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Sum of three dice is eleven the sample space of one toss of three dice is:
$\Omega = \left \{ (1,1,1), ..., (6,6,6) \right \}$
so there are $6^3 = 216$ possible outcomes.
What is the probability to obtain an outcome where the sum of its three components is equal to 11?
I've considered the possible value can assume dice without a particular position and then I have considered the permutations to include every position:
$(6,4,1), 3! = 6 \\ (6,3,2), 3! = 6 \\ (5,5,1), \frac{3!}{2!} = 3 \\ (5,4,2), 3! = 6 \\ (5,3,3), \frac{3!}{2!} = 3 \\ (4,4,3), \frac{3!}{2!} = 3$
so I've summed up obtaining $27$ and the probability would be $\frac{27}{216}$
Now, consider if I have to do this same passages for sums from 3 to 18, it is very exhausting.
So, my question is: Is there any "faster" way to do that?
|
For the favorable cases: Since the dice can assume integer values between $1$ and $6$, we wish to find the number of solutions in the positive integers of the equation
$$x_1 + x_2 + x_3 = 11 \tag{1}$$
subject to the restrictions that $x_k \leq 6$ for $1 \leq k \le 3$.
A particular solution of equation 1 corresponds to the placement of two addition signs in the ten spaces between successive ones in a row of eleven ones. For instance,
$$1 1 1 1 1 + 1 1 + 1 1 1 1$$
corresponds to the solution $x_1 = 5$, $x_2 = 2$, and $x_3 = 4$. Hence, the number of solutions of equation 1 in the positive integers is the number of ways we can select two of the ten spaces between successive ones in a row of eleven ones in which to place addition signs, which is
$$\binom{10}{2}$$
More generally, the equation
$$x_1 + x_2 + \cdots + x_k = n$$
has
$$\binom{n - 1}{k - 1}$$
solutions in the positive integers since we must choose which $k - 1$ of the $n - 1$ spaces between successive ones in a row of $n$ ones will be filled with addition signs.
However, we have counted solutions in which one of the variables exceeds $6$. Notice that it is not possible for two of the variables to exceed $6$ simultaneously since $2 \cdot 7 = 14 > 11$.
Suppose $x_1 > 6$. Let $x_1' = x_1 - 6$. Then $x_1'$ is a positive integer. Substituting $x_1' + 6$ for $x_1$ in equation 1 yields
\begin{align*}
x_1' + 6 + x_2 + x_3 & = 11\\
x_1' + x_2 + x_3 & = 5 \tag{2}
\end{align*}
Equation 2 is an equation in the positive integers with
$$\binom{5 - 1}{3 - 1} = \binom{4}{2}$$
solutions. By symmetry, there are an equal number of solutions in which $x_2 > 6$ or $x_3 > 6$. Hence, the number of solutions we must exclude is
$$\binom{3}{1}\binom{4}{2}$$
Hence, the number of permissible solutions is
$$\binom{10}{2} - \binom{3}{1}\binom{4}{2}$$
Addendum: You asked about the sums from $3$ to $18$. If $3 \leq n \leq 8$, then the number of solutions of the equation
$$x_1 + x_2 + x_3 = n \tag{3}$$
in the positive integers with $x_k \leq 6$ for $1 \leq k \leq 3$ is
$$\binom{n - 1}{3 - 1} = \binom{n - 1}{2}$$
since it is not possible for one of the variables to exceed $6$.
Notice also that by symmetry, the number of solutions with sum $3$ (all ones) is equal to the number of solutions with sum $18$ (all sixes), the number of solutions with sum $4$ (two ones and a two) is equal to the number of solutions with sum $17$ (two fives and a six), and so forth. Hence, knowing the number of solutions for $3 \leq n \leq 8$ also tells us the number of solutions for $13 \leq n \leq 18$.
The following argument shows that the number of solutions with sum $n$ is equal to the number of solutions with sum $21 - n$. If $y_k = 7 - x_k$, then $1 \leq x_k \leq 6 \implies 1 \leq y_k \leq 6$. Moreover, substituting $7 - y_k$ for $x_k$ for $1 \leq k \leq 3$ in equation 3 yields
\begin{align*}
7 - y_1 + 7 - y_2 + 7 - y_3 & = n\\
21 - y_1 - y_2 - y_3 & = n\\
-21 + y_1 + y_2 + y_3 & = -n\\
y_1 + y_2 + y_3 & = 21 - n
\end{align*}
By an argument similar to that given above for $n = 11$, the number of solutions of the equation
$$x_1 + x_2 + x_3 = 9$$
in positive integers not exceeding six is
$$\binom{8}{2} - \binom{3}{1}\binom{2}{2} = \binom{8}{2} - \binom{3}{1}$$
By symmetry, this is also the number of solutions for $n = 12$.
Also, by symmetry, the number of solutions for $n = 10$ is equal to the number of solutions for $n = 11$.
|
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|
How to solve $ 2^{\vert x + 1 \vert} - 2^x = \vert 2^x - 1\vert + 1 $ for $ x $ I am new to modulus and inequalities , I came across this problem:
$ 2^{\vert x + 1 \vert} - 2^x = \vert 2^x - 1\vert + 1 $ for $ x $
How to find $ x $ ?
|
If $x\geq 0$, then
*
*$|x+1| = x+1$
*$|2^x-1| = 2^x-1$
So the equation becomes
$$2^{x+1}-2^x=2^x-1+1\\
2^{x+1} - 2\cdot 2^x=0\\
2^{x+1}-2^{x+1}=0\\
0=0$$
so it is always true.
There are two more cases to do: $-1\leq x<0$ and $x<-1$.
|
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How many positive integer cubes are divisors of $1!\cdot 2! \cdot 3!\cdot 4! \cdot 5!\cdot 6!\cdot 7!\cdot 8!$? What I have tried:
$$N= 1!\cdot 2! \cdot 3!\cdot 4! \cdot 5!\cdot 6!\cdot 7!\cdot 8!\\
= 1^8\cdot 2^7\cdot 3^6\cdot 4^5\cdot 5^4\cdot 6^3\cdot 7^2\cdot 8^1 \\
= 2^{23}\cdot 3^9\cdot 5^4 \cdot 7^2$$
|
You're off to a good start.
Hint 1: Let $n^3$ be a positive integers cube. Suppose $n$ has the prime factorization $n = p_1^{a_1} p_2^{a_2} \cdots p_n^{a_n}$. What is the prime factorization of $n^3$? In general, what can you say about the exponents of the prime factors of a perfect cube?
Hint 2: Let's simplify the problem a bit and count the number of positive integer cubes that divide $3^9 5^4$. If $n^3$ divides $3^9 5^4$, then it can only have $3$ and $5$ as prime factors. By Hint 1, the possible exponents for $3$ are $0$, $3$, $6$ and $9$. Similarly, the possible exponents for $5$ are $0$ and $3$. So there are $4 \cdot 2 = 8$ total positive integer cubes that divide $3^9 \cdot 5^4$. Can you generalize this to your number?
|
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In how many ways can we permute the digits $2,3,4,5,2,3,4,5$ if identical digits must not be adjacent?
In how many ways can we permute the digits $2,3,4,5,2,3,4,5$ if identical digit must not be adjacent?
I tried this by first taking total permutation as $\dfrac{8!}{2^4}$
Now $n_1$ as $22$ or $33$ or $44$ or $55$ occurs differently
$N_1 = \left(^7C_1\times \dfrac{7!}{8}\right)$
And $n_2 = \left(^4C_1 \times 4!\right)$
Using the inclusion-exclusion principle I got:
$\dfrac{8!}{16}-\left(^7C_1\times\dfrac{7!}{8}\right)+\left(^4C_1\times4!\right)$
But answer was wrong
Please help me solve the question
This question is from combinatorics and helpful for RMO
|
You need a few more inclusion-exclusion steps to complete this approach.
Without constraints, you do indeed have $\dfrac {8!}{2^4} = 2520 $ arrangements.
Then there are $\dfrac {7!}{2^3} = 630$ cases where a $22$ is found in the arrangement, and similarly for the other digits.
Then there are $\dfrac {6!}{2^2} = 180$ cases where both a $22$ and a $33$ are found, and similarly for other pairs, etc.
So by inclusion-exclusion, we have to subtract the paired cases then add back the double-paired cases, then subtract off triple-paired again and finally add in the cases where all digits appear in pairs.
$$\frac {8!}{2^4} - \binom 41\frac {7!}{2^3} + \binom 42\frac {6!}{2^2} -
\binom 43\frac {5!}{2} + \binom 44\frac {4!}{1} \\[3ex]
=2520 -4\cdot 630 +6\cdot 180-4\cdot60 + 24 = 864$$
[Sharp eyes might notice that $\frac {8!}{2^4} = \binom 41\frac {7!}{2^3}$, shortening the calculation process.]
|
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Solving system of differential equations to evaluate $x(t)$ I have to solve the following system of equations and find $x(t)$
$$\frac{dx}{dt} =a\cos(y)-b$$
$$x\frac{dy}{dt}=-a\sin(y)$$
Initial conditions:
*
*$t=0,x=x_o, y=\frac{\pi}{2}$
By dividing equations and integration, i obtained $x(y)$:
$$x=x_o \csc(y) \left|\csc(y)-\cot(y)\right|^{\frac{b}{a}}$$
Now I cannot proceed to find $x(t)$.
|
Too long for a comment.
Domains of my math competence are far from differential equations, so I can miss some subtleties. I assume that $a$ and $b$ are constants. Given a differentiable function $u=u(t)$ by $u’$ I shall denote its derivative with respect to $t$.
Differentiating the first equation we obtain
$x’’=-a\sin y\cdot y’.$
Then
$xx’’=-a\sin y\cdot xy'=a^2\sin^2 y=a^2-(x’+b)^2=a^2-(x’)^2-2bx’-b^2$
$(xx’)’=xx’’+(x’)^2=a^2 -b^2-2bx’$
Integrating with respect to $t$ we obtain
$xx’=(a^2 -b^2)t-2bx+C$,
This equation looks simpler than the initial system and maybe it already can be solved by standard methods.
Also we can find $C$ as follows. Applying the initial conditions we obtain
$x(0)x’(0)=(a^2 -b^2)\cdot 0 -2bx(0)+C$,
but $x(0)=x_o$ and $x’(0)=a\cos y(0)-b= a\cos\frac\pi2-b=-b$, so $-bx_o=-2bx_o+C$,
and $C=bx_o$.
Update. According to player100 comment, we have to calculate $\int\frac {x}{x^2-x-A}dx$. But $$x^2-x-A= x^2-x+\frac{b^2-a^2}{4b^2}.$$ The discriminant of this polynomial is $$1-4\frac{b^2-a^2}{4b^2}=\frac{a^2}{b^2}.$$ Thus
$$x^2-x-A=\left(x-\frac{b+a}{2b}\right)\left(x-\frac{b-a}{2b}\right).$$
If $a=0$ then
$$\frac x{x^2-x-A}=\frac{x}{(x-\frac 12)^2}=\frac{1}{x-\frac 12}+\frac{1}{2(x-\frac 12)^2}. $$
So
$$\int\frac {x}{x^2-x-A}dx=\int \frac{dx}{x-\frac 12}dx+\int\frac{dx}{2(x-\frac 12)^2}dx=\ln\left(x-\frac 12\right)-\frac 1{2x-1}+C’$$
(provided $x>\frac 12$).
Then
$$e^{-\int\frac {x}{x^2-x-A}dx}=C’’\frac 1{x-\frac 12}e^{\frac 1{2x-1}}.$$
If $a\ne 0$ then
$$\frac x{x^2-x-A}=\frac{ab+b^2}{2abx-a^2-ab}+
\frac{ab-b^2}{2abx+a^2-ab}.$$
So
$$\int\frac {x}{x^2-x-A}dx=$$ $$\int \frac{ab+b^2}{2abx-a^2-ab}dx+
\int\frac{ab-b^2}{2abx+a^2-ab}dx=$$ $$\frac{ab+b^2}{2ab}\ln(2abx-a^2-ab)+\frac{ab-b^2}{2ab}\ln(2abx+a^2-ab)+C'.$$
(provided the denominators are positive)
Then
$$e^{-\int\frac {x}{x^2-x-A}dx}=
C''\left(2abx-a^2-ab\right)^{-\frac{ab+b^2}{2ab}}\left(2abx+a^2-ab\right)^{-\frac{ab-b^2}{2ab}}.$$
|
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$\sin(40^\circ)<\sqrt{\frac{3}7}$
Prove without using of calculator, that $\sin40^\circ<\sqrt{\frac{3}7}$.
My attempt.
Since $$\sin(40^\circ)=2\sin(20^\circ)\cos(20^\circ)<2\sin(20^\circ)$$
$$=2\sin(60^\circ-40^\circ)=\sqrt{3} \cos(40^\circ)-\sin(40^\circ),$$
$$2\sin(40^\circ)<\sqrt{3} \cos(40^\circ).$$
Hence, $$4\sin^2(40^\circ)<3\cos^2(40^\circ)=3(1-\sin^2(40^\circ))$$
$$7\sin^2(40^\circ)<3$$
$$\sin(40^\circ)<\sqrt{\frac{3}7}$$
Is there another way to prove this inequality?
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Let $\sin\theta=\sqrt{\frac37}$, so that $\cos\theta=\sqrt{\frac47}$ and $$\sin(\theta-30°)=\sqrt{\frac37}\sqrt{\frac34}-\sqrt{\frac47}\sqrt{\frac14}=\frac1{\sqrt{28}}.$$
Then by the triple angle formula
$$\sin(3\theta-90°)=\frac3{\sqrt{28}}-\frac4{28}\frac1{\sqrt{28}}=\frac{10}{7\sqrt 7}\approx0.54,$$
while
$$\sin(3\cdot40°-90°)=\frac12.$$
The shift by $90°$ is to stay in a monotonic section of the sinusoid.
|
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Question based on inequality with $\sum_{i=1}^{n}x_{i}=1$ Question:
If $n$ be given positive integer,let $x_{i}\ge 0$ such that $x_{1}+x_{2}+\cdots+x_{n}=1$. Find the maximum of the value $$x_{1}+x_{1}x_{2}+x_{1}x_{2}x_{3}+\cdots+x_{1}x_{2}x_{3}x_{4}\cdots x_{n}.$$
Try: (1):when $n=1$,it is clear equal $1$.
(2):when $n=2$, the condtion is $x_{1}+x_{2}=1$, then
$$x_{1}+x_{1}x_{2}=x_{1}+x_{1}(1-x_{1})=2x_{1}-x^2_{1}=-(x_{1}-1)^2+1\le 1$$
when $n=3$,the condtion $x_{1}+x_{2}+x_{3}=1$,then
$$x_{1}+x_{1}x_{2}+x_{1}x_{2}x_{3}=x_{1}(1+x_{2}+x_{2}x_{3})\le x_{1}(1+x_{2})(1+x_{3})\le \left(\dfrac{x_{1}+1+x_{2}+x_{3}+1}{3}\right)^3=1$$
when $x_{1}=1,x_{2}=x_{3}=0$
For $n\ge 4$,How find it ?
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By AM-GM we obtain:
$$x_{1}+x_{1}x_{2}+x_{1}x_{2}x_{3}+\cdots+x_{1}x_{2}x_{3}x_{4}\cdots x_{n}$$
$$\leq x_1(1+x_2)\cdots(1+x_n)\leq\left(\frac{n-1+x_1+\cdots+x_n}{n}\right)^n=1.$$
The equality occurs for $x_2=\cdots=x_n=0$ and $x_1=1$, which says that $1$ is a maximal value.
|
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Help understanding why converting repeating decimal to fraction works I recently learned how to convert a repeating decimal like $3.424242...$ to a fraction. I was however wondering why that actually works. After reading a few resources, I tried to understand it from a previous question on this site.
The explanation goes like this:
Let $x=y.a_1a_2\ldots a_m b_1b_2\ldots b_p b_1b_2\ldots b_p \ldots$, where $y\in \mathbb N$.
Then $10^m x=t+f$, where $t\in \mathbb N$ and $f=0.b_1b_2\ldots b_p b_1b_2\ldots b_p \ldots$ .
That makes sense so far for me. The following part is where I get lost:
Now, $10^p f=b+f$, where $b=(b_1b_2\ldots b_p)_{10}\in \mathbb N$.
So, $f=\dfrac{b}{10^p-1}$
Thus $x=\dfrac{t+\dfrac{b}{10^p-1}}{10^m}=\dfrac{t(10^p-1)+b}{10^m(10^p-1)}$ is a quotient of two natural numbers.
Why do we talk about $10^p f=b+f$ now instead of the $10^m x=t+f$ from the previous lines?
|
In the equation $$10^m x = t + f$$ we have two "annoying" variables, $f$ and $t$ (annoying in the sense that they are not written as a fraction). The two variables are $f$ and $x$.
Our goal is to transform $x$ into a fraction, and since $t$ is already a fraction, the equation shows that we will be able to write $x$ as a fraction if we can figure out a way to write $f$ as a fraction. So, our next goal is to discover an equation for $f$ that involves only $f$ and whole numbers.
Which is exactly what the equation $$10^p f = b+f$$ is.
By the way, there is another way of looking at the transformation, and that involves knowing that
$$x=y.a_1a_2\dots a_mb_1b_2\dots b_p b_1b_2\dots b_p\dots =y+\frac{a_1a_2\dots a_m}{10^m} + \frac{b_1}{10^{m+1}} + \frac{b_2}{10^{m+2}} + \cdots$$
You can reduce this to
$$x=y+a' + \frac{1}{10^{m+1}}\left(\frac{b_1}{10}+\frac{b_2}{10^2} + \cdots + \frac{b_p}{10^p} + \frac{b_1}{10^{p+1}}+\dots+\frac{b_p}{10^{p+p}} + \dots\right)$$
where $a'=\frac{a_1a_2\dots a_m}{10^m}$. Now define $b=\frac{b_1}{10}+\frac{b_2}{10^2}+\cdots+\frac{b_p}{10^p}$ and you get
$$x=y+a'+\frac{1}{10^{m+1}}\left(b + \frac{1}{10^p} b + \frac{1}{10^{2p} }b + \cdots\right)
\\x=y+a'+\frac{b}{10^{m+1}}\sum_{k=0}^\infty\left(\frac{1}{10^p}\right)^k$$
and using the fact that $\sum_{k=0}^\infty r^k = \frac{1}{1-r}$ you get
$$x=y+a'+\frac{b}{10^{m+1}}\frac{1}{1-10^p}$$
All the numbers in the expression above are fractions, so the result is also a fraction.
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Prove $\frac{a(a^2+2)}{3}$ is an integer for all integer $a\geqslant 1$ If $\frac{a(a^2+2)}{3}$ then $3\mid{a(a^2+2)}$.
By induction:
Lets define the set, $S=\left\{a\in N:a\geqslant1, 3\mid a(a^2+2) \right\}$
If $a=1$ then, $1\in S$
So we have to prove that if $k(k^2+2)=3m$ then $(k+1)((k+1)^2+2)=3n$ with $m,n\in Z$
If $k(k^2+2)=3m$ then,
$\begin{align*}k(k^2+2)+3(k^2+k+1)=&3m+3(k^2+k+1)\\=&3(m+k^2+k+1)\end{align*}$
Also,
$\begin{align*}k(k^2+2)+3(k^2+k+1)=&k^3+2k+3k^2+3k+3\\=&k^3+2k^2+k^2+2k+3k+3\\=&k^2(k+2)+k(k+2)+3(k+1)\\=&(k+2)(k^2+k)+3(k+1)\\ =&k(k+2)(k+1)+3(k+1)\\=&(k+1)(k(k+2)+3)\\ =&(k+1)(k^2+2k+1+2)\\ =&(k+1)((k+1)^2+2)\\ =&3(m+k^2+k+1) \end{align*}$
where $n=m+k^2+k+1$
Therefore,
$3\mid(k+1)((k+1)^2+2)$
Can I do it simplier using induction?
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Yes, we can give a simpler and more conceptual inductive proof. Notice that
$\qquad\qquad a(a^2+2)\, =\, a(a^2\!-\!1 + 3)\, =\, \color{#0a0}{(a-1)a(a+1)} + 3a$
so it suffices to show that one of any $\rm\color{#0a0}{3\ consecutive\ integers}$ is divisible by $3$
This has a simple inductive proof. Note that shifting such a sequence by one simply replaces the old least element $\,\:\color{#C00}n\,$ by the new greatest element $\,\color{#C00}{n+3}$
$$ \begin{array}{} \:\color{#C00}n & n+1 & n+2 \\
\to & n+1 & n+2 & \color{#C00}{n+3} \end{array}$$
Since $ \: \color{#C00}n\equiv \color{#C00}{n+3} \pmod{\! 3},\,$ the shift does not change the set of remainders $\bmod 3$ of the elements. Thus the remainders remain the same as in the base case $ \ 0,1,2\: =\: $ all possible remainders mod $ \,3.\,$ Therefore the sequence has an element with remainder $\,0,\,$ i.e. an element divisible by $ \,3.$
Remark $ $ The same method works to show that a sequence of $d$ consecutive integers contains a multiple of $d.\,$ Alternatively this can be proved by using division with remainder (which has a natural proof by induction), which is closely connected to the above method.
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Prove that $ \sin \frac{A}{2} \sin \frac{B}{2} \sin \frac{C}{2} \le \frac{1}{8}$ Prove that $ \sin \frac{A}{2} \sin \frac{B}{2} \sin \frac{C}{2} \le \frac{1}{8}$ when $A,B,C$ are the angles in a triangle.
I need a help on this question, one can enter $A=B=C$ and get the maximum value $\frac{1}{8}$, but that is not a solution, but hit and trial. Can this conclusion be derived using maximum/mininum concept?
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$$\sin\frac{\alpha}{2}\sin\frac{\beta}{2}\sin\frac{\gamma}{2}=\frac{r}{4R}\leq\frac{1}{8}.$$
Another way.
Since $f(x)=\ln\sin\frac{x}{2}$ is a concave function, by Jensen and AM-GM we obtain:
$$\sin\frac{\alpha}{2}\sin\frac{\beta}{2}\sin\frac{\gamma}{2}\leq\sin^2\frac{\frac{\alpha}{2}+\frac{\beta}{2}}{2}\sin\frac{\gamma}{2}=$$
$$=\frac{1}{2}\left(1-\sin\frac{\gamma}{2}\right)\sin\frac{\gamma}{2}\leq\frac{1}{2}\left(\frac{1-\sin\frac{\gamma}{2}+\sin\frac{\gamma}{2}}{2}\right)^2=\frac{1}{8}.$$
Done!
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How to calculate $\lim_{x\to\infty}\frac{7x^4+x^2 3^x+2}{x^3+x 4^x+1}$?
$$\lim_{x\to\infty}\frac{7x^4+x^2 3^x+2}{x^3+x 4^x+1}$$
I can't seem to find away to get rid of the $3^x$ and $4^x$ and then resolve it.
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Numerator : $7x^4 + x^2 3^x + 2 \lt 3x^2 3^x$
for $x\gt 4$ (say).
Denominator: $x^3 + x4^x +1 \gt x4^x$
for $x \gt 0.$
$0 \lt \dfrac{7x^4 + x^2 3^x +3}{x^3 +4^x +1} \lt $
$ \dfrac{3x^23^x}{x4^x} = 3x (\dfrac{3}{4})^x =
3x(\dfrac{ 4}{3})^{-x} $.
Note: $y := (\dfrac{4}{3})^{-x} = e^{-x \log(4/3)}$.
Let $a:= \log(4/3) \gt 0.$
Using :
$\star) \lim_{x \rightarrow \infty} x^k e^{-x} = 0$, $ k \in \mathbb{N}$:
$z:= ax$;
$3xe^{-ax} = (3/a) (ax)e^{-ax} = (3/a)ze^{-z}$.
Altogether :
$0\le \lim_{x \rightarrow \infty} \dfrac{7x^4 +3x^2 x^3 +2}{x^3 + x4^x + 1} $
$\lt \lim_{z \rightarrow \infty}(3/a)ze^{-z} = 0$.
|
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|
An olympic mathematics problem regarding Cauchy-Schwarz This question was asked in Turkish National Maths Olympiad in 2008.
For all $xy=1$ we have $((x+y)^2+4)\cdot ((x+y)^2-2) \ge A\cdot(x-y)^2$.
What is the maximum value $A$ can get?
My efforts regarding this problem;
$(x+y)^2-8 \ge A\cdot(x-y)^2$
Using the property $xy=1$ ;
$x^2+y^2-6\ge A\cdot (x^2+y^2-2)$
$\sqrt{\dfrac{x^2+y^2}{2}} \ge\sqrt{\dfrac{A\cdot(x^2+y^2-2)}{2}}$
Therefore $\sqrt{\dfrac{A\cdot(x^2+y^2-2)}{2}}=\dfrac{x+y}{2}$
Although moving further that doesn't work I applied Cauchy-Schwarz by the way if I didn't mention it before.
How should I proceed?
What are you suggestions?
|
Note that your first expression:
$$(x+y)^2-8\ge A\cdot (x-y)^2$$
must be:
$$(x+y)^4+2(x+y)^2-8\ge A\cdot (x-y)^2.$$
Alternative solution: You can also denote:
$$(x+y)^2=z; \ \ \ (x-y)^2=(x+y)^2-4xy=z-4.$$
Then:
$$(z+4)(z-2)\ge A\cdot (z-4) \Rightarrow$$
$$z^2+(2-A)z+4A-8\ge 0$$
This inequality is true for all $z$ (all $x,y$) if:
$$(2-A)^2-4(4A-8)\le 0 \Rightarrow (A-2)(A-18)\le 0 \Rightarrow 2\le A\le 18.$$
Hence the maximum value of $A$ is $18$.
P.S. Was there any other condition, like $x\ne y$? Otherwise if $x=y=1$, the left hand side is $16$, while the right hand side is $0$, implying $A$ can be any number, hence no max.
|
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|
Upper Bound for $\frac{\ln(1+nx)}{1+x^2\ln(n)}\leq \frac{1+\ln(x)+\ln(2)}{x^2}$ I was reading a solution for an analysis problem and they argued that $$\frac{\ln(1+nx)}{1+x^2\ln(n)}\leq \frac{1+\ln(x)+\ln(2)}{x^2}$$ for $x\geq 1$ and $n\geq 1$, $x$ a real number and $n$ a natural number. Why is that?
|
First the identity definitely doesn't hold for $n=1$ so let's assume $n\geq 2$, then we have
$$
\begin{align*}
\ln(1 + nx) &= \ln(nx) +\ln(1+1/nx) \\
&\leq \ln(nx) + 1/nx =\frac{ x^2 \ln (nx) + x/n }{x^2}.
