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How can $\frac{x^3-4x^2+4x}{x^2-4}$ be both $0$ and "undefined" when $x = 2$? Suppose I have a function defined as $$F(x)= \frac{x^3-4x^2+4x}{x^2-4}$$
Now I want to find the value of $F(2)$. I can do it in 2 ways:
*
*Put $x=2$ and solve the function. It will give:
$$F(2)=\frac{0}{0}$$ which is not defined.
*Solve $F(x)$ first and then put $x=2$.
$$F(x)= \frac{x(x-2)^2}{(x-2)(x+2)}=\frac{x(x-2)}{x+2}$$
It will give $${F(2)=\frac{0}{4}}$$ which is zero.
How can zero equal not defined?
|
"Solve $F(x)$" is the wrong terminology. "Simplify $F(x)$" fits better. One solves problems; one solves equations; one evaluates or sometimes simplifies expressions.
$\dfrac 5 0$ is undefined because there is no number $x$ for which $0x = 5.$
But $\dfrac 0 0$ is undefined because there are many numbers $x$ for which $0x=0$, rather than just one such number.
A basic fact of algebra is that if you plug a number into a polynomial in a variable $x$ and get $0$, then $x$ minus that number is a factor of the polynomial. For example, suppose
$$
f(x) = x^3 -7x^2 + 5x + 21
$$
so that
$$
f(3) = 0.
$$
We conclude that
$$
x^3 -7x^2 + 5x + 21 = (x-3)(\cdots\cdots\cdots).
$$
You still have to do some work to find the other factor, and you get
$$
x^3 -5x^2 + 4x + 6 = (x-3)(x^2 - 4x - 7).
$$
(Factoring $x^2-2x-2$ further will not concern us for now.)
Now suppose you divide this by another polynomial that is $0$ when $x=3$; for example $g(x) = x^2 - 4x + 3.$ Since $g(3)=0$, you conclude that $g(x) = (x-3)(\cdots\cdots\cdots)$, and when you find the other factor you've got $ g(x) = (x-3)(x-1).$
Now look at $\dfrac{f(x)}{g(x)},$ and see that $\dfrac {f(3)}{g(3)} = \dfrac 0 0$ is undefined.
Now simplify:
$$
\frac{f(x)}{g(x)} = \overbrace{ \frac{x^3 - 7x^2 + 5x + 21}{ x^2 - 4x + 3} = \frac{x^2-4x-7}{x-1}}^{\text{when } x\ne3} {} \underbrace{ {} = \frac{-10}{2} = -5}_{\text{when }x=3}.
$$
Are we saying that $\dfrac 0 0 = -5$? Are we saying $\dfrac{f(x)}{g(x)} = -5$ when $x=3$?
No, we're not, because one of the "equals" signs is true when $x\ne3$ and the other when $x=3$, so logically we cannot conclude that $\dfrac{f(3)}{g(3)} = -5.$
However, we can conclude that $\dfrac{f(x)}{g(x)}$ can be made as close to $-5$ as desired by making $x$ close enough, but not equal, to $3$. And that is expressed by saying $\dfrac{f(x)}{g(x)}$ approaches $-5$ as $x$ approaches $-3$, or by saying $\dfrac{f(x)}{g(x)} \to -5$ as $x\to-3$, or by saying $\lim\limits_{x\to3} \dfrac{f(x)}{g(x)} = -5.$
|
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|
How can we factorize $a{b^2} + {a^2}b + {a^2}c + a{c^2} + {b^2}c + b{c^2} + 2abc$? How can we factorize $a{b^2} + {a^2}b + {a^2}c + a{c^2} + {b^2}c + b{c^2} + 2abc$?
|
You just need to do a little rearrangement :
$$a{b^2} + {a^2}b + {a^2}c + a{c^2} + {b^2}c + b{c^2} + 2abc$$
$$a^2b+a^2c+b^2c+bc^2+ab^2+abc+abc+ac^2$$
$$a^2(b+c)+bc(b+c)+ab(b+c)+ac(b+c)$$
$$(b+c)(a^2+bc+ab+ac)$$
$$(b+c)(a(a+b)+c(a+b))$$
$$(b+c)(a+b)(a+c)$$
$$(a+b)(b+c)(c+a)$$
*And we are done !!!!!
|
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|
Summation of series with $r^{th}$ term $\frac{1}{\sqrt{a+rx} +\sqrt{a+(r-1)x}} $
Find the summation of series with $r^{th}$ term $t_r$ given by $\displaystyle \frac{1}{\sqrt{a+rx} +\sqrt{a+(r-1)x}} $
Working:
$$t_r=\frac{1}{\sqrt{a+rx} +\sqrt{a+(r-1)x}}$$
$$t_r=\frac{\sqrt{a+rx} -\sqrt{a+(r-1)x}}{a+rx-(a+(r-1)x)}$$
$$t_r=\frac{\sqrt{a+rx} -\sqrt{a+(r-1)x}}{x}$$
How do I calculate the summation $S_n$, from the $r^{th}$ term?
|
Continue from your working $$S_{n} = \sum^{n}_{r=1}t_{r} = \frac{1}{x}\sum^{n}_{r=1}\bigg[\sqrt{a+rx}-\sqrt{a+(r-1)x}\bigg]$$
Now Using Telescopis Sum (expanding summation.)
$$S_{n} = \frac{1}{x}\bigg[\left(\sqrt{a+x}-\sqrt{a}\right)+(\sqrt{a+2x}-\sqrt{a+x})+\cdots \cdots +(\sqrt{a+nx}-\sqrt{a+(n-1)x)}\bigg]$$
So we get $$S_{n} = \frac{1}{x}(\sqrt{a+nx}-\sqrt{a}) = \frac{(\sqrt{a+nx}-\sqrt{a})\cdot (\sqrt{a+nx}+\sqrt{a})}{\sqrt{a+nx}+\sqrt{a}} = \frac{n}{\sqrt{a}+\sqrt{a+nx}}$$
|
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|
maclaurin series of $f(x)=xe^{-x}$ I'm not sure how to do: $$f(x)=xe^{-x}$$I did this one so far:
$$ f(x) = x^2 \cos x = x^2 \left( 1 - \dfrac{x^2}{2!} + \dfrac{x^4}{4!} - \dfrac{x^6}{6!} +\dfrac{x^8}{8!} \dots \right) = x^2 - \dfrac{x^4}{2!} + \dfrac{x^6}{4!} - \dfrac{x^8}{6!} +\dfrac{x^{10}}{6!} \dots $$
i know that the maclaurin series for $f(x)=e^x= 1 + x+ \dfrac{x^2}{2!} + \dfrac{x^3}{3!} - \dfrac{x^4}{4!}\dots$
|
From
$$e^x= 1 + x+ \dfrac{x^2}{2!} + \dfrac{x^3}{3!} + \dfrac{x^4}{4!}+\cdots$$
one gets that
$$
f(x)=e^{-x}= 1 - x+ \dfrac{x^2}{2!} - \dfrac{x^3}{3!} + \dfrac{x^4}{4!}+\cdots
$$ giving
$$
f(x)=xe^{-x}= x - x^2+ \dfrac{x^3}{2!} - \dfrac{x^4}{3!} + \dfrac{x^5}{4!}+\cdots
$$ or
$$
f(x)=xe^{-x}=\sum_{n=0}^\infty(-1)^n \frac{x^{n+1}}{n!}.
$$
|
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|
Calculation of some limit I'm having problems with limits like that $ \lim_{x β 0} \left(\frac{1- \left(\cos x\right)^{\sin x}} {x^3}\right)$. I need to calculate it in pretty rigorous way.
I can use Taylor and write $\cos x^{\sin x}$ as $\left(1+o\left(x^2\right)\right)^{x + o\left(x^2\right)}$ but I have no idea what to do with such expression.
|
You can compose polynomial expansions. We'll use expansions at order $3$.
By definition, $\;\cos x^{\sin x}=\mathrm e^{\sin x\ln(\cos x)}$.
Now $\;\cos x=1-\dfrac{x^2}2+o(x^3)$, and $\;\ln(1-u)=-u-\dfrac{u^2}2-\dfrac{u^3}3-\dotsm$, so it will be enough to use the expansion of $\;\ln(1-u)$ at order $1$:
$$\ln(\cos x)=-\frac{x^2}2+o(x^3),$$
$$\text{so}\quad\sin x\,\ln(\cos x)=\Bigl(x-\frac{x^3}6+o(x^3)\Bigr)\Bigl(-\frac{x^2}2+o(x^3)\Bigr)=-\frac{x^3}2+o(x^3).$$
Now expand $\;\mathrm e^u$ at order $1$:
$$1-\mathrm e^{\sin x\ln(\cos x)}=1-\Bigl(1-\frac{x^3}2+o(x^3)\Bigr)=\frac{x^3}2+o(x^3)\sim_0\frac{x^3}2,$$
and finally
$$\frac{1-\cos x^{\sin x}}{x^3}\sim_0\frac{\dfrac{x^3}2}{x^3}=\frac12.$$
|
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|
Prove $2^{1/3} + 2^{2/3}$ is irrational What's the nice 'trick' to showing that the following expression is irrational?
$2^{1/3} + 2^{2/3}$
|
Let $x=\sqrt[3]{2}+\sqrt[3]{4}$ then raising to the $3^{rd}$ power and expanding the binomial on the right:
$$x^3 = (\sqrt[3]{2})^3 + 3 \cdot \sqrt[3]{2} \cdot \sqrt[3]{4} \cdot (\sqrt[3]{2}+\sqrt[3]{4}) + (\sqrt[3]{4})^3 = 6 x + 6$$
By the rational root theorem, the equation $x^3-6x-6=0$ can only have divisors of $6$ as rational roots, but it's easily verified that none of the divisors is in fact a root. Therefore $x$ is irrational.
|
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|
How many three-digit numbers can be generated from 1, 2, 3, 4, 5, 6, 7, 8, 9, such that the digits are in ascending order? Solution for the same is given below
Numbers starting with 12 β 7 numbers
Numbers starting with 13 β 6 numbers;
14 β 5, 15 β 4, 16 β 3, 17 β 2, 18 β 1.
Thus total number of
numbers starting from 1 is given by the sum of 1 to 7 = 28.
Number of numbers starting from 2- would be given by the sum of 1 to 6 = 21
Number of numbers starting from 3- sum of 1 to 5 = 15
Number of numbers starting from 4 β sum of 1 to 4 = 10
Number of numbers starting from 5 β sum of 1 to 3 = 6
Number of numbers starting from 6 = 1 + 2 = 3
Number of numbers starting from 7 = 1
Thus a total of: 28 + 21 + 15 + 10 + 6 + 3 + 1 = 84 such numbers.
suggest some trick to approach without thinking about this gibberish given above?
|
Here is a trick:
*
*An increasing arrangement must contain unique digits
*There are $\binom{9}{3}$ combinations of $3$ out of $9$ unique digits
*Each combination has exactly $1$ increasing arrangement
*Hence the number of increasing arrangements is $\binom{9}{3}$
|
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|
Prove or disprove $(a^2+b^2)(b^2+c^2)(c^2+a^2)\leq(a+b)(b+c)(c+a)$ Prove or disprove:
If $a, b, c \in (0, \infty)$ such that $a^2+b^2+c^2=3$ then
$$(a^2+b^2)(b^2+c^2)(c^2+a^2)\leq(a+b)(b+c)(c+a).$$
All my attempts to prove inequality have been unsuccessful. Maybe someone has an idea. Thank you very much!
|
It's wrong! Try $c\rightarrow0^+$ and $a=b\rightarrow\sqrt{1.5}$
|
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|
If $ \alpha_i, i=0,1,2...n-1 $ be the nth roots of unity, the $\sum_{i=0}^{n-1} \frac{\alpha_i}{3- \alpha_i}$ is equal to? If $ \alpha_i, i=0,1,2...n-1 $ be the nth roots of unity, the $\sum_{i=0}^{n-1} \frac{\alpha_i}{3- \alpha_i}$ is equal to?
A) $ \frac{n}{3^n-1} $
B) $ \frac{n-1}{3^n-1} $
C) $ \frac{n+1}{3^n-1} $
D) $ \frac{n+2}{3^n-1} $
Attempt: I know that $(3- \alpha_0)(3- \alpha_1)....(3-\alpha_{n-1})= (3^n-1)/2 $
But I have no clue about the numerator. Adding 3 and subracting 3 from the numberator would make the fraction simpler, but I would still have to sum up $ 1/(3- \alpha_i) $
|
(For future reference.) Introducing
$$f(z) = \frac{z}{3-z} \frac{nz^{n-1}}{z^n-1}$$
we get
$$S_n = \sum_{q=0}^{n-1}
\frac{\exp(2\pi i q/n)}{3-\exp(2\pi i q/n)}
= \sum_{q=0}^{n-1} \mathrm{Res}_{z=\exp(2\pi i q/n)} f(z).$$
Residues sum to zero so we have
$$S_n + \mathrm{Res}_{z=3} f(z) +
\mathrm{Res}_{z=\infty} f(z) = 0$$
Observe that
$$\mathrm{Res}_{z=3} f(z) =
- \frac{n3^n}{3^n-1}$$
and
$$\mathrm{Res}_{z=\infty} f(z) =
-\mathrm{Res}_{z=0} \frac{1}{z^2} f(1/z)
\\ = -\mathrm{Res}_{z=0}
\frac{1}{z^2} \frac{1/z}{3-1/z} \frac{n/z^{n-1}}{1/z^n-1}
\\ = -\mathrm{Res}_{z=0}
\frac{1}{z^2} \frac{1}{3z-1} \frac{nz}{1-z^n}
\\ = -\mathrm{Res}_{z=0}
\frac{1}{z} \frac{1}{3z-1} \frac{n}{1-z^n}
= n.$$
Hence
$$S_n - \frac{n3^n}{3^n-1} + n = 0$$
or
$$S_n = \frac{n3^n-n(3^n-1)}{3^n-1}
= \frac{n}{3^n-1}.$$
|
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|
Can someone please explain why this is the first move when solving this integral. The integral which I am solving is $$\int_0^c\sqrt{c^2-x^2}~\mathrm dx$$ and the first thing which they suggest doing is setting $x=(\sqrt{a})/(\sqrt{b}) \sin u$. I understand everything after this but I do not comprehend where and why this step came to be.
|
If you must do this by a method other than recognizing this as the area of one quarter of a circle, then the expression $\sqrt{c^2-x^2}$ suggests using the substitution
\begin{align}
x & = c\sin\theta \\
dx & = c\cos\theta\,d\theta \\[10pt]
\sqrt{c^2 -x^2} & = \sqrt{c^2 - c^2\sin^2\theta} = c\sqrt{1-\sin^2\theta} \\[5pt]
& = c\sqrt{\cos^2\theta} = c \cos\theta.
\end{align}
As $x$ goes from $0$ to $c$, then $\sin\theta$ goes from $0$ to $1$, so $\theta$ goes from $0$ to $\pi/2$. So we get
$$
\int_0^c \sqrt{c^2-x^2}\,dx = \int_0^{\pi/2} c^2\cos^2\theta\,d\theta.
$$
One way of evaluating the integral is by using the trigonometric identity
$$
\cos^2\theta = \frac 1 2 + \frac 1 2 \cos(2\theta).
$$
There is also a way to do it without finding any antiderivatives: Observe that by symmetry, since $\cos\theta = \sin(\frac\pi2 - \theta)$, we have
$$
\int_0^{\pi/2} \cos^2\theta\,d\theta = \int_0^{\pi/2} \sin^2\theta\,d\theta
$$
and then find the sum of those two integrals by using the identity $\cos^2\theta+\sin^2\theta=1$.
|
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|
Prove that $\int J_5(x) dx= -J_4(x) - \frac {4}{x} J_3(x) - \frac {8}{x^2}J_2(x) +c$
Prove that $$\int J_5(x) dx= -J_4(x) - \frac {4}{x} J_3(x) - \frac
{8}{x^2}J_2(x) +c$$ where $J_n (x)$ is the Bessel function of first kind
and order $n $.
My attempt:
I know the following recurrence relations:
$$\frac {d}{dx}\left[x^nJ_n (x)\right]=x^nJ_{n-1} (x)$$
$$\frac {d}{dx}\left[x^{-n}J_n (x)\right]=-x^{-n}J_{n+1} (x)$$
$$2J'_n(x)=J_{n-1} (x)-J_{n+1} (x)$$
$$2\frac {n}{x}J_n(x)=J_{n-1} (x)+J_{n+1} (x)$$
But I cannot figure out how to get the above expression. Please help.
|
Following Alex R's comment, I am proceeding as follows:
$$\int J_5(x) dx$$
$$=\int \frac {x^4}{x^4} J_5(x) dx$$
$$=\int x^4 \cdot x^{-4} J_5(x) dx$$
Integrating by parts, we get
$$x^4 \cdot \int x^{-4} J_5(x) dx - \int \left\{\frac {d}{dx}(x^4) \cdot \int x^{-4} J_5(x) dx \right\} dx$$
$$=x^4 \cdot (-1) \cdot x^{-4} J_4(x) - \int \left\{4x^3 \cdot (-1) \cdot x^{-4} J_4(x)\right\} dx$$
$$=-J_4(x) +4 \int \left\{x^3 \cdot x^{-4} J_4(x)\right\} dx$$
$$=-J_4(x) +4 \int \left\{x^2 \cdot x^{-3} J_4(x)\right\} dx$$
Similarly, again integrating by parts, we get
$$=-J_4(x) +4 x^2 \cdot \int x^{-3} J_4(x) dx - 4\cdot \int \left\{\frac {d}{dx}(x^2) \cdot \int x^{-3} J_4(x) dx\right\} dx$$
$$=-J_4(x) +4 x^2 \cdot (-1) \cdot x^{-3} J_3(x) - 4\cdot \int \left\{2x \cdot (-1) \cdot x^{-3} J_3(x)\right\} dx$$
$$=-J_4(x) - \frac {4}{x} J_3(x) + 8\cdot \int \left\{x^{-2} \cdot J_3(x)\right\} dx$$
$$=-J_4(x) - \frac {4}{x} J_3(x) - 8\cdot \int \left\{x^{-2} \cdot J_2(x)\right\}$$
$$=-J_4(x) - \frac {4}{x} J_3(x) - \frac {8}{x^2} J_2(x) +c$$
Hence the relation is proved.
Thanks Alex R for the great hint.
|
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Diophantine equation of prime numbers Determine all pairs of prime numbers $(p,q)$ that satisfy the equation
$p^3-q^3=pq^3-1$.
I can easily understand that
$p>q$, $q^3=p^2-p+1$, $q^3\equiv 1 \pmod{p}$, $p^3\equiv -1 \pmod{q}$.
But I don't know how to go on. (I know that p=19 and q=7 works)
|
$p^2 - p + 1 = q^3$
$p(p-1) = (q-1)(q^2+q+1)$.
As you noticed, $p>q$, so $q-1$ is coprime with $p$. So $q-1|p-1$, or instead we can write
$k(q-1)=p-1$ and $kp=q^2 + q + 1$ for some integer $k>1$.
Get $p$ from the first equation and plug it to the second one:
$k(kq - k + 1) = q^2 + q + 1$
$q^2 + (1-k^2)q +1-k+k^2=0$
So $q$ can only be an integer if discriminant of above equation is a square:
$\Delta = k^4 - 6k^2 + 4k - 3$
But $(k^2 - 3)^2 < \Delta < (k^2 - 2)^2$ for $k>3$.
So we only need to consider $k=2, 3$. For $k=2$, $\Delta$ is not a square. For $k=3$, $\Delta=36$, the quadratic equation has solutions $q=1, 7$. $q=7$ gives $p=19$.
So the only solution for $p, q$ primes is $(19, 7)$.
|
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|
What is wrong with the "proof" for $\ln(2) =\frac{1}{2}\ln(2)$? I have got a question which is as follows:
Is $\ln(2)=\frac{1}{2}\ln(2)$??
The following argument seems suggesting that the answer is yes:
We have the series $1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\cdots$
which has a mathematically determined value $\ln(2)=0.693$.
Now, let's do some rearrangement:
$$1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\frac{1}{5}-\frac{1}{6}+\frac{1}{7}-\frac{1}{8}+\frac{1}{9}-\frac{1}{10}+\frac{1}{11}-\frac{1}{12}......$$
$$
(1-\frac{1}{2})-\frac{1}{4}+(\frac{1}{3}-\frac{1}{6})-\frac{1}{8}+(\frac{1}{5}-\frac{1}{10})-\frac{1}{12}.......$$
$$\frac{1}{2}-\frac{1}{4}+\frac{1}{6}-\frac{1}{8}+\frac{1}{10}-\frac{1}{12}......$$
$$\frac{1}{2}(1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\frac{1}{5}-\frac{1}{6}+\frac{1}{7}-\frac{1}{8}+\frac{1}{9}-\frac{1}{10}+\frac{1}{11}-\frac{1}{12}......)$$
$$\frac{1}{2}\ln(2).$$
I know that mathematics can't be wrong, and I have done something wrong. But here is my question: where does the argument above go wrong?
|
The series $\sum\limits_{k=1}^n\frac{(-1)^{k+1}}k$ is convergent but not absolutely convergent, i.e. the sum itself has a limit, but $\sum_{k=1}^\infty\left\lvert\frac{(-1)^{k+1}}{k}\right\rvert=+\infty$. Therefore, by Riemann-Dini theorem, for any extended real numbers $\alpha\le \beta\in\Bbb R\cup\{-\infty,\infty\}$ there is a bijective function $f:\Bbb N^+\to\Bbb N^+$ such that $$\alpha=\liminf_{n\to\infty}\sum_{k=1}^n \frac{(-1)^{f(k)}}{f(k)}\le\limsup_{n\to\infty}\sum_{k=1}^n \frac{(-1)^{f(k)}}{f(k)}=\beta$$
You happen to have described, more or less explicitly, a bijective function $f$ which goes well with $\alpha=\beta=\ln\sqrt2$, which happens to be the example(s) on Wikipedia too.
|
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General solution to $(\sqrt{3}-1)\cos x+(\sqrt{3}+1)\sin x=2$ $(\sqrt{3}-1)\cos x+(\sqrt{3}+1)\sin x=2$ is said to have a general solution of $x=2n\pi\pm\frac{\pi}{4}+\frac{\pi}{12}$.
My Approach:
Considering the equation as
$$
a\cos x+b\sin x=\sqrt{a^2+b^2}\Big(\frac{a}{\sqrt{a^2+b^2}}\cos x+\frac{b}{\sqrt{a^2+b^2}}\sin x\Big)=\sqrt{a^2+b^2}\big(\sin y.\cos x+\cos y.\sin x\big)=\sqrt{a^2+b^2}.\sin(y+x)=2
$$
$\frac{a}{\sqrt{a^2+b^2}}=\sin y$ and $\frac{b}{\sqrt{a^2+b^2}}=\cos y$.
$$
{\sqrt{a^2+b^2}}=\sqrt{8}=2\sqrt{2}\\\tan y=a/b=\frac{\sqrt{3}-1}{\sqrt{3}+1}=\frac{\frac{\sqrt{3}}{2}.\frac{1}{\sqrt{2}}-\frac{1}{2}.\frac{1}{\sqrt{2}}}{\frac{\sqrt{3}}{2}.\frac{1}{\sqrt{2}}+\frac{1}{2}.\frac{1}{\sqrt{2}}}=\frac{\sin(\pi/3-\pi/4)}{\sin(\pi/3+\pi/4)}=\frac{\sin(\pi/3-\pi/4)}{\cos(\pi/3-\pi/4)}=\tan(\pi/3-\pi/4)\implies y=\pi/3-\pi/4=\pi/12
$$
Substituting for $y$,
$$
2\sqrt{2}.\sin(\frac{\pi}{12}+x)=2\implies \sin(\frac{\pi}{12}+x)=\frac{1}{\sqrt{2}}=\sin{\frac{\pi}{4}}\\\implies \frac{\pi}{12}+x=n\pi+(-1)^n\frac{\pi}{4}\implies x=n\pi+(-1)^n\frac{\pi}{4}-\frac{\pi}{12}
$$
What's going wrong with the approach ?
|
Hint:
For $(\tan A-\tan45^\circ)\cos x+(\tan A+\tan45^\circ)\sin x=\sec A$
$\cos(x+A+45^\circ)=\cos45^\circ$
|
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|
Find all integer solutions $(x,y)$ such that $2x^2 + y^2 = 3x^2y$
Find all integer solutions $(x,y)$ such that $2x^2 + y^2 = 3x^2y$.
