Q
stringlengths 70
13.7k
| A
stringlengths 28
13.2k
| meta
dict |
|---|---|---|
If $a,b\in\mathbb R\setminus\{0\}$ and $a+b=4$, prove that $(a+\frac{1}{a})^2+(b+\frac{1}{b})^2\ge12.5$. If $a,b\in\mathbb R\setminus\{0\}$ and $a+b=4$, prove that $$\left(a+\frac{1}{a}\right)^2+\left(b+\frac{1}{b}\right)^2\ge12.5$$
I could expand everything: $$a^2+2+\frac{1}{a^2}+b^2+2+\frac{1}{b^2}\ge12.5$$
Subtract $4$ from both sides: $$a^2+\frac{1}{a^2}+b^2+\frac{1}{b^2}\ge8.5$$
We could use AM-GM here ($a^2,\frac{1}{a^2},b^2,\frac{1}{b^2}$ are all positive), but obviously it wouldn't do anything useful. And we still have to use the fact that $a+b=4$ somehow.
I've tried substituting $b$ with $4-a$, but after clearing the denominators and simplifying we don't quite see anything useful, just a random 6th degree polynomial.
The polynomial is actually: $$2a^6-24 a^5+103.5 a^4-188 a^3+122 a^2-8 a+16\ge 0$$
How could I solve this? We can't use calculus by the way. Thanks.
| Without loss of generality, we can assume $a\ge b$, so let's write $a=2+x$, $b=2-x$ with $0\le x$. The inequality is clearly satisfied if $a\ge4$, so we need only worry about the range $0\le x\lt2$. Plugging this into the OP's inequality and simplifying like crazy, we find we need only prove
$$f(x)=x^2+{(4+x^2)\over(4-x^2)^2}\ge{1\over4}\quad\text{for }0\le x\lt2$$
Now it does not require calculus to see that the function $f(x)$ is increasing on the interval $0\le x\lt2$: The $x^2$ term is clearly increasing, and so is the quotient term, since the numerator $4+x^2$ is increasing and the denominator $(4-x^2)^2$ is decreasing. Finally, since $f(0)={1\over4}$, we can conclude that the inequality holds.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/616695",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 2,
"answer_id": 1
} |
Calculation of $\lambda$, If $x^2+2(a+b+c)\cdot x+3\lambda \cdot (ab+bc+ca) = 0$ has real roots Let $a,b,c$ be the sides of a $\triangle$ where $a\neq b\neq c$ and $\lambda\in \mathbb{R}$. If the roots of the equation
$$x^2+2(a+b+c)\cdot x+3\lambda \cdot (ab+bc+ca) = 0$$ are real , Then which one is Right.
$\bf{Options}::$ $\displaystyle (a)\;\; \lambda <\frac{4}{3}\;\;\;\;\;\; (b)\; \lambda >\frac{4}{3}\;\;\;\;\;\; (c)\; \lambda \in \left(\frac{1}{3},\frac{5}{3}\right)\;\;\;\;\;\; (d)\;\; \lambda \in \left(\frac{4}{3},\frac{5}{3}\right)$
$\bf{My\; Try}::$ If given equation has real roots , Then its $\bf{Discriminant}\geq 0$
So $$\displaystyle 4(a+b+c)^2-12\lambda\cdot (ab+bc+ca)\geq 0$$
$$\displaystyle (a+b+c)^2-3\lambda\cdot (ab+bc+ca)\geq 0$$
$$\displaystyle 3\lambda\leq \frac{(a+b+c)^2}{(ab+bc+ca)} = \frac{a^2+b^2+c^2}{(ab+bc+ca)}+2$$
Now I did not understand how can i solve after that
Help Required
Thanks
| Use that $(a,b,c)$ forms a triangle, hence $$a \leq b+c \\ b \leq c+a \\ c\leq a+b$$ showing $$2(ab+bc+ca) = a(b+c)+b(c+a)+c(a+b) \geq a^2+b^2+c^2.$$
Thus $\lambda \leq \frac{4}{3}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/618036",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Taylor expansion of $e^{\cos x}$ I have to find the 5th order Taylor expansion of $e ^{\cos x}$. I know how to do it by computing the derivatives of the function, but the 5th derivative is about a mile long, so I was wondering if there is an easier way to do it.
I'd appreciate any help.
| You can start with the Taylor expansion of $\cos x$ and when you expand
$\exp(\cos x)$, you just throw away terms you know won't affect the final result.
For lower order Taylor expansion, the derivation is actually pretty short and straight forward.
$$\begin{align}
\exp(\cos x)
&= \exp\left(1 - \frac{x^2}{2} + \frac{x^4}{24} + O(x^6)\right)\\
&= e\left[1 + \left(-\frac{x^2}{2} + \frac{x^4}{24} + O(x^6)\right) + \frac{1}{2}\left( -\frac{x^2}{2} + O(x^4)\right)^2 + O(x^6) \right]\\
&= e\left[ 1 - \frac{x^2}{2} + \left(\frac{1}{24} + \frac{1}{8}\right) x^4 \right] + O(x^6)\\
&= e\left[ 1 - \frac{x^2}{2} + \frac{x^4}{6} \right] + O(x^6)
\end{align}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/620065",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 6,
"answer_id": 1
} |
evaluation of $\lim_{x\rightarrow \infty}\frac{\ln x^n-\lfloor x \rfloor }{\lfloor x \rfloor} = $ (1) $\displaystyle \lim_{x\rightarrow \infty}\frac{\ln x^n-\lfloor x \rfloor }{\lfloor x \rfloor} = $, where $n\in \mathbb{N}$ and $\lfloor x \rfloor = $ floor function of $x$
(2)$\displaystyle \lim_{x\rightarrow \infty}\left({\sqrt{\lfloor x^2+x \rfloor }-x}\right) = , $where $\lfloor x \rfloor = $ floor function of $x$
$\bf{My\; Try}::$ for (1) one :: We can write as $\displaystyle \lim_{x\rightarrow \infty}\frac{n\cdot \ln x-\lfloor x \rfloor }{\lfloor x \rfloor}$
and we can say that when $x\rightarrow \infty$, Then $\lfloor x\rfloor \rightarrow x$
So $\displaystyle \lim_{x\rightarrow \infty}\frac{n\cdot \ln(x)-x}{x} = n\lim_{x\rightarrow \infty}\frac{\ln (x)}{x}-1$
Now Let $\displaystyle L = \lim_{x\rightarrow \infty}\frac{\ln(x)}{x}{\Rightarrow}_{L.H.R} =\lim_{x\rightarrow \infty}\frac{1}{x} = 0$
So $\displaystyle \lim_{x\rightarrow \infty}\frac{n\cdot \ln x-\lfloor x \rfloor }{\lfloor x \rfloor} = n\cdot 0-1 =-1$
$\bf{My\; Try}::$ for (2)nd one::we can say that when $x\rightarrow \infty$, Then $\lfloor x^2+x\rfloor\rightarrow (x^2+x)$
So $\displaystyle \lim_{x\rightarrow \infty}\left({\sqrt{x^2+x}-x}\right) = \lim_{x\rightarrow \infty}\frac{\left({\sqrt{x^2+x}-x}\right)\cdot \left({\sqrt{x^2+x}+x}\right)}{\left({\sqrt{x^2+x}+x}\right)}$
$\displaystyle \lim_{x\rightarrow \infty}\frac{x}{\left(\sqrt{x^2+x}+x\right)} = \frac{1}{2}$
Now my doubt is can we write when $x\rightarrow \infty$, Then $\lfloor x\rfloor \rightarrow x$
and when $x\rightarrow \infty$, Then $\lfloor x^2+x\rfloor\rightarrow (x^2+x)$
please clear me
Thanks
| Since $x-1\leq \lfloor x\rfloor \leq x$ we have $$ \frac{n\log{x}}{x}\leq \frac{n\log{x}}{\lfloor x\rfloor}\leq \frac{n\log{x}}{x-1}$$ for $x>1$. If the left and right converge to zero as $x\rightarrow \infty$ then the center does by the squeeze lemma. However, the left and right are of the form $\frac{\infty}{\infty}$ so by L'Hospitals rule they converge to zero. Your limit is the middle part minus $1$ as $x\rightarrow \infty$ which is minus $1$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/620152",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 2
} |
How prove this stronger than Weitzenbock's inequality:$(ab+bc+ac)(a+b+c)^2\ge 12\sqrt{3}\cdot S\cdot(a^2+b^2+c^2)$ In $\Delta ABC$,$$AB=c,BC=a,AC=b,S_{ABC}=S$$
show that
$$(ab+bc+ac)(a+b+c)^2\ge 12\sqrt{3}\cdot S\cdot(a^2+b^2+c^2)$$
I know this Weitzenböck's_inequality
$$a^2+b^2+c^2\ge 4\sqrt{3}S$$
But my inequality is stronger than this Weitzenbock's inequality.
my try:
let the semiperimeter, inradius, and circumradius be $s,r,R$ respectively
$$a+b+c=2s,ab+bc+ac=s^2+4Rr+r^2,S=rs$$
$$\Longleftrightarrow (s^2+4Rr+r^2)4s^2\ge 12\sqrt{3}\cdot rs[4s^2-2(s^2+4Rr+r^2)]$$
$$\Longleftrightarrow s^3+4Rrs+r^2s\ge 6\sqrt{3}rs^2-24\sqrt{3}Rr^2-6\sqrt{3}r^3$$
and use this Gerretsen inequality:
$$r(16R-5r)\le s^2\le 4R^2+4Rr+3r^2$$
and Euler inequality
$$R\ge 2r$$
But seems is not usefull,
Thank you
| Your inequality is equivalent to $12\sqrt{3}S \le f(a,b,c)$ where
$$f(a,b,c)=\frac{(a+b+c)^2(ab+bc+ac)}{a^2+b^2+c^2}.$$
One link of Hadwiger's inequality (credit to Macavity for the link) says that $12\sqrt{3}S \le g(a,b,c)$ where, defining as in the statement there $Q=(a-b)^2+(b-c)^2+(c-a)^2,$ $g$ is defined as
$$g(a,b,c)=(a+b+c)^2-2Q.$$
This latter inequality implies yours, as we will argue. It is also strictly stronger for any nonequilateral triangle.
Now since $f,g$ are each symmetric and homogeneous we may assume that $a=1,b=s,c=t$ and that $s,t\ge 1.$ (This amounts to taking $a$ as the shortest side.) The $g$ bound is better if we show $f-g \ge 0.$ We have after algebra (I used maple here) an expression whose numerator factors into two second degree terms in $s,t.$
$$f(1,s,t)-g(1,s,t)=\frac{AB}{s^2+t^2+1},$$
where $A,B$ are factors to be analyzed. It turns out that
$$A=(s-t)^2+(s-1)(t-1), \\ B=3(s-t)^2+(2s-1)(2t-1)+2.$$
Since $s,t \ge 1$ the factor $B$ is positive in any case, while the factor $A$ is nonnegative. Note that the only way for $A$ to be zero is if $s=t$ because of the square term, and then $s=t=1$ since the term $(s-1)(t-1)=(s-1)^2$ occurs as a summand. This gives, assuming the two bounds $f,g$ agree on a given triangle, the case of an equilateral triangle.
Conclusion: Your inequality does indeed hold, but it seems more involved to state than the (known) Hadwiger inequality which is in fact stronger.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/621099",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
if range of $f(x) = \frac{x^2+ax+b}{x^2+2x+3}$ is $[-5,4]$. Then $a$ and $b$ are If Range of $\displaystyle f(x) = \frac{x^2+ax+b}{x^2+2x+3}$ is $\left[-5,4\; \right]$ for all $\bf{x\in \mathbb{R}}$. Then values of $a$ and $b$.
$\bf{My\; Try}::$ Let $\displaystyle y=f(x) = \frac{x^2+ax+b}{x^2+2x+3} = k$,where $k\in \mathbb{R}$.Then $\displaystyle kx^2+2kx+3k=x^2+ax+b$
$\Rightarrow (k-1)x^2+(2k-a)x+(3k-b) = 0$
Now we will form $2$ cases::
$\bf{\bullet}$ If $(k-1)=0\Rightarrow k=1$, Then equation is $(2-a)x+(3-b)=0$
$\bf{\bullet}$ If $(k-1)\neq 0\Rightarrow k\neq 1$ means either $k>1$ or $k<1$
How can i solve after that
Help Required
Thanks
| Note that $x^2+2x+3\gt 0 $ for any $x\in\mathbb R$.
Hence, we have
$$-5(x^2+2x+3)\le x^2+ax+b\le 4(x^2+2x+3).$$
You can simplify this, and you have two quadratic inequality which will be easy to solve.
You'll have two inequality.
$$6x^2+(a+10)x+b+15\ge0$$
$$3x^2+(8-a)x+12-b\ge0.$$
This leads that each discriminant has to be equal or greater than $0$. So, we have
$$(a+10)^2-4\cdot 6\cdot (b+15)\le 0$$
$$(8-a)^2-4\cdot 3\cdot (12-b)\le 0.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/621512",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 0
} |
Integral: $I=\int\limits_{0}^{1}\dfrac{x^2dx}{\sqrt{3+2x-x^2}}$ Evaluate: $\displaystyle I=\int\limits_{0}^{1}\dfrac{x^2dx}{\sqrt{3+2x-x^2}}$
| As $\displaystyle\frac{x^2}{3+2x-x^2}=\frac{x^2}{2^2-(x-1)^2}$ and we are dealing with definite integral
I would like to use Trigonometric substitution $\displaystyle x-1=2\sin u$ from the very start as we don't need to get back to $x$ again
When $\displaystyle x=0,\sin u=-\frac12\implies u=-\frac\pi6$ and when $\displaystyle x=1,\sin u=0\implies u=0$ based on the definition of the principal value of sine inverse
$\displaystyle\implies\cos u>0 $
$$\int_0^1\frac{x^2}{\sqrt{3+2x-x^2}}dx=\int_0^{-\frac\pi6}\frac{(1+2\sin u)^2}{+2\cos u}2\cos udu=\int_0^{-\frac\pi6}(1+2\sin u)^2du$$
Now, $\displaystyle(1+2\sin u)^2=1+4\sin u+4\sin^2u=1+4\sin u+2(1-\cos2u)$ (using $\cos2A=1-2\sin^2A$)
$\displaystyle\implies(1+2\sin u)^2=3+4\sin u-2\cos2u$
Can you take it home from here?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/623754",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Find all integer solutions: $x^4+x^3+x^2+x=y^2$ Find all integer solutions of the following equation:
$$x^4+x^3+x^2+x=y^2$$
| EDIT : The case (1) in my answer has a big mistake. The case (2) is true.
$(x,y)=(0,0)$ is one of the solutions. In the following, suppose that $(x,y)\not=(0,0).$
We have
$$x(x+1)(x^2+1)=y^2.$$
First, we know that $x$ and $x+1$ are coprime, and that $x$ and $x^2+1$ are coprime, too. This leads that there is an integer $k$ such that $x=k^2.$ (see the right hand side)
In the following, we break up the answer into two cases.
(1) The case that $x+1$ and $x^2+1$ are not coprime.
Letting $$x+1=ps,x^2+1=pt\ \ (\text{$p$ is a prime number, $s,t\in\mathbb Z$})$$
we have
$$(ps-1)^2+1=pt\iff p(t-ps^2+2s)=2\Rightarrow p=2.$$
So, we know we only have the following possibilities :
$$x+1=0,\pm1, \pm2^k\Rightarrow x=-1,0,-2,-1\pm2^k.$$
However, we have $x\not=-1, x\not=0, x\not=-2$, because $x$ has to be a square of an integer (and $x\not=0$). So, we have $$x=-1\pm2^k.$$ In the same argument, we have $$x^2=-1\pm2^l.$$ Note that $k,l\ge1\in\mathbb Z.$
Hence, we have
$$(-1\pm2^k)^2=-1\pm2^l\iff 2^{2k}\mp2^k+2=\pm2^l\iff 2^{2k-1}\mp2^{k-1}+1=\pm2^{l^1}.$$
Now, if $k-1\ge1$ and $l-1\ge1$, then we know that the left hand side is odd, and the right hand side is even, which is a contradiction.
So, we have "$k=1$ or $l=1$".
This leads $x=\pm2^k-1=\pm2^1-1\Rightarrow x=1,-3\Rightarrow x=1.$ ($x^2$ has to be a square.) Also, this leads $x^2=\pm2^l-1=\pm2^l-1\Rightarrow x^2=1,-3\Rightarrow x=1.$
So, this case leads $(x,y)=(1,\pm 2).$
(2) The case that $x+1$ and $x^2+1$ are coprime.
In this case, we know any two of $x,x+1,x^2+1$ are coprime. So, this leads that each of $x, x+1,x^2+1$ has to be a square number of an integer. However, it is impossible that $x$ and $x+1$ are both square numbers except $x=0$. However, we suppose that $x\not=0$ at the top. Hence, this case has no solution.
Now we reach a conclusion that the solutions of the given equation is only the followings:
$$(x,y)=(0,0),(1,2),(1,-2).$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/624792",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
"answer_id": 3
} |
Prove $ \cos\left(\frac{2\pi}{5}\right) + \cos\left(\frac{4\pi}{5}\right) + \cos\left(\frac{6\pi}{5}\right)+ \cos\left(\frac{8\pi}{5}\right) = -1 $ $$
\cos\left(\frac{2\pi}{5}\right) + \cos\left(\frac{4\pi}{5}\right) + \cos\left(\frac{6\pi}{5}\right)+ \cos\left(\frac{8\pi}{5}\right) = -1
$$
Wolframalpha shows that it is a correct identity, although I can't prove it.
I've tried to use the formula
$$ \cos(z) = \frac{e^ {iz} - e^ {iz}}{2} $$ but without any satisfying result.
This exercise is from chapter on series.
EDIT: I corrected a mistake in the formula I wanted to use.
| Why not use trig?
for $$\cos\left(\frac{2\pi}{5}\right)+ \cos\left(\frac{4\pi}{5}\right) + \cos\left(\frac{6\pi}{5}\right) + \cos\left(\frac{8\pi}{5}\right) =-1.$$
First of notice that if $a = \frac{2\pi}{5}$ then we have
$$\cos(a) + \cos(2a) + \cos(3a)+ \cos(4a) =-1.$$
Knowing the identity that $$\cos(2a) = 2\cos^2(a) -1$$ we can directly derive:
$$\cos(3a) = 4 \cos^3(a)-3\cos(a)$$ and
$$\cos(4a) = 8 \cos^4(a) - 8\cos^2(a) +1$$
Also we know that $\cos\left(\frac{2\pi}{5}\right) = \frac{1}{5^{1/2} + 1}$ which is $\color{red}{1/2 1/ \phi}$ the golden ratio
Yes there is a lot arithmetic, but once you have gotten to the arithmetic the problem is solved...just use care in expanding the powers. I worked it out and proved the problem, using the above scheme. Look once you know the key trig identity for $\cos(2a)$ which is $\cos( a+a)$ you can very simply derive the other identities for $\cos(3a) = \cos(2a + a)$ and $\cos(4a) = \cos(3a + a)$ or $\cos( 2a + 2a)$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/625101",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 8,
"answer_id": 7
} |
Prove that $\lim\limits_{n \rightarrow \infty} \left(\frac{23n+2}{4n+1}\right) = \frac{23}{4} $. My attempt:
We prove that $$\displaystyle \lim\limits_{n \rightarrow \infty} \left(\frac{23n+2}{4n+1}\right) = \frac{23}{4} $$
It is sufficient to show that for an arbitrary real number $\epsilon\gt0$, there is a $K$
such that for all $n\gt K$, $$\left| \frac{23n+2}{4n+1} - \frac{23}{4}
\right| < \epsilon. $$
Note that $$ \displaystyle\left| \frac{23n+2}{4n+1} - \frac{23}{4} \right| = \left| \frac{-15}{16n+4} \right| $$ and for $ n > 1 $ $$ \displaystyle \left| \frac{-15}{16n+4} \right| = \frac{15}{16n+4} < \frac{1}{n}. $$
Suppose $ \epsilon \in \textbf{R} $ and $ \epsilon > 0 $. Consider $ K = \displaystyle \frac{1}{\epsilon} $. Allow that $ n > K $. Then $ n > \displaystyle \frac{1}{\epsilon} $. So $ \epsilon >\displaystyle \frac{1}{n} $.
Thus
$$ \displaystyle\left| \frac{23n+2}{4n+1} - \frac{23}{4} \right| = \left| \frac{-15}{16n+4} \right| = \frac{15}{16n+4} < \frac{1}{n} < \epsilon. $$ Thus $$ \displaystyle \lim\limits_{n \rightarrow \infty} \left(\frac{23n+2}{4n+1}\right) = \frac{23}{4}. $$
Is this proof correct? What are some other ways of proving this? Thanks!
| Your proof is basically correct, but I would encourage you to practice a bit on articulating exactly what you mean. Where you say
It is sufficient to show that $$\left|
\frac{23n+2}{4n+1} - \frac{23}{4} \right| < \epsilon $$
you mean to say something like
It is sufficient to show that for all $\epsilon\gt0$, there is a $K$
such that for all $n\gt K$, $$\left| \frac{23n+2}{4n+1} - \frac{23}{4}
\right| < \epsilon $$
As is, the $\epsilon$ comes out of nowhere and there's no stated restriction on $n$, so the inequality that is "sufficient" to show could be trivially true or patently false.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/631462",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 1
} |
Evaluating $\prod_{n=2}^\infty {n^3-1\over n^3+1}$
The value of the infinite product $$P = \frac 79 \times \frac{26}{28} \times \frac{63}{65} \times \cdots \times \frac{n^3-1}{n^3+1} \times \cdots$$ is
(A) $1$
(B) $2/3$
(C) $7/3$
(D) none of the above
I wrote first 6 terms and tried to cancel out but did not get any idea what will be the last term
I did one same kind of problem $\prod (1-{1\over k^2}),k\ge2$ whose answer is $1\over 2$
here I think the answer will be $D$, none of this?
| HINT:
Observe that $$n^3-1=(n-1)(n^2+n+1)\text{ and }n^3+1=(n+1)(n^2-n+1)$$
Again, $\displaystyle (n+1)^2-(n+1)+1=\cdots=n^2+n+1$
Just set a few values of $n$ to find the surviving terms
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/634973",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 2
} |
Probability that the second roll comes up yellow given the first roll was purple. A bag contains $20$ dice. $5$ of the dice have entirely purple sides, $7$ of the dice have $2$ purple and $4$ yellow sides, and $8$ of the dice have $3$ purple and $3$ yellow sides. If you randomly pick a die, roll it, and observe that the roll comes up purple, what is the probability that if you roll the same die again, the roll comes up yellow?
Update I have tried the following: The probability that you pick die 1 and roll a purple is $5/20*6/6=5/20$; the probability you pick die 2 and roll a purple is $7/20*2/6=7/60$; and the probability you pick die 3 and roll a purple is $8/20*3/6=1/5$. The sum of these probabilities is $5/20+7/60+1/5=17/30$. Now the probability that the second roll is yellow given the first is purple is given by: $(5/20)\div(17/30)*0+(7/60)\div(17/30)*(4/6)+(1/5)\div(17/30)*(3/6)=.31$. This is what I think is right; can someone verify it or point out where it is wrong if it is?
| The probability of picking the first type of die with all purple sides equals $\Pr(T=1) = \frac{5}{20} = \frac{1}{4}$. The probability of picking the second type of die with 2 purple and 4 yellow sides equals $\Pr(T=2) = \frac{7}{20}$. The probability of picking the third type of die with 3 purple and 3 yellow sides equals $\Pr(T=3)=\frac{8}{20} = \frac{2}{5}$.
