Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
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Prove that $a^2 - b^2 + c^2 - d^2 \ge (a - b + c - d)^2$ In thinking about a base case in this problem, I came up with the following question.
Given real numbers $a \ge b \ge c \ge d \ge 0$, prove that the following holds:
$a^2 - b^2 + c^2 - d^2 \ge (a - b + c - d)^2 \tag{A}$
My attempt:
After simplification, this reduces to proving the inequality:
$\underbrace{ab}_{(1)} + \underbrace{bc}_{(2)} + \underbrace{cd + da}_{(3)} \ge \underbrace{ac}_{(1)} + \underbrace{bd}_{(2)} + \underbrace{b^2 + d^2}_{(3)} \tag{B}$
I tried to attack pairs of terms individually. This gave pairs $(1)$ and $(2)$ which satisfied the $\ge$ relation, since $ab \ge ac \implies b \ge c$ true and $bc \ge bd \implies c \ge d$ true.
But then I got stuck at proving pair $(3)$ also satisfied the $\ge$ relation. That is $cd + da \ge b^2 + d^2 \tag{C}$
It turned out that $(C)$ doesn't hold in general. For example $(a, b, c, d) = (5, 4, 3, 2)$ gives $3 \cdot 2 + 2 \cdot 5 \ge 4^2 + 2^2 \implies 16 \ge 20$ false.
So, my strategy was incorrect. I would appreciate if anyone could show me the right approach for proving either (A) or (B).
Update: See this for a generalization of this problem.
| A standard trick for a problem like this is to make a variable substitution which simplifies the problem in some way; hopefully you hit on a substitution which makes the problem more amenable. One idea for such a substitution is to simplify the condition that $a\geq b\geq c\geq d\geq 0$, and we can do this by defining new variables representing the difference between consecutive original variables:
\begin{align}
x &= a - b \\
y &= b - c \\
z &= c - d.
\end{align}
Using the new variables, we find that the condition becomes $x,y,z,d\geq0$. Next, replace $a$, $b$, and $c$ in the inequality to be proved (we're working backwards here).
The lefthand side becomes
\begin{align}
a^2 - b^2 + c^2 - d^2 &=
(a-b)(a+b) + (c-d)(c+d) \\
&= x\cdot [(x+y+z+d) + (y+z+d)] + z\cdot [(z + d) + d] \\
&= x^2 + 2xy + 2xz + 2xd + z^2 + 2zd \\
&= (x^2 + 2xz + z^2) + 2xy + 2xd + 2zd.
\end{align}
The righthand side becomes
$$
(a-b + c-d)^2 = (x + z)^2 = x^2 + 2xz + z^2.
$$
With the new variables, your inequality is obvious! In particular, we see that equality holds if and only if $xy + xd + zd = 0$, that is, if and only if one of the following is true:
*
*$x=0$ and either $z=0$ or $d=0$
*$y=0$ and $d=0$.
In other words, these conditions say that
*
*$a=b$ and either $c=d$ or $d=0$
*$b=c$ and $d=0$.
| {
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"timestamp": "2023-03-29T00:00:00",
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Compositions - Fruit Salad I'm asked to find $s(n)$ which is the number of ways to make a fruit salad with $n$ pieces of fruit, given that we must use strawberries by the half-dozen, an odd number of apples, between 2 and 7 bananas and at most 1 pineapple.
I started evaluating the ordinary generating function $S(X)$. I found
$$S(X)=\frac{1}{1-x^6}\frac{x}{1-x^2}\frac{x^2-x^8}{1-x}(1+x)=\frac{1}{1-x^6}\frac{x}{1-x}\frac{x^2-x^8}{1-x}$$
However, I don't manage to find a closed formula for $s(n)$ after. I know that I'm supposed to "evalute $S(X)$ at $x^n$" but I don't really understand how to do it.
Any help would be appreciated.
Thanks.
| You can write:
\begin{align}
S(z) &= \frac{1}{1 - z^6} \frac{z}{1 - z^2} \frac{z^2 - z^8}{1 - z} (1 + z) \\
&= \frac{1}{1 - z^6} \frac{z}{(1 - z) (1 + z)}
\frac{z^2 (1 - z^6)}{1 - z} (1 + z) \\
&= \frac{z^3}{(1 - z)^2} \\
&= z^3 \sum_{k \ge 0} \binom{-2}{k} (-1)^k z^k \\
&= z^3 \sum_{k \ge 0} \binom{k + 1}{1} z^k \\
&= z^3 \sum_{k \ge 0} (k + 1) z^k
\end{align}
I'm sure you can take it from here.
| {
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Which performance enhancement is better? Suppose you are given the task of improving the performance of a program consisting of three parts. Part $A$ requires 20% of the overall run time, part $B$ requires 30%, and part $C$ requires 50%.
You determine that for \$1000 you could either speed up part $B$ by a factor of 3.0 or part $C$ by a factor of 1.5. Which choice would maximize performance?
I think if you speed up part $B$ by a factor of 3.0, then you only spend 10% of your time on that, saving 20% time. But if you speed up part $C$ by a factor of 1.5, then you spend 25% time and save 25% time.
So is speed up part $C$ the right answer since you save 25% vs 20%?
Thanks
| Let $x$ be the runtime of your program.
Then
\begin{align}
t(A) :&= a = 0.2 x\\
t(B) :&= b = 0.3 x\\
t(C) :&= c = 0.5 x\\
\end{align}
Speed something up by a factor of 2 (double as fast) means to reduce the runtime to half of the time it needed before. The thought behind that is that you can do twice as much in the same time.
Similar, I think when you say "speed up part $B$ by a factor of 3.0" it means reducing the runtime to $\frac{1}{3}$ of the time before. So a speed up of the factor of 1.5 means the new runtime is $\frac{1}{1.5} = \frac{2}{3}$ of the original runtime.
So the question is
$$a + \frac{1}{3}b + c < a + b + \frac{3}{2}c$$
or
$$a + \frac{1}{3}b + c > a + b + \frac{3}{2}c$$
Now resubstitute:
\begin{align}
0.2x + \frac{1}{3} \cdot 0.3x + 0.5x &= \frac{1}{2}x + \frac{1}{10}x\\
&= \frac{6}{10}x\\
&< \frac{5}{10}x + \frac{1}{3} x\\
&= \frac{1}{2}x + \frac{1}{3} x\\
&= 0.5x + \frac{2}{3} \cdot \frac{1}{2} x\\
&= 0.2x + 0.3x + \frac{2}{3} \cdot 0.5 x\\
&= a + b + \frac{2}{3}c
\end{align}
So the answer is: Reduce the execution time (= speed up) part $B$.
| {
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Generating functions for compositions Let $g(n)$ be the number of compositions of n where each part is an odd number. Let $h(n)$ number of compositions of $n$ where each part is either 1 or 2. Using the ordinary generating functions $G(x)$ and $H(x)$, show that $g(n) = h(n-1)$
| Let $\mathcal{O}$ denote the class of all odd numbers, so that it has generating function $O(z) = z + z^3 + z^5 + \dots = \dfrac{z}{1-z^2}$. Then the class of compositions into odd parts is
$$\mathcal{G} = \operatorname{S\scriptsize EQ}(\mathcal{O}) \implies G(z) = \frac{1}{1-O(z)} = \frac{1}{1-\frac{z}{1-z^2}} = \frac{1-z^2}{1-z-z^2}$$
where $G(z) = \sum_{n \ge 0} g(n) z^n$ is the generating function for $\mathcal{G}$.
Similarly, let $\mathcal{C}$ denote the class containing just the numbers $1$ and $2$ (so $C(z) = z+z^2$), then the class of compositions into parts equal to $1$ and $2$ is
$$\mathcal{H} = \operatorname{S\scriptsize EQ}(\mathcal{C}) \implies H(z) = \frac{1}{1-C(z)} = \frac{1}{1-z-z^2}$$
where $H(z) = \sum_{n \ge 0} h(n) z^n$ is the generating function for $\mathcal{H}$.
Now you want to prove that $g(n) = h(n-1)$ for $n \ge 1$, or equivalently that $g(n+1) = h(n)$ for $n \ge 0$. We have
$$\sum_{n \ge 0}g(n+1)z^n = \frac{G(z) - g(0)}{z} = \frac1z \left( \frac{1-z^2}{1-z-z^2} - 1\right) = \frac{1}{1-z-z^2} = H(z)$$
which proves the assertion.
| {
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Proving that if $a,b > 1$, then $5^a - 3^b=16$ has only one solution with $a=2$ and $b=2$ This may be one of those problems that is easy to state but very hard to prove. I don't know.
I have tried to show that there is only one solution but I have not made much progress.
Here's what I have:
If there exists $a,b > 2$, then:
$$5^2(5^{a-2} - 1) = 3^2(3^{b-2}-1)$$
$a-2 = u(3-1)(3) = 6u$ where $u \ge 1$
$b-2 = v(5-1)(5) = 20v$ where $v \ge 1$
So, as I understand it, the problem comes down to showing that for all $u,v$:
$$5^2(2)(1 + 5 + 5^2 + \cdots + 5^{6u-1}) \ne 3^2(1 + 3 + 3^2 + \cdots + 3^{20v-1})$$
I can't figure out any step after this. Is my thinking wrong? Is there $a,b > 2$ where $5^a - 3^b = 16$
Thanks very much,
-Larry
| Consider everything modulo $3$. Then,
$$(-1)^a \equiv 1 \pmod 3$$
From this, $a$ must be even, i.e. $a = 2m$ for some integer $m \ge 1$.
Now, consider everything modulo $5$. Then,
$$-(-2)^b \equiv 1 \pmod 5$$
From this (why?), $b$ must even, i.e. $b = 2n$ for some $n \ge 1$.
So we rewrite:
$$5^{2m} - 3^{2n} = 16$$
$$(5^m - 3^{n})(5^m + 3^{n}) = 2^4$$
Now, let $5^m - 3^{n} = 2^x$ and $5^m + 3^{n} = 2^{x + k}$ for some integers $x,k \ge 0$ s.t. $2x + k = 4$. Then, adding both up,
$$2\cdot5^m = 2^x + 2^{x + k} = 2^x(1 + 2^k)$$
$$5^m = 2^{x - 1}(1 + 2^k)$$
But if $x - 1 > 0$, then both sides will be even, which is false since $5^m$ is odd. So $x - 1 = 0 \implies x = 1 \implies k = 4 - 2 = 2$. Therefore, we have
$$5^m - 3^{n} = 2$$
$$5^m + 3^{n} = 8$$
The rest is trivial (Hint: sum up both, or subtract one from another).
| {
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An almost impossible limit The following limit appeared in a qualification exam: Find the limit of
$$\lim_{x \to 0} \left( \frac{\tan (\sin (x))-\sin (\tan (x))}{x^7} \right).$$
I ended up doing it in Mathematica, is there any other way?
Thanks in advance!
| I have not been able to find an elementary solution which avoids Taylor's series but I think we can provide much simpler approach in terms of calculations while following the Taylor's series approach.
Let $\tan x = t$ so that $\sin x = t/\sqrt{1 + t^{2}} = s$ and then $$\tan(\sin x) - \sin (\tan x) = \tan s - \sin t$$ Since $(\tan x)/x \to 1$ as $x \to 0$ we can replace the $x^{7}$ in denominator in the question by $t^{7}$. So the desired limit is equal to $$L = \lim_{t \to 0}\frac{\tan s - \sin t}{t^{7}}$$ where $s = t/\sqrt{1 + t^{2}}$. We will now need the expansion $$\tan s = s + \frac{s^{3}}{3} + \frac{2s^{5}}{15} + \frac{17s^{7}}{315} + \cdots$$ And then we have the tedious algebra $$\begin{aligned}A &= s + \frac{s^{3}}{3} + \frac{2s^{5}}{15}\\
&= \frac{t}{\sqrt{1 + t^{2}}} + \frac{t^{3}}{3(1 + t^{2})\sqrt{1 + t^{2}}} + \frac{2t^{5}}{15(1 + t^{2})^{2}\sqrt{1 + t^{2}}}\\
&= \frac{15t(1 + t^{2})^{2} + 5t^{3}(1 + t^{2}) + 2t^{5}}{15(1 + t^{2})^{2}\sqrt{1 + t^{2}}}\\
&= \frac{22t^{5} + 35t^{3} + 15t}{15(1 + t^{2})^{2}\sqrt{1 + t^{2}}}\\
&= \frac{22t^{5} + 35t^{3} + 15t}{15}\cdot(1 + t^{2})^{-5/2}\\
&= \left(t + \frac{7t^{3}}{3} + \frac{22t^{5}}{15}\right)\cdot\left(1 - \frac{5t^{2}}{2} + \frac{35t^{4}}{8} - \frac{105t^{6}}{16} + \cdots\right)\\
&= t + t^{3}\left(\frac{7}{3} - \frac{5}{2}\right) + t^{5}\left(\frac{35}{8} - \frac{35}{6} + \frac{22}{15}\right) + t^{7}\left(-\frac{105}{16} + \frac{245}{24} - \frac{22}{6}\right) + \cdots\\
&= t - \frac{t^{3}}{6} + \frac{t^{5}}{120} - \frac{t^{7}}{48} + \cdots\\
&= t - \frac{t^{3}}{3!} + \frac{t^{5}}{5!} - \frac{t^{7}}{48} + \cdots\end{aligned}$$ Note how nicely the first three terms match with those in expansion of $\sin t$.
The above tedious algebra was the only part which dealt with calculation of coefficients in a certain power series. One can see that it involves very less calculation and the series for $(1 + t^{2})^{-5/2}$ needs to be calculated till 3 terms only (ignoring the obvious first term 1).
Now the limit $L$ can be calculated very easily as follows $$\begin{aligned}L &= \lim_{t \to 0}\frac{\tan s - \sin t}{t^{7}}\\
& = \lim_{t \to 0}\frac{\tan s - A}{t^{7}} + \frac{A - \sin t}{t^{7}}\\
&= \lim_{t \to 0}\frac{\tan s - A}{s^{7}}\cdot\left(\frac{s}{t}\right)^{7} + \frac{A - \sin t}{t^{7}}\\
&= \frac{17}{315} - \frac{1}{48} + \frac{1}{7!} = \frac{1}{30}\end{aligned}$$ In the above we have used the expansion of $\sin t$ and $\tan s$ and we can see that the appropriate number of terms cancel because of the term $A$ and only the terms containing $s^{7}$ and $t^{7}$ are significant while the terms with higher powers of $s, t$ tend to $0$ and do not contribute to the limit. Also we have used the fact that as $t \to 0$ the variable $s \to 0$ and $s/t = 1/\sqrt{1 + t^{2}} \to 1$.
Note: An approach which avoids use of Taylor series and uses LHR is presented here (but it is really laborious).
| {
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question on surds i already asked this question but the answer I got did not match the one in the book $$\sqrt{ 3x }= x + \sqrt {3}$$
Give x in the form
$$A \sqrt {B} + C $$
Can you show me how this is done step by step.
The answer I have in the book is:
$$\frac {1}{2} \sqrt{3} + \frac {3}{2} $$
this is where I got stuck:
$$ \frac {x^2 +2x \sqrt{3} +3}{3x} $$
| $\sqrt{3x}=x+\sqrt{3}$
$3x=x^2+2x\sqrt{3}+3$
$x^2+(2\sqrt{3}-3)x+3=0$
$D=(2\sqrt{3}-3)^2-4 \times 1 \times 3=12-12\sqrt{3}+9-12=9-12\sqrt{3}<0$
The equation nas no real solutions.
you can check that the answer you have is not the solution at all.
| {
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Green balls and Red balls, probability problem I'm studying for my exam and I came across the following draw without replacement problem :
$N$ boxes filled with red and green balls.
The box $r$ contains $r-1$ red balls and $N-r$ green balls.
We pick a box at random and we take $2$ random balls inside it, without putting them back.
1) What is the probability that the second ball is green ?
2) What is a probability that the second ball is green, knowing the first one is green.
I don't know where to start, all those dependance (to $r$ and $N$) are blowing my mind.
I don't know if I should concider it as a Binomial law (with Bernoulli : ball = green, $n=2, p = ?$) or go with the formula $$p(X=k)=\frac{C_{m}^{k} C_{N-m}^{n-k}}{C_{N}^{n}}$$
or something else...
Could someone advise me ?
| All of the boxes contain $N - 1$ balls. This is just a complicated conditional probability problem. Lets look at a single box with $r$ red balls and $g$ green balls. What would the probability be of getting green on the second? Well it depends on whether or not you draw a red or green first. If you draw a red first, then there are $\left.p(\text{second green } \right| \text{ first red}) = \frac{g}{r + g - 1}$. However, if you draw a green ball first then you have one less green to choose from giving: $\left.p(\text{second green } \right| \text{ first green}) = \frac{g - 1}{g + r - 1}$. So what are the chances of each condition happening? $p(\text{first red}) = \frac{r}{g + r}$ and $p(\text{first green}) = \frac{g}{r + g}$. Therefore we can finally write:
\begin{align}
p(\text{second green}) =& \left.p(\text{second green } \right| \text{ first red})p(\text{first red}) + \left.p(\text{second green } \right| \text{ first green})p(\text{first green})\\
=& \frac{r}{r + g}\frac{g}{r+g-1} + \frac{g}{r + g}\frac{g-1}{r+g-1} = \frac{g(r + g - 1)}{(r + g)(r + g - 1)} = \frac{g}{r + g}
\end{align}
Not surprising that drawing the second green has just as good of a chance of being green as the first pick.
Therefore for each of the $N$ boxes you need to compute $p(\text{second green})$ (which is just the probability of drawing a green on the first try). Now the condition is that we choose box $r$ which has $p(\text{second green}) = p(\text{first green}) = \frac{N - r}{N - 1}$. The probability of choosing box $r$ among $N$ boxes is just $\frac{1}{N}$ which gives:
$$
p(\text{second green}) = \sum_1^N \frac{1}{N}\frac{N - r}{N - 1} = \frac{1}{N(N - 1)}\sum_1^r (N - r)
$$
The first sum is very easy (you're just summing the same number, $N$, $N$ times) $\sum_1^N N = N\cdot N = N^2$. The second part is easy if you remember the sum of the first $n$ consecutive integers is $\sum_1^n i = \frac{n(n + 1)}{2}$. So this gives:
$$
p(\text{green}) = \frac{N^2 - \frac{N(N + 1)}{2}}{N(N - 1)} = \frac{2N^2 - N^2 - N}{2N(N - 1)} = \frac{N^2 - N}{2\left(N^2 - N\right)} = \frac{1}{2}
$$
For part $2$), we actually already computed that above: $\left.p(\text{second green }\right|\text{ first green}) = \frac{g - 1}{g + r - 1}$. But now you need to sum over the condition that it could be any of the $N$ boxes (edit: However, the last box, box $N$, has $0$ green balls (and thus seeing green first means it definitely wasn't this box. So we should only sum over the first $N - 1$ boxes and divide by $N - 1$, not $N$.):
\begin{align}
\left.p(\text{second green }\right|\text{ first green}) =& \sum_1^{N - 1}
\frac{1}{N - 1}\frac{N - r - 1}{N - 2} \\
=& \frac{N(N - 1) - (N - 1) - \frac{N(N - 1)}{2}}{N(N - 2)} \\
=& \frac{2N(N - 1) - 2(N - 1) - N(N - 1)}{2(N - 1)(N - 2)}\\
=& \frac{N(N - 1) - 2(N - 1))}{2(N - 1)(N - 2)} \\
=& \frac{(N - 1)(N - 2)}{2(N - 1)(N - 2)} \\
=& \frac{1}{2}
\end{align}
This is only valid for $N > 2$ (since if $N = 1$ there are no balls in each box and if $N = 2$ there is only one ball in each box). This result just confirms that drawing balls are independent events.
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How many surjective functions are there from $A=${$1,2,3,4,5$} to $B=${$1,2,3$}? I want to find how many surjective functions there are from the set $A=${$1,2,3,4,5$} to the set $B=${$1,2,3$}? I think the best option is to count all the functions ($3^5$) and then to subtract the non-surjective functions. However, I'm not sure how can I count these functions.
Thanks
| Pedestrian approach:
How many surjective functions from:
$A$ ={ $1, 2, 3, 4, 5$} to $B$= {$a, b, c$} ?
1) Let $3$ distinct elements of $A$ be mapped onto $a, b$, or $c$.
There are
$\binom{5}{3} = 10$,
ways to pick $3$ elements from $5$.
There are $3$ ways to map these elements onto $a,b$, or $c$.
Altogether $3×10 = 30$ ways.
Each choice leaves $2$ spots in $B$ empty; $2$ ways of filling the vacant spots with the $2$ remaining elements of $A$
Combining: $2×30 = 60$ ways of generating a surjectice map with $3$ elements mapped onto $1$ element of $B$.
2) $2$ elements of $A$ are mapped onto $1$ element of $B$, another $2$ elements of $A$ are mapped onto another element of $B$, and the remaining element of $A$ is mapped onto the remaining element of $B$.
The mapping looks, for example, like :
$( ||, |, || )$.
Let's start with the single element:
$5$ ways to choose an element from $A$, $3$ ways to map it to $a,b$ or $c$. Altogether: $5×3 =15$ ways.
$2$ vacant spots remain to be filled with $2$ elements of $A$ each.
$4$ elements are left in $A$, the number of ways of choosing $2$ of the remaining $4$: $ \binom{4}{2} = 6.$
To avoid double counting fix any one empty spot of $B$ (there are $2$).
There are $6$ ways to put $2$ numbers in this spot, the remaining open spot is taken care of with the remaining $2$ numbers of $A$ automatically.
Altogether there are $15×6 = 90$ ways of generating a surjective function that maps $2$ elements of $A$ onto $1$ element of $B$, another $2$ elements of $A$ onto another element of $B$, and the remaining element of $A$ onto the remaining element of $B$.
Combining: There are 60 + 90 = 150 ways.
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Calculate for $(1+\tan 20^\circ)(1+\tan 25^\circ)$. Help me with my works I have no idea what I am doing here,
I started with $\tan 20^\circ=\tan(45^\circ-25^\circ)=(1-\tan 25^\circ)/(1+\tan 25^\circ)$
I am sure the work I have shown so far are ok, but how do you get $1+\tan 20^\circ=2/(1+\tan 25^\circ)$ from that?
| It's a consequence of the following trigonometric identity
\begin{equation*}
\color{blue}{\left( 1+\tan a\right) \left( 1+\tan b\right) =2+\tan a+\tan b-\dfrac{\tan
a+\tan b}{\tan \left( a+b\right) }.}\tag{1}
\end{equation*}
On the one hand we rewrite
\begin{equation*}
\tan (a+b)=\frac{\tan a+\tan b}{1-\tan a\tan b}
\end{equation*}
as
\begin{equation*}
\color {blue}{\tan a\tan b}=1-\frac{\tan a+\tan b}{\tan \left( a+b\right) }.\tag{2}
\end{equation*}
On the other hand setting $x=\tan a$ and $y=\tan b$ in the algebraic identity
\begin{equation*}
xy=\left( 1+x\right) \left( 1+y\right) -1-x-y
\end{equation*}
yields:
\begin{equation*}
\color {blue}{\tan a\tan b}=\left( 1+\tan a\right) \left( 1+\tan b\right) -1-\tan a-\tan b.\tag{3}
\end{equation*}
If we equate $(3)$ to $(2)$, then we get
\begin{equation*}
\left( 1+\tan a\right)
\left( 1+\tan b\right) -1-\tan a-\tan b=1-\frac{\tan a+\tan b}{\tan \left( a+b\right) },
\end{equation*}
from which $(1)$ follows. For $a=20^{{}^\circ},b=25^{{}^\circ}$ we obtain
\begin{eqnarray*}
\left( 1+\tan 20{{}^\circ}\right) \left( 1+\tan 25{{}^\circ}\right)
&=&2+\tan 20{{}^\circ}+\tan 25{{}^\circ}-\frac{\tan 20{{}^\circ}+\tan 25{{}^\circ}}{\tan \left( 45{{}^\circ}\right) } \\
&=&2+\tan 20{{}^\circ}+\tan 25{{}^\circ}-\frac{\tan 20{{}^\circ}+\tan 25{{}^\circ}
}{1} \\
&=&2.\tag {4}
\end{eqnarray*}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/745929",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Roots of Taylor's series.
