^{2k+1}}{\binom{2k}{k}} $$
$$\stackrel{\text{Mathematica}}{=} \frac{\pi^2}{8}-\frac{\pi^2}{8}+ \frac{1}{2}\sinh^{-1}(1)^2 $$
$$I_1= \frac{1}{2}\sinh^{-1}(1)^2 $$
$$I_2=\int_0^\frac{\pi}{2} \frac{ x\sin(x) dx}{1+\cos^2(x)}= \sum_{k \ge 0}(-1)^k \int_0^\frac{\pi}{2} x \sin(x) \cos^{2k}(x)dx$$
$$= \sum_{k \ge 0}(-1)^k \frac{1}{2k+1}\int_0^\frac{\pi}{2} \cos^{2k+1}(x)dx$$
$$= \sum_{k \ge 0}(-1)^k \frac{1}{2k+1}\int_0^\frac{\pi}{2} \sin^{2k+1}(x)dx$$
$$I_2=\frac{\pi^2}{8}- \frac{1}{2} \sinh^{-1}(1)^2$$
Thus
$$J=\sinh^{-1}(1)^2- \frac{\pi^2}{4}.$$
|
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|
How to show that $ \frac{\cos x}{1 - \sin x} - \tan x = \sec x$? Question is: verify the identity:
$$
\frac{\cos x}{1 - \sin x} - \tan x = \sec x.
$$
How do I show that the left side equals the right?
I changed $\tan x$ into $\sin x/\cos x$ but didn't get anywhere.
Please help.
|
$\frac{\cos x}{1-\sin x} - \tan x = \sec x$
$\displaystyle\frac{\cos x}{1-\sin x} - \frac{\sin x}{\cos x} = \sec x$
$$(\cos^2 x - \sin x (1 - \sin x)) / (1-\sin x) (\cos x) = \sec x$$
$$(\cos^2 x - \sin x + \sin^2 x) / (1-\sin x)\cos x = \sec x$$
$$\begin{array}{lll}
(1 - \sin x) / ((1-\sin x)\cos x) &=& \sec x\\
1/\cos x &=& \sec x\\
1/\sec x &=& \sec x
\end{array}$$
|
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"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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|
Infinite Series :$ \sum_{n=0}^\infty \frac{\Gamma \left(n+\frac{1}{2} \right)\psi \left(n+\frac{1}{2} \right)}{n! \left(n+\frac{3}{2}\right)^2}$ Prove that:
$$\sum_{n=0}^\infty \frac{\Gamma \left(n+\frac{1}{2} \right)\psi \left(n+\frac{1}{2} \right)}{n! \left(n+\frac{3}{2}\right)^2} = \frac{-\pi^{\frac{3}{2}}}{12}\left( \pi^2+6\gamma(1-2\log 2)-12\log 2\right)$$
where $\gamma$ is Euler-Mascheroni Constant and $\psi(z)$ is the Digamma Function.
|
Begin with
$$\sum_{n=0}^\infty \frac{\Gamma(n+1-y)}{n!}\frac{1}{x+n} = \frac{\pi \Gamma(x)}{\sin(\pi y) \Gamma(x+y)}$$
Differentiating with respect to $x$ gives
$$-\sum_{n=0}^\infty \frac{\Gamma(n+1-y)}{n!}\frac{1}{(x+n)^2} = \frac{\pi \Gamma(x)}{\sin(\pi y) \Gamma(x+y)} \left\{ \psi(x)-\psi(x+y)\right\}$$
Now, differentiate with respect to $y$:
$$\sum_{n=0}^\infty \frac{\Gamma(n+1-y)\psi(n+1-y)}{n!}\frac{1}{(n+x)^2}=\frac{-\pi \Gamma(x)}{\sin(\pi y)\Gamma(x+y)}\left[ \{ \sin(\pi y)\psi(x+y)+\pi\cos(\pi y)\}\{\psi(x)-\psi(x+y)\}+\psi_1(x+y)\right]$$
Putting $x=\frac{3}{2}$ and $y=\frac{1}{2}$, gives the desired result.
$$\sum_{n=0}^\infty \frac{\Gamma \left(n+\frac{1}{2} \right)\psi \left(n+\frac{1}{2} \right)}{n! \left(n+\frac{3}{2}\right)^2} = \frac{-\pi^{\frac{3}{2}}}{12}\left( \pi^2+6\gamma(1-2\log 2)-12\log 2\right)$$
If you have any other method, please enlighten me.
|
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|
Prove that a set of matrices is a subspace I'm self studying linear algebra and now I'm starting with proofs and so on. I found this exercise and this is the way I prove it. I think it's correct but I'm not sure I mean, what do you think?
Is the set of matrices
$
\begin{pmatrix}
x && x+y \\
x-y && y
\end{pmatrix}
$ where $x, y \in R $ a subspace of $M_{2\times 2}$
I know that for the set to be a subspace it needs to be closed under vector addition and under scalar multiplication. So:
Given $M_1, M_2 \in M_{2\times 2}, x,y, \in R$. Then $M_1 + M_2$ will be:
$$\begin{align}
\begin{pmatrix}
x_1 && x_1+y_1\\
x_1-y_1 && y_1
\end{pmatrix}
&+
\begin{pmatrix}
x_2 && x_2+y_2\\
x_2-y_2 && y_2
\end{pmatrix}
=
\\&
\begin{pmatrix}
(x_1 + x_2) && (x_1+x_2)+(y_1+y_2)\\
(x_1+x_2)-(y_1+y2) && (y_1 + y_2)
\end{pmatrix}
\end{align}$$
Which has the same structure so it's closed under vector addition.
Now, Given $M \in M_{2\times 2}, x,y,r \in R$. Then $rM_1$ will be:
$$
r\begin{pmatrix}
x && x+y\\
x-y && y
\end{pmatrix}
=
\begin{pmatrix}
rx && r(x+y)\\
r(x-y) && ry
\end{pmatrix}
$$
Which, given that $x,y,r \in R$ is also closed under vector multiplication.
So yes. The set is a subspace of $M_{2\times 2}$
|
Note
$$
\begin{pmatrix}
x && x+y \\
x-y && y
\end{pmatrix}
=
x \begin{pmatrix}
1 && 1 \\
1 && 0
\end{pmatrix}
+
y \begin{pmatrix}
0 && 1 \\
-1 && 1
\end{pmatrix}
$$
so it is the subspace generated by the two matrices
$$
\begin{pmatrix}
1 && 1 \\
1 && 0
\end{pmatrix},
\qquad
\begin{pmatrix}
0 && 1 \\
-1 && 1
\end{pmatrix}.
$$
|
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|
Proof: Two non identical circles have at most 2 same points I'm struggeling with an analytic proof for the fact, that two different circles have at most 2 same points. (I try to solve it analytical, because geometrical I already prooved it).
I tried to start with the equations
$r_1^2=(x-a_1)^2+(y-b_1)^2$ and $r_2^2=(x-a_2)^2+(y-b_2)^2$, further
$r_1^2=x^2-2xa_1+a_1^2+y^2-2yb_1+b_1^2$ and the same for the 2nd equation.
Then I get
\begin{eqnarray*}
r_1^2&=&x^2-2xa_1+a_1^2+y^2-2yb_1+b_1^2\\
r_2^2&=&x^2-2xa_2+a_2^2+y^2-2yb_2+b_2^2\\
0&=&x^2-2xa_1+a_1^2+y^2-2yb_1+b_1^2-r_1^2\\
0&=&x^2-2xa_2+a_2^2+y^2-2yb_2+b_2^2-r_2^2\\
&\Rightarrow& x^2-2xa_1+a_1^2+y^2-2yb_1+b_1^2-r_1^2=x^2-2xa_2+a_2^2+y^2-2yb_2+b_2^2-r_2^2\\
&\Leftrightarrow& 0=-2xa_1+a_1^2-2xb_1+b_1^2-r_1^2+2xa_2-a_2^2+2yb_2-b_2^2+r_2^2
\end{eqnarray*}
But now I don't know how to move on. Can someone give a hint?
|
A circle is defined as the locus of a degree 2 polynomial.
The intersection of two circles is then, the roots of a degree 2 polynomial
There are at most $n$ roots of a degree $n$ polynomial
|
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|
A question about an equilateral triangle Suppose that $\triangle ABC$ is an equilateral triangle. Let $D$ be a point inside the triangle so that $\overline{DA}=13$, $\overline{DB}=12$, and $\overline{DC}=5$. Find the length of $\overline{AB}$.
|
The Wikipedia article on equilateral triangles quotes the following theorem from "Curious Properties of the Circumcircle and Incircle of an Equilateral Triangle," by Prithwijit De (http://ms.appliedprobability.org/data/files/Abstracts%2041/41-1-7.pdf):
Theorem: Let $ABC$ be an equilateral triangle with side $s$, and let $P$ be a point in the plane of the triangle with distances $p$, $q$, and $r$ to $A$, $B$, and $C$, respectively. Then $$3(p^4+q^4+r^4+s^4) = (p^2+q^2+r^2+s^2)^2{\textrm.}$$
Solving $3(5^4+12^4+13^4+s^4) = (5^2+12^2+13^2+s^2)^2$ for $s$ gives $s=\sqrt{169\pm 60\sqrt3}$. Because $P$ is inside the triangle, $s>13$, which allows only $s=\sqrt{169+ 60\sqrt3}$, the same solution others have obtained.
|
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|
triple integral to calculate volume Let $V$ be the volume of the $3$-dimensional structure bounded by the paraboloid $z=1−x^2−y^2$, planes $x\ge0, y\ge0$ and $z\ge0$ and by the cylinder $x^2+y^2−x\le0$. If $V=a\pi/b$, where $a$ and $b$ are coprime positive integers, what is the value of $a+b$?
if we consider $x^2+y^2=r^2,x=r\cos x,y =r\sin x$, we get $0\le z\le 1-r^2, 0\le r\le\cos x$ and$0\le x\le\pi/2$.
Is anything wrong with the limits?
|
The cross-section of this volume is the intersection of the circles $(x-1/2)^2+y^2=1$ and $x^2+y^2=1-z$. Draw the picture to see that the intersection breaks into two pieces: that bounded on top by the first circle, and that by the second. The integral is
$$\int_0^1 dz \: \left [ \int_0^{1-z} dx \: \int_0^{\sqrt{1/4 - (x-1/2)^2}} dy + \int_{1-z}^{\sqrt{1-z}} dx \: \int_0^{\sqrt{1-z - x^2}} dy \right ]$$
|
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|
Show $ \frac{x}{xy+x+1}+\frac{y}{yz+y+1}+\frac{z}{zx+z+1}=1 $ given $xyz = 1$ Please help me prove the equality: If $xyz=1$, prove that
$$
\frac{x}{xy+x+1}+\frac{y}{yz+y+1}+\frac{z}{zx+z+1}=1
$$
|
Note that $$\dfrac{x}{xy+x+1} = \dfrac{xz}{xyz+xz + z} = \dfrac{xz}{xz+z+1}$$
Similarly, we get
$$\dfrac{y}{yz+y+1} = \dfrac1{z+1+\dfrac1y} = \dfrac1{z+1+xz}$$
Now add the three terms and finish it off by canceling the numerator and denominator.
|
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|
finding the real values of $x$ such that : $x=\sqrt{2+\sqrt{2-\sqrt{2+x}}}$ How to find the real values of $x$ such that : $$x=\sqrt{2+\sqrt{2-\sqrt{2+x}}}$$
|
Here's an approach leading to a "closed formula".
The two innermost square roots are defined only, if $x\in[-2,2]$, so we can write
$x=2\cos\theta$ for some $\theta\in[0,\pi]$. My solution relies on the trig identities
$$
\sqrt{\frac{1+\cos\alpha}2}=\cos\frac\alpha2
$$
and
$$
\sqrt{\frac{1-\cos\alpha}2}=\sin\frac\alpha2
$$
that hold for all $\alpha\in[0,\pi]$ (outside this range we may get differing signs).
Thus
$$
\sqrt{2+x}=\sqrt{2+2\cos\theta}=2\cos\frac\theta2,
$$
and then
$$
\sqrt{2-\sqrt{2+x}}=\sqrt{2-2\cos\frac\theta2}=2\sin\frac\theta4=2\cos(\frac\pi2-\frac\theta4).
$$
Going on we get
$$
2\cos\theta=x=\sqrt{2+\sqrt{2-\sqrt{2+x}}}=\sqrt{2+2\cos(\frac\pi2-\frac\theta4)}=2\cos(\frac\pi4-\frac\theta8).
$$
In the interval $[0,\pi]$ cosine is injective, so we can conclude that
$$
\theta=\frac\pi4-\frac\theta8\Leftrightarrow\theta=\frac{2\pi}9.
$$
The only real solution is thus
$$
x=2\cos\frac{2\pi}9\approx1.53209.
$$
|
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|
confirm which one is correct? Let $f(z)=-(x^2+y^2)^{1/2}$ and $\Delta=\frac{\partial^2}{\partial x^2}+\frac{\partial^2}{\partial y^2}$. Help to confirm which one is correct for $\Delta f$; this or $(x^2+y^2)^{3/2}-\frac{2}{(x^2+y^2)^{1/2}}$?
thankyou.
EDIT: sorry, I meant above. Just one minus sign more.
|
This is an application of the chain rule.
$$\nabla^2\!f = \frac{\partial^2f}{\partial x^2} + \frac{\partial^2f}{\partial y^2} . $$
If $f(x,y) = \sqrt{x^2+y^2}$ then the first order partial derivatives are:
$$\begin{array}{ccc}
\frac{\partial\! f}{\partial x} &=& \frac{x}{\sqrt{x^2+y^2}} \\
\frac{\partial\! f}{\partial y} &=& \frac{y}{\sqrt{x^2+y^2}}
\end{array}$$
Differentiating a second time, gives:
$$\begin{array}{ccc}
\frac{\partial^2 f}{\partial x^2} &=& \frac{y^2}{(x^2+y^2)^{3/2}} \\
\frac{\partial^2 f}{\partial y^2} &=& \frac{x^2}{(x^2+y^2)^{3/2}}
\end{array}$$
It follows that the Laplacian is given by
$$\begin{array}{ccc}
\nabla^2f &=& \frac{y^2}{(x^2+y^2)^{3/2}} + \frac{x^2}{(x^2+y^2)^{3/2}} \\
&=& \frac{x^2+y^2}{(x^2+y^2)^{3/2}} \\
&\equiv& \frac{1}{\sqrt{x^2+y^2}}
\end{array}$$
|
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|
Finding polynomial given the remainders Question: Find a polynomial $f(x) \in \mathbb{Q} (x)$ of minimal degree that has both the following properties:
When $f(x)$ is divided by $(x-1)^2$, the remainder is $2x$; and when $f(x)$ is divided by $(x-2)^3$, the remainder is $3x$.
Answer provided: $f(x)=(x-2)^3 \cdot (4x-3)+3x$
Work so far: Okay, so I know this problem shouldn't be difficult, but I've been stumped. I know that this is probably a simple application of the division algorithm, where $f(x)=q(x) \cdot d(x)+r(x)$ but I can't seem to get an answer. The polynomial's minimum degree should be $4$, intuitively.
So I have that $f(x)=(x-1)^2 \cdot q_1(x) + 2x$ and $f(x)=(x-2)^3 \cdot q_2(x)+3x$. Then I know that for the second equation, $q_2(x)$ should be of form $(ax+b)$ for some $a,b \in \mathbb{Q}$, also I also applied the remainder theorem to find that $f(1)=4$ and that $f(2)=6$, and so
$f(1)=4=(-1)^3 \cdot (a+b) +3(1) \rightarrow a+b=1$
so it makes sense that the answer is $q_2(x)=4x-3$, but I don't know how to get there after finding $a+b=1$.
|
It might be worth noticing that this can be phrased in terms of congruences
$$
\begin{cases}
f(x) \equiv 2x \pmod{(x-1)^2}\\
f(x) \equiv 3x \pmod{(x-2)^3}.
\end{cases}
$$
There is definitely a solution because of CRT, since $(x-1)^2$ and $(x-2)^3$ are clearly coprime.
The solution is of course the same the others have given, I just would like to illustrate how you get it using the general theory of system of congruences.
Now use the extended Euclidean algorithm on $(x-1)^2$ and $(x-2)^3$ to get
$$
1 = (x-1)^2 \cdot (3x^2 -14 x + 17) + (x-2)^3 \cdot (-3x+2).
$$
So applying the general method for solving a system of congruences, write
$$
3 x - 2 x = x = (x-1)^2 \cdot (3x^3 -14 x^2 + 17 x) + (x-2)^3 \cdot (-3x^2+2x),
$$
so that
$$
3 x + (x-2)^3 \cdot (3x^2-2x) = 2 x + (x-1)^2 \cdot (3x^3 -14 x^2 + 17 x)
=
3 x^5-20 x^4+48x^3-48x^2+19 x
$$
is a solution. This is not minimal, because you can still take the remainder with respect to the lcm of $(x-1)^2$ and $(x-2)^3$ (which is their product in this case), so one gets
$$
4 x^4-27 x^3+66 x^2-65x+24.
$$
|
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|
For which $n$ is $n\sigma(n)\equiv 2 \pmod {\phi(n)}$? How to find all of $n \in \Bbb N$ such that: $$n\sigma(n)\equiv 2 \pmod {\phi(n)}$$
$\sigma(n)$ is summation of all distinct divisors of $n$
For $p$ prime we have: $p(p+1)=p^2+p\equiv 2 \pmod {p-1}$
but how to prove for composite number : $4,6,22$ is only solution.
|
Note that $d \mid \phi(n)$ implies that $n\sigma(n) \equiv 2 \pmod{d}$.
If $p^2 \mid n$ for some odd prime $p$, then $p \mid \phi(n)$. Thus $0 \equiv n\sigma(n) \equiv 2 \pmod{p}$, a contradiction. Thus $p^2 \nmid n$ for any odd prime $p$.
If $8 \mid n$, then $4 \mid \phi(n)$. Thus $0 \equiv n\sigma(n) \equiv 2 \pmod{4}$, a contradiction. Thus $8 \nmid n$.
If $4 \|n$, then write $n=4m$, where $m$ is odd. If $m>1$, then $2 \mid \phi(m)$ so $4 \mid 2\phi(m)=\phi(n)$. Thus $0 \equiv n\sigma(n) \equiv 2 \pmod{4}$, a contradiction. Therefore $m=1$, so $n=4$, which is a solution.
Otherwise $n$ is squarefree.
If $pq \mid n$ for distinct odd primes $p, q$, then write $n=pqm, \gcd(m, pq)=1$. Then $\sigma(n)=\sigma(pq)\sigma(m)=(p+1)(q+1)m$, and $\phi(n)=\phi(n)=\phi(pq)\phi(m)=(p-1)(q-1)\phi(m)$. Thus $4 \mid \sigma(n), \phi(n)$, so $0 \equiv n\sigma(n) \equiv 2 \pmod{4}$, a contradiction.
Therefore we have $n=1, 2, p$ or $2p$, where $p$ is an odd prime. If $n=1, 2$, then $\phi(n)=1$, so we trivially have $n\sigma(n) \equiv 2 \pmod{1}$. If $n=p$, then $\phi(p)=p-1, \sigma(p)=p+1$, so $p\sigma(p)=p(p+1) \equiv 2 \pmod{p-1}$. Therefore $n=1, 2, p$ are solutions.
Suppose that $n=2p$. Then $\phi(n)=p-1$ and $\sigma(n)=1+2+p+2p=3(p+1)$. Then we have $2 \equiv 2p\sigma(p)=6p(p+1) \equiv 12 \pmod{p-1}$, so $(p-1) \mid 10$. Since $p$ is an odd prime, we get $p=3, 11$. This gives $n=6, 22$, which are indeed solutions.
Therefore all solutions are given by $n=1, 4, 6, 22$ and $n=p$, $p$ prime.
|
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"timestamp": "2023-03-29T00:00:00",
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|
Evaluate $\sum_{n=1}^{\infty}\frac{1}{n^3+3 n^2+2 n}$ Summing this series from $0$ to $\infty$, the result is $\frac{1}{4}$. I tried a lot, but I could not get this result. I think it´s wrong.
