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Prove the inequality $4S \sqrt{3}\le a^2+b^2+c^2$ Let a,b,c be the lengths of a triangle, S - the area of the triangle. Prove that
$$4S \sqrt{3}\le a^2+b^2+c^2$$
|
Let $\triangle$ denote the area.
$\triangle=\sqrt{s(s-a)(s-b)(s-c)}$
Take $b+c-a=x$
$c+a-b=y$ and $a+b-c=z$.
$\dfrac{x}{2}=(s-a)$
$\dfrac{y}{2}=(s-b)$
$\dfrac{z}{2}=(s-c)$
$\dfrac{x+y+z}{2}=s$
Now use AM-GM inequality.
$\dfrac{x+y+z}{4} \ge ((xyz)(x+y+z))^\frac{1}{4}$
$\dfrac{2}{x+y+z} \le \dfrac{1}{((xyz)(x+y+z))^\frac{1}{4}}$
$\dfrac{4}{(x+y+z)^2} \le \dfrac{1}{((xyz)(x+y+z))^\frac{1}{2}} \implies \dfrac{4}{(x+y+z)^2} \le \dfrac{1}{\triangle}$
|
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|
How can I determine limit: $\lim\limits_{x\to4}\frac{x-4}{x-\sqrt[]{x}-2}$? How can I determine limit:
$$\lim_{x\to4}\frac{x-4}{x-\sqrt[]{x}-2}$$
Thanks
|
If you are unfamiliar with L'Hopital's rule, this is what you can do.
$$ \lim_{x \rightarrow 4} \frac{ x - 4 } { x - \sqrt{x} - 2} = \lim_{x \rightarrow 4} \frac{ (\sqrt{x} - 2) ( \sqrt{x} + 2) } { (\sqrt{x} -2) ( \sqrt{x} +1) }= \lim_{x \rightarrow 4} \frac{ ( \sqrt{x} + 2) } { ( \sqrt{x} +1) } = \frac{ 4}{3}.$$
|
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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."
|
A general way to solve such problems is to set up a system of recurrences. Call $a_n$, $b_n$ and $c_n$ the number of strings of length $n$ of interest that end in 0, 1, 2 respectively. Then:
$$
\begin{align*}
a_{n + 1} &= a_n + b_n + c_n \\
b_{n + 1} &= a_n + b_n + c_n \\
c_{n + 1} &= a_n + b_n
\end{align*}
$$
It is also $a_1 = b_1 = c_1 = 1$. We are interested in $s_n = a_n + b_n + c_n$,
which coincidentally is just $a_{n + 1}$.
Define ordinary generating functions:
$A(z) = \sum_{n \ge 0} a_{n + 1} z^n$ and similarly $B(z)$, $C(z)$. By properties of ordinary generating functions:
$$
\begin{align*}
\frac{A(z) - a_1}{z} &= A(z) + B(z) + C(z) \\
\frac{B(z) - a_1}{z} &= A(z) + B(z) + C(z) \\
\frac{C(z) - a_1}{z} &= A(z) + B(z)
\end{align*}
$$
The solution to this linear system of equations is:
$$
\begin{align*}
A(z) &= \frac{1 + z}{1 - 2 z - 2 z^2} \\
B(Z) &= \frac{1 + z}{1 - 2 z - 2 x^2} \\
C(z) &= \frac{1}{1 - 2 z - 2 z^2}
\end{align*}
$$
We are interested in $a_{n +1}$. Spliting the expresison for $A(z)$ into partial fractions:
$$
A(z) = - \frac{2 - \sqrt{3}}{2 \sqrt{3}} \cdot \frac{1}{1 - (1 - \sqrt{3}) z}
+ \frac{2 + \sqrt{3}}{2 \sqrt{3}} \cdot \frac{1}{1 - (1 + \sqrt{3}) z}
$$
Two geometric series.
|
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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
|
You missed $(-1)^k\left(\frac{3}{2}\right)^{k}:$ $$f'(x)=-\frac{3}{2} \sum_{k=0}^{\infty}\left(-\frac{3}{2}x^2\right)^k=\frac{3}{2} \sum_{k=0}^{\infty}(-1)^{k+1}\left(\frac{3}{2}x^2\right)^k= \sum_{k=0}^{\infty}(-1)^{k+1}\left(\frac{3}{2}\right)^{k+1} x^2.$$
Then
$$
f(x)=\int{f'(x)\,dx}= \sum_{k=0}^{\infty}(-1)^{k+1}\left(\frac{3}{2}\right)^{k+1} \frac{x^{2k+1}}{2k+1}+C.$$
From $f(0)=\arctan(-1)=-\dfrac{\pi}{4}$ we have $C=\dfrac{\pi}{4}.$
|
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|
Evaluate $\int_0^{2\pi} \frac{d\theta}{(1-a\cos(\theta)+a^2)}$ Evaluate $\displaystyle \int_0^{2\pi} \frac{d\theta}{(1-a\cos(\theta)+a^2)}$
Super general. I get to a step: $\displaystyle \frac{2}{i}$ multiplied by Path integral $\displaystyle \frac{z}{[(2-a)z^2 + 2(a^2 z) + a]}.$
No idea if I'm on the right track. Maybe distribute the $i$? Wondering if I can get some help.
|
We have
$$I(a) = \int_{0}^{2 \pi} \dfrac{dx}{1+a^2-a \cos(x)} = \dfrac1{1+a^2} \int_0^{2\pi} \dfrac{dx}{1 - \dfrac{a}{1+a^2} \cos(x)}$$
Note that we always have $b = \dfrac{a}{1+a^2} < 1$. Hence, we can write
$$\dfrac1{1 - b \cos(x)} = \sum_{k=0}^{\infty}b^k \cos^k(x)$$
Now we have
$$\int_0^{2\pi} \dfrac{dx}{1 - b \cos(x)} = \sum_{k=0}^{\infty} \int_0^{2 \pi} b^k \cos^k(x)dx = \sum_{k=0}^{\infty} b^{2k} \int_0^{2 \pi} \cos^{2k}(x)dx (\because\text{odd terms integrate to }0)$$
We now have from here that
$$\int_0^{2 \pi} \cos^{2k}(x)dx= \dbinom{2k}k \dfrac{2 \pi}{4^k}$$
This gives us
$$\int_0^{2\pi} \dfrac{dx}{1 - b \cos(x)} = 2 \pi \left(\sum_{k=0}^{\infty} \dbinom{2k}k \left(\dfrac{b^2}4\right)^k\right) = \dfrac{2 \pi}{\sqrt{1-b^2}}$$
where we made use of the fact that the Taylor series of $\dfrac1{\sqrt{1-x}}$ is
$$\dfrac1{\sqrt{1-x}} = \sum_{k=0}^{\infty} \dbinom{2k}k \left(\dfrac{x}4\right)^k$$
Hence, we now have
$$I(a) = \dfrac1{1+a^2} \dfrac{2 \pi}{\sqrt{1-\left(\dfrac{a}{1+a^2}\right)^2}} = \dfrac{2 \pi}{\sqrt{1+a^2+a^4}}$$
|
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Show abelian groups of order 3240? Show how to get all abelian groups of order $2^3 \cdot 3^4 \cdot 5$.
I just started learning this and was wondering how you would do this?
Is this correct?
$2^3 \cdot 3^4 \cdot 5 = 3240$. Therefore the number of abelian groups of order $3240$ is $3 \cdot 4 = 12$.
Is this the entire proof or do we need to do the table showing divisors like this?
Divisors:
$2^3 \cdot 3^4 \cdot 5$
$2^2 \cdot 2 \cdot 3^4 \cdot 5$
$2^2 \cdot 2 \cdot 3^3 \cdot 3 \cdot 5$
$2 \cdot 2 \cdot 2 \cdot 3^3 \cdot 3 \cdot 5$
$2 \cdot 2 \cdot 2 \cdot 3^2 \cdot 3 \cdot 3 \cdot 5$
$2 \cdot 2 \cdot 2 \cdot 3 \cdot 3 \cdot \cdot 3 \cdot 5$
$2 \cdot 2 \cdot 2 \cdot 3^4 \cdot 5$
$2^2 \cdot 2 \cdot 3^2 \cdot 3 \cdot 3 \cdot 5$
$2^3 \cdot 3 \cdot 3 \cdot 3 \cdot 3 \cdot 5$
$2^3 \cdot 3^2 \cdot 3 \cdot 3 \cdot 5$
$2^3 \cdot 3^3 \cdot 3 \cdot 5$
|
You have the right idea, but there are 5 abelian groups of order $3^4$, not 4. You can have:
*
*$\def\zt{\Bbb Z_3}\Bbb Z_{81}$
*$ \Bbb Z_{27}\times\zt$
*$\Bbb Z_{9}\times\Bbb Z_{9}$
*$\Bbb Z_{9}\times\zt\times\zt$
*$\zt\times\zt\times\zt\times\zt$
These correspond to the 5 ways (not 4) that you can express 4 as a sum of positive integers: $4, 3+1, 2+2, 2+1+1, $ and $1+1+1+1$, respectively. (Similarly, there are not 5 but 7 abelian groups of order $3^5$.)
I would observe this, list the groups of order $2^3$ and $3^4$, and then say that there were $3\cdot5\cdot1 = 15$ abelian groups of order $2^3\cdot3^4\cdot5$, without listing all 15, but on the other hand listing them couldn't hurt. If you do list them, do it methodically, not all mixed up as you did above, so that you (and the grader) can be certain you didn't omit any or list any twice.
I agree with vadim123 that you should cite the theorem by name, particularly since the whole point of this exercise is to show that you know the fundamental classification theorem of finite abelian groups. Find out what it is called in your text, and call it that.
|
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|
Solving this 3-degree polynomial I'm trying to factor the following polynomial by hand:
$-x^3 + 9x^2 - 24x + 20 = 0$
The simplest I could get is:
$-x^2(x-9) - 4(5x+5) = 0$
Any ideas on how I could go ahead and solve this by hand? This seems pretty tough.
|
You have not a polynomial, rather an equation, whose LHS is the cubic polynomial $$-x^{3}+9x^{2}-24x+20.\tag{0}$$
To get rid of the negative coefficient of $x^3$ multiply $(0)$ by $-1$
$$
\begin{equation*}
-\left( -x^{3}+9x^{2}-24x+20\right) =x^{3}-9x^{2}+24x-20.\tag{1}
\end{equation*}
$$
By trial and error$^1$, inspection, or per DonAntonio's hint (Rational root test), we find that $x=5$ is a root: $5^{3}-9\left( 5\right)
^{2}+24\left( 5\right) -20=0$. As a consequence we can divide $(1)$ by $x-5$, using polynomial long division or Ruffini's rule to obtain a quadratic polynomial, since the remainder must be zero
$$
\begin{equation*}
\frac{x^{3}-9x^{2}+24x-20}{x-5}= x^{2}-4x+4 =\left( x-2\right)\tag{2}
^{2}.
\end{equation*}
$$
So
$$
\begin{equation*}
-x^{3}+9x^{2}-24x+20=-\left( x-5\right) \left( x-2\right) ^{2},\tag{3}
\end{equation*}
$$
whose roots are $x=5$ (single root) and $x=2$ (double root).
If you start with the root $x=2$, you get the same result:
$$\frac{x^{3}-9x^{2}+24x-20}{x-2}=x^{2}-7x+10=\left( x-2\right) \left(
x-5\right).\tag{2'} $$
$^1$ Trial and error is a fundamental method of solving problems.
|
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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.
|
You can use two nested inductions.
In order to show the equality $\forall n, 1 + (1+2) + (1+2+3) + \ldots (1+2+\ldots+n) = \frac 1 6 n (n+1) (n+2)$,
you check that it is true for $n=1$, and then you are left to show the equality $\forall n, \frac 1 6 n (n+1) (n+2) + (1+2+ \ldots+n+(n+1)) = \frac 1 6 (n+1) (n+2) (n+3)$,
which can be simplified to $\forall n, (1+2+ \ldots +n+(n+1)) = \frac 1 6 (n+1) (n+2) [(n+3)-n] = \frac 1 2 (n+1) (n+2) $.
Now, you use induction again to prove this. You check that it is true for $n=1$, and are left to show the equality $\forall n, \frac 1 2 (n+1)(n+2) + (n+2) = \frac 1 2 (n+2)(n+3)$. Which should be straightforward to prove.
Alternately you can use a single strong induction : check that it is true for $n=0,1$, and for $n\ge 1$, assume the equality is true for $n-1$ and $n$, and show that it is true for $n+1$, by using $1 + (1+2) + (1+2+3) + \dots + (1+2+\ldots n+1) = 2*(1 + (1+2) + \ldots (1+2+\ldots + n)) - (1+(1+2) + \ldots + (1+2+\ldots +(n-1)) + (n+1)$.
Replace everyone with the corresponding $\frac 1 6 k(k+1)(k+2)$, and then it should be straightforward.
|
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|
Complex analytic function and Cauchy-Riemann conditions question here again asking for some guidance
Let me state the problem and working on and then my question,
So I have two functions:
$$u(x, y) = \frac{x(1+x)+y^2}{(1+x)^2 + y^2}$$
and
$$v(x, y) = \frac{y}{(1+x)^2 + y^2}$$
and I'm being asked if they can serve as real and imaginary parts respectively of some analytic function of $z = x + iy$
So what I thought I had to do was check for the Cauchy-Riemann conditions, so I started with the first condition
$$\frac{du( x,y )}{dx} = \frac{dv( x,y )}{dy}$$
and I got $x(2(1+x)^2 -2x^2 -4x -2) = 0$
does that mean they don't satisfy the equation? does it? should I check for the other condition?
Thank you!
|
The Cauchy-Riemann equations on a pair functions $u(x,y)$ and $v(x,y)$ are the two equations:
$\displaystyle \frac{du}{dx} = \frac{dv}{dy}$ and $\frac{du}{dy} = -\frac{dv}{dx}$.
Checking these two equations for the given functions $u(x,y)$ and $v(x,y)$ gives that
$\displaystyle \frac{du}{dx} = \frac{1+ 2x + x^2 - y^2}{(1 + 2x + x^2 + y^2)^2}$ and
$\displaystyle \frac{dv}{dy} = \frac{1+ 2x + x^2 - y^2}{(1 + 2x + x^2 + y^2)^2}$ so the first equation holds. For the second one note that $\displaystyle \frac{du}{dy} = \frac{2(1+x)y}{(1 + 2x + x^2 + y^2)^2}$ and $\displaystyle \frac{dv}{dx} = -\frac{2(1+x)y}{(1 + 2x + x^2 + y^2)^2}$ so the second equation also holds. You can derive the appropriate conclusion yourself.
|
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Differentiating $\;y = x a^x$ My attempt:
$$\eqalign{
y &= x{a^x} \cr
\ln y &= \ln x + \ln {a^x} \cr
\ln y &= \ln x + x\ln a \cr
{1 \over y}{{dy} \over {dx}} &= {1 \over x} + \left(x \times {1 \over a} + \ln a \times 1\right) \cr
{1 \over y}{{dy} \over {dx}} &= {1 \over x} + \left({x \over a} + \ln a\right) \cr
{1 \over y}{{dy} \over {dx}} &= {1 \over x} + \left({{x + a\ln a} \over a}\right) \cr
{1 \over y}{{dy} \over {dx}} &= {{a + {x^2} + ax\ln a} \over {ax}} \cr
{{dy} \over {dx}} &= {{a + {x^2} + ax\ln a} \over {ax}} \times x{a^x} \cr
{{dy} \over {dx}} &= {{x{a^{x + 1}} + {x^3}{a^x} + {a^{x + 1}}{x^2}\ln a} \over {ax}} \cr
{{dy} \over {dx}} &= {a^x} + {x^2}{a^{x - 1}} + {a^x}x\ln a \cr
{{dy} \over {dx}} &= {a^x}\left(1 + {x^2}{a^{ - 1}} + x\ln a\right) \cr
{{dy} \over {dx}} &= {a^x}\left({{a + {x^2} + x\ln a} \over a}\right) \cr} $$
However the answer in the back of the book is:
$${dy \over dx} = a^x (1 + x\ln a)$$
What have I done wrong?
|
Proceeding with your method in approaching the problem, but recalling that we can take $a$ as representing some constant: that is, $\;$ '$a$' $\;$ is a value that is NOT a function of $x$, and so does not get differentiated with respect to $x$. Similarly, so we can take $\ln a$ to be a constant.
So in your third line, when we see $\,x\ln a,\,$ we will treat it no differently than we would treat, say, $x\ln 3$; that is, $\;\dfrac{d}{dx}\left(x\ln a\right) = \dfrac{d}{dx}\Big((\ln a)\cdot x\Big) = \ln a.\;\;$ (And we will use this when moving from the third line to the fourth line below).
In contrast, $\,y\,$ IS a function of $x$, so you're correct that $\;\dfrac{d}{dx}(\ln y) = \dfrac 1y \cdot \dfrac{dy}{dx}$:
$$\eqalign{
& y = x{a^x} \\
& \ln y = \ln x + \ln {a^x}\\
& \ln y = \ln x + x\ln a \\
& {1 \over y}{{dy} \over {dx}} = {1 \over x} + \ln a \\
& {{dy} \over {dx}} = \left({1 \over x} + \ln a\right)\times x{a^x} \\
& {{dy} \over {dx}} = a^x + xa^x\cdot \ln a \\
& {{dy} \over {dx}} = a^x(1 + x\ln a)\cr}$$
|
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Is the sum of two singular matrices also singular? If $A$ and $B$ are $n \times n$ singular matrices, is $A+B$ also singular?
|
The sum of two singular $n × n$ matrices may be non-singular.
For example, consider two $2 × 2$ matrix $A=\begin{pmatrix}
1 & 1 \\
1 & 1
\end{pmatrix}$ and $B=\begin{pmatrix}
1 & -1 \\
-1 & 1
\end{pmatrix}$
Here $\det A = 1-1=0$ and $\det B=1-1=0$
So both the matrix $A \quad \text{and} \quad B$ are singular matrices.
Now $A+B =\begin{pmatrix}
2 & 0 \\
0 & 2
\end{pmatrix}$
$\det (A+B)=4\neq 0$
Hence $A + B$ is non-singular.
|
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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?
|
It's really easy to compute $\left\lfloor \frac{\sqrt a + b}{c}\right\rfloor$ for integers $a,b,c$. Just use the fact that
$$\left\lfloor \frac{r + b}{c}\right\rfloor = \left\lfloor \frac{\lfloor r \rfloor + b}{c}\right\rfloor$$
for real $r$ and integers $b,c$.
Here, $\lfloor\sqrt{14}\rfloor = 3$, so $x = \left\lfloor\frac{6}{5}\right\rfloor = 1$.
|
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help in solving a recursion EDIT: Turns out that the given solution has an error in it.
I have the following recursion question and following that is my answer. The problem is it doesn't seem to agree with the marking scheme. Can you help me by pointing out where i have gone wrong?
Question:
Use substitution to solve the recurrence equation $a_n = a_{n-1}+n^2$ for $n\ge1$, given $a_0 = 7$
My answer
step 1: $a_n = a_{n-1}+n^2$
step 2: $a_n = a_{n-2}+(n-1)^2+n^2$
step 3: $a_n = a_{n-3}+(n-2)^2+(n-1)^2+n^2$
step i :$a_n = a_{n-i}+\sum_{j=0}^{i-1} (n-j)^2$
since
$n-i=0$ then
$n=i$
therefore:
$a_n=a_0+\sum_{j=0}^{n-1} (n-j)^2$
$a_n=7+\sum_{j=0}^{n-1} (n-j)^2$
End of my answer
Now the marking scheme has something totally different which i don't understand
So again, where am i going wrong?
|
So u have the the the difference between the nth term and (n-1)th term is n^2
And the zeroth term is 7,
Thus f(0) = 7, f(1) = 8, f(2) = 12, f(3) = 21
Now we know the difference is quadratic in the term number so the function itself will be (via integration) cubic. Thus it has form
F(n) = an^3 + bn^2 + cn + d
We know d = 7 from the initial case of F(0) = 7
So substituting each of the values from above into the equation we have:
F(1) = a(1)^3 + b(1)^2 + c(1) = 8
F(2) = a(2)^3 + b(2)^2 + c(2) = 12
F(3) = a(3)^3 + b(3)^2 + c(3) = 21
Thus:
a + b + c + 7 = 8
8a + 4b + 2c + 7 = 12
27a + 9b + 3c + 7 = 21
This is a a system of 3 linear equations in 3 unknowns (which simplifies to):
a + b + c = 1
8a + 4b + 2c = 5
27a + 9b + 3c = 14
It has a matrix representation of:
1 1 1 1
8 4 2 5
27 9 3 14
Which when put into reduced row echelon form (rref): is
1 0 0 1/3
0 1 0 1/2
0 0 1 1/6
Corresponding to a = 1/3, b = 1/2, c = 1/6
Therefore we conclude that for our sequence:
F(n) = 1/3(n^3) + 1/2(n^2) + 1/6(n) + 7
Hopefully that made sense :)
|
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|
A series: $1-\frac{1}{2}-\frac{1}{3}-\frac{1}{4}+\frac{1}{5}-\frac{1}{6}+\frac{1}{7}+\cdots$ Denote $$b_1=1,b_{n}=b_{n-1}-\dfrac{S(b_{n-1})}{n},(n>1 )\tag1$$
where $S(x)=1$ if $x>0,S(x)=-1$ if $x<0$, and $S(0)=0.$
So $b_n=1,\dfrac{1}{2},\dfrac{1}{6},-\dfrac{1}{12},\dfrac{7}{60},-\dfrac{1}{20},\dfrac{13}{140},-\dfrac{9}{280},\dfrac{199}{2520},-\dfrac{53}{2520}…$
For example: $b_7=1-\dfrac{1}{2}-\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{5}-\dfrac{1}{6}+\dfrac{1}{7}=\dfrac{13}{140}.$
We can prove that $$\lim_{n \rightarrow \infty} b_n=0,$$
and $$\sum_{i=0}^2{S(b_{n+i})}=±1. (n>1) \tag2$$
Question 1: Can we determine $S(b_n)$ for very large $n$ (such as $n=10^{100}$)?
Question 2: Prove or disprove that if $n>3$, then
$$\left|\sum_{i=4}^n{S(b_i)}\right|\leq 1.\tag3$$
I have checked it for $n<4 \cdot 10^4.$Has this problem been studied before?Thanks in advance!
|
I can't answer question 1, but I do have an answer for question 2. In particular, the fact that
$$
\left|\sum_{i=4}^n S(b_i)\right| \leq 1
$$
is an immediate consequence of the following theorem.
Theorem. For all $n\geq 2$, the terms $b_{2n}$ and $b_{2n+1}$ have opposite signs.
Proof: It is easy to check that the theorem holds for $2\leq n\leq 7$. Therefore, it suffices to show that
$$
|b_{2n}| < \dfrac{1}{2n+1}
$$
for all $n\geq 8$. To do so, we shall prove the much better bound
$$
|b_{2n}| \;\leq\; \frac{1}{2n-1}-\frac{1}{2n}
$$
for all $n\geq 8$. (Note that this difference is always less than $1/(2n+1)$.)
We proceed by induction on $n$. The base cases is $n=8$, for which
$$
|b_{16}| \;=\; \frac{547}{144144} \;\leq\; \frac{1}{15}-\frac{1}{16}.
$$
Now suppose that $|b_{2n}|\leq \dfrac{1}{2n-1}-\dfrac{1}{2n}$. Without loss of generality, we may assume that $b_{2n}$ is positive. Then $b_{2n} \leq \dfrac{1}{2n+1}$, so
$$
b_{2n+1} \;=\; b_{2n} \,-\, \frac{1}{2n+1}
$$
is negative, and therefore
$$
b_{2n+2} \,=\, b_{2n+1} + \frac{1}{2n+2} \,=\, b_{2n} - \frac{1}{2n+1} + \frac{1}{2n+2}.
$$
But $b_{2n} \leq \dfrac{1}{2n-1}-\dfrac{1}{2n} \leq 2\left(\dfrac{1}{2n+1} - \dfrac{1}{2n+2}\right)$, so it follows that
$$
|b_{2n+2}| \;\leq\; \frac{1}{2n+1} - \frac{1}{2n+2}.\tag*{$\square$}
$$
This proof was based on the observation that $b_n$ is "small" when $n$ is even, and "large" when $n$ is odd. There are similar patterns for other powers of $2$: the term $b_n$ tends to be "small" when $n$ is $2$ mod $4$, and when $n$ is $6$ mod $8$.
