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Number of natural numbers less than million with sum of the digits equal to $12$ Number of natural numbers less than million with sum of the digits equal to $12$
My Try: obviously we need 6 places to filled with digits $0$ to $9$ such that sum of the digits is $12$
so the required number is number of non negative integral solutions of
$$x_1+x_2+x_3+x_4+x_5+x_6=12$$ which is nothing but $\binom{17}{5}$
Is this correct approach?
|
We seek the number of solutions of the equation in the nonnegative integers
$$x_1 + x_2 + x_3 + x_4 + x_5 + x_6 = 12 \tag{1}$$
subject to the restrictions that $x_k \leq 9$ for $1 \leq k \leq 6$.
As you determined, if there were no restrictions, equation 1 has
$$\binom{12 + 6 - 1}{6 - 1} = \binom{17}{5}$$
solutions in the nonnegative integers.
Suppose $x_1 > 9$. Then $x_1' = x_1 - 10$ is a nonnegative integer. Substituting $x_1' + 10$ for $x_1$ in equation 1 yields
\begin{align*}
x_1' + 10 + x_2 + x_3 + x_4 + x_5 + x_6 & = 12\\
x_1' + x_2 + x_3 + x_4 + x_5 + x_6 & = 2 \tag{2}
\end{align*}
Equation 2 is an equation with
$$\binom{2 + 6 - 1}{6 - 1} = \binom{7}{5}$$
solutions in the nonnegative integers. By symmetry, there are an equal number of solutions for each of the six variables that could exceed $9$. Hence, the number of solutions of equation 1 that do not satisfy the restrictions is
$$\binom{6}{1}\binom{7}{5}$$
so the number of natural numbers less than a million that have digit sum $12$ is
$$\binom{17}{5} - \binom{6}{1}\binom{7}{5}$$
|
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|
Polynomial : $P(a)P(a^2)P(a^3)P(a^4)$ Let $P(x) = x^3 + 2x^2+3x+4$ and $a$ be the root of equation $x^4+x^3+x^2+x+1=0$.
Find the value of $P(a)P(a^2)P(a^3)P(a^4)$
Is my answer correct ?
Since root of equation $x^4+x^3+x^2+x+1=0$ is the $5^{th}$ primitive root of 1,
so $a, a^2, a^3, a^4$ are roots of $x^4+x^3+x^2+x+1=0$
$P(a)P(a^4)=(a^3+2a^2+3a+4)(\frac{1}{a^3}+\frac{2}{a^2}+\frac{3}{a}+4)= 15+5a^4+5a$
Similarly, $P(a^2)P(a^3)=15+5a^3+5a^2$
$P(a)P(a^2)P(a^3)P(a^4)=(15+5a^4+5a)(15+5a^3+5a^2)=125$
|
Suppose $P(x)=x^3+2x^2+3x+4=(x-p)(x-q)(x-r)$ for some $p,q,r\in\mathbb{C}$. Then, $$\prod_{j=1}^4\,P\left(a^j\right)=Q(p)\,Q(q)\,Q(r)\,,$$
where $Q(x):=x^4+x^3+x^2+x+1$. Now, $$Q(x)=(x-1)\,P(x)+5\,.$$
Thus, $Q(p)=Q(q)=Q(r)=5$.
This is actually quite a nice technique. Let $P(x)$ and $Q(x)$ be two nonconstant polynomials in $x$ over a field $K$. Write $\bar{P}$ and $\bar{Q}$ for the leading coefficients of $P(x)$ and $Q(x)$, respectively. Suppose that $t_1,t_2,\ldots,t_n$ are the roots of $Q(x)$ in the algebraic closure of $K$ (with multiplicities). Then, $$\prod_{j=1}^n\,P\left(t_j\right)=(-1)^{mn}\,\frac{\bar{P}^n}{\bar{Q}^m}\,\prod_{i=1}^m\,Q\left(s_i\right)\,,\tag{1}$$
where $s_1,s_2,\ldots,s_m$ are the roots of $P(x)$in the algebraic closure of $K$ (with multiplicities).
If $m>n$, then write
$$P(x)=Q(x)\,A(x)+B(x)$$
for some $A(x),B(x)\in K[x]$ with $B(x)$ having degree less than $n$. Clearly, we have $$\prod_{j=1}^n\,P\left(t_j\right)=\prod_{i=1}^m\,B\left(t_i\right)\,,\tag{2}$$
Then, we can use the paragraph below using $B(x)$ in place of $P(x)$ to simplify things even more, provided that $B(x)$ is nonconstant (or nonlinear).
If $m \leq n$, then we write $Q(x)=P(x)\,U(x)+V(x)$ for some polynomials $U(x),V(x)\in K[x]$, with $V(x)$ having degree less than $m$. Then, from $(1)$, $$\prod_{j=1}^n\,P\left(t_j\right)=(-1)^{mn}\,\frac{\bar{P}^n}{\bar{Q}^m}\,\prod_{i=1}^m\,V\left(s_i\right)\,.\tag{3}$$
In fact, we can play this game again between $V(x)$ and $P(x)$ to make further degree reductions, provided that $V(x)$ is nonconstant (or nonlinear).
For example, let $P(x)=x^5+x^2+2x+1$ and $Q(x)=x^3-3x-2$. Then, we see that $P(x)=A(x)\,Q(x)+B(x)$ with $A(x)=x^2+3$ and $B(x)=3x^2+11x+7$. If $t_1$, $t_2$, and $t_3$ are the roots of $Q(x)$, then
$$\prod_{j=1}^3\,P\left(t_j\right)=\prod_{j=1}^3\,B\left(t_j\right)\,.$$
Note that $Q(x)=B(x)\,U(x)+V(x)$, where $U(x)=\frac{x}{3}-\frac{11}{9}$ and $V(x)=\frac{73}{9}x+\frac{59}{9}$. Using $(3)$, we obtain
$$\prod_{j=1}^3\,B\left(t_j\right)=(-1)^{2\cdot 3}\,\frac{3^3}{1^2}\,\prod_{i=1}^2\,V\left(r_i\right)\,,$$
where $r_1$ and $r_2$ are the roots of $B(x)$. That is,
$$\prod_{j=1}^3\,P\left(t_j\right)=3^3\,\prod_{i=1}^2\,V\left(r_i\right)=3^2\left(\frac{73}{9}\right)^2\left(-\frac{59}{73}-r_1\right)\left(-\frac{59}{73}-r_2\right)=3^2\left(\frac{73}{9}\right)^2\,B\left(-\frac{59}{73}\right)\,.$$
That is, $\prod_{j=1}^3\,P\left(t_j\right)=41$. This can be easily verified as $t_1=-1$, $t_2=-1$, and $t_3=2$, so that $P\left(t_1\right)=-1$, $P\left(t_2\right)=-1$, and $P\left(t_3\right)=41$. (I picked an easy-to-check example, of course, so this reduction method is actually more complicated than simply computing the values of $P\left(t_j\right)$'s.)
|
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|
Evaluating $\frac{1}{2 \pi i} \int_{|z|=3} \frac{e^{\pi z}}{z^2(z^2+2z+2)}dz$ I need some help with Complex Analysis:
To evaluate the integral
$$\frac{1}{2 \pi i} \int_{|z|=3} \frac{e^{\pi z}}{z^2(z^2+2z+2)}dz$$
Here is what I tried:
So, $f(z)=\frac{e^{\pi z}}{z^2(z^2+2z+2)}$ has 3 singularities, $z_0 = 0$, $z_0= -1+i $ and $z_0=-1-i$.
And all of them are interior to the contour.
$\Rightarrow$ We need to find the residues at all the 3 points.
$$Res_{z_0=0}(f(z)) = Res_{z_0=0} \left (\frac{e^{\pi z}}{z^2(z^2+2z+2)} \right)\\=\frac{1}{2}(\pi-1)$$
$$Res_{z_0=-1+i}(f(z)) = Res_{z_0=-1+i} \left (\frac{e^{\pi z}}{z^2(z^2+2z+2)} \right)\\=-\frac{e^{-\pi}}{4}$$
$$Res_{z_0=-1-i}(f(z)) = Res_{z_0=-1-i} \left (\frac{e^{\pi z}}{z^2(z^2+2z+2)} \right)\\=-\frac{e^{-\pi}}{4}$$
$$\Rightarrow \frac{1}{2\pi i}\int_{|z|=3} \frac{e^{\pi z}}{z^2(z^2+2z+2)}dz = \frac{1}{2 \pi i} \times \left[ 2\pi i \left(\frac{1}{2}(\pi -1) - \frac{e^{-\pi}}{4} - \frac{e^{-\pi}}{4}\right)\right]\\ = \left( \frac{1}{2}(\pi-1)-\frac{e^{-\pi}}{2}\right)\\ = \frac{\pi-1-e^{-\pi}}{2}$$
But I was told that this was wrong. I couldn't find any mistakes for my work. Can any of you check to see if my work is valid? Or, if it is wrong, how else can I evaluate this? Any helps or comments would be appreciated. Can someone provide a valid solution (or a different approach)to evaluate this integral? Because some comments say it is wrong. But, some say it is correct. I am confused....
|
So checking the residues...
The pole at $z=0$ is double, so (in my opinion) it's easier to find the residue via the series expansion than with the explicit formula. Factoring the denominator, the integrand is $\frac{e^{\pi z}}{z^2(z+1-i)(z+1+i)}$. The series for $e^{\pi z}$ is just $1 + \pi z + \frac{\pi^2 z^2}{2!} + ...$ The other factors except for $\frac{1}{z^2}$ are of the form $\frac{1}{a + z}$, which has the series representation $\frac{1}{a} - \frac{z}{a^2}+...$ So the integrand, as a product of series, is $\frac{1}{z^2}(1 + \pi z + ...)(\frac{1}{1-i} - \frac{z}{(1-i)^2} + ...)(\frac{1}{1+i} - \frac{z}{(1+i)^2} + ...)$ Since we're looking for the residue, the only term we care about is $z^{-1}$. Since $\frac{1}{z^2}$ is the only factor with negative exponents, we only need the first two terms of each series. Finding the terms that would have order $-1$, we find the coefficient to be $\frac{\pi}{(1+i)(1-i)} - \frac{1}{(1-i)^2(1+i)} - \frac{1}{(1+i)^2(1-i)}$. The first term is $\frac{\pi}{2}$, and the second and third terms combine to $-\frac{1}{2(1+i)} - \frac{1}{2(1-i)} = -\frac{1-i}{4} - \frac{1+i}{4} = -\frac{1}{2}$. So the first residue is (unless I've also made a horrible mistake) $\frac{\pi - 1}{2}$, as you found.
Now for the next two residues: since these are both simple poles, they can be evaluated easily with the explicit formula. The second residue should be $\frac{e^{\pi z}}{z^2(z+1-i)}$ evaluated at $z = -1 - i$, and the third is similar. Plugging in, the second residue is $\frac{e^{\pi (-1 - i)}}{(-1-i)^2(-1-i+1-i)} = \frac{e^{-\pi}e^{-i\pi}}{(2i)(-2i)}$ which is $-\frac{e^{-\pi}}{4}$. The third is similar, and I also found it also matches what you have.
Maybe we both made the same mistakes, but your evaluation looks right to me.
|
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Find all values of $x$ at which $P(x)=x^4-4x^3+22x^2-36x+18$ is a perfect square Find all values of positive integer $x$ at which the following expression is perfect square
$$P(x)=x^4-4x^3+22x^2-36x+18$$
I tried to assume $P= (x^2+ax+b)^2$ ; and comparing the cofactors , get that $a= -2 ; b= 9$
, but when expand the $x^2-2x+9$, I didn't get the same as $P$
What is wrong in my work?
|
As Jyrki Lahtonen mentioned in comments, my bounds can be improved to
$$(x^2-2x-14)^2 < P(x) < (x^2-2x-13)^2,$$
or in expanded form:
$$196 + 56 x - 24 x^2 - 4 x^3 + x^4 < 18 - 36 x - 22 x^2 - 4 x^3 + x^4 < 169 + 52 x - 22 x^2 - 4 x^3 + x^4$$
for sufficiently large values of $x$ ($x \ge 48$ in first inequality, $x \ge -1$ in the second one). But these bounds are actually squares of adjacent integers, so after inspection of finitely many cases ($x = 1, 2, \ldots, 47$) we can surelz say that $P(x)$ is never a square.
|
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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?
|
Note that $\sin x$ is concave function when $0\leq x\leq \dfrac{\pi}{2}$. Then by Jensen's inequality on concave function, we have:
$$\frac{\sin \dfrac{A}{2}+\sin \dfrac{B}{2}+\sin \dfrac{C}{2}}{3}\leq \sin \dfrac{A+B+C}{2\cdot 3}=\sin\dfrac{\pi}{6}=\frac{1}{2}$$
Now by AM-GM inequality $$\frac{\sin \dfrac{A}{2}+\sin \dfrac{B}{2}+\sin \dfrac{C}{2}}{3}\geq \Big(\sin \frac{A}{2} \sin \frac{B}{2} \sin \frac{C}{2}\Big)^{\dfrac{1}{3}}$$
Hence $$\sin \frac{A}{2} \sin \frac{B}{2} \sin \frac{C}{2} \le \frac{1}{8}$$
Equality holds when $A=B=C=\dfrac{\pi}{3}$
|
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Finding $a + b + c$ for the $101$st number of the form $3^a + 3^b + 3^c$ when these numbers are listed in increasing order $a,b,c\neq0$ & $a,b,c$ are distinct & $a,b,c\in \mathbb{N}$
numbers in form of
$3^a+3^b+3^c$
If we order them in increasing order, what is the sum of
$a+b+c=?$
for the $101$st such number?
There are some things I need to clarify about the question a number is represented by $3^a+3^b+3^c$, not $3^a,3^b,3^c$. These do not represent an individual number.
That being said we could accept $a>b>c$ and then acknowledge that $a,b,c\in \left\{ 0,1,2,3,\cdots,n \right\}$. There are $\dbinom{n+1}{3}$ choices possible for $n$. Therefore, $n$ must give a number greater than $101$. I have chosen $\dbinom{10}{3}=120$, so $n=9$. But I also used $\dbinom{9}{3}=84$ to know which number is $3^9+3^1+3^0$ (it is the $84$th). I couldn't spot how I would proceed in getting the $101$st number? What do you propose?
|
Without loss of generality we can assume that $a > b > c$.
Let's consider the lexicographic ordering;
we will write $(a,b,c) > (a',b',c')$,
if one of the following occures:
*
*$a > a'$ ; or
*$a=a' $ and $b > b'$; or
*$a=a' $ and $b=b' $ and $c > c'$.
Remark:
Let's consider two ordered triples $(a,b,c) , \ (a',b',c')$;
i.e. $a > b > c$ and $a' > b' > c'$.
Then:
$$
3^a+3^b+3^c > 3^{a'}+3^{b'}+3^{c'}
\Longleftrightarrow
(a,b,c) > (a',b',c').
$$
As you have been said;
$3^9+3^1+3^0$ is greater than $3^a+3^b+3^c $
for every $\{ a,b,c \} \subset \{ 0, 1, 2, ..., 8 \}$.
Also notice that there are $84 = {9 \choose 3}$ choices for $\{ a,b,c \}$ ;
so $3^9+3^1+3^0$ is in the $85^{\text{th}}$-place.
$3^9+3^6+3^0$ is greater than $3^9+3^b+3^c $
for every $\{ b,c \} \subset \{ 0, 1, 2, ..., 5 \}$.
Also notice that there are $15 = {6 \choose 2}$ choices for $\{b,c \}$ ;
so $3^9+3^6+3^0$ is in the $1+15+84=100^{\text{th}}$-place.
Finally one can easilly check that $3^9+3^6+3^1$
is in the $101^{\text{th}}$-place.
How to find the number, at the $m^{\text{th}}$-place?
An effective algorithm for the above problem is as follows:
*
*Find greatest integer $a \in \mathbb{N}_0$ such that
${a \choose 3} \leq m-1$.
*Find greatest integer $b \in \mathbb{N}_0$ such that
${b \choose 2} \leq m-{a \choose 3}-1$.
*Find greatest integer $c \in \mathbb{N}_0$ such that
${c \choose 1} \leq m-{a \choose 3}-{b \choose 2}-1$;
i.e $c=m-{a \choose 3}-{b \choose 2}-1$.
One can check that $3^a+3^b+3^c$
is in the $1+{c \choose 1}+{b \choose 2}+{a \choose 3}=m^{\text{th}}$-place.
|
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|
maybe $(2^m-1)(3^n-1)$ can never be a perfect square? Conjecture：
For any positive integer $m,n$, show that $(2^m-1)(3^n-1)$ is never a perfect square?
maybe this is old problem？ I find some $m,n$ the problem is right ,so How to prove it?
The conjecture was based on my solution to this following (different) problem :
Problem: >Find all non-negative integers $m$ and $n$, such that $(2^n-1) \cdot (3^n-1)=m^2$
Solution:
Assume that $n > 0$ is a solution. First, note that $2 \mid n$, otherwise the exponent of $2$ in LHS in odd. From this (by lifting) we deduce that $n$ must be divisible by $3$ as well.
Let $n=6k$. Then $$ (2^{6k}-1)(3^{6k}-1) \equiv (2^k-1)(16^k-1) \equiv (2^k-1)^2 (2^k+1)(4^k+1) \pmod{31} $$ must be a quadratic residue. We claim this implies $5 \mid k$. Indeed, for $k \equiv 1,2,3,4 \pmod5$ we get $3 \cdot 5$, $5 \cdot 17$, $5 \cdot 13$, $17 \cdot 9$ are non-quadratic residues mod $31$, since of the primes here, only $5$ is a quadratic residue.
So let $n = 10c$. Then we have that $\left( (2^{10})^c - 1 \right)\left( (3^{10})^c - 1 \right)$ is a perfect square. But now by lifting the exponent on the prime $11$ we get a contradiction, because $\nu_{11}(2^{10}-1) = 1$ and $\nu_{11}(3^{10}-1) = 2$, thus the exponent of $11$ in the original is $3+2\nu_{11}(c)$.
|
Another partial solution, is to note that all squares of integers are 0 1 or 4 mod 8, $2^m-1$ is either 1,3 or 7 mod 8, and $3^n-1$ is either 0 or 2 mod 8.
$$1\cdot0\equiv0\bmod8;
1\cdot2\equiv2\bmod8;
3\cdot0\equiv0 \bmod8;
3\cdot2\equiv6\bmod8;
7\cdot0\equiv0\bmod8;
7\cdot2\equiv6\bmod8$$
None of these are 1 or 4 mod 8, so the only possibility left is 0 mod 8. when $3^n-1\equiv0\bmod8$ it's always one less than a square. so that part doesn't help. it then comes down can the square free parts for a square because a non primitive is the product of two or more squares. so either it's a primitive square in which case $2^m-1 =3^n-1$ which means there's a power of two that's a power of three, contradiction. So if it's not a primitive square it must factor into squares. since it takes $3^n-1\equiv0\bmod8$ for there to be a chance, and this factors as $(3^{n\over 2}+1)\cdot(3^{n\over2}-1)$ these must then half to combine with the factors of $2^m-1$ to make squares.
|
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${2^\left|(x+2)\right|}$ - $\left|2^{x+1} -1 \right| $=$ 2^{x+1}+1$ .What is the Minimum value of ${x}$? I try to use logarithm but I cannot be applied on the equation.
So how can I solve this equation.
Any help will be appreciate.
|
If $x \in [-1 , \infty)$,
then we have $|x+2|=x+2$ and $|2^{x+1}-1|=2^{x+1}-1$;
so in this interval, the equation has this form:
$$
\ \ \ \ \
2^{x+2}-(2^{x+1}-1)=
2^{x+1}+1
\Longleftrightarrow
2.2^{x+1}-2^{x+1}+1=
2^{x+1}+1
\Longleftrightarrow
\\
(2-1).2^{x+1}+1=
2^{x+1}+1
\Longleftrightarrow
2^{x+1}+1=
2^{x+1}+1
;
\ \ \ \ \
\checkmark
\checkmark
\checkmark
$$
so we can conclude that
every $x \in [-1 , \infty)$ satisfies the equation.
If $x \in (\infty , -1]$, then we have
$%% |x+2|=-(x+2)$
$|2^{x+1}-1|=2^{x+1}-1$;
so in this interval, the equation has this form:
$$
2^{|x+2|}-\Big(-(2^{x+1}-1)\Big)=
2^{x+1}+1
\Longleftrightarrow
2^{|x+2|}=2
\Longleftrightarrow
|x+2|=1;
$$
which gives us the new solution $x=-3 \in [-1 , \infty)$.
So by the above we can conclude that all solutions are the following:
$$
\{ -3 \}
\cup
[-1 , \infty)
.
$$
|
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|
Prove by Induction: $1- \frac{1}{2}+ \frac{1}{3}- \frac{1}{4} + \ldots + \frac{(-1)^n}{n}$ is always positive How do you prove by induction that $1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\ldots+\frac{(-1)^n}{n}$ is always positive.
The base case works out and the inductive step is n=k, so $1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\ldots+\frac{(-1)^k}{k}$ holds true for the statement. I am stuck on the n = k+1.
if any one can give an idea, it would help alot.
|
Proving that $1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\ldots+\frac{(-1)^n}{n}$ is always positive is equivalent to proving that:
$1+\frac{1}{3}+\dots +\frac{1}{2n-1} > \frac{1}{2}+\frac{1}{4}+\dots +\frac{1}{2n}$
For all $n\geq 1$
Clearly this is true when $n=1$. Assuming this is true for $n=k$, how could you show this is true for $n=k+1$?
|
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|
Find $\int\frac{\tan^2(x)}{\sqrt{x}} \, dx$ I can not find the next antiderivative
$$\displaystyle\int\dfrac{\tan^2(x)}{\sqrt{x}}\,dx$$
I tried with integration by parts the second integral, please help me.
My try
$$
\begin{aligned}
\int \dfrac{\sec^2(x)-1}{\sqrt{x}}\,dx
=\int\dfrac{\sec^2(x)}{\sqrt{x}}\,dx-\int\dfrac{dx}{\sqrt{x}}=\int\dfrac{\sec^2(x)}{\sqrt{x}}\,dx-2\cdot\sqrt{x}+C\end{aligned}
$$
Thank you so much.
|
$\int\dfrac{\tan^2x}{\sqrt x}~dx$
$=\int\dfrac{\sec^2x-1}{\sqrt x}~dx$
$=\int\dfrac{\sec^2x}{\sqrt x}~dx-\int\dfrac{1}{\sqrt x}~dx$
$=\int\dfrac{1}{\sqrt x}~d(\tan x)-2\sqrt x$
$=\dfrac{\tan x}{\sqrt x}-\int\tan x~d\left(\dfrac{1}{\sqrt x}\right)-2\sqrt x$
$=\dfrac{\tan x}{\sqrt x}-2\sqrt x-\int\tan u^2~d\left(\dfrac{1}{u}\right)$ (Let $u=\sqrt x$)
$=\dfrac{\tan x}{\sqrt x}-2\sqrt x+\int\dfrac{\tan u^2}{u^2}~du$
$=\dfrac{\tan x}{\sqrt x}-2\sqrt x+\int\sum\limits_{n=0}^\infty\dfrac{8}{(2n+1)^2\pi^2-4u^4}~du$ (use Mittag-Leffler Expansion of tangent)
$=\dfrac{\tan x}{\sqrt x}-2\sqrt x+\int\sum\limits_{n=0}^\infty\dfrac{8}{((2n+1)\pi+2u^2)((2n+1)\pi-2u^2)}~du$
$=\dfrac{\tan x}{\sqrt x}-2\sqrt x+\int\sum\limits_{n=0}^\infty\dfrac{4}{(2n+1)\pi((2n+1)\pi+2u^2)}~du+\int\sum\limits_{n=0}^\infty\dfrac{4}{(2n+1)\pi((2n+1)\pi-2u^2)}~du$
$=\dfrac{\tan x}{\sqrt x}-2\sqrt x+\sum\limits_{n=0}^\infty\dfrac{2\sqrt2}{(2n+1)\pi\sqrt{(2n+1)\pi}}\tan^{-1}\dfrac{\sqrt2u}{\sqrt{(2n+1)\pi}}+\sum\limits_{n=0}^\infty\dfrac{2\sqrt2}{(2n+1)\pi\sqrt{(2n+1)\pi}}\tanh^{-1}\dfrac{\sqrt2u}{\sqrt{(2n+1)\pi}}+C$
$=\dfrac{\tan x}{\sqrt x}-2\sqrt x+\sum\limits_{n=0}^\infty\dfrac{2\sqrt2}{(2n+1)\pi\sqrt{(2n+1)\pi}}\tan^{-1}\dfrac{\sqrt{2x}}{\sqrt{(2n+1)\pi}}+\sum\limits_{n=0}^\infty\dfrac{2\sqrt2}{(2n+1)\pi\sqrt{(2n+1)\pi}}\tanh^{-1}\dfrac{\sqrt{2x}}{\sqrt{(2n+1)\pi}}+C$
|
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|
Complex transformation $ w = z + \frac{c}{z} $ where $ |z| = 1 $ I am to show that if $ w = z + \frac{c}{z} $ and $ |z| = 1 $, then $w$ is an ellipse, and I must find its equation.
