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Prove by induction that for every $n \ge 0$ the term $2^{2n+1} - 9n^2 + 3n -2$ is divisible by $54$ What I get is $4 \cdot 54k + 27(n(n-1)) - 2$.
I understand first two terms but $-2$ is giving me a headache. I suppose to be true every term should be divisible by $54$. Or did I make wrong conclusion?
P. S: Is there mathjax tutorial?
| When $n$ is even, each term of $2^{2n+1} - 9n^2 + 3n -2$ is even. When $n$ is odd, $-9n^2 + 3n$ is odd + odd = even, hence $2^{2n+1} - 9n^2 + 3n -2$ is even.
Now $2 \cdot 4^{n} - 9n^2 + 3n -2 \equiv 2 \cdot 1^n - 2 \equiv 0 \pmod 3$. Now let $n = 3k$, so $2 \cdot 64^k - 9(3k)^2 + 9k - 2 \equiv$ $ 2\cdot 1^k - 9(3k)^2 + 9k - 2 \equiv 0 \pmod 9$. Letting $k = 9m$, $2 \cdot (64^9)^m - 9(9m)^2 + 81m - 2$ $ \equiv 2 \cdot 1^m - 9(9m)^2 + 81m - 2 \equiv 0 \pmod {27}$. This is because $\phi(27) = 27(1 - 1/3) = 18$, thus $64^{18} \equiv 1 \pmod {27} \implies 64^9 \equiv ±1 \pmod {27}$. But from before, $64^k \equiv 1 \pmod 9$ for $k \in \mathbb Z^+$, so it is congruent to $1$ and not $-1$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2584789",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Evaluating $\int\frac{x^2+1}{(x^2-2x+2)^2}dx$ - Section 7.3, #28 in Stewart Calculus 8th Ed. I'd like to evaluate $$I=\int\frac{x^2+1}{(x^2-2x+2)^2}dx.$$
According to WolframAlpha, the answer is $$\frac{1}{2}\bigg(\frac{x-3}{x^2-2x+2}-3\tan^{-1}(1-x)\bigg)+C.$$
To ensure that my answer is correct, I'd like to match it with this answer.
First I complete the square by writing the denominator as $((x-1)^2+1)^2$, then use a trig-sub $x-1=\tan\theta$ so $dx=\sec^2\theta d\theta.$
Then $$I=\int\frac{(\tan\theta +1)^2+1}{\sec^4\theta}\sec^2\theta d\theta\\=\int\frac{\tan^2\theta+2\tan\theta+2}{\sec^2\theta}d\theta\\=\int\sin^2\theta +2\sin\theta\cos\theta+2\cos^2\theta d\theta\\=\int\bigg(\frac{1}{2}-\frac{\cos(2\theta)}{2}\bigg)+\sin(2\theta)+(1+\cos(2\theta))d\theta\\=\int\frac{3}{2}+\frac{\cos(2\theta)}{2}+\sin(2\theta)d\theta\\=\frac{3\theta}{2}+\frac{\sin(2\theta)}{4}-\frac{\cos(2\theta)}{2}+C\\=\frac{3\theta}{2}+\frac{\sin\theta \cos\theta}{2}-\frac{1-2\sin^2\theta}{2}+C.$$
Here's a picture corresponding to my trig-sub.
From the picture, I conclude $$I=\frac{3\tan^{-1}(x-1)}{2}+\frac{1}{2}\frac{x-1}{\sqrt{x^2-2x+2}}\frac{1}{\sqrt{x^2-2x+2}}- \frac{1-2\frac{(x-1)^2}{x^2-2x+2}}{2}+C\\=\frac{-3\tan^{-1}(1-x)}{2}+\frac{x-1}{2(x^2-2x+2)}-\frac{1}{2}+\frac{(x-1)^2}{x^2-2x+2}+C.$$
Although the $\arctan$ portion of the answer matches Wolfram, the remaining portion does not (since there is a square in the numerator after finding a common denominator). Where am I making the mistake?
I've also tried using the identity $\cos(2\theta)= 2\cos^2\theta-1$ instead of $\cos(2\theta)= 1-2\sin^2\theta$, but the same issue occurs.
| Your work is correct; the only difference between your antiderivative and the one calculated by WolframAlpha is a constant of integration, $1/2$. You may verify that $$\frac{(x-1)^2}{x^2-2 x+2}+\frac{x-1}{2 \left(x^2-2 x+2\right)}-\frac{1}{2} = \frac{x^2-x-1}{2 \left(x^2-2 x+2\right)},$$ and it follows that $$\frac{x^2-x-1}{2 \left(x^2-2 x+2\right)}-\frac{x-3}{2 \left(x^2-2 x+2\right)} = \frac{1}{2}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2585274",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Calculate the determinant of $A-5I$ Question
Let $ A =
\begin{bmatrix}
1 & 2 & 3 & 4 & 5 \\
6 & 7 & 8 & 9 & 10 \\
11 & 12 & 13 & 14 & 15 \\
16 & 17 & 18 & 19 & 20 \\
21& 22 & 23 & 24 & 25
\end{bmatrix}
$.
Calculate the determinant of $A-5I$.
My approach
the nullity of $A$ is $3$, so the algebraic multiplicity of $\lambda = 0$ is $3$, i.e. $\lambda_1 = \lambda_2 = \lambda_3 = 0.$
Now trace($A$) = $\lambda_4 + \lambda_5 = 1+6+13+19+25 = 65$
Then det($A-5I$) = $(\lambda_1-5)(\lambda_2-5)(\lambda_3-5)(\lambda_4-5)(\lambda_5-5)=(-5)^3(\lambda_4\lambda_5 - 5 \times 65 + 25)$
We need to calculate the value of $\lambda_4 \lambda_5$, which includes sum of lots of determinant of $2 \times 2$ matrices.
Is there any fast way to calculate the determinant?
| The calculation of $\lambda_4 \lambda_5$ (which in this case is equivalent to the cubic term of the characteristic polynomial) is indeed given by the sum of principal $2\times2$ subdeterminants
$$\lambda_4 \lambda_5 = \sum_{i=2}^5 \sum_{j=1}^{i-1} \left|\begin{matrix}a_{ii}&a_{ij}\\a_{ji}&a_{jj}\end{matrix}\right|.$$
In this case we actually have the easy formula $a_{ij} = 5i+j-5$, so the above subdeterminant simplifies to $-5(i-j)^2$. We can of course evaluate the double sum algebraically, but it's just as easy to regroup the sum according to the value of $i-j$:
$$\sum_{1\le j < i \le 5} (i-j)^2 = 4\cdot 1^2 + 3\cdot 2^2 + 2\cdot 3^2 + 1\cdot4^2 = 4 + 12 + 18 + 16 = 50.$$
Thus $\lambda_4 \lambda_5 = -5(50) = -250$ and $\det(A-5I) = (-5)^3 (-250-5\cdot 65+25) = 68750$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2585742",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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How many $3$-digit numbers can be formed using the digits $ 2,3,4,5,6,8 $ such that the number contains the digits $5$ and repetitions are allowed? The solution I have is by counting the complement which gives an answer of $91$. But I think that it should be solved as follows-
Let the $3$-digit number be denoted by $3$ boxes. We can put the digit $5$ in one of the three boxes and we can put fill remaining two boxes with $6$-digits. Hence answer is $6\cdot 6\cdot 3= 108$. Help!
| The total numbers that can be formed using these $6$ digits is $6^3$. We find the number of numbers in which there is no $5$. Therefore numbers without the digit $ 5$ is $5^3$. Therefore the numbers with at least one $5$ in the number is $6^3-5^3=91$
Hence answer is $91$
Or you can do it the other way using cases:
1) Number contains only one $5$ and all letters are different:
Number of such numbers = $^5C_2. 3!=60$
2) Numbers with only one $5$ and other letters alike:
Number of such numbers=$^5C_1.\frac {3!}{2!}=15$
3)Numbers with two $5's$ and one other number:
Number of such numbers=$^5C_1.\frac {3!}{2!}=15$
4)Numbers with three $5's= 1$
Hence the total number of numbers satisfying the given condition =$60+15+15+1=91$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2586564",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Solving $\cos x + \cos 2x - \cos 3x = 1$ with the substitution $z = \cos x + i \sin x$ I need to solve
$$\cos x+\cos 2x-\cos 3x=1$$
using the substitution$$z= \cos x + i \sin x $$
I fiddled around with the first equation using the double angle formula and addition formula to get
$$\cos^2 x+4 \sin^2x\cos x-\sin^2 x=1$$
which gets me pretty close to something into which I can substitute $z$, because $$z^2= \cos^2 x-\sin^2 x+2i\sin x\cos x$$
I have no idea where to go from there.
| There are already some good solutions, but the following solution does use the OP's substitution.
I assume that $x$ is real, then
$$z=\cos(x)+i \sin(x), z^*=\cos(x)-i \sin(x)=z^{-1},$$ so the equation is equivalent to the following:
$$z+z^{-1}+z^2+z^{-2}-z^3-z^{-3}=2.$$
Let us use the following substitution: $q=z+z^{-1}$, so $q^2=z^2+2+z^{-2}$ and $q^3=z^3+3 z+3 z^{-1}+z^{-3}$, so the equation is equivalent to the following:
$$q+q^2-2-q^3+3q=2$$ or $$q^3-q^2-4 q+4=0.$$
The roots of this cubic equation are integer, so it is easy to find them.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2586848",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Probability of chosing at least two given letters out of three in sequence of five. I am self studying elementary Probability Theory, and I am not sure on how to solve this exercise, taken from the book Understanding Probability by Henk Tijms.
For the upcoming drawing of the Bingo Lottery, five extra prizes have
been added to the pot. Each prize consists of an all-expenses paid
vacation trip. Each prize winner may choose from among three possible
destinations $A$, $B$ and $C$. The three destinations are equally
popular. The prize winners choose their destination indipendently of
each other. Calculate the probabilty that at least one of the
destinations $A$ and $B$ will be chosen. Also, calculate the
probability that not each of the three destinations will be chosen.
Regarding the first question, at first I thought that the probability that at least one of the destinations $A$ and $B$ is chosen should be the complementary probability that all the five winners choose the destination $C$, so:
$$P(A\cup B)=1-P(C)=1-1/3^5=\frac{242}{243}.$$
I would like to compute the same probability through the inclusion-exclusion rule,
$$P(A\cup B) = P(A)+P(B)-P(A\cap B).$$
In this case it should be $P(A)=P(B)$, and to compute $P(A)$ I consider first the sequences of five letters containing exactly one time the letter $A$, which are $\binom{5}{1}$ and have probability $\frac13 (\frac23)^4$.
Similarly, the sequences of five letters containing exactly two $A$s are $\binom{5}{2}$ and have probability $(\frac13)^2(\frac23)^3$. In this way, I get
$$P(A)=\sum_{k=1}^5 \binom{5}{k}\left(\frac13\right)^k\left(\frac23\right)^{5-k}.$$
To compute the probability that both the destinations $A$ and $B$ are chosen, I start by computing the probability that $A$ and $B$ are chosen one time each, and the remaining three winners choose $C$. In this case we have $\binom{5}{2}$ ways of forming the string $ABCCC$, which has probability $1/3^5$. Next, consider the two sequences $AABCC$ and $ABBCC$. The number of sequences are $2\cdot\binom{5}{3}$ and the probability is again $1/3^5$.
Summing up, I have
$$P(A\cap B)=\frac{1}{3^5}\sum_{k=2}^5 \binom{5}{k}(k-1).$$
However, when I compute the numerical values for $P(A\cup B)$, I get a number greater than one so I should be making a mistake somewhere.
| For your calculation of $P(A\cap B)$, it appears you're missing some cases.
Here's one way to organize the count . . .
$$
\begin{array}
{|c|c|c|c|}
A&B&C&\text{count}&\text{times}\\ \hline
1&4&0&{\large{\binom{5}{1}}}{\large{\binom{4}{4}}}=5&2\\
2&3&0&{\large{\binom{5}{2}}}{\large{\binom{3}{3}}}=10&2\\ \hline
1&3&1&{\large{\binom{5}{1}}}{\large{\binom{4}{3}}}{\large{\binom{1}{1}}}=20&2\\
2&2&1&{\large{\binom{5}{2}}}{\large{\binom{3}{2}}}{\large{\binom{1}{1}}}=30&1\\ \hline
1&2&2&{\large{\binom{5}{1}}}{\large{\binom{4}{2}}}{\large{\binom{2}{2}}}=30&2\\ \hline
1&1&3&{\large{\binom{5}{1}}}{\large{\binom{4}{1}}}{\large{\binom{3}{3}}}=20&1\\ \hline
\end{array}
$$
The entries in the "times" column are symmetry correction factors. For example, the count for the triple $(A,B,C)=(1,4,0)$ needs to be multiplied by $2$, so as to include the count for the symmetrically equivalent triple $(A,B,C)=(4,1,0)$.
Thus, the total count is
$$(5)(2)+(10)(2)+(20)(2)+(30)(1)+(30)(2)+(20)(1) =180$$
hence
$$P(A\cap B) = \frac{180}{3^5} = \frac{20}{27}$$
so then
\begin{align*}
P(A\cup B) &=P(A)+P(B) - P(A\cap B)\\[6pt]
&=2\left(\frac{211}{3^5}\right)- \frac{20}{27}\\[6pt]
&=\frac{242}{243}
\end{align*}
as expected.
For the second problem, first find the probability that all $3$ destinations are chosen.
As shown in the table below, there are only two cases . . .
$$
\begin{array}
{|c|c|c|c|}
A&B&C&\text{count}&\text{times}\\ \hline
1&1&3&{\large{\binom{5}{1}}}{\large{\binom{4}{1}}}{\large{\binom{3}{3}}}=20&3\\ \hline
1&2&2&{\large{\binom{5}{1}}}{\large{\binom{4}{2}}}{\large{\binom{2}{2}}}=30&3\\ \hline
\end{array}
$$
Thus, the total count is
$$(20)(3)+(30)(3)=150$$
hence
$$P(A\cap B \cap C) = \frac{150}{3^5} = \frac{50}{81}$$
so the probability that not all $3$ destinations are chosen is
$$1- \frac{50}{81} = \frac{31}{81}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2588051",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
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} |
Evaluate $\int (2x+3) \sqrt {3x+1} dx$ Evaluate $\int (2x+3) \sqrt {3x+1} dx$
My Attempt:
Let $u=\sqrt {3x+1}$
$$\dfrac {du}{dx}= \dfrac {d(3x+1)^\dfrac {1}{2}}{dx}$$
$$\dfrac {du}{dx}=\dfrac {3}{2\sqrt {3x+1}}$$
$$du=\dfrac {3}{2\sqrt {3x+1}} dx$$
| Hint. One may find $a,b \in \mathbb{R}$ such that
$$
(2x+3) \sqrt {3x+1}=\color{red}{a}\cdot(3x+1)^{3/2}+\color{red}{b}\cdot(3x+1)^{1/2}.
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2588847",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 0
} |
Error while integrating reciprocal of irreducible quadratic $\int{\frac{1}{ax^2 + bx + c}}{dx} = ?$ I am trying to derive a formula for indefinite integral of reciprocal of irreducible real quadratic.
$$\int{\frac{1}{ax^2 + bx + c}}{dx} = ?$$, $b^2 - 4ac \lt 0$. According to WolframAlpa it should come out as: $$\frac{2\arctan{\frac{2ax+b}{\sqrt{4ac - b^2}}}}{\sqrt{4ac - b^2}} + C$$ Instead I get $$\frac{\arctan{\frac{x + \frac{b}{2a}}{\sqrt{\frac{c}{a} - (\frac{b}{2a})^2}}}}{a(\frac{c}{a} - (\frac{b}{2a})^2)^\frac{3}{2}} + C$$ which, no matter how I try to simplify it, doesn't give the right answer. Here are my steps, what am I doing wrong?
1) Reduce to a monic quadratic and complete the square: $$\int{\frac{1}{ax^2 + bx + c}}{dx} = \frac{1}{a}\int{\frac{1}{x^2 + \frac{b}{a}x + \frac{c}{a}}}{dx} = \frac{1}{a}\int{\frac{1}{(x + \frac{b}{2a})^2 + \frac{c}{a} - (\frac{b}{2a})^2}}{dx}$$
3) Substitute $x = u - \frac{b}{2a}$, $u = x + \frac{b}{2a}$, $k = \frac{c}{a} - (\frac{b}{2a})^2$: $$\frac{1}{a}\int{\frac{1}{(x + \frac{b}{2a})^2 + \frac{c}{a} - (\frac{b}{2a})^2}}{dx} = \frac{1}{a}\int{\frac{1}{u^2 + k}}{du}$$
4) Substitute $u = v\sqrt{k}$, $v = \frac{u}{\sqrt{k}}$, $du = \frac{1}{\sqrt{k}}\hphantom ddv$:
$$\frac{1}{a}\int{\frac{1}{u^2 + k}}{du} = \frac{1}{a}\int{\frac{\frac{1}{\sqrt{k}}}{kv^2 + k}}{dv} = \frac{1}{a}\int{\frac{1}{k\sqrt{k}(v^2 + 1)}}{dv} = \frac{1}{ak^\frac{3}{2}}\int{\frac{1}{v^2 + 1}}{dv}$$
5) Use common integral $$\frac{1}{ak^\frac{3}{2}}\int{\frac{1}{v^2 + 1}}{dv} = \frac{\arctan{v}}{ak^\frac{3}{2}} + C$$
6) Substitute back $$\frac{\arctan{v}}{ak^\frac{3}{2}} + C = \frac{\arctan{\frac{u}{\sqrt{k}}}}{ak^\frac{3}{2}} + C = \frac{\arctan{\frac{x + \frac{b}{2a}}{\sqrt{\frac{c}{a} - (\frac{b}{2a})^2}}}}{a(\frac{c}{a} - (\frac{b}{2a})^2)^\frac{3}{2}} + C$$
| The error is in step 4: if $u=\sqrt kv$, then $\mathrm du=\sqrt k\mathrm dv$, and not $\frac1{\sqrt k}\,\mathrm dv$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2591781",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Prove that $(-1)^n\sum_{k=0}^{n}{{{n+k}\choose{n}}2^k}+1=2^{n+1}\sum_{k=0}^{n}{{{n+k}\choose{n}}(-1)^k}$ Define:
$$A_n:=\sum_{k=0}^{n}{{n+k}\choose{n}} 2^k,\quad{B_n}:=\sum_{k=0}^{n}{{n+k}\choose{n}}(-1)^k$$
I've found that (based on values for small $n$) this identity seems to be true:
$${\left(-1\right)}^nA_n+1=2^{n+1}B_n$$
However, I'm stuck on trying to find a proof. Any ideas? Thanks!
| The striking symmetry of this binomial identity might lead us to ask what are the roles of $2$ and $-1$ and are there similar identities like this one? Here is a slight generalisation using essentially the same approach as @MarkoRiedel did in his nice answer.