Indeed, the theorem above appears to be just the first in a sequence of theorems about the signs of the $b_n$. Then next two are
Theorem. For all $n\geq 4$, the terms $b_{4n+2}$ and $b_{4n+4}$ have opposite signs.
Theorem. For all $n\geq 2$, the terms $b_{8n+6}$ and $b_{8n+10}$ have opposite signs.
The first of these follows from the fact that
$$
|b_{4n+2}| \;\leq\; \frac1{4 n - 1} - \frac1{4 n} - \frac1{4 n + 1} + \frac1{4 n + 2}
$$
for $n\geq 5$. The second follows from the fact that
$$
|b_{8n+6}| \;\leq\; \frac1{8 n - 1} - \frac1{8 n} - \frac1{8 n + 1} + \frac1{8 n + 2} - \frac{1}{8n+3} + \frac{1}{8n+4} +\frac{1}{8n+5} - \frac{1}{8n+6}
$$
for all $n\geq 2$.
What seems to be happening is that it gets "caught" in these patterns, e.g. if the $k$th term happens to be unusually small (on the order of $1/n^{2^n}$), then the same will be true of the $(k+2^n)$'th term. However, I don't see any way to predict when it gets "caught", or what the parity will be (i.e. why 6 mod 8 instead of 2 mod 8), so I don't have any way of predicting the sign of $b_n$ for $n=10^{100}$.
|
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|
To find the x and y-intercepts of the line $ax+by+c=0$ Please check if I've solved the problem in the correct way:
The problem is as follows:
Find the points at which the line $ax+by+c=0$ crosses the x and y-axes. (Assume that $a \neq 0$ and $b \neq 0$.
My solution:
We have to find the x and y-intercepts of the line. At the 'x-intercept' the ordinate must be equal to $0$ and at the 'y-intercept' the abscissa must be equal to $0$.
Now we solve the equation $ax+by+c=0$ for $y$:
$ax + by + c=0$
$ax + by = -c$
$by = -ax -c$
$y = \frac {-ax-c}{b}$
$\because x = 0$ at y-intercept,
$\therefore y = \frac {-a(0)}{b} -\frac{c}{b}$
$y = -\frac cb$.
The point at which the line crosses the y-axis is $(0,-\frac cb)$
Now we solve the equation $ax+by+c=0$ for $x$:
$ax+by+c=0$
$ax+by=-c$
$ax = -by-c$
$x = \frac {-by-c}{a}$
$\because y = 0$ at x-intercept
$\therefore x = \frac {-b(0)}{a} -\frac{c}{a}$
$x = -\frac ca$
The point at which the line crosses the x-axis is $(-\frac ca,0)$
|
Your work is exemplary, Samama. You know your definitions well, and your answers are entirely correct.
A nice "shortcut" is to take advantage of what you already know:
*
*the $x$ intercept is the value of $x$ when $y = 0$, and
*the $y$ intercept is the value of $y$ when $x = 0$.
*$ax + by + c = 0 \iff ax + by = -c$
x intercept
We can use $\;ax + by = -c\;$ to solve for $x$ when $y = 0$, by plugging in $0$ for $y,\;$ right at the start:
$$ax + \underbrace{by}_{y = 0} = -c \iff ax = - c \iff x = \frac {-c}{a}$$
y intercept
And we can do the same to solve for $y$ when $x = 0$, by plugging in $0$ for $x,\,$ right at the start:
$$\underbrace{ax}_{x = 0} + by = -c \;\;\iff\;\; by = -c \;\;\iff \;\; y = \frac{-c}{b}$$
|
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|
Proving that $x - \frac{x^3}{3!} < \sin x < x$ for all $x>0$
Prove that $x - \frac{x^3}{3!} < \sin(x) < x$ for all $x>0$
This should be fairly straightforward but the proof seems to be alluding me.
I want to show $x - \frac{x^3}{3!} < \sin(x) < x$ for all $x>0$. I recognize this shouldn't be too difficult but perhaps finals have fried my brain.
|
A possible solution without use a Taylor series.Observe that:
$\\ \\ \displaystyle \sin(3\gamma)=\sin(2\gamma)\cos(\gamma)+\sin(\gamma)\cos(2\gamma)=2\sin(\gamma)\cos^2(\gamma)+\sin(\gamma)(1-2\sin^2(\gamma))=2\sin(\gamma)(1-\sin^2\gamma)+\sin(\gamma)(1-2\sin^2(\gamma))=3\sin(\gamma)-4\sin^3(\gamma)\Rightarrow \sin^3(\gamma)=\frac{1}{4}\left(3\sin(\gamma)-\sin(3\gamma)\right)$
\begin{equation}
\sin^3(\gamma)=\frac{1}{4}\left(3\sin(\gamma)-\sin(3\gamma)\right)
\end{equation}
Do it $\displaystyle \gamma=\frac{\phi}{3^k}$:
\begin{equation}
\sin^3\left(\frac{\phi}{3^k}\right)=\frac{1}{4}\left(3\sin\left(\frac{\phi}{3^k}\right)-\sin\left(\frac{\phi}{3^{k-1}}\right)\right)
\end{equation}
Multiplying by $\displaystyle 3^{k-1}$:
\begin{equation}
3^{k-1}\sin^3\left(\frac{\phi}{3^k}\right)=\frac{1}{4}\left(3^{k}\sin\left(\frac{\phi}{3^k}\right)-3^{k-1}\sin\left(\frac{\phi}{3^{k-1}}\right)\right)
\end{equation}
Applying summation on both sides of equality, we will have:
$\\ \\ \displaystyle \sum_{k=1}^{n}3^{k-1}\sin^3\left(\frac{\phi}{3^k}\right)=\sum_{k=1}^{n}\frac{1}{4}\left(3^{k}\sin\left(\frac{\phi}{3^k}\right)-3^{k-1}\sin\left(\frac{\phi}{3^{k-1}}\right)\right)= \frac{1}{4}\left(\left(3\sin\left(\frac{\phi}{3^k}\right)-\sin\left(\phi\right)\right)+\left(3^{2}\sin\left(\frac{\phi}{3^k}\right)-3\sin\left(\frac{\phi}{3}\right)\right)+...+\left(3^{n}\sin\left(\frac{\phi}{3^n}\right)-3^{n-1}\sin\left(\frac{\phi}{3^{n-1}}\right)\right)\right) =\frac{1}{4}\left(3^{n}\sin\left(\frac{\phi}{3^n}\right)-\sin(\phi)\right)\Rightarrow \sum_{k=1}^{n}3^{k-1}\sin^3\left(\frac{\phi}{3^k}\right)=\frac{1}{4}\left(3^{n}\sin\left(\frac{\phi}{3^n}\right)-\sin(\phi)\right)\\ \\$
Take the limit:
\begin{equation*}
\lim_{n\rightarrow \infty}\sum_{k=1}^{n}3^{k-1}\sin^{3}\left(\frac{\phi}{3^{k}}\right)=\frac{1}{4}\left(\phi-\sin(\phi)\right)
\end{equation*}
Notice, on the other hand, using the inequality $ \displaystyle \sin x \leq x $ and using the infinite arithmetic progression formula, follows:
$\\ \displaystyle \sin\left(\frac{\phi}{3^{k}}\right) \leq \frac{\phi}{3^{k}}\Rightarrow \sin^{3}\left(\frac{\phi}{3^{k}}\right) \leq \frac{\phi^3}{3^{3k}} \Rightarrow 3^{k-1}\sin^{3}\left(\frac{\phi}{3^{k}}\right) \leq \frac{\phi^3}{3^{2k+1}}\Rightarrow \sum_{k=1}^{\infty}3^{k-1}\sin^{3}\left(\frac{\phi}{3^{k}}\right) \leq \sum_{k=1}^{\infty} \frac{\phi^3}{3^{2k+1}}=\frac{\phi^3}{3}\sum_{k=1}^{\infty} \frac{1}{3^{2k}}=\frac{\phi^3}{3\times 8}\Rightarrow \sum_{k=1}^{\infty}3^{k-1}\sin^{3}\left(\frac{\phi}{3^{k}}\right) \leq \frac{\phi^3}{3\times 8} \Rightarrow \frac{1}{4}\left(\phi-\sin(\phi)\right) \leq \frac{\phi^3}{3\times 8} \Rightarrow \phi-\frac{\phi^3}{6}\leq \sin(\phi) $
|
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|
Find max of $x^7+y^7+z^7$
Find max of $x^7+y^7+z^7$ where $x+y+z=0$ and $x^2+y^2+z^2=1$
I tried to use the inequality:$$\sqrt[8]{\frac {x^8+y^8+z^8} 3}\ge\sqrt[7]{\frac {x^7+y^7+z^7} 3}$$ but stuck
|
Here is an elementary way to proceed, which may not be wholly convenient, but which illustrates some interesting techniques.
If $x,y,z$ are the roots of a cubic $t^3-s_1t^2+s_2t-s_3=0$, then we have$$s_1=x+y+z=0$$$$2s_2=2xy+2yz+2zx=(x+y+z)^2-(x^2+y^2+z^2)=-1$$
Thus we have (for some $u$) $$2t^3-t-u=0$$
This gives also (multiply by $t^r$) $$2t^{r+3}-t^{r+1}-ut^r=0$$
Now let $a_r=x^r+y^r+z^r$, and substitute $x, y, z$ successively into the last equation and add the three together to obtain:$$2a_{r+3}-a_{r+1}-ua_r=0 \text{ with }a_0=3, a_1=0, a_2=1$$
Then you can express $a_7$ in terms of $u$ using the recurrence. The constraint on $u$ is that all the roots of the cubic must be real. Combining these two pieces of information you can get a handle on your question.
|
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|
finding nth term Let
3,8,17,32,57 . . . . .
be a pattern.How do we find the nth number?My brains are completely jammed,I am tired.I do not even recognize the pattern.I calculated a few ways,but all I want is a little hint,not the whole solution.
|
If you want the smallest degree polynomial that gives these numbers (which is only one of the infinitely many possible explanations for them), we can use the calculus of finite differences:
$$\fbox{3}\quad 8\quad 17\quad 32\quad 57 \\
\fbox{5}\quad 9\quad 15\quad 25\\
\fbox{4}\quad 6\quad 10\\
\fbox{2} \quad 4\\
\fbox{2}$$
Define
$$\begin{align*}
f(n)&=3\binom{n}{0}
+5\binom{n}{1}+4\binom{n}{2}+2\binom{n}{3}+2\binom{n}{4}\\\\
&=3 + \frac{19 n}{6} + \frac{23 n^2}{12} - \frac{n^3}{6} + \frac{n^4}{12}
\end{align*}$$
Then
$$f(0)=3,\quad f(1)=8,\quad f(2)=17,\quad f(3)=32,\quad f(4)=57$$
|
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|
Partial fraction decomposition with a nonrepeated irreducible quadratic factor I'm trying to do a partial fraction decomposition on the following rational eqn with a nonrepeated irreducible quadratic factor:
$$\dfrac{-28x^2-92}{(x-4)^2(x^2+1)}$$
I've broken it down into an identiy: $-28x^2 -92 = A((x-4)(x^2-1))+B(x^2+1)+(Cx+D)(x-4)^2$
and solved for B by setting x to 4 ($B-12$), distributed the -12 and moved the resulting quadratic to the left side, leaving me with:
$$12x^2-28x-80 = A((x-4)(x^2-1)) + (Cx+D)(x-4)^2$$
Now I want to solve for A by setting x to something that will make the coefficient on $Cx+D$ zero, but I'm stumped - setting it to 4 also gets rid of my A. What do I do? Or have I messed up somewhere along the way?
|
This is what you have done...
$$\dfrac{-28x^2-92}{(x-4)^2(x^2+1)}
= \dfrac{A}{(x-4)} + \dfrac{B}{(x-4)^2} + \dfrac{Cx+D}{x^2+1}$$
$$-28x^2 -92 = A(x-4)(x^2+1) + B(x^2+1) + (Cx+D)(x-4)^2$$
Letting $x=4$, we get
$$-540 =17B$$
$$B = -\dfrac{540}{17}$$
What I would do now...
Let $B = -\dfrac{540}{17}$ and simplify.
\begin{align}
17(-28x^2 -92) &= 17A(x-4)(x^2+1) - 540(x^2+1) + 17(Cx+D)(x-4)^2 \\
64(x^2-16) &= 17A(x-4)(x^2+1) + 17(Cx+D)(x-4)^2 \\
64(x+4) &= 17A(x^2+1) + 17(Cx+D)(x-4) \\
\end{align}
Again, let $x=4$...
$$512 = 289A$$
$$A = \dfrac{512}{289}$$
Finally. let $A = \dfrac{512}{289}$ and simplify one more time...
\begin{align}
64(x+4) &= \dfrac{512}{17}(x^2+1) + 17(Cx+D)(x-4) \\
1088(x+4) &= 512(x^2+1) + 289(Cx+D)(x-4) \\
-64(8x+15)(x-4) &= 289(Cx+D)(x-4) \\
-64(8x+15) &= 289(Cx+D) \\
-\dfrac{512x}{289} - \dfrac{960}{289} &= Cx+D
\end{align}
$$C = -\dfrac{512}{289}$$
$$D = - \dfrac{960}{289}$$
|
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|
Circular motion trig We have $x_P = -2 + 4 \cos (-\pi t)$ and $y_P = 1 + 4 \sin ( - \pi t)$ with $t$ in seconds.
We have to find the coordinates of the intersection with the y-axis. So I use trig and I eventually end up with:
$ t = -\dfrac{1}{3} + k \cdot 2$ and $ t = \dfrac{1}{3} + k \cdot 2$
The correction model does this too, and then they just say 'the corrosponding points are ...' and give the 2 points. But what I don't understand is how this can be correct. If we fill in for k, we get 6 points within $[0, 2\pi]$. This would imply 6 intersections, right? What am I missing here?
p.s. - $t$ would be 1/3, 7/3, 10/3, 5/3, 11/3, 17/3
|
$$\text{If }t=2k+ \frac13,$$
$$\sin(-\pi t)=-\sin \pi t=-\sin\left(2k\pi+\frac\pi3\right)=-\sin\left(\frac\pi3\right)=-\frac{\sqrt3}2$$
$$\text{If }t=2k- \frac13,$$
$$\sin(-\pi t)=-\sin \pi t=-\sin\left(2k\pi-\frac\pi3\right)=-\sin\left(-\frac\pi3\right)=\sin\left(\frac\pi3\right)=+\frac{\sqrt3}2$$
So, the two coordinates of the intersection with the y-axis are $\left(0,1+4\cdot\pm \frac{\sqrt3}2 \right)=\left(0,1\pm2\sqrt3\right)$
Alternatively, as $-1\le \cos(-\pi t)\le 1\implies-3\le x_p\le 2$
If $x_p$ assumes any value in the above range $=c$(say),
$-2+ 4\cos(-\pi t)=c\implies \cos(-\pi t)=\frac{c+2}4$
$$\implies \sin (-\pi t)=\pm\frac{\sqrt{\left(4^2-(c+2)^2\right)}}4=\pm\frac{\sqrt{\left(12-c^2-4c\right)}}4$$
$$\implies y_p=1+4\left( \pm\frac{\sqrt{\left(12-c^2-4c\right)}}4\right)=1\pm \sqrt{12-c^2-4c}$$
So, the intersections will be $\left(c,1\pm \sqrt{12-c^2-4c}\right)$
|
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|
Evaluating $\int_0^\infty \frac{\log (1+x)}{1+x^2}dx$ Can this integral be solved with contour integral or by some application of residue theorem?
$$\int_0^\infty \frac{\log (1+x)}{1+x^2}dx = \frac{\pi}{4}\log 2 + \text{Catalan constant}$$
It has two poles at $\pm i$ and branch point of $-1$ while the integral is to be evaluated from $0\to \infty$. How to get $\text{Catalan Constant}$? Please give some hints.
|
If you're still interested in approaches that use contour integration, consider the function $$f(z) = \frac{\log(1+z) \log(-z)}{1+z^{2}}.$$
Using the principal branch of the logarithm, there is a branch cut along $[0,\infty)$ and a branch cut along $(-\infty, -1]$.
Then integrating counterclockwise around a keyhole contour deformed around the branch cuts (see here for a picture),
$$ \begin{align} &\int_{\infty}^{0} \frac{\log(1+x) \big(\log(x) + i \pi \big)}{1+x^{2}} \ dx + \int_{0}^{\infty} \frac{\log(1+x) \big(\log (x) - i \pi \big)}{1+x^{2}} \ dx \\ &+\int_{-\infty}^{-1} \frac{\big(\log|1+x| + i \pi \big) \log(-x)}{1+x^{2}} \ dx + \int_{-1}^{-\infty} \frac{\big(\log|1+x| - i \pi \big) \log(-x)}{1+x^{2}} \ dx \\ &= - 2 \pi i \int_{0}^{\infty} \frac{\log(1+x)}{1+x^{2}} \ dx + 2 \pi i \int_{1}^{\infty} \frac{\log(x)}{1+x^{2}} \ dx \\ &= 2 \pi i \big( \text{Res} [f(z), i] + \text{Res} [f(z), -i] \big) \\ &= 2 \pi i \left(\frac{\log(1+i) \log(-i)}{2i} + \frac{\log(1-i)\log(i)}{-2i} \right) \\ &= 2 \pi i \left( - \frac{\pi}{4} \log(2)\right) . \end{align}$$
Therefore,
$$ \begin{align} \int_{0}^{\infty} \frac{\log(1+x)}{1+x^{2}} \ dx &= \frac{\pi}{4} \log(2) + \int_{1}^{\infty} \frac{\log (x)}{1+x^{2}} \ dx \\ &= \frac{\pi}{4} \log(2) - \int_{1}^{0} \frac{\log(\frac{1}{u})}{1+ (\frac{1}{u})^{2}} \frac{1}{u^{2}} \ du \\ &= \frac{\pi}{4} \log(2) - \int^{1}_{0} \frac{\log u}{1+u^{2}} \ du \\ &= \frac{\pi}{4} \log(2) + G . \end{align}$$
|
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|
How can I show this field extension equality? How can I show this field extension equality $\mathbb{Q}(\sqrt{2}, \sqrt{3}) = \mathbb{Q}(\sqrt{2} + \sqrt{3})$?
|
The easier direction is $\mathbb{Q}(\sqrt{2} + \sqrt{3} ) \subseteq \mathbb{Q}(\sqrt{2}, \sqrt{3}).$ Can you tell us why this is true?
Now we have $\mathbb{Q} \subseteq \mathbb{Q}(\sqrt{2} + \sqrt{3} ) \subseteq \mathbb{Q}(\sqrt{2}, \sqrt{3}).$
Since $\sqrt{2}$ is a root of the degree 2 polynomial $X^2-2$ which is irreducible over $\mathbb{Q}$ (because if it were reducible, it would have a linear factor corresponding to a rational root which the rational root theorem rules outs) we have $[\mathbb{Q}(\sqrt{2}): \mathbb{Q}]=2.$
Now $\sqrt{3}$ is a root of the degree 2 polynomial $X^2-3.$ Let's check this is irreducible over $\mathbb{Q}(\sqrt{2}).$ It is reducible if and only if it has a root in $\mathbb{Q}(\sqrt{2}).$ Again by the rational root theorem, it has no roots in $\mathbb{Q}.$ So if there is a root $a+b\sqrt{2} \in \mathbb{Q}(\sqrt{2})$ it must have $b\neq 0$ and satisfy $(a+b\sqrt{2})^2-3 = a^2+2ab\sqrt{2}+2b-3=0.$ If $a=0$ then $2b^2=3$ so $b=\pm \sqrt{3/2}$ which is rational if and only if $2b = \pm \sqrt{6}$ is rational. It is not hard to show that $\sqrt{n}$ is rational if and only if $n$ is a perfect square which $6$ is not, so $a+b\sqrt{2}$ is a root of $X^2-3$ then $a$ and $b$ must both be non-zero. But then we can rearrange to get $\sqrt{2} = \dfrac{3-2b+a^2}{2ab}$ and the right hand side is rational while the left is not. So $X^2-3$ is irreducible over $\mathbb{Q}(\sqrt{2})$ so $[\mathbb{Q}(\sqrt{2})(\sqrt{3}):\mathbb{Q}(\sqrt{2})]=2.$
Thus $$[\mathbb{Q}(\sqrt{2},\sqrt{3}):\mathbb{Q}] = [\mathbb{Q}(\sqrt{2},\sqrt{3}):\mathbb{Q}(\sqrt{2})][\mathbb{Q}(\sqrt{2}):\mathbb{Q}] = 2\cdot 2=4.$$
Now we find the minimal polynomial of $\sqrt{2}+\sqrt{3}$ over $\mathbb{Q}.$ The natural guess is $$(X-\sqrt{2}-\sqrt{3})(X-\sqrt{2}+\sqrt{3})(X+\sqrt{2}-\sqrt{3})(X+\sqrt{2}+\sqrt{3})=X^4+2X^2+25.$$ Let's check it is irreducible to confirm. Gauss's lemma allows us to simply check it is irreducible over $\mathbb{Z}.$ The rational root theorem gives $\pm 1, \pm 5, \pm 25$ as the only possibilities for integer roots, none of which work. So any factorization of $X^2+2X^2+25$ has no linear factors, and thus is of the form $$X^4+2X^2+25 = (X^2+aX+b)(X^2+cX+d)$$ where $a,b,c,d\in \mathbb{Z}.$ Equating coefficients of powers of $X$ gives $a=-c, ac+b+d=2, ad+bc=0$ and $bd=25.$ If $a=-c=0$ then $b+d=2$ and $bd=25$ which implies $4 = (b+d)^2 = b^2+2bd + d^2 = b^2+d^2+50 \implies b^2+d^2<0.$ Thus $a=-c\neq 0$ and from $ad+bc=0$ we get $b=d$ and from $bd=25$ we conclude $b=d=\pm 5.$ This makes $ac+b+d=2$ becomes $-a^2 = 2\pm 10$ which is impossible.
Thus $$ [\mathbb{Q}(\sqrt{2}+\sqrt{3}):\mathbb{Q}]=4.$$
Putting everything together, $$ 4 = [\mathbb{Q}(\sqrt{2},\sqrt{3}):\mathbb{Q}] = [\mathbb{Q}(\sqrt{2},\sqrt{3}):\mathbb{Q}(\sqrt{2}+\sqrt{3})][\mathbb{Q}(\sqrt{2}+\sqrt{3}):\mathbb{Q}]=[\mathbb{Q}(\sqrt{2},\sqrt{3}):\mathbb{Q}(\sqrt{2}+\sqrt{3})]\cdot 4
$$
so $$[\mathbb{Q}(\sqrt{2},\sqrt{3}):\mathbb{Q}(\sqrt{2}+\sqrt{3})]=1$$ which implies $\mathbb{Q}(\sqrt{2},\sqrt{3})=\mathbb{Q}(\sqrt{2}+\sqrt{3}).$
|
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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?
|
Everytime there is a complicated cube root or square root, there is a simplification that can occur.
For example, you can compute $(35+18i\sqrt 3)^{1/3}$ approximately in $\Bbb C$, or you can magically realize that $(-1-2i\sqrt 3)^3 = 35+18i\sqrt 3$ and so you can pick $Q = -1-2i\sqrt 3$ (don't try to find this cube root algebraically).