Previously, I have solved transformation questions by finding the modulus of the transformation in either the form $ w = f(z) $ or $ z = f(w) $. However, I think the part stumping me here is that the transformation has both $z$ and $z^{-1}$ in it.
Attempt with $w = f(z)$:
$$ w = x+iy + \frac{c}{x+iy} $$
$$= x+iy + \frac{c(x-iy)}{x^2 + y^2} $$
$$= x+iy+c(x-iy)$$
[as $|z| = 1 \implies x^2 + y^2 = 1^2$]
However, from here I am unable to work out how to proceed.
I also tried to find $z=f(w)$:
$$ z^2 - zw + c = 0 $$
$$ (z - \frac{w}{2})^2 + c - \frac{w^2}{4} = 0 $$
But I cannot see how this is of any use either.
|
$z$ is on the unit circle; let $z=e^{i \theta}$ so
\begin{eqnarray*}
w= (1+c) \cos( \theta) +i (1-c) \sin(\theta)
\end{eqnarray*}
which gives $ x= (1+c) \cos( \theta) , y= (1-c) \sin(\theta)$ and considered in cartesian coordinates
\begin{eqnarray*}
\frac{x^2}{ (1+c)^2} + \frac{y^2}{ (1-c)^2} =1 .
\end{eqnarray*}
|
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|
Prove that no perfect number of the form $3^m 5^n 7^k$ exists. For all perfect numbers $N$, $\sigma (N) = 2N$, where $\sigma$ is the divisor sigma function.
Let $s$ be a perfect number of the form $3^m 5^n 7^k$, where $m,n,k \geq 1$ are integers.
Then $\sigma (s)= \sigma (3^m 5^n 7^k)$
$ =\sigma (3^m) \sigma (5^n) \sigma (7^k)$ since $3, 5,$ and $7$ are coprime to each other.
$ =\left(\frac{3^{m+1}-1}{2}\right)\left(\frac{5^{m+1}-1}{4}\right)\left(\frac{7^{k+1}-1}{6}\right)$
$ =2(3^m 5^n 7^k)$ since $s$ is a perfect number.
$\implies 9 (3^m 5^n 7^k) = 3^{m+1} 5^{n+1}+3^{m+1} 7^{k+1} + 5^{n+1} 7^{k+1} - 3^{m+1}-5^{n+1} - 7^{k+1}-1$ after some algebra.
This is as far as I got using this method. Any and all help would be appreciated.
|
Note that the Question stipulates the exponents considered are $m,n,k\ge 1$. As @lulu points out, the cases where one of the exponents is zero can (if desired) be ruled out by this previous Question.
The following is a simplification of the proof that an odd perfect number cannot be divisible by $105$ found here, as previously linked under the older Question to that effect.
$N= 3^m 5^n 7^k$ is a perfect number if and only $S(N)$, the sum of all divisors of $N$ (including itself and $1$), equals $2N$.
Since $N$ is odd, it must be that $S(N)=2N$ is not divisible by $4$. Now:
$$ \frac{S(N)}{N} = \left(1+\frac{1}{3}+\ldots+\frac{1}{3^m}\right)
\left(1+\frac{1}{5}+\ldots+\frac{1}{5^n}\right)
\left(1+\frac{1}{7}+\ldots+\frac{1}{7^k}\right) $$
Since $m=1$ would give $\left(1+\frac{1}{3}+\ldots+\frac{1}{3^m}\right)=\frac{4}{3}$ and $k=1$ would give $\left(1+\frac{1}{7}+\ldots+\frac{1}{7^k}\right)=\frac{8}{7}$, either would imply $S(N)$ is divisible by $4$, contradicting our observation above.
Knowing thus $m,k\ge 2$, we get a contradiction:
$$ \begin{align*} 2 = \frac{S(N)}{N}
&\ge \left(1+\frac{1}{3}+\frac{1} {3^2} \right)
\left(1+\frac{1}{5}\right)
\left(1+\frac{1}{7}+\frac{1}{7^2}\right) \\
&= \frac{13}{9} \frac{6}{5} \frac{57}{49} = \frac{4446}{2205} \gt 2 \end{align*} $$
|
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|
Show that ${\sum^{n}_{k=1}} \frac{1}{n+k} \le \frac{3}{4}$
Prove that $\displaystyle{\sum^{n}_{k=1}} \frac{1}{n+k} \le \frac{3}{4}$ for each positive integer $n$.
My work. I think that i have to use induction, but i can't see how... What i did:
$$f(n)=\displaystyle{\sum^{n}_{k=1}} \frac{1}{n+k} \implies k(n)=f(n+1)-f(n)=\frac{1}{2n+1}-\frac{1}{2n+2}.$$
Now we have
$$f(n+1)=\displaystyle{\sum^{n}_{k=1}} \left ( \frac{1}{2k+1} \right )- \displaystyle{\sum^{n}_{k=1}} \left ( \frac{1}{2k+2} \right )$$
But now I have no more ideas.
|
Here there are two different proofs.
1) Note that since $1/(1+x)$ is decreasing and positive for $x\geq 0$ then
$$\sum^{n}_{k=1} \frac{1}{n+k}=\frac{1}{n}\sum^{n}_{k=1} \frac{1}{1+\frac{k}{n}}\leq
\sum^{n}_{k=1}\int_{(k-1)/n}^{k/n} \frac{dx}{1+x}
= \int_0^1 \frac{dx}{1+x}=\ln(2)<\frac{3}{4}.$$
2) For $n\geq 1$, show by induction the stronger inequality (which turns out also in Andreas' answer)
$$\sum^{n}_{k=1} \frac{1}{n+k} \le \frac{3}{4}-\frac{1}{4n}.$$
In fact, it holds for $n=1$ and for $n\geq 1$,
\begin{align*}
\sum^{n+1}_{k=1} \frac{1}{n+1+k}
&=\sum^{n+2}_{j=2} \frac{1}{n+j}=\sum^{n}_{j=1} \frac{1}{n+j}-\frac{1}{n+1}+\frac{1}{2n+1}+\frac{1}{2n+2}\\
&\leq \frac{3}{4}-\frac{1}{4n}-\frac{1}{n+1}+\frac{1}{2n+1}+\frac{1}{2n+2}\\
&= \frac{3}{4}-\frac{1}{4(n+1)}-\frac{1}{4n(n+1)(2n+1)}
\leq \frac{3}{4}-\frac{1}{4(n+1)}.
\end{align*}
|
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|
How to find the one real root of $(x-1)(x-2)(x-3) + 1$, manually? The polynomial
$$ (x-1)(x-2)(x-3) + 1 $$
has one root (I have seen it by a plot in LaTeX), $0 < x_{root} < 1$. So I presume that the polynomial can be rewritten as :
$$ (x-1)(x-2)(x-3) + 1 = (x - a)^{3} $$
but this is not possible, since
$$ (x-1)(x-2)(x-3)+1 = x^{3} - 6x^{2} + 11x -5 $$
$$ (x-a)^{3} = x^{3} - 3ax^{2} + 3a^{2}x -a^{3} $$
How to find the form of the polynomial without the remainder? (also
manually, without numerical method)
Thanks before.
|
We can solve $ (x-1)(x-2)(x-3) + 1 = 0 $ for $x \in \mathbb{R}$ by manipulating it in a quadratic equation with some convenient variable substitutions as follows:
Expand the polynomial
$$ x^3-6 x^2+11 x-5 = 0 $$
Substitute $ y = x - 2 $ to get rid of the quadratic term
$$ (y+2)^3 - 6(y+2)^2 + 11(y+2) - 5 = y^3 - y + 1 = 0 $$
Let $c \in \mathbb{R}$ be a constant we yet don't know its value but will determine a suitable value later. Substitute $ y = z + \frac{c}{z} $
$$ (z + \frac{c}{z})^3 - (z + \frac{c}{z}) + 1 = 0 $$
Multiply both sides by $z^3$ and group by $z$
$$ z^6 + z^4(3c-1) + z^3 + z^2c(3c-1)+c^3 = 0 $$
Note that by setting $ c = \frac{1}{3} $ we get rid of both the quadratic and the quartic terms yielding
$$ z^6 + z^3 + c^3 = 0 $$
To make computations easier make $ k = c^3$. Now substitute $ u = z^3 $ and we get a quadratic equation in $u$
$$ u^2 + u + k = 0 $$
Can you carry on from here? Let me know if you get stuck.
|
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|
Use the AGM inequality to find the maximum Use the AGM inequality to find the maximum of $(5+\sqrt{x^4+1}) \cdot (9-\sqrt{x^4 + 1})$.
$$ab \le (\frac{a+b}{2})^2$$
I don't know how to relate this inequality to find the maximum of that.
|
Take $a=5+\sqrt{x^4+1}, b=9-\sqrt{x^4+1}$. Then:
$$ab\le 49$$
Equality occurs when $x=\pm\sqrt[4]{3}$.
Because:
$$ab=(5+\sqrt{x^4+1})(9-\sqrt{x^4+1})\le \left(\frac{(5+\sqrt{x^4+1})+(9-\sqrt{x^4+1})}{2}\right)^2=7^2=49.$$
|
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|
Prove that $\int \frac{x^m dx}{\ln x}=\ln(\ln x)+\frac{(m+1)\ln x}{1}+\frac{(m+1)^2 \ln ^2 x}{1 \times 2^2}+\cdots $ Prove that
$$\int \frac{x^m dx}{\ln x}=\ln(\ln x)+\frac{(m+1)\ln x}{1}+\frac{(m+1)^2 \ln ^2 x}{1 \times 2^2}+\frac{(m+1)^3 \ln^3 x}{1 \times 2 \times 3^2}+\cdots \infty $$
My Try:
I started with $$I=\int \frac{x^m}{\ln x}=\int \frac{x^{m+1} dx}{x \ln x}$$
Now by using parts taking $u=x^{m+1}$ and $v=\frac{1}{x \ln x}$ we get
$$I=x^{m+1} \times \ln(\ln x)-(m+1)\int x^m \ln (\ln x)dx$$
can i have any clue to proceed
|
With substitution $x=e^u$:
\begin{align}
\int \frac{x^m}{\ln x}dx
&=\int\dfrac{1}{u}e^{u(m+1)}du \\
&=\int\dfrac{1}{u}\sum_{n=0}^\infty \dfrac{u^n(m+1)^n}{n!}du \\
&=\int\left(\dfrac{1}{u}+\sum_{n=1}^\infty \dfrac{u^{n-1}(m+1)^n}{n!}du\right) \\
&=\ln u+\sum_{n=1}^\infty \dfrac{u^n(m+1)^n}{n.n!} \\
&=\ln\ln x+\sum_{n=1}^\infty \dfrac{\ln^nx(m+1)^n}{n.n!}
\end{align}
|
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|
What's the maximum value of $\frac{2}{a^2+1}-\frac{2}{b^2+1}+\frac{3}{c^2+1}$ given that: $abc+a+c=b$
Given that: $abc+a+c=b$. What's the maximum value of $$\frac{2}{a^2+1}-\frac{2}{b^2+1}+\frac{3}{c^2+1}.$$
(DO NOT use trigonometric methods)
|
For $a=\frac{1}{\sqrt2}$, $b=\sqrt2$ and $c=\frac{1}{2\sqrt2}$ we'll get a value $\frac{10}{3}$.
We'll prove that it's a maximal value.
Indeed, the condition gives $c(ab+1)=b-a$.
If $ab=-1$ then $a=b$ and $a^2+1=0$, which is impossible.
Thus, $ab\neq-1$, $c=\frac{b-a}{ab+1}$ and we need to prove that
$$\frac{2}{1+a^2}-\frac{2}{1+b^2}+\frac{3}{1+\left(\frac{b-a}{1+ab}\right)^2}\leq\frac{10}{3}$$ or
$$\frac{2(b^2-a^2)}{(1+a^2)(1+b^2)}+\frac{3(1+ab)^2}{(1+a^2)(1+b^2)}\leq\frac{10}{3}$$ or
$$(a^2+4)b^2-18ab+16a^2+1\geq0,$$
for which it's enough to prove that
$$81a^2-(a^2+4)(16a^2+1)\leq0,$$
which is
$$(2a^2-1)^2\geq0.$$
Done!
I got a value $\frac{10}{3}$ and values $a$, $b$ and $c$ by the following way.
Let $k$ be a maximal value of our expression.
Thus, the following inequality is true for all values of $a$ and $b$.
$$\frac{2(b^2-a^2)}{(1+a^2)(1+b^2)}+\frac{3(1+ab)^2}{(1+a^2)(1+b^2)}\leq k$$ or
$$(ka^2-3a^2+k-2)b^2-6ab+ka^2+2a^2+k-3\geq0.$$
Thus, $k\geq3$ and since we have a quadratic inequality of $b$, we need that the equality
$$9a^2-(ka^2-3a^2+k-2)(ka^2+2a^2+k-3)\leq0$$ or
$$(k^2-k-6)a^4+2(k^2-3k-2)a^2+k^2-5k+6\geq0$$
will be true for all value of $a$.
But the last inequality is a quadratic inequality of $a^2$, which says that we need
$$(k^2-3k-2)^2-(k^2-k-6)(k^2-5k+6)\leq0$$ or
$$3k-10\geq0,$$
which gives a value $\frac{10}{3}$ and from here we can get the case of the equality occurring.
|
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|
$x\le y \le z$; $\text{ }x \cdot y \cdot z = 1 $; Prove that $(x+1) \cdot(z+1)>3$ Good morning,
Assuming that $x$,$y$,$z$ are positive real numbers and $x \le y \le z$ and $ x \cdot y \cdot z = 1$.
How can I prove that $(x+1) \cdot (z+1) > 3$?
I know that $x$ has to be less than $1$ and $z$ greater than $1$ and I have already tried to rearrange that term but this didn't help me.
So do you have any ideas how to prove that?
|
Let $y < z$
Set $y_1 = z_1 = \sqrt{yz}$
$(x+1)(z+1)> (x+1)(z_1+1)$
So, you must prove you task only for $z = y$ => $z=\frac{1}{\sqrt{x}}$, $x \le 1$
$(x+1)(z+1) = 1 + x + \sqrt{x} + \frac{1}{\sqrt{x}} \ge 3+x$
(by $t + \frac{1}{t} \ge 2$ )
|
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|
Find the center of the octagon, complex plane The problem given read as follows,
Two consecutive vertices of a regular convex octagon are $(1,2)$ and $(3,-2)$. Find the center of the octagon.
I would like to know if my answer is correct. Here's what I have thought:
I've drawn the complex plane and set the affixes given accordingly (I've called them $z_{1} = (1,2)$ and $z_{2}=(3,-2)$).
Since $z_{1}, z_{2}, ... z_{8}$ compose a regular octagon, the angle between $z_{1}$ and $z_{2}$ must be $\frac{\pi}{4}$. So I have an angular relation at my disposal.
Say $z_{0} = x+iy$ is the center of the polygon, then I have two vectors, namely $\vec{v} = \vec{z_{0} z_{2}}$ and $\vec{u} = \vec{z_{0} z_{1}}$ and since I have an angle relation between them, I can write:
$\vec{u} = \vec{v}.e^\frac{i \pi}{4} \iff z_{1}-z_{0} = (z_{2} - z_{0}).cis \left(\frac{\pi}{4} \right) (*)$
Substituting with my known affixes, I get:
$1+2i-z_{0}=(3-2i-z_{0}).e^\frac{i \pi}{4}$
Working out some algebra, I've got:
$z_{0} = \frac{3e^{\frac{i \pi}{4}} - 1}{e^\frac{i \pi}{4} - 1} + i\frac{-2-2 e^\frac{i \pi}{4}}{e^\frac{i \pi}{4} - 1}$
Converting from polar form to trigonometric form and doing a lengthy algebra, I've got:
$\boxed{z_{0} = 2 + i \frac{\sqrt{2}}{2-\sqrt{2}}}$
I believe that my reasoning is correct, although substituting my result at $(*)$ I get:
$\left(1 - \sqrt{2}, \frac{4-3\sqrt{2}}{2-\sqrt{2}}\right) \neq \left(\frac{3 \sqrt{2} - 2}{2}, \frac{1-2 \sqrt{2}}{2-\sqrt{2}}\right)$
I would appreciate if someone could help me figure out my mistake.
Thanks, in advance.
|
$z_0=x+i y;z_1=1+2 i;z_2=3-2 i$
$z_1-z_0=e^{\frac{i \pi }{4}} (z_2-z_0)$
$-e^{\frac{i \pi }{4}} (-x-i y+(3-2 i))-x-i y+(1+2 i)=0$
$
\left\{
\begin{array}{c}
\frac{x}{\sqrt{2}}-x-\sqrt{2}-\frac{y}{\sqrt{2}}-\frac{3}{\sqrt{2}}+1=0 \\
\frac{x}{\sqrt{2}}+\frac{y}{\sqrt{2}}-y+\sqrt{2}-\frac{3}{\sqrt{2}}+2=0 \\
\end{array}
\right.
$
$x=-2 \sqrt{2},y=-1-\sqrt{2}$
but there is also another possibility, as you can see in the
picture
$z_2-z_0=e^{\frac{i \pi }{4}} (z_1-z_0)$
which leads to the other solution
$x=2 \left(2+\sqrt{2}\right),y=1+\sqrt{2}$
Hope this is useful
|
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|
Solve the equation $18x^2-18x \sqrt{x}-17x-8 \sqrt{x}-2=0$
Solve the following equation. $$18x^2-18x \sqrt{x}-17x-8 \sqrt{x}-2=0.$$
Taking $\sqrt{x}=t$ we get equivalent equation $18t^4 -18t^3 - 17t^2-8t-2=0$.
From this point I have tried to factor it , write RHS as sum of two squares and its variants but nothing seem to work. Then putting the original equation in wolfram alpha I got solution $x=\frac{2}{9}(7+2 \sqrt{10})$. Can anyone suggest a method to solve it without wolfram alpha or any such computer method. Thanks in advance.
|
Let $\sqrt{x}=\frac{t}{6}$.
Thus, we need to solve
$$18\cdot\frac{t^4}{1296}-18\cdot\frac{t^3}{216}-17\cdot\frac{t^2}{36}-8\cdot\frac{t}{6}-2=0$$ or
$$t^4-6t^3-34t^2-96t-144=0$$ or for all real $k$
$$(t^2-3t+k)^2-9t^2-k^2-2kt^2+6kt-34t^2-96t-144=0$$ or
$$(t^2-3t+k)^2-((2k+43)t^2-(6k-96)t+k^2+144)=0.$$
Now, we'll choose a value of $k$ such that
$$(2k+43)t^2-(6k-96)t+k^2+144=(at+b)^2,$$
for which we need
$$9(k-16)^2-(2k+43)(k^2+144)=0.$$
Easy to see that $k=-9$ is valid.
Id est, we need to solve
$$(t^2-3t-9)^2-(25t^2+150t+225)=0$$ or
$$(t^2-3t-9)^2-25(t+3)^2=0$$ or
$$(t^2-3t-9-5(t+3))(t^2-6t-9+5(t+3))=0$$ or
$$(t^2-8t-24)(t^2+2t+6)=0$$ or
$$t^2-8t-24=0$$ or
$$(t-4)^2=40,$$ which gives
$$6\sqrt{x}=4+\sqrt{40}$$ or
$$\sqrt{x}=\frac{2+\sqrt{10}}{3}$$ or
$$x=\frac{14+4\sqrt{10}}{9}.$$
|
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|
How to prove $8^{2^n} - 5^{2^n}$ is divisible by $13$ where $n\in\mathbb{N}-\{0\}$ with induction? So the question is as above is shown:
How to prove $8^{2^n} - 5^{2^n}$ is divisible by $13$?
$n=1: 8^2 - 5^2 = 39$ is a multiple of $13$.
I.H.: Suppose $8^{2^n} - 5^{2^n}$ is divisible by $13$ is true $\forall n \in \mathbb{N}\setminus\{0\}$.
I got stuck on this $n= m + 1$ case.
$8^{2^{m+1}} - 5^{2^{m+1}}= 8^{2^m\cdot 2}- 5^{2^m\cdot 2}$
Can somebody please help me?
|
To complete the proof with the following identity $ A^2 - B^2 = (A + B) (A-B) $.
I get
$\begin{align}\\
\underline {\text{n= m+1:}} \\
8^{2^n} -5^{2^n} & = 8^{2^{m+1}} -5^{2^{m+1}} \\
& = (8^{2^m} + 5^{2^m})(8^{2^m} -5^{2^m})\\
& = (8^{2^m} + 5^{2^m})\cdot \text{some multiple of 13}
\end{align}$
Thus concluding that the entire expression is divisible by 13.
|
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|
Proving pointwise convergence I'm having trouble proving: $$f_n(x) = \frac{1-n^2x^2}{(1+n^2x^2)^2}$$
converges pointwise to zero for $x \in \,[-1,0)\cup(0,1]$.
My attempt for finding : $N(\epsilon,x)$
\begin{align*}
\left|\frac{1-n^2x^2}{(1+n^{2}x^{2})} - 0\right|<\epsilon &\iff \frac{|1-n^2x^2|}{(1+n^2x^2)^2} < \epsilon\\
\end{align*}
Since,
\begin{align*}
\frac{|1-n^2x^2|}{(1+n^2x^2)^2} &< \frac{|1-n^2x^2|}{n^4x^4}\\
\end{align*}
we can then find $\epsilon$ such that:
\begin{align*}
\frac{|1-n^2x^2|}{n^4x^4}<\epsilon
\end{align*}
I get stuck after this step. Does anyone know how to prove this without using $$\lim_{n \rightarrow \infty}f_n(x)$$
|
HINT
Let $x \in [-1,0) \cup (0,1]$ and $\epsilon>0$
\begin{align*}
|f_n(x)| \leq \frac{|1-n^2x^2|}{n^4x^4}\leq\frac{1}{n^4x^4}+\frac{n^2x^2}{n^4x^4}=\frac{1}{n^4x^4}+\frac{1}{n^2x^2}
\end{align*}
Exists $n_1 \in \mathbb{N}$ such that for all n$\geq n_1$ we have $\frac{1}{n^4x^4} \leq \epsilon\Rightarrow n_1 \geq \frac{1}{\sqrt[4]{\epsilon}x^4}$
Exists $n_2 \in \mathbb{N}$ such that for all n$\geq n_2$ we have $\frac{1}{n^2x^2} \leq \epsilon\Rightarrow n_2 \geq \frac{1}{\sqrt{\epsilon}x^2}$
Take $N \geq \max\{n_1,n_2\}$ and you are done.
|
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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|
Solve the following nonlinear system of equations:$ xy+xz+yz=12 , xyz=2+x+y+z$
Solve the following system of equations in $\Bbb R^+$:
$$
\left\{
\begin{array}{l}
xy+yz+xz=12 \\
xyz=2+x+y+z\\
\end{array}
\right.
$$
I did as follows.
First equation yields to: $(x+y+z)^2-x^2-y^2-z^2=24$. Now, using second equation we get: $x^2y^2z^2-4xyz=20+x^2+y^2+z^2$ . Here I stopped...
|
On one hand, we have $$(xy+yz+zx)^2 \geq 3xyz(x+y+z)$$, then one has $$-8 \leq x+ y +z \leq 6.$$
On other hand, we have $$(x+y+z)^2 \geq 3(xy+yz+zx) = 36,$$
then $$x+y+z \geq 6$$ or $$x+y+z \leq -6.$$
So, you have $x+y+z = 6$ or $-8 \leq x+y+z \leq -6$ (but x, y, z is non-negative!).
