Q
stringlengths 70
13.7k
| A
stringlengths 28
13.2k
| meta
dict |
|---|---|---|
Critique on a proof by induction that $\sum\limits_{i=1}^n i^2= n(n+1)(2n+1)/6$? I need to make the proof for this
1:$$1^2 + 2^2 + 3^2 + ... + n^2=\frac{(n(n+1)(2n+1))}{6}$$
By mathematical induction I know that,
If P(n) is true for $n>3^2$ then P(k) is also true for k=N and also P(k+1) must be true.
Then,
2:$$1^2 + 2^2 + 3^2 + ... + k^2=\frac{(k(k+1)(2k+1))}{6}$$
Substitution:
3:$$1^2 + 2^2 + 3^2 + ... +n^2+ (n+1)^2=\frac{((n+1)((n+1)+1)(2(n+1)+1))}{6}$$
Questions
a: Do anyone see notation errors in this post?
b: It is right so far?
|
For the third step, notice that
$$ 1^2 + 2^2 + 3^2 + ... +n^2+ (n+1)^2 = \frac{(n(n+1)(2n+1))}{6}+(n+1)^2 $$
$$ = \frac{(n(n+1)(2n+1))}{6}+\frac{6(n+1)^2}{6}$$
$$ = \frac{(n+1)(n(2n+1)+6(n+1))}{6}=\dots $$
Can you finish it?
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/472292",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
}
|
Determine the minimum of $a^2 + b^2$ if $a,b\in\mathbb{R}$ are such that $x^4 + ax^3 + bx^2 + ax + 1 = 0$ has at least one real solution I just wanted the solution, a hint or a start to the following question.
Determine the minimum of $a^2 + b^2$ if $a$ and $b$ are real numbers for which
the equation
$$x^4 + ax^3 + bx^2 + ax + 1 = 0$$
has at least one real solution.
Thanks in advance.
|
Hint: Your quartic is palindromic. Divide through by $x^2$. We get the equation
$$x^2+ax+b+\frac{a}{x}+\frac{1}{x^2}=0.$$
Make the substitution $x+\frac{1}{x}=t$. Then $x^2+\frac{1}{x^2}=t^2-2$, and we arrive at the equation
$$t^2+at+b-2=0.$$
Imagine solving the quadratic. We will get two not necessarily real values of $t$. The original equation has at least one real solution if and only if $t$ is real and has absolute value $\ge 2$. This is because, for example, for $x$ positive, the minimum value of $x+\frac{1}{x}$ is $2$.
Added, almost to the answers: There are two cases to consider, $a\ge 0$ and $a\lt 0$. Look at the case $a\ge 0$. The roots of the quadratic in $t$ are
$$\frac{-a\pm \sqrt{a^2-4(b-2)}}{2}.$$
The root of largest absolute value is given by choosin the $-$ in $\pm$. We want this root to be $\le -2$. After some manipulation, that gives the inequality
$$\sqrt{a^2-4(b-2)}\ge 4-a.$$
Squaring both sides and simplifying, we get $2a\ge b+ 2$. The minimum value of $a^2+b^2$ is attained when $2a=b+2$. That will give you the answer that has $a\ge 0$. The answer for negative $a$ is obtained by a similar calculation. The relevant root is then $t=\frac{-a+\sqrt{a^2-4(b-2)}}{2}$.
Remark: I feel somewhat guilty at bashing the problem with standard machinery. But routine calculation does work. There is undoubtedly a simple way. However, simple can take time.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/474507",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 3,
"answer_id": 0
}
|
A cyclic inequality $\sum\limits_{cyc}{\sqrt{3x+\frac{1}{y}}}\geqslant 6$ Given positive real numbers $x,y,z$ satisfying $x+y+z=3$, prove that $$ \sqrt{3x+\frac{1}{y}}+\sqrt{3y+\frac{1}{z}}+\sqrt{3z+\frac{1}{x}}\geqslant 6. $$
|
Remark: @chenbai gave a nice solution using the Buffalo Way (BW).
Here, I give an alternative solution using BW:
By Holder, we have
\begin{align}
\left(\sum_{\mathrm{cyc}} \sqrt{3x + 1/y}\right)^2 \sum_{\mathrm{cyc}} \frac{(6x + y + 3z)^3}{3x + 1/y}
\ge \left(\sum_{\mathrm{cyc}} (6x + y + 3z)\right)^3.
\end{align}
It suffices to prove that
$$\left(\sum_{\mathrm{cyc}} (6x + y + 3z)\right)^3 \ge 36 \sum_{\mathrm{cyc}} \frac{(6x + y + 3z)^3}{3x + 1/y}.$$
After homogenization, it suffices to prove that, for all $x, y, z > 0$,
$$\left(\sum_{\mathrm{cyc}} (6x + y + 3z)\right)^3 \ge 12(x+y+z) \sum_{\mathrm{cyc}} \frac{(6x + y + 3z)^3}{3x + \frac{(x+y+z)^2}{9y}}.$$
The Buffalo Way kills it. We are done.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/475842",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "12",
"answer_count": 1,
"answer_id": 0
}
|
Solve the following system of simultaneous congruences: \begin{gather}
3x\equiv1 \pmod 7 \tag 1\\
2x\equiv10 \pmod {16} \tag 2\\
5x\equiv1 \pmod {18} \tag 3
\end{gather}
Hi everyone, just a little bit stuck on this one. I think I am close, but I must be getting tripped up somewhere. Here is what I have so far:
from (2), $2x=10+16k \implies x=5+8k$
Putting this into (1):
\begin{align*}
3(5+8k) \equiv 1 \pmod 7
&= 15+24k \equiv 1 \pmod 7 \\
&= 24k \equiv -14 \pmod 7
\end{align*}
By co-prime cancellation, I get $12k\equiv -7 \pmod 7$
And since $12k \equiv 5k \pmod 7 \implies 5k \equiv -2k \pmod 7$ and $-7 \equiv 0 \pmod 7$, we conclude that
$ -2k \equiv 0 \pmod 7 \implies -2k = 7l$ for some integer $l$.
Multiplying by $-4 \implies 8k = -28l$
It follows that, $x = 5 + 8k = 5-28l \implies x \equiv 5 (mod \space -28) $
So now, solving (1), (2) and (3) is equivalent to solving:
$x \equiv 5 (mod \space -28) $ (4)
$5x\equiv1 (mod\space 18)$ (3)
Then substitute $x = 5-28l$ into (3),
$5(5-28l) \equiv 1 (mod \space 18)$
= $25 - 140l \equiv 1 (mod \space 18)$
= $140l \equiv 24 (mod \space 18) $
And since $140l \equiv 14l (mod \space 18) \implies 14l \equiv -4l (mod \space 18)$ and $24 \equiv 6 (mod \space 18)$
we have, $-4l \equiv 6 (mod \space 18) \implies -4l = 6 + 18M$ for some integer M.
Multiplying by 7 $\implies -28l = 42 + 126M$
Finally, substituting this back into x, $x = 5-28l \implies x = 5+42+126M = 47+126M \implies x \equiv 47 (mod \space 126)$
But when I substitute $x = 47$ back into my original equations, it works for (1) and (3), but fails for (2).
Can anyone tell me where I went wrong? Many thanks!!
|
This is not an answer, but a guide to a simplification. The first congruence is fine. Note that the second congruence is equivalent to $x\equiv 5\pmod{8}$. Any solution of this congruence must be odd.
Now look at the third congruence. Note that as long as we know that $x$ is odd, we automatically have $5x\equiv 1\pmod{2}$. So in the presence of the second congruence, the third one can be replaced by $5x\equiv 1\pmod{9}$.
Thus we are looking at the congruences $3x\equiv 1 \pmod{7}$, $x\equiv 5\pmod{8}$, and $5x\equiv 1\pmod{9}$. Now the moduli are pairwise relatively prime. Relatively prime moduli are easier to handle, there is less risk of error.
There will be a unique solution modulo $(7)(8)(9)$. In particular, your final modulus of $126$ cannot be right.
It is probably worthwhile to simplify the congruences still further. Note that the congruence $3x\equiv 1\pmod{7}$ has the solution $x\equiv 5\pmod{7}$. And the congruence $5x\equiv 1\pmod{9}$ is equivalent to $x\equiv 2\pmod{9}$.
We got lucky, ended up with $x\equiv 5\pmod{7}$ and $x\equiv 5\pmod{8}$, which has as only solution $x\equiv 5\pmod{56}$.
So we are trying to solve $x\equiv 5\pmod{56}$, $x\equiv 2\pmod{9}$.
One can find a solution by a short search. Or else we want $x=56a+5=9b+2$. That gives $9b=56a+3$, so $3$ must divide $a$, say $a=3c$. We arrive at $56c+1=3b$. Clearly $c=1$ works, so $a=168$. The solution is therefore $x\equiv 173\pmod{(56)(9)}$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/476293",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
}
|
Solve $(a^2-1)(b^2-1)=\frac{1}4 ,a,b\in \mathbb Q$ Does the equation $(a^2-1)(b^2-1)=\dfrac{1}4$ have solutions $a,b\in \mathbb Q$?
I search $0<p<1000,0<q<1000$, where $a=\dfrac{p}q$, but no solutions exist. I wonder is this equation solvable?
|
Here are some observations which are too extensive really for a comment, and which reduce the search space somewhat, and which may assist in establishing a contradiction.
You can rewrite your equation as $$(ab+1)^2=\frac 14+(a+b)^2$$
Since this is a rational equation, clearing denominators gives a primitive pythagorean triple $$p^2=q^2+r^2$$ with $ab+1=\cfrac p{2q}$ and $a+b=\cfrac r{2q}$ whence $a$ and $b$ are roots of the quadratic $$2qx^2-rx+p-2q=0$$
This has rational solutions only if its discriminant $r^2-8q(p-2q)$ is a square. Now $r^2=p^2-q^2$ so this becomes $$p^2-8qp+15q^2=(p-4q)^2-q^2=s^2$$ This gives us a second pythagorean triple, and also $r^2-s^2=8q(p-2q)$.
Note that $p-2q$ or $p-4q$ may be negative.
To summarise - a solution implies two related pythagorean triples, the first of which can be taken primitive
$$p^2=q^2+r^2$$$$(p-4q)^2=q^2+s^2 \text{ or }(4q-p)^2=q^2+s^2$$
[continued]
The second of these is then also primitive, because $(p-4q,q)=(p,q)=1$
We know that a primitive pythagorean triple is of the form $m^2+n^2, 2mn, m^2-n^2$ with $m,n$ coprime and one even and one odd.
If $p=m^2+n^2$ and $4q-p=v^2+w^2$ then $4q=m^2+n^2+v^2+w^2$, but the right-hand side is the sum of two even squares and two odd squares, and is congruent to $2$ mod $4$. So this case is impossible.
So we must have $p=m^2+n^2$ and $p-4q=v^2+w^2\gt0$, and in passing we note that $p\gt 5q$ since $(p-4q)^2\gt q^2$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/476375",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "14",
"answer_count": 2,
"answer_id": 0
}
|
Finding minima and maxima of $\sin^2x \cos^2x$ I am trying to find the turning points of $\sin^2x \cos^2x$ in the range $0<x<\pi$.
So far I have:
$\begin{align} f'(x) & =\cos^2x \cdot 2\sin x \cdot \cos x + \sin^2x \cdot 2\cos x \cdot -\sin x \\
& = 2 \cos^3x \sin x - 2\sin^3x \cos x \\
& = 2 \sin x \cos x (\cos^2x - \sin^2x) \\
& = 2 \sin x \cos x (\cos x + \sin x) (\cos x - \sin x)
\end{align}$
And, after lots of simplification, I think I've found that:
$f''(x) = 2 \left[\left( \cos^2x-\sin^2x \right)^2 - 4\sin^2x \cos^2 x\right]$
My questions are:
*
*How can I evaluate $0 = \cos x + \sin x$ and $0 = \cos x - \sin x$ without resorting to graph plotting?
*Are there trigonometric identities that I could have used to simplify either derivative?
|
It s easier to deal with this form
$$ f(x) = \sin^2(x)\cos^2(x)=\frac{\sin^2(2x)}{4} $$
$$ \implies f'(x) = \sin(2x)\cos(2x)=\frac{\sin(4x)}{2}. $$
Now, you should be able to finish the problem.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/477636",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 0
}
|
Let $X$ be the number on the first ball drawn and $Y$ the larger of the two numbers draw. Find the joint discrete density function of $X$ and $Y$. Consider a sample of size 2 drawn without replacement from an urn containing three balls, numbered 1,2, and 3. Let $X$ be the number on the first ball drawn and $Y$ the larger of the two numbers draw
(a) Find the joint discrete density function of $X$ and $Y$.
$$\begin{array}{c|cc|c}
x/y & 2 & 3 & f_x(x)\\
\hline
\\ 1 & 1/6 & 1/6 & 2/6
\\ 2 & 1/6 & 1/6 & 2/6
\\ 3 & 0 & 2/6 & 2/6
\\
\hline
\\ f_y(y) & 2/6 & 4/6 &
\end{array}
$$
(b) Find $P[X=1|Y=3]$
$$P(X=1|Y=3)=\frac{P(X=1,Y=3)}{P(Y=3)}=\frac{1}{4}$$
(c) Find $cov[X,Y]$
$E[XY]=(1)(2)(\frac{1}{6})+(1)(3)(\frac{1}{6})+(2)(2)(\frac{1}{6})+(2)(3)(\frac{1}{6})+(3)(3)(\frac{2}{6})=\frac{33}{6}$
$E[X]=(1)(\frac{2}{6})+(2)(\frac{2}{6})+(3)(\frac{2}{6})=2$
$E[Y]=(2)(\frac{2}{6})+(3)(\frac{4}{6})=\frac{16}{6}$
$cov[X,Y]=\frac{33}{6}-(2)(\frac{16}{6})=\frac{1}{6}$
|
The solution is correct. Congratulations.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/478997",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
}
|
After what interval in degrees or radians do sine, cosine and tangent values repeat? Between $0$ to $2π$, I have noticed that $\sin x$, $\cos x$ and $\tan x$ values repeat for different values of $x$.
For example, $\sin 30 = \sin 150$
What exactly is the interval between two successive values of $x$ such that the value of $\sin x$, $\cos x$ or $\tan x$ are equal?
|
$1:$
Let $\sin x=\sin y$
Applying $\sin C-\sin D=2\sin\frac{C-D}2\cos\frac{C+D}2,$
we have $2\sin\frac{x-y}2\cos\frac{x+y}2=0$
If $\sin\frac{x-y}2=0, \frac{x-y}2=n180^\circ\implies x-y=n180^\circ$ where $n$ is any integer
If $\cos\frac{x+y}2=0, \frac{x+y}2=(2m+1)90^\circ\implies x+y=(2m+1)180^\circ$ where $m$ is any integer
$2:$
Apply $\cos C-\cos D=-2\sin\frac{C-D}2\sin\frac{C+D}2,$
$3:$
$\displaystyle\tan x=\tan y\iff \frac{\sin x}{\cos x}=\frac{\sin y}{\cos y}$
$$\implies \sin x\cos y-\cos x\sin y=0\implies \sin(x-y)=0$$
$\implies x-y=r180^\circ$ where $r$ is any integer
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/480052",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
}
|
Showing Whether a Sequence is Bounded Above or Not I am trying to solve the following problem about a sequence:
Consider the sequence ${a_n}$ where $a_n = 1 + \frac{1}{1 \cdot 3} + \frac {1}{1 \cdot 3 \cdot 5} + \frac {1}{1 \cdot 3 \cdot 5 \cdot 7} + ... + \frac{1}{1 \cdot 3 \cdot ... \cdot (2n-1)}.$ Decide whether ${a_n}$ is bounded above or not, and prove your answer is correct.
I started to solve it in the following manner:
Let $a_n \geq a_{n+1} \Rightarrow \frac{1}{1 \cdot 3 \cdot ... \cdot (2n-1)} \geq \frac{1}{1 \cdot 3 \cdot ... \cdot (2n-1) \cdot 2(n+1)-1} \Rightarrow \frac{1}{1 \cdot 3 \cdot ... \cdot (2n-1)} \geq \frac{1}{1 \cdot 3 \cdot ... \cdot (2n-1) \cdot 2n+1}.$ Multiplying by the reciprocal of $a_n$, we have $1 \geq \frac{1}{2n+1}$ which shows that $a_n$ is increasing.
This is where I am stuck since I do not know how to proceed in showing it is bounded from what I have so far. Any assistance or criticism is welcome.
I am using the following textbook: Introduction to Analysis by Arthur Mattuck.
|
Hint: Note that
$$a_n\le 1+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\cdots+\frac{1}{2^{n-1}}\lt 2.$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/480266",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 0
}
|
$g(x+y)+g(x)g(y)=g(xy)+g(x)+g(y)$ for all $x,y$. Find all functions $g:\mathbb{R}\to\mathbb{R}$ with $g(x+y)+g(x)g(y)=g(xy)+g(x)+g(y)$ for all $x,y$.
I think the solutions are $0, 2, x$. If $g(x)$ is not identically $2$, then $g(0)=0$. I'm trying to show if $g$ is not constant, then $g(1)=1$. I have $g(x+1)=(2-g(1))g(x)+g(1)$. So if $g(1)=1$, we can show inductively that $g(n)=n$ for integer $n$. Maybe then extend to rationals and reals.
|
Let $P(x,y)$ denote $g(x+y) + g(x) g(y) =g(xy) + g(x) + g(y) $. The important statements that we'd consider are
$$ \begin{array} { l l l }
P(x,y) & : g(x+y) + g(x) g(y) =g(xy) + g(x) + g(y) & (1) \\
P(x,1) & : g(x+1) + g(x) g(1) = 2 g(x) + g(y) & (2) \\
P(x, y+1) & : g(x+y+1) + g(x) g(y+1) = g(xy+x) + g(x) +g(y+1) & (3)\\
\end{array}
$$
Let $g(1) = A$. With $x=y=1$, we get $g(2) + A^2 = 3A$.
With $x=y=2$, we get $g(4) + g(2)^2 = g(4) + 2g(2) $. Hence either $g(2) = 0$ or $g(2) = 2$.
Case 1: If $g(2) = 0$, then $A^2 - 3A = 0$ so $A=0$ or $A=3$.
Case 1a: (This case is incomplete) If $A=0$:
$(2)$ gives us $g(x+1) = 2g(x)$.
$(3)$ gives us $g(x+y+1) +g(x)g(y+1)= g(xy+y) + g(x)+g(y+1)$.
Applying the above gives $2g(x+y)+g(x)2g(y) = g(xy+x) + g(x) + 2g(y)$, or that $2g(x+y) + 2g(x)g(y) = g(xy+x) + g(x) + 2g(y)$.
$(1)$, combined with the above gives us $g(xy+x) = 2g(xy) + g(x) $.
Substitute $y=1$ in the above identity, we get $g(2x) = 3g(x)$.
(Incomplete here)
Comparing with the first equation, we thus have $g(x) = 0 $.
It is easy to check that $g(x) = 0$ is a solution.
Case 1b: If $A=3$:
$(2)$ gives us $g(x+1) = - g(x) + 3$.
$P(x,2)$ gives us $g(x+2) = g(2x) + g(x)$.
From the previous identity, $g(x+2) = -g(x+1) + 3 = g(x)$. Hence $g(2x) = 0$ for all $x$. However, with $x = \frac{1}{2}$, we have a contradiction. There is no solution in this case.
Case 2: If $g(2) = 2$, then $A^2 - 3A +2 = 0 $ so $A= 1$ or $A=2$.
Case 2a: If $A = 1$:
$P(0,1) $ gives us $g(1) + g(0) = g(0) + g(0) + g(1)$ so $g(0) = 0 $.
$(2)$ gives us $g(x+1) = g(x) + g(1)$.
$(3)$ gives us $g(x+y+1) + g(x) g(y+1) = g(xy+x) + g(x) + g(y+1)$. Using the above, this gives us $ g(x+y) + g(x) g(y) = g(xy+x) + g(x) + g(y) $.
$(1)$ gives us $g(x+y) + g(x) g(y) = g(xy) + g(x) + g(y)$. Hence $g(xy+x) = g(xy) + g(x) $.
Now, for non-zero $a, b$, let $x = b$ and $ y = \frac{b}{a}$. This gives us $g(a+b) = g(a) + g(b) $. It is clear that this equation also holds if $a$ or $b$ equals 0, since $g(0) = 0 $. Thus, we have $g(x+y) = g(x) + g(y) $.
This also yields $g(xy) = g(x) g(y)$. With these 2 equations, the only solution is $g(x) = x$. (slight work here, but this is standard in functional equations).
Case 2b: If $A=2$:
$(2)$ gives us $g(x+1) +2g(x) = g(x) + g(x) + g(1)$, so $g(x) = 2$ for all $x$. This is clearly a solution.
In conclusion, we only have the 3 solutions $g(x) = 0$ or $g(x) = 2$ or $g(x) = x$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/481673",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "21",
"answer_count": 6,
"answer_id": 0
}
|
Compute $ \int_{-\infty}^{\infty} \frac{x^2}{(1+x^2)^2} dx$ Compute $$ \int_{-\infty}^{\infty} \frac{x^2}{(1+x^2)^2} dx$$
Of course we have $$ \int_{-\infty}^{\infty} \frac{x^2}{(1+x^2)^2} dx = 2 \int_{0}^{\infty} \frac{x^2}{(1+x^2)^2} dx = 2 \int_{0}^{\infty} \left( \frac{x}{1+x^2} \right) ^2 dx = \lim_{ A \to \infty } \int_{0}^{A} \left( \frac{x}{1+x^2} \right) ^2 dx $$
I think that firstly I should compute $ \int \left( \frac{x}{1+x^2} \right) ^2 dx $ but I don't have idea.
|
$$\frac{\pi}{t}=\int_{-\infty}^{\infty} \frac{dx}{t^2x^2+1}\Rightarrow\frac{\pi}{t^2}=\int_{-\infty}^{\infty}\frac{2tx^2}{(t^2x^2+1)^2}\,dx\Rightarrow \frac{\pi}{2}=\int_{-\infty}^{\infty}\frac{x^2}{(x^2+1)^2}\,dx$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/483180",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 7,
"answer_id": 1
}
|
How do I write this sum in summation notation? $$\sum_{n=?}^\infty \left(\frac{x^n}{?}\right) = \frac{x^0}{1} + \frac{x^1}{x^2 -1}+\frac{x^2}{x^4 - x^2 +1}+\frac{x^3}{x^6 -x^4 + x^2 -1}+\frac{x^4}{x^8-x^6 +x^4 - x^2 +1}+\cdots$$
I am pretty sure I have the numerator of the summation, $x^n$ correct, but don't know how to write the denominator because of the alternating signs in each term. As for the starting point, $n=?$, I think its going to be either $0$ or $1$, depending on the denominator - I don't want to be kicked from this site by attempting to divide by $0$ .
|
The denominators alternate signs, but they are of the form
$$\sum_{k=0}^m (-1)^k x^{2 k} = \frac{(-1)^{m+1} x^{2(m+1)}-1}{x^2+1}$$
The sum may then be written as
$$(x^2+1) \sum_{n=0}^{\infty} (-1)^n \frac{x^n}{(-1)^{n+1} x^{2(n+1)}-1} = -(x^2+1) \sum_{n=0}^{\infty} \frac{x^n}{x^{2(n+1)}+(-1)^n}$$
I should note that convergence is achieved when $|x|<1$ when $x$ is real.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/483327",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 3,
"answer_id": 0
}
|
Evaluate $\lim_{x\to0} \frac{6}{x}-\frac{42}{x^2+7x}$ Evaluate $\lim_{x\to0} \frac{6}{x}-\frac{42}{x^2+7x}$
I'm I want to say that you cross multiply to get the same denominator, but I could be wrong.
Please Help!!
|
You are correct that a common denominator is necessary:
\begin{align}
\frac 6 x - \frac {42} {x^2 + 7x} &= \frac 6 x - \frac {42} {x(x + 7)}\\
&= \frac{6(x + 7) - 42}{x(x + 7)} \\
&= \frac{6x}{x(x + 7)} \\
&= \frac{6}{x + 7}
\end{align}
for $x \ne 0$. Do you see how to compute the limit now?
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/485185",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 0
}
|
How prove $\lim_{n\to \infty}\frac{n}{\ln{(\ln{n}})}\left(1-a_{n}-\frac{n}{\ln{n}}\right)=-1$? Let equation $x^n+x=1$ have positive root $a_{n}$.
Show that
$$\lim_{n\to \infty}\frac{n}{\ln{(\ln{n}})}\left(1-a_{n}-\dfrac{n}{\ln{n}}\right)=-1$$
some hours ago,it prove that $$\lim_{n\to \infty}\frac{n}{\ln{n}}(1-a_{n})=1$$
How prove this limit $\displaystyle\lim_{n\to \infty}\frac{n}{\ln{n}}(1-a_{n})=1$
this problem is from china paper(1993):http://www.cnki.com.cn/Article/CJFDTotal-YEKJ199301013.htm
and this paper poof is very ugly,so I want see other nice methods,I kown these can use Incremental estimation analysis,Thank you
|
I'd like to post a separate answer here in which I'll use a very different method to obtain the result. The search for this method was prompted by this comment on the previous question where the OP asks for a bound of the form $f(n) \leq a_n \leq g(n)$ with $f(n),g(n) \to 1$ as $n \to \infty$. The bound I obtained turned out to be sufficient not only for that question but for this one as well.
Some inequalities for the exponential function.
In this answer robjohn uses Bernoulli's inequality to establish the monotonicity of some familiar sequences which converge to $e$. Using this technique I'll show that
If $y$ is fixed with $0 < y < n$ then
$$
\left(1-\frac{y}{n}\right)^n \text{ increases monotonically to } e^{-y}
\tag{1}
$$
and if $y>0$ is fixed then
$$
\left(1-\frac{y}{n+y}\right)^n \text{ decreases monotonically to } e^{-y}
\tag{2}
$$
as $n \to \infty$.
The limits can be established in the usual way so I'll just show monotonicity.