This should help you with the rest, and feel free to see what happens when you replace $Q$ with one of the other two cube roots.
|
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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 both sides you should get $9(x+13) = (x+9)^2$ which rearranges to
\[x^2+9x-36 = 0 \ ,\]
which has the solutions
\[x_{1,2} = -\frac 92 \pm \sqrt{\frac{81}{4}+36} = \frac{-9\pm 15}{2} \: ,\]
i.e. $\begin{cases}x_1 = 3 \\ x_2 = -12\end{cases}$. By putting these into the original equation $3\sqrt{x+13} = x+9$ you realize that $x_2 = -12$ is not a solution, but $x_1 = 3$ is.
|
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|
Plane geometry tough question $\triangle ABC$ is right angled at $A$. $AB=20, CA= 80/3, BC=100/3$ units. $D$ is a point between $B$ and $C$ such that the $\triangle ADB$ and $\triangle ADC$ have equal perimeters. Determine the length of $AD$.
|
Add up the the sides of $\triangle ADB$ and $\triangle ADC$ and you get their perimeters and then equate them:
$x+\frac{100}{3}-y+\frac{80}{3}=x+y+20$
Solve for $y$ to get $y=20$
and then for $x$ using the cosine rule we get
$x=\sqrt{20^2+20^2-2(20)(20)\cos\left(\tan^{-1}\frac{80}{20\times3}\right)}$
$x=8\sqrt{5}$
|
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|
$\frac{a}{1-a}+\frac{b}{1-b}+\frac{c}{1-c}+\frac{d}{1-d}+\frac{e}{1-e}\ge\frac{5}{4}$ I tried to solve this inequality:
$$\frac{a}{1-a}+\frac{b}{1-b}+\frac{c}{1-c}+\frac{d}{1-d}+\frac{e}{1-e}\ge\frac{5}{4}$$
with
$$a+b+c+d+e=1$$
I am stuck at this. I don't want the full solution, a hint would be enough.
|
Hint:
You have $1-a=b+c+d+e$ and
$a=1-(b+c+d+e)$
$\dfrac{1-(b+c+d+e)}{b+c+d+e}=\dfrac{1}{b+c+d+e}-1$ and $b+c+d+e=1-a$
Using AM $\ge$ HM
$(\dfrac{1}{1-a}+\dfrac{1}{1-b}+\dfrac{1}{1-c}+\dfrac{1}{1-d}+\dfrac{1}{1-e}) \ge \dfrac{5^2}{1+1+1+1+1-(a+b+c+d+e)} $
|
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|
Proving that either $2^n-1 $ or $ 2^n+1$ is not prime Not sure what approach to take with this:
Prove that at least $2^n-1 $ or $ 2^n+1$ is composite $\forall$ $n>2$
|
As $n>2,2^n-1>3$
If $n$ is even, $=2m$(say),
$2^n-1=2^{2m}-1=4^m-1$ which is divisible by $4-1=3$ as $(a-b)$ divides $a^n-b^n$ for integer $n\ge 0$
If $n$ is odd, $=2m+1$(say),
$2^{2m+1}+1=2^{2m+1}+1^{2m+1}$ which is divisible by $2+1=3$ as $(a+b)$ divides $a^{2m+1}+b^{2m+1}$ for integer $m\ge 0$
Alternatively,
$(2^n-1)(2^n+1)=4^n-1$ is divisible by $4-1=3,$
but $2^n+1>2^n-1>3$
References : Fermat Prime 1,2, Mersenne Prime1,2
|
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Solving summation $2n+2^2(n-1)+2^3(n-2)+....+2^n$ Can anyone help me with this summation? I tried to use the geometric series on this, but I can't use that.
$$2n+2^2(n-1)+2^3(n-2)+....+2^n$$
I am trying to do this for studying algorithms. Can we get a closed form for this ?
|
This does have a nice closed form.
We can write your original sum, $2n+2^2(n-1)+2^3(n-2)+....+2^n$, as
$(2+2^2+2^3+...2^n)+(2+2^2+2^3+...2^{n-1})+2+2^2+2^3+...2^{n-2})+...+2$, because this sum will have $n$ $2$'s, $n-1$ $2^2$'s, and so on. By the geometric series formula, this sum simplifies to $(2^{n+1}-2) + (2^n-2) + (2^{n-1}-2)+...+(2^2-2)$, which itself simplifies (again by the geometric series formula) to $2^{n+2}-4-2n$.
|
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|
$1+1+1+1+1+1+1+1+1+1+1+1 \cdot 0+1 = 12$ or $1$? Does the expression $1+1+1+1+1+1+1+1+1+1+1+1 \cdot 0+1$ equal $1$ or $12$ ?
I thought it would be 12 this as per pemdas rule:
$$(1+1+1+1+1+1+1+1+1+1+1+1)\cdot (0+1) = 12 \cdot 1 = 12$$
Wanted to confirm the right answer from you guys. Thanks for your help.
|
Multiplication takes priority over addition, so this
$1+1+1+1+1+1+1+1+1+1+1+1\times0+1$
becomes:
$1+1+1+1+1+1+1+1+1+1+1+0+1$
Now add everything which gives you 12.
|
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|
Why dividing by zero still works Today, I was at a class. There was a question:
If $x = 2 +i$, find the value of $x^3 - 3x^2 + 2x - 1$.
What my teacher did was this:
$x = 2 + i \;\Rightarrow \; x - 2 = i \; \Rightarrow \; (x - 2)^2 = -1 \; \Rightarrow \; x^2 - 4x + 4 = -1 \; \Rightarrow \; x^2 - 4x + 5 = 0 $. Now he divided $x^3 - 3x^2 + 2x - 1$ by $x^2 - 4x + 5$.
The quotient was $x + 1$ and the remainder $x - 6$. Now since $\rm dividend = quotient\cdot divisor + remainder$, he concluded that $x^3 - 3x^2 + 2x - 1 = x-6$ since the divisor is $0$.
Plugging $2 + i$ into $x - 6$, we get $-4 + i$.
But my question is, how was he able to divide by zero in the first place? Why did dividing by zero work in this case?
|
The division was a division in the polynomial ring $\mathbb{C}[x]$
by a non-zero polynomial
For example: by dividing $$x^{2}+2x+1$$ by $$x+1$$ which is a non-zero
polynomial we get $$(x+1)(x+1)=x^{2}+2x+1$$ since the quotient is
$x+1$ and the reminder is $0$. The result is an equality
of polynomials, and it is valid if you set any element in the field
in those polynomials, even $x=-1$.
|
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$X$ is an odd number, $Y$ is a natural number more than 36. If $\frac{1}{X}+\frac{2}{Y}=\frac{1}{18}$, find the set $(X,Y)$? $X$ is an odd number, $Y$ is a natural number more than 36. If $\frac{1}{X}+\frac{2}{Y}=\frac{1}{18}$, find the set $(X,Y)$ ?
Re arranging the given equation, we have,
$\frac{2}{Y}=\frac{X-18}{18*X}$
$\frac{1}{Y}=\frac{X-18}{36*X}$
$Y=\frac{36*X}{X-18}$
$Y=\frac{1*2*2*3*3*X}{X-18}$
Now we know that $X$ is an odd number, therefore $X-18$ must also be an odd number.
By visual inspection we see that, $X-18$ can take values of 1, 3 and 9
i.e. $X-18 = 1 = 3^{0}$
and $X=19$
$X-18 = 3 = 3^{1}$
i.e, $X=21$
$X-18 = 9 = 3^{2}$
i.e, $X=27$
when I see the pattern for $X-18$ to be $3^{0}, 3^{1}, 3^{2}$ I'm tempted to try for $3^{3}, 3^{4},...$
when $X-18=3^3$ we have $X=45$
when $X-18=3^4$ we have $X=99$
I get the answer but whats the reasoning behind this ?
Is there another way to solve this question ?
|
As you have $Y=\frac{36X}{X-18},$
So, $Y=\frac{36(X-18)+36\cdot 18}{X-18}=36+\frac{3^4\cdot2^3}{X-18}$
also as, $Y>0, X>18$
and as you have identified $X-18=3^r ,0\le r\le 4 $
Alternatively
HINT:
So, $Y=\frac{36X}{X-18}$ which is even $=2Z$(say)
So, we have $$\frac1 X+\frac1 Z=\frac1{18}\implies XZ-18(X+Z)=0$$
$$\implies (X-18)(Z-18)=18^2\implies Z-18=\frac{18^2}{X-18}\text{ which is an integer }$$
Clearly, $(Z-18)(X-18)\ne0\implies X,Z>18$
|
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|
Quadratic Equation find the value of $\lambda$ when other roots are given in restriction Problem :
If $\lambda$ be an integer and $\alpha, \beta$ be the roots of $4x^2-16x+\lambda$=0 such that $ 1 < \alpha <2$ and $2 < \beta <3$, then find the possible values of $\lambda$
My approach :
The roots $\alpha, \beta = \frac{16 \pm \sqrt{256-16\lambda}}{8}$
$\Rightarrow \alpha, \beta = \frac{4 \pm \sqrt{16-\lambda}}{2}$
$\Rightarrow \alpha, \beta = \frac{4 \pm \sqrt{16-\lambda}}{2}$
$1 < \frac{4 \pm \sqrt{16-\lambda}}{2} < 2$
Also $ 2 < \frac{4 \pm \sqrt{16-\lambda}}{2} < 3$
Please suggest further..Thanks..
|
Using Vieta's formulas, $\alpha+\beta=4$ and $\alpha \beta=\frac{\lambda}4\implies \lambda=4\alpha \beta $
If $1< \alpha<2 \iff 1< 4-\beta <2\iff -1>\beta-4>-2\iff 3>\beta>2$
So, one condition implies the other
Now, $\lambda=4\alpha \beta=4\alpha(4-\alpha)=16-(2\alpha-4)^2$
As $1< \alpha<2 \implies 0>2\alpha-4>-2 \implies0<(2\alpha-4)^2<4 $
$\implies 0>-(2\alpha-2)^2>-4 \implies 16>16-(2\alpha-2)^2>12 $
|
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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
|
An equivalent identity is that $x+y+z = 0$ implies $x^3+y^3+z^3-3xyz = 0$.
So suppose $x+y+z = 0$.
Then the determinant $ \begin{vmatrix} x & z & y\\ y & x & z\\ z & y & x \end{vmatrix}$ must be zero, because the sum of the elements of each column is zero.
Expanding the determinant, we have the required result.
|
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|
Find the length of the parametric curve Find the length of the parametric curve defined by:
$x=t+\dfrac{1}{t}$
and
$y=\ln{t^2}$
on the interval
$(1 \le t \le 4)$.
|
$$\begin{align*}s&=\int_1^4\sqrt{\left(\frac{dx}{dt}\right)^2+\left(\frac{dy}{dt}\right)^2}dt\\&=\int_1^4\sqrt{\left(1-\frac1{t^2}\right)^2+\left(\frac2t\right)^2}dt\\&=\int_1^4\sqrt{1-\frac2{t^2}+\frac1{t^4}+\frac4{t^2}}dt\\&=\int_1^4\sqrt{\left(\frac1{t^2}\right)^2+2\left(\frac1{t^2}\right)+1}dt\\&=\int_1^4\sqrt{\left(\frac1{t^2}+1\right)^2}dt\\&=\int_1^4\left(\frac1{t^2}+1\right)dt\\&=\left[-\frac1t+t\right]_1^4\\&=-\frac14+4+1-1\\&=\frac{15}4\end{align*}$$
|
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How to find $\int_0^{2\pi} \frac{dt}{1+2\cos(t)}$ The problem is
$$\int_0^{2\pi} \frac{dt}{1+2\cos(t)}.$$ I know it is equal to
$$\int\limits_{|z|=1}\frac{2dz}{i(1+z)^2}$$
but I don't know how I should calculate the last integral.
|
If you write
$$
\begin{equation*}
z=e^{it}\qquad \left( 0\leq t\leq 2\pi \right),
\end{equation*}
$$
since
$$
\begin{equation*}
\cos t=\frac{z+z^{-1}}{2},\qquad dt=\frac{dz}{iz},
\end{equation*}
$$
the integral takes the form
$$
\begin{eqnarray*}
\int\limits_{|z|=1}\left( \frac{1}{1+2\frac{z+z^{-1}}{2}}
\right) \frac{1}{iz}dz &=&\int\limits_{|z|=1}\frac{1}{i}\times
\frac{1}{z^{2}+z+1}dz \\
&=&\int\limits_{|z|=1}\frac{dz}{i\left( z+\frac{1-i\sqrt{3}}{2}
\right) \left( z+\frac{1+i\sqrt{3}}{2}\right) }
\end{eqnarray*}
$$
and not
$$\int\limits_{|z|=1}\frac{2dz}{i(1+z)^2}.$$
Since $\cos \frac{2\pi }{3}=\cos \frac{4\pi }{3}=-\frac{1}{2}$, hence $1+2\cos \frac{2\pi }{3}=1+\cos \frac{4\pi }{3}=0$, the given integral has singularities at $t\in\{\frac{2\pi, }{3},\frac{4\pi }{3}\}$. We thus split it as follows:
$$
\begin{equation*}
\int_{0}^{\frac{2\pi }{3}}\frac{dt}{1+2\cos t}+\int_{\frac{2\pi }{3}}^{\frac{4\pi }{3}}\frac{dt}{1+2\cos t}+\int_{\frac{4\pi }{3}}^{2\pi }\frac{dt}{1+2\cos t}.
\end{equation*}
$$
Each integral is a divergent improper integral of the second kind. Using the Weirstrass substitution$^1$ $x=\tan \frac{t}{2}$, for instance the first integral becomes
$$
\begin{eqnarray*}
I_{1} &=&\int_{0}^{\frac{2\pi }{3}}\frac{1}{1+2\cos t}dt,\qquad x=\tan \frac{t}{2} \\
&=&\int_{0}^{\sqrt{3}}\frac{2}{\left( 1+2\frac{1-x^{2}}{1+x^{2}}\right)
\left( 1+x^{2}\right) }\,dx \\
&=&\int_{0}^{\sqrt{3}}\frac{2}{3-x^{2}}\,dx,\qquad x=\sqrt{3}u \\
&=&\frac{2\sqrt{3}}{3}\int_{0}^{1}\frac{1}{1-u^{2}}\,du \\
&=&\left. \frac{2\sqrt{3}}{3}\operatorname{arctanh}u\right\vert _{0}^{1}=\infty .
\end{eqnarray*}
$$
--
$^1$ The Weierstrass substitution is a universal standard substitution to evaluate an integral of a rational fraction in $\sin t,\cos t$, i.e. a rational fraction of the form
$$R(\sin t,\cos t)=\frac{P(\sin t,\cos t)}{Q(\sin t,\cos t)},$$
where $P,Q$ are polynomials in $\sin t,\cos t$
$$
\begin{equation*}
\tan \frac{t }{2}=x,\qquad t =2\arctan x,\qquad dt =\frac{2}{1+x^{2}}dx
\end{equation*},
$$
which converts the integrand into a rational function in $x$. We know from trigonometry (see this answer) that
$$\cos t =\frac{1-\tan ^{2}\frac{t }{2}}{1+\tan ^{2}\frac{
t}{2}}=\frac{1-x^2}{1+x^2},\qquad \sin t =\frac{2\tan \frac{t }{2}}{1+\tan ^{2}
\frac{t }{2}}=\frac{2x}{1+x^2}.$$
|
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|
Is my proof correct? $2^n=x^2+23$ has an infinite number of (integer) solutions. This is how I tried to prove it. Is it correct? Thanks!!
$2^n = x^2+23$
$x^2$ must be odd, therefore $x^2 = 4k+1$, where $k \in \mathbb{N}$.
$2^n=4k+24$
$k=2(2^{n-3}-3)$
Since $x^2=4k+1$,$ \ \ \ \ $ $k_1 = \frac{x_1^2-1}{4}$
and $k_2=\frac{(x_1+2)^2-1}{4}=\frac{x_1^2+4x_1+3}{4}$
If we substitute $x_1^2=4k_1 + 1$, we end up with:
$k_2=k_1 + \sqrt{4k_1+1} + 1$
Therefore, finding solutions of $k=2(2^{n-3} - 3)$ is comparable to
finding the solutions of $k+\sqrt{4k+1}+1=2(2^{n-3}-3)$
Let $p=2(2^{n-3}-3)$
Therefore, $(\sqrt{4k+1})^2=(p-k-1)^2$, so
$(p-k)^2 = 2(p+k)$
Since $p$ has an infinite number of solutions $(p-k)^2=2(p+k)$ also has an infinite number of solutions, which implies the the original does also.
|
This seems wrong to me.
You went from
$k_1 = \frac{x_1^2-1}{4}$
to $x_1^2 = 4 k_1^2+1$.
Somehow the $k_1$ got squared.
This is not a correct proof.
Also, having to download that big image is a pain.
Please learn how to enter math in $\LaTeX$.
|
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|
The angle at which a circle and a hyperbola intersect?
$x^2 - 2y^2 = 4$
$ (x-3)^2 + y^2 = 25 $
How do you calculate the angle at which a circle and a hyperbola intersect?
If I express $y^2$ from the first equation and apply it to the second equation, I get the following:
$y^2 = -2 + \frac{x^2}{2}$
$(x-3)^2 + -2 + \frac{x^2}{2} = 25$ ... $x^2 - 4x - 12 = 0$
$x_1 = 6, x_2 = -2 \implies y_1 = 4, y_{1'} = -4, y_2 = 0$
Now, for the points $(6,4)$ I could calculate the line equation which intersects the circle and the hyperbola: $(6-3)(x-3) + (4-0)(y-0) = 25 \implies y = -\frac{3x}{4} + \frac{17}{2}$
I calculated this because I thought I could apply the formula $\tan\phi = |\frac{k_2 - k_1}{1 + k_1k_2}|$
|
HINT:
$(1):$ From the article $151$ of The elements of coordinate geometry (Loney), the gradient $(m_1)$ of $x^2+y^2+2gx+2fy+c=0$ is $=-\frac{x_1+g}{y_1+f}$ where $(x_1,y_1)$ is the given point on the circle
and from the Article $305,262$ the gradient $(m_2)$ of $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$ is $\frac{b^2\cdot x_2}{a^2\cdot y_2}$ where $(x_2,y_2)$ is the given point on the hyperbola
$(2):$ Alternatively, find the gradients of each curve at $(6,4)$ and at $(-2,0)$ applying first order derivative
The acute angle between the curves will be $$\arctan\left|\frac{m_1-m_2}{1+m_1m_2} \right|$$
|
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|
Cyclotomic integers: Why do we have $x^n+y^n=(x+y)(x+\zeta y)\dots (x+\zeta ^{n-1}y)$? Why do we have the factorization $$x^n+y^n=(x+y)(x+\zeta y)\dots (x+\zeta ^{n-1}y)$$ for $\zeta$ a primitive $n^{\text{th}}$ root of unity where $n$ is an odd prime?
|
If $y = 0$, the factorisation reads $x^n = \underbrace{x\,.x \dots x}_{n\ \text{times}}$.
Consider $y\neq 0$. If $\zeta \in \mathbb{C}$ is a primitive $n^{\text{th}}$ root of unity then $\zeta^n = 1$ and $\zeta^k \neq 1$ for $k = 1, \dots, n-1$. In particular, $\zeta, \zeta^2, \dots, \zeta^{n-1}, \zeta^n=1$ are all distinct. Now consider $x^n + y^n$ as a polynomial in $x$ over the complex numbers (treat $y$ as a fixed constant). Then by the Fundamental Theorem of Algebra, $x^n + y^n$ has precisely $n$ zeroes. For each $k = 1, \dots, n$, $x = -\zeta^ky$ is a zero as
\begin{align*}
x^n + y^n &= (-\zeta^ky)^n + y^n\\
&= (-1)^n(\zeta^k)^ny^n + y^n\\
&= -(\zeta^n)^ky^n + y^n\qquad \text{as $n$ is odd, $(-1)^n = -1$}\\
&= -y^n + y^n\\
&= 0.
\end{align*}
As $y \neq 0$, $x = -\zeta^ky$ for $k=1, \dots, n$ are all distinct, so they are all of the zeroes of $x^n + y^n$. By the factor theorem, each zero corresponds to a linear factor, so we have
\begin{align*}
x^n + y^n &= a(x+\zeta y)(x+\zeta^2 y)\dots(x+\zeta^{n-1}y)(x+\zeta^ny)\\
&= a(x+\zeta y)(x+\zeta^2 y)\dots(x+\zeta^{n-1}y)(x+y)\\
&= a(x+y)(x+\zeta y)(x+\zeta^2 y)\dots(x+\zeta^{n-1}y)
\end{align*}
for some $a \in \mathbb{C}$. Comparing the coefficients of $x^n$, we deduce that $a = 1$ and arrive at the desired factorisation
$$x^n+y^n = (x+y)(x+\zeta y)(x+\zeta^2 y)\dots(x+\zeta^{n-1}y).$$
|
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|
About Rayleigh's formula How to use $\displaystyle j_n(x)=(-1)^nx^n\left(\frac{1} {x} \frac{d} {dx}\right)^n \left(\frac{\sin x}{x}\right)$?
for example, to find $j_3(x)$
|
The formula you provided uses a compact way of writing operators.
For example, at $n = 2$, the term $\displaystyle \left(\frac{1}{x} \frac{d}{dx}\right)^2$ should be understood to mean the operator
$$\displaystyle \left(\frac{1}{x} \frac{d}{dx}\right)^2=\frac{1}{x} \frac{d}{dx}\frac1x \frac d{dx}.$$
You can see the values of Rayleigh's formula which are Spherical Bessel functions on the Wiki, but lets calculate some of them so you can get a feel for how this operator works.