\end{align*}
$$
Now
\begin{align*}
x^2\ln(nx) + x/n &= \left[ 1+x^2\ln(n) \right] \left( \frac{\ln(n)}{\ln(n) + x^{-2} } + \frac{\ln(x) }{\ln(n) + x^{-2} } + \frac{1}{nx(\ln(n) + x^{-2} )} \right)
\end{align*}
giving
\begin{align*}
\frac{ \ln( 1 + nx) }{ 1 + x^2 \ln(n) } &\leq \frac{1}{x^2}\left( \frac{x^2 \ln(n) }{x^2 \ln(n) + 1 } + \frac{x^2 \ln(x) }{x^2 \ln(n) + 1 } +\frac{x^2 }{nx ( x^2 \ln(n) + 1 )}\right) \\
&= \frac{\ln(n) }{x^2 \ln(n) + 1 } + \frac{\ln(x) }{x^2 \ln(n) + 1 } +\frac{1}{nx ( x^2 \ln(n) + 1 )} \\
&\leq
\frac{1}{x^2} \left( 1 + \frac{\ln(x)}{\ln(n)} + \frac{1}{n \ln (n) }\right)
\end{align*}
For $n \geq 3$ this last bound is actually tighter than the one you are trying to prove, and in fact your original inequality fails for $n=2$ but I don't have an immediate assertion of that fact. Will come back to it.
Just to finish the answer off for the case $n=2$ the original inequality is equivalent to
$$
\frac{ \ln(1 + 2x) }{ 1 + \ln(2 x) } \leq \frac{1 + x^2 \ln(2) }{x^2 } = \frac{1}{x^2} + \ln(2), \qquad x \geq 1.
$$
which is true for $x = 1$, but
$$
\lim_{x \rightarrow \infty}\frac{\ln(1 + 2x)}{1 + \ln(2x) } = 1, \qquad \lim_{x \rightarrow \infty} \frac{1}{x^2} + \ln(2) =\ln(2) < 1,
$$
and so the claim is false for $n=1, 2$.
|
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|
Finding the Coefficient of $(x^3 + 2y^2)^n$ containing $x^{18} y^{12}$ I was helping somebody when I scratched my head because of this question. It goes like
this:
If the middle term of the expansion of $(x^3 + 2y^2)^n$ is $C x^{18} y^{12},$ find C.
My work:
I let $u = x^3$ and $v = 2y^2.$ Then doing this: $$(u)^6 = (x^3)^6 \space and \space \space \left(\frac{v}{2} \right)^6 = (y^2)^6$$
we get $u^6 = x^{18}$ and $\left(\frac{v^6}{64} \right) = y^{12}$
We now conclude that the expression $(u + v)^n$ has a term $(u^6)\left(\frac{v^6}{64} \right)$ along its expansion when $u = x^3$ and $v = 2y^2$ We need to find its equivalent term of $(u^6)\left(\frac{v^6}{64} \right)$ when we go back to dealing with $(x^3 + 2y^2)^n.$
Everybody knows that in the binomial expansion of $(u + v)^n,$ in each term, the sum of the exponents of $u$ and $v$ is $n.$ and there are $n+1$ terms.
With that in mind, the sum of the particular term $(u^6)\left(\frac{v^6}{64} \right)$ is $n = 12$ and the number of terms in that particular expansion is $12+1 = 13.$ Since the problem asks for the coefficient of the middle term $C x^{18} y^{12},$ we need to find its middle term. Turns out, in the binomial expansion containing $13$ terms, the middle term would
be the $7$th term.
Now looking for for the expression of the $7$th term:
$$nth \space term = C(n, r-1) u^{n-r+1} v^{r-1}$$
$$expression \space of \space 7th \space term = C(12, 7-1) (x^3)^{12-7+1} (2y^2)^{7-1}$$
$$ = (924) (x^3)^{6} (2y^2)^{6}$$
$$ = (924) (x^{18}) (2)^{6}(y^2)^{6}$$
$$ = (924) (x^{18}) (64) (y^2)^{6}$$
$$ = (924) (x^{18}) (64) (y^{12})$$
$$ = 59136 x^{18} y^{12}$$
Therefore, we conclude that $C = 59136.$
Lastly, the equivalent term of $(u^6)\left(\frac{v^6}{64} \right)$ from $(u + v)^n$, when we go back to dealing with $(x^3 + 2y^2)^n$, is $59136x^{18}y^{12}$
I've done my best but I couldn't verify it. Is my solution correct?
|
The general term of the expansion $(x^3+2y^2)^n$ is
$$\binom{n}{r}2^{n-r}x^{3r}y^{2n-2r}$$
For the term of $x^{18}y^{12}$, take $r=6$ and such that $n=12$
So the coefficient, $C=\binom{12}{6}2^{12-6}=59136$
|
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|
Solving a System of Two Differential Equations and getting the wrong answer Problem:
Solve the following system of differential equations.
\begin{eqnarray*}
5x' + y' - 3x + y &=& 0 \\
4x' + y' - 3x &=& -3t \\
\end{eqnarray*}
Answer:
To solve this, we use the operator method.
\begin{eqnarray*}
(5D-3)x + (D+1)y &=& 0 \\
(D-3)x + Dy &=& -3t \\
D(5D-3)x + D(D+1)y &=& 0 \\
(D+1)(D-3)x + D(D+1)y &=& (D+1)(-3t) = -3t - 3 \\
D(5D-3)x - (D+1)(D-3)x &=& 3t + 3 \\
(5D^2 - 3D)x - (D^2 - 2D + 3)x &=& 3t + 3 \\
(4D^2 - D - 3)x &=& 3t + 3 \\
\end{eqnarray*}
Now to solve this equation, we need to find both the complementary solution and the particular solution. We call the complementary solution $x_c$ and the particular solution $x_p$. To find the complementary solution, we
set up the characteristic equation.
\begin{eqnarray*}
4m^2 - m - 3 &=& 0 \\
m &=& \frac{1 \pm \sqrt{1 - 4(4)(-3)} }{2(4)} = \frac{1 \pm \sqrt{1+ 48} }{8} \\
m &=& \frac{1 \pm 7 }{8} \\
m &=& 1 \,\, \text{or} \,\, m = -\frac{3}{4} \\
x_c &=& c_1 e^{x} + c_2 e^{-\frac{3}{4}} \\
\end{eqnarray*}
Now we need to find $x_p$.
\begin{eqnarray*}
x_p &=& At + B \\
x'_p &=& A \\
x''_p &=& 0 \\
4(0) - A - 3(At + B) &=& 3t + 3 \\
-3A &=& 3 \\
A &=& -1 \\
-A - 3B &=& 3 \\
-(-1) - 3B &=& 3 \\
-3B &=& 3 - 1 = 2\\
B &=& -\frac{2}{3} \\
x_p &=& -t -\frac{2}{3} \\
\end{eqnarray*}
The books answer is:
\begin{eqnarray*}
x &=& c_1 e^{t} + c_2 e ^{3t} + t + \frac{7}{3} \\
y &=& - c_1 e^{t} - 3c_2e^{3t} + 3t + 1 \\
\end{eqnarray*}
Since my answer for $x$ is going to be different from the book, I conclude that I went wrong some where. Where did I go wrong?
Bob
|
from the second equation we obtain:
$$y'=-3t+3x-4x'$$ plugging this in the first equation:
$$x'-y=3t$$ differentiating this with respect to $t$:
$$x''-y'=3$$ and again we have
$$x''-(-3t+3x-4x')=3$$ this is easy to solve.
|
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|
An isosceles triangle with sides... An isosceles triangle with sides $|AB|=6$, $|BC|=|AC|=5$. Calculate the height of the triangle, against the side $BC$.
I drew the following:
By the figure I get that $|CD|=\sqrt{25-x^2}$ and $|BD|=5-\sqrt{25-x^2}.$ Using the pythagorean theorem on $\triangle ABD,$ I get $$36=x^2+\left(5-\sqrt{25-x^2}\right)^2 \Longleftrightarrow x^2 = 25 - \frac{49}{25}.$$
Here is where the trouble starts. I'm not allowed to have a calculator to solve this, but I have to simplify the RHS and check if it is a perfect square. What I did with the RHS was this:
$$25-\frac{49}{25}=\frac{25^2-7^2}{25}=\frac{(25-7)(25+7)}{25}=\frac{18\cdot 32}{25}=\frac{3\cdot3\cdot2\cdot2^5}{5^2}=\frac{3^2(2^3)^2}{5^2}=\left(\frac{24}{5}\right)^2.$$
Thus the only positive solution is $x=\frac{24}{5},$ which indeed is correct. However time is of the essence here, doing that last simplification on the RHS took a while, on probably on a test, I'd just give up thinking I'm wrong because it's not supposed to be that much of arithmetic on these questions.
Is there any simpler way to solve this problem?
|
An easy way to solve this problem is to compute the area of the triangle using Heron's formula. In particular, $s = (6+5+5)/2=8$, and $A = \sqrt{8(8-6)(8-5)^2}=12$. But $A=\frac{5x}{2}$. Therefore, $x = \frac{24}{5}$.
|
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|
Prove that the expression $5^{2n+1} \cdot 2^{n+2} + 3^{n+2} \cdot 2^{2n+1}$ is divisible by 19.
Prove that the expression
$$5^{2n+1} * 2^{n+2} + 3^{n+2} * 2^{2n+1}$$
is divisible by $19$.
I'll skip the basis step (as I have done last time) but I can conclude that it's only divisible by 19 for integers n ≥ 0 (or whole numbers).
II.
Assume that
$$5^{2k+1} * 2^{k+2} + 3^{k+2} * 2^{2k+1}$$
is divisible by 19. Then,
$$5^{2k+3} * 2^{k+3} + 3^{k+3} * 2^{2k+3}$$
is divisible by 19.
Now this is where I get lost, I try to "dismember" the expression to get
$$5^{2k}* 5^3 * 2^k * 2^3 + 3^k * 3^3 * 2^{2k} * 2^3$$
I also try to get it similar to to the assumption to make use of the said assumption yielding
$$5^{2k}* 5 * 5^2 * 2^k * 2^2 * 2 + 3^k * 3^2 * 3 * 2^{2k} * 2 * 2^2$$
$$5^{2k+1} * 5^2 * 2^{k+2} * 2 + 3^{k+2} * 3 * 2^{2k+1} * 2^2$$
$$5^{2k+1} * 25 * 2^{k+2} * 2 + 3^{k+2} * 3 * 2^{2k+1} * 2^2$$
$$50 * 5^{2k+1} * 2^{k+2} + 12 * 3^{k+2} * 2^{2k+1}$$
And this is where I get lost.. : (
Am I missing out something? Had I done it wrong? The number 19 is prime which makes it hard to handle for me. Thanks!
EDIT : After some pondering, I answered it this way :
$$50 * 5^{2k+1} * 2^{k+2} + 12 * 3^{k+2} * 2^{2k+1}$$
I realized that 50 can be written as 38 + 12 (and 38 is a multiple of 19) Hence,
$$ 38 + 12 * 5^{2k+1} * 2^{k+2} + 12 * 3^{k+2} * 2^{2k+1} $$
Factoring out 12, I get :
$$ 38 + 12(5^{2k+1} * 2^{k+2} + 3^{k+2} * 2^{2k+1}) $$
38 is divisible by 19 and the long expression is divisible by 19 (per the assumption) and qed.
Is this correct ?
|
Hint:
$
5^{2n+1} 2^{n+2} + 3^{n+2} 2^{2n+1}
\\= 20\cdot 50^n + 18 \cdot 12^n
\\= 19(50^n + 12^n) + 50^n - 12^n
\\= 19(50^n + 12^n) + (2\cdot 19 +12)^n - 12^n
$
|
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|
About using the lagrange multiplier. Today I have seen a question like the following;
$x+y+z=5$ and $xy+yz+xz=3$ and $x,y,z \in \mathbb{R}^+$
What is the maximum value $x$ can get?
Now it is pretty obvious that question is solvable using many simple elementary methods like say $(x+y+z)^2=x^2+y^2+z^2+2(xy+yz+xz)=25$
$25=x^2+y^2+z^2+6 \rightarrow x^2+y^2+z^2=19$
Applying Cauchy-Schwarz $(y^2+z^2)(1+1)\geq (z+y)^2$
$2(19-x^2)\ge(5-x)^2$
$x_{max}=\dfrac{13}{3}$
However I would also like to solve it by the lagrange multiplier, here are my efforts
$f(x,y,z,k)=x+5-(y+z)+k(x+y+z-5)$
$f_k'=0$
$f_x'=k+1$
$f_y'=k-1$
$f_z'=k-1$
$k=1,-1$ and $x$ goes nowhere near $\dfrac{13}{3}$
So my question is where am I wrong and how can I fix it?
Besides are there maxima minima questions that can't be solved with lagrange,
is this one of them?
|
We will use the quadratic equation approach to determine the min and max values of $x$. We have: $y+z = 5-x$, and $yz + x(y+z) = 3\implies 3- x(5-x)= yz\implies x^2-5x+3= yz\le \dfrac{(y+z)^2}{4}= \dfrac{(5-x)^2}{4}\implies 4x^2-20x+12 \le 25-10x+x^2\implies 3x^2-10x-13 \le 0 \implies (3x -13)(x + 1)\le 0\implies -1 \le x \le \dfrac{13}{3}\implies x_{\text{min}} = -1, x _{\text{max}} = \dfrac{13}{3}$ .
|
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|
Let $(x, y, z)\in \Bbb R$, prove that $\big(\frac{2x-y}{x-y}\big)^2+\big(\frac{2y-z}{y-z}\big)^2 +\big(\frac{2z-x}{z-x}\big)^2 \geqslant 5$
Olympiad Inequation
Let $x$, $y$ and $c$ be distinct real numbers. Prove that:
$$\left(\frac{2x-y}{x-y}\right)^2+\left(\frac{2y-z}{y-z}\right)^2+\left(\frac{2z-x}{z-x}\right)^2 \geqslant 5.$$
This is an assignment I got from a teacher. I've been baffled by the negative sign in the denominator.
Could you please provide insight as to how would one resolve this inequality please? (Do not solve yet, I've yet to turn in this assignment, I just want to get some tips)
|
We want to minimize
$$
\begin{align}
&\underbrace{\left(\frac{mx-ny}{x-y}\right)^2}_{\left(\frac{m+n}2+\frac{m-n}2\frac{x+y}{x-y}\right)^2}+\underbrace{\left(\frac{my-nz}{y-z}\right)^2}_{\left(\frac{m+n}2+\frac{m-n}2\frac{y+z}{y-z}\right)^2}+\underbrace{\left(\frac{mz-nx}{z-x}\right)^2}_{\left(\frac{m+n}2+\frac{m-n}2\frac{z+x}{z-x}\right)^2}\tag{1}
\end{align}
$$
Letting
$$
u=\frac{x+y}{x-y}\quad v=\frac{y+z}{y-z}\quad w=\frac{z+x}{z-x}\tag{2}
$$
transforms minimizing $(1)$ into minimizing
$$
\left(\frac{m+n}2+\frac{m-n}2u\right)^2+\left(\frac{m+n}2+\frac{m-n}2v\right)^2+\left(\frac{m+n}2+\frac{m-n}2w\right)^2\tag{3}
$$
where
$$
\underbrace{\log\left(\frac{u-1}{u+1}\right)}_{\log\left(\frac yx\right)}+\underbrace{\log\left(\frac{v-1}{v+1}\right)}_{\log\left(\frac zy\right)}+\underbrace{\log\left(\frac{w-1}{w+1}\right)}_{\log\left(\frac xz\right)}=0\tag{4}
$$
Thus, for all variations that maintain $(4)$; i.e.
$$
\frac{\delta u}{u^2-1}+\frac{\delta v}{v^2-1}+\frac{\delta w}{w^2-1}=0\tag{5}
$$
we want to minimize $(3)$; i.e.
$$
\left(\frac{m+n}2+\frac{m-n}2u\right)\delta u+\left(\frac{m+n}2+\frac{m-n}2v\right)\delta v+\left(\frac{m+n}2+\frac{m-n}2w\right)\delta w=0\tag{6}
$$
Orthogonality requires that to satisfy $(6)$ for all variations that satisfy $(5)$ there be a $\lambda$ so that
$$
\begin{align}
\lambda
&=\left(u^2-1\right)\left(\frac{m+n}2+\frac{m-n}2u\right)\\
&=\left(v^2-1\right)\left(\frac{m+n}2+\frac{m-n}2v\right)\\
&=\left(w^2-1\right)\left(\frac{m+n}2+\frac{m-n}2w\right)
\end{align}\tag{7}
$$
That is, $u,v,w$ are roots of
$$
x^3+\frac{m+n}{m-n}x^2-x-\frac{m+n+2\lambda}{m-n}=0\tag{8}
$$
The sum of the roots of $(8)$ is
$$
-\frac{m+n}{m-n}\tag{9}
$$
and the sum of the squares of the roots of $(8)$ is
$$
\left(\frac{m+n}{m-n}\right)^2+2\tag{10}
$$
Applying $(9)$ and $(10)$ to $(3)$ yields that the minimum is
$$
3\frac{(m+n)^2}4-\frac{m^2-n^2}2\frac{m+n}{m-n}+\frac{(m-n)^2}4\left[\left(\frac{m+n}{m-n}\right)^2+2\right]=\bbox[5px,border:2px solid #C0A000]{m^2+n^2}\tag{11}
$$
|
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|
Let $y= \frac{x^2+3x+1}{x^2+x+1} \forall x\in R$. Find the range of $y$
Let $y= \dfrac{x^2+3x+1}{x^2+x+1} \forall x\in R$. Find the range of $y$
I have reached:
$x^2(y-1) + x(y-3) + (y-1) = 0$
I also know that the denominator of $y= f(x)$ is greater than $0$.
How do I continue from here? I am unable to find a suitable condition to proceed.
Edit:
vertex of function in the numerator is at $(\frac{-3}{2}, \frac{-5}{4})$
and of the function in the denominator is at: $(\frac{-1}{2},\frac{3}{4})$
but that ain't much helpful, is it?
|
$$y=1+\frac{2x}{x^2+x+1}$$
The function $y(x)$ is continuous and differentiable in $\Bbb R$. Moreover
$\lim_{x\to\pm\infty}y(x)=1$, so $y(x)$ is bounded. This means that the range of $y$ is an interval $I$.
Then
$$y'=\frac{2x^2+2x+2-4x^2-2x}{(x^2+x+1)^2}=\frac{2-2x^2}{(x^2+x+1)^2}$$
which vanishes at $x=\pm1$.
Compute $y(1)$ and $y(-1)$ to find the endpoints of $I$.
|
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|
Logarithmic problem If $\log_{16} 15 =a$ and $\log_{12} 18 =b$, then show that $$\log_{25} 24 = \frac{5-b}{16a-8ab-4b+2}.$$
|
Let $x=\log _{25}24$ so that $25^x=24$
From $16^a=15$ we get $$2^{8a x}=16^{2a x}=15^{2 x}= 3^{2 x}5^{2x}=3^{2x}24=2^33^{2x+1}$$
$$2^{8a x-3}=3^{2x+1}$$
From $12^b=18$ we get
$$2^{2b-1}=3^{2-b}$$
Multiplying these two equations we have
$$(8a x-3)(2-b)=(2b-1)(2x+1)$$
Solve for $x$
|
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|
Proof verification: Determine whether $f(x)$ is one-to-one or not
$$f:\mathbb{R}\mapsto\mathbb{R}
\text{ defined as } f(x)=2x^3+3x-4 \\
\underline{\textit{proof(by contradiction)}}\\
\text{By definition a function is called one-to-one if}\\
f(x)=f(y),\text{ } x,y\in\mathbb{R} \text{ implies that x=y.}\\
\text{Suppose for the sake of contradiction }f(x)=f(y)\text{ but }x\neq y.\\ \text{Therefore } 2x^3+3x-4 = 2y^3+3y-4\\
2x^3-2y^3+3x-3y=0\\
2(x-y)(x^2+xy+y^2)+3(x-y)=0\\
(x-y)\left [2(x^2+xy+y^2)+3)\right ]=0\\
\text{which implies that either } x=y\text{ or }2(x^2+xy+y^2)+3=0 \\
\text{which contradicts to the assumption that } x\neq y \Rightarrow \Leftarrow $$
Is there anything missing in my proof or there is a better solution? I got confused at the either or part because $x=y$ my or may not be true. My logic is given above, if there is something I misunderstand please give your feedback.
$\textbf{FINAL EDIT}$
Here is the continuation of the proof that it is never true that the right factor $x^2+xy+y^2=-\frac{3}{2}$.
$$\text{We know that } x^2\geq0 \text{ and } y^2\geq0\Rightarrow x^2+y^2\geq 0 \text{ and that there are two cases for }x,y.\\
\textbf{case 1: }\text{If nonzero x and y have the same sign then we are done since that implies that }\\
x^2+xy+y^2>0\\
\textbf{case 2: } \text{If nonzero x and y have differing signs }\\ xy<0
\text{ and also we know that } (x+y)^2>0\\
\Rightarrow x^2+2xy+y^2>0 \\
\Rightarrow x^2+xy+y^2>-xy\\
\text{ Since } xy<0 \Rightarrow -xy>0 \Rightarrow x^2+xy+y^2>0 \\
\text{Therefore it is never true that } x^2+xy+y^2=-\frac{3}{2}$$
Is my proof complete now?
|
$$f(x) = f(y)$$
$$\Rightarrow2x^3 + 3x - 4 = 2y^3 + 3y - 4$$
$$\Rightarrow 2(x^3-y^3) + 3(x-y) = 0$$
$$\Rightarrow 2(x-y)(x^2+xy+y^2) + 3(x-y) = 0$$
$$\Rightarrow (x-y)(2x^2 + 2xy + 2y^2 +3) = 0$$
$$\Rightarrow x = y$$
The last step can be justified by saying that the right factor is never $0$. I leave this as an exercise for you.
An alternative solution:
Note that $f'(x) = 6x^2 + 3 = 3(2x^2 +1) > 0$, such that $f$ is strictly increasing and therefore injective.
|
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|
Evaluating $\max(ab+bc+ac)$
Let a,b,c be real numbers such that $a+2b+c=4$. What is the value of $\max(ab+bc+ac)$
My attempt:
Squaring both the sides:
$a^2 +4b^2+c^2+2ac+4bc+4ab=16$
Then I tried factoring after bringing 16 to LHS but couldn't. It's not even a quadratic in one variable or else I could have directly found the maximum.
How do I proceed?
|
From the given condition $\,2b=4-a-c\,$, then:
$$\require{cancel}
\begin{align}
2(ab+bc+ca) &= 2b(a+c)+2ac \\
&= (a+c)(4-a-c)+2ac= \\
&= 4a - a^2 - \bcancel{ac} + 4c - \bcancel{ac} - c^2 + \bcancel{2ac} = \\
&= \color{red}{8} -a^2 + 4a \color{red}{-4} - c^2 +4c \color{red}{-4} = \\
&= 8 - (a-2)^2 - (c-2)^2 \\
&\le 8
\end{align}
$$
Equality is attained for $a=c=2\,$, so the upper bound is in fact a maximum.
|
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|
Evaluate $\lim_{x\rightarrow \infty} (\sqrt{x^2 +ax} - \sqrt{x^2 +bx})$ Evaluate $$\lim_{x\rightarrow \infty} (\sqrt{x^2 +ax} - \sqrt{x^2 +bx})$$
I tried the following:
$$\lim_{x\rightarrow \infty} (\sqrt{x^2 +ax} - \sqrt{x^2 +bx}) \cdot \frac{\sqrt{x^2 +ax} + \sqrt{x^2 +bx}}{\sqrt{x^2 +ax} + \sqrt{x^2 +bx}}$$
But ended up with
$$\lim_{x\rightarrow \infty} \frac{ax-bx}{\sqrt{x^2 +ax} + \sqrt{x^2 +bx}}$$
Which I'm not sure what to do with.
|
$$\lim_{x\rightarrow \infty} (\sqrt{x^2 +ax} - \sqrt{x^2 +bx}) \cdot \frac{\sqrt{x^2 +ax} + \sqrt{x^2 +bx}}{\sqrt{x^2 +ax} + \sqrt{x^2 +bx}}
=(a-b)\lim_{x\rightarrow \infty}\frac{x}{\sqrt{x^2 +ax} - \sqrt{x^2 +bx}}=(a-b)\lim_{x\rightarrow \infty}\frac{x}{\vert{x}\vert\left(\sqrt{1+\frac{a}{x}}+\sqrt{1+\frac{b}{x}}\right)}=\frac{(a-b)}{2}\lim_{x\rightarrow \infty}\frac{x}{\vert x\vert}=\frac{a-b}{2}.$$
Using that $\vert x\vert=\sqrt{x^{2}}$.
|
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|
$66$ points in $100$ shots. I just received a probability problem from a friend via a text and by the time I took to read it, I was sent a solution as well - which is confusing.. The question goes like this..
A person shoots basketball 100 times. First time he scores a point and second time he misses it. For the following shots, the probability of him scoring a point is equal to number of points scored before this shot divided by number of shots taken before this shot, for ex: if he is into his 21st shot and he has scored 13 points in the first 20 shots then the probability of him scoring in 21st shot is 13/20. What is the probability of scoring 66 points in the 100 shots (including the first two)
The confusing solution:
$\dfrac1{(n-1)} \ge \dfrac1{99}$
Can someone please explain the logic behind this?
|
Let $P(x, n)$ be the probability that you have $x$ points after $n$ throws. We are given that $$P(1,2) = 1$$
You are also given information about the probability of scoring at any step, if you know the number of points previously scored. In particular:
$$P(\text{scoring from } x-1, n-1 ) = \frac{x-1}{n-1}$$
$$P(\text{scoring from } x, n-1 ) = \frac{x}{n-1}$$
Now, note that in order to have $x$ points after $n$ throws, we can either have $x-1$ points after $n-1$ throws, then score a point, or have $x$ points after $n-1$ throws, then not score a point.
We can write this as follows:
$$P(x,n) = P(x-1,n-1)P(\text{scoring from } x-1, n-1 ) + P(x,n-1)(1 -
P(\text{scoring from } x, n-1 ))$$
This is the same as
$$P(x,n) = P(x-1,n-1)\left(\frac{x-1}{n-1}\right) + P(x,n-1)\left(1 -\left(\frac{x}{n-1}\right)\right)$$
$$P(x,n) = P(x-1,n-1)\left(\frac{x-1}{n-1}\right) + P(x,n-1)\left(\frac{n-1 - x}{n-1}\right)$$
Call this last result "potato", because we will refer to it later as such.
Statement:
For all $n > 1$ and $1 \leq x \leq n-1$,
$$P(x,n) = \frac{1}{n-1}$$
Proof:
Our base case is $n=2$, for which we are given that $$P(1,2)=1$$
For the inductive step, we assume that for some $k \geq 2$, the following is true for all $x$ between $1$ and $k-1$, inclusive: $$P(x,k) = \frac{1}{k-1}$$
Now suppose $y$ satisfies $2 \leq y \leq k-1$, and we wish to find $P(y, k+1)$. By the previous result "potato", we can write
$$P(y,k+1) = P(y-1,k)\left(\frac{y-1}{k}\right) + P(y,k)\left(\frac{k - y}{k}\right)$$
But we know from our inductive assumption that $P(y,k) = P(y-1,k) = 1/(k-1)$. Therefore,
$$P(y,k+1) = \left(\frac{1}{k-1}\right)\left(\frac{y-1}{k}\right) + \left(\frac{1}{k-1}\right)\left(\frac{k - y}{k}\right)$$
$$P(y,k+1) =\frac{y-1}{k(k-1)} + \frac{k-y}{k(k-1)}$$
$$P(y,k+1) = \frac{k-1}{k(k-1)}$$
$$P(y,k+1) = \frac{1}{k}$$
We are finished for the cases when $2 \leq y \leq k-1$, but we still have to consider the edge cases $y=1$ and $y=k$.
