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Trying to simplify $\frac{\sqrt{8}-\sqrt{16}}{4-\sqrt{2}} - 2^{1/2}$ into $\frac{-5\sqrt{2}-6}{7}$ I'm asked to simplify $\frac{\sqrt{8}-\sqrt{16}}{4-\sqrt{2}} - 2^{1/2}$ and am provided with the solution $\frac{-5\sqrt{2}-6}{7}$
I have tried several approaches and failed. Here's one path I took:
(Will try to simplify the left hand side fraction part first and then deal with the $-2^{1/2}$ later)
$\frac{\sqrt{8}-\sqrt{16}}{4-\sqrt{2}}$
The root of 16 is 4 and the root of 8 could be written as $2\sqrt{2}$ thus:
$\frac{2\sqrt{2}-4}{4-\sqrt{2}}$
Not really sure where to go from here so I tried multiplying out the radical in the denominator:
$\frac{2\sqrt{2}-4}{4-\sqrt{2}}$ = $\frac{2\sqrt{2}-4}{4-\sqrt{2}} * \frac{4+\sqrt{2}}{4+\sqrt{2}}$ = $\frac{(2\sqrt{2}-4)(4+\sqrt{2})}{16-2}$ =
(I become less certain in my working here)
$\frac{8\sqrt{2}*2(\sqrt{2}^2)-16-4\sqrt{2}}{14}$ = $\frac{8\sqrt{2}*4-16-4\sqrt{2}}{14}$ = $\frac{32\sqrt{2}-16-4\sqrt{2}}{14}$ = $\frac{28\sqrt{2}-16}{14}$
Then add back the $-2^{1/2}$ which can also be written as $\sqrt{2}$
This is as far as I can get. I don't know if $\frac{28\sqrt{2}-16}{14}-\sqrt{2}$ is still correct or close to the solution. How can I arrive at $\frac{-5\sqrt{2}-6}{7}$?
| \begin{align}
\frac{\sqrt{8}-\sqrt{16}}{4-\sqrt{2}} - 2^{1/2} & = \frac{2\sqrt{2}-4}{4-\sqrt{2}}\cdot \frac{4+\sqrt{2}}{4+\sqrt{2}} - \sqrt{2} \\
& = \frac{4\sqrt{2}-12}{14} - \sqrt{2} \\
& = \frac{2\sqrt{2}-6}{7} - \sqrt{2} \\
& = \frac{2\sqrt{2}-6 -7 \sqrt{2}}{7}\\
& = \frac{-5\sqrt{2} -6 }{7}
\end{align}
| {
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Polynomial Long Division Confusion (simplifying $\frac{x^{5}}{x^{2}+1}$) I need to simplify \begin{equation}
\frac{x^{5}}{x^{2}+1}
\end{equation}
by long division in order to solve an integral.
However, I keep getting an infinite series:
\begin{equation}
x^{3}+x+\frac{1}{x}-\frac{1}{x^{3}}+...
\end{equation}
| Why wouldn't you get an infinite series?
If $x^2 + 1$ doesn't divide evenly into $x^5$ (which it doesn't) you will get a remainder. Just like with numbers with remainders if you try to continue you will get a decimal. Here if try to divide into the remainder you will get an expression with a negative power.
The thing is when you get a remainder that can be divided into any further.... you stop. And you let it be simply a remainder.
$\frac {x^5}{x^2+ 1} = \frac {x^5 + x^3}{x^2 + 1} -\frac {x^3}{x^2 + 1}$
$= x^3 - \frac {x^3 + x}{x^2 + 1} + \frac {x}{x^2+ 1} =$
$x^3 - x + \frac {x}{x^2 + 1}$.
Now we can't divide any further as the degree of the denominator ($x^2 + 1$) is $2$ and that is larger than the degree of that numerator ($x$). So we are done.
$\frac {x^5}{x^2 +1} = x^3 - x +\frac {x}{x^2 +1}$.
Put another way: $x^5 = (x^3 - x)(x^2 + 1) + x$.
$x$ is .... just a remainder you cant do any thing with.
It is exactly like.
$\frac {249}{7} = \frac {210 + 39}{7} = \frac {210}7 + \frac {39}7=$
$30 + \frac {35 + 4}{7} = 30 + \frac {35}7 + \frac 47=$
$30 + 5 + \frac 47 = 35\frac 47$.
We've divided as far as we can go.
If you tried to go further we would get decimals:
$30 + 5 + \frac {40}{7*10} = 30 + 5 + {35 + 5}{70} = $
$30 + 5 + \frac 5{10} + \frac 5{70} =30 + 5 + \frac 5{10} + \frac {50}{700} =$
$30 + 5 + \frac 5{10} + \frac 7{100} + \frac 1{1000} + ......$.
$= 35.571428571428571428571428571429.....$
But we weren't asked to go that for and as we aren't masochists.... we stopped at $35\frac 47$.
$
| {
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What is the expected number of randomly generated numbers in the range [a, b] required to reach a sum $\geq X$? We are generating random numbers (integers) in the range $[a, b]$. All values are equally likely.
We will continue to generate random numbers in this range, and add up successive values until their combined sum is greater than or equal to a set number $X$.
What is the expected number of rolls to reach at least $X$?
Example:
a = 1000
b = 2000
X = 5000
Value 1: 1257 (total sum so far = 1257)
Value 2: 1889 (total sum so far = 3146)
Value 3: 1902 (total sum so far = 5048; all done)
So it took $3$ rolls to reach $\geq5000$. Intuitively, we can say that it will not take more than $5$ rolls if each roll is $1000$. We can also say that it will not take less than $3$ rolls if each roll was $2000$. So it stands to reason that in the example above, the expected number of rolls lies somewhere between $3$ and $5$.
How would this be solved in the general case for arbitrary values $[a, b]$ and $X$? It's been quite a while since I last took statistics, so I've forgotten how to work with discrete random variables and expected value.
| The expected number is dominated by the term based on the mean $\frac{2X}{a+b}$ - but we can do better than that. As $X\to\infty$,
$$E(n) = \frac{2X}{a+b}+\frac{2(a^2+ab+b^2)}{3(a^2+2ab+b^2)}+o(1)$$
Why? Consider the excess. In the limit $X\to\infty$, the frequency of hitting a particular point approaches a uniform density of $\frac{2}{b+a}$. For points in $[X,X+a)$, this is certain to be the first time we've exceeded $X$, so the density function for that first time exceeding $X$ is approximately $\frac{2}{b+a}$ on that range. For $t\in[X+a,X+b)$, the probability that it's the first time exceeding $X$ is $\frac{X+b-t}{b-a}$; we need the last term added to be at least $t-X$. Multiply by $\frac{2}{b+a}$, and the density function for the first time exceeding $X$ is approximately $\frac{2(X+b-t)}{b^2-a^2}$ on $[X+a,X+b)$. For $t<X$ or $t\ge X+b$, of course, the density is zero. Graphically, this density is a rectangular piece followed by a triangle sloping to zero. Now, the expected value of the first sum to be greater than $X$ is approximately
$$\int_{X}^{X+a}\frac{2t}{b+a}\,dt+\int_{X+a}^{X+b}\frac{2t(X+b-t)}{b^2-a^2}\,dt$$
$$=\frac{2a(2X+a)}{2(b+a)}+\frac{2(b-a)(2X+a+b)(X+b)}{2(b^2-a^2)}-\frac{2(b-a)((X+a)^2+(X+a)(X+b)+(X+b)^2)}{3(b^2-a^2)}$$
$$=\frac{6aX+3a^2+6X^2+3(a+3b)X+3ab+3b^2-6X^2-6(a+b)X-2(a^2+ab+b^2)}{3(b+a)}$$
$$=\frac{3(a+b)X+a^2+ab+b^2}{3(b+a)}=X+\frac{a^2+ab+b^2}{3(a+b)}$$
Divide by the mean $\frac{a+b}{2}$ of each term in the sum, and we have an asymptotic expression for $E(n)$.
This is not a rigorous argument. If I wanted to do that, I'd bring out something like the moment-generating function.
| {
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"timestamp": "2023-03-29T00:00:00",
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If $ f(x) = x^3+\frac{3x}{4}-\frac{3x^2}{2}+\frac{7}{8}.$ Then $ \int^{\frac{3}{4}}_{\frac{1}{4}}f(f(x))dx$
If $\displaystyle f(x) = x^3+\frac{3x}{4}-\frac{3x^2}{2}+\frac{7}{8}.$ Then $\displaystyle \int^{\frac{3}{4}}_{\frac{1}{4}}f(f(x))dx$
Try: Given $\displaystyle f(x) =x^3+\frac{3x}{4}-\frac{3x^2}{2}+\frac{7}{8}. $
Then $\displaystyle f(1-x) = (1-x)^3+\frac{3}{4}(1-x)-\frac{3}{2}(1-x)^2+\frac{7}{8}$
$$f(1-x) = -x^3+\frac{3x^2}{2}-\frac{3x}{4}+\frac{9}{8}$$
$$f(x)+f(1-x) = 2$$
Let $$I = \int^{\frac{3}{4}}_{\frac{1}{4}}f(f(x))dx = \int^{\frac{3}{4}}_{\frac{1}{4}}f(f(1-x))dx$$
I did not understand how to solve from there
could some help me to solve it
| Hint
Note that $$f(x)={(x-{1\over 2})^3}+1$$and $$\int^{\frac{3}{4}}_{\frac{1}{4}}f(f(x))dx=\int^{\frac{1}{4}}_{-\frac{1}{4}}f(f(u+{1\over 2}))du=\int^{\frac{1}{4}}_{-\frac{1}{4}}f(u^3+1)du$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Induction proof for stirling of first kind. I have to show that for every stirling number of the first kind $\forall n \geq 2 : s_{n,n-2} = \frac{1}{24}n(n-1)(n-2)(3n-1) $ is true.
I've started like this:
Base case: Let $n$ be $n=2$, then per defintion
$s_{n,0} = 0$.
Since we have $s_{2,2-2} = s_{2,0} = 0$ and $\frac{1}{24}2(2-1)(2-2)(3*2-1) = \frac{1}{24}2(2-1)(0)(3*2-1) = 0$ the base case is true.
Now we assume $\forall n \in \mathbb{N} $ with $ n \geq 2$
that $s_{n,n-2} = \frac{1}{24}n(n-1)(n-2)(3n-1) $ is true.
We also now, that there exists a recursive formula for the stirling numbers of the first kind which will help us for the induction.
It says that $s_{n,k} = s_{n-1,k-1}+(n-1)s_{n-1,k} $.
Now let $n \rightarrow n+1$.
$s_{n+1,n-1} = s_{n,n-2}+n*s_{n,n-1} $.
If we insert our assumption, that leaves us with
$s_{n+1,n-1} = \frac{1}{24}n(n-1)(n-2)(3n-1)+n*s_{n,n-1} $.
If we insert $n=n+1$ into $\frac{1}{24}n(n-1)(n-2)(3n-1)$ we get
$\frac{1}{24}n+1(n)(n-1)(3n)$, so that's what we would like after the induction.
The big problem now is that I can'T get rid of $s_{n,n-1}$. I've seen a similar question for the stirling numbers of second kind which simply say that $S_{n,n-1} = $$n \choose 2$ but I don't know if this is true for the first kind and neither did I understand how we come to that conclusion. Can anyone help me?
| Induction step:
$\begin{aligned}s\left(n+1,n-1\right) & =ns\left(n,n-1\right)+s\left(n,n-2\right)\\
& =\frac{1}{2}n^{2}\left(n-1\right)+\frac{1}{24}n\left(n-1\right)\left(n-2\right)\left(3n-1\right)\\
& =\frac{1}{24}n\left(n-1\right)\left[12n+\left(n-2\right)\left(3n-1\right)\right]\\
& =\frac{1}{24}n\left(n-1\right)\left(n+1\right)\left(3n+2\right)\\
& =\frac{1}{24}\left(n+1\right)n\left(n-1\right)\left(3n+2\right)
\end{aligned}
$
In the second equality we use the equality $s(n,n-1)=\frac12n(n-1)$ and also the induction hypothesis.
| {
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Calculate $\lim_{x\to 0}\left(\frac{3\sin(x)-3x \cos(x)}{x^3}\right)^\frac{1}{x}$ This is an exercise from today's exam and I think I did something wrong:
Calculate the limit:
$$\lim_{x\to 0}\left(\frac{3\sin(x)-3x\cos(x)}{x^3}\right)^{1/x}.$$
To tackle this problem I used equivalent infinitesimals: $1-\cos(x) \sim \frac{x^2}{2}$ as $(x\to 0)$ and $\sin(x)\sim x$ as $(x\to 0)$.
Then our limit:
$$\begin{align}
\lim_{x\to 0}\left(\frac{3\sin(x)-3x\cos(x)}{x^3}\right)^{1/x}&=\lim_{x\to 0}\left(\frac{3x-3x\cos(x)}{x^3}\right)^{1/x}\\
&=\lim_{x\to 0}\left(\frac{3x(1-\cos(x))}{x^3}\right)^{1/x}\\
&=\lim_{x\to 0}\left(\frac{3x \frac{x^2}{2}}{x^3}\right)^{1/x} \\
&=\lim_{x\to 0}\left(\frac{3}{2}\right)^{1/x}
\end{align}$$
Now I split the limit to see what the effect of negative $x$ will be:
$\lim_{x\to 0^+}\left(\frac{3}{2}\right)^{1/x}=+\infty$
$\lim_{x\to 0^-}\left(\frac{3}{2}\right)^{1/x}=0$
I graphed the function and the limit seems to be 1 from both sides.
| $$3\sin{x}-3x\cos{x}=3\left(x-\frac{x^3}{6}+\frac{x^5}{120}-...\right)-3x\left(1-\frac{x^2}{2}+\frac{x^4}{24}-...\right)=$$
$$=x^3-\frac{1}{10}x^5+...,$$
which says that your limit is equal to $1$.
| {
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Find all positive triples of positive integers a, b, c so that $\frac {a+1}{b}$ , $\frac {b+1}{c}$, $\frac {c+1}{a}$ are also integers. Find all positive triples of positive integers a, b, c so that $\frac {a+1}{b}$ , $\frac {b+1}{c}$, $\frac {c+1}{a}$ are also integers.
WLOG, let a$\leqq b\leqq c$,
| If $a \le b$ but $b|a+1$ then $b \le a+1$ so either $a = b$ or $b=a+1$.
If $a = b$ then $b|b+1$ and $a=b=1$. The you $c|b+1=2$ and $b|c+1$. So either $c=1$ or $c = 2$.
So so far we have $(1,1,1)$ or $(1,1,2)$
If $b = a+1$ then $c|a+2$ so $c \le a+2$ but $a < b=a+1 \le c\le a+2 $ so either $c = b = a+1$ or $c= a+2$.
If $c= b =a+1$ then we have $b|a+1 = b$ and $c=b|b+1$ and $a=b-1|c+1=b+1$. So $b=1$ but then $a=0$ and that's a contradiction.
If $c=a+2$ and $b= a+1$ we have: $b|a+1=b$; $c|b+1 =c$ and $a|c+1 = a+3$. This means $a|3$ and $a =1$ or $a =3$.
So we have $(1,2,3)$ or $(3,4,5)$.
To double check:
$(a,b,c) = (1,1,1)\implies $ $b|a+1, c|b+1, a|c+1$ all translate to $1|2$. which is true
$(a,b,c) = (1,1,2) \implies b=1|a+1=2; c=2|b+1=2; a=1|c+1=3$. All true.
$(a,b,c) = (1,2,3)\implies b=2|a+1=2; c=3|b+1=3; a=1|c+1 = 4$. All true.
$(a,b,c) = (3,4,5)\implies b=4|a+1=4; c=5|b+1=5; a=3|c+1=6$. All true.
| {
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Find the sum $\sqrt{5+\sqrt{11+\sqrt{19+\sqrt{29+\sqrt{41+\cdots}}}}}$ Okay so this can be written as $$\sqrt{5+\sqrt{(5+6)+\sqrt{(5+6+8)+\sqrt{(5+6+8+10)+\sqrt{(5+6+8+10+12)\cdots}}}}}$$
Putting it as $y$ and squaring both sides doesn't seem to help, and I don't know what else can be done.
| Maybe works, $$3=\sqrt{3^{2}}=\sqrt{5+4}=\sqrt{5+\sqrt{16}}=\sqrt{5+\sqrt{11+5}}$$
$$=\sqrt{5+\sqrt{11+\sqrt{25}}}=\sqrt{5+\sqrt{11+\sqrt{19+6}}}=\sqrt{5+\sqrt{11+\sqrt{19+\sqrt{36}}}}$$
$$=\sqrt{5+\sqrt{11+\sqrt{19+\sqrt{29+7}}}}=\sqrt{5+\sqrt{11+\sqrt{19+\sqrt{29+\sqrt{49}}}}}=\sqrt{5+\sqrt{11+\sqrt{19+\sqrt{29+\sqrt{41+8}}}}}=\ldots$$
This is just a beautiful way of writing 3.
| {
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Why is $(x^{2n+1} - (2n+1)x^{n+1} + (2n+1)x^n - 1)/(x-1)^3$ irreducible? Consider the following polynomial
$$
f_n(x)=x^{2n+1} - (2n+1)x^{n+1} + (2n+1)x^n - 1
$$
Try the first $n$, I find that the $(x-1)^3$ is its factor:
$$
f_1(x)=(x-1)^3,\\
f_2(x)=(x-1)^3 (x^2+3x+1),\\
f_3(x)=(x-1)^3 (x^4+3x^3+6x^2+3x+1),\\
\cdots
$$
and that this is the complete factorization of $f_n(x)$ in $\mathbb{Z}[x]$.
But how can you prove that factoring out $(x-1)^3$ gives the complete factorization over $\mathbb{Z}$? That is, how can you prove that $f_n(x)/(x-1)^3$ is irreducible over $\mathbb{Z}$?
PS: I use Mathematica to test $n\leqslant 1000$, all is correct.
| Here is a proof in the special case that $2n+1$ is prime (and $n>1$).
Let $g_n(x)=f_n(x)/(x-1)^3$ and let $h_n(x)=g_n(x+1)$. By dividing $f_n(x)$ by $x-1$ three times using long division, we find $$g_n(x)=\binom{2}{2}x^{2n-2}+\binom{3}{2}x^{2n-3}+\dots+\binom{n}{2}x^n+\binom{n+1}{2}x^{n-1}+\binom{n}{2}x^{n-2}+\dots+\binom{3}{2}x+\binom{2}{2}.$$ Now note that $$f_n(x)-(x-1)^{2n+1}=x^{2n+1} - 1 - (x-1)^{2n+1} + (2n+1)(x^n-x^{n+1})$$ is divisible by $2n+1$ since every term of $x^{2n+1}-1-(x-1)^{2n+1}$ is divisible by $2n+1$ (the coefficients have the form $\pm\binom{2n+1}{k}$ for $0<k<2n+1$ which is divisible by $2n+1$ since it is prime). It follows that $h_n(x)-x^{2n-2}$ is also divisible by $2n+1$. That is, every term except the leading term of $h_n(x)$ is divisible by $2n+1$. Moreover, the constant term of $h_n(x)$ is
\begin{align*}
g_n(1) &= \binom{2}{2}+\binom{3}{2}+\dots+\binom{n}{2}+\binom{n+1}{2}+\binom{n}{2}+\dots+\binom{3}{2}+\binom{2}{2} \\
&=\binom{n+2}{3}+\binom{n+1}{3} \\
&=\frac{n(n+1)(2n+1)}{6}.
\end{align*}
(Alternatively, $g_n(1)$ can be computed as $f_n'''(1)/6$.) In particular, the constant term of $h_n(x)$ is not divisible by $(2n+1)^2$. So, by Eisenstein's criterion for the prime $2n+1$, $h_n(x)$ is irreducible over $\mathbb{Z}$, and hence so is $g_n(x)=h_n(x-1)$.
(This approach sometimes also works when $2n+1$ is not prime; for instance, $h_4(x)$ turns out to still satisfy Eisenstein's criterion for the prime $3$. However, $h_7(x)$ does not satisfy Eisenstein's criterion for any prime.)
| {
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Probability of $2$ pairs on $5$ dice We throw $5$ dice. What is the probability of getting $2$ pairs ?
My solution says it is $$\frac{6\cdot 5\cdot 4\cdot 5! }{6^5\cdot 2\cdot 2!\cdot 2!},$$ where as for me it's $$\frac{6\cdot 5\cdot 4\cdot 5!}{6^5\cdot 2!\cdot 2!}.$$
I do as follows: Throwing $5$ dice is the same thing as throwing one die $5$ times. What we want is $AABBC$.
We have $6$ possibilities for $A$, $5$ possibilities for $B$ and 4 possibilities for C. Since the order doesn't count we have to multiply by $\frac{5!}{2!2!}$. At the end, I get
$$\frac{6\cdot 5\cdot 4 \cdot 5!}{6^5\cdot 2!\cdot 2!}.$$
What's wrong in my argument?
| Let's take a simple example: two dice and you want them to be different values. As a check, this is the probability they are not the same so is $1-\dfrac16=\dfrac56$
Taking your approach, you would say the probability should be $\dfrac{6\cdot 5 \cdot 2!}{6^2\cdot 1!\cdot 1!} = \dfrac{10}{6}$, which is clearly wrong. The problem is that you have double counted the possibilities. For example you have treated choosing $A=3, B=4$ as distinct from choosing $A=4, B=3$, but then have counted the ordered pattern $A,B$ as distinct from $B,A$; so $3,4$ gets counted twice (as does $4,3$)
There are two ways of avoiding this problem (you should only use one of them, as they are alternatives)
*
*You could say there are ${6 \choose 2}=\dfrac{6 \times 5}{2}$ ways of choosing $A$ and $B$
*Having chosen $A$ and $B$, you could divide the number of orderings by $2!$ since $A$ must appear before $B$
This transfers across to your particular question and would get you the correct answer
| {
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Is it true that $\forall n \in \Bbb{N} : (\sum_{i=1}^{n} a_{i} ) (\sum_{i=1}^{n} \frac{1}{a_{i}} ) \ge n^2$ , if all $a_{i}$ are positive?
If $\forall i \in \Bbb{N}: a_{i} \in \Bbb{R}^+$ , is it true that $\forall n \in \Bbb{N} : \big(\sum_{i=1}^{n}a_{i}\big) \big(\sum_{i=1}^{n}
\frac{1}{a_{i}}\big) \ge n^2$ ?
I have been able to prove that this holds for $n=1$ , $n=2$, and $n=3$ using the following lemma:
Lemma 1: Let $a,b \in \Bbb{R}^+$. If $ab =1$ then $a+b \ge 2$
For example, the case for $n=3$ can be proven like this:
Let $a,b,c \in \Bbb{R}^+$. Then we have:
$(a+b+c)\big(\frac{1}{a} + \frac{1}{b} + \frac{1}{c}\big) = 1 + \frac{a}{b} + \frac{a}{c} + \frac{b}{a} + 1 + \frac{b}{c} + \frac{c}{a} + \frac{c}{b} + 1 $
$= 3 + \big(\frac{a}{b} + \frac{b}{a}\big) + \big(\frac{a}{c} + \frac{c}{a}\big) + \big(\frac{b}{c} + \frac{c}{b}\big) $
By lemma 1, $\big(\frac{a}{b} + \frac{b}{a}\big) \ge 2$, $ \big(\frac{a}{c} + \frac{c}{a}\big) \ge 2$ and $\big(\frac{b}{c} + \frac{c}{b}\big) \ge 2$ , therefore:
$3 + \big(\frac{a}{b} + \frac{b}{a}\big) + \big(\frac{a}{c} + \frac{c}{a}\big) + \big(\frac{b}{c} + \frac{c}{b}\big) \ge 3 + 2 + 2 +2 = 9 = 3^2 \ \blacksquare $
However I'm not sure the generalized version for all natural $n$ is true. I can't come up with a counterexample and when I try to prove it by induction I get stuck.
Here is my attempt:
Let $P(n)::\big(\sum_{i=1}^{n}a_{i}\big) \big(\sum_{i=1}^{n} \frac{1}{a_{i}}\big) \ge n^2$
Base case: $\big(\sum_{i=1}^{1}a_{i}\big) \big(\sum_{i=1}^{1} \frac{1}{a_{i}}\big) = a_{1} \frac{1}{a_{1}} = 1 = 1^2$ , so $P(1)$ is true.
Inductive hypothesis: I assume $P(n)$ is true.
Inductive step:
$$\left(\sum_{i=1}^{n+1}a_{i}\right) \left(\sum_{i=1}^{n+1} \frac{1}{a_{i}}\right) = \left[\left(\sum_{i=1}^{n}a_{i}\right) + a_{n+1}\right] \left[\left(\sum_{i=1}^{n} \frac{1}{a_{i}}\right) + \frac{1}{a_{n+1}}\right]$$
$$=\left(\sum_{i=1}^{n}a_{i}\right) \left[\left(\sum_{i=1}^{n} \frac{1}{a_{i}}\right) + \frac{1}{a_{n+1}}\right] + a_{n+1} \left[\left(\sum_{i=1}^{n} \frac{1}{a_{i}}\right) + \frac{1}{a_{n+1}}\right]$$
$$=\left(\sum_{i=1}^{n}a_{i}\right)\left(\sum_{i=1}^{n} \frac{1}{a_{i}}\right) +\left(\sum_{i=1}^{n}a_{i}\right) \frac{1}{a_{n+1}} + a_{n+1} \left(\sum_{i=1}^{n} \frac{1}{a_{i}}\right) +a_{n+1} \frac{1}{a_{n+1}}$$
$$=\left(\sum_{i=1}^{n}a_{i}\right)\left(\sum_{i=1}^{n} \frac{1}{a_{i}}\right) +\left(\sum_{i=1}^{n}a_{i}\right) \frac{1}{a_{n+1}} + a_{n+1} \left(\sum_{i=1}^{n} \frac{1}{a_{i}}\right) +1 $$
$$\underbrace{\ge}_{IH} n^2 + \left(\sum_{i=1}^{n}a_{i}\right) \frac{1}{a_{n+1}} + a_{n+1} \left(\sum_{i=1}^{n} \frac{1}{a_{i}}\right) + 1$$
And here I don't know what to do with the $\big( \sum_{i=1}^{n}a_{i} \big) \frac{1}{a_{n+1}} + a_{n+1} \big(\sum_{i=1}^{n} \frac{1}{a_{i}}\big)$ term.
Is this inequality true? If it is, how can I prove it? If it isn't, can anyone show me a counterexample?
| HINT:
The Cauchy-Schwarz Inequality
\begin{eqnarray*}
(x_1^2+ \cdots +x_n^2) (y_1^2+ \cdots +y_n^2) \geq (x_1 y_1+ \cdots +x_n y_n)^2
\end{eqnarray*}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3081320",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 0
} |
Find the area of parallelogram and its missing vertex Given three radius-vectors: $OA(5; 1; 4), OB(6;2;3), OC(4;2;4)$, find the missing vertex $D$ and calculate the area of obtained parallelogram.
My attempt:
Firstly, we are to find the vectors which form the parallelogram. $$AB = OA - OB = (-1; -1; 1)$$
$$AC=OA-OC=(1; -1; 0)$$
Since the parallelogram is spanned by $AB, AC$, cross product and the norm of resulting vector should be then calculated.
\begin{align*}
&AB \times AC=
{\begin{vmatrix}
\vec{i} & \vec{j} & k \\
-1 & -1 & 1 \\
1 & -1 & 0
\end{vmatrix}} = \sqrt{1^2 + 2^2 + 1^2} = \sqrt{6}
\end{align*}
$D$ is equidistant from $E$ and $B$, where $E$ is the point of diagonal's intersection.
$$x_E = \frac{x_A+x_C}{2} = 9/2$$
$$y_E = \frac{y_A+y_C}{2} = 3/2$$
$$z_E = \frac{z_A+z_C}{2} = 4$$
\begin{cases} x_B - x_E = x_E - x_D\\ y_B - y_E = y_E - y_D \\ z_B - z_E = z_E - z_D \end{cases}
After plugging and solving, the final answer would be $\sqrt{6}$ and $D(3;1;5).$
| A simpler way to obtain $D$ is $D = C+AB$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3081821",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"answer_id": 0
} |
What's the answer to $\int \frac{\cos^2x \sin x}{\sin x - \cos x} dx$? I tried solving the integral $$\int \frac{\cos^2x \sin x}{\sin x - \cos x}\, dx$$ the following ways:
*
*Expressing each function in the form of $\tan \left(\frac{x}{2}\right)$, $\cos \left(\frac{x}{2}\right)\,$ and $\,\sin \left(\frac{x}{2}\right)\,$ independently, but that didn't go well for me.
*Multiplying and dividing by $\cos^2x$ or $\sin^2x$.
*Expressing $\cos^2x$ as $1-\sin^2x$ and splitting the integral, and I was stuck with $\int \left(\frac{\sin^3x}{\sin x - \cos x}\right)\, dx$ which I rewrote as $\int \frac{\csc^2x}{\csc^4x (1-\cot x) } dx,\,$ and tried a whole range of substitutions only to fail.