We have (I am writing these as they are typically written - like on the Wiki page):
$$\displaystyle j_n(x)=(-1)^nx^n\left(\frac{1} {x} \frac{d} {dx}\right)^n \left(\frac{\sin x}{x}\right)$$
So:
$\displaystyle j_0(x) = (-1)^0x^0\left(\frac{1} {x} \frac{d} {dx}\right)^0 \left(\frac{\sin x}{x}\right) = \frac{\sin x}{x}$
$\displaystyle j_1(x) = (-1)^1x^1\left(\frac{1} {x} \frac{d} {dx}\right)^1 \left(\frac{\sin x}{x}\right) = (-1)^1x^1\left(\frac{1} {x} \frac{x \cos x - \sin x}{x^2}\right) = \frac{\sin x}{x^2} - \frac{\cos x}{x}$
$\displaystyle j_2(x) = (-1)^2x^2\left(\frac{1} {x} \frac{d} {dx}\right)^2 \left(\frac{\sin x}{x}\right)$
$\displaystyle = x^2 \left(\frac{1} {x} \frac{d} {dx} \frac{1} {x} \frac{d} {dx}\right) \left(\frac{\sin x}{x}\right)$
$\displaystyle = x^2 \left(\frac{1} {x} \frac{d} {dx} \frac{1} {x} \right) \left(\frac{x \cos x - \sin x}{x^2}\right)$
$\displaystyle = x^2 \left(\frac{1} {x} \frac{d} {dx} \right) \left(\frac{x \cos x - \sin x}{x^3}\right) $
$\displaystyle = x^2 \left(\frac{1} {x} \right) \left(\frac{(3-x^2) \sin x - 3x \cos x}{x^4}\right)$
$\displaystyle = \left(\frac{3}{x^2} -1 \right) \frac{\sin x}{x} - \frac{3 \cos x}{x^2}$
$\displaystyle j_3(x) = (-1)^3x^3\left(\frac{1}{x} \frac{d} {dx}\right)^3 \left(\frac{\sin x}{x}\right) = -x^3 \left(\frac{1}{x} \frac{d}{dx} \frac{1}{x} \frac{d}{dx} \frac{1}{x} \frac{d}{dx} \right)\left(\frac{\sin x}{x}\right) $
After some calculations, we arrive at:
$$j_3(x) = \left(\frac{15}{x^3} - \frac{6}{x} \right) \frac{\sin x}{x} -\left(\frac{15}{x^2} - 1 \right) \frac{\cos x}{x}$$
I will let you work that last one, but if you get stuck, just reply and I will add the details!
|
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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 $\det A$?
|
Add all the columns to the first and we find
$$ B=\begin{pmatrix} a+3 & 1 & 1 & 1 \\ a+3 & a & 1 & 1\\ a+3 & 1 & a & 1\\ a+3 & 1 & 1 & a \end{pmatrix}$$
then subtract the first row from the other we find
$$ C=\begin{pmatrix} a+3 & 1 & 1 & 1 \\ 0 & a-1 & 0 & 0\\ 0 & 0 & a-1 & 0\\ 0 & 0 & 0 & a-1 \end{pmatrix}$$
hence
$$\det A_a=\det C=(a+3)(a-1)^3$$
and note that the elementary operation preserve the rank so
*
*if $a\neq -3$ and $a\neq 1$ then $\mathrm{rank}A_a=4$
*if $a=1$ then $\mathrm{rank}A_1=1$
*if $a=-3$ then $\mathrm{rank}A_{-3}=3$
|
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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.
|
$$\text{As }\frac1{(2r-1)2r}=\frac{2r-(2r-1)}{(2r-1)2r}=\frac1{2r-1}-\frac1{2r},$$
$$\text{we can write }\frac{1}{1\cdot 2}+\frac{1}{3\cdot 4}+\frac{1}{5\cdot 6}+...+\frac{1}{(2n-1)\cdot 2n}$$
$$=\left(\frac11-\frac12\right)+\left(\frac13-\frac14\right)+\cdots+\left(\frac1{2n-1}-\frac1{2n}\right)$$
$$=\left(\frac11+\frac12-2\cdot\frac12\right)+\left(\frac13+\frac14-2\cdot\frac14\right)+\cdots+\left(\frac1{2n-1}+\frac1{2n}-2\cdot\frac1{2n}\right)$$
$$=\frac11+\frac12+\frac13+\frac14+\cdots+\frac1{n-1}+\frac1n+\frac1{n+1}+\cdots+\frac1{2n-1}+\frac1{2n}-2\left(\frac12+\frac14+\frac16+\cdots+\frac1{2n}\right)$$
$$=\frac11+\frac12+\frac13+\frac14+\cdots+\frac1{n-1}+\frac1n+\frac1{n+1}+\cdots+\frac1{2n-1}+\frac1{2n}-\left(1+\frac12+\frac13+\cdots+\frac1n\right)$$
$$=\frac1{n+1}+\frac1{n+2}+\cdots+\frac1{2n-1}+\frac1{2n}$$
|
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|
Determine whether a multi-variable limit exists $\lim_{(x,y)\to(0,0)}\frac{\cos x-1-\frac{x^2}2}{x^4+y^4}$ I need to determine whether the next limit exists:
$$\lim_{(x,y)\to(0,0)}f(x,y)=\lim_{(x,y)\to(0,0)}\frac{\cos x-1-\frac{x^2}2}{x^4+y^4}$$
Looking at the numerator $(-1-\frac{x^2}2)$ it immediately reminds me of maclaurin series of $\cos$, Where:
$$\cos x=1-\frac{x^2}{2!}+\frac{x^4}{4!}+...=1-\frac{x^2}{2!}+\frac{x^4}{4!}+O(x^6)$$
So we can simplify the expression:
$\begin{align} \lim_{(x,y)\to(0,0)}\frac{\cos x-1-\frac{x^2}2}{x^4+y^4}&=\lim_{(x,y)\to(0,0)}\frac{1-\frac{x^2}{2!}+\frac{x^4}{4!}+O(x^6)-1-\frac{x^2}2}{x^4+y^4}\\
&=\lim_{(x,y)\to(0,0)}\frac{-x^2+\frac{x^4}{4!}+O(x^6)}{x^4+y^4} \end{align}$
This limit does not exist.
So, after solving it the 'hard' way, I was thinking:
If I place $y=mx$ and then try to solve the limit I get:
$$\lim_{(x,y)\to(0,0)}\frac{\cos x-1-\frac{x^2}2}{x^4(1+m^4)}=\infty$$
Is that enough to prove that the limit does not exists? I mean, If I find at least one limit that does not 'converge' is it enough to say that a multi-variable function limit doesn't exist?
|
I think you can assume the following limit as well to show that it doesn't exist:
$$\lim_{n\to\infty}f(x,y_n),~~y_n=\pi/n$$
|
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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!
|
Use L'Hospital's Rule. Since plugging in $x=1$, gives you indeterminate form, take the derivative of the numerator and the derivative of the denominator, and try the limit again.
$\lim_{x\to 1}\frac{(x^2+3)^{\frac{1}{2}}-2}{x^2-1}\implies$ (Via L
Hospital's Rule...) $\lim_{x\to 1}\frac{\frac{1}{2}(x^2+3)^{-\frac{1}{2}}(2x)}{2x}=\frac{\frac{1}{2}(1^2+3)^{-\frac{1}{2}}(2(1))}{2(1)}=\frac{\frac{1}{2}(4)^{-\frac{1}{2}}(2)}{2}=\frac{1}{2}(4)^{-\frac{1}{2}}=(\frac{1}{2})(\frac{1}{4})^{-\frac{1}{2}}=(\frac{1}{2})(\frac{1}{2})=\frac{1}{4}$
|
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|
Proving a formula for $\int_0^\infty \frac{\log(1+x^{4n})}{1+x^2}dx $ if $n=1,2,3,\cdots$ I came across the formula
$$\int_0^\infty \frac{\log \left(1+x^{4n} \right)}{1+x^2}dx = \pi \log \left\{2^n \prod_{k=1 ,\ k \text{ odd}}^{2n-1} \left(1+\sin \left( \frac{\pi k}{4n}\right) \right)\right\} $$
where $n=1,2,3,4,\cdots$
I checked it for a few values of $n$ and it seems to give correct results.
Please help me prove this result.
|
Evaluate
$J=\int_0^\infty\frac{\ln(1+x^{4n})}{1+x^2} \,dx $
with
$1+x^{4n} = \prod_{k=1}^{2n}(1+e^{i\pi\frac{2n+1-2k}{2n} }x^2)
$
and
$\int_0^\infty \frac{\ln(1+r x^2)}{1+x^2}dx= \pi\ln(1+r^{\frac12})
$
\begin{align}
J& =\int_0^\infty\frac{dx}{1+x^2} \sum_{k=1}^{2n}
\ln (1+e^{i\pi\frac{2n+1-2k}{2n} }x^2)
= \pi\sum_{k=1}^{2n} \ln (1+e^{i\pi\frac{2n+1-2k}{4n} })\\
&=2\pi\sum_{k=1}^{n} \ln \left(2\cos\frac{(2n+1-2k)\pi}{8n} \right)
=2\pi\sum_{j=1,odd}^{2n-1} \ln \left(2\cos\frac{(2n-j)\pi}{8n}\right) \\
&=\pi \sum_{j=1,odd}^{2n-1} \ln \left(4\cos^2\frac{(2n-j)\pi}{8n} \right)
=\pi \sum_{j=1,odd}^{2n-1} \ln \left(2(1+\sin\frac{\pi j}{4n} )\right)\\
& =\pi \ln \bigg( 2^n \prod_{j=1,odd}^{2n-1} \left(1+\sin \frac{\pi j}{4n}\right)\bigg)
\end{align}
|
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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.
$$\lim_{x\to0}\frac{\sin(x^2+\frac{1}{x})-\sin\frac{1}{x}}{x}$$
|
Using the identity
$$
\sin(A)-\sin(B)=2\sin\left(\frac{A-B}{2}\right)\cos\left(\frac{A+B}{2}\right)
$$
we get
$$
\begin{align}
\lim_{x\to0}\frac{\sin\left(x^2+\frac1x\right)-\sin\left(\frac1x\right)}{x}
&=\lim_{x\to0}\frac{2\sin\left(\frac{x^2}{2}\right)\cos\left(\frac{x^2}{2}+\frac1x\right)}{x}\\
&=\lim_{x\to0}x\frac{\sin\left(\frac{x^2}{2}\right)\cos\left(\frac{x^2}{2}+\frac1x\right)}{\frac{x^2}{2}}\\
&=\lim_{x\to0}x\cos\left(\frac{x^2}{2}+\frac1x\right)\lim_{x\to0}\frac{\sin\left(\frac{x^2}{2}\right)}{\frac{x^2}{2}}\\[12pt]
&=0\cdot1
\end{align}
$$
|
{
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"timestamp": "2023-03-29T00:00:00",
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|
Find the natural number $n$ satisfy the condition Find the natural number $n$ satisfy the condition
$$\dfrac{1}{2}C_{2n}^1 - \dfrac{2}{3} C_{2n}^2 + \dfrac{3}{4} C_{2n}^3 - \dfrac{4}{5} C_{2n}^4 + \cdots - \dfrac{2n}{2n+1} C_{2n}^{2n} =\dfrac{1}{2013}.$$
|
HINT:
$$\frac r{r+1}\binom {2n}r=\binom {2n}r-\frac{(2n)!}{r!(2n-r)!(r+1)}$$
$$=\binom {2n}r-\frac1{2n+1}\cdot \binom{2n+1}{r+1}$$
$$\text{So, }\sum_{1\le r\le 2n}\frac r{r+1}(-1)^{r-1}\binom {2n}r$$
$$=\sum_{1\le r\le 2n}(-1)^{r-1}\binom{2n}r-\frac1{2n+1}\cdot \sum_{1\le r\le 2n}(-1)^{r-1}\binom{2n+1}{r+1}$$
$$\text{ Now, }\sum_{1\le r\le 2n}(-1)^{r-1}\binom{2n}r=1-\left(\binom{2n}0-\binom{2n}1+\binom{2n}2-\cdots+\binom{2n}{2n}\right)=1-(1-1)^{2n}=1$$
$$\text{ and } \sum_{1\le r\le 2n}(-1)^{r-1}\binom{2n+1}{r+1}$$
$$=\binom{2n+1}0-\binom{2n+1}1+\binom{2n+1}2-\cdots+\binom{2n+1}{2n}-\binom{2n+1}{2n+1}-\binom{2n+1}0+\binom{2n+1}1$$
$$=(1-1)^{2n+1}-\binom{2n+1}0+\binom{2n+1}1=2n+1-1=2n$$
So, the result will be $$1-\frac{2n}{2n+1}=\frac1{2n+1}$$
|
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|
Compute this limit $\lim_{x\to0}\frac{\sin(x^2+\frac{1}{x})-\sin\frac{1}{x}}{x}$ using L'Hôpital's rule I have asked this problem before, but I can't understand the explanation, I couldn't understand how the sin multiply for cos, and too multiply for A + and - B: $$\sin(A)-\sin(B)=2\sin\left(\frac{A-B}{2}\right)\cos\left(\frac{A+B}{2}\right)$$ and I don't understand in this step how/why the $A-B$ and $A+B$ was replaced by $\frac{x^2}{2}$ and $\frac{x^2}{2}+\frac{1}{x}$ :
$$\lim_{x\to0}\frac{\sin\left(x^2+\frac1x\right)-\sin\left(\frac1x\right)}{x}= \lim_{x\to0}\frac{2\sin\left(\frac{x^2}{2}\right)\cos\left(\frac{x^2}{2}+\frac1x\right)}{x}.$$
|
It's easier if you use the plain old addition formula:
$$\sin{\left(x^2+\frac{1}{x}\right)} = \cos{x^2}\, \sin{\frac{1}{x}} + \sin{x^2} \,\cos{\frac{1}{x}}$$
which behaves as
$$\left(1-\frac12 x^2\right) \sin{\frac{1}{x}} + x^2 \cos{\frac{1}{x}}$$
so that
$$\frac{\sin{\left(x^2+\frac{1}{x}\right)} - \sin{\frac{1}{x}}}{x} \sim \frac{x^2 \cos{\frac{1}{x}} - (1/2)x^2 \sin{\frac{1}{x}}}{x}$$
which goes to zero as $x \to 0$.
|
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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 $a^2 b^2 c^2 \le \frac{1}{54}$.
|
Several good answers have been posted. Just feeling like adding a single variable solution with a geometric twist.
The set of points $(a,b,c)$ is the intersection of the unit sphere and the plane $T:x+y+z=0$ with normal vector $(1,1,1)$. IOW a circle of radius one in the plane $T$ centered at the origin. An occasionally useful factoid is that those points are parametrized by the formula
$$
(a,b,c)=K(\cos t, \cos(t+\frac{2\pi}3), \cos(t-\frac{2\pi}3)),
$$
with the constant $K=\sqrt{2/3}$. Some of you may have seen the relevant trig identities in high school physics in connection with three-phase power: the total power is constant (=> on the unit sphere) and the instantaneous sum of the voltages is zero (=> on the plane $T$).
Anyway, here
$$
bc=K^2\cos(t+\frac{2\pi}3)\cos(t-\frac{2\pi}3)=K^2(\cos^2t-\frac34)
$$
by basic trig identities. Therefore
$$
abc=K^3\frac{4\cos^3t-3\cos t}4=\frac{K^3\cos3t}4.
$$
Hence
$$
(abc)^2\le \frac{K^6}{16}=\frac{8/27}{16}=\frac1{54}.
$$
|
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|
How to compute $\sum_{n=0}^{\infty}\frac{(2n+1)}{n!}x^{2n+1}$? I don't know how to compute $\sum_{n=0}^{\infty}\frac{(2n+1)}{n!}x^{2n+1}$,appreciate any help!
Is there any general rule for solving such problems?
|
$$\text{As } e^x=\sum_{0\le r<\infty}\frac{x^r}{r!} $$
$$\text{and }\frac{(2n+1)}{n!}x^{2n+1}=2x^3 \cdot \frac{(x^2)^{n-1}}{(n-1)!}+x\cdot \frac{(x^2)^n}{n!}$$
$$\text{So,} \sum_{n=0}^{\infty}\frac{(2n+1)}{n!}x^{2n+1}$$
$$=2x^3 \cdot \sum_{n=0}^{\infty}\frac{(x^2)^{n-1}}{(n-1)!}+x\cdot \sum_{n=0}^{\infty}\frac{(x^2)^n}{n!}$$
$$=2x^3 \cdot \sum_{m=0}^{\infty}\frac{(x^2)^m}{m!}+x\cdot \sum_{n=0}^{\infty}\frac{(x^2)^n}{n!}\text{ putting } n-1=m \text{ and using} \frac1{(-1)!}=0 $$
$$=2x^3 \cdot e^{x^2}+x\cdot e^{x^2}$$
|
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|
Ramanujan-Sums... How do they do? Prove that:
$\displaystyle \frac{1}{e^{2\pi}-1}+\frac{2}{e^{4\pi}-1}+\frac{3}{e^{6\pi}-1}+...=\frac{1}{24}-\frac{1}{8\pi}$
I would like to see different ways of solving this beautiful sum, whoever is encouraged? :)
Thanks to all.
|
This problem is already solved in Summing $\frac{1}{e^{2\pi}-1} + \frac{2}{e^{4\pi}-1} + \frac{3}{e^{6\pi}-1} + \cdots \text{ad inf}$ and has a nice solution there.
But then the simplest proof comes from Ramanujan. He uses the transformation formula for "eta" function and logarithmic differentiation. If we define $\eta(q)$ by
$$\eta(q) = q^{1/12}(1 - q^{2})(1 - q^{4})(1 - q^{6})\cdots$$ then we have $$\frac{\eta(e^{-\pi\sqrt{n}})}{\eta(e^{-\pi/\sqrt{n}})} = n^{-1/4}$$ or in other words $$n^{1/4}e^{-\pi\sqrt{n}/12}(1 - e^{-2\pi\sqrt{n}})(1 - e^{-4\pi\sqrt{n}})\cdots = e^{-\pi/(12\sqrt{n})}(1 - e^{-2\pi/\sqrt{n}})(1 - e^{-4\pi/\sqrt{n}})\cdots$$
Then if we differentiate the above equation logarithmically with respect to $n$ we get
$$n\left\{1 - 24\left(\frac{1}{e^{2\pi\sqrt{n}} - 1} + \frac{2}{e^{4\pi\sqrt{n}} - 1}+ \cdots\right)\right\} + \left\{1 - 24\left(\frac{1}{e^{2\pi/\sqrt{n}} - 1} + \frac{2}{e^{4\pi/\sqrt{n}} - 1}+ \cdots\right)\right\} = \frac{6\sqrt{n}}{\pi}$$
Putting $n = 1$ in above equation we get $$1 - 24\left(\frac{1}{e^{2\pi} - 1} + \frac{2}{e^{4\pi} - 1} + \frac{3}{e^{6\pi} - 1}\right) = \frac{3}{\pi}$$
and thus the proof is completed. The above derivation is coming straight from Ramanujan's most famous and amazing paper "Modular Equations and Approximations to $\pi$".
|
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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 $$\left(\frac{a}{b+1}\right)^2 +\left(\frac{b}{a+1}\right)^2$$
|
$$a+b=-3,ab=1$$
$$(a+b)^2=a^2+b^2+2ab\implies a^2+b^2=7$$
$$(a+b)^3=a^3+b^3+3ab(a+b)\implies a^3+b^3=-18$$
$$(a^2+b^2)^2=a^4+b^4+2(ab)^2\implies a^4+b^4=47$$
$${\left(\dfrac {a}{b+1}\right)}^2+{\left(\dfrac {b}{a+1}\right)}^2$$
$$\dfrac {a^2(a+1)^2+b^2(b+1)^2}{((b+1)(a+1))^2}$$
$$\dfrac {a^4+2a^3+a^2+b^4+2b^3+b^2}{(ab+a+b+1)^2}$$
$$\dfrac {a^4+b^4+2(a^3+b^3)+a^2+b^2}{(1-3+1)^2}$$
$${47+2(-18)+7}$$
$$18$$
There is one alternate as in comment @mark suggest:
$$\dfrac {a^4+2a^3+a^2+b^4+2b^3+b^2}{(ab+a+b+1)^2}$$
$${a^4+3a^3+a^2-a^3+b^4+3b^3+b^2-b^3}$$
$${a^2(a^2+3a+1)-a^3+b^2(b^2+3b+1)-b^3}$$
since $a^2+3a+1=b^2+3b+10=0$
$$-a^3-b^3$$
$$18$$
|
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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 can find quickly $S$ if you note that $$5+2\sqrt{13}=\left(\frac{1+\sqrt{13}}{2}\right)^3$$ and similarly for $5-2\sqrt{13}$.
|
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|
Prove $\int_0^{\infty}\! \frac{\mathbb{d}x}{1+x^n}=\frac{\pi}{n \sin\frac{\pi}{n}}$ using real analysis techniques only I have found a proof using complex analysis techniques (contour integral, residue theorem, etc.) that shows $$\int_0^{\infty}\! \frac{\mathbb{d}x}{1+x^n}=\frac{\pi}{n \sin\frac{\pi}{n}}$$ for $n\in \mathbb{N}^+\setminus\{1\}$
I wonder if it is possible by using only real analysis to demonstrate this "innocent" result?
Edit
A more general result showing that
$$\int\limits_{0}^{\infty} \frac{x^{a-1}}{1+x^{b}} \ \text{dx} = \frac{\pi}{b \sin(\pi{a}/b)}, \qquad 0 < a <b$$
can be found in another math.SE post
|
In general, let $$y=\dfrac{1}{1+x^b}\quad\Rightarrow\quad x=\left(\dfrac{1-y}{y}\right)^{\frac1b}\quad\Rightarrow\quad dx=-\left(\dfrac{1-y}{y}\right)^{\frac1b-1}\ \dfrac{dy}{by^2}\ ,$$ then
\begin{align}
\int_0^\infty\dfrac{x^{a-1}}{1+x^b}\ dx&=\int_0^1 y\left(\dfrac{1-y}{y}\right)^{\large\frac{a-1}b}\left(\dfrac{1-y}{y}\right)^{\large\frac1b-1}\ \dfrac{dy}{by^2}\\&=\frac1b\int_0^1y^{\large1-\frac{a}{b}-1}(1-y)^{\large\frac{a}{b}-1}\ dy,
\end{align}
where the integral in RHS is Beta function.
$$
\text{B}(x,y)=\int_0^1t^{\ x-1}\ (1-t)^{\ y-1}\ dt=\frac{\Gamma(x)\cdot\Gamma(y)}{\Gamma(x+y)}.
$$
Hence
\begin{align}
\int_0^\infty\dfrac{x^{a-1}}{1+x^b}\ dx&=\frac1b\int_0^1y^{1-\frac{a}{b}-1}(1-y)^{\frac{a}{b}-1}\ dy\\&=\frac1b\cdot\Gamma\left(1-\frac{a}{b}\right)\cdot\Gamma\left(\frac{a}{b}\right)\\&={\color{blue}{\frac{\pi}{b\sin\left(\frac{a\pi}{b}\right)}}}.
\end{align}
The last part uses Euler's reflection formula for Gamma function provided $0<a<b$. Thus
$$
\int_0^\infty\dfrac{1}{1+x^n}\ dx=\color{blue}{\frac{\pi}{n\sin\left(\frac{\pi}{n}\right)}}.
$$
QED
|
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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?
|
$$\Longleftrightarrow \dfrac{a^4}{ab^2-abc+ac^2}+\dfrac{b^4}{bc^2-abc+ba^2}+\dfrac{c^4}{ca^2-abc+cb^2}\ge a+b+c$$
by cauchy-Schwarz we have
$$\left[ \dfrac{a^4}{ab^2-abc+ac^2}+\dfrac{b^4}{bc^2-abc+ba^2}+\dfrac{c^4}{ca^2-abc+cb^2}\right][bc(b+c)+ac(a+c)+ab(a+b)-3abc]\ge (a^2+b^2+c^2)^2$$
$$\Longleftrightarrow (a^2+b^2+c^2)^2\ge (a+b+c)[(bc(b+c)+ac(a+c)+ab(a+b)-3abc]$$
let $a=\min{\{a,b,c\}}$
$$\Longleftrightarrow a^2(a-b)(a-c)+(b^2+c^2+bc-ab-ac)(b-c)^2\ge 0$$
By Done
|
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|
Integration trig substitution $\int \frac{dx}{x\sqrt{x^2 + 16}}$ $$\int \frac{dx}{x\sqrt{x^2 + 16}}$$
With some magic I get down to $$\frac{1}{4} \int\frac{1}{\sin\theta} d\theta$$
Now is where I am lost. How do I do this? I tried integration by parts but it doesn't work.
|
UPDATE 2 The first integral is not $\int \frac{dx}{x\sqrt{x^{2}-16}}$ but $\int
\frac{dx}{x\sqrt{x^{2}+16}}$ (as noticed by julien), because
\begin{eqnarray*}
I &=&\int \frac{dx}{x\sqrt{x^{2}+16}},\qquad x=\tan \theta ,dx=\sec
^{2}\theta d\theta \\
&=&\int \frac{\sec ^{2}\theta }{\left( \tan \theta \right) 4\sec \theta }%
\,d\theta =\int \frac{\sec \theta }{4\tan \theta }\,d\theta \\
&=&\int \frac{\sec \theta }{4\tan \theta }\,d\theta =\int \frac{1}{4\sin
\theta }\,d\theta \text{.}
\end{eqnarray*}
Use the Weierstrass substitution $$t=\tan\frac{\theta}{2}.$$
Then
$$\int \frac{1}{\sin \theta }\,d\theta =\int \frac{2}{\frac{2t}{1+t^{2}}
\left( 1+t^{2}\right) }\,dt=\int \frac{1}{t}\,dt=\ln \left\vert t\right\vert
+C=\ln \left\vert \tan \frac{\theta }{2}\right\vert +C.$$
Comment: The Weierstrass substitution is a universal standard substitution to evaluate an integral of a rational fraction in $\sin \theta,\cos \theta$, i.e. a rational fraction of the form
$$R(\sin \theta,\cos \theta)=\frac{P(\sin \theta,\cos \theta)}{Q(\sin \theta,\cos \theta)},$$
where $P,Q$ are polynomials in $\sin \theta,\cos \theta$
$$
\begin{equation*}
\tan \frac{\theta }{2}=t,\qquad\theta =2\arctan t,\qquad d\theta =\frac{2}{1+t^{2}}dt
\end{equation*},
$$
which converts the integrand into a rational function in $t$. We know from trigonometry that
$$\cos \theta =\frac{1-\tan ^{2}\frac{\theta }{2}}{1+\tan ^{2}\frac{
\theta}{2}}=\frac{1-t^2}{1+t^2},\qquad \sin \theta =\frac{2\tan \frac{\theta }{2}}{1+\tan ^{2}
\frac{\theta }{2}}=\frac{2t}{1+t^2}.$$
Proof. A possible proof is the following one, which uses the double-angle formulas and the identity $\cos ^{2}\frac{\theta}{2}+\sin ^{2}\frac{\theta}{2}=1$:
$$
\begin{eqnarray*}
\cos \theta &=&\cos ^{2}\frac{\theta}{2}-\sin ^{2}\frac{\theta }{2}=\frac{\frac{\cos ^{2}
\frac{\theta}{2}-\sin ^{2}\frac{\theta}{2}}{\cos ^{2}\frac{\theta}{2}}}{\frac{\cos ^{2}
\frac{\theta}{2}+\sin ^{2}\frac{\theta}{2}}{\cos ^{2}\frac{\theta}{2}}}=\frac{1-\tan ^{2}
\frac{\theta}{2}}{1+\tan ^{2}\frac{\theta}{2}}, \\
&& \\
\sin \theta &=&2\sin \frac{\theta}{2}\cos \frac{\theta}{2}=\frac{\frac{2\sin \frac{\theta}{2}
\cos \frac{\theta}{2}}{\cos ^{2}\frac{\theta}{2}}}{\frac{\cos ^{2}\frac{\theta}{2}+\sin ^{2}
\frac{\theta}{2}}{\cos ^{2}\frac{\theta}{2}}}=\frac{2\tan \frac{\theta}{2}}{1+\tan ^{2}
\frac{\theta}{2}}.
