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Integer as sum of 6 positive squares Can every integer be written as sum of exactly $6$ positive squares?
I am also curious to know when an integer can be written as sum of exactly 8 squares.
I know the problem is related to Waring-Hilbert Theorem, but I couldnn't find anything specific.
If it is not possible for every integer then what type of integers can be written as sum of 6 (and 8) squares?
Edit: It appearss to me that I have to be more 'specific'. All squares are non-zero. And by every integer, I mean, integers after a large number, so that 1, 2, 3 ... does not become counterexample!
|
Here's a way to show that every sufficiently large integer is a sum of eight positive squares.
Legendre's three-square theorem guarantees (among much else) that every number congruent to $3$ mod $8$ is the sum of three squares, and it's easy to see that all three squares must be odd, hence nonzero. Any sufficiently large integer $N$ can be reduced to a number $k\equiv3$ mod $8$ by subtracting three squares, each from the set $\{1,4,16\}$ and either $5=1^2+2^2$ or $17=1^2+4^2$. That is, noting that $\{1,4,16\}\equiv\{0,1,4\}$ mod $8$ and $\{5,17\}\equiv\{1,5\}$ mod $8$, we see that
$$\begin{align}
\{0,1,4\}+\{0,1,4\}+\{0,1,4\}+\{1,5\}
&\equiv\{0,1,2,4,5\}+\{0,1,4\}+\{1,5\}\\
&\equiv\{0,1,2,3,4,5,6\}+\{1,5\}\\
&\equiv\{0,1,2,3,4,5,6,7\}\mod 8
\end{align}$$
which covers all the residue classes mod $8$. Thus we can write
$$N=a^2+b^2+c^2+m^2+n^2+x^2+y^2+z^2$$
with $a,b,c\in\{1,2,4\}$, $m^2+n^2\in\{5,17\}$, and $x^2+y^2+z^2\equiv3$ mod $8$.
For example, $68-16-16-16-17=3$, so
$$68=4^2+4^4+4^2+1^2+4^2+1^1+1^1+1^1$$
In fact, since $16+16+16+17=65$ is the largest that need ever be subtracted from $N$ to get a remainder congruent to $3$ mod $8$, the only numbers that cannot be expressed as the sum of eight positive squares must all be less than $65$. It shouldn't be too hard to find the complete list (in particular the largest number not expressible as the sum of eight positive squares), but I don't see any really easy way to do it that doesn't boil down to a boatload of arithmetic. Maybe someone else can.
Remark: This argument almost works for six positive squares as well. That is, $\{0,1,4\}=\{0,1,4\}+\{0,1,4\}\equiv\{0,1,2,3,4,5,6\}$ mod $8$ lacks only a $7$. This implies any (sufficiently large) $N\not\equiv2$ mod $8$ can be written in the form $a^2+b^2+c^2+x^2+y^2+z^2$ with $a,b,c\in\{1,2,4\}$ and $x^2+y^2+z^2\equiv3$ mod $8$. (The set $\{5,17\}$ doesn't help here, since $\{0,1,4\}+\{1,5\}\equiv\{1,2,5,6\}$ lacks $7$ as well.) Maybe some separate argument could account for the missing residue class.
|
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"timestamp": "2023-03-29T00:00:00",
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|
How to solve $\lim_{x\to1}=\frac{x^2+x-2}{1-\sqrt{x}}$? let $f(x)=\dfrac{x^2+x-2}{1-\sqrt{x}}$
How do I solve this limit?
$$\lim_{x\to1}f(x)$$
I can replace the function with its content
$$\lim_{x\to1}\dfrac{x^2+x-2}{1-\sqrt{x}}$$
Then rationalizing the denominator
$$\lim_{x\to1}\dfrac{x^2+x-2}{1-\sqrt{x}}\cdot\dfrac{1+\sqrt{x}}{1+\sqrt{x}}$$
With $(a + b)(a - b) = a^2 - b^2$, I can remove the irrational denominator.
$$\lim_{x\to1}\dfrac{(x^2+x-2)(1+\sqrt{x})}{1-x}$$
I the multiply the two parenthesis
$$\lim_{x\to1}\dfrac{\sqrt{x} \cdot x^2+\sqrt{x}\cdot x- \sqrt{x} \cdot 2 + x^2 + x - 2}{1-x}$$
I'm not sure where to continue to solve this limit.
|
Alternatively you can alwasy factor the hard way:
$\frac {x^2 + x - 2}{-\sqrt x + 1}=$
$\frac {x^{\frac 32}(\sqrt{x} - 1) + x^{\frac 32} + x - 2}{-\sqrt x + 1}=$
$-x^{\frac 32} + \frac {x(\sqrt x-1) + 2x - 2}{-\sqrt x+ 1}=$
$-x^{\frac 32} -x + \frac {2\sqrt x(\sqrt x -1) +2\sqrt x - 2}{-\sqrt x +1}=$
$-x^{\frac 32} - x - 2\sqrt{x}-2$
So $\lim\limits_{x\to 1} \frac {x^2 + x - 2}{-\sqrt x + 1}=\lim\limits_{x\to 1}(-x^{\frac 32} - x - 2\sqrt{x}-2) = -(1^{\frac 32}) - 1 - 2\sqrt 1 - 2= -1-1-2-2 = -6$.
Which is kind of difficult and hard and one ought to have seen and made use of $x^2 +x -2= (x+2)(x-1)$ and somehow noticed that $(x -1) = (\sqrt x+1)(\sqrt x -1)$ to make it easier.
But sometimes we don't see things and it's nice to know that brute force can work when we don't.
........
Or if we just let $y = \sqrt x$ then $\lim_{x\to 1} \frac {x^2 + x - 2}{1-\sqrt x }= \lim\limits_{y^2 \to 1;y > 0}\frac {y^4 + y^2 - 2}{1-y}$ and psychologically that seems easier. We just factor $y^4 + y^2 -2 = (y^2 + 2)(y^2 - 1) = (y^2+2)(y+1)(y-1)$ and the $y-1$s cancel out.
|
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|
Factoring $(x+y+z)^3 xyz - (xz+xy+yz)^3$ and $(x-a)^3(b-c)^3+(x-b)^3(c-a)^3+(x-c)^3(a-b)^3$ I am trying to factor the expressions $(x+y+z)^3 xyz - (xz+xy+yz)^3$ and $(x-a)^3(b-c)^3+(x-b)^3(c-a)^3+(x-c)^3(a-b)^3$.
I am rather stuck though. Is there a general method for going about this? I always find myself having to guess which is not so useful here. I notice in the first one that if $x=y=z$ then the polynomial is zero. I'm not sure how useful this is though in the case of three variables.
|
Note that
$$(x+y+z)^3xyz-(xz+xy+yz)^3=(x^2-yz)(y^2-xz)(z^2-xy)$$
and
$$(x-a)^3(b-c)^3+(x-b)^3(c-a)^3+(x-c)^3(a-b)^3=3 (b-a) (c-a) (a-x) (b-c) (b-x) (c-x)$$
|
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Find all functions such that $f(1+xf(y))=yf(x+y)$ where $x,y \in R^+$
Find all functions run over positive real numbers such that
$f(1+xf(y))=yf(x+y)$ where $x,y\in R^+$
MY ANSWER:
Putting $x=y=0$,we get,
$f(1)=0$
Putting $x=0$ we get,
$f(1)=yf(y)$
or,$yf(y)=0$
or,$f(y)=0$ (since $y\ne 0$., $y \in \mathbb R^+$)
Hence,$f(x)=0$ is the solution.
Is my answer and solution collect? If not then please tell me the proper answer and solution and where I have made the mistake!!
|
For reference, here is CanVQ's solution taken from here.
Replacing $x$ by $\frac{x}{f(y)}$ in $(1),$ we have
$$f(1+x)=y\cdot f\left(\frac{x}{f(y)}+y\right),\quad \forall x ,\,y \in \mathbb R^+. \quad (2)$$
Next, we replace $y$ by $\frac{f(1+x)}{y}$ in $(2)$ to get
$$y=f\left(\frac{x}{f\left(\frac{f(1+x)}{y}\right)}+\frac{f(1+x)}{y}\right),\quad \forall x,\,y \in \mathbb R^+. \quad (3)$$
From $(3),$ it follows that $f$ is surjective. Now, we will prove that $f$ is decreasing. Replacing $x$ by $x+z$ in $(1),$ we get
$$f\big(1+(x+z)\cdot f(y)\big)=y\cdot f(x+y+z),\quad \forall x,\,y,\,z \in \mathbb R^+. \quad (4)$$
Replacing $y$ by $y+z$ in $(1),$ we also have
$$f\big(1+x\cdot f(y+z)\big)=(y+z)\cdot f(x+y+z),\quad \forall x,\,y,\,z \in \mathbb R^+. \quad (5)$$
Dividing $(4)$ for $(5),$ side by side, we obtain
$$\frac{f\big(1+(x+z)\cdot f(y)\big)}{f\big(1+x\cdot f(y+z)\big)}=\frac{y}{y+z},\quad \forall x,\,y,\,z \in \mathbb R^+. \quad (6)$$
Now, assume that there exists a pair $(y,\,z)$ such that $f(y+z)>f(y).$ In this case, by choosing $x=\frac{z\cdot f(y)}{f(y+z)-f(y)}$ vào $(6),$ we obtain $y=y+z,$ which is a contradiction. So we must have
$$f(y+z) \le f(y),\quad \forall y,\,z \in \mathbb R^+. \quad (7)$$
Now, we will prove that $f$ is injective. Assume that there are two numbers $a,\,b$ such that $f(a)=f(b).$ Replacing $y=a$ and $y=b$ in $(1)$ respectively in $(1),$ we get
$$a\cdot f(x+a)=b\cdot f(x+b),\quad \forall x\in \mathbb R^+. \quad (8)$$
From this, it follows that
$$1+a(y-1)\cdot f(x+a)=1+b(y-1)\cdot f(x+b),\quad \forall x, \, y \in \mathbb R^+,\, y>1. \quad (9)$$
Plugging this into $f$ and using $(1),$ we get
$$(x+a)\cdot f(x+ay)=(x+b)\cdot f(x+by),\quad \forall x,\,y \in \mathbb R^+ ,\, y>1. \quad (10)$$
From $(10),$ it follows that, for any $x,\,y,\,z \in \mathbb R^+,\, y>1,$
$$1+(xz+az)\cdot f(x+ay)=1+(xz+bz)\cdot f(x+by). \quad (11)$$
Again, we plug this into $f$ and using $(1).$ It follows that
$$(x+ay)\cdot f(x+ay+az+xz)=(x+by)\cdot f(x+by+bz+xz)\quad (12)$$
for any $x,\,y,\,z \in \mathbb R^+$ and $ y>1.$ On the other hand, according to $(10),$ we also have
$$\big[(x+xz)+a\big]\cdot f\big( (x+xz)+a(y+z)\big)=\big[(x+xz)+b\big] \cdot f\big((x+xz)+b(y+z)\big),$$
or
$$(x+xz+a)\cdot f(x+ay+az+xz)=(x+xz+b)\cdot f(x+ay+az+xz). \quad (13)$$
Dividing $(12)$ for $(13),$ side by side, we obtain
$$\frac{x+ay}{x+xz+a}=\frac{x+by}{x+xz+b},\quad \forall x,\,y,\,z \in \mathbb R^+,\, y>1. \quad (14)$$
It is easy to deduce that $a=b$ here, so $f $ is injective. Now, replacing $x=y=1$ in $(1)$ with notice that $f $ is injective, we have $f(1)=1.$ Since $f$ is strictly decreasing ($f$ is decreasing and injective), we have
$$f(x)<1,\quad \forall x>1.\quad (15)$$
Now, let us consider the case $y>x.$ Replacing $y$ by $y-x$ in $(1)$ and using the above remark, we get
$$f(y)=\frac{f\big(1+x\cdot f(y-x)\big)}{y-x}<\frac{1}{y-x},\quad \forall x,\,y \in \mathbb R^+,\, x<y. \quad (16)$$
In $(16),$ we let $x\to 0^+$ and obtain
$$f(y) \le \frac{1}{y},\quad \forall y >0. \quad (17)$$
Next, replacing $x$ by $x-1$ and $y$ by $f(y)$ in $(3),$ we get
$$y=\frac{x-1}{f\left(\frac{f(x)}{f(y)}\right)}+\frac{f(x)}{f(y)},\quad \forall x ,\, y \in \mathbb R^+,\, x >1. \quad (18)$$
Since $f\left(\frac{f(x)}{f(y)}\right) \le \frac{f(y)}{f(x)},$ from $(18),$ we deduce that
$$y\ge \frac{(x-1)\cdot f(x)}{f(y)}+\frac{f(x)}{f(y)},$$
or
$$y\cdot f(y) \ge x \cdot f(x),\quad \forall x,\,y \in \mathbb R^+,\, x>1. \quad (19)$$
Changin the position of $x$ and $y$ in $(19),$ we also have
$$x\cdot f(x) \ge y\cdot f(y),\quad \forall x,\,y \in \mathbb R^+,\, y>1. \quad (20)$$
From the inequalities $(19)$ and $(20),$ we can easily deduce that
$$f(x)=\frac{k}{x},\quad \forall x>1. \quad (21)$$
Now, taking $x=1$ in $(1)$ and using $(21),$ we have
$$\frac{1}{1+f(y)}=\frac{y}{1+y},$$
or
$$f(y)=\frac{1}{y},\quad \forall y\in \mathbb R^+. \quad (22)$$
Clearly, the function $f(x)=\frac{1}{x}$ satisfies our equation.
|
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Prove that $\frac{x^2y}{1+x^4+y^2}$ has no global minimum I am not sure how to approach this problem. The usual methods do not work to find a minimum.
I can see that, but how to show that there must not exist a minimum?
|
$$f(x,y) = \frac{x^2y}{1+x^4+y^2}$$
$$\frac{\partial f}{\partial x} = \frac{2xy(1+x^4+y^2) - 4x^5y}{(1+x^4+y^2)^2}$$
$$\frac{\partial f}{\partial y} = \frac{x^2(1+x^4+y^2) - 2x^2y^2}{(1+x^4+y^2)^2}$$
$$\frac{\partial f}{\partial x} = 0 \implies 1+x^4+y^2 = 2x^4 \lor x=0 \lor y=0$$
$$\frac{\partial f}{\partial y} = 0 \implies 1+x^4+y^2 = 2y^2 \lor x=0$$
One check that when $x=0$ or $y=0$, the function is zero, so it is not the global minimum.
Otherwise, on solving the equations we have $y=x^2$, and $1+2x^4=2x^4 \implies 1 = 0$.
|
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Show that $\sin (x/2)+2\sin (x/4)\leq3\sqrt{3}/2$ for all $x \in [0,2\pi]$
Given that $$f (x)=\sin (x/2)+2\sin (x/4)$$
Show that $f (x)\leq3\sqrt{3}/2$ for all $x \in [0,2\pi]$.
My try
$f (x)=\sin (x/2)+2\sin (x/4)$
$f' (x)=\frac {1}{2}\cos(x/2)+\frac {1}{2}\cos (x/4)$
$f' (x)=\frac {1}{2}\cos(x/2)+\frac {1}{2}\cos (x/4)$
$f' (x)=\frac {1}{2}(\cos(x/2)+\cos (x/4))$
$f' (x)=\cos(3x/4)\cos (x/4))$
$f'(x)=0$ $\implies$ $\cos(3x/4)\cos (x/4))=0$
$\cos(3x/4)=0\:$ or $\:\cos (x/4)=0$
$3x/4= \pi/2\:$ and so $\:x= 2\pi/3$ or $ x=2\pi$
$x/4= \pi/2\:$ and so $\:x= 2\pi $
Is my work ok ? What should I do now?
|
By AM-GM we obtain:
$$\sin\frac{x}{2}+2\sin\frac{x}{4}=2\sin\frac{x}{4}\left(1+\cos\frac{x}{4}\right)=\frac{2}{\sqrt3}\sqrt{\left(3-3\cos\frac{x}{4}\right)\left(1+\cos\frac{x}{4}\right)^3}\leq$$
$$\leq\frac{2}{\sqrt3}\sqrt{\left(\frac{3-3\cos\frac{x}{4}+3\left(1+\cos\frac{x}{4}\right)}{4}\right)^4}=\frac{3\sqrt3}{2}$$
|
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Solving $\cos(3x) = \cos(2x)$ I'm struggling with solving given trigonometric equation
$$\cos(3x) = \cos(2x)$$
Let's take a look at the trigonometric identities we can use:
$$\cos(2x) = 2\cos^2-1$$
and
$$\cos(3x) = 4\cos^3(x) -3\cos(x)$$
Plugging into the equation and we have that
$$4\cos^3(x) -3\cos(x) = 2\cos^2(x)-1$$
$$4\cos^3(x) -3\cos(x) - 2\cos^2(x)+1= 0$$
Recalling $t = \cos (x)$,
$$4t^3-2t^2-3t +1 = 0$$
Which is a cubic equation. Your sincerely helps will be appreciated.
Regards!
|
The equality $\cos(3x)=\cos(2x)$ is obviously true when $x=0$ and thus when $t=1.$ Therefore the polynomial
$$
4t^3-2t^2-3t +1
$$
has $t=1$ as one of its zeros. Consequently it can be factored:
$$
4t^3-2t^2-3t +1 = (t-1)(\cdots\cdots\cdots)
$$
The other zeros are those of a quadratic polynomial, written here as $(\cdots\cdots\cdots).$
|
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Finding limit using Euler number $$\lim_{x\rightarrow \infty }\left ( 1+\frac{3}{x+2} \right )^{3x-6}$$
I've tried to factor and simplfy the expression. I got:
$${\left ( 1+\frac{3}{x+2} \right )^{\frac{1}{x+2}}}^{3({x^2-4})}$$
I set $x$ to $1/t$ I get:
$${\left ( 1+\frac{3}{\frac{1}{t}+2} \right )^{\frac{1}{\frac{1}{t}+2}}}^{3 \left({\frac{1}{t}^2-4} \right)}$$
then I am left with:
$$\left ( e^{3} \right )^{3\left(\frac{1}{t^2}-4\right)}$$ which I get by using Euler number.
The answer is $e^9$, but clearly the answer I get is $(e^9)^{\text{expression}}$ which is not equal to the answer.
|
$\lim_\limits{x\to\infty}\left(1+\frac {3}{x+2}\right)^{3x-6}$
$y = x+2$
$\lim_\limits{y\to\infty}\left(1+\frac {3}{y}\right)^{3y-12}\\
\lim_\limits{y\to\infty}\left(1+\frac {3}{y}\right)^{3y}\lim_\limits{y\to\infty}\left(1+\frac {3}{y}\right)^{-12}$
Let't attack these separately
$\lim_\limits{y\to\infty}\left(1+\frac {3}{y}\right)^{-12} = 1$
As $y$ gets to be large $\frac {3}{y}$ becomes effectively $0,$ and the limit goes to 1.
$\lim_\limits{y\to\infty}\left(\left(1+\frac {3}{y}\right)^{y}\right)^3\\
\lim_\limits{y\to\infty}\left(1+\frac {3}{y}\right)^{y} = e^3$
$\lim_\limits{y\to\infty}\left(\left(1+\frac {3}{y}\right)^{y}\right)^3 = (e^3)^3 = e^9$
|
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What is the number of ways in which I can select 17 balls from an urn containing 30 balls? I have an urn that contains 30 balls with 10 coloured white, 10 coloured black and the remaining coloured red. Each ball is numbered, from 1 to 10: that is I have red balls numbered 1 to 10, black balls numbered 1 to 10 and white balls numbered 1 to 10. In how many ways can I select 17 balls such that I select a minimum of 4 red balls, 4 black balls and 4 white balls?
I need the quickest method to find the above out. There are long methods of solving it where I find the number of scenarios that violate the condition, such as those scenarios with 3 red balls or no black balls; these I subtract from the total number of ways of choosing 17 balls from the urn.
But is there a quicker way to solve this question?
(EDIT: All balls are now numbered. Please accept my apologies.)
(I will say this: The above problem is not a home-work problem from school.)
|
There are 21 composition of 17:
$17 = 4+4+9 , 17 = 4+5+8, .... , 17=9+4+4$
However, some of have the same structure are equivalent and we may take 3 or 6 of them at a time:
$ 3 \binom{10}{4} \binom{10}{4} \binom{10}{9} +
3 \binom{10}{5} \binom{10}{5} \binom{10}{7} +
3 \binom{10}{6} \binom{10}{6} \binom{10}{5} +
6 \binom{10}{4} \binom{10}{5} \binom{10}{8} +
6 \binom{10}{4} \binom{10}{6} \binom{10}{7} $
|
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Showing that $4b^2+4b = a^2+a$ has no non-zero integer solutions? The problem is
Show that $4b^2+4b = a^2+a$ has no integer solutions where none of $a, b$ are zero.
I have a solution but I think there must be some better ways:
My Solution:
$$4b^2+4b = a^2+a$$
$$(2b+a)(2b-a)+4b-a= 0$$
Now letting $x = 2b + a$ and $y = 2b-a$, we see that $x+y = 4b$. Substituting,
$$xy+\dfrac {x+y}{2}+y=0$$
$$2xy+x+3y=0$$
From this we see that $y|x$, so we can substitute $x = ky$ for some integer $k$
$2ky^2+ky+3y = 0$
$$k = \dfrac {3}{2y+1}$$
From here we get that $y \in \{-2, -1 , 1 \}$, and each of the cases can be checked individually.
|
Left side looks almost like $(2b+1)^2$, indeed adding $1$ we can write $(2b+1)^2=a^2+a+1$. It is known that right side can be a perfect square only in few cases - see Integral value of $n$ that makes $n^2+n+1$ a perfect square.
Specifically, when $a>0$, the right side lies between two consecutive squares: $a^2<a^2+a+1<(a+1)^2,$
so it cannot be itself a square. Similarly for $a<-1$, we have $(a+1)^2< a^2+a+1 < a^2$, same reasoning applies. So it only remains to check $a=-1$.
|
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|
Why doesn't the quadratic equation contain $2|a|$ in the denominator? When deriving the quadratic equation as shown in the Wikipedia article about the quadratic equation (current revision) the main proof contains the step:
$$
\left(x+{\frac {b}{2a}}\right)^{2}={\frac {b^{2}-4ac}{4a^{2}}}
$$
the square root is taken from both sides, so why is
$$\sqrt{4a^2} = 2a$$
in the denominator and not
$$ \sqrt{4a^2} = 2\left |a \right | $$
Could somebody explain this to me? Thank you very much
|
If you put
$x
=\dfrac{-b\pm\sqrt{b^2-4ac}}{2a}
$
into
$ax^2+bx+c$,
since
$x^2
=\dfrac{b^2\mp2b\sqrt{b^2-4ac}+(b^2-4ac)}{4a^2}
=\dfrac{2b^2-4ac\mp2b\sqrt{b^2-4ac}}{4a^2}
$
you get
$\begin{array}\\
ax^2+bx+c
&a\dfrac{2b^2-4ac\mp2b\sqrt{b^2-4ac}}{4a^2}
+b\dfrac{-b\pm\sqrt{b^2-4ac}}{2a}+c\\
&=\dfrac{2b^2-4ac\mp2b\sqrt{b^2-4ac}}{4a}
+\dfrac{-2b^2\pm 2b\sqrt{b^2-4ac}}{4a}+c\\
&=\dfrac{2b^2-4ac\mp2b\sqrt{b^2-4ac}-2b^2\pm 2b\sqrt{b^2-4ac}+4ac}{4a}\\
&=0\\
\end{array}
$
If you use $|a|$,
it won't work
since you can't
combine the terms.
|
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|
For the differential equation given by $(x^2-y^2)dx+3xydy=0$ the general solution is $(x^2+2y^2)^3 =cx^2$
For the differential equation given by $$(x^2-y^2)dx+3xy \ dy=0$$ the general soultion is $$(x^2+2y^2)^3 =cx^2$$ Check this solution by differentiating and rewriting to get the original function.
I solved for $\frac{dy}{dx}$ and got $$\frac{dy}{dx}=\frac{cx-3x(x^2+2y^2)^2} {6y(x^2+2y^2)^2}$$ I was told to solve for $c$ and multiply by $dx$. but I cannot figure out how to get this to the original equation. My teacher said to use implicit differentiation so I solved for $\frac{dy}{dx}$.
|
The hint says you should solve for $c$ before differentiating, so let's do that
$$ \frac{(x^2+2y^2)^3}{x^2} = c $$
Performing implicit differentiation on this form will eliminate $c$
$$ \frac{3x^2(x^2+2y^2)^2\left(2x + 4y\frac{dy}{dx}\right)-2x(x^2+2y^2)^3}{x^4} = 0 $$
Now you can solve for $\frac{dy}{dx}$ and finish the job
|
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|
Reciprocal solutions of a differential equation
I need to show that if $a$ is a constant and $b(x)$ is a function, then
$$y''+\frac{b'(x)}{b(x)}y'-\frac{a^2}{[b(x)]^2}y=0$$
has a pair of linearly independent solutions which are reciprocal and then find them.
I would have thought I could just substitute in $y=u(x)+\dfrac{1}{u(x)}$ but I seem to get nowhere.
Anyway, we have $y'=u'-\dfrac{u'}{u^2}$ and $y''=u''-\dfrac{u''}{u^2}+\dfrac{2(u')^2}{u^3}$. Therefore, we get $$u''-\frac{u''}{u^2}+\frac{2(u')^2}{u^3}+(\frac{b'(x)}{b(x)})(u'-\frac{u'}{u^2})-(\frac{a^2}{[b(x)]^2})(u(x)+\frac{1}{u(x)})=0\,.$$
tidying gives
$$-\frac{u''}{u^2}+\frac{2(u')^2}{u^3}+(\frac{b'(x)}{b(x)})(-\frac{u'}{u^2})-(\frac{a^2}{[b(x)]^2})(\frac{1}{u(x)})=0\,.$$
multiply by $-u^2$ to get
$$u''-\frac{2(u')^2}{u}+(\frac{b'(x)}{b(x)})(u')+(\frac{a^2}{[b(x)]^2})(u(x))=0\,.$$
|
You got so close. Suppose that $y_1=u$ is a solution, then $y_2=1/u$ must also be a solution. Substituting this, we obtain
$$ u'' -2\frac{(u')^2}{u} + \frac{b'}{b}u' + \frac{a^2}{b^2}u = 0 $$
But we already know that
$$ u'' + \frac{b'}{b}u' = \frac{a^2}{b^2}u $$
From the given assumption. Therefore
$$ 2\frac{a^2}{b^2}u - 2\frac{(u')^2}{u} = 0 \implies \frac{u'}{u} = \pm\frac{a}{b} $$
Integrating both sides gives
$$ u(x) = c\exp\left(\pm \int \frac{a}{b(x)} dx\right) $$
where $c$ is some arbitrary constant. Taking the positive and negative signs to be distinct solutions, we can put together the general solution
$$ y(x) = c_1 \exp\left(\int \frac{a}{b(x)}dx \right) + c_2 \exp\left(-\int \frac{a}{b(x)}dx\right) $$
which is indeed a linear combination of two reciprocal functions.
