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Number of real roots of $3^x+4^x=2^x+5^x$ with proof This equation
$$3^x+4^x=2^x+5^x$$
has two obvious real roots. The question is if it has more real roots than two. A proof is required in any case.
|
We consider the equation
$$3^x+4^x=2^x+5^x~~~~(1)$$
Use Lagranges Mean Value Theorem (LMVT) for the function $f(t)=t^x$ for two intervals $(2,3)$ and $(4,5)$. So
$$\frac{3^x-2^x}{3-2}=xt_1^{x-1}, ~~~t_1 \in (2,3)~~~~(2)$$
and
$$\frac{5^x-4^x}{5-4}=xt_2^{x-1}, ~~~t_2 \in (4,5)~~~~(3).$$
By equating (2) and (3), we get (1) and
$$xt_1^{x-1} = xt_2^{x-1}, ~~~t_1 \ne t_2\Rightarrow x=0~ \mbox{or}~ x=1.$$
Hence Eq. (1) can have only two real roots.
|
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|
Eigenvalues with no eigenvectors Can a matrix (example 2x2) with 2 distinct eigenvalues have no eigenvetors?
(since, the A-3I as well as A-1I both are coming out as invertible with -3 and -3 with only 0 vector in the nullspace)?
|
By definition, a matrix has eigenvalue $\lambda$ and eigenvector $\mathbf{x}$ when $A \mathbf{x}=\lambda \mathbf{x}$ holds for some $\lambda$ and a non-zero vector $\mathbf{x}$.
To address the matrix example in the comments ...
$$A=\begin{pmatrix} 2 & 1\\ 1 & 2 \end{pmatrix}$$
The eigenvalues are $\lambda_1 =1$ and $\lambda_2=3$.
The eigenvectors are given by
$$\begin{pmatrix} 2 & 1\\ 1 & 2 \end{pmatrix}
\begin{pmatrix} x \\ y \end{pmatrix}= \begin{pmatrix} x \\ y \end{pmatrix}$$
$$\mathbf{x}_1 = \begin{pmatrix} \phantom{-}1 \\ -1 \end{pmatrix}$$
and
$$\begin{pmatrix} 2 & 1\\ 1 & 2 \end{pmatrix}
\begin{pmatrix} x \\ y \end{pmatrix}= \begin{pmatrix} 3x \\ 3y \end{pmatrix}$$
$$\mathbf{x}_2 = \begin{pmatrix} 1 \\ 1 \end{pmatrix}$$
|
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|
Asymptotic expansion of maximum of function. Denoted $M(n)$ by maximum of
$$e^{-t}-\left(1-\frac{t}{n}\right)^n \ \ \ t\in [0,n]$$
how to calculate the asymptotic expansion of $M(n)$ as $n\rightarrow \infty$?
|
If we consider the function
$$f(t)=e^{-t}-\left(1-\frac{t}{n}\right)^n $$ its first derivative is given by
$$f'(t)=-e^{-t}+\left(1-\frac{t}{n}\right)^{n-1} $$ It cancels at a value
$$t_*=n+(n-1)\, W\left(-\frac{ n}{n-1}e^{-\frac{n}{n-1}}\right)$$ where $W(.)$ is Lambert function.
So $M(n)=f(t_*)$.
Using successive Taylor expansions, we have
$$e^{-\frac{n}{n-1}}=\frac{1}{e}-\frac{1}{e n}-\frac{1}{2 e n^2}-\frac{1}{6 e
n^3}+O\left(\frac{1}{n^4}\right)$$
$$-\frac{ n}{n-1}e^{-\frac{n}{n-1}}=-\frac{1}{e}+\frac{1}{2 e n^2}+\frac{2}{3 e
n^3}+O\left(\frac{1}{n^4}\right)$$
$$ W\left(-\frac{ n}{n-1}e^{-\frac{n}{n-1}}\right)=-1+\frac{1}{n}+\frac{1}{3 n^2}-\frac{37}{72n^3}+O\left(\frac{1}{n^4}\right)$$
$$t_*=2-\frac{2}{3 n}-\frac{61}{72 n^2}+O\left(\frac{1}{n^3}\right)$$
$$n \log\left(1-\frac{t_*}{n}\right)=-2-\frac{4}{3 n}-\frac{35}{72 n^2}+O\left(\frac{1}{n^3}\right)$$
$$\left(1-\frac{t_*}{n}\right)^n=\frac{1}{e^2}-\frac{4}{3 e^2 n}+\frac{29}{72 e^2
n^2}+O\left(\frac{1}{n^3}\right)$$ from which
$$M(n)=f(t_*)=\frac{2}{e^2 n}+\frac{2}{3 e^2 n^2}+O\left(\frac{1}{n^3}\right)=\frac{2}{e^2 n}\left(1+\frac{1}{3 n}+O\left(\frac{1}{n^2}\right)\right)$$
So, your result is
$$\color{blue}{M(n)=\frac{2}{e^2 n}+O\left(\frac{1}{n^2}\right)}$$
It has been checked numerically.
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|
Probability that $2$ dice selected from $3$ rolled dice will have sum $7$ If I roll $3$ dice ($6$-sided) and then proceed to choose $2$ of them (discarding the 3rd), what is the probability of the total of those $2$ dice totalling $7$?
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Let's say the dice are blue, green, and red. Each outcome can be represented by the ordered triple $(b, g, r)$. Since there are six choices for each entry, there are $6^3 = 216$ possible outcomes. Assuming the dice are fair, they are equally likely to occur.
Method 1: The favorable outcomes are those in which at least one pair of dice selected from the three rolled dice have sum $7$.
There are $\binom{3}{2} = 3$ ways to select a pair of dice. Say we select the blue die and the green die. There are six ways for that pair of dice to have a sum of $7$: $(1, 6), (2, 5), (3, 4), (4, 3), (5, 2), (6, 1)$, where each ordered pair is of the form $(b, g)$. For each such pair, there are six possible outcomes for the red die (or, in general, the other die), giving $3 \cdot 6 \cdot 6$ favorable outcomes.
However, we have counted each outcome in which there are two pairs of dice that have sum $7$ twice, once for each way we could have designated one of the pairs as the pair that has sum $7$. For this to occur, two of the dice must show the same number and the third die must show the difference between $7$ and the number that appears on the other two dice. There are $\binom{3}{2} = 3$ ways for exactly two dice to display the same number and $6$ numbers these two dice could display. For any such outcome, there is only one possible outcome for the third die. Hence, there are $3 \cdot 6$ such outcomes.
Thus, the number of favorable cases is $3 \cdot 6 \cdot 6 - 3 \cdot 6 = 108 - 18 = 90$.
Therefore, the probability that when three dice are rolled that a pair of dice may be selected with sum $7$ is
$$\frac{3 \cdot 6 \cdot 6 - 3 \cdot 6}{6^3} = \frac{90}{216} = \frac{5}{12}$$
as InterstellarProbe found.
Method 2: The favorable outcomes are those in which at least one pair of dice selected from the three rolled dice have sum $7$.
There are two possibilities. Either each die shows a different outcomes or two of the three dice show the same outcome. Since $7$ is not a multiple of $3$, it is not possible for all three dice to show the same outcome.
Each die shows a different outcome: There are $\binom{3}{2} = 3$ ways for two of the dice to have sum $7$ and $6$ ways for those two dice to have sum $7$. There are four possible outcomes for the other die. Hence, there are $3 \cdot 6 \cdot 4$ such outcomes.
Two of the dice show the same outcome: There are $\binom{3}{2} = 3$ ways for exactly two of the three dice to show the same outcome and six possible outcomes those two dice could show. The other die must show the difference between $7$ and the outcome on those two dice. Hence, there are $3 \cdot 6$ such outcomes.
That yields a total of $3 \cdot 6 \cdot 4 + 3 \cdot 6 = 90$ favorable outcomes.
Hence, the probability that a pair of the three dice has sum $7$ is
$$\frac{3 \cdot 6 \cdot 4 + 3 \cdot 6}{6^3} = \frac{90}{216} = \frac{5}{12}$$
in agreement with the answer we obtained above.
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|
Solve $3x(2+\sqrt{9x^2 + 3}) + (4x-2)(\sqrt{x^2 - x +1}+1) = 0$ for $x\in \mathbb{R}$
Solve the equation in $ \mathbb{R}$: $$3x(2+\sqrt{9x^2 + 3}) + (4x-2)(\sqrt{x^2 - x +1}+1) = 0$$
I've been tried to solve this question for 3 hours, but can't find out any answers.
Just like I running in the maze, if I represent $\sqrt{9x^2 + 3} = A$, $ \sqrt{x^{2}-x+1}= B$
Finally we've got $2A^{2} -15 = 18B^2 + 9\sqrt{4B^2 +1}$ , which is not help me to find relation between $A$ and $B$ anymore.
I just wonder if we have a nice solution approaches to the problem.
I appreciate any suggestion and help. Thank you.
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It's obvious that for $x\leq0$ our equation has no real roots and since
$$\left(4x-2)\sqrt{x^2-x+1}\right)'=\frac{8x^2-8x+5}{\sqrt{x^2-x+1}}>0,$$ we see that $$10x-2+3x\sqrt{9x^2+3}+(4x-2)\sqrt{x^2-x+1}$$ increases for $x>0$ and our equation has one real root maximum.
But $\frac{1}{5}$ is a root and we are done!
|
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|
distribution of digits in prime numbers I am curious about this: suppose we consider all numbers in base $b$ such that the number of digits $n$ in this range is the same ( eg, in base $10$ it could be $10-to-99$ for $n=2$, or $100-to-999$ for $n=3$, etc; leading digit is non-zero), for the prime numbers in this range, if I were to choose a prime number at random can I expect the distribution of the digits of my prime to be uniform random? That is, $\frac{n}{b}$.
Thank you.
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In an odd base, an odd number always has an odd number of odd digits.
Proof
$$\begin{eqnarray}(2n+1)+(2p+1)=2q
\\
(2r+1)+(2s)=(2t+1)
\\
(2u)+(2v)=(2w)\end{eqnarray}$$
By $$(x+y)+z=x+(y+z)\tag{4}$$ and $$a+c+d=a+d+c=c+a+d=c+d+a=d+a+c=d+c+a\tag{5}$$ we have $$\underbrace{(2h+1)+\cdots +(2h+1)}_{\text{2i+1 times}}=2e+1\implies (2i+1)(2h+1)=(2e+1)$$ $$ 123456789_b=1\cdot b^8+2\cdot b^7+3\cdot b^6+4\cdot b^5+5\cdot b^4 +6\cdot b^3+7\cdot b^2+8\cdot b^1 +9\cdot b^0$$ that in an odd (2h+1) base b, that all odd digits create odd summands. it follows from (1),(2)(3), that the even (2u) digits create even (2w) summands regardless of base.
even (2v) bases luck out in that $b^0=1\quad b\neq 0$. Otherwise, they couldn't represent odd (2r+1) numbers at all.
There are other things like all primes greater than 3 being 1 or 5 on division by 6, that can play with things. in a $6k+1$ base, then the last digit being $6j+4$, will force the rest of the digits to represent a number of forms $6l+1$ or $6m+3$ etc.
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|
The integral $\int_{0}^{\infty} \frac{\cot^{-1}\sqrt{1+x^2}}{\sqrt{1+x^2}}dx =\frac{\pi}{2}\ln (1+\sqrt{2})$ At Mathematica the numerical value of the integral
$$\int_{0}^{\infty} \frac{\cot^{-1}\sqrt{1+x^2}}{\sqrt{1+x^2}} dx$$
equals 1.3844.., which is nothing but $\frac{\pi}{2}\ln (1+\sqrt{2})=z$. Also, one of its transformed forms is evaluated to be $z$ by Mathematica. The question is: How to do it by hand?
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Let
$$I=\int_{0}^{\infty} \frac{\cot^{-1}\sqrt{1+x^2}}{\sqrt{1+x^2}}dx =\int_{0}^{\infty} \frac{\tan^{-1}(1/\sqrt{(1+x^2})}{\sqrt{1+x^2}} dx$$
Let us use the integral representation of $$\frac{\tan^{-1}(1/\sqrt{1+x^2})}{\sqrt{1+x^2}}=\int_{0}^{1} \frac{dt}{1+x^2+t^2}$$
$$I=\int_{0}^{\infty} \int_{0}^{1} \frac {dt dx}{1+x^2+t^2}=\int_{0}^{1}\frac{dt}{\sqrt{1+t^2}} \tan^{-1} \left.\frac{x}{\sqrt{1+t^2}}\right|_{0}^{\infty} =\frac{\pi}{2} \ln (1+\sqrt{2}).$$
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Solve $x^{3} = 6+ 3xy - 3 ( \sqrt{2}+2 )^{{1}/{3}} , y^{3} = 9 + 3xy(\sqrt{2}+2)^{{1}/{3}} - 3(\sqrt{2}+2)^{{2}/{3}}$ Solve the system of equations for $x,y \in \mathbb{R}$
$x^{3} = 6+ 3xy - 3\left ( \sqrt{2}+2 \right )^{\frac{1}{3}} $
$ y^{3} = 9 + 3xy(\sqrt{2}+2)^{\frac{1}{3}} - 3(\sqrt{2}+2)^{\frac{2}{3}}$
I just rearranged between those equations and get $ \frac{y^{3}-9}{x^{3} -6} = (\sqrt{2}+2)^{\frac{1}{3}}$ then I don't know how to deal with it.
Please give me a hint or relevant theorem to solve the equation.
Thank you, and I appreciate any help.
Furthermore I get an idea
how about we subtract two equation and get
$y^{3}-x^{3} = 3 + 3xy((\sqrt{2}+2)^{\frac{1}{3}} -1) - (3(\sqrt{2}+2)^{\frac{2}{3}} - 3(\sqrt{2}+2)^{\frac{1}{3}})$
$(y-x)(x^2+xy+y^2)= 3[1-((\sqrt{2}+2)^{\frac{1}{3}}-1)(\sqrt{2}+2)^{\frac{1}{3}}-xy)]$
$y-x = 3$ and $x^2 +xy+y^2 =[1-((\sqrt{2}+2)^{\frac{1}{3}}-1)(\sqrt{2}+2)^{\frac{1}{3}}-xy)] $
or,
$y-x = [1-((\sqrt{2}+2)^{\frac{1}{3}}-1)(\sqrt{2}+2)^{\frac{1}{3}}-xy)] $ and $x^2 +xy+y^2 = 3$
Am I on the right track?
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import numpy as np
import matplotlib.pyplot as plt
a = pow(2+np.sqrt(2),1/3)
y, x = np.ogrid[-10:10:100j, -10:10:100j]
plt.contour(x.ravel(), y.ravel(), x**3-3*x*y+3*a, [6], colors='r')
plt.contour(x.ravel(), y.ravel(), y**3-3*a*x*y+3*a*a, [9])
plt.grid()
plt.show()
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|
Integrate $\int_a^\infty\frac{\sqrt{x^2-a^2}}{\sinh x}\,dx$ I wonder whether the following integral ($a\in\mathbb{R}$, $a>0$)
$$S(a)=\int_a^\infty\frac{\sqrt{x^2-a^2}}{\sinh x}\,dx$$
admits a closed form (perhaps using some known special functions).
The integral representation of $K_1(z)$ gives just
$$S(a)=2a\sum\limits_{n=0}^{\infty}\frac{K_1\big((2n+1)a\big)}{2n+1}$$
which doesn't lead me to anything meaningful.
An alternative form comes from contour integration:
$$\frac{S(a\pi)}{\pi^2}=\frac{1-2a}{4}+\sum_{n=1}^{\infty}(-1)^{n-1}(\sqrt{n^2+a^2}-n).$$
For a context, this is what I arrive at in this answer.
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Let's denote:
$$F(a)=\frac{1}{2}+\frac{d}{da} \frac{S(a\pi)}{\pi^2} \tag{1}$$
Remember for later that $S(0)= \frac{\pi^2}{4}$.
From the second expression in the OP we obtain:
$$F(a)=a \sum_{n=1}^{\infty} \frac{(-1)^{n-1}}{\sqrt{n^2+a^2}} \tag{2}$$
In this answer I found that:
$$\frac{F(a)}{a}=\log 2 +\sum_{k=1}^\infty \frac{(-1)^k (2k)!}{k!^2}\left(1-\frac{1}{2^{2k}} \right) \zeta(2k+1) \left( \frac{a}{2} \right)^{2k} \tag{3}$$
and
$$\frac{F(a)}{a}=\log 2 +\int_0^\infty \frac{J_0 (a x)-J_0 (a x/2)}{e^x-1}dx \tag{4}$$
Integrating (1) from $0$ to $a$, we have:
$$\int_0^a F(a') da'=\frac{a}{2}+\frac{S(a\pi)}{\pi^2}-\frac{1}{4}$$
On the other hand (3) gives us:
$$\int_0^a F(a') da'=\frac{\log 2}{2} a^2+2\sum_{k=1}^\infty \frac{(-1)^k (2k)!}{k!^2 (k+1)}\left(1-\frac{1}{2^{2k}} \right) \zeta(2k+1) \left( \frac{a}{2} \right)^{2k+2}$$
This gives us explicit Taylor series for $S(a)$:
$$S(a)= \frac{\pi^2}{4}-\frac{\pi a}{2}+\frac{\log 2}{2} a^2+ \frac{a^2}{2} \sum_{k=1}^\infty \frac{(-1)^k (2k)!}{k!^2 (k+1)}\left(1-\frac{1}{2^{2k}} \right) \zeta(2k+1) \left( \frac{a}{2 \pi} \right)^{2k} \tag{5}$$
While this is not a closed form, it still may prove useful.
From (4) we obtain:
$$\int_0^a F(a') da'=\frac{\log 2}{2} a^2+a \int_0^\infty \frac{J_1 (a x)-2 J_1 (a x/2)}{x(e^x-1)}dx$$
Which gives us an additional integral form for the function:
$$S(a)= \frac{\pi^2}{4}-\frac{\pi a}{2}+\frac{\log 2}{2} a^2+ \pi a \int_0^\infty \frac{J_1 (a x)-2 J_1 (a x/2)}{x(e^{\pi x}-1)}dx \tag{6}$$
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How many numbers with 6 digits can be formed with the digits 1,2,3,4,5 such that the digit 2 appears every time at least three times? How many numbers with 6 digits can be formed with the digits $1,2,3,4,5$ such that the digit $2$ appears every time at least three times.
My try:
Total numbers: $5^6$
Numbers in which 2 doesn't appear: $4^6$
Numbers in which 2 appear once : $6\cdot4^5$
Numbers in which 2 appear twice : $13\cdot4^4$
So my result is: $5^6-4^6- 6 \cdot4^5-13\cdot4^4=2057 $ but the right answer is $1545$ How solve it ?
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...or the other way around: $\binom{6}{3}4^3 + \binom{6}{4}4^2 + \binom{6}{5}6^1 +1$. Keep in mind $\binom{6}{4} = \binom{6}{2}$. It's a term longer, but gives you an opportunity to compute sm binomial coefficients.
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Solving 5th degree polynomial Solve the equation, $$6x^5 + 5x^4 − 51x^3 + 51x^2 − 5x − 6 = 0$$(hints: the pattern of the coefficients)
How I attempted this problem is to use the rational root theorem to obtain the factors. The possible roots are $$±(1, 1/2, 1/3, 1/6, 2, 2/3, 3, 3/2, 6)$$
By substituting the values inside the equation, only the following roots satisfy the equation:
$$x=1,\:x=\frac{3}{2},\:x=\frac{2}{3}$$
Henceforth, I came up with the following based on the above roots:
$$\left(x-1\right)\left(2x-3\right)\left(3x-2\right)$$
Clearly expanding them would give me a 3rd degree polynomial as follows:
$$6x^3 - 19x^2 + 19x - 6$$
Using polynomial division where I divided the original 5th degree equation with the above equation,
I obtained the following equation:
$$x^2+4x+1$$
Now, solving the above equation using quadratic formula, I am able to get the roots. Hence, I have obtained all the roots of the solution.
However, looking at the hint that asks me to observe the pattern of the coefficients, I think that the method used to arrive at my solution might have been long-winded (it was indeed tedious plugging in the rational .roots one-by-one into the equation to see which returns value of 0). Is there something that I'm missing in the way I approached the question?
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The "pattern" is that w hen you reverse the order of the coefficients you get the negative of the original polynomial:
$6x^5+5x^4-51x^3+51x^2-5x-6\to-6x^5-5x^4+51x^3-51x^2+5x+6$
When the polynomial has this pattern you can render it this way:
$6x^5+5x^4-51x^3+51x^2-5x-6=6\color{blue}{(x^5-1)}+5\color{blue}{(x^4-x)}-51\color{blue}{(x^3-x^2)}$
where the blue factors are all multiples of $x-1$ forcing $x=1$ to be a root. When you divide out this factor (using the above rearrangement to help things along) you get
$6x^5+5x^4-51x^3+51x^2-5x-6=(x-1)\color{blue}{(6x^4+11x^3-40x^2+11x+6)}$
where the blue quartic factor now reads exactly the same when the coefficients are reversed. This arrangement, in an even degree polynomial, forces a "symmetric" factorization into quadratics:
$6x^4+11x^3-40x^2+11x+6=6(x^2+ax+1)(x^2+bx+1)$
Multiplying the right side out and matching terms with either odd power of $x$ gives $a+b=11/6$, and matching the terms with $x^2$ gives $ab=-26/3$. From the known sum and product we infer that $a$ and $b$ solve
$w^2-(11/6)w-(26/3)=0$
$6w^2-11w-52=0$
Thereby the roots for $a$ and $b$ are $4$ and $-13/6$, from which we now have
$6x^5+5x^4-51x^3+51x^2-5x-6=6(x-1)(x^2+4x+1)(x^2-(13/6)x+1)=(x-1)(x^2+4x+1)(6x^2-13x+6)$
and the remaining quadratic factors are solved by conventional methods.
It looks more complicated, but note that no trial and error is needed.
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Find generating function for the pattern Denote $p(n)$ is the numbers of any size rectangle count on these pattern
Problem :
Find generating function that $p(n)$ is coefficient of function.
I try to read out for some $p(n)$ and get
$p(1)=1$
$p(2)=2\binom{3}{2}-1 = 5$
$p(3)=(2\binom{4}{2}-1) + \binom{3}{2}^{2} - (2\binom{3}{2}-1) = 15$
$p(4)= (2\binom{5}{2}-1) + (2\binom{3}{2}\binom{4}{2}-\binom{3}{2}^{2}) - (2\binom{4}{2}-1) = 35$ and so on
We found that $p(n) = \frac{1}{3}(2n^{3}-3n^{2}+7n-3)$
The problem is I cannot find a generating function that has $p(n)$ as a coefficient.
Please give me some hint or any advice. Thank you and I appreciate for any helps.
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We derive a recurrence relation for $p(n), n\geq 1$ and calculate from it the generating function
\begin{align*}
P(x)&=\sum_{j=1}^\infty p(j)x^j\\
&=p(1)x+p(2)x^2+p(3)x^3+p(4)x^4\cdots\\
&=x+5x^2+15x^3+35x^4+\cdots
\end{align*}
In order to go from $p(n-1)$ to $p(n)$ it is convenient to add squares at the diagonal. See the graphic below.
In each step we add a new square at the diagonal. The rectangles we count in one step do not interfere with rectangles from former steps, since the new square is always part of these rectangles. In fact this way we partition the rectangles we count when going from $p(n-1)$ to $p(n)$, so that we can simply sum them up. In general we have for $n\geq 1$:
\begin{align*}
\color{blue}{p(n)}&\color{blue}{=p(n-1)+\sum_{j=1}^{n}j(n+1-j)\qquad\qquad n\geq 2}\tag{1}\\
\color{blue}{p(1)}&\color{blue}{=1}
\end{align*}
We simplify (1) by using the summation formulas
$$\sum_{j=1}^nj=\frac{1}{2}n(n+1), \qquad\qquad\sum_{j=1}^nj^2=\frac{1}{6}n(n+1)(2n+1)$$ and obtain
\begin{align*}
\color{blue}{p(n)}&\color{blue}{=p(n-1)+\frac{1}{6}n(n+1)(n+2)\qquad\qquad n\geq 2}\tag{2}\\
\color{blue}{p(1)}&\color{blue}{=1}
\end{align*}
We obtain from (2) the generating function $P(x)$ as follows
\begin{align*}
P(x)&=\sum_{n=1}^\infty p(n)x^n=x+\sum_{n=2}^\infty p(n)x^{n}\\
&=x+\sum_{n=2}^\infty \left(p(n-1)+\frac{1}{6}n(n+1)(n+2)\right)x^n\\
&=x+\sum_{n=1}^\infty p(n)x^{n+1}+\frac{1}{6}x\sum_{n=2}^\infty n(n+1)(n+2)x^{n-1}\\
&=x+\sum_{n=1}^\infty p(n)x^{n+1}+\frac{1}{6}xD_x\sum_{n=2}^\infty (n+1)(n+2)x^{n}\tag{3}\\
&=x+\sum_{n=1}^\infty p(n)x^{n+1}+\frac{1}{6}xD_x^2\sum_{n=2}^\infty(n+2)x^{n+1}\\
&=x+xP(x)+\frac{1}{6}xD_x^3\sum_{n=2}^\infty x^{n+2}\tag{4}\\
&=x+xP(x)+\frac{1}{6}xD_x^3\frac{x^4}{1-x}\\
&=x+xP(x)+\frac{x}{(1-x)^4}-x\\
\color{blue}{P(x)}&\color{blue}{=\frac{x}{(1-x)^5}}\tag{5}
\end{align*}
In (3) we use the differential operator $D_x=\frac{d}{dx}$ and continue successively till we get a geometric series in (4). It is convenient to use the coefficient of operator $[x^n]$ to denote the coefficient of $x^n$.