Can anybody help me?
|
$$\frac{1}{n(n+1)(n+2)}=\frac{1}{n+2} \frac{1}{n(n+1)}=\frac{1}{n+2}\left( \frac{1}{n}- \frac{1}{n+1} \right)
=\frac{1}{n(n+2)}-\frac{1}{(n+1)(n+2)}=\frac{1}{2}\frac{1}{n}-\frac{1}{n+1}+\frac{1}{n+2}-\frac{1}{n+1}$$
|
{
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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|
Positive integer $2\times 2$ matrix with two positive integer square root matrices So suppose I have this matrix:
$$M=\begin{pmatrix}a & b \\ c & d \end{pmatrix}$$
I am sure that every entry is a positive integer. I am trying to figure out if it has 2 square roots in which every entry is also a positive integer. I took a look at http://en.wikipedia.org/wiki/Square_root_of_a_2_by_2_matrix . It basically says that, setting $s = \pm \sqrt{\det M}$ and $t = \pm \sqrt{a + d + 2s}$ you get that the square roots of $M$ are all of the form $$R = \frac{1}{t} \begin{pmatrix}a+s & b \\ c & d+s\end{pmatrix}$$
So, in order for every entry of $R$ to be a positive integer, am I correct in thinking $\det{M}$ must be a perfect square, $a + d \gt 2s$, $a>s$, $d>s$, $t>0$, t is an integer, and all enteries must be divisible by t? Or are there wholes in these conditions?
|
The Wikipedia article is not very clearly written. Let us derive the result by ourselves. Suppose
$$
M=\begin{pmatrix}A&B\\C&D\end{pmatrix}
=R^2={\underbrace{\begin{pmatrix}a&b\\c&d\end{pmatrix}}_{R}}^2
=\begin{pmatrix}a^2+bc&b(a+d)\\c(a+d)&bc+d^2\end{pmatrix}.
$$
Then $A+D+2(\det R) = (a+d)^2$ and
$$
\begin{pmatrix}A+\det R&B\\C&D+\det R\end{pmatrix}
=(a+d)\begin{pmatrix}a&b\\c&d\end{pmatrix}.
$$
Therefore, if $A+D+2\delta\neq0$ for some $\delta\in\{-\sqrt{\det M},\sqrt{\det M}\}$, then
$$
\begin{pmatrix}a&b\\c&d\end{pmatrix}
=\frac{1}{\pm\sqrt{A+D+2\delta}}
\begin{pmatrix}A+\delta&B\\C&D+\delta\end{pmatrix}\tag{1}
$$
is a (perhaps complex) square root of $M$. Also, if $A+D+2\delta\neq0$ for both $\delta=-\sqrt{\det M}$ and $\delta=\sqrt{\det M}$, then all square roots of $M$ are given by $(1)$.
When $A+D+2\delta=0$ for some $\delta\in\{-\sqrt{\det M},\sqrt{\det M}\}$, $M$ can have other square roots that do not take the form of $(1)$. For instance, as pointed out by Samuel, the identity matrix has infinitely many square roots: apart from $\pm I$, there are also square roots of the form
$$
\begin{pmatrix}a&b\\c&-a\end{pmatrix}
$$
where $a\in\mathbb{C}$ is arbitrary and $bc=1-a^2$. (Putting $a=0$ and $b=c=1$ gives the well known one; my favourite one is obtained by taking $a=b=1$ and $c=0$.)
Now, go back to your question. If $M=R^2$ for some positive integer matrix $R$, then $\det M=(\det R)^2$ must be a perfect square.
Also, we have
$$A+D+2\sqrt{\det M}>A+D-2\sqrt{\det M}> A+D-2\sqrt{AD}=(\sqrt{A}-\sqrt{D})^2 \ge0.$$
Therefore, every positive square root matrix of $M$, if any, takes the form of $(1)$.
It follows that for each $\delta\in\{-\sqrt{\det M},\sqrt{\det M}\}$ that produces an integer square root matrix from $(1)$, $\sqrt{A+D+2\delta}$ must be an integer that divides $A+\delta,\,B,\,C,\,D+\delta$. Note that it may not divide $A$ or $D$ (so your claim that it divides all entries of $M$ is false). For a counterexample, consider
$$
M=\begin{pmatrix}2&5\\5&17\end{pmatrix}
=\begin{pmatrix}1&1\\1&4\end{pmatrix}^2,\ \sqrt{\det M}=3.
$$
We have $\sqrt{A+D+2\sqrt{\det M}}=5$ divides $A+3=B=C=5$ and $D+3=20$, but not $A=2$ or $D=17$. This counterexample also shows that the condition $A>\sqrt{\det M}$ does not necessarily hold. By permuting the rows and columns of $M$, it follows that the condition $D>\sqrt{\det M}$ also is not necessarily true.
|
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"timestamp": "2023-03-29T00:00:00",
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|
Interesting log sine integrals $\int_0^{\pi/3} \log^2 \left(2\sin \frac{x}{2} \right)dx= \frac{7\pi^3}{108}$ Show that
$$\begin{aligned} \int_0^{\pi/3} \log^2 \left(2\sin \frac{x}{2}
\right)dx &= \frac{7\pi^3}{108} \\ \int_0^{\pi/3}x\log^2
\left(2\sin\frac{x}{2} \right)dx &= \frac{17\pi^4}{6480}\end{aligned}$$
*
*I can solve $\displaystyle \int_0^\pi \log^2 \left(2\sin \frac{x}{2}
\right)dx $ but I don't know what to do if the limits are from $0$ to $\pi/3$.
*I have no idea what to do if the integrand contains an $x$.
*I feel that the Polylogarithm function will be involved however I don't know how it can be implemented here.
It would be really great if someone could take the initiative to prove these.
|
The best way I see to do the simpler integral (the one without the $x$ in front) is to substitute $u=2 \sin{(x/2)}$ and expand the resulting integrand in a series. To wit, upon doing the substitution, we get
$$\int_0^{\pi/3} dx \: \log^2{[2 \sin{(x/2)}]} = \int_0^1 du \frac{\log^2{u}}{\sqrt{1-u^2/4}}$$
Note that
$$\frac{1}{\sqrt{1-u^2/4}} = \sum_{k=0}^{\infty} \frac{1}{2^{4 k}} \binom{2 k}{k} u^{2 k}$$
Then the integral becomes
$$\sum_{k=0}^{\infty} \frac{1}{2^{4 k}} \binom{2 k}{k} \int_0^1 du\: u^{2 k} \log^2{u}$$
The integral on the right may be done through integration by parts; the process is very interesting, but I leave it to the reader to get the nifty result that
$$\int_0^1 du\: u^{2 k} \log^2{u}=\frac{2}{(2 k+1)^3}$$
Therefore the evaluation of the integral becomes an evaluation of the following sum:
$$2 \sum_{k=0}^{\infty} \frac{1}{2^{4 k}} \binom{2 k}{k}\frac{1}{(2 k+1)^3}$$
To evaluate this sum, define
$$f(z) = 2 \sum_{k=0}^{\infty} \frac{1}{2^{4 k}} \binom{2 k}{k}\frac{z^{2 k+1}}{(2 k+1)^3}$$
The desired integral is $f(1)$. To derive an equation for $f$, recall the binomial series above:
$$\frac{d}{dz}\left [z \frac{d}{dz} \left [ z \frac{d}{dz} f(z)\right ] \right ] = 2 \sum_{k=0}^{\infty} \frac{1}{2^{4 k}} \binom{2 k}{k} z^{2 k} = \frac{2}{\sqrt{1-z^2/4}}$$
The resulting integrations are elementary except for the last one, which results in a nasty generalized hypergeometric function, something like
$$2 _4F_3\left(\frac{1}{2},\frac{1}{2},\frac{1}{2},\frac{1}{2},\frac{3}{2},\frac{3}{2},\frac{3}{2},\frac{1}{4}\right)$$
which in fact does numerically check out.
|
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|
bernoulli equations1 Question:
Solve the following bernoulli equations.
$$xy'+y=y^{-2}$$
I solving:
$$w=y^1+2$$ $$,$$
$$\frac{w'}{1+2}+1w=1$$
$$\frac{-w'}{3}=-1$$
$$\frac{1}{3}(\frac{-1}{3}w'-w)=-1$$
$$d(\frac{-1}{3}w)=-1$$
$$\frac{-1}{3}w=-1+C$$
$$= \left\{
\begin{array}{l l}
w=9 & \quad \\\
w=y^3 & \quad \ \end{array} \right.$$
I think I got the wrong solved.
Please help me.
|
First observe that if $x = 0$ then $y = 1$ is the solution.
Next, assume $x \neq 0$. The equation can be written as
$$y^2y' + \frac{y^3}{x} = \frac{1}{x}.$$
Put $w := y^3$ then $w' = 3y^2y'$ and the equation becomes linear:
$$\frac{w'}{3} + \frac{w}{x} = \frac{1}{x}.$$
This can be solved by separation of variables. Therefore the answer is
$$
y = \begin{cases}
1 & x = 0 \\
\sqrt[3]{1 + \frac{c}{x^3}} & x \neq 0
\end{cases}
$$
for some positive constant $c$.
|
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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|
What is the remainder when $25^{889}$ is divided by 99? What is the remainder when $25^{889}$ is divided by 99 ?
$25^3$ divided by $99$ gives $26$ as a remainder.
$25*(25^3)$ divided by $99$ gives (remainder when $25*26$ is divided by $99$) as a remainder.
i.e. $25*(25^3)$ divided by $99$ gives $56$ as a remainder.
$(25^3)*(25^3)$ divided by $99$ gives (remainder when $26*26$ is divided by $99$) as a remainder.
i.e. $(25^3)*(25^3)$ divided by $99$ gives $82$ as a remainder.
|
$\rm mod\ 9\!:\ 25^3\equiv (-2)^3\equiv -8\equiv 1\:\Rightarrow\:n = 25^{889}\equiv 25^{889\ mod \ 3}\equiv 25\equiv \color{#0A0}{-2}$
$\rm mod\ 11\!:\ 25^5\equiv 5^{10}\equiv 1\:\Rightarrow\:n = 25^{889}\equiv 25^{889\ mod\ 5}\equiv 25^4 \equiv 3^4\equiv (-2)^2\equiv \color{#C00}4$
$\rm mod\ 9\!:\ \color{#0A0}{{-}2} \equiv n\equiv \color{#C00}4\!+\!11k\equiv 4\!+\!2k\:\Rightarrow\:k\equiv -3\equiv\color{blue}{6}\:\Rightarrow\:n = 4\!+\!11(\color{blue}{6}\!+9\,j) = 70+99\,j$
|
{
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|
Integral Solution Technique Could somebody explain the processes in the following integral solution?
$$\int_{0}^{\infty}f(x)dx, $$
$\text{ where } f(x) = \frac{x^{3}}{\theta^{2}}e^{-x^{2}/(2\theta^{2})}$
$$\int_{0}^{\infty}\frac{x^{3}}{\theta^{2}}e^{-x^{2}/(2\theta^{2})}dx \dots (1)$$
$$= \int_{0}^{\infty}x^{2}e^{-x^{2}/(2\theta^{2})}d\frac{x^{2}}{2\theta^{2}} \dots (2)$$
$$= 2\theta^{2}\int_{0}^{\infty}ye^{-y}dy \dots (3)$$
$$= 2\theta^{2}\Gamma(2) \dots (4)$$
$$= 2\theta^{2} \dots (5)$$
More specifically, whilst it seems that the substitution $y = x^{2}/2\theta^{2}$ is made in line 2 (and made further explicit in line 3), why does the $x^{3}$ term become $x^{2}$ and why does the $\theta^{2}$ term in the denominator vanish?
Finally, how does the integral in line 3 evaluate to the gamma function with an argument of 2?
|
Doesn't answer the question, but I'd do:
$$
\int_0^\infty x^3 e^{- x^2 / 2 \theta^2} dx
$$
As Caran-d'Ache says, try to simplify the exponent. Try $y = x^2 / 2 \theta^2$,
that is $x = \theta \sqrt{2 y}$. When $x = 0$, $y = 0$; $x = \infty$ gives $y = \infty$. So :
$$
\begin{align*}
d x
&= \theta \sqrt{2} \frac{d y}{\sqrt{y}} \\
x^3 e^{- x^2 / 2 \theta^2}
&= \left( \theta \sqrt{2 y} \right)^3 e^{-y} \\
\int_0^\infty x^3 e^{- x^2 / 2 \theta^2} dx
&= \int_0^\infty \left( \theta \sqrt{2 y} \right)^3 e^{-y}
\cdot \theta \sqrt{2} \frac{d y}{\sqrt{y}} \\
&= 2 \theta \int_0^\infty y e^{-y} d y
\end{align*}
$$
For the remaining integral, use integration by parts, $u = y$, $d v = e^{-y} dy$, so $du = dy$, $v = - e^{-y}$:
$$
\int_0^\infty y e^{-y} dy
= \left. - y e^{-y} \right|_0^\infty + \int_0^\infty e^{-y} d y = 1
$$
And the final result is:
$$
\int_0^\infty x^3 e^{- x^2 / 2 \theta^2} dx
= 2 \theta
$$
|
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|
Prove $\frac{1}{bc + cd + da -1} + \frac{1}{ab + cd + da -1} + \frac{1}{ab + bc + da -1} + \frac{1}{ab + bc + cd -1} \le 2$ If $a,b,c,d > 0$ and $abcd = 1$, prove that the following inequality holds:
$$\frac{1}{bc + cd + da -1} + \frac{1}{ab + cd + da -1} + \frac{1}{ab + bc + da -1} + \frac{1}{ab + bc + cd -1} \le 2$$
The problem I had is because it says that I need to prove that it's less than or equal to 2. With this I can't use any better known inequality (Cauchy-Schwarz, Holder, Minkowski...), because using that I'll prove that $\frac{1}{bc + cd + da -1} + \frac{1}{ab + cd + da -1} + \frac{1}{ab + bc + da -1} + \frac{1}{ab + bc + cd -1}$ is bigger than something, and even if I prove that that new inequality is true, that doesn't mean that the original inequallity holds.
Other my idea was to split the 2 in $4 \times \frac{1}{2}$ and i'll get:
$$\frac{1}{bc + cd + da -1} + \frac{1}{ab + cd + da -1} + \frac{1}{ab + bc + da -1} + \frac{1}{ab + bc + cd -1} \le \frac{1}{2} + \frac{1}{2} + \frac{1}{2} + \frac{1}{2}$$
Now I need to prove that:
$$\frac{1}{bc + cd + da -1} \le \frac{1}{2}$$
$$\frac{1}{ab + cd + da -1} \le \frac{1}{2}$$
$$\frac{1}{ab + bc + da -1} \le \frac{1}{2}$$
$$\frac{1}{ab + bc + cd -1} \le \frac{1}{2}$$
or:
$$bc + cd + da -1 \ge 2$$
$$ab + cd + da -1 \ge 2$$
$$ab + bc + da -1 \ge 2$$
$$ab + bc + cd -1 \ge 2$$
But if all inequalites hold then $a=b=c=d=1$ and that's as far I can go.
|
By inequality of arithmetic and geometric means, $bc+da\ge 2\sqrt{abcd}=2$, so
$$\frac{1}{bc+cd+da-1}\le\frac{1}{1+cd}.\tag{1}$$
Similarly,
$$\frac{1}{ab+bc+da-1}\le\frac{1}{1+ab}.\tag{2}$$
$(1)+(2)$ gives
$$\frac{1}{bc+cd+da-1}+\frac{1}{ab+bc+da-1}\le\frac{1}{1+cd}+\frac{1}{1+ab}=\frac{2+ab+cd}{1+ab+cd+abcd}=1.\tag{3}$$
Similarly,
$$\frac{1}{ab+cd+da-1}+\frac{1}{ab+bc+cd-1}\le 1.\tag{4}$$
The conclusion follows from $(3)+(4)$.
|
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|
Find minimum of $\frac{a+3c}{a+2b+c}+\frac{7a+6b+3c}{a+b+2c}+\frac{c-a}{2a+b+c}$ for non-negative reals
Let $a, b, c\ge 0$, not all zero. Find the minimum of
$$N = \frac{a+3c}{a+2b+c}+\frac{7a+6b+3c}{a+b+2c}+\frac{c-a}{2a+b+c}. $$
|
Hint: Let $x,y,z$ be the denominators of the three fractions, and express $a,b,c$ in terms of $x,y,z$. Then write $N$ in terms of $x,y,z$. After some simplification we will get terms like $\frac{x}{z},\frac{x}{y},\frac{y}{x},\frac{y}{z},\cdots$Then consider inequality $\frac{x}{y}+\frac{y}{x}\geqslant2$ if $xy$ is positive and analogously $\frac{x}{z}+\frac{z}{x}\geqslant2$ and so on...
|
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"timestamp": "2023-03-29T00:00:00",
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|
is this relation true? i found this relation in old notes , it's mentioned with no proof , and i want to know if it's true or false
the relation says ,
$$ \sum_{k=1}^n n^6 = \frac{n^7}{7} + \frac{n^6}{2} + \frac{5}{2} n^5 - \frac{15}{2}n^4 - \frac{31}{6}n^3 + \frac{11}{21}n - 8 $$
is this relation true or not ?
and how can we prove it ?
i think that using induction " if it is true " will be so so hard !
so , any other ideas ?
|
Use $$(r+1)^{n+1}-r^{n+1}=\sum_{0\le s\le n}\binom {n+1}s r^s$$
to find $\sum_{1\le t \le n}r^s$ for $s=1,2,3,5,6$
For $s=0, \sum_{1\le r \le n}r^0=\sum_{1\le r \le n}1=n$
For $n=2, (r+1)^2-r^2=2r+1$
Put $r=1,2,\cdots,n-1,n$ and add to get $$(n+1)^2-1^2=2\sum_{1\le s\le n}r+\sum_{1\le s\le n}1$$
$$\implies 2\sum_{0\le s\le n}r=(n+1)^2-(n+1)\implies \sum_{0\le s\le n}r=\frac{n(n+1)}2$$
For $n=2+1=3,(r+1)^3-r^3=3r^2+3r+1$
Put $r=1,2,\cdots,n-1,n$ and add to get $$(n+1)^3-1^3$$
$$=3\sum_{1\le s\le n}r^2+3\sum_{1\le s\le n}r+\sum_{1\le s\le n}1$$
$$\implies 3\sum_{0\le s\le n}r^2=(n+1)^3-1-3\sum_{1\le s\le n}r-\sum_{1\le s\le n}1$$
$$=(n+1)^3-1-3\frac{n(n+1)}2-n=\frac{n(n+1)(2n+1)}2$$
and so on
|
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"timestamp": "2023-03-29T00:00:00",
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|
Ordered triples solution to system of equations How many ordered triples $(x,y,z)$ of integer solutions are there to the following system of equations?
$$
\begin{align}
x^2+y^2+z^2&=194 \\
x^2z^2+y^2z^2&=4225
\end{align}
$$
|
We have $x^2 + y^2 = \dfrac{4225}{z^2}$. Plugging this into the first equation, we get that
$$z^2 + \dfrac{4225}{z^2} = 194 \implies z^2 + \left(\dfrac{65}z\right)^2 = 194$$ Once you have this you should be able to proceed and get the answer.
|
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|
Variable change in integral I need to calculate the integral $$\int_0^{2\pi}\sqrt{1-\cos\,x}dx$$ but when I try the substitution with $u = 1-\cos\,x$ both integration limits goes to 0.
The same happens using the t-substitution.
|
Hint: Note that
\begin{align}
1-\cos(x)
=
&
1-\cos\left(2\cdot \frac{x}{2}\right)
\\
=
&
1+\sin^2\left(\frac{x}{2}\right)-\cos^2\left(\frac{x}{2}\right)
\\
=
&
\sin^2\left(\frac{x}{2}\right)+\sin^2\left(\frac{x}{2}\right)
\\
=
&
2\sin^2\left(\frac{x}{2}\right)
\end{align}
implies
$$
\int_{0}^{2\pi}\sqrt[2\,]{1+\cos(x)}\,\mathop{d}x=
2\cdot\int_{0}^{2\pi}\left|\sin\left(\frac{x}{2}\right)\right|\,\mathop{d}x
$$
|
{
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|
Choosing teams with minimum number of boys and girls. I need to find different number of teams I can make with 6 people that needs to have at least 2 girls and 2 boys. There are 8 girls and 12 boys. So the way I think is I need to find total number of different teams that I can make first and subtract all boys team, all girls team, one girl 5 boys team, 5 girls one boy team.
$$\binom{20}{6} - \binom{12}{6} - \binom{8}{6} - \left(\binom{8}{5} \binom{12}{1}\right) - \left(\binom{8}{1} \binom{12}{5}\right)$$
$$\binom{20}{6} - \binom{12}{6} - \binom{8}{6} - \left( \binom{8}{5} \binom{12}{1}\right) - \left( \binom{8}{1} \binom{12}{5}\right) $$
|
if boys divided into such as Tom,Jack,Jerry....then
$$X={20\choose 6} - {8\choose6}-{12\choose6}-{8\choose5}{12\choose1}-{12\choose5}{8\choose1}=30800 $$
else it's a naive problem
just 3
|
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|
How find this $a_{1}+a_{2}+\cdots+a_{500}=b_{1}+b_{2}+\cdots+b_{500}$? Let $A=\{1^2,2^2,3^2,\cdots,1000^2\}$. How to prove :
There exist $A_{1}=\{a_{1},a_{2},a_{3},\cdots,a_{500}\}\subset A$, $A_{2}=\{b_{1},b_{2},\cdots,b_{500}\}\subset A$, such that $A_{1}\bigcup A_{2}=A,A_{1}\bigcap A_{2}=\varnothing$, and $a_{1}+a_{2}+\cdots+a_{500}=b_{1}+b_{2}+\cdots+b_{500}$ ?
|
Note that
$$(k+1)^2+(k+2)^2 = k^2+(k+3)^2\quad -4.$$
Therefore,
$$\tag1(k+1)^2+(k+2)^2+(k+4)^2+(k+7)^2 = k^2+(k+3)^2+(k+5)^2+(k+6)^2.$$
Now partition $A$ into $125$ groups of $8$ consecutive numbers each and distribute each such group among $A_1$ and $A_2$ accoring to $(1)$. This works because $1000$ is a multiple of $8$.
|
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Strategies to prove inequalities with interval notation How to prove a inequalities with interval notation, for example:
Find minimum of $a^3+b^3+c^3$ with $a,b,c \in [-1;\infty), a^2+b^2+c^2=9$
|
Clearly $a, b, c$ cannot be all negative.
