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Show that, $2\arctan\left(\frac{1}{3}\right)+2\arcsin\left(\frac{1}{5\sqrt2}\right)-\arctan\left(\frac{1}{7}\right)=\frac{\pi}{4}$ Show that,
$$2\arctan\left(\frac{1}{3}\right)+2\arcsin\left(\frac{1}{5\sqrt2}\right)-\arctan\left(\frac{1}{7}\right)=\frac{\pi}{4}$$
There is a mixed of sin and tan, how can I simplify this to $\frac{\pi}{4}$
We know the identity of $\arctan\left(\frac{1}{a}\right)+\arctan\left(\frac{1}{b}\right)=\arctan\left(\frac{a+b}{ab-1}\right)$
|
It would be easier to attack if you would substitute $\tan^{-1}{\frac17}\;$ for $\sin^{-1}{\frac1{5\sqrt2}}\;$.
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|
How to compute $\int_0^1\frac{\ln(x)}{1+x^5}dx$?
Let $\phi$ denote the golden ratio $\phi=\frac{1+\sqrt5}{2}$.
How can I prove this sum?
$$\sum_{n=0}^{\infty}(-1)^n\left[\frac{\phi}{(5n+1)^2}+\frac{\phi^{-1}}{(5n+2)^2}-\frac{\phi^{-1}}{(5n+3)^2}-\frac{\phi}{(5n+4)^2}\right]=\left(\frac{2\pi}{5}\right)^2$$
My try:
Change the sums into an integrals:
$\sum_{n=0}^{\infty}\frac{(-1)^n}{(5n+1)^2}=\int_0^1\frac{-\ln(x)}{1+x^5}dx$
Can somebody give a hint how on to integrate this integral.
Try substitution by letting $u=\ln(x)$ is not working and integration by part is making it more complicated than before. What kind of substitution should I be using?
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Consider the function $f(x)=\frac12(3x^2-1)$ for $x\in(-1,1)$ and its periodic extension with period $2$. Since $f(x)=f(-x)$ we can write
$$f(x)=\sum_{k=0}^{\infty}a_n\cos n\pi x$$
$$\int_{-1}^1\frac12(3x^2-1)dx=0=2a_0$$
$$\begin{align}\int_{-1}^1\frac12(3x^2-1)\cos n\pi x\,dx&=\left[\frac1{n\pi}\frac12(3x^2-1)\sin n\pi x+\frac{3x}{n^2\pi^2}\cos n\pi x-\frac3{n^3\pi^3}\sin n\pi x\right]_{-1}^1\\
&=\frac{6(-1)^n}{n^2\pi^2}=a_n\end{align}$$
So
$$f(x)=\frac6{\pi^2}\sum_{n=1}^{\infty}\frac{(-1)^n}{n^2}\cos n\pi x$$
Note that $\cos\left(n\pi\left(1-\frac15\right)\right)=(-1)^n\cos\frac{n\pi}5$. So
$$\begin{align}f\left(\frac45\right)&=\frac{23}{50}=\frac6{\pi^2}\sum_{n=1}^{\infty}\frac1{n^2}\cos\frac{n\pi}5\\
&=\frac6{\pi^2}\left(\frac{\sigma}2+\sum_{n=1}^{\infty}\frac1{25n^2}(-1)^n\right)\\
&=\frac6{\pi^2}\left(\frac{\sigma}2+\frac1{25}\left(-\frac12\right)\frac{\pi^2}6\right)\end{align}$$
Here $\sigma$ is the topical sum, where the coefficient of $1/n^2$ is $2\cos n\pi/5$ except where $n=5k$ where the $\cos 5k\pi/5=(-1)^k$ has been omitted. Thus
$$\sigma=2\frac{\pi^2}6\left(\frac{23}{50}+\frac1{50}\right)=\frac{4\pi^2}{25}$$
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Prove that $\displaystyle \sum_{k=1}^n \bigg(\dfrac{1}{k}+\dfrac{2}{k+n}\bigg ) \leq \ln(2n) + 2 -\ln(2)$
Prove that $$\displaystyle \sum_{k=1}^n \bigg(\dfrac{1}{k}+\dfrac{2}{k+n}\bigg ) \leq \ln(2n) + 2 -\ln(2).$$
I was thinking of using mathematical induction for this. That is,
We prove by induction on $n$. The case $n=1$ holds trivially since $2 \leq 2$. Now assume the result holds for some $m$. Then by assumption we know that $$\displaystyle \sum_{k=1}^{m+1} \bigg(\dfrac{1}{k}+\dfrac{2}{k+m}\bigg ) \leq \ln(2m) + 2 -\ln(2)+\dfrac{1}{m+1}+\dfrac{2}{2m+1}. $$
We must relate this somehow to $\ln(2(m+1)) + 2 -\ln(2)$.
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A non-induction version, just for diversification. Let's note $$S_n=\sum_{k=1}^{n} \frac{1}{k}$$
which is $$S_n=\ln(n)+\gamma +\varepsilon_n$$
Now
$$\sum_{k=1}^{n}\left ( \frac{1}{k} + \frac{2}{k+n} \right )= \sum_{k=1}^{n} \frac{1}{k} + \sum_{k=1}^{n} \frac{2}{k+n} $$
$$ = 1 + \frac{1}{2} + \frac{1}{3} + ... + \frac{1}{n} + 2\left ( \frac{1}{n+1} + \frac{1}{n+2} + ... + \frac{1}{2n} \right ) $$
$$=S_n+2S_{2n}-2S_n=\ln(n)+\gamma +\varepsilon_n + 2\ln(2n)+2\gamma +2\varepsilon_{2n} - 2\ln(n)-2\gamma -2\varepsilon_n$$
$$=\ln(n)+\gamma + 2\ln2 + 2\varepsilon_{2n} -\varepsilon_n < $$
$$<\ln(n)+1.96352 + 2\varepsilon_{2n} -\varepsilon_n$$
And $$2\varepsilon_{2n} -\varepsilon_n \sim \frac{2}{4n} - \frac{1}{2n}$$
i.e. first $n=1..4$ can be verified "manually".
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How to solve this equation algebraically Solve the following simultaneous equations on the set of real numbers:
\begin{cases}x^2 + y^3 = x+1 \\ x^3+y^2=y+1\end{cases}
Thanks for helping!
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Subtract the two equations to get
$(x^2 -y^2) -(x^3 -y^3)-(x-y)=0$
$\Rightarrow (x-y)(x+y-x^2-y^2 -xy-1)=0$
So, either $x-y=0$ or $x+y-x^2-y^2 -xy-1=0$
*
*For the first case,
We have $x^3+x^2-x-1=(x-1)(x^2 -1)=(x-1)^2 (x+1)=0 \Rightarrow x=1 $ or $x=-1$
So, $x=y=1$ or $x=y=-1$
*And for the second case,
$x+y-x^2-y^2 -xy-1=-\frac{1}{2} \cdot [(x-1)^2 +(y-1)^2 +(x+y)^2]=0$
$\Rightarrow x=1, y=1 $ and $x=-y$, which is not possible at the same time.
So, this case yields no solutions.
Thus the only solutions are $x=y=1$ and $x=y=-1$
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Number of polynomials which are divisible by $x+1$ Let $a,b,c,d$ be four integers (not necessarily distinct) in the set ${1,2,3,4,5}$ . The number of polynomials $f(x)=x^4+ax^3+bx^2+cx+d$ which are divisible by $x+1$ are:
$(A)$ Between 55 and 65
$(B)$ Between 65 and 85
$(C)$ Between 86 and 105
$(D)$ More than 105
I see that it will be divisible when $f(-1)=0$
i.e. $1-a+b-c+d=0$ but how should I proceed further ?
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On approach would be to brute-force the result by counting. Notice that you've reached the necessary and sufficient condition $a+c = 1+b+d$.
Since $1\le a,b,c,d \le 5$, then $a+c \in \{2, \dots, 10\}$, so $b+d \in \{1, \dots, 9\}$. Notice that $b+d=1$ is impossible, because it would force either $b$ or $d$ to be $0$ which is not allowed. Thus, $a+c \ne 2$.
For $n \ge 1$ natural, let $S_n$ be the number of ways of writing it as $a+c$, with $1 \le a,c \le 5$ and counting both $a+c$ and $c+a$ (i.e. order does matter).
*
*Every natural number $1 \le n \le 6$ can be written as $1 + (n-1) = 2 + (n-2) = \dots = (n-1) + 1$, so $S_n = n-1$ for $1 \le n \le 6$.
*For $7 \le n \le 10$ we can perform a case-by-case analysis: $7 = 2 + 5 = 3 + 4 = 4 + 3 = 5 + 2$, so $S_7 = 4$. $8 = 3 + 5 = 4 + 4 = 5 + 3$, so $S_8 = 3$. $9 = 4 + 5 = 5 + 4$, so $S_9 = 2$. $10 = 5 + 5$, so $S_{10} = 1$.
*No natural number $n \ge 11$ can be written in the desired way because at least one summand would be $\ge 6$, so $S_n = 0$ for $n \ge 6$.
The number of such polynomials, then, is $\sum \limits _{n = 3} ^{10} S_n S_{n-1}$ because $n$ plays the role of $a+c$ and $n-1$ plays the role of $b+d$. This produces
$$\sum \limits _{n = 3} ^6 S_n S_{n-1} + S_7 S_6 + S_8 S_7 + S_9 S_8 + S_{10} S_9 = \\
\sum \limits _{n = 3} ^6 (n-1) (n-2) + 4 \cdot 5 + 3 \cdot 4 + 2 \cdot 3 + 1 \cdot 2 = 40 + 20 + 12 + 6 + 2 = 80 .$$
The correct answer, then, is B.
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Prove that a complex equation has a solution of module 1
Prove that the equation
$$z^n + z + 1=0 \ z \in \mathbb{C}, n \in \mathbb{N} \tag1$$
has a solution $z$ with $|z|=1$ iff $n=3k +2, k \in \mathbb{N} $.
One implication is simple: if there is $z \in \mathbb{C}, |z|=1$ solution for (1) then $z=cos \alpha + i \cdot sin\alpha$ and $|z + 1|=1$. It follows $cos\alpha=-\frac 1 2$ etc.
The other implication is the one I failed to prove.
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If $z^n + z + 1=0 $ then $z^n = -(1+z)$ and $$|z|^n = |1+z|.$$
If $|z| = 1$, then $z = \cos a + i \sin a$, with $a \in [0, 2\pi]$. Moreover:
$$1^n = |1 + \cos a + i \sin a| \Rightarrow \sqrt{(1+\cos a)^2 + \sin^2 a} = 1 \Rightarrow \\
1 + \cos^2 a + 2 \cos a + \sin^2 a = 1\Rightarrow
\cos a = -\frac{1}{2}\\ \Rightarrow a = \frac{2\pi}{3}\vee a = \frac{4\pi}{3}.$$
Let's plug $z = \cos a + i \sin a$ in the starting equation. We get:
$$\cos (na) + i \sin(na) + \cos (a) + i \sin(a) + 1 = 0 \Rightarrow\\
\begin{cases}
\cos(na) + \cos(a)+ 1 &= 0\\
\sin(na) + \sin(a) &= 0
\end{cases}$$
Let's work on the second equation, which becomes $\sin(a) =-\sin(na)$.
If $a=\frac{2\pi}{3} $, then $na = \frac{4\pi}{3}+2\pi k \vee na = \frac{5\pi}{3}+2\pi k$.
First case:
$$\frac{2\pi}{3}n = \frac{4\pi}{3}+2\pi k \Rightarrow
\frac{2\pi}{3}n = \frac{2\pi}{3}(2 + 3k) \Rightarrow
n = 2+3k$$
Second case:
$$\frac{2\pi}{3}n = \frac{5\pi}{3}+2\pi k \Rightarrow
\frac{2\pi}{3}n = \frac{2\pi}{3}\left(\frac{5}{2} + 3k\right) \Rightarrow
n = \frac{5}{2} + 3k.$$
The last one can't be satisfied since both $n$ and $k$ are integer.
If $a=\frac{4\pi}{3} $, then $na = \frac{\pi}{3}+2\pi k \vee na = \frac{2\pi}{3}+2\pi k$.
First case:
$$\frac{4\pi}{3}n = \frac{\pi}{3}+2\pi k \Rightarrow
\frac{4\pi}{3}n = \frac{4\pi}{3}\left(\frac{1}{4} + \frac{3}{2}k\right) \Rightarrow
n = \frac{1}{4} + \frac{3}{2}k.$$
The last one can't be satisfied since both $n$ and $k$ are integer.
Second case:
$$\frac{4\pi}{3}n = \frac{2\pi}{3}+2\pi k \Rightarrow
\frac{4\pi}{3}n = \frac{4\pi}{3}\left(\frac{1}{2} + \frac{3}{2}k\right) \Rightarrow
n = \frac{1}{2} + \frac{3}{2}k.$$
The last one can't be satisfied since both $n$ and $k$ are integer.
Finally, the only feasible case is:
$$a = \frac{2\pi}{3}$$
and
$$n = 2 + 3k.$$
Notice that in this case also the equation $\cos(na) + \cos(a)+ 1 = 0$ is satisfied. Indeed:
$$\cos\left((2+3k)\frac{2\pi}{3}\right) + \cos\left(\frac{2\pi}{3}\right) + 1 = 0 \Rightarrow \\
\cos\left(\frac{4\pi}{3} + 2\pi k\right) + \cos\left(\frac{2\pi}{3}\right) + 1 = 0 \Rightarrow \\
\cos\left(\frac{4\pi}{3} \right) + \cos\left(\frac{2\pi}{3}\right) + 1 = 0 \Rightarrow \\
-\frac{1}{2}-\frac{1}{2} + 1 = 0.$$
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Find $\frac{1}{7}+\frac{1\cdot3}{7\cdot9}+\frac{1\cdot3\cdot5}{7\cdot9\cdot11}+\cdots$ upto 20 terms Find $S=\frac{1}{7}+\frac{1\cdot3}{7\cdot9}+\frac{1\cdot3\cdot5}{7\cdot9\cdot11}+\cdots$ upto 20 terms
I first multiplied and divided $S$ with $1\cdot3\cdot5$
$$\frac{S}{15}=\frac{1}{1\cdot3\cdot5\cdot7}+\frac{1\cdot3}{1\cdot3\cdot5\cdot7\cdot9}+\frac{1\cdot3\cdot5}{1\cdot3\cdot5\cdot7\cdot9\cdot11}+\cdots$$
Using the expansion of $(2n)!$
$$1\cdot3\cdot5\cdots(2n-1)=\frac{(2n)!}{2^nn!}$$
$$S=15\left[\sum_{r=1}^{20}\frac{\frac{(2r)!}{2^rr!}}{\frac{(2(r+3))!}{2^{r+3}(r+3)!}}\right]$$
$$S=15\cdot8\cdot\left[\sum_{r=1}^{20}\frac{(2r)!}{r!}\cdot\frac{(r+3)!}{(2r+6)!}\right]$$
$$S=15\sum_{r=1}^{20}\frac{1}{(2r+5)(2r+3)(2r+1)}$$
How can I solve the above expression? Or is there an simpler/faster method?
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Alpha finds the sum to be $\frac {32}{129}$, giving an answer to your question of $1$. The way to do it by hand is telescoping the series.
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Show that $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} -\cdots $ converge $$1-\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{4}+\dfrac{1}{5}+\dfrac{1}{6}-\dfrac{1}{7}
-\dfrac{1}{8}-\dfrac{1}{9}-\dfrac{1}{10} ... $$
I added parentheses for each sub-sequence with the same sing.
so i got :
$$1-(\dfrac{1}{2}+\dfrac{1}{3})+(\dfrac{1}{4}+\dfrac{1}{5}+\dfrac{1}{6})-(\dfrac{1}{7}
+\dfrac{1}{8}+\dfrac{1}{9}+\dfrac{1}{10}) ... $$
I want to show that the new sequence is a leibniz sequence and by that conclude that is converge.
I managed to show that each pair of parentheses is greater than:
$$\dfrac{2}{n+1} $$
Cant find a way to proceed.
Thanks for helping.
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So we have to deal with
$$ S=\sum_{k\geq 1}(-1)^{k+1} \sum_{n=\binom{k}{2}+1}^{\binom{k+1}{2}}\frac{1}{n} = \sum_{k\geq 2} (-1)^k A_k $$
and to prove convergence it is enough to show that $\{A_k\}_{k\geq 1}$ is decreasing (from some point on) and convergent to zero. The last claim is straightfoward to prove, since $A_k\geq 0$ but
$$A_k\leq \frac{k}{\binom{k}{2}+1}\sim\frac{2}{k}. $$
The first claim can be easily proved by induction or a convexity argument:
$$ A_k = H_{\binom{k+1}{2}}-H_{\binom{k}{2}}, $$
hence $\{A_k\}$ is decreasing iff $k\mapsto H_{\binom{k}{2}}$ is a midpoint-concave function. Since:
$$\begin{eqnarray*} H_n=\sum_{k=1}^{n}\frac{1}{k} &=& \sum_{k=1}^{n}\left(\frac{1}{k}-\log\frac{k+1}{k}\right)+\sum_{k=1}^{n}\log\frac{k+1}{k}\\&=&\gamma+\log(n+1)-\sum_{k>n}\left(\frac{1}{k}-\log\frac{k+1}{k}\right)\end{eqnarray*}$$
and $\left(\frac{1}{k}-\log\frac{k+1}{k}\right)$ is bounded between $\frac{1}{2k(k+1)}$ and $\frac{1}{2k^2}$, we have:
$$ H_{\binom{k+1}{2}}-H_{\binom{k}{2}} = \log\frac{k^2+k+2}{k^2-k+2}+O\left(\frac{1}{k^4}\right)$$
so $x\mapsto H_{\binom{k}{2}}$ is a midpoint-concave function from some point on.
We may also notice that the value of our series depends on the integral:
$$ I =\int_{0}^{1}\frac{1+2\sum_{k\geq 1}(-1)^k x^{\binom{k}{2}}}{x-1}\,dx.
$$
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Integrate $ \int\frac{1+x^2}{(1-x^2)(\sqrt{1+x^4})}dx $ $$ \int\frac{1+x^2}{(1-x^2)(\sqrt{1+x^4})}dx $$
I thought of substituting $ x-\frac{1}{x} $ as $t$ but it gets stuck midway. I am close but I think I need to sustitute something else here.
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Given expression is also equal to:
$$\int\frac{1+\frac{1}{x^2}}{(x-\frac{1}{x})(\sqrt{\frac{1}{x^2}+x^2})}dx$$Which further reduces to
$$\int\frac{1+\frac{1}{x^2}}{(x-\frac{1}{x})\sqrt{(x-\frac{1}{x})^2-2}}dx$$Now let $x-\frac{1}{x}=t$ The rest is evident.
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Error in solving $\int \sqrt{1 + e^x} dx$ . I want to solve this integral for $1 + e^x \ge 0$
$$\int \sqrt{1 + e^x} dx$$
I start by parts $$\int \sqrt{1 + e^x} dx = x\sqrt{1 + e^x} - \int x \frac{e^x}{2\sqrt{1 + e^x}} dx $$
Substitute $\sqrt{1 + e^x} = t \implies dt = \frac{e^x}{2\sqrt{1 + e^x}} dx$
So I remain with $$x\sqrt{1 + e^x} - \int \ln((t-1)(t+1)) dt = x\sqrt{1 + e^x} + 2t +(-t+1)\ln(t-1)+ (-t+1)\ln(t+1) $$
that is
$$(x+2)\sqrt{1 + e^x} + (-\sqrt{1 + e^x} + 1) \ln(+\sqrt{1 + e^x} - 1) + (-\sqrt{1 + e^x} + 1) \ln(+\sqrt{1 + e^x} + 1) + C $$
But the solution should be $$2\sqrt{1 + e^x} + \ln(+\sqrt{1 + e^x} - 1) - \ln(+\sqrt{1 + e^x} + 1) $$
Where is the mistake hidden?
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HINT:
Let $\sqrt{1+e^x}=y\implies e^x=y^2-1\implies e^x\ dx=2y\ dy$
$$\int\sqrt{1+e^x}\ dx=\int\dfrac{2y^2}{y^2-1}dy$$
Now $y^2=y^2-1+1$
$$\dfrac{2y^2}{y^2-1}=2+\dfrac{y+1-(y-1)}{y^2-1}=2+\dfrac1{y-1}-\dfrac1{y+1}$$
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$\left(x+\sqrt{x^2+1}\right)\left(y+\sqrt{y^2+1}\right)=1$. Find $(x+y)$.
We know that $\left(x+\sqrt{x^2+1}\right)\left(y+\sqrt{y^2+1}\right)=1$. Find the expression $(x+y)$.
My work so far:
$$\left(x+\sqrt{x^2+1}\right)\left(y+\sqrt{y^2+1}\right)=1$$
$$\color{red}{\left(x-\sqrt{x^2+1}\right)\cdot}\left(x+\sqrt{x^2+1}\right)\left(y+\sqrt{y^2+1}\right)=\color{red}{\left(x-\sqrt{x^2+1}\right)\cdot}1$$
$$-\left(y+\sqrt{y^2+1}\right)=\left(x-\sqrt{x^2+1}\right)$$
$$-\color{red}{\left(y-\sqrt{y^2+1}\right)\cdot}\left(y+\sqrt{y^2+1}\right)=\color{red}{\left(y-\sqrt{y^2+1}\right)\cdot}\left(x-\sqrt{x^2+1}\right)$$
$$1=\left(x-\sqrt{x^2+1}\right)\left(y-\sqrt{y^2+1}\right)$$
I need help here.
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$t\mapsto t+\sqrt{t^2+1}$ is a strictly increasing function, hence for each $x$ there is at most one $y$ that makes the equation true. On the other hand, $$(x+\sqrt{x^2+1})(-x+\sqrt{x^2+1})=(x^2+1)-x^2=1$$ suggests that $y=-x$ is a valid such choice. We conclude that $x+y=0$.
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continued fraction $F(x)$ that is a generating function of central binomial coefficients Given the following continued fraction
$$F(x) =\cfrac{1}{x+\cfrac{2^2(2^2-1)}{6x+\cfrac{3^2(3^2-1)}{12x+\cfrac{4^2(4^2-1)}{20x+\cfrac{5^2(5^2-1)}{30x+\ddots}}}}}=\frac{1}{\sqrt{x^2+4}}$$
Then $$\frac{1}{x}F\left(\frac{1}{x}\right)=\sum^{\infty}_{n=0}(-1)^{n} \binom{2n}{n} x^{2n}$$
Where $\binom{2n}{n}$ are central binomial coefficients
How do we prove that the given continued fraction is a generating function of central binomial coefficients?
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$F(x)$ can be rewritten as $\displaystyle\;\frac{1}{\frac{2}{P(x)} - x}$ where
$\displaystyle\;\def\CF{\mathop{\LARGE\mathrm K}}
P(x) = \cfrac{1\cdot 2}{1\cdot 2 x + \cfrac{ (1 \cdot 2)(2\cdot 3)}{2\cdot 3 x + \cfrac{(2\cdot 3)(3\cdot 4)}{3\cdot 4 x + \ddots} }}
$.
The CF $P(x)$ has the form $
\displaystyle\;
\CF_{\ell=1}^{\infty} \frac{\alpha_\ell\gamma_{\ell-1}}{\beta_\ell}
$
where
$\gamma_0 = 1$ and
$\displaystyle\;
\begin{cases}
\alpha_\ell &= \ell (\ell+1)\\
\beta_\ell &= \ell (\ell+1) x\\
\gamma_\ell &= \ell (\ell+1)
\end{cases}
$ for $\ell > 0$.
In general, CF of this form is invariant when we scale all $\alpha_\ell, \beta_\ell, \gamma_\ell$ by same factor for all $\ell > 0$.
If we scale $\alpha_\ell, \beta_\ell, \gamma_\ell$ by $\ell(\ell+1)$ for all $\ell > 0$, we find
$$P(x) = \CF_{\ell=1}^{\infty} \frac{1}{x} = \cfrac{1}{x + \cfrac{1}{x + \cfrac{1}{x + \ddots}}}$$
The CF at RHS is well known. It is not hard to verify its convergents have the form:
$$
\CF_{\ell=1}^{n} \frac{1}{x} = \lambda_{+}\lambda_{-}\frac{\lambda_{-}^n - \lambda_{+}^n}{\lambda_{-}^{n+1} - \lambda_{+}^{n+1}}
\quad\text{ where }\quad
\lambda_{\pm} = \frac{-x \pm \sqrt{x^2+4}}{2}$$
When $x > 0$, we have $|\lambda_{-}| > |\lambda_{+}|$.