We finally obtain from (5) for $n\geq 1$
\begin{align*}
[x^n]P(x)&=[x^n]\frac{x}{(1-x)^5}\\
&=[x^{n-1}]\sum_{j=0}^\infty\binom{-5}{j}(-x)^j\\
&=[x^{n-1}]\sum_{j=0}^\infty\binom{j+4}{4}x^5\\
&=\binom{n+3}{4}\\
&\,\,\color{blue}{=\frac{1}{24}n(n+1)(n+2)(n+3)}
\end{align*}
|
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"timestamp": "2023-03-29T00:00:00",
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|
Splitting a population, probability of 2 people landing in same subpopulation.
$10$ people has been split into $3$ groups $A,B,C$ of $5,3,2$ people respectively. What is the probability that $2$ predetermined people $x,y$ land in the same group?
My attempt:
There is $\binom{10}{5, 3, 2} = \frac{10!}{5!3!2!}$ ways to split the population into mentioned above subpopulations.
Let's count all the partitions in which $x,y\in A$.
There's $\binom{10-2}{3}$ ways to choose the rest of people in $A$, $\binom{5}{3}$ ways to choose people in $B$ from the remaining 5, and $\binom{2}{2}$ ways to choose people in $C$. Repeating the above for the cases $x,y\in B$ and $x,y \in C$ we get:
$$P(A)=\frac{ \binom{8}{3}\binom{5}{3} +\binom{8}{1}\binom{7}{5}+\binom{8}{0}\binom{8}{5}}{\binom{10}{5,3,2}},$$
which is wrong.
The given answer is
$$P(A)=\frac{ \binom{8}{3}\binom{2}{2} +\binom{8}{1}\binom{2}{2}+\binom{2}{2}}{\binom{10}{5,3,2}},$$
and I'm having a hard time engineering it back. Could anyone point where I made mistake?
|
I also think that you are correct. I get the same answer by just partitioning the remaining eight people ...
$$N(A)= \binom8{3,3,2}+\binom8{5,1,2}+\binom8{5,3,0}
\\= \binom{8}{3}\binom{5}{3} +\binom{8}{1}\binom{7}{5}+\binom{8}{0}\binom{8}{5}
\\=784$$
|
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|
Prove $x+y+z \ge xy+yz+zx$ Given $x,y,z \ge 0$ and $x+y+z=4-xyz$ Then Prove that
$$x+y+z \ge xy+yz+zx$$
My try:
Letting $x=1-a$, $y=1-b$ and $z=1-c$ we get
$$(1-a)+(1-b)+(1-c)+(1-a)(1-b)(1-c)=4$$
$$-(a+b+c)-(a+b+c)+ab+bc+ca-abc=0$$
$$ab+bc+ca-abc=2(a+b+c)$$
Where $a, b,c \le 1$
is there a clue here?
|
Also, we can use $uvw$ here.
Indeed, let $x+y+z=3u$, $xy+xz+yz=3v^2$ and $xyz=w^3$.
Thus, the condition does not depend on $v^2$ and it's enough to prove our inequality for a maximal value of $v^2$, which happens for equality case of two variables.
Let $y=x$.
Thus, the condition gives $z=\frac{4-2x}{1+x^2},$ where $0\leq x\leq2$ and we need to prove that
$$2x+\frac{4-2x}{1+x^2}\geq x^2+2x\cdot\frac{4-2x}{1+x^2}$$ or
$$(2+x)(2-x)(x-1)^2\geq0$$ and we are done!
|
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"timestamp": "2023-03-29T00:00:00",
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|
Solving the following limit without L'Hospital's rule: $\lim_{x\to 0} \frac{\sin(x^2+2)-\sin(x+2)}{x} $
I have been trying to solve the following limit $$\lim_{x\to 0} \frac{\sin(x^2+2)-\sin(x+2)}{x}.$$
I came across the right answer as shown by the steps below, but I would to check if the steps are correct or if someone has a more straightforward solution.
So applying the sum formula for sine and doing simple algebra we have:
$$\lim_{x\to 0} \frac{\cos{2} \,(\sin{x^2}-\sin{x})}{x} - \frac{\sin{2} (\cos{x^2}-\cos{x})}{x} .$$
The first limit is easy to evaluate and is equal to $-\cos{2}$. However, the second limit is harder, as it follows:
$$\lim_{x\to 0}\frac{\sin{2} (\cos{x^2}-\cos{x})}{x} .$$
I came across a solution by using the following sum-to-product identity:
$$\cos{A}-\cos{B}=-2\sin{\Big(\frac{A+B}{2}\Big)} \sin{\Big(\frac{A-B}{2}\Big)}$$
Setting $A=x^2$ and $B=x$, we have that
$$\cos{x^2}-\cos{x}=-2\sin{\Big(\frac{x^2+x}{2}\Big)} \sin{\Big(\frac{x^2-x}{2}\Big)}$$
This is my only point of concern whether I applied the identity correctly. The rest of it flows more easily:
$$\lim_{x\to 0}\frac{\sin{2} (\cos{x^2}-\cos{x})}{x} = \lim_{x\to 0} \frac{-2\sin{2}\,\sin{\Big(\frac{x^2+x}{2}\Big)} \sin{\Big(\frac{x^2-x}{2}\Big)}}{x}$$
$$= -2 \sin{2} \lim_{x\to 0} \frac{\sin{\Big(\frac{x^2+x}{2}\Big)}}{x} \lim_{x\to 0} \sin{\Big(\frac{x^2-x}{2}\Big)} $$
The first limit can be solved as it follows:
$$\lim_{x\to 0} \frac{\sin{\Big(\frac{x^2+x}{2}\Big)}}{x} = \lim_{x\to 0} \frac{\sin{\Big(\frac{x^2+x}{2}\Big)} \Big(\frac{x^2+x}{2}\Big)}{x \Big(\frac{x^2+x}{2}\Big)} = 1 \cdot \lim_{x \to 0} \frac{x^2 + x}{2x} = \frac{1}{2} $$
The second limit is equal to zero
$$\lim_{x\to 0} \sin{\Big(\frac{x^2-x}{2}}\Big)=0$$
|
You can just sum-to-product it from the beginning and use $\lim_{x\rightarrow 0}\sin(x)/x = 1$:
\begin{align}
\lim_{x\rightarrow 0}\frac{\sin(x^2+2)-\sin(x+2)}{x} &= \lim_{x\rightarrow 0}\frac{2\cos[2+x(x+1)/2]\sin[x(x-1)/2]}{x}
\\ & = \cos(2)\lim_{x\rightarrow 0}\frac{\sin[x(x-1)/2]}{x/2}
\\ &= -\cos(2)\lim_{x\rightarrow 0}\frac{\sin[x(x-1)/2]}{x(x-1)/2}
\\ &= -\cos(2)
\end{align}
|
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"timestamp": "2023-03-29T00:00:00",
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|
Find all extrema of a complicated trigonometric function Problem
Find all local extrema for
$$f(x) = \frac{\sin{3x}}{1+\frac{1}{2}\cos{3x}}$$
Attempted solution
My basic approach is to take the derivative, set the derivative equal to zero and solve for x.
Taking the derivative with the quotient rule and a few cases of the chain rule for the trigonometric functions with a final application of the Pythagorean identity:
$$f'(x) = \frac{(1+\frac{1}{2}\cos{3x})(3\cos{3x})+1.5\sin{3x}\sin {3x}}{(1+\frac{1}{2}\cos{3x})^2} = \frac{3\cos 3x+1.5\cos^2 3x + 1.5\sin^2 3x}{(1+\frac{1}{2}\cos 3x)^2} = \frac{3 \cos 3x + 1}{(1+\frac{1}{2}\cos 3x)^2}$$
Putting it equal to zero and solving for x:
$$3\cos 3x + 1 = 0 \Rightarrow x = \frac{\arccos{\Big(-\frac{1}{3}\Big)}}{3} = \frac{\pi}{6} + \frac{2\pi n}{3}$$
...however the expected answer is $\pm\frac{2\pi}{9} + \frac{2\pi n}{3}$
So I must have gone wrong somewhere.
|
It should be $$f'(x) = \frac{(1+\frac{1}{2}\cos{3x})(3\cos{3x})-\frac{3}{2}\sin{3x}\sin {3x}}{(1+\frac{1}{2}\cos{3x})^2}.$$
I like the following way.
Let $x=\frac{2\pi}{9}.$
Thus, we get a value $\frac{2}{\sqrt3}.$
We'll prove that it's a maximal value.
Indeed, we need to prove that
$$\frac{\sin3x}{2+\cos3x}\leq\frac{1}{\sqrt3}$$ or
$$\sqrt3\sin3x-\cos3x\leq2,$$ which is true by C-S:
$$\sqrt3\sin3x-\cos3x\leq\sqrt{((\sqrt3)^2+(-1)^2)(\sin^23x+\cos^23x)}=2.$$
By the same way we can get a minimal value.
I got $-\frac{2}{\sqrt3},$ which occurs for $x=-\frac{2\pi}{9}.$
|
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|
Calculate the maximum value of $x^3y + y^3z + z^3x$ where $x + y + z = 4$ and $x, y, z \ge 0$.
Given non-negatives $x, y, z$ such that $x + y + z = 4$. Calculate the maximum value of $$\large x^3y + y^3z + z^3x$$
As an assumption, the maximum value is $27$, occured when $(x, y, z) = (0, 1, 3)$.
I have a guess about a working-in-process prove. Let $y$ be the median of $x, y, z$.
$$\iff (zx - yz)(y^2 - z^2) \ge 0 \iff y^2zx - y^3z - z^3x + yz^3 \ge 0$$
$$\iff x^3y + y^3z + z^3x \le x^3y + y^2zx + yz^3 = y(x^3 + xyz + z^3)$$
And further down the line is what I haven't accomplished yet.
|
Now, use AM-GM:
$$y(x^3+xyz+z^3)\leq y(x+z)^3=27y\left(\frac{x+z}{3}\right)^3\leq27\left(\frac{y+3\cdot\frac{x+z}{3}}{4}\right)^4=27.$$
|
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"timestamp": "2023-03-29T00:00:00",
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|
Inequality with Fibonacci numbers $ \sum \limits_{k=1}^{2n+1} (-1)^{k+1} F_k \: \text{arccot} F_k < \frac{\pi+1-\sqrt{5}}{2}$
Prove that $$ \sum \limits_{k=1}^{2n+1} (-1)^{k+1} F_k \: \text{arccot} F_k < \frac{\pi+1-\sqrt{5}}{2}$$ holds for all $n \in \mathbb{N}$.
(The Fibonacci sequence, defined by the recurrence $F_1 = F_2 = 1$ and $\forall n \in \mathbb{N},$ $F_{n+2} = F_{n+1} + F_n$)
My work. I proved that the sequence $a_n=\sum \limits_{k=1}^{2n+1} (-1)^{k+1} F_k \: \text{arccot} F_k \;$ is increasing. Therefore inequality cannot be proved by induction.
|
Not a full answer.
Let's group even and odd terms in the sum:
$$S_n=\sum_{k=1}^{2n+1} (-1)^{k+1} F_k \operatorname{arccot} F_k= \\ = \sum_{l=0}^{n-1} (F_{2l+1} \operatorname{arccot} F_{2l+1}-F_{2l+2} \operatorname{arccot} F_{2l+2}) +F_{2n+1} \operatorname{arccot} F_{2n+1}$$
It's easy to prove that $F_{2n+1} \operatorname{arccot} F_{2n+1} < 1$ and:
$$\lim_{n \to \infty} F_{2n+1} \operatorname{arccot} F_{2n+1}=1$$
So we only need to find the limit for the rest of the sum.
Denote:
$$P_n=\sum_{l=0}^{n-1} (F_{2l+1} \operatorname{arccot} F_{2l+1}-F_{2l+2} \operatorname{arccot} F_{2l+2})$$
The first terms is $0$, it's easy to check. Also, this sum converges to a single limit, so we can denote:
$$P=\sum_{l=1}^\infty \left(F_{2l+1} \arctan \frac{1}{ F_{2l+1} } -F_{2l+2} \arctan \frac{1}{ F_{2l+2} } \right)$$
We need to prove that ($\varphi$ - Golden ratio):
$$P= \frac{\pi}{2}-\varphi$$
In this post it is shown how to prove the following identity:
$$\frac{\pi}{2} = 2 \sum_{l=1}^\infty \arctan \frac{1}{ F_{2l+1} }$$
So we are left to prove that:
$$Q=\sum_{l=1}^\infty \left(F_{2l+2} \arctan \frac{1}{ F_{2l+2}}- (F_{2l+1}-2) \arctan \frac{1}{ F_{2l+1} } \right)= \varphi$$
Numerically this works. The proof eludes me.
I've spent a few hours trying different methods, but haven't had much luck so far.
Some useful identities:
$$\arctan \frac{1}{a} = \int_0^1 \frac{a dt}{t^2+a^2}$$
$$\arctan \frac{1}{a} = \sum_{k=0}^\infty \frac{(-1)^k}{2k+1} \frac{1}{a^{2k+1}}$$
$$F_n= \frac{1}{\sqrt{5}} \left(\varphi^n -\frac{(-1)^n}{\varphi^n} \right)$$
Notice that we can write the odd terms as:
$$F_{2l+1}=\frac{2}{\sqrt{5}} \cosh ((2l+1) \alpha)$$
$$F_{2l+2}=\frac{2}{\sqrt{5}} \sinh ((2l+2) \alpha)$$
Where $$\alpha= \log \varphi$$
Let's not forget the special property of the golden ratio:
$$\varphi-1=\frac{1}{\varphi}$$
There's also a more general arctangent identity listed on this page:
$$\arctan \frac{1}{ F_{2l} }= \sum_{j=l}^\infty \arctan \frac{1}{ F_{2j+1} }$$
|
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|
Find all $x$ such that $\sin x = \frac{4}{5}$ and $\cos x = \frac{3}{5}$.
Let
$$
\left\{
\begin{array}{c}
\sin x = \frac{4}{5} \\
\cos x = \frac{3}{5}
\end{array}
\right.
$$
Find all of the possible values for $x$.
My try: By dividing the equations we obtain $\tan x = \frac{4}{3}$ and then $$x = \arctan\frac{4}{3} + k\pi$$
But WolframAlpha gives
$$x = 2k\pi + 2\arctan\frac{1}{2}$$
Using $\arctan(x)+\arctan(y) = \arctan\left(\frac{x+y}{1-xy}\right)$, we get $$2\arctan\frac{1}{2} = \arctan\frac{4}{3}$$
but the answers are different still.
Why does this happen? And what is the correct answer?
|
$\cos x=\dfrac35$
$\implies x=2n\pi\pm\arccos\dfrac35$
As for $x>0,\arccos x=\arcsin\sqrt{1-x^2}=\arctan\dfrac{\sqrt{1-x^2}}x$
$\implies x=2n\pi\pm\arcsin\dfrac45$ where $n$ is any integer
Now , $\sin x=\sin\left(2n\pi\pm\arcsin\dfrac45\right)=\pm\dfrac45$
But $\sin x=+\dfrac45$
$$\implies x=2n\pi+\arcsin\dfrac45=2n\pi+\arccos\dfrac35=2n\pi+\arctan\dfrac43$$
Further if $2y=\arctan\dfrac43,\dfrac43=\tan2y=\dfrac{2\tan y}{1-\tan^2y}$
Solve the quadratic equation in $\tan y$ to find the values to be $\dfrac12,-2$
But as $0<2y<\dfrac\pi2, y=\arctan\dfrac12$
which can be validated using Inverse trigonometric function identity doubt: $\tan^{-1}x+\tan^{-1}y =-\pi+\tan^{-1}\left(\frac{x+y}{1-xy}\right)$, when $x<0$, $y<0$, and $xy>1$
|
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|
Alternative methods for solving a system of one linear one non linear simultaneous equations Take the equations $$x+y=5$$ $$x^2 + y^2 =13$$
The most basic method to solve this system is to first express the linear equation in terms of one of the variables and then sub that into the non-linear equation.
But I am curious if there are other methods to solve such a system ?
|
$\DeclareMathOperator{\lcm}{lcm}$
Compute the Gröbner basis of your system. Let us start by writing this with zeroes on the right of the equals signs. \begin{align*}
0 &= x+y-5 \\
0 &= x^2 + y^2 - 13 \text{.}
\end{align*}
We pick a variable ordering. Let us choose $x < y$. (The given system is unchanged by the exchange of the variables $x$ and $y$, so we get the same computation, but with the variables swapped, if we choose the other ordering.) We compute the first $s$-polynomial. We need the LCM of the leading terms
$$ \lcm(x, x^2) = x^2 $$
and using this we get
\begin{align*}
0 &= \frac{x^2}{x}(x+y-5) - \frac{x^2}{x^2}(x^2 + y^2 - 13) \\
&= x^2 + xy - 5x -(x^2 + y^2 - 13) \\
&= xy - y^2 -5x + 13 \text{.}
\end{align*}
Now $\lcm(xy, x) = xy$ and
\begin{align*}
0 &= \frac{xy}{x}(x+y-5) - \frac{xy}{xy}(xy - y^2 -5x + 13) \\
&= xy + y^2 - 5y -(xy - y^2 -5x + 13) \\
&= 2y^2 +5x -5y -13
\end{align*}
and since we have a relation for $x$ and $y$ both of degree $1$, \begin{align*}
0 &= 2y^2 +5x - 5y - 13 -5(x+y-5) \\
&= 2y^2 +5x - 5y - 13 -5x -5y + 25 \\
&= 2y^2 -10y + 12 \\
&= 2(y^2 - 5y + 6) \text{,}
\end{align*}
and since twice a thing is zero means the thing is zero, we have
$$ y^2 - 5y + 6 = 0 \text{.} $$
Our collection of expressions which evaluate to zero is then (sorting by decreasing total degree, then according to the order we picked for the variables)
\begin{align*}
x^2 + y^2 - 13 &= 0 \\
xy - y^2 -5x + 13 &= 0 \\
y^2 - 5y + 6 &= 0 \\
x+y-5 &= 0 \text{.}
\end{align*}
Notice that in degree $2$ we slowly decreased the degree of the dependence on $x$ until we were left with a polynomial in $y$ alone. Solving that polynomial, $y = 2$ or $y = 3$. Then the collection becomes (by specializing the value of $y$ and appending a final equation for that value of $y$) either
\begin{align*}
x^2 - 9 &= 0 \\
-3x + 9 &= 0 \\
0 &= 0 \\
x-3 &= 0 \\
y -2 &= 0 \text{,}
\end{align*}
giving the solution $(x,y) = (3,2)$, or
\begin{align*}
x^2 - 4 &= 0 \\
-2x + 4 &= 0 \\
0 &= 0 \\
x-2 &= 0 \\
y - 3 &= 0 \text{,}
\end{align*}
giving the solution $(x,y) = (2,3)$.
|
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|
Given $(x-1)^3+3(x-1)^2-2(x-1)-4=a(x+1)^3+b(x+1)^2+c(x+1)+d$, find$(a,b,c,d)$
Given $(x-1)^3+3(x-1)^2-2(x-1)-4=a(x+1)^3+b(x+1)^2+c(x+1)+d$,
find$(a,b,c,d)$
my attempt:
$$(x+1)=(x-1)\frac{(x+1)}{(x-1)}$$
but this seems useless?
I want to use synthetic division but I don't know how
|
It's $$(x+1-2)^3+3(x+1-2)^2-2(x+1-2)-4=(x+1)^3-3(x+1)^2-2(x+1)+4.$$
Can you end it now?
|
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|
What is the coefficient of $x^3$ in expansion of $(x^2 - x + 2)^{10}$
What is the coefficient of $x^3$ in expansion of $(x^2 - x + 2)^{10}$.
I have tried:
$$\frac{10!}{(3!\times7!)} \times (-x + 2)^7 \times (x^2)^3 $$
But got an incorrect answer $-15360$.
|
By the multinomial formula,
$$(x^2-x+2)^{10}=\sum_{a+b+c=10}\begin{pmatrix}10!\\\!a!\,b!\;c!\!\end{pmatrix}x^{2a}(-1)^bx^b 2^c,$$
and obtaining $x^3$ means $2a+b=3$, whence $c-a=7$, so the possibilities aren't so many:
*
*either $a=0$, so $b=3$, $c=7$;
*or $a=1$, so $b=1$, $c=8$.
Therefore, the coefficient will be
$$ -2^7\biggl(\frac{10!}{3!\,7!}+2\frac{10!}{8!}\biggr)=-38\mkern 2mu400.$$
|
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|
How can one integrate $\int\frac{1}{(x+1)^4(x^2+1)} dx$?
How can one integrate $\displaystyle\int\frac{1}{(x+1)^4(x^2+1)}\ dx$?
Attempt:
I tried partial fraction decomposition (PFD) and got lost. The method of u-substitution didn't work for me either.
What else can I do? Can one calculate the integral without PFD?
|
Here is a secure and faster method when the fraction has a pole of comparatively high order:
*
*If the pole is not $0$, as is the case here, perform the substitution $u=x+1$ and express the other factors in function of $u$. We have to take care of $x^2+1$. The method of successive divisions yields $x^2+1=u^2-2u+2$, so we have
$$\frac 1{(x+1)^4(x^2+1)}=\frac1{u^4}\cdot\frac 1{2-2u+u^2}.$$
*Perform the division of $1$ by $2-2u+u^2$ along increasing powers of $u$, up to order $4$:
$$\begin{array}{r}
\phantom{\frac12}\\
\phantom{u}\\
2-2u+u^2\Big(
\end{array}\begin{array}[t]{&&rr@{}rrrrr}
\frac12&{}+\frac 12 u&{}+\frac 14u^2 \\
%\hline
1 \\
-1&{}+u&{}-\frac12u^2 \\\hline
&u&{}-\frac12u^2 \\
&-u& +u^2 &{}-\frac12u^3\\
\hline
&&&\frac12u^2&{}-\frac12u^3 \\
&&&-\frac12u^2&{}+\frac12u^3&-\frac14u^4 \\
\hline
&&&&&-\frac14u^4
\end{array} $$
*This yields the equality:
$$1=(2-2u+u^2)\bigl(\tfrac12+\tfrac 12 u+\tfrac 14u^2\bigr)-\tfrac14u^4,$$
whence the partial fractions decomposition:
$$\frac 1{u^4(2-2u+u^2)}=\frac1{2u^4}+\frac 1{2u^3} u+\frac 1{4u^2}-\frac1{4(2-2u+u^2)},$$
or with $x$ :
$$\frac 1{(x+1)^4(x^2+1)}=\frac1{2(x+1)^4}+\frac 1{2(x+1)^3} +\frac 1{4(x+1)^2}-\frac1{4(x^2+1)}.$$
|
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|
How to show the matrix has Rank $\le 5$
I want to show that the following matrix has Rank $\le 5$.
The matrix is
\begin{bmatrix}
2&1&1&1&0&1&1&1\\
1&2&1&1&1&0&1&1\\
1&1&2&1&1&1&0&1\\
1&1&1&2&1&1&1&0\\
0&1&1&1&2&1&1&1\\
1&0&1&1&1&2&1&1\\
1&1&0&1&1&1&2&1\\
1&1&1&0&1&1&1&2
\end{bmatrix}
I found that there is a submatrix in the matrix which has rank $ =4$ given by
$[2,1,1,1],[1,2,1,1],[1,1,2,1],[1,1,1,2]$.
I need to show the given matrix has at least 3 zero rows in order to show that Rank $\le 5$..
But I dont know how to show it.
Can someone help.
|
Call column $j$ of this matrix $C_j$, and denote by $\def\2{\mathbf 2}\2$ the column vector with all $8$ entries equal to $2$. Note that for $j=1,2,3,4$ one has $C_j+C_{j+4}=\2$. Thus $C_6=\2-C_2=C_1+C_5-C_2$, and similarly $C_7=C_1+C_5-C_3$ and $C_8=C_1+C_5-C_4$. The last three columns being in the span of the first five, the rank of the matrix is at most$~5$.
|
{
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|
Find $n$ if $\frac{9^{n+1}+4^{n+1}}{9^n+4^n} = 6$ Find $n$ if $$\frac{9^{n+1}+4^{n+1}}{9^n+4^n} = 6$$
In this video they show a shortcut and say $n=-1/2$ without any explanation.
Key observation here is that the geometric mean of $9$ and $4$ is $6$.
It seems numerator and denominator are partial sums of geometric series, but I don't know how to proceed. Any help?
|
This is a possible way:
$$\frac{9^{n+1}+4^{n+1}}{9^n+4^n}=6$$
Now: $$3^{2(n+1)}+2^{2(n+1)}=3\cdot 2\cdot 2^{2n}+2\cdot 3\cdot 3^{2n}$$
I obtain: $$3^{2(n+1)}+2^{2(n+1)}=3\cdot2^{2n+1}+2\cdot3^{2n+1}$$
In other words: $$3^{2n+1}(3-2)=2^{2n+1}(3-2)$$
The solution is $n=-\frac{1}{2}$.
|
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|
Find positive integers $a$ such that there are exactly distinct $2014$ positives integers $b$ satisfied $2 \le \dfrac{a}{b} \le 5$.
Find positive integers $a$ such that there are exactly distinct $2014$ positives integers $b$ satisfied $2 \le \dfrac{a}{b} \le 5$.
This is another problem which I couldn't solve in the test my teacher gave us.
Here are my thoughts.
$2 \le \dfrac{a}{b} \le 5 \iff 2b \le a \le 5b$.
Let the possible $2014$ positive integers $b$ be $b_1 < b_2 < \cdots < b_{2013} < b_{2014}$
$ \implies b_{2014} - b_1 \le 2013$.