Having picked the type of die, the outcome of the second roll is independent of the first outcome. Hence:
$$
\Pr(\mathcal{O}_2 = Y) = \Pr(T=1) \cdot \frac{0}{6} + \Pr(T=2) \frac{4}{6} + \Pr(T=3) \frac{3}{6} = \frac{13}{30} = \frac{2}{5} + \frac{1}{30} = 0.4(3)
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/636171",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Solve $12x\equiv9\pmod{15}$ Question:
Solve $12x\equiv9\pmod{15}$
My try:
$\gcd(12,15)=3$ so it has at least $3$ solutions.
Now
$15=12\times1+3\\
3=15-12\times 1\\
3=15+2\times(-1)\\
\implies9=15\times3+12\times(-3)\\
\implies12\times(-3)\equiv9\pmod{15}$
So $x\equiv-3\pmod{15}$
Am I correct?
| $\begin{eqnarray}{\bf Hint}\quad 12x\equiv 9\!\!\!\pmod{15} &\iff& 15\mid 12x-9\\ &\iff& \dfrac{12x-9^{\phantom I}}{15} = \dfrac{4x-3}5\in\Bbb Z\\ \\ &\iff& 5\mid 4x-3\\ \\ &\iff& 4x\equiv3\!\!\!\pmod 5 \end{eqnarray}$
Hence $\,{\rm mod}\ 5\!:\,\ 3 \equiv 4x\equiv -x\ \Rightarrow\ x \equiv -3\equiv 2\ $ hence
$$\ x = 2 + 5n = \ldots,-3,2,7,12,17\ldots \equiv\, 2,7,12\!\!\pmod{15}$$
Remark $\ $ Notice in particular how one needs to cancel the gcd $(12,9) = 3$ from the modulus too, and how this is explained above by putting the associated fraction into lowest terms.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/637041",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
(Highschool Pre-calculus) Solving quadratic via completing the square I'm trying to solve the following equation by completing the square:
$x^2 - 6x = 16$
The correct answer is -6,1. This is my attempt:
$x^2 - 6x = 16$
$(x - 3)^2 = 16$
$(x - 3)^2 = 25$
$\sqrt(x -3)^2 = \sqrt(25)$
$x - 3 = \pm5$
$x =\pm5 - 3$
$x = -8,2$
I did everything according to what I know,but my answer was obviously wrong. Any help is appreciated. Thanks
| I am just going to do the thing first and we will talk about what you did after.
Note that $x^2-6x\sim (x-3)^2$. In fact
$$\begin{align}(x-3)^2=x^2-6x+9&=(x^2-6x)+9\\\Rightarrow x^2-6x&=(x-3)^2-9.\end{align}$$
Therefore if
$$x^2-6x=16$$ then
$$\begin{align}(x-3)^2-9&=16
\\\Rightarrow (x-3)^2&=25
\\ \Rightarrow x-3&=\pm5
\\ \Rightarrow x&=8\text{ or }-2.
\end{align}$$
The problems with what you did are as follows:
*
*If $x^2-6x=16$ it does not follow that $(x-3)^2=16$.
*You don't need to take the square roots of both sides necessarily. If $x^2=a$ then $x=\pm \sqrt{a}$ is a perfectly good implication.
*At the end you have two equations... $x-3=5$ and $x-3=-5$. To get rid of the three, remember you want $x=\dots$ add three to both sides, e.g.
$$\begin{align}
x-3&=5
\\\Rightarrow x-3+3&=5+3
\\\Rightarrow x+0&=8
\\ \Rightarrow x&=8.
\end{align}$$
You can do this because if two quantities are equal, and you add the same thing to both of them, then they are still equal.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/641599",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 1
} |
How to contruct such a sequence of rational? How to order all rational numbers from $(0,1)$ in a sequence $(x_n)_{n=1}^\infty$ in such a way that $$|x_n-x_k| \geq \frac{1}{(n+1)^2}$$ for $k<n$ ?
| It seems that $\dfrac{1}{(n+1)^2}$ can be improved. If we do the first thing that comes to our mind:
$$
\frac{1}{2},\,\,\,\frac{1}{3},\frac{2}{3},\,\,\,\frac{1}{4},\frac{3}{4},\,\,\,
\frac{1}{5},\frac{2}{5},\frac{3}{5},\frac{4}{5},\,\,\,
\frac{1}{6},\frac{5}{6},\,\,\,\frac{1}{7},\frac{2}{7},\frac{3}{7},\frac{4}{7},
\frac{5}{7},\frac{6}{7},\ldots
$$
then $x_n$ is a fraction with denominator at $n$, and if $k<n$, then $x_k$ has numerator either $n+1$ or less than $n+1$. If its numerator is $n+1$, then $|x_k-x_n|\ge\frac{1}{n+1}$, whereas if it is less than $n$, then
$$
|x_n-x_k|\ge\frac{1}{n(n+1)}.
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/642638",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
} |
Prove $\int \limits_0^b x^3 = \frac{b^4}{4} $ by considering partitions $[0, b]$ in $n$ equal subinvtervals. I was given this question as an exercise in real analysis class. Here is what I came up with. Any help is appreciated!
Prove $\int \limits_0^b x^3$ = $\frac{b^4}{4} $ by considering partitions [0, b] in $n$ equal subinvtervals.
Consider $f$ on the interval $[0, b]$ where $b > 0$. For a partition $ P = \{0=t_0 < t_1 < ...< t_n = b\}$ we have:
$U(f, P) = $$\sum_{k=1}^n t^3_k (t_k - t_1) $
If we choose $t_k = \frac{kb}{n} $ then we can say
$U(f, P) = $$\sum_{k=1}^n \frac{k^3b^3}{n^3}\ . \frac{b}{n}$
=> $U(f, P) = \frac{b^4}{n^4} \sum_{k=1}^n k^3$
=> $U(f, P) = \frac{b^4}{n^4}\ . \frac{n^2(n + 1)}4$
so $L(f) \ge \frac{b^4}{n^4}$ . Therefore $f(x) = x^3$ is integrable on $[0, b]$ and $\int \limits_0^b x^3$ = $\frac{b^4}{4} $.
| You can consider the Riemann sums. For instance, the left Riemann sum
$$ \sum_{i=0}^{n-1} f(x_i) \Delta x_i = \sum_{i=0}^{n-1} \left(\frac{b}{n}i\right)^3\frac{b}{n} = \frac{b^4}{n^4}\sum_{i=0}^{n-1} i^3 = \frac{b^4}{n^4}\left({n}^{4}/4 - {n}^{3}/2+ {n}^{2}/4\right)\longrightarrow_{n\to \infty} \frac{b^4}{4} $$
To find the last sum, see here.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/645932",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Find the volume between two surfaces
Find the volume between $z=x^2$ and $z=4-x^2-y^2$
I made the plot and it looks like this:
It seems that the projection over the $xy$-plane is an ellipse, because if $z=x^2$ and $z=4-x^2-y^2$ then $2x^2+y^2=4$ which means that $\displaystyle\frac{x^2}{(\sqrt{2})^2}+\displaystyle\frac{y^2}{2^2}=1$
Stewart define a region I if it is of the kind $\{(x,y,z):(x,y)\in D, u_1(x,y)\leq z \leq u_2(x,y)\}$.
I believe that the region I'm asked for cab be describe setting $D=\{(x,y): \displaystyle\frac{x^2}{(\sqrt{2})^2}+\displaystyle\frac{y^2}{2^2}=1\}$ and then $E=\{(x,y,z):(x,y)\in D, x^2\leq z \leq 4-x^2-y^2\}$.
Can the volume $V(E)$ be computed by $\displaystyle\int\displaystyle\int_D\displaystyle\int_{x^2}^{4-x^2-y^2}z\;dz$?
Or maybe considering that $0\leq \sqrt{x} \leq \sqrt{2}$ and $0 \leq y \leq \sqrt{4-x^2-y^2}$ could I compute the volume calculating $\displaystyle\int_0^{\sqrt{2}}\displaystyle\int_0^{\sqrt{4-2x^2}}\displaystyle\int_{x^2}^{4-x^2-y^2}z\;dzdydx ?$
The issue with the approach above is that doesn't seem too easy after the first two integrals, because:
$\displaystyle\int_0^{\sqrt{2}}\displaystyle\int_0^{\sqrt{4-2x^2}}\displaystyle\int_{x^2}^{4-x^2-y^2}z\;dz = \displaystyle\frac{1}{2} \displaystyle\int_0^{\sqrt{2}}\displaystyle\int_0^{\sqrt{4-2x^2}} [(4-x^2-y^2)^2-x^2]\; dydz \\
=\displaystyle\frac{1}{2} \displaystyle\int_0^{\sqrt{2}} \left(16y-7x^2y-\frac{8}{3}y^3+\frac{2}{3}x^2y^3+x^4y+\frac{1}{5}y^5\right)\rvert_0^{\sqrt{4-2x^2}}\;dx$ which after the substitution seems hard to evaluate.
Is there an easy way?
| I agree with Fantini that you do not need the $z$ inside the integral. Generally, when you see a lot of squares, especially $x^2+y^2$, then you should be thinking about switching to polar/cylindrical coordinates. This will get the answer for this integral, though since the boundary is an ellipse it is not quite so straight-forward an integral. The equation of the ellipse becomes
$2r^2\cos^2\theta+r^2\sin^2\theta=4 \Longrightarrow r^2 = \dfrac{4}{\sin^2\theta+2\cos^2\theta} = \dfrac{4}{1+\cos^2\theta}$
So I get the following result:
$\begin{align}
\iint 4 -2x^2 -y^2 \, dA &= \iint 4-(x^2+y^2) - x^2 dA \\
& = 4\int_0^{\pi/2} \hspace{-5pt} \int_0^{\sqrt{\frac{4}{1+\cos^2\theta}}} (4-r^2-r^2\cos^2\theta) \, r \, dr \, d \theta \\
&= 4\int_0^{\pi/2} \hspace{-5pt} \int_0^{\sqrt{\frac{4}{1+\cos^2\theta}}} (4r-r^3-r^3\cos^2\theta) \, dr \, d \theta \\
&= 4\int_0^{\pi/2} \hspace{-5pt} (2r^2-\frac{1}{4}r^4-\frac{1}{4}r^4\cos^2\theta)\bigg\vert_0^{\sqrt{\frac{4}{1+\cos^2\theta}}} \, d \theta \\
& = 4\int_0^{\pi/2} \dfrac{8}{1+\cos^2\theta} - \dfrac{4}{(1+\cos^2\theta)^2} - \dfrac{4\cos^2\theta}{(1+\cos^2\theta)^2} \, d \theta \\
& = 4\int_0^{\pi/2} \dfrac{8+8\cos^2\theta}{(1+\cos^2\theta)^2} - \dfrac{4+4\cos^2\theta}{(1+\cos^2\theta)^2} \, d \theta \\
& = \int_0^{\pi/2} \dfrac{16+16\cos^2\theta}{(1+\cos^2\theta)^2} \, d \theta \\
& = \int_0^{\pi/2} \dfrac{16}{1+\cos^2\theta} \, d \theta \\
&= \int_0^{\pi/2} \dfrac{16}{\sin^2\theta+2\cos^2\theta} \, d \theta\\
& = \int_0^{\pi/2} \dfrac{16}{\sin^2\theta+2\cos^2\theta}\cdot\dfrac{\frac{1}{\cos^2\theta}}{\frac{1}{\cos^2\theta}} \, d \theta \\
&= \int_0^{\pi/2} \dfrac{16\sec^2\theta}{\tan^2\theta+2} \, d \theta \\
&= \int_0^{\infty} \dfrac{16}{u^2+(\sqrt{2})^2} \, d u \\
&= \dfrac{16}{\sqrt{2}} \tan^{-1}\left(\dfrac{u}{\sqrt{2}}\right) \bigg\vert_0^{\infty} = \dfrac{16}{\sqrt{2}}\cdot\dfrac{\pi}{2} = 4\sqrt{2}\pi
\end{align}$
Note that I've done a bit of hand-waving at the end to deal with the improper integral and that the above solutions agrees with a computer calculation of the answer.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/650264",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 2,
"answer_id": 1
} |
how to find $ \lim\limits_{x \to \infty} \left(\sqrt{x^2 +1} +\sqrt{4x^2 + 1} - \sqrt{9x^2 + 1}\right)$ How can I find this?
$ \lim\limits_{x \to \infty} \left(\sqrt{x^2 +1} +\sqrt{4x^2 + 1} - \sqrt{9x^2 + 1}\right)$
| Here is another tack. If $a > 0$,
$${\sqrt{a^2 x^2 + 1} - ax } = {1\over{\sqrt{a^2 x^2 + 1} + ax }}= O\left
({1\over x}\right). $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/651148",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 1
} |
Find the value of the expression The expression $ax^2 + bx + 1$ takes the values $1$ and $4$ when $x$ takes the values $2$ and $3$ respectively. How can we find the value of the expression when $x$ takes the value of $4$?
| Let $f(x) = ax^2 + bx + 1$
Plug the values in $f(x)$ and we have:
$$f(2) = a(2)^2 + 2b + 1 = 4a + 2b + 1 = 1$$
$$f(3) = a(3)^2 + 3b + 1 = 9a + 3b + 1 = 4$$
Then just solve the following system of equations to find the values for $a$ and $b$, which is fairly easy:
\begin{cases} 4a + 2b = 0\\ 9a + 3b = 3 \end{cases}
Then just plug $x=4$ in $f(x)$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/652269",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
Proving that $\frac{e^x + e^{-x}}2 \le e^{x^2/2}$
Prove the following inequality:
$$\dfrac{e^x + e^{-x}}2 \le e^{\frac{x^2}{2}}$$
This should be solved using Taylor series.
I tried expanding the left to the 5th degree and the right site to the 3rd degree, but it didn't help.
Any tips?
| $$\frac{e^x+e^{-x}}{2} = \frac{\left(1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \cdots \right) + \left(1 - x + \frac{x^2}{2!} - \frac{x^3}{3!} + \cdots \right)}{2} = \frac{2 + 2 \frac{x^2}{2!} + 2 \frac{x^4}{4!} + \cdots}{2}$$
$$ = 1 + \frac{x^2}{2} + \frac{x^4}{24} + \frac{x^6}{720}+ \cdots$$
$$e^{\frac{x^2}{2}} = 1 + \frac{x^2}{2} + \frac{\left(\frac{x^2}{2}\right)^2}{2!} + \frac{\left(\frac{x^2}{2}\right)^3}{3!} + \cdots = 1 + \frac{x^2}{2} + \frac{x^4}{8} + \frac{x^6}{48} + \cdots$$
Basically, it comes down to comparing $\displaystyle \frac{1}{(2n)!}$ and $\displaystyle \frac{1}{2^n n!}$, because they are the coefficients for $\displaystyle \frac{e^x + e^{-x}}{2}$ and $e^{\frac{x^2}{2}}$, respectively. We note that $$\frac{1}{(2n)!} = \frac{1}{1 \cdot 2 \cdot 3 \cdots (2n-1) \cdot (2n)}$$ while $$\frac{1}{2^n n!} = \frac{1}{2 \cdot 4 \cdot 6 \cdots (2n-2) \cdot (2n)}$$
Because the denominator for the second one is obviously smaller than that of the first, the second fraction is bigger than the first. This applies to every $n$, and therefore $$\frac{e^x+e^{-x}}{2} \le e^{\frac{x^2}{2}}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/654839",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 3,
"answer_id": 1
} |
Integrating $\int_0^\infty\frac{1}{1+x^6}dx$ $$I=\int_0^\infty\frac{1}{1+x^6}dx$$
How do I evaluate this?
| Without complex analysis nor special functions:
$$x^6+1=(x^2+1)(x^4-x^2+1)=(x^2+1)(x^2+\sqrt3x +1)(x^2-\sqrt3x+1).$$
($x=\pm i$ are obviously roots and $(x^4-x^2+1)$ is biquadratic)
And now you have the following partial fraction decomposition:
$${1\over x^6+1} =
{Ax+B\over x^2+1} + {Cx+D\over x^2+\sqrt3x +1} + {Ex+F\over x^2 -\sqrt3x + 1}.$$
$\cdots$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/658637",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 7,
"answer_id": 5
} |
Sum from $k=1$ to $n$ of $k^3$ $$\sum_{k=1}^n k^3 = \left(\frac{1}{2}n(n+1) \right)^2$$
I want to prove this using induction. I start with $(\frac{n}{2}(n+1))^2 + (n+1)^3$ and rewrite $(n+1)^3$ as $(n+1)(n+1)^2$, then factor out an $(n+1)^2$ from the expression:
$(n+1)^2((\frac{n}{2})^2 + (n+1))$
I'm confused where to go from here though.
| $(n+1)^2((\frac{n}{2})^2 + (n+1))=(n+1)^2\frac{n^2+4n+4}{4}=(n+1)^2\frac{(n+2)^2}{2^2}= \left(\frac{1}{2}(n+1)((n+1)+1) \right)^2$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/658956",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Prove trig identity: $\csc \tan - \cos = \frac{\sin^2}{\cos}$ I keep hitting seeming dead-ends.
\begin{align*}
\csc\ x \tan\ x - \cos\ x &= \left(\frac{1}{\sin\ x}\right)\left(\frac{\sin\ x}{\cos\ x}\right) - \cos\ x \\
&= \frac{\sin\ x}{(\sin\ x)(\cos\ x)} - \cos\ x \\
&= \frac{\sin\ x}{(\sin\ x)(\cos\ x)} - \frac{(\cos\ x)(\sin\ x)(\cos\ x)}{(\sin\ x)(\cos\ x)} \\
&= \frac{(\cos^2 x)(\sin\ x)}{(\sin\ x)(\cos\ x)} \\
&= \cos\ x
\end{align*}
Thank you!
| Your method was also not wrong,
L.H.S=$\frac{sinx}{sinxcosx}$-$\frac{cosxsinxcosx}{sinxcosx}$
= $\frac{sinx-cos^2xsinx}{sinxcosx}$
= $\frac{sinx(1-cos^2)}{sinxcosx}$
= $\frac{1-cos^2}{cosx}$
= $\frac{sin^2}{cosx}$=R.H.S
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/661495",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Finding $\lim_{x \to 0} \frac {a\sin bx -b\sin ax}{x^2 \sin ax}$ witouth L'Hopital, what is my mistake? I was working on this question.
$\lim_{x \to 0} \dfrac {a\sin bx -b\sin ax}{x^2 \sin ax}$
$\lim_{x \to 0} \dfrac {1}{x^2} \cdot \lim_{x \to 0} \dfrac { \frac {1}{abx}}{\frac {1}{abx}} \cdot \dfrac {a\sin bx -b\sin ax}{\sin ax}$
$\lim_{x \to 0} \dfrac {1}{x^2} \cdot \dfrac {\lim_{x \to 0} \frac {\sin bx}{bx}- \lim_{x \to 0}\frac{\sin ax}{ax}}{\frac 1b \lim_{x \to 0} \frac {\sin ax}{ax}}$
$b \lim_{x \to 0} \dfrac {1}{x^2} \cdot \dfrac {b-a}{a}$
It seems like this limit does not exist, but if you apply L'Hopital's rule you seem to get an answer. What is wrong with what I did?
| A good way is to use Taylor expansion. You know what is Taylor series for $sin(x)$. So, develop accordingly the different terms.
For the numerator, you will have $\frac{1}{6} x^3 \left(a^3 b-a b^3\right)+O\left(x^4\right)$ and for the denominator $a x^3+O\left(x^4\right)$. So, the result should be $$\frac{1}{6} \left(a^2 b-b^3\right)$$ you can still simplify to $$\frac{1}{6} b (a-b) (a+b)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/661612",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 3
} |
Prove that $\frac{1}{1*3}+\frac{1}{3*5}+\frac{1}{5*7}+...+\frac{1}{(2n-1)(2n+1)}=\frac{n}{2n+1}$ Trying to prove that above stated question for $n \geq 1$. A hint given is that you should use $\frac{1}{(2k-1)(2k+1)}=\frac{1}{2}(\frac{1}{2k-1}-\frac{1}{2k+1})$. Using this, I think I reduced it to $\frac{1}{2}(\frac{1}{n^2}-(\frac{1}{3}+\frac{1}{5}+\frac{1}{7}+\cdots+\frac{1}{2n+1}))$. Just not sure if it's correct, and what to do with the second half.
| Hint: Multiply and divide by $2$.
What you have done is correct. All you need to observe is terms cancel out in pairs starting from $2nd$ term.
$$\frac{1}{2}(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...)$$
So we are left with $1st$ and the last terms.
$$\frac{1}{2}(1-\frac{1}{2n+1})$$
$$\frac{n}{2n+1}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/661701",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
A formula for a sequence which has three odds and then three evens, alternately We know that triangular numbers are 1, 3, 6, 10, 15, 21, 28, 36... where we have alternate two odd and two even numbers. This sequence has a simple formula $a_n=n(n+1)/2$.
What would be an example of a sequence, described by a similar algebraic formula, which has three odds and then three evens, alternately?
Ideally, it would be described by a polynomial of low degree.
| How about the sequence
$$a_n=\frac{1+(-1)^{\lfloor n/3\rfloor}}{2},\qquad \begin{array}{c|c|c|c|c|c|c|c|c|c}
n \strut& 0 & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8\\\hline
a_n \strut& 1 & 1 & 1 & 0 & 0 & 0 & 1 & 1 & 1
\end{array}\;\cdots$$
or even simpler, the sequence
$$a_n=\left\lfloor \frac{n}{3}\right\rfloor+1,\qquad \begin{array}{c|c|c|c|c|c|c|c|c|c}
n \strut& 0 & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8\\\hline
a_n \strut& 1 & 1 & 1 & 2 & 2 & 2 & 3 & 3 & 3
\end{array}\;\cdots$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/665722",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
How to tackle a recurrence that contains the sum of all previous elements? Say I have the following recurrence:
$$T(n) = n + T\left(\frac{n}{2}\right) + n + T\left(\frac{n}{4}\right) + n + T\left(\frac{n}{8}\right) + \cdots +n + T\left(\frac{n}{n}\right) $$
where $n = 2^k$, $k \in \mathbb{N} $ and $T(1) = 1$.
simplified to:
$$T(n) = n \log_2n + \sum_{i=1}^{\log_2n}T\left(\frac{n}{2^i}\right) $$
The Master's theorem is not applicable; neither is the Akra-Bazzi method since $k = \log_2$ is not a constant.