We consider $a\in\mathbb{C}$ instead of $2$ resp. $-1$ and obtain
\begin{align*}
\color{blue}{\sum_{k=0}^n}&\color{blue}{\binom{n+k}{n}a^k}\tag{0}\\
&=\sum_{k=0}^n\binom{-n-1}{k}(-a)^k\tag{1}\\
&=\sum_{k=0}^n[z^k]\frac{1}{(1-az)^{n+1}}\tag{2}\\
&=[z^0]\frac{1}{(1-az)^{n+1}}\sum_{k=0}^n\frac{1}{z^k}\tag{3}\\
&=[z^0]\frac{1}{(1-az)^{n+1}}\cdot\frac{1-z^{n+1}}{z^n(1-z)}\tag{4}\\
&=[z^n]\frac{1}{(1-az)^{n+1}(1-z)}\tag{5}\\
&=\frac{(-1)^n}{a^{n+1}}\mathrm{Res}_{z=0}\underbrace{\frac{1}{z^{n+1}\left(z-\frac{1}{a}\right)^{n+1}(z-1)}}_{f(z)}\tag{6}\\
&=\left(-\frac{1}{a}\right)^{n+1}
\left(\mathrm{Res}_{z=\frac{1}{a}}+\mathrm{Res}_{z=1}+\mathrm{Res}_{z=\infty}\right)f(z)\tag{7}
\end{align*}
Comment:
*
*In (1) we use $\binom{p}{q}=\binom{p}{p-q}$ and $\binom{-p}{q}=\binom{p+q-1}{q}(-1)^q$.
*In (2) and (3) we apply the coefficient of operator and use the rule $[z^{p-q}]A(z)=[z^p]z^qA(z)$.
*In (4) we use the finite geometric series formula.
*In (5) we apply the same rule as we did in (3) and skip the term $z^{n+1}$ since it does not contribute to $[z^n]$.
*In (6) we use the representation with residual at $z=0$.
*In (7) we use the fact that computing the residue at zero is minus the sum of the residuals at the other poles and infinity.
Calculating the residuals gives
\begin{align*}
\mathrm{Res}_{z=\frac{1}{a}}f(z)&=\frac{1}{n!}\left[\frac{d^n}{dz^n}\left(\frac{1}{z^{n+1}(z-1)}\right)\right]_{z=\frac{1}{a}}\\
&=\frac{1}{n!}\left[\sum_{q=0}^n\binom{n}{q}\frac{d^q}{dz^q}\left(\frac{1}{z^{n+1}}\right)\frac{d^{n-q}}{dz^{n-q}}\left(\frac{1}{z-1}\right)\right]_{z=\frac{1}{a}}\\
&=\frac{1}{n!}\left[\sum_{q=0}^n\binom{n}{q}(-1)^q\frac{(n+q)!}{q!}\frac{1}{z^{n+1+q}}
(-1)^{n-q}(n-q)!\frac{1}{(z-1)^{n-q+1}}\right]_{z=\frac{1}{a}}\\
&=(-1)^n\left(\frac{a^2}{1-a}\right)^{n+1}\sum_{q=0}^n\binom{n+q}{q}(1-a)^q\\
\mathrm{Res}_{z=1}f(z)&=\frac{1}{\left(1-\frac{1}{a}\right)^{n+1}}\\
&=\left(\frac{a}{a-1}\right)^{n+1}\\
\mathrm{Res}_{z=\infty}f(z)&=\mathrm{Res}_{z=0}\left(-\frac{1}{z^2}f\left(\frac{1}{z}\right)\right)\\
&=-\mathrm{Res}_{z=0}\left(\frac{1}{z^2}\cdot\frac{z^{n+1}}{\left(\frac{1}{z}-\frac{1}{a}\right)^{n+1}\left(\frac{1}{z}-a\right)}\right)\\
&=-\mathrm{Res}_{z=0}\frac{z^{2n+1}a^{n+1}}{(a-z)^{n+1}(1-z)}\\
&=0
\end{align*}
Putting all together in (7) gives
\begin{align*}
&\left(-\frac{1}{a}\right)^{n+1}
\left(\mathrm{Res}_{z=\frac{1}{a}}+\mathrm{Res}_{z=1}+\mathrm{Res}_{z=\infty}\right)f(z)\\
&\quad=\left(-\frac{1}{a}\right)^{n+1}\left((-1)^n\left(\frac{a^2}{1-a}\right)^{n+1}\sum_{k=0}^n\binom{n+k}{k}(1-a)^k
+\left(\frac{a}{a-1}\right)^{n+1}+0\right)\\
&\quad\color{blue}{=-\left(\frac{a}{1-a}\right)^{n+1}\sum_{k=0}^n\binom{n+k}{n}(1-a)^k+\left(\frac{1}{1-a}\right)^{n+1}}\tag{8}
\end{align*}
We finally conclude from (0) and (8)
The following is valid for non-negative integer $n$ and $a\in\mathbb{C}$
\begin{align*}
\color{blue}{(1-a)^{n+1}\sum_{k=0}^n\binom{n+k}{n}a^k
=1-a^{n+1}\sum_{k=0}^n\binom{n+k}{n}(1-a)^k}
\end{align*}
Some special cases:
$a=-1, a=2$ (OP's identity):
\begin{align*}
2^{n+1}\sum_{k=0}^n\binom{n+k}{n}(-1)^k=1+(-1)^n\sum_{k=0}^n\binom{n+k}{n}2^k
\end{align*}
$a=\frac{1}{2}$:
\begin{align*}
\sum_{k=0}^n\binom{n+k}{k}\frac{1}{2^k}=2^n
\end{align*}
$a=\frac{1}{3}, a=\frac{2}{3}$:
\begin{align*}
\left(\frac{2}{3}\right)\sum_{k=0}^n\binom{n+k}{n}\frac{1}{3^k}
=1-\frac{1}{3^{n+1}}\sum_{k=0}^n\binom{n+k}{n}\left(\frac{2}{3}\right)^k
\end{align*}
| {
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"url": "https://math.stackexchange.com/questions/2593862",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Problem with factorials: find the least n for which (n + 16)!(n+20)! ends with a number of zeros divisible by 2016 I have been working on this problem during the last few days but I can't find a good solution.
| Number of zeros ended $n!$ gives formula
$$p = \left[\frac{n}{5^1}\right] + \left[\frac{n}{5^2}\right]+... + \left[\frac{n}{5^k}\right]$$
At first check $n=4040$:
$$\left[\frac{4040}{5^1}\right] + \left[\frac{4040}{5^2}\right]+ \left[\frac{4040}{5^3}\right]+ \left[\frac{4040}{5^4}\right]+ \left[\frac{4040}{5^5}\right]=808+161+32+6+1=1008$$
and $n+4=4044$:
$$\left[\frac{4044}{5^1}\right] + \left[\frac{4044}{5^2}\right]+ \left[\frac{4044}{5^3}\right]+ \left[\frac{4044}{5^4}\right]+ \left[\frac{4044}{5^5}\right]=808+161+32+6+1=1008.$$
For $n=4024$ product $(n+16)!\cdot (n+20)!=4040!\cdot 4044!$ ends with a 2016 zeros
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Solving $\frac{1}{|x+1|} < \frac{1}{2x}$ Solving $\frac{1}{|x+1|} < \frac{1}{2x}$
I'm having trouble with this inequality. If it was $\frac{1}{|x+1|} < \frac{1}{2}$, then:
If $x+1>0, x\neq0$, then
$\frac{1}{(x+1)} < \frac{1}{2} \Rightarrow x+1 > 2 \Rightarrow x>1$
If $x+1<0$, then
$\frac{1}{-(x+1)} < \frac{1}{2} \Rightarrow -(x+1) > 2 \Rightarrow x+1<-2 \Rightarrow x<-3$
So the solution is $ x \in (-\infty,-3) \cup (1,\infty)$
But when solving $\frac{1}{|x+1|} < \frac{1}{2x}$,
If $x+1>0, x\neq0$, then
$\frac{1}{x+1} < \frac{1}{2x} \Rightarrow x+1 > 2x \Rightarrow x<1 $
If $x+1<0$, then
$\frac{1}{-(x+1)} < \frac{1}{2x} \Rightarrow -(x+1) > 2x \Rightarrow x+1 < -2x \Rightarrow x<-\frac{1}{3}$
But the solution should be $x \in (0,1)$.
I can see that there can't be negative values of $x$ in the inequality, because the left side would be positive and it can't be less than a negative number. But shouldn't this appear on my calculations?
| First observe this inequation is defined for $x\ne 0,-1$ and that it implies $x>0$. Now you can square both sides to remoave the absolute value:
$$\frac1{|x+1|}<\frac1{2x}\iff|x+1|>2x\iff(x+1)^2>4x^2\iff 0>3x^2-2x-1.$$
One obvious root of the quadratic polynomial is $x=1$, and the other root is negative since their product is $-\dfrac13$.
So $3x^2-2x-1<0\iff -\dfrac13<x<1$. Taking into account the condition on$x$, we obtain
$$0<x<1.$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Find $f(2^{2017})$ The function $f(x)$ has only positive $f(x)$. It is known that $f(1)+f(2)=10$, and $f(a+b)=f(a)+f(b) + 2\sqrt{f(a)\cdot f(b)}$. How can I find $f(2^{2017})$?
The second part of the equality resembles $(\sqrt{f(a)}+\sqrt{f(b)})^2$, but I still have no idea what to do with $2^{2017}$.
| $f(2) = f(1+1) = f(1) + f(1) + 2 \sqrt{f(1) \cdot f(1)} = 4f(1)$
Together with $f(1) + f(2) = 10$, this gives $f(1) = 2$ and $f(2) = 8$.
Then, $f(n+1) = f(n) + 2\sqrt{2f(n)} + 2 = (\sqrt{f(n)}+\sqrt2)^2$, i.e. $\sqrt{f(n+1)} = \sqrt{f(n)} + \sqrt2$
This tells us that $\sqrt{f}$ is an arithmetic sequence with common difference $\sqrt2$.
With $\sqrt{f(1)} = \sqrt2$ and $\sqrt{f(2)} = 2\sqrt2$, we obtain $\sqrt{f(n)} = n\sqrt2$, i.e. $f(n) = 2n^2$.
In particular, $f(2^{2017}) = 2 \cdot (2^{2017})^2 = 2^{4035}$.
| {
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"timestamp": "2023-03-29T00:00:00",
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Factoring the polynomial $3(x^2 - 1)^3 + 7(x^2 - 1)^2 +4x^2 - 4$ I'm trying to factor the following polynomial:
$$3(x^2 - 1)^3 + 7(x^2 - 1)^2 +4x^2 - 4$$
What I've done:
$$3(x^2 -1)^3 + 7(x^2-1)^2 + 4(x^2 -1)$$
Then I set $p=x^2 -1$ so the polynomial is:
$$3p^3 + 7p^2 + 4p$$
Therefore: $$p(3p^2 + 7p + 4)$$
I apply Cross Multiplication Method: $$p(p+3)(p+4)$$
I substitute $p$ with $x^2-1$:
$$(x^2-1)(x^2-1+3)(x^2-1+4)$$
$$(x-1)(x+1)(x^2-2)(x^2-3)$$
I don't know if I've done something wrong or if I have to proceed further and how. The result has to be: $x^2(3x^2+1)(x+1)(x-1)$. Can you help me? Thanks.
| Hint: $y = x^2-1$:
$3(x^2 - 1)^3 + 7(x^2 - 1)^2 +4(x^2 - 1) = 3y^3+7y^2+4y = y(3y^2+7y+4) = y(y+1)(3y+4)$
| {
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"timestamp": "2023-03-29T00:00:00",
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$P(X=x, Y=y)= e^{-2}\binom{x}{y} \frac{3}{4}^{y}\frac{1}{4}^{x-y}\frac{2^{x}}{x!}$ ; Then $E(Y)$? Let the random variables $X$ and $Y$ have the joint probability mass function
$$P(X=x, Y=y)= e^{-2}\binom{x}{y} \dfrac{3}{4}^{y}\frac{1}{4}^{x-y}\frac{2^{x}}{x!} ; \ \ y=0,1,...x ; x = 0,1,2...$$
Then $E(Y)$?
I tried to simplify this but couldn't. I can see one part is of binomial other is of Poisson with parameter $2$ but they are not independent so that is not gonna help me. How should i simplify it ?
| If it's too hard to find the marginals, write
$$\begin{align}
\mathbb{E}[Y] &= \sum_{\text{all x}}\sum_{\text{all }y}y \cdot \mathbb{P}(X = x, Y = y)\\
&=\sum_{x=0}^{\infty}\sum_{y=0}^{x}y\cdot e^{-2}\binom{x}{y}\left(\dfrac{3}{4}\right)^y\left(\dfrac{1}{4}\right)^{x-y}\dfrac{2^x}{x!} \\
&= \sum_{x=0}^{\infty}e^{-2}\dfrac{2^x}{x!}\left[\sum_{y=0}^{x}y\binom{x}{y}\left(\dfrac{3}{4}\right)^y\left(\dfrac{1}{4}\right)^{x-y}\right]
\end{align}$$
The sum
$$\sum_{y=0}^{x}y\binom{x}{y}\left(\dfrac{3}{4}\right)^y\left(\dfrac{1}{4}\right)^{x-y}$$
is the expected value of a binomial distribution based on $x$ trials and "success" probability $3/4$. Hence,
$$\sum_{y=0}^{x}y\binom{x}{y}\left(\dfrac{3}{4}\right)^y\left(\dfrac{1}{4}\right)^{x-y} = x\left(\dfrac{3}{4}\right)$$
Thus,
$$\mathbb{E}[Y]=\sum_{x=0}^{\infty}e^{-2}\dfrac{2^x}{x!}x\dfrac{3}{4}=\dfrac{3}{4}\sum_{x=0}^{\infty}x\cdot\dfrac{e^{-2}2^x}{x!}$$
The sum
$$\sum_{x=0}^{\infty}x\cdot\dfrac{e^{-2}2^x}{x!}$$
is the expected value of a Poisson distribution with parameter $2$, hence we have
$$\mathbb{E}[Y] = \dfrac{3}{4} \cdot 2 = \dfrac{3}{2}\text{.}$$
| {
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Prove by induction: $\sum\limits_{k=1}^n (-1)^{k+1}k^2 = \frac{(-1)^{n+1}n(n+1)}{2}$ The whole problem has been translated from German, so apologies if I made any mistakes. Thank you for taking the time to help!
So this is a problem from my math book
Prove that $\sum\limits_{k=1}^n (-1)^{k+1}k^2 = \frac{(-1)^{n+1}n(n+1)}{2}$ for all $n\ \varepsilon \ N$
and I have come as far as this:
*
*let $n_0 = 1$. Then $\sum\limits_{k=1}^n (-1)^{k+1}k^2 = (-1)^{1+1}1^2 = 1 = \frac{(-1)^{1+1}1(1+1)}{2} = 1 $
*For one $n\ \varepsilon \ N$ $\sum\limits_{k=1}^n (-1)^{k+1}k^2 = \frac{(-1)^{n+1}n(n+1)}{2}$ is true.
*Now we have to show that $\sum\limits_{k=1}^{(n+1)} (-1)^{k+1}k^2 = \frac{(-1)^{(n+1)+1}(n+1)((n+1)+1)}{2} = \sum\limits_{k=1}^{(n+1)} (-1)^{k+1}k^2 = \frac{(-1)^{(n+2)}(n+1)((n+2)}{2}$ for all $n\ \varepsilon \ N$
From here, I did this: $\sum\limits_{k=1}^{(n+1)} (-1)^{k+1}k^2 = \sum\limits_{k=1}^{(n)} (-1)^{k+1}k^2 + (-1)^{(n+2)}(n+1)^2 = \frac{(-1)^{n+1}n(n+1)}{2} + (-1)^{(n+2)}(n+1)^2 = \frac{(-1)^{n+1}n(n+1)}{2} + \frac{2((-1)^{(n+2)}(n+1)^2)}2$
at which point I got stuck.