We have that $2b_{2014} \le a \le 5b_1 \implies 2b_{2014} \le 5b_1$
$\implies \left\{ \begin{align} 2(b_1 + 2013) &\le 5b_1\\ 2b_{2014} &\le 5(b_{2014} - 2013) \end{align} \right.$ $\iff \left\{ \begin{align} 1342 &\le b_1\\ 3355 &\le b_{2014} \end{align} \right.$
$\implies b_1 = 1342, b_2 = 1343, \cdots, b_{2013} = 3354, b_{2014} = 3355$.
$\implies a = 2b_{2014} = 5b_1 = 6710$.
Is this the correct solution and is there any retouching needed?
|
Your solution isn't quite correct, as I discuss at the end. Instead, as trisct's comment indicates, it's easier to adjust your inequality to bound the $b$ values by multiples of $a$. In particular, with
$$2 \le \frac{a}{b} \le 5 \tag{1}\label{eq1B}$$
First, multiply the left & middle parts by $\frac{b}{2}$ to get
$$b \le \frac{a}{2} \tag{2}\label{eq2B}$$
Next, multiply the middle & right parts of \eqref{eq1B} by $\frac{b}{5}$ to get
$$\frac{a}{5} \le b \tag{3}\label{eq3B}$$
You can combine \eqref{eq2B} and \eqref{eq3B} into one set of inequalities of
$$\frac{a}{5} \le b \le \frac{a}{2} \tag{4}\label{eq4B}$$
For there to be exactly $2014$ values of $b$ (with all of them being consecutive) requires that the difference between the right & left sides of \eqref{eq4B} be at least $2013$, but less than $2015$. Thus, you get
$$2013 \le \frac{a}{2} - \frac{a}{5} = \frac{3a}{10} \lt 2015 \tag{5}\label{eq5B}$$
Multiplying all parts by $\frac{10}{3}$ gives
$$6710 \le a \lt 6716\;\frac{2}{3} \tag{6}\label{eq6B}$$
This gives up to $7$ possible values of $a$. However, it's possible some of them may not give the correct number of values of $b$, so you should check each one. To make it a bit simpler, let $a = 6710 + c$, with $0 \le c \le 6$. Then \eqref{eq4B} becomes
$$1342 + \frac{c}{5} \le b \le 3355 + \frac{c}{2} \tag{7}\label{eq7B}$$
For $c = 0$, you get $1342 \le b \le 3355$, so there's $2014$ values of $b$. However, for $c = 1$, you get $1342.2 \le b \le 3355.5$, so there's only $2013$ values of $b$ in this case. Continuing, you'll find that $c = 2,3$ work, while $c = 4,5,6$ don't. Thus, the final set of values of $a$ which work are $\{6710,6712,6713\}$.
Note this includes your solution of $6710$, and also includes $2$ others you didn't find. I see two issues with your written solution. First, you tried using $2b_{2014} \le a \le 5b_1$. Note the inequality is for each value of $b$, so you can't just use the largest on the left and the smallest on the right. Next, you have $2b_{2014} \le 5(b_{2014} - 2013) \iff b_{2014} \le 3355$. However, the correct result from this is $3355 \le b_{2014}$ instead.
|
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|
Evaluating $\int ^\infty _{0} \frac{x\ln x}{(1+x^2)^2} \,dx$ My first instinct was to evaluate the indefinite form of the integral, which I did by substituting $x=\tan t$, therefore yielding
\begin{align}
\int \frac{x\ln x}{(1+x^2)^2} \,dx
&= \int \frac{\tan t \sec^2 t \ln\tan t}{(1+\tan^2 t)^2} \,dt
&& \text{by substitution} \\
&= \int \tan t \cos^2 t \ln \tan t \,dt \\
&= \int \sin t \cos t \ln \tan t \,dt \\
&= -\frac{1}{2} \cos^2 t \ln \tan t + \frac{1}{2} \int \cot t \, dt
&& \text{by parts} \\
&= -\frac{1}{2} \cos^2 t \ln \tan t +\frac{1}{2} \ln \sin t + k
\end{align}
I run into a wall when I introduce the limits of the integral, since I get
\begin{align}
\int ^\infty _{0} \frac{x\ln x}{(1+x^2)^2} \,dx
&= \bigg[ -\frac{1}{2} \cos^2 t \ln \tan t +\frac{1}{2} \ln \sin t \bigg] ^\frac{\pi}{2} _0
\end{align}
I'm not too sure how to evaluate the limit of the final equation as $t \rightarrow 0$. I feel like the solution is something very trivial, but I can't quite put my finger on what I'm forgetting.
|
Another way:
Integrate by parts
$$\int\ln x\cdot\dfrac x{(1+x^2)^2}\ dx=\ln x\int\dfrac x{(1+x^2)^2}\ dx-\int\left(\dfrac{d(\ln x)}{dx}\int\dfrac x{(1+x^2)^2}\ dx\right)dx$$
$$=-\dfrac{\ln x}{2(1+x^2)}+\int\dfrac{dx}{2x(1+x^2)}$$
Again $$\int\dfrac{dx}{x(1+x^2)}=\int\dfrac{(x^2+1-x^2)\ dx}{x(1+x^2)}=\int\dfrac{dx}x-\int\dfrac{x\ dx}{1+x^2}=\dfrac12\ln\dfrac{x^2}{1+x^2}$$
|
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|
Does $-\frac{1}{2}+\frac{1}{3}-\frac{1}{5}+\frac{1}{7}+\frac{1}{11}+\frac{1}{13}+\frac{1}{17}+\frac{1}{19}+\frac{1}{23}+\dots$ converge? Let $d(m)$ be the number of positive divisors of $m$ [including $1$ and $m$]. Let $p_k$ be the $k^\text{th}$ prime number. Consider the series
$$\sum_{k=1}^{\infty }\frac{(-1)^{d(p_{k}-1)}}{p_{k}}=-\frac{1}{2}+\frac{1}{3}-\frac{1}{5}+\frac{1}{7}+\frac{1}{11}+\frac{1}{13}+\frac{1}{17}+\frac{1}{19}+\frac{1}{23}+\dots$$
Does the series converge? If yes, then what is its value of
convergence?
Any help/hint will be appreciated. THANKS!
|
Since $d(n)$ is odd iff $n$ is a square, your sum is equal to the sum of reciprocals of primes, minus twice the sum of reciprocals of primes of the form $n^2+1$. But we have
$$\sum_{p=n^2+1\text{ for some $n$}}\frac{1}{p}<\sum_{n=1}^\infty\frac{1}{n^2+1}<\infty,$$
so your sum diverges to infinity.
|
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|
Finding the limit of the sequence Let $a ∈ R$ Consider $x_{1} = a, x_{2} =
(1+a)/2$
, and by induction $x_{n} :=
(1+x_{n−1})/2$
What is the limit $?$
By replacing $x_{n}$ and $x_{n-1}$ by $l$, we get the limit $l=1$.
So limit should be $1$.
Also the nth term can be represented by $x_{n} = ( a+ 1 + 2 + 2^{2}+......+ 2^{n-2})/ 2^{n-1}$
And again the limit is $1$
I want to know, why does the value of $a$ doesn't affect the limit$?$
Does this sequence always converge$?$
And also what can I say about the monotonicity of the sequence$?$
|
As you've already determined,
$$\begin{equation}\begin{aligned}
x_n & = \frac{a + 1 + 2 + 2^2 + \ldots + 2^{n-2}}{2^{n-1}} \\
& = \frac{a - 1}{2^{n-1}} + \frac{2 + 2 + 2^2 + \ldots + 2^{n-2}}{2^{n-1}} \\
& = \frac{a - 1}{2^{n-1}} + \frac{2^{n-1}}{2^{n-1}} \\
& = \frac{a - 1}{2^{n-1}} + 1
\end{aligned}\end{equation}\tag{1}\label{eq1}$$
Since $\frac{1}{2^{n-1}} \to 0$ as $n \to \infty$, the first term goes to $0$ regardless of what $a$ is, with the value of $a - 1$ only affecting how fast the first term goes to $0$. This also shows the sequence always converges to a limit of $1$.
As for the sequence's monotonicity, this depends on the value of $a - 1$. If it's $0$, the sequence is a constant of $1$, while if it's positive, then the sequence is monotonically decreasing, and if it's negative, then the sequence is monotonically increasing.
|
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|
$\epsilon - N$ proof of $\sqrt{4n^2+n} - 2n \rightarrow \frac{1}{4}$ I have the following proof for $\lim_{n\rightarrow\infty} \sqrt{4n^2+n} - 2n = \frac{1}{4}$ and was wondering if it was correct. Note that $\sqrt{4n^2+n} - 2n = \frac{n}{\sqrt{4n^2+n} + 2n}$.
$$\left|\frac{n}{\sqrt{4n^2+n} + 2n} - \frac{1}{4}\right| \\
= \left|\frac{2n - \sqrt{4n^2+n}}{4(\sqrt{4n^2+n} + 2n)}\right|=\left|\frac{\sqrt{4n^2+n} - 2n}{4(\sqrt{4n^2+n} + 2n)}\right|\\
= \left|\frac{n}{4(\sqrt{4n^2+n} + 2n)^2}\right| \leq \left|\frac{n}{4(4n)^2}\right| = \left|\frac{n}{64n^2}\right| \\
= \left|\frac{1}{64n}\right| < \epsilon \\
\implies n>\frac{1}{64\epsilon}$$
|
You want to show that $\lim_{n\rightarrow\infty} \sqrt{4n^2+n} - 2n = \frac{1}{4}$. To do this, I would split up the analysis into scratch work and the formal proof.
For the scratch work, you need to find a suitable upper bound. You have done this by showing
\begin{align}
\left|\frac{n}{\sqrt{4n^2+n} + 2n} - \frac{1}{4}\right| &
= \left|\frac{2n - \sqrt{4n^2+n}}{4(\sqrt{4n^2+n} + 2n)}\right|\\&=\left|\frac{\sqrt{4n^2+n} - 2n}{4(\sqrt{4n^2+n} + 2n)}\right|\\&
= \left|\frac{n}{4(\sqrt{4n^2+n} + 2n)^2}\right| \\&\leq \left|\frac{n}{4(4n)^2}\right| \\&= \left|\frac{n}{64n^2}\right| \\&
= \left|\frac{1}{64n}\right| \\&< \epsilon
\end{align}
which means that $n>\frac{1}{64\epsilon}$ is the upper bound.
For the formal proof:
Let $\epsilon>0$ (you need to fix $\epsilon$ as a small positive constant). It follows from $\frac{1}{\epsilon}>0$ that $\frac{1}{64\epsilon}>0$. Then by the Archimedean property there exists a $N\in\mathbb N$ such that $N>\frac{1}{64\epsilon}$. Then if $n\geq N > \frac{1}{64\epsilon}$, we have
$$\left|\frac{n}{\sqrt{4n^2+n} + 2n} - \frac{1}{4}\right|\leq \frac{1}{64n}<\epsilon $$
where the last inequality follows from
$$n\geq N > \frac{1}{64\epsilon} \implies n>\frac{1}{64\epsilon} \implies\epsilon > \frac{1}{64n}$$
|
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|
If $x=\cot6^\circ\cot42^\circ$ and $y=\tan66^\circ\tan78^\circ$, then determine the ratio of $x$ and $y$
If $x=\cot6^\circ\cot42^\circ$ and $y=\tan66^\circ\tan78^\circ$, then
A) $2x=y$
B) $x=2y$
C) $x=y$
D) $2x=3y$
I seriously don’t know where to start. I don’t need the complete answer, but a starting statement which would give me the direction to solve it would be helpful.
Thanks a lot.
|
as @Ross said
$x = \cot 6 \tan 48 , y = \cot 24 \cot 12$ then we look at
$$\frac{x}{y} = \dfrac{\cot 6 \tan 48 }{\cot 24 \cot 12} = \frac{\cot 6 \tan 24}{\cot 48 \cot 12}$$
$$\cot 6 \tan 24 = \frac{\cos 6 \sin 24}{\sin 6 \cos 24}= \frac{\sin 30 + \sin 18}{\sin 30 - \sin 18} = \frac{1 + 2\sin 18 }{1 - 2\sin 18} $$
I used
$$\sin a \cos b = 0.5 ( \sin (a+b) + \sin (a-b)) \\ \cos a \sin b = 0.5 ( \sin (a+b) - \sin (a-b))$$
You need also
$$\cos a \cos b = 0.5 ( \cos (a+b) + \cos (a-b) ) \\ \sin a \sin b = 0.5 ( \cos (a-b) - \cos (a+b))$$
to simplify the Denominator can you proceed ?
|
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|
How do I evaluate $\lim _{x\to 0}\left(\frac{x-\sin x}{x\sin x}\right)$ without using L'Hopital or series?
How do I evaluate $\lim _{x\to 0}\left(\frac{x-\sin x}{x\sin x}\right)$ without using L'Hopital or series?
I've tried expanding the variable such as $x = 2y$ or $x = 3y$, but seemed to still get stuck.
|
Try the following
$$L=\lim \dfrac{x-x\cos\frac{x}{2}+x\cos\frac{x}{2}-\sin x}{x\sin x}$$ then use the $\sin(x)=2\sin \frac{x}{2}\cos\frac{x}{2}$ now the first two terms limit is zero and the second two terms have a relation with the limit $L$
$$\lim \frac{x \cos\frac{x}{2} -2 \sin\frac{x}{2} \cos\frac{x}{2}}{2x \sin\frac{x}{2} \cos\frac{x}{2}}$$
$$\lim \frac{2 \cos\frac{x}{2}\left( \frac{x}{2}- \sin\frac{x}{2}\right)}{2x \sin\frac{x}{2} \cos\frac{x}{2}}$$
$$\lim \frac{1}{2}\frac{\frac{x}{2}-\sin \frac{x}{2}}{\frac{x}{2}\sin\frac{x}{2}}$$ and the last limit is $1/2L$ so $L=1/2L$ hence $L=0$ by this method you can solve something like
$\frac{\sin x -x}{x^3}$ but you need to add and subtract a certain term can figure it out ?
|
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|
Help differentating $f(x) = \sqrt\frac{x^2-1}{x^2+1}$ The equation I'm trying to differentiate is, $ f(x) = \sqrt\frac{x^2-1}{x^2+1}$ and I know the answer is meant to be
$$=\frac{\frac{x\sqrt {x^2+1}}{\sqrt {x^2-1}}-\frac{x\sqrt {x^2-1}}{\sqrt {x^2+1}}}{x^2+1}$$
But when I do the working out I get this
$$=\frac{(x^2-1)^\frac{1}{2}}{(x^2+1)^\frac{1}{2}}$$
$$=\frac{\frac{1}{2}(x^2-1)^\frac{-1}{2}\cdot2x\cdot(x^2+1)^\frac{1}{2}-(x^2-1)^\frac{1}{2}\cdot\frac{1}{2}(x^2+1)^\frac{-1}{2}\cdot2x}{x^2+1}$$
simplify
$$=\frac{x(x^2-1)^\frac{-1}{2}\cdot(x^2+1)^\frac{1}{2}-(x^2-1)^\frac{1}{2}\cdot x(x^2+1)^\frac{-1}{2}}{x^2+1}$$
$$=\frac{\frac{\sqrt {x^2+1}}{x\sqrt {x^2-1}}-\frac{\sqrt {x^2-1}}{x\sqrt {x^2+1}}}{x^2+1}$$
As you can see two of my $x$'s are in the wrong location, and I just can't figure out what I'm doing wrong. Any help as to what steps I'm doing wrong or missing would be much appreciated.
|
Going from your second last line to the last line, you put the $x$ factors into the denominator. Instead, you need to leave them in the numerator. This is because the power of $\frac{-1}{2}$ only apply to the expressions in the brackets, but the $x$ factors are outside of the brackets, so they are not affected. Thus, for example, with the first term in the numerator, you have
$$\begin{equation}\begin{aligned}
x(x^2-1)^\frac{-1}{2}\cdot(x^2+1)^\frac{1}{2} & = x\left((x^2-1)^\frac{-1}{2}\cdot(x^2+1)^\frac{1}{2}\right) \\
& = x\left(\frac{\sqrt {x^2+1}}{\sqrt {x^2-1}}\right) \\
& = \frac{x\sqrt {x^2+1}}{\sqrt {x^2-1}}
\end{aligned}\end{equation}\tag{1}\label{eq1}$$
|
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|
Getting different answers for an integral: $\frac{1}{2}x-\frac{3}{2}\ln{|x+2|}+C$ vs $\frac{1}{2}x-\frac{3}{2}\ln{|2x+4|}+C$ Problem:
$$\int\frac{1}{2}-\frac{3}{2x+4}dx$$
Using two different methods I am getting two different answers and have trouble finding why.
Method 1:
$$\int\frac{1}{2}-\frac{3}{2x+4}dx$$
$$\int\frac{1}{2}-\frac{3}{2}\left(\frac{1}{x+2}\right)dx$$
$$\int\frac{1}{2}dx-\frac{3}{2}\int\frac{1}{x+2}dx$$
$$\frac{1}{2}x-\frac{3}{2}\int\frac{1}{x+2}dx$$
$$x+2=u$$
$$dx=du$$
$$\frac{1}{2}x-\frac{3}{2}\int\frac{1}{u}du$$
$$\frac{1}{2}x-\frac{3}{2}\ln{|x+2|}+C$$
Method 2:
$$\int\frac{1}{2}-\frac{3}{2x+4}dx$$
$$\int\frac{1}{2}-3\left(\frac{1}{2x+4}\right)dx$$
$$\int\frac{1}{2}dx-3\int\frac{1}{2x+4}dx$$
$$\frac{1}{2}x-3\int\frac{1}{2x+4}dx$$
$$2x+4=u$$
$$dx=\frac{du}{2}$$
$$\frac{1}{2}x-\frac{3}{2}\int\frac{1}{u}du$$
$$\frac{1}{2}x-\frac{3}{2}\ln{|2x+4|}+C$$
|
$\ln|2x+4|=\ln|2(x+2)|=\ln|2|+\ln|x+2|=\ln|x+2|+C$
|
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|
Evaluating $\lim_{x \to -\infty} \frac{1}{-\sqrt{\frac{1}{x^6}}\sqrt{x^6+4}}$ Problem:
$$\lim_{x \to -\infty} \frac{1}{-\sqrt{\frac{1}{x^6}}\sqrt{x^6+4}}$$
$$\lim_{x \to -\infty} \sqrt{\frac{1}{x^6}}=0$$ so...
$$\lim_{x \to -\infty} \frac{1}{-\sqrt{\frac{1}{x^6}}\sqrt{x^6+4}}=\frac{1}{0}$$
The answer is $-1$ and I know how to get that answer. Where is the mistake in this method though?
|
Just as $\sqrt{1/x^6}$ goes to $0$, so does $\sqrt{x^6+4}$ go to $\infty$. You cannot substitute just one of these radicals and then simplify, and their unsimplified product is the indeterminate form $0\cdot\infty$ and so cannot be handled directly.
|
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|
What is the greatest possible radius of a circle that passes through the points (1, 2) and (4, 5), whose interior is contained Q1?
What is the greatest possible radius of a circle that passes through the points (1, 2) and (4, 5) and whose interior is contained in the first quadrant of the coordinate plane?
I drew approximate diagrams of 3 circles I could think of that satisfy the points criteria:
1) Points represents diameter(This completely satisties problem criteria, and its radius is $\frac{3\sqrt{2}}{2}$)'
2 & 3(which do not apparently work) are below:
Yet, my first answer is incorrect. What am I missing?
|
Consider the $2$ points to be $A(1,2)$ and $B(4,5)$. The center of any circle passing through these $2$ points must be perpendicular bisector of $AB$. The slope of $AB$ is $\frac{5-2}{4-1} = 1$, so the slope of the perpendicular bisector is the negative reciprocal, i.e., $-1$. Also, the midpoint of $AB$ is $M(\frac{1+4}{2},\frac{5+2}{2}) = M(\frac{5}{2},\frac{7}{2})$. Thus, if the perpendicular bisector line's formula is of the form $y = mx + b$, with $m = -1$, you get $\frac{7}{2} = -\frac{5}{2} + b \implies b = 6$. Thus, the perpendicular bisector line's formula is
$$y = -x + 6 \tag{1}\label{eq1}$$
Consider a point $C(t, -t + 6)$ along the line in \eqref{eq1} to be the center point of a circle through $AB$. Let $r$ be the radius of this circle. For the entire circle to be in the first quadrant requires that
$$r \le t \implies r^2 \le t^2 \tag{2}\label{eq2}$$
and
$$r \le -t + 6 \implies r^2 \le (-t + 6)^2 = t^2 - 12t + 36 \tag{3}\label{eq3}$$
Next, note the lengths of $AC$ and $BC$ are equal to each other and to $r$. Consider just $AC$. This give the equation, when the distance is squared, of
$$\begin{equation}\begin{aligned}
r^2 & = (t - 1)^2 + (-t + 6 - 2)^2 \\
r^2 & = (t - 1)^2 + (t - 4)^2 \\
r^2 & = t^2 - 2t + 1 + t^2 - 8t + 16 \\
r^2 & = 2t^2 - 10t + 17
\end{aligned}\end{equation}\tag{4}\label{eq4}$$
From \eqref{eq2}, this gives
$$2t^2 - 10t + 17 \le t^2 \implies t^2 - 10t + 17 \le 0 \tag{5}\label{eq5}$$
and, from \eqref{eq3}, \eqref{eq4} gives
$$2t^2 - 10t + 17 \le t^2 - 12t + 36 \implies t^2 + 2t - 19 \le 0 \tag{6}\label{eq6}$$
The maximum radius occurs where \eqref{eq5} or \eqref{eq6} is $0$. With \eqref{eq5}, the roots are $t = 5 \pm 2\sqrt{2}$. With $r = t = 5 - 2\sqrt{2}$, you have $-t + 6 \gt r$, so eqref{eq3} doesn't hold. Since $t = 5 + 2\sqrt{2} \gt 6$ means $-t + 6 \le 0$, so it's not a valid root. With \eqref{eq6}, the roots are $t = -1 \pm 2\sqrt{5}$. Since $t \gt 0$, the only valid root is $t = -1 + 2\sqrt{5}$, so $r = -t + 6 = 7 - 2\sqrt{5}$ (and \eqref{eq2} also holds), with this being the maximum radius.
|
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|
How can we not use Muirhead's Inequality for proving the following inequality? There was a question in the problem set in my math team training homework:
Show that $∀a, b, c ∈ \mathbb{R}_{≥0}$ s.t. $a + b + c = 1, 7(ab + bc + ca) ≤ 2 + 9abc.$
I used Muirhead's inequality to do the question (you can try out yourself):
By Muirhead's inequality, $$\begin{align}7(ab+bc+ca)&=7(a+b+c)(ab+bc+ca)\\&=21abc+6\big(a^2b+b^2c+c^2a+ab^2+bc^2+ca^2\big)+\sum_{sym}a^2b\\&\le21abc+6\big(a^2b+b^2c+c^2a+ab^2+bc^2+ca^2\big)+\sum_{sym}a^3\\&=2(a+b+c)^3+9abc\\&=2+9abc\end{align}$$As $(3,0,0)$ majorizes $(2,1,0)$.
Is above proof correct? Also, can we find a proof not using Muirhead's inequality?
Any help is appreciated!
|
Your proof looks good.
Here's an unsophisticated alternative proof, using only elementary algebra . . .
We don't even need $a,b,c$ to be nonnegative.
As shown below, if $a,b,c\;$are real numbers such that $a+b+c=1$, and if at least one of $a,b,c\;$is between $-1$ and ${\large{\frac{7}{9}}}$ inclusive, then the inequality holds.
Without loss of generality, assume $-1\le a\le {\large{\frac{7}{9}}}$.
Replacing $c\;$by $1-a-b$, we get
\begin{align*}
&(2+9abc)-7(ab+bc+ca)\\[4pt]
=\;&\bigl(2+9ab(1-a-b)\bigr)-7\bigl(ab+(1-a-b)(a+b)\bigr)\\[4pt]
=\;&(7-9a)b^2+\bigl((7-9a)(a-1)\bigr)b+(2+7a^2-7a)\\[4pt]
\end{align*}
which is nonnegative since:
*If $a={\large{\frac{7}{9}}}$, it evaluates to ${\large{\frac{64}{81}}}$.$\\[10pt]$
*If $-1\le a < {\large{\frac{7}{9}}}$, then regarded as a quadratic in the variable $b$, its leading coefficient is positive, and its discriminant
$$
\bigl((7-9a)(a-1)\bigr)^2-4(7-9a)(2+7a^2-7a)
$$
factors as
$$
(a+1)(1-3a)^2(9a-7)
$$
which is nonpositive.
|
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|
Laurent Expansion of a Composition (with the inverse) Suppose $f(z) = a_1z + a_0 + O(\frac{1}{z})$ as $|z| \rightarrow \infty$. Here $a_1 > 0$. Let $W(z) = z + \frac{1}{z}$.