After $k+1$ throws, in order to have only $1$ point, we must miss every throw from the third to the $(k+1)^\text{th}$. The probability of missing the third throw is $1/2$. Given this, the probability of missing the fourth throw is $2/3$. Given this, the probability of missing the fifth throw is $3/4$, and so on. We get a telescoping product:
$$P(1, k+1) = \prod_{j=1}^{k-1} \frac{j}{j+1} = \frac{1}{2} \cdot \frac{2}{3} \cdot \frac{3}{4} \cdots \frac{k-2}{k-1} \cdot \frac{k-1}{k} = \frac{1}{k}$$
Similarly, after $k+1$ throws, in order to have exactly $k$ points, we must get a point on every throw from the third to the $(k+1)^\text{th}$. The probability of getting a point on the third throw is $1/2$. Given this, the probability of getting a point on the fourth throw is $2/3$. Given this, the probability of getting a point on the fifth throw is $3/4$, and so on. We once again have a telescoping product:
$$P(k, k+1) = \prod_{j=1}^{k-1} \frac{j}{j+1} = \frac{1}{2} \cdot \frac{2}{3} \cdot \frac{3}{4} \cdots \frac{k-2}{k-1} \cdot \frac{k-1}{k} = \frac{1}{k}$$
This completes the inductive step, so we've finally shown that for all $n >1$ and $1 \leq x \leq n-1$, $$\boxed{P(x,n) = \frac{1}{n-1}\,}$$
In particular, your question asked for $$P(66, 100) = \frac{1}{99}$$
|
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|
Is writing equalities within an equation abuse of notation? I'll occasionally write equalities within parentheses or sqrt signs to make my steps more compact.
E.g.:
$$ r = \frac{3}{4}\sqrt[4]{\frac{7}{9}\cdot\frac{16}{7}=\frac{16}{9}} = \frac{3}{4}\sqrt{\frac{4}{3}}=\sqrt{\frac{3}{4}} = \frac{\sqrt{3}}{2} $$
I always assumed this was pretty clear, and an acceptable (if unusual) way to use equalities.
However, I was recently told that "you can't take the root of a truth value", which is certainly true.
Is my notation confusing? Does it classify as abuse of notation?
Is the below correct?
$$
r = \left(\frac{7}{9} \cdot \frac{16}{7} = \frac{16}{9}\right) \implies r = \top
$$
|
Your notation is very non-standard and will be confusing to most people. Many people use curly braces to denote intermediate results:
$$
r = \frac{3}{4}\underbrace{\sqrt[4]{\frac{7}{9}\cdot\frac{16}{7}}}_{\sqrt[4]{\frac{16}{9}}} = \frac{3}{4}\sqrt{\frac{4}{3}}=\sqrt{\frac{3}{4}} = \frac{\sqrt{3}}{2}
$$
|
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|
Sum of terms $1^2-2^2+3^2-...$ Prove that: $$1^2-2^2+3^2-4^2+...+(-1)^{n-1}n^2=\frac{(-1)^{n-1}n(n+1)}{2}$$
I proved it by induction but is there any other way to solve it?
If it was not a proof but rather a question like find the term,how to solve it?
I realized that alternate terms were under same sign but can't understand whether to take $\frac{n}{2}$ odd and even terms[if n is even] or $\frac{n+1}{2}$ odd terms and $\frac{n-1}{2}$ even terms[if n is odd].
I thought of this $$1^2-2^2+3^2-4^2+5^2=1^2+{(1+2)}^2+{(1+4)}^2-2^2-4^2$$
$$=1^2+1^2+1^2+2(0+2+4)+2^2+4^2-2^2-4^2$$
But then how to generalize??
|
For example, if $n$ is odd, $n=2k-1$ then we obtain:
$$LS=\frac{n(n+1)(2n+1)}{6}-8\cdot\frac{k(k-1)(2k-1)}{6}=$$
$$=\frac{n(n+1)(2n+1)}{6}-8\cdot\frac{\frac{n+1}{2}\cdot\frac{n-1}{2}\cdot n}{6}=\frac{n(n+1)}{2}$$
|
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|
How to compute $S_{2016}=\sum\limits_{k=1}^{2016}\left(\sum\limits_{n=k}^{2016}\frac1n\right)^2+\sum\limits_{k=1}^{2016}\frac1k$? I came across a question asking the value of the following sum:
\begin{align}
\left(1+ \frac{1}{2}+\frac{1}{3}+\cdots +\frac 1{2015}+\frac{1}{2016}\right)^2 \\
+\left(\frac{1}{2}+\frac{1}{3}+\cdots +\frac 1{2015}+\frac{1}{2016}\right)^2 \\
+ \left(\frac{1}{3}+\cdots +\frac 1{2015}+\frac{1}{2016}\right)^2 \\[5 pt]
+\cdots \qquad\quad \vdots\qquad\qquad\\[5 pt]
+ \left(\frac{1}{2015}+\frac{1}{2016}\right)^2 \\
+ \left(\frac{1}{2016}\right)^2\\
+ \left(1+ \frac{1}{2}+\frac{1}{3}+\cdots +\frac 1{2015}+\frac{1}{2016}\right)\;
\end{align}
I can not find a good way to solve it. Any ideas?
Edit: That is, with no dots, $$S_{2016}=\sum_{k=1}^{2016}\left(\sum_{n=k}^{2016}\frac1n\right)^2+\sum_{k=1}^{2016}\frac1k$$
|
(new solution - a bit shorter)
$$\require{cancel}\begin{align}
&\;\;\;\sum_{j=1}^n\left(\sum_{r=j}^n\frac 1r\right)^2+\sum_{j=1}^n\frac 1j\\
&=\sum_{j=1}^n\left(\sum_{r=j}^n\sum_{s=j}^n\frac 1{rs}\right)+\sum_{j=1}^n\frac 1j\\
&=\sum_{j=1}^n\left[2\sum_{r=j}^n\sum_{s=r}^n\frac 1{rs}-\sum_{r=j}^n\frac 1{r^2}\right]+\sum_{j=1}^n\frac 1j\\
&=2\sum_{j=1}^n\sum_{r=j}^n\sum_{s=r}^n\frac 1{rs}-\sum_{j=1}^n\sum_{r=j}^n\frac 1{r^2}+\sum_{j=1}^n\frac 1j\\
&=2\sum_{r=1}^n\sum_{j=1}^r\sum_{s=r}^n\frac 1{rs}-\sum_{r=1}^n\sum_{j=1}^r\frac 1{r^2}+\sum_{r=1}^n\frac 1r
&&\scriptsize(1\le j\le r\le n)\\
&=\color{lightgrey}{2\sum_{r=1}^n\sum_{s=r}^nr\cdot \frac 1{rs}-\sum_{r=1}^nr\cdot \frac 1{r^2}+\sum_{r=1}^n \frac 1r}\\
&=2\sum_{r=1}^n\sum_{s=r}^n\frac 1s\cancel{-\sum_{r=1}^n\frac 1r}+\cancel{\sum_{r=1}^n\frac 1r}\\
&=2\sum_{s=1}^n\sum_{r=1}^s\frac 1s
&&\scriptsize(1\le r\le s\le n)\\
&=2\sum_{s=1}^n1\\
&=\color{red}{2n}
\end{align}$$
Putting $n=2016$ gives the solution as $2\times 2016=\color{red}{4032}$.
(earlier solution below)
$$\begin{align}
\left(1+\frac 12+\frac 13+\cdots+\frac 1{n-1}+\frac 1n\right)\;\, \\
+\left(1+\frac 12+\frac 13+\cdots+\frac 1{n-1}+\frac 1n\right)^2\\
+\left(\frac 12+\frac13+\cdots+\frac1{n-1}+\frac1n\right)^2\\
+\left(\frac13+\cdots+\frac1{n-1}+\frac1n\right)^2\\
+\cdots\qquad \qquad\vdots\quad \quad\\
+\left(\frac 1{n-1}+\frac 1n\right)^2\\
+\left(\frac 1n\right)^2\\
&=\sum_{j=1}^n \frac 1j+\sum_{j=1}^n\left(\sum_{r=j}^n\frac 1r\right)^2\\\
&=\sum_{j=1}^n\frac 1j+\sum_{j=1}^n\left(\sum_{r=j}^n \frac 1{r^2}+2\sum_{j\le r<s}^n\frac 1{rs}\right)\\
&=\sum_{j=1}^n\frac 1j+\sum_{j=1}^n\sum_{r=j}^n\frac 1{r^2}+2\sum_{j=1}^n\sum_{r=j}^n\sum_{s=r+1}^n\frac 1{rs}\\
&=\sum_{j=1}^n\frac 1j+\sum_{r=1}^n\sum_{j=1}^r\frac 1{r^2}+2\sum_{r=1}^{n-1}\sum_{j=1}^r\sum_{s=r+1}^n\frac 1{rs}
&&\scriptsize(1\le j\le r\le n)\atop{\scriptsize(1\le j\le r<s\le n)}\\
&=\sum_{j=1}^n\frac 1j+\sum_{r=1}^nr\cdot \frac 1{r^2}+2\sum_{r=1}^{n-1}\sum_{s=r+1}^nr\cdot \frac 1{rs}\\
&\color{lightgrey}{=\sum_{j=1}^n \frac 1j+\sum_{r=1}^n \frac 1r+2\sum_{r=1}^{n-1}\sum_{s=r+1}^n\frac 1s}\\
&=2\sum_{r=1}^n \frac 1r+2\sum_{r=1}^{n-1}\sum_{s=r+1}^n\frac 1s\\
&=2\sum_{r=0}^{n-1}\sum_{s=r+1}^n\frac 1s\\
&=2\sum_{s=1}^n\frac 1s\sum_{r=0}^{s-1}1\\
&=2\sum_{s=1}^n\frac 1s\cdot s\\
&=\color{red}{2n}\end{align}$$
Putting $n=2016$ gives the solution as $2\times 2016=\color{red}{4032}$.
|
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|
Prove this sequence using induction Prove the following formula for all positive intergers $n$ using induction:
$1^2+3^2+5^2+...+(2n-1)^2=(n(2n-1)(2n+1))/3$
I have pretty gotten through the whole induction, but something seems wrong with my algebra. Can someone provide an answer to correct my algebra?
Induction step:
$1^2+3^2+...+(2n-1)^2+(2n+1)^2=(n+1)(2(2n+1)-1)(2(n+1)+1)/3$
$n(2n-1)(2n+1)/3+3(2n+1)^2/3=(n+1)(2(2n+1)-1)(2(n+1)+1)/3$
$4n(2n-1)(2n+1)/3=(n+1)(2n+3)/3$
What is wrong?
|
\begin{eqnarray*}
1^2+3^2+5^2+\cdots+(2n-1)^2=\frac{(2n-1)n(2n+1)}{3}
\end{eqnarray*}
So
\begin{eqnarray*}
1^2+3^2+5^2+\cdots+(2n-1)^2 + (2n+1)^2 &=& \frac{(2n-1)n(2n+1)}{3} + (2n+1)^2 \\
&=& \frac{(2n+1)}{3} \underbrace{\left( n(2n-1) +3(2n+1) \right)}_{2n^2+5n+3} \\
&=& \frac{(2n+1)(n+1)(2n+3)}{3}.
\end{eqnarray*}
|
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|
Find minimum possible values of the leading coefficient of quadratic equation let the polynomial $$f(x) = ax^2- bx+ c$$ (where a, b, are positive integers). If $f(p)=f(q)=0$ where $0<p<q<1$ then what is the minimum possible value of a.
I used some basic inequalities to get the following -
$$1) b<2a$$
$$2) a>c$$
$$3) a+c>b$$
But couldn't reach for the inequality for a.
Edit1- The answer is 5
|
You are on the right track. First of all, you don't really need condition $1)$ since it is implied by $2)$ and $3)$. On the other hand, you need to add condition $b^2>4ac$, for example $2x^2-x+1$ satisfies your conditions but has no real roots.
Now, you can exploit the fact that $a,b,c$ are positive integers to get better bounds:
\begin{align}
a&\geq c + 1\\
a+c&\geq b + 1\\
b^2&\geq 4ac + 1
\end{align}
which can be rewritten as $$\max\{b-c+1,c+1\}\leq a\leq \frac{b^2-1}{4c}.$$
Since $\frac{b^2-1}{4c}\geq c + 1$, we get that $b\geq 2c + 1$ and from there that $a\geq b - c + 1\geq c + 2$.
Iterate the procedure, i.e. $\frac{b^2-1}{4c}\geq c + 2$ implies $$b\geq \sqrt {4c^2+8c + 1} > 2c + 1$$ and again, since we are working with integers, $b\geq 2c + 2$, or $a\geq b - c + 1\geq c + 3$.
Rinse and repeat, $\frac{b^2-1}{4c}\geq c + 3$ implies $$b\geq \sqrt{4c^2+12c + 1}> 2c + 2,$$
i.e. $b\geq 2c + 3$ which implies $a\geq b - c + 1\geq c + 4$.
Notice that repeating the same thing will not give us $$b\geq \sqrt{4c^2+16c + 1} > 2c + 3$$ so we cannot improve the bound $a\geq c + 4$. But, this bound gives us plenty, since it implies that $a\geq 5$.
Finally, to see that $a = 5$ works, you just need to find an example, for instance, $5x^2 -5x + 1$ gives roots $x_{1,2} = \frac{1}{10}(5\pm\sqrt 5)$.
|
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|
$(a-3)^3+(b-2)^3+(c-2)^3=0$, $a+b+c=2$, $a^2+b^2+c^2=6$. Prove that at least one from $a,b,c$ is 2.
Assume that $\{a,b,c\} \subset \Bbb R$, $(a-3)^3+(b-2)^3+(c-2)^3=0$, $a+b+c=2$, $a^2+b^2+c^2=6$.
Prove that at least one of the numbers $a, b, c\ $ is 2.
This is from a list of problems used for training a team for a math olympics. I tried to use known Newton identities and other symmetric polynomial results but without success (perhaps a wrong approach). Sorry if it is a duplicate. Hints and answers are always welcomed.
Edit: There is a problem with the original statement of the question in the original source. Under these assumptions it is impossible to have $a, b, c$ with value 2, as spotted in the comments and proved by the answers below
|
Suppose that $a=2$. Then $b+c=0$ and hence $b=-c$. The third equation gives $b^2+c^2=2$ hence $2b^2=2$ and $b^2=1$. If $b=1$ then $c=-1$ and vice versa. Then $(2-3)^3+(1-2)^3+(-1-2)^3=-1-1-27=-29$. The same thing happens when $b=-1$ and $c=1$.
If $b=2$, then $a=-c$ again which leaves with $a$ being either $1$ or $-1$. Either way the first equation doesn't hold. Similar thing happens when $c=2$
|
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|
Probability about oranges There are $15$ oranges in each bag.
Each orange has a probability of $10$% to be bad-looking.
Supermarkets only sell the bags that have $12$ or more good-looking oranges.
What percentage of all the bags will be sold?
|
Compute:
$$
\ \ \ \
{15 \choose 3} \cdot Pr(3 \ \text{bad-look oranges})
\cdot Pr(12 \ \text{good-look oranges})
\\ +
{15 \choose 2} \cdot Pr(2 \ \text{bad-look oranges})
\cdot Pr(13 \ \text{good-look oranges}) \\ +
{15 \choose 1} \cdot Pr(1 \ \text{bad-look oranges})
\cdot Pr(14 \ \text{good-look oranges}) \\ +
Pr(15 \ \text{good-look oranges}) \\ =
{15 \choose 3} \cdot (\dfrac{1}{10})^3 \cdot (\dfrac{9}{10})^{12} +
{15 \choose 2} \cdot (\dfrac{1}{10})^2 \cdot (\dfrac{9}{10})^{13} +
{15 \choose 1} \cdot (\dfrac{1}{10}) \cdot (\dfrac{9}{10})^{14} +
(\dfrac{9}{10})^{15}
$$
|
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|
Induction proof of $F(n)^2+F(n+1)^2=F(2n+1)$, where $F(n)$ is the $n$th Fibonacci number.
Let $F(n)$ denotes the $n$th number in Fibonacci sequence. Then for all $n\in\mathbb{N}$,
$$F(n)^2+F(n+1)^2=F(2n+1).$$
I know how to prove it by using the formula
$$F(n)=\frac{\left(\frac{1+\sqrt5}{2}\right)^n-\left(\frac{1-\sqrt5}{2}\right)^n}{\sqrt{5}},$$
but is there a way to prove it by induction?
I am year 12 standard so please don't go too deep.
|
\begin{align} &F(x)=F(x-1)+F(x-2) = F(2)F(x-1)+F(1)F(x-2). \ \\ \ \\ &F(x-1)=F(x-2)+F(x-3) \\ &\Rightarrow F(x)=F(2)(F(x-2)+F(x-3))+F(1)F(x-2)=(F(1)+F(2))F(x-2) \\ &+F(2)F(x-3)=F(3)F(x-2)+F(2)F(x-3). \ \\ \ \\ &F(x-2)=F(x-3)+F(x-4). \\ &\Rightarrow F(x)=F(3)(F(x-3)+F(x-4))+F(2)F(x-3)=(F(2)+F(3))F(x-3) \\ &+F(3)F(x-4)=F(4)F(x-3)+F(3)F(x-4). \\ &\cdot \\ &\cdot \\ &\cdot \\ &\therefore F(x)=F(k+1)F(x-k)+F(k)F(x-k-1). \\ &x=2k+1; \ F(2k+1)=F(k)^2+F(k+1)^2. \\ \ \\ &\therefore F(x)^2+F(x+1)^2=F(2x+1).\end{align}
|
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|
Determine for what values of $x$ the given series converges The given series is
$\sum_{n=1}^∞
(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\frac{1}{6}+\frac{1}{7}+...+\frac{1}{n}
)x^{n} $.
I tried it by using Cauchy Root Test as follows-
Let
$y=\lim_{n\to\infty}(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\frac{1}{6}+\frac{1}{7}+...+\frac{1}{n} )^{1/n}$,
then by taking logarithm both sides,we get
$\log(y)=\lim_{n\to\infty}\frac{1}{n}\log(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\frac{1}{6}+\frac{1}{7}+...+\frac{1}{n} )$
Since, $\log(0)=-\infty$. So, $\log(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\frac{1}{6}+\frac{1}{7}+...+\frac{1}{n} )=-\infty$.
Applying,L'Hôpital's rule,we get
$\log(y)=\lim_{n\to\infty}-\frac{(\frac{1}{n^2}+\frac{1}{(n-1)^2}+\frac{1}{(n-2)^2}+...+...+... )}{(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\frac{1}{6}+\frac{1}{7}+...+\frac{1}{n} )}$.(Please Check this step!!)
Since,$\sum_{k=1}^\infty\frac{1}{k^2}=\lim_{n\to\infty}\sum_{k=1}^n\frac{1}{k^2}=\frac{\pi^2}{6}$ & $\sum_{k=1}^\infty\frac{1}{k}=\lim_{n\to\infty}\sum_{k=1}^n\frac{1}{k}=\infty$.So,$\log(y)=0\implies y=1$.
Now let $a_n=(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\frac{1}{6}+\frac{1}{7}+...+\frac{1}{n} )x^{n} $. Then,$$(a_n)^{1/n}=\lim_{n\to\infty}(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\frac{1}{6}+\frac{1}{7}+...+\frac{1}{n} )^{1/n}x$$
$\implies \lim_{n\to\infty}\vert (a_n)^{1/n}\vert=1.x$.By ,Cauchy root test the given series converges if $\lim_{n\to\infty}(a_n)^{1/n}<1$.Hence, the given series converges if $\vert x\vert<1$.
I NEED TO KNOW WHETHER MY SOLUTION IS CORRECT OR NOT?
|
There are two serious problems. The first one is your assertion that
$$
\log\Bigl(1+\frac12+\dots+\frac1n\Bigr)=-\infty.
$$
It is well known (harmonic series) that
$$
\lim_{n\to\infty}\Bigl(1+\frac12+\dots+\frac1n\Bigr)=+\infty,
$$
so that
$$
\lim_{n\to\infty}\log\Bigl(1+\frac12+\dots+\frac1n\Bigr)=+\infty.
$$
The second one is about the use of L'Hôpital's rule. It is incorrect by two reasons: 1) you have a sequence, not a function and 2) the index of summation is also a variable.
There are several ways to solve the problem. For example, from the inequality
$$
1\le1+\frac12+\dots+\frac1n\le n
$$
it follows that
$$
\lim_{n\to\infty}\Bigl(1+\frac12+\dots+\frac1n\Bigr)^{1/n}=1.
$$
Or you can use the well known fact that
$$
1+\frac12+\dots+\frac1n\sim\log n.
$$
|
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|
Find $\lim\limits_{x\to 0} \frac{\sqrt{\ 1+x} - \sqrt{\ 1-x}}{\sqrt[3]{\ 1+x} - \sqrt[3]{\ 1-x}}$ This problem
$$\lim_{x\to 0} \frac{\sqrt{\ 1+x} - \sqrt{\ 1-x}}{\sqrt[3]{\ 1+x} - \sqrt[3]{\ 1-x}}$$
is from Silverman's "Modern Calculus and Analytical Geometry" Section 22, #16d. I've been struggling on it for a while and can't figure out what to do besides trying to multiply by the conjugate and/or substitution but it doesn't work out. What do you all think? Keep in mind you can't use L'Hopital's rule, only elementary math.
|
Let
$$a=(x+1)^{\frac{1}{6}}$$
$$b=(1-x)^{\frac{1}{6}}$$
Our equation becomes,
$$\lim_{x\to 0} \frac{a^3-b^3}{a^2-b^2}$$
$$\lim_{x\to 0} \frac{(a-b)(a^2+ab+b^2)}{(a-b)(a+b)}$$
$$\lim_{x\to 0} \frac{a^2+ab+b^2}{a+b}$$$$\lim_{x\to 0} \frac{(x+1)^{\frac{1}{3}}+(1-x^2)^{\frac{1}{6}}+(1-x)^{\frac{1}{3}}}{(x+1)^{\frac{1}{6}}+(1-x)^{\frac{1}{6}}}$$$$\frac{1+1+1}{1+1}$$$$\frac{3}{2}$$
|
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|
Binomial expansion of complex variable For the binomial expansion $ (1+x)^n = p_0 + p_1x+p_2x^2+...,$ where
$p_i$ refer to the binomial coefficients.
Need to substitute the cube roots of unity (1, w, $w^2$) for the variable x; with w = $\dfrac{-1+i\sqrt{3}}{ 2},$
find the sums:\begin{align} (a) p_0 + p_3 + p_6 + ..., \\ (b) p_1 + p_4 + p_7 + ..., \\ (c) p_2 + p_5 + p_8 + ..., \\
\end{align}
I have substituted the values as follows:
$(d)\;2^n = $$p_0+p_1+p_2+ p_3 + p_4 + p_5+...$
$(e)\;(1+w)^n = $$p_0+p_1.w+p_2.w^2+p_3+p_4.w+p_5.w^2+...$
$(f)\;(1+$$w^2$$)^n =(-w)^n = $$p_0+p_1.w^2+p_2.w+p_3+p_4.^2+p_5.w^2+...$
On adding (d), (e), (f); get the non-multiple of 3 terms vanished by the sum property of the cube roots of unity (1+w+$w^2$=0). Rest (non-zero) terms have each $p_i$ terms' coefficient being 3 (1+1+1=3) for all such terms.
(i) $(2^n + (1+w)^n + (-w)^n)/3 = p_0 + p_3 + p_6 + ...$
Geometrically/algebraically plotting the points (1+w), (1+$w^2$); we get:
1+w=$\dfrac{1+i\sqrt{3}}{2},$ 1+$w^2$=-w=$\dfrac{1-i\sqrt{3}}{2}.$
Obtaining the trigonometrical, & then exponential forms for the same, get:
(ii) 1+w=$\dfrac{1+i\sqrt{3}}{2}$= $cis(\dfrac{n\pi}{3})$= $e^\dfrac{in\pi}{3}$
(iii) -w =$\dfrac{1-i\sqrt{3}}{2}$= $cis(\dfrac{-n\pi}{3})$= $e^\dfrac{-in\pi}{3}$
Adding $e^\dfrac{in\pi}{3}$ + $e^\dfrac{-in\pi}{3}$ we get:
2.$\dfrac{e^\dfrac{in\pi}{3} + e^\dfrac{-in\pi}{3}}{2} = 2.\dfrac{cos(n\pi)}{3}$
Hence, get (i) modified as:
(iv) $p_0 + p_3 + p_6 + ... = \dfrac{2^n + 2cos(\dfrac{n\pi}{3})}{3}$
This solves only the part (a) of the question.
For the parts (b) & (c), there is no such closed form expression possible for me.
|
Ill do part 2. Since all coefficients of $x^{1+3k}$ are to be retained, we multiply by $x^2$ to make the power $x^{3+3k}$, ie a multiple of $3$.
$$(1+x)^n = a_o + a_1 x + a_2 x^2...a_nx^n$$
$$x^2(1+x)^n = a_ox^2+a_1x^3+a_2 x^4 ... a_nx^{n+2}$$
Now substituting cube roots of unity in above equation leaves required terms intact ($z^3 = 1$, where $z$ is cube root of unity).
All the other terms vanish, ( $1+z+z^2 =0$ if $z$ is cube root of unity)
Let $P(x) = x^2 (1+x)^n$. Then on we have:
$$\begin{align} a_1+a_4+a_7 ... &= \dfrac{P(1) + P(\omega) + P(\omega^2)}{3} \\
&= \dfrac{2^n + \omega^2(1+\omega)^n + \omega(1+\omega^2)^n}{3}
\end{align}$$
|
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|
Find the area of a triangle using the sides lengths
The sides lengths of a triangle $a,$ $b$ and $c$ verify: $$\sqrt{a-24} + b^2 +|c-12\sqrt{3} |+144=12b.$$ The task is to find the area of the triangle.
I'm trying to apply the heron's formula: $$\dfrac{\sqrt{(a+b+c)(a+b-c)(a+c-b)(b+c-a)}} {4}.$$
How do i get to heron's formula from what i know? Or it can be solved in another way?
|
$$\sqrt{a-24} + b^2 +|c-12\sqrt{3} |+144-12b=\sqrt{a-24} + (b-6)^2 +|c-12\sqrt{3} |+108>0.$$
Thus, we have no any triangle with your given.