This implies
$$\begin{align}
P(x) &= \lim_{n\to\infty} \CF_{\ell=1}^{n} \frac{1}{x} = \lambda_{+} = \frac{\sqrt{x^2+4} - x}{2}\\
\implies
F(x) &= \frac{1}{\frac{4}{\sqrt{x^2+4}-x} - x}
= \frac{1}{(\sqrt{x^2+4}+x) - x} = \frac{1}{\sqrt{x^2+4}}
\end{align}
$$
Recall for any $\alpha \in \mathbb{C}$ and $|z| < 1$, we have
$$\frac{1}{(1-z)^{\alpha}} = \sum_{k=0}^\infty \frac{(\alpha)_k}{k!} z^k$$
where $(\alpha)_k = \alpha(\alpha+1)\cdots(\alpha+k-1)$. When $\alpha = \frac12$, the coefficient for $z^k$ becomes
$$\frac{(\frac12)_k}{k!} = \frac{1}{k!}\prod_{j=0}^{k-1}\left(j + \frac12\right)
= \frac{(2k-1)!!}{2^k k!} = \frac{(2k)!}{4^k(k!)^2} = \frac{1}{4^k} \binom{2k}{k}$$
From this, we find
$$\frac{1}{\sqrt{1-4z}} = \sum_{k=0}^\infty \binom{2k}{k} z^k
\quad\implies\quad
\frac1x F\left(\frac1x\right)
= \frac{1}{\sqrt{1+4x^2}}
= \sum_{k=0}^\infty (-1)^k \binom{2k}{k} x^{2k}$$
|
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|
Discriminant formula issue Could you please explain to me, how to get the formula of discriminant ? How can I visualize it, any articles, lectures?
I can memorize it $b^2 - 4ac$ But, want to understand it.
Thanks.
|
Consider the polynomial
$$ax^2+bx+c$$
First I would like to make this polynomial a multiple of a polynomial with its first term a perfect square and the second term even. To do this, we multiply by $\frac{4a}{4a}.$
$$\begin{align} \frac{1}{4a}(4a^2x^2 + 4abx+4ac) &= \frac{1}{4a}((2ax)^2+2b(2ax)+4ac) \\
& = \frac{1}{4a}((2ax)^2+2b(2ax)+b^2-b^2+4ac)\\
& = \frac{1}{4a}(\bigg((2ax)^2+2b(2ax)+b^2\bigg)-(b^2-4ac))\\
& = \frac{1}{4a}((2ax+b)^2-(b^2-4ac))\\
& = \frac{1}{4a}((2ax+b)^2-\sqrt{b^2-4ac}^2)\\
& = \frac{1}{4a}(2ax+b+\sqrt{b^2-4ac})(2ax+b-\sqrt{b^2-4ac})\\
\end{align}$$
You can easily see from this that when the discriminant is zero, the second and third factors both have the same root. When the discriminant is positive, the roots are different. And when the discriminant is negative, there are no real roots.
|
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|
Decouple a system of two second order differential equations I have a system of second-order differential equations that I want to decouple. they are,
$\ddot{x} = \frac{\omega_1^2}{2} x + \omega_2 \dot{y}$
and
$\ddot{y} = \frac{\omega_1^2}{2} y - \omega_2 \dot{x}$
I am thinking that I should use some transformation, but it just isn't clear in my head yet!
Thanks!
|
Differentiate the first wrt $t$ to gain an expression for $\ddot y$:
$\dddot{x} = \frac{\omega_1^2}{2} \dot x + \omega_2 \ddot{y}$
Substitute $\ddot{y} = \frac{\omega_1^2}{2} y - \omega_2 \dot{x}$ to get:
$\dddot{x} = \frac{\omega_1^2}{2} \dot x + \frac{\omega_1^2 \omega_2}{2} y - \omega_2^2 \dot{x}$
Rearrange: $\frac{\omega_1^2 \omega_2}{2} y =\dddot{x} - \frac{\omega_1^2}{2} \dot x + \omega_2^2 \dot{x}$
Differentiate: $\frac{\omega_1^2 \omega_2}{2} \dot y =\ddddot{x} + \frac{2\omega_2^2 -\omega_1^2}{2} \ddot x$
Recall that $\ddot{x} = \frac{\omega_1^2}{2} x + \omega_2 \dot{y} \Rightarrow \omega_2 \dot{y}=\ddot{x} - \frac{\omega_1^2}{2} x$
Thus: $\frac{\omega_1^2}{2} \left (\ddot{x} - \frac{\omega_1^2}{2} x \right ) =\ddddot{x} + \frac{2\omega_2^2 -\omega_1^2}{2} \ddot x$
... which becomes $\ddddot{x} + \frac{2\omega_2^2 -\omega_1^2}{2} \ddot x - \frac{\omega_1^2}{2} \left (\ddot{x} - \frac{\omega_1^2}{2} x \right )=0 $
or $\ddddot{x} + \left(\omega_2^2 -\omega_1^2 \right ) \ddot x - \frac{\omega_1^4}{2}x =0 $
Auxiliary equation $\lambda^4+p\lambda^2-q=0$ where $p=\left(\omega_2^2 -\omega_1^2 \right )$ and $q=\omega_1^4$
$\lambda^2={-p+\sqrt{p^2+4q} \over 2}$ or $\lambda^2={-p-\sqrt{p^2+4q} \over 2}$
$\lambda_1=\sqrt{{-p+\sqrt{p^2+4q} \over 2}}$
$\lambda_2=-\sqrt{{-p+\sqrt{p^2+4q} \over 2}}$
$\lambda_3=\sqrt{{-p-\sqrt{p^2+4q} \over 2}}$
$\lambda_4=-\sqrt{{-p-\sqrt{p^2+4q} \over 2}}$
|
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|
Find the missing digits in the expansion of $34!$ If $34!=\overline{295232799cd96041408476186096435ab000000}$ then find the value of $a,b,c$ and $d.$
My Attempt: I can find that $b=0$ because it has seven five integers.
$\lfloor{\frac{34}{5}}\rfloor+\lfloor{\frac{34}{25}}\rfloor=6+1=7$
Also I think $a$ can be found using divisibility rule of $8$ that means $\overline{35a}$ is divisible by $8$ which gives us $a=2$. But I'm stuck in finding $c$ and $d$. Using the divisibility rule of $9$ I get an equation. And using divisibility rule of $11$ I get another equation but when I solve them I get two values for $c$ and $d$.
Note: calculator is not allowed.
|
This is a partial solution...
Use all the divisibility rules.
But first, you can notice that $32!=2^31\times5^7\times\cdots$, so there are $7$ zeros at the end of the number. So $b=0$, and $a\ne 0$.
$34!$ is divisible by $9$, so:
$$4+a+c+d=0\pmod 9.$$
It's divisible by $7$, so:
$$000-000+5a0-643+609-618+847-140+604-cd9+799+327-952+2=0\pmod 7,$$
so
$$835+5a0-cd9=0\pmod 7.$$
It's divisible by $11$, so:
$$2-2+5-2+9-7+2-3+9-9+7-9+d-c+4-0+6-0+4-1+7-4+1-7+4-8+8-1+6-9+0-6+3-4+6-0+a-5=0\pmod {11},$$
so
$$6+a-c+d=0\pmod {11}.$$
It's also divisible by $13$, so you can look at the alternate sum of three digits from the left.
And so on...
Try to look at this article about divisibility rules.
I hope this helps!
|
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|
If $a^3=1$, is $G$ abelian? If $G$ is a group that satisfies $a^3=1$ for every $a\in G$, then is $G$ abelian?
This is an exercise I found in Jacobson's Basic Algebra. It is analogous to the question: If $G$ is a group that satisfies $a^2=1$ for every $a\in G$, then $G$ is abelian. I tried to multiply on both sides of $ab$ ($a,b\in G$) by some appropriate ($b***$) and ($***a$) three times to yield $ba$ but failed. Nor could I give a counterexample. Can someone give me an answer to the question and some hints to solve it?
|
The multiplicative group of matrices
$$G = \left\{ \begin{pmatrix}1&a&b\\0&1&c\\0&0&1\end{pmatrix} \middle\vert\, a,b,c \in \mathbb{F}_3\right\} \subset \operatorname{Mat}(3 \times 3, \mathbb{F}_3)$$
is a counterexample (it is isomorphic to the group in Joanpemo's answer).
For all those who are not willing to check this example on there own, here are the arguments:
*
*Every element $g \in G$ has characteristic polynomial $\chi_g(\lambda) = \lambda^3 - 1$, and thus has order $3$ by the Cayley-Hamilton theorem.
*The elements $\begin{pmatrix}1&1&0\\0&1&0\\0&0&1\end{pmatrix}$ and $\begin{pmatrix}1&0&0\\0&1&1\\0&0&1\end{pmatrix}$ do not commute:
$$ \begin{pmatrix}1&1&0\\0&1&0\\0&0&1\end{pmatrix} \cdot \begin{pmatrix}1&0&0\\0&1&1\\0&0&1\end{pmatrix} = \begin{pmatrix}1&1&1\\0&1&1\\0&0&1\end{pmatrix} $$
$$ \begin{pmatrix}1&0&0\\0&1&1\\0&0&1\end{pmatrix} \cdot \begin{pmatrix}1&1&0\\0&1&0\\0&0&1\end{pmatrix} = \begin{pmatrix}1&1&0\\0&1&1\\0&0&1\end{pmatrix}$$
|
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|
Using induction to prove for $n ≥ 1, $ $1 \times 5+2\times6+3\times7 +\cdots +n(n + 4) = \frac 16n(n+1)(2n+13).$ This is a very interesting problem that I came across in an old textbook of mine. So I know its got something to do with mathematical induction, which yields the shortest, simplest proofs, but other than that, the textbook gave no hints really and I'm really not sure about how to approach it. Any guidance hints or help would be truly greatly appreciated. Thanks in advance :) So anyway, here the problem goes:
Prove that for $n ≥ 1,$
$1 \times 5+2\times6+3\times7 +\cdots +n(n + 4) = \frac 16n(n+1)(2n+13).$
|
Base Case: For $n = 1$, we have:
$$
1 \times 5 = 5 = \frac{1}{6}(1)(1 + 1)(2(1) + 13)
$$
which works.
Inductive Hypothesis: Assume that the claim holds for $n' = n - 1$, where $n \geq 2$.
It remains to show that the claim holds for $n' = n$. Indeed, observe that:
\begin{align*}
&1 \times 5 +\cdots + n(n + 4) & \\
&= [1 \times 5 + \cdots + (n - 1)(n + 3)] + n(n + 4) &\text{since } n \geq 2 \\
&= \frac{1}{6}(n - 1)(n)(2n + 11) + n(n + 4) &\text{by the inductive hypothesis} \\
&= \frac{1}{6}n [(n - 1)(2n + 11) + 6(n + 4)] \\
&= \frac{1}{6}n [(2n^2 + 9n - 11) + (6n + 24)] \\
&= \frac{1}{6}n [2n^2 + 15n + 13] \\
&= \frac{1}{6}n (n + 1)(2n + 13) \\
\end{align*}
as desired. $~~\blacksquare$
|
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|
When one of these two equations has integral sol, why p=3 (mod 4)? Let p is not 7 and p be an odd prime.
AND one of two equations
$x^2+7y^2=p, x^2-7y^2=p$
has integral sol (x,y). then show that $p \equiv3 \pmod4$
what i have done is this, in $\pmod7$, two equations are $x^2 \equiv p \pmod7$
and $p$ can be only $1$ or $2$ or $4$ by testing from $1$ to $7$
|
Suppose $x^2+7y^2=p$. Applying modulo 4:
$$x^2+3y^2=p \pmod 4 \tag1$$ then $$x^2-y^2=p \pmod 4 \tag2$$ Because $p,4$ are coprime from (2) we get $x \ne y \pmod 4$.
Because $a^2 \in \{0, 1\} \pmod 4 \ \forall a$, we have $x^2=0,y^2=1$ or $x^2=1,y^2=0$.
From the first case we get $p = 3 \pmod 4$ and from the second case $p=1 \pmod 4$.
Both cases are possible:
1) $x=2, y=1$ then $p=11$ and $11 = 3 \pmod 4$
2) $x=1,y=2$ then $p=29$ and $29 = 1 \pmod 4$
Therefore your claim $p \equiv3 \pmod4$ is false.
|
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|
Find the volume of the region enclosed by $x^2+y^2+z^2=2$ and $x^2+y^2=z$.
Find the volume of the region enclosed by $x^2+y^2+z^2=2$ and $x^2+y^2=z$.
I tried to solve the problem above by doing a change of variables to the spherical coordinate system, that is,
$x= \rho \cos \theta \sin \varphi$
$y = \rho \sin \theta \sin \varphi$
$z= \rho \sin \varphi$
But I'm struggling with the range of each variable, $\rho, \theta$ and $\varphi$. What I did is set $\theta$ from $0$ to $2 \pi$, $\varphi$ from $\pi /4$ to $3 \pi /4$ and $\rho$ from $0$ to $\sqrt{2}$, so I get
$$v= \int_0^{\sqrt{2}}\left( \int_0^{2 \pi} \left( \int_{\frac{\pi}{4}}^{\frac{3 \pi}{4}} \rho ^2 \sin \varphi d\varphi \right) d\theta \right)d \rho = \frac{8 \pi}{3}$$
Is this result right? If not, can you suggest me how to proceed? Thanks in advance!
EDIT: After some thinking, I've arrived to the following:
Using Tom-Tom suggestion in the comment that I should split the figure in two parts and change to cylimdral coordinates, let's call the two volumes $S_1$ and $S_2$, where $S_1$ is the upper part (from $z=1$ to $z=\sqrt{2}$) and $S_2$ the rest of it.
For $S_1$, let's first fix the variable $\rho$ which in this case goes from $0$ to $1$. Now, $z$ goes from $0$ to the sphere defined by $x^2+y^2+z^2=2$, which in cylimdral coordinates is $z^2= 2- \rho^2$, or $z=\sqrt{2-\rho^2}$. And finall, $\theta$, which moves from $0$ to $2 \pi$. So then,
$$v(S_1)=\int_{S_1} 1 = \int_0^{2\pi} \int_0^1 \int_1^{\sqrt{2-\rho^2}} \rho \ dz \ d\rho \ d\theta = 2\pi \int_0^1 \rho (\sqrt{2-\rho^2}-1)\ d\rho = 2\pi \left(\frac{\sqrt{8}}{3}-\frac{5}{6}\right)$$
Mow, for $S_2$, we fix the variable $z$ first, that in this case goes from $0$ to $1$. $\rho$ moves from $0$ to the paraboloid defined by $z=x^2+y^2$, which in cylindral coordinates is $\rho = \sqrt{z}$ and $\theta$ goes fro $0$ to $2\pi$ as before. So
$$v(S_2)=\int_{S_2} 1= \int_0^1 \int_0^{2\pi} \int_0^{\sqrt{z}} \rho \ d\rho \ d\theta \ dz = \frac{\pi}{2}$$
And since $S_1$ and $S_2$ intersect on a circle, whic has null measure in $\mathbb{R}^3$,
$$v(S_1 \cup S_2)= v(S_1) + v(S_2)$$
Is these reasoning correct or I'm stil messing it up with the change of coordinates?
|
We will assume that the problem refers to the region
above the paraboloid $z=x^2+y^2$ and inside the sphere $x^2+y^2+z^2=2$.
Since the volume can be found by integrating the area of the horizontal slices of the region,
and each of the horizontal slices is a circular disc,
$\displaystyle V=\int_0^1\pi(r(z))^2dz+\int_1^{\sqrt{2}}2\pi(r(z))^2dz=\int_0^1\pi z\;dz+\int_1^{\sqrt{2}}\pi(2-z^2)dz$
$\displaystyle\hspace{.16 in}=\frac{\pi}{2}+\pi\left(\frac{4}{3}\sqrt{2}-\frac{5}{3}\right)=\frac{\pi}{6}\left(8\sqrt{2}-7\right)$
|
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|
Extreme values of $\frac{(a+b+c)(2ab+2bc+2ca-a^2-b^2-c^2)}{abc}$ Let $a,b,c$ be side lengths of a triangle. What are the minimum and maximum of $$\frac{(a+b+c)(2ab+2bc+2ca-a^2-b^2-c^2)}{abc}?$$
When $a=b=c$, the value is $9$. In addition, we can write $a=x+y,b=y+z,c=z+x$ since they are side lengths of a triangle. The expression becomes
$$\frac{8(x+y+z)(xy+yz+xz)}{(x+y)(y+z)(z+x)}.$$
|
If $a=b=1$ and $c\rightarrow2^-$ so $$\frac{(a+b+c)(2ab+2bc+2ca-a^2-b^2-c^2)}{abc}\rightarrow8,$$
but $$\frac{(a+b+c)(2ab+2bc+2ca-a^2-b^2-c^2)}{abc}>8$$ it's
$$\sum_{cyc}\left(-a^3+a^2b+a^2c-\frac{2}{3}abc\right)>0$$ or
$$(a+b-c)(a+c-b)(b+c-a)>0,$$
which says that the minimum does not exist and
$$\inf\frac{(a+b+c)(2ab+2bc+2ca-a^2-b^2-c^2)}{abc}=8$$
Let $a=b=c$.
Hence, $$\frac{(a+b+c)(2ab+2bc+2ca-a^2-b^2-c^2)}{abc}=9,$$
but $$\frac{(a+b+c)(2ab+2bc+2ca-a^2-b^2-c^2)}{abc}\leq9$$ it's
$$\sum_{cyc}(a^3-a^2b-a^2c+abc)\geq0,$$
which is Schur.
Thus, $$\max\frac{(a+b+c)(2ab+2bc+2ca-a^2-b^2-c^2)}{abc}=9.$$
|
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|
Confusion in rotation matrix - rotation about $y$ axis The rotation matrix about y axis should look like
$$\left[
\begin{array}{ccc}
\cos\frac{\pi}{2} & 0 &\sin\frac{\pi}{2}\\
0 & 1 & 0\\
-\sin\frac{\pi}{2} & 0 &\cos\frac{\pi}{2}\\
\end{array}
\right]
=
\left[
\begin{array}{ccc}
0 & 0 &1\\
0 & 1 & 0\\
-1& 0 &0\\
\end{array}
\right]
$$
Now, I pictured this rotation, and it should look like rotation about y axis, counterclockwise
I drew $(1,1,1)$ vector in original coordinate system, and in the new coordinate system, the vector should be $(-1,1,1)$
BUT! the matrix calculation gives otherwise.
$$
\left[
\begin{array}{ccc}
0 & 0 &1\\
0 & 1 & 0\\
-1& 0 &0\\
\end{array}
\right]
\left[
\begin{array}{c}
1\\
1\\
1
\end{array}
\right]
=\left[
\begin{array}{c}
1\\
1\\
-1
\end{array}
\right]
$$
Am I missing something?
Okay, it seems like I should rotate vector instead of coordinate system.
Then, How can you explain rotation about z axis?
The rotation matrix is
$$
\left[
\begin{array}{ccc}
\cos\frac{\pi}{2} &\sin\frac{\pi}{2} & 0\\
-\sin\frac{\pi}{2} &\cos\frac{\pi}{2} & 0\\
0 & 0 & 1
\end{array}
\right]
=
\left[
\begin{array}{ccc}
0 & 1 &0\\
-1 & 0 & 0\\
0& 0 &1\\
\end{array}
\right]
$$
But... calculation is correct.
$$
\left[
\begin{array}{ccc}
0 & 1 &0\\
-1 & 0 & 0\\
0& 0 &1\\
\end{array}
\right]
\left[
\begin{array}{c}
1\\
1\\
1
\end{array}
\right]
=\left[
\begin{array}{c}
1\\
-1\\
1
\end{array}
\right]
$$
IF you rotate the vector, then in the picture, it should be $(-1,1,1)$
Oh. god.... I think I am looking at the wrong source.
My book tells
well, clearly, googling gives me
http://what-when-how.com/the-3-d-global-spatial-data-model/rotation-matrix-derivation-the-3-d-global-spatial-data-model/
I think I solved my case. Thank you all.
|
I believe that you are viewing the problem backwards. The matrix
$$
A=\begin{pmatrix}
0 & 0 & 1 \\
0 & 1 & 0 \\
-1 & 0 & 0
\end{pmatrix}
$$
rotates a vector into a new, rotated frame, according to what you are describing with your images. However, what you are asking is, "What matrix brings me from the rotated frame back to the original frame", which would be the inverse of $A$, i.e.
$$
A^{-1} = \begin{pmatrix}
0 & 0 & -1 \\
0 & 1 & 0 \\
1 & 0 & 0
\end{pmatrix}.
$$
|
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|
Decomposition of a rational function in partial fractions I am trying to decompose the following rational function: $\dfrac{1}{(x^2-1)^2}$ in partial fractions (in order to untegrate it later).
I have notices that $(x^2-1)^2 = (x+1)^2(x-1)^2$
Therefore $\exists A, B, C, D$ s.t: $\dfrac{1}{(x^2-1)^2} = \dfrac{1}{(x-1)^2(x+1)^2} = \dfrac{Ax+B}{(x+1)^2}+\dfrac{Cx+D}{(x-1)^2}$
We then have $Ax+B =\left. \dfrac{1}{(x-1)^2} \right\vert _{x=-1} \implies B-A=1/4$
Same thing for Cx+D: $Cx+D =\left. \dfrac{1}{(x+1)^2} \right\vert _{x=1} \implies C+D=1/4 $
How do I find A, B, C, D from here?
|
We have
$$1=(x-1)^2(Ax+B)+(x+1)^2(Cx+D)$$
You already have
$$B-A=C+D=\frac 14$$
Substituting $x=0$ gives
$$1=B+D$$
Substituting $x=2$ gives
$$1=2A+B+9(2C+D)$$
Solving these gives
$$A=\frac 14,\quad B=\frac 12,\quad C=-\frac 14,\quad D=\frac 12$$
|
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|
Finding $\lim_{x\to 0} \frac{\ln(x+4)-\ln(4)}{x}$ without L'hôpital I have this $\lim_{}$.
$$\lim_{x\to 0} \frac{\ln(x+4)-\ln(4)}{x}$$
Indetermation:
$$\lim_{x\to 0} \frac{\ln(0+4)-\ln(4)}{0}$$
$$\lim_{x\to 0} \frac{0}{0}$$
Then i started solving it:
$$\lim_{x\to 0} \frac{\ln\frac{(x+4)}{4}}{x}$$
$$\lim_{x\to 0} \frac{1}{x}\phantom{2} .\phantom{2} \ln\lim_{x\to 0} \frac{(x+4)}{4}$$
$$\lim_{x\to 0} \frac{1}{x}\phantom{2} .\phantom{2} \ln\lim_{x\to 0} (\frac{x}{4}+1)$$
$$\lim_{x\to 0} \frac{1}{x}\phantom{2} .\phantom{2} \ln\lim_{x\to 0} (1 + \frac{x}{4})$$
$$\ln\lim_{x\to 0} (1 + \frac{x}{4})^\frac{1}{x}$$
Then multiply the power, by 4.
$$\ln\lim_{x\to 0} (1 + \frac{x}{4})^\frac{1.4}{x.4}$$
$$\ln \begin{bmatrix}\lim_{x\to 0} (1 + \frac{x}{4})^\frac{4}{x} \end{bmatrix}^\frac{1}{4} $$
$$ \ln \phantom{2}\mathcal e^\frac{1}{4} = \frac{1}{4}$$
Until here is fine, but someone's telling me that it can be made like this:
$$\lim_{x\to 0} \frac{\ln(x) + \ln(4) -\ln(4)}{x}$$
$$\lim_{x\to 0} \frac{\ln(x)}{x} = 1$$
Alright, it uses logarithmic property to separate the expression, but i said it can't be, because of the indetermination, and $\frac{1}{4} $ is closer of $0$ than $1$. There's another reasonable explanation for this ? Why it can't be or when can it ?
And, there's a trick to way out quickly with this limit, keeping L'Hôspital aside ?
|
You want
$\lim_{x\to 0} \frac{\ln(x+4)-\ln(4)}{x}
$.
One of the definitions
of $\ln$ is that
$\ln(x)
=\int_1^x \frac{dt}{t}
$.
Therefore
$\ln(x+4)-\ln(4)
=\int_4^{x+4} \frac{dt}{t}
$.
If
$x > 0$
then,
since
$\frac14
\ge \frac{1}{t}
\ge \frac1{x+4}
$
for
$4 \le t \le 4+x$,
$\frac{x}{4}
\ge \int_4^{x+4}\frac{dt}{t}
\ge \frac{x}{x+4}
$,
so that
$\frac{1}{4}
\ge \frac1{x}\int_4^{x+4}\frac{dt}{t}
\ge \frac{1}{x+4}
$.
Therefore
$\lim_{x \to 0^+}\frac1{x}\int_4^{x+4}\frac{dt}{t}
=\frac14
$.
We can similarly show that
$\lim_{x \to 0^-}\frac1{x}\int_4^{x+4}\frac{dt}{t}
=\frac14
$.
Therefore the limit is
$\frac14$.
|
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|
Show that $x$ and $Q x$ are equidistant $$Q= \begin{bmatrix} \cos x & -\sin x\\ \sin x & \cos x\end{bmatrix}$$
Given x belongs to $\mathbb{R^2}$, show $Qx$ and $x$ are equidistant.
I've tried dot producting $Qx$ and seeing whether they are equal. I just can't seem to get it.
|
Of course, @thanasissdr answer is the well educated one, but arguing more naïvely:
As written in my comment:
$$d((a,b),(0,0)=\sqrt{(a-0)^2+(b-0)^2}=\sqrt{a^2+b^2}=\|(a,b)\|$$
Thus we compare the norms.