We have
$$
\begin{align}
\frac{\left(1-\frac{y}{n+1}\right)^{n+1}}{\left(1-\frac{y}{n}\right)^n} &= \left(\frac{n+1-y}{n+1}\right)^{n+1} \left(\frac{n}{n-y}\right)^n \\
&= \frac{n-y}{n} \left(\frac{n+1-y}{n+1} \cdot \frac{n}{n-y}\right)^{n+1} \\
&= \frac{n-y}{n} \left(1 + \frac{y}{(n-y)(n+1)}\right)^{n+1} \\
&> \frac{n-y}{n} \left(1 + \frac{y(n+1)}{(n-y)(n+1)}\right) \\
&= 1,
\end{align}
$$
proving the first statement, and
$$
\begin{align}
\frac{\left(1-\frac{y}{n+y}\right)^n}{\left(1-\frac{y}{n+1+y}\right)^{n+1}} &= \left(\frac{n}{n+y}\right)^n \left(\frac{n+1+y}{n+1}\right)^{n+1} \\
&= \frac{n+y}{n} \left(\frac{n}{n+y} \cdot \frac{n+1+y}{n+1}\right)^{n+1} \\
&= \frac{n+y}{n} \left(1 - \frac{y}{(n+y)(n+1)}\right)^{n+1} \\
&> \frac{n+y}{n} \left(1 - \frac{y(n+1)}{(n+y)(n+1)}\right) \\
&= 1,
\end{align}
$$
proving the second.
Bounding the positive root.
Let $x$ be a root of the equation $x^n + x - 1 = 0$ with $0 < x < 1$, the existence of which is guaranteed by the intermediate value theorem. Let
$$
x = 1 - \frac{y}{n},
$$
so that $0 < y < n$. We then have
$$
\begin{align}
0 &= \left(1-\frac{y}{n}\right)^n + 1 - \frac{y}{n} - 1 \\
&= \left(1-\frac{y}{n}\right)^n - \frac{y}{n} \\
&< e^{-y} - \frac{y}{n}
\end{align}
$$
by $(1)$. Rearranging this gives
$$
ye^y < n
$$
which implies that
$$
y < W(n)
$$
and hence
$$
x > 1 - \frac{W(n)}{n}.
\tag{3}
$$
Similarly let
$$
x = 1 - \frac{y}{n+y}
$$
so that $y>0$. We have
$$
\begin{align}
0 &= \left(1-\frac{y}{n+y}\right)^n + 1 - \frac{y}{n+y} - 1 \\
&= \left(1-\frac{y}{n+y}\right)^n - \frac{y}{n+y} \\
&> e^{-y} - \frac{y}{n+y}
\end{align}
$$
by $(2)$, from which we deduce that
$$
ye^y > n+y > n.
$$
Thus
$$
y > W(n)
$$
and so
$$
x < 1 - \frac{W(n)}{n + W(n)}.
\tag{4}
$$
By combining $(3)$ and $(4)$ we get
$$
1 - \frac{W(n)}{n} < x < 1 - \frac{W(n)}{n + W(n)}.
\tag{5}
$$
Calculating the desired limit.
We can simplify the right-hand side of $(5)$ slightly by applying the inequality $(1+t)^{-1} \geq 1-t$, valid for $t > -1$, to get
$$
\begin{align}
x &< 1 - \frac{W(n)}{n + W(n)} \\
&= 1 - \frac{W(n)}{n} \cdot \frac{1}{1 + \frac{W(n)}{n}} \\
&\leq 1 - \frac{W(n)}{n} \left(1 - \frac{W(n)}{n}\right) \\
&= 1 - \frac{W(n)}{n} + \frac{W(n)^2}{n^2}.
\end{align}
$$
Using this, inequality $(5)$ becomes
$$
1 - \frac{W(n)}{n} < x < 1 - \frac{W(n)}{n} + \frac{W(n)^2}{n^2}.
\tag{6}
$$
Let's get a handle on the Lambert $W$ function here. In this answer I derived the inequalities
$$
\log n - \log\log n < W(n) < \log n - \log\log n - \log\left(1 - \frac{\log\log n}{\log n}\right)
\tag{7}
$$
and $W(n) < \log n$ which hold for $n>e$.
Substituting the first of these, the left-hand side of $(6)$ becomes
$$
\begin{align}
x &> 1 - \frac{W(n)}{n} \\
&> 1 - \frac{\log n}{n} + \frac{\log\log n}{n} + \frac{\log\left(1 - \frac{\log\log n}{\log n}\right)}{n},
\end{align}
$$
which rearranges to
$$
\frac{n}{\log\log n} \left(1-x-\frac{\log n}{n}\right) < -1 - \frac{\log\left(1-\frac{\log\log n}{\log n}\right)}{\log\log n}.
\tag{8}
$$
Mirroring this, the right-hand side of $(6)$ becomes
$$
\begin{align}
x &< 1 - \frac{W(n)}{n} + \frac{W(n)^2}{n^2} \\
&< 1 - \frac{\log n}{n} + \frac{\log\log n}{n} + \frac{(\log n)^2}{n^2},
\end{align}
$$
so that
$$
\frac{n}{\log\log n} \left(1-x-\frac{\log n}{n}\right) > -1 - \frac{(\log n)^2}{n \log\log n}.
\tag{9}
$$
Combining $(8)$ and $(9)$ yields
$$
- \frac{(\log n)^2}{n \log\log n} < \frac{n}{\log\log n} \left(1-x-\frac{\log n}{n}\right) + 1 < - \frac{\log\left(1-\frac{\log\log n}{\log n}\right)}{\log\log n}.
$$
I learned this method from this answer by Qiaochu Yuan.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/485408",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 4,
"answer_id": 2
}
|
Expression of an Integer as a Power of 2 and an Odd Number (Chartrand Ex 5.4.2[a])
Let $n$ be a positive integer. Show that every integer $m$ with $ 1 \leq m \leq 2n $ can be expressed as $2^pk$, where $p$ is a nonnegative integer and $k$ is an odd integer with $1 \leq k < 2n$.
I wrote out some $m$ to try to conceive the proof. I observed:
$\bbox[5px,border:1px solid grey]{\text{Case 1: $m$ odd}}$ Odd numbers $\neq 2p$, thus the only choice is to put $p = 0$ and $k = m$.
$\bbox[5px,border:1px solid grey]{\text{Case 2: $m$ even and power of 2}}$ Then $p$ can be determined by division or inspection to "square with" the power of $2$. This requires $k = 1$. Is an explicit formula for $p$ necessary?
$\bbox[5px,border:1px solid grey]{\text{Case 3: $m$ even and NOT A power of 2}}$
$\begin{array}{cc}
\\
\boxed{m = 6}: 6 = 2^1 \cdot 3 \qquad \qquad & \boxed{m = 10}: 10 = 2^1 \cdot 5 \qquad \qquad & \boxed{m = 12}: 12 = 2^2 \cdot 3 \\
\boxed{m = 14}: 14 = 2 \cdot 7 \qquad \qquad & \boxed{m = 18}: 18 = 2^1 \cdot 9 \qquad \qquad & \boxed{m = 20}: 20 = 2^2 \cdot 5\\
\boxed{m = 22}: 22 = 2 \cdot 11 \qquad \qquad & \boxed{m = 24}: 24 = 2^3 \cdot 3 \qquad \qquad & \boxed{m = 26}: 26 = 2^1 \cdot 13
\end{array}$
Solution's Proof by Contradiction: $\color{#0073CF}{\Large{\mathbb{[}}}$Let $p$ be the greatest nonnegative integer
such that $2^p | m. \color{#0073CF}{\Large{\text{]}}}$ $\color{red}{\Large{\text{[}}}$ Then $m= 2^pk$ for some positive integer $k$. Necessarily $k$ is odd,
for otherwise this would contradicts the definition of $p$. $\color{red}{\Large{\text{]}}}$
$\Large{\text{1.}}$ Could someone please expound on the answer? It differs from my work above?
$\Large{\text{2.}}$ Is there a general formula/pattern for Case $3$?
I referenced 1. Source: Exercise 5.42(a), P125 of Mathematical Proofs, 2nd ed. by Chartrand et al
|
$p$ is the greatest integer such that $2^p$ divides $m$.
This means that :
*
*there exists an integer $k$ such that $m = 2^p k$
*for any integer $q$, if there is an integer $l$ such that $m = 2^q l$, then $p \ge q$.
Suppose the $k$ you get by 1. is even. Then $k = 2l$ for some integer $l$. Hence $m = 2^p(2l) = 2^{p+1}l$. Now apply 2. with $q = p+1$ : you get that $p \ge p+1$. This is a contradiction, so $k$ must be odd.
For example, let $m = 24$. It turns out that $24 = 2^3 * 3$, so $2^3$ divides $24$. $3$ is the greatest integer $p$ such that $2^p$ divides $m$ (you can indeed check that $16,32,64,128,\ldots$ don't divide $24$) and so we find that indeed, $3$ is odd and $24 = 2^3 * 3$ was the decomposition we wanted.
The solution gave you not an algebraic formula for $p$ (as you may have wanted), but a logical formula for $p$, which is the next best thing, as it tells you exactly which $p$ you have to take.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/486395",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 5,
"answer_id": 2
}
|
Integrating $\cos^3 (x) \, dx$ I am wondering whether I integrated the following correctly.
*
*$\int \cos^3 x \, dx$
I did
\begin{align}
\int \cos^3 x \, dx &= \int \cos(x)(1-\sin^{2}x) \, dx \\
&= \int \cos(x)-\sin^{2}x \cos x \, dx \\
&= \sin(x)-\frac{u^{3}}{3} + c, \quad(u=\sin(x)) \\
&= \sin(x)-\dfrac{\sin^{3}x}{3}+c
\end{align}
2.$\int \sin^{3}x \cos^{2}x\,dx$
\begin{align}
\int \sin^{3}x \cos^{2}x\,dx &= \int(1-\cos^2x)(\cos^2x)\sin(x)\,dx \\
&= \int \cos^2x\sin(x)-\cos^4x\sin(x)\,dx, \quad u=\cos(x) \\
&= \dfrac{u^3}{3}-\dfrac{u^5}{5}+c
\end{align}
And plug in my u.
|
You can even put the formula of $$\cos^{3}x = \frac{\cos3x + 3 \cos x}{4}$$
And then integrate it where you get the answer as
$$ \frac{\sin 3x}{12} + \frac{3 \sin x}{4} + C$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/486849",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
}
|
integral part of surd
$(\sqrt{a}+b)^n=N+f$ where $f \in (0,1)$
$(\sqrt{a}+b)^{n+2} =M+g$ where $g \in (0,1)$
Given that $0<\sqrt{a}-b<1$ and $(a,b)$
belongs to integers, then
*
*If $n$ is odd, $f>g$
*If $n$ is even, $f<g$
How to prove/disprove it?
|
Since $a$ and $b$ are integers (and $a > 0$ is not a square, otherwise $0 < \sqrt{a}-b < 1$ would not hold), we have
$$\begin{align}
(b+\sqrt{a})^n + (b-\sqrt{a})^n &= \sum_{k=0}^n \binom{n}{k}b^{n-k}a^{k/2} + \sum_{k=0}^n (-1)^k\binom{n}{k}b^{n-k}a^{k/2}\\
&= \sum_{k=0}^n \bigl(1 + (-1)^k\bigr)\binom{n}{k}b^{n-k}a^{k/2}\\
&= 2\sum_{m=0}^{\lfloor n/2\rfloor} \binom{n}{2m}b^{n-2m}a^m,
\end{align}$$
so $(b+\sqrt{a})^n + (b-\sqrt{a})^n$ is an integer. From $0 < \sqrt{a}-b < 1$ it follows that $0 < \lvert b-\sqrt{a}\rvert^n < 1$ and $(b-\sqrt{a})^n$ is positive when $n$ is even, and negative when $n$ is odd.
Hence for odd $n$, we have
$$(\sqrt{a}+b)^n = \underbrace{(b+\sqrt{a})^n + (b-\sqrt{a})^n}_{N(n)} + \underbrace{(\sqrt{a}-b)^n}_{f(n)}$$
and with the notations of the problem, $g = f(n+2) = (\sqrt{a}-b)^2\cdot f(n) = (\sqrt{a}-b)^2\cdot f < f$.
For even $n$ we have
$$(\sqrt{a}+b)^n = \underbrace{(b+\sqrt{a})^n + (b-\sqrt{a})^n - 1}_{N(n)} + \underbrace{1-(\sqrt{a}-b)^n}_{f(n)},$$
and in the notation of the problem $g = 1-(\sqrt{a}-b)^{n+2} > 1 - (\sqrt{a}-b)^n = f$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/489646",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 1,
"answer_id": 0
}
|
How find this inequality $\sqrt{a^2+64}+\sqrt{b^2+1}$ let $a,b$ are positive numbers,and such $ab=8$ find this minum
$$\sqrt{a^2+64}+\sqrt{b^2+1}$$
My try:
and I find when $a=4,b=2$,then
$$\sqrt{a^2+64}+\sqrt{b^2+1}$$ is minum $5\sqrt{5}$
it maybe use Cauchy-Schwarz inequality
Thank you
|
As noted by Mher Safaryan, your inequality is equivalent to
$$
(1+a)\sqrt{b^2+1} \geq 5\sqrt{5} \tag{1}
$$
or in other words,
$$
(1+a)^2(1+b^2) \geq 125 \tag{2}
$$
or equivalently,
$$
(1+\frac{8}{b})^2(1+b^2) \geq 125 \tag{3}
$$
We are now done, because of the identity
$$
(1+\frac{8}{b})^2(1+b^2)-125=\frac{(b-2)^2\bigg(b^2+20b+16\bigg)}{b^2}
$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/491394",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
}
|
Pre calculus fraction simplify question Simplify:
$$\frac{\frac{16x^4}{81} - y^4}{\frac{2x}{3} + y}$$
Wolfram alpha confirms the answer from the answer sheet: Wolframalpha answer
|
Let $u=\frac{2x}{3}$ and notice that $u^{4}=\big(\frac{2x}{3}\big)^{4}=\frac{2^{4}x^{4}}{3^{4}}=\frac{16x^{4}}{81}$.
So we get:
$\frac{\frac{16x^{4}}{81}-y^{4}}{\frac{2x}{3}+y}=\frac{u^{4}-y^{4}}{u+y}=\frac{(u^{2}-y^{2})(u^{2}+y^{2})}{u+y}=\frac{(u-y)(u+y)(u^{2}+y^{2})}{u+y}=(u-y)(u^{2}+y^{2})=(\frac{2x}{3}-y)(\frac{4x^{2}}{9}+y^{2})$
$=(\frac{1}{3}(2x-3y))(\frac{1}{9}(4x^{2}+9y^{2}))=\frac{1}{27}(2x-3y)(4x^{2}+9y^{2})=\frac{1}{27}(8x^{3}-12x^{2}y+18xy^{2}-27y^{3})$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/491975",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 0
}
|
How to get all solutions to equations with square roots I would like to find all solutions to
$$b-a\sqrt{1+a^2+b^2}=a^2(ab-\sqrt{1+a^2+b^2})$$
$$a-b\sqrt{1+a^2+b^2}=b^2(ab-\sqrt{1+a^2+b^2})$$
I found some solutions. For example, $a = 1, b = \pm i$ and $b = 1 , a = \pm i$. How can I find all solutions?
|
The first equation becomes
$$b(1-a^3)=a(1-a)\sqrt{1+a^2+b^2},$$
the second
$$a(1-b^3)=b(1-b)\sqrt{1+a^2+b^2}.$$
If $a=0$, then $b=0$.
Clearly $a=b=1$ is a solution. Let $a=1$ and $b\neq1$. Then from the second equation
$$1+b+b^2=b\sqrt{2+b^2}$$
we derive the solutions $b=\pm i$.
Let $a$ and $b$ not equal 0 or 1. Then we have
$$b(1+a+a^2)=a\sqrt{1+a^2+b^2},$$
and
$$a(1+b+b^2)=b\sqrt{1+a^2+b^2}.$$
Consider now $1+a+a^2=0\iff a=\frac{-1\pm i\sqrt{3}}{2}$. In this case $1+a^2+b^2$ must be 0, this gives together with $1+a+a^2=0$ the condition $b^2=-a$, from where we get four more solutions.
Now lets divide one equation by the other. An easy compution yields
$$(b-a)(a+b+ab)=0.$$
Now if you multiply the equations, you'll get
$$ab(1+ab)(a+b+ab)=0\iff (1+ab)(a+b+ab)=0.$$
So either $a+b=-ab$ or $a=b$ and $ab=-1$, that is $a=b=i$.
So we're done. (Remind that if $(a,b)$ is a solution, so is $(b,a)$.
Michael
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/492434",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
}
|
Solve $x(x+1)=y(y+1)(y^2+2)$ for $x,y$ over the integers Solve $$x(x+1)=y(y+1)(y^2+2)$$ , for $x,y$ over the integers
|
Here's a solution for positive $x$ and $y$.
I will show that
the only solutions
for positive $x$ and $y$ are
$(x, y) = (2, 1)$ and $(11, 3)$.
$x(x+1) = y(y+1)(y^2+2)
=y(y^3+y^2+2y+2)
=y^4+y^3+2y^2+2y
$
Multiplying by 4,
$(2x+1)^2-1
=4y^4+4y^3+8y^2+8y
$
or
$(2x+1)^2
=4y^4+4y^3+8y^2+8y+1
$
My goal is to show algebraically
that this polynomial in $y$
is between two consecutive squares
for large enough $y$,
and then examine the remaining cases.
$(2y^2+y)^2
=4y^4+4y^3+y^2
$.
$\begin{align}
(2y^2+y+1)^2
&=4y^4+4y^3+y^2
+2(2y^2+y)+1 \\
&=4y^4+4y^3+y^2 +4y^2+2y+1 \\
&=4y^4+4y^3+5y^2+2y+1 \\
\end{align}
$.
$\begin{align}
(2y^2+y+2)^2
&=4y^4+4y^3+y^2 +4(2y^2+y)+4 \\
&=4y^4+4y^3+y^2 +8y^2+4y+4 \\
&=4y^4+4y^3+9y^2+4y+4 \\
\end{align}
$.
For
$(2x+1)^2$
to be between these consecutive squares,
we need
$5y^2+2y+1
<8y^2+8y+1
<9y^2+4y+4
$.
The first inequality is true
for $y \ge 1$.
For the second inequality to be true,
we need
$8y^2+8y+1
<9y^2+4y+4
$
or
$y^2-4y+3
> 0
$
or
$(y-2)^2-1
> 0$.
This is true for
$y \ge 4$,
so the equation has no solution for $y \ge 4$.
If $y = 3$,
the equation is
$x(x+1) = 3(4)(11)$
and this is true for
$x=11$
(surprise!).
If $y = 2$,
the equation is
$x(x+1) = 2(3)(4)=24$
which has no solution.
If $y = 1$,
the equation is
$x(x+1) = 1(2)(3)$
and this is true for
$x=2$.
Therefore
the only solutions
for positive $x$ and $y$ are
$(x, y) = (2, 1)$ and $(11, 3)$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/492581",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
}
|
Consider the funcition and find an expression Consider the function $f(x) = \frac{2}{x-1}$ and find an expression for $\frac{f(x+h)-f(x)}{h}$ and simplify where he cannot be equal to $0$.
|
$\require{cancel}$
$$f(x) = \frac 2{x - 1}$$
$$\frac{f(x + h) - f(x)}{h} = \frac{\dfrac{2}{(x + h) - 1} - \dfrac {2}{x - 1}}{h}$$
Multipy numerator and denominator by $(x + h - 1)(x - 1)$
$$\begin {align}\frac{\dfrac{2}{(x + h) - 1} - \dfrac {2}{x - 1}}{h} \cdot \frac{ \dfrac{(x + h - 1)(x - 1)}{1}}{(x + h - 1)(x - 1)} & = \frac{2(x - 1) - 2(x + h - 1)}{h(x+h - 1)(x - 1)}\\ \\ & = \dfrac{-2\cancel{h}}{\cancel{h}(x+h + 1)(x-1)}\end{align}$$
After canceling the common factor, you're done.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/494599",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
}
|
Find $(a,b)$ if $x^2-bx+a = 0, x^2-ax+b = 0$ both have distinct positive integers roots
If $x^2-bx+a = 0$ and $x^2-ax+b = 0$ both have distinct positive integers roots, then what is $(a,b)$?
My Try:
$$\displaystyle x^2-ax+b = 0\Rightarrow x = \frac{a\pm \sqrt{a^2-4b}}{2}$$
So here $a^2-4b$ is a perfect square.
Similarly
$$\displaystyle x^2-bx+a = 0\Rightarrow x = \frac{b\pm \sqrt{b^2-4a}}{2}$$
So here $b^2-4a$ is a perfect square.
But I did not understand how can I solve after that.
|
Would a=b=4 work? In that case the (double) solution in both cases is 2 and that's positive
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/494679",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 3,
"answer_id": 2
}
|
Recurrence relation problem, need help:) I´m stuck on a problem. Can anyone help me?
The problem: Find the recurrence relation to
$$a_n=a_{n-1}+2a_{n-2}+\cdots+(n-1)a_1+na_0\;(\text{for }n\ge 1),\\a_0=1$$
I guess I have to compare $a_n-a_{n-1}$ with $a_{n-1}-a_{n-2}$?
|
Use generating functions. Define $A(z) = \sum_{n \ge 0} a_n z^n$, and write your recurrence as:
$\begin{equation*}
a_{n + 1}
= \sum_{0 \le k \le n} (k + 1) a_{n - k}
\end{equation*}$
Multiply the recurrence by $z^n$ and sum over $n \ge 0$, recognize some sums and use $a_0 = 1$:
$\begin{align*}
\sum_{n \ge 0} a_{n + 1} z^n
&= \sum_{n \ge 0} z^n \sum_{0 \le k \le n} (k + 1) a_{n - k} \\
\frac{A(z) - a_0}{z}
&= \left( \sum_{n \ge 0} (n + 1) z^n \right)
\cdot \left( \sum_{n \ge 0} a_n z^n \right) \\
\frac{A(z) - 1}{z}
&= \left( \frac{d}{d z} \sum_{n \ge 0} z^n \right) \cdot A(z) \\
&= \left( \frac{d}{d z} \frac{1}{1 - z} \right) \cdot A(z) \\
&= \frac{A(z)}{(1 - z)^2}
\end{align*}$
Solving for $A(z)$, as partial fractions:
$\begin{align*}
A(z)
&= \frac{(1 + z)^2}{1 - 3 z + z^2} \\
&= \frac{\sqrt{5}}{5} \cdot \frac{1}{1 - z \frac{3 + \sqrt{5}}{2}}
- \frac{\sqrt{5}}{5} \cdot \frac{1}{1 - z \frac{3 - \sqrt{5}}{2}}
+ 1
\end{align*}$
We are after the coefficient of $z^n$, the expression is just geometric series:
$\begin{equation*}
[z^n] A(z)
= \frac{\sqrt{5}}{5}
\cdot \left(
\left( \frac{3 + \sqrt{5}}{2} \right)^n
- \left( \frac{3 - \sqrt{5}}{2} \right)^n
\right)
+ [n = 0]
\end{equation*}$
Here $[\mathit{condition}]$ is Iverson's convention: $1$ if the condition is true, $0$ if it is false.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/496516",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
}
|
Proving $\sqrt[3]{\sqrt[3]{2}-1}=\sqrt[3]{\frac19}-\sqrt[3]{\frac29}+\sqrt[3]{\frac49}$ I found the following two relational expressions in a book without any additional information:
$$\sqrt{\sqrt[3]{5}-\sqrt[3]{4}}=\frac13(\sqrt[3]{2}+\sqrt[3]{20}-\sqrt[3]{25})$$
$$\sqrt[3]{\sqrt[3]{2}-1}=\sqrt[3]{\frac19}-\sqrt[3]{\frac29}+\sqrt[3]{\frac49}$$
Wolfram tells these are true, but I can't prove at all. Can anyone help?
|
let
$$\sqrt[3]{\dfrac{1}{3}}=x,\sqrt[3]{\dfrac{2}{3}}=y$$
then
$$\sqrt[3]{\dfrac{1}{9}}-\sqrt[3]{\dfrac{2}{9}}+\sqrt[3]{\dfrac{4}{9}}=x^2-xy+y^2$$
then
$$(x+y)(x^2-xy+y^2)=x^3+y^3=\dfrac{1}{3}+\dfrac{2}{3}=1$$
$$\Longrightarrow x^2-xy+y^2=\dfrac{1}{x+y}=\sqrt[3]{3}(\sqrt[3]{1}+\sqrt[3]{2})^{-1}=\sqrt[3]{\sqrt[3]{2}-1}$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/498325",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
}
|
question on separable differential equations I'm trying to solve this equation but always got stuck in a term. How can I separate the $x$,$y$ variables in the equation?
$$(y^2 -yx^2)dy + (y^2 + xy^2)dx =0$$
|
For the ODE $(y^2-yx^2)~dy+(y^2+xy^2)~dx=0$ , I don't think you can separate the $x$ and $y$ variables directly, instead you can only solve it like this:
$(y^2-yx^2)~dy+(y^2+xy^2)~dx=0$
$y^2(x+1)~dx=(yx^2-y^2)~dy$
$(x+1)\dfrac{dx}{dy}=\dfrac{x^2}{y}-1$
Let $u=x+1$ ,
Then $x=u-1$
$\dfrac{dx}{dy}=\dfrac{du}{dy}$
$\therefore u\dfrac{du}{dy}=\dfrac{(u-1)^2}{y}-1$
$u\dfrac{du}{dy}=\dfrac{u^2-2u+1}{y}-1$
$u\dfrac{du}{dy}=\dfrac{u^2}{y}-\dfrac{2u}{y}+\dfrac{1}{y}-1$
This belongs to an Abel equation of the second kind.