What strategy can I use to find a closed form solution? I have a feeling that the closed form is $T(n) = \sum_{i=0}^{\log_2n}\left[j\frac{n}{2^i} \log_2 \left(\frac{n}{2^i} \right)\right] + 1 $ where $j = \max\left(1, 2^{i-1}\right)$ but would like a proof.
| As an additional comment on this, note that if we define $T(n)$ for all $n$ and not just powers of two with $T(0)=0$ and $T(1)=1$ like this
$$T(n) = \sum_{k=1}^{\lfloor \log_2 n \rfloor} T(\lfloor n/2^k \rfloor)
+ n \lfloor \log_2 n \rfloor,$$
and the binary representation of $n$ is given by
$$n = \sum_{k=0}^{\lfloor \log_2 n \rfloor} d_k 2^k$$
then the exact formula for all $n$ where $n\ge 2$ is
$$T(n) = 2^{\lfloor \log_2 n \rfloor - 1} + \sum_{j=0}^{\lfloor \log_2 n \rfloor}
(\lfloor \log_2 n \rfloor -j )
[z^j] \frac{1-z}{1-2z} \sum_{k=j}^{\lfloor \log_2 n \rfloor} d_k 2^{k-j}$$
because $$\frac{1}{1-z-z^2-z^3-\cdots} = \frac{1-z}{1-2z}.$$
This simplifies to
$$2^{\lfloor \log_2 n \rfloor - 1} + \lfloor \log_2 n \rfloor \times n +
\sum_{j=1}^{\lfloor \log_2 n \rfloor}
(\lfloor \log_2 n \rfloor -j )
[z^j] \frac{1-z}{1-2z} \sum_{k=j}^{\lfloor \log_2 n \rfloor} d_k 2^{k-j}$$
Now note that for $j\ge 1$
$$[z^j] \frac{1-z}{1-2z} = 2^{j-1}$$ so this in turn again simplifies, this time to
$$2^{\lfloor \log_2 n \rfloor - 1} + \lfloor \log_2 n \rfloor \times n +
\sum_{j=1}^{\lfloor \log_2 n \rfloor}
(\lfloor \log_2 n \rfloor -j )
2^{j-1} \sum_{k=j}^{\lfloor \log_2 n \rfloor} d_k 2^{k-j}$$
which is
$$2^{\lfloor \log_2 n \rfloor - 1} + \lfloor \log_2 n \rfloor \times n +
\frac{1}{2} \sum_{j=1}^{\lfloor \log_2 n \rfloor}
(\lfloor \log_2 n \rfloor -j )
\sum_{k=j}^{\lfloor \log_2 n \rfloor} d_k 2^k.$$
Now for an upper bound consider a string of one digits giving
$$T(n)\le 2^{\lfloor \log_2 n \rfloor - 1} + \lfloor \log_2 n \rfloor \times n +
\frac{1}{2} \sum_{j=1}^{\lfloor \log_2 n \rfloor}
(\lfloor \log_2 n \rfloor -j ) (2^{\lfloor \log_2 n \rfloor+1}-2^j)
\\= 2^{\lfloor \log_2 n \rfloor - 1} + \lfloor \log_2 n \rfloor \times n +
\frac{1}{2} (\lfloor \log_2 n \rfloor +1)
\left(\lfloor \log_2 n \rfloor 2^{\lfloor \log_2 n \rfloor}
- 2^{\lfloor \log_2 n \rfloor+1} + 2\right).$$
For a lower bound consider a one digit followed by zeros giving
$$T(n)\ge 2^{\lfloor \log_2 n \rfloor - 1} + \lfloor \log_2 n \rfloor \times n +
\frac{1}{2} \sum_{j=1}^{\lfloor \log_2 n \rfloor}
(\lfloor \log_2 n \rfloor -j ) 2^{\lfloor \log_2 n \rfloor}
\\ = 2^{\lfloor \log_2 n \rfloor - 1} + \lfloor \log_2 n \rfloor \times n +
\frac{1}{4} \lfloor \log_2 n \rfloor(\lfloor \log_2 n \rfloor - 1)
2^{\lfloor \log_2 n \rfloor}.$$
The upper and lower bounds are attained and cannot be improved upon.
Selecting the dominant terms from the two bounds we finally get a complexity of
$$\Theta\left((\lfloor \log_2 n \rfloor)^2\times 2^{\lfloor \log_2 n \rfloor}\right)
= \Theta\left((\log_2 n)^2 2^{\log_2 n}\right)
= \Theta\left(n \times (\log_2 n)^2\right).$$
A similar calculation was done at this MSE link.
Addendum. The sequence of values of $T(2^j)$ using our definition is
$$3, 12, 40, 120, 336, 896, 2304, 5760, 14080, 33792,\ldots$$
which agrees (as it ought to) with the formula found by @Ragnar
$$T(2^j) = 2^{j-2}\times(2+3j+j^2).$$
(This is just a re-write of our lower bound, which to give credit was posted second.)
Remark as of Sat Feb 8 20:37:05 CET 2014. The upper limit of the outer sum can be replaced by $\lfloor \log_2 n\rfloor-1$ but the formulas are correct as stated.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/667929",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
} |
How prove this $|ON|\le \sqrt{a^2+b^2}$ let ellipse $M:\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}=1$,and there two point $A,B$ on $\partial M$,and the point $C\in AB$ ,such $AC=BC$,and the Circle $C$ is directly for the AB circle,for any point $N$ on $\partial C$,
show that
$$|ON|\le\sqrt{a^2+b^2}$$
my try: let
$$A(x_{1},y_{1}),B(x_{2},y_{2}),C(\dfrac{x_{1}+x_{2}}{2},\dfrac{y_{1}+y_{2}}{2})$$
then
$$\dfrac{x^2_{1}}{a^2}+\dfrac{y^2_{1}}{b^2}=1,\dfrac{x^2_{2}}{a^2}+\dfrac{y^2_{2}}{b^2}=1$$
so the circle $C$ equation is
$$\left(x-\dfrac{x_{1}+x_{2}}{2}\right)^2+\left(y-\dfrac{y_{1}+y_{2}}{2}\right)^2=r^2$$
where
$$4r^2=(x_{1}-x_{2})^2+(y_{1}-y_{2})^2$$
let $N(c,d)$,then
$$\left(c-\dfrac{x_{1}+x_{2}}{2}\right)^2+\left(d-\dfrac{y_{1}+y_{2}}{2}\right)^2=r^2$$
How prove
$$c^2+d^2\le a^2+b^2?$$
Thank you
| Following Tian's observation and taking
$$A:(a\cos\theta,b\sin\theta),$$
$$B:(a\cos\phi,b\sin\phi)$$
we have to prove
$$4\|OC\|^2+8\|OC\|\|AC\|+4\|AC\|^2\leq 4a^2+4b^2,\tag{1} $$
or:
$$a^2(\cos\theta+\cos\phi)^2+b^2(\sin\theta+\sin\phi)^2+a^2(\cos\theta-\cos\phi)^2+b^2(\sin\theta-\sin\phi)^2+2\sqrt{\left(a^2(\cos\theta+\cos\phi)^2+b^2(\sin\theta+\sin\phi)^2\right)\cdot\left(a^2(\cos\theta-\cos\phi)^2+b^2(\sin\theta-\sin\phi)^2\right)}\leq 4a^2+4b^2,\tag{2}$$
equivalent to:
$$\sqrt{\left(a^2(\cos\theta+\cos\phi)^2+b^2(\sin\theta+\sin\phi)^2\right)\cdot\left(a^2(\cos\theta-\cos\phi)^2+b^2(\sin\theta-\sin\phi)^2\right)}\leq a^2(\sin^2\theta+\sin^2\phi)+b^2(\cos^2\theta+\cos^2\phi),\tag{3}$$
and to:
$$a^4\left(\sin^2\theta-\sin^2\phi\right)^2+b^4\left(\cos^2\theta-\cos^2\phi\right)^2+a^2b^2\left(\sin(2\theta)-\sin(2\phi)\right)^2\leq a^4(\sin^2\theta+\sin^2\phi)^2+b^4(\cos^2\theta+\cos^2\phi)^2+2a^2b^2(\sin^2\theta+\sin^2\phi)(\cos^2\theta+\cos^2\phi),\tag{4}$$
that further reduces to:
$$4a^2 b^2(\sin\theta\cos\theta-\sin\phi\cos\phi)^2\leq 4a^4\sin^2\theta\sin^2\phi+4b^4\cos^2\theta\cos^2\phi+2a^2b^2(\sin^2\theta\cos^2\theta+\sin^2\phi\cos^2\phi+\sin^2\theta\cos^2\phi+\sin^2\phi\cos^2\theta),\tag{5} $$
and finally to:
$$2a^2 b^2(\sin^2\theta-\sin^2\phi)(\cos^2\theta-\cos^2\phi)\leq 4(a^2\sin\theta\sin\phi+b^2\cos\theta\cos\phi)^2\tag{6} $$
that is trivial since the LHS is non-positive while the RHS is non-negative.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/668125",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 0
} |
Proof for the formula of sum of arcsine functions $ \arcsin x + \arcsin y $ It is known that
\begin{align}
\arcsin x + \arcsin y =\begin{cases}
\arcsin( x\sqrt{1-y^2} + y\sqrt{1-x^2}) \\\quad\text{if } x^2+y^2 \le 1 &\text{or} &(x^2+y^2 > 1 &\text{and} &xy< 0);\\
\pi - \arcsin( x\sqrt{1-y^2} + y\sqrt{1-x^2}) \\\quad\text{if } x^2+y^2 > 1&\text{and} &0< x,y \le 1;\\
-\pi - \arcsin( x\sqrt{1-y^2} + y\sqrt{1-x^2}) \\\quad\text{if } x^2+y^2 > 1&\text{and} &-1\le x,y < 0.
\end{cases}
\end{align}
I tried to prove this myself, have no problem in getting the 'crux' $\arcsin( x\sqrt{1-y^2} + y\sqrt{1-x^2})$ part of the RHS, but face trouble in checking the range of that 'crux' under the given conditions.
| Hope the graph explains the x,y domain ( |x|< 1, |y|< 1 ) and range.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/672575",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "16",
"answer_count": 4,
"answer_id": 1
} |
How can I find all integers $x≠3$ such that $x−3|x^3−3$ How can I find all integers $x≠3$ such that $x−3|x^3−3$?
I tried expand $x^3−3$ as a sum but I couldn't find a way after that.
| I am sitting with this exact problem right now. I used a more brute approach to solving the problem but I think that I am missing some solutions. Could someone take a look at my solution?
Consider $$\dfrac{x^{3}-3}{x-3}=\cdots =1-\dfrac{x(x+1)(x-1)}{x-3}.$$ Therefore $$x-3\vert x^{3}-3 \Leftrightarrow x-3\vert x~~or~~x-3\vert x+1~~or~~x-3\vert x-1.$$ Now, we may consider each case by themselves and combine each solution set from these to solve the problem.
$(i)$ Consider $$\dfrac{x}{x-3}=1+\dfrac{3}{x-3}$$ and $x-3$ divides $3$ exactly when $x\in A= \{0,4,6\}$.
$(ii)$ Consider $$\dfrac{x+1}{x-3}=1+\dfrac{4}{x-3}$$ and $x-3$ divides $4$ exactly when $x\in B=\{-1,4,5,7\}$.
$(iii)$ Lastly consider $$\dfrac{x-1}{x-3}=1+\dfrac{2}{x-3}$$ and $x-3$ divides $2$ exactly when $x\in C=\{1,4,5\}$.
Answer: $x-3$ divides $x^{3}-3$ when $x\in A\cup B\cup C=\{-1,0,1,4,5,6,7\}$.
Is this correct?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/672854",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
} |
Solving trig equation with two ways gives two different answers It is given that $x$ satisfies
$$ 3\sin x + 4\cos x = 5 $$
We have to find the value(s) of $\tan (x/2) $. My approach is this:
$$\begin{align} &3\sin x + 4\cos x = 5 \\
&\implies \dfrac 3 5\sin x + \dfrac 4 5 \cos x = 1 \\
&\implies \cos ( x - \arccos(\dfrac 4 5) ) = 1 \\
&\implies x = 2n\pi + \arccos(\dfrac 4 5) \\
&\implies x/2 = n\pi + 0.5\arctan(\dfrac 3 4)\\
&\implies \tan(x/2) = + \tan(0.5 \arctan(\dfrac 3 4)) = y \;\; \text{ (say)}
\end{align}$$
Now, $$ \dfrac{2y}{1-y^2} = \tan (\arctan(\dfrac 3 4 ) ) = \dfrac 3 4 $$
Which gives me $ y = 3, -\dfrac 1 3 $
However, the solution given does something like this:
$$\begin{align} &3\sin x + 4\cos x = 5 \\
&\implies \dfrac 3 5 \dfrac{ 2t } { 1 + t^2 } + \dfrac 4 5 \dfrac{ 1 - t^2 } { 1 + t^2 } = 1 \;\;\;\;(t = \tan (x/2))\\
&\implies (3t-1)^2 = 0 \implies t = \dfrac 1 3
\end{align}$$
Which one is correct and why?
| There is a sign error. The solution of $ \dfrac{2y}{1-y^2} = \dfrac 3 4 $ is $ y = -3, \dfrac 1 3 $.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/673670",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
How can you find the cubed roots of $i$? I am trying to figure out what the three possibilities of $z$ are such that
$$
z^3=i
$$
but I am stuck on how to proceed. I tried algebraically but ran into rather tedious polynomials. Could you solve this geometrically? Any help would be greatly appreciated.
| You can solve this geometrically if you know polar coordinates.
In polar coordinates, multiplication goes $(r_1, \theta_1) \cdot (r_2, \theta_2) = (r_1 \cdot r_2, \theta_1 + \theta_2)$, so cubing goes $(r, \theta)^3 = (r^3, 3\theta)$. The cube roots of $(r, \theta)$ are $\left(\sqrt[3]{r}, \frac{\theta}{3}\right)$, $\left(\sqrt[3]{r}, \frac{\theta+2\pi}{3}\right)$ and $\left(\sqrt[3]{r}, \frac{\theta+4\pi}{3}\right)$ (recall that adding $2\pi$ to the argument doesn't change the number). In other words, to find the cubic roots of a complex number, take the cubic root of the absolute value (the radius) and divide the argument (the angle) by 3.
$i$ is at a right angle from $1$: $i = \left(1, \frac{\pi}{2}\right)$. Graphically:
A cubic root of $i$ is $A = \left(1, \frac{\pi}{6}\right)$. The other two are $B = \left(1, \frac{5\pi}{6}\right)$ and $\left(1, \frac{9\pi}{6}\right) = -i$.
Recalling basic trigonometry, the rectangular coordinates of $A$ are $\left(\cos\frac{\pi}{6}, \sin\frac{\pi}{6}\right)$ (the triangle OMA is rectangle at M). Thus, $A = \cos\frac{\pi}{6} + i \sin\frac{\pi}{6} = \frac{\sqrt{3}}{2} + i\frac{1}{2}$.
If you don't remember the values of $\cos\frac{\pi}{6}$ and $\sin\frac{\pi}{6}$, you can find them using geometry. The triangle $OAi$ has two equal sides $OA$ and $Oi$, so it is isoceles: the angles $OiA$ and $OAi$ are equal. The sum of the angles of the triangle is $\pi$, and we know that the third angle $iOA$ is $\frac{\pi}{2} - \frac{\pi}{6} = \frac{\pi}{3}$; therefore $OiA = OAi = \dfrac{\pi - \frac{pi}{3}}{2} = \dfrac{\pi}{3}$. So $OAi$ is an equilateral triangle, and the altitude AN is also a median, so N is the midpoint of $[Oi]$: $\sin\frac{\pi}{6} = AM = ON = \frac{1}{2}$. By the Pythagorean theorem, $OM^2 + AM^2 = OA^2 = 1$ so $\cos\frac{\pi}{6} = \sqrt{1 - \left(\frac{1}{2}\right)^2} = \dfrac{\sqrt{3}}{2}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/674621",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
"answer_count": 7,
"answer_id": 1
} |
Find the value of $\lim_{x \to - \infty} \left( \sqrt{x^2 + 2x} - \sqrt{x^2 - 2x} \right)$ I am stuck on this. I would like the algebraic explanation or trick(s) that shows that the equation below has limit of $-2$ (per the book). The wmaxima code of the equation below.
$$
\lim_{x \to - \infty} \left( \sqrt{x^2 + 2x} - \sqrt{x^2 - 2x} \right)
$$
I've tried factoring out an $x$ using the $\sqrt{x^2} = |x|$ trick. That doesn't seem to work. I get $1 - 1 = 0$ for the other factor meaning the limit is zero...but that's obviously not correct way to go about it :(
Thanks.
| The direct approach of just factoring an $|x|$ from each piece is not fruitful: It leads to
$$ |x| \Big(\sqrt{1 + 2/x} - \sqrt{1 - 2/x}\Big)$$
The first term grows, and the second term tends to $0$, so there's a balance between them.
Multiply top and bottom by the conjugate to find that
\begin{align*}
\sqrt{x^2 + 2x} - \sqrt{x^2 - 2x} &= \Big(\sqrt{x^2 + 2x} - \sqrt{x^2 - 2x}\Big) \left(\frac{\sqrt{x^2 + 2x} + \sqrt{x^2 - 2x}}{\sqrt{x^2 + 2x} + \sqrt{x^2 - 2x}}\right) \\
&= \frac{4x}{\sqrt{x^2 + 2x} + \sqrt{x^2 - 2x}} \\
&= \frac{-4}{\sqrt{1 + 2/x} + \sqrt{1 - 2/x}}
\end{align*}
since $\sqrt{x^2} = |x| = -x$ for $x < 0$. Can you finish it from here?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/675516",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
Incongruent solutions to $7x \equiv 3$ (mod $15$) I'm supposed to find all the incongruent solutions to the congruency $7x \equiv 3$ (mod $15$)
\begin{align*}
7x &\equiv 3 \mod{15} \\
7x - 3 &= 15k \hspace{1in} (k \in \mathbb{Z}) \\
7x &= 15k+3\\
x &= \dfrac{15k+3}{7}\\
\end{align*}
Since $x$ must be an integer, we must find a pattern for $k$ that grants this. We know that $\frac{k+3}{7}$ must be equal to some integer, say $m$. Solving for $k$, we have $k=4+7m$.
Substituting this into our value for $x$, we get:
\begin{align*}
x & = \dfrac{15(4+7m) + 3}{7}.\\
&= \dfrac{63}{7} + \frac{105m}{7}.\\
&= 9+15m.
\end{align*}
So, $x = 9+15m, m\in \mathbb{Z}.$
So, is this what I was looking for? I'm not exactly sure what is meant by incongruent solutions.
| Let us take your equation $7x=15k+3$ and multiply both sides by $13$.
We get $91x=195k+39$. From where $(6\cdot 15+1)x=15\cdot 13k+2\cdot 15+9$, or what is the same $x=15(13k-6x+2)+9$. Since $k$ was certain integer we could say that $K:=13k-6x+2$ is certain integer and get $x=15K+9$.
So, all solutions leave the same remainder $9$ after division by $15$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/675867",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 1
} |
Trigonometric Series Proof I am posed with the following question:
Prove that for even powers of $\sin$:
$$ \int_0^{\pi/2} \sin^{2n}(x) dx = \dfrac{1 \cdot 3 \cdot 5\cdots (2n-1)}{2 \cdot 4 \cdot 6 \cdots 2n} \times \dfrac{\pi}{2} $$
Here is my work so far:
*
*Proof by induction
$P(1) \Rightarrow n = 2 \Rightarrow $
$$\int_0^{\pi/2} \sin^4(x) dx = \dfrac{1 \cdot 3}{2 \cdot 4} \times \dfrac{\pi}{2} $$
$$ \frac{3\pi}{16} = \frac{3 \pi}{16} $$
Base Case Succeeds. (The left hand side evaluated by MAPLE)
$P(n+1) \Rightarrow n = n + 1 \Rightarrow$
$$ \int_0^{\pi/2} \sin^{2n + 1}(x) dx = \int_0^{\pi/2} \sin^{2n} x \sin x dx$$
$$ = \int_0^{\pi/2} (1 - \cos^2x)^n \sin x dx$$
Let $u = \cos x, du = - \sin x dx \Rightarrow$
$$ -\int_0^{\pi/2} (1-u^2)^n du$$
Now I am stuck and unsure what to do with this proof ... Any help would be greatly appreciated
(I also tried using the reduction formulas to no avail)
EDIT
I have completed the proof.
I am posting this here for any other people who may have the same question...
We will prove by induction that $\forall n \in 2 \mathbb{N}_{>0}$
\begin{align*}
\int_0^{\pi/2} \sin^{2n} x dx & = \frac{1 \times 3 \times 5 \times \cdots \times (2n - 1)}{2 \times 4 \times 6
\times \cdots \times 2n} \times \frac{\pi}{2} \tag{1}
\end{align*}
With $k = 1$ as our base case, we have
\begin{align*}
\frac{1}{2}x - \frac{1}{4} \sin{2x} \bigg|_{0}^{\pi / 2} & = \frac{\pi}{4} \tag{67}
\\ \frac{1}{2} \times \frac{\pi}{2} & = \frac{\pi}{4}
\end{align*}
Let $n \in 2\mathbb{N}_{>0}$ be given and suppose (1) is true for $k = n$. Then
\begin{align*}
\int_0^{\pi/2} \sin^{2n+2} x dx & = - \frac{1}{2n+2} \cos^{2n-1}x \sin x \bigg|_{0}^{\pi / 2} + \frac{2n+1}{2n+2} \int_0^{\pi/2} \sin^{2n} x dx \tag{67} \\
& = \frac{2n+1}{2n+2} \int_0^{\pi/2} \sin^{2n} x dx \\
& = \frac{1 \times 3 \times 5 \times \cdots \times (2n + 1)}{2 \times 4 \times 6
\times \cdots \times {(2n+2)}} \times \frac{2n+2}{2n+1} \times \frac{\pi}{2} \\
& = \frac{1 \times 3 \times 5 \times \cdots \times (2n - 1)}{2 \times 4 \times 6
\times \cdots \times 2n} \times \frac{\pi}{2} \\
& = \int_0^{\pi/2} \sin^{2n} x dx
\end{align*}
Thus, (1) holds for $k = n + 1$, and by the principle of induction, it follows that that (1) holds for all even numbers. $\square$
| This proof makes use of the gamma and beta functions. We have the basic identity $$B(x, y) = \frac{\Gamma(x)\Gamma(y)}{\Gamma(x + y)} $$
Note that $$B(x, y) = \int_0^1 t^x(1 - t)^y \,dt $$
by definition. Put $x = \sin^2\theta$. Thus we have $$\int_{0}^{\pi/2}\sin^{2m + 1}\theta\cos^{2n + 1}\theta \,\,d\theta = \frac{\Gamma(m)\Gamma(n)}{2\Gamma(m + n)}$$
Note that since we were originally dealing with two independent variables, we need to maintain those after making this substitution, hence the $m$ and $n$.
Now put $2k = 2m + 1$ and $2n + 1 = 0$. This gives the following result:
$$\int_{0}^{\pi/2}\sin^{2k}\theta \,d\theta = \frac{\Gamma[\frac{1}{2}(2k + 1)]\Gamma(1/2)}{2\Gamma(k + 1)} $$
The denominator is simply $2\cdot k!$ and the numerator is $\pi\cdot k!\cdot \binom{k - 1/2}{k}$
Expanding the binomial coefficient gives the required result.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/680340",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
An inequality for sides of a triangle Let $ a, b, c $ be sides of a triangle and $ ab+bc+ca=1 $. Show
$$(a+1)(b+1)(c+1)<4 $$
I tried Ravi substitution and got a close bound, but don't know how to make it all the way to $4 $.