*I tried to substitute $\frac{1}{ \sin x - \cos x}$, $\frac{\sin x}{ \sin x - \cos x}$, $\frac{\cos x \sin x}{ \sin x - \cos x}$ and $\frac{\cos^2x \sin x}{\sin x - \cos x},$ independently, none of which seemed to work out.
*I expressed the denominator as $\sin\left(\frac{\pi}{4}-x\right)$ and tried multiplying and dividing by $\sin\left(\frac{\pi}{4}+x\right)$, and carried out some substitutions. Then, I repeated the same with $\cos\left(\frac{\pi}{4}+x\right)$. Neither of them worked.
| try it with $$\sin(x)=\frac{2t}{1+t^2}$$
$$\cos(x)=\frac{1-t^2}{1+t^2}$$
$$dx=\frac{2dt}{1+t^2}$$
You will get this integral $$\int \frac{4 t
\left(t^2-1\right)^2}{\left(t^2+1
\right)^3 \left(t^2+2 t-1\right)}dt$$
and this is $$\int \left(1/2\,{\frac {-t+1}{{t}^{2}+1}}+{\frac {-4\,t+4}{ \left( {t}^{2}+1
\right) ^{3}}}+1/2\,{\frac {t+1}{{t}^{2}+2\,t-1}}+{\frac {2\,t-4}{
\left( {t}^{2}+1 \right) ^{2}}}\right)
dt$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3082128",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 6,
"answer_id": 3
} |
Cauchy Schwarz - Finding minimum of a function For $x, y , z$ in real numbers, If $2x+y+z=5$, then what is the min value of $x^2 + y^2 + z^2 - 2x + 4y + 6$. This is a weekend brain teaser in the 2nd week of Calc 3.
| This problem can be solved by using elementary math only (no derivatives, no Lagrange multipliers, no C-S ineqaulity etc).
We have to minimize the expression:
$$f=x^2+y^2+z^2-2x+4y+6=(x-1)^2+(y+2)^2+z^2+1$$
Introduce substitution:
$$u=x-1, \ \ v=y+2, \ \ w=z$$
The expression that has to be minimized now becomes:
$$f=u^2+v^2+w^2+1\tag{1}$$
...with the following constraint:
$$2u+v+w=5\tag{2}$$
Make one more subsitution:
$$v=x-y, \ \ w=x+y$$
Replace that into (1) and (2) and you get:
$$f=u^2+2x^2+2y^2+1\tag{3}$$
...and the constraint becomes:
$$2u+2x=5\tag{4}$$
The point is: contraint (4) give us the possibility to minimize (3) by simply choosing $y=0$. Now the problem is much simpler. We have to minimize:
$$f=u^2+2x^2+1\tag{5}$$
...with constraint:
$$u+x=\frac52\tag{6}$$
From (6):
$$x=\frac 52-u$$
...so (7) becomes:
$$f=u^2+2(\frac 52 - u)^2+1=3(u-\frac53)^2+\frac{31}{6}$$
So the minimum value is $f_{min}=\frac{31}{6}$ and it is reached for $u=\frac53$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3082755",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 0
} |
$\left\lfloor \frac{a-b}{2} \right\rfloor + \left\lceil \frac{a+b}{2} \right\rceil = a$ when $a,b$ are integers? Let $a$ and $b$ be positive integers.
If $b$ is even, then we have $$\left\lfloor \frac{a-b}{2} \right\rfloor + \left\lceil \frac{a+b}{2} \right\rceil = a$$
I think the equality also hold when $b$ is odd. What could be a proof for it?
| As $s:=a+b$ and $a-b$ have the same parity, we can write
$$\left\lfloor\frac{s-2b}2\right\rfloor+\left\lceil\frac s2\right\rceil=\left\lfloor\frac s2\right\rfloor+\left\lceil\frac s2\right\rceil-b=s-b=a.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3087071",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 5,
"answer_id": 3
} |
If $\cos^6 (x) + \sin^4 (x)=1$, find $x$ if $x\in [0, \dfrac {\pi}{2}]$ If $\cos^6 (x) + \sin^4 (x)=1$, find $x$ if $x\in [0, \dfrac {\pi}{2}]$
My attempt:
$$\cos^6 (x) + \sin^4 (x)=1$$
$$\cos^6 (x) + (1-\cos^2 (x))^{2}=1$$
$$\cos^6 (x) + 1 - 2\cos^2 (x) + \cos^4 (x) = 1$$
$$\cos^6 (x) + \cos^4 (x) - 2\cos^2 (x)=0$$
| Let $\cos x = c, \sin x = s$.
$c^6 + s^4 = 1$
$c^6 + s^4 = c^2 + s^2$
$c^2(c^4-1)+s^2(s^2-1) = 0$
$c^2(c^4-1)-s^2c^2=0$
$c^2(c^4-1-s^2) = 0$
$c^2((c^2+1)(c^2-1)-s^2) = 0$
$-s^2c^2(c^2+2) = 0$
The only real zeroes occur when $\sin x = 0$ or $\cos x = 0$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3087249",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 3
} |
What is the value of $\operatorname{arctan} \left(-\frac{12}{5}\right)$? I am trying to find the value of $\sin(\operatorname{arctan}(-12/5))$ manually (without a calculator). I know I need to solve the inner portion I need to find the angle at which $\operatorname{tan}(\theta) = -12/5$. But this is not one of the easy angles to simply look up the value.
I also know it is a $5, 12, 13$ right triangle, both because I recognize the numbers and because $x^2 +y^2 = r^2$ and $r = 13$. But I don't know how to find that angle and can't find a good example.
| Let $\theta = \arctan \frac {-12}{5}$
$\tan \theta = \frac {-12}{5} = \frac {\sin \theta}{\cos \theta}$ which means there there is a right triangle with angle $\theta$ and the opposite side = $r\sin \theta = -12$ ($-12$ meaning extends below the $x$ axis) and an adjacent side $r \cos \theta = 5$ where $r$ is the hypotenuse.
So $\sin \theta = \frac {-12}r$ where $r$ is the hypotenuse = $\sqrt {(-12)^2 + 5^2} = 13$.
So $\sin \theta = \frac {-12}{13}$.
....
In general:
$\sin (\arcsin \frac ab) = \frac {a}{\sqrt{a^2 + b^2}}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3090234",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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} |
Find all integer solutions to $ x^3 - y^3 = 3(x^2 - y^2) $ The objective is to find all solutions to
$$ x^3 - y^3 = 3(x^2 - y^2) $$
where $x,y \in \mathbb{Z}$.
So far I've got one pair of solution. Try $(x, y)=(0,0)$:
$$ 0^3-0^3=3(0^2-0^2) \\
0=0 \qquad \text{equation satisfied}$$
Another try $ (x, y) = (x, x)$ then
$$ x^3 - x^3 = 3(x^2 -x^2) \\
0 = 0 \qquad \text{equation satisfied}$$
But the above are just particular solutions. To find all $(x, y)$ pairs, I tried:
$$\begin{aligned}
(x-y)(x^2+xy+y^2) &= 3(x+y)(x-y) \\
x^2+xy+y^2 &= 3(x+y) \end{aligned}\\
\begin{aligned}
x^2+xy+y^2-3x-3y &= 0 \\
x^2+x(y-3)+y(y-3) &=0 \\
x^2 + (y-3)(x+y) &= 0 \end{aligned}$$
What now? Am I on the right track?
| So first you factorize the expression:
$$(x-y)(x^2+xy+y^2-3x-3y)=0$$
$$(x-y)(x^2+x(y-3)+(y^2-3y))=0$$
When the product of two expressions are $0$ either of them is $0$
So first factor is $(x-y)=0$ giving the solution pair $x=y$ so for all such pairs the equation will be 0.
Second factor can be treated as a quadratic equation in terms of either $x$ or $y$,
$$x^2+x(y-3)+(y^2-3y)=0$$
Where you consider each $y$ as a constant and solve the equation to get:
$$x=\frac{-(y-3)\pm\sqrt{(y-3)^2-4(y^2-3y)}}{2}$$
$$x=\frac{-(y-3)\pm\sqrt{3(9-y^2)}}{2}$$
For $x$ to be real the quantity inside the square root should be positive, moreover you need the integer solutions for $x$ and $y$ so the only possibility that remains is if the discriminant becomes a perfect square, So $y=0,\pm1,\pm2,\pm3$ but you also want it to be a perfect square so only $y=\pm3$ which gives:
$$x = 0,3 $$
Final solutions:
All integers for which $x=y$,
$(0,3)$,
$(3,-3)$
You can also solve the equation for $y$, hope this helps .....
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3090345",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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On a multisection of a series Let $x$ be a real number and $\lfloor{x}\rfloor$ denote the greatest integer less than or equal to $x$. Let $r$ be a positive integer. Let $a$ and $n$ be nonnegative integers such that $n-a+1\geq r$. Prove that
$$ \sum_{k=a}^{n} f(k) = \sum_{j=0}^{r-1} \sum_{k=\lfloor{\frac{a+r-1-j}{r}}\rfloor}^{\lfloor{\frac{n-j}{r}}\rfloor} f(rk+j).$$
I couldn't arrive at any proof.
Also it is noted that for a fixed $r$, this identity partitions the sum modulo remainder classes. The number of remainder classes is indexed by $j$. Could you explain this note?
I generally have trouble with series whose indices involves greatest integer function, are there any fast techniques to solve them?
| This is an answer based on series manipulation. In order to ease things we make a two-step approach. At first we show the identity is valid in case $a=0$ and then the generalisation for integral $a\geq 0$ follows easily.
Step 1: $a=0$
The following is valid for integral $1\leq r\leq n$:
\begin{align*}
\sum_{k=0}^nf(k)=\sum_{j=0}^{r-1}\sum_{k=0}^{\left\lfloor\frac{n-j}{r}\right\rfloor}f(rk+j)\tag{1}
\end{align*}
Observe the simple lower limit $k=0$ of the inner sum. Setting $a=0$ we have
\begin{align*}
\left\lfloor\frac{a+r-1-j}{r}\right\rfloor=\left\lfloor\frac{r-1-j}{r}\right\rfloor=0
\end{align*}
since $0\leq \frac{r-1-j}{r}<1$ if $0\leq j<r$.
At first we consider (1) when $n$ is an integer multiple of $r$. In this case we see
\begin{align*}
\sum_{k=0}^nf(k)&=\sum_{k=0}^{\frac{n}{r}-1}\sum_{j=0}^{r-1}f(rk+j)+f(n)\\
&=\sum_{j=0}^{r-1}\sum_{k=0}^{\frac{n}{r}-1}f(rk+j)+f(n)\tag{2}
\end{align*}
The multisection in (2) corresponds to a partitioning of $[0,n]\cap\mathbb{Z}$ into $r$ equally sized, right half-open intervals $$[rk,rk+r)\cap\mathbb{Z}, \qquad\qquad 0\leq k<\frac{n}{r}$$ of length $\frac{n}{r}$ together with the right-most point $\{n\}$.
Now we consider $0\leq r\leq n$. In this more general setting we have
\begin{align*}
n=r\left\lfloor\frac{n}{r}\right\rfloor+l\qquad\qquad 0\leq l <\left\lfloor\frac{n}{r}\right\rfloor\tag{3}
\end{align*}
We obtain from (2) and (3)
\begin{align*}
\color{blue}{\sum_{k=0}^n f(k)}&=\sum_{j=0}^{r-1}\sum_{k=0}^{\left\lfloor\frac{n}{r}\right\rfloor-1}f(rk+j)
+\sum_{j=0}^l f\left(r\left\lfloor\frac{n}{r}\right\rfloor+j\right)\tag{4}\\
&=\sum_{j=0}^{l}\sum_{k=0}^{\left\lfloor\frac{n}{r}\right\rfloor}f(rk+j)
+\sum_{j=l+1}^{r-1}\sum_{k=0}^{\left\lfloor\frac{n}{r}\right\rfloor-1}f(rk+j)\tag{5}\\
&=\sum_{j=0}^{r-1}\,\,\ \sum_{k=0}^{\left\lfloor\frac{n}{r}\right\rfloor+\left\lfloor\frac{l-j}{r}\right\rfloor}f(rk+j)\tag{6}\\
&=\sum_{j=0}^{r-1}\sum_{k=0}^{\left\lfloor\frac{r\lfloor n/r\rfloor +l-j}{r}\right\rfloor} f(rk+j)\tag{7}\\
&\,\,\color{blue}{=\sum_{j=0}^{r-1}\sum_{k=0}^{\left\lfloor\frac{n-j}{r}\right\rfloor} f(rk+j)}\tag{8}\\
\end{align*}
and the claim (1) follows.
Comment:
*
*In (4) we use $\left\lfloor\frac{n}{r}\right\rfloor$ instead of $\frac{n}{r}$ and instead of the single value $f(n)$ we have now to consider a sum ranging over $0\leq j\leq l$ according to (3) resulting in the right-hand sum.
*In (5) we split the left-hand of (4) into two double sums with index region $0\leq j\leq l$ and $l<j\leq r-1$. This way we can merge the right-hand sum of (4) into the left-hand sum. Note the upper limit of the inner sum from the left-hand sum is $\left\lfloor\frac{n}{r}\right\rfloor$ (and no longer $\left\lfloor\frac{n}{r}\right\rfloor-1$).
*In (6) we collect both double sums of (5) into one double sum with a small trick. We add
\begin{align*}
\left\lfloor\frac{l-j}{r}\right\rfloor=
\begin{cases}
0&\quad 0\leq j\leq l\\
-1&\quad l< j\leq r-1
\end{cases}
\end{align*}
*In (7) we observe $m:=\left\lfloor\frac{n}{r}\right\rfloor$ is an integer and we use the identity
$$m+\left\lfloor\frac{l-j}{r}\right\rfloor=\left\lfloor\frac{rm+l-j}{r}\right\rfloor.$$
*In (8) we use $n=r\left\lfloor\frac{n}{r}\right\rfloor+l$ from (3).
Step 2: $a\geq 0$
We obtain from (8)
\begin{align*}
\color{blue}{\sum_{k=a}^nf(k)}
&=\sum_{k=0}^n f(k)-\sum_{k=0}^{a-1} f(k)\\
&=\sum_{j=0}^{r-1}\sum_{k=0}^{\left\lfloor\frac{n-j}{r}\right\rfloor} f(rk+j)
-\sum_{j=0}^{r-1}\sum_{k=0}^{\left\lfloor\frac{a-1-j}{r}\right\rfloor} f(rk+j)\\
&=\sum_{j=0}^{r-1}\ \sum_{k=\left\lfloor\frac{a-1-j}{r}\right\rfloor+1}^{\left\lfloor\frac{n-j}{r}\right\rfloor} f(rk+j)\\
&\,\,\color{blue}{=\sum_{j=0}^{r-1}\ \sum_{k=\left\lfloor\frac{a+r-1-j}{r}\right\rfloor}^{\left\lfloor\frac{n-j}{r}\right\rfloor} f(rk+j)}\\
\end{align*}
and OPs claim follows.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Evaluate $\int_{0}^{1}\frac{1+x+x^2}{1+x+x^2+x^3+x^4}dx$ Evaluate $$I=\int_{0}^{1}\frac{(1+x+x^2)}{1+x+x^2+x^3+x^4}dx$$
My try:
We have:
$$1+x+x^2=\frac{1-x^3}{1-x}$$
$$1+x+x^2+x^3+x^4=\frac{1-x^5}{1-x}$$
So we get:
$$I=\int_{0}^{1}\frac{1-x^3}{1-x^5}dx$$
$$I=1+\int_{0}^{1}\frac{x^3(x^2-1)}{x^5-1}dx$$
Any idea from here?
| Your integral $$ \int_{0}^{1}\frac{1-x^3}{1-x^5}dx$$ is similar to that of the digamma function $$-\gamma+\int_{0}^{1}\frac{1-t^s}{1-t}dt=\psi(s + 1) $$Maybe the change of variable $t=x^5$ could be helpful.
| {
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"url": "https://math.stackexchange.com/questions/3090916",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
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How to find the value $\sum\limits_{n=1}^{\infty}\frac{(-1)^{n}}{2n-1}$ converges to? We have been finding the value of sums related to the Harmonic series by using that
$\sum_{n=1}^{N}\frac{1}{n}= ln(N)+\gamma + \varepsilon_N$.
When I try to calculate $\sum_{n=1}^{\infty}\frac{(-1)^{n}}{2n-1}$ using that method there is a point where I don´t know how to continue.
I have read that its value is $\frac{\pi}{4}$ but I don´t understand how is that obtained and I would like to know if it´s possible to solve it using the method I know.
This is what I did:
$S_N=\sum_{n=1}^{N}\frac{(-1)^n}{2n-1}$
$S_{2N}=\sum_{n=1}^{2N}\frac{(-1)^n}{2n-1}=-1+\frac{1}{3}-\frac{1}{5}+...-\frac{1}{4N-3}+\frac{1}{4N-1}=$
$=1+\frac{1}{3}+\frac{1}{5}+...+\frac{1}{4N-1}-2(1+\frac{1}{5}+\frac{1}{9}+...+\frac{1}{4N-3})=\sum_{n=1}^{2N}\frac{1}{2n-1}-2\sum_{n=1}^{N}\frac{1}{4n-3}$
.
$\sum_{n=1}^{2N}\frac{1}{2n-1}=1+\frac{1}{3}+\frac{1}{5}+...+\frac{1}{4N-3}+\frac{1}{4N-1}=$
$=1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{4N-1}+\frac{1}{4N}-(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{4N})=\sum_{n=1}^{4N}\frac{1}{n}-\frac{1}{2}\sum_{n=1}^{2N}\frac{1}{n}$
.
But I don´t know what to do with the second one. I tried to write it like this:
$\sum_{n=1}^{N}\frac{1}{4n-3}=1+\frac{1}{5}+\frac{1}{9}+...+\frac{1}{4N-3}=$
$=1+\frac{1}{3}+\frac{1}{5}+...+\frac{1}{4N-3}+\frac{1}{4N-1}-(\frac{1}{3}+\frac{1}{7}+...\frac{1}{4N-1})=\sum_{n=1}^{2N}\frac{1}{2n-1}-\sum_{n=1}^{N}\frac{1}{4n-1}$
Wich didn´t help because I find $\sum_{n=1}^{N}\frac{1}{4n-1}$ as difficult to solve as $\sum_{n=1}^{N}\frac{1}{4n-3}$ and I have no idea of what else to do.
| Indeed, this sum requires more than playing with the asymptotic development of the harmonic series.
I suggest that you consider “instead” $$S_n=\sum_{k=0}^n{\frac{(-1)^n}{2n+1}}=\int_0^1{\sum_{k=0}^n{(-x^2)^n}\,dx},$$ and then prove that $$\left|S_n-\int_0^1{\frac{1}{1+x^2}\,dx}\right| \leq n^{-1}.$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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perfect square of summation of odd numbers This is not homework , and I am old enough to be your father :-).We know the summation of odd numbers results in perfect squares , like $1 + 3 = 4 $, $1+ 3 + 5 = 9$ and so on.
My question is , if we start from a random odd number (say $315$) , how many odd numbers after $315$ results in a perfect square ? For what odd number $(2n-1)$ does :- $315 + 317 + 319 + .....(2n-1) = x^2 $ occur for $ n,x \in \mathbb{N} $ , and $(2n-1)> 315$ ?
| Lets suppose that the random odd number in which we stop is $2k+1$. We known that the sum:
$$1+3+5+\cdots +(2k+1) = k^2$$
Consider two cases:
Case 1: k is even
If this is the case then que can write $k = 2n$. Follows to see that $k^2 = (2n)^2$ and,
$$(2n)^2+(n^2-1)^2 = (n^2+1)^2$$
We know that:
$$1+3+4+\cdots+(2(n^2+1)+1)$$
Hence is follows that:
$$(n^2+1)^2-(2n)^2 = (2k+3) + (2k+5)+\cdots + (2(n^2+1)+1) = (n^2-1)^2$$
To sum up: if k is even, then the sum from $2k+3$ to $2(\frac{k^2}{4} + 1)+1$ going only from odd to odd gives a perfect square.
Case 2: k is odd
We can express k as $2n+1$ and then it can easily be seen that:
$$(2n+1)^2+(2n^2+2n)^2 = (2n^2+2n+1)^2$$
Hence, by the same reasoning than before:
$$(2k+3)+(2k+5)+\cdots + (2(2n^2+2n+1)+1) = (2n^2+2n)^2$$
Meaning that the sum that goes from $2k+3$ to $2\left(2\left(\frac{k-1}{2}\right)^2+2\left(\frac{k-1}{2}\right)+1\right)+1$ is a perfect square.
This provides an answer for any odd number, but it is not the smallest possible answer. I will try to figure out how to get the smallest possible one.
Hope I was helpful
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Find limit (type 0/0) I'm struggling to find the limit
$$I = \lim_{x\to 0}\frac{2\sqrt{1-x} - \sqrt[3]{8-x}}{x}$$
What I was trying:
$$ I = \lim_{x\to 0}\frac{1-x + 2\sqrt{1-x} + 1 - (1-x) - 1 - \sqrt[3]{8-x}}{x} $$
$$ = \lim_{x\to 0}\frac{(\sqrt{1-x}+1)^2 - (2-x)- \sqrt[3]{8-x}}{x} \qquad \quad $$
$$ = \lim_{x\to 0}\frac{(\sqrt{1-x}+1)^2 - (2-x+ \sqrt[3]{8-x})}{x} \qquad \qquad $$
$$ = \lim_{x\to 0}\frac{(\sqrt{1-x}+1)^2}{x} - \lim_{x\to 0}\frac{(2-x+ \sqrt[3]{8-x})}{x} \qquad $$
Thank all of you for your answers.
| I would substitute $$a=\sqrt[3]{8-x}$$ then you term is given by $$\frac{2\sqrt{a^3-7}-a}{8-a^3}$$ and then multiply numerator and denominator by $$2\sqrt{a^3-7}+a$$ and then you will get
$$\frac{(a-2)(14+7a+4a^2)}{-(a-2)(a^2+2a+4)(2\sqrt{a^3-7}+a)}$$
| {
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"question_score": "2",
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How to solve $2x(4-x)^{\frac{-1}{2}}-3\sqrt{4-x}=0$ for $x$? I'm struggling to figure out how to solve $2x(4-x)^{\frac{-1}{2}}-3\sqrt{4-x}=0$ for $x$.
The answer is $x = \frac{12}{5}$, but I am getting $x = \frac{-35}{9}$
My steps are:
$2x(4-x)^{\frac{-1}{2}}-3\sqrt{4-x}=0$
*
*make exponent postive
$2x\frac{1}{\sqrt{4-x}} - 3\sqrt{4-x} = 0$
*
*square everything
$4x^2\frac{1}{4-x} - 9(4-x) =0$
*
*combine into one fraction
$\frac{1-9(4-x)[4x^2(4-x)]}{4x^2(4-x)} = 0$
*
*remove like terms
$1-9(4-x)=0$
$1-36-9x = 0$
$x = \frac{-35}{9}$
| This step is wrong
square everything
$4x^2\frac{1}{\sqrt{4-x}} - 9(4-x) =0$
Once you square everything you get (i.e. $a=b$, then $a^2=b^2$, then $a^2-b^2=0$)
$$4x^2\frac{1}{4-x} - 9(4-x) =0$$
Now multiply by $(4-x)$, we get
$$4x^2 - 9(4-x)^2 =0$$
or
$$(2x)^2 - [3(4-x)]^2 =0$$
which gives
$$(2x -12 + 3x)(2x +12 - 3x)=0$$
Finally we get
$$(5x -12 )(12 - x)=0$$
Obviously $12=x$ is not accepted, but $x = \frac{12}{5}$ is.
| {
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"timestamp": "2023-03-29T00:00:00",
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Factorise polynomial with real and complex roots How would you go about finding the roots of the polynomial:
$$x^4 +5x^3+4x^2+6x-4=0.$$
I attempt to form two quadratics e.g.
$$(x^2+ax+b)(x^2+cx+d)$$
and then tried to expand, collect like terms, and solve for the coefficients simultaneously, but this results in a hard system of equations.
I was wondering if there was a better approach to this question.
Any help would be greatly appreciated.
| I like the following way.
For all real value of $k$ we have:
$$x^4+5x^3+4x^2+6x-4=\left(x^2+\frac{5}{2}x+k\right)^2-\left(\left(2k+\frac{9}{4}\right)x^2+(10k-6)x+k^2+4\right).$$
Now, we'll choose a value of $k$, for which $2k+\frac{9}{4}>0$ and $$(5k-3)^2-\left(2k+\frac{9}{4}\right)(k^2+4)=0.$$
Easy to see that $k=0$ is valid.
Thus,$$x^4+5x^3+4x^2+6x-4=\left(x^2+\frac{5}{2}x\right)^2-\left(\frac{9}{4}x^2-6x+4\right)=$$
$$=\left(x^2+\frac{5}{2}x\right)^2-\left(\frac{3}{2}x-2\right)^2=(x^2+x+2)(x^2+4x-2).$$
Can you end it now?
| {
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"timestamp": "2023-03-29T00:00:00",
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How many ways can 26 letters be divided into 13 pairs? The question is: a plug board has 26 letters, and there are 13 cables. The cables connect all possible pair of letters.
How many possible configurations does the plug board provide? In other words, how many ways can 26 letters be divided into 13 pairs?
The result I found by calculation and some logic is:
$$25\cdot 23\cdot 21\cdot 19\cdot 17\cdot 15\cdot 13\cdot 11\cdot 9\cdot 7\cdot 5\cdot 3 = 7 905 853 580 625.$$
Is this correct?
How is the mathematical algorithm and formula to calculate this and similar problems?
| Another approach is: the first pair can be selected in ${26\choose 2}$ ways, but we are not interested in the position of the pair among the $13$ pairs, hence we divide it by ${13\choose 1}$:
$$\frac{{26\choose 2}}{{13\choose 1}}$$
Similarly, for other pairs.
At the end we get:
$$\frac{{26\choose 2}}{{13\choose 1}}\cdot \frac{{24\choose 2}}{{12\choose 1}}\cdot\frac{{22\choose 2}}{{11\choose 1}}\cdot\frac{{20\choose 2}}{{10\choose 1}}\cdot \frac{{18\choose 2}}{{9\choose 1}}\cdot\frac{{16\choose 2}}{{8\choose 1}}\cdot\frac{{14\choose 2}}{{7\choose 1}}\cdot\frac{{12\choose 2}}{{6\choose 1}}\cdot\frac{{10\choose 2}}{{5\choose 1}}\cdot\frac{{8\choose 2}}{{4\choose 1}}\cdot\frac{{6\choose 2}}{{3\choose 1}}\cdot\frac{{4\choose 2}}{{2\choose 1}}\cdot\frac{{2\choose 2}}{{1\choose 1}},$$
which is what you have.
| {
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"timestamp": "2023-03-29T00:00:00",
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Inequality $\frac{a_1}{1^2}+\frac{a_2}{2^2}+...+\frac{a_n}{n^2}\ge\frac{1}{1}+\frac{1}{2}+...+\frac{1}{n}$ Suppose $a_i$ are dinstinct positive integers $\forall1\le i\le n$. Prove that
$$\frac{a_1}{1^2}+\frac{a_2}{2^2}+...+\frac{a_n}{n^2}\ge\frac{1}{1}+\frac{1}{2}+...+\frac{1}{n}$$
My approach:
I will procede by proof by contradiction. If $a_1,a_2,...,a_n$ is not a permutation of $1,2,...,n$, then one can reduce $a_k (\exists a_k>n)$ to a value $<n$ and $>1$ So that $a_1,a_2,...,a_n$ is a permutation of $1,2,...,n$. From now on assume $a_1,a_2,...,a_n$ is a permutation of $1,2,...,n$. Now, by the rearrangement inequality, $$\frac{a_1}{1^2}+\frac{a_2}{2^2}+...+\frac{a_n}{n^2}\ge\frac{1}{1^2}+\frac{2}{2^2}+...+\frac{n}{n^2}=\text{LHS}$$ with the rearrangement sequence being $\{a_1,a_2,...,a_n\} \text{and} \{\frac{1}{1^2},\frac{1}{2^2},...,\frac{1}{n^2}\}$.
Is this proof correct? Is there a better way of doing it?
| I think your proof is correct, but I think it's better to get it without contradiction.
$(1,2,...,n)$ and $\left(\frac{1}{1^2},\frac{1}{2^2},...,\frac{1}{n^2}\right)$ have an opposite ordering.