We can rearrange the given equation to $$y^2 = x^2(3y-2)\tag1$$ Thus $3y-2$ must be a perfect square and so $3y-2 = k^2$.
How can we continue?
|
We note that $x^2\mid 3x^2y$, so $x^2\mid 2x^2+y^2$, then $x^2\mid y^2$, which implies that $x\mid y$. Let's call $y=kx$, if we replace this into your equation $(1)$ we get $k^2x^2=x^2(3kx-2)$, then $k^2=3kx-2$, which lead us to the quadratic equation $k^2-3xk+2=0$. Using the formula for quadratic equations we deduce that $$k=\frac{3x\pm \sqrt{9x^2-8}}{2}.$$
So in order to have $k\in\mathbb{Z}$ we must have $9x^2-8=z^2$, for some $z\in \mathbb{Z}$, so $(3x+z)(3x-z)=8$, therefore $3x+z$ and $3x-z$ are both divisors of $8$ with the same parity. Hence $3x+z=\pm 4$ and $3x-z=\pm 2$, or $3x+z=\pm 2$ and $3x-z=\pm 4$.
For example if $3x+z=4$ and $3x-z=2$, then $x=1$ and $z=1$, which gives us $k=1$ or $k=2$, and thus $y=1$ or $y=2$. The other cases are similar.
|
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|
Distributing $8$ different articles among $7$ boys
Problem Statement:-
Find the number of ways in which $8$ different articles can be distributed among $7$ boys, if each boy is to receive at least one article.
Attempt at a solution:-
First, start by numbering the boys from $1$ to $7$, which can be done in $7!$ ways.
After the boys have been numbered then we resume our task of distributing the articles to the boys. As, each boy is supposed to at least get a single article.
Hence,
$\text{Ways of giving $1$ article to the $1^{\text{st}}$ boy= $8\choose1$}\\
\text{Ways of giving $1$ article to the $2^{\text{nd}}$ boy= $7\choose1$}\\
\text{Ways of giving $1$ article to the $3^{\text{rd}}$ boy= $6\choose1$}\\
\vdots\\
\text{Ways of giving $1$ article to the $7^{\text{th}}$ (and the last) boy= $2\choose1$}$
Now remains the last article, which can go to any one boy of the $7$ boys in ${{7}\choose{1}}=7$ ways
So, the number of ways of distributing $8$ articles among $7$ boys$=$
$$7!\times \left( \binom{8}{1} \times \binom{7}{1} \times \binom{6}{1} \times \binom{5}{1} \times \binom{4}{1} \times \binom{3}{1} \times \binom{2}{1}\right)\times \binom{7}{1}$$
But, the arrangement of the two things that the $1^{\text{st}}$ boy gets is of no relevance so we gotta remove the consideration of the two objects being permuted among themselves.
Hence, the total the number of ways of distributing $8$ articles among $7$ boys$=$
$$\frac{7!\times \left( \binom{8}{1} \times \binom{7}{1} \times \binom{6}{1} \times \binom{5}{1} \times \binom{4}{1} \times \binom{3}{1} \times \binom{2}{1}\right)\times \binom{7}{1}}{2!}$$
The textbook gives the answer to the question as
$$7\times \left( \binom{8}{2} \times \binom{6}{1} \times \binom{5}{1} \times \binom{4}{1} \times \binom{3}{1} \times \binom{2}{1} \times \binom{1}{1}\right)$$
Where am I going wrong
|
You have permuted both the boys and the articles, which is incorrect.
I'd prefer to work it out as
[ Choose "lucky" boy ] $\times$ [ Permute articles ]
$= \dbinom71 \times \dfrac{8!}{2!1!1!1!1!1!1!}$
which, of course, could be abbreviated as $ 7\times \dfrac{8!}{2!}$
PS:
On closer scrutiny, I find that the book answer you give is also incorrect !
|
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|
Prove that $\vec{AI}.\vec{BH}=0$ First I'm sorry I don't know how to draw the triangle on LaTeX.
We consider $\triangle ABC$ isoscele in $A$. We denote $O$ the midpoint of $[BC]$ and $H$ the orthogonal projection of $O$ on $(AC)$. We also note $I$ the midpoint of $[OH]$.
Is it a good idea to use Apollonius theorem in $\triangle AOH$ and $\triangle HBC$ ?
In that case we will have :
$AO^2+AH^2=\frac{1}{2}OH^2+2AI^2$ and $HB^2+HC^2=\frac{1}{2}BC^2+2OH^2$ and maybe find an expression without $OH^2$ will help.
Thanks in advance !
|
Since the title invites a vector proof, note that by construction:
$$
\begin{align}
& \overrightarrow{BO}=\overrightarrow{OC} && \text{since } O \text{ is the midpoint of } BC \tag{1} \\
& \overrightarrow{AO} \cdot \overrightarrow{BC}=0 && \text{since } \triangle ABC \text{ is isosceles, so } AO \perp BC\tag{2} \\
& \overrightarrow{OH} \cdot \overrightarrow{AC}=0 && \text{since } OH \perp AC \tag{3} \\
& \overrightarrow{AI} = \frac{1}{2}(\overrightarrow{AO}+\overrightarrow{AH}) && \text{since } I \text{ is the midpoint of } OH \tag{4}
\end{align}
$$
Then:
$$
\require{cancel}
\begin{align}
2 \overrightarrow{AI} \cdot \overrightarrow{BH} &= (\overrightarrow{AO}+\overrightarrow{AH})\cdot(\overrightarrow{BO}+\overrightarrow{OH}) && \text{by } (4) \\
&= \cancel{\overrightarrow{AO}\cdot\overrightarrow{BO}} + \overrightarrow{AO}\cdot\overrightarrow{OH}+\overrightarrow{AH}\cdot\overrightarrow{BO}+\bcancel{\overrightarrow{AH}\cdot\overrightarrow{OH}} && \text{by } (2), (3) \\
&= (\cancel{\overrightarrow{AH}}+\overrightarrow{HO})\cdot\overrightarrow{OH} + (\bcancel{\overrightarrow{AO}}+\overrightarrow{OH})\cdot\overrightarrow{OC} && \text{by } (2),(3),(1) \\
&= -\overrightarrow{OH}\cdot\overrightarrow{OH} + \overrightarrow{OH}\cdot(\overrightarrow{OH}+\cancel{\overrightarrow{HC}}) && \text{by } (3)\\
&= 0
\end{align}
$$
|
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|
Prove that $\sum\limits_{cyc}(4a^6+5a^5b)\geq\frac{(a+b+c)^6}{27}$ Let $a$, $b$ and $c$ be real numbers. Prove that:
$$4(a^6+b^6+c^6)+5(a^5b+b^5c+c^5a)\geq\frac{(a+b+c)^6}{27}$$
I tried SOS, uvw and more, but without success.
|
The following proof is true if they are non-negative reals.
$4(x^3+y^3+z^3)+15xyz\geq (x+y+z)^3$- this directly follows from Schur. So $$4(a^6+b^6+b^6)+5(a^5b+b^5c+c^5a)\geq 4(a^6+b^6+c^6)+15a^2b^2c^2\geq (a^2+b^2+c^2)^3\geq\dfrac{(a+b+c)^6}{27}$$
|
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|
Is something wrong with this precalculus question? If $x^{2}-4x+6=0$, then what can be the value of $1-\frac{4}{3x}+\frac{2}{x^{2}}$?
My answer is $1-\frac{4}{3x}+\frac{2}{x^{2}}=\frac{3x^{2}-4x+6}{3x^{2}}=\frac{3x^{2}-x^{2}}{3x^{2}}=\frac{2}{3}$ for $x\neq 0$. But according to the book answer is 2. What is the point i miss?
|
$3x^2-4x+6=2x^2+(x^2-4x+6)=2x^2$
You're right. Note that $x$ is complex and not real, but apart from this, the value of the other expression is $2/3$.
|
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|
How to prove that trace$(ABA^{-1}B^{-1})$=$3$ If $A,B$ are two $3 \times 3$ square matrices and trace(A) is defined as the sum of all diagonal elements. trace$(ABA^{-1}B^{-1})$=$3$
I could easily verify the above for the identity matrix.But I couldn't generalise it.
Please help me in this regard.thanks.
|
Counterexample:
$$\begin{align}
A&=\begin{pmatrix}1&1&0\\0&1&0\\0&0&1\end{pmatrix}
&B&=\begin{pmatrix}1&0&0\\1&1&0\\0&0&1\end{pmatrix}
\\A^{-1}&=\begin{pmatrix}1&-1&0\\0&1&0\\0&0&1\end{pmatrix}
&B^{-1}&=\begin{pmatrix}1&0&0\\-1&1&0\\0&0&1\end{pmatrix}
\end{align}$$
results in
$$ABA^{-1}B^{-1}=\begin{pmatrix}3&-1&0\\1&0&0\\0&0&1\end{pmatrix}$$
with trace $4$.
|
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|
Common proof for $(1+x)(1+x^2)(1+x^4)...(1+x^{2^n})=\dfrac{1-x^{2^{n+1}}}{1-x} $ I'm asking for an alternative (more common?) proof of the following equality, more specifically an alternative proof for the inductive step:
$$(1+x)(1+x^2)(1+x^4)...(1+x^{2^n})=\dfrac{1-x^{2^{n+1}}}{1-x} (x\neq 1)$$
This is how I proved it:
Basecase: substitute $1$ for $n$, everything works out.
Inductive step: assume that $$\prod _{i=1}^{n}(1+x^i)=\dfrac{1-x^{2^{n+1}}}{1-x}$$
then $$\prod _{i=1}^{n+1}(1+x^i)=\dfrac{1-x^{2^{n+1}}}{1-x} (1+x^{n+1})$$
Let $a$, $b$ and $c$ be positive real numbers, then $\dfrac {c}{a}=b\Leftrightarrow ab=c$, thus
$$\dfrac {\left( \dfrac{1-x^{2^{n+2}}}{1-x}\right)}{\left( \dfrac{1-x^{2^{n+1}}}{1-x}\right)}=1-x^{2^{n+1}} \Leftrightarrow \dfrac{1-x^{2^{n+1}}}{1-x} (1+x^{n+1}) = \dfrac{1-x^{2^{n+2}}}{1-x}$$
$$\dfrac {\left( \dfrac{1-x^{2^{n+2}}}{1-x}\right)}{\left( \dfrac{1-x^{2^{n+1}}}{1-x}\right)}=\dfrac {1-x^{2^{n+2}}}{1-x^{2^{n+1}}}$$
Applying polynomial division, we see that indeed $\dfrac {1-x^{2^{n+2}}}{1-x^{2^{n+1}}} = 1+x^{2^{n+1}}$.
Thus $\dfrac{1-x^{2^{n+1}}}{1-x} (1+x^{n+1}) = \dfrac{1-x^{2^{n+2}}}{1-x}$.
However, the exercise was in a chapter on binomial coefficients and pascal's triangle, furthermore we didn't mention polynomial division in class. Which makes me think that there was another solution that I was "supposed" to see. How was I supposed to prove it?
|
A Proof with a Bit of Field Theory
Over a field of characteristic $2$, $1+x^{2^r}=(1+x)^{2^r}$ and $1-x=1+x$, so the LHS is $$
\begin{align}
\prod_{i=0}^n\,\left(1+x^{2^i}\right)
&=\prod_{i=0}^n\,(1+x)^{2^i}=(1+x)^{\sum\limits_{i=0}^n\,2^i}=(1+x)^{2^{n+1}-1}
\\
&=\frac{(1+x)^{2^{n+1}}}{1+x}=\frac{1+x^{2^{n+1}}}{1+x}=\frac{1-x^{2^{n+1}}}{1-x}\,.
\end{align}$$
Over a field of characteristic not equal to $2$, we may assume to be algebraically closed, hence every nonconstant polynomial has roots in the field. The polynomial $1-x^{2^{n+1}}$ has the derivative $-2^{n+1}\,x^{2^{n+1}-1}$, which is nonzero at any root of $1-x^{2^{n+1}}$. Thus, $1-x^{2^{n+1}}$ has only simple roots, one of which is $1$. Therefore, $\frac{1-x^{2^{n+1}}}{1-x}$ is the product of $x-\omega$, where $\omega\neq 1$ are other roots of $1-x^{2^{n+1}}$. However, if $r$ is the smallest positive integer such that $\omega^{2^r}=1$, then $\omega^{2^{r-1}}=-1$, or $\omega$ is a root of $1+x^{2^{r-1}}$. Furthermore, all the roots of $1+x^{2^{r-1}}$, for $r=1,2,\ldots,n+1$, are clearly roots of $1-x^{2^{n+1}}$ and they are not equal to $1$. Finally, if $r$ and $s$ are distinct positive integers, then we can easily show that $\gcd\left(1+x^{2^{r-1}},1+x^{2^{s-1}}\right)=1$ (well, the roots of $1+x^{2^{r-1}}$ are all elements $\omega$ in the field such that $\omega^k=1$ if and only if $2^{r}$ divides $k$). Hence, $\prod\limits_{r=1}^{n+1}\,\left(1+x^{2^{r-1}}\right)=\prod\limits_{i=0}^n\,\left(1+x^{2^i}\right)$ is a factor of $\frac{1-x^{2^{n+1}}}{1-x}$. As both of these polynomials have the same degree and are monic, we conclude that $$\prod\limits_{i=0}^n\,\left(1+x^{2^i}\right)=\frac{1-x^{2^{n+1}}}{1-x}\,.$$
|
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|
prove $(a^3+1)(b^3+1)(c^3+1)\ge 8$ let $a,b,c\ge 0$ and such $a+b+c=3$ show that
$$(a^3+1)(b^3+1)(c^3+1)\ge 8$$
My research:It seem use Holder inequality,so
$$(a^3+1)(b^3+1)(c^3+1)\ge (abc+1)^3$$
Use AM-GM
$$abc\le\dfrac{(a+b+c)^3}{27}=1$$
what? then I think this method is wrong
|
Let $f(x,y,z) = (x^3 +1)(y^3 +1)(z^3 +1)$ subject to the constraint $g(x,y,z) = x + y + z - 3 =0$ and $ x,y,z \ge 0$.
Then using lagrange multipliers we have
*
*$f_x = 3x^2(y^3 +1)(z^3 +1)= g_x= \lambda$
*$f_y = 3y^2(x^3 +1)(z^3 +1)= g_y = \lambda$
*$f_z = 3z^2(x^3 +1)(y^3 +1)=g_z = \lambda$
*$g(x,y,z) = x + y + z - 3 =0$
From the first three equations we see $x^2(y^3 +1) = y^2(x^3 +1) = z^2(y^3 +1)$. So then $x =z$, since $x, z \ge 0$. Then clearly $x=y=z$, so plugging into our formula for $g$, we see $x=y=z=1$, so $(1,1,1)$ is a critical point.
Now, $f(1,1,1) = 8$. To see this is a minimum, we see $(0,0,3)$ lies on our curve, and $f(0,0,3) = 28 \gt f(1,1,1) = 8.$
We conclude that $(a^3 +1)(b^3 +1)(c^3 +1) \ge 8$.
|
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|
What is the remainder left after dividing $1! + 2! + 3! + ... + 100!$ by $5$? I have this question as a homework.
What is the remainder left after dividing $1! + 2! + 3! + \cdots + 100!$ by $5$?
I tried this: I noticed that every $n! \equiv 0 \pmod{5}$ for every $n\geq 5$.
For $n < 5$:
$$\begin{align*}
1! &\equiv 1 \pmod{5} \\
2! &\equiv 2 \pmod{5} \\
3! &\equiv 1 \pmod{5} \\
4! &\equiv 4 \pmod{5} \\
\end{align*}$$
So $1! + 2! + \cdots + 100! \equiv 8 \equiv 3 \pmod{5}$.
Therefore the remainder is $3$.
Am I thinking properly?
Thanks a lot.
|
To complement the other answer in less theory-heavy terms, you can view it as:
$$1! + 2! + 3! + 4! + 5! (1 + 6 + [6 \times 7] + \dots + [6 \times \dots \times 100])$$
The final term is clearly divisible by $5$; so we just need to look at $1! + 2! + 3! + 4!$ and its remainder, as you did.
|
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|
Choose $a, b$ so that $\cos(x) - \frac{1+ax^2}{1+bx^2}$ would be as infinitely small as possible on ${x \to 0}$ using Taylor polynomial $$\cos(x) - \frac{1+ax^2}{1+bx^2} \text{ on } x \to 0$$
If $\displaystyle \cos(x) = 1 - \frac{x^2}{2} + \frac{x^4}{4!} - \cdots $
Then we should choose $a, b$ in a such way that it's Taylor series is close to this.
However, I'm not sure how to approach this. I tried to take several derivates of second term to see its value on $x_0 = 0$, but it becomes complicated and I don't see general formula for $n$-th derivative at point zero to find $a$ and $b$.
|
Note you just need to compute the Taylor expansion of
\begin{align}
g(x)&=\cos x+bx^2\cos x-1-ax^2 \\[6px]
&=1-\frac{x^2}{2!}+\frac{x^4}{4!}-\frac{x^6}{6!}
+bx^2-\frac{bx^4}{2!}+\frac{bx^6}{4!}-1-ax^2+o(x^6)\\[6px]
&=\left(-\frac{1}{2}+b-a\right)x^2+
\left(\frac{1}{24}-\frac{b}{2}\right)x^4+
\left(-\frac{1}{6!}+\frac{b}{4!}\right)x^6+o(x^6)
\end{align}
Therefore
$$
\begin{cases}
a-b=-\dfrac{1}{2} \\[6px]
b=\dfrac{1}{12}
\end{cases}
$$
|
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|
If $abc+1=0$, verify that' If $abc+1=0$, prove that:
$\frac {1}{1-a-b^{-1}}+\frac {1}{1-b-c^{-1}} +\frac {1}{1-c-a^{-1}}=1$.
My Attempt:
$abc+1=0$
$abc=-1$.
Now,
$$L.H.S=\frac {1}{1-a-b^{-1}}+\frac {1}{1-b-c^{-1}}+\frac {1}{1-c-a^{-1}}$$
$$=\frac{b}{b-ab-1} + \frac {1}{1-b-c^{-1}} + \frac {c^{-1}}{c^{-1}-1-(ca)^{-1}}$$
$$=\frac{b}{b+c^{-1} -1} +\frac{1}{1-b-c^{-1}} +\frac{c^{-1}}{c^{-1}-1+b}$$
What should I do next?
|
You are on the correct path. Your steps are all correct. Proceeding from there, just note that
$$\frac{b}{b+c^{-1} -1} +\frac{1}{1-b-c^{-1}} +\frac{c^{-1}}{c^{-1}-1+b}$$
$$=\frac{b+c^{-1}}{b+c^{-1} -1} -\frac{1}{b+c^{-1}-1}$$
$$=\frac{b+c^{-1}-1}{b+c^{-1} -1}$$
$$=1$$
Hope this helps you.
|
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|
Short technique of tackling or another method $\int_{0}^{\infty}{2x\over (x^4+2x^2+1)+\sqrt{x^4+2x^2+1}}dx=\ln{2}$ Prove that,
$$I=\int_{0}^{\infty}{2x\over (x^4+2x^2+1)+\sqrt{x^4+2x^2+1}}dx=\ln{2}$$
I try:
$x^4+2x^2+1=(x^2+1)^2$
$$\int_{0}^{\infty}{2x\over (x^2+1)(x^2+2)}dx$$
Let $u=x^2+1$, $du=2xdx$
$$\int_{1}^{\infty}{1\over u(u+1)}du$$
$$\int_{1}^{\infty}\left({1\over u}-{1\over u+1}\right)du$$
$$I=\ln{1}-\ln{{1\over2}}=\ln{2}$$
Is there another short way of integrating this integral?
|
Set $x=\tan\theta$, we have
$$I=\int_{0}^{\infty}{2x\over (x^2+1)(x^2+2)}dx=\int_{0}^{\frac{\pi}{2}}\frac{2\tan\theta(1+\tan^2\theta)}{(1+\tan^2\theta)(2+\tan^2\theta)}d\theta$$
thus
$$I=\int_{0}^{\frac{\pi}{2}}\frac{\sin2\theta}{1+\cos^2\theta}d\theta=-\ln(1+\cos^2\theta)\Big{|}_{0}^{\frac{\pi}{2}}=\ln(2)$$
|
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How to compute the dimension of $\Bbb C[x,y]/I$ on $\Bbb C$? The problem is compute the dimension of $\Bbb C[x,y]/I$ over $\Bbb C$ as vector space where $I=\langle(x+2)^2,(x+2)(y+1),(y+1)^3\rangle$. $\Bbb C[x,y]$ is the polynomial ring over $\Bbb C$.
I have tried in this way.
For $f(x,y)$ in $\Bbb C[x,y]$ it has the form $f(x)=(a_{0}+a_{1}(x+2)+β¦a_{i}(x+2)^i+β¦a_{n}(x+2)^n)(b_{0}+b_{1}(y+1)+β¦b_{j}(y+1)^j+β¦b_{m}(x+2)^m))=[a_{0}+a_{1}(x+2)][(b_{0}+b_{1}(y+1)+b_{2}(y+1)^2]+f_{1}(x,y)(x+2)^2+f_{2}(x,y)(y+1)^3$
And $[a_{0}+a_{1}(x+2)][(b_{0}+b_{1}(y+1)+b_{2}(y+1)^2]=a_{0}b_{0}+a_{1}b_{0}(x+2)+a_{0}b_{1}(y+1)+a_{0}b_{2}(y+1)^2+f_{3}(x,y)(x+2)(y+1)$
So $f(x,y)=a_{0}b_{0}+a_{1}b_{0}(x+2)+a_{0}b_{1}(y+1)+a_{0}b_{2}(y+1)^2$ in $\Bbb C[x,y]/I$ .
The basis are {$1,x+2,y+1,(y+1)^2$}.Any polynomial's coordinate in $\Bbb C[x,y]/I$ is $(a_{0}b_{0},a_{1}b_{0},a_{0}b_{1},a_{0}b_{2})$
The remain is the dimension for $(a_{0}b_{0},a_{1}b_{0},a_{0}b_{1},a_{0}b_{2})$ in $\Bbb C^4$ .I am not very sure for it.4 or 5,or other?
|
A basis of $C[x,y]$ is $\{(x+2)^m(y+1)^n: m,n\ge 0\}$.
In $C[x,y]/I$, $(x+2)^2$ and $(y+1)^3$ are identified with $1$, hence, just because of them the basis reduces to
$$
(x+2)^m(y+1)^n, \quad m=0,1,\,\,n=0,1,2.
$$
But as $(x+2)(y+1)$ is also identified with $1$, then we are reduced to
$$
1, x+2, y+1, (y+2)^2.
$$
Hence
$$
\dim C[x,y]/I=4.
$$
|
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|
Why do I first need to bring $-4x$ into the numerator in $\lim_{x\to \infty} 4x^2/(x-2) - 4x$ I tried solving the question in the title as follows:
$$\lim_{x\to \infty} \frac{4x^2}{x-2} - 4x \to 4x - 4x \to 0$$
However, apparently that first step ($\to 4x - 4x$) was wrong, and I should first have brought the second $4x$ into the numerator.