If $a, b, c$ are all non-negative, then by Power mean inequality $(\frac{a^3+b^3+c^3}{3})^2 \geq (\frac{a^2+b^2+c^2}{3})^3=27$, so $a^3+b^3+c^3 \geq 9\sqrt{3}$ with equality when $a=b=c=\sqrt{3}$.
If exactly 1 of $a, b, c$ is negative, then WLOG assume that $c$ is negative, so $a, b$ are non-negative. Then $c \geq -1$, so $c^2 \leq 1$ so $a^2+b^2 \geq 8$. By Power mean inequality $(\frac{a^3+b^3}{2})^2 \geq (\frac{a^2+b^2}{2})^3 \geq (\frac{8}{2})^3=64$, so $a^3+b^3+c^3 \geq 2\sqrt{64}+c^3 \geq 16-1=15$, with equality when $a=b=2, c=-1$. (and permutations)
If exactly 2 of $a, b, c$ are negative, then WLOG assume that $b, c$ are negative and $a$ is non-negative. We have $b, c \geq -1$, so $b^2, c^2 \leq 1$, so $a^2 \geq 7$, so $a^3+b^3+c^3 \geq (\sqrt{7})^3-1-1=7\sqrt{7}-2$, with equality when $a=\sqrt{7}, b=c=-1$. (and permutations)
Thus $a^3+b^3+c^3 \geq \min(9\sqrt{3}, 15, 7\sqrt{7}-2)=15$, with equality when $a=b=2, c=-1$ and permutations.
|
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How to factorise $x^6 + 1$? How do I factorise this? I already said that
$$x^6 + 1 = 0 \implies (x^3)^2 + 1 = 0$$
so we then get
$$(x^3)^2 = - 1$$
and then there are no real roots. Similar thing happens if I try it as $(x^2)^3$.
How would I then go about factorising this? I know its possible as
$$(x^6 + 1) = (x^2 + 1)(x^4 - x^2 + 1).$$
|
If $x^6+1=0$, then $x^3$ must be either $i$ or $-i$ (because $(x^3)^2=x^6=-1$). Hence, one must find the cube roots of $\pm i$. The numbers $i$ and $-i$ are easily seen to be cube roots of each other.
Now, $(a+bi)^3=a^3+3a^2bi+3a(bi)^2+(bi)^3=a^3+3a^2bi+3ab^2i^2+b^3i^3=a^3+3a^2bi-3ab^2-b^3i=(a^3-3ab^2)+(3a^2b-b^3)i$. If this were to equal $i$, then the real part $a^3-3ab^2$ would have to be zero. Hence, either $a=0$ or $a^2-3b^2=0$. If $a=0$, then one gets the solution $-i$. For the other two solutions, substitute $a=\pm b\sqrt{3}$ into the equation $3a^2b-b^3=1$ to get $3(3b^2)b-b^3=1$. Simplifying the left side, one gets $8b^3=1$. The only real solution for $8b^3=1$ is $b=\frac{1}{2}$.
Hence, the other two cube roots of $i$ (besides $-i$) are $\frac{1}{2}\sqrt{3}+\frac{1}{2}i$ and $-\frac{1}{2}\sqrt{3}+\frac{1}{2}i$. Conjugating those two cube roots of $i$, one gets $\frac{1}{2}\sqrt{3}-\frac{1}{2}i$ and $-\frac{1}{2}\sqrt{3}-\frac{1}{2}i$ as the two cube roots of $-i$ besides $i$.
The fully factorized form of $x^6+1$ (over the complex numbers) is therefore $(x-i)(x+i)(x-\frac{1}{2}\sqrt{3}-\frac{1}{2}i)(x+\frac{1}{2}\sqrt{3}-\frac{1}{2}i)(x-\frac{1}{2}\sqrt{3}+\frac{1}{2}i)(x+\frac{1}{2}\sqrt{3}+\frac{1}{2}i)$.
|
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How to find the eigenvalues and Jordan canonical form of this matrix Question:
let $a_{i,j}\in R,A=(a_{i,j})_{n\times n} $,and
$a_{i,j}=\begin{cases}
1&i+j\in\{n,n+1\}\\
0&i+j\notin\{n,n+1\}
\end{cases}$
that's meaning:
$$A=\begin{bmatrix}
0&0&0&\cdots&0&1&1\\
0&0&0&\cdots&1&1&0\\
0&0&\cdots&1&1&0&0\\
\vdots&\vdots&\vdots&\vdots&\vdots&\vdots&\vdots\\
0&1&1&\cdots&0&0&0\\
1&1&0&\cdots&0&0&0\\
1&0&0&\cdots&0&0&0
\end{bmatrix}_{n\times n}$$
Problem (1): Find the Jordan canonical form of $A$.
I know this matrix Jordan is
$$diag(\lambda_{1},\lambda_{2},\cdots,\lambda_{n})$$
where $\lambda_{i}$ is eigenvalue
But this problem key find the eigenvalue is hard,
Thank you.maybe this problem is not easy,But I hope see someone can solve it.Thank you very much!
|
Although Ewan Delanoy was more quick than I, :-) I’ll continue user1551‘s answer.
It seems the following. Denote the determinant of the $n\times n$ matrix $xI-A$ as $q_n(x)$. Then we can calculate the first polynomials:
$q_1=x-1$
$q_2=x^2-x-1$
$q_3=x^3-x^2-2x+1$
$q_4=x^4-x^3-3x^2+2x+1$
$q_5=x^5-x^4-4x^3+3x^2+3x-1$
$q_6=x^6-x^5-5x^4+4x^3+6x^2-3x-1$
$q_7=x^7-x^6-6x^5+5x^4+10x^3-6x^2-4x+1$
Investigating decomposition of the determinant $|xI-A|$ by the first row, we obtain the following recurrence:
$q_n=(x^2-1)q_{n-2}-xq_{n-3}$ for each $n\ge 4$.
Adding for the completeness the polynomial $q_0\equiv 1$, we can easily prove the recurrence
$q_n=xq_{n-1}-q_{n-2}$ for each $n\ge 2$.
The Google search by these polynomials suggests that roots of the polynomial $q_n$ are $x_{k}=2\cos\frac{2k+1}{2n+1}\pi $, where $k=0,\dots,n-1$ (but not always for $k=n$).
Geometric interpretation (and Ewan Delanoy’s solution :-) ) suggest to try to take as $q_n$ a sum $s_n(x)=(-1)^n\sum_{j=-n}^n \cos jt,$ where $t=\arccos (-x/2)$.
We can check that $s_n=q_n$ for all $n\le 3.$ Let $n\ge 2$. Then
$$s_n(x)+s_{n-2}(x)= (-1)^n\sum_{j=-n}^n \cos jt+(-1)^{n-2}\sum_{j=-n+2}^{n-2} \cos jt=
(-1)^n\left(\sum_{j=-1}^1 \cos jt + 2\sum_{j=2}^n \left( \cos jt+\cos (j-2)t \right)-\sum_{j=0}^0 \cos jt \right)= (-1)^n\left(-x+2\sum_{j=2}^n 2\cos (j-1)t\cos t \right)=
(-1)^n\left(-x-2x\sum_{j=1}^{n-1} \cos jt \right)=s_{n-1}x.$$
That is $s_n(x)+s_{n-2}(x)=xs_{n-1}x$. Therefore $s_n=q_n$ for each $n$. At last, we show that $s=s_n(2\cos\frac{2k+1}{2n+1}\pi)=0$ for $k=0,\dots,n-1$. But in this case $$s=
(-1)^{n-1}\sum_{j=-n}^n \cos j\left(\frac{2k+1}{2n+1}+1\right)\pi$$ is a multiple of the sum of the abscissas of vertices of a right $(2n+1)/d$-gon $P$ centered at the zero of the complex plane, where $d$ is a greatest common divisor of the numbers $k+n+1$ and $2n+1$. The rotational symmetry of $P$ implies that $se^{2d\pi i/(2n+1)}=s$. Therefore $s=0$.
PS. The Google search by “x^5-x^4-4x^3+3x^2+3x-1” yielded me only two links. One of then to *.jp site, and the other to this MSE question. The recurrent relations and the roots of the polynomials $q_n$ are similar to these of Chebyshev (and cyclotomic) polynomials, so I added relative tags to the question. But the general trigonometric formula for polynomials $q_n$ is much more complex than the trigonometric formula for Chebyshev polynomials.
|
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Inequality: $\frac{a^2+b^2}{(a+b)^4}+\frac{b^2+c^2}{(b+c)^4}+\frac{c^2+d^2}{(c+d)^4}+\frac{d^2+a^2}{(d+a)^4} \leq \frac{1}{abcd}$ If $ab+bc+cd+da\leq 8$ and $a,b,c,d \in \mathbb{R}_{+},$how can I prove the following inequality :
$$\frac{a^2+b^2}{(a+b)^4}+\frac{b^2+c^2}{(b+c)^4}+\frac{c^2+d^2}{(c+d)^4}+\frac{d^2+a^2}{(d+a)^4} \leq \frac{1}{abcd}$$
I use : $$\frac{a+b}{2}\geq \frac{2ab}{a+b}$$ or $$\frac{1}{(a+b)^2} \leq \frac{1}{4ab}$$
$$\frac{1}{(a+b)^4} \leq \frac{1}{16a^2b^2}$$
$$\frac{a^2+b^2}{(a+b)^4}\leq \frac{a^2+b^2}{16a^2b^2}$$
but from here I have no idea.
thanks :)
|
that is not a hard question !
$\frac{a^2+b^2}{(a+b)^4}\leq \frac{a^2+b^2}{16a^2b^2}$$\Longrightarrow$
$\frac{a^2+b^2}{(a+b)^4}+\frac{b^2+c^2}{(b+c)^4}+\frac{c^2+d^2}{(c+d)^4}+\frac{d^2+a^2}{(d+a)^4} $$=\frac{1}{8}$$(\frac{1}{ab}+\frac{1}{bc}+\frac{1}{cd}+\frac{1}{ad})=\frac{1}{8}\cdot{\frac{ab+bc+cd+ad}{abcd}}$
$\leq \frac{1}{abcd}$
|
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Find all solutions: $x^2 + 2y^2 = z^2$ I'm use to finding the solutions of linear Diophantine equations, but what are you suppose to do when you have quadratic terms? For example consider the following problem:
Find all solutions in positive integers to the following Diophantine equation
$x^2 + 2y^2 = z^2$
I'd usually start by finding the gcd and use some other tricks, but I'm not sure how to approach this type of problem
|
We can approach this in the same manner as Pythagorean Triples.
Let's only look for primitive solutions, $\gcd(x,y,z)=1$. Since
$$
z^2-x^2=2y^2
$$
$z$ and $x$ must have the same parity. That means both $z-x$ and $z+x$ are even so $y$ must be even. Therefore, for the triple to be primitive, $x$ and $z$ must be odd. Let $y=2ab$ where $(a,b)=1$ and $b$ is odd. Then, since one of $z+x$ or $z-x$ must be $2\bmod4$ and the other must be $0\bmod4$,
$$
\overbrace{2y^2}^{8a^2b^2}=\overbrace{(z+x)}^{4a^2}\overbrace{(z-x)}^{2b^2}\qquad\text{smaller factor is }2\bmod4
$$
or
$$
\overbrace{2y^2}^{8a^2b^2}=\overbrace{(z+x)}^{2b^2}\overbrace{(z-x)}^{4a^2}\qquad\text{larger factor is }2\bmod4
$$
Solving for $x$ and $z$ gives $x=2a^2-b^2$ if the smaller factor is $2\bmod4$, or $x=b^2-2a^2$ if the larger factor is $2\bmod4$. That is,
$$
x=\left|\,2a^2-b^2\right|,y=2ab,z=2a^2+b^2
$$
where $(a,b)=1$ and $b$ is odd.
For example,
$$
\begin{array}{c|cc}
a\backslash b\!\!&1&3&5&7\\\hline
1&(1,2,3)&(7,6,11)&(23,10,27)&(47,14,51)\\
2&(7,4,9)&(1,12,17)&(17,20,33)&(41,28,57)\\
3&(17,6,19)&\text{n/a}&(7,30,43)&(31,42,67)\\
4&(31,8,33)&(23,24,41)&(7,40,57)&(17,56,81)\\
5&(49,10,51)&(41,30,59)&\text{n/a}&(1,70,99)\\
6&(71,12,73)&\text{n/a}&(47,60,97)&(23,84,121)\\
7&(97,14,99)&(89,42,107)&(73,70,123)&\text{n/a}
\end{array}
$$
|
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|
Proving existence of a square-free sequence I found this problem and a solution sketch in a MathOverflow answer, and I thought it was nice enough to deserve more attention and a properly written solution.
Problem:
Prove that for each natural number $n$, there is some natural number $r$ for which the $n$ integers $r+1^2,r+2^2,…r+n^2$ are all square-free.
|
Let $p_1, p_2, \cdots , p_k$ be all the prime numbers less than $n^2$ (for some $k$). For any $K > k$, let $p_{k+1}, p_{k+2}, \cdots , p_K$ be the next prime numbers after that (larger than $n^2$).
For $1 \le i \le k$, there is at least one value of $r$ $\pmod {p_i^2}$ that satisfies $r + 1^2, r+2^2, \cdots , r+n^2 \not \equiv 0$ (i.e. there is at least one viable possible remainder for $r$.) To see this, note that the squares modulo $p_i^2$ do not include $p_i$ itself, thus setting $r \equiv -p_i$ is sufficient.
For $k+1 \le i \le K$, the numbers $1^2, 2^2, \cdots , n^2$ are all distinct $\pmod {p_i}$, because $p_i > p_{k+1} > n^2$. Therefore, there are exactly $(p_i^2 - n)$ viable possibilities for $r \pmod {p_i}$.
Now let $N = p_1^2p_2^2p_3^2\cdots p_K^2$.
By the Chinese remainder theorem, there are at least
$$(1)(1) \cdots (1)(p_{k+1}^2 - n)(p_{k+2}^2 - n)\cdots (p_{K}^2 - n)$$
viable options for $r$ modulo $N$ (by which we mean values of $r$ modulo $N$ that cause $r + 1^2, r + 2^2, \cdots r+n^2$ to be free from squares of all the primes $p_1, \cdots ,p_K$).
For any $K$, we define
$$
x_K := \frac{\text{Number of viable options for } r \text{ mod } N}{N}
$$
which evaluates to
$$
\frac{(p_{k+1}^2 - n)(p_{k+2}^2 - n)\cdots (p_{K}^2 - n)}{p_1^2p_2^2p_3^2\cdots p_K^2}
=
\frac{1}{p_1}\frac{1}{p_2} \cdots \frac{1}{p_k} \left(1 - \frac{n}{p_{k+1}^2}\right)\cdots \left(1 - \frac{n}{p_K^2}\right)
$$
The infinite product $\lim_{K \to \infty} \left(x_K \right)$ converges to a positive number, and the viable options for $r$ modulo $N$ for the first $K+1$ primes are a subset of the viable options for $r$ modulo $N$ for the first $K$ primes for any $K$.
Therefore, there are infinitely many values of $r$ which work for every $K$.
In particular, there is at least one positive integer $r$ such that $r + 1^2, r+2^2, \cdots r+n^2$ are squarefree.
|
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Strings and Substrings So here is one of the last homeworks we are doing in my Discrete math class. It seems like it should be simple but I am really stuck. Any help would be greatly appreciated.
*
*Find the ordinary generating series with respect to length
for the strings in $\{0,1,2\}^*$ having no "$22$" substring.
*Find a recurrence satisfied by the coefficients of the generating
series.
*Find an explicit formula for the number of such strings of length n,
where n is a non-negative integer."
|
You can uniquely generate the set of strings using $(0|1|20|21)^∗(2|ϵ)$, hence your generating function is $\frac{1}{1 - (x + x + x^2 + x^2)} \cdot (1 + x) = \frac{1 + x}{1 - 2(x + x^2)}$.
Let $\frac{1 + x}{1 - 2(x + x^2)} = a_0 + a_1x + a_2x^2 + \cdots$. Then by multiplying both sides by $1 - 2(x + x^2)$ and looking at the terms with the appropriate powers, we get $1 = 1a_0$ so $a_0 = 1$; $1 = 1a_1 - 2a_0 = 1a_1 - 2$ so $a_1 = 3$; and for $n \geq 2$, $0 = 1a_n - 2a_{n - 1} - 2a_{n - 2}$, so $a_n = 2a_{n - 1} + 2a_{n - 2}$. Note this checks out since there is one such string of length 0 (the empty string), and 3 such strings of length 1. You can also check that $a_2 = 2a_1 + 2a_0 = 8$ checks out since there are 8 such strings of length 2.
Since you have the linear recurrence $a_{n + 2} - 2a_{n + 1} - 2a_n = 0$, you have to look at the roots of $x^2 + x + 2$, which are $1 + \sqrt{3}$ and $1 - \sqrt{3}$. Hence $a_n = A(1 + \sqrt{3})^n + B(1 - \sqrt{3})^n$ for some constants $A$ and $B$. By plugging in $n = 0$ and $n = 1$ you can solve $A = \frac{\sqrt{3} + 2}{2\sqrt{3}}$ and $B=\frac{\sqrt{3} - 2}{2\sqrt{3}}$, hence $a_n = \frac{(\sqrt{3} + 2)(1 + \sqrt{3})^n + (\sqrt{3} - 2)(1 - \sqrt{3})^n}{2\sqrt{3}}$.
|
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|
expand $ \arctan\left(\frac{3x+2}{3x-2}\right)$ into pwer series, find radius of convergence (check solution) I would be grateful if someone could check what I've worked out:
$$ f(x)=\arctan\left(\frac{3x+2}{3x-2}\right)\implies f'(x)=\frac{1}{1+(\frac{3x+2}{3x-2})^2}\cdot \frac{3(3x-2)-3(3x+2)}{(3x-2)^2}$$
$$=\frac{(3x-2)^2}{(3x-2)^2+3x+2)^2}\cdot \frac{-12}{(3x-2)^2}=-\frac{3}{2}\cdot \frac{1}{1+(\frac{3}{2}x)^2}$$
$$=-\frac{3}{2} \cdot \frac{1}{1+(\frac{3}{2}x)^2}=-\frac{3}{2} \sum_{k=0}^{\infty}\left(-\frac{3}{2}x^2\right)^k $$
Which implies $$f(x)=\int f'(x)dx=-\frac{3}{2} \sum_{k=0}^{\infty}\left(-\frac{3}{2} \frac{x^{2k+1}}{2k+1}\right)$$
Radius of convergence:
$$\Big|(\frac{3}{2}x)^2\Big|<1 \Rightarrow -\frac{2}{3}<x<\frac{2}{3}$$
Is this correct? Thank you in advance
|
Here is what you are missing,
$$ f(x)=\int f'(x)dx=-\frac{3}{2} \sum_{k=0}^{\infty}\left((-1)^k\frac{3}{2} \frac{x^{2k+1}}{2k+1}\right) +C $$
$$ \implies f(0) = \arctan(-1) = 0 + C \implies C=-\frac{\pi}{4}. $$
|
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|
Distributing identical objects to identical boxes We have 6 identical things to be distributed in 4 identical boxes such that empty boxes are allowed the find the number of ways to distribute the things ?
|
The result is the number of partitions of 6 into a sum of 4 integral summands $\geq 0$.