Hope this helps.
|
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|
Find the length of AB in Triangle ABC In $\Delta ABC , m \angle A = 2 m \angle C$ , side $BC$ is 2 cm longer than side $AB$ . $AC = 5 $What is $AB$ ?
Well I thought you can use trigonometry or Complete Pythagoras theorem , but I don't really know how to apply it
|
Let $|AB|=c$, $|BC|=a=c+2$,
$|AC|=b=5$,
$\angle BCA=\gamma$,
$\angle CAB=\alpha=2\gamma$
By the sine rule we have
\begin{align}
\frac{\sin\alpha}{a}
&=
\frac{\sin\beta}{b}
=\frac{\sin\gamma}{c}
,\\
\frac{\sin2\gamma}{c+2}
&=
\frac{\sin(\pi-3\gamma)}{5}
=\frac{\sin\gamma}{c}
.
\end{align}
By the rules based on
componendo and dividendo,
\begin{align}
\frac{\sin2\gamma}{c+2}
&=
\frac{\sin\gamma}{c}
=\frac{\sin2\gamma-\sin\gamma}{c+2-c}
=\frac{\sin2\gamma-\sin\gamma}{2}
,\\
\frac{\sin(3\gamma)}{5}
&=
\frac{\sin2\gamma-\sin\gamma}{2}
,
\end{align}
\begin{align}
2\sin(3\gamma)
-
5\sin2\gamma+5\sin\gamma
&=0
,\\
8\sin\gamma\cos^2\gamma-2\sin\gamma
-10\sin\gamma\cos\gamma+5\sin\gamma
&=0
,\\
8\cos^2\gamma
-10\cos\gamma+3
&=0
,
\end{align}
so $\cos\gamma$
must be either $\tfrac12$ or $\tfrac34$.
It follows that possible values for $\gamma$
are $60^\circ$ or $\arccos\tfrac34\approx41.41^\circ$.
But since $\beta=180^\circ-3\gamma$,
$\gamma=60^\circ$ results in $\beta=0$, a degenerate case,
so the only suitable choice is
\begin{align}
\cos\gamma&=\tfrac34
,\\
\frac{\sin\gamma}c&=
\frac{\sin2\gamma-\sin\gamma}2
,\\
c&=\frac{2\sin\gamma}{\sin2\gamma-\sin\gamma}
,\\
&=
\frac{2\sin\gamma}{2\sin\gamma\cos\gamma-\sin\gamma}
=
\frac{2}{2\cos\gamma-1}
=
\frac{2}{2\cdot\tfrac34-1}
=4.
\end{align}
Thus $\triangle ABC$ has sides $4,5$ and $6$cm.
|
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|
Doubt with an inequality involving integrals I have to prove this inequality about integrals. I did it but I'm not sure if my arguments were correct. Take a look:
Prove:
$$0\leq\int_{-1}^1\frac{1-x^2}{x^4+1}dx\leq\int_{-1}^1\frac{2+x^4}{x^4+1}dx$$
My try is this:
Let $f(x)=\frac{1-x^2}{x^4+1}$ with $f:[-1,1]\to\mathbb{R}$ and $g(x)=\frac{2+x^4}{x^4+1}$ with $g:[-1,1]\to\mathbb{R}$. So we have to prove: $f(x)\leq g(x)$ for all $x\in[-1,1]$.
Clearly we only need to prove that $1-x^2\leq2+x^4$ because the denominator is always a positive number in $[-1,1]$. So we get:
$$0\leq x^2(x^2+1)+1=(x^2+1)(x^2+\frac{1}{x^2+1})$$
Therefore we need:
$$x^2+1\geq0 \land x^2+\frac{1}{x^2+1}\geq0$$
$$\lor x^2+1\leq0 \land x^2+\frac{1}{x^2+1}\leq0$$
The first pair of inequalities are always true for any $x\in\mathbb{R}$, because they involve positive numbers. And the second pair is the empty set by the same reason.
Therefore $f(x)\leq g(x)$ for any $x\in\mathbb{R}$ in particular for $x\in[-1,1]$ so we have: $\int_{-1}^1\frac{1-x^2}{x^4+1}dx\leq\int_{-1}^1\frac{2+x^4}{x^4+1}dx$
The part of the zero is easy because we only need to prove: $0\leq1-x^2$, that is $x\leq|1|$, $-1\leq x\leq1$.
Is the first part correct or it is "forced"? Thanks!
|
$$\int\limits_{-1}^1\frac{2+x^4}{x^4+1}dx-\int\limits_{-1}^1\frac{1-x^2}{x^4+1}dx=\int\limits_{-1}^1\frac{x^4+x^2+1}{x^4+1}dx>0$$
$$\int\frac{1-x^2}{x^4+1}dx=-\int\frac{1-\frac{1}{x^2}}{x^2+\frac{1}{x^2}}dx=-\int\frac{1}{\left(x+\frac{1}{x}\right)^2-2}d\left(x+\frac{1}{x}\right)=$$
$$=\frac{1}{2\sqrt2}\int\left(\frac{1}{x+\frac{1}{x}+\sqrt2}-\frac{1}{x+\frac{1}{x}-\sqrt2}\right)d\left(x+\frac{1}{x}\right)=$$
$$=\frac{1}{2\sqrt2}\ln\frac{x+\frac{1}{x}+\sqrt2}{x+\frac{1}{x}-\sqrt2}\frac{1}{2\sqrt2}+C=\frac{1}{2\sqrt2}\ln\frac{x^2+\sqrt2x+1}{x^2-\sqrt2x+1}+C.$$
Thus, $$\int\frac{1-x^2}{x^4+1}dx=\frac{1}{\sqrt2}\ln\frac{2+\sqrt2}{2-\sqrt2}=\frac{1}{\sqrt2}\ln\frac{\sqrt2+1}{\sqrt2-1}=\sqrt2\ln(\sqrt2+1)>0.$$
|
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|
Finding probability of ball being white using Bayes theorem? From an urn containing 3 white and 5 black balls, 4 balls are transferred into an empty urn. From this urn 2 balls are taken and they both happen to be white. They are then replaced. What is the probability that the third ball taken from the same urn will be white?
The answer seems to be 7/12, but my answer is coming to be 15/28.
Here is my approach so far:
Since 2 balls were found to be white and replaced already, at least 2 balls out of 4 are white, so probability will be >50%.
The two possibilities are: 2 white + 2 black, or 3 white + 1 black.
So the probability of another ball being white will be: 3/4x1/7 + 2/4x6/7 = 15/28
|
Let $X$ denote the number of white balls initially transferred into the urn from which you are drawing balls. Then, before you have drawn any balls, $X$ has a hypergeometric distribution: $$\Pr[X = x] = \frac{\binom{3}{x}\binom{5}{4-x}}{\binom{8}{4}}, \quad x \in \{0, 1, 2, 3\}.$$ After you draw two balls from this urn and note they are white, then it becomes clear that $X \in \{2, 3\}$: We are assured at least two of the four balls is white, and as there were only three white balls from which the second urn is populated, it is impossible to have more than three in the second urn. This much you have noted. However, what is the posterior probability of $X = 2 \mid Y = 2$? Well, Bayes theorem gives us $$\Pr[X = 2 \mid Y = 2] = \frac{\Pr[Y = 2 \mid X = 2]\Pr[X = 2]}{\Pr[Y = 2]}.$$ The numerator is simple: If there were only two white balls in the second urn, the chance that you draw both of them is $1/6$, since there is only one way to pick both white balls out of the $\binom{4}{2}$ ways to take two balls without replacement. So the numerator is simply $$\Pr[Y = 2 \mid X = 2]\Pr[X = 2] = \frac{1}{6} \cdot \frac{\binom{3}{2}\binom{5}{2}}{\binom{8}{4}} = \frac{1}{14}.$$ The denominator is trickier, since we must calculate $\Pr[Y = 2 \mid X = 3]$. In such a case, we can reason that the only way we can fail to obtain $2$ white balls is if one of the balls we pick is the black one, thus $$\Pr[Y = 2 \mid X = 3] = 1 - \frac{3}{6} = \frac{1}{2}.$$ Thus $$\Pr[Y = 2] = \Pr[Y = 2 \mid X = 2]\Pr[X = 2] + \Pr[Y = 2 \mid X = 3]\Pr[X = 3] \\ = \frac{1}{14} + \frac{1}{2}\cdot\frac{\binom{3}{3}\binom{5}{1}}{\binom{8}{4}} = \frac{3}{28}.$$ Therefore, $$\Pr[X = 2 \mid Y = 2] = \frac{1/14}{3/28} = \frac{2}{3},$$ and $$\Pr[X = 3 \mid Y =2] = \frac{1}{3}.$$
We then easily compute that the probability of obtaining a white ball from a single draw from this urn is $$\frac{2}{3}\cdot\frac{2}{4} + \frac{1}{3}\cdot\frac{3}{4} = \frac{7}{12}.$$
|
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|
In how many ways can a committee of $5$ members be formed from $4$ women and $6$ men such that at least $1$ woman is a member of the committee.
In how many ways can a committee of $5$ members be formed from $4$ women and $6$ men such that at least $1$ woman is a member of the committee.
I know that the correct answer is:
$\binom{10}{5} - \binom{6}{5} = 246$
In hindsight (after reading the solution guide), that all makes sense (All Groups - Just those with $5$ Men). However, I can't figure out why my original solution doesn't work. I would choose $1$ of the women, and then choose any $4$ from the remaining $9$ people. This gives us:
$\binom{4}{1}\binom{9}{4} = 504$
Can anyone explain why that doesn't work? Thank you for your help!
|
The number of groups with exactly $k$ of the four women and $5 - k$ of the six men is
$$\binom{4}{k}\binom{6}{5 - k}$$
so the number of groups with exactly $k$ women is
\begin{align*}
\sum_{k = 1}{4} \binom{4}{k}\binom{6}{5 - k} & = \binom{4}{1}\binom{6}{4} + \binom{4}{2}\binom{6}{3} + \binom{4}{3}\binom{6}{2} + \binom{4}{4}\binom{6}{1}\\
& = 4 \cdot 15 + 6 \cdot 20 + 4 \cdot 15 + 1 \cdot 6\\
& = 60 + 120 + 60 + 6\\
& = 246
\end{align*}
as you found more easily by subtracting the number of groups of five people with no women from the number of groups of five people that can be formed without restriction.
Suppose the women are Angela, Brenda, Charlotte, and Denise and that the men are Edward, Frank, George, Henry, Ivan, and Jeffrey. By designating a particular woman as the representative of the women in the group, you count each group the number of times as the number of women in the group. For instance, consider the group consisting of the four women and Edward. You count it in each of the following ways:
\begin{array}{c c}
\text{designated woman} & \text{additional members of the group}\\ \hline
\text{Angela} & \text{Brenda, Charlotte, Denise, Edward}\\
\text{Brenda} & \text{Angela, Charlotte, Denise, Edward}\\
\text{Charlotte} & \text{Angela, Brenda, Denise, Edward}\\
\text{Denise} & \text{Angela, Brenda, Charlotte, Edward}
\end{array}
Notice that
$$\color{red}{\binom{1}{1}}\binom{4}{1}\binom{6}{4} + \color{red}{\binom{2}{1}}\binom{4}{2}\binom{6}{3} + \color{red}{\binom{3}{1}}\binom{4}{3}\binom{6}{2} + \color{red}{\binom{4}{1}}\binom{4}{4}\binom{6}{1} = 504$$
where the factor $\color{red}{\binom{k}{1}}$ is the number of ways of designating one of the $k$ women in the group as the representative of the women in the group.
|
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|
Prove that $5^{2^n} - 1$ is divisible by $2^{n+1}$ for all $n ≥ 1$ I made the following using induction:
If $n=1$ then the proposition is true: $5^{2^1} - 1=24$ is divisible by $2^{1+1} = 4$
Now I suppose that for a natural number $k$, $5^{2^k} - 1$ is divisible by $2^{k+1}$ is true. And I want to prove (using this) that the proposition is true for $n=k+1$ but I don't know how to do this.
I appreciate the help you give me.
|
First assume that $5^{2^k} - 1$ is divisible by $2^{k + 1}$ so $2^{k + 1}\,|\,5^{2^k} - 1$. Now consider
$$5^{2^{k + 1}} - 1 = 5^{2\cdot 2^k} - 1 = (5^{2^k})^2 - 1 = (5^{2^k} - 1)(5^{2^k} + 1)$$
Now we need to show that $2^{(k + 1) + 1} = 2^{k + 2}$ divides $(5^{2^k} - 1)(5^{2^k} + 1)$. Our inductive assumption allows to know that $2^{k + 1}\,|\, 5^{2^k} - 1$. Now we just need to show that $2 \,|\, 5^{2^k} + 1$. Well $5^{2^k}$ will always be an odd number, so adding one makes it even and hence $2 \,|\, 5^{2^k} + 1$.
Since $2$ and $2^{k + 1}$ divide $5^{2^k} + 1$ and $5^{2^k} - 1$, then $2\cdot 2^{k + 1} = 2^{k + 2} \,| 5^{2^{k + 1}} - 1\,$.
By induction then $2^{n+ 1}\,|\, 5^{2^n} - 1$ for all $n \in \mathbb{N}$.
|
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|
Determine Minimum Value.
Find the minimum value of
$$\frac{(x+1/x)^6-(x^6+1/x^6)-2}{(x+1/x)^3+(x^3+1/x^3)}$$
for $x>0$.
When $x=1$, $$\frac{(x+1/x)^6-(x^6+1/x^6)-2}{(x+1/x)^3+(x^3+1/x^3)}=6$$
I tried to plot some points on a graph and I observed that the minimum value is $6$.
Any hints would be sufficient.
Thanks
I think differentiation would be really complicated
|
A nice trick:
As $x>0$, $$f(x)=\frac{\left(x+\cfrac1x\right)^6-\left(x^6+\cfrac1{x^6}\right)-2}{\left(x+\cfrac1x\right)^3+\left(x^3+\cfrac1{x^3}\right)}=\frac{6x^4+15x^2+18+\cfrac{15}{x^2}+\cfrac6{x^4}}{2x^3+3x+\cfrac3x+\cfrac2{x^3}}$$ giving $$f(x)=\frac{6x^8+15x^6+18x^4+15x^2+6}{2x^7+3x^5+3x^3+2x}=\frac{6x^8+9x^6+9x^4+6x^2}{2x^7+3x^5+3x^3+2x}+\frac{6x^6+9x^4+9x^2+6}{2x^7+3x^5+3x^3+2x}$$ which boils down to $$f(x)=3x+\frac3x.$$
|
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|
Correct way of finding $\delta $ for $\lim_{x \to a} \sqrt{x} = \sqrt{a}$ Prove: $\lim_{x \to a} \sqrt{x} = \sqrt{a}$ using $\epsilon$-$\delta$.
My solution:
We have that $0 < |x-a| < \delta $.
Also, $ |\sqrt{x} - \sqrt{a}| < \epsilon $
$ \therefore -\epsilon < \sqrt{x} - \sqrt{a} < \epsilon $
$ \therefore \sqrt{a} -\epsilon < \sqrt{x} < \sqrt{a} + \epsilon $
$ \therefore (\sqrt{a} -\epsilon)^2 < x < (\sqrt{a} + \epsilon)^2 $
$ \therefore (\sqrt{a} -\epsilon)^2 - a < x - a < (\sqrt{a} + \epsilon)^2 - a $
$ \therefore \epsilon^2 - 2\sqrt{a}\epsilon < x - a < \epsilon^2 + 2\sqrt{a}\epsilon $
$ \delta = \min(|\epsilon^2 - 2\sqrt{a}\epsilon|,|\ \epsilon^2 + 2\sqrt{a}\epsilon|) $
By letting $ \delta = \min(|\epsilon^2 - 2\sqrt{a}\epsilon|,|\ \epsilon^2 + 2\sqrt{a}\epsilon|) $, we get that $|\sqrt{x}-\sqrt{a}|<\epsilon$ if $0 < |x-a| < \delta $.
Thus, $\lim_{x \to a} \sqrt{x} = \sqrt{a}$.
Question: is it correct to find a $ \delta $ using the method above?
I've seen a bunch of other methods used to find $\delta$ but I'm totally confused what is the "right" way of doing it. If anyone could explain this or give some other examples, I would appreciate it.
Thank you a lot!
|
The OP's logic can be salvaged. One problem is they assume on their third $\therefore$ that $u \lt v$ implies $u^2 \lt v^2$ without knowing for sure that $u \ge 0$.
Here is a 'logic patch':
If $a = 0$, simple algebra shows that $\delta = \varepsilon^2$ works.
To show continuity when $a \gt 0$, we can take a few easy steps:
$\tag 1 |\sqrt{x} - \sqrt{a}| < \varepsilon $
$ \therefore -\varepsilon < \sqrt{x} - \sqrt{a} < \varepsilon $
$ \therefore \sqrt{a} -\varepsilon < \sqrt{x} < \sqrt{a} + \varepsilon $
Now to get rid of $\sqrt x$ and replace it with $x$, we would like to run the 'square it' on our expression, but to do that we must assume that $\sqrt{a} -\varepsilon$ in non-negative. i.e. $\varepsilon \le \sqrt{a}$. But that is not a restriction at all, so we proceed:
$ \therefore (\sqrt{a} -\varepsilon )^2 < x < (\sqrt{a} + \varepsilon )^2 $
$ \therefore -2 \sqrt{a} \, \varepsilon + \varepsilon^2 < x - a < 2 \sqrt{a} \, \varepsilon + \varepsilon^2$
We are trying to find our $\delta \gt 0$ 'setup':
$\tag 2 -2 \sqrt{a} \, \varepsilon + \varepsilon^2 \le -\delta < x - a < \delta \le 2 \sqrt{a} \, \varepsilon + \varepsilon^2$
so that $\text{(2)}$ implies $\text{(1)}$ (we can use our developed "$\therefore$-logic-chain" in reverse).
Since $\varepsilon \le \sqrt{a}$ here, we can define our $\delta \gt 0$ with
$\tag 3 \delta = 2 \sqrt{a} \, \varepsilon - \varepsilon^2$
and so by definition the left side of $\text{(2)}$ 'clicks'. For the right side, it is easy to see that $\delta \le 2 \sqrt{a} \, \varepsilon + \varepsilon^2$ since
$\quad (2 \sqrt{a} \, \varepsilon - \varepsilon^2) \le (2 \sqrt{a} \, \varepsilon + \varepsilon^2) \text{ iff } 2 \varepsilon^2 \ge 0$
So we can use the OP's logic to find a $\delta$ given any starting $\varepsilon$, where if necessary we pound $\varepsilon$ down first to $\sqrt a$.
|
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|
Minimum $e$ where $a,b,c,d,e$ are reals such that $a+b+c+d+e=8$ and $a^2+b^2+c^2+d^2+e^2=16$ I have a question about this 1978 USAMO problem:
Given that $a,b,c,d,e$ are real numbers such that $a+b+c+d+e=8$ and $a^2+b^2+c^2+d^2+e^2=16$, find the maximum value that $e$ can attain.
I had the following solution:
Let $a+b+c+d=x$. Then $x+e=8\implies e=8-x$. Also, $\frac{a^2+1}{2}+\frac{b^2+1}{2}+\frac{c^2+1}{2}+\frac{d^2+1}{2}\geq (a+b+c+d)=x$ by AM-GM inequality. Hence, $a^2+b^2+c^2+d^2\geq 2x-4$.
Now we have $2x-4+e^2\leq a^2+b^2+c^2+d^2+e^2=16$. Substituting $x=8-e$, we get $e^2-2e-4\leq 0$. We can easily calculate that the lowest value that $e$ can attain is $1 -\sqrt{5}$.
However, the answer given on the internet is $\frac{16}{5}$. Where am I going wrong?
EDIT $1$ -- Is this a case of how the value $1-\sqrt{5}$ can never be attained by $e$, although the inequality is true?
EDIT $2$ -- It seems that we need to find the maximum. By my method, I’ve found the maximum to be $1+\sqrt{5}$. This is greater than $\frac{16}{5}$. Have I found a sharper inequality?
|
By the RMS-AM inequality:
$$
\dfrac{a^2+b^2+c^2+d^2}{4} \ge \left(\dfrac{a+b+c+d}{4}\right)^2 = \dfrac{(8-e)^2}{16} \tag{1}
$$
Therefore:
$$
e^2 = 16-(a^2+b^2+c^2+d^2) \le 16 - \dfrac{(8-e)^2}{4} \;\;\iff\;\; e(5e-16) \le 0 \tag{2}
$$
Equality is in fact attained for $\,e = \dfrac{16}{5}\,$ and $\,a=b=c=d\,$.
Where am I going wrong?
Besides miscopying the problem as a "minimum" rather than "maximum", the equality case in your first step ("by AM-GM") is attained for $\,a=b=c=d=1\,$, but that's not necessarily where the global maximum for $\,e\,$ is attained.
[ EDIT ] Condition $\,(2) \iff e \in \left[0, \,16/5\right]\,$, which establishes both the minimum $\,e=0\,$ and maximum $\,e=16/5\,$, both attained when the other variables are equal $\,a=b=c=d\,$ so that the RMS-AM inequality $\,(1)\,$ becomes an equality. This should answer both the question referenced in the title (about the maximum $e$) as well as the posted variation (about the minimum $e$).
|
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|
If the minimum and the maximum values of $y = (x^2-3x+c)/(x^2+3x+c)$ If the minimum and the maximum values of $y = (x^2-3x+c)/(x^2+3x+c)$ are $7$ and $1/7$ , then the value of $c$ is?
I cross multiplied the equation and tried to find it's discriminant, but I don't think it gets me anywhere. A little hint would be appreciated!
|
Solution
First, we should constrain the value of $c$ such that $x^2+3x+c \neq 0$ for all $x \in \mathbb{R}.$ Otherwise, there necessarily exists at least one infinite discontinuity for $y=f(x)$, and if so, there exists no minimum or maximum value for $y$. For this purpose, let $\Delta=9-4c<0$, i.e. $c>\dfrac{9}{4},$ which is enough.
Under this constraint, $y=f(x)$ is continuous over $(-\infty,+\infty)$. It's clear that the maximum and minimum value given can only be reached at the local extremum point.
Now,notice that $$y'=f'(x)=\frac{6(x^2-c)}{(x^2+3x+c)^2}.$$Let $y'=0$. Then $$x=\pm\sqrt{c}.$$
Hence,$$f(\sqrt{c})=\frac{2\sqrt{c}-3}{2\sqrt{c}+3}=1-\frac{6}{2\sqrt{c}+3}<1,~~~~~f(-\sqrt{c})=\frac{2\sqrt{c}+3}{2\sqrt{c}-3}=1+\frac{6}{2\sqrt{c}-3}.$$
Hence, $f(\sqrt{c})=\dfrac{1}{7},$ then $$c=4.$$
Finally,we may verify that $c=4$ could satisfy all the conditions. We are done.
|
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|
Error Correcting Polynomial Code I apologize for the screenshot, it's quite a long passage. I've linked the textbook below, page number 428-429, example 16.4.3 http://faculty.uml.edu/klevasseur/ads2
I constructed the power table for $GF(8)$, but honestly I'm extremely confused on the two things being asked of me:
Suppose the four message bits are $1, 0, 0, 1$. What seven bits should be transmitted?
Suppose the seven received bits are $1, 0, 1, 0, 1, 0, 1$. What was the original four-bit message?
Any tips on how to go about this would be great. TIA.
|
$\newcommand{\b}{\beta}$We bear in mind $\b^3 = \b + 1$.
Given $1001$, we need to reduce $1 \cdot \b^3 + 0 \cdot \b^4 + 0 \cdot \b^5 + 1 \cdot \b^6$.
We compute $\b^6+\b^3 = \b^4+\b^3+\b^3 = \b^4 = \b^2 + \b$, so we transmit $0111001$.
We received $1010101$.
We check if $\b^0+\b^2+\b^4+\b^6 = 0$:
$\b^0+\b^2+\b^4+\b^6 = \b^0+\b^2+\b^4+\b^4+\b^3 \\ = \b^0+\b^2+\b^3 = \b^0+\b^2+\b^1+\b^0 = \color{red}{\b^2+\b^1}$
So we need to check which power of $\beta$ gives $\b^2+\b^1$: well, $\b^2+\b^1 = \b(\b+1) = \b^4$.
So our original message is $0001$ instead of $0101$ ($0101$ is the last 4 bits of the received code).
|
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|
Triples $(x, y, z)$ that satisfy a set of equations
Suppose that $a$ is a fixed (but unknown) real number, with $a^2 \neq 1$. Determine all triples $(x, y, z)$ of real numbers that satisfy the system of equations:
$x + y + z = a$
$xy + yz + xz = -1$
$xyz = -a$
I've tried making substitutions but don't seem to be able to make much progress. Another thing I noticed is that:
$(x + y + z) = a \implies$
$(x + y + z)^2 = a^2 \implies$
$x^2 + y^2 + z^2 + 2(xy + yx + xz) = a^2 \implies$
$x^2 + y^2 + z^2 + 2(-1) = a^2 \implies$
$x^2 + y^2 + z^2 = a^2 + 2$
but I don't know if that actually is useful. Any suggestions on how to move forward?
|
A more systematic way: Use the third equation to eliminate $z$ in the second equation,
$$xy-a/x -a/y = -1$$
$$z = -a/xy \Rightarrow x+y-\frac{a}{xy} = a$$
Now spot $-a/x - a/y = -a\frac{x+y}{xy}$.