I am looking for a non-calculus solution (no Lagrange multipliers).
Do you know how to do it?
| WOLOG assume that $a\geq b\geq c>0$. The constraints the problem imposes are $c=\frac{1-ab}{a+b}$, $b+c>a$. Equivalently, we have $\frac{1-ab}{a+b}+b > a\geq b$ and $b\geq \frac{1-ab}{a+b}$. These inequalities yield $$0\leq a^2-b^2<1-ab<1\quad (1)$$ and $$ab\geq \frac{1-b^2}{2}.\quad(2)$$
Note that $(1)$ implies $0<b\leq a<1$. Now, suppose that $a+b\leq 1$. Then from $(2)$ we have $2b\geq 2b(a+b)\geq 1+b^2$, which is not possible since $b<1$. Thus, we must have $$a+b>1.\quad (3)$$ Invoking $(3)$ we have
$$(1-a)(1-b)>0\\
\implies ab>a+b-1\\
=\left\vert a+b-1\right\vert\\
\implies a^2b^2>(a+b-1)^2\\
\iff 1-a^2b^2<2(a+b)-(a+b)^2\\
\iff a+b+(1+ab)\frac{1-ab}{a+b}<2\\
\iff a+b+c +abc<2\\
\implies (1+a)(1+b)(1+c)<4.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/681433",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 3,
"answer_id": 0
} |
Trying to get a bound on the tail of the series for $\zeta(2)$ $$\frac{\pi^2}{6} = \zeta(2) = \sum_{k=1}^\infty \frac{1}{k^2}$$ I hope we agree.
Now how do I get a grip on the tail end $\sum_{k \geq N} \frac{1}{k^2}$ which is the tail end which goes to zero?
I want to show that $\sqrt{x}\cdot \mathrm{tailend}$ is bounded as $x \to \infty$.
All this to show that $x\cdot \mathrm{tailend} = \mathcal O\sqrt{x}$
The purpose is to get the asymptotic formula for the distribution of square free integers.
p.269 Exercise 8 Stewart and Tall.
| The tail is
$$\frac{1}{N^2}+\frac{1}{(N+1)^2}+\frac{1}{(N+2)^2}+\cdots.$$
For $N\gt 1$ this is less than
$$\frac{1}{(N-1)(N)}+ \frac{1}{(N)(N+1)}+ \frac{1}{(N+1)(N+2)}+\cdots.$$
Note that $\frac{1}{t(t+1)}=\frac{1}{t}-\frac{1}{t+1}$. Using this we find that the tail is less than
$$\left(\frac{1}{N-1}-\frac{1}{N}\right)+ \left(\frac{1}{N}-\frac{1}{N+1}\right)+ \left(\frac{1}{N+1}-\frac{1}{N+2}\right)+\cdots.$$
Open up the brackets and note the massive cancellation (telescoping). We find that the tail is less than $\frac{1}{N-1}$.
Another way: Draw a picture of the curve $y=\frac{1}{x^2}$. Note that the area of the rectangle of height $\frac{1}{n^2}$ that goes from $x=n-1$ to $x=n$ is $\frac{1}{n^2}$.
This rectangle is entirely below the curve on the interval from $n-1$ to $n$. Thus
the tail is less than
$$\int_{N-1}^\infty \frac{dx}{x^2}.$$
Evaluate the integral. We get $\frac{1}{N-1}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/685435",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 8,
"answer_id": 2
} |
Relations and Combinatorics exercise Be $A=\{1,2,3,\ldots,10\}$ Determine how many equivalence relations can be defined in $A$ with exactly two equivalence classes.
Determine how many equivalence relations can be defined in $A$ with exactly three equivalence classes.
What I did for two is:
$${10\choose1}+{10\choose2}+\cdots+{10\choose9}=\sum_{i=1}^9{10\choose i}=2^{10}-2$$
For three:
$$\text{ #(All the posible Classes)} - {10\choose1}{9\choose0}-{10\choose0}=2^{10}-11$$
is this ok? is there a easier way to do think it?
| If the chosen subset is $\{1,2,3\}$, then it's the SAME equivalence relation as if the chosen subset is $\{4,5,6,7,8,9,10\}$. Therefore when you add $\dbinom{10}{3}+\dbinom{10}{7}$, you're counting each of those equivalence relations twice.
Similarly with $\dbinom{10}{1}+\dbinom{10}{9}$ and with $\dbinom{10}{2}+\dbinom{10}{8}$ and with $\dbinom{10}{4}+\dbinom{10}{6}$.
A different sort of thing happens with $\dbinom{10}{5}$. That one term alone counts each of $126$ equivalence relations twice: $\{1,2,3,4,5\}$ corresponds to the SAME equivalence relation as $\{6,7,8,9,10\}$.
So the answer is
$$\dfrac{\sum_{k=1}^9 \binom{10}{k}}{2} = \dbinom{10}{1} + \dbinom{10}{2} + \dbinom{10}{3} + \dbinom{10}{4} + \frac{\dbinom{10}{5}}{2} = 511$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/688831",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Show that $(x^2-yz)^3+(y^2-zx)^3+(z^2-xy)^3-3(x^2-yz)(y^2-zx)(z^2-xy)$ is a perfect square and find its square root. Show that $(x^2-yz)^3+(y^2-zx)^3+(z^2-xy)^3-3(x^2-yz)(y^2-zx)(z^2-xy)$ is a perfect square and find its square root.
My work:
Let, $x^2-yz=a,y^2-zx=b,z^2-xy=c$. So, we can have,
$a^3+b^3+c^3-3abc=(a+b+c)(a^2+b^2+c^2-ab-bc-ca)$
$(a+b+c)(a^2+b^2+c^2-ab-bc-ca)=\dfrac12[(x-y)^2+(y-z)^2+(z-x)^2]\cdot\dfrac12[(a-b)^2+(b-c)^2+(c-a)^2]$
Now, I got into a mess. I have got two products with sum of three squares which I cannot manage nor can I show this to be a perfect square. Please help.
| Use the formula $$a^3+b^3+c^3-3abc=(a+b+c)(a^2+b^2+c^2-ab-bc-ca)=\frac{1}{2}(a+b+c)((a-b)^2+(b-c)^2+(c-a)^2)$$ and then notice that $a-b=x^2-y^2+zx-yz=(x-y)(x+y+z)$ This will lead you to get the answer as $(x^3+y^3+z^3-3xyz)^2$ as pointed out by Ewan Delanoy.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/691078",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 2,
"answer_id": 1
} |
Finding J-invariant of Legendre form of Elliptic Curve PROBLEM: Put the Legendre equation $y^2 = x(x − 1)(x − λ)$ into Weierstrass
form and use this to show that the j-invariant is
j = $2^8\frac{(λ2 − λ + 1)^3}{λ^2(λ − 1)^2}$ .
Recall:
Weierstrass equation form: E: y^2 = x^3 + Ax +B
and
J(E) = 1728$\frac{4A^3}{4A^3+27B^2}$
Attempt:
$y^2 = x^3 + (-\lambda-1)x^2 + \lambda x$
Now I'm having trouble transforming this into Weierstass form. What would be the way to go for transforming this equation? $x^2 = x_1$ would not work. Idk..
| In general, if a curve is given by $Y^2=X^3+AX^2+BX+C$, a change of variables $Y=y$ and $X=x-A/3$ will provide a model for the curve of the form $y^2=x^3+A'x+B'$.
The reason is that
$$X^3+AX^2+\cdots = (x-A/3)^3 + A(x-A/3)^2+\cdots$$
and the coefficient in $x^2$ is given by $3(-A/3)+A=0$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/692309",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
To find the measure of the given angle
In triangle $ABC$ angle $\widehat C=60°$. $(AD)$ and $(BE)$ are perpendicular on $(BC)$ and $(AC)$ respectively. $M$ is the midpoint of $[AB]$. How to find the measure of angle $\widehat{EMD}$ in degrees?.
| Here is a heavyweight, straightforward solution.
Putting: $AB=c$, $AC=b$ and $BC=a$.
Let us see $\triangle DMB$ first:
*
*$BM=\frac{c}{2}$
*$BD=c\cos B$ [why?]
So, applying the cosine rule on $DMB$, we have $DM=\sqrt{(\frac{c}{2})^2+c^2\cos ^2B-c^2\cos ^2B}=\frac{c}{2}$. Similarly, doing this on $\triangle MAE$, yields $ME=\frac{c}{2}$.
On $\triangle DCE$:
*
*$CD=b\cos C$ [why?]
*$CE=a\cos C$
So, applying the cosine rule on $DCE$, we have:
$$DE=\sqrt{b^2\cos ^2C +a^2\cos ^2C-2ab\cos ^3C}=\cos C\sqrt{b^2 +a^2-2ab\cos C}=c\cos C=\frac{c}{2}$$ So, $ME=DM=DE=\frac{c}{2}$. Thus, $MED$ is an equilateral triangle.
This proof tells us more about the question. We have shown that $DMB$ is isosceles for ever acute triangle, not necessarily when $\angle C$ is $60^{\circ}$. However, it is equilateral if and only if $\angle C= 60^{\circ}$, since $\cos C = \frac{1}{2}$, only when $\angle C= 60^{\circ}$, when $C$ is an angle of a triangle .
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/693734",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Prove the trigonometric identity $\sin^4{x} = \frac{3-4\cos{2x}+\cos{4x}}{8}$ I need to show the steps to prove this identity:
$$\sin^4{x} = \frac{3-4\cos{2x}+\cos{4x}}{8}$$
I know that $\cos{2x}=\cos^2{x}-\sin^2{x}$. From there I do not know what to do.
The solution should look like:
$$\sin^4{x}=sin^4{x}$$
I need to prove the right side equals the left side.
| Let $ z = \cos(\theta) +i \sin(\theta)$
We can then find through De Moive's formula and the fact that $\cos -\theta = \cos \theta$ and $\sin -\theta = - \sin \theta$:
$ z - \frac1z \equiv 2i\sin\theta$
Thus: ($i^4=1$)
$ (z-\frac1z)^4 \equiv 16\sin^4\theta$
Expanding the RHS:
$ (z-\frac1z)^4 \equiv z^4 - 4z^2 + 6 - 4z^{-2} + z^{-4}$
Then subsituting $z$ back into the equation. (The $\sin$s will cancel out)
$ (z-\frac1z)^4 \equiv 2\cos4\theta - 8\cos2\theta + 6$
Finally dividing by 16 to get $\sin^4\theta$:
$\sin^4\theta = \frac {\cos4\theta - 4\cos2\theta + 3} 8$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/696593",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 4
} |
Determinant algebra If $A$ and $B$ are $4 \times 4$ matrices with $\det(A) = −2$, $\det(B) = 3$, what is $\det(A+B)$?
At first I approached the problem that $\det(A+B) = \det(A) + \det(B)$ but this general rule would not hold true, so I do not know how to approach the problem from here.
| The number $\det(A+B) $ is not determined by the determinants of $A$ and $B$. For instance if
$$
A=\begin{bmatrix} 1&0&0&0\\0&1&0&0\\0&0&1&0\\0&0&0&-2\end{bmatrix}, \ \
B=\begin{bmatrix} -1&0&0&0\\0&-1&0&0\\0&0&1&0\\0&0&0&3\end{bmatrix}
$$
then $A$ and $B$ are as required, with $\det A+B=0$. While if$$
A=\begin{bmatrix} 1&0&0&0\\0&1&0&0\\0&0&1&0\\0&0&0&-2\end{bmatrix}, \ \
B=\begin{bmatrix} 1&0&0&0\\0&1&0&0\\0&0&1&0\\0&0&0&3\end{bmatrix}
$$
then $\det A+B=8$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/697259",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Help with a simple derivative I am trying to solve $\dfrac {6} {\sqrt {x^3+6}}$
and so far I made it to $6(x^3+6)^{-\frac 1 2}$
then I continued and now I have $(x^3+6)^{- \frac 3 2} * 3x^2$
and I cannot figure out what how to find the constant that should be before the parenthesis.
| $\dfrac{d}{dx} \left(\dfrac{6}{\sqrt{x^3+6}}\right) = 6 \dfrac{d}{dx}\left(\dfrac{1}{\sqrt{x^3+6}}\right) = 6\cdot -\dfrac{1}{2}\left(x^3+6\right)^{\frac{-3}{2}}\cdot 3x^2 = -9x^2\left(x^3+6\right)^{\frac{-3}{2}}$ via chain rule.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/699607",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
} |
How to determine the eigenvectors of this matrix? I have some problems to determine the eigenvectors of a given matrix:
The matrix is:
$$
A = \left( \begin{array}{ccc}
1 & 0 &0 \\
0 & 1 & 1 \\
0 & 0 & 2
\end{array} \right)
$$
I calculated the eigenvalues first and got $$ \lambda_1 = 1, \lambda_2 = 2, \lambda_3 = 1$$
There was no problem for me so far. But I do not know how to determine the eigenvectors. The formula I have to use is
$$ (A-\lambda_i E)u=0, \lambda_i = \{1,2,3\}, u\ is\ eigenvector$$
When I determined the eigenvector with $ \lambda_2=2$ there was not a problem. I got the result that $x_3 = variable$ and $x_2 = x_3$, so:
$$
EV_2= \left( \begin{array}{ccc}
0 \\
\beta \\
\beta
\end{array} \right) \ \beta\ is\ variable,\ so\ EV = span\{\left( \begin{array}{ccc}
0 \\
1 \\
1
\end{array} \right)\}
$$
But when I used $ \lambda_1 = \lambda_3 = 1 $, I had to calculate:
$$
\left( \begin{array}{ccc}
0 & 0 &0 \\
0 & 0 & 1 \\
0 & 0 & 1
\end{array} \right) *
\left( \begin{array}{ccc}
x_1 \\
x_2 \\
x_3
\end{array} \right)
=0
$$
what in my opinion means that $x_3 = 0 $ and $x_1$ and $x_2$ are variable, but not necessarily the same as in the case above, so $ EV_{1,3} = \left( \begin{array}{ccc}
\alpha \\
\beta \\
0
\end{array} \right) $
What does that mean for my solution? is it
$$
EV_{1,3} = span\{\left( \begin{array}{ccc}
1 \\
0 \\
0
\end{array} \right),
\left( \begin{array}{ccc}
0 \\
1 \\
0
\end{array} \right),
\left( \begin{array}{ccc}
1 \\
1 \\
0
\end{array} \right)\}
$$
What exactly is now my solution in this case for the eigenvectors $ \lambda_1, \lambda_3 $? In university we just had one variable value in the matrix so I don't know how to handle two of them being different.
| Every linear combination of $EV_{1}=\pmatrix{1\\0\\0}$ and $EV_3=\pmatrix{0\\1\\0}$ is a eigenvector with eigenvalue $1$.
$EV_{1,3} = span\{\left( \begin{array}{ccc}
1 \\
0 \\
0
\end{array} \right),
\left( \begin{array}{ccc}
0 \\
1 \\
0
\end{array} \right),
\left( \begin{array}{ccc}
1 \\
1 \\
0
\end{array} \right)\}$
is the same as $EV_{1,3} = span\{\left( \begin{array}{ccc}
1 \\
0 \\
0
\end{array} \right),
\left( \begin{array}{ccc}
0 \\
1 \\
0
\end{array} \right)\}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/701673",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
} |
Prove $\frac {1} {1+x^2} = \sum^{n-1}_{j=0} (-1)^j x^{2j} + (-1)^n \frac {x^{2n}} {1+x^2}, x\in \mathbb R, n \in \mathbb N$ by induction Prove $\frac {1} {1+x^2} = \sum^{n-1}_{j=0} (-1)^j x^{2j} + (-1)^n \frac {x^{2n}} {1+x^2}, x\in \mathbb R, n \in \mathbb N$ by induction.
$n = 1$: $\sum^{(1)-1}_{j=0} (-1)^j x^{2j} + (-1)^n \frac {x^{2n}} {1+x^2} = 1 -\frac {x^{2}} {1+x^2} = \frac {1} {1+x^2}$.
Now assume the identity is true for $n \le k$.
We show the identity holds for $n=k+1$:
$\sum^{(k+1)-1}_{j=0} (-1)^j x^{2j} + (-1)^{k+1} \frac {x^{2(k+1)}} {1+x^2} = \sum^{k}_{j=0} (-1)^j x^{2j} + (-1)^kx^{2k} + (-1)^{k+1} \frac {x^{2(k+1)}} {1+x^2} = \frac {1} {1+x^2} + (-1)^kx^{2k} + (-1)^{k+1} \frac {x^{2(k+1)}} {1+x^2}$
where I've used the induction hypotheses in the last equality. However I cannot show that this sum equals $\frac {1} {1+x^2}$.
Thanks for your time.
| The equation in question is
$\dfrac{1}{1 + x^2} = \sum_{j = 0}^{n - 1} (-1)^j x^{2j} + (-1)^n \dfrac{x^{2n}}{1 + x^2} \tag{1}$
which, upon multiplication by $1 + x^2$ is seen to be equivalent to
$1 = (1 + x^2) \sum_{j = 0}^{n - 1} (-1)^j x^{2j} + (-1)^n x^{2n} \tag{2}$
or
$1 - (-1)^n x^{2n} = (1 + x^2) \sum_{j = 1}^{n - 1} (-1)^j x^{2j}; \tag{3}$
we prove (3) by induction on $n$. It is easy to see that, for $n = 1$, (3) reduces to the identity
$1 + x^2 = 1 + x^2, \tag{4}$
since the sum on the right of (3) clearly takes the value $1$ for $n = 1$. Suppose then that (3) holds for some positive integer $k$:
$1 - (-1)^k x^{2k} = (1 + x^2) \sum_{j = 1}^{k - 1} (-1)^j x^{2j}; \tag{5}$
we subtract $(-1)^{k + 1}x^{2(k + 1)}$ from, and add $(-1)^k x^{2k}$ to, each side of (5), yielding
$1 - (-1)^{k + 1} x^{2(k + 1)} = (1 + x^2) \sum_{j = 1}^{k - 1} (-1)^j x^{2j} + (-1)^k x^{2k} - (-1)^{k + 1}x^{2(k + 1)} ; \tag{6}$
examining the last two terms on the right-hand side of (6), we see that
$(-1)^k x^{2k} - (-1)^{k + 1}x^{2(k + 1)} = (-1)^k x^{2k} (1 - (-1)x^2) = (-1)^k x^{2k} (1 + x^2), \tag{7}$
and thus we have
$(1 + x^2) \sum_{j = 1}^{k - 1} (-1)^j x^{2j} + (-1)^k x^{2k} - (-1)^{k + 1}x^{2(k + 1)}$
$= (1 + x^2) \sum_{j = 1}^{k - 1} (-1)^j x^{2j} + (-1)^k x^{2k} (1 + x^2) = (1 + x^2)\sum_{j = 1}^k (-1)^j x^{2j}, \tag{8}$
which shows that (6) becomes
$1 - (-1)^{k + 1} x^{2(k + 1)} = (1 + x^2)\sum_{j = 1}^k (-1)^j x^{2j}, \tag{9}$
completing the inductive proof of (3). If we now add $(-1)^n x^{2n}$ to each side of (3), and then divide by $(1 + x^2)$, we obtain
$\dfrac{1}{1 + x^2} = \sum_{j = 1}^{n -1} (-1)^j x^{2j} + \dfrac{(-1)^n x^{2n}}{1 + x^2}, \tag{10}$
thus establising (1). QED.
Hope this helps. Cheerio,
and as always,
Fiat Lux!!!
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/702834",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Where is the mistake in this limit calculation? I got this limit:
$$\lim_{x\to1}\frac{\sqrt[3]{x}-1}{\sqrt[4]{x}-1} \implies \lim_{x\to1}\frac{\frac{x-1}{\sqrt[3]{x²}+\sqrt[3]{x}+1}}{\sqrt[4]{x}-1} \implies \lim_{x\to1}\frac{x-1}{\sqrt[3]{x²}+\sqrt[3]{x}+1}*\frac{1}{\sqrt[4]{x}-1}*\frac{\sqrt[4]{x}+1}{\sqrt[4]{x}+1} \implies \lim_{x\to1}\frac{\sqrt[4]{x}+1}{\sqrt[3]{x²}+\sqrt[3]{x}+1} \xrightarrow[x\to 1]{}\frac{2}{3}.$$
But in the book the answer is $$ \frac{4}{3}.$$
I cannot find the mistake in my calculation.
| If fractional exponents are difficult to work with, you can also try a variable substitution: let $x = u^{12}$, so that $\sqrt[3]{x} = u^4$ and $\sqrt[4]{x} = u^3$. Then we have $$\frac{\sqrt[3]{x} - 1}{\sqrt[4]{x} - 1} = \frac{u^4 - 1}{u^3 - 1} = \frac{(u-1)(u+1)(u^2+1)}{(u-1)(u^2+u+1)} = \frac{(u+1)(u^2+1)}{u^2+u+1}.$$ And since $x \to 1$ implies $u \to 1$, the limit is simply $$\frac{(1+1)(1^2+1)}{1^2+1+1} = \frac{4}{3},$$ as claimed.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/703251",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 1
} |
Why doesn't this calculation work? I want to find some closed form for $\gcd(x^3+1,3x^2 + 3x + 1)$ but get $7$ which is not always true.
| First
$$
3(x^3+1)-(3x^2+3x+1)(x-1)=2x+4
$$
and
$$
2(3x^2+3x+1)-3(2x+4)(x-1)=14
$$
Thus, we have
$$
(3x^2-6x+5)(3x^2+3x+1)-9(x-1)(x^3+1)=14
$$
and thus, $(x^3+1,3x^2+3x+1)\mid14$. Since $3x^2+3x+1=6\binom{x+1}{2}+1$, it is always odd. Thus, we can improve the statement to
$$
(x^3+1,3x^2+3x+1)\mid7
$$
If we look mod $7$, we see that the gcd is $7$ when $x\equiv5\pmod7$ and $1$ otherwise.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/703506",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 0
} |
Solve inequality: $\frac{2x}{x^2-2x+5} + \frac{3x}{x^2+2x+5} \leq \frac{7}{8}$ Rational method to solve $\frac{2x}{x^2-2x+5} + \frac{3x}{x^2+2x+5} \leq \frac{7}{8}$ inequality?