\end{eqnarray*}
$$
Another possible substitution is the Euler substitution
$$
\begin{equation*}
\sqrt{x^{2}+16}=t+x.
\end{equation*}
$$
Then
$$
\begin{eqnarray*}
I &=&\int \frac{dx}{x\sqrt{x^{2}+16}}=\int \frac{2}{t^{2}-16}\,dt \\
&=&\int \frac{1}{4\left( t-4\right) }-\frac{1}{4\left( t+4\right) }dt=\frac{1
}{4}\ln \left\vert \frac{t-4}{t+4}\right\vert +C \\
&=&\frac{1}{4}\ln \left\vert \frac{\sqrt{x^{2}+16}-x-4}{\sqrt{x^{2}+16}-x+4}
\right\vert +C.
\end{eqnarray*}
$$
|
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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!
|
We can write
$$
\begin{align}
x & = \sqrt{1+2\sqrt{2+3\sqrt{3+4\sqrt{4+5\sqrt{5+\cdots}}}}} \\
& = \sqrt{1+2\sqrt{2}\sqrt{1+\frac{3\sqrt{3}}{2}\sqrt{1+\frac{4\sqrt{4}}{3}\sqrt{1+\frac{5\sqrt{5}}{4}\sqrt{1+\cdots}}}}} \\
&< \sqrt{1+2\sqrt{2}\sqrt{1+\frac{3\sqrt{3}}{2}\sqrt{1+3\sqrt{1+4\sqrt{1+5\sqrt{1+\cdots}}}}}}
\end{align}
$$
since
$$
\frac{n\sqrt{n}}{n-1}<n-1 \quad \mathrm{for~n>3}
$$
Then using the Ramanujan identity @ShuhaoCao cited
$$
1+n = \sqrt{1+n\sqrt{1+(n+1)\sqrt{1+(n+2)\sqrt{1+\cdots}}}}
$$
it follows that $x$ converges and we can get upper bounds
$$
x < \sqrt{1+2\sqrt{2}\sqrt{1+\frac{3\sqrt{3}}{2}\cdot (1+3)}} = 3.247\cdots \\
x< \sqrt{1+2\sqrt{2}\sqrt{1+\frac{3\sqrt{3}}{2}\sqrt{1+\frac{4\sqrt{4}}{3}\sqrt{1+\frac{5\sqrt{5}}{4}\cdot (1+5)}}}} = 3.159\cdots
$$
and so forth.
Refining these upper bounds and the lower bounds described by @ShuhaoCao
$$
a_n = \sqrt{1+2\sqrt{2+3\sqrt{3+\cdots\sqrt{n+(n+1)(n+3)}}}} \\
b_n = \sqrt{1+2\sqrt{2}\sqrt{1+3\sqrt{3}\sqrt{1+\cdots\sqrt{1+\frac{(n+1)^{3/2}}{n}(n+2)}}}} \\
a_n<x<b_n \quad n>1
$$
by computing $a_{1000},b_{1000}$ we can find
$$
x = 3.083355141830694458051142580088\cdots
$$
with a range of $<10^{-100}$.
|
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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??
|
HINT:
$$a^4+b^4+c^4=(a^2+b^2+c^2)^2-2(a^2b^2+b^2c^2+c^2a^2)$$
Now $$a^2b^2+b^2c^2+c^2a^2=(ab+bc+ca)^2-2abc(a+b+c)$$
$$2(ab+bc+ca)=(a+b+c)^2-(a^2+b^2+c^2)$$ we can find $ab+bc+ca$ from here
$$a^3+b^3+c^3-3abc$$
$$=(a+b)^3-3ab(a+b)+c^3-3abc$$
$$=(a+b+c)\{(a+b)^2-(a+b)c+c^2\}-3ab(a+b+c)$$
$$=(a+b+c)\{(a+b)^2+c^2-3ab\}$$
$$= (a+b+c)\{(a+b+c)^2-3(ab+bc+ca)\}$$ we can find $abc$ from here
|
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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??
|
$$z=2x^2 + y^2 +2xy -2x +2y +2$$
$$\implies 2x^2+2x(y-1)+y^2+2y+2-z=0$$
As $x$ is real, the discriminant of the above Quadratic equation must be $\ge0$
$$\implies \{2(y-1)\}^2-4\cdot 2\cdot(y^2+2y+2-z)\ge0 $$
$$\implies (y-1)^2- 2\cdot(y^2+2y+2-z)\ge0 $$
$$\implies 2z\ge y^2+6y+3$$
$$\text{Now, }y^2+6y+3=(y+3)^2-6\ge -6$$
|
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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.
|
You were on the right track!
$x\left(z^3-y^3\right)+y\left(x^3-z^3\right)+z\left(y^3-x^3\right)$
$=x\left(z^3-x^3 + x^3-y^3\right)+y\left(x^3-z^3\right)+z\left(y^3-x^3\right)$
$=\left(z^3-x^3\right)(x-y) + \left(x^3-y^3\right) (x - z)$
$=(z-x)(x-y)[z^2 + x^2 + zx -x^2 -y^2 -xy]$
$=(z-x)(x-y)[z^2 -y^2 + zx -xy]$
$=(z-x)(x-y)(z-y)[z + y + x]$
|
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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 $c=2a-d$.
|
Vieta's formulas $\Rightarrow \ \ $ $a+b=2c$, $c+d=2a$ $\ \ \Rightarrow \ \ $ $a+b+c+d=2(a+c)$ $\ \Rightarrow \ $ $a+c=b+d$.
Denote
$$m = \dfrac{a+c}{2}=\dfrac{b+d}{2}, \qquad p=\dfrac{c-a}{2}\color{gray}{=m-a=c-m}.$$
Then
$$
\left\{
\begin{array}{r}
a = m-p, \quad b=m+3p, \\
c = m+p, \quad d=m-3p.
\end{array}
\right.
$$
Vieta's formulas $\Rightarrow \ $ $ab=-5d$, $\ \ $ $cd=-5b$ $\ \ \Rightarrow$
$$
\left\{
\begin{array}{r}
(m-p)(m+3p)=-5m+15p, \\
(m+p)(m-3p)=-5m-15p;
\end{array}
\right.
$$
$$
\left\{
\begin{array}{r}
m^2+2mp-3p^2=-5m+15p, \\
m^2-2mp-3p^2=-5m-15p;
\end{array}
\right.
$$
$$
\left\{
\begin{array}{c}
m^2-3p^2=-5m, \\
2mp=15p.
\end{array}
\right.
$$
Since $a,b,c,d$ are distinct, then $p\ne 0$, then $2m=15$, then
$$\color{#660011}{\Large{a+b+c+d=4m=30}}.$$
Note:
$3p^2=m^2+5m=\dfrac{375}{4}$ $\ \ \Rightarrow \ \ $ $p =\pm \dfrac{5\sqrt{5}}{2}$.
|
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|
Taylor series expansion for $f(x)=\sqrt{x}$ for $a=1$ I seem to be stuck defining an alternating sequence of terms in this series because $f^{(0)}(x)=f(x)$ is positive, as well as $f'(x)$, but then every other term starting with $f''(x)$ is negative. How can I define $f^{(n)}(x)$ given this?
\begin{array}{ll}
f(x)=x^{\frac{1}{2}} & f(1)=1 \\
f'(x)=\frac{1}{2}\cdot x^{-\frac{1}{2}} & f'(1)=\frac{1}{2} \\
f''(x)=(-1)^1\cdot\left(\frac{1}{2}\right)^{2}\cdot x^{-\frac{3}{2}} & f''(1)=(-1)^1\cdot\left(\frac{1}{2}\right)^{2} \\
f'''(x)=(-1)^2\cdot 3\cdot\left(\frac{1}{2}\right)^{3}\cdot x^{-\frac{5}{2}} & f'''(1)=(-1)^2\cdot3\cdot\left(\frac{1}{2}\right)^{3} \\
f^{(4)}(x)=(-1)^3\cdot3\cdot 5\cdot\left(\frac{1}{2}\right)^{4}\cdot x^{-\frac{7}{2}} & f^{(4)}(1)=(-1)^3\cdot3\cdot 5\cdot\left(\frac{1}{2}\right)^{4} \\
f^{(n)}(x)=(-1)^{n-1}\left(\frac{1}{2}\right)^{n}\cdot x^{\frac{1-2n}{2}} & f^{(n)}(1)=(-1)^{n-1}\left(\frac{1}{2}\right)^{n}
\end{array}
I thought I had the right answer until I realized that I'd be defining $f(x)$ to be negative.
|
Well you're not defining $f(x)$ to be negative, you're finding that certain derivatives of $f(x)$ are negative - if you think about the graph of this $f(x)$ it decreasingly increases; that is, it is increasing, but the rate at which it increases is decreasing, so it should seem to me that in the least, the second derivative should be negative, which is what your formula suggests!
Was in comments then realized this was a long comment :P
EDIT: So thanks to my oversight I wasn't too helpful. Hopefully this edit will provide more help
So if $f(x) = x^{1/2}$ then...
$$
f'(x) = \frac{1}{2} x^{-1/2} \\
f''(x) = \frac{-1}{4} x^{-3/2} \\
f'''(x) = \frac{3}{8} x^{-5/2} \\
f^{(4)} (x) = \frac{-15}{16} x^{-7/2} \\
f^{(5)} (x) = \frac{105}{32} x^{-9/2}
$$
so, at least to me, it appears as though you get $f^{(0)} (x) = x^{1/2}$, $f^{(1)} (x) = \frac{1}{2} x^{-1/2}$ and $f^{(n)} (x) = (-1)^{n+1} \frac{(2n - 3)(2n - 5) \cdots (3)(1)}{2^n} x^{\frac{1-2n}{2}} : n \ge 2$. At least for me (I can be wrong though) I don't see any outstanding pattern other than the one I just said. Hope this helps!
|
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|
Integral of $\frac{2}{x^3-x^2}$ How can I integrate $\dfrac2{x^3-x^2}$? Can someone please give me some hints? Thanks a lot!
|
$$\int \frac2{x^3-x^2}\, dx =\frac2{x^2(x-1)}= \int \left(\frac{A}x+\frac{B}{x^2}+\frac{C}{x-1}\right)\,dx$$
$$Ax(x-1) + B(x-1) + Cx^2 = 2$$
We choose values for $x$ that "zero out" some of the terms, to simplifying the computation of the unknown values $A, B, C$:
$$x = 1 \implies A(1)(1 - 1) + B(1-1) + C(1^2) = 2 \iff 0 + 0 + C = 2\quad \checkmark$$
$$x = 0 \implies B(-1) = 2 \implies B = -2\quad \checkmark$$
$$x = 2, C = 2, B = -2 \implies 2A - 2 + 8 = 2\iff A = -2\quad \checkmark$$
$$ \int \left(\frac{A}x+\frac{B}{x^2}+\frac{C}{x-1}\right)\,dx = \int \left(-\frac{2}x -\frac{2}{x^2}+\frac{2}{x-1}\right)\,dx$$ $$ =-2 \int \frac{dx}x \quad -\quad 2\int \frac{dx}{x^2}\quad + \quad 2 \int \frac{dx}{x-1}$$
|
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|
Find five prime factors of $3^{140 }- 1$ I tried to simplify $3^{140}$ but I couldn't go past $81^{35}$, any help would be greatly appreciated.
|
$$a^n - b^n = (a-b) ( a^{n-1} + a^{n-2} b + \cdots + b^{n-1})$$
Using $n = 140$, we get that $3 - 1$ is a factor, which gives $2$ as a factor.
Using $n = 35$, we get that $3^4 - 1$ is a factor, which gives $5$ as a factor.
Using $n = 28$, we get that $3^5 - 1$ is a factor, which gives $11$ as a factor.
Using $n = 20$, we get that $3^7-1$ is a factor, which gives $1093$ as a factor.
Using $n = 14$, we get that $3^{10} -1 $ is a factor, which gives $61$ as a factor.
|
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|
Prove that : $\sqrt[4]{(a^2+1)b}+\sqrt[4]{(b^2+1)c}+ \sqrt[4]{(c^2+1)a} \le 3\sqrt[4]{2}.$ For $a,b,c\in\mathbb{R}^+$ and $a+b+c=3$ .
Prove that : $\sqrt[4]{(a^2+1)b}+\sqrt[4]{(b^2+1)c}+ \sqrt[4]{(c^2+1)a} \le 3\sqrt[4]{2}.$
|
Oops, give a wrong answer in previous edit. Here is the corrected version.
Notice both $\sqrt[4]{x}$ and $\sqrt[4]{x^2+1}$ are strictly increasing function in $x$.
The list of numbers $(\sqrt[4]{a},\sqrt[4]{b},\sqrt[4]{c})$ are in same sorted order as $(\sqrt[4]{a^2+1},\sqrt[4]{b^2+1},\sqrt[4]{c^2+1})$.
By Rearrangement inequality, we have:
$$\begin{align}&\sqrt[4]{(a^2+1)b}+\sqrt[4]{(b^2+1)c}+ \sqrt[4]{(c^2+1)a}\\
\le & \sqrt[4]{(a^2+1)a}+\sqrt[4]{(b^2+1)b}+ \sqrt[4]{(c^2+1)c}\tag{*1}
\end{align}$$
Notice
$$\frac{d^2}{dx^2} \sqrt[4]{(x^2+1)x} =
-\frac{3\,{\left( x^2 -1\right) }^{2}}{16\,{x}^{\frac{7}{4}}\,{\left( {x}^{2}+1\right) }^{\frac{7}{4}}} \le 0$$
$\sqrt[4]{(x^2+1)x}$ is a concave function in $x$ and by Jensen's inequality, we have:
$$
\text{R.H.S of } (*1)
\le 3 \left[ \left( \left(\frac{a+b+c}{3}\right)^2 + 1\right) \left(\frac{a+b+c}{3}\right)\right]^{\frac14}
= 3\sqrt[4]{2}
$$
|
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|
tough algebric problem? I wanted to know how can i prove that if
$xy+yz+zx=1$, then
$$ \frac{x}{1+x^2}+\frac{y}{1+y^2}+\frac{z}{1+z^2}
= \frac{2}{\sqrt{(1+x^2)(1+y^2)(1+z^2)}}$$
I did let $x=\tan A$, $y=\tan B$, $z=\tan C$
given $xy+yz+zx =1$ we have $\tan A \tan B+ \tan B \tan C+\tan C \tan A=1$
$\tan C(\tan A+\tan B)=1-\tan A \tan B$, or $\tan(A+B)=\tan(\pi/2 -C)$ we have $A+B+C=\pi/2$.
what to do now?
Any help appreciated.
thanks.
|
HINT:
$$\frac x{1+x^2}=\frac {\tan A}{1+\tan^2A}=\frac{2\sin A\cos A}2=\frac{\sin2A}2$$
Now , $$\sin2A+\sin2B+\sin2C=2\sin(A+B)\cos(A-B)+2\sin C\cos C$$
$$=2\sin\left(\frac\pi2-C\right)\cos(A-B)+2\sin C\cos C$$
$$=2\cos C\{\cos(A-B)+\cos(A+B)\}$$ as $\sin C=\sin\{\frac\pi2-(A+B)\}=\cos(A+B)$
$$\implies \sin2A+\sin2B+\sin2C=2\cos C\cdot2\cos A\cos B$$
and $$\frac1{\sqrt{1+x^2}}=\frac1{\sec A}=\cos A$$
|
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|
Prove that $2^n < \binom{2n}{n} < 2^{2n}$ Prove that $2^n < \binom{2n}{n} < 2^{2n}$. This is proven easily enough by splitting it up into two parts and then proving each part by induction.
First part: $2^n < \binom{2n}{n}$. The base $n = 1$ is trivial. Assume inductively that some $k$ satisfies our statement. The inductive step can be proved as follows.
$2^k < \binom{2k}{k} \implies 2^{k+1} < 2\binom{2k}{k} = \frac{2(2k)!}{k!k!} = \frac{2(2k)!(k + 1)}{k!k!(k + 1)} = \frac{2(2k)!(k+2)}{(k+1)!k!}<\frac{(2k)!(2k+2)(2k+1)}{(k+1)!k!(k+1)} = \binom{2(k+1)}{k+1}$
Second part: $2^{2n} > \binom{2n}{n}$. Again, the base is trivial. We can assume that for some $k$ our statement is satisfied and prove that inductive step as follows:
$2^{2k} > \binom{2k}{k} \implies 2^{2k + 2} > 2^2\binom{2k}{k} = \frac{2\cdot2(2k)!}{k!k!} = \frac{2\cdot2(2k)!(k+1)(k+1)}{k!k!(k+1)(k+1)} = \frac{(2k)!(2k+2)(2k+2)}{(k+1)!(k+1)!} > \frac{(2k)!(2k+1)(2k+2)}{(k+1)!(k+1)!} = \binom{2(k+1)}{k+1}$
Is there a non-inductive derivation for the inequality?
|
For the first part, we don't need induction as $$\binom {2n}n=\prod_{0\le r\le n-1}\frac{2n-r}{n-r}=2\prod_{1\le r\le n-1}\frac{2n-r}{n-r}>2\cdot 2^{n-1}$$ and $2n-r>2(n-r)$ for $r>0$
|
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|
On $x^3+y^3+z^3 = 1$ and a Pell equation Given,
$$(1-ac+bc)^3 + (a+c^2-ac^3)^3 + (ac^3-b-c^2)^3 = 1\tag{1}$$
where,
$$a,b,c,r = 12qrt,\;\; 3(q-r)(3q+r)t,\;\; 3s^2t^2,\;\; p-18qs^3t^3$$
then $(1)$ holds true if $p,q,s,t$ satisfies,
$$p^2-3(108s^6t^6-1)q^2=s\tag{2}$$
For $s=1$, this gives the Pell equation,
$$p^2-3(108t^6-1)q^2=1$$
which, starting with fundamental solution $p_1, q_1 = 216t^3-1,\; \pm12t^3$, gives an infinite family of polynomial parameterizations to $(1)$.
Question: Other than square $s$ and $s=3$, is there any other integer $s$ such that $(2)$ has a non-trivial solution in the integers?
|
Yes, there are other integers $s$. Just put $t=0$, so the equation becomes $p^2+3q^2=s$. Then pick any integers $p$ and $q$, and most likely $p^2+3q^2$ won't be either a square or $3$. For instance, if $p=2$ and $q=1$ then $s=7$.
If you require that $t$ be nonzero, then there are still solutions to (2) for other integers $s$ besides squares and $3$, for instance $s=12$ works with $t=1$, $q=2$, and $p=62208$.
|
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|
Matrix determinant using Laplace method I have the following matrix of order four for which I have calculated the determinant using Laplace's method.
$$
\begin{bmatrix}
2 & 1 & 3 & 1 \\
4 & 3 & 1 & 4 \\
-1 & 5 & -2 & 1 \\
1 & 3 & -2 & -1 \\
\end{bmatrix}
$$
Finding the determinant gives me $-726$. Now if I check the result at Wolfram Alpha, it says the result is $-180$ (Because there are no zeros in the matrix, expand with respect to row one) so it uses only the first row to calculate the determinant of the matrix.
My question is: Why it uses only the first row to find the determinant?
|
The Laplace development can be performed with respect to any row or column. Let's see when developing with respect to the first row:
\begin{align}
\det\begin{bmatrix}
2 & 1 & 3 & 1 \\
4 & 3 & 1 & 4 \\
-1 & 5 & -2 & 1 \\
1 & 3 & -2 & -1
\end{bmatrix}={}&
(-1)^{1+1}\cdot 2 \cdot
\det\begin{bmatrix}
3 & 1 & 4 \\
5 & -2 & 1 \\
3 & -2 & -1
\end{bmatrix}+{}\\
&(-1)^{1+2}\cdot 1 \cdot
\det\begin{bmatrix}
4 & 1 & 4 \\
-1 & -2 & 1 \\
1 & -2 & -1
\end{bmatrix}+{}\\
&(-1)^{1+3}\cdot 3\cdot
\det\begin{bmatrix}
4 & 3 & 4 \\
-1 & 5 & 1 \\
1 & 3 & -1
\end{bmatrix}+{}\\
&(-1)^{1+4}\cdot1\cdot
\det\begin{bmatrix}
4 & 3 & 1 \\
-1 & 5 & -2 \\
1 & 3 & -2
\end{bmatrix}
\end{align}
Now you can go on by computing the determinants of the four $3\times3$ matrices with the same (or another) method. The final result is indeed $-180$.
When one row or column has many zeroes it's convenient to use that one, but any row or column can be used.
|
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|
$n$ and $n^5$ have the same units digit? Studying GCD, I got a question that begs to show that $n$ and $n^5$ has the same units digit ...
What would be an idea to be able to initiate such a statement?
testing
$0$ and $0^5=0$
$1$ and $1^5=1$
$2$ and $2^5=32$
In my studies, I have not got "mod", please use other means, if possible of course.
I demonstrated in a previous period that
$$2|n^5-n$$because $$n^5-n=(n+1)n(5n^4+5n+5)$$,
and$$5|n^5-n$$By Fermat's Little Theorem
Only I do not understand what should happen to the units of the two numbers are equal ... What must occur?
|
We know if unit digits of two numbers are same, their difference is divisible by 10 and vice versa.
Method $1a:$
Using Fermat's Little Theorem $n^5-n\equiv0\pmod 5$
and $n^5-n=n(n^4-1)=n(n^2-1)(n^2+1)=n(n-1)(n+1)(n^2+1)$ which is divisible by $n(n-1)$ which is always even
$\implies 2|(n^5-n)$ and we have $5|(n^5-n)$
$\implies n^5-n$ is divisible by lcm $(2,5)=10$
Method $1b:$
As $10=2\cdot5,$
using Fermat's Little Theorem, we have $$n^5-n\equiv0\pmod 5\text{ and } n^2-n\equiv0\pmod 2$$
Now, lcm $(n^5-n,n^2-n)=n(n^4-1,n-1)=n(n^4-1)$ as $(n-1)|(n^4-1)$
$\implies $lcm $(n^5-n,n^2-n)=n^5-n$ which is divisible by $5,2$ hence by lcm$(2,5)=10$
Method $2:$
Alternatively,
$$n^5-n=n(n^4-1)=n(n^2-1)(n^2+1)=n(n^2-1)(n^2-4+5)$$
$$=n(n^2-1)(n^2-4)+5n(n^2-1)$$
$$=\underbrace{(n-2)(n-1)n(n+1)(n+2)}_{\text{ product of }5\text{ consecutive integers }}+5\cdot \underbrace{(n-1)n(n+1)}_{\text{ product of }3\text{ consecutive integers }}$$
Now, we know the product $r$ consecutive integers is divisible by $r!$ where $r$ is a positive integer
So, $(n-2)(n-1)n(n+1)(n+2)$ is divisible by $5!=120$ and $(n-1)n(n+1)$ is divisible by $3!=6$
$$\implies n^5-n\equiv0\pmod{30}\equiv0\pmod{10}$$
|
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|
Show the sequence $(1 - \frac{1}{n})^{-n}$ is decreasing. How do you show the sequence $(1 - \frac{1}{n})^{-n}$ is decreasing?