So the first equation can be written as $$a(x+y) = (xy)^{2}+(xy)$$
Getting rid of $x+y$, the second equation becomes
$$(xy)^{2}+(xy)-a^{2}/(xy) = a^{2}$$
Aha! This is a cubic equation in $w=xy$. Precisely,
$$w^{3}+w^{2}-a^{2}w-a^{2} = w^{2}(w+1)-a^{2}(w+1) = (w^{2}-a^{2})(w+1)=0$$
We have three possibilities: $w=a$,$w=-a$ and $w=-1$.
Note that $zw=-a$, so we can just consider the possibilities $z=1,-1,a$. If $z$ is known, then
$$y=a-z-x \Rightarrow x(a-z-x) +z(a-z) = -1$$
which is a quadratic in $z$. The rest is enumeration, good luck!
|
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|
Evaluate the Determinants A? Evaluate the Determinants
$$A=\left(\begin{matrix}
1 & 1 & 0 & 0 & 0\\
-1 & 1 & 1 & 0 & 0\\
0 & -1& 1 & 1 & 0\\
0& 0 & -1 & 1 & 1\\
0 & 0 & 0 & -1 & 1\\
\end{matrix}\right)$$
My attempst : I was thinking abouts the
Schur complement https://en.wikipedia.org/wiki/Schur_complement
$\det \begin{pmatrix}
A&B\\
C&D
\end{pmatrix}= \det (A-BD^{-1}C)\det D
$
As i not getting How to applied this formula and finding the Determinant of the Given question,,
Pliz help me or is There another way to find the determinant of the matrix.
Thanks u
|
After adding the first line to the second one, your matrix becomes$$\begin{pmatrix}1 & 1 & 0 & 0 & 0\\ 0 & 2 & 1 & 0 & 0\\ 0 & -1& 1 & 1 & 0\\ 0& 0 & -1 & 1 & 1\\ 0 & 0 & 0 & -1 & 1\end{pmatrix}$$and therefore$$\det A=\begin{vmatrix}2 & 1 & 0 & 0\\ -1& 1 & 1 & 0\\ 0 & -1 & 1 & 1\\ 0 & 0 & -1 & 1\end{vmatrix}=2\begin{vmatrix}1 & \frac12 & 0 & 0\\ -1& 1 & 1 & 0\\ 0 & -1 & 1 & 1\\ 0 & 0 & -1 & 1\end{vmatrix}.$$Can you take it from here?
|
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|
Difficulty finding Lagrange multiplier because of $\leq$
Let $f: \mathbb R^3 \to \mathbb R$ be defined by $$f(x,y,z)=x-y+z$$
and $$E:=\{(x,y,z)\in \mathbb R^{3} \mid x^2+2y^2+2z^2\leq1\}$$ Find the extrema of $f$ on $E$.
Path: I have already proven that $E$ is compact and then defined $g_\lambda:=f-\lambda h$ where $h:\mathbb R^{3} \to \mathbb R, h(x,y,z)=x^2+2y^2+2z^2-1$ and $\lambda \in \mathbb R$
Setting $0=\nabla g_{\lambda}(x,y)=\begin{pmatrix}
1-2\lambda x\\
-1-4\lambda y\\
1-4\lambda z
\end{pmatrix}$ while $x^2+2y^2+2z^2\leq1$(*)
It follows that: $\frac{1}{2}=\lambda x, -\frac{1}{4}=\lambda y,\frac{1}{4}=\lambda z$
And this is where I get stuck: and using (*) does not help that much as all we gain from it is that: $x\leq 1, y\leq\frac{1}{\sqrt{2}}$ and $z\leq\frac{1}{\sqrt{2}}$
We could of course look at $(0,0,1),(0,\frac{1}{\sqrt{2}},0),(0,0\frac{1}{\sqrt{2}})$
But what about all the points in between? Aren't there so many points to cover in order to assess the extrema. Is there any standardized way about going about this?
That is why the "$\leq$" is causing trouble.
|
You can handle the inequality condition introducing a slack variable $\epsilon$ such that
$$
E\to x^2+2y^2+z^2+\epsilon^2 = 1
$$
and now the lagrangian formulation gives
$$
L(x,y,z,\lambda,\epsilon) = f(x,y,z) + \lambda(x^2+2y^2+z^2+\epsilon^2 -1)
$$
and the stationary conditions give
$$
\nabla L = \left\{
\begin{array}{rcl}
2 \lambda x+1=0 \\
4 \lambda y-1=0 \\
2 \lambda z+1=0 \\
\epsilon ^2+x^2+2 y^2+z^2-1=0 \\
2 \epsilon \lambda =0 \\
\end{array}
\right.
$$
and solving gives
$$
\left[
\begin{array}{cccccc}
x & y & z & \lambda & \epsilon & f\\
\sqrt{\frac{2}{5}} & -\frac{1}{\sqrt{10}} & \sqrt{\frac{2}{5}} & -\frac{\sqrt{\frac{5}{2}}}{2} & 0 & \sqrt{\frac{5}{2}} \\
-\sqrt{\frac{2}{5}} & \frac{1}{\sqrt{10}} & -\sqrt{\frac{2}{5}} & \frac{\sqrt{\frac{5}{2}}}{2} & 0 & -\sqrt{\frac{5}{2}} \\
\end{array}
\right]
$$
Analyzing the solution we conclude that it is at the ellipsoid surface because $\epsilon = 0$
|
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|
What can we say about the series $\sum_{n=1}^{\infty}n\cdot (1-\cos(\frac{a}{n}))$ Series $\sum_{n=1}^{\infty}n\cdot (1-\cos(\frac{a}{n}))$
$$\sum_{n=1}^{\infty}n\cdot(1-\cos(\frac{a}{n}))=\sum_{n=1}^{\infty}n\cdot[1- \{1-\frac{a^2}{2!\cdot n^2} + O(\frac{1}{n^4})\}] = \sum_{n=1}^{\infty} \frac{a^2}{2n} + O(\frac{1}{n^3})$$
Can we conclude with Limit comparison test with $\sum \frac1n$, that the given series is divergent?
Also can we say anything about its oscillatory nature?
|
Another way, using
$\frac{2x}{\pi}
\le \sin(x)
\le x$
for
$0 \le x \le \pi/2$:
$\begin{array}\\
\sum_{n=1}^{m}n(1-\cos(\frac{a}{n}))
&=\sum_{n=1}^{m}n(2\sin^2(\frac{a}{2n}))\\
&=2\sum_{n=1}^{m}n\sin^2(\frac{a}{2n})\\
&\ge 2\sum_{n=1}^{m}n(\frac{2a}{2\pi n})^2
\qquad\text{for }\frac{a}{2n} \le \frac{\pi}{2}
\text{ or }n \ge \frac{a}{\pi} \\
&\ge 2\sum_{n=\lceil \frac{a}{\pi} \rceil}^{m}n(\frac{a}{\pi n})^2\\
&=\frac{2a^2}{\pi^2}\sum_{n=\lceil \frac{a}{\pi}\rceil}^{m}\frac1{n}\\
\end{array}
$
and this diverges
by comparison with
$\sum \frac1{n}$.
|
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|
Series Convergence of Harmonic Means
Let $\{x_n\}$ be a sequence of real numbers such that $0< x_1 <x_2$. If $$x_n= \frac{2}{\frac{1}{x_{n-1}}+\frac{1}{x_{n-2}}}, $$then show that
$$\lim_{n\to\infty}x_n=\frac{3x_1x_2}{2x_1+x_2}.$$
|
Solution
Since $$x_n = \frac{2}{\dfrac{1}{x_{n-1}} + \dfrac{1}{x_{n-2}}},$$
then $$ \frac{2}{x_{n}}=\frac{1}{x_{n-1}} + \frac{1}{x_{n-2}}.$$
Thus, $$2y_n=y_{n-1}+y_{n-2},$$ where $y_n=\dfrac{1}{x_n}$.
From $(1)$, we may obtain $$y_n-y_{n-1} = -\frac{1}{2}\left(y_{n-1}-y_{n-2}\right),$$ for $n=3,4,\cdots.$
Hence，$$y_n-y_{n-1}=(y_2-y_1)\left(-\frac{1}{2}\right)^{n-2}.$$
Thus, $$y_n-y_1=\sum_{k=2}^n(y_k-y_{k-1})=\sum_{k=2}^n(y_2-y_1)\left(-\frac{1}{2}\right)^{k-2}=(y_2-y_1)\cdot \frac{2}{3}\left[1-\left(-\frac{1}{2}\right)^{n-1}\right]$$
Thus, $$y_n=y_1+(y_2-y_1)\cdot \frac{2}{3}\left[1-\left(-\frac{1}{2}\right)^{n-1}\right].$$
Let $n \to \infty$. Then$$y_n \to y_1+(y_2-y_1)\cdot \frac{2}{3}=\frac{1}{3}y_1+\frac{2}{3}y_2=\frac{1}{3x_1}+\frac{2}{3x_2}=\frac{2x_1+x_2}{3x_1x_2}.$$
As a result, $$x_n=\frac{1}{y_n} \to \frac{3x_1x_2}{2x_1+x_2}.$$
|
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|
Area of a quadrilateral inside a square In the image, the segments inside the square go from a vertex to the middle point of the opposite side. If the length of the sides of the square is $1$, the area of $ABCD$ is?
Any hints?
I tried some things matching the bigger triangles but i can't get the area of $ABCD$.
|
If we move the figure to a $xy$ axis, we can solve this problem by finding the line equations and intersection points. Consider this image:
Since we know the point coordinates in the image, we can easily find the line equation of lines 1, 2 and 3 (we can substitute the point values in the slope-intercept form equation $y = ax + b$ and find the equation of each line).
For line 1, the line equation is $y = 2x$.
For line 2, the line equation is $y = \frac{2 - x}{2}$
For line 3, the line equation is $y = \frac{1 - x}{2}$
To find point A, we have to find the intersection point of lines 1 and 3:
$$
2x = \cfrac{1 - x}{2} \Rightarrow x = \cfrac{1}{5} \;and \; y = \cfrac{2}{5} \Rightarrow A = \left(\cfrac{1}{5}, \cfrac{2}{5}\right)
$$
We can do the same with lines 1 and 2 in order to find point B:
$$
2x = \cfrac{2 - x}{2} \Rightarrow x = \cfrac{2}{5} \; and \; y = \cfrac{4}{5} \Rightarrow B = \left(\cfrac{2}{5}, \cfrac{4}{5}\right)
$$
The side of the square (we can call it $L$) will then be the distance $\overline{AB}$:
$$
L = \overline{AB} = \sqrt{(x_B - x_A)^2 + (y_B - y_A)^2 } = \sqrt{\cfrac{1}{25} + \cfrac{4}{25}} = \sqrt{\cfrac{1}{5}}
$$
Finally, the area of the square will be:
$$
Area = L^2 = \cfrac{1}{5}
$$
|
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|
Minimum value of $\frac{b+1}{a+b-2}$
If $a^2 + b^2= 1 $ and $u$ is the minimum value of the $\dfrac{b+1}{a+b-2}$, then find the value of $u^2$.
Attempt:
Then I tried this way: Let $a= bk$ for some real $k$.
Then I got $f(b)$ in terms of b and k which is minmum when $b = \dfrac{2-k}{2(k+1)}$ ... then again I got an equation in $k$ which didn't simplify.
Please suggest an efficient way to solve it.
|
Note that $$u=\frac{b+1}{\sqrt{1-b^2}+b-2}$$ so $$\frac{du}{db}=\frac{1\big(\sqrt{1-b^2}+b-2\big)-(b+1)\left(-\frac{2b}{\sqrt{1-b^2}}+1\right)}{\big(\sqrt{1-b^2}+b-2\big)^2}$$ and setting to zero gives $$-3\sqrt{1-b^2}+b+1=0\implies 1-b^2=\frac{b^2+2b+1}9\implies 5b^2+b-4=0$$ and we see that $b=4/5,-1$ are roots.
Checking second derivatives, we have that $4/5$ is a minimum.
Hence $$u^2=\left(\frac{\frac45+1}{\frac35+\frac45-2}\right)^2=9.$$
Note that the negative root ($-3/5$) is also possible, but that yields a lower value of $u^2$ since $$\bigg|-\frac35+\frac45-2\bigg|>\bigg|\frac35+\frac45-2\bigg|$$
|
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|
Minimizing in 3 variables
Find the least possible value of the fraction $\dfrac{a^2+b^2+c^2}{ab+bc}$, where $a,b,c > 0$.
My try:
$a^2+b^2+c^2 = (a+c)^2 - 2ac + b^2$,
$= (a+c)/b +b/(a+c) -2ac/b(a+c)$
AM > GM
$3\sqrt[3]{-2ac/b(a+c)}$
And I cant somehow move on.
|
Write
$$\frac{2a^2+b^2+b^2+2c^2}{2}=\frac{2a^2+b^2}{2}+\frac{b^2+2c^2}{2}\geq \sqrt{2}ab+\sqrt{2}bc$$
|
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|
Trigonometry and quadratics : Possible mismatch? There’s this problem I came across, gives me an invalid answer by using general quadratic formula. Wonder why?
$2\sin^2{x} -5\cos{x} -4 =0 $
Here’s what I did:
$2\sin^2{x} -5\cos{x} -4 =0 $
$2(1-\cos^2{x}) - 5 \cos{x} - 4 = 0$
$2 \cos^2{x} + 5 \cos{x} + 2 =0$
This is a quadratic function of $\cos x$, thus,
$\cos{x} = (-5 + 3)/4$ or $\cos{x}= (-5 - 3)/4$
But, the answer given is $\cos{x}=-\frac{1}{2} $ and WolframAlpha says the same but doesn’t show how to do it.
What did I do wrong?
Update: Thank you very much, everyone. Turns out that I wrote the squareroot of 9 as squareroot of 3. My bad
|
Yes from here (you are right) we have:
$$2\sin^2x -5\cos x -4 =0 \implies 2\cos^2 x+5\cos x +2=0$$
that is by quadratic formula (here was your mistake):
$$\cos x = \frac{-5\pm \sqrt{25-16}}{4}= \frac{-5\pm 3}{4}\implies \cos x=-\frac12$$
|
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|
Calculate $\lim_{n\to\infty} \sqrt{5n^2+4}~-~\sqrt{5n^2+n}$
Calculate:$$\lim_{n\to\infty} \sqrt{5n^2+4}~-~\sqrt{5n^2+n}$$
Even though I know how to handle limits like this, I would be interested in other ways to approach to tasks similar to this. My own solution will be at the bottom.
My own solution
$$\lim_{n\to\infty} \sqrt{5n^2+4}~-~\sqrt{5n^2+n}~\frac{\sqrt{5n^2+4}~+~\sqrt{5n^2+n}}{\sqrt{5n^2+4}~+~\sqrt{5n^2+n}}$$
$$\lim_{n\to\infty} \frac{5n^2+4~-~(5n^2+n)}{\sqrt{5n^2+4}~+~\sqrt{5n^2+n}}~=~\lim_{n\to\infty} \frac{4-n}{n\left(\sqrt{5+\frac4{n^2}}~+~\sqrt{5+\frac1{n}}\right)}$$
$$=~\frac{-1}{2\sqrt{5}}~=~-\frac{\sqrt{5}}{10}$$
|
Two alternative ideas:
(1) Write
$$
\sqrt{5n^2+4} - \sqrt{5n^2+n} =
\sqrt 5n(\sqrt{1 + 4/5n^2} - \sqrt{1 + 1/5n})
$$
and use Taylor ($\sqrt{1 + x} = 1 + x/2 - x^2/8 + \cdots$).
(2) Let be $f(x) = \sqrt x$. By the Mean Value Theorem, for some $c_n\in(5n^2 + 4,5n^2 + n)$
$$
\sqrt{5n^2 + 4} - \sqrt{5n^2 + n} =
f'(c_n)((5n^2 + 4) - (5n^2 + n)) =
-\frac{4 - n}{2\sqrt{c_n}} =
-\frac{4/n - 1}{2\sqrt{c_n/n^2}},
$$
and by sqeezing $c_n/n^2\to 5$.
|
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|
How to prove that $\pi=\pi$? I am trying to prove that indeed $\pi=\pi$. More precisely, that:
$$6\sum_{n=0}^\infty \frac{(2n)!}{4^n (n!)^2 (2n+1) 2^{2n+1}}=\pi$$
Where the definition of $\pi:$
$$\pi=4\sum_{n=1}^\infty\frac{(-1)^{(n+1)}}{(2n-1)}$$
Using the epsilon delta definition, we should prove:
$$\forall \epsilon_+\exists\delta\forall k(k>\delta\rightarrow|3+6\sum_{n=1}^k \frac{(2n)!}{4^n (n!)^2 (2n+1) 2^{2n+1}}-4\sum_{n=1}^k\frac{(-1)^{(n+1)}}{(2n-1)}|<\epsilon)$$
Re-arranging the sum, it can be written as:
$$\sum_{n=1}^k\frac{(-1)^{(n+1)} 4^{n+1} (n!)^2 (2n+1) 2^{(2n+1)}-6(2n)!(2n-1)}{(4n^2-1) 4^n (n!)^2 2^{2n+1} }-3$$
But from this point it's difficult to go on and find delta as a function of epsilon. What techniques should be used to prove the identity? Looking at a graph I think it's clear that $\delta=\text{ceil}(\frac{1} {\epsilon})$ would satisfy the criteria, but proving it is a other matter.
|
Consider the expression
\begin{eqnarray*}
\sum_{n=0}^{\infty} \binom{2n}{n} \frac{x^{2n+1}}{2n+1}.
\end{eqnarray*}
Now use
\begin{eqnarray*}
\frac{x^{2n+1}}{2n+1}= \int_0^x x^{2n} dx
\end{eqnarray*}
and
\begin{eqnarray*}
\sum_{n=0}^{\infty} \binom{2n}{n} x^{2n} = \frac{1}{\sqrt{1-4x^2}}.
\end{eqnarray*}
Invert the order of the plum & integral and the expression becomes
\begin{eqnarray*}
\int_0^x \frac{1}{\sqrt{1-4x^2}} dx.
\end{eqnarray*}
This integral can be done by the substition $2x=\sin( \theta)$ to give
\begin{eqnarray*}
\frac{1}{2}\sin^{-1} (2x).
\end{eqnarray*}
Now substitute $x^2=1/8$ and we have
\begin{eqnarray*}
\sum_{n=0}^{\infty} \frac{1}{2n+1}\binom{2n}{n} \frac{1}{8^{n}}= \frac{\pi}{2 \sqrt{2}}.
\end{eqnarray*}
|
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|
Help summing the telescoping series $\sum_{n=2}^{\infty}\frac{1}{n^3-n}$. I know a priori that the series $$\sum_{n=2}^{\infty}\frac{1}{n^3-n}$$ converges. However, I am tasked with summing the series by treating it as a telescoping series.
By partial fraction decomposition, the series can be written as: $$\sum_{n=2}^{\infty}\left(\frac{1}{2(n+1)}+\frac{1}{2(n-1)}-\frac{1}{n}\right)$$
$s_n=\sum_{n=2}^{\infty}\left(\frac{1}{2(n+1)}+\frac{1}{2(n-1)}-\frac{1}{n}\right)=\left(\frac{1}{6}+\frac{1}{2}-\frac{1}{2}\right)+\left(\frac{1}{8}+\frac{1}{4}-\frac{1}{3}\right)+...+\left(\frac{1}{2(n+1)}+\frac{1}{2(n-1)}-\frac{1}{n}\right)$
I then grouped the terms by their position in each partial sum,
First terms: $\frac{1}{6},\frac{1}{8},\frac{1}{10},\frac{1}{12}...$
Second terms: $\require\cancel{\cancel{\frac{1}{2}}},\cancel{\frac{1}{4}},\cancel{\frac{1}{6}},\cancel{\frac{1}{8}}...$
Third terms: $\cancel{-\frac{1}{2}},-\frac{1}{3},\cancel{-\frac{1}{4}},-\frac{1}{5}...$
Cancelling leaves the series: $$\sum_{n=3}^{\infty}\frac{1}{2n}-\sum_{n=1}^{\infty}\frac{1}{2n+1}$$
However I'm stuck here since I see a divergent harmonic series summed with another harmonic series but I know the original series in question is convergent to $\frac{1}{4}$. What can I do? I suspect my error can be fixed somehow by adjusting the bounds of the sums...? Thanks in advance.
|
Alt. hint: it is not necessary to do a full partial fractions decomposition, it's enough to telescope:
$$
\frac{1}{n^3-n} = \frac{1}{2}\frac{(n+1) - (n-1)}{(n-1)n(n+1)} = \frac{1}{2}\left(\frac{1}{(n-1)n} - \frac{1}{n(n+1)}\right)
$$
|
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|
Evaluating $\lim_{x\to0} \frac{\cos x - \cos 3x}{\sin 3x^2 - \sin x^2}$
$$ \lim_{x\to0}\frac{\cos x-\cos (3x)}{\sin (3x^2)-\sin (x^2)} $$
Is there a simple way of finding the limit?
I know the long one: rewrite it as
$$ -\lim_{x\to 0}\frac{\cos x-\cos(3x)}{\sin(3x^2)}\cdot\frac{1}{1-\dfrac{\sin(3x^2)}{\sin(x^2)}} $$
and then find both limits in separately applying L'Hospital's rule several times. The answer is $2$.
|
Use the Maclaurin series
$$\cos x = 1-1/2 \; x^2+1/24 \; x^4 +O(x^6)$$
$$\cos 3x = 1-9/2 \; x^2+27/8 \; x^4 +O(x^6)$$
$$\sin x^2 = x^2-1/6\; x^6+O(x^8)$$
$$\sin 3x^2 = 3x^2-9/2 \; x^6+O(x^8)$$
then quotient is
$$\frac{-1/2 + 9/2}{3-1} + O(x^2)$$
and therefore the limit is 2.
|
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|
Maximum minus minimum of $c$ where $a+b+c=2$ and $a^2+b^2+c^2=12$ Let $a,b,$ and $c$ be real numbers such that
$a+b+c=2 \text{ and } a^2+b^2+c^2=12.$
What is the difference between the maximum and minimum possible values of $c$?
$\text{(A) }2\qquad \text{ (B) }\frac{10}{3}\qquad \text{ (C) }4 \qquad \text{ (D) }\frac{16}{3}\qquad \text{ (E) }\frac{20}{3}$
As I was reading the solution for this problem, I noticed that it said to use Cauchy–Schwarz inequality. I know what this inequality is (dot product of two vectors < vector 1*vector 2), but I don't understand how it can be applied in this situation. Thanks!
https://artofproblemsolving.com/wiki/index.php?title=2013_AMC_12B_Problems/Problem_17
|
We wish to find the values of $c$ such that the following simultaneous equations have real solutions to $a$ and $b$:
$$\begin{cases}a+b&=&2-c\\a^2+b^2&=&12-c^2\end{cases}$$
Labelling the equations as $(1)$ and $(2)$ respectively, $(1)^2 - (2)$ gives $2ab
= -8-4c+2c^2$, so $(a-b)^2 = a^2+b^2 - 2ab = (12-c^2) - (-8-4c+2c^2) = 20+4c-3c^2$.
If that number is non-negative, then $a-b = \pm\sqrt{20+4c-3c^2}$, which together with $(1)$ gives two sets of solutions.
So it remains to solve $20 + 4c - 3c^2 \ge 0$, no Cauchy-Schwarz needed.
|
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|
Application of Chebyschev inequality
I want to prove the inequality above. On the extreme ends we get a clear application of AM-GM, and I want to use the chebyschev inequality for the middle but was having trouble.
My Attempt:
Since Chebyschevs inequalty is
We can square the left hand side of our inequality with the term to its right to get
$$\frac{a+b+c}{3}\cdot \frac{a+b+c}{3}\geq \frac{ca+b^2+ca}{3}$$
My problem
I would've wanted to get $\dfrac{ab+bc+ca}{3}$ on the right instead. Did I apply the inequality wrong, or does it follow that $\dfrac{ab+bc+ca}{3}$ is less than or equal to $\dfrac{a+b+c}{3}\cdot \dfrac{a+b+c}{3}$?
|
As you've noted here the use of the Chebyshev's Inequality doesn't help much. However we can make use of the AM-GM inequality.
Squaring both sides and expanding them we get that the inequality is equivalent to: $(a+b+c)^2 \ge 3(ab+bc+ca)$, which is equivalent to $a^2 + b^2 + c^2 \ge ab + bc + ca$.
The last equality is true, as we have that $a^2 + b^2 \ge 2ab$, $b^2 + c^2 \ge 2bc$ and $c^2 + a^2 \ge 2ac$. Add the inequalities to get the final answer.
|
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|
What is the domain of the function $f(x)=\sin^{-1}\left(\frac{8(3)^{x-2}}{1-3^{2(x-1)}}\right)$?
What is the domain of the function
$f(x)=\sin^{-1}\left(\frac{8(3)^{x-2}}{1-3^{2(x-1)}}\right)$?
I started using the fact $-\frac{\pi}{2}\le f(x) \le \frac{\pi}{2}\implies-1 \le\frac{8(3)^{x-2}}{1-3^{2(x-1)}}\le1$.Now,on dissecting it into two cases.
$CASE (1): -1 \le\frac{8(3)^{x-2}}{1-3^{2(x-1)}}$
$CASE (2): \frac{8(3)^{x-2}}{1-3^{2(x-1)}}\le1$
The calculations is inboth cases are bewidering,that's why i'm not showing it.
I need someone who can help me in solving this.
|
Hint:
Let $3^{x-2}=a\implies a>0$ for real $x$
$$-1\le\dfrac{8a}{1-9a^2}\le1$$
$$\dfrac{8a}{1-9a^2}\le1\iff0\le\dfrac{9a^2+8a-1}{9a^2-1}=\dfrac{(9a-1)(a+1)}{(3a-1)(3a+1)}$$
As $a.0,$ we need $\dfrac{9a-1}{3a-1}\ge0$
$\implies$ either $9a-1=0\iff a=?$ or $ a>$max$\left(\dfrac19,\dfrac13\right)$ or $a<$min$\left(\dfrac19,\dfrac13\right)$
$\implies x<-2$ or $x\ge-1$
Similarly for $$-1\le\dfrac{8a}{1-9a^2}\iff\dfrac{8a+1-9a^2}{1-9a^2}\ge0$$
$$\iff\dfrac{(9a+1)(a-1)}{(3a+1)(3a-1)}\le0$$
As $a.0,$ we need $\dfrac{a-1}{3a-1}\ge0$
Can you take it from here?
|
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|
If the roots of the equation $x^2 + 3x -1 = 0$ also satisfy the equation $x^4 + ax^2 + bx + c = 0$ find the value of $a + b + 4c + 100$
If the roots of the equation $x^2 + 3x -1 = 0$ also satisfy the equation $x^4 + ax^2 + bx + c = 0$ find the value of $a + b + 4c + 100$
I tried really hard but the most I could get is the sum of the roots of the second equation is $3$
Please could someone solve this please !