I tried to lead fractions to a common denominator, but I think that this way is wrong, because I had fourth-degree polynomial in the numerator.
| I think I'm found nice solving method. If $x\neq 0$:
$$\frac{2x}{x^2-2x+5} + \frac{3x}{x^2+2x+5} \leq \frac{7}{8} \equiv \frac{2}{x-2+\frac{5}{x}} + \frac{3}{x+2+\frac{5}{x}} \leq \frac{7}{8}$$
Let $t=x+\frac{5}{x}$.
$$\frac{2}{t-2} + \frac{3}{t+2} \leq \frac{7}{8}$$
$$\frac{(t-6)(7t+2)}{(t-2)(t+2)} \geq 0$$
$t \in (-\infty; -2) \cup [-\frac{2}{7}; 2) \cup [6; \infty)$
And finally solved the following inequalities:
$$x + \frac{5}{x} < -2$$
$$-\frac{2}{7}\leq x + \frac{5}{x} < 2$$
$$x + \frac{5}{x} \geq 6$$
I got $x \in (-\infty; 1] \cup [5; \infty)$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/708201",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
linear algebra exercise problem, please help me with this linear combination To express each equation of this system $$x_1-x_3=0 \qquad x_2+x_3=0$$ as a linear combination of this system $$-x_1+x_2+4x_3=0 \qquad x_1+3x_2+8x_3=0 \qquad \frac{1}{2}x_1+x_2+\frac{5}{2}x_3=0$$
but I found $\begin{pmatrix}-a+b+\frac{c}{2}\\ a+3b+c\\4a+8b+\frac{5c}{2} \end{pmatrix}= \begin{pmatrix} 1\\0\\-1\end{pmatrix}$ and $\begin{pmatrix}-a+b+\frac{c}{2}\\ a+3b+c\\4a+8b+\frac{5c}{2} \end{pmatrix}= \begin{pmatrix} 0\\1\\1\end{pmatrix}$
are incosistent , then how can we make the above said linear combination? please help me with best regards
| let's call the equations in the basis set as follows:
$$\begin{align}
E1 &: -x_1+x_2+4x_3 &=0 \\
E2 &: x_1+3x_2+8x_3 &=0 \\
E3 &: \frac{1}{2}x_1+x_2+\frac{5}{2}x_3 &=0
\end{align}$$
Now, we want to find linear combinations of these three equations that give each of the following equations:
$$\begin{align}
R1 &: x_1 - x_3 &=0 \\
R2 &: x_2 + x_3&=0 \\
\end{align}$$
notice that $\frac{-1}{8}E1 + \frac{3}{8} E2 = E3$ so, E3 is a linear combination of $E1, E2$. It adds nothing…it is redundant. So, we can work with the first two equations.
After some simple algebraic manipulations, I get...
$$\frac{-3}{4}E1 + \frac{1}{4} E2 = R1$$
The second result, R2 is not a linear combination of the equations E2, E2 and E3.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/708392",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Two Dimensional Delta Epsilon Proof I was dawdling in some 2D delta epsilon examples, and I was wondering how to prove that the limit of $x^2+2xy+y^2$ has limit 3 as $(x,y)\rightarrow(1,1)$, using epsilon delta.
| Using Cauchy-Schwarz inequality twice:
$(x^2 + 2xy + y^2 - 4)^2 = ((x+y)^2 - 2^2)^2 = (((x-1) + (y-1))((x-1) + (y-1) + 4))^2$ <=
$2*((x-1)^2 + (y-1)^2)*18((x-1)^2 + (y-1)^2 + 1) = S$ .
So let epsilon e > 0 be given, we need to choose delta = d > 0 so that the square-root of S is less than epsilon e. First we want:
$(x-1)^2 + (y-1)^2 < 3$
so we can take square-root and get a whole number. So we force d to be less than 3^(1/2). So we also want that $12*((x-1)^2 + (y-1)^2)^(1/2) < e$
==>
$((x-1)^2 + (y-1)^2)$^(1/2) < e/12. So naturally, we choose
delta d = min { e/12, 3^(1/2) } > 0.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/710576",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Order of operations when using evaluation bar Suppose we have the function
\begin{align*}
f(x) = \sin(x)
\end{align*}
with first derivative
\begin{align*}
\frac{d}{dx}f(x) = \cos(x).
\end{align*}
If we evaluate $f'(x)$ at $x=0$, the result depends on whether you evaluate $f(0)$ or differentiate $f(x)$ first.
\begin{align*}
\displaystyle \frac{d}{dx}f(x)\mid_{x = 0} = \cos(x)\mid_{x = 0} = 1\\
\displaystyle \frac{d}{dx}f(x)\mid_{x = 0} = \frac{d}{dx}f(0) = \frac{d}{dx}0 = 0
\end{align*}
First question: Does this mean the following two statements are not equivalent?
\begin{align*}
\displaystyle \left(\frac{d}{dx}f(x)\right)\mid_{x = 0}\\
\displaystyle \frac{d}{dx}\left(f(x)\mid_{x = 0}\right)
\end{align*}
Second question: if so, which of the following is true, and why?
\begin{align*}
\displaystyle \frac{d}{dx}f(x)\mid_{x = 0} =
\left(\frac{d}{dx}f(x)\right)\mid_{x = 0}
\end{align*}
or
\begin{align*}
\displaystyle \frac{d}{dx}f(x)\mid_{x = 0} = \frac{d}{dx}\left(f(x)\mid_{x = 0}\right)
\end{align*}
| *
*These are not equivalent indeed. $$\displaystyle \frac{d}{dx}\left(f(x)\mid_{x = 0}\right)$$will always be 0.
*For that reason, this notation always means the first form.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/712261",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
How prove this $\frac{\cos{x}-\sin{y}}{\sin{x}-\cos{y}}=\frac{1-2\cos{x}}{1-2\sin{x}}$ let $\dfrac{\pi}{2}<x<\dfrac{3\pi}{2},0<y<\dfrac{\pi}{2}$,and such
$$\dfrac{1-\sin{x}}{1-\cos{x}}=\dfrac{1-\sin{y}}{1-\cos{y}}$$
show that
$$\dfrac{\cos{x}-\sin{y}}{\sin{x}-\cos{y}}=\dfrac{1-2\cos{x}}{1-2\sin{x}}$$
my idea:
$$\Longleftrightarrow (\cos{x}-\sin{y})(1-2\sin{x})=(\sin{x}-\cos{y})(1-2\cos{x})$$
But I fell this follow can't solve this problem,Thank you
| I'm sure that there is at least one sublimer solution than this as this is not how the problem came along , but its legitimate.
If we cross multiply the relation to be proved, we find
$$\sin x+\sin y-(\cos x+\cos y)=-2\cos(x+y)$$
Using Prosthaphaeresis & Double Angle Formulas,
$$2\cos\frac{x-y}2\left(\sin\frac{x+y}2-\cos\frac{x+u}2\right)=-2\left(\cos^2\frac{x+u}2-\sin^2\frac{x+u}2\right)$$
Now we can safely cancel $\displaystyle\cos\frac{x+y}2-\sin\frac{x+y}2$ as for $\displaystyle\cos\frac{x+u}2-\sin\frac{x+y}2=0\implies\tan\frac{x+y}2=1$
$\displaystyle\iff\frac{x+y}2=n\pi+\frac\pi4\iff x+y=2n\pi+\frac\pi2$ where $n$ is any integer
But, this is not possible due to the given ranges of $x,y$
So, we need $\displaystyle\cos\frac{x-y}2=\cos\frac{x+y}2+\sin\frac{x+y}2$
$\displaystyle\implies\sin\frac{x+y}2=\cos\frac{x-y}2-\cos\frac{x+y}2$
$\displaystyle\implies\sin\frac x2\cos\frac y2+\cos\frac x2\sin\frac y2=2\sin\frac x2\sin\frac y2$
$\displaystyle\implies\cot\frac x2+\cot\frac y2=2\ \ \ \ (1) $
$$\text{Now, }\frac{1-\sin x}{1-\cos x}=\frac{1-2\sin\frac x2\cos\frac x2}{2\sin^2\frac x2}=\frac12\csc^2\frac x2-\cot\frac x2$$
$$=\frac12\left(1+\cot^2\frac x2\right)-\cot\frac x2=\frac{\cot^2\dfrac x2-2\cot\dfrac x2+1}2$$
Method $\#1:$
So from the given condition, if we set $$\frac{1-\sin x}{1-\cos x}=\frac{1-\sin y}{1-\cos y}=K$$
$\displaystyle\implies\cot\frac x2,\cot\frac y2$ will be roots of
$$\frac{\cot^2\dfrac u2-2\cot\dfrac u2+1}2=K\iff\cot^2\dfrac u2-2\cot\dfrac u2+1-2K=0$$
$\displaystyle\implies\cot\frac x2+\cot\frac y2=\frac21$ which is same as $(1)$
Method $\#2:$
So from the given condition, $\displaystyle\frac{\cot^2\dfrac x2-2\cot\dfrac x2+1}2=\frac{\cot^2\dfrac y2-2\cot\dfrac y2+1}2$
$\displaystyle\implies \left(\cot\dfrac x2-\cot\dfrac y2\right)\left(\cot\frac x2+\cot\frac y2-2\right)=0$
If $\displaystyle\cot\dfrac x2-\cot\dfrac y2=0,\tan\frac x2=\tan\frac y2\implies \dfrac x2=\dfrac y2+m\pi\iff x=2m\pi+y$ where $m$ is any integer
But, this is not possible due to the given ranges of $x,y$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/712800",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Prove that $2^x \cdot 3^y - 5^z \cdot 7^w = 1$ has no solutions Prove that $$2^x \cdot 3^y - 5^z \cdot 7^w = 1$$ has no solutions in $\mathbb{Z}^+$, if $y\ge 3$.
| Here's a proof without Størmer's theorem, relying purely on (a lot of) modular arithmetic:
Observation 1: $z$ is odd.
Reducing mod $3$ shows that
$$1=2^x\cdot3^y-5^z\cdot7^w\equiv-(-1)^z\pmod{3}.$$
Observation 2: $x=2$ and $y\equiv2\pmod{12}$.
Reducing mod $8$ shows that if $x\geq3$ then
$$1=2^x\cdot3^y-5^z\cdot7^w\equiv-5^z\cdot7^w\pmod{8},$$
which implies that $z$ is even, a contradiction, hence $x\leq 2$. Reducing mod $5$ and $7$ shows that
$$1\equiv2^x\cdot3^y\equiv2^{x-y}\pmod{5}
\qquad\text{ and }\qquad
1\equiv2^x\cdot3^y\equiv3^{2x+y}\pmod{7},$$
which tells us that $x-y\equiv0\pmod{4}$ and $2x+y\equiv0\pmod{6}$. It follows that $y$ is even and hence also $x$ is even. Because $x\leq2$ we find that $x=2$ and hence that $y\equiv2\pmod{12}$.
Observation 3: $w=1$.
Reducing mod $8$ shows that
$$1\equiv4\cdot3^y-5^z\cdot7^w\equiv4-5\cdot7^w\pmod{8},$$
because $y$ is even and $z$ is odd, and so $w$ is also odd. Reducing mod $49$ shows that if $w\geq2$ then
$$1=4\cdot3^y-5^z\cdot7^w\equiv4\cdot3^y\pmod{49},$$
which implies that $y\equiv32\pmod{42}$. Then reducing mod $43$ shows that
$$1=4\cdot3^y-5^z\cdot7^w\equiv9-5^z\cdot7^w\pmod{43},$$
and hence $5^z\cdot7^w\equiv8\pmod{43}$. But $5$, $7$ and $8$ are all quadratic nonresidues mod $43$, and $z$ and $w$ are odd, a contradiction, hence $w=1$.
Observation 4: $y=2$.
Reducing mod $27$ shows that if $y\geq3$ then
$$1=4\cdot3^y-5^z\cdot7\equiv-5^z\cdot7\pmod{27},$$
which implies that $z\equiv13\pmod{18}$. Then $5^z\equiv17\pmod{19}$ so reducing mod $19$ then shows that
$$1=4\cdot3^y-5^z\cdot7\equiv4\cdot3^y-17\cdot7\equiv4\cdot3^y-5\pmod{19},$$
and so $4\cdot3^y\equiv6\pmod{19}$. But $4\cdot3^y\equiv5,16,17\pmod{19}$ because $y\equiv2\pmod{6}$, a contradiction, hence $y\leq2$. Because $y$ is even we find that $y=2$.
Conclusion: The only solution to
$$2^x\cdot3^y-5^z\cdot7^w=1,$$
in the positive integers is $(x,y,z,w)=(2,2,1,1)$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/713432",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Solve congruence system $\ x\equiv m_i-1 \pmod{m_i}\,$ for $\,i = 1,\ldots, k$ Find natural number $x$ so that
$$\begin{align}x&\equiv 9\pmod{10}\\ x&\equiv8\pmod9\\ &\ \ \vdots\\ x&\equiv 1\pmod2\end{align}$$
| \begin{align}
x &\equiv 9 \pmod{10} \\
x &\equiv 8 \pmod 9 \\
x &\equiv 7 \pmod 8 \\
x &\equiv 6 \pmod 7 \\
x &\equiv 5 \pmod 6 \\
x &\equiv 4 \pmod 5 \\
x &\equiv 3 \pmod 4 \\
x &\equiv 2 \pmod 3 \\
x &\equiv 1 \pmod 2 \\
\end{align}
Is equivalent to
\begin{align}
x &\equiv -1 \pmod{10} \\
x &\equiv -1 \pmod 9 \\
x &\equiv -1 \pmod 8 \\
x &\equiv -1 \pmod 7 \\
x &\equiv -1 \pmod 6 \\
x &\equiv -1 \pmod 5 \\
x &\equiv -1 \pmod 4 \\
x &\equiv -1 \pmod 3 \\
x &\equiv -1 \pmod 2 \\
\end{align}
which is equivalent to
$$x \equiv -1 \mod{\operatorname{lcm}\{2,3,4,5,6,7,8,9,10\}}$$
$$x \equiv -1 \pmod{2520}$$
$$x \equiv 2519 \pmod{2520}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/714678",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Prove that $\sqrt[3]{\sec\frac{2\pi }{7}} + \sqrt[3]{\sec\frac{4\pi }{7}} + \sqrt[3]{\sec\frac{8\pi }{7}} = \sqrt[3]{8-6\sqrt[3]{7}}$ Prove that:
$$\sqrt[3]{\sec\frac{2\pi }{7}} + \sqrt[3]{\sec\frac{4\pi }{7}} + \sqrt[3]{\sec\frac{8\pi }{7}} = \sqrt[3]{8-6\sqrt[3]{7}}$$
Thank you!
Avdiu...
| $$
\sec\frac{2\pi}{7},\qquad
\sec\frac{4\pi}{7},\qquad
\sec\frac{8\pi}{7}
$$
are the three zeros of the polynomial $z^3+4z^2-4z-8$ .
The cube-roots of those zeros are the zeros of the polynomial $y^9+4y^6-4y^3-8$ . Using as a hint the right-hand side above, we can factor this polynomial to get a polynomial of degree 6 with no real roots, and a polynomial of degree 3 with three real roots
$$
y^3+(6\cdot 7^{1/3} - 8)^{1/3}y^2+\left(
\frac{(6\cdot 7^{1/3}-8)^{2/3}(3-4\cdot 7^{1/3}-3\cdot 7^{2/3})}{25}
\right)y-2
$$
and therefore the sum of its zeros is, indeed,
$$
-(6\cdot 7^{1/3}-8)^{1/3}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/716009",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
What is the sum of $1^4 + 2^4 + 3^4+ \dots + n^4$ and the derivation for that expression What is the sum of $1^4 + 2^4 + 3^4+ \dots + n^4$ and the derivation for that expression using sums $\sum$ and double sums $\sum$$\sum$?
| The sum can be derived via combinatorial argument.
Count the number of quintuples $(x_1,x_2,x_3, x_4,x_5)$ with the property $x_5> \max\limits_{i\in S}$ from the set $S=\{1,2,...,n,n+1\}$,
then as $x_5$ varies from $2$ to $n+1$, the positions $(x_1,x_2,x_3, x_4)$
the positions can be filled in $\sum\limits_{i=1}^n i^4$ ways.
Counting it according to the cases: $5 = 4 + 1 = 2 + 2 + 1 = 3 + 1 + 1 = 2+1+1+1=1+1+1+1+1$ (where the numbers in partitions indicate the number of equal components in that partition, i.e. for example $5=4+1$ corresponds to the case $x_1=x_2=x_3=x_4<x_5$; and $2+2+1$ corresponds to the case two pairs have equal value but the value of the pair are different ,.. and so on). Count the number of ways in which each of the partitioning can happen: $1,14,36,24$ respectively (in the same order as I have presented the partitions).
You get the number of ways to be: ${n+1\choose 2}+14{n+1\choose 3}+36{n+1 \choose 4}+24{n+1\choose 5}$
Therefore, $\sum\limits_{i=1}^n i^4 = {n+1\choose 2}+14{n+1\choose 3}+36{n+1 \choose 4}+24{n+1\choose 5} = (1/5)n^5 + (1/2)n^4 + (1/3)n^3 - (1/30)n$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/718939",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 6,
"answer_id": 2
} |
Solving the functional equation $f\big(4f(x)-3x\big)=x$ Find all functions $f:[0,+\infty)\to [0,+\infty)$ such that $f(x)\geq \frac{3x}{4}$ and
$$f\big(4f(x)-3x\big)=x,\forall x\in[0,+\infty)$$
| Note that $4f(x) - 3x\geq 0$ for all $x\geq 0$.
Given a lower bound of the form $f(x) \geq a x$ with $a>0$ for $x \geq 0$. Let $x > 0$ and set $y = 4f(x) - 3x \geq 0$. Then
$$f(y) = x \geq a y = a(4f(x) - 3x) = 4af(x) - 3 ax,$$
so $f(x) \leq \frac{(3a + 1)x}{4a}$.
Similarly, given an upper bound of the form $f(x)\leq ax$ with $a>0$ for $x \geq 0$, we may again let $x > 0$ and $y = 4f(x) - 3x$ to get
$$f(y) = x \leq ay = 4af(x) - 3ax.$$
Thus, we get a lower bound $f(x) \geq \frac{(3a + 1)x}{4a}$.
Set $a_0 = \frac{3x}{4}$ and define inductively $a_{n+1} = \frac{3a_n + 1}{4a_n} = \frac{3}{4} + \frac{1}{4a_n}$ for $n > 0$. Note that $a_n > 0$ for all $n$ and that, by the above calculations, for all $x\geq 0$, $a_nx\leq f(x)\leq a_{n+1} x$ for all even $n$. We claim that
$$\lim_{n\to \infty} a_n = 1.$$
Applying the recurrence relation twice yields
$$a_{n + 2} = \frac{3\frac{3a_n + 1}{4a_n} + 1}{4\frac{3a_n + 1}{4a_n}} = \frac{13a_n + 3}{12a_n + 4} = 1 + \frac{a_n - 1}{a_n + 4}.$$
From this one can check that $a_{n-2}\leq a_n\leq 1$ when $n$ is even and that $a_{n-2}\geq a_n\geq 1$ when $n$ is odd. In particular, the sequences $\{a_n\mid n\text{ even}\}$ and $\{a_n\mid n\text{ odd}\}$ are bounded monotone and thus convergent. Let $a^-$, resp. $a^+$, be the limit of the even, resp. odd sequence, so that $a^-\leq 1\leq a^+$. Both $a^+$ and $a^-$ must satisfy
$$a = 1 + \frac{a - 1}{a + 4},$$
i.e., $a^2 + 2a - 3 = (a + 2)(a - 1) = 0$. As $a^\pm$ are positive, the only possibility is $a = 1$.
As noted above, we have
$$a_n x \leq f(x) \leq a_{n+1} x$$
for all $n$ even, so $f(x) = x$ is the only possibility.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/720604",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Definite integral by u-substitution (1/u^2), u given $$\int_{-3}^0 \frac{-8x}{(2x^2+3)^2}dx; u=2x^2+3$$
I need help solving this integral -- I'm completely bewildered. I've attempted it many times already and I don't know what I am doing wrong in my work, and I seem to be having the same problem with other definite integrals with the format $\frac{1}{u^2}$.
Here is how I've worked it out without success:
$$du=4xdx$$
$$-2\int_{21}^3\frac{1}{u^2}du=-2\int_{21}^3{u^{-2}}du$$
$$-2(\frac{-1}{2(3)^2+3}-(\frac{-1}{2(21)^2+3}))$$
$$-2(\frac{-1}{21}+\frac{1}{885})=\frac{192}{2065}$$
According to WA and the answer key to my homework I should have gotten $\frac{4}{7}$, a far cry from my answer. What am I doing wrong? (I am new to StackExchange by the way -- I apologize for poor formatting or tagging on my part)
| Since the derivative of $2x^2 + 3$ is almost the numerator, we set $u = 2x^2 + 3$. Then $du = 4x dx$ or $dx = du/(4x)$. Then take the integral so,
$$-\int \frac{8x}{(2x^2 + 3)^2}dx = -\int \frac{2}{u^2} du = \frac{2}{u} + C = \frac{2}{2x^3 + 3} + C.$$
Therefore
$$-\int_{-3}^0 \frac{8x}{(2x^2 + 3)^2}dx = 2\left[\frac{1}{2x^2 + 3}\right]_{-3}^0 = 2\left(\frac{1}{3} - \frac{1}{21}\right) = 2\left(\frac{7}{21} - \frac{1}{21}\right) = \frac{4}{7}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/720695",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Rationalization of $\frac{2\sqrt{6}}{\sqrt{2}+\sqrt{3}+\sqrt{5}}$
Question:
$$\frac{2\sqrt{6}}{\sqrt{2}+\sqrt{3}+\sqrt{5}}$$ equals:
My approach:
I tried to rationalize the denominator by multiplying it by $\frac{\sqrt{2}-\sqrt{3}-\sqrt{5}}{\sqrt{2}-\sqrt{3}-\sqrt{5}}$. And got the result to be (after a long calculation):
$$\frac{\sqrt{24}+\sqrt{40}-\sqrt{16}}{\sqrt{12}+\sqrt{5}}$$
which is totally not in accordance with the answer, $\sqrt{2}+\sqrt{3}-\sqrt{5}$.
Can someone please explain this/give hints to me.
| $$\frac{2\sqrt{6}}{\sqrt{2}+\sqrt{3}+\sqrt{5}}=\frac{2\sqrt{6}}{\sqrt{2}+\sqrt{3}+\sqrt{5}}\cdot\frac{\sqrt{2}+\sqrt{3}-\sqrt{5}}{\sqrt{2}+\sqrt{3}-\sqrt{5}}$$
$$=\frac{2\sqrt{6}\cdot(\sqrt{2}+\sqrt{3}-\sqrt{5})}{(\sqrt{2}+\sqrt{3})^2-5}=\frac{2\sqrt{6}\cdot(\sqrt{2}+\sqrt{3}-\sqrt{5})}{5+2\sqrt{6}-5}$$
$$=\frac{2\sqrt{6}\cdot(\sqrt{2}+\sqrt{3}-\sqrt{5})}{2\sqrt{6}}$$
Cancelling those $2\sqrt{6}$ you would end up with required result...
Can you see it at least now?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/720867",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "12",
"answer_count": 4,
"answer_id": 1
} |
How to prove/show that the sequence $a_n=\frac{1}{\sqrt{n^2+1}+n}$ is decreasing? How to prove/show that the sequence $a_n=\frac{1}{\sqrt{n^2+1}+n}$ is decreasing?