Thus, $x+y+z = 6$ and $x=y=z$.
|
{
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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|
For what values of $a$ does the system have infinite solutions? Find the solutions. The system is
$$\left\{
\begin{array}{rcr}
x+ay+z & = & 1 \\
ax+y+z & = & 1+a \\
x-y+z & = & 2+a
\end{array}
\right.$$
After row reducing I got
$$\left\{
\begin{array}{rcr}
x+ay+z & = & 1 \\
-(1+a)y+0 & = & 1+a \\
(1-a)z& = & 2-a + (1-a)(1+a)
\end{array}
\right.$$
In order to have an infinite set of solutions, I want to have $0=0$ in the last row, meaning that $a=1\Rightarrow 0=1,$ which is not what I want. Have I made any arithmetical mistake? Did this computation times now.
|
The system reduces to
$\left[\begin{matrix}
1 & 0 & 1 & 1+a\\
0 & -1 & 0 & 1 \\
0 & 0 & 1-a & 1+(1+a)^2
\end{matrix}\right]$ which means $z=\frac{1+(1+a)^2}{1-a}$
It implies for $a$ any value other than $1$ the system can have infinite solutions.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2427722",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 3
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|
limit of $\frac{1-\sin x+\cos x }{x-\frac{\pi }{2}}$ how can i show that
$$\lim _{x\to \frac{\pi }{2}}\left(\frac{1-\sin \:x\:+\cos \:x\:}{x-\frac{\pi }{2}}\right)=-1$$
$$\left(\frac{1-\sin \:x\:+\cos \:x\:}{x-\frac{\pi }{2}}\right)=\left(\frac{1+\cos \:x -\sin x}{x-\frac{\pi }{2}}\right)$$
any help thanks
|
$$\lim _{x\to \frac{\pi }{2}}\left(\frac{1-\sin \:x\:+\cos \:x\:}{x-\frac{\pi }{2}}\right)=\\
\lim _{x\to \frac{\pi }{2}}\left(\frac{1-\sin \:x\:}{x-\frac{\pi }{2}}\right)+\lim _{x\to \frac{\pi }{2}}\left(\frac{\cos \:x\:}{x-\frac{\pi }{2}}\right)=\\
\lim _{x\to \frac{\pi }{2}}\left(\frac{\sin(\frac{\pi}{2})-\sin x}{x-\frac{\pi }{2}}\right)+\lim _{x\to \frac{\pi }{2}}\left(\frac{\sin(\frac{\pi}{2}- x)}{x-\frac{\pi }{2}}\right)=\\
\lim _{x\to \frac{\pi }{2}}\left(\frac{2\sin(\frac{\frac{\pi}{2}-x}{2}).\cos(\frac{\frac{\pi}{2}+x}{2})}{-(\frac{\pi }{2}-x)}\right)+\lim _{x\to \frac{\pi }{2}}\left(\frac{\sin(\frac{\pi}{2}- x)}{-(\frac{\pi }{2}-x)}\right)=\\$$
so $$2(\frac{1}{-2})\lim _{x\to \frac{\pi }{2}}\cos(\frac{\frac{\pi}{2}+x}{2})+\underbrace{\lim _{x\to \frac{\pi }{2}}\left(\frac{\sin(\frac{\pi}{2}- x)}{-(\frac{\pi }{2}-x)}\right)}_{-1}=\\-1(0)+(-1)=-1$$
|
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"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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|
Defining a piecewise function using restricted operations Question
Can the piecewise function
$$f(x) = \begin{cases}
0 & \text{if $x > 0$} \\
1 & \text{if $x = 0$} \\
0 & \text{if $x < 0$} \\
\end{cases}$$
be defined using only the operations $+ , -, *, /, |\cdots|, \max$, and $\min$?
What I have tried
I can define the first and last pieces: $0$ if $x > 0$ or $x < 0$ with
$$1 - \frac{x}{x}$$
But this will fail with a division by $0$ in the case where $x = 0$
$$1 - \frac{0}{0}$$
I can fix the division error by forcing a 1 on the bottom.
$$a(x) = 1 - \frac{x}{\max(1, x) \min(-1, x)}$$
This works for most negatives and $0$ and fails when $-1 < x < 0$ and $x > 0$. When $x > 0$, $a(x) = 2$. Fixing this requires another max to check a number is positive. Defining $b(x)$ to be $2$ when $x > 0$ and $0$ when $x = 0$ or $x <= -1$
$$b(x) = 2\frac{\max(0, x)}{\max(1, x) \min(-1, x)}$$
Combining them to get
$$c(x) = a(x) - b(x) = 1 - \frac{x - 2\max(0, x)}{\max(1, x) \min(-1, x)}$$
This mess is what I want except when $-1 < x < 0$ and $0 < x < 1$. This is as far as I have gotten.
|
You asked how to define your function with an infinite number of operations
from your list.
Here it is defined by means of an infinite sum:
$$
f(x) =
\max(0,1-|x|)
+ \frac12\sum_{n=1}^\infty (\min(1,|2^n x+1|) + \min(1,|2^n x-1|) - 2).
$$
The first part of this, $\max(0,1-|x|),$ is zero except for $-1<x<1,$
where its graph is an isoceles triangle whose vertices are
$(-1,0),$ $(1,0),$ and $(0,1)$.
Each of the terms of the sum is zero except for
$-\frac{2}{2^n}<x<\frac{2}{2^n},$ where its graph consists of two
isoceles triangles with vertices at
$\left(-\frac{2}{2^n},0\right),$ $(0,0),$ and
$\left(-\frac{1}{2^n},-1\right)$ (first triangle)
and at $(0,0),$ $\left(\frac{2}{2^n},0\right),$ and
$\left(\frac{1}{2^n},-1\right)$ (second triangle).
The function
$$
f_m(x) =
\max(0,1-|x|)
+ \frac12\sum_{n=1}^m (\min(1,|2^n x+1|) + \min(1,|2^n x-1|) - 2)
$$
defined by taking only a partial sum is zero everywhere except
for $-\frac{1}{2^m}<x<\frac{1}{2^m},$
where its graph is an isoceles triangle with vertices
$\left(-\frac{1}{2^n},0\right),$ $\left(\frac{1}{2^n},0\right),$ and
$(0,1).$
So for any $x\neq 0,$ $f_m(x) = 0$ whenever $m$ is large enough,
whereas $f_m(0) = 1$ for all $m.$
|
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
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|
substitution in $y' = \cos(y-x)$ equation I have one simple DE with separable variables to test substitutions skill:
$$\begin{align*}
y' &= \cos(y-x) \\
\frac{dy}{dx} &= \cos(y-x) \\
t &= y -x \\
\frac{dy}{dx} &= \cos t
\end{align*}$$
But what should I do next? I mean I do not know if there's an algorithm to handle substitutions in DEs or no
|
$y-x=u$ derive $y'-1=u'\to y'=u'+1$
The equation becomes
$u'+1=\cos u$
$\dfrac{du}{dx}=\cos u-1$
$\dfrac{du}{\cos u-1}=dx$
$\int dx=x+C$
$\cos u=\dfrac{1-t^2}{1+t^2}$ substituting $t=\tan\frac{u}{2}$
$u=2\arctan t$ and $du=\dfrac{2dt}{1+t^2}$
$$\int \frac{1}{\cos u-1} \, du=\int \frac{2}{\left(\frac{1-t^2}{1+t^2}-1\right) \left(t^2+1\right)} \, dt=\int -\frac{1}{t^2} \, dt=\dfrac{1}{t}=\cot\frac{u}{2}$$
$x+C=\cot\dfrac{u}{2}$
$x+C=\cot\dfrac{y-x}{2}$
$y=x+2\,\text{arccot}(x+C)$
Hope this helps
|
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|
Find the total number of coefficients in the expansion of $(x^2-x +1)^{22}$ that are divisible by $6$. Find the total number of coefficients in the expansion of $(x^2-x +1)^{22}$ that are divisible by $6$.
So I am really stuck on this one.. I immediately defeated when I saw that I need to compute $\binom {22} {a,b,c}$ where $a,b,c$ is the coefficients of $x^2,-x,1$. So what should be the next step I need to take?
|
We provide a method for determining when a binomial coefficient is a multiple of a prime $p$.
We first prove the following:
Lemma: Let $p$ be a prime and let $m,n$ be natural numbers. Let $n = lp+t$ and $m = kp+s$, where $0 \leq t < p$ and $0 \leq s < p$. Then
\begin{align*}
\binom{n}{m} = \binom{l}{k}\binom{t}{s}\mod p
\end{align*}
Proof: Observe first that for $0 < r < p$, $\binom{p}{r}$ is divisible by $p$. Hence in
$$ (1+X)^p - (1+X^p) = \binom{p}{1}X + \binom{p}{2}X^2+\cdots + \binom{p}{p-1}X^{p-1} $$
all the coefficients are divisible by $p$.
Now, consider
\begin{align*}
P(X) &= (1+X)^{lp+t} - (1+X)^t (1+X^p)^{l} \\
&= (1+X)^t\left\{(1+X)^{lp} - (1+X^p)^l \right\} \\
&= (1+X)^t \left\{(1+X)^p - (1+X^p)\right\}\left\{(1+X)^{p(l-1)} + \cdots + (1+X^p)^{l-1}\right\}
\end{align*}
As observed before, $P(X)$ is a multiple of $p$.
Consider the coefficient of $X^{kp+s}$ in $P(X)$. This equals
\begin{align*}
\binom{lp+t}{kp+s} - \binom{t}{s}\binom{l}{k}
\end{align*}
and hence all the coefficients are multiples of $p$. Note that this holds even when $k = 0$ or $l=0$. This completes the proof of the lemma.
Applying the Lemma repeatedly, it follows that if
\begin{align*}
n &= t_1+t_2p+t_3p^2+\cdots + t_rp^{r-1} \\
m &= s_1+s_2p+s_3p^2+\cdots + s_rp^{r-1}
\end{align*}
then
\begin{equation*}
\binom{n}{m} = \binom{t_1}{s_1}\binom{t_2}{s_2}\cdots\binom{t_{r-1}}{s_{r-1}} \mod p
\end{equation*}
As observed, $x^2-x+1 = (x+1)^2 \pmod 3$ and hence we can find which binomials $\binom{44}{k}$ are multiples of 3.
Here, $44 = 3^3 + 3^2 + 2\cdot 3 + 2 $. If $k = s_1\cdot 3^3 + s_2 \cdot 3^2 + s_3 \cdot 3 + s_4$, where $0 \leq s_i \leq 2$, $\binom{44}{k}$ is a multiple of 3 if and only if $s_2 = 2$ and note that $s_1 \neq 2$ since $k \leq 44$. This gives that for $k = 18, 19, 20, 21, 22, 23, 24, 25, 26$, $\binom{44}{k}$ is a multiple of 3.
Of these we need to eliminate 18, 22 and 26. We do this as follows:
Since $x^2-x+1 = x^2+x+1 \pmod 2$, we have
$$(x^2-x+1)^{22} = \left(\frac{1-x^3}{1-x}\right)^{22} \pmod 2$$
Again, using the Lemma we obtain upto $\mod\, 2$,
$$(1-x^3)^{22} = 1 + x^6 + x^{12} + x^{18} + x^{48} + x^{54} + x^{60} + x^{66}$$
and again $mod\, 2$, $(1-x)^{-22}$ contains only even powers of $x$. Now,
we eliminate $18, 22, $ and $26$ as follows:
The coefficient of $x^{18}$ is obtained by multiplying the terms containing $(1, x^{18}), (x^6, x^{12}), (x^{12}, x^6)$ and hence the coefficient is sum of three odd numbers and hence is odd.
The coefficient of $x^{22}$ is obtained by multiplying the terms containing $(1, x^{22}), (x^6, x^{16}), (x^{18}, x^4)$ and hence the coefficient is sum of three odd numbers and hence is odd.
The coefficient of $x^{26}$ is obtained by multiplying the terms containing $(1, x^{26}), (x^6, x^{20}), (x^{12}, x^{14})$ and hence the coefficient is sum of three odd numbers and hence is odd.
Thus the only coefficients that are multiples of 6 are those of $x^n$ where $n = 19, 20, 21, 23, 24, 25$.
|
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|
Yet another family of hypergeometric sums that has a closed form solution. Let $m \ge 2$ and $j\ge 0$ be integers. Now, let $0 < z < \frac{(m-1)^{m-1}}{m^m}$ be a real number. Consider a following sum:
\begin{equation}
{\mathfrak S}^{(m,j)}(z) := \sum\limits_{i=0}^\infty \binom{m \cdot i + j}{i}\cdot z^i
\end{equation}
By using both my answer to Closed form solutions for a family of hypergeometric sums. and by generalizing the approach from my answer to About the identity $\sum\limits_{i=0}^{\infty}\binom{2i+j}{i}z^i=\frac{B_2(z)^j}{\sqrt{1-4z}}$ I have derived the following results:
\begin{eqnarray}
{\mathfrak S}^{(m,0)}(z) &=& \frac{x \left(1-z x^{m-1}\right)}{1-m z x^{m-1}}\\
{\mathfrak S}^{(m,1)}(z) &=& \frac{m}{m-1} \cdot\frac{ x \left(1-z x^{m-1}\right)}{ \left(1-m z x^{m-1}\right)}-\frac{1}{m-1}\cdot \left(\frac{x-1}{x z}\right)^{\frac{1}{m-1}}\\
{\mathfrak S}^{(m,2)}(z) &=& \frac{m^2}{(m-1)^2}\cdot \frac{x \left(1-z x^{m-1}\right)}{ \left(1-m z x^{m-1}\right)}
-\frac{(m-2) }{(m-1)^2}\cdot \left(\frac{x-1}{x z}\right)^{\frac{1}{m-1}}+\\
&&
-\frac{1}{(m-1)^2}\cdot \left(\frac{x-1}{x z}\right)^{\frac{2}{m-1}} \cdot \frac{((m-1) x+2) }{ x}\\
{\mathfrak S}^{(m,3)}(z) &=& \frac{m^3}{(m-1)^3}\cdot \frac{x \left(1-z x^{m-1}\right)}{ \left(1-m z x^{m-1}\right)}-
\frac{(m-2)(2 m-3)}{2(m-1)^3}\cdot \left(\frac{x-1}{x z}\right)^{\frac{1}{m-1}}+\\
&&-\frac{m-3}{(m-1)^3}\cdot \left(\frac{x-1}{x z}\right)^{\frac{2}{m-1}}\cdot \frac{((m-1) x+2)}{x}+\\
&&-\frac{1}{2(m-1)^3}\cdot \left(\frac{x-1}{x z}\right)^{\frac{3}{m-1}}\cdot
\frac{2(m-1)^2 x^2+6(m-1)x+3(m+2)}{x^2}
\end{eqnarray}
Here $x:=x(z)$ is obtained in the following way. Out of the solutions of the equation:
\begin{equation}
1-x+z \cdot x^m=0
\end{equation}
we choose the one that is the closest to unity.
Now the question is how does the result look like for arbitrary $j \ge 2$.
|
Here I am using the results provided above by Markus Scheuer. Firstly I provide a closed form expression for the quantity $B_m(z)$. We have:
\begin{eqnarray}
B_m(z)&=& \sum\limits_{i=0}^\infty \binom{m\cdot i}{i} \cdot \underbrace{\frac{1}{(m-1) i+1}}_{\int\limits_0^1 \theta^{(m-1) i} d\theta} \cdot z^i\\
&=& \int\limits_0^1 \frac{x\cdot(1-z\theta^{m-1}\cdot x^{m-1})}{1-m z \theta^{m-1} \cdot x^{m-1}} d\theta \\
&\underbrace{=}_{z \theta^{m-1}=(\xi-1)/\xi^m}& \frac{1}{z^{1/(m-1)} \cdot(m-1)}\int\limits_0^{x-1} \frac{\xi^{1/(m-1)-1}}{(1+\xi)^{m/(m-1)}} d\xi\\
&=& \left( \frac{x-1}{z \cdot x}\right)^{1/(m-1)} = x
\end{eqnarray}
where $1-x+ z \cdot x^m=0$.
Therefore using the result quoted by Markus Scheuers' we have:
\begin{equation}
{\mathfrak S}^{(m,j)}(z) = \frac{(\frac{x-1}{z\cdot x})^{j/(m-1)}}{1-m+m \cdot (\frac{z \cdot x}{x-1})^{1/(m-1)}} = \frac{x^{j+1}}{(1-m) \cdot x+m}
\end{equation}
In particular if $m=1,2$ the result reads:
\begin{eqnarray}
\left\{ {\mathfrak S}^{(m,j)}\right\}_{m=1}^2 = \left\{ (\frac{1}{1-z})^{j+1}, \frac{(\frac{1-\sqrt{1-4 z}}{2 z})^j}{\sqrt{1-4 z}}\right\}
\end{eqnarray}
as it should be.
|
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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??
|
hint: A classic trick is for example $1^2 - 2^2 + 3^2 - 4^2 = (1^2 +2^2+3^2+4^2) - 2(2^2+4^2)= S_4 - 2^3(1^2+2^2)= S_4 - 8S_2$. Now you can generalize this with $4,2$ are replaced by $2n, n$. Can you finish it? If the last term is odd, then isolate it and use up to the preceeding term which is even. So it can always be done this way...
|
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|
Simplify $\sin{(\arctan{2}-\arcsin{\frac{1}{\sqrt{10}}})}.$ First I denote $x=\arctan{2}$ and $y=\arcsin{\frac{1}{\sqrt{10}}}$ and then use the addition formula for sine:
$$\sin{(x-y)}=\sin{x}\cos{y}-\cos{x}\sin{y}=\sin{x}\cos{y}-\cos{x}\cdot \frac{1}{\sqrt{10}}.$$
Now I use the fact that $\cos{y}=1-\sin^2{y}$ which gives
$$\cos{y}=1-\sin^2{\left(\arcsin{\frac{1}{\sqrt{10}}}\right)}=1-\frac{1}{10}=\frac{9}{10}$$
So I have reduced the problem to the following
$$\sin{x}\cdot \frac{9}{10}-\cos{x}\cdot \frac{1}{\sqrt{10}}.$$
Here I'm stuck. I cant just divide this expression by $(\sin{x}\cos{x})$ because I'd still be left with sine and cos terms and I'd also change the value of the expression. Everything now boils down to compute $\sin(\arctan{2})$ and $\cos{\arctan2}.$ I also tried rewriting $\cos(\arctan{2})$ as $1-\sin^2(\arctan{2}),$ but to no avail.
Any suggestions on
*
*how to proceed from where I left;
*how tocompute this by means more effective;
*both of the above.
|
Put $\arctan 2 = x$ then $\tan x= 2$, so $\sin x = {2\over \sqrt{5}}$ and $\cos x = {1\over \sqrt{5}}$.
Let $\arcsin {1\over \sqrt{10}}= y$ so $\sin y = {1\over \sqrt{10}}$, thus $\cos y = {3\over \sqrt{10}}$.
Now:
$$\sin{(\arctan{2}-\arcsin{\frac{1}{\sqrt{10}}})} = \sin (x-y) = \sin x \cos y- \cos x \sin y= {\sqrt{2}\over 2} $$
|
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"timestamp": "2023-03-29T00:00:00",
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|
Determinant of a symmetric zero-diagonal matrix I am trying to write an algorithm which computes determinants of $n\times n$ real-symmetric matrices of the form
$$\mathbf S = \begin{pmatrix}
0 & a_{1,1} & a_{1,2} & a_{1,3} & \cdots &a_{1,n-1}\\
a_{1,1} & 0 & a_{2,1} & a_{2,2} & \cdots& a_{2,n-2}\\
a_{1,2} & a_{2,1} & 0 & a_{3,1} & \cdots& a_{3,n-3}\\
a_{1,3} & a_{2,2} & a_{3,1} & 0& \cdots& a_{4,n-4}\\
\vdots & \vdots & \vdots& \vdots& \ddots &\vdots\\
a_{1,n-1}&a_{2,n-2}& a_{3,n-3} & a_{4,n-4} & \dots &0
\end{pmatrix}$$
which are to be encoded in the form $\{\{a_{1,1},a_{1,2},\dots,a_{1,n-1}\},\{a_{2,1},a_{2,2},\dots,a_{2,n-2}\},\dots,\{a_{n-1,1}\}\}$.
For example, the encoding {{1,2,3},{4,5},{6}} corresponds to the $4\times 4$ matrix
$$ \begin{pmatrix}
0&1&2&3\\
1&0&4&5\\
2&4&0&6\\
3&5&6&0
\end{pmatrix}.
$$
Is there a way of efficiently finding the eigenvalues of $\mathbf S$, or, alternatively, finding the number of distinct eigenvalues, by taking advantage of its symmetric properties?
The only way I've managed so far is to compute is to evaluate $\det(\mathbf S - \mu \mathbf I)$ by treating $a_{i,j} = a_{j,i}$, but this is still $O(\frac{n^2}{2})$, i.e. $O(n^2)$.
|
You won't be able to do anything that's asymptotically better than an algorithm for computing general determinants (which is $O(n^3)$ the easy way, though there are algorithms that are $O(n^k)$ for $k \approx 2.373$). The reason for this is that a special case of your problem is finding the determinant of a block matrix of the form
$$ \begin{bmatrix} 0 & A \\ A^T & 0\end{bmatrix} = \begin{bmatrix} 0 & 0 & \cdots & 0 &a_{1,n/2} & a_{1,n/2+1} & \dots & a_{1,n-1} \\
0 & 0 & \cdots & 0 & a_{2,n/2} & a_{2,n/2+1} & \dots & a_{2,n-1} \\
\vdots & \vdots & \ddots & \vdots & \vdots & \vdots & \ddots & \vdots \\
0 & 0 & \cdots & 0 & a_{n/2,n/2} & a_{n/2, n/2+1} & \cdots & a_{n/2, n-1} \\
a_{1,n/2} & a_{2,n/2} & \cdots & a_{n/2,n/2} & 0 & 0 & \cdots & 0 \\
a_{1,n/2+1} & a_{2,n/2+1} & \cdots & a_{n/2, n/2+1} & 0 & 0 & \cdots & 0 \\
\vdots & \vdots & \ddots & \vdots & \vdots & \vdots & \ddots & \vdots \\
a_{1,n-1} & a_{2,n-1} & \cdots & a_{n/2,n-1} & 0 & 0 & \cdots & 0\end{bmatrix}$$
and this determinant simplifies to $(-1)^n \det(A)^2$, so computing it is as hard as computing the determinant of $A$, an arbitrary $\frac n2 \times \frac n2$ matrix.
|
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|
Evaluating $\int_0^\infty dq\, \frac{\sin(qr)}{r} \left(\frac{q}{\sqrt{q^2+m^2}}-1\right)$ I am trying to evaluate, at least in the limit $|x|\to0$,
$$
F_m(r)\equiv\int \frac{d^3q}{(2\pi)^3}\left(\frac{1}{2\sqrt{\mathbf q^2-m^2}}-\frac{1}{2|\mathbf q|}\right) e^{i\mathbf q \cdot \mathbf x}\,.
$$
Going to spherical coordinates and letting $r=|\mathbf x|$ yields
$$
\frac{1}{(2\pi)^2}\int_0^\infty dq\, \frac{\sin(qr)}{r} \left(\frac{q}{\sqrt{q^2+m^2}}-1\right)\,,
$$
which is absolutely integrable since the term between round brackets behaves as $-m^2/2q^2$ for large $q$.
Since
$$\begin{aligned}
\left|\int_0^\infty dx\, \sin(qr) \left(\frac{q}{\sqrt{q^2+m^2}}-1\right)\right|&=
\int_0^\infty dq |\sin(qr)|\left(1-\frac{q}{\sqrt{q^2+m^2}}\right)\\
& \le \int_0^\infty dq \left(\frac{q}{\sqrt{q^2+m^2}}-1\right)\\
&=m<\infty
\end{aligned}$$
we can conclude that, by dominated convergence,
$$
\lim_{r\to0}\int_0^\infty dx\, \sin(qr) \left(\frac{q}{\sqrt{q^2+m^2}}-1\right)=0\,;
$$
but unfortunately this does not help since it gives an indefinite expression for
$$
\lim_{r\to0} F_m(r)= \left[\frac{0}{0}\right]\,.
$$
I tried applying l'Hospital's rule but then we get
$$
\int_0^\infty dq\, \cos(qr) \left(\frac{q^2}{\sqrt{q^2+m^2}}-q\right)\,.
$$
Alternatively, I tried using $\sin(qr)=\Im e^{iq r}$ and integrating by parts formally (we can think that $r$ acquires a small positive imaginary part which we then send to zero), which gives
$$\begin{aligned}
F_m(r)&=\frac{1}{(2\pi)^2} \Re \int_0^\infty dq\, e^{iqr}\left(q-\sqrt{q^2+m^2}\right)
\end{aligned}$$
But still, I'm not getting anywhere. Any suggestion? :)
|
With $q=mt$ and $x=rm$,
\begin{align}
I&=\int_0^\infty dq\, \frac{\sin(qr)}{r} \left(\frac{q}{\sqrt{q^2+m^2}}-1\right)\\
&=\frac{m}{r}\int_0^\infty \sin (xt) \left( \frac{t}{\sqrt{1+t^2}}-1 \right)\,dt
\end{align}
We integrate by parts,
\begin{align}
I&=\frac{m}{r}\left[\left.-\frac{\cos (xt)}{x}\left( \frac{t}{\sqrt{1+t^2}}-1 \right)\right|_0^\infty+\frac{1}{x}\int_0^\infty\frac{\cos (xt)}{\left( 1+t^2 \right)^{\tfrac{3}{2}}} \right]\,dt\\
&=\frac{m}{r}\left[-\frac{1}{x}+K_1(x)\right]
\end{align}
$K_1(.)$ is the modified Bessel function of the second kind and we used its Basset's integral representation.
Then,
\begin{equation}
I=\frac{m}{r}K_1(rm)-\frac{1}{r^2}
\end{equation}
Its series expansion for $r\to 0^+$ is then
\begin{equation}
I\sim \frac{m^2}{2}\left( \gamma-\frac{1}{2}+\ln \frac{mr}{2} \right)+\frac{m^4r^2}{16}\left(\gamma-\frac{5}{4}+\ln\frac{mr}{2} \right)+...