Show that there is exactly one value of x which satisfies the equation $$2\cos^2 (x^3+x)=2^x+2^{-x} $$
I solved this using Taylor's series:
$$2^x+2^{-x}=2\{1+\frac {x^2 \{\ln2\}^2}{2!}+\frac {x^4 \{\ln2\}^4}{4!}....\} $$
$$\cos^2 (x^3+x)=1+\cos (2x^3+2x)=2-\frac{{2x^3+2x}^2}{2!}.... $$
Equating the two gives
$$ x^2\{{\ln2}^2+.....\}$$
$$\implies x=0 $$
But I can't seem to be able to find an argument justifying the existence if only one root. If I said that as the other term is an infinite series hence it's roots are undefined, would that be correct?
| As for real $x, 2^{\pm x}>0$
$$\frac{2^x+2^{-x}}2\ge \sqrt{2^x\cdot 2^{-x}}=1\ \ \ \ (1)$$
So, we have $$2\cos^2(x^3+x)=2^x+2^{-x}\ge2$$
$$\iff \sin^2(x^3+x)\le0$$
For real $x,\displaystyle\sin^2(x^3+x)\ge0\implies$ the relation will hold iff
$\displaystyle2^x+2^{-x}=2$ and $\displaystyle\sin^2(x^3+x)=0\iff\sin(x^3+x)=0$
But, the equality in $(1)$ will hold iff $\displaystyle2^x=2^{-x}\iff 2^{2x}=1$
Can you take it home from here?
Another way:
If $\displaystyle a+a^{-1}=2\cos\theta, a^2-2a\cos\theta+1=0$
$\displaystyle\implies a=\cos\theta\pm i\sin\theta$
If $a$ is real, $\sin\theta$ must be $\displaystyle0\implies a=\cos\theta=\pm1$
Again, if $\displaystyle a>0, a=1\implies \theta=2n\pi$ where $n$ is any integer
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/751770",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Need help with basic factoring equation I'm just trying to brush up on my factoring of quadratic equations.
$$\frac1{x+3} + \frac1{x^2 + 5x +6}$$
$$\frac1{x+3} + \frac1{(x+2)(x+3)}$$
$$\frac{(x+2)(x+3) + (x+3)}{(x+2)(x+2)(x+3)}$$
Then I'm stuck. I think you can factor out the $(x+2)(x+3)$ from top and bottom?
| $$\begin{align}
&{1\over{x+3}}+{1\over{x^2+5x+6}}\to\\
&{1\over{x+3}}+{1\over{(x+2)(x+3)}}\to\\
&{x+2+1\over{(x+2)(x+3)}}\to\\
&{x+3\over{(x+2)(x+3)}}\to\\
&1\over{x+2}
\end{align}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/755869",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Find three numbers given their sum, product and sum of their squares Given three unknown positive integers. Is it possible to find the three numbers if we are given their
Sum->(a+b+c) = X
Product-> (abc) = Y
Sum of Squares-> (a^2 + b^2 + c^2) = Z
| Here is a method you can try. Let
$s_1=a+b+c$,
$s_2=a^2+b^2+c^2$,
$p_2=ab+bc+ac$,
$p_3=abc$
$a, b, c$ are the roots of the cubic $$0=(x-a)(x-b)(x-c)=x^3-s_1x^2+p_2x-p_3$$
We don't know $p_2$ but can calculate it using $s_1^2=s_2+2p_2$, and then solve the cubic to find $a,b,c$.
| {
"language": "en",
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"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
How find this $x^3-5x+10=2^y$ let $x,y$ is positive integer,and such
$$x^3-5x+10=2^y$$
find all $x,y$.
since $$x=1\Longrightarrow 1^3-5+10=6$$ can't
$$x=2,2^3-5\cdot 2+10=8=2^3$$
so $x=2,y=3$
$$x=3,LHS=27-15+10=22$$
$$x=4,LHS=64-20+10=54$$
$$x=5,LHS=125-25+10=110$$
$$x=6,LHS=216-30+10=236$$
$$\cdots$$
I find $$(x,y)=(2,3)$$
I only find $x\le 7$ this solution.
maybe this have other solution.and This problem is from Mathematical olympiad problems
Thank you
| Some experimenting reveals that $\pmod{7}$ is the way to go:
$$x^3-5x+10=2^y$$
\begin{align}
\begin{array}{|c|c|}
\hline
x \pmod{7} & x^3-5x+10 \pmod{7} \\
\hline
0 & 3 \\
\hline
1 & 6 \\
\hline
2 & 1 \\
\hline
3 & 1 \\
\hline
4 & 5 \\
\hline
5 & 5 \\
\hline
6 & 0 \\
\hline
\end{array}
&
\begin{array}{|c|c|}
\hline
y \pmod{3} & 2^y \pmod{7} \\
\hline
0 & 1 \\
\hline
1 & 2 \\
\hline
2 & 4 \\
\hline
\end{array}
\end{align}
We see that we must have $3 \mid y$, so let $y=3z$, then
$$x^3-5x+10=2^{3z}=(2^z)^3$$
This means that the LHS is a perfect cube, which can't happen for large values of $x$ as we may bound it between consecutive perfect cubes. Indeed, for $x \geq 3$, we have
$$(x-1)^3<x^3-5x+10<x^3$$
(reduces to $0<3x^2-8x+11=x(3x-8)+11$ and $10<5x$ which are true for $x \geq 3$)
Thus there are no solutions for $x\geq 3$. It is now straightforward to check that $x=1$ fails and $x=2$ gives the only positive integer solution $(2, 3)$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/756355",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 0
} |
If $a^2+b^2=1$ where $a,b>0$ then find the minimum value of $(a+b+{1\over{ab}})$ If $a^2+b^2=1$ where $a,b>0$ then find the minimum value of $(a+b+{1\over{ab}})$
This can be easily done by calculas but is there any way to do do this by algebra
| Edit: The last line is incorrect.
In light of @Lemur's answer, it will suffice to show that
$$
a + b \geq \sqrt{2}
$$
where $a,b > 0, a^2 + b^2 = 1$ is enforced.
To see this, note that by AMGM, $ab \leq \frac{a^2 + b^2}{2} = \frac{1}{2}$, and so
$$
(a + b)^2 = a^2 + b^2 + 2 ab \geq 1 + 2 ab \geq 1 + 1 = 2
$$
Thus $a + b \geq \sqrt{2}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/756702",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 4
} |
Find $4\cos\theta-3\sin\theta$, given that $4\sin \theta +3\cos \theta = 5$ Another problem that I already wasted hours on.
Given
$$4\sinθ +3\cosθ = 5$$
Find
$$4\cosθ -3\sinθ$$
Help me guys (PS:I'm not that good in maths)
| Its simple:
Given
$$4\sin\theta + 3\cos\theta=5$$
Squaring on both sides gives
$$16\sin^2\theta +24\sin\theta \cos\theta+9\cos^2\theta=25$$
$$16-16\cos^2\theta+24\sin\theta \cos\theta+9-9\sin^2\theta=25$$
$$16\cos^2\theta-24\sin\theta \cos\theta+9\sin^2\theta=0$$
$$(4\cos\theta-3\sin\theta)^2=0$$
$$4\cos\theta-3\sin\theta=0$$
Hope it helps!.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/757497",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "21",
"answer_count": 14,
"answer_id": 12
} |
Series $\sum_{n=0}^\infty (-1)^n \frac{x^{4n+1}}{4n+1}$ Does anyone know the sums of the following two series?
$$\sum_{n=0}^\infty (-1)^n \frac{x^{4n+1}}{4n+1}$$
$$\sum_{n=1}^\infty (-1)^{n+1} \frac{x^{4n-1}}{4n-1}$$
I encounter such series in my work.
| Well, to attack the first sum, if we call it $f(x)$, then
$$f'(x) = \sum_{n=0}^{\infty} (-1)^n x^{4 n} = \frac1{1+x^4}$$
so that
$$f(x) = \int_0^x \frac{dt}{1+t^4}$$
You can use partial fractions to deduce that
$$\begin{align}f(x) &= \frac1{2 \sqrt{2}}\int_0^x dt \, \left [\frac{t+\sqrt{2}}{t^2+\sqrt{2} t+1}-\frac{t-\sqrt{2}}{t^2-\sqrt{2} t+1} \right ] \\ &= \frac1{2 \sqrt{2}} \int_0^x dt \left [\frac{t+1/\sqrt{2}+1/\sqrt{2}}{(t+1/\sqrt{2})^2+1/2}-\frac{t-1/\sqrt{2}-1/\sqrt{2}}{(t-1/\sqrt{2})^2+1/2} \right ] \\ &= \frac1{2 \sqrt{2}} \left [\arctan{(\sqrt{2} x+1)} - \frac{\pi}{4} + \arctan{(\sqrt{2} x-1)} + \frac{\pi}{4} \right ]\\ &+ \frac1{4 \sqrt{2}} \left [\log{(x^2+\sqrt{2} x+1)} - \log{(x^2-\sqrt{2} x+1)} \right ]\\ &=\frac1{2 \sqrt{2}} \arctan{\left (\frac{\sqrt{2} x}{1-x^2}\right )}+\frac1{4 \sqrt{2}} \log{\left (\frac{x^2+\sqrt{2} x+1}{x^2-\sqrt{2} x+1} \right )}\end{align} $$
The second sum is obviously similar, except that you are evaluating
$$g(x) = \int_0^x dt \frac{t^2}{1+t^4} $$
and, using similar steps, i.e., partial fractions, you get
$$\begin{align}g(x) &= \frac1{2 \sqrt{2}}\int_0^x dt \, \left [\frac{t}{t^2-\sqrt{2} t+1}-\frac{t}{t^2+\sqrt{2} t+1} \right ] \\ &= \frac1{2 \sqrt{2}} \int_0^x dt \left [\frac{t-1/\sqrt{2}+1/\sqrt{2}}{(t-1/\sqrt{2})^2+1/2}-\frac{t+1/\sqrt{2}-1/\sqrt{2}}{(t+1/\sqrt{2})^2+1/2} \right ]\\ &= \frac1{2 \sqrt{2}} \arctan{\left (\frac{\sqrt{2} x}{1-x^2}\right )}-\frac1{4 \sqrt{2}} \log{\left (\frac{x^2+\sqrt{2} x+1}{x^2-\sqrt{2} x+1} \right )} \end{align} $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/758616",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
A different type binomial expansion problem Suppose we have $$(1+x+x^2)^n = a_0 + a_1 x + a_2 x^2 + \cdots + a_{2n} x^{2n}.$$
What will be the value of $a_0^2 - a_1^2 + a_2^2 - \cdots + a_{2n}^2$?
The answer is $a_n$, but I can't solve it.
See, what I've done is substitute $x$ as $-\frac{1}{x}$ and I've got:
${\frac{(x^2-x+1)}{x^2}}^n = a_0 - \frac{a_1}{x} + \frac{a_2}{x^2}+...$
I've got the alternating signs but I can't get the squares of the numbers.
| Let $(1+x+x^2)^n=\sum_ka_kx^k$. Then:
\begin{align}
(1+x^2+x^4)^n&=(1-x^{-1}+x^{-2})^n(1+x+x^2)^nx^{2n}\\
\sum_ja_jx^{2j}&=\sum_k(-1)^ka_kx^{-k}\sum_ja_jx^jx^{2n}\\
&=\sum_j\sum_k(-1)^ka_ka_jx^{2n+j-k}\\
&=\sum_j\sum_k(-1)^ka_ka_{k+j-2n}x^j\\
\end{align}
The $x^{2n}$ coefficient on the left side is $a_n$; the same coefficient on the right side is $\sum_k(-1)^ka_k^2$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/762762",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 0
} |
Any one help me solve Algebraically equation $\sqrt{x^2+x+1}+ x^{3}= \sqrt{2x+2}+x^{2}+x$ I can't solve this. Can any one show me detail solution
$$
\sqrt{x^2+x+1}+x^3=\sqrt{2x+2}+x^2+x
$$
| This can be solved easily by observing the aid equation once you rearrange the terms.
$$
\sqrt{x^2 + x + 1} - \sqrt{2x+2} = -x^3 + x^2 + x
$$
$$
\sqrt{x^2 + x + 1} - \sqrt{2x+2} = -x(x^2 -x -1)
$$
Notice that on the LHS, the difference between squares of the terms in under the root is RHS/$x$
Let's multiply and divide by the conjugate on the LHS
$$
\sqrt{x^2 + x + 1} - \sqrt{2x+2} \times \frac{(\sqrt{x^2 + x + 1} + \sqrt{2x+2})}{\sqrt{x^2 + x + 1} + \sqrt{2x+2}} = -x(x^2 -x -1)
$$
$$
\frac{x^2 -x -1}{\sqrt{x^2 + x + 1} + \sqrt{2x+2}}= -x(x^2 -x -1)
$$
Rearranging a few terms,
$$
\frac{x^2 -x -1}{\sqrt{x^2 + x + 1} + \sqrt{2x+2}} + x(x^2 -x -1) = 0
$$
$$
(x^2 -x -1) \times \left ( \frac{1}{\sqrt{x^2 + x + 1} + \sqrt{2x+2}} + x \right ) = 0
$$
Now, either of $ (x^2 -x -1) , \Bigl ( \frac{1}{\sqrt{x^2 + x + 1} + \sqrt{2x+2}} + x \Bigr)$ could be trivial.
We know that $x \geq -1$ (Constraint placed on $ \sqrt{2x + 2}$ ) making $ \frac{1}{\sqrt{x^2 + x + 1} + \sqrt{2x+2}} + x = 0$ at $ x = -1$
The other solutions lie in $ x^2 -x -1 = 0$, which can be solved as a standard quadratic polynomial.
These 3 values will be your solution.
Edit : There is a another solution inside the equation, which can be found by solving $ \Bigl ( \frac{1}{\sqrt{x^2 + x + 1} + \sqrt{2x+2}} + x \Bigr) = 0$ completely.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/764508",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Expression generating $\left( \frac{3}{10}, \, \frac{3}{10} + \frac{33}{100}, \, \frac{3}{10} + \frac{33}{100} + \frac{333}{1000}, \dots \right)$ I'm looking for a closed-form expression (in terms of $n$), that will give the sequence
$$
(s_n) = \left( \frac{3}{10}, \, \frac{3}{10} + \frac{33}{100}, \, \frac{3}{10} + \frac{33}{100} + \frac{333}{1000}, \dots \right).
$$
Can anyone think of one? I made a related post to this question several minutes ago but I realized I was interpreting the sequence wrong.
| Let
$$S=0.3+0.33+0.333+0.3333+\cdots \text{ to $n$ terms}$$
$$=\underbrace{\frac {3}{10}+\frac {33}{100}+\frac {333}{1000} + \frac {3333}{10000}+\cdots}_n$$
$$=3\left(\frac {1}{10}+\frac {11}{100}+\frac {111}{1000} + \frac {1111}{10000}+\cdots\right)$$
$$=3\left(\frac {1}{10^1}+\frac {11}{10^2}+\frac {111}{10^3} + \frac {1111}{10^4}+\cdots\right)$$
$$=3\left(\frac {\sum_{k=0}^010^k}{10^1}+\frac {\sum_{k=0}^110^k}{10^2}+\frac {\sum_{k=0}^210^k}{10^3} + \frac {\sum_{k=0}^310^k}{10^4}+\cdots+\frac {\sum_{k=0}^{(n-1)}10^k}{10^n}\right)$$
So $$\frac{1}{10^{n}}\left(\sum_{k=0}^{n-1}10^k\right)= \frac{1}{10^{n}}\left(\frac{10^{n}-1}{9}\right) = \frac{1}{9}\left(1 - \frac{1}{10^n}\right) = \frac{1}{9}\left(1 - 10^{-n}\right)$$
Therefore
\begin{align}
S &= 3\left(\sum_{k=1}^n\left[\frac{1}{9}\left(1 - 10^{-k}\right)\right]\right)\\
&= \frac{3}{9}\sum_{k=1}^n\left(1 - 10^{-k}\right)\\
&= \frac{1}{3} \left( \sum_{k=1}^n 1 - \sum_{k=1}^n10^{-k}\right)\\
&= \frac{1}{3} \left( n- \sum_{k=1}^n10^{-k}\right)\\
&= \frac{1}{3} \left( n- \frac{1}{9}\left(1 - 10^{-n}\right)\right)
\end{align}
$$S = \frac{1}{27}\left(10^{-n}+ 9n -1\right) $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/765660",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 3
} |
Find the axis of rotation of a rotation matrix by inspection (NOT by solving $Kv=v$) $$K=\
\begin{pmatrix}
0 & 0 & 1\\
-1 & 0 & 0\\
0 & -1 & 0
\end{pmatrix}$$
Find the axis of rotation for the rotation matrix $K$ by INSPECTION.
This is from my other thread click here to view it
Everything you see below is me finding the axis of rotation by solving $Kv=v$. Just to show you how much working it requires:
Noting that the axis of rotation consists of vectors that remain unmoved. That is a vector $v$ satisfying $Kv = v$. Or, $Kv - Iv=0$ where $I$ is the $3\times3$ identity matrix. For matrix $K$ after solving the homogeneous equations given by $(K-I)v=0$ and showing the working:
$(K-I)v=0$
So
$$K-I=\
\begin{pmatrix}
0 & 0 & 1\\
-1 & 0 & 0\\
0 & -1 & 0
\end{pmatrix}-\begin{pmatrix}
1 & 0 & 0\\
0 & 1 & 0\\
0 & 0 & 1
\end{pmatrix}=\
\begin{pmatrix}
-1 & 0 & 1\\
-1 & -1 & 0\\
0 & -1 & -1
\end{pmatrix}$$
therefore
$$\begin{pmatrix}
-1 & 0 & 1\\
-1 & -1 & 0\\
0 & -1 & -1
\end{pmatrix}v=0$$
writing out the components for $v$ gives
$$\begin{pmatrix}
-1 & 0 & 1\\
-1 & -1 & 0\\
0 & -1 & -1
\end{pmatrix}\begin{pmatrix}
x \\
y \\
z
\end{pmatrix}=0$$
Multiplying out gives three equations
$-x+z=0$
$-x-y=0$
$-y-z=0$
Since
$$
v=\begin{bmatrix}x\\y\\z\end{bmatrix}
$$
Here's the solution parametrically in terms of $x$
\begin{align*}
z&= x\\
y&=-x\\
x&=x
\end{align*}
Hence the axis of rotation is given by the line
$$
\begin{bmatrix}
x\\-x\\x
\end{bmatrix}=x\begin{bmatrix}1\\-1\\1\end{bmatrix}\quad x\in\Bbb R
$$
That is, the axis of rotation is
$$
\operatorname{Span}\left\{\begin{bmatrix}1\\-1\\1\end{bmatrix}\right\}
$$
As you can see this was a lot of work so i would be so grateful if someone could please explain in simple english how to get the answer: $$
\operatorname{Span}\left\{\begin{bmatrix}1\\-1\\1\end{bmatrix}\right\}
$$
by using Inspection?
Many thanks to all that helped so far particularly Brian Fitzpatrick in the last thread
| I can offer a remark on how you could simplify in some sense the calculation, but I'm not sure whether you would call the final result "inspection" and not "calculation".
Sometimes it easier to do mental calculations with matrices by thinking of the result of matrix multiplication by a vector as a linear combination of the matrix rows. That is,
$$ \begin{pmatrix}
a_{11} & a_{12} & a_{13} \\
a_{21} & a_{22} & a_{23} \\
a_{31} & a_{32} & a_{33}
\end{pmatrix} \cdot
\begin{pmatrix}
x \\
y \\
z
\end{pmatrix} = x \begin{pmatrix} a_{11} \\ a_{21} \\ a_{31} \end{pmatrix} + y \begin{pmatrix} a_{12} \\ a_{22} \\ a_{32} \end{pmatrix} + z \begin{pmatrix} a_{13} \\ a_{23} \\ a_{33} \end{pmatrix}. $$
The result of the multiplication is what you get if you multiply $x$ by the first row, add to it $y$ multiplied by the second row and then $z$ by the third row.
In your case, we have
$$ \begin{pmatrix}
0 & 0 & 1 \\
-1 & 0 & 0 \\
0 & -1 & 0
\end{pmatrix} \cdot
\begin{pmatrix}
x \\
y \\
z
\end{pmatrix} = x \begin{pmatrix} 0 \\ -1 \\ 0 \end{pmatrix} + y \begin{pmatrix} 0 \\ 0 \\ -1 \end{pmatrix} + z \begin{pmatrix} 1 \\ 0 \\ 0 \end{pmatrix} = \begin{pmatrix} z \\ -x \\ -y \end{pmatrix} = \begin{pmatrix} x \\ y \\ z \end{pmatrix}. $$
After some practice, you can do this in your head and then note that the first coordinate of the result is going to be $z$ and you want $z = x$. So you can take $z = x = 1$. Then, the second coordinate should be $-x = y$, so you can take $y = -1$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/766565",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 3
} |
Solve the initial value problem x'=Ax $A =
\begin{array}{cc}
-2 & 1 \\
5 & -4 \\
\end{array}$
$x(0) =
\begin{array}{cc}
1\\
3\\
\end{array}$
I undertsand that this is an eigenvector problem, and I got the values of $ -3+ \sqrt{6}$ and $ -3- \sqrt{6}$. I am unable to calculate the eigenvectors from here.
| Using the eigenvalue $\lambda = -3 - \sqrt{6}$, we find $[A-\lambda I]v_1 = 0$ and arrive at a RREF of:
$$\left(
\begin{array}{cc}
1 & \frac{1}{5} \left(-1+\sqrt{6}\right) \\
0 & 0 \\
\end{array}
\right)v_1 = 0$$
We choose $b = 1, a = \dfrac{1}{5} \left(1-\sqrt{6}\right)$, so:
$$v_1 = \left(\dfrac{1}{5} \left(1-\sqrt{6}\right), 1\right)$$
Since we have conjugate eigenvalues, we can write the eigenvector for the second eigenvalue as:
$$v_2 = \left(\frac{1}{5} \left(1+\sqrt{6}\right),1\right)$$
You can now write:
$$x(t) = c_1 ~e^{\lambda_1 t}~v_1 + c_2~ e^{\lambda_2 t}~v_2$$
Use the IC to find the constants.
Your final solution should be:
$x(t) = \dfrac{-4 e^{\left(-3-\sqrt{6}\right) t}+\sqrt{6} e^{\left(-3-\sqrt{6}\right) t}+4 e^{\left(-3+\sqrt{6}\right) t}+\sqrt{6} e^{\left(-3+\sqrt{6}\right) t}}{2 \sqrt{6}} =\dfrac{1}{3} e^{-3 t} \left(2 \sqrt{6} \sinh \left(\sqrt{6} t\right)+3 \cosh \left(\sqrt{6} t\right)\right) \\y(t) =\dfrac{-2 e^{\left(-3-\sqrt{6}\right) t}+3 \sqrt{6} e^{\left(-3-\sqrt{6}\right) t}+2 e^{\left(-3+\sqrt{6}\right) t}+3 \sqrt{6} e^{\left(-3+\sqrt{6}\right) t}}{2 \sqrt{6}} = \dfrac{1}{3} e^{-3 t} \left(\sqrt{6} \sinh \left(\sqrt{6} t\right)+9 \cosh \left(\sqrt{6} t\right)\right)$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
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} |
Calculation of $\int_{0}^{1}\frac{x-1+\sqrt{x^2+1}}{x+1+\sqrt{x^2+1}}dx$ Calculation of $\displaystyle \int_{0}^{1}\frac{x-1+\sqrt{x^2+1}}{x+1+\sqrt{x^2+1}}dx$
$\bf{My\; Try::}$ Let $x=\tan \psi\;,$ Then $\displaystyle dx = \sec^2 \psi$
So Integral convert into $\displaystyle \int_{0}^{\frac{\pi}{4}}\frac{\tan \psi-1+\sec \psi}{\tan \psi+1+\sec \psi}d\psi = \int_{0}^{\frac{\pi}{4}}\frac{\tan \psi+\sec \psi - 1}{\tan \psi+\sec \psi+1}d\psi$
Now Multiply both Numerator and Denominator by $\left(\tan \psi+\sec \psi-1\right)$
So $\displaystyle \int_{0}^{\frac{\pi}{4}}\frac{\tan \psi+\sec \psi-1}{\tan \psi+\sec \psi+1} \times \frac{\tan \psi+\sec \psi-1}{\tan \psi+\sec \psi-1}d\psi = \int_{0}^{\frac{\pi}{4}}\frac{\left(\tan \psi+\sec \psi-1\right)^2}{\left(\tan \psi+\sec \psi\right)^2-1}d\psi$
Now I did not understand how can i solve after that
Help required
Thanks.
| Setting $x=\tan2y$ $$F=\frac{\tan2y+\sec2y-1}{\tan2y+\sec2y+1}=\frac{\sin2y+1-\cos2y}{\sin2y+1+\cos2y}$$
Using Double Angle formula, $$F=\frac{2\sin y\cos y+2\sin^2y}{2\sin y\cos y+2\cos^2y}=\frac{2\sin y(\cos y+\sin y)}{2\cos y(\sin y+\cos y)}=\tan y$$ assuming $\cos y+\sin y\ne0\iff\tan y\ne-1\implies x=\tan2y\ne\infty$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 6,
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Definite Integral of $1+\sqrt{9-x^2}$? Right, so I'm to find the definite integral (interpreting it as an area)...