How do I obtain that the composite function $L(z) = W \circ f^{-1}(z)$ has the expansion $\frac{z}{a_1} - \frac{a_0}{a_1} + O(\frac{1}{z})$ as $|z| \rightarrow \infty$
|
Suppose $f(z) = a_1z + a_0 + O(\frac{1}{z})$ as $|z| \rightarrow \infty$ with $a_1>0$. Consider $g(z) = \frac{1}{f(\frac{1}{z})}$. Then $g(0) = 0$ and $g'(0) \neq 0$. So $g$ is analytic near $0$ and has the form $g(z) = g_1z + g_2z^2 + O(z^3)$ near $0$.
$$
\frac{1}{f\left(\frac{1}{z}\right)} = g(z)
$$
$$
\therefore g_1 = \frac{1}{a_1} \text{ and } g_2 = -\frac{a_0}{a_1^2}$$
Now as the comment (above) suggests, for $g(z) = \frac{1}{f\left(\frac{1}{z}\right)}$ we have $g^{-1} = \frac{1}{f^{-1}\left(\frac{1}{z}\right)}$ and $g^{-1}$ is also analytic at $0$. So $g^{-1}(z)$ is of the form $b_1z + b_2z^2 + O\left(z^3\right)$ near $0$.
$$
(g^{-1}(g(z)))' = 1
$$
$$
\left( \frac{1}{a_1} - \frac{2a_0}{a_1^2} (b_1z + b_2z^2)) + O(z^3)\right)\left( b_1 + 2b_2z + O(z^3)\right) = 1
$$
Which gives us $b_1 = a_1$ and $b_2 = a_0a_1$
Therefore we get that $\frac{1}{f^{-1}(\frac{1}{z})} = a_1z + a_0a_1z^2 + O(z^3)$.
Now since ord$_0 f^{-1}\left(\frac{1}{z}\right)) = -1$ we have $f^{-1}\left(\frac{1}{z}\right)$ is of the form $\frac{c_{-1}}{z} + c_o + c_1z + O(z^2)$ near $0$. Thus
$$
\left( \frac{c_{-1}}{z} + c_0 + c_1z + O(z^2)\right)(a_1z + a_0a_1z^2 + O(z^3) = 1
$$
Which yields, $c_1 = \frac{1}{a_1}$ and $c_0 = -\frac{a_0}{a_1}$. So, $f^{-1}\left(\frac{1}{z}\right) = \frac{1}{a_1z} - \frac{a_0}{a_1} + O(z)$.
Now, $L(z) = W \circ f^{-1}(z) = f^{-1}(z) + \frac{1}{f^{-1}(z)}$. Both of these can be found by taking the reciprocals inside $f^{-1}\left(\frac{1}{z}\right)$ and $\frac{1}{f^{-1}\left(\frac{1}{z}\right)}$.
|
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|
Evaluate $\int_{0}^{2\pi} \frac{cos \theta}{2 + cos \theta} d\theta$ using the residue theorem My attempt to a solution for $I = \int_{0}^{2\pi} \frac{cos \theta}{2 + cos \theta} d\theta$ is as follows.
On the unit circle we have $z=e^{i\theta} \implies dz = izd\theta \iff d\theta = \frac{dz}{iz}$, and furthermore $cos \theta = \frac{e^{i\theta} + e^{-i\theta}}{2} = \frac{z+1/z}{2}.$
By this variable change and the residue theorem, the integral becomes, $$I = \int_{|z|=1}^{} \frac{(z+1/z)\frac{1}{2}}{(2+\frac{z+1/z}{2})} \frac{dz}{iz} = \frac{1}{i} \int_{|z|=1}^{} \frac{z^2+1}{z(z^2+4z+1)}dz = \frac{1}{i} \int_{|z|=1}^{} \frac{z^2+1}{z(z+2-\sqrt{3})(z+2+\sqrt{3})}dz = 2\pi i \ \sum_{j=1}^{2}\text{Res}\left[\frac{z^2+1}{z(z+2-\sqrt{3})(z+2+\sqrt{3})}, z_j\right],$$
where, on the unit disc, the integrand $f(z) = \frac{z^2+1}{z(z+2-\sqrt{3})(z+2+\sqrt{3})}$, has a simple pole at $z_1=0$, and one at $z_2=-2+\sqrt{3}$.
Evaluating the residues we obtain $\text{Res}\left[f(z),0\right] = \lim_{z \to 0} \left[\frac{z^2+1}{(z+2-\sqrt{3})(z+2+\sqrt{3})}\right] = \frac{1}{(2-\sqrt{3})(2+\sqrt{3})} = 1$, and $\text{Res}\left[f(z),-2+\sqrt{3}\right] = \lim_{z \to -2+\sqrt{3}} \left[\frac{z^2+1}{(z+2-\sqrt{3})(z+2+\sqrt{3})}\right] = \frac{(-2+\sqrt{3})^2 +1}{(-2+\sqrt{3})(-2+\sqrt{3} + 2 + \sqrt{3}))} = \frac{8-4\sqrt{3}}{6-4\sqrt{3}} = \frac{4-2\sqrt{3}}{3-2\sqrt{3}}.$
Finally we obtain, $$I = 2\pi i \left(1 + \frac{4-2\sqrt{3}}{3-2\sqrt{3}}\right) = \frac{2\pi i (7-4\sqrt{3})}{3-\sqrt{3}}.$$
However the solution in the text claims the answer to be $I = \pi(1-\frac{2}{\sqrt{3}}).$ And I do not doubt the validity of the texts solution since I agree with how they derived it. However they solved it using a different way which I would never apply my self.
All help is appreciated. Thanks!
|
$$I = \int_{|z|=1}^{} \frac{(z+1/z)\frac{1}{2}}{(2+\frac{z+1/z}{2})} \frac{dz}{iz} = \frac{1}{i} \int_{|z|=1}^{} \frac{z^2+1}{z(z^2+4z+1)}dz = \frac{1}{i} \int_{|z|=1}^{} \frac{z^2+1}{z(z+2-\sqrt{3})(z+2+\sqrt{3})}dz = 2\pi \color{red}{i} \ \sum_{j=1}^{2}\text{Res}\left[\frac{z^2+1}{z(z+2-\sqrt{3})(z+2+\sqrt{3})}, z_j\right]$$
Note that you have an extra factor of $i$ in the last step ($2\pi i \cdot \dfrac{1}{i} = 2\pi$, not $2\pi i$). Also, by rationalizing the denominator, $$\frac{4 - 2\sqrt{3}}{3 - 2\sqrt{3}} = \frac{(4 - 2\sqrt{3})(3 + 2\sqrt{3})}{3^2 - (2\sqrt{3})^2} = \frac{12 - 6\sqrt{3} + 8\sqrt{3} - 12}{9-12} = \frac{2\sqrt{3}}{-3}$$ which is the same as $-\dfrac{2}{\sqrt{3}}$. So the answer should be $$I = 2\pi\left(1 + \frac{4 - 2\sqrt{3}}{3 - 2\sqrt{3}}\right)=2\pi\left(1 - \frac{2}{\sqrt{3}}\right)$$
|
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|
Locus of mid point of $AB$
If the family of lines $tx+3y-6=0.$ where $t$ is variable intersect the lines $x-2y+3=0$ and $x-y+1=0$ at point $A$ and $B.$ Then locus of mid point of $AB$ is
what i try
Intersection of line $tx+3y-6=0$ and $x-2y+3=0$ is
$\displaystyle A\bigg(\frac{3}{3+2t},\frac{6+3t}{3+2t}\bigg)$
and Intersection of $tx+3y-6=0$ and $x-y+1=0$ is
$\displaystyle B\bigg(\frac{3}{3+t},\frac{6+t}{3+t}\bigg)$
Let locus of mid point of $AB$ is at $M(h,k)$
Then
$\displaystyle h=\frac{1}{2}\bigg(\frac{3}{3+2t}+\frac{3}{3+t}\bigg)$ and $\displaystyle k = \frac{1}{2}\bigg(\frac{6+3t}{3+2t}+\frac{6+t}{3+t}\bigg)$
How do i eliminate $t$ in an easy way
or please help me is any other easier ways to solve it thanks
|
Hint:
$$2k=\dfrac{6+3t}{3+2t}+\dfrac{6+t}{3+t}$$
$$2k-\dfrac32-1=\dfrac{6+3t}{3+2t}-\dfrac32+\dfrac{6+t}{3+t}-1=\dfrac{3}{2(3+2t)}+\dfrac3{3+t}$$
$$2h=\dfrac3{3+t}+\dfrac3{3+2t}$$
Solve the simultaneous equations for $\dfrac1{3+2t},\dfrac1{3+t}$
Finally use $$2(3+t)-(3+2t)=3$$
Optionally we can simplify the result
|
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|
Maclaurin expansion of $\arccos(1-2x^2)$ Maclaurin expansion of $\arccos(1-2x^2)$
This is what I tried.
$f'(x)=2(1-x^2)^{-1/2} \\
f''(x)=2(1-x^2)^{-3/2}+3 \cdot 2 x^2(1-x^2)^{-5/2} \\
f^{(3)}(x)=18x(1-x^2)^{-5/2}+2\cdot 3\cdot 5x^3(1-x^2)^{-7/2} \\
f^{(4)}(x)=18(1-x^2)^{-5/2}+180x^2(1-x^2)^{-7/2}+2\cdot 3\cdot 5\cdot 7x^4(1-x^2)^{-9/2}$
From this I get $f'(0)=2 $, $f''(0)=0 $, $f^{(3)}(0)=2 $, $f^{(4)}(0)=0 $
But I don't know how to find a general term. Can this be solved in easier steps?
|
$arcos(1-2x^2) = \frac{\pi}{2} \sum\limits_{k=0}^{\infty} \frac{(1-2x^2)^{1+2k}(\frac{1}{2})k}{k!+2k\times k!}$
$\forall |1-2x^2|<1$
|
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|
How to solve two inequalities that which simultaneous answers I have two inequalities here, and I must find the answer for both of them simultaneously (joint answer):
$\left\{
\begin{aligned}
\dfrac{2}{x-3} \gt \dfrac{5}{x+6} \\
\dfrac{1}{3} \lt \dfrac{1}{x-2}
\end{aligned}
\right.$
Please give me some hints and I'll do the rest. But if you have full time you can show me how to do it. Anyway, I can't find anything like this question on the Internet. I can solve one inequality, but two inequalities with a joint answer is new to me.
|
Let us consider the inequalties
$(1) \quad \dfrac{2}{x-3} \gt \dfrac{5}{x+6} $
and
$(2) \quad \dfrac{1}{3} \lt \dfrac{1}{x-2} $.
Then compute the set $L_1$ of the solutions of $(1)$ and the set $L_2$ of the solutions of $(2)$.
The set of solutions of
$\left\{
\begin{aligned}
\dfrac{2}{x-3} \gt \dfrac{5}{x+6} \\
\dfrac{1}{3} \lt \dfrac{1}{x-2}
\end{aligned}
\right.$
is given by $L_1 \cap L_2.$
|
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|
Closed form for the series: $\sum_{n=1}^{\infty}x^{n^2}$ Let $$S=\sum_{n=1}^{\infty}x^{n^2}, \quad|x|<1.$$ It is convergent and the sum is definitely less than $1/(1-x).$
Is there any closed form for $S$ ?
|
There is no known closed form of
$$
S(q):=\sum_{n=1}^{\infty}q^{n^2}, \qquad|q|<1.
$$ in terms of elementary functions. A related function, called the Jacobi theta function,
$$
\begin{align}
\vartheta(z; \tau) &= \sum_{n=-\infty}^\infty \exp \left(\pi i n^2 \tau + 2 \pi i n z\right) \\
&= 1 + 2 \sum_{n=1}^\infty \left(e^{\pi i\tau}\right)^{n^2} \cos(2\pi n z) \\
&= \sum_{n=-\infty}^\infty q^{n^2}\eta^n
\end{align}
$$
has been studied showing very interesting properties. By setting $q=e^{-x \pi}$ and
\begin{align}
\varphi(e^{-\pi x}) = \vartheta(0; ix) = \sum_{n=-\infty}^\infty e^{-x \pi n^2}=2S(q)+1
\end{align} one may prove that
\begin{align}
\varphi\left(e^{-\pi} \right) &= \frac{\sqrt[4]{\pi}}{\Gamma\left(\frac34\right)},
\\\\
\varphi\left(e^{-2\pi}\right) &= \frac{\sqrt[4]{6\pi+4\sqrt2\pi}}{2\Gamma\left(\frac34\right)},
\\\\
\varphi\left(e^{-3\pi}\right) &= \frac{\sqrt[4]{27\pi+18\sqrt3\pi}}{3\Gamma\left(\frac34\right)},
\\\\
\varphi\left(e^{-4\pi}\right) &= \frac{\sqrt[4]{8\pi}+2\sqrt[4]{\pi}}{4\Gamma\left(\frac34\right)},
\\\\
\varphi\left(e^{-5\pi}\right) &= \frac{\sqrt[4]{225\pi+ 100\sqrt5 \pi}}{5\Gamma\left(\frac34\right)},
\\\\
\varphi\left(e^{-6\pi}\right) &= \frac{\sqrt[3]{3\sqrt{2}+3\sqrt[4]{3}+2\sqrt{3}-\sqrt[4]{27}+\sqrt[4]{1728}-4}\cdot\sqrt[8]{243{\pi}^2}}{6\sqrt[6]{1+\sqrt6-\sqrt2-\sqrt3}{\Gamma\left(\frac34\right)}}.
\end{align}
|
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|
Find value for $k$ such that $(x^2-k)$ is a factor for $f(x)=2x^4+(3k-4)x^3+(2k^2-5k-5)x^2+(2k^3-2k^2-3k-6)x+6$ Find value for $k$ such that $(x^2-k)$ is a factor for, $$f(x)=2x^4+(3k-4)x^3+(2k^2-5k-5)x^2+(2k^3-2k^2-3k-6)x+6$$
My Try
Since $x^2-k=0$ when we substitute $x=\pm k$ to $f(x)$ it should be equal to $0.$
But this gives a polynomial of k where it has $\sqrt{k}$ terms as well. Is my approach correct or is there a simpler way? Please any hint would be highly appreciated.
|
Just do the Euclidean division of the two polynomials. You get
$$f_k(x)=\left(x^2-k\right)\left(2x^2+(3k-4)x+2k^2-3k-5\right)+(2k^3+k^2-7k-6)x+2k^3-3k^2-5k+6$$
And we want the remainder
$$(2k^3+k^2-7k-6)X+(2k^3-3k^2-5k+6)$$
to be the null polynomial. This is possible for the common roots if any of
$$\begin{align}2k^3+k^2-7k-6=&(2k+3)(k+1)(k-2)\\
2k^3-3k^2-5k+6=&(2k+3)(k-1)(k-2)\end{align}$$
|
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|
Four balls are randomly dropped into four boxes Four balls are randomly dropped into four boxes, where any ball is equally likely to fall into
each box.
For a fixed $k = 0, 1, 2, 3$, let $A_k$ denote the event that exactly $k$ boxes are empty. Compute
$P(A_k)$ for each $k = 0, 1, 2, 3$.
I have computed $P(A_0)$ and $P(A_3)$ which are $\frac{4!}{4^4}=\frac{3}{32}$ and $\frac{4}{4^4}=\frac{1}{64}$ recpectively. But I don't know how to find $P(A_1)$ and $P(A_2)$. Could anybody help me?
|
I'll write out the computation "by hand" for $A_2$. There are slicker answers, but this doesn't require any clever observations.
First we drop the first ball, and it doesn't matter where it goes. Then we drop the second ball. There is a $1/4$ chance it lands in the same place as the first, and a $3/4$ chance otherwise. So
$$
P(A_2) = \frac{1}{4} P(A_2 | \text{same}) + \frac{3}{4} P(A_2 | \text{different}).
$$
If they landed in different places, then we don't care where balls 3 and 4 land as long as they land in the boxes already occupied. The probability of this is $1/2$ for each, and the events are independent, so
$$
P(A_2) = \frac{1}{4} P(A_2 | \text{same}) + \frac{3}{4} \cdot\frac{1}{2} \cdot\frac{1}{2}.
$$
If the first two balls landed in the same place, then drop the third ball. Either it lands in the first place again, with probability $1/4$, or lands in a second place. If it lands in a second place, then we just need the fourth ball to land in one of the two already occupied.
$$
P(A_2) = \frac{1}{4} \left(\frac{1}{4} P(A_2| 3 \text{same}) + \frac{3}{4} \cdot \frac{1}{2} \right) + \frac{3}{4} \cdot\frac{1}{2} \cdot\frac{1}{2}.
$$
If the first three balls all landed in the same place, then we need the fourth ball to land somewhere else, with probability $3/4$. So we have
$$
P(A_2) = \frac{1}{4} \left(\frac{1}{4} \cdot\frac{3}{4} + \frac{3}{4} \cdot \frac{1}{2} \right) + \frac{3}{4} \cdot\frac{1}{2} \cdot\frac{1}{2} = \frac{3 + 6 + 12}{64} = \frac{21}{64}.
$$
|
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"timestamp": "2023-03-29T00:00:00",
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|
Simplifying expression, is it possible? To show that,
$$
(7 + 50^{1/2})^{1/3} + (7 - 50^{1/2})^{1/3} = 2
$$
I am aware of the way where we can "guess" and come up with the following:
7 + 50^(1/2) = (1 + 2^(1/2))^3
7 - 50^(1/2) = (1 - 2^(1/2))^3
Hence simplifying the expression. But can we do it without the guesswork?
|
We can use also the following identity.
$$a^3+b^3+c^3-3abc=(a+b+c)(a^2+b^2+c^2-ab-ac-bc).$$
Let $a=\sqrt[3]{7-5\sqrt2},$ $b=\sqrt[3]{7+5\sqrt2}$ and $c=\sqrt[3]{7-5\sqrt2}+\sqrt[3]{7+5\sqrt2}$.
Thus, since $a+b-c=0,$ we obtain:
$$7-5\sqrt2+7+5\sqrt2-c^3+3\sqrt[3]{(7-5\sqrt2)(7+5\sqrt2)}c=0$$ or
$$c^3+3c-14=0$$ or
$$c^3-2c^2+2c^2-4c+7c-14=0$$ or
$$(c-2)(c^2+2c+7)=0$$ or
$$c=2$$ and we are done!
|
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|
Inequality $3\sin^{-1}\left(\frac{\pi^{1/3}}{2^{2/3}}\right)\leq a+b+c$ with some conditions I have a new problem create by myself :
Let $0<a,b,c<\frac{\pi}{2}$ such that $\sin(a)\sin(b)\sin(c)=\frac{\pi}{4}$ then we have :
$$3\sin^{-1}\left(\frac{(\pi)^{\frac{1}{3}}}{2^{\frac{2}{3}}}\right)\leq a+b+c$$
I have tried to use the logarithm function :
$$\ln(\sin(a))+\ln(\sin(b))+\ln(\sin(c))=\ln(\frac{\pi}{4})$$
And use the concavity of $\ln(\sin(x))$ we get :
$$\ln(\sin(a))+\ln(\sin(b))+\ln(\sin(c))\leq 3\ln\left(\sin\left(\frac{a+b+c}{3}\right)\right)$$
Now it's easy to conclude .
My question have you an alternative way ?
Thanks a lot !
|
We can use the Tangent Line method here:
$$\sum_{cyc}\left(a-\arcsin\sqrt[3]{\frac{\pi}{4}}\right)=\sum_{cyc}\left(a-\arcsin\sqrt[3]{\frac{\pi}{4}}-\frac{\ln\sin{a}-\ln\sqrt[3]{\frac{\pi}{4}}}{\sqrt{\sqrt[3]{\frac{16}{\pi^2}}-1}}\right)\geq0$$
because $$a-\arcsin\sqrt[3]{\frac{\pi}{4}}-\frac{\ln\sin{a}-\ln\sqrt[3]{\frac{\pi}{4}}}{\sqrt{\sqrt[3]{\frac{16}{\pi^2}}-1}}\geq0$$ for any $0<a<\frac{\pi}{2}.$
Indeed, let $f(a)=a-\arcsin\sqrt[3]{\frac{\pi}{4}}-\frac{\ln\sin{a}-\ln\sqrt[3]{\frac{\pi}{4}}}{\sqrt{\sqrt[3]{\frac{16}{\pi^2}}-1}}.$
Thus, easy to see that $$f\left(\arcsin\sqrt[3]{\frac{\pi}{4}}\right)=0,$$
$$f'\left(\arcsin\sqrt[3]{\frac{\pi}{4}}\right)=0$$ and for all $0<a<\frac{\pi}{2}$ $$f''\left(a\right)>0.$$
|
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|
Express $\cos 4x$ in terms of only $\sin 2x$ and hence solve the trigonometric equation The question is to express $\cos 4x$ in terms of only $\sin 2x$ and hence solve the trigonometric equation with the restriction of $\theta \in (0^{\circ}, 135^{\circ})$
$$\frac{\cos 5 \theta}{\sin \theta} + \frac{\sin 5 \theta}{\cos \theta} = 2.$$
I have come up to this point:
$$\cos 4x = \cos (2x + 2x) = \cos 2x \cos 2x - \sin 2x \sin 2x = (\cos(2x))^2 - (\sin(2x))^2.$$
Now I only need to express $\cos 2x$ in terms of $\sin 2x$, but for all quadrants. Regarding the trigonometric equation, I have done these steps:
\begin{align}
\frac{\cos(5x)\cos(x) + \sin(5x)\sin(x)}{\sin(x)\cos(x)}=2
&\implies \cos(5x-x)=2\sin(x)\cos(x) \\
&\implies \cos(4x)=\sin(2x).
\end{align}
So I believe that when I express $\cos(4x)$ in terms of $\sin 2x$ I continue with solving the equation.
|
Rearrange
$$\dfrac{\cos 5 \theta}{\sin \theta} + \dfrac{\sin 5 \theta}{\cos \theta} = 2$$
to get
$$\cos(4\theta) = \sin(2\theta)$$
Then, use $\cos 2x=1-2\sin^2 x$
$$1-2\sin^2(2\theta)= \sin(2\theta)$$
$$\sin(2\theta) = -1, \space \sin(2\theta)=1/2$$
$$\theta = 15^\circ,\space \theta=75^\circ$$
|
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|
Alligator population Hints only. I feel like I am so close. Population growth: The time rate change of an alligator population $P$ in a swamp is proportional to the square of $P$. The swamp contained a dozen alligators in $1988$ and $2$ dozen in $1998$
$$\frac{dp}{dt} = kp^2$$
$$\int \frac{dp}{p^2} = \int k$$
$$ \frac{p^{-2+1}}{-2+1} = tk + C$$
$$ -\frac{1}{p} = tk+C$$
Use $P(0) = 12$ to solve for C.
$$-\frac{1}{12} = C$$
$$p(t) = -\frac{1}{tk - \frac{1}{12}}$$
Use $P(10)=24$ to solve for k:
$$24 = -\frac{1}{10k-\frac{1}{12}}$$
$$10k -\frac{1}{12} = -\frac{1}{24}$$
$$k = \frac{1}{240}$$
Now solve for $t$ with $P(t)=48$
$$48=-\frac{1}{\frac{t}{240}-\frac{1}{12}}$$
$$\frac{t}{240} - \frac{1}{12} = -\frac{1}{48}$$
$$\frac{t}{20}-1=-3$$
$$\frac{t}{20}=-2$$
$$t=-40$$
Which is obviously wrong. I assume that t should in-fact represent 1 year even though our information is given in increments of 10. I am very sure my set-up is correct and I integrated correctly...
|
You have correctly derived that
$$P(t) = -\frac{1}{tk - \frac{1}{12}}\tag{1}$$
and
$$k = \frac{1}{240}\tag{2}$$
so that
$$P(t) = -\frac{1}{\frac{t}{240} - \frac{1}{12}}\tag{3}$$
Your mistake happens when solving for $P(t)=48$. Starting off at
$$48=-\frac{1}{\frac{t}{240}-\frac{1}{12}}\implies\frac{t}{240} - \frac{1}{12} = -\frac{1}{48}$$
is correct. However
$$\frac{t}{240} - \frac{1}{12} = -\frac{1}{48}\implies \frac{t}{20}-1=-3$$
is incorrect. Multiplying all three terms by $12$ would produce
$$\frac{t}{240} - \frac{1}{12} = -\frac{1}{48}\implies \frac{t}{20}-1=-\frac{1}{4}$$
which you could then multiply again by $20$ to form
$$\frac{t}{20}-1=-\frac{1}{4}\implies t-20=5$$
so that $t=15$.
|
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|
What is the method to factor $x^3 + 1$? In the solution to a problem, it's stated that
We see that $x^3+1=(x+1)(x^2-x+1)$.
Why is this, and what method can I use for similar problems with different coefficients?
The full problem is
Find the remainder when $x^{81}+x^{48}+2x^{27}+x^6+3$ is divided by $x^3+1$.
|
For the full problem, working modulo $x^3+1$ we have $$x^3=-1\implies x^{3n}=(-1)^n\implies x^{81}+x^{48}+2x^{27}+x^6+3=-1+1-2+1+3=2.$$
So no such factorisation is needed. But when it is needed, here's how to do it. By the factor theorem, the fact that $(-1)^3+1=0$ implies $x-(-1)=x+1$ is a factor. So try $x^3+1=(x+1)(ax^2+bx+c)$. We know $a=1$ from the $x^3$ coefficient and $c=1$ from the constant term, and the $x^2$ coefficient tells us $0=a+b\implies b=-1$.
|
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Finding $\displaystyle \lim_{x \to 2}\frac{\sqrt{x^3+1}-\sqrt{4x+1}}{\sqrt{x^3-2x}-\sqrt{x+2}}$ I came across this question.
Evaluate the limit $$ \lim_{x \to 2}\frac{\sqrt{x^3+1}-\sqrt{4x+1}}{\sqrt{x^3-2x}-\sqrt{x+2}}$$
I tried rationalizing the denominator, substitution, yet nothing seems to cancel out with the denominator. I don't think we are supposed to use squeeze theorem or L'Hopital rule for this.