Thus, by Rearrangement
$$\frac{a_1}{1^1}+\frac{a_2}{2^2}+...+\frac{a_n}{n^2}\geq\frac{a'_1}{1^1}+\frac{a'_2}{2^2}+...+\frac{a'_n}{n^2}\geq$$
$$\geq\frac{1}{1^1}+\frac{2}{2^2}+...+\frac{n}{n^2}=1+\frac{1}{2}+...+\frac{1}{n}.$$
Here $(a_1',a_2',...,a_n')$ is a permutation of $(1,2,...,n)$.
| {
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Prove that $\lim_{(x,y)\to (0,0)}\frac{x^{2}+xy+y^{2}}{\sqrt{x^{2}+y^{2}}}=0$.
Prove that $\displaystyle \lim_{(x,y)\to (0,0)}\frac{x^{2}+xy+y^{2}}{\sqrt{x^{2}+y^{2}}}=0$.
This has been my rough work so far, but I am not sure how to go further or if I am doing it the wrong way...
$\displaystyle|f(x,y)-L|=\left |\frac{x^{2}+xy+y^{2}}{\sqrt{x^{2}+y^{2}}} \right | \leq \frac{|x^{2}+xy+y^{2}|}{\big|\sqrt{x^{2}+y^{2}}\big|}\cdot \frac{\sqrt{x^{2}+y^{2}}}{\sqrt{x^{2}+y^{2}}} = \frac{\big(\sqrt{x^{2}+y^{2}}\big)|x^{2}+xy+y^{2}|}{|x^{2}+y^{2}|}$
I wanted the square root on top to help determine what I should set my $\delta$ as. Any ideas on how to finish this or do it in a better way?
| You may also procced as follows using GM-QM (inequality between geometric and quadratic mean):
*
*$\sqrt{ab}\leq \sqrt{\frac{a^2+b^2}{2}}$
So, you get
\begin{eqnarray*} \left| \frac{x^2+xy+y^2}{\sqrt{x^2+y^2}}\right|
& = & \left| \frac{x^2+y^2}{\sqrt{x^2+y^2}} + \frac{xy}{\sqrt{x^2+y^2}}\right| \\
& \leq & \sqrt{x^2+y^2} + \frac{|xy|}{\sqrt{x^2+y^2}}\\
& \stackrel{GM-QM}{\leq} & \sqrt{x^2+y^2} + \frac{\left(\sqrt{\frac{x^2+y^2}{2}} \right)^2}{\sqrt{x^2+y^2}} \\
& = & \frac{3}{2}\sqrt{x^2+y^2} \\
& \stackrel{(x,y)\to (0,0)}{\longrightarrow} & 0
\end{eqnarray*}
| {
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How to show $\sum_{i=1}^{\lceil L/2\rceil}\frac{\binom{2i-2}{i-1}}{i}2^{-2i}=1/2-2^{-L-1}\binom{L}{(L-1)/2}$? By numerical evaluation, it seems that the following identity holds. For any odd positive integer $L$,
$$\sum_{i=1}^{\lceil L/2\rceil}\frac{\binom{2i-2}{i-1}}{i}2^{-2i}=1/2-2^{-L-1}\binom{L}{(L-1)/2}.$$
I was trying to expand everything out but didn't have a good luck. I guess there should be a more clever way using some binomial identity.
| When looking at OPs identity the left-hand side smells like a telescoping sum where we have something like the first and last entry at the right hand side, and indeed we can transform it accordingly. We start with some preliminary arrangements. Shifting the index $i$ by one to start with $i=0$ we want to show
\begin{align*}
\sum_{i=0}^{\lceil L/2\rceil -1}\frac{1}{i+1}\binom{2i}{i}2^{-2i-2}=\frac{1}{2}-\frac{1}{2^{L+1}}\binom{L}{\frac{L-1}{2}}\tag{1}
\end{align*}
Since $L$ is assumed to be an odd positive integer, we set $L=2K+1$ in (1), multiply both sides with $4$ and show
The following is valid for non-negative integer $K$:
\begin{align*}
\sum_{i=0}^K\frac{1}{4^i(i+1)}\binom{2i}{i}=2-\frac{1}{4^K}\binom{2K+1}{K}\tag{2}
\end{align*}
We obtain
\begin{align*}
\color{blue}{\sum_{i=0}^K}&\color{blue}{\frac{1}{4^i(i+1)}\binom{2i}{i}}\\
&=1+\sum_{i=1}^K\frac{1}{4^i}\left(\binom{2i}{i}-\binom{2i}{i-1}\right)\tag{3}\\
&=1+\sum_{i=1}^K\frac{1}{4^i}\left(4\binom{2i-1}{i-1}-\binom{2i+1}{i}\right)\tag{4}\\
&=1+\sum_{i=1}^K\frac{1}{4^{i-1}}\binom{2i-1}{i-1}-\sum_{i=1}^K\frac{1}{4^i}\binom{2i+1}{i}\tag{5}\\
&=1+\sum_{i=0}^{K-1}\frac{1}{4^{i}}\binom{2i+1}{i}-\sum_{i=1}^K\frac{1}{4^i}\binom{2i+1}{i}\tag{6}\\
&=1+1-\frac{1}{4^K}\binom{2K+1}{K}\tag{7}\\
&\,\,\color{blue}{=2-\frac{1}{4^K}\binom{2K+1}{k}}
\end{align*}
and OPs claim follows.
Comment:
*
*In (3) we use the binomial identity $\frac{1}{i+1}\binom{2i}{i}=\binom{2i}{i}-\binom{2i}{i-1}$ which might be familiar in context with the Catalan numbers.
*In (4) we use the identity
\begin{align*}
\binom{2i}{i}-\binom{2i}{i-1}&=2\binom{2i}{i}-\binom{2i}{i}-\binom{2i}{i-1}\\
&=2\binom{2i}{i}-\binom{2i+1}{i}\\
&=4\binom{2i-1}{i-1}-\binom{2i+1}{i}
\end{align*}
*In (5) we split the sum.
*In (6) we shift the index of the left-hand sum by one to start with $i=0$.
*In (7) we cancel terms due to the telescoping sum.
| {
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Formula for partial sum of binoms: $\sum_{k=0}^{\lfloor n/2 \rfloor} \binom nk \binom mk$ I am dealing with some expressions containing combinatoric numbers. Does anybody know a formula for this?
$$\displaystyle\sum_{k=0}^{\left\lfloor \dfrac{n}{2} \right\rfloor} \binom{n}{k}\binom{m}{k}$$
| Not simple and we need to consider three cases since we do not know the parity of $n$ and what $\left[ \dfrac{n}{2} \right]$ means (floor or ceiling function).
*
*Suppose that
$n$ is odd $(n=2p+1)$ and that $\left[ \dfrac{n}{2} \right]=\left\lfloor \frac{n}{2}\right\rfloor=p$, the expression is
$$S_p=\sum_{k=0}^{p} \binom{2p+1}{k}\binom{m}{k}$$ and the result is given by $$S_p=\frac{\Gamma (m+2 p+2)}{\Gamma (m+1) \Gamma (2 (p+1))}-\binom{2 p+1}{p+1}
\binom{m}{p+1} \, _3F_2(1,-p,-m+p+1;p+2,p+2;1)$$
*Suppose that $n$ is odd $(n=2p+1)$ and that $\left[ \dfrac{n}{2} \right]=\left\lceil \frac{n}{2}\right\rceil=p+1$, the expression is
$$S_p=\sum_{k=0}^{p+1} \binom{2p+1}{k}\binom{m}{k}$$ and the result is given by
$$S_p=\frac{\Gamma (m+2 p+2)}{\Gamma (m+1) \Gamma (2 (p+1))}-\binom{2 p+1}{p+2}
\binom{m}{p+2} \, _3F_2(1,1-p,-m+p+2;p+3,p+3;1)$$
*If $n$ is even $(n=2p)$, whatever $\left[ \dfrac{n}{2} \right]$ could mean, the expression is
$$S_p=\sum_{k=0}^{p} \binom{2p}{k}\binom{m}{k}$$ and the result is given by
$$S_p=\frac{\Gamma (m+2 p+1)}{\Gamma (m+1) \Gamma (2 p+1)}-\binom{2 p}{p+1} \binom{m}{p+1}
\, _3F_2(1,1-p,-m+p+1;p+2,p+2;1)$$ In all cases, appears the generalized hypergeometric function.
| {
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"timestamp": "2023-03-29T00:00:00",
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recurrence relation limit $x_{n+2}=(a+b)\cdot x_{n+1}-ab\cdot x_{n}$ Let $0<b<a$ and $(x_{n})_{n\in \mathbb{N}}$ with $x_{0}=1, \ x_{1}=a+b$
$$x_{n+2}=(a+b)\cdot x_{n+1}-ab\cdot x_{n}$$
a) If $0<b<a$ and $L=\lim_{n\rightarrow \infty }\frac{x_{n+1}}{x_{n}}$ then $L= ?$ ($L=a$ right answer )
b) If $0<b<a<1$ and $L=\lim_{n\rightarrow \infty }\sum_{k=0}^{n}x_{k}$ then $L= ?$ ($L=\frac{1}{(1-a)(1-b)}$ right answer)
My try:
First, I tried to write $x_{2},x_{3}...$ but I didn't get too far.I obtained $x_{2}=(a+b)^{2}-ab$ and $x_{3}=(a+b)^{2}-ab-ab(a+b)$.
I tried to write the characteristic equation like:
$x_{n+2}-(a+b)x_{n+1}-abx_{n}=0$ so the equation is $r^{2}-(a+b)r-ab=0$.The discriminant is positive so $x_{n}=c\cdot r_{1}^{n}+d\cdot r_{2}^{n}$ where $c$ and $d$ are the solutions of the system: $\begin{cases}
c\cdot r_{1}^{1}+d\cdot r_{2}^{1}=x_{1} \\
c\cdot r_{1}^{2}+d\cdot r_{2}^{2}=x_{2}
\end{cases}$ but the calculations are too complicated and I get stuck.
| Hint: Try rewriting your recurrence as follows:
$$
x_{n+2} = (a+b)x_{n+1} - abx_n \\
x_{n+2} = ax_{n+1}+bx_{n+1} - abx_n \\
x_{n+2} - bx_{n+1} = a(x_{n+1} - bx_n)
$$
So you get:
$$
x_{n+2} - bx_{n+1} = a(x_{n+1} - bx_n) = a^2(x_{n} - bx_{n-1}) = \cdots
$$
Can you proceed from here?
| {
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What is the value of $m+n$ if $\frac{m}{n}$ is the radius of the smallest of the three circles?
Circles of radii 5, 5, 8 and $\frac{m}{n} $(the smallest circle) are
mutually externally tangent to all circles, where $m$ and $n$ are
relatively prime positive integers. Find $m + n$.
Source: Bangladesh Math Olympiad 2017 Junior category.
I can not figure the radius of the smallest circle. Is there any formula for an internal point in a isosceles triangle which can help me to solve this math?
| The triangle has a base side of length $10$ and two sides of length $13$ each. The hight this trinagle is $h = \sqrt {13^2-5^2} = 12$. This height consists of the radius of the large circle, the radius $r=\frac mn$ of the smallest circle and the hight of another isoceles triangle with the same base side and two other sides of lengths $5+r$, which is $\sqrt {(5+r)^2-5^2} = \sqrt{10r+r^2}$. Therefore, we have
$$
\begin{align}
& \quad 12 = 8 + r + \sqrt{10r+r^2}\\
\Leftrightarrow & \quad 4-r = \sqrt {10r+r^2}\\
\Leftrightarrow & \quad (4-r)^2 = 10r + r^2\\
\Leftrightarrow & \quad 16 - 8r + r^2 = 10r + r^2\\
\Leftrightarrow & \quad 18r = 16\\
\Leftrightarrow & \quad r = \frac{16}{18} = \frac89\\
\end{align}
$$
Thus $m=8$ and $n=9$, so $n+m=17$.
| {
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Show that the equation represents a sphere, find center and radius $2x^2 + 2y^2 + 2z^2 = 8x - 24z + 1$
So here's my guess:
$2x^2 - 8x + 2y^2 + 2z^2 + 24z = 1$
Get all variables on one side
$2x^2 - 8x + 16 + 2y^2 + 2z^2 + 24z + 144 = 1 + 16 + 144$ Complete square
$(2x - 8)^2 + 2y^2 + (2z^2 + 24)^2 = 161$ Stopped here
Is it possible to complete the square with the $y$ value too? Would it just be 1?
| Here's a way to do it: divide through by $2$ to get
$$ x^2+y^2+z^2=4x-12z+\frac{1}{2}$$
$$ x^2-4x+y^2+z^2 +12 z=\frac{1}{2}.$$
Now, to complete the square for $x^2-4x$ we need to add $4$ to both sides. To complete the square for $z^2+12z$ we need to add $36$ to both sides. In total we add $40$ to both sides to get
$$ x^2-4x+4+y^2+z^2+12z+36=\frac{81}{2}$$
$$ (x-2)^2+y^2+(z+6)^2=\frac{81}{2}.$$
As we know, this is the equation for a sphere of radius $9/\sqrt{2}$ centred at $(2,0,-6)$.
| {
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"url": "https://math.stackexchange.com/questions/3104569",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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How do we use the picture to get the polynomial? Calculate the solution $p\in \mathbb{P}^4$ off the following interpolatin exercise:
\begin{align*}&p(0)=2 , \ p'(0)=3, \ p''(0)=1 \\ &p(1)=2, \ p'(1)=0\end{align*}
$$$$
I have done the following:
The polynomial that we are looking for is of the form \begin{equation*}p(x)=ax^4+bx^3+cx^2+dx+e\end{equation*}
The first two derivatives are \begin{align*}&p'(x)=4ax^3+3bx^2+2cx+d \\ &p''(x)=12ax^2+6bx+2c\end{align*}
Using the given conditions we get the system:
\begin{align*}\begin{cases}p(0)=2 \\ p'(0)=3 \\ p''(0)=1 \\ p(1)=2 \\ p'(1)=0\end{cases} \Rightarrow \begin{cases}e=2 \\ d=3 \\ 2c=1 \\ a+b+c+d+e=2 \\ 4a+3b+2c+d=0\end{cases} \Rightarrow \begin{cases}e=2 \\ d=3 \\ c=\frac{1}{2} \\ b=-10 \\ a=\frac{13}{2}\end{cases}\end{align*}
Therefore we get the polynomial \begin{equation*}p(x)=\frac{13}{2}x^4-10x^3+\frac{1}{2}x^2+3x+2\end{equation*}
$$$$
Now I saw that there is an easier way to solve this using the following picture, but I haven't really understood how we use that. Do you have an idea?
| You may solve this directly with Taylor:
$$p(x) =\sum_{k=0}^4\frac{p^{(k)}(0)}{k!}x^k$$
Pluggin in the given values you get
$$p(x) = p(0) + p'(0)x + \frac{p''(0)}{2}x^2 + \frac{p^{(3)}(0)}{6}x^3 + \frac{p^{(4)}(0)}{24}x^4 $$ $$= 2 + 3x + \frac{1}{2}x^2 + ax^3+bx^4$$
Now, use the other two conditions to find $a$ and $b$:
$$0 = p'(1) =3+1+3a+4b$$
$$2 = p(1) = 2+3+\frac{1}{2} + a + b$$
$$\Rightarrow a=-10, b=\frac{13}{2}$$
$$\Rightarrow p(x) = 2 + 3x + \frac{1}{2}x^2 -10x^3+\frac{13}{2}x^4$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3107684",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
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Find $\int^{\pi/2}_0 \frac{\sin^{n-2}(x)}{(1+\cos x)^n}\mathrm dx$
Finding
$$I=\int_0^{\pi/2}\frac{\sin^{n-2}(x)}{(1+\cos x)^n}\mathrm dx$$
What I tried:
\begin{align*}
I &= \int^{\pi/2}_0 \left(\frac{\sin x}{1+\cos x}\right)^n\csc^2(x)\mathrm dx \\
&= \int^{\pi/2}_0 \tan^n(x/2)\csc^2(x)\mathrm dx
\end{align*}
But I don't know how to proceed further. Could you please help?
| Note that the $\csc^2(x)$ term can be expressed in terms of $\tan\left(\frac x2\right)$ aswell. To be precise we got that
\begin{align*}
\csc^2(x)=\left(\frac1{\sin(x)}\right)^2=\left(\frac1{2\sin\left(\frac x2\right)\cos\left(\frac x2\right)}\right)^2=\left(\frac{\frac1{\cos^2\left(\frac x2\right)}}{2\frac{\sin\left(\frac x2\right)}{\cos\left(\frac x2\right)}}\right)^2
=\left(\frac{1+\tan^2\left(\frac x2\right)}{2\tan\left(\frac x2\right)}\right)^2
\end{align*}
Using this and further noticing that $\frac{\mathrm d}{\mathrm dx}\tan\left(\frac x2\right)=\frac12\left(1+\tan^2\left(\frac x2\right)\right)$ we may enforce the substition $\tan\left(\frac x2\right)=u$ to obtain
\begin{align*}
I_n=\int_0^{\pi/2}\tan^n\left(\frac x2\right)\csc^2(x)\mathrm dx&=\int_0^{\pi/2}\tan^n\left(\frac x2\right)\left(\frac{1+\tan^2\left(\frac x2\right)}{2\tan\left(\frac x2\right)}\right)^2\mathrm dx\\
&=\frac12\int_0^{\pi/2}\tan^{n-2}\left(\frac x2\right)\left(1+\tan^2\left(\frac x2\right)\right)\left[\frac12\left(1+\tan^2\left(\frac x2\right)\right)\mathrm dx\right]\\
&=\frac12\int_0^1 u^{n-2}(1+u^2)\mathrm du\\
&=\frac12\left[\frac{u^{n-1}}{n-1}+\frac{u^{n+1}}{n+1}\right]_0^1\\
&=\frac12\left[\frac1{n-1}+\frac1{n+1}\right]
\end{align*}
$$\therefore~I_n~=~\int_0^{\pi/2}\tan^n\left(\frac x2\right)\csc^2(x)\mathrm dx~=~\frac n{n^2-1}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3107807",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
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A sum of series problem with alternating sign of terms I came across a problem that requires me to find the sum of a series. The term of the series $T_n$ is given by
$$T_n = (-1)^{\frac{n(n+1)}2}n^2$$
Sum till $4n$ terms is to be found.
Writing down the first few terms:
$$-(1)^2-(2)^2+3^2+4^2-(5)^2-(6)^2+7^2+8^2+ \ldots$$
This can be arranged into a sum of difference between squares:
$$(3^2-1^2)+(4^2-2^2)+(7^2-5^2)+(8^2-6^2)+\ldots$$
$$=2\times(4+6+12+14+20+\ldots)$$
$$=4\times(2+3+6+7+10\ldots)$$
In the last step, the expression inside the parentheses contains $n$ terms.
I am unable to proceed from here. How do I find the sum of the series inside the parentheses? A hint would be much appreciated.
This is a question from Resonance DLPD Algebra, Page #30, Question #2.
| $$\begin{align}
&\;\;\;\ \sum_{r=1}^{4n}(-1)^{\frac {r(r+1)}2}r^2\\
&=\sum_{r=1}^n (4r)^2+(4r-1)^2-(4r-2)^2-(4r-3)^2\\
&=\sum_{r=1}^n \underbrace{(4r)^2-(4r-2)^2}+\underbrace{(4r-1)^2-(4r-3)^2}\\
&=\sum_{r=1}^n \qquad 2(8r-2)\;\;\;+\qquad 2(8r-4)\\
&=\sum_{r=1}^n4(8r-3)\\
&=4\sum_{r=1}^n (8r-4+1)\\
&=4\sum_{r=1}^n 4(2r-1)+1\\\\
&=4(4n^2+n)\\\\
&=4n(4n+1)\end{align}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3110394",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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How to compute remainder of division of $P(x)$ by $x^2 -3x+2$?
The remainder of division of $P(x)$ by $x^2−1$ is $2x+1$, and the remainder of division of the same polynomial by $x^2−4$ is $x+4$. Compute the remainder of division of $P(x)$ by $x^2−3x+2$.
I will translate these into math equations
$$P(x) = (x^2-1)Q(x)+ 2x+1$$
$$P(x) = (x^2 -4)R(x)+x+4$$
And let
$$f(x) = P(x) $$
We're asked to find the remainder when this polynomial is divided by $x^2 -3x+2$. So, there are two equations, which is why I'm confused with what to use in the equation $f(x) = P(x)$. What am I missing here?
Regards
| You want to find $r(x)$ where the degree of $r$ is less than 2 in the following equation:
$$P(x) = (x^2-3x+2)S(x) + r(x) = (x-1)(x-2)S(x) + r(x)$$
You also have the two formulas you wrote above (representing the polynomial remainders you are given). If you plug in $x=1$ and $x=2$, you have two different values of $x$ that make the $S(x)$ term zero. So, $P(1) = r(1)$ and $P(2) = r(2)$.
From your first equation, you also know:
$$P(1) = (1^2-1)Q(1)+2(1)+1 = 3$$
From your second equation, you have:
$$P(2) = (2^2-4)R(2)+(2)+4 = 6$$
Since $r(x) = ax+b$ (since your quotient has degree 2, your remainder will have a maximum degree of 1)
You have $r(1)=3=a+b$ and $r(2) = 6 = 2a+b$.
This gives $a=3, b=0$.
Plugging in gives:
$$P(x) = (x^2-3x+2)S(x)+3x$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3110509",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Find out the value of the integral $\int_{-2}^{2} \lfloor x^2-1\rfloor dx$ Find out the value of the integral $$\int_{-2}^{2} \lfloor x^2-1\rfloor dx$$ where $[x]$ denotes the floor function (i.e., $[x]$ is the greatest integer $\le x$.)
My attempt ..... $$\int_{-2}^2 \lfloor x^2 – 1\rfloor dx = 2\int_0^2 \lfloor x^2-1\rfloor dx$$ Because $\lfloor x^2 – 1\rfloor$ is even. $$2\int_0^2 \lfloor x^2-1\rfloor dx =\\ 2\int_0^1 \lfloor x^2-1\rfloor dx+2\int_1^{\sqrt{2}} \lfloor x^2-1\rfloor dx +2\int_{\sqrt{2}}^{\sqrt{3}} \lfloor x^2-1\rfloor dx + 2\int_{\sqrt{3}}^2 \lfloor x^2-1\rfloor dx$$
But how to evaluate this or am I wrong in the whole assumption?
| Since $f$ is even we have $$ I/2 =\int_0^2[x^2-1]\;dx=$$ $$=\int_0^1-1\;dx+\int_1^\sqrt{2}0\;dx+\int_\sqrt{2}^\sqrt{3}1\;dx+\int_\sqrt{3}^22\;dx=...$$
$$=(-1)+0+(\sqrt{3}-\sqrt{2})+(4-2\sqrt{3})=3-\sqrt{3}-\sqrt{2}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3111444",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 1,
"answer_id": 0
} |
Decide all solutions to the following trigonometric equation: $\sin \left( 4\,x \right) =\cos \left( 3\,x \right).$ in the interval $[0, 2π[ $. Decide all solutions to the following trigonometric equation:
$$
\sin \left( 4\,x \right) =\cos \left( 3\,x \right).
$$
in the interval $[0, 2π[$.
I start by expanding $\sin \left( 4\,x \right)$ and $\cos \left( 3\,x \right)$ to get to get that
$\sin \left( 4\,x \right) =4\,\sin \left( x \right) \cos \left( x
\right) \left( 1-2\, \left( \sin \left( x \right) \right) ^{2}
\right)$
$\cos \left( 3\,x \right) =\cos \left( x \right) \left( 1-4\, \left(
\sin \left( x \right) \right) ^{2} \right)$.
I cancel $\cos \left( x \right)$ in both equations above and get that
$4\,\sin \left( x \right) \left( 1-2\, \left( \sin \left( x \right)
\right) ^{2} \right) =1-4\, \left( \sin \left( x \right) \right) ^{2}$
which is equivalent to
$8\, \left( \sin \left( x \right) \right) ^{3}-4\, \left( \sin \left(
x \right) \right) ^{2}-4\,\sin \left( x \right) +1=0$.
Now I use the substitution $u=\sin \left( x \right)$ to get to
$8\,{u}^{3}-4\,{u}^{2}-4\,u+1=0$ ⟺ $u \left( u-1 \right) \left( 2\,u+1 \right) =-1/4$.
Now, I'm not so sure I'm on the right track. To solve the cubic equation for $u$, seem to lead to a real nasty piece of a solution. Appreciate any help!
| It'd be much easier to use the identities
$\sin \theta = \sin(2k \pi + \theta)$
$\cos \theta = \cos (\pm \theta +2k\pi)$
$\sin \theta = \cos (\frac \pi 2 - \theta); \cos(\theta) = \sin (\frac \pi 2-\theta)$.
So solutions occur when
$4x = \pm (\frac \pi 2 - 3x) + 2k\pi$
$\pm 3x = \frac \pi 2 - 4x + 2k \pi$ so
$7x = \frac \pi 2 + 2k \pi$ or
$x = -\frac \pi 2 + 2k\pi$ or
$x = \frac \pi 2 + 2k \pi$ or
$x = \frac {(4k+1) \pi}{14}$ or $x= \frac {3\pi} 2$ or $x = \frac {\pi} 2$.
So $x = \frac 1{14}\pi, \frac 5{14}\pi, \frac 12 \pi, \frac 9{14} \pi, \frac {13}{14} \pi, \frac {17}{14}\pi, \frac 32\pi,$ or $ \frac {25}{14}\pi$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Stuck at proving whether the sequence is convergent or not I have been trying to determine whether the following sequence is convergent or not. This is what I got:
Exercise 1: Find the $\min,\max,\sup,\inf, \liminf,\limsup$ and determine whether the sequence is convergent or not:
$X_n=\sin\frac{n\pi}{3}-4\cos\frac{n\pi}{3}$
I wrote down a few cases:
$X_1 = \sin\frac{\pi}{3}-4\cos\frac{\pi}{3}= \frac{\sqrt3-4}{2} $
$X_2 = \sin\frac{2\pi}{3}-4\cos\frac{2\pi}{3}= \frac{\sqrt3+4}{2} $
$X_3 = \sin\frac{3\pi}{3}-4\cos\frac{3\pi}{3}=4 $
$X_4 = \sin\frac{4\pi}{3}-4\cos\frac{4\pi}{3}= \frac{4-\sqrt3}{2} $
$X_5 = \sin\frac{5\pi}{3}-4\cos\frac{5\pi}{3}= \frac{4-\sqrt3}{2} $
$X_6 = \sin\frac{6\pi}{3}-4\cos\frac{6\pi}{3}= -4 $
So as found above, $\min = -4$, $\max = 4$, $\inf = -4$, $\sup = 4$, $\liminf = -4$, $\limsup = 4$.
Let's check whether its convergent or not:
The sequence is bounded as stated above so lets check if its decreasing or increasing.
$X_n \geq X_{n+1}$
$\sin\frac{n\pi}{3}-4\cos\frac{n\pi}{3} \geq\sin\frac{(n+1)\pi}{3}-4\cos\frac{(n+1)\pi}{3}$
$\sin\frac{n\pi}{3} - \sin\frac{(n+1)\pi}{3} \geq -4\cos\frac{(n+1)\pi}{3} + 4\cos\frac{n\pi}{3}$
I used trigonometrical identity for $\sin\alpha+\sin\beta$ and $\cos\alpha-\cos\beta$ :
$-\cos\frac{\pi(2n+1)}{6} \geq 4\sin\frac{\pi(2n+1)}{6}$
What should I do next? I am stuck here.
Thanks, and sorry if I made mistakes.
| It happens that your sequence is a cyclic sequence. More precisely, its first six terms are$$-2+\frac{\sqrt3}2,2+\frac{\sqrt3}2,4,2-\frac{\sqrt3}2,-2-\frac{\sqrt3}2,\text{ and }-4$$and then it repeats itself again and again. Therefore:
*
*$\max\{X_n\,|\,n\in\mathbb N\}=\sup\{X_n\,|\,n\in\mathbb N\}=\limsup_nX_n=4$;
*$\min\{X_n\,|\,n\in\mathbb N\}=\inf\{X_n\,|\,n\in\mathbb N\}=\liminf_nX_n=-4$;
*the sequence diverges.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3114035",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 5,
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Why is $(5\sqrt{5p}-3\sqrt{5q})(5\sqrt{5p}+3\sqrt{5q}) \equiv 5(5p-3q)(5p+3q)$? I was working on the difference of two squares, $125p^2-45q^2$
Writing my answer, $$(5\sqrt{5}p-3\sqrt{5}q)(5\sqrt{5}p+3\sqrt{5}q),$$ onto Pearson, I got a popup that said my answer was equivalent to the correct answer but in incorrect form. Apparently, the correct answer is $$5(5p-3q)(5p+3q).$$
Why is that? I'm failing to see the intuition here.
| $$125p^2-45q^2=5(25p^2-9q^2)=5(5p+3q)(5p-3q)$$
Your answer was incorrect because it should have been
$$125p^2-45q^2=(\sqrt{125}p^2+\sqrt{45}q^2)(\sqrt{125}p^2-\sqrt{45}q^2)=(5\sqrt{5}p+3\sqrt{5}q)(5\sqrt{5}p-3\sqrt{5}q)=5(5p+3q)(5p-3q)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3117487",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Evaluate $\int_0^1 x\arcsin\left(\sin\left(\frac 1x\right)\right) dx$
Evaluate $$\int_0^1 x\arcsin\left(\sin\left(\frac 1x\right)\right) dx=\int_1^{\infty} \frac {\arcsin(\sin x)}{x^3}dx$$
I tried to use Feynman's trick in the second one as
$$I(a)=\int_1^{\infty} \frac {\arcsin(\sin ax)}{x^3}dx$$
We have $I(0)=0$
Now differentiating both sides leads me to an absurd integral as $$I'(a)=\int_1^{\infty} \frac {\cos ax }{x^2\vert \cos ax \vert}dx$$
I am now unable proceed further.