The possibilities are
$$
6 = 6 + 0 + 0 + 0 \\
= 5 + 1 + 0 + 0 \\
= 4 + 2 + 0 + 0 \\
= 4 + 1 + 1 + 0 \\
= 3 + 3 + 0 + 0 \\
= 3 + 2 + 1 + 0 \\
= 3 + 1 + 1 + 1 \\
= 2 + 2 + 2 + 0 \\
= 2 + 2 + 1 + 1
$$
so the answer is $9$.
|
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Sum of the first $n$ triangular numbers - induction Question:
Prove by mathematical induction that $$(1)+(1+2)+(1+2+3)+\cdots+(1+2+3+\cdots+n)=\frac{1}{6}n(n+1)(n+2)$$ is true for all positive integers n.
Attempt:
I did the the induction steps and I got up to here:
$$RTP:\frac{1}{6}n(n+1)(n+2)+(1+2+3+\cdots+n+(n+1))=\frac{1}{6}(n+1)(n+2)(n+3)$$
Where do I go from here?
Thank you very much.
|
$$\frac{1}{6}n(n+1)(n+2)+(1+2+3+\cdots+n+1)=$$
$$=\frac{1}{6}n(n+1)(n+2)+\frac{(n+1)(n+2)}{2}=$$
$$=\frac{1}{6}n(n+1)(n+2)+\frac{3(n+1)(n+2)}{6}=$$
$$=\frac{1}{6}(n+1)(n+2)(n+3)$$
|
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|
Continued fraction for $\sqrt{14}$ I'm referencing this page: An Introduction to the Continued Fraction, where they explain the algebraic method of solving the square root of $14$.
$$\sqrt{14} = 3 + \frac1x$$
So, $x_0 = 3$, Solving for $x$, we get
$$x = \frac{\sqrt{14} + 3}5$$
However, in the next step, how do we get the whole number $x_1$ = 1?
$$\frac{\sqrt{14} + 3}5 = 1 + \frac1x$$
My understanding is we would substitute for $x$ in the original equation for $\sqrt{14}$ whereas $$\sqrt{14} = 3 + \frac1{\frac{\sqrt{14} + 3}5}$$
Then substitute the $\sqrt{14}$ again here for $x = \frac{\sqrt{14} + 3}5$ to get the $x_1$ of the continued fraction? Am I just getting the algebra at this point wrong or am I botching steps?
|
$\sqrt{14}=3+\sqrt{14}-3=3+\frac{1}{\frac{\sqrt{14}+3}{5}}\implies x_0 = 3$
$\frac{\sqrt{14}+3}{5}=\frac{6+\sqrt{14}-3}{5}=1+\frac{\sqrt{14}-2}{5}=1+\frac{1}{\frac{\sqrt{14}+2}{2}} \implies x_1 = 1$
$\frac{\sqrt{14}+2}{2}=\frac{5+\sqrt{14}-3}{2}=2+\frac{\sqrt{14}-2}{2}=2+\frac{1}{\frac{\sqrt{14}+2}{5}} \implies x_2 = 2$
etc.
|
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|
How do I factorize equations of the form $x^2 + Bxy + Cy^2 = 0$ Given equation
$$ x^2 + Bxy + Cy^2 = 0. $$
I want to factorize it in the form
$$ (x + my)(x + ny) = 0. $$
What are the values of $m$ and $n$ in terms of $B$ and $C$?
I tried writing the equation as
$$ x^2 + Bxy + Cy^2 = (x + kCy)(x + \dfrac{1}{k}y) = 0. $$
In this case the following equation must be satisfied:
$$ kC + \dfrac{1}{k} = B $$
I solved it for $k$ to find
$$ k = \dfrac{B \mp \sqrt{B^2 - 4C}}{2C}. $$
That makes
$$ x^2 + Bxy + Cy^2 = (x + \dfrac{B \mp \sqrt{B^2 - 4C}}{2}y)(x + \dfrac{2C}{B \mp \sqrt{B^2 - 4C}}y) = 0. $$
Which implies that
$$ \begin{array}{rcl}
m &=& \dfrac{B \mp \sqrt{B^2 - 4C}}{2} \\
n &=& \dfrac{2C}{B \mp \sqrt{B^2 - 4C}},
\end{array} $$
and there are at most four $(m,n)$ pairs.
Is my solution correct? Are there any (other) solution methods for this factorization?
|
Nice solution! Please note that Glen O has answered your question very well, and what comes below does not. Its purpose is to give a slightly different view of the matter, and to connect it with other things that you know.
Let us factor $4x^2+4Bxy+4Cy^2$. Completing the square, we get $(2x+By)^2-(B^2-4C)y^2$. If $B^2-4C\lt 0$, there is no factorization over the reals. But if we are willing to work with complex non-real numbers, we can go on. If we are unwilling, assume that $B^2-4C\ge 0$.
Let $D=B^2-4C$. We are looking at $(2x+By)^2-(\sqrt{D}y)^2$. This is a difference of squares, and factors as
$$(2x+By+\sqrt{D}y)(2x+By-\sqrt{D}y).$$
But we had multiplied by $4$, so let's undo that by dividing each term by $2$. We get
$$\left(x-\frac{B+\sqrt{D}}{2}y\right)\left(x+\frac{B+\sqrt{D}}{2}y\right).$$
Remark: Note that in particular we have derived the Quadratic Formula. For an alternate approach, let's use the Quadratic Formula. Divide our quadratic by $y^2$, not worrying about possible division by $0$, and temporarily let $w=\frac{x}{y}$. Then our quadratic is $y^2(w^2+Bw+C)$. The quadratic equation $w^2+Bw+C=0$ has the roots $\frac{-B\pm\sqrt{B^2-4C}}{2}$, and therefore the polynomial $w^2+Bw+C$ factors as
$$w^2+Bw+C=\left(w-\frac{-B+\sqrt{B^2-4C}}{2}\right) \left(w-\frac{-B-\sqrt{B^2-4C}}{2}\right) .$$
Now replace $w$ by $\frac{x}{y}$, and multiply each term by $y$.
|
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|
Closed form for $\sum_{n=-\infty}^{\infty} \frac{1}{n^4+a^4}$ I've been working with series this week and came across a couple that have been bugging me.
I'm looking for the closed form of:
$$
J(a)=\sum_{-\infty}^{\infty} \frac {1} {n^4+a^4}
$$
As with the other one I posted, solving for the closed form seems like a futile effort as nothing involving differentiation, integration, or DE form leads to a known series solution, such as $e^{x}$, $ln(x)$, etc. I'm running out of hair here. I'll thank anyone who's willing to help.
|
Rather "magical" approach :
Recall the Mittag-Leffler meromorphic expansion of the $\cot$ ($\coth$, resp.) function :
$$\sum_{n=-\infty}^{\infty}\frac{1}{n^2-a^2} = -\frac{\pi\cot\pi a}{a} $$
holding for any complex $a$ (exept the poles obviously) - this fact we will use later as change $a\rightarrow a i$.
From this we have
$$\sum_{n=-\infty}^{\infty}\frac{1}{(n\!-\!a)^2\!-\!b^2} = \frac{1}{2b}\sum_{n=-\infty}^{\infty}\frac{a-b}{n\!-\!(a\!-\!b)^2}\!+\!\frac{a+b}{n\!-\!(a\!+\!b)^2} =\frac{\pi}{2b}\left[\cot\pi (a\!-\!b)\!-\!\cot\pi (a\!+\!b)\right]\tag{1}$$
Similarly we can show
$$\sum_{n=-\infty}^{\infty}\frac{n+a}{(n\!+\!a)^2\!-\!b^2}\!-\!\frac{n-a}{(n\!-\!a)^2\!-\!b^2} = \pi\left[\cot\pi (a\!+\!b)\!+\!\cot\pi (a\!-\!b)\right]\tag{2}$$
Start with
$$n^4\!+\!a^4\!=\!n^4\!+\!2a^2n^2\!+\!a^4\!-\!2a^2n^2\!=\!\left(n^2\!+\!a^2\right)^2\!-\!\left(\sqrt{2}an\right)^2\!=\!\left(n^2\!+\!na\sqrt{2}\!+\!a^2\right)\left(n^2\!-\!na\sqrt{2}\!+\!a^2\right)$$
Then, using partial fraction expansion :
$$\begin{align*}&\sum_{n=-\infty}^{\infty}\frac{1}{n^4+a^4}=\frac{1}{2\sqrt{2}a^3}\sum_{n=-\infty}^{\infty}\frac{n+\sqrt{2}a}{n^2+\sqrt{2}an+a^2}-\frac{n-\sqrt{2}a}{n^2-\sqrt{2}an+a^2} = \\ \\
&\frac{1}{2\sqrt{2}a^3}\sum_{n=-\infty}^{\infty}\frac{n+\frac{a}{\sqrt{2}}+\frac{a}{\sqrt{2}}}{\left(n+\frac{a}{\sqrt{2}}\right)^2+\left(\frac{a}{\sqrt{2}}\right)^2}-\frac{n-\frac{a}{\sqrt{2}}-\frac{a}{\sqrt{2}}}{\left(n-\frac{a}{\sqrt{2}}\right)^2+\left(\frac{a}{\sqrt{2}}\right)^2}\end{align*} $$
By (1) and (2) we immidiatelly conclude this equals :
$$\frac{\pi}{2\sqrt{2}a^3}\left[\cot \left(\pi a\frac{1+i}{\sqrt{2}}\right)\!+\!\cot \left(\pi a\frac{1-i}{\sqrt{2}}\right)\!-i\cot \left(\pi a\frac{1-i}{\sqrt{2}}\right)\!+i\cot \left(\pi a\frac{1+i}{\sqrt{2}}\right)\right] $$
Denote $\alpha := \pi a(1-i)/\sqrt{2}$, then $\bar{\alpha}=\pi a(1+i)/\sqrt{2}$, and since $\cot\bar{\alpha}=\overline{\cot\alpha}$ meromorphic :
$$\sum_{n=-\infty}^{\infty}\frac{1}{n^4+a^4}=\frac{\pi}{2\sqrt{2}a^3}\left[\cot\bar{\alpha} + \cot\alpha -i\cot\alpha+i\cot\bar{\alpha}\right]=\frac{\pi}{\sqrt{2}a^3}\left[\Re\cot \alpha + \Im\cot \alpha\right]\tag{3} $$
Now a bit of half angle trigonometry, from wiki for example :
$$\cot\left(\frac{A-B}{2}\right)=\frac{\sin A-\sin B}{\cos B - \cos A}$$
This implies, since $\sin ix=i \sinh x$ and $\cos ix = \cosh x$ :
$$\cot \alpha=\frac{\sin\left(\sqrt{2} \pi a\right)+i\sinh\left(\sqrt{2} \pi a\right)}{\cosh\left(\sqrt{2} \pi a\right) -\cos\left(\sqrt{2} \pi a\right)}$$
And therefore by $(3)$ :
$$\sum_{n=-\infty}^{\infty} \frac{1}{n^4+a^4} = \frac{\pi}{\sqrt{2} \, a^3} \frac{\sinh\left(\sqrt{2} \pi a\right)+\sin\left(\sqrt{2} \pi a\right)}{\cosh\left(\sqrt{2} \pi a\right) -\cos\left(\sqrt{2} \pi a\right)}$$
|
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|
Examine the continuity of $f(x)=x^2+\frac{x^2}{(1+x^2)}+\frac{x^2}{(1+x^2)^2}+...+ \frac{x^2}{(1+x^2)^n}+....$ at $x=0$
Examine the continuity of $$f(x)=x^2+\frac{x^2}{(1+x^2)}+\frac{x^2}{(1+x^2)^2}+...+ \frac{x^2}{(1+x^2)^n}+....$$ at $x=0$
I tried to solve the problem using $$\lim_{x \to0^+}f(x)=\lim_{x \to0^-}f(x)=f(0)$$ and I got both limits to be $0$. I think there is an error somewhere since my book says it is discontinuous.
Can you help me out?
Thankyou
|
Note that it's a geometric series, as Haral Hanche-Olsen commented. You can write $f(x)$ as
$$f(x) = x^2 \frac{1-\frac{1}{(1+x^2)^{n+1}}}{1-\frac{1}{1+x^2}}=x^2\frac{(1+x^2)^{n+1}-1}{x^2(1+x^2)^{n}} = \frac{(1+x^2)^{n+1}-1}{(1+x^2)^{n}}$$
This function is continuous at $x=0$ in my sense...
|
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|
Laurent Series Expansion Give the Laurent series development of the function $f(z)=\frac{1}{z(z-1)(z-2)}$ in the three rings $A_1=\{z:|z|<1\}$, $A_2=\{z:1<|z|<2\}$ and $A_3=\{z:2<|z|\}$.
I have gotten the partial fraction form $f(z)=\frac{1}{2z}-\frac{1}{z-1}+\frac{1}{z-2}$
For $A_1$, I think it's the standard Taylor expansion about 0 and I can do this for the second and third terms:
$-\frac{1}{z-1}=(1-z)^{-1}=\sum^\infty_{n=0}z^n$ and $\frac{1}{2(z-2)}=\frac{1}{4}\sum^\infty_{n=0}(\frac{z}{2})^n$. How do I get the Taylor expansion of the first term?
I foresee similar problems if I work out the expansion for $A_2$ and $A_3$ as well, so a general method is appreciated. Thanks!
|
First of all, the partial fraction decomposition is
$$
f(z) = \frac{1}{2z} - \frac{1}{z - 1} + \frac{1}{2(z - 2)}.
$$
The Laurent series converges on an annulus, so $A_1$ must not contain the origin:
$$
A_1 = \left\{z : 0 < |z| < 1 \right\}.
$$
Then, you can just use the usual Taylor expansion around $z = 0$:
$$
\begin{align}
f(z) &= \frac{1}{2z} + \frac{1}{1 - z} - \frac{1}{4(1 - \tfrac{z}{2})} \\
&= \tfrac{1}{2}z^{-1} + \sum_{n=0}^\infty z^n - \tfrac{1}{4} \sum_{n=0}^\infty \left(\frac{z}{2}\right)^n \\
&= \tfrac{1}{2}z^{-1} + \sum_{n=0}^\infty \left(1 - \frac{1}{2^{n+2}}\right)z^n \\
&= \sum_{n=-1}^\infty \left(1 - \frac{1}{2^{n+2}}\right)z^n \\
&= \tfrac{1}{2}z^{-1} + \tfrac{3}{4} + \tfrac{7}{8}z + \tfrac{15}{16}z^2 + \cdots
\end{align}
$$
There is actually an easier way to expand the series (around $z = 0$) which generalizes nicely for the other regions:
$$
f(z) = \frac{1}{z} \cdot \frac{1}{(z - 1)(z - 2)} = \frac{1}{z} \cdot \left( \frac{1}{1 - z} - \frac{1}{2 - z} \right).
$$
Each of these is a geometric series that can be expanded inside or outside its radius of convergence.
$$
\frac{1}{1 - z} = \begin{cases}
\phantom{-}\sum_{n = 0}^\infty z^n, &|z| < 1, \\
-\sum_{n = 1}^\infty \frac{1}{z^n}, &|z| > 1.
\end{cases}
$$
Similarly,
$$
\frac{1}{2 - z} = \frac{1}{2(1 - \tfrac{z}{2})}
= \begin{cases}
\phantom{-}\frac{1}{2}\sum_{n = 0}^\infty \left(\frac{z}{2}\right)^n
= \frac{1}{2}\sum_{n = 0}^\infty \frac{z^n}{2^n}
= \sum_{n = 0}^\infty \frac{z^n}{2^{n + 1}}, &|z| < 2, \\
-\frac{1}{2}\sum_{n = 1}^\infty \frac{1}{\left(\frac{z}{2}\right)^n}
= -\frac{1}{2}\sum_{n = 1}^\infty \frac{2^n}{z^n}
= -\sum_{n = 1}^\infty \frac{2^{n - 1}}{z^n}, &|z| > 2.
\end{cases}
$$
Now, for each of the three regions, choose the appropriate expansion, and combine them. For instance, on $A_2$, where $1 < |z| < 2$,
$$
\begin{align}
f(z) &= \frac{1}{z} \left(-\sum_{n = 1}^\infty \frac{1}{z^n}
- \sum_{n = 0}^\infty \frac{z^n}{2^{n + 1}} \right) \\
&= -\sum_{n = 1}^\infty \frac{1}{z^{n + 1}}
- \sum_{n = 0}^\infty \frac{z^{n - 1}}{2^{n + 1}}.
\end{align}
$$
|
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|
conic sections, ellipse A particle is travelling clockwise on the elliptical orbit given by
$$\displaystyle \frac{x^2}{100} + \frac{y^2}{25} = 1$$
The particle
leaves the orbit at the point $(-8, 3)$ and travels in a straight line tangent to the ellipse.
At which point will the particle cross the $y$-axis?
|
I like to work with homogeneous coordinates.
The ellipse is represented by
$$C = \begin{bmatrix} \frac{1}{100} & 0 & 0 \\
0 & \frac{1}{25} & 0 \\
0 & 0 & -1 \end{bmatrix}$$
the 3×3 matrix such that
$$ \left. \begin{pmatrix} x & y & 1 \end{pmatrix} \begin{bmatrix} \frac{1}{100} & 0 & 0 \\
0 & \frac{1}{25} & 0 \\
0 & 0 & -1 \end{bmatrix} \begin{pmatrix} x \\ y \\ 1 \end{pmatrix} =0 \right\} \frac{x^2}{100} + \frac{y^2}{25} -1 =0 $$
The tangent point is $P=(-8,3,1)$ (you verify tangency by $P\cdot C P=0$)
The tangent line is $L=C P = (-\frac{2}{25}, \frac{3}{25}, -1)$ with equation
$$ \left. (x,y,1) (-\frac{2}{25}, \frac{3}{25}, -1)=0 \right\} -2x+3y-25=0 $$
You can work out the rest.
|
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|
Solve $\sqrt[3]{x+10}-\sqrt[3]{x-10}=2$.
Solve $\sqrt[3]{x+10}-\sqrt[3]{x-10}=2$.
I tried cubing the both sides but things then go very ugly. Are there simpler way to solve it? Thanks.
p.s. The answers are $\pm 6\sqrt 3$.
|
$\displaystyle\sqrt[3]{x+10}-\sqrt[3]{x-10}=2---(1)$
Let $y=x+10$
$(1)$ becomes
$\displaystyle\sqrt[3]{y}-\sqrt[3]{y-20}=2$
$\displaystyle\sqrt[3]{y}=2+\sqrt[3]{y-20}$
Cube both sides to get
$\displaystyle{y}=8+12 (y-20)^{\frac{1}{3}}+6(y-20)^{\frac{2}{3}}+y-20$
$\displaystyle{0}=-12+ 6 (y-20)^{\frac{1}{3}}(2+(y-20)^{\frac{1}{3}})$
$\displaystyle{12}= 6 (y-20)^{\frac{1}{3}}(y)^{\frac{1}{3}}$
$\displaystyle{2}= (y-20)^{\frac{1}{3}}(y)^{\frac{1}{3}}$
Cube both sides
$\displaystyle{8}= (y-20)(y)$
$\displaystyle{y^2-20y-8}= 0$
Solving the above equation yields $y=10\pm 6\sqrt{3}$ which implies $x=\pm 6\sqrt{3}$.
|
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|
Proving two equations involving the greatest common divisor
Show or prove that $$\gcd \left(\frac{a^{2m}-1}{a+1} ,a + 1\right )=\gcd(a + 1 , 2m),$$
and that
$$\gcd \left(\frac{a^{2m + 1}+1}{a+1} , a + 1\right)=\gcd(a + 1 , 2m + 1).$$
|
Let integer $d\ne0$ divides $a+1$ i.e., $a+1=c\cdot d$(say) where $c\ne0$ is some integer
$\implies a=c\cdot d-1$
$$\gcd \left(\frac{a^{2m}-1}{a+1} ,a + 1\right )$$
$$=\gcd \left(\frac{(c\cdot d-1)^{2m}-1}{c\cdot d} ,c\cdot d\right )$$
$$=\gcd \left( (c\cdot d)^{2m-1}-\binom{2m}1(c\cdot d)^{2m-2}+ \binom{2m}2(c\cdot d)^{2m-3}+\cdots+\binom{2m}{2m-2}(c\cdot d)-\binom{2m}{2m-1},c\cdot d\right)$$
$$=\gcd\left( c\cdot d\{(c\cdot d)^{2m-2}-\binom{2m}1(c\cdot d)^{2m-3}+ \binom{2m}2(c\cdot d)^{2m-4}+\cdots+\binom{2m}{2m-2}\}-2m,c\cdot d\right)\text{ as } \binom{2m}{2m-1}=\binom{2m}{2m-(2m-1)}=\binom{2m}1=2m$$
$$=\gcd(-2m, c\cdot d)\text{ as }\gcd(p+kq,q)=\gcd(p,q)$$
$$=\gcd(2m, a+1)$$
Can you solve the second one now?
|
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|
On Sixth Powers $x_1^6+x_2^6+\dots+x_6^6 = z^6$ Fourteen years ago, in 1999 (has it been that long?) Merignac started a search for,
$$x_1^6+x_2^6+\dots+x_6^6 = \color{red}z^6$$
using the five congruence classes,
$$\begin{aligned}
&b^6(x_1^6+x_2^6+\dots+x_5^6)+(1x_6)^6+(42x_7)^6 = z^6\\
&b^6(x_1^6+x_2^6+\dots+x_5^6)+(2x_6)^6+(21x_7)^6 = z^6\\
&b^6(x_1^6+x_2^6+\dots+x_5^6)+(3x_6)^6+(14x_7)^6 = z^6\\
&b^6(x_1^6+x_2^6+\dots+x_5^6)+(6x_6)^6+(7x_7)^6 = z^6\\
&b^6(x_1^6+x_2^6+\dots+x_4^6)+(21x_5)^6+(14x_6)^6+(6x_7)^6 = z^6
\end{aligned}$$
where $b =42$. Primitive integer solutions are known for all five classes, but none has one $x_i = 0$ with $\color{red}z<730000$. (See Further work section of The Smallest Solutions to the Diophantine Equation $a^6+b^6+c^6+d^6+e^6=x^6+y^6$.)