\end{equation}
|
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|
Can I compute $\frac{dy}{dx}$ of $y=x\sqrt{3x+1}\sqrt{x+1}$ by taking $ln$ on both sides My teacher has suggested that to compute $\frac{dy}{dx}$ of $y=x\sqrt{3x+1}\sqrt{x+1}$, it's better to take $ln$ on both sides of the equation $y=x\sqrt{3x+1}\sqrt{x+1}$, and try taking derivative of logarithms on both sides, and then solve for $\frac{dy}{dx}$. But I am confused because when you consider the function $y=x\sqrt{3x+1}\sqrt{x+1}$, $y \lt 0$ for $-1/3 \lt x \lt 0$, so $ln(y)$ is not defined for $-1/3 \lt x \lt 0$, i.e. we cannot take logarithm for entire domain in which $y=x\sqrt{3x+1}\sqrt{x+1}$ is defined for. In this case, can we still compute $\frac{dy}{dx}$ by taking $ln$ on both sides, like my teacher is suggesting?
Thank you,
|
Multiply both sides of the equation by $-1$,
$$-y=-x\sqrt{3x+1}\sqrt{x+1}$$
Now that both sides are positive you can take the logarithm:
$$\ln(-y)=\ln(-x)+\frac{1}{2}\ln(3x+1)+\frac{1}{2}\ln(x+1)$$
Then take the derivative with respect to $x$ (note that $\ln(-x)'=\frac{-1}{-x}=\frac{1}{x}$)
$$\frac{y'}{y}=\frac{1}{x}+\frac{3}{2(3x+1)}+\frac{1}{2(x+1)}$$
$$y'=x\sqrt{3x+1}\sqrt{x+1}\left(\frac{1}{x}+\frac{3}{2(3x+1)}+\frac{1}{2(x+1)}\right)$$
|
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|
How to find values such that the curves $y=\frac{a}{x-1}$ and $y=x^2-2x+1$ intersect at right angles? Problem:
Find all values of $a$ such that the curves $y = \frac{a}{x-1}$ and $y = x^2-2x+1$ intersect at right angles.
My attempt:
First, I set the two curves equal to each other:
$ \frac{a}{x-1} = x^2 - 2x + 1 $
$ \frac{a}{x-1} = (x-1)^2 $
$ \frac{a}{x-1}(x-1) = (x-1)^2(x-1) $
$ a = (x-1)^3 $
$ \sqrt[3] a = x-1 $
$ \sqrt[3] a + 1 = x $
I found the derivative of the first curve:
$ y = \frac{a}{x-1} $
$ y = a(x-1)^{-1} $
$ y' = -a(x-1)^{-2} $
$ y' = \frac{-a}{(x-1)^{2}} $
Then I found the derivative of the second curve:
$ y = x^2 - 2x + 1 $
$ y' = 2x - 2 $
Next, I multiplied them together and set them equal to -1:
$ \frac{-a}{(x-1)^{2}} ⋅ 2x - 2 = -1 $
$ \frac{-2ax + 2a}{(x-1)^2} = -1 $
$ \frac{-2a(x-1)}{(x-1)^2} = -1 $
$ \frac{-2a}{x-1} = -1 $
$ \frac{-2a}{x-1} ⋅ (x-1) = -1(x-1) $
$ -2a = -x+1 $
$ a = \frac{-x+1}{-2} $
Now I am unsure how to finish the problem. Do I substitute what I found for $x$ into $ a = \frac{-x+1}{-2} $?
But my problem shows that there are two possible answers for $a$ so I am confused.
Any help would be appreciated!
Thank you in advance!
|
Then,
$$\displaystyle a=\frac{X-1}2
a=(x-1)^3((\text{eq}
\frac{x-1}2=\frac{(x-1)^3}2
=\frac{(x-2)^2}2
2x^2-4x+2=1$$
So,
$$X=\frac{2+\sqrt 2}2, \frac{2-\sqrt 2}2$$
Put $x$ in eq $1$:
$$a= 2^{-\frac32}= \sqrt \frac 42$$
|
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|
Complex quintic equation Given the equation $x^5=i$, I need to show by both algebraic and trigonometrical approaches that
$$\cos18^{\circ}=\frac{\sqrt{5+2\sqrt5}}{\sqrt[5]{176+80\sqrt5}}$$
$$\sin18^{\circ}=\dfrac1{\sqrt[5]{176+80\sqrt5}}$$
Trying by trigonometric approach,
$x^5$ = i $\;\;\;\;$ -- eqn. (a)
=> x = $i\sin(\dfrac{\pi}{2} +2k\pi)$ => $i\sin(\pi\dfrac{4k + 1}{2}) $
Taking the value of k=0, for getting the principal root of 18$^{\circ}$, have x = $i\sin(\dfrac{\pi}{10}) $
Solving algebraically, the solution approach is : $(a+bi)^5$ = i $\;\;\;\;$ -- eqn. (b)
=> $a^5 + 5ia^4b -10a^3b^2 -10ia^2b^3 +5ab^4 +ib^5$
Separating the real & imaginary parts:
$a^5 -10a^3b^2 +5ab^4=0$$\;\;\;$ -- eqn. (c); $\;\;\;\;$$5a^4b -10ia^2b^3+b^5=1$$\;\;\;$ -- eqn. (d)
Solving (c), we have : $a(a^4 -10a^2b^2 +5b^4)=0$$\;\;\;$ -- eqn. (c);
Either $a$ = $0$, or $(a^4 -10a^2b^2 +5b^4)=0$$\;\;$ -- eqn. (c'),
dividing both sides by $b^4$, and having c = a/b, $(c^4 -10c^2 +5)=0$$\;\;$ -- eqn. (c''),
having d = $c^2$, get : $(d^2 -10d +5)=0$$\;\;$ -- eqn. (c'''), with factors as : d =$5\pm 2\sqrt5$
finding value of c for the two values, get square roots of the two values for d.
//Unable to proceed any further with (c''').
Only root of significance, from eqn. (c) is $a = 0$.
Taking eqn.(d), and substituting $a = 0$, we get:$\;\;\;b^5$=1 => $b =1$
//Unable to prove any of the two values for $\sin18^{\circ}$, or $\cos18^{\circ}$
|
This problem seems a bit wonky to me because there are simpler expressions for these functions and easier ways to prove them, but let's roll with it.
Once you render $c^2=5\pm2\sqrt5$, take its square roots to get four roots
$c=a/b=\pm\sqrt{5\pm2\sqrt5}$
with the $\pm$ signs independent. Now from DeMoivre's Theorem we know that the complex root we are seeking is $\cos18°+i\sin18°$, so $a/b$ must be the cotangent of $18°$, therefore greater than $1$. Only by choosing both $\pm$ signs as positive can we get that result; $\sqrt{5-2\sqrt5}$ is too small with $2\sqrt5=\sqrt{20}$ lying between $4$ and $5$.
So, $a=(\sqrt{5+2\sqrt5})b$. Plug that into the imaginary part of $(a+bi)^5$ you computed above and equate this to $1$, which gives $1/b^5=176+80\sqrt5$. This will lead easily to the final claims.
|
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|
Seeking the maximal parameter value s.t. two-variable inequality still holds Consider the expression
$$\frac 12\left(\frac{a^2}b+\frac{b^2}a\right)$$
in two variables $\,a,b\,$ residing in $\,\mathbb R^{>0}$.
The arithmetic mean $\,\frac{a+b}2\,$ is a lower bound for it
$\big[$ one has $\,a^2b\le(2a^3+b^3)/3\,$ by AM-GM, do the same with $\,ab^2$, then add and divide by $2ab\,$$\big]$,
but $\,\max\{a,b\}\,$ is not
$\big[$ choose $\,(a,b)=(3,2)\,$ for instance, the expression then evaluates to $\,2\tfrac{11}{12}\,\big]$.
Both arithmetic mean ($x=1$) and the maximum ($x=\infty$) are instances of the Hölder mean
$$\left(\frac{a^x+b^x}2\right)^{\frac 1x}\quad\text{with }\; x\,\in\,\{-\infty\}\cup\mathbb R\cup\{\infty\}$$
aka Power mean, known to be strictly increasing with $\,x\,$ if $\,a\ne b$, and my question is:
What is the maximal value of $\,x\,$ such that
$$\left(\frac{a^x+b^x}2\right)^{\frac 1x}\;\leq\;\frac 12\left(\frac{a^2}b+\frac{b^2}a\right)$$
holds for all $\,a,b>0\,$?
Let's pick up the specific "max counter-example" $\,(a,b)=(3,2)\,$.
The following plot screen-shot displays the zero, by courtesy of WolframAlpha:
Returning to the general case we can at least state that $\,x_{max}\geqslant 5\,$:
After taking the fifth power of the corresponding expression and clearing denominators (to arrive at the LHS as of below) I could find a
Certificate of positivity
$$\begin{eqnarray}
\left(a^3 + b^3\right)^5 -16a^5b^5\left(a^5 +b^5\right)\; & =\;\left(a+b\right)\left(a-b\right)^2\big[a^{12} + a^{11}b +2a^{10}b^2 +4a^9b^3 +8a^8b^4 \\
& +8a^6b^6 +8a^4b^8 +4a^3b^9 + 2a^2b^{10} +ab^{11} +b^{12}\big] \\
& + 3a^3b^3\left(a^2+b^2\right)\left(a+b\right)^3\left(a-b\right)^4
\end{eqnarray}$$
It shows also that the equality case holds iff $\,a=b$.
|
Let $a=b-1$, so $\frac{1}{2}(\frac{a^2}{b}+\frac{b^2}{a}) =b+\frac{1}{2 (b-1)}+\frac{1}{2 b}-\frac{1}{2} < b-\frac{1}{3} $ true for $b>6.54138$.
Now $ \lim \limits_{x \to \infty} (\frac{a^x+b^x}{2})^{\frac{1}{x}} =(\frac{(b-1)^x+b^x}{2})^{\frac{1}{x}} = ((b-1)^x +b^x)^{\frac{1}{x}} *0.5^{1/x} = (b^x(1+(1-1/b)^x))^{\frac{1}{x}}*0.5^{\frac{1}{x}} = (b^x)^{\frac{1}{x}} (1+(1-1/b)^x)^{\frac{1}{x}} *0.5^x = b *(1+(1-\frac{1}{b})^x)^{\frac{1}{x}} *0.5^\frac{1}{x} = b$. assuming $b>2$.
So from $x_0$ every $x \geq x_0$ will have that $(\frac{(b-1)^x+b^x}{2})^{\frac{1}{x}} > b-\frac{1}{3}$.
And its not just $a=b-1$ , any $a$ you choose between $0.9b <a <b$, will have such $x_0$ that $x\geq x_0$ your inequality does not hold.
For instance $b=500$ so $\frac{1}{2}(\frac{a^2}{b}+\frac{b^2}{a}) = 499.502002 <499 \frac{2}{3} = 500-\frac{1}{3}$
Yet for all $x \geq 723 $ the term $(\frac{(b-1)^x+b^x}{2})^{\frac{1}{x}} \geq 499 \frac{2}{3}$.
The inequality is false when $a <b$ and $a$ "near" $b$ , and $x$ is sufficiently large.
|
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|
How to prove closed form for a binomial sum? Let $l\ge 0$ be an integer.
In the process of solving A tough series related with a hypergeometric function with quarter integer parameters we have discovered the following identity:
\begin{eqnarray}
\frac{1}{4} \sum\limits_{j=0}^{l-1} \sum\limits_{\begin{array}{r} p=-l-1\\p\neq -1 \end{array}}^{l-1}
\binom{-3/2}{j} \binom{-1/2}{l-1-j} \frac{\binom{2j+1}{p+j+1} - \binom{2j+1}{p+j+2}}{2^{2 j+1}}\cdot \frac{(-1)^{p+l}}{p+1}\cdot \left( 1+(-1)^p\right)=\frac{(1/2)^{(l)}}{l!}-1
\end{eqnarray}
Is it possible to prove that identity in some other way, i.e. without resorting to the calculations in the aforementioned question?
|
This question is actually easier than I thought and it can be solved using elementary methods. First of all note that :
\begin{equation}
\binom{2j+1}{p+j+1}- \binom{2j+1}{p+j+2}= \binom{2j+1}{p+j+1} \left(1-\frac{j-p}{p+j+2}\right)= \binom{2j+1}{p+j+1}\cdot \frac{2p+2}{p+j+2}
\end{equation}
Therefore the sum over $p$ gives:
\begin{eqnarray}
\sum\limits_{p=-l-1,p\neq-1}^{l-1} \frac{\binom{2j+1}{p+j+1}-\binom{2j+1}{p+j+2}}{2^{2j+1}}\cdot \frac{(-1)^{p+l}}{p+1} \cdot (1+(-1)^p)=\\
\sum\limits_{p=-l-1,p\neq-1}^{l-1} \frac{\binom{2j+1}{p+j+1}}{2^{2j+1}}\cdot \frac{2p+2}{p+j+2}\cdot \frac{(-1)^{p+l}}{p+1} \cdot (1+(-1)^p)=\\
\frac{1}{2^{2 j}}\cdot \frac{(-1)^l}{2j+2}\sum\limits_{p=-l-1}^{l-1} \binom{2j+2}{p+j+2} \cdot (1+(-1)^p)=\\
\frac{1}{2^{2 j}}\cdot \frac{(-1)^l}{2j+2}\sum\limits_{p=0}^{j+l+1}
\binom{2j+2}{p} \cdot (1+(-1)^{p+j})= \frac{(-1)^l}{j+1} \cdot 2
\end{eqnarray}
Therefore the sum in question reads:
\begin{eqnarray}
lhs&=& \frac{1}{4} \sum\limits_{j=0}^{l-1} \binom{-3/2}{j} \binom{-1/2}{l-1-j} \cdot \frac{(-1)^l}{j+1} \cdot 2\\
&=&-\sum\limits_{j=1}^l \binom{-1/2}{j} \binom{-1/2}{l-j} \cdot (-1)^l\\
&=&-1+\sqrt{\pi}(-1)^l \frac{1}{l! (-1/2+l)!}\\
&=&-1+\frac{(l-1/2)!}{l!(-1/2)!}\\
&=&-1+\frac{(1/2)^{(l)}}{l!}
\end{eqnarray}
|
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|
Prove that $a^2 \equiv b^2 $ mod $p$ implies that $a \equiv \pm b$ mod $p$ Prove that $a^2 \equiv b^2 $ (mod $p$) implies that $a \equiv \pm b$ (mod $p$). Where $p$ is a prime number.
So I know that $a^2 \equiv b^2 $ (mod $p$) implies $p|(a^2 -b^2)$ which implies that $a^2 -b^2 = mp$ for some $m \in \mathbb{Z}$. So $a^2 =b^2 +mp$.
But I am not sure where to go from here.
|
Just to be different (but essentially exactly the same):
Obviously if $a\equiv \pm b \mod p$ then $a^2 \equiv b^2 \mod p$. ($a = kp \pm b\implies a^2 = (k^2p \pm 2b)*p + b^2$.)
There are $p$ possible integers, $m$ so that $0\le m < p$ and for all integers $a$, $a \equiv m \mod p$ for one of these.
Let $0 \le n < m \le \frac {p-1}2$ and let $d = m-n>0$. Then $m^2 = (n+d)^2 = n^2 + 2nd + d^2$ so $n^2 \equiv m^2 \mod p \implies p|2nd+d^2=d(2n+d)=d(m+d)$.
But $d< p$ and $m+d<p$ so the prime factors of $d(m+d)$ are less than $p$ and $p|d(m+d)$ is impossible.
Now for any two $a,b$ there are for possibilities:
Either $a \equiv m \mod p$ for some $0\le m \le \frac{p-1}2$ or $a \equiv m \mod p$ for some $\frac{p-1}2 < m < p$. If the latter then $-m \equiv p-m \mod p$ and $0 < p-m < \frac {p-1}2$. So $\pm a\equiv m \mod p$ for some $m;0\le m \le \frac{p-1}2$.
Likewise $\pm b \equiv n \mod p$ for some $n;0\le n \le \frac{p-1}2$.
The if $a^2 \equiv b^2 \mod p$ then $m = n$ and thus $\pm a \equiv \pm b \mod p$ and $a \equiv \pm b \mod p$.
.....
Just as the other answers assuming if $p|(a+b)(a-b)$ then $p|a+b$ or $p|a-b$ (Euclid's lemma), this answer assumes if $p|N$ then $N$ has a unique prime factorization and $p$ is one of the factors, and further if $N$ has factors, that $p$ is a factor of one of the factors (Unique prime factorization theorem).
These are essentially the same assumption.
|
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|
Proof that $\binom{n + 3}{4} = n + 3 \binom{n + 2}{4} - 3 \binom{n + 1}{4} + \binom{n}{4}$. I was trying to count the number of equilateral triangles with vertices in an regular triangular array of points with n rows. After putting the first few rows into OEIS, I saw that this was described by A000332: $\binom{n}{4} = n(n-1)(n-2)(n-3)/24$.
A000332 has the comment:
Also the number of equilateral triangles with vertices in an equilateral triangular array of points with n rows (offset 1), with any orientation.
The linked solution is insightful, but the proof is very mechanical.
I tried to write an inductive proof, but I'm unable to come up with a nice way to prove the final identity (in particular for $n \geq 4$):
$$\binom{n + 3}{4} = n + 3 \binom{n + 2}{4} - 3 \binom{n + 1}{4} + \binom{n}{4}$$.
Wolfram|Alpha admits that these are indeed equal, and this could certainly be shown by writing everything as a polynomial.
However, I was hoping to find some nice combinatorial identities that give some intuition as to why this equality holds.
|
$$\begin{align}
\binom {n+3}4-3\binom {n+2}4+3\binom {n+1}4-\binom n4
&=\sum_{r=0}^3(-1)^{r+1}\binom 3r\binom {n+r}4\\
&=\sum_{r=0}^3(-1)^{r+1}\binom 3r\binom {n+r}{n+r-4}\\
&=\sum_{r=0}^3(-1)^{r+1}\binom 3r (-1)^{n+r-4}\binom {-5}{n+r-4}\tag{*}\\
&=(-1)^{n-3}\sum_{r=0}^3\binom 3{3-r}\binom {-5}{n+r-4}\\
&=(-1)^{n-1}\binom {-2}{n-1}\tag{**}\\
&=(-1)^{n-1}\cdot (-1)^{n-1}\binom n{n-1}\tag{*}\\
&=\binom n1\\
&=n\\\\
\binom {n+3}4&=n+3\binom {n+2}4-3\binom {n+1}4+\binom n4\;\color{red}\blacksquare
\end{align}$$
*using upper negation
**using the Vandermonde Identity
|
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|
If we know range of a function , how can construct range of other function?
If we know $\frac{-1}{2}\leq \frac{x}{x^2+1}\leq \frac{1}{2}$ how can obtain the range of the function below?
$$y=\frac{4 x}{9x^2+25}$$
The problem is what's the range with respect to $\frac{-1}{2}\leq \frac{x}{x^2+1}\leq \frac{1}{2}$
|
We have
\begin{eqnarray*}
- \frac{1}{2} \leq \frac{x}{x^2+1} \leq \frac{1}{2}
\end{eqnarray*}
Let $x= \frac{3z}{5}$
\begin{eqnarray*}
- \frac{1}{2} \leq \frac{\frac{3z}{5}}{(\frac{3z}{5})^2+1} \leq \frac{1}{2} \\
- \frac{1}{2} \leq \frac{15z}{9z^2+25} \leq \frac{1}{2}
\end{eqnarray*}
Now multiply by $ \frac{4}{15}$
\begin{eqnarray*}
- \frac{2}{15} \leq \frac{4z}{9z^2+25} \leq \frac{2}{15}.
\end{eqnarray*}
|
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|
Inequality involving sum and modulus of complex numbers Let $z_1,z_2,...,z_n$ be complex numbers, prove the inequality:
$$\frac{\vert\sum_{k=1}^n z_k\vert}{1+\vert\sum_{k=1}^n z_k\vert} \le \sum_{k=1}^n \frac{\vert z_k \vert}{1+\vert z_k \vert}$$
|
For the base case $n=2$, let $a= \mid z_1 \mid $ and $b=\mid z_2 \mid $
\begin{eqnarray*}
&ab(a+b+2) & \geq & 0 \\
& ab(a+b)+b(2a+b) +b & \\ &+ab(a+b)+a(a+2b) +a & \geq & ab(a+b)+a(a+b)+b(a+b) +a+b \\
&a(1+b)(1+a+b)+b(1+a)(1+a+b) & \geq & (a+b)(1+a)(1+b) \\
& \frac{a}{1+a}+\frac{b}{1+b} \geq \frac{a+b}{1+a+b}
\end{eqnarray*}
The inductive step is established by ...
\begin{eqnarray*}
\sum_{k=1}^{n+1} \frac{\vert z_k \vert}{1+\vert z_k \vert} &=&\sum_{k=1}^{n} \frac{\vert z_k \vert}{1+\vert z_k \vert} +\frac{\vert z_{n+1} \vert}{1+\vert z_{n+1} \vert} \\
&\geq & \frac{\vert\sum_{k=1}^n z_k\vert}{1+\vert\sum_{k=1}^n z_k\vert}+\frac{\vert z_{n+1} \vert}{1+\vert z_{n+1} \vert} \\
&\geq & \frac{\vert\sum_{k=1}^n z_k\vert+\vert\ z_{n+1} \vert}{1+\vert\sum_{k=1}^n z_k\vert +\vert\ z_{n+1} \vert} \\
\end{eqnarray*}
|
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|
Then find the value of ... Let $a, \: b, \: c$ be three variables which can take any value (Real or Complex). Given that $ab + bc + ca = \frac{1}{2}$; $a + b + c = 2$; $abc = 4$. Then find the value of $$\frac{1}{ab + c - 1} + \frac{1}{bc + a - 1} + \frac{1}{ac + b - 1}$$
I always get stuck in such problems. Please give a detailed solution and also give me an approach that I can use to solve other such kind of problems.
|
$$\sum_{cyc}\frac{1}{ab+c-1}=\frac{\sum\limits_{cyc}(ab+c-1)(ac+b-1)}{\prod\limits_{cyc}(ab+c-1)}=$$
$$=\frac{\sum\limits_{cyc}(a^2bc+a^2b+a^2c-ab-2a+1)}{a^2b^2c^2+abc-1+\sum\limits_{cyc}(a^3bc+a^2b^2-a^2bc-a^2b-a^2c+a)}=$$
$$=\frac{2\cdot4+2\cdot\frac{1}{2}-3\cdot4-\frac{1}{2}-2\cdot2+3}{4^2+4-1+4\cdot\left(2^2-2\cdot\frac{1}{2}\right)+\left(\frac{1}{2}\right)^2-2\cdot4\cdot2-2\cdot\frac{1}{2}+3\cdot4+2}=-\frac{18}{113}.$$
|
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|
If $f(1)=1$, then is it true that $f(n)=n$ for all $n \in \mathbb{N}\cup\{0\}$. Let $f:\mathbb{N}\cup\{0\}\to\mathbb{N}\cup\{0\}$ be a function which satisfies $f(x^2+y^2)=f(x)^2+f(y)^2$ for all $x,y \in\mathbb{N}\cup\{0\}$. It' easy to see that $f(0)=0$ and $f(1)=0$ or $f(1)=1$. Suppose let's assume that $f(1)=1$. Then it's very easy to see that $f(2)=2$ and since $f(2)=2$, we can see that $f(4)=4$ and $f(5)=5$ because $5=2^2+1^2$. To see $f(3)=3$, note that $$25=f(5^2)=f(3)^2+f(4)^2 \implies f(3)=3$$ One can even see that $f(6)=6$, $f(7)=7$ and so on. Is it generally true that $f(n)=n$ for all $n$?
Edit. If anyone wondering how $f(7)=7$ here is the way. Note that $f$ satisfies $f(n^2)=f(n)^2$. One can see that $$25^{2}=24^2+7^2 \implies 25^{2}=f(24)^{2}+f(7)^{2}$$ We have to show $f(24)=24$. For this note that $$26^2 = 24^2+10^2 \implies f(26)^{2} = f(24)^{2}+f(10)^{2}$$ But $f(26)=26$ because $26=5^2+1^2$ and $f(10)=10$ since $10=3^{2}+1^{2}$. In short if $n=x^2+y^2$ for some $x,y$ and $f(1)=1$, then we can see that $f(n)=n$.
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Let $N$ be the smallest number with $f(N)\neq N$.
Notice that $f(N^2+a^2)=f(N)^2 + a^2 \neq N^2 + a^2$ for all $a<N$.
On the other hand if we can write $N^2+a^2$ as $b^2 + c^2$ with $b,c<N$ then $f(b^2+c^2)=f(b)^2+f(c)^2=b^2+c^2$.
We now prove that for large $N$ we can always find $0\leq a,b,c < N$ such that $N^2+a^2=b^2+c^2$ or $N^2-c^2=(a+b)(a-b)$.
Letting $c=(N-k)$ we get $k(2N-k)=(a+b)(a-b)$.
Checking with $k=1,2,3$ we'll find a value such that $2N-k$ that is a multiple of $3$, this will give rise to a factorization $(\frac{2N-K}{3})(3k)$ which can be realized in the way $(a+b)(a-b)$ with $0\leq a,b<N$.
The first value that makes $N$ large enough is the first value such that $3\times 3 \leq (2N-3)$ which is $6$.