Also If I keep in mind that I need value of integral at $a=1$ and so break the integral in intervals as $$\left(1,\frac {\pi}{2}\right) ;\left(\frac {\pi}{2},\frac {3\pi}{2}\right) ; \left(\frac {3\pi}{2},\frac {5\pi}{2}\right) $$
and so on then this gives me the value of integral as 0. But I suspect it isn't correct because one online software suggests it's value to be $\frac {1}{2}$
Moreover the signum function is always reminding me of the Grandi's series which also in some definition equals $1/2$
Any help would be greatly appreciated.
| Write $I_n = \left[\left(n-\frac{1}{2}\right)\pi, \left(n+\frac{1}{2}\right)\pi\right]$ and notice that the behavior of $\arcsin\sin x$ on $I_n$ depends on the value of $n$. In particular, $\arcsin\sin x$ is linear on each interval $I_n$ with the slope $(-1)^n$.
So, we can compute the integral as
\begin{align*}
\int_{1}^{\infty} \frac{\arcsin\sin x}{x^3} \, \mathrm{d}x
&= \left[ - \frac{\arcsin\sin x}{2x^2} \right]_{1}^{\infty}
+ \int_{1}^{\infty} \frac{(\arcsin\sin x)'}{2x^2} \, \mathrm{d}x \\
&= \frac{1}{2} + \int_{1}^{\infty} \frac{1}{2x^2} \left( \sum_{n=0}^{\infty} (-1)^n \mathbf{1}_{I_n}(x) \right) \, \mathrm{d}x \\
&= \frac{1}{2} + \sum_{n=0}^{\infty} (-1)^n \int_{I_n\cap[1,\infty)} \frac{1}{2x^2} \, \mathrm{d}x \\
&= \frac{1}{2} + \left[ - \frac{1}{2x} \right]_{1}^{\frac{\pi}{2}} + \sum_{n=1}^{\infty} (-1)^n \left[ - \frac{1}{2x} \right]_{\left(n-1/2\right)\pi}^{\left(n+1/2\right)\pi} \\
&= 1 - \sum_{n=1}^{\infty} \frac{(-1)^{n-1}}{\left( n - \frac{1}{2} \right) \pi} \\
&= 1 - \frac{1}{2}
= \frac{1}{2}.
\end{align*}
Or, more concisely,
\begin{align*}
\int_{1}^{\infty} \frac{\arcsin\sin x}{x^3} \, \mathrm{d}x
&= \left[ - \frac{\arcsin\sin x}{2x^2} \right]_{1}^{\infty}
+ \int_{1}^{\infty} \frac{(\arcsin\sin x)'}{2x^2} \, \mathrm{d}x \\
&= \frac{1}{2} + \left[ -\frac{(\arcsin\sin x)'}{2x} \right]_{1}^{\infty}
+ \int_{1}^{\infty} \frac{(\arcsin\sin x)''}{2x} \, \mathrm{d}x \\
&= 1 + \int_{1}^{\infty} \left( \sum_{n=1}^{\infty} \frac{(-1)^n}{x} \delta\left(x - \frac{2n-1}{2}\pi \right) \right) \, \mathrm{d}x \\
&= 1 + \sum_{n=1}^{\infty} \frac{(-1)^n}{\left(\frac{2n-1}{2}\right)\pi}
= 1 - \frac{1}{2}
= \frac{1}{2}.
\end{align*}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3120350",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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Convergence of the Arithmetic mean of a sequence taking values in $\{0, 1\}$ Let $(a_i)_{i \in \mathbb{N}}$ be a sequence taking values in $\{0, 1\}$. Now define $x_n$ as the Arithmetic mean of the first $n$ elements of $(a_i)_{i \in \mathbb{N}}$. So let
$$x_n = \frac{1}{n}\sum_{i = 1}^{n}a_i.$$
Can we find such a sequence for which $x_n$ diverges (when $n \to \infty$)? I know that $0 \leq x_n \leq 1$ for all $n \in \mathbb{N}$, but can we be certain that the limit actually exists?
| This is not the case. Take for example $a_n=1$ for certain consecutive $n$ and otherwise $a_n=0$ such that the mean of the sequence wiggles around between a third and two third. To let this happen, the sequence needs to contain longer and longer sequences of consecutive $1$s and $0$s.
To be precise one could take the following sequence
$$ a_n = \begin{cases} 1, & \text{if there exists an }\ m:\ 2^{2m-1}\leq n \leq 2^{2m} \\ 0 ,& \text{otherwise}. \end{cases} $$
In this case,
$$ \frac{1}{2^{2m}}\sum_{n=1}^{2^{2m}}a_n = \frac{1}{2^{2m}}\sum_{j=1}^{m}\sum_{n=2^{2j-1}}^{2^{2j}} 1 = \frac{1}{2^{2m}} \sum_{j=1}^{m} 2^{2j-1}= \frac{1}{2^{2m+1}} \sum_{j=1}^{m} 4^j = \frac{1}{2^{2m+1}}\cdot\frac{4^{m+1}-1}{4-1} = \frac{2^{2m+2} - 1}{2^{2m+1}\cdot3} \to \frac{2}{3} $$
but
$$ \frac{1}{2^{2m-1}}\sum_{n=1}^{2^{2m-1}}a_n = \frac{1}{2^{2m-1}}\sum_{j=1}^{m-1}\sum_{n=2^{2j-1}}^{2^{2j}} 1 = \frac{1}{2^{2m-1}} \sum_{j=1}^{m-1} 2^{2j-1}= \frac{1}{2^{2m}} \sum_{j=1}^{m-1} 4^j = \frac{1}{2^{2m}}\cdot\frac{4^m - 1}{4-1} = \frac{2^{2m} - 1}{2^{2m}\cdot 3} \to \frac{1}{3} $$
And thus the limit $\lim_{N\to\infty}\frac{1}{N}\sum_{n=1}^{N} a_n$ is undefined.
| {
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"timestamp": "2023-03-29T00:00:00",
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calculation of $\int^{1}_{0}\frac{1}{\sqrt{x-x^2}(x^2+3x+2)}dx$
Calculation of $$\int^{1}_{0}\frac{1}{\sqrt{x-x^2}(x^2+3x+2)}dx$$
My Try: Let $$I = \int^{1}_{0}\frac{1}{\sqrt{x-x^2}(x+1)(x+2)}dx$$
Put $x=\sin^2 \theta$ and $dx = 2\sin \theta \cos \theta$ and changing limits
So $$I = \int^{\frac{\pi}{2}}_{0}\frac{2}{(1+\sin^2 \theta)(2+\sin^2 \theta)}d\theta$$
Could someone help me to solve it , Thanks
| Hint. Note that
$$\frac{2}{(1+\sin^2 \theta)(2+\sin^2 \theta)}
=\frac{2}{1+\sin^2 \theta}-\frac{2}{2+\sin^2 \theta}.$$
Moreover, for $k>0$,
$$\frac{1}{k+\sin^2 \theta}=\frac{\frac{1}{k}d(\tan \theta)}{1+\frac{(k+1)\tan^2 \theta}{k}}.$$
Finally let $t=\frac{\sqrt{k+1}\tan \theta}{\sqrt{k}}.$
| {
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Find $Tr(B)+Tr(C)$ If $B,C$ are $2 \times 2 $ matrices with integer entries such that:
$$\begin{bmatrix}
-1 &1 \\
0& 2
\end{bmatrix}=B^3+C^3$$
Find value of $Tr(B)+Tr(C)$
My try:
Taking trace on both sides we get:
$$Tr(B^3)+Tr(C^3)=1$$
Any clue from here?
| Let $\displaystyle A=\begin{bmatrix}
-1 &1 \\
0& -2
\end{bmatrix}$ and $\displaystyle I =\begin{bmatrix}1&0 \\0&1\end{bmatrix}$.
Then Using Characteristic Equation(Caly -Hamilton Theorem)
$|A-\lambda I|=0\Rightarrow \begin{vmatrix}
-1-\lambda && 1& \\
0& & -2-\lambda\\
\end{vmatrix}=0$
$\Rightarrow \lambda^2+3\lambda+2=0\Rightarrow A^2+3A+2I=O$
$$ \Rightarrow A^3+3A^2+2A=O$$
$\Rightarrow A=A^3+3A^2+3A+I^3-I^3$
$$\Rightarrow A=(A+I)^3-I^3=B^3+C^3$$
So we have $$B=A+I\;\;\;,\;\;\;C=-I$$
| {
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List of asymptotic expressions for $\sum_r r^\beta \ln r$? Expression for $\beta =1$ in $\sum_r r^\beta \ln r$
I recently had the following idea to use the below identity:
$$ (1!2! 3! \dots n!) (12^2 3^3 4^4 \dots n^n) = n!^{n+1}$$
Dividing both sides by $n^{n(n+1)/2}$ and using $1+2+3 \dots + n = \frac{n(n+1)}{2}$:
$$ (1!2! 3! \dots n!) (\frac{1}{n}(\frac{2}{n})^2 (\frac{3}{n})^3 (\frac{4}{n})^4 \dots (\frac{n}{n})^n) = \frac{n!^{n+1}}{n^{n(n+1)/2}}$$
Raising both sides to power $1/n$:
$$ (1!2! 3! \dots n!)^{\frac{1}{n}} ((\frac{1}{n})^{\frac{1}{n}}(\frac{2}{n})^{\frac{2}{n}} (\frac{3}{n})^{\frac{3}{n}} (\frac{4}{n})^{\frac{4}{n}} \dots (\frac{n}{n})^{\frac{1}{n}}) = \frac{n!^{1+1/n}}{n^{(n+1)/2}}$$
Taking $\ln$ both sides:
$$ \sum_{r=1}^n \ln(r!) \frac{1}{n} + \sum_{r=1}^n \frac{r}{n} \ln(\frac{r}{n}) = \ln(\frac{n!^{1+1/n}}{n^{(n+1)/2}}) $$
Dividing both sides with $1/n$:
$$ \sum_{r=1}^n \ln(r!) \frac{1}{n^2} + \sum_{r=1}^n \frac{r}{n} \ln(\frac{r}{n})\frac{1}{n} = \frac{1}{n} \ln(\frac{n!^{1+1/n}}{n^{(n+1)/2}}) $$
In the limit $n \to \infty $ then $ \sum_{r=1}^n \frac{r}{n} \ln(\frac{r}{n})\frac{1}{n} \to \int_0^1 x \ln x dx = -1/4 $. Hence,
$$ \sum_{r=1}^n \ln(r!) \frac{1}{n^2} - \frac{1}{4}\sim \frac{1}{n} \ln(\frac{n!^{1+1/n}}{n^{(n+1)/2}}) $$
Thus,
$$ \sum_{r=1}^n \ln(r!)\frac{1}{n^2} \sim (1+ \frac{1}{n} ))( n \ln(n) -n + O(\ln n)) - \frac{(n+1)}{2}\ln(n) + \frac{1}{4} $$
Using Stirling's approximation on the R.H.S:
$$ \sum_{r=1}^n \ln(r!)\frac{1}{n^2} \sim \frac{n-1}{2} \ln n + \ln n -\frac{3}{4} + O(\frac{\ln n}{n}) $$
Using Stirling's approximation on the L.H.S:
$$ \sum_{r=1}^n(r \ln r + r + O(\ln r))\frac{1}{n^2} \sim \frac{n-1}{2} \ln n + \ln n -\frac{3}{4} + O(\frac{\ln n}{n}) $$
Simplifying both sides:
$$ \sum_{r=1}^n r \ln r \sim n^2 \frac{n-1}{2} \ln n + n^2\ln n -\frac{(5n+2)n}{4} + O(n \ln n + \ln n!) $$
Question
I did not consider the error when converting to an integral when I did the step "In the limit $n \to \infty $ then $ \sum_{r=1}^n \frac{r}{n} \ln(\frac{r}{n})\frac{1}{n} \to \int_0^1 x \ln x dx = -1/4 $" .. What is the error? Is there a list of asymptotic expressions for $\sum r^\beta \ln r$ without using $ \sum r^\beta \leq \sum r^\beta \ln r \leq \sum r^{\beta + 1}$?
| By the Euler-Maclaurin summation formula,
$$\sum_{k=1}^nk^\beta\ln k\approx\frac{n^{\beta+1}}{\beta+1}\log n-\frac{n^{\beta+1}-1}{(\beta+1)^2}+\frac{n^\beta\ln n}2+\frac{\beta n^{\beta-1}\ln n-n^{\beta-1}+1}{12}+\cdots$$
| {
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Prove that any number that is not a power of 2 can be expressed as a sum of two or more consecutive positive integers 1) Prove that any number that is not a power of $2$ can be expressed as a sum
of two or more consecutive positive integers, but that this is not possible for powers of $2$.
2) Which numbers can be expressed as a sum of two or more consecutive
odd positive integers?
I have found many questions the first part but am unsure on how to solve the second part of the question. My guess is using $(n+1)(n+3)...$ and so on but am unsure how to use this.
| 1. Assume $2^m;\ m\ge 2$ can be expressed as the sum of consecutive integers $n+1,n+2,\dots n+k$. Then it is the difference between the sum of the first $k$ integers and the sum of the first $n$ integers. $$2^m=\frac{k(k+1)}{2}-\frac{n(n+1)}{2}=\frac{k^2-n^2+k-n}{2}\Rightarrow 2^{m+1}=(k+n)(k-n)+(k-n)=(k+n+1)(k-n)$$ Now $(k+n)$ and $(k-n)$ have the same parity, so $(k+n+1)$ and $(k-n)$ must have different parity, i.e.one of them must be odd. But $2^{m+1}$ has no odd factors, so the assumption is false and $2^m$ cannot be expressed as the sum of consecutive integers.
Any odd number $2a+1$ can be expressed as the sum of two consecutive integers, $2a+1=a+(a+1)$
Any even number $b$ not a perfect power of two has at least one odd prime factor: $b=2^m\prod q_i$ where $q_i$ are odd primes, not necessarily distinct.
Case 1: $b$ contains more than one odd prime factor: $i\ge 2$. Let $q_1=\min(q_i); j=\frac{q_1-1}{2}$. Also, $s=\frac{b}{q_1}$. The consecutive integers $s-j, s-j+1,\dots s,\dots s+j-1,s+j$ sum to $b$ as the sequence contains $q_1$ integers of average value $s$, and $s\cdot q_1=b$. This approach will certainly work for the smallest odd prime factor of $b$, and may work for some larger odd prime factors of $b$ as well.
Case 2: $b=2^mq$. Then $b$ is the sum of $2^{m+1}$ consecutive integers: $$b=((\frac{q}{2}-\frac{2^{m+1}-1}{2})+(\frac{q}{2}-\frac{2^{m+1}-1}{2}+1)+\dots +(\frac{q}{2}-\frac{2^{m+1}-1}{2}+(2^{m+1}-1)))$$ Notice that each term features the difference between two half integers, and so results in an integer. This sequence contains $2^{m+1}$ integers with an average value of $$\frac{(\frac{q}{2}-\frac{2^{m+1}-1}{2})+(\frac{q}{2}-\frac{2^{m+1}-1}{2}+(2^{m+1}-1))}{2}=\frac{q}{2}$$ Thus the sum is $2^{m+1}\cdot \frac{q}{2}=2^mq=b$.
2. Any even number which is a multiple of $4$ can be expressed as the sum of two consecutive odd integers: $4a=(2a-1)+(2a+1)$.
Any odd composite number $b$ has at least two odd prime factor: $b=\prod q_i$ where $q_i$ are odd primes, not necessarily distinct, and $i\ge 2$. Let $q_1=\min(q_i); j=\frac{q_1-1}{2}$. Also, let $s=\frac{b}{q_1}$. Note that $s$ is odd. The consecutive odd integers $s-2j, s-2(j+1),\dots s,\dots s+2(j-1),s+2j$ sum to $b$ as the sequence contains $q_i$ integers of average value $s$, and $s\cdot q_1=b$.
If an odd prime number could be expressed as the sum of two or more consecutive odd integers, it would be equal to the number of such consecutive odd integers times their average value, which would be equal to the least plus the greatest divided by $2$. Note that the average of two odd numbers is always an integer. Thus the prime number would have factors, which is a contradiction. So odd prime numbers cannot be expressed as the sum of two or more consecutive odd numbers.
This leaves even numbers which are twice odd numbers, which cannot be expressed as the sum of two or more consecutive odd numbers. The easiest way to see this is by reference to the other answers which point out that numbers which can be expressed as the sum of consecutive odd numbers are always the difference of squares, and numbers which are twice odd numbers cannot be so expressed. If $b=2(2a+1)=4a+2$, then $b\equiv 2 \mod 4$. But $\forall n: n^2\equiv 0,1 \mod 4$ so $n_1^2-n_2^2 \equiv 0,1,3 \mod 4$
As an example of how an odd number can be represented as the sum of consecutive integers or consecutive odd integers in multiple ways, consider $$255=3\cdot 5\cdot 17=(127+128)=(84+85+86)=(49+50+51+52+53)=(7+8+9+10+11+12+13+14+15+16+17+18+19+20+21+22+23)=(83+85+87)=(47+49+51+53+55)$$
| {
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Existence of closed form solution of the below DE I have a problem finding an analytical closed form solution of the following DE.
$ \displaystyle \left( \frac{y(x)+2}{3} \right) \frac{dy}{dx} + y(x) +\frac{1}{x} \left( a y(x)^2 + b y(x) -c \right) =0. \\
\{a,b,c\} \in \mathbb{R}, \qquad \{a,b,c\}>0.
$
I encountered this differential equation in my research in fluid dynamics. I failed to reduce it to some standard form of DE for which solutions are possible. Mathematica and Maple are unable to find analytical solutions for this.
Can anyone point out any method that can be used to know if a solution for this DE exists?
And if a solution exists, can anyone point out a method to solve the DE?
| Hint:
Approach $1$:
This belongs to an Abel equation of the second kind.
Let $u=y+2$ ,
Then $\dfrac{du}{dx}=\dfrac{dy}{dx}$
$\therefore\dfrac{u}{3}\dfrac{du}{dx}+u-2+\dfrac{1}{x}(a(u-2)^2+b(u-2)-c)=0$
$\dfrac{u}{3}\dfrac{du}{dx}+u-2+\dfrac{1}{x}(au^2-(4a-b)u+4a-2b-c)=0$
$u\dfrac{du}{dx}=-\dfrac{3au^2}{x}-3\left(1-\dfrac{4a-b}{x}\right)u+6-\dfrac{3(4a-2b-c)}{x}$
Let $u=x^{-3a}v$ ,
Then $\dfrac{du}{dx}=x^{-3a}\dfrac{dv}{dx}-3ax^{-3a-1}v$
$\therefore x^{-3a}v\left(x^{-3a}\dfrac{dv}{dx}-3ax^{-3a-1}v\right)=-\dfrac{3ax^{-6a}v^2}{x}-3\left(1-\dfrac{4a-b}{x}\right)x^{-3a}v+6-\dfrac{3(4a-2b-c)}{x}$
$x^{-6a}v\dfrac{dv}{dx}-3ax^{-6a-1}v^2=-3ax^{-6a-1}v^2-3\left(1-\dfrac{4a-b}{x}\right)x^{-3a}v+6-\dfrac{3(4a-2b-c)}{x}$
$x^{-6a}v\dfrac{dv}{dx}=-3\left(1-\dfrac{4a-b}{x}\right)x^{-3a}v+6-\dfrac{3(4a-2b-c)}{x}$
$v\dfrac{dv}{dx}=-3(x^{3a}-(4a-b)x^{3a-1})v+6x^{6a}-3(4a-2b-c)x^{6a-1}$
Approach $2$:
$\dfrac{y+2}{3}\dfrac{dy}{dx}+y+\dfrac{ay^2+by-c}{x}=0$
$y+\dfrac{ay^2+by-c}{x}=-\dfrac{y+2}{3}\dfrac{dy}{dx}$
$\left(x+ay+b-\dfrac{c}{y}\right)\dfrac{dx}{dy}=-\left(\dfrac{1}{3}+\dfrac{2}{3y}\right)x$
This belongs to an Abel equation of the second kind.
Let $u=x+ay+b-\dfrac{c}{y}$ ,
Then $x=u-ay-b+\dfrac{c}{y}$
$\dfrac{dx}{dy}=\dfrac{du}{dy}-a-\dfrac{c}{y^2}$
$\therefore u\left(\dfrac{du}{dy}-a-\dfrac{c}{y^2}\right)=-\left(\dfrac{1}{3}+\dfrac{2}{3y}\right)\left(u-ay-b+\dfrac{c}{y}\right)$
$u\dfrac{du}{dy}-\left(a+\dfrac{c}{y^2}\right)u=-\left(\dfrac{1}{3}+\dfrac{2}{3y}\right)u+\dfrac{(y+2)(ay^2+by-c)}{3y^2}$
$u\dfrac{du}{dy}=\dfrac{(3a-1)y^2-2y+3c}{3y^2}u+\dfrac{(y+2)(ay^2+by-c)}{3y^2}$
| {
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Another way to compute $\lim\limits_{x\to+\infty}x^2\log\left(\frac{x^2+1}{x^2+3}\right)$ I need to compute as a title a limit with $x\to+\infty$. The only way I found to obtain a result is to use the L'Hôpital's rule:
$$\lim\limits_{x\to+\infty}x^2\log\bigg(\frac{x^2+1}{x^2+3}\bigg)=\lim\limits_{t\to 0}\frac{1}{t^2}\log\bigg(\frac{1+t^2}{1+3t^2}\bigg)\stackrel{H}{=}\lim\limits_{t\to 0}\frac{1}{2t}\bigg(\frac{2t}{1+t^2}-\frac{6t}{1+3t^2}\bigg)=\lim\limits_{t\to 0}\bigg(\frac{1}{1+t^2}-\frac{3}{1+3t^2}\bigg)=-2$$
It seems the result is the difference between the two functions inside the log, but how this can be possible? Is there another way to compute these type of limits? I mean a calculus way without using theorems.
| You can write
$$x^2 \ln \left( \frac{x^2+1}{x^2+3} \right) = x^2 \ln \left( 1- \frac{2}{x^2+3} \right) \sim x^2 \times \frac{-2}{x^2+3}$$
So the limit is $-2$.
| {
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Geometric center of convex set of positive semidefinite matrices Consider the set of $d\times d$ matrices that are positive semidefinite and have unit trace. This is a convex set, $S$. Is it possible to think of a geometric center of this set? The criterion for geometric center is an element $\rho$ such that if $\rho + X\in S$, then $\rho - X\in S$
For $d = 2$, it is the identity matrix. For $d>2$, it is not the identity matrix. Indeed $$\begin{pmatrix} 1 & 0 & 0\\ 0 & 0 &0 \\ 0 & 0 & 0 \end{pmatrix} = \frac{1}{3}\begin{pmatrix} 1 & 0 & 0\\ 0 & 1 &0 \\ 0 & 0 & 1 \end{pmatrix} + \begin{pmatrix} 2/3 & 0 & 0\\ 0 & -1/3 &0 \\ 0 & 0 & -1/3 \end{pmatrix} \in S$$ but $$\frac{1}{3}\begin{pmatrix} 1 & 0 & 0\\ 0 & 1 &0 \\ 0 & 0 & 1 \end{pmatrix} - \begin{pmatrix} 2/3 & 0 & 0\\ 0 & -1/3 &0 \\ 0 & 0 & -1/3 \end{pmatrix} \notin S$$
So is it possible to find such an element for $d>2$ or if not, why not?
| When $d\ge3$, there cannot be any geometric centre. Since the trace of a matrix and the property of being a centre are preserved by unitary similarity, we may consider only the case where $\rho$ is a nonnegative diagonal matrix.
If $\rho=\frac1nI$, then a similar example to yours shows that $\rho$ is not a centre.
If $\rho$ contains two different diagonal entries $a>b$ instead, $\rho$ is also not a centre because
$$
\pmatrix{a\\ &b\\ &&c\\ &&&\ddots}
+\pmatrix{-\frac{a+b}2\\ &\frac{a+b}2\\ &&0\\ &&&\ddots}
=\pmatrix{\frac{a-b}2\\ &\frac{a+3b}2\\ &&c\\ &&&\ddots}\succeq0
$$
but
$$
\pmatrix{a\\ &b\\ &&c\\ &&&\ddots}
-\pmatrix{-\frac{a+b}2\\ &\frac{a+b}2\\ &&0\\ &&&\ddots}
=\pmatrix{\frac{3a+b}2\\ &\frac{b-a}2\\ &&c\\ &&&\ddots}\nsucceq0.
$$
| {
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Find $\lim\limits_{x\to0^+}x(\lfloor \frac{1}{x}\rfloor+\lfloor \frac{2}{x}\rfloor+\cdots+\lfloor \frac{k}{x}\rfloor), \, k \in \mathbb N$. $$M:=x\left(\left\lfloor \frac{1}{x} \right\rfloor+\left\lfloor \frac{2}{x}\right\rfloor+\cdots+\left\lfloor \frac{k}{x}\right\rfloor\right),\, k \in \mathbb N.$$
Using $\lfloor y \rfloor=y-\{y\}$,
$$M=x\sum_{i=1}^{k}\left(\frac{i}{x}-\left\{\frac{i}{x}\right\}\right)=\frac{k(k+1)}{2}-x \sum_{i=1}^{k}\left\{\frac{i}{x}\right\}$$
Since $0\leq\{y\}<1$, the coefficient of $x$ is finite and thus, $$\lim_{x\to0^+}M=\frac{k(k+1)}{2}$$
Is this correct?
| Your proof is correct. Here is another way to compute this limit.
Note that $\frac{i}{x}-1 < \left \lfloor \frac{i}{x} \right \rfloor \le \frac{i}{x}$,$\forall i=\overline{1,k}$.
Therefore, $\frac{k(k+1)}{2x} -k < \left \lfloor \frac{1}{x} \right \rfloor + \left \lfloor \frac{2}{x} \right \rfloor +...+ \left \lfloor \frac{k}{x} \right \rfloor \le \frac{k(k+1)}{2x}$.
Hence, $\frac{k(k+1)}{2} -kx < x\left(\left \lfloor \frac{1}{x} \right \rfloor + \left \lfloor \frac{2}{x} \right \rfloor +...+ \left \lfloor \frac{k}{x} \right \rfloor \right) \le \frac{k(k+1)}{2}$.
By the squeeze theorem we get that your limit equals $\frac{k(k+1)}{2}$ as you proved.
| {
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Factor $2x^3 + 3x^2 - 2x$ I am having the most difficult time factoring this equation out. I should have paid more attention in precalculus in high school :(
$2x^3 + 3x^2 - 2x = x(2x^2 + 3x - 2) = x(2x-1)(x+2)$
How do you get the $(2x-1)(x+2)$ out of $(2x^2+3x-2)$?
Quadratic equation didn't work, I can't take out a common variable, I am out of options in my head
| You can always use the quadratic equation to factor a quadratic polynomial $Q(x)$.
If the solutions to $Q(x)=0$ are $x=r$ and $x=s$, then the polynomial can be factored as
$$Q(x)=k(x-r)(x-s)$$
where $k$ is the leading coefficient of $Q(x)$.
In your case, $Q(x)=2x^2+3x-2$ and the quadratic formula says the solutions to $Q(x)=0$ are
$$x=\frac{-3 \pm\sqrt{3^2-4(2)(-2)}}{2(2)}=\frac{-3 \pm 5}{4}$$
So the solutions are $x=\frac12$ and $x=-2$.