$\int^0_{-3}(1+\sqrt{9-x^2})dx$
How do I go about doing this?
I am to specifically use the following theorem to work it out...
If $f$ is integrable on $[a,b]$ then...
$\int^b_a f(x)\,dx = \lim_{n\rightarrow \infty}\sum^n_{i=1}f(x_i)\Delta x$
where $\Delta x = \frac{b-a}{n}$ and $x_i = a + i\Delta x$
I keep getting halfway through and getting stuck with $\lim_{n\rightarrow\infty}\frac{3\sqrt{3}}{n^3}\sum^n_{i=1}\sqrt{4n^2-3i}$
Is this correct? How do I work on from here, if so? The answer given in the book is $3 + \frac{9}{4}\pi$.
Any ideas?
| The expression you are having trouble with seems to be incorrect.
You have $\mathrm{f}(x) = 1+\sqrt{9-x^2}$. Since $a=-3$ and $b=0$, you have $\displaystyle{\Delta x = \frac{3}{n}}$ and $\displaystyle{x_k = -3+\frac{3k}{n}}$.
Applying your definition:
\begin{eqnarray*}
\mathrm{f}(x_k) &=& 1+\sqrt{9-\left(-3+\frac{3k}{n}\right)^2} \\ \\ \\
&=&1+\sqrt{18\frac{k}{n}-9\frac{k^2}{n^2}} \\ \\ \\
&=&1+\frac{3}{n}\sqrt{2kn-k^2}
\end{eqnarray*}
From this, we see that
\begin{eqnarray*}
\mathrm{f}(x_k)\Delta x &=& \frac{3}{n}\left(1+\frac{3}{n}\sqrt{2kn-k^2}\right) \\ \\ \\
&=& \frac{3}{n}+\frac{9}{n^2}\sqrt{2kn-k^2}
\end{eqnarray*}
Then the final expression should be
$$\lim_{n\to\infty}\left[\sum_{k=1}^n \left(\frac{3}{n}+\frac{9}{n^2}\sqrt{2kn-k^2}\right)\right]$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "1",
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"answer_id": 1
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Show that $\int_0^\pi \log^2\left(\tan\frac{ x}{4}\right)dx=\frac{\pi^3}{4}.$ Hi I am trying to prove the relation
$$
I:=\int_0^\pi \log^2\left(\tan\frac{ x}{4}\right)dx=\frac{\pi^3}{4}.
$$
I tried expanding the log argument by using $\sin x/ \cos x=\tan x,$ and than used $\log(a/b)=\log a-\log b$, I get
$$
I=\int_0^\pi \left( \log \sin \frac{x}{4}-\log\cos \frac{x}{4}\right)^2dx.
$$
We can distribute this out
$$
\int_0^\pi \log^2 \sin \frac{x}{4}dx +\int_0^\pi \log^2\cos \frac{x}{4}dx-2\int_0^\pi\log \sin \frac{x}{4}\log \cos \frac{x}{4}dx.
$$
Now I am stuck at how to solve these. Thanks.
| Letting $x\mapsto \frac{1}{x},$ we have
$$
I=\int_0^\pi \ln ^2\left(\tan \frac{x}{4}\right) d x=4 \int_0^{\frac{\pi}{4}} \ln ^2(\tan x) d x
$$
By my post,
$$\int_0^{\frac{\pi}{4}} \ln ^2(\tan x) d x= \frac{\pi^3}{16} ,$$
we can conclude that
$$
\boxed {I=\frac{\pi^3}{4}}
$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "26",
"answer_count": 4,
"answer_id": 3
} |
How to integrate $\frac{\sqrt{x}}{1-\sqrt{x}}$? How to integrate $\frac{\sqrt{x}}{1-\sqrt{x}}$?
I tried by using integration by parts, but always got sucked. Should be very easy...
| $$\int \frac{ \sqrt{x}}{1 - \sqrt{x}} = - \int \frac{\sqrt{x}}{\sqrt{x}-1} = - \int \frac{ \sqrt{x}-1+1}{\sqrt{x}-1}= - \int dx - \int \frac{dx}{\sqrt{x}-1}$$
But, $ d( \sqrt{x} - 1) = \frac{1}{2 \sqrt{x} } dx $. Hence
$$\int \frac{dx}{\sqrt{x}-1} = 2 \int \frac{\sqrt{x} d( \sqrt{x}-1)}{\sqrt{x}-1}=2 \int \frac{(\sqrt{x}-1+1) d( \sqrt{x}-1)}{\sqrt{x}-1} = 2 \int d( \sqrt{x}-1) + 2 \int \frac{ d(\sqrt{x}-1)}{\sqrt{x}-1} = 2 (\sqrt{x}-1) + 2 \ln(\sqrt{x}-1)$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 3
} |
Integration by parts with log function $$\int2x^3\ln(1-x^2)\;dx$$
How do I integrate this equation?
Thanks.
| Use substitution rule. Let:\begin{align}u&=1-x^2\\x^2&=u+1\\du&=2x \, dx\end{align}
Then \begin{align}\int 2x^3 \ln(1-x^2) \, dx &= \int x^2\ln(1-x^2) \,2x \, dx \\
&=\int(u+1)\ln (u) \, du\end{align}
Now use integration by parts:
\begin{align}a=u+1 &\implies da=du\\
db=\ln(u) \, du &\implies b = u \ln(u)-u \end{align}
So \begin{align} \int 2x^3 \ln(1-x^2) \, dx &= \int(u+1) \ln (u) \, du \\ &= \int a \, db \\
&=ab-\int b \, da \\
&= (u+1) (u \ln(u)-u) - \int (u \ln(u)-u) \, du \\
&=(u+1)(u \ln(u)-u)- [\frac{1}{4}u^2(2 \log(u)-3)] + C \\
&=(2-x^2)[(1-x^2)\ln(1-x^2)-(1-x^2)]\\&\, \, \, \, \, \, \, \,-[\frac{1}{4}(1-x^2)^2(2\log(1-x^2)-3)]+C\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/773718",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 0
} |
Inequality problem about sides of a triangle and the semiperimeter Let $a,b,c$ the sides of a triangle and $s$ be the semi perimeter. Then show that
$$
a^2+b^2+c^2 > \frac{36}{35}(a^2+\frac{abc}{s})
$$
I tried it doing in many ways using some changes but cannot help my cause.
| The inequality seems really loose to me, and this approach shows that it is indeed loose. ( which makes me somewhat doubt the inequality is what we want to show)
We want to show that
$$ 35 b^2 +35 c^2 > a^2 + 36 abc/s .$$
We have $b+c > a$ and so $ 2s > 2a $ and so $abc/s < bc $. We will show that
$$ 35 b^2 +35 c^2 > a^2 + 36 bc .$$
This is true because $a < b+c $ so $a^2 < 2 b^2 + 2c^2$, which gives us
$$ 18(b-c)^2 +17 b^2 +17 c^2 > a^2$$
In fact, with the above, we can show that
$$
a^2+b^2+c^2 > \frac{6}{5}(a^2+\frac{abc}{s})
$$
the 'equality condition' occurs when $a=2b=2c$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/777876",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Find Polynomial coefficients How can I find the real coefficients a,b if the polynomial
$ P(X)=X^4-5X^3+8X^2+aX+b $ is divisible by $ Q(X)=(X-1)^2 $
So if 1 is a solution, I get a+b=-4; where should i look for other roots?
| Using synthetic division:
\begin{array}{rrrrrrr}
& | & 1 & -5 & 8 & a & b \\
1 & | & 0 & 1 & -4 & 4 & a+4 \\
& + & --- & --- & --- & --- & ----- \\
& & 1 & -4 & 4 & a+4 & a+b+4 \\
\end{array}
So $a+b+4 = 0$.
Dividing again by $(x-1)$ :
\begin{array}{rcrrrr}
& | & 1 & -4 & 4 & a+4 \\
1 & | & 0 & 1 & -3 & 1 \\
& + & --- & --- & --- & --- \\
& & 1 & -3 & 1 & a+5 \\
\end{array}
So $a+5= 0$, hence $a = -5$ and $b=1$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/778466",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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How to evaluate $\frac{1}{2\pi }\int_{-\pi }^{\pi }\dfrac{\sin n\theta }{\sin\theta }d\theta $? How to evaluate the integral given below?
$$\dfrac{1}{2\pi }\int_{-\pi }^{\pi }\dfrac{\sin n\theta }{\sin\theta }d\theta $$
| Since the integrand is $ 2 \pi$-periodic,
$$ \begin{align} \frac{1}{2\pi } \int_{-\pi }^{\pi } \frac{\sin n\theta }{\sin\theta } \ d\theta &= \frac{1}{2 \pi} \int_{-\pi/2}^{3 \pi/2} \frac{\sin n \theta}{\sin \theta} \ d \theta \\ &=\frac{1}{2 \pi} \int_{-\pi/2}^{3 \pi/2} \text{Im} \frac{e^{in \theta}}{\sin \theta} \ d \theta \\ &= \frac{1}{2\pi} \text{Im} \ \text{PV} \int_{-\pi/2}^{3 \pi /2} \frac{e^{in \theta}}{\sin \theta \ } \ d \theta \tag{1} \\ &= \frac{1}{\pi} \text{Im} \ i \ \text{PV} \int_{-\pi/2}^{3 \pi /2} \frac{e^{in \theta}}{e^{i \theta}-e^{-i \theta}} \ d \theta . \end{align} $$
Now let $z=e^{i \theta}$.
Then
$$ \begin{align} \frac{1}{2\pi } \int_{-\pi }^{\pi } \frac{\sin n\theta }{\sin\theta } \ d\theta &= \frac{1}{\pi} \text{Im}\ i \ \text{PV} \int_{|z|=1} \frac{z^{n}}{z-z^{-1}}\frac{dz}{iz} \\ &= \frac{1}{\pi} \text{Im} \ \text{PV} \int_{|z|=1} \frac{z^{n}}{z^{2}-1} \ dz \end{align}$$
where the unit circle is indented around the simple poles at $z=1$ and $z=-1$.
Therefore,
$$ \frac{1}{2\pi } \int_{-\pi }^{\pi } \frac{\sin n\theta }{\sin\theta } \ d\theta= \frac{1}{\pi } \text{Im} \left(i \pi \ \text{Res} \left[\frac{z^{n}}{z^{2}-1},1 \right] + i \pi \ \text{Res} \left[\frac{z^{n}}{z^{2}-1},-1 \right]\right) .$$
If $n=2k+1$ (that is, if $n$ is odd),
$$ \frac{1}{2\pi }\int_{-\pi }^{\pi }\frac{\sin (2k+1)\theta }{\sin\theta } \ d\theta = \frac{1}{\pi } \text{Im} \left[i \pi \left(\frac{1}{2} \right) +i \pi \left(\frac{1}{2} \right) \right] = 1.$$
And if $n=2k$,
$$ \frac{1}{2\pi }\int_{-\pi }^{\pi } \frac{\sin 2k \theta }{\sin\theta } \ d\theta = \frac{1}{\pi } \text{Im} \left[i \pi \left(\frac{1}{2} \right) +i \pi \left(-\frac{1}{2} \right) \right] = 0 .$$
$ $
$(1)$ On the interval $[- \frac{\pi}{2}, \frac{3 \pi}{2} ]$, $\frac{e^{in \theta}}{\sin \theta}$ has a simple pole at $0$ and a simple pole at $\pi$.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 5,
"answer_id": 1
} |
$\int_{0}^{\infty}\frac{x}{x^3+1}dx$ =? So guys I have this improper integral $\int_{0}^{\infty}\frac{x}{x^3+1}dx$. I checked that it converges by $ \int_{0}^{1}\frac{x}{x^3+1}dx + \int_{1}^{\infty}\frac{x}{x^3+1}dx $ and using the $\frac{c}{(x-a)^\lambda} $ and $\frac{1}{x^\lambda} $ criteria. But for finding the value I have trouble.I tried $ \int_{0}^{\infty}\frac{x}{(x+1)(x^2-x+1)}dx$ and $\int_{0}^{\frac{\pi}{2}}\frac{tg\alpha}{tg\alpha^3+1}dx$ but it gets me nowhere. Any ideas on solving it?
| One thing that simplifies the computation is splitting the integral and substituting $y = x^{-1}$ in one part:
$$\begin{align}
\int_0^\infty \frac{x}{x^3+1}\,dx &= \int_0^1\frac{x}{x^3+1}\,dx + \int_1^\infty \frac{x}{x^3+1}\,dx\\
&= \int_0^1 \frac{x}{x^3+1}\,dx + \int_1^0 \frac{y^{-1}}{y^{-3}+1}\,d(y^{-1})\\
&= \int_0^1 \frac{x}{x^3+1}\,dx + \int_0^1 \frac{1}{y^3+1}\,dy\\
&= \int_0^1 \frac{dx}{x^2 -x+1}.
\end{align}$$
This can be seen to be related to an $\arctan$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/782531",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Pie Integral $\int_0^1 \log\frac{(x+\sqrt{1-x^2})^2}{(x-\sqrt{1-x^2})^2} \frac{x\, dx}{1-x^2}=\frac{\pi^2}{2}.$ Hi I am trying to show this astonishing result$$
\int_0^1 \log\frac{\big(x+\sqrt{1-x^2}\big)^2}{\big(x-\sqrt{1-x^2}\big)^2} \frac{x\, dx}{1-x^2}=\frac{\pi^2}{2}.
$$
Note we can to use $\ln(a/b)=\ln a-\ln b$ but that didn't help me much . After this I obtained integrals of the form
$$
\int_0^1 \log \big[\big(x\pm\sqrt{1-x^2}\big)^2\big] \frac{x\, dx}{1-x^2}
$$
which I am not sure how to handle. Thanks.
| Let
$$$$
\begin{align}
I(a)&=\int_0^1 2\, \log\frac{\big(x+a\, \sqrt{1-x^2}\big)}{\big(x-a\, \sqrt{1-x^2}\big)} \frac{x\, dx}{1-x^2} \tag 1\\
\therefore \frac{\partial}{\partial a}I(a) &= 2\, \int_0^1 \left(\frac{\sqrt{1-x^2}}{\big(x+a\, \sqrt{1-x^2}\big)}+\frac{\sqrt{1-x^2}}{\big(x-a\, \sqrt{1-x^2}\big)}\right) \frac{x}{1-x^2}\, dx\\
&= \int_0^{\pi/2} \frac{4\, \tan{(t)}^2}{\tan{(t)}^2-a^2}\, dt \hspace{100pt}\text{(subst. $x=\sin{t}$)}\\
&= \int_0^{\pi/2} \frac{4\, \tan{(t)}^2}{(\tan{(t)}^2-a^2)(1+\tan{(t)}^2)} \sec{(t)}^2 \, dt\\
&= \int_0^{\infty} \frac{4\, y^2}{(y^2-a^2)(1+y^2)} \, dy \hspace{82pt}\text{(subst. $\tan{t}=y$)}\\
&= \int_0^{\infty} \frac{1}{1+a^2}\left(\frac{2a}{y-a}-\frac{2a}{y+a}+\frac{4}{1+y^2}\right)\, dy\\
&= \frac{1}{1+a^2}\left(a\, \log\left(\frac{y-a}{y+a}\right)^2 +4\arctan{y}\right)\Bigg|_0^\infty\\
&= \frac{2\, \pi}{a^2+1}\\
\implies I(a)&=2\,\pi\arctan{a}+C \tag 2
\end{align}
Putting $a=0$ in $(1)$ and $(2)$, we find that $C=0$
Hence, $$I(a)=2\,\pi\arctan{a}$$
and the required integral $$I(1)=\frac{\pi^2}{2}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/782964",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "14",
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Limit when x approaches a number $\ne \infty$ I was rushing through a analysis task book when I suddenly encountered
$$\text{Find }~\lim_{x \to 3}\frac{x^2-9}{x-3}~\text{ by using the definition of limits.}$$
I thought, as the term equals to x+3, the limit should be 6.
Here's my beginning: I have to show, that $$\forall \epsilon >0~\exists ~X\in \mathbb N :\forall x \gt X : |a_x-a|<\epsilon$$
So $|x+3-6|=|x-3|<\epsilon$ is true because as x is approaching 3, |x-3| is approaching 0.
I just don't know how to wrap it up formally.
Can I say: $|x-3|=|3-x|\overset{as ~x~ is~ positive}{=}3-x<\epsilon,$
so choose $X=3-\epsilon$ ?
| Since you want to prove the limit, $L=6$, you need to show $\forall\epsilon>0,\ \exists\delta>0$ such that $|x-3|<\delta\implies \left|\dfrac{x^2-9}{x-3}-6\right|<\epsilon$. It would be a good idea to start with the right side of the $\implies$ and work backwards, to show you can actually find such a $\delta$ that makes this statement true. After you find it, then write out the proof. Now, to find a suitable $\delta$:
Since $\left|\dfrac{x^2-9}{x-3}-6\right|=\left|\dfrac{x^2-9-6(x-3)}{x-3}\right|=\left|\dfrac{x^2-6x+18}{x-3}\right|=\left|\dfrac{(x-3)^2}{x-3}\right|=|x-3|<\epsilon.$ Since $|x-3|<\delta$ by assumption, then just let $\delta=\epsilon.$
Now to start the proof:
Given $\epsilon>0$, let $\delta=\epsilon$. Then $$|x-3|<\epsilon\implies\left|\dfrac{(x-3)^2}{x-3}\right|=\left|\dfrac{x^2-6x+9}{x-3}\right|=\left|\dfrac{x^2-6x-9+18}{x-3}\right|=\left|\dfrac{x^2-9-6(x-3)}{x-3}\right|=\left|\dfrac{x^2-9}{x-3}-6\frac{(x-3)}{x-3}\right|=\left|\dfrac{x^2-9}{x-3}-6\right|<\epsilon$$ which proves that $\lim\limits_{x\to 3}\frac{x^2-9}{x-3}=6$ and we are done.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/783059",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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If $y =\sqrt{5+\sqrt{5-\sqrt{5+ \cdots}}}$, what is the value of $y^2-y$? If $$y =\sqrt{5+\sqrt{5-\sqrt{5 + \cdots}}},$$
what is the value of $y^2-y$ ?
I am unable to get the clue due to these alternative signs of plus and minus please help on this thanks...
| Hint:
$$\begin{align}
(y^2 -5)^2 - 5 &= -y \\
y^4 - 10y^2 + 25 - 5 + y &= 0 \\
y^4 -10y^2 + y + 20 &= 0.
\end{align}
$$
Edit: To elaborate a bit on the comments given below: you have produced a depressed quartic equation. As explained on the Wikipedia article you can try to factor this into quadratics
$$
y^4 -10y^2 + y + 20 = (y^2 +py + q)(y^2 + ry + s).
$$
All you have to do is find $p,q,r$, ands $s$. You get
$$\begin{align}
y^4 -10y^2 + y + 20 &= (y^2 +py + q)(y^2 + ry + s)\\
&= y^4 + (p + r)y^3 + (q + s + pr)y^2 + (ps + qr)y + qs.
\end{align}
$$
Now put coefficients equal to each other and solve ...
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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Find the indefinite integral of...? I have spent well over an hour trying to solve this equation:
$$\int\cos^4{x}\sin^3{x}\, dx$$
I have tried substituting u as $\cos{x}$, $\sin{x}$, $\cos^4{x}$, and $sin^3{x}$ to no avail. How can I solve this?
| Through "brute force", you can just write out the values...there are several identities you need to get there:
$$
\sin(a)\sin(b) = \frac{\cos(a - b)-\cos(a + b)}{2} = \frac{\cos(b - a)-\cos(a + b)}{2} \\
\cos(a)\cos(b) = \frac{\cos(a - b)+\cos(a + b)}{2} = \frac{\cos(b - a) + \cos(a + b)}{2} \\
\sin(a)\cos(b) = \frac{\sin(a + b) + \sin(a - b)}{2}
$$
You can verify all of those properties from the angle addition properties:
\begin{align}
\cos(a + b) =&& \cos(a)\cos(b) - \sin(a)\sin(b) \\
\cos(a - b) =& \cos(a)\cos(-b) - \sin(a)\sin(-b)= &\cos(a)\cos(b) + \sin(a)\sin(b)\\
\sin(a +b) =&& \sin(a)\cos(b) + \sin(b)\cos(a) \\
\sin(a - b) =& \sin(a)\cos(-b) + \sin(-b)\cos(a) =&\sin(a)\cos(b) - \sin(b)\cos(a)
\end{align}
This gives:
\begin{align}
\cos^4(x)sin^3(x) = &\cos(x)\left(\sin(x)\cos(x)\right)^3 \\
=& \cos(x)\frac{\sin^3(2x)}{2^3} = \cos(x)\sin(2x)\sin^2(2x) \\
=& \frac{1}{8}\cos(x)\sin(2x)\frac{1 - \cos(4x)}{2} \\
\cos(x)\sin(2x) =& \frac{\sin(3x) + \sin(x)}{2} \\
\cos^4(x)\sin^3(x) = & \left.\left.\frac{1}{32}\right(\sin(3x) - \sin(3x)\cos(4x) + \sin(x) - \sin(x)\cos(4x)\right) \\
=& \left.\left.\frac{1}{32}\right(\sin(3x) - \sin(x)\right) - \left.\left.\frac{1}{64} \right(\sin(7x) + \sin(x) + \sin(5x) + \sin(3x)\right) \\
=& \left.\left.\frac{1}{64}\right(-3\sin(x) + \sin(3x) - \sin(5x) - \sin(7x)\right)
\end{align}
This then gives the integral easily as:
$$
\int\cos^4(x)\sin^3(x)dx = \left.\left.\frac{1}{64}\right(3\cos(x) - \frac{1}{3}\cos(3x) + \frac{1}{5}\cos(5x) + \frac{1}{7}\cos(7x)\right) + C
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/784245",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Points on $(x^2 + y^2)^2 = 2x^2 - 2y^2$ with slope of $1$ Let the curve in the plane defined by the equation: $(x^2 + y^2)^2 = 2x^2 - 2y^2$
How can i graph the curve in the plane and determine the points of the curve where $\frac{dy}{dx} = 1$.
My work:
First i found the roots of this equation with a change of variable $z = y^2$ and get:
and then i tried to graph the point $ x - y$ and $x + y $ but i stuck i can't graph this and find the point where the derivative is 1. Some help please.
| You've already got some good suggestions on the graphing part using polar coordinates (if you are familiar with that). For the other, implicitly differentiate the given relation to obtain ${dy\over dx}$:
\begin{align*}
(x^2 + y^2)^2 &= 2x^2 - 2y^2\\
2(x^2+y^2)(2x+2y{dy\over dx})&=4x-4y{dy\over dx}\\
2x^3+2x^2y{dy\over dx}+2xy^2+2y^3{dy\over dx}&=2x-2y{dy\over dx}\\
(2x^2y+2y^3+2y){dy\over dx}&=2x-2x^3-2xy^2\\
{dy\over dx}&={2x-2x^3-2xy^2\over 2x^2y+2y^3+2y}\\
{dy\over dx}&={x(1-x^2-y^2)\over y(x^2+y^2+1)}\\
\end{align*}
Now solve ${dy\over dx}=1$ in terms of $x$ and $y$, keeping in mind you can use the original expression to simplify things along the way. Edit: This was much more involved than I anticipated; see below.