So we just need to prove $f(n)=n$ holds for $n\in \{0,1,2,\dots,6\}$, we prove them in this order:
$0$ (clearly)
$1$ (clearly)
$2$ (sum of two squares)
$4$ (a square)
$5$ (sum of $1^2+2^2$)
$3$ ($3^2+4^2=5^2$)
$8$ ($2^2+2^2$)
$10$ ($1^2+3^2$)
$6$ ($6^2+8^2=10^2$)
|
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|
Limit of $\sqrt{x^2+3x}+x$ when $x\to-\infty$ Limit of $ \lim_{x\to -\infty}(\sqrt{x^2+3x}+x)$, I know that the final answer is $-3/2$, my question is about Wolfram Alpha step by step solution:
$$x+\sqrt{x^2+3x}=\frac{(x+\sqrt{x^2+3x})(x-\sqrt{x^2+3x})}{x-\sqrt{x^2+3x}}$$
$$=-\frac{3x}{x-\sqrt{x^2+3x}}$$
$$\lim_{x\to-\infty}-\frac{3x}{x-\sqrt{x^2+3x}}$$
$$\lim_{x\to-\infty}-\frac{3x}{x-\sqrt{x^2+3x}}=-3$$
$$\lim_{x\to-\infty}\frac{x}{x-\sqrt{x^2+3x}}=-3\lim_{x\to-\infty}\frac{x}{x-\sqrt{x^2+3x}}$$
$$\frac{x}{x-\sqrt{x^2+3x}}=\frac{1}{1-\frac{\sqrt{x^2+3x}}{x}}$$
$$-3\lim_{x\to-\infty}\frac{x}{x\left(1-\frac{\sqrt{x^2+3x}}{x}\right)}$$
To prepare the product $\frac{1}{1-\frac{\sqrt{x^2+3x}}{x}}$ for solution by l'Hopital's rule, write it as $\frac{x}{x\left(1-\frac{\sqrt{x^2+3x}}{x}\right)}$
$$-3\lim_{x\to-\infty}\frac{x}{x\left(1-\frac{\sqrt{x^2+3x}}{x}\right)}$$
Is it correct to use L'Hopital here like Wolfram did?
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You are right to render
$\sqrt{x^2+3x}+x=\frac{3x}{\sqrt{x^2+3x}-x}$
The next step is to complete the square on $x^2+3x$ getting
$x^2+3x=(x+(3/2))^2-(9/4)$
Use this to show that for negative $x$ with $x<-3$ (why?):
$-x-(3/2)<\sqrt{x^2+3x}<-x$
and then
$-\frac{3x}{2x+(3/2)}<\frac{3x}{\sqrt{x^2+3x}-x}<-(3/2)$
Now get the limit from the squeeze theorem.
|
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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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Hint: Let $y=e^{\frac25t}$ and use $\sinh^2t+1=\cosh^2t$.
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|
A difficult integral $I=\int_0^1\sqrt{1+\sqrt{1-x^2}}\frac{dx}{1+x^2}$ How to prove
$$I=\int_0^1\sqrt{1+\sqrt{1-x^2}}\frac{dx}{1+x^2}=\sqrt{\sqrt{2}+1}\arctan\sqrt{\sqrt{2}+1}-\frac{1}{2}\sqrt{\sqrt{2}-1}\ln(1+\sqrt{2}+\sqrt{2+2\sqrt{2}})$$
$$ I=\int_0^{\pi/4}\sqrt{1+\sqrt{1-\tan^2y}}dy=\int_0^{\pi/4}\sqrt{{cosy}+\sqrt{\cos2y}}\frac{dy}{\sqrt{cosy}} $$
put $$x=tany$$
But how to calculate this integral?
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$\displaystyle I=\int_0^1\sqrt{1+\sqrt{1-x^2}}\frac{dx}{1+x^2}\,dx$
Perform the change of variable $y=\sqrt{1-x^2}$,
$\begin{align}I&=\int_0^1 \frac{x}{(2-x^2)\sqrt{1-x}}\,dx\tag{1}\\
&=\Big[\sqrt{\sqrt{2}-1} \cdot\text{arctanh}\left(\sqrt{\sqrt{2}-1}\sqrt{1-x}-\sqrt{\sqrt{2}+1}\cdot\arctan\left(\sqrt{\sqrt{2}+1}\sqrt{1-x}\right)\right)\Big]_0^1\\
&=\boxed{\sqrt{\sqrt{2}+1}\cdot\arctan\left(\sqrt{\sqrt{2}+1}\right)-\sqrt{\sqrt{2}-1}\cdot\text{arctanh}\left(\sqrt{\sqrt{2}-1}\right)}
\end{align}$
Addendum:
if you want to transform the integral into an integral whose integrand is a rational fraction,
perform the change of variable $y=\sqrt{1-x}$ in $(1)$
$\begin{align}I=\int_0^1 \dfrac{2(x^2-1)}{x^4-2x^2-1}\,dx\end{align}$
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|
Proof that $(abc)^2 = \frac{(c^6-a^6-b^6)}{3}$ We are given the length of the sides of a right triangle T, where $a \leq b \leq c$. We are asked to prove if $T$ is a right triangle. then $(abc)^2 = (c^6-a^6-b^6)/3$.
What I tried:
I tried substituting with the Pythagorean theorem $a^2 + b^2 = c^2$.
Where I am stuck:
I am unable to get $\frac{1}{3}$ or $3$ anywhere in my simplification. Is there some kind of identity that I am missing?
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with $$c^2=a^2+b^2$$ the left-hand side is given by
$$(ab\sqrt{a^2+b^2})^2$$ and the other side
$$\frac{(\sqrt{a^2+b^2})^6-a^6-b^6}{3}$$
can you compute this?
expanding the right-hand side we get $$a^4b^2+a^2b^4$$ and so is the left-hand side.
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|
Finding integral solutions of $x+y=x^2-xy+y^2$
Find integral solutions of $$x+y=x^2-xy+y^2$$
I simplified the equation down to
$$(x+y)^2 = x^3 + y^3$$
And hence found out solutions $(0,1), (1,0), (1,2), (2,1), (2,2)$ but I dont think my approach is correct . Is further simplification required? Is there any other method to solve this? I am thankful to those who answer!
|
$x+y=x^2-xy+y^2
$.
Since this is
symmetrical in
$x$ and $y$,
we can assume that
$x \le y$.
If $x=y$,
this becomes
$2x = x^2$,
so
$x=0$ or $x=2$.
If $x < y$,
then
$2y
\gt x+y
=(x-y/2)^2+3y^2/4
\ge 3y^2/4
$.
There is no solution if
$y \le 0$.
If $ > 0$,
$2 \gt 3y/4
$
or
$y \lt 8/3$
so
$y \le 2$.
If $y=2$,
then
$x=0$ or $x=1$;
of there,
only $x=1$ works.
If $y=1$,
then $x=0$ works.
|
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|
Finding equation of circle under the given geometric conditions
Finding the equation of the circle which touches the pair of lines
$7x^2 - 18xy +7 y^2 = 0$ and the circle $x^2 + y^2 - 8x -8y = 0$ and
contained in the given circle??
My attempt
The centre of required circle would lie on angle bisector of the pair of lines ie $x=y$.
Assuming circle to be $(x-h)^2+(y-h)^2=r^2$
Now $2(h-8)^2=r^2$ ( distance between the extreme of larger circle and center of contained circle,)
I am unable to frame a second equation . One way would be to calculate the angle between pair of straight lines and use it to find a relation between $r$ and $h$.
However I was looking for a better solution or suggestion ?
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The angle bisector of the two lines $y=\dfrac{1}{7} \left(9 \pm4 \sqrt{2} \right)x$ is the line $y=x$
The wanted circle has then centre $H(k,k)$ and its radius is the distance from the given lines $\left(9+4 \sqrt{2}\right) x-7 y=0$
$$r_k=\frac{\left|\left(9+4 \sqrt{2}\right) k-7 k\right|}{\sqrt{\left(9+4 \sqrt{2}\right)^2+49}}=\frac{\left|2 \left(1+2 \sqrt{2}\right) k\right|}{12+3 \sqrt{2}}$$
The wanted circle must also be tangent to the given circle $x^2-8 x+y^2-8 y=0$ having centre $C(4,4)$ and radius $R=4\sqrt{2}$
A circle is internally tangent to another when the distance of the centres is equal to the difference of the radii (in absolute value)
Therefore we must have $CH=R-r_k$
that is $$\sqrt{(4-k)^2+(4-k)^2}=4\sqrt{2}-\frac{\left|2 \left(1+2 \sqrt{2}\right) k\right|}{12+3 \sqrt{2}}$$
which simplified, noticing that $k>4$ becomes
$$\sqrt{2}(k-4)=\frac{2 \left(1+2 \sqrt{2}\right) (12-k)}{3 \left(4+\sqrt{2}\right)}$$
After a look to the conditions we can say that $k>4$ so the previous equation becomes
$$\left(12+3 \sqrt{2}\right) \sqrt{2} (k-4)=2 \left(1+2 \sqrt{2}\right) k$$
Solution is $k=6$ and the wanted circle has equation
$(x-6)^2+(y-6)^2=8\to \color{red}{x^2+y^2-12x-12y+64=0}$
|
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|
Find the perfect numbers of the product of two primes, $2^p-1$ and $2^{p-1}$ A number $n\in N$ Show that if $p$ is a prime, such that $2^p - 1$ is also a prime, a Mersenne prime that is, then $n = 2^{p-1}(2^p-1)$ is a perfect number.
So I know that $n$ must be divisible by, $2^p-1$, $2^{p-1}$ and $1$ (the proper divisors of $n$), which sum up to be $2^p-1 + 2^{p-1} + 1 = 2^p + 2^{p-1}$. I'm stumped on where to go from here, since the sum isn't equal to $n$.
|
Assuming $2^p-1$ is prime number then $2^{p-1} (2^p -1)$ is perfect number if
$\sigma(2^{p-1} (2^p -1)) = \sigma(2^{p-1}) \sigma(2^p-1) = 2^p \sigma(2^{p-1}) = 2^p (2^{p}-1) = 2 * 2^{p-1} (2^{p}-1) $ twice the number we start with so its perfect number.
Done.
|
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|
Prove that there exists a $c$ such that $f(c-1/8)=f(c+1/8)$ Suppose that the function $f$ is continuous on $[0,1]$ and $f(0)=f(1)$, prove that there exists a number $c$ such that $$f(c-\frac{1}{8})=f(c+\frac{1}{8})$$
I am not sure how to prove its existence, but what I did was to find the possible range that $c$ is inside:
$0\le c-\frac{1}{8} \le1$ and $0\le c+\frac{1}{8} \le1$, and solving both inequalities gives: $$\frac{1}{8}\le c\le\frac{7}{8}$$
So far I have learnt the Extreme Value Theorem, Rolle's Theorem, Mean Value Theorem, Fermat's Theorem.
Appreciate it if you can help me on what should I do!
|
Let $g:[\frac{1}{8},\frac{7}{8}]\to \mathbb{R}$ be defined by
$\displaystyle{g(x) = f(x+{\small{\frac{1}{8}}}) - f(x-{\small{\frac{1}{8}}})}$.
The goal is to show that $g(c) = 0$, for some $c$.
If $g(\frac{k}{8})=0$, for some $k \in \{1,3,5,7\}$, then let $c=\frac{k}{8}$, and we're done.
Suppose then that $g(\frac{1}{8}),g(\frac{3}{8}),g(\frac{5}{8}),g(\frac{7}{8})$ are all nonzero.
\begin{align*}
\text{Then}\;&
g({\small{\frac{1}{8}}})+
g({\small{\frac{3}{8}}})+
g({\small{\frac{5}{8}}})+
g({\small{\frac{7}{8}}})
\\[4pt]
=\;\,&
\left(f({\small{\frac{2}{8}}}) - f(0)\right) +
\left(f({\small{\frac{4}{8}}}) - f({\small{\frac{2}{8}}})\right) +
\left(f({\small{\frac{6}{8}}}) - f({\small{\frac{4}{8}}})\right) +
\left(f(1) - f({\small{\frac{6}{8}}})\right)
\\[4pt]
=\;\,&f(1) - f(0)\\[4pt]
=\;\,&0\\[4pt]
\end{align*}
Since $g(\frac{1}{8}),g(\frac{3}{8}),g(\frac{5}{8}),g(\frac{7}{8})$ sum to zero, and since they're all nonzero, it follows that at least one of them is positive, and least one is negative.
Since $f$ is continuous on $[\frac{1}{8},\frac{7}{8}]$, so is $g$.
Hence, by the Intermediate Value Theorem, we must have $g(c) = 0$, for some $c$.
This completes the proof.
|
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|
Proving $\frac{x+1}{y+1}$ is equal to or less than $\frac{x-1}{y-1}$ I know that $\frac{x+1}{y+1}$ is equal to (when $x = y$) or less than $\frac{x-1}{y-1}$. Suppose we cross multiply, then
*
*$xy - x+ y -1$ is less than or equal to $xy + x - y -1$
So,
*$y- x$ is less than or equal to $x- y$.
But when $y$ is greater than $x$ then the inequality is wrong.
Where am I wrong?
|
If $ 1 \leq y \leq x $ then $ y-x \leq x-y $. Now add $xy-1$ to both sides and factorise to get
\begin{eqnarray*}
\underbrace{xy+y-x-1}_{(x+1)(y-1)} \leq \underbrace{xy+x-y-1}_{ (x-1)(y+1)} \\
\frac{x+1}{y+1} \leq \frac{x-1}{y-1}.
\end{eqnarray*}
|
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|
Find all natural number(s) $n$ such that $55\mid n^2 + 3n + 1$
Find all natural number(s) $n$ such that $55\mid n^2 + 3n + 1$
I see that $n^2+3n+1 =n^2-2n+1+5n\equiv n^2-2n+1=(n-1)^2 \pmod{5}$
I also see that $n^2+3n+1=n^2+3n-10+11=(n-2)(n+5)+11\equiv(n-2)(n+5) \pmod{11}$
After that what?
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from what you deduced we have $n\equiv 1 \bmod 5$ and $n\equiv 2$ or $n\equiv 6 \bmod 11$.
This tells us $n\equiv 46$ or $n\equiv 6 \bmod 55$.
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Solve Recurrence relation -:$T(n)=T(n-1)+6n^{2}+2n$
Solve Recurrence relation -:$T(n)=T(n-1)+6n^{2}+2n$
Base case $T(1)=8$
$T(n)=T(n-1)+6n^{2}+2n$
$T(n-1)=T(n-2)+6(n-1)^{2}+2(n-1)............(1)$
$T(n-2)=T(n-3)+6(n-2)^{2}+2(n-2)...........(2)$
Substituting $(1)\,\, \text{and} \,\,(2) \text{in our question},$
$T(n)=T(n-k)+6(n-k)^{2}+2(n-k)+......6(n-2)^{2}+2(n-2)+6(n-1)^{2}+2(n-1)+6n^{2}+2n$
$T(n)=T(n-k)+6((n-k)^{2}+....+(n-2)^{2}+(n-1)^{2}+n^{2})+2((n-k)+...(n-2)+(n-1)+2n)$
finding $k$ using base case.
$T(n-k)=8$
$\Rightarrow n-k=1$
$\Rightarrow k=n-1$
putting the value of $k$ in our question.
$T(n)=T(n-k)+6((n-n+1)^{2}+....+(n-2)^{2}+(n-1)^{2}+n^{2})+2((n-n+1)+...(n-2)+(n-1)+n)$
$T(n)=T(1)+6(1^{2}+2^{2}+3^{3}+....n^{2})+2*(1+2+3..+n)$
$$T(n)=8+6\frac{n*(n+1)(2n+1)}{6}+2* \frac{n*(n+1)}{2}$$
$$T(n)=n*(n+1)(2n+1+1)+8$$
$$T(n)=n*(2n+2)*(n+1)+8$$
But annsweris given as $$T(n)=n*(2n+2)*(n+1)$$
Am i wrong?
Please help
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Since the equation contains a quadratic in $n$, it is legitimate to suppose that
$$T_n=a+bn+c n^2+d n^3$$ Just replace in the equation and group terms to end with
$$(3 d-6) n^2+ (2 c-3 d-2)n+(b-c+d)=0$$ Since this must hold for all $n$, then
$$3d-6=0 \qquad 2c-3d-2=0 \qquad b-c+d=0$$ that is to say $b=2$, $c=4$, $d=2$.
After simplification, then $$T_n=a+2 n (n+1)^2$$ and $a$ will be fixed by any condition.
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If $a+b+c=0$ prove that $ (a^2+b^2+c^2) \times (a^2+b^2+c^2) = 2(a^4+b^4+c^4)$
If $a+b+c=0$, for $a,b,c \in\mathbb R$, prove
$$(a^2+b^2+c^2) \times (a^2+b^2+c^2) = 2(a^4+b^4+c^4)$$
What is a good way to do this?
This question came from answering this slightly harder question. Those answers were somewhat hard to understand for me. To get something easier to digest I made a very similar but easier (lower exponent) question.
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$$(a^2+b^2+c^2)^2-2(a^4+b^4+c^4)=$$
$$=\sum_{cyc}(2a^2b^2-a^4)=(a+b+c)(a+b-c)(a+c-b)(b+c-a)=0.$$
Because
$$\sum_{cyc}(2a^2b^2-a^4)=4a^2b^2-(a^4+b^4+c^4-2a^2c^2-2b^2c^2+2a^2b^2)=$$
$$=(2ab)^2-(a^2+b^2-c^2)^2=(2ab-a^2-b^2+c^2)(2ab+a^2+b^2-c^2)=$$
$$=(c^2-(a-b)^2)((a+b)^2-c^2)=(c-a+b)(c+a-b)(a+b-c)(a+b+c).$$
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Solve identity: $\frac {\sin x}{1-\cos x} = \frac {1+\cos x}{\sin x}$ $$\frac {\sin x}{1-\cos x} = \frac {1+\cos x}{\sin x}$$
The only way I can see of doing this is by cross multiplying but isn't that not allowed when trying to prove something?
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Set $t=\tan \frac x2$ so that $\sin x = \cfrac {2t}{1+t^2}$ and $\cos x = \cfrac {1-t^2}{1+t^2}$ (Weierstrass substitution) or $$(1+t^2)\sin x =2t; (1+t^2)\cos x=1-t^2$$
Since $1+t^2\ge 1 \gt 0$ we can multiply the original fractions by $1=\cfrac {1+t^2}{1+t^2}$ to obtain $$\frac {\sin x}{1-\cos x}=\frac {2t}{2t^2}=\frac 2{2t}=\frac {1+t^2+1-t^2}{2t}=\frac {1+\cos x}{\sin x}$$
The only issue is cancelling a factor of $\frac tt$ which is undefined when $t=0$, but then one of the original fractions reduces to $\frac 00$.
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Proving that: $ | a + b | + |a-b| \ge|a| + |b|$ I am trying to prove this for nearly an hour now:
$$
\tag{$\forall a,b \in \mathbb{R}$}| a + b | + |a-b| \ge|a| + |b|
$$
I'm lost, could you guys give me a tip from where to start, or maybe show a good resource for beginners in proofs ?
Thanks in advance.
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Very Simple Trick: We have that
\begin{split} (|a|-|b|)^2 +2|a^2-b^2| \ge 0&\Longleftrightarrow & a^2+b^2 -2|a||b|+ 2|a +b||a-b| \ge 0\\
&\Longleftrightarrow & a^2+b^2 + 2|a +b||a-b|\ge 2|a||b|\\
&\Longleftrightarrow& \color{red}{2a^2+2b^2} + 2|a +b||a-b|\ge \color{red}{a^2+b^2}+2|a||b|\\
&\Longleftrightarrow& (|a +b|+|a-b|)^2 \ge (|a|+|b|)^2\\
&\Longleftrightarrow& |a +b|+|a-b| \ge |a|+|b|\end{split}
Given that $$\color{red}{ (|a +b|+|a-b|)^2 = 2a^2+2b^2 + 2|a +b||a-b|}$$
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How much smaller is the set of ratios than the set of ordered pairs? For integers $a, b$ with $0 < a < N$ and $0 < b < N$ for some integer $N$, let $\{(a, b)\}$ be the set of all their ordered pairs and let $\{\frac{a}{b}\}$ be the set of all their ratios.
Clearly, $|\{\frac{a}{b}\}| < |\{(a, b)\}|$ (the number of possible ratios is smaller than the number of possible ordered pairs), since for example $\frac{2}{1} = \frac{4}{2}$ but $(2, 1) \neq (4, 2)$.
My question is: how much smaller is $|\{\frac{a}{b}\}|$ than $|\{(a, b)\}|$, as $N \to \infty$?
Here's my approach so far: for any given $b$, we can determine how many $a$s get "eliminated." For example, if $b=2$, then $a=2, 4, 6, \ldots$ are all eliminated (because $\frac{2}{2}=\frac{1}{1}, \frac{4}{2}=\frac{2}{1}, \frac{6}{2}=\frac{3}{1}, \ldots$). So when $b=2$, $|\{\frac{a}{b}\}| = (1 - \frac{1}{2}) |\{(a, b)\}|$. The first few such ratios $r$ are:
$b = 1 \implies r = 1$
$b = 2 \implies r = \frac{1}{2}$
$b = 3 \implies r = \frac{1}{3}$
$b = 4 \implies r = \frac{1}{2}$
$b = 5 \implies r = \frac{1}{5}$
$b = 6 \implies r = \frac{1}{2} + \frac{1}{3} - \left(\frac{1}{2}\right) \left(\frac{1}{3}\right) = \frac{2}{3}$
$b = 7 \implies r = \frac{1}{7}$
In other words, we find the prime factors $p_1, p_2, \ldots$ of $b$ and that reduces $|\{\frac{a}{b}\}|$ by $\frac{1}{p_1} + \frac{1}{p_2} + \ldots - \left(\frac{1}{p_1}\right) \left(\frac{1}{p_2}\right) - \ldots$
So, we want to know the average $r$ for all $b$ (as $b \to \infty$). Does this help us?
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I will use inclusive bounds, so $1 \leq a, b \leq n$.
$|\{\frac{a}{b}\}|$ is the number of coprime pairs $1 \leq a, b \leq n$. This is OEIS sequence A018805. For large $n$ it's noted that we have $\text{A018805(n)} \approx \dfrac{6}{\pi^2} n^2$.
$|\{a, b\}|$ is the number of pairs $1 \leq a, b \leq n$. This is simply $n^2$.
So their asymptotic density is the fraction of these two, giving simply $\dfrac{6}{\pi^2} \approx 0.60793$.
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What is the sum of $E=\frac{1}{3}+\frac{2}{3^{2}}+\frac{3}{3^{3}}+\frac{4}{3^{4}}+...$ What is the right way to assess this problem?
$$E=\frac{1}{3}+\frac{2}{3^{2}}+\frac{3}{3^{3}}+\frac{4}{3^{4}}+...$$
To find the value of the sum I tried to use the fact that it could be something convergent like a geometric series. However it does not seem to be the case as there is no common ratio between the terms. Since the numerator is increasing from $1,2,3,...$ makes it impossible to find a ratio. What should I do?.
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$$\sum _{k=1}^{\infty } \frac{k}{3^k}=\frac34$$
To prove this I'll use generating functions
Consider $$f(x)=\sum _{k=1}^{\infty } \frac{ x^k}{3^k}=\sum _{k=1}^{\infty } \left(\frac{ x}{3}\right)^k=\frac{1}{1-x/3}=\frac{3}{3-x}\quad(*)$$
$$f'(x)=\frac{1}{3}\sum _{k=1}^{\infty } k\left(\frac{ x}{3}\right)^{k-1}$$
To make the index back to $k$ multiply both sides by $x$ so that
$$xf'(x)=\frac{x}{3}\sum _{k=1}^{\infty } k\left(\frac{ x}{3}\right)^{k-1}=\sum _{k=1}^{\infty } k\left(\frac{ x}{3}\right)^{k}$$
The sum is then $xf'(x)$ for $x=1$
$$\sum _{k=1}^{\infty } k\,\frac{1}{3^k}$$
On the other side $(*)$ we have $f'(x)=\dfrac{3}{(3-x)^2}$
that is $xf'(x)=\dfrac{3 x}{(3-x)^2}$
The sum $\sum _{k=1}^{\infty } \frac{k}{3^k}$ is $xf'(x)$ for $x=1$, which gives $\dfrac{3}{4}$
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Proving inequality.. Question:
$$
f(a,b)= \begin{cases} ja & \text{if } b \le 2 \\ jb & \text{if } a \le 2 \\ jab + f(d, \frac{b}{2}) + f(a-d, \frac{b}{2}) & \text{if } a,b >2\end{cases}
$$ where $\ j$ is a positive constant and $ \ 1 \le d<a$.
Prove using some form of induction that $f(a,b) = \Theta(ab)$ or $ f(a,b) = \Theta(a^2b^2)$. Only one is correct.
My attempt:
I think that $ f(a,b) = \Theta(ab)$. We first have to show that $ f(a,b) = O(ab)$ and $ \ f(a,b) = \Omega(ab)$.