Follow the method in http://eqworld.ipmnet.ru/en/solutions/ode/ode0126.pdf:
Let $u=yv$ ,
Then $\dfrac{du}{dy}=y\dfrac{dv}{dy}+v$
$\therefore yv\left(y\dfrac{dv}{dy}+v\right)=yv^2-2v+\dfrac{1}{y}-1$
$y^2v\dfrac{dv}{dy}+yv^2=yv^2-2v+\dfrac{1}{y}-1$
$y^2v\dfrac{dv}{dy}=-2v+\dfrac{1}{y}-1$
$v\dfrac{dv}{dy}=-\dfrac{2v}{y^2}+\dfrac{1}{y^3}-\dfrac{1}{y^2}$
Let $t=\dfrac{2}{y}$ ,
Then $y=\dfrac{2}{t}$
$\dfrac{dv}{dy}=\dfrac{dv}{dt}\dfrac{dt}{dy}=-\dfrac{2}{y^2}\dfrac{dv}{dt}$
$\therefore-\dfrac{2v}{y^2}\dfrac{dv}{dt}=-\dfrac{2v}{y^2}+\dfrac{1}{y^3}-\dfrac{1}{y^2}$
$v\dfrac{dv}{dt}=v+\dfrac{1}{2}-\dfrac{1}{2y}$
$v\dfrac{dv}{dt}=v+\dfrac{1}{2}-\dfrac{t}{4}$
$\dfrac{dv}{dt}=1-\dfrac{t-2}{4v}$
Luckily this becomes a first-order homogeneous ODE.
Let $w=\dfrac{v}{t-2}$ ,
Then $v=(t-2)w$
$\dfrac{dv}{dt}=(t-2)\dfrac{dw}{dt}+w$
$\therefore(t-2)\dfrac{dw}{dt}+w=1-\dfrac{1}{4w}$
$(t-2)\dfrac{dw}{dt}=-\dfrac{4w^2-4w+1}{4w}$
$\dfrac{4w}{(2w-1)^2}~dw=-\dfrac{dt}{t-2}$
$\int\dfrac{4w}{(2w-1)^2}~dw=-\int\dfrac{dt}{t-2}$
$\ln(2w-1)-\dfrac{1}{2w-1}=-\ln(t-2)+c_1$
$(2w-1)e^{-\frac{1}{2w-1}}=\dfrac{c_2}{t-2}$
$\dfrac{2v-t+2}{t-2}e^{-\frac{t-2}{2v-t+2}}=\dfrac{c_2}{t-2}$
$\dfrac{yv+y-1}{y-1}e^\frac{y-1}{yv+y-1}=\dfrac{Cy}{y-1}$
$(u+y-1)e^\frac{y-1}{u+y-1}=Cy$
$(x+y)e^\frac{y-1}{x+y}=Cy$
Checking:
$\left(\dfrac{1}{y}\dfrac{dx}{dy}-\dfrac{x}{y^2}\right)e^\frac{y-1}{x+y}+\left(\dfrac{x}{y}+1\right)\left(\dfrac{1}{x+y}-\dfrac{y-1}{(x+y)^2}\left(\dfrac{dx}{dy}+1\right)\right)e^\frac{y-1}{x+y}=0$
$\dfrac{1}{y}\dfrac{dx}{dy}-\dfrac{x}{y^2}+\dfrac{1}{y}-\dfrac{y-1}{y(x+y)}\left(\dfrac{dx}{dy}+1\right)=0$
$\left(\dfrac{1}{y}-\dfrac{y-1}{y(x+y)}\right)\left(\dfrac{dx}{dy}+1\right)=\dfrac{x}{y^2}$
$\dfrac{x+1}{y(x+y)}\left(\dfrac{dx}{dy}+1\right)=\dfrac{x}{y^2}$
$(x+1)\left(\dfrac{dx}{dy}+1\right)=\dfrac{x(x+y)}{y}$
$(x+1)\dfrac{dx}{dy}+x+1=\dfrac{x^2}{y}+x$
$(x+1)\dfrac{dx}{dy}=\dfrac{x^2}{y}-1$ , correct!
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/498683",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
}
|
Improving bound on $\sqrt{2 \sqrt{3 \sqrt{4 \ldots}}}$ An old challenge problem I saw asked to prove that $\sqrt{2 \sqrt{3 \sqrt{4 \ldots}}} < 3$. A simple calculation shows the actual value seems to be around $2.8$, which is pretty close to $3$ but leaves a gap. Can someone find $C < 3$ and prove $\sqrt{2 \sqrt{3 \sqrt{4 \ldots}}} < C$?
|
Take a logarithm from the both sides and use $\log n\leq n-5$ for $n\geq 7$ and we have:
$$
\log C= \log \sqrt{2 \sqrt{3 \sqrt{4 \ldots}}} = \frac{\log 2}{2}+\frac{\log 3}{2^2}+\frac{\log 4}{2^3}+\frac{\log 5}{2^4}+...\\
\leq \frac{\log 2}{2}+...+\frac{\log 6}{2^5}+\frac{2}{2^6}+\frac{3}{2^7}+...\\
\leq 0.92 +\frac{1}{16}\sum_{i=1}^{\infty} \frac{i}{2^i}=1.045
$$
which gives $C<2.8434$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/498774",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 6,
"answer_id": 1
}
|
integration by parts !!!! PD: I did a little change in the denominator !!!!
I need to solve this integral using integration by parts.
$\displaystyle\int\frac{x\,dx}{\sqrt{(a^2+b^2)+(x-c)^2}}$
Thanks!
PS: I know that I can to do:
$\displaystyle\int\frac{x\,dx}{\sqrt{(a^2+b^2)+(x-c)^2}}=\int\frac{(x-c)\,dx}{\sqrt{(a^2+b^2)+(x-c)^2}}+c\int\frac{dx}{\sqrt{(a^2+b^2)+(x-c)^2}}$
but according to the book it is easier using integration by part.
|
Let $x-c=\sqrt{a^2+b^2} \tan \theta,$ then $d x=\sqrt{a^2+b^2} \sec ^2 \theta d \theta$ and
$$
\begin{aligned}
\int \frac{x d x}{\sqrt{\left(a^2+b^2\right)+(x-c)^2}} &=\int \frac{c+\sqrt{a^2+b^2 \tan \theta}}{\sqrt{a^2+b^2 \sec \theta}} \sqrt{a^2+b^2 \sec ^2 \theta} d \theta\\
&=\int\left(c+\sqrt{a^2+b^2} \tan \theta\right) \sec \theta d \theta \\
&=c \ln |\sec \theta+\tan \theta|+\sqrt{a^2+b^2} \sec \theta+C\\
&=c \ln \left|\frac{x-c+\sqrt{a^2+b^2+(x-c)^2}}{\sqrt{a^2+b^2}}\right|+\sqrt{a^2+b^2+(x-c)^2}+C
\end{aligned}
$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/499901",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
}
|
Convergence of $x_{n}=\frac{n^2}{\sqrt{n^{6}+1}}+\frac{n^2}{\sqrt{n^{6}+2}}+\cdots+\frac{n^2}{\sqrt{n^{6}+n}}$ Let $$x_{n}=\frac{n^2}{\sqrt{n^{6}+1}}+\frac{n^2}{\sqrt{n^{6}+2}}+\frac{n^2}{\sqrt{n^{6}+3}}+...+\frac{n^2}{\sqrt{n^{6}+n}}.$$ Then $(x_{n})$ converges to
(A)$1$
(B)$0$
(C)$\frac{1}{2}$
(D)$\frac{3}{2}$
My Try:
$$\lim_{n\to \infty}x_{n}=\frac{n^2}{n^{3}}(\frac{1}{\sqrt{1+\frac{1}{n^6}}}+\frac{1}{\sqrt{1+\frac{2}{n^{6}}}}+...+\frac{1}{\sqrt{1+\frac{1}{n^5}}})=0.$$
Am i right?
|
$$
x_{n}=\frac{n^2}{\sqrt{n^{6}}}+\frac{n^2}{\sqrt{n^{6}}}+\cdots+\frac{n^2}{\sqrt{n^{6}}} = n(\frac{n^2}{\sqrt{n^{6}}}) = \frac{n^3}{n^3} = 1
$$
This is how I solve this.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/500235",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
}
|
Finding the equation of an ellipse centered at (-1,3) and through the points (1,3) and (-1,4) I've been taking a computer graphics class and as a review we had some questions about some geometric problems. This one I can not seem to figure out though it seems that it should be straight forward.
I know that the standard equation of an ellipse is
x^2/a^2 + y^2/b^2 = 1
From this we shift to its stated center and the equation becomes
(x+1)^2/a^2 + (y-3)^2/b^2 =1
But from here I am at a loss for where to go from here...
|
Substitute into your second equation (corrected with $b^2$ in the second denominator) the $x, y$ values for each of the points on the ellipse $(1, 3), (-1, 4)$, and you'll obtain two equations in two unknowns, from which you can solve for $a^2$ and $b^2$.
That is, evaluate $$\dfrac{(x+1)^2}{a^2} + \dfrac{(y-3)^2}{b^2} =1$$
at (1) $(1, 3)$ and again at (2) $(-1, 4)$. That will give you two equations in $a^2, b^2$, for which you can solve simultaneously.
$$\dfrac{(1 + 1)^2}{a^2} + \dfrac{(3-3)^2}{b^2} = 1 \implies \dfrac 4{a^2} = 1 \implies a^2 = 4\tag{1}$$
$$\dfrac{(-1 + 1)^2}{a^2} + \dfrac{(4-3)^2}{b^2} = 1 \implies \dfrac 1{b^2} = 1 \implies b^2 = 1\tag{2}$$
Now, use the general equation with your new-found values for $a^2, b^2$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/500395",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
}
|
Show $\binom{k+1}{r}+\binom{k+1}{r+1} = \binom{k+2}{k+1} $ I have been attempting to show $$\binom{k+1}{r}+\binom{k+1}{r+1} = \binom{k+2}{r+1} $$ and my work is
$$\binom{k+1}{r}+\binom{k+1}{r+1} = \frac{(k+1)!}{r!((k+1)-r)!} + \frac{(k+1)!}{(r+1)!((k+1)-(r+1))!}$$
$$=\frac{(k+1)!(r+1)}{(r+1)!((k+1)-r)!} + \frac{(k+1)!
\color{red}{(k-r)}}{(r+1)!((k+1)-(r+1))!\color{red}{(k-r)}}$$
$$=(k+1)!\frac{k+1}{(r+1)!((k+1)-r)!}$$
which appears correct. But I am having trouble seeing the final step; where does the $(k+2)$ come from?
Somewhere I've made a mistake but I just don't see it.
|
We have $\binom{k+1}{r} = \frac{(k+1)!}{(k+1-r)!r!}$ and $\binom{k+1}{r+1} = \frac{(k+1)!}{(k+1-(r+1))!(r+1)!} = \frac{(k+1)!}{(k-r)!(r+1)!}$. Both have a common factor of $\frac{(k+1)!}{(k-r)!r!}$ so let's factor that out to get
$$\binom{k+1}{r}+\binom{k+1}{r+1} = \frac{(k+1)!}{(k-r)!r!}\left(\frac{1}{k+1-r}+\frac{1}{r+1}\right).$$
Making a common denominator, we have
$$\binom{k+1}{r}+\binom{k+1}{r+1} = \frac{(k+1)!}{(k-r)!r!}\cdot\frac{r+1+k+1-r}{(k+1-r)(r+1)}.$$
Notice we can simplify the numerator some to get $k+2$. Rewriting we have
$$\binom{k+1}{r}+\binom{k+1}{r+1} = \frac{(k+1)!}{(k-r)!r!}\cdot\frac{k+2}{((k+1)-r)(r+1)}.$$
Notice that we have $(k-r)!(k+1-r) = (k+1-r)! = ((k+2)-(r+1))!$. So we get
$$\binom{k+1}{r}+\binom{k+1}{r+1} =\frac{(k+2)(k+1)!}{((k+2)-(r+1))!(r+1)r!} = \frac{(k+2)!}{((k+2)-(r+1))!(r+1)!}.$$
Thus,
$$\binom{k+1}{r}+\binom{k+1}{r+1} = \binom{k+2}{r+1}.$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/500802",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
}
|
How many eight-digit numbers can be formed with the numbers 2, 2, 2, 3, 3, 3, 3, 4, 4? I know how to determine the problem when we form nine-digit numbers that is $\frac{9!}{(3!\cdot 4!\cdot 2!)}$. But what about eight-digit numbers?
|
We need to omit exactly one of the numbers. So any $8$-digit number induces the multiset of numbers:
\begin{align*}
& \{2,2,3,3,3,3,4,4\}, \\
& \{2,2,2,3,3,3,4,4\}, \text{or } \\
& \{2,2,2,3,3,3,3,4\}.
\end{align*}
These give rise to
\begin{align*}
\binom{8}{2,4,2}+\binom{8}{3,3,2}+\binom{8}{3,4,1} &= 420+560+280\\
&= 1260
\end{align*}
possible $8$-digit numbers.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/501131",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
}
|
How do I reach this form I just want to check with you guys how to do this correctly:
I have to reduce:
$$ \frac{1}{N}+3=\frac{2R+2}{5R+2}$$
...to the following form:
$$ N=\frac{aR+b}{cR+d}$$
Any leads on how I reach this form? Should I flip all terms, to get N=, then bring 1/3 to the other side and subtract it there?
Thank you in advance.
|
First, subtract $3$ from each side of the equation. Then find a common denominator: $$\begin{align} \frac{1}{N}+3=\frac{2R+2}{5R+2} &\iff \frac{1}{N} =\frac{2R+2}{5R+2} -3 \\ \\ & \iff \dfrac 1N = \dfrac{(2R+2) - 3(5R+2)}{5R + 2} \\ \\ & \iff \frac 1N = \frac{-13 R- 4}{5R + 2}\end{align}$$
Simplify, and "flip"/invert":
$$\iff N =\frac {5R + 2}{-13 R- 4} = \frac {-5R - 2}{13 R + 4} =-\frac {5R + 2}{13 R+ 4}$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/501196",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
}
|
Find all natural numbers n such that $7n^3 < 5^n$ Find all natural numbers $n$ such that $7n^3 < 5^n$.
I drew a graph which showed that $n \geq 4$
wolfram
How can I prove that? I guess I need to use induction with the base case $n=4$?
But I am stuck because the induction hypothesis uses a $\geq$ sign so I can not substitute...
|
First, check for $n = 4$:
$$7n^3 = 448 < 625 = 5^n.$$
Now, assume that it is true for all numbers strictly less than $n$. Let us check for $n$:
\begin{align*}
7n^3 &= 7((n-1)+1)^3 = 7(n-1)^3 + 7 \cdot 3 \cdot (n-1)^2 + 7 \cdot 3 \cdot (n-1) + 7. %\\
%&< 4 \cdot 7 (n-1)^3 < 4 \cdot 5^{n-1} < 5 \cdot 5^{n-1} = 5^n.
\end{align*}
Here, use that $3(n-1)^k < (n-1)^3$ for $k \in \{1,2\}$, which comes from the fact that $n > 4$, so $n - 1 > 3$. Also, $7 < (n-1)^3$. If you get stuck, ask.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/501695",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
}
|
How to prove: $\lim_{n\to\infty}\left(\int_{0}^{\frac{\pi}{2}}\left\vert\frac{\sin{(n+1)x}}{\sin{x}}\right\vert dx-\frac{2\ln{n}}{\pi}\right)$ show that
$$\mathop {\lim }\limits_{n \to \infty } \left( {\int\limits_0^{\frac{\pi }{2}} {\left\vert\frac{{\sin \left( {2n + 1} \right)x}}{{\sin x}}\right\vert\,dx - \frac{{2\ln n}}{\pi }} } \right) = \frac{{6\ln 2}}{\pi } + \frac{{2\gamma }}{\pi } + \frac{2}{\pi }\sum\limits_{k = 1}^\infty {\frac{1}{{2k + 1}}\ln \left( {1 + \frac{1}{k}} \right)}\cdots (1) $$
I can prove $(1)$ it exsit it.and also it is well kown that
$$I_{n}=\int_{0}^{\frac{\pi}{2}}\dfrac{\sin{(2n+1)x}}{\sin{x}}dx=\dfrac{\pi}{2}$$
proof:$$I_{n}-I_{n-1}=\int_{0}^{\frac{\pi}{2}}\dfrac{\sin{(2n+1)x}-\sin{(2n-1)x}}{\sin{x}}dx=2\int_{0}^{\frac{\pi}{2}}\cos{(2nx)}dx=0$$
so
$$I_{n}=I_{n-1}=\cdots=I_{0}=\dfrac{\pi}{2}$$
But I can't prove $(1)$,Thank you
|
Notice for any continuous function $f(x)$ on $[0,\frac{\pi}{2}]$, we have:
$$\lim_{n\to\infty} \int_0^{\frac{\pi}{2}} \Big|\sin((2n+1)x)\Big| f(x) dx = \frac{2}{\pi}\int_0^{\frac{\pi}{2}} f(x) dx$$
Apply this to $\frac{1}{\sin x} - \frac{1}{x}$, we get
$$\lim_{n\to\infty} \int_0^{\frac{\pi}{2}} \Big|\sin((2n+1)x)\Big| \Big(\frac{1}{\sin x} - \frac{1}{x} \Big) dx
= \frac{2}{\pi} \int_0^{\frac{\pi}{2}} \Big(\frac{1}{\sin x} - \frac{1}{x} \Big) dx\\
= \frac{2}{\pi} \left[\log\left(\frac{\tan(\frac{x}{2})}{x}\right)\right]_0^{\frac{\pi}{2}}
= \frac{2}{\pi} \left[\log\frac{2}{\pi} - \log{\frac12}\right] = \frac{2}{\pi} \log\frac{4}{\pi}
\tag{*1}$$
So it suffices to figure out the asymptotic behavior of following integral:
$$\int_0^{\frac{\pi}{2}} \frac{|\sin((2n+1)x)|}{x} dx
= \int_0^{\pi(n+\frac12)} \frac{|\sin x|}{x} dx = \int_0^{\pi n} \frac{|\sin x|}{x} dx + O(\frac{1}{n})
$$
We can rewrite the rightmost integral as
$$\int_0^{\pi} \sin x \Big( \sum_{k=0}^{n-1} \frac{1}{x+k\pi} \Big) dx
= \int_0^1 \sin(\pi x) \Big( \sum_{k=0}^{n-1} \frac{1}{x+k} \Big) dx\\
= \int_0^1 \sin(\pi x) \Big( \psi(x+n) - \psi(x) \Big) dx
\tag{*2}
$$
where $\displaystyle \psi(x) = \frac{\Gamma'(x)}{\Gamma(x)}$ is the
digamma function.
Using following asymptotic expansion of $\psi(x)$ for large $x$:
$$\psi(x) = \log x - \frac{1}{2x} + \sum_{k=1}^{\infty}\frac{\zeta(1-2k)}{x^{2k}}$$
It is easy to verify
$$\int_0^1 \sin(\pi x)\psi(x+n) dx = \frac{2}{\pi} \log n + O(\frac{1}{n})\tag{*3}$$.
Substitute $(*3)$ into $(*2)$ and combine it with $(*1)$, we get
$$\lim_{n\to\infty} \left(\int_0^{\frac{\pi}{2}} \left|\frac{\sin((2n+1)x)}{\sin x}\right| dx - \frac{2}{\pi} \log n\right) = \frac{2}{\pi} \log\frac{4}{\pi} - \int_0^1 \sin(\pi x)\psi(x) dx \tag{*4}$$
To compute the rightmost integral of $(*4)$, we first integrate it by part:
$$\int_0^1 \sin(\pi x)\psi(x) dx = \int_0^1 \sin(\pi x)\,d\log\Gamma(x) =
-\pi\int_0^1 \cos(\pi x)\log\Gamma(x) dx
$$
We then apply following result$\color{blue}{^{[1]}}$
Kummer (1847) Fourier series for $\log\Gamma(x)$ for $x \in (0,1)$
$$\log\Gamma(x) = \frac12\log\frac{\pi}{\sin(\pi x)} + (\gamma + \log(2\pi))(\frac12 - x) + \frac{1}{\pi}\sum_{k=2}^{\infty}\frac{\log k}{k}\sin(2\pi k x)$$
Notice
*
*$\displaystyle \int_0^1 \cos(\pi x)\log \frac{\pi}{\sin(\pi x)} dx = 0\quad$ because of symmtry.
*$\displaystyle \int_0^1 \cos(\pi x)\Big(\frac12 - x\Big) dx = \frac{2}{\pi^2}$
*$\displaystyle \int_0^1 \cos(\pi x)\sin(2\pi k x) dx = \frac{4k}{(4k^2-1)\pi} $
We can evaluate RHS of $(*4)$ as
$$\begin{align}
\text{RHS}_{(*4)} = & \frac{2}{\pi}\log\frac{4}{\pi} + \pi \left[
\Big(\gamma + \log(2\pi)\Big)\frac{2}{\pi^2}
+ \frac{4}{\pi^2}\sum_{k=2}^{\infty}\frac{\log k}{4k^2-1}
\right]\\
= & \frac{2}{\pi}\left[\log 8 + \gamma + \sum_{k=2}^{\infty}\log k \left(\frac{1}{2k-1}-\frac{1}{2k+1}\right) \right]\\
= & \frac{6\log 2}{\pi} + \frac{2\gamma}{\pi} + \frac{2}{\pi}\sum_{k=1}\frac{\log(1+\frac{1}{k})}{2k+1}
\end{align}$$
Notes
$\color{blue}{[1]}$ For more infos about Kummer's Fourier series, please see
following paper by Donal F. Connon.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/501984",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 1,
"answer_id": 0
}
|
How do I get the integral of $\frac{1}{(x^2 - x -2)}$ I'm working with this problem $$ \int \frac{1}{x^2 - x - 2}$$
I'm thinking I break up the bottom so that it looks like this $$\int \frac{1}{(x-2)(x+1)} $$
Then I do $$x^2 - x -2 = \frac{A}{x-2} + \frac{B}{x+1} $$
Multiple both sides by the common denominator and come out with $$ A(x+1) + B(x-2) = x^2 - x - 2 $$
Which equals $$Ax + A + Bx - 2B = x^2 - x - 2$$
Or $$ (A+B)x + (A-2B) = x^2 - x - 2$$
After that I tried to get values for my A and B but it doesn't seem right since I don't have anything for the $x^2$
Did I mess up somewhere?
|
You need to do $$\frac 1{x^2-x-2}=\frac A{x-2}+\frac B{x+1}$$
Clear fractions (multiply both sides by $x^2-x-2$) to obtain $$1=A(x+1)+B(x-2)$$
You should be able to do it from there. Easy way - set $x=2$, $x=-1$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/502641",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
}
|
Trig. Indefinite Integral $\int\frac{\tan x +\tan ^3 x}{1+\tan^3 x}dx$ $\displaystyle \int\frac{\tan x +\tan ^3 x}{1+\tan^3 x}dx$
$\underline{\bf{My \; Try}}$:: Let $\tan x = t$. Then $\sec^2 xdx = dt\Rightarrow \displaystyle dx = \frac{1}{1+\tan^2 t}dt\Rightarrow dx = \frac{1}{1+t^2}dt$
So $\displaystyle \int\frac{t+t^3}{1+t^3}\cdot \frac{1}{1+t^2}dt = \int\frac{t}{1+t^3}dt$
Now My Question is can we solve the Given Integral without Using Partial fraction Method
If Yes How can I solve
plz Help me , Thanks
|
Addendum:
Evaluating $ \displaystyle{\int\frac{t}{t^3 + 1}\,\mathrm{d}t} $ without partial fractions:
$$
\begin{aligned}
\int\frac{t}{t^3 + 1}\,\mathrm{d}t&=\frac{1}{2}\int\frac{t+1+t-1}{t^3+1}\,\mathrm{d}t\\
&=\frac{1}{2}\int\frac{t+1}{(t+1)(t^2-t+1)}\,\mathrm{d}t-\frac{1}{2}\int\frac{t^2 - t + 1 - t^2}{t^3 + 1}\,\mathrm{d}t\\
&=\frac{1}{2}\int\frac{\mathrm{d}t}{\left(t-1/2\right)^2 + \left(\sqrt{3}/2\right)^2} - \frac{1}{2}\int\frac{\mathrm{d}t}{t+1} + \frac{1}{6}\int\frac{3t^2}{t^3 + 1}\,\mathrm{d}t\\
&=\frac{1}{\sqrt{3}}\arctan\left(\frac{2t-1}{\sqrt{3}}\right)-\frac{1}{2}\log\left|t+1\right| + \frac{1}{6}\log\left|t^3 + 1\right| + C.
\end{aligned}
$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/503647",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
}
|
Find $\lim_{(x,y) \to (0,0)} \frac{\sin{(x^3 + y^5)}}{x^2 + y^4}$. Prove your result. Find $\lim_{(x,y) \to (0,0)} \frac{\sin{(x^3 + y^5)}}{x^2 + y^4}$. Prove your result.
I've attempted to apply the Squeeze Theorem as such:
$\frac{-1}{x^2 + y^4} \leq \frac{\sin{(x^3 + y^5)}}{x^2 + y^4} \leq \frac{1}{x^2 + y^4}$. Clearly, though, the leftmost and rightmost functions of this inequality tend to $\infty$ as $(x,y) \to 0$, so this result does not help.
|
$$
\left|\frac{\sin(x^{2+\color{red}{n}} + y^{4+\color{blue}{m}})}{x^2 + y^4}\right|
\leqslant\left|\frac{x^{2+\color{red}{n}} + y^{4+\color{blue}{m}}}{x^2 + y^4}\right|
\leqslant\frac{|x|^\color{red}{n}\cdot x^2+|y|^\color{blue}{m}\cdot y^4}{x^2+y^4}
\leqslant|x|^\color{red}{n}+|y|^\color{blue}{m}
$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/505391",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
}
|
Polynomial Question Find polynomials $A(x)$ and $B(x)$ such that $A(x)P(x) + B(x)Q(x) = x + 1$ for all $x$ where $P(x) = x^4 - 1$ and $Q(x) = x^3 + x^2$. I'm stumped on this question. I know that I'm supposed to apply the extended version of Euclid's algorithm for polynomials but I'm unsure of how to do that. I thought about trying to create some kind of linear system but guessing arbitrary coefficients but that wouldn't work as $A$ and $B$ don't have fixed degrees.
|
By using Extended Euclidean Algorithm,
$$\begin{align}
x^4-1 =& \left(x^3+x^2\right)(x-1) +x^2-1\\
x^3+x^2 =& \left(x^2-1\right)(x+1) + x+1\\
x^2-1 =& \left(x+1\right)(x-1) + 0
\end{align}$$
And so $x+1$ is the GCD of $P(x)$ and $Q(x)$. Now, we substitute previous remainders to the second-to-last line:
$$\begin{align}
x+1 =& \left(x^3+x^2\right) - \left(x^2-1\right)(x+1)\\
=& \left(x^3+x^2\right) - \left[\left(x^4-1\right) - \left(x^3+x^2\right)(x-1)\right](x+1)\\
=& \left(x^3+x^2\right)\left[1+(x-1)(x+1)\right] - \left(x^4-1\right)(x+1)\\
=& \left(-x-1\right)\left(x^4-1\right) + x^2\left(x^3+x^2\right)\\
\end{align}$$
Therefore, $A(x) = -x-1$ and $B(x)=x^2$ is a pair of solution.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/509402",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
}
|
How prove this $\frac{1}{2a^2-6a+9}+\frac{1}{2b^2-6b+9}+\frac{1}{2c^2-6c+9}\le\frac{3} {5}\cdots (1)$ let $a,b,c$ are real numbers,and such $a+b+c=3$,show that
$$\dfrac{1}{2a^2-6a+9}+\dfrac{1}{2b^2-6b+9}+\dfrac{1}{2c^2-6c+9}\le\dfrac{3}
{5}\cdots (1)$$
I find sometimes,and I find this same problem:
let $a,b,c$ are real numbers,and such $a+b+c=3$,show that
$$ \frac{1}{5a^2-4a+11}+\frac{1}{5b^2-4b+11}+\frac{1}{5c^2-4c+11}\leq\frac{1}{4} $$
and this problem have some methods,you can see:http://www.artofproblemsolving.com/Forum/viewtopic.php?f=52&t=223910&start=20
and I like this can_hang2007 methods and Honey_S methods,But for $(1)$ I can't prove it.Thank you
|
A comment to complete the other proof.