A look at the solutions at the back of the book told me that the next step is to factor (n+1) and $(-1)^{(n+2)}$, which resulted into this:
$\frac{(-1)^{n+2}(n+1)((-1)n+2(n+1))}{2} = \frac{(-1)^{n+2}(n+1)(-n+2n+2)}{2} = \frac{(-1)^{n+2}(n+1)(n+2)}{2}$
Which makes sense, but here is my question: How on earth do I get from $\frac{(-1)^{n+1}n(n+1)}{2} + \frac{2((-1)^{(n+2)}(n+1)^2)}2$ to $\frac{(-1)^{n+2}(n+1)((-1)n+2(n+1))}{2}$? I have been trying for quite a while and I just can't seem to figure it out. Any help is greatly appreciated!
| This is obviously true for $n=1$. Assume it's true for any positive integer $n$
$$\sum\limits_{k=1}^n (-1)^{k+1}k^2 = \frac{(-1)^{n+1}n(n+1)}{2}$$
we must show that:
$$\sum\limits_{k=1}^{n+1} (-1)^{k+1}k^2 = \frac{(-1)^{n+2}(n+1)(n+2)}{2}$$
from the other side we have:
$$\sum\limits_{k=1}^{n+1} (-1)^{k+1}k^2 =\sum\limits_{k=1}^n (-1)^{k+1}k^2+(-1)^{n+2}(n+1)^2 $$
by substitution from our induction assumption we obtain:
$$\sum\limits_{k=1}^{n+1} (-1)^{k+1}k^2=\frac{(-1)^{n+1}n(n+1)}{2}+(-1)^{n+2}(n+1)^2=\frac{(-1)^{n+2}(-n^2-n+2n^2+4n+2)}{2}=\frac{(-1)^{n+2}(n^2+3n+2)}{2}=\frac{(-1)^{n+2}(n+1)(n+2)}{2}$$
which is what we wnated to show.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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If $n \in \mathbb{N}\setminus\{1\}$ then $\gcd(n^2-1,3n+1)=1$ if and only if $n$ is even If $n\in\mathbb N\setminus\{1\}$ then $\gcd(n^2-1,3n+1)=1$ if and only if $n$ is even.
Can somebody help me prove this problem?
| If $d$ divides both $n^2-1,3n+1$
$d$ must divide $n(3n+1)-3(n^2-1)=n+3$
$d$ must divide $3(n+3)-(3n+1)=8$
$d$ must divide $(n+3,8)= \begin{cases}8&\mbox{if} n+3=8m\iff n\equiv5\pmod8 \\
4& \mbox{if }n+3=4(2t+1)\iff n\equiv1\pmod8\\2& \mbox{if } n+3=2(2r+1)\iff n\equiv3\pmod4\\1& \mbox{if } n+3=2s+1\iff n\equiv0\pmod2\end{cases}$
where $m,t,r,s$ are arbitrary integers
| {
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"timestamp": "2023-03-29T00:00:00",
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Prove that $\sum\limits_{cyc}\frac{a}{a^{11}+1}\leq\frac{3}{2}$ for $a, b, c > 0$ with $abc = 1$
Let $a$, $b$ and $c$ be positive numbers such that $abc=1$. Prove that:
$$\frac{a}{a^{11}+1}+\frac{b}{b^{11}+1}+\frac{c}{c^{11}+1}\leq\frac{3}{2}.$$
I tried homogenization and the BW (https://artofproblemsolving.com/community/c6h522084),
but it does not work.
Indeed, let $a=\frac{x}{y}$, $b=\frac{y}{z}$, where $x$, $y$ and $z$ are positives.
Hence, $c=\frac{z}{x}$ and we need to prove that $$\sum_{cyc}\frac{xy^{10}}{x^{11}+y^{11}}\leq\frac{3}{2},$$
which has a problem around $(x,y,z)=(7,5,6)$.
For these values $$\frac{3}{2}-\sum_{cyc}\frac{xy^{10}}{x^{11}+y^{11}}=0.0075...$$
I tried also TL, uvw, C-S, Lagrange multipliers and more, but without success.
Also, Vasc's Theorems don't help.
Also, the following method does not help here. Find the maximum of the expression
Because the inequality $\frac{x}{x^{11}+1}\leq\frac{3(a^9+1)}{4(a^{18}+a^9+1)}$ is wrong.
| This is probably wrong, but it might provide some ideas.
First observe that $a^nb^nc^n\leq3\;\;\forall n$, which is trivial by AM-GM.
Then, when $x\geq y$, then $\frac1x\leq\frac1y$.
First, expand to get $$a(b^{11}+1)(c^{11}+1)+b(a^{11}+1)(c^{11}+1)+c(a^{11}+1)(b^{11}+1)\over(a^{11}+1)(b^{11}+1)(c^{11}+1)$$
so
$$a+b+c+ab(a^{10}+b^{10})+bc(b^{10}+c^{10})+ac(a^{10}+c^{10})+abc(a^{10}b^{10}+b^{10}c^{10}+a^{10}c^{10})\over(a^{11}+1)(b^{11}+1)(c^{11}+1)$$
By the first observation $a+b+c\geq3$. Also, by AM-GM, $a^{10}+b^{10}\geq\frac2{c^5}$. And since $ab=\frac1c$, we get
$$3+2\left(\frac1{a^6}+\frac1{b^6}+\frac1{c^6}\right)+(a^{10}b^{10}+b^{10}c^{10}+a^{10}c^{10})\over(a^{11}+1)(b^{11}+1)(c^{11}+1)$$
Then, by GM-HM, $$1\geq{3\over{(a^{10}b^{10}+b^{10}c^{10}+a^{10}c^{10})\over a^{10}b^{10}c^{10}}}$$ so $$(a^{10}b^{10}+b^{10}c^{10}+a^{10}c^{10})\geq3$$
Thus we get $$6+2\left(\frac1{a^6}+\frac1{b^6}+\frac1{c^6}\right)\over(a^{11}+1)(b^{11}+1)(c^{11}+1)$$
And similarly to above, with GM-HM we get $$\frac1{a^6}+\frac1{b^6}+\frac1{c^6}\geq3$$
Thus we get $$12\over(a^{11}+1)(b^{11}+1)(c^{11}+1)$$
And expanding the denominator we get $$12\over2+(a^{10}b^{10}+b^{10}c^{10}+a^{10}c^{10})+(a^{11}+b^{11}+c^{11})$$
And now I'm not sure what to do. Hope this helps.
| {
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"timestamp": "2023-03-29T00:00:00",
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Prove that $ \frac{1}{1+n^2} < \ln(1+ \frac{1}{n} ) < \frac {1}{\sqrt{n} }$ For $n >0$ ,
Prove that $$ \frac{1}{1+n^2} < \ln(1+ \frac{1}{n} ) < \frac {1}{\sqrt{n}}$$
I really have no clue. I tried by working on $ n^2 + 1 > n > \sqrt{n} $ but it gives nothing.
Any idea?
| Note that
$$\frac{1}{1+n^2} < \ln\left(1+ \frac{1}{n} \right)< \frac {1}{\sqrt{n}}\iff\frac{n^2}{1+n^2} < n\ln\left(1+ \frac{1}{n} \right)^n < \frac {n^2}{\sqrt{n}}=n\sqrt n$$
which is true, indeed for $n=1$ the given inequality is true and for $n>1$
$$\frac{n^2}{1+n^2} <1< n\ln\left(1+ \frac{1}{n} \right)^n\leq n\log e=n< n\sqrt n$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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Find the integral $\int\sqrt{\frac{1-\sqrt{x}}{1+\sqrt{x}}}\,dx.$ The question is $$\int\sqrt{\frac{1-\sqrt{x}}{1+\sqrt{x}}}\,dx.$$
I have tried to multiply both numerator and denominator by $1-\sqrt{x}$ but can't proceed any further, help!
| $\int \sqrt{\frac{1-\sqrt{x}}{1+\sqrt{x}}} dx$ = $\int \sqrt{\frac{\left( 1-\sqrt{x} \ \right)\left( 1-\sqrt{x}\right)}{\left( 1+\sqrt{x} \ \right)\left( 1-\sqrt{x}\right)}} dx$ = $\int \ $$\frac{1-\sqrt{x}}{\sqrt{1-\ x}} dx$ \ = $\int \ \frac{dx}{\sqrt{1-\ x}} \ -$$\int \ \frac{\sqrt{x} dx}{\sqrt{1-\ x}} \ =A\ -\ B$
$A.\ Let\ u\ =\ sin^{2} x;\ \ du\ =\ 2sinxcosxdx$
$ \begin{gathered}
=2\int \ \frac{sinxcosxdx}{\sqrt{1-\ sin^{2} x}} \ =\ 2\int \ \frac{sinxcosxdx}{cosx} =2\int sinxdx\ =-2cosx\ +\ C\ =-2\left(\sqrt{1-x}\right)\\
\end{gathered}$
$B.\ Let\ u\ =\ sin^{2} x;\ \ du\ =\ 2sinxcosxdx$
$\int \ \frac{\sqrt{sin^{2} x}}{\sqrt{1-\ sin^{2} x}} 2sinxcosxdx\ =\ \int \ \frac{sinx}{\sqrt{cos^{2} x}} 2sinxcosxdx\ =\ 2\int \ \frac{sinx}{cosx} sinxcosxdx\ =2$$\int sin^{2} xdx$
$=2\left(\frac{1}{2} x-\frac{1}{4} sin( 2x) +C\right) =x-\frac{1}{2} sin( 2x) +C\ =x-sinxcosx+C=sin^{-1}\sqrt{x} -\sqrt{x( 1-x)} +C$
$\int \sqrt{\frac{1-\sqrt{x}}{1+\sqrt{x}}} dx\ =\ $$-2\sqrt{1-x} \ -\ $$sin^{-1}\sqrt{x} +\sqrt{x( 1-x)} +C$
| {
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Solve $ \int_\frac{-π}{3}^{\frac{π}{3}}\frac{\cos^2(x)-\sin^2(x)}{\cos^2(x)}dx$ I came across this question in my textbook and have been trying to solve it for a while but I seem to have made a mistake somewhere.
$$ \int_\frac{-π}{3}^{\frac{π}{3}}\frac{\cos^2(x)-\sin^2(x)}{\cos^2(x)}dx$$ and here is what I did. First I simplified the equation as $$ \int_\frac{-π}{3}^{\frac{π}{3}}(1-\tan^2(x))dx=x|_{\frac{-π}{3}}^{\frac{π}{3}}-\int_\frac{-π}{3}^{\frac{π}{3}}(\tan^2(x))dx$$
Then I simplified $\tan^2(x)\equiv\frac{\sin^2(x)}{\cos^2(x)}, \sin^2(x)\equiv1-\cos^2(x)$ so it becomes, $\tan^2(x)\equiv\frac{1-\cos^2(x)}{\cos^2(x)}=\frac{1}{\cos^2(x)}-1$ making the overall integral
$$\int_\frac{-π}{3}^{\frac{π}{3}}(1-\tan^2(x))dx=x|_{\frac{-π}{3}}^{\frac{π}{3}}-\int_\frac{-π}{3}^{\frac{π}{3}}(\frac{1}{\cos^2(x)}-1)dx$$
$$=x|_{\frac{-π}{3}}^{\frac{π}{3}}-\int_\frac{-π}{3}^{\frac{π}{3}}\frac{1}{\cos^2(x)}dx-\int_\frac{-π}{3}^{\frac{π}{3}}1dx=x|_{\frac{-π}{3}}^{\frac{π}{3}}-x|_{\frac{-π}{3}}^{\frac{π}{3}}-\int_\frac{-π}{3}^{\frac{π}{3}}\frac{1}{\cos^2(x)}dx$$
I know that $\int\frac{1}{\cos^2(x)}dx=\tan(x)+c$ but that's off by heart and not because I can work it out. Since $x|_{\frac{-π}{3}}^{\frac{π}{3}}-x|_{\frac{-π}{3}}^{\frac{π}{3}}=0$, the final equation becomes
$$\int_\frac{-π}{3}^{\frac{π}{3}}\frac{\cos^2(x)-\sin^2(x)}{\cos^2(x)}dx=\tan(x)|_{\frac{-π}{3}}^{\frac{π}{3}}=2 \sqrt3$$
Is what I did correct because I feel like I've made a mistake somewhere but can't find it. Also why does $\int\frac{1}{\cos^2(x)}dx=\tan(x)+c$.
EDIT - Made an error in $\int\frac{1}{\cos^2(x)}dx=\tan^2(x)+c$, it's actually $\int\frac{1}{\cos^2(x)}dx=\tan(x)+c$.
| Note that after your words: making the overall integral, because of the minus sign in $$\frac{1}{\cos^2 x} - 1 =\tan^2 x$$ we should have: $$x\bigg \lvert_{-\pi/3}^{\pi/3} - \int_{-\pi/3}^{\pi/3} \frac1{\cos^2 x} \, dx \color{red}{+} \int_{-\pi/3}^{\pi/3} 1\, dx$$
By the way, if you want to solve your integral quickly, note that we have: $$\int_{-\pi/3}^{\pi/3} \frac{\cos^2 x - \sin^2 x} {\cos^2 x} \, dx$$ $$=\int_{-\pi/3}^{\pi/3} 1-\tan^2 x \, dx$$ $$=\int_{-\pi/3}^{\pi/3} 1-(\sec^2 x - 1)\, dx$$ $$ =\int_{-\pi/3}^{\pi/3} 2-\sec^2 x\, dx$$ $$= 2\left[\frac{2\pi}3\right] - \tan x \bigg \lvert_{-\pi/3}^{\pi/3}$$ $$=2\left[\frac{2\pi}3-\sqrt 3\right]$$
| {
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"timestamp": "2023-03-29T00:00:00",
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A recursive divisor function Question:
Function definition:
$$f(1)=1$$
$$f(p)=p$$ where $p$ is a prime, and
$$f(n)=\prod {f(d_n)}$$ where $d_n$ are the divisors of $n$ except $n$ itself.
End result:
The end result of the function is when all divisors have been reduced to primes or 1.
Example:$$f(12)=f(2)f(3)f(4)f(6)=f(2)f(3)f(2)f(2)f(3)=f(2)^3f(3)^2=72$$
Question parts:
(a) Find a general formula for $f(a^n)$ where $a$ is a prime and $n$ is a natural number.
(b) Find a general formula for $f(a^nb^m)$ (following same notation). [Note: $a$ and $b$ are unique primes. $n$ and $m$, however, may be equal.]
Attempts at solutions:
(a) We have solved it. The solution is:
$a^{2^{n-2}}$ if $n≥2$,
$a$ if $n=1$.
(b) As of yet, none of us (me and my colleagues) have come up with a solution. We have solved the special cases
$$f(ab^m)=a^{2^{m-1}} \times b^{(2^{m-2})(m+1)}$$
$$f(a^2b^m)=a^{(2^{m-1})(m+2)} \times b^{(2^{m-2})(m^2+5m+2)/2}$$
$$f(a^3b^m)=a^{(2^{m-1})(m^2+7m+8)/2} \times b^{(2^{m-2})(m^3+12m^2+29m+6)/6}$$
Update 1: $f(a^4b^m)$ has been solved as well.
$$f(a^4b^m)=a^{(2^{m-1})(m^3+15m^2+56m+48)/6} \times b^{(2^{m-2})(m^4+22m^3+131m^2+206m+24)/24}$$
An answer to the above questions is needed. A general formula for $f(n)$ is appreciated, along with an explanation.
| The polynomials in the exponent of $b$ can be written
$${m+1\choose1}\\{m+3\choose2}-{2\choose1}\\
{m+5\choose3}-{3\choose1}{m+3\choose1}\\
{m+7\choose4}-{4\choose1}{m+5\choose2}+{4\choose2}$$
The values at $m=0,1,2$ are $1,2^k,2^{k-1}(k+2)$ so I predict the next two polynomials are
$${m+9\choose5}-{5\choose1}{m+7\choose3}+{5\choose2}{m+5\choose1}\\
{m+11\choose6}-{6\choose1}{m+9\choose4}+{6\choose2}{m+7\choose2}-{6\choose3}$$
The polynomial for $a$ is the polynomial for $b$ coming from $m+1$ and $n−1.$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2613617",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
Convergence of the sequence $ \sqrt {n-2\sqrt n} - \sqrt n $ Here's my attempt at proving it:
Given the sequence $$ a_n =\left( \sqrt {n-2\sqrt n} - \sqrt n\right)_{n\geq1} $$
To get rid of the square root in the numerator:
\begin{align}
\frac {\sqrt {n-2\sqrt n} - \sqrt n} 1 \cdot \frac {\sqrt {n-2\sqrt n} + \sqrt n}{\sqrt {n-2\sqrt n} + \sqrt n} &= \frac { {n-2\sqrt n} - \ n}{\sqrt {n-2\sqrt n} + \sqrt n} = \frac { {-2\sqrt n}}{\sqrt {n-2\sqrt n} + \sqrt n} \\&= \frac { {-2}}{\frac {\sqrt {n-2\sqrt n}} {\sqrt n} + 1}
\end{align}
By using the limit laws it should converge against:
$$
\frac { \lim_{x \to \infty} -2 } { \lim_{x \to \infty} \frac {\sqrt {n-2\sqrt n}}{\sqrt n} ~~+~~\lim_{x \to \infty} 1}
$$
So now we have to figure out what $\frac {\sqrt {n-2\sqrt n}}{\sqrt n}$ converges against:
$$
\frac {\sqrt {n-2\sqrt n}}{\sqrt n} \leftrightarrow \frac { {n-2\sqrt n}}{ n} = \frac {1-\frac{2\sqrt n}{n}}{1}
$$
${\frac{2\sqrt n}{n}}$ converges to $0$ since:
$$
2\sqrt n = \sqrt n + \sqrt n \leq \sqrt n ~\cdot ~ \sqrt n = n
$$
Therefore $~\lim_{n\to \infty} a_n = -1$
Is this correct and sufficient enough?
| It can be much shorter with asymptotic analysis:
\begin{align}
\sqrt {n-2\sqrt n\,} - \sqrt n&=\sqrt n\biggl(\sqrt{1-\frac2{\smash[b]{\sqrt n}\,}}-1\biggr)=\sqrt n\biggl(1-\frac1{\sqrt n}+o\biggl(\frac1{\sqrt n}\biggr)-1\biggr)\\[1ex]
&=-1+o(1) \to -1.