If you mean $$\sqrt{a-24} + b^2 +|c-12\sqrt{3} |+144=24b$$ then we get $$(a,b,c)=(24,12,12\sqrt3)$$ and now you can use the Heron's formula.
|
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|
Show that $\limsup \frac{\epsilon_1+\cdots +\epsilon_n}{n}\geq 0$？ Suppose $a_n$ is a sequence such that $a_n \downarrow 0$,
$\epsilon_n=-1$ or $1$ for all $n$, the series $\sum a_n$ diverges but the series $\sum_{n=1}^\infty \epsilon_n a_n$ converges. Is it true that $\limsup \frac{\epsilon_1+\cdots +\epsilon_n}{n}\geq 0$?
Question history: Originally, this question is about the assertion $\limsup \frac{\epsilon_1+\cdots +\epsilon_n}{n}=0$, which has been proved wrong by some wonderful answers.
Note $a_n$ is non-increasing to zero here. Some people forget this assumption.
|
The statement is not true. Using Michael's construction, but simplified gives:
$$\epsilon_n = \{1, -1, -1, 1, 1, 1,1, \dots\}$$
$$\epsilon_n = (-1)^{\lfloor \log_2 n\rfloor}$$
After $1$, $7$, $31$, $\dots$, $2^{2k+1} - 1$ numbers we hit a maximum. This maximum is exactly:
$$\sum_{i=0}^{2k} (-2)^i = \frac{1}{3}(2^{2k+1} + 1)$$
So at this maximum our $\sum e_n/n = \frac{1}{3}\dfrac{2^{2k+1} + 1}{2^{2k+1} - 1}$. So for large $n$ our $\limsup \sum e_n/n = \frac{1}{3}$.
$$a_n = \{\frac1{1\cdot 2^0}, \frac1{2\cdot 2^1}, \frac1{2\cdot 2^1}, \frac1{3\cdot 2^2}, \frac1{3\cdot 2^2}, \frac1{3\cdot 2^2}, \frac1{3\cdot 2^2}, \frac1{4\cdot 2^3}, \dots\}$$
$$a_n = \{\frac11, \frac1{4}, \frac14, \frac1{12}, \frac1{12}, \frac1{12}, \frac1{12}, \frac1{32}, \dots\}$$
$$a_n = \frac{1}{(\lfloor\log_2 n\rfloor + 1)2^{\lfloor\log_2 n\rfloor}}$$
$a_n$ is the harmonic series, with the $k$th element repeated $2^{k-1}$ times, divided by $2^{k-1}$. $\sum a_n$ diverges because the harmonic series does. Then, $\epsilon_n a_n$ is the alternating harmonic series, with the $k$th element repeated $2^{k-1}$ times, divided by $2^{k-1}$. The sum of this series converges just like the alternating harmonic series does:
$$\sum_{n=1}^\infty \epsilon_na_n = \sum_{k=0}^\infty 2^{k}\cdot (-1)^{k} \cdot \frac{1}{(k + 1)2^{k}} = \sum_{k=0}^\infty \frac{(-1)^k}{k+1} = \ln(2) \tag*{$\blacksquare$}$$
|
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|
Why is $\sqrt[3]{18+5\sqrt{13}} + \sqrt[3]{18-5\sqrt{13}} = 3$?
How to show that $$\sqrt[3]{18+5\sqrt{13}} + \sqrt[3]{18-5\sqrt{13}} = 3?$$
This equality comes from solving $$t^3 - 15 t - 4 = 0$$ using Cardanos fomula and knowing the solution $t_1=4$.
I have attempted multiplying the whole thing with $(\sqrt[3]{18+5\sqrt{13}})^2 - (\sqrt[3]{18-5\sqrt{13}})^2$, but no success. Then I have solved for one cubic root and put all to the third power. Also no success.
|
Let $(a + b\sqrt{13})^3 = (18 + 5\sqrt{13})$ for $a, b \in \Bbb Q$
Expanding the LHS gives,
$$(a^3 + 39 ab^2 - 18 ) +\sqrt{13}(3a^2 b + 13 b^3 - 5) = 0$$,
From this we get,
$$\begin{cases}a^3 + 39 ab^2 - 18 = 0 \\ 3a^2 b + 13 b^3 - 5 = 0\end{cases}$$
Solving the system give $ a = \dfrac 32$ and $ b = \dfrac12$
Therefore
$$\sqrt[3]{(18 + 5\sqrt{13})} = \dfrac 32 +\dfrac12\sqrt{13}$$
Similarly,
$$\sqrt[3]{(18 - 5\sqrt{13})} = \dfrac 32 -\dfrac12\sqrt{13}$$
Hence the sum is $3$.
|
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|
If $u_{n+1} = u_n^2 - u_n + 1$ and $u_0=a>1$, show that $\sum\frac{1}{u_n} \rightarrow \frac{1}{a-1}$ If $(u_n)\in\mathbb{R}^\mathbb{N},u_{n+1} = u_n^2 - u_n + 1$ and $u_0=a>1$, show that $\sum\dfrac{1}{u_n} \rightarrow \dfrac{1}{a-1}$.
I don't even know how to prove that the series converges
|
The key observation, which kills theproblem quite instantly, is the following: for every $n$ it holds the relation $$\frac{1}{u_n} = \frac{1}{u_n - 1} - \frac{1}{u_{n + 1} - 1}$$ Indeed we have $$\frac{1}{u_n - 1} - \frac{1}{u_{n + 1} - 1} = \frac{1}{u_n - 1} - \frac{1}{u_n(u_n - 1)} = \frac{u_n - 1}{u_n(u_n - 1)} = \frac{1}{u_n}$$
Hence we have a telescopic sum: $$\sum_{n = 0}^{N} \frac{1}{u_n} = \sum_{n = 0}^N \left(\frac{1}{u_n - 1} - \frac{1}{u_{n + 1} - 1}\right) = \\ = \left(\frac{1}{u_0 - 1} - \frac{1}{u_1 - 1}\right) + \left(\frac{1}{u_1 - 1} - \frac{1}{u_2 - 1}\right) + \cdots + \left(\frac{1}{u_N - 1} - \frac{1}{u_{N + 1} - 1}\right) = \\ = \frac{1}{a - 1} + \frac{1}{u_{N + 1} - 1}$$ Now it is pretty easy to prove that $\displaystyle \lim_{n \rightarrow \infty} u_n = \infty$ and thus $$\lim_{N \rightarrow \infty} \sum_{n = 0}^N \frac{1}{u_i} = \lim_{N \rightarrow \infty} \left(\frac{1}{a - 1} + \frac{1}{u_{N + 1} - 1}\right) = \frac{1}{a - 1}$$
So the series converges to $\dfrac{1}{a - 1}$.
|
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|
Find the number of solutions of x+y+z=17? Find the number of solutions of $x+y+z=17$ where $2\le x\le 5, 3\le y \le 6, 4\le z\le7$.
My approach:
The number of solutions with the indicated constraints is the coefficient of $x^{17}$ in the expansion of
($x^2+x^3+x^4+x^5)(x^3+x^4+x^5+x^6)(x^4+x^5+x^6+x^7)$
I have changed the above polynomial to
$x^9(1+x+x^2+x^3)^3$
Now $x^{17}=x^9*x^8$
So I must now find the coefficient of $x^8$ in the expansion of ($1+x+x^2+x^3)^3$
$x_1+x_2+x_3=8$
This is equal to $C(3+8-1,8)=45$
However it takes into values of $x,y,z$ greater than $3$.
So I must subtract those combinations where either of $x_1,x_2,x_3$ is greater than $3$.
Let's suppose $x_1\ge 4$.Then
$x_1+x_2+x_3=4$
Solutions=$15$.
Similarly for $x_2,x_3$ we get $15$ solutions each.
Total=$45$.
Now we must consider the case when more than one of $x_1+x_2+x_3\ge 4$.
For this we have $3$ solutions.
Total solutions=$48$.
Now I must subtract these from original solutions of $45$. This gives answer $=-3$.
But the correct answer is $3$.
What is wrong in this approach?
|
I double counted some solutions. The 3 solutions which I added to 45 to get 48 must be subtracted from 45 to get 42. This is because these 3 solutions were counted twice.
|
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|
Find the matrix of a linear transformation If T : $\mathbb R^{3}$$\mapsto$$\mathbb R^{3}$ is a linear transformation such that
T $\begin{pmatrix}
1 \\
0 \\
0 \\
\end{pmatrix}$ = $\begin{pmatrix}
3 \\
1 \\
4 \\
\end{pmatrix}$, $T$ $\begin{pmatrix}
0 \\
1 \\
0 \\
\end{pmatrix}$ = $\begin{pmatrix}
-1 \\
-1\\
3\\
\end{pmatrix}$ , $T$ $\begin{pmatrix}
0 \\
0 \\
1 \\
\end{pmatrix}$ = $\begin{pmatrix}
4 \\
-3 \\
-1\\
\end{pmatrix}$ then $T$ $\begin{pmatrix}
-5\\
4 \\
4 \\
\end{pmatrix}$ = $\begin{pmatrix}
? \\
? \\
? \\
\end{pmatrix}$
I do not know where to start with this question, I tried doing RREF of the transformation numbers but I feel that is wrong.
|
To expand on my comment above since @Gregory already gave the answer using one method:
Given information on how $T$ acts on the standard basis, knowing $T\begin{bmatrix}1\\0\\0\end{bmatrix}=\begin{bmatrix}3\\1\\4\end{bmatrix}$ and so on, this tells you exactly how to represent $T$ as a matrix. The columns of $T$ are quite simply the corresponding results of $T$ being applied to the standard basis vectors.
We get then $T=\begin{bmatrix}3&-1&4\\1&-1&-3\\4&3&-1\end{bmatrix}$ and so all that remains to your problem is completing the matrix arithmetic necessary.
|
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|
Determine the limit of a sequence using squeeze theorem $$ \left(\frac{\left(1+2\left(-1\right)^x\left(x+4x^2\right)\right)}
{5+7x^3}\right)\left(\sin \left(\frac{\left(3x\pi \right)}{7}\right)\ +\cos
\left(\frac{1}{x}\right)\right)
$$
I have to find the limit of this equation. I am confused by the (-1)^x and how to factor that out. Thanks for the help
|
Noting
$$ \bigg|\left(\frac{\left(1+2\left(-1\right)^x\left(x+4x^2\right)\right)}
{5+7x^3}\right)\left(\sin \left(\frac{\left(3x\pi \right)}{7}\right)\ +\cos
\left(\frac{1}{x}\right)\right)\bigg|\le\frac{2\left(1+2\left(x+4x^2\right)\right)}
{5+7x^3}
$$
and
$$ \lim_{x\to\infty}\frac{2\left(1+2\left(x+4x^2\right)\right)}
{5+7x^3}=0$$
one has
$$ \lim_{x\to\infty}\left(\frac{\left(1+2\left(-1\right)^x\left(x+4x^2\right)\right)}
{5+7x^3}\right)\left(\sin \left(\frac{\left(3x\pi \right)}{7}\right)\ +\cos
\left(\frac{1}{x}\right)\right)=0. $$
|
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|
The manager chooses 2 guests at random. Find the probability that they are both staying in the same room. There are $25$ rooms in a hotel; $3$ with $0$ guests, $4$ with $1$ guest each, $12$ with $2$ guests each, $5$ with $3$ guests each and $1$ with $4$ guests.
Total rooms: $25$
Total guests: $47$
Q. The manager chooses $2$ guests at random. Find the probability that they are both staying in the same room.
I got
$$12\left(\frac{2}{47} \cdot \frac{1}{46}\right) + 5\left(\frac{3}{47} \cdot \frac{2}{46}\right) + 1\left(\frac{4}{47} \cdot \frac{3}{46}\right) = \frac{33}{1081}$$
Somebody explain this to me.
|
There are
$$\binom{47}{2}$$
ways for the manager to select two of the forty-seven guests in the hotel.
In order to select two guests from the same room, the manager must select both guests from a room with two people or two of the three guests in a room with three people or two of the four guests in a room with four people. Since there are $12$ rooms with two people, $5$ rooms with three people, and one room with four people, the manager can do this in
$$12\binom{2}{2} + 5\binom{3}{2} + \binom{4}{2}$$
ways.
Hence, the desired probability is
$$\frac{12\dbinom{2}{2} + 5\dbinom{3}{2} + \dbinom{4}{2}}{\dbinom{47}{2}}$$
Note: I assume you meant to write
$$12\left(\frac{2}{47} \cdot \frac{1}{\color{red}{46}}\right) + 5\left(\frac{3}{47} \cdot \frac{2}{46}\right) + 1\left(\frac{4}{47} \cdot \frac{3}{46}\right)$$
in which case you would have obtained the correct probability.
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Does $c^{a} + c^{b} = a^{c} + b^{c}$ when $a \ne b$? Right now I'm working on a computer algorithm that manipulates a binary tree data structure. Within this tree, there are many "gaps"; i.e., nodes that don't produce children nodes, thus resulting in gaps that become increasingly large as the tree grows.
That being said, when attempting to account for these gaps, I describe a cross section of the tree as a collection of sequences of nodes that either do or don't exist, and represent the length of each sequence as a power of $2$. When doing so, I stumbled upon something that I found to be interesting..
Let $a, b$ and $c \in \mathbb{Z}^+$. For a given value of $c$, does there exist an $a$ and $b$ s.t. $c^{a} + c^{b} = a^{c} + b^{c}$ when $a \ne b$?
Can someone please show this to be true/not true? And if this is true for some values of $a, b,$ and $c$, is it possible to generalize this behavior?
Or, if this has already been studied, can someone please provide a link to relevant literature? Thanks in advance.
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The only triples of positive integers $(a, b, c)$ for which $a \neq b$ and $c^a + c^b = a^c + b^c$, up to interchange of $a$ and $b$, are $(1, 3, 2)$, $(2, 4, 2)$, and $(2, 4, 4)$. Allowing $a = b$ adds the the solutions $(4, 4, 2)$ and $(2, 2, 4)$ as well as the trivial solutions $a = b = c$.
For $c$ fixed, $a^c + b^c$ grows polynomially in $a$ and $b$, whereas $c^a + c^b$ grows exponentially. So there should be few solutions except when $a$ is large compared to $c$ and $b$ is small, or vice versa. To eliminate these possibilities, we need three lemmas providing (rather loose) bounds on expressions of the form $c^a - a^c$.
Lemma 1: For $3 \leq k < n$, $$k^n - n^k > k^\theta$$ where $\theta = 1$ for $k = 3$ and $\theta = k$ for $k \geq 4$.
Proof: Rearrange to $(k^{n-\theta} - 1) k^\theta > n^k$, take logarithms to get $\ln (k^{n-\theta} - 1) + \theta \ln k > k \ln n$, and further rearrange to $$\ln (k^{n-\theta} - 1) + (\theta - k) \ln k > k \ln \frac{n}{k}.$$
The LHS is $\ln (k^{n-k} - k^{\theta-k})$, and the RHS is less than $n-k$ (because $\ln x \leq x-1$ with equality only for $x=1$), so it sufficies to prove $$\ln (k^{n-k} - k^{\theta-k}) > n-k$$
or, taking exponentials and rearranging,
$$e^{n-k} ((k/e)^{n-k} - 1) > k^{\theta - k}.$$ Since $k \geq 3 > e$, both terms on the LHS are positive and increasing with $n$, so it suffices to prove the case where $n = k+1$, i.e., $k - e > k^{\theta - k}$. If $k \geq 4$, taking $\theta = k$ gives $k - e > 1$; if $k = 3$, taking $\theta = 1$ gives $3 - e > 1/9$, both manifestly true. QED
Lemma 2: For $n \geq 5$, we have $2^n - n^2 \geq 9$.
Proof: The bound holds for $n = 5$, and the difference $2^{n+1} - 2^n = 2^n$ between consecutive powers of 2 is greater than the difference $(n+1)^2 - n^2 = 2n+1$ between consecutive squares (and thus $2^n - n^2$ is strictly increasing) for $n \geq 5$. QED
Corollary: The only pairs of integers $(k, n)$ for which $k^n \leq n^k$ and $2 \leq k < n$ are $(2, 3)$ and $(2, 4)$.
Lemma 3: Again with $3 \leq k < n$, we have $k^n - n^k > n$.
Proof: We'll prove that we can pick some constant $c > 1 + n^{1-k}$ to get an inequality of the form $k^n > c n^k$. Since $1 + n^{1-k}$ decreases with increasing $n$, it suffices to show that there's some possible $c > 1 + (1+k)^{1-k}$.
Take logarithms of $k^n > c n^k$ to get $n \ln k > \ln c + k \ln n$ and rearrange to $$(n-k) \ln k > \ln c + k \ln \frac{n}{k}.$$ As $\ln x \leq x-1$, it suffices to prove $$(n-k) \ln k > \ln c + (n-k)$$ or, rearranging, $$(n-k) (\ln k - 1) > \ln c.$$ The LHS of this inequality increases with $n$, so it suffices to prove the case $n = k + 1$, which holds as long as $c < k/e$.
Now we need to find some $c$ to satisfy $1 + (1+k)^{1-k} < c < k/e$; that is, we need to show that $1 + (1+k)^{1-k} < k/e$. For $k = 3$, this inequality becomes $17/16 < 3/e$, which is true. As the LHS $1 + (1+k)^{1-k}$ decreases for increasing $k$ (as the base $1+k$ increases and the negative exponent $1-k$ decreases) and the RHS $k/e$ increases, the inequality also holds for all greater $k$. QED
Now we return to the main question. Consider five cases, depending on the value of $c$.
Case 1 ($c = 1$): $c^a + c^b = a^c + b^c$ becomes $a + b = 2$, which has no solutions in distinct positive integers.
Case 2 ($c = 2$): The equation is $2^a + 2^b = a^2 + b^2$. At least one of $a$ or $b$ has to be in $\{2, 3, 4\}$, as otherwise each term on the LHS exceeds the corresponding term on the RHS. Suppose without loss of generality that $a \in \{2, 3, 4\}$. If $a = 2$ then $b = 4$ and vice versa, as $2^n = n^2$ is solved only for $2$ and $4$. If $a = 3$, then we have $2^b - b^2 = 1$, which is solved only for $b = 1$ (by Lemma 2).
Case 3 ($c = 3$): The equation is $3^a + 3^b = a^3 + b^3$. But $3^n \geq n^3$ for every integer $n$, with equality only if $n = 3$ (by testing $n = 1, 2, 3$ and then using Lemma 1 for $n \geq 4$), so this case is impossible.
Case 4 ($c = 4$): The equation is $4^a + 4^b = a^4 + b^4$. $4^n = n^4$ only for $n = 2, 4$ (giving $(a, b) = (2, 4)$ as a solution), and $4^n < n^4$ only for $n = 3$, so there are no solutions with $a$ and $b$ both than $4$. A solution with $a = 3$ must have $b$ such that $4^b - b^4 = 3^4 - 4^3 = 17$. But $b = 1, 2, 3, 4$ can be eliminated individually, and for $b > 4$ we have $4^b - b^4 > 4^4 = 64$ by Lemma 1, so this is impossible.
Case 5 ($c \geq 5$): Assume without loss of generality that $a < b$. For any positive $n$, $c^n > n^c$ if and only if $n > c$ or $n = 1$, and $c^n = n^c$ only if $n = c$, so either $a=1$ or $a < c < b$.
Sub-case a ($a = 1$). We must have $b$ (necessarily less than $c$) such that $b^c - c^b = c - 1$. This is impossible if $b \geq 3$ by Lemma 3, which holds that $b^c - c^b \geq c$, and $b = 2$ can be eliminated by parity: $2^c - c^2$ is even if and only if $c$ is even, but $c - 1$ is even if and only if $c$ is odd.
Sub-case b ($2 \leq a < c < b$). Rearrange $a^c + b^c = c^a + c^b$ as $a^c - c^a = c^b - b^c$. The LHS is less than $a^c < c^c$, but by Lemma 1, the RHS is strictly greater than $c^c$. QED
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An integral of a rational function with high degrees: Evaluate $\int_{-\infty}^{\infty}\frac{(x^4 + 1)^2}{x^{12} + 1}dx$.
Calculate: $$\int_{-\infty}^{\infty}\frac{(x^4 + 1)^2}{x^{12} + 1}dx.$$
What I have tried is to divide both numerator and denominator with $x^4 + 1$ and then get two following integrals, because of parity of the integrand (nothing else worked for me): $$2\int_0^\infty \frac{x^4 + 1}{(x^4 - \sqrt3x^2 + 1)(x^4 + \sqrt3x^2 + 1)}dx = $$ $$ = \int_0^\infty \frac1{x^4 - \sqrt3x^2 + 1}dx + \int_0^\infty\frac1{x^4 + \sqrt3x^2 + 1}dx = $$ $$ = \int_0^\infty \frac1{(x^2 - \frac{\sqrt3}2)^2 + \frac14}dx + \int_0^\infty \frac1{(x^2 + \frac{\sqrt3}2)^2 + \frac14}dx.$$
I don't see what would be continuation of this. Any help is appreciated.
Thank you for any help. Appreciate it.
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Hint: The difference of two squares $a^2-b^2$ can be factored as $(a-b)(a+b)$. A sum of two squares $a^2+b^2$ can be thought of as $a^2-(ib)^2$...
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In how many ways can a student score exactly $100$ points on four $50$ point exams? greater than $100$? A student takes up $4$ exams with $50$ points each. In how many ways can he score exactly $100$? Similarly, in how many ways can he score greater than $100$?
My try:
I cannot seem to fill in the $4$ possibilities. The range depends on the constraint of equating to $100$.
Thanks for any help!
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The scores he can get are the coefficients of the different powers of $x$ in the expansion
\begin{align*}
(1+x+\cdots + x^{50})^4 &= \left(\frac{1-x^{51}}{1-x}\right)^4 \\
&= (1-4x^{51} + 6X^{102} - 4X^{153} + X^{204})\left(1+ 4x + \frac{4\cdot 5}{1\cdot 2}x^2 + \frac{4\cdot 5 \cdot 6}{3!}x^3 + \cdots +\binom{n+3}{3}x^n+\cdots \right)
\end{align*}
Hence the number of ways in which he can get exactly 100 is
$$ \binom{103}{3} - 4 \binom{52}{3} = 88451$$
Second part has already been solved in the other solutions posted.
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Rationalize a fraction. Rationalize the denominator $\frac{1}{2^{\frac{1}{3}} + 3^{\frac{1}{3}} + 4^{\frac{1}{3}}}$. Is there a short solution for this task ? Thanks in advance.
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First, use the identity
$$(a+b+c)(a^2 + b^2 + c^2 - ab - ac - bc) \;\; = \;\; a^3 + b^3 + c^3 - 3abc$$
with $a = \sqrt[3]{2}$ and $b = \sqrt[3]{3}$ and $c = \sqrt[3]{4}.$
Multiplying both the numerator and denominator of your fraction by the numerical form of $a^2 + b^2 + c^2 - ab - ac - bc$ will give you
$$ \frac{\text{stuff}}{2 + 3 + 4 - 3\sqrt[3]{24}} \;\; = \;\; \frac{\text{stuff}}{9 - 6\sqrt[3]{3}} $$
Now use the identity
$$(a-b)(a^2 + ab + b^2) = a^3 \, – \, b^3$$
with $a = 9$ and $b = 6\sqrt[3]{3}.$
Multiplying both the numerator and denominator of the displayed fraction above by the numerical form of $a^2 + ab + b^2$ will give you
$$ \frac{(\text{stuff})(\text{other stuff})}{9^3 - (6\sqrt[3]{3})^3} \;\; = \;\; \frac{(\text{stuff})(\text{other stuff})}{81}$$
At this point you simply need to multiply out (stuff)(other stuff). This is a bit tedious, but not really excessively so --- it'll be polynomial with $6$ terms multiplied by a polynomial with $3$ terms, for a total of $18$ "FOIL multiplications" that need to be carried out.
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Find the area of a triangle inscribed in the ellipse (Unicamp Mathematics Olympiad) Let $\zeta$ denote the Carteasian region given by $\zeta = \{(x,y): \frac{x^2}{4} + \frac{y^2}{9} = 1\}$. Let $A = (0,3), B = (x,y), C = (z,w)$ be points such that $A,B,C \in \zeta%$ and $\Delta ABC$ is equilateral. What is the area of $\Delta ABC$?
My approach: Since $B, C$ are homogeneous, I tried to do enough algebra so everything "cuts out". I used Gauss' formula for the area of a triangle given tree points, the fact that ${AB}^2 = {BC}^2 = {AC}^2$ and used the explicit equation of $\zeta$, but it ended up being too much algebra and I couldn't find an explicit result.
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$A=(0,3)$ is a vertex of the given ellipse. If $ABC$ is equilateral, by symmetry we have that $B$ and $C$ share the same $y$-coordinate, hence they are points of the form
$$ B=\left(-2\sqrt{1-\frac{y^2}{9}},y\right),\qquad C=\left(2\sqrt{1-\frac{y^2}{9}},y\right) $$
with $y\in(-3,3)$. In order that $ABC$ really is equilateral, we must have
$$ 3-y = \sqrt{3}\cdot 2\sqrt{1-\frac{y^2}{9}} $$
hence $y=-\frac{3}{7}$ and $BC=\frac{16}{7}\sqrt{3}$. It follows that $[ABC]=\color{red}{\frac{192}{49}\sqrt{3}}$.
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Find the arclength of the function $ x = \frac{1}{3} y^{\frac{-3}{2}} + \frac{1}{7}y^{\frac{7}{2}}$ for $1\leq y \leq 25$. I'm trying to find the arclength of the function $ x = \frac{1}{3} y^{\frac{-3}{2}} + \frac{1}{7}y^{\frac{7}{2}}$ for $1\leq y \leq 25$.
I can find the equation for the length pretty easily but I'm looking at thow to solve for the actual length. It looks like a very complex integral so I'm assuming I made a mistep or theres some easy reduction I can make.
After determining the area of an incredibly small section of the function:
$$ds = \sqrt{\left(\frac{dx}{dy}\right)^2 + 1}$$
$$\frac{dx}{dy} = \frac{-y^{\frac{-5}{2}}}{2} + \frac{y^{\frac{5}{2}}}{2} $$
$$ds = \sqrt{\left(\frac{-y^{\frac{-5}{2}}}{2} + \frac{y^{\frac{5}{2}}}{2}\right)^2 + 1}$$
This leaves me with the integral
$$\int{\sqrt{\left(\frac{-y^{\frac{-5}{2}}}{2} + \frac{y^{\frac{5}{2}}}{2}\right)^2 + 1}} dy$$
I do still have to calculate from 1 - 25 but I like to plug in after I solve my integral. Anyway, I can't tell how to solve this, but I have a feeling I need to play with the squared term. Perhaps a substitution or maybe the reciprocals simplify into something. If anyone has any tips I appreciate it!