We try to show that it is enough to consider the inequality for only positive $a,b,c$. Assume $a\leq b\leq c$.
$$
\sum_{cyc}\dfrac{1}{2a^2-6a+9}=
\sum_{cyc}\dfrac{1}{(b+c)^2+a^2}=
\sum_{cyc}\dfrac{1}{9-2ab-2ac}
$$
Now it can be seen that if $a\leq 0$ and $b\leq 0$, one can choose $(-a,-b,c)$ instead and decrease the denominator and hence increase the whole sum (similar for the case where $a<0$).
In other words, for each triple $(a,b,c)$ for which at least one of $a,b,c$ is negative, one can find another triple with whole positive elements such that the sum is increased.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/509580",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 5,
"answer_id": 3
}
|
Partial Fractions-Hows I seem to have serious problem understanding entry points of Partial fractions. I would like to decompose the following:
$$\dfrac{x^4-8}{x^2+2x}$$.
My workings. Please help me judge if I am getting the concept or completely lost:
I first simply the denominator ${x^2+2x}$ to become $x(x+2)$.
Using $x(x+2)$ as LCD I create my partial fractions in form:
$$\dfrac{x^4-8}{x^2+2x}=\dfrac{A}{x}+\dfrac{B}{x+2}$$
Multiply both sides by LCD, I get:
$$x^4-8=Ax+2A+Bx$$
Collecting like terms: $x^4-8 =Ax+Bx+2A$.
Simplified with coeff. to: $$(x^3)x-8=(A+B)x+2A$$
When I match with coefficients, I get $$
A+B = x^3$$ and $$2A=-8 \therefore A=-4 $$
substituting to get value of B: $B=(x^3-(-4))$ i.e. $B=(x^3+4)$
My Solution:
$$\dfrac{x^4-8}{x^2+2x}=\dfrac{-4}{x}+\dfrac{x^3+4}{x+2}$$
|
You missed the first step: long division on polynomials to make the numerator of smaller degree than the denominator.
After long division, you should get $$x^2-2x+4+\frac{-8x-8}{x^2+2x}$$ and then you proceed as in your initial attempt, i.e. $$\frac{-8x-8}{x^2+2x}=\frac{A}{x}+\frac{B}{x+2}$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/511072",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
}
|
Prove that $1^2 3^2 5^2 \cdots (p-4)^2 (p-2)^2 \equiv (-1)^{(p+1)/2} \pmod p$. Let $p$ be an odd prime number. Prove that $$1^2 3^2 5^2 \cdots (p-4)^2 (p-2)^2 \equiv (-1)^{(p+1)/2} \pmod p.$$ I know I can use Wilson's Theorem somehow. It would make sense if I could show that all the factors $(p-2k)^2$ is congrunt to $-1 \pmod p$ because then the product would result in congruent to $(-1)^{(p+1)/2}\pmod p$. But I'm not quite sure how to go about that. Can you help?
EDIT: Okay, I think I have it now. So, using that $1^2 \equiv (-1) \cdot 1 \cdot (p-1)$, $3^2\equiv (-1) \cdot 3 \cdot (p-3)$ etc., we get $$1^2 3^2 5^2 \cdots (p-4)^2 (p-2)^2 \equiv (-1) \cdot 1 \cdot (p-1) \cdot (-1) \cdot 3 \cdot (p-3) \cdots (-1) \cdot (p-4) \cdot (4) \cdot (-1) \cdot (p-2) \cdot 2 \pmod p \Leftrightarrow \\ 1^2 3^2 5^2 \cdots (p-4)^2 (p-2)^2 \equiv (-1)^{(p-1)/2}(p-1)! \equiv (-1)^{(p+1)/2} \pmod p.$$ Correct?
|
I will give you an answer that does not use the 'trick' $a^2\equiv(-1)\cdot a\cdot(p-a)$, which was given in a comment above. (Ok, this is less beautiful but, honestly, this was the first thing I thought and I think I would not have found the 'trick'.)
I will use the fact that $2\cdot4\cdot6\cdots2n=2^n\cdot n!$, hoping to prove the congruency by rewriting $1^2\cdot3^2\cdots(p-2)^2$ as the multiplicative inverse of $2^2\cdot4^2\cdots(p-1)^2$. (Note that Wilson's theorem says this is indeed its inverse.)
It remains to compute $2^2\cdot4^2\cdots(p-1)^2\pmod p$, which is $2^{p-1}\left[\left(\frac{p-1}2\right)!\right]^2\equiv\left[\left(\frac{p-1}2\right)!\right]^2\pmod p$, using the fact $2\cdot4\cdot6\cdots2n=2^n\cdot n!$.
The following identity (sometimes called a generalisation of Wilson's theorem) can easily be deduced from Wilson's theorem:
$$(k-1)!\cdot(p-k)!\equiv(-1)^k\pmod p,$$
if $k\leq p$ and $p$ is odd. (To see how, write $(p-k)!=(p-k)(p-(k+1))(p-(k+2))\cdots(p-(p-1))\equiv(-1)^{p-k}\cdot k\cdot(k+1)\cdots(p-1)$. Also note that the case $k=1$ is the original theorem.)
Plugging in $k=\frac{p+1}2$ gives us
$$\left[\left(\frac{p-1}2\right)!\right]^2\equiv(-1)^{\frac{p+1}2}\pmod p,$$ which is exactly what we were looking for.
Our final answer is the multiplicative inverse of $(-1)^{\frac{p+1}2}$ which is clearly $(-1)^{\frac{p+1}2}$ since $(-1)^{\frac{p+1}2}\cdot(-1)^{\frac{p+1}2}\equiv1\pmod p$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/511170",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 1,
"answer_id": 0
}
|
$n!+k$ is never any power of any prime number if $n\ge 6$ and $2\le k\le n$? Question : Is the following true?
"If $n!+k$ is a power of a prime number, then it is one of $2!+2, 3!+2, 3!+3, 4!+3, 5!+5$ where $n,k\in\mathbb Z$ satisfy $n\ge 2$ and $2\le k\le n$."
Motivation : The following is well known :
1. A sequence $(n+1)!+k\ (k=2,3,\cdots,n+1)$ does not have any prime number for any $n\in\mathbb N$.
I've just got the following :
2. A sequence $\{(n+1)!+(n+1)\}!+(n+1)!+k\ (k=2,3,\cdots,n+1)$ does not have any power of any prime number for any $n\in\mathbb N$.
After thinking about these sequences, I reached the above expectation. However, I can neither prove this expectation is true nor find any counterexample. Can you help?
|
Suppose that
$$n!+k=p^r\ \ (r\ge 2).$$
Since it is obvious that $k$ is a power of $p$, let $k=p^s\ (1\le s\lt r).$ However, if $s\gt 1,$ then $n\ge k=p^s\gt p$ leads that $p^{s+1}$ is a divisor of $n!$. Noting that $s+1\le r$ and that $p^{s+1}$ is a divisor of $p^r$, this leads that $p^{s+1}$ is a divisor of $p^r-n!=p^s$, which is a contradiction. Hence, we get $s=1, k=p\le n$. By the same argument as above, we get $n\lt 2p$.
By the way, the number of a prime factor $2$ included in $n!\ (n\ge 2)$ is $\sum_{i=1}^{\infty}\lfloor{\frac{n}{2^i}}\rfloor$ where $\lfloor x\rfloor$ is the largest integer not greater than $x$, so we get
$$\begin{align}\sum_{i=1}^{\infty}\lfloor\frac{n}{2^i}\rfloor\ge\lfloor\frac n2\rfloor+\lfloor\frac n4\rfloor\ge\left(\frac n2-\frac 12\right)+\left(\frac n4-\frac 34\right)=\frac{3n-5}{4}\qquad(1)\end{align}$$
On the other hand, letting $r-1=2^mu$ where $m$ is non-negative integer and $u$ is odd, we get
$$p^{r-1}-1=(p^{2^m})^u-1=(p^{2^m}-1)\{(p^{2^m})^{u-1}+\cdots+(p^{2^m})^2+p^{2^m}+1\}.$$
Since $\{\ \}$ is odd, we get $e(m)=\sum_{i=1}^{\infty}\lfloor{\frac{n}{2^i}}\rfloor$ where $e(m)$ represents the number of a prime factor $2$ included in $p^{2^m}-1$. Since $p^{2^m}-1=(p^{2^{m-1}}+1)(p^{2^{m-1}}-1)$, we know that $e(m)=1+e(m-1)$. (This is because $p^{2^s}+1\not\equiv 0$ (mod $4$) for any $s\in\mathbb N$.) This leads
$$\begin{align}e(m)\le e(1)+m-1\ \ (m\ge 0)\qquad(2)\end{align}$$
Let's consider $e(1)$. Letting $p=2l+1\ (l\in\mathbb N)$, since $p^2-1=(p+1)(p-1)=(2l+2)(2l)$, we get $2^{e(1)}=\frac{2^2(l+1)l}{q}$. Since $l\le q$, we get $2^{e(1)}\le 2^2(l+1)=2(p+1)\le 2(n+1)$, which leads
$$\begin{align}e(1)\le \log_2(n+1)+1\qquad(3)\end{align}$$
Let's consider $m$. Since $u$ is odd, noting $r-1=2^mu,$ we get $2^m\le r-1\iff m\le \log_2(r-1).$ Since $p^r\lt p+2p^n\le 3p^n\le p^{n+1}$, we get $r\le n$. Here I used $n!\le 2\left(\frac n2\right)^n\lt 2p^n$ since $n\lt 2p$. Hence we get
$$\begin{align}m\le \log_2(n-1)\qquad(4)\end{align}$$
From $(1)(2)(3)(4)$, we get
$$\frac{3n-5}{4}\le \log_2(n^2-1).$$
To make calculations easier, let's consider $\frac{3n-5}{4}\lt \log_2n^2$. Then, we know that $n$ must satisfy the following condition :
$$3n-5\lt 8\log_2n,\ n\ge 3.$$
Observing the function $f(x)=8\log_2x-3x+5=8\frac{\log_ex}{\log_e2}-3x+5\ (x\ge 3),$ we get $3\le n\le 10.$ Observing every $n$, we know that $2!+2, 2!+2, 3!+3,4!+3,5!+5$ are the only such examples. Now the proof is completed.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/511958",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
}
|
Use mathematical induction to prove that 9 divides $n^3 + (n + 1)^3 + (n + 2)^3$; Looking for explanation, I already have the solution. I have the solution for this but I get lost at the end, here's what I have so far.
basis $n = 0$; $9 \mid 0^3 + (0 + 1)^3 + (0 + 2)^2 ?$
$9 \mid 1 + 8$ = true
Induction: Assume $n^3 + (n + 1)^3 + (n + 2)^3 = k * 9$ // Why set it equal to $k * 9$? I know it works but why not just make the assumption => $n^3 + (n + 1)^3 + (n + 2)^3$ for some $n = k \ge 0$
Then, //and here's where I get lost
$(n + 1)^3 + (n + 2) + (n + 3)^3 = k * 9 + [(n + 3)^3 - n^3] = 9 (k + n^2 + 3n + 3)$
.
I've done similar examples but none like this. What am I not seeing?
|
You assume $\;9\mid\left[n^3+(n+1)^3+(n+2)^3\right]\;$ , and now you want to prove that also
$$9\mid\left[(n+1)^3+(n+2)^3+(n+3)^3\right]\;\;,\;\;\text{but}$$
$$(n+1)^3+(n+2)^3+(n+3)^3=3n^3+18n^2+42n+36=$$
$$=\left(3n^3+9n^2+15n+9\right)+9n^2+27n+27$$
and you can see the left part between the parentheses is divisible by nine by the inductive hypothesis, whereas the second part is obviously divisible by nine, too...
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/512072",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
}
|
About the area of integer-edge-length triangles Let $a,b,c$ be three edge lengths of a triangle whose area is $S$.
Then, here is my question.
Question : Supposing that $a,b,c$ are natural numbers, then does there exists $(a,b,c)$ such that $S=6k$ for any $k\in\mathbb N$?
Motivation : I've known the following fact:
Fact : If $a,b,c,S$ are natural numbers, then $S$ is a multiple of $6$.
Proof : By Heron's formula, we get
$$\begin{align}16S^2=(a+b+c)(-a+b+c)(a-b+c)(a+b-c).\qquad(\star)\end{align}$$
Hence, we know that in mod $2$
$$a+b+c\equiv -a+b+c\equiv a-b+c\equiv a+b-c.$$
If all of these are odd, we reach a contradiction. So we know that all of these are even. Here, letting
$$x=\frac{-a+b+c}{2}, y=\frac{a-b+c}{2}, z=\frac{a+b-c}{2},$$
then, getting $\frac{a+b+c}{2}=x+y+z,$ we know
$$(\star)\iff S^2=(x+y+z)xyz.$$
By considering in mod $3$, we know that $S^2$ can be divided by $3$, which means that $S$ can be divided by $3$.
By considering in mod $4$, we know that $S^2$ can be divided by $4$, which means that $S$ can be divided by $2$.
Now we know that $S$ can be divided by $6$ as desired. Hence, the proof is now completed.
This got me interested in the above question, but I'm facing difficulty. Can anyone help?
|
Theorem: There is no integer-sided triangle whose area is 18 units.
Proof: Assume such a triangle exists. Then there are integers $x, y, z$ such that $18^2 = xyz(x + y + z)$. Assume without loss of generality that $x \ge y \ge z$. Since $y \ge 1$ and $z \ge 1$, we have $x(x + 2) \le 18^2$ which certainly forces $x < 18$. This leaves us with only 17 possibilities for $x$ and hence $17 \times 18 / 2 = 153$ possibilities for the pair $(x, y)$; for each of these there is at most one $z$ which will work, and a moment's computer calculation shows that none of the 153 cases leads to a solution.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/513379",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
}
|
Critique of Solution "Find all integers $n$ such that the quadratic $7x^2 + nx - 11$ can be expressed as the product of two linear factors with integer coefficients.
In a quadratic in the form of $ax^2+bx+c$, the product of the roots of the quadratic equal the constant ($c$). Since there can only be integer coefficients, the only factors of $c$ ($11$) are $-1$ and $11$, and $1$ and $-11$. So we can put these in.
$(7x+1)(x-11)=$
$7x^2+x-77x-11=$
$7x^2-76x-11$
$N$ can be $-76$.
$(7x-1)(x+11)=$
$7x^2-x+77x-11=$
$7x^2+76x-11$
$N$ can be $76$. There is also another change you can make with the coefficient of $x^2$. You can move it over to the other side, so now we have....
$(x+1)(7x-11)=$
$7x^2+7x-11x-11=$
$7x^2-4x-11$
$N$ can be $-4$.
$(x-1)(7x+11)=$
$7x^2-7x+11x-11=$
$7x^2+4x-11.$
$N$ can be $4$.
So now we have these values of $n$: $-76$, $76$, $-4$, $4$.
Basically I am looking for a critique of my solution if it is correct.
|
I think it is correct, but it would be easier to use that the discriminant must be a square number, i.e. $n^2+308$ must be a square number. This has the solutions $n=\pm 4, \pm 76$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/513763",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
}
|
Functional equation $f(a+b)^3-f(a)^3-f(b)^3=3f(a)f(b)f(a+b)$ Find all of the functions defined on the set of integers and receiving the integers value, satisfying the condition:
$$f(a+b)^3-f(a)^3-f(b)^3=3f(a)f(b)f(a+b)$$
for each pair of integers $(a,b)$.
|
Not cleaned. I just wrote it as I thought about it.
$A^3+B^3+C^3-3ABC=(A+B+C)(A+wB+w^2C)(A+w^2B+wC)$. So in our case
$$\left(f(a+b)-f(a)-f(b)\right)\left(f(a+b)-wf(a)-w^2f(b)\right)\left(f(a+b)-w^2f(a)-wf(b)\right)=0.$$
If for some $a,b$ we have $f(a+b)-wf(a)-w^2f(b)=0$ then $f(a+b)-wf(a)-w^2f(b)$ is divisible by $w^2+w+1$. Using long division we get $f(a+b)-wf(a)-w^2f(b)=(w^2+w+1)(-f(b))+(f(b)-f(a))w+f(a+b)-f(b)$. From where $f(b)=f(a)$ and $f(a+b)=-f(b)$.
So we are getting that, for each $a,b$, either:
$$f(a+b)=f(a)+f(b)$$
or $$f(a)=f(b)=-f(a+b).$$
We clearly get that $f(0)=0$.
If for some $x$, $f(x)=f(-x)$ then we get $f(x)=f(-x)=0$, otherwise we get $f(x)=-f(-x)$. If for some $x$, $f(x)=0$, then $f(nx)=0$.
If $f(1)=0$, then $f(a+1)=f(a)$. So, $f$ is constant equal to zero.
If $f(1)=r\neq0$ then $$f(a+1)=f(a)+r$$. Therefore $f(x)=rx$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/517749",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
}
|
Calculate the integral $\int_{0}^{2\pi}\frac{1}{a^{2}\cos^2t+b^{2}\sin^{2}t}dt$, by deformation theorem. I want to prove:
$$\int_{0}^{2\pi}\frac{1}{a^{2}\cos^2t+b^{2}\sin^{2}t}dt=\frac{2\pi}{ab}$$
by the deformation theorem of complex variable.
Then I consider a parameterization $\gamma:[0,2\pi]\rightarrow A$, traveled in the opposite direction of the clock hand, of the ellipse (in $\mathbb{C}$):
$$\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$$
I thought of searching for a function $f:A\rightarrow \mathbb{C}$ (analytic in $A$) and a curve $\lambda:[0,2\pi]\rightarrow A$ (homotopic to $\gamma$) and use the deformation theorem:
$$\int_{\gamma}f = \int_{\lambda}f$$
But I can not find a function $f(z)$ that meets what I need, can you help me to find the f?
P.D: Necessarily I have to use the theorem of deformation
|
Let the ellipse $\Gamma$
$$ \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$$
be parameterized by
$$ z = a\cos\theta + ib\sin\theta$$
with $\theta\in[0,2\pi].$
Consider the integral
$$ \int_\Gamma \frac{1}{z} dz.$$
This is equal to
$$ \int_0^{2\pi} \frac{1}{a\cos\theta + ib\sin\theta}
(-a\sin\theta + ib\cos\theta) \; d\theta\\ =
\int_0^{2\pi}
\frac{a\cos\theta-ib\sin\theta}{a^2\cos^2\theta + b^2\sin^2\theta}
(-a\sin\theta + ib\cos\theta) \; d\theta.$$
which is
$$\int_0^{2\pi}
\frac{(b^2-a^2)\sin\theta\cos\theta + iab(\cos^2\theta+\sin^2\theta)}
{a^2\cos^2\theta + b^2\sin^2\theta} \; d\theta
\\=\int_0^{2\pi}
\frac{(b^2-a^2)\sin\theta\cos\theta + iab}
{a^2\cos^2\theta + b^2\sin^2\theta} \; d\theta.$$
This implies that
$$ \frac{1}{ab} \Im \int_\Gamma \frac{1}{z} dz =
\int_0^{2\pi} \frac{1} {a^2\cos^2\theta + b^2\sin^2\theta} \; d\theta.$$
Now apply the deformation theorem, turning the ellipse into a circle
round the origin of radius one, getting
$$ \frac{1}{ab} \Im \int_\Gamma \frac{1}{z} dz =
\frac{1}{ab} \Im \int_{|z|=1} \frac{1}{z} dz =
\frac{1}{ab} \Im (2\pi i) = \frac{2\pi}{ab}.$$
Verification. We can verify this result using the Cauchy residue theorem directly. Put $z=e^{i\theta}$ so that $dz = i e^{i\theta} \; d\theta$ and $d\theta = 1/(iz) \; dz$ to get
$$\int_0^{2\pi} \frac{1} {a^2\cos^2\theta + b^2\sin^2\theta} \; d\theta
= \int_{|z|=1} \frac{4}{a^2(z+1/z)^2 - b^2(z-1/z)^2} \frac{1}{iz} \; dz.$$
This gives
$$\frac{1}{i}\int_{|z|=1} \frac{4z}{a^2(z^2+1)^2 - b^2(z^2-1)^2} \; dz$$
This has the following four simple poles:
$$\rho_{0,1} = \pm i\sqrt{\frac{a-b}{a+b}}
\quad\text{and}\quad
\rho_{2,3} = \pm i\sqrt{\frac{a+b}{a-b}}.$$
Now suppose that $a>b>0$ (the other cases are treated similarly). This leaves only $\rho_{0,1}$ inside the unit circle.
The residues are easy to calculate:
$$\mathrm{Res}\left(\frac{4z}{a^2(z^2+1)^2 - b^2(z^2-1)^2}; z=\rho_{0,1} \right)
= \left.\frac{4z}{2a^2(z^2+1)2z - 2b^2(z^2-1)2z}\right|_{z=\rho_{0,1}}
\\ = \left.\frac{2}{2a^2(z^2+1) - 2b^2(z^2-1)}\right|_{z=\rho_{0,1}}
= \left.\frac{1}{(a^2-b^2)z^2 + (a^2+b^2)}\right|_{z=\rho_{0,1}}.$$
Note that $$\rho_{0,1}^2 = - \frac{a-b}{a+b}$$ giving for the residues
$$ \frac{1}{-(a^2-b^2)(a-b)/(a+b) + (a^2+b^2)}
= \frac{1}{-(a-b)^2 + (a^2+b^2)} = \frac{1}{2ab}.$$
Therefore the value of the integral is
$$2\pi i \times \frac{1}{i} \times
\left(\frac{1}{2ab} + \frac{1}{2ab}\right)= \frac{2\pi}{ab},$$
QED.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/518173",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "12",
"answer_count": 2,
"answer_id": 1
}
|
$z$-transform of $1/n$ How can one calculate the $z$-transform of:
$x(n) = \frac{1}{n}$ , where $n \geq 1$? I have searched for table entries, then got stuck while trying to do it with the definition of $z$-transform (summation).
|
Good Question
$$\frac{u(n-1)}{n} \rightleftharpoons ??$$
We know that Z transform is defined as :
$$X(z) = \sum_{n=-\infty}^{\infty}x(n)z^{-n}$$
$$u(n) \rightleftharpoons \frac{z}{z-1}$$
By the use of time shifting property
$$x(n-1) \rightleftharpoons z^{-1} X(z)$$
$$u(n-1) \rightleftharpoons \frac{1}{z-1}$$
Differentiate the General formula of X(z) with respect to z
$$\frac{dX(z)}{dz} = -\sum_{n=-\infty}^{\infty}nx(n)z^{-n-1}$$
$$\frac{zdX(z)}{dz} = -\sum_{n=-\infty}^{\infty}nx(n)z^{-n}$$
$$nx(n) \rightleftharpoons -\frac{zdX(z)}{dz}$$
Now we want to find out the Z transform of $$\frac{u(n-1)}{n}$$
$$x(n) = \frac{u(n-1)}{n} \rightleftharpoons X(z)$$
$$n x(n) \rightleftharpoons -\frac{zdX(z)}{dz}$$
$$n [\frac{u(n-1)}{n}] \rightleftharpoons -\frac{zdX(z)}{dz}$$
$$u(n-1) \rightleftharpoons -\frac{zdX(z)}{dz}$$
Means
$$-\frac{zdX(z)}{dz} = \frac{1}{z-1}$$
$$\frac{dX(z)}{dz} = -\frac{1}{z(z-1)}$$
$$\frac{dX(z)}{dz} = [\frac{1}{z}] + [\frac{1}{z-1}]$$
$$X(z) = ln(\frac{z}{z-1})$$
Hence
$$\frac{u(n-1)}{n} \rightleftharpoons ln(\frac{z}{z-1})$$
Now we have some interesting series for natural log
$$\frac{1}{z} + \frac{1}{2z^2} + \frac{1}{3z^3} + ...... = ln(\frac{z}{z-1})$$
For z = 2
$$\frac{1}{2} + \frac{1}{2\times2^2} + \frac{1}{3\times2^3} + ...... = ln(2)$$
For $$z = i^2 = -1$$
$$\frac{1}{1} - \frac{1}{2} + \frac{1}{3} + ...... = ln(2)$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/518677",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
}
|
finding range of function of three variables Three real numbers $x$, $y$, $z$ satisfy the following conditions.
$x^{2}+y^{2}+z^{2}=1~$, $~y+z=1$
Find the range of $~x^{3}+y^{3}+z^{3}~$ without calculus.