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2614626",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 1
} |
Does the series $\sum 2^n \sin(\frac{\pi}{3^n})$ converge?
Check if $$\sum_{n = 1}^{\infty}2^n \sin\left(\frac{\pi}{3^n}\right)$$ converges.
I tried to solve this by using the ratio test - I have ended up with the following limit to evaluate:
$$\lim_{n \to \infty} \left(\frac{2\sin\left(\frac{\pi}{3 \cdot 3^n} \right)}{\sin \left(\frac{\pi}{3^n} \right)} \right)$$
And now - I am stuck and don't know how to proceed with this limit. Any hints?
| If you want to stick with your approach using the ratio test, you can (although Olivier's answer is more direct). Caveat: I'll detail every step of the derivation, which is not actually necessary for a proof.
We will only rely on elementary arguments, specifically the fact that $\sin'(0)=\cos 0 = 1$ — which is equivalent to $$\lim_{x\to 0} \frac{\sin x}{x} = 1 \,.\tag{1}
$$
From there, you can write
$$
\lim_{n \to \infty} \frac{2\sin\left(\frac{\pi}{3 \cdot 3^n} \right)}{\sin \left(\frac{\pi}{3^n} \right)}
= \lim_{n \to \infty} 2\cdot \frac{1}{3}\cdot \frac{\sin\left(\frac{\pi}{3 \cdot 3^n} \right)}{\frac{\pi}{3 \cdot 3^n}}\cdot\frac{\frac{\pi}{3^n}}{\sin \left(\frac{\pi}{3^n} \right)}
= \frac{2}{3}\lim_{n \to \infty} \frac{\sin\left(\frac{\pi}{3 \cdot 3^n} \right)}{\frac{\pi}{3 \cdot 3^n}}\cdot\left(\frac{\sin \left(\frac{\pi}{3^n} \right)}{\frac{\pi}{3^n}}\right)^{-1} \tag{2}
$$
and, by (1), we get
$$\begin{align*}
\lim_{n \to \infty} \frac{2\sin\left(\frac{\pi}{3 \cdot 3^n} \right)}{\sin \left(\frac{\pi}{3^n} \right)}
&=
\frac{2}{3}\lim_{n \to \infty} \frac{\sin\left(\frac{\pi}{3 \cdot 3^n} \right)}{\frac{\pi}{3 \cdot 3^n}}\cdot \lim_{n \to \infty} \left(\frac{\sin \left(\frac{\pi}{3^n} \right)}{\frac{\pi}{3^n}}\right)^{-1}
=\frac{2}{3}\lim_{n \to \infty} \frac{\sin\left(\frac{\pi}{3 \cdot 3^n} \right)}{\frac{\pi}{3 \cdot 3^n}}\cdot \left(\lim_{n \to \infty} \frac{\sin \left(\frac{\pi}{3^n} \right)}{\frac{\pi}{3^n}}\right)^{-1} \\
&= \frac{2}{3}\cdot 1\cdot 1^{-1} = \boxed{\frac{2}{3}}
\end{align*}$$
and you can conclude with the ratio test.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2615300",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 2
} |
Evaluate limit containing $\sum{n^6}$ Evaluate:
$$\lim_{n\to\infty}{\frac{1^6+2^6+3^6+\ldots+n^6}{(1^2+2^2+3^2+\ldots+n^2)(1^3+2^3+3^3+\ldots+n^3)}}$$
I can solve the denominator as:
$$\frac{n(n+1)(2n+1)}{6}\cdot\frac{n^2(n+1)^2}{4}$$
$$n^7\cdot\frac{(1+\frac{1}{n})(2+\frac{1}{n})}{6}\cdot\frac{(1+\frac{1}{n})}{4}$$
$$=\frac{n^7}{12}$$
How can I reduce the numerator?
| Using Cesaro-Stolz (or limit of a Riemann sum) one can easily prove that $$\lim_{n\to\infty} \frac{1^p+2^p+\dots +n^p} {n^{p+1}}=\frac{1}{p+1}, p>-1$$ For the current question divide the numerator and denominator by $n^7=n^3\cdot n^4$ and then the limit is equal to $1/7/((1/3)(1/4))=12/7$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2616313",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 5,
"answer_id": 4
} |
The complex equation $x^3 = 9 + 46i$ has a solution of the form $a + bi$ where $a,b\in \mathbb Z$. Find the value of $a^3 + b^3$ The complex equation $x^3 = 9 + 46i$ has a solution of the form $a + bi$ where $a,b\in \mathbb Z$. Find the value of $a^3 + b^3$ .
| $\mathbb{Z}[i]$ is a Euclidean domain, hence a UFD. Since the norm of $9+46i$ is $9^2+46^2=13^3$, $(a+bi)^3=9+46i$ implies $a^2+b^2=13$. It follows that $a,b\in\{-3,-2,2,3\}$ and to check that the only solution is $a=-3$ and $b=2$ is straightforward.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2617729",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Computing the norms of a non-principal ideal in $\Bbb Z[\sqrt{-5}]$? $$ 2 x^2 + 2 xy + 3 y^2 $$
Apparantly these are the norms of the non-principal ideals of $\Bbb Z[\sqrt{-5}]$.
Why is that ?
How is that computed ?
Apparantly the ideals of prime norm in this ring have norms $2$, $5$ and $p$
with $p\equiv1,3,7,9\pmod{20}$.
I assume this follows from quadratic reciprocity and the norm of the ring being $x^2 + 5 y^2$.
Is that correct ?
Apparantly the principal ideals cannot have norms
$\equiv3,7$ and the non-principal ones cannot have norms $\equiv1,9$.
Again , why is that ?
I think I need to understand ideals better. These questions will help me.
| 0. Number field properties and ideal class
Let $w = \sqrt{-5}$, $K = \mathbb Q(w)$ and $R = \mathbb Z[w]$. We can compute the Minkowski's bound as $M_K <3$, then checking the ideal factorization of $(2)$ we find that $(2) = (2, w+1)^2 = J^2$ and that $J$ is non-principal. We remark that its norm is $N(J)=2$.
Hence the number field $K$ has class number $2$ and we can think of the ideal class as having two representatives: $P(K)$ the principal ideal group and $P(K)J$. In particular, multiplication by $J$ switches the principality of a given ideal $I$ since
$$P(K)J\cdot J=P(K)J^2=P(K)(2)=P(K)$$
1. Norms of ideals in $R$
For a principal ideal $(a+bw)\subseteq R$, the norm is straightforward:
$$ N(a+bw) = a^2+5b^2$$
So we are mostly concerned about $N(I)$ for a non-principal ideal $I\subseteq R$. From previous paragraph, $IJ$ is principal so we have $IJ=(a+bw)$ for some $a,b\in\mathbb Z$. Therefore using the multiplicative property of norm:
$$
2N(I)=N(J)N(I)=N(IJ)=N(a+bw)=a^2+5b^2
$$
Taking modulo 2, we see that
$$
a\equiv b \pmod 2 \implies a = b+2u
$$
for some integer $u$. Now
$$
\begin{align*}
2N(I) &= a^2+5b^2 = (b+2u)^2+5b^2 = 4u^2+4ub+6b^2\\
N(I) &= 2u^2 + 2ub + 3b^2
\end{align*}
$$
giving us the required form.
2. Prime norms
We have see that $N(J) = N(2,w+1)=2$. Norm $5$ comes from $N(w)=-w^2 = 5$. Ideals are either principal or not, so it comes down to solving for prime $p$:
$$
\begin{align*}
p &= x^2+5y^2\\
p &= 2x^2+2xy+3y^2
\end{align*}
$$
For the first, $p\equiv x^2\pmod 5$ so since $p$ is prime we must have $p\equiv 1$ or $4$ mod $5$. Using quadratic reciprocity:
$$
\left(\frac{5}{p}\right) = \left(\frac{p}{5}\right) = 1
$$
so $5$ is a square mod $p$. But then
$$
-1 \equiv (\sqrt 5 y x^{-1})^2 \pmod p
$$
$-1$ is a square iff $p\equiv 1 \pmod 4$, so combining both we get
$$
p \equiv 1, 9\pmod {20}
$$
For the other case, we can try to transform into the first case and use what we found. The obvious way is to multiply by 2:
$$
\begin{align*}
2p &= 4x^2 + 4xy + 6 y^2 = (2x+y)^2 + 5y^2 \\
2p &\equiv (2x+y)^2 \pmod 5
\end{align*}
$$
Now since $2$ is a quadratic non-residue, this time round we require $p\equiv 2$ or $3$ mod $5$. Using quadratic reciprocity again, we have
$$
\left(\frac{5}{p}\right) = \left(\frac{p}{5}\right) = -1
$$
Now taking modulo $p$, we have
$$
\begin{align*}
(2x+y)^2 + 5y^2 &\equiv 0 \pmod p \\
((2x+y)y^{-1})^2 &\equiv -5 \pmod p
\end{align*}
$$
Since $-5$ is a square and we have seen earlier that $5$ is not, hence $-1$ is a quadratic non-residue. This is possible iff $p\equiv 3 \pmod 4$. Now combining both we have
$$
p \equiv 3, 7\pmod{20}
$$
Therefore combining both sets of solutions the primes must be
$$
p\equiv 1,3,7,9 \pmod {20}
$$
Notice that $1,9\pmod{20}$ came from $p=x^2+5y^2$, which is from principal ideals. On the other hand $3,7\pmod{20}$ came from $p=2x^2+2xy+3y^2$, which is for non-principal ideals. This explains your last question: principal ideals cannot go to the $\{3,7\}$ class and vice-versa.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2618496",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Squared Summation(Intermediate Step) I am studying Economics and are trying to get a firm grasp of summation rules and applications. Looking into the following relation,
$~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~$$\sum_{k=1}^nk^2=\frac{n(n+1)(2n+1)}{6}$
The following "trick" is given below, to understand the above.
$~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~$$\sum_{k=1}^n[(k+1)^3-k^3]=n^3 + 3n^2+3n$
As I expand the left-handside of the equation for a given sequence $S=[1^2, 2^2, 3^3,..., n^3]$, the following leads to the right-hand side of the equation,
$(2^3 - 1^3) + (3^3 - 2^3) + (4^3 - 3^3) + .... + (n^3 - (n - 1)^3) + ((n+1)^3 - n^3) = (n + 1)^3 - 1^3 = n^3 + 3n^2 + 3n$
Understanding the intermediate steps, ie. the above expanding, Im struggling with the intuition so to speak. Which kind of mentality should I have had applied on the second equation, in order get to the right hand side, without expanding it?
Any help, with some mathematical explanation is highly appreciated.
Thank you!
| You need to know also that $1+2+...+n=\frac{n(n+1)}{2}$ and use your work:
$$n^3+3n^2+3n=3x+3\cdot\frac{n(n+1)}{2}+n,$$ which gives which you wish.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2619374",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
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Proving that the sequence $\{\frac{3n+5}{2n+6}\}$ is Cauchy. I'm not quite sure how to tackle these kinds of questions in general, but I tried something that I thought could be right. Hoping to be steered in the right direction here!
Let $\{\frac{3n+5}{2n+6}\}$ be a sequence of real numbers. Prove that this sequence is Cauchy.
Proof:
We want to establish that $\forall_{\epsilon>0}\exists_{{n_0}\in{\mathbb{N}}}\forall_{n,m\geq n_0}\big(|f(n)-f(m)|\big)<\epsilon.$
Suppose $n>m$ without loss of generality. We then know that $\frac{3n+5}{2n+6}>\frac{3m+5}{2m+6}$ and thus that $\frac{3n+5}{2n+6}-\frac{3m+5}{2m+6}>0$ such that $\frac{3n+5}{2n+6}-\frac{3m+5}{2m+6}=|\frac{3n+5}{2n+6}-\frac{3m+5}{2m+6}|=|f(n)-f(m)|.$
Let us work out the original sequence:
$\frac{3n+5}{2n+6}-\frac{3m+5}{2m+6}=\frac{(3n+5)(2m+6)-(3m+5)(2n+6)}{(2n+6)(2m+6)} = \frac{8(n-m)}{(n2+6)(2m+6)}<\frac{8(n-m)}{nm}= 8(\frac{1}{n}- \frac{1}{m}).$
We know that $\frac{1}{n}<\frac{1}{m}$ as $n>m$ and that $\frac{1}{n}\leq\frac{1}{n_0}, \frac{1}{m}\leq\frac{1}{n_0}$ for $n,m\geq n_0$.
This means that $\frac{1}{n}-\frac{1}{m}\leq \frac{1}{n_0}- \frac{1}{m}\leq\frac{1}{n_0}$, and thus $8(\frac{1}{n}- \frac{1}{m})\leq \frac{8}{n_0}$.
Let $\epsilon=\frac{8}{n_0}$, as it only depends on $n_0$ it can become arbitrarily small.
Then the following inequality holds: $\frac{3n+5}{2n+6}-\frac{3m+5}{2m+6}<8(\frac{1}{n}- \frac{1}{m})\leq \frac{8}{n_0}$.
So: $\frac{3n+5}{2n+6}-\frac{3m+5}{2m+6}<\epsilon$, and thus the sequence is Cauchy.$\tag*{$\Box$}$
| $|\frac{3n+5}{2n+6}-\frac{3m+5}{2m+6}|=\frac{|n-m|}{(2n+6)(2m+6)}\le\frac{8|n-m|}{nm} =8|\frac{1}{n}-\frac{1}{m}|$.
Can you proceed ?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2620493",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Proof explanation about matrix for block. Prove:
$$\det\left[\begin{array}[cc]\\A&C\\
0&B\end{array}\right]=\det(A)\det(B)$$
Proof:
$$A = Q_A R_A, \quad B = Q_B R_B$$
be QR decompositions of $A$ and $B$. Then
\begin{align*}
\det \begin{bmatrix} A & C \\ 0 & B \end{bmatrix} &= \det \begin{bmatrix} Q_A R_A & Q_A Q_A^T C \\ 0 & Q_B R_B \end{bmatrix} = \det \left( \begin{bmatrix} Q_A \\ & Q_B \end{bmatrix} \begin{bmatrix} R_A & Q_A^T C \\ 0 & R_B \end{bmatrix} \right) \\
&= \det \begin{bmatrix} Q_A \\ & Q_B \end{bmatrix} \det \begin{bmatrix} R_A & Q_A^T C \\ 0 & R_B \end{bmatrix} = \det Q \det R,
\end{align*}
where
$$Q := \begin{bmatrix} Q_A \\ & Q_B \end{bmatrix}, \quad R := \begin{bmatrix} R_A & Q_A^T C \\ 0 & R_B \end{bmatrix}.$$
Notice that $R$ is (upper) triangular, so its determinant is equal to the product of its diagonal elements, so
$$\det R = \det \begin{bmatrix} R_A & 0 \\ 0 & R_B \end{bmatrix}.$$
Combining what we have,
\begin{align*}
\det \begin{bmatrix} A & C \\ 0 & B \end{bmatrix} &= \det Q \det R = \det \begin{bmatrix} Q_A \\ & Q_B \end{bmatrix} \det \begin{bmatrix} R_A \\ & R_B \end{bmatrix} \\
&= \det Q_A \det Q_B \det R_A \det R_B = \det (Q_AR_A) \det (Q_B R_B) \\
&= \det A \det B.
\end{align*}
I don't understand this line:$$\det \begin{bmatrix} Q_A \\ & Q_B \end{bmatrix} \det \begin{bmatrix} R_A \\ & R_B \end{bmatrix} \\
= \det Q_A \det Q_B \det R_A \det R_B$$
Can explain me that detail?
| Determinants multiply.
With $A$ square $m$ by $m,$ and $B$ square $n$ by $n,$ also $C$ $m$ by $n,$
$$
\left(
\begin{array}{c|c}
I_m & 0 \\ \hline
0 & B
\end{array}
\right)
\left(
\begin{array}{c|c}
A & C \\ \hline
0 & I_n
\end{array}
\right) =
\left(
\begin{array}{c|c}
A & C \\ \hline
0 & B
\end{array}
\right)
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2621480",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Find the area of underneath the curve $ \ f(x)=4-3x \ $ over the interval $ \ [2,4] \ $ by dividing the interval into $ \ n \ $ Find the area of underneath the curve $ \ f(x)=4-3x \ $ over the interval $ \ [2,4] \ $ by dividing the interval into $ \ n \ $ equal subintervals and taking limit.
Answer:
The length of the interval $ \ [2,4] \ $ is
$ \ 4-2=2 \ $ ,
Thus subinterval size $ =\Delta x=\frac{2}{n} , \ $
Then the required area is given by
$ \lim_{n \to \infty} \sum_{i=1}^{n} f(x_i) \Delta x \\ = \lim_{n \to \infty} \sum_{i=1}^{n} [4-3 (2+i \frac{2}{n} )] \frac{2}{n}
\\ = -\lim_{n \to \infty} \sum_{i=1}^{n} \left[2+i \frac{6}{n}\right] \frac{2}{n} \\ \\ = -\large \lim_{n \to \infty} \left(\frac{4}{n}+12 \{\frac{1}{n^2} + \frac{2}{n^2} +..........+\frac{n}{n^2}\} \right) \\ =\large 0- \lim_{n \to \infty} \frac{12}{n^2} (1+2+3+.....+n) \\ = -\large \lim_{n \to \infty} \frac{12}{n^2} \cdot \frac{n(n+1)}{2} \\ =-6 $
Thus the required area $ \ =6 \ $
But $ \ \int_{2}^{4} (4-3x) dx=-10 \ $ i.e., the area is $ \ 10 \ $
Why such difference in approximation above.
I thought I have done something wrong in calculation .