Here's the situation (note ${dy\over dx}$ is not defined at the origin):
So how to find these coordinates? As noted in another answer, the original relation can be expressed in polar coordinates as $r^2=2\cos(2t)$. Recall that the slope of the tangent line in polar coordinates is given by
$$
{dy\over dx}={{dr\over dt}\sin t+r\cos t\over {dr\over dt}\cos t-r\sin t},
$$
so setting ${dy\over dx}=1$ and using $r^2=2\cos(2t)\implies {dr\over dt}={-2\sin(2t)\over r}$, we get
\begin{align*}
{dy\over dx}&=1\\
{dr\over dt}\sin t+r\cos t&={dr\over dt}\cos t-r\sin t\\
{-2\sin(2t)\over r}\sin t+r\cos t&={-2\sin(2t)\over r}\cos t-r\sin t\\
-2\sin(2t)\sin t+r^2\cos t&=-2\sin(2t)\cos t-r^2\sin t\\
r^2&={2\sin(2t)[\sin t-\cos t]\over \cos t+\sin t}\\
2\cos(2t)&={2\sin(2t)[\sin t-\cos t]\over \cos t+\sin t}\\
\end{align*}
Multiply on the right by ${\cos t-\sin t\over \cos t-\sin t}$, use $\cos^2t-\sin^2t=\cos(2t)$ and $\sin(2t)=2\sin t\cos t$ and rearrange to get
\begin{align*}
\sin(2t)&=-(\sin t+\cos t)^2\\
2\sin t\cos t&=-1-2\sin t\cos t\\
4\sin t\cos t&=-1\\
\sin(2t)&=-{1\over 2}\\
t&={7\pi\over 12},\ {11\pi\over 12}.
\end{align*}
But $r^2=2\cos(2t)$ so, $t={11\pi/12}\implies r^2=2\cos(11\pi/6)=\sqrt{3}$. Thus, the polar coordinates of this point are $(3^{1/4},11\pi/12)$ which in rectangular coordinates is $$(3^{1/4}\cos(11\pi/12),3^{1/4}\sin(11\pi/12))\approx (-1.27,0.34)$$ as shown in the figure above.
(The fourth quadrant point of tangency can be found similarly.)
| {
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"source": "stackexchange",
"question_score": "4",
"answer_count": 5,
"answer_id": 1
} |
Inverse Laplace Transform of $ \left(\frac{1-s^{1/2}}{s^2}\right)^2$ I found this question in my N.P Bali's Engineering Mathematics 7th Edition.
I could not find any solved questions related to this.
How can I find the Inverse Laplace Transform of :
$\left({1-s^{1/2} \over s^{2}}\right)^2$
I know only to find the inverse Laplace of $s$ when the power is a whole number.
The answer to the question is given as : $\dfrac{t^3}{6} + \dfrac{t^2}{2} - \dfrac{16t^\frac{5}{2}}{15 \pi^{\frac{1}{2}}}$
I want to know how to get it.
| The partial fraction expansion yields;
$$\left(\dfrac{1-s^{1/2}}{s^{2}}\right)^2 = -\dfrac{2}{s^{7/2}}+\dfrac{1}{s^4}+\dfrac{1}{s^3} $$
Using this table of Laplace Transforms (item $6$ and item $3$ (twice)) yields:
$$\mathscr{L}^{-1} \left(-\dfrac{2}{s^{7/2}}+\dfrac{1}{s^4}+\dfrac{1}{s^3}\right) = -\dfrac{2 \times 2^3}{1 \times 3 \times 5 ~ \sqrt{\pi}}t^{5/2} +\dfrac{t^3}{6} + \dfrac{t^2}{2} = -\dfrac{16}{15~ \sqrt{\pi}}t^{5/2} +\dfrac{t^3}{6} + \dfrac{t^2}{2}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/785981",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Summation of Fibonacci numbers $F_n$ with $n$ odd vs. even Compare the summation below:
$$\begin{align}
\smash[b]{\sum_{i=1}^n F_{2i-1}}&=F_1+F_3+F_5+\cdots+F_{2n-1}\\
&=1+2+5+\cdots+F_{2n-1}\\
&=F_{2n}\\
\end{align}
$$
with this one:
$$\begin{align}
\smash[b]{\sum_{i=1}^n F_{2i}}&=F_2+F_4+F_6+\cdots+F_{2n}\\
&=1+3+8+\cdots+F_{2n}\\
&=F_{2n+1}-1\\
\end{align}$$
When I first discovered these patterns I was amazed. Naively I had thought that an every-other-number sum of Fibonacci numbers would be the same pattern whether the parity of their indices was odd or even, but I was wrong! Why is the above true, where the summation of odd-indexed Fibonacci numbers is another Fibonacci number, but the even-indexed sum is a Fibonacci number minus 1?
| Here is a late answer using my favorite Fibonacci technique.
The matrix formulation for the Fibonacci sequence is well worth knowing and easily proved:
$$
\begin{pmatrix}1&1\\1&0\end{pmatrix}^n=
\begin{pmatrix}F_{n+1}&F_n\\F_n&F_{n-1}\end{pmatrix}
$$
Let
$
A=\begin{pmatrix}1&1\\1&0\end{pmatrix}
$.
Since the second column of $A^{2k}$ is $\begin{pmatrix}F_{2k}\\F_{2k-1}\end{pmatrix}$,
the second column of
$$
S = A^2 + A^4 + \cdots + A^{2n}
$$
contains exactly the sums we're interested in.
We have $(A^2-I)S=A^{2n+2}-A^2$.
Since $A^2=A+I$ (the Fibonacci recurrence!), we get $AS=A^{2n+2}-A^2$ and so
$$
S=A^{2n+1}-A
=\begin{pmatrix}F_{2n+2}&F_{2n+1}\\F_{2n+1}&F_{2n}\end{pmatrix}
-\begin{pmatrix}1&1\\1&0\end{pmatrix}
=\begin{pmatrix}*&F_{2n+1}-1\\*&F_{2n}\end{pmatrix}
$$
as claimed. This also explains where the $-1$ comes from.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/787341",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 5,
"answer_id": 3
} |
Calculating integral I need some help to solve this integral:
$$\int(x^2+y^2)^{-\frac32} \mathrm dx$$
Thank you.
| Put $x = y \tan t$ and $dx = y \sec^2 t\ dt$, then
$$
\begin{align}
\int\frac{dx}{\sqrt{(x^2+y^2)^3}}&=\int\frac{y \sec^2 t\ dt}{\sqrt{(y^2\tan^2 t+y^2)^3}}\\
&=\int\frac{y \sec^2 t\ dt}{y^3\sec^3 t}\\
&=\frac{1}{y^2}\int\cos t\ dt\\
&=\frac{\sin t}{y^2}+C\\
&=\frac{x}{y^2\sqrt{x^2+y^2}}+C.
\end{align}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/787879",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Prove that $n$ is divisible by $6$ Problem:
Let $x^2+mx+n$ and $x^2+mx-n$ give integer roots where $(m,n)$ are integers. Show that $n$ is divisible by $6$
My attempt:
Since the roots are integers then the discriminants of both the equations should be perfect squares.
Let $a=\sqrt{m^2-4n}$ and $b=\sqrt{m^2+4n}$, then $(ab)^2=m^4-16n^2$ where $(a,b,m,n)$ are all integers. I am stuck here...
| We have $a^2 + b^2 = 2 m^2$, whose solution in the integers is
$$a = x^2+2xy-y^2, b = y^2 + 2xy - x^2, m = x^2 + y^2, \text { for } x, y \in \mathbb{N}$$
Hence, $ 4n = m^2 - a^2 = 4 xy (x-y) ( x+y)$
It is clear that $ xy (x-y)(x+y)$ must be a multiple of 6.
Proof of classification: (I'm slightly surprised I can't find a derivation on this site, but I'm bad at searching)
If $a$ is even, then clearly $b$ is even, and thus so is $m$, and we can then divide out by 2. Hence, we may assume that $a$ and $b$ are odd, so $ a = 2p-1, b = 2q-1$.
Observe that $m^2 = ( p-q) ^2 + (p+q-1)^2$, hence by the classification of pythagorean triples, we have
$$ m = x^2 + y^2, (p-q) = x^2 - y^2, (p+q-1) = 2xy$$
In other words,
$$a = x^2+2xy-y^2, b = y^2 + 2xy - x^2, m = x^2 + y^2, \text { for } x, y \in \mathbb{N}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/788993",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 1,
"answer_id": 0
} |
Solve $\sin(2\theta) -\tan(\theta) = 0 \ $ for $ 0\leq \theta \leq 2\pi $ I want to use the fact that $$\sin(2\theta) = \frac{2\tan(\theta)}{1 + \tan^2(\theta)}$$
to solve $\sin(2\theta) -tan(\theta) = 0 \ $ for $ 0\leq \theta \leq 2\pi$
My solution:
$\frac{2tan(\theta)}{1 + tan^2(\theta)} - tan(\theta) = 0 $
so $\frac{2tan(\theta) - (1 + tan^2(\theta)) tan(\theta)}{1 + \tan^2(\theta)} = 0$
so $ 2tan(\theta) - (1 + tan^2(\theta)) tan(\theta)= 0$
$ \implies \tan^3(\theta) + \tan(\theta) = 0$ $\implies \tan(\theta) [\tan^2(\theta) + 1] = 0$ $\implies \tan(\theta) = 0 \;\textrm{or}\; \tan^2(\theta) + 1 =0 $ since $\theta$ must be real.
Then we solve $\tan(\theta) = 0$ $\implies$ $\theta = n\pi, \ \ $ $ n \in Z \ \ $ so $\theta = \pi,2\pi$
| Another way to solve:
Since identity $\sin (2\theta)=\dfrac{2\tan\theta}{1+\tan^2\theta}$, equation $\sin (2\theta)-\tan\theta=0$ is equivalent to
\begin{align}
\left(\dfrac{2}{1+\tan^2\theta}-1\right)\tan\theta&=0\;\;\;\text{ or}\\
\left(\dfrac{1-\tan^2\theta}{1+\tan^2\theta}\right)\tan\theta&=0\;\;\;\text{ or}\\
\frac{(1+\tan\theta)(1-\tan\theta)\tan\theta}{1+\tan^2\theta}&=0
\end{align}
It follows $\tan\theta=\pm 1$ or $\tan\theta=0$. Thus $\theta=0,\frac{\pi}{4},\frac{3\pi}{4},\pi,\frac{5\pi}{4},\frac{7\pi}{4},2\pi$ are the solutions in $[0,2\pi]$.
| {
"language": "en",
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"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 1
} |
Proving some trig identities. Stuck on two questions.
First one: $\tan^2x - \sin^2x = (\sin^2 x)(\tan^2 x)$
Tried solving that with the right side but wasn't able to.
Second one: $\csc x / \sec x = \cot x$
I tried solving this one with the left side doing,
L.S = $\csc x / \sec x $
L.S = $(1 / \sin x) / (1 / \cos x) $
L.S = $\sin x / \cos x\quad$ [(1's cancel out]
L.S = $\tan x$
Don't know what I'm doing wrong here.
For reference, here is the trig identity worksheet we're supposed to use.
http://imgur.com/TXxaYUQ
| *
*$$
\begin{align}
\tan^2x - \sin^2x&=\frac{\sin^2x}{\cos^2x}-\sin^2x\quad;\ \text{where}\ \tan x=\frac{\sin x}{\cos x}\\
&=\sin^2x\left(\frac{1}{\cos^2x}-1\right)\\
&=\sin^2x\left(\frac{1}{\cos^2x}-\frac{\cos^2x}{\cos^2x}\right)\\
&=\frac{\sin^2x}{\cos^2x}\left(1-\cos^2x\right)\\
&=\color{blue}{\tan^2x \cdot \sin^2x}\quad;\ \text{where}\ \sin^2x + \cos^2x=1.
\end{align}
$$
$$$$
*$$
\begin{align}
\frac{\csc x}{\sec x}&=\dfrac{\dfrac{1}{\sin x}}{\dfrac{1}{\cos x}}\\
&=\frac{1}{\color{red}{\sin x}}\times\frac{\color{red}{\cos x}}{1}\\
&=\frac{\cos x}{\sin x}\\
&=\color{blue}{\cot x}.
\end{align}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/790517",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 0
} |
Solving a fractional quadratic equation problem by completing the square I have the following problem to solve using the method of completing the square.
$$2x^2-3x-1 = 0$$
Here is where I've gotten to so far on this problem.
$$2x^2-3x = 1$$
$$x^2-\frac{3}{2}x = \frac{1}{2}$$
$$x^2-\frac{3}{2}x +\frac{9}{16} = \frac{1}{2}+\frac{9}{16}$$
$$x^2-\frac{3}{2}x +\frac{9}{16} = \frac{17}{16}$$
$$(x -\frac{3}{4})^2 = \frac{17}{16}$$
The book that I'm using provides the answer, but not how the answer was arrived at. I can't figure out how to get from the last step that I completed to here:
$$x=\frac{3±\sqrt{17}}{4}$$
Thanks in advance for any assistance that you can provide.
| You made a mistake in the last line. It should read
$(x-\frac{3}{4})^2=\frac{17}{16}$...
This gives $(x-\frac{3}{4})^2 - \frac{17}{16} = 0$. Using the third binomial formula, we obtain $(x-\frac{3}{4} + \sqrt \frac{17}{16})(x-\frac{3}{4} - \sqrt \frac{17}{16}) = 0$.
Remembering that a product is zero iff one of its coefficients is zero, you have the answer.
Edit:
Consider $x=\frac{3}{4} + \sqrt \frac{17}{16}$ which is a solution to your equation.
Remember that for any two numbers $a,b$ $\sqrt{a b}=\sqrt{a} \sqrt{b}$. The same is true for quotients (change $b$ to $\frac{1}{c}$).
Then we get $x=\frac{3}{4} + \frac{\sqrt{17}}{\sqrt{16}} = \frac{3}{4} + \frac{\sqrt{17}}{4} = \frac{3 + \sqrt{17}}{4}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/790892",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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How prove this $\frac{r}{R}\ge\frac{AM}{AX}$ Let acute $\triangle ABC$ have largest angle $\angle A$, and let $R,r$ be the circumradius and inradius respectively. Let $M$ be the midpoint of $BC$, and $X$ be the intersection of the tangents at $B,C$ to the circumcircle of $\triangle ABC$
Show that:
$$\frac{r}{R}\ge\frac{|AM|}{|AX|}$$
Here is the original chinese version:
My method:
So now I have the following problem:
let $x,y,z>0$,and let
$$R=\dfrac{\sqrt{(x^2+y^2+z^2+2xz)(x^2+y^2+z^2-2xz)}}{2y},a=\sqrt{x^2+y^2}$$
$$r=\dfrac{2yz}{2z+\sqrt{x^2+y^2+z^2-2xz}+\sqrt{x^2+y^2+z^2+2xz}}$$
$$b=\sqrt{\dfrac{(x^2+y^2)(x^2+y^2+z^2+2xz)(x^2+y^2+z^2-2xz)}{(x^2+y^2-z^2)^2}}$$
Show that
$$\dfrac{r}{R}\ge\dfrac{a}{b}$$
I think this problem has other methods. If you find other methods, can you explain?
Thank you very much
| Let us identify the plane with the set of complex numbers $\Bbb{C}$, with $\Gamma$ being the unit circle. (So that $R=1$). The complex number representing a point will be denoted by the same, but small, corresponding letter.
Now, $X$ is the image of $M$ under the inversion with respect to $\Gamma$, hence
$$
m=\frac{b+c}{2},\qquad x=\frac{1}{\overline{m}}.
$$
Since $|a|=|b|=|c|=1$, we conclude that
$$
AX=|a-x|=|\overline{a}-\overline{x}|=\left\vert\frac{1}{a}-\frac{1}{m}\right\vert=\frac{|a-m|}{|m|}=\frac{AM}{OM}
$$
where $O$ is the center of $\Gamma$. Thus
$$
\frac{AM}{AX}=OM=\cos A \tag{1}
$$
On the other hand, it is well-known that
$$
\frac{r}{R}=4\sin \frac{A}{2}\sin \frac{B}{2} \sin \frac{C}{2} \tag{2}
$$
Thus
$$\eqalign{
\frac{r}{R}-\frac{AM}{AX}&=4\sin \frac{A}{2}\sin \frac{B}{2} \sin \frac{C}{2} -1+2\sin^2 \frac{A}{2}\cr
&=2\sin \frac{A}{2}\left(2\sin \frac{B}{2} \sin \frac{C}{2}+\sin \frac{A}{2}\right) -1\cr
&=2\sin \frac{A}{2}\left(2\sin \frac{B}{2} \sin \frac{C}{2}+\cos \frac{B+C}{2}\right) -1\cr
&=2\cos \frac{B+C}{2} \cos \frac{B-C}{2} -1\cr
&=\cos B+\cos C -1~\buildrel{\rm(1)}\over{\geq}~ 2\cos\left(\frac{B+C}{2}\right)-1\cr
&\geq2\sin\frac{A}{2}-1~\buildrel{\rm(2)}\over{\geq}~ 2\sin 30^\circ-1=0
}
$$
where $(1)$ follows from the concavity of $x\mapsto \cos x$ on $[0,\pi/2]$, and $(2)$ follows from the fact that $A$ is the largest angle in the acute triangle $\triangle ABC$.$\qquad\square$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/791263",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Length of Curve $6xy=3+x^4$ Question :
Find the length of the curve $6xy=3+x^4$ between $x=1$ and $x=2$.
Answer = 17/12
I have tried this:
I obtain a different answer. Where did I do wrongly?
Thank you for your help.
| Your integration is not correct. You should simplify the integrand before proceeding. We have $$\mathscr{L}=\int_{1}^{2}\sqrt{1+\left(\frac{x^{4}-1}{2x^2}\right)^{2}}dx=\int_{1}^{2}\sqrt{\frac{(x^{4}+1)^2}{4x^4}}dx\\=\int_{1}^{2}\frac{x^{4}+1}{2x^2}dx=\frac{1}{2}\int_{1}^{2} x^2\,dx+\frac{1}{2}\int_{1}^{2} \frac{1}{x^2}dx\\=\frac{1}{6}x^3\biggr|_{1}^{2}-\frac{1}{2}\frac{1}{x}\biggr|_{1}^{2}=\frac{17}{12}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/791329",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Find limit of this recursive sequence $$
a_0=0,\ a_1=2,\ a_{n+1}=\sqrt{2 - \frac{a_{n-1}}{a_n}} \\
\lim_{n\to\infty}2^na_n\ =\ ?
$$
| Let's prove by induction that
$$
a_n=2\sin\left(\frac{\pi}{2^n}\right)
$$
Let $\theta_n=\pi/2^n$. Clearly, $2\sin(\pi)=0=a_0$ and $2\sin(\frac{\pi}{2})=2=a_1$. Furthermore,
$$
a_{n+1}=\sqrt{2-\frac{a_{n-1}}{a_n}}=2\sqrt{\frac{1-\frac{a_{n-1}}{2a_n}}{2}}
$$
But
$$
\frac{a_{n-1}}{2a_n}=\frac{\sin \theta_{n-1}}{2\sin\theta_n}=
\frac{\sin 2\theta_n}{2\sin\theta_n}=
\frac{2\sin \theta_n\cos\theta_n}{2\sin\theta_n}=\cos\theta_n
$$
and
$$
\cos2\alpha=1-2\sin^2\alpha
$$
$$
\sin\alpha=\sqrt{\frac{1-\cos 2\alpha}{2}}
$$
Hence,
$$
a_{n+1}=2\sqrt{\frac{1-\cos\theta_n}{2}}=
2\sin\frac{\theta_n}{2}=2\sin\theta_{n+1}
$$
Using Taylor series expansion,
$$
a_n=2\theta_n+O\left(\theta_n^3\right)=
\frac{\pi}{2^{n-1}}+O\left(8^{-n}\right)
$$
so
$$
2^n a_n=2\pi+O\left(4^{-n}\right)\to 2\pi
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/795306",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "14",
"answer_count": 2,
"answer_id": 0
} |
Integral $\int_0^1 \log \frac{1+ax}{1-ax}\frac{dx}{x\sqrt{1-x^2}}=\pi\arcsin a$ Hi I am trying to solve this integral $$
I:=\int_0^1 \log\left(\frac{1+ax}{1-ax}\right)\,\frac{{\rm d}x}{x\sqrt{1-x^2}}=\pi\arcsin\left(a\right),\qquad
\left\vert a\right\vert \leq 1.
$$
It gives beautiful result for $a = 1$
$$
\int_0^1 \log\left(\frac{1+ x}{1-x}\right)\,\frac{{\rm d}x}{x\sqrt{1-x^2}}
=\frac{\pi^2}{2}.
$$
I tried to write
$$
I=\int_0^1 \frac{\log(1+ax)}{x\sqrt{1-x^2}}dx-\int_0^1 \frac{\log(1-ax)}{x\sqrt{1-x^2}}dx
$$
If we work with one of these integrals we can write
$$
\sum_{n=1}^\infty \frac{(-1)^{n+1} a^n}{n}\int_0^1 \frac{x^{n-1}}{\sqrt{1-x^2}}dx-\sum_{n=1}^\infty \frac{a^n}{n}\int_0^1 \frac{x^{n-1}}{\sqrt{1-x^2}}dx,
$$
simplifying this I get an infinite sum of Gamma functions. which i'm not sure how to relate to the $\arcsin$ Thanks.
| View $I$ as a function of $a$, differentiate under integral sign and let
$x = \sin\theta$, we have
$$\begin{align}
I'(a) &= \int_0^1 \left( \frac{x}{1+ax} - \frac{-x}{1-ax}\right) \frac{dx}{x\sqrt{1-x^2}}
= \int_{-1}^1 \frac{dx}{(1+ax)\sqrt{1-x^2}}\\
&= \int_{-\pi/2}^{\pi/2} \frac{d\theta}{1+a\sin\theta}
= \frac12 \int_0^{2\pi}\frac{d\theta}{1+a\sin\theta}
= \frac12 \int_0^{2\pi}\frac{d\theta}{1+a\cos\theta}
\end{align}
$$
Introduce $z = e^{i\theta}$ and convert above integral to a contour integral over the unit
circle in $z$, we get
$$I'(a) = \frac{1}{2i}\oint_{|z|=1} \frac{dz}{z+\frac{a}{2}(z^2+1)}
= \frac{1}{ai}\oint_{|z|=1} \frac{dz}{(z - \lambda_{+})(z - \lambda_{-})}
$$
where $\displaystyle\;\lambda_{\pm} = -\frac{1}{a} \pm \sqrt{\frac{1}{a^2}-1}.\;$
When $|a| \le 1$, only the root $\lambda_{+}$ lies inside the unit circle, we have
$$I'(a) = \frac{1}{ai}\frac{2\pi i}{\lambda_{+} - \lambda_{-}}
= \frac{2\pi}{2a\sqrt{\frac{1}{a^2}-1}} = \frac{\pi}{\sqrt{1-a^2}}
$$
Since $I(0) = 0$, we get
$$I(a) = \pi \int_0^a \frac{dt}{\sqrt{1-t^2}} = \pi \arcsin(a)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/795493",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "17",
"answer_count": 5,
"answer_id": 2
} |
Integral comparison test $$ \int_a^b \frac{x+1}{\sqrt[4]{x^5+x^2}} \,dx $$
I want to know this integral converges or not when $(a, b) = (0, 1), (1, \infty)$. I was thinking of using the comparison test, but I can't think of anything to compare it to.