It would mean the world to me
|
Hint: The solutions of your polynomial are given by
$$x_{1,2}=-\frac{3}{2}\pm\frac{\sqrt{13}}{2}$$
plugging $x_1$ into the other equation we get
$$b\sqrt{13}+3c\sqrt{13}-3\sqrt{13}+2a+3b+11c+11=0$$
the rest is for you!
|
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|
determine whether the following series converge or diverge? determine whether the following series converge or diverge ?
$a)$ $\sum_{n=1}^{\infty} (\frac {n}{n+1})^{n(n+1)}$
$b)$ $\sum_{n=1}^{\infty} (\frac {n^2 +1 }{n^2 +n +1})^{n^2}$
My attempts : for $a)$ $(\frac {n}{n+1})^{n(n+1)}= (1- \frac{1}{n+1})^{n(n+1)}= e^{-n}$, now $\sum_{n=1}^{\infty}\frac{1}{e^n}$ is converge
For $b )$ $(\frac {n^2 +1 }{n^2 +n +1})^{n^2}=( 1- \frac{n}{n^2+n+1})^{n. n}$...after that i can not able to procedd Further,,,,
Any hints/ solution will be appreciated
thanks in advance
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By root test we have that
$$\sqrt[n]{a_n}=\left(\frac {n}{n+1}\right)^{n+1}=\left(1-\frac {1}{n+1}\right)^{n+1}\to \frac1e$$
$$\sqrt[n]{b_n}=\left(\frac {n^2 +1 }{n^2 +n +1}\right)^{n}=\left[\left(1-\frac {n}{n^2+n+1}\right)^{\frac{n^2+n+1}{n}}\right]^{\frac{n^2}{n^2+n+1}} \to \frac1e$$
therefore $\sum a_n$ and $\sum b_n$ both converge.
|
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|
Prove that: $xy\sqrt{z}+yz\sqrt{x}+zx\sqrt{y}\geq x+y+z$ Let $x$,$y$ and $z$ are positive and $$\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\leq 3$$
Prove that: $$xy\sqrt{z}+yz\sqrt{x}+zx\sqrt{y}\geq x+y+z$$
The things I have done so far
$$3\geq \sum \limits_{cyc}\frac{1}{x}\geq \frac{9}{\sum \limits_{cyc}x}\Rightarrow \sum \limits_{cyc}x\geq 3$$
Then, I tried to use AM-GM and Cauchy -Schwarz but without success.
|
We need to prove that
$ xy\sqrt{z}+yz\sqrt{x}+xz\sqrt{y}\geq x+y+z $
$\Leftrightarrow \frac{xy\sqrt{z}+yz\sqrt{x}+xz\sqrt{y}}{xyz}\geq \frac{x+y+z}{xyz}$
$\Leftrightarrow \frac{1}{\sqrt{x}}+\frac{1}{\sqrt{y}}+\frac{1}{\sqrt{z}}\geq \frac{1}{xy}+\frac{1}{yz}+\frac{1}{xz}$
$\Leftrightarrow \frac{1}{x^{2}}+\frac{1}{y^{2}}+\frac{1}{z^{2}}+\frac{2}{\sqrt{x}}+\frac{2}{\sqrt{y}}+\frac{2}{\sqrt{z}}\geq \frac{1}{x^{2}}+\frac{1}{y^{2}}+\frac{1}{z^{2}}+\frac{2}{xy}+\frac{2}{yz}+\frac{2}{xz}$
$\Leftrightarrow \frac{1}{x^{2}}+\frac{1}{y^{2}}+\frac{1}{z^{2}}+\frac{2}{\sqrt{x}}+\frac{2}{\sqrt{y}}+\frac{2}{\sqrt{z}} \geq (\frac{1}{x}+\frac{1}{y}+\frac{1}{z})^{2}$
We have: $3\geq \frac{1}{x}+\frac{1}{y}+\frac{1}{z}\Rightarrow 3(\frac{1}{x}+\frac{1}{y}+\frac{1}{z})\geq (\frac{1}{x}+\frac{1}{y}+\frac{1}{z})^{2}$
Now, we need to prove:
$\frac{1}{x^{2}}+\frac{1}{y^{2}}+\frac{1}{z^{2}}+\frac{2}{\sqrt{x}}+\frac{2}{\sqrt{y}}+\frac{2}{\sqrt{z}}\geq 3(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}) $
Using Cauchy's inequality, we have:
$\frac{1}{x^{2}}+\frac{2}{\sqrt{x}}=\frac{1}{x^{2}}+\frac{1}{\sqrt{x}}+\frac{1}{\sqrt{x}}\geq 3\sqrt[3]{\frac{1}{x^{2}}.\frac{1}{\sqrt{x}}.\frac{1}{\sqrt{x}}}=\frac{3}{x}$
Similar, we have:
$\frac{1}{y^{2}}+\frac{2}{\sqrt{y}}\geq \frac{3}{y};\frac{1}{z^{2}}+\frac{2}{\sqrt{z}}\geq \frac{3}{z}$
So, $\frac{1}{x^{2}}+\frac{1}{y^{2}}+\frac{1}{z^{2}}+\frac{2}{\sqrt{x}}+\frac{2}{\sqrt{y}}+\frac{2}{\sqrt{z}}\geq 3(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}) $
|
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|
Evaluate $I_n=\int_0^{\pi/2} \frac{1}{\left( a\cos^2x+b\sin^2x\right)^n} \, dx$ I would like to evaluate (using elementary methods if possible) : (for $a>0,\ b>0$)
$$
I_n=\int_0^{\pi/2} \frac{1}{( a\cos^2x+b\sin^2x)^n} \, dx,\quad \ n=1,2,3,\ldots
$$
I thought about using $u=\tan(x)$ or $u=\frac{\pi}{2}-x$ but did not work. wolfram alpha evaluates the indefinite integral but not definite integral???
|
This is not an answer but it is too long for a comment.
For the antiderivative, Wolfram Alpha (and other CAS) return, for $$J_n(x)=(n-1)(a-b) I_n(x)$$s the messy expression
$$2^{n-1} \csc (2 x) \sqrt{\frac{(a-b) \sin ^2(x)}{a}} \sqrt{\frac{(b-a) \cos ^2(x)}{b}}$$ $$
((a-b) \cos (2 x)+a+b)^{1-n}$$ $$
F_1\left(1-n;\frac{1}{2},\frac{1}{2};2-n;\frac{a+b+(a-b) \cos (2 x)}{2
b},\frac{a+b+(a-b) \cos (2 x)}{2 a}\right)$$ where appears
The problem is that both $J_n\left(\frac{\pi }{2}\right)$ and $J_n\left(0\right)$ result in indeterminate forms and that the limits need to be worked.
These would be
$$J_n\left(\frac{\pi }{2}\right)=-\frac{\sqrt{\pi } \sqrt{1-\frac{a}{b}} \sqrt{1-\frac{b}{a}}\, b^{1-n}\, \Gamma (2-n)}{2
\, \Gamma \left(\frac{3}{2}-n\right)}\, _2F_1\left(\frac{1}{2},1-n;\frac{3}{2}-n;\frac{b}{a}\right)$$
$$J_n\left(0\right)=\frac{\sqrt{\pi } \sqrt{1-\frac{a}{b}} \sqrt{1-\frac{b}{a}} \,a^{1-n}\, \Gamma (2-n)}{2\,
\Gamma \left(\frac{3}{2}-n\right)}\, \, _2F_1\left(\frac{1}{2},1-n;\frac{3}{2}-n;\frac{a}{b}\right)$$
Have fun !
|
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|
Find the quotient and the remainder of $(n^6-7)/(n^2+1)$
Given that $n$ belong to $\mathbb{N}$.
Find the quotent and the remainder of $(n^6-7)/(n^2+1)$.
So I tried to divide them up and got a negative expression $(-n^4-7)$.
How to continue?
Or what can be done differently?
How to find the quotent and the remainder?
|
We have $n^6-7 = (n^2+1)q(n)+an+b$, with $a,b \in \mathbb Z$, because $n^2+1$ is monic.
Plugging $n=\pm i$, we get
$$
-8 = \pm ai+b
$$
which implies $a=0$ and $b=-8$. Therefore,
$$
q(n) = \frac{n^6-7+8}{n^2+1} = \frac{n^6+1}{n^2+1} = n^4 - n^2+ 1
$$
Thus,
$$
n^6-7 = (n^2+1)(n^4 - n^2+ 1)-8
$$
|
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|
Evaluate: $u =\int_0^\infty\frac{dx}{x^4 +7x^2+1}$ Evaluate:
$u =\displaystyle\int_0^\infty\dfrac{dx}{x^4 +7x^2+1}$
Attempt:
$$u = \int_0^\infty \dfrac{dx}{\left(x^2+ \left(\dfrac{7 - \sqrt {45}}{2}\right)\right)\left(x^2+ \left(\dfrac{7 + \sqrt {45}}{2}\right)\right)}$$
$$u = \int_0^\infty \dfrac{dx}{(x^2+a^2)(x^2+b^2)}$$
After partial fraction decomposition and simplifying I get:
$u = \dfrac{\pi}{2(a+b)}$
But answer is $\frac \pi 6$.
Where have I gone wrong?
|
Hint:
$$\dfrac2{x^4+ax^2+1}=\dfrac{1-1/x^2}{x^2+a+1/x^2}+\dfrac{1+1/x^2}{x^2+a+1/x^2}$$
$$x^2+1/x^2=(x-1/x)^2+?=(x+1/x)^2-2$$
|
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|
How to prove that $\sqrt[3]{20+14\sqrt{2}}+\sqrt[3]{20-14\sqrt{2}}=4$? Using the Cardano formula, one can show that $\sqrt[3]{20+14\sqrt{2}}+\sqrt[3]{20-14\sqrt{2}}$ is a real root of the depressed cubic
$f(x)=x^3-6x-40$. Actually, one can show by the calculating the determinant that this is the only real root. On the other hand, by the rational root theorem, one can see that the possible rational root must be a factor of 40 and one can check that $f(4)=0$. Therefore, by uniqueness of the real root, one must have
$$
\sqrt[3]{20+14\sqrt{2}}+\sqrt[3]{20-14\sqrt{2}}=4\tag{1}
$$
I had the observation above when I solved the cubic equation $x^3-6x-40=0$. My question is as follows:
without referring to the unique real root of the cubic, can we show (1) directly?
[An attempt.]
When taking the cube on both sides of (1) and simplifying further, I ended up with (1) again.
|
Can we express $\sqrt[3]{20+14\sqrt2}$ as $a+b\sqrt2$, with $a,b\in\mathbb Z$? In other words, are thre integers $a$ and $b$ such that $(a+b\sqrt2)^3=20+14\sqrt2$? Note that\begin{align}(a+b\sqrt2)^3=20+14\sqrt2&\iff a^3+3\sqrt2a^2b+6ab^2+2\sqrt2b^3=20+14\sqrt2\\&\iff a^3+6ab^2+(3a^2b+2b^3)\sqrt2=20+14\sqrt2.\end{align}Therefore, it is enough to have$$a^3+6ab^2=20(\iff a(a^2+6b^2)=20)\text{ and }3a^2b+2b^3=14(\iff b(3a^2+2b^2)=14).$$It is easy to see that you can take $a=2$ and $b=1$. Therefore, $\sqrt[3]{20+14\sqrt2}=2+\sqrt2$ and it is now easy to see that $\sqrt[3]{20-14\sqrt2}=2-\sqrt2$. Therefore$$\sqrt[3]{20+14\sqrt2}+\sqrt[3]{20-14\sqrt2}=2+\sqrt2+2-\sqrt2=4.$$
|
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|
Find the roots of $3x^3-4x-8$ It is given that $\alpha$, $\beta$ and $\gamma$ are the roots of the polynomial $3x^3-4x-8$.
I have been asked to calculate the value of $\alpha^2 + \beta^2 + \gamma^2$.
However I am unsure how to find these roots, seeing as though I haven't been given a root to start with.
I began by identifying that
$\alpha + \beta + \gamma = 0$
$\alpha\beta + \alpha\gamma + \beta\gamma = -4/3$
$\alpha\beta\gamma = 8/3$
However I am unsure how to continue to find $\alpha^2 + \beta^2 + \gamma^2$.
|
$ x(3x^2-4)=8$
Squaring we get $x^2(3x^2-4)^2=8^2$
Let $x^2=y\implies y(3y-4)^2=64\iff9y^3-24y^2+16y-64=0$
whose roots are $a^2,b^2,c^2$
$\implies a^2+b^2+c^2=\dfrac{24}9=?$
|
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|
Differential equation: $(2x^2 + 3y^2 -7)x dx = (3x^2 + 2y^2 -8 )y dy $
$(2x^2 + 3y^2 -7)x dx = (3x^2 + 2y^2 -8 )y dy $
Attempt:
After expanding, everything is neat except: $3x^2y dy - 3y^2 x dx$
I can't convert it to exact differential.
Also, there's weird symmetry in the equation wrt the coefficients of $x^2$ and $y^2$. But not sure how to utilise that symmetry.
As an attempt, though, I reached this:
$\dfrac{d(x^2 - y^2 -1)}{2(x^2- y^2 -1)}= \dfrac{y dy}{(2x^2 + 3y^2 -7)}$ which is not useful at all.
Answer given is:
$(x^2 +y^2 - 3)=(x^2 + y^2 - 1)^5C$
|
Given $(2x^2+3y^2-7)x\ dx=(3x^2+2y^2-8)y\ dy$
Let us take $$X=x^2$$$$Y=y^2$$and we get$$\dfrac{dY}{dX}=\dfrac{2X+3Y-7}{3X+2Y-8}$$Again let us consider $$X=p+a$$$$Y=q+b$$$$\implies\dfrac{dq}{dp}=\dfrac{2p+2a+3q+3b-7}{3p+3a+2q+2b-8}=\dfrac{2p+3q}{3p+2q}$$From $2a+3b-7=0$ and $3a+2b-8=0$ we get $\implies a=2,b=1$
$$\dfrac{dq}{dp}=\dfrac{2p+3q}{3p+2q}\mbox{ is a homogeneous equation. So, the change of variable is given by}$$$$q(p)=p\ u(p)\implies u+s\dfrac{du}{dp}=\dfrac{2+3u}{3+2u}$$$$s\dfrac{du}{dp}=\dfrac{2+3u}{3+2u}-u=\dfrac{2-2u^2}{3+2u}$$$$\dfrac{dp}{p}=\dfrac{3+2u}{2-2u^2}du$$$$\ln|p|=\int\dfrac{3+2u}{2-2u^2}\ du=\dfrac14\ln|u+1|-\dfrac54-\ln|u-1|+C$$
$$s=C\dfrac{(u+1)^{\frac14}}{(u-1)^{\frac54}}\implies(q-p)^{\frac54}=C(q+p)^{\frac14}$$
$$(q-p)^5=C(q+p)\implies(q-p)^5-C(q+p)=0$$$$p=X-2\mbox{ }\mbox{ and }q=Y-1$$$$\implies(Y-X+1)^5-C(Y+X-3)=0$$Therefore,$$\boxed{(x^2 +y^2 - 3)=(x^2 + y^2 - 1)^5C}$$
|
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"timestamp": "2023-03-29T00:00:00",
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|
Solve $ \binom{a}{2} + \binom{b}{2} = \binom{c}{2} $ with $a,b,c \in \mathbb{Z}$ I am trying to solve the Diophantine equation:
$$ \binom{a}{2} + \binom{b}{2} = \binom{c}{2} $$
Here's what it looks like if you expand, it's variant of the Pythagorean triples:
$$ a \times (a-1) + b \times (b-1) = c \times (c-1) $$
I was able to find solutions by computer search but this could have also been checked using the Hasse principle.
\begin{eqnarray*}
\binom{3}{2}+ \binom{3}{2}&=& \binom{4}{2} \\ \\
\binom{5}{2}+ \binom{10}{2}&=& \binom{11}{2} \\ \\
\binom{15}{2}+ \binom{15}{2}&=& \binom{21}{2}
\end{eqnarray*}
and many others. Is there a general formula for the $(a,b,c) \in \mathbb{Z}^3$ that satisfy this integer constraint.
>>> N = 25
>>> f = lambda a : a*(a-1)/2
>>> X = [(a,b,c,f(a) + f(b) - f(c)) for a in range(N) for b in range(N) for c in range(N)]
>>> [(x[0],x[1],x[2]) for x in X if x[3] == 0 and x[0] > 1 and x[1] > 1 and x[2] > 1]
[( 3, 3, 4), ( 4, 6, 7) , ( 5, 10, 11), ( 6, 4, 7), ( 6, 7, 9),
( 6, 15, 16), ( 7, 6, 9) , ( 7, 10, 12), ( 7, 21, 22), ( 9, 11, 14),
(10, 5, 11), (10, 7, 12), (10, 14, 17), (10, 22, 24), (11, 9, 14),
(12, 15, 19), (12, 21, 24), (13, 18, 22), (14, 10, 17), (15, 6, 16),
(15, 12, 19), (15, 15, 21), (15, 19, 24), (18, 13, 22), (19, 15, 24),
(21, 7, 22), (21, 12, 24), (22, 10, 24)]
|
After some algebra you can rewrite the equation as
$$
a(a-1)=(c-b)(c+b-1)
$$
Note that $c-b$ and $c+b-1$ have opposite parity. Conversely, if $p$ and $q$ are integers with opposite parity, then we can find integers $b,c$ such that $p=c-b$, $q=c+b-1$: namely, $b=\frac{q+1-p}{2}$ and $c=\frac{q+1+p}{2}$.
That is to say, all solutions to the equation can be obtained in the following way:
*
*Choose some number $a$. Factor $a(a-1)$ into two numbers $p,q$ of opposite parity. Since $a(a-1)$ is always even, this just means $p$ and $q$ can't both be even.
*Take $b=\frac{q+1-p}{2}$, $c=\frac{q+1+p}{2}$.
For example, we could take $a=16$. Then $a(a-1)=240$, which factors into two numbers with opposite parity in four different ways:
\begin{align}
1&\cdot 240\\
3 &\cdot 80\\
5 &\cdot 48\\
15 &\cdot 16
\end{align}
These give four essentially different solutions to the equation:
\begin{align}
\binom{16}{2}+\binom{120}{2}&=\binom{121}{2}\\
\binom{16}{2}+\binom{39}{2}&=\binom{42}{2}\\
\binom{16}{2}+\binom{22}{2}&=\binom{27}{2}\\
\binom{16}{2}+\binom{1}{2}&=\binom{16}{2}
\end{align}
A few notes:
*
*Swapping $p,q$, or replacing them by $-p,-q$, gives a solution which is essentially the same if you remember that $\binom{n}{2}=\binom{1-n}{2}$.
*The symmetry between $a$ and $b$ means that this method will produce each solution twice. For example, the third solution above ($\binom{16}{2}+\binom{22}{2}=\binom{27}{2}$) could also have been found by taking $a=22$, $p=11$, $q=42$.
|
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"language": "en",
"url": "https://math.stackexchange.com/questions/2883759",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
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|
Partial fraction of $\frac{2s+12}{ (s^2 + 5s + 6)(s+1)} $ then inverse transform it
Find the inverse Laplace transform of
$$\mathcal{L}^{-1}\left(\frac{2s+12}{ (s^2 + 5s + 6)(s+1)}\right)$$
I recognise I need to use partial fractions to solve it and that is where I got stuck. Here’s my working,
After factoring the denominator, I got a case of non repeating linear factors
$\dfrac{2s+12}{ (s+2)(s+3)(s+1)} = \dfrac{A}{s+2} + \dfrac{B}{s+3} + \dfrac{C}{s+1} $
$2s + 12 = A (s+3)(s+1) + B (s+2)(s+1) + C (s+2)(s+3) $
$2s + 12 = (As^2 + Bs^2 + Cs^2) + (4As + 3Bs + 5Cs) + (3A + 2B+6C) $
So...
$A + B + C = 0$
$ 4A + 3B + 5C = 2$
$3A + 2B + 6C = 12$
how do I solve this complicated equations ? This is where i got stuck and cannot continue.
|
There is a shortcut way here, in this simple case. Let to mutiply the relation to $\color{red}{s+1}$, then
$$\frac{2s+12}{ (s+2)(s+3)(s+1)} \color{red}{(s+1)}= \frac{A\color{red}{(s+1)}}{s+2} + \frac{B\color{red}{(s+1)}}{s+3} + \frac{C\color{red}{(s+1)}}{s+1}$$
or
$$\frac{2s+12}{ (s+2)(s+3)} = \frac{A\color{red}{(s+1)}}{s+2} + \frac{B\color{red}{(s+1)}}{s+3} + C$$
now set $s=-1$ then $C=\dfrac{10}{2}$.
Can you proceed?
|
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"timestamp": "2023-03-29T00:00:00",
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|
Roots of unity and large expression
Let $\omega$ be a complex number such that $\omega^5 = 1$ and $\omega \neq 1$. Find
$$\frac{\omega}{1 + \omega^2} + \frac{\omega^2}{1 + \omega^4} + \frac{\omega^3}{1 + \omega} + \frac{\omega^4}{1 + \omega^3}.$$
I have tried combining the first and third terms & first and last terms. Here is what I have so far:
\begin{align*}
\frac{\omega}{1 + \omega^2} + \frac{\omega^2}{1 + \omega^4} + \frac{\omega^3}{1 + \omega} + \frac{\omega^4}{1 + \omega^3}
&= \frac{\omega}{1 + \omega^2} + \frac{\omega^4}{1 + \omega^3} + \frac{\omega^2}{1 + \omega^4} + \frac{\omega^3}{1 + \omega} \\
&= \dfrac{\omega(1+\omega^3) + \omega^4(1+\omega^2)}{(1+\omega^2)(1+\omega^3)} + \dfrac{\omega^2(1+\omega) + \omega^3(1+\omega^4)}{(1+\omega^4)(1+\omega)} \\
&= \dfrac{\omega + 2\omega^4 +\omega^6}{1+\omega^2 + \omega^3 + \omega^5} + \dfrac{\omega^2 + 2\omega^3 + \omega^7}{1+\omega + \omega^4 + \omega^5} \\
&= \dfrac{2\omega + 2\omega^4}{2+\omega^2 + \omega^3} +
\dfrac{2\omega^2 + 2\omega^3}{2+\omega+\omega^4}
\end{align*}
OR
\begin{align*}
\frac{\omega}{1 + \omega^2} + \frac{\omega^2}{1 + \omega^4} + \frac{\omega^3}{1 + \omega} + \frac{\omega^4}{1 + \omega^3}
&= \frac{\omega}{1 + \omega^2} + \frac{\omega^3}{1 + \omega} + \frac{\omega^4}{1 + \omega^3} + \frac{\omega^2}{1 + \omega^4} \\
&= \dfrac{\omega(1+\omega) + \omega^3(1+\omega^2)}{(1+\omega)(1+\omega^2)} + \dfrac{\omega^2(1+\omega^3) + \omega^4(1+\omega^4)}{(1+\omega^3)(1+\omega^4)} \\
&= \dfrac{\omega + \omega^2 + \omega^3 + \omega^5}{1+\omega + \omega^2 + \omega^3} + \dfrac{\omega^2 + \omega^4 + \omega^5 + \omega^8}{1 + \omega^3 + \omega^4 + \omega^7} \\
&= \dfrac{2\omega+\omega^2+\omega^3}{1+\omega+\omega^2+\omega^4} + \dfrac{1+\omega+\omega^2+\omega^4}{1+2\omega^3+\omega^4}
\end{align*}
|
$\frac{w}{1+w^2 }+\frac{w^3 }{1+w} +\frac{w^2 }{1 +w^4}+\frac{w^4}{1+w^3 }=\frac{w+w^2 +w^3 +w^5}{(1+w)(1+w^2) }+\frac{w^2 +w^5+w^4+w^8}{(1+w^4)(1+w^3) }=\frac{w+w^2 +w^3 +w^5}{w+w^2 +w^3 +1}+\frac{w^2 +1+w^4+w^3 }{1+w^3 +w^4+w^2} =2$
Because $w^5 =1\Rightarrow w^7=w^2$ and $w ^8 = w^3$
|
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"language": "en",
"url": "https://math.stackexchange.com/questions/2886460",
"timestamp": "2023-03-29T00:00:00",
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|
Is there a smarter way to differentiate the function $f(x) = \sin^{-1} \frac{2x}{1+x^2}$?
Given $f(x) = \sin^{-1} \frac{2x}{1+x^2}$,
Prove that $$f'(x) = \begin{cases}\phantom{-}\frac{2}{1+x^2},\,|x|<1 \\\\ -\frac{2}{1+x^2},\,|x|>1 \end{cases}$$
Obviously the standard approach would be to use the chain rule and simplify from there.
But I noticed that some of these expressions are familiar, specifically, from the tangent half-angle formulae:
If $x = \tan \frac \theta 2$, then $\sin \theta = \frac{2x}{1+x^2}$ and $\frac{d\theta}{dx} = \frac{2}{1+x^2}$.