Kindly examine my work and help me out
| There is a mistake in one of your steps:
$$ \lim_{n \to \infty} \sum_{i=1}^{n} f(x_i) \Delta x = \lim_{n \to \infty} \sum_{i=1}^{n} \left\{\left[4-3\left(2+i \frac{2}{n}\right)\right] \frac{2}{n} \right\}= -\lim_{n \to \infty} \sum_{i=1}^{n} \left[2+i \frac{6}{n}\right] \frac{2}{n} \\ \\ = -\large \lim_{n \to \infty} \left(\color{blue}{n\cdot\frac{4}{n}}+12 \left[\frac{1}{n^2} + \frac{2}{n^2} +..........+\frac{n}{n^2}\right] \right) \\ =- \lim_{n \to \infty} \left(4+\frac{12}{n^2} \sum_{k=1}^{n}k\right) =- 4 -\large \lim_{n \to \infty} \frac{12}{n^2} \cdot \frac{n(n+1)}{2} =\color{red}{-10} $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2621856",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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is there away to give $= \sqrt[3]{5 + 10i } + \sqrt[3]{5- 10i}$ question is there away to give
$$= \sqrt[3]{5 + 10i }
+ \sqrt[3]{5- 10i}$$
in just reals?
I am thinkting maybe use some clever algebra raise it to like 3 and magically imaginary terms dissapear. Or expressed the complex numbers in polar form but idk
Background (might delete if this is not really the question)
Let $$ m(x)=x^3-15x-10 $$
Find the roots of $m(x)$ using the cubic formula and show that they are all real
Using Howell complex approach
Lets change variables $$ \begin{aligned}
&z^3-15z-10
\\=&z^3+0z^2-15z-10
\end{aligned}$$
so, general form is denoted as
$$ z^3+a_2z^2+a_1z+a_0=0$$
so $$\begin{aligned}
a_2&=0
\\a_1&=-15
\\a_0&=-10
\end{aligned}$$
we get by substituting that $z=x-a_2/3$. $a_2=0$ its already depressed cubic but so theses steps are unnecesary for this one problem but will use them for other problems
$$ x^3+bx+c$$
where $$\begin{aligned}
b&=a_1-\frac{a_2^2}{3}
=-15-\frac{0^2}{3}=-15
\\c&=\frac{-a_1a_2}{3}+\frac{2}{27}a_2^3+a_0
=\frac{-(-15)(0)}{3}+\frac{2}{27}(0)^3+-10=-10
\end{aligned} $$
our depressed cubic is of $$ x^3-15x-10 $$
Ferro-Tartaglia Formula
$$ x=\sqrt[3]{\frac{-c}{2} + \sqrt{\frac{c^2}{4} +\frac{b^3}{27}}}\
+ \sqrt[3]{\frac{-c}{2} - \sqrt{\frac{c^2}{4} -\frac{b^3}{27}}}\ $$
in our case $$
\begin{aligned}
x&=\sqrt[3]{\frac{10}{2} + \sqrt{\frac{100}{4} +\frac{(-15)^3}{27}}}\
+ \sqrt[3]{\frac{10}{2} - \sqrt{\frac{100}{4} +\frac{(-15)^3}{27}}}\
\\&= \sqrt[3]{5 + \sqrt{25 -125}}
+ \sqrt[3]{5- \sqrt{25 -125}}
\\ &= \sqrt[3]{5 + 10i }
+ \sqrt[3]{5- 10i}
\end{aligned}
$$
| Note that $$ \sqrt[3]{5 + 10i } + \sqrt[3]{5- 10i}=
Z+ \bar Z =2\operatorname{Re}(Z)$$
Thus it is a real number.
In order to find the real part of Z, we may write it in polar form and get $$\operatorname{Re}(Z) = \sqrt 5 \cos(\theta /3)$$ where $$\theta =\cos^{-1} (\frac {\sqrt 5}{5})$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2623575",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 3
} |
Find the value of $\frac{a+b+c}{d+e+f}$ Given real numbers $a, b, c, d, e, f$, such that:
$a^2 + b^2 + c^2 = 25$
$d^2 + e^2 + f^2 = 36$
$ad + be + cf = 30$
What is the value of $\frac{a+b+c}{d+e+f}$?
I've tried combining equations in several ways but haven't gotten very far. Any hints would be appreciated.
| By C-S $$(a^2+b^2+c^2)(d^2+e^2+f^2)\geq(ad+be+cf)^2,$$
where the equality occurs for $(a,b,c)||(d,e,f)$, which we got.
Let $(a,b,c)=k(d,e,f).$
Thus, $5=|k|6$ and $|k|=\frac{5}{6}$ and
$$\frac{a+b+c}{d+e+f}=\frac{5}{6}$$ or
$$\frac{a+b+c}{d+e+f}=-\frac{5}{6}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2623782",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
} |
Find the x for the minimum area of triangle Triangle area = f(x) = $$\frac{{(4+a^2)}^2}{-4a}$$
So, to find min area, i take the derivative and equals to 0
$$f'(x) = 16 -8a^2-3a^4$$
$$0 = 16 -8a^2-3a^4$$
How to factorize it? Thanks
| At the risk of being pedantic, even if you arrive to the correct equation, you must know that your derivative if wrong.
$$f(x)=-\frac{\left(a^2+4\right)^2}{4 a}\implies f'(x)=-a^2-4+\frac{\left(a^2+4\right)^2}{4 a^2}=-\frac{3 a^2}{4}+\frac{4}{a^2}-2$$ What you wrote, as $f'(x)$ is in fact $4a^2 f'(x)$ assuming that $a\neq 0$.
Now, to get the solution of $f'(x)=0$, you need to solve $$-\frac{3 a^2}{4}+\frac{4}{a^2}-2=0$$ Let $x=a^2$ to make the equation
$$-\frac{3 x}{4}+\frac{4}{x}-2=0\implies 3 x^2+8 x-16=0$$ which is a quadratic in $x$; its solutions are
$x_1=-4$ and $x_2=\frac 43$. $x_1$ must be discarded since $x=a^2>0$ and the only root is then $a^2=x_2=\frac 43$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2626014",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
If $a,b,c$ be in Arithmetic Progression, If $a,b,c$ be in Arithmetic Progression, $b,c,a$ in Harmonic Progression, prove that $c,a,b$ are in Geometric Progression.
My Attempt:
$a,b,c$ are in AP so
$$b=\dfrac {a+c}{2}$$
$b,c,a$ are in HP so
$$c=\dfrac {2ab}{a+b}$$
Multiplying these relations:
$$bc=\dfrac {a+c}{2} \dfrac {2ab}{a+b}$$
$$=\dfrac {2a^2b+2abc}{2(a+b)}$$
$$=\dfrac {2a^2b+2abc}{2a+2b}$$
| Since $a,b,c$ are in AP we have $a=b-x$ and $c=b+x$ for some $x$ and since $b,c,a$ are in HP we have $$(b+x)(2b-x) = 2(b-x)b$$
Solwing this we get: $x=0$ or $x=3b$. So in the later case we get $a=-2b$ and $c=4b$ and so $$a^2=4b^2= bc$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2626517",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 0
} |
Number of answers : $f(x)=f^{-1}(x)$
let $f(x)= 1+\sqrt{x+k+1}-\sqrt{x+k} \ \ k \in \mathbb{R}$
Number of answers :
$$f(x)=f^{-1}(x) \ \ \ \ :f^{-1}(f(x))=x$$
MY Try :
$$y=1+\sqrt{x+k+1}-\sqrt{x+k} \\( y-1)^2=x+k+1-x-k-2\sqrt{(x+k+1)(x+k)}\\(y-1)^2+k-1=-2\sqrt{(x+k+1)(x+k)}\\ ((y-1)^2+k-1)^2=4(x^2+x(2k+1)+k^2+k)$$
now what do i do ?
| Let $y=f(x)=f^{-1}(x)$ and then
\begin{eqnarray}
y=1+\sqrt{x+k+1}-\sqrt{x+k},\tag{1}\\
x=1+\sqrt{y+k+1}-\sqrt{y+k}.\tag{2}
\end{eqnarray}
Subtracting (1) from (2) gives
\begin{eqnarray}
x-y&=&(\sqrt{y+k+1}-\sqrt{x+k+1})-(\sqrt{y+k}-\sqrt{x+k})\\
&=&\frac{y-x}{\sqrt{y+k+1}+\sqrt{x+k+1}}-\frac{y-x}{\sqrt{y+k}+\sqrt{x+k}}
\end{eqnarray}
and hence
$$ (x-y)\left[\frac{1}{\sqrt{y+k+1}+\sqrt{x+k+1}}-\frac{1}{\sqrt{y+k}+\sqrt{x+k}}\right]=0.$$
Thus one has either
$$ x=y $$
or
$$ \frac{1}{\sqrt{y+k+1}+\sqrt{x+k+1}}=\frac{1}{\sqrt{y+k}+\sqrt{x+k}}\tag{3}. $$
Since (3) does holds, one must have $x=y$ and hence
$$ x=1+\sqrt{x+k+1}-\sqrt{x+k}. $$
It is easy to see that $x$ is the real roots of the following equation
$$x(x^3-8x^2+12x-4)-4k(x+1)^2=0.\tag{4}$$
The number of answers depends on the real solutions of (4), either 1,2,3 or 4, depending on $k$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2627131",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Recursion of polynomials I have the following equation on the functions $f_i$ for $i>1$, defined in $[0,1/4)$
$$f_i(x) = f_{i-1}(x) + \frac{x}{1-4x}f_{i-2}(x)$$
And assume that $f_0$ equals $c(x)$, the generating function of the Catalan numbers, and $f_1(x)=(c(x)-1)/x$.
Any idea on how can I get a closed form if it exists, or differential equation, or relation to well known polynomials?
| Testing the form $f_i = g^i$, we get $g^2 = g + \frac{x}{1-4x} \implies g_{\pm} = \frac12 \pm \frac1{2\sqrt{1-4x}}$
So letting $f_i =a g_+^i + b g_-^i$, we get $c = a+b$ and
$$\frac{c-1}x = \frac{a+b}2 + \frac{a-b}{2\sqrt{1-4x}} \implies a - b = \frac{\sqrt{1-4x}}x(2c - 2-x c) $$
From which we should get
$$f_i = \frac12\left(c+ \frac{\sqrt{1-4x}}x(2x-2-xc) \right)\left(\frac12 + \frac1{2\sqrt{1-4x}} \right)^i + \frac12\left(c - \frac{\sqrt{1-4x}}x(2x-2-xc) \right)\left(\frac12 - \frac1{2\sqrt{1-4x}} \right)^i $$
Assuming the working's all right, does the form help?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2630029",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 0
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Find the limit of $\lim\limits_{x\to0^+}\frac{1}{e} \frac{e- e^ \frac{\ln (1+x)}{x}}{x}$ as $x$ Find the limit of $\frac{1}{e} \frac{e- e^ \frac{\ln (1+x)}{x}}{x}$ as $x$ approaches right of zero.
The answer is $\frac{1}{2}$ but I keep getting 1. Here's what I have:
Since $\lim_{x\to0^+} \frac{\ln (1+x)}{x}$ is of indeterminate form (0/0), we can apply L Hôpitals.
So now I have
$\lim_{x\to0^+}\frac{1}{e} \frac{e- e^ \frac{1}{1+x}}{x}$, which is also of indeterminate form (0/0).
So by L Hôpitals, $\lim_{x\to0^+}\frac{1}{e} \frac{{- e^ \frac{1}{1+x} \frac{-1}{(1+x)^2}}}{x}$ which is equal to 1.
But as I said, the answer is $\frac12$. Can you tell me where did I go wrong? Thanks!
| Given that $$\ln (1+x)=x-\frac{x^2}{2} +o(x^2)$$ we have $$\frac{1}{e} \frac{e- e^ \frac{\ln (1+x)}{x}}{x} = \frac{1- e^ \frac{\ln (1+x)-x}{x}}{x} = \frac{1- e^{ \frac{-x}{2}+o(x)} }{x} =\frac{1- 1+\frac{x}{2}+o(x) }{x} = \frac{1}{2}+o(1)\to\frac{1}{2}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2631343",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 2
} |
limit without expansion I was solving this limit instead of using sum of expansion i did this and got zero but answer is 1/5 by expansion why this is wrong
$$\lim_{n\rightarrow \infty } \frac{1^4 + 2^4 + \ldots + n^4}{n^5}$$
$$\lim_{n\rightarrow \infty } \frac{n^4( \frac{1^4}{n^4} + \frac{2^4}{n^4} +\ldots + 1 )}{n^5}$$
$$= \lim_{n\rightarrow \infty } \frac{( \frac{1^4}{n^4} + \frac{2^4}{n^4} +\ldots + 1 )}{n}$$
$$= \frac{(0 + 0 +0 \ldots + 1 )}{n} = 0$$ this is how i got ,
| If
$m > 1$ then
$k^m
\lt \int_k^{k+1} x^m dx
\lt (k+1)^m
$.
Summing the left inequality
from $1$ to $n-1$,
$\begin{array}\\
\sum_{k=1}^{n-1}k^m
&\lt \sum_{k=1}^{n-1}\int_k^{k+1} x^m dx\\
&= \int_1^{n} x^m dx\\
&= \dfrac{x^{m+1}}{m+1}\big|_1^{n}\\
&< \dfrac{n^{m+1}-1}{m+1}+(n+1)^m\\
&< \dfrac{n^{m+1}}{m+1}\\
\text{so}\\
\sum_{k=1}^{n}k^m
&< \dfrac{n^{m+1}}{m+1}+n^m\\
\text{so}\\
\dfrac1{n^{m+1}}\sum_{k=1}^{n}k^m
&< \dfrac{1}{m+1}+\dfrac1{n}\\
\end{array}
$
Similarly,
summing the right inequality
from $0$ to $n-1$,
$\begin{array}\\
\sum_{k=0}^{n-1}(k+1)^m
&\gt \sum_{k=0}^{n-1}\int_k^{k+1} x^m dx\\
&= \int_0^{n} x^m dx\\
&= \dfrac{x^{m+1}}{m+1}\big|_0^{n}\\
&= \dfrac{n^{m+1}}{m+1}\\
\text{so}\\
\sum_{k=1}^{n}k^m
&> \dfrac{n^{m+1}}{m+1}\\
\text{so}\\
\dfrac1{n^{m+1}}\sum_{k=1}^{n}k^m
&> \dfrac{1}{m+1}\\
\end{array}
$
Therefore
$0
\lt \dfrac1{n^{m+1}}\sum_{k=1}^{n}k^m
- \dfrac{1}{m+1}
\lt \dfrac1{n}
$
so
$\lim_{n \to \infty} \dfrac1{n^{m+1}}\sum_{k=1}^{n}k^m
=\dfrac1{m+1}$.
Note:
Nothing original here,
though I took pains
to make the result
come about
as painlessly as possible.
| {
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"url": "https://math.stackexchange.com/questions/2632317",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 3
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Calculate floor of $\frac{1}{x_1 + 1} + \frac{1}{x_2 + 1} + ... + \frac{1}{x_{100} + 1}$ There is a recurrence sequence $x_1 = \frac{1}{2}$, $x_{n+1} = x_n^2 + x_n$. How much is floor of $\frac{1}{x_1 + 1} + \frac{1}{x_2 + 1} + ... + \frac{1}{x_{100} + 1}$? Floor is an integer part of a real number.
| The instructive hints of @AchilleHui deserve an answer by its own.
Given the recurrence relation
\begin{align*}
x_{n+1}&=x_n(x_n+1)\tag{1}\\
x_1&=\frac{1}{2}\\
\end{align*}
we derive from (1) a telescoping representation of the reciprocal values via
\begin{align*}
\frac{1}{x_{n+1}}&=\frac{1}{x_n(x_n+1)}
=\frac{1}{x_n}-\frac{1}{x_n+1}
\end{align*}
We obtain
\begin{align*}
\sum_{j=1}^{100}\frac{1}{x_j+1}&=\sum_{j=1}^{100}\left(\frac{1}{x_j}-\frac{1}{x_{j+1}}\right)\\
&=\sum_{j=1}^{100}\frac{1}{x_j}-\sum_{j=2}^{101}\frac{1}{x_j}\tag{2}\\
&=\frac{1}{x_1}-\frac{1}{x_{101}}\\
&=2-\frac{1}{x_{101}}\tag{2}
\end{align*}
From $x_1=\frac{1}{2}$ and (1) we get $x_2=\frac{3}{4}, x_3=\frac{21}{16}>1$ and it follows again from (1) $x_n>1$ with $n\geq 3$.