To compare to 4th powers, one can primitively solve,
$$a^4(x_1^4+x_2^4+x_3^4)+x_4^4 = z^4$$
for $a=10$ or $a=20$ with the smallest being,
$$10^4(24^4+34^4+43^4)+599^4 = 651^4$$
$$20^4(19^4+83^4+94^4)+4907^4 = 4949^4$$
Question: Is $b=42$ overly restrictive? Can it be reduced to just $b = 21$? (As the fourth power example shows, $a = 10$ has a smaller solution. If so, then maybe they overshot one in the $z<730000$ range.)
|
The restriction to $b = 42$ is correct because any solution of the Diophantine equation
$$x_1^6+x_2^6+\dots+x_6^6 = z^6$$
would have to meet constraints associated with each of the prime factors of 42:
a) If $a \not\equiv 0 \pmod{2}$ then $a^6 \equiv 1 \pmod{8}$;
b) If $a \not\equiv 0 \pmod{3}$ then $a^6 \equiv 1 \pmod{9}$;
c) If $a \not\equiv 0 \pmod{7}$ then $a^6 \equiv 1 \pmod{7}$.
(These can be found on this page of Meyrignac's website, albeit applied to a different 6th power equation.) To meet these constraints, a (primitive) solution must have $x_i \equiv 0 \pmod{2}$ for 5 of the 6 left hand terms, and similarly for mod 3 and mod 7. Hence a solution must have $x_i^6 \equiv 0 \pmod{42}$ for at least 3 of the 6 left hand terms.
There is also a restriction associated with 13 arising from the fact that:
d) If $a \not\equiv 0 \pmod{13}$ then $a^6 \equiv 1$ or $-1 \pmod{13}$.
This implies that a solution must have an odd number of terms (which could include $z$) congruent to 0 mod 13 and, among the remaining terms, an appropriate balance of sixth powers congruent to 1 and to -1 mod 13. To combine this restriction with those relating to 2, 3 and 7 would be quite tedious but it can be said that the five congruence classes are not restrictive enough to exclude all cases which are impossible for reasons of congruence.
|
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Floor of log equation $S=\left(\lfloor\log_{10}(x)\rfloor+1\right)x - \frac{10^{\lfloor\log_{10}(x)\rfloor+1}-10}{9}$ I must find 'x' and I don't know how to solve the following equation.
Does it have a solution? How can I solve it?
$$
S=\left(\lfloor\log_{10}(x)\rfloor+1\right)x - \frac{10^{\lfloor\log_{10}(x)\rfloor+1}-10}{9}
$$
$$S,x\in\mathbb N$$
|
For an integer $k\ge 0$, if $10^k \le x < 10^{k+1}$, then $\lfloor \log_{10}x\rfloor = k$ so
$$
S(x) = (k+1)x -\frac{10^{k+1}-10}{9}=(k+1)x-\frac{10}{9}(10^k-1)
$$
So, for example, if $1\le x \le 9$, then $S(x) = x$. If $10\le x \le 99$, then $S=2x-10$. If $100\le x\le 999$, then $S=3x-110$, and so on.
The first few $S$ values are
$$
\begin{array}
&x = & 1 & 2 &\dots & 9 & 10 & 11 & 12 & \dots & 99 & 100 & 101 & 102 & \dots 999\\
S = & 1 & 2 &\dots & 9 & 10 & 12 & 14 & \dots & 188 & 190 & 193 & 196 & \dots 2887\\
\end{array}
$$
|
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What am I doing wrong in these quartic formula calculations? I was a bit surprised that there is a general formula for the roots of a quartic equation, so I decided to test Wikipedia's version of it myself. To my surprise, I have arrived at a correct answer only once in about five attempts, using only integer coefficients of factorizable polynomials.
I am testing the formula for the equation $(x-1)(x-2)(x-3)(x-4)=0$, which should obviously return 1, 2, 3, and 4. However, my calculations tell a different story:
*
*Expanding gives $x^4 - 10x^3 + 35x^2 - 50x + 24$. Therefore, $a = -10, b = 35, c = -50, d = 24$.
*$u$ is therefore equal to $\frac{3(-10)^2 - 8(35)}{12} = \frac{300 - 280}{12} = \frac{20}{12} = \frac{5}{3}$.
*$\Delta_0$ is $35^2 - 3(-10)(-50) + 12(24) = 1225 - 1500 + 288 = 13$.
*$\Delta_1$ is $2(35^3) - 9(-10)(35)(-50) + 27(-10)^2(24) + 27(-50)^2 - 72(35)(24)$, which works out to 70.
*The discriminant is $-\frac{\Delta_1^2 - 4\Delta_0^3}{27} = -\frac{-3888}{27} = 144$. Because it is positive, the equation must have either four real roots or four complex roots.
*$Q$ is equal to $\sqrt[3]{\frac{70 + \sqrt{-3888}}{2}} = \sqrt[3]{\frac{70 + 36i\sqrt{3}}{2}} = \sqrt[3]{35 + 18i\sqrt{3}}$.
*$v$ is $\frac{\left(\sqrt[3]{35 + 18i\sqrt{3}}\right)^2 + 13}{3\sqrt[3]{35 + 18i\sqrt{3}}}$. After rationalizing the denominator it becomes $\frac{1 + \sqrt[3]{35 - 18i\sqrt{3}}}{3}$.
*$w$ might make things even more complicated, but mercifully its numerator (and therefore value) is zero.
Placing these values into one of the final expressions for the formula gives
$$\frac{1}{4}\left(10 + 2\sqrt{\frac{6 + \sqrt[3]{35 - 18i\sqrt{3}}}{3}} \pm 2\sqrt{\frac{9 - \sqrt[3]{35 - 18i\sqrt{3}}}{3}}\right),$$
a number which is neither rational nor even fully real, and the other two roots are the same.
I expected that the error would become self-evident as I typed this question (which took a while, as you can imagine) — but it has not. So what is wrong with these calculations?
|
My error was made in rationalizing the denominator of $v$. I multiplied both parts of the fraction by $Q^2$, but rendered the fraction as
$$\frac{35 + 18i\sqrt{3} + 13\left(\sqrt[3]{35 + 18i\sqrt{3}}\right)^2}{3(35 + 18i\sqrt{3})}$$
(The first $35 + 18i\sqrt{3}$ should be $(35 + 18i\sqrt{3})\sqrt[3]{35 + 18i\sqrt{3}}$.) Following from this, we can multiply both parts of the fraction by $35 - 18i\sqrt{3}$:
$$\frac{2197\sqrt[3]{35 + 18i\sqrt{3}} + 13(35 - 18i\sqrt{3})\left(\sqrt[3]{35 + 18i\sqrt{3}}\right)^2}{6561}$$
And, simplifying, we find that the true value of $v$ is
$$\frac{\sqrt[3]{35 + 18i\sqrt{3}} + \sqrt[3]{35 - 18i\sqrt{3}}}{3}$$
Or, to put it more simply, $\frac{7}{3}$. (This is what Wolfram Alpha evaluates as the decimal — I have no idea how to obtain this from the cube roots; any help in that respect will be appreciated.)
The expressions for the roots are then
$$\frac{1}{2}(5 + 2 \pm 1), \frac{1}{2}(5 - 2 \pm 1)$$
Which yield the expected 1, 2, 3, 4.
|
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"url": "https://math.stackexchange.com/questions/396489",
"timestamp": "2023-03-29T00:00:00",
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|
Solve for $x$, $3\sqrt{x+13} = x+9$ Solve equation: $3\sqrt{x+13} = x+9$
I squared both sides and got $9 + x + 13 = x^2 + 18x + 81$
I then combined like terms $x^2 + 17x + 59 = 0$
I then used the quadratic equation $x= -\frac{17}2 \pm \sqrt{\left(-\frac{17}2\right)^2-59}$
However, the answer is 3
|
When you square the LHS, it should be $9(x+13)$ not $9+x+13$.
|
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|
System of two simple modular equations $$x \equiv -7 \mod 13$$
$$x \equiv 39 \mod 15$$
I need to find the smallest x for which these equations can be solved. I've been always doing this using Chinese Reminder Theorem, but it seems that it doesn't work here, I'm not sure why though. Let's see:
$N = 13*15 = 195$
$N_{1} = \frac{195}{13} = 13$
$N_{2} = \frac{195}{15} = 15$
$GCD(13, 13) = 1*13 + 0*13$
$GCD(15,15) = 1*15 + 0*15$
$$x = 1 * (-7) * 13 + 1 * 39 * 15 = 480$$
The answer however is supposed to be $x = 84$...What's wrong?
|
The way I usually go about these is to solve the two systems
$$
\mbox{$\begin{align}
x_{13}&\equiv1\pmod{13}\\
x_{13}&\equiv0\pmod{15}
\end{align}$}
\qquad\text{and}\qquad
\begin{align}
x_{15}&\equiv0\pmod{13}\\
x_{15}&\equiv1\pmod{15}
\end{align}
$$
Then combine these to get the solution to the equation given. These can both be solved at once using the Euclid-Wallis Algorithm:
$$
\begin{array}{r}
&&1&6&2\\\hline
1&0&1&-6&13\\
0&1&-1&7&-15\\
15&13&2&1&0\\
\end{array}
$$
This says that $(-6)15+(7)13=1$ which immediately gives $x_{13}\equiv(-6)15\pmod{195}$ and $x_{15}\equiv(7)13\pmod{195}$. The solution to the original question is
$$
\begin{align}
x
&\equiv-7x_{13}+39x_{15}\\
&\equiv84\pmod{195}
\end{align}
$$
Caveat: in the case of two equations as above, this only requires one application of the Euclidean algorithm. However, in the case of three or more equations, we need to apply the Euclidean algorithm once per modulus. For example, to solve
$$
\begin{align}
x&\equiv 1\pmod{3}\\
x&\equiv 2\pmod{5}\\
x&\equiv 4\pmod{7}
\end{align}
$$
we would need to solve each of
$$
\mbox{$\begin{align}
x_3&\equiv1\pmod{3}\\
x_3&\equiv0\pmod{35}
\end{align}$}
\quad\text{and}\quad
\mbox{$\begin{align}
x_5&\equiv1\pmod{5}\\
x_5&\equiv0\pmod{21}
\end{align}$}
\quad\text{and}\quad
\begin{align}
x_7&\equiv1\pmod{7}\\
x_7&\equiv0\pmod{15}
\end{align}
$$
using the Euclid-Wallis Algorithm.
|
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|
Rearranging a series. Consider the following two series.
$$\sum a_n = 1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\frac{1}{5}+ \cdots$$ $$\sum b_n = 1+\frac{1}{3}-\frac{1}{2}+\frac{1}{5}+\frac{1}{7}-\frac{1}{4}+ \cdots$$
I was able to say that $a_n = (-1)^{n+1} \frac{1}{n}$ while $b_n = \frac{1}{4n-3}+\frac{1}{4n-1}-\frac{1}{2n}$, which was obtained after rearranging the numbers.
I understand up to the fact that if a series is not absolutely convergent, rearranging the sequence will change the limit of the sum.
I haven't seen the proof yet, but I have two questions.
1), I want to prove that the second series converges. The ratio test was inconclusive, so trying to use the root test I got the following. Can someone check my work ? Is this a natural approach ?
$$\sqrt[n]{b_n} =\sqrt[n] {\frac{8n-3}{2n(4n-1)(4n-3)}} \le \sqrt[n]{\frac{8}{32n^2}}$$
so $$\lim_{n \to \infty}\sqrt[n]{b_n}=1$$
which is inconclusive, again.
2), Assuming that convergence was known, how can we calculate the limit of the sum ?
|
Since
$$
\small
b_n=\frac{8n-3}{2n(4n-1)(2n-3)}\sim\frac{1}{2n^2}
$$
The series $\sum b_n$ absolutely converges. To find its sum we need to work a little. Note that
$$
\small
\begin{align}
\sum\limits_{k=1}^n b_k
&=\left(1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\ldots-\frac{1}{4n}\right)+\color{green}{\left(\frac{1}{2n}+\frac{1}{2n+1}+\frac{1}{2n+3}+\ldots+\frac{1}{4n-1}\right)}\\
&=\left(1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\ldots-\frac{1}{4n}\right)\color{green}{+\frac{1}{2n}}\color{green}{+\frac{1}{2n+1}}\color{red}{+\frac{1}{2n+2}}\color{green}{+\frac{1}{2n+3}}\color{red}{+\frac{1}{2n+4}}\color{green}{+\ldots+\frac{1}{4n-1}}\color{red}{+\frac{1}{4n}}\color{red}{-\left(\frac{1}{2n+2}+\frac{1}{2n+4}+\ldots+\frac{1}{4n}\right)}\\
&=\left(1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\ldots-\frac{1}{4n}\right)\color{green}{+\left(\frac{1}{2n}+\ldots+\frac{1}{4n}\right)}\color{red}{-\frac{1}{2}\left(\frac{1}{n+1}+\frac{1}{n+2}+\ldots+\frac{1}{2n}\right)}\\
\end{align}
$$
Thus
$$
\small
\begin{align}
\sum\limits_{k=1}^\infty b_k
&=\lim\limits_{n\to\infty}\left(\sum\limits_{k=1}^{4n}\frac{(-1)^{k-1}}{k}+\sum\limits_{k=2n}^{4n}\frac{1}{k}-\frac{1}{2}\sum\limits_{k=n+1}^{2n}\frac{1}{k}\right)\\
&=\lim\limits_{n\to\infty}\sum\limits_{k=1}^{4n}\frac{(-1)^{k-1}}{k}+\lim\limits_{n\to\infty}\left(\ln\frac{4n}{2n}+o(n)-\frac{1}{2}\left(\ln\frac{2n}{n+1}+o(n)\right)\right)\\
&=\ln 2+\ln 2-\frac{1}{2}\ln 2=\frac{1}{2}\ln 8
\end{align}
$$
|
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|
Solve system of first order differential equations I have to solve differential systems like this:
$$
\left\{
\begin{array}{c}
x' = 3x - y + z \\
y' = x + 5y - z \\
z' = x - y + 3z
\end{array}
\right.
$$
Until now I computed the eigenvalues $k = \{2,4,5\}$ by solving the equation resulted from this determinant of this matrix:
$$
\begin{pmatrix}
3 && -1 && 1 \\
1 && 5 && -1 \\
1 && -1 && 3
\end{pmatrix} - kI_3 =
\begin{pmatrix}
3-k && -1 && 1 \\
1 && 5-k && -1 \\
1 && -1 && 3-k
\end{pmatrix}
$$
I don't know what to do next.
NOTE: This is an example but from it I want to learn the method to solve any system of this kind. I learn better from particular examples than directly from generalization.
|
To find the eigenvalues of: $$A =\begin{bmatrix}3 & -1 & 1\\1 & 5 & -1\\1 & -1 &3\end{bmatrix},$$
we set up $|A - \lambda I| = 0$ and solve the characteristic polynomial, so we have:
$$|A -\lambda I| = \begin{bmatrix} 3-\lambda & -1 & 1\\1 & 5-\lambda & -1\\1 & -1 &3 -\lambda \end{bmatrix} = 0$$
From this, we get the characteristic polynomial as: $$-\lambda^3+11 \lambda^2-38 \lambda+40 = -(\lambda-5) (\lambda-4) (\lambda-2)= 0$$
This gives us three eigenvalues: $λ_1 = 2, λ_2 = 4$ and $λ_3 = 5$
To find the eigenvectors, we set up and solve $[A - \lambda_i I]v_i = 0$, so lets do $\lambda_1 = 2$. We have:
$$[A - \lambda_i I]v_i = \begin{bmatrix} 3-2 & -1 & 1\\1 & 5-2 & -1\\1 & -1 &3 -2 \end{bmatrix}v_1 = \begin{bmatrix} 1 & -1 & 1\\1 & 3 & -1\\1 & -1 & 1 \end{bmatrix}v_1 = 0$$
This leads to the row-reduced-echelon form:
$$\begin{bmatrix} 1 & 0 & \dfrac{1}{2}\\0 & 1 & -\dfrac{1}{2}\\0 & 0 & 0 \end{bmatrix}v_1 = 0$$
This gives us a solution of:
$b = \dfrac{1}{2}c \rightarrow c = 2 \rightarrow b = 1$, and $a = -\dfrac{1}{2} c \rightarrow a = -1$.
Thus the eigenvector for the eigenvalue $\lambda_1 = 2$ is $v_1 = (-1, 1, 2)$.
If we repeat this process two more times for the other two eigenvalues, we end up with the eigenvalue/eigenvector pairs:
$$\lambda_1 = 2, ~v_1 = (-1, 1, 2)$$
$$\lambda_2 = 4, ~v_2 = (1, 0, 1)$$
$$\lambda_3 = 5, ~v_3 = (1, -1, 1)$$
Note that different approaches are needed when we get into repeated eigenvalues, but that is for another problem.
Try to derive those last two eigenvectors and report back if they do not work out.
To write out the solution for the three equation, we would use a linear combination of the the eigenvalues, eigenvectors and some unknown constants, as:
$$X(t) = \begin{bmatrix} x(t) \\ y(t) \\ z(t) \end{bmatrix} = c_1e^{2t} \begin{pmatrix} -1 \\ 1 \\ 2 \end{pmatrix} + c_2 e^{4t} \begin{pmatrix} 1 \\ 0 \\ 1 \end{pmatrix} + c_3 e^{5t} \begin{pmatrix} 1 \\ -1 \\ 1 \end{pmatrix}$$
|
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|
Convergence/Divergence of a particular infinite nested radical Is it known if the following infinite nested radical converges or diverges (for $n \in \mathbb N$)?:
$$R(n) = \sqrt{n+\sqrt{(n+1)+\sqrt{(n+2)+ \cdots}}}$$
I recently became interested in these problems when reading about Ramanujan's (famous?) formula: $$\sqrt{1+2\sqrt{1+3\sqrt{1+4\sqrt{\cdots}}}} = 3$$ Which he showed to be a special case of: $$\sqrt{ax+(n+a)^2+x\sqrt{a(x+n)+(n+a)^2+(x+n)\sqrt{\cdots}}} = x+n+a$$
My formula does not fit into that of Ramanujan and the farthest I am able to get is:
$$R^2(n) = n + R(n+1)$$ which I prefer to rewrite as $$R(n+1) = R^2(n) - n$$
This is well...not far at all. Knowing $R(0)$ would supply me with all values of $R(n)$ and so I end up needing to solve $$\sqrt{\sqrt{1+\sqrt{2+ \sqrt{3+\cdots}}}}$$
I feel that the general method for solving these problems is currently beyond me without any instruction. Would anyone be able to offer hints and get me a little bit further down the path?
|
If you accept
$$
\phi=\sqrt{1+\sqrt{1+\sqrt{1+\sqrt{\cdots}}}}=\sqrt{1+\phi}=\frac{1+\sqrt{5}}{2}
$$
then
$$
\begin{align}
\phi < R(1) &< \sqrt{1+\sqrt{2+\sqrt{2^2+\sqrt{2^4+\sqrt{2^8+\cdots}}}}}\\
&=\sqrt{1+\sqrt{2}\phi}
\end{align}
$$
so
$1.615<R(1)<1.815$ and as you say you can get the other $R(n)$ from there.