EDIT: Problem solved! (I think)
In the answer to the question I can make the simplification
$$\int{\sqrt{\left(\frac{y^{\frac{5}{2}} + y^{\frac{-5}{2}}}{2}\right)^2}} dy$$
$$\int{\left(\frac{y^{\frac{5}{2}} + y^{\frac{-5}{2}}}{2}\right)} dy$$
$$\frac{1}{2}\left(\int{y^{\frac{5}{2}}}dy + \int{y^{\frac{-5}{2}}}dy\right)$$
$$\frac{y^{\frac{7}{2}}}{7} + \frac{1}{-3y^{\frac{3}{2}}}$$
We plug in our bounds here and the answer is $F(25) - F(1)$
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$$\int \sqrt{1+(x'(y))^2}\,dy=\int \sqrt{1+\left(\frac{y^{5/2}}{2}-\frac{1}{2 y^{5/2}}\right)^2}\,dy=\int\sqrt{\frac{1+\left(y^5-1\right)^2}{4 y^5}}\,dy=$$
$$=\int\frac{1}{2} \sqrt{\frac{\left(y^5+1\right)^2}{y^5}}\,dy=\frac12\int\frac{y^5+1}{\sqrt{y^5}}\,dy=\frac12\int\left(y^{-5/2}+y^{5/2}\right)\,dy=\frac{y^{7/2}}{7}-3 y^{-3/2}+C$$
The arc length is $$\left[\frac{y^{7/2}}{7}-3 y^{-3/2}\right]_1^{25}=\frac{29297368}{2625}\approx 11160.9$$
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How would one prove that for all positive real $a$ and $b$, $\sqrt{a+b} \neq \sqrt{a} + \sqrt{b}$? I think that the best way to prove this would be to prove by contradiction. Am I right?
If so, can I just assume that $\sqrt{a+b} = \sqrt{a} + \sqrt{b}$ and say $a = 1$ and $b = 1$ and show that $\sqrt{1+1} \neq \sqrt{1} + \sqrt{1}$, or is that not enough?
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Let $a,b \in \mathbb R^+$ and suppose for the sake of contradiction that $\sqrt{a+b}=\sqrt{a}+\sqrt{b}$. Squaring both sides we get $$a+b = a+b+2\sqrt{ab}$$
$$\implies 2\sqrt{ab}=0 \implies ab=0$$ a contradiction as $a,b \neq 0$. Therefore $\sqrt{a+b} \neq \sqrt{a}+\sqrt{b}$ anytime $a$ and $b$ are nonzero.
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Solving the ODE: $\frac{1}{r} \frac{d}{d r} \left( r \frac{d f}{d r} \right) + \left( a - b e^{r^2} \right) f = c+ d e^{- r^2} $ I'm trying hard to solve this:
$$\frac{1}{r} \frac{d}{d r} \left( r \frac{d f}{d r} \right) + \left( a - b e^{r^2} \right) f = c+ d e^{- r^2} $$
where $r$ ranges between $0$ and $\infty$, $a$ and $b$ are positive constants, $c$ and $d$ may have either sign.
Any of you is able to handle this?
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Hint:
$\dfrac{1}{r}\dfrac{d}{dr}\left(r\dfrac{df}{dr}\right)+\left(a-be^{r^2}\right)f=c+de^{-r^2}$
$\dfrac{d^2f}{dr^2}+\dfrac{1}{r}\dfrac{df}{dr}+\left(a-be^{r^2}\right)f=c+de^{-r^2}$
Let $s=r^2$ ,
Then $\dfrac{df}{dr}=\dfrac{df}{ds}\dfrac{ds}{dr}=2r\dfrac{df}{ds}$
$\dfrac{d^2f}{dr^2}=\dfrac{d}{dr}\left(2r\dfrac{df}{ds}\right)=2r\dfrac{d}{dr}\left(\dfrac{df}{ds}\right)+2\dfrac{df}{ds}=2r\dfrac{d}{ds}\left(\dfrac{df}{ds}\right)\dfrac{ds}{dr}+2\dfrac{df}{ds}=2r\dfrac{d^2f}{ds^2}2r+2\dfrac{df}{ds}=4r^2\dfrac{d^2f}{ds^2}+2\dfrac{df}{ds}$
$\therefore4r^2\dfrac{d^2f}{ds^2}+2\dfrac{df}{ds}+2\dfrac{df}{ds}+\left(a-be^{r^2}\right)f=c+de^{-r^2}$
$4r^2\dfrac{d^2f}{ds^2}+4\dfrac{df}{ds}+\left(a-be^{r^2}\right)f=c+de^{-r^2}$
$4s\dfrac{d^2f}{ds^2}+4\dfrac{df}{ds}+(a-be^s)f=c+de^{-s}$
Let $f=e^{ks}g$ ,
Then $\dfrac{df}{ds}=e^{ks}\dfrac{dg}{ds}+ke^{ks}g$
$\dfrac{d^2f}{ds^2}=e^{ks}\dfrac{d^2g}{ds^2}+ke^{ks}\dfrac{dg}{ds}+ke^{ks}\dfrac{dg}{ds}+k^2e^{ks}g=e^{ks}\dfrac{d^2g}{ds^2}+2ke^{ks}\dfrac{dg}{ds}+k^2e^{ks}g$
$\therefore4s\left(e^{ks}\dfrac{d^2g}{ds^2}+2ke^{ks}\dfrac{dg}{ds}+k^2e^{ks}g\right)+4\left(e^{ks}\dfrac{dg}{ds}+ke^{ks}g\right)+(a-be^s)e^{ks}g=c+de^{-s}$
$4s\left(\dfrac{d^2g}{ds^2}+2k\dfrac{dg}{ds}+k^2g\right)+4\left(\dfrac{dg}{ds}+kg\right)+(a-be^s)g=ce^{-ks}+de^{-(k+1)s}$
$4s\dfrac{d^2g}{ds^2}+4(2ks+1)\dfrac{dg}{ds}+(4k^2s+a+4k-be^s)g=ce^{-ks}+de^{-(k+1)s}$
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Two-variable limit of $\lim_{(x,y)\to(0,0)}\frac{\sin(x^4+y^4)}{x^2+y^2}$ $$\lim_{(x,y)\to(0,0)}\frac{\sin(x^4+y^4)}{x^2+y^2}$$
I tried to bound it with $\frac{\sin((x^2+y^2)^2)}{x^2+y^2}$ and using polar coordinates with $x = r\cos\theta$ and $y = r\sin\theta$, but neither of the approaches provided any results. I know that the limit exists and is equal to 0, so tricks with different paths won't work. Should I use the squeeze theorem, or is there another solution?
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Use $|\sin t|\leq |t|$ then
$$\Big|\frac{\sin(x^4+y^4)}{x^2+y^2}\Big|\leq\frac{x^4+y^4}{x^2+y^2}\leq\frac{x^4+y^4+2x^2y^2}{x^2+y^2}= x^2+y^2$$
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Can the inequality $\frac{a+b}{c} + \frac{b+c}{a} + \frac{c+a}{b} \geq 6$ be proved with differentiation? $$\frac{a+b}{c} + \frac{b+c}{a} + \frac{c+a}{b} \geq 6,\quad \text{with}\quad a,b,c > 0$$
I could do it with letting $x=\frac{a}{b}$, $y=\frac{b}{c}$, $z=\frac{c}{a}$, but I wonder if it is solvable somehow with differentiation.
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Consider you that you look for the minimum of function $$\Phi=\frac{a+b}{c} + \frac{b+c}{a} + \frac{c+a}{b} \tag 1$$ Compute the partial derivatives
$$\frac{\partial\Phi}{da}=-\frac{b+c}{a^2}+\frac{1}{b}+\frac{1}{c}\tag 2$$
$$\frac{\partial\Phi}{db}=-\frac{a+c}{b^2}+\frac{1}{a}+\frac{1}{c}\tag 3$$
$$\frac{\partial\Phi}{dc}=-\frac{a+b}{c^2}+\frac{1}{a}+\frac{1}{b}\tag 4$$ and say that all of them are equal to $0$.
From $(2)$ solve the quadratic for $b$. The two roots are
$$b_1=\frac {a^2} c\qquad \text{and}\qquad b_2=-c\implies b=\frac {a^2} c$$ since $a,b,c$ are positive.
Plug $b=\frac {a^2} c$ in $(3)$ to get $$\frac{(a+c) \left(a^3-c^3\right)}{a^4 c}=0\implies c=a\implies b=c=a$$
Plug in $(4)$ to get $0=0$.
Replace in $(1)$ and get $6$.
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Improper integral depending on parameter Find the value of the constant $C$ for which the integral $$\int \limits_{0}^{\infty}\left (\dfrac{1}{\sqrt{x^2+4}}-\dfrac{C}{x+2}\right)dx$$ converges. Evaluate the integral for this value of $C$.
I have some difficulties with above problem. I know some methods such as $x+2 \sim x$ and $\sqrt{x^2+4}\sim x$ for $x \to \infty$. But I would like to see the rigorous proof.
Can anyone please show it?
Would very thankful for that.
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First we have,
$\int_0^\infty \frac{1}{\sqrt{x^+4}}-\frac{C}{x+2}$
$=\int_0^\infty \frac{x+2-C\sqrt{x^2+4}}{(x+2)(\sqrt{x^2+4}}$
$=\int_0^\infty \frac{x+2-Cx\sqrt{1+\frac{4}{x^2}}}{(x^2+2x)(\sqrt{1+\frac{4}{x^2}}})$
$=\int_0^\infty \frac{x(1-C\sqrt{1+\frac{4}{x^2}})+1}{x^2\sqrt{1+\frac{4}{x^2}}+x\sqrt{1+\frac{4}{x^2}}}$
For this to converge we must have the coefficient of the x in the numerator going to 0 as $x\to\infty$(otherwise the integral will diverge by limit comparison with $\int\frac{1}{x}$).SO
$\lim_{x\to\infty}(1-C\sqrt{1+\frac{4}{x^2}})=1-C$ $\Rightarrow C=1$
|
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|
Group with $[a,b][c,d] \ne [f,g]$ I guess that multiplication of commutators isn't equal to commutator of some elements. But I don't know how to find such group. Because my task became so : I tried to get four elements (for example in $GL_{2}$) , consider their product of commutators and then I stuck because it become difficult to prove that there is no such elements $f$ and $g$ : product = $[f,g]$. Any ideas?
|
In general, $[G,G] \not = \{[g,h] :g,h \in G\}$. We shall illustrate this via $\operatorname{SL}(2,\mathbb{R})$, the group of real $2 \times 2$ matrices with determinant 1.
First we show $-I$ is not a commutator. Suppose, to the contrary, that $A^{-1}B^{-1}AB = -I$ for some $A,B \in SL(2,\mathbb{R})$. Then considering traces yields,
$$ \operatorname{tr}(A) =\operatorname{tr} (B^{-1}AB) = \operatorname{tr}(-A)$$
And therefore,
$$
A=\left( \begin{array}{cc} a_{11}&a_{12} \\ a_{21}&-a_{11}\\ \end{array} \right) \quad \text{for some} \quad a_{11},a_{12},a_{21} \in \mathbb{R}
.$$
Since $\det(A)= -{a_{11}}^2 -a_{12}a_{21} =1$ we have that $a_{12} \neq 0 \neq a_{21}$ and $A^2 =-I$. That is $\left( \begin{array}{c} 1 \\ 0\\ \end{array} \right)$ and $A \left( \begin{array}{c} 1 \\ 0\\ \end{array} \right)$ are linearly independen; with respect to this basis,
$
A=\left( \begin{array}{cc} 0&-1 \\ 1&0\\ \end{array} \right)
.$
Thus,
$$
-\left( \begin{array}{cc} b_{11}&b_{12} \\ b_{21}&b_{22}\\ \end{array} \right)= -B = ABA^{-1} = \left( \begin{array}{cc} b_{22}&-b_{21}\\ -b_{12}&b_{11} \\\end{array} \right),
$$
and wee see that $b_{11} = b_{22}$ and $b_{12} = b_{21}$. So $\det(B) = -{ b_{11}}^2 -{ b_{12}}^2 \leq 0$, contrary to $B \in \operatorname{SL}(2,\mathbb{R})$. A contradiction!
It remains to show that $-I \in \left[ \operatorname{SL}(2,\mathbb{R}), \operatorname{SL}(2,\mathbb{R}) \right]$. To this end verify that,
\begin{align*}
X&= \left( \begin{array}{cc} 1&0 \\ 1&1\\ \end{array} \right) =\left[ \left( \begin{array}{cc} 1&-1 \\0&1\\ \end{array} \right) , \left( \begin{array}{cc} \sqrt{2}&0 \\0&1 \over \sqrt{2}\\ \end{array} \right)\right] \\
Y&=\left( \begin{array}{cc} 1&-1 \\ 0&1\\ \end{array} \right) =\left[ \left( \begin{array}{cc} 1&0 \\1&1\\ \end{array} \right) , \left( \begin{array}{cc} 1 \over\sqrt{2}&0 \\0&\sqrt{2}\\ \end{array} \right)\right]
.
\end{align*}
And since $(XY)^3 = -I$ we are done.
This answer is based on Tom Goodwill's answer on MathOverflow.
|
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|
is there a name for a series wich alternates + and - terms? Is there a specific name for a geometric series such as this?
$1-\frac{c (os\theta)}{k}+\frac{c^2}{k^2}-\frac{c^3}{k^3}+\frac{c^4}{k^4}-....$
How can we identify it if positive terms are even or odd?
Also, does its definition change if the numerator follows a more complex pattern such as:
$1-\frac{c (os\theta)}{k}+\frac{c^2}{k^2}-\frac{c^3+c^2}{k^3}+\frac{c^4+c^3}{k^4}-....$
|
It's the Taylor expansion of $\frac 1 {1+ \frac c k}$
$$1-\frac{c}{k}+\frac{c^2}{k^2}-\frac{c^3}{k^3}+\frac{c^4}{k^4}-....=\sum_{n=0}^{\infty}(-1)^n( \frac c k)^n=\sum_{n=0}^{\infty}( \frac {-c} k)^n=\frac 1 {1+ \frac c k}$$
|
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|
Limit of a function using Taylor series
How can I verify that
$\lim_{x \to 0} \dfrac{\cos{2x} - \sqrt{1-4x^2}}{2x \sin x^3} = \frac{4}{3}$?
I've tried with Taylor so that:
$2x\sin x^3 \sim 4x^4 - \frac{2}{3}x^{10}$
$\cos2x \sim 1- 2x^2$
$\sqrt{1-4x^2} \sim 1-2x^2-2x^4$
But it keeps me giving the wrong result, so the question is, am I using wrong the Taylor series?
When approximating the function and plugging them in the limit should them all be of the same order?
A well-detailed explanation is more than welcome.
|
Hint. First of all $2x\sin(x)^3=2x^4+o(x^4)$. Moroever, here you need a longer expansion for $\cos(2x)$:
$$\cos(2x)=1-\frac{(2x)^2}{2}+\frac{(2x)^4}{4!}+o(x^4)=1-2x^2+\frac{2x^4}{3}+o(x^4).$$
Hence
$$\frac{\cos(2x) - \sqrt{1-4x^2}}{2x \sin x^3}=
\frac{1-2x^2+\frac{2x^4}{3} - (1-2x^2-2x^4)+o(x^4)}{2x^4+o(x^4)}
$$
Do you mind to give another try?
P.S. I suggest the use of the little-o notation. It will help you to understand whether a Taylor expansion is sufficient for the purpose.
|
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|
Lagrange multiplier to function $x^2+y^2+z^2$ Use Lagrange multipliers to find the maximum and minimum values of the function subject to the given condition:
$$f(x,y,z)=x^2+y^2+z^2; \quad x^4+y^4+z^4=1$$
My solution: As we do in Lagrange multipliers I have considered $\nabla f=\lambda \nabla g$ where $g(x,y,z)=x^4+y^4+z^4$ and the last equation is equivalent to the system of equations $$\begin{cases}
2x=4\lambda x^3 \\
2y=4\lambda y^3 \\
2z=4\lambda z^3
\end{cases}$$
After dividing into $2$ and multiplying to $x,y$ and $z$, respectively we get: $$\begin{cases}
x(1-2\lambda x^2)=0 \\
y(1-2\lambda y^2)=0 \\
z(1-2\lambda z^2)=0
\end{cases}$$
Considerong the first equation we get two cases: $x=0$ or $1-2\lambda x^2=0$
After that I am stuck. How to rule out or consider each case?
Can anyone demonstrate it clearly?
Would be very thankful for help
|
$x^2+y^2+z^2 \ge x^4+y^4+z^4 = 1$, and equality occurs when $x = 0,y = 0, z = \pm 1$ or permutations of them. Also by Cauchy-Schwarz inequality: $x^2+y^2+z^2 \le \sqrt{3(x^4+y^4+z^4)} = \sqrt{3}$ with equality occurs when $x = y = z = \pm \dfrac{1}{\sqrt[4]{3}}$ . Thus we can conclude that the min $= 1$ and the max $= \sqrt{3}$ .
|
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|
Prove $\frac{1}{2} + \frac{1}{2(u+1)^2} - \frac{1}{\sqrt{1+2u}} \geq 0$ for $u \geq 0$ This inequality provides a tight lower bound to $\sqrt{1+2u}$ for $u\geq 0$ without a radical. I was trying to solve it by squaring the radical and cross-multiplying and repeated differentiation of the resulting expression, I wonder if there is a quicker solution. Thanks.
|
It's
$$\sqrt{2u+1}(u^2+2u+2)\geq2(u+1)^2$$ or $$(2u+1)(u^2+2u+2)^2\geq4(u+1)^4$$ or
$$u^3(2u^2+5u+4)\geq0.$$
Done!
Also, we can use the following way.
Let $\sqrt{2u+1}=x$.
Thus, $x\geq1$ and we need to prove that
$$\frac{1}{2}+\frac{1}{2\left(\frac{x^2-1}{2}+1\right)^2}\geq\frac{1}{x}$$ or
$$1+\frac{4}{(x^2+1)^2}\geq\frac{2}{x}$$ or
$$x(x^2+1)^2+4x\geq2(x^2+1)^2$$ or
$$x^5+2x^3+x+4x\geq2x^4+4x^2+2$$ or
$$x^5-2x^4+2x^3-4x^2+5x-2\geq0$$ or
$$x^5-2x^4+x^3+x^3-2x^2+x-2x^2+4x-2\geq0$$ or
$$(x-1)^2(x^3+x-2)\geq0,$$
which is obvious for $x\geq1$.
Done againe!
|
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|
Crazy Functional Equation - Possibly related to trig identities? $p(2x)=p(x)p(-x)-p(x-1)p(1-x)+4x-1$ I'm trying to find a set of nontrivial (identity function is not allowed, and neither are constant functions) continuous solutions to the functional equation
$$p(2x)=p(x)p(-x)-p(x-1)p(1-x)+4x-1$$
...and, to be honest, I'm not quite sure where to go with this. This functional equation looks vaguely trigonometric to me, so I tried making some functional substitutions, like
$$p(x)=\cos(\psi(x))$$
$$p(x)=\cos^2(\psi(x))$$
$$p(x)=\psi(x)\cos(x)$$
$$p(x)=\psi(x)\cos^2(x)$$
...but I got nowhere with that. Any ideas?
|
One can take trial functions, say
$$p_{n}(x) = a_{0} + a_{1} \, x + a_{2} \, x^2 + \cdots + a_{n} \, x^n$$
and attempt to find the values of the coefficients.
As an example consider the case of $n=2$, or $p_{2}(x) = a_{0} + a_{1} \, x + a_{2} \, x^2.$ By expanding terms one will find that
\begin{align}
a_{0} + 2 a_{1} x + 4 a_{2} x^2 &= (-1 - 3 a_{0} a_{2} + a_{1} a_{2} + a_{1}^2 - 2 a_{2}^2) + 2 \, x \, (2 + 2a_{0} a_{2} - a_{1}^2 + 2 a_{2}^2) \\
& \hspace{5mm} + x^2 \, (a_{0} - a_{1} - 5 a_{2}) \, a_{2} + 4 \, a_{2}^2 \, x^3
\end{align}
Since $x^3$ only appears on one side of the equation then the coefficient is zero, or $a_{2} = 0$. This yields
\begin{align}
a_{0} + 2 a_{1} x &= (-1 + a_{1}^2) + 2 \, x \, (2 - a_{1}^2).
\end{align}
Now $a_{0} = a_{1}^2 - 1$ and $a_{1} = 2 - a_{1}^2$ which yields $a_{1} = \{1, -2\}$ and the set of values $(a_{0}, a_{1}) \in \{ (0,1), (3,-2)\}$. For $p_{2}(x)$ the two solutions are
$$p_{2}(x) = \begin{cases} x \\ 3- 2x \end{cases}.$$
Taking the case of $p_{1}(x) = a_{0} + a_{1} \, x$ then one finds that
$$p_{1}(x) = \begin{cases} x \\ 3- 2x \end{cases}.$$
Leaving the details for the interested it would be presumable to say that the solutions to this equation are:
Given the equation
$$p(2x)=p(x)p(-x)-p(x-1)p(1-x)+4x-1$$
then the solutions are
$$p(x) = \begin{cases} x \\ 3- 2x \end{cases}.$$
|
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|
Polynomial sum: $x^n+x^{n-1}y+x^{n-2}y^2+x^{n-3}y^3+\dots+xy^{n-1}+y^n$ Find sum of the expression,
$$x^n+x^{n-1}y+x^{n-2}y^2+x^{n-3}y^3+\dots+xy^{n-1}+y^n$$
where $x,y$ are real numbers and $n$ is a natural number.
|
Hint: It’s a geometric progression with common ratio $\dfrac{y}x$.
|
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|
Prove the existence of complex matrix that satisfies the following conditions. B is a 2 by 2 complex matrix. $B=0$ or $B^2 \neq 0$. Prove that a complex matrix C exists such that $C^2=B$.
I cannot understand what this problem wants me to do. "$B=0$ or $B^2 \neq 0$". Should I deal with the two cases? It would be great if somebody can give me some hints or solutions.
|
Ok, if $B=0$ then take $C=0$ and you are done, so $B^2 \neq 0$.
Since $B$ is complex matrix it has Jordan normal form. There exists invertable matrix $S$ such that $B=SXS^{-1}$ where X has one of 2 possible forms:
$X=\begin{pmatrix} \lambda_1 & 0 \\ 0 & \lambda_2 \end{pmatrix} $ or $X=\begin{pmatrix} \lambda & 1 \\ 0 & \lambda \end{pmatrix} $.
In the first case $\sqrt{X}=\begin{pmatrix} \sqrt{\lambda_1} & 0 \\ 0 & \sqrt{\lambda_2} \end{pmatrix}$ and $C = S \sqrt{X}S^{-1}$.
In the second case $\lambda \neq 0$, otherwise $B^2=SXS^{-1}SXS^{-1}=SX^2S^{-1}=0$.
So $\sqrt{X}=\begin{pmatrix} \sqrt{\lambda} & \frac{1}{2 \sqrt{\lambda}} \\ 0 & \sqrt{\lambda} \end{pmatrix}$ and $C = S \sqrt{X}S^{-1}$.
And, of course, $C^2 = S \sqrt{X}S^{-1} S \sqrt{X}S^{-1} =SXS^{-1}=B$.
|
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|
Expansion of Binomial Coefficient $(1+x+x^2)^n$ Suppose $n$ is a natural number and consider expansion:
$$\left(1+x+x^2\right)^n=\sum_{r=0}^{2n} \ a_r x^r$$
Find $\ a_0+ \ a_3+ \ a_6+ \ a_9\ldots$
I used different method, but could not arrive at the answer.
|
Note: I think the nice comment from @labbhattacharjee is worth an answer by its own. In order to obtain a formula for
\begin{align*}
a_0+a_3+a_6+\cdots
\end{align*}
with $(1+x+x^2)^n=\sum_{r=0}^{2n}a_rx^n$ it is convenient to consider the third roots of unity
\begin{align*}
1,\,\omega_1=\frac{-1+i\sqrt{3}}{2},\,\omega_2=\frac{-1-i\sqrt{3}}{2}
\end{align*}
which are the zeros of
\begin{align*}
1-x^3=(1-x)(1+x+x^2)
\end{align*}
We observe that following holds
\begin{align*}
1+\omega_1+\omega_1^2&=0\\
1+\omega_2+\omega_2^2&=0\tag{1}\\
\omega_1^2&=\omega_2\\
1&=\omega_1^3=\omega_2^3
\end{align*}
In order to obtain a formula for $a_0+a_3+a_6+\cdots$ we use the relationship (1) to filter out the coefficients $a_r$ with $r$ not a multiple of $3$. In order to do so, we evaluate the polynomial $(1+x+x^2)^{n}$ at $x=1,\omega_1$ and $\omega_2$.