We finally get from (2)
\begin{align*}
\color{blue}{\left\lfloor 2-\frac{1}{x_{101}} \right\rfloor = 1}
\end{align*}
Note: Since $x_1=\frac{1}{2}$ and $x_{n+1}\approx x_n^2$ we have $x_{101}\approx \frac{1}{2^{2^{100}}}$. So, the difference of $2-\frac{1}{x_{101}}$ to the value $2$ is very small.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2632611",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
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Evaluating a limit using Taylor series How do I evaluate the limit
$$ \lim_{x \to 0}\frac{e^{x^2-x} -1 + x - \alpha x^2 + x^4 \log x }{\cosh (2x) -1 + x^4 \sin (1/x^2)}$$
depending on $\alpha, $using Taylor series? I know I consider
$e^t = 1 +t +\frac{t^2}{2} + o(t^2)$
$ \log(1+t) = t - \frac{t^2}{2} + \frac{t^3}{3} + o(t^3)$
$\cosh(t) = 1+ \frac{t^2}{2} + \frac{t^4}{24} + o(t^4)$
$\sin(t) = t - \frac{t^3}{6} + \frac{t^5}{120} + o(t^5)$
and substitute. I know why it is possibile and to justify this method. Though, I am not able to solve this limit... Thanks!
| Note that
$$\begin{cases}x^3 \cdot x \sin (1/x^2)\to 0 \implies x^4\sin (1/x^2)=o(x^3)\\\\
x^3 \cdot x\log x\to 0\implies x^4\log x=o(x^3)\\\\
e^{x^2-x} -1 =x^2-x+\frac{(x^2-x)^2}{2}+o(x^2)=-x+\frac32x^2+o(x^2)\\\\
\cosh 2x=1+2x^2+o(x^2)\end{cases}$$
thus
$$\frac{e^{x^2-x} -1 -\alpha x^2 + x^4 \log x }{\cosh (2x) -1 + x^4 \sin (1/x^2)}
=\frac{-x+\frac32x^2+o(x^2) - \alpha x^2 + o(x^3) }{1+2x^2+o(x^2) -1 + o(x^3)}
=\frac{-x+\left(\frac32-\alpha \right)x^2 +o(x^2) }{2x^2+o(x^2)}=\frac{-\frac1x+\left(\frac32-\alpha \right) +o(1) }{2+o(1)}\to\pm\infty $$
therefore
$$\lim_{x \to 0}\frac{e^{x^2-x} -1 - \alpha x^2 + x^4 \log x }{\cosh (2x) -1 + x^4 \sin (1/x^2)}$$
does not exist.
| {
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"url": "https://math.stackexchange.com/questions/2633021",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Prove by induction that $(k + 2)^{k + 1} \leq (k+1)^{k +2}$
Prove by induction that $$ (k + 2)^{k + 1} \leq (k+1)^{k +2}$$ for $ k > 3 .$
I have been trying to solve this, but I am not getting the sufficient insight.
For example, $(k + 2)^{k + 1} = (k +2)^k (k +2) , (k+1)^{k +2}= (k+1)^k(k +1)^2.$
$(k +2) < (k +1)^2 $ but $(k+1)^k < (k +2)^k$ so what I want would clearly not be immediate from using something like If $ 0 < a < b, 0<c<d $ then $0 < ac < bd $. THe formula is valid for n = 4, So if it is valid for $n = k$ I would have to use
$ (k + 2)^{k + 1} \leq (k+1)^{k +2} $ somewhere in order to get that $ (k + 3)^{k + 2} \leq (k+2)^{k +3} $ is also valid. This seems tricky.
I also tried expanding $(k +2)^k $ using the binomial formula and multiplying this by $(k + 2)$, and I expanded $(k+1)^k$ and multiplied it by $(k + 1)^2 $ term by term. I tried to compare these sums, but it also gets tricky. I would appreaciate a hint for this problem, thanks.
| For $k=4$ it's true.
Let $(k+2)^{k+1}\leq(k+1)^{k+2}.$
Thus, $$((k+2)^2)^{k+1}\leq(k+1)^{k+2}(k+2)^{k+1}$$ or
$$((k+1)(k+3)+1)^{k+1}\leq(k+1)^{k+2}(k+2)^{k+1},$$ which gives
$$((k+1)(k+3))^{k+1}\leq(k+1)^{k+2}(k+2)^{k+1}$$ or
$$(k+3)^{k+1}\leq(k+1)(k+2)^{k+1}.$$
Thus, $$(k+3)^{k+2}\leq(k+3)(k+1)(k+2)^{k+1}=$$
$$=(k^2+4k+3)(k+2)^{k+1}\leq(k+2)^2(k+2)^{k+1}=(k+2)^{k+3}$$ and we are done!
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2633720",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
Integral of $\int{\frac{\cos^4x + \sin^4x}{\sqrt{1 + \cos 4x}}dx}$ How do we integrate the following?
$\int{\frac{\cos^4x + \sin^4x}{\sqrt{1 + \cos 4x}}dx}$ given that $\cos 2x \gt 0$
I tried to simplify this, but I cannot seem to proceed further than the below form:
$\int{\frac{\sec2x}{\sqrt{2}}dx + \sqrt{2}\int{\frac{\sin^2x\cos^2x}{\cos2x}dx}}$
$\implies \frac{1}{2\sqrt2}\log |\sec 2x + \tan 2x| + \sqrt{2}\int{\frac{\sin^2x\cos^2x}{\cos^2x-\sin^2x}dx} + C$
The answer that I'm supposed to get is:
$\frac{x}{\sqrt2}+C$
Please help, thanks!
| If $\cos 2x > 0$ and the integrand is wrong as Rudr Pratap Singh points out, then:
\begin{align}
\frac{\cos^4 x-\sin^4 x}{\sqrt{1+\cos 4x}}
&= \frac{(\cos^2x+\sin^2x)(\cos^2x-\sin^2x)}{\sqrt{2\cos^2 2x}} \\
&= \frac{(1)(\cos 2x)}{\sqrt2|\cos2x|} \\
&= \frac 1 {\sqrt2}.
\end{align}
Hence,
$$\int \frac{\cos^4x-\sin^4x}{\sqrt{1+\cos 4x}} \ \text dx = \frac x {\sqrt2} +c,$$
as desired by O.P.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2634477",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 4
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Some chain rule questions. Answer checks I have these problems and I just wanted to make sure I was doing Chain Rule correctly and that no further simplification was possible:
*
*$$y = (2x^3 + 5)^4$$
$$ so \frac{dy}{dx} = 4(2x^3+5)^3 * 6x^2$$
$$ = 24x^2(2x^3 + 5)^3$$
*$$f(x) = (5x^6 + 2x^3)^4$$
$$f'(x) = 4(5x^6 + 2x^3)^3 *(30x^5 + 6x^2)$$
3.
$$f(x) = (1 + x + x^2)^{99}$$
$$f'(x) = 99(1 + x + x^2)^{98} * (1 + 2x)$$
4.
$$f(x) = \sqrt{5x + 1}$$
$$f'(x) = \frac{1}{2} * (5x+1)^{-\frac{1}{2}} * 5$$
$$\frac{2.5}{\sqrt{5x + 1}}$$
*I could use some help on this one:
$$f(x) = (2x-3)^4 * (x^2+x+1)^5$$
I started with this:
$$ f'(x) = (2x-3)^4 * 5(x^2+x+1)^4 * (2x+1) + (x^2 +x+1)^5 * 4(2x-3)^3 * 2$$
But how do I simplify from here?
| My comment was getting too long so I'll post a short answer.
For number $5$, as mentioned in the comments you should use the product rule. Recall that:
$$\frac{d}{dx}f(x)g(x)=\frac{df(x)}{dx}g(x)+f(x)\frac{dg(x)}{dx}.$$
Now, set $$\begin{aligned}&f(x)=(2x-3)^4\\&g(x)=(x^2+x+1)^5\end{aligned}$$
and differentiate both (using the chain rule as you did on the previous exercises)
$$\begin{aligned}&\frac{df(x)}{dx}=8(2x-3)^3\\&\frac{dg(x)}{dx}=5(2x+1)(x^2+x+1)^4\end{aligned}$$
so
$$\frac{d}{dx}\left((2x-3)^4(x^2+x+1)^5\right)=8(2x-3)^3(x^2+x+1)^5+5(2x+1)(2x-3)^4(x^2+x+1)^4$$
and by simplifying furthermore, you obtain
$$(2 x-3)^3 \left(x^2+x+1\right)^4 \left(28 x^2-12 x-7\right).$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Closed form of $u_n$ with $u_{0} =1$ and $u_n=\frac{1}{3}u_{n-1}-1$? Let $u_n$ be a sequence defined by recurrence relation as : $u_{0} =1$ and $u_n=\frac{1}{3}u_{n-1}-1$, I'm not familiar to look for the closed form of $u_n $ , I have tried to find $u_n $ with a function of $n$ but I don't succeed because it's not defined as an arithmetic progression or a geometric progression. Then, is there any simple way to get it closed form ?
Thank for any help
| You can express $u_n$ through $u_0$ sequentially through $u_{n-1}$, $u_{n-2}$ and so on
$$u_n =
\frac{1}{3} \cdot u_{n-1} - 1$$
Substitute $u_{n-1}$
$$u_n = \frac{1}{3} \left( \frac{1}{3} \cdot u_{n-2} - 1 \right) - 1 =
\frac{1}{3^2} \cdot u_{n-2} - \frac{1}{3} - 1$$
Then $u_{n-2}$ and so on
$$u_n =
\frac{1}{3^2} \left( \frac{1}{3} \cdot u_{n-3} - 1 \right) - \frac{1}{3} - 1 =
\frac{1}{3^3} \cdot u_{n-3} - \frac{1}{3^2} - \frac{1}{3} - 1 =
\dotsc$$
Finally
$$u_n = \frac{1}{3^n} - \sum \limits_{k = 1}^n \frac{1}{3^{n-k}}$$
To check this formula we can use mathematical induction.
The base case:
$$n = 1: u_1 = \frac{1}{3} - 1$$
The inductive step:
$$u_{n+1} = \frac{1}{3} \cdot u_n - 1 =
\frac{1}{3} \left( \frac{1}{3^n} - \sum \limits_{k=1}^n \frac{1}{3^{n-k}} \right) - 1 =
\frac{1}{3^{n+1}} - \sum \limits_{k=1}^n \frac{1}{3^{n+1-k}} - 1$$
We can substitute
$$1 = \frac{1}{3^{n+1 - k}}$$
where $k = n+1$, thus
$$u_{n+1} = \frac{1}{3^{n+1}} - \sum \limits_{k=1}^{n+1} \frac{1}{3^{n+1-k}}$$
So the formula is right
| {
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$a_n $ is a positive integer for any $n\in \mathbb {N} $. Let $(a_n)_{n\geq 1}$ be a sequence defined by $a_{n+1}=(2n^2+2n+1)a_n-(n^4+1 )a_{n-1} $.
$a_1=1$, $a_2=3$.
I have to show that $a_n $ is a positive integer for any $n\in \mathbb {N}, n\geq 1$.
I tried to prove it by induction but it doesn't work.
| It can be proven that for $n\ge 4$, $$n^2+\frac 32n\lt \frac{a_{n+1}}{a_n}\lt n^2+3n\tag1$$
From $(1)$, we have, for $n\ge 4$, $$\frac{a_{n+1}}{a_n}\gt 0$$
Since we have
$$a_1=1,\quad a_2=3,\quad a_3=22,\quad a_4=304,\quad a_5=6810$$
it follows that $a_n$ is a positive integer for all $n\ge 1$.
Now, let us prove that $(1)$ holds for $n\ge 4$.
Proof :
Let us prove $(1)$ by induction on $n$.
We see that $(1)$ holds for $n=4$ since
$$n^2+\frac 32n=22,\quad \frac{a_{n+1}}{a_n}=22+\frac{61}{152},\quad n^2+3n=28$$
Suppose that $(1)$ holds for some $n\ (\ge 4)$.
Then, we have
$$\begin{align}&\frac{a_{n+2}}{a_{n+1}}-\left((n+1)^2+\frac 32(n+1)\right)
\\\\&=2(n+1)^2+2(n+1)+1-\frac{a_n}{a_{n+1}}\left((n+1)^4+1\right)-\left((n+1)^2+\frac 32(n+1)\right)
\\\\&\gt 2(n+1)^2+2(n+1)+1-\frac{(n+1)^4+1}{n^2+\frac 32n}-\left((n+1)^2+\frac 32(n+1)\right)
\\\\&=\frac{n^2 - n - 8}{2 n (2 n + 3)}
\\\\&\gt 0\end{align}$$
and
$$\begin{align}&(n+1)^2+3(n+1)-\frac{a_{n+2}}{a_{n+1}}
\\\\&=(n+1)^2+3(n+1)-\left((2(n+1)^2+2(n+1)+1)-\frac{a_n}{a_{n+1}}\left((n+1)^4+1\right)\right)
\\\\&\gt (n+1)^2+3(n+1)-(2(n+1)^2+2(n+1)+1)+\frac{(n+1)^4+1}{n^2+3n}
\\\\&=\frac{2 n^2 + n + 2}{n (n + 3)}
\\\\&\gt 0\qquad\blacksquare\end{align}$$
| {
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} |
Reciprocal sum with 2017 Find 2017 distinct postive integers ($a_1, a_2, \dots, a_{2017}$) such that
$$\sum_{i=1}^{2017}\dfrac{1}{a_i}=\dfrac{2017}{1000}$$
That is what I know: $1=1/2+1/3+1/6$. From that we can find infinity many ways to reach 1 as sum of reciprocals of distinct positive integers.
| You can also start with
$$ \frac{2017}{1000}=\frac{1}{1}+\frac{1}{2}+\frac{1}{4}+\frac{1}{5}+\frac{1}{15}+\frac{1}{3000} $$
or
$$ \frac{2017}{1000}=\frac{1}{1}+\frac{1}{2}+\frac{1}{3}+\frac{1}{6}+\frac{1}{60}+\frac{1}{3000} $$
then expand $\frac{1}{3000}$ into a sum of $2012$ egyptian fractions by exploiting $\frac{1}{n}=\frac{1}{n+1}+\frac{1}{n(n+1)}$ multiple times, or
$$\frac{1}{3000}=\frac{1}{3000}\underbrace{\left(\sum_{k=1}^{2010}\frac{1}{2^k}+\frac{1}{3\cdot 2^{2009}}+\frac{1}{6\cdot 2^{2009}}\right)}_{2012\text{ terms}}$$
The number of possible solutions for this problem is probably gargantuan.
You know I've always liked that word gargantuan? I so rarely have an opportunity to use it in a sentence.
| {
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Suppose that $a$ and $b$ are positive integers such that $ab$ divides $a + b$. Suppose that $a$ and $b$ are positive integers such that $ab$ divides $a + b$.
Prove that $a = b$, then prove that $a$ is either $1$ or $2$.
I was thinking using the theorem:
If $n|a$ and $n|b$,then $n|(ax+by)$ for any $x,y\in \mathbb Z$.
Then I can derive that $ab$ divides $a$ and $b$. I could not go further. However, this doesn't seem like can help me solve the question I had.
| $abx=(a+b)$, where $x$ is greater than or equal to $1$ ($x$ cannot be $0$)
$$x = \frac{1}{a} + \frac{1}{b}$$
This implies that $x$ must be $1$ or $2$.
($2$ is the greatest value possible for sum of two reciprocals of natural numbers)
so if $x=1$, then $\frac{1}{a}$ and $\frac{1}{b}$ must both be $\frac{1}{2}$ and $a=b=2$.
so if $x=2$ then $\frac{1}{a}$ and $\frac{1}{b}$ must both be $1$ and $a=b=1$
| {
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Solve $a^3 -b^3 -c^3=3abc, a^2=2(b+c)$ in natural numbers.
Solve $a^3 -b^3 -c^3=3abc, a^2=2(b+c)$ in natural numbers.
Substituting $a=\sqrt{2(b+c)}$ in the cubic equation, we get:
$2\sqrt{2}(b+c)^{\frac{3}{2}} - b^3 -c^3 = 3\sqrt{2(b+c)}bc$
$2\sqrt{2}(b+c)\sqrt{b+c} - b^3 -c^3 = 3\sqrt{2(b+c)}bc$
Not able to proceed further, although I think that binomial expansion of $\sqrt{b+c}$ can yield some more steps. But, am confused about applying that too.
| Hint:
$$0=a^3-b^3-c^3-3abc = \frac12(a-b-c)[(a+b)^2+(b-c)^2+(c+a)^2]$$
So we have two cases:
Case 1:
$a=b+c \implies a^2=2a \implies a \in \{0, 2\}$.
You should be able to fnd out $b+c$ bounded in this case...
OR Case 2:
$a+b=b-c=c+a=0$, which again reduces cases considerably among natural numbers.
| {
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} |
Inequality concerning complex numbers and their modulus properties If $\left| z^3+\frac{1}{z^3} \right| \leq 2$, prove that $\left| z+\frac{1}{z} \right| \leq 2$.
My approach:
Let $z=x+yi$, where $x, y \in \mathbb{R}$ and $i=\sqrt{-1}$ is the imaginary unit.
Factor the expression$$\left| z^3+\frac{1}{z^3} \right| = \left| \Big(z+\frac{1}{z}\Big)\Big(z^2+\frac{1}{z^2}-1\Big)\right|=\left|z+\frac{1}{z}\right|\left|z^2+\frac{1}{z^2}-1\right|$$
Now use that fact that $$\left|\sum_{i=1}^{k}z_i\right| \leq \sum_{i=1}^{k}\left| z_i \right|$$
Hence, $$\left|z^2+\frac{1}{z^2}-1\right| \leq \left|z^2\right|+\left|\frac{1}{z^2}\right| - 1$$
But $$\left|z^2\right|+\left|\frac{1}{z^2}\right| = x^2+y^2 + \frac{1}{x^2+y^2} \leq 2$$
Which implies $$\left|z^2\right|+\left|\frac{1}{z^2}\right|-1 \leq \implies \left|z^2+\frac{1}{z^2}-1\right| \leq 1$$
Which implies $$\left| z + \frac{1}{z}\right| \leq 2$$
Did I miss something or is there a better way to prove the inequality?
Thank you :)
| $$2\ge|a^3+b^3|=|(a+b)^3-3ab(a+b)|$$
$$\ge|a+b|^3-3|ab(a+b)|$$
| {
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Find Area bounded by Inverse of $f(x)=x^3+3x+2$ between $x=-2$ and $x=6$ Find Area bounded by Inverse of $f(x)=x^3+3x+2$ between $x=-2$ and $x=6$
My Try:
we need $$I=\int_{-2}^{6} f^{-1}(x) dx$$
Use substitution $x=f(t)$, then limits will change to $-1$ and $1$
So
$$I=\int_{-1}^{1}f^{-1}(f(t)) f'(t)dt=\int_{-1}^{1}tf'(t)dt=\int_{-1}^{1}t(3t^2+3)dt$$
Now since we need Unsigned area, Required area is
$$I=2\int_{0}^{1}t(3t^2+3)dt=\frac{9}{2}$$
But Answer is $\frac{5}{4}$...
| Firstly note that as $f'(x) = 3x^2 + 3 > 0$ for all $x$, $f$ is an increasing function and so its inverse exists for all $x$. Denote this inverse by $f^{-1} (x)$.