So my question is: can this observation be used to construct a more elegant proof?
|
We have
$$ \frac{t}{1+\sqrt{1-t^2}} \Bigg \rvert_{t = \frac{2x}{1+x^2}} = \frac{2x}{1+x^2 + \lvert 1-x^2\rvert} = \begin{cases} x &\!\!\!, |x|<1 \\ \frac{1}{x} &\!\!\!, |x|>1 \end{cases} \, .$$
Therefore we can use
$$ \arcsin(t) = 2 \arctan\left(\frac{t}{1+\sqrt{1-t^2}}\right) \, ,$$
which follows from the half-angle formula, to obtain
$$ f(x) = \begin{cases} 2\arctan(x) &\!\!\!, |x|<1 \\ 2\arctan\left(\frac{1}{x}\right) &\!\!\!, |x|>1 \end{cases} = \begin{cases} 2\arctan(x) &\!\!\!, |x|<1 \\ \operatorname{sgn}(x)\pi -2\arctan(x) &\!\!\!, |x|>1 \end{cases} \, .$$
|
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"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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|
Prove that $\frac{1}{2}+ \sum_{k=1}^n \cos (k\theta) = \sin((n+ \frac{1}{2})\theta)/2\sin\frac{\theta}{2}$ Suppose $\sin \frac{\theta}{2} \neq 0$ . Prove that
$$\frac{1}{2}+ \sum_{k=1}^n \cos (k\theta) = \frac{\sin[(n+ \frac{1}{2})\theta]}{2\sin\frac{\theta}{2}}$$
The question also give the hint,
$$z=\cos\theta + i\sin\theta = e^{i\theta}$$
$$\sum_{k=1}^n z^k = z + z^2 + \cdots + z^n = \frac{z(z^n-1)}{z-1}$$
What I did was to change z into polar form and applied double angle formula to remove $\cos\theta$, but I have no idea what to do after that and it is still $\sum_{k=1}^n z^k = \sum_{k=1}^n \cos(k\theta)+i\sin(k\theta)$, not exactly $\sum_{k=1}^n \cos(k\theta)$ that I am supposed to prove
|
\begin{equation}
\cos k \theta
=
\frac{e^{jk\theta} + e^{-jk\theta}}{2}
\end{equation}
Let's use $\sum\limits_{k=0}^{N-1 }r^k= \frac{1-r^N}{1-r} $
\begin{equation}
\sum \cos k \theta
=
\frac{1}{2}
\sum\limits_{k=1}^{N}
e^{jk\theta} +
\frac{1}{2}
\sum\limits_{k=1}^{N}
e^{-jk\theta}
=
\frac{1}{2}
\big(
\frac{1 - e^{-j (N+1) \theta}}{1 - e^{j \theta}}-1 +
\frac{1 - e^{j (N+1) \theta}}{1 - e^{-j \theta}}-1
\big)
\end{equation}
So
\begin{equation}
\frac{1}{2}
+
\sum \cos k \theta
=
\frac{1}{2}
\big(
-1+
\frac{1 - e^{j (N+1) \theta}}{1 - e^{j \theta}} +
\frac{1 - e^{-j (N+1) \theta}}{1 - e^{-j \theta}}
\big)
\end{equation}
So
\begin{equation}
\frac{1}{2}
+
\sum \cos k \theta
=
\frac{1}{2}
\big(
\frac{-(1-e^{j \theta})(1-e^{-j \theta})
+
(1 - e^{j (N+1) \theta})(1-e^{-j \theta})
+
(1 - e^{-j (N+1) \theta})(1-e^{j \theta})
}
{(1-e^{j \theta})(1-e^{-j \theta})}
\big)
\end{equation}
that is
\begin{equation}
\frac{1}{2}
+
\sum \cos k \theta
=
\frac{1}{2}
\big(
\frac{
-2 + e^{j \theta} + e^{- j\theta}
+
1 - e^{-j \theta} - e^{j (N+1) \theta} + e^{j N \theta}
+
1 - e^{j \theta} - e^{- j (N+1) \theta }+ e^{-jN \theta}
}
{(1-e^{j \theta})(1-e^{-j \theta})}
\big)
\end{equation}
that is
\begin{equation}
\frac{1}{2}
+
\sum \cos k \theta
=
\frac{1}{2}
\big(
\frac{
- (e^{j (N+1) \theta} +e^{- j (N+1) \theta })+ (e^{j N \theta} + e^{-jN \theta})
}
{(1-e^{j \theta})(1-e^{-j \theta})}
\big)
\end{equation}
i.e.
\begin{equation}
\frac{1}{2}
+
\sum \cos k \theta
=
\frac{1}{2}
\big(
\frac{
- 2 \cos (N+1) \theta+ 2 \cos N \theta
}
{4 \sin^2 \frac{\theta}{2}}
\big)
\end{equation}
Using $\cos a - \cos b = 2 \sin \frac{1}{2}(a+b) \sin \frac{1}{2} (b-a)$, we get
\begin{equation}
\frac{1}{2}
+
\sum \cos k \theta
=
\frac{1}{2}
\big(
\frac{
4 \sin \frac{ \theta}{2}
\sin \frac{ \theta}{2} (2N + 1)
}
{4 \sin^2 \frac{\theta}{2}}
\big)
=
\frac{1}{2}
\frac{\sin \frac{ \theta}{2} (2N + 1) }{\sin \frac{ \theta}{2}}
\end{equation}
|
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"timestamp": "2023-03-29T00:00:00",
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|
Solve $8\sin x=\frac{\sqrt{3}}{\cos x}+\frac{1}{\sin x}$
Solve
$$8\sin x=\dfrac{\sqrt{3}}{\cos x}+\frac{1}{\sin x}$$
My approach is as follow
$8 \sin x-\frac{1}{\sin x}=\frac{\sqrt{3}}{\cos x}$
On squaring we get
$64 \sin^2 x+\frac{1}{\sin^2 x}-16=\frac{3}{\cos^2 x}$
$(64\sin^4 x-16\sin^2 x+1)(1-\sin^2 x)=3 \sin^2 x$
Solving and re-arranging we get
$-64\sin^6 x+80\sin^4 x-20\sin^2 x+1=0$
Using the substitution $\sin^2 x=t$
$-64t^3+80t^2-20t+1=0$
I am not able to solve it from hence forth
|
Nice job so far.
Notice that every exponent is even, so you can further substitute $t^2=y$, to obtain a cubic equation in $y$, which you can solve by the Cardano formula.
|
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|
The remainder when $x^2$ $+ ax +4$ is divided by $x-4$ is four times the remainder when the same expression is divided by $x-1$. Find the value of $a$ According to my textbook, the answer to this is $-2$.
Isn't this true for any value of a though?
I started by doing the following:
$4a+20=4x$
$a+5 = x$
I'm not sure how they got $-2$ from this, as both equations are equivalent.
|
(This is not an answer, but it is too long for a comment.)
Let $f(x) = x^2 + ax + 4$.
I can't make any sense of what you claim to get on dividing $f(x)$ by $x-4$, "$(x-4)(x+4+a)(4a+20)$" or on dividing $f(x)$ by $x-1$, "$(x-1)(x+1+a)(a+5)$". You also claim the first is the second multiplied by four, which is evidently false. Why? The expressions you write must contain "$a^2$"s, but $f(x)$ does not. \begin{align*}
(x-4)&(x+4+a)(4a+20) \\ &= 4 a x^2 + 20 x^2 + 4 a^2 x + 20 a x - 16 a^2 - 144 a - 320 \\
(x-1)&(x+1+a)(a+5) \\ &= a x^2 + 5 x^2 + a^2 x + 5 a x - a^2 - 6 a - 5
\end{align*}
It is also straightforward to see that the first is not the second multiplied by four. In particular, neither $x+4+a = 4(x+1+a)$ nor $x+4+a = x+1+a$.
|
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|
Find the maximum value of $a+b$ The question:
Find the maximum possible value of $a+b$ if $a$ and $b$ are different non-negative real numbers that fulfill
$$a+\sqrt{b} = b + \sqrt{a}$$
Without loss of generality let us assume that $a\gt b$. I rearrange the equation to get $$a - \sqrt{a} = b - \sqrt{b}$$
If $f(x)= x - \sqrt{x},$ then we are trying to solve $f(a)=f(b).$ Using some simple calculus I found that the turning point of $f(x)$ is $(\frac{1}{4}, -\frac{1}{4})$. Hence $0 \le b \le \frac{1}{4}$ and $\frac{1}{4} \le a \le 1$. From here, I have no idea how to proceed.
I used trial and error to find that when $a$ increases, the value of $a+b$ increases as well. Hence I hypothesise that $a+b$ is at a maximum when $a=1$ and $b=0$, which implies that $a+b=1$ is a maximum. Can anyone confirm this?
|
As
$$
a-b=(\sqrt{a}-\sqrt{b})(\sqrt{a}+\sqrt{b})=\sqrt{a}-\sqrt{b}
$$
then with $a\ne b$ it gives $\sqrt{a}+\sqrt{b}=1$. Now you have to maximize
$$
(\sqrt{a})^2+(\sqrt{b})^2
$$
subject to $\sqrt{a}+\sqrt{b}=1$.
|
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|
Show that $3y''+4xy'-8y=0$ has an integral which is a polynomial in x. Deduce the general solution.
Show that $3y''+4xy'-8y=0$ has an integral which is a polynomial in x. Deduce the general solution.
In this problem, i tried in direct method. But i could not do. So, i did in following way
Since order is 2, i am assuming a polynomial of degree 2.
Let $ y=ax^2+bx+c$ be the solution. By substitution, i get Solution as $y= (4/3)x^2+c$, c is some constant. Pls correct me if am wrong. Any other better way, pls suggest
|
If $y(x)=a_nx^n+a_{n-1}x^{n-1}+\ldots+a_1x+a_0$ is a solution with $a_n\neq 0$, then note that $3\,y''(x)$ is of degree $n-2$, but $4x\,y'(x)$ and $-8\,y(x)$ are of the same degree $n$. As $3\,y''(x)+4x\,y'(x)-8\,y(x)=0$, the leading terms of $4x\,y'(x)$ and $-8\,y(x)$ must cancel. This proves that
$$4n\,a_n\,x^n-8\,a_n\,x^n=0\,,\text{ whence }n=2\,.$$
Thus, we may assume that $y(x)=x^2+bx+c$ is a solution. Now,
$$0=3\,y''(x)+4x\,y'(x)-8\,y(x)=3\cdot 2+4x\,(2x+b)-8\,(x^2+bx+c)=-4b\,x+(6-8c)\,,$$
so $b=0$ and $c=\dfrac{3}{4}$. This means
$$y(x)=x^2+\frac34$$
is a solution.
To find the general solution, we suppose that $y(x)=\left(x^2+\dfrac34\right)\,z(x)$ satisfies the differential equation. Plugging this in to get
$$3\,\left(x^2+\frac34\right)\,z''(x)+\Biggl(12x+4x\,\left(x^2+\frac34\right)\Biggr)\,z'(x)=0\,.$$
In other words,
$$z''(x)+\left(\frac{4}{3}\,x+\frac{4x}{x^2+\frac34}\right)\,z'(x)=0\,,$$
or
$$\frac{\text{d}}{\text{d}x}\,\left(\left(x^2+\frac34\right)^2\,\exp\left(\frac{2}{3}\,x^2\right)\,z'(x)\right)=0\,.$$
Ergo,
$$z'(x)=A'\,\left(\frac{\exp\left(-\frac{2}{3}\,x^2\right)}{\left(x^2+\frac{3}{4}\right)^2}\right)\text{ for some constant }A'\,.$$
In conclusion,
$$y(x)=A'\,\left(x^2+\frac34\right)\,\int_0^x\,\frac{\exp\left(-\frac{2}{3}\,t^2\right)}{\left(t^2+\frac{3}{4}\right)^2}\,\text{d}t+B\,\left(x^2+\frac{3}{4}\right)$$
for some constant $B$. We may write
$$y(x)=\small A\,\Biggl(\sqrt{\frac{2\pi}{3}}\,\left(x^2+\frac34\right)\,\text{erf}\left(\sqrt{\frac23}\,x\right)+x\,\exp\left(-\frac23\,x^2\right)\Biggr)+B\,\left(x^2+\frac{3}{4}\right)\,,$$
where $A:=\frac{2}{3}\,A'$ and $\text{erf}$ is the error function:
$$\text{erf}(x)=\frac{2}{\sqrt{\pi}}\,\int_0^x\,\exp\left(-t^2\right)\,\text{d}t\,.$$
We also have $A=\dfrac{1}{2}\,y'(0)$ and $B=\dfrac{4}{3}\,y(0)$. That is,
$$y(x)=\small y(0)\,\left(\frac{4}{3}\,x^2+1\right)+y'(0)\,\Biggl(\frac{\sqrt{6\pi}}{8}\,\left(\frac43\,x^2+1\right)\,\text{erf}\left(\sqrt{\frac23}\,x\right)+\frac12\,x\,\exp\left(-\frac23\,x^2\right)\Biggr)\,.$$
Interestingly, we may write
$$\left(\frac{\text{d}}{\text{d}x}+\frac{4x}{3}+\frac{2x}{x^2+\frac34}\right)\,\left(\frac{\text{d}}{\text{d}x}-\frac{2x}{x^2+\frac34}\right)=\frac{\text{d}^2}{\text{d}x^2}+\frac{4x}{3}\,\frac{\text{d}}{\text{d}x}-\frac{8}{3}\,.$$
|
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|
How to calculate the integral $\int \frac{a\tan^2{x}+b}{a^2\tan^2{x}+b^2} dx$? How to calculate the integral $\displaystyle\int \dfrac{a\tan^2{x}+b}{a^2\tan^2{x}+b^2} dx$?
It seems like a $\arctan$ of what else... but I can not work it out.
I have previously asked how to calculate the definite integral between $0$ and $\pi$ here, but the answers there only use complex analysis, so they are of no help for the indefinite integral.
|
Note that:
$$\begin{align}
\left(a+b\right)\left(a\tan^2{x}+b\right) & =a^2\tan^2{x}+b^2+ab\left(1+\tan^2{x}
\right) \\[5pt] & = \color{blue}{a^2\tan^2{x}+b^2}+ab\sec^2x\end{align}$$
So we then have:
$$\begin{align}\int \frac{a\tan^2{x}+b}{\color{blue}{a^2\tan^2{x}+b^2}} \,\mbox{d}x
& = \frac{1}{a+b}\int\frac{\color{blue}{a^2\tan^2{x}+b^2}+ab\sec^2x}{\color{blue}{a^2\tan^2{x}+b^2}} \,\mbox{d}x \\[8pt]
& = \frac{1}{a+b}\int\left( 1 + \frac{ab\sec^2x}{a^2\tan^2{x}+b^2} \right) \,\mbox{d}x \tag{$\star$}\\[8pt]
& = \frac{1}{a+b}\Bigl( x+\arctan\left( \tfrac{a}{b}\tan x\right) \Bigr) +C
\end{align}$$
Where $(\star)$ is a standard integral, or follows after $t=\tfrac{a}{b}y$, here with $y=\tan x$:
$$\int\frac{y(x)'}{a^2y(x)^2+b^2}\,\mbox{d}x = \frac{1}{ab}\arctan\left(\tfrac{a}{b}y\right)+C$$
|
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|
Solve the system $x+y=\sqrt{4z-1},\ x+z=\sqrt{4y-1},\ z+y=\sqrt{4x-1}$
Find all real numbers $x,\ y,\ z$ that satisfy $$x+y=\sqrt{4z-1},\ x+z=\sqrt{4y-1},\ z+y=\sqrt{4x-1}$$
First natural move would be rewriting the system as:
$$x^{2}+y^{2}+2xy=4z-1$$
$$x^{2}+z^{2}+2xz=4y-1$$
$$z^{2}+y^{2}+2zy=4x-1$$
Thus,
$$2x^{2}+2y^{2}+2z^{2}+2xy+2yz+2zx-4x-4y-4z+3=0$$
And I tried to write the polynomial above as the sum of complete squares. Is this possible or do you have other ideas?
|
Note that these equations are defined only for $x,y,z\ge 1/4$.
Consider the first two equations
$$x+y=\sqrt{4z-1}$$
$$x+z=\sqrt{4y-1}$$
Subtracting these two equations gives us
$$y-z=\sqrt{4z-1}-\sqrt{4y-1}$$
If $y\gt z$, then the LHS is positive but the RHS is negative, so this cannot be. Similarly, if $y\lt z$, then the LHS is negative but the RHS is positive, which is also impossible. Thus, we have that
$$y-z=\sqrt{4z-1}-\sqrt{4y-1}=0$$
or
$$y=z$$
We may show by similar reasoning that $x=y$, and so $x=y=z$.
Then we have that
$$2x=\sqrt{4x-1}$$
or
$$4x^2-4x+1=0$$
Which has one solution:
$$x=\frac{1}{2}$$
This gives us the only solution to your system:
$$x=y=z=\frac{1}{2}$$
NOTE: This problem is trivialized by the fact that both $x$ and $\sqrt{4x-1}$ are increasing functions. The following system of equations might be more interesting:
$$x+y=\frac{1}{\sqrt{4z-1}}$$
$$x+z=\frac{1}{\sqrt{4y-1}}$$
$$y+z=\frac{1}{\sqrt{4x-1}}$$
|
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|
Find out when $(x + y)^5 = x^5 + y^5$ (Spivak's Calculus Book) I know that I can see that this match with $y = 0$, $x = 0$ and $y = -x$.
But, the author of the book says:
"Hint: From the assumption $(x + y)^5 = x^5 + y^5$ you should be able to derive the equation $x^3 + 2x^2y + 2xy^2 + y^3 = 0$ if $xy \neq 0$. This implies that $(x + y)^3 = x^2y + y^2x = xy(x + y)$"
Then, I realize that $(x + y)^3 - xy(x + y) = x^3 + 2x^2y + 2xy^2 + y^3$.
That's why if $(x + y)^3 - xy(x + y) = 0$ then $(x + y)^3 = xy(x + y)$.
Then, I did the same with $(x + y)^5$
$x^5 + 5x^4y +9x^3y^2 + 9x^2y^3 + 5xy^4 + y^5 = 0$ if $xy \neq 0$
Then $(x + y)^5 - x^2y^2(x + y) = 0$
$(x + y)^5 = x^2y^2(x + y)$ Then I can prove that its true when $y = -x$
But I still don't get it why I have to did this and what the author want to tell me doing this.
|
Hint: From the assumption $(x + y)⁵ = x⁵ + y⁵$ you should be able to derive the equation $x³ + 2x²y + 2xy² + y³ = 0$ if $xy \neq 0$.
Using the binomial expansion:
$$\require{cancel}
\begin{align}
(x+y)^5=x^5+y^5 \;\;&\iff\;\; \cancel{x^5} + 5x^4y +10x^3y^2 + 10x^2y^3 + 5xy^4 + \bcancel{y^5} = \cancel{x^5}+\bcancel{y^5} \\
&\iff\;\; 5xy\,(x^3+2x^2y+2xy^2+y^3) = 0
\end{align}
$$
Assuming that $\,xy\ne 0\,$, this implies $\,x^3+2x^2y+2xy^2+y^3 = 0\,$
But I still don't get it [...] what the author want to tell me doing this
Note that $\,x^3+2x^2y+2xy^2+y^3 = (x + y) (x^2 + x y + y^2)\,$, and the only (real) zeros occur for $\,x+y=0\,$ since the quadratic factor is always strictly positive if $\,xy \ne 0\,$.
|
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|
Sum of roots of equation $x^4 - 2x^2 \sin^2(\displaystyle {\pi x}/2) +1 =0$ is My try:
$$x^4-2x^2\sin^2(\frac{\pi x}{2})+1=0\\x^4+1=2x^2\left (1-\cos^2\left(\frac{\pi x}{2}\right)\right )\\(x^2-1)^2=-2x^2\cos^2\left(\frac{\pi x}{2}\right)\\(x^2-1)^2+2x^2\cos^2\left(\frac{\pi x}{2}\right)=0\\x^2-1=0\,\text{and}\, 2x^2\cos^2\left(\frac{\pi x}{2}\right)=0$$
I am stuck , I am confused now what to do now
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If $f(x)\ge0$ and $g(x)\ge0$, for all $x$, then
$$
f(x)+g(x)=0
\qquad\text{if and only if}\qquad
f(x)=0\text{ and }g(x)=0
$$
Take $f(x)=(x^2-1)^2$. This equals zero only at $-1$ and $1$.
If $g(x)=2x^2\cos^2\bigl(\frac{\pi x}{2}\bigr)$, is $g(1)=0$ or $g(-1)=0$? No other values of $x$ can make $f(x)+g(x)=0$.
A different strategy is to note that $0\le\sin^2(\pi x/2)\le1$, so that $-2x^2\le-2x^2\sin^2(\pi x/2)$ and therefore
$$
x^4-2x^2+1\le x^4-2x^2\sin^2(\frac{\pi x}{2})+1
$$
If the right-hand side is $0$, then also $x^4-2x^2+1=(x^2-1)^2$ must be $0$, which implies $x=1$ or $x=-1$. Then it's just a matter of checking whether these values are solutions of the given equation.
|
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|
Stirling type formula for Sum on $\ln(n)^2$ Is there a similar formula like the Stirling one on the sum over $\ln(n)$ (take logarithms on its factorial representation):
$$\sum_{n=1}^N \ln(n) = N \ln(N)-N+\ln(N)/2+\ln(2\pi)/2+\mathcal{O}(\ln(N)/N)$$
but on the sum over its squares?
$$\sum_{n=1}^N (\ln(n))^2$$
I already advanced on getting good approximation on asymptotics integrating $\ln(n)^2$ and arrive to correct terms till $\mathcal{O}(N)$ order. But further advance is becoming hard for me in $\mathcal{O}(\ln(N))$ terms.
I am specially interested in $\mathcal{O}(1)$ term.
|
We can use the Euler-Maclaurin formula to obtain the asymptotic expansion
$$ f_N (a) \equiv \sum \limits_{n=1}^N n^a \sim \zeta(-a) + \sum \limits_{k=0}^\infty \frac{B_k^*}{a+1} {a+1 \choose k} N^{a+1-k} $$
for $N \in \mathbb{N}$ and $a \in \mathbb{R} \setminus \{-1\}$ . Here $(B_k^*)_{k\in \mathbb{N}_0}$ are the Bernoulli numbers with $B_1^* = \frac{1}{2}$. We want to find an asymptotic expansion for
$$f_N''(0) = \sum \limits_{n=1}^N \ln^2 (n) \, .$$
Obviously, the contribution of the terms with $k \geq 2$ vanishes as $N \to \infty$ . More precisely, their leading term is $\mathcal{O} (\ln^2 (N) /N)$ . Therefore we have
\begin{align}
\sum \limits_{n=1}^N \ln^2 (n) &\sim \frac{\mathrm{d}^2}{\mathrm{d}a^2} \left[\zeta(-a) + \frac{N^{a+1}}{a+1} + \frac{1}{2} N^{a} \right] \Bigg\rvert_{a=0} + \mathcal{O} \left(\frac{\ln^2 (N)}{N}\right) \\
&\sim N \ln^2 (N) - 2 N \ln(N) + 2N + \frac{1}{2} \ln^2 (N) + \zeta''(0) + \mathcal{O} \left(\frac{\ln^2 (N)}{N}\right)
\end{align}
as $N \to \infty$ .
$\zeta''(0)$ can be found using the series expansions
\begin{align}
(2\pi)^s &= 1 + \ln(2 \pi) s + \frac{1}{2} \ln^2 (2 \pi) s^2 + \mathcal{O}(s^3) \, , \\
\sin \left(\frac{\pi s}{2}\right) &= \frac{\pi}{2} s - \frac{\pi^3}{48} s^3 + \mathcal{O}(s^5) \, , \\
\Gamma(1-s) &= 1 - \gamma s + \frac{6 \gamma^2 + \pi^2}{12} s^2 + \mathcal{O}(s^3) \, , \\
\zeta(1-s) &= - \frac{1}{s} + \gamma + \gamma_1 s + \mathcal{O}(s^2)
\end{align}
near $s = 0$ in the functional equation
\begin{align}
\zeta(s) &= \frac{1}{\pi} (2\pi)^s \sin \left(\frac{\pi s}{2}\right) \Gamma(1-s) \zeta(1-s)\\
&= - \frac{1}{2} - \frac{1}{2} \ln (2 \pi) s - \frac{1}{2} \left[\frac{\pi^2}{24} + \frac{1}{2} \ln^2(2\pi) - \frac{1}{2} \gamma^2 - \gamma_1 \right] s^2 + \mathcal{O}(s^3)\, .
\end{align}
We obtain $\zeta(0) = -\frac{1}{2}$ , $\zeta'(0) = -\frac{1}{2} \ln(2 \pi)$ and
$$ \zeta''(0) = - \left[\frac{\pi^2}{24} + \frac{1}{2} \ln^2(2\pi) - \frac{1}{2} \gamma^2 - \gamma_1 \right] \, . $$
|
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|
Prove $\ \frac{z-1}{z+1} $ is imaginary no' iff $\ |z| = 1 $ Let $\ z \not = -1$ be a complex number. Prove $\ \frac{z-1}{z+1} $ is imaginary number iff $\ |z| = 1 $
Assuming $\ |z| = 1 \Rightarrow \sqrt{a^2+b^2} = 1 \Rightarrow a^2+b^2 = 1 $ and so
$$\ \frac{z-1}{z+1} = \frac{a+bi-1}{a+bi+1} = \frac{a-1+bi}{a+1+bi} \cdot \frac{a+1-bi}{a+1-bi} = \frac{(a^2+b^2)-1+2bi}{(a^2+b^2)+1 +2a} = \frac{bi}{1+a} $$
and therefore $\ \frac{z-1}{z+1}$ is imaginary
now let me assume $\ \frac{z-1}{z+1} $ is imaginary number, how could I conclude that $\ |z| =1 $ I really can't think of any direction..
Thanks
|
Recall that $w\in \mathbb{C}$ is purely imaginary $\iff w=-\bar w$, then
$$\frac{z-1}{z+1}=-\overline{\left(\frac{z-1}{z+1}\right)} =-\frac{\bar z-1}{\bar z+1} \iff (z-1)(\bar z+1)=-(\bar z-1)(z+1)$$
$$z\bar z+z-\bar z-1=-z\bar z+z-\bar z+1$$
$$2z\bar z=2 \iff|z|^2=1$$
|
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|
If $q\equiv 3 \pmod 4$ then is it true ${q+1 \choose 2}$ is semiprime if and only if $q=3$?
Question: If $q\equiv 3 \pmod 4$ then is it true ${q+1 \choose 2}$ is semiprime if and only if
$q=3.$
I am certain the answer to the question is yes, and I believe I have worked out a solution - although I am not certain it is correct; whence the solution-verification tag. I feel it is ok to also ask if there are solutions that are "deeper" - maybe using better applications of modular arithmetic or making use of group theory ?