This means $Q(x)$ factors as $2(x-\frac12)(x+2)$. It is convenient to combine the leading constant with the first factor to get rid of the fraction, giving
$$\boxed{Q(x)=(2x-1)(x+2)}.$$
Addendum:
It may be worth remembering how the quadratic formula is derived in the first place. Believe it or not, it's through factoring (!). Here's how. Suppose we have a quadratic equation (assume $a\neq0$, otherwise it's really just a linear equation):
$$ax^2+bx+c=0$$
We work towards the goal of building the constant discriminant $b^2-4ac$ on the right side:
$$ax^2+bx = \boxed{-c}\tag{subtract $c$ from both sides}$$
$$4a^2x^2 + 4abx = \boxed{-4ac}\tag{multiply both sides by $4a$}$$
$$4a^2x^2 + 4abx +b^2= \boxed{b^2-4ac}\tag{add $b^2$ to both sides}$$
When you do this, the left side is always a perfect square polynomial (the square of a binomial). You can then factor the left side:
$$\overbrace{(2ax+b)(2ax+b)}^{4a^2x^2 + 4abx +b^2\textrm{ factored}}=
\overbrace{b^2-4ac}^{\textrm{constant discriminant}}$$
Assuming that $b^2-4ac$ isn't negative, we can write $b^2-4ac=(\sqrt{b^2-4ac})^2$ so
$$(2ax+b)^2 = (\sqrt{b^2-4ac})^2$$
$$(2ax+b)^2 - (\sqrt{b^2-4ac})^2 = 0$$
which can be factored, being a difference of squares, as
$$(2ax+b - \sqrt{b^2-4ac})(2ax+b + \sqrt{b^2-4ac})=0$$
Dividing both sides by $2a$ twice, and then multiplying by $a$, we have
$$a\left(x+\frac{b- \sqrt{b^2-4ac}}{2a}\right)\left(x+\frac{b+ \sqrt{b^2-4ac}}{2a}\right)=0$$
$$a\left(x-\left[\frac{-b+ \sqrt{b^2-4ac}}{2a}\right]\right)\left(x-\left[\frac{-b- \sqrt{b^2-4ac}}{2a}\right]\right)=0$$
The left side is indeed of the form $k(x-r)(x-s)$, where $k$ is the leading coefficient of $Q(x)$ (that is, $a$) and $r,s$ are the solutions $\tfrac{-b\pm \sqrt{b^2-4ac}}{2a}$from the quadratic formula.
If the discriminant is negative, the same formula works but the square root will produce an imaginary number. The quadratic will still factor, but the factors will contain complex numbers. In other words, if $b^2-4ac<0$, we can write
$$\pm\sqrt{b^2-4ac} = \pm i\sqrt{-(b^2-4ac)}.$$
| {
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A mistake on computing $\int \frac{dx}{\sqrt{x+1}+1}$ I have to find $\int \frac{dx}{\sqrt{x+1}+1}$. This was my attempt, which is wrong and I cannot find where exactly is the mistake.
First I write $\frac{1}{\sqrt{x+1}+1}=\frac{\sqrt{x+1}-1}{x}=\frac{\sqrt{x+1}}{x}-\frac{1}{x}$, therefore $\int \frac{dx}{\sqrt{x+1}+1}=\int \frac{\sqrt{x+1}}{x}dx-\log\left (x\right )$, so I only have to deal with $\int \frac{\sqrt{x+1}}{x}dx$.
Setting $u=\sqrt{x+1}$, we obtain $du=\frac{1}{2\sqrt{x+1}}dx$, therefore $dx=2udu$ and $x=u^2-1$, so
$\int\frac{\sqrt{x+1}}{x}dx=\int\frac{2u^2}{u^2-1}du=\int \left(2+\frac{2}{u^2-1}\right)du=2\sqrt{x+1}+\int \left(\frac{1}{u-1}-\frac{1}{u+1}\right)du=$
$=2\sqrt{x+1}+\log \left (\sqrt{x+1}-1\right )-\log \left (\sqrt{x+1}+1\right )=2\sqrt{x+1}+\log\left ( \frac{\sqrt{x+1}-1}{\sqrt{x+1}+1} \right )$
But then $\int \frac{dx}{\sqrt{x+1}+1}=2\sqrt{x+1}+\log\left ( \frac{\sqrt{x+1}-1}{\sqrt{x+1}+1} \right )-\log\left (x\right )=2\sqrt{x+1}+\log\left ( \frac{x+2-2\sqrt{x+1}}{x^2} \right )$, which is not what I am supposed to obtain. The actual answer is $2\sqrt{x+1}-2\log\left ( 1+\sqrt{x+1} \right )$.
Where is my mistake?
(I already know a correct way to solve it, I just want to know where I am committing a mistake).
| Note that $$\ln\frac{x+2-2\sqrt{x+1}}{x^2}=\ln\frac{(\sqrt{x+1}-1)^2}{x^2}=2\ln\frac{\sqrt{x+1}-1}{x}$$
if we further simplify,
$$2\ln(\frac{\sqrt{x+1}-1}{x}\cdot\frac{\sqrt{x+1}+1}{\sqrt{x+1}+1})
=2\ln\frac{1}{\sqrt{x+1}+1}=-2\ln(\sqrt{x+1}+1)$$
You did nothing wrong.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3132934",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 1
} |
If nine coins are tossed, what is the probability that the number of heads is even? If nine coins are tossed, what is the probability that the number of heads is even?
So there can either be 0 heads, 2 heads, 4 heads, 6 heads, or 8 heads.
We have $n = 9$ trials, find the probability of each $k$ for $k = 0, 2, 4, 6, 8$
$n = 9, k = 0$
$$\binom{9}{0}\bigg(\frac{1}{2}\bigg)^0\bigg(\frac{1}{2}\bigg)^{9}$$
$n = 9, k = 2$
$$\binom{9}{2}\bigg(\frac{1}{2}\bigg)^2\bigg(\frac{1}{2}\bigg)^{7}$$
$n = 9, k = 4$
$$\binom{9}{4}\bigg(\frac{1}{2}\bigg)^4\bigg(\frac{1}{2}\bigg)^{5}$$
$n = 9, k = 6$
$$\binom{9}{6}\bigg(\frac{1}{2}\bigg)^6\bigg(\frac{1}{2}\bigg)^{3}$$
$n = 9, k = 8$
$$\binom{9}{8}\bigg(\frac{1}{2}\bigg)^8\bigg(\frac{1}{2}\bigg)^{1}$$
Add all of these up:
$$=.64$$ so there's a 64% chance of probability?
| There are two cases here:
*
*There's an even number of heads: 0, 2, 4, 6 or 8 heads
*There's an odd number of heads: 1, 3, 5, 7 or 9 heads
But notice that having an odd number of heads means having an even number of tails (e.g. 5 heads means 4 tails), so the second case is the same as:
*
*There's an even number of tails: 0, 2, 4, 6 or 8 tails
Since heads and tails are equally probable, we can, by symmetry, see that these two cases have the same probability. Therefore each must have probability $1/2$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3134991",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "54",
"answer_count": 13,
"answer_id": 6
} |
Find range of $x$ satisfying $\left \lfloor \frac{3}{x} \right \rfloor+\left \lfloor \frac{4}{x} \right \rfloor=5$ Find range of $x$ satisfying $$\left \lfloor \frac{3}{x} \right \rfloor +\left \lfloor \frac{4}{x} \right \rfloor=5$$ Where $\lfloor\cdot\rfloor$ is the floor function
My try:
As far as domain of LHS is concerned we have $x \ne 0$ and since RHS is positive, we have $x \gt 0$
Now since LHS is sum of two positive integers, let us suppose:
$$\left \lfloor \frac{3}{x} \right \rfloor=m$$ and
$$\left \lfloor \frac{4}{x} \right \rfloor=5-m$$
Thus we have:
$$ m \le \frac{3}{x} \lt m+1$$
$$5-m \le \frac{4}{x} \lt 6-m$$
Adding both we get:
$$5 \le \frac{7}{x} \lt 7$$
$\implies$
$$1 \lt x \le \frac{7}{5}$$
Hence $$x \in (1, 1.4]$$
But answer in book is given as $$x \in (1,\frac{4}{3})$$
What went wrong?
| Your method is fine, but you dropped the inequalities prematurely. Indeed, by comparing the extreme members, we must fulfill
$$\max\left(\dfrac m3,\dfrac{5-m}4\right)<\min\left(\dfrac{m+1}3,\dfrac{6-m}4\right)$$
which is only possible with $m=2$.
Then
$$\max\left(\dfrac 23,\frac34\right)\le\frac1x<\min\left(1,1\right)$$ constrains $x$ to be smaller than $\dfrac43$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3137417",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 3
} |
Solve the inequality $x^2 - 3 > 0$ For the inequality $x^2 - 3 > 0$, we have
\begin{align} x^2 - 3 & = (x+\sqrt 3)(x- \sqrt 3) > 0 \end{align}
Therefore,
\begin{align} x > -\sqrt 3 \end{align}
and
\begin{align} x > \sqrt 3 \end{align}
But, this is clearly wrong as we should get $x < -\sqrt 3$ and $x > \sqrt 3$ as the two intervals. What have I done wrong?
| You have $3$ intervals to consider: $(-\infty,-\sqrt3), (-\sqrt3,\sqrt3)$ and $(\sqrt3,\infty)$. The sign of the product $(x-\sqrt3)(x+\sqrt3)$ is constant on each interval. (That's by continuity, since the function $f(x)=x^2-3$ must pass through zero to change sign.)
Use a test point in each interval: \begin{align} (-\infty,-\sqrt3): f(x)&\gt0\\ (-\sqrt3,\sqrt3): f(x)&\lt0\\ (\sqrt3,\infty): f(x)&\gt0\end{align}
So actually you get $x\lt -\sqrt3 \color{blue}{\text{ or }} x\gt\sqrt3 $.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3139705",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 1
} |
Proof that $a\nabla^2 u = bu$ is the only homogenous second order 2D PDE unchanged/invariant by rotation Looking for feedback and maybe simpler intuition for my proof of the theorem, shown below
The statement of the theorem:
Theorem
Among all second-order homogeneous PDEs in two dimensions with constant coefficients, show that the only ones that do not change under a rotation of the coordinate
system (i.e., are rotationally invariant), have the form
$$a\nabla^2u = bu $$
Proof:
The general PDE of those conditions is written as:
$$a_1u_{xx} + 2a_2u_{xy} + a_3 u_{yy} + b_1u_x +b_2u_y +cu = 0$$
A counter-clockwise rotation of a point $x,y$ can be given by the rotation matrix is given by the figure below,
where with some basic geometry, we can derive
$$x' = \|{\mathbf{v}}\|\cos\left(\theta + \tan^{-1}\left(\frac{y}{x}\right)\right) = x\cos\theta - y\sin\theta$$
$$y' = \|{\mathbf{v}}\|\sin\left(\theta + \tan^{-1}\left(\frac{y}{x}\right)\right) = x\sin\theta + y\cos\theta$$
This can be summarized in a matrix transformation
$$\begin{bmatrix} x' \\ y' \end{bmatrix} =
\begin{bmatrix} \cos\theta & -\sin\theta \\ \sin\theta & \cos\theta \end{bmatrix} \begin{bmatrix}x\\y\end{bmatrix}$$
giving the map:
\begin{align*}
x\mapsto x'=x\cos\theta - y\sin\theta \\
y\mapsto y'=x\sin\theta+y\cos\theta
\end{align*}
From here we find derivatives of our new coordinates:
$$ \frac{\partial x'}{\partial x} = \cos\theta \quad \frac{\partial y'}{\partial x}=\sin\theta$$
$$\frac{\partial x'}{\partial y} = -\sin\theta \quad \frac{\partial y'}{\partial y} = \cos\theta$$
Now the first derivatives of $u(x',y')$ with respect to $x,y$:
\begin{align*}
u_x = \frac{\partial u}{\partial x} = \frac{\partial u}{\partial x'} \frac{\partial x'}{\partial x} + \frac{\partial u}{\partial y'} \frac{\partial y'}{\partial x} = u_{x'}\cos\theta + u_{y'} \sin\theta \\
u_y = \frac{\partial u}{\partial x} = \frac{\partial u}{\partial x'} \frac{\partial x'}{\partial y} + \frac{\partial u}{\partial y'} \frac{\partial y'}{\partial y} = -u_{x'}\sin\theta + u_{y'} \cos\theta
\end{align*}
And then the second derivatives:
\begin{align*}
&u_{xx} = u_{x'x'} \cos^2\theta +2u_{x'y'}\sin\theta\cos\theta+ u_{y'y'} \sin^2\theta \\
&u_{xy} = -u_{x'x'} \cos\theta\sin\theta - u_{y'x'} \sin^2\theta + u_{x'y'} \cos^2\theta + u_{y'y'}\sin\theta\cos\theta \\
&u_{yy}= u_{x'x'} \sin^2\theta -2u_{x'y'}\sin\theta\cos\theta+ u_{y'y'} \cos^2\theta
\end{align*}
Substituting into the general PDE and rearranging factors of partial derivatives, can be written as
$$ \color{blue}{\widetilde{a_1}u_{x'x'} + \widetilde{a_2}u_{x'y'} + \widetilde{a_3} u_{y'y'} + \widetilde{b_1}u_{x'} +\widetilde{b_2}u_{y'} +\widetilde{c}u = 0} $$
where:
\begin{align*}
&\widetilde{a_1} = a_1 \cos^2\theta -2a_2\cos\theta\sin\theta + a_3\sin^2\theta \\
&\widetilde{a_2} = (a_1-a_3)\sin 2\theta + 2a_2 \cos 2\theta \\
&\widetilde{a_3} = a_1 \sin^2\theta + 2a_2\sin\theta\cos\theta +a_3\cos^2\theta \\
&\widetilde{b_1} = b_1\cos\theta - b_2\sin\theta \\
&\widetilde{b_2} = b_1\sin\theta + b_2\cos\theta \\
&\widetilde{c} = c
\end{align*}
Since we require rotational invariance, the original equation and the transformed PDE must have the same value,namely $0$, on all of $u$. This can be written as
$$\small{a_1u_{xx} + 2a_2u_{xy} + a_3 u_{yy} + b_1u_x +b_2u_y +cu = \widetilde{a_1}u_{x'x'} + \widetilde{a_2}u_{x'y'} + \widetilde{a_3} u_{y'y'} + \widetilde{b_1}u_{x'} +\widetilde{b_2}u_{y'} +\widetilde{c}u} $$
whence:
\begin{align}
\tag{1} a_1& = a_1 \cos^2\theta -2a_2\cos\theta\sin\theta + a_3\sin^2\theta&\\
\tag{2} 2a_2& = (a_1-a_3)\sin 2\theta + 2a_2 \cos 2\theta& \\
\tag{3} a_3& = a_1 \sin^2\theta + 2a_2\sin\theta\cos\theta +a_3\cos^2\theta& \\
\tag{4} b_1& = b_1\cos\theta - b_2\sin\theta&\\
\tag{5} b_2& = b_1\sin\theta + b_2\cos\theta&\\
\tag{6} c &= c&
\end{align}
Excluding the trivial case where $\{a_i\},\{b_i\},c = 0$, we can make several conclusions. Note that the deductions below are made with the understanding that any arbitrary angle $\theta$ must be valid, hence it is erroneous to apply $\theta =0$ in order to reach equality.
*
*can only be true when $a_1=a_3$ and $a_2=0$,
*implies $a_2=0$ and $a_1=a_3$,
*like (1) is only true when $a_1=a_3$ and $a_2=0$,
*is true when $b_1=b_2=0$
*like (4) is true when $b_1=b_2=0$,
*implies $c\in \mathbb{R}$ is valid.
All together we then know $a1=a3$, $a_2=b_1=b_3=0$ and $c=c$.
Returning to the rotated PDE we now know:
$$\widetilde{a_1} = a_1,
\widetilde{a_2} = 0,
\widetilde{a_3} = a_1 ,
\widetilde{b_1} = 0,
\widetilde{b_2} = 0,
\widetilde{c} = c
$$
So the PDE under a rotation, $u(x',y')$, becomes
\begin{align*}
&a_1u_{x'x'} + a_1u_{y'y'} +cu = 0\\
\Rightarrow& a_1(u_{x'x'}+u_{y'y'}) = -cu \\
\Rightarrow& a\nabla^2{u} = bu
\end{align*}
where we have chosen $a_1=a, -c = b$ for all $a,b\in\mathbb{R}$. This is in terms of the new rotation $u(x',y')$, and so it remains to show that $\nabla^2{u(x,y)} = \nabla^2{u(x',y')}$, consider from the derivatives before:
\begin{align*}
&u_{xx} = u_{x'x'} \cos^2\theta +2u_{x'y'}\sin\theta\cos\theta+ u_{y'y'} \sin^2\theta \\
&u_{yy}= u_{x'x'} \sin^2\theta -2u_{x'y'}\sin\theta\cos\theta+ u_{y'y'} \cos^2\theta
\end{align*}
Sum them together
$$u_{xx}+u_{yy}= u_{x'x'} (\sin^2\theta+\cos^2\theta) -2u_{x'y'}\sin\theta\cos\theta+ 2u_{x'y'}\sin\theta\cos\theta+ u_{y'y'} (\cos^2\theta+\sin^2\theta)=u_{x'x'}+u_{y'y'}$$
hence, $\nabla^2{u(x,y)} = \nabla^2{u(x',y')}$ as required.
Therefore a rotation applied to any second order homogeneous 2D PDE with constant coefficents will transform to a PDE of the form $a\nabla^2{u} = bu$ under the rotated coordinate $x',y'$, which we have shown to be equivalent under the regular coordinates $x,y$. This is the only PDE that is invariant under rotation. $$\tag*{$\blacksquare$}$$
Additional Remark
Also was just curious about rotational invariant functions and operators. Anything that solves laplaces equation ($\nabla^2=0$) is called a harmonic function, and satisfies properties such as the mean value property and the maximal principle. I assumed at first harmonic functions meant they were radial, but I think it is more along the lines of being symmetric?
The laplacian is rotationally invariant but the laplace equation has some solutions which are radial (rotationally invariant) and some which are not. I also noticed that the converse is not true, i.e, a radial function does not imply $\nabla^2 = 0$, like $f(x,y)=x^2+y^2$
See bounty remark below
| We say that a linear operator $L$ is rotationally invariant if and only if $L$ commutes with the orthogonal group, i.e.
$[L, O] = 0$ for every $O \in \text{O}(n)$.
Hence what you are proving is that if $L$ is a second order linear operator then
\begin{align}
LO[f](x) = L[f(O x)] = [Lf](O x) = OL[f](x)
\end{align}
if and only if $L = a\Delta-bI$. Moreover, this is equivalent to showing
\begin{align}
L[f](x, y) = O^{-1}LO[f](x, y)
\end{align}
for every function $f$, that is, $L$ remains fixed under the conjugation action of orthogonal transformations.
Example:
Let us look at an example. Consider $f(x, y) = x e^y$ and $L=\Delta$. Observe
\begin{align}
O[f]=&\ f(\cos\theta x - \sin\theta y, \sin\theta x+\cos\theta y) \\
=&\ (\cos\theta x-\sin\theta y)e^{\sin\theta x+\cos\theta y}
\end{align}
where
\begin{align}
O
=
\begin{pmatrix}
\cos\theta & -\sin\theta\\
\sin\theta & \cos\theta
\end{pmatrix}.
\end{align}
Then we see that
\begin{align}
g(x, y):=LO[f](x, y)= e^{\sin\theta x+\cos\theta y}(x\cos\theta-y\sin\theta)
\end{align}
and finally
\begin{align}
O^{-1}[g](x, y) =&\ g(\cos\theta x+\sin\theta y, -\sin\theta x+\cos\theta y)\\
=&\ e^{\sin\theta\cos\theta x+\sin^2\theta y-\sin\theta\cos\theta x+\cos^2\theta y}(\cos^2\theta x+\sin\theta \cos\theta y+\sin^2\theta x-\sin\theta\cos\theta y)\\
=&\ xe^y.
\end{align}
Hence
\begin{align}
O^{-1}LO [f](x, y) = xe^y.
\end{align}
Also, note that $\Delta f =x e^y$. Thus, $L[f](x, y) = O^{-1}LO[f](x, y)$.
Radial Function: In fact the only radial harmonic solution defined on the entire $xy$-plane are constants. This is a simple consequence of the mean-value identity and the maximum principle for harmonic function. Hence $L$ being rotationally invariant doesn't mean
\begin{align}
f(Ox) = f(x) \text{ for all } O \in \text{O}(2)\ \ \implies \ \ \Delta f =0.
\end{align}
Last Remark: Unfortunately, I don't think there are much easier ways to show the only rotationally invariant second order differential operators are given by $L=a\Delta-bI$ other than direct computation.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3140725",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "13",
"answer_count": 2,
"answer_id": 0
} |
calculate the surface area of the part of a cylinder $ x^2 + (y-1)^2 = 1 $ that is inside the sphere $ x^2 + y^2 + z^2 = 4 $. calculate the surface area of the part of a cylinder $ x^2 + (y-1)^2 = 1 $ that is inside the sphere
$ x^2 + y^2 + z^2 = 4 $.
my trial :
The Domain of integration on the YZ plane is :
solving :
(*) $ x^2 + (y-1)^2 = 1 $
$ x^2 + (y)^2 + z^2 \leq 4 $
We get : $ 0 \leq z \leq \sqrt{4-2y}$
and $ -2 \leq y \leq 2 $
$||\nabla{Cylinder(x,y,z)}|| = ||(2x,2(y-1),0)|| = 2 ~$ see(*)
S = $2~\int_{-2}^{2}\int_{0}^{\sqrt{4-2y}}~~2~dydz$ ( by symmetry *2)
I also tried solving by parmetrization and i didn't get the same answer i really need HELP is this way alright ? or is there something wrong
| Let $\gamma : [0, 2\pi] \to \mathbb{R}^2, \gamma(t) = (\cos t, 1+\sin t)$ be the parameterization of the circle $x^2+(y-1)^2 = 1$ in the $xy$-plane.
The height of the cylinder over a point $(x,y)$ on this circle is given by $z = \sqrt{4-x^2-y^2}$ so the area is
$$A = 2\int_\gamma z\,d\gamma = 2\int_\gamma \sqrt{4-x^2-y^2}\,d\gamma = 2\int_0^{2\pi}\sqrt{4-\cos^2t -(1+\sin t)^2}\,dt = 2\int_0^{2\pi}\sqrt{2(1-\sin t)}\,dt$$
We can solve this integral by noting that $1-\sin t = \left(\sin\frac{t}2 - \cos\frac{t}2\right)^2$ so we get
$$A = 16$$
Is this result what you are getting?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3141139",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Evaluate $\int \frac{x^2 + x + 1}{(x+1)^2(x+2)}dx$ via partial fractions $\int \frac{x^2 + x + 1}{(x+1)^2(x+2)}dx$
= $ \int \frac{Ax+B}{(x+1)} + \frac{Cx+B}{(x+1)^2} + \frac{Dx+E}{x+2}$
= $\int (Ax+B)(x+1)(x+2) + (Cx+B)(x+2) + (Dx+E)(x+1)^2$
= $ \int Ax^3 + 3Ax^2 + 2Ax + Bx^2 + 3Bx + 2B + Cx^2 + 2Cx + Bx + 2B + Dx^3 + 2Dx^2 + Dx + Ex^3 + 2Ex^2 + E$
= $ \int (A + D + E)x^3 + (3A + B + C + 2D + 2E)x^2 + (2A + 2C + 4B + D)x + (4B + E)$
Turn into matrix, find reduced row echelon form to solve system of equations:
$\begin{bmatrix}
0 & 0 & 0 & 0 & 0 & 0 \\
1 & 0 & 0 & 1 & 1 & 0 \\
3 & 1 & 1 & 2 & 2 & 1 \\
2 & 4 & 2 & 1 & 0 & 1 \\
0 & 4 & 0 & 0 & 1 & 1 \\
\end{bmatrix}$
This is where things go wrong, apparently this doesn't reduce down properly :( and I have been relying on RREF to solve systems of equations everytime up until now.
I am also confused where the $x^2 + x + 1$ is supposed to go exactly. I know I put it into the system of equations later but until then, I feel like I kinda just ignored it and left it out of all my work up until then (is that okay?).
Other note: I recognize that this is a proper rational fraction so no long division is nesecary and that this has irreducible factors that are repeated so that is why I split them up into the partial fractions up above in that manner. Did I miss any intermittent steps that made the RREF turn out wrong? I am not sure where I went wrong thus far either
| It must be $$\frac{x^2+x+1}{(x+1)^2(x+2)}=\frac{A}{x+1}+\frac{B}{(x+1)^2}+\frac{C}{x+2}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3141438",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 0
} |
Evaluate $\int \frac{x+4}{x^2 + 2x + 5}$ I am having issues with this integral. I am not sure if it is irreducible or not. I can't use the quadratic formula, but I can rearrange the integral to be $\int \frac{x+4} {(x^2 + 2x + 1) + 4}$, but I don't know how to deal with the $+4$.
Here is my work treating the quadratic equation as irreducible with no repeating factors.
$\int \frac{x+4}{x^2 + 2x + 5}$
= $\frac {Ax + B} {x^2 + 2x + 5} $
= $Ax+B(x^2 + 2x + 5)$
= $Ax^3 + 2Ax^2 + 5Ax + Bx^2 + 2Bx + 5B$
= $Ax^3 + (2A + 2B)x^2 + (5A + 2B)x + 5B$
However I get stuck trying to solve my system of equations. This leads me to believe that I did the partial fractions improperly.
$\begin{bmatrix}
1 & 0 & 0 \\
2 & 2 & 0 \\
5 & 2 & 1 \\
0 & 5 & 4 \\
\end{bmatrix}$
| \begin{align}
\frac{x+4}{x^2+2x+5} &= \frac{x+4}{(x+1)^2 +4} \\
&=\frac{\color{blue}{x+1}}{\color{blue}{(x+1)}^2 +4} +\frac{3}{\color{blue}{(x+1)}^2 +4} \\
&=\frac{\color{blue}{u}}{\color{blue}{u}^2 +4} +\frac{3}{\color{blue}{u}^2 +4}
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3143709",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 2
} |
A triangle has sides x, x+4, and 3x−5. What is the possible range of x?
A triangle has sides $x, x+4,$ $and$ $3x−5$. What is the possible range of x?
The answer to this is not:
$2x-9$ $<$ $x$ $<$ $4x-1$
Which is gotten from:
--> $(3x-5)-(x+4)<x<3x-5+x+4$
--> $3x-5-x-4 < x < 4x-1$
--> $5x-9 < x < 4x-1$
So what is the answer?
The correct answer has to be in:
__ < x < __
| Hint Consider using the triangle inequality
$$a+b>c$$
Thus you have that $$x+(x+4)>3x-5\iff 9>x\tag{1}$$
$$x+(3x-5)>x+4\iff 3x>9\iff x>3$$ Hence
$$x\in(3, 9)\text{ or equivalently }\color{red}{3<x<9}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3144286",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
If the equation $\sin^2x-a\sin x+b=0$ has only one solution in $(0,\pi)$, then what is the range of $b$?
If the equation $\sin^2(x)-a\sin(x)+b=0$ has only one solution in $(0,\pi)$, then what is the range of $b$?
What I try:
$$\displaystyle \sin x=\frac{a\pm \sqrt{a^2-4b}}{2}\in\bigg(0,1\bigg)$$
for one real solution $a^2=4b$.
How do I solve it?
| To $x$ to have only one solution in the range $(0,\pi)$, then $x$ must be $\pi\over 2$, according to the unit circle for trigonometry.
Since, $\sin{\frac{\pi}{2}} = 1$,
$$\sin^2 x - a\sin x +b = 1-a+b =0$$
$$\implies a-b=1 \to\text{eq.1} $$
From what you have tried we get,
$$a^2 =4b \to \text{eq.2} $$
Thus, to satisfy eqs. 1&2 is,
From eq.1 , we get,
$$a=1+b$$
Subtituting into eq.2,
$$(1+b)^2=4b$$
$$1+2b+b^2=4b$$
$$(b-1)^2=0$$
$$\therefore b=1$$
and $a=2$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3144983",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Question about partial fractions with irreducible quadratic factors Given this rational function: $$\frac{-4x^4-2x^3-26x^2-8x-44}{(x+1)(x^2 +3)^2}$$
The decomposition would look like this: $$\frac{A}{x+1} + \frac{Bx+C}{(x^2+3)} + \frac{Dx+E}{(x^2+3)^2}$$
And the final answer would be: $$\frac{-4}{x+1} - \frac{2}{(x^2+3)} - \frac{2}{(x^2+3)^2}$$
But, if you were to set it up like this: $$\frac{A}{x+1} + \frac{B}{(x^2+3)} + \frac{C}{(x^2+3)^2}$$
You end up with the same answer:
$$\frac{-4}{x+1} - \frac{2}{(x^2+3)} - \frac{2}{(x^2+3)^2}$$
I want to know why this is.