Prove $ \ f(a,b) = O(ab)$:
We must show there exist positive constants $c$ and $ n_0$ such that $ \ f(a,b) \le cab$ for $a,b \ge n_0$.
Base cases:
$a=1$: $f(a,b) = jb$ by definition. If we let $a=b$ then $jb \le cb^2$.
$b=1$: $f(a,b) = ja$ by definition. If we let $a=b$ then $ja \le ca^2$.
$a=2$: $f(a,b) = jb$ by definition. If we let $a=b$ then $jb \le cb^2$.
$b=2$: $f(a,b) = ja$ by definition. If we let $a=b$ then $ja \le ca^2$.
I don't understand how to do the induction step. What should I assume in my induction hypothesis and how should I complete the proof?
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We assume that the domain of $f$ is a set $\Bbb R^2_+$ of pairs $(a,b)$ of non-negative real numbers and the third row of the definition means that there exists $d$ such that the claim holds. Remark that $f(a,b)=O(ab)$ fails, for instance for the sequence $a_n=\frac 1n$ and $b_n=n^2$, because then $a_nb_n=n$, whereas $f(a_n,b_n)=jn^2$. So the problem conclusion should be corrected too.
We use induction by $n$ to show that $\left(2-\frac 3{2^{n}}\right)jab-3jb\le f(a,b)\le \left(2-\frac 3{2^{n}}\right)jab+ja+jb$, where $n$ is the smallest positive integer such that $b\le 2^n$. For $n=1$ this follows from the definition of $f$. Assume that we have already proved the inequality for $n\ge 1$ and assume that $2^{n}<b\le 2^{n+1}$. Then $n$ is the smallest positive number such that $\frac b2\le 2^n$. If $a\le 2$ then $f(a,b)=jb$ and it is easy to check that the inequality holds. If $a>2$ then by the inductive assumption
$$f(a,b)=jab + f\left(d, \frac{b}{2}\right) + f\left(a-d, \frac{b}{2}\right)\ge $$ $$jab + \left(2-\frac 3{2^{n}}\right)jd\frac{b}{2} + \left(2-\frac 3{2^{n}}\right)j(a-d)\frac{b}{2}-3j\frac{b}{2}-3j\frac{b}{2} =$$ $$\left(2-\frac 3{2^{n+1}}\right)jab-3jb.$$
$$f(a,b)=jab + f\left(d, \frac{b}{2}\right) + f\left(a-d, \frac{b}{2}\right)\le $$ $$jab + \left(2-\frac 3{2^{n}}\right)jd\frac{b}{2} + \left(2-\frac 3{2^{n}}\right)j(a-d)\frac{b}{2}+jd+j\frac{b}{2}+j(a-d)+j\frac{b}{2}=$$ $$\left(2-\frac 3{2^{n+1}}\right)jab+ja+jb.$$
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How to find the maximum and minimum values of $\frac{8x(x^2-1)}{(x^2+1)^2}$ algebraically? The function is $f(x) = \frac{8x(x^2-1)}{(x^2+1)^2}$.
I have tried using calculus, only to fail.
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Suppose $b$ is in the range of the given function. Then equation
$$ \frac{8x(x^2-1)}{(x^2+1)^2} = b $$
$$ \frac{8x^2(x-{1\over x})}{x^2(x+{1\over x})^2} = b $$
Mark $t = x-{1\over x}$ and $t$ takes all real values. Then we have
$$ b= \frac{8t}{t^2+4} $$ If $t$ is positive then $b\leq 2$ since $4t\leq t^2+4$ or $(t-2)^2\geq 0$ is true. If $t$ is negative then $b\geq -2$. So the range is $[-2,2]$ since $b(t)$ is continious function.
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Prove by induction that $1^1\cdot 2^2\cdot \dots\cdot n^n\leq \left(\frac{2n+1}{3}\right)^{\frac{n(n+1)}{2}}.$ How can I prove by induction that for every natural number $n$,
$$1^1\cdot 2^2\cdot \dots\cdot n^n\leq \left(\frac{2n+1}{3}\right)^{\frac{n(n+1)}{2}}.$$
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The inductive proof requires the Bernoulli inequality (using $3$ terms).
\begin{eqnarray*}
(2n+3)^{n+2}=(2n+1+2)^{n+2} \geq \\(2n+1)^{n+2} +2(n+2) (2n+1)^{n+1} + \frac{(n+2)(n+1)}{2} 2^2 (2n+1)^{n}=(2n+1)^n (10n^2+20n+9) > (3(n+1))^2 (2n+1)^n.
\end{eqnarray*}
So we have $(2n+3)^{n+2} \geq (3(n+1))^2 (2n+1)^n$ & the result will follow with a bit of algebra.
The inductive step requires
\begin{eqnarray*}
1 \times 2^2 \cdots n^n \times(n+1)^{n+1} \leq ( \frac{2n+1}{3})^{\frac{n(n+1)}{2}} (n+1)^{n+1} \leq ( \frac{2n+3}{3})^{\frac{(n+1)(n+2)}{2}}
\end{eqnarray*}
So we need to show
\begin{eqnarray*}
( \frac{2n+1}{2n+3})^{\frac{n(n+1)}{2}} \leq ( \frac{2n+3}{3(n+1)})^{(n+1)}
\end{eqnarray*}
Take the $n+1$ root and use the previously derived result.
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How many solutions does the equation $a + b + c = 10$ if $(1, 4, 5)$ and $(1, 5, 4)$ are not considered distinct?
Question: What if we consider (1,4,5) and (1,5,4) as non-distinct possibilities, then what should we do?
$${{9}\choose{2}}-2\cdot\frac{{9}\choose{2}}{3}$$
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Since $10$ is not a multiple of $3$, it is not possible for all three numbers to be the same.
However, it is possible for two numbers of the same. Since each number is positive, the repeated number must be $1$, $2$, $3$, or $4$. There are $\binom{3}{2}$ ways to choose the locations of the repeated number. Choosing the repeated number determines the third number. There are
$$4\binom{3}{2} = 12$$
such ordered positive integer solutions of the equation $a + b + c = 10$. We count each unordered solution of this type three times, depending on where the single number is located.
That leaves $36 - 12 = 24$ solutions of the equation $a + b + c = 10$ in which all three numbers are distinct. Since $a, b, c$ can be arranged in $3! = 6$ ways, we count each unordered solution of this type $6$ times.
Hence, the number of unordered positive integer solutions to the equation
$a + b + c = 10$ is
$$\frac{12}{3} + \frac{24}{6} = 4 + 4 = 8$$
As a check, we list them.
$$\{1, 2, 7\}, \{1, 3, 6\}, \{1, 4, 5\}, \{2, 3, 5\}, \{1, 1, 8\}, \{2, 2, 6\}, \{3, 3, 4\}, \{2, 4, 4\}$$
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How to solve $\sqrt{x} + \sqrt{x-\sqrt{1-x}} = 1$ The equation is $$\sqrt{x} + \sqrt{x-\sqrt{1-x}} = 1$$
I cannot find any way to simplify this, other than squaring repeatedly, which is, btw, again not simplifying. Also tried substitutions like $1-x = t^2$ and trigonometric substitutions. None seem to work.
By trial and error, I find one root to be $\dfrac{16}{25}$. Is there a general way to treat this equation?
EDIT
The domain of definition of this function seems to be quite small, some subset of $(0,1)$. Also we know that there is only one root as the function is strictly increasing.
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$$\sqrt{x} + \sqrt{x-\sqrt{1-x}} = 1$$
$$\sqrt{x-\sqrt{1-x}} = 1 -\sqrt{x}$$
$$x-\sqrt{1-x} = (1 -\sqrt{x})^2$$
$$x-\sqrt{1-x} = x- 2\sqrt{x} +1$$
$$\sqrt{1-x} = 2\sqrt{x} +1$$
$$1-x = (2\sqrt{x} +1)^2$$
$$1-x = 1 + 4 \sqrt{x} + 4 x$$
$$- 5x= 4 \sqrt{x} $$
$$- \frac{5}{4}x= \sqrt{x} $$
$$\left(- \frac{5}{4}x\right)^2= x $$
$$\frac{25}{16}x^2= x $$
$$\frac{25}{16}= \frac{1}{x} $$
$$\frac{16}{25}= x $$
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Test the convergence of the series $\sum [(n^3+1)^{\frac{1}{3}}- n]$
Test the convergence of the series $\sum [(n^3+1)^{\frac{1}{3}}- n]$
I have no idea which test to use here because of this complicated expression.
|
$n^{th}$ term of the series is
= $ (n^3+1)^{\frac{1}{3}}- n$
= $ n(1+\frac{1}{n^3})^{\frac{1}{3}}- n$
= $ n(1+\frac{1}{3n^3} + other terms)- n$ . . . . . . . .(other terms goes to 0 as n approaches $\infty$)
= $ \frac{1}{3n^2}$ + other terms
Now as $n\to\infty$ ,our $n^{th}$ term can be shown with following inequality
= $ \frac{1}{3n^2}$+other terms$ \le \frac{1}{n^2} < \frac{1}{n}$
Since $n^{th}$ term of series is less than $\frac{1}{n}$, the series is convergent
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Rationalize denominator: $\frac{1}{\sqrt[3]{5}+\sqrt[3]{1}+\sqrt[3]{6}}$ $$\frac{1}{\sqrt[3]{5}+\sqrt[3]{1}+\sqrt[3]{6}}$$
So this is what I thought: the square root of 1 is obviously one, so I have $1^3 +(\sqrt[3]{5} + \sqrt[3]{6})$. In my head I see that this is the first part for the sum of cubes formula. I multiplied with the rest of the formula so I can get $1^3 +(\sqrt[3]{5} + \sqrt[3]{6})^3$ in the denominator. Now when I try to do this bracket I have a problem, what do I do with this? $$11+3\sqrt[3]{25 \cdot 6}+3\sqrt[3]{5 \cdot 36}$$
How do I get rid of the square roots?
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Use the identity: $$\dfrac{1}{\sqrt[3]{x}+\sqrt[3]{y}+\sqrt[3]{z}} = \dfrac{\sqrt[3]{x^2}+\sqrt[3]{y^2}+\sqrt[3]{z^2}-\sqrt[3]{xy}-\sqrt[3]{yz}-\sqrt[3]{zx}}{x+y+z - 3\sqrt[3]{xyz}}$$
and then use the difference of cube formula. Pretty sure there is no easier way to do this.
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Number of sequences that have at least five consecutive positions in which the numbers are in increasing order? A sequence of numbers is formed from the numbers $1, 2, 3, 4, 5, 6, 7$ where all $7!$ permutations are equally likely. What is the probability that anywhere in the sequence there will be, at least, five consecutive positions in which the numbers are in increasing order?
I approached this problem in the following way, but I am wondering if there is a better way, since my approach is quite complicated.
My Approach: There are three possibilities: a sequence have $7$ consecutive positions in which numbers increase, have $6$ consecutive positions in which numbers increase, and $5$ consecutive positions in which numbers increase.
There is only $1$ sequence that have $7$ consecutive positions. Namely, the sequence $(1,2,3,4,5,6,7)$.
There are $12$ sequences that have $6$ consecutive positions. Namely, we choose $1$ number from $(1,2,3,4,5,6,7)$, and move it to either sides. As an illustration, if we choose $3$, then we can get $(3,1,2,4,5,6,7)$ or $(1,2,4,5,6,7,3)$.
Now consider when there are $5$ consecutive positions in which numbers increase. We choose $2$ numbers that are not in the increasing subsequence.
If $1$ and $7$ are not chosen, we can place them in front of the subsequence, of after. For example, if we choose $(2,5)$, then we will have $(2,5,1,3,4,6,7)$,$(5,2,1,3,4,6,7)$, $(1,3,4,6,7,2,5)$ and $(1,3,4,6,7,5,2)$. This is $\binom{5}{2}\times4$.
Then I'm not sure how to proceed when we choose $1$ and/or $7$?
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Both myself and @N.F.Taussig used the following approach, although I'd like to see if it could be generalised to increasing runs of arbitrary length.
Define set $S_{i,j}$ as the set of permutations of $[7]$ with an increasing run between position $i$ and $j$ inclusive. Then by inclusion-exclusion the desired success count is
$$\begin{align}&(|S_{1,5}|+|S_{2,6}|+|S_{3,7}|) -\\ (|S_{1,5}\cap S_{2,6}| + |S_{1,5}&\cap S_{3,7}|+|S_{2,6}\cap S_{3,7}|)+ |S_{1,5}\cap S_{2,6}\cap S_{3,7}|\tag{1}\end{align}$$
Clearly
$$|S_{1,5}|=|S_{2,6}|=|S_{3,7}|=\binom{7}{5}2!\tag{2}$$
since we choose $5$ of the $7$ numbers to go in increasing order in positions $1$ to $5$, $2$ to $6$ or $3$ to $7$ and the remaining $2$ numbers can go in the remaining $2$ spots in $2!$ ways.
Also
$$S_{1,5}\cap S_{2,6}=S_{1,6}$$
$$\implies |S_{1,5}\cap S_{2,6}|=|S_{1,6}|=\binom{7}{6}1!\tag{3}$$
and
$$S_{1,5}\cap S_{3,7}=S_{1,7}$$
$$\implies |S_{1,5}\cap S_{3,7}|=|S_{1,7}|=\binom{7}{7}0!\tag{4}$$
and
$$S_{2,6}\cap S_{3,7}=S_{2,7}$$
$$\implies |S_{2,6}\cap S_{3,7}|=|S_{2,7}|=\binom{7}{6}1!\tag{5}$$
and
$$S_{1,5}\cap S_{2,6}\cap S_{3,7}=S_{1,7}$$
$$\implies |S_{1,5}\cap S_{2,6}\cap S_{3,7}|=|S_{1,7}|=\binom{7}{7}0!\tag{6}$$
using similar reasoning to $(2)$ in each case.
Putting the results of $(2)$, $(3)$, $(4)$, $(5)$ and $(6)$ into $(1)$ gives:
$$\text{success count}=3\binom{7}{5}2!-\left(2\binom{7}{6}1!+\binom{7}{7}0!\right)+\binom{7}{7}0!=112$$
Then since there are $7!$ permutations the desired probability is:
$$\text{probability of an increasing run of length $\ge 5$}=\frac{112}{7!}=\frac{1}{45}\tag{Answer}$$
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Justify the recurrence relation $a_{n+2}=2a_{n+1}+a_n$ and find $a_n$ Let $a_n$ be the number of ways to color the squares of a $1$ x $n$ chessboard using the colors red, white, and blue, so that no red square is adjacent to a white square. Justify the relation $a_{n+2}=2a_{n+1}+a_n$ (for certain $n$), and then find $a_n$.
My Attempt:
.If the first square is blue, then the number of ways of coloring the other $n-1$ squares is $a_{n-1}$
.If the first square is red, then the next square must be blue (1 option), and the rest can be filled in $a_{n-2}$ ways
.If it is white, then it is also $a_{n-2}$
This means the recurrence relation is $a_n=a_{n-1}+2a_{n-2}$ for $n\ge2$
By inspection, $a_0=1$ and $a_1=3$
Not sure where to go from here.
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I'm also learning recurrence relations as well so here's my take on it :) (please let me know if this makes sense)
I got a recurrence relation of $$a_n = a_{n-1} + 2a_{n-2} + 2\times 3^{n-2}$$
for $n\geq 2$ with $a_0 = 1, a_1 = 3$.
Here's how I derived it:
Case 1: The first block is blue. The remaining board is filled up in $a_{n-1}$ ways.
Case 2: The first block is red.
Case 2.1: If the second block is blue, then the remaining is filled in $a_{n-2}$ ways.
Case 2.2: The second block is also red. We count the number of ways the first block is red, the second block is red by counting the total number of unrestricted ways, less the number of ways the second block is not red.
The total number of ways is $3^{n-1}$.
The number of ways the second block is not red is $2\times 3^{n-2}$.
So, this case has $3^{n-1} - 2\times 3^{n-2} = 3^{n-2}$ ways.
So case 2 has $a_{n-2} + 3^{n-2}$ ways
Case 3: The first block is white.
This is the same as case 2.
So the number of ways is also $a_{n-2} + 3^{n-2}$ ways.
So,
$$a_n = a_{n-1} + 2a_{n-2} + 2\times 3^{n-2}.$$
To solve this, we can use the Method of Undetermined Coefficients. Basically, educated guessing.
The solution will be in the form of
$$a_n = h_n + p_n$$
where $h_n$ is the homogenous solution and $p_n$ is the particular solution.
For the homogenous solution:
$$h_n - h_{n-1} - 2h_{n-2} = 0,$$
the characteristic equation is
$$r^2 - r - 2 = 0$$
which has roots $r=2,-1$.
So $$h_n = A(2)^n + B(-1)^n$$
for arbitrary constants $A,B$.
The particular solution $p_n$ will be in the form
$$p_n = a\times 3^{n-2}$$
for some constant $a$.
So
$$p_n - p_{n-1} - 2p_{n-2} = a(3^{n-2} - 3^{n-3} - 2\times 3^{n-4}) = 3^{n-2} \\ a(9 - 3 - 2) = 9 \\ a = \frac{9}{4}.$$
So, $$p_n = \frac{9}{4} \times 3^{n-2}$$
So,
$$a_n = A(2)^n + B(-1)^n + \frac{9}{4} 3^{n-2}$$
which you can solve for $A,B$ using the initial conditions.
|
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Factoring a quartic expression
We have a function $$f(x)=x^{\frac{2}{3}}-\frac{2}{x^{\frac{1}{3}}}+1$$ and let line segment $AB$ be represented by $g(x)$ such that $$g(x)=x-4$$
Find the greatest positive real solution representing the intersection between the two lines.
The first step is obviously to let them equal each other, and the other steps will follow logically, they are demonstrated here
$$x-4=x^{\frac{2}{3}}-\frac{2}{x^{\frac{1}{3}}}+1$$ $$x-5=\frac{x-2}{x^{\frac{1}{3}}}$$ $$x^{\frac{1}{3}}=\frac{x-2}{x-5}$$ $$x=\frac{x^3-6x^2+12x-8}{x^3-15x^2+75x-125}$$ $$x^4-15x^3+75x^2-125x=x^3-6x^2+12x-8$$ $$x^4-16x^3+81x^2-137x+8=0$$
and here I'm very much stuck. By the factor theorem we know that for $(x-a)$ to a factor, $a$ must be a factor of $8$, i.e. $1,2,4,-1,-2$ or $-4$, but I've tried every one of these values and haven't found a value of $a$ such that $(x-a)$ is a factor of the latest expression.
The graph of the two equations is given here, and the point of intersection clearly falls at the intersection of two grid lines, so I know the solution is an integer, but under examination conditions I obviously won't have a graphing tool this sophisticated, so I'm eager to know how to proceed from here algebraically.
Any help is appreciated, thank you.
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You are very close, just haven't tried $a=8$ yet.
\begin{align}
f(x)&=x^4 -16x^3+81x^2-137x+8 \\
f(8) &= 8^4 - 16 \cdot8^3 + 81 \cdot8^2 - 137\cdot8+8 \\
&= 4096-8192+5184-1096+8\\
&= 0
\end{align}
Then we know $(x-8)$ is a factor of $f(x)$.
\begin{align}
f(x)&=(x-8)(x^3-8x^2+17x-1) \\
\end{align}
Used polynomial division to find $(x^3-8x^2+17x-1)$.
Let $p(x) = x^3-8x^2+17x-1$
You can then prove that $p(x)$ has no integer solutions. Your end answer should be $x=8$.
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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
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HINT
If $4^k + 6k -10$ is divisible by $18$, then $4(4^k + 6k -10)$ is divisible by $18$ as well.
Subtract the resulting expression from the expression you get when plugging in $k+1$:
$$4^{k+1} +6(k+1)-10 - 4(4^k + 6k -10) = $$
$$4 \cdot 4^k +6k+6-10 - 4 \cdot 4^k -24k + 40 = $$
$$-18k+36$$
Since $-18k+36$ is clearly divisible by $18$, and since $4(4^k + 6k -10)$ is divisible by $18$, it follows that $4^{k+1} + 6(k+1) -10$ is divisible by $18$ as well.
|
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Minimum of an expression involving radicals: $\sqrt{\frac{2a}{a+b}} + \sqrt{\frac{2b}{b+c}} + \sqrt{\frac{2c}{c+a}}$ The link below is a MSE question discussing the maximum value of the expression
$$ \sqrt{\frac{2a}{a+b}} + \sqrt{\frac{2b}{b+c}} + \sqrt{\frac{2c}{c+a}} $$
which is $3$.
Prove inequality $\sqrt{\frac{2a}{b+a}} + \sqrt{\frac{2b}{c+b}} + \sqrt{\frac{2c}{a+c}} \leq 3$
I thought about the minimum of the same expression over nonnegative reals.
My reasoning was as follows :
Using the fact that $\sqrt{X+Y+Z} \le \sqrt X + \sqrt Y + \sqrt Z$, and the fact $u + v \le u+v+w$ we can already assert that the expression to minimize is $\ge \sqrt 2$
Now taking $a = 0$, $b\to 0^+$ and $c$ arbitrary we see that the expression can take values arbitrarily close to $\sqrt 2$. Thus $\sqrt 2 $ is the infimum. but It seems not to be a minimum while the continuous function we have must have a minimum on the compact sphere (homogenous expression). Is there anything wrong with this approach ?
Then, What would the result be if we restrict ourselves to positive reals.
Thanks.
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For $b=\epsilon^2$ and $c=\epsilon$, where $\epsilon\rightarrow0^+$ our expression is closed to $\sqrt2$.
We'll prove that for positive variables it's an infimum.
Indeed, we need to prove that
$$\sum_{cyc}\sqrt{\frac{a}{a+b}}\geq1,$$ which is true by Holder:
$$\left(\sum_{cyc}\sqrt{\frac{a}{a+b}}\right)^2\sum_{cyc}a^2(a+b)\geq(a+b+c)^3.$$
Thus, it remains to prove that
$$(a+b+c)^3\geq\sum_{cyc}a^2(a+b)$$ or
$$\sum_{cyc}(2a^2b+3a^2c+2abc)\geq0,$$ which is obvious.
Done!
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Find $\lim_{x\rightarrow0}\frac{3^x-5^x}{4^x-10^x}$ Find $$\lim_{x\rightarrow0}\frac{3^x-5^x}{4^x-10^x}$$
My work so far:
$$\lim_{x\rightarrow0}\frac{3^x-5^x}{4^x-10^x}=\frac{\ln3-\ln5}{\ln4-\ln10}$$
Is correct?
Add:
I used $a^x\sim 1+x\ln a$ for $x\rightarrow 0$
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The expression equals
$$\frac{(3^x-3^0) - (5^x-5^0)}{(4^x - 4^0)-(10^x-10^0)}.$$
Divide top and bottom by $x=x-0.$ Then by definition of the derivative, the desired limit equals
$$ \frac{(3^x)'(0) - (5^x)'(0)}{(4^x)'(0)- (10^x)'(0)}= \frac{\ln 3 - \ln 5}{\ln 4 - \ln 10}.$$
|
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Can someone explain the steps of this Partial fraction decomposition?
My thoughts:
$$\frac{(Ax + B)}{(x^2+1)} + \frac{(Cx + D)}{(x^2+4)} = \frac{x}{(x^2+1)(x^2+4)}$$
I combined the left terms
Set the numerator of the combined left term to "x" which is the numerator of the right term
I got
$$4Ax + Cx = x$$
I am not sure what to do next
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I’m not entirely sure what you did to get there, but here are the steps to decomposing your partial fraction. You can compare and see where went wrong.
Starting off with$$\frac x{(1+x^2)(4+x^2)}=\frac {Ax+B}{1+x^2}+\frac {Cx+D}{4+x^2}$$We get rid of the fractions to see$$x=(Ax+B)(4+x^2)+(Cx+D)(1+x^2)$$Now, we set $x^2=-4$ to get rid of one of the expressions. Thus$$x=-3(Cx+D)\implies x=-3Cx-3D$$So $C=-\tfrac 13$ and $D=0$. Similarly, with the other expression, set $x^2=-1$ and we find that $A=\tfrac 13$ and $B=0$. Hence$$\frac x{(1+x^2)(4+x^2)}=\frac {x}{3(1+x^2)}-\frac x{3(4+x^2)}$$
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Probability that length of Randomly chosen chord of a circle Find Probability that length of Randomly chosen chord of a circle lies between $\frac{2}{3}$ and $\frac{5}{6}$ of its diameter.
My try: I assumed unit circle with center origin. Let two randomly chosen distinct points be $A(\cos \alpha, \sin \alpha)$ and $B(\cos \beta, \sin \beta)$
Length of the chord is $$p=2\sin \left(\frac{\alpha-\beta}{2}\right)$$
Now we have to find Probability that
$$\frac{4}{3} \le 2\sin \left(\frac{\alpha-\beta}{2}\right) \le \frac{5}{3}$$
can i have any clue here?
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The distance between $\alpha$ and $\beta$ cannot be negative.