We have:
$$\|(a,b)\|=\sqrt{a^2+b^2}$$
and
\begin{align*}\|Q(x,y)\|& =\|(\cos(x)a+\sin(x)b,-\sin(x)a+\cos(x)b)\|\\ & =\sqrt{(\cos(x)a+\sin(x)b)^2+(-\sin(x)a+\cos(x)b)^2}\\ & =\sqrt{\cos^2(x)a^2+\sin^2(x)b^2+2\sin(x)\cos(x)ab\cos^2(x)b^2+\sin^2(x)a^2\color{red}{-}2\sin(x)\cos(x)ab}\\ & =\sqrt{[\sin^2(x)+ \cos^2(x)]a^2+[\sin^2(x)b^2+\cos^2(x)]b^2}\\ & =\sqrt{a^2+b^2}\end{align*}
as claimed, where we have used the classic identity $\sin^2(x)+\cos^2(x)=1$.
Note: Any matrix $Q$ which fulfills this property is called orthogonal matrix. This is equivalent to the above cited criterion $Q^\top Q=1$.
|
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|
Simplifying derivative result I am doing the derivative of
$$f(x) = \frac{x^2 -4x +3}{x^2-1}$$
So my result is the following
$$f'(x) = \frac{4x^2 -8x +4}{(x^2-1)^2}$$
I am sure the answer is correct, but in my solutions book and In Wolfram Alpha they simplify until
$$f'(x) = \frac{4}{(x+1)^2}$$
And I don't know why, which steps are they doing?
|
You are correct. Now note that
$$\frac{4x^2-8x+4}{(x^2-1)^2}=\frac{4(x^2-2x+1)}{((x-1)(x+1))^2}=\frac{4(x-1)^2}{(x-1)^2(x+1)^2}=\frac{4}{\color{red}{(x+1)^2}}$$
This is not $$\frac{4}{(x^2-1)^2}$$
|
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|
Solving Chinese Remainder Theorem Algebraically I am doing a practice problem for my final which asks:
Solve the following Chinese Remainder Theorem:
$$
x \equiv 2 \pmod{3}, \\
x \equiv 3 \pmod{5}, \\
x \equiv 5 \pmod{7}, \\
x \equiv 7 \pmod{11} \\
x \equiv 11 \pmod{13}
$$
From the first I can conclude that $x = 3k + 2$ for some $k \in \mathbb{Z}$.
Now I can apply that to the second equation which gives $ 3k+2 \equiv 3 \pmod{5}.$
Then I get lost here. Do I subtract $2$ and solve $ 3k \equiv 1 \pmod{5}$?
I don't have a solid understanding of solving the Chinese Remainder Theorem algebraically in general.
|
\begin{align}
x &\equiv 2 \pmod 3 \\
x &= 2 + 3a \\
\hline
x &\equiv 3 \pmod 5 \\
2+3a &\equiv 3 \pmod 5 \\
3a &\equiv 1 \pmod 5 \\
a &\equiv 2 \pmod 5 \\
a &= 2 + 5b \\
x &= 2 + 3(2 + 5b)\\
x &= 8 + 15b \\
\hline
x &\equiv 5 \pmod 7 \\
8 + 15b &\equiv 5 \pmod 7 \\
1 + b &\equiv 5 \pmod 7 \\
b &\equiv 4 \pmod 7 \\
b &= 4 + 7c \\
x &= 8 + 15(4 + 7c) \\
&\text{and so on...}
\end{align}
|
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|
find all primes $p$ and $q$ such that $p \cdot q | 2^p + 2^q$
I have to find all prime numbers $p,q$ such that $p\cdot q | 2^p + 2^q$.
I don't know from what I have to start.
|
If $p=q$, then, $p^2|2^{p+1}$. So, $p=q=2$.
Now,suppose $p\ne q$.
If $p=2$, we have $q|2+2^{q-1}$. By Fermat's Little Theorem(FLT), $q=3$. Similarly, $q=2,p=3$ is another solution.
Now, suppose $p$ and $q$ are odd primes. Let $p-1=2^ms$ and $q-1=2^nr$, where $r$ and $s$ odd numbers.
$2^q\equiv -2^p\equiv -2\pmod p$. So, $2^{(q-1)}\equiv -1\pmod p$ and $2^{2(q-1)}\equiv 1\pmod p$. Let $k$ be the order of $2$ in $\bmod p$. Then, $k|2(q-1)=2^{n+1}r$. So, $k=2^{n_0}r_0$ for some $n_0\le n+1$ and $r_0|r$. On the other hand, $k\not\lvert (q-1)$, thus, $n_0> n$.
So, $k=2^{n+1}r_0$. However, by FLT, $k|p-1$, too. So, $m\ge n+1$.
However, $2^p\equiv -2^q\equiv -2\pmod q$. So, $2^{(p-1)}\equiv -1\pmod q$ and $2^{2(p-1)}\equiv 1\pmod q$. Let $l$ be the order of $2$ in $\bmod q$. Then, $l|2(p-1)=2^{m+1}s$. So, $l=2^{m_0}s_0$ for some $m_0\le m+1$ and $s_0|s$. On the other hand, $l\not\lvert (p-1)$, thus, $m_0> m$.
So, $l=2^{m+1}s_0$. However, by FLT, $l|q-1$, too. So, $n\ge m+1$.
However, we get
$$n\ge m+1\ge n+2$$
Contradiction.Thus, $(2,2),(2,3),(3,2)$ are the only solutions.
|
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|
Integrate $\int \frac {x^{2}} {\sqrt {x^{2}-16}}dx$ $\displaystyle\int \dfrac {x^{2}} {\sqrt {x^{2}-16}}dx$
Effort 1:
Let be $x=4\sec u$
$dx=4.\sin u.\sec^2u.du$
Then integral;
$\displaystyle\int \dfrac {\sec^2u \; .4.\sin u.\sec^2u.du} {\sqrt {16\sec^2u-16}}=\displaystyle\int \sec^3.du$
After I didn't nothing.
Effort 2:
$\displaystyle\int \dfrac {x^{2}dx} {\sqrt {x^{2}-16}}$
Let's doing integral by parts;
$du=\dfrac{x}{\sqrt{x^2-16}}$
$v=x$
$\displaystyle\int \dfrac {x^{2}dx} {\sqrt {x^{2}-16}}=x.\sqrt{x^2-16}-\displaystyle\int\sqrt{x^2-16}dx$
We have $\quad \displaystyle\int\sqrt{x^2-16}dx$
let be $\quad x=4\sec j$
$dx=4\dfrac{\sin j}{\cos^2 j}dj$
and;
$\displaystyle\int\sqrt{x^2-16}\;dx=16.\displaystyle\int\dfrac{\sin j}{\cos j}\dfrac{\sin j}{\cos^2 j}dj=16.\displaystyle\int \sec j.tan^2j.dj$
After I didn't nothing.
|
$\newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle}
\newcommand{\braces}[1]{\left\lbrace\, #1 \,\right\rbrace}
\newcommand{\bracks}[1]{\left\lbrack\, #1 \,\right\rbrack}
\newcommand{\dd}{\mathrm{d}}
\newcommand{\ds}[1]{\displaystyle{#1}}
\newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
\newcommand{\half}{{1 \over 2}}
\newcommand{\ic}{\mathrm{i}}
\newcommand{\iff}{\Leftrightarrow}
\newcommand{\imp}{\Longrightarrow}
\newcommand{\ol}[1]{\overline{#1}}
\newcommand{\pars}[1]{\left(\, #1 \,\right)}
\newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
\newcommand{\root}[2][]{\,\sqrt[#1]{\, #2 \,}\,}
\newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
\newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert}$
With $\ds{\quad t \equiv x - \root{x^{2} - 16}\quad\imp\quad
x = {t^{2} + 16 \over 2t}}$:
\begin{align}
\color{#f00}{\int{x^{2} \over \root{x^{2} - 16}}\,\dd x} & =
\int\pars{-\,{64 \over t^{3}} - {8 \over t} - {t \over 4}}\,\dd t =
{32 \over t^{2}} - 8\ln\pars{t} - {1 \over 8}\,t^{2}
\\[3mm] & =
{32 \over \pars{x - \root{x^{2} - 16}}^{2}} -
8\ln\pars{x - \root{x^{2} - 16}} - {1 \over 8}\pars{x - \root{x^{2} - 16}}^{2}
\\[3mm] & =
{1 \over 8}\,\bracks{\pars{x + \root{x^{2} - 16}}^{2} -
\pars{x - \root{x^{2} - 16}}^{2}} - 8\ln\pars{x - \root{x^{2} - 16}}
\\[3mm] & =
\color{#f00}{\half\,x\root{x^{2} - 16} - 8\ln\pars{x - \root{x^{2} - 16}}}
\end{align}
|
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|
Sum of combinatorics sequence $\binom{n}{1} + \binom{n}{3} +\cdots+ \binom{n}{n-1}$ I need to find sum like $$\binom{n}{1} + \binom{n}{3} +\cdots+ \binom{n}{n-1},\qquad \text{ for even } n$$
Example:
Find the sum of $$\binom{20}{1} + \binom{20}{3} +\cdots+ \binom{20}{19}=\ ?$$
|
If you know your Pascal's triangle, you know that
$$\binom{20}{1} + \binom{20}{3} +..+ \binom{20}{19}= \sum_{k=0}^{19} \binom{19}{k}$$
|
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|
Find the $k$ such that $2^{(k-1)n+1}$ does not divide $\frac{(kn)!}{n!}$. Find all positive integers $k$ such that for any positive integer $n$, $2^{(k-1)n+1}$ does not divide $\frac{(kn)!}{n!}$.
From olympiad problem
I'm curious So far no one to solve this problem,Maybe it is very difficult May be other reasons? But I really want to know how to solve this problem more easy?
|
The number of factors $2$ in a number $a!$ is:
$$
\left\lfloor \frac a 2 \right\rfloor
+
\left\lfloor \frac a 4 \right\rfloor
+
\left\lfloor \frac a 8 \right\rfloor
+ \cdots
$$
So we have to find all $k$ such that for every $n$, we have
$$
kn - n + 1 >
\sum_{x=1} \left\lfloor \frac{kn}{2^x} \right\rfloor
-
\sum_{x=1} \left\lfloor \frac{n}{2^x} \right\rfloor
$$
Now, to save typing, I define $F(a) = \sum_{x=1} \left\lfloor \frac{a}{2^x} \right\rfloor$. So how much is $F(a)$ compared to $a$?
Lemma: $F(2^a) = 2^a-1$
Proof: We have $F(2^a) = 2^{a-1} + 2^{a-2} + \cdots + 1 = 2^a-1$.
$\blacksquare$
Lemma: If $b < 2^a$, then $F(2^a + b) = F(2^a) + F(b)$.
Proof: For every $x < a$, $\frac{2^a}{2^x}$ is an integer and thus
$$
\left\lfloor \frac{2^a + b}{2^x} \right\rfloor
=
\left\lfloor \frac{2^a}{2^x} \right\rfloor
+
\left\lfloor \frac{b}{2^x} \right\rfloor
$$
For every $x \geq a$, we have $\left\lfloor \frac{b}{2^x} \right\rfloor = 0$, so the equation holds as well.
$\blacksquare$
A consequence of this is that $a - F(a)$ is the number of ones in the binary representation of $a$.
So $kn - n + 1 > F(kn) - F(n)$ means that $kn - F(kn) \geq n - F(n)$, which means that the binary representation of $kn$ has at least as many ones as the one of $n$.
So it's obvious that $k = 2^a$ always works. I'll think of some reasons why other $k$s don't work shortly.
|
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|
Prove that $a_1+\frac{a_2^2}{a_1+a_2}+\frac{a_3^2}{a_1+a_2+a_3}>b_1+\frac{b_2^2}{b_1+b_2}+\frac{b_3^2}{b_1+b_2+b_3}.$ Suppose $a_1>a_2>a_3>0$ and $b_1>b_2>b_3>0$ and $a_1>b_1,a_2>b_2,a_3>b_3$.
I want to prove that
$$a_1+\frac{a_2^2}{a_1+a_2}+\frac{a_3^2}{a_1+a_2+a_3}>b_1+\frac{b_2^2}{b_1+b_2}+\frac{b_3^2}{b_1+b_2+b_3}.$$
It doesn't appear to fit nicely with any standard inequality. A brute force simplification doesn't seem to work either. Has anyone seen a clean approach to a problem like this that might work?
|
Let $f(a,b,c)=a+\dfrac{b^2}{a+b}+\dfrac{c^2}{a+b+c}$.
Suppose $a_1 \geq a_2 \geq a_3$ and $b_1 \geq b_2 \geq b_3$ and $a_1 \geq b_1,a_2 \geq b_2,a_3 \geq b_3$.
I prove $f(a_1,a_2,a_3) \geq f(a_1,a_2,b_3) \geq f(a_1,b_2,b_3) \geq f(b_1,b_2,b_3)$.
The first inequality follows from the monotonicity of $\dfrac{x^2}{r+x}$ for any $r>0$.
The second is: if $a \geq b \geq B \geq c$, then
$\dfrac{b^2}{a^2+b}+\dfrac{c^2}{a+b+c}>\dfrac{B^2}{a^2+B}+\dfrac{c^2}{a+B+c}$,
which is equivalent to:
$\dfrac{a(b+B)+bB}{(a+b)(a+B)} \geq \dfrac{c^2}{(a+b+c)(a+B+c)}$,
which is easily seen to be true.
The third inequality is: if $a \geq A \geq b \geq c$, then $f(a,b,c) \geq f(A,b,c)$, which is equivalent to:
$1 \geq \dfrac{b^2}{(A+b)(a+b)}+\dfrac{c^2}{(a+b+c)(A+b+c)}$,
which is again seen to be true.
Each of the inequalities is tight only when the corresponding variables are equal; hence if in addition we have $a_i>b_i$ for some $i$, then we get $f(a_1,a_2,a_3)>f(b_1,b_2,b_3)$.
|
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|
Prove $a^2+b^2+c^2 \ge a+b+c$ if $abc=1$, and $a$, $b$, $c$ are positive real numbers
Prove $a^2+b^2+c^2 \ge a+b+c$ if $abc=1$, and $a$, $b$, $c$ are positive real numbers
It is in the exercises of the AM-GM inequality chapter of a book, and that is why I believe it will be solved by that. Can anyone give me a proof using that or otherwise, too?
|
We use AM-GM three times
$$\frac{a^2 +a^2+a^2+a^2 +b^2+c^2}{6} \geq a (abc)^{1/3}$$
$$\frac{b^2 +b^2+b^2+b^2 +a^2+c^2}{6} \geq b (abc)^{1/3}$$
$$\frac{c^2 +c^2+c^2+c^2 +b^2+a^2}{6} \geq c (abc)^{1/3}$$
Summing these inequalities and dividing by 6 gives
$$a^2+b^2+c^2 \geq (a+b+c)(abc)^{1/3}$$
Now using that $abc=1$, we conclude the desired inequality.
|
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|
Restricted equality involving prime numbers Given three real numbers such that $a + b + c = 0$, it can be proved that
\begin{align*}
\frac{a^{5} + b^{5} + c^{5}}{5} & = \frac{a^{3} + b^{3} + c^{3}}{3}\cdot \frac{a^{2} + b^{2} + c^{2}}{2}\\
\frac{a^{7} + b^{7} + c^{7}}{7} & = \frac{a^{5} + b^{5} + c^{5}}{5}\cdot \frac{a^{2} + b^{2} + c^{2}}{2}
\end{align*}
Thence I would like to ask: given three real numbers under the same restriction as above and prime numbers $p_{2} = p_{1} + 2$, for which of them does the following equation hold:
\begin{align*}
\frac{a^{p_{2}} + b^{p_{2}} + c^{p_{2}}}{p_{2}} & = \frac{a^{p_{1}} + b^{p_{1}} + c^{p_{1}}}{p_{1}}\cdot \frac{a^{2} + b^{2} + c^{2}}{2}
\end{align*}
Thank you in advance.
|
$p=3,5$ are the only possible solutions.
To check this, substitute $a=2, \ b,c=-1$.
Then
$\displaystyle \frac{2^{p+2}-2}{p+2}=3\frac{2^p-2}{p}$.
This equation can (after some effort) be rewritten as $2^p(p-6)=-4p-12$.
The left hand side can only be negative if $p<6$.
|
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|
Area of the triangle formed by circumcenter, incenter and orthocenter Lets say we have $\triangle$$ABC$ having $O,I,H$ as its circumcenter, incenter and orthocenter. How can I go on finding the area of the $\triangle$$HOI$.
I thought of doing the question using the distance (length) between $HO$,$HI$ and $OI$ and then using the Heron's formula, but that has made the calculation very much complicated. Is there any simple way to crack the problem?
|
A área do triângulo HOI do triângulo ABC de lados a=|BC|, b=|AC| e c=|AB| é dada por:
$AreaHOI=|\frac{1}{4}.\frac{\left|\begin{array}{} a & a^3 & 1 \\ b & b^3 & 1 \\ c & c^3 & 1 \\ \end{array} \right| }{\left|\begin{array}{} 0 & a^2 & b^2 &1\\ a^2 & 0 & c^2&1 \\ b^2 & c^2 & 0&1 \\1&1&1&0 \end{array} \right| }|$
Dedução: Seja o triângulo ABC, com o vértice A na origem do Sistema Cartesiano e C no semi eixo positivo x. Assim:
$A=(0,0)$
$B=(\frac{-a^2+b^2+c^2}{2b},\frac{2S}{b})$
$C=(b,0)$
$S=\sqrt{p(p-a)(p-b)(p-c)}$
$p=\frac{a+b+c}{2}$
$AreaHOI=3.AreaGIO$
$AreaGIO=\frac{1}{2}\left|\begin{array}{} xG & yG & 1 \\ xI & yI & 1 \\ xO & yO & 1 \\ \end{array} \right| $
"Obtendo as coordenadas de G, I e O através de suas respectivas fórmulas e fazendo substituições obtemos:"
$AreaHOI=3.\frac{1}{2}\left|\begin{array}{} \frac{-a^2+3b^2+c^2}{6b} & \frac{2S}{3b} & 1 \\ \frac{-a+b+c}{2} & \frac{2S}{a+b+c} & 1 \\ \frac{b}{2} & \frac{b(a^2-b^2+c^2)}{8S} & 1 \\ \end{array} \right| $
"Desenvolvendo, substituindo S, p e fatorando obtemos:"
$AreaHOI=\left|\frac{(a-b)(a-c)(b-c)(a+b+c)}{16S} \right| $
"Substituindo as expressões por determinantes (S no formato de área de Cayley-Menger), obtemos:"
$AreaHOI=|\frac{1}{4}.\frac{\left|\begin{array}{} a & a^3 & 1 \\ b & b^3 & 1 \\ c & c^3 & 1 \\ \end{array} \right| }{\left|\begin{array}{} 0 & a^2 & b^2 &1\\ a^2 & 0 & c^2&1 \\ b^2 & c^2 & 0&1 \\1&1&1&0 \end{array} \right| }|$
|
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|
Evaluate $\int \frac{1}{3+4\tan x} \, dx$ I'm am trying to evaluate $$\int \frac{1}{3+4\tan x} \, dx.$$
I tried to use the universal trig substitution, that is
$$t = \tan{x \over 2}, \quad \tan x = {2t \over 1-t^2}, \quad dx = {2 \over t^2 + 1}dt$$
and after manipulation I got the integral to be
$$\int \frac{2(1 - t^2)}{-3t^4 + 8t^3 + 8t +3} \, dt.$$
Is there a more clever approach to this problem or do I have to continue and face partial fractions?
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Note that $$\frac{1}{3 + 4 \tan x} = \frac{\cos x}{3 \cos x + 4 \sin x} = f(x).$$
Next, observe that $$\log (3 \cos x + 4 \sin x) + C = \int \frac{4 \cos x - 3 \sin x}{3 \cos x + 4 \sin x} \, dx = \int g(x) \, dx,$$ via the obvious substitution $u = 3 \cos x + 4 \sin x$. Now the goal is to find a linear combination of the integrands $f$ and $g$ such that for some real-valued scalar constants $A$, $B$, $$A f(x) + B g(x) = 1.$$ Well, this is simple: we have $$A \cos x + B(4 \cos x - 3 \sin x) = 3 \cos x + 4 \sin x,$$ and collecting like terms, we obtain $$A + 4B = 3, \quad -3B = 4,$$ or $$A = \frac{25}{3}, \quad B = -\frac{4}{3}.$$ Therefore, $$x = \int 1 \, dx = \int A f(x) + B g(x) \, dx = A \int f(x) \, dx + B \int g(x) \, dx,$$ or $$\int f(x) \, dx = \frac{3}{25} \left( x + \frac{4}{3} \log (3 \cos x + 4 \sin x) \right) + C,$$ and the rest is trivial.
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|
Prove that $\Gamma(-k+\frac{1}{2})=\frac{(-1)^k 2^k}{1\cdot 3\cdot 5\cdots(2k-1)}\sqrt{\pi}$. I was able to prove that
$$
\Gamma\left (k+\frac{1}{2} \right )=\frac{1\cdot 3\cdot 5\cdots(2k-1)}{2^k}\sqrt{\pi}.\tag{$k\geq 1$}$$
using the Legendre's duplication formula. But I can't do the same to
$$\Gamma\left ( -k+\frac{1}{2} \right )=\frac{(-1)^k 2^k}{1\cdot 3\cdot 5\cdots(2k-1)}\sqrt{\pi}.\tag{$k\geq 1$}$$
If possible, I'd like you to give a hint. If it is not possible to use Legendre's duplication formula, then I tried this way, for $n\geq 1$,
\begin{align*}
\Gamma\left ( -n+\frac{1}{2} \right )&=\left ( -n-\frac{1}{2} \right )\Gamma\left ( -n-\frac{1}{2} \right )\\
&=\dots\\
&=\left ( \frac{1}{2} \right )\left ( \frac{3}{2} \right )\cdots\left ( \frac{-2n-3}{2} \right )\left ( \frac{-2n-1}{2} \right )\Gamma\left ( \frac{1}{2} \right )
\end{align*}
and not anymore. How do I count how many factors there are to right side beside $\Gamma(1/2)$?
|
If you are willing to accept that for $z \notin \Bbb Z$
$$\Gamma (z) \Gamma (1-z) = \frac \pi {\sin \pi z}$$
then letting $z = \frac 1 2 - n$ leads to
$$\Gamma \left( \frac 1 2 - n \right) \Gamma \left( \frac 1 2 + n \right) = \frac \pi {\sin \left( \frac \pi 2 - n \pi \right)} = \frac \pi {\cos n \pi} = (-1)^n \pi$$
and since you already know
$$\Gamma \left( n + \frac 1 2 \right) = \frac {1 \cdot 3 \cdot 5 \cdot \cdots \cdot (2n-1)} {2^n} \sqrt{\pi}$$
then
$$\Gamma \left( \frac 1 2 - n \right) = \frac {(-1)^n 2^n \sqrt \pi} {1 \cdot 3 \cdot 5 \cdot \cdots \cdot (2n-1)} .$$
Alternatively, your own attempt based upon $\Gamma (z) = (z-1) \Gamma (z-1)$ is fine, too, because
$$\Gamma \left( \frac 1 2 \right) = \left( \frac 1 2 - 1 \right) \Gamma \left( \frac 1 2 - 1 \right) = \\
\left( \frac 1 2 -1 \right) \left( \frac 1 2 - 2 \right) \Gamma \left( \frac 1 2 - 2 \right) = \dots \\
\underbrace { \left( \frac 1 2 -1 \right) \left( \frac 1 2 - 2 \right) \dots \left( \frac 1 2 - n \right) } _{n \text{ factors}} \Gamma \left( \frac 1 2 - n \right) = \\
\underbrace { \left( - \frac 1 2 \right) \left( - \frac 3 2 \right) \dots \left( - \frac {2n-1} 2 \right) } _{n \text{ factors}} \Gamma \left( \frac 1 2 - n \right) = (-1)^n \frac {1 \cdot 3 \cdot \dots \cdot (2n-1)} {2^n} \Gamma \left( \frac 1 2 - n \right)$$
whence the desired result follows.
|
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|
Solve Sum of Arccos I am working through a situation with trying to fit an equilateral triangle into a square, and I have boiled it down to the following equation:
$$\arccos\left(\frac{\frac{L}{2}+x}{S}\right) + \arccos\left(\frac{\frac{L}{2}-x}{S}\right) = \frac{2}{3}\pi\;\mathrm{rad}$$
I need to solve this equation for $S$. After researching it for a couple of hours, I am still lost on how to resolve the two $\arccos$ terms. Could someone please show me how to go about solving this equation for $S$?
I would really appreciate it!
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Notice that: $\cos(A+B)=\cos A \cos B - \sin A \sin B$
In your example $A=\arccos(\frac{\frac{L}{2}+x}{S})$, and $B=\arccos(\frac{\frac{L}{2}-x}{S})$.