My idea:
*
*$n^2<(n+1)^2 /+1$
*$n^2+1<(n+1)^2+1/ \sqrt{}$
*$\sqrt{n^2+1}<\sqrt{(n+1)^2+1}/+n$
*$\sqrt{n^2+1}+n<\sqrt{(n+1)^2+1}+n$
And now I'm stuck since if I add 1 to the both sides, I don't know how to move it from the right side without also moving it from the left side.
| You have to prove $$\frac{1}{\sqrt{n^2+1}+n}>\frac{1}{\sqrt{(n+1)^2+1}+n+1}$$
i.e., $$\sqrt{n^2+1}+n< \sqrt{(n+1)^2+1}+n+1$$
i.e., $$\sqrt{n^2+1}< \sqrt{(n+1)^2+1}+1$$
i.e., $$n^2+1< (n+1)^2+1+1+2(\sqrt{(n+1)^2+1})$$
i.e., $$n^2< (n+1)^2+1+2(\sqrt{(n+1)^2+1})$$
i.e., $$0 <2n+1+1+2(\sqrt{(n+1)^2+1})$$
i.e., $$0 <n+(\sqrt{(n+1)^2+1})$$
Can you conclude this and fill gaps in between?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/722171",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 3
} |
Proving an inequality using induction Use induction to prove the following:
$\frac{1}{1}+\frac{1}{2}+\frac{1}{3}+\ldots+\frac{1}{2^n}\geq1+\frac{n}{2}$
What would the base case be? Would it still be $n=0$
so $\frac{1}{1}+\frac{1}{2}\geq 1+\frac{0}{2}$, which holds true.
then how would you prove for $n$ and $n+1$ to prove the proof with induction?
| Base case for $n = 0$: $\frac{1}{1} \geq 1$
Inductive step: Assume $\displaystyle \sum_{k = 1}^{2^n} \frac{1}{k} \geq 1 + \frac{n}{2}$
Then, after substituting the inductive step, what we want to prove is $\displaystyle \sum_{k = 2^n + 1}^{2^{n+1}} \frac{1}{k} \geq \frac{1}{2}$
This is clear to show by a comparison:
$\displaystyle \sum_{k = 2^n + 1}^{2^{n+1}} \frac{1}{k} \geq 2^n \left( \frac{1}{2^{n+1}} \right) = \frac{1}{2}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/725474",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
How to evaluate $\int_0^1\frac{1+x^4}{1+x^6}\,dx$ $$\int_0^1\frac{1+x^4}{1+x^6}\,dx$$
Can anyone help me solve the question? I am struggling with this.
| $$ \begin{align} \int_{0}^{1} \frac{x^{4}+1}{x^{6}+1} \ dx &= \int_{0}^{1} \frac{x^{4}+1+x^{2}-x^{2}}{x^{6}+1} \ dx \\ &= \int_{0}^{1} \frac{x^{4}-x^{2}+1}{x^{6}+1} \ dx + \int_{0}^{1}\frac{x^{2}}{x^{6}+1} \ dx \\ &= \int_{0}^{1} \frac{x^{4}-x^{2}+1}{(x^{2}+1)(x^{4}-x^{2}+1)} \ dx + \int_{0}^{1}\frac{x^{2}}{(x^{3})^{2}+1} \ dx \\ &= \int_{0}^{1} \frac{1}{x^{2}+1} \ dx + \frac{1}{3} \int_{0}^{1} \frac{1}{u^{2}+1} \ du \\ &= \frac{4}{3} \int_{0}^{1} \frac{1}{x^{2}+1} \ dx \\&= \frac{4}{3} \left(\frac{\pi}{4} \right) \\ &= \frac{\pi}{3} \end{align}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/725592",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
} |
Equation with floor function How would one solve an equation with a floor function in it:
$$a - (2x + 1)\left\lfloor{\frac {a - 2x(x + 1)}{2x + 1}}\right\rfloor - 2x(x + 1) = 0$$
$a$ is a given and can be treated as a natural number, and all $x$ other than integers can be discarded. At least one non-trivial solution would be sufficient.
Maybe an algorithmic method could be used?
| Rearranging we get,
$$\frac{a - 2x(x + 1)}{(2x + 1) } =\left\lfloor{\frac {a - 2x(x + 1)}{2x + 1}}\right\rfloor $$
This means,$\frac{a - 2x(x + 1)}{(2x + 1) }$ is an integer.
$$\frac{a - 2x(x + 1)}{(2x + 1) }=k$$
$$a=2xk+k+2x^2+2x$$
$$2x^2+2(k+1)x+k-a=0$$
Applying the quadratic formula,
$$x = \frac{-2(k+1)\pm\sqrt{4(k+1)^2-8(k-a)}}{4}$$
$a$ is known, now we substitute integers in place of $k$ to obtain solutions.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/726195",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
How to find characteristic polynomial of this matrix?
Let, $A=\begin{bmatrix} 4&0&1&0\\1&1&1&0\\0&1&1&0 \\0&0&0&4 \end{bmatrix}$. Knowing that $4$ is one of its eigenvalues, find the characteristic polynomial of $A$.
Well if $4$ is an eigenvalues of $A$, one should have $|A-4I_{4}|=0$ .
And so,
$\begin{vmatrix} 0&0&1&0\\1&-3&1&0\\0&1&-3&0 \\0&0&0&0 \end{vmatrix}=0$
It's clear that the previous equation is true (the determinant of $(A-4I_{4})=0$). Now that the factor $(\lambda-4)$ was pull out, one gets a new matrix by removing the null row and null column.
$A'=\begin{bmatrix} 0&0&1\\1&-3&1\\0&1&-3&\end{bmatrix}$
The characteristic polynomial of $A'$ will be a $3^{th}$ degree polynomial, which product with $(\lambda-4)$ equals to a $4^{th}$ degree polynomial.
Now, in order of finding the characteristic polynomial of $A'$ one must to solve the characteristic equation:
$\begin{vmatrix} -\lambda&0&1\\1&-3-\lambda&1\\0&1&-3-\lambda&\end{vmatrix}=0$
My doubt is on finding this determinant. I already tryed Laplace's transformations in order to make null row or a column, but I couldn't do it.
Can you give me a clue? Thanks.
| ...and how to find the charateristic polynomial of the original matrix, $A$
$$\begin{align}\mathrm{det}(A - \lambda \mathrm{I}) &= 0 \tag{1}\\
\begin{vmatrix}(4-\lambda)&0&1&0 \\
1&(1-\lambda)&1&0\\
0&1&(1-\lambda)&0\\
0&0&0&(4-\lambda) \end{vmatrix} &= 0 \tag{2}\\
(4-\lambda)\begin{vmatrix}(4-\lambda)&0&1\\
1&(1-\lambda)&1\\
0&1&(1-\lambda) \tag{3}\\
\end{vmatrix} &= 0 \\
(4-\lambda) \left[(4-\lambda) \left[(1-\lambda)^2 -1 \right] + 1\right] &= 0\tag{4} \\
(4-\lambda) \left[(4-\lambda)(\lambda^2 -2\lambda) + 1\right] &= 0\tag{5} \\
(4-\lambda) \left[(4\lambda^2 -8\lambda -\lambda^3+2\lambda^2) + 1\right] &= 0 \tag{6}\\
(4-\lambda)(-\lambda^3 + 6\lambda^2 - 8\lambda +1 ) &= 0 \tag{7}\\
(\lambda -4)(\lambda^3 - 6\lambda^2 + 8\lambda - 1)&= 0\tag{8}\\
\end{align}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/727972",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 3
} |
How to write $x=2\cos(3t) y=3\sin(2t)$ in rectangular coordinates? How would I write the following in terms of $x$ and $y$? I think I use the inverse $\cos$ or $\sin$?
$$x=2\cos(3t)\,, \quad y=3\sin(2t)$$
| We have $\cos(3t)=\frac{x}{2}$. Using the identity $\cos 2u=2\cos^2(u)-1$, we get
$$\cos(6t)=2\left(\frac{x}{2}\right)^2-1.\tag{1}$$
We also have $\sin(2t)=\frac{y}{3}$. Using the not so well-known identity $\sin(3u)=3\sin u-4\sin^3 u$, we get
$$\sin(6t)=3\left(\frac{y}{3}\right)-4\left(\frac{y}{3}\right)^3.\tag{2}$$
Now use (1) and (2), and the fact that $\cos^2(6t)+\sin^2(6t)=1$ to eliminate $t$.
We get the remarkably ugly equation
$$\left( 2\left(\frac{x}{2}\right)^2-1 \right)^2+\left( 3\left(\frac{y}{3}\right)-4\left(\frac{y}{3}\right)^3 \right)^2=1.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/728135",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
} |
Prove $\int\limits_{0}^{\pi/2}\frac{dx}{1+\sin^2{(\tan{x})}}=\frac{\pi}{2\sqrt{2}}\bigl(\frac{e^2+3-2\sqrt{2}}{e^2-3+2\sqrt{2}}\bigr)$
Prove the following integral
$$I=\int\limits_{0}^{\frac{\pi}{2}}\dfrac{dx}{1+\sin^2{(\tan{x})}}=\dfrac{\pi}{2\sqrt{2}}\left(\dfrac{e^2+3-2\sqrt{2}}{e^2-3+2\sqrt{2}}\right)$$
This integral result was calculated using Mathematica and
I like this integral. But I can't solve it.
My idea:
Let $$\tan{x}=t\Longrightarrow dx=\dfrac{1}{1+t^2}dt$$
so
$$I=\int\limits_{0}^{\infty}\dfrac{dt}{1+\sin^2{t}}\cdot \dfrac{1}{1+t^2}$$
then I can't proceed. Can you help me? Thank you.
| I hope nobody cares that i exhume this question, but i found it interesting that it is possible to obtain this integral by a relativly straightforward contour integration method.
Observe that,following the question opener and using parity, that we can rewrite the integral as
$$
\frac{1}{2}\int^{\infty}_{-\infty}\frac{1}{1+t^2}\frac{1}{1+\sin^2(t)}
$$
It is now easy to show that the poles are
$$
t_{\pm}=\pm i\\
t_{n\pm}=\pi n\pm i \text{arcsinh(1)}
$$
so we have two isolated poles and the rest lies on two straight lines paralell to the real axis.
Because the integrand interpreted as a complex function converges as $|z|\rightarrow\infty$ we can choose a semicircle closed in the upper half plane as an integration contour. We find
$$
I=\pi i\sum_{n=-\infty}^{\infty}\text{res}(t_{n+})+\pi i \text{res}(t_{+})
$$
Where the residues are given by
$$
\text{res}(t_{+})=\frac{i}{2}\frac{1}{2 \sinh^2(1)-1}\\
\text{res}(t_{n+})=\frac{-i}{2\sqrt{2}}\frac{1}{1+(n \pi+i \text{arcsinh(1)} )^2}
$$
Therefore the integral reduces to the following sum
$$
I=\frac{\pi}{2\sqrt{2}} \sum_{n=-\infty}^{\infty} \frac{1}{1+(n \pi+i \text{arcsinh(1)})^2} -\frac{\pi}{2}\frac{1}{2 \sinh^2(1)-1}
$$
Using a partial fraction decomposition together with the Mittag-Leffler expansion of $\coth(x)$, this can be rewritten as
$$
I=\frac{\pi}{4\sqrt{2}} \sum_{n=-\infty}^{\infty} \frac{-i}{-i+n \pi+ \text{arcsinh(1)}}+ \frac{i}{i+n \pi+ i\text{arcsinh(1)}}-\frac{\pi}{2}\frac{1}{2 \sinh^2(1)-1}=\\
\frac{\sqrt{2} \pi}{8} \left( \coth \left(1-\text{arcsinh(1)}\right)+ \coth \left(1+\text{arcsinh(1)}\right)\right)-\frac{\pi}{2}\frac{1}{2 \sinh^2(1)-1}\\
$$
Or
$$
I\approx 1.16353
$$
Which matches the claimed result.
One can also compute this explicitly noting that $\text{arcsinh(1)}=\log(1+\sqrt{2})$ (*). But this is rather tedious so i just leave this step to the reader and conclude that
$$
I=\frac{e^2+3-2\sqrt{2}}{e^2-3+2\sqrt{2}}
$$
Appendix
Just to give some details of the last part of the calculations:
Using (*) the part stemming from the sum is
$$
\frac{\pi}{4\sqrt{2}}\left(\frac{ \frac{1+\sqrt{2}}{e}+\frac{e}{1+\sqrt{2}}}{ \frac{e}{1+\sqrt{2}}-\frac{1+\sqrt{2}}{e}}+\frac{e \left(1+\sqrt{2}\right)+\frac{1}{1+\sqrt{2} e} }{\left(1+\sqrt{2}\right) e-\frac{1}{\left(1+\sqrt{2}\right) e}}\right)=\\
\frac{\left(e^4-1\right) \pi }{2 \sqrt{2} \left(1-6 e^2+e^4\right)}
$$
The part of the single pole gives
$$
\frac{\pi }{2 \left(\left(\frac{e}{2}-\frac{1}{2 e}\right)^2-1\right)}=\frac{2 e^2 \pi }{1-6 e^2+e^4}
$$
Adding both terms and factorizing then yields the desired result
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/730876",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "52",
"answer_count": 3,
"answer_id": 1
} |
Find the least value of x which when divided by 3 leaves remainder 1, ... A number when divided by 3 gives a remainder of 1; when divided by 4, gives a remainder of 2; when divided by 5, gives a remainder of 3; and when divided by 6, gives a remainder of 4. Find the smallest such number.
How to solve this question in 1 min?
Any help would be appreciated. :)
| So we need $\displaystyle x=3a+1=4b+2=5c+3=6d+4$
which can also the written as $\displaystyle x=3(a+1)-2=4(b+1)-2=5(c+1)-2=6(d+1)-2$
So, we need to find $x$ such the remainder $=-2$ for the divisors $3,4,5,6$
Now, the smallest number which is divisible by $3,4,5,6$ is lcm$(3,4,5,6)=60$
So, $60m-2$ (where $m$ is an integer) will leave $-2$ as remainder
Find proper $m$ for the minimum positive value of $x$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/731299",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 1
} |
cyclic subgroup elements I'm having hard time finding elements of the cyclic subgroup $\langle a\rangle$ in $S_{10}$, where $a = (1\ 3\ 8\ 2\ 5\ 10)(4\ 7\ 6\ 9)$
This is my attempt:
\begin{align}
a^2 &= (1\ 8\ 5\ 10)(4\ 6\ 9) \\
a^3 &= (1\ 3\ 5\ 10)(4\ 7\ 9\ 6) \\
a^4 &= (1\ 5\ 10)(4\ 9\ 7) \\
a^5 &= (1\ 3\ 8\ 2\ 10)(7\ 6) \\
a^6 &= (1\ 8\ 10)(4\ 6\ 9) \\
a^7 &= (1\ 3\ 10)(4\ 7\ 9\ 6) \\
a^8 &= (1\ 10)(4\ 9\ 7) \\
a^9 &= (1\ 3\ 8\ 2\ 5\ 10)(7\ 6) \\
a^{10} &= (1\ 8\ 5)(4\ 6\ 9) \\
a^{11} &= (1\ 3\ 5\ 10)(4\ 7\ 9) \\
a^{12} &= (1\ 5)(4\ 9\ 7\ 6)
\end{align}
I suspect I already went wrong somewhere. I understand I need to get to $e = (1)$ at some point. Is there a way to check and make sure there are no mistakes when you calculate this?
| Your calculations look wrong. Keep in mind that $a = (1\ 3\ 8\ 2\ 5\ 10)(4\ 7\ 6\ 9)$ is the map $a:\{1,2,\dots,10\}\to\{1,2,\dots,10\}$ given by
$$
a : 1\mapsto 3, 3\mapsto 8, 8 \mapsto 2, 2\mapsto5, 5\mapsto 10, 10\mapsto 1, 4\mapsto 7, 7\mapsto 6, 6 \mapsto 9, 9\mapsto 4.
$$
Applying this map twice yields
$$
a^2 : 1\mapsto 8, 8\mapsto 5, 5\mapsto 1, 3\mapsto 2, 2\mapsto 10, 10\mapsto 3, 4\mapsto 6, 6\mapsto 4, 7\mapsto 9, 9 \mapsto 7,
$$
which is written in cycle notation $a^2 = (1\ 8\ 5)(3\ 2\ 10)(4\ 6)(7\ 9)$.
For $a^3$ you have to apply $a$ three times and get $a^3 = (1\ 2)(3\ 5)(8\ 10)(4\ 9\ 6\ 7)$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/731792",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
How to calculate the diagonal matrix The question says: find the eigenvalues and corresponding eigenvectors of the matrix $A$. This I could do. But then it says: hence find a non-singular matrix $P$ and a diagonal matrix $D$ such that $A + A^2 + A^3 = PDP^{-1}$ , where
$$
A =\begin{pmatrix}
6 & 4 & 1 \\
-6 & -1 & 3 \\
8 & 8 & 4 \\
\end{pmatrix}
$$
The eigenvalues are $-1, 2, 8$ with corresponding eigenvectors $\begin{pmatrix}
4 \\
-9 \\
8 \\
\end{pmatrix}$,
$\begin{pmatrix}
5 \\
-6 \\
4 \\
\end{pmatrix}$,
$\begin{pmatrix}
1 \\
0 \\
2 \\
\end{pmatrix}$
I cannot calculate $A + A^2 + A^3$ numerically, so how can I do this?
| You can find $Q$ such that $A=QDQ^{-1}$ with $D$ diagonal, right? Then show that $$A+A^2+A^3=Q(D+D^2+D^3)Q^{-1}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/731995",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Proof by induction that $n^3 + (n + 1)^3 + (n + 2)^3$ is a multiple of $9$. Please mark/grade. What do you think about my first induction proof? Please mark/grade.
Theorem
The sum of the cubes of three consecutive natural numbers is a multiple of 9.
Proof
First, introducing a predicate $P$ over $\mathbb{N}$, we rephrase the theorem as follows.
$$\forall n \in \mathbb{N}, P(n)
\quad \text{where} \quad
P(n) \, := \, n^3 + (n + 1)^3 + (n + 2)^3 \text{ is a multiple of 9}$$
We prove the theorem by induction on $n$.
Basis
Below, we show that we have $P(n)$ for $n = 0$.
$$0^3 + 1^3 + 2^3 = 0 + 1 + 8 = 9 = 9 \cdot 1$$
Inductive step
Below, we show that for all $n \in \mathbb{N}$, $P(n) \Rightarrow P(n + 1)$.
Let $k \in \mathbb{N}$. We assume that $P(k)$ holds.
In the following, we use this assumption to show that $P(k + 1)$ holds.
By the assumption,
there is a $i \in \mathbb{N}$ such that
$i \cdot 9 = k^3 + (k + 1)^3 + (k + 2)^3$.
We use this fact in the following equivalent transformation.
The transformation turns the sum of cubes in the first line,
for which we need to show that it is a multiple of 9,
into a product of 9 and another natural number.
$(k + 1)^3 + (k + 2)^3 + (k + 3)^3 \\
= (k + 1)^3 + (k + 2)^3 + k^3 + 9k^2 + 27k + 27 \\
= k^3 + (k + 1)^3 + (k + 2)^3 + 9k^2 + 27k + 27 \quad | \text{ using the induction hypothesis} \\
= 9i + 9k^2 + 27k + 27 \\
= 9 \cdot i + 9 \cdot k^2 + 9 \cdot 3k + 9 \cdot 3 \\
= 9 \cdot (i + k^2 + 3k + 3)$
We see that the above product has precisely two factors: 9 and another natural number.
Thus the product is a multiple of 9.
This completes the induction.
| Formulation, base case, inductive hypothesis, inductive step, it all looks good. :)
One might also conclude with a clarifying statement about what has been done - that the hypothesis is true for all $n \in \Bbb N$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/732445",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "19",
"answer_count": 3,
"answer_id": 2
} |
Taylor series of $\sqrt{1+x}$ using sigma notation I want help in writing Taylor series of $\sqrt{1+x}$ using sigma notation
I got till
$1+\frac{x}{2}-\frac{x^2}{8}+\frac{x^3}{16}-\frac{5x^4}{128}+\ldots$ and so on.
But I don't know what will come in sigma notation.
| If $f(x) = \sqrt{1+x} = (1+x)^\frac{1}{2}$, then $f'(x) = \frac{1}{2}(1+x)^{-\frac{1}{2}}$ and $f''(x) = \frac{1}{2}\frac{-1}{2}(1+x)^{-\frac{3}{2}}$. In general, $$
\frac{d}{dx}(1 + x)^{\frac{-n}{2}} = \frac{-n}{2}(1 + x)^{\frac{-n-2}{2}} \text{,}
$$
and therefore the $n$-th derivative of $f(x) = \sqrt{1+x}$ is (for $n > 1$) $$
\frac{1}{2}\underbrace{\frac{-1}{2}\frac{-3}{2}\ldots\frac{-2n+3}{2}}_{n-1\textrm{ terms}}(1+x)^{-\frac{-2n+1}{2}} \text{.}
$$
Evaluating at $x=0$ yields (again for $n > 1$) $$
f^{(n)}(0) = (-1)^{n-1}\frac{1\cdot 3 \cdot \ldots \cdot (2n-3)}{2^n}
$$
and the taylor series around $x=0$ is thus $$
\sum_{n=0}^\infty x^n\frac{f^{(n)}(0)}{n!} = 1 + x\frac{1}{2} - x^2\frac{1}{8} + \sum_{n=3}^\infty x^n (-1)^{n-1}\frac{1\cdot 3 \cdot \ldots \cdot (2n-3)}{n!2^n} \text{.}
$$
Cutting this off at $x^4$ yields $$
\sqrt{1+x} \approx 1 + x\frac{1}{2} - x^2\frac{1}{8} + x^3\frac{1\cdot 3}{3!2^3} - x^4\frac{1\cdot 3\cdot 5}{4!2^4}
= 1 + x\frac{1}{2} - x^2\frac{1}{8} + x^3\frac{1}{16} - x^4\frac{5}{128}
$$
so your initial coefficients are correct.
You can write the series in terms of factorials by using that $1\cdot 3 \cdots\ldots\cdot (2k+1) = \frac{(2k+1)!}{2 \cdot 4\cdot\ldots\cdot (2k)}$ and that $2 \cdot 4\cdot\ldots\cdot (2k) = (2\cdot1)\cdot(2\cdot 2)\cdot(2\cdot 3)\cdot\ldots\cdot(2k) = k!2^k$. Overall, $$
1\cdot 3 \cdots\ldots\cdot (2k+1) = \frac{(2k+1)!}{k!2^k}
$$
and therefore ($k = n - 2$) $$
\frac{1\cdot 3 \cdot \ldots \cdot (2n-3)}{n!2^n}
= \frac{(2n-3)!}{n!2^n(n-2)!2^{n-2}}
= \frac{(2n-3)!}{n!(n-2)!2^{2n-2}} \text{,}
$$
so the taylor series becomes $$
\sum_{n=0}^\infty x^n\frac{f^{(n)}(0)}{n!} = 1 + x\frac{1}{2} - x^2\frac{1}{8} + \sum_{n=3}^\infty x^n (-1)^{n-1}\frac{(2n-3)!}{n!(n-2)!2^{2n-2}} \text{.}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/732540",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 3,
"answer_id": 0
} |
Cartesian to polar coordinates - Integration $$\iint_R \frac{1}{1+x^2+y^2} \,dA$$
$$R=\left\{(r,\theta):1\le r\le 2,0\le \theta \le \pi\right\}$$
limits of outer integral are $0$ to $\pi$ and inner integral are $1$ to $2$. I wanted to confirm if i did the problem right.