My question is not how I need to solve the question, as I know that now. My question is why what I did was wrong, as I lack any intuition for it, and it seems a mystery to me.
|
Hint
You can use long division first and get $$\frac{4x^2}{x-2}=4 x+8+\frac{16}{x}+\cdots$$
The other way could be $$\frac{4x^2}{x-2}=4\frac{x^2}{x-2}=4\frac{x^2-4x+4+4x-4}{x-2}=4\frac{(x-2)^2+4(x-1)}{x-2}=4\left(x-2+4\frac{x-1}{x-2} \right)=4\left(x-2+4\frac{x-2+1}{x-2} \right)=4\left(x-2+4\left(1+\frac{1}{x-2}\right) \right)=$$ $$4\left(x+2+\frac 4{x-2}\right)=4x+8+\frac {16}{x-2}$$
|
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|
How to differentiate $y=\ln(x+\sqrt{1+x^2})$? I'm trying to differentiate the equation below but I fear there must have been an error made. I can't seem to reconcile to the correct answer. The problem comes from James Stewart's Calculus Early Transcendentals, 7th Ed., Page 223, Exercise 25.
Please differentiate $y=\ln(x+\sqrt{1+x^2})$
My Answer:
Differentiate using the natural log rule:
$$y'=\left(\frac{1}{x+(1+x^2)^{1/2}}\right)\cdot\left(x+(1+x^2)^{1/2}\right)'$$
Now to differentiate the second term, note the chain rule applied and then simplification:
$$\left(x+(1+x^2)^{1/2}\right)'=1+\frac{1}{2}\cdot(1+x^2)^{-1/2}\cdot(2x)$$
$$1+\frac{1}{2}\cdot(1+x^2)^{-1/2}\cdot(2x)=1+\frac{x}{(1+x^2)^{1/2}}$$
Our expression is now:
$$y'=\left(\frac{1}{x+(1+x^2)^{1/2}}\right)\cdot\left(1+\frac{x}{(1+x^2)^{1/2}}\right)$$
Distribute the left term across the two right terms for my result:
$$y'=\left(\frac{1}{x+(1+x^2)^{1/2}}\right)+\left(\frac{x}{\left(x+(1+x^2)^{1/2}\right)\left(1+x^2\right)^{1/2}}\right)$$
$$y'=\left(\frac{1}{x+(1+x^2)^{1/2}}\right)+\left(\frac{x}{\left(x(1+x^2)^{1/2}\right)+(1+x^2)^{1}}\right)$$
At this point I can see that if I simplify further by adding the fractions I'll still have too many terms, and it will get awfully messy. The answer per the book (below) has far fewer terms than mine. I'd just like to know where I've gone wrong in my algebra. Thank you for your help. Here's the correct answer:
$$y'=\frac{1}{\sqrt{1+x^2}}$$
|
Let $u=x+\sqrt{1+x^2}$. Then
$$u'=1+\left[\frac{1}{2}(1+x^2)^{-\frac{1}{2}}\cdot 2x\right]=1+\frac{x}{\sqrt{1+x^2}}=\frac{\sqrt{1+x^2}+x}{\sqrt{1+x^2}}.$$
Thus,
$$y'=\frac{1}{u}\cdot u'=\frac{1}{x+\sqrt{1+x^2}}\cdot \frac{\sqrt{1+x^2}+x}{\sqrt{1+x^2}}=\frac{1}{\sqrt{1+x^2}}.$$
|
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|
Find coefficient of $x^n$ in $(1+x+2x^2+3x^3+.....+nx^n)^2$ Find coefficient of $x^n$ in
$(1+x+2x^2+3x^3+.....+nx^n)^2$
My attempt:Let $S=1+x+2x^2+3x^3+...+nx^n$
$xS=x+x^2+2x^3+3x^4+...+nx^{n+1}$
$(1-x)S=1+x+x^2+x^3+....+x^n-nx^{n+1}-x=\frac{1-x^{n+1}}{1-x}-nx^{n+1}-x$
$S=\frac{1}{(1-x)^2}-\frac{x}{1-x}=\frac{1-x+x^2}{(1-x)^2}$. (Ignoring terms which have powers of x greater than $x^n$)
So one can say that
coefficient of $x^n$ in $(1+x+2x^2+3x^3+.....+nx^n)^2$
=coefficient of $x^n$ in $(1-x+x^2)^2(1-x)^{-4}$
Is there a shorter way.
|
Concluding my attempt
Let $S=1+x+2x^2+3x^3+...+nx^n$
$xS=x+x^2+2x^3+3x^4+...+nx^{n+1}$
$(1-x)S=1+x+x^2+x^3+....+x^n-nx^{n+1}-x=\frac{1-x^{n+1}}{1-x}-nx^{n+1}-x$
$S=\frac{1}{(1-x)^2}-\frac{x}{1-x}=\frac{1-x+x^2}{(1-x)^2}$. (Ignoring terms which have powers of x greater than $x^n$)
So one can say that
coefficient of $x^n$ in $(1+x+2x^2+3x^3+.....+nx^n)^2$
=coefficient of $x^n$ in $(1-x+x^2)^2(1-x)^{-4}$
=coefficient of $x^n$ in $(1-2x+3x^2-2x^3+x^4)(1-x)^{-4}$
=$\binom{n+3}{3}-2\binom{n+2}{3}+3\binom{n+1}{3}-2\binom{n}{3}+\binom{n-1}{3}$
=$\frac{n^3+11n}{6}$.
which is too long.
|
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|
Choosing one type of ball without replacement. Suppose I have $9$ balls, among which $3$ are green and $6$ are red. What is the probability that a ball randomly chosen is green?
It is $\dfrac{3}{9}=\dfrac{1}{3}$.
If three balls are randomly chosen without replacement, then what is the probability that the three balls are green?
Is it $\dfrac{3}{9}\times\left\{\dfrac{2}{8}+\dfrac{3}{8}\right\}\times\left\{\dfrac{1}{7}+\dfrac{2}{7}+\dfrac{3}{7}\right\}=\dfrac{90}{504}$?
But in hypergeometric distribution formula, it is
$$f(X=3)=\dfrac{\binom{3}{3}\binom{9-3}{3-3}}{\binom{9}{3}}=\dfrac{1}{84}.$$
|
The probability that the first ball selected is green is $P_1 = \frac {3}{9} $. After selecting a green balls, there are now $2$ green and a total of $8$ balls. Now the probability of selecting a green balls is $P_2 = \frac {2}{8} $. Now there are a total of $7$ balls and a green ball. Now the probability is $P_3=\frac {1}{7} $. Thus, $$P_{\text {req}} = P_1 \times P_2 \times P_3 = \frac {3}{9} \times \frac {2}{8} \times \frac {1}{7} = \frac {1}{84} $$ which is the same as that got using the hypergeometric distribution formula. Hope it helps.
|
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|
Show that:$\sum\limits_{n=1}^{\infty}{n\over (4n^2-1)(16n^2-1)}={1\over 12}(1-\ln{2})$ Show that
$$\sum_{n=1}^{\infty}{n\over (4n^2-1)(16n^2-1)}={1\over 12}(1-\ln{2})$$
My try:
We split into partial decomposition
$$n={A\over 2n-1}+{B\over 2n+1}+{C\over 4n-1}+{D\over 4n+1}$$
Setting $n={1\over 2}$, ${-1\over2}$ we have $A={1\over3}$ and $B={-1\over 3}$
Finding C and D is a bit tedious
I wonder what is the closed form for
$$\sum_{n=1}^{\infty}{1\over an+b}=F(a,b)?$$
This way is not a good approach. Can anyone help me with a better approach to tackle this problem? Thank you.
|
If we consider
$$ f(x)=\frac{x}{(4x^2-1)(16x^2-1)} $$
we may compute its partial fraction decomposition through the residue theorem:
$$ f(x) = \frac{1}{24}\left(\frac{1}{x-\tfrac{1}{2}}+\frac{1}{x+\tfrac{1}{2}}\right)-\frac{1}{24}\left(\frac{1}{x-\tfrac{1}{4}}+\frac{1}{x+\tfrac{1}{4}}\right)$$
and that leads to:
$$\begin{eqnarray*} \sum_{n\geq 1}f(n) &=& \frac{1}{6}\sum_{n\geq 1}\left(\frac{1}{4n-2}+\frac{1}{4n+2}-\frac{1}{4n-1}-\frac{1}{4n+1}\right) \\&=&\frac{1}{6}\int_{0}^{1}\sum_{n\geq 1}\left(x^{4n-3}+x^{4n+1}-x^{4n-2}-x^{4n}\right)\,dx\\&=&\frac{1}{6}\int_{0}^{1}\frac{x(1-x)(1-x^3)}{1-x^4}\,dx\\&=&\frac{1}{6}\left(\int_{0}^{1}(1-x)\,dx-\int_{0}^{1}\frac{(1-x)^2}{1-x^4}\,dx\right)\\&=&\frac{1}{6}\left(\int_{0}^{1}(1-x)\,dx-\int_{0}^{1}\frac{dx}{x+1}+\frac{1}{2}\int_{0}^{1}\frac{2x}{x^2+1}\,dx\right)\\&=&\frac{1-\log 2}{12}.\end{eqnarray*}$$
|
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|
From complex rotation matrix to real matrix Let's consider $R_{n}$ to be an $n \times n$ real rotation matrix. $R_{n}$ can be diagonalized with a unitary matrix $U_{n}$ to $D_{n}$, which is composed of blocs of
$ \left( \begin{array}{ccc}
e^{j\theta} & 0 \\
0 & e^{-j\theta} \end{array} \right) $ and eventually ones.
But what I want is to have a real rotation matrix decomposition, with blocs of $ \left( \begin{array}{ccc}
\cos\theta & -\sin\theta \\
\sin\theta & \cos\theta \end{array} \right)$ and ones.
How do I get the change of basis matrix that allows me to get such a decomposition ? Can I derive it from the $U_{n}$ matrix and the eigenvalues (the $\theta$s) ?
|
Notice that $$\begin{pmatrix}\cos\theta & -\sin\theta \\ \sin\theta & \cos\theta\end{pmatrix} = \begin{pmatrix}\tfrac{1}{\sqrt{2}} & -\tfrac{j}{\sqrt{2}} \\ -\tfrac{j}{\sqrt{2}} & \tfrac{1}{\sqrt{2}}\end{pmatrix} \begin{pmatrix}e^{j\theta} & 0\\ 0 & e^{-j\theta}\end{pmatrix} \begin{pmatrix}\tfrac{1}{\sqrt{2}} & \tfrac{j}{\sqrt{2}} \\ \tfrac{j}{\sqrt{2}} & \tfrac{1}{\sqrt{2}}\end{pmatrix}.$$
So we can define $W_n$ to be a block diagonal matrix with the same block structure as $D_n$ such that for each $\begin{pmatrix}e^{j\theta} & 0\\ 0 & e^{-j\theta}\end{pmatrix}$ block in $D_n$, the corresponding block in $W_n$ is $\begin{pmatrix}\tfrac{1}{\sqrt{2}} & \tfrac{j}{\sqrt{2}} \\ \tfrac{j}{\sqrt{2}} & \tfrac{1}{\sqrt{2}}\end{pmatrix}$, and for each $\begin{pmatrix}1\end{pmatrix}$ block in $D_n$, the corresponding block in $W_n$ is also $\begin{pmatrix}1\end{pmatrix}$.
Then, $W_n^*D_nW_n = W_n^*U_n^*R_nU_nW_n$ will be block diagonal with several blocks of the form $\begin{pmatrix}\cos\theta & -\sin\theta \\ \sin\theta & \cos\theta\end{pmatrix}$ and then several ones.
|
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|
Angles in triangles are in AP, find:$\dfrac{a}{c}\sin 2A+\dfrac{c}{a}\sin 2C$
If angles of a $\triangle{ABC}$ are in A.P. then find the value of : $$\dfrac{a}{c}\sin 2A+\dfrac{c}{a}\sin 2C$$ where $a,b,c$ are sides and $A,B,C$ are angles.
My attempts:
Let $\angle A =x-d , \angle B=x, \angle C=x+d\ \implies \angle B=60^{o}$.
So one possibility is following:
*
*$\angle A=30^{o}$
*$\angle C=90^{o}$
By sine law: $\dfrac{a}{\sin A}=\dfrac{b}{\sin B}=\dfrac{c}{\sin C}=2R$
$\implies \dfrac{a}{c}\sin 2A+\dfrac{c}{a}\sin 2C=\dfrac{\sin^{2}A \sin 2A+ \sin^{2} C \sin 2C}{\sin A \sin C} $ By value of $\angle A\ ,\angle C$.
Answer comes to be $\dfrac{\sqrt{3}}{4}$.
Is it correct, if not then where I go wrong? And I don't want to prove this by an assumption that $\angle A=30, \angle C=90$. How to do this without assuming angles.
Edit as suggested by @rohan :
original question is follows:
If angles of a $\triangle{ABC}$ are in A.P. then find the value of : $$\dfrac{a}{c}\sin 2C+\dfrac{c}{a}\sin 2A$$ where $a,b,c$ are sides and $A,B,C$ are angles.
I earnestly apologise for wasting your precious time, it was not deliberate.
|
I think there is a typo in your question. It should be to find the value of $\frac{a}{c}\sin 2C + \frac{c}{a}\sin 2A$.
Assuming the case it is so, then the problem simplifies to: $$\frac{\sin A}{\sin C}(2\sin C\cos C) + \frac{\sin C}{\sin A}(2\sin A\cos A) = 2\sin A\cos C + 2\sin C\cos A = 2\sin (A+C) = 2\sin 2B = 2\times \frac{\sqrt{3}}{2} =\sqrt{3}$$
The problem with the question as it is is the fact that it allows multiple possibilities.
|
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|
Can I express $x^7y+xy+x+1$ by a repeated use of operation $xy+x+y+1$? A certain calculator can only give the result of $xy+x+y+1$ for any two real numbers $x$ and $y$.
How to use this calculator to calculate $x^7y+xy+x+1$ for any given $x$ and $y$?
When $x$ and $y$ are equal, it will give $(x+1)^2$. But I cannot proceed beyond that.
|
Follow these steps:
*
*Give $(x,x)$ to the calculator to get $x^2 + 2x + 1$ and subtract $2x + 1$ from this. You can even get $2x$ from the calculator by giving in $(x,1)$ to get $2x + 2$.
*Now, using prev result, give $(x^2, x^2)$ to get $x^4+ 2x^2 + 1$. Use previous result to obtain $x^4$.
*Using previous result, give $(x^4, x^2)$ to get $x^6 + x^4 + x^2 + 1$. Subtract values to get $x^6$.
*Give $(x^6, x)$ to get $x^7 + x^6 + x + 1$ and from this extract $x^7$.
*Give $(x^7, y)$ to get $x^7y + x^7 + y + 1$. From this pull out the $x^7y$ to get $x^7y$
*Give $(x,y)$ to get $xy + x + y + 1$. From this pull out the $y$ to get $xy + x + 1$
*Add 5. and 6. to get $x^7y + xy + x + 1$.
This question makes little sense. Are there more stringent rules on what you aren't allowed to do?
|
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|
Find $AC: CB$ in $\triangle XYZ$ Problem:
In $\triangle XYZ$, $XY = 4$, $YZ = 7$, and $XZ = 9$. Let $M$ be the midpoint of $\overline{XZ}$, and let $A$ be the point on $\overline{XZ}$ such that $\overline{YA}$ bisects angle $XYZ$. Let $B$ be the point on $\overline{YZ}$ such that $\overline{YA} \perp \overline{AB}$. Let $\overline{AB}$ meet $\overline{YM}$ at $C$. Find $AC: CB$.
Attempt:
We know that $XM$ = $MZ$ = $\dfrac{9}{2}$, and by the Angle-Bisector theorem,
$$\dfrac{4}{x} = \dfrac{7}{9-x}$$
$$\implies 36-4x = 7x$$
$$\implies x= \dfrac{36}{11}=XA$$
Therefore, $AZ$ = $\dfrac{63}{11}$, and $AM$ = $\dfrac{9}{2} - \dfrac{36}{11}= \dfrac{27}{22}.$
Also, by the Menelaus theorem,
$$\dfrac{ZM}{MA} \times \dfrac{AC}{CB} \times \dfrac{YB}{YZ} = 1$$
$$\implies \dfrac{11}{3} \times \dfrac{AC}{CB} \times \dfrac{YB}{7} = 1$$
From here, I got stuck. I'm wondering if one could use mass points, since the problem wants to find the ratio of lengths, and not specific side lengths. Any help is appreciated!
|
(Wrong!!! This is the case for $YB\perp AB.$)
Following your attempts, we get $\frac{11}{3}\times\frac{AC}{CB}\times\frac{YB}{7} = 1.$ Hence it remains to find the length of $YB.$ Let $K$ be the point on $YZ$ such that $XK\perp YZ.$ Then we can use the area of the triangle to find $XK$:
$$\sqrt{10(10-9)(10-7)(10-4)}=\frac{7\times XK}{2}.$$
Therefore $XK=\frac{12\sqrt{5}}{7}.$
Then since $\triangle XKZ\sim\triangle ABZ,$ we can get
$$AB = XK \times\frac{63/11}{9}=\frac{12\sqrt{5}}{11}$$
and obtain $BZ = \sqrt{(\frac{63}{11})^2-(\frac{12\sqrt{5}}{11})^2}=\frac{57}{11}.$
Then our $YB=7- \frac{57}{11} = \frac{20}{11},$ so the answer should be
$$\frac{AC}{CB} = \frac{3}{11}\times\frac{7}{YB} = \frac{21}{20}.$$
|
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|
Is it possible to find the sum of the infinite series $1/p + 2/p^2 + 3/p^3 + \cdots + n/(p^n)+\cdots$, where $p>1$? Is it possible to find the sum of the series:
$$\frac{1}{p} + \frac{2}{p^2} +\frac{3}{p^3} +\dots+\frac{n}{p^n}\dots$$
Does this series converge? ($p$ is finite number greater than $1$)
|
Let:
$
S = {1\over p} + {2\over p^2} + {3 \over p^3} + \cdots \\
\implies {S\over p} = {1\over p^2} + {2\over p^3} + {3\over p^4} + \cdots.
$
Hence,
$
S\left ( 1 - {1\over p} \right ) = {1\over p} + {1\over p^2} + {1\over p^3} + \cdots \\
S\left ( 1 - {1\over p} \right ) = \frac{1}{p}\frac{1}{1-{1\over p}} = \frac{1}{p-1}\\
\implies S = \frac{p}{(p-1)^2}.
$
The series converges for $p>1$.
|
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|
Find the number of three elements sets of positive integers $\{a,b,c\}$ such that $a\times b\times c=2310$. Find the number of three elements sets of positive integers $\{a,b,c\}$ such that $a\times b\times c=2310$.
My attempt:
Now, $2310=2\times 3\times 5\times 7\times 11=a\times b\times c$
$a=2^{x_{1}}3^{y_{1}}5^{z_{1}}7^{w_{1}}11^{t_{1}}$
$b=2^{x_{2}}3^{y_{2}}5^{z_{2}}7^{w_{2}}11^{t_{2}}$
$c=2^{x_{3}}3^{y_{3}}5^{z_{3}}7^{w_{3}}11^{t_{3}}$
where $x_{1}+x_{2}+x_{3}=1$; $x_{1},x_{2},x_{3}$ being non-negative integers.
Number of solutions of this equation clearly $3$ and similarly for other variables as well.
So,number of solutions $=3^5$.
But answer given is $40$ .
Am I misinterpreting the question
|
I got a combinatorial solution.
$2310=2 \times 3 \times 5 \times 7 \times 11$
$\dfrac{\dbinom{5}{1}\dbinom{4}{1}\dbinom{3}{3}}{2!}=10\\
\dfrac{\dbinom{5}{2}\dbinom{3}{1}\dbinom{2}{2}}{2!}=15\\
\dbinom{5}{3}\dbinom{2}{2}=10\\
\dbinom{5}{4}\dbinom{1}{1}=5$
Total: $40$
|
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|
Relation between inverse tangent and inverse secant I've been working on the following integral
$$\int\frac{\sqrt{x^2-9}}{x^3}\,dx,$$
where the assumption is that $x\ge3$. I used the trigonometric substitution $x=3\sec\theta$,which means that $0\le\theta<\pi/2$. Then, $dx=3\sec\theta\tan\theta\,dx$, and after a large number of steps I achieved the correct answer:
$$\int\frac{\sqrt{x^2-9}}{x^3}\,dx=\frac16\sec^{-1}\frac{x}{3}-\frac{\sqrt{x^2-9}}{2x^2}+C$$
I was able to check my answer using Mathematica.
expr = D[1/6 ArcSec[x/3] - Sqrt[x^2 - 9]/(2 x^2), x];
Assuming[x >= 3, FullSimplify[expr]]
Which returned the correct response:
Sqrt[-9 + x^2]/x^3
Mathematica returns the following answer:
Integrate[Sqrt[x^2 - 9]/x^3, x, Assumptions -> x >= 3]
-(Sqrt[-9 + x^2]/(2 x^2)) - 1/6 ArcTan[3/Sqrt[-9 + x^2]]
Which I can write to make more clear.
$$-\frac16\tan^{-1}\frac{3}{\sqrt{x^2-9}}-\frac{\sqrt{x^2-9}}{2x^2}+D$$
Now, you can see that part of my answer is there, but here is my question. How can I show that
$$\frac16\sec^{-1}\frac{x}{3}\qquad\text{is equal to}\qquad -\frac16\tan^{-1}\frac{3}{\sqrt{x^2-9}}$$
plus some arbitrary constant? What identities can I use? Also, can anyone share the best web page for inverse trig identities?
Update: I'd like to thank everyone for their help. The Trivial Solution's suggestion gave me:
$$\theta=\sec^{-1}\frac{x}{3}=\tan^{-1}\frac{\sqrt{x^2-9}}{3}$$
Then the following identity came to mind:
$$\tan^{-1}x+\tan^{-1}\frac1x=\frac{\pi}{2}$$
So I could write:
\begin{align*}
\frac16\sec^{-1}\frac{x}{3}-\frac{\sqrt{x^2-9}}{2x^2}
&=\frac16\tan^{-1}\frac{\sqrt{x^2-9}}{3}-\frac{\sqrt{x^2-9}}{2x^2}\\
&=\frac16\left(\frac{\pi}{2}-\tan^{-1}\frac{3}{\sqrt{x^2-9}}\right)-\frac{\sqrt{x^2-9}}{2x^2}\\
&=\frac{\pi}{12}-\frac16\tan^{-1}\frac{3}{\sqrt{x^2-9}}-\frac{\sqrt{x^2-9}}{2x^2}
\end{align*}
Using Olivier's and Miko's thoughts, I produced this plot in Mathematica.
Plot[{1/6 ArcSec[x/3] - Sqrt[x^2 - 9]/(
2 x^2), -(1/6) ArcTan[3/Sqrt[x^2 - 9]] - Sqrt[x^2 - 9]/(
2 x^2)}, {x, -6, 6},
Ticks -> {Automatic, {-\[Pi]/12, \[Pi]/12}}]
Which shows that the two answers differ by $\pi/12$, but only for $x>3$.
|
Hint. Applying the chain rule, for $x>3$, one may check that
$$
\left(\frac16\sec^{-1}\frac{x}{3}\right)'=\frac{1}{2x\sqrt{9-x^2}}
$$$$
\left(-\frac16\tan^{-1}\frac{3}{\sqrt{x^2-9}}\right)'=\frac{1}{2x\sqrt{9-x^2}}
$$ since
$$
\lim_{x \to 3^+}\left(\frac16\sec^{-1}\frac{x}{3}\right)=0,\quad \lim_{x \to 3^+}\left(-\frac16\tan^{-1}\frac{3}{\sqrt{x^2-9}}\right)=-\frac{\pi}{12}
$$ then
$$
\frac16\sec^{-1}\frac{x}{3}=-\frac16\tan^{-1}\frac{3}{\sqrt{x^2-9}}+\frac{\pi}{12},\quad x>3.
$$
|
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|
Identifying the digits of $37 \cdot aaaa\ldots a$. With a calculator, I have noticed that the integer $37$ multiplied with some particular numbers yields numbers with some structures.