I understand that the binomial theorem should be used here but I don't see how we can use it to prove that $a_{n+1} < a_n$.
I will rewrite the sequence as,
\begin{align*}
(1 - \frac{1}{n})^{-n} &= (\frac{n-1}{n})^{-n} \\
&= (\frac{n}{n-1})^n \\
&= (1 + \frac{1}{n-1})^n
\end{align*}
Then I can apply binomial theorem to it.
This is as far as I got now.
|
Ultimately, you want to show that$^{(1)}$ for $n\geqslant 2$ $${\left( {1 + \frac{1}{{\left( {n - 1} \right)\left( {n + 1} \right)}}} \right)^n} > 1 + \frac{1}{n}$$
Using the Binomial Theorem, the left hand side is $${\left( {1 + \frac{1}{{\left( {n - 1} \right)\left( {n + 1} \right)}}} \right)^n} > 1 + \frac{n}{{\left( {n - 1} \right)\left( {n + 1} \right)}}$$
Can you show the right hand side is $>1+n^{-1}$? Note that $$1 + \frac{n}{{\left( {n + 1} \right)\left( {n - 1} \right)}} - \left( {1 + \frac{1}{n}} \right) = \frac{1}{{n\left( {n - 1} \right)\left( {n + 1} \right)}} > 0\;;\; \text{if }\;n\geqslant 2$$
$(1)$ $$\begin{align}
{\left( {1 - \frac{1}{n}} \right)^{ - n}} &> {\left( {1 - \frac{1}{{n + 1}}} \right)^{ - n - 1}} \\
{\left( {\frac{{n - 1}}{n}} \right)^{ - n}} &> {\left( {\frac{n}{{n + 1}}} \right)^{ - n}}{\left( {\frac{n}{{n + 1}}} \right)^{ - 1}} \\
{\left( {\frac{{n - 1}}{n}} \right)^{ - n}} &> {\left( {\frac{n}{{n + 1}}} \right)^{ - n}}\left( {1 + \frac{1}{n}} \right) \\
{\left( {\frac{n}{{n - 1}}} \right)^n}{\left( {\frac{n}{{n + 1}}} \right)^n} &> 1 + \frac{1}{n} \\
{\left( {\frac{{{n^2}}}{{{n^2} - 1}}} \right)^n} &> 1 + \frac{1}{n} \\
{\left( {1 + \frac{1}{{{n^2} - 1}}} \right)^n} &> 1 + \frac{1}{n} \end{align} $$
|
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|
Finding $\frac{a+b}{a-b}$ such that $a^2+b^2=6ab$ For $a,b > 0$ such that $a^2+b^2=6ab$ .How to find $\frac{a+b}{a-b}$
|
$$(a-b)^2=4 a b = (a+b)^2-(a-b)^2$$
which means that
$$1 = \left ( \frac{a+b}{a-b}\right)^2 - 1$$
or
$$\frac{a+b}{a-b} =\pm \sqrt{2}$$
depending whether $a > b$ or not.
|
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|
show that $\int_{0}^{\infty} \frac {\sin^3(x)}{x^3}dx=\frac{3\pi}{8}$ show that
$$\int_{0}^{\infty} \frac {\sin^3(x)}{x^3}dx=\frac{3\pi}{8}$$
using different ways
thanks for all
|
In this answer, the more general integral
$$
\int_0^\infty\left(\frac{\sin(x)}{x}\right)^n\,\mathrm{d}x
$$
is calculated.
Your integral is that integral for $n=3$.
A Different Way
In a fashion similar to this answer, we will use the equation
$$
\frac{\mathrm{d}^2}{\mathrm{d}x^2}\frac{\sin^3(kx)}{k^3}
=\frac{9\sin(3kx)-3\sin(kx)}{4k}\tag{1}
$$
and the series for $0\lt x\le\pi$,
$$
\sum_{k=1}^\infty\frac{\sin(kx)}{k}=\frac{\pi-x}{2}\tag{2}
$$
Using $(2)$, we get
$$
\begin{align}
\sum_{k=1}^\infty\frac{9\sin(3kx)-3\sin(kx)}{4k}
&=\frac94\frac{\pi-3x}{2}-\frac34\frac{\pi-x}{2}\\
&=\frac{3\pi}{4}-3x\tag{3}
\end{align}
$$
Integrating from $0$ twice to back out the derivatives taken in $(1)$ yields
$$
\sum_{k=1}^\infty\frac{\sin^3(kx)}{k^3}=\frac{3\pi}{8}x^2-\frac12x^3\tag{4}
$$
Set $x=1/n$ and multiply by $n^2$ to get
$$
\sum_{k=1}^\infty\frac{\sin^3(k/n)}{k^3/n^3}\frac1n=\frac{3\pi}{8}-\frac1{2n}\tag{5}
$$
and $(5)$ is a Riemann sum for
$$
\int_0^\infty\frac{\sin^3(x)}{x^3}\,\mathrm{d}x=\frac{3\pi}{8}\tag{6}
$$
|
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|
Fractional part of rational power arbitrary small I think that $\{a^n\}$ (where $\{x\}$ is $x \pmod 1$), where $a$ is fixed rational greater than 1 and $n$ is positive integer, is dense in $[0,1]$ is unsolved. However what about $\{a^n\}$ is arbitrary small for some $n$ ($a$ is fixed rational as well).
|
I will show that
if $a = 1+\sqrt{2}$
then the limit points of
$\{a^n\}$
are $0$ and $1$.
I know that this doesn't tell anything
about rational $a$,
but this might be of use.
Note that this method can show this
for $a$ and $b$ roots of
$x^2-2ux-v = 0$
where $u$ and $v$ are
positive integers such that
$v < 2u+1$.
This case is $u=v=1$;
$u=1, v=2$ also works.
If $a = \sqrt{2}+1$
and
$b = 1-\sqrt{2}$
then
$ab = -1$
and
$a+b = 2$.
Therefore $a$ and $b$
are the roots of
$x^2-2x-1 = 0$.
If $u_n = a^n+b^n$,
then
$u_0 = 2,
u_1 = 2$.
Since
$a^{n+2}
=a^n(a^2)
=a^n(2a+1)
=2a^{n+1}+a^n$
and similarly for $b$,
$\begin{array}\\
u_{n+2}
&=a^{n+2}+b^{n+2}\\
&=2a^{n+1}+a^n+2b^{n+1}+b^n\\
&=2u_{n+1}+u_n\\
\end{array}
$
Therefore
$u_n$ is a
positive integer for all $n$.
Since
$|b| < 1$,
$|b^n| \to 0$.
Since
$b < 0$,
$b^{2n} > 0$
and
$b^{2n+1} < 0$.
Therefore,
since
$a^n = u_n - b^n$,
$\{a^{2n}\}
=\{u_{2n}-b^{2n}\}
=1-b^{2n}
$
so
$\{a^{2n}\}
\to 1
$
and
$\{a^{2n+1}\}
=\{u_{2n+1}-b^{2n+1}\}
=\{u_{2n+1}+|b^{2n+1}\}
=|b^{2n+1}|
$
so
$\{a^{2n+1}\}
\to 0
$.
Therefore
the limits points of
$\{a^n\}$
are $0$ and $1$.
|
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"timestamp": "2023-03-29T00:00:00",
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|
Existence of linear mapping I am studying for an exam in linear algebra and I am having trouble solving the following:
Do linear mappings $\phi : \mathbb{R}^2 \rightarrow \mathbb{R}^2$ with the following properties exist?
$1)$ $\phi_1 \begin{pmatrix} 2 \\ 0 \end{pmatrix} = \begin{pmatrix} 0 \\ 1 \end{pmatrix} $, $\phi_1 \begin{pmatrix} 1 \\ 1 \end{pmatrix} = \begin{pmatrix} 5 \\ 2 \end{pmatrix} $, $\phi_1 \begin{pmatrix} 1 \\ 2 \end{pmatrix} = \begin{pmatrix} 2 \\3 \end{pmatrix} $
$2)$ $\phi_2 \begin{pmatrix} 2 \\ 3 \end{pmatrix} = \begin{pmatrix} 0 \\0\end{pmatrix} $, $\phi_2 \begin{pmatrix} 0 \\ 0 \end{pmatrix} = \begin{pmatrix} 2\\ 3 \end{pmatrix} $
$3)$ $\phi_3 \begin{pmatrix} 1 \\ -1 \end{pmatrix} = \begin{pmatrix} 1\\ 0 \end{pmatrix} $, $\phi_3 \begin{pmatrix} -2 \\ 5 \end{pmatrix} = \begin{pmatrix} -3 \\ 1 \end{pmatrix} $, $\phi_3 \begin{pmatrix} 2 \\ 1 \end{pmatrix} = \begin{pmatrix} 1 \\1 \end{pmatrix} $
I know that the following properties have to hold for a linear mapping:
*
*$f(\mathbf{x}+\mathbf{y}) = f(\mathbf{x})+f(\mathbf{y})$
*$f(\alpha \mathbf{x}) = \alpha f(\mathbf{x})$
*$f(0) = 0 $
I conclude that $2)$ is a not a linear mapping since $\phi(0)$ is not mapped to $0$.
But how shall I proceed with the others?
|
Hint: A linear transformation is completely determined by how it acts on a basis for your vector space (a linearly independent and spanning set), and a linear transformation can send basis vectors to wherever it wants (you can send them anywhere). Show these fact if you haven't yet!!
For $\phi_1$, note that $\begin{pmatrix} 2 \\ 0 \end{pmatrix}$ and $\begin{pmatrix} 1 \\ 1 \end{pmatrix}$ form a basis for $\Bbb R^2$, so how $\phi_1$ acts on all of $\Bbb R^2$ if $\phi_1$ is a linear map. There is a unique linear map extending values of $\phi_1$ on this basis to all of $ \Bbb R^2$. Could $\phi_1$ be that map based on its 3rd known value?
A similar proof works for $\phi_3$.
|
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"timestamp": "2023-03-29T00:00:00",
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|
Find the sum : $\sin^{-1}\frac{1}{\sqrt{2}}+\sin^{-1}\frac{\sqrt{2}-1}{\sqrt{6}}+\sin^{-1}\frac{\sqrt{3}-\sqrt{2}}{\sqrt{12}}+\cdots$ Problem :
Find the sum of :
$$\sin^{-1}\frac{1}{\sqrt{2}}+\sin^{-1}\frac{\sqrt{2}-1}{\sqrt{6}}+\sin^{-1}\frac{\sqrt{3}-\sqrt{2}}{\sqrt{12}}+\cdots$$
My approach :
Here the $n$'th term is given by :
$$t_n = \sin^{-1}\left[\frac{\sqrt{n}-\sqrt{n-1}}{\sqrt{n}\sqrt{n+1}}\right]$$
From now how to proceed further please suggest thanks....
|
Hint: Use the fact that $\sin^{-1}a+\sin^{-1}b = \sin^{-1}(a\sqrt{1-b^2}+b\sqrt{1-a^2})$. The first two terms then give you $\sin^{-1}(\sqrt{2/3})$. Then applying the same identity with this term and the third term gives you $\sin^{-1}(\sqrt{3/4})$, and...
Edit: Induction step gives some headache, so let me write it for completeness. Suppose the sum of the first $n$ terms are $a = \sin^{-1}\sqrt\frac{n}{n+1}$. We show that the sum of the first $n+1$ terms are $\sin^{-1}\sqrt{\frac{n+1}{n+2}}$. Let $$b = \frac{\sqrt{n+1}-\sqrt{n}}{\sqrt{(n+1)(n+2)}}$$ denote the $(n+1)$th term. Then, it is sufficient to prove that $c = a\sqrt{1-b^2}+b\sqrt{1-a^2} = \sqrt\frac{n+1}{n+2}$. Indeed, we have
\begin{align}
c &= \sqrt\frac{n}{n+1}\sqrt{1-\frac{(\sqrt{n+1}-\sqrt{n})^2}{(n+1)(n+2)}}+\frac{\sqrt{n+1}-\sqrt{n}}{\sqrt{(n+1)(n+2)}}\sqrt{1-\frac{n}{n+1}}\\
& = \sqrt\frac{n}{n+1}\sqrt{\frac{n^2+n+1+2\sqrt{n(n+1)}}{(n+1)(n+2)}}+\frac{\sqrt{n+1}-\sqrt{n}}{(n+1)\sqrt{(n+2)}}
\end{align}
Noting that the nested radical $\sqrt{n^2+n+1+2\sqrt{n(n+1)}}$ is equal to $1+\sqrt{n(n+1)}$, we obtain
$$c = \sqrt n \frac{1+\sqrt{n(n+1)}}{(n+1)\sqrt{(n+2)}}+\frac{\sqrt{n+1}-\sqrt{n}} {(n+1)\sqrt{(n+2)}} = \sqrt\frac{n+1}{n+2}$$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/454523",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "12",
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|
How can prove this $\sum_{k=0}^{p}\binom{p}{k}(\pm i\sqrt{3})^k\equiv 1\pm (i\sqrt{3})^p-p\sum_{k=1}^{p-1}\frac{(\mp i\sqrt{3})^k}{k}\pmod{p^2}?$
For any prime $p>3$ show that
$$p\sum_{j=0}^{p-1}\dfrac{(-3)^j}{2j+1}\equiv \left(\dfrac{p}{3}\right)\pmod{p^2}$$
where $\left(\dfrac{p}{3}\right)$ denotes the Legendre symbol.
This is proof: we have
$$(1\pm i\sqrt{3})^p=2^pe^{\pm i\pi p/3}=2^p(\cos{(\pi p/3)}\pm i\sin{(\pi p/3)}=2^{p-1}\left(1\pm i\left(\dfrac{p}{3}\right)\sqrt{3}\right)$$
On the other hand we have
\begin{align*}
(1\pm i\sqrt{3})^p=\sum_{k=0}^{p}\binom{p}{k}(\pm i\sqrt{3})^k&\equiv 1\pm (i\sqrt{3})^p-p\sum_{k=1}^{p-1}\dfrac{(\mp i\sqrt{3})^k}{k}\\
&\equiv 1\pm i\sqrt{3}(-3)^{(p-1)/2}-S_{0}\pm i\sqrt{3}S_{1}\pmod{p^2}
\end{align*}
where
$$S_{0}=p\sum_{j=1}^{\frac{p-1}{2}}\dfrac{(-3)^j}{2j},S_{1}=p\sum_{j=0}^{\frac{p-3}{2}}\dfrac{(-3)^j}{2j+1}$$
then
$$S_{0}\equiv 1-2^{p-1}\pmod{p^2}, S_{1}\equiv 2^{p-1}\left(\dfrac{p}{3}\right)-(-3)^{(p-1)/2}\pmod{p^2}$$
so
\begin{align*}
p\sum_{j=0}^{p-1}\dfrac{(-3)^j}{2j+1}&=p\sum_{j=0}^{(p-3)/2}\dfrac{(-3)^j}{2j+1}+(-3)^{(p-1)/2}+(-3)^{(p-1)/2}p\sum_{j=1}^{(p-1)/2}\dfrac{(-3)^j}{p+2j}\\
&\equiv S_{1}+(-3)^{(p-1)/2}+(-3)^{(p-1)/2}S_{0}\\
&\equiv (2^{p-1}-1)\left(\left(\dfrac{p}{3}\right)-(-3)^{(p-1)/2}\right)+\left(\dfrac{p}{3}\right)\\
&\equiv \left(\dfrac{p}{3}\right)\pmod{p^2}
\end{align*}
where use
$$p|(2^{p-1}-1),2|\left(\left(\dfrac{p}{3}\right)-(-3)^{(p-1)/2}\right)$$
My question:
(1)：why $$\sum_{k=0}^{p}\binom{p}{k}(\pm i\sqrt{3})^k\equiv 1\pm (i\sqrt{3})^p-p\sum_{k=1}^{p-1}\dfrac{(\mp i\sqrt{3})^k}{k}\pmod{p^2}?$$
(2):
why
$$S_{0}\equiv 1-2^{p-1}\pmod{p^2}, S_{1}\equiv 2^{p-1}\left(\dfrac{p}{3}\right)-(-3)^{(p-1)/2}\pmod{p^2}?$$
Thank you someone can solve my two problem,Thank you very much,and this
$$p\sum_{j=0}^{p-1}\dfrac{(-3)^j}{2j+1}\equiv \left(\dfrac{p}{3}\right)\pmod{p^2}$$
have other methods? Thank you
|
Now,I have konw why
$$S_{0}\equiv 1-2^{p-1}\pmod {p^2},S_{1}\equiv 2^{p-1}\left(\frac{p}{3}\right)-(-3)^{(p-1)/2}\pmod {p^2}$$
Because use
$$(1\pm i\sqrt{3})^p=2^{p-1}(1\pm i\left(\frac{p}{3}\right)\sqrt{3})$$
and other hand we have
$$(1\pm i\sqrt{3})^p\equiv 1\pm i\sqrt{3}(-3)^{(p-1)/2}-S_{0}\pm i\sqrt{3}S_{1}\pmod{p^2}$$
so we have done
|
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"url": "https://math.stackexchange.com/questions/455369",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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|
Finding the limit $\lim \limits_{n \to \infty}\ (\cos \frac x 2 \cdot\cos \frac x 4\cdot \cos \frac x 8\cdots \cos \frac x {2^n}) $ This limit seemed quite unusual to me as there aren't any intermediate forms or series expansions which are generally used in limits. Stuck on this for a while now .Here's how it goes :
$$
\lim \limits_{n \to \infty}
\left[\cos\left(x \over 2\right)\cos\left(x \over 4\right)
\cos\left(x \over 8\right)\ \cdots\ \cos\left(x \over 2^{n}\right)\right]
$$
|
Hint $$\begin{align}{\sin x}&=2^1\sin\frac x 2 \cos\frac x2\\{}\\\sin x& =2^2\sin \frac x4\cos\frac x 4\cos \frac x 2\\{}\\\sin x& =2^3\sin \frac x8\cos \frac x8\cos\frac x 4\cos \frac x 2\\{}\\\cdots\;&=\hspace{2cm }\cdots\end{align} $$
One further hint
$$\sin x = {2^n}\sin \frac{x}{{{2^n}}}\prod\limits_{k = 1}^n {\cos \frac{x}{{{2^k}}}} $$
You'll need $\dfrac{\sin x}x\to 1$ as $x\to 0$.
Final spoiler:
$$\lim\limits_{n \to \infty } \prod\limits_{k = 1}^n \cos {\frac{x}{2^k}} = \lim \limits_{n \to \infty } \frac{\sin x}{x}\left( \frac{\sin {2^{ - n}x}}{2^{ - n}x} \right)^{ - 1} = \frac{\sin x}{x}$$
|
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"url": "https://math.stackexchange.com/questions/455995",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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|
Prove: $\frac{1}{1^2} +\frac{1}{2^2} + \cdots + \frac{1}{n^2} + \cdots = \sum_{n=1}^\infty \frac{1}{n^2} < 2$ While I don't doubt that this question is covered somewhere else I can't seem to find it, or anything close enough to which I can springboard. I however am trying to prove $$\frac{1}{1^2} +\frac{1}{2^2} + \cdots + \frac{1}{n^2} + \cdots = \sum_{n=1}^\infty \frac{1}{n^2} < 2$$ by induction.
I have seen it many times and proved it before but can't remember what it was I did. I see that for the first two terms $n = 1, n=2$ I get:
for $n = 1$, $\frac{1}{1^2} = 1 < 2$
for $n = 2$, $\frac{1}{1^2} + \frac{1}{2^2} = \frac{5}{4} < 2$
Now I am stumped, I know I want to show this works for the $n+1$ term and am thinking, let the series $\sum_{n=1}^\infty \frac{1}{n^2} = A(n)$ Then look to show the series holds for $A(n+1)$ But $A(n+1) = A(n) + \frac{1}{(n+1)^2}$ But now what? If I tried $A(n+1) - A(n) = \frac{1}{(n+1)^2}$ , but would have to show that this is less than $2 - A(n)$. I am stuck.
Thanks for your thoughts,
Brian
|
One way is to look at the lower Riemann sum with width $1$ under the graph of $f(x) = {1 \over x^2}$, from $x = 1$ to $x = \infty$. Since ${1 \over x^2}$ is decreasing, the lower
Riemann sum will be $\sum_{n=2}^{\infty} f(n) = \sum_{n=2}^{\infty} {1 \over n^2}$, which must be less than the integral $\int_1^{\infty} {1 \over x^2}\,dx = 1$. Thus adding $1$ to this inequality gives the result you seek.
|
{
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"timestamp": "2023-03-29T00:00:00",
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|
Splitting Stacks Nim A game is played with two players and an initial stack of $n$ pennies ($n\geq 3$). The players take turns choosing one of the stacks of pennies on the table and splitting it into two stacks. When a player makes a move that causes all the stacks to be of height $1$ or $2$ at the end of his or her turn, that player wins. Which starting values of $n$ are wins for each player?
For all the values I've tested, if it's even the first player wins and if it's odd the second player wins. If it's even, then the piles must be split into two evens or two odds. If it's two evens, the next player wins one of the splits recursively but the first player wins the other one (so the first player wins). If it's odd, the opposite happens (the second player recursively 'loses' both piles). If it's odd, then it must be split into one odd and one even pile, so the second player can choose to lose the odd pile and so win the even pile. BUT this doesn't take into account the fact that when you have a pile of $1$s and $2$s, that is considered finished for a pile of $n$, but when there is more than one pile, then you can still split the $2$s into $1$s and so delay your turn.
So what strategy works for all values that guarantees a win for one of the players?
|
Here for reference is a table of nim-values for every nontrivial position with less than 12 pennies. Piles of size 1 have been omitted.
A nim-value of 0 indicates a position that is a win for the player who moves to it; any other value indicates a position that is a win for the next player to move.
$$
\begin{array}{lr}
3 & 1 \\
\hline\\
4 & 2 \\
\hline\\
3,2 & 2 \\
5 & 0 \\
\hline\\
4,2 & 1 \\
3,3 & 0 \\
6 & 2 \\
\hline\\
3,2,2 & 1 \\
5,2 & 2 \\
4,3 & 2 \\
7 & 0 \\
\hline\\
4,2,2 & 2 \\
3,3,2 & 2 \\
6,2 & 0 \\
5,3 & 0 \\
4,4 & 0 \\
8 & 1 \\
\hline\\
3,2,2,2 & 2 \\
5,2,2 & 0 \\
4,3,2 & 0 \\
7,2 & 1 \\
3,3,3 & 0 \\
6,3 & 1 \\
5,4 & 1 \\
9 & 0 \\
\hline\\
4,2,2,2 & 1 \\
3,3,2,2 & 0 \\
6,2,2 & 2 \\
5,3,2 & 1 \\
4,4,2 & 2 \\
8,2 & 0 \\
4,3,3 & 1 \\
7,3 & 0 \\
6,4 & 0 \\
5,5 & 0 \\
10 & 1 \\
\hline\\
3,2,2,2,2 & 1 \\
5,2,2,2 & 2 \\
4,3,2,2 & 2 \\
7,2,2 & 0 \\
3,3,3,2 & 1 \\
6,3,2 & 0 \\
5,4,2 & 0 \\
9,2 & 1 \\
5,3,3 & 0 \\
4,4,3 & 0 \\
8,3 & 1 \\
7,4 & 1 \\
6,5 & 1 \\
11 & 0 \\
\end{array}
$$
|
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"timestamp": "2023-03-29T00:00:00",
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|
$2+2 = 5$? error in proof $$\begin{align} 2+2 &= 4 - \frac92 +\frac92\\
&= \sqrt{\left(4-\frac92\right)^2} +\frac92\\
&= \sqrt{16 -2\times4\times\frac92 +\left(\frac92\right)^2} + \frac92\\
&= \sqrt{16 -36 + \left(\frac92\right)^2} +\frac92\\
&= \sqrt {-20 +\left(\frac92\right)^2} + \frac92\\
&= \sqrt{25-45 +\left(\frac92\right)^2} +\frac92\\
&= \sqrt {5^2 -2\times5\times\frac92 + \left(\frac92\right) ^2} + \frac92\\
&= \sqrt {\left(5-\frac92\right)^2} +\frac92\\
&= 5 + \frac92 - \frac92 \\
&= 5\end{align}$$
Where did I go wrong
|
$$\sqrt{a^2}=|a|\not=a$$
Watch the signs.
|
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|
Nested Radical of Ramanujan I think I have sort of a proof of the following nested radical expression due to Ramanujan for $x\ge 0$.
$$\large x+1=\sqrt{1+x\sqrt{1+(x+1)\sqrt{1+(x+2)\sqrt{1+\cdots}}}}$$ for $ x\ge -1$
I just want to know if my proof is okay or there is a flaw, and if there is one I request to give some suggestions to eliminate them. Thank you. The proof is the following:
Proof: Let us define $$ a_n(x)=\underbrace{\sqrt{1+x\sqrt{1+(x+1)\sqrt{1+(x+2)\sqrt{1+\cdots}}}}}_{n \ \mbox{terms}}$$ for $x\ge 0$ so that $$ a_1(x)=\sqrt{1+x},\ a_2(x)=\sqrt{1+x\sqrt{1+(x+1)}},\ a_3(x)=\cdots$$
and so on. Since $x\ge 0$ each one of the $a_n$ is defined (I am taking only the positive square root). Also, we note that $$a_{n+1}^2(x)=1+xa_{n}(x+1)$$
Now, we note that $\{a_n(x)\}$ is an increasing sequence and that $$a_n(x)<x+1$$ $\forall n\ge 1$, this is easy to prove by induction as below:
For $n=1$, $a_1(x)=\sqrt{1+x}<1+x$ since $x\ge 0\Rightarrow 1+x\ge 1$. SO it is true for $n=1$. Similarly, the truth can be proved for $n>1$.