We obtain
\begin{align*}
\color{blue}{3^n}&=(1+1+1)^n=a_0+a_1+a_2+a_3+a_4+a_5+a_6+\cdots\\
\color{blue}{0}&=(1+\omega_1+\omega_1^2)^n=a_0+a_1\omega_1+a_2\omega_1^2+a_3+a_4\omega_1+a_5\omega_1^2+a_6+\cdots\\
\color{blue}{0}&=(1+\omega_2+\omega_2^2)^n\\
&=(1+\omega_1^2+\omega_1)^n=a_0+a_1\omega_1^2+a_2\omega_1^1+a_3+a_4\omega_1^2+a_5\omega_1^1+a_6+\cdots
\end{align*}
Adding up these three equations we obtain
\begin{align*}
\color{blue}{3^n+0+0}&=3a_0+a_1(1+\omega_1+\omega_1^2)+a_2(1+\omega_1^2+\omega_1)\\
&\qquad +3a_3+a_4(1+\omega_1+\omega_1^2)+a_5(1+\omega_1^2+\omega_1)\\
&\qquad +3a_6+\cdots\\
&=3a_0+3a_3+3a_6+\cdots\\
&\color{blue}{=3(a_0+a_3+a_6+\cdots)}
\end{align*}
We finally conclude
\begin{align*}
\color{blue}{a_0+a_3+a_6+\cdots=3^{n-1}\qquad\qquad n\geq 1}
\end{align*}
|
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Solve this : $\,\displaystyle{\frac{dy}{dx}} = \cfrac{2xy \,e^{(x/y)^2}}{y^2(1+e^{(x/y)^2})+2x^2e^{(x/y)^2}}$
Solve this : $\,\cfrac{dy}{dx} = \cfrac{2xy \,e^{(x/y)^2}}{y^2(1+e^{(x/y)^2})+2x^2e^{(x/y)^2}}$
I tried to solve it using the homogeneous equation method:
$$y=vx\\ \cfrac{dy}{dx}=v\,+ x\cfrac{dv}{dx}$$
$\implies v\,+x\cfrac{dv}{dx} = \cfrac{2v\,e^{1/v^2}}{v^2(1+e^{1/v^2})+2e^{1/v^2}}$
$\implies x\cfrac{dv}{dx}= \cfrac{-v^2(1+e^{1/v^2})}{v^2(1+e^{1/v^2})+2e^{1/v^2}}$
$\implies \cfrac{v^2(1+e^{1/v^2})+2e^{1/v^2}}{v^2(1+e^{1/v^2})}dv=-\cfrac{dx}{x}$
$\implies \left[1 + \cfrac{2e^{1/v^2}}{v^2(1+e^{1/v^2})}\right]dv = -\cfrac{dx}{x}$
after this I have no clue, please give hint to solve the LHS.
|
$$\, \cfrac { dy }{ dx } =\cfrac { 2xy\, e^{ (x/y)^{ 2 } } }{ y^{ 2 }(1+e^{ (x/y)^{ 2 } })+2x^{ 2 }e^{ (x/y)^{ 2 } } } \\ \, \cfrac { dy }{ dx } =\cfrac { 2\, e^{ (x/y)^{ 2 } } }{ \frac { y }{ x } (1+e^{ (x/y)^{ 2 } })+2\frac { x }{ y } e^{ (x/y)^{ 2 } } } \\ y=xt\\ \frac { dy }{ dx } =x\frac { dt }{ dx } +t\\ x\frac { dt }{ dx } +t=\frac { 2{ e }^{ \frac { 1 }{ { t }^{ 2 } } } }{ t\left( 1+{ e }^{ \frac { 1 }{ { t }^{ 2 } } } \right) +\frac { 2 }{ t } { e }^{ \frac { 1 }{ { t }^{ 2 } } } } =\frac { 2{ te }^{ \frac { 1 }{ { t }^{ 2 } } } }{ { t }^{ 2 }\left( 1+{ e }^{ \frac { 1 }{ { t }^{ 2 } } } \right) +2{ e }^{ \frac { 1 }{ { t }^{ 2 } } } } \\ x\frac { dt }{ dx } =\frac { 2{ te }^{ \frac { 1 }{ { t }^{ 2 } } } }{ { t }^{ 2 }\left( 1+{ e }^{ \frac { 1 }{ { t }^{ 2 } } } \right) +2{ e }^{ \frac { 1 }{ { t }^{ 2 } } } } -t=\frac { -{ t }^{ 3 }-{ t }^{ 3 }{ e }^{ \frac { 1 }{ { t }^{ 2 } } } }{ { t }^{ 2 }\left( 1+{ e }^{ \frac { 1 }{ { t }^{ 2 } } } \right) +2{ e }^{ \frac { 1 }{ { t }^{ 2 } } } } \\ \int { \frac { { t }^{ 2 }\left( 1+{ e }^{ \frac { 1 }{ { t }^{ 2 } } } \right) +2{ e }^{ \frac { 1 }{ { t }^{ 2 } } } }{ { t }^{ 3 }\left( 1+{ e }^{ \frac { 1 }{ { t }^{ 2 } } } \right) } dt } =-\int { \frac { dx }{ x } } \\ \int { \frac { dt }{ t } -\int { \frac { d\left( { 1+e }^{ \frac { 1 }{ { t }^{ 2 } } } \right) }{ \left( 1+{ e }^{ \frac { 1 }{ { t }^{ 2 } } } \right) } } } =-\ln { \left| x \right| +C } \\ \ln { \left| t \right| -\ln { \left| 1+{ e }^{ \frac { 1 }{ { t }^{ 2 } } } \right| = } } -\ln { \left| x \right| +C } \\ $$
$$\ln { \left| \frac { y }{ \left( 1+e^{ (x/y)^{ 2 } } \right) } \right| } =C\\ y=C\left( 1+e^{ (x/y)^{ 2 } } \right) \\ \\ \\ \\ $$
|
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|
Proving that $\frac{6n^2+3n}{2n^2-5}$ converges to 3. Prove the following:
$$\lim_{n\to \infty}\frac{6n^2+3n}{2n^2-5}=3$$
Here's my solution:
Consider the function $\frac{6n^2+3n}{2n^2-5}$.
Then $$|\frac{6n^2+3n}{2n^2-5}-3|= |\frac{6n^2+3n}{2n^2-5}-\frac{6n^2+15}{2n^2-5}|=|\frac{3n-15}{2n^2-5}|$$
The upper bound for the numerator is $3n-15 \le 3n$ for all $n$ and the lower bound for the denominator is $2n^2-5 \le 4n^2$ if $n \le 2$.
It follows that
$|\frac{3n-15}{2n^2-5}| \le \frac{3n}{4n^2}$. Meaning $|\frac{3n-15}{2n^2-5}| \le \frac{3}{4n}$.
By the Archimedean Property, $\frac{3}{4n} \le \frac{3}{4N} \lt \epsilon$
Therefore $(\frac{6n^2+3n}{2n^2-5})\rightarrow 3$.
$\blacksquare$
I feel like I'm missing a step after showing that $\frac{3}{4n} \le \frac{3}{4N} \lt \epsilon$.
Can $\frac{3}{4n} \le \frac{3}{4N} \lt \epsilon$ be simplified more so that it resembles $\frac{1}{n} \le \frac{1}{N} \lt \epsilon$ or is it fine how it is?
Please advise.
|
$$\left|\frac{6n^2+3n}{2n^2-5}-3\right|= \left|\frac{6n^2+3n}{2n^2-5}-\frac{6n^2\color{red}-15}{2n^2-5}\right|=\left|\frac{3n\color{red}+15}{2n^2-5}\right|$$
Hence upper bound for the numerator when $n> 15$, is $|3n+15| \leq 4n.$
A lower bound for the denominator for $n \geq 3$, is $|2n^2-5| \geq n^2$
It follows that $\left|\frac{3n+15}{2n^2-5}\right|\leq \frac4n$ for $n>15$.
Now, by archimedian property, $\forall \epsilon>0 , \exists N>0, \frac{4}{N}<\epsilon$, now let $n>\max(N, 15)$. and we have
$$\left|\frac{3n+15}{2n^2-5}\right|\leq \frac4n \leq \frac4N<\epsilon$$
|
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|
Weird integral found in textbook I found the following integral in a calculus textbook.
$$\int_{-1}^1\frac{\mathrm d x}{\sqrt{4-x^2}}$$
What is $\mathrm d x$ doing up there and how do I fix this??? I didn't even know it was possible to write an integral like this. Much help needed.
|
The $dx$ is a symbol, that tells you by which variable to integrate. But it is often treated like a factor. Different ways to write this integral would be
$$ \int_{-1}^1 \frac{dx}{\sqrt{4-x^2}} = \int_{-1}^1 \frac{1}{\sqrt{4-x^2}}dx = \int_{-1}^1 dx \frac{1}{\sqrt{4-x^2}}. $$
The first variant is often used to save space, the last variant is often used in a physics context.
In order to solve this, we can use a substitution. $\sqrt{4-x^2}$ has actually something to do with a circle, so we choose a substitution that involves trig functions. In this case $x=2\sin t$ would be a good choice. We see that for $t=-\frac\pi6$ it is $-1=2\sin(-\frac\pi6)$ and for $t=\frac\pi6$ it is $1=2\sin(\frac\pi6)$. Also $2\sin t$ is monotonous on the interval $(-\frac\pi6,\frac\pi6)$. We have done this to find out the limits of our substituted integral.
Now we need the derivative of $x$ with respect to $t$:
$$ \frac{dx}{dt} = 2\cos t.$$
Since we want to substitute $t$ for $x$, we treat $dx$ and $dt$ symbolically as numbers in order to solve for $dx = 2\cos t\;dt$.
Now we plug everything together:
\begin{align*}
\int_{-1}^1\frac{dx}{\sqrt{4-x^2}} &= \int_{-\frac\pi6}^{\frac\pi6} \frac{2\cos t\;dt}{\sqrt{4 - (2\sin t)^2}} \\
&= \int_{-\frac\pi6}^{\frac\pi6} \frac{2\cos t\;dt}{\sqrt{4(1 - \sin^2 t)}} \\
&= \int_{-\frac\pi6}^{\frac\pi6} \frac{2\cos t\;dt}{2\sqrt{\cos^2 t}},
\end{align*}
where we have used the trigonometric pythogoras $1=\sin^2+\cos^2$. Now we see that for $t\in(-\frac\pi6,\frac\pi6)$ we have $\cos t \geq 0$ and thus $\sqrt{\cos^2t} = \cos t$. Going on:
\begin{align*}
\int_{-1}^1\frac{dx}{\sqrt{4-x^2}}
&= \int_{-\frac\pi6}^{\frac\pi6} \frac{2\cos t\;dt}{2\sqrt{\cos^2 t}} \\
&= \int_{-\frac\pi6}^{\frac\pi6} \frac{\cos t\;dt}{\cos t} \\
&= \int_{-\frac\pi6}^{\frac\pi6} dt
= \int_{-\frac\pi6}^{\frac\pi6} 1 dt \\
&= [t]_{-\frac\pi6}^{\frac\pi6} = \frac\pi6 - (-\frac\pi6) = 2\frac\pi6 = \frac\pi3.
\end{align*}
|
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|
Does $n+1$ divides $\binom{an}{bn}$? Suppose that $a>b>0$ be integers. Is it true that for an integer $n>2$ that
$$n+1|\binom{an}{bn}$$
or is there a counter example. Certainly i think the right hand side would reduce to
$$\frac{an(an-1)(an-2)...((a-1)n+1)}{n(n-1)(n-2)...2\cdot 1}$$
But I'm not seeing how this could reduce better to show there is a factor of $n+1$ left.
Examples show this is true for small n; for example
$$\binom{9}{6}=\binom{3\cdot 3}{2\cdot 3}=\frac{9\cdot8\cdot7}{3\cdot 2\cdot 1}=4(3\cdot 7)$$
$$\binom{16}{8}=\binom{4\cdot 4}{2\cdot 4}=\frac{16\cdot15\cdot...\cdot 10\cdot 9}{8\cdot 7\cdot...\cdot2\cdot 1}=5(2\cdot 3^2\cdot11\cdot 13)$$
|
There is a counterexample.
Take $(n,a,b)=(3,5,1)$.
$\binom{5\times 3}{1\times 3}=455$ is not divisible by $4$.
|
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|
Prove that $\forall n \in \mathbb{N}: 4^n + 6n - 10$ is divisible by $18$
Prove that $\forall n \in \mathbb{N}: 4^n + 6n - 10$ is divisible by $18$
Base case: for $n = 1: 4^1 +6\cdot 1 - 10 = 0$ is divisible by 18.
Inductive Assumption: Assume that for for some $k \in \mathbb{N} :4^k +6k-10$
Proving that $4^{k+1}+6(k+1)-10$ is divisible by 18
$4^{k+1}+6(k+1)-10= 4^k \cdot 4 + 6k + 6 - 10$ $$= \color{green}{4^k +6k-10} + 3
\cdot 4^k + 6$$
The first term is divisible by 18 according to the inductive assumption and I have to find a way to manipulate the second term to be divisible by 18.
There is a similar question posted, but the answers for it are substituting another expression $18m$ instead of manipulating $4^{k+1}+6(k+1)-10$. I want to know how to solve this without substituting
|
\begin{align}3\cdot 4^k+6 &= 6(2\cdot4^{k-1}+1) \\
&=6(3\cdot4^{k-1}+1-4^{k-1})\\
&=6(3\cdot4^{k-1}+(1-2^{k-1})(1+2^{k-1}))\\
&=6(3\cdot4^{k-1}-(2^{k-1}-1)(1+2^{k-1}))\end{align}
Notice that $2^{k-1}$ is not divisible by $3$, hence either $(2^{k-1}-1)$ is divisible by $3$ or $(2^{k-1}+1)$ is divisible by $3$.
Hence $(3\cdot4^{k-1}-(2^{k-1}-1)(1+2^{k-1}))$ is divisble by $3$ and $6(3\cdot4^{k-1}-(2^{k-1}-1)(1+2^{k-1}))$ is divisible by $18$.
Remark: Using modulo arithmetic makes the task much easier.
|
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|
Show that $ a \equiv 1 \pmod{2^3 } \Rightarrow a^{2^{3-2}} \equiv 1 \pmod{2^3} $ Show that $ a \equiv 1 \pmod{2^3 } \Rightarrow a^{2^{3-2}} \equiv 1 \pmod{2^3} $
Show that$ a \equiv 1 \pmod{2^4 } \Rightarrow a^{2^{4-2}} \equiv 1 \pmod{2^4} $
Answer:
$ a \equiv 1 \pmod{2^3} \\ \Rightarrow a^2 \equiv 1 \pmod{2^3} \\ \Rightarrow a^{2^{3-2}}=a^{2^1} \equiv 1 \pmod{2^3} $
Am I right?
|
Yes, it is correct.
In general, if $a \equiv 1 \pmod m$,then we have $a^n \equiv 1 \pmod m$ for any $n$.
|
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|
let $f(x) = \left\lbrace\begin{array}{ll} x^2+3x, &x \geq1\\ x^2-3x+6, & x<1 \end{array}\right.$ then find the limit… Let
$f(x) =
\begin{cases}
x^2+3x, &x \geq1\\[2ex]
x^2-3x+6, & x<1
\end{cases}$
Then find the
$$\lim_{h \to0} \frac{f(1)-f(1-h^2)}{h^2}=?$$
My Try :
$$f(1)=4$$
$$f(1-h^2)=(1-h^2)^2-3(1-h^2)+6=h^4-h^2+4=h^2(h^2-1)+4$$
So we have :
$$\lim_{h \to0} \frac{h^2(h^2-1)}{h^2}=-1$$
it is right ?
|
Yes, it's right (with a minor slip, check your algebra). For computing $f(1)$ you have to use the “upper branch”, so
$$
f(1)=1^2+3\cdot 1=4
$$
whereas, since $h^2>0$ when computing the limit, $1-h^2<1$ and $f(1-h^2)$ requires following the “lower branch”, so
\begin{align}
f(1-h^2)
&=(1-h^2)^2-3(1-h^2)+6\\[4px]
&=1-2h^2+h^4-3+3h^2+6\\[4px]
&=4+h^2+h^4
\end{align}
Then
\begin{align}
f(1)-f(1-h^2)
&=4-4-h^2-h^4\\[4px]
&=-h^2-h^4
\end{align}
and
$$
\frac{f(1)-f(1-h^2)}{h^2}=-1-h^2
$$
so the limit is $-1$.
|
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|
show that:for every $x\in S$, there exsit $y,z,w\in S$, such $x=y^2+z^2+w^2$ Define $S=\{a+b\cdot\dfrac{-1+\sqrt{3}i}{2}|a,b\in Z\}$. Show that for every $x\in S$, there exit $y,z,w\in S$,such
$$x=y^2+z^2+w^2$$where $i$ such $i^2=-1$
My attempt:
Since
$$\left(a+b\dfrac{-1+\sqrt{3}i}{2}\right)^2=\left(a^2-ab-\dfrac{1}{2}b^2\right)+\dfrac{\sqrt{3}}{2}b(a-b)i$$
|
Let $\omega = e^{\frac{2\pi}{3}i} = \frac{-1 + \sqrt{3}i}{2}$ be the primitive cubic root of unity. It satisfies the identities
$$1 + \omega + \omega^2 = 0\quad\text{ and }\quad \omega^3 = 1$$
For any $x = a + b\omega \in S$, define $r$ according to following table.
$$\begin{array}{|cc:c:l|}
\hline
a & b & r & x + r^2\\
\hline
\text{even} & \text{ even } & 0 & a + b\omega\\
\text{even} & \text{ odd } & \omega^2 & a + (b+1)\omega\\
\text{odd} & \text{even} & 1 & (a+1) + b\omega\\
\text{odd} & \text{ odd} & \omega &(a-1) + (b-1)\omega\\
\hline
\end{array}$$
As one can see, independent of the parity of $a,b$, we have $x + r^2 = a' + b'\omega$ for some even $a'$ and $b'$. This means we can find a $\mu \in S$ such that $x + r^2 = 2\mu$.
Now for any $p, q \in S$, define $y, z, w$ by
$$
\begin{cases}
y &= p + q - r\\
z &= p\omega + q\omega^2\\
w &= p\omega^2 - q\omega
\end{cases}
$$
It is easy to see
$$\begin{align} y^2 + z^2 + w^2
&= (p+q)^2 + (p\omega+q\omega^2)^2 + (p\omega^2 - q\omega)^2 - 2r(p+q) + r^2\\
&= (p^2 + 2pq + q^2) + (p^2\omega^2 + 2pq + q^2\omega) + (p^2\omega -2pq + q^2\omega^2) - 2r(p+q) + r^2\\
&= p^2(1+\omega^2+\omega) + 2pq(1+1-1) + q^2(1+\omega+\omega^2)-2r(p+q)+r^2\\
&= 2pq -2r(p+q)+r^2\\
&= 2(p-r)(q-r)-r^2
\end{align}
$$
If we choose $q = r+1$ and $p = r+\mu$, this becomes
$$y^2 + z^2+w^2 = 2\mu - r^2 = (x+r^2)-r^2 = x$$
This is precisely the decomposition we seek.
|
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|
Solve $\lfloor \sqrt x +\sqrt{x+1}+\sqrt{x+2}\rfloor=x$ $$\lfloor \sqrt x +\sqrt{x+1}+\sqrt{x+2}\rfloor=x$$
I tried to solve this equation.
First thing is $\lfloor \sqrt x +\sqrt{x+1}+\sqrt{x+2}\rfloor \in \mathbb{Z} $ so $x \in \mathbb{Z}$
second $$\sqrt x +\sqrt{x+1}+\sqrt{x+2} \geq \sqrt 0 +\sqrt{0+1}+\sqrt{0+2} \\\to x \in \mathbb{N}$$ so we can check $x=1,2,3,4,5,6,7,8,9,\ldots$ by a MATLAB program. I checked the natural numbers to find solution. I found $x=8,9$ worked here.
Now my question is about somehow an analytical solving of the equation, or another idea. Can any one help me? Thanks in advance.
|
First solve the real inequalities
$$
x \le \sqrt{x}+\sqrt{x+1}+\sqrt{x+2} < x+1
\tag{1}$$
Solution of $x = \sqrt{x}+\sqrt{x+1}+\sqrt{x+2}$ is numerically $9.8956$ and solution of $x = \sqrt{x}+\sqrt{x+1}+\sqrt{x+2}=x+1$ is numerically $7.9813$. So solution of (1) is
$$
7.9813 < x \le 9.8956
$$
Finally, assume $x$ is an integer. We get $x=8$ or $x=9$.
|
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|
Prove that $1^2 + 2^2 + \cdots + n^2 = \frac{n(n+1)(2n+1)}{6}$ without using induction. I have to deduce the following formula $$1^2 + 2^2 + \cdots + n^2 = \frac{n(n+1)(2n+1)}{6},$$ while using the given formula $$\binom{k}{0}+\binom{k+1}{1}+\cdots+\binom{k+r}{r}=\binom{k+r+1}{r}$$
I tried to find values for $k$, such that $\binom{k}{0}=1^2$ etc. but that didn't work. Does anybody have a push in the right direction? Thanks!
|
When I was (much) younger I wanted to find the area of a segment of parabola so I needed the sum of the squares and I found the formula by myself in this way.
Inspired by the famous $1+2+\ldots+n=\dfrac{n(n+1)}{2}$
I supposed that the sum of the square could be a third degree polynomial in $n$
$P(n)=an^3+bn^2+cn$
Then I plugged the first $3$ values for $n$ getting
$
\left\{
\begin{array}{l}
a+b+c=1 \\
8a+4b+2c=5 \\
27 a + 9 b + 3 c=14\\
\end{array}
\right.
$
which gives $a=\frac13;\;b=\frac12;\;c=\frac16$
and therefore
$$P(n)=\frac13 n^3+\frac12 n^2+\frac16 n=\frac16 (2 n^3+3 n^2+n)=\frac16 n(n+1)(2n+1)$$
It is not elegant, but in 1977 there was no wikipedia :)
Hope it can be useful
|
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|
Show that $\zeta_K(2) = \frac{\pi^4}{48 \sqrt{2}} $, with $K = \mathbb{Q}(\sqrt{2})$ For the real number field $K = \mathbb{Q}(\sqrt{2})$ the ring of integers is $\mathcal{O}_K = \mathbb{Z}[\sqrt{2}] $. We can solve Pell's equation and so there are units $(1 - \sqrt{2})^k$ with $k \in \mathbb{Z}$. One can show that $K$ is a:
*
*Euclidean domain
*principal ideal domain
*unique factorization domain
Taking their word for it, there a norm $N_K: a + b \sqrt{2} \mapsto |a^2 - 2b^2 | $ and we can write the Dedekind Zeta function:
$$ \zeta_K(s) = \sum_{(a,b) \in \mathbb{Z}^2} \frac{1}{(a^2 - 2b^2)^k} = \frac{1}{1 - 2^{-s}} \prod_{p = \pm 1 (8)} \frac{1}{(1 - p^{-s})^2} \prod_{p = \pm 1 (8)} \frac{1}{(1 - p^{-2s})} $$
If we plug in $s = 2$ I found the numerical result stated without proof. And I'm starting to migrate the existing proofs over $\mathbb{Z}$ to the ring $\mathbb{Z}[\sqrt{2}]$:
$$ \zeta_K(2) = \sum_{x \in \mathbb{Z}[\sqrt{2}]} \frac{1}{N(x)^2} = \sum_{(a,b) \in \mathbb{Z}^2} \frac{1}{\big(a^2 - 2b^2\big)^2} = \frac{\pi^4}{48 \sqrt{2}} $$
The definition does not quite make sense over numbers, which explains the shift to ideals. We have $ \mathcal{O}(K)= \mathbb{Z}[\sqrt{2}$ and
$$ \zeta_K(2) = \sum_{I \subseteq \mathcal{O}(K)} \frac{1}{N_{K/\mathbb{Q}}(I)^2} = \sum_{(a,b) \in \mathbb{Z}^2} \frac{1}{\big(a^2 - 2b^2\big)^2} = \frac{\pi^4}{48 \sqrt{2}} $$
The Euler product is the product over prime ideals:
$$ \zeta_K(2) = \prod_{P \subseteq \mathcal{O}(K)} \frac{1}{1- N_{K/\mathbb{Q}}(P)^{-2} } = \frac{1}{1 - 2^{-2}} \prod_{p = \pm 1 (8)} \frac{1}{(1 - p^{-2})^2} \prod_{p = \pm 1 (8)} \frac{1}{(1 - p^{-4})}
\stackrel{?}{=}
\sum_{(a,b) \in \mathbb{Z}^2} \frac{1}{(a^2 - 2b^2)^2} $$
|
We have that
$$ L(\chi_8,s) = \sum_{n\geq 0}\left[\frac{1}{(8n+1)^s}-\frac{1}{(8n+3)^s}-\frac{1}{(8n+5)^s}+\frac{1}{(8n+7)^s}\right]$$
equals
$$\prod_{p\equiv \pm 1\!\!\pmod{8}}\left(1-\frac{1}{p^s}\right)^{-1}\prod_{p\equiv \pm 3\!\!\pmod{8}}\left(1+\frac{1}{p^s}\right)^{-1} $$
by Euler's product. On the other hand
$$ L(\chi_8,1) = \int_{0}^{1}\sum_{n\geq 0} x^{8n}(1-x^2-x^4+x^6)\,dx = \int_{0}^{1}\frac{1-x^2}{1+x^4}\,dx = \frac{1}{\sqrt{2}}\log(1+\sqrt{2}) $$
and similarly
$$ L(\chi_8,2)=\int_{0}^{1}\frac{x^2-1}{x^4+1}\log(x)\,dx = \frac{\pi^2}{8\sqrt{2}}.$$
Since $\zeta_K(2) = \zeta(2)\cdot L(\chi_8,2)$ by Euler's product, $\zeta_K(2)=\frac{\pi^4}{48\sqrt{2}}$ is proved.
|
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|
A polynomial can not be the product of two polynomials of degree 2 and 3 I want to show that
the polynomial $x^5+x^2-1$ in $\Bbb{Z}/2\Bbb{Z}[x]$ can not be written can a product of a polynomial of degree $2$ and an other one of degree $3$.
For that I wrote $x^5-x^2+1=(x^2+bx+c)(x^3+dx^2+ex+f)=x^5+(b+d)x^4+(c+e+bd)x^3+(f+cd+be)x^2+(bf+ce)x+cf$
where $a,b,c,d,e,f\in \Bbb{Z}/2\Bbb{Z}$
then by identification we have
$b+d=0 \\
c+e+bd=0\\
f+cd+be=-1\\
bf+ce=0\\
cf=1$
but I didn't find a contradiction, can you please help me? Thanks
|
Let us show that the polynomial $x^5-x^2+1$ is irreducible in $\Bbb Z[x]$.
We khnow that if a polynomial in reducible in $\Bbb Z[x]$ then it is reducible in $\Bbb{Z}/p\Bbb{Z}[x]$ for any prime number $p$.
Therefore if we show for example that $x^5+x^2+1$ is irreducible in $\Bbb{Z}/2\Bbb{Z}[x]$ then $x^5-x^2+1$ is irreducible in $\Bbb Z[x]$.
So let us show that $x^5+x^2+1$ is irreducible in $\Bbb{Z}/2\Bbb{Z}[x]$ :
Since $0^5+0^2+1=1\not=0$ and $1^5+1^2+1=1\not=0$ then we conclude that $x^5+x^2+1$ does not have any linear factor in $\Bbb{Z}/2\Bbb{Z}[x]$.
Now we need to show that $x^5+x^2+1$ can not be written as a product of a polynomials of degree 2 and 3.
For that let us suppose that there exist $a,b,c,d,e,f\in \Bbb{Z}/2\Bbb{Z}$
such that $\begin{align*}
x^5+x^2+1&=(x^2+bx+c)(x^3+dx^2+ex+f)\\&=x^5+(b+d)x^4+(c+e+bd)x^3+(f+cd+be)x^2+(bf+ce)x+cf\end{align*}$
then by identification we have
$$\left\{
\begin{array}{ll}
b+d=0 \\
c+e+bd=0\\
f+cd+be=1\\
bf+ce=0\\
cf=1 \end{array}
\right.$$
First off, the line $cf=1$ tells us that $c = f = 1$. Now insert that into the rest of the equations:
$$\left\{
\begin{array}{ll}b+d=0 \\
1+e+bd=0\\
1+d+be=1\\
b+e=0\end{array}
\right.$$
From the first and last equations, we get that $d = b = e$, so we insert that into the final two equations, and we get
$$\left\{
\begin{array}{ll}
1+b+b^2=0\\
1+b+b^2=1\end{array}
\right.$$
and there we have a contradiction.
|
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|
$\int_{0}^{\pi\over 2}\cos(x)\ln(\tan{x}){\mathrm dx\over 1+\tan(x)}=\int_{0}^{\pi\over 2}\cos(3x)\ln(\tan{x}){\mathrm dx\over 1+\tan(x)}=-\ln{2}?$
$$I=\int_{0}^{\pi\over 2}\cos(x)\ln(\tan{x}){\mathrm dx\over 1+\tan(x)}$$
$$J=\int_{0}^{\pi\over 2}\cos(3x)\ln(\tan{x}){\mathrm dx\over 1+\tan(x)}$$
How to show that $I=J=-\ln{2}?$
For Integral $I$
${\cos{x}\over 1+\tan{x}}={1-\sin^2{x}\over \sin{x}+\cos{x}}$
$$I=\int_{0}^{\pi\over 2}{\ln{\tan{x}}}{\mathrm dx\over \sin{x}+\cos{x}}-\int_{0}^{\pi\over 2}{\sin^2{x}}\ln(\tan{x}){\mathrm dx\over \sin{x}+\cos{x}}=I_1-I_2$$
$$$$
$u=\tan{x}$ then $\mathrm dx=\cos^2{x}\mathrm du$
$$I_1=\int_{0}^{\infty}{\ln{u}}{\mathrm du\over (1+u)\sqrt{1+u^2}}$$
$$I_2=\int_{0}^{\infty}{u^2\ln{u}}{\mathrm du\over (1+u)(1+u^2)\sqrt{1+u^2}}$$
|
As an alternative to the solution provided by Jack D'Aurizio, I will find a value to the integral $I$ by picking up where you left off.