Can someone give me a hint in the right direction?
|
$${{\sqrt{x^3+1}-\sqrt{4x+1} \over \sqrt{x^3-2x} - \sqrt{x+2}}
= \left({\sqrt{x^3+1}-\sqrt{4x+1} \over \sqrt{x^3-2x} - \sqrt{x+2}} \right)
\left( {\sqrt{x^3+1}+\sqrt{4x+1} \over \sqrt{x^3+1}+\sqrt{4x+1}} \right)
\left({\sqrt{x^3-2x} + \sqrt{x+2} \over \sqrt{x^3-2x} + \sqrt{x+2}} \right)
=\left({x^3-4x \over x^3-3x-2}\right)
\left({\sqrt{x^3-2x}+\sqrt{x+2} \over \sqrt{x^3+1}+\sqrt{4x+1}} \right)
=\left({x(x+2) \over (x+1)^2}\right)
\left({\sqrt{x^3-2x}+\sqrt{x+2} \over \sqrt{x^3+1}+\sqrt{4x+1}} \right)}$$
At $x=2$, above simplified to:
$\displaystyle \left({2 \times 4 \over 3 \times 3}\right)
\left({2+2 \over 3+3} \right) = \left({8 \over 9}\right) \left({2 \over 3}\right) = {16 \over 27}$
|
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|
if $4n^2-1=3m^2$ has a positive integer solution, show that $2n-1$ always square Let $n\in \mathbb{N}$, if $4n^2-1=3m^2$ has a positive integer solution, show that $2n-1$ is a perfect square.
For example $n=1$, it is clear.
and $n=13$ because $$ 4\cdot 13^2-1=3\cdot 15^2$$ and $$2n-1=25=5^2$$
|
$a^2-1=3b^2$ is a Pell equation, $a^2-3b^2=1$. The solutions correspond to
$$
a_k + b_k \sqrt3 = (2+\sqrt3)^{k}
$$
We have
$$
a_{k+2}=4a_{k+1}-a_k
, \quad
a_0 = 1
, \quad
a_1 = 2
$$
because $2+\sqrt3$ is a root of $x^2 - 4 x + 1$.
Therefore
$$
a_k \bmod 2 = 1,0,1,0,\dots
\\
a_k \bmod 3 = 1,2,1,2,\dots
$$
Write $4n^2-1=3m^2$ as $(2n)^2-3m^2=1$. Therefore, $2n=a_{2k+1}$ and so $2n \equiv 2 \bmod 3$.
Write $4n^2-1=3m^2$ as $(2n-1)(2n+1)=3m^2$. Since $3$ divides $2n+1$ and $2n-1$ and $2n+1$ are coprime, we must have that $2n-1$ is square.
|
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|
Cubic diophantine equation with a prime $x^3 + y^3 + z^3 - 3xyz = p$. Question: Find all triple positive integers $(x, y, z)$ so that
$$x^3 + y^3 + z^3 - 3xyz = p,$$
where $p$ is a prime number greater than $3$.
I have tried the following: The equation factors as
$$(x + y + z) (x^2 + y^2 + z^2-xy-yz-zx) = p.$$
Since $x + y + z> 1$, we must have $x + y + z = p$ and
$$x^2 + y^2 + z^2-xy-yz - zx = 1.$$
The last equation is equivalent to
$$(x-y)^2 + (y-z)^2 + (z-x)^2 = 2.$$
Without loss of generality you can assume that $x ≥ y ≥ z$, we have $xy ≥ 1$ and $xz ≥ 2$, implying
$$(xy)^2 + (yz)^2 + (zx)^2 ≥ 6> 2.$$
Who can help me and correct me, thank you.
|
Strangely enough, the solution is finite.
for the equation:
$$X^3+Y^3+Z^3-3XYZ=q=ab$$
If it is possible to decompose the coefficient as follows: $4b=k^2+3t^2$
Then the solutions are of the form:
$$X=\frac{1}{6}(2a-3t\pm{k})$$
$$Y=\frac{1}{6}(2a+3t\pm{k})$$
$$Z=\frac{1}{3}(a\mp{k})$$
Thought the solution is determined by the equation Pell, but when calculating the sign was a mistake. There's no difference, but the amount should be. Therefore, the number of solutions of course.
I may be wrong, though. We still need to check other options.
|
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|
Expression of $x^n+\frac1{x^n}$ by $x+\frac1{x}$ where $n$ is a positive odd number. There was a problem in a book:
Denote that $y=x+\dfrac{1}{x}$, express $x^7+\dfrac{1}{x^7}$ using $y$.
It's not a hard question, but I find a special sequence:
$x+\dfrac{1}{x}=y\\x^3+\dfrac{1}{x^3}=\left(x+\dfrac{1}{x}\right)^3-3\left(x+\dfrac{1}{x}\right)=y^3-3y \\ x^5+\dfrac{1}{x^5}=\left(x+\dfrac{1}{x}\right)^5-5\left(x^3+\dfrac{1}{x^3}\right)-10\left(x+\dfrac{1}{x}\right)=y^5-5\left(y^3-3y\right)-10y=y^5-5y^3+5y\\x^7+\dfrac{1}{x^7}=\left(x+\dfrac{1}{x}\right)^7-7\left(x^5+\dfrac{1}{x^5}\right)-21\left(x^3+\dfrac{1}{x^3}\right)-35\left(x+\dfrac{1}{x}\right)=y^7-7\left(y^5-5y^3+5y\right)-21\left(y^3-3y\right)-35y=y^7-7y^5+14y^3-7y$
I find that the coefficient has some relationship between the Pascal Triangle, such as $y^7-7y^5+14y-7y=y^7-7 \binom{2}{0}y^5+7\binom{2}{1}y^3-7\binom{2}{2}y$. That's strange but funny! However, I can't really prove this, or show that it is false. Hope there is someone who can answer me. Thank you!
|
Krechmar's `A problem book in Algebra' gives an identity if $x+z=p$ and $xz=q$, then
$$x^n+z^n=p^n-\frac{n}{1}p^{n-2} q+\frac{n(n-3)}{1.2}p^{n-4} q^2+...+(-1)^k \frac{n(n-k-1)(n-k-2)....(n-2k+1)}{k!} p^{n-2k} q^k+...$$
Take $z=1/x$ then $q=1$ and letting $x+\frac{1}{x}=y$ we can write
$$f_n(x)=x^n+\frac{1}{x^n}=\sum_{k=0}^{n/2} A_{n,k} ~y^{n-2k},...(1)$$
where $$A_{n,k}=(-1)^k \frac{n(n-k-1)(n-k-2)....(n-2k+1)}{k!}....(2)$$
We can re-write
$A_{n,0}=1$ and $$A_{n,k}=(-1)^k\frac{n}{k}{n-k-1 \choose k-1}, ~ 0<k <n/2 ......(3)$$
Fpr $n=5$ we get $A_{5,0}=1, A_{5,1}= -5, A_{5,2}=5.$
For $n=7$ we get $A_{7,0}=1, A_{7,1}=-7, A_{7,2}= 14, A_{7,3}=-7$
For $n=8$, we get $A_{8,0}=1, A_{8,1}=-8, A_{8,2}= 20, A_{8,3}=-16, A_{8,4}=2.$
So $$x^8+1/x^8=y^8-8y^6+20y^4-16y^2+2,~ y=x+1/x.$$
Similarly $$x^9+1/x^9=y^9-9y^7+27y^5-30y^3+9y.$$
Finally, we would like to assert that the expansion (1) along with (2) or (3) is the required generalization which works for both even and odd $n$.
|
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|
Formulas for Sequences Removing Multiples of 2, 3, and 5 First off, I am a programmer so please excuse if some of the terms I use are not the correct mathematical terms. I was working on devising a function to improve one of my prime number generation algorithms. With this in mind, I first set out to find the formulas for a sequence removing multiples of 2 and 3:
\begin{array}{c|c}
x&y\\ \hline
0&5\\ \hline
1&7\\ \hline
2&11\\ \hline
3&13\\ \hline
4&17\\ \hline
5&19\\ \hline
6&23\\ \hline
7&25\\ \hline
8&29\\ \hline
\vdots&\vdots
\end{array}
The equations that I came up for for this sequence are as follows:
$$y = 3x + 5 - x \bmod 2$$
$$x = \left\lfloor\frac{y - 5 + [y \bmod 3 \neq 0]}{3}\right\rfloor$$
After this, I tried to do the same for a sequence removing multiples of 2, 3, and 5:
\begin{array}{c|c}
x&y\\ \hline
0&7\\ \hline
1&11\\ \hline
2&13\\ \hline
3&17\\ \hline
4&19\\ \hline
5&23\\ \hline
6&29\\ \hline
7&31\\ \hline
8&37\\ \hline
9&41\\ \hline
10&43\\ \hline
11&47\\ \hline
12&49\\ \hline
13&53\\ \hline
\vdots&\vdots
\end{array}
While I think I found an equation to get $y$ from a value of $x$, I cannot find a way to get the value of $x$ from a given value $y$.
$$y = 4x + 7 - 2\left\lfloor\frac{1}{8}x\right\rfloor - 2\left[\{2, 3, 6\} \ \text{contains}\ (x \bmod 8)\right] - 4\left[\{4, 5, 7\} \ \text{contains}\ (x \bmod 8)\right]$$
$$x =\ ?$$
I am wondering if an equation that produces the corresponding value of $x$ for a given value of $y$ for the aforementioned sequence exists, and if indeed it does, what the equation is.
|
This isn’t exactly what you’re looking for, but it’s definitely relevant, and too long for a comment. The formula
$$\bigg\lfloor\frac{n}{2}\bigg\rfloor + \bigg\lfloor\frac{n}{3}\bigg\rfloor + \bigg\lfloor\frac{n}{5}\bigg\rfloor - \bigg\lfloor\frac{n}{6}\bigg\rfloor - \bigg\lfloor\frac{n}{10}\bigg\rfloor - \bigg\lfloor\frac{n}{15}\bigg\rfloor + \bigg\lfloor\frac{n}{30}\bigg\rfloor $$
is equal to the number of positive integers less than or equal to $n$ that aren’t divisible by $2,3,$ or $5$. Perhaps you can use this to find an answer to your question?
|
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|
Given $p^2+q^2+r^2+1=s+\sqrt{p+q+r-s}$, solve for $s$
The real numbers $p$, $q$, $r$, $s$ satisfy
$$p^2+q^2+r^2+1=s+\sqrt{p+q+r-s}$$
Find the value of $s$.
I don't even know how to start. Thanks!
|
Let $t = \sqrt{p+q+r-s}$, the equation is equivalent to
$$\begin{align} & p^2+q^2+r^2+1 = (p+q+r-t^2) + t\\
\iff & \left(p - \frac12\right)^2+\left(q - \frac12\right)^2+\left(r - \frac12\right)^2 + \left(t - \frac12\right)^2 = 0\\
\implies & p = q = r = t = \frac12\\
\implies & s = p + q + r - t^2 = \frac54\end{align}$$
|
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If $a$, $b$, $c$ are three positive integers such that $a^3+b^3=c^3$ then one of the integer is divisible by $7$ Let on contrary that none of the $a$, $b$, $c$ is divisible by $7$. Then either $a^3\equiv b^3\pmod{7}$ or $b^3\equiv c^3\pmod{7}$ or $c^3\equiv a^3\pmod{7}$.
Now how to go further?
|
$7$ is prime. This implies that the multiplicative group $\mathbb{Z}_7^*$ is cyclic and of order $6$. It follows that the image of the action $x\mapsto x^3$ is a subgroup of order $6/3=2$. So there are only two values for $x^3$ mod $7$.
(Plus a third value, $0$, since $0$ was excluded from $\mathbb{Z}_7^*$. But we are assuming none of the three variables are $0$ mod $7$.)
It's easy to see those values are $1$ and $-1\equiv6$. Now, can you solve this equation mod $7$: $$A+B\equiv C$$ using only the values $1$ and $-1$?
In general, this fact about how powers of integers mod $p$ can sometimes only take certain values can be useful. For example, consider $a^5+b^5=c^5$ and look at $11$ in place of $7$. It's the same situation, since $5$ divides $11-1$.
Or consider $a^4+b^4=c^{12}+7$ If you look at that one mod $13$, the $12$th power can only be $0$ or $1$. But the $4$th powers are so restricted that they cannot sum to $7$ or $8$.
|
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|
Solutions to $\left( \frac{1+3x}{1+2x} \right)^{\frac{1+x}{x}}=2$? I am looking for all non-negative real solutions to
$$
\left( \frac{1+3x}{1+2x} \right)^{\frac{1+x}{x}}=2.
$$
Numerically it seems that there is a unique solution $x \approx 0.4256$. Any ideas how to prove/find it?
|
Starting from Andrew Chin's answer and taking logarithms, we need to solve
$$(A-1) \log \left(1+\frac{1}{A}\right)=\log(2)$$
Building the simple $[1,1]$ Padé approximant around $A=4$ gives
$$(A-1) \log \left(1+\frac{1}{A}\right)\sim\frac{3 \log \left(\frac{5}{4}\right)+\frac{ \left(18+800 \log
^2\left(\frac{5}{4}\right)-201 \log \left(\frac{5}{4}\right)\right)}{40 \left(20
\log \left(\frac{5}{4}\right)-3\right)}(A-4) } {1+\frac{13 }{40 \left(20 \log \left(\frac{5}{4}\right)-3\right)}(A-4) }$$ Then one linear equation in $A$ the approximate solution of which being
$$A=4+\frac{40 \log \left(\frac{128}{125}\right) \left(20 \log
\left(\frac{5}{4}\right)-3\right)}{18+\log \left(\frac{5}{4}\right) \left(800
\log \left(\frac{5}{4}\right)-201\right)-13 \log (2)}\approx 4.349417$$ while the "exact" solution is $\approx 4.349409$.
Then $x$.
We could go further building, around $A=4$, the $[1,n]$ Padé approximant of
$$(A-1) \log \left(1+\frac{1}{A}\right)-\log(2)$$ Setting the numerator equal to $0$ would give exact (but nasty) expressions. The numerical values are reported below
$$\left(
\begin{array}{cc}
n & A_{(n)} \\
0 & 4.324246310 \\
1 & 4.349417080 \\
2 & 4.349408445 \\
3 & 4.349409126 \\
4 & 4.349409072 \\
5 & 4.349409076
\end{array}
\right)$$
Edit
It could be interesting to notice that, assuming that $A$ is large
$$(A-1) \log \left(1+\frac{1}{A}\right)=1-\frac{3}{2 A}+\frac{5}{6 A^2}+O\left(\frac{1}{A^3}\right)$$ which gives as an estimate
$$A=\frac{9+\sqrt{3 (40 \log (2)-13)}}{12 (1-\log (2))} \approx 4.24922$$
|
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|
how can I find the Side length Two squares inside an equilateral Triangle?
Question: Figure shows an equilateral triangle with side length equal to $1$ . Two squares of side length a and $2a$ placed side by side just fit inside the triangle as shown.
Find the exact value of $a$.
Its an Assessment question from edX course "A-Level Mathematics Course 1" and I am supposed to use skills that I learnt in Indices and surds,Inequalities and The Factor Theorem.
I have tried finding the height of triangle and then use similar triangles to find the right triangle length still No luck.
I am just looking for food for thought or very small hints thats all.
|
I struggled with this too. But the info is along the bottom. The triangle is equilateral so all angles are $60°$. On the left there is a right-angled triangle - let's call its base $x$.
Triangle 1: Angle = $60°$, opposite = $a$, and adjacent = $x$
On the right there is another right-angled triangle and its base is $1-3a-x$.
Triangle 2: Angle = $60°$, opposite = $2a$, and adjacent = $1 - 3a - x$
From triangle 1:
$$\tan60° = \frac{a}{x} \implies x = \frac{a}{\sqrt{3}}$$
From triangle 2:
$$\tan60° = \frac{2a}{1-3a-x}$$
Equalising $\tan60°$ and substituting for $x$:
$$\frac{2a}{1-3a-x} = \frac{a}{x} \implies \frac{2a}{1-3a-\frac{a}{\sqrt{3}}} = \frac{a}{\frac{a}{\sqrt{3}}}$$
Simplify:
$$\frac{2}{1-3a-\frac{a}{\sqrt{3}}} = \frac{\sqrt{3}}{a}$$
$$\frac{2a}{1-3a-\frac{a}{\sqrt{3}}} = \sqrt{3}$$
$$2a = \sqrt{3}(1-3a-\frac{a}{\sqrt{3}})$$
$$3a +3a\sqrt{3}= \sqrt{3}$$
$$a= \frac{\sqrt{3}}{(3 +3\sqrt{3})}$$
Rationalise the Denominator
$$a= \frac{\sqrt{3}(3 -3\sqrt{3})}{(3 +3\sqrt{3})(3 -3\sqrt{3})}$$
$$a= \frac{3 -\sqrt{3}}{6}$$
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|
Equation of circle touching three circles, two of which are intersecting Find the equation of the circle which is tangentially touching three given circles: $x^2+y^2=49$, $x^2+(y-3.5)^2=49/4$, and $y^2+(x-3.5)^2=49/4$.
By tangentially i mean, it touches the smaller two circle externally and the larger one internally.
The problem would have been much easier had the three given circles been tangent to each other but the smaller two of them intersect, making finding the radius of the circle in question much difficult for me.
Well, I should write the general equation of the circle and equate sum of radius with the distance between the centre with the three given equations.
I am not getting the correct answer with this approach.
Please help.
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Say the center of the circle is at $(h,k)$, and it’s radius is $r$.
From the fact that it is touching the “smaller” circles, the distances between its center and their centers are equal to the sum of the radii of the two chosen circles.
This gives us the following equations:
$$\Bigl(h-\frac 72\Bigr)^2 + k^2 = \Bigl(r + \frac 72\Bigr)^2…(1)$$
$$h^2 + \Bigl(k-\frac 72\Bigr)^2 = \Bigl(r + \frac 72\Bigr)^2…(2)$$
Solving them simultaneously gives us this relation:
$$h=k…(3)$$
Now, since it touches the “large” circle internally, the distance between their centers is the difference between their radii.
So, we have
$$h^2 + k^2 = (7-r)^2…(4)$$
Solving $(2), (3) and (4)$, it is now a trivial task to find the required circles:
$$\left(x-\frac{14}{3\sqrt{2}-1}\right)^2+\left(y-\frac{14}{3\sqrt{2}-1}\right)^2=\left(\frac{7\left(\sqrt{2}-1\right)}{3\sqrt{2}-1}\right)^2$$
and
$$\left(x+\frac{14}{3\sqrt{2}+1}\right)^2+\left(y+\frac{14}{3\sqrt{2}+1}\right)^2=\left(\frac{7\left(\sqrt{2}+1\right)}{3\sqrt{2}+1}\right)^2$$
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If $a_{k}=2^{2^k}+2^{-2^k}$ then evaluate $\prod_{k=1}^\infty\left(1-\frac{1}{a_{k}}\right)$ If $$a_{k}=2^{2^k}+2^{-2^k}$$ then evaluate $$\prod_{k=1}^\infty\left(1-\frac{1}{a_{k}}\right)$$
I tried using Sophie-Germaine Identity about factorisation for $x^4+4$ but it did not work
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Let $2^{2^{-k}}=b, 2^{2^k}=\dfrac1b $
$1-\dfrac1{a_k}=\dfrac{b^2-b+1}{b^2+1}$
$a_{k+1}=(2^{2^k})^2+(2^{2^{-k}})^2=b^2+\dfrac1{b^2}=\dfrac{b^4+1}{b^2}$
$1-\dfrac1{a_{1+k}}=\dfrac{b^4-b^2+1}{b^4+1}$
Observe that $$(b^2-b+1)(b^2+b+1)=(b^2+1)^2-b^2=b^4+b^4+1$$
and $$(1+b^2)(1-b^2)=1-b^4$$
$$\implies\prod_{m=k}^n\left(1-\dfrac1{a_m}\right)=\dfrac{1-b^2}{1+b+b^2}\cdot\dfrac{1-b^{2^n}+b^{2^{n+1}}}{1-b^{2^{n+1}}}$$ as for $1-b^2\ne0$
Observe that $n\to\infty, b^{2^n}=0$
Here $k=1,4b=1$
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Integration of a rational function involving a quadratic
Evaluate the integral: $$\int\frac{3x}{(1-4x-2x^2)^2}\ dx$$
Here is my work:
*
*Complete the square on the denominator: $$(1-4x-2x^2)=(1-2(2x+x^2+1-1))=(3-2(x+1)^2)$$
*Insert back into denominator. Use substitution $u=x+1$; $x=u-1$; $dx=du$. $$\int\frac{3(u-1)}{(3-2u^2)^2}\ du$$
*Split the integral into two: $$\int\frac{3u}{(3-2u^2)^2}\ du\ -\int\frac3{(3-2u^2)^2}\ du$$
*Integrate the first integral: $$\int\frac{3u}{(3-2u^2)^2}\ du$$
$v=(3-2u^2)$
$\frac{dv}{du}=-4u$
$\frac{-dv}4=u\ du$
$$\frac{-1}4\int\frac{3}{v^2}\ dv=\frac{-3}{4}\int\frac{1}{v^2}\ dv=\frac1{4v}$$$$=\frac{1}{4(3-2u^2)}=\frac1{4(3-2(x+1)^2)}$$
*
*Integrate the second integral: $$-\int\frac3{(3-2u^2)^2}\ du$$
Use sine trig substitution: $u=\frac{\sqrt3}{\sqrt2}\sin\theta$; $\frac{\sqrt2}{\sqrt3}du=\cos\theta\ d\theta$
$$-\frac{\sqrt2}{\sqrt3}\int\frac{3\cos\theta}{(3-2(\frac{\sqrt3}{\sqrt2}\sin\theta)^2)^2}\ d\theta=-\frac{\sqrt2}{\sqrt3}\int\frac{3\cos\theta}{(3-2(\frac{3}{2}\sin^2\theta))^2}\ d\theta$$
$$=-\frac{\sqrt2}{\sqrt3}\int\frac{3\cos\theta}{(3-3\sin^2\theta))^2}\ d\theta=-\frac{\sqrt2}{3\sqrt3}\int\frac{\cos\theta}{(\cos^2\theta))^2}\ d\theta=-\frac{\sqrt2}{3\sqrt3}\int\frac{1}{(\cos^3\theta)}\ d\theta$$$$=-\frac{\sqrt2}{3\sqrt3}\int\sec^3\theta\ d\theta=\frac{\sqrt2}{3\sqrt3}\sec\theta\tan\theta+\ln|\sec\theta+\tan\theta|$$
Besides for the last resubstitution which I still have to do, I also seem to have a different leading fraction for my second integral than the answer key. Where did I go wrong?
|
Hint:
This formula should be known by heart:
$$\int\!\frac {\mathrm d x}{a^2-x^2}=\frac 1a\,\operatorname{argtanh}\left(\frac xa\right)=\frac 1{2a} \ln\left(\frac{a+x}{a-x}\right).$$
(Similar to $\;\displaystyle\int\!\frac {\mathrm d x}{a^2+x^2}=\frac 1a\,\arctan\left(\frac xa\right) )$
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If $n$ numbers are generated, what is the probability that the product of all those numbers is a multiple of 10? A computer generates random numbers from the set $\{1,2,3,4,5,6,7,8,9\}$ (each has equal probability). If $n$ numbers are generated (with replacement), what is the probability that the product of all those numbers is a multiple of 10?
Attempt:
Let us see easier case, when $n=4$. The total number of possible random numbers can be counted as the number of non-negative solutions to
$$ x_{1} + x_{2} + ... + x_{8} + x_{9} = 4$$
This is because, for example: $\{1,2,3,9\}$ can be seen as $x_{1}=x_{2}=x_{3}=x_{9}=1$ and all others are 0, then again $\{5,5,2,4\}$ as $x_{5}=2, x_{2}=x_{4}=1$ and the others 0. The number of solutions is $\binom{8+4}{4}$.
Now we have to count the total number of possibilities when the product of the generated numbers is multiple of 10. If it is multiple 10, then it must contain a 5 and at least one from $\{2,4,6,8\}$. We can count by the following:
*
*2 generated numbers are $5$ and $2$, with no restriction for the other 2. (note that this can contain $4,6,8$). This will be the same as counting nonnegative solutions: $x_{1} + ... + x_{9} = 2$. Count is $\binom{8+2}{8}$
*2 generated numbers are $5$ and $4$, but this time there is restriction for the other 2: it cannot contain 2 because this was counted earlier. This is the number of nonegative solutions for: $x_{1} + x_{3} + ... + x_{9} = 2$. Count is $\binom{7+2}{7}$
*2 generated numbers are $5$ and $6$, but restriction for the other 2: cannot contain a 2 or a 4. Similarly we consider: $x_{1} + x_{3} + x_{5} + ... +x_{9} = 2$. Count is $\binom{6+2}{6}$
*2 generated numbers are $5$ and $8$, restriction for other 2: cannot contain a 2, 4, or 6. Similarly consider: $x_{1} + x_{3} + x_{5} + x_{7} + x_{8} + x_{9} = 2$. Count is $\binom{5+2}{5}$
So when $n=4$, the probability of interest is $\frac{\binom{8+2}{8} + \binom{7+2}{7} + \binom{6+2}{6} + \binom{5+2}{5}}{\binom{8+4}{8}}$
I want to double check my approach, and also looking for better answers.
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You are assuming that getting $2,4,3,5$ is the exact same as getting $5,2,4,3$ and that somehow the probability of getting $\{2,3,4,5\}$ in any of the 24 orders is the exact same probability of getting $\{3,3,4,5\}$ any of its 12 orders, or getting $\{4,4,4,\}$ in its one order.
You are assuming that getting a set $\{2,3,4,5\}$ is equally likely as getting a set $\{5,5,5,5\}$. And therefore you are counting (with great difficulty) the ways to get different sets of numbers.
But each order is as equally likely as any other as there are $24$ ways to order $\{2,3,4,5\}$ but only $12$ ways to order $\{3,3,4,5\}$ there is no reason at all do do this.
Each $(a_1,a_2,a_3,a_4)$ tuple counts and each order is a seperate option.