I solved this problem only with Lagrange-Multiplier and wonder if there exist other methods.
|
to find max value, we only consider $f(u)=1-3u^2+2\sqrt{2}u^3, u=\sqrt{y(1-y)}, 0 \le u \le \dfrac{1}{2}$.we try first to see if it's increasing or decreasing function. $f(0)=1, f(\dfrac{1}{2})=1-\dfrac{3-\sqrt{2}}{4} < 1$, so we guess it is a decreasing function. to prove this, let $ u_1>u_2$
$f(u_1)-f(u_2)=2\sqrt{2}u_1^3-2\sqrt{2}u_2^3-3u_1^2+3u_2^2=(u_1-u_2)(2\sqrt{2}(u_1^2+u_2^2+u_1u_2)-3(u_1+u_2))$
since $u_1-u_2>0$, it remains $(2\sqrt{2}(u_1^2+u_2^2+u_1u_2)-3(u_1+u_2)) <0 \iff 2\sqrt{2}(u_1^2+u_2^2+u_1u_2) < 3(u_1+u_2) \iff 2\sqrt{2}(u_1^2+u_2^2+\dfrac{u_1^2+u_2^2}{2}) <3(u_1+u_2) \iff (u_1-\sqrt{2}u_1^2)+(u_2-\sqrt{2}u_2^2) >0 \iff u_1(1-\sqrt{2}u_1)>0 \iff (u_1\le \dfrac{1}{2}) and ( 1-\sqrt{2}u_1>0) \implies f(u_1)< f(u_2)$
so the max is $1$ when $u=0 \implies x=y=0$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/519218",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
}
|
Proof by induction that $\sum_{i=1}^n \frac{i}{(i+1)!}=1- \frac{1}{(n+1)!}$ Prove via induction that $\sum_{i=1}^n \frac{i}{(i+1)!}=1- \frac{1}{(n+1)!}$
Having a very difficult time with this proof, have done pages of work but I keep ending up with 1/(k+2). Not sure when to apply the induction hypothesis and how to get the result $1- \frac{1}{(n+2)!}$. Please help!
thanks guys, youre the greatest!
|
You can also prove this by power series manipulation (generating functions). Note first that changing the sum by starting at $i=0$ instead of $i=1$ doesn't change its value. Compute as follows:
\begin{align}\sum_{n=0}^\infty \sum_{i=1}^n \frac{i}{(i+1)!}x^{n+1} &= \sum_{n=0}^\infty \sum_{i=0}^n \frac{i}{(i+1)!}x^{n+1}
\\\\&= \sum_{i=0}^\infty \sum_{n=i}^\infty \frac{i}{(i+1)!}x^{n+1}
\\\\ &= \sum_{i=0}^\infty \frac{i}{(i+1)!} \left( \sum_{n=-1}^\infty x^{n+1} - \sum_{n=-1}^{i-1} x^{n+1}\right)
\\\\&= \sum_{i=0}^\infty \frac{i}{(i+1)!} \left(\frac{1}{1-x}-\frac{1-x^{i+1}}{1-x} \right)
\\\\&= \sum_{i=0}^\infty \frac{i}{(i+1)!}\frac{x^{i+1}}{1-x}
\\\\&= \frac{x^2}{1-x} \sum_{i=0}^\infty \frac{ix^{i-1}}{(i+1)!}
\\\\&= \frac{x^2}{1-x} \frac{d}{dx} \sum_{i=0}^\infty \frac{x^i}{(i+1)!}
\\\\&= \frac{x^2}{1-x} \frac{d}{dx} \frac{e^x-1}{x}
\\\\&= \frac{x^2}{1-x} \frac{xe^x-(e^x-1)}{x^2}
\\\\&=\frac{1}{1-x}-e^x
\\\\&=\sum_{n=0}^\infty x^n - \sum_{n=0}^\infty \frac{x^n}{n!}
\\\\&= \sum_{n=0}^\infty \left(1-\frac{1}{n!}\right)x^n
\\\\&= \sum_{n=1}^\infty \left(1-\frac{1}{n!}\right)x^n
\\\\&= \sum_{n=0}^\infty \left(1-\frac{1}{(n+1)!}\right)x^{n+1} \end{align}
Coefficients of like powers of $x$ in two equal power series must be equal, proving that $$\sum_{i=1}^n \frac{i}{(i+1)!} = 1-\frac{1}{(n+1)!},$$ as desired.
This can all be justified either as formal power series manipulation or as calculations with absolutely convergent series for $|x| < 1.$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/521861",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 4,
"answer_id": 3
}
|
How can I prove that one of $n$, $n+2$, and $n+4$ must be divisible by three, for any $n\in\mathbb{N}$ Intuitively it's true, but I just can't think of how to say it "properly".
Take for example, my answer to the following question:
Let $p$ denote an odd prime. It is conjectured that there are infinitely many twin primes $p$, $p+2$. Prove that the only prime triple $p$, $p+2$, $p+4$ is the triple $3,\ 5,\ 7$.
And my solution:
Given an odd integer $n$, between the three integers $n$, $n+2$ and $n+4$, one of them must be divisible by $3$... Three possible cases are $n=3k$, $n+2=3k$, and $n+4=3k$. The only such possible $k$ that makes $n$ prime is $k=1$. In this case, given an odd prime $p$, either $p=3$, $p+2=3$, or $p+4=3$. This would imply that $p=3$, $p=1$, or $p=-1$. The only of these three that is prime is $p=3$, therefore the only three evenly distributed primes are $3$, $5$, and $7$.
Is there a "better" way that I can assert that one of the integers is divisible by 3? This feels too weak.
|
A somewhat silly answer:
You probably know that $\displaystyle \sum_{i=1}^n i = \frac{n(n+1)}{2}$ and $\displaystyle \sum_{i=1}^n i^2 = \frac{n(n+1)(2n+1)}{6}$; if not these can be proven easily via induction. Consider then
\begin{align*}
\sum_{i=1}^n (i^2 + 3i + 1) &= \frac{n(n+1)(2n+1)}{6} + 3\frac{n(n+1)}{2} + n \\
&= \frac{(2n^3 + 3n^2 + n) + 9(n^2+n) + 6n}{6} \\
&= \frac{2n^3 +12 n^2 + 16 n}{6} \\
&= \frac{n^3 + 6n^2 + 8n}{3} \\
&= \frac{n(n+2)(n+4)}{3}
\end{align*}
On the left hand side, you have a sum of integers, so the right hand side is an integer. Hence $3 | n(n+2)(n+4)$, and since $3$ is prime it divides one of the three factors.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/522585",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "16",
"answer_count": 14,
"answer_id": 1
}
|
Residue at infinity of $f(z)=z^3\cos\big(\frac{1}{z-2}\big)$ I have trouble with the residue of:
$f(z)=z^3\cos\left(\frac{1}{z-2}\right)$ at $z = \infty$.
I tried to solve it at $z=0$ but it turns out that I was wrong while $z=0$ is not a pole.
I must solve it at $z=2$ but I'm stuck.
Any suggestion will be much appreciated.
|
The residue at infinity of a function $f$ holomorphic in a punctured neighbourhood of $\infty$ is the residue in $0$ of the function
$$g(z) = -\frac{1}{z^2}f\left(\frac1z\right).$$
For $f(z) = z^3\cos \frac{1}{z-2}$, that becomes the residue in $0$ of
$$-\frac{1}{z^5}\cos \frac{z}{1-2z}.$$
Expanding $\frac{z}{1-2z}$ into a geometric series and inserting the result into the Taylor series of $\cos$ yields
$$\begin{align}
\frac{z}{1-2z} &= z + 2z^2 + 4z^3 + O(z^4)\\
\cos (z + 2z^2 + 4z^3 + O(z^4)) &= 1 - \frac12(z+2z^2+4z^3)^2 + \frac{1}{24}z^4 + O(z^5)\\
&= 1 - \frac12 z^2 - 2z^3 - \frac{143}{24}z^4 + O(z^5),
\end{align}$$
so the residue is $\frac{143}{24}$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/522728",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
}
|
How to prove this inequality $\left|x\sin{\frac{1}{x}}-y\sin{\frac{1}{y}}\right|<2\sqrt{|x-y|}$? For any real numbers $x,y\neq 0$,show that
$$\left|x\sin{\dfrac{1}{x}}-y\sin{\dfrac{1}{y}}\right|<2\sqrt{|x-y|}$$
I found this problem when I dealt with this problem. But I can't prove it. Maybe the constant $2$ on the right hand side can be replaced by the better constant $\sqrt{2}$?
Thank you.
|
We'll first assume that $0 \le x < y$.
I need three kinds of estimates here.
*
*$\left| x \sin \frac{1}{x} - y \sin \frac{1}{y} \right| \le x + y$
*$\left| x \sin \frac{1}{x} - y \sin \frac{1}{y} \right| \le 2$
*$\left| x \sin \frac{1}{x} - y \sin \frac{1}{y} \right| \le \frac{y}{x} - 1$
Estimates 1 and 2 are trivial: $\left| x \sin \frac{1}{x} \right| \le \min(x, 1)$
Estimate 3 is proved as follows.
*
*First we show that $\left| (x \sin \frac{1}{x})^\prime \right| \le \frac{1}{x}$. Indeed, $(x \sin \frac{1}{x})^{\prime\prime} = - \frac{1}{x^3} \sin \frac{1}{x}$, so local maxima and minima of $(x \sin \frac{1}{x})^\prime$ are located at $\frac{1}{\pi n}, n \in \mathbb{N}$, and its values there are $\frac{(-1)^n}{\pi n}$, so that $\sup_{z \ge x} \left| (z \sin \frac{1}{z})^\prime \right| \le \frac{1}{\pi n} \le \frac{1}{x}$, where $\frac{1}{\pi n}$ is the smallest one in $[x,+\infty)$.
*Now $\left| x \sin \frac{1}{x} - y \sin \frac{1}{y} \right| \le \intop_x^y \frac{1}{z} dz = \log \frac{y}{x} \le \frac{y}{x} - 1$
Now let's find the three regions corresponding to:
*
*$x + y \le C \sqrt{y - x}$
*$2 \le C \sqrt{y - x}$
*$\frac{y}{x} - 1 \le C \sqrt{y - x}$
and choose the constant $C$ in such a way that these regions cover $\{0 < x < y\}$. Clearly, 2 is bounded by a line, 1 and 3 are bounded by parabolas. Miraculously, $C = 2$ is the unique value when the three boundaries intersect at a single point, namely $(x,y) = (1/2, 3/2)$...
Anyway, let's rewrite our regions for $C = 2$:
*
*$y-x \ge \frac{1}{4} (x+y)^2$
*$y-x \ge 1$
*$y-x \le 4 x^2$
So to cover the whole space we have to prove $y-x \ge \frac{1}{4} (x+y)^2$ on $\{4 x^2 \le y-x \le 1\}$, for which it is sufficient to consider just the two boundary cases, namely $y-x = 1$ and $y-x = 4 x^2$, since a quadratic polynomial with positive leading term attains maximum on the boundary of a segment. And these cases are easy to verify. Indeed, the inequality in terms of $x$ and $z := y-x$ looks like $x^2 + xz + \frac{1}{4} z^2 \le z$; for $z=1$ it's equivalent to $(x + \frac{1}{2})^2 \le 1$, which is true, since $4 x^2 \le 1$; for $z = 4 x^2$ it's equivalent to $3 x^2 \ge 4 x^3 + 4 x^4$, which follows once again from $4 x^2 \le 1$ (since $x^2 \ge 2 x^3$ and $x^2 \ge 4 x^4$).
Now the case $x < 0 < y$ is even simpler. We only need analogs of estimates 1 and 2:
*
*$\left| x \sin \frac{1}{x} - y \sin \frac{1}{y} \right| \le |x| + |y| \le C \sqrt{|x| + |y|}$ whenever $|x| + |y| \le C^2 = 4$
*$\left| x \sin \frac{1}{x} - y \sin \frac{1}{y} \right| \le 2 \le C \sqrt{|x| + |y|}$ whenever $|x| + |y| \ge (\frac{2}{C})^2 = 1$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/523297",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "17",
"answer_count": 2,
"answer_id": 1
}
|
Prove the following determinant identities without expanding the determinants a)
$$\begin{vmatrix}
\sin^2 x & \cos^2 x & \cos 2x \\
\sin^2 y & \cos^2 y & \cos 2y \\
\sin^2 z & \cos^2 z & \cos 2z \\
\end{vmatrix} = 0;$$
$$\begin{vmatrix}
\sin^2 x & \cos^2 x & \cos^2x-\sin^2x \\
\sin^2 y & \cos^2 y & \cos^2y-\sin^2y \\
\sin^2 z & \cos^2 z & \cos^2z-\sin^2z \\
\end{vmatrix} = 0;$$
$$\begin{vmatrix}
\cos^2 x & \cos^2 x & \cos^2x-\sin^2x \\
\cos^2 y & \cos^2 y & \cos^2y-\sin^2y \\
\cos^2 z & \cos^2 z & \cos^2z-\sin^2z \\
\end{vmatrix} = 0;$$
b)
$$\begin{vmatrix}
1 & a & p+c \\
1 & p & c+a \\
1 & c & a+p \\
\end{vmatrix} = 0,$$
$$\begin{vmatrix}
1 & a & 1\cdot(a+p+c) \\
1 & p & 1\cdot(p+c+a) \\
1 & c & 1\cdot(c+a+p)
\end{vmatrix} =
(a + p + c)\cdot
\begin{vmatrix}
1 & a & 1 \\
1 & p & 1 \\
1 & c & 1
\end{vmatrix} = 0.$$
I am stuck on a), any hints and help is appreciate, and please check if b) is correct.
|
Remember that $\cos 2x = \cos^2 x - \sin^2 x$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/526447",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
}
|
Convert boolean expression into SOP and POS Convert the following expression into SOP (sum of products) and POS (product of sums) canonical forms using boolean algebra method:
$(ac + b)(a + b'c) + ac$
Attempt at solution:
$(ac + b)(a + b'c) + ac$
*
*$(a + b)(c + b)(a + b')(a + c) + ac$
*$...$
*$...$
I'm stuck at this point. Any help would be greatly appreciated. Thanks.
|
One way to get the SoP form starts by multiplying everything out, using the distributive law:
$$\begin{align*}
(ac+b)(a+b'c)+ac&=ac(a+b'c)+b(a+b'c)+ac\\
&=aca+acb'c+ba+bb'c+ac\\
&=ac+ab'c+ab+ac\\
&=ac+ab'c+ab\;.
\end{align*}$$
Then make sure that every term contains each of $a,b$, and $c$ by using the fact that $x+x'=1$:
$$\begin{align*}
ac+ab'c+ab&=ac(b+b')+ab'c+ab(c+c')\\
&=abc+ab'c+ab'c+abc+abc'\\
&=abc+ab'c+abc'\;.
\end{align*}$$
Alternatively, you can make what amounts to a truth table for the expression:
$$\begin{array}{cc}
a&b&c&ac+b&b'c&a+b'c&ac&(ac+b)(a+b'c)+ac\\ \hline
0&0&0&0&0&0&0&0\\
0&0&1&0&1&1&0&0\\
0&1&0&1&0&0&0&0\\
0&1&1&1&0&0&0&0\\
1&0&0&0&0&1&0&0\\
1&0&1&1&0&1&1&1\\
1&1&0&1&1&1&0&1\\
1&1&1&1&0&1&1&1
\end{array}$$
Now find the rows in which the expression evaluates to $1$; here it’s the last three rows. For a product for each of those rows; if $x$ is one of the variables, use $x$ if it appears with a $1$ in that row, and use $x'$ if it appears with a $0$. Thus, the last three rows yield (in order from top to bottom) the terms $ab'c$, $abc'$ and $abc$.
You can use the truth table to get the PoS as well. This time you’ll use the rows in which the expression evaluates to $0$ — in this case the first five rows. Each row will give you a factor $x+y+z$, where $x$ is either $a$ or $a'$, $y$ is either $b$ or $b'$, and $z$ is either $c$ or $c'$. This time we use the variable if it appears in that row with a $0$, and we use its negation if it appears with a $1$. Thus, the first row produces the sum $a+b+c$, the second produces the sum $a+b+c'$, and altogether we get
$$(a+b+c)(a+b+c')(a+b'+c)(a+b'+c')(a'+b+c)\;.\tag{1}$$
An equivalent procedure that does not use the truth table is to begin by using De Morgan’s laws to negate (invert) the original expression:
$$\begin{align*}
\Big((ac+b)(a+b'c)+ac\Big)'&=\Big((ac+b)(a+b'c)\Big)'(ac)'\\
&=\Big((ac+b)'+(a+b'c)'\Big)(a'+c')\\
&=\Big((ac)'b'+a'(b'c)'\Big)(a'+c')\\
&=\Big((a'+c')b'+a'(b+c')\Big)(a'+c')\\
&=(a'b'+b'c'+a'b+a'c')(a'+c')\\
&=a'b'(a'+c')+b'c'(a'+c')+a'b(a'+c')+a'c'(a'+c')\\
&=a'b'+a'b'c'+a'b'c'+b'c'+a'b+a'bc'+a'c'+a'c'\\
&=a'b'+a'b'c'+b'c'+a'b+a'bc'+a'c+a'c'\\
&=a'b'+b'c'+a'b+a'(c+c')\\
&=a'b+b'c'+a'b+a'\\
&=b'c'+a'\;,
\end{align*}$$
where in the last few steps I used the absorption law $x+xy=x$ a few times. Now find the SoP form of this:
$$\begin{align*}
b'c'+a'&=b'c'(a+a')+a'(b+b')(c+c')\\
&=ab'c'+a'b'c'+a'b(c+c')+a'b'(c+c')\\
&=ab'c'+a'b'c'+a'bc+a'bc'+a'b'c+a'b'c'\\
&=ab'c'+a'b'c'+a'bc+a'bc'+a'b'c\;.
\end{align*}$$
Now negate (invert) this last expression, and you’ll have the PoS form of the original expression:
$$\begin{align*}
(ab'c'&+a'b'c'+a'bc+a'bc'+a'b'c)'\\
&=(ab'c')'(a'b'c')'(a'bc)'(a'bc')'(a'b'c)'\\
&=(a'+b+c)(a+b+c)(a+b'+c')(a+b'+c)(a+b+c')\;,
\end{align*}$$
which is of course the same as $(1)$, though the factors appear in a different order.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/526520",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
}
|
Factorise $y^2 -3yz -10z^2$ How do I solve this question? I have looked at the problem several times. However, I cannot find a viable solution. I believe that it is a perfect square trinomial problem.
|
I made up a method with no guesswork at all, Factoring Quadratics: Asterisk Method
and my earlier answer at How to factor the quadratic polynomial $2x^2-5xy-y^2$?
We require that $b^2 - 4ac$ be a perfect square, and (positive) $\delta = \sqrt {b^2 - 4ac},$ otherwise factoring is impossible anyway. Then: we may simply take $$ a_1 = \gcd \left( a,
\frac{(b + \delta)}{2} \right) ; \; \; \; \; \; a_2 = \frac{a}{a_1} $$ without paying attention to any prime factorizations and
$$ a x^2 + b x y + c y^2 = \; \left(a_2x+ \left( \frac{b + \delta}{2a_1} \right) y \right) \; \; \left(a_1x+ \left( \frac{b - \delta}{2a_2} \right) y \right) \; $$ in integers.
You have $a=1, b= -3, c=-10, b^2 - 4 a c = 49, \delta = 7.$ So $a_1 = 1, a_2 = 1.$ Then $(b+\delta)/ (2 a_1) = 4/2 = 2,$ and $(b-\delta)/ (2 a_2) = -10/2 = -5,$ so $x^2 - 3 x y - 10 y^2 = (x + 2 y)(x - 5 y).$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/526684",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 3
}
|
If $x+y+z=\pi/2$, $\sin{x}\sqrt{1-\sin x}+\sin y\sqrt{1-\sin y}+\sin z\sqrt{1-\sin z} \ge 4\sqrt{2} \sin{x}\sin{y}\sin{z}(\sin{x}+\sin{y}+\sin{z})$
Let: $x,y,z >0$ and $ x+y+z=\dfrac{\pi}{2}$ then prove:
$$\sin{x}\sqrt{1-\sin{x}}+\sin{y}\sqrt{1-\sin{y}}+\sin{z}\sqrt{1-\sin{z}}
\ge 4\sqrt{2} \sin{x}\sin{y}\sin{z}(\sin{x}+\sin{y}+\sin{z})$$
I had a solution with Lagrange multipliers from this problem. But it is ugly and I'm thinking of a method without Lagrange multipliers but failed. I hope to see a nice solution.
|
Note that $$\sqrt{1-\sin x} = \sqrt{2}\sin\left(\frac{\pi}{4}-\frac{x}{2}\right) = \sqrt{2}\sin\left(\frac{y+z}{2}\right) \geq \frac{1}{2}\sqrt{2}\left( \sin y + \sin z\right).$$ Therefore $$\sin x \sqrt{1-\sin x}+\sin y \sqrt{1-\sin y} + \sin z \sqrt{1-\sin z} \geq \sqrt{2}\left(\sin x \sin y+ \sin x \sin z + \sin y \sin z\right) \\ = \sqrt{2}\sin x \sin y \sin z\left(\frac{1}{\sin x} + \frac{1}{\sin y} + \frac{1}{\sin z} \right) \\ \geq \sqrt{2}\sin x \sin y \sin z \frac{3}{\sin\frac{\pi}{6}}=6 \sqrt{2}\sin x \sin y \sin z$$ where the last inequality follows from the convexity of $\frac{1}{\sin}$. By concavity of $\sin$ $$\sin x + \sin y + \sin z \leq 3 \sin \frac{\pi}{6} = \frac{3}{2} $$ and your inequality follows.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/526864",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
}
|
Limit in form of 0/0
If I multiply by I get zero/2x anyway.
What manipulation needed to get 2/3?
|
$$
\lim_{x\to 0}\frac{\sqrt[3]{1+x}-\sqrt[3]{1-x}}{x}=\lim_{x\to 0}\frac{\sqrt[3]{1+x}-\sqrt[3]{1-x}}{x}\cdot\frac{\sqrt[3]{(1+x)^2}+\sqrt[3]{1-x^2}+\sqrt[3]{(1-x)^2}}{x(\sqrt[3]{1-x^2}+\sqrt[3]{1-x^2}+\sqrt[3]{(1-x)^2}}=
$$
$$
=\lim_{x\to 0}\frac{1+x-1+x}{x(\sqrt[3]{1-x^2}+\sqrt[3]{(1-x^2}+\sqrt[3]{(1-x)^2}}=\lim_{x\to 0}\frac{2x}{x\left(\sqrt[3]{(1+x)^2}+\sqrt[3]{1-x^2}+\sqrt[3]{(1-x)^2}\right)}
$$
$$
=\lim_{x\to 0}\frac{2}{\sqrt[3]{(1+x)^2}+\sqrt[3]{1-x^2}+\sqrt[3]{(1-x)^2}}=\lim_{x\to 0}\frac{2}{\sqrt[3]{(1+0)}^2+\sqrt[3]{1-0^2}+\sqrt[3]{(1-0)^2}}=\frac{2}{3}
$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/527254",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 2
}
|
Closed form for $\int_0^\infty\frac{\sqrt{x+\sqrt{x^2+1}}}{\sqrt{x\phantom{|}}\sqrt{x^2+1}}e^{-x}dx$ Is it possible to evaluate this integral in a closed form?
$$\int_0^\infty\frac{\sqrt{x+\sqrt{x^2+1}}}{\sqrt{x\phantom{|}}\sqrt{x^2+1}}e^{-x}dx$$
|
Hint:
$\int_0^\infty\dfrac{\sqrt{x+\sqrt{x^2+1}}}{\sqrt{x}\sqrt{x^2+1}}e^{-x}~dx$
$=\int_0^\infty\dfrac{\sqrt{\sinh x+\sqrt{\sinh^2x+1}}}{\sqrt{\sinh x}\sqrt{\sinh^2x+1}}e^{-\sinh x}~d(\sinh x)$
$=\int_0^\infty\dfrac{\sqrt{\sinh x+\cosh x}}{\sqrt{\sinh x}\cosh x}e^{-\sinh x}\cosh x~dx$
$=\int_0^\infty\dfrac{e^{-\sinh x}\sqrt{e^x}}{\sqrt{\dfrac{e^x-e^{-x}}{2}}}dx$
$=\sqrt2\int_0^\infty e^{-\sinh x}\sqrt{\dfrac{e^x}{e^x-e^{-x}}}~dx$
$=\sqrt2\int_0^\infty\dfrac{e^{x-\frac{e^x-e^{-x}}{2}}}{\sqrt{e^{2x}-1}}dx$
$=\sqrt2\int_0^\infty\dfrac{e^{-\frac{e^x}{2}+\frac{1}{2e^x}}}{\sqrt{e^{2x}-1}}d(e^x)$
$=\sqrt2\int_1^\infty\dfrac{e^{-\frac{x}{2}+\frac{1}{2x}}}{\sqrt{x^2-1}}dx$
$=\sqrt2\int_1^\infty e^{-\frac{x}{2}+\frac{1}{2x}}~d(\cosh^{-1}x)$
$=\sqrt2\int_0^\infty e^{-\frac{\cosh x}{2}+\frac{1}{2\cosh x}}~dx$
$=\sqrt2\int_0^\infty e^\frac{1-\cosh^2x}{2\cosh x}~dx$
$=\sqrt2\int_0^\infty e^{-\frac{\sinh^2x}{2\cosh x}}~dx$
$=\sqrt2\int_0^\infty e^{-\frac{x^2}{2\sqrt{x^2+1}}}~d(\sinh^{-1}x)$
$=\sqrt2\int_0^\infty\dfrac{e^{-\frac{x^2}{2\sqrt{x^2+1}}}}{\sqrt{x^2+1}}dx$
$=\sqrt2\int_\infty^0\dfrac{e^{-\frac{1}{2x^2\sqrt{\frac{1}{x^2}+1}}}}{\sqrt{\dfrac{1}{x^2}+1}}d\left(\dfrac{1}{x}\right)$
$=\sqrt2\int_0^\infty\dfrac{e^{-\frac{1}{2x\sqrt{x^2+1}}}}{x\sqrt{x^2+1}}dx$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/527438",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "25",
"answer_count": 2,
"answer_id": 1
}
|
Prove that $\sum\limits_{k=0}^{m}\binom{m}{k}\binom{n+k}{m}=\sum\limits_{k=0}^{m}\binom{n}{k}\binom{m}{k}2^k$ Prove that
$$\sum_{k=0}^{m}\binom{m}{k}\binom{n+k}{m}=\sum_{k=0}^{m}\binom{n}{k}\binom{m}{k}2^k$$
What should I do for this equation? Should I focus on proving $\binom{m}{k}\binom{n+k}{m}=\binom{n}{k}\binom{m}{k}2^k$?
|
This can also be done using a basic complex variables technique.