For the first of your integrals $I_1$, if we write
$$I_1 = \int^1_0 \frac{\ln x}{(1 + x) \sqrt{1 + x^2}} \, dx + \int^\infty_1 \frac{\ln x}{(1 + x) \sqrt{1 + x^2}} \, dx,$$
letting $x \mapsto 1/x$ in the second of these integrals we see that
$$\int^\infty_1 \frac{\ln x}{(1 + x) \sqrt{1 + x^2}} \, dx = -\int^1_0 \frac{\ln x}{(1 + x) \sqrt{1 + x^2}} \, dx.$$
Thus $I_1 = 0$.
For the second of your integrals, we begin again by writing
$$I_2 = \int^1_0 \frac{x^2 \ln x}{(1 + x)(1 + x^2) \sqrt{1 + x^2}} \, dx + \int^\infty_1 \frac{x^2 \ln x}{(1 + x)(1 + x^2) \sqrt{1 + x^2}} \, dx.$$
In the second of these integrals if we again let $x \mapsto 1/x$, then
\begin{align*}
I_2 &= \int^1_0 \frac{x^2 \ln x}{(1 + x)(1 + x^2) \sqrt{1 + x^2}} \, dx - \int^1_0 \frac{\ln x}{(1 + x)(1 + x^2) \sqrt{1 + x^2}} \, dx\\
&= \int^1_0 \frac{(x^2 - 1) \ln x}{(1 + x)(1 + x^2) \sqrt{1 + x^2}} \, dx = \int^1_0 \frac{(x - 1) \ln x}{(1 + x^2)^{3/2}} \, dx.
\end{align*}
Surprisingly (perhaps to some), this last integral can be found in elementary terms. To find it we write
$$\int \frac{(x - 1) \ln x}{(1 + x^2) \sqrt{1 + x^2}} = \int \frac{x \ln x}{(1 + x^2)^{3/2}} \, dx - \int \frac{\ln x}{(1 + x^2)^{3/2}} \, dx = I_\alpha - I_\beta.$$
For the first integral $I_\alpha$, integrating by parts we have
\begin{align*}
I_\alpha &= \int \frac{x}{(1 + x^2)^{3/2}} \cdot \ln x \, dx\\
&= -\frac{\ln x}{\sqrt{1 + x^2}} + \int \frac{dx}{x \sqrt{1 + x^2}} \, dx\\&= -\frac{\ln x}{\sqrt{1 + x^2}} - \ln (\sqrt{1 + x^2} + 1) + \ln x + C_\alpha.
\end{align*}
(The integral that results after applying integration by parts can be found by applying a substitution of $x = \tan \theta$, for example)
For the second integral $I_\beta$, integrating by parts again gives
\begin{align*}
I_\beta &= \int \frac{1}{(1 + x^2)^{3/2}} \cdot \ln x \, dx\\
&= \frac{x \ln x}{\sqrt{1 + x^2}} - \int \frac{dx}{\sqrt{1 + x^2}}\\
&= \frac{x \ln x}{\sqrt{1 + x^2}} - \sinh^{-1} x + C_\beta.
\end{align*}
So
$$\int \frac{(x - 1) \ln x}{(1 + x^2) \sqrt{1 + x^2}} \, dx = -\frac{(x + 1)\ln x}{\sqrt{1 + x^2}} - \ln (\sqrt{1 + x^2} + 1) + \ln x + \sinh^{-1} x + C.$$
Now, on applying the limits of integration we have
\begin{align*}
I_2 &= \int^1_0 \frac{(x - 1) \ln x}{(1 + x^2) \sqrt{1 + x^2}} \, dx = -\ln (1 + \sqrt{2}) + \ln 2 + \sinh^{-1} (1)\\ &= \ln \left (\frac{2}{1 + \sqrt{2}} \right ) + \ln (1 + \sqrt{2}) = \ln 2,
\end{align*}
giving
$$I = I_1 - I_2 = 0 - \ln 2 = -\ln 2,$$
as expected.
|
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"timestamp": "2023-03-29T00:00:00",
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|
Finding the limit of a function . How can I calculate the following limit:
\begin{equation*}
\lim_{x \rightarrow a}
\frac{\sqrt{x} - \sqrt{a} + \sqrt{x-a} }{\sqrt{x^2 - a^2}}
\end{equation*}
I feel that I should multiply by the conjugate, but which conjugate?
|
Note that the above limit doesn't make sense for $x < a$, since the square root in the numerator has a negative argument, but I avoid writing this.
$$
\frac{\sqrt x - \sqrt a + \sqrt{x-a}}{\sqrt{x^2-a^2}} = \frac{1}{\sqrt{x+a}}\left(\frac{\sqrt x - \sqrt a}{\sqrt{x-a}} + 1\right)
$$
Finally, $\lim_{x \to a} \frac{\sqrt x - \sqrt a}{\sqrt{x-a}} = 0$. This is because we can write this as $\frac{\sqrt x - \sqrt a}{x-a} \times \frac{x-a}{\sqrt{x-a}}$. The limit of the first term exists, since it is the derivative of the square root function at $a$. The second term is just $\sqrt{x-a}$, whose limit is zero. So the limit of the product exists and is zero.
Finally,
$$ \bbox[yellow,5px,border:2px solid red]
{
\lim_{x \to a} \frac{1}{\sqrt{x+a}} \left(\frac{\sqrt x - \sqrt a}{\sqrt{x-a}} + 1\right) = \frac{1}{\sqrt{2a}}
}
$$
since $\sqrt{x+a} \to \sqrt{2a}$ as $x \to a$.
|
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"timestamp": "2023-03-29T00:00:00",
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|
Product of quadratic residues in terms of primitive root Let $a$ be a primitive root for prime $p(\geq 3)$. Show that the product of all non-zero quadratic residues is
congruent to $a^\frac{p^2−1}{4}$ and that the product of quadratic nonresidues is congruent to $a^\frac{(p−1)^2}{4}$ modulo $p$.
I know that the product of the non-zero residues is $(1^2).(2^2)...((p-1)/2)^2$, and that every number from $1, 2, ... p-1$ can be written as $a, a^2, a^3, ..., a^{p-1}$, if $a$ is a primitive root. I think I'm just missing one link to convert the product of some form of powers of $a$ to the final answer
|
Hint: If $a$ is a primitive root mod $p$, $p$ is an odd prime, then all the non-zero quadratic residues mod $p$ are exactly $a^2,a^4,\ldots,a^{p-1}$.
Proof: $a$ is not a quadratic residue because if $a\equiv b^2\pmod{p}$, then by Fermat's little theorem $a^{\frac{p-1}{2}}\equiv b^{p-1}\equiv 1 \pmod {p}$, contradiction. $a^{2k}\equiv (a^k)^2\pmod{p}$. If $a^{2k+1}\equiv c^2\pmod{p}$, then $a\equiv (c\cdot a^{-k})^2\pmod{p}$, contradiction.
Edit: here's another proof. $x^2\equiv y^2\pmod p$ $\iff p\mid x^2-y^2$ $=(x-y)(x+y)$ and by Euclid's lemma $\iff (x\equiv y\pmod p$ or $x\equiv -y\pmod p)$. So there are exactly $\frac{p-1}{2}$ non-zero quadratic residues mod $p$ and they are $(\pm 1)^2, (\pm 2)^2,\ldots$ $, \left(\pm \frac{p-1}{2}\right)^2$ mod $p$. If $a^{2i}\equiv a^{2j}\pmod p$, $i> j$, $i,j\in\{1,2,\ldots, \frac{p-1}{2}\}$, then $a^{2(i-j)}\equiv 1\pmod p$, contradiction. $a^{2k}\equiv (a^k)^2\pmod p$. There are $\frac{p-1}{2}$ different non-zero quadratic residues in the list $a^2, a^4,\ldots, a^{p-1}$.
|
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|
What is the simplest way to compute :$\int^{\pi}_0\bigl(\frac{\sin(x)}{5-4\cos(x)}\bigr)^2dx=\frac{\pi}{24}$ I am trying to compute the following integral.
$$\int^{\pi}_0\biggl(\frac{\sin(x)}{5-4\cos(x)}\biggr)^2dx$$
My attempt was to expand the integrand and make use of the standard change of variables $t =\tan x/2$. But it turns out to be a lengthy and exhausting computations. I can put all details here since it is not pleasant.
at the end I got the answer:
$$\int^{\pi}_0\biggl(\frac{\sin(x)}{5-4\cos(x)}\biggr)^2dx =\frac{\pi}{24}$$
Maybe it is not correct so do not trust this result at 100%
I would like to know if there is an easiest way or trick that quickly leads to the answer?
|
Let $z=e^{ix}$ and then
\begin{eqnarray}
&&\int^{\pi}_0\bigg(\frac{\sin(x)}{5-4\cos(x)}\bigg)^2dx\\
&=&\frac12\int^{\pi}_{-\pi}\bigg(\frac{\sin(x)}{5-4\cos(x)}\bigg)^2dx\\
&=&\frac12\int_{|z|=1}\bigg(\frac{\frac{z-\frac1z}{2i}}{5-4\frac{z+\frac1z}{2}}\bigg)^2\frac{dz}{iz}\\
&=&\frac12\int_{|z|=1}\bigg(\frac{z^2-1}{2i(5z-2z^2-2)}\bigg)^2\frac{dz}{iz}\\
&=&-\frac1{8i}\int_{|z|=1}\bigg(\frac{z^2-1}{2z^2-5z+2}\bigg)^2\frac{dz}{z}\\
&=&-\frac1{8i}\int_{|z|=1}\frac{(z^2-1)^2}{z(2z-1)^2(z-2)^2}dz\\
&=&-\frac1{32i}\int_{|z|=1}\frac{(z^2-1)^2}{z(z-\frac{1}{2})^2(z-2)^2}dz\\
&=&-\frac1{32i}\cdot2\pi i\left(\text{Res}\bigg(\frac{(z^2-1)^2}{z(z-\frac{1}{2})^2(z-2)^2},z=0\bigg)+\text{Res}\bigg(\frac{(z^2-1)^2}{z(z-\frac{1}{2})^2(z-2)^2},z=\frac12\bigg)\right)\\
&=&-\frac\pi{16}(1-\frac{5}{3})\\
&=&\frac{\pi}{24}.
\end{eqnarray}
|
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|
Partial fractions decomposition of ${\frac{2x}{(x+2)^2}}$ Express in partial fraction form: $\displaystyle{\frac{2x}{(x+2)^2}}$
I think is $\displaystyle{\frac{2x}{(x+2)^{2}} = \frac{A}{x+2}+\frac{B}{(x+2)^2}}$
However when identifying $A$ and $B$, I'm not sure how to calculate A.
E.g. $$2x = A\cdot (x+2) + B$$
Substitute $x=-2$
$2\cdot(-2)$ = $A\cdot (2-2) +B$
$-4 = B$
In other questions there is always another factor to multiply by at this stage.
|
Starting from (as you almost wrote)
$$\frac{2x}{(x+2)^2} = \frac{A}{x+2}+\frac{B}{(x+2)^2},$$
rewrite as
$$\frac{2x}{(x+2)^2} = \frac{A(x+2)+B}{(x+2)^2}
= \frac{Ax+(2A+B)}{(x+2)^2}.$$
From this, one sees that $2x = Ax + (2A+B)$, so that $A=2$ and $2A+B=0$ and thus $B=-4$ (as you correctly derived).
An alternative method, using substitution, starts again from
$$\frac{2x}{(x+2)^2} = \frac{A(x+2)+B}{(x+2)^2}$$
so that $2x = A(x+2)+B$. Substituting $x=-2$ gives $B=-4$ and thus $2x = A(x+2)-4$. Now substitute any value other than $-2$ (say $0$) for $x$, giving $0 = 2A-4$, so that $A=2$.
|
{
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"timestamp": "2023-03-29T00:00:00",
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|
Sum of series $\sum_{\underset{m \neq n}{n = 1}}^{\infty} \frac{1}{m^2 - n^2}$ I was trying to solve this series and I have an exam in a week. I can't understand how to find its sum although I managed to rework it by transforming $\frac{1}{m^2 - n^2}$ into $\frac{1}{2m}(\frac{1}{m+n} - \frac{1}{n-m})$.
I'm sure there is a way, as my book explicitly confirms this as being convergent and having a sum.
Any help is appreciated.
|
I get
$-\frac{3}{4m^2}
$.
$\begin{array}\\
s(m)
&=\sum_{\underset{m \neq n}{n = 1}}^{\infty} \frac{1}{m^2 - n^2}\\
&=\sum_{\underset{m \neq n}{n = 1}}^{\infty} \frac{1}{2m}(\frac{1}{m+n} + \frac{1}{m-n})\\
&= \frac{1}{2m}\sum_{\underset{m \neq n}{n = 1}}^{\infty}(\frac{1}{m+n} + \frac{1}{m-n})\\
&= \frac{1}{2m}(\sum_{n = 1}^{m-1}(\frac{1}{m+n} + \frac{1}{m-n})
+\sum_{n = m+1}^{\infty}(\frac{1}{m+n} + \frac{1}{m-n}))\\
&= \frac{1}{2m}(t(m)+u(m))\\
t(m)
&=\sum_{n = 1}^{m-1}(\frac{1}{m+n} +\frac{1}{m-n})\\
&= \sum_{n = 1}^{m-1}\frac{1}{m+n} +\sum_{n = 1}^{m-1}\frac{1}{m-n}\\
&= \sum_{n = m+1}^{2m-1}\frac{1}{n} + \sum_{n = 1}^{m-1}\frac{1}{n}\\
&=H_{2m-1}-H_m+H_{m-1}\\
&=H_{2m-1}-\dfrac1{m}\\
u(m)
&=\sum_{n = m+1}^{\infty}(\frac{1}{m+n} + \frac{1}{m-n})\\
&=\lim_{k \to \infty}\sum_{n = m+1}^{k}(\frac{1}{m+n} + \frac{1}{m-n})\\
&=\lim_{k \to \infty}\left(\sum_{n = m+1}^{k}\frac{1}{m+n} +\sum_{n = m+1}^{k}\frac{1}{m-n}\right)\\
&=\lim_{k \to \infty}\left(\sum_{n = m+1}^{k}\frac{1}{m+n} -\sum_{n = m+1}^{k}\frac{1}{n-m}\right)\\
&=\lim_{k \to \infty}\left(\sum_{n = 2m+1}^{k+m}\frac{1}{n} - \sum_{n = 1}^{k-m}\frac{1}{n}\right)\\
&=\lim_{k \to \infty}\left(\sum_{n = 2m+1}^{k-m}\frac{1}{n}+\sum_{n = k-m+1}^{k+m}\frac{1}{n} - (\sum_{n = 1}^{2m}\frac{1}{n}+\sum_{n = 2m+1}^{k-m}\frac{1}{n})\right)\\
&=\lim_{k \to \infty}\left(\sum_{n = k-m+1}^{k+m}\frac{1}{n} - \sum_{n = 1}^{2m}\frac{1}{n}\right)\\
&=-H_{2m}
\qquad\text{since }\sum_{n = k-m+1}^{k+m}\frac{1}{n}<\frac{2m}{k-m+1} \to 0\\
\text{so}\\
s(m)
&=\frac1{2m}(H_{2m-1}-\dfrac1{m}-H_{2m})\\
&=\frac1{2m}(-\dfrac1{m}-\dfrac1{2m})\\
&=-\frac{3}{4m^2}\\
\end{array}
$
|
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"timestamp": "2023-03-29T00:00:00",
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|
Find the integral $\int_{0}^{1} f(x)dx$ for $f(x)+f(1-{1\over x})=\arctan x\,,\quad \forall \,x\neq 0$. Suppose that $f:\mathbb{R}\rightarrow\mathbb{R}$ such that
$$f(x)+f\left(1-{1\over x}\right)=\arctan x\,,\quad \forall \,x\neq 0$$
Find $$\int_{0}^1 f(x)\,dx$$
My Attempt :
Replace $x$ by $1/x$ in given equation
$$f\left({1\over x}\right)+f(1-x)=\arctan {1\over x}$$Add both equations
$$f(x)+f\left(1-{1\over x}\right)+f\left({1\over x}\right)+f(1-{x})=\arctan x\,+\arctan {1\over x}$$Rearranging thenm gives
$$f(x)+f(1-x)+f\left({1\over x}\right)+f\left(1-{1\over x}\right)={\pi\over2}$$Now it seems to me that $f(x)=f\left({1\over x}\right)$ Am I correct here? (I don't have proof though)
$$f(x)+f(1-x)={\pi\over 4}$$ $$\int_0^1 f(x)\,dx =\int_0^1f(1-x)\, dx={\pi\over 8}$$
I'm not sure about my assumption. Thank you
|
An Attempt: Let
\begin{equation}
f(x) + f(1 - \frac{1}{x}) = arctan(x) \tag{1}
\end{equation}
Integrating:
\begin{equation}
\int_0^1 f(x) dx + \int_0^1 f(1 - \frac{1}{x}) dx = \int_0^1 arctan(x) dx
\end{equation}
\begin{equation}
\int_0^1 f(x) dx + \int_0^1 f(1 - \frac{1}{x}) dx = \frac{1}{4}(\pi-\ln 4) \tag{2}
\end{equation}
Now using one of your results:
$$f(\frac{1}{x}) + f(1 - x) = arctan(\frac{1}{x})$$
and integrating and plugging in value for the RHS
\begin{equation}
\int_0^1 f(\frac{1}{x}) dx + \int_0^1 f(1 - x) dx = \frac{1}{4}(\pi+\ln 4) \tag{3}
\end{equation}
Now using another of your results:
$$ f(x) + f(1 - \frac{1}{x}) + f(\frac{1}{x}) + f(1 - x) = \frac{\pi}{2}$$
Integrating:
$$
\int_0^1 f(x) dx + \int_0^1 f(1 - \frac{1}{x}) dx +\int_0^1 f(\frac{1}{x}) dx + \int_0^1 f(1 - x) dx =\frac{\pi}{2}$$
Using the fact that
$$\int_0^1 f(x) dx = \int_0^1 f(1 - x) dx \tag{a}$$
we have:
$$
2\int_0^1 f(x) dx + \int_0^1 f(1 - \frac{1}{x}) dx +\int_0^1 f(\frac{1}{x}) dx =\frac{\pi}{2}
$$
Now substituting the value of $\displaystyle \int_0^1 f(\frac{1}{x}) dx $ from (3) into above result and using (a):
$$
\int_0^1 f(x) dx + \int_0^1 f(1 - \frac{1}{x}) dx =\frac{\pi}{2} - \frac{1}{4}(\pi+\ln 4) \tag{4}
$$
Leads no where. General procedure is fine but didn't work.
|
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|
When $f(x)=\frac{-b^2m-ba+ax}{-mx-bm-a}$ is an integer $a,b,m,x$ are positive integers.
For which $x>0$ is $f(x)$ an integer?
$$f(x)=\frac{-b^2m-ba+ax}{-mx-bm-a}$$
I been trying to play with it, I changed it to:
$$\frac{b^2m-a\left(b+x\right)}{a+m\left(b+x\right)}$$
And then I been trying to say:
$$a+m\left(b+x\right)| b^2m-a\left(b+x\right) $$
$$a+m\left(b+x\right)| (a+m\left(b+x\right))(b+x)$$
So
$$a+m\left(b+x\right)| b^2m-a\left(b+x\right)+ (a+m\left(b+x\right))(b+x)$$
$$a+m\left(b+x\right)| m\left(\left(b+x\right)^2+b^2\right)$$
But I don't see how it helps, so please help me.
|
For every even $x>0$ we have that $f(x)$ is not an integer for all $a,b,m$. To see this, chose an even $x$ and take $a=b=m=x$. Then
$$
f(x)=\frac{-x^3}{-2x^2-x}=\frac{x^2}{2x+1}.
$$
Because $x^2$ is even, and $2x+1$ is odd, the fraction cannot be an integer. Similarly, for $x$ odd, take $a=b=x$ and $m=x+1$ to see that
$$
f(x)=\frac{x^2+x}{2x+3},
$$
which cannot be an integer for $x$ odd.
|
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|
Limit of $f(x)$ when $x$ goes to zero Let $f(x) = \frac{1 + \tan x + \sin x - \cos x}{\sin^2 x + x^3}$ . Find value of $\lim_{x \to 0} f(x)$ if it exists . I can solve it using L'Hospital's Rule and Taylor series but I'm looking for another way suing trigonometric identities .
|
\begin{align}
\dfrac{1-\cos x+\sin x+\tan x}{\sin^2x+x^3}
&= \dfrac{2\sin^2\frac{x}{2}+\sin x\dfrac{1+\cos x}{\cos x}}{\sin^2x+x^3} \\
&= \dfrac{2\dfrac{\sin^2\frac{x}{2}}{x^2}+\dfrac{\sin x}{x}\dfrac{1+\cos x}{x\cos x}}{\dfrac{\sin^2x}{x^2}+x} \\
&\to\dfrac{\dfrac12+1\times\infty}{1+0}=\infty
\end{align}
as $x\to0$.
|
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|
Easiest way to show that $\int_0^{2\pi}\frac{\sin x\,dx}{\left|\sin x\right| +\left|\cos x\right|}=0$ What's the easiest way to show that $$\int_0^{2\pi}\frac{\sin x \,dx}{\left|\sin x\right| +\left|\cos x\right|}=0$$
I was thinking about to change the interval of integration and to show that the function is odd.
$\displaystyle y=x-\pi \Rightarrow \int_{-\pi}^\pi \frac{-\sin y \, dy}{\left|\sin y\right| + \left|\cos y\right|}=0$ (I guess). Does it seem legit?
|
$$\int_0^{2\pi}\frac{\sin x}{|\sin x| +|\cos x|}dx=$$
$$=\int_0^{\frac{\pi}{2}}\tfrac{\sin x}{\sin x +\cos x}dx+\int_{\frac{\pi}{2}}^{\pi}\tfrac{\sin x}{\sin x -\cos x}dx+\int_{\pi}^{\frac{3\pi}{2}}\tfrac{\sin x}{-\sin x-\cos x}+\int_{\frac{3\pi}{2}}^{2\pi}\tfrac{\sin x}{-\sin x +\cos x}dx=$$
$$=\int_0^{\frac{\pi}{2}}\tfrac{\sin x}{\sin x +\cos x}dx+\int_0^{\frac{\pi}{2}}\tfrac{\sin \left(x+\frac{\pi}{2}\right)}{\sin \left(x+\frac{\pi}{2}\right) -\cos \left(x+\frac{\pi}{2}\right) }dx+$$
$$+\int_0^{\frac{\pi}{2}}\tfrac{\sin(x+\pi)}{-\sin (x+\pi)-\cos(x+\pi)}dx+\int_0^{\frac{\pi}{2}}\tfrac{\sin\left( x+\frac{3\pi}{2}\right)}{-\sin\left( x+\frac{3\pi}{2}\right) +\cos\left( x+\frac{3\pi}{2}\right)}dx=$$
$$=\int_0^{\frac{\pi}{2}}\tfrac{\sin x}{\sin x +\cos x}dx+\int_0^{\frac{\pi}{2}}\tfrac{\cos{x}}{\cos{x}+\sin{x} }dx+$$
$$+\int_0^{\frac{\pi}{2}}\tfrac{-\sin{x}}{\sin{x}+\cos{x}}dx+\int_0^{\frac{\pi}{2}}\tfrac{-\cos{x}}{\cos{x} +\sin{x}}dx=0.$$
|
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|
Solve $\sqrt{n^3+1}+\sqrt{n^3}>10^3$ I was given this task:
"Find $n_0$ so that for all $n > n_0$:
$$\frac{1}{\sqrt{n^3+1}+\sqrt{n^3}}<\ 10^{-3}$$
This should be equal to:
$$\sqrt{n^3+1}+\sqrt{n^3}>10^3$$
But this is where I am already stuck. Wolfram Alpha gave me this solution for n:
$$n\ >\ \frac{9}{100}\left(\frac{37037}{2}\right)^{\frac{2}{3}}$$
And this is equal to $n_0$ = 62
But how did wolfram Alpha get that solution?
|
$$\sqrt{n^3+1}+\sqrt{n^3}\gt 10^3$$
is equivalent to
$$\sqrt{n^3+1}\gt 10^3-\sqrt{n^3}$$
For $n$ such that $10^3-\sqrt{n^3}\ge 0$, i.e. $n\le 10^2$, the both sides are non-negative, so squaring the both sides gives
$$n^3+1\gt (10^3-\sqrt{n^3})^2,$$
i.e.
$$\sqrt{n^3}\gt \frac{10^6-1}{2\times 10^3}$$
So, we get
$$n\gt \left(\frac{10^6-1}{2\times 10^3}\right)^{2/3}=\frac{9}{100}\left(\frac{37037}{2}\right)^{2/3}$$
|
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|
Condititon for roots of quartic to be real and two be coincident We wish the roots of the following quartic to be real and and distinct, but two roots should be equal. Eg. Roots should be $a,b,c,c$ where $a,b,c$ are Real and distinct.
$$[x^2-2mx-4(m^2+1)][x^2-4x-2m(m^2+1)]$$
We have to find values of $m$ corresponding to this condition.
I observed that discriminant of first quadratic is positive, and discriminant of second quadratic is $0$ at $m=-1$. BUT when $m=-1$, then the two quadratic have a common root! So the roots are $-4,1,1,1$. This is not required.
Now I think if we find common root by subtracting quadratics, then we get desired value of $m$. On subtracting quadratic I got:
$$(m-2)x = (m^2+1)(m-2)$$
Meaning either $m=2$ or $x=m^2+1$. Still none lead to answer.
The answer is $m=3$.
|
Let $f(x)=x^2-2mx-4(m^2+1),g(x)=x^2-4x-2m(m^2+1)$.
Also, let $c\in\mathbb R$ be the double root of $f(x)g(x)$.
We have three cases to consider :
Case 1 : $x=c$ is a double root of $f(x)$
Case 2 : $x=c$ is a double root of $g(x)$
Case 3 : $f(c)=g(c)=0$
*
*Case 1 : If $x=c$ is a double root of $f(x)$, then we have to have $(-2m)^2-4\times 1\times (-4(m^2+1))=0$, but there are no such $m\in\mathbb R$.
*Case 2 : If $x=c$ is a double root of $g(x)$, then solving $(-4)^2-4\times 1\times (-2m(m^2+1))=0$ gives $m=-1$. Then, we have $f(x)=(x+4)(x-2),g(x)=(x-2)^2$ which don't satisfy our condition.