$p = 2|\sin \frac {\beta - \alpha}{2}| = p = 2\sin \frac {|\beta - \alpha|}{2}$
Now, $\beta - \alpha$ is uniformly distributed. Let $\theta = \beta - \alpha.$ (In fact, we can fix $\alpha = 0$ and get the same result.) And rather than dealing with the absolute value, put theta on the interval $[0,\pi)$
$\frac 43 < 2 |\sin\frac {\theta}{2}| < \frac 53$
$2\arcsin \frac {4}{6} < \theta < 2\arcsin \frac {5}{6}$
This gives us a range of $\theta$ equal to $2(\arcsin \frac 56 - \arcsin \frac 46)$
And we must divide this by possible values of $\theta.$
$\frac {2}{\pi}(\arcsin \frac 56 - \arcsin \frac 46)$
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What is the exact value of $\int^\infty_0\frac{\sin^2 x}{x^{5/2}}\,dx$
I would like to get the exact value of the following integral.
$$\int^\infty_0\frac{\sin^2 x}{x^{5/2}}\,dx$$
I was able to prove the convergence as well. But I don't how to compute its exact value. I tried with the Residue Theorem of the complex function
$$z\mapsto \frac{\sin^2 z}{z^{5/2}}$$
But I could not move further.
|
Performing the change of variables $2u = x^2$ together with two integration by parts, we get, $$ \int_0^\infty \cos(x^2)dx = \frac{1}{\sqrt{2}}\int^\infty_0\frac{\cos(2x)}{\sqrt{x}}\,dx\\=\frac{1}{2\sqrt{2}}\underbrace{\left[\frac{\sin 2 x}{x^{1/2}}\right]_0^\infty}_{=0} +\frac{1}{4\sqrt{2}} \int^\infty_0\frac{\sin 2 x}{x^{3/2}}\,dx\\=
\frac{1}{4\sqrt{2}}\underbrace{\left[\frac{\sin^2 x}{x^{1/2}}\right]_0^\infty}_{=0} +\frac{3}{8\sqrt{2}} \int^\infty_0\frac{\sin^2 x}{x^{5/2}}\,dx$$
Hence $$\int^\infty_0\frac{\sin^2 x}{x^{5/2}}\,dx = \frac{8\sqrt2}{3}\int_0^\infty \cos(x^2)dx = \frac{4\sqrt \pi}{3}$$
Since See Here, $$\int_0^\infty \cos(x^2)dx= \sqrt\frac\pi8$$
or
How to prove only by Transformation that: $ \int_0^\infty \cos(x^2) dx = \int_0^\infty \sin(x^2) dx $
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Proving that $\int_0^\infty e^{-a^2x^2} \cos (bx) dx = \frac {\sqrt \pi}{2a}e^{\frac{-b^2}{4a^2}}$ Any clue regarding this:
$$\int_0^\infty e^{-a^2x^2} \cos (bx) dx = \frac {\sqrt \pi}{2a}e^{\frac{-b^2}{4a^2}}$$
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Let
$$I(b) = \int^\infty_0 e^{-a^2 x^2} \cos (b x) \, dx, \quad a > 0.$$
Integrating by parts gives
$$I(b) = \frac{2a^2}{b} \int^\infty_0 x e^{-a^2 x^2} \sin (bx) \, dx.$$
Also, differentiating $I(b)$ with respect to the parameter $b$ we have
$$I'(b) = - \int^\infty_0 x e^{-a^2 x^2} \sin (bx) \, dx = - \frac{b}{2a^2} I(b).$$
Solving this first-order differential equation yields
$$I(b) = K e^{-\frac{b^2}{4a^2}},$$
where $K$ is a constant to be determined. To find this constant, setting $b = 0$ one has
$$I(0) = \int^\infty_0 e^{-(ax)^2} \, dx = \frac{\sqrt{\pi}}{2a} \cdot \frac{2}{\sqrt{\pi}} \int^\infty_0 e^{-u^2} \, du = \frac{\sqrt{\pi}}{2a} \cdot \text{erf} (\infty) = \frac{\sqrt{\pi}}{2a}.$$
So $K = \sqrt{\pi}/(2a)$ and we have
$$I(b) = \int^\infty_0 e^{-a^2 x^2} \cos (b x) \, dx = \frac{\sqrt{\pi}}{2a} e^{-\frac{b^2}{4a^2}}, \quad a > 0,$$
as required.
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Finding exact solutions of an equation I have the question "Find the exact solutions of the following equation, giving your answer in terms of i and simplifying any surds (where possible):
2X^2 + 8X + 9 = 0"
Here is my attempt is this correct ?
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Looks correct to me. Here's how I would do it though:
$$
2x^2 + 8x + 9 = 0\implies\\
x^2 + 4x + \frac{9}{2} = 0\implies\\
x^2+2\cdot x\cdot 2+2^2-2^2 +\frac{9}{2} = 0\implies\\
(x+2)^2=\frac{8}{2}-\frac{9}{2}\implies\\
(x+2)^2=-\frac{1}{2}\implies\\
x_{1,2}+2=\pm\sqrt{-\frac{1}{2}}\implies\\
x_{1,2}=-2\pm\sqrt{\frac{1}{2}\cdot (-1)}\implies\\
x_{1,2}=-2\pm\sqrt{\frac{1}{2}\cdot i^2}\implies\\
x_{1,2}=-2\pm\sqrt{\frac{1}{2}}\cdot\sqrt{i^2}\implies\\
x_{1,2}=-2\pm\frac{1}{\sqrt{2}}i\implies\\
x_{1,2}=-2\pm\frac{\sqrt{2}}{2}i
$$
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estimate the maximum error of taylor approximation of $\ln(x)$ find the error bound using the Lagrange Remainder:
$$\ln(x)\approx (x-1)-\frac{(x-1)^2}{2}+\frac{(x-1)^3}{3}, |x-1|<\dfrac{1}{64}$$
My attempt:
\begin{align}
R_3(x)&=\dfrac{f^{(4)}(c)}{4!}(x-1)^4\\
f^{4}(c)&=\dfrac{-6}{(x-1)^4}\\
\implies R_3(x)&=\dfrac{-6}{4!}\dfrac{(x-1)^4}{(c-1)^4}\\
\implies |R_3(x)|&\leq \dfrac{6}{4!}\dfrac{(\frac{1}{64})^4}{(\frac{-1}{64})^4}=.25
\end{align}
Is this correct?
|
Note that,
\begin{equation}
f(x) = log(x) \implies f^{(4)}(x) = \frac{-6}{x^4}
\end{equation}
Therefore,
\begin{equation}
R_3(x) = -\frac{6}{4!c^4}(x-1)^4\quad; \quad \text{where } c \text{ is in between 1 and x}
\end{equation}
Since $c$ depends on $x$, first find an upper bound as;
\begin{equation}
|R_3(x)| \leq
\begin{cases}
\frac{6}{4!}(x-1)^4 \quad &if \quad x>1\\
\frac{6}{4!x^4}(x-1)^4 \quad &if \quad x<1
\end{cases}
\end{equation}
Now we can bound for $|x-1|<1/64$,
\begin{equation}
|R_3(x)| \leq
\begin{cases}
\frac{6}{4!64^4} \quad &if \quad x>1\\
\frac{6}{4!(1-1/64)^464^4} \quad &if \quad x<1
\end{cases}
\end{equation}
Here is a plot done by wolframalpha.com.
You can visaully see that we need two different bounds for $x<1$ and $x>1$. If you need a single bound, $x<1$ case will work.
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|
Elementary method to solve this equation I have to describe ''the sign of the root(s)''$$\sqrt[3]{3+\sqrt{x}}+\sqrt[3]{3-x}-\sqrt[3]{6}=0$$ for k-11 students , who did not learned derivation.
I can solve the equation by taking $f(x)=\sqrt[3]{3+\sqrt{x}}+\sqrt[3]{3-x}-\sqrt[3]{6}$ also by graphing https://www.desmos.com/calculator/u0pbqp1hvu
but , is there a simple trick to show the sign of roots(s) .
Remark:by derivation or graphing we can see 1 positive root exists .
thanks in advance.
|
Since $$a^3+b^3+c^3-3abc=(a+b+c)(a^2+b^2+c^2-ab-ac-bc)$$ and since
$$3+\sqrt{x}=3-x=-6$$ is impossible,
our equation is equivalent to
$$3+\sqrt{x}+3-x-6+3\sqrt[3]{6(3+\sqrt{x})(3-x)}=0$$ or
$$(\sqrt{x}-x)^3+162(3+\sqrt{x})(3-x)=0,$$
which after substitution $\sqrt{x}=t$ gives
$$t^6-3t^5+3t^4+161t^3+486t^2-486t-1458=0$$ and the last equation has unique non-negative root $t_1=1.731...$
|
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|
Find the number of real roots for $x+\sqrt{a^2-x^2}=b$, $a>0$, $b>0$, as a function of $a$ and $b$
Given: (1) $x+\sqrt{a^2-x^2}=b$, $(a,b,x)\subset \mathbb R$, $a>0$, $b>0$.
Find: number of roots for (1), given possible values for $a$ and $b$.
This is a question from a book for the preparation for math contests.
It states as final answer: (a) 1 root if $b<a$; and (b) 2 roots if $a<b<a\sqrt{2}$.
I'm having difficulties on finding this answer. I don't know whether it is correct or perhaps I'm not finding the right approach.
I started moving $x$ in (1) to the left, to get
$$\sqrt{a^2-x^2}=b-x$$
Before proceeding with squaring both sides, I saved 2 needed conditions for checking the final solution (c1) $a^2-x^2\ge 0$ and (c2) $b-x\ge 0$. Then squaring both sides, we get:
$$a^2-x^2=b^2+x^2-2bx\Leftrightarrow 2x^2-2bx+(b^2-a^2)=0$$
with discriminant $\triangle$ defined by:
$$\triangle=4(2a^2-b^2)$$
From this it is easy to see that a condition for 2 roots is (c3) $\sqrt{2}a>b,$ and for 1 root is (c4) $\sqrt{2}a=b,$ as $a>0$ and $b>0$. Then I find the roots as $$x=\frac{2b\pm \sqrt{\triangle}}{4}=\frac{b\pm \sqrt{2a^2-b^2}}{2}$$
From this point, I can't see a way to reach the stated answer, if it is right.
Full solutions or helpful hints are welcome. Sorry if it is a duplicate.
|
You want to study the function
$$
f(x)=x+\sqrt{a^2-x^2}
$$
defined over $[-a,a]$. We have $f(-a)=-a$, $f(a)=a$; moreover
$$
f'(x)=1-\frac{x}{\sqrt{a^2-x^2}}
$$
for $x\in(-a,a)$. The derivative can only vanish where
$$
x=\sqrt{a^2-x^2}
$$
so $x\ge0$ and $x^2=a^2-x^2$, that is, $x=a/\sqrt{2}$. Note that
$$
f(a/\sqrt{2})=\frac{a}{\sqrt{2}}+\frac{a}{\sqrt{2}}=a\sqrt{2}
$$
and that $a/\sqrt{2}$ is a point of absolute maximum for $f$.
Therefore the equation $x+\sqrt{a^2-x^2}=b$ has
*
*no solution for $b<-a$
*one solution for $-a\le b\le a$
*two solutions for $a<b<a\sqrt{2}$
*one solution for $b=a\sqrt{2}$
*no solutions for $b>a\sqrt{2}$
An algebraic solution.
The equation $\sqrt{a^2-x^2}=b-x$ has solutions only for $x\le b$. Then you can square: $a^2-x^2=b^2-2bx+x^2$ or $2x^2-2bx+b^2-a^2$. The roots of this equation are
$$
\frac{b-\sqrt{2a^2-b^2}}{2}
\qquad\text{and}\qquad
\frac{b+\sqrt{2a^2-b^2}}{2}
$$
There is no solution for $2a^2-b^2<0$, that is, $b>a\sqrt{2}$.
There will be two solutions when $b<a\sqrt{2}$ and
$$
\frac{b+\sqrt{2a^2-b^2}}{2}\le b
$$
that is,
$$
\sqrt{2a^2-b^2}\le b
$$
or $b\ge a$. Hence, $a\le b<a\sqrt{2}$.
There will be one solution when the largest root is greater than $b$, that is $b<a$, but also the smallest root is $\le b$, that is
$$
\frac{b-\sqrt{2a^2-b^2}}{2}\le b
$$
which is always satisfied when $b>0$.
When the discriminant is $0$, then there is a single solution.
|
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|
When $f(x) = \frac{ax + b}{a -x + 1}$ is an integer Given that $a$ and $b$ are positive integers. and $$f(x) = \frac{ax + b}{a -x + 1}$$ is an integer, what integer values can x have?
If I could only somehow move $x$ from numerator to the denominator I would be able to solve this by factoring the numerator.
|
Since
$$a-x+1\mid ax+b \;\;\;{\rm and} \;\;\;a-x+1\mid a(a-x+1)$$
we have $$a-x+1\mid (ax+b)+(a^2-ax+a)= a^2+a+b$$
So $x=a-d+1$ where $d$ divides $a^2+a+b$.
|
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|
Inverse Laplace transform of $f(s)=\frac{1}{s\sinh^2(c\sqrt{s})}$ I want to calculate inverse Laplace transform of $f(s)=\frac{1}{s\sinh^2(c\sqrt{s})}$, where $s$ the is Laplacian variable.
The function has one pole at $s=0$, for which the residue can be easily found, and infinitely many poles at $c\sqrt{s} = n\pi i$. I have some problems finding the residues of those poles because seem to be of order of $2$? How can I solve the second part of the problem?
Thanks in advance.
|
$$f(s)=\frac{1}{s \sinh ^2\left(c \sqrt{s}\right)}$$
Integrate f(s) as a function a:
$$\text{Int}(f(a,s))=\int \frac{1}{s \sinh ^2\left(a c \sqrt{s}\right)} \, da=-\frac{\coth \left(a c \sqrt{s}\right)}{c
s^{3/2}}+\text{c1}
$$
Using this identity and substituting:
$$\coth (s)=\frac{1}{s}+\sum _{k=1}^{\infty } \frac{2 s}{\pi ^2 \left(\frac{s^2}{\pi ^2}+k^2\right)}$$
$$\text{Int}(f(a,s))=-\frac{1}{a c^2 s^2}-\sum _{k=1}^{\infty } \frac{2 a}{\pi ^2 s \left(k^2+\frac{a^2 c^2 s}{\pi
^2}\right)}+\text{c1}$$
Using Inverse Laplace Transfrom:
$$\mathcal{L}_s^{-1}[\text{Int}(f(a,s))](t)=\mathcal{L}_s^{-1}\left[-\frac{1}{a c^2 s^2}-\sum _{k=1}^{\infty } \frac{2 a}{\pi ^2 s
\left(k^2+\frac{a^2 c^2 s}{\pi ^2}\right)}+\text{c1}\right](t)$$
$$\text{Int}(F(a,t))=-\frac{t}{a c^2}-\sum _{k=1}^{\infty } \left(\frac{2 a}{k^2 \pi ^2}-\frac{2 a e^{-\frac{k^2 \pi ^2 t}{a^2
c^2}}}{k^2 \pi ^2}\right)+\text{c1} \delta (t)$$
We differentiate for a and a=1:
$$\frac{\partial \text{Int}(F(a,t))}{\partial a}=\frac{\partial }{\partial a}\left(-\frac{t}{a c^2}-\sum _{k=1}^{\infty }
\left(\frac{2 a}{k^2 \pi ^2}-\frac{2 a e^{-\frac{k^2 \pi ^2 t}{a^2 c^2}}}{k^2 \pi ^2}\right)+\text{c1} \delta (t)\right)$$
$$F(t)=\frac{t}{c^2}-\sum _{k=1}^{\infty } \frac{2}{k^2 \pi ^2}+\sum _{k=1}^{\infty } \frac{2 e^{-\frac{k^2 \pi ^2 t}{c^2}}}{k^2
\pi ^2}+\sum _{k=1}^{\infty } \frac{4 e^{-\frac{k^2 \pi ^2 t}{c^2}} t}{c^2}$$
$$F(t)=-\frac{1}{3}-\frac{t}{c^2}+\frac{2 t \vartheta _3\left(0,e^{-\frac{\pi ^2 t}{c^2}}\right)}{c^2}+\sum _{k=1}^{\infty }
\frac{2 e^{-\frac{k^2 \pi ^2 t}{c^2}}}{k^2 \pi ^2}$$
where $\vartheta _3\left(0,e^{-\frac{\pi ^2 t}{c^2}}\right)$: is Jacobi Theta function
|
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|
A problem about the dimension of the intersection of two subspaces I'm trying to solve the problem below.
$U$ and $W$ are subspaces of polynomials over $\mathbb{R}$.
$U = Span(t^3 + 4t^2 - t + 3, t^3 + 5t^2 + 5, 3t^3 + 10t^2 -5t + 5)$
$W = Span(t^3 + 4t^2 + 6, t^3 + 2t^2 - t + 5, 2t^3 + 2t^2 -3t + 9)$
What is $dim(U \cap W)$?
I have solved it using the fact that $dim(U) + dim(W) - dim(U \cap W) = dim(U \cup W)$, but was wondering how to solve it without using this fact.
In order to find $dim(U \cap W)$, I first try and find $U \cap W$.
Clearly if $v \in U \cap W$, then $$\alpha_1(t^3 + 4t^2 - t + 3) +\alpha_2(t^3 + 5t^2 + 5) +\alpha_3(3t^3 + 10t^2 -5t + 5) = \beta_1(t^3 + 4t^2 + 6) + \beta_2(t^3 + 2t^2 - t + 5) + \beta_3(2t^3 + 2t^2 -3t + 9)$$ for some $\alpha_1, \alpha_2, \alpha_3, \beta_1, \beta_2, \beta_3 \in \mathbb{R}$.
Using this fact, you can reduce a system of linear equations to work out that:
$\alpha_1 + 5\alpha_3 - \beta_2 - 3\beta_3 = 0$
$\alpha_2 -2 \alpha_3 + 2\beta_2 + 6\beta_3 = 0$
$\beta_1 + 2\beta_2 + 5\beta_3 = 0$
But I don't know where to go from here.
Any help would be greatly appreciated.
|
Consider a polynomial $p(t)=at^{3}+bt^{2}+ct+d$ belonging to $U$ and $V$ at the same time and solve two separated systems to find the conditions for $p(t)$ to be in $U$ and beside find the coditions for $p(t)$ to be in $V$ and once you find those two list of conditions consider a system on the unkowns $a,b,c$.
$$p(t)=\alpha_1(t^3 + 4t^2 - t + 3) +\alpha_2(t^3 + 5t^2 + 5) +\alpha_3(3t^3 + 10t^2 -5t + 5)$$
$$ p(t)=\beta_1(t^3 + 4t^2 + 6) + \beta_2(t^3 + 2t^2 - t + 5) + \beta_3(2t^3 + 2t^2 -3t + 9)$$
then
$$at^{3}+bt^{2}+ct+d=\alpha_1(t^3 + 4t^2 - t + 3) +\alpha_2(t^3 + 5t^2 + 5) +\alpha_3(3t^3 + 10t^2 -5t + 5)$$
$$at^{3}+bt^{2}+ct+d=\beta_1(t^3 + 4t^2 + 6) + \beta_2(t^3 + 2t^2 - t + 5) + \beta_3(2t^3 + 2t^2 -3t + 9)$$
this implies that
For $U$ you need to find the conditions on $a,b$ and $c$ for the following system to have solution:
$$at^{3}+bt^{2}+ct+d=\alpha_1(t^3 + 4t^2 - t + 3) +\alpha_2(t^3 + 5t^2 + 5) +\alpha_3(3t^3 + 10t^2 -5t + 5)$$
$$A_{U}=\left(\begin{array}{ccc|c}
1&1&3&a\\
4&5&10&b\\
-1&0&-5&c\\
3&5&5&d
\end{array}\right)$$
For $V$ you need to find the conditions on $a,b$ and $c$ for the following system to have solution:
$$at^{3}+bt^{2}+ct+d=\beta_1(t^3 + 4t^2 + 6) + \beta_2(t^3 + 2t^2 - t + 5) + \beta_3(2t^3 + 2t^2 -3t + 9)$$
$$A_{V}=\left(\begin{array}{ccc|c}
1&1&2&a\\
4&2&2&b\\
0&-1&-3&c\\
6&5&9&d
\end{array}\right)$$
The row reduced echelon form of $A_{U}$ will give you the conditions for a polynomial to be in $U$. The same for $A_{V}$. These conditions are linear equations in $a,b,c$ that you can put in a linear system of equations in the variables $a,b,c$ to finally find the conditions for $p(t)$ to be in the intersection of $U$ and $V$.
|
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|
To find the values of a for $f$ has no critical number Determine the values of $a$ for which $$f(x)=(a^2+a-6)\cos{2x}+(a-2)x+\cos{1}$$has no critical number
My attempt:
A critical number of $f$ is a number $c$ in the domain of $f$ such that $f'(c)=0$ or $f'(c)$ does not exist
$$f'(x)=-2(a^2+a-6)\sin{2x}+(a-2)$$
So,
$$f'(0)=0$$
gives $a=2$
But how do I find the other values?
|
We need
$$f'(x)=-2(a^2+a-6)\sin{2x}+(a-2)=0$$
Then
$$f'(x)=-2(a+3)(a-2)\sin{2x}+(a-2)$$
Because we want that $f$ has no critical points, we need that $f'(x)$ is never $0$. But $-2(a+3)(a-2)\sin{2x}$ can be $0$ (since $\sin{2x}=0$ an infinite number times in the domain).
So really, we need that $(a-2)$ is never $0$, so that even when $-2(a+3)(a-2)\sin{2x}=0$, no critical point is achieved. And $(a-2)$ is never $0$ as long as $x \not = 2$.
Furthermore, we need to exclude values of $a$ such that $2(a+3)(a-2)\sin{2x} = (a-2)$. Divide by $(a-2)$ to get $$2(a+3)\sin{2x} = 1$$
Solve for $a$ to get $$a = \frac 12 (\csc(2x) - 6)$$ Which means that if $a$ can be expressed in this form for some $x$, then $f$ will have critical points. So we need that $a \not = \frac 12 (\csc(2x) - 6)$ for all $x$.
For ease of notation, let $g(x) = \frac 12 (\csc(2x) - 6)$. The range of $g$ is every real number except for those between $- \frac 72$ and $- \frac 52$, exclusive. So if $a$ is in the excluded range, it will never equal $g(x)$. In other words, if $a \in (- \frac 72, - \frac 52)$, then we have satisfied the criterion that $2(a+3)(a-2)\sin{2x} \not = (a-2)$.
Combining these results gives us that $f$ has no critical points when $a \in (- \frac 72, - \frac 52)$ and $a \not = 2$, which is the same as saying
$$a \in (- \frac 72, - \frac 52)$$
|
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|
If $A$ and $B$ are $2\times 2$ matrices and $AB=0$ then $A=0$ or $B=0$?
Is it true that if $A$ and $B$ are $2\times 2$ matrices and $AB=0$ then $A=0$ or $B=0$. Prove it, or prove the contrary.
I tried saying that if:
$$A= \begin{pmatrix}
0 & 0 \\
0 & 0 \\
\end{pmatrix}\quad\text{and}\quad B= \begin{pmatrix}
e & f \\
g & h \\
\end{pmatrix}.$$
The product will be 0. But, if $A$ is non zero, i tried doing something like this:
$$A= \begin{pmatrix}
a & b \\
c & d \\
\end{pmatrix},\quad
B = \begin{pmatrix}
e & f \\
g & h \\
\end{pmatrix}
\implies
AB = \begin{pmatrix}
ae+bg & af+bh \\
ce+dg & cf+dh \\
\end{pmatrix}$$
is equal to \begin{pmatrix}
0 & 0 \\
0 & 0 \\
\end{pmatrix}
But then, how should i proceed? I don't know if these are the right steps to follow, i've tried searching, but i didn't find the right examples.
|
For example you may try with
$$A=B=\begin{bmatrix}
0 & 1 \\
0 & 0 \\
\end{bmatrix}.$$
|
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|
What is the number of 5 digit numbers divisible by 3? What is the number of $5$ digit numbers divisible by $3$ using the digits
$0,1,2,3,4,6,7$ and repetition is not allowed?
|
the rule of three says that a number is divisible by three the sum of the digits is divisible by three. SO if one five digit number is divisible by three then any arrangement of its digits is divisible by three.
So it's a matter of finding the ways of picking $5$ out of $7$ numbers that add to a multiple of $3$ and multiplying by the number of ways we can arrange $5$ digits.
If $0$ is among the $5$ the number can't start with $0$ so there are four choice for the fist digit and $4!$ for the rest so $4*4!$ total. If $0$ is not among the $5$ there are $5!$ ways or arranging them.
Picking $5$ numbers that add to a multiple of three is a matter of picking the $2$ that we don't need. $0+1+2 +3+4+6+7= 23\equiv 2 \mod 3$ so this is matter of picking two numbers that add to $2 \mod 3$ (so the remaining $5$ will add to $0 \mod 3$. That is to say, $2$ that add to $2,5, 8,$ or $11$.
So those pairs are $(0,2),(1,4)(2,3),(1,7),(2,6),(4,7)$.