Solution To Question: Since $q\equiv 3 \pmod 4$ then there exist a positive number $m\in\mathbb{N}$ such that $q=4m+3$. Then after substitution, ${q+1 \choose 2}={4m+4 \choose 2}=\frac{(4m+4)(4m+3)}{2}=(2m+2)(4m+3);$ which has even parity. In particular if $p$ is a prime number I can write $(2m+2)(4m+3)=2p.$ Division by two yields $\left(\frac{2m+2}{2}\right)\left(4m+3\right)=p.$ Recall a prime number has as divisors the numbers $1$ and itself. Note that $4m+3\neq 1$ for positive $m$ and so $\frac{2m+2}{2}=1;$ which implies $m=0$ in which case $4m+3=3.$
On the other hand if ${q+1 \choose 2}=6$ then explicitly $\frac{q(q+1)}{2}=6$ and so $q^2+q-12=0.$ Equivalently $(q-3)(q+4)=12;$ which has roots $q=3$ or $q=-4.$ We require the positive root $q=3.$ This completes the solution to the question. $\blacksquare$
|
You have $q=4m+3$, so $m=\frac{q-3}{4}$
\begin{align}
{q+1 \choose 2}={4m+4 \choose 2}&=(2m+2)(4m+3)=2(m+1)(4m+3)\\
&=2q(m+1)
\end{align}
When $m=0$ we have ${3+1 \choose 2}=6=2\cdot3$ as required. Now you always get $2$ and $q$ as two distinct factors along with $m+1$. Now $m+1$ is never equal to $q$, and equals $2$ when $m=1$, i.e., when $q=7$, so ${7+1 \choose 2}=28=2^2\cdot7$, which is not a semiprime. After this $m$ is distinct from $2$ or $q$ and the theorem is proved.
|
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|
Find a formula for $\left(\begin{smallmatrix} -4 & -15 \\ 2 & 7 \end{smallmatrix}\right)^n$ We're going to consider the matrix $\begin{pmatrix} -4 & -15 \\ 2 & 7 \end{pmatrix}.$
(a) Let $\mathbf{P} = \begin{pmatrix} 2 & 5 \\ 1 & 3 \end{pmatrix}$. Find the $2 \times 2$ matrix $\mathbf{D}$ such that
$\mathbf{P}^{-1} \mathbf{D} \mathbf{P} = \begin{pmatrix} -4 & -15 \\ 2 & 7 \end{pmatrix}.$
(b) Find a formula for $\mathbf{D}^n,$ where $\mathbf{D}$ is the matrix you found in part (a).
(You don't need to prove your answer, but explain how you found it.)
(c) Using parts (a) and (b), find a formula for $\begin{pmatrix} -4 & -15 \\ 2 & 7 \end{pmatrix}^n.$
I have completed part a (thanks for the helpful hints), but I am confused on part b. I solved for a few powers of $\mathbf{D}$ to find a pattern. So far I have if $\mathbf{D} = \begin{pmatrix} a & b \\ c & d \end{pmatrix}$ then $\mathbf{D}^{n} = \begin{pmatrix} -4^n & ? \\ 0 & -19^n \end{pmatrix}.$ I don't know if I'm solving for this correctly and as you can see, I'm not sure how to find the $b$ part of $\mathbf{D}^{n}.$ Thanks again!
|
Let consider
$$\begin{pmatrix} 3 & -5 \\ -1 & 2 \end{pmatrix}\begin{pmatrix} x & y \\ z & w \end{pmatrix}\begin{pmatrix} 2 & 5 \\ 1 & 3 \end{pmatrix}=\begin{pmatrix} -4 & -15 \\ 2 & 7 \end{pmatrix}\\$$
$$\iff
D=\begin{pmatrix} x & y \\ z & w \end{pmatrix}=\begin{pmatrix} 2 & 5 \\ 1 & 3 \end{pmatrix}\begin{pmatrix} -4 & -15 \\ 2 & 7 \end{pmatrix}\begin{pmatrix} 3 & -5 \\ -1 & 2 \end{pmatrix}=$$
$$=\begin{pmatrix} 2 & 5 \\ 1 & 3 \end{pmatrix}\begin{pmatrix} 3 & -10 \\-1 & -3 \end{pmatrix}=\begin{pmatrix} -4 & -5 \\ 0 & -19 \end{pmatrix}$$
|
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|
$ABCD$ a tetrahedron. $P$ and $Q$ be points on the sides $AB$ and $CD$ respectively. What is the locus of the midpoint of the line segment $PQ$? Let $ABCD$ a tetrahedron. $P$ and $Q$ be points on the sides $AB$ and $CD$ respectively. What is the locus of the midpoint of the line segment $PQ$?
The locus is the set of points $X$ satisfying the condition that $X$ is the midpoint of the line segment $PQ$. How do I go about finding and describing this set?
|
Denote points of tetrahedron with $A(0, 0, 0)$, $B(a, 0, 0)$, $C(\frac{a}2,\frac{a\sqrt 3}{2}, 0)$, $D(\frac{a}2,\frac{a\sqrt3}{6},\frac{a\sqrt6}3)$
Arbitrary points $P\in AB$ and $Q\in CD$ have the following coordinates:
$$P(pa,0,0), Q(\frac{a}2, \frac{a\sqrt3}{6}(3-2q),\frac{aq\sqrt6}3)$$
...with:
$$p,q\in[0,1]$$
Coordinates of the midpoint are:
$$M(\frac{a}4(2p+1), \frac{a\sqrt3}{12}(3-2q),\frac{aq\sqrt6}6)$$
or:
$$x=\frac{a}4(2p+1)\tag{1}$$
$$y=\frac{a\sqrt3}{12}(3-2q),\space z=\frac{aq\sqrt6}6\tag{2}$$
Obviously, you can pick the value for $x$ independently of $y,z$.
The range for $x$ is:
$$x_{p=0}=\frac{a}4,\space x_{p=1}=\frac{3a}4,$$
For any given $x$ the locus of points is given with (2) which is actually a straight line going from:
$$y_{q=0}=a\frac{\sqrt{3}}4, \space z_{q=0}=0$$
...to:
$$y_{q=1}=a\frac{\sqrt{3}}{12}, \space z_{q=1}=a\frac{\sqrt 6}6$$
So the locus of midpoints is actually a rectangle with the following coordinates:
$$(\frac{a}4, a\frac{\sqrt{3}}4, 0),\space(\frac{a}4, a\frac{\sqrt{3}}{12}, a\frac{\sqrt 6}6)$$
$$(\frac{3a}4, a\frac{\sqrt{3}}4, 0),\space(\frac{3a}4, a\frac{\sqrt{3}}{12}, a\frac{\sqrt 6}6)$$
|
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|
Prove that $3^{16} -33$ and $3^{15} +5$ is divisible by 4 by means of binomial theorem This is a question that I found in a textbook:
Given that $p=q+1$, $p$ and $q$ are integers, then show that
$p^{2n} - 2nq-1$ is divisible by $q^2$ given that $n$ is a positive integer.
By taking a suitable value of $n$, $p$ and $q$, show that $3^{16}-33$ and $3^{15}+5$ are divisible by 4.
My proof:
$$p^{2n}-2nq-1=(1+q)^{2n}-2nq-1$$
$$=[1+2nq+\frac{(2n)(2n-1)}{2!} q^2+\frac{2n(2n-1)(2n-2)}{3!}q^3+...]-2nq-1$$
$$=n(2n-1)q^2+\frac{2}{3} n(2n-1)(n-1)q^3+...$$
$$=q^2[n(2n-1)+\frac{2}{3} n(2n-1)(n-1)q+...]$$
Hence the expansion has a common factor $q^2$
Taking $n=8$ and $p=3$, and given that $p=q+1, q=2$, by substitution,
$$3^{16} -33=4[120+1120+...]$$
By factoring a 3:
$$3(3^{15}-11)=4[120+1120+...]$$
Dividing both sides by 3 and adding 15 to both sides:
$$3^{15} +5=4[\frac{1}{3}(120+1120+...)+4]$$
Then it is proven that it is also divisible by 4.
The only problem I have with the proof is that how do I know that each term in the brackets $(120+1120+...)$ are divisible by 3?
|
Use the Euclid lemma in last implication:
$$\begin{eqnarray}
3(3^{15}-11)=4[120+1120+...]&\implies &3\mid 4[120+1120+...] \\
&\implies & 3\mid 120+1120+...
\end{eqnarray}$$
|
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|
Show that $\iint_S (x^2+y^2) dA = 9 \pi /4$ In exam it was asked to show that
$$\iint_S x^2+y^2 dA = 9 \pi /4$$ for $$S = {\{(x, y, z) | x>0, y>0,3>z>0, z^2 = 3(x^2 + y^2)}\}$$
I have tried many times but I don't get the $9 \pi /4$.
$$\begin{align}
\iint_S\sqrt{1+f_x^2+f_y^2}\,dA &=\int_0^{\sqrt3}\int_0^{2\pi} r^2\sqrt{1+36r^2}\,r\,d\theta\,dr\\
&=2\pi\,\int_0^\sqrt3 r^3\sqrt{1+36r^2}\,dr = 2 \pi \dfrac{\left(36r^2+1\right)^\frac{3}{2}\left(54r^2-1\right)}{9720}\Bigg|_0^{\sqrt{3}} \\&= 2 \pi\dfrac{161{\cdot}109^\frac{3}{2}+1}{9720} \ne 9 \pi/4 \end{align}$$
Where I did it wrong?
|
Your problem arises from using the fact that $f_x = 6x$ and $f_y = 6y$. Indeed this isn't the case, as $z^2 = 3(x^2+y^2)$, not $z$. Thus you would have:
$$z=f(x,y) = \sqrt{3(x^2+y^2)}$$
$$f_x = \frac{\sqrt{3}x}{\sqrt{x^2+y^2}} \quad \quad f_y = \frac{\sqrt{3}y}{\sqrt{x^2+y^2}}$$ Then we have:
$$\iint_S (x^2+y^2)\sqrt{1+f_x^2 + f_y^2}dxdy = \iint_S (x^2+y^2)\sqrt{1+\frac{3x^2+3y^2}{x^2+y^2}} dxdy = \iint_S 2(x^2+y^2) dxdy$$
Now use polar coordinates and compute the integral. But be careful that $\theta \in \left(0,\frac \pi2\right)$, as $x,y$ are positive.
|
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|
prove that $\lim _{x\to 1}f(x)=3$ if $f(x)=\frac{x^3-1}{x-1}$ Prove that $\lim _{x\to 1}f(x)=3$ where $f:(0,\infty)\to \mathbb{R}$ is given by $f(x)=\frac{x^3-1}{x-1}$.
I proved by definition of the limit
$$|f(x)-3|=\left|\frac{x^3-1}{x-1}-3\right|=|x^2+x-2|=|x-1||x+2|\leq |x-1|(|x|+2)$$
how to processed from this
|
$x^3-1=(x-1)(x^2+x+1)$.
Hence $\lim_{x\to1}\frac{x^3-1}{x-1}=\lim_{x\to1}x^2+x+1=1^2+1+1=3$.
|
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|
Find $\lim_{x\to \infty}\left({x \over x^2+1}+{x \over x^2+2}+\cdots +{x\over x^2+x}\right)$ Find $$\lim_{x\to \infty}\left({x \over x^2+1}+{x \over x^2+2}+\cdots +{x\over x^2+x}\right)$$ , without using squeeze theorem.
I have done the solution as below using squeeze theorem ...
$$Let \left[\left({x \over x^2+1}+{x \over x^2+2}+\cdots +{x\over x^2+x}\right)\right]=f(x)\implies \\ \left({x \over x^2+x}+{x \over x^2+x}+\cdots +{x\over x^2+x}\right)\lt f(x)\lt \left({x \over x^2+1}+{x \over x^2+1}+\cdots +{x\over x^2+1}\right) \\ {x^2 \over x+x^2}\lt f(x) \lt {x^2\over 1+x^2}\\ \text{applying limit on both sides }\\ \implies\lim_{x\to \infty}{x^2 \over x+x^2}= \lim_{x\to \infty}{x^2\over 1+x^2}=1\\ \implies \lim_{x\to \infty}f(x)=1$$
Can we do this without squeeze theorem?
|
By geometric series we have
$$\frac x {x^2+k}=\frac1x\frac 1 {1+k/x^2}=$$$$=\frac1x\left(1-\frac{k}{x^{2}}+\left(-\frac{k}{x^{2}}\right)^2+\ldots\right)=\frac1x-\frac{k}{x^{3}}+\frac{k^2}{x^{5}}+\ldots$$
and therefore
$$\sum_{k=1}^x \frac x {x^2+k}=\sum_{k=1}^x \left(\frac1x-\frac{k}{x^{3}}+\frac{k^2}{x^{5}}+\ldots\right)=1-\frac{\sum_{k=1}^xk}{x^{3}}+\frac{\sum_{k=1}^xk^2}{x^{5}}+\ldots $$
$$\sim 1-\frac{x^2}{2x^{3}}+\frac{x^3}{3x^{5}}+\ldots\to 1$$
indeed recall that by Faulhaber's formula
$$\sum_{k=1}^xk^p \sim \frac{x^{p+1}}{p+1}$$
|
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|
How to prove that for all natural $n$, $133|11^{n+2}+12^{2n+1}$
How to prove that for every natural $n$, the number
$$11^{n+2} +12^{2n+1}$$
is divisible by $133$?
I tried Induction method, so assuming that
$$
n=k \implies
A = (11^{k+2} +12^{2k+1} ) \pmod{133} \equiv 0
$$
Then trying to prove that
$$
n=k+1 \implies
B = (11^{k+3} +12^{2k+3} ) \pmod{133} \equiv 0
$$
For this I wanted to split $B$ into multiplication with at least one multiple of $A$. But no luck.
Any Ideas?
|
$\displaystyle 11^{k + 3} + 12^{2k + 3} = 11 \times 11^{k + 2} +
\underbrace{\qquad 144\qquad}_{\displaystyle =\ 133 + 11} \times 12^{2k + 1} =
11\left(\color{red}{11^{k + 2} + 12^{2k + 1}}\right) + \color{red}{133} \times 12^{2k + 1}$
|
{
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "2",
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|
$\sqrt{n+\sqrt[3]{n+\sqrt[3]{n+\cdots \cdots \cdots }}}$ is a natural number, is
The number of natural number $n\leq 50$ such that
$\displaystyle \sqrt{n+\sqrt[3]{n+\sqrt[3]{n+\cdots \cdots \cdots }}}$ is a natural number, is
Try: Let $\displaystyle x=\sqrt[3]{n+\sqrt[3]{n+\cdots \cdots \cdots }}$
So $\displaystyle x=\sqrt[3]{n+x}\Rightarrow x^3=n+x\Rightarrow x^3-x-n=0$
could some help me how to solve it, Thanks
|
By the rational root theorem, a solution of $$x^3-x-n=0$$ must be a divisor of $-n.$ Since $n\leq50$ such a solution can't be very large. By definition of $x$ we have $x\geq0$. If $x=4,$ then $x^3 =64$ and there's already no hope. So $x$ must be one of $0, 1, 2, 3.$
$$\begin{align}
x&=0\implies n=0\\
x&=1\implies n=0\\
x&=2\implies n=6\\
x&=3\implies n=24
\end{align}$$
Now you have to test these combinations of $n$ and $x$ to see $n+x$ is a perfect square. The only ones that work are the ones where $n=0$ and obviously $x=1$ is wrong in that case. So we are left with $n=0,$ which isn't very interesting.
I suspect that there really is supposed to be a cube root on the outside, and the answer is $n=24, x=3.$
|
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|
Solve $a^2 - 2b^2 - 3 c^2 + 6 d^2 =1 $ over integers $a,b,c,d \in \mathbb{Z}$ Are we able to completely solve this variant of Pell equation?
$$ x_1^2 - 2x_2^2 - 3x_3^2 + 6x_4^2 = 1 $$
This has an interpretation as is related to the fundamental unit equation of $\mathbb{Q}(\sqrt{2}, \sqrt{3}) = \mathbb{Q}[x,y]/(x^2 - 2, y^2 - 3)$ as well as various irreducible quartics. Is this the same as solving three separate Pell equations?
\begin{eqnarray*}
x^2 - 2y^2 &=& 1 \\
x^2 - 3y^2 &=& 1 \\
x^2 + 6y^2 &=& 1 \tag{$\ast$}
\end{eqnarray*}
Our instinct suggests there should be three degrees of freedom here, and setting different variables to zero we could find three two of these (the third equation has no solutions over $\mathbb{R}$). Does that generate all the solutions?
Perhaps I should remark this quadratic form is also a determinant
$$ a^2 - 2b^2 - 3c^2 + 6d^2 = \det \left[
\begin{array}{cc}
a + b \sqrt{2} & c - d \sqrt{2}\\
3(c + d \sqrt{2}) & a - b \sqrt{2}
\end{array}
\right]$$
This might not even contain Oscar's solution. Extending Keith's solution We could have:
$$
\left[
\begin{array}{cc}
a + b \sqrt{2} & c - d \sqrt{2}\\
3(c + d \sqrt{2}) & a - b \sqrt{2}
\end{array}
\right] =
\left[
\begin{array}{cc}
3 + 1 \sqrt{2} & 2 - 1 \sqrt{2}\\
3(2 + 1 \sqrt{2}) & 3 - 1 \sqrt{2}
\end{array}
\right]^n
$$
|
$$x^2-2y^2-3z^2+6q^2=1$$
Use what any decision $a^2-2b^2=1$ and $c^2-2d^2-3k^2+6t^2=1$
$$x=ac\pm{2bd}$$
$$y=ad\pm{bc}$$
$$z=ak\pm{2bt}$$
$$q=at\pm{bk}$$
|
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|
Calculate $A^5 - 27A^3 + 65A^2$, where $A$ is the matrix defined below. If $A=\begin{bmatrix} 0 & 0 & 1 \\
3 & 1 & 0 \\
-2&1&4\end{bmatrix}$, find $A^5 - 27A^3 + 65A^2$
$$A=\begin{bmatrix} 0 & 0 & 1 \\
3 & 1 & 0 \\
-2&1&4\end{bmatrix}$$
Let $\lambda$ be its eigenvalue, then
$$(A-\lambda I) = \begin{bmatrix} 0-\lambda & 0 & 1 \\
3 & 1-\lambda & 0 \\
-2&1&4-\lambda\end{bmatrix}$$
$$|A-\lambda I| = -(\lambda)^3 + 5(\lambda)^2 - 6(\lambda) +5$$
Using Cayley-Hamilton theorem
$$A^3-5^2+6A-5=0$$
How do I use this find $A^5 - 27A^3 + 65A^2$?
|
To add to the above answer, using $A^3 = 5A^2 - 6A + 5$ to simplify further $A^2(5A^2 -33A + 70) = A(5(5A^2 - 6A + 5) - 33A^2 + 70A) = A(-8A^2 + 40A + 25) = -8(5A^2 -33A + 70) + 40A^2 + 25A = 289A - 560$
|
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|
Solve: $\log_3(5(2^2+3^2)(2^4+3^4)(2^8+3^8)(2^{16}+3^{16})(2^{32}+3^{32})+2^{64})=?$
$$\log_3(5(2^2+3^2)(2^4+3^4)(2^8+3^8)(2^{16}+3^{16})(2^{32}+3^{32})+2^{64})=?$$
At a first glance, it seems that I need to do this:
$$\log_3((2+3)(2^2+3^2)(2^4+3^4)(2^8+3^8)(2^{16}+3^{16})(2^{32}+3^{32})+2^{64})=?$$
But afterwards, I can't find an algebric manipulation that will lead me to the solution, I took the exercise out of one of the entry tests of TAU.
|
Since
\begin{align*}
&\phantom{==}(3-2)(3+2)(3^2+ 2^2)(3^4+2^4)\cdots(3^{32}+2^{32})\\
&= (3^2-2^2)(3^2 + 2^2)(3^4+2^4)\cdots(3^{32}+2^{32})\\
&= (3^4-2^4)(3^4+2^4)(3^8+2^8)\cdots(3^{32}+2^{32})\\
&= \cdots\\
&= 3^{64}-2^{64},
\end{align*}
the result is $\log_3(3^{64}) = 64$. The trick is $(a-b)(a+b)= a^2 - b^2$.
|
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|
knowing: $ \tan x=2-\sqrt{3}$ , obtain: $ \cos 2x$
Knowing: $$ \tan x=2-\sqrt{3} $$
Obtain: $$\cos2x$$
I tried converting $\tan x$ into it's sinus and cosine form and trying to square both sides to try to get the form of:
$$\cos^2x-\sin^2x$$
But I can't really get to this form without having extra expressions of sine or cosine, any ideas how to start this properly?
Taken out of one of the entry tests to Maths in TAU.
Solution:
$$\cos2x=\cos^2x-\sin^2x=\frac{\cos^2x-\sin^2x}{\sin^2x+\cos^2x}:\frac{\cos^2x}{\cos^2x}=\frac{1-\tan^2x}{1+\tan^2x}$$
$$ \frac{1-\tan^2x}{1+\tan^2x}=\frac{-6+4\sqrt{3}}{8-4\sqrt{3}}=\frac{-3+2\sqrt{3}}{4-2\sqrt{3}} $$
|
Draw a right triangle with the opposite side being $2 - \sqrt3$, and the adjacent side being $1$ so that $\tan x = 2 - \sqrt{3}$. Then the hypotenuse is $\sqrt{1^2 + (2 - \sqrt{3})^2} = \sqrt{1 + 4 - 4\sqrt3 + 3} = \sqrt{8 - 4 \sqrt{3}}$, and so $\cos^2 x = \frac{1}{8 - 4 \sqrt{3}}$ as $\cos x$ is adjacent/hypotenuse.
Now $\cos 2x = 2 \cos^2 x - 1$, which is $\frac{1}{4 - 2 \sqrt3} - 1$ or $\frac{4 + 2 \sqrt3}{4} - \frac{4}{4} = \frac{\sqrt 3}{2}$.
|
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|
How to prove $b \ll a (a^{2}b^2)^{2.1/k} $ In the paper linear forms in the logarithms of real algebraic numbers close to 1, it is
written on page $9$ that
It follows from (13) and Lemma 2 with $\eta=0.1 $ we have that,
$$b \ll a (a^{2}b^2)^{2.1/k} $$
The image of page 9 :
How do we prove $b \ll a (a^{2}b^2)^{2.1/k} $?
My Confusion : Note, if $k$ increases then ${2.1/k}$ goes close to $0$, so it is confusing
how the inequality holds true.
|
We know,
$ \left|\frac{a+1}{a}- \left(\frac{xz}{y^2}\right)^k\right|\leq \frac{1}{b} \cdots (13)$
Here, $xz= ((a+1)(ab^2+1))^\frac{1}{k}, y^2= (ab+1)^\frac{2}{k}.$
So, $ \left|\frac{a+1}{a}- \frac{(xz)^k}{(ab+1)^2} \right|\leq \frac{1}{b} \cdots (a)$
Since, $(ab+1)>ab\implies(ab+1)^\frac{2}{k}>(ab)^\frac{2}{k}$
$\implies \frac{1}{(ab+1)^\frac{2}{k}}<\frac{1}{(ab)^\frac{2}{k}} \implies \frac{(xz)^k}{(ab+1)^\frac{2}{k}}<\frac{(xz)^k}{(ab)^\frac{2}{k}}$
$\implies -\frac{(xz)^k}{(ab+1)^\frac{2}{k}}> -\frac{(xz)^k}{(ab)^\frac{2}{k}}$ [since, the inequality sign
changes when both sides are multiplied by $-1$.]
$\implies-\frac{(xz)^k}{(ab)^\frac{2}{k}} < -\frac{(xz)^k}{(ab+1)^\frac{2}{k}}$ [changing sides]
$\implies (\frac{a+1}{a}) -\frac{(xz)^k}{(ab)^\frac{2}{k}} < (\frac{a+1}{a}) -\frac{(xz)^k}{(ab+1)^\frac{2}{k}}$
$\implies \left|\frac{a+1}{a}- \frac{(xz)^k}{(ab)^\frac{2}{k}} \right| \leq \left|\frac{a+1}{a}- \frac{(xz)^k}{(ab+1)^\frac{2}{k}} \right|$
Using inequality (a), we deduce-
$\left|\frac{a+1}{a}- \frac{(xz)^k}{(ab)^\frac{2}{k}} \right| \leq \frac{1}{b} \cdots (b)$
By inspection, we see- $ \left|(\frac{a+1}{a})^\frac{1}{k} \right |\leq \left|\frac{a+1}{a} \right | $
$\implies \left|(\frac{a+1}{a})^\frac{1}{k}- \frac{(xz)^k}{(ab)^\frac{2}{k}}\right |\leq \left|\frac{a+1}{a}- \frac{(xz)^k}{(ab)^\frac{2}{k}} \right | $
[subtracting $\frac{(xz)^k}{(ab)^\frac{2}{k}}$ on both sides]
Using inequality (b), we deduce-
$\left|(\frac{a+1}{a})^\frac{1}{k}- \frac{(xz)^k}{(ab)^\frac{2}{k}}\right |\leq \frac{1}{b}\ \cdots (c) $
To use Lemma 2, we need to re-parameterize $p, q$ of Lemma 2
according to inequality $(13)$, so
let, $p= (xz)^k, q= (ab)^\frac{2}{k} $, then-
$ \left|(1+\frac{1}{a})^\frac{1}{k}- \left(\frac{p}{q}\right)\right| > \frac{c_5}{aq^{\epsilon +2 }} $ [ Lemma 2]
$\implies \left|(1+ \frac{1}{a})^\frac{1}{k}- \frac{(xz)^k}{(ab)^\frac{2}{k}} \right| > \frac{c_5}{a(ab)^{\frac{2(\epsilon +2 )}{k}}} $
$\implies \frac{1}{b} > \frac{c_5}{a(ab)^{\frac{2(\epsilon +2 )}{k}}} $ [ using inequality (c)]
$\implies a(a^2b^2)^{\frac{2.1}{k}} >bc_5 $ [$\epsilon =0.1$]
$\implies a(a^2b^2)^{\frac{2.1}{k}} >b \implies b \ll_k a(a^2b^2)^{\frac{2.1}{k}} $
|
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|
Evaluate $\lfloor n / \lfloor n / \lfloor \sqrt n \rfloor \rfloor \rfloor$ for positive integers $n$. I'm considering:
$$
\left\lfloor {n \over \lfloor n / \lfloor \sqrt n \rfloor\rfloor} \right\rfloor \:\:\:\: \forall n \in \mathbf N^+
$$
which seems to be $\lfloor \sqrt n \rfloor$.