The solution follows from the fact that $\cos(\arccos(x))=x$
A specific solution:
$\arccos\left(\frac{\frac{L}{2}+x}{S}\right) + \arccos\left(\frac{\frac{L}{2}-x}{S}\right) = \frac{2}{3}\pi\;\mathrm{rad}$
From here, take the cosine of both sides:
$\cos(\arccos\left(\frac{\frac{L}{2}+x}{S}\right) + \arccos\left(\frac{\frac{L}
{2}-x}{S}\right))=\cos(\frac{2π}{3})$
Using the addition formula above, we find that:
$\cos(\arccos\left(\frac{\frac{L}{2}+x}{S}\right))\cdot\cos(\arccos\left(\frac{\frac{L}
{2}-x}{S}\right)))-\sin(\arccos\left(\frac{\frac{L}{2}+x}{S}\right))\cdot\sin(\arccos\left(\frac{\frac{L}
{2}-x}{S}\right)))=-\frac{1}{2}$
Notice that, by the Pythagorean relationship:
$\sin^2(\arccos(x))+\cos^2(\arccos(x))=1$
Therefore: $\sin(\arccos(x))=\sqrt{1-x^2}$.
Hence, the above equation can be rewritten as:
$(\frac{\frac{L}{2}+x}{S})\cdot(\frac{\frac{L}{2}-x}{S})-\sqrt{1-(\frac{\frac{L}{2}+x}{S})^2}\cdot\sqrt{1-(\frac{\frac{L}{2}-x}{S})^2}=-\frac{1}{2}$
Or, equivalently:
$\frac{L^2-4x^2}{4S^2}-\sqrt{(1-(\frac{\frac{L}{2}+x}{S})^2)(1-(\frac{\frac{L}{2}-x}{S})^2)}=-\frac{1}{2}$
And:
$L^2-4x^2-\sqrt{(-L^2-4Lx+4S^2-4x^2)(-L^2+4Lx+4S^2-4x^2)}=-2S^2$
WolframAlpha tells me (it's probably solvable, but to save time):
$x=±\frac{1}{2}\sqrt{3}\sqrt{S^2-L^2}$
|
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|
Determine matrix of linear map Linear map is given through:
$\phi\begin{pmatrix} 3 \\ -2 \end{pmatrix} =\begin{pmatrix} -3 \\ -14 \end{pmatrix} $
$\phi\begin{pmatrix} 3 \\ 0 \end{pmatrix} =\begin{pmatrix} -9 \\ -6 \end{pmatrix}$
Determine matrix $A$ linear map.
Here I have solution but I dont understand how to get it.
$A=\begin{pmatrix} -3 & -3 \\ -2 & 4 \end{pmatrix} $
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Let's go through the fundamental ideas here from the beginning:
If you multiply a matrix $$A=\pmatrix{a_{11} & a_{12} & \cdots & a_{1m} \\ a_{21} & a_{22} & \cdots & a_{2m} \\ \vdots & \vdots & \ddots & \vdots \\ a_{n1} & a_{n2} & \cdots & a_{nm}}$$ by the vector $e_1 = \pmatrix{1 \\ 0 \\ \vdots \\ 0}$, what do you get? Let's see:
$$Ae_1 = \pmatrix{a_{11} & a_{12} & \cdots & a_{1m} \\ a_{21} & a_{22} & \cdots & a_{2m} \\ \vdots & \vdots & \ddots & \vdots \\ a_{n1} & a_{n2} & \cdots & a_{nm}}\pmatrix{1 \\ 0 \\ \vdots \\ 0} = \pmatrix{a_{11} + 0a_{12} + \cdots + 0a_{1m} \\ a_{21} + 0a_{22} + \cdots + 0a_{2m} \\ \vdots \\ a_{n1} + 0a_{n2} + \cdots + 0a_{nm}} = \pmatrix{a_{11} \\ a_{21} \\ \vdots \\ a_{n1}}$$
Notice anything about this vector? It's just the first column of $A$. That's interesting.
I claim that if you multiply $A$ by $e_2 = \pmatrix{0 \\ 1 \\ 0 \\ \vdots \\ 0}$, you'd get the second column of $A$. And if you multiplied $A$ by $e_i$, which is a column vector with zeroes everywhere except for a 1 in the $i$th row, then you'd get the $i$th column of $A$. Try it out and see what you get.
So, with that in mind, we can see that any matrix $A$ can be thought of as $$\pmatrix{Ae_1 & Ae_2 & \cdots & Ae_m}$$
That may not seem very useful but let's consider your function $\phi$. If we could figure out what $\phi(e_1)$ and $\phi(e_2)$ are then the matrix representation of $\phi$ would just be $$\pmatrix{\phi(e_1) & \phi(e_2)}$$
So let's see if we can do that. We know that $\phi\begin{pmatrix} 3 \\ -2 \end{pmatrix} =\begin{pmatrix} -3 \\ -14 \end{pmatrix}$ and $\phi\begin{pmatrix} 3 \\ 0 \end{pmatrix} =\begin{pmatrix} -9 \\ -6 \end{pmatrix}$ and we know that $\phi$ is linear. So let's use that linearity. The second equation almost looks like $\phi\pmatrix{1 \\ 0}$, so let's start with that one:
$$\phi\pmatrix{3 \\ 0} = \phi\left[3\pmatrix{1 \\ 0}\right] = 3\phi\pmatrix{1 \\ 0} = \pmatrix{-9 \\ -6} \\ \implies \phi\pmatrix{1 \\ 0} = \frac 13\pmatrix{-9 \\ -6} = \pmatrix{-3 \\ -2}$$
Using that, let's figure out $\phi\pmatrix{0 \\ 1}$:
$$\phi\pmatrix{3 \\ -2} = \phi\left[3\pmatrix{1 \\ 0} -2\pmatrix{0 \\ 1}\right] = 3\phi\pmatrix{1 \\ 0}-2\phi\pmatrix{0 \\ 1} = \pmatrix{-9 \\ -6}-2\phi\pmatrix{0 \\ 1} = \pmatrix{-3 \\ -14} \\ \implies \phi\pmatrix{0 \\ 1} = -\frac 12\left[\pmatrix{-3 \\ -14}-\pmatrix{-9 \\ -6}\right] = \pmatrix{-3 \\ 4}$$
So then we can see that $$\bbox[5px,border:2px solid red]{A=\pmatrix{-3 & -3 \\ -2 & 4}}$$
|
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|
Prove: if $|x-1|<\frac{1}{10}$ so $\frac{|x^2-1|}{|x+3|}<\frac{1}{13}$
Prove: $$|x-1|<\frac{1}{10} \rightarrow \frac{|x^2-1|}{|x+3|}<\frac{1}{13}$$
$$|x-1|<\frac{1}{10}$$
$$ -\frac{1}{10}<x-1<\frac{1}{10}$$
$$ \frac{19}{10}<x+1<\frac{21}{10}$$
$$|x+1|<\frac{19}{10}$$
Adding 4 to both sides of $$ -\frac{1}{10}<x-1<\frac{1}{10}$$ gives:
$$\frac{39}{10}<x+3<\frac{41}{10}$$
$$|x+3|<\frac{39}{10}$$
Plugging those results in $$\frac{|x-1|*|x+1|}{|x+3|}<\frac{1}{13}$$
We get: $$\frac{\frac{1}{10}*\frac{19}{10}}{\frac{39}{10}}<\frac{1}{13}$$
$$\frac{19}{390}<\frac{1}{13}$$ Which is true, is this proof is valid as I took the smallest intervals, like $|x+3|<\frac{39}{10}$ and not $|x+3|<\frac{41}{10}$?
|
Easier method: use the inequalities
$$|a+b|\le|a|+|b|\quad\hbox{and}\quad |a+b|\ge|a|-|b|\ .$$
If
$$|x-1|<\frac1{10}$$
then
$$|x+1|=|(x-1)+2|\le|x-1|+2<\frac{21}{10}$$
and
$$|x+3|=|4+(x-1)|\ge4-|x-1|>\frac{39}{10}\ .$$
Therefore
$$\frac{|x^2-1|}{|x+3|}=|x-1|\frac{|x+1|}{|x+3|}<\frac1{10}\frac{21/10}{39/10}=\frac{21}{390}<\frac{30}{390}=\frac1{13}\ .$$
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|
Tips for integrating $\int \frac{dx}{1+\cos(x)}$ I tried the following
$$
\int \frac{dx}{1+\cos(x)}=\int \frac{1-\cos(x)}{1-\cos^2(x)}\,dx=
\int \frac{1-\cos(x)}{\sin^2(x)}\,dx\\
=\int \frac{1}{\sin^2(x)}\,dx-\int \frac{\cos(x)}{\sin^2(x)}\,dx=\int \csc^2x\,dx-\int \cot(x)\csc(x)\,dx
$$
Which I looked up and found to be equal to
$$
\csc(x)-\cot(x)+c
$$
Do I need to memorize the final identity, or is there a more elegant way to evaluate this integral?
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Here is a method similar to C. Dubussy's solution except no substitution needs to be made.
Recalling $1 = \cos^2 \frac{x}{2} + \sin^2 \frac{x}{2}$ and $\cos x = \cos^2 \frac{x}{2} - \sin^2 \frac{x}{2}$, the integral may be written as
\begin{align*}
\int \frac{dx}{1 + \cos x} &= \int \frac{dx}{\left (\cos^2 \frac{x}{2} + \sin^2 \frac{x}{2}\right ) + \left (\cos^2 \frac{x}{2} - \sin^2 \frac{x}{2} \right )}\\
&= \int \frac{dx}{2 \cos^2 \frac{x}{2}}\\
&= \frac{1}{2} \int \sec^2 \frac{x}{2} \, dx\\
&= \tan \frac{x}{2} + \cal{C}
\end{align*}
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|
Prove that $\sin \theta=\frac{3 \sin \alpha+\sin^3 \alpha}{1+3\sin^2 \alpha}$ using given condition If $$\tan(\frac{\pi}{4}+\frac{\theta}{2})=\tan^3(\frac{\pi}{4}+\frac{\alpha}{2})$$, then prove that $$\sin \theta=\frac{3 \sin \alpha+\sin^3 \alpha}{1+3\sin^2 \alpha}$$
I tried using the fact that
$\frac{\cos A}{1-\sin A}=\tan(\frac{\pi}{4}+\frac{A}{2})$ but now not able to eliminate $\cos \alpha$ and $\cos \theta$. How should I proceed?
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Notations and assumptions
$$
a=\tan{\theta\over2},\quad b=\tan{\alpha\over2},\quad p=\sin{\theta},\quad q=\sin\alpha
$$
Consequence of our notation
$$
{1+a\over1-a}=\left({1+b\over1-b}\right)^3\cdots\spadesuit
$$
An identity
$$
\sin z=2\sin{z\over2}\cos{z\over2}=2\tan{z\over2}\cos^2{z\over2}={2\tan{z\over2}\over\sec^2{z\over2}}={2\tan {z\over2}\over1+\tan^2{z\over2}}\cdots\heartsuit
$$
Using $\heartsuit$ and
$$
{x\over y}={u\over v}\implies {y+x\over y-x}={v+u\over v-u}\cdots\diamondsuit
$$
we get to
$$
{1+\sin z\over1-\sin z}=\left({1+\tan{z\over2}\over1-\tan{z\over2}}\right)^2\cdots\clubsuit
$$
Using $\clubsuit$ we get
$$
{1+p\over1-p}=\left({1+a\over1-a}\right)^2,\quad
{1+q\over1-q}=\left({1+b\over1-b}\right)^2$$
Using our $\spadesuit$ we get
$${1+p\over1-p}=\left({1+q\over1-q}\right)^3
$$
Using $\diamondsuit$ we get the required identity
$$
-{1\over p}={(1-q)^3+(1+q)^3\over(1-q)^3-(1+q)^3}\implies p={(1+q)^3-(1-q)^3\over(1+q)^3+(1-q)^3}={3q+q^3\over1+3q^2}
$$
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|
A solvable quintic with the root $x=(\sqrt[5]{p}+\sqrt[5]{q})^5$ - what are the other roots? I derived a two parameter quintic equation with the root:
$$x=(\sqrt[5]{p}+\sqrt[5]{q})^5,~~~~~p,q \in \mathbb{Q}$$
$$\color{blue}{x^5}-5(p+q)\color{blue}{x^4}+5(2p^2-121pq+2q^2)\color{blue}{x^3}-5(2p^3+381p^2q+381pq^2+2q^3)\color{blue}{x^2}+\\ + 5(p^4-121p^3q+381p^2q^2-121pq^3+q^4)\color{blue}{x}-(p+q)^5=0 \tag{1}$$
I'm unsure what the other roots are.
My thoughts: this is really a $25$ degree equation, it has $25$ roots.
Since there are 5 fifth degree roots, it makes sence, that we have:
$$x=y^5$$
$$y_{kn}=w_{5k}\sqrt[5]{p}+w_{5n}\sqrt[5]{q},~~~~~k,n=\{1,2,3,4,5\}$$
With $w_{5k}$ - $5$th roots of unity and $\sqrt[5]{p},\sqrt[5]{q}$ - principal roots.
Is this correct?
If we have $25$ different roots of the $25$th degree equation, what about the quintic equation $(1)$?
Edit
Another possible way - we can probably write:
$$x=p(1+\sqrt[5]{q/p})^5=q(1+\sqrt[5]{p/q})^5$$
Then we have $5$ possible roots inside the bracket.
Is this the correct way to get all the roots of $(1)$?
|
I see what you are getting at. Since there are actually five solutions $\alpha$ to the equation,
$$\alpha^5 = p$$
then it makes sense that $p^{1/5}+q^{1/5} = z$ should be the root of a $25$th deg eqn, correct? This is essentially given by your equation slightly modified as,
Eq.1:
$$z^{25}-5(p+q)z^{20}+5(2p^2-121pq+2q^2)z^{15}-5(2p^3+381p^2q+381pq^2+2q^3)z^{10}+\\ + 5(p^4-121p^3q+381p^2q^2-121pq^3+q^4)z^5-(p+q)^5=0 $$
However, since you raised your sum to a fifth power $x = \big(p^{1/5}+q^{1/5}\big)^5 = z^5$, you lost $20$ of the roots,
Eq.2:
$$x^{5}-5(p+q)x^{4}+5(2p^2-121pq+2q^2)x^{3}-5(2p^3+381p^2q+381pq^2+2q^3)x^{2}+\\ + 5(p^4-121p^3q+381p^2q^2-121pq^3+q^4)x-(p+q)^5=0 $$
But there is another way to get a quintic out of $(1)$: one can factor it. This can be done by using appropriately chosen $p,q$, namely,
$$p=\tfrac{-b+\sqrt{b^2+4}}{2}, \quad q=\tfrac{-b-\sqrt{b^2+4}}{2}$$
and $(1)$ simplifies as,
$$z^{25} + 5 b z^{20} + 5 (125 + 2 b^2) z^{15} + 5 b (-375 + 2 b^2) z^{10} + 5 (625 + 125 b^2 + b^4) z^5 + b^5 = 0$$
This factors into a $5$-deg and $20$-deg $P(z)$, easily done by Wolfram Alpha or Mathematica,
$$(z^5+5z^3+5z+b)\, P(z) = 0$$
with the $5$-deg as the familiar DeMoivre quintic. Thus, the $25$ possible sums of,
$$z = p^{1/5}+q^{1/5} = \Big(\tfrac{-b+\sqrt{b^2+4}}{2}\Big)^{1/5} +\Big(\tfrac{-b-\sqrt{b^2+4}}{2}\Big)^{1/5}$$
solve the DeMoivre quintic, as well as a $20$th deg eqn you can see by clicking on the WA link above.
|
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|
Eliminate the parameter given $x = \tan^{2}\theta$ and $y=\sec\theta$ $x = \tan^{2} (\theta)$ and $y = \sec (\theta)$
knowing that $\tan^{2} (\theta) = (\tan (\theta))^2 = \dfrac{\sin^{2}\theta}{\cos^{2}\theta}$
and that $\sec(\theta) = \dfrac{1}{\cos(\theta)}$ $\to$ $y=\dfrac{1}{\cos(\theta)}$
For $y$ we can get an even more simplified expression that will be useful for substituting into $x$ by multiplying both sides by $\cos(\theta)$ and then dividing both sides by $y$ to obtain: $\cos(\theta)=\frac{1}{y}$
With the trig simplified throught identities know we can now use more identities that become obvious and then simplify to a final expression through substitution into the x component:
$x = \dfrac{1-(\cos(\theta))^{2}}{(\cos(\theta))^{2}}$
Now substituting $\cos(\theta)$ for $\frac{1}{y}$ we get:
$x = \dfrac{1-(\frac{1}{y})^2}{(\frac{1}{y})^2}$
which simplifies to:
$x = \dfrac{1-\frac{1}{y^2}}{\frac{1}{y^2}}$
which can be simplified to:
$x = \dfrac{1-(\frac{1}{y^2})}{1} \cdot \dfrac{y^2}{1} \to y^2 - 1$
So removing the parameter we get:
$x= y^2 - 1$
I feel very confident that this answer is correct but I would like verification.
|
Recall the circular identity $$\sin^2 \theta + \cos^2 \theta = 1.$$ Dividing both sides by $\cos^2 \theta$ gives $$\tan^2 \theta + 1 = \sec^2 \theta.$$ Consequently, $$x + 1 = y^2.$$
|
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|
Evaluate $\int_0^\infty \frac{dx}{x^2+2ax+b}$ For $a^2<b$, is there an identity of evaluating the following integral?
$$\int_0^\infty \frac{dx}{x^2+2ax+b}$$
What about:
$$\int_0^\infty \frac{dx}{(x^2+2ax+b)^2}$$
My attempt is using partial fractions and completing the square, but I still failed to obtain a nice result.
|
For the first one:
$$(x+a)^2=x^2+2ax+a^2$$
So
$$(x+a)^2+b-a^2=x^2+2ax+b$$
$$=(b-a^2)\left(\frac{(x+a)^2}{b-a^2}+1 \right)$$
$$=(b-a^2) \left( \left( \frac{x+a}{\sqrt{b-a^2}} \right)^2+1 \right)$$
For $b-a^2 \neq 0$. In which case, for evaluating the integral, we enforce the substitution $\tan {u}=\frac{x+a}{\sqrt{b-a^2}}$ and proceed from there. Or the substitution $t=\frac{x+a}{\sqrt{b-a^2}}$, $\sqrt{b-a^2} dt=dx$:
$$\int_{0}^{\infty} \frac{1}{(b-a^2) \left( \left( \frac{x+a}{\sqrt{b-a^2}} \right)^2+1 \right)} dx$$
$$=\frac{\sqrt{b-a^2}}{b-a^2} \int_{\frac{a}{\sqrt{b-a^2}}}^{\infty} \frac{1}{t^2+1} dt$$
$$=\frac{\sqrt{b-a^2}}{b-a^2}\left(\frac{\pi}{2}-\arctan (\frac{a}{\sqrt{b-a^2}}) \right)$$
$$=\frac{\text{arccot} (\frac{a}{\sqrt{b-a^2}})}{\sqrt{b-a^2}}$$
The second integral can be found by the method already mentioned, i.e. finding:
$$\frac{\partial}{\partial b} \left(-\frac{\text{arccot} (\frac{a}{\sqrt{b-a^2}})}{\sqrt{b-a^2}} \right)$$
In fact if we take:
$$\begin{equation*} I_n = \int^{\infty}_{0}\frac{1}{(x^2 + 2ax + b)^n} \mathrm{d}x\end{equation*}$$
Like zain did,
Then for $n \geq 2$:
$$I_n=\frac{\partial^{n-1}}{\partial b} \left(\frac{(-1)^{n-1}}{(n-1)!}\frac{\text{arccot} (\frac{a}{\sqrt{b-a^2}})}{\sqrt{b-a^2}} \right)$$
Because:
$$\frac{\partial^{n}}{\partial x} \frac{1}{x+c}=\frac{(-1)^nn!}{(x+c)^{n+1}}$$
Thus:
$$I_n(a,b)=\frac{(-1)^{n-1}}{(n-1)!}\frac{\partial^{n-1}}{\partial b_0} \left(\frac{\text{arccot} (\frac{a_0}{\sqrt{b_0-a_0^2}})}{\sqrt{b_0-a_0^2}} \right) \biggr \rvert_{(a,b)}$$
|
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"timestamp": "2023-03-29T00:00:00",
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|
Prove the equation has infinitely many solutions
Let $a,b$ be coprime integers. Show that the equation $ax^2+by^2=z^3$ has an infinite set of solutions $(x,y,z)$ with $x,y,z \in \mathbb{Z}$ and $x,y$ mutually coprime (in each solution).
I thought of doing like $x = a^4$ and $y = b^4$, but then we get $a^9+b^9 = z^3$, which has no solutions by Fermat's Last Theorem. I was trying to get it into a Pythagorean form such as $x^2+y^2 = z^2$, which we know has an infinite number of solutions, but I wasn't able to do that.
|
$$ (A x^2 + B y^2)^3 = A (A x^3 - 3 B x y^2)^2 + B (3 A x^2 y - B y^3)^2 $$
We have
$$ U = (Ax^2 - 3By^2), \; \; \; V = (3Ax^2 - B^2). $$
To begin with, we certainly need the restrictions
$$ \gcd(A,B) = 1, \gcd(x,y) = 1, \gcd(A,y) = 1, \gcd(B,x) = 1. $$
Getting there: the prime $2.$ If $A$ is odd but $B$ even, we must have $x$ odd. If $A$ is even but $B$ odd, we must have $y$ odd.
If $A,B$ are both odd, we must have $x + y$ odd.
NEXT: the prime $3:$ just a minute... If $A$ is prime to $3$ but $3 | B,$ we must have $x$ prime to $3.$ If $B$ is prime to $3$ but $3 | A,$ we must have $y$ prime to $3.$ If both $A,B$ are prime to $3,$ we need nothing more than $\gcd(x,y) = 1,$ insofar as we already know $x,y$ are not both divisible by $3.$
Next we consider $$ \gcd(xU, yV). $$
In particular, we consider a prime $p \neq 2,3,$ such that $$ p | \gcd(xU, yV). $$
I see five cases: I am going to use the symbol $\perp$ to mean does not divide. Nope, we can use backslash nmid giving $\nmid $
(I) if $p|x,$ then $p \nmid y, $ $p \nmid B,$ so $p \nmid yV.$
(II) if $ p | y,$ then $p \nmid x,$ $p \nmid A,$ so $p \nmid xU$
(III) if $p | A,$ then $ p \nmid y,$ $ p \nmid B,$ so $p \nmid yV$
(IIII) if $p | B,$ then $ p \nmid x,$ $ p \nmid A,$ so $p \nmid xU$
(IIIII) if $p | U$ and $p | V,$ then
$$ A x^2 \equiv 3 B y^2 \pmod p $$ and
$$ B y^2 \equiv 3 A x^2 \pmod p. $$
Therefore
$$ A x^2 \equiv 9 A x^2 \pmod p, $$
$$ 8 A x^2 \equiv 0 \pmod p. $$
after case III, we conclude $x \equiv 0 \pmod p.$ After case IIII, we conclude $y \equiv 0 \pmod p.$ That is, $p | \gcd(x,y),$ which contradicts $\gcd(x,y) = 1.$
Apparently this was written first by Euler in 1770, where he complains that not all solutions can be found this way. Indeed, if the class number is divisible by three, and we have some form $e x^2 + f xy + g y^2$ whose cube in the class group is the identity, then the cube of any number primitively represented by this form is then represented primitively by the principal form. I get it: from his example, we can also use any form whose square in the class group is the identity. Such forms are traditionally called ambiguous. His example is $2x^2 - 5 y^2.$ Well, something to think about.
|
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|
Given MGF of X, find MGF of $ Y=X_1\dot\ X_2 \dot\ X_3$ Let $X_1$, $X_2$, $X_3$ be a random sample from a discrete distribution with probability funciton
$p(0)= 1/3$
$p(1) = 2/3$
Calculate moment generating function, $M(t)$, of $Y=$$X_1$$X_2$$X_3$
My Work
$M_x(t) = \frac{1}{3} + \frac{2}{3}e^t$
then $E[e^{t(X_1X_2X_3)}]$
$=E[e^{tX_1}+e^{tX_2}+e^{tX_3}]$
$= E[e^{tX_1}]+E[e^{tX_2}]+E[e^{tX_3}]$
$=3(\frac{1}{3} + \frac{2}{3}e^t)$
$=1+2e^t$
However, $M_x(0)=3\neq1$, so this must be wrong, but why?
|
If $X_1\,,\, X_2$ and $X_3$ be independent , then
$$P(Y=0)=\left( \begin{matrix}
3 \\
1 \\
\end{matrix} \right){{\left( \frac{1}{3} \right)}^{1}}{{\left( \frac{2}{3} \right)}^{2}}+\left( \begin{matrix}
3 \\
2 \\
\end{matrix} \right){{\left( \frac{1}{3} \right)}^{2}}{{\left( \frac{2}{3} \right)}^{1}}+\left( \begin{matrix}
3 \\
3 \\
\end{matrix} \right){{\left( \frac{1}{3} \right)}^{3}}
$$
$$P(Y=0)=\frac{12}{27}+\frac{6}{27}+\frac{1}{27}=\frac{19}{27}$$
$$P(Y=1)=\left( \begin{matrix}
3 \\
3 \\
\end{matrix} \right){{\left( \frac{2}{3} \right)}^{3}}=\frac8{27}$$
$$M_Y(t) = \operatorname{E}[e^{tY}] = \frac8{27}e^t + \frac{19}{27}$$
|
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|
How to find this function, and what method to use? The function is $f(x-\frac{1}{x})= x^3-\frac{1}{x^3}$ and they are asking us to find out what $f(-x)$ is?
|
A general method for solving equations of the type:
$$f(g(x))=z(x)$$
Let:
$$u=x-\frac{1}{x}$$
Here we let $u$ be equal to $g(x)$.
$$ux=x^2-1$$
$$x^2-ux-1=0$$
Using the quadratic formula we have:
$$x=\frac{u \pm \sqrt{u^2+4}}{2}$$
Note in a way we found an inverse for $x-\frac{1}{x}$ even though it does not have an inverse function. Now we have:
$$f(u)=(\frac{u \pm \sqrt{u^2+4}}{2})^3-\frac{1}{(\frac{u \pm \sqrt{u^2+4}}{2})^3}$$
And:
$$f(x)=(\frac{x \pm \sqrt{x^2+4}}{2})^3-\frac{1}{(\frac{x \pm \sqrt{x^2+4}}{2})^3}$$
Which can be simplified down to get the result $f(x)=x^3+3x$.
$$f(-x)=(\frac{-x \pm \sqrt{x^2+4}}{2})^3-\frac{1}{(\frac{-x \pm \sqrt{x^2+4}}{2})^3}$$
Again this can be simplified to get your desired result $f(-x)=-x^3-3x$.
|
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"url": "https://math.stackexchange.com/questions/1853364",
"timestamp": "2023-03-29T00:00:00",
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|
Maximize $k=x^2+y^2$ Subject to $x^2-4x+y^2+3=0$ Question
Let $x$ and $y$ be real numbers satisfying the equation $x^2-4x+y^2+3=0$. Find the maximum and minimum values of $x^2+y^2$.