Edit: You can use this technique for any $n$:
$$
\phi_n=\sqrt{n+\sqrt{n+\sqrt{n+\sqrt{\cdots}}}}=\sqrt{n+\phi_n}=\frac{1+\sqrt{4n+1}}{2} \\
\phi_n < R(n) < \sqrt{n+\sqrt{\frac{n+1}{n}}\phi_n} \\
n+\frac{1}{2}+\sqrt{n+\frac{1}{4}} < R^2(n) < n+\frac{1}{2}\sqrt{1+\frac{1}{n}}+\sqrt{n+\frac{5}{4}+\frac{1}{4n}}
$$
|
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|
Find volume using double integrals? Question: Use double integral to find the volume of the solid enclosed by the spheres $x^2+y^2+z^2=1$ and $x^2+y^2+(z-1)^2=1$
Alright so I tried to doing this by myself and I'm not sure if this is right. Could someone check over my work?
Curve of intersection:
\begin{align*}
x^2 + y^2 + z^2 &= x^2 + y^2 + (z - 1)^2\\
\implies z^2 &= z^2 - 2z + 1\\
\implies z &= 1/2.
\end{align*}
So, the curve of intersection is $x^2 + y^2 = 3/4$ with $z = 1/2$.
By symmetry, it suffices to double the volume between $z = 1/2$ and $z = 1$.
$x^2 + y^2 + (z-1)^2 = 1$ is above $x^2 + y^2 + z^2 = 1$ when $z > 1/2$.
Solving for (positive) $z$ yields
$z = \sqrt{1 - x^2 - y^2}$ and $z = 1 + \sqrt{1 - x^2 - y^2}$.
Hence, the volume equals
\begin{align*}
V &= 2 \iint \left[(1 + \sqrt{1 - x^2 - y^2}) - \sqrt{1 - x^2 - y^2}\right]\, dA\\ \\
&= 2 \int_0^{2\pi}\!\int_0^{\sqrt{3/4}}\!r\,dr \,d\theta, \qquad\textrm{via polar coordinates}\\ \\
&= 2 \int_0^{2\pi} \!\left[(1/2)r^2\right]_{r = 0}^{\sqrt{3/4}}\, d\theta\\ \\
&= \int_0^{2\pi}\!3/4\,d\theta\\ \\
&= 3\pi/2.
\end{align*}
|
Besides to @Shuhao's approach you can use the Spherical coordinates as well. In fact the limits in this system would be:
$$r|_{\frac{1}{2\cos(\theta)}}^1,~~\theta|_0^{2\pi},~~\phi|_0^{\pi/3}$$
Note that two creatures intersects at $z=1/2$.
|
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"timestamp": "2023-03-29T00:00:00",
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|
Find the largest number having this property. The $13$-digit number $1200549600848$ has the property that for any $1 \le n \le 13$, the number formed by the first $n$ digits of $1200549600848$ is divisible by $n$ (e.g. 1|2, 2|12, 3|120, 4|1200, 5|12005, ..., 13|1200549600848 using divisor notation).
Question 1: Find the largest computed number having this property.
Question 2: Is there a theoretical upper bound on the largest possible number with this property?
Edit: Added Question 2 as I believe it is more insightful as compared to brute force computer calculations.
|
Let $n_k = 10^{k - 1} a_1 + 10^{k - 2} a_2 + \cdots + a_k$ for positive integer $k$. Also,
$$\begin {eqnarray}
n_{k + 1} & = & 10^k a_1 + 10^{k - 1} a_2 + \cdots + 10 a_k + a_{k + 1} \nonumber \\ & = & 10 n_k + a_{k + 1}. \tag{1}
\end {eqnarray}$$
Since $k | n_k$, we have, by $(1)$, that
$$\begin {equation*}
\begin {cases}
k + 1 |10 n_k + a_{k + 1}, \\
k | n_k, \tag{2} \\
0 \leqslant a_{k + 1} \leqslant 9.
\end {cases}
\end {equation*}$$
Solving $(2)$ we have $$k + 1 | 10 r_k - a_{k + 1},$$ where $r_k = n_k/k$. Thus,
$$\begin {equation*}
\begin {cases}
a_k \equiv 10 r_{k - 1} \pmod {k}, \\
0 \leqslant a_k \leqslant 9. \tag{3} \\
\end {cases}
\end {equation*}$$
Let $S_k$ be the set of solutions of $(3)$. Then, the number of such numbers is $$\sum_{\substack {k \geqslant 1 \\ n_k \in S_k}} 1 = \sum_{k = 1}^{\infty} |S_k|.$$
Empirical data shows that for some $k$ the set $S_k$ is empty. We can even state a far stronger claim, that the set $S_k$ is empty for infinitely many $k$.
Lemma. There exists a natural number $N$ such that $S_k$ is empty for all $k \geqslant N$.
Proof. Let $N$ be the smallest integer such that $S_N$ is empty, that is, $n_N$ does not exist. Observe that if $S_N$ is empty, then all $S_k$ are also empty for $k > N$, since if $n_N$ does not exist, $n_k$ for $k > N$ cannot exist, either.
Necessary and sufficient condition for minimality of non-existence of $n_N$ is that $n_{N - 1}$ exists, but that $N$ never divides $10 r_{N - 1} - a_N$. We can guarantee that if and only if when we write $10 r_{N - 1} = N t + b$, where $t$ and $b$ are positive integers and $b \leqslant N - 1$, then $$10 \leqslant b \leqslant N - 1.$$ Indeed, then we would have $10 r_{N - 1} - a_N = N t + b - a_N$ and $0 \leqslant a \leqslant 9$, which gives $$N t + 1 \leqslant N t + b - a_N = 10 r_{N - 1} - a_N \leqslant N t + N - 1,$$ which is never divisible by $N$.
Suppose that $N$ is odd. Denote by $\langle x \rangle$ the smallest positive integer that a positive integer $x$ gives modulo $N$. Let $m$ be the smallest positive integer such that $10 m \equiv 1 \pmod {N}$, which exists, since $N$ is odd. Then, $10 m r_{N - 1} \equiv r_{N - 1} \pmod {N}$. On the other hand, we have $10 m r_{N - 1} \equiv m b \pmod {N}$. Hence, $$r_{N - 1} \equiv m b \pmod {N}.$$ Since $10 \leqslant b \leqslant N - 1$, we have $$1 = \langle 10 m \rangle \leqslant \langle m b \rangle = \langle r_{N - 1} \rangle \leqslant N - m.$$ Since $n_{N - 1} \equiv -r_{N - 1} \pmod {N}$, we further have $$m \leqslant \langle n_{N - 1} \rangle \leqslant N - 1.$$
to be continued...
|
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|
Calculating $\sum^{10}_{k=1}\left(\sin\frac{2k\pi}{11}+i\cos\frac{2k\pi}{11}\right)$
Find the value of $$\sum^{10}_{k=1}\left (\sin\left (\frac{2k\pi}{11} \right )+i\cos\left (\frac{2k\pi}{11}\right ) \right)$$
My approach:
Since $\cos\theta + i\sin\theta = e^{i\theta}$, we can write the given equation as:
$$\begin{align*}
&i \left \{\sum^{10}_{k=1} \left (\cos\frac{2k\pi}{11} -i\sin\frac{2k\pi}{11} \right ) \right \}\\
= &i \left \{\sum^{10}_{k=1}\left (e^{-i\frac{2k\pi}{11}} \right ) \right \} \tag{i}
\end{align*}$$
Solving the index part only which is
$$\begin{align*}
-i\frac{2k\pi}{11} &= -i\frac{2\pi}{11}(1+2+3+\cdots+10) \quad (\text{putting the values of } k)\\
&= -i\frac{2\pi}{11}( 55) \quad \left(\text{By applying sum of first $n$ natural numbers} = \frac{n(n+1)}{2}\right )\\
&=-i10\pi
\end{align*}$$
Putting this value in $(\text{i})$ we get:
$e^{-i10\pi} = i\cos10\pi = i.$
But the answer is $-i$. Please suggest where I went wrong… Thanks..
|
You went wrong when you turned the sum into a product (which you'd have to do in order to sum the indices like that).
Instead, multiply (i) by $1=e^{i2\pi}$ and reindex to get $$i\sum_{k=1}^{10}e^{i2\pi\left(1-\frac{k}{11}\right)}=i\sum_{k=1}^{10}e^{\frac{i2\pi}{11}},$$ then substitute $\omega=e^{\frac{i2\pi}{11}}$ to get $$i\sum_{k=1}^{10}\omega^k.$$ All that's left is to use the fact that $$\sum_{k=0}^{10}\omega^k=0.$$ (Why is this true?)
|
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|
How to prove $(a-b)^3 + (b-c)^3 + (c-a)^3 -3(a-b)(b-c)(c-a) = 0$ without calculations I read somewhere that I can prove this identity below with abstract algebra in a simpler and faster way without any calculations, is that true or am I wrong?
$$(a-b)^3 + (b-c)^3 + (c-a)^3 -3(a-b)(b-c)(c-a) = 0$$
Thanks
|
We know $$(x+y+z)^3=x^3+y^3+z^3+3(x+y)(y+z)(z+x)$$
If $x+y+z=0, x^3+y^3+z^3=-3(x+y)(y+z)(z+x)=-3(-z)(-x)(-y)=3xyz$
Alternatively if $x+y+z=0,$
$x^3+y^3+z^3=(x+y)^3-3xy(x+y)+z^3=(-z)^3-3xy(-z)+z^3=3xyz$
Put $x=a-b,y=b-c,z=c-a$
|
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|
Volume bounded by sphere $x^2+y^2+z^2=a^2$ and cylinder $x^2+y^2=a|x|$ What is the volume bounded by the sphere $x^2+y^2+z^2=a^2$ and the cylinder $x^2+y^2=a|x|$?
The answer can be in terms of the value $a$ (or $r$). Does someone know how to do this?
Thank you in advance!
|
The figure is symmetric, with equal volume in each of the eight octants, so we focus on the first octant, and multiply by $8$. Let's work in cylindrical coordinates. In that case, our function is $z=\sqrt{a^2-r^2}$, and our region of integration is bounded by
$$0\le\theta\le\frac{\pi}{2}\\0\le r\le a\cos\theta$$
The volume is then:
$$8\int_0^{\frac{\pi}{2}}\int_{0}^{a\cos\theta}\int_{0}^{\sqrt{a^2-r^2}}rdzdrd\theta=8\int_0^{\frac{\pi}{2}}\int_{0}^{a\cos\theta}r\sqrt{a^2-r^2}drd\theta=\frac{2a^3\pi}{3}$$
Recall that the volume element in cylindrical coordinates is $rdzdrd\theta$. The above integral is then evaluated by the substitution $x=a^2-r^2$.
Notice that our answer is half of the volume of the sphere, so there is an equal volume inside of the cylinder bounded by the sphere as there is outside of the cylinder bounded by the sphere.
|
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|
How to integrate $\frac{x+2}{x^2+2x+2}$ with the substitution method This is what I got so far:
$$\frac{x+2}{x^2+2x+2} = \frac{x+2}{(x+1)^2+1} = \int \frac{x}{(x+1)^2+1} + 2 \times \int \frac{1}{(x+1)^2+1}$$
I know the last integral $= \arctan(x+1) + c$, where $c$ is a constant, but I don't know how to integrate the first one, i.e $\int \frac{x}{(x+1)^2+1}$.
Thanks for your help.
EDIT: MY SOLUTION
I split the fraction: $\frac{x+1}{x^2+2x+2} + \frac{1}{x^2+2x+2}$
$$\int \frac{x+1}{(x+1)^2+1}\;dx \implies (x+1)^2+1 = u$$
$$\implies \int\frac{(x+1)}{u} \text{ and } \frac{du}{2(x+1)}$$
$$=\frac{1}{2}\int \frac{1}{u}\,du$$
$$=\frac{1}{2}\ln(u)$$
$$=\frac{1}{2}\ln(x^2+2x+2)$$
so finally the last integral $\int \frac{1}{x^2+2x+2}\,dx = \int \frac{1}{(x+1)^2+1}\,dx = \arctan(x+1)$
and now the solution = $\frac{1}{2}\ln(x^2+2x+2) + \arctan(x+1) + C$
|
HINT: You want one of the integrals to integrate to give a logarithmic function. Take this into account when you split up the fractions and you should be fine.
|
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|
The system of equations $x^2 + y^2 - x - 2y = 0$ and $x + 2y = c$ I have
$(1.) \quad x^2 + y^2 - x - 2y = 0 \\
(2.) \quad x + 2y = c$
Solving for $y$ in $(2.)$ gives
$y = (c - x) / 2$
Is there a way to simplify equation $(1.)$?
Because at the end I arrive at
$c^2 - 2x - c = 0$
and can't proceed. Need to get typical form of square equation $Ax^2 + Bx + C = 0$.
The solution for $c$ is $0$ or $5$.
EDIT 1:
For what real values of the parameter do the common solutions of the following pairs of simultaneous equations became identical?
(g) $ \quad x^2 + y^2 - x - 2y = 0, \quad x + 2y = c$, Ans. c = 0 or 5.
The process is to solve for $y$, then to substitute that into second equation.
From that we get A, B and C. Delta being $B^2 - 4AC$ we can get parameter.
So
$y = (c - x) / 2$
$x^2 + y^2 - x -2y = 0$
$x^2 + ((c-x)/2)^2 - x - 2 * ((c-x)/2) = 0$
and i got lost...
|
You're intersecting a circle with the straight lines parallel to $x+2y=0$.
The circle can be written, by completing the squares, as
$$
x^2 -2\frac{1}{2} x + \frac{1}{4} + y^2 - 2y + 1 = \frac{1}{4}+1
$$
or, writing the squares,
$$
\biggl(x-\frac{1}{2}\biggr)^2+(y-1)^2=\frac{5}{4}
$$
Thus the lines you're looking for must have distance $\sqrt{5}/2$ from the center $(1/2,1)$.
The formula for the distance of a point from a line gives then
$$
\frac{\left|\frac{1}{2}+2\cdot1-c\right|}{\sqrt{1^2+2^2}}=\frac{\sqrt{5}}{2}
$$
that becomes
$$
2\left|\frac{5}{2}-c\right|=5.
$$
This in turn gives the two equations
$$
5-2c=5,\qquad -5+2c=5
$$
that have solutions, respectively, $c=0$ and $c=5$.
You can also solve the system by substitution. From the second equation you get $x=c-2y$ and, substituting in the quadratic equation,
$$
(c-2y)^2+y^2-(c-2y)-2y=0
$$
Developing and simplifying, you get
$$
5y^2-4cy+c^2-c=0
$$
that has coincident roots when the discriminant is zero:
$$
(4c)^2-4\cdot 5(c^2-c)=0
$$
that becomes
$$
-4c^2+20c=0
$$
that is, $c^2-5c=0$. The solutions are $c=0$ and $c=5$.
|
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|
How find $a$ such that $x^2-\sqrt{a-x}=a$ has exactly two real solutions Consider the equation
$$ x^2-\sqrt{a-x}=a.$$ I wish to determine the values of $a$ for which the above equation has exactly two real solutions (for $x$).
My idea:
$$a-x=(x^2-a)^2=x^4-2ax^2+a^2\Longrightarrow f(x)=x^4-2ax^2+x+a^2-a=0$$ and we must have
$$a-x\ge 0$$
$$x^2= a+\sqrt{a-x}\ge a,$$
so $f(x)=x^4-2ax^2+x+a^2-a=0$ has only real solution $x$, and this solution $x^2\ge a, x\le a.$ But can I use this to find the possible values of $a$?
then we have $f''(x)=0 ?$
$$\Longrightarrow f'(x)=4x^3-4ax+1,\Longrightarrow f''(x)=12x^2-4a=0$$
then $$12x^2=4a\Longrightarrow a=3x^2>0$$
so $$ a^2\le \sqrt{\dfrac{a}{3}}\le a$$
$$\Longrightarrow \dfrac{1}{3}\le a\le \sqrt[3]{\dfrac{1}{3}}$$
my reslut is true?and I think this problem have other nice methods.
I hope someone can write the final results，because I don't know the correct result.Thank you everyone
|
Continuing from the moment of squaring both parts of the equation, we will solve the equation with respect to the parameter $a$. From $x^4-2ax^2+x-a+a^2=0$ we have:
$$a^2-\left(2x^2+1\right)a+\left(x^4+x\right)=0 \Rightarrow \left(a-x^2-x\right)(a-x^2+x-1)=0.$$
Now find $x$ in terms of $a$:
$$x=\frac{1\pm \sqrt{4a-3}}{2} \text{ or } x=\frac{-1\pm\sqrt{4a+1}}{2}.$$
Plugging this values in the initial equation we will get that $x=\dfrac{-1-\sqrt{4a+1}}{2}$ and $x=\dfrac{1+\sqrt{4a-3}}{2}$ are solutions. Then for $a\in\left[-\dfrac{1}{4}; 0\right]$ we have two solutions. When $a\in\left(0; 1\right)$ only one solution, because one of the root on this interval is bigger than $a$. For $a\in\left[1;+\infty\right)$ we have two solutions.
Answer: $a\in\left[-\dfrac{1}{4};0\right]\cup\left[1;+\infty\right).$
|
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|
How to determine the rank and determinant of $A$? let $A$ be $$A_{a} = \begin{pmatrix} a & 1 & 1 & 1 \\ 1 & a & 1 & 1\\ 1 & 1 & a & 1\\ 1 & 1 & 1 & a \end{pmatrix}$$
How can I calculate the rank of $A$ by the Gauss' methode and $\det A$?
|
We know that this matrix is equivalent to $$B=\begin{pmatrix}a&0&0&1-a\\1&a-1&0&0\\1&1-a&a-1&0\\1&0&1-a&a-1\end{pmatrix}$$
and $$\det(B)=a(a-1)^3-(1-a)((1-a)^2-(a-1)((1-a)-(a-1))=(a-3)(a-1)^3$$ and $\text{rank}(A)=4$ for all values of $a$ except $1$ and $3$.
|
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|
proving $\frac{1}{1\cdot 2}+\frac{1}{3\cdot 4}+\frac{1}{5\cdot 6}+\cdots+$ Without Induction i proved that:
$$
\begin{align}
& {} \quad \frac{1}{1\cdot 2}+\frac{1}{3\cdot 4}+\frac{1}{5\cdot 6}+\cdots+\frac{1}{(2n-1)\cdot 2n} \\[10pt]
& =\frac{1}{n+1}+\frac{1}{n+2}+\frac{1}{n+1}+\cdots+\frac{1}{2n}\text{ for }n\in \mathbb{N}
\end{align}
$$
by induction.
i wonder if it can be done without using induction. if so, i'll appreciate if someone could show how.
thanks.
|
We have
$$\frac{1}{2k(2k-1)}=\frac{1}{2k-1}-\frac{1}{2k}=\frac{1}{2k-1}+\frac{1}{2k}-\frac{1}{k}$$
so
$$\sum_{k=1}^n\frac{1}{2k(2k-1)}=\sum_{k=1}^n\left(\frac{1}{2k-1}+\frac{1}{2k}\right)-\sum_{k=1}^n\frac{1}{k}=\sum_{k=1}^{2n}\frac{1}{k}-\sum_{k=1}^n\frac{1}{k}=\sum_{k=n+1}^{2n}\frac{1}{k}$$
|
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|
Evaluating $\lim_{x\to1}\frac{\sqrt{x^2+3}-2}{x^2-1}$? I tried to calculate, but couldn't get out of this:
$$\lim_{x\to1}\frac{x^2+5}{x^2 (\sqrt{x^2 +3}+2)-\sqrt{x^2 +3}}$$
then multiply by the conjugate.
$$\lim_{x\to1}\frac{\sqrt{x^2 +3}-2}{x^2 -1}$$
Thanks!
|
Let $t=x-1$ so $x=t+1$ and since $(1+y)^\frac{1}{2}\sim_0 1+\frac{y}{2}$ and $y^2=_0o(y)$ then we find
$$\lim_{x\to 1}\frac{\sqrt{x^2+3}-2}{x^2-1}=\lim_{t\to 0}\frac{\sqrt{t^2+2t+4}-2}{t^2+2t}=\lim_{t\to 0}2\frac{\sqrt{\frac{t^2+2t}{4}+1}-1}{t^2+2t}=\lim_{t\to 0}2\frac{t/4}{2t}=\frac{1}{4}$$
|
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|
Find all $n$ such that $\sigma(n)=12$ Let $ \sigma (n) = \sum_{k|n}^{}{k} $. I need to solve $\sigma(n)=12$.