For instance, let $aaaa\ldots a$ be a natural number of $n$ identical digits. Then, $ 37 \cdot aaaa\ldots a$ is a number with $n+1$ or $n+2$ digits of the form
$$\underbrace{4\cdot a}_{1\text{ or }2}~ \underbrace{aaa\ldots a}_{n-2} ~\underbrace{7\cdot a}_{2^{*}}.$$
$*$ if $a=1$, then $7 \cdot a$ is the sequence of digits $07$.
I am wondering whether the result above can be proven using some number-theory tool.
Thanks in advance!
|
Nice observation!
This follows because $aaa\cdots a = a \cdot 111 \cdots 1$ and
$37 \cdot 11 = 407$
$37 \cdot 111 = 4107$
$37 \cdot 1111 = 41107$
$\cdots$
Indeed, let $u_n = 111 \cdots 1$ ($n$ ones). Then $u_n = \dfrac{10^n-1}{9}$ and
$$
\begin{align}
37 \cdot 111\cdots 1 \quad (n \text{ ones}) &=
37u_n\\&= 36u_n + u_n\\
&= 4\cdot9\cdot u_n+10u_{n-2}+11\\
&=4(10^n-1)+10u_{n-2}+4+7\\
&=4\cdot10^n+10u_{n-2}+7\\
&=4111\cdots107 \quad (n-2 \text{ ones})
\end{align}
$$
|
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|
Integral of $2x^2 \sec^2{x} \tan{x}$ I've been trying for a while to find $\int{( 2x^2 \sec^2{x} \tan{x} )} dx$, using integration by parts.
I always end up getting a more complicated integral in the second part of the equation.
For example:
$$ \int{( 2x^2 \sec^2{x} \tan{x} )} dx =
\\ 2x^2 \tan^2x - \int{\tan{x} \cdot \frac{d}{dx}(2x^2 \tan{x})}
\\ \frac{d}{dx}(2x^2 \tan{x})=4x\tan{x} + 2x\sec^2{x} \rightarrow
\\ 2x^2 \tan^2x - \int{4x \tan^2{x}+2x\tan{x}\sec^2{x}}
$$
I've tried integrating with different value for $u$ and $v$, such as:
$$ 1:( 2x^2 \sec^2{x} \tan{x} ),
\\ \tan{x} : 2x^2 \sec^2{x},
\\ 2x^2: \sec^2{x} \tan{x},
\\ \sin{x}: 2x^2 \sec^3{x} $$ etc, however, haven't succeeded.
|
Use integration by parts with $u=x^2$ and $v=\tan^2(x)$. Then, we have
$$\int 2x^2 \tan(x)\sec^2(x)\,dx=x^2\tan^2(x)-2\int x\tan^2(x)\,dx$$
Continue with a subsequent integration by parts with $u=x$ and $v=\tan(x)-x$ to obtain
$$\begin{align}
\int 2x^2 \tan(x)\sec^2(x)\,dx&=x^2\tan^2(x)-2\left(x\tan(x)-x^2-\int(\tan(x)-x)\,dx\right)\\\\
&=x^2\tan^2(x)-2x\tan(x)+x^2-2\log(\cos(x))+C\\\\
&=x^2\sec^2(x)-2(x\tan(x)+\log(\cos(x)))+C
\end{align}$$
|
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|
Decompose $X^8 + X^7 + X^6 + X^4 + 1$ over $\mathbb{Z}_2$ Is there a clever way to decompose $f(X) = X^8 + X^7 + X^6 + X^4 + 1$ into irreducible factors over $\mathbb{Z}_2$? Or to see that it is irreducible itself?
This is what I have thought about so far: $f(0) \ne 0$ and $f(1) \ne 0$, so $f$ does not have linear factors. It also does not have quadratic factors, because the only irreducible quadratic polynomial over $\mathbb{Z}_2$, $X^2 + X + 1$, does not divide $f$. I could continue finding all irreducible polynomials of degree 3 and 4 and carrying out the corresponding polynomial divisions.
Is there a more clever solution?
|
Note that $(x^8+x^7+x^6+x^4+1)(x+1)=x^9 + x^6 + x^5 + x^4 + x + 1$.
Thus, if $\alpha^5=1$, then $\alpha$ is a root of the latter polynomial. That is, the latter polynomial is divisible by $x^5+1=(x^4 + x^3 + x^2 + x + 1)(x+1)$. But this means that
$x^4 + x^3 + x^2 + x + 1$ is a factor of your original polynomial. Now polynomial division leads us to
$$x^8+x^7+x^6+x^4+1=(x^4 + x + 1)(x^4 + x^3 + x^2 + x + 1).$$
It is easy to see that both fators on the RHS are irreducible over $\mathbb{Z}_2$. In fact, they have no roots in $\mathbb{Z}_2$ and are not divisible by $x^2+x+1$.
|
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|
show that $a_{n+874}=a_{n}$,if such $a_{n+2}=\left\lceil \frac{4}{3}a_{n+1}-a_{n}+0.5\right\rceil$
Let the sequence $\{a_{n}\}$ be such that $a_{1}=1, a_{2}=100$, and $$a_{n+2}=\left\lceil \dfrac{4}{3}a_{n+1}-a_{n}+0.5\right\rceil$$
Prove that the sequence $\{a_{n}\}$ is periodic.
I have used a computer and found the periodic is $T=874$, but how to prove it?
|
N.B. This is not complete answer, but just extrapolation of the clue provided by @SangchulLee.
First of all we can rewrite the recurrence as:
$$a_{n+2}=\frac{4a_{n+1}}{3}-a_{n}+v_{n+1}$$
where
$$v_{n+1}=\begin{cases}
1 & a_{n+1} = 0 \bmod 3\\
\frac{2}{3} & a_{n+1} = 1 \bmod 3 \\
\frac{4}{3} & a_{n+1} = 2 \bmod 3
\end{cases}$$
Now we define
$$A_{n} = \begin{bmatrix}
a_{n+1} & a_{n} \\
a_{n+2} & a_{n+1}
\end{bmatrix}$$
$$B = \begin{bmatrix}
0&1 \\
-1&\frac{4}{3}
\end{bmatrix}$$
$$U_1 = \begin{bmatrix}
0&0 \\
0&1
\end{bmatrix}$$
$$U_2 = \begin{bmatrix}
0&0 \\
1&0
\end{bmatrix}$$
This means
$$A_{n+1} = BA_{n}+v_{n+1}U_1+v_{n+2}U_2$$
Then we can write
$$\det(A_{n+1}-v_{n+2}U_2) = \det(A_{n}+v_{n+1}B^{-1}U_1) $$
Simplifying this one obtains
$$a_{n+2}^2-\frac{4a_{n+2}a_{n+1}}{3}+a_{n+1}^2-v_{n+1}(a_{n+2}+a_{n+1}) = a_{n+1}^2-\frac{4a_{n+1}a_n}{3}+a_n^2-v_{n+1}(a_{n+1}+a_n)$$
Notice that the equation $x^2-4xy/3+y^2-v(x+y)=C$ represents an ellipse for a given constant $C$ and $v\in\{1,2/3,4/3\}$
I was not able to proceed from here! Hope this helps although its a little late!
|
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|
Prove that $3 \mid (a+b+c+d)$
Given $a,b,c,d \in \mathbb{Z}$ satisfying $a^3+b^3 = 2(c^3-8d^3)$, prove that $3 \mid (a+b+c+d)$.
I first factorized $a^3+b^3$ to get $a^3+b^3 = (a+b)(a^2-ab+b^2)$. I wasn't sure how to use the right-hand side to get $a+b+c+d$. How can we prove that $3 \mid (a+b+c+d)$?
|
$x^3\equiv x\bmod3$, which you can prove by trial and error in this case. This should reduce it down to $a+b\equiv2c-d\bmod3$. Adding $c+d$ to both sides yields $a+b+c+d\equiv3c\equiv0\bmod3$, which means $3|(a+b+c+d)$.
|
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|
Hint in integration $\int\frac{x^{2}}{\left(x\cos x-\sin x \right )\left( x\sin x+\cos x \right )}\,\mathrm{d}x$ In the following integration
$$\int \frac{x^{2}}{\left ( x\cos x-\sin x \right )\left ( x\sin x+\cos x \right )}\, \mathrm{d}x$$
I tried alot. But does not get any proper start.
Can anybody provide me a hint.
|
Hint. One may observe the following trick
$$
\begin{align}
\frac{x^2}{(x \cos x-\sin x)(x\sin x+\cos x)}&=\frac{x\cos x(x \cos x-\sin x)+x\sin x(x\sin x+\cos x)}{(x \cos x-\sin x)(x\sin x+\cos x)}
\\\\&=\frac{x\cos x}{x\sin x+\cos x}+\frac{x \sin x}{x \cos x-\sin x}
\\\\&=\frac{(x\sin x+\cos x)'}{(x\sin x+\cos x)}-\frac{(x \cos x-\sin x)'}{(x \cos x-\sin x)}
\end{align}
$$ then one may conclude.
|
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|
There is a single pair $(a,b)$ such that $x^2+2(1+a)x+(3a^2+4ab+4b^2+2)=0$. For this pair, what is $a+b$?
There is a single pair $(a,b)$ such that $x^2+2(1+a)x+(3a^2+4ab+4b^2+2)=0$. For this pair, what is $a+b$?
I know how to find rational roots now because of my last question asked, but what about real roots? What do I do?
|
$$0=x^2+2(1+a)x+(3a^2+4ab+4b^2+2) =\\(x+1+a)^2+(3a^2+4ab+4b^2+2-1-2a-a^2)
=\\(x+1+a)^2+(2a^2+4ab+4b^2+1-2a)
=\\(x+1+a)^2+(a^2+4ab+4b^2)+(a^2-2a+1) \\
=(x+1+a)^2+(a+2b)^2+(a-1)^2
$$
Since
$$(x+1+a)^2+(a+2b)^2+(a-1)^2=0$$
you get
$$x+1+a=0\\
a+2b=0\\
a-1=0$$
|
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|
How to integrate $\int\sqrt{\frac{4-x}{4+x}}$? Let
$$g(x)=\sqrt{\dfrac{4-x}{4+x}}.$$
I would like to find the primitive of $g(x)$, say $G(x)$.
I did the following: first the domain of $g(x)$ is $D_g=(-4, 4]$. Second, we have
\begin{align}
G(x)=\int g(x)dx &=\int\sqrt{\dfrac{4-x}{4+x}}dx\\
&=\int\sqrt{\dfrac{(4-x)(4-x)}{(4+x)(4-x)}}dx\\
&=\int\sqrt{\dfrac{(4-x)^2}{16-x^2}}dx\\
&=\int\dfrac{4-x}{\sqrt{16-x^2}}dx\\
&=\int\dfrac{4}{\sqrt{16-x^2}}dx-\int\dfrac{x}{\sqrt{16-x^2}}dx\\
&=\int\dfrac{4}{\sqrt{16(1-x^2/16)}}dx+\int\dfrac{-2x}{2\sqrt{16-x^2}}dx\\
&=\underbrace{\int\dfrac{1}{\sqrt{1-(x/4)^2}}dx}_{\text{set $t=x/4$}}+\sqrt{16-x^2}+C\\
&=\underbrace{\int\dfrac{4}{\sqrt{1-t^2}}dt}_{\text{set $t=\sin \theta$}}+\sqrt{16-x^2}+C\\
&=\int\dfrac{4\cos\theta}{\sqrt{\cos^2\theta}}d\theta+\sqrt{16-x^2}+C\\
\end{align}
So finally, I get
$$G(x)=\pm\theta+\sqrt{16-x^2}+C'.$$
With wolframalpha I found some different answer. Could you provide any suggestions?
Also, multiplying by $4-x$ is it correct at the beginning? because I should say then that $x\neq 4$.
|
Substitute $x=4\frac{1-y^2}{1+y^2}$ so that $y=\sqrt{\frac{4-x}{4+x}}$:
$$
\begin{align}
\int\sqrt{\frac{4-x}{4+x}}\,\mathrm{d}x
&=4\int y\,\mathrm{d}\frac{1-y^2}{1+y^2}\\
&=-16\int\frac{y^2}{\left(1+y^2\right)^2}\,\mathrm{d}y\\
&=-16\int\frac{\tan^2(\theta)}{\sec^2(\theta)}\,\mathrm{d}\theta\\
&=-16\int\sin^2(\theta)\,\mathrm{d}\theta\\
&=-16\int\frac{1-\cos(2\theta)}2\,\mathrm{d}\theta\\
&=-8\theta+4\sin(2\theta)+C\\
&=-8\theta+8\frac{\tan(\theta)}{\sec^2(\theta)}+C\\
&=-8\tan^{-1}\left(\sqrt{\frac{4-x}{4+x}}\right)+\sqrt{16-x^2}+C
\end{align}
$$
|
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|
Let $f(x) = x^3-\frac{3}{2}x^2+x+\frac{1}{4}.$ Then the value of $ \int^{3/4}_{1/4}f(f(x))\mathrm dx$ If $\displaystyle f(x) = x^3-\frac{3}{2}x^2+x+\frac{1}{4}.$ then the value of $\displaystyle \int^{\frac{3}{4}}_{\frac{1}{4}}f(f(x))\mathrm dx$
$\displaystyle \int^{\frac{3}{4}}_{\frac{1}{4}}(f(x))^3-1.5(f(x))^2+f(x)+0.25 \mathrm dx$
could some help me with this, thanks
|
Apply kings rule, I=int(f(f(1-x))) =integral( -x3 +1.5x2 -x +0.75)
Add the two 2I = int(1)
I= 0.25.
|
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|
A simple proof by induction $P(n) = \frac{6^{2n} - 3^n}{11} \in \mathbb{N}$ Verify by induction that $P(n) = \frac{6^{2n} - 3^n}{11} \in \mathbb{N} \quad \forall n \ge 1 \in \mathbb{N}$
Basis: $P(1) \Rightarrow \frac{33}{11} \in \mathbb{N}$.
Induction: if the statement holds for some $n$ $\Rightarrow$ holds for $n+1$
I would appreciate some small hints because I'm stuck at $P(n+1) = \frac{6^{2(n+1)} - 3^{n+1}}{11}$
|
We have $P(n)=\frac{6^{2n}-3^n}{11} \in \mathbb{N}$.
You have proven that the base case is true.
Now, assume true for $n=k$.
$$6^{2k}-3^k=11p \tag{1}$$
Where $p \in \mathbb{N}$.
True for $n=k+1$.
$6^{2(k+1)}-3^{k+1}=6^{2k+2}-3^{k+1}=36 \cdot 6^{2k}-3 \cdot 3^k \tag{2}$
Now, here comes the trick.
We can substitute from $6^{2k}$ from equation $(1)$ into equation $(2)$.
$$36 \cdot (11p+3^k)-3 \cdot 3^k=11\cdot 36p+36\cdot 3^k-3 \cdot 3^k=11 \cdot 36p-33 \cdot 3^k \tag{3}$$
Since $11\cdot 36p$ is obviously divisible by $11$, and $33 \cdot 3^k$ can be divided by $11$ to give $3 \cdot 3^k$, the whole of equation $(3)$ is divisible by $11$.
Hence, we are done.
|
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|
Solve the system of equations ... Solve the system of equations :
(EDIT : The problem does not say anything about the nature of $x$ and $y$ (integer, natural number ,..etc.) )
$4xy + 4(x^2 + y^2) + {\frac {3} { (x+y) ^ 2 } } = \frac {85} {3} $
$2x + {\frac {1} {x+y}} = \frac {13} {3}$
I do not know how to approach these types of problems. I tried finding value of $\frac {1} {x+y}$ in terms of $x$ and $y$, but it complicates the problem even more.
Can anyone provide a pointer to what should be done ?
|
Less elegant than Anurag A's answer and using brute force.
Considering the equations $$4xy + 4(x^2 + y^2) + {\frac {3} { (x+y) ^ 2 } } = \frac {85} {3}\tag 1$$
$$2x + {\frac {1} {x+y}} = \frac {13} {3}\tag 2$$ extract $y$ from $(2)$; this gives $$y=\frac{-6 x^2+13 x-3}{6 x-13}\tag 3$$ Replace $(3)$ in $(1)$ an simplify to get $$\frac{8 (x-2) (3 x-2) \left(24 x^2-118 x+149\right)}{(13-6 x)^2}=0\tag 4$$ and the quadratic term does not show real solution (so,two real roots and two complex conjugate roots for $x$).
The solutions of $(4)$ are simple; when you have them, go back to $(3)$ for the corresponding $y$'s.
|
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|
Positive pairs of integral values satisfying $2xy β 4x^2 +12x β 5y = 11$ The number of positive pairs of integral values of $(x, y)$ that solves
$2xy β 4x^2 +12x β 5y = 11$ is?
I rearranged it to $(2x-5)(y+1-2x)=6$, which took quite a bit of time.
So it can be $2*3$ , $3*2$, $6*1$ or $1*6$ which gives us 2 possible positive integral pairs. Answer: 2.
Is there a faster way to do similar problems?
|
First of all, $6=\pm1\cdot\pm6=\pm1\cdot\pm6=\pm2\cdot\pm3=\pm3\cdot\pm2=\pm6\cdot\pm1$
We've got the following equations:
$$\begin{cases} 2x = 5+a\\-2x+y = -1+b\end{cases}$$
We can rearrange it to:
$$\begin{cases} x = 2.5+\frac{a}{2}\\y = 4+a+b\end{cases}$$
We want $x$ and $y$ to be integer, so $a$ must be an odd number and $b$ must be then an integer number. Also we want $ab=6$.
There are 4 pairs $[a,b]$ satisfying these conditions: $[\pm 1, \pm 6]$ and $[\pm 3, \pm2]$
We have then 4 different pairs $[x,y]$:
$$\left\{[2.5\pm \frac{1}{2}, 4 \pm 1 \pm 6], [2.5\pm \frac{3}{2}, 4 \pm 3 \pm 2] \right\} =\\
=\left\{[3,11],[2,-3],[4,9],[1,-1]\right\}$$
We also want $[x,y]$ to be a pair of positive integers, so from our pairs we pick two satisfying this condition:
$$\left\{[3,11],[4,9]\right\}$$
|
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|
Any solution for $\iiint\frac{x^2+2y^2}{x^2+4y^2+z^2}\,dv$ I tried to solve this triple integral but couldn't integrate the result.
$$\iiint\frac{x^2+2y^2}{x^2+4y^2+z^2}\,dv$$ and the surface to integrate in is $$x^2+y^2+z^2\le1$$
Is there any way to transform the integral into polar coordinates?
|
Notice that:
$$\iiint \frac{x^2+2y^2}{x^2+4y^2+z^2} \mbox{d}v =
\iiint \frac{x^2+4y^2+z^2-2y^2-z^2}{x^2+4y^2+z^2} \mbox{d}v
= \iiint 1-\frac{z^2+2y^2}{x^2+4y^2+z^2} \mbox{d}v $$
But by symmetry $x \leftrightarrow z$, we have:
$$\iiint \frac{x^2+2y^2}{x^2+4y^2+z^2} \mbox{d}v
= \iiint \frac{z^2+2y^2}{x^2+4y^2+z^2} \mbox{d}v $$
So:
$$\iiint \frac{x^2+2y^2}{x^2+4y^2+z^2} \mbox{d}v = \frac{1}{2}
\iiint 1 \mbox{d}v = \frac{1}{2}\frac{4}{3}\pi = \frac{2}{3}\pi$$
|
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|
Definite integral of $\int_0^{1}x^{7}\sqrt{\frac{1+x^{2}}{1-x^{2}}}dx$. When i integrate $\int_{0}^{1}x^{7}\sqrt{\frac{1+x^{2}}{1-x^{2}}}dx$, putting
$t^{2}=1-x^{2}$ then integration $\int{(1-t^{2})}^{3}\sqrt{t^{2}+2}dt$, but i don't understand how to proceed . please someone help me. Thank you.
|
Substitute $\text{u}=\frac{1}{1-x^2}$:
$$\mathcal{I}=\int x^7\cdot\sqrt{\frac{1+x^2}{1-x^2}}\space\text{d}x=\frac{1}{2}\int\frac{\left(\text{u}-1\right)^3\cdot\sqrt{2\text{u}-1}}{\text{u}^5}\space\text{d}\text{u}$$
Now, substitute $\text{v}=\sqrt{2\text{u}-1}$:
$$\mathcal{I}=2\int\frac{\text{v}^2\cdot\left(\text{v}^6-3\text{v}^4+3\text{v}^2-1\right)}{\left(1+\text{v}^2\right)^5}\space\text{d}\text{v}$$
And now you do partial fraction decomposition:
$$\frac{\text{v}^2\cdot\left(\text{v}^6-3\text{v}^4+3\text{v}^2-1\right)}{\left(1+\text{v}^2\right)^5}=\frac{1}{1+\text{v}^2}-\frac{7}{\left(1+\text{v}^2\right)^2}+\frac{18}{\left(1+\text{v}^2\right)^3}-\frac{20}{\left(1+\text{v}^2\right)^4}+\frac{8}{\left(1+\text{v}^2\right)^5}$$
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|
Limit of type $\frac{0}{0}$: $\lim_{x\to0} \frac{\sin(x^n)-\sin^n(x)}{x^{n+2}}$ $$\lim_{x\to 0}\ \frac{\sin\left(x^n\right)-\sin^n\left(x\right)}{x^{n+2}} ~~~~~~~ \mbox{for} ~~~ n \geq 2$$
It's a multiple choice problem and the answer is $\dfrac{n}{6}$.
I tried it for $n=2$ and i got the answer $\dfrac{2}{6}$ which fits, but i had to apply L'Hospital multiple times and it was kinda annoying.
I wonder if this can be solved for the general case (without using Taylor) or even for $n=2$ but in a simpler way.
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Write it
$$\frac{\sin (x^n) - (\sin x)^n}{x^{n+2}} = \underbrace{x^{2n-2}}_{\to 0}\underbrace{\frac{\sin (x^n) - x^n}{x^{3n}}}_{\to - \frac{1}{6}} + \frac{x^n - (\sin x)^n}{x^{n+2}},$$
and then
$$\frac{x^n - (\sin x)^n}{x^{n+2}} = \frac{x-\sin x}{x^3}\sum_{k = 0}^{n-1} \frac{x^{n-1-k}(\sin x)^k}{x^{n-1}}$$
using $a^n - b^n = (a-b)(a^{n-1} + a^{n-2}b + \dotsc + ab^{n-2} + b^{n-1})$. With $\lim\limits_{x\to 0} \frac{\sin x}{x} = 1$, the limit of the last sum is easily determined as $n$.
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|
Help me to calculate $\int_{0}^{\pi/4}\sqrt{1+\tan x}\,\mathrm dx$ How to calculate $$\int_{0}^{\pi/4}\sqrt{1+\tan x}\,\mathrm dx$$
My attempt:
Let
$$I=\int_{0}^{\pi/4}\sqrt{1+\tan x}\,\mathrm dx$$
substitute $\sqrt{1+\tan x}=t$,then
$$I=\int_{1}^{\sqrt{2}}\frac{2t^{2}}{t^{4}-2t^{2}+2}\,\mathrm dt=\int_{1}^{\sqrt{2}}\frac{2}{t^{2}-2+\dfrac{2}{t^{2}}}\,\mathrm dt$$
but I got stuck here for a long time.Any idea?