Now, since $f(0) = 2$, then $f^{-1} (2) = 0$. As $f^{-1}(x)$ is an increasing function for all $x$ on its domain, we see that $f^{-1}(x) < 0$ for $x < 2$ and $f^{-1} (x) > 0$ for $x > 2$. The required area $A$ between the curve $f^{-1} (x)$, the $x$-axis, and the lines $x = -2$ and $x= 6$ will therefore be given by
$$A = \left |\int_{-2}^2 f^{-1} (x) \, dx \right | + \int_2^6 f^{-1} (x) \, dx.$$
To find the integral containing the inverse function the following result of
$$\int_a^b f(x) \, dx + \int_{f(a)}^{f(b)} f^{-1} (x) \, dx = b f(b) - a f(a),$$
will be used.
Firstly, by inspection we note that
\begin{align*}
f(-1) = -2 &\Rightarrow f^{-1} (-2) = -1\\
f(0) = 2 &\Rightarrow f^{-1} (2) = 0\\
f(1) = 6 &\Rightarrow f^{-1} (6) = 1
\end{align*}
So
\begin{align*}
\int_{-2}^2 f^{-1} (x) \, dx &= 0 \cdot f(0) - (-1) \cdot f(-1) - \int_{-1}^0 (x^3 + 3x + 2) \, dx\\
&= -2 - \left [\frac{x^4}{4} + \frac{3x^2}{2} + 2x \right ]_{-1}^0\\
&= -2 - \frac{1}{4} = -\frac{9}{4}
\end{align*}
And
\begin{align*}
\int_{2}^6 f^{-1} (x) \, dx &= 1 \cdot f(1) - 0 \cdot f(0) - \int_0^1 (x^3 + 3x + 2) \, dx\\
&= 6 - \left [\frac{x^4}{4} + \frac{3x^2}{2} + 2x \right ]_0^1\\
&= 6 - \frac{15}{4} = \frac{9}{4}.
\end{align*}
So the required area will be
$$A = \left |-\frac{9}{4} \right | + \frac{9}{4} = \frac{9}{2} \, \text{units}^2.$$
| {
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Complex number ABC Conjecture Regarding the abc conjecture, I can't find records for Gaussian integers.
Let $rad(a) = \lVert{\prod{prime factors(a)}}\rVert$.
For relatively prime $(a,b,c)$ with $a+b=c$, define quality as
$Q = \frac{\log{c}}{\log(rad(a b c))}$.
For Gaussian integers, the highest quality I can find in initial searching is
$Q = 1.65169:\ -1 +\ (2 + i)^4 =\ i(1+2 i)(1+i)^7 : $
$Q = 1.50515:\ 1 +\ (2 + i)^2 =\ -(1+i)^5 : $
$Q = 1.34527:\ 1 +\ (8 + 7 i)^2 =\ -i(1+2 i)^2(1+i)^9 : $
$Q = 1.23754:\ -3^4 +\ (4 + i)^4 =\ -i(1+2 i)(2+ i)^2(1+i)^9 : $
$Q = 1.21850:\ (1+2 i)^2 +\ (4+i)^4 =\ -(1+i)^2 (3+2 i)^2 (2+i)^3 : $
$Q = 1.21684:\ -1 +\ (4+i)^4 =\ -i(1+i)^8(1+2 i)(2+i)(2+3 i) : $
Can anyone beat that first item, or supply other complex a+b=c sums with $Q>1.22$?
| For equation $a+b=c$ $$i(1+i)\;+\;(2+3i)^4\;=\;i(3+2i)^4$$
we have
$$Q = \frac{\log\left(13^2\right)}{\log\left(13\sqrt{2}\right)}\approx 1.761929.$$
| {
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Prove that $2^{5n + 1} + 5^{n + 2} $ is divisible by 27 for any positive integer My question is related to using mathematical induction to prove that $2^{5n + 1} + 5^{n + 2} $ is divisible by 27 for any positive integer.
Work so far:
(1) For n = 1:
$2^{5(1) + 1} + 5^{(1) + 2} = 26 + 53 = 64 + 125 = 189$
Check if divisible by $27$: $189$ mod $27$ = $0$
As no remainder is left, the base case is divisible by $27$.
(2) Assume $n = k$, then $2^{5k + 1} + 5^{k + 2} = 27k$
(3) Prove that this is true for n = k + 1:
$$2^{5(k + 1) + 1} + 5^{(k + 1) + 2} $$
$$= 2^{5k + 5 + 1} + 5^{k + 1 + 2} $$
$$ = 32 * 2^{5k + 1} + 5 * 5^{k + 2}$$
$$= ? $$
I know I am supposed to factor out 27 somehow, I just cant seem to figure it out. Any help would be appreciated.
| Here is a different take, for fun.
Let $x_n = 2^{5n + 1} + 5^{n + 2} = 2\cdot 32^n+25\cdot 5^n$.
Since $32$ and $5$ are the roots of an equation $x^2=ax+b$ (*), we have $x_{n+2}=ax_{n+1}+bx_n$.
The result follows by induction because the base cases $27 \mid x_0 = 27$ and $27 \mid x_1 = 189$ are easily checked.
(*) where $a=32+5, b=32\cdot 5$. These values are not important. What matters is that the recurrence is linear with integer coefficients.
| {
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "2",
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Why $K^0 = \{0\}$? I am reading "Linear Algebra" by Takeshi SAITO.
Why $n \geq 0$ instead of $n \geq 1$?
Why $K^0 = \{0\}$?
Is $K^0 = \{0\}$ a definition or not?
He wrote as follows in his book:
Let $K$ be a field, and $n \geq 0$ be a natural number.
$$K^n = \left\{\begin{pmatrix}
a_{1} \\
a_{2} \\
\vdots \\
a_{n}
\end{pmatrix} \middle| a_1, \cdots, a_n \in K \right\}$$
is a $K$ vector space with addition of vectors and scalar multiplication.
$$\begin{pmatrix}
a_{1} \\
a_{2} \\
\vdots \\
a_{n}
\end{pmatrix} +
\begin{pmatrix}
b_{1} \\
b_{2} \\
\vdots \\
b_{n}
\end{pmatrix} = \begin{pmatrix}
a_{1}+b_{1} \\
a_{2}+b_{2} \\
\vdots \\
a_{n}+b_{n}
\end{pmatrix}\text{,}$$
$$c \begin{pmatrix}
a_{1} \\
a_{2} \\
\vdots \\
a_{n}
\end{pmatrix} =
\begin{pmatrix}
c a_{1} \\
c a_{2} \\
\vdots \\
c a_{n}
\end{pmatrix}\text{.}$$
When $n = 0$, $K^0 = 0 = \{0\}$.
| It is a convention, which you can take as a definition.
Since $K^n$ is an $n$-dimensional vector space when $n$ is a positive integer, we would like to have $K^0$ to denote a zero-dimensional vector space, which would be $\{0\}$.
| {
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"source": "stackexchange",
"question_score": "1",
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Trigonometric Inequality $\sin{A}+\sin{B}-\cos{C}\le\frac32$ To prove
$$\sin{A}+\sin{B}-\cos{C}\le\frac32$$
Given $A+B+C=\pi$ and $A,B,C>0$
I have managed to convert LHS to $$1-4\cos{\frac C2}\sin{\frac{A+C-B}2}\cos{\frac{A-B-C}2}$$ but that clearly isn't very helpful
One other conversion was $$\sin A+\sin B+\cos A\cos B-\sin A\sin B$$ but not sure how to proceed.
Any hints will be appreciated
| Like In $ \triangle ABC$ show that $ 1 \lt \cos A + \cos B + \cos C \le \frac 32$ OR $ \cos {A} \cos {B} \cos {C} \leq \frac{1}{8} $
Let $y=\sin A+\sin B-\cos C=2\cos\dfrac C2\cos\dfrac{A-B}2-2\cos^2\dfrac C2+1$
$$\iff2\cos^2\dfrac C2-2\cos\dfrac C2\cos\dfrac{A-B}2+y-1=0$$
As $\cos\dfrac C2$ is real,the discriminant $\ge0$
$$4\cos^2\dfrac{A-B}2-8(y-1)\ge0\iff8y\le8+4\cos^2\dfrac{A-B}2\le8+4$$
| {
"language": "en",
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How to solve the equation $x^2+2=4\sqrt{x^3+1}$? From the Leningrad Mathematical Olympiad, 1975:
Solve $x^2+2=4\sqrt{x^3+1}$.
In answer sheet only written $x=4+2\sqrt{3}\pm \sqrt{34+20\sqrt{3}}$.
How to solve this?
| I think the "fastest" way to deal with this is surely to square both members, to get
$$x^4 -16x^3 +4x^2 - 12 = 0$$
Now you can treat is as the quartic equation it is, via Ferrari's method for quartic equations.
I will write you down the general procedure for a quartic of the form
$$ax^4 + bx^3 + cx^2 + dx + e = 0$$
And leave you the easy maths behind.
Define the following quantities
$$P = 8ac - 3b^2$$
$$Q = b^3 + 8da - 4abc$$
$$R = c^2 - 3bd + 12 ae$$
$$S = 64a^3e - 16a^2c^2 + 16ab^2c - 16 a^2bd - 3b^4$$
Define other elements
$$U = \frac{P}{8a^2}$$
$$V = \frac{Q}{8a^3}$$
$$Y = 2c^3 - 9bcd + 27b^2e - 27ad^2 - 27ace$$
$$Z = \sqrt[3]{\frac{Y + \sqrt{Y^2 - 4R^2}}{2}}$$
$$W = \frac{1}{2} \sqrt{-\frac{2U}{3} + \frac{1}{3a} \left( Z + \frac{R}{Z}\right) }$$
Final Solutions (note: only two of them will be real, in your case)
$$x_{1,\ 2} = -\frac{b}{4a} - W \pm \sqrt{-4W^2 - 2U + \frac{V}{W}}$$
$$x_{3,\ 4} = -\frac{b}{4a} + W \pm \sqrt{-4W^2 - 2U - \frac{V}{W}}$$
NOTICE In your case $\color{red}{d = 0}$ which simplifies a bit the resolvent.
| {
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How many sub-square matrices does a square matrix have and is there a simple formula for it? Consider an $n \times n$ matrix $M$. I want to find the determinant for ALL sub-square matrices of $M$. There may be a better way but my method is to find all sub-square matrices and check them individually.
*
*How many sub-square matrices does a square matrix have and is there a simple formula for it?
The other problem is, after checking several videos I am not sure what counts as a sub-square matrix. I thought I could use the formula for the sum of squares, i.e.
$$S(n) = \frac{n(n+1)(2n+1)}{6}$$
But this gives sixteen $(1 \times 1)$ matrices, nine $(2 \times 2)$ matrices, four $(3 \times 3)$ matrices and one $(4 \times 4)$ matrices, i.e. a total of $30$ sub-square matrices.
\begin{pmatrix}
2 & 3 & 1 & 1 \\
1 & 2 & 3 & 1 \\
1 & 1 & 2 & 3 \\
3 & 1 & 1 & 2
\end{pmatrix}
But, for example, I can find thirty-six $(2 \times 2)$ matrices by observation, i.e. six for any pair of rows. For example, the first two rows give:
\begin{equation}
\begin{pmatrix}
2 & 3 \\
1 & 2
\end{pmatrix}
\begin{pmatrix}
2 & 1 \\
1 & 3
\end{pmatrix}
\begin{pmatrix}
2 & 1 \\
1 & 1
\end{pmatrix}
\begin{pmatrix}
3 & 1 \\
2 & 3
\end{pmatrix}
\begin{pmatrix}
3 & 1 \\
2 & 1
\end{pmatrix}
\begin{pmatrix}
1 & 1 \\
3 & 1
\end{pmatrix}
\end{equation}
If I do the same for rows $1$ and $3$, $1$ and $4$, $2$ and $3$, $2$ and $4$, and $3$ and $4$, I get thirty-six sub-square matrices by considering only $(2 \times 2)$ matrices. So now I am wondering if my way of counting sub-square matrices is wrong.
Any ideas?
| There’s 9 2x2 matrices if you only count ones that are touching. 3 for each row.
2 3 _ 3 1 _ 1 1 _ 1 2 _ 2 3 _ 3 1 _ 1 1 _ 1 2 _ 2 3
1 2 _ 2 3 _ 3 1 _ 1 1 _ 1 2 _ 2 3 _ 3 1 _ 1 1 _ 1 2
| {
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Algorithm complexity using iteration method I want to find the complexity of
$$T(n) = T\left(\frac{n}{2}\right) + n \left(\sin\left(n-\frac{π}{2}\right) +2\right)$$ by iteration method.
Assume $T(1) = 1$.
\begin{align*}
T(n) &= T\left(\frac{n}{2}\right) + n \left(\sin\left(n-\frac{π}{2}\right) +2\right)\\
&= T\left(\frac{n}{2^2}\right) + \frac{n}{2} \left(\sin\left(\frac{n}{2}-\frac{π}{2}\right) +2\right) + n \left(\sin\left(n-\frac{π}{2}\right) +2\right)\\
&= T\left(\frac{n}{2^3}\right) + \frac{n}{2^2} \left(\sin\left(\frac{n}{2^2}-\frac{π}{2}\right) +2\right) + \frac{n}{2} \left(\sin\left(\frac{n}{2}-\frac{π}{2}\right) +2\right)\\
&\mathrel{\phantom{=}}{}+ n \left(\sin\left(n-\frac{π}{2}\right) +2\right)\\
&=\cdots\\
&= T\left(\frac{n}{2^k}\right)+ n \sum_{i = 0}^{k-1} \frac{1}{2^i} \sin\left(\frac{n}{2^i} - \frac{π}{2}\right) + 2n \sum_{i = 0}^{k-1} \frac{1}{2^i}\\
&= T\left(\frac{n}{2^k}\right)+ n \sum_{i = 0}^{k-1} \frac{1}{2^i} \sin\left(\frac{n}{2^i} - \frac{π}{2}\right) + 2n \frac{1-(\frac{1}{2})^{k}}{1-\frac{1}{2}}.
\end{align*}
Now, how to I can simplify the second summation?
Thanks.
| We know that: $sin(x-\frac{\pi}{2})=-sin(x)$
For the second sum ($S_2$):
$\displaystyle S_2=n \sum_{i = 0}^k \frac{1}{2^i} \sin(\frac{n}{2^i} - \frac{π}{2})=-n \sum_{i = 0}^k \frac{1}{2^i} \sin(\frac{n}{2^i})=\Im ( -n \sum_{i = 0}^k \frac{1}{2^i} e^{j(\dfrac{n}{2^i})} ) =-n \Im( \sum_{i = 0}^k \frac{e^{j(\dfrac{n}{2^i})}}{2^i} )$
This is also a geometric sequence. Note that $\Im$ stands for imaginary part of the number and $\Im (e^{xj})=sin(x)$.
$$\displaystyle\to S_2= -n \Im ( \sum_{i = 0}^k \frac{e^{j(\dfrac{n}{2^i})}}{2^i} )=-n \Im(\dfrac{1-\frac{e^{j(\dfrac{n}{2})(k+1)}}{2^{(k+1)}}}{1-\frac{e^{j(\dfrac{n}{2})}}{2}})$$
We now multiply numerator and denominator by the conjugate of denominator to separate the real and imaginary part:
$$\to S_2=-n \Im \bigg(\displaystyle\dfrac{1-\frac{e^{j(\dfrac{n}{2})(k+1)}}{2^{(k+1)}}}{1-\frac{e^{j(\dfrac{n}{2})}}{2}} \times \frac{1+\frac{e^{j(\dfrac{n}{2})}}{2}}{1+\frac{e^{j(\dfrac{n}{2})}}{2}} \bigg)=-n\Im \bigg(\dfrac{1+\frac{e^{j(\dfrac{n}{2})}}{2}-\frac{e^{j(\dfrac{n}{2})(k+1)}}{2^{(k+1)}}-\frac{e^{j(\dfrac{n}{2})(k+2)}}{2^{(k+2)}}}{1-\dfrac{e^{nj}}{2}}\bigg)=-n\Im \bigg(\dfrac{1+\frac{e^{j(\dfrac{n}{2})}}{2}-\frac{e^{j(\dfrac{n}{2})(k+1)}}{2^{(k+1)}}-\frac{e^{j(\dfrac{n}{2})(k+2)}}{2^{(k+2)}}}{1-\dfrac{(-1)^n}{2}}\bigg)=-\dfrac{n}{1-\dfrac{(-1)^n}{2}}\Im \bigg({1+\frac{e^{j(\dfrac{n}{2})}}{2}-\frac{e^{j(\dfrac{n}{2})(k+1)}}{2^{(k+1)}}-\frac{e^{j(\dfrac{n}{2})(k+2)}}{2^{(k+2)}}}\bigg)
$$
$\displaystyle\to S_2=-\dfrac{n}{1-\dfrac{(-1)^n}{2}} \bigg( \dfrac{\sin(\dfrac n2)}{2} - \dfrac{\sin(\dfrac {n}{2} (k+1))}{2^{k+1}} - \dfrac{\sin(\dfrac {n}{2} (k+2))}{2^{k+2}} \bigg)$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2652742",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
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On the cubic generalization $(a^3+b^3+c^3+d^3)(e^3+f^3+g^3+h^3 ) = v_1^3+v_2^3+v_3^3+v_4^3$ for the Euler four-square We are familiar with the Euler Four-Square identity,
$$(a^2+b^2+c^2+d^2)(e^2+f^2+g^2+h^2 ) = u_1^2+u_2^2+u_3^2+u_4^2$$
where,
$$u_1 = ae-bf-cg-dh\\
u_2 = af+be+ch-dg\\
u_3 = ag-bh+ce+df\\
u_4 = ah+bg-cf+de$$
or the product of two sums of four squares is itself a sum of four squares.