Thanks in advance.
| For $(a,b)=(0,1)$ notice that
$$\frac{x+1}{\sqrt[4]{x^{5}+x^{2}}}\le\frac{x+1}{\sqrt[4]{x^{2}}}=\frac{x+1}{\sqrt{x}}=\sqrt{x}+\frac{1}{\sqrt{x}}$$
and for $(a,b)=(1,\infty)$ notice that
$$\frac{x+1}{\sqrt[4]{x^{5}+x^{2}}}\ge\frac{x}{\sqrt[4]{x^{5}+x^{2}}}\ge\frac{x}{\sqrt[4]{2x^{5}}}=\left(\frac{1}{2}\right)^{\frac{1}{4}}\frac{x}{x^{\frac{5}{4}}}=\left(\frac{1}{2}\right)^{\frac{1}{4}}x^{1-\frac{5}{4}}=\left(\frac{1}{2}\right)^{\frac{1}{4}}x^{-\frac{1}{4}}.$$
Notice that for $x$ near $0$ we have $$\frac{x+1}{\sqrt[4]{x^{5}+x^{2}}}\approx\frac{1}{\sqrt[4]{x^{2}}}=x^{-\frac{2}{4}}=x^{-\frac{1}{2}}$$
and that for large $x$ we have $$\frac{x+1}{\sqrt[4]{x^{5}+x^{2}}}\approx\frac{x}{x^{\frac{5}{4}}}=x^{-\frac{1}{4}}.$$ So these estimates are not unexpected.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/799851",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Simple Integral $\int_0^\infty (1-x\cot^{-1} x)dx=\frac{\pi}{4}$. Hellow I am trying to prove this result.
$$
I:=\int_0^\infty (1-x\cot^{-1} x)dx=\frac{\pi}{4}.
$$
The indefinite integral exists for this integral.
The function $\cot^{-1} x$ is the arc-cotangent function, not the multiplicative inverse. Note, we can not just break the integral up into two pieces because we will have problems with divergence.
Using the relation
$$
x\cot^{-1} x=x \tan^{-1} \frac{1}{x},\to \quad I=\int_0^\infty \left(1-x\tan^{-1} \frac{1}{x}\right)dx
$$
may be of help to some but didn't help me. I am not sure if I am missing a clever substitution, perhaps integration by parts will work, I obtained using this method
$$
I=(x-x^2\cot^{-1} x)\big|^\infty_0-\int_0^\infty \frac{x^2}{x^2+1}dx+\int_0^\infty x\cot^{-1} x \, dx
$$
but this is clearly a problem since we have divergence issue now.
The indefinite integral is given by
$$
\int (1-x\cot^{-1} x ) dx =\frac{1}{2}\left(x+\tan^{-1} x-x^2 \cot^{-1} x\right)
$$
but I am unable to prove this result. Thanks.
| Let $y=\cot^{-1}x$, then $x=\cot y\;\Rightarrow\;dx=d(\cot y)$. The integral turns out to be
$$
\int(1-y\cot y)\ d(\cot y)=\int\ d(\cot y)-\int y\cot y\ d(\cot y).$$
The second integral in the RHS can be solved using IBP by taking $u=y$ and $dv=\cot y\ d(\cot y)$, we obtain $du=dy$ and $v=\dfrac12\cot^2 y$. Hence
$$
\int y\cot y\ d(\cot y)=\frac12y\cot^2 y-\frac12\int\cot^2 y\ dy,
$$
where
$$
\int\cot^2 y\ dy=\int(\csc^2 y-1)\ dy=-\cot y-y+C.
$$
Therefore
$$
\begin{align}
\int(1-y\cot y)\ d(\cot y)&=\cot y-\frac12y\cot^2 y-\frac12\cot y-\frac12y+C\\
&=\frac12(\cot y-y\cot^2 y-y)+C\\
\int (1-x\cot^{-1} x)\ dx&=\frac12\left(x-x^2\cot^{-1}x-\cot^{-1}x\right)+C.
\end{align}
$$
Actually, the result of integral is as the same result as yours Jeff. To make sure, using the identity $\tan^{-1}x+\cot^{-1}x=\dfrac\pi2$ for $x>0$ yields
$$
\begin{align}
\int (1-x\cot^{-1} x)\ dx&=\frac12\left(x-x^2\cot^{-1}x\right)-\frac12\cot^{-1}x+C\\
&=\frac12\left(x-x^2\cot^{-1}x\right)-\frac12\cot^{-1}x+\frac\pi4+K\\
&=\frac12\left(x-x^2\cot^{-1}x\right)+\frac12\left(\frac\pi2-\cot^{-1}x\right)+K\\
&=\frac12\left(x-x^2\cot^{-1}x\right)+\frac12\tan^{-1}x+K\\
&=\frac12\left(x+\tan^{-1}x-x^2\cot^{-1}x\right)+K.\qquad\blacksquare
\end{align}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/800989",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 5,
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} |
The $n$'th derivative of $x^x$ I want to know the $n$'th derivative of $f(x)=x^x$.
Then, I'll calculate $f(0)$ with Taylor expansion of $f(x)$ on $a=1$.
Here is my answer, but it is unfinished.
The derivative of $f(x)=x^x$
$$\begin{align}
f'(x)&=x^x(\log x+1)\\
f''(x)&=x^x(\log x+1)^2+x^{x-1}\\
f'''(x)&=x^x(\log x+1)^3+3x^{x-1}(\log x+1)-x^{x-2}\\[5pt]
f(x)^{(4)}&=x^x(\log x+1)^4+4x^{x-1}(\log x+1)^2-4x^{x-2}(\log x+1)+3x^{x-2}+2x^{x-3}\\
f(x)^{(5)}&=x^x(\log x+1)^5+10x^{x-1}(\log x+1)^3-10x^{x-2}(\log x+1)^2+15x^{x-2}(\log x+1)\\&\quad+10x^{x-3}(\log x+1)-10x^{x-3}-6x^{x-4}\\
f(x)^{(6)}&=x^x(\log x+1)^6+15x^x(\log x+1)^4-20x^{x-2}(\log x+1)^3+45x^{x-2}(\log x+1)^2\\&\quad+30x^{x-3}(\log x+1)^2-50x^{x-3}(\log x+1)+15x^{x-3}-46x^{x-4}(\log x+1)\\&\quad+40x^{x-4}+24x^{x-5}
\end{align}$$
Taylor expansion of $f(x)=x^x$ in $a=1$
$$\begin{align}
f(x)&=\sum_{i=0}^{n-1}\frac{f^{(i)}(1)}{i!}\\[5pt]
&\qquad=\frac1{0!}+\frac1{1!}(x-1)+\frac2{2!}(x-1)^2+\frac3{3!}(x-1)^3+\frac8{4!}(x-1)^4+\frac{12}{5!}(x-1)^5\\&\qquad+\frac{54}{6!}(x-1)^6+\cdots
\end{align}$$
| I tried the simplest approach to test myself. Let $u=x^{x-1}=e^{(x-1)\ln x}$, then $u=1+\sum_{n=1}^{\infty}\frac{1}{n!}(x-1)^n(\sum_{s=1}^{\infty}\frac{(-1)^{s-1}}{s}(x-1)^s)^n$ and expanding the binomial we have
$$u=1+\sum_{n=1}^{\infty}\sum_{k=1}^{\infty}\frac{1}{n!}C_{n,k}(x-1)^{n+k}$$
where
$$C_{n,k}=\sum_{r_1+r_2+...+r_n=k}^{r_i\geq 1,\forall i}\frac{(-1)^{k-n}}{r_1r_2...r_n}.$$
Letting $n+k=m$, we have $u=1+\sum_{m=2}^\infty a_m(x-1)^m$ where
$$a_m=\sum_{n=1}^{\lfloor m/2\rfloor}\frac{C_{n,m-n}}{n!}.$$
Then we have
$$y=x^x=1+(x-1)+(x-1)^2+\sum_{m=3}^{\infty}(a_m+a_{m-1})(x-1)^m.$$
$a_2=C_{1,1}=1$, $a_3=C_{1,2}=-\frac12$, $a_4=C_{1,3}+\frac{C_{2,2}}{2}=\frac 13+\frac12=\frac56$, $a_5=C_{1,4}+\frac{C_{2,3}}{2}=-\frac 14+-\frac12=-\frac34$, $a_6=C_{1,5}+\frac{C_{2,4}}{2}+\frac{C_{3,3}}{6}=\frac15+\frac{11}{24}+\frac16=\frac{33}{40},.....$ and hence
$$y=1+(x-1)+(x-1)^2+\frac{1}{2}(x-1)^3+\frac{1}{3}(x-1)^4+\frac{1}{12}(x-1)^5+\frac{3}{40}(x-1)^6+...$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/802256",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 6,
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} |
How to integrate $\int_{0}^{1} \int_{0}^{\pi} \int_{0}^{\pi} r^2 \sin\theta \sqrt{1 - r^2\cos^2\theta - r^2\sin^2\theta} \,d\phi\, d\theta \,dr$
Find the center of mass of the hemispherical region $W$ defined by the inequalities $x^2 + y^2 + z^2 \leq 1$ and $z \geq 0$ with unity density.
By symmetry we know the $x$ and $y$ coordinates will be $0$ so we just need the $z$ coordinate. The volume of a hemisphere is just $\frac{2}{3} \pi$ from the formula for the volume of a sphere, so we just need to compute $\iiint_{W}^{} z \, dz \, dy \, dz$.
This formula along with using the same limits for the integration of the volume of a hemisphere, we have: $\int_{0}^{1} \int_{0}^{\pi} \int_{0}^{\pi} r^2 \sin\theta z \, d\phi \, d\theta \, dr$. $z = \sqrt{1 - x^2 - y^2} = \sqrt{1 - r^2\cos^2\theta - r^2\sin^2\theta}$, so we substitute that in and get: $\int_{0}^{1} \int_{0}^{\pi} \int_{0}^{\pi} r^2 \sin\theta \sqrt{1 - r^2\cos^2\theta - r^2\sin^2\theta} \, d\phi \, d\theta \, dr$.
Is there an easy way to integrate this?
| The right way is to use cylindrical coordinates:
$$
Z=\frac{\int_0^1z (\pi(1-z^2))dz}{\int_0^1 (\pi(1-z^2))dz}=\frac{3}{2}\int_0^1(z-z^3)dz=\frac{3}{8}.
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/803303",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
How prove $a_{n}=[\sqrt{2}n]+[\sqrt{5}n]$ Contains infinitely even numbers.
let sequence $$a_{n}=[\sqrt{2}n]+[\sqrt{5}n]$$
where $[x]$ is the largest integer not greater than $x$
show that $\{a_{n}\}$ Contains infinitely even numbers.
also I guess contains infinitely odd numbers.
before I have ask this How prove this sequence $S_{n}=[2^n\cdot \sqrt{2}],n\in N$ contains infinitely many composite numbers
I found :
$$a_{2}=[2\sqrt{2}]+[2\sqrt{5}]=2+4=6$$
$$a_{3}=[3\sqrt{2}]+[3\sqrt{5}]=4+6=10$$
$$a_{5}=[5\sqrt{2}]+[5\sqrt{5}]=7+11=18$$
$$a_{7}=[7\sqrt{2}]+[7\sqrt{5}]=9+15=24$$
$$a_{8}=[8\sqrt{2}]+[8\sqrt{5}]=11+17=28$$
$$a_{9}=[9\sqrt{2}]+[9\sqrt{5}]=12+20=32$$
$$a_{10}=[10\sqrt{2}]+[10\sqrt{5}]=14+22=36$$
$$a_{12}=[12\sqrt{2}]+[12\sqrt{5}]=16+26=42$$
$$a_{14}=[14\sqrt{2}]+[14\sqrt{5}]=19+31=50$$
$$a_{15}=[15\sqrt{2}]+[15\sqrt{5}]=21+33=54$$
and so on
this problem is my found it. It seem this is interesting problem,and How prove it?
Thank you
| When you increase $n$ by one, $\lfloor \sqrt 2 n \rfloor$ increases by either $1$ (with frequency $2-\sqrt 2$ ) or $2$ (with frequency $\sqrt 2 - 1$), while $\lfloor\sqrt 5n \rfloor$ does by either $2$ (with frequency $3-\sqrt 5$) or $3$ (with frequency $\sqrt 5 - 2$).
If your sequence only contains finitely many odd (or even) terms, then it has to increase by $4$ at every step past a certain point.
This means that whenever $\lfloor\sqrt2 n\rfloor$ increases by $2$ (resp. $1$), $[\sqrt 5 n\rfloor$ increases by $2$ (resp. $3$).
However, the frequency at which those increments occur are all different. Therefore this is impossible, and so you must obtain infinitely many even / odd terms .
You can make this proof more precise and show that there can't be more than $4$ consecutive terms of the same parity.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/805248",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
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} |
Evaluate $\int_0^1\frac{x\ln x}{(1+x^2)^2}\ dx$ $$\int_0^1\frac{x\ln x}{(1+x^2)^2}\ dx
$$
Help me please. I don't know any ways of solution. Thank you.
| Rewrite:
$$
\int\frac{x\ln x}{(1+x^2)^2}\ dx=\frac14\int\frac{2x}{1+x^2}\cdot\ln x\cdot\frac{2\ dx}{1+x^2}.
$$
Now, consider Weierstrass substitution: $x=\tan\dfrac t2$, $\sin t=\dfrac{2x}{1+x^2}$, $\cos t=\dfrac{1-x^2}{1+x^2}$, and $dt=\dfrac{2\ dx}{1+x^2}$. The integral turns out to be
$$
\frac14\int\sin t\cdot \ln \tan\frac t2\ dt.
$$
Use IBP by taking $u=\ln \tan\dfrac t2\;\Rightarrow\;du=\dfrac12\dfrac{\sec^2\frac t2}{\tan\frac t2}\ dt=\dfrac{dt}{\sin t}$ and $dv=\sin t\ dt\;\Rightarrow\;v=-\cos t$. We obtain
\begin{align}
\frac14\int\sin t\cdot \ln \tan\frac t2\ dt&=\frac14\left[-\cos t\cdot\ln \tan\dfrac t2+\int\dfrac{\cos t}{\sin t}\ dt\right]\\
&=\frac14\left[-\cos t\cdot\ln \tan\dfrac t2+\int\dfrac{1}{\sin t}\ d(\sin t)\right]\\
&=\frac14\left[-\cos t\cdot\ln \tan\dfrac t2+\ln\sin t\right]+C\\
&=\frac14\left[-\dfrac{1-x^2}{1+x^2}\cdot\ln x+\ln\dfrac{2x}{1+x^2}\right]+C\\
&=\frac14\left[\dfrac{x^2\ln x-\ln x}{1+x^2}+\ln2+\ln x-\ln\left(1+x^2\right)\right]+C\\
&=\frac14\left[\dfrac{x^2\ln x-\ln x+\left(1+x^2\right)\ln x}{1+x^2}+\ln2-\ln\left(1+x^2\right)\right]+C\\
&=\frac14\left[\dfrac{2x^2\ln x}{1+x^2}-\ln\left(1+x^2\right)\right]+\frac{\ln2}{4}+C\\
\int\frac{x\ln x}{(1+x^2)^2}\ dx&=\frac14\left[\dfrac{2x^2\ln x}{1+x^2}-\ln\left(1+x^2\right)\right]+K.
\end{align}
Thus
$$
\int_0^1\frac{x\ln x}{(1+x^2)^2}\ dx=\frac14\left[\dfrac{2x^2\ln x}{1+x^2}-\ln\left(1+x^2\right)\right]_{x=0}^1=\large\color{blue}{-\frac{\ln2}{4}}.
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/805617",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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"answer_id": 0
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What is a simple way of computing the following fraction? Compute the value of the expression:
$$\frac{(10^4+324)(22^4+324)(34^4+324)(46^4+324)(58^4+324)}{(4^4+324)(16^4+324)(28^4+324)(40^4+324)(52^4+324)}$$
| Note that
$$x^4+324=(x^2-6x+18)(x^2+6x+18).$$
Hence
$$
\begin{align*}
\frac{(n+6)^4+324}{n^4+324}
&= \frac{((n+6)^2-6(n+6)+18)((n+6)^2+6(n+6)+18)}{(n^2-6n+18)(n^2+6n+18)} \\
&= \frac{(n^2+6n+18)((n+6)^2+6(n+6)+18)}{(n^2-6n+18)(n^2+6n+18)} \\
&=\dfrac{(n+6)^2+6(n+6)+18}{n^2-6n+18}
\end{align*}$$
which means that most of the terms cancel and we are left with
$$\begin{align*}
&\frac{(10^4+324)(22^4+324)(34^4+324)(46^4+324)(58^4+324)}{(4^4+324)(16^4+324)(28^4+324)(40^4+324)(52^4+324)} \\
&\qquad = \frac{(52+6)^2+6(52+6)+18}{4^2-6(4)+18} = \frac{3730}{10}=373
\end{align*}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/805782",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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$\int_{0}^{\infty}\int_{1}^{\infty}\frac{x^2-y^2}{(x^2+y^2)^2}dxdy$ diverges? I want someone to review my proof that $$\int_{1}^{\infty}\int_{0}^{\infty}\frac{x^2-y^2}{(x^2+y^2)^2}dxdy$$ does not converge.
To make things easier, I said let's look at the entire first quadrant and then subtract the integral over the small rectangle that we added.
move to polar coordinates:
$$\int_{0}^{\frac{\pi}{2}}\int_{o}^{\infty}r^2\frac{(\cos^2\theta-\sin^2\theta)}{r^3}drd\theta=\int_{0}^{\frac{\pi}{2}}\int_{o}^{\infty}\frac{\cos^2\theta-\sin^2\theta}{r}drd\theta=\int_{0}^{\frac{\pi}{2}}\int_{o}^{\infty}\frac{1-2\sin^2\theta}{r}drd\theta\leq\int_{0}^{\frac{\pi}{2}}\int_{o}^{\infty}\frac{1}{r}drd\theta$$
The inequality part is true since $-2\sin^2 x$ is always negative.
So:
$$\int_{0}^{\frac{\pi}{2}}\int_{o}^{\infty}\frac{1}{r}drd\theta=\frac{\pi}{2}\ln(r)|_0^{\infty}=\frac{\pi}{2}\ln(\frac{\infty}{0})=\frac{\pi}{2}\ln({\infty})=\infty$$
So over the entire first quadrant it diverges.
Is there a point to checking the small rectangle that we added? I mean, even if it diverges, the answer would still diverge.
| As an iterated integral, this integral converges. Indeed
$$ \int_{0}^{\infty} \frac{x^{2} - y^{2}}{(x^{2} + y^{2})^{2}} \, dx = \left[ - \frac{x}{x^{2} + y^{2}} \right]_{0}^{\infty} = 0. $$
So the whole integral also converges to 0. But it does not converge absolutely. Indeed,
$$ \int_{0}^{\infty} \left|\frac{x^{2} - y^{2}}{(x^{2} + y^{2})^{2}}\right| \, dx = \int_{0}^{y} \frac{y^{2} - x^{2}}{(x^{2} + y^{2})^{2}} \, dx + \int_{y}^{\infty} \frac{x^{2} - y^{2}}{(x^{2} + y^{2})^{2}} \, dx = \frac{1}{y} $$
and therefore
$$ \int_{1}^{\infty} \int_{0}^{\infty} \left|\frac{x^{2} - y^{2}}{(x^{2} + y^{2})^{2}}\right| \, dxdy = \infty. $$
Much frustrating fact is that if you switch the order of integration, you get
$$ \int_{1}^{\infty} \frac{x^{2} - y^{2}}{(x^{2} + y^{2})^{2}} \, dy = \left[ \frac{y}{x^{2} + y^{2}} \right]_{1}^{\infty} = - \frac{1}{x^{2}+1} $$
and thus the whole inegral becomes
$$ \int_{0}^{\infty}\int_{1}^{\infty} \frac{x^{2} - y^{2}}{(x^{2} + y^{2})^{2}} \, dydx = -\frac{\pi}{2}. $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/808085",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 2,
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Solve the following equation in positive integers $x$ and $y$ What are the solutions in positive integers of the equation:
$${1+2^x+2^{2x+1}=y^2}$$
I tried to factorize the equation but it didn't help much.
Clearly $y $ is an odd integer. Substituting $y =2n+1$, we get
$2^x+2^{2x+1}=(2n)\cdot{(2n+2)}$
$\Rightarrow (2^{x-2})\cdot(1+2^{x+1})=(n)\cdot(n+1)$
Which is the product of 2 consecutive integers. Does it help? I don't know.
| $1+2^x+2^{2x+1}=y^2$ or $1+2^{(x-1)+1}+2^{2(x-1)+3}=y^2$ or $1+2*2^{(x-1)}+8*2^{2(x-1)}=y^2$. Now, let $p=2^{(x-1)}$, then, $1+2p+p^2+7p^2=y^2$ or $(p+1)^2=y^2-7p^2$ , this a Rational Pell equation, now, it can be shown that any such Pell equation has solutions; $(r^2-d)^2=(r^2+d)^2-d(2r)^2$. Hence, by comparison, $d=7$, $p(=2^{(x-1)})=2r$, $y=r^2+7$ and $p+1=r^2-7$, from these equations we get; $8=r^2-2r=r(r-2)$. Hence, $r=4=2^{(x-2)}$. $x=4$, $y=23$. This is the only non-trivial integer solution.
| {
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"timestamp": "2023-03-29T00:00:00",
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Application of Definite Integral I been solving this problem for an hour, but I cant get the correct answer. Please help me with this.
| If you split the desired violet area into $2$ equals parts by drawing the line $x=\frac32$, you can evaluate it. In fact you only need to solve the system below
\begin{cases}
\int_0 ^\frac32 \sqrt{4-x^2} = A+S\\
\int_0 ^2 \sqrt{4-x^2} = A+2S\\
\end{cases}
letting $S=$ half of the violet area, and $A =$ the sector of the first circle obtained by removing the entire violet area. By solving the system you get
\begin{cases}
\left[\frac x2\sqrt{4-x^2}+2\arcsin\left(\frac x2 \right)\right]_0 ^\frac32 = A+S\\
\left[\frac x2\sqrt{4-x^2}+2\arcsin\left(\frac x2 \right)\right]_0 ^2 = A+2S\\
\end{cases}
\begin{cases}
\frac 34\sqrt{4-\frac{9}{16}}+2\arcsin\left(\frac 34 \right)= A+S\\
2\arcsin\left(1\right)= A+2S\\
\end{cases}
\begin{cases}
\frac{3\sqrt{55}}{16}+2\arcsin\left(\frac 34 \right)= A+S\\
\pi = A+2S\\
\end{cases}
$$\Rightarrow S=A+2S-(A+S)=\pi-\left(\frac{3\sqrt{55}}{16}+2\arcsin\left(\frac 34 \right)\right)\approx 0.0549\Rightarrow \\\text{Violet Coloured Area}\approx 2\cdot 0.0549 \approx 0.109$$
| {
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Show that $f\in \mathcal{R}[0,1]$ Let $f$ be defined by
$$f(x) =\left\{\frac{1}{n}, \frac{1}{n+1} \lt x \le \frac{1}{n} \right\}$$ and $f(x) = 0$ when $x=0$ and $n \in N$. Show that $f$ is integrable and $$\int_{0}^{1}f(x)dx=\frac{\pi^2}{6}-1.$$
| Let $\epsilon > 0$ be given, and choose $n$ such that $n > \dfrac{1}{\epsilon}$, consider the partition $P = \{0, \frac{1}{n}, \frac{1}{n-1},..., \frac{1}{2}, 1\}$, then:
On $[0,\frac{1}{n}]$, $M - m = \dfrac{1}{n} - 0 = \dfrac{1}{n}$,
On $(\frac{1}{n}, \frac{1}{n-1}]$, $M - m = \dfrac{1}{n-1} - \dfrac{1}{n-1} = 0$,
On $(\frac{1}{n-1}, \frac{1}{n-2}]$, $M - m = \dfrac{1}{n-2} - \dfrac{1}{n-2} = 0$, and
...
On $(\frac{1}{2}, 1]$, $M - m = 1 - 1 = 0$. Thus:
$U(f,P) - L(f,P) = \dfrac{1}{n^2} < \dfrac{1}{n} < \epsilon$, proving $f$ is integrable.
A Riemann sum for $f$ using $M_i$ for each subinterval is:
$\dfrac{1}{n}\cdot \dfrac{1}{n} + \dfrac{1}{n-1}\cdot \left(\dfrac{1}{n-1} - \dfrac{1}{n}\right) + \dfrac{1}{n-2}\cdot \left(\dfrac{1}{n-2} - \dfrac{1}{n-1}\right) + ... + 1\cdot \left(1 - \dfrac{1}{2}\right) = \displaystyle \sum_{k=1}^n \dfrac{1}{k^2} - 1 + \dfrac{1}{n} \to \dfrac{\pi^2}{6} - 1$, as $n \to \infty$.
| {
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Evaluating the integral with trigonometric integrand While solving another problem I have come across this integral which I am unable to evaluate. Can someone please evaluate the following integral? Thank you.
$$\int_0^{2\pi}\frac{1}{(2+\cos\theta)^2}\,d\theta.$$
| Here's the contour integration approach.
$$\int_{0}^{2 \pi} \frac{1}{(2+ \cos \theta)^{2}} \ d \theta = \int_{0}^{2 \pi} \frac{1}{(2+\frac{e^{i \theta} + e^{- i \theta}}{2})^{2}} \ d \theta$$
Let $z=e^{i \theta}$.