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Hint:
\begin{align*}
I&=\int_{1}^{\sqrt{2}}\frac{2t^{2}}{t^{4}-2t^{2}+2}\,\mathrm{d}t\\
&=\int_{1}^{\sqrt{2}}\frac{\sqrt{2}+t^{2}+\left ( t^{2}-\sqrt{2} \right )}{t^{4}-2t^{2}+2}\,\mathrm{d}t \\
&=\int_{1}^{\sqrt{2}}\frac{\displaystyle\frac{\sqrt{2}}{t^{2}}+1}{t^{2}-2+\displaystyle\frac{2}{t^{2}}}\,\mathrm{d}t+\int_{1}^{\sqrt{2}}\frac{1-\displaystyle\frac{\sqrt{2}}{t^{2}}}{t^{2}-2+\displaystyle\frac{2}{t^{2}}}\,\mathrm{d}t \\
&=\frac{1}{\sqrt{2\sqrt{2}-2}}\int_{1}^{\sqrt{2}}\frac{1}{\left ( \frac{\displaystyle t-\displaystyle\frac{\sqrt{2}}{t}}{\displaystyle\sqrt{2\sqrt{2}-2}} \right )^{2}+1}\,\mathrm{d}\left (\frac{t-\displaystyle\frac{\sqrt{2}}{t}}{\displaystyle\sqrt{2\sqrt{2}-2}} \right ) \\
&~~~+\int_{1}^{\sqrt{2}}\frac{1}{\left ( t+\displaystyle\frac{\sqrt{2}}{t}-\sqrt{2\sqrt{2}+2} \right )\left ( t+\displaystyle\frac{\sqrt{2}}{t}+\sqrt{2\sqrt{2}+2} \right )}\,\mathrm{d}\left ( t+\frac{\sqrt{2}}{t} \right )
\end{align*}
then you can take it from here.
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|
Two fair dice are rolled. What is the conditional probability that at least one lands on 6 given that the dice land on different numbers? $F =$ The probability that the dice land on different numbers
$F = {(1,2),(1,3),(1,4),(1,5),(1,6),(2,1),(2,3),(2,4),(2,5),(2,6),(3,1),(3,2),(3,4),(3,5),(3,6),(4,1),(4,2),(4,3),(4,5),(4,6),(5,1),(5,2),(5,3),(5,4),(5,6),(6,1),(6,2),(6,3),(6,4),(6,5)}$
$F = \frac{30}{36} = \frac{5}{6}$
$E =$ The event that at least one lands on 6
$E = {(1,6),(2,6),(3,6),(4,6),(5,6),(6,1),(6,2),(6,3),(6,4),(6,5),(6,6)}$
$E = \frac{11}{36}$
What is the probability of E, when given that F has occurred?
$P = \frac{P(EF)}{P(F)}$
$P = \frac{1-P(EF^c)}{P(F)}$
$P = \frac{\frac{11}{36}}{\frac{5}{6}}$
$P = \frac{11}{30}$
This answer was incorrect it is $\frac{1}{3}$ I think there is an issue in identifying $P(EF)$, what should that be and why? Identifying F is easy because that is what is given to have occurred.
Added: Is $EF$ equivalent to $E \cap F$?
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The mistake in your work is that while the question asks us to find that the numbers on the dice are different, in writing the sample space for event $E $, you have also included $(6,6) $ where the numbers are not different. Correct that and we will get, $$P=\frac{\frac{10}{36}}{\frac {5}{6}} =\frac {1}{3} $$ Hope it helps.
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|
If $q^k n^2$ is an odd perfect number with Euler prime $q$, does $I(n^2) \geq 5/3$ imply $k=1$? Let $\sigma(x)$ denote the sum of the divisors of $x \in \mathbb{N}$. Denote the abundancy index of $y \in \mathbb{N}$ by $I(y)=\sigma(y)/y$.
If $\sigma(N)=2N$, then $N$ is said to be perfect.
Euler proved that every odd perfect number has the form $q^k n^2$ where $q$ is prime (called the Euler prime) with $q \equiv k \equiv 1 \pmod 4$ and $\gcd(q,n)=1$.
If $k=1$, then since $q \geq 5$, then we have
$$I(q)I(n^2)=2 \Longleftrightarrow I(n^2)=\dfrac{2q}{q+1} \geq \dfrac{5}{3}.$$
If $k>1$, then we can use the bound
$$I(q^k) < \dfrac{q}{q-1} \leq \dfrac{5}{4}$$
so that
$$I(n^2)=\dfrac{2}{I(q^k)} > \dfrac{2(q-1)}{q} \geq \dfrac{8}{5}.$$
Here is my question:
If $q^k n^2$ is an odd perfect number with Euler prime $q$, does $I(n^2) \geq 5/3$ imply $k=1$?
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We consider three cases:
Case 1 $I(n^2) = 5/3$
This implies that
$$2 - I(n^2) = \frac{2n^2 - \sigma(n^2)}{n^2} = 2 - \frac{5}{3} = \frac{1}{3},$$
which implies that $(2n^2 - \sigma(n^2)) \mid n^2$, since
$$\frac{n^2}{2n^2 - \sigma(n^2)} = 3$$
then holds. This means that $n^2$ is deficient-perfect, which is true if and only if $k=1$ (Dris [2017], pages 17 to 18).
Case 2 $I(n^2) < 5/3$
Since $k = 1 \implies I(n^2) \geq 5/3$ holds, by the contrapositive, we have
$$I(n^2) < 5/3 \implies k > 1.$$
Thus, $k > 1$.
Case 3 $I(n^2) > 5/3$
Since we have the inequality
$$I(n^2) \leq \frac{2q}{q+1}$$
with equality if and only if $k=1$, then we have $q>5$ under this case.
Next, since $q$ is the special/Euler prime (satisfying $q \equiv k \equiv 1 \pmod 4$ and $\gcd(q,n)=1$), then $q>5$ implies that $q \geq 13$.
So we now have
$$I(q^k) < \frac{q}{q-1} \leq \frac{13}{12}$$
and therefore that
$$I(n^2) = \frac{2}{I(q^k)} > \frac{24}{13}.$$
From the other answer, we have
$$I(n^2) = \dfrac{2{q^k}(q - 1)}{q^{k+1} - 1} = 2 - 2\dfrac{q^k - 1}{q^{k+1} - 1}.$$
Thus, we obtain
$$\frac{2}{13} > 2 - I(n^2) = 2\dfrac{q^k - 1}{q^{k+1} - 1}$$
$$q^{k+1} - 1 > 13q^k - 13$$
$$q^{k+1} - 13q^k + 12 > 0.$$
This last inequality does not force $k=1$, and neither proves that $k>1$.
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|
How do you compute the $\gcd(1+n+n^2,1+n+n^2+s+2ns+s^2)$ I would like to prove the following claim which I think is true:
Claim: Let $n,$ $m$ and $s$ be positive numbers. Fix $s$, then for every positive number $n$ the $\gcd(1+n+n^2,1+n+s+n^2+2ns+s^2)$ will be equal to a divisor of $1+5s^2+s^4.$
For example for every positive number $n$ if we set $s=8$ the $\gcd(1+n+n^2,73+n+16n+n^2)=1,7,631$ or $4417$. We can see that $1+5*8^2+8^4=4417$ and $4417=7*631$.
I came to the claim by moving numbers around in GAP. If it is wrong a counter example would be awesome.
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Take $n=1$ and $s=3$. Then
$$\displaylines{
1+n+n^2=3\cr
1+n+n^2+s+2ns+s^2=21\cr
\gcd(3,21)=3\cr
1+5s^2+s^4=127\cr
3\ \hbox{is not a factor of}\ 127\ .\cr}$$
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How to compute the sum $\sum_{n=0}^\infty \tfrac{n^2}{2^n}$?
How to find this sum :
$\sum_{n=0}^\infty \dfrac{n^2}{2^n}$
$\sum_{n=0}^\infty \dfrac{n^2}{2^n}=\dfrac{1}{2}+\dfrac{4}{4}+\dfrac{9}{8}+\dfrac{16}{16}+\dfrac{25}{32}+\dfrac{36}{64}+\dfrac{49}{128}+\dots$
Now $\sum_{n=0}^\infty \dfrac{n}{2^n}\leqslant \sum_{n=0}^\infty \dfrac{n^2}{2^n}$
And I know that $\sum_{n=0}^\infty \dfrac{n}{2^n}=2$.
But how to find this sum ? I am confused.Please give some hints.
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Interesting... let's look at it.
$S=\frac{1}{2}+\frac{4}{4}+\frac{9}{8}+...$
$\frac{S}{2}=\frac{1}{4}+\frac{4}{8}+\frac{9}{16}+...$
$\frac{S}{2}=\frac{1}{2}+\frac{3}{4}+\frac{5}{8}+...$
$\frac{S}{4}=\frac{1}{4}+\frac{3}{8}+\frac{5}{16}+...$
$\frac{S}{4}=\frac{1}{2}+\frac{2}{4}+\frac{2}{8}+...=\frac{1}{2}+1=\frac{3}{2}$
So $S=6.$
EDIT: According to the comments this is wrong. I'll check over it (?)
EDIT2: Apparently that solution (which is for $\Sigma_{n=0}^{\infty}\frac{n}{2^n}$) gives 2 for the original summand. I can't read, oops. :\
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|
Help for series calculation $\sum_{n\ge1} \frac{1}{4n^3-n}$ I want to find the following series.
$$\sum_{n\ge1} \frac{1}{4n^3-n}$$
So, fisrt of all, patial fraction : $$\frac{1}{4n^3-n}=\frac{1}{2n+1}+\frac{1}{2n-1}-\frac{1}{n}.$$
Next, I consider the geometric series $$\sum_{n\ge1} x^{2n-2}+x^{2n}-x^{n-1}=\frac{1}{1-x^2}+\frac{x^2}{1-x^2}-\frac{1}{1-x}\tag{1}$$ where ${-1\lt x\lt1}$.
Then, I integrate both side of (1) from $0$ to $1$:
$$\sum_{n\ge1}\frac{1}{2n-1}+\frac{1}{2n+1}-\frac{1}{n}=\int_0^1 \frac{x^2-x}{1-x^2}dx$$
Finally, I get the value that is equal to $ln(2)-1$ by improper integral but it's less than $0$. What did I do wrong? All help is appreciated.
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Following a comment of mine above:
*
*without checking too carefully, the issue in your step most likely resides in the "I integrate both sides of (1) from 0 to 1." You then integrate the LHS termwise -- why can you do that? The interval of convergence of your power series is $1$, so it is not obvious you can swap $\sum_n$ and $\int_0^1$. And, as it turns out, worse than "not obvious" it appears to be wrong.
*Now, another approach to compute the sum: starting with your $(1)$, we write, for $N\geq 1$,
$$\begin{align}
\sum_{n=1}^N a_n &= \sum_{n=1}^N \frac{1}{2n+1}+\sum_{n=1}^N \frac{1}{2n-1}-\sum_{n=1}^N \frac{1}{n}\\
&= \sum_{n=1}^N \left(\frac{1}{2n+1}+\frac{1}{2n}\right)+\sum_{n=1}^N \left(\frac{1}{2n-1}+\frac{1}{2n}\right)-2\sum_{n=1}^N \frac{1}{n} \\
&= \sum_{n=2}^{2N+1} \frac{1}{n}+\sum_{n=1}^{2N} \frac{1}{n}-2\sum_{n=1}^N \frac{1}{n} \\
&= H_{2N+1}-1+H_{2N}-2H_N \\
&= \ln(2N+1)+\ln(2N)-2\ln N - 1 +o(1)\\
&= 2\ln 2 - 1 + \ln(N+\frac{1}{2})+ \ln(N+\frac{1}{2})-2\ln N +o(1)\\
&= 2\ln 2 - 1 + \ln\frac{N+\frac{1}{2}}{N}+ \ln\frac{N-\frac{1}{2}}{N} +o(1)\\
&\xrightarrow[N\to\infty]{} \boxed{2\ln 2 -1}
\end{align}$$
where $H_N = \sum_{n=1}^N \frac{1}{n}$ denotes the $N$-th Harmonic number, which satisfies
$$
H_N = \ln N + \gamma + o(1)
$$
when $N\to\infty$ ($\gamma$ being Euler's constant). We also used that by continuity of the logarithm, $\ln\frac{N\pm\frac{1}{2}}{N} \xrightarrow[N\to\infty]{} \ln 1=0$.
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What is the Laurent expansion of $f(z)=\frac{1}{(z-1)(z-2)(z-3)}$ for $1<|z|<2$ Find the Laurent expansion for the function $f(z)$ in the domains $i),ii)$
$f(z)=\frac{1}{(z-1)(z-2)(z-3)}$
$i)$ $0<|z|<1$
$ii)$ $1<|z|<2$
What I did for the first:
$f(z)=\frac{1}{(z-1)(z-2)(z-3)}=\frac{-1}{(z-2)}+\frac{1}{2(z-3)}-\frac{1}{2}\sum_{n=0}^\infty z^n$
Since $|z|<1$
Now, I don't know if this is correct, and also, I don't know how to do $ii)$
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1. Laurent expasions of $\; f_1(z)=\dfrac{1}{z-1}$
$$\frac{1}{z-1}=\begin{cases} -\dfrac{1}{1-z}=-\displaystyle\sum_{n=0}^{+\infty}z^n,\quad |z|<1,\quad (L_1) \\\dfrac{1}{z}\dfrac{1}{1-\dfrac{1}{z}}= \dfrac{1}{z}\displaystyle\sum_{n=0}^{+\infty}\left(\frac{1}{z}\right)^n=\sum_{n=0}^{+\infty}\frac{1}{z^{n+1}},\quad |z|>1,\quad (L_2)\end{cases}$$
2. Laurent expasions of $\;f_2(z)=\dfrac{1}{z-2}$
$$\frac{1}{z-2}=\begin{cases} -\dfrac{1}{2}\dfrac{1}{1-\dfrac{z}{2}}=-\dfrac{1}{2}\displaystyle\sum_{n=0}^{+\infty}\left(\dfrac{z}{2}\right)^n=-\displaystyle\sum_{n=0}^{+\infty}\dfrac{z^n}{2^{n+1}},\quad |z|<2,\quad (L_3) \\\dfrac{1}{z}\dfrac{1}{1-\dfrac{2}{z}}= \dfrac{1}{z}\displaystyle\sum_{n=0}^{+\infty}\left(\frac{2}{z}\right)^n=\sum_{n=0}^{+\infty}\frac{2^n}{z^{n+1}},\quad |z|>2,\quad (L_4)\end{cases}$$
3. Laurent expasions of $\;f_3(z)=\dfrac{1}{z-3}$
$$\frac{1}{z-3}=\begin{cases} -\dfrac{1}{3}\dfrac{1}{1-\dfrac{z}{3}}=-\dfrac{1}{3}\displaystyle\sum_{n=0}^{+\infty}\left(\dfrac{z}{3}\right)^n=-\displaystyle\sum_{n=0}^{+\infty}\dfrac{z^n}{3^{n+1}},\quad |z|<3,\quad (L_5) \\\dfrac{1}{z}\dfrac{1}{1-\dfrac{3}{z}}= \dfrac{1}{z}\displaystyle\sum_{n=0}^{+\infty}\left(\frac{3}{z}\right)^n=\sum_{n=0}^{+\infty}\frac{3^n}{z^{n+1}},\quad |z|>3,\quad (L_6)\end{cases}$$
4. Laurent expasions of $\;f(z)=\dfrac{1}{(z-1)(z-2)(z-3)}=\dfrac{1}{2}f_1(z)-\dfrac{1}{2}f_2(z)+\dfrac{1}{2}f_3(z).$
(i) $\;f(z)=\dfrac{1}{2}(L_1)-\dfrac{1}{2}(L_3)+\dfrac{1}{2}(L_5)\quad \text{ valid for }0<|z|<1$
(ii) $\;f(z)=\dfrac{1}{2}(L_2)-\dfrac{1}{2}(L_3)+\dfrac{1}{2}(L_5)\quad \text{ valid for }1<|z|<2$
Etc. for other possible regions.
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Derivative of arcsin In my assignment I need to analyze the function
$f(x)=\arcsin \frac{1-x^2}{1+x^2}$
And so I need to do the first derivative and my result is:
$-\dfrac{4x}{\left(x^2+1\right)^2\sqrt{1-\frac{\left(1-x^2\right)^2}{\left(x^2+1\right)^2}}}$
But in the solution of this assignment it says
$f'(x)=-\frac{2x}{|x|(1+x^2)}$
I don't understand how they get this. I checked my answer on online calculator and it is the same.
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Use that $(1+x^2)^2\sqrt{1-\frac{(1-x^2)^2}{(1+x^2)^2}}=(1+x^2)\sqrt{(1+x^2)^2-\frac{(1+x^2)^2(1-x^2)^2}{(1+x^2)^2}}=(1+x^2)\sqrt{((1+x^2)+(1-x^2))((1+x^2)-(1-x^2))}=(1+x^2)\sqrt{2\cdot2x^2}$
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Find all possible positive integer $n$ such that $3^{n-1} + 5^{n-1} \mid 3^n + 5^n $
Find all possible positive integers $n$ such that $3^{n-1} + 5^{n-1} \mid 3^n + 5^n $.
I don't know how to start with. Any hint or full solution will be helpful.
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Let $d$ be a positive integer such that $d$ divides both $3^{n-1}+5^{n-1}$ and $3^{n}+5^{n}$.
then d divides $5\cdot (3^{n-1}+5^{n-1})-( 3^{n}+5^{n} )= 3^{n-1}\cdot 2$, and
$3\cdot(3^{n-1}+5^{n-1})-( 3^{n}+5^{n} )= -5^{n-1}\cdot 2$
Thus, as $\gcd(5,3)=1$, it follows that $d$ is either $1$ ou $2$.
Since $3^{n-1}+5^{n-1}$ divides $3^{n}+5^{n}$, we must have
$d=3^{n-1}+5^{n-1}=2$, which gives $n=1$.
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Explain why $(aβb)^2 = a^2 βb^2$ if and only if $b = 0$ or $b = a$. This is a question out of "Precalculus: A Prelude to Calculus" second edition by Sheldon Axler. on page 19 problem number 54.
The problem is Explain why $(aβb)^2 = a^2 βb^2 $ if and only if $b = 0$ or $b = a$.
So I started by expanding $(aβb)^2$ to $(aβb)^2 = (a-b)(a-b) = a^2 -2ab +b^2$. To Prove that $(aβb)^2 = a^2 βb^2 $ if b = 0 I substituted b with zero both in the expanded expression and the original simplified and I got $(aβb)^2 = (a-0)^2 = (a-0)(a-0) = a^2 - a(0)-a(0)+0^2 = a^2$ and the same with $a^2 -2ab +b^2$ which resulted in $a^2 - 2a(0) + 0^2 = 2a$ or if I do not substite the $b^2$ I end up with $a^2 + b^2$. That's what I got when I try to prove the expression true for $b=0$.
As for the part where $b=a$, $(aβb)^2 = (a-b)(a-b) = a^2-2ab+b^2$, if a and b are equal, let $a=b=x$ and I substite $a^2-2ab+b^2 = x^2-2(x)(x) + x^2 = x^2-2x^2+x^2 = 1-2+1=0$ I do not see where any of this can be reduced to $a^2-b^2$ unless that equals zero......I do see where it holds but I do not see how would a solution writting out look.After typing this it seems a lot clearer but I just can't see how to phrase a "solution".
P.S: This is my first time asking a question here so whatever I did wrong I am sorry in advance and appreciate the feedback.
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we have $$(a-b)^2=a^2-b^2$$ we know that $$(a-b)^2=a^2+b^2-2ab$$ and if we use the first equation we get
$$2b^2-2ab=0$$ thus we get $$b=0$$ or $$b=a$$
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How can I find $\sum_{cyc}\sin x\sin y$ $x$, $y$ & $z$ are real number such that
$$\frac{\sin{x}+\sin{y}+\sin{z}}{\sin{(x+y+z)}}=\frac{\cos{x}+\cos{y}+\cos{z}}{\cos{(x+y+z)}}=2$$
find the value
$$\sum_{cyc}\sin x\sin y$$
All help would be appreciated.
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I used the Blue's beautiful idea.
Let $e^{ix}=a$, $e^{iy}=b$ and $e^{iz}=c$.
Hence, $\sin{x}=\frac{a-\frac{1}{a}}{2i}=\frac{a^2-1}{2ai}$, $\cos{x}=\frac{a^2+1}{2a}$, $\sin{y}=\frac{b^2-1}{2bi}$, $\sin{x}=\frac{c^2-1}{2ci}$ and $\cos{x}=\frac{c^2+1}{2c}$.
Thus, $\sum\limits_{cyc}\sin{x}=2\sin(x+y+z)$ gives $\sum\limits_{cyc}(a^2bc-ab)=2(a^2b^2c^2-1)$ and
$\sum\limits_{cyc}\cos{x}=2\cos(x+y+z)$ gives $\sum\limits_{cyc}(a^2bc+ab)=2(a^2b^2c^2+1)$ or
$ab+ac+bc=2$ and $a+b+c=2abc$.
Thus, $$-\sum\limits_{cyc}\sin{x}\sin{y}=\sum_{cyc}\frac{(a^2-1)(b^2-1)}{4ab}=\frac{\sum\limits_{cyc}c(a^2-1)(b^2-1)}{4abc}=$$
$$=\frac{abc(ab+ac+bc)-\sum\limits_{cyc}(a^2b+a^2c)+a+b+c}{4abc}=\frac{abc(ab+ac+bc)-(a+b+c)(ab+ac+bc)+3abc+a+b+c}{4abc}=\frac{1}{2}-1+\frac{3}{4}+\frac{1}{2}=\frac{3}{4}$$
|
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|
The ratio of their $n$-th term. The sum of $n$ terms of two arithmetic series are in the ratio of $(7n+ 1) : (4n+ 27)$. We have to find the ratio of their $n$-th term.
οΏ½
I tried to find the ratio by using the formula of summation of A.P.
But it becomes too long due to many variables that is $a_1,a_2,d_1,d_2$
|
Let $a_1$, $a_2$ be the first terms and $d_1$, $d_2$ the common differences of the given APs. Then, their sum of $n$ terms are given by: $S_{n}=\frac{n}{2}\left[2 a_{1}+(n-1) d_{1}\right]$ and $S_{n}^{\prime}=\frac{n}{2}\left[2 a_{2}+(n-1) d_{2}\right]$
On dividing, we get:
$$
\frac{2 a_{1}+(n-1) d_{1}}{2 a_{2}+(n-1) d_{2}}=\frac{(7 n+1)}{(4 n+27)}
$$
$\therefore$ The ratio of their 11 th terms $=\frac{a_{1}+10 d_{1}}{a_{2}+10 d_{2}}=\frac{2 a_{1}+20 d_{1}}{2 a_{2}+20 d_{2}}$
$=\frac{2 a_{1}+(21-1) d_{1}}{2 a_{2}+(21-1) d_{2}}=\frac{(7 \times 21+1)}{(4 \times 21+27)}=\frac{148}{111} \quad[$ Putting $n=21$ in (iii) ]
Hence, the required ratio is $148: 111$.
|
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|
Analytic Proof of alternating series Having some problems figuring out the following proof. If we have a series $a_n$ that is both nonnegative and decreasing. Then if we consider the corresponding alternating series $$\sum (-1)^{n+1}a_n$$ I need to prove that the sequence of odd partial sums $$s_{2n+1}= \sum _{k=1}^{2n+1} (-1)^{k+1}a_k $$ is bounded and decreasing.
So if we start to write out some terms of this series we see the following,
$(-1)^2a_1+(-1)^3a_2+(-1)^5a_4+...+(-1)^{2n+2}a_k$
So the terms seem to always equal 1 but how do I show that this is bounded and decreasing?
|
Let
$A_n
=\sum_{k=1}^n (-1)^{k+1}a_k
$.