There are $9^4$ possible tuples.
That's it.
So there are $9^n$ ways to pick $n$ numbers.
If you have at least one even and at least one $5$ you will have a multiple of $10$.
There are $4^n$ ways to have no evens or $5$s at all. And there are $5^n$ ways to have no evens but maybe or maybe not have $5$s. ANd there are $8^n$ ways to have no $5$s but maybe or maybe not have evens.
By inclusion exclusion:
Number of ways to not have a even and a 5= Number of ways to not have an even + Number of ways to not have a 5 - (due to double counting) Number of ways to not have both=
$5^n + 8^n - 4^n$.
So probability of not being a multiple of $10$ is $\frac {5^n + 8^n - 4^n}{9^n}$
So probability of being a multiple of $10$ is $1 - \frac {5^n + 8^n - 4^n}{9^n}$
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Calculate the coordinates of points on a circle in 3D space Finding the $(x,y)$ coordinates of points along the circumference of a circle in 2D space is fairly easy.
x = r * cos() + Xc
y = r * sin() + Yc
r = radius of circle
(Xc, Yc) = coordinates of circle center
= current angle
I am looking for similar equations for a circle in 3D space. Keep in mind I haven't had geometry in 20 years and we didn't cover 3D geometry. So please go easy on me and explain any notation you use.
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Think on the intersection between an sphere $x^2+y^2+z^2= r^2$ and a plane $a x + b y + c z = 0$
Solving for $x,y$ we get
$$
x = \frac{-a c z\mp\sqrt{b^2 \left(r^2 \left(a^2+b^2\right)-z^2 \left(a^2+b^2+c^2\right)\right)}}{a^2+b^2}\\
y = \frac{-b^2 c z\pm a \sqrt{b^2 \left(r^2
\left(a^2+b^2\right)-z^2 \left(a^2+b^2+c^2\right)\right)}}{b \left(a^2+b^2\right)}
$$
now $z$ is such that $r^2 \left(a^2+b^2\right)-z^2 \left(a^2+b^2+c^2\right) \ge 0$ or
$$
-\frac{r\sqrt{a^2+b^2}}{\sqrt{a^2+b^2+c^2}}\le z \le \frac{r\sqrt{a^2+b^2}}{\sqrt{a^2+b^2+c^2}}
$$
so finally we have a parametric representation
$$
x = x(z)\\
y = y(z)\\
$$
for
$$
-\frac{r\sqrt{a^2+b^2}}{\sqrt{a^2+b^2+c^2}}\le z \le \frac{r\sqrt{a^2+b^2}}{\sqrt{a^2+b^2+c^2}}
$$
NOTE
we assumed $a,b,c$ all non null. The cases in which with one of them, or two are null, are trivial because the plane equation is simpler.
Attached the solution for $a=1,b=1,c=1,r=1$ and $-\sqrt{\frac 23}\le z\le \sqrt{\frac 23}$
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Given three positive numbers $a,b,c$. Prove that $\sum\limits_{sym}\frac{a+b}{c}\geqq 2\sqrt{(\sum\limits_{sym}a)(\sum\limits_{sym}\frac{a}{bc}})$ .
(A problem due to Mr. Le Khanh Sy). Given three positive numbers $a, b, c$. Prove that
$$\sum\limits_{sym}\frac{a+ b}{c}\geqq 2\sqrt{(\sum\limits_{sym}a)(\sum\limits_{sym}\frac{a}{bc}})$$
I'm eagerly interested in learning one method which assumes $c\not\equiv {\rm mid}(\!a, b, c\!)$. But if $c\equiv {\rm mid}(\!a, b, c\!)$:
$$2\sqrt{(\!\frac{a}{bc}+ \frac{b}{ca}+ \frac{c}{ab}\!)(\!a+ b+ c\!)}\leqq c(\!\frac{a}{bc}+ \frac{b}{ca}+ \frac{c}{ab}\!)+ \frac{a+ b+ c}{c}= \frac{a+ b}{c}+ \frac{a}{b}+ \frac{b}{a}+ 1+ \frac{c^{2}}{ab}$$
We need to prove
$$\begin{align} \frac{a+ b}{c}+ \frac{a}{b}+ \frac{b}{a}+ 1+ \frac{c^{2}}{ab}\leqq \frac{a+ b}{c}+ \frac{b+ c}{a} & + \frac{c+ a}{b}\Leftrightarrow 1+ \frac{c^{2}}{ab}\leqq \frac{c}{a}+ \frac{c}{b}\Leftrightarrow \\ & \Leftrightarrow (\frac{c}{a}- 1)(\frac{c}{b}- 1)\leqq 0\Leftrightarrow \frac{(c- a)(c- b)}{ab}\leqq 0 \end{align}$$
Who can teach me what would we do if $c\not\equiv {\rm mid}(\!a, b, c\!)$ ? I am goin' to set a bounty, thank u so much
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After squaring of the both sides we need to prove that
$$\sum_{sym}(a^4b^2-a^4bc+a^3b^3-2a^3b^2c+a^2b^2c^2)\geq0,$$ which is true by Muirhead and Schur.
|
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Solving $3\sin(2x+45^\circ)=2\cos(x+135^\circ)$ for $x$ between $0^\circ$ and $360^\circ$ Please find the value of $x$ in degree from this equation, with explanation
$$3\sin(2x+45^\circ)=2\cos(x+135^\circ)$$
For $x$ between $0^\circ$ and $360^\circ$.
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Your simplification of
$$3\sin(2x+45^\circ)=2\cos(x+135^\circ)$$
obtaining the left hand side of
$$3\sin2x\cos 45^\circ+3\cos 2x\sin 45^\circ =2\cos x\cos135^\circ-2\sin x\sin135^\circ$$
was very well done.
In fact, $\cos135^\circ=-\cos45^\circ$ and $\sin135^\circ=\sin45^\circ$ and so you can cancel down to
$$3\sin2x+3\cos 2x=-2\cos x-2\sin x$$
Using the double angle formulae you mention gives
$$6\sin x\cos x+6\cos^2 x-3=-2\cos x-2\sin x.$$
Check that you can follow these steps and that should suffice for your tutor until more work has been done on this topic.
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Differentiation using l´Hopital I need to use L´Hopital's rule with this functions:
$$\lim_{x\rightarrow\frac{\pi}{2}} (1-\sin(x))^{\cos(x)}$$
$$\lim_{x\rightarrow\frac{\pi}{4}} (\tan(x))^{\tan(2x)}$$
I take the exponent down using the properties of logarithms and then make the denominator like: $\lim_{x\rightarrow\frac{\pi}{2}} \frac{\cos(x)}{\frac{1}{\ln(1-\sin(x))}}$ but I still get stuck.
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Without using L'Hospital:
$$\begin{align}1) \ \lim_{x\rightarrow\frac{\pi}{2}} (1-\sin(x))^{\cos(x)}&=\lim_{x\rightarrow\frac{\pi}{2}} (1-\sin(x))^{\cos(x)}\cdot 1=\\
&=\lim_{x\rightarrow\frac{\pi}{2}} (1-\sin(x))^{\cos(x)}\cdot \lim_{x\rightarrow\frac{\pi}{2}} (1+\sin(x))^{\cos(x)}=\\
&=\lim_{x\rightarrow\frac{\pi}{2}} (1-\sin^2(x))^{\cos(x)}=\\
&=\lim_{x\rightarrow\frac{\pi}{2}} (\cos^2(x))^{\cos(x)}\stackrel{(1)}=\\
&=1.\\
2) \ \lim_{x\rightarrow\frac{\pi}{4}} (\tan(x))^{\tan(2x)}&=\lim_{x\rightarrow\frac{\pi}{4}} (1+\tan(x)-1)^{\frac{2\tan(x)}{(1-\tan x)(1+\tan x)}}=\\
&=\lim_{x\rightarrow\frac{\pi}{4}} (1+\tan(x)-1)^{-\frac{1}{\tan x-1}}\stackrel{(2)}=\\
&=e^{-1}.\end{align}$$
Note:
$$\begin{align}\lim_\limits{x\to 0} x^x&=1 \quad (1)\\
\lim_\limits{x\to 0} (1+x)^{1/x}&=e \quad (2) \end{align}$$
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|
Differentiating $f(x)=\frac{3(1-\sin x)}{2\cos x}$ I'm having trouble to solve the following derivative. I tried to apply the quotient rule, but I cannot get the same result as my textbook.
The function:
$$f(x)=\frac{3(1-\sin x)}{2\cos x}$$
The result I supposed to get:
$$f'(x)=\frac{3}{2}\sec x(\tan x-\sec x)$$
I have tried this by applying the quotient rule:
$$f(x) =\frac{3(1-\sin x)}{2\cos x}=\frac{3-3\sin x}{2\cos x}$$
\begin{align*}
f'(x) & =\frac{-3\cos x \cdot 2\cos x - (3- \sin x) \cdot -2\sin x}{2\cos^2x}\\
& =\frac{-6\cos^2x-(-6\sin x+6\sin^2x)}{2\cos^2x}\\
& =\frac{-6\cos^2x+6\sin x+6\sin^2x}{2\cos^2x}
\end{align*}
Is someone able to show me step by step how to get the above result?
Thank you in advance for all of your help!
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$$f(x)=\frac{3(1-\sin x)}{2\cos x} =\frac {3}{2} (\frac {1-\sin x}{\cos x})$$
Apply the quotient rule $$ f'(x) =\frac {3}{2}( \frac {-\cos x (\cos x)+\sin x (1-\sin x)}{\cos ^2 x})=$$
$$\frac {3}{2}( \frac {\sin x -1}{\cos ^2 x})=$$
$$\frac {3}{2}( \frac {1}{\cos x} \frac {\sin x -1}{\cos x})=$$
$$\frac {3}{2} \sec x(\tan x - sec x)$$
|
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Find the thousandth number in the sequence of numbers relatively prime to $105$.
Suppose that all positive integers which are relatively prime to $105$ are arranged into a increasing sequence: $a_1 , a_2 , a_3, . . . .$ Evaluate $a_{1000}$
By inclusion exclusion principle I found the following equation:
$n - \left( \left\lfloor \dfrac {n} {3} \right\rfloor + \left\lfloor \dfrac {n} {5} \right\rfloor + \left\lfloor \dfrac {n} {7} \right\rfloor \right) + \left( \left\lfloor \dfrac {n} {15} \right\rfloor + \left\lfloor \dfrac {n} {21} \right\rfloor + \left\lfloor \dfrac {n} {35} \right\rfloor \right) - \left\lfloor \dfrac {n} {105} \right\rfloor=1000$
I tried to solve the equation by setting $n=105k+r ; 0 \leq 0<105$. But this method is so tiring and lengthy to solve. I thought to invoke inequalities like $a-1 <\lfloor a \rfloor \leq a$, but I couldn't do as the equation contains positive as well as negative terms .Please provide me any other method to solve such equations involving floor function. Any help would be appreciated.
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The number of integers $n \in \{1,2,3,\dots, 105\}$ that are relatively prime to $105$ is
$$\phi(105) = \phi(3 \times 5 \times 7) = 2 \times 4 \times 6 = 48$$
We know that $\gcd(105A + n, 105) = \gcd(n, 105)$.
Since $1000 = 20\times 48 + 40$, we know that there are $20 \times 48 = 960$ numbers relatively prime to $105$ in the interval $[1, 2100]$ (where $2100 = 20 \times 105$).
$86$ is the $40^{\text{th}}$ number in that range that is relatively prime to $105$.
So the thousandth number relatively prime to $105$ is $2100 + 86 = 2186$.
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Given $0 < x \le 1 \le y$. Calculate the minimum value of $\frac{x}{y + 1} + \frac{y}{x + 1} + \frac{x^2 + 3y^2 + 2}{6xy(xy + 1)}$.
Given $0 < x \le 1 \le y$. Calculate the minimum value of $$\large \dfrac{x}{y + 1} + \dfrac{y}{x + 1} + \dfrac{x^2 + 3y^2 + 2}{6xy \cdot (xy + 1)}$$
We have that $$\frac{x^2 + 3y^2 + 2}{6xy \cdot (xy + 1)} \ge \left[ \begin{align} \frac{2 \cdot (x + 3y + 2)^2}{12xy \cdot (6xy + 6)}\\ \frac{2 \cdot (2xy + 4y)}{12xy \cdot (xy + 1)} \end{align} \right. = \frac{8 \cdot y(\sqrt{x} + 2)^2}{9 \cdot (3xy + 1)^2}$$
But that's all I have.
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Show that $$\frac{x}{y+1}+\frac{y}{x+1}+\frac{x^2+3y^2+2}{6xy(xy+1)}\geq \frac{3}{2}$$
The equal sign holds if $x=y=1$
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Roots in equation In the equation
$\sqrt{x-2} - \sqrt{6x-11} + \sqrt{x+3} =0$
I got the roots of $x$ being $6$ and $7\sqrt{3}$.
Considering the graph shows only $6$ as being a valid solution, how should I go as figuring this out in the equation itself?
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Let $x-2=p^2,x+3=q^2;p,q\ge0$
As $x+3>x-2, \dfrac pq<1\ \ (1)$
$6x-11=a(x-2)+b(x+3)$
$x=2\implies1=5b$
$x+3=0\implies6(-3)-11=a(-3-2)$
$$\implies p+q=\sqrt{\dfrac{q^2+29p^2}5}$$
$$25(p+q)^2=q^2+29p^2$$
$$2p^2-25pq+12q^2=0$$
$$\dfrac pq=\dfrac{25\pm\sqrt{25^2-4\cdot2\cdot12}}{2\cdot2}=\dfrac{25\pm23}4$$
$\implies \dfrac pq =\dfrac12$ by $(1)$
Now take square in both sides and replace the values of $p,q$
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Geometry Rotation and Trigonometry
$A, B, C$ and $D$ are on a line such that $AB=BC=CD$. Also, $P$ is a point on a circle with $BC$ as a diameter. Find $\tan\angle{APB} \cdot \tan\angle{CPD}$.
Let $O$ be the center of $(BPC)$. Let $P$' be the point of intersection of $PO$ and $(BOC)$ again. Then $C$ is the centroid of $PP'D$ since $OD$ is a median and $CD=2OD$. By symmetry,
$$\tan\angle{APB} \cdot \tan\angle{CPD}=\tan\angle{CP'D} \cdot \tan\angle{CPD}$$
This all looks promising and hopefully, it will help (or maybe it is misleading).
Thanks!
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Let $A\equiv(0,0),B\equiv(1,0),C\equiv(2,0)$ and $D\equiv(3,0)$. Also let $\angle PBC=\theta$
$PB=\cos \theta$. Also, slope of $BP=\tan\theta$.
So, co-ordinates of $P\equiv(1+\cos^2\theta,\sin\theta \cos\theta)$
So, slope of $AP=\frac{\tan\theta}{2+\tan^2\theta}$
$tan\alpha=\frac{\tan\theta-\frac{\tan\theta}{2+\tan^2\theta}}{1+\frac{tan^2\theta}{2+\tan^2\theta}}=\frac{\tan\theta}{2}$
Slope of $CP=-\cot\theta$
$$Slope(PD)=\frac{\sin\theta\cos\theta}{\cos^2\theta-2}=-\frac{\tan\theta}{1+2\tan^2\theta}$$
similarly $\tan\beta=\frac{1}{2\tan\theta}$
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Markov Chain: Calculating Expectation Reach a Certain Set of States Suppose I have a Markov chain $Z_k$ with $6$ states, as depicted below:
The probability of moving from one node to a neighboring node is $1/2$. For example, the probability of moving from node $1$ to node $2$ is $1/2$ and the probability of moving from node $1$ to node $6$ is $1/2$ etc.
Suppose $P(Z_0=1)=1$. That is we start state $1$. We need to compute two things:
*
*Compute the expected time we first reach the bottom of our pyramid (states $3$, $4$, or $5$). That is compute $E[T_B]$ where $T_B=min\{j: Z_j \in \{3,4,5\}\}$
My attempt: I try listing out all the possibilities
I can go from:
*
*I can go from State $1-2-3$. Time is $2$ when base is reached and probability $\frac{1}{2} \cdot \frac{1}{2}$
*$1-2-1-2-3$. Time is $4$ when base is reached. Probability of occurring is $\frac{1}{16}$.
*$1-2-1-2-1-2-3$. Time is $6$ when base is reached. Probability of occurring is $\frac{1}{64}$
*$1-2-1-2-1-2-1-2-3$. Time is $8$ when base is reached. Probability of occurring is $\frac{1}{64}$.
etc...
Thus my conclusion for these types of sequences expected value is:
$2 \cdot \frac{1}{4}+ 4*\frac{1}{16}+6*\frac{1}{64}+8*\frac{1}{256} ...$
$\sum_{k=1}^{\infty} 2k \cdot (\frac{1}{2})^{2k}=\frac{8}{9}$
*Now, I can also go from State $1-6-5$. Time is $2$ when base is reached and probability $\frac{1}{2} \cdot \frac{1}{2}$
*$1-6-1-6-5$. Time is $4$ when base is reached. Probability of occurring is $\frac{1}{16}$.
*$1-6-1-6-1-6-5$. Time is $6$ when base is reached. Probability of occurring is $\frac{1}{64}$
*$1-6-1-6-1-6-1-6-5$. Time is $8$ when base is reached. Probability of occurring is $\frac{1}{64}$.
Thus my conclusion for these types of sequences expected value is:
$2 \cdot \frac{1}{4}+ 4*\frac{1}{16}+6*\frac{1}{64}+8*\frac{1}{256} ...$
$\sum_{k=1}^{\infty} 2k \cdot (\frac{1}{2})^{2k}=\frac{8}{9}$
But there are more possibilities:
*$1-2-1-6-5.$ The time is $4$ probability $1/16$
*$1-2-1-6-1-6-5$. The time is $6$ probability $1/64$
10.$1-2-1-6-1-6-1-6-5 etc..$. The time is $8$ with probability $1/256$.
$\sum_{k=2}^{\infty} 2k \cdot (\frac{1}{2})^{2k}=\frac{7}{18}$
Still more possibilities....:
Too many possibilities (unfortunately gave up) as its like the Markov Chain restarts when we go back to $1$. Couldn't figure it out. Please let me know what I should do and thank you for the help
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For $i\in\{1,\dots,6\}$, let $m_i = E(T_B | Z_t = i)$ be the expected number of steps until reaching $\{3,4,5\}$, starting from state $i$. Trivially, $m_3=m_4=m_5=0$.
To obtain a system of linear constraints, apply first-step analysis (conditioning on the first step out of state $i$). For $i=1$, we have
\begin{align}
m_1 &= \sum_{j\in\{2,6\}} E(T_B | Z_t = 1, Z_{t+1} = j) P(Z_{t+1} = j | Z_t=1)\\
&= \sum_{j\in\{2,6\}} (1 + E(T_B | Z_{t+1} = j)) \frac{1}{2} \\
&= \frac{1}{2}\sum_{j\in\{2,6\}} 1 + \frac{1}{2} \sum_{j\in\{2,6\}} E(T_B | Z_{t+1} = j) \\
&= 1+\frac{1}{2}\sum_{j\in\{2,6\}} m_j.
\end{align}
For $i=2$, we have
\begin{align}
m_2 &= \sum_{j\in\{1,3\}} E(T_B | Z_t = 2, Z_{t+1} = j) P(Z_{t+1} = j | Z_t=2) \\
&= \sum_{j\in\{1,3\}} (1 + E(T_B | Z_{t+1} = j)) \frac{1}{2} \\
&= \frac{1}{2} \sum_{j\in\{1,3\}} 1 + \frac{1}{2} \sum_{j\in\{1,3\}} E(T_B | Z_{t+1} = j) \\
&= 1+\frac{1}{2}\sum_{j\in\{1,3\}} m_j\\
&= 1+\frac{1}{2}(m_1+0).
\end{align}
The $i=6$ case is similar. So we obtain the following linear constraints:
\begin{align}
m_1 &= 1 + (m_2+m_6)/2 \\
m_2 &= 1 + m_1/2 \\
m_6 &= 1 + m_1/2
\end{align}
By substituting the last two constraints into the first one, we obtain
$$m_1 = 1 + (1 + m_1/2 + 1 + m_1/2)/2 = 2 + m_1/2,$$
which implies that $m_1 = 4$.
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$\sqrt{2} +\sqrt[3]{2}$ is irrational
Prove that $\sqrt{2} +\sqrt[3]{2}$ is irrational.
My attempt: Suppose $\sqrt{2} +\sqrt[3]{2}$ is rational then $\exists$ $x\in \mathbb{Q}$ such that $$\sqrt{2} +\sqrt[3]{2}=x$$ Rewriting the above equation as $$x-\sqrt{2}=\sqrt[3]{2}$$ cubing the above equation gives us $$x^3-3x^2\sqrt{2}+6x-2\sqrt{2}=2$$ This implies that $$\sqrt{2}=\frac{x^3+6x-2}{3x^2+2}\in\mathbb{Q}$$ but this is absurd . Therefore ,$\sqrt{2} +\sqrt[3]{2}$ is irrational. Does this look good ? Can this be done any other way.
This question is from chapter 2 of Spivak's Calculus. In there he's given a hint which says "start by working with the 6th power " of this expression.. I don't see how that helps . Can you show me how it can be done using the hint the Author has provided ? Thank you
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The OP shows great ingenuity with their (now fixed up) answer.
You can also go with Spivak's hint:
From Spivak's Exercise 18.a we know that if $u$ satisfies
$\tag 1 u^n + a_{n-1}u^{n-1} + \dots + a_0 = 0$
for integers $a_{n-1}, \dots, a_0$, then $u$ is either irrational or an integer.
Let
$\quad u = 2^{\frac{2}{6}} + 2^{\frac{3}{6}}$
It can be show that $u$ isn't an integer (see next section).
Moreover, it can be shown (by solving a system of linear equations) that
$\tag 2 u^6 -6u^4-4u^3+12u^2 -24u - 4 = 0$
Since
$\quad \sqrt{2} \lt \frac{3}{2}$
$\quad \sqrt[3]{2} \lt \frac{4}{3}$
we can write
$\quad \sqrt{2} + \sqrt[3]{2} \lt \frac{3}{2} + \frac{4}{3} \lt 3$
Since
$\quad \sqrt{2} \gt \frac{5}{4}$
$\quad \sqrt[3]{2} \gt \frac{5}{4}$
we can write
$\quad \sqrt{2} + \sqrt[3]{2} \gt \frac{5}{4} + \frac{5}{4} \gt 2$
So $\sqrt{2} + \sqrt[3]{2}$ can't be an integer.
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|
How to find the basis (and cartesian equation(s)) of a sum of two vectorial subspaces? I have the following vectorial subspaces :
U = span $\left(\begin{pmatrix} 2 \\ 0 \\ 1 \\ -2 \end{pmatrix},\begin{pmatrix} 3 \\ 6 \\ 9 \\ -12 \end{pmatrix} \right)$ and V = span $\left(\begin{pmatrix} 0 \\ 2 \\ 1 \\ 0 \end{pmatrix},\begin{pmatrix} -1 \\ 1 \\ 0 \\ 1 \end{pmatrix} \right)$
To calculate the basis of U + V should have use this method :
$\left(\begin{array}{cccc|c} 2 & 3 &0 & -1& 0 \\ 0 & 6 &2&1 & 0 \\ 1 & 9 &1&0&0 \\ -2 & -12&0&1 & 0 \end{array}\right)$ which generate this solution after reduction $\left(\begin{array}{cccc|c} 1 & 0 &0 & -\frac{1}{2}& 0 \\ 0 & 1 & 0 & 0 & 0 \\ 0 & 0 & 1 & \frac{1}{2} & 0 \\ 0 & 0 & 0 & 0 & 0 \end{array}\right)$ $\sim$ $\begin{pmatrix} 1 & 0 & 0 & -\frac{1}{2} \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & \frac{1}{2} \end{pmatrix}$
Now I don't understand how to have cartesian equation(s) of this hyperplane. Can someone help me understanding this?
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Since it is a subspace, it pass the origin, so has an equation like:
$$ax+by+cz+dt=0$$
Now put $4$ points in that and solve a system of linear equations to find $a,b,c,d$.
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Change of basis linear transformation Let $S : M_{2,2} → M_{2,2}$ be the linear transformation defined by $S(A) = A + A^T$
.
Consider the bases for $M_{2,2}$ given by:
$$B = \left\{\begin{bmatrix} 1&0\\0&0 \end{bmatrix}, \begin{bmatrix} 0&1\\0&0 \end{bmatrix}, \begin{bmatrix} 0&0\\1&0 \end{bmatrix}, \begin{bmatrix} 0&0\\0&1 \end{bmatrix} \right\}$$
and
$$c = \left\{\begin{bmatrix} 1&0\\0&0 \end{bmatrix}, \begin{bmatrix} 2&-1\\0&0 \end{bmatrix}, \begin{bmatrix} 0&0\\1&1 \end{bmatrix}, \begin{bmatrix} -3&0\\0&2 \end{bmatrix} \right\}$$
(a) Determine the transition matrix $P_{B,C}$ from C to B.
(b) Determine the transition matrix $P_{C,B}$ from B to C.
This question is very different from the types I have encountered thus far so please help me solve it, I have been struggling for a few hours now.
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We have that by matrix representation of the same vector $v$ with respect to the two basis
*
*$v=M_Bv_B$
*$v=M_Cv_C$
with
$$M_B=\left(\begin{array} &1 & 0 & 0 & 0 \\ 0 & 1 &0 &0 \\ 0& 0 &1&0\\ 0 & 0 &0 & 1 \end{array} \right), \quad M_C=\left(\begin{array} &1 & 2 & 0 & -3 \\ 0 & -1 &0 &0 \\ 0& 0 &1&0\\ 0 & 0 &1 & 2 \end{array} \right)$$
therefore
$$M_Bv_B=M_Cv_C \implies v_B=M_B^{-1}M_Cv_C \quad \land \quad v_C=M_C^{-1}M_Bv_B$$
that is $P_{C,B}=M_C^{-1}M_B$ and $P_{B,C}=M_B^{-1}M_C$.