My work
Let $k=x^2+y^2$
Therefore, $x^2-4x+y^2+3=0$ ---> $k-4x+3=0$ .
What do I do next? How do I find an expression in terms of $k$ that I can maximize?
|
$x^2 + y^2$ is essentially the square of the distance of a point $(x, y)$ from the origin. Now, the question is asking us to find two points $(x_{1}, y_{1})$ and $(x_{2}, y_{2})$ which lie on the circle and are at a minimum and maximum distance from the origin.
It makes sense that these two points will lie on the line joining the center of the circle and the origin. I won't be proving this here, but it should be pretty obvious upon drawing a rough diagram.
For the given circle
$$
x^2 - 4x + y^2 + 3 = 0
$$
the center is $(2,0)$. Hence, we draw a line passing through this point and the origin (the red line in the below diagram).
The red line is
$$
y = 0
$$
Now we find the point of intersection between this line and the circle.
$$
x^2 - 4x + 3 = 0 \\
(x - 3)(x - 1) = 0 \\
x = 1, 3
$$
Since $y = 0$, the maximum and minimum value of $x^2 + y^2$ is $1$ and $9$.
|
{
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"timestamp": "2023-03-29T00:00:00",
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|
Integral value of $n$ that makes $n^2+n+1$ a perfect square. Find all integers $n$ for which $n^2+n+1$ is a perfect square.
By hit and trial we get $n=-1,0$ but could someone suggest any genuine approach as how to approach this problem?
|
We have: $n^2 + n + 1 = m^2 \implies n^2 + n + (1-m^2) = 0 \implies \triangle = 1^2 - 4(1)(1-m^2) = k^2 \implies 1-4(1-m^2) = k^2 \implies 4m^2 - k^2 = 3 \implies (2m+k)(2m-k) = 3\implies 2m+k = 3, 2m-k = 1$ or $2m+k = -1, 2m-k = -3$. Either case gives $n^2 + n = 0 \implies n = -1, 0$.
|
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"timestamp": "2023-03-29T00:00:00",
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|
Prove that $\sqrt{a^2+3b^2}+\sqrt{b^2+3c^2}+\sqrt{c^2+3a^2}\geq6$ if $(a+b+c)^2(a^2+b^2+c^2)=27$ Let $a$, $b$ and $c$ be non-negative numbers such that $(a+b+c)^2(a^2+b^2+c^2)=27$. Prove that:
$$\sqrt{a^2+3b^2}+\sqrt{b^2+3c^2}+\sqrt{c^2+3a^2}\geq6$$
A big problem here around $(a,b,c)=(1.6185...,0.71686...,0.4926...)$.
In this case we get $\sum\limits_{cyc}\sqrt{a^2+3b^2}-6=0.000563...$.
My trying.
Let $a^2+3b^2=4x^2$, $b^2+3c^2=4y^2$ and $c^2+3a^2=4z^2$, where $x$, $y$ and $z$ are non-negatives.
Hence, we need to prove that
$$\sum\limits_{cyc}\sqrt{x^2-3y^2+9z^2}\leq\frac{\sqrt7(x+y+z)^2}{\sqrt{3(x^2+y^2+z^2)}}$$
Let $k$ and $m$ be non-negatives, for which
$x-ky+mz>0$, $y-kz+mx>0$, $z-kx+my>0$ and $1-k+m>0$.
By C-S $\left(\sum\limits_{cyc}\sqrt{x^2-3y^2+9z^2}\right)^2\leq(1-k+m)(x+y+z)\sum\limits_{cyc}\frac{x^2-3y^2+9z^2}{x-ky+mz}$.
Thus, it remains to prove that
$$(1-k+m)\sum\limits_{cyc}\frac{x^2-3y^2+9z^2}{x-ky+mz}\leq\frac{7(x+y+z)^3}{3(x^2+y^2+z^2)}$$
It's a sixth degree, but I didn't find a values of $k$ and $m$, such that the last inequality will be true.
By this way we can prove that $\sum\limits_{cyc}\sqrt{a^2+2b^2}\geq3\sqrt3$ is true, but it's not comforting.
Also I tried to use Holder, but without success.
Thank you!
|
There seems to be bugs in the segment after where I marked "***". Needed for check.
When I saw the form $\sqrt{a^2+3b^2}$ I thought of the absolute value of a complex number.
So let
$$u=a+\sqrt{3}bi$$
$$v=b+\sqrt{3}ci$$
$$w=c+\sqrt{3}ai$$
And now what you want to prove becomes
$$|u|+|v|+|w|\geq6$$
$$u+v+w=(1+\sqrt{3}i)(a+b+c)$$
$$|u|^2+|v|^2+|w|^2=4(a^2+b^2+c^2)$$
$$(u+v+w)^2(|u|^2+|v|^2+|w|^2)$$
$$=(1+\sqrt{3}i)^2(a+b+c)^2\cdot4(a^2+b^2+c^2)$$
$$=4(1+\sqrt{3}i)^2(a+b+c)^2(a^2+b^2+c^2)$$
$$=4(1+\sqrt{3}i)^2\cdot27$$
$$|u+v+w|^2(|u|^2+|v|^2+|w|^2)=|4(1+\sqrt{3}i)^2\cdot27|=4\cdot27\cdot4$$
Now I thought I should separate $|u+v+w|$ to $|u|+|v|+|w|$ so that all the elements in the expression were independent $|u|$, $|v|$, $|w|$, and its form would be closer to the inequality we want to prove.
So I used the triangle inequality,
$$|u|+|v|+|w|\geq|u+v|+|w|\geq|u+v+w|$$
$$(|u|+|v|+|w|)^2(|u|^2+|v|^2+|w|^2)\geq|u+v+w|^2(|u|^2+|v|^2+|w|^2)=4\cdot27\cdot4$$
Let $x=|u|,\ \ y=|v|,\ \ z=|w|$
Then the problem becomes,
$$\mathrm{if\ \ }(x+y+z)^2(x^2+y^2+z^2)\geq4\cdot27\cdot4$$
$$\mathrm{prove\ that\ \ }x+y+z\geq6$$
Proof:
let $k$ be a number so that $x+y+z\geq k\geq0$ is always true.
A known formula:
$$(x-y)^2+(y-z)^2+(z-x)^2=3(x^2+y^2+z^2)-(x+y+z)^2\geq0$$
***Then
$$3(x^2+y^2+z^2)\geq(x+y+z)^2\geq k^2$$
$$\implies (x+y+z)^2(x^2+y^2+z^2)\geq k^2\cdot\frac{k^2}{3}=\frac{k^4}{3}$$
But it's already known that $(x+y+z)^2(x^2+y^2+z^2)\geq4\cdot27\cdot4$ has to be true. So to let $$(x+y+z)^2(x^2+y^2+z^2)\geq\frac{k^4}{3}$$ be always true,
$$\frac{k^4}{3}\leq4\cdot27\cdot4$$
$$k\leq6$$
This proof doesn't need $a,b,c\geq0$. It just needs them to be real numbers.
|
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"timestamp": "2023-03-29T00:00:00",
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|
Statement regarding primes $ \le n$ Following is the statement I believe is true, but can't prove.
Let $n$ be a natural. Let the primes less than equal to $\sqrt{n}$ be $p_1,p_2,...,p_k$. Let $\alpha_i$ be the greatest natural number such that $p_i^{\alpha_i} \le n$. Out of these $k$ primes let us take $2 \le r \le k$ primes namely $p_{j_1},p_{j_2},...,p_{j_r}$. Then there exist no natural numbers $b_1,b_2,..,b_r$ such that
$$ \sum_{i=1}^{i=r}p_{j_i}^{\alpha_{j_i}} < \prod_{i=1}^{i=r}p_{j_i}^{b_i} \le n ...(1)$$
Example. Let $n=10$ then $k=2$, ie. we have $2$ primes less than $\sqrt{10}$ namely $2,3$ with the values of $\alpha$ being $3,2$ respectively. Now let us take $r=2$ primes, say $2$ and $3$, ie. take both the primes. The summation part of equation $(1)$ becomes $2^3+3^2=17$. The product in equation $(1)$ can only be $6$ in this case and hence clearly $6 < 17$. Thus there is no values of $b$ for which $(1)$ is true in this case.
PS: This question is in relation to this problem max-sum co-prime set. I would appreciate only hints.
|
Let $n = 124$, $p_1 = 2$, $p_2 = 5$. Then $\alpha_1 = 6$, since
$$2^6 = 64 \leq 124 < 2^7 = 128$$
and $\alpha_2 = 2$ since
$$5^2 = 25 \leq 124 < 5^3 = 125.$$
Now, $2^6 + 5^2 = 89$. Choose $b_1 = 2$ and $b_2 = 2$. Then
$$89 < 2^2\cdot 5^2 = 100 \leq 124 = n.$$
The answer below was for an earlier version of the question, before it was edited.
Let $n = 21$, take the primes $p_1 = 3$ and $p_2 = 7$. Then $\alpha_1 = 2$ since $3^2 = 9 \leq 21 < 3^3 = 27$ and $\alpha_2 = 1$ since $7 \leq 21 < 7^2 = 49$. Take $b_1 = b_2 = 1$. Then
$$9 + 7 = 16 < 3 \cdot 7 = 21 \leq 21 = n$$
|
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|
How to prove: $\sum_{k=m+1}^{n} (-1)^{k} \binom{n}{k}\binom{k-1}{m}= (-1)^{m+1}$ Show that if $m$ and $n$ are integers with $0\leq m<n$ then $$\sum_{k=m+1}^{n} (-1)^{k} \binom{n}{k}\binom{k-1}{m}= (-1)^{m+1}$$
Attempts:
*
*$(-1)^{k}\binom{n}{k}$ is the coefficient of $x^{k}$ in the expansion of $(1-x)^{n}$
*And $\binom{k-1}{m}$ is the coefficient of $x^{m}$ in the expansion of $(1+x)^{k-1}$.
Thats all what I could come up with.
|
For those who enjoy integrals here is another approach using the
Egorychev method as presented in many posts by @FelixMarin and also by
@MarkusScheuer.
Suppose we seek to verify that
$$\sum_{k=m+1}^n (-1)^k {n\choose k} {k-1\choose m} = (-1)^{m+1}.$$
First proof. Introduce
$${k-1\choose m} = {k-1\choose k-1-m} =
\frac{1}{2\pi i}
\int_{|z|=\epsilon}
\frac{1}{z^{k-m}} (1+z)^{k-1} \; dz.$$
Now clearly when $k\le m$ this vanishes so we may lower the limit in
the sum to $k=0.$ We obtain
$$\frac{1}{2\pi i}
\int_{|z|=\epsilon}
\frac{z^m}{1+z}
\sum_{k=0}^n {n\choose k} (-1)^k \frac{(1+z)^k}{z^k} \; dz
\\ = \frac{1}{2\pi i}
\int_{|z|=\epsilon}
\frac{z^m}{1+z}
\left(1-\frac{1+z}{z} \right)^n \; dz
\\ = \frac{1}{2\pi i}
\int_{|z|=\epsilon}
\frac{z^m}{1+z}
\frac{(-1)^n}{z^n} \; dz
\\ = \frac{1}{2\pi i}
\int_{|z|=\epsilon}
\frac{(-1)^n}{z^{n-m} (1+z)} \; dz
= (-1)^n (-1)^{n-m-1} = (-1)^{m+1}.$$
Second proof. Introduce
$${k-1\choose m} = {k-1\choose k-1-m} =
\frac{1}{2\pi i}
\int_{|z|=\epsilon}
\frac{1}{z^{m+1}} \frac{1}{(1-z)^{k-m}} \; dz.$$
Now for $1\le k\le m$ this is $[z^m] (1-z)^{m-k} = 0$ so we may again
extend the summation back to $k=0$, taking care of the value at $k=0$
which is $(-1)^m.$ We obtain
$$-(-1)^m + \frac{1}{2\pi i}
\int_{|z|=\epsilon}
\frac{1}{z^{m+1}} (1-z)^m
\sum_{k=0}^n {n\choose k} \frac{(-1)^k}{(1-z)^k}
\; dz
\\ = -(-1)^m + \frac{1}{2\pi i}
\int_{|z|=\epsilon}
\frac{1}{z^{m+1}} (1-z)^m
\left(1-\frac{1}{1-z}\right)^n
\; dz
\\ = -(-1)^m + \frac{1}{2\pi i}
\int_{|z|=\epsilon}
\frac{1}{z^{m-n+1}} \frac{(-1)^n}{(1-z)^{n-m}} \; dz.$$
Note however that we have $m\lt n$ so the contribution from the
integral vanishes, once more leaving just $$(-1)^{m+1}.$$
|
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|
On the proof $\tan 70°-\tan 20° -2 \tan 40°=4\tan 10°$ I am currently studying in class 10 and I am unable to do this problem. $$\tan 70 ° -\tan 20° -2 \tan 40° =4\tan 10°$$ Can anybody please help me.
Thanks!
|
\begin{align*}
\tan 70-\tan 20 & =\tan 70-\cot 70\\
& =\tan 70-\frac{1}{\tan 70}\\
& = \frac{\tan^2 70-1}{\tan 70}=-\frac{1-\tan^2 70}{\tan 70}\\
& = -\frac{2(1-\tan^2 70)}{2\tan 70}=-\frac{2}{\frac{2\tan 70}{1-\tan^2 70}}\\
& = \frac{-2}{\tan 140}= \frac{-2}{-\tan 40}= \frac{2}{\tan 40}=2\cot 40
\end{align*}
Now, $$ \tan 70-\tt-2\tan 40 =2\cot 40-2\tan 40=-2(\tan 40-\cot 40) $$
\begin{align*}
-2(\tan 40-\cot 40)= -2\left( \tan 40-\frac{1}{\tan 40} \right)
\end{align*}
Now do the same as above to get the result. You will get $ 4\cot 80=4\tan 10. $
|
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|
integrate $\int \frac{x^4-16}{x^3+4x^2+8x}dx$
$$\int \frac{x^4-16}{x^3+4x^2+8x}dx$$
So I first started with be dividing $p(x)$ with $q(x)$ and got:
$$\int x-4+\frac{8x^2+32x-16}{x^3+4x^2+8x}dx=\frac{x^2}{2}-4x+\int \frac{8x^2+32x-16}{x^3+4x^2+8x}dx$$
Using partial sum I have received:
$$\int \frac{8x^2+32x-16}{x^3+4x^2+8x}dx=8\int -\frac{1}{4x}+\frac{5x}{4(x^2+4x+8)}+\frac{5}{x^2+4x+8} dx=-2ln|x|+8(\frac{5}{4}\int\frac{x}{(x^2+4x+8)} +5\int \frac{1}{x^2+4x+8})=-2ln|x| +10\int\frac{x}{(x^2+4x+8)}dx +40\int \frac{1}{x^2+4x+8}dx$$
How do I continue from here?
|
$\displaystyle \int \dfrac{1}{x^2+4x+8}dx$
How do I continue from here?
Observe that by writing
$$
x^2+4x+8=(x+2)^2+4
$$ and by making the change of variable
$$
t=2(x+2), \quad dx=\frac12dt, \quad x^2+4x+8=4(t^2+1)
$$ you are led to evaluate
$$
\int \dfrac{1}{x^2+4x+8}dx=\frac18\int\frac1{t^2+1}dt
$$ which is classic.
$\displaystyle \int \dfrac{x}{x^2+4x+8}dx$
How do I continue from here?
One may write
$$
\int \dfrac{x}{x^2+4x+8}dx= \frac12\int \dfrac{2x+4}{x^2+4x+8}dx-2\int \dfrac{1}{x^2+4x+8}dx
$$that is $$
\int \dfrac{x}{x^2+4x+8}dx=\frac12 \int \dfrac{(x^2+4x+8)'}{x^2+4x+8}dx-2\int \dfrac{1}{x^2+4x+8}dx
$$ then conclude with the first part.
|
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Integrate $\int\frac{x+1}{\sqrt{1-x^2}} \; dx$ without using trigonometric substitution I want to solve:
$$\int\frac{x+1}{\sqrt{1-x^2}} \; dx$$
I know how to solve this using trigonometric substitution, but how can I solve the integral in an other way ?
|
Given that the answer for $\int\frac1{\sqrt{1-x^2}}dx$ will be $\sin^{-1}x$ you can't hope to solve the integral without any clue about trig functions. However, there is an easy way to deduce the answer just knowing that sine (and cosine) are the solutions to the harmonic differential equation
$$
\frac{d^2 y}{dx^2} = -y \implies y = A\sin(x+\delta)
$$
(We are actually going to use this insight with $x$ and $y$ swapped.)
Start from $y= \int\frac1{\sqrt{1-x^2}}dx \implies \frac{dy}{dx}= \frac1{\sqrt{1-x^2}} \implies \frac{dx}{dy} = \sqrt{1-x^2}$
Now differentiate with respect to $y$ again, using the chain rule, with the knowledge that $\frac{dx}{dy} = \sqrt{1-x^2}$:
$$\frac{d}{dy} \sqrt{1-x^2} = \frac{-2x}{2\sqrt{1-x^2}} \frac{dx}{dy}=\frac{-2x}{2\sqrt{1-x^2}} \sqrt{1-x^2} = -x$$.
So if $y= \int\frac1{\sqrt{1-x^2}}dx $ then $x = A \sin(y+\delta)
$, or more properly, if $y= \int_0^x\frac1{\sqrt{1-t^2}}dt $ then $x = A \sin(y+\delta)$.
Now when $x=0$, the limits in the integral are identical so $\delta$ must be zero. (We know $A$ cannot be zero; that would make the integral identically zero.) We are left with the probelm of determining $A$.
From the integrand, we can read off that at zero, $d\frac{dy}{dx} = \frac{dx}{dy} = 1$. Since the slope of $\sin y$ at the origin is $1$, the multiplicative constant $A$ must be $1$.
Thus if $y= \int\frac1{\sqrt{1-x^2}}dx$ then $x=\sin y$ or $y = \sin^{-1}x$.
|
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|
inequality $\sqrt{\cos x}>\cos(\sin x)$ for $x\in(0,\frac{\pi}{4})$ How can I prove the inequality $\sqrt{\cos x}>\cos(\sin x)$ for $x\in(0,\frac{\pi}{4})$ ?
The derivative of $f(x):=\sqrt{\cos x}-\cos(\sin x)$ is very unpleasant, so the standard method is probably not be the right choice...
|
The only way I can think about is Taylor expansions (tedious but doable) as Henning Makholm commented. $$\cos(x)=1-\frac{x^2}{2}+\frac{x^4}{24}-\frac{x^6}{720}+O\left(x^8\right)$$ $$\sqrt{\cos(x)}=1-\frac{x^2}{4}-\frac{x^4}{96}-\frac{19 x^6}{5760}+O\left(x^8\right)$$ $$\sin(x)=x-\frac{x^3}{6}+\frac{x^5}{120}-\frac{x^7}{5040}+O\left(x^8\right)$$ $$\cos(\sin(x))=1-\frac{x^2}{2}+\frac{5 x^4}{24}-\frac{37 x^6}{720}+\frac{457
x^8}{40320}+O\left(x^9\right)$$ $$\sqrt{\cos(x)}-\cos(\sin(x))=\frac{x^2}{4}-\frac{7 x^4}{32}+\frac{277 x^6}{5760}+O\left(x^8\right)$$ $$\sqrt{\cos(x)}-\cos(\sin(x))=\frac{x^2}4 \left(1-\frac{7 x^2}{8}+ \frac{277 x^4}{1440}\right)+O\left(x^8\right)$$ There is no real root for the quadratic in $x^2$ inside brackets.
|
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|
Evaluate $\int \frac{\sin^4 x}{\sin^4 x +\cos^4 x}{dx}$ $$\int \frac{\sin^4 x}{\sin^4 x +\cos^4 x}{dx}$$
$$\sin^2 x =\frac{1}{2}{(1- \cos2x)}$$
$$\cos^2 x =\frac{1}{2}{(1+\cos2x)}$$
$$\int \frac{(1- \cos2x)^2}{2.(1+\cos^2 2x)}{dx}$$
$$\frac{1}{2} \int \left[1-\frac{2 \cos2x}{1+\cos^22x}\right] dx$$
What should I do next ?
Please also tell me alternative way to do this .
|
HINT:
$$\cos^22x=1-\sin^22x$$
Set $\sin2x=y$
Otherwise,
$$\dfrac{\sin^4x}{\sin^4x+\cos^4x}=\dfrac1{1+\cot^4x}$$
Let $\cot^2x=u\implies dx=-\dfrac{du}{1+u^2}$
Method$\#1:\dfrac2{(1+u^4)(1+u^2)}=\dfrac{1+u^4+1-u^4}{(1+u^4)(1+u^2)}=?$
Method$\#2:$ Writing $u^2=y$
$$\dfrac1{(1+y^2)(1+y)}=\dfrac{Ay+B}{1+y^2}+\dfrac C{1+y}$$
$$\iff1=(Ay+B)(1+y)+C(1+y^2)=y^2(C+A)+y(B+A)+B+C$$
$C+A=0\iff C=-A, B+A=0\iff B=-A$ and $1=B+C=-2A$
$$\implies\dfrac1{(1+y^2)(1+y)}=\dfrac{-y+1}{2(1+y^2)}+\dfrac1{2(1+y)}$$
In either case, finally $\dfrac{1-u^2}{1+u^4}=\dfrac{\dfrac1{u^2}-1}{\dfrac1{u^2}+u^2}$
and $\displaystyle\int\left(\dfrac1{u^2}-1\right)=?$ and $\dfrac1{u^2}+u^2=\left(u+\dfrac1u\right)^2-2$
|
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|
Proving that $x^{16} > 5$ when given a polynomial of degree $15$. I am unable to prove the following
If $x^{15}-x^{13}+x^{11}-x^9+x^7-x^5+x^3-x = 7$ prove that $x^{16} > 15$.
|
Notice that the left hand is equal to
$(x^3-x)(x^{16}-1)/(x^4-1)$; hence the equation will be transfer into:
$(x^{16}-1)=7(x^4-1)/(x^3-x)=7(x^2+1)/x$; notice that $(x^2+1)/x$ is graeter than 2, hence the right hand is greater than 7*2, therefor $x^{16}>15$.
|
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If $n$ is a positive integer, then $(-2^n)^{-2} + (2^{-n})^2 = 2^{-2n+1}$ I'm not sure why $$(-2^n)^{-2} + (2^{-n})^2=2^{-2n+1}$$
I have been going over this equation for a while now, noticing, and have successfully got quite far in the equation, finding that
$$ (-2^n)^{-2} + (2^{-n})^2 \implies \frac{1}{(-2^n)^2} + \frac{1}{(2^n)^2} $$
which I think then becomes $ \dfrac{2}{2^{2n}}$
But then I get stuck.
|
So, we've got $\dfrac{1}{-2^n \cdot -2^n} + \dfrac{1}{2^n \cdot 2^n} = \dfrac{1}{2^n \cdot 2^n} + \dfrac{1}{2^n \cdot 2^n} = \dfrac{2}{2^{2n}}$ since adding two of the same things up is the same as multiplying by $2$.
Alright, so you have $\frac{2}{2^{2n}}$. That's good work! Now use the property that $\frac{a}{b} = ab^{-1}$, since $\frac{1}{b} = b^{-1}$. Different notations, same thing.
Now, that means you have $\frac{2}{2^{2n}} = 2\cdot 2^{-2n}$. We're almost there: remember that $a^b \cdot a^c = a^{b+c}$, this can be seen by writing $a^b \cdot a^c = \underbrace{a \cdot a \cdots a}_{b \, \text{times}} \cdot \underbrace{a \cdot a \cdots a}_{c \, \text{times}} = \underbrace{a \cdot a \cdots a}_{(b+c) \, \text{times}} = a^{b+c}$.