Suppose we seek to verify that
$$\sum_{k=0}^m {m\choose k} {n+k\choose m}
= \sum_{k=0}^m {m\choose k} {n\choose k} 2^k.$$
Introduce the two integral representations
$${n+k\choose m}
= \frac{1}{2\pi i}
\int_{|z|=\epsilon} \frac{1}{z^{m+1}} (1+z)^{n+k} \; dz$$
and
$${n\choose k}
= \frac{1}{2\pi i}
\int_{|z|=\epsilon} \frac{1}{z^{k+1}} (1+z)^n \; dz$$
This gives the following integral for the sum on the LHS
$$\frac{1}{2\pi i}
\int_{|z|=\epsilon}
\sum_{k=0}^m {m\choose k}
\frac{1}{z^{m+1}} (1+z)^{n+k} \; dz
\\ = \frac{1}{2\pi i}
\int_{|z|=\epsilon} \frac{(1+z)^n}{z^{m+1}}
\sum_{k=0}^m {m\choose k}
(1+z)^k \; dz
\\ = \frac{1}{2\pi i}
\int_{|z|=\epsilon} \frac{(1+z)^n}{z^{m+1}} (2+z)^m \; dz.$$
We get the following integral for the sum on the RHS
$$\frac{1}{2\pi i}
\int_{|z|=\epsilon}
\sum_{k=0}^m {m\choose k} 2^k
\frac{1}{z^{k+1}} (1+z)^n \; dz
\\ = \frac{1}{2\pi i}
\int_{|z|=\epsilon} \frac{(1+z)^n}{z}
\sum_{k=0}^m {m\choose k} 2^k
\frac{1}{z^k} \; dz
\\ = \frac{1}{2\pi i}
\int_{|z|=\epsilon} \frac{(1+z)^n}{z}
\left(1 + \frac{2}{z}\right)^m \; dz
\\ = \frac{1}{2\pi i}
\int_{|z|=\epsilon} \frac{(1+z)^n}{z}
\frac{(z+2)^m}{z^m} \; dz.$$
We can stop here without further evaluation because the integrals for
LHS and RHS are seen to be the same.
A trace as to when this method appeared on MSE and by whom starts at this
MSE link.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/531772",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 4,
"answer_id": 2
}
|
Prove by induction the following inequality for all n∈N $\frac1{\sqrt{1}} + \frac1{\sqrt{2}}+\frac1{\sqrt{3}}+...+\frac1{\sqrt{x}}\ge {\sqrt{x}}$
I proved the basic case: and realize it is equal to 1, but I have absolutely no idea how to create prove the left and right side using the induction hypothesis.
Please help :) I've solved equality's before using induction, but this is the first time I've done an inequality, and although I'm sure the process is fairly similar, I think the square root is messing me up.
|
Let $s_n$ represent the sum up to $n$, and suppose that $s_n \ge \sqrt{n}$ for induction. We want to show that $s_{n + 1} \ge \sqrt{n + 1}$; to this end, note
\begin{align*}
s_{n + 1} - s_n &= \frac{1}{\sqrt{n + 1}}
\end{align*}
Therefore,
\begin{align*}
s_{n + 1} &= \frac{1}{\sqrt{n + 1}} + s_n \\
&\ge \frac{1}{\sqrt{n + 1}} + \sqrt{n} \\
&= \frac{1 + \sqrt{n}\sqrt{n + 1}}{\sqrt{n + 1}} \\
&\ge \frac{1 + \sqrt{n} \sqrt{n}}{\sqrt{n + 1}} \\
&= \frac{n + 1}{\sqrt{n + 1}} = \sqrt{n + 1}
\end{align*}
as desired. The first inequality is by the induction hypothesis, and the second is just by noting that
$$\sqrt{n + 1} \ge \sqrt{n}$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/533961",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
}
|
Prove that $1<\frac{1}{n+1}+\frac{1}{n+2}+...+\frac{1}{3n+1}$ Prove that $1<\dfrac{1}{n+1}+\dfrac{1}{n+2}+...+\dfrac{1}{3n+1}$.
By using the Mathematical induction. Suppose the statement holds for $n=k$.
Then for $n=k+1$. We have $\dfrac{1}{k+2}+\dfrac{1}{k+3}+...+\dfrac{1}{3k+1}+\dfrac{1}{3k+2}+\dfrac{1}{3k+3}+\dfrac{1}{3k+4}=(\dfrac{1}{k+1}+\dfrac{1}{k+2}+\dfrac{1}{k+3}+...+\dfrac{1}{3k+1})+(\dfrac{1}{3k+2}+\dfrac{1}{3k+3}+\dfrac{1}{3k+4}-\dfrac{1}{k+1})$
we know $\dfrac{1}{k+1}+\dfrac{1}{k+2}+\dfrac{1}{k+3}+...+\dfrac{1}{3k+1}>1$
What can we do for $(\dfrac{1}{3k+2}+\dfrac{1}{3k+3}+\dfrac{1}{3k+4}-\dfrac{1}{k+1})$?
|
using $maxima$ I get
(%i1) 1/(3*k+2)+1/(3*k+3)+1/(3*k+4)-1/(k+1),ratexpand;
2
(%o1) -------------------------
3 2
27 k + 81 k + 78 k + 24
(%i2) solve(denom(%),[k]);
4 2
(%o2) [k = - -, k = - 1, k = - -]
3 3
For $k=0$ the denominator is greater than $0$, so it is greater than $0$ for all $ k \gt -\frac{2}{3}$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/534117",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 4,
"answer_id": 3
}
|
$\frac{1}{z} \prod_{n=1}^{\infty} \frac{n^2}{n^2 - z^2} = \frac{1}{z} + 2z\sum_{n=1}^{\infty} \frac{(-1)^n}{z^2-n^2}$? I am trying to show that
$$\frac{1}{z} \prod_{n=1}^{\infty} \frac{n^2}{n^2 - z^2} = \frac{1}{z} + 2z\sum_{n=1}^{\infty} \frac{(-1)^n}{z^2-n^2}$$
This question stems from the underlying homework problem, which asks to show
$$ \frac{\pi}{\sin(\pi z)} = \frac{1}{z} + 2z\sum_{n=1}^{\infty} \frac{(-1)^n}{z^2-n^2}, $$
to which I am at my wits end. I have a couple of identities on hand, namely
$$ \pi \cot (\pi z) = \frac{1}{z} + \sum_{n \in \mathbb{Z}; n \neq 0} \frac{1}{z - n} + \frac{1}{n} $$
and
$$ \frac{\sin (\pi z)}{\pi} = z \prod_{n=1}^{\infty} \left( 1 - \frac{z^2}{n^2} \right) $$
and
$$ \frac{\pi^2}{\sin^2 (\pi z)} = \sum_{n \in \mathbb{Z}} \frac{1}{(z - n)^2} $$
I've tried fooling around with these identities and am getting nowhere. Any hints or suggestions would be greatly appreciated.
|
Let $ \displaystyle f(z) = \frac{\pi}{\sin \pi z} - \frac{1}{z}$.
Then according to the Mittag-Leffler pole expansion theorem, $$ \frac{\pi}{\sin \pi z} - \frac{1}{z} = f(0) + \sum_{n=1}^{\infty} \text{Res}[f,n] \Big( \frac{1}{z-n} + \frac{1}{n} \Big) + \sum_{n=1}^{\infty} \text{Res}[f,-n] \Big( \frac{1}{z+n} - \frac{1}{n} \Big)$$
$$ = \sum_{n=1}^{\infty} (-1)^{n} \Big( \frac{1}{z-n} + \frac{1}{n} \Big) + \sum_{n=1}^{\infty} (-1)^{n} \Big( \frac{1}{z+n} - \frac{1}{n} \Big)$$
$$ = \sum_{n=1}^{\infty} (-1)^{n} \Big(\frac{1}{z-n} + \frac{1}{z+n} \Big) = \sum_{n=1}^{\infty} (-1)^{n} \frac{2z}{z^{2}-n^{2}}$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/535116",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 2
}
|
Prove that there exists only 2 solutions for $x^2 \equiv 9 \pmod {p^k}$, ($p$ an odd prime > 3 and $x$ a natural number < $n$) It appears that the only two solutions are always $3$ and $p^k-3$, I want to prove this, here has been my approach, I think I am close but just missing something, would really appreciate any help!!!
Suppose,
$b^2 \equiv 9 \pmod {p^k}$ for ($k \ge 1$)
$\implies b^2 = 9 + cp^k $ for some integer $c$
I want to find $r$ s.t. $(b+rp^k)^2 \equiv 9 \pmod {p^{k+1}}$
(I.e. since $b \equiv \pm 3$, I want $r \equiv \pm 1$ )
By squaring i get,
$(b+rp^k)^2 \equiv b^2 + 2brp^k + r^2p^{2k} \equiv 9 \pmod {p^{k+1}}$
Substituting $b^2 = 9 + cp^k$ yields,
$9 + cp^k+2brp^k+r^2p^{2k} \equiv 9 \pmod {p^{k+1}}$
$ \implies (c+2br)p^k+r^2p^{2k} \equiv 0 \pmod {p^{k+1}}$
So I need $r$ s.t.
$c+2br \equiv 0 \pmod p$
Since $b \equiv \pm 3$
$ \implies 2(-3)r \equiv -c \pmod p$ or $2(3)r \equiv c \pmod p$
Now I'm kind of stuck, I'm not sure where I need to go from here. I think that if $b\equiv-3$ then I want $r=1$ and if $b\equiv3$ then I want $r=-1$...
I would really appreciate some help! I think I am close!
|
Let $p\gt 3$. Note that $x^2\equiv 9\pmod{p^k}$ if and only if $p^k$ divides $(x-3)(x+3)$. But $p$ cannot divide both $x-3$ and $x+3$, else it would divide their difference $6$. So $(x-3)(x+3)$ is divisible by $p^k$ if and only if $p^k$ divides $x-3$ or $p^k$ divides $x+3$. Thus there are two solutions, $x\equiv 3\pmod{p^k}$ and $x\equiv -3\pmod{p^k}$. Thse are clearly incongruent modulo $p^k$.
Remark: The situation is more complicated when $p=2$ or $p=3$. Note for example that the congruence $x^2\equiv 9\pmod{2^3}$ has $4$ solutions. So does the congruence $x^2\equiv 9\pmod{3^2}$. A complete analysis for these special primes is not difficult.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/537885",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
}
|
How to find value of $x+y+z+u+v+w$ let $x,y,z,u,v,w$ be positive integer numbers,and such
$$1949(xyzuvw+xyzu+xyzw+xyvw+xuvw+zuvw+xy+xu+xw+zu+zw+vw+1)=2004(yzvw+yzu+yzw+uvw+y+u+w)$$
Find this value of
$$x+y+z+u+v+w=?$$
My try:
maybe use
$$(x+1)(y+1)(z+1)(u+1)(v+1)(w+1)=(xyz+xy+yz+xz+x+y+z+1)(uvw+uv+uw+vw+u+v+w+1)$$
|
This is not a solution but a little progress towards the solution pointed out in Ewan Delanoy's comment to show that $x=u=1$ and $y \ge 36$.
Let each side of original equation be $2004 \times 1949 \times k$. Multiplying the parentheses on right-hand-side with $x$ and subtracting from the parentheses on left-hand-side gives $xyzvw(u−1)+xyvw+zuvw+zu+zw+vw+1=(2004−1949x)k$ . The left-hand-side is positive so right-hand-side must be positive so $x=1$.
If $x=1$,
$yzuvw+yzu+yzw+yvw+uvw+zuvw+y+u+w+zu+zw+vw+1=2004k$,
$yzvw+yzu+yzw+uvw+y+u+w=1949k$.
Since $z\ge 1$, first eqn yields $y(zuvw+zu+zw+vw)+2(uvw+u+w)+y+vw+1 \le 2004k$ and dividing by second eqn yields $\frac{y(zuvw+zu+zw+vw)+2(uvw+u+w)+y+vw+1}{y(zvw+zu+zw+1)+uvw+u+w} \le \frac{2004}{1949}$ which implies $\frac{zuvw+zu+zw+vw}{zvw+zu+zw+1} < \frac{2004}{1949}$ or $1949uzvw+1949vw<2004zvw+55(zu+zw)+2004$.
Substituting $zu \le uzvw$, $zw \le zvw$, and $1 \le vw$, $1949uzvw+1949vw<2059zvw+55uzvw+2004vw$ or $1894uz<2059z+55$ which precludes $u \ge 2$.
With $x=u=1$, the two equations reduce to
$yzvw+yz+yzw+yvw+vw+zvw+y+w+z+zw+vw+2=2004k$,
$yzvw+yz+yzw+vw+y+w+1=1949k$
and the difference of the two equations is
$yvw+zvw+z+zw+vw+1=55k$.
The ratio of the last two equations is
$\frac{(yz+1)(vw+w+1)+y}{(z+1)(vw+w+1)+yvw-w} = \frac{1949}{55}$.
But $y \le yvw - w$ unless $v=w=1$. Even if $v=w=1$, $y/(yvw - w) \le 1959/55$ so
$\frac{yz+1}{z+1} \ge \frac{1949}{55}$ so $y \ge 36$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/538037",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 2,
"answer_id": 0
}
|
Sketch curve $y = (4x^3-2x^2+5)/(2x^2+x-3)$ I'm trying to sketch the curve
$$
y = (4x^3-2x^2+5)/(2x^2+x-3).
$$
I tried to find the first and second derivative but I don't know how to find the roots of these.
\begin{align}
y' &= \frac{-5-8 x-38 x^2+8 x^3+8 x^4}{(-3+x+2 x^2)^2}
&&\text{Fermat Theorem}\\
0 &= -5-8 x-38 x^2+8 x^3+8 x^4
&&\leftarrow\text{Stuck}\\
y'' &= \frac{34+276 x-24 x^2+64 x^3}{(-3+x+2 x^2)^3} \\
0 &= (-5-8 x-38 x^2+8 x^3+8 x^4)(-3+x+2 x^2)
&&\leftarrow\text{Stuck}
\end{align}
I am currently learning Calculus I.
|
You don't need the second derivative.
*
*Factorise the denominator to find vertical asymptotes.
*Ask yourself what happens when $x \to \pm \infty$ to find the horizontal asymptotes.
*Put $x=0$ to find the $y$-intercept.
*Look for the roots of the numerator to give you the $x$-interecepts.
*Solve $y'=0$ to give yourself the turning points.
Once you have all of these the sketch should be straightforward.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/538232",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
}
|
Problem involving trigonometry and cubics One of my teachers proposed me the following problem:
$$\text{If } (3\sec x+\csc x)\sin x=5\cos^2 x\text{, calculate } z=\tan x+\sec x$$
I started by manipulating
$$3\tan x +1=5\cos^2 x$$
$$\sec^2 x(3\tan x +1)=5$$
$$(1+\tan^2 x)(3\tan x +1)=5$$
$$3\tan^3x +\tan^2 x + 3\tan x -4=0$$
$$3a^3 +a^2 + 3a -4=0$$
Unfortunately, this doesn't have any rational zeroes. Replacing $a=b-1/9$, we get
$$3b^3+26b/9-1051/243=0$$
Unfortunately again, the linear term is positive, so we can't solve using cosines and arcocosines directly.Alternatively, this can be seen as $(3a^3 +a^2 + 3a -4)'=(3a+1/3)^2+26/9>0$, so it is strictly increasing, therefore it has only one real root, therefore we can't use trigonometric solutions. Of course there is a solution using radicals, but it goes really ugly and I am quite convinced that was not the intent of the problem. Is there a good solution to this? Or was the problem ill-posed because of a typo?
|
Maybe this will be helpful for you. It seems the following.
Put $y=\pi/2-x$. Then $\cos x=\sin y$, $\sin x=\cos y$ and $\frac 1z=\frac {\cos x}{1+\sin x}=\frac {\sin y}{1+\cos y}=\tan\frac y2$. Then $\sin y=\frac {2z}{z^2+1}$,
$\cos y=\frac {z^2-1}{z^2+1}$ and
$$\frac{3(z^2-1)}{z^2+1}+\frac{2z}{z^2+1}=5\left(\frac {2z}{z^2+1}\right)^3,$$
or $3z^6+2z^5+3z^4-36z^3-3z^2+2z-3=0$. Mathcad yieds that four roots of this equation are not real and other two are $z_1\simeq –.50330721831204403873$ and $z_2\simeq 1.9868580533252213137$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/538715",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
}
|
Proof of $(\forall x)(x^2+4x+5 \geqslant 0)$ $(\forall x)(x^2+4x+5\geqslant 0)$ universe is $\Re$
I went about it this way
$x^2+4x \geqslant -5$
$x(x+4) \geqslant -5$
And then I deduce that if $x$ is positive, then $x(x+4)$ is positive, so it's $\geqslant 5$
If $ 0 \geqslant x \geqslant -4$, then $x(x+4)$ is also $\geqslant -5$.
If $ x < -4$, then $x(x+4)$ will be negative times negative = positive, so obviously $\geqslant -5$
I'm wondering if there's an easier to solve this problem? My way seems a little clunky.
|
One of the best ways in general to handle questions about quadratics is via the process of completing the square: that is, turning a quadratic of the form $P(x) = x^2+bx+c$ into one of the form $Q(x) = (x+r)^2+s$. To do this, note that the latter can be written as $x^2+2rx+r^2+s$, and we can equate terms; this means that $r$ must be equal to $\frac b2$, and then since $s+r^2=c$, we must have $s=c-r^2=c-\frac{b^2}{4}$ (and if you were to use an 'a' term for your quadratic and pursued the algebra along this path a little further, you'd derive the famous quadratic formula).
In the case at hand, we have $b=4$ so we know that $r=\frac b2=2$, and we get that $x^2+4x+5$ = $(x+2)^2+s$ = $x^2+4x+4+s$, so $s=1$ and the quadratic can be written in the form $(x+2)^2+1$. In this form it's immediate that it's positive for all $x$ (since $(x+2)^2\geq 0$ for all $x$ and $1\gt 0$). What's more, we can see that its minimum value is acheived where $(x+2)^2$ is at a minimum - that is, when $x+2=0$, or in other words when $x=-2$; thus, the vertex of the parabola given by $y=P(x) = x^2+4x+5$ is at $(x,y) = (-2, P(-2)) = (-2, 1)$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/539375",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 6,
"answer_id": 5
}
|
If $a^3 + b^3 +3ab = 1$, find $a+b$
Given that the real numbers $a,b$ satisfy $a^3 + b^3 +3ab = 1$, find $a+b$.
I tried to factorize it but unable to do it.
|
The three solutions of the equation $a^3+b^3+3ab=1$ are
$$b_1=1-a,$$
$$b_2,b_3 = \frac{1}{2}\left(a-1\pm i\sqrt{3}(1+a)\right).$$
If $a\neq -1$, since $a,b\in\mathbb{R}$, we must have $a+b=a+b_1=1$. If
$a=-1$, then the imaginary part of $b_2,b_3$ vanishes and we find two solutions for $(a,b)$: $(-1,-1)$ and $(-1,2)$ so $a+b$ is -2 or 1 respectively.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/540084",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 9,
"answer_id": 4
}
|
how to solve $x^2 \equiv -1 \pmod{13}$ Knowing that $p$ is prime and if $p \equiv 1 \pmod 4$, then $\left(\left(\frac{p-1}{2}\right)!\right)^2 \equiv -1 \pmod p$; how do I solve $x^2 \equiv -1 \pmod{13}$?
|
Observe that $$p-r\equiv-r\pmod p$$
putting $r=1,2,\cdots,\frac{p-1}2$ and multiplying we get
$$\prod_{\frac{p+1}2\le r\le p-1 }r\equiv(-1)^{\frac{p-1}2} \prod_{1\le s\le \frac{p-1}2 }s$$
$$\prod_{\frac{p+1}2\le r\le p-1 }r\equiv\prod_{1\le s\le \frac{p-1}2 }s\text{ as } \frac{p-1}2 \text{ is even}$$
Multiplying by $\displaystyle \prod_{1\le s\le \frac{p-1}2 }s\text{ as } \frac{p-1}2, $ we get
$$\left(\prod_{1\le s\le \frac{p-1}2 }s\right)^2\equiv \prod_{1\le r\le p-1 }r=(p-1)!\equiv-1\pmod p$$ using Wilson's Theorem
But $\displaystyle \prod_{1\le s\le \frac{p-1}2 }s=\left(\frac{p-1}2\right)!$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/542487",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
}
|
$\sin ^6x+\cos ^6x=\frac{1}{8}\left(3\cos 4x+5\right)$, Any quick methods? How to prove the following equation by a quick method?
\begin{eqnarray}
\\\sin ^6x+\cos ^6x=\frac{1}{8}\left(3\cos 4x+5\right)\\
\end{eqnarray}
If I use so much time to expand it and take extra care of the calculation process, I can find the answer. Since it is so easy to get mistakes, I have to try it 3 time to obtain the correct answer. Therefore, I wonder there are other easy ways to deal with this question. Is it right?
Thank you for your attention
|
To flesh out Thomas's comment into another answer: start with the factorization $x^3-y^3 = (x-y)(x^2+xy+y^2)$; this can be derived easily from the finite geometric series. Replacing $y$ with $-y$ yields $x^3+y^3 = (x+y)(x^2-xy+y^2)$, and then replacing $x$ with $x^2$ and $y$ with $y^2$ yields $x^6+y^6 = (x^2+y^2)(x^4-x^2y^2+y^4)$. Now, plugging in $x=\cos\theta, y=\sin\theta$ and using $\cos^2\theta+\sin^2\theta=1$ yields $\cos^6\theta+\sin^6\theta = \cos^4\theta-\cos^2\theta\sin^2\theta+\sin^4\theta$. From here, the simplest route is through an approach similar to lab bhattacharjee's answer: $\cos^4\theta-\cos^2\theta\sin^2\theta+\sin^4\theta$ $=(\cos^4\theta+2\cos^2\theta\sin^2\theta+\sin^4\theta)-3\cos^2\theta\sin^2\theta$ $=(\cos^2\theta+\sin^2\theta)^2-3\cos^2\theta\sin^2\theta$ $=1-3\cos^2\theta\sin^2\theta$; and now things proceed exactly as in that answer, using the double-angle formulas for sin and cos in turn.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/543030",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 1
}
|
How can I Prove $\frac{2xy}{x+y}\leq \sqrt{xy}\leq \frac{x+y}{2}$ for $x,y>0$ prove that
$\frac{2xy}{x+y}\leq \sqrt{xy}\leq \frac{x+y}{2}$
I have tried to develop $(x+y)^2=$ and to get to an expression that must be bigger than those above
Thanks!
|
Start by noticing that since $(x-y)^2 \geq 0$ then $x^2 - 2xy +y^2 \geq 0$ which implies that $x^2+y^2 \geq 2xy$. Now we attack these inequalities one at a time.
To show the left inequality, we can show $2xy \leq (x+y)\sqrt{xy}$, and since everything is positive we can square both sides and further reduce the problem to showing $4x^2 y^2 \leq(x^2+2xy+y^2)(xy)$. Then, since $x^2+y^2 \geq 2xy$ we have $(x^2+2xy+y^2)(xy) \geq 4xy(xy) = 4x^2y^2$, which is exactly what you wanted to show.
To show the right inequality we will square both sides again and reduce the problem to showing $xy \leq \frac{1}{4}(x^2 + 2xy +y^2)$. To do so we again use that $x^2+y^2 \geq 2xy$. Therefore, $\frac{1}{4}(x^2+2xy+y^2) \geq \frac{1}{4}(4xy) = xy$ which is again the inequality you wanted to show.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/543253",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
"answer_id": 3
}
|
Finite Sum $\sum\limits_{k=1}^{m-1}\frac{1}{\sin^2\frac{k\pi}{m}}$
Question : Is the following true for any $m\in\mathbb N$?
$$\begin{align}\sum_{k=1}^{m-1}\frac{1}{\sin^2\frac{k\pi}{m}}=\frac{m^2-1}{3}\qquad(\star)\end{align}$$
Motivation : I reached $(\star)$ by using computer. It seems true, but I can't prove it. Can anyone help?
By the way, I've been able to prove $\sum_{n=1}^{\infty}\frac{1}{n^2}=\frac{{\pi}^2}{6}$ by using $(\star)$.
Proof : Let
$$f(x)=\frac{1}{\sin^2x}-\frac{1}{x^2}=\frac{(x-\sin x)(x+\sin x)}{x^2\sin^2 x}.$$
We know that $f(x)\gt0$ if $0\lt x\le {\pi}/{2}$, and that $\lim_{x\to 0}f(x)=1/3$. Hence, letting $f(0)=1/3$, we know that $f(x)$ is continuous and positive at $x=0$. Hence, since $f(x)\ (0\le x\le {\pi}/2)$ is bounded, there exists a constant $C$ such that $0\lt f(x)\lt C$. Hence, substituting $x={(k\pi)}/{(2n+1)}$ for this, we get
$$0\lt \frac{1}{\frac{2n+1}{{\pi}^2}\sin^2\frac{k\pi}{2n+1}}-\frac{1}{k^2}\lt\frac{{\pi}^2C}{(2n+1)^2}.$$
Then, the sum of these from $1$ to $n$ satisfies
$$0\lt\frac{{\pi}^2\cdot 2n(n+1)}{(2n+1)^2\cdot 3}-\sum_{k=1}^{n}\frac{1}{k^2}\lt\frac{{\pi}^2Cn}{(2n+1)^2}.$$
Here, we used $(\star)$. Then, considering $n\to\infty$ leads what we desired.
|
Using the partial fractions identity, it's possible to split the summation into two
$$ \frac{2}{\sin^2 \theta} = \frac{1}{1-\cos \theta} + \frac{1}{1+\cos \theta}$$
Now our sum over the nth roots
$$ (\star) =\sum_{k=1}^{m-1} \frac{1}{\sin^2 \frac{\pi k }{n}}
= \frac{1}{2}\sum_{k=1}^{m-1} \frac{1}{1-\cos \frac{\pi k }{n}}
+ \frac{1}{2}\sum_{k=1}^{m-1} \frac{1}{1+\cos \frac{\pi k }{n}}
$$
If we can find the right polynomial $P(x)$, we can find our sum using the identity for the logarithmic derivative. We need to plug in $x=1$ and $x=-1$ and subtract.
$$ \frac{P'(x)}{P(x)} =\sum \frac{1}{x-r_i} \text{ then }
\sum \frac{1}{1-r_i} + \sum \frac{1}{1+r_i}
= \frac{P'(1)}{P(1)} - \frac{P'(-1)}{P(-1)} $$
My first guess is the Chebyshev polynomials satisfy $T_n(\cos \theta) = \cos n \theta $.