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Integrate $\sin^4(x)$ Consider the integral
$$\int \sin^4(x)dx$$
Now I could separate the $\sin^4(x)$ into two $\sin^2(x)$ terms and the use power reducing formula
$$\int \sin^2(x)\sin^2(x)dx $$
$$\int \frac{1-\cos(2x)}{2}*\frac{1-\cos(2x)}{2}dx $$
$$\int \frac{(1-\cos(2x))^2}{4} dx$$
$$\int\frac{1}{4}-\frac{2\cos(2x)}{4}+\frac{\cos^2(x)}{4}dx $$
$$\frac{1}{4} \int1-8\cos(2x)+\cos^2(2x)dx $$
Now would it be good to rewrite that $\cos^2(2x)$ as $1-\sin^2(2x)$ and then integrate or should I also separate the terms into their own integrands?
The answer I am looking for which is $\frac{3}{8}x -\frac{1}{4}\sin(2x)+\frac{1}{32}\sin(4x)+C$.
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Surely your $8$ should be a $2$ (not to mention each $\cos^2x$ should be $\cos^22x$), giving$$\frac14\int(1-2\cos 2x+\cos^22x)dx=\frac18\int(3-4\cos 2x+\cos 4x)dx=\frac{1}{32}(12x-8\sin 2x+\sin 4x)+C.$$
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Compute $\sum_{n=1}^{\infty} \frac{ H_{n/2}}{(2n+1)^3}$ How to prove that
$$S=\displaystyle \sum_{n=1}^{\infty} \frac{ H_{n/2}}{(2n+1)^3} \quad=\quad \frac{\pi^2G}{4}-\frac{21\zeta(3)\ln(2)}{8}+\frac{\pi^4}{64}+\frac{\Psi^{(3)}(\frac{1}{4})}{512}- \frac{\Psi^{(3)}(\frac{3}{4})}
{512}$$
This problem was proposed by @Ahmad Bow but unfortunately it was closed as off-topic and you can find it here.
Any way, I tried hard on this one but no success yet. here is what I did:
Using the identity
$$H_{n/2}=H_n-n\int_0^1 x^{n-1}\ln(1+x)\ dx, \quad x\mapsto x^2$$
$$H_{n/2}=H_n-2n\int_0^1 x^{2n-1}\ln(1+x^2)\ dx$$
We can write
$$S=\sum_{n=0}^\infty\frac{H_n}{(2n+1)^3}-\int_0^1\frac{\ln(1+x^2)}{x}\sum_{n=0}^\infty \frac{2nx^{2n}}{(2n+1)^3}\ dx$$
where
\begin{align}
\sum_{n=0}^\infty \frac{2nx^{2n}}{(2n+1)^3}&=\frac1x\sum_{n=0}^\infty \frac{x^{2n+1}}{(2n+1)^2}-\frac1x\sum_{n=0}^\infty \frac{x^{2n+1}}{(2n+1)^3}\\
&=\frac1{2x}\sum_{n=0}^\infty \frac{x^{n+1}}{(n+1)^2}(1+(-1)^n-\frac1{2x}\sum_{n=0}^\infty \frac{x^{n+1}}{(n+1)^3}(1+(-1)^n\\
&=\frac1{2x}\sum_{n=1}^\infty \frac{x^{n}}{n^2}(1-(-1)^n-\frac1{2x}\sum_{n=1}^\infty \frac{x^{n}}{n^3}(1-(-1)^n\\
&=\frac1{2x}\left(\operatorname{Li}_2(x)-\operatorname{Li}_2(-x)-\operatorname{Li}_3(x)+\operatorname{Li}_3(-x)\right)
\end{align}
Therefore
$$S=\sum_{n=0}^\infty\frac{H_n}{(2n+1)^3}-\frac12\int_0^1\frac{\ln(1+x^2)}{x^2}\left(\operatorname{Li}_2(x)-\operatorname{Li}_2(-x)-\operatorname{Li}_3(x)+\operatorname{Li}_3(-x)\right)\ dx$$
The sum can be done using the following identity
$$ \sum_{n=1}^{\infty} \frac{H_{n}}{ (n+a)^{2}}= \left(\gamma + \psi(a) \right) \psi_{1}(a) - \frac{\psi_{2}(a)}{2} \, , \quad a >0.$$
Differentiate both sides with respect to $a$ then set $a=1/2$ we get
$$\sum_{n=0}^\infty\frac{H_n}{(2n+1)^3}=\frac{45}{32}\zeta(4)-\frac74\ln2\zeta(3)$$
and the question here is how to calculate the the remaining integral or a different way to tackle the sum $S$ ? Thanks
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Cornel's way to make it easy. Replace the harmonic number in the numerator by Digamma function, using that $H_{n/2}= \psi(n/2+1)+\gamma$, and then splitting the series using the parity, we have
$$ S=\sum_{n=1}^{\infty} \frac{ \psi(n/2+1)+\gamma}{(2n+1)^3}=\sum_{n=1}^{\infty} \frac{ \psi(n+1)+\gamma}{(4n+1)^3}+\sum_{n=1}^{\infty} \frac{ \psi(n+1/2)+\gamma}{(4n-1)^3}$$
$$=\sum_{n=1}^{\infty} \frac{H_n}{(4n+1)^3}+\sum_{n=1}^{\infty} \frac{ 2H_{2n}-H_n-2\log(2)}{(4n-1)^3}$$
$$=\sum_{n=1}^{\infty} \frac{H_n}{(4n+1)^3}-\sum_{n=1}^{\infty} \frac{H_n}{(4n-1)^3}-2\log(2)\sum_{n=1}^{\infty} \frac{1}{(4n-1)^3}+2\sum_{n=1}^{\infty} \frac{H_{2n}}{(4n-1)^3},$$
and since the first two series are straightforward using Cornel's Master Theorem of Series from A master theorem of series and an evaluation of a cubic harmonic series, which is also given in the book, (Almost) Impossible Integrals, Sums, and Series, and then noting that
$$\sum_{n=1}^{\infty} \frac{H_{2n}}{(4n-1)^3}=\frac{1}{2}\left(\sum_{n=1}^{\infty} \frac{H_{n}}{(2n-1)^3}-\sum_{n=1}^{\infty} (-1)^{n-1} \frac{H_{n}}{(2n-1)^3}\right),$$
where for the first series we can use the same mentioned master theorem, and then the second one is already known in the form $\displaystyle \sum_{n=1}^{\infty} (-1)^{n-1} \frac{H_{n}}{(2n+1)^3}$ (it's easy to rearrange the series according to our needs), and you may find its value here together with a solution in comments, we're done.
End of story.
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Showing $\lim_{x\to 0}\frac{\sin(x^2\sin\frac{1}{x})}{x}=0$ $$\lim_{x\to 0}\frac{\sin(x^2\sin\frac{1}{x})}{x}=\lim_{x\to 0 }\frac{x^2\sin\frac{1}{x}}{x}=\lim_{x\to 0} x\sin\frac{1}{x}=0$$
Is this solution right?
Thank you very much!
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A simple proof:
$$ \lim_{x\rightarrow0} \frac{\sin(x^2\sin(\frac{1}{x}))}{x} = \lim_{x\rightarrow0} \frac{\sin(x^2\sin(\frac{1}{x}))}{x} \cdot\frac{x}{x}\frac{\sin(\frac{1}{x})}{\sin(\frac{1}{x})} = \lim_{x\rightarrow0} \frac{\sin(x^2\sin(\frac{1}{x}))}{x^2\sin(\frac{1}{x})} \cdot x\sin(\frac{1}{x}) $$
The term $\frac{\sin(x^2\sin(\frac{1}{x}))}{x^2\sin(\frac{1}{x})} $ goes to $1$ as $ x \rightarrow 0 $ by the famous identity $\lim_{f(x)\rightarrow0}\frac{\sin(f(x))}{f(x)} = 1 $ as the term $ \sin(1/x) $ is bounded between [-1,1] and the term $ x^2 $ goes to zero, so overall the dummy function $ f(x) = x^2\sin(\frac{1}{x}) \rightarrow 0 $ as $ x \rightarrow 0 $. We now have:
$$ \lim_{x\rightarrow0} x\sin(\frac{1}{x}) = 0$$
as again the term $ \sin(1/x) $ is bounded between [-1,1].
I hope the answer is satisfying.
|
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|
Prove that if $z$ is uni modular then $\frac{1+z}{1 + \bar z}$ is equal to $z$. The expression can be written as
$$\frac{1+z}{\overline {1+z}}$$
Since $z \cdot \overline z=|z|^2$
$$\overline{1+z}= \frac{1}{1+z}$$
As $|z|=1$
So it will become $(1+z)^2$
But the answer is $z$. What am I doing wrong?
|
You are asserting that if $z\cdot \overline z=1$ then $(1+\overline z)(1+z)=1,$ too. That is not true. A simple case is $z=1.$
Being unimodular means $z\cdot \overline{z}=1.$ Then $\overline z = z^{-1}.$
So $$\frac{1+z}{1+\overline{z}}=\frac{1+z}{1+z^{-1}}=\frac{1+z}{1+z^{-1}}\cdot \frac z z=\frac{z(1+z)}{z+1}=z$$
Note, the left side is not defined in the case $z=-1.$
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|
Determining whether three values can be the lengths of a triangle or not
Let $x,y,z$ be three distinct real positive numbers, Determine whether or not the three real numbers $$ \left| \frac{x}{y} - \frac{y}{x}\right| ,\left| \frac{y}{z} - \frac{z}{y}\right |, \left| \frac{z}{x} - \frac{x}{z}\right| $$ can be the lengths of the sides of a triangle.
Hello! I hope everybody is doing well. Can anybody help with the above problem?
I am not able to approach this problem.
WLOG $x > y > z$. Hence $\frac{x}{y} > 1, \frac{y}{z} > 1 $ and $\frac{x}{z} > 1$. Let $a=\frac{x}{y},b=\frac{y}{z},ab=\frac{x}{z}$ and hence we want to show whether $a - \frac{1}{a}$, $b - \frac{1}{b}$ and $ab - \frac{1}{ab}$ can form the sides of a triangle. If I could somehow know the smallest or the greatest side and could work on it using the Triangle Inequality we could be done.
Any help would be appreciated. Thanks.
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The hint.
Let $x>y>z.$
Thus, show that
$$\frac{x^2-z^2}{xz}>\frac{x^2-y^2}{xy}$$ and
$$\frac{x^2-z^2}{xz}>\frac{y^2-z^2}{yz},$$ which says that it's enough to check
$$\frac{x^2-y^2}{xy}+\frac{y^2-z^2}{yz}>\frac{x^2-z^2}{xz}$$ or
$$\sum_{cyc}\frac{x^2-y^2}{xy}>0$$ or
$$\sum_{cyc}(x^2z-x^2y)>0$$ or
$$(x-y)(y-z)(z-x)>0,$$ which is wrong.
|
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|
Number of possible polynomials
Let $a,b,c,d$ be four integers (not necessarily distinct) in the set $\{1,2,3,4,5\}$. Find the number of polynomials of the form $x^4+ ax^3 + bx^2 + cx +d$ which is divisible by $x+1$.
My Try:
Let $f(x) = x^4+ ax^3 + bx^2 + cx +d$, then $f(-1) = 0$. Thus $1+ (b+d) = c+a$. On counting cases I got 80 permissible cases. Is there a way to solve the above equation $1+ (b+d) = c+a$?
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The number of solutions of $a-b+c-d=1$ for $a,b,c,d\in\{1,2,3,4,5\}$ can be counted as the coefficient of $x$ in the Laurent series
$$ \left(x+x^2+x^3+x^4+x^5\right)^2\left(\frac{1}{x}+\frac{1}{x^2}+\frac{1}{x^3}+\frac{1}{x^4}+\frac{1}{x^5}\right)^2,$$
which is also the coefficient of $x$ in
$$ \left(\frac{1}{x^4}+\frac{2}{x^3}+\frac{3}{x^2}+\frac{4}{x}+5+4x+3x^2+2x^3+x^4\right)^2, $$
so it is given by
$$ 2\cdot 1+3\cdot 2+4\cdot 3+5\cdot 4+4\cdot 5+3\cdot 4+2\cdot 3+1\cdot 2=2\sum_{k=1}^{4}k(k+1)=80 $$
as claimed.
|
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|
quick way to count Suppose we have 10 sticks with length 1-10, respectively. Pick three from them, how many triangles can we form?
I counted one by one and got 50. Is there a quick way? Any help would be appreciated.
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While bruteforcing you can notice a pattern.
$$\begin{array}{c|c}
Odd&Even\\
\hline
4=3+1<3+\{2\}&5=4+1<4+\{2,3\}\\
\hline
6=5+1<5+\{2,3,4\},&7=6+1<6+\{2,3,4,5\},\\ \qquad \quad \ 4+\{3\}&
\qquad \quad \ 5+\{3,4\}\\
\hline
8=7+1<7+\{2,3,4,5,6\},&9=8+1<8+\{2,3,4,5,6,7\}, \\
\qquad \quad \ \ 6+\{3,4,5\},&\qquad \quad \ 7+\{3,4,5,6\}\\
\quad \ 5+\{4\}&\quad \ 6+\{4,5\}\\
\hline
10=9+1<9+\{2,3,4,5,6,7,8\}, \\
\qquad \quad \ \ 8+\{3,4,5,6,7\}\\
\quad \ \ 7+\{4,5,6\}\\
6+\{5\}\\
\end{array}$$
Hence, the cardinalities of the sets:
$$\sum_{i=1}^{4}\sum_{j=1}^i (2j-1)+\sum_{i=1}^3\sum_{j=1}^i (2j)=\\
\sum_{i=1}^4 (2\cdot \frac{1+i}{2}\cdot i-i)+\sum_{i=1}^3 (2\cdot \frac{1+i}{2}\cdot i)=\\
\sum_{i=1}^4 i^2+\sum_{i=1}^3 (i^2+i)=\\
\frac{4\cdot 5\cdot 9}{6}+\frac{3\cdot 4\cdot 7}{6}+\frac{1+3}{2}\cdot 3=\\
30+14+6=50.$$
The general formula for $n$:
$$\large{\sum_{i=1}^{\lfloor \frac{n}{2}-1\rfloor}\sum_{j=1}^{i} (2j-1)+\sum_{i=1}^{\lfloor \frac{n}{2}-1\rfloor}\sum_{j=1}^i (2j)}$$
|
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|
Maximum value of $a+b+c$ in an inequality Given that $a$, $b$ and $c$ are real positive numbers, find the maximum possible value of $a+b+c$, if
$$a^2+b^2+c^2+ab+ac+bc\le1.$$
From the AM-GM theorem, I have
$$a^2+b^2+c^2+ab+ac+bc\geq 6\sqrt[6]{a^4b^4c^4} = 6\sqrt[3]{a^2b^2c^2} \\
6\sqrt[3]{a^2b^2c^2} \le1 \\
a^2b^2c^2 \le \frac{1}{216} \\
abc \le \frac{\sqrt{6}}{36}$$
However, I don't know where to go from here.
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$$4(a+b+c)^2+(a-b)^2+(b-c)^2+(c-a)^2=6(a^2+b^2+c^2+ab+bc+ca)$$
and so $$4(a+b+c)^2+(a-b)^2+(b-c)^2+(c-a)^2\le 6$$
Each of $(a-b)^2,(b-c)^2,(c-a)^2$ is non-negative and so
$$4(a+b+c)^2\le 6.$$
Therefore $a+b+c\le \sqrt \frac{3}{2}$.
|
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Evaluation of a digamma series involving golden-ratio Let $\varphi =\frac{1}{2} \left(\sqrt{5}+1\right), a=\tan \left(\frac{\sqrt{5} \pi }{2}\right)$, then how can one prove
$$\sum _{n=1}^{\infty } \frac{\psi ^{(0)}(n+\varphi)-\psi ^{(0)}\left(n-\frac{1}{\varphi}\right)}{n^2+n-1}=\frac{\pi ^2 a^2}{\sqrt{5}}+\frac{4 \pi a}{5}+\frac{\pi ^2}{2 \sqrt{5}}$$
Note that $n^2+n-1=(n+\varphi) \left(n-\frac{1}{\varphi}\right)$. Maybe we should consider the generalized sum i.e. $\sum _{n=1}^{\infty } \frac{\psi ^{(0)}(n+t)-\psi ^{(0)}(n+s)}{(n+s) (n+t)}$? Any help will be appreciated.
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We can use the representation
\begin{equation}
\psi(a)-\psi(b)=(a-b)\sum_{p=0}^\infty\frac{1}{(p+a)(p+b)}
\end{equation}
to express the series
\begin{align}
S(s,t)&=\sum _{n=1}^{\infty } \frac{\psi (n+t)-\psi (n+s)}{(n+s) (n+t)}\\
&=(t-s)\sum _{n=1}^{\infty }\sum_{p=0}^\infty\frac{1}{(p+n+t)(p+n+s)} \frac1{(n+s) (n+t)}\\
&=(t-s)\sum _{n=1}^{\infty }\sum_{k=n}^\infty\frac{1}{(k+t)(k+s)} \frac1{(n+s) (n+t)}
\end{align}
The case $t=\phi, s=-1/\phi$ is explicitely solved in this question.
For the general case, as shown in the answers, and with $f(n)=1/\left((n+s) (n+t) \right)$, we have
\begin{equation}
\left[\sum_{n=1}^\infty f(n)\right]^2=2\sum_{n=1}^\infty \sum_{k=n}^{\infty} f(n)f(k) - \sum_{n=1}^{\infty}f(n)^2
\end{equation}
then,
\begin{equation}
S(s,t)=\frac{t-s}{2}\left[\sum _{n=1}^{\infty }\frac1{(n+s) (n+t)}\right]^2+\frac{t-s}{2}\sum _{n=1}^{\infty }\frac1{(n+s)^2 (n+t)^2}
\end{equation}
With
\begin{equation}
\frac1{(n+s)^2 (n+t)^2}=\frac{1}{(t-s)^2}\left[\frac{1}{(n+s)^2}+\frac{1}{(n+t)^2}\right]+\frac{2}{(t-s)^3}\left[\frac{1}{n+t}-\frac{1}{n+s}\right]
\end{equation}
and using again the representations in terms of the polygamma functions given above, we obtain
\begin{align}
S(s,t)=\frac{1}{2(t-s)}&\left[\psi(t+1)-\psi(s+1)\right]^2\\
&+\frac{1}{2(t-s)}\left[\psi^{(1)}(t+1)+\psi^{(1)}(s+1)\right]\\
&-\frac{1}{(t-s)^2}\left[\psi(t+1)-\psi(s+1)\right]
\end{align}
|
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|
Show that $3^{-n}$ have the interesting property that one half of their repeating binary string is the inverse of the other.
$3^{-n}$ have the interesting property that one half of their repeating binary string is the inverse of the other. Prove it!
$3^{-1}=\overline{0\color{red}{1}}_2$
$3^{-2}=\overline{000\color{red}{111}}_2$
$3^{-3}=\overline{000010010\color{red}{111101101}}_2$
$\ldots$
These are the binary representations - add a minus sign to the left hand side and the right hand side is the 2-adic representation.
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Lemma: For any sequence of $k$ digits $A$, where $B$ is the opposite sequence of digits: $$0.\overline{AB}_2 = \frac{A + 1}{2^k + 1}.$$
Proof: Let $x = 0.\overline{AB}$. Then
$$
2^{k} x = A.\overline{BA}.
$$
Therefore,
$$
x + 2^k x = A.\overline{11\ldots 1} = A + 1 \quad \implies \quad x = \frac{A + 1}{2^k + 1}.
$$
So we have to show that $3^{-n} = \frac{1}{3^n}$ has this form, $\frac{A+1}{2^k + 1}$, where $A$ is between $0$ and $2^{k} - 1$. Because $3^{-n}$ is definitely between $0$ and $1$, this is equivalent to just finding $k$ such that $3^n \mid 2^{k} + 1$.
From the pattern in the examples in your question, we guess that we should pick
$$
k = 3^{n-1}.
$$
So we show by induction that
$$
3^n \mid 2^{3^{n-1}} + 1.
$$
Base case: $3^1 \mid 2 + 1 = 3$.
Inductive step:
Let's assume that $3^n \mid 2^{3^{n-1}} + 1$, for some $n$. Specifically, let
$2^{3^{n-1}} = a \cdot 3^n - 1$.
Then
\begin{align*}
2^{3^n} + 1
&= \left(2^{3^{n-1}}\right)^3 + 1 \\
&= \left(a \cdot 3^n - 1\right)^3 + 1 \\
&= a^3 3^{3n} - 3 a^2 3^{2n} + 3 a 3^n - 1 + 1 \\
&\equiv 0 - 0 + 0 - 1 + 1 \pmod{3^{n+1}} \\
&= 0.
\end{align*}
So $2^{3^n} + 1$ is divisible by $3^{n+1}$, and the induction is complete.
|
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|
Number theory problem on divisors!
For any positive integer $n$, let $d(n)$ denote the number of positive divisors of $n$ (including $1$ and itself). Determine all positive integers $k$ such that $$\frac{d\left(n^2\right)}{d(n)} = k$$ for some $n$.
Please help me solve this number theory problem. I have tried a lot and found two values of k, $1$ and $3$, but I don't know if there are some others.
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All odd numbers and only them are representable in the form $d(n^2)/d(n)$. But the proof is tricky!
It is clear that no even number is representable as the numerator is always odd.
For $k$ odd we proceed by induction, clearly $1 = d(1^2)/d(1)$ so $1$ is representable. Now suppose that $k>1$ and that all odd integers up to $k-2$ are representable then if $2^a$ is the largest power of $2$ that divides $k+1$, write
$$ k = 2^at-1 $$
with $t$ odd and $a \ge 1$, then
$$ (2^a-1)k = 2^{a+1}(2^{a-1}t-\tfrac{t+1}2)+1 = 2^{a+1}b+1$$
whith $b = 2^{a-1}t-\tfrac{t+1}2$. Now we see that $2b+1 = (2^a-1)t$ and so:
$$ \frac{2^{a+1}b+1}{2^ab+1}\cdot \frac{2^{a}b+1}{2^{a-1}b+1}\cdot \dots \cdot \frac{2^2b + 1}{2b+1} = \frac{(2^a-1)k}{(2^a-1)t} = \frac{k}{t}$$
As $t < k$ is odd, using the induction hypothesis we can find $m$ such that
$$ t = \frac{d(m^2)}{d(m)} $$
But then chosing primes $p,q,\dots,s$ coprime with $m$ we find that
$$ n = p^{2^{a}b}q^{2^{a-1}b}\dots s^{2b} m $$
verifies
$$ \frac{d(n^2)}{d(n)} = \frac{d(p^{2^{a+1}b}q^{2^{a}b}\dots s^{2^2 b})}{d(p^{2^{a}b}q^{2^{a-1}b}\dots s^{2 b})} \cdot \frac{d(m^2)}{d(m)} = \frac{k}t\cdot t = k$$
and the proof is complete!.
|
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|
Prove that $\sqrt[n]{n} < (1 + \frac{1}{\sqrt{n}})^2$ for all n in the naturals. I need to prove that $\sqrt[n]{n} < (1 + \frac{1}{\sqrt{n}})^2$ for all n in the naturals.
I started by using Bernoulli's inequality:
$(1+\frac{2}{\sqrt{n}}) < (1 + \frac{1}{\sqrt{n}})^2$
I can say that:
$(1+\frac{2}{\sqrt{n}}) = (1+\frac{2\sqrt{n}}{n})$
I can also subtract the one and divide by 2 on the left side without changing the inequality (because it makes it even smaller): $(\frac{\sqrt{n}}{n}) < (1 + \frac{1}{\sqrt{n}})^2$
But now I am stuck...
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Well for natural (positive) $n$ then $\sqrt[n]{n} \le (1+ \frac 1{\sqrt n})^2 \iff$
$n \le (1 + \frac 1{\sqrt n})^{2n}$
So $(1+\frac 1{\sqrt n})^{2n} \ge 1 + \frac {2n}{\sqrt n}$ and dang... that's not enough.
But lets go one more term.
Remember the reason $(1 + b)^n \ge 1+ nb$ is becase $(1 + b)^n = 1 + nb + C_2b^2 + C_3b^3 + ..... + b^n$ and $C_kb^k \ge 0$. (Note if $b$ is large and $n>0$, in particular if $b \ge 1$, this is $(1+b)^n \ge 1 + nb$ is a very large discrepancy in the inequality).
So $(1+\frac 1{\sqrt n})^{2n} \ge 1 + \frac {2n}{\sqrt n} + {2n \choose 2}\frac 1{n}$.
So ${2n \choose 2} = \frac {2n(2n-1)}2 = n(2n-1)$ so
$(1 + \frac 1{\sqrt n})^{2n} \ge 1 +\frac{2n}{\sqrt n} + n(2n-1)\frac 1n =$
$1 + 2\sqrt n + 2n- 1 = 2(\sqrt n + n) > n$.
.... which if we did that correctly is not even close.
|
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|
Find all the numbers and aware A $3$-digit number $n$ is said and aware if the last $ 3$ digits of $n ^ 2$ are the same digits of $n$ and in the same order. Find all the numbers and aware
I solved it with some nasty casework:
We must find all integers $0\leq n < 1000$ such that $n^k \equiv n \pmod{1000}$ for any integer $k$. Actually we only need to check this statement for $k = 2$ because the rest will follow by induction.