Applying this to your case with $a = 2$, $b = 1$, $c = -2n$ we have $2 \cdot 2^{-2n} = 2^{1 - 2n} = 2^{-2n + 1}$.
|
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|
If $\sin x + \sin y = 1$ and $\cos x + \cos y = 0$, solve for $x$ and $y$
*
*$\sin x + \sin y = 1$
*$\cos x + \cos y = 0$
Any valid pair of $(x, y)$ is fine, as the restrictions on the board in the image below are obscured.
I got the question from chapter 26 of a comic called Yamada-kun.
How can I solve this equation?
|
$$\sin x + \sin y = 1\Rightarrow 2\sin\frac{x+y}{2}\cos \frac{x-y}{2}=1\\\cos x+\cos y=0\Rightarrow 2\cos\frac{x+y}{2}\cos \frac{x-y}{2}=0$$
It follows $$ \cos \frac{x+y}{2}=0\text{ and } \cos \frac{x-y}{2}=0\iff $$ The first ($\cos \frac{x+y}{2}=0$) gives, from the first given equation,
$$\begin{cases}x+y=\pi\\x-y=\frac{2\pi}{3}\end{cases}\iff(x,y)=(\frac{\pi}{6},\frac{5\pi}{6})$$
The second ($\cos \frac{x-y}{2}=0$) is discarded by incompatibility.
Thus the solutions are given by $$(x,y)=( \frac{\pi}{6}+2m\pi,\space \frac{5\pi}{6}+2n\pi)$$
It is not worthless to see the graphic solutions: all intersections of the red lattice, with blue closed curves.
|
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|
Evaluate the expression $\sqrt{6-2\sqrt5} + \sqrt{6+2\sqrt5}$ $$\sqrt{6-2\sqrt5} + \sqrt{6+2\sqrt5}$$
Can anyone tell me the formula to this expression.
I tried to solve in by adding the two expression together and get $\sqrt{12}$ but as I insert each expression separately in calculator the answer is above $\sqrt{12}$.
|
Set
$$t=\sqrt{6-2\sqrt5} + \sqrt{6+2\sqrt5}$$
$$t^2=6-2\sqrt5+2\sqrt{(6-2\sqrt5)(6+2\sqrt5)}+6+2\sqrt5$$
$$t^2=12+2\sqrt{36-20}=12+2(4)=20$$
$$t^2=20\implies t=2\sqrt5$$
|
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|
Proof that $P(x)=x-\frac{1}3 x^3+\frac{1}{5}x^5-\frac{1}{7}x^7+\cdots$ has radius of convergence $1$
Proof that
$P(x)=x-\frac{1}{3}x^3+\frac{1}{5}x^5-\frac{1}{7}x^7+\cdots$ has
radius of convergence $1$
First of all, I need to convert this to a series:
$$\sum_{k=1}^\infty \frac{x^{2k-1}(-1)^k}{1-2k}$$
(I hope this is correct so far?)
On this series, you use ratio test, count the limit and then take the reciprocal of the limit which will hopefully equal 1:
$$\lim_{k\rightarrow\infty}\left (\frac{x^{2(k+1)-1}(-1)^{k+1}}{1-2(k+1)}: \frac{x^{2k-1}(-1)^k}{1-2k}\right ) = \lim_{k\rightarrow\infty}\frac{\left(x^{2k+1}(-1)^{k+1}\right)(1-2k)}{(-2k-1) (x^{2k-1}(-1)^k)}$$
$$=\lim_{k\rightarrow\infty}\left(\frac{x^{2k+1}}{x^{2k-1}}\right ) \cdot \left(\frac{(-1)^{k+1}}{(-1)^k}\right)\cdot\left(\frac{1-2k}{-2k-1}\right)$$
$$=\lim_{k\rightarrow\infty}\left (x^{2k+1-2k+1} \right )\cdot\left ((-1)^{k+1-k} \right )\cdot\left (\frac{k(\frac{1}{k}-2)}{k(-2-\frac{1}{k})} \right )$$
$$=\lim_{k\rightarrow\infty}x^2\cdot((-1)^k)\cdot\left (\frac{\frac{1}{k}-2}{-2-\frac{1}{k}} \right )=-x^2\cdot\left(\frac{0-2}{-2-0} \right) = -x^2$$
$$R=-\frac{1}{x^2}$$
After all, I don't get 1 as result... :(
Where is my mistake? I cannot imagine it's as complicated and long as I did?
Or maybe, at the end, I ignore the $x^2$? Then result would be 1!?
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Suppose $\{a_n\}$ is the sequence of terms of this series. Then it is clear that $\lim \sup (|a_n|)^{1/n} = 1 $ so that the radius of convergence is $1$.
|
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Which answer is correct? Finding the limit of a radical as $x$ approaches infinity. When I take $$\lim_{x \to -∞} \sqrt{x^2+7x}+x,$$ I multiply by the conjugate over the conjugate to get
$$\lim_{x \to -∞}\frac{7x}{\sqrt{x^2+7x}-x},$$ and multiply by either $\frac{\frac{1}{x}}{\frac{1}{x}}$ or $\frac{\frac{1}{-x}}{\frac{1}{-x}}$ to get an undefined answer or $\frac{-7}{2}.$
My teacher's solution involves multiplying by $\frac{\frac{1}{-x}}{\frac{1}{-x}}:$
$$=\lim_{x \to -∞}\frac{-7}{\sqrt{x^2/x^2+7x/x^2}+1}$$
$$=-\frac{7}{\sqrt{1+0}+1}$$
$$=\frac{-7}{2}$$
However, I multiplied
by $\frac{\frac{1}{x}}{\frac{1}{x}}$ and got the following:
$$\lim_{x \to -∞}\frac{7}{\sqrt{x^2/x^2+7x/x^2}-1}$$
$$\frac{7}{\sqrt{1+0}-1}$$
$$\frac{7}{0}$$
Which is undefined.
Why does multiplying by what is essentially $1$ cause different answers in general, and in particular for evaluating limits?
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$$\lim_{x \to -\infty} \sqrt{x^2+7x}+x=$$
$$=\lim_{x \to -\infty} |x|\sqrt{1+\frac{7}{x}}+x=$$
$$=\lim_{x \to -\infty} -x\sqrt{1+\frac{7}{x}}+x=$$
$$=\lim_{x \to -\infty} -x\Big(\sqrt{1+\frac{7}{x}}-1\Big)=$$
$$=\lim_{x \to -\infty} -x\Big(1+\frac{7}{2x}+O\Big(\frac{1}{x^2}\Big)-1\Big)=$$
$$=\lim_{x \to -\infty} -\frac{7}{2}+O\Big(\frac{1}{x}\Big)=-\frac{7}{2}$$
|
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|
Minimum value of $f(x) = \sqrt{x^2 + (1-x)^2} + \sqrt{(1-x)^2 +(1+x)^2}$ If $f(x) = \sqrt{x^2 + (1-x)^2} + \sqrt{(1-x)^2 +(1+x)^2}$
Find the minimum value of the function
I tried using the AMGM inequality and differentiation but didn't know how to solve it any ideas?
This is from a math competition. ( I would like to see the most efficient way as I think differentiation in a math competition is not that efficient)
Using Mogjals comment and using the AM-GM inequality , setting $A = \sqrt{x^2 + (1-x)^2}$ and $B=\sqrt{(1-x)^2 +(1+x)^2}$
then
$A+B \geq 2\sqrt{AB}$
$$ \sqrt{x^2 + (1-x)^2} + \sqrt{(1-x)^2 +(1+x)^2} \geq 2\sqrt{2}\sqrt{x^2+1}\sqrt{2x^2-2x+1}$$
With equality if and only $A=B$ so $x=-\frac{1}{2}$
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A posible way: Let $A,B,C$ a triangle and $AD$ the height. The problem is $AD=1-x$, $BD=x$, $CD=1+x$ and you want minimize $BA+AC$, but note that the area is fixed ($(1-x)(2x+1)/2$), then this sum is minime if the triangle is right in A.
|
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|
For which $a$, equation $4^x-a2^x-a+3=0$ has at least one solution.
Find all values of $a$ for which the equation $4^x-a2^x-a+3=0$ has at least one solution.
$\bf{My\; Try::}$ We can write it as $$2^{x}-a-\frac{a}{2^x}+\frac{3}{2^x}=0$$
So $\displaystyle \left(2^x+\frac{3}{2^x}\right)=a\left(1+\frac{1}{2^x}\right).$
Now for the existance of solution $\displaystyle 2^x+\frac{3}{2^x}\geq 2\sqrt{3}$ Using $\bf{A.M\geq G.M}$
So $\displaystyle a\left(1+\frac{1}{2^x}\right)\geq 2\sqrt{3}\Rightarrow a\geq \sqrt{3}\cdot \frac{2^x}{2^x+1}\geq \sqrt{3}$
But answer given as $a\geq 2,$ please explain me whats wrong with that, Thanks
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Put $t = 2^x$. The equation is $t^2-at-a+3 = 0$. This quadratic should have a positive root. Hence $a^2 + 4(a-3) \geq 0$ which gives $(a+6)(a-2) \geq 0$. Thus $a \geq 2$.
|
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|
$x^4 - 4x^3 + 6x^2 - 4x + 1 = 0$ Determine all the possibilities for rational roots of the polynomial $x^4 - 4x^3 + 6x^2 - 4x + 1 = 0$. Then determine how many of the real roots of the polynomial may be positive and how many may be negative. Factor the polynomial to confirm your results.
The answer is possible rational roots: $+-1$; number of possible real roots - positive: four or two or zero, negative: zero; actual roots: $x = 1, 1, 1, 1$ (a quadruple root).
Using the rational root theorem, you divide the factors of the constant, $1$, by the factors of the lead coefficient, also a 1. That step gives you only two different possibilities for rational roots: $1$ and $-1$.
The signs change four times in the original polynomial, indicating $4$ or $2$ or $0$ positive real roots. Replacing each $x$ with $-x$, you get $x^4 + 4x^3 + 6x^2 + 4x + 1 = 0$. The signs never change. The polynomial is the fourth power of the binomial $(x - 1)$, so it factors into $(x - 1)^4 = 0$, and the roots are $1, 1, 1, 1$. There are four positive roots (all the same number, of course).
Can someone explain, the factorization of the polynomial? I do not understand, how it factors into $(x - 1)^4$.
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Knowing that x=1 is a solution we know that $(x-1) $ is a factor so we force it to factor out.
$x^4-4x^3+6x^2-4x+1=$
$(x-1)x^3-3x^3 +6x^2-4x+1=$
$(x-1)x^3-(x-1)3x^2+3x^2-4x+1=$
$(x-1)x^3-(x-1)3x^2 +(x-1)3x -x +1=$
$(x-1)x^3-(x-1)3x^2+(x-1)3x-(x-1)=$
$(x-1)[x^3-3x^2+3x-1]=$
Now, we may not know that 1 is a double root yet. But if $x=1$ then $x^3-3x^2+3x-1=0$ so we do know that $x-1$ will factor out again. So we force it.
$(x-1)[(x-1)x^2-2x^2+3x-1]=$
$(x-1)[(x-1)x^2-(x-1)2x+x-1]=$
$(x-1)[(x-1)x^2-(x-1)2x+(x-1)]=$
$(x-1)(x-1)[x^2-2x+1]=$
I'll pretend it's not obvious at this point.
We test $x =1$ so $x^2-2x+1=0$ so we know $(x-1) $ must factor a third time. We force it out but by this point it should look like something very clear.
$(x-1)^2 [(x-1)x-x+1]=$
$(x-1)^2 [(x-1)x-(x-1)]=$
$(x-1)^2 (x-1)[x-1]=$
$(x-1)^4$
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Two tangents to the hyperbola $4x^2-y^2=36$ intersect at the point $(0,4)$. Find the coordinates for the points on the hyperbola for this to occur. The question is two tangents to the hyperbola $4x^2-y^2=36$ intersect at the point $(0,4)$. Find the coordinates for the points on the hyperbola for this to occur.
My attempt:
Consider the line $y=mx+c$ as it passes through $(0,4)$ then $c=4$ --> Condition 1
Plug in $y=mx+c$
$$ 4x^2-(mx+c)^2=36$$
$$4x^2-36=m^2x^2+2mcx+c^2$$
$$(m^2-4)x^2+(2mc)x^2+c^2+36=0$$
For tangency $b^2-4ac=0$
Hence
$$ 4m^2c^2-4(c^2+36)(m^2-4)=0$$
$$ 16c^2-144m^2+576=0$$
But $c=4$ due to the line condition hence
$$ m = \pm \frac{2\sqrt{13}}{3}$$
So the lines are
$y= \frac{2\sqrt{13}}{3}x + 4$ and $y= -\frac{2\sqrt{13}}{3}x + 4$
How to find the coordinates for the points on the hyperbola for this to occur tho?
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You already have
$$(m^2-4)x^2+(2mc)x^2+c^2+36=0$$
Solving this gives
$$x=\frac{-mc}{m^2-4}=\frac{-4m}{m^2-4}$$
So, for $m=\pm\frac{2\sqrt{13}}{3}$, the coordinates we want are
$$\color{red}{\left(\mp\frac{3}{2}\sqrt{13},-9\right)}$$
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How to show $\sum_{k=0}^{n}\binom{n+k}{k}\frac{1}{2^k}=2^{n}$
How does one show that $$\sum_{k=0}^{n}\binom{n+k}{k}\frac{1}{2^k}=2^{n}$$ for each nonnegative integer $n$?
I tried using the Snake oil technique but I guess I am applying it incorrectly. With the snake oil technique we have $$F(x)= \sum_{n=0}^{\infty}\left\{\sum_{k=0}^{n}\binom{n+k}{k}\frac{1}{2^k}\right\}x^{n}.$$ I think I have to interchage the summation and do something. But I am not quite comfortable in interchanging the summation. Like after interchaging the summation will $$F(x)=\sum_{k=0}^{n}\sum_{n=0}^{\infty}\binom{n+k}{k}\frac{1}{2^k}x^{n}?$$ Even if I continue with this I am unable to get the correct answer.
*
*How does one prove this using the Snake oil technique?
*A combinatorial proof is also welcome, as are other kinds of proofs.
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Suppose we seek to verify that
$$\sum_{k=0}^n {n+k\choose k} \frac{1}{2^k} = 2^n.$$
In the following we make an effort to use a different set of integrals
from the answer by @MarkusScheuer, for variety's sake, even if this is
not the simplest answer.
The difficulty here lies in the fact that the binomial coefficients on
the LHS do not have an upper bound for the sum wired into them. We use
an Iverson bracket to get around this:
$$[[0\le k\le n]]
= \frac{1}{2\pi i}
\int_{|w|=\gamma} \frac{w^k}{w^{n+1}} \frac{1}{1-w} \; dw.$$
Introduce furthermore
$${n+k\choose k} =
\frac{1}{2\pi i}
\int_{|z|=\epsilon} \frac{1}{z^{n+1}}
\frac{1}{(1-z)^{k+1}} \; dz.$$
With the Iverson bracket in place we can let the sum range to
infinity, getting
$$\frac{1}{2\pi i}
\int_{|w|=\gamma} \frac{1}{w^{n+1}} \frac{1}{1-w}
\frac{1}{2\pi i}
\int_{|z|=\epsilon} \frac{1}{z^{n+1}}
\frac{1}{1-z}
\sum_{k\ge 0} \frac{w^k}{(1-z)^k} \frac{1}{2^k}
\; dz\; dw.$$
This converges when $|w| < |2(1-z)|.$ We require $\gamma \lt 2(1-\epsilon)$ or $\epsilon \lt 1-\gamma/2.$ Simplifying we have
$$\frac{1}{2\pi i}
\int_{|w|=\gamma} \frac{1}{w^{n+1}} \frac{1}{1-w}
\frac{1}{2\pi i}
\int_{|z|=\epsilon} \frac{1}{z^{n+1}}
\frac{1}{1-z}
\frac{1}{1-w/(1-z)/2}
\; dz\; dw
\\ = \frac{1}{2\pi i}
\int_{|w|=\gamma} \frac{1}{w^{n+1}} \frac{1}{1-w}
\frac{1}{2\pi i}
\int_{|z|=\epsilon} \frac{1}{z^{n+1}}
\frac{1}{1-z-w/2}
\; dz\; dw.$$
The pole at $z=1-w/2$ is outside the contour due to the requirements
on convergence, so we may use the negative of the residue there,
getting
$$\frac{1}{2\pi i}
\int_{|w|=\gamma} \frac{1}{w^{n+1}} \frac{1}{1-w}
\frac{1}{(1-w/2)^{n+1}} \; dw.$$
This could have been obtained by inspection, bypassing the Iverson
bracket. Now put $w (1-w/2) = v$ so that $w = 1-\sqrt{1-2v}$ (this
branch maps $w=0$ to $v=0$) to get (here we have $v=w-\cdots$ so the image of $|w|=\gamma$ makes one turn around the origin and may be deformed to a circle $|v|=\gamma'$)
$$\frac{1}{2\pi i}
\int_{|v|=\gamma'} \frac{1}{v^{n+1}}
\frac{1}{\sqrt{1-2v}} \frac{1}{\sqrt{1-2v}} \; dv
\\ = \frac{1}{2\pi i}
\int_{|v|=\gamma'} \frac{1}{v^{n+1}}
\frac{1}{1-2v} \; dv = 2^n.$$
This is the claim. Note that we may take $\gamma' \lt \gamma - \frac{1}{2} \gamma^2.$
Observe that
$$\mathrm{Res}_{z=\infty} \frac{1}{z^{n+1}} \frac{1}{1-z-w/2}
= - \mathrm{Res}_{z=0} \frac{1}{z^2} z^{n+1} \frac{1}{1-w/2-1/z}
\\ = - \mathrm{Res}_{z=0} z^{n} \frac{1}{z(1-w/2)-1} = 0.$$
This was an interesting exercise showing how the choice of contour for
convergence influences the computation. The branch of $\sqrt{1-2v}$
that was used has the branch cut on $[1/2, \infty).$
|
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|
Multiplicate a number per an elevated parenthesis I have a little problem in calculating the second derivative of a function, because I don't know how to compute this operation:
$(2x)*(x+2)^2$
I have to calculate the $(x+2)^2$ in this way: $(x^2+4+4x)$ and then multiply every term per $(2x)$ or there is another and faster way?
Thanks
|
Use the product rule $(fg)' = f'g + g'f$ with $f = 2x$ and $g = (x+2)^2$ so you get $$(2x(x+2)^2)' = (2x)'(x+2)^2 + ((x+2)^2)'(2x)$$
giving $$2(x+2)^2 + 4x(x+2) = 2(x+2)(x+2+2x) = 2(x+2)(3x+2)$$ and then differentiating again with the product rule with $f = x+2$ and $g = 3x+2$ gives the derivative of $(x+2)(3x+2)$ as $$(x+2)'(3x+2) + (3x+2)'(x+2) = 3x+2+3(x+2)$$
and accounting for the factor of $2$ gives you what you want: $4(3x+4)$
|
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"timestamp": "2023-03-29T00:00:00",
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|
Simplifying Ramanujan-type Nested Radicals Ramanujan found many awe-inspiring nested radicals, such as...
$$\sqrt{\frac {1+\sqrt[5]{4}}{\sqrt[5]{5}}}=\frac {\sqrt[5]{16}+\sqrt[5]{8}+\sqrt[5]{2}-1}{\sqrt[5]{125}}\tag{1}$$$$\sqrt[4]{\frac {3+2\sqrt[4]{5}}{3-2\sqrt[4]{5}}}=\frac {\sqrt[4]{5}+1}{\sqrt[4]{5}-1}\tag{2}$$$$\sqrt[3]{\sqrt[5]{\frac {32}{5}}-\sqrt[5]{\frac {27}{5}}}=\frac {1+\sqrt[5]{3}+\sqrt[5]{9}}{\sqrt[5]{25}}\tag{3}$$$$\sqrt[3]{(\sqrt{2}+\sqrt{3})(5-\sqrt{6})+3(2\sqrt{3}+3\sqrt{2})}=\sqrt{10-\frac {13-5\sqrt{6}}{5+\sqrt{6}}}\tag{4}$$$$\sqrt[6]{4\sqrt[3]{\frac {2}{3}}-5\sqrt[3]{\frac {1}{3}}}=\sqrt[3]{\sqrt[3]{2}-1}=\frac {1-\sqrt[3]{2}+\sqrt[3]{4}}{\sqrt[3]{9}}\tag{5}$$
And there's more!
Question: Is there a nice algebraic way to denest each radical such as above?
For me, I've only been able to prove such identities by raising both sides to the appropriate exponents and use Algebra to simplify them. But sometimes, that can be very difficult for identities such as $(1)$.
|
Considering
$$(\sqrt[4]{a} \pm \sqrt[4]{b})^{4}=
a+b+6\sqrt{ab} \pm 4\sqrt[4]{ab}(\sqrt{a}+\sqrt{b})$$
Making a factor in the form of $\sqrt{m}+\sqrt{n}$,
$$\displaystyle \frac{6\sqrt{ab}}{a+b}=\frac{\sqrt{a}}{\sqrt{b}}
\implies \frac{a}{b}=5$$
Then
$$\left( \frac{\sqrt[4]{5}+1}{\sqrt[4]{5}-1} \right)^{4}=
\frac{(\sqrt{5}+1)(6+4\sqrt[4]{5})}
{(\sqrt{5}+1)(6-4\sqrt[4]{5})}=
\frac{3+2\sqrt[4]{5}}
{3-2\sqrt[4]{5}}$$
Note on the symmetric roles of $a$ and $b$:
$$\displaystyle \frac{6\sqrt{ab}}{a+b}=\frac{\sqrt{b}}{\sqrt{a}}
\implies \frac{b}{a}=5$$
which gives the same result.
Hence, there's no other cases similar to $(2)$.
|
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|
Problem in the solution of a trigonometric equation $\tan\theta + \tan 2\theta+\tan 3\theta=\tan\theta\tan2\theta\tan3\theta$
I needed to solve the following equation:
$$\tan\theta + \tan 2\theta+\tan 3\theta=\tan\theta\tan2\theta\tan3\theta$$
Now, the steps that I followed were as follows.
Transform the LHS first:
$$\begin{split}
\tan\theta + \tan 2\theta+\tan 3\theta
&= (\tan\theta + \tan 2\theta)
+ \dfrac{\tan\theta + \tan 2\theta}
{1-\tan\theta\tan2\theta} \\
&= \dfrac{(\tan\theta + \tan 2\theta)(2-\tan\theta\tan2\theta)}
{1-\tan\theta\tan2\theta}
\end{split}$$
And, RHS yields
$$\begin{split}
\tan\theta\tan2\theta\tan3\theta
&= (\tan\theta\tan2\theta)\dfrac{\tan\theta + \tan 2\theta}
{1-\tan\theta\tan2\theta}
\end{split}$$
Now, two terms can be cancelled out from LHS and RHS, yielding the equation:
$$
\begin{split}
2-\tan\theta\tan2\theta &= \tan\theta\tan2\theta\\
\tan\theta\tan2\theta &= 1,
\end{split}$$
which can be further reduced as:
$$\tan^2\theta=\frac{1}{3}\implies\tan\theta=\pm\frac{1}{\sqrt3}$$
Now, we can yield the general solution of this equation:
$\theta=n\pi\pm\dfrac{\pi}{6},n\in Z$. But, setting $\theta=\dfrac{\pi}{6}$ in the original equation is giving one term $\tan\dfrac{\pi}{2}$, which is not defined.
What is the problem in this computation?
|
You concluded from
$$\dfrac{(\tan\theta + \tan 2\theta)(2-\tan\theta\tan2\theta)}{1-\tan\theta\tan2\theta}=(\tan\theta\tan2\theta)\dfrac{\tan\theta + \tan 2\theta}{1-\tan\theta\tan2\theta}$$
(by “canceling” $\dfrac{\tan\theta + \tan 2\theta}{1-\tan\theta\tan2\theta}$ from both sides) that
$$2-\tan\theta\tan2\theta=\tan\theta\tan2\theta.$$
The correct conclusion is that either $\dfrac{\tan\theta + \tan 2\theta}{1-\tan\theta\tan2\theta}=0$ or $2-\tan\theta\tan2\theta=\tan\theta\tan2\theta$. And you still have to check your solutions and keep only those for which the original equation is defined.
Canceling terms doesn’t give you an equivalent equation (one with the same solution set). If you cancel zero, you can lose solutions. If you cancel something undefined, you can introduce wrong solutions.
For example, consider the equation $$\frac{x}{1-x}=\frac{1}{1-x}.$$ If you “cancel” $1-x$ from each side, you get $x=1$, but that is not a solution, because the equation is not defined when $x=1$. And consider the equation
$${x}({1-x})=2({1-x}).$$ If you “cancel” $1-x$, the only solution to what’s left is $x=2$, and you lose the other solution ($x=1$) to the equation, because for that value, you canceled zero from each side.