I thought that for irreducible quadratic factors, you needed a linear term in the numerator. Is this purely coincidence that it worked with constant terms in the numerator?
| I would consider this a coincidence.
$$\frac{-4x^4-2x^3-26x^2-8x-4\mathbf{5}}{(x+1)(x^2 +3)^2},$$
for example, requires the linear term.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3146263",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "12",
"answer_count": 4,
"answer_id": 1
} |
Integrate $\int \frac{dx}{\sqrt{x^2-9}}$ by trig substitution $$
\begin{align}
x = 3\sec\theta, dx &= 3\sec\theta\tan\theta d\theta\\\\
\int \frac{dx}{\sqrt{x^2-9}} &= \int \frac{3\sec\theta\tan\theta d\theta}{\sqrt{(3\sec\theta)^2 - 3^2}} \\\\
& = \int \frac{3\sec\theta\tan\theta d\theta}{\sqrt{3^2(\sec^2\theta -1)}} \\\\
&= \int \frac{3\sec\theta\tan\theta d\theta}{\sqrt{3^2\tan^2\theta}} = \int \sec\theta\\\\
&= \ln|\sec\theta + \tan\theta| + C = \ln| \frac{x}{3} + \frac{\sqrt{x^2-9}}{3}|
\end{align}
$$
However, wolphram alpha says the answer is $\ln |x+ \sqrt{x^2-9}$
I am wondering how did it get rid of the 3 in the denominator? This is pretty much how I got my answer:
$$
x = 3\sec\theta \\
\frac{x}{3} = \sec\theta \\
\frac{\sqrt{x^2-9}}{3} = \tan\theta
$$
| Note that
$$\ln\left| \frac{x}{3} + \frac{\sqrt{x^2-9}}{3}\right| = \ln\left|x+\sqrt{x^2-9}\right|-\ln 3$$
and $\ln3$ is a constant. Since $C$ is an arbitrary constant, you can define a new constant $C_1 = C - \ln3$ and the answer is
$$\ln\left|x+\sqrt{x^2-9}\right| + C_1$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3146542",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
Differentiate $11x^5 + x^4y + xy^5=18$ I am not sure how to differentiate $11x^5 + x^4y + xy^5=18$. I have a little bit of experience with implicit differentiation, but I'm not sure how to handle terms where both variables are multiplied together.
I have tried
$$\frac{d}{dx}(11x^5 + x^4y+xy^5) = \frac{d}{dx}(18)$$
$$\frac{d}{dx}(11x^5)+\frac{d}{dx}(x^4y) + \frac{d}{dx}(xy^5)=0$$
differentiating each term
$$\frac{d}{dx} (11x^5)=11\frac{d}{dy}(5x^4) = 55x^4$$
$$\frac{d}{dx} (x^4y) = [4x^3 \cdot y] + [1 \cdot x^4] = 4yx^3+x^4$$
$$\frac{d}{dx}(xy^5) = [1 \cdot y^5] + [x \cdot 5y^4] = y^5 + 5xy^4$$
finding $\frac{dy}{dx}$
$$\frac{dy}{dx}=\frac{-x}{y} = \frac{-[4yx^3+x^4] + [y^5+5xy^4]}{55x^4}$$
According to the website I'm using, "WebWork", this is wrong.
| For each term involving both $x$ and $y$, differentiate the $x$ and $y$ parts separately and combine the two using the product rule:
$$(x^4y)'=x^4(y)'+(x^4)'y=x^4\frac{dy}{dx}+4x^3y$$
$$(xy^5)'=x(y^5)'+(x)'y^5=5xy^4\frac{dy}{dx}+y^5$$
Thus
$$55x^4+x^4\frac{dy}{dx}+4x^3y+5xy^4\frac{dy}{dx}+y^5=0$$
$$(x^4+5xy^4)\frac{dy}{dx}=-(55x^4+4x^3y+y^5)$$
$$\frac{dy}{dx}=-\frac{55x^4+4x^3y+y^5}{x^4+5xy^4}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3147423",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 5,
"answer_id": 1
} |
All the roots of $5\cos x - \sin x = 4$ in the interval $0^{\circ} \leq x \leq 360^{\circ}$? This is a problem that I stumbled upon in one of my books.
Representing $5\cos x - \sin x$ in the form $R\cos(x + \alpha)$ (as demanded by the question):
$
\rightarrow R = \sqrt{5^2 + ({-}1)^2} = \sqrt{26}\\
\rightarrow R\cos x \cos \alpha - R\sin x \sin \alpha = 5\cos x - \sin x \\
\rightarrow ➊\hspace{0.25cm}5 = \sqrt{26}\cos \alpha \\
\rightarrow ➋\hspace{0.25cm}{-}1 = \sqrt{26}\sin \alpha \\
$
So here are my question(s): $\\$
• Why is only ➊ working out?
• When I find one solution in the interval, which is 27°, why is another solution like $(360 - x)$ not working out (in this case 333°)?
• I see 310.4° as a solution to this equation in my book. How do you get to 310.4° from 27° (which is the solution I got)? Which $\cos$ identity am I missing? And surprisingly, my book doesn't include 27° as a solution!
| It seems that you know why $R=\sqrt{26}$,
Hence $5\cos x - \sin x$ can be expressed as $\sqrt{26}(\frac{5}{\sqrt{26}}\cos x - \frac{1}{\sqrt{26}}\sin x)$
$$\text{Let, } \cos\alpha = \frac{5}{\sqrt{26}} \text{ and } \sin\alpha = \frac{1}{\sqrt{26}}$$
Hence, $5\cos x - \sin x$ can now be expressed as $\sqrt{26}\cos (x + \alpha)$.
To solve your initial equation, we can now write it as,
$$\sqrt{26}\cos (x + \alpha) =4$$
$$\cos (x + \alpha) = \frac{4}{\sqrt{26}}$$
$$x +\alpha = \cos^{-1} \frac{4}{\sqrt{26}}$$
If $\cos\alpha = \frac{5}{\sqrt{26}}$, then $\alpha = \cos^{-1} \frac{5}{\sqrt{26}}$,
$$x = \cos^{-1} \frac{4}{\sqrt{26}} - \alpha \rightarrow x = \cos^{-1} \frac{4}{\sqrt{26}} - \cos^{-1} \frac{5}{\sqrt{26}}$$
$$\therefore x = 27^o \text{ or } 321.40^o$$
| {
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"url": "https://math.stackexchange.com/questions/3148923",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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} |
$e^{\frac{1}{x}} < 1 + \frac{1}{x-1} $ I want to prove that $e^{1/x} < 1 + \frac{1}{x-1}$ for $x > 1$.
The first thing I tried is differentiating $f(x) = e^{1/x} - 1 - \frac{1}{x-1}$: this gives
$$ \frac{1}{x^2} \left( \left(1 + \frac{1}{x-1}\right)^2 - e^{\frac{1}{x}} \right) $$
If I could show that $\left(1 + \frac{1}{x-1} \right)^2 > e^\frac{1}{x} $ for $x>1$, then $f(x)$ would be increassing, and since $ \lim_{x \to \infty} f(x) = 0$ this would mean that $f(x) < 0$ for $x>1$. However, proving that inequality is very similar to the first one, and still involves bounding above $e^\frac{1}{x}$.
The other thing I tried is considering $f(x) = e^{x-1} - (x-1) - 1$ which is increasing for $x > 1$. Then $f(\frac{1}{x})$ is decreasing, so $e^{\frac{1}{x}-1} - \frac{1}{x}$ is decreasing; However this also does not seem to help too much.
Any ideas?
| Let $t=1/x$, then $e^{1/x} < 1 + \frac{1}{x-1}$ for $x>1$ is equivalent to
$e^t -1< +\frac{t}{1-t}$ for $0<t<1$.
We have $e^t-1=t+ \frac{t^2}{2!}+\frac{t^3}{3!}+.....$ and $\frac{t}{1-t}=t+t^2+t^3+...$for $|t|<1.$
Can you proceed ?
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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} |
Hint: Showing $f_{n}(x):=\frac{n\sqrt{x} \sin{(x)}}{1+nx^{2}}$ converges in $L^{p}$ to $x^{-\frac{3}{2}}\sin{(x)}$ Let $f_{n}:]0, \infty[\to \mathbb R$ and $f_{n}(x):=\frac{n\sqrt{x} \sin{(x)}}{1+nx^{2}}$
Show that $f_{n}$ converges in $L^{p}$ to $f$ where $f(x):=x^{-\frac{3}{2}}\sin{(x)}$ and $p \in [1,2[$
My idea:
\begin{align}
\lVert f_{n}-f\rVert_{p}^{p}&=\int_{0}^{\infty}\left\lvert\frac{n\sqrt{x} \sin{(x)}}{1+nx^{2}}-x^{-\frac{3}{2}}\sin{(x)}\right\rvert^{p}dx\leq\int_{0}^{\infty}\left\lvert\frac{n\sqrt{x}}{1+nx^{2}}-x^{-\frac{3}{2}}\right\rvert^{p}dx \\
&=\int_{0}^{\infty}\left\lvert\frac{n\sqrt{x}-x^{-\frac{3}{2}}(1+nx^{2})}{1+nx^{2}}\right\rvert^{p}dx=\int_{0}^{\infty}\left\lvert\frac{n\sqrt{x}-x^{-\frac{3}{2}}+nx^{-\frac{1}{2}}}{1+nx^{2}}\right\rvert^{p}dx
\end{align}
Am I on the right track? How do I continue from here?
| First notice that
$$\frac{n\sqrt{x}\sin x}{1+nx^2} = \frac{\sqrt{x}\sin x}{\frac1n+x^2} \xrightarrow{n\to\infty} x^{-3/2}\sin x$$
pointwise.
Then notice that
$$\left|\frac{\sqrt{x}\sin x}{\frac1n+x^2}\right|^p \le \frac1{x^{p/2}}\chi_{(0,1]}(x) + \frac1{x^{3p/2}}\chi_{[1,\infty)}(x) =: g(x) \in L^1(0,\infty)$$
and also $|x^{-3/2}\sin x|^p$ is dominated by the same integrable function $g$.
Hence
$$\left|\frac{\sqrt{x}\sin x}{\frac1n+x^2} - x^{-3/2}\sin x\right|^p\le 2^p g(x)$$
so Lebesgue dominated convergence theorem yields
$$\lim_{n\to\infty}\int_0^\infty \left|\frac{\sqrt{x}\sin x}{\frac1n+x^2} - x^{-3/2}\sin x\right|^p\,dx = \int_0^\infty\lim_{n\to\infty} \left|\frac{\sqrt{x}\sin x}{\frac1n+x^2} - x^{-3/2}\sin x\right|^p\,dx = \int_0^\infty 0\,dx = 0$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3150949",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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integral $C(a,b)=\int_0^{2\pi}\frac{xdx}{a+b\cos^2 x}$ I am looking for other methods to find the general integral
$$C(a,b)=\int_0^{2\pi}\frac{xdx}{a+b\cos^2x}$$
To do so, I first preformed $u=x-\pi$:
$$C(a,b)=\int_{-\pi}^{\pi}\frac{xdx}{a+b\cos^2x}+\pi\int_{-\pi}^{\pi}\frac{dx}{a+b\cos^2x}$$
The sub $x\mapsto -x$ provides
$$\int_{-\pi}^\pi \frac{xdx}{a+b\cos^2x}=0$$
and with symmetry,
$$C(a,b)=2\pi\int_0^\pi \frac{dx}{a+b\cos^2x}$$
Then we use $t=\tan(x/2)$:
$$C(a,b)=4\pi\int_0^\infty \frac1{a+b\left[\frac{t^2-1}{t^2+1}\right]^2}\frac{dt}{t^2+1}$$
$$C(a,b)=\frac{4\pi}{a+b}\int_0^\infty \frac{x^2+1}{x^4+2\frac{a-b}{a+b}x^2+1}dx$$
Then we consider
$$N_s(k)=\int_0^\infty\frac{x^{2s}}{x^4+2kx^2+1}dx$$
Then $x\mapsto 1/x$ gives
$$N_s(k)=N_{1-s}(k)$$
and with $s=0$:
$$C(a,b)=\frac{8\pi}{a+b}\int_0^\infty \frac{dx}{x^4+2kx^2+1}\qquad k=\frac{a-b}{a+b}$$
Then with
$$x^4+(2-c^2)x^2+1=(x^2+cx+1)(x^2-cx+1)$$
we have that
$$\begin{align}
u(c)=N_0\left(\frac{2-c^2}2\right)=&\frac1{4c}\int_0^\infty\frac{2x+c}{x^2+cx+1}dx-\frac1{4c}\int_0^\infty\frac{2x-c}{x^2-cx+1}dx\\
&+\frac14\int_0^\infty\frac{dx}{x^2+cx+1}+\frac14\int_0^\infty\frac{dx}{x^2-cx+1}
\end{align}$$
The first two integrals vanish and we have
$$u(c)=\frac14I(1,c,1)+\frac14I(1,-c,1)$$
where
$$I(a,b,c)=\int_0^\infty\frac{dx}{ax^2+bx+c}=\frac2{\sqrt{4ac-b^2}}\left[\frac\pi2-\arctan\frac{b}{\sqrt{4ac-b^2}}\right]$$
So
$$4u(c)=\frac2{\sqrt{4-c^2}}\left[\frac\pi2-\arctan\frac{c}{\sqrt{4-c^2}}+\frac\pi2-\arctan\frac{-c}{\sqrt{4-c^2}}\right]$$
$$u(c)=\frac{\pi}{2\sqrt{4-c^2}}$$
$$N_0(a)=\frac\pi{2\sqrt{2+2a}}$$
And then
$$C(a,b)=\frac{8\pi}{a+b}N_0\left(\frac{a-b}{a+b}\right)$$
$$C(a,b)=\frac{4\pi^2}{\sqrt{a^2+ab}}$$
How else can you prove this? Have fun ;)
As it turns out, we may be missing a factor of $1/2$, so we may actually have
$$C(a,b)=\frac{2\pi^2}{\sqrt{a^2+ab}}$$
an issue discussed in the comment section of @Song's answer.
| Putting $t = 2\pi - x$ we get $$C(a, b) = \int_{0}^{2\pi}\frac{2\pi - t}{a+b\cos^2t}\,dt= 2\pi\int_{0}^{2\pi}\frac{dt}{a + b\cos^2t} - C(a, b)$$ so that $$C(a, b) = \pi\int_{0}^{2\pi}\frac{dx}{a+b\cos^2x}$$ Since the integrand satisfies equation $f(2\pi - x) = f(x)$ it follows that $$C(a, b)=2\pi\int_{0}^{\pi}\frac{dx}{a+b\cos^2x}$$ and using the same argument again we get $$C(a,b)=4\pi\int_{0}^{\pi/2}\frac{dx}{a+b\cos^2x}=8\pi\int_{0}^{\pi/2}\frac{dx}{2a+b+b\cos 2x}$$ Putting $2x=t$ we get $$C(a, b) = 4\pi\int_{0}^{\pi}\frac{dt}{2a+b + b\cos t}=\frac{4\pi^2}{\sqrt{(2a+b)^2-b^2}}=\frac{2\pi^2}{\sqrt{a^2+ab}}$$ where we have used the standard integral formula $$\int_{0}^{\pi}\frac{dx}{A+B\cos x}=\frac{\pi}{\sqrt{A^2-B^2}}, A>|B|$$ which can be proved using the substitution $$(A + B\cos x)(A - B\cos t) = A^2-B^2$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3154937",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 2
} |
Find the area of $y = 12-x^2, y=x^2-6$ I found the x-intercepts via my graphing calculator to be $(-3,3)$ and $(3,3)$ and from it I formed the Area equation,
$$A = \int_{-3}^{3} (12-x^2 - x^2 - 6)dx = \int_{-3}^{3}(-2x^2+6)dx \\
= -\frac{2}{3}x^3 + 6x \bigg]_{-3}^{3}$$
$$A = \bigg(3\frac{2}{3}\cdot 3^3 + 6(3) \bigg) - \bigg(-\frac{2}{3} \cdot -3^3 + 6(-3)\bigg) \\
= 0 - 0$$
I calculated the integral and solved for the definite integral, but I got 0....I did the calculations correctly, but the answer doesn't seem to make sense for what the graphs look like.
Did I set the problem up wrong? or something?
Edit: I think I need to split the integral bounds up to $\int_{-3}^{0}$ and $\int_{0}^{3}$ to split the positive/negative area portions
| $$
12-x^2-\color{red}(x^2-6\color{red})=\color{red}{18}-2x^2\implies A=\left.18x-2\frac{x^3}3\right]_{-3}^3=72.
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3156121",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Fewest number of marbles in a bag such that drawing the probability of drawing 2 blue marbles is $\frac{1}{6}$.
Two marbles are randomly selected without replacement from a bag
containing blue and green marbles. Probability of drawing both blue is
$\frac{1}{6}$. If three marbles are drawn, then the probability all
three are blue is $\frac{1}{21}$. What is the fewest number of marbles
that must have been in the bag before any were drawn?
I use the criteria to make two equations:
$$\frac{x}{y}\cdot\frac{x-1}{y-1}=\frac{1}{6}$$
for drawing 2 marbles$$\frac{x}{y}\cdot\frac{x-1}{y-1}\cdot\frac{x-2}{y-2}=\frac{1}{21}$$
Where $x$ is the number of blue marbles and $y$ is the total number of marbles.
Simplifying these equations gives:
$$6x^2-6x=y^2-y$$ for the first equation and $$21x^3-63x^2+42x=y^3-3y^2+2y$$
I can't seem to solve this set of equations.
Is my approach correct? If so, how should I continue? If not, how would one solve this problem?
Thanks! Your help is appreciated!
Max0815
| Let's look at the second equation:
$$\underbrace{\dfrac{x}{y}\cdot \dfrac{x-1}{y-1}}_{\dfrac{1}{6}}\cdot \dfrac{x-2}{y-2} = \dfrac{1}{21}$$
The first two terms are the same as the LHS of your first equation. So, we can substitute.
$$\dfrac{x-2}{y-2} = \dfrac{6}{21}$$
Reducing the RHS to lowest terms gives:
$$\dfrac{x-2}{y-2} = \dfrac{2}{7}$$
There you have $x-2=2, y-2=7$
So, $x=4, y=9$ should be the smallest possible values for which the problem should hold true (if there were smaller values, it would contradict $\dfrac{2}{7}$ being in lowest terms).
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3157655",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
How can I simplify my boolean expression further I have the following boolean expression that I want to simplify
$$B\cdot D+ \overline{A\cdot B\cdot D} + \overline{B}\cdot C\cdot \overline{D}$$
Here is what I have been able to due so far
$$B\cdot D+ \overline{A} + \overline{B} + \overline{D} + \overline{B} \cdot \overline{D}\cdot C$$
I know that the answer is suppose to be
$$\overline{A} + \overline{B}\cdot B $$
How can I simplify my initial expression any further thank you very much any help.
| Factorising
$$\overline{A}+B\cdot D + \overline{B}+\overline{D} \cdot (1+\overline{B}\cdot C)$$
$$=\overline{A}+B\cdot D + \overline{B}+\overline{D} \cdot 1$$
$$=\overline{A}+B\cdot D + \overline{B}+\overline{D}$$
Then by the absorption law,
$$B\cdot D + \overline{B}=D + \overline{B}$$
So this simplifies to
$$=\overline{A}+D + \overline{B}+\overline{D}$$
$$=\overline{A} + \overline{B}+D+\overline{D}$$
$$=\overline{A} + \overline{B}+1$$
$$=1$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3158312",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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How can I block diagonalise this matrix? I have this matrix:
$$A = \left(
\begin{array}{cccc}
0 & 0 & 0 & -1 \\
1 & 0 & 0 & -1 \\
0 & 1 & 0 & -1 \\
0 & 0 & 1 & -1 \\
\end{array}
\right)$$
and I would like to show that it can be block diagonalised into :
$$ B = \left(
\begin{array}{cccc}
\cos 2\pi/5 & -\sin 2\pi/5 & 0 & 0 \\
\sin 2\pi/5 & \cos 2\pi/5 & 0 & 0 \\
0 & 0 & \cos 4\pi/5 & -\sin 4\pi/5 \\
0 & 0 & \sin 4\pi/5 & \cos 4\pi/5\\
\end{array}
\right) = \left(
\begin{array}{cccc}
\frac{1}{4} \left(-1+\sqrt{5}\right) & -\sqrt{\frac{5}{8}+\frac{\sqrt{5}}{8}} & 0 & 0 \\
\sqrt{\frac{5}{8}+\frac{\sqrt{5}}{8}} & \frac{1}{4} \left(-1+\sqrt{5}\right) & 0 & 0 \\
0 & 0 & \frac{1}{4} \left(-1-\sqrt{5}\right) & -\sqrt{\frac{5}{8}-\frac{\sqrt{5}}{8}} \\
0 & 0 & \sqrt{\frac{5}{8}-\frac{\sqrt{5}}{8}} & \frac{1}{4} \left(-1-\sqrt{5}\right) \\
\end{array}
\right)$$
What is the general procedure?
| When you diagonalise a matrix over the complex numbers, you start by finding its eigenvalues, which may be complex. If the matrix is diagonalisable, then we can expect a basis of eigenvectors corresponding to these eigenvalues. If we compute the transformation $x \mapsto Ax$ in terms of this basis of eigenvectors, we get a complex diagonal matrix, similar to $A$.
To block-diagonalise $A$ over the reals, again find all the complex eigenvalues and a basis of eigenvectors. Suppose $\alpha + i \beta$ is a non-real eigenvalue, with corresponding eigenvector $v + i w$, where $v, w$ are vectors with real components. One can easily verify that $\alpha - i \beta$ is another eigenvalue with eigenvector $v - i w$. We can form a basis of eigenvectors such that, if $v + iw$ is in the basis (where $v$ and $w$ are real vectors), then so is $v - iw$.
Now, form a real basis like so: replace the conjugate eigenvector pairs $v \pm i w$ with the vectors $v, w$. If $v + iw$ has an eigenvalue $\alpha + i \beta$, then
\begin{align*}
Av &= A \left(\frac{(v + iw) + (v - iw)}{2}\right) \\
&= \frac{(\alpha + i\beta)(v + iw) + (\alpha - i\beta)(v - iw)}{2} \\
&= \frac{\alpha v + i \beta v + i \alpha w - \beta w + \alpha v - i \beta v - i \alpha w - \beta w}{2} \\
&= \alpha v - \beta w.
\end{align*}
Similar calculation reveals
$$Aw = \beta v + \alpha w.$$
So, when computing the matrix for $x \mapsto Ax$ with this basis, you'll find a block diagonal form, where each block is either a $1 \times 1$ block containing a real eigenvalue, or a $2 \times 2$ block of the form
$$\begin{pmatrix} \alpha & \beta \\ -\beta & \alpha \end{pmatrix},$$
where $\alpha \pm \beta i$ is a non-real eigenvalue of $A$.
| {
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"url": "https://math.stackexchange.com/questions/3158856",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
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} |
Existence of 3 natural numbers that divide each other when squared and have 1 taken away from them Does there exist natural numbers, $a,b,c > 1$, such that;
$a^2 - 1$ is divisible by $b$ and $c$,
$b^2 - 1$ is divisible by $a$ and $c$ and
$c^2 - 1$ is divisible by $a$ and $b$.
| No such numbers exist.
First, observe that $a,b,c$ are pairwise coprime: For example, since $a$ divides $b^2-1$, any prime $p$ that divides $a$ also divides $b^2 - 1$; hence $p$ does not divide $b^2$ and also does not divide $b$. Thus $a$ and $b$ are coprime; and likewise for the other pairs.
Since $b,c$ are coprime and each divide $a^2 - 1$, so does their product $bc$. Since all the quantities are positive, that implies $bc \le a^2 - 1 \lt a^2$. So we have three strict inequalities:
$$
\begin{align}
bc &\lt a^2\\
ac &\lt b^2\\
ab &\lt c^2
\end{align}
$$
Multiplying these inequalities together yields
$$
a^2 b^2 c^2 \lt a^2 b^2 c^2
$$
which is impossible.
| {
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"timestamp": "2023-03-29T00:00:00",
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"answer_id": 0
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Shell integration to find volume Find the volume generated by rotating the region bounded by the given curves about the y-axis.
$$y = \sqrt[3]{x}, y = 0, x = 1$$
Here is my work:
$$\begin{align}
V &= \int_{0}^{1}2\pi(x)\sqrt[3]{x}dx = 2\pi\int_{0}^{1}x^\frac{4}{3}dx \\
&= 2\pi \bigg[\frac{3}{6}x^\frac{6}{3} \bigg]_{0}^{1} = 2\pi\bigg[ \frac{1}{2}x^\frac{1}{2} \bigg]_{0}^{1}\\
&= \pi
\end{align}$$
However the answer is $\frac{6\pi}{7}$. It looks like you have to do integration by parts to integrate to get the answer. Why can't I "simplify" $x$ and $\sqrt[3]{x}$ by multiplying them together and adding the exponents? That seems to be reasonable in my mind, yet the result isn't the same
| The setup is correct. The only thing that you got wrong is the antiderivative of the function $x^{\frac{4}{3}}$:
$$\int x^{\frac{4}{3}}\,dx=\frac{1}{\frac{4}{3}+1}x^{\frac{4}{3}+1}+C=
\frac{1}{\frac{7}{3}}x^{\frac{7}{3}}+C=
\frac{3x^{\frac{7}{3}}}{7}+C.$$
$$
V=2\pi\int_{0}^{1}x\sqrt[3]{x}\,dx=
2\pi\int_{0}^{1}x^{\frac{4}{3}}\,dx=
2\pi\left[\frac{3}{7}x^{\frac{7}{3}}\right]_{0}^{1}=\\
2\pi\left(\frac{3}{7}1^{\frac{7}{3}}-\frac{3}{7}0^{\frac{7}{3}}\right)=
2\pi\left(\frac{3}{7}-0\right)=\frac{6\pi}{7}\ cubic\ units.
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3164400",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Solving $\int \frac{\sqrt{(x-2)^3}}{(\sqrt{x+1})^2} dx$ This is the last indefinite integral I am attempting before proceeding to definite integrals study. Any ideas how to solve it?
$$\int \frac{\sqrt{(x-2)^3}}{(\sqrt{x+1})^2} dx$$
My attempt: (not the longest one... :P)
$$\int \frac{\sqrt{(x-2)^3}}{(\sqrt{x+1})^2} dx = \int \frac{\sqrt{(x-2)^3}}{x+1} dx = \left(\left(u=x-2,du=dx \right)\right) =\int\frac{\sqrt{u^3}}{u+3}du$$
But that substitution does not really make it simpler for me. If the expression inside the numerator's root was of the form $ax^2 + bx + c$ then I could apply one of Euler's substitution. But it's polynomial of third power.
How to deal with it?
| So you have brought the ques down to solving integral of u^(3/2)/(u+3)
Now simply put u= z^2 so after taking constants out it becomes
z^4/(z^2+3)= (z^4 - 9 + 9)/(z^2+3)
=(z^2 - 3)(z^2+3)/(z^2+3) + 9/(z^2+3)
= z^2 - 3 + 9/(z^2+3)
Here, z^2 - 3 is very easy to solve. For 9/(z^2+3) simply put z= sqrt(3).tan(u) and the integral becomes elementary.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3167804",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
What is wrong with this solution of find the least value of $ \sec^6 x +\csc^6 x + \sec^6 x\csc^6 x$ What is wrong with this solution of find the least value of $ \sec^6 x +\csc^6 x + \sec^6 x\csc^6 x$
They all are positive terms so arithmetic mean is greater than equal to geometric mean.
$$ \sec^6 x +\csc^6 x + \sec^6 x\csc^6 x\geq 3( \sec^6 x \csc^6 x \sec^6 x\csc^6 x)^\frac{1}{3} $$
$$ \sec^6 x +\csc^6 x + \sec^6 x\csc^6 x \geq 3( \sec x \csc)^4 $$
$$
\sec^6 x +\csc^6 x + \sec^6 x\csc^6 x\geq
\frac{3 * 2^4}{\sin ^4 2x} $$
Clearly least value is 48, but something is wrong here, as the answer is 80, if I use other methods.
| In your way you proved that the minimal value is greater than $48$.
It's true, but the equality does not occur, which says that $48$ is not a minimal value.
The right solution can be the following, for example.
Let $\sin^2x\cos^2x=t.$
Thus, by AM-GM
$$t\leq\left(\frac{\sin^2x+\cos^2x}{2}\right)^2=\frac{1}{4}.$$
The equality occurs for $x=45^{\circ},$ which gives a value $80$.
We'll prove that it's a minimal value.
Indeed, we need to prove that
$$\frac{\sin^4x-\sin^2x\cos^2x+\cos^4x}{\sin^6x\cos^6x}+\frac{1}{\sin^6x\cos^6x}\geq80$$ or
$$\frac{1-3\sin^2x\cos^2x}{\sin^6x\cos^6x}+\frac{1}{\sin^6x\cos^6x}\geq80$$ or
$$\frac{2}{t^3}-\frac{3}{t^2}\geq80$$ or
$$80t^3+3t-2\leq0$$ or
$$80t^3-20t^2+20t^2-5t+8t-2\leq0$$ or
$$(4t-1)(20t^2+5t+2)\leq0,$$ which is obvious.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3168377",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 2
} |
Sum of combinations - convergence I am trying to evaluate the sum of combinations for a specific probability problem, but I am kind of stuck.
Let A have toss a coin $n+1$ times and B toss another coin $n$ times. What is the probability that A has more heads than B. (the probability of heads is $P = 1/2$). All experiments are independent.