*Case 3 : If $f(c)=g(c)=0$, then from $0=f(c)-g(c)=-2(m-2)(c-m^2-1)$, we have $m=2$ or $c=m^2+1$. If $m=2$, then $f(x)=g(x)$ which don't satisfy our condition. If $c=m^2+1$, then by Vieta's formulas, $x=-4$ is a root of $f(x)$, so $f(-4)=0\implies m=-1,3$. We already see that $m\not=-1$. If $m=3$, then we have $f(x)=(x-10)(x+4),g(x)=(x-10)(x+6)$ which satisfy our condition.
Therefore, the answer is $\color{red}{m=3}$.
|
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|
Proof explanation: finding the coefficient of $(r+1)$th term in the expansion of $\left(1-6x\right)^{-\frac{1}{2}}$? Here is the answer of the math..
$\displaystyle\left(1-6x\right)^{-\frac{1}{2}}$
$\displaystyle=\frac{\left(-\frac{1}{2}\right)\left(-\frac{1}{2}-1\right)....\left(-\frac{1}{2}-r+1_{ }\right)}{r!}\left(-6x\right)^r$
$\displaystyle=\frac{\left(-1\right)^r\left(\frac{1}{2}\right)\left(\frac{1}{2}+1\right)\left(\frac{1}{2}+2\right)....\left(r-1+\frac{1}{2}\right)}{r!}\left(-1\right)^r.\:x^r.\:2^r.3^r$
$\displaystyle=\left(-1\right)^{2r}\:\frac{1.3.5.7.....\left(2r-1\right)}{2^r\:.\:r!}\:\:x^r.\:2^r.3^r$
$\displaystyle=\frac{\left\{1.3.5.7.....\left(2r-1\right)\right\}\left(2.4.6....2r\right)}{r!\:\left(2.4.6.....2r\right)}\:\:x^r.\:3^r$
$\displaystyle=\frac{1.2.3.4.....2r}{r!\:2^r\:\left(1.2.3.\:....r\right)}\:\:x^r.\:3^r$
$\displaystyle=\frac{\left(2r\right)!\:\:.\:3^r}{2^r\:.\:\left(r!\right)^2}\:\:x^r$
$\displaystyle=\left(\frac{3}{2}\right)^r\:\frac{\left(2r\right)!\:}{\:\left(r!\right)^2}\:.\:\:x^r$
I can't understand the 4th and 6th line of this math . Please explain me that lines..
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We obtain
\begin{align*}
&\color{blue}{\frac{\left(-1\right)^r\left(\frac{1}{2}\right)\left(\frac{1}{2}+1\right)\left(\frac{1}{2}+2\right)\cdots\left(r-1+\frac{1}{2}\right)}{r!}\left(-1\right)^rx^r2^r3^r}\\
&\qquad=\frac{\left(-1\right)^r\left(\frac{1}{2}\right)\left(\frac{1+2}{2}\right)\left(\frac{1+4}{2}\right)\cdots\left(\frac{2r-2+1}{2}\right)}{r!}\left(-1\right)^rx^r2^r3^r\tag{1}\\
&\qquad=\frac{\left(-1\right)^{2r}\left(\frac{1}{2}\right)\left(\frac{3}{2}\right)\left(\frac{5}{2}\right)\cdots\left(\frac{2r-1}{2}\right)}{r!}x^r2^r3^r\tag{2}\\
&\qquad\color{blue}{=(-1)^{2r}\frac{1\cdot3\cdot5\cdots(2r-1)}{2^rr!}x^r2^r3^r}\tag{3}\\
&\qquad=\frac{1\cdot3\cdot5\cdots(2r-1)\cdot 2\cdot 4\cdot 6\cdots (2r)}{r!2\cdot 4\cdot 6\cdots (2r)}x^r3^r\tag{4}\\
&\qquad=\frac{(2r)!}{r!(2\cdot1)\cdot (2\cdot 2)\cdot (2\cdot 3)\cdots (2r)}x^r3^r\tag{5}\\
&\qquad\color{blue}{=\frac{(2r)!}{r!2^r1\cdot 2\cdot 3\cdots r}x^r3^r}\tag{6}\\
\end{align*}
Common:
*
*In (1) we use a common denominator $2$ for the factors in the numerator.
*In (2) we do a simplification and collect the factors $(-1)^r$.
*In (3) we observe there are $r$ factors $\frac{1}{2}$ which can be written as $2^r$ in the denominator.
*In (4) we see odd factors $1,3,5,\ldots,2r-1$ in the numerator. Since we wish to use the more compact notation of factorials we expand numerator and denominator with corresponding even factors. We also use that an even number of factors $(-1)$ give $1$.
*In (5) we write the numerator using factorial notation and observe that each even factor in the denominator can be written a $2\cdot k$.
*In (6) we factor out $r$ factors $2$ giving $2^r$.
|
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|
prove that $a^{25}-a$ is divisible by 30 I'm trying to prove that $30|a^{25}-a$.
First of all I said that it equals
$a(a^3-1)(a^3+1)(a^6+1)(a^{12}+1)$, so that it must be divisible by 2.
Now I want to show it can be divided by $3,6$ using Fermat's little theorem.
I need a bit of direction since I'm kind of lost.
|
You need divisibility by $2,3,5$.
$a$ and $a^3-1$ are of different parity, so one of them is odd, one is even. Thie product is of course even.
$a^3\equiv a\pmod{3}$, so one of numbers $a^3-1$, $a$, $a^3+1$ is always divisible by $3$.
Now, for $5$.
*
*The case $5|a$ is trivial.
*$a\equiv 1\pmod{5}\implies a^3\equiv 1\pmod{5}\implies 5|a^3-1$
*$a\equiv 2\pmod{5}\implies a^6+1\equiv 65\equiv 0\pmod{5}\implies 5|a^6+1$
*$a\equiv 3\pmod{5}\implies a^6\equiv 9^3\equiv (-1)^3\equiv -1\pmod{5}\implies 5|a^6+1$
*$a\equiv 4\pmod{5}\implies a\equiv -1\pmod{5}\implies a^3\equiv -1\pmod{5}\implies 5|a^3+1$
|
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|
Particular solution of non-homogeneous recurrence relation can somebody help me with my homework, please? I have to solve this:
$a_{n}$ = $-a_{n-1}$ + $12a_{n-2}$ - 10n + 13 + $7.3^{n}$
$a_{0} = 3$, $a_{1}=24$.
I know to solve this (homogeneous equation):
$a_{n}$ = $-a_{n-1}$ + $12a_{n-2}$
$a_{n}$ = $x^{n}$
$x^{n}$ = -$x^{n-1}$ + $12.x^{n-2}$
$x^{n}$ + $x^{n-1}$ - $12.x^{n-2}$ = 0
(x + 4)(x - 3) = 0
$t_n$ = $(-4)^{n}.\alpha$ + $3^{n} \beta$
How can I find particular solution? ....using A,B,C... because i don't know solve this with sum (I found here something with sum, but we did't use sum on course).
Thanks everyone for your help :)
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You have:
$$a_n = -a_{n-1} + 12a_{n-2} - 10n + 13 + 7\cdot 3^n$$
$$a_{n+1} = -a_{n} + 12a_{n-1} - 10(n+1) + 13 + 7\cdot 3^{n+1}$$
Subtracting the first from the second equation you get:
$$a_{n+1} = 13a_{n-1} - 12a_{n-2} - 10 + 14 \cdot 3^n$$
Repeat the same trick to get rid of the constant and you have:
$$a_{n+2} = a_{n+1} + 13a_n - 25a_{n-1} + 12a_{n-2} + 28\cdot 3^n$$
Now subtract three times the above equation from:
$$a_{n+3} = a_{n+2} + 13a_{n+1} - 25a_{n} + 12a_{n-1} + 28\cdot 3^{n+1}$$
You will get a homogeneous linear recurrence relation
|
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|
How can I calculate $\alpha=\arccos\left(-\frac{1}{4}\right)$ without using a calculator? How can I calculate $\alpha$, without using a calculator?
$\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad$
I know $x = -\frac{1}{4} \implies y= \frac{\sqrt{15}}{4}, $ now how can I calculate $$\arccos\left(-\frac{1}{4}\right) = \alpha,\quad \arcsin\left(\frac{\sqrt{15}}{4}\right) = (180° - \alpha),$$ without using a calculator? How did the Greeks to calculate the angle?
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You can read the binary digits of $\arccos(x)/\pi$ off the signs of $2\cos(2^kx)$, which is an easy to compute sequence defined recursively with $x_{n+1} = x_n^2-2$.
More precisely, you put a $1$ digit when the product of the signs so far is negative, and a $0$ otherwise :
$\begin{matrix}x_0 & -1/2 & - & - \\
x_1 &-7/4 & - & + \\
x_2 & 17/16 & + & + \\
x_3 & -223/256 & - & - \end{matrix}$
Now this starts getting hard because squareing $3$ digits number is a lot of hard work, so let me roughly approximate the fractions with $2$ digit numerators and denominators.
$\begin{matrix} -23/25 & & & \le x_3 \le & & & -11/13 & - & - \\
-11/8 & \le & -217/169 & \le x_4 \le & -721/625 & \le& -8/7 & - & + \\
-34/49 & \le & -34/49 & \le x_5 \le & -7/64 & \le & -7/64 & - & - \\
-2 & \le & -8143/4096 & \le x_6 \le & -3646/2401 & \le & -36/25 & - & + \\
4/63 & \le & 46/625 & \le x_7 \le & 2 & \le & 2 & + & + \\
\end{matrix}$
And now this is too imprecise to continue.
So far I got the cumulative sign sequence $(-,+,+,-,+,-,+,+)$ and so the angle is between $(2^{-1}+2^{-4}+2^{-6})\pi$ and $(2^{-1}+2^{-4}+2^{-6}+2^{-8})\pi$
In degrees you replace $\pi$ with $180$, so those are $104.06\ldots$ and $104.77\ldots$
The recurrence follows from the addition formula :
$2\cos(2x) = 2\cos^2(x)-2\sin^2(x) = 4\cos^2(x)-2 = (2\cos(x))^2-2$
Suppose you call $a_n \in [0 ; \pi]$ the angle whose cosine is $2x_n$.
If $x_n\ge 0$ then $a_n \in [0 ; \pi/2] $ and then $a_{n+1} = 2a_n$, so the binary digits of $a_n/\pi$ are $.0$ followed with the binary digits of $a_{n+1}/\pi$
If $x_n \le 0$ then $a_n \in [\pi/2 ; \pi]$ and then $a_{n+1} = 2\pi-2a_n$, so the binary digits of $a_n/\pi$ are $.1$ followed with the inverted binary digits of $a_{n+1}/\pi$
Thus $a_{n+1} = \pm 2 a_n \mod {2\pi}$, and by induction, $a_n = \pm 2^n a_0 \pmod {2\pi}$ where the sign
depends on the parity of the number of negative $x_k$ encountered for $0 \le k < n$. The $n$th digit is $0$ if and only if $2^n a_0 \in [0 ; \pi] \pmod {2\pi}$, which means $\pm a_n \in [0;\pi] \pmod {2\pi}$ with the same sign. But since $a_n \in [0;\pi]$, the digit is $0$ if the sign was $+$ and it is $1$ is the sign was $-$.
And so the $n$th binary digit correspond to the parity of the number of negative cosines encountered for $0 \le k < n$.
|
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If roots of $x^6=p(x)$ are given then choose the correct option If $x_1,x_2,x_3,x_4,x_5,x_6$ are real positive roots of the equation $x^6=p(x)$ where $P(x)$ is a 5 degree polynomial where $\frac{x_1}{2}+\frac{x_2}{3}+\frac{x_3}{4}+\frac{x_4}{9}+\frac{x_5}{8}+\frac{x_6}{27}=1$ and $p(0)=-1$, then choose the correct option(s):
$(A)$ $x_5-x_1=x_3x_4$
$(B)$ Product of roots of $P(x)=0$ is $\frac{6}{53}$
$(C)$ $x_2,x_4,x_6$ are in Geometric Progeression
$(D)$ $x_1,x_2,x_3$ are in Arithmetic Progression
Now
$x_1,x_2,x_3,x_4,x_5,x_6$ are real positive roots of the equation $x^6=p(x)$, so I wrote it as
$x^6-p(x)=(x-x_1)(x-x_2)(x-x_3)(x-x_4)(x-x_5)(x-x_6)$ but I am not getting how to use the condition $\frac{x_1}{2}+\frac{x_2}{3}+\frac{x_3}{4}+\frac{x_4}{9}+\frac{x_5}{8}+\frac{x_6}{27}=1$ to get the answer. Could someone please help me with this?
|
$P(0)=-1$ gives $x_1 x_2 x_3 x_4 x_5 x_6 =1$ Now apply AM-GM to $ x_1/2+ x_2/3+ x_3/4+ x_4/9+ x_5/8 + x_6/27 =1$
\begin{eqnarray*}
1 = \frac{x_1}{2 }+ \frac{x_2}{3 }+ \frac{x_3}{ 4}+ \frac{x_4}{9 }+ \frac{x_5}{8 } + \frac{x_6}{27 } \geq \frac{6 \sqrt[6]{x_1 x_2 x_3 x_4 x_5 x_6}}{6} = 1.
\end{eqnarray*}
For this bound to attained each of the terms in the sum must be $1/6$ so we have
\begin{eqnarray*}
x_1= \frac{1}{3} ,x_2= \frac{1}{2} ,x_3= \frac{2}{3} ,x_4= \frac{3}{2} ,x_5= \frac{4}{3} ,x_6= \frac{9}{2} .
\end{eqnarray*}
Quick check $x_5-x_1=1=x_3 x_4$.
$\sum x_i =53/6$ so (B) is also true. ($p(x)=(x_1+ \cdots+x_6)x^5-\cdots-1$).
$x_2,x_4,x_6 = 1/2,3/2,9/2$ are in geometric progression (common ratio $3$).
$x_1,x_2,x_3 = 1/3,1/2,2/3$ are in arithematic progression ( common difference $1/6$).
Thus all $\color{red}{4}$ statements are true.
|
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|
Factorize $x+{1\over 2}x^2+{1\over 3}x^3+ \cdot\cdot$ by $1-x$ I want to derive explicit formula for given recursive relation below:
$$a_{n+1} = (n + 1)a_n + n!$$
for $n ≥ 0$ and $a_0 = 0$.
I had exploited $EGF$, resulting in:
$$g(x)\cdot(1-x) = x+{1\over 2}x^2+{1\over 3}x^3+ \cdot\cdot$$
Thus to derive the explicit formula of $a_n$, I am thinking about how I can manage the $RHS$ to be factorized by $1-x$ so that I can have $a_n$ for corresponding $x^n/n!$
Any advice to proceed further?
I yet haven't took the abstract algebra class where I seemingly guess I could have more chance to be familiar to polynomial series.
|
We can obtain the numbers $a_n (n\geq 1, a_0=0)$ by calculating the coefficients of the exponential generating function
\begin{align*}
g(x)=a_1x+a_2\frac{x^2}{2!}+a_3\frac{x^3}{3!}+a_4\frac{x^4}{4!}+\cdots
\end{align*}
We obtain
\begin{align*}
\color{blue}{g(x)}&=\left(x+\frac{1}{2}x^2+\frac{1}{3}x^3+\frac{1}{4}x^4+\cdots\right)\frac{1}{1-x}\\
&=\left(x+\frac{1}{2}x^2+\frac{1}{3}x^3+\frac{1}{4}x^4+\cdots\right)\left(1+x+x^2+x^3+x^4+\cdots\right)\tag{1}\\
&=x+\left(1+\frac{1}{2}\right)x^2+\left(1+\frac{1}{2}+\frac{1}{3}\right)x^3
+\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}\right)x^4+\cdots\tag{2}\\
&=\sum_{n=1}^\infty H_nx^n\tag{3}\\
&=\sum_{n=1}^\infty\color{blue}{n!H_n}\frac{x^n}{n!}\tag{4}
\end{align*}
We conclude the numbers are $$\color{blue}{a_n=n!H_n \qquad n\geq 1}$$.
Comment:
*
*In (1) we use the geometric series expansion.
*In (2) we multiply the series of the right-hand side and collect the terms with equal powers of $x$. We observe the coefficients are
\begin{align*}
1,\,1+\frac{1}{2},\,1+\frac{1}{2}+\frac{1}{3},\,1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4},\cdots
\end{align*}
which are the harmonic numbers denoted with $H_n$.
*In (3) we use the sigma notation for brevity.
*In (4) we write the series as exponential generating series to better see the coefficients $a_n$.
Note: If we write the recursion for $a_n$ as
\begin{align*}
\frac{a_{n+1}}{(n+1)!}&=\frac{a_n}{n!}+\frac{1}{n+1}\qquad\qquad n\geq 1\\
a_0&=0
\end{align*}
then the solution $a_n=n!H_n$ might be seen easily.
|
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|
Condition for the inclination of generators through the principal elliptic section of a hyperboloid
Question: Show that the generators through any one of the ends of an equi-conjugate diameter of the principal elliptic section of the hyperboloid $\frac{x^2}{a^2}+\frac{y^2}{b^2}-\frac{z^2}{c^2} = 1$ are inclined to each other at an angle of $60 ^\circ$ if $a^2+b^2=6c^2$.
Solution: I took the point on the diameter as $(a\cos\theta,b\sin\theta,0)$ . So the two generators through this point are
$$\begin{align}
\frac{x-a\cos\theta}{a\sin\theta}=\frac{y-b\sin\theta}{-b\cos\theta}=\frac{z}{\pm c}
\end{align}$$
So the direction ratios of two generators are $(a\sin\theta,-b\cos\theta,c)$ and $(a\sin\theta, -b\cos\theta,-c)$ respectively.
$$\begin{align}
\cos\theta &=\frac{l_1l_2+m_1m_2+n_1n_2}{\sqrt{{l_1}^2+{m_1}^2+{n_1}^2}\sqrt{{l_2}^2+{m_2}^2+{n_3}^2}} \\
\implies \cos\theta &=\frac{a^2\sin^2\theta+b^2\cos^2\theta-c^2}{a^2\sin^2\theta+b^2\cos^2\theta+c^2}
\end{align}$$
Putting $\theta = 60^\circ$ and solving I got $\,3a^2+b^2 = 12c^2$
I am wondering whether the question is wrong or I missed anything or my approach itself is wrong. Can some one please help?
|
Continuing your solution,
$$\begin{align}
\cos\phi &=\frac{l_1l_2+m_1m_2+n_1n_2}{\sqrt{{l_1}^2+{m_1}^2+{n_1}^2}\sqrt{{l_2}^2+{m_2}^2+{n_3}^2}} \\
\implies \cos\phi &=\frac{a^2\sin^2\theta+b^2\cos^2\theta-c^2}{a^2\sin^2\theta+b^2\cos^2\theta+c^2}
\end{align}$$
Putting $\phi = 60^\circ$ and solving, we get
$$a^2sin^2\theta+b^2cos^2\theta= 3c^2$$
Equiconjugate diameters of ellipse implies $\theta=\pi/4, 3\pi/4$. Put this to get the answer.
|
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|
How to prove that $\langle 3+8\sqrt{2},7\rangle = \langle 3+\sqrt{2}\rangle$ in the ring $\mathbb{Z}[\sqrt{2}]$? How to prove that the ideals $\langle 3+8\sqrt{2},7\rangle$ and $\langle 3+\sqrt{2}\rangle$ are equal in the ring $\mathbb{Z}[\sqrt{2}]$?
I tried using the factors and reducing to the form of the other and vice versa, but it did not work.
If there was not two factors involved in the first I would've tried the associativity of generators.
A hint would be appreciated to find out.
|
Hint: show that each contains the generators of the other.
Solution:
Note that $7=(3+\sqrt{2})(3-\sqrt{2})$, so $7 \in (3+\sqrt{2})$, hence $7\sqrt{2}$, and $3+\sqrt{2}+7\sqrt{2}=3+8\sqrt{2}\in(3+\sqrt{2})$. Thus $(7,3+8\sqrt{2})\subset (3+\sqrt{2})$. Conversely, since $7\in(7,3+8\sqrt{2})$, $3+8\sqrt{2}-7\sqrt{2} = 3+\sqrt{2}\in (7,3+8\sqrt{2})$. Hence $(3+\sqrt{2}) \subset (7,3+8\sqrt{2})$ as well. Thus $(7,3+8\sqrt{2})=(3+\sqrt{2})$.
|
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How to solve this recurrence relation? $f(1) = a; f(2) = b; f(x) = 2f(x-1)-f(x-2)+2;$ How to solve this recurrence relation?
*
*$f(1) = a;$
*$f(2) = b;$
*$f(x) = 2f(x-1)-f(x-2)+2;$
Where $a$ and $b$ are positive integers
I want to find $f(x)$ representation with $a$ and $b$ only
Also I am sorry, I really not familiar with it, I got into this as part of my development project
|
By combining all the above clues, we have
\begin{equation}
f(n) = (n-1)b - (n-2)a + (n-1)(n-2)
\end{equation}
To verify this, it is clear that $f(1) = a$ and $f(2) = b$.
Also the simple calculation reveals that
\begin{align}
2f(x-1) - f(x-2) +2 &= 2[(x-2)b - (x-3)a + (x-2)(x-3)] \\
&\hspace{0.2cm} ...- [(x-3)b - (x-4)a + (x-3)(x-4)] + 2 \\
&= (x-1)b - (x-2)a + (x-1)(x-2) = f(x).
\end{align}
|
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|
Finding value of $c+d$
At a point $A(1,1)$ on the ellipse , equation of tangent is $y=x.$ If one of the
foci of the ellipse is $(0,-2)$ and coordinate of center of ellipse is $(c,d)$.
Then find value of $c+d$ (given length of major axis is $4\sqrt{10} unit$)
Attempt : assuming one foci is at $S_{1}(0,-2)$ and other is at $S_{2}(\alpha,\beta)$ and $A(1,1)$ be a point on ellipse. then $AS_{1}+AS_{2} = 4\sqrt{2}$
$\sqrt{10}+\sqrt{(1-\alpha)^2+(1-\beta)^2} = 4\sqrt{2}$
could some help me to solve it , thanks
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The slope of the normal at $(1,1)$ is $-1$ and the normal bisects the angle between $S_1A, S_2A$. Since the slope of $AS_1$ is 3, slope of normal is $-1$, slope of $AS_2$ is $\frac{1}{3}$. The equation of $AS_2$ is $y-1 = \frac{1}{3}(x-1)$. Thus $\beta -1 = \frac{1}{3}(\alpha-1)$. Substitute this in the equation $AS_1+AS_2 = 4\sqrt{10}$ to get $|1-\alpha| = 9$ giving $\alpha = 10, -8$. We can now obtain $\beta$. There are two possible ellipses. When $\alpha = -8$, we get $\beta = -2$ and the center of the ellipse is $(-4, -2)$ and $c+d = -6$. When $\alpha = 10$, $\beta = 4$ and center is $(5,1)$ and $c+d = 6$. Thus $c+d = \pm 6$.
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"timestamp": "2023-03-29T00:00:00",
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|
Proving ${\sum_{n=1}^\infty {1\over F_n}} <4$ I'm trying to prove the sum of Fibonacci numbers' reciprocals is less than 4, which is:
$${\sum_{n=1}^\infty {1\over F_n}} <4$$
It makes me confused because the only information I know about Fibonacci numbers that might be useful are its recurrence relation and general formula. But when dealing with reciprocals, I found the info hard to use.
I also thought of induction: maybe turning this into:
$${\sum_{n=1}^\infty {1\over F_n}} <4-A$$
where A is related to $F_n$. But this method also seems to be not working.
Could anyone please give me some hints?
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This solution uses only simple arithmetic and some recursion.
First we prove that $F_{n+1}-\dfrac{3}{2}F_n\gt 0$ for all $n\ge 4$. Simplify the left hand part:
$$
F_{n+1}-\dfrac{3}{2}F_n = F_{n}+F_{n-1}-\dfrac{3}{2}F_n = F_{n-1}-\dfrac{1}{2}F_n = F_{n-1}-\dfrac{1}{2}(F_{n-1}+F_{n-2}) = \dfrac{1}{2}(F_{n-1}-F_{n-2}) = \dfrac{1}{2}F_{n-3}
$$
Since $F_{n-3}$ is positive for all $n\ge 4$, the statement is true.
Now rearrange the inequality: $F_{n+1}\gt\dfrac{3}{2}F_n$
Invert both sides: $\dfrac{1}{F_{n+1}}\lt\dfrac{2}{3}\dfrac{1}{F_n}$
This holds for all $n\ge 4$, so we can recursively expand the right-hand side:
$$
\dfrac{1}{F_{n+1}}\lt\dfrac{2}{3}\dfrac{1}{F_n}\lt\left(\dfrac{2}{3}\right)^2 \dfrac{1}{F_{n-1}}\lt\cdots\lt\left(\dfrac{2}{3}\right)^{n-3} \dfrac{1}{F_4}
$$
This provides an upper bound for the sum since each term is smaller than a corresponding term in a convergent geometric series. It could also be used as an alternative proof of convergence for Barry Cipra's excellent answer, but using finite descent rather than induction.
The rest just involves calculating this upper bound. First a partial sum of $\dfrac{1}{F_n}$ where $n>4$:
$$
\sum\limits_{n=5}^\infty \dfrac{1}{F_n} \lt \sum\limits_{k=1}^\infty \left(\dfrac{2}{3}\right)^{k} \dfrac{1}{F_4}=\dfrac{1}{3}\cdot\dfrac{2}{3}\sum\limits_{k=0}^\infty \left(\dfrac{2}{3}\right)^{k} = \dfrac{2}{9}\cdot\dfrac{1}{1-\frac{2}{3}} = \dfrac{2}{3}
$$
Then to complete the proof we add the first four terms to both sides:
$$
\sum\limits_{n=1}^\infty \dfrac{1}{F_n} = \dfrac{1}{F_1}+\dfrac{1}{F_2}+\dfrac{1}{F_3}+\dfrac{1}{F_4}+\sum\limits_{n=5}^\infty \dfrac{1}{F_n} \lt 1+1+\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{2}{3} = \frac{7}{2} \lt 4
$$
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Prove that any field $F$ containing $\sqrt{a}+\sqrt{b}$ also contains $\sqrt{a}$ and $\sqrt{b}$.
Prove that any field $F$ containing $\sqrt{a}+\sqrt{b}$ also contains $\sqrt{a}$ and $\sqrt{b}$.
I started by taking $(\sqrt{a}+\sqrt{b})^2=2\sqrt{ab}+a+b$ and want to conclude that $\sqrt{ab}\in F$ but am unsure of this. Then I took $\sqrt{ab}(\sqrt{a}+\sqrt{b})=a\sqrt{b}+\sqrt{a}b\rightarrow b\sqrt{a}+a\sqrt{b}\in F$ and I know this one is because it is of the form $x\sqrt{a}+y\sqrt{b}\in F$. Thus $1\cdot \sqrt{a}+0\cdot \sqrt{b}=\sqrt{a}\in F$ and$ 0\cdot \sqrt{a}+1\cdot \sqrt{b}=\sqrt{b}\in F$. I think I have shown what is needed.
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Hint: $\sqrt{a}-\sqrt{b} = (a-b)(\sqrt{a}+\sqrt{b})^{-1}$
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