Then $a_n(x)$ converges to $$l(x)=\sup_{n}a_n(x)\le x+1$$ Now I make the following claim:
Claim: $l(x)=x+1\quad \forall i\ge 0$
Proof: Fix $x$. Let $l(x)<x+1$. Then, $l(x)=x+1-\epsilon$ for some $\epsilon>0$. Now, I claim that there must be a $n$ such that $$x+1-a_n(x)<\epsilon$$, and if that is true then $$a_n(x)>x+1-\epsilon=l(x)$$ which is a contradiction since $$l(x)=\sup_{n}a_n(x)$$ and then it implies that $$l(x)=x+1$$ To prove my claim it requires $$x+1-a_n(x) < \epsilon$$ Now, \begin{align}
x+1-a_n(x) = & x+1-\sqrt{1+xa_{n-1}(x+1)} \\
\ =& \frac{(x+1)^2-({1+xa_{n-1}(x+1)})}{x+1+\sqrt{1+xa_{n-1}(x+1)} } \\
\ =& x\frac{(x+1)+1-a_{n-1}(x+1)}{x+1+\sqrt{1+xa_{n-1}(x+1)} } \\
\ <& \frac{x}{x+2}((x+1)+1-a_{n-1}(x+1))\\
\ <& \frac{x}{x+2}\cdot\frac{x+1}{x+3}((x+2)+1-a_{n-2}(x+2))\\
\ <& \frac{x}{x+n-1}\cdot \frac{x+1}{x+n} ((x+n-1)+1-a_{1}(x+n-1))\\
\ <& \frac{x}{x+n-1}\cdot \frac{x+1}{x+n}(x+n-\sqrt{x+n})\\
\ <& \frac{x(x+1)}{x+n-1}
\end{align}
Now, if one is able to find $n$ such that $$\frac{x(x+1)}{x+n-1}<\epsilon
\Rightarrow x< \frac{-(1-\epsilon)+\sqrt{(1-\epsilon)^2+4\epsilon(n-1)}}{2}$$ then we're done.
Now from the upper bound it seems that there always exists some $n$ that satisfies this requirement. Hence the claim is proved.
|
Variant of your proof, for some clarity.
Define $a_1(x)=\sqrt{1+x}$ and $a_{n+1}(x)=\sqrt{1+xa_n(x+1)}$.
Define $b_n(x)=1+x-a_n(x)$. By your proof, we know that $b_n(x)$ is decreasing and bounded below by zero. You want to show that $b_n(x)\to 0$, and then you are done.
Now (this is pretty much exactly your proof, but it is made clearer by have $b_n$ defined):
$$\begin{align}b_{n+1}(x) &= 1+x - a_{n+1}(x)
\\&=\frac{(1+x)^2-a_{n+1}(x)^2}{1+x+a_{n+1}(x)}
\\&=\frac{1+2x+x^2-(1+xa_n(x+1))}{1+x+a_{n+1}(x)}
\\&= \frac{xb_n(x+1)}{1+x+a_{n+1}(x)}
\\&\leq\frac{x}{2+x}b_{n}(x+1)
\end{align}$$
By induction (for $n>k\geq 1$) you can show that:
$$b_n(x) \leq \frac{x(x+1)}{(x+k)(x+k+1)}b_{n-k}(x+k)$$
Therefore, for $k=n-1$, we get:
$$0\leq b_n(x)\leq \frac{x(x+1)}{(x+n-1)(x+n)}b_1(x+n-1) \leq \frac{x(x+1)}{x+n}$$
Therefore, $b_n(x)\to 0$, and hence $a_n(x)\to x+1$.
|
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|
what are real and imaginary part of this expression I have $M:=\sqrt{\frac{a\cdot(b+ic)}{de}}$ and all variables $a,b,c,d,e$ are real. Now I am looking for the real and imaginary part of this, but this square root makes it kind of hard.
|
$$\sqrt{\frac{a(b+ic)}{de}}=\sqrt{\frac{a}{de}}\cdot\sqrt{b+ic}$$
Let $$\sqrt{b+ic}=x+iy$$
$$\implies b+ic=(x+iy)^2=x^2-y^2+2xyi$$
Equating the real & the imaginary parts, $b=x^2-y^2, c=2xy$
So, $b^2+c^2=(x^2-y^2)^2+(2xy)^2=(x^2+y^2)^2\implies x^2+y^2=\sqrt{b^2+c^2}$
We have $$x^2-y^2=b$$
$$\implies 2x^2=\sqrt{b^2+c^2}+b\implies x^2=\frac{\sqrt{b^2+c^2}+b}2$$
$$\implies x=\pm\frac{\sqrt{\sqrt{b^2+c^2}+b}}{\sqrt2}$$
and $$\implies y^2=x^2-b=\frac{\sqrt{b^2+c^2}-b}2$$
$$\implies y=\pm\frac{\sqrt{\sqrt{b^2+c^2}-b}}{\sqrt2}$$
Now, the sign of $y=$ sign of $x\cdot$ sign of $c$
|
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|
What would have been our number system if humans had more than 10 fingers? Try to solve this puzzle. Try to solve this puzzle:
The first expedition to Mars found only the ruins of a civilization.
From the artifacts and pictures, the explorers deduced that the
creatures who produced this civilization were four-legged beings with
a tentatcle that branched out at the end with a number of grasping
"fingers". After much study, the explorers were able to translate
Martian mathematics. They found the following equation:
$$5x^2 - 50x + 125 = 0$$
with the indicated solutions $x=5$ and $x=8$. The value $x=5$ seemed
legitimate enough, but $x=8$ required some explanation. Then the explorers
reflected on the way in which Earth's number system developed, and found
evidence that the Martian system had a similar history. How many fingers would
you say the Martians had?
$(a)\;10$
$(b)\;13$
$(c)\;40$
$(d)\;25$
P.S. This is not a home work. It's a question asked in an interview.
|
Many people believe that since humans have $10$ fingers, we use base $10$. Let's assume that the Martians have $b$ fingers and thus use a base $b$ numbering system, where $b \neq 10$ (note that we can't have $b=10$, since in base $10$, $x=8$ shouldn't be a solution). Then since the $50$ and $125$ in the equation are actually in base $b$, converting them to base $10$ yields $5b+0$ and $1b^2 + 2b + 5$, so we now have:
$$
5x^2-(5b)x + (b^2+2b+5)=0
$$
Since $x=5$ is a solution, substitution yields:
$$ \begin{align*}
5(5)^2-(5b)(5) + (b^2+2b+5) &= 0 \\
b^2-23b+130 &= 0 \\
(b-10)(b-13) &= 0 \\
b&=10,13
\end{align*} $$
Since we know that $b\neq10$, we conclude that the Martians must have $13$ fingers. Indeed, this makes sense, because if $50$ and $125$ are in base $13$, then converting them to base $10$ yields $5(13)=65$ and $1(13)^2+2(13)+5=200$, so our equation becomes:
$$ \begin{align*}
5x^2-65x+200 &= 0 \\
x^2-13x+40&= 0 \\
(x-5)(x-8)&= 0 \\
x&= 5,8 \\
\end{align*} $$
as desired.
|
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How Many License Numbers Consist of $4$ digits and $4$ letters, where one letter or digit is repeated? If license numbers consist of four letters and four numbers how many different licenses can be created having at least one letter or digit repeated?
$$26 \cdot 26 \cdot 26 \cdot 26 \cdot 10 \cdot 10 \cdot 10 \cdot 10 = 4569760000$$
Is this right?
|
Assuming the licences are of the form $\text{LLLLNNNN}$ where $\text{L}$ and $\text{N}$ denote letters and numbers respectively:
Number of licences with no letters or numbers repeated:
$26\cdot 25\cdot 24\cdot 23\cdot 10\cdot 9\cdot 8\cdot 7$.
So, number of licences with at least one letter or number repeated:
$\text{Total number of licences} - 26\cdot 25\cdot 24\cdot 23\cdot 10\cdot 9\cdot 8\cdot 7$.
Where the total number of licences is $26^4\cdot 10^4$ which is what you calculated.
Assuming the letters and numbers can appear in any order:
You can obtain this answer by multiplying the answer obtained above by $\dfrac{8!}{4!4!}$ which is the number of permutations of $8$ things in which $4$ objects are of one kind and $4$ objects are of the other kind. This is the same as the answer given by Omnitic.
Please note that as pointed out by Omnitic, the following does not restrict the number of letters and digits to $4$:
Assuming the numbers or letters can be filled in any order (and are not restricted to be $4$):
The licences have $8$ places which can be filled in with a number or a letter. Thus each place can have one of $26+10=36$ values.
Now, number of licences with no letters or numbers repeated: $36\cdot 35\cdot 34\cdot\cdot\cdot 30\cdot29$.
So, number of licences with at least one letter or number repeated:
$\text{Total number of licences} - 36\cdot 35\cdot 34\cdot\cdot\cdot 30\cdot29$
The total number is given by $36^8$.
|
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show that if $ n-1$ and $n+1$ are both primes and $n>4$, then $\phi(n) \leq n/3$ Show that if $n-1$ and $n+1$ are both primes and $n>4$, then $\phi(n)$ is less than or equal to $n/3$
I tried a few cases
If $n=6$, $n-1=5$, and $n+1=7$ then $~\phi(6)=2=n/3$
If $n=12$, $n-1=11$, and $n+1=13$ then $~\phi(12)=4=n/3$
$\phi(n)=n(1-1/p_1)(1-1/p_2)...(1-1/p_k)$
|
You need to show that $$\phi(n)\leq \frac{n}{3}$$
If $n>4$, with $n-1$ and $n+1$ both being primes,
Now since $n-1$ is a prime $p =(n-1)> 3$
And therefore odd, it follows that, $$(n-1)+1=n, \text{ is even}$$
Likewise we know that at least one of $3$ successive integers, must be divisible by $3$.
Now looking at the integers $\{ n-1,n,n+1 \}$ we see that $n-1$ and $n+1$ are primes
So there only divisors are $1$ and themselves
Thus if $3$ divides $n-1$ or $n+1$, we must have either $3=n-1$ or that $3=n+1$
But we know that $n-1>3$ and $n+1>5$, so we see neither of these can happen
And thus $3$ must divide $n$
So we have that $3\mid n$ and $2\mid n$
Now we need to show that $$\phi(n)\leq \frac{n}{3}$$
Or equivalently $$\frac{\phi(n)}{n} \leq \frac{1}{3}$$
Or $$\prod_{p\mid n}(1-\frac{1}{p})\leq \frac{1}{3}$$
But since $2\mid n$ and $3\mid n$ we can re-write this as,
$$\frac{1}{3}\prod_{p\mid n}_{p\ne 2}_{p\ne 3}(1-\frac{1}{p})=(1-\frac{1}{3})(1-\frac{1}{2})\prod_{p\mid n}_{p\ne 2}_{p\ne 3}(1-\frac{1}{p})\leq \frac{1}{3}$$
$$\iff \frac{1}{3}\prod_{p\mid n}_{p\ne 2}_{p\ne 3}(1-\frac{1}{p})\leq \frac{1}{3}$$
$$\iff \prod_{p\mid n}_{p\ne 2}_{p\ne 3}(1-\frac{1}{p})\leq 1$$
Which is clearly true,
Thus if $n>4$, with $n-1$ and $n+1$ both being primes
$$\phi(n)\leq \frac{n}{3}$$
As required.
|
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Using the definition of the derivative prove that if $f(x)=x^\frac{4}{3}$ then $f'(x)=\frac{4x^\frac{1}{3}}{3}$ So I have that $f'(x)=\lim_{h \to 0} \frac{f(x+h)-f(x)}{h}$
and know that applying
$f(x)=x^\frac{4}{3}=f\frac{(x+h)^\frac{4}{3} -x^\frac{4}{3}}{h}$
but am at a loss when trying to expand
$(x+h)^\frac{4}{3}$
|
You can use the generalized binomial coeficients to solve this problem, which gives you
$(x + h)^\frac{4}{3} = x^\frac{4}{3}(1 + \frac{4h}{3x} + \frac{4h^2}{18x^2} + \ldots)$
It will go on forever, but since you divide by $h$, most will tend to zero, and you'll be left with $x^\frac{4}{3} \frac{4}{3x} = \frac{4}{3}x^\frac{1}{3}$.
|
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|
Easy trigonometry question How is
$$\frac{2 \sin x}{(1+ \cos x)^2}= \tan\left(\frac{x}{2}\right)\sec^2\left(\frac{x}{2}\right)\;?$$
It should be easy.
But somehow I don't get it.
Can you help me with this?
|
$$\frac{2 \sin x}{(1+ \cos x)^2}=\frac{4 \sin \frac{x}{2}\cos \frac{x}{2}}{(1+ 2\cos^2 \frac{ x}{2}-1)^2}=\frac{ \sin \frac{x}{2}\cos \frac{x}{2}}{( \cos^4 \frac{ x}{2})}=\tan\frac{x}{2}\sec^2\frac{x}{2}$$
|
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|
Solving recurrence equation using generating function method I've been trying to solve the following equation intuitively (I only know the method if there are minuses in the equation - $a_{n-1}, a_{n-2}...$).
$$a_{n+2}=4a_{n+1}-4a_{n}$$
$$a_{0}=3$$
$$a_{1}=8$$
$$
\begin{align}
A(x)&=\sum\limits_{n>=0}a_{n}x^{n} \\
&= \sum\limits_{n>=0}(a_{n+1}-\frac{1}{4}a_{n+2})x^{n} \\
&= \sum\limits_{n>=0}(a_{n+1})x^{n}-\frac{1}{4}\sum\limits_{n>=0}(a_{n+2})x^{n} \\
&= \sum\limits_{n>=1}(a_{n})x^{n+1}-\frac{1}{4}\sum\limits_{n>=2}(a_{n})x^{n+2} \\
&= \frac{1}{x}\sum\limits_{n>=1}(a_{n})x^{n}-\frac{1}{4x^{2}}\sum\limits_{n>=2}(a_{n})x^{n} \\
&= \frac{1}{x}[\sum\limits_{n>=0}(a_{n})x^{n} - 3]-\frac{1}{4x^{2}}[\sum\limits_{n>=0}(a_{n})x^{n} - 3 - 8x] \\
&= \frac{1}{x}[A(x) - 3]-\frac{1}{4x^{2}}[A(x) - 3 - 8x]
\end{align}
$$
So I get
$$A(x)=\frac{-4x+3}{4x^{2}-4x+1}$$
Is this correct? I'm asking because the answer to this question according to the source I got it from is $x^{2}-4x+4$ as the denominator...
|
Not sure where you went wrong, so I'll start from scratch with a different method:
Let $A(x)=\sum_{n=0}^\infty a_nx^n$. Note that $a_n$ satisfies $a_{n+2}-4a_{n+1}+4a_n=0$. Let $p(x)=1-4x+4x^2$. We then define
$$
B(x) = \sum_{n=0}^\infty b_nx^n=p(x)A(x)
$$
Note that for $n\geq2$, we have $b_n=a_{n+2}-4a_{n+1}+4a_n=0$. So, we simply have
$$
p(x)A(x) = b_0+b_1 x
$$
Thus,
$$
A(x) = \frac{b_0+b_1 x}{1-4x+4x^2}=\frac{a_0+(a_1-4a_0) x}{1-4x+4x^2}=\frac{3-4x}{1-4x+4x^2}
$$
It seems you have the right answer, and that your source has a typo.
|
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Compute $\int x^2 \cos \frac{x}{2} \mathrm{d}x$ I am trying to compute the following integral:
$$\int x^2 \cos \frac{x}{2} \mathrm{d}x$$
I know this requires integration by parts multiple times but I am having trouble figuring out what to do once you have integrated twice. This is what I have done:
Let $u = \cos \frac{x}{2}$ and $\mathrm{d}u = \frac{-\sin\left(\dfrac{x}{2}\right)}{2} \mathrm{d}x$ and $\mathrm{d}v = x^2$ and $v= \frac{x^3}{3}$.
\begin{align}
&\int x^2 \cos \frac{x}{2} \\
&\cos \frac{x}{2} \cdot \frac{x^3}{3} - \int \frac{x^3}{3} \cdot \frac{-\sin\left(\dfrac{x}{2}\right)}{2} \mathrm{d}x
\end{align}
So now I integrate $\int \frac{x^3}{3} \cdot \frac{-\sin\left(\dfrac{x}{2}\right)}{2} \mathrm{d}x$ to get:
\begin{align}
\frac{-\sin\left(\dfrac{x}{2}\right)}{2} \cdot \frac{x^4}{12} - \int \frac{x^4}{12} \cdot \frac{\cos x}{2}
\end{align}
Now, this is where I get stuck. I know if I continue, I will end up with $\frac{-\sin\left(\dfrac{x}{2}\right)}{2}$ again when I integrate $\cos \frac{x}{2}$. So, where do I go from here?
Thanks!
|
let S be the integrate... That is, S x"2cosx/2 dx........let u=x"2 and dv=cosx/2dx. so that, du=2x and v=2sinx/2 dx. Therefore, S x"2cosx/2 dx= x"2(2sinx/2)-S 2sinx/2(2x) =2x"2sinx/2-S 2x.2sinx/2 dx = 2x"2sinx/2-4 S xcosx/2 dx. Apply another integration. S xcosx/2 dx. let u=x and dv=cosx/2dx. so that du=(1)dx=dx and v=2sinx/2 dx. Hence, S xcosx/2 dx=x(2cosx/2)-S 2sinx/2 dx. combining all formulas we get. S x"2cosx/2dx=2x"2sin(x/2)-2(4xcos(x/2)-S 2sin(x/2) dx). That is, S x"2sin(x/2)+8xcos(x/2)-16sin(x/2) dx = 2x"2sin(x/2)+8xcos(x/2)-16sin(x/2)+C. SO, (2x"2-16)sin(x/2)+8xcos(x/2)+C. I hope this help. By O'john
|
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Prove that $\sum_{n=1}^{\infty} \frac{1}{2^n - 1}$ is convergent and $\sum_{n=1}^{\infty} \frac{1}{2^n - 1} < 2$. Prove that $\sum_{n=1}^{\infty} \frac{1}{2^n - 1}$ is convergent and $\sum_{n=1}^{\infty} \frac{1}{2^n - 1} < 2$.
I'm not sure, but I suppose that $$\sum_{n=1}^{\infty} \frac{1}{2^n - 1} < \sum_{n=0}^{\infty} \frac{1}{2^n} = 2$$
I don't know how can I prove it exactly, but we have:
$$\sum_{n=1}^{\infty} \frac{1}{2^n - 1} = 1+ \frac{1}{3} + \frac{1}{7} + \frac{1}{15}+... \\
\sum_{n=0}^{\infty} \frac{1}{2^n} = 1+ \frac{1}{2} + \frac{1}{4} + \frac{1}{8}+...$$
Because
$$1 \le 1 \\ \frac{1}{3} < \frac{1}{2} \\ \frac{1}{7} < \frac{1}{4} \\ \frac{1}{15} < \frac{1}{8} \\ \mbox{and so on}$$
we have
$$\sum_{n=1}^{\infty} \frac{1}{2^n - 1} = 1+ \frac{1}{3} + \frac{1}{7} + \frac{1}{15}+... < 1+ \frac{1}{2} + \frac{1}{4} + \frac{1}{8}+... = \sum_{n=0}^{\infty} \frac{1}{2^n}$$
Could you tell me, does it work? Maybe exist more formal proof?
|
This is a formalization of your proof
$$
\sum\limits_{n=1}^\infty\frac{1}{2^n-1}
=1+\sum\limits_{n=2}^\infty\frac{1}{2^n-1}
=1+\sum\limits_{n=1}^\infty\frac{1}{2^{n+1}-1}
<1+\sum\limits_{n=1}^\infty\frac{1}{2^n}=2
$$
|
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How prove that $10(a^3+b^3+c^3)-9(a^5+b^5+c^5)\le\dfrac{9}{4}$ let $a,b,c\ge 0$, such that $a+b+c=1$, prove that
$$10(a^3+b^3+c^3)-9(a^5+b^5+c^5)\le\dfrac{9}{4}$$
This problem is simple as 2005, china west competition problem
$$10(a^3+b^3+c^3)-9(a^5+b^5+c^5)\ge 1$$
see:(http://www.artofproblemsolving.com/Forum/viewtopic.php?p=362838&sid=00aa42b316d41e251e24e658594fcc51#p362838)
for 2005 china west problem we have two methods (at least)
solution 1：
note
$$(a+b+c)^3=a^3+b^3+c^3+3(a+b)(b+c)(a+c)$$
$$(a+b+c)^5=a^5+b^5+c^5+5(a+b)(b+c)(a+c)(a^2+b^2+c^2+ab+bc+ac)$$
then
$$\Longleftrightarrow 10[1-3(a+b)(b+c)(a+c)]-9[1-5(a+b)(b+c)(a+c)(a^2+b^2+c^2+ab+bc+ac)]\ge 1$$
$$\Longleftrightarrow 3(a+b)(b+c)(a+c)(a^2+b^2+c^2+ab+bc+ac)-2(a+b)(b+c)(a+c)\ge 0$$
$$\Longleftrightarrow 3(a^2+b^2+c^2+ab+bc+ac)\ge 2=2(a+b+c)^2$$
$$ a^2+b^2+c^2-ab-bc-ac\ge 0$$
It's Obviously.
solution 2:
$$10(a+b+c)^2(a^3+b^3+c^3)-9(a^5+b^5+c^5)-(a+b+c)^5\ge0$$
it is equivalent to
$$15(a+b)(b+c)(c+a)(a^2+b^2+c^2-ab-bc-ca)\ge0$$
But for
$$10(a^3+b^3+c^3)-9(a^5+b^5+c^5)\le\dfrac{9}{4}$$
and for this equality I think
$$10a^3-9a^5\le p (a-1/3)+q$$
and let
$$f(x)=10x^3-9x^5\Longrightarrow f'(x)=30x^2-45x^4\Longrightarrow p=f'(1/3)=\dfrac{25}{9}$$
$$q=f(1/3)=\dfrac{1}{9}$$
so if we can prove this
$$10a^3-9a^5\le\dfrac{25}{9}(a-1/3)+\dfrac{1}{9}$$
These methods I can't work, can someone help deal it. Thank you
|
We need to prove that
$$40(a^3+b^3+c^3)(a+b+c)^2-36(a^5+b^5+c^5)\leq9(a+b+c)^5$$ and since the last inequality is homogeneous already,
it's enough to prove this inequality for all non-negatives $a$, $b$ and $c$.
Now, let $a+b+c=3u$, $ab+ac+bc=3v^2$ and $abc=w^3$.