Probably the following might be of use: if $n={p_1}^{a_1}{p_2}^{a_2}...{p_s}^{a_s}$ then $\sigma(n)=\frac{{p_1}^{a_1+1}-1}{p_1-1}\frac{{p_2}^{a_2+1}-1}{p_2-1}...\frac{{p_s}^{a_s+1}-1}{p_s-1}$.
|
Note that
$$
\begin{align}
\sigma (1) &= 1\\
\sigma (2) &= 1+2 = 3\\
\sigma (3) &= 1+3 = 4\\
\sigma (4) &= 1+2+4 = 7\\
\sigma (5) &= 1+5 = 6\\
\sigma (6) &= 1+2+3+6 = 12\\
\sigma (7) &= 1+7 = 8\\
\sigma (8) &= 1+2+4+8 = 15\\
\sigma (9) &= 1+3+9 = 13\\
\sigma (10) &= 1+2+5+10 = 18\\
\sigma (11) &= 12
\end{align}
$$
so the only solutions are $n = 6$ and $n = 11$.
|
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Probability and arithmetic sequence question Three numbers are chosen at random from $1,2,...2n$ with $n > 1$. Show that the probability that the numbers are in
A.P. is $\frac{3}{4n-2}$
I don't know how to deal with this. Please help.
|
The total number of $3$-number combinations from $1$ to $2n$ is $\binom{2n}{3}$. Now consider $1$ as the first taken number. Then, the $3$-number A.P. with the greatest common difference will occur if the then highest chosen number is $2n-1$ as common difference is $\dfrac{(2n-1)-(1)}{2}=n-1$. Hence the combination of numbers in A.P. with greatest common difference and with $1$ chosen as one of the numbers is $\{1,n,2n-1\}$. Now we start to narrow our common difference until it becomes one and notice that with $1$ as the first chosen number, there are a total of $n-1$ combinations of $3$ numbers forming an A.P. Repeat the process for $2$ and you'll notice that the number of combinations of $3$ numbers forming an A.P. is again $n-1$. However for $3$ it becomes $n-2$. For any $k$ chosen as the first number, there are $n-1-\left\lfloor \frac{k-1}{2}\right\rfloor$ such possible combinations. Sum it all up to $2n-2$ (since the last choice for $k$ is $2n-2$ for which there is only one A.P. combination, namely $\{2n-2,2n-1,2n\}$), you get $\displaystyle\sum_{k=1}^{2n-2}n-1-\left\lfloor \frac{k-1}{2}\right\rfloor=n(n-1)$.
Then
\begin{align}
Pr&=\frac{n(n-1)}{\binom{2n}{3}}\\&=\frac{n(n-1)}{\frac{2n!}{3!(2n-3)!}}\\&=\frac{3!n(n-1)}{2n(2n-1)(2n-2)}\\&=\frac{3}{4n-2}
\end{align}
|
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|
Distributing two distinct objects to identical boxes Hiii,
I've been struck with a problem which deals with the distribution of two distinct objects such that p of one type and q of other type into three identical boxes.
As if it were only one object with q copies i'd have used integer partitioning, and if all objects were distinct then could use Stirling Number of second kind. But what about this mixed case...?
Thnx!
|
For the sake of completeness I am posting an answer using the Polya Enumeration Theorem. As was pointed out, the group acting on the boxes is $S_3$ with cycle index
$$Z(S_3) = \frac{1}{6} a_1^3 + \frac{1}{2} a_1 a_2 + \frac{1}{3} a_3.$$
The pattern repertoire is given by
$$ \frac{1}{1-X}\frac{1}{1-Y}.$$
Substituting into the cycle index it now follows that the desired value is
$$ [X^p Y^q] \left(
\frac{1}{6} \frac{1}{(1-X)^3}\frac{1}{(1-Y)^3} +
\frac{1}{2} \frac{1}{1-X}\frac{1}{1-Y} \frac{1}{1-X^2}\frac{1}{1-Y^2} +
\frac{1}{3} \frac{1}{1-X^3}\frac{1}{1-Y^3} \right).$$
Now the first term (see the formula at the end) contributes
$$ \frac{1}{6} {p+2 \choose 2} {q+2 \choose 2}.$$
The second term combines the initial segment of the series for $1/(1-X^2)$ of degree at most $p$ with a term from $1/(1-X)$ and the initial segment of the series for $1/(1-Y^2)$ of degree at most $q$ with a term from $1/(1-Y)$ and hence contributes
$$ \frac{1}{2}
\left(\sum_{m=0}^{\lfloor p/2\rfloor} 1 \right)
\left(\sum_{m=0}^{\lfloor q/2\rfloor} 1 \right) =
\frac{1}{2} \lfloor (p+2)/2\rfloor \lfloor (q+2)/2\rfloor$$
and the third term contributes $1/3$ if $p$ and $q$ are both divisible by three because $1/(1-X^3)$ only contains terms of degree divisible by three and $1/(1-Y^3)$ also only contains terms of degree divisible by three.
This confirms the first answer that was posted, namely
$$\frac{1}{6} {p+2 \choose 2} {q+2 \choose 2}
+ \frac{1}{2} \lfloor (p+2)/2\rfloor \lfloor (q+2)/2\rfloor
+ \frac{1}{3} [[p\equiv 0 (3); q\equiv 0 (3)]].$$
Here we have made use several times of the fact that
$$[z^k] \frac{1}{(1-z)^m} = {k+m-1\choose m-1}.$$
|
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How to calculate $ \int_{0}^{\infty} \frac{ x^2 \log(x) }{1 + x^4} $? I would like to calculate $$\int_{0}^{\infty} \frac{ x^2 \log(x) }{1 + x^4}$$ by means of the Residue Theorem. This is what I tried so far: We can define a path $\alpha$ that consists of half a half-circle part ($\alpha_r$) and a path connecting the first and last point of that half circle (with radius $r$) so that we have $$ \int_{-r}^{r} f(x) dx + \int_{\alpha_r} f(z) dz = \int_{\alpha} f(z) dz = 2 \pi i \sum_{v = 1}^{k} \text{Res}(f;a_v) $$ where $a_v$ are zeros of the function $\frac{x^2 \log(x) }{1+x^4}$.
If we know $$\lim_{r \to \infty} \int_{\alpha_r} f(z) dz = 0 \tag{*} $$ then we know that $$\lim_{r \to \infty} \int_{-r}^{r} f(x) dx = \int_{-\infty}^{\infty} f(x) dx = 2 \pi i \sum_{v=1}^{k} \text{Res}(f;a_v) $$ and it becomes 'easy'.
Q: How do we know (*) is true?
|
$$\underbrace{\dfrac{x^2 \log(x)}{1+x^4} dx \to \dfrac1{x^2} \dfrac{\log(1/x)}{1+1/x^4} \dfrac{-dx}{x^2}}_{x \to 1/x}$$
Hence,
\begin{align}
I = \int_0^{\infty} \dfrac{x^2 \log(x)}{1+x^4} dx & = \int_0^1 \dfrac{x^2 \log(x)}{1+x^4} dx + \int_1^{\infty} \dfrac{x^2\log(x)}{1+x^4}dx\\
& = \int_0^1 \dfrac{x^2 \log(x)}{1+x^4} dx - \int_0^1 \dfrac{\log(x)}{1+x^4}dx\\
& = \int_0^1 \dfrac{x^2-1}{1+x^4} \log(x) dx
\end{align}
\begin{align}
\int_0^1 \dfrac{x^2 \log(x)}{1+x^4} dx & = \int_0^1 \sum_{k=0}^{\infty}(-1)^k x^{4k+2} \log(x)dx\\
& = \sum_{k=0}^{\infty}(-1)^k \int_{0}^{\infty} x^{4k+2} \log(x)dx\\
& = \sum_{k=0}^{\infty} \dfrac{(-1)^{k+1}}{(4k+3)^2}
\end{align}
\begin{align}
\int_0^1 \dfrac{\log(x)}{1+x^4} dx & = \int_0^1 \sum_{k=0}^{\infty}(-1)^k x^{4k} \log(x)dx\\
& = \sum_{k=0}^{\infty}(-1)^k \int_{0}^{\infty} x^{4k} \log(x)dx\\
& = \sum_{k=0}^{\infty} \dfrac{(-1)^{k+1}}{(4k+1)^2}
\end{align}
$$I = \sum_{k=0}^{\infty} (-1)^k \left(\dfrac1{(4k+1)^2} - \dfrac1{(4k+3)^2}\right) = \dfrac{\pi^2}{8\sqrt2}$$
which can be obtained from the identity that the PolyLogarithmic function satisfies
$$\text{Li}_n(e^{2 \pi ix}) + (-1)^n \text{Li}_n(e^{-2 \pi ix}) = - \dfrac{(2\pi i)^n}{n!}B_n(x)$$
You can see the posts here and here for more details.
|
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|
How to determine whether this series convergent or divergent? Does
$$
\dfrac{7}{19}+\dfrac{7}{19}\sqrt{\dfrac{7}{19}}+\dfrac{7}{19}\sqrt{\dfrac{7}{19}}\sqrt[3]{\dfrac{7}{19}}+\cdots+\dfrac{7}{19}\sqrt{\dfrac{7}{19}}\sqrt[3]{\dfrac{7}{19}}\cdots\sqrt[n]{\dfrac{7}{19}}+\cdots
$$
converge or diverge?
The following is my idea:
use
$1+\dfrac{1}{2}+\dfrac{1}{3}+\cdots+\dfrac{1}{n}>\ln{n}$
$$
\sum_{n=1}^{\infty}\left(\dfrac{7}{19}\right)^{(1+1/2+\cdots+1/n)}
< \sum_{n=1}^{\infty}\left(\dfrac{7}{19}\right)^{\ln{n}}
= \sum_{n=1}^{\infty}n^{-\ln(19/7)}
$$
But $p=\ln{\dfrac{19}{7}}<1$, becasue $\dfrac{19}{7}\approx 2.71428<e=2.71828$
I guess the series is divergent because I use
$$1+1/2+\cdots+1/n\approx \ln{n}, n\to\infty$$
to find $$\sum_{n=1}^{\infty} (1/x)^{1+1/2+1/3+\cdots+1/n}$$ is convergent only if $x>e$.
So, my question is: how do I determine whether
$$
\dfrac{7}{19}+\dfrac{7}{19}\sqrt{\dfrac{7}{19}}+\dfrac{7}{19}\sqrt{\dfrac{7}{19}}\sqrt[3]{\dfrac{7}{19}}+\cdots+\dfrac{7}{19}\sqrt{\dfrac{7}{19}}\sqrt[3]{\dfrac{7}{19}}\cdots\sqrt[n]{\dfrac{7}{19}}+\cdots
$$
converges or diverges? Thank you
|
It can be used Raabe's test.
We have series
$$\sum_{n=1}^{\infty} a_n, \qquad \mathrm{where } \quad
a_n = \left(\dfrac{7}{19}\right)^{1+\frac{1}{2}+\cdots+\frac{1}{n}}.
$$
We wil construct value $R_n= n \left( \dfrac{a_{n+1}}{a_n}-1\right)$. Denote
$R=\lim\limits_{n\to\infty} R_n$.
If $R<-1$, then series converges.
If $R>-1$, then series diverges.
$R_n = n\left(\left(\dfrac{7}{19}\right)^{\frac{1}{n+1}}-1\right)= n\left( \exp(\frac{1}{n+1}\ln\frac{7}{19})-1\right)$.
Using Taylor series for $\exp$, we get
$$R_n = n \sum_{j=1}^{\infty} \dfrac{ \left(\frac{1}{n+1} \ln\frac{7}{19}\right)^j }{j!}.$$
So, $R=\lim\limits_{n\to\infty} R_n = \ln\frac{7}{19} >\ln \frac{1}{e}=-1$. (because
$\frac{19}{7}<e$).
Actually, $R \approx -0,99852883011112715490367468844.\;$
So, series diverges.
|
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|
How can evaluate $\lim_{x\to0}\frac{\sin(x^2+\frac{1}{x})-\sin\frac{1}{x}}{x}$ I don't know if I apply for this case sin (a-b), or if it is the case of another type of resolution, someone with some idea without using derivation or L'Hôpital's rule? Thank you.
$$\lim_{x\to0}\frac{\sin(x^2+\frac{1}{x})-\sin\frac{1}{x}}{x}$$
|
Perhaps not an elegant proof but I considered this.
Use a Taylor series for $\sin$.
$\sin(x)=x+ a_1x^3 + a_2x^5 + ...$
Let $A$ be the value of the limit.
Then we get
$A=\dfrac{x^2+\frac{1}{x}-\frac{1}{x}+a_1(x^2+\frac{1}{x})^3-a_1(\frac{1}{x})^3+\ldots}{x} = \dfrac{x^2+a_1(x^2+\frac{1}{x})^3-a_1(\frac{1}{x})^3)+\ldots}{x}$.
This equals $A =\dfrac{x^2}{x}+\dfrac{a_1((x^3+1)^3-1)}{x^4}+\dfrac{a_2((x^3+1)^5-1)}{x^6}+\ldots$
Now use big $O$ notation to rewrite as follows : $A=\dfrac{O(x^2)}{x}+\dfrac{a_1O(x^9)}{x^4}+\dfrac{a_2O(x^{15})}{x^6}+\ldots$ hence $A=0 + a_1 0+a_20 + \ldots =0$. To justify that infinite sum notice that each limit $O(x^a)/x^b$ goes faster to $0$ than $x$ does.
|
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|
intersection of two subsets of vectors and its geometrical meaning E and F is a subset of $ \mathbb R^3$.
How to compute $ P = E \cap F $ ?
What is the geometrical meaning of E, F and P ?
$ E =\left( \begin{array}{ccc}
1 \\
2 \\
-3 \end{array} \right)+\alpha\left( \begin{array}{ccc}
1 \\
0 \\
1 \end{array} \right)+\beta \left( \begin{array}{ccc}
1 \\
1 \\
1 \end{array} \right) $
$ F =\left( \begin{array}{ccc}
-2 \\
1 \\
4 \end{array} \right)+\gamma\left( \begin{array}{ccc}
0 \\
1 \\
1 \end{array} \right)+\delta \left( \begin{array}{ccc}
-1 \\
3 \\
2 \end{array} \right) $
|
To find the intersection, the analytic way would be to write out both planes as equations of the form $ax + by + cz + d = 0$, equate the left hand sides of each plane's equations with each other, and solve for x, y, and z.
|
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|
$a+b+c =0$; $a^2+b^2+c^2=1$. Prove that: $a^2 b^2 c^2 \le \frac{1}{54}$ If a,b,c are real numbers satisfying $a+b+c =0; a^2+b^2+c^2=1$.
Prove that $a^2 b^2 c^2 \le \frac{1}{54}$.
|
There are three real numbers, and only two sign(negative, non-negative),So there are two real numbers with same sign,then without loss generality, Assume ab $\ge$ 0
$1 = a^2+b^2+c^2 = a^2+b^2+(-a-b)^2 = (a^2+b^2+ab) * 2 ≥ 3*ab * 2$
So
$ab ≤ \dfrac{1}{6}$
$a^2 b^2 c^2 = a^2b^2(a^2+b^2+2ab)=a^2b^2(\dfrac{1}{2}+ab)≤\left(\dfrac{1}{6} \right)^2 *(\dfrac{1}{2}+\dfrac{1}{6})=\dfrac{1}{54}$
Hi, does someone give me any tips about expression format?
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|
Limit involving power tower: $\lim\limits_{n\to\infty} \frac{n+1}n^{\frac n{n-1}^\cdots}$ What is the value of the following limit?
$$\large \lim_{n \to \infty} \left(\frac{n+1}{n}\right)^{\frac{n}{n-1}^{\frac{n-1}{n-2}^{...}}}$$ In general what do limits of infinite decreasing numbers strung together in familiar ways approach?
|
For $n\ge 1$, let $$a_n=\left(\frac{n+1}{n}\right)^{\frac{n}{n-1}^{\frac{n-1}{n-2}^{...}}}$$ where the tower stops when we reach $2/1=2$, so $a_1=2$ and $$a_{n+1}=\left(\frac{n+2}{n+1}\right)^{a_n}.$$
Note that each $a_n$ is (strictly) larger than $1$, and that the sequence is decreasing from $n=2$ on: First, $a_2=9/4>2>1.911>a_3$. Next, if $a_{n+1}<a_n$, then $$a_{n+2}=\left(\frac{n+3}{n+2}\right)^{a_{n+1}}<\left(\frac{n+2}{n+1}\right)^{a_{n+1}}<\left(\frac{n+2}{n+1}\right)^{a_n}=a_{n+1}. $$
Since the sequence is decreasing and bounded below by $1$, it follows that $\lim_n a_n=a$ exists, and satisfies $a\ge 1$.
It remains to argue that $a=1$. To see this, note that $a_{n+1}=b_n^{a_n}$, where $\displaystyle b_n=\frac{n+2}{n+1}$, so $a=\lim_n a_{n+1}=(\lim_n b_n)^{\lim_n a_n}=1^a=1$.
|
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|
Orthogonal basis for this indefinite symmetric bilinear form Find an orthogonal basis for the bilinear form over $\mathbb{R}$ given by $(\mathbf{x}, \mathbf{y})\mapsto\mathbf{x}^{t}A\mathbf{y}$ where $A=\begin{bmatrix} 1 & 4 & 4\\ 4 & 4 & 10\\ 4 & 10 & 16 \end{bmatrix}$.
I'm not sure if this is as easy as using Gram-Schmidt, or if there is another way. I used Gram-Schmidt and obtained very complicated fractions in my vectors, so I have a feeling that this is wrong...
|
PREFACE: this is called Hermite reduction. It is, for example, Theorem 23 in The Arithmetic Theory of Quadratic Forms by Burton W. Jones, pages 56-59 primarily. On page 58 you clearly see an upper triangular matrix as the change of variables. The W stands for Wadsworth.
I get
$$ x^2 + 4 y^2 + 16 z^2 + 20 y z + 8 z x + 8 x y = (x + 4 y + 4 z)^2 - 12 \left( y + \frac{z}{2} \right)^2 + 3 z^2, $$ so
$$ x^2 + 4 y^2 + 16 z^2 + 20 y z + 8 z x + 8 x y = (x + 4 y + 4 z)^2 - 3 \left( 2y + z \right)^2 + 3 z^2. $$
Note that this was very, very little work. Nowhere near finding eigenvectors.
This is called Hermite reduction.
The relevant matrix equation is
$$
\left( \begin{array}{ccc}
1 & 0 & 0 \\
4 & 2 & 0 \\
4 & 1 & 1
\end{array}
\right)
\left( \begin{array}{ccc}
1 & 0 & 0 \\
0 & -3 & 0 \\
0 & 0 & 3
\end{array}
\right)
\left( \begin{array}{ccc}
1 & 4 & 4 \\
0 & 2 & 1 \\
0 & 0 & 1
\end{array}
\right)
=
\left( \begin{array}{ccc}
1 & 4 & 4 \\
4 & 4 & 10 \\
4 & 10 & 16
\end{array}
\right).
$$
Either that, or they really want the inverses, as in
$$
\left( \begin{array}{ccc}
2 & 0 & 0 \\
-4 & 1 & 0 \\
-4 & -1 & 2
\end{array}
\right)
\left( \begin{array}{ccc}
1 & 4 & 4 \\
4 & 4 & 10 \\
4 & 10 & 16
\end{array}
\right)
\left( \begin{array}{ccc}
2 & -4 & -4 \\
0 & 1 & -1 \\
0 & 0 & 2
\end{array}
\right)
=
\left( \begin{array}{ccc}
4 & 0 & 0 \\
0 & -12 & 0 \\
0 & 0 & 12
\end{array}
\right).
$$
Either way, less work than eigenvectors.
EDDDIITTTT: @copper.hat has confirmed that the eigenvalues are horrible, I get the characteristic polynomial as $$ \lambda^3 - 21 \lambda^2 - 48 \lambda + 36 $$ with three irrational real roots, so this is the Casus Irrededucibilis and so the eigenvectors are horrible. Which tells me they really want Hermite.
=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=
=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=
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A question on differentiation : Let $$f(x)=\sin^{-1}(2x\sqrt{1-x^2})$$
I found out $f'(x)$ in three methods and got three different answers !