Now we can apply the Chinese Remainder Theorem:
For the factor of 8, we easily check by hand that $n^2 \equiv n\pmod{8}$ iff $n\equiv 0\pmod{8}$ or $n\equiv 1\pmod{8}$.
As for the other factor of $125$, we also check that $n^2 \equiv n\pmod{5}$ iff $n\equiv 0\pmod{5}$ or $n\equiv 1\pmod{5}$. Of the integers $n$ with $n \equiv 0 \pmod{5}$, the only integers with $n^2 \equiv n\pmod{25}$ are those with $n\equiv 0\pmod{25}$; similarly, of the integers $n$ with $n \equiv 1 \pmod{5}$, the only integers with $n^2 \equiv n\pmod{25}$ are those with $n\equiv 1\pmod{25}$ (because when we write $n = 5k + 1$, we then find that $n^2 - n \equiv 5k\pmod{25}$, so that $k\equiv 0\pmod{5}$). Of the integers $n$ with $n\equiv 0\pmod{25}$, we know that $n^2 \equiv n\pmod{125}$ only when $n\equiv 0\pmod{125}$; similarly, of the integers $n$ with $n \equiv 1 \pmod{25}$, the only integers with $n^2 \equiv n\pmod{125}$ are those with $n\equiv 1\pmod{125}$ (because when we write $n = 25k + 1$, we then find that $n^2 - n \equiv 25k\pmod{125}$, so that $k\equiv 0\pmod{5}$). Thusly the only solutions to the congruence $n^2 \equiv n\pmod{125}$ are those with $n\equiv 0\pmod{125}$ or $n\equiv 1\pmod{125}$.
So now we know that there are exactly four such integers: $n = 0$ (which corresponds to $n\equiv 0\pmod{8}, n\equiv 0\pmod{125}$), $n = 625$ (which corresponds to $n\equiv 1\pmod{8}, n\equiv 0\pmod{125}$), $n = 376$ (which corresponds to $n\equiv 0\pmod{8}, n\equiv 1\pmod{125}$), and $n = 1$ (which corresponds to $n\equiv 1\pmod{8}, n\equiv 1\pmod{125}$). Now we are done.
We remark in passing that this approach can be applied to other moduli besides $1000$, as long as the modulus is prime-factorized.
Is there a shorter or more enjoyable solution?
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There is one very nice thing we can use. It is not a coincidence that 0 and 1 were the only numbers that worked for both modulo 8 and 125. We can prove that quite easily, actually. Let $p^k$ be a prime power. Then:
$$
n^2=n\pmod{p^k}
\iff p^k\mid n^2-n = n(n-1)
\iff p^k \mid n \lor p^k \mid n-1
$$
$$
\iff n\in\{0,1\} \pmod{p^k}
$$
(Do you see why the second "if and only if" is valid?)
With this, we can easily solve any problem asking $n^2=n \pmod m$. Just find the prime factorisation $m=p_1^{\alpha_1}\cdots p_k^{\alpha_k}$. Then $n=0$ or $n=1$ modulo each of the prime powers, giving $2^k$ solutions, which we can find with the Chinese Remainder Theorem.
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|
Find the minimum of :$P=a+b+c-ab-bc-ca$ soluLet $a,b,c$ be positive real numbers and $a+b+c+abc=4$.
We can rewrite the first equation as $a+b+c=4-abc.$ Then,
\begin{align*}
P&=a+b+c-ab-bc-ca\\&=(4-abc)-ab-bc-ca\\&=4-abc-ab-bc-ca-(a+b+c+1)+(a+b+c+1)\\&=4-(abc+ab+bc+ca+a+b+c+1)+a+b+c+1\\&=4-(a+1)(b+1)(c+1)+a+b+c+1\\&=5+a+b+c-(a+1)(b+1)(c+1)\\&=5+(4-abc)-(a+1)(b+1)(c+1)\\&=9-abc-(a+1)(b+1)(c+1).
\end{align*}It seems that both $abc$ and $(a+1)(b+1)(c+1)$ are maximized when $a=b=c,$ which would give $a=b=c=1$ and then $P=9-(1)(1)(1)-(2)(2)(2)=9-(1+8)=0.$ However, I'm not sure how to prove that statement.
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This problem can be taken down with some heavy machinery. From the initial condition, obtain $c=\frac{4-a-b}{ab+1}$. We now want to minimize $$f(a,b)=a+b-ab+\left(\frac{a+b-4}{ab+1}\right)(a+b-1),$$ where $a$, $b$ are on the region bounded by the x and y axes, and the line $x+y\leq 4$. Since $f$ is continuous, it must have a minimum on this region.
If this minimum lies on a border, then one of $a$, $b$, $c$ must be zero. In this case, taking WLOG $c=0$, the initial condition becomes $a+b=4$, and our value to minimize, $a+b-ab$. By AM-GM, this last expression attains a minimum value of $0$ when $a=b=2$.
Otherwise, if the minimum lies inside the triangle, both partial derivatives at that point must be zero. Churning out these derivatives (which is not that hard, since $f$ is symmetric), and setting to zero, we find that either $$a=\frac{\sqrt{5}+1}{2}\text{ or }a=\frac{4-2b}{b^2+1},$$and viceversa for $b$. However, $f\left(\frac{\sqrt{5}+1}{2},b\right)=f\left(a,\frac{\sqrt{5}+1}{2}\right)=\frac{7-3\sqrt{5}}{2}>0$, which as we've already established, is not the minimum. This implies $$a=\frac{4-2b}{b^2+1},\,\,b=\frac{4-2a}{a^2+1}\Rightarrow$$ $$a=\frac{4-2\left(\frac{4-2a}{a^2+1}\right)}{\left(\frac{4-2a}{a^2+1}\right)^2+1}\Rightarrow$$ $$a^3-5a^2+3a+1=0\Rightarrow$$ $$(a-1)(a^2-4a-1)=0.$$ This all implies $a=b=c=1$, which also gives a value of $0$ on the expression to minimize.
In conclusion, the minimum is $0$, which is attained at $(a,b,c)=(0,2,2)$, $(2,0,2)$, $(2,2,0)$, $(1,1,1)$.
|
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|
Calculate the approximate value I am supposed to calculate the approximate value of $$\cos 151^\circ$$ My idea was that I can divide it in the form: $$\cos 90^\circ+ 61^\circ= \cos \frac{\pi}{2}+ \left (\frac{\pi}{3} +\frac{\pi}{180} \right )$$ Then I use the derivation for cosx:
$$-\sin \frac{\pi}{2}\left ( \frac{\pi}{3} +\frac{\pi}{180} \right )=-\frac{61\pi}{180}+61^\circ$$
But I guess, it is not correct.
Can anyone help me?
|
One way is to use derivatives.
Let $y=f(x) = \cos x$
$dy = -\sin x dx$
Let $x=150^\circ = \frac{2\pi}{3} , dx = 1^\circ\approx 0.0174$
$\cos x = \cos150^\circ \approx -0.8660, \sin x = \sin 150^\circ =0.5$
$y+dy = f(x+dx) = \cos x -\sin x dx = \cos(150^\circ)-\sin(150^\circ)\times0.0174 \approx-0.8747$
$\implies\cos(151^\circ) \approx -0.8747$
|
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Closed form of $\sum^{\infty}_{n=1} \dfrac{1}{n^a{(n+1)}^a}$ where $a$ is a positive integer Recently, I bought a book about arithmetic. I saw a question is like that:
Given that $\dfrac{1}{1^2}+\dfrac{1}{2^2}+\dfrac{1}{3^2}+\cdots=\dfrac{\pi^2}{6}$, find the value of $$\dfrac{1}{1^32^3}+\dfrac{1}{2^33^3}+\dfrac{1}{3^34^3}+\cdots$$
Then, I did this:
$$\sum_{n=1}^\infty \dfrac{1}{n(n+1)}=\sum_{n=1}^\infty \left(\dfrac{1}{n}-\dfrac{1}{n+1}\right)=1 \\ \sum_{n=1}^\infty \dfrac{1}{n^2(n+1)^2}=\sum_{n=1}^\infty \dfrac{n^2+(n+1)^2-2n(n+1)}{n^2(n+1)^2}\\=\sum_{n=1}^\infty \left[\dfrac{1}{n^2}+\dfrac{1}{(n+1)^2}-\dfrac{2}{n(n+1)}\right]=\zeta(2)+\zeta(2)-1-2=\dfrac{\pi^2}{3}-3 \\ \sum_{n=1}^\infty \dfrac{1}{n^3(n+1)^3}=\sum_{n=1}^\infty \dfrac{(n+1)^3-n^3-3n(n+1)}{n^3(n+1)^3} \\= \sum_{n=1}^\infty\left[\dfrac{1}{n^3}-\dfrac{1}{(n+1)^3}-\dfrac{3}{n^2(n+1)^2}\right]=1-3\left(\dfrac{\pi^2}{3}-3\right)=10-\pi^2$$
Though I finished the problem, but I found something interesting, so I wanted to discover more. I let $f(a)=\sum_{n=1}^\infty \dfrac{1}{n^a(n+1)^a}$. Then, I found something special:
$f(4)=2\zeta(4)-2f(3)-2f(2)-1 \\ f(5)=1-5f(4)-5f(3) \\ f(6)=2\zeta(6)-6f(5)-9f(4)-2f(3) \\ f(7)= 1-7f(6)-14f(5)-7f(4)$
It seems to have a sequence, but I can't tell what it is. If anyone knows what is the sequence or even the closed form of $f(a)$, please tell me. Thank you very much!
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Let $h(z)= \frac1{(z+1)^k},g(z)=\frac{1}{z^k}$ then
$$\frac{1}{z^k(z+1)^k} -\frac{\sum_{m=0}^{k-1} \frac{h^{(m)}(0)}{m!} z^m}{z^k}- \frac{\sum_{m=0}^{k-1} \frac{g^{(m)}(-1)}{m!} (z+1)^m}{(z+1)^k}$$
is a rational function with no pole thus it is a polynomial and since it vanishes at $\infty$ it is $0$.
$\frac{h^{(m)}(0)}{m!} =(-1)^m {m+k-1 \choose m}$, $\frac{g^{(m)}(-1)}{m!} = (-1)^{k}{m+k-1 \choose m}$
Thus $$\sum_{n=1}^\infty \frac{1}{n^k(n+1)^k}= \sum_{m=0}^{k-2}\zeta(k-m) {m+k-1 \choose m} ((-1)^m+ (-1)^{k})-\sum_{m=0}^{k-1} (-1)^{k}{m+k-1 \choose m}$$
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"url": "https://math.stackexchange.com/questions/3412007",
"timestamp": "2023-03-29T00:00:00",
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|
Given the following function how many solutions to $f(x)=0$ are there? I have the following function:
$f: \mathbb{R} \rightarrow \mathbb{R}$
$f(x)=9^x-5^x-4^x$
And I have to find the number of real soltutions (so not necessarily the solutions themselves, just how many are there) for $f(x)=0$ and $f(x)-2 \sqrt{20^x}=0$.
For $f(x)=0$ this is what I did:
$9^x-5^x-4^x=0$
I divided by $9^x$,
$1- \bigg (\dfrac{5}{9} \bigg)^x - \bigg (\dfrac{4}{9} \bigg)^x = 0$
$\bigg (\dfrac{5}{9} \bigg)^x + \bigg (\dfrac{4}{9} \bigg)^x = 1$
Since the left-hand side is the sum of two strictly decreasing functions, I conculded that the left-hand side is strictly decreasing ($1$). So the equation can have at most $1$ solution. By pure guessing, I found that $x=1$ is a solution to the equation and because of ($1$) it is the only solution. So, $f(x)=0$ has only one solution ($x=1$ to be precise, but this is not necessary). I think I got this right, but if someone finds a mistake, please let me know.
My real trouble is at the second part of the problem, where I have to find the number of solutions for:
$9^x-5^x-4^x - 2 \sqrt{20^x} = 0$
I don't see how should I approach this. Any help will be appreciated.
|
Yes we have that
$$f(x)=9^x-5^x-4^x=0 \iff \left(\frac49\right)^x+\left(\frac59\right)^x=1$$
and since
$$ g(x)=\left(\frac49\right)^x+\left(\frac59\right)^x\implies g'(x)=-\left(\frac49\right)^x\log \left(\frac94\right)-\left(\frac95\right)^x\log \left(\frac95\right)<0$$
$f(x)$ is strictly decreasing and
*
*$\lim_{x\to \infty} f(x)=0$
*$\lim_{x\to -\infty} f(x)=\infty$
by IVT we have exactly one solution for $f(x)=1$.
For the second equation we can use that
$$f(x)=9^x-5^x-4^x - 2 \sqrt{20^x}=(3^x)^2-((\sqrt 5)^x+(\sqrt 4)^x)^2=\\=(3^x+2^x+(\sqrt 5)^x)(3^x-2^x-(\sqrt 5)^x)=0$$
that is $3^x-2^x-(\sqrt 5)^x=0$ which can be studied in the same way.
|
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"timestamp": "2023-03-29T00:00:00",
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|
Evaluate $\left(1+\sin\frac{\pi}{5}+i\cos\frac{\pi}{5}\right)^5+i\left(1+\sin\frac{\pi}{5}-i\cos\frac{\pi}{5}\right)^5$
Evaluate
$$\left(1+\sin\frac{\pi}{5}+i\cos\frac{\pi}{5}\right)^5+i\left(1+\sin\frac{\pi}{5}-i\cos\frac{\pi}{5}\right)^5$$
I did this by $$\left(1+\sin\frac{\pi}{5}+i\cos\frac{\pi}{5}\right)^5=\left(1+\cos\frac{3\pi}{10}+i\sin\frac{3\pi}{10}\right)^5$$ and get $0$
Does anyone have another idea?
Thanks
|
$$\frac{1+\sin x+i\cos x}{1+\sin x-i\cos x}=\frac{(1+\sin x+i\cos x)^2}{(1+\sin x)^2+\cos^2 x}$$
$$=\frac{2(1+\sin x)(\sin x+i\cos x)}{2(1+\sin x)}=\cos\left(\frac\pi2-x\right)+i\sin\left(\frac\pi2-x\right).$$
Apply de Movire's Theorem, you can get
$$\left(\frac{1+\sin x+i\cos x}{1+\sin x-i\cos x}\right)^n=\cos n\left(\frac{\pi }{2}-x\right)+i\sin n\left(\frac{\pi}{2}-x\right).$$
In youe situation, take $x=\pi/5$.
$$\left(1+\sin\frac{\pi}{5}+i\cos\frac{\pi}{5}\right)^5+i\left(1+\sin\frac{\pi}{5}-i\cos\frac{\pi}{5}\right)^5$$
$$=\left(\frac{1+\sin\frac{\pi}{5}+i\cos\frac{\pi}{5}}{1+\sin\frac{\pi}{5}-i\cos\frac{\pi}{5}}\right)^5+i=(-i)^5+i=0.$$
|
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|
Prove that $\sum \frac{x}{x^2+7}\le \frac{3}{8}$ Let $x,y,z>0$ such that $xy+yz+xz=3$. Show that $$\frac{x}{x^2+7}+\frac{y}{y^2+7}+\frac{z}{z^2+7}\le \frac{3}{8}$$
We have: $$x+y+z\ge \sqrt{3\left(xy+yz+xz\right)}=3\rightarrow \frac{3}{8\left(x+y+z\right)}\le \frac{3}{8\cdot 3}$$
Then i will prove $$\sum \frac{x}{3x^2+7\left(xy+xz+yz\right)}\le \frac{3}{8\left(x+y+z\right)}$$
But it's wrong. I tried to use uvw and done. So i need another idead without uvw
|
Also, the Tangent Line method helps.
Let $x=\sqrt3\tan\alpha,$ $y=\sqrt3\tan\beta$ and $z=\sqrt3\tan\gamma,$ where $\{\alpha,\beta,\gamma\}\subset\left(0,\frac{\pi}{2}\right).$
Thus, $$\sum_{cyc}\tan\alpha\tan\beta=1.$$ It follows that $$\gamma=\arctan\left(\frac{1-\tan(\alpha)\tan(\beta)}{\tan\alpha+\tan\beta}\right)=\arctan(\cot(\alpha+ \beta))=\arctan\left(\tan\left(\frac\pi2-\alpha-\beta\right)\right),$$ which gives
$$\alpha+\beta+\gamma=\frac{\pi}{2}$$ and we obtain:
\begin{split}\frac{3}{8}-\sum_{cyc}\frac{x}{x^2+7}&=\sqrt3\sum_{cyc}\left(\frac{1}{8\sqrt3}-\frac{\tan\alpha}{3\tan^2\alpha+7}\right)\\
&=\sqrt3\sum_{cyc}\left(\frac{1}{8\sqrt3}-\frac{\tan\alpha}{3\tan^2\alpha+7}+\frac{1}{8}\left(\alpha-\frac{\pi}{6}\right)\right)\geq0.\end{split}
Proof of the last inequality: For $x\in\left[0,\frac{\pi}{2}\right]$, let
$$f(x)= \frac{1}{8\sqrt3}-\frac{\tan x}{3\tan^2 x+7}+\frac{1}{8}\left(x-\frac{\pi}{6}\right).$$
Then $$f'(x)=\frac18+\frac{\sec^2(x)\cdot (-7 + 3 \tan^2(x))}{(7 + 3 \tan^2(x))^2}=\frac18-\frac{(5\cos(2x)+2)\cdot\sec^4(x)}{\big((2\cos(2x)+5)\cdot\sec^2(x)\big)^2}=\frac18-\frac{5\cos(2x)+2}{(2\cos(2x)+5)^2}.$$
Hence, for $x\in(0,\pi/2)$, we have $f'(x)=0$ iff $x=\frac\pi6$. Since $f(\pi/6)=0$ and $f(0)>0$, $f(\pi/2)>0$, we conclude $f\geq 0$.
|
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|
Continued fraction expansion of $\sqrt{d^2+1}$ How can I show that the continued fraction expansion of $\sqrt{d^2+1}$ is $[d, \overline{2d}]$?
And why is the fundamental unit of $\mathbb Q(\sqrt{d^2+1})$ equivalent to $d+\sqrt{d^2+1}$, if $d^2+1$ is squarefree?
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Partial answer/hint. From
$$\sqrt{d^2+1}-d=\frac{1}{\sqrt{d^2+1}+d} \Rightarrow
\sqrt{d^2+1}=d+\frac{1}{d+\sqrt{d^2+1}}\Rightarrow\\
\sqrt{d^2+1}=\color{red}{d}+\frac{1}{\color{red}{2d}+\frac{1}{d+\sqrt{d^2+1}}} \Rightarrow\\
\sqrt{d^2+1}=\color{red}{d}+\frac{1}{\color{red}{2d}+\frac{1}{\color{red}{2d}+\frac{1}{d+\sqrt{d^2+1}}}}$$
|
{
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"url": "https://math.stackexchange.com/questions/3427567",
"timestamp": "2023-03-29T00:00:00",
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|
Find all $n≥1$ natural numbers such that : $n^{2}=1+(n-1)!$ Problem :
Find all $n≥1$ natural numbers such that : $n^{2}=1+(n-1)!$
My try :
$n=1$ we find : $1=1+1$ $×$
$n=2$ we find : $4=1+1$ $×$
$n=3$ we find : $9=1+2$ $×$
$n=4$ we find : $16=1+6$ $×$
$n=5$ we find : $25=1+24$ $√$
Now how I prove $n=5$ only the solution ?
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If $n\ge 6$, dividing $n^2-1=(n-1)!$ by $n-1$ we get
$$n+1=(n-2)!\ge(n-2)(n-3)(n-4)=n^3-9n^2+26n-24$$
Define
$$\begin{align}f(n)&=n^3-9n^2+25n-25\\
&=(n-5)(n^2-4n+5)\\
&=(n-5)((n-2)^2+1)\end{align}$$
and $f(n)>0$ for $n\ge 6$.
|
{
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"timestamp": "2023-03-29T00:00:00",
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|
Find the units digit of $572^{42}$ The idea of this exercise is that you use the modulus to get the right answer.
What I did was:
$$572\equiv 2\pmod {10} \\
572^2 \equiv 2^2 \equiv 4\pmod{10} \\
572^3 \equiv 2^3 \equiv 8\pmod{10} \\
572^4 \equiv 2^4 \equiv 6\pmod{10} \\
572^5 \equiv 2^5 \equiv 2\pmod{10} \\
572^6 \equiv 2^6 \equiv 4\pmod{10} \\
(...)$$
I can see that this goes 2,4,8,6 and then repeats. I remember that the gist of the exercise is to find the remainder based on this repetition. How do I do that? I know that $572^{42} \equiv 2^{42}\equiv ? \pmod {10}$. How do I simplify that 42 and answer this using that repetition?
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Consider $2^k$ modulo $10$.
You will notice that it enters a repeating pattern: $2,4,8,6,2,4,8,6,2,\dots$
Where $2^k\pmod {10} \equiv \begin{cases}2&\text{when}~k\equiv 1\pmod 4\\ 4&\text{when}~k\equiv 2\pmod 4\\8&\text{when}~k\equiv 3\pmod 4\\6&\text{when}~k\equiv 0\pmod 4\end{cases}$
Now, consider the exponent in this case. It is $42$. Since the pattern of $2^k$ depends on $k\pmod 4$, we see that $42\equiv2 \pmod 4 $. Therefore, $2^{42}\equiv 4 \pmod {10}$ which implies that ${572}^{42}\equiv 4 \pmod {10}$.
|
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"timestamp": "2023-03-29T00:00:00",
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|
If a variable chord of the hyperbola subtend a right angle at the centre, find the radius of the circle it is tangent to
If a variable line $x\cos\alpha+y\sin\alpha=p$ which is a chord of the hyperbola $\dfrac{x^2}{a^2}-\dfrac{y^2}{b^2}=1$, $(b>a)$ subtend a right angle at the centre of hyperbola,then prove that it always touches a fixed circle whose radius is $\dfrac{ab}{\sqrt{b^2-a^2}}$
$x\cos\alpha+y\sin\alpha=p$ touches(tangent to) a fixed circle $x^2+y^2=p^2$ of radius $p$. So I need to find $p$.
Let $A$,$B$ be the points where the given lines intersects the hyperbola, ie. $\angle AOB=90^\circ$
How do I proceed further and find the radius of the required circle ?
Note: I really do not want to use some existing shortcut formula here
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Let $A=(x_1,y_1)$ and $B=(x_2,y_2)$ be the intersections of line $\cos+\sin=$ with the hyperbola. We have $\angle AOB=90°$ if $x_1x_2+y_1y_2=0$. Substituting there
$$y={p-x\cos\alpha\over\sin\alpha}$$
we thus obtain:
$$
\tag{*}
x_1x_2-(x_1+x_2)p\cos\alpha+p^2=0.
$$
But $x_1$ and $x_2$ are the solutions of the equation
$$
{x^2\over a^2}-{(p-x\cos\alpha)^2/\sin^2\alpha\over b^2}=1
$$
which can be rearranged to
$$
(b^2\sin^2\alpha-a^2\cos^2\alpha)x^2+2a^2p\cos\alpha\ x-a^2(p^2+b^2\sin^2\alpha)=0.
$$
From this equation we get:
$$
x_1+x_2=-a^2{2p\cos\alpha\over b^2\sin^2\alpha-a^2\cos^2\alpha},
\quad
x_1x_2=-a^2{p^2+b^2\sin^2\alpha\over b^2\sin^2\alpha-a^2\cos^2\alpha}.
$$
Substituting these into (*) dependence on $\alpha$ cancels out and we can solve for $p$.
|
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|
Find the number of natural solutions for $x_1 +x_2 + \cdots + x_k = n$, with $ x_i \notin 3\mathbb{N}$. Find the number of natural solutions for $$x_1 +x_2 + \cdots + x_k = n,$$ with the constraints $x_i \notin 3\mathbb{N}$ for $i=1,2,\dots,k$.
My Attempt:
the generating function of the equation is:
$f(x) =(x+x^2)(1+x^3 +x^6+\cdots +x^{3k} +\cdots)$
Now I know that for $g(x) = 1+x+x^2+\dots +x^k +\cdots = \ \sum_{k=0}^\infty x^k = \frac{1}{1-x}$
does that mean that $f(x) =(x+x^2)(1+x^3 +\cdots +x^{3k} +\cdots)= \sum_{k=0}^\infty (x^{3k})\cdot(x+x^2) = \frac{(x+x^2)}{1-x^{3k}}$
But How do I Continue with my function?
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The generating function is
$$
\left(\frac{x+x^2}{1-x^3}\right)^k\tag1
$$
since each variable can take values in the exponents of
$$
\frac{x+x^2}{1-x^3}=x+x^2+x^4+x^5+x^7+x^8+\dots\tag2
$$
Compute the coefficient of $x^n$ in $(1)$:
$$
\begin{align}
\left[x^n\right]\left(\frac{x+x^2}{1-x^3}\right)^k
&=\left[x^{n-k}\right]\left(\frac{1+x}{1-x^3}\right)^k\\
&=\left[x^{n-k}\right]\sum_{i=0}^\infty\binom{k}{i}x^i\sum_{j=0}^\infty(-1)^j\binom{-k}{j}x^{3j}\\
&=\sum_{j=0}^\infty\binom{k}{n-k-3j}(-1)^j\binom{-k}{j}\\
&=\sum_{j=0}^\infty\binom{k}{n-k-3j}\binom{j+k-1}{j}\tag3
\end{align}
$$
That is, the coefficient of $x^n$ in $(1)$ is
$$
\sum_{j=0}^{\left\lfloor\frac{n-k}3\right\rfloor}\binom{k}{n-k-3j}\binom{j+k-1}{j}\tag4
$$
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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.