Thus, since our inequality is fifth degree, we need to prove that $f(w^3)\geq0,$ where $f$ is a linear function.
But the linear function gets a minimal value for the extreme value of $w^3$,
which happens in the following cases.
*
*$w^3=0$.
Let $c=0$ and $b=1$.
We obtain:
$$(a^2-4a+1)^2\geq0;$$
2. Two variables are equal.
Let $b=c=1$.
We obtain:
$$9(a+2)^5\geq40(a^3+2)(a+2)^2-36(a^5+2)$$ or
$$a^5-14a^4+40a^3+128a^2+80a+8)\geq0,$$ which is true by AM-GM:
$$a^5-14a^4+40a^3+128a^2+80a+8\geq$$
$$\geq a^2\left(3\left(\frac{a^3}{3}\right)+40a+128-14a^2\right)\geq a^4\left(5\sqrt[5]{\left(\frac{1}{3}\right)^3\cdot40\cdot128}-14\right)\geq0.$$
Done!
|
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|
Find the value of $\textrm{cosec}^2\left(\frac\pi7\right) +\textrm{cosec}^2\left(\frac{2\pi}7\right)+\textrm{cosec}^2\left(\frac{4\pi}7\right)$ What is the value of $$\textrm{cosec}^2\left(\frac\pi7\right) +\textrm{cosec}^2\left(\frac{2\pi}7\right)+\textrm{cosec}^2\left(\frac{4\pi}7\right) \qquad\qquad ? $$
I tried to write $\textrm{cosec}^2\left(\frac{4\pi}7\right)$ as $\textrm{cosec}^2\left(\frac{3\pi}7\right)$. Then converted in $\sin$... But in vain.. Is there any other approach?
|
$$\cos\frac{2\pi}{7}+\cos\frac{4\pi}{7}+\cos\frac{6\pi}{7}=\frac{2\sin\frac{\pi}{7}\cos\frac{2\pi}{7}+2\sin\frac{\pi}{7}\cos\frac{4\pi}{7}+2\sin\frac{\pi}{7}\cos\frac{6\pi}{7}}{2\sin\frac{\pi}{7}}=$$
$$=\frac{\sin\frac{3\pi}{7}-\sin\frac{\pi}{7}+\sin\frac{5\pi}{7}-\sin\frac{3\pi}{7}+\sin\frac{7\pi}{7}-\sin\frac{5\pi}{7}}{2\sin\frac{\pi}{7}}=-\frac{1}{2},$$
$$\cos\frac{2\pi}{7}\cos\frac{4\pi}{7}+\cos\frac{2\pi}{7}\cos\frac{6\pi}{7}+\cos\frac{4\pi}{7}\cos\frac{6\pi}{7}=$$
$$=\frac{1}{2}\left(\cos\frac{6\pi}{7}+\cos\frac{2\pi}{7}+\cos\frac{6\pi}{7}+\cos\frac{4\pi}{7}+\cos\frac{4\pi}{7}+\cos\frac{2\pi}{7}\right)=$$
$$=\cos\frac{2\pi}{7}+\cos\frac{4\pi}{7}+\cos\frac{6\pi}{7}=-\frac{1}{2}$$ and
$$\cos\frac{2\pi}{7}\cos\frac{4\pi}{7}\cos\frac{6\pi}{7}=\frac{8\sin\frac{2\pi}{7}\cos\frac{2\pi}{7}\cos\frac{4\pi}{7}\cos\frac{8\pi}{7}}{8\sin\frac{2\pi}{7}}=\frac{\sin\frac{16\pi}{7}}{8\sin\frac{2\pi}{7}}=\frac{1}{8}.$$
Id est,
$$\frac{1}{\sin^2\frac{\pi}{7}}+\frac{1}{\sin^2\frac{2\pi}{7}}+\frac{1}{\sin^2\frac{4\pi}{7}}=$$
$$=2\left(\frac{1}{1-\cos\frac{2\pi}{7}}+\frac{1}{1-\cos\frac{4\pi}{7}}+\frac{1}{1-\cos\frac{6\pi}{7}}\right)=$$
$$=\frac{2\left(\left(1-\cos\frac{2\pi}{7}\right)\left(1-\cos\frac{4\pi}{7}\right)+\left(1-\cos\frac{2\pi}{7}\right)\left(1-\cos\frac{6\pi}{7}\right)+\left(1-\cos\frac{4\pi}{7}\right)\left(1-\cos\frac{6\pi}{7}\right)\right)}{\left(1-\cos\frac{2\pi}{7}\right)\left(1-\cos\frac{4\pi}{7}\right)\left(1-\cos\frac{6\pi}{7}\right)}=$$
$$=\frac{2\left(3-2\left(-\frac{1}{2}\right)+\left(-\frac{1}{2}\right)\right)}{1-\left(-\frac{1}{2}\right)+\left(-\frac{1}{2}\right)-\frac{1}{8}}=8.$$
|
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On the Pell-like $Ax^2-By^2 = 1$ This is connected to the post, Mere coincidence? (prime factors). I was looking at NeuroFuzzy's dataset and noticed the line,
{{{1, {4, 2}}, {1, 4, 2, 4, 2}, 23762}}
It seems this could be generalized. Is it true that given the equation,
$$(n+1)x^2-ny^2 = 1\tag{1}$$
then its solutions are given by,
$$\frac{y_1}{x_1} = 1+\cfrac{1}{2n+\cfrac{1}{2}} = \frac{4n+3}{4n+1}\tag{2}$$
$$\frac{y_2}{x_2} = 1+\cfrac{1}{2n+\cfrac{1}{2+\cfrac{1}{2n+\cfrac{1}{2}}}} = \frac{16n^2+20n+5}{16n^2+12n+1} \tag{3}$$
and so for all $x_i, y_i$? I assume it is connected to the fact that,
$$\sqrt{\frac{n+1}{n}} = 1+\cfrac{1}{2n+\cfrac{1}{2+\cfrac{1}{2n+\cfrac{1}{2+\ddots}}}}\tag{4}$$
and truncating $(4)$ at the right periodic points, correct?
|
Here
$$
\left( \begin{array}{c}
x \\
y
\end{array}
\right)
$$
will be a solution of $$ (n+1) x^2 - n y^2 = 1. $$
We have an "automorph" or generator of the automorphism group or isometry group or orthogonal group of the indefinite binary quadratic form depicted; the form is $ (n+1) x^2 - n y^2. $
$$ A =
\left( \begin{array}{cc}
2n+1 & 2 n \\
2n+2 & 2n+1
\end{array}
\right) ,
$$ and
$$
\left( \begin{array}{cc}
2n+1 & 2 n \\
2n+2 & 2n+1
\end{array}
\right)
\left( \begin{array}{c}
1 \\
1
\end{array}
\right) =
\left( \begin{array}{c}
4n+1 \\
4n+3
\end{array}
\right),
$$
$$
\left( \begin{array}{cc}
2n+1 & 2 n \\
2n+2 & 2n+1
\end{array}
\right)
\left( \begin{array}{c}
4n+1 \\
4n+3
\end{array}
\right) =
\left( \begin{array}{c}
16n^2 +12n+1 \\
16 n^2 + 20 n + 5
\end{array}
\right),
$$
$$
\left( \begin{array}{cc}
2n+1 & 2 n \\
2n+2 & 2n+1
\end{array}
\right)
\left( \begin{array}{c}
16n^2 +12n+1 \\
16 n^2 + 20 n + 5
\end{array}
\right) =
\left( \begin{array}{c}
64 n^3 + 80 n^2 + 24 n + 1 \\
64 n^3 + 112 n^2 + 56 n + 7
\end{array}
\right),
$$
$$
\left( \begin{array}{cc}
2n+1 & 2 n \\
2n+2 & 2n+1
\end{array}
\right)
\left( \begin{array}{c}
64 n^3 + 80 n^2 + 24 n + 1 \\
64 n^3 + 112 n^2 + 56 n + 7
\end{array}
\right) =
\left( \begin{array}{c}
256 n^4 + 448 n^3 + 240 n^2 + 40 n + 1 \\
256 n^4 + 576 n^3 + 432 n^2 + 120 n + 9
\end{array}
\right),
$$
and so on forever. Except for $\pm$ sign these are all.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/470734",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 1,
"answer_id": 0
}
|
Prove by induction $\frac{n^n}{3^n}$\dfrac{n^n}{3^n}<n!<\dfrac{n^n}{2^n}$
The case $n!<\dfrac{n^n}{2^n}$ is easier.
|
Another way this might be viewed is by rearranging the inequalities as
$$2^n \ < \ \frac{n^n}{n!} \ \ \text{and} \ \ \frac{n^n}{n!} \ < \ 3^n , $$
the entirety of which holds for $ \ n \ = \ 6 \ , $ as already mentioned by user84413 . The "induction step" for the ratio is then
$$\frac{(n+1)^{n+1}}{(n+1)!} \ = \ \frac{(n+1)}{(n+1)} \ \cdot \ \frac{(n+1)^{n}}{n!} \ = \ \left( \frac{n+1}{n} \right)^n \ \cdot \ \frac{n^{n}}{n!} \ . $$
We know that this first factor produces a number between 2 and 3 for $ \ n \ \ge \ 1 \ $ (in fact, it's a familiar statement that $ \ \lim_{n \rightarrow \infty} \ \left( \frac{n+1}{n} \right)^n = \ e \ )^{*} \ , $
$^{*}$ indeed, some regard this as the defining equation for $ \ e \ $
so we can write
$$2^{n+1} \ = \ 2 \ \cdot \ 2^n \ < \ 2 \ \cdot \frac{n^n}{n!} \ \le \ \left( \frac{n+1}{n} \right)^n \ \cdot \ \frac{n^{n}}{n!} \ = \ \frac{(n+1)^{n+1}}{(n+1)!} \ $$
[the equation being true for $ \ n = 1 \ $ ]
and
$$ \frac{(n+1)^{n+1}}{(n+1)!} \ = \ \left( \frac{n+1}{n} \right)^n \ \cdot \ \frac{n^{n}}{n!} \ < \ 3\ \cdot \frac{n^{n}}{n!} \ < \ 3 \ \cdot \ 3^n \ = \ 3^{n+1} \ . $$
Thus, for $ \ n \ \ge \ 6 \ , $ we find $ \ 2^n \ < \ \frac{n^n}{n!} \ < \ 3^n \ , \ $ equivalent to the original inequality
$$ \ \frac{n^n}{3^n} \ < \ n! \ < \ \frac{n^n}{2^n} \ . \ $$
$$\\$$
Note that in so doing, we have also proven that $ \ \left(\frac{1}{3}\right)^n < \ \frac{n!}{n^n} \ < \ \left(\frac{1}{2}\right)^n \ , \ $ and hence, by the "Squeeze Theorem", that $ \ \lim_{n \rightarrow \infty} \ \frac{n!}{n^n} \ = \ 0 \ . $ The induction work is similar to that required in applying the Ratio Test to demonstrate the absolute convergence of $ \ \sum_{n=1}^{\infty} \ \frac{n!}{n^n} \ $ (and so also the divergence of $ \ \sum_{n=1}^{\infty} \ \frac{n^n}{n!} \ ) \ . $
ADDENDUM: It might seem, from the preceding discussion, that we ought to get something like an equation out of this by just using $ \ e \ $ ; but, in fact we find that $ \ \frac{n^n}{e^n} < \ n! \ . \ $ This touches on the Stirling approximation, which requires some additional non-linear factors in order to get closer to accurate values of $ \ n! \ $ for large $ \ n \ . $
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/472220",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 2
}
|
Find the value of the the term The sequence $a_1,a_2,a_3,\ldots$ satisfies $a_1=1$, $a_2=2$, and $$a_{n+2}=\frac2{a_{n+1}}+a_n\;;$$ find the value of
$$\frac{a_{2012}2^{2009}}{2011}$$
|
As lab bhattacharjee shows, $a_{n+2}a_{n+1}=2n+2$, and thus,
$$
\begin{align}
\frac{a_{n+2}}{a_n}
&=\frac{a_{n+2}}{a_{n+1}}\frac{a_{n+1}}{a_n}\\
&=\frac{n+1}{n}
\end{align}
$$
Therefore,
$$
\begin{align}
\frac{a_{2n}}{a_2}
&=\frac{2n-1}{2n-2}\frac{2n-3}{2n-4}\cdots\frac32\\
&=\color{#C00000}{\frac1{4^{n-1}}\frac{(2n-1)!}{(n-1)!(n-1)!}}\\
&=\frac{n}{2^{2n-1}}\binom{2n}{n}
\end{align}
$$
and since $a_2=2$,
$$
a_{2n}=\frac{n}{4^{n-1}}\binom{2n}{n}
$$
I believe this verifies lab's hint (in red).
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/472989",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
}
|
How to find $\sum_{n=0}^{\infty}(-1)^{n}\left(\frac{1}{n+1}-\frac{1}{n+2}+\frac{1}{n+3}-\cdots\right)^{2}$ Find the value
$$\sum\limits_{n=0}^{\infty}(-1)^{n}\left(\frac{1}{n+1}-\frac{1}{n+2}+\frac{1}{n+3}-\cdots\right)^{2}.$$
This problem is from this and I am interested in this problem, but I can't solve it.
Here is my idea:
$$(-1)^{k-1}\dfrac{1}{n+k}=(-1)^{-n}\int_{0}^{-1}x^{n+k-1}dx$$
so
\begin{align*}\sum\limits_{n=0}^{\infty}(-1)^{n}\left(\frac{1}{n+1}-\frac{1}{n+2}+\frac{1}{n+3}-\cdots\right)^{2}&=\sum_{n=0}^{\infty}(-1)^n\left(\sum_{k=1}^{\infty}(-1)^n\int_{0}^{-1}x^{n+k-1}dx\right)^2\\
&=\sum_{n=0}^{\infty}(-1)^n\left(\int_{0}^{-1}\sum_{k=1}^{\infty}x^{n+k-1}dx\right)^2\\
&=\sum_{n=0}^{\infty}(-1)^n\left(\int_{0}^{-1}\dfrac{x^n}{1-x} dx \right)^2.
\end{align*}
|
You can take that last expression and turn it into an integral that gets you a result that agrees with both Mathematica and the answer linked to (and not really quite explained).
Your last sum may be rewritten as
$$\begin{align}\sum_{n=0}^{\infty} (-1)^n \int_0^{-1} dx \frac{x^n}{1-x} \, \int_0^{-1} dy \frac{y^n}{1-y}&= \int_0^{-1} dx \frac{1}{1-x} \, \int_0^{-1} dy \frac{1}{1-y} \sum_{n=0}^{\infty} (-1)^n x^n y^n \\ &= \int_0^{-1} dx \frac{1}{1-x} \, \int_0^{-1} dy \frac{1}{1-y} \frac{1}{1+x y} \\&= \int_0^{-1} dx \frac{1}{1-x} \, \int_0^{-1} dy \left (\frac{1}{1+x}\frac{1}{1-y}+ \frac{x}{1+x} \frac{1}{1+x y}\right )\\ &= \int_0^{-1} dx \frac{\log{(1-x)}-\log{2}}{1-x^2} \\ &= \frac{\pi^2}{24}\end{align}$$
ADDENDUM
That last integral may be evaluated by substituting $x=1-2 u$ as follows:
$$\begin{align}\int_0^{-1} dx \frac{\log{(1-x)}-\log{2}}{1-x^2} &= -\frac12 \int_{1/2}^1 du \frac{\log{u}}{u-u^2} \\ &= -\frac12 \int_{1/2}^1 du \frac{\log{u}}{u}-\frac12 \int_{1/2}^1 du \frac{\log{u}}{1-u} \\ &= \frac12 \log^2{2} + \frac12 \text{Li}_2\left(\frac12\right) \\ &= \frac12 \log^2{2} + \frac{\pi^2}{24} - \frac12 \log^2{2}\end{align}$$
The result follows.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/473996",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "19",
"answer_count": 1,
"answer_id": 0
}
|
How many factors does 6N have? Given a number $2N$ having 28 factors another number $3N$ having 30 factors, then find out the number of factors of $6N$.
|
Let $p_i$, be primes in increasing order. Consider the prime factorization of $n$.
$$ n = 2^{a_1} \times 3^{a_2} \times \prod p_i^{a_i}, $$
where $a_i$ are non-negative integers. For this question, let the indexing run over $i\geq 3$.
You are given that
$$ (a_1 + 2) \times (a_2 +1 ) \times \prod (a_i + 1) = 28 $$
and
$$( a_1 + 1) \times (a_2 + 2) \times \prod(a_i + 1) = 30 $$
The factors of 28 are $1, 2, 4, 7, 14, 28$. The factors of 30 are $1, 2, 3, 5, 6, 10, 15, 30$.
A quick check shows that the possible values of $a_1$ are 2 or 5.
If $a_1 = 2$, then we must have
$$ \frac{ a_2 +1 } { a_2 + 2} = \frac{7}{10},$$
which has no positive integer solution.
If $a_1 = 5$, then we must have
$$\frac{a_2 + 1} { a_2 + 2} = \frac{4}{5}, $$
which has a positive integer solution $a_2 = 3$. This implies that $\prod (a_i+1) = 1$. Hence, the number of factors of $6n$ is
$$( a_1 + 2) \times (a_2 + 2) \times \prod(a_i + 1) = 7 \times 5 \times 1 = 35 $$
In this case, $n = 2^5 \times 3^3$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/475002",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
}
|
Differentiate $x \sqrt{1+y}+y \sqrt{1+x}=0$
If $x \sqrt{1+y}+y \sqrt{1+x}=0$, prove that $(1+x^2)\frac{dy}{dx}+1=0.$
The answer I got is $$\frac{dy}{dx}= -\frac{2 \sqrt{1+x} \sqrt{1+y}+y}{x+2 \sqrt{1+x}\sqrt{1+y}}$$ but I cannot simplify it further.
Please provide your assistance.
|
$$x\sqrt{1+y} = -y\sqrt{1+x}$$
squaring both sides
$$ x^2(1+y) = y^2(1+x)$$
simplifying
$$x^2 - y^2 = xy(y - x)$$
$$x+y = -xy$$
$$y =\frac{-x}{1+x}$$
further take derivatives.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/475824",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
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|
How can we determine the number of terms which we have to take in a series to get a particular accurate? As I remember , two days ago , there was a question ( here ) asks for calculating this limit
$\displaystyle \lim \limits_{x\rightarrow \infty } \frac{x^3}{e^x}$ and the question was answered .
of course this is an easy limit , we can calculate it using l'hopital rule .
Now , if I expressed $\displaystyle e^x = 1+x+\frac{x^2}{2!} + \frac{x^3}{3!} + \ldots = \sum\limits_{i=0}^{\infty}\frac{x^i}{i!}$ and substitute back an approximation for $e^x$.
If we approximated $e^x$ for 3 terms then the limit is $\infty$ , if we approximated it for 4 terms then the limit is 6 but if we approximated for 5 terms then we get the right answer which is 0 .
Now , How can we know that approximation for 5 terms will work while approximation for 4 or 3 terms would fail ? Is there any test or analytical method to determine this ?
I used the geometric approach to see the reason behind that . here is the sketch for those approximations and the original function which illustrate and show why approximation for 3 and 4 terms fail . but I ask for Analytical approach as using the sketch of approximations is not rigorous and practical in general .
|
It took me a while to figure out what the question was asking. It seems the question is why
$$
\begin{align}
\lim_{x\to\infty}\frac{x^3}{1+x+\frac{x^2}{2}}&=\infty\tag{1}\\
\lim_{x\to\infty}\frac{x^3}{1+x+\frac{x^2}{2}+\frac{x^3}{6}}&=6\tag{2}\\
\lim_{x\to\infty}\frac{x^3}{1+x+\frac{x^2}{2}+\frac{x^3}{6}+\frac{x^4}{24}}&=0\tag{3}\\
\end{align}
$$
while
$$
\lim_{x\to\infty}\frac{x^3}{e^x}=0\tag{4}
$$
In actuality, $e^x$ grows faster than any power of $x$. We see why in the example of $x^3$ above. First of all, for $x\ge0$
$$
e^x\ge1+x+\frac{x^2}{2}+\frac{x^3}{6}+\dots+\frac{x^n}{n!}\tag{5}
$$
for any $n$.
Thus, $(5)$ (substituting $n\mapsto n+1$) implies
$$
\begin{align}
\lim_{x\to\infty}\frac{x^n}{e^x}
\le&\lim_{x\to\infty}\frac{x^n}{1+x+\frac{x^2}{2}+\frac{x^3}{6}+\dots+\frac{x^n}{n!}+\frac{x^{n+1}}{(n+1)!}}\\
=&\lim_{x\to\infty}\frac1{\frac1{x^n}+\frac1{x^{n-1}}+\frac1{2x^{n-2}}+\frac1{6x^{n-3}}+\dots+\frac1{n!}+\frac{x}{(n+1)!}}\\
\le&\lim_{x\to\infty}\frac{(n+1)!}{x}\\[14pt]
=&0\tag{6}
\end{align}
$$
If we'd used one fewer term in the denominator of $(6)$, we would get
$$
\begin{align}
\lim_{x\to\infty}\frac{x^n}{e^x}
\le&\lim_{x\to\infty}\frac{x^n}{1+x+\frac{x^2}{2}+\frac{x^3}{6}+\dots+\frac{x^{n-1}}{(n-1)!}+\frac{x^n}{n!}}\\
=&\lim_{x\to\infty}\frac1{\frac1{x^n}+\frac1{x^{n-1}}+\frac1{2x^{n-2}}+\frac1{6x^{n-3}}+\dots+\frac1{x(n-1)!}+\frac1{n!}}\\[4pt]
=&n!\tag{7}
\end{align}
$$
because we are using a polynomial of the same order as the numerator to approximate $e^x$ near $x=\infty$.
If we'd used two fewer terms in the denominator of $(6)$, we would get
$$
\begin{align}
\lim_{x\to\infty}\frac{x^n}{e^x}
\le&\lim_{x\to\infty}\frac{x^n}{1+x+\frac{x^2}{2}+\frac{x^3}{6}+\dots+\frac{x^{n-2}}{(n-2)!}+\frac{x^{n-1}}{(n-1)!}}\\
=&\lim_{x\to\infty}\frac1{\frac1{x^n}+\frac1{x^{n-1}}+\frac1{2x^{n-2}}+\frac1{6x^{n-3}}+\dots+\frac1{x^2(n-2)!}+\frac1{x(n-1)!}}\\
=&\lim_{x\to\infty}x\cdot\lim_{x\to\infty}\frac1{\frac1{x^{n-1}}+\frac1{x^{n-2}}+\frac1{2x^{n-3}}+\frac1{6x^{n-4}}+\dots+\frac1{x(n-2)!}+\frac1{(n-1)!}}\\
=&\lim_{x\to\infty}x\cdot(n-1)!\\[9pt]
=&\infty\tag{8}
\end{align}
$$
because we are using a polynomial of a lower order than the numerator to approximate $e^x$ near $x=\infty$.
However, given $(6)$, $(7)$, and $(8)$, we can easily conclude that
$$
\lim_{x\to\infty}\frac{x^n}{e^x}\le\min(0,n!,\infty)=0\tag{9}
$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/476875",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
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|
$\tan A + \tan B + \tan C = \tan A\tan B\tan C\,$ in a triangle I want to prove this (where each angle may be negative or greater than $180^\circ$):
When $A+B+C = 180^\circ$ \begin{equation*} \tan A + \tan B + \tan C = \tan A\:\tan B\:\tan C. \end{equation*}
We know that
\begin{equation*}\tan(A+B) = \frac{\tan A+\tan B}{1-\tan A\tan B}\end{equation*} and that \begin{equation*}\text{and that}~A+B = 180^\circ-C.\end{equation*}
Therefore $\tan(A+B) = -\tan C.$
From here, I got stuck.
|
$$A+B=180^\circ-C\\
\tan(A+B)=\tan(180^\circ-C)\\
\frac{\tan A+\tan B}{1-\tan A\tan B} =-\tan C\\
\tan A +\tan B +\tan C =\tan A \tan B \tan C.$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/477364",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "15",
"answer_count": 11,
"answer_id": 10
}
|
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.