Then
$$ \begin{align}\int_{0}^{2 \pi} \frac{1}{(2+ \cos \theta)^{2}} \ d \theta &= \int_{|z|=1} \frac{1}{(2+\frac{z+z^{-1}}{2})^{2}} \frac{dz}{iz} \\ &= \frac{4}{i}\int_{|z|=1} \frac{z}{(z^{2}+4z+1)^{2}} \ d z \\ &=\frac{4}{i} \int_{|z|=1}\frac{z}{[(z-\sqrt{3}+2)(z+\sqrt{3}+2)]^{2}} \ dz \end{align}$$
The only pole inside the unit circle is at $z= \sqrt{3}-2$.
Therefore,
$$ \begin{align} \int_{0}^{2 \pi} \frac{1}{(2+ \cos \theta)^{2}} \ d \theta &= 2 \pi i \frac{4}{i} \text{Res} \left[ \frac{z}{[(z-\sqrt{3}+2)(z+\sqrt{3}+2)]^{2}}, \sqrt{3}-2 \right] \\ &= 8 \pi \lim_{z \to \sqrt{3}-2} \frac{d}{dz} \frac{z}{(z+\sqrt{3}+2)^{2}}
\\ &= 8 \pi \lim_{z \to \sqrt{3}-2} \frac{(z+\sqrt{3}+2)^{2}-2z(z+\sqrt{3}+2)}{(z+\sqrt{3}+2)^{4}} \\ &= 8 \pi \lim_{z \to \sqrt{3}-2} \frac{-z+ \sqrt{3} + 2}{(z+\sqrt{3}+2)^{3}} \\ &= 8 \pi \frac{4}{24 \sqrt{3}} = \frac{4 \pi}{3 \sqrt{3}} \end{align}$$
| {
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Laurent series for $\frac1{z^2+1}$ I have this problem:
Find the Laurent series around $z=0$, for $\dfrac{10}{(z+2)(z^2+1)}$ in the region $1<|z|<2$.
I did partial fractions and found this: $\dfrac{2}{z+2}-\dfrac{2z-4}{z^2+1}$,
then I have to know what's the Laurent series for $\dfrac{1}{z^2+1}$ to solve the problem.
Do you know that series? Thank you.
| The answer by Santosh is not correct.
The partial fraction decomposition of $f(z)$ is
$$
f(z) = \frac{2}{z+2} - \frac{1-2i}{z+i} - \frac{1+2i}{z-i}
$$
Since we are looking for Laurent series in the annulus, $1<\lvert z\rvert < 2$, we have that $\frac{1}{\lvert z\rvert} < 1$ and $\frac{\lvert z\rvert}{2}<1$.
\begin{align}
\frac{2}{z+2} - \frac{1-2i}{z+i} - \frac{1+2i}{z-i} &=
\frac{1}{1+\frac{z}{2}} - \frac{1-2i}{z}\frac{1}{1+\frac{i}{z}} - \frac{1+2i}{z}\frac{1}{1-\frac{i}{z}}\\
&=\sum_{n=0}^{\infty}(-1)^n\biggl[\Bigl(\frac{z}{2}\Bigr)^n-\frac{1-2i}{z^{n+1}}(i)^n\biggr]-(1+2i)\sum_{n=0}^{\infty}\frac{(i)^n}{z^{n+1}}
\end{align}
| {
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Simplifying $(1/x-1/5 )/( 1/x^2-1/25)$ How do I get $\frac{5x}{x+5}$ from simplifying the following?
$$\frac{(\frac{1}{x}-\frac{1}{5} )}{( \frac{1}{x^2}-\frac{1}{25})}$$
My work:
I multiplied the top and bottom by the LCD: $25x^2$ (to get the same denominators).
Then I got: $\frac{25x-5x}{25-x^2}$
Then I got this for my answer: $\frac{20}{-(x-5)(x+5)}$
But the real answer is: $\frac{5x}{x+5}$
So my questions is how to I get from $\frac{20}{-(x-5)(x+5)}$ to $\frac{5x}{x+5}$?
| Easier: $\frac{1}{x^2} - \frac{1}{25} = \big( \frac{1}{x} -\frac{1}{5} \big)\big(\frac{1}{x} + \frac{1}{5} \big)$ and the first term happily cancels out, now sum two fractions and you are done.
| {
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Logarithm Fraction Contest Math Question The question is as follows:
If $\dfrac{\log_ba}{\log_ca}=\dfrac{19}{99}$ then $\dfrac{b}{c}=c^k$. Compute $k$.
| Rewrite
$$
\begin{align}
\dfrac{\log_ba}{\log_ca}&=\dfrac{19}{99}\\
\dfrac{\frac{\log a}{\log b}}{\frac{\log a}{\log c}}&=\dfrac{19}{99}\\
\dfrac{\log a}{\log b}\cdot\dfrac{\log c}{\log a}&=\dfrac{19}{99}\\
\dfrac{\log c}{\log b}&=\dfrac{19}{99}\\
99\log c&=19\log b\\
\log c^{99}&=\log b^{19}\\
c^{99}&=b^{19}\tag1
\end{align}
$$
and
$$
\frac{b}{c}=c^k\quad\Rightarrow\quad b=c^{k+1}\quad\Rightarrow\quad b^{19}=c^{19(k+1)}.\tag2
$$
Substituting $(1)$ to $(2)$ yields
\begin{align}
c^{99}=c^{19(k+1)}\quad\Rightarrow\quad 19(k+1)&=99\\
k+1&=\frac{99}{19}\\
k&=\frac{99}{19}-1\\
&=\large\color{blue}{\frac{80}{19}}.
\end{align}
| {
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Congruency by completing the square $x^2+x+1\equiv 0\mod 49$
We have the ring isomorphism $\mathbb{Z}/49\mathbb{Z}\to\mathbb{Z}/7\mathbb{Z}\times\mathbb{Z}/7\mathbb{Z}$.
Consider $x^2+x+1\equiv 0\mod 7$
I usually solve these polynomial congruencies using the 'complete the square' method. How would I go about doing that in this case?
EDIT:
$x^2+x+1\equiv x^2-6x+8\equiv(x-3)^2\equiv 1\mod 7$
I get solutions $x=4$ and $x=2$ $\mod 7$.
Now I tried to use Hemel's lifting, so,
We have $f(x)=x^2+x+1$ thus $f'(x)=2x+1$.
In the case of $x\equiv 2\mod 7$, we have $f(2)=7$, $f'(2)=5$; as $7\nmid f'(2)$, we can lift to a unique solutions by solving $f'(2)\equiv -\frac{f(2)}{7}\mod 7$, i.e. $5t\equiv -1\mod 7$.
Clearly $t\equiv 4\mod 7$, leading to the solution $x\equiv 2+4\cdot 7\equiv 30\mod 49$.
In the case of $x\equiv 4\mod 7$, we have $f(4)=21$, $f'(4)=9$; as $7\nmid f'(4)$, we can lift to a unique solutions by solving $f'(4)t\equiv -\frac{f(4)}{7}\mod 7$ or $9t\equiv -\frac{9}{7}\mod 7$...and this is where I'm stuck, assuming I'm even doing it correctly.
| Mod 7, we have $$x^2+x+1\equiv x^2-6x+8=0$$ or $$(x-3)^2=1$$
Since $7$ is prime, each coprime number has either two square roots, or none. Hence $x-3=1$ or $x-3=-1$, i.e. $x=4$ and $x=2$ are the two solutions mod $7$.
| {
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$4 \sin 72^\circ \sin 36^\circ = \sqrt 5$ How do I establish this and similar values of trigonometric functions?
$$
4 \sin 72^\circ \sin 36^\circ = \sqrt 5
$$
| A key observation is $36 + 72 + 72 = 180$. This means $$\sin 72 = \sin (36 + 72) = \sin 36 \cos 72 + \cos 36 \sin 72$$
Using the identity $\sin 72 = 2\sin 36 \cos 36$, we get
$$2\sin 36 \cos 36 = \sin 36 \cos 72 + 2 \sin 36 \cos^2 36$$
Now notice $\sin 36$ appears everywhere, so we can divide by it...
$$2\cos 36 = \cos 72 + 2\cos^2 36$$
Use the identity $\cos 72 = 2 \cos^2 36 - 1$, we get
$$2\cos 36 = 4\cos^2 36 - 1$$
So $\cos36$ is one of the solutions to $2x = 4x^2 - 1$. Using the quadratic formula, you see that the solutions are $$\frac {2 \pm \sqrt{20}}4 = \frac {1 \pm \sqrt 5}2$$
But you know that $\cos 36$ must be positive, so you know the correct value. Now you can get $\sin 72$ from any identity you wish, and multiply.
How to get such values "in general": This is hard. The reason $36$ and $72$ are easy values is that a small integer multiple of them gives you a multiple of $180$. If you wanted to compute such a value for let's say $\sin 1$, it would be a very nasty expression, as you would have to iterate many times until you reach a "basic" angle like $180$ (or even $30$).
| {
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Applications of the identity $ab + \left(\frac{a+b}{2} - b\right)^2 = \left(\frac{a+b}{2}\right)^2$ I am reading Euclid's elements
I found the algebraic identity
$ab + \left(\frac{a+b}{2} - b\right)^2 = \left(\frac{a+b}{2}\right)^2$
I ponder on usage of this identity for $2$ hours.
but I can't click anything.
$a^2 + b^2 = c^2$ can be used when you want to know the direction between $2$ coordinates.
Any example involving this identity?
| If I understand properly $$ab + \left(\frac{a+b}{2} - b\right)^2 =ab+\left(\frac{a-b}{2} \right)^2=\frac{1}{4} \Big(4ab+(a-b)^2\Big)=\frac{1}{4} (a+b)^2=\left(\frac{a+b}{2}\right)^2$$
| {
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Another inequality to prove involving arbitrary reals Let $a,b$ be non-zero reals such that $ab\ge \frac{1}{a}+\frac{1}{b}+3$ then prove the following inequality :
$$ \sqrt[3]{ab}\ge \frac{1}{\sqrt[3]{a}}+\frac{1}{\sqrt[3]{b}}$$
This one stumped me completely as usual. A solution would be welcome.
| [Clearly we can substitute all the variables with their cubes, which I will do so to avoid writing so many cube-roots.]
For any $a,b < 0$:
$a b > 0 > \frac{1}{a} + \frac{1}{b}$
For any $a > 0$ and $b \ne 0$:
If $a b < \frac{1}{a} + \frac{1}{b}$:
Let $b' > b$ such that $a b' = \frac{1}{a} + \frac{1}{b'}$ and ( $b' < 0$ if $b < 0$ ), which is possible because:
$f : x \mapsto a x - \frac{1}{a} - \frac{1}{x}$ is continuous except at $0$
$f(b) < 0$ and $f(0^-) = \infty$ and $f(\infty) = \infty$
Thus the claim follows by Intermediate value theorem
Then $a^3 b^3 < a^3 b'^3 = ( \frac{1}{a} + \frac{1}{b'} )^3 = \frac{1}{a^3} + \frac{1}{b'^3} + 3 \frac{1}{ab'}( \frac{1}{a} + \frac{1}{b'} ) = \frac{1}{a^3} + \frac{1}{b'^3} + 3 < \frac{1}{a^3} + \frac{1}{b^3} + 3$
Therefore $a b \ge \frac{1}{a} + \frac{1}{b}$ if $a^3 b^3 \ge \frac{1}{a^3} + \frac{1}{b^3} + 3$
Likewise for $b > 0$ and $a \ne 0$
Therefore ( $a b \ge \frac{1}{a} + \frac{1}{b}$ if $a^3 b^3 \ge \frac{1}{a^3} + \frac{1}{b^3} + 3$ ) for any $a,b \ne 0$
[I should add that this solution was found by plotting the two inequalities using Graph and seeing that they were equivalent. The solution is essentially to show that equality for the simpler one implies equality for the other, and then to move within the regions of inequality towards the equality case.]
| {
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With how many ways can we choose $9$ balls of a box? With how many ways can we choose $9$ balls of a box that contains $12$ balls, of which $3$ are green, $3$ are white, $3$ are blue and $3$ are red?
$$$$
I have done the following:
$x_1=\# \text{ green balls that we choose }$
$x_2=\# \text{ white balls that we choose }$
$x_3=\# \text{ blue balls that we choose }$
$x_4=\# \text{ red balls that we choose }$
$$x_1+x_2+x_3+x_4=9, \ \ \ \ 0 \leq x_i \leq 3$$
$$\\$$
The number of solutions of the equation: $x_1+x_2+x_3+x_4=9, \ \ \ x_i \geq 0 \ \ \ $ is
$$\binom{9+4-1}{3}=\binom{12}{3}$$
Then I take the cases I don't want, to subtract them from the number above.
$x_1>3 \Rightarrow x_1 \geq 4$
$y_1=x_1-4 \geq 0$
So we become the equation:
$$y_1+4+y_2+y_3+y_4=9, \ \ \ y_i \geq 0 \ \ \ \ \Rightarrow y_1+y_2+y_3+y_4=5, \ \ \ y_i \geq 0$$
And the number of solutions are:
$$\binom{4+5-1}{3}=\binom{8}{3}$$
We do the same for $i=2,3,4$.
$$\\$$
Finally the solution is:
$$\binom{12}{3}- 4\cdot \binom{8}{3}=-4$$
That cannot be right..
What have I done wrong??
| Generating function solution: Each color is represented by:
$$
1 + z + z^2 + z^3 = \frac{1 - z^4}{1 - z}
$$
All four colors are then:
$$
\left( \frac{1 - z^4}{1 - z} \right)^4
$$
and you want 9 balls, i.e., the coefficient of $z^9$, ellipses are terms that don't affect the result:
\begin{align}
[z^9] \left( \frac{1 - z^4}{1 - z} \right)^4
&= [z^9] \frac{(1 - z^4)^4}{(1 - z)^4} \\
&= [z^9] (1 - 4 z^4 + 6 z^8 - \ldots) (1 - z)^{-4} \\
&= [z^9] (1 - z)^{-4} - 4 [z^5] (1 - z)^{-4} + 6 [z] (1 - z)^{-4} \\
&= (-1)^9 \binom{-4}{9}
- 4 \cdot (-1)^5 \binom{-4}{5}
+ 6 \cdot (-1)^1 \binom{-4}{1}
\end{align}
Use of the generalized binomial theorem finishes this off:
$$
\binom{-m}{k} = (-1)^k \binom{k + m - 1}{m - 1}
$$
for $m \in \mathbb{N}$. You get 20 as result.
| {
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Trying to evaluate integral$\int_0^\infty x \sqrt{1-e^{-x}}\,e^{-x}dx$ I am trying to integrating
$$
\int_0^\infty x \sqrt{1-e^{-x}}\,e^{-x}dx\equiv I
$$
but cannot get the answer, I would like a proof not a numerical answer. My attempt at proof:
$$
y=\sqrt{1-e^{-x}}\\
y(0)=0, \ y(\infty)=1\\
y^2=1-e^{-x}\\
2ydy=e^{-x}dx\\
e^{-x}=1-y^2\\
x=\ln\frac{1}{1-y^2} \rightarrow\\
I=2\int_0^1 y^2\ln\frac{1}{1-y^2} dy=\\
-2\int_0^1y^2\ln(1-y^2)\,dy=\\2\int_0^1y^2\sum_{k=1}^\infty \frac{y^{2k}}{k}=\\
2\sum_{k=1}^\infty \frac{1}{k}\int_0^1 y^{2(k+1)}dy=\\
2\sum_{k=1}^\infty \frac{1}{k(3+2k)}=\\
2\sum_{k=1}^\infty \left(\frac{1}{3k}-\frac{2}{3(2k+3)} \right)
$$
but this diverges because $\sum_k\frac{1}{k}\to\infty$? Mistakes I made...
Please help if can on doing the sum or integral. Thank you, Grazie
| Using the well-Known Series formula for the Digammafunction http://en.m.wikipedia.org/wiki/Digamma_function we have $$2\sum_{k=1}^{\infty}\frac{1}{n(2n+3)}=\frac{2}{3}\sum_{n=0}^{\infty}\frac{\frac{5}{2}-1}{(n+1)(n+\frac{5}{2})}=\frac{2}{3}(\gamma +\psi(\frac{5}{2}))$$
Recall the recurrence relation formula for Digamma, i.e $$\psi(x+1)=\psi(x)+\frac{1}{x}$$ hence it reduces to $$\frac{2}{3}(\gamma +\frac{8}{3}+\psi(\frac{1}{2}))$$ It is pretty simple to calculate that $\psi(\frac{1}{2})=-\gamma -2\log2$ So your sum finally evaluates to $$\frac{2}{3}(\frac{8}{3}-2\log2)=\frac{4}{3}(\frac{4}{3}-\log2)$$
| {
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Find the limit of $ \lim_{x \to 7} \frac{\sqrt{x+2}-\sqrt[3]{x+20}}{\sqrt[4]{x+9}-2} $ I need to evaluate the limit without using L'Hopital's rule.
$$\Large
\lim_{x \to 7} \frac{\sqrt{x+2}-\sqrt[3]{x+20}}{\sqrt[4]{x+9}-2}
$$
| Sorry to add a second answer, dont upvote if you dont like it, but this answer is different from my first and is essentially the same as some of the other answers, but I wanted to point a different formulation that is a little to long for a comment.
One can use the limit
$$\lim\limits_{y \to 0} \frac{(y+k)^{\alpha}-k^{\alpha}}{y}=\alpha k^{\alpha -1}$$
then with the substitution, $y=x-7$ we can write the expression as
$$\frac{\left( \frac{\sqrt{y+9}-3}{y} \right) -\left( \frac{\sqrt[3]{y+27}-3}{y} \right)}{\frac{\sqrt[4]{y+16}-2}{y} }$$
and this gives,
$$\frac{\frac{1}{2}9^{-\frac{1}{2}}-\frac{1}{3}27^{-\frac{2}{3}} }{\frac{1}{4}16^{-\frac{3}{4}}}=\frac{112}{27}$$
| {
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Limit $\lim_{x\rightarrow\infty}1-x+\sqrt{2+2x+x^2}$ $$\lim_{x\rightarrow\infty}1-x+\sqrt{2+2x+x^2}=\lim_{x\rightarrow\infty}1-x+x\sqrt{\frac{2}{x^2}+\frac 2x+1}=\lim_{x\rightarrow\infty}1=1\neq2$$ as Wolfram Alpha state. Where I miss something?
| You need to Taylor expand the square root part as well:
Recall that $\sqrt{1+x} = 1 + \frac{1}{2}x + o(x)$ for $x \to 0$
$\sqrt{2/x^2+2/x+1} = (1+\frac{1}{2}2/x+o(1/x)) = 1+1/x + o(1/x)$ as $x \to \infty$
Now you get:
$1-x+x\sqrt{2/x^2+2/x+1} = 1-x+x[1+1/x + o(1/x)] = 2 + o(1) \to 2$ as $x \to \infty$
| {
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Proving by induction: $ \frac{1\cdot3\cdot5\cdot \ldots \cdot (2n-1)}{1\cdot2\cdot3\cdot\ldots\cdot n} \leq 2^n $ WTS $ \frac{1\cdot3\cdot5\cdot \ldots \cdot (2n-1)}{1\cdot2\cdot3\cdot\ldots\cdot n} \leq 2^n $ for all natural $n$.
Have checked $P_1$, and assumed $P_k$.
Trying the following argument:
$P_{k+1} = \frac{1\cdot3\cdot5\cdot \ldots \cdot (2k-1)\cdot(2k+1)}{1\cdot2\cdot3\cdot\ldots\cdot k \cdot (k+1)} = \frac{1\cdot3\cdot5\cdot \ldots \cdot (2k-1)}{1\cdot2\cdot3\cdot\ldots\cdot k}\cdot \frac{2k+1}{k+1} \leq 2^k \cdot\frac{2k+1}{k+1} \leq 2^{k+1}\cdot \frac{2k+1}{k+1}$ using $P_k$ and that these are all positive numbers.
Now dividing through by $\frac{2k+1}{k+1}$ would give the desired result.
I just feel like I've done something wrong here. Can anyone tell me if I've overlooked anything?
| Here is an alternate solution that gives a bit more insight.
$\frac{1 \cdot 3 \cdots (2n-1)}{n!} = \frac{(2n)!}{2^n(n!)^2}$ so we have to prove $2^{2n} \ge \frac{(2n)!}{(n!)^2} = \dbinom{2n}{n}$.
However, $2^{2n} = \sum_{k = 0}^{2n} \dbinom{2n}k > \dbinom{2n}n$ and the conclusion follows. Obviously, the inequality is very weak.
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"language": "en",
"url": "https://math.stackexchange.com/questions/822526",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
Integration by parts - $\int \ln (2x+1) \text{dx}$ Use integration by parts to find
$\int \ln (2x+1) \text{dx}$.
So far I have:
$$x\ln(2x+1)-\int\dfrac{2x}{2x+1}dx+c$$
Using integration by substitution to find the integral
$$u=2x+1\Rightarrow\text{du}=2\text{dx}$$
$$\int\dfrac{2x}{2x+1}\cdot\dfrac{1}{2}\text{du}=\int xu^{-1}$$
$$=\int \left(\dfrac{u}{2}-\dfrac{1}{2}\right)u^{-1}\text{du}=\int\left[\dfrac{1}{2}-\dfrac{1}{2}u^{-1}\right]\text{du}$$
$$=\dfrac{1}{2}x-\dfrac{1}{2} \ln \left|2x+1\right|$$
Looking at the answer in the back, this is wrong.
The answer is $x \ln(2x+1)-x+\dfrac{1}{2}\ln(2x+1)+c$.
What have I done wrong?
| You are right till $$I = x\ln(2x+1) - \int\dfrac{2xdx}{2x+1}.$$
The last integral can be calculated like this:
$$\int\dfrac{2xdx}{2x+1} =\int \dfrac{2x+1 - 1}{2x+1}$$
$$\int dx - \int \dfrac{1}{(2x+1)}$$
Substituting this you get the desired integral:
$$ I = x\ln(2x+1) -x +\frac{ \ln(2x+1)}{2}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/823162",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 6,
"answer_id": 2
} |
Find a closed form for $\sum_{k=0}^{n} k^3$ Find a closed form for $\sum_{k=0}^{n} k^3$.
I would appreciate ideas for approaching questions like this in general as well.
Thanks.
| Consider the polynomial
$$\sum_{k=0}^n(z+1)^k=\frac{(z+1)^{n+1}-1}{z}=(n+1)+\frac{(n+1)n}2z+\frac{(n+1)n(n-1)}6z^2+\frac{(n+1)n(n-1)(n-2)}{24}z^3...$$
when $z=0$, it takes the value $$\color{blue}{\sum_{k=0}^n1=n+1}.$$
Derive on $z$:
$$\sum_{k=0}^nk(z+1)^{k-1}=\frac{(n+1)n}2+\frac{2(n+1)n(n-1)}6z+\frac{3(n+1)n(n-1)(n-2)}{24}z^2...$$
when $z=0$, it takes the value $$\color{blue}{\sum_{k=0}^nk=\frac{(n+1)n}2}.$$
Multiply by $z+1$:
$$\sum_{k=0}^nk(z+1)^{k}=\frac{(n+1)n}2+\frac{3n(n+1)+2(n+1)n(n-1)}6z+\frac{8(n+1)n(n-1)+3(n+1)n(n-1)(n-2)}{24}z^2...$$
Derive on $z$:
$$\sum_{k=0}^nk^2(z+1)^{k-1}=\frac{(n+1)n(2n+1)}6+\frac{(n+1)n(n-1)(3n+2)}{12}z...$$
when $z=0$, it takes the value $$\color{blue}{\sum_{k=0}^nk^2=\frac{(n+1)n(2n+1)}6}.$$
Multiply once more by $z+1$ and derive to get the third order sum. The constant term is
$$\frac{(n+1)n(2n+1)}6+\frac{(n+1)n(n-1)(3n+2)}{12}=\frac{3(n+1)n(n^2+n)}{12},$$
giving:
$$\color{blue}{\sum_{k=0}^nk^3=\frac{(n+1)^2n^2}4}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/823897",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 1
} |
Probability of getting the sum of two numbers selected from two ranges as odd. Consider two integers N and M.