To summarize, if $E\cdot A=E\cdot B$, where $E$, $A$, and $B$ are expressions, solutions occur when both
*
*$E\cdot A$ and $E\cdot B$ are defined
*Either $E=0$ or $A=B$ (or both).
|
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|
Find the general solution for: $\frac{dy}{dx} + \frac{y}{x}=\frac{1}{x^2}$
Find the general solution for: $$\frac{dy}{dx} + \frac{y}{x}=\frac{1}{x^2} $$
I don't want to solve this using an integrating factor, I wanted to try solve this with a substitution $y=vx$
With the sub of $y=vx$ it implies that $dy = xdv+vdx$ and $\frac{y}{x}= v$
Hence the differential equation is transformed into
$$ \frac{xdv + vdx}{dx} + v =\frac{1}{x^2}$$
$$ \Leftrightarrow \frac{xdv}{dx} + 2v = \frac{1}{x^2}$$
$$ \Leftrightarrow \frac{xdv}{dx} = \frac{1}{x^2} - 2v $$
$$ \Leftrightarrow \frac{dv}{dx} = \frac{1}{x^3} - \frac{2v}{x}$$
Now I am stuck how should I continue? Is this substitution even correct for this question?
|
Try the natural substitution $y(x)=x^n f(x)$.
Note that
$y' = x^n f' + n x^{n-1}f$.
Therefore,
$x^n f' + n x^{n-1}f + x^{n-1}f = 1/x^2$
or
$$x^n f' + (n+1)x^{n-1}f = \frac{1}{x^2}.$$
This differential equation simplifies significantly if we let $n=-1$, which we are free to do.
Then
$f'/x = 1/x^2$
or
$$f' = \frac{1}{x}.$$
Integrate to find $f$.
The solution to the original differential equation is $y = f/x$.
|
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|
What is the $\lim_\limits{x \to 0} \frac{\cos(x)-1+\frac{x^2}{2}}{x^4}$? What is the limit of
$\lim_\limits{x \to 0} \frac{\cos(x)-1+\frac{x^2}{2}}{x^4}$?
I attempted the problem via L^Hopital's Rule so I rewrote it as
$$y=\frac{\cos(x)-1+\frac{x^2}{2}}{x^4}$$
then took the natural of both sides
$$\ln(y)=\ln(\frac{\cos(x)-1+\frac{x^2}{2}}{x^4})$$
then using the properties of logarithms I came to the conclusion that
$$\ln(y)= \ln(\cos(x)-1+\frac{x^2}{2})+\ln(x^4)$$
$$\ln(y)= \ln(\cos(x)-1+\frac{x^2}{2})+4\cdot \ln(x)$$
So then I took the limit as the $\ln(y)$ approaches 0.
$$\lim_\limits {x \to 0} (\ln(\cos(x)-1+\frac{x^2}{2})+4\cdot \ln(x))$$
Here I used L'Hopital's Rule and got
$$\lim_\limits {x \to 0} \frac{-\sin(x)+x}{\cos(x)-1+x^2}+\frac{4}{x}$$
then got a common denominator
$$\lim_\limits {x \to 0} \frac{-\sin(x)+x+4 \cdot (\cos(x)-1+x^2)}{x\cdot (\cos(x)-1+x^2)}$$
I clearly made a mistake somewhere because the denominator is 0. The answer by the way is $\frac{1}{24}$. I have no idea how to arrive at that conclusion.
|
Using Taylor series $$\cos(x)=1-\frac{x^2}{2}+\frac{x^4}{24}-\frac{x^6}{720}+O\left(x^8\right)$$ So $$\cos(x)-1+\frac{x^2}{2}=\frac{x^4}{24}-\frac{x^6}{720}+O\left(x^8\right)$$ $$\frac{\cos(x)-1+\frac{x^2}{2}}{x^4}=\frac{1}{24}-\frac{x^2}{720}+O\left(x^4\right)$$ which shows the limit and how it is approached.
|
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|
Determinant of a $4 \times 4$ matrix $A$ and $(\det(A))^5$ Calculate $\det(A)$ and $\det(A)^5$:
$$A= \begin{bmatrix}a&a&a&a\\a&b&b&b\\a&b&c&c\\a&b&c&d\end{bmatrix}$$
I found $\det A$ with Laplace expansion: $$a(-abc+b^2c+a c^2-b c^2+a b d-b^2 d-a c d+b c d) .$$
But how can I determine $\det A^5$ easily/fast? I know that there is the possibility to use $\det(A^5)=\det(A)^5)$, but this is too long for given time to resolve the problem.
|
If calculations are done mentally but not jotted down, I think the quickest way to calculate $\det A$ is to note that
\begin{align}
A&=a(e_1+e_2+e_3+e_4)(e_1+e_2+e_3+e_4)^T\\
&+(b-a)(e_2+e_3+e_4)(e_2+e_3+e_4)^T\\
&+(c-b)(e_3+e_4)(e_3+e_4)^T\\
&+(d-c)e_4e_4^T.
\end{align}
Therefore $A$ has the $LDL^T$ decomposition
$$
A=\pmatrix{1&0&0&0\\ 1&1&0&0\\ 1&1&1&0\\ 1&1&1&1}
\pmatrix{a\\ &b-a\\ &&c-b\\ &&&d-c}
\pmatrix{1&1&1&1\\ 0&1&1&1\\ 0&0&1&1\\ 0&0&0&1}
$$
and $\det A=a(b-a)(c-b)(d-c)$. And as noted by the others, $\det A^5$ is just $(\det A)^5$.
The above decomposition has the additional merit that it gives also the signature of $A$. In particular, $A$ and $A^5$ are positive semidefinite if and only if $d\ge c\ge b\ge a\ge0$.
|
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|
If $n$ is an integer then $4$ does not divide $n^2 - 3$ Having some trouble proving this. Was going to attack with using the contrapositive of this statement but can't seem to show that $n$ isn't an integer.
|
You just need to look at two cases: $n$ is odd and $n$ is even.
*
*If $n$ is even, then $n \equiv 0 \textrm{ or } 2 \pmod 4$ but $n^2 \equiv 0 \pmod 4$ regardless, which means that $n^2 - 3 \equiv 1 \pmod 4$, and therefore $$\frac{n^2 - 3}{4} = m + \frac{1}{4},$$ where $m$ is some integer. For example, $n = 10$ gives us $24.25$.
*If $n$ is odd, then $n \equiv 1 \textrm{ or } 3 \pmod 4$ but $n^2 \equiv 1 \pmod 4$ regardless, which means that $n^2 - 3 \equiv 2 \pmod 4$, and therefore $$\frac{n^2 - 3}{4} = m + \frac{1}{2},$$ where $m$ is some integer. For example, $n = 11$ gives us $29.5$.
|
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"timestamp": "2023-03-29T00:00:00",
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|
How to prove that $\frac{a^ab^b}{a!b!}\le\frac{(a+b)^{(a+b)}}{(a+b)!}$? As stated in the question. Thank you!
$$\frac{a^ab^b}{a!b!}\le\frac{(a+b)^{(a+b)}}{(a+b)!}$$
|
Base case: Assume $a=b=0$. Then $\frac{0^00^0}{0!0!}\le\frac{(0+0)^{(0+0)}}{(0+0)!}$ goes to $1 \le 1$, which is true.
Inductive case: Suppose $a=b$. Increase $b$ to equal $a+k$ for positive $k$. Then $\frac{a^a(a+k)^{a+k}}{a!(a+k)!}\le\frac{(a+a+k)^{(a+a+k)}}{(a+a+k)!}$ rearranges to $\frac{(2a+k)!}{a!(a+k)!} \le \frac{(2a+k)^{2a+k}}{a^a(a+k)^{a+k}}$ at which point you can compare each piece: $(2a+k)! \le (2a+k)^{2a+k}$, $a! \le a^a$, and $(a+k)! \le (a+k)^{a+k}$.
The same can be said for setting $a$ equal to $b+k$.
|
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|
Solve $x \equiv 32 \pmod{81}$ and $x \equiv 59 \pmod{64}$.
Solve $x \equiv 32 \pmod{81}$ and $x \equiv 59 \pmod{64}$.
$32 + 81k = 59 + 64n \implies 81k - 64n = 27$
$17k \equiv 27 \pmod{64}$.
$64 = 3(17) + 13$
$17 = 1(13) + 4$
$13 = 3(4) + 1$
So $1 = 13 - 3(4) = 13 - 3[17 - 13] = 4(13) - 3(17) = 4(64 - 3*17) - 3*17 =
4(64) - 15(17)$. Thus $k \equiv 1 \pmod{64} \implies k = 1 + 64y$ so we have $n = \frac{54 + 5184y}{64}$
But this is not possible . Help?
EDIT $k\equiv 43 \pmod{64}$ thus,
$k = 43 + 64y$ thus $x = 32 + 81(43 + 64y) = 3515 + 5824y$ so then $x \equiv 3515 \pmod{5824}$.
|
$x=32+81k\equiv 32+17k \equiv 59 \pmod {64}$
$k\equiv \frac{27}{17}\equiv\frac{27}{81}\equiv \frac{1}{3}\equiv \frac{129}{3}\equiv43 \pmod {64}$
$x=32+81(43+64k')=3515+5184k'$
|
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"timestamp": "2023-03-29T00:00:00",
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|
Composition relation of R1 ∘ R2
Let $R_1$ and $R_2$ be the relations on $\{1, 2, 3, 4, 5\}$ defined by
$$R_1 = \{(1,1),(2,3),(2,4),(3,5),(5,2),(5,5)\}$$
$$R_2 = \{(1,1),(2,2),(2,3),(2,5),(4,3),(5,5)\}$$
The answer for this is below but I'm not sure how they arrived at this answer.
Answer:
$$R_2 \circ R_1 = \{(1,1),(2,3),(2,4),(2,5),(4,5),(5,5)\}$$
What I got:
$$R_2 \circ R_1 = \{(1,1),(2,3),(2,4),(2,5),(2,2),(4,5),(5,5)\}$$
|
An alternative is through matrix representations of relations ($a_{ij}=1$ if $(i,j)$ is present in the relation, $0$ otherwise) with composition of relations replaced by matrix product (in the same order as in the composition, with boolean addition convention: $1+1=1$). This can be very useful on a computer. Here, we have :
$$\begin{pmatrix}1&0&0&0&0\\0&1&1&0&1\\0&0&0&0&0\\0&0&1&0&0\\0&0&0&0&1\end{pmatrix}
\times \begin{pmatrix}1&0&0&0&0\\0&0&1&1&0\\0&0&0&0&1\\0&0&0&0&0\\0&1&0&0&1\end{pmatrix}=\begin{pmatrix}1&0&0&0&0\\0&1&1&1&1\\0&0&0&0&0\\0&0&0&0&1\\0&1&0&0&1\end{pmatrix}$$
|
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|
Find the values for $k$ such that the system of equations: $x^2+y^2=k$ and $y^2=4(x-2)$ has no solutions, a unique solution and two solutions?
Question Find the values for $k$ such that the system of equations: $x^2+y^2=k$ and $y^2=4(x-2)$ has no solutions, a unique solution and two solutions
My attempt:
If we try to solve this problem by substituiton the second equations into the first we get:
$$x^2+4(x-2)=k$$
which has discriminant $48+4k$. Thus the equation has $0,1$ and $2$ solutions when $k<-12$, $k=-12$ and $k>-12$ respectively.
However I notice this doesn't match the picture where it's clear the nature of the solutions changes when $k=4$ , why is this?
|
The equation $y^2=4(x-2)$ implies that $x\geq2$. That means the curve defined by $y^2=4(x-2)$ (a parabola) has no point for $x<2$. So, if you are looking for its intersection with another curve (here $x^2+y^2=k$) you should look over $x>2$.
For example, if the goal is having no intersection you should find $k$ such that the equation$$x^2+4(x-2)=k$$has no roots over $x>2$. This is equivalent to find $k$ such that the greatest root of the equation happens before $x=2$. Thus by solving the quadratic equation we have $$x^*=\frac{1}{2}\left(-4+\sqrt{48+4k}\right)<2$$ which leads to$$k<4$$
|
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|
Prove $\frac{a}{b+c}+\frac{b}{c+d}+\frac{c}{d+a}+\frac{d}{a+b} \ge \frac{1}{2}$ where $a,b,c,d \ge 0$ Prove $\frac{a}{b+c}+\frac{b}{c+d}+\frac{c}{d+a}+\frac{d}{a+b} \ge \frac{1}{2}$ where $a,b,c,d \ge 0$
My attempt:I used two ways but I get to a wrong answer.
My first way:We know that $\frac{a}{b}+\frac{b}{a} \ge 2$ where $a,b \ge 0$
Then:
$\frac{a}{b+c}+\frac{b+c}{a}+\frac{b}{c+d}+\frac{c+d}{b}+\frac{c}{d+a}+\frac{d+a}{c}+\frac{d}{a+b}+\frac{a+b}{d} \ge 8$
And:
$\frac{b+c}{a}+\frac{c+d}{b}+\frac{d+a}{c}+\frac{a+b}{d}=\frac{b}{a}+\frac{c}{a}+\frac{c}{b}+\frac{d}{b}+\frac{d}{c}+\frac{a}{c}+\frac{a}{d}+\frac{b}{d}=$
$\frac{a}{c}+\frac{c}{a}+\frac{b}{d}+\frac{d}{b} \ge 4$
$+$
$\frac{b}{a}+\frac{c}{b}+\frac{d}{c}+\frac{a}{d}\ge 4\sqrt{\frac{b}{a}*\frac{c}{b}*\frac{d}{c}*\frac{a}{d}}=4$
Then:
$\frac{a}{b+c}+\frac{b+c}{a}+\frac{b}{c+d}+\frac{c+d}{b}+\frac{c}{d+a}+\frac{d+a}{c}+\frac{d}{a+b}+\frac{a+b}{d} \ge 8$
And:
$\frac{b+c}{a}+\frac{c+d}{b}+\frac{d+a}{c}+\frac{a+b}{d}=\frac{b}{a}+\frac{c}{a}+\frac{c}{b}+\frac{d}{b}+\frac{d}{c}+\frac{a}{c}+\frac{a}{d}+\frac{b}{d}\ge 8$
Then we will get:
$\frac{a}{b+c}+\frac{b}{c+d}+\frac{c}{d+a}+\frac{d}{a+b} \ge 0$
Which is not true.
My second way:I don't have enough time then I just explain it.
I used caushy-shuartz and I get:
$\frac{a}{b+c}+\frac{b}{c+d}+\frac{c}{d+a}+\frac{d}{a+b} \ge 2$
|
$\sum\limits_{cyc}\frac{a}{b+c}=\sum\limits_{cyc}\frac{a^2}{ab+ac}\geq\frac{(a+b+c+s)^2}{\sum\limits_{cyc}(ab+ac)}\geq2$
Because the last inequality it's $(a-c)^2+(b-d)^2\geq0$. Done!
|
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|
Area enclosed by the curve $5x^2+6xy+2y^2+7x+6y+6=0$ We have to find the area enclosed by the curve
$$5x^2+6xy+2y^2+7x+6y+6=0.$$
I tried and I got that it is an ellipse, and I know its area is $\pi ab$ where $a$ and $b$ are the semiaxis lengths of the ellipse.
But I am unable to find the value of $a$ and $b$.
|
For the curve $ax^2+2hxy+by^2+2gx+2fy+c = 0$, the center is given by
\begin{align*}
ax+hy+g &= 0\\
hx+by+f &= 0
\end{align*}
and the equation of the ellipse with the axes through the center parallel to the axes is
\begin{align*}
ax^2+2hxy+by^2 + c' = 0
\end{align*}
where $c' = gx_c + fy_c + c$, where $(x_c, y_c)$ is the center.
|
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|
Prove that $\frac{a}{b} < \frac{a+c}{b+d} <\frac{ c}{d}$ I'm reading a introductory book on mathematical proofs and I am stuck on a question.
Let $a, b, c, d$ be positive real numbers, prove that if $\frac{a}{b} < \frac{c}{d}$, then $\frac{a}{b} < \frac{a+c}{b+d} <\frac{ c}{d}$.
|
HINT: Note that
$$\begin{align*}
\frac{a+c}{b+d}-\frac{a}b&=\frac{a+c}{b+d}\cdot\frac{b}b-\frac{a}b\cdot\frac{b+d}{b+d}\\
&=\frac{b(a+c)-a(b+d)}{b(b+d)}\\
&=\frac{ab+bc-ab-ad}{b(b+d)}\\
&=\frac{bc-ad}{b(b+d)}\;.
\end{align*}$$
You’d like to show that this difference is positive, so you’d like to know that $bc>ad$. Use the fact that $\frac{a}b<\frac{c}d$ to show this.
The other inequality can be handled similarly.
|
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|
Complex number fraction Find the real and imaginary number:
$$\frac{1}{z}=\frac{2}{2+j3}+\frac{1}{3-j2}$$
How do I invert $\frac{1}{z}$ to $z$ so that I can start solving it?
|
Solve a more general way, assume $q\in\mathbb{C}$:
$$q=\Re[z]+\Im[z]i$$
So, we get:
$$\frac{1}{q}=\frac{\overline{q}}{q\overline{q}}=\frac{\overline{q}}{|q|^2}=\frac{\Re[q]-\Im[q]i}{\Re^2[q]+\Im^2[q]}=\frac{\Re[q]}{\Re^2[q]+\Im^2[q]}-\frac{\Im[q]}{\Re^2[q]+\Im^2[q]}\cdot i$$
Now, in your case:
$$\frac{2}{2+3i}+\frac{1}{3-2i}=\frac{2(3-2i)+2+3i}{(2+3i)(3-2i)}=\frac{8-i}{12+5i}=$$
$$\frac{(8-i)\cdot\overline{12+5i}}{(12+5i)\cdot\overline{12+5i}}=\frac{91-52i}{12^2+5^2}=\frac{91-52i}{169}=\frac{7-4i}{13}\to\color{red}{z=\frac{13}{7-4i}}$$
For $z$ we find:
$$z=\frac{13}{7-4i}=\frac{13\cdot\overline{7-4i}}{(7-4i)\cdot\overline{7-4i}}=\frac{13(7+4i)}{7^2+4^2}=\frac{7+4i}{5}$$
So:
*
*$$\Re[z]=\Re\left[\frac{7+4i}{5}\right]=\frac{7}{5}$$
*$$\Im[z]=\Im\left[\frac{7+4i}{5}\right]=\frac{4}{5}$$
|
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|
Proof verification: $a_1=2$, $a_{n+1}=3+\frac{a_n}{2}$ is increasing and bounded For base case, $a_2=3+\frac{a_1}{2}=4>a_1$
For $n=k$, let $a_{k+1}>a_k$
Adding 3 on both sides after dividing by 2,
$$3+\frac{a_{k+1}}{2}>3+\frac{a_k}{2}$$
$$a_{k+2}>a_{k+1}$$
Hence the sequence is increasing.
How can I show that the sequence is bounded?
|
$$\begin{align}
a_{n+1}&=3+\frac {a_n}2\\
a_{n+1}-6&=\frac 12 \left(a_n-6\right)\\
b_n&=\frac 12 b_{n-1}=\cdots =\frac 1{2^{n-1}}b_1&&(b_n=a_n-6)\\
a_n-6&=\frac 1{2^{n-1}}(\underbrace{\;a_1\;}_{=2}-6)\\
\color{red}{a_n}&\color{red}{=6-\frac 8{2^n}}
\end{align}$$
which is increasing and bounded (at $6$) as $n\to \infty$.
|
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|
Neglecting Terms Based On Their Relevant Magnitude If we have an expression like:
$$\frac{1}{l}[ \frac{1}{a} - \frac{1}{a+l}]$$
and we know that $l<<a$ then we can convert the above expression to
$$ \frac{1}{a(a+l)}$$
and then neglect the l from the denominator. So we end up with:
$$ \frac{1}{a^2} $$
But this is a very easy case. A more complicated case whould be this:
$$ \frac{1}{ \sqrt{(β-α-l)^2+γ^2}}-\frac{1}{ \sqrt{(β-α)^2+γ^2}} $$
Where $ l<<β-l-a$
My question is what rules do we follow to make those neglections?
I assume these hold:
*
*$a+l \approx a$
*$\frac{a}{l} \to \infty$
*$\frac{l}{a} \approx 0$
Present examples if possible.
|
Hint. One may write, by the Taylor series expansion, as $\dfrac{l}{a} \to 0$,
$$
\frac{1}{l}\left[ \frac{1}{a} - \frac{1}{a+l}\right]=\frac{1}{a(a+l)}=\frac1{a^2}\cdot\frac{1}{1+\dfrac{l}{a}}=\frac1{a^2}\left[1-\frac{l}{a}+O\left(\frac{l^2}{a^2}\right)\right]=\frac1{a^2}+O\left(\frac{l}{a^3}\right). \tag1
$$ Similarly, as $\dfrac{l}{β-α} \to 0$, one has
$$
\begin{align}
& \frac{1}{ \sqrt{(β-α-l)^2+γ^2}}-\frac{1}{ \sqrt{(β-α)^2+γ^2}}
\\\\&=\left(\frac{1}{ \sqrt{(β-α)^2+γ^2-2l\cdot(β-α)+l^2}}-\frac{1}{ \sqrt{(β-α)^2+γ^2}}\right)
\\\\&=\frac{1}{\sqrt{(β-α)^2+γ^2}}\left[\frac{1}{\sqrt{1-2\dfrac{l\cdot(β-α)}{(β-α)^2+γ^2}+\dfrac{l^2}{(β-α)^2+γ^2}}}-1\right]
\\\\&=\frac{1}{\sqrt{(β-α)^2+γ^2}}\left[1+\dfrac{l\cdot(β-α)}{(β-α)^2+γ^2}+O\left(\dfrac{l^2}{(β-α)^2}\right)-1\right]
\\\\&=\frac{1}{\sqrt{(β-α)^2+γ^2}}\cdot O\left(\dfrac{l}{β-α}\right)
\\\\&=O\left(\dfrac{l}{β-α}\right) \tag2
\end{align}
$$ where we have used the standard Taylor series expansion, as $x \to 0$,
$$
\frac1{\sqrt{1-x}}=1+\frac{x}2+O(x^2).
$$
|
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|
What is the remainder when the product of the primes between 1 and 100 is divided by 16? The product of all the prime numbers between 1 and 100 is equal to $P$. What is the remainder when $P$ is divided by 16?
I have no idea how to solve this, any answers?
|
Skip the first prime $2$ and look for the product modulo $8$.
The twenty-four odd primes $<100$ are
$$3,5,7,11,13,17, 19,23,29,31,37,41, 43,47,53,59,61,67, 71,73,79,83,89,97. $$
Modulo $8$, these are
$$3,5,7,3,5,1,3,7,5,7,5,1,3,7,5,3,5,3,7,1,7,3,1,1$$
so their product is $$1^53^75^67^6\equiv 3\pmod 8 $$
(where we might profit from using $x^2\equiv 1\pmod 8$ for odd $x$).
Thus $P\equiv 6\pmod{16}$.
|
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|
Find all pairs $(a,b)$ of positive integers, such that $\frac{b^2+ab+a+b-1}{a^2+ab+1}$ is integer.
Find all positive integers $a$ and $b$ such that $$\frac{b^2+ab+a+b-1}{a^2+ab+1}$$ is integer.
My work so far:
If $a=1, b\in\mathbb N$ then
$$\frac{b^2+ab+a+b-1}{a^2+ab+1}=\frac{b^2+2b}{b+2}=b \in \mathbb Z$$
|
You have $a(a+b)+1=a^2+ab+1$ divides $$\left(b^2+ab+a+b-1\right)+\left(a^2+ab+1\right)=(a+b)(a+b+1)\,.$$ Because $\gcd\big(a(a+b)+1,a+b\big)=1$, we must have $a(a+b)+1\mid a+b+1$. The rest should be easy.
|
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|
Find $\frac{x^2}{y^2}$ + $\frac{y^2}{x^2}$ If $\frac{x}{y}$ + $\frac{y}{x}$ = 3 Find $\frac{x^2}{y^2}$ + $\frac{y^2}{x^2}$
Any Ideas on how to begin ?
|
Hint: Look at:
$$\left(\frac{x}{y} + \frac{y}{x}\right)^2=3^2=9$$
Why to look at this? We know if we expand we will have a $(\frac{x}{y})^2=\frac{x^2}{y^2}$ term and a $(\frac{y}{x})^2=\frac{y^2}{x^2}$ term so this might be worth a try.
|
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|
If $a,b,c$ are in Arithmetic Progression, find the other two vertices of the triangle. A triangle $ABC$ is given where vertex $A$ is $(1,1)$ and the orthocentre is $(2,4)$. Also sides $AB$ and $BC$ are members of the family of the lines $ax+by + c = 0$, where $a,b,c$ are in Arithmetic Progression..Find the other two vertices of the triangle.
Using information about Arithmetic Progression and vertex $A$, I have deduced that
$ax-a=0$. But now I am able to proceed further and use coordinates of orthocentre. Could someone help me with this?
|
The arithmetic progression condition amounts to $a+c=2b$.
Now, consider $AB$. Because $A$ is on this line, we have $a+b+c=0$, and as you deduced, we get $ax-a=0\implies x=1$. Thus, the $x$-coordinate of point $B$ must be 1 as well i.e. $B=(1,m)$.
Then, consider point $C$. Because $AB$ is on the line $x=1$, and the orthocenter is at $(2,4)$, $C$ must lie on the line passing through $(2,4)$ that is perpendicular to $x=1$ i.e. the line $y=4$. Thus, $C=(n,4)$.