The roots of $T_m(x) = 0$ are $x = \cos \frac{2\pi k}{m}$ with $k = 0,1,2, \dots, m-1$.
Instead, we need a polynomial with roots of $x = \cos \frac{\pi k}{m}$ which is the derivative $\boxed{P(x)=T'_m(x)}$ (Chebyshev polynomial of second kind)
In the case of the Chebyshev polynomial I got these two expressions for the values of the derivative at $x=1$:
Let $T(\cos \theta) = \cos n \theta$. Then set $\theta = 0$:
$$ T(1) = 1 $$
If we take the derivative (whose zeros have roots exactly where we want)
$$\boxed{ \sin \theta \; T'(\cos \theta) = -n \sin n \theta} \hspace{0.25in}\text{ so that }\hspace{0.25in}
\boxed{\displaystyle T'(\cos \theta) = -\frac{n \sin n \theta}{\sin \theta} \bigg|_{\theta=0}= -n^2}$$
and for the second derivative, try the quotient rule:
\begin{eqnarray} T''(\cos \theta) &=& n \cdot \frac{n \cos n \theta \sin \theta - \sin n \theta \cos \theta}{\sin^3 \theta} \\
&=\bigg|_{\theta \approx 0}& n \cdot \frac{n (1 - n^2 \theta^2 /2)(\theta - \theta^3 /6) - (n \theta - (n\theta)^3/6)(1 - \theta^2 /2)}{\theta^3} \\ &=\bigg|_{\theta \approx 0}& \frac{n^2 - n^4 }{3}\end{eqnarray}
If we set $\theta = \pi$, just get a $(-1)^n$ factor. In general your sum is
$$ (\star) = \frac{T'(1)}{T''(1)} = \frac{T'(-1)}{T''(-1)} =
\frac{1}{2}\left( \frac{T'(1)}{T''(1)} + \frac{T'(-1)}{T''(-1)}\right) = \frac{\tfrac{1}{3}(n^4 - n^2)}{n^2} = \frac{n^2 - 1 }{3}$$
COMMENTS It is surprisingly hard to pin all the details down. See also: Sum of the reciprocal of sine squared
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/544228",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "49",
"answer_count": 8,
"answer_id": 6
}
|
Inequality $\frac{a}{b} + \frac{b}{c} + \frac{c}{a} \geq a + b+ c$ when $abc = 1$
If $a,b,c > 0$ are such that $abc=1$, then
$$ \frac{a}{b} + \frac{b}{c} + \frac{c}{a} \geq a + b+ c. $$
I would be pleased if you give me a hint. Thanks in advance.
|
$$a+b+c =$$
$$\{ \mbox{multiply by 1: maybe } abc\mbox{, maybe }\sqrt{abc}, ... \}$$
$$= \frac{a}{\sqrt[3]{abc}}+\frac{b}{\sqrt[3]{abc}}+\frac{c}{\sqrt[3]{abc}}=$$
$$
=\sqrt[3]{\frac{a}{b} \cdot \frac{a}{b} \cdot \frac{b}{c}}
+\sqrt[3]{\frac{b}{c} \cdot \frac{b}{c} \cdot \frac{c}{a}}
+\sqrt[3]{\frac{c}{a} \cdot \frac{c}{a} \cdot \frac{a}{b}}
\le
$$
$$
\{\mbox{ GM}\le\mbox{AM }\}
$$
$$
\le \frac{\frac{a}{b}+\frac{a}{b}+\frac{b}{c}}{3}
+ \frac{\frac{b}{c}+\frac{b}{c}+\frac{c}{a}}{3}
+ \frac{\frac{c}{a}+\frac{c}{a}+\frac{a}{b}}{3} =
$$
$$
=\frac{a}{b}+\frac{b}{c}+\frac{c}{a}.
$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/546685",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 1,
"answer_id": 0
}
|
Let $a,b,c \in \mathbb{R^+}$, does this inequality holds $\frac{a}{na + kb} + \frac{b}{nb+kc} + \frac{c}{nc + ka} \ge \frac{3}{k+n}$? Does the following statement/inequality holds for $a,b,c \in \mathbb{R^+}$?
$$\frac{a}{na + kb} + \frac{b}{nb+kc} + \frac{c}{nc + ka} \ge \frac{3}{k+n}$$
I've been thinking for hours and I couldn't find a solution to this inequality. I tried to use Cauchy-Schwarz Inequality for fraction:
$$\frac{x^2}{a} + \frac{y^2}{b} + \frac{z^2}{c} \ge \frac{(z+x+y)^2}{a+b+c}$$
But I only get:
$$LHS \ge \frac{(\sqrt{a} + \sqrt{b} + \sqrt{c})^2}{(n+k)(a+b+c)}$$
And then I can't prove that:
$$(\sqrt{a} + \sqrt{b} + \sqrt{c})^2 \ge 3(a+b+c)$$
$$a + b + c + 2\sqrt{ab} + 2\sqrt{ac} + 2\sqrt{bc} \ge 3(a+b+c)$$
$$\sqrt{ab} + \sqrt{ac} + \sqrt{bc} \ge a+b+c$$
Which simplifyies to:
$$\sqrt{a}(\sqrt{b} - \sqrt{a}) + \sqrt{b}(\sqrt{c} - \sqrt{b}) + \sqrt{c}(\sqrt{a} - \sqrt{c}) \ge 0$$
But if $a\neq b \neq c$, one of the terms must be negative, so nothing here.
And at last I plug it into Wolfram Alpha, and for some random numbers $n$ and $k$ it's true. I wonder maybe the fact that the LHS is cyclic we make the LHS to have minimal values when $a=b=c$?
|
By applying CBS is obtained
$$\frac{a^2}{na^2+kab} + \frac{b^2}{nb^2+kbc} + \frac{c^2}{nc^2+kca} \ge \frac{(a+b+c)^2}{n(a^2+b^2+c^2)+k(ab+bc+ca)}.$$
But
$$\frac{(a+b+c)^2}{n(a^2+b^2+c^2)+k(ab+bc+ca)}\ge \frac{3}{n+k}$$
is equivalent to
$$(k-2n)(a^2+b^2+c^2-ab-bc-ca)\ge 0$$
which is true under the additional condition
$$k\ge2n.$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/549149",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
}
|
Generalizing the sum of consecutive cubes $\sum_{k=1}^n k^3 = \Big(\sum_{k=1}^n k\Big)^2$ to other odd powers We have,
$$\sum_{k=1}^n k^3 = \Big(\sum_{k=1}^n k\Big)^2$$
$$2\sum_{k=1}^n k^5 = -\Big(\sum_{k=1}^n k\Big)^2+3\Big(\sum_{k=1}^n k^2\Big)^2$$
$$2\sum_{k=1}^n k^7 = \Big(\sum_{k=1}^n k\Big)^2-3\Big(\sum_{k=1}^n k^2\Big)^2+4\Big(\sum_{k=1}^n k^3\Big)^2$$
and so on (apparently).
Is it true that the sum of consecutive odd $m$ powers, for $m>1$, can be expressed as sums of squares of sums* in a manner similar to the above? What is the general formula?
*(Edited re Lord Soth's and anon's comment.)
|
This is not an answer but does not fit in a comment-box.
Another way to describe the relation between the sums and the sums of squares of various sums-of-like-powers makes use of the Eulerian-numbers, which converts the expressions for the sums-of-like-powers into polynomials. The sums of like powers can be expressed as
$$\small \begin{array} {rclllll}
s_0(n) = & 1 \cdot \binom{n}{1} \\
s_1(n) = & & 1 \cdot \binom{n+1}{2} \\
s_2(n) = & & 1 \cdot \binom{n+1}{3}& +1 \cdot \binom{n+2}{3} \\
s_3(n) = & & 1 \cdot \binom{n+1}{4}& +4 \cdot \binom{n+2}{4} & +1 \cdot \binom{n+3}{4} \\
s_4(n) = & & 1 \cdot \binom{n+1}{5}& +11 \cdot \binom{n+2}{5} & +11 \cdot \binom{n+3}{5} & +1 \cdot \binom{n+4}{5} \\
s_5(n) = & & 1 \cdot \binom{n+1}{6}& +26 \cdot \binom{n+2}{6}& +66 \cdot \binom{n+3}{6} & +26 \cdot \binom{n+4}{6} & +1 \cdot \binom{n+5}{6} \\
\vdots & &\vdots
\end{array}
$$
This looks a bit more regular than the expressions by Bernoulli-polynomials and possibly the relations between the sums-expressions of odd orders $s_{2k+1}(n)$ can be written using these polynomials nicely and with more ease than with the Bernoulli-polynomials, don't know.
The first of your equality would then be expressed as
$$ \begin{array} {rcl} \sum_{k=1}^n k^3 &=&\left(\sum_{k=1}^n k \right)^2 \\ s_3(n) &=& s_1(n)^2 \\
1 \cdot \binom{n+1}{4} +4 \cdot \binom{n+2}{4} +1 \cdot \binom{n+3}{4} &=& \left( 1 \cdot \binom{n+1}{2}\right)^2 \\
\end{array}$$
and which must then be expanded to reproduce the equality (but I've not the time at the moment to do this in the required length...)
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/549823",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "52",
"answer_count": 4,
"answer_id": 1
}
|
Find the Laurent series for $f(z) = (z^2 - 4)/(z-1)^2 $ for $z=1$ What I understand is that we have to expand $f(z$) in the positive and negative powers of $(z-1)$.
Hence I tried factorizing the numerator $(z^2-4)=(z+2)(z-2)$ , which can then be written in terms of $(z-1)$ as:
$(z-1-1)(z-1+3)/(z-1)^2$ . however i cannot expand it using the geometric series expansion:
$$1/(1-z) = 1 + z + z^2 + \cdots$$
Help.
|
Use
$$z^2-4 = (z-1+1)^2 -4 = (z-1)^2+ 2 (z-1)-3$$
so that
$$f(z) = -\frac{3}{(z-1)^2} + \frac{2}{z-1}+1$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/550014",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
}
|
Find the equation of the tangent line to the curve $y = \sqrt{x}$ at the point $(9,3)$. Find the equation of the tangent line to the curve $y = \sqrt{x}$ at the point $(9,3)$.
I did
$$\frac{f(x+h)-f(x)}{h}=\frac{\sqrt{9+h}-3}{h} \frac{\sqrt(9+h)+3}{\sqrt(9+h)+3} = \frac{9+h-3}{h\sqrt{9+h}+3}= \frac{6}{\sqrt{9+h}+3}= \frac{6}{6}=1.$$
|
To follow up on Oria Gruber's answer:
If
$y = \sqrt x, \tag{1}$
then
$x = y^2, \tag{2}$
so
$dx/dy = 2y = 2\sqrt x, \tag{3}$
whence
$dy/dx = (dx/dy)^{-1} = \frac{1}{2 \sqrt x}; \tag{4}$
thus when $x = 9$ we obtain
$dy/dx = 1/6, \tag{5}$
so the equation for the desired tangent line is
$y - 3 = \frac{1}{6}(x - 9). \tag{6}$
The derivative $dx/dy$ of $x$ with repect to $y$ may be easily had using the definition
$f'(x) = \lim_{h \to 0}((f(x + h) -f(x) / h), \tag{7}$
with which the OP seems familiar.
Hope this helps. Cheerio,
and as always
Fiat Lux!!!
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/553494",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
}
|
Laplace differential equation Can somebody help me work out $2y''+y'-y=\mathrm{e}^{3t}$, y(0)=2 and y'(0)=0 with the method of Laplace? I got
\begin{align*}
Y(s)&=\frac{1}{(s-3)(2s^2+s-1)}+ \frac{2+4s}{(s-3)(2s^2+s-1)}\\
&=-\frac{4}{15}\frac{1}{2s-1}+\frac{1}{12}\frac{1}{s+1}+\frac{1}{20}\frac{1}{s-3}-\frac{1}{6}\frac{1}{s+1}-\frac{16}{15}\frac{1}{2s-1}+\frac{7}{10}\frac{1}{s-3}\\
&=-\frac{2}{15}\mathrm{e}^{0.5t}+\frac{1}{12}\mathrm{e}^{-t}+\frac{1}{20}\mathrm{e}^{3t}-\frac{1}{6}\mathrm{e}^{-t}-\frac{8}{15}\mathrm{e}^{0.5t}+\frac{7}{10}\mathrm{e}^{3t}
\end{align*}
Is this correct?
|
I think the answer is $$ L(y)=\frac{-\exp(t)}{6}+\frac{\exp(3t)}{4}+\frac{2\exp(-.5t)}{21}.$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/554187",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
}
|
How prove $\sum\frac{1}{2(x+1)^2+1}\ge\frac{1}{3}$ let $x,y,z>0$ and such $xyz=1$ show that
$$\dfrac{1}{2(x+1)^2+1}+\dfrac{1}{2(y+1)^2+1}+\dfrac{1}{2(z+1)^2+1}\ge\dfrac{1}{3}$$
My try: I will find a value of the $k$ such
$$\dfrac{1}{2(x+1)^2+1}\ge\dfrac{1}{9}+k\ln{x}$$
note $\ln{x}+\ln{y}+\ln{z}=0$,so
$$\sum_{cyc}\dfrac{1}{2(x+1)^2+1}\ge\dfrac{1}{3}+k(\ln{x}+\ln{y}+\ln{z})=\dfrac{1}{3}$$
so let
$$f(x)=\dfrac{1}{2(x+1)^2+1}-k\ln{x}-\dfrac{1}{9}$$
$$\Longrightarrow f'(x)=\dfrac{-4x-4}{(2x^2+4x+3)^2}-\dfrac{k}{x}$$
let $f'(1)=0\Longrightarrow k=-\dfrac{8}{81}$
so
$$f'(x)=\dfrac{-4x-4}{(2x^2+4x+3)^2}+\dfrac{8}{81x}=\dfrac{4(x-1)(8x^3+40x^2+15x-18)}{81x(2x^2+4x+3)^2}$$
so note when $1>x>\dfrac{1}{2}$ then
$$f'(x)=\dfrac{4(x-1)(8x^3+40x^2+15x-18)}{81x(2x^2+4x+3)^2}<0$$
$x>1,f'(x)>0$
so
$$f(x)\ge f(1)=0$$
so if $x,y,z>\dfrac{1}{2}$ we have prove done.
But for other case,How prove it? Thank you
|
$z=\dfrac{1}{xy}$, put in LHS and and clean the denominators, we have:
edit:
LHS-RHS=$ 9y^2x^4-8y^3x^3+2y^2x^3+9y^4x^2+2y^3x^2-9y^2x^2-8yx^2-8y^2x+2yx+9 \ge0 \iff $
$4y^2x^4-8yx^2+4\ge 0,\\4y^4x^2-8y^2x+4\ge0,\\5y^2x^4+5y^4x^2\ge 10x^3y^3,\\2y^2x^3+2y^3x^2\ge 4(xy)^{\frac{5}{2}} \iff\\ LHS \ge 2x^3y^3+4(xy)^{\frac{5}{2}}-9x^2y^2+2xy+1=2t^6+4t^5-9t^4+2t^2+1=(t-1)^2(2t^4+8t^3+5t^2+2t+1) \ge0, t=\sqrt{xy}$
the "=" will hold when $xy=1,x=y,y^2x=yx^2=1 \implies x=y=z=1$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/555497",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 0
}
|
Is $\int\frac{\cos^5x\sin^3x}{1+\cos2x}dx = \frac{\sin^4x}{8}-\frac{\sin^6x}{12} +C$ or $\frac{\cos^6x}{12}-\frac{\cos^4x}{8} +C$? $\int\dfrac{\cos ^5x\sin ^3x}{1+\cos 2x}dx = \dfrac{\sin ^4x}{8}-\dfrac{\sin ^6x}{12} +C$ or $\dfrac{\cos ^6x}{12}-\dfrac{\sin ^4x}{8} +C$?
$\int\dfrac{\cos ^5x\sin ^3x}{1+\cos 2x}dx$ = $\dfrac{1}{2}\int{\cos ^3x\sin ^3x} dx$
1) $\dfrac{1}{2}\int{\cos ^3x\sin ^3x} dx$
= $\dfrac{1}{2}\int{\cos ^2x\sin ^3x\cos x} dx$
= $\dfrac{1}{2}\int{(1-\sin ^2x)\sin ^3x} d\sin x$
= $\dfrac{1}{2}\int{\sin ^3x-\sin ^5x} d\sin x$
= $\dfrac{\sin ^4x}{8}-\dfrac{\sin ^6x}{12} +C$
2) $\dfrac{1}{2}\int{\cos ^3x\sin ^3x} dx$
= $\dfrac{1}{2}\int{\cos ^3x\sin ^2x\sin x} dx$
= $-\dfrac{1}{2}\int{(1-\cos ^2x)\cos ^3x} d\cos x$
= $-\dfrac{1}{2}\int{\cos ^3x-\cos ^5x} d\cos x$
= $\dfrac{\cos ^6x}{12}-\dfrac{\cos ^4x}{8} +C$
Which one is correct? or both are correct?
|
You can differentiate and simplify your two answers to see that they are both correct. If you take the first answer, expand it using the identity $\sin^2 x = 1-\cos^2 x$, and simplify, you will get the second answer plus an additional constant. Thus the two expressions (ignoring the $C$'s) differ by a constant.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/555578",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
}
|
Find the first three terms of the Maclaurin Series
Determine using multiplication/division of power series (and not via WolframAlpha!) the first three terms in the Maclaurin series for $y=\sec x$.
I tried to do it for $\tan(x)$ but then got kind of stuck. For our homework we have to do it for the $\sec(x)$. It is kind of tricky. Help would be awesome!
Thanks!
Taylor series for $\tan(x)$:
\begin{align*}
\tan (x)
&=\frac{\sin(x)}{\cos(x)}\\
&=\frac{x-\frac {x^3}6+\frac{x^5}{120}-\cdots}{1-\frac{x^2}2+\frac{x^4}{24}-\cdots}\\
&=x+\frac{x^3}3+\frac{2x^5}{15}+\cdots
\end{align*}
|
$\sec(x)=\frac{1}{\cos x}$. The three first terms are $1,x^2$ and $x^4$. Then we write: $$\cos x=1-\frac{x^2}2 + \frac{x^4}{24} + o(x^4)$$
Putting $$u=-\frac{x^2}2 + \frac{x^4}{24}=-\frac{x^2}{2}\left(1-\frac{x^2}{12}\right)$$ we have:
$$\sec(x)=(1+u)^{-1} =\operatorname{Tronc}_4 (1 -u +u^2-u^3+u^4)=\operatorname{Tronc}_4 (1 -u +u^2)$$
Since: $\operatorname{Tronc}_4(u)=u$ and $\operatorname{Tronc}_4(u^2)=\frac{x^4}{4}$ you can finish:
$$\sec x = 1+\frac{x^2}{2}-\frac{x^4}{24} + \frac{x^4}4 + o(x^4)= 1+\frac{x^2}{2} + \frac{5x^4}4 + o(x^4) $$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/556423",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
}
|
Check my solution: Find $\lim_{x\to+\infty}(\sqrt{(x+a_1)(x+a_2)}-x)$ Find $\lim_{x\to+\infty}(\sqrt{(x+a_1)(x+a_2)}-x)$
First I rationalized the numerator,
$$
\begin{align*}
&\lim_{x\to+\infty}(\sqrt{(x+a_1)(x+a_2)}-x) \cdot \frac{\sqrt{(x+a_1)(x+a_2)}+x}{\sqrt{(x+a_1)(x+a_2)}+x} \\
=&\lim_{x\to+\infty}\frac{x(a_1+a_2)+a_1a_2}{\sqrt{(x+a_1)(x+a_2)}+x} \\
=&\lim_{x\to+\infty}\frac{(a_1+a_2)+(a_1a_2)/x}{\sqrt{(\frac{1}{x}+\frac{a_1}{x^2})(\frac{1}{x}+\frac{a_2}{x^2})}+1} \\
=& a_1+a_2
\end{align*}
$$
Some calculations in my calculator tells me my answer is likely to be wrong.
I think the real answer is $\frac{a_1+a_2}{2}$.
Can anyone find my mistake or drop me a hint?
|
The mistake happens when you divide numerator and denominator by $x$. Inside the square root this means division by $x^{2}$ and so you should divide one factor $(x + a_{1})$ by $x$ and another factor $(x + a_{2})$ by another $x$ so that after the division the denominator will be $$\sqrt{\left(1 + \frac{a_{1}}{x}\right)\left(1 + \frac{a_{2}}{x}\right)} + 1$$ and this tends to $2$ as $x \to \infty$. You have mistakenly divided both factors inside square root by $x^{2}$ so effectively this means that the square root expression has been divided by $x^{2}$ instead of $x$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/560938",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
}
|
last three digits of $7^{100}-3^{100}$ How can I find the las three digits of $7^{100}-3^{100}$ ? I know one way is to use $7^{100}=(10-3)^{100}=\sum_{k=0}^n{100 \choose k}10^{100-k}(-3)^k$ but I'm totally stuck...
Thanks
|
Well, note that for $0\le k\le 97,$ we have that $1000$ is readily a factor of $\binom{100}{k}10^{100-k}(-3)^k,$ so the last three digits of all of those numbers will be $0$s. Hence, the last three digits of $7^{100}$ will be the last three digits of $$\binom{100}{98}10^{2}(-3)^{98}+\binom{100}{99}10^1(-3)^{99}+\binom{100}{100}10^{0}(-3)^{100},$$ that is, of $$\binom{100}{98}10^{2}(-3)^{98}+\binom{100}{99}(-3)^{99}+3^{100}.$$ Hence, the last three digits of $7^{100}-3^{100}$ will be the last three digits of $$\binom{100}{98}10^{2}(-3)^{98}+\binom{100}{99}10^1(-3)^{99}.$$
Since $\binom{100}{98}=50\cdot99$ and $\binom{100}{99}=100$, the last three digits of this number are easily found.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/562670",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 0
}
|
$2(1+abc)+\sqrt2(1+a^2)(1+b^2)(1+c^2)\ge(1+a)(1+b)(1+c)$ for real numbers $a,b,c$ $a,b,c$ reel sayılar için ;
$$2(1+abc)+\sqrt2(1+a^2)(1+b^2)(1+c^2)\ge(1+a)(1+b)(1+c)$$
Olduğunu gösteriniz.
Translation:1 For real numbers $a,b,c$, show that:
$$2(1+abc)+\sqrt2(1+a^2)(1+b^2)(1+c^2)\ge(1+a)(1+b)(1+c)$$
1With some help from Google translate
|
let $$u=a+b+c,v=ab+bc+ac,w=abc$$
it suffices to show that
$$2(u^2+v^2+w^2-2wu-2v+1)\ge (u+v-w-1)^2$$
this inequality is equivalent to
$$u^2+v^2+w^2-2uv+2vw-2uw+2u-2v-2w+1\ge 0$$
or
$$(u-v-w+1)^2\ge 0$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/565300",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
}
|
find all the values of a and b so that the system has a) no solution b) 1 solution c) exactly 3 solutions and 4) infinitely many solutions $$\begin{cases}
&x &- &y &+ &2z &= 4 \\
&3x &- &2y &+ &9z &= 14 \\
&2x &- &4y &+ &az &= b
\end{cases}
$$
I know that $a$ and $b$ has to either equal to something or not in order to satisfy the $4$ conditions stated above.
my matrices looked like
$$
\begin{bmatrix}
1 & -1 & 2 \\
0 & 1 & 3 \\
0 & 0 & a+12 \\
\end{bmatrix}
\begin{bmatrix}
x & \\
y &\\
z & \\
\end{bmatrix}=
\begin{bmatrix}
6 & \\
2 & \\
b+8 & \\
\end{bmatrix}
$$
|
Apply Gaussian Elimination to the reduced matrix.
$$
\pmatrix{1 & -1 & 2 & 4\\3 & -2 & 9 & 14\\2 & -4 & a & b} \to
\pmatrix{1 & -1 & 2 & 4\\0 & 1 & 3 & 2\\0 & -2 & a-4 & b-8} \to
\pmatrix{1 & -1 & 2 & 4\\0 & 1 & 3 & 2\\0 & 0 & a+2 & b+4}
$$
So
*
*if you want a unique solution, $\frac{b+4}{a+2}$ must be defined, i.e. $a \neq -2$.
*if you want infinite number of solutions, $a=-2$ and $b = -4$ removes the last equation.
*if you want no solutions, $a=-2$ and $b \neq -4$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/567213",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 1,
"answer_id": 0
}
|
How show $8|[(\sqrt[3]{n}+\sqrt[3]{n+2})^3]+1$ let $n$ be postive integer numbers,and such $n>2$,show that
$$
8\,\,\left.\right\vert\,\,\left\lfloor%
\left(\vphantom{\Large A}\sqrt[3]{n\,} + \sqrt[3]{n + 2\,}\,\right)^{3}%
\right\rfloor + 1
$$
where $\left\lfloor x\right\rfloor$ is is the largest integer not greater than $x$,
My try:
since
$$(a+b)^3=a^3+b^3+3ab(a+b)$$
then
$$(\sqrt[3]{n}+\sqrt[3]{n+2})^3=n+n+2+3\sqrt[3]{n^2+2n}(\sqrt[3]{n}+\sqrt[3]{n+2})$$
so
$$[(\sqrt[3]{n}+\sqrt[3]{n+2})^3]=2n+2+[3\sqrt[3]{n^2+2n}(\sqrt[3]{n}+\sqrt[3]{n+2})]$$
then I can't,Thank you
|
Claim For $n >2$ we have
$$\left\lfloor%
\left(\vphantom{\Large A}\sqrt[3]{n\,} + \sqrt[3]{n + 2\,}\,\right)^{3}%
\right\rfloor =8n+7$$
This is equivalent to
$$8n+7 \leq
\left(\vphantom{\Large A}\sqrt[3]{n\,} + \sqrt[3]{n + 2\,}\,\right)^{3}< 8n+8 \Leftrightarrow \\
8n+7 \leq 2n+2+[3\sqrt[3]{n^2+2n}(\sqrt[3]{n}+\sqrt[3]{n+2})] < 8n+8 \Leftrightarrow \\
6n+5 \leq3\sqrt[3]{n^2+2n}(\sqrt[3]{n}+\sqrt[3]{n+2}) < 6n+6 $$
Hopefully this part is right: By Jensen
$$\sqrt[3]{n}+\sqrt[3]{n+2}< 2 \sqrt[3]{\frac{n+n+2}{2}}=2\sqrt[3]{n+1}$$
also
$$3\sqrt[3]{n^2+2n}<3\sqrt[3]{n^2+2n+1}=3\sqrt[3]{(n+1)^2}\,.$$
Multiplying these two inequalities yields the upperbound.