One number is selected from a range of 1 to N.
Another number is selected from a range of 1 to M.
What is the probability that the sum of the two numbers is odd?
Note:
I figured out that, whenever either N or M is even, the probability will be 1/2. But what if both or any one of them is odd.
| $A \equiv \text{Number selected from the range $1\ldots N$ is odd}\\
B \equiv \text{Number selected from the range $1\ldots M$ is odd}
$
Now $P(A) = \frac 1 N\left\lceil\frac N 2\right\rceil,\ P(\bar{A}) = \frac{1}{N}\left\lfloor \frac N 2 \right\rfloor$
and $P(B) = \frac 1 M\left\lceil\frac M 2\right\rceil,\ P(\bar{B}) = \frac{1}{M}\left\lfloor \frac M 2 \right\rfloor$
Then the probability that the sum of the two numbers is odd is
$P(A)P(\bar{B}) + P(\bar{A})P(B)\\
= \dfrac 1 N\left\lceil\dfrac N 2\right\rceil \dfrac{1}{M}\left\lfloor \dfrac M 2 \right\rfloor + \dfrac{1}{N}\left\lfloor \dfrac N 2 \right\rfloor \dfrac 1 M \left\lceil\dfrac M 2\right\rceil\\
= \dfrac{1}{MN}\left( \left\lceil\dfrac N 2\right\rceil \left\lfloor \dfrac M 2 \right\rfloor + \left\lfloor \dfrac N 2 \right\rfloor \left\lceil\dfrac M 2\right\rceil \right)
$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/825911",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
What is the value of the following? $3^{302} \mod 5.$ I have to choose from
a. 0
b. 1
c. 2
d. 3
e. 4
I think its e. 4 because
$$3^{302} = 3^{300} \cdot 3^2 = 3^{4\cdot 75} \cdot 3^2 = (3^4)^{75} \cdot 3^2.$$
Applying Fermat's Little Theorem to turn $3^4$ into $1$, then evaluate and solve:
$$(3^4)^{75} \cdot 3^2 \equiv 1^{75} \cdot 9 \equiv 1 \cdot 9\equiv 9 \equiv 4 \pmod 5$$
Can you please confirm that the answer is e. 4. Thank you.
| Hint $\ {\rm mod}\ 5\!:\,\ \color{#c00}{3^2\equiv\, -1}\ \Rightarrow\ 3^{2k}\equiv (\color{#c00}{3^2})^k\equiv (\color{#c00}{-1})^k\ $ [$\,\equiv -1\ $ if $\,k\,$ is odd, as in your case]
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/828145",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
} |
Help w/ Partial Fraction Decomposition I need some help figuring out how to decompose $\displaystyle\frac{1}{x^4+1}$ into partial fractions.
This is what I have done so far: $$\frac{1}{x^4+1} = \frac{1}{(x^2 - \sqrt{2}x + 1)(x^2 + \sqrt{2}x + 1)}$$
From there, I do not know how to decompose it into partial fractions, or if I even set it up correctly.
| My favorate partial fraction decomposition.
$$\frac{1}{x^4+1}=\frac{1}{2\sqrt{2}}\left(\frac{x+\sqrt{2}}{x^2+\sqrt{2}x+1}
-\frac{x-\sqrt{2}}{x^2-\sqrt{2}x+1}
\right)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/828598",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
How is $2\sum_{n=2}^{\infty}\frac{1}{(n-1)(n+1)}=\frac{6}{4}$ calculated? $$2\sum_{n=2}^{\infty}\frac{1}{(n-1)(n+1)}=\frac{6}{4}$$
I cant figure out why this is $\frac64$. I try to use telescopic series without success.
| Since OP is apparently aware this is a telescoping series, here's the computation.
$$
2\sum_{n=2}^{\infty}\frac{1}{(n-1)(n+1)} =
\sum_{n=2}^{\infty}\frac{2}{(n-1)(n+1)}=
\sum_{n=2}^{\infty}\left(\frac{1}{n-1}-\frac{1}{n+1}\right) =\\
\left[
\left(1 - \frac 13\right) + \left(\frac 12 - \frac 14\right)
+ \left(\frac 13 - \frac 15\right) + \cdots
\right] = \\
\left[
1 + \frac 12 + \left(\frac 13 - \frac 13\right) + \left(\frac 14 - \frac 14\right) + \cdots
\right] = \frac 32
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/829220",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 3
} |
Inf and Sup for ratio and root test of convergent series I would like to use the ratio and root test on the following series:
$$s = 1/2 + 1/3 + (1/2)^2 + (1/3)^2 + \ldots = a1 + a2 + a3 + \ldots$$
where $a_2$ is $\left(\frac{1}{2} \right)^2 + \left(\frac{1}{3} \right)^2 $ for example.
I know we have a sum of two geometric series so the sum will be convergent but I'd like to find the following results
$\lim_{n \to \infty} \inf\left(\frac{a_{n+1}}{a_n}\right) = 0 $
$\lim_{n \to \infty} \sup\left(\frac{a_{n+1}}{a_n}\right) = +\infty $
$\lim_{n \to \infty} \inf \sqrt[n]{a_n} = \frac{1}{\sqrt{3}} $
$\lim_{n \to \infty} \sup \sqrt[n]{a_n} = \frac{1}{\sqrt{2}} $
How to calculate such supremum of infimum ?
I know that $\frac{(a_{n+1})}{(a_n)} = \frac{(3^{(n+1)} + 2^{(n+1)})}{(6\cdot(3^n+2^n))}$ and $(a_n) = \frac{3^n + 2^n}{3^n \cdot 2^n}$. But what to do afterwards ?
How to get these calculations ?
| Hint: Since $a_n=2^{-n}+3^{-n}$, we get
$$
\begin{align}
\frac{a_{n+1}}{a_n}
&=\frac{2^{-n-1}+3^{-n-1}}{2^{-n}+3^{-n}}\\
&=\frac{1/2+1/3(2/3)^n}{1+(2/3)^n}
\end{align}
$$
and
$$
\begin{align}
a_n^{1/n}
&=\left(2^{-n}+3^{-n}\right)^{1/n}\\
&=\frac12\left(1+(2/3)^n\right)^{1/n}
\end{align}
$$
To get the limits listed in the question, you may have meant to define
$$
a_n=\left\{\begin{array}{}
2^{-k}&\text{if }n=2k-1\\
3^{-k}&\text{if }n=2k
\end{array}\right.
$$
Then
$$
\frac{a_{n+1}}{a_n}=\left\{\begin{array}{}
(2/3)^k&\text{if }n=2k-1\\
1/2(3/2)^k&\text{if }n=2k
\end{array}\right.
$$
and
$$
a_n^{1/n}=\left\{\begin{array}{}
2^{-k/(2k-1)}&\text{if }n=2k-1\\
3^{-k/(2k)}&\text{if }n=2k
\end{array}\right.
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/831446",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Find a generating function for $\sum_{k=0}^{n} k^2$
Find a generating function for $\sum_{k=0}^{n} k^2$
I know my solution is wrong, but why?
My solution:
If $F(x)$ generates $\sum_{k=0}^{n} k^2$ then $F(x)(1-x)$ generates $k^2$.
$\frac{x}{(1-x)^4}: \left\{ 0,1,4,9,16,25... \right\}$
$\frac{x}{(1-x)^3}: \left\{ 0,1,3,5,7,9... \right\}$
$\frac{x}{(1-x)^2}: \left\{ 0,1,2,2,2,2... \right\}$
$\frac{x}{1-x}: \left\{ 0,1,1,1,1,1... \right\}$
$x: \left\{ 0,1,0,0,0,0... \right\}$
So, $F(x)=\frac{x}{(1-x)^5}$ generates $\sum_{k=0}^{n} k^2$
| Just consider:
\begin{align}
\sum_{n \ge 0} z^n
&= \frac{1}{1 - z} \\
\sum_{n \ge 0} n^2 z^n
&= z \frac{\mathrm{d}}{\mathrm{d} z}
\left( z \frac{\mathrm{d}}{\mathrm{d} z} \frac{1}{1 - z} \right) \\
&= \frac{z + z^2}{(1 - z)^3} \\
\sum_{n \ge 0} \left( \sum_{0 \le k \le n} k^2 \right) z^n
&= \frac{z + z^2}{(1 - z)^4}
\end{align}
Just for the heck of it:
\begin{align}
\sum_{0 \le k \le n} k^2
&= [z^n] \frac{z + z^2}{(1 - z)^4} \\
&= [z^{n - 1}] (1 - z)^{-4} + [z^{n - 2}] (1 - z)^{-4} \\
&= (-1)^{n - 1} \binom{-4}{n - 1} + (-1)^{n - 2} \binom{-4}{n - 2} \\
&= \binom{n - 1 + 4 - 1}{4 - 1} + \binom{n - 2 + 4 - 1}{4 - 1} \\
&= \binom{n + 2}{3} + \binom{n + 1}{3} \\
&= \frac{(n + 2) (n + 1) n + (n + 1) n (n - 1)}{3!} \\
&= \frac{(2 n + 1) (n + 1) n}{6}
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/832858",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 1
} |
Partial fraction (doubt) I have this partial fraction
$$\displaystyle\frac{1}{(2+x)^2(4+x)^2}$$
I tried to resolve using this method:
$$\displaystyle\frac{A}{2+x}+\displaystyle\frac{B}{(2+x)^2}+\displaystyle\frac{C}{4+x}+\displaystyle\frac{D}{(4+x)^2}$$
$$1=A(2+x)(4+x)^2+B(4+x)^2+C(4+x)(2+x)^2+D(2+x)^2$$
When x=-2
$$1=B(4-2)^2$$
$$B=\displaystyle\frac{1}{4}$$
When x=-4
$$1=D(2-4)^2$$
$$D=\displaystyle\frac{1}{4}$$
When x=0
$$1=A(2)(16)+B(16)+C(4)(4)+D(4)$$
$$1=A(32)+B(16)+C(16)+D(4)$$
Replacing the values for B y D
$$1=A(32)+4+C(16)+1$$
$$1-4-1=A(32)+c(16)$$
$$-4=A(32)+C(16)$$
How I can get the values of $A$ and $D$?
| If you clear fractions you get $$A(2+x)(4+x)^2+B(4+x)^2+C(4+x)(2+x)^2+D(2+x)^2=1$$
The easy ones are $x=-2$ which gives $B=\frac 14$ and $x=-4$ which gives $D=\frac 14$
Now equate the coefficients of $x^3$ on each side to give $$A+C=0$$ and set $x=0$ (or equivalently equate constant coefficients), to give $$32A+16B+16C+4D=1$$ which becomes $$16A+4+0+1=1$$
Where the $0$ comes from $16(A+C)=0$
I think it is the equating coefficients of $x^3$ - or using a value like $x=1$ which you've missed - you need four data points to identify the four unknown values.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/833170",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 0
} |
How Find the sum $\sum_{n=2}^{\infty}\left(\sum_{j=1}^{n}\frac{1}{(2j-1)^2} \binom{2n}{n}\frac{1}{2^{2n}(2n+1)}\right)$
Prove or disprove
$$I=\sum_{n=2}^{\infty}\left(1+\dfrac{1}{3^2}+\dfrac{1}{5^2}+\cdots+\dfrac{1}{(2n-1)^2}\right)
\binom{2n}{n}\dfrac{1}{2^{2n}(2n+1)}=\dfrac{\pi^3}{48}-\dfrac{1}{6}$$
My try: since
$$\binom{2n}{n}\dfrac{1}{2^{2n}(2n+1)}=\dfrac{(2n)!}{(n!)^2\cdot 2^{2n}(2n+1)}=\dfrac{(2n-1)!!}{2^n\cdot n!(2n+1)}$$
so
$$I=\sum_{n=2}^{\infty}\left(\sum_{j=1}^{n}\dfrac{1}{(2j-1)^2}\cdot\dfrac{(2n-1)!!}{2^n\cdot n!(2n+1)}\right)$$
Then I can't continue.Maybe this sum have closed form,Thank you
|
Lemma. For $x\in[-1,1]$, we have
$$
\frac{1}{6}(\arcsin x)^3=\sum_{n=1}^\infty a_nx^{2n+1}
$$
with
$$a_n=\left(\sum_{j=1}^n\frac{1}{(2j-1)^2}\right)\binom{2n}{n}\frac{1}{2^{2n}(2n+1)}.$$
Proof. First, we will suppose that $x \in(-1,1)$. Recall that,
$$\arcsin x= \sum_{n=0}^\infty \binom{2n}{n}\frac{x^{2n+1}}{2^{2n}(2n+1)}\tag{1}$$
Now, let $f(x)=\frac{1}{6}(\arcsin x)^3$. Clearly,
$$\sqrt{1-x^2}f'(x)=\frac{1}{2}(\arcsin x)^2,$$
hence
$$(1-x^2)f''(x)-x f'(x)= \arcsin x \tag{2}$$
Now, if $\sum_{n=0}^\infty a_nx^{2n+1}$ (with $a_0=0,a_1=1/6$,) is the power series expansion of $f$ then
replacing in $(2)$ we get
$$
\sum_{n=0}^\infty((2n+3)(2n+2)a_{n+1}-(2n+1)^2a_n)x^{2n+1}=\arcsin x
$$
comparing with $(1)$ we see that
$$
(2n+3)(2n+2)a_{n+1}-(2n+1)^2a_n=\binom{2n}{n}\frac{1}{2^{2n}(2n+1)}
$$
This is equivalent to
$$
2^{2n}\frac{(2n+3)(2n+2)}{(2n+1)\binom{2n}{n}}a_{n+1}-\frac{(2n+1)2^{2n}}{\binom{2n}{n}}a_n=\frac{1}{(2n+1)^2}
$$
or
$$
2^{2n+2}\frac{(2n+3)}{\binom{2n+2}{n+1}}a_{n+1}-\frac{(2n+1)2^{2n}}{\binom{2n}{n}}a_n=\frac{1}{(2n+1)^2}
$$
adding these equalities for $n=0,1,\ldots,m-1$ we get
$$
2^{2m}\frac{(2m+1)}{\binom{2m}{m}}a_{m}= \sum_{n=0}^{m-1}\frac{1}{(2n+1)^2}
$$
which is the desired formula for $a_m$. Finally, noting that $a_n=\mathcal{O}(n^{-3/2})$, we conclude that the power series $\sum a_mx^{2m+1}$ does converge as
$x$ tend to $+1$ or $-1$. So by Abel's theorem we conclude that this power series
converges to $f(x)$ also for $x=1$ and $x=-1$, and the lemma follows.$\qquad\square$
In particular, taking $x=1$ we obtain
$$
\frac{\pi^3}{48}=\sum_{n=1}^\infty\left(\sum_{j=1}^n\frac{1}{(2j-1)^2}\right)\binom{2n}{n}\frac{1}{2^{2n}(2n+1)}
$$
taking away the first term, we obtain $I=\frac{\pi^3}{48}-\frac{1}{6}$.
Also, taking $x=\frac{1}{2}$ we get
$$
\frac{\pi^3}{648}=\sum_{n=1}^\infty\left(\sum_{j=1}^n\frac{1}{(2j-1)^2}\right)\binom{2n}{n}\frac{1}{2^{4n}(2n+1)}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/833680",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Find the intersections of the functions I have $f(x)=-x^2+4$ a parabola and $g(x)=\sqrt{(4-x^2})$ a semi circle with a raduis of $2$
if I say $g(x)=f(x)$ and solve for $x$. I should find the points at which $x$ intercepts
$\sqrt{4-x^2}=-x^2+4$
then
$4-x^2=x^4-8x^2+16\Rightarrow
-x^4+7x^2-12=0$
this is as far as i got, how do I continue from here?
| I would use the quadratic formula.
if you set $f(x) = g(x)$,
you get $x^4 - 7x^2 + 12 = 0$
factoring, we get $(x^2-4)*(x^2-3) = 0$
So $x^2 -4 = 0$,
$x = 2, -2$
Then we have $x^2 - 3$ $= 0$,
$x$ = $\sqrt{3},$ $-\sqrt{3}$
But now we must check for extraneous solutions.
Seeing as there are none, our answers are $x$ = $\sqrt{3}, -\sqrt{3}, 2, -2$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/833928",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
Consider convergence of series: $\sum_{n=1}^{\infty}\sin\left[\pi\left(2+\sqrt{3}\right)^n\right]$ Consider convergence of series:
$$\sum_{n=1}^{\infty}\sin\left[\pi\left(2+\sqrt{3}\right)^n\right]$$
My tried:
We have
$$\sum_{n=1}^{\infty }\sin(\pi (2+\sqrt{3})^{n})=\sum_{n=1}^{\infty}\sin\left(\pi[(2+\sqrt{3})^{n}+(2-\sqrt{3})^{n}]-\pi(2-\sqrt{3})^{n}\right)\, (*)$$
Because $$(2+\sqrt{3})^{n}=\sum_{k=0}^{n}C_{n}^{k}2^{n-k}3^{\frac{k}{2}}$$
$$(2-\sqrt{3})^{n}=\sum_{k=0}^{n}(-1)^{k}C_{n}^{k}2^{n-k}3^{\frac{k}{2}}$$
Hence $$(2+\sqrt{3})^{n}+(2-\sqrt{3})^{n}=\left\{\begin{matrix} 0&,k=2l+1 \\ m\in N&,k=2l \end{matrix}\right.$$
$$\Rightarrow (1)=\sum_{n=1}^{\infty}\sin\left(m\pi-\pi(2-\sqrt{3})^{n}\right)=\sum_{n=1}^{\infty}(-1)^{m+1}\sin\frac{\pi}{(2+\sqrt{3})^{n}}<\sum_{n=1}^{\infty}\sin\frac{\pi}{(2+\sqrt{3})^n}$$
$\sum_{n=1}^{\infty}\sin\frac{\pi}{(2+\sqrt{3})^n}$ converge
Hence series is converge.
True or False?
| As you noticed,
$$\sin(\pi(2+\sqrt{3})^n+\pi(2-\sqrt{3})^n)=0$$
so$$\sin(\pi(2+\sqrt{3})^n)=-\dfrac{\sin(\pi(2-\sqrt{3})^n)\cos(\pi(2+\sqrt{3})^n)}{\cos(\pi(2-\sqrt{3})^n)}$$
or for $n$ large enough
$$\begin{cases} \sin(\pi(2-\sqrt{3})^n)=\pi(2-\sqrt{3})^n+o(\pi(2-\sqrt{3})^n)\\ \cos(\pi(2-\sqrt{3})^n)\ge \frac12 \end{cases}$$
and
$$|\cos(\pi(2+\sqrt{3})^n)|\le1$$
Therefore
$$|n^2\sin(\pi(2+\sqrt{3})^n)|\le 2\pi n^2(2-\sqrt{3})^n+o(n^2(2-\sqrt{3})^n)$$
$$|n^2\sin(\pi(2+\sqrt{3})^n)|\le 2\pi \exp(2\ln(n)-n\ln(2-\sqrt{3}))+o(n^2(2-\sqrt{3})^n)\to 0$$
so your serie converge.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/835554",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 3,
"answer_id": 1
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If $I_n =\int \cot^nx\ dx$ then $I_0 +I_1 +2(I_2+I_3+ \cdots I_8) +I_9+I_{10}= $? If $\displaystyle I_n =\int \cot^nx\ dx$ then find :
$I_0 +I_1 +2(I_2+I_3+ \cdots I_8) +I_9+I_{10} $ = ?
My approach :
$I_n = \displaystyle\int \cot^{n-2} \cot^2x dx$
$\Rightarrow I_n = \displaystyle\int \cot^{n-2} (\csc^2x -1)dx$
$\Rightarrow I_n = \displaystyle\int (\cot^{n-2} \csc^2x -\cot^{n-2} )dx$
$\Rightarrow I_n =\displaystyle \int( \cot^{n-2} \csc^2x) dx- I_{n-2} $
$\Rightarrow I_n +I_{n-2} =\displaystyle \int( \cot^{n-2} \csc^2x) dx$
I am not getting how to integrate the RHS. now please guide thanks.
| Consider using the integral
\begin{align}
I_{n} = \int \cot^{n}(x) \ dx
\end{align}
to evaluate the integral $J = I_{0} + I_{1} + 2(I_{3} + \cdots + I_{8}) + I_{9} + I_{10}$. Let $P$ be the integrand of $J$ for which
\begin{align}
P &= 1 + \cot^{1}(x) + 2( \cot^{2}(x) + \cdots + \cot^{8}(x)) + \cot^{9}(x) + \cot^{10}(x) \\
&= \sum_{k=0}^{10} \cot^{k}(x) + \cot^{2}(x) \ \sum_{k=0}^{6} \cot^{k}(x) \\
&= (1 + \cot^{2}(x)) \ \frac{1-\cot^{9}(x)}{1-\cot(x)} \\
&= \frac{1-\cot^{9}(x)}{(1-\cot(x)) \ \sin^{2}(x)}.
\end{align}
Now,
\begin{align}
J &= \int \frac{1-\cot^{9}(x)}{(1-\cot(x)) \ \sin^{2}(x)} \ dx \\
&= - \int \frac{1 - u^{9}}{1-u} \ du \\
&= - \sum_{k=1}^{9} \frac{u^{k}}{k} \\
J &= - \sum_{k=1}^{9} \frac{\cot^{k}(x)}{k}.
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/837337",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 4,
"answer_id": 1
} |
Easier Proof of $\sin{3\theta} + \sin\theta = 2\sin{2\theta}\cos\theta$ I am curious to see whether anybody can give me a proof that takes less steps.
Here is how I did it:
$$\sin{3\theta} + \sin\theta = 2\sin{2\theta}\cos\theta$$
LHS $$\eqalign{\sin(2\theta + \theta) + \sin\theta &= \sin2\theta\cos\theta + \cos2\theta\sin\theta + \sin\theta\\
&= \sin2\theta\cos\theta + (\cos^2\theta - \sin^2\theta)\sin\theta + \sin\theta\\
&= \sin2\theta\cos\theta + \sin\theta(\cos^2\theta - \sin^2\theta + 1)\\
&= \sin2\theta\cos\theta + \sin\theta(2\cos^2\theta)\\
&= \sin2\theta\cos\theta + 2\sin\theta\cos^2\theta\\
&= \sin2\theta\cos\theta + \cos\theta(\sin\theta\cos\theta + \sin\theta\cos\theta)\\
&= \sin2\theta\cos\theta + \cos\theta(\sin2\theta)\\
&= \sin2\theta\cos\theta + \cos\theta(\sin2\theta)\\
&= 2\sin2\theta\cos\theta.}$$
| Yes, you can differentiate both sides and check whether they are equal: $$\eqalign{
\dfrac{\mathrm d}{\mathrm d\theta}\big[\sin3\theta+\sin\theta\big]&\overset?=\dfrac{\mathrm d}{\mathrm d\theta}\big[2\sin2\theta\cos\theta\big] \\
3\cos3\theta+\cos\theta&\overset?= 2\big[\tfrac12( \cos\theta+3\cos3\theta)\big]\tag{$\overset{\rm Chain}{\underset{\sf rule}{}}\overset{\,+}{}\overset{\rm Prod.}{\underset{\sf rule}{}}$}\\
3\cos3\theta+\cos\theta&\overset?= \cos\theta+3\cos3\theta. \quad\checkmark
}$$
So you can conclude that they're the same.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/837430",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 2
} |
Extrema of $f(x)=\frac{\sin (5x)} 5 - \frac{2\sin(3x)} {3} + \sin (x)$. (a) I need help in finding maxima and minima of the following funcion:
$$f(x)=\frac{\sin (5x)} 5 - \frac{2\sin(3x)} {3} + \sin (x)$$
therefore I need to find the roots of $f'(x)=\cos(5x)-2\cos(3x)+\cos(x)$.