Now, let $X$ denote the numbers of heads A has and $Y$ the number of heads B has. So:
$$ \begin{eqnarray} P(X>Y) &= \sum_{y=0}^{n}P(X>y|Y=y) P(Y=y) \\
&= \sum_{y=0}^{n} P(X>y|Y=y) {n \choose y} \left(\frac{1}{2}\right)^n \\
&= \sum_{y=0}^{n} \sum_{x=y+1}^{n+1} {n+1 \choose x} {n \choose y} \left(\frac{1}{2}\right)^{2n+1} \\
&= \left(\frac{1}{2}\right)^{2n+1} \sum_{y=0}^{n} \sum_{x=y+1}^{n+1} {n+1 \choose x} {n \choose y} \end{eqnarray} $$
I know this sum must result $2^{2n}$, because the probability must be $1/2$. Yet I am stuck. Any advices?
Thanks! :)
|
On the one hand we obtain
\begin{align*}
\color{blue}{\sum_{y=0}^n\sum_{x=y+1}^{n+1}\binom{n+1}{x}\binom{n}{y}}
&=\sum_{y=0}^n\left(2^{n+1}-\sum_{x=0}^y\binom{n+1}{x}\right)\binom{n}{y}\tag{1}\\
&\,\,\color{blue}{=2^{2n+1}-\sum_{y=0}^n\sum_{x=0}^y\binom{n+1}{x}\binom{n}{y}}\tag{2}
\end{align*}
on the other hand we obtain
\begin{align*}
\color{blue}{\sum_{y=0}^n\sum_{x=y+1}^{n+1}\binom{n+1}{x}\binom{n}{y}}
&=\sum_{y=0}^n\sum_{x=n+1-y}^{n+1}\binom{n+1}{x}\binom{n}{y}\tag{3}\\
&=\sum_{y=0}^n\sum_{x=-y}^0\binom{n+1}{x+n+1}\binom{n}{y}\tag{4}\\
&=\sum_{y=0}^n\sum_{x=0}^y\binom{n+1}{n+1-x}\binom{n}{y}\tag{5}\\
&\,\,\color{blue}{=\sum_{y=0}^n\sum_{x=0}^y\binom{n+1}{x}\binom{n}{y}}\tag{6}\\
\end{align*}
Comparing (2) and (6) we observe
\begin{align*}
2^{2n+1}-\sum_{y=0}^n\sum_{x=0}^y\binom{n+1}{x}\binom{n}{y}&=\sum_{y=0}^n\sum_{x=0}^y\binom{n+1}{x}\binom{n}{y}\\
\color{blue}{\sum_{y=0}^n\sum_{x=0}^y\binom{n+1}{x}\binom{n}{y}}&\color{blue}{=2^{2n}}
\end{align*}
and the claim follows.
Comment:
*
*In (1) and (2) we use $\sum_{j=0}^n\binom{n}{j}=2^n$.
*In (3) we change the order of summation $y\to n-y$ and we use $\binom{p}{q}=\binom{p}{p-q}$.
*In (4) we shift the index $x$ by $n+1$.
*In (5) we substitute $x$ with $-x$.
*In (6) we use again $\binom{p}{q}=\binom{p}{p-q}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3168900",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
coefficient $x ^ n$ in development The advisor asks to verify that the coefficient of
$$x^n$$
in the development of:
$$(1+x)^{2n}+x(1+x)^{2n−1}+x2(1+x)^{2n−2}+......+x^n(1+x)^n$$
is equal to
$$\binom{2n+1}{n}$$
I tried for summations but not.
I did with the denominator change too, but I can not even,how can i match the 2 questions
| You can explicitly evaluate your sum and extract the coefficient afterward. Recall that $$(y-x) (y^n + xy^{n-1} + \cdots + x^{n-1}y + x^n) = y^{n+1} - x^{n+1}$$ Letting $y = 1+x$ (so $y-x = 1$) gives
$$(1+x)^n + x(1+x)^{n-1} + \cdots + x^{n-1}(1+x) + x^n = (1+x)^{n+1} - x^{n+1}.$$
Then multiplying by $(1+x)^n$ gives your expression
$$(1+x)^{2n} + x(1+x)^{2n-1} + \cdots + x^n (1+x)^n = (1+x)^{2n+1} - (1+x)^n x^{n+1}.$$
The coefficient of $x^n$ of $(1+x)^nx^{n+1}$ is $0$, while the coefficient of $x^n$ in $(1+x)^{2n+1}$ is $${ 2n+1 \choose n}$$ by the binomial theorem.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3169345",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 3
} |
$x^2 + y^2+xy = 1$ , then find the minimum of $x^3 y + xy^3 +4$ x and y belongs to real numbers. $ x^2 + y^2+xy = 1 $. then find the minimum value of $x^3 y + xy^3 +4$.
I assume $ x = r \sin (w)$ and $ y = r\cos(w) $. $ x^3 y + xy^3 +4 = L $ which give me $ \frac{2}{3} \le r^2 \le 2 $ I am stuck after that.Its my Humble request to help me after that.
| Hint: $x^2+y^2=1-xy$, so that $x^3y+xy^3+4=xy(x^2+y^2)+4=xy(1-xy)+4$. Substitute $z=xy$. What do you end up with?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3171407",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 0
} |
Recurrence Relation: $B(n) = 5\cdot B(n/3) + c(n^2)$ I'm working on deriving this recurrence relation in the form $O(n^d)$ for some value of d:
$B(n) = 5\cdot B(\frac{n}{3}) + cn^2$
The initial condition is:
$B(1) = c$
I'm having trouble incorporating the n^2 into my derivation. This is what I have so far:
We set $n = 3^k$
$5^k(1 + \frac{5}{3} + (\frac{5}{3})^2 + ... + (\frac{5}{3})^k$
Which leads to $3^k\cdot (\frac{10}{3})^k$ -> $5^k = 5^{log_3n}$ -> $log_{3}n = k$
Not sure how to derive the final running time to be $O(n^d)$. Any help would be appreciated!
| \begin{equation}
\begin{split}
B(n) &= 5B(\frac{n}{3}) + cn^2 \\
&= 5(5B(\frac{n}{3^2}) + c(\frac{n}{3})^2) + cn^2 \\
&= 5^2B(\frac{n}{3^2}) + c[5(\frac{n}{3})^2 + n^2] \\
&= 5^2(5B(\frac{n}{3^3}) + c(\frac{n}{3^2})^2) + c[5(\frac{n}{3})^2 + n^2] \\
&= 5^3B(\frac{n}{3^3}) + c[5^2(\frac{n}{3^2})^2 +5(\frac{n}{3})^2 + n^2] \\
&= \vdots \\
&= 5^kB(\frac{n}{3^k}) + c[5^{k-1}(\frac{n}{3^{k-1}})^2 + \ldots +5(\frac{n}{3})^2 + n^2]\\
&= 5^kB(\frac{n}{3^k}) + cn^2 \sum_{i=0}^{k-1} 5^i(\frac{1}{3^i})^2 \\
&= 5^kB(\frac{n}{3^k}) + cn^2 \sum_{i=0}^{k-1} (\frac{5}{3^2})^i \\
&= 5^kB(\frac{n}{3^k}) + cn^2 \frac{1-(\frac{5}{9})^k}{1- \frac{5}{9}}
\end{split}
\end{equation}
But $B(1) = c$ hence for $k=\log_3 n$
$$B(n) = 5^{\log_3 n}c + \frac{9}{4} cn^2 (1-(\frac{5}{9})^{\log_3 n})$$
It seems that the dominating term is $O(n^2)$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3171538",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Why partial fraction decomposition of $\frac{1}{s^2(s+2)}$ is $\frac{A}{s}+\frac{B}{s^2}+\frac{C}{(s+2)}$? Can someone please explain why: $$\frac{1}{s^2(s+2)}=\frac{A}{s}+\frac{B}{s^2}+\frac{C}{(s+2)}$$
And not:$$\frac{1}{s^2(s+2)}=\frac{A}{s^2}+\frac{B}{(s+2)}$$
I'm a bit confused where the extra s term comes from in the first equation.
| You can do it step-by-step:
$$\begin{align}\frac{1}{s^2(s+2)}&=\frac1s\cdot \frac{1}{s\cdot (s+2)}=\\
&=\frac1s\cdot \left(\frac As+\frac{B}{s+2}\right)=\\
&=\frac A{s^2}+\frac1s\cdot \frac B{s+2}=\\
&=\frac{A}{s^2}+\frac{C}{s}+\frac{D}{s+2}.\end{align}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3172683",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 6,
"answer_id": 4
} |
What are the integer coeffcients of a cubic polynomial having two particular properties? Let $f(x) = x^3 + a x^2 + b x + c$ and $g(x) = x^3 + b x^2 + c x + a\,$ where $a, b, c$ are integers and $c\neq 0\,$. Suppose that the following conditions hold:
*
*$f(1)=0$
*The roots of $g(x)$ are squares of the roots of $f(x)$.
I'd like to find $a, b$ and $c$.
I tried solving equations made using condition 1. and relation between the roots, but couldn't solve.
The equation which I got in $c$ is $c^4 + c^2 +3 c-1=0$ (edit: eqn is wrong).
Also I was able to express $a$ and $b$ in terms of $c$. But the equation isn't solvable by hand.
| The conditions 1 and 2 imply that
*
*$\;c\,$ must be $\,0\,$ or $\,-1$,
*$\;a= -c^2$, and $\:b= c^2-c-1\,$.
Condition 1 gives $0=f(1)=1+a+b+c=g(1)\,.\,$ Hence both $f$ and $g$ have a zero at $1$ and factor as
$$f(x) \,=\, (x-1)\big(x^2 + (a+1)x -c\big)\\[1.5ex]
g(x) \,=\, (x-1)\big(x^2 -(a+c)x -a\big)\,.$$
Denote the roots of the quadratic factor of $f$ by $x_1$ and $x_2$. Condition 2 says that the roots of $g$ are contained in $\{1,x_1^2,x_2^2\}$. By Vieta's formula one gets
$$\,-a \,=\, x_1^2x_2^2=(-c)^2 \,=\, c^2\,,\;\text{thus}\;\;
b \,=\, -a-c-1 \,=\, c^2-c-1\,.\tag{1}$$
Note that condition 2 remains true when restricted to the quadratic factors of $f$ and $\,g$. These are
$$q_f \,=\,x^2 + \left(1-c^2\right)x -c\tag{2}\\
q_g \,=\,x^2 -c\,(1-c)x +c^2$$
when written in terms of $\,c$.
It is shown next that condition 2 cannot hold if $c\neq 0\,$ or $\,-1$.
*
*Assume $c\geqslant 1$. Then $q_f(0)=-c<0$, and the roots $x_1,x_2$ of $q_f$ are real and distinct. By Vieta's formula regarding $q_g$ one reaches the contradiction $0<x_1^2+x_2^2=c(1-c)\leq 0\,$.
*Assume $c\leq -2\,$. Then the discriminant $\left(1-c^2\right)^2+4c$ in $(2)$ is positive, and we run into the same contradiction as before.
So we are left with the two solutions (with condition 2 obviously satisfied)
*
*$a=0,b=-1,c=0\:$ which was ruled out a priori
then $f(x)=x(x+1)(x-1)\:$ and $\:g(x)=x^2(x-1)$
*$a=-1,b=1,c=-1\:$
where $f(x)=\left(x^2+1\right)(x-1)\:$ and $\:g(x)=(x+1)^2(x-1)$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3173125",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
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$h^2 = x^2 + (x+1)^4$ This is a question I had of finding an exact value of $h$ as I was interested in the process and techniques used. The question was one to do with Pythagoras and the math shown is the question in $a^2 + b^2 = c^2$ form.
The question was what is the size of the hypotenuse of a right-angled triangle if one side is a size $x$ and the other is a size $x^2 + 2x + 1$ (which was rewritten as $(x + 1)^2$). I got an answer from Wolfram Alpha of a value for $h$ that does not include $x$, only numbers.
Wolfram Input:
Wolfram Answer:
What would be the process of finding the exact value of h as to where there is no sign of x in the h = equation?
P.S. If the process is simply too lengthy or tedious, feel free to take this question down.
| \begin{align}
c^2 &= a^2 + b^2 \\[1em]
h^2 &= x^2 + \left[(x+1)^2\right]^2 \\
h^2 &= x^2 + (x+1)^4 \\
h^2 &= x^2 + x^4 + 4x^3 + 6x^2 + 4x +1\\
h^2 &= x^4 + 4x^3 + 7x^2 + 4x +1\\
h &= \sqrt{x^4 + 4x^3 + 7x^2 + 4x +1}
\end{align}
You probably put into Wolfram Alpha some other task - you wanted the square root, and you obtained the root (= solution) of the system of equations with unknowns $h$ and $x$.
Addendum:
If you wanted from Wolfram Alpha the result of my computing, enter into it
solve(h^2 = x^2 + (x+1)^4, h)
(h at the end determines which you want to compute), and you will obtain
$$h = \pm \sqrt{x^4 + 4x^3 + 7x^2 + 4x +1}$$
If you want the result for a particular $x$, enter its value, e. g.
h^2 = x^2 + (x+1)^4, x=1
to obtain plot
and solutions
$$\begin{array}{l}{h=-\sqrt{17}, \quad x=1} \\ {h=\sqrt{17}, \quad x=1}\end{array}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3175766",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Implicit differentiation of an equation of a hyperbola Prove that an equation of the tangent line to the graph of the hyperbola :
$(x^2/a^2) - (y^2/b^2) = 1$
at the point ($x_0$, $y_0$) is
$x x_0/a^2 - y y_0/b^2 = 1$ (1)
I implicitly differentiated the equation and then found the gradient by substituting in the points to get the gradient ( $b^2x_0/a^2y_0$) and use the points, plug it into $y-y_1=m(x-x_1)$
But I don't know how to rearrange it to get to (1). Please help me!
| \begin{align}
y-y_0 & = \frac{b^2x_0}{a^2y_0}(x-x_0)
\\ a^2y_0(y-y_0)& = (b^2x_0)(x-x_0)
\\ a^2y_0y-a^2y_0^2& = b^2x_0x-b^2x_0^2
\\ b^2x_0^2 -a^2y_0^2& = b^2x_0x-a^2y_0y
\end{align}
$(x_0,y_0)$ satisfaces that $\frac{x_0^2}{a^2}-\frac{y_0^2}{b^2}=1$ doing the operations, $b^2x_0^2 -a^2y_0^2=a^2b^2$, then:
\begin{align}
\\ b^2x_0^2 -a^2y_0^2& = b^2x_0x-a^2y_0y
\\ a^2b^2& = b^2x_0x-a^2y_0y
\\ 1& = \frac{xx_0}{a^2}-\frac{yy_0}{b^2}
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3177674",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Unitriangular matrix group over primary finite field I have unitriangular matrix group over a field $F_q$.
It consists of the following elements:
$$E + A = \left[\begin{matrix}
1 & a12 & ... & a1n\\
0 & 1 & a23 & ...\\
... & ... & ... & ... \\
0 & 0 & ... & 1\\
\end{matrix}\right]$$
I'm trying to prove that $(E+A)^{p^{n-1}}=E$, where $n$ is a size of matrix.
So far, I found that $(E+A)^{p}=E + A^p$ if we use Newton's binomial theorem.
So, $(E+A)^{p^{n-1}}=E + A^{p^{n-1}}$.
But why $A^{p^{n-1}} = 0$?
I have a hint, that for matrix $A^{p}$ first secondary diagonal is $0$, but I also don't have any idea how to prove that.
| Try to use induction on $n$. Note that you can think $A = A_{n}$ as a block matrix
$$
A = A_{n} = \begin{pmatrix} A_{n-1} & v_{n-1} \\ \mathbf{0}_{1, n-1} & 0 \end{pmatrix}
$$
where $v_{n-1} \in \mathbb{F}_{p}^{n-1}$. Then it is not hard to check that
$$
A^{m} = \begin{pmatrix} A_{n-1}^{m} & A_{n-1}^{m-1}v_{n-1} \\ \mathbf{0}_{1, n-1} & 0 \end{pmatrix}
$$
for any $m\geq 1$, so
$$
A^{p^{n-2}} = \begin{pmatrix} A_{n-1}^{p^{n-2}} & A_{n-1}^{p^{n-2}-1}v_{n-1} \\ 0^{T} & 0 \end{pmatrix} = \begin{pmatrix} \mathbf{0}_{n-1, n-1} & A_{n-1}^{p^{n-2}-1}v_{n-1} \\ \mathbf{0}_{1, n-1} & 0 \end{pmatrix}
$$
By the way, we can apply the same argument for the same matrix with different blocks
$$
A = \begin{pmatrix} 0 & w_{n-1}^{T} \\ \mathbf{0}_{n-1, 1} & A_{n-1}' \end{pmatrix}
$$
gives us
$$
A^{p^{n-2}} = \begin{pmatrix} 0 & w_{n-1}^{T}(A_{n-1}')^{p^{n-2}-1} \\ \mathbf{0}_{n-1, 1} & \mathbf{0}_{n-1, n-1}\end{pmatrix}
$$
This proves that $A^{p^{n-2}}$ has a form of
$$
A^{p^{n-2}} = \begin{pmatrix} \mathbf{0}_{1, n-1} & a_{1, n} \\ \mathbf{0}_{n-1, n-1} & \mathbf{0}_{n-1, 1} \end{pmatrix}
$$
and we get $A^{p^{n-1}} = (A^{p^{n-2}})^{p} = \mathbf{0}_{n, n}$.
When $n = 3$, we have
$$
A = \begin{pmatrix} 0 & a_{12} & a_{13} \\ 0 & 0 & a_{23} \\ 0 & 0 & 0 \end{pmatrix} = \begin{pmatrix} A_{2} & v_{2} \\ \mathbf{0}_{1, 2} & 0 \end{pmatrix}
$$
where
$$
A_{2} = \begin{pmatrix} 0 & a_{12} \\ 0 & 0 \end{pmatrix}, \quad v_{2} = \begin{pmatrix} a_{13} \\ a_{23} \end{pmatrix}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3177911",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
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} |
Find the minimum of $(1+a^2)(1+b^2)(1+c^2)$ where $a,b,c\geq 0$
Find the minimum of:
$$(1+a^2)(1+b^2)(1+c^2) \ \ \ a,b,c\geq 0$$
Knowing that $$ab+bc+ac=27$$
I tried my best using QM-AM-GM inequalities, Cauchy-Schwarz, etc. I tried also to do it with partial derivatives but it's too long. Thank you for your time :)
| We note that $(1+a^2)(1+b^2)(1+c^2)$ is a symmetric function with respect to $a$, $b$, and $c$. We rewrite it by using the elementary symmetric polynomials $a+b+c$, $ab+bc+ca$ and $abc$:
$$\begin{align}
(1+a^2)(1+b^2)(1+c^2)&=(ab+bc+ca-1)^2+(a+b+c-abc)^2\\
&=26^2+(a+b+c-abc)^2\geq 26^2=676
\end{align}$$
where the condition $ab+bc+ac=27$ has been applied.
Hence the minimum is $676$ as soon as we show that there are $a,b,c\geq 0$ such that $ab+bc+ac=27$ and $a+b+c=abc$.
Can you take it from here?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3179770",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
} |
Prove that for all $x\in \mathbb R$ $, \arctan x=\frac{\pi}{2}-\arccos(\frac{x}{\sqrt{1+x^{2}}})$
Prove that for all $x\in \mathbb R$, $$\arctan x=\frac{\pi}{2}-\arccos \left(\frac{x}{\sqrt{1+x^{2}}}\right)$$
From Lagrange form of Taylor's theorem I have:$$\arctan x+\arccos\left(\frac{x}{\sqrt{1+x^{2}}}\right)=x-\frac{c_{1}x^{2}}{\sqrt{1+c_{1}^{2}}}+\frac{\pi}{2}-x-\frac{c_{2}x^{2}}{2\sqrt{(1-c_{2}^{2})^{3}}}=-\frac{c_{1}x^{2}}{\sqrt{1+c_{1}^{2}}}+\frac{\pi}{2}-\frac{c_{2}x^{2}}{2\sqrt{(1-c_{2}^{2})^{3}}}$$I know that: $$\left(-\frac{c_{1}x^{2}}{\sqrt{1+c_{1}^{2}}}-\frac{c_{2}x^{2}}{2\sqrt{(1-c_{2}^{2})^{3}}}\right) \rightarrow 0$$But I don't know how to show that: $$\left(-\frac{c_{1}x^{2}}{\sqrt{1+c_{1}^{2}}}-\frac{c_{2}x^{2}}{2\sqrt{(1-c_{2}^{2})^{3}}}\right) = 0$$ Can you help me?
| By definition, $\arctan x=\theta$ means
$$\tan\theta=x\quad\hbox{and}\quad -\frac\pi2<\theta<\frac\pi2\ .$$
Let $\theta$ be the RHS of your equation. We have
$$-1<\frac{x}{\sqrt{1+x^2}}<1\quad
\Rightarrow\quad 0<\arccos\frac{x}{\sqrt{1+x^2}}<\pi\quad
\Rightarrow\quad -\frac\pi2<\theta<\frac\pi2\ ,$$
so the second part of the condition is true. The fact that $0<\arccos\frac{x}{\sqrt{1+x^2}}<\pi$ also means that
$$\sin\Bigl(\arccos\frac{x}{\sqrt{1+x^2}}\Bigr)
=\sqrt{1-\cos^2\Bigl(\arccos\frac{x}{\sqrt{1+x^2}}\Bigr)}$$
and not the negative of this square root, so
$$\eqalign{\tan\theta
&=\cot\arccos\frac{x}{\sqrt{1+x^2}}\cr
&=\frac{\cos(\arccos(x/\sqrt{1+x^2}))}{\sin(\arccos(x/\sqrt{1+x^2}))}\cr
&=\frac{x/\sqrt{1+x^2}}{\sqrt{1-\frac{x^2}{1+x^2}}}\cr
&=x\ .\cr}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3183120",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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In how many ways can we place $3$ red and $4$ blue balls into $3$ indistinguishable boxes so that no box is empty? In how many ways can we place $3$ red and $4$ blue balls into $3$ indistinguishable boxes so that no box is empty?
What if the boxes can be empty?
| Using the Polya Enumeration Theorem we get for the two cases using the
cycle index of the symmetric group with empty boxes
$$Q_1 = [R^3 B^4] Z\left(S_3;
(1+R+R^2+R^3)\times(1+B+B^2+B^3+B^4)\right)$$
and without
$$Q_2 = [R^3 B^4] Z\left(S_3; -1 +
(1+R+R^2+R^3)\times(1+B+B^2+B^3+B^4)\right).$$
Now the cycle index is
$$Z(S_3) = 1/6\,{a_{{1}}}^{3}+1/2\,a_{{2}}a_{{1}}+1/3\,a_{{3}}.$$
Doing the substitution we find
$$\bbox[5px,border:2px solid #00A000]{
Q_1 = 28 \quad\text{and}\quad Q_2 = 18.}$$
If we want to do these by hand, here is an example. We use the
alternate form
$$[R^3 B^4] Z\left(S_3;
\frac{1}{1-R}\frac{1}{1-B}\right).$$
We get from the first term of the cycle index
$$[R^3 B^4] \frac{1}{6}
\frac{1}{(1-R)^3}\frac{1}{(1-B)^3}
= \frac{1}{6} {3+2\choose 2} {4+2\choose 2} = 25.$$
We get from the second term
$$[R^3 B^4] \frac{1}{2}
\frac{1}{1-R^2}\frac{1}{1-B^2}
\frac{1}{1-R}\frac{1}{1-B}
= \frac{1}{2} (1+1)\times (1+1+1) = 3.$$
Here we have e.g. for the coefficient on $B^4$ the possibilities
$(B^2)^2 (B^1)^0,$ $(B^2)^1 (B^1)^2$ and $(B^2)^0 (B^1)^4.$
At last we get from the third term
$$[R^3 B^4] \frac{1}{3}
\frac{1}{1-R^3}\frac{1}{1-B^3} = 0.$$
Add these to obtain $25+3=28.$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3183575",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
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} |
Find Taylor's expansion around $ x=2$ for the function $f(x)=x^5+x^4+2x+1 $
Find Taylor's expansion around $ x=2$ for the function $f(x)=x^5+x^4+2x+1 $
My try:
$$f(2)=53$$
$$f^{(1)}(x)=5x^4+4x^3+2 \Rightarrow f^{(1)}(2)=114$$
$$f^{(2)}(x)=20x^3+12x^2 \Rightarrow f^{(2)}(2)=208$$
$$f^{(3)}(x)=60x^2+24x \Rightarrow f^{(3)}(2)=288$$
$$f^{(4)}(x)=120x+24 \Rightarrow f^{(4)}(2)=264$$
$$f^{(5)}(x)=120 \Rightarrow f^{(5)}(2)=120$$
$$f^{(n)}(x)=0, n\ge6 \Rightarrow f^{(n)}(2)=0$$That is why I have: $$f(x)=53+114(x-2)+\frac{208}{2!}(x-2)^2+\frac{288}{3!}(x-2)^3+\frac{264}{4!}(x-2)^4+\frac{120}{5!}(x-2)^5$$ I checked this answer in Mathematica and I get that this is true. However I also checked diagrams for $f(x)=x^5+x^4+2x+1 $ and $f(x)=53+114(x-2)+\frac{208}{2!}(x-2)^2+\frac{288}{3!}(x-2)^3+\frac{264}{4!}(x-2)^4+\frac{120}{5!}(x-2)^5$ and I get that there are the same when $x\ge2$.Why this diagrams are different? I think that these diagrams should be the same in all domain.
| You have polynomials on both sides of $$x^5+x^4+2x+1=53+114(x-2)+\frac{208}{2!}(x-2)^2+\frac{288}{3!}(x-2)^3+\frac{264}{4!}(x-2)^4+\frac{120}{5!}(x-2)^5$$
Thus the two sides match on the entire range of $(-\infty, \infty).$
You do not have to worry about the interval of convergence of either side.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3186691",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
maximum value of $\sum (a-b)^2$
If $a^2+b^2+c^2=5$ and $a,b,c \in \mathbb{R},$ find the maximum
value of $(a-b)^2+(b-c)^2+(c-a)^2$.
My Try: $(a-b)^2+(b-c)^2+(c-a)^2=2(a^2+b^2+c^2)-2(ab+bc+ac)$
$$=10-2(ab+bc+ac)$$
Now this implies $ab+bc+ac\le 5$. So
$$(a-b)^2+(b-c)^2+(c-a)^2 = 10-2(ab+bc+ac) \geq 20.$$
Could some help me to find max of $(a-b)^2+(b-c)^2+(c-a)^2$? Thanks.
| $$\sum_{cyc}(a-b)^2=3(a^2+b^2+c^2)-(a+b+c)^2\leq15.$$
The equality occurs for $a+b+c=0,$ which says that we got a maximal value.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3191049",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Proving $a^{ab}+b^{ab}\leq a^{a^2}+b^{b^2}$ and an identity . I was working on this Prove that $a^{ab}+b^{bc}+c^{cd}+d^{da} \geq \pi$ when I have discovered the following identity :
$$\Bigg|\Big(\frac{1}{2}\Big)^{\frac{x}{8}}\pm\Big(\frac{1}{4}\Big)^{\frac{x}{8}}\Bigg|=\Bigg|\Big(\frac{1}{2}\Big)^{\frac{x}{4}}\pm\Big(\frac{1}{4}\Big)^{\frac{x}{16}}\Bigg|$$
With $x$ a real number such that $x\neq 0$
Related to this identity we can have this inequality :
Let $a,b>0$ such that $a+b=\frac{3}{4}$ and $\frac{1}{2}\geq a\geq b\geq \frac{1}{4}$ then we have :
$$a^{ab}+b^{ab}\leq a^{a^2}+b^{b^2}$$
I can prove a part of this inequality using log-majorization but I would have a full alternative proof . Furhtermore this inequality is very precise (we have two equality case) and I don't think that reduce to a one variable inequality is a good idea .
So my questions are :
How to prove this two facts ?
Thanks in advance for your time .
Ps: The sign in the identity are the same .
| The identity is trivial, but the inequality is not trivial.
I cannot see any important connection between these two statements.
The following section is just a hint.
Let $~f(x) := x^{x(\frac{3}{4}-x)} + (\frac{3}{4}-x)^{x(\frac{3}{4}-x)}~$ and $~g(x) := x^{x^2} + (\frac{3}{4}-x)^{ (\frac{3}{4}-x)^2}~$
with $~\frac{1}{4}\leq x\leq\frac{1}{2}~$ .
It’s $~f(\frac{3}{4}-x) = f(x)~$ , $~g(\frac{3}{4}-x) = g(x)~$
and $~g(x_0) = f(x_0)~$ for $~x_0\in\{\frac{1}{4},\frac{3}{8},\frac{1}{2}\}~$ .
So we are looking for a proof of $~f(x)\leq g(x)~$ .