Is there any way to prove or disprove?
|
Let $n$ be a real number such that $n\geq 1$, and $m:=\left\lfloor\sqrt{n}\right\rfloor$. Then, we have
$$m^2\leq n< (m+1)^2=m^2+2m+1\leq m^2+3m\,.$$
Thus,
$$m\leq \frac{n}{m}< m+3\,.$$
Set $k:=\left\lfloor\dfrac{n}{m}\right\rfloor$, so that $k\in\{m,m+1,m+2\}$. If $k=m$, then $m^2\leq n<m(m+1)$. That is,
$$m\le\frac{n}{k}<m+1\,,\text{ whence }\left\lfloor\frac{n}{k}\right\rfloor=m\,.$$
If $k=m+1$, then $m(m+1)\leq n<m(m+2)$. Thus,
$$m\leq \frac{n}{k}<\frac{m(m+2)}{m+1}=\frac{(m+1)^2-1}{m+1}<m+1\,,$$
so $\left\lfloor\dfrac{n}{k}\right\rfloor=m$. If $k=m+2$, then $m(m+2)\leq n<(m+1)^2$. Hence,
$$m\leq \frac{n}{k}<m+\frac{1}{m+2}<m+1\,,\text{ making }\left\lfloor\frac{n}{k}\right\rfloor=m\,.$$
In other words,
$$\left\lfloor\frac{n}{\left\lfloor\frac{n}{\left\lfloor\sqrt{n}\right\rfloor}\right\rfloor}\right\rfloor=\left\lfloor\sqrt{n}\right\rfloor$$
for all real numbers $n\geq 1$.
|
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|
Finding a closed form for $\sum_{n = 1}^{\infty} \frac{1}{(1 + x^2)^{n}}$ So I got the sum
$$\sum_{n = 1}^{\infty} \frac{1}{(1 + x^2)^{n}}$$
and I want a closed form.
Can I just do
$$\frac{1}{1 - r} = \frac{1}{1 - \frac{1}{1 + x^2}} = \frac{1}{x^2} + 1 $$
for a closed form?
|
Another way of solving it is by bringing it to the standard form, then you can use the formula for infinite geometric series considering $x \ne 0$
take $n=m+1$, so you have:
$$ \sum_{m = 0}^{\infty} \frac{1}{(1 + x^2)^{m+1}}= \frac{1}{(1 + x^2)} \sum_{m = 0}^{\infty} \frac{1}{(1 + x^2)^{m}}=\frac{1}{(1 + x^2)}\cdot \frac{(1 + x^2)}{ x^2}=\frac{1}{x^2}$$
|
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|
If $x>0$ real number and $n>1$ integer, then $(1+x)^n>\frac{1}{2}n(n-1)x^2$ If $x>0$ real number and $n>1$ integer, then $(1+x)^n>\frac{1}{2}n(n-1)x^2$.
Is there a way to prove it without using the Binomial Theorem? Is it possible to use the Bernoulli's Inequality to prove it? If yes, please show me.
I tried to prove it by induction on $n$, where $x>0$ is fixed. But it doesn't let me reach the conclusion. Here's what I did:
Fix $x>0$. We instead prove that
\begin{equation*}
\frac{2(1+x)^n}{n(n-1)}> x^2
\end{equation*}
for all $n>1$. Clearly, this is true for $n=2$. Assume this statement holds for $n=k$, for some $k\geq 2$. We have
\begin{equation*}
\frac{2(1+x)^{k+1}}{(k+1)k}=\frac{2(1+x)^k}{k(k-1)}\frac{(x+1)(k-1)}{k+1}>\frac{(x+1)(k-1)}{k+1}x^2.
\end{equation*}
Since $y\mapsto \frac{y-1}{y+1}$ is increasing on $\mathbb{R}\setminus \{-1\}$, we get $\frac{y-1}{y+1}\geq \frac{1}{3}$ for all $y\geq 2$. Therefore,
\begin{equation*}
\frac{(x+1)(k-1)}{k+1}x^2\geq \frac{1}{3}(x+1)x^2>\frac{1}{3}x^2.
\end{equation*}
As you can see, I can not reach $>x^2$ instead of $>\frac{1}{3}x^2$.
|
As an alternative we can proceed by induction for the stronger
$$(1+x)^{n}>1+nx+\frac12n(n-1)x^2>\frac12n(n-1)x^2$$
that is for the induction step
$$(1+x)^{n+1}=(1+x)(1+x)^{n}\stackrel{Ind.Hyp.}>1+nx+\frac12n(n-1)x^2+x+nx^2+\frac12n(n-1)x^3>$$$$\stackrel{?}>1+(n+1)x+\frac12(n+1)nx^2$$
therefore we need to prove that
$$1+nx+\frac12n(n-1)x^2+x+nx^2+\frac12n(n-1)x^3\stackrel{?}>1+(n+1)x+\frac12(n+1)nx^2$$
which is true indeed
$$\frac12n(n-1)x^2+nx^2+\frac12n(n-1)x^3\stackrel{?}>\frac12(n+1)nx^2$$
$$\frac12\left(n(n-1)+2n\right)x^2+\frac12n(n-1)x^3\stackrel{?}>\frac12(n+1)nx^2$$
$$\frac12\left(n^2+n\right)x^2+\frac12n(n-1)x^3\stackrel{?}>\frac12(n+1)nx^2$$
$$\frac12(n+1)nx^2+\frac12n(n-1)x^3\stackrel{?}>\frac12(n+1)nx^2$$
$$\frac12n(n-1)x^3>0$$
for $n>1$.
|
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|
Value of 'a' for which $f(x)=\frac{x^3}{3} +\frac{x^2}{2} +ax+b $ is injective Let $f(x)=\frac{x^3}{3} +\frac{x^2}{2} +ax+b \forall x \in R$.
Find the least value of a for which $f(x)$ is injective function.
My approach , for $f(x)$ to be injective either f(x) should be increasing or decreasing function.'
$Y=f'(x)=x^2+x+a$ [Increasing function]
If $Y>0$, then $x^2+x+a>0$, it is possible when $1-4a<0$ or $a>\frac{1}{4}$
If $Y<0$, then $x^2+x+a<0$, it is possible when $1-4a \ge 0$ or $a \le \frac{1}{4}$
The options are (A) $\frac{1}{4}$ which is the correct option, the other given options are (B) $1$, (C) $\frac{1}{2}$, (D) $\frac{1}{8}$
Though my optin is correct but i have one doubt when the equality=$0$, then Y=$0$ hence it is neither increasing nor decreasing function.
|
Your idea of computing $f'$ is fine, but your computation is wrong. In fact$$f'(x)=x^2+x+a=\left(x+\frac12\right)^2+a-\frac14.$$Can you take it from here?
|
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|
Ask a about hard integral of $\int_{0}^{\infty} \log x \log (\frac{a^2}{x^2}+1) \log(\frac{b^2}{x^2}+1)dx$ I want to evaluate the integral:
$$I(a,b)=\int_{0}^{\infty} \log x \log (\frac{a^2}{x^2}+1) \log(\frac{b^2}{x^2}+1)dx$$
Attempt:$$\frac{\partial ^2I}{\partial a\partial b}=4ab\int_{0}^{\infty}\frac{\log x}{(a^2+x^2)(b^2+x^2)}dx=\frac{4ab}{b^2-a^2}\int_{0}^{\infty}\log x\left(\frac{1}{a^2+x^2}-\frac{1}{b^2+x^2}\right)dx$$
$$=\frac{4ab}{b^2-a^2}\frac{\pi}{2}\left(\frac{\log a}{a}-\frac{\log b}{b}\right)=\frac{2\pi(b\log a-a\log b)}{b^2-a^2}$$
Then $$I(a,b)=2\pi\int_{0}^{b}\int_{0}^{a}\frac{(y\log x-x\log y)}{y^2-x^2}dxdy$$
But this integral very hard to solve,can anyone help me,thank you!
|
One way is to replace $\ln x$ with $x^p$, then the integrand becomes a product of two linear Meijer G-functions after the change of variables $t = 1/x^2$. We obtain
$$I(p) = \int_0^\infty x^p
\ln \left(1 + \frac {a^2} {x^2} \right)
\ln \left(1 + \frac {b^2} {x^2} \right) dx = \\
\frac 1 2 \int_0^\infty t^{(-3-p)/2}
G_{2, 2}^{1, 2} \left( a^2 t \middle| {1, 1 \atop 1, 0} \right)
G_{2, 2}^{1, 2} \left( b^2 t \middle| {1, 1 \atop 1, 0} \right) dt = \\
\frac {a^{1+p}} 2 G_{4, 4}^{3, 3} \left( \frac {b^2} {a^2} \middle|
{1, 1, \frac {1+p} 2, \frac {3+p} 2 \atop
1, \frac {1+p} 2, \frac {1+p} 2, 0} \right),$$
which is expressible in terms of the Lerch transcendent. Then
$$\int_0^\infty \ln x
\ln \left(1 + \frac {a^2} {x^2} \right)
\ln \left(1 + \frac {b^2} {x^2} \right) dx =
I'(0) = \\
-\pi \left( \frac {a \omega (1-\omega)} 2
\Phi \!\left( \omega^2, 2, \frac 1 2 \right) + \frac {\pi^2 b} 2 - \\
(2b(1 - \ln b) - (a-b) \ln(1-\omega)) \ln \omega -
(a+b) (\ln(a b) - 2) \ln(1+\omega) \right), \\
0 < b < a, \quad\omega = \frac b a.$$
|
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|
Polar form of $z^2+1$, $z\in \mathbb C$ Let $z\in \mathbb C$. Can we write $z^2+1 = re^{i\theta}$, for some $r\in \mathbb R$ and $\theta \in (-\pi, \pi]$?
We have $z^2+1 = r^2e^{i2\theta}+e^{i0}$ or we can write $z^2+1 = x^2-y^2+1 + i2xy$. We can use formulas to calculate $r$ and $\theta$ of $z^2+1$, but the process seems to be long. From the geometric point of view, I think $z^2+1$ is stretching the length of $z$ to $r^2$ and rotating the angle $\theta$ to $2\theta$, then shifting the vector to the right by a unit.
My ultimate goal is to calculate $\frac{d}{dz}Log(z^2+1)$. I think we can write $Log(z^2+1)$ in terms of $x$ and $y$, then use the formula $f'(z) = u_x + iv_x$, but I would like to see if I can use $f'(z) = e^{-i\theta}(u_r + iv_r)$.
|
First of all, notice that $z^{2} + 1 = (z-i)(z+i)$. Thus we have:
\begin{align*}
|z^{2} + 1| = |z-i||z+i|
\end{align*}
If we agree that $z = \rho e^{i\theta}$, we get
\begin{align*}
z - i = \rho\cos(\theta) + i(\rho\sin(\theta) - 1) \Rightarrow |z - i|^{2} & = \rho^{2}\cos^{2}(\theta) + (\rho\sin(\theta) - 1)^{2}\\
& = \rho^{2} - 2\rho\sin(\theta) + 1
\end{align*}
Similarly, we have:
\begin{align*}
z + i = \rho\cos(\theta) + i(\rho\sin(\theta) + 1) \Rightarrow |z+i|^{2} & = \rho^{2}\cos^{2}(\theta) + (\rho\sin(\theta) + 1)^{2}\\
& = \rho^{2} + 2\rho\sin(\theta) + 1
\end{align*}
From whence we obtain:
\begin{align*}
|z^{2} + 1|^{2} = |z-i|^{2}|z+i|^{2} = (\rho^{2} + 1)^{2} - 4\rho^{2}\sin^{2}(\theta)
\end{align*}
Based on this, it is possible to find the argument of $z^{2} + 1$ as well, since we have
\begin{align*}
\begin{cases}
\displaystyle\cos(\theta_{1}) = \frac{\mathrm{Re}(z-i)}{|z - i|} = \frac{\rho\cos(\theta)}{\sqrt{\rho^{2} - 2\rho\sin(\theta)+1}}\\
\displaystyle\cos(\theta_{2}) = \frac{\mathrm{Re}(z+i)}{|z+i|} = \frac{\rho\cos(\theta)}{\sqrt{\rho^{2}+2\rho\sin(\theta)+1}}
\end{cases}
\end{align*}
\begin{align*}
\therefore z^{2} + 1 = \sqrt{(\rho^{2}+1)^{2} - 4\rho^{2}\sin^{2}(\theta)}\times e^{i\alpha}\,\,\text{where}\,\,\alpha = \theta_{1} + \theta_{2}
\end{align*}
|
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|
Tricks to find a closed form $x_n$ of recurrence relation $x_{n+1} = \frac{1}{1+x_n}$ where $x_1 = a > 0$, $n \in \mathbb N$ With the help of you guys I've been able to learn how to solve various recurrence equations, but today I came across one I couldn't handle:
Find a closed form of the sequence $\{x_n\}$ of recurrence relation (where $n \in \mathbb N$):
$$
\begin{cases} x_1 = a > 0\\x_{n+1} = \dfrac{1}{1+x_n}\end{cases}
$$
I've already made several attempts which include:
*
*Trying to infer the pattern by expanding the continuous fraction from bottom up. No result
*Expressing $x_{n+2}$ and multiply/add/subtract/divide expressions for $x_{n+1}$ and $x_{n+2}$. No result
*Suppose that $x_{n}$ has a rational form $k_n \over p_n$ and playing with the equations in different ways. No result
I believe there should be some clever trick to untangle this, but apparently i'm too dumb to see it.
The answer is known to be a fraction involving the Golden Ratio $\frac{\sqrt5 - 1}{2}$ so maybe this may help.
Update
Using @AnotherJohnDoe's hint we may perform the following transformation:
Let $x_n = \frac{t_n}{t_{n+1}}$, then:
$$
\frac{t_{n+1}}{t_{n+2}} = \frac{1}{1+\frac{t_n}{t_{n+1}}} = \frac{t_{n+1}}{t_n + t_{n+1}}
$$
Which defines a recurrence relation in terms of $t$:
$$
t_{n+2} - t_{n+1} - t_n = 0
$$
The characteristic equation for this recurrence has two root so the closed form of $\{t_n\}$ is:
$$
t_n = C_1\cdot\left({1+\sqrt5\over 2}\right)^n + C_2 \cdot \left({1-\sqrt5 \over 2}\right)^n
$$
Let $\phi_1 = \frac{1+\sqrt5}{2}$ and $\phi_2 = \frac{1-\sqrt5}{2}$:
$$
t_n = C_1\cdot \phi_1^n + C_2\cdot\phi_2^n \\
t_{n+1} = C_1\cdot \phi_1^{n+1} + C_2\cdot\phi_2^{n+1}
$$
Then:
$$
x_n = \frac{t_n}{t_{n+1}} \\
x_n = \frac{C_1\cdot \phi_1^n + C_2\cdot\phi_2^n}{C_1\cdot \phi_1^{n+1} + C_2\cdot\phi_2^{n+1}}
$$
So using the initial conditions $x_1 = a$ and $x_2 = {1\over 1+a}$:
$$
\begin{align}
a &= \frac{C_1\cdot \phi_1 + C_2\cdot\phi_2}{C_1\cdot \phi_1^2 + C_2\cdot\phi_2^2} \\
{1\over 1+a} &= \frac{C_1\cdot \phi_1^2 + C_2\cdot\phi_2^2}{C_1\cdot \phi_1^3 + C_2\cdot\phi_2^3}
\end{align}
$$
But this system doesn't seem to have solutions.
|
When stuck, there's always the option of computing the first few terms of the sequence, to see if a pattern appears.
Computing the first $6$ terms, we get
\begin{align*}
x_1&=a\\[4pt]
x_2&=\frac{1}{1+a}\\[4pt]
x_3&=\frac{1+a}{2+a}\\[4pt]
x_4&=\frac{2+a}{3+2a}\\[4pt]
x_5&=\frac{3+2a}{5+3a}\\[4pt]
x_6&=\frac{5+3a}{8+5a}\\[4pt]
\end{align*}
so it appears to be the case that
$$x_n=\frac{F_{n-1}+aF_{n-2}}{F_n+aF_{n-1}}$$
where $F_n$ is the $n$-th Fibonacci number (using $F_{-1}=1$ and $F_{0}=0$).
The proof by induction is straightforward . . .
For $n=1$, we have
$$\frac{F_0+aF_{-1}}{F_1+aF_0}=\frac{a}{1}=a=x_1$$
so the base case is verified.
Suppose the claim holds for some positive integer $n$.$\;$Then
\begin{align*}
x_{n+1}&=\frac{1}{1+x_n}\\[4pt]
&=\frac{1}{1+{\Large{\frac{F_{n-1}+aF_{n-2}}{F_n+aF_{n-1}}}}}\\[4pt]
&=\frac{F_n+aF_{n-1}}{(F_{n-1}+F_n)+a(F_{n-2}+F_{n-1})}\\[4pt]
&=\frac{F_n+aF_{n-1}}{F_{n+1}+aF_n}\\[4pt]
\end{align*}
which completes the induction.
|
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|
Choosing an interval of the CDF to find each quartile I have a random variable $X$ which has the following CDF:
$$F(y) = \left\{\begin{array}{ll}
0 & : y \lt 0\\
\frac{y}{30} & : 0 \le y \lt 20\\
\frac{2}{3} + \frac{y-20}{60} & : 20 \le y \lt 40\\
1 & : y \ge 40
\end{array}
\right.$$
To find the median of $X$, I know I need to plug $q_2$ into the CDF and set it equal to 0.5:
$$ F(q_2) = 0.5 $$
and solve for $q_2$. But which interval of the CDF should I use? My intuition tells me it would be the third interval since 20 falls in the middle of 0 and 40 and this is the interval that would be used for $F(20)$, but I have a feeling this is wrong. Which one should I use for $q_2$, as well as $q_1$ and $q_3$?
|
If your random variable $X$ has the cdf
$$F(y) = \left\{\begin{array}{ll}
0 & : y \lt 0\\
\frac{y}{30} & : 0 \le y \lt 20\\
\frac{2}{3} + \frac{y-20}{60} & : 20 \le y \lt 40\\
1 & : y \ge 40
\end{array}
\right.$$
you're supposed to find the inverse so you get $F(y) = \frac{y}{30} $ for $y$. then the inverse is $30y$. If plug in $F^{-1}(.5) = \frac{1}{2}30 = 15$
I'll just note you switch from $X$ to $y$. You can also see it has $\frac{2}{3} + \frac{y-20}{60} $. If you plug in $y=20$ you get $\frac{2}{3}$
|
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|
Classifying singular points of $\frac{\sin(z^2)}{z^3-\frac{\pi}{4}z^2}$
I am trying to classify the singular points of the function $$f(z)=\frac{\sin(z^2)}{z^3-\frac{\pi}{4}z^2}.$$
My attempt:
$$f(z)=\frac{\sin(z^2)}{z^3-\frac{\pi}{4}z^2}=\frac{\sin(z^2)}{z^2\left(z-\frac{\pi}{4}\right)}.$$
Hence the singular points are $z=0,\frac{\pi}{4}$. For classification:
$z=\frac{\pi}{4}$ is a simple pole as $\left(z-\frac{\pi}{4}\right)$ is a simple zero.
For $z=0$, we can see that $z^2\left(z-\frac{\pi}{4}\right)$ is a zero of order $2$. Also,$\ \sin(z^2)$ is a zero of order $2$ for $z=0$. Hence, $z=0$ is a removable singularity.
I am unsure about the $z=0$ case, particularly if $\sin(z^2)$ is a zero of order $2$ for $z=0$.
|
It is correct.
Also you can notice that
$$\lim_{z\to 0} \frac{\sin(z^2)}{z^2\left(z-\frac{\pi}{4}\right)}= \lim_{z\to 0} \frac{\sin(z^2)}{z^2} \cdot \lim_{z\to 0}\frac1{z-\frac\pi4} = 1 \cdot \frac1{-\frac\pi4} = -\frac4\pi$$
so $z = 0$ is a removable singularity.
|
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|
Prove by induction that, for all positive integers n, the following inequality holds: (1-2-1)(1-2-2)...(1-2-n) ≥ 1/4 + 2-(n+1)
My solution so far:
After verifying the base case for n = 1, I began the inductive step:
(1-2-1)(1-2-2)...(1-2-(k+1)) ≥ 1/4 + 2-(k+2)
Subtracting 1/4 + 2-(k+1) on the RHS as it's an inequality:
(1-2-1)(1-2-2)...(1-2-(k+1)) ≥ 1/4 + 2-(k+2) - (1/4 + 2-(k+1))
1 - 2-(k+1) ≥ 2-(k+2) - 2-(k+1)
Taking common denominator on RHS (skipped a few algebraic steps):
(2k+1 - 1)/2k+1 ≥ 1/(2k+2) - 1/(2k+1)
Taking common denominator on LHS resulting in:
(2k+1 - 1)/2k+1 ≥ - 1/(2-(k+1))
It's my first time learning about induction so I'm wondering if the last step is sufficient for proving the proposition since the numerator on the LHS is greater than the numerator on the RHS, and they have the exact same denominator, or is this just the completely wrong approach? Any hints and/or tips would be helpful. Thanks.
|
I'd avoid starting the induction step from what you need to prove; rather
$$
\left(1-\frac{1}{2}\right)\left(1-\frac{1}{2^2}\right)\dots
\left(1-\frac{1}{2^k}\right)\left(1-\frac{1}{2^{k+1}}\right)\ge
\left(\frac{1}{4}+\frac{1}{2^{k+1}}\right)\left(1-\frac{1}{2^{k+1}}\right)
$$
by the induction hypothesis.
The right hand side can be rewritten
$$
\left(\frac{1}{4}+\frac{1}{2^{k+1}}\right)\left(1-\frac{1}{2^{k+1}}\right)=
\frac{1}{4}+\frac{1}{2^{k+1}}-\frac{1}{2^{k+3}}-\frac{1}{2^{2k+2}}
$$
and now we can try to see whether
$$
\frac{1}{4}+\frac{1}{2^{k+1}}-\frac{1}{2^{k+3}}-\frac{1}{2^{2k+2}}
\ge
\frac{1}{4}+\frac{1}{2^{k+2}}
$$
that is,
$$
\frac{1}{2^{k+1}}\ge\frac{1}{2^{k+2}}+\frac{1}{2^{k+3}}+\frac{1}{2^{2k+2}}
$$
If we multiply both sides by $2^{k+1}$ we obtain the equivalent
$$
1\ge\frac{1}{2}+\frac{1}{4}+\frac{1}{2^{k+1}}
$$
that is,
$$
\frac{1}{4}\ge\frac{1}{2^{k+1}}
$$
or $2^{k+1}\ge 4$ that's true as soon as $k\ge1$. Therefore
$$
\left(1-\frac{1}{2}\right)\left(1-\frac{1}{2^2}\right)\dots
\left(1-\frac{1}{2^k}\right)\left(1-\frac{1}{2^{k+1}}\right)\ge
\left(\frac{1}{4}+\frac{1}{2^{k+1}}\right)\left(1-\frac{1}{2^{k+1}}\right)
\ge
\frac{1}{4}+\frac{1}{2^{k+2}}
$$
|
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|
Finding value of $\int^{2\pi}_{0}\frac{\sin^2(x)}{a-b\cos x}dx$ without contour Integration
Finding value of $\displaystyle \int^{2\pi}_{0}\frac{\sin^2(x)}{a-b\cos x}dx\;\;, a>b>0$ without contour Integration
Try: Let $\displaystyle I =\int^{2\pi}_{0}\frac{\sin^2 x}{a-b\cos x}dx=2\int^{\pi}_{0}\frac{\sin^2 x}{a-b\cos x}dx$
Using Integration by parts
$$ I =\frac{2}{b}\bigg[\sin (x)\cdot \ln|a-b\cos x|\bigg]\bigg|^{\pi}_{0}-\frac{2}{b}\int^{\pi}_{0}\cos x\cdot \ln|a-b\cos x|dx$$
$$I=-\frac{2}{b}\int^{\pi}_{0}\cos x\cdot \ln(a-b\cos x)dx$$
Could some help me to solve it, Thanks in advance
|
$$I=\int_0^{2\pi}\frac{\sin^2(x)}{a-b\cos(x)}dx=\int_0^\pi\frac{\sin^2(x)}{a-b\cos(x)}dx+\int_\pi^{2\pi}\frac{\sin^2(x)}{a-b\cos(x)}dx$$
now using substitution:
$$t=\tan\frac{x}{2}$$
$$dx=\frac{2}{1+t^2}dt$$
$$\sin(x)=\frac{2t}{1+t^2}$$
$$\cos(x)=\frac{1-t^2}{1+t^2}$$
because of the discontinuities of the function, we must first change the limits. Now our integral becomes:
$$I=\int_{-\pi}^\pi\frac{\sin^2(x)}{a-b\cos(x)}dx=\int_{-\infty}^\infty\frac{\left(\frac{2t}{1+t^2}\right)^2}{a-b\left(\frac{1-t^2}{1+t^2}\right)}.\frac{2}{1+t^2}dt=2\int_{-\infty}^\infty\frac{2t}{(1+t^2)^2\left(a(1+t^2)-b(1-t^2)\right)}dt=2\int_{-\infty}^\infty\frac{2t}{(1+t^2)^2\left((a-b)+(a+b)t^2\right)}dt$$
then by using PFD this can be evaluated
|
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|
Evaluating $\mathcal{P}\int_0^\infty \frac{dx}{(1 + a^2x^2)(x^2 - b^2)}$ I am trying to evaluate this integral
$$\mathcal{P}\int_0^\infty \frac{dx}{(1 + a^2x^2)(x^2 - b^2)}$$
with $\mathcal{P}$ the principal value and $a,b>0$.
I already know the answer to be $$ - \frac{a\pi}{2(1 + a^2b^2)}$$
after fiddling with Mathematica putting numbers for $a$ and $b$.
The poles of this function are at $x=\pm b, \pm \frac{i}{a^2}$ so I cannot find a proper contour.
Any help in getting the answer will be appreciated!
|
Thanks to both of you for the help!
So indeed, this contour is a good one to evaluate this integral.
The residues at $b$ and $-b$ just cancel each other.
The residue at $i/a$ gives $$ - \frac{a}{2i\left( a^2 b^2 +1 \right)}$$
Thus we have
$$ 2\mathcal{P}\int_0^\infty \frac{dx}{(1 + a^2x^2)(x^2 - b^2)} = 2\pi i \left(- \frac{a}{2i\left( a^2 b^2 +1 \right)} \right) = \frac{-\pi a}{1 + a^2b^2}$$
$$ \mathcal{P}\int_0^\infty \frac{dx}{(1 + a^2x^2)(x^2 - b^2)} = -\frac{\pi}{2}\frac{a}{1 + a^2b^2}$$
|
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|
In how many ways can ten indistinguishable cookies be distributed to four friends with restrictions?