(b) I need to find the minima of $$f(x)=4x+\frac{9\pi^2} x + \sin x.$$
How can I find the roots of $$f'(x)=4-\frac{9\pi^2 }{x^2} + \cos x?$$
Thank you.
| $$\begin{align}
\cos5x-2\cos3x+\cos x&=\cos5x-\cos 3x + \cos x-\cos 3x\\
&=-2\sin4x\sin x+2\sin2x\sin x\\
&=2\sin x(-\sin 4x+\sin2x)\\
&=-4\sin x\sin3x\cos x
\end{align}$$
Now you only have to equal $0$ to each factor.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/840793",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Compute the indefinite integral $I=\int y^{-a}(1−y)^{b-1} dy$ or $I=\int_{d}^1 y^{-a}(1−y)^{b-1} dy$ I need to calculate the indefinite integral $I=\int y^{-a}(1−y)^{b-1} dy$, where $a$, $b$ are REAL NUMBERS and $b>0$.
(my goal is to determine the definite integral $I=\int_{d}^1 y^{-a}(1−y)^{b-1} dy$, where $d<1$).
Thanks !
| \begin{align}
\int_d^1 {y^{ - a} \left( {1 - y} \right)^{b - 1} dy} = \left\{ \begin{array}{l}
\int_d^1 {y^{r - 1} \left( {1 - y} \right)^{b - 1} dy} ,\,\,\,\,\,\,a < 0,\,\,\text{setting}\,\,a = 1 - r,r > 1 \\
\frac{1}{b}\left( {1 - d} \right)^b ,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,a = 0 \\
\int_d^1 {y^{r - 1} \left( {1 - y} \right)^{b - 1} dy} ,\,\,\,\,\,\,a > 0,\,\,\text{setting}\,\,a = 1 - r,r < 1 \\
\end{array} \right.
\end{align}
So that, you can continue with incomplete beta function.
Another method:
Case 1:If $a=0$ and $b>0$, then $$I=
\int_d^1 {\left( {1 - y} \right)^{b - 1} dy} = \frac{1}{b}\left( {1 - d} \right)^b
$$
Case 2:If $b=0$ and $a>0$ and $a\ne1$, then $$
\int_d^1 {y^{ - a} dy} = \frac{{1 - d^{1 - a} }}{{1 - a}}
$$
Case 3:If $b>0$ with $b\ne1$ and $a>0$, $-1<d<1$, then the binomial series
\begin{align}
\left( {1 + x} \right)^\alpha = \sum\limits_{k = 0}^\infty {\left( \begin{array}{l}
\alpha \\
k \\
\end{array} \right)x^k } , \alpha \in \mathbb{R}-\{0\},\,\,\, \forall x \in (-1,1),
\end{align}
where $$
\left( \begin{array}{l}
\alpha \\
k \\
\end{array} \right) = \frac{{\alpha \left( {\alpha - 1} \right)\left( {\alpha - 2} \right) \cdots \left( {\alpha - \left( {k - 1} \right)} \right)}}{{k!}}$$
may be used by setting $\alpha:=b-1$, $(b\ne1)$, the interval $(d,1)\subseteq (-1,1)$, we have
\begin{align}
\left( {1 - y} \right)^{b - 1} = \sum\limits_{k = 0}^\infty {\left( { - 1} \right)^k \left( {\begin{array}{*{20}c}
{b - 1} \\
k \\
\end{array}} \right)y^k }
\end{align}
Multiplying both sides by $y^{ - a} $ and then integrating both sides with respect to $y$, we get
\begin{align}
&\int_d^1 {y^{ - a} \left( {1 - y} \right)^{b - 1} dy}
\\
&= \int_d^1 {\left( {\sum\limits_{k = 0}^\infty {\left( { - 1} \right)^k \left( {\begin{array}{*{20}c}
{b - 1} \\
k \\
\end{array}} \right)y^{k - a} } } \right)dy}
\\
&= \sum\limits_{k = 0}^\infty {\left( { - 1} \right)^k \left( {\begin{array}{*{20}c}
{b - 1} \\
k \\
\end{array}} \right)\left( {\int_d^1 {y^{k - a} dy} } \right)}
\\
&= \sum\limits_{k = 0}^\infty {\left( { - 1} \right)^k \left( {\begin{array}{*{20}c}
{b - 1} \\
k \\
\end{array}} \right)\frac{{1 - d^{k - a + 1} }}{{k - a + 1}}}, \,\,\,\, \text{with} \,\,
a \ne k + 1
\end{align}
where, we used the fact that the series converges uniformly for all $y \in (d,1)\subseteq (-1,1)$.
We note that the condition $a\ne k+1$ is equivalent to say that $a \notin \mathbb{N}$ i.e., $a\in\mathbb{R}-\mathbb{N}$. One more point, If $d<-1$ the series diverges so that we cannot integrate. All these additional conditions should be added.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/849577",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
How to calculate $\int\frac{1}{x + 1 + \sqrt{x^2 + 4x + 5}}\ dx$? How to calculate $$\int\frac{1}{x + 1 + \sqrt{x^2 + 4x + 5}}dx?$$ I really don't know how to attack this integral. I tried $u=x^2 + 4x + 5$ but failed miserably. Help please.
| $$\int\frac{1}{x+1+\sqrt{x^{2}+4x+5}}dx=\int\frac{(x+1)-\sqrt{x^{2}+4x+5}}{-2x-4}$$
$$=\frac{-1}{2}\int\frac{x+1}{x+2}dx-\frac{1}{2}\int\frac{\sqrt{x^{2}+4x+5}}{x+2}dx$$
The first integral can be dealt with but noticing:
$$\int\frac{x+1}{x+2}dx=\int1dx-\int\frac{1}{x+2}dx$$
The second integral is dealt with as follows:
$$\int\frac{\sqrt{x^{2}+4x+5}}{x+2}dx=\int\frac{\sqrt{(x+2)^{2}+1}}{x+2}dx$$
Let $x+2=\tan(u)$ then:
$$\int\frac{\sec^{3}(u)}{\tan(u)}du=\int\sec^{2}(u)\csc(u)=\tan(u)\csc(u)+\int\csc(u)du$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/850500",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 4,
"answer_id": 0
} |
Upper bound for $\frac{x^{2k+2}}{1-x^2} $ My book says that if $x \in (-0.5, 0.5)$, then
$$\frac{x^{2k+2}}{1-x^2} \leq \frac{1}{3\cdot 4^k}$$
I don't understand how to get such upper bound?
Take $k=0 \Rightarrow \frac{x^{2}}{1-x^2} \leq \frac{0.25}{0.75} = \frac{1}{3}$
But how to figure out the general form?
| Note that if $|x| < .5 = \frac 1 2$, then
$$\frac{x^{2k + 2}}{1 - x^2} = \frac{x^2}{1 - x^2} \cdot (x^2)^k \le \frac 1 3 (x^2)^k \le \frac 1 3 \cdot \left(\frac 1 4\right)^k$$
using your previous estimate.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/851487",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
$ \Big(\dfrac{x^7+y^7+z^7}{7}\Big)^2=\Big(\dfrac{x^5+y^5+z^5}{5}\Big)^2\cdot\Big(\dfrac{x^4+y^4+z^4}{2}\Big) $ I have a question. I tried so to solve it, but there is a problem.
that is i don't have any idea to findout how can i work with degrees 4,5,7 ...
this is the problem :
let $ x , y $ and $ z $ three real numbers such $ x+y+z = 0 $.
prove : $ \Big(\dfrac{x^7+y^7+z^7}{7}\Big)^2=\Big(\dfrac{x^5+y^5+z^5}{5}\Big)^2\cdot\Big(\dfrac{x^4+y^4+z^4}{2}\Big) $
Please think and write your solutions! ; )
| HINT:
Let $x,y,z$ be the roots of $\displaystyle t^3+bt+c=0\ \ \ \ (1)$
$\displaystyle\implies xy+yz+zx=b, xyz=-c$
Multiplying $(1)$ by $t^n\ne0$
$\displaystyle \implies t^{n+3}+bt^{n+1}+ct^n=0$
$\displaystyle \implies\sum x^{n+3}=-b\sum x^{n+1}-c\sum x^n$
$\displaystyle n=1\implies \sum x^4=-b\sum x^2-c\sum x =-b[(\sum x)^2-2(xy+yz+zx)]=-b(-2b)=2b^2$
From $\displaystyle n=0\implies \sum x^3=-b\sum x-3c=-3c$
$\displaystyle n=2\implies \sum x^5=-b\sum x^3-c\sum x^2 =-b(-3c)-c[(\sum x)^2-2(xy+yz+zx)]=3bc-c(-2b)=5bc$
Can you take home from here?
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "13",
"answer_count": 3,
"answer_id": 2
} |
Max. and Min. value of $|z|$ in $\left|z+\frac{2}{z}\right| = 2\,$ If $z$ is a complex no. such that $\displaystyle \left|z+\frac{2}{z}\right| = 2\,$ Then find max. and min. value of $\left|z\right|$.
$\bf{My\; Try:}$ Given $\displaystyle \left|z+\frac{2}{z}\right| = 2\Rightarrow \left|z+\frac{2}{z}\right|^2 = 2^2=4$.
So $\displaystyle \left(z+\frac{2}{z}\right)\cdot \left(\bar{z}+\frac{2}{\bar{z}}\right) = 4\Rightarrow \left|z\right|^2+2\left(\frac{z}{\bar{z}}+\frac{\bar{z}}{z}\right)+\frac{1}{|z|^2} = 4$.
Now how can I find the max. and min. values of $|z|$?
Help me please.
Thanks
| Here is one approach:
Let $z=r e^{i\theta}$, then we want to compute the $\min,\max$ of
$I=\{ r \mid \, |r e^{i2\theta} + {2 \over r}| = 2 \}$. We see that $r \in I$ iff $|r \cos ( 2 \theta)+ {2 \over r} +i r \sin ( 2 \theta)|^2 = r^2 +{4 \over r^2} + 4 \cos(2 \theta)= 4$ for some $\theta$.
If we let $f(r) = 1-{1 \over 4} (r^2 +{4 \over r^2} )$, we see that $I = f^{-1} [-1,1]$.
The function $f$ is strictly concave, $\lim_{r \downarrow 0} f(r) = \lim_{r \to \infty} f(r) = -\infty$, $f$ has a maximum value of $0$ at $r=\sqrt{2}$. Hence $I = [r_0,r_1]$ where $r_k$ solve $f(r_k) = -1$.
Solving $f(r) = -1$ yields the equation $r^2 +{4 \over r^2} = 8$ which
has positive solutions $r = \sqrt{2} \sqrt{2 \pm \sqrt{3}}$.
Hence the minimum value of $|z|$ is $\sqrt{2} \sqrt{2 - \sqrt{3}}$ and the maximum value if $\sqrt{2} \sqrt{2 + \sqrt{3}}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/853992",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 2
} |
Extrema and inflection points of the function $y = \cos^2(x) - \cos(x)$ Please help me find the extrema and inflection points of the function $y = \cos^2(x) - \cos(x)$.
So far: $$y'=-2 \cos(x)\sin(x)+\sin(x)$$ $$y'' = 2\sin^2(x) - 2\cos^2(x) + \cos(x)$$
$y' = 0$ when $x = 0, \pi, \pi/3, 5\pi/3$
I plugged these values into $y''$ to find min/max which are as follows:
$x = 0, \pi$ : MAX
$x = \pi/3$, $5\pi/3$ : MIN
The issue I have is finding the inflection points, i used the quadratic formula to solve for $\cos(x)$ by converting (via trig ID's) $y'' = 2\sin^2(x) + 2\cos^2(x) + \cos(x)$ to $y'' = -2\cos^2(x) + \cos(x) = 0$ which gave me $\cos(x) = 0, 1/2$, and $x = \pi/2, 3\pi/2, \pi/3, 5\pi/3$
The solution shows that the inflection points are at $x = 32$ deg $32', 126$ deg $23', 233$ deg $37'$, and $327$ deg $28' $
Can you please help by showing how these inflection points were solved for? Where did I make my mistake? I have no idea how they got those degrees.
| You changed from addition to subtraction of the first two terms in $y''$:
Recall that $$y'' = 2sin^2(x) - 2cos^2(x) + cos(x)$$
$$y'' = 2(1-\cos^2 x) - 2 \cos^2(x) + \cos x \\
= -4\cos^2 x + \cos x + 2 \iff 4\cos^2 x - \cos x - 2$$
Try solving for $\cos x$ now, since we have a quadratic in $\cos x$. I suggest using the quadratic formula: $\cos x = \frac{1 \pm \sqrt{33}}{8}$. You can get an approximation by using $\cos^{-1}(\cos(x))$ for each of the two solutions for $\cos x$.
| {
"language": "en",
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"source": "stackexchange",
"question_score": "2",
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} |
$\frac{2t-t^2}{t+2} \cdot (\frac{5t}{t-2} - \frac{2t}{t-2} )$ Simplify:
$$\frac{2t-t^2}{t+2} \cdot \left(\frac{5t}{t-2} - \frac{2t}{t-2} \right)$$
*
*I first subtracted the parenthesis because the denominator is equal. I then got:
$$\frac{2t-t^2}{t+2} \cdot \frac{3t}{t-2}$$
*
*Then I was lost. I tried multiplying by $t+2$ and $t-2$ on either sides. I tried multiplying $3t$ with $-1$. To make $t+2$ on both sides. Again I didn't get the answer. I have so many different calculations that I'm lost.
None of these seem correct. Or I multiplied incorrect but I doubt that. Is there a trick for doing these? I'm wasting lots of time on just one simplification. Just when I think I'm progressing I'm stuck again.
| $$\begin{align}
\frac{2t-t^2}{t+2}\cdot\left(\frac{5t}{t-2}-\frac{2t}{t-2}\right) &= \frac{2t-t^2}{t+2}\cdot \frac{3t}{t-2} \\
&= \frac{3t\left(2t-t^2\right)}{(t+2)(t-2)} \\
&= \frac{6t^2-3t^3}{(t+2)(t-2)} \\
&= \frac{-3t^2(t-2)}{(t+2)(t-2)} \\
&= -\frac{3t^2}{t+2} \\
\end{align}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/858182",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Evaluating $\sum_{n=0}^{\infty}{\sin^{3}\left(3^{n}\right) \over 3^{n}}$
How do I find the sum of this series?
$$\sum_{n=0}^{\infty}{\sin^{3}\left(3^{n}\right) \over 3^{n}}$$
Hints in the right direction would be appreciated.
| Hint: Use this
$\sin^3x=\frac{3}{4}\sin x - \frac{1}{4}\sin3x. $
Solution:
Let $a_n=\frac{\sin^33^n}{3^n}$ and $b_n=\frac{\sin3^n}{3^n}$. So from the identity mentioned above, one simply has that
$
a_n = \frac{1}{4}(3b_n -3b_{n+1}).
$
Therefore,
$$
\sum_{n\ge 0} \frac{\sin^33^n}{3^n} = \sum_{n\ge 0} a_n = \sum_{n\ge 0} \frac{1}{4}(3b_n -3b_{n+1}) = \frac{3}{4}b_0 = \frac{3\sin 1}{4}.
$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 2,
"answer_id": 1
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The sum of three consecutive cubes numbers produces 9 multiple I want to prove that $n^3 + (n+1)^3 + (n+2)^3$ is always a $9$ multiple
I used induction by the way.
I reach this expression: $(n+1)^3 + (n+2)^3 + (n+3)^3$
But is a lot of time to calculate each three terms, so could you help me to achieve the induction formula
Thanks in advance
| There is one really elegant and simple way of proving this by means of induction.
It goes as following:
Define the problem as: $n^3 + (n+1)^3 + (n+2)^3 = 9\lambda$
Test for $n=1$; $1^3+2^3+3^3=9\lambda$; $36=4*9$
Assume that for some $n \in N$ -> $n^3 + (n+1)^3 + (n+2)^3 = 9\lambda$
Test for $n+1$
$
(n+1)^3+(n+2)^3+(n+3)^3 = 9\lambda
$
$
(n+1)^3+(n+2)^3+(n+3)(n^2+6n+9)=9\lambda
$$(n+1)^3+(n+2)^3+n^3+9n^2+27n+27$
According to our assumption $n^3+(n+1)^3+(n+2)^3$ is divisible by 9 so we can put
$9k$ instead giving us:
$9k + 9n^2+27n+27 = 9(k+n^2+3n+3) = \lambda9$
$Q.E.D$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/859572",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 7,
"answer_id": 2
} |
Find the Value of Integral Find the Value of $$\begin{align}I=\int_{0}^{1}\frac{\ln(x)\,dx}{1-x^2}\end{align}$$
I have tried like this: We have $$\begin{align}2I=\int_{0}^{1}\frac{\ln(x^2)\,dx}{1-x^2}=\int_{0}^{1}\frac{\ln(1-(1-x^2))\,dx}{1-x^2}\end{align}$$ So
$$2I=\begin{align}\int_{0}^{1}\frac{-(1-x^2)-\frac{(1-x^2)^2}{2}-\frac{(1-x^2)^3}{3}-\cdots }{1-x^2}\end{align}=$$
I need help from here..
| Integrating by parts,
$$ \begin{align} \int \frac{\ln (x)}{1-x^{2}} \ dx &= \ln(x) \ \text{arctanh} (x) - \int \frac{\text{arctanh}(x)}{x} \ dx \\ &= \ln (x) \ \text{arctanh} (x) - \frac{1}{2} \int \frac{\ln (1+x)}{x} \ dx + \frac{1}{2} \int \frac{\ln (1-x)}{x} \ dx \\ &= \ln(x) \ \text{arctanh}(x) + \frac{1}{2} \text{Li}_{2}(-x) - \frac{1}{2} \text{Li}_{2}(x) + C \end{align}$$
where $\text{Li}_{2}(x)$ is the dilogarithm function.
Then
$$ \begin{align} \int_{0}^{1} \frac{\ln (x)}{1-x^{2}} \ dx &= \frac{1}{2} \Big( \text{Li}_{2}(-1) - \text{Li}_{2}(1) \Big) \\ &= \frac{1}{2} \Big( - \frac{\zeta(2)}{2} - \zeta(2) \Big) \\ &= \frac{1}{2} \left(- \frac{\pi^{2}}{12}-\frac{\pi^{2}}{6} \right) \\ &= -\frac{\pi^{2}}{8} . \end{align}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/860091",
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"source": "stackexchange",
"question_score": "4",
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} |
Find the maximum value of $ \sqrt{x^4-3x^2-6x+13} - \sqrt{x^4-x^2+1} $ If $x\in\mathbb{R}$ find the maximum value of
$$ \sqrt{x^4-3x^2-6x+13} - \sqrt{x^4-x^2+1} $$
I tried this:
Let $$y= \sqrt{x^4-3x^2-6x+13} - \sqrt{x^4-x^2+1}$$
For maxima $\frac{dy}{dx}=0$ and $\frac{d^2y}{dx^2} < 0$. However, the equation $\frac{dy}{dx}=0$ (after simplifying and clearing the square roots) came out to be a nine degree equation which gave me a nightmare! Moreover, simplifying the derivative was also a tedious task. I found this question in my book in the chapter on theory of equation. I can't think of an algebraic solution. Please Help!
Thanks!
| since
$$\sqrt{(x^2-2)^2+(x-3)^2}-\sqrt{(x^2-1)^2+(x-0)^2}$$
let $$P(x,x^2),A(3,2),B(0,1)$$
so
$$|PA|-|PB|\le |AB|=\sqrt{10}$$
if and only is $A,P,B$ on a line.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/860305",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "12",
"answer_count": 1,
"answer_id": 0
} |
Find expression for : $ S_n =\sum_{i=1}^{n} \frac{i}{i^4+i^2+1} $ I want to find a formula for the sum of this series using its general term.
How to do it?
Series
$$
S_n = \underbrace{1/3 + 2/21 + 3/91 + 4/273 + \cdots}_{n \text{ terms}}
$$
General Term
$$
S_n = \sum_{i=1}^{n} \frac{i}{i^4+i^2+1}
$$
| Write it out:
$$
\begin{eqnarray}
\sum_{i=1}^\infty \frac{i}{i^4 + i^2 + 1}
&=& \sum_{i=1}^\infty \left( \frac{2}{4i^2 - 4i + 4}
- \frac{2}{4i^2 + 4i + 4}\right)\\
&=& \sum_{i=1}^\infty \left( \frac{2}{\Big(2i-1\Big)^2 + 3}
- \frac{2}{\Big( 2i + 1\Big)^2 + 3}\right)\\
&=& \frac{2}{\Big( 2 - 1\Big)^2 + 3} + \sum_{i=1}^\infty \left( \frac{2}{\Big(2i+1\Big)^2 + 3}
- \frac{2}{\Big( 2i + 1\Big)^2 + 3}\right)\\
&=& \frac{1}{2}.
\end{eqnarray}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/865442",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 2
} |
Simplify expression $(x\sqrt{y}- y\sqrt{x})/(x\sqrt{y} + y\sqrt{x})$ I'm stuck at the expression: $\displaystyle \frac{x\sqrt{y} -y\sqrt{x}}{x\sqrt{y} + y\sqrt{x}}$.
I need to simplify the expression (by making the denominators rational) and this is what I did:
$$(x\sqrt{y} - y\sqrt{x}) \times (x\sqrt{y} - y\sqrt{x}) = (\sqrt{y} - \sqrt{x})^2$$
Divided by
$$(x\sqrt{y} + y\sqrt{x}) \times (x\sqrt{y} - y\sqrt{x} ) = (x\sqrt{y})^2$$
So I'm left with $\displaystyle \frac{(\sqrt{y} - \sqrt{x})^2}{(x\sqrt{y})^2}$.
This answer is incorrect. Can anyone help me understand what I did wrong? If there is a different approach to solve this it will also be much appreciated. Please explain in steps.
| I am assuming your ambiguous notation begins with the task of simplifying:
$$\frac{x\sqrt y - y\sqrt x}{x\sqrt y + y\sqrt x}.$$
Assuming I'm correct, then we can rationalize the denominator (get rid of the factors with square roots), as follows:
Multiply the numerator and denominator by $(x\sqrt{y}-y\sqrt{x})$ to get a difference of squares. Recall that $$(a+b)(a-b) = a^2 - b^2.$$ If you carry out this multiplication, you'll have $$\dfrac{(x\sqrt{y}-y\sqrt{x})^2}{x^2y-xy^2}= \dfrac{x^2y - 2xy\sqrt{xy} + xy^2}{x^2y-xy^2}\; =\; \frac{xy(x-2\sqrt{xy} + y)}{xy(x-y)}\;= \; \frac{x-2\sqrt{xy} + y}{x-y}$$
You seemed to have the right idea, looking at your strategy, to multiply numerator and denominator by $x\sqrt y - y\sqrt x$, but you miscalculated.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/869037",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
} |
find a polynomial whose roots are inverse of squares of roots of $x^3+px+q$ Question is :
Given a polynomial $f(x)=x^3+px+q\in \mathbb{Q}[x]$ find a polynomial whose roots are inverse of sqares of roots of $f(x)$
Supposing $a,b,c$ as roots of $f(x)$ we have :
*
*$a+b+c=0$
*$ab+bc+ca=p$
*$abc=-q$
Now i need to know what
*
*$\dfrac{1}{a^2}+\dfrac{1}{b^2}+\dfrac{1}{c^2}$
*$\dfrac{1}{a^2}\cdot\dfrac{1}{b^2}+\dfrac{1}{b^2}\cdot\dfrac{1}{c^2}+\dfrac{1}{c^2}\cdot\dfrac{1}{a^2}$
*$\dfrac{1}{(abc)^2}$
All i have to do is use $(a+b+c)^2$ formula and others and conclude what those sums,products are.. I am fairly comfortable with that...
But then, this question was from a Galois theory course.. So, i some how guess there is a better way to do this...
Can some one suggest something..
| *
*$x^3+px-q$ has roots $-a,-b,-c$.
*$(x^3+px+q)(x^3+px-q)$ has roots $a,b,c,-a,-b,-c$.
*$x^3+2px^2+p^2x-q^2$ has roots $a^2,b^2,c^2$.
*$1+2px+p^2x^2-q^2x^3$ has roots $1/a^2,1/b^2,1/c^2$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/872264",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 0
} |
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Fractions in Questions and Answers
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