Then
$\begin{array}\\
A_{n+2}
&=\sum_{k=1}^{n+2} (-1)^{k+1}a_k\\
&=\sum_{k=1}^{n} (-1)^{k+1}a_k+(-1)^{n+2}a_{n+1}+(-1)^{n+3}a_{n+2}\\
&=A_n+(-1)^{n+2}(a_{n+1}-a_{n+2})\\
\end{array}
$
Putting $2n$
for $n$,
$\begin{array}\\
A_{2n+2}
&=A_{2n}+(-1)^{2n+2}(a_{2n+1}-a_{2n+2})\\
&=A_{2n}+(a_{2n+1}-a_{2n+2})\\
&>A_{2n}
\qquad\text{since }a_{2n+1}>a_{2n+2}\\
\end{array}
$
Putting $2n+1$
for $n$,
$\begin{array}\\
A_{2n+3}
&=A_{2n+1}+(-1)^{2n+3}(a_{2n+2}-a_{2n+3})\\
&=A_{2n+1}-(a_{2n+2}-a_{2n+3})\\
&<A_{2n+1}
\qquad\text{since }a_{2n+2}>a_{2n+3}\\
\end{array}
$
Therefore
the even terms are increasing
and
the odd terms are decreasing.
To show that
the odd terms are bounded below,
$\begin{array}\\
A_{2n+1}
&=\sum_{k=1}^{2n+1} (-1)^{k+1}a_k\\
&=\sum_{k=1}^{2n} (-1)^{k+1}a_k+(-1)^{2n+2}a_{2n+1}\\
&=\sum_{k=1}^{n} ((-1)^{2k}a_{2k-1}+(-1)^{2k+1}a_{2k})+a_{2n+1}\\
&=\sum_{k=1}^{n} (-1)^{2k}(a_{2k-1}-a_{2k})+a_{2n+1}\\
&=\sum_{k=1}^{n} (a_{2k-1}-a_{2k})+a_{2n+1}\\
&> 0\\
\end{array}
$
since all terms are positive.
To show that
the even terms are bounded above,
$\begin{array}\\
A_{2n+1}-A_{2n}
&=(-1)^{2n+2}a_{2n+1}\\
&=a_{2n+1}\\
&> 0\\
\end{array}
$
so
$A_{2n}
< A_{2n+1}
$.
Since the odd terms
are bounded below and decreasing,
the even terms are
bounded above.
|
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|
If $N = \frac{{n_1}(n_1+1)}{2}\cdot{d_1}$, is it possible to have $N = \frac{{n_2}(n_2+1)}{2}\cdot{d_2}$? If $$N = \frac{{n_1}(n_1+1)}{2}\cdot{d_1},$$
is it possible to have $$N = \frac{{n_2}(n_2+1)}{2}\cdot{d_2}?$$
Here, $d_i, n_i \in \mathbb{N}$, $d_i > 1$ and $n_1 \neq n_2$.
That is,
Question 1
If a number can be represented as a nontrivial multiple of a triangular number, can it then be represented as a nontrivial multiple of another triangular number?
Added January 27 2017
and
Question 2
If such a number can be so represented (as in Question 1), must it necessarily be (a nontrivial multiple of or) the least common multiple of the triangular numbers
$$T(n_1)=\dfrac{{n_1}(n_1 + 1)}{2}$$
and
$$T(n_2)=\dfrac{{n_2}(n_2 + 1)}{2}?$$
Thanks to users Alex Macedo, dxiv, and Stahl for their initial comments!
|
Let $\dfrac{n_1(n_1+1)}{n_2(n_2+1)} = \dfrac{d_2}{d_1}$
It follows that
$\dfrac{{n_1}(n_1+1)}{2}\cdot{d_1} = \dfrac{{n_2}(n_2+1)}{2}\cdot{d_2}$
Example $T(6) = 21$ and $T(8)=36$. So $\dfrac{T(6)}{T(8)} = \dfrac{7}{12}$
Hence
$\dfrac{{6}(6+1)}{2}\cdot{12} = \dfrac{{8}(8+1)}{2}\cdot{7}$
If $n_1$ and $d_1$ are given {It would have been nice if you were a bit more explicit}, then you want to find $n$ and $d$ that solve
$\dfrac{{n}(n+1)}{2}\cdot{d} = \dfrac{n_1(n_1+1)}{2}\cdot{d_1}$
Which can be expressed as
$n^2 + n - \dfrac{n_1(n_1+1) d_1}{d} = 0$
and can be solved by inspection.
Example
\begin{align}
\dfrac{4(4+1)}{2}\cdot{3} &= \dfrac{{n}(n+1)}{2}\cdot{d} \\
n^2 + n - \dfrac{60}{d} &= 0 \\
n &= \dfrac{-1 + \sqrt{1+\dfrac{240}{d}}}{2} \\
\end{align}
The possible integer values of $\dfrac{240}{d}$ are
$$\{1, 2, 3, 4, 5, 6, 8, 10, 12, 15, 16, 20, 24, 30, 40, 48, 60, 80, 120, 240\}$$
We note that $1+\{3, 8, 15, 24, 48, 80, 120\}$ are perfect squares.
Which leads to $d \in \{80, 30, 16, 10, 5, 3, 2\}$
and $n \in \left\{\dfrac 12, 1, \dfrac 32, 2, 3, 4, 5\right\}$.
Removing the fractions, we end up with
$(n,d) = \{(1,30), (2,10), (3,5), (4,3), (5,2) \}$
|
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|
Find the value of $a^4-a^3+a^2+2$ when $a^2+2=2a$ Find the value of $a^4-a^3+a^2+2$ when $a^2+2=2a$
My Attempt,
$$a^4-a^3+a^2+2=a^4-a^3+2a$$
$$=a(a^3-a^2+2)$$
What's next?
|
Hint $\ $ We seek the value of $\,f(a)\,$ given that $\, g(a) = a^2-2a+2 = 0.\,$ By the Polynomial Division Algorithm we can write $\ f = q\,g + r\,$ so $\, g=0\,\Rightarrow\, f = r = f\bmod g.\,$ So it suffices to compute the remainder $\, r = (f\bmod g).\,$ Doing so easily yields $\,f = r = 0.$
Because $\, g = 0\,$ we can perform all arithmetic modulo $g.\,$ This is the point of modular arithmetic - when we are interested only in the remainder then we can ignore the quotients. This is exactly what is done (implicitly) in Stefan4024's answer.
|
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|
Angle between two lines explanation I gave two lines $q_1 = 2x - y + 2 = 0$ and $q_2 = x + 2y - 3 = 0$.
They have vectors $n_1 = (2,1)$ and $n_2 = (1,2)$.
When I have to find angle between them I must apply that formula:
$$\cos \theta = \frac{2\cdot 1 + (-1)\cdot2}{\sqrt{4 + 1}\sqrt{1 + 4}} = 0 \Rightarrow \theta = 90^\circ$$
Replacing $n_1$ and $n_2$ is clear to me, but can you please explain how I get $90^\circ$ from solving the formula. Thank you.
Where those $\sqrt{4 + 1}\sqrt{1 + 4}$ comes from?
If $ \theta $ is $90^ \circ$ at $0$, how much degrees it will be if it is $\dfrac{1}{5}$?
|
Let $\theta$ be an angle.
$\cos \theta = \frac{2.1 + (-1).2}{\sqrt 5. \sqrt 5}$
$\cos \theta = \frac{2 - 2}{\sqrt 5. \sqrt 5}$
$\cos \theta = \frac{0}{5}$
$\cos \theta = 0$
$\cos \theta = \cos 90Β°$
$\theta = 90Β°$
|
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|
Integral of product of two error complementary functions (erfc) Could you please help me to show that the integral
$$
\int_0^{\infty} \mathrm{erfc}(ax) \, \mathrm{erfc}(bx)\, \mathrm{d}x
$$
is equal to
$$
\frac{1}{ab\sqrt{\pi}} (a+b-\sqrt{a^2+b^2}),
$$
where
$$
\mathrm{erfc}(y)=\frac{2}{\sqrt{\pi}} \int_y^{\infty} \exp(-t^2)\, \mathrm{d} t.
$$
I have tried to expand the integral as
$$
\frac{4}{\pi }\int_0^{\infty} \int_{ax}^{\infty} \int_{bx}^{\infty} \exp(-t^2 -s^2) \, \mathrm{d}s \, \mathrm{d}t \, \mathrm{d} x
$$
but I could not come up with the right change of variables. Any ideas on how to proceed?
Thank you in advance!
|
A multiple integral approach. Since
$$
\int_{0}^{\infty}x^2 e^{-\alpha x^2}\mathrm{d}x \overset{\alpha x^2 \to x}{=} \frac{1}{2\alpha^{\frac{3}{2}} }\int_{0}^{\infty} x^{\frac{3}{2}-1}e^{-x} \mathrm{d}x = \frac{1}{2\alpha^{\frac{3}{2}} } \Gamma\left(\frac{3}{2}\right) = \frac{\sqrt{\pi}}{4\alpha^{\frac{3}{2}} }
$$
We get
\begin{align}
\int_0^\infty \operatorname{erfc}(ax) \operatorname{erfc}(bx)\, \mathrm{d}x & =\int_0^\infty \left( \frac{2}{\sqrt{\pi}}\int_1^\infty ax e^{-a^2x^2z^2} \, \mathrm{d}z\right)\left( \frac{2}{\sqrt{\pi}}\int_1^\infty bx e^{-b^2x^2y^2}\, \mathrm{d}y\right) \mathrm{d}x\\
& =\frac{4ab}{\pi} \int_1^\infty \int_1^\infty \int_0^\infty x^2 e^{-(a^2z^2 +b^2y^2)x^2} \, \mathrm{d}x\, \mathrm{d}z\, \mathrm{d}y\\
& =\frac{ab}{\sqrt{\pi}} \int_1^\infty \int_1^\infty \frac{1}{\left(a^2z^2 +b^2y^2 \right)^{\frac32}} \, \mathrm{d}z\, \mathrm{d}y\\
& \overset{z\to \tan(\theta)by/a}{=}\frac{ab}{\sqrt{\pi}} \int_1^\infty \frac{1}{ab^2y^2} - \frac{1}{y^2b^2\sqrt{y^2b^2+a^2}} \, \mathrm{d}y\\
& =\frac{ab}{\sqrt{\pi}}\left(\frac{1}{ab^2} - \frac{1}{b^2}\int_1^\infty \frac{1}{y^2\sqrt{b^2y^2+a^2}} \, \mathrm{d}y\right)\\
& \overset{y\to\tan(\varphi)a/b }{=}\frac{a}{b\sqrt{\pi}}\left(\frac{1}{a} - \frac{\sqrt{a^2+b^2} - b}{a^2} \right)\\
& = \frac{a+b - \sqrt{a^2+b^2}}{ab\sqrt{\pi}}
\end{align}
|
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|
Find a formula for $\cos(5x)$ in terms of $\sin(x)$ and $\cos(x)$ I was asked to find a formula for $\cos(5x)$ in terms of $\sin(x)$ and $\cos(x)$.
I tried to use the formula $\cos(5x) + i\sin(5x) = (\cos(x)+i\sin(x))^5
$ and what I get is
$16i\sin^5(x) - 20i\sin^3(x) + 5i\sin(x) + \cos(x) + 16 \sin^4(x) \cos(x) - 12 \sin^2(x) \cos(x)$
But how do I deal with the $i\sin(5x)$? Because I only can use $\sin(x)$ and $\cos(x)$ to express $\cos(5x)$. Thank you for your help!
|
Your way is right. But I think it should be
$$(\cos x + i\sin x)^5 = \cos^5 x + 5i\cos^4x\sin x - 10\cos^3x\sin^2 x - 10i\cos^2x\sin^3x + 5\cos x\sin^4x +i\sin^5x$$
So, you have
$$\cos 5x = \cos^5x-10\cos^3x\sin^2x+5\cos x\sin^4x$$
$$\sin 5x = \sin^5x-10\cos^2x\sin^3x+5\cos^4 x\sin x$$
|
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|
How do I solve $(x+1)^5 +36x+36 = 13(x+1)^3$? I tried $$(x+1)^5 + 36 x + 36 = 13 (x +1)^3\\
(x+1)^5 + 36(x+1) = 13 (x +1)^3\\
(x+1)^4 +36 = 13 (x+1)^2
$$
But, don't understand how to solve further. Can somebody show step by step please. Thanks!
|
It's a simple problem. Well
$$(x+1)^5+36x+36=13(x+1)^3$$
$$\implies (x+1)^5β13(x+1)^3+36(x+1) = 0$$
Let $y=x+1$
$$\implies y^5-13y^3+36y = 0$$
$$\implies y(y^4-13y^2+36)= 0$$
$$\implies y(y^2-9)(y^2-4)= 0$$
$$\implies y = 0,\pm 2,\pm 3$$
$$\implies x+1 = 0,\pm 2,\pm 3$$
$$\implies x = -4, -3, -1, 1, 2$$
and weβre done.
|
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|
Prove the $\frac {2r+5}{r+2}$ is always a better approximation of $\sqrt {5}$ than $r$.
Prove the $\frac {2r+5}{r+2}$ is always a better approximation of $\sqrt {5}$ than $r$.
SOURCE : Inequalities (Page Number 4 ; Question Number 207)
I tried a lot of approaches, but without success.
I rewrote $\frac {2r+5}{r+2}$ as $2 + \frac {1}{r+2}$.
$\sqrt {5} \approx2.2360679775$
Equating $\frac {1}{r+2}$ and $0.2360679774$ , I get $r=2.23606797929$.
So, $\frac {2r+5}{r+2}$ is a still a better approximation than $r$.
How to proceed ?
Any hints/ideas/pointers ?
|
Rather than doing manipulations with $\sqrt{5}$ it is better to use just the rationals. We can see that $$\left(\frac{2r+5}{r+2}\right)^{2}-5=\frac{4r^{2}+20r+25-5r^{2}-20r-20}{(r+2)^{2}}=\frac{5-r^{2}}{(r+2)^{2}}$$ Since $r>0$ it follows that $(2r+5)/(r+2)$ is a better approximation to $\sqrt{5}$ than $r$, but in a different direction.
The real benefit comes from iterating this procedure two times to get $$\dfrac{2\cdot\dfrac{2r+5}{r+2}+5}{\dfrac{2r+5}{r+2}+2}=\frac{9r +20}{4r+9}$$ so that $(9r+20)/(4r+9)$ is a better approximation to $\sqrt{5}$ and that too in same direction.
|
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|
How is the infinite sum of a series calculated by symbolic math? I wonder how Wolfram can solve this series and provide the solution symbolically:
$$\sum_{k=1}^\infty\frac 1{(2k-1)^4}$$
In this particular case I know how to use a Fourier series on a triangle function to get the result by employing Parseval's theorem, but this is only a particular example. The proof for $\sum_{k=1}^{k=\infty} \frac{1}{k^2} = \frac{\pi^2}{6}$ was found by Euler and uses a taylor series of a special function.
But is there a recipe working correctly for each possible series? I cannot imagine that such algorithm exists. But how can Wolfram do it?
|
I do not know how symbolic programs evaluate series in closed form. But I thought it might be instructive to see one methodology to solve the problem of interest. It is to that end we now proceed.
Note that $\zeta(4)=\sum_{n=1}^\infty\frac{1}{n^4}$. Next, we can write $\zeta(4)$ as
$$\begin{align}
\zeta(4)&=\sum_{n=1}^\infty\frac{1}{n^4}\\\\\
&=\sum_{n=1}^\infty\frac{1}{(2n-1)^4}+\sum_{n=1}^\infty\frac{1}{(2n)^4}\tag 1
\end{align}$$
by pairing even and odd terms. But $\sum_{n=1}^\infty\frac{1}{(2n)^4}=\frac1{16}\zeta(4)$. Using this in $(1)$ and solving for the term of interest reveals
$$\sum_{n=1}^\infty\frac{1}{(2n-1)^4}=\frac{15}{16}\zeta(4)$$
FINDING $\displaystyle \zeta(4)$:
To find $\zeta(4)$, we can use contour integration. Let $f(z)=\frac{\pi \cot(\pi z)}{z^4}$ and $C$ be a circular contour of radius $N+1/2$. Then, we have
$$\begin{align}
\oint_C \frac{\pi \cot(\pi z)}{z^4}\,dz&=2\pi i \sum_{n=-N}^N \text{Res}\left(\frac{\pi \cot(\pi z)}{z^4}, z=n\right)\\\\
&=2\pi i \sum_{|n|\ge 1} \frac1{n^4}+2\pi i \text{Res}\left(\frac{\pi \cot(\pi z)}{z^4}, z=0\right) \tag 2
\end{align}$$
As $N\to \infty$, the integral in $(2)$ approaches $0$. Hence, from $(2) we have
$$\sum_{n=1}^\infty \frac1{n^4}=-\frac12 \text{Res}\left(\frac{\pi \cot(\pi z)}{z^4}, z=0\right) \tag 3$$
EVALUATING THE RESIDUE AT $0$:
Noting that $z=0$ is a fifth-order pole, we could calculate the residue by applying the formula
$$\text{Res}\left(\frac{\pi \cot(\pi z)}{z^4}, z=0\right)=\frac1{4!}\lim_{z\to 0}\frac{d^4 (\pi z \cot(\pi z))}{dz^4}
$$
Rather than pursue this approach, we expand the integrand as
$$\begin{align}
\frac{\pi\cot(\pi z)}{z^4}&= \frac{1-\frac12(\pi z)^2+\frac{1}{24}(\pi z)^4+O(z^6)}{ z^5\left(1-\frac16 (\pi z)^2+\frac{1}{120}(\pi z)^4+O(z^6)\right)}\\\\
&=\frac{1}{z^5}\left(1-\frac12(\pi z)^2+\frac{1}{24}(\pi z)^4+O(z^6)\right)\left(1+\frac16 (\pi z)^2+\frac{7}{360}(\pi z)^4+O(z^6)\right)
\end{align}$$
The residue is the coefficient on the term $z^{-1}$. Therefore,
$$\text{Res}\left(\frac{\pi \cot(\pi z)}{z^4}, z=0\right)=\left(-\frac1{12}+\frac{1}{24}+\frac{1}{360}\right)\,\pi^4=-\frac{\pi^4}{45} \tag 4$$
Using $(4)$ in $(3)$ we find that $\zeta(4)=\frac{\pi^4}{90}$. And finally, we have
$$\sum_{n=1}^\infty\frac{1}{(2n-1)^4}=\frac{\pi^4}{96}$$
And we are done!
|
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|
Find all conditions for $x$ that the equation $1\pm 2 \pm 3 \pm 4 \pm \dots \pm n=x$ has a solution. Find all conditions for $x$ so that the equation $1\pm 2 \pm 3 \pm 4 \pm \dots \pm 1395=x$ has a solution.
My attempt: $x$ cannot be odd because the left hand side is always even then we have $x=2k(k \in \mathbb{N})$ also It has a maximum and minimum
$1-2-3-4-\dots-1395\le x \le 1+2+3+4+\dots +1395$
But I can't show if these conditons are enough or make some other conditions.
|
The solution for $n$ as in the title is as follows:
Let $p = (n-3)(n+4)/2$ then the solutions for x are $-p-4, -p, -p+2, \ldots,p-2, p, p +2, p+4$ where from the term $-p$ on we go to the term $p$ each time adding $2$. So only the first and the last two terms are irregular. We can verify:
*
*for $n = 4$ we have $-8, -4, -2, 0, 2, 4, 6, 10$ with p = $(4-3)(4+4/2 = 4)$
*for $n = 5$ we have $ -13, -9, -7, -5, -3, -1, 1, 3, 5, 7, 9, 11, 15$ with $p = (5-3)(5+4)/2 = 9$
*for $n = 1395$ we will have $-973708, -973704, -973702, \text{ each time add 2 , 973704, 973706, 973710}$ with $p = (1935-3)(1935+4)/2 = 973704$
Proof: we have to show that, having established the series for $n$ then we obtain the new series by adding and subtracting $n+1$ to all previous numbers. Indeed we have that $p' = (n'-3(n'-4)/2 = p + n + 1$ for $n' = n+1$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2127922",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "13",
"answer_count": 9,
"answer_id": 7
}
|
What is $x$, if $3^x+3^{-x}=1$? I came across a really brain-racking problem.
Determine $x$, such that $3^x+3^{-x}=1$.
This is how I tried solving it:
$$3^x+\frac{1}{3^x}=1$$
$$3^{2x}+1=3^x$$
$$3^{2x}-3^x=-1$$
Let $A=3^x$.
$$A^2-A+1=0$$
$$\frac{-1Β±\sqrt{1^2-4\cdot1\cdot1}}{2\cdot1}=0$$
$$\frac{-1Β±\sqrt{-3}}{2}=0$$
I end up with
$$\frac{-1Β±i\sqrt{3}}{2}=0$$
which yields no real solution. And this is not the expected answer.
I'm a 7th grader, by the way. So, I've very limited knowledge on mathematics.
EDIT
I made one interesting observation.
$3^x+3^{-x}$ can be the middle term of a quadratic equation:
$$3^x\cdot\frac{1}{3^x}=1$$
$$3^x+3^{-x}=1$$
|
Note that the function $f(x)=3^x+3^{-x}$ has first derivative $f'(x)=3^x-3^{-x}$ and second derivative $f''(x)=3^x+3^{-x}$. Since $f''(x)>0$ always, this is a convex function that attains a global minimum wherever $f'(x)=0$, so wherever $3^x=3^{-x}$. This occurs at $x=0$, in which the function takes value $f(0)=3^0+3^0=1+1=2$. Therefore, $f(x)\geq2$ for all $x\in\mathbb{R}$ and there does not exist any $x$ such that $3^x+3^{-x}=1$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2128506",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 5,
"answer_id": 4
}
|
Conditional probability and calculation if $P(F|E)=3P(F|E^c)$
$P(E)=0.3$
$P(F)=0.4$
Then $P(E|F)=??$
How do i use the conditional probability equation that has been given in the question
|
From the first equality you wrote,
$$P(F|E) = \frac{P(F\cap E)}{P(E)} = \frac{P(F\cap E)}{.3} = 3P(F|E^{c}) = 3\frac{P(F\cap E^{c})}{P(E^{c})} = 3\frac{P(F\cap E^{c})}{.7} $$
We can rearrange some terms to get
$$P(F\cap E) = \frac{3*.3}{.7} P(F\cap E^{c}) = \frac{9}{7}P(F\cap E^{c})$$
Now, since $P(F) = P(F\cap E) + P(F\cap E^{c})$, substitute the above equality into this one to get
$$\frac{2}{5} = .4 = P(F) = P(F\cap E) + P(F\cap E^{c}) = \frac{9}{7} P(F\cap E^{c}) + P(F\cap E^{c}) = \frac{16}{7}P(F\cap E^{c})$$
Therefore $P(F\cap E^{c}) = \frac{\frac{2}{5}}{\frac{16}{7}} = \frac{14}{80} = \frac{7}{40}$. Now
$$P(F\cap E) = \frac{2}{5} - \frac{7}{40} = \frac{9}{40}$$
By definition, $P(E|F) = \frac{P(E\cap F)}{P(F)}$, which will be $\frac{\frac{9}{40}}{\frac{2}{5}} = \frac{45}{80} = \frac{9}{16}$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2130601",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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|
Find that solution satisfying $Ο(1) = 3Ο(0)$ for $y' + 5y = 2$. Find that solution satisfying $Ο(1) = 3Ο(0)$ for the following second order linear ordinary differential equation:
$y' + 5y = 2$
I found the solution to be $Ο(x) = \frac{2}{5} + ce^{5x}$.