Of the sets of five remaining, $(1,3,4,6,7), (0,2,3,6,7), etc.$ only one has no $0$; $(1,3,4,6,7)$ and $5!$ ways to arrange them. The remaining $5$ have $0$s and so $4*4!$ ways to arange them. So there are $5*(4*4!) + 5! = 4*5! +5! = 5*5! = 5*120 = 600$ possible numbers.
|
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|
Divisibility theorem based proof for any square mod 4 being either 0 or 1 Kindly help me in understanding the below proof for given statement:
If $n$ is a square, then leaves a remainder $0$ or $1$ when divided by 4.
Proof: The divisibility theorem states that for two integers $a,b$ with $b>0$, then there is a unique pair of integers $q$ and $r$ such that $a =qb +r$ and $0\le r <b$.
Let the number $n= a^2$, for $a$ being an integer. If $b$(divisor)$=4$, then $a = 4q +r$, where $r \in \{0, 1, 2, 3\}$, so that $n = (4q+r)^2 = 16q^2 + 8qr + r^2$.
If $r=0$, then $n=4(4q^2 +2qr) +0$; if $r=1$, then $n=4(4q^2 +2qr)+1$; if $r=2$, then $n =4(4q^2 +2qr+1) +0$; and if $r =3$, then $n = (4q^2 +2qr+2)+1$. In each case , the remainder is $0$ or $1$.
The difficulty lies in the last para. where according to me it should be:
If $r=0$, then $n=4(4q^2 +2qr) +0$; if $r=1$, then $n=4(4q^2 +2qr)+1$; if $r=2$, then $n =4(4q^2 +2qr) +2$; and if $r =3$, then $n = (4q^2 +2qr)+3$.
|
Any integer is of the form $2n$ or $2n-1$ (i.e odd or even)
Now sq. of any number will be in the form $$(2n)^2 =4p or (4n-1)^2 =4q +1$$
So we find when that if $n$ is a square, then leaves a remainder $0$ or $1$ when divided by $4$.
|
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|
Find value of $a_{2012}$ A sequence $\left\{a_n\right\}$ is defined as:
$a_1=1$, $a_2=2$ and
$$a_{n+1}=\frac{2}{a_n}+a_{n-1}$$ $\forall$ $n \ge 2$
Find $a_{2012}$
My Try:
we have
$$a_{n+1}-a_{n-1}=\frac{2}{a_n}$$
$$a_n a_{n+1}-a_{n-1}a_n=2 \tag{1}$$
Replacing $n$ with $n-1$ we get
$$a_{n-1} a_{n}-a_{n-2}a_{n-1}=2 \tag{2}$$ adding $(1)$ and $(2)$
we get
$$a_n a_{n+1}-a_{n-2}a_{n-1}=4 \tag{3}$$
Again replace $n$ with $n-1$ in $(3)$ and adding with $(1)$ we get
$$a_n a_{n+1}-a_{n-2}a_{n-3}=6 \tag{4}$$ Again replace $n$ with $n-1$ in $(4)$ and adding with $(1)$ we get
$$a_n a_{n+1}-a_{n-3}a_{n-4}=8 \tag{5}$$
Continuing the process we get
$$a_na_{n+1}-a_{n-2010}a_{n-2011}=4022$$
Now in above equation put $n=2012$ we get
$$a_{2012}a_{2013}-a_1a_2=4022$$ $\implies$
$$a_{2012}a_{2013}=4024$$
Any further clue?
|
You already have $$a_n a_{n+1}-a_{n-1}a_n=2$$
So if you set the new sequence $b_k=a_{k+1}a_{k}$ then you get
$$b_k-b_{k-1}=2$$ so,
$b_k$ is an arithmetic sequence with $b_1=a_1a_2=2$ and step $2$. Then
$$b_k=2+2(k-1)=2k$$
then
$$a_{2}a_{1}=2$$
$$4=a_{3}a_{2}$$
$$a_{4}a_{3}=6$$
$$8=a_{5}a_{4}$$
$$...$$
$$2\cdot 2010=a_{2011}a_{2010}$$
$$a_{2012}a_{2011}=2\cdot2011$$
Multiply every equation and get
$$4\cdot8\cdot12\cdot...2020\cdot (a_{2012}\cdot a_1)=2\cdot6\cdot10\cdot...2022$$
$$a_{2012}=\frac{2\cdot6\cdot10\cdot...2022}{4\cdot8\cdot12\cdot...2020}$$
|
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|
>Does $\{f_n\}$ converge pointwise to a function on $[0,\infty)?$ Let $g_n(x)=\sin^2(x+\frac{1}{n}),x\in (0,\infty)$
and $f_n(x)=\int _0^x g_n(t)\, dt$.
Does $\{f_n\}$ converge pointwise to a function on $[0,\infty)?$
I try to show that $$\int_0^x \sin^2\left(t +\frac1n\right) \, dt = \frac{x}{2} - \frac{1}{4}\sin\left(2x+\frac2n\right) + \frac{1}{4} \sin \left(\frac2n\right),$$
and
$$\int_0^x \sin^2(t) \, dt = \frac{x}{2} - \frac{1}{4}\sin(2x) .$$
Hence,
$$|f_n(x) - f(x)| \leqslant \frac{1}{4}\left|\sin\left(2x+\frac2n\right) - \sin(2x) \right|+ \frac{1}{4} \left|\sin\left(\frac2n\right)\right| \leqslant \frac{1}{2n} + \frac{1}{4}\left|\sin\left(\frac2n\right)\right|$$
We have uniform convergence on $[0,\infty)$.
From this proof i can conclude that $\{f_n\}$ doesnot converge pointwise to a
function $f$ on $[0,\infty)$. so My answer is no.......IS it correct ?
|
So $\dfrac{1}{2n}+\dfrac{1}{4}|\sin(2/n)|\leq\dfrac{1}{2n}+\dfrac{1}{4}\cdot\dfrac{2}{n}=\dfrac{1}{n}\rightarrow\infty$ uniformly in $x$, so the convergence is uniform and hence pointwise by fixing $x$, and then taking $n\rightarrow\infty$ to $|f_{n}(x)-f(x)|<\epsilon$.
|
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|
Limit of the $n$th root of the product How one can evaluate $$\lim_{n\to\infty} \frac{\left( \prod_{k=1}^n (n^2+k^2) \right)^{1/n}}{n^2}?$$ I was unable to found any trick to do the computation.
|
$$\frac{\left( \prod_{k=1}^n (n^2+k^2) \right)^{1/n}}{n^2}=\frac{\left( \prod_{k=1}^n (n^2(1+\frac{k^2}{n^2}) \right)^{1/n}}{n^2}= \frac{n^2 \left( \prod_{k=1}^n (1+\frac{k^2}{n^2} \right)^{1/n}}{n^2}$$
$$\log\frac{n^2 \left( \prod_{k=1}^n (1+\frac{k^2}{n^2} \right)^{1/n}}{n^2}= \sum_{k=1}^n \frac{1}{n}\log (1+\frac{k^2}{n^2})$$
$$\int_0^1 \log(1+x^2)dx =\lim \sum_{k=1}^n \frac{1}{n}\log (1+\frac{k^2}{n^2})$$
The last integral can be evaluated by parts: it equals $-2 + \frac{\pi}{2}+\log(2)\approx 0.2639435$.
|
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|
Evaluate $\int \frac{x+1}{x^2+2x}dx$ in two different ways I am trying to evaluate
$$\int \frac{x+1}{x^2+2x}dx$$
Sounds like $$\int \frac{x+1}{x^2+2x}dx=\int \frac{(x+1)}{(x+1)^2-1}dx\\u=x+1 \implies\int\frac{u}{u^2-1}du\\t=u^2 \implies \frac{1}{2}\int\frac{1}{t-1}dt $$ Which is also known as $$\frac{1}{2}\ln|t-1|=\frac{1}{2}\ln|x^2+2x|$$But before I tried that, I had tried $$\int \frac{x+1}{x^2+2x}dx=\int \frac{1}{x+2}dx+\int \frac{1}{x^2+2x}dx\\=\ln|x+2| + \int \frac{1}{(x+1)^2-1}dx$$ Recalling that $\int \frac{1}{u^2-1}du=-\tanh^{-1}(u)$ or $-\coth^{-1}(u)$, $$\ln|x+2| + \int \frac{1}{(x+1)^2-1}dx=\ln|x+2|-\tanh^{-1}(x+1)$$ or $\ln|x+2|-\coth^{-1}(x+1)$ on the different parts of the domain.
Why did I get two different answers for the two different methods? Thanks.
Edit: They are not different. Just from the logarithmic definition of $\coth^{-1}$ and $\tanh^{-1}$ they are the same thing.
|
You can use substitution to solve. I will be a little bit faster. In fact, let $u=x^2+2x$. Then $du=2(x+1)dx$ and so
$$ \int\frac{x+1}{x^2+2x}dx=\frac12\int\frac{du}{u}=\frac12\ln|u|+C=\frac12\ln|x^2+2x|+C.$$
|
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|
Combinatorics and Matrices Find the number of $4\times4$ matrices such that $|a_{ij}| = 1 \forall i,j\in[1,4]$ , and sum of every row and column is zero.
I tried 'counting' the number of matrices that satisfy the above conditions, that is, elements are $1$ or $-1$ and sum of every row and column is zero.
In the attempt to generate a recursion I started off with a $2\times2$ matrix, for which case the answer is $2$. (First element is 1 or -1, other elements are decided accordingly)
However, this method becomes cumbersome and mathematically disappointing for $3x3$ and larger matrices.
Could someone please explain the method, or post a solution to the problem?
Is it possible to generalise the result to an nxn matrix?
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The first row has two $1$s and two $-1$s. There are $\binom{4}{2} = 6$ ways they can be arranged in the first row. We'll count the number of matrices with first row $(1,1,-1,-1)$, then multiply by $6$ to account for all the other arrangements of the first row.
The second row can be one of the three following things: $(1,1,-1,-1)$, or $(-1,-1,1,1)$, or $(1,-1,1,-1)$. In the third case this counts for $4$ possibilities: the first two columns might be switched, or the last two columns might be switched. We'll count the number of matrices in each of these three cases (multiplying the third cases's count by $4$, to account for the switches).
Case 1: The matrix looks like
$$
\begin{pmatrix}
1 & 1 & -1 & -1 \\
1 & 1 & -1 & -1 \\
* & * & * & * \\
* & * & * & *
\end{pmatrix}
$$
There is only one possible matrix:
$$
\begin{pmatrix}
1 & 1 & -1 & -1 \\
1 & 1 & -1 & -1 \\
-1 & -1 & 1 & 1 \\
-1 & -1 & 1 & 1 \\
\end{pmatrix}
$$
Case 2: The matrix looks like
$$
\begin{pmatrix}
1 & 1 & -1 & -1 \\
-1 & -1 & 1 & 1 \\
* & * & * & * \\
* & * & * & *
\end{pmatrix}
$$
In the last two rows, there is precisely one $-1$ in each column and two $-1$s in each row. There are $\binom{4}{2}=6$ ways to choose the locations of two $-1$s in the third row; then the fourth row is determined (it has $-1$s in the complementary positions).
Case 3: The matrix looks like
$$
\begin{pmatrix}
1 & 1 & -1 & -1 \\
1 & -1 & 1 & -1 \\
* & * & * & * \\
* & * & * & *
\end{pmatrix}
$$
The first and last columns are determined:
$$
\begin{pmatrix}
1 & 1 & -1 & -1 \\
1 & -1 & 1 & -1 \\
-1 & * & * & 1 \\
-1 & * & * & 1
\end{pmatrix}
$$
There are two solutions: the remaining block can be $\begin{pmatrix} 1 & -1 \\ -1 & 1 \end{pmatrix}$ or the opposite. That gives two solutions when the second row is as displayed, but because of the possibility of switching the first two or last two columns, we count $8$ solutions.
Subtotal: With this first row, we have $1+6+8=15$ matrices.
Total: There are $6$ equivalent arrangements of the first row, so we have $6 \cdot 15 = 90$ matrices in total.
It seems unlikely that this kind of approach could generalize to larger matrices.
|
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|
Probability of rolling two dice with either showing a six We roll two dice, one red and one green. Under each assumption below, what is the probability that the roll is double sixes?
a) The red die shows six.
b) At least one of the dice shows a six.
I think this is pretty straight forward:
For a) it is $\dfrac{1}{6}$
For b) it is $\dfrac{6+6+6-2}{36} \quad$ ($6$ for only red, $6$ for only green, $6$ for both green and red have sixes, $2$ redundant cases)
Please correct me if I'm wrong.
|
I am suspicious of your second computation. I think it would be easier to think in terms of inclusion-exclusion and independence:
\begin{align}
P(\text{at least one 6})
&= P(\color{red}{\text{red 6}} \lor \color{green}{\text{green 6}})\\
&= P(\color{red}{\text{red 6}}) + P(\color{green}{\text{green 6}}) - P(\color{red}{\text{red 6}} \land \color{green}{\text{green 6}}) && (\text{inclusion-exclusion}) \\
&= P(\color{red}{\text{red 6}}) + P(\color{green}{\text{green 6}}) - P(\color{red}{\text{red 6}})\cdot P(\color{green}{\text{green 6}}) && (\text{independence}) \\
&= \frac{1}{6} + \frac{1}{6} - \frac{1}{6}\cdot\frac{1}{6} \\
&= \frac{1}{6} + \frac{1}{6} - \frac{1}{36} \\
&= \frac{6 + 6 - 1}{36} \\
&= \frac{11}{36}.
\end{align}
Alternatively, you can just write out the entire sample space (albeit in an organized manner):
\begin{array}{r|rrrrrr}
& \color{green}{1} & \color{green}{2} & \color{green}{3} & \color{green}{4} & \color{green}{5} & \color{green}{6} \\\hline
\color{red}{1} & \square & \square & \square & \square & \square & \blacksquare \\
\color{red}{2} & \square & \square & \square & \square & \square & \blacksquare \\
\color{red}{3} & \square & \square & \square & \square & \square & \blacksquare \\
\color{red}{4} & \square & \square & \square & \square & \square & \blacksquare \\
\color{red}{5} & \square & \square & \square & \square & \square & \blacksquare \\
\color{red}{6} & \blacksquare & \blacksquare & \blacksquare & \blacksquare & \blacksquare& \blacksquare \\
\end{array}
Here, the black squares represent "good" outcomes (i.e. at least one 6), while the white squares represent "bad" outcomes. There are a total of 36 outcomes, 11 of which are good, thus
$$ P(\text{at least one 6}) = \frac{\text{good outcomes}}{\text{all outcomes}} = \frac{11}{36},$$
as above.
$\color{white}{\text{Boy, I hope no one is red/green colorblind...}}$
|
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How to find matrix exponential $e^A$ I have the matrix $$A =\begin{pmatrix} 0 & 1 \\ - 1 & 0 \end{pmatrix}$$
and I have to find $e^A$
I've found two complex-conjugate eigenvalues $\lambda_{1,2} = \pm i$
so substracting $\lambda_1 = i$ from the matrix's diagonal I got:
$$A_1 = \begin{pmatrix} -i & 1 \\-1 & i \end{pmatrix}$$
and therefore. to find eigenvector I have to solve the system:
$$A_1 = \begin{pmatrix} -i & 1 \\-1 & i \end{pmatrix} \begin{pmatrix} x_1 \\ x_2 \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \end{pmatrix}$$
so the first eigenvector is $h_1 = \begin{pmatrix}1 \\ i\end{pmatrix}$
and the second one is $h_2 = \begin{pmatrix}1 \\ -i\end{pmatrix}$
so the general solution is $$x(t) = C_1e^{it}\begin{pmatrix} 1 \\i\end{pmatrix} + C_2e^{-it}\begin{pmatrix} 1 \\-i\end{pmatrix}$$
I know that now I have to solve two Cauchy's problems for the standard basis $\mathbb{R}^2$ with vectors $v_1 = (1, 0)$ and $v_2 = (0,1)$ But I do not know how to approach it for complex numbers
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Let$$T=\begin{pmatrix}1&1\\i&-i\end{pmatrix};$$its columns are the eigenvectors that you found. Then$$T^{-1}.A.T=\begin{pmatrix}i&0\\0&-i\end{pmatrix}.$$Therefore,$$T^{-1}.A^n.T=\begin{pmatrix}i&0\\0&-i\end{pmatrix}^n=\begin{pmatrix}i^n&0\\0&(-i)^n\end{pmatrix}$$and so$$T^{-1}.e^A.T=\sum_{n=0}^\infty\frac1{n!}\begin{pmatrix}i^n&0\\0&(-i)^n\end{pmatrix}=\begin{pmatrix}e^i&0\\0&e^{-i}\end{pmatrix}.$$Therefore,\begin{align}e^A&=T.\begin{pmatrix}e^i&0\\0&e^{-i}\end{pmatrix}.T^{-1}\\&=\begin{pmatrix}\frac{e^i+e^{-i}}2&\frac{e^i-e^{-i}}{2i}\\-\frac{e^i-e^{-i}}{2i}&\frac{e^i+e^{-i}}2\end{pmatrix}\\&=\begin{pmatrix}\cos 1&\sin 1\\-\sin1&\cos1\end{pmatrix}.\end{align}
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|
Prove $5^n + 2 \cdot 3^n - 3$ is divisible by 8 $\forall n\in \mathbb{N}$ (using induction)
Prove $5^n + 2 \cdot 3^n - 3$ is divisible by 8 $\forall n\in \mathbb{N}$
Base case $n = 1\to 5 + 6 - 3 = 8 \to 8 \mid 8 $
Assume that for some $n \in \mathbb{N}\to 8 \mid 5^n + 2 \cdot 3^n - 3$
Showing $8 \mid 5^{n+1} + 2 \cdot 3^{n+1} - 3$
$$5^{n+1} + 2 \cdot 3^{n+1} - 3$$ $$5\cdot 5^n + 2\cdot 3\cdot 3^n - 3$$ $$ (5^n + 2\cdot 3^n - 3) + 4\cdot 5^n + 2\cdot 2\cdot3^n $$ $$ 5\cdot(5^n + 2\cdot 3^n - 3) + 4\cdot 5^n + 2\cdot 2\cdot3^n - 4\cdot(5^n + 2\cdot 3^n - 3)$$
$$ [5\cdot(5^n + 2\cdot 3^n - 3)] - [4\cdot 3^n - 12]$$
The first term divides by 8 but I am not sure how to get the second term to divide by 8.
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If $f(m)=5^m+2\cdot3^m-3,$
$$f(n+2)-f(n)=5^n(5^2-1)+2(3^2-1)3^n$$ which is clearly divisible by $8$
$\implies8|f(n+2)\iff8|f(n)$
Now establish the base cases $f(0),f(1)$
|
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|
If $z=\tan(\frac{x}{2})$, show that $\sin(x)=\frac{2z}{1+z^2}$ So as the title states I've got the following problem:
If $z=\tan(\frac{x}{2})$, show that $\sin(x)=\frac{2z}{1+z^2}$
So I'd guess that this probably involves the formula for half-angles, but that is is a dead-end.
Any suggestions?
Thank you in advance!
|
$$
\sin x = \overbrace{\sin\left( 2\cdot \frac x 2 \right) = 2\sin\left( \frac x 2 \right) \cos\left( \frac x 2 \right)}^{\large\text{the double-angle formula for the sine}}$$
If $z = \tan \frac x 2,$ then what are $\sin \frac x 2$ and $\cos \frac x 2\,$?
In the first quadrant, we can say $\tan = \dfrac \sin \cos$ and $\sin^2+\cos^2 = 1,$ and thereby conclude that
$$
\sin = \frac \tan {\sqrt{1+\tan^2}} \text{ and } \cos = \frac 1 {\sqrt{1+\tan^2}}
$$
so you have
$$
2\cdot \frac z {1+z^2} \cdot \frac 1 {\sqrt{1+z^2}} = \frac {2z}{1+z^2}.
$$
In the fourth quadrant, the denominator will be the same, but the sign of the tangent will agree with the sign of the sine.
And in the other two quadrants, the whole function just repeats.
|
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|
Factor $z^5 + z +1 =0$ The problem stated in the title is:
Factor $z^5 + z + 1 = 0$
(naturally without the use of computers or calculators)
How do I go about solving this? Is there a more systematic approach than simply guessing a root and then applying polynomial division etc. ? Thank you for any help!
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If $z^5 + z + 1 = f(z)g(z)$, then:
*
*$\deg f =1, \deg g=4$, or
*$\deg f =2, \deg g=3$
In both cases, you can try to determine the coefficients. Perhaps simplify things by first trying monic polynomials with independent term $1$.
The second case gives
$$
(z^2+a z+1)(z^3+bz^2+c z+ 1) =
z^5 + (a + b)z^4 + (a b + c + 1)z^3 + (a c + b + 1)z^2 + (a + c)z + 1
$$
Forcing the coefficients of $z^4,z^3,z^2$ to be zero and $a+c=1$, we get
$$z^5 + z + 1 = (z^2 + z + 1) (z^3 - z^2 + 1)$$
|
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|
How many possibilities are there for two full houses to be dealt to two players in one game? Imagine dealing cards from a classic 52 card deck to two poker players.
How many possibilities are there for both of them to be dealt a full house(three cards in same rank and two cards of another rank) in same round?
As I know totally ${13 \choose 1}{4 \choose 3}{12 \choose 1}{4 \choose 2}$ possibilities for one player (AAABB). According to that, I tried to solve my problem as shown below: $${13 \choose 1}{4 \choose 3}{12 \choose 1}{4 \choose 2}{11 \choose 1}{4 \choose 2} + {13 \choose 1}{4 \choose 3}{12 \choose 1}{4 \choose 3}{11 \choose 1}{4 \choose 2}$$ for two players ((AAABB)(AAACC) or (AAABB)(CCCBB)). Is my solution correct or am I missing something?
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There are two possibilities. Either both players receive two cards of the same rank or they do not.
Both players receive two cards of the same rank: There are $\binom{13}{1}$ ways to choose the rank from which player A receives three cards and $\binom{4}{3}$ ways to choose three cards of that rank. That leaves $\binom{12}{1}$ possible ways to choose the rank from which player B receives three cards and $\binom{4}{3}$ ways to choose three cards of that rank. There are $\binom{11}{1}$ ways to choose the rank from which both players receive two cards. There are $\binom{4}{2}$ ways to select which two of the four cards are received by player A. Hence, there are
$$\binom{13}{1}\binom{4}{3}\binom{12}{1}\binom{4}{3}\binom{11}{1}\binom{4}{2}$$
such deals.
The players receive two cards from different ranks: The number of ways to select three cards of one rank for each player is the same as above. There are $\binom{11}{1}$ ways to select the rank from which player A receives two cards and $\binom{4}{2}$ ways to select two of the four cards of that rank. There are $\binom{10}{1}$ ways to select the rank from which player B receives two cards and $\binom{4}{2}$ ways to select two of the four cards of that rank. Hence, there are
$$\binom{13}{1}\binom{4}{3}\binom{12}{1}\binom{4}{3}\binom{11}{1}\binom{4}{2}\binom{10}{1}\binom{4}{2}$$
such deals.
Since the two cases are mutually exclusive and exhaustive, the number of deals in which both players receive a full house is found by adding the results for the above cases.
|
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|
Show that the sequence $a_1=1$, $a_2=2$, $a_{n+2} = (a_{n+1}+a_n)/2$ converges by showing it is Cauchy
Show that the sequence $a_1=1$, $a_2=2$, $a_{n+2} = (a_{n+1}+a_n)/2$ converges by showing it is Cauchy.
My work :
Need to show that for every $\epsilon \gt 0$ there exist $N$ such that $n,m\ge N \implies | a_n - a_m| \lt\epsilon$.
$$|a_n-a_m| = \dfrac{1}{2}|(a_{n-1} + a_{n-2}) - ( a_{m-1} + a_{m-2})|$$
I feel triangle inequality might be helpful here, but really not sure how to link it to the $\epsilon$. Appreciate any help...
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A slightly unorthodox approach to this problem (as I have nothing else to add to Robert's spot on answer) is to solve the recurrence using (for example) characteristic polynomials, which for
$a_{n+2}=\frac{a_{n+1}+a_n}{2}$ is:
$$2x^2-x-1=0$$
which has the following roots $x_1=1$ and $x_2=-\frac{1}{2}$ and the general form
$$a_n=C_1\cdot x_1^n+C_2\cdot x_2^n=C_1+C_2\cdot \left(-\frac{1}{2}\right)^n$$
Using the initial condition $a_1=1, a_2=2$ we have
$$\left\{\begin{matrix}
1=C_1-\frac{C_2}{2}\\
2=C_1+\frac{C_2}{4}
\end{matrix}\right.$$
leading to $C_1=\frac{5}{3}$ and $C_2=\frac{4}{3}$ or
$$a_n=\frac{5}{3}+\frac{4}{3}\cdot \left(-\frac{1}{2}\right)^n$$
and
$$\lim\limits_{n\rightarrow\infty}a_n=\frac{5}{3}$$
|
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Subsets and Splits
Fractions in Questions and Answers
The query retrieves a sample of questions and answers containing the LaTeX fraction symbol, which provides basic filtering of mathematical content but limited analytical insight.