You have ten indistinguishable cookies that you’ve baked for four of your friends Jianbei, Sione, Sina and Julia. You want to give away all ten of your cookies to your friends, and for each friend to get at least one cookie. You also know that Sione wants an even number of cookies, so he can share them with a friend. In how many ways can you give away your cookies?
I'm having difficulty going about this. Firstly, I thought of breaking down the problem into three separate parts:
Sione has $2$ cookies $\implies$ split $8$ between $3$
Sione has $4$ cookies $\implies$ split $6$ between $3$
Sione has $6$ cookies $\implies$ split $4$ between $3$
From here, I'm a bit stuck on how to find the total, any hints or advice on formula? I got $80$ first time round, but this seems so wrong.
So using the $3$ separate parts above:
$$\text{Total} = 8C3 + 6C3 + 4C3 = \frac{8 \cdot 7 \cdot 6}{3 \cdot 2 \cdot 1} + \frac{6 \cdot 5 \cdot 4}{3 \cdot 2 \cdot 1} + \frac{4 \cdot 3 \cdot 2}{3 \cdot 2 \cdot 1} = 56 + 20 + 4 = 80$$
However, I think this may be wrong as the cookies are all identical e.g. Jianbei getting $2$ cookies and then Sione getting $2$ cookies is the same as Sione getting $2$ cookies and then Jianbei getting $2$ cookies, so I'm thinking maybe this formula would work: $$\frac{n!}{k!(n-k)!}$$
|
Your strategy is sound, but your answer is indeed incorrect.
Case 1: Sione receives two cookies.
Let $x_i$, $1 \leq i \leq 4$, be the number of cookies distributed to the $i$th friend. Let $x_4$ denote the number of cookies given to Sione. Since a total of ten cookies are distributed and Sione receives two,
\begin{align*}
x_1 + x_2 + x_3 + 2 & = 10\\
x_1 + x_2 + x_3 & = 8 \tag{1}
\end{align*}
Since each friend receives at least one cookie, equation 1 is an equation in the nonnegative integers. A particular solution of equation 1 corresponds to the placement of two addition signs in the seven spaces between successive ones in a row of eight ones.
$$1 \square 1 \square 1 \square 1 \square 1 \square 1 \square 1 \square 1$$
For instance, if we place an addition sign in the third and seventh boxes, we obtain
$$1 1 1 + 1 1 1 1 + 1$$
which corresponds to the solution $x_1 = 3$, $x_2 = 4$, and $x_3 = 1$. The number of such solutions is the number of ways we can select two of the seven spaces in which to place an addition sign, which is
$$\binom{7}{2}$$
A particular solution of the equation
$$x_1 + x_2 + x_3 + \cdots + x_n = k$$
in the positive integers corresponds to the placement of $n - 1$ addition signs in the $k - 1$ spaces between successive ones in a row of $k$ ones. Therefore, the number of such solutions is
$$\binom{k - 1}{n - 1}$$
since we must choose which $n - 1$ of those $k - 1$ spaces will receive an addition sign.
Case 2: Sione receives four cookies.
Let the variables be assigned as above. Since Sione receives four cookies,
\begin{align*}
x_1 + x_2 + x_3 + 4 & = 10\\
x_1 + x_2 + x_3 & = 6 \tag{2}
\end{align*}
Since each friend receives at least one cookie, equation 2 is an equation in the positive integers with
$$\binom{6 - 1}{3 - 1} = \binom{5}{2}$$
solutions.
Case 3: Sione receives six cookies.
Let the variables be assigned as above. Since Sione receives six cookies,
\begin{align*}
x_1 + x_2 + x_3 + 6 & = 10\\
x_1 + x_2 + x_3 & = 4 \tag{3}
\end{align*}
Since each friend receives at least one cookie, equation 3 is an equation in the positive integers with
$$\binom{4 - 1}{3 - 1} = \binom{3}{2}$$
solutions.
Total: Since these cases are mutually exclusive and exhaustive, you can distribute the cookies in
$$\binom{7}{2} + \binom{5}{2} + \binom{3}{2} = 21 + 10 + 3 = 34$$
ways.
|
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|
Help with proof by induction and divisibility I have a question about induction (I'm a little fuzzy on it).
*
*Prove $3n+1<n^2$ for all integers $n\geq 4$.
*Assume $k^2+k$ is an even integer for any $k\in\mathbb{Z}$. Prove $n^3-n$ is divisible by 6 for all $n\in\mathbb{N}$.
For 1, I have: Base Case: Let $n=4$. Evaluate the inequality, returning $13<16$. Thus, the base case is valid. Induction step: Suppose that, given $k\geq 4$, $3k+1<k^2$ is true for $n=k$. Then, $3(k+1)+1<(k+1)^2$. I am stuck from here.
For 2, I have: Base case: Let $n=1$ (the first natural number in this case). Evaluate the expression $n^3-n$ as $0$. Now I am stuck. Am I supposed to evaluate at $n=2$? Induction step: take $k$ to be true such that $k^3-k$ is divisible by 6 (am I allowed to do this?). Then, $(k+1)^3-(k+1)$, or $k^3+3k^2+3k+1-(k+1)$, or $(k^3-k)+3(k^2+k)$. Now I am stuck.
Help is appreciated!
|
1)
$3(k+1)+1 = 3k + 4$ and $3k+1 < k^2$ so $3k + 4 < k^2 +3$.
And $(k+1)^2 = k^2 + 2k + 1$. ANd $k\ge 4$ so $2k + 1 \ge 9$.
So $k^2 + 3 < k^2 + 9 \le k^2 + 2k +1 = (k+1)^2$
2)
$n^3 - n = 0$ if $n = 1$. And $0$ is divisible by $6$ (because $0 = 6*0$) so, no, you are not stuck.
But if you want to do $n =2$ for backup go ahead. $n^3 -n = 2^3 - 2=6$ and $6$ is divisible by $6$.
And for the induction step. If $k^3 - k$ is divisible by $6$ then $(k+1)^3 - (k+1) = k^3 + 3k^2 + 3k + 1 - k - 1 = (k^3 - k) + 3(k^2 + k)$. You know that $k^3 -k$ is divisible by $6$ so all you have to do is prove $3(k^2 + k)$ is divisible by $6$. It's divisible by $3$ so all you have to do is prove $k^2 + k$ is even. Which you can do be induction if you want. Or but taking cases if $k$ is even or odd.
|
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|
Distance between the Orthocenter and the Lemoine Point (Symmedian Point) I need help to solve this problem. I have no idea how to solve.
Is there previously solved copy of this problem?
Thank you very much.
Let $H$ be the orthocenter of a triangle $ABC$ with circumradius $R$. Show that the distance $HK$ between $H$ and the Lemoine point $K$ of the triangle $ABC$ satisfies
$$HK^2=4R^2-\frac{a^4+b^4+c^4}{a^2+b^2+c^2}-\frac{3a^2b^2c^2}{\left(a^2+b^2+c^2\right)^2}\,,$$
where $a:=BC$, $b:=CA$, and $c:=AB$.
|
You can solve this question by analytic geometry, using two lemmas:
Lemma 1: In any triangle ABC these relations hold true:
$$a^2+{AH}^2=b^2+{BH}^2=c^2+{CH}^2=4R^2$$
Lemma 2: If $A=(x_1,y_1)$, $B=(x_2,y_2)$, $C=(x_3,y_3)$ are the coordinates of vertices of triangle ABC, then the coordinates of its Lemoine point are $$K=({a^2x_1+b^2x_2+c^2x_3\over a^2+b^2+c^2},{a^2y_1+b^2y_2+c^2y_3\over a^2+b^2+c^2})$$
Now, if you put the origin of cartesian orthogonal axes at orthocenter H,
$$HK^2=({a^2x_1+b^2x_2+c^2x_3\over a^2+b^2+c^2})^2 +({a^2y_1+b^2y_2+c^2y_3\over a^2+b^2+c^2})^2,$$
$$(a^2+b^2+c^2)^2HK^2=(a^2x_1+b^2x_2+c^2x_3)^2 +(a^2y_1+b^2y_2+c^2y_3)^2,$$
$$(a^2+b^2+c^2)^2HK^2=a^4(x_1^2+y_1^2)+b^4(x_2^2+y_2^2)+c^4(x_3^2+y_3^2)+(2x_1x_2+2y_1y_2)a^2b^2+(2x_1x_3+2y_1y_3)a^2c^2+(2x_2x_3+2y_2y_3)b^2c^2,$$
$$(a^2+b^2+c^2)^2HK^2=a^4{AH}^2+b^4{BH}^2+c^4{CH}^2+({AH}^2+{BH}^2-c^2)a^2b^2+({AH}^2+{CH}^2-b^2)a^2c^2+({BH}^2+{CH}^2-a^2)b^2c^2,$$
$$(a^2+b^2+c^2)^2HK^2=(a^4+a^2b^2+a^2c^2){AH}^2+(b^4+a^2b^2+b^2c^2){BH}^2+(c^4+a^2c^2+b^2c^2){CH}^2-3a^2b^2c^2$$
Having done that, we use lemma 1 stated above and after further algebraic simplification we finally get
$${HK}^2=4R^2-{a^4+b^4+c^4\over a^2+b^2+c^2}-{3a^2b^2c^2\over (a^2+b^2+c^2)^2}$$
QED.
|
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|
Proving $7|13\cdot 6^{n}+8\cdot 13^{n}$ for all natural numbers. I would like to verify whether my proof is correct. The answer sheet used a much more intuitive and logical approach but I think mine is correct also.
To prove: $7|13\cdot 6^{n}+8\cdot 13^{n}$ for all natural numbers
Proof: We proceed by induction and show the base case holds.
$13+8=21 $. Since 7 divides 21 the base case holds.
We assume $7|13\cdot 6^{m}+8\cdot 13^{m}$ and we need to show $7|13\cdot6^{m+1}+8\cdot 13^{m+1}$.
Since, $7|13\cdot 6^{m}+8\cdot 13^{m}$, we have, $13\cdot 6^{m}+8\cdot 13^{m} =7x, \space x \in \mathbb{N}$.
We rewrite $13\cdot 6^{m+1}+8\cdot 13^{m+1}$ as $6(13\cdot 6^m)+13(8\cdot 13^m)$ and notice, $13\cdot 6^m=7x-8\cdot13^m$.
We substitute and find,
$6(7x-8\cdot13^m)+13(8\cdot13^m)=42x-48\cdot13^m+104\cdot13^m = 42x+56\cdot13^m=7(6x+8\cdot13^m)$
Since $6x+8\cdot13^m \space \in \mathbb{N}$
$\\ \therefore $ By the principle of mathematical induction, $7|13\cdot 6^{n}+8\cdot 13^{n}$ for all natural numbers $n \space \blacksquare$.
|
Rewrite $13\cdot 6^n+8\cdot 13^n$ in base $7$:
$$16\cdot 6^n+11\cdot 16^n$$
Note that $6\cdot 6=51$ in base $7$.
The first term ends with $1$ if $n$ is even and ends with $6$ if $n$ is odd. Conversely, the second term ends with $1$ if $n$ is even and ends with $6$ if $n$ is odd. Therefore, the sum ends with $0$.
|
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|
Find $\sum_{n=1}^\infty\frac{2^{f(n)}+2^{-f(n)}}{2^n}$, where $f(n)=\left[\sqrt n +\frac 12\right]$ denotes greatest integer function
Question: Let $f(n)=\left[\sqrt n +\dfrac 12\right]$, where $[\cdot]$ denotes greatest integer function, $\forall n\in \Bbb N$. Then,
$$\sum_{n=1}^\infty\frac{2^{f(n)}+2^{-f(n)}}{2^n}={\color{red}?}$$
My solution :
I tried with trapping gif function into its lower value and
upper values using $x-1< [x] <x$,
but i am not getting any result.
|
Note that $f(n) = \left\lfloor \sqrt{n}+\tfrac{1}{2}\right\rfloor = k$ iff $k^2-k+\tfrac{1}{4} \le n < k^2+k+\tfrac{1}{4}$, i.e. $k^2-k+1 \le n \le k^2+k$.
Therefore, we have:
\begin{align*}
\sum_{n = 1}^{\infty}\dfrac{2^{f(n)}+2^{-f(n)}}{2^n} &= \sum_{k = 1}^{\infty}\sum_{f(n) = k}\dfrac{2^{f(n)}+2^{-f(n)}}{2^n}
\\
&= \sum_{k = 1}^{\infty}\sum_{n = k^2-k+1}^{k^2+k}\dfrac{2^k+2^{-k}}{2^n}
\\
&= \sum_{k = 1}^{\infty}(2^k+2^{-k})\cdot\sum_{n = k^2-k+1}^{k^2+k}\dfrac{1}{2^n}
\\
&= \sum_{k = 1}^{\infty}(2^k+2^{-k})\left(2^{-(k^2-k)}-2^{-(k^2+k)}\right)
\\
&= \sum_{k = 1}^{\infty}\left(2^{-k^2+2k}+2^{-k^2}-2^{-k^2}-2^{-k^2-2k} \right)
\\
&= \sum_{k = 1}^{\infty}\left(2^{-(k-1)^2+1} - 2^{-(k+1)^2+1}\right)
\\
&= \sum_{i = 0}^{\infty}2^{-i^2+1} - \sum_{j = 2}^{\infty}2^{-j^2+1}
\\
&= 2^{-0^2+1}+2^{-1^2+1}
\\
&=3
\end{align*}
|
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|
Finding the $2n+1$ th derivative of $\frac{y^{2n+1}xy}{1-x^2y^2}$ with respect to $x$. $f(x,y) = \frac{y^{2n+1}xy}{1-x^2y^2}$.
I made the following table:
\begin{align}
& 2n+1 = 1 \implies f^{(1)} = \frac{1!y^2(1+x^2y^2)}{(1-x^2y^2)^2}\\
& 2n+1 = 3 \implies f^{(3)} = \frac{3!y^6(x^4y^4+6x^2y^2+1)}{(1-x^2y^2)^4} \\
& 2n+1 = 5 \implies f^{(5)} = \frac{5!y^{10}(x^6y^6+15x^4y^4+15x^2y^2+1)}{(1-x^2y^2)^6} \\
& 2n+1 = 7 \implies f^{(7)} = \frac{7!y^{14}(x^8y^8+28x^6y^6+70x^4y^4+28x^2y^2+1)}{(1-x^2y^2)^8}
\end{align}
I think $f^{(2n+1)}$ has the shape:
\begin{align}
f^{(2n+1)} = \frac{(2n+1)!y^{2(2n+1)}P(x,y)}{(1-x^2y^2)^{2n+2}}
\end{align}
The problem is finding $P(x,y)$ above.
|
You may write $z = xy\Rightarrow z'(x) = y$.
$$f(x) = \frac{y^{2n+1}xy}{1-x^2y^2} = y^{2n+1}\color{blue}{\frac{z}{1-z^2}} $$
$$\Rightarrow f(x) = g(z(x)) \Rightarrow f'(x) = g'(z)\cdot z'(x) = g'(z)\cdot y \Rightarrow \boxed{f^{(k)}(x) = g^{(k)}(z)\cdot y^k}$$
$$\color{blue}{\frac{z}{1-z^2} = -\frac{1}{2}\left( \frac{1}{z+1}+ \frac{1}{z-1} \right)}$$
$$\Rightarrow f^{(2n+1)}(x) = g^{(2n+1)}(z)\cdot y^{2n+1} = -\frac{y^{2(2n+1)}}{2}\left( \frac{(-1)^{2n+1}\cdot (2n+1)!}{(z+1)^{2n+2}}+ \frac{(-1)^{2n+1}\cdot (2n+1)!}{(z-1)^{2n+2}} \right) $$ $$= \frac{y^{2(2n+1)}\cdot (2n+1)!}{2}\left( \frac{1}{(z+1)^{2n+2}}+ \frac{1}{(z-1)^{2n+2}} \right) = \frac{y^{2(2n+1)}\cdot (2n+1)!}{2}\left( \frac{1}{(xy+1)^{2n+2}}+ \frac{1}{(xy-1)^{2n+2}} \right)$$
|
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|
Dual Numbers and Automatic Differentiation I have the following equation. I would like to solve this at the point x = a, but using dual numbers.
$$f\left(x\right) = \dfrac{1}{x} + \sin\left(\dfrac{1}{x}\right)$$
Now, the derivative of this function is below, and that is the function we'd like to reach in our answer.
$$-\dfrac{1}{x^2} - \dfrac{\cos\left(\dfrac{1}{x}\right)}{x^2}$$
Here is what I have tried, and I am finding it difficult to simplify beyond that.
$$\begin{align} f\left(x\right) = \dfrac{1}{x} + \sin\left(\dfrac{1}{x}\right)\\
&= \frac{1}{a^{\prime}+1\epsilon}+ sin(\dfrac{1}{a+1\epsilon})\\ &= \frac{a-1\epsilon}{(a-1\epsilon)(a+1\epsilon)} + sin(\frac{a-1\epsilon}{(a-1\epsilon)(a+1\epsilon)} )\end{align}$$
I do this using the following identities:
*
*$$\sin\left(\alpha + \beta \epsilon\right) = \sin\left(\alpha\right) + \cos\left(\alpha\right)\beta\epsilon$$
*$$\dfrac{1}{\alpha + \beta\epsilon} = \dfrac{\alpha - \beta\epsilon}{\left(\alpha + \beta\epsilon\right)\left(\alpha - \beta\epsilon\right)}.$$
However, I am not certain how to simplify beyond that painfully obvious first step. Any hints?
|
Break your problem into steps. Let's say $f(x) = g(x) + h(x)$ then $f' = g' + h'$ (omitting $x$). In general even with dual variables you should use the chain rule.
First solve the derivative of $1/x$ at $x=a$.
$$\begin{align} \frac{1}{a + \epsilon} = \frac{1}{a} \frac{1}{1+ \epsilon/a} = \frac{1}{a} \big(1 - \frac{\epsilon}{a} + O(\epsilon^2) \big)\end{align}.$$ The $O(\epsilon^2)$ terms become zero. so you get $-1/a^2$
Next find the derivative of $sin(y)$. $\sin(y+\epsilon) = \sin(y)+cos(y)\epsilon\ $ so the derivative of $\sin(y)$ is $\cos(y)$. Now put $y=1/x$, and use chain rule.
|
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|
Finding range of expression $f(x,y)=x^2+y^2$
Finding range of $f(x,y)=x^2+y^2$ subjected to
the condition $2x^2+6xy+5y^2=1$ without Calculus
Try: Let $k=x^2+y^2,$ Then $\displaystyle k=\frac{x^2+y^2}{2x^2+6xy+5y^2}$
Now put $\displaystyle \frac{y}{x}=t(x\neq 0)$ and $t\in \mathbb{R}$
$$k=\frac{1+t^2}{2+6t+5t^2}\Rightarrow (5k-1)t^2+6kt+(2k-1)=0$$
For real $t$ its Discriminant $\geq 0$
$$36k^2-4(5k-1)(2k-1)\geq 0$$
$$k^2-7k+1\leq 0$$
$$\Rightarrow k\in \bigg[\frac{7-3\sqrt{5}}{2}\;\;,\frac{7+3\sqrt{5}}{2}\bigg]$$
I have a question . can i solve it using Inequality like
$(a+b)^2\geq 0$
or $(a-b)^2\geq 0$ , explain me thanks
|
Yes, of course. We can make it.
For example, we need to prove that
$$x^2+y^2\geq\frac{7-3\sqrt5}{2}(2x^2+6xy+5y^2)$$ or
$$(2\sqrt5-4)x^2-2(7-3\sqrt5)xy+(5\sqrt5-11)y^2\geq0$$ or
$$2x^2-2(7-3\sqrt5)(2+\sqrt5)xy+(5\sqrt5-1)(2+\sqrt5)y^2\geq0$$ or
$$2x^2-2(-1+\sqrt5)xy+(3-\sqrt5)y^2\geq0$$ or
$$4x^2-4(-1+\sqrt5)xy+(6-2\sqrt5)y^2\geq0$$ or
$$(2x-(\sqrt5-1)y)^2\geq0.$$
The second inequality we can prove by the similar way.
|
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|
Solve $9^x-2^{x+\frac{1}{2}}=2^{x+\frac{7}{2}}-3^{2x-1}$ $$9^x-2^{x+\frac{1}{2}}=2^{x+\frac{7}{2}}-3^{2x-1}.$$
The equation states solve for $x$.
What I first did was put like bases together.
$$3^{2x}+3^{2x-1}= 2^{x+\frac{7}{2}}+ 2^{x+\frac{1}{2}}.$$
Then I factored $3^{2x}$ and $2^x$
$$3^{2x}(1+\frac{1}{3})=2^x(2^{\frac{7}{2}}+2^{\frac{1}{2}}),$$
then I got
$$\frac{3^{2x}}{2^x}=9\sqrt{2}.$$
From here I took $\log$s, but the answer wasn't nice.
What to do?
|
If you let $x=u+{1\over2}$, you can get rid of the pesky square roots: the expression simplifies to
$$9^{u+1/2}-2^{u+1}=2^{u+4}-3^{2u}$$
or
$$3\cdot9^u-2\cdot2^u=16\cdot2^u-9^u$$
This simplifies first to $4\cdot9^u=18\cdot2^u$ and then to $9^{u-1}=2^{u-1}$, which clearly implies $u=1$, i.e., $x=3/2$.
|
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|
Purchasing Donuts
At closing time at a Dunkin Donuts store they still have $10$ vanilla,
$20$ custard, $24$ cinnamon, and $30$ chocolate donuts available.
Donuts of the same kind are regarded as identical.
(a) How many ways are there to purchase $8$ donuts?
(b) How many ways are there to purchase $15$ donuts with at least
three of each kind?
(c) How many ways are there to purchase $15$ donuts?
Part (a):
We have four different flavor donuts to choose from. We would like to purchase $8$ of them. So this is a stars and bars problem. We have $8$ stars and $3$ bars. So this is $\binom{8+3}{3}=\binom{11}{3}=165$ ways.
Part (b):
Step $1$: Put three of each kind in the box.
Now we need to purchase $3$ more donuts. This is again, stars and bars.
$\binom{3+3}{3}=\binom{6}{3}=20$ ways.
Part (c):
I know we need to be careful because there are only $10$ vanilla donuts left.
This will be a stars and bars problem.
How do I go about this problem?
|
(c) How many ways are there to purchase $15$ donuts?
The question can be formulated as:
$$x_1+x_2+x_3+x_4=15,\\
0\le x_1\le 10,\\
0\le x_2\le 20,\\
0\le x_3\le 24,\\
0\le x_4\le 30,$$
which is equivalent to:
$$x_1+x_2+x_3+x_4=15,\\
0\le x_1\le 10,\\
0\le x_2\le 15,\\
0\le x_3\le 15,\\
0\le x_4\le 15.$$
which is the difference between (1):
$$x_1+x_2+x_3+x_4=15,\\
0\le x_1\le 15,\\
0\le x_2\le 15,\\
0\le x_3\le 15,\\
0\le x_4\le 15,$$
and (2):
$$x_1+x_2+x_3+x_4=15,\\
11\le x_1\le 15,\\
0\le x_2\le 15,\\
0\le x_3\le 15,\\
0\le x_4\le 15,$$
(1):${15+4-1\choose 4-1}={18\choose 3}$.
(2):
$$x_1+x_2+x_3+x_4=15,\\
11\le x_1\le 15,\\
0\le x_2\le 15,\\
0\le x_3\le 15,\\
0\le x_4\le 15,$$
which is equivalent to (let $x_1=y_1+11$):
$$y_1+x_2+x_3+x_4=4,\\
0\le y_1\le 4,\\
0\le x_2\le 4,\\
0\le x_3\le 4,\\
0\le x_4\le 4,$$
which is: ${4+4-1\choose 4-1}={7\choose 3}$.
Thus, the final answer is: ${18\choose 3}-{7\choose 3}=781$.
|
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|
Prove $f(x)=x^n(1-x)\lt \frac{1}{ne}$ for all $n\in \mathbb{N},x\in(0,1)$
Prove $f(x)=x^n(1-x)\lt \frac{1}{ne}$ for all $n\in \mathbb{N},x\in(0,1)$.
My try: $f'(x)=nx^{n-1}-(n+1)x^n=x^{n-1}(n-(n+1)x)=0\Rightarrow x=\frac{n}{n+1} $, hence $\max_{x\in(0,1)} f(x)=f(\frac{n}{n+1})=(1-\frac{1}{n+1})^n\cdot\frac{1}{n+1}$. I know $(1-\frac{1}{n+1})^n\to e^{-1}$. But I don't know how to prove $(1-\frac{1}{n+1})^n\lt \frac{1}{e}$.
|
\begin{equation}
\max_{x\in(0,1)} f(x)=f(\frac{n}{n+1})=(1-\frac{n}{n+1})^n\cdot\frac{n}{n+1}
=
\frac{n}{(n+1)^{n+1}}
\tag{1}
\end{equation}
Now
\begin{equation}
\frac{n}{(n+1)^{n+1}} - \frac{1}{ne}
=
\frac{1}{n}( \frac{n^2}{(n+1)^{n+1}} - \frac{1}{e})
=
\frac{1}{n}( \frac{en^2 - (n+1)^{n+1}}{e(n+1)^{n+1}})
\tag{2}
\end{equation}
Sign depends on $en^2 - (n+1)^{n+1}$. It is easy to see that for $n=1$, $en^2 - (n+1)^{n+1} < 0$. We can say that for $n \geq 2$,
\begin{equation}
\begin{split}
en^2 - (n+1)^{n+1} &< e(n+1)^2 - (n+1)^{n+1} \\ &= (n+1)^2(e - (n+1)^{n-1}) <0
\end{split}
\end{equation}
So for all $n \geq 1$, we have that $en^2 - (n+1)^{n+1} < 0$. Going back to $(2)$, we get $\frac{n}{(n+1)^{n+1}} - \frac{1}{ne} < 0$, which using $(1)$, gives
\begin{equation}
\max_{x\in(0,1)} f(x)
<
\frac{1}{ne}
\end{equation}
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2959146",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
}
|
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.