My answeR: $(1/2)\ln(5/2)\pi$
| Let $r=\sqrt{x^2+y^2}$ and $dA=dxdy =rdrd\theta$, then
$$\int \int_{R}\frac{dA}{1+x^2+y^2}=\int_{\theta=0}^{\theta=\pi}\int_{r=1}^{r=2}\frac{r drd\theta}{1+r^2}= $$
$$=\int_{\theta =0}^{\theta =\pi}\left(\frac{1}{2} \int_{r=1}^{r=2}\frac{2r dr}{1+r^2}\right)d\theta=\int_{\theta=0}^{\theta=\pi}d\theta \cdot \left.\frac{1}{2}\ln(1+r^2)\right|_{r=1}^{r=2}= $$
$$=\pi\frac{1}{2} (\ln(1+4)-\ln(1+1))=\frac{\pi}{2}\ln\frac{5}{2}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/735949",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Demonstrate that $\displaystyle \frac{(2n - 2)!!}{(2n - 3)!!} \simeq 1.7 \sqrt{n}$ As in the title, I know that
$\displaystyle \frac{(2n - 2)!!}{(2n - 3)!!} = \frac{(2n - 2)(2n - 4)\cdots 4 \cdot 2}{(2n - 3)(2n - 5) \cdots 3 \cdot 1} \simeq 1.7 \sqrt{n}$
Could you give some hint to prove it?
(should I look the series expansion of $\sqrt{n}$?)
Thank you anyway!
| Use the Stirling formula:
$$\frac{(2n - 2)(2n - 4)\cdots 4 \cdot 2}{(2n - 3)(2n - 5) \cdots 3 \cdot 1}
=\frac{(2n - 2)^2(2n - 4)^2\cdots 4^2 \cdot 2^2}
{(2n - 2)!}
\\=
2^{2n-2}
\frac{(n - 1)^2(n - 2)^2\cdots 2^2 \cdot 1^2}
{(2n - 2)!}
\\=2^{2n-2}
\frac{((n - 1)!)^2}{(2n - 2)!}
\sim 2^{2n-2}\frac{(n-1)^{2n-2}e^{-2n+2}2\pi n}
{(2n - 2)^{2n-2}e^{-2n+2}\sqrt{2\pi 2n}}=
{\sqrt{\pi n}}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/738107",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 0
} |
How find the $a_{10}+a_{2014}$ if $a_{n+1}=\frac{8}{5}a_{n}+\frac{6}{5}\sqrt{4^n-a^2_{n}}$
The sequence $\{a_{n}\}$ satisfies $a_{0}=1$,and
$$a_{n+1}=\dfrac{8}{5}a_{n}+\dfrac{6}{5}\sqrt{4^n-a^2_{n}},n\ge 0$$
Find the $a_{10}+a_{2014}$.
My idea: since
$$5a_{n+1}-8a_{n}=6\sqrt{4^n-a^2_{n}},n\ge 0$$
so
$$25a^2_{n+1}-80a_{n+1}a_{n}+64a^2_{n}=36(4^n-a^2_{n})$$
$$\Longrightarrow 25a^2_{n+1}-80a_{n+1}a_{n}+100a^2_{n}=36\cdot 4^n\tag{1}$$
and
$$25a^2_{n}-80a_{n}a_{n-1}+100a^2_{n-1}=36\cdot 4^{n-1}$$
$$\Longrightarrow 100a^2_{n}-320a_{n}a_{n-1}+400a^2_{n-1}=36\cdot 4^n\tag{2}$$
then $(1)-(2)$,we have
$$25(a^2_{n+1}-16a^2_{n-1})=80a_{n}(a_{n+1}-4a_{n-1})$$
so
$$(a_{n+1}-4a_{n-1})(5a_{n+1}-16a_{n}+20a_{n-1})=0$$
so
$a_{n+1}=4a_{n-1}$,
or
$5a_{n+1}=16a_{n}-20a_{n-1}$
and I find this ugly,But I fell very ugly.maybe have other methods,it is said can find the $a_{n}$ close form.
$$a_{n}=2^n\sin{x_{n}}?$$.
But I can't.Thank you very much
| We have $a_0=1, a_1=\frac{8}{5}, a_2=4, a_3=\frac{32}{5}$. This leads us to see the pattern $$a_{2n}=2^{2n}, a_{2n+1}=\frac{2^{2n+3}}{5}$$ This is easily proven by induction. Thus $a_{10}+a_{2014}=2^{10}+2^{2014}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/739247",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
} |
Integrating partial fractions I have
$\int{\frac{2x+1}{x^2+4x+4}}dx$
Factorising the denominator I have
$\int{\frac{2x+1}{(x+2)(x+2)}}dx$
From there I split the top term into two parts to make it easier to integrate
$\int{\frac{2x+1}{(x+2)(x+2)}}dx$ = $\int{\frac{A}{(x+2)}+\frac{B}{(x+2)}}dx$
=$\int{\frac{A(x+2)}{(x+2)^2}+\frac{B(x+2)}{(x+2)^2}}dx$
Therefore
$2x+1 = A(x+2) +B(x+2)$
This is where I would normally use a substitution method to eliminate either the A term or B term by letting x = something like -2, which would get rid of the A and usually leave me with the B term to solve. However since they are the same I'm not sure what to do.
I've been told to try evaluate the co-efficients, but am not sure how.
| You want to try a split like
$$\frac{2 x+1}{(x+2)^2} = \frac{A}{x+2} + \frac{B x}{(x+2)^2}$$
Then $A+B=2$ and $2 A=1$. The decomposition is then
$$\frac{2 x+1}{(x+2)^2} = \frac12 \frac1{x+2} + \frac{3}{2} \frac{x}{(x+2)^2}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/741861",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 1
} |
Optimize function on $x^2 + y^2 + z^2 \leq 1$ Optimize $f(x,y,z) = xyz + xy$ on $\mathbb{D} = \{ (x,y,z) \in \mathbb{R^3} : x,y,z \geq 0 \wedge x^2 + y^2 + z^2 \leq 1 \}$. The equation $\nabla f(x,y,z) = (0,0,0)$ yields $x = 0, y = 0, z \geq 0 $ and we can evaluate $f(0,0,z) = 0$.
Now studying the function on the boundary $x^2 + y^2 + z^2 = 1$ gets really hairy. I tried replacing $x$ with $\sqrt{1 - y^2 - z^2}$ in order to transform $f(x,y,z)$ into a two-variable function $g(y,z)$ and optimize it on $y^2 + z^2 \leq 1$ but $g(y,z)$ is a pain to differentiate. I then tried spherical coordinates which really did not make it any much easier.
Got any suggestions on how to tackle it?
| $f=xy(z+1) \le \dfrac{x^2+y^2}{2}(z+1)=\dfrac{(z+1)(k-z^2)}{2},k\le1$
$f'_z=\dfrac{k-z^2-2z^2-2z}{2}=0,z=\dfrac{-2 \pm \sqrt{4+12k}}{6}=\dfrac{-1 \pm \sqrt{1+3k}}{3}$
for max: $z=\dfrac{-1 + \sqrt{1+3k}}{3},f_{max}=\dfrac{(2+\sqrt{1+3k})(6k-1+2\sqrt{1+3k})}{2*27}\le \dfrac{32}{2*27}$ ,
when $k=1$ get max becasue $6k-1+2\sqrt{1+3k}$ is mono increasing funtion
$z= \dfrac{1}{3}, x=y=\dfrac{2}{3}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/742031",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
Area bounded by two circles $x^2 + y^2 = 1, x^2 + (y-1)^2 = 1$ Consider the area enclosed by two circles: $x^2 + y^2 = 1, x^2 + (y-1)^2 = 1$
Calculate this area using double integrals:
I think I have determined the region to be $D = \{(x,y)| 0 \leq y \leq 1, \sqrt{1-y^2} \leq x \leq \sqrt{1 - (1-y)^2}\}$
Now I can't seem to integrate this. Is this region wrong? Should the integral just be $\int_0^1 \int_{\sqrt{1-y^2}}^{\sqrt{1- (1-y)^2}} dx dy$?
Do I need to convert this to polar form?
| Solve the system: $x^2 + y^2 = 1 = x^2 + (y - 1)^2$ we have: $x =\dfrac{\sqrt{3}}{2}$ and $y = \dfrac{1}{2}$, and $x = -\dfrac{\sqrt{3}}{2}$ and $y = \dfrac{1}{2}$. So the set up for the integral is:$\int_\frac{-\sqrt{3}}{2}^{\frac{\sqrt{3}}{2}} \int_{1 -\sqrt{1 - x^2}}^{\sqrt{1 - x^2}} 1dydx$ which can be easily computed.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/742949",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
AM-GM inequality On the wikipedia page on
"Nesbit's inequality", the fifth proof ends as follows:
$$ \frac{x+z}{y}+\frac{y+z}{x}+\frac{x+y}{z}\geq 6$$
which is true, by AM-GM inequality.
I am wondering if the inequality is obvious / immediate from just looking at it and how you see this immediately without resorting to something like the following proof:
\begin{align*}
\frac{x+z}{y}+\frac{y+z}{x}+\frac{x+y}{z} &\geq
2\cdot\left(\frac{\sqrt{xz}}{y}+\frac{\sqrt{yz}}{x}+\frac{\sqrt{xy}}{z}\right) \\
&\geq
6\cdot\left(\frac{\sqrt{xz}}{y} \cdot \frac{\sqrt{yz}}{x} \cdot\frac{\sqrt{xy}}{z}\right)^{\frac13} \\
&=6\cdot(1)^{\frac13}=6
\end{align*}
| $$ \frac{x+z}{y}+\frac{y+z}{x}+\frac{x+y}{z}\geq 6$$ is equivalent to
$$ \frac{x^2 z+z^2x + y^2 z+z^2 y + x^2 y + y^2 x}{6}\geq xyz$$
which is true by the AM-GM inequality:
$$ \frac{x^2 z+z^2x + y^2 z+z^2 y + x^2 y + y^2 x}{6}\geq \sqrt[6]{x^6y^6z^6}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/746977",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Best strategy for rolling $20$-sided and $10$-sided dice There is a $20$-sided (face value of $1$-$20$) die and a $10$-sided (face value of $1$-$10$) dice. $A$ and $B$ roll the $20$ and $10$-sided dice, respectively. Both of them can roll their dice twice. They may choose to stop after the first roll or may continue to roll for the second time. They will compare the face value of the last rolls. If $A$ gets a bigger (and not equal) number, $A$ wins. Otherwise, $B$ wins. What's the best strategy for $A$? What's $A$'s winning probability?
I know this problem can be solved using the indifference equations, which have been described in details in this paper by Ferguson and Ferguson. However, this approach is complicated and it’s easy to make mistakes for this specific problem. Are there any other more intuitive methods?
Note: $A$ and $B$ roll simultaneouly. They don't know each other's number until the end when they compare them with one another.
| $B$ should reroll anything $5$ or below as he is likely to improve, while with a $6$ he is more likely to get worse. He then has $\frac 1{20}$ chance of getting each number $1$ to $5$ and $\frac 3{20}$ chance of any number $6$ to $10$.
$A$ can then compute the chance that each number wins. Any number $11$ through $20$ is a guaranteed win, so clearly should be accepted. If he rolls a second die, each number is equally probable, and he will win $\frac 12$ (that he rolls $11-20$) plus $\frac 1{20} \cdot \frac {17}{20}$ (that he rolls a $10$ and wins) plus $\frac 1{20} \cdot \frac {14}{20}$ (that he rolls a $9$ and wins) plus ... $\frac 1{20} \cdot \frac {4}{20}$ (that he rolls a $5$ and wins) plus ... The total is
$$\frac 12+\frac 1{20}\left(\frac{17+14+11+8+5+4+3+2+1}{20}\right)=\frac{53}{80}$$
He should therefore accept a $9$ or higher as these have a higher winning chance than a random roll. His winning chance is then $\frac 12$ (that the first roll is $11-20$) plus $\frac 1{20} \cdot \frac {17}{20}$ (that he rolls a $10$ and wins) plus $\frac 1{20} \cdot \frac {14}{20}$ (that he rolls a $9$ and wins) plus $\frac8{20}\cdot \frac {53}{80}$ (that he rolls $8$ or less and wins with the second roll). The total is
$$\frac 12+\frac {17+14}{20^2}+\frac 8{20}\cdot \frac {53}{80}=\frac {1348}{1600}=\frac {337}{400}=0.8425$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/747664",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 0
} |
Convergence of $ \sum_{n=1}^{\infty} (\frac{n^2+1}{n^2+n+1})^{n^2}$
Find if the following series converge: $$\displaystyle \sum_{n=1}^{\infty} \left(\frac{n^2+1}{n^2+n+1}\right)^{n^2}$$
What I did:
$$a_n=\left(\frac{n^2+1}{n^2+n+1}\right)^{n^2}$$
$$b_n=\frac {1}{(n+1)^{n^2}}$$
$$\forall n\ge 1 : a_n < b_n \ \Rightarrow \ \displaystyle \sum_{n=1}^{\infty}\left(\frac{n^2+1}{n^2+n+1}\right)^{n^2}<\displaystyle \sum_{n=1}^{\infty}\frac {1}{(n+1)^{n^2}} \approx \displaystyle \sum_{n=1}^{\infty} \frac 1 {n^2}$$
$\displaystyle \sum_{n=1}^{\infty}b_n$ converges like $\displaystyle \sum_{n=1}^{\infty}\frac 1 {n^2}$ so $\displaystyle \sum_{n=1}^{\infty}a_n$ converges as well.
| This is likely what I'd have a Calculus student do. Use the root test:
$$\lim_{n \rightarrow \infty} \left| \left(\frac{n^2+1}{n^2+n+1}\right)^{n^2}\right|^\frac{1}{n} = \lim_{n \rightarrow \infty} \left( \frac{n^2+1}{n^2+n+1}\right)^n$$
$$\begin{array} \\
\lim_{n \rightarrow \infty} \ln \left(\left( \frac{n^2+1}{n^2+n+1}\right)^n\right) &= \lim_{n \rightarrow \infty} n \ln \left( \frac{n^2+1}{n^2+n+1}\right) \\
&= \lim_{n \rightarrow \infty} \frac{\ln \left( \frac{n^2+1}{n^2+n+1}\right)}{\frac{1}{n}} \\
&= \lim_{n \rightarrow \infty} \frac{\left( \frac{n^2+n+1}{n^2+1}\right)\left(\frac{(n^2+n+1)(2n)-(n^2+1)(2n+1)}{(n^2+n+1)^2}\right)}{-\frac{1}{n^2}} \\
&= \lim_{n \rightarrow \infty} -\frac{n^2(2n^3+2n^2+2n - 2n^3 - n^2 - 2n - 1)}{(n^2+1)(n^2+n+1)} \\
&= \lim_{n \rightarrow \infty} -\frac{n^2(n^2-1)}{(n^2+1)(n^2+n+1)} = -1
\end{array}$$
Now since the limit of $\ln (f(n)) = -1$, we have that the limit of $f(n) = \frac{1}{e} < 1$, and the series is absolutely convergent.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/748110",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Finding correlation coefficient Two dice are thrown. $X$ denotes number on first die and $Y$ denotes maximum of the numbers on the two dice. Compute the correlation coefficient.
| We outline a basically computational approach, leaving details to you. The mean of $X$ is easily found to be $\frac{7}{2}$. For the variance of $X$, we first find $E(X^2)$, which is
$$\frac{1}{6}(1^2+2^2+3^2+4^2+5^2+6^2).$$
Thus $\text{Var}(X)=\frac{91}{6}-\left(\frac{7}{2}\right)^2=\frac{35}{12}$.
Next we find the mean and variance of $Y$. The probability that the maximum is $1$ is $\frac{1}{36}$. We have $Y=2$ if our tosses are $(1,2)$, $(2,1)$, or $(2,2)$. Thus $\Pr(Y=2)=\frac{3}{36}$. We have $Y=3$ if our tosses are $(1,3)$, $(2,3)$, $(3,3)$, $(3,1)$, or $(3,2)$, for a probability of $\frac{5}{36}$. Similarly, we can see that $\Pr(X=4)=\frac{7}{36}$, and $\Pr(X=5)=\frac{9}{36}$ and $\Pr(X=6)=\frac{11}{36}$. Now we can compute $E(Y)$ and $E(Y^2)$, and then $\text{Var}(Y)$.
The correlation coefficient is
$$\frac{\text{Cov}(X,Y)}{\sqrt{\text{Var}(X)\text{Var}(Y)}}.$$
We still need to find $\text{Cov}(X,Y)$, which is $E(XY)-E(X)E(Y)$.
To find $E(XY)$, we compute
$$\frac{1}{6}\left(E(XY|X=1)+E(XY|X=2)+\cdots+E(XY|X=6)\right).$$
We do a sample calculation, say $E(XY|X=3)$. If $X=3$, the maximum is $3$ with probability $\frac{3}{6}$ (second toss any of $1$, $2$, or $3$) and $4$, $5$, or $6$ each with probability $\frac{1}{6}$. So the mean of the maximum, given $X=3$, is $3\cdot\frac{1}{6}+4\cdot\frac{1}{6}+5\cdot\frac{1}{6}+6\cdot\frac{1}{6}$, which is $3$, and therefore $E(XY|X=3)=9$.
Remark: Alternatively, we could write down the full $6\times 6$ joint distribution table for $(X,Y)$, and do the computations using that table.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/749680",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
How do I solve this definite integral: $\int_0^{2\pi} \frac{dx}{\sin^{4}x + \cos^{4}x}$? $$\int_0^{2\pi} \frac{dx}{\sin^{4}x + \cos^{4}x}$$
I have already solved the indefinite integral by transforming $\sin^{4}x + \cos^{4}x$ as follows:
$\sin^{4}x + \cos^{4}x = (\sin^{2}x + \cos^{2}x)^{2} - 2\cdot\sin^{2}x\cdot\cos^{2}x = 1 - \frac{1}{2}\cdot\sin^{2}(2x) = \frac{1 + \cos^{2}(2x)}{2}$,
and then using the $\tan(2x) = t$ substitution. But if I do the same with the definite integral, both bounds of the integral become $0$.
| Let $z=e^{i x}$; then $dx = -i dz/z$ and the integral is equal to
$$-i 8 \oint_{|z|=1} dz \frac{z^3}{z^8 + 6 z^4+1}$$
By the residue theorem, the integral is then equal to $i 2 \pi$ times the sum of the residues at each pole inside the unit circle. The residue at each pole $z_k$ is equal to
$$-i 8 \frac{z_k^3}{8 z_k^7 + 24 z_k^3} = -i \frac1{z_k^4+3}$$
Each pole $z_k$ inside the unit circle satisfies $z_k^4+3=2 \sqrt{2}$, so the integral is therefore
$$2 \pi \frac{4}{2 \sqrt{2}} = 2 \sqrt{2} \pi$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/754750",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 4,
"answer_id": 2
} |
Find the solutions of: $\sin x+\cos x=\sin^2 x+0.5\sin{2x}$ Find the solutions of: $\sin x+\cos x=\sin ^2 x+0.5\sin{(2x)}$
How can I find the solutions ?
| Your equation, as you wrote it, is not true for all $x \in \mathbb{R}$, nor even for most of them. Consider $x=0$, in which case the left hand side evaluates to $1$ and the right hand side equals $0$.
We can still find what values of $x$, if any, satisfy your equation.
Recall that
$$\sin (2x) = 2\sin x \cos x$$
Substitute this into your equation, to get
$$\sin x+\cos x=\sin ^2 x+ (0.5 \cdot 2)(\sin x \cos x)$$
$$\sin x+\cos x=\sin ^2 x+ \sin x \cos x $$
$$\sin x+\cos x= (\sin x)(\sin x + \cos x)$$
This equation is true when either
$$\sin x = 1$$
or
$$\sin x + \cos x = 0$$
The first condition occurs when $x=2\pi n + \frac{\pi}{2}$ for some $n \in \mathbb{Z}$. The second condition holds whenever $x= \pi n - \frac{\pi}{4}$ for some $n \in \mathbb{Z}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/755999",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Show that $(x + 1)^{(2n + 1)} + x^{(n + 2)}$ can be divided by $x^2 + x + 1$ without remainder I am in my pre-academic year. We recently studied the Remainder sentence (at least that's what I think it translates) which states that any polynomial can be written as $P = Q\cdot L + R$
I am unable to solve the following:
Show that $(x + 1)^{(2n + 1)} + x^{(n + 2)}$ can be divided by $x^2 + x + 1$ without remainder.
| Hint: try to check if complex roots of $x^2+x+1$ are also roots for $(x+1)^{2n+1}+x^{n+2}$.
Also there is another way to prove it. Let see that $((x+1)^2-x)\cdot (x+1)^{2n-1}$ also divided by $x^2+x+1$. So we just need to prove that $x\cdot (x+1)^{2n-1}+x^{n+2}$ is divided by $x^2+x+1$. In this way we can come to prove that $(x^k\cdot (x+1)^{2n+1-2\cdot k}+x^{n+2}), \dots ,$ $(x^n(x+1)+x^{n+2})$ are all divided by $x^2+x+1$. But it's true, because $x^n(x+1)+x^{n+2} = x^n(x^2+x+1) $
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/757702",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 5,
"answer_id": 1
} |
How to prove $x^3$ is strictly increasing I am trying to use $f(x)=x^3$ as a counterexample to the following statement.
If $f(x)$ is strictly increasing over $[a,b]$ then for any $x\in (a,b), f'(x)>0$.
But how can I show that $f(x)=x^3$ is strictly increasing?
| Consider $a^3-b^3$, where $a\gt b$. We want to show this is positive.
We have
$$a^3-b^3=(a-b)(a^2+ab+b^2).$$
Note that $a^2+ab+b^2$ is positive unless $a=b=0$. There are many ways to do this. For example,
$$a^2+ab+b^2=\frac{1}{2}\left(a^2+b^2+(a+b)^2\right).$$
More conventionally, complete the square. We have $4(a^2+ab+b^2)=(2a+b)^2+3b^2$.
Remark: If we are in a calculus mood, note that
$$\frac{a^3-b^3}{a-b} =3c^2$$
for some $c$ between $a$ and $b$. This argument breaks down if $c=0$. But that can only happen when $0$ is in the interval $(b,a)$. That means $b$ is negative and $a$ is positive, making the inequality $a^3\gt b^3$ obvious.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/758158",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 6,
"answer_id": 1
} |
Quadratic equations and inequalities $\sqrt{4n+1}<\sqrt{n} + \sqrt{n+1}<\sqrt{4n+2}$ and $[\sqrt{n}+\sqrt{n+1}] = [\sqrt{4n+1}]$ For every positive integer $n$, prove that
$$\sqrt{4n+1}<\sqrt{n} + \sqrt{n+1}<\sqrt{4n+2}$$
Hence or otherwise, prove that $[\sqrt{n}+\sqrt{n+1}] = [\sqrt{4n+1}]$, where $[x]$ denotes the greatest integer not exceeding $x$.
This question was posed to me in class by my teacher. Since we are discussing quadratic equations. I can only imagine that this question is related to that topic.
Actually, by squaring the terms on both sides of the inequality, the first part of the question is solved easily. It is the second half that is causing me trouble.
Clearly we have to show, that if $x <\sqrt{4n+1} < x+1$ where $x$ is a natural number, then $x <\sqrt{n} + \sqrt{n+1} < x+1$, but how?