Now, since $BC$ lies on a line of the form $ax+by+c=0$, we plug in $B$ to get $a+bm+c=0$ Recalling that $a+c=2b$, the first equation becomes $b(m+2)=0$, and from this we must have $m=-2$ or $b=0$. But $b=0$ forces $a+c=0$, and thus gives a degenerate triangle, so $m=-2$.
So $B=(1,-2)$, $A=(1,1)$, and $C=(n,4)$. The slope of the line through $B$ and $C$ is $\frac{6}{n-1}.$ The line with perpendicular to $BC$ passing through $A$ then, is $$y-1=\frac{1-n}{6}(x-1).$$ This must pass through the orthocenter, $(2,4)$, so we find $$3=\frac{1-n}{6}\cdot 1\implies n=-17.$$
Thus, the triangle is $A=(1,1), B=(1,-2), C=(-17,4)$.
|
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|
Maxmum distance from a point to a sphere Which point of the sphere $x^{2} + y^{2} + z^{2} = 1$ is at the maximum distance from the point $\left(\, 2,1,3\, \right)$ ?.
I know that the point is outside the sphere. Should we proceed with maxima and minima concept ?.
|
WLOG any point on the sphere $(\cos u,\sin u\cos t,\sin u\sin t)$
$$D^2=(\cos u-2)^2+(\sin u\cos t-1)^2+(\sin u\sin t-3)^2$$
$$2^2+1^2+3^2+\cos^2u+\sin^2u(\cos^2t+\sin^2t)-4\cos u-2\sin u(\cos t+3\sin t)$$
$$=2^2+1^2+3^2+\cos^2u+\sin^2u-4\cos u-2\sin u(\cos t+3\sin t)$$
$\cos t+3\sin t=\sqrt{1^2+3^2}\cos\left(t-\arccos\dfrac1{\sqrt{1^2+3^2}}\right)$
$\implies-\sqrt{10}\le\cos t+3\sin t\le\sqrt{10}$
$\implies-2\sqrt{10}|\sin u|\le-2\sin u(\cos t+3\sin t)\le2\sqrt{10}|\sin u|$
$$D^2-(2^2+1^2+3^2+1)\le-4\cos u+2\sqrt{10}|\sin u|\le\sqrt{4^2+(2\sqrt{10})^2}=2\sqrt{14}$$
|
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|
Proving $\left|\sqrt2-(a/b)\right|\geq1/(3b^2)$ This is Problem 4.26 on p.58 of The Theory of Algebraic Numbers by Harry Pollard and Harold G. Diamond (Dover edition).
Prove that $\left|\sqrt2-\dfrac ab\right|\geq\dfrac1{3b^2}$ for all positive integers $a,b$.
Here's what I've done so far. Split into cases:
*
*(i) $\dfrac ab<\sqrt2$
*(ii) $\dfrac ab>\sqrt2+\dfrac13$
*(iii) $\sqrt2<\dfrac ab<\sqrt2+\dfrac13$.
(i) I followed the method used by the authors in proving a similar result in that chapter. Let $f(x)=x^2-2$; this is the minimum polynomial of $\sqrt2$ in $\mathbb Q$ so $f\left(\sqrt2\right)=0$. By the mean-value theorem:
$$\frac{f\left(\sqrt2\right)-f\left(\dfrac ab\right)}{\sqrt2-\dfrac ab}\ =\ f'(\xi)$$
for some $\dfrac ab<\xi<\sqrt2$. We have $f'(\xi)=2\xi<2\sqrt2<3$, and so
$$\left|f\left(\frac ab\right)\right|\ <\ 3\left|\sqrt2-\frac ab\right|$$
But
$$\left|f\left(\frac ab\right)\right|=\left|\frac{a^2}{b^2}-2\right|=\frac{|a^2-2b^2|}{b^2}\geq\frac1{b^2}$$
Hence
$$\left|\sqrt2-\frac ab\right|\ \geq\ \frac1{3b^2}$$
as required.
(ii) Done immediately as
$$\left|\sqrt2-\frac ab\right|=\frac ab-\sqrt2\,> \frac13\geq\frac1{3b^2}$$
(iii) This is where I'm stuck.
So I've basically done much of the hard work above and the final piece is all I need to complete the jigsaw. I would be grateful for some help. Thanks.
|
Prove that $$\left|\sqrt2-\dfrac ab\right|>\dfrac1{3b^2}$$ for integers $a$ and $b$ with $b\neq 0$.
Without loss of generality, assume that $b>0$. Note that
$$\left|\sqrt{2}-\frac{a}{b}\right|=\frac{\left|a^2-2b^2\right|}{b(a+\sqrt{2}b)}\,.\tag{*}$$
If $a<0$, then
$$\left|\sqrt{2}-\frac{a}{b}\right|>\sqrt{2}>\frac{1}{3b^2}\,.$$
If $0\leq a\leq b$, then, using (*),
$$\left|\sqrt{2}-\frac{a}{b}\right|\geq\frac{b^2}{b(b+\sqrt{2}b)}\geq \frac{1}{1+\sqrt{2}}>\frac{1}{3}\geq\frac{1}{3b^2}\,.$$
If $a\geq \frac{3}{2}b$ and $b>1$, then
$$\left|\sqrt{2}-\frac{a}{b}\right|\geq\frac{3}{2}-\sqrt{2}>\frac{1}{12}\geq\frac{1}{{3b^2}}\,.$$
If $a>1$ and $b=1$, then
$$\left|\sqrt{2}-\frac{a}{b}\right|\geq 2-\sqrt{2}>\frac{1}{3}\geq\frac{1}{3b^2}\,.$$
If $b<a<\frac{3}{2}b$, then, using (*),
$$\left|\sqrt{2}-\frac{a}{b}\right|\geq \frac{1}{b\left(a+\sqrt{2}b\right)}>\frac{1}{b\left(\frac{3}{2}b+\sqrt{2}b\right)}=\frac{1}{\left(\frac{3}{2}+\sqrt{2}\right)b^2}>\frac{1}{3b^2}\,.$$
Indeed, the smallest constant $\gamma>0$ such that $$\left|\sqrt{2}-\frac{a}{b}\right|\geq \frac{1}{\gamma\,b^2}$$ for all integers $a$ and $b$ with $b\neq 0$ is $\gamma=\frac{3}{2}+\sqrt{2}$. The equality holds if and only if $a=3$ and $b=2$.
Related Observation
Conjecture. Let $\gamma_b:=\max\,\left\{\frac{1}{b^2\,\left|\sqrt{2}-\frac{a}{b}\right|}\,\Big|\,a\in\mathbb{Z}\right\}$ for each $b\in\mathbb{N}$. Define $$\Gamma:=\liminf_{b\to\infty}\,b\cdot\gamma_b\,.$$ Then, $$\Gamma=\inf\,\left\{b\cdot\gamma_b\,\big|\,b\in\mathbb{N}\right\}=2\,.$$
After checking all positive integers $b\leq 1500$, I find that $$\Gamma\leq 1+\frac{29}{41}\sqrt{2}\lesssim 2.0003\,.$$ Below is a plot illustrating the credibility of this conjecture. Th horizontal axis is $b$. The fuzzy blue line shows $\gamma_b$ versus $b$, and the nice orange line is $\dfrac{2}{b}$ versus $b$ for $b=1,2,\ldots,1500$.
What is very strange is $\Gamma$ seems to be equal to $2$ as well if $\sqrt{2}$ is replaced by $\sqrt{3}$, $\sqrt{5}$, $\sqrt{6}$, $\sqrt{7}$, and $\sqrt{8}$. Could $\Gamma=2$ universally hold when $\sqrt{d}$ replaces $\sqrt{2}$ for any non-square positive integer $d$? I tried replacing $\sqrt{2}$ by $\frac{1+\sqrt{5}}{2}$, $\sqrt[3]{2}$, $\pi$, $\text{e}$, and $\ln(3)$ as well, and it looks like $\Gamma=2$ still holds.
On the other hand, $$\limsup\limits_{b\to\infty}\,\gamma_b=2\sqrt{2}\,.$$ If $\sqrt{2}$ is replaced by $\sqrt{d}$ for any non-square positive integer $d$, then $$\limsup\limits_{b\to\infty}\,\gamma_b=2\sqrt{d}\,.$$
|
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find integers a and b such $x^2-x-1$ divides $ax^{17}+bx^{16}+1 = 0$
find integers a and b such $x^2-x-1$ divides $ax^{17}+bx^{16}+1 = 0$
By really long division i got :-
$$Q=ax^{15} + (a+b)x^{14} + \dots +(610a + 377b) $$
$$R = x(987a+610b)+1+610a+377b$$
since remainder is $0 $,
$$987a+610b = 0$$
$$1+610a+377b = 0$$
from which i got $a = -610, b = 987$
but from wolfram alpha the remainder of $-610x^{17}+987x^{16}+1 \over x^2-x-1$ is $x-1$
Somebody please show me where i went wrong ?
Thanks.
|
Still another approach (and a quite compact one). Since
$$ \frac{1}{1-x-x^2}=\sum_{n\geq 0}F_{n+1} x^n \tag{1}$$
it is not difficult to write down the Taylor series of $\frac{1+bx^{16}+a x^{17}}{1+x-x^2} $:
$$\frac{1+bx^{16}+a x^{17}}{1+x-x^2}=\sum_{n\geq 0}(-1)^n\left(F_{n+1} x^n + bF_{n+1} x^{n+16} + a F_{n+1} x^{n+17}\right) \tag{2}$$
If $1+x-x^2$ is a divisor of $1+bx^{16}+ax^{17}$, the RHS of $(2)$ is a polynomial with degree $\leq 15$, hence the coefficients of $x^{16}$ and $x^{17}$ have to be zero. That translates into:
$$ F_{17}+b = 0,\qquad F_{18}+b-a = 0\tag{3}$$
from which $(a,b)=(F_{16},-F_{17})=\color{red}{(987,-1597)}$ readily follows. Now we just have to check that, with this choice, for any $m\geq 18$ the coefficient of $x^m$ in the RHS of $(2)$ is actually zero. That is easy by induction.
|
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Find value of x, where $\frac{3+\cot 80^{\circ} \cot 20^{\circ}}{\cot 80^{\circ}+\cot 20^{\circ}}=\cot x^{\circ}$ $$\frac{3+\cot 80^{\circ}\cot 20^{\circ}}{\cot 80^{\circ}+\cot 20^{\circ}}=\cot x^{\circ}$$ Then find $x$
My Try:
Using $\cot 80=\tan 10$ and $\cot 20=\frac{1}{\tan 20}$
we have LHS as $$\frac{3+\frac{\tan 10^{\circ}}{\tan 20^{\circ}}}{\tan 10^{\circ}+\frac{1}{\tan 20^{\circ}}}$$
assuming $y=10$ and using $\tan 2y=\frac{2 \tan y}{1-\tan ^2 y}$ we get
LHS as
$$\frac{3+\frac{\tan y}{\tan 2y}}{\tan y+\frac{1}{\tan 2y}}=\frac{\tan y(7-\tan ^2 y)}{1+\tan ^2 y}$$
how to proceed further?
|
Let $\alpha = 20^\circ, \,\, u = \tan(\alpha)$, then
$$ LHS = \frac{1 + 3\tan(80^\circ)\tan(20^\circ)}{\tan(80^\circ) + \tan(20^\circ)} = \frac{1 + 3\tan(\alpha + 60^\circ)\tan(\alpha)}{\tan(\alpha + 60^\circ) + \tan(\alpha)} .$$
notice that
$$ \tan(\alpha + 60^\circ) = \frac{\sqrt{3} + \tan{\alpha}}{1 - \sqrt{3} \tan{\alpha}} = \frac{\sqrt{3}+u}{1-u\sqrt{3}},$$
thus the LHS becomes
$$ LHS = \frac{1 + 3u\cdot \frac{\sqrt{3}+u}{1-u\sqrt{3}}}{u + \frac{\sqrt{3}+u}{1-u\sqrt{3}}} = \frac{1 - u\sqrt{3} + 3u\sqrt{3} + 3u^2}{u - u^2\sqrt{3}+ u + \sqrt{3}} = \frac{u\sqrt{3} + 1}{\sqrt{3} - u} = \cot(60^\circ - \alpha) = \cot(40^\circ).$$
|
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"timestamp": "2023-03-29T00:00:00",
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|
Find a law, and prove it by induction. From the following equalities, I'm supposed to find a law and prove it by induction.
$1=1$; $1-4=-(1+2)$; $1-4+9=1+2+3$, and $1-4+9-16=-(1+2+3+4)$.
I supposed the law was $\sum^{n}_{k=1} (-1)^{k+1}k^2=(-1)^{n+1}\sum^n_{k=1} k$.
Let's say it's valid for a $n\in \mathbb{N}$. For $n+1$, I have $\sum^{n+1}_{k=1} (-1)^{k+1}k^2=\sum^{n}_{k=1} (-1)^{k+1}k^2+(-1)^{n+2}(n+1)^2$, which by the induction hypothesis it's equal to
$(-1)^{n+1}\sum^n_{k=1} k+(-1)^{n+2}(n+1)^2$.
I'm stuck here. What's the 'trick'?
Any help would be appreciated.
|
although this is not an inductive proof it is worth noting, in passing, that this rather pretty result follows from the fact that the difference of the squares of two consecutive integers is equal to their sum. thus, for example:
$$
9^2-8^2 = 9+8\\
7^2-6^2 = 7+6\\
5^2-4^2 = 5+4\\
3^2-2^2= 3+2\\
1^2-0^2= 1+0
$$
adding these equations gives
$$
9^2-8^2+7^2-6^2+5^2-4^2+3^2-2^2+1^2 = 9+8+7+6+5+4+3+2+1
$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1901310",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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|
Evaluate $\sum\limits_{k=0}^n(-1)^{n+k}{n\choose k}{ {n+k}\choose n} \frac{1}{k+2}$ I am trying to evaluate the following sum :
$S=\displaystyle\sum_{k=0}^n(-1)^{n+k}{n\choose k} {{n+k}\choose n} \frac{1}{k+2}$
which is the same as
$\displaystyle\sum_{k=0}^n(-1)^{n+k}{n\choose k} {{n+k}\choose k} \frac{1}{k+2}$
My approach so far has been aimed at converting this sum to a known form (given as a standard result in Integrals and series by Prudnikov et.al.)
$\displaystyle\sum_{k=1}^n(-1)^{k+1}{n\choose k} {{n+k}\choose k} \frac{1}{k}=\sum_{k=1}^n\frac{1}{k}$
I thought of combining the like terms ${n\choose k} {{n+k}\choose k}$ in the two series but that again depends on whether $n$ is even or odd. Any suggestion to proceed further ?
|
$$
\begin{align}
&\sum_{k=0}^n(-1)^{n+k}\binom{n}{k}\binom{n+k}{n}\frac1{k+2}\\[6pt]
&=\sum_{k=0}^n(-1)^{n+k}\binom{n}{k}\binom{n+k}{k}\left(\frac1{k+1}-\frac1{(k+1)(k+2)}\right)\tag{1}\\[6pt]
&=\frac1{n+1}\sum_{k=0}^n(-1)^{n+k}\binom{n+1}{k+1}\binom{n+k}{k}\\
&-\frac1{(n+1)(n+2)}\sum_{k=0}^n(-1)^{n+k}\binom{n+2}{k+2}\binom{n+k}{k}\tag{2}\\[6pt]
&=\frac{(-1)^n}{n+1}\sum_{k=0}^n\binom{n+1}{n-k}\binom{-n-1}{k}\\
&-\frac{(-1)^n}{(n+1)(n+2)}\sum_{k=0}^n\binom{n+2}{n-k}\binom{-n-1}{k}
\tag{3}\\[6pt]
&=\frac{(-1)^n}{n+1}\binom{0}{n}-\frac{(-1)^n}{(n+1)(n+2)}\binom{1}{n}\tag{4}\\[6pt]
&=\bbox[5px,border:2px solid #C0A000]{\left\{\begin{array}{}
\frac12&\text{if }n=0\\
\frac16&\text{if }n=1\\
0&\text{if }n\ge2
\end{array}\right.}\tag{5}
\end{align}
$$
Explanation:
$(1)$: $\frac1{k+2}=\frac1{k+1}-\frac1{(k+1)(k+2)}$ and $\binom{n+k}{n}=\binom{n+k}{k}$
$(2)$: $\frac1{k+1}\binom{n}{k}=\frac1{n+1}\binom{n+1}{k+1}$ applied once to the first sum and twice to the second
$(3)$: $\binom{n+j}{k+j}=\binom{n+j}{n-k}$ and $\binom{n+k}{k}=(-1)^k\binom{-n-1}{k}$
$(4)$: Vandermonde's Identity
$(5)$: evaluate at $n=0$ and $n=1$, and for $n\ge2$, $\binom{0}{n}=\binom{1}{n}=0$
|
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"timestamp": "2023-03-29T00:00:00",
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|
cardinality of cartesian product of the infinite set of natural numbers One of the problems in my discrete math course states that we need to prove that $\mathcal{N}\times\mathcal{N}$ is countable specifically when there's a function $f:\mathcal{N}\times\mathcal{N}\to\mathcal{N}$ defined as follows:
$$f(a,b)=\frac{1}{2}(a+b+1)(a+b)+a$$
where $a,b \in \mathcal{N}$. The solution uses the function definition in order to prove that the function is bijective as shown below:
$$f(a,b+1)=\frac{1}{2}(a+b+1+1)(a+b+1)+a=$$
$$=\frac{1}{2}(a+b+1)(a+b)+a+\frac{1}{2}(a+b+1)\cdot2$$
How is the transition achieved? I tried all kinds of arithmetics and couldn't arrive to the expression after the equal sign.
|
How is the transition achieved?
$$f(a,b+1)=\color{red}{ \dfrac{1}{2}}(\color{green}{a+b+1}\color{blue}{+1})(a+b+1)+a=$$
$$=\color{red}{ \dfrac{1}{2}}(\color{green}{a+b+1})(a+b+1) + \color{red}{\dfrac{1}{2}}(\color{blue}{+1})(a+b+1)+a$$
$$=\color{red}{ \dfrac{1}{2}}(\color{green}{a+b+1})(a+b\color{orange}{+1}) + \color{red}{\dfrac{1}{2}}(\color{blue}{+1})(a+b+1)+a$$
$$=\color{red}{ \dfrac{1}{2}}(\color{green}{a+b+1})(a+b) + \underbrace{\color{red}{ \dfrac{1}{2}}(\color{green}{a+b+1})(\color{orange}{+1}) + \color{red}{\dfrac{1}{2}}(\color{blue}{+1})(a+b+1)}_{\color{purple}{\text{two copies}}}+a$$
$$=\color{red}{ \dfrac{1}{2}}(\color{green}{a+b+1})(a+b) +\color{purple}{ \dfrac{1}{2}(a+b+1)\cdot 2}+a$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1906620",
"timestamp": "2023-03-29T00:00:00",
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|
Prove the inequality using induction Prove $3^n \geq 2n^2 +1$ for $n = 1,2,\ldots$ using induction
This is what I have so far
Base case - $n=1,3^1 \geq 2^1 + 1 = 3$ true
Induction step - Assume true for some n, then,
$ 3*3^n \geq 3*(2n^2 +1)=6n^2 +3 $
I have to somehow manipulate and show its $\geq 2*(n+1)^2 +1$
|
First, show that this is true for $n=1$:
$3^1\geq2\cdot1^2+1$
Second, assume that this is true for $n$:
$3^n\geq2n^2+1$
Third, prove that this is true for $n+1$:
$3^{n+1}=$
$3\cdot\color\red{3^n}\geq$
$3\cdot(\color\red{2n^2+1})=$
$6n^2+3=$
$2n^2+4n^2+3\geq$
$2n^2+4n+3=$
$2n^2+4n+2+1=$
$2(n^2+2n+1)+1=$
$2(n+1)^2+1$
Please note that the assumption is used only in the part marked red.
|
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|
Contradictory result when testing Linear independence using Gaussian elimination Consider a set of vectors - (2,3,1) , (1,-1,2) and (7,3,8).
I want to find if its linearly dependent or independent.
Putting it as:
\begin{equation}
2a + b + 7c = 0 \\
3a - b + 3c = 0 \\
a + 2b + 8c = 0
\end{equation}
If I use Gaussian elimination of equations to calculate row echelon form of the matrix:
\begin{bmatrix}
2&1&7&0 \\
3&-1&3&0 \\
1&2&8&0 \\
\end{bmatrix}
I get:
\begin{bmatrix}
1&0&0&0 \\
0&1&0&0 \\
0&0&1&0 \\
\end{bmatrix}
which means a = b = c = 0
*
*So, as per
Determine if vectors are linearly independent
the vectors are linearly independent.
*But I know that they are linearly dependent, with a = 2, b = 3, c =
-1
Why is there such a contradiction? did i do something procedurally wrong?
Note: I just want to understand the mistake in above procedure, dont want an alternate solution like using determinant etc (unless this procedure itself is totally wrong!)
|
There's no contradiction: you reduced echelon form is wrong. Let me do it again:
\begin{align*}
&\begin{bmatrix}
2&1&7\\
3&-1&3\\
1&2&8
\end{bmatrix}
\rightsquigarrow
\begin{bmatrix}
1&2&8\\
2&1&7\\
3&-1&3
\end{bmatrix}
\rightsquigarrow
\begin{bmatrix}
1&2&8\\
0&-3&-9\\
0&-7&-21
\end{bmatrix}
\rightsquigarrow
\begin{bmatrix}
1 & 2 & 8\\
0 & 1 & 3\\
0 & 1 & 3
\end{bmatrix}
\rightsquigarrow
\begin{bmatrix}
1 & 0 & 2\\
0 & 1 & 3\\
0 & 0 & 0
\end{bmatrix},\\
&\text{whence the solutions: }\hspace{3.5em}\begin{bmatrix}a\\b\\c\end{bmatrix} =t\begin{bmatrix}2\\3\\-1\end{bmatrix}.
\end{align*}
|
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"timestamp": "2023-03-29T00:00:00",
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|
Find $f^{(100)}(x)$ where $f(x)=\frac{1}{4x^2-1}$ Find $f^{(100)}(x)$ where $f(x)=\frac{1}{4x^2-1}$. I found first,second and third derivative:
$$f'(x)=\frac{-8x}{(4x^2-1)^2}$$
$$f''(x)=\frac{96x^2+8}{(4x^2-1)^3} $$
$$f'''(x)=\frac{-384x(4x^2+1)}{(4x^2-1)^4}$$
I can't seem to find any rule between them. Anyone has any ideas or hints?
|
$$
f(x) = \frac{1}{2}\left(\frac{1}{2x-1} - \frac{1}{2x+1}\right)
$$
Now if $g(x) = \frac{1}{2x-1} = (2x-1)^{-1}$, then check (by induction) that
$$
g^{(n)}(x) = (-2)^nn!(2x-1)^{-(n+1)}
$$
and if $h(x)=(2x+1)^{-1}$, then
$$
h^{(n)}(x) = (-2)^nn!(2x+1)^{-(n+1)}
$$
Hence,
$$
f^{(n)}(x) = \frac{(-2)^nn!}{2} \left(\frac{1}{(2x-1)^{(n+1)}} - \frac{1}{(2x+1)^{(n+1)}}\right)
$$
Now take $n=100$.
|
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"timestamp": "2023-03-29T00:00:00",
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|
How to prove that $[2^n \sqrt{2}],[2^{n+1} \sqrt{2}],\ldots, [2^{2n} \sqrt{2}]$ contains at least one even numbers for every integer $n\ge 1$? I tried as follows:
If not, denote $x=\{2^{n-1}\sqrt{2}\}$, then
$$1-\frac{1}{2^{n+1}}<x<1.$$
Denote $y=[2^{n-1}\sqrt{2}+1]$ and assume $|\sqrt{2}-p/q|<1/q^2$, then
$$\frac{2^{n-1}p}{q}-\frac{2^{n-1}}{q^2}<y<\frac{2^{n-1}p}{q}+\frac{2^{n-1}}{q^2}+\frac{1}{2^{n+1}}.$$
But it doesn't work.
|
By a basic trick of Diophantine approximation we have, for all positive integers $p,q$, the inequality
$$
\left|\frac pq-\sqrt2\right|\cdot\left|\frac pq+\sqrt2\right|=\left|\frac{p^2}{q^2}-2\right|\ge\frac1{q^2}.
$$
In particular, if $3/2>p/q>\sqrt2$, we get the estimate
$$
\left|\frac pq-\sqrt2\right|\ge\frac1{3q^2}.
$$
Therefore, with $M=[2^{n-1}\sqrt2]$ we get, using $x=\{2^{n-1}\sqrt2\}$,
$$
1-x=2^{n-1}(\frac{M+1}{2^{n-1}}-\sqrt2)\ge\frac1{3\cdot2^{n-1}}>\frac1{2^{n+1}}.
$$
Therefore $x<1-2^{-(n+1)}$, so your contrapositive case cannot happen.
|
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"timestamp": "2023-03-29T00:00:00",
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|
Subsets and Splits
Fractions in Questions and Answers
The query retrieves a sample of questions and answers containing the LaTeX fraction symbol, which provides basic filtering of mathematical content but limited analytical insight.