Now how do I find a particular solution satisfying $Ο(1) = 3Ο(0)?$ Please help me with this. In the previous part, I was asked to find the solution satisfying $Ο(1) = 2$ which I found to be $Ο(x) = \frac{2}{5} + \frac{8}{5}e^5e^{-5x}$ if that helps.
|
Firstly it should be $$y'=2-5y\\ \frac { dy }{ 2-5y } =dx\\ \int { \frac { dy }{ 2-5y } } =\int { dx } \\ \int { \frac { d\left( 5y-2 \right) }{ 5y-2 } =-5\int { dx } } \\ \ln { \left| 5y-2 \right| =-5x+C } \\ 5y-2={ Ce }^{ -5x }\\$$
$$ \phi \left( x \right) =\frac { 2 }{ 5 } +{ Ce }^{ -5x }$$
then we are given $Ο(1) = 3Ο(0)$ so $$\frac { 2 }{ 5 } +C{ e }^{ -5 }=\frac { 6 }{ 5 } +3C\\ C\left( { e }^{ -5 }-3 \right) =\frac { 4 }{ 5 } \\ C=\frac { 4 }{ 5 } \left( { e }^{ -5 }-3 \right) $$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2130925",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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}
|
A closed form for $\sum _{j=0}^{\infty } -\frac{\zeta (-j)}{\Gamma (j)}$ Is there a closed form for
$$\sum _{j=0}^{\infty } -\frac{\zeta (-j)}{\Gamma (j)}$$
where $\zeta (-j)$ Zeta function and $\Gamma (j)$ Gamma function.
I tried everything, but I still can not solve it. Any Ideas?
|
The reflection formula for the Riemann zeta function gives
$$
\begin{align}
\zeta(-k)
%&=\zeta(k+1)\frac{\Gamma\!\left(\frac{k+1}2\right)}{\pi^{\frac{k+1}2}}\frac{\pi^{-\frac{k}2}}{\Gamma\!\left(-\frac{k}2\right)}\frac{\Gamma\!\left(1+\frac{k}2\right)}{\Gamma\!\left(1+\frac{k}2\right)}\\
%&=\zeta(k+1)\frac{\Gamma\!\left(\frac{k+1}2\right)\Gamma\!\left(\frac{k+2}2\right)}{\pi^{\frac{2k+3}2}}\sin(-k\pi/2)\\
&=\zeta(k+1)\frac{\Gamma(k+1)}{2^k\pi^{k+1}}\sin(-k\pi/2)\tag1
\end{align}
$$
If $k$ is an even integer, $\sin(-k\pi/2)=0$. This matches the zeroes of the zeta function at the negative even integers. Let $k=2j-1$.
$$
\zeta(1-2j)
=(-1)^j\frac{\zeta(2j)}{\pi^{2j}}\frac{(2j-1)!}{2^{2j-1}}\tag2
$$
Therefore,
$$
\begin{align}
-\sum_{j=1}^\infty\frac{\zeta(1-2j)}{\Gamma(2j-1)}
&=\sum_{j=1}^\infty(-1)^{j-1}\frac{\zeta(2j)}{\pi^{2j}}\frac{2j-1}{2^{2j-1}}\\
&=\sum_{k=1}^\infty\sum_{j=1}^\infty(-1)^{j-1}\frac{\color{#C00}{4j}-\color{#090}{2}}{(2k\pi)^{2j}}\\
&=\sum_{k=1}^\infty\left(\color{#C00}{4\frac{4k^2\pi^2}{\left(4k^2\pi^2+1\right)^2}}-\color{#090}{2\frac1{4k^2\pi^2+1}}\right)\\
&=\sum_{k=1}^\infty\left(2\frac1{4k^2\pi^2+1}-4\frac1{\left(4k^2\pi^2+1\right)^2}\right)\tag3
\end{align}
$$
Using the formula from this answer
$$
\sum_{k=1}^\infty\frac1{k^2+x^2}
=\frac{\pi x\coth(\pi x)-1}{2x^2}\tag4
$$
and its derivative times $-\frac1{2x}$
$$
\sum_{k=1}^\infty\frac1{\left(k^2+x^2\right)^2}
=\frac{\pi^2x^2\operatorname{csch}^2(\pi x)+\pi x\coth(\pi x)-2}{4x^4}\tag5
$$
we get
$$
\sum_{k=1}^\infty\frac1{4k^2\pi^2+1}
=\frac{\frac12\coth\left(\frac12\right)-1}{2}\tag6
$$
and
$$
\sum_{k=1}^\infty\frac1{\left(4k^2\pi^2+1\right)^2}
=\frac{\frac14\operatorname{csch}^2\left(\frac12\right)+\frac12\coth\left(\frac12\right)-2}{4}\tag7
$$
Applying $(6)$ and $(7)$ to $(3)$ yields
$$
\bbox[5px,border:2px solid #C0A000]{-\sum_{k=1}^\infty\frac{\zeta(-k)}{\Gamma(k)}=1-\frac14\operatorname{csch}^2\left(\frac12\right)}\tag8
$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2133424",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 2
}
|
Antiderivative of a product of a rational function and a square root. Let $0<x<1$ and $p$ be a non-negative integer.
Consider a following definite integral:
\begin{equation}
{\mathbb I}_p(x):=\int\limits_1^x \frac{(1-v^2)^p}{v^{2p+2}} \cdot \sqrt{\frac{1-v}{1+v}} dv
\end{equation}
We have computed it in a following way. Firstly we substituted for the square root and reduced the problem to integrating a rational function. Then we decomposed the resulting rational function into simple fractions and by integrating terms by term we completed the computation. The result reads:
\begin{equation}
{\mathbb I}_p(x) = (-1)^p 2 \binom{p-\frac{1}{2}}{-\frac{1}{2}} \text{arctanh}(\sqrt{\frac{1-x}{1+x}})+\frac{\sqrt{1-x^2}}{x^{2 p+1}} \cdot\sum\limits_{l=0}^{2 p} {\mathcal C}_{l,p} x^l
\end{equation}
where the coefficients read:
\begin{eqnarray}
&&{\mathcal C}_{l,p} := \sum _{q=0}^{\left\lfloor \frac{l}{2}\right\rfloor }
\frac{\sqrt{\pi } 2^{-l+2 q+1} (-l+2 p+q)! \binom{4 p+2}{l-2 q} \binom{-l+2 p+2 q+1}{2 q+1} }
{\left(-q-\frac{3}{2}\right)! (-l+2 p+2 q+1) (-l+2 p+2 q+1)!}
\, _3F_2\left(-p-1,l-4 p-2 q-2,2 q-l;-2 p-1,-2 p-\frac{1}{2};1\right)\\
&&=
\left(
\begin{array}{c}
\{-1\} \\
\left\{-\frac{1}{3},\frac{1}{2},\frac{1}{3}\right\} \\
\left\{-\frac{1}{5},\frac{1}{4},\frac{2}{5},-\frac{5}{8},-\frac{1}{5}\right\} \\
\left\{-\frac{1}{7},\frac{1}{6},\frac{3}{7},-\frac{13}{24},-\frac{3}{7},\frac{11}{16},\frac{1}{7}\right\} \\
\left\{-\frac{1}{9},\frac{1}{8},\frac{4}{9},-\frac{25}{48},-\frac{2}{3},\frac{163}{192},\frac{4}{9},-\frac{93}{128},-\frac{1}{9}\right\} \\
\left\{-\frac{1}{11},\frac{1}{10},\frac{5}{11},-\frac{41}{80},-\frac{10}{11},\frac{171}{160},\frac{10}{11},-\frac{149}{128},-\frac{5}{11},\frac{193}{256},\frac{1}{11}\right\} \\
\end{array}
\right)
\end{eqnarray}
Above we have given numerical values of those coefficients for $p=0,\dots,5$ (rows) and $l=0,\dots,2 p$ (columns). Now the question is can we come up with some closed form solution for the coefficients. Note that even at the first glance we see that in case $l=0$ and $l=p$ there is a closed form. What about other values of $l$?
|
Let us complete the calculation given above by Felix Marin. Clearly that result above is correct however it can be simplified further and what's more important expressed through elementary functions. We demonstrate here how is this done. Firstly we use the series expansion of the incomplete beta function. We have:
\begin{equation}
{\mathbb I}_p(x) = -2^{2 p+1} \left(\frac{1-x}{1+x} \right)^{p+3/2} \cdot \sum\limits_{n=0}^\infty \frac{(2+2 p)^{(n)}}{(p+3/2+n) n!} \cdot \left(\frac{1-x}{1+x}\right)^n
\end{equation}
Now, the point is that the coefficient under the sum (in front of the $n$th power)is a rational function in $n$ and as such can be decomposed into simple fractions. In fact the decomposition will consist of a polynomial of order $2 p$ and one fraction only. To be precise we have:
\begin{equation}
\frac{(2+2 p)^{(n)}}{(p+3/2+n) n!} = \frac{1}{2 p+1} \sum\limits_{j=0}^{2 p} \frac{(-p-1/2)^{(j)}}{(2p-j+1)^{(j)}} \cdot \binom{n+2 p+1}{2 p-j} - \binom{p-1/2}{2 p} \cdot \frac{1}{2 n+2 p+3}
\end{equation}
The only thing we need to do now is to multiply both sides of the above by $y^n$ (here $y:=(1-x)/(1+x)$) and sum over $n=0,1,\cdots,\infty$. Here we use the following, easily derived by means of elementary methods, identities:
\begin{eqnarray}
\sum\limits_{n=0}^\infty \binom{n+2 p+1}{2p-j} y^n &=& \sum\limits_{l=0}^{2p-j} \binom{2p+1}{j+l+1} \cdot \frac{y^l}{(1-y)^{l+1}}\\
\sum\limits_{n=0}^\infty \frac{y^n}{(2 n+2 p+3)} &=& \frac{1}{y^{p+3/2}} \left[ \text{arctanh}(\sqrt{y}) - \sum\limits_{j=0}^p \frac{(\sqrt{y})^{2 j+1}}{2j+1} \right]
\end{eqnarray}
The final result is then as follows:
\begin{eqnarray}
&&{\mathbb I}_p(x)=\\
&&2 (-1)^p \binom{p-1/2}{-1/2} \text{arctanh}(\sqrt{y})-\\
&&\frac{2^{2p+1} y^{p+3/2}}{2p+1}\sum\limits_{l=0}^{2 p} \binom{2p+1}{l+1}\, _3F_2\left(1,l-2 p,-p-\frac{1}{2};l+2,-2 p;1\right) \frac{y^l}{(1-y)^{l+1}}\\
&&-\sqrt{y} \binom{p-1/2}{-1/2} (-1)^p \sum\limits_{j=0}^p \frac{y^j}{2j+1}
\end{eqnarray}
There are two things to be noted. Firstly, this is an elementary function. Secondly, we can already see that the first term on the right hand side is identical with that in the formulation of the problem. The remaining terms need to be simplified further and then the unknown coefficients will be extracted. We will complete this task as soon as possible.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2133670",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
}
|
How to do this question that talks about dependency of x Let $x > 0$. Prove that the value of the following expression doesn't depend on x
$$\int_{0}^{x} \frac{1}{1+t^2} dt + \int_{0}^{\frac{1}{x}} \frac{1}{1+t^2}dt$$
Attempt:
Left: f'(x) = $\frac{1}{1+x^2}$
Right: f'(x) = $\frac{1}{1+(\frac{1}{x})^2} -\frac{1}{x^2}$
$= \frac{1}{(1+\frac{1}{x^2})} - \frac{1}{x^2}$
$=\frac{x^2}{1+x^2} - \frac{1}{x^2}$
$=\frac{x^4 - x^2 - 1}{1+x^2}$
Yeah I don't know what I am doing, I tried to remove the integral but failed miserably
|
To answer the question as-is after having been heavily edited, let for $x \gt 0\,$:
$$f(x) = \int_{0}^{x} \frac{1}{1+t^2} dt + \int_{0}^{\frac{1}{x}} \frac{1}{1+t^2}dt$$
Then, using the Leibniz integral rule:
$$
f'(x) = \frac{1}{1+x^2} \,-\, \frac{1}{x^2} \cdot \frac{1}{1+ \cfrac{1}{x^2}} = \frac{1}{1+x^2} \,-\, \frac{1}{1+x^2} \,=\, 0
$$
Thus $f'(x)=0\,$, so $f(x)$ is a constant, and therefore does not depend on $x$.
P.S. Note to the OP:
Right: f'(x) = $\frac{1}{1+(\frac{1}{x})^2} -\frac{1}{x^2}$
This looks like you attempted to use Leibniz' rule, but misapplied it. The $-\,\frac{1}{x^2}$ derivative of the upper bound is multiplied with, not added to, the end value of the function being integrated.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2134430",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 1
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|
Find the maximum and minimum value of $\cos^2x-6\sin(x)\cos(x)+3\sin^2x+2$ Find the maximum and minimum value of $$\cos^2x-6\sin(x)\cos(x)+3\sin^2x+2$$.
i simplified and reach to expression as follows :
$5 + 2\sin(x)[\sin(x)-3\cos(x)]$. How do i go from here?
Thanks
|
The expression can be simplified to $$\cos^2 x + \sin^2 x +2\sin^2 x +2 -3\sin 2x $$ $$=3 -3\sin 2x + 2\frac {1-\cos 2x}{2} $$ $$=4 - 3\sin 2x - \cos 2x $$
Now we know that $$-\sqrt{a^2 +b^2} \leq a\sin \alpha + b \cos \alpha \leq \sqrt {a^2 +b^2} $$
Hope you can take it from here.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2134658",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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|
Find minimal value of $abc$ if the quadratic equation $ax^2-bx+c = 0$ has two roots in $(0,1)$ If $$ ax^2-bx+c = 0 $$ has two distinct real roots in (0,1) where a, b, c are natural numbers then find the minimum value of product abc ?
|
Since $a,b,c $ are positive the roots are trivially greater than 0.
What remains is to solve the inequality:
$\frac{b + \sqrt{b^2-4ac}}{2a} <1$
This reduces to $ a+c>b$
But the roots being real and distinct we have $b^2 >4ac$
Combining both we have :
$a^2 + c^2 + 2ac > b^2 > 4ac$
$b^2 > 4ac$ tells us $b> 2$ (why?)
$ a^2 + c^2+ 2ac > 4ac $ tells us $a \neq c$
Checking small cases we get $(a,b,c) =(5,5,1)$ where $abc =25$
EDIT:
Checking "small" cases is not informative, so adding an explanation:
Keeping in mind $a+c>b$, the minimum value of $ac$ occurs when $a=b$ and $c=1$. So for given $b$, the minim of $abc$ is $b^2$. The smallest value of $b$ which agrees the inequality $b^2>4b$ is 5 (as $ac=b$). Hence the corresponding minimum value is $5^2$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2134787",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 2,
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|
Evaluate the integral $\int \frac{x^2(x-2)}{(x-1)^2}dx$
Find $$\int \frac{x^2(x-2)}{(x-1)^2} dx .$$
My attempt:
$$\int \frac{x^2(x-2)}{(x-1)^2}dx = \int \frac{x^3-2x^2}{x^2-2x+1}dx $$
By applying polynomial division, it follows that
$$\frac{x^3-2x^2}{x^2-2x+1} = x + \frac{-x}{x^2-2x+1}$$
Hence $$\int \frac{x^3-2x^2}{x^2-2x+1}dx = \int \left(x + \frac{-x}{x^2-2x+1}\right) dx =\int x \,dx + \int \frac{-x}{x^2-2x+1} dx \\ = \frac{x^2}{2} + C + \int \frac{-x}{x^2-2x+1} dx $$ Now using substitution $u:= x^2-2x+1$ and $du = (2x-2)\,dx $ we get $dx= \frac{du}{2x+2}$.
Substituting dx in the integral:
$$\frac{x^2}{2} + C + \int \frac{-x}{u} \frac{1}{2x-2} du =\frac{x^2}{2} + C + \int \frac{-x}{u(2x-2)} du $$
I am stuck here. I do not see how using substitution has, or could have helped solve the problem. I am aware that there are other techniques for solving an integral, but I have been only taught substitution and would like to solve the problem accordingly.
Thanks
|
$$\int\frac{-x}{x^2-2x+1} dx=\int\frac{1-x}{(1-x)^2} dx+\int\frac{-1}{(1-x)^2} dx=-\ln(1-x)+\frac{1}{1-x}$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2135003",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 7,
"answer_id": 5
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|
Bessel to integral form Can someone tell me how the Bessel function be this form.
$$\frac{2}{Ο}\int_0^1 \frac{\cos (xt)}
{\sqrt{1-t^2}} dt = J_0(x)$$
|
$\newcommand{\dif}{\mathrm d}$
$$\begin{align*} \int_0^1 \frac{\cos (xt)}{\sqrt{1-t^2}} \dif t
&= \int_0^1 \sum_{j=0}^\infty \frac{(-1)^j (xt)^{2j}}{(2j)!\sqrt{1-t^2}} \dif t \\
&= \sum_{j=0}^\infty \frac{(-1)^j x^{2j}}{(2j)!} \int_0^1 \frac{t^{2j} \dif t}{\sqrt{1-t^2}}
\end{align*}$$
Observe that (this part needs some justification) by some identities of the $\Gamma$ function,
$$ \int_0^1 \frac{t^{2j} \dif t}{\sqrt{1-t^2}} = \frac{\sqrt\pi \Gamma(j+1/2)}{2\Gamma(j+1)} = \frac{\sqrt\pi}{j!}\frac{(2j)!\sqrt\pi}{4^j j! } = \pi\frac{(2j)!}{2 \cdot 4^j (j!)^2}
$$
Hence
$$\sum_{j=0}^\infty \frac{(-1)^j x^{2j}}{(2j)!} \int_0^1 \frac{t^{2j} \dif t}{\sqrt{1-t^2}} = \frac{\pi}{2}\sum_{j=0}^\infty \frac{(-1)^j x^{2j}}{(2j)!} \frac{(2j)!}{2 \cdot 4^j (j!)^2} = \frac{\pi}{2} \sum_{j=0}^\infty \frac{(-1)^j}{(j!)^2} \left( \frac x 2 \right)^{2j} = \frac{\pi}{2} \mathrm J_0(x)
$$
A good exercise would be to use the completeness of $\mathrm L^{\!1}$ to justify the switch of the order of summation.
Justification of the integral $ \int_0^1 \frac{t^{2j} \dif t}{\sqrt{1-t^2}}$:
Via the diffeomorphism $s \mapsto \sqrt s$, and the $\mathrm B$ identity $\mathrm B(x,y) \Gamma(x+y)=\Gamma(x) \Gamma(y)$,
$$\begin{align*} \int_0^1 \frac{t^{2j}}{\sqrt{1-t^2}} \dif t
&= \int_0^1 \frac{1}{2s^{1/2}}\frac{s^j}{\sqrt{1-s}} \dif s \\
&= \frac12 \int_0^1 s^{j-1/2} (1-s)^{-1/2} \dif s \\
&= \frac12\mathrm B(j+1/2,1/2) \\
&= \frac{\sqrt\pi \Gamma(j+1/2)}{2\Gamma(j+1)}
\end{align*}$$
|
{
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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|
Divergence of $\vec{f} = \frac{1}{r^2} \hat{r}$ Why am I getting zero divergence of function $\vec{f} = \frac{1}{r^2} \hat{r}$, where $r$ is the distance from the origin and $\hat{r}$ is the unit vector in the radial direction. The divergence of this function over a sphere of radius $R$, which includes the origin.
$$\nabla \cdot f = \frac{1}{r^2} \frac{\partial}{\partial r}(r^2 f_r) = \frac{1}{r^2} \frac{\partial}{\partial r}(r^2 \frac{1}{r^2}) = 0$$
|
$\hat{r} = (x,y,z)/\sqrt{x^2 + y^2 + z^2}$
$$
F(x,y,z) = \frac{1}{(x^2 + y^2 + z^2)^{3/2}} (x,y,z) \\
\frac{\partial}{\partial x} F = \frac{\partial}{\partial x} \frac{x}{(x^2 + y^2 + z^2)^{3/2}} = \frac{-2x^2 + y^2 + z^2}{(x^2 + y^2 + z^2)^{5/2}} \\
\frac{\partial}{\partial y} F = \frac{\partial}{\partial y} \frac{y}{(x^2 + y^2 + z^2)^{3/2}} = \frac{x^2 -2y^2 + z^2}{(x^2 + y^2 + z^2)^{5/2}} \\
\frac{\partial}{\partial z} F = \frac{\partial}{\partial z} \frac{z}{(x^2 + y^2 + z^2)^{3/2}} = \frac{x^2 + y^2 -2z^2}{(x^2 + y^2 + z^2)^{5/2}} \\
$$
Putting together :
$$
\nabla\cdot F = \frac{\partial}{\partial x} F + \frac{\partial}{\partial y} F + \frac{\partial}{\partial z} F = \frac{0}{(x^2 + y^2 + z^2)^{5/2}} = 0
$$
|
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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|
Difficult integral with trigonometric substitution: $\int\frac{\sqrt{1-x}}{\sqrt x}dx$ This is the integral I am talking about
$$\int\frac{\sqrt{1-x}}{\sqrt x}dx$$
As you can tell I tried substitution $u = \sqrt x$, and from there I went to $u = \sin{\theta}$ like here.
$$\newcommand{\dd}{\; \mathrm{d}}
\int \frac{\sqrt{1-x}}{\sqrt x} \dd x=
\begin{vmatrix} u=\sqrt x \\ \dd u = \frac1{2\sqrt x}\end{vmatrix} =
2\int \sqrt{1-u^2} \dd u =
\begin{vmatrix} u=\sin\theta \\ \theta = \arcsin u \\ \dd u = \cos\theta \dd\theta \end{vmatrix} =
2\int \sqrt{1-\sin^2\theta} \cos\theta \dd\theta =
2\int \cos^2\theta \dd\theta=
2\int \frac{1+\cos2\theta}2 \dd\theta =
\int \dd\theta + \int\cos2\theta \dd\theta =
\theta + \frac12 \sin2\theta + C =
\arcsin u + \frac12 \sin(2\arcsin u) + C =
\arcsin \sqrt x + \frac12 \sin(2\arcsin \sqrt x) + C
$$
Anyways, you can see I entered this answer and was wrong and I have no idea why. I even tried to go back and substitute in cosine instead of sine since in this case they were equivalent and I still got the answer wrong, any help is appreciated thank you!
|
$$
I:=\int \frac{\sqrt{1-x}}{\sqrt{x}} d x
$$
Since $0 <x<1$, therefore let $x=\sin ^{2}\theta$, then $d x=2 \sin \theta \cos \theta d\theta.$
$$\begin{aligned}
I &=\int \frac{\sqrt{\cos ^{2} \theta}}{\sqrt{\sin ^{2} \theta}} \cdot 2 \sin \theta \cos \theta d \theta =\int 2 \cos ^{2} \theta d\theta =\int(1+\cos 2 \theta) d \theta \\&=\theta+\frac{\sin 2 \theta}{2}+C=\theta+\sin \theta \cos \theta +C =\boxed{\sin ^{-1} \sqrt{x}+\sqrt{x(1-x)}+C}\end{aligned}
$$
|
{
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"timestamp": "2023-03-29T00:00:00",
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|
Does $\sqrt{a+b}\leq \sqrt{a}+\sqrt{b}$? In the case of $$\sqrt{(x_n-\ell_1)+(y_n-\ell_2)}\leq \sqrt{(x_n-\ell_1)^2} + \sqrt{(y_n-\ell_2)^2} = |x_n-\ell_1|+|y_n-\ell_2|$$
it is true, if we take the rise the two sides in the power of $2$ we get:
\begin{align}
& (x_n-\ell_1)+(y_n-\ell_2)\leq \left( \sqrt{(x_n-\ell_1)^2}+\sqrt{(y_n-\ell_2)^2} \,\right)^2 \\[10pt]
= {} &{(x_n-\ell_1)+(y_n-\ell_2)}+\sqrt{(x_n-\ell_1)+(y_n-\ell_2)}
\end{align}
Does the same is true for $$\sqrt{a+b}\leq \sqrt{a}+\sqrt{b} \text{ ?}$$
doesn't $\sqrt{a+b}$ is a product of $3$ components? $\sqrt{a}+\sqrt{b}+\text{(something positive)}$ and therefore
$$\sqrt{a+b}\geq \sqrt{a}+\sqrt{b} \text{ ?}$$
|
Quite simply, the answer to your question from the title is "yes", if we assume that $a$ and $b$ are non-negative real numbers.