For the lowerbound we know
$$3\sqrt[3]{n^2+2n}(\sqrt[3]{n}+\sqrt[3]{n+2}) =3\sqrt[3]{n^3+2n^2}+3\sqrt[3]{n^3+4n^2+4n}$$
Now, I think you can prove:
$$3\sqrt[3]{n^3+4n^2+4n} \geq 3n+3.5$$
and for $n > 2$ we have
$$3\sqrt[3]{n^3+2n^2} > 3n+1.5$$
Much Simpler Solution
By AM-GM
$$\left( \sqrt[3]{n\,} + \sqrt[3]{n + 2\,} \right)^3 > 8 \sqrt{n^2+2n\,} >8n+7$$
with the last inequality being true for $n >3$. The case $n=3$ is trivial to check.
As Calvin pointed, by Jensen we have
$$\left(\vphantom{\Large A}\sqrt[3]{n\,} + \sqrt[3]{n + 2\,}\,\right)^{3} < \left(2\vphantom{\Large A}\sqrt[3]{n+1\,} \right)^{3} =8n+8$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/567745",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
}
|
Finding $\lim\limits_{x\to 0} (\sqrt{4+x}-1)^{1/(e^x-1)}$ I have to find the limit of $(\sqrt{4+x}-1)^{1/(e^x-1)}$ as $x\to0$ without de l'Hopital's rule and only with notable limits
|
$\ln \left( \sqrt{4+x}-1 \right)^{\frac{1}{e^x-1}}= \frac{x}{e^x-1} \frac{\ln \left(1+(\sqrt{4+x}-2)) \right)}{x} = \frac{x}{e^x-1} \frac{\ln \left(1+ \frac{x}{2+\sqrt{4+x}} \right)}{x} = \frac{x}{e^x-1} \frac{\ln \left(1+ \frac{x}{2+\sqrt{4+x}} \right)}{\frac{x}{2+\sqrt{4+x}}} \frac{1}{2+\sqrt{4+x}} \rightarrow 1 \cdot 1 \cdot \frac{1}{4}$
Therefore $\lim_{x\to 0} \left( \sqrt{4+x}-1\right)^\frac{1}{e^x-1} = e^\frac{1}{4}$
I used well known limits $\lim_{x\to 0} \frac{e^x-1}{x}=1=\lim_{x\to 0} \frac{\ln (1+x)}{x}$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/568299",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 3
}
|
Find a primitive of $x^2\sqrt{a^2 - x^2}$ I'm stuck on this problem for the last while now and any help would be appreciated. I need to find the indefinite integral of $$\int x^2\sqrt{a^2-x^2} dx$$
and show that it equals $$\frac x8(2x^2-a^2)\sqrt{a^2-x^2}+{a^4\over 8}\arcsin{x\over a}+c$$
I've tried using the substitution $x=a\sin t$ and $x=a\sec t$ Both have looked promising but I just can't seem to finish it out.
Thanks in advance for any help.
|
We will assume that $a>0$ for the following.
We have
$$\int x^2\sqrt{a^2-x^2} \,dx.$$
Let
$$
\begin{align*}
x&=a\sin{t} \\
dx &= a\cos{t} \, dt \\
a^2-x^2&=a^2-a^2\sin{t}\\
&=a^2\left( 1-\sin^2{t} \right)\\
&=a^2\cos^2{t}.
\end{align*}
$$
We substitute and integrate,
$$
\begin{align*}
&\int a^2\sin^2t \cdot \sqrt{a^2\cos^2t}\,\cdot a \cos t \,dt \\
=&a^4\int\sin^2 t \cos^2 t \, dt \\
=&a^4 \int\left( \sin t \cdot \cos t \right)^2 \, dt \\
=&a^4 \int\left( \frac{1}{2}\sin(2t) \right)^2 \, dt \\
=&\frac{a^4}{4}\int \sin^2(2t)\, dt \\
=& \frac{a^4}{4} \int \frac{1-\cos(4t)}{2} \, dt \\
=& \frac{a^4}{8} \int \left(1-\cos(4t)\right) \, dt \\
=& \frac{a^4}{8} \left( t-\frac{1}{4}\sin(4t) \right)+c.
\end{align*}
$$
The back substitution will be simpler if we have single angled trig solutions, and so we can reduce,
$$
\begin{align*}
\sin(4t) &= 2\sin(2t)\cos(2t) \\
&=2\left( 2\sin t \cdot \cos t \left( \cos^2 t- \sin^2 t \right)\right) \\
&= 4\sin t \cos^3 t-4\sin^3 t \cos t.
\end{align*}
$$
Hence our integral is
$$\frac{a^4}{8}\left( t- \sin t \cos^3 t + \sin^3 t \cos t \right)+c.$$
For the back substitution, we have that
$$x=a\sin t,$$
and so
$$t=\sin^{-1}\left(\frac{x}{a}\right).$$
For the remaining part, we draw a right triangle with angle $t$, opposite side $x$, hypotenuse $a$, and it follows that the adjacent side will be $\sqrt{a^2-x^2}$.
We use the definition of $t$ and read straight from the right triangle to back substitute,
\begin{align*}
& \frac{a^4}{8}\left( t- \sin t \cos^3 t + \sin^3 t \cos t \right) +c \\
=& \frac{a^4}{8}\left( \sin^{-1}\left(\frac{x}{a}\right)-\left(\frac{x}{a}\right)\left( \frac{\sqrt{a^2-x^2}}{a} \right)^3 +\left( \frac{x}{a} \right)^3\frac{\sqrt{a^2-x^2}}{a} \right) +c \\
=&\frac{a^4}{8}\left( \sin^{-1}\left(\frac{x}{a}\right) -\left(\frac{x}{a}\right)\frac{\sqrt{a^2-x^2}}{a}\left( \frac{a^2-x^2}{a^2}-\frac{x^2}{a^2} \right) \right) +c \\
=& \frac{a^4}{8}\left( \sin^{-1}\left(\frac{x}{a}\right)-\frac{x\sqrt{a^2-x^2}}{a^2}\left( \frac{a^2-2x^2}{a^2} \right) \right) +c \\
=&\frac{x}{8}\left( 2x^2-a^2 \right)\sqrt{a^2-x^2}+\frac{a^4}{8}\sin^{-1}\left(\frac{x}{a}\right)+c.
\end{align*}
This is the desired form.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/575933",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 6,
"answer_id": 0
}
|
Interesting determinant: Let $A$ be an $n$ by $n$ matrix with entries $a_{i,j}$ given that $a_{i,j}=2$ if $i=j$ Let $A$ be an $n$ by $n$ matrix with entries $a_{i,j}$ given that $a_{i,j}=2$ if $i=j$, $a_{i,j}=1$ if $i-j\equiv\pm2\pmod n$, and $a_{i,j}=0$ otherwise.
Find $\det A$.
It seems that the determinant is 4 for all odd $n$. What about even $n$?
|
Note that we have the degenerate cases $n=1, 2, 4$, where there are less than $3$ non-zero elements in each row/column.
When $n=1, 2, 4$ respectively, we get $\det(A)=2, 4, 9$.
Let us now consider $n \not =1, 2, 4$, so that each row/column has one $2$ and two $1$s and no other non-zero element.
Observe that $A$ is a circulant matrix, for which the determinant is easy to compute.
We may verify that $A$ has eigenvectors
$$\begin{pmatrix} 1 \\ \omega^j \\ \omega^{2j} \\ \ldots \\ \omega^{(n-1)j} \end{pmatrix}$$
for each $j=0, 1, \ldots , n-1$ where $\omega=e^{\frac{2 \pi i}{n}}$ is a $n$th root of unity. The corresponding eigenvectors are $2+\omega^{2j}+\omega^{(n-2)j}$.
Now the $n$ eigenvectors given above ($j=0, 1, \ldots , n-1$) are exactly the roots of the characteristic polynomial of $A$, so
$$\det(A-tI)=\chi_A(t)=(-1)^n\prod_{j=0}^{n-1}{(t-(2+\omega^{2j}+\omega^{(n-2)j}))}$$
Putting $t=0$ gives
\begin{align}
\det(A)& =\prod_{j=0}^{n-1}{(2+\omega^{2j}+\omega^{(n-2)j})} \\
&=\prod_{j=0}^{n-1}{(\omega^j+\omega^{-j})^2} \\
&=4\prod_{j=1}^{n-1}{(\omega^j+\omega^{-j})^2} \\
&=4\prod_{j=1}^{n-1}{\omega^{-2j}}\prod_{j=1}^{n-1}{(\omega^{2j}+1)^2} \\
&=4\omega^{-(n-1)n}\left(\prod_{j=1}^{n-1}{(\omega^{2j}+1)}\right)^2 \\
&=4\left(\prod_{j=1}^{n-1}{(\omega^{2j}+1)}\right)^2
\end{align}
It remains to compute $\prod_{j=1}^{n-1}{(\omega^{2j}+1)}$.
Notice that if $n$ is odd, then $\omega^2, \omega^4, \ldots , \omega^{2(n-1)}$ are simply $\omega, \omega^2, \ldots , \omega^{n-1}$ in some order. Furthermore, we know that $\omega^j$ are the roots of $\frac{x^n-1}{x-1}=x^{n-1}+\ldots +x+1$, so $$x^{n-1}+\ldots +x+1=\prod_{j=1}^{n-1}{(x-\omega^j)}$$
Putting $x=-1$ then gives
$$1=\prod_{j=1}^{n-1}{(-1-\omega^j)}=(-1)^{n-1}\prod_{j=1}^{n-1}{(1+\omega^j)}=(-1)^{n-1}\prod_{j=1}^{n-1}{(1+\omega^{2j})}$$
Thus when $n$ is odd, $\det(A)=4$.
When $n=2m$ is even, we have
$$\prod_{j=1}^{n-1}{(1+\omega^{2j})}=2\left(\prod_{j=1}^{m-1}{(1+\omega^{2j})}\right)^2$$
Now $\omega^{2j}$ are roots of $x^{m-1}+ \ldots +x+1$, so
$$x^{m-1}+ \ldots +x+1=\prod_{j=1}^{m-1}{(x-\omega^{2j})}$$
Putting $x=-1$ gives $$(-1)^{m-1}\prod_{j=1}^{m-1}{(1+\omega^{2j})}= \begin{cases} 1 & 2 \nmid m \\ 0 & 2\mid m \end{cases}$$
Thus $$\det(A)=4\left(2\left(\prod_{j=1}^{m-1}{(1+\omega^{2j})}\right)^2\right)^2=\begin{cases} 16 & 2\nmid m \\ 0 & 2 \mid m \end{cases}$$
All in all,
$$\det(A)= \begin{cases} 2 & n=1 \\ 4 & n=2 \\ 9 &n=4 \\ 4 & 2 \nmid n, n>1\\ 16 & n \equiv 2 \pmod{4}, n>2 \\ 0 & 4 \mid n, n>4 \end{cases}$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/576551",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
}
|
Showing that $\int_0^{\pi/4} \frac{1-\cos{16x}}{\sin{2x}}\,\mathrm{d}x=\frac{176}{105}$ Wolfram Alpha tells me that $$\int_0^{\pi/4} \frac{1-\cos{16x}}{\sin{2x}}\,\mathrm{d}x=\frac{176}{105}$$
What are some quick/elegant ways of proving this?
|
$$1-\cos(16x) = 2\sin^2(8x) = 8 \sin^2(4x) \cos^2(4x) = 32 \sin^2(2x) \cos^2(2x) \cos^2(4x)$$
Hence,
$$I = \int_0^{\pi/4}32 \sin(2x) \cos^2(2x) \cos^2(4x) dx = \int_0^{\pi/4}32 \sin(2x) \cos^2(2x) (2\cos^2(2x)-1)^2 dx$$
Now let $\cos(2x) = t$ and compute.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/577553",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
}
|
Integrate: $ \int_0^\infty \frac{\log(x)}{(1+x^2)^2} \, dx $ without using complex analysis methods Can this integral be solved without using any complex analysis methods: $$ \int_0^\infty \frac{\log(x)}{(1+x^2)^2} \, dx $$
Thanks.
|
Using twice the result that $\int_0^{\infty} \frac{\ln x}{1+x^2} d x=0$, we have
$$
\begin{aligned}
\int_0^{\infty} \frac{\ln x}{\left(1+x^2\right)^2} d x&=\int_0^{\infty} \frac{\left(1+x^2-x^2\right) \ln x}{\left(1+x^2\right)^2} d x \\
&=\int_0^{\infty} \frac{\ln x}{1+x^2} d x-\int_0^{\infty} \frac{x^2 \ln x}{\left(1+x^2\right)^2} d x \\
&=\frac{1}{2} \int_0^{\infty} x \ln x d\left(\frac{1}{1+x^2}\right) \\
&=\frac{1}{2} \int_0^{\infty} \frac{1-\ln x}{1+x^2} d x \\
&=\frac{1}{2} \int_0^{\infty} \frac{d x}{1+x^2} \\
&=\frac{\pi}{4}
\end{aligned}
$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/581155",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "21",
"answer_count": 12,
"answer_id": 11
}
|
How prove this $\frac{a}{(bc-a^2)^2}+\frac{b}{(ca-b^2)^2}+\frac{c}{(ab-c^2)^2}=0$ let $a,b,c\in \mathbb{R}$, if such
$$\dfrac{1}{bc-a^2}+\dfrac{1}{ca-b^2}+\dfrac{1}{ab-c^2}=0$$
show that
$$\dfrac{a}{(bc-a^2)^2}+\dfrac{b}{(ca-b^2)^2}+\dfrac{c}{(ab-c^2)^2}=0$$
Does this problem has nice methods?
My idea:let
$$(ca-b^2)(ab-c^2)+(bc-a^2)(ab-c^2)+(bc-a^2)(ca-b^2)=0$$
then I can't. Thanks
|
$$0 = \left( \dfrac{1}{bc-a^2}+\dfrac{1}{ca-b^2}+\dfrac{1}{ab-c^2} \right)\left( \dfrac{a}{bc-a^2}+\dfrac{b}{ca-b^2}+\dfrac{c}{ab-c^2} \right) =$$ $$ \dfrac{a}{(bc-a^2)^2}+\dfrac{b}{(ca-b^2)^2}+\dfrac{c}{(ab-c^2)^2} + \frac{a(ca-b^2) +a(ab-c^2) + b(bc-a^2) +b(ab-c^2)+c(bc-a^2)+c(ca-b^2)}{(bc-a^2)(ca-b^2)(ab-c^2)}$$ $$=\dfrac{a}{(bc-a^2)^2}+\dfrac{b}{(ca-b^2)^2}+\dfrac{c}{(ab-c^2)^2}$$
The last equation is true since the numerator of the big fraction is $0$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/581884",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 0
}
|
Solve $3^x + 28^x=8^x+27^x$ The equation $3^x + 28^x=8^x+27^x$ has only the solutions $x=2$ and $x=0$? If yes, how to prove that these are the only ones?
|
Since for all sufficiently large $x$ the expression $28^x + 3^x - 8^x - 27^x$ is positive and increasing, there cannot be roots $x$ beyond that point.
The question is how to pin down where the expression becomes positive and increasing.
Now $28^x$ grows fastest, by a factor of $28$ when $x$ is incremented by 1. This suggests a strategy of dividing through by $28^x$ to get a simpler expression to estimate.
Define $f(x) = 1 + \left(\frac{3}{28}\right)^x - \left(\frac{8}{28}\right)^x - \left(\frac{27}{28}\right)^x$. As noted in the Question, $f(2) = 0$. By standard calculus techniques:
$$ f'(x) = -\left( \log \frac{28}{3} \right)\left(\frac{3}{28}\right)^x
+ \left( \log \frac{7}{2} \right)\left(\frac{2}{7}\right)^x
+ \left( \log \frac{28}{27} \right)\left(\frac{27}{28}\right)^x $$
If $x \ge 2$, we can prove the derivative is positive:
$$ f'(x) = \left(\frac{2}{7}\right)^x \left[ \left(\log \frac{7}{2}\right)
- \left(\log \frac{28}{3}\right)\left(\frac{3}{8}\right)^x \right]
+ \left(\log \frac{28}{27}\right)\left(\frac{27}{28}\right)^x $$
$$ \ge \left(\frac{2}{7}\right)^x \left[ \left(\log \frac{7}{2}\right)
- \left(\log \frac{28}{3}\right)\left(\frac{3}{8}\right)^2 \right]
+ \left(\log \frac{28}{27}\right)\left(\frac{27}{28}\right)^x $$
and by direct computation the quantity in square brackets is positive, approx. $0.938664$.
Thus $f(x) \gt 0 $ for all $x \gt 2$. After checking $f(1)$ we know the only nonnegative integer solutions are the two identified in the Question, $x=0,2$.
As clarified, however, we need to determine all real solutions $x$. We can do this almost by inspection, calling upon an extension of Descartes Rule of Signs due to Laguerre(1883). Setting $z=3^x \gt 0$ we can rewrite the problem as finding positive real roots of the function:
$$ g(z) = z^{\log_3 28} - z^3 - z^{\log_3 8} + z $$
The number of sign changes is two, and by the version Thm. 3.1 in the PDF linked above, there are at most two positive real roots. Since we know $z=1,9$ are roots, these are the only positive real roots (from which it follows $x=0,2$ are the only real roots of the original problem).
It should be evident from this brief discussion that a much broader application of the principle can be made, e.g. like the generalization @IvanLoh has proved.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/581947",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
"answer_count": 4,
"answer_id": 1
}
|
$(a^2+b^2)\mid (ax+by)$, implies $\gcd(x^2+y^2,a^2+b^2)>1$ Suppose $a,b,x,y$ are natural number such that $(a^2+b^2)\mid (ax+by)$, prove that $\gcd(x^2+y^2,a^2+b^2)>1$. Can anyone give me a hint to solve the problem? I am rather stuck, thanks!
|
First notice that $$(a^2+b^2)(x^2+y^2)=(ax+by)^2+(bx-ay)^2$$
If $a=da_1$ and $b=db_1$ where $(a_1, b_1) = 1$ the equation can be rewritten as $$(a_1^2+b_1^2)(x^2+y^2)=(a_1x+b_1y)^2+(b_1x-a_1y)^2$$
Suppose now that $p$ is a prime such that $p \mid a_1^2+b_1^2$. Since we have $$d^2(a_1^2+b_1^2)=(a^2+b^2)\mid (ax+by)=d(a_1x+b_1y)\rightarrow d(a_1^2+b_1^2) \mid (a_1x+b_1y)$$ we would get that $p \mid a_1x + b_1y$ and hence from the equation it follows that $p \mid (b_1x-a_1y)$. So we would have that $$p \mid y(a_1x+b_1y) + x(b_1x-a_1y)=b_1(x^2+y^2)$$ Notice that $p \not \mid b_1$ since otherwise we would have that $p \mid b_1$ and from $p \mid a_1^2+b_1^2$ we would get that $p \mid a_1$ and that would effectively mean that $(a_1, b_1) \not = 1$. So we are left with the option of $p \mid x^2+y^2$. Since $p \mid d^2(a_1^2+b_1^2)= a^2+b^2$ we get that $(a^2+b^2, x^2+y^2)$ is at least $p$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/582991",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
}
|
Ring theory: Ideals being equal Question: Prove directly, without gcd computations, the following equalities of ideals.
(i) $(5, 7) = (1)$ in $\Bbb{Z}$ (of course (1) = $\Bbb Z$).
(ii) $(15, 9) = (3)$ in $Z$.
(iii) $(X^3 −1,X^2 −1) = (X −1) \text{ in } \Bbb{Q} [X]$
Attempted solution:
Try to show that $(1)$ is a subset of $(5,7)$ and that $(5,7)$ is a subset of $(1)$
First thing is $(5,7) = 5x + 7y$ with $x,y$ integers. Then by the closure under addition of the integers $(5,7)$ is a subset of $(1)$ as $(1)$ is the integers
Next to show $(5,7)$ is a subset of $(1)$ I try to say if $x$ belongs to integers, $x=5q + 7p, 1=5q + 7p$, then $x=x . 1= x(5q + 7 p)$ which is $5q' +7p'$, which shows that if $x$ is an integer then it's also in $(5,7)$
|
For (i) you have that $1 = 5x + 7y$ for some $x,y \in \Bbb{Z}$. This can be done using the Euclidean algorithm. One solution is $x = 3$, $y=-2$. Thus any integer $z$ can be written as $z = z\cdot1 = z\cdot(5\cdot3 - 14\cdot2)$, and we have $\Bbb{Z} = (1) = (5,7)$.
For (ii), we use basically the same process, but show that $3$ can be written as $15x+9y$, since both $15$ and $9$ can obviously be written as a multiple of $3$. Euclidean algorithm once again gives $x=2,y=-3$.
For (iii), we have that $X^3-1 = (X-1)(X^2+X+1)$, and $X^2-1 = (X-1)(X+1)$, thus $(X^3-1,X^2-1) \subset (X-1)$. For the other direction we want some $f,g \in \Bbb{Q}[X]$ such that $(X^3-1)f + (X^2-1)g = X-1$. We use the same Euclidean algorithm, but use polynomial division instead of integer division and get $f = 1$ and $g=-X$. This gives $(X-1) \subset (X^3 -1,X^2-1)$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/583793",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
}
|
How solve equation:$y^4+4y^2x-11y^2+4xy-8y+8x^2-40x+52=0$ Let $x,y\in \mathbb{R}$, solve this follow equation:$$y^4+4y^2x-11y^2+4xy-8y+8x^2-40x+52=0$$
My try: Since
$$8x^2+4x(y^2+y-10)+y^4-11y^2-8y+52=0$$
then I can't. Maybe this problem have other nice methods.Thank you
|
$$8x^2+x\left( 4y^2+4y-40\right) +y^4-11y^2-8y+52=0$$
This is quadratic in $x$, thus
$$x_{1,2}=\frac{-4y^2-4y+40\pm \sqrt{\left( 4y^2+4y-40\right)^2-32\left( y^4-11y^2-8y+52\right)}}{16}$$
Now, use
$$\left( 4y^2+4y-40\right)^2-32\left( y^4-11y^2-8y+52\right)=-16(y-2)^2(y+1)^2$$
to show that the equation has solutions if and only if $y \in \{-1,2\}$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/584154",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
}
|
Why can't I simply use algebra to solve this inequality? Consider the inequality:
$\frac{(x+3)(x-5)}{x(x+2)}\geq 0$
Why can't I simply multiply both sides by $x(x+2)$ and get $(x+3)(x-5)\geq 0$ ?
Which would yield: $x^2-2x-15\geq 0$ and I could then use the quadratic formula to derive the answer..?
This seems algebraically correct but I assume that is because I am misunderstanding something fundamental about inequalities.
|
if we assume that $x \gt 0$ then $(x+2) \gt 0$ so
$$\frac{(x+3)(x-5)}{x(x+2)} \ge 0 \Rightarrow (x+3)(x-5) \ge 0$$
but if $x \cdot (x+2) < 0$
$$\frac{(x+3)(x-5)}{x(x+2)} \ge 0 \Rightarrow (x+3)(x-5) \le 0$$
and if $x = 0$ or $x = -2$ then $\frac{(x+3)(x-5)}{x(x+2)}$ is undefined.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/584433",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 6,
"answer_id": 2
}
|
Compute $a(x)b(x)+c(x)$ in $\mathrm{GF}(2^4)$ where the irreducible generator polynomial $x^4+x+1$.
Let the coefficients of
$$a(x) = x^3+x^2+1,$$
$$b(x) = x^2+x+1,$$
$$c(x) = x^3+x^2+x+1$$
be in $\mathrm{GF}(2)$.
Compute $a(x)b(x)+c(x)$ in $\mathrm{GF}(2^4)$ where the irreducible generator polynomial $x^4+x+1$.
This is the exact question in a criptography lesson midterm.
I have calculated the $a(x)b(x)+c(x) = x^5 + 2x^4 + 3x^3 + 3x^2 + 2x + 1$. As I know in a $\mathrm{GF}(2^4)$, coefficients can't be greater than 15. So it is ok for a intermediate result. Now what to do with $x^4+x+1$?
Can you help me?
|
I assume that Pablo's interpretation of your question is correct, and answer accordingly.
I just added a self-answered question giving a useful discrete logarithm table. I rely heavily on that.
Below $\gamma$ will denote the coset $x+I$ in the quotient ring $GF(2)[x]/I$, where the ideal $I=\langle x^4+x+1\rangle$. That table then tells us that
$$
\begin{aligned}
a(x)+I&=x^3+x^2+1+I=\gamma^3+\gamma^2+1=\gamma^{13},\\
b(x)+I&=x^2+x+1+I=\gamma^2+\gamma+1=\gamma^{10},\\
c(x)+I&=x^3+x^2+x+1+I=\gamma^3+\gamma^2+\gamma+1=\gamma^{12}.
\end{aligned}
$$
Thus
$$
a(x)b(x)+I=\gamma^{13}\cdot\gamma^{10}=\gamma^{23}=\gamma^{15+8}=\gamma^8=\gamma^2+1.
$$
Therefore
$$
a(x)b(x)+c(x)+I=(\gamma^3+\gamma^2+\gamma+1)+(\gamma^2+1)=x^3+x+I.
$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/587000",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
}
|
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.