Let $~\displaystyle G(x) := \frac{g(\frac{1}{2})-g(x)}{(x-\frac{1}{4})(\frac{1}{2}-x)}~$ and $~\displaystyle F(x) := \frac{f(\frac{1}{2})-f(x)}{(x-\frac{1}{4})(\frac{1}{2}-x)}~$ .
Then, equivalent to $~f(x)\leq g(x)~$, we should proof:
$$G(x) \leq G\left(\frac{3}{8}\right) = F\left(\frac{3}{8}\right) \leq F(x)$$
This means that we have to find out that $~G(x)~$ has it’s (only) maximum
and $~F(x)~$ has it’s (only) minimum at $~x=\frac{3}{8}~$ :
$F'(\frac{3}{8})=0~$ and $~F''(x)>0$
$G'(\frac{3}{8})=0~$ and $~G''(x)<0$
With numerical estimates, we can conclude that $~F''(x)>4~$ and $~G''(x)<-2~$ for $~\frac{1}{4}<x<\frac{1}{2}~$ , but it's better to do that by a computer program than by hand.
Note:
Taylor series of $~G''(x)~$ at $~x=\frac{3}{8}~$ is about:
$-2.65756 - 25.0727\left(x-\frac{3}{8}\right)^2 - 121.808\left(x-\frac{3}{8}\right)^4 - 792.222\left(x-\frac{3}{8}\right)^6 $
$\hspace{2cm} -~5121.11\left(x-\frac{3}{8}\right)^8 - 32824.3\left(x-\frac{3}{8}\right)^{10} - O\left(\left(x-\frac{3}{8}\right)^{12}\right)$
Taylor series of $~F''(x)~$ at $~x=\frac{3}{8}~$ is about:
$4.71384 + 122.309\left(x-\frac{3}{8}\right)^2 + 1114.49\left(x-\frac{3}{8}\right)^4 + 8974.12\left(x-\frac{3}{8}\right)^6 $
$\hspace{1.7cm} +~68027.7\left(x-\frac{3}{8}\right)^8 + 502855\left(x-\frac{3}{8}\right)^{10} + O\left(\left(x-\frac{3}{8}\right)^{12}\right)$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3192378",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 0
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How can this equation be true? I'm currently stuck on an equation that no matter what numbers I insert to it is true, but I can't prove that is true. The equation is:
$$\frac{1}{\sqrt{1-\frac{\frac{(u+v)^2}{(1+\frac{vu}{c^2})^2}}{c^2}}}=(1+\frac{vu}{c^2})*\frac{1}{\sqrt{1-\frac{u^2}{c^2}}}*\frac{1}{\sqrt{1-\frac{v^2}{c^2}}}$$
So I was hoping someone could show me how this equation is actually true. The only thing I have been able to derive so far is: $$\frac{1}{\sqrt{1-\frac{\frac{(u+v)^2}{(1+\frac{vu}{c^2})^2}}{c^2}}}=(1+\frac{vu}{c^2})*\frac{1}{\sqrt{1+\frac{vu+vc+uc}{c^2}}}*\frac{1}{\sqrt{1+\frac{vu-vc-uc}{c^2}}}$$
| $$\frac{1}{\sqrt{1-\frac{\frac{(u+v)^2}{(1+\frac{vu}{c^2})^2}}{c^2}}}=(1+\frac{vu}{c^2})*\frac{1}{\sqrt{1-\frac{u^2}{c^2}}}*\frac{1}{\sqrt{1-\frac{v^2}{c^2}}} \iff$$
$${\sqrt{1-\frac{\frac{(u+v)^2}{(1+\frac{vu}{c^2})^2}}{c^2}}}=\frac{1}{(1+\frac{vu}{c^2})}*{\sqrt{1-\frac{u^2}{c^2}}}*{\sqrt{1-\frac{v^2}{c^2}}} \iff$$
$$ 1-\frac{\frac{(u+v)^2}{(1+\frac{vu}{c^2})^2}}{c^2}=\frac{1}{(1+\frac{vu}{c^2})^2}*(1-\frac{u^2}{c^2})*(1-\frac{v^2}{c^2}) \iff $$
$$ c^2-\frac{(u+v)^2}{(1+\frac{vu}{c^2})^2}=\frac{c^2}{(1+\frac{vu}{c^2})^2}*(1-\frac{u^2}{c^2})*(1-\frac{v^2}{c^2}) \iff $$
$$ c^2(1+\frac{vu}{c^2})^2-(u+v)^2=c^2*(1-\frac{u^2}{c^2})*(1-\frac{v^2}{c^2}) \iff $$
$$ c^4(1+\frac{vu}{c^2})^2-c^2(u+v)^2=c^4*(1-\frac{u^2}{c^2})*(1-\frac{v^2}{c^2}) \iff$$
$$ (c^2+vu)^2-c^2(u+v)^2=(c^2-u^2)(c^2-v^2)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3192890",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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Proving or disproving $12\mid x$ given $x^2+2\mid y^2-2$
Let $x$, $y$ be positive integers such that $x^2+2\mid y^2-2$. Prove or disprove that $12\mid x$.
This conclusion comes when I was dealing with another problem, and I feel it is right because when $x=12$ and $y=32$,
$$\frac{32^2-2}{12^2+2}=7.$$
Let me explain the origin of this example when $x=12$ can consider $\sqrt{146k+2}\in \Bbb Z$, I try $k=1,2,\cdots$,when $k=7$ is such it
but when $x=24$, I can't find $y\le 48$ no example such it,because I want to find $k$ such $\sqrt{478k+2}\in \Bbb Z$
| If $3\nmid x$ then $$ x^2\equiv 1\pmod 3\implies x^2+2\equiv 0\pmod 3$$ so $$ 3\mid y^2-2$$ which means $2$ is a square modulo 3. But this does not hold so $3\mid x$.
If $x$ is odd, then $x^2+2 = 4k(k+1)+3$, so there exists prime $p= 8l\pm 3$ such that $p\mid x^2+2$, but then $p\mid y^2-2$ and this is not true since by Legendre symbols we have $$\Big({2\over p}\Big) = (-1)^{p^2-1\over 8} =-1 $$
So $2\mid x$.
Added after Mike Bennett comment. So if $x=4k+2$ then $$ x^2+2\equiv 6\pmod 8$$ so again we have $p= 4l+3$ such that $p\mid x^2+2$, so $-2$ is a square modulo $p$. But since $p\mid y^2-2$, number $2$ is also square modulo $p$ so we have $$\Big({2\over p}\Big)= 1$$
Since Legendre symbol is multiplicative we have: $$ 1= \Big({-2\over p}\Big) =\Big({2\over p}\Big)\cdot \Big({-1\over p}\Big) = 1\cdot (-1)^{p-1\over 2}= -1$$
A contradiction. So $4\mid x$ and we are done.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3200458",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 1,
"answer_id": 0
} |
Primality test for numbers of the form $N=4 \cdot 3^n-1$ Can you provide proof or counterexample for the claim given below?
Inspired by Lucas-Lehmer primality test I have formulated the following claim:
Let $P_m(x)=2^{-m}\cdot((x-\sqrt{x^2-4})^m+(x+\sqrt{x^2-4})^m)$ . Let $N= 4 \cdot 3^{n}-1 $ where $n\ge3$ . Let $S_i=S_{i-1}^3-3 S_{i-1}$ with $S_0=P_9(6)$ . Then $N$ is prime if and only if $S_{n-2} \equiv 0 \pmod{N}$ .
You can run this test here .
Numbers $n$ such that $4 \cdot 3^n-1$ is prime can be found here .
I was searching for counterexample using the following PARI/GP code:
CE431(n1,n2)=
{
for(n=n1,n2,
N=4*3^n-1;
S=2*polchebyshev(9,1,3);
ctr=1;
while(ctr<=n-2,
S=Mod(2*polchebyshev(3,1,S/2),N);
ctr+=1);
if(S==0 && !ispseudoprime(N),print("n="n)))
}
| This is a partial answer.
This answer proves that if $N$ is prime, then $S_{n-2}\equiv 0\pmod N$.
Proof :
First of all, let us prove by induction on $i$ that
$$S_i=s^{2\cdot 3^{i+2}}+t^{2\cdot 3^{i+2}}\tag1$$
where $s=\sqrt 2-1,t=\sqrt 2+1$ with $st=1$.
We see that $(1)$ holds for $i=0$ since
$$S_0=P_9(6)=(3-2\sqrt 2)^9+(3+2\sqrt 2)^9=s^{2\cdot 3^2}+t^{2\cdot 3^2}$$
Supposing that $(1)$ holds for $i$ gives
$$\begin{align}S_{i+1}&=S_i^3-3S_i
\\\\&=(s^{2\cdot 3^{i+2}}+t^{2\cdot 3^{i+2}})^3-3(s^{2\cdot 3^{i+2}}+t^{2\cdot 3^{i+2}})
\\\\&=s^{2\cdot 3^{i+3}}+t^{2\cdot 3^{i+3}}\qquad\square\end{align}$$
Using $(1)$ and $N=4\cdot 3^n-1$, we get
$$S_{n-2}=s^{2\cdot 3^{n}}+t^{2\cdot 3^{n}}=s^{(N+1)/2}+t^{(N+1)/2}$$
So, we have, by the binomial theorem,
$$\begin{align}&S_{n-2}^2-2
\\\\&=s\cdot s^{N}+t\cdot t^{N}
\\\\&=(\sqrt 2-1)(\sqrt 2-1)^N+(\sqrt 2+1)(\sqrt 2+1)^N
\\\\&=\sqrt 2\sum_{i=0}^{N}\binom Ni(\sqrt 2)^i((-1)^{N-i}+1^{N-i})
\\&\qquad\quad +\sum_{i=0}^{N}\binom Ni(\sqrt 2)^i(1^{N-i}-(-1)^{N-i})
\\\\&=\sum_{j=1}^{(N+1)/2}\binom N{2j-1}2^{j+1}+\sum_{j=0}^{(N-1)/2}\binom{N}{2j}2^{j+1}\end{align}$$
Using that $\binom{N}{m}\equiv 0\pmod N$ for $1\le m\le N-1$, we get
$$S_{n-2}^2-2\equiv 2^{\frac{N+1}{2}+1}+2=4\cdot 2^{\frac{N-1}{2}}+2\pmod N\tag 2$$
Here, since
$$4\cdot 3^{2k}-1\equiv 4\cdot 9^k-1\equiv 3\pmod 8$$
and
$$4\cdot 3^{2k+1}-1\equiv 12\cdot 9^k-1\equiv 3\pmod 8$$
we see that
$$N\equiv 3\pmod 8$$
from which we have
$$2^{\frac{N-1}{2}}\equiv \left(\frac 2N\right)=(-1)^{(N^2-1)/8}=-1\pmod N\tag3$$
where $\left(\frac{q}{p}\right)$ denotes the Legendre symbol.
From $(2)(3)$, we have
$$S_{n-2}^2-2\equiv 4\cdot (-1)+2\pmod N$$
from which
$$S_{n-2}\equiv 0\pmod N$$
follows.$\quad\blacksquare$
| {
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "5",
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Integral $\int_0^\infty \frac{\ln(1+x+x^2)}{1+x^2}dx$
Prove that$$I=\int_0^\infty \frac{\ln(1+x+x^2)}{1+x^2}dx=\frac{\pi}{3}\ln(2+\sqrt 3)+\frac43G$$
I've found this integral in my notebook and perhaps I encountered it before since it looks quite familiar.
Anyway I thought it's quite a trivial integral so I'm gonna solve it quickly, but I am having some hard time to finish it. I went on with Feynman's trick:
$$I(a)=\int_0^\infty \frac{\ln((1+x^2)a+x)}{1+x^2}dx\Rightarrow I'(a)=\int_0^\infty \frac{dx}{a+x+ax^2}$$
$$=\frac1a\int_0^\infty \frac{dx}{\left(x+\frac{1}{2a}\right)^2+1-\frac{1}{4a^2}}=\frac{1}{a}\frac{1}{\sqrt{1-\frac{1}{4a^2}}}\arctan\left(\frac{x+\frac{1}{2a}}{\sqrt{1-\frac{1}{4a^2}}}\right)\bigg|_0^\infty$$$$=\frac{\pi}{\sqrt{4a^2-1}}-\frac{2}{\sqrt{4a^2-1}}\arctan\left(\frac{1}{\sqrt{4a^2-1}}\right)=\frac{2\arctan\left(\sqrt{4a^2-1}\right)}{\sqrt{4a^2-1}}$$
We can prove easily via the substitution $x\to \frac{1}{x}$ that $I(0)=0$ so we have that:
$$I=I(1)-I(0)=2\int_0^1 \frac{\arctan\left(\sqrt{4a^2-1}\right)}{\sqrt{4a^2-1}}da$$
Now I thought about two substitutions:
$$ \overset{a=\frac12\cosh x}=\int_{\operatorname{arccosh}(0)}^{\operatorname{arccosh}(2)} \arctan(\sinh x)dx$$
$$\overset{a=\frac12\sec x}=\int_{\operatorname{arcsec}(0)}^{\frac{\pi}{3}}\frac{x}{\cos x}dx$$
But in both cases the lower bound is annoying and I think I am missing something here (maybe obvious).
So I would love to get some help in order to finish this.
Edit: We can apply once again Feynman's trick. First consider: $$I(t)=\int_0^1 \frac{2\arctan(t\sqrt{4a^2-1})}{\sqrt{4a^2-1}}da\Rightarrow I'(t)=2\int_0^1 \frac{1}{1+t^2(4a^2-1)}da$$
$$=\frac{1}{t\sqrt{1-t^2}}\arctan\left(\frac{2at}{\sqrt{1-t^2}}\right)\bigg|_0^1=\frac{1}{t\sqrt{1-t^2}}\arctan\left(\frac{2t}{\sqrt{1-t^2}}\right)$$
So once again we have $I(0)=0$, so $I=I(1)-I(0)$.
$$\Rightarrow I=\int_0^1\frac{1}{t\sqrt{1-t^2}}\arctan\left(\frac{2t}{\sqrt{1-t^2}}\right)dt\overset{t=\sin x}=\int_0^\frac{\pi}{2}\frac{\arctan(2\tan x)}{\sin x}dx$$
At this point Mathematica can evaluate the integral to be:
$$I=\frac{\pi}{3}\ln(2+\sqrt 3)+\frac43G$$
I didn't try the last integral yet, but I am thinking of Feynman again $\ddot \smile$.
Edit 2: Found that I already was on it some time ago, and actually posted it here, which means I have solved it before using Feynman's trick, but right now I can't remember how I did it.
So given the circumstances I am positive that it can be solved starting with my approach, but if you have any other ways then feel free to share it.
| One more by Feynman’s Technique
Parameterising our integral
$$I=\int_{0}^{\infty} \frac{\ln \left(1+x+x^{2}\right)}{1+x^{2}} d x$$
by the integral
$$
I(a)=\int_{0}^{\infty} \frac{\ln \left(1+2 x \sin a+x^{2}\right)}{1+x^{2}} d x,
$$
where $a\in [-\frac{\pi}{2}, \frac{\pi}{2}]. $
Differentiating $I(a)$ w.r.t. $a$ yields
$$
\begin{aligned}
I^{\prime}(a) &=\int_{0}^{\infty} \frac{2 x \cos a}{\left(1+x^{2}\right)\left(1+2 x \sin a+x^{2}\right)} d x \\
&=\cot a\int_{0}^{\infty}\left(\frac{1}{1+x^{2}}-\frac{1}{1+2 x \sin a+x^{2}}\right) d x \\
&=\cot a\left[\tan ^{-1} x-\frac{1}{\cos a} \tan ^{-1}\left(\frac{x+\sin a}{\cos a}\right)\right]_{0}^{\infty} \\
&=\cot a\left[\frac{\pi}{2}-\frac{1}{\cos a}\left(\frac{\pi}{2}-a\right)\right]
\end{aligned}
$$
Integrating $I’(a)$ back to $I(a)$, we have
$$
\begin{aligned}
I\left(\frac{\pi}{6}\right)- \underbrace{I(0)}_{=\pi\ln 2} &=\int_{0}^{\frac{\pi}{6}} \cot a\left[\frac{\pi}{2}-\frac{1}{\cos a}\left(\frac{\pi}{2}-a\right)\right] d a \\
&=\frac{\pi}{2} \underbrace{ \int_{0}^{\frac{\pi}{6}}\left(\cot a-\frac{1}{\sin a}\right) d a}_{=\ln \left(\frac{2+\sqrt{3}}{4}\right)} + \underbrace{\int_{0}^{\frac{\pi}{6}} \frac{a}{\sin a} d a}_{K}
\end{aligned}
$$
$$
\begin{aligned}
K &=\int_{0}^{\frac{\pi}{6}} \frac{a}{\sin a} d a=\int_{0}^{\frac{\pi}{6}} a\, d\left[\ln \left(\tan \frac{a}{2}\right)\right] \\
&=\left[a \ln \left(\tan \frac{a}{2}\right)\right]_{0}^{\frac{\pi}{6}}-\int_{0}^{\frac{\pi}{6}} \ln \left(\tan \frac{a}{2}\right) d a \\
&=\frac{\pi}{6} \ln \left(\tan \frac{\pi}{12}\right)-2 \int_{0}^{\frac{\pi}{12}} \ln (\tan a) d a \\
&=-\frac{\pi}{6} \ln (2+\sqrt{3})+\frac{4}{3} G,
\end{aligned}
$$
where $G$ is the Catalan’s constant and the last integral from post.
Now we can conclude that
$$
\boxed{I=I\left(\frac{\pi}{6}\right)=\frac{\pi}{3} \ln (2+\sqrt{3})+\frac{4}{3} G}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3205906",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "12",
"answer_count": 4,
"answer_id": 3
} |
Solve the equation $\frac{7x^2 - x + 4}{\sqrt{3x^2 - 1} + \sqrt{x^2 - x} - x\sqrt{x^2 + 1}} = 2\sqrt 2$.
Solve the equation $\dfrac{7x^2 - x + 4}{\sqrt{3x^2 - 1} + \sqrt{x^2 - x} - x\sqrt{x^2 + 1}} = 2\sqrt 2$ over the reals.
Remember when I posted a question almost identical to this one?
Well, turned out my teacher actually mistyped the question. But even when the problem is correctly typed, we couldn't still figure it out throughout the lesson. (And remember the solution must be able to be executed in a test setting.)
| Using the Bunyakovsky and AM - GM inequalities for $(\sqrt{3x^2 - 1}, \sqrt{x^2 - x}, \sqrt{x^2 + 1})$, $(1, 1, -x)$ and $(2x^2 + 4)$, $(5x^2 - x)$, we have that $$(\sqrt{3x^2 - 1} + \sqrt{x^2 - x} - x\sqrt{x^2 + 1})^2 \le (1 + 1 + x^2)(3x^2 - 1 + x^2 - x + x^2 + 1)$$
$$ = (x^2 + 2)(5x^2 - x) = \dfrac{(2x^2 + 4)(5x^2 - x)}{2} \le \dfrac{(7x^2 - x + 4)^2}{8}$$
The equality sign occurs when $\sqrt{3x^2 - 1} = \sqrt{x^2 - x} = -\dfrac{\sqrt{x^2 + 1}}{x}$ and $2x^2 + 4 = 5x^2 - x$.
From the second condition, we have that $3x^2 - x - 4 = 0 \implies \left [ \begin{align} x &= -1\\ x &= \dfrac{4}{3} \end{align} \right.$.
Plugging in $x = -1$, we have that $\sqrt{3x^2 - 1} - \sqrt{x^2 - x} = \sqrt{3 \cdot 1^2 - 1} - \sqrt{1^2 + 1} = 0$.
And $-\dfrac{2\sqrt{x^2 + 1}}{x} - \sqrt{3x^2 - 1} = \dfrac{2\sqrt{1^2 + 1}}{1} - \sqrt{3 \cdot 1^2 - 1} = 0$
Plugging in $x = \dfrac{4}{3}$, we have that $\sqrt{3x^2 - 1} - \sqrt{x^2 - x} = \sqrt{3 \cdot \left(\dfrac{4}{3}\right)^2 - 1} - \sqrt{\left(\dfrac{4}{3}\right)^2 - \dfrac{4}{3}} \ne 0$.
So $x = -1$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3206499",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Evaluate $\int _0^{\infty }\frac{x^6}{\left(x^4+a^4\right)^2}dx$ The function
$$f\left(z\right)=\frac{z^6}{\left(z^4+a^4\right)^2}$$
Has the following poles of order 2:
$$ z(k)=a \exp\left( \frac{\left(2k+1\right)}4 i\pi \right)$$
$f$ is even, therefore: $$\int _0^{+\infty }\frac{x^6}{\left(x^4+a^4\right)^2}dx =\frac{1}{2}\int _{-\infty }^{+\infty \:}\frac{x^6}{\left(x^4+a^4\right)^2}dx$$
$$\int _0^{+\infty }\frac{x^6}{\left(x^4+a^4\right)^2}dx=i\pi \sum _k\:Res\left(f,\:z\left(k\right)\right)$$
$$Res\left(f,\:z\left(k\right)\right)=\lim _{z\to z\left(k\right)}\left(\frac{1}{\left(2-1\right)!}\left(\frac{d}{dz}\right)^{2-1}\frac{z^6\left(z-z\left(k\right)\right)^2}{\left(z^4+a^4\right)^2}\right)$$
$$z^4+a^4=z^4-z_k^4\implies\dfrac{z^6(z-z_k)^2}{(z^4+a^4)^2}=\dfrac{z^6}{(z^3+z_k z^2+z_k^2 z+z_k^3)^2}$$
$$Res\left(f,\:z_k\right)=\lim _{z\to \:z_k}\left(\frac{d}{dz}\left(\frac{z^6}{\left(z^3+z_kz^2+z_k^2z+z_k^3\right)^2}\right)\right)$$
$$Res\left(f,\:z_k\right)=\frac{2z_kz^5\left(z^2+2z_kz+3z_k^2\right)}{\left(z^3+z_kz^2+z_k^2z+z_k^3\right)^3}=\frac{2z_k^6\cdot 6z_k^2}{\left(4z_k^3\right)^3}$$
$$Res\left(f,\:z_k\right)=\frac{12z_k^8}{64z_k^9}=\frac{3}{16z_k}$$
$$\int _0^{+\infty }\frac{x^6}{\left(x^4+a^4\right)^2}dx=\frac{3i\pi }{16a}\sum _{k=0}^n\:e^{-\frac{\left(2k+1\right)}{4}i\pi }$$
We consider only the residues within the upper half plane, that is to say those corresponding to $k=0$ and $k=1$.
$$\int _0^{+\infty \:}\frac{x^6}{\left(x^4+a^4\right)^2}dx=\frac{3i\pi \:}{16a}\left(e^{-\frac{i\pi }{4}\:\:}+e^{-\frac{3i\pi \:}{4}\:\:}\right)$$
$$\int _0^{+\infty \:}\frac{x^6}{\left(x^4+a^4\right)^2}dx=\frac{3i\pi \:}{16a}\left(\frac{\sqrt{2}}{2}\:-i\frac{\sqrt{2}}{2}-\frac{\sqrt{2}}{2}-i\frac{\sqrt{2}}{2}\right)$$
$$\int _0^{+\infty \:}\frac{x^6}{\left(x^4+a^4\right)^2}dx=\frac{3\pi \sqrt{2}\:}{16a}$$
| $\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
\newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace}
\newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
\newcommand{\dd}{\mathrm{d}}
\newcommand{\ds}[1]{\displaystyle{#1}}
\newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
\newcommand{\ic}{\mathrm{i}}
\newcommand{\mc}[1]{\mathcal{#1}}
\newcommand{\mrm}[1]{\mathrm{#1}}
\newcommand{\pars}[1]{\left(\,{#1}\,\right)}
\newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
\newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
\newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
\newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$
\begin{align}
&\bbox[10px,#ffd]{\int_{0}^{\infty}{x^{6} \over
\pars{x^{4} + a^{4}}^{2}}\,\dd x}
\\[5mm] \stackrel{x/\verts{a}\ \mapsto\ x}{=}\,\,\,&
{1 \over \verts{a}}\int_{0}^{\infty}{x^{6} \over
\pars{x^{4} + 1}^{2}}\,\dd x
\\[5mm] \stackrel{\mrm{I.B.P.}}{=}\,\,\,&\
-\,{1 \over 4\verts{a}}\int_{x\ =\ 0}^{x\ \to\ \infty}x^{3}
\,\dd\pars{1 \over x^{4} + 1}
\\[5mm] = &\
{1 \over 4\verts{a}}\int_{0}^{\infty}
{1 \over x^{4} + 1}\,\pars{3x^{2}}\,\dd x
\\[5mm] = &\
{3 \over 4\verts{a}}\int_{0}^{\infty}
{\dd x \over x^{2} + 1/x^{2}}\,\dd x
\\[5mm] = &\
{3 \over 4\verts{a}}\int_{0}^{\infty}
{\dd x \over \pars{x - 1/x}^{2} + 2}
\\[5mm] = &\
{3 \over 8\verts{a}}\left[\int_{0}^{\infty}
{\dd x \over \pars{x - 1/x}^{2} + 2}\right.
\\[2mm] & \phantom{{3 \over 8\verts{a}}\,\,}
\left. + \int_{\infty}^{0}
{-\dd x/x^{2} \over \pars{1/x - x}^{2} + 2}
\right]
\end{align}
In the last line, the last integral is equivalent to the first one: It just arises from a $\ds{x \mapsto 1/x}$ change of variable.
Then,
\begin{align}
&\bbox[10px,#ffd]{\int_{0}^{\infty}{x^{6} \over
\pars{x^{4} + a^{4}}^{2}}\,\dd x}
\\[5mm] = &\
{3 \over 8\verts{a}}\int_{0}^{\infty}
{1 + 1/x^{2} \over \pars{x - 1/x}^{2} + 2}\,\dd x
\\[5mm] \stackrel{x - 1/x\ \mapsto\ x}{=}\,\,\, &\
{3 \over 8\verts{a}}\int_{-\infty}^{\infty}
{\dd x \over x^{2} + 2}
\\[5mm] = &\
{3 \over 8\verts{a}}\,{1 \over 2}\,\root{2}
\int_{-\infty}^{\infty}{\dd x/\root{2} \over
\pars{x/\root{2}}^{2} + 1}
\\[5mm] \stackrel{x/\root{2}\ \mapsto\ x}{=}\,\,\, &\
{3\root{2} \over 16\verts{a}}\
\underbrace{\int_{-\infty}^{\infty}{\dd x \over x^{2} + 1}}
_{\ds{=\ \pi}}\ =\
\bbx{{3\root{2}\pi \over 16}\,{1 \over \verts{a}}}
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3206701",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 0
} |
Relation defined on the set of real numbers by xRy when $x^2 + y^2 = 1$. Show whether or not R is reflexive, symmetric, antisymmetric or transitive.
Let R be the relation defined on the set of real numbers by $xRy$
whenever $x^2 + y^2 = 1.$ Show whether or not $R$ is reflexive,
symmetric, antisymmetric or transitive.
All right so I think I've got this right and just want to make sure.
*
*Symmetric - Yes for all $x$ and $y$, if $x^2 + y^2 = 1$ then $1 = x^2 + y^2 $
*Reflexive - Yes for all $x$ and $y$, if $x^2 + y^2 = x^2 +y^2$
*Anti-Symmetric - Yes for all $x$ and $y$, if $x^2 + y^2 = 1$ and $1 = x^2 + y^2$ then $x^2 + y^2 = 1$ is true.
*Transitive - Yes, if $x^2 + y^2 = x_2^2 + y_2^2 $ and $x^2 + y^2 = 1$ then $x_2^2 + y_2^2 =1$
I think I've done all these right, unless I've misunderstood the definitions for the relations.
| You misunderstood. Instead of using the $R$ notation, say that $x \sim y$ if and only if $x^2+y^2=1$. So
*
*Not reflexive, as we have elements $x$ such that $x \not\sim x$. For example, $1 \not\sim 1$, as $1^2+1^2 = 2 \neq 1$.
*It is symmetric, because if $x \sim y$, then $x^2+y^2 = 1$. So $y^2+x^2=1$ says that $y \sim x$.
*Not transitive for reasons similar to reflexivity. For example, $1 \sim 0$ and $0 \sim 1$, but $1 \not\sim 1$.
And so on...
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3210272",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
$a, b$ are positive whole numbers such that $a + b = a/b + b/a$. How many possible values can $a^4 + b^4$ be? $a, b$ are positive whole numbers such that $a + b = a/b + b/a$. How many possible values can $a^4 + b^4$ be?
I tried using $(a^2 + b^2)^2$ but I don't know what to do after.
| You have $a^2-ba^2+ba-b^2=0$ then $a=\dfrac{-b\pm\sqrt{b^2-4(1-b)(-b^2)}}{2(1-b)}$ so that $5-4b$ must be a square. The only possible value is $b=1$ thus $a=1$ and $a^4+b^4=2$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3211657",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
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Fractions in Questions and Answers
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