Since both of the sides are positive, we can square the entire equation, and we get a trivial inequality: $$ \sqrt{a+b} \leq \sqrt{a} + \sqrt{b} \iff a+b \leq a+b+2\sqrt{ab} \iff 0 \leq 2\sqrt{ab}. $$
|
{
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"timestamp": "2023-03-29T00:00:00",
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|
Particle starts at $(0,-3)$ and moves clockwise around origin on graph $x^2+y^2=9$, find parametric equation Question
particle starts at $(0,-3)$ and moves clockwise around origin on graph $x^2+y^2=9$, revolve in $9$ seconds find parametric equation in term of $t$.
What I've done so far:
I first thought that the graph ought to be $x^2+y^2=9$
so then I say that $x=\frac{9}{2}\cos{t}=x$ and $y=\frac{9}{2}\sin{t}$
but then this particle in this graph travels CCW so then I change it to :
$x=\frac{9}{2}\cos{t}=x$ and $y=\frac{9}{2}\sin{-t}$
but then I found out that when I plug in $t=0$, I do not get -3
How do I phase shift this parametric equation so that it satisfies the fact that the particle starts at $(0.-3)$?
|
Let's start with the value of $\theta$, in radians, as a function of time. Because the particle is travelling at a constant velocity, revolving in $9$ seconds, $\dfrac{d\theta}{dt} = -\dfrac{2\pi}{9}$. The minus sign is because the particle is travelling clockwise. Then
$\theta(t) = -\dfrac{2\pi}{9}t + \theta_0$. Since the angle of the point $(-3,0)$ is $-\pi$, we must have
$-\pi = \theta(0) = \theta_0$
So
$\theta(t) = -\dfrac{2\pi}{9}t - \pi$.
So
$(x,y) = \left(3 \cos\left( -\dfrac{2\pi}{9}t - \pi\right),
3 \sin\left( -\dfrac{2\pi}{9}t - \pi\right)
\right)
= \left( -3\cos\left( \dfrac{2\pi}{9}t\right),
3\sin\left( \dfrac{2\pi}{9}t\right)
\right)$
|
{
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"timestamp": "2023-03-29T00:00:00",
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|
To prove that prove that $cos^8 \theta sec^6 \alpha , \frac{1}{2 } ,sin^8 \theta cosec^6 \alpha $ are in A.P If $\cos^4 \theta \sec^2 \alpha , \frac{1}{2 } ,\sin^4 \theta \csc^2 \alpha $ are in A.P ,
then prove that $\cos^8 \theta \sec^6 \alpha , \frac{1}{2 } ,\sin^8 \theta \csc^6 \alpha $ are in A.P
Now i have reached upto
$1=\cos^4 \theta \sec^2 \alpha + \sin^4 \theta \csc^2 \alpha$.
By completing square i have $(\sin^2 \theta \cot\alpha - \cos^2 \theta \tan \alpha)^2=0$ so i get $\tan \theta= \pm \tan \alpha$
How do i proceed?
Thanks
|
We have $\cos^4\theta\sec^2\alpha+\sin^4\theta\csc^2\alpha=1$
Set $\cos^2\theta=a,\sin^4\theta=(1-a)^2$ to solve for $a$ to find $a=\cos^2\alpha$
|
{
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"timestamp": "2023-03-29T00:00:00",
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|
$\lim_{x\to2}\frac{\sqrt{1+\sqrt{x+2}-\sqrt3}}{x-2}$ $\lim_{x\to2}\frac{\sqrt{1+\sqrt{x+2}-\sqrt3}}{x-2}$
My try:
$\lim_{x\to2}\frac{\sqrt{1+\sqrt{x+2}-\sqrt3}}{x-2}=$
$\lim_{x\to2}\frac{\sqrt{1+\sqrt{x+2}-\sqrt3}}{x-2}\times\frac{\sqrt{1+\sqrt{x+2}+\sqrt3}}{\sqrt{1+\sqrt{x+2}+\sqrt3}}=$
$\lim_{x\to2}\frac{\sqrt{x+2\sqrt{x+2}}}{(x-2)\sqrt{1+\sqrt{x+2}+\sqrt3}}$
I am stuck here.
|
Following StackTD's comment
let $\sqrt{1+\sqrt{x+2}}-\sqrt3=y\implies y\to0$
and $1+\sqrt{x+2}=(y+\sqrt3)^2=y(y+2\sqrt3)+3$
$$x+2=(y^2+2\sqrt3y+2)^2=y^2(y+2\sqrt3)^2+4y(y+2\sqrt3)+4$$
$$\lim_{x\to2}\frac{\sqrt{1+\sqrt{x+2}}-\sqrt3}{x-2}=\lim_{y\to0}\dfrac y{y^2(y+2\sqrt3)^2+4y(y+2\sqrt3)}=?$$
|
{
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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|
differential linear equation of order one $(2xy+x^2+x^4)\,dx-(1+x^2)\,dy=0$ I have no idea how to solve it. Should be linear equation of order one since I am passing through this chapter, but I can't put into the form of $$y'+P(x)y=Q(x)$$
Here is the equation: $$(2xy+x^2+x^4)\,dx-(1+x^2)\,dy=0$$
It is not exact since partial derivatives are not equal.
Any help would be appreciated.
|
$$(2xy+x^2+x^4)\,dxβ(1+x^2)\,dy=0\\
(1+x^2) \frac {dy}{dx} = (2xy+x^2+x^4)\\
y' - \frac {2x}{1+x^2} y = x^2$$
Integrating factor
$$e^{\int \frac {-2x}{1+x^2}} = e^{-\ln(1+x^2)} = \frac {1} {(1+x^2)}$$
$$\frac {1}{1+x^2}y = \int \frac {x^2}{1+x^2}\\
\frac {1}{1+x^2}y = x - \arctan x + C\\
y = x^3 + x - (1+x^2)\arctan x + C(1+x^2)$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2143195",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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|
Solving for varible with use of ln Looking for some help with trying to solve for $x$ in the following equation,
$$0=5-re^{-x^2}$$
The solution is,
$$x^2=\ln\frac{r}{5}$$
|
\begin{align*}
\implies 5 = re^{-x^2} \iff \frac{5}{r} = e^{-x^{2}} \iff \ln\frac{5}{r} = -x^{2} \iff -\ln\frac{5}{r} = x^2 \iff \ln \frac{r}{5} = x^{2}
\end{align*}
Note: $\ln a^{n} = n\ln a$. In this case, $n=-1$, i.e. $\ln a^{-1} = -\ln a$.
|
{
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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|
Related Rates: Angular Velocity The figure shows a rotating wheel with radius 40 cm and a connecting rod AP with length 1.2 m. The pin P slides back and forth along the x-axis as the wheel rotates counterclockwise at a rate of 360 revolutions per minute.
*
*Find the angular velocity of the connecting rod, $\frac{d\alpha}{dt}$, in radians per second, when $\theta = \frac{\pi }{3}$.
*Express the distance $x = \left|OP \right|$ in terms of $\theta$.
*Find an expression for the velocity of the pin P in terms of $\theta$.
I have found a solution for all three questions, but my answer to QUESTION 1 is different from the given solution from the textbook and I am struggling to figure out why.
QUESTION 1
I use the law of sines to solve this.
$$\frac{\sin\alpha }{OA} = \frac{\sin\theta }{AP}$$
from which I find that:
$$\sin\alpha =\frac{OA}{AP}\sin\theta =\frac{40}{120}\sin\theta =\frac{1}{3}\sin\theta$$
$$\alpha = \arcsin\left(\frac{1}{3}\sin\theta \right)$$
$$\frac{d\alpha }{dt} = \frac{1}{\sqrt{1 - \frac{{\sin\theta }^{2}}{9}}} \cdot \frac{1}{3}\cos\theta \cdot \frac{d\theta }{dt}$$
and this is equal to $$\frac{d\alpha }{dt} = \frac{3\pi }{\sqrt{33}}$$ when $$\theta = \frac{\pi }{3}$$
The problem here is that according to the textbook the right answer is $\frac{4\pi \sqrt{3}}{\sqrt{11}}$ which I quite do not understand. Furthermore I am not supposed, at this stage, to know the derivative of inverse functions including arcsin. Is there any other way to solve this without using the arcsin funcion?
|
We have
$$
\sin(\theta) = \frac{s}{r} \\
\sin(\alpha) = \frac{s}{l}
$$
such that equating on $s$ we get
$$
r \sin(\theta) = l \sin(\alpha) \iff \\
\alpha = \arcsin\left( \frac{r}{l} \sin(\theta) \right)
$$
Then the time derivative is
$$
\dot{\alpha}
= \frac{1}{\sqrt{1-\left( \frac{r}{l} \sin(\theta)\right)^2}}
\frac{r}{l}\cos(\theta) \dot{\theta}
$$
with $r=0.4\,\text{m}$, $l=1.2\,\text{m}$, $\dot{\theta}=6\cdot 2\pi\,\text {rad}/\text{s}$. Then for $\theta=\pi/3$ we get:
$$
\sin(\pi/3) = \sqrt{3}/2 \\
\cos(\pi/3) = 1/2 \\
\dot{\alpha} =
\frac{1}{\sqrt{1-(\frac{1}{3} \sqrt{3}/2)^2}}
\frac{1}{3} \cdot \frac{1}{2} \cdot 12 \pi
= \sqrt{\frac{36}{33}} 2 \pi
= \sqrt{\frac{9}{33}} 4\pi
= \sqrt{\frac{3}{11}} 4\pi
$$
Alternative without $\arcsin$:
$$
\sin(\alpha) = \frac{r}{l} \sin(\theta)
$$
Differentiating
$$
\cos(\alpha) \dot{\alpha} = \frac{r}{l} \cos(\theta) \dot{\theta} \iff \\
\dot{\alpha}
= \frac{r}{l} \frac{\cos(\theta) \dot{\theta}}{\cos(\alpha)}
= \frac{r}{l} \frac{\cos(\theta) \dot{\theta}}{\sqrt{1-\sin(\alpha)^2}}
= \frac{r}{l} \frac{\cos(\theta) \dot{\theta}}{\sqrt{1-(\frac{r}{l} \sin(\theta))^2}}
$$
|
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|
Linear Equations system The system
$$\begin{cases}x-y+3z=-5\\5x+2y-6z=\alpha \\2x-y+\alpha z = -6 \end{cases}$$
for which $\alpha$ values the linear equation system:
*
*has no solution
*has one solution
*has more than one solution
I started to do Gauss elimination on it, but i have no idea what i am looking for and how to approach this, I'm stuck with the Gauss elimination.
My work so far:
\begin{align}
\left(\begin{array}{rrr|r}
1 & -1 & 3 & -5 \\
5 & 2 & 6 & \alpha \\
2 & -1 & \alpha & -6
\end{array}\right)
&\leadsto \left(\begin{array}{rrr|r}
1 & -1 & 3 & -5 \\
0 & 7 & -9 & \alpha + 25 \\
2 & -1 & \alpha & -6
\end{array}\right) \\
&\leadsto \left(\begin{array}{rrr|r}
1 & -1 & 3 & -5 \\
0 & 7 & -9 & \alpha+25 \\
0 & 1 & \alpha-6 & 4
\end{array}\right) \\
&\leadsto \left(\begin{array}{rrr|r}
1 & 0 & \alpha + 3 & -1 \\
0 & 7 & -9 & \alpha+25 \\
0 & 1 & \alpha-6 & 4
\end{array}\right) \\
\end{align}
|
Gaussian elimination:
\begin{align}
\left[\begin{array}{ccc|c}
1 & -1 & 3 & -5 \\
5 & 2 & -6 & \alpha \\
2 & -1 & \alpha & -6
\end{array}\right]
&\to
\left[\begin{array}{ccc|c}
1 & -1 & 3 & -5 \\
0 & 7 & -21 & \alpha+25 \\
0 & 1 & \alpha-6 & 4
\end{array}\right]
&&\begin{aligned} R_2&\gets R_2-5R_1 \\ R_3&\gets R_3-2R_1\end{aligned}
\\[6px]&\to
\left[\begin{array}{ccc|c}
1 & -1 & 3 & -5 \\
0 & 1 & -3 & (\alpha+25)/7 \\
0 & 1 & \alpha-6 & 4
\end{array}\right]
&&R_2\gets \tfrac{1}{7}R_2
\\[6px]&\to
\left[\begin{array}{ccc|c}
1 & -1 & 3 & -5 \\
0 & 1 & -3 & (\alpha+25)/7 \\
0 & 0 & \alpha-3 & (3-\alpha)/7
\end{array}\right]
&&R_3\gets R_3-R_2
\end{align}
If $\alpha\ne3$, the system has unique solution.
If $\alpha=3$, the system has infinitely many solutions.
|
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|
Without using calculus, how can I find the formula for the area of the shaded region in the attached figure in terms of a and b? Note that the center of the larger circle is on the edge of the smaller.
|
If we consider the centers of both the circles on y axis, origin as the center of the bigger circle, then let the two circles intersect at $A=(x_1,y_1), B=(x_2,y_2)$
Equation of the bigger circle is $x^2+y^2=b^2...(I)$
Equation of the smaller circle $(x-h)^2+(y-k)^2=a^2$, here $h=0, k=-a$
$\therefore x^2+(y+a)^2=a^2...(II)$
$(II)-(I)$ gives $y=\frac{-b^2}{2a}$ and $x=\pm \frac{-b\sqrt{4a^2-b^2}}{2a}$
$\therefore x_1=-\frac{b\sqrt{4a^2-b^2}}{2a}$ and $x_2=\frac{b\sqrt{4a^2-b^2}}{2a}$
It can be proved that $AB=2CB$ and $\angle OCB=90$ by Pythagoras theorem.
$\ell(AB)=\frac{b}{a}\sqrt{4a^2-b^2}$
angle subtended by $AB$ at the center of the bigger circle$=\phi=2\cdot \sin^{-1}\left(\frac{\sqrt{4a^2-b^2}}{2a}\right)$
angle subtended by $AB$ at the center of the smaller circle$=\phi'=2\cdot \sin^{-1}\left(\frac{b\sqrt{4a^2-b^2}}{2a^2}\right)$
The isosceles triangles $AO'O$ and $BO'O$, both with equal sides length$=a$ and base$=b$ and area$=\frac{b}{4}\sqrt{4a^2-b^2}$
Area of the crescent shape region or segment = area of sector of the smaller circle formed by $AO$ or $BO$ - combined area of isosceles triangles $AO'O$ and $BO'O$
$=\frac{a^2}{2}\cdot 2 \sin^{-1}\left( \frac{b\sqrt{4a^2-b^2}}{2a^2} \right)-\left( \frac{b}{2}\sqrt{4a^2-b^2} \right)$
area of the bigger circle sector$=\frac{b^2}{2}\cdot 2 \sin^{-1} \left( \frac{\sqrt{4a^2-b^2}}{2a} \right)$
$\therefore$ area of the required region $\color{red}{= \pi b^2 - b^2\cdot \sin^{-1} \left( \frac{\sqrt{4a^2-b^2}}{2a} \right)-\left( a^2\cdot \sin^{-1}\left( \frac{b\sqrt{4a^2-b^2}}{2a^2} \right)-\left( \frac{b}{2}\sqrt{4a^2-b^2} \right)\right)}$
To check if the formula is correct, put $b=\sqrt2 a$, so that points $O'$ and $C$ coincide and the $\angle AOB=90$ and the required area is (area of bigger circle) - (area of sector $AOB$) - (area of two segments or crescent shaped regions), which is
$\frac{3\pi b^2}{4}-\frac{\pi a^2}{2}+\frac{b}{2}\sqrt{4a^2-b^2}=a^2(\pi+1)$
if we substitute $b=\sqrt2 a $ in the $\color{red}{red}$ formula, we get the same result
Check using calculus: $(b=\sqrt2 a)$
In the diagram consider only the region marked I, to the right of the dashed line
area I=$\int_0^a (\sqrt{b^2-y^2}-a)dy = \dfrac{\pi a^2}{4}-\dfrac{a^2}{2}$
In the diagram consider only the region marked II, to the left of the dashed line (note that axis are shifted vertically down by $a$ units)
area II=$\int_0^a(a-\sqrt{a^2-y^2})dy = a^2-\dfrac{\pi a^2}{4}$
area I + area II = $\dfrac{a^2}{2}$, two such areas = $a^2$, area of upper half circle with radius $ b = \dfrac{\pi b^2}{2}=\pi a^2$
adding everything together we get the required area = $\pi a^2+a^2=a^2(\pi+1)$, $\color{red}{voila!}$
|
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|
Determine all possible real values of h so that the equation has 4 real roots The equation is as following: $$x(x+1)(x+h)(x+1+h)=h^2$$
I did the multiplication and got the expression: $$x^4+2(h+1)x^3+(h+1)^2x^2+hx+h=0$$
From now on I do not know how to proceed.
|
At $h = 0$ you have double roots at $x=-1$ and $x = 0$
$x(x+1)(x+h)(x+1+h)=h^2\\
u = x+ \frac h2$
$(u-\frac h2)(u+\frac h2)(u+1-\frac h2)(u+1+\frac h2)=h^2\\
(u^2-\frac {h^2}4)((u+1)^2-\frac {h^2}{4})=h^2$
$(u^2-\frac {h^2}4)((u+1)^2-\frac {h^2}{4})$ has its maximum at $u = -\frac 12$
If the max $(u^2-\frac {h^2}4)((u+1)^2-\frac {h^2}{4})< h^2$ then there are complex roots for that value of $h.$
$(\frac 14 - \frac {h^2}{4})^2 > h^2\\
|1 - h^2|> 4|h|$
$(h^2 + 4h - 1)(h^2 - 4h - 1) < 0$
$h = \pm 2\pm\sqrt 5)$
$h\in(-\sqrt 5-2,2-\sqrt 5)\cup 0\cup(\sqrt 5-2,\sqrt 5+2)$ there are only 2 real roots.
Outside of these intervals there are 4
|
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|
Find the remainder of $\frac{3^{11}-1}{2}$ divided by $9$ I have the following question:
Find the remainder of $$\frac{3^{11}-1}{2}$$ divided by $9$
I tried to reformat the question:
$$\frac{3^{11} -1 } {2} \times \frac{1}{9} = \frac{3^{11}-1}{18}$$
Since $3^2 = 9$
$$\frac{3^2(3^9) -1}{3^2 \times2}$$
I don't know where to go next. Anyway, this is one of my many attempts to solve this question, and most of them ends with a complicated solution. I don't want to use modular arithmetic for this question. A hint or anything will help me.
|
We have that $$\frac {3^{11}-1}{2} = \frac {3-1}{2}(3^{10} + 3^9 + \cdots + 3^2+3+1) \tag {1} $$
We know that $3^2 =9 \equiv 0 \mod 9$ and similarly $3^3 =3 (3^2) \equiv 0 \mod 9$ and so on. Thus, $(1)$ reduces to, $$1 (0 + 0 + 0 \cdots 0 + 3 +1 ) \equiv 4 \mod 9$$
Hope it helps.
|
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About the formulas to determine the integral ideal with specific norm (1) For $\mathbb{Q}(\sqrt{10})$, the ring of integers is $\mathbb{Z}[\sqrt{10}]$, and to determine the integral ideal with norm $3$, we can use the formula below:
By $X^2 β 10\equiv (X + 1)(X β 1)\text{ mod 3}$, then we get the conjugate ideals with norm 3: $(3, \sqrt{10}+1), (3, \sqrt{10}-1)$.
(2) For $\mathbb{Q}(\sqrt{17})$ the ring of integers is $\mathbb{Z}[\alpha]$, where $\alpha=\frac{1+\sqrt{17}}{2}$ and to determine the integral ideal with norm $2$, we can use the formula below:
$0=(\alpha-1/2)^2-17/4=\alpha^2-\alpha-4$ $X^2 β X β 4 β‘ X(X + 1) \text{ mod 2}$ then we get the conjugate ideals with norm 2: $(2,\alpha),(2,\alpha+1)$.
I am confused with the reason of these two formulas, where are these formulas from. Actually, I can't find any reference to illustrate these formulas. Can anybody explain these two formulas? Thank you!
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This answer, despite its length, is not a complete answer either, but I think it'll give you some insights. Your question feels like another one I've answered before, although I think the earlier question was more concerned with numbers than with ideals.
Regardless, I feel that the excessive early emphasis on defining $\alpha$ (sometimes $\theta$ or $\omega$, but I think the latter should be reserved for a specific ring) only serves to confuse students.
However, I am not certain the formulas you were given are correct, but I could be wrong, as I've approached this topic from a different, far less systematic path than yours.
Given a squarefree integer $d$, the norm function is defined for any number in $\mathbb Q(\sqrt d)$, whether or not that number is also in the ring of algebraic integers. Namely, if $a$ and $b$ are rational real numbers, then $N(a + b \sqrt d) = a^2 - db^2$. And if $a + b \sqrt d$ is an algebraic integer, then its norm is an ordinary integer.
For example, consider the number $$\frac{3 + \sqrt{10}}{2}.$$ Then $$N\left(\frac{3 + \sqrt{10}}{2}\right) = \left(\frac{3}{2}\right)^2 - 10\left(\frac{1}{2}\right)^2 = \frac{9}{4} - \frac{10}{4} = -\frac{1}{4},$$ which is not an integer.
In fact, for a number $a + b\sqrt{10}$ to be an algebraic integer, $a$ and $b$ must both be ordinary integers. From there it is easy to see that $a - 10b = \pm 3$ has no solutions, hence no number in this domain can have a norm of 3, and so the ideal with that norm can't be a principal ideal.
The smallest positive integer multiple of 3 that is a possible norm in this domain is 6, but 9 works, too, giving us the ideals $\langle 3, 1 - \sqrt{10} \rangle$ and $\langle 3, 1 + \sqrt{10} \rangle$.
Now let's go to $\mathbb Q(\sqrt{17})$, and consider the number $$\frac{3 + \sqrt{17}}{2}.$$ Then $$N\left(\frac{3 + \sqrt{17}}{2}\right) = \left(\frac{3}{2}\right)^2 - 17\left(\frac{1}{2}\right)^2 = \frac{9}{4} - \frac{17}{4} = -\frac{8}{4} = -2,$$ which is an integer.
Since the fundamental unit of this domain is of norm 1 rather than norm $-1$, a number with a norm of 2 is impossible. (Oops, silly mistake, there's $29 + 7\sqrt{17}$ divided by 2, and infinitely many others. I stand by everything else.)
But with ideals, we don't have to worry about signs as much, and so $$\langle 2 \rangle = \left\langle \frac{3 - \sqrt{17}}{2} \right\rangle \left\langle \frac{3 + \sqrt{17}}{2} \right\rangle,$$ a product of principal ideals.
Of course if you want to interpret those in terms of $$\alpha = \frac{1 + \sqrt{17}}{2},$$ you can, like so:
$$\frac{3}{2} - \frac{\sqrt{17}}{2} = \frac{4}{2} - \left(\frac{1}{2} + \frac{\sqrt{17}}{2}\right) = 2 - \alpha$$ and $$\frac{3}{2} + \frac{\sqrt{17}}{2} = \frac{2}{2} + \left(\frac{1}{2} + \frac{\sqrt{17}}{2}\right) = 1 + \alpha.$$
You should recognize the latter number as a co-generator of one of the ideals you found: $\langle 2, 1 + \alpha \rangle$. But since $(2 - \alpha)(1 + \alpha) = -2$, the ideal $\langle 2 \rangle$ is contained in the ideal $\langle 1 + \alpha \rangle$, and therefore 2 is a redundant generator in $\langle 2, 1 + \alpha \rangle$.
As for the ideal $\langle 2, \alpha \rangle$, that can also be boiled down to a principal ideal, namely $\langle \gcd(2, \alpha) \rangle$, but I don't think that's a prime ideal, since, for starters, $N(\alpha) = -4$.
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"language": "en",
"url": "https://math.stackexchange.com/questions/2147678",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
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