1) Putting $x=\cos\theta$, we get $f(x)=2\cos^{-1}x$, on differentiating this we get
$$f'(x)=\frac{-2}{\sqrt{1-x^2}}$$
2) Putting $x=\sin\theta$, we get $f(x)=2\sin^{-1}x$, on differentiating we get
$$f'(x)=\frac{2}{\sqrt{1-x^2}}$$
This is contradicting with the previous result that we arrived at by substituting $x=\cos\theta$. Can someone help me out with the correct solution for this problem!
|
HINT:
As the principal value of $\sin^{-1}x$ lies $\in [-\frac\pi2, \frac\pi2]; \pi\le 2\sin^{-1}x\le \pi$
So, $$\sin^{-1}(2x\sqrt{1-x^2})=\begin{cases}
2\sin^{-1}x &\mbox{if } -\frac\pi4\le \sin^{-1}x\le \frac\pi4 \\
-\pi-2\sin^{-1}x & \mbox{if } -\frac\pi2\le \sin^{-1}x< -\frac\pi4 \\
\pi-2\sin^{-1}x & \mbox{if } \frac\pi4< \sin^{-1}x\le \frac\pi2 \end{cases} $$
As the principal values of $\cos^{-1}x$ lies $\in [0,\pi], 0\le 2\cos^{-1}x\le 2\pi,$
So, $$\sin^{-1}(2x\sqrt{1-x^2})=\begin{cases} 2\cos^{-1}x &\mbox{if } 0\le \cos^{-1}x< \frac\pi4 \iff \frac\pi4< \sin^{-1}x\le \frac\pi2 \\
2\cos^{-1}x-2\pi & \mbox{if } \frac\pi4\le \cos^{-1}x< \frac{3\pi}4\iff -\frac\pi2\le \sin^{-1}x< -\frac\pi4\\
\pi-2\cos^{-1}x & \mbox{if } \frac{3\pi}4\le \cos^{-1}x\le \frac\pi2\iff -\frac\pi4\le \sin^{-1}x\le \frac\pi4 \end{cases} $$
Now differentiate for each region
|
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|
Trigonometric identity: $\frac {\tan\theta}{1-\cot\theta}+\frac {\cot\theta}{1-\tan\theta} =1+\sec\theta\cdot\csc\theta$ I have to prove the following result :
$$\frac {\tan\theta}{1-\cot\theta}+\frac {\cot\theta}{1-\tan\theta} =1+\sec\theta\cdot\csc\theta$$
I tried converting $\tan\theta$ & $\cot\theta$ into $\cos\theta$ and $\sin\theta$.
That led to a huge expression which I wasn't able to simplify.
Please help!!!
|
You are on the right track.
writing $\tan\theta$ as$ \dfrac {\sin\theta}{\cos\theta}$ and $\cot\theta$ as $ \dfrac {\cos\theta}{\sin\theta} $, we get
$ \dfrac {\frac {\sin\theta}{\cos\theta} }{1-\frac {\cos\theta}{\sin\theta} }+\dfrac {\frac {\cos\theta}{\sin\theta} }{1-\frac {\sin\theta}{\cos\theta} }$
$= \dfrac {\sin^2\theta}{cos\theta\cdot(\sin\theta-\cos\theta)} + \dfrac {\cos^2\theta}{\sin\theta\cdot(\cos\theta-\sin\theta)}$ (how?)
$= \dfrac {\sin^2\theta}{\cos\theta\cdot(\sin\theta-\cos\theta)} - \dfrac {\cos^2\theta}{\sin\theta\cdot(\sin\theta-\cos\theta)}$
$=\dfrac{1}{(\sin\theta-\cos\theta)}\big(\dfrac {\sin^2\theta}{\cos\theta}-\dfrac {\cos^2\theta}{\sin\theta})$
$=\dfrac{1}{(\sin\theta-\cos\theta)}\big(\dfrac {\sin^3\theta-\cos^3\theta}{\sin\theta\cdot\cos\theta})$
$=\dfrac{\sin\theta-\cos\theta}{\sin\theta-\cos\theta}\dfrac{\big(\sin^2\theta+\sin\theta\cdot\cos\theta+\cos^2\theta)}{\sin\theta\cdot\cos\theta}$(how?)
$=1\cdot \dfrac{1+\sin\theta\cdot\cos\theta}{\sin\theta\cdot\cos\theta}$ (why?)
which is
$1+\sec\theta\cdot\csc\theta$
QED.
|
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|
Prove that $\sqrt{7}^{\sqrt{8}}>\sqrt{8}^{\sqrt{7}}$ show that
$$\sqrt{7}^{\sqrt{8}}>\sqrt{8}^{\sqrt{7}}$$
and I found
$$LHs-RHS=0.017\cdots$$
I have post this interesting problem Prove $\left(\frac{2}{5}\right)^{\frac{2}{5}}<\ln{2}$
can someone suggest any other nice method? Thank you everyone.
|
A solution from a greek math forum.
$\frac{lne^{2}-ln7}{e^{2}-7} < \frac{1}{7} \Leftrightarrow \frac{1}{ln7} < \frac{7}{21-e^{2}}<\frac{7}{13,6}$
$\frac{ln8-lne^{2}}{8-e^{2}} >\frac{1}{8} \Leftrightarrow \frac{1}{ln8} < \frac{8}{24-e^{2}}<\frac{8}{16,6}$
So,
$ \frac{1}{ln7}+ \frac{1}{ln8} < \frac{7}{13,6}+ \frac{8}{16,6}= \frac{225}{225,76}<1$.
$\int_{ln7}^{ln8} \frac{1}{x} dx < \frac{(\frac{1}{ln7}+\frac{1}{ln8}) \cdot (ln8-ln7)}{2} < ln \sqrt{8}-ln\sqrt{7}=\int_{\sqrt{7}}^{\sqrt{8}} \frac{1}{x} dx$.
So,
$\int_{ln7}^{ln8} \frac{1}{x} dx < \int_{\sqrt{7}}^{\sqrt{8}} \frac{1}{x} dx\Leftrightarrow $
$ln\frac{ln8}{ln7} < ln \frac{\sqrt{8}}{\sqrt{7}} \Leftrightarrow \sqrt{8}ln7 > \sqrt{7} ln8 \Leftrightarrow$
$ \sqrt{7}^{\sqrt{8}}> \sqrt{8}^{\sqrt{7}}$.
|
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|
Evaluating $\int{{x^2 -1}\over{x^3 \sqrt{2x^4-2x^2+1}}} \mathrm dx$ How to evaluate:
$$\int{{x^2 -1}\over{x^3 \sqrt{2x^4-2x^2+1}}} \mathrm dx$$
|
HINT:
First put $x^2=y$ in
$$\int{{x^2 -1}\over{x^3 \sqrt{2x^4-2x^2+1}}} \mathrm dx=\int{{x^2 -1}\over{2x^4 \sqrt{2x^4-2x^2+1}}} \mathrm 2xdx$$
to get $$\int \frac{y-1}{2y^2\sqrt{2y^2-2y+1}}dy$$
Now $2y^2-2y+1=\frac{4y^2-4y+2}2=\frac{(2y-1)^2+1}2$
Put $2y-1=\tan\theta$
|
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|
If $a,b$ are roots for $x^2+3x+1=0$.Calculating $(\frac{a}{b+1})^2 +(\frac{b}{a+1})^2$ If $a,b$ are roots for the equation $x^2+3x+1=0$.How to calculate $$\left(\frac{a}{b+1}\right)^2 +\left(\frac{b}{a+1}\right)^2$$
|
We note that the expression given is a symmetric rational function of the roots of the original polynomial, so can be expressed in terms of the coefficients of the original polynomial. We then use the basic facts we know about the sum and product of the roots, and the fact that both $a$ and $b$ satisfy the polynomial to achieve successive simplification of otherwise unwieldy terms. Here are some hints for a way through.
Note that $a+b=-3$ and $ab=1$ so that $(a+1)(b+1)=ab+(a+b)+1=-3+2=-1$
Use this to put the whole thing over a common denominator and simplify.
Note also that $a^2(a+1)^2=a^2(a^2+2a+1)=a^2(a^2+3a+1-a)=-a^3$
You might also need to use $a^2+b^2=(a+b)^2-2ab$
Since the full answer closest in spirit to this has been deleted, note first $(a^2+b^2) = (-3)^2-2=7$. Then put the desired expression over common denominator $(a+1)^2(b+1)^2=1$ to obtain:$$a^2(a+1)^2+b^2(b+1)^2=-a^3-b^3=3(a^2+b^2)+(a+b)=21-3=18$$ [see comment for middle step]
|
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|
Find $S$ where $S=\sqrt[3] {5+2 \sqrt {13}}+\sqrt[3]{5-2 \sqrt {13}}$, why am I getting an imaginary number? $\large S=\sqrt[3] {5+2 \sqrt {13}}+\sqrt[3]{5-2 \sqrt {13}}$
Multiplying by conjugate:
$\large S=\dfrac {-3}{\sqrt[3] {5+2 \sqrt {13}}-\sqrt[3]{5-2 \sqrt {13}}}$
From the original:
$\large S-2\sqrt[3]{5-2 \sqrt {13}}
=\sqrt[3] {5+2 \sqrt {13}}-\sqrt[3]{5-2 \sqrt {13}}$
Substituting:
$\large S=\dfrac{-3}{S-2\sqrt[3]{5-2 \sqrt {13}}}$
This leads to a quadratic equation in $\large S$ which I checked in wolframalpha and I got imaginary solutions. Why does this happen? I am not looking for an answer telling me how to solve this problem, I just want to know why this is wrong. Thanks.
|
You are not right: the conjugate to $S$ is
$$
\left[\sqrt[3] {5+2 \sqrt {13}}\right]^2-\sqrt[3]{5+2 \sqrt {13}}\sqrt[3] {5-2 \sqrt {13}}+\left[\sqrt[3]{5-2 \sqrt {13}}\right]^2.
$$
|
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Proving the inequality $\frac{a^3}{b^2-bc+c^2}+\frac{b^3}{a^2-ac+c^2}+\frac{c^3}{a^2-ab+b^2}\geq a+b+c$ I am trying to prove the following inequality
For all positive numbers $a$, $b$ and $c$ we have
$$\dfrac{a^3}{b^2-bc+c^2}+\dfrac{b^3}{a^2-ac+c^2}+\dfrac{c^3}{a^2-ab+b^2}\geq a+b+c$$
I can probably solve this by reducing it to Schur's inequality.
However, is there any other method?
|
$\dfrac{a^3}{b^2-bc+c^2}+\dfrac{b^3}{a^2-ac+c^2}+\dfrac{c^3}{a^2-ab+b^2}\geq a+b+c \iff \dfrac{a^3(b+c)}{b^3+c^3}+\dfrac{b^3(a+c)}{a^3+c^3}+\dfrac{c^3(b+a)}{b^3+a^3} \geq a+b+c \iff a^9b+a^9c+b^9a+b^9c+c^9a+c^9b \ge a^7b^3+a^7c^3+b^7a^3+b^7c^3+c^7a^3+c^7b^3$
lemma: $a^n+b^n\ge a^{n-1}b+b^{n-1}a \ge a^{n-2}b^2+b^{n-2}a^2 \ge ...$
$a^n+b^n\ge a^{n-1}b+b^{n-1}a \iff (a-b)(a^{n-1}-b^{n-1})\ge 0 \iff (a-b)^2(a^{n-2}+a^{n-3}b+a^{n-4}b^2+....ab^{n-3}+b^{n-2}) \ge 0$
which is true ,when $a=b$ the "=" holds.
so we have $a^9b+b^9a \ge a^8b^2+a^8b^2 \ge a^7b^3+b^7a^3$
QED.
|
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Finding the value of $\sqrt{1+2\sqrt{2+3\sqrt{3+4\sqrt{4+5\sqrt{5+\dots}}}}}$ Is it possible to find the value of
$$\sqrt{1+2\sqrt{2+3\sqrt{3+4\sqrt{4+5\sqrt{5+\dots}}}}}$$
Does it help if I set it equal to $x$? Or I mean what can I possibly do?
$$x=\sqrt{1+2\sqrt{2+3\sqrt{3+4\sqrt{4+5\sqrt{5+\dots}}}}}$$
$$x^2=1+2\sqrt{2+3\sqrt{3+4\sqrt{4+5\sqrt{5+\dots}}}}$$
$$x^2-1=2\sqrt{2+3\sqrt{3+4\sqrt{4+5\sqrt{5+\dots}}}}$$
$$\frac{x^2-1}{2}=\sqrt{2+3\sqrt{3+4\sqrt{4+5\sqrt{5+\dots}}}}$$
$$\left(\frac{x^2-1}{2}\right)^2=2+3\sqrt{3+4\sqrt{4+5\sqrt{5+\dots}}}$$
$$\left(\frac{x^2-1}{2}\right)^2-2=3\sqrt{3+4\sqrt{4+5\sqrt{5+\dots}}}$$
$$\vdots$$
I don't see it's going anywhere. Help appreciated!
|
In the spirit of Ramanujan, let
$$G(x)=\sqrt{x-1+x\sqrt{(x+n-1)+(x+n)\sqrt{(x+2n-1)+(x+2n)\sqrt{\ldots}}}}$$
and note that by setting $x=2$ and $n=1$ we recover our nested radical. After squaring,
$$G(x)^{2}=(x-1)+xG(x+n)$$
This is a slightly more complicated functional than Ramanujan encountered and I'm not sure how to solve it, but hopefully someone else can provide more insight.
|
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How to solve these three equations? If α ,β ,γ are three numbers s.t.:
$\ α^ \ $ + $\ β \ $ + $ γ \ $ = −2
$\ α^2 \ $ + $\ β^2 \ $ + $ γ^2 \ $ = 6
$\ α^3 \ $ + $\ β^3 \ $ + $ γ^3 \ $ = −5,
then $\ α^4 \ $ + $\ β^4 \ $ + $ γ^4 \ $ is equal to ??
I tried out substituting the values of each equation to one other ...but it became very
complex .. I also remember some crammers rule for this ..using matrices??
Is that the way??
|
Let $A_{n}=a^n+b^n+c^n$. Then we have
$$A_{n+3}=(a+b+c)A_{n+2}-(ab+bc+ac)A_{n+1}+abcA_{n}$$
and
$$2(ab+bc+ac)=(a+b+c)^2-(a^2+b^2+c^2)=4-6=-2$$
$$a^3+b^3+c^3=(a+b+c)^3-3(a+b+c)(ab+bc+ac)+3abc$$
Then we can easily find $abc$.
|
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|
show that the function $z = 2x^2 + y^2 +2xy -2x +2y +2$ is greater than $-3$ Show that the function
$$z = 2x^2 + y^2 +2xy -2x +2y +2$$
is greater than $-3$
I tried to factorize but couldn't get more than $(x-1)^2 + (x+y)^2 +(y-1)^2 - (y)^2$.
Is there any another way to factorize or another method??
|
Here is a way of doing it that requires no previous knowledge of 3D analytic geometry or calculus:
You are trying to show $z = 2x^2+y^2+2xy-2x+2y+2 \gt -3$, or, (more usefully) that $z+3 = 2x^2+y^2+2xy-2x+2y+5 \gt 0$
$$z+3 = 2x^2+y^2+2xy-2x+2y+5 \\= 2x^2+(2y-2)x+(y^2+2y+5)$$
Now we can use the quadratic formula on it as if it were just a 2nd degree polynomial in $x$.
$$x = \frac{2-2y\pm \sqrt{4(y-1)^2 - 8(y^2-2y+5)}}{2} = 1-y\pm(y+3)i$$
This is clearly only a real value when $y=-3$, so a point of intersection will only occur if $y=-3$. Another interpretation is that the vertex of the elliptic paraboloid is at $y=-3$.
We now substitute $y=-3$ into our original equation to find the parabola that represents the minimum value of the function WRT $y$:
$$z+3 = 2x^2+y^2+2xy-2x+2y+5 = 2x^2 - 8x + 8$$
We find that this has a double-root at $x=2$ (can be obtained by using quadratic formula again, although its much simpler this time).
So, we have determined that the vertex of this elliptic paraboloid is at $(x,y) = (2, -3)$. We can now evaluate the function at this point: $z+3 = 2x^2+y^2+2xy-2x+2y+5 = 0$, or $z=-3$. Clearly, the original proposition, $z>-3$, is not correct.
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{
"language": "en",
"url": "https://math.stackexchange.com/questions/436679",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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|
Factor Equation Help me with this,
Question: factor $x^3y-x^3z+y^3z-xy^3+xz^3-yz^3$.
Solution:
$$\begin{eqnarray}&=&x^3y-x^3z+y^3z-xy^3+xz^3-yz^3\\
&=&x\left(z^3-y^3\right)+y\left(x^3-z^3\right)+z\left(y^3-x^3\right)\\
&=&x\left[(z-y)\left(z^2+zy+y^2\right)\right]+y\left[(x-z)\left(x^2+xz+z^2\right)\right]+z\left[(y-x)\left(y^2+xy+x^2\right)\right]\end{eqnarray}$$
This expression is quite simple at first glance, but I stuck up again in that line. I appreciate any help.
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If $E(x,y,z)=x^3y-x^3z+y^3z-xy^3+xz^3-yz^3=x^3(y-z)+y^3(z-x)+z^3(x-y)$
Observe that $E=0$ if $x=y\implies x-y$ divides $f$
Similarly, $y-z,z-x$ divide $f$ $\implies E=(x-y)(y-z)(z-x)F$
As $E$ is Alternating polynomial wrt $x,y,z$
and "the ratio of two alternating polynomials is a symmetric function" (Source : wiki, Proof Article# $4$ of this)
$F$ must be Symmetric polynomial of degree $1$ in $x,y,z$
So, $E=(x-y)(y-z)(z-x)\cdot k(x+y+z)$ where $k$ is independent of $x,y,z$
Comparing the coefficients of $x^3y, 1=-k\implies k=-1$
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{
"language": "en",
"url": "https://math.stackexchange.com/questions/438088",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
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|
$a$ and $b$ are the roots of quadratic equation $x^2 -2cx-5d=0$ and $c$ and $d$ are the roots of quadratic equation $x^2 -2ax-5b=0 $ Let $a,\,b,\,c,\,d$ be distinct real numbers and $a$ and $b$ are the roots of quadratic equation $x^2 -2cx-5d=0$ and $c$ and $d$ are the roots of quadratic equation $x^2 -2ax-5b=0$. Then find the value of $a+b+c+d$.
I could only get $2$ equations that
$a=2c-b$ and $c=2a-d$.
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Let the two polynomials be $$
p(x) = x^2 - 2cx -5d \\
q(x) = x^2 - 2ax - 5b
$$
You also know that $$
p(x) = (x-a)(x-b) = x^2 - (a+b)x + ab \\
q(x) = (x-c)(x-d) = x^2 - (c+d)x + cd
$$
You might try playing around with these two forms. For example, you can take the product of the polynomials and equate the coefficients for each power of $x$.
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{
"language": "en",
"url": "https://math.stackexchange.com/questions/438492",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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|
Prove that $n^k < 2^n$ for all large enough $n$ If $k\ge 2$ is an integer,
prove by elementary means
(no calculus or limits)
that
there is a $N(k)$
such that
$n^k < 2^n$
for all integers $n \ge N(k)$.
Give an explicit form for $N(k)$.
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Write the inequality as
$n < 2^{n/k}$.
Suppose this is true for $n$.
I want to find conditions such that
this is also true for $n+1$.
By assumption,
$n+1
= n(1+\frac1{n})
< 2^{n/k}(1+\frac1{n})
$
and this is
less than
$2^{(n+1)/k}$
if
$2^{n/k}(1+\frac1{n})
< 2^{(n+1)/k}$
or
$2^{1/k} > 1+\frac1{n}$.
I showed in another solution
(from $(1+\frac{x}{k})^k < \frac1{1-x}$
for $0 < x < 1$,with $x = \frac12$)
that
$2^{1/k} > 1+\frac1{2k}$,
so, if $n \ge 2k$
and
$2^n > n^k$,
then
$2^m > m^k$ for
all $m \ge n$.
We now need to find an initial $n \ge 2k$
such that $2^n > n^k$.
Let's try $n = ak$ for some $a$.
$2^{ak} > (ak)^k
\iff 2^a > ak
$.
Almost, but not quite.
Let's try $n = k^2$.
$2^{k^2} > (k^2)^k
\iff 2^{k} > k^2
$.
This is easy by induction.
$2^5 > 5^2$,
and, if $2^k > k^2$ and $k \ge 5$,
$$(k+1)^2 = k^2+2k+1
= k^2(1+\frac{2}{k}+\frac1{k^2})
< 2^k(1+\frac{2}{5}+\frac1{25})
< 2^k(2)
= 2^{k+1}
$$
So, by this nested induction,
if $k \ge 5$,
if $n \ge k^2$
(so that, certainly, $n \ge 2k$),
$2^n > n^k$.
For $k = 2, 3,$ and $4$,
respective values of
$n \ge 2k$ such that
$2^n > n^k$ are
$5$ ($2^5 > 5^2$),
$10$ ($2^{10} > 10^3$),
and
$17$ ($2^{17} = 131072> 17^4=83521$).
These values of $n$ happen to be
$k^2+1$,
so the final result is:
If $n$ and $k$ are integers
and $k \ge 2$
and
$n \ge k^2+1$,
then
$2^n > n^k$.
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{
"language": "en",
"url": "https://math.stackexchange.com/questions/439026",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 4,
"answer_id": 1
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|
Subsets and Splits
Fractions in Questions and Answers
The query retrieves a sample of questions and answers containing the LaTeX fraction symbol, which provides basic filtering of mathematical content but limited analytical insight.