I am in high school, so please use techniques appropriate for my level.
| Clearly $\lfloor\sqrt{4n+1}\rfloor\le\lfloor\sqrt n+\sqrt{n+1}\rfloor\le \lfloor\sqrt{4n+2}\rfloor$. Thus the claim could only be wrong if $\lfloor\sqrt{4n+1}\rfloor< \lfloor\sqrt{4n+2}\rfloor$, i.e. if there exist an ineger $m$ with $\sqrt {4n+1}<m\le\sqrt {4n+2}$, equivalently: with $4n+1<m^2\le 4n+2$. But we cannot have $m^2=4n+2$ because the latter is even but not divisible by $4$, whereas $m^2$ is either odd or divisible by $4$. Thus the integer $m^2$ is between $4n+1$ and $4n+2$, which is absurd.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/760330",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
How to prove uniform convergence for sequences $f_n = x^n(1-x), f_n = \frac {n^3x} {1+n^4x^2}$ and $ f_n = \sqrt {n} xe^{-nx^2}$ on $[0,1]$
How to prove uniform convergence for sequences $f_n = x^n(1-x), f_n = \frac {n^3x} {1+n^4x^2}$ and $ f_n = \sqrt {n} xe^{-nx^2}$ on $[0,1]$
I've already shown that the following sequence of functions converge pointwise to $0$ on $[0,1]$:
$$f_n = \frac {2x} {1+nx}$$
$$f_n = x^n(1-x)$$
$$f_n = \frac {n^3x} {1+n^4x^2}$$
$$f_n = \sqrt {n} xe^{-nx^2}$$
For $f_n = \frac {2x} {1+nx}$ I have $\frac {2x} {1+nx} =\frac {2} {1/x+n} \le \frac 2 n \le \epsilon$, so I can choose $N$ large enough and the inequality will hold for all $x \in [0,1]$.
However, I don't see how to prove whether or not the other three functions converge uniformly. I'm not sure I'm supposed to use majorant series, since this exercise is a basic one.
Could someone point out how to prove uniform convergence for the three sequences in question ?
| I will show the 1st function $f_n(x) = x^n\cdot (1 - x)$ to converge uniformly on $[0, 1]$. $f_n'(x) = nx^{n-1} - (n+1)x^n = 0 \iff x = 0, \dfrac{n}{n+1}$. So $\displaystyle \sup_{x \in [0, 1]} |f_n(x) - 0| = f_n\left(\frac{n}{n+1}\right) = \dfrac{1}{n+1}\cdot \dfrac{1}{(1 + \frac{1}{n})^n} \to 0$ as $n \to \infty$.
For the second function $f_n(x) = \dfrac{xn^3}{1 + n^4x^2}$, we have:
$f_n'(x) = \dfrac{n^3 - x^2n^7}{(1 + n^4x^2)^2} = 0 \iff n^3 - x^2n^7 = 0 \iff x = \dfrac{1}{n^2}$. So $\displaystyle \sup_{x \in [0,1]} |f_n(x) - 0| = f_n\left(\frac{1}{n^2}\right) = \dfrac{n}{1 + n^2} \to 0$ as $n \to \infty$.
Surprisingly, the 3rd function does not converge uniformly on $[0, 1$. To see this, $f_n(x) = \sqrt{n}xe^{-nx^2}$ has $f_n'(x) = \sqrt{n}e^{-nx^2}\cdot \left(1 - 2nx^2\right) = 0 \iff x = \dfrac{1}{\sqrt{2n}}$. So $\displaystyle \sup_{x \in [0,1]}|f_n(x) - 0| = f_n\left(\frac{1}{\sqrt{2n}}\right) = \dfrac{1}{\sqrt{2e}}$. So $\displaystyle \lim_{n \to \infty} \sup_{x \in [0,1]}|f_n(x) - 0| = \dfrac{1}{\sqrt{2e}} > 0$, proving this function not converging uniformly to $0$ on $[0,1]$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/762777",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
For which $a$ is a matrix $A$ diagonalizable? Say I have a matrix $A_a$ with
$$A_a:= \left(\begin{array}{c}
2 & a+1 & 0 \\
-a & -3a & -a \\
a & 3a+2 & a+2
\end{array}\right)$$
I was wondering if there was an easy way to determine for which $a$ the matrix would be diagonalizable.
I tried to determine its eigenvalues first.
$det(A_a - \lambda I_3) = 0$ gave me a rather complex formula, from where on I don't know how to continue.
$0 = 7a \lambda - 6a - 4 \lambda + 4 \lambda^2 - 2a\lambda^2 - \lambda^3 - a^2\lambda + 2a^2$
| Given:
$$A_a:= \left(\begin{array}{c}
2 & a+1 & 0 \\
-a & -3a & -a \\
a & 3a+2 & a+2
\end{array}\right)$$
We find the eigenvalues using $|A_a - \lambda I| = 0$, yielding:
$$-a^2 \lambda +2 a^2-2 a \lambda ^2+7 a \lambda -6 a-\lambda ^3+4 \lambda ^2-4 \lambda = 0 \implies (\lambda-2)(\lambda - (1-a-\sqrt{a+1}))(\lambda - (1-a+\sqrt{a+1}))=0$$
This gives us the three eigenvalues:
$$\lambda_{1,2,3} = 2,-a-\sqrt{a+1}+1,-a+\sqrt{a+1}+1$$
Hint: try the cases $a = 0$ and $a = -1$ (one is good and the other is not)! Do you see why those two values were chosen? Think algebraic multiplicity of the eigenvalues.
Generally, the eigenvectors are found by $[A-\lambda_i I]v_i = 0$, so for $\lambda_1 = 2$, the RREF is:
$$\left(
\begin{array}{ccc}
1 & 0 & 1 \\
0 & 1 & 0 \\
0 & 0 & 0 \\
\end{array}
\right)v_1=0$$
This gives $v_1 = (-1,0,1)$.
For $\lambda_2 = -a-\sqrt{a+1}+1$, the RREF is:
$$\left(
\begin{array}{ccc}
1 & 0 & \frac{-a+\sqrt{a+1}-1}{a} \\
0 & 1 & 1 \\
0 & 0 & 0 \\
\end{array}
\right)v_2=0$$
This gives $v_2 = \left(-\dfrac{-a+\sqrt{a+1}-1}{a},-1,1\right)$.
We can immediately write (due to conjugates) $v_3 = \left(-\dfrac{-a-\sqrt{a+1}-1}{a},-1,1\right)$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/762859",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Find all integer solutions of $x^4 + 2x^3 + 2x^2 + 2x +5 = y^2$ Find all integer solutions to $x^4 + 2x^3 + 2x^2 + 2x +5 = y^2$.
I'm in a dead end. I've transformed the expression in the following state:
$(x^2+1)(x+1)^2 = y^2 -4$
I couldn't see anyway in which I could work with this expression in this state, so I continued into writing $(x^2+x+2)^2 - y^2 = 4x(x+1)$. Now I'm trying to use mod 8 and trying with different modulo. Please suggestions?
not entire solution tho...
| Hint: $(x^2+x)^2 < y^2 < (x^2+x+1)^2$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/765147",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Last digits, numbers Can anyone please help me?
1) Find the last digit of $7^{12345}$
2) Find the last 2 digits of $3^{3^{2014}}$.
Attempt: 1)
By just setting the powers of $7$ we have $7^1 = 7$, $7^2=49$, $7^3=343$, $7^4 = 2401$, $7^5 = 16807$, $7^6 = 117649$, $\dots$
After the power of $4$, the last digits will repeat. Then by noticing the pattern the digits will end in $7,9,3$ and $1$.
Then we can divide the exponent $(12345)$ by $4$ since this is the cycle that makes it repeat. Then $12345 : 4$ has remainder $1$. So $7^1 = 7$ is the unit digit to $7^{12345}$.
So the last digit is $7$.
I know how to do it like this, the problem does not state how to find the last digit, but I know it has something do do with Euler's theorem.
for part $2$) I don't know how to start. Can anyone please help me?
Thank you for the help.
| First of all, finding last $n$ digits in Base $10$ essentially finding modulo $\displaystyle10^n$
Using Carmichael function, $\displaystyle3^{\lambda(100)}\equiv1\pmod{100}$ as $(3,100)=1$
$$\implies3^{3^{2014}}\equiv3^{3^{2014}\pmod{\lambda(100)}}\pmod{100}$$
Now $\displaystyle\lambda(100)=20,$ so we need to find $\displaystyle3^{2014}\pmod{20}$
and by observation $\displaystyle3^4=81\equiv1\pmod{20}$ or using Carmichael function $\displaystyle\lambda(20)=4\implies3^4\equiv1\pmod{20}$
and as $\displaystyle2014\equiv2\pmod4\implies3^{2014}\equiv3^2\pmod{20}\equiv9$
$$\implies3^{3^{2014}}\equiv3^9\pmod{100}$$
Now $\displaystyle3^9=3\cdot3^8=3\cdot(3^2)^4=3(10-1)^4=3(1-10)^4=3\left[1-\binom4110+\binom4210^2-\binom4310^3+10^4\right]$
$\displaystyle\equiv3\left(1-40\right)\pmod{100}\equiv-117\equiv83$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/765530",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
$-5| 2+4x | = -32(x+3/4)- | x | + 1$ This was my attempt:
$$-5| 2+4x | = -32\left(x+\frac34\right)- | x | + 1\\
\implies|2+4x|=\frac{-32x-24- | x | + 1}{-5}\\
\implies2+4x=\pm \frac{-32x+-24- x + 1}{-5}\\
\implies4x=\pm \frac{-33x+-23 }{-5}-2\\
\implies-20x=\pm (-33x-23)+10\\
\implies-20x= -33x-13\text{ or} -20x=33x+33\\
\implies13x= -13\text{ or} -53x=33\\
\implies x=-1\text{ or} x=-\frac{33}{53}$$
Neither of these answers works when checking.
Using a graphing calculator I get an answer of $\frac{-11}{17}$ which does check, but I don't know how to get that answer using algebraic means.
| You made a mistake between the second and the third line:
$$|2+4x|=\frac{-32x-24- | x | + 1}{-5}\\
2+4x=\pm \frac{-32x-24- \color{red}{x} + 1}{-5}\\$$ instead it should have been $$2+4x=\pm \frac{-32x-24- \color{red}{|x|} + 1}{-5}.\\$$
Going on you get
$$-10-20x=\pm(-32x-24- |x| + 1)$$
Now you should treat the 2 cases ($x\ge0$ and $x\lt0$) differently:
\begin{cases}
x\ge0 \to -10-20x=\pm(-32x-24- x+ 1)\\
x\lt0 \to -10-20x=\pm(-32x-24+x+ 1)
\end{cases}\begin{cases}
x\ge0 \to x=-\frac{33}{13} \lor -\frac{13}{53} \text{ solutions not acceptable}\\
x\lt0 \to x=-\frac{13}{11} \lor x=-\frac{11}{17}
\end{cases}
So the final solutions are $$x=-\frac{13}{11} \lor x=-\frac{11}{17}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/765951",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Is this a valid method of finding magnitude of complex fraction If I have a complex fraction $\dfrac{a+bi}{c+di}$ and I want the magnitude, then will it be $\left|\dfrac{a+bi}{c+di}\right|=\dfrac{|a+bi|}{|c+di|}$?
Scratch that ... I just found the answer on another page; however, I'm still unclear why it's true?
| $\frac{a+bi}{c+di} = \frac{a+bi}{c+di} * \frac{c-di}{c-di} = i (\frac{b c}{c^2+d^2}-\frac{a d}{c^2+d^2})+\frac{a c}{c^2+d^2}+\frac{b d}{c^2+d^2}$. At this point, you should be able to get the magnitude easily. Yes, it'll be cumbersome computation wise, but that should be it.
Suppose $e = \frac{b c}{c^2+d^2}-\frac{a d}{c^2+d^2}$ and $f = \frac{a c}{c^2+d^2}+\frac{b d}{c^2+d^2}$
Then, $\|f + ei\| = \sqrt{f^2+e^2} = \sqrt{\frac{(bc-ad)^2}{(c^2+d^2)^2} + \frac{(ac+bd)^2}{(c^2+d^2)^2}} = \sqrt{\frac{2(a^2d^2+b^2c^2)}{(c^2+d^2)^2}}$ and you could take it from there.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/766841",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 3,
"answer_id": 0
} |
solve this problem of trigonometry. It is given :
$$\sin(A-B)/\sin B = \sin(A + Y)/\sin (Y)$$
We have to prove $$\cot B - \cot Y = \cot(A + Y) + \cot(A - B).$$
Please help me solving this. I have tried to solve this by analyzing $$\cot B - \cot Y$$
From the given equation,we get $$\frac{\sin A \cos B - \cos A \sin B}{\sin B} = \frac{\sin A \cos Y + \cos A \sin Y}{\sin Y} $$ or
$$\sin A \cot B - \cos A = \sin A \cot Y + \cos A$$ or
$$\sin A(\cot B - \cot Y) = 2 \cos A $$ or
$$\cot B - \cot Y = 2\cot A $$
but we have to prove $$\cot B - \cot Y = \cot(A + Y) + \cot(A - B).$$
I couldn't proceed further.
| OK, you started with:
$$
{\sin(A−B)\over \sin B}={\sin(A+Y)\over\sin Y}
$$
Which led you to:
$$
\cot B−\cot Y=2\cot A
$$
Let $U = A - B$, which imples $B = A - U$.
Let $V = -(A+Y)$, which implies $Y = -(A+V)$.
Substitute these values into the first equation to get:
$$
{\sin U \over \sin (A - U)} = {\sin (-V) \over \sin(-(A+V))}
$$
Note that $\sin (-x) = - \sin (x)$, so the negative signs on the right half of the equation factor out, then cancel each other. Now invert the equations to yield:
$$
{\sin (A - U) \over \sin U} = {\sin (A + V) \over \sin V}
$$
This equation is exactly the same as the first equation, but with U, V instead of B, Y. Thus, using the same logic, you can conclude:
$$
\cot U - \cot V = 2 \cot A
$$
Substitute the definitions for $U$ and $V$:
$$
\cot U - \cot V = \cot (A - B) - \cot (-(A+Y)) = \cot (A - B) + \cot (A + Y).
$$
This is also equal to $2 \cot A$, so we can combine with the second equation above to yield:
$$
\cot B - \cot Y = \cot(A+Y) + \cot(A-B).
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/767108",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
why $\int \frac{1}{1+\sin(x)+\cos(x)}dx = \ln\left | \tan(\frac{x}{2})+1 \right |+const.$? This is how I solve $$\int \frac{1}{1+\sin(x)+\cos(x)}dx$$, but I got the wrong answer, and the correct answer is $$\ln\left | \tan(\frac{x}{2})+1 \right |+\text{const}.$$
How to solve this？
| \begin{align}
\sin x &= \frac{2\tan(x/2)}{1 + \tan(x/2)} \\
\cos x &= \frac{1-\tan^2x}{1+ \tan^2x} \\
1+\sin x+\cos x &= \frac{2 + 2\tan(x/2)}{1+\tan(x/2)} \\
I &= \int\frac{dx}{1+\sin x+\cos x} \\
&= \int\frac{\sec(x/2)}{2+\tan(x/2)} dx
\end{align}
Let $z=\tan(x/2)$, then $dz=\frac12\sec(x/2)dx$ and
\begin{align}
I &= \int \frac22 \frac{dz}{1+z} \\
&= \log|z+1| +c \\
&= \log(\tan(x/2)+1) + c
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/769114",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Find the eigen values and vectors of the matrix Find the eigenvalues and eigenvectors of the matrix $A$:
$$A = \begin{bmatrix}-2 & 2 & -3\\2 & 1 & -6\\-1 & -2 & 0\\\end{bmatrix}.$$
$$A - \lambda I = \begin{bmatrix}-2-\lambda & 2 & -3\\2 & 1-\lambda & -6\\-1 & -2 & -\lambda\\\end{bmatrix}\\
\det(A-\lambda I)=(-2-\lambda)[(-1-\lambda)(-\lambda)-12]-2(-2\lambda-6)-3(-4-(-1+\lambda))\\
=(-2-\lambda)[(-\lambda+\lambda^2)-12]+4\lambda+12+9+3\lambda\\
=2\lambda-2\lambda^2+\lambda^2-\lambda^3+24+12\lambda+4\lambda+12+9+3\lambda\\
=-\lambda^3-\lambda^2+21\lambda+45$$
After factoring....... $\lambda=5,\lambda=-3$
When $\lambda=5$,
$$\begin{bmatrix}-7 & 2 & -3\\2 & -4 & -6\\-1 & -2 & -5\\\end{bmatrix}\begin{bmatrix}x\\y\\z\end{bmatrix}=\begin{bmatrix}0\\0\\0\end{bmatrix}$$
$$-7x+2y-3z=0,\\
2x-4y-6z=0,\\
-x-2y-5z=0.$$
I am stuck here, i have no idea what to do next. I hope someone can help
Please and thanks
| This is easier to proceed if we permute the matrix:
$$
\pmatrix{2& -7& -3\\
-4& 2& -6\\
-2&-1&-5} \rightarrow \pmatrix{2& -7& -3\\
0& -12& -12\\
0&-8&-8}
$$
This clearly leads to the equation:
$$
-12x - 12z = 0 \rightarrow z = -x \\
\text{using the first equation, }2y - 7x - 3z = 0 \rightarrow y = \frac{7x + 3z}{2} = \frac{7x - 3x}{2} = 2x
$$
This gives the eigenvector as $\langle x, 2x, -x\rangle \leadsto \langle1, 2, -1\rangle$, or as a unit vector: $\left\langle \pm\sqrt{\frac{1}{6}}, \pm \sqrt{\frac{2}{3}}, \mp \sqrt{\frac{1}{6}}\right\rangle$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/769783",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
How do I solve this square root problem? I need to solve the following problem:
$$\frac{\sqrt{7+\sqrt{5}}}{\sqrt{7-\sqrt{5}}}=\,?$$
| $\frac{7+\sqrt{5}}{7-\sqrt{5}}=\frac{7+\sqrt{5}}{7-\sqrt{5}}\cdot\frac{7+\sqrt{5}}{7+\sqrt{5}}=\frac{(7+\sqrt{5})^{2}}{49-5}=\frac{(7+\sqrt{5})^{2}}{44}$
Then you are taking the root of the quotient (note for $a,b >0$ $\sqrt{\frac{a}{b}}=\frac{\sqrt{a}}{\sqrt{b}}$) and you get
$\frac{7+\sqrt{5}}{\sqrt{44}}=\frac{7+\sqrt{5}}{2\sqrt{11}}=\frac{\sqrt{11}(7+\sqrt{5})}{22}$
None of the answers proposed is correct: we can use the squared value we have calculated
$\frac{(7+\sqrt{5})^{2}}{44}=\frac{3}{11}+\frac{\sqrt{35}}{22}$
As you can see it is not rational, so you exclude $1$ and $2$
Then $(6 \pm \sqrt{35})^2= 36+35 \pm 12 \sqrt{35}$ and you can see that both of them are incorrect.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/770259",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
Why does $\int^{ab}_{a} \frac{1}{x} dx = \int^{b}_{1} \frac{1}{t} dt$? I can't understand how the integral having limits from $a$ to $ab$ in Step 1 is equivalent to the integral having limits from $1$ to $b$. I'm a beginner here. Please explain in detail.
\begin{align*}
\ln(ab) = \int^{ab}_{1} \frac{1}{x} dx &= \int^{a}_{1} \frac{1}{x} dx + \int^{ab}_{a} \frac{1}{x} dx\\
&= \int^{a}_{1} \frac{1}{x} dx + \int^{b}_{1} \frac{1}{at} d(at)\\
&= \int^{a}_{1} \frac{1}{x} dx + \int^{b}_{1} \frac{1}{t} dt\\
&= \ln(a) + \ln(b).
\end{align*}
| Your are asking why
$$ \int_{a}^{ab} \frac{1}{x} dx = \int_{1}^b \frac{1}{t} dt $$
Well, it follows by substituting $x = at \implies dx = a dt $.
Now, the limits in the first integral are $x =a$ to $x=ab$. Hence, if $x =a $, then $ t = \frac{a}{a} = 1 $. and if $ x = ab $ , then $ t = \frac{ab}{a} = b$ which are your new limits of integration.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/771015",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
} |
A 3-minute algebra problem I have just taken advance math test level 2 and there are several problems that have been bugging me. This is the first question:
If $x,y>0$, then determine the value of $x$ that satisfies the system
of equations: \begin{align} x^2+y^2-xy&=3\\ x^2-y^2+\sqrt{6}y&=3\\
\end{align}
I can answer this problem using a 'standard' algebra but it takes time more than 3 minutes. For the sake of curiosity, is there a way to answer this problem less than 3 minutes?
| Substracting second equation to the first you obtain $$2y^2-xy-\sqrt{6}y=0$$
Then, factoring $$y(2y-x-\sqrt{6})=0$$
Thus $2y-x-\sqrt{6}=0$ cause $x,y$ are greater than $0$. So $y=\frac{x+\sqrt{6}}{2}$, putting this expression into the second equation you have
\begin{align}
x^2-\left(\frac{x+\sqrt{6}}{2}\right)^2+\sqrt{6}\left(\frac{x+\sqrt{6}}{2}\right)&=3 \\
x^2-\frac{x^2+2\sqrt{6}x+6}{4}+\frac{\sqrt{6}}{2}x+3&=3\\
\frac{3}{4}x^2-\frac{3}{2}&=0\\
\end{align}
Whose positive root is $x=\sqrt{2}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/771551",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 3,
"answer_id": 0
} |
Multivariable Chain Rule Does anyone know how to use the multivariable chain rule to solve the following problem?
If $G(x^2+y^2, x+y)=(7x+3y, x+5\,y)$, knowing that $G:\mathbb{R}^2\rightarrow \mathbb{R}^2$, find $(G^{-1})'(24,8)$
| Let $f:\mathbb{R}^2\longrightarrow\mathbb{R}^2$ defined by $(x,y)\mapsto(7x+3y,x+5y)$, observe
$$G^{-1}\circ f(x,y)=(x^2+y^2,x+y)$$
applying Chain Rule it follows
\begin{align}
DG^{-1}[f(x,y)]Df(x,y)&=\begin{bmatrix}2x&2y\\1&1\end{bmatrix}\\
DG^{-1}[f(x,y)]\begin{bmatrix}7&3\\1&5\end{bmatrix}&=\begin{bmatrix}2x&2y\\1&1\end{bmatrix}
\end{align}
Then, putting $x=3,y=1$ and since $f(3,1)=(24,8)$ we get
\begin{align}
DG^{-1}(24,8)\begin{bmatrix}7&3\\1&5\end{bmatrix}&=\begin{bmatrix}6&2\\1&1\end{bmatrix}\\
DG^{-1}(24,8)&=\begin{bmatrix}6&2\\1&1\end{bmatrix}\begin{bmatrix}7&3\\1&5\end{bmatrix}^{-1}\\
&=\begin{bmatrix}6&2\\1&1\end{bmatrix}\begin{bmatrix}\frac{5}{32}&-\frac{3}{32}\\-\frac{1}{32}&\frac{7}{32}\end{bmatrix}\\
&=\begin{bmatrix}\frac{7}{8}&-\frac{1}{8}\\\frac{1}{8}&\frac{1}{8}\end{bmatrix}
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/771751",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Subsets and Splits
No saved queries yet
Save your SQL queries to embed, download, and access them later. Queries will appear here once saved.