Tinkering about, I came across a cubic version,
$$(a^3+b^3+c^3+d^3)(e^3+f^3+g^3+h^3 ) = 6^{-3} (w_1^3+w_2^3+w_3^3+w_4^3)$$
where,
$$w_1= 9 + a^3 + b^3 + c^3 + d^3 + e^3 + f^3 + g^3 + h^3\\
w_2= -9 - a^3 - b^3 - c^3 - d^3 + e^3 + f^3 + g^3 + h^3\\
w_3= 9 - a^3 - b^3 - c^3 - d^3 - e^3 - f^3 - g^3 - h^3\\
w_4= -9 + a^3 + b^3 + c^3 + d^3 - e^3 - f^3 - g^3 - h^3$$
Q: Does the cubic version have any number theoretic implications, like the set of sums of four cubes is closed under multiplication? Or is it just an interesting curiosity?
| Summarizing some comments and adding a bit on top:
The identity is a special case of
$$24ABC = (\underbrace{A+B+C}_{w_1})^3 + (\underbrace{-A+B-C}_{w_2})^3
+ (\underbrace{-A-B+C}_{w_3})^3 + (\underbrace{A-B-C}_{w_4})^3
\tag{1}$$
where setting $C=9$ turns the left-hand side into $6^3AB$.
The settings
$$\begin{align}
A &= a^3 + b^3 + c^3 + d^3
\\ B &= e^3 + f^3 + g^3 + h^3
\end{align}$$
are not necessary for $(1)$ but lead to an interesting specialization.
To make the right-hand side of $(1)$ a symbolic integer multiple of $6^3$,
we might want to require that each $w_i$ is divisible by $6$.
Straightforward calculations show that this happens if and only if both $A$
and $B$ are divisible by $3$ and $A+B$ is odd.
In other words,
$$\{A,B\}=\{6m,6n+3\}\quad\text{for some}\quad m,n\in\mathbb{Z}$$
Conversely, for every choice of $m,n\in\mathbb{Z}$ there exist
representations of $A$ and $B$ as sums of four cubes, such as
$$\begin{align}
6m &= (m+1)^3 + (m-1)^3 + (-m)^3 + (-m)^3 \tag{2}
\\ 6n+3 &= n^3 + (-n + 4)^3 + (2n - 5)^3 + (-2n + 4)^3 \tag{3}
\end{align}$$
taken from the Alpertron
hyperlinked in Dietrich Burde's comment.
Specializing your identity to that scenario gives
$$18\,m\,(2n+1) =
(2 + m + n)^3 + (-1 - m + n)^3 + (1 - m - n)^3 + (-2 + m - n)^3 \tag{4}$$
which may give much smaller solutions for representations of $18q$ than $(2)$,
particularly if $q$ has an odd divisor near $\sqrt{2|q|}$ which we can use for
$2n+1$.
Examples:
$$\begin{array}{r|rr|rrr}
18q & (2)\quad\text{with} & m
& (4)\quad\text{with} & m, & n
\\\hline
18 & 4^3 + 2^3 + (-3)^3 + (-3)^3 & 3
& 3^3 + (-2)^3 + 0^3 + (-1)^3 & 1 & 0
\\ 504 & 85^3 + 83^3 + (-84)^3 + (-84)^3 & 84
& 9^3 + (-2)^3 + (-6)^3 + (-1)^3 & 4 & 3
\end{array}$$
The following points may be worth mentioning:
*
*Replacing $2$ with $0$ in the right-hand side of $(4)$ turns the left-hand
side into $6\,m\,(2n-1)$ which is more widely applicable.
*Replacing $2$ with $1$ in the right-hand side of $(4)$ turns the left-hand
side into $24mn$, an easy specialization of $(1)$.
*$(1)$ itself, where applicable, might give solutions
with even smaller absolute maximum of the numbers involved, particularly if
$A,B,C$ can be chosen to be pairwise close.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2653086",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 1,
"answer_id": 0
} |
Trigonometry Exact Value using Half Angle Identity I have a quick question regarding a little issue.
So I'm given a problem that says "$\tan \left(\frac{9\pi}{8}\right)$" and I'm supposed to find the exact value using half angle identities. I know what these identities are $\sin, \cos, \tan$. So, I use the tangent half-angle identity and plug-in $\theta = \frac{9\pi}{8}$ into $\frac{\theta}{2}$. I got $\frac{9\pi}{4}$ and plugged in values into the formula based on this answer. However, I checked my work with slader.com and it said I was wrong. It said I should take the value I found, $\frac{9\pi}{4}$, and plug it back into $\frac{\theta}{2}$. Wouldn't that be re-plugging in the value for no reason? Very confused.
| Because the period of tangent is $\pi$,
$\tan \dfrac{9 \pi}{8} = \tan \dfrac{\pi}{8}$
You could just look this up, but its pretty easy to derive.
$$ \tan \frac x2
= \frac{\sin \frac x2}{\cos \frac x2}
= \frac{2 \sin \frac x2 \ \cos \frac x2}{1 + 2\cos^2 \frac x2 - 1}
= \frac{\sin x}{1 + \cos x}
= \frac{1 - \cos x}{\sin x}$$
Since $\sin \frac{\pi}{4} = \cos \frac{\pi}{4} = \frac{1}{\sqrt 2}$
$$ \tan \dfrac{9 \pi}{8}
= \tan \frac{\pi}{8}
= \tan \left( \frac 12 \cdot \frac{\pi}{4} \right)
= \frac{1 - \cos \frac{\pi}{4}}{\sin \frac{\pi}{4}}
= \frac{1-\frac{1}{\sqrt 2}}{\frac{1}{\sqrt 2}}
= \sqrt 2 - 1$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2655018",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
} |
Why limit for this function exists as $x$ approaches $0$? $$f(x) = \frac{\sqrt{x^{2}+100} - 10}{x^{2}} $$
As $$\lim_{x \to 0.01} f(x)=0.05$$ but, as we keep decreasing the value of $x$ to $0$, $$\lim_{x \to 0.000001} f(x) = 0.000000$$ So why this function's Limit exist?
| Without using the definition of the derivate:
$$\frac{\sqrt{x^2+100}-10}{x^2}=$$
Let's multiply the top and the bottom by $\sqrt{x^2+100}+10$ to get rid of the square root in the top:
$$\frac{\sqrt{x^2+100}-10}{x^2}\frac{\sqrt{x^2+100}+10}{\sqrt{x^2+100}+10}=$$
$$\frac{\left(\sqrt{x^2+100}-10\right)\left(\sqrt{x^2+100}+10\right)}{x^2\left(\sqrt{x^2+100}+10\right)}=$$
$$\frac{\left(\sqrt{x^2+100}\right)^2-(10)^2}{x^2\left(\sqrt{x^2+100}+10\right)}=$$
$$\frac{x^2+100-100}{x^2\left(\sqrt{x^2+100}+10\right)}=$$
$$\frac{x^2}{x^2\left(\sqrt{x^2+100}+10\right)}=$$
$$\frac{1}{\sqrt{x^2+100}+10}\to \frac{1}{\sqrt{100}+10}=\frac{1}{20}$$
Or you can also use Taylor-series:
$$\frac{\sqrt{x^2+100}-10}{x^2}=$$
$$\frac{10}{x^2}\left(\sqrt{1+\frac{x^2}{100}}-1\right)=$$
$$\frac{10}{x^2}\left(1+\frac{x^2}{200}+O(x^4)-1\right)=$$
$$\frac{10}{x^2}\left(\frac{x^2}{200}+O(x^4)\right)=$$
$$\frac{1}{20}+O(x^2)\to\frac{1}{20}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2658222",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Finding value of product of Cosines
Finding $$\left(\frac{1}{2}+\cos \frac{\pi}{20}\right)\left(\frac{1}{2}+\cos \frac{3\pi}{20}\right)\left(\frac{1}{2}+\cos \frac{9\pi}{20}\right)\left(\frac{1}{2}+\cos \frac{27\pi}{20}\right)$$
My Try: $$\left(\frac{1}{2}+\cos \frac{\pi}{20}\right)\left(\frac{1}{2}+\cos \frac{3\pi}{20}\right)\left(\frac{1}{2}+\sin \frac{\pi}{20}\right)\left(\frac{1}{2}-\sin\frac{3\pi}{20}\right)$$
So we have $$\left(\frac{1}{2}+\cos \frac{\pi}{20}\right)\left(\frac{1}{2}+\sin\frac{\pi}{20}\right)\left(\frac{1}{2}+\cos \frac{3\pi}{20}\right)\left(\frac{1}{2}-\sin\frac{3\pi}{20}\right)$$
Could some help me to solve it, Thanks in Advanced
| Let $\xi=\exp\left(\frac{2\pi i}{40}\right)$. The given product equals
$$ \frac{1}{16}\left(1+\xi+\xi^{-1}\right)\left(1+\xi^3+\xi^{-3}\right)\left(1+\xi^9+\xi^{-9}\right)\left(1+\xi^{27}+\xi^{-27}\right)$$
or
$$ \frac{1}{16\xi\xi^3\xi^9\xi^{27}}\cdot\frac{\xi^3-1}{\xi-1}\cdot\frac{\xi^9-1}{\xi^3-1}\cdot\frac{\xi^{27}-1}{\xi^9-1}\cdot\frac{\xi^{81}-1}{\xi^{27}-1}$$
or (by telescopic property and the fact that $\xi^{81}=\xi$)
$$ \frac{1}{16 \xi^{1+3+9+27}}=\frac{1}{16\xi^{40}}=\color{red}{\frac{1}{16}}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2659008",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 5,
"answer_id": 4
} |
Find $\int \sqrt{2+\tan x} dx$ I am trying to get the answer $\int \sqrt{2+\tan x} dx$
What I did was to write $2+\tan x$ as $\frac{2\cos x+\sin x}{\cos x}$ but then find no way. Is there any simple form or should I proceed some other way ?
| If you enjoy complex numbers, let us do as in the linked post
$$\sqrt{2+\tan (x)}=u^2 \implies x=-\tan ^{-1}\left(2-u^2\right)\implies dx=\frac{2 u}{1+\left(2-u^2\right)^2}\,du$$
$$\int \sqrt{2+\tan (x)}\, dx=\int \frac{2 u^2}{1+\left(2-u^2\right)^2}\,du$$
$$1+\left(2-u^2\right)^2=u^4-4 u^2+5=(u^2-(2-i))(u^2-(2+i))$$ Now, partial fraction decomposition
$$\frac{2 u^2}{1+\left(2-u^2\right)^2}=\frac{1+2 i}{u^2-(2-i)}+\frac{1-2 i}{u^2-(2+i)}$$ where you find almost classical integrals.
$$\int \frac{2 u^2}{1+\left(2-u^2\right)^2}\,du=i \left(\sqrt{-2-i} \tan ^{-1}\left(\frac{u}{\sqrt{-2-i}}\right)-\sqrt{-2+i} \tan
^{-1}\left(\frac{u}{\sqrt{-2+i}}\right)\right)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2661168",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Find all $z$ s.t $|z|=1$ and $|z^2 + \overline{z}^2|=1$ Here's my attempt. Let $z=x+i\ y$, then $$z^2=x^2-y^2+i\ 2xy$$ and $$\bar z^2=x^2-y^2-i\ 2xy$$
Then, $$z^2+\bar z^2=2x^2-2y^2$$ so $$1=|z^2+\bar z^2|=\sqrt{(2x^2)^2+(-2y^2)^2}$$
Simpliflying the expression above, we get $$1=4x^4+4y^4$$
which gives us $$\frac14=x^4+y^4$$. I am stuck here. Is it wrong approach? is there an easier one?
| Use exponential form: since $|z|=1$, $z=\mathrm e^{i\theta}$, so the second equation can be rewritten as
$$1=\bigl|z^2+\bar z^2\bigr|^2=\bigl(z^2+\bar z^2\bigr)^2=\bigl(\mathrm e^{2i\theta}+\mathrm e^{-2i\theta}\bigr)^2=\mathrm e^{4i\theta}+\mathrm e^{-4i\theta}+2=2\cos4\theta+2,$$
so finally we have to solve
$$\cos4\theta=-\frac12\iff 4\theta\equiv =\pm\frac{2\pi}3\pmod{2\pi}\iff\theta\equiv\pm\frac\pi6\mod{\frac\pi2}.$$
This corresponds the $8$ different complex numbers in $\mathbf U$:*
$$z=\mathrm e^{\pm\tfrac{k\pi}6},\quad k=1,2,4,5.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2662090",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 4
} |
finding an $r$-simple polar coordinate paramtrization of the area bounded by $y \geq 12$ and inside the circle $x^2+(y-9)^2=9^2$ I want to find an $r$-simple polar coordinate paramtrization of the area bounded by $y \geq 12$ and inside the circle $x^2+(y-9)^2=9^2$, meaning find two constants $a$ and $b$ and two functions $f$ and $g$ such that this area is given by $a \leq \theta \leq b$ and $f(\theta) \leq r \leq g(\theta)$
Solving $x^2+(y-9)^2=9^2$ where $y=12$ gives me $x=6\sqrt{2}$ and $x=-2\sqrt{6}$. But I'm not sure if I've interpreted the question right, as it seems like $\theta$ should be between $0$ and $\pi$ if we're talking about angles. But in that case I don't understand how I'm supposed to find the functions $f$ and $g$...
any ideas?
| First things first, the region above the line $y = 12$ in polar coordinates is
$$ r\sin\theta \ge 12 \implies r \ge 12\csc\theta $$
Second, for the region inside the circle
$$ x^2 + (y-9)^2 \le 9^2 $$
$$ x^2 + y^2 - 18y \le 0 $$
$$ r^2 - 18r\sin\theta \le 0 $$
$$ r \le 18\sin\theta $$
Combining them together gives you the bounds for $r$
$$ 12\csc\theta \le r \le 18\sin\theta $$
As for the angles, note that the two points of intersection are $(\pm 6\sqrt{2}, 12)$, corresponding with $r = 6\sqrt{6}$ and $\tan\theta = \pm\sqrt{2}$, or $\theta = \tan^{-1}\sqrt{2}$ (first quadrant) and $\theta = \pi - \tan^{-1}\sqrt{2}$ (fourth quadrant). So the bounds for $\theta$ are
$$ \tan^{-1}\sqrt{2} \le \theta \le \pi - \tan^{-1}\sqrt{2} $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2663047",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
For real numbers $x,y$ satisfy the condition $x^2+5y^2-4xy+2x-8y+1=0$. Find the Maximum and Minimum Value of the expression $A=3x-2y$. For real numbers $x,y$ satisfy the condition $x^2+5y^2-4xy+2x-8y+1=0$.Find the Maximum and Minimum Value of the expression $A=3x-2y$.
| by the Lagrange Multiplier method we get $$A\le 5(1+\sqrt{6})$$ for $$x=\frac{1}{5}(15+11\sqrt{6})$$,$$y=\frac{10}{11}+\frac{4}{55}(15+11\sqrt{6})$$
and $$A\geq -5(-1+\sqrt{6})$$ for $$x=\frac{1}{5}(15-11\sqrt{6}),y=-\frac{2}{5}(-5+2\sqrt{6})$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2663469",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Is a proof using modular arithmetic in a question like this valid? It's been two years or so since I've finished my math undergrad (and I'm doing something non-math related now, unfortunately), so I apologize if what is to follow isn't a very good question!
Prove that for all Integers $n$, $n(n + 1)(2n + 1)$ will always be divisible by 6.
I can do that using induction, but I wanted to try a different way. Does it work to use modular arithmetic in the following way?
Let $f(n) = n(n+1)(2n+1) = 2n^3 + 3n^2 + n$. All we need to show is that $f(n)$ is divisible by both $2$ and $3$ for any choice of $n$.
Evaluate mod $2$.
$f(n) = n(n+1)(2n+1) = n(n+1)(0 + 1)$ mod $2 = n(n+1)$ mod $2$. Two consecutive numbers; one of them must be even, and so $f(n)$ is divisible by $2$.
Evaluate mod $3$
There are three possible residues for n modulo $3$: $0, 1,$ or $2$.
If the residue is $0$, then $f(n)$ is divisible by $3$.
If the residue is $1$, then $f(n) = n(n+1)(2n+1) = 1(1+1)(2*1+1) = 1(2)(3) = 0$ mod $3$.
If the residue is $2$, then $f(n) = 2(2+1)(2*2+1) = 2(3)(2) = 0$ mod $3$.
In any case, $f(n)$ is divisible by $3$.
Since $f(n)$ is divisible by $2$ and by $3$, it is divisible by $6$. The result follows.
Thank you!
| Yes, your reasoning is valid.
If a number is divisible by $2$ and $3$, it is divisible by $6$ and each case can be proven using your method.
Trivia:
$$\sum_{i=1}^n i^2= \frac{n(n+1)(2n+1)}{6}$$
hence,$$n(n+1)(2n+1)= 6 \sum_{i=1}^n i^2$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2668673",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "27",
"answer_count": 4,
"answer_id": 3
} |
Alternative way to calculate the sequence Given $a>0$ that satisfies $4a^2+\sqrt 2 a-\sqrt 2=0$.
Calculate $S=\frac{a+1}{\sqrt{a^4+a+1}-a^2}$.
Attempt:
There is only one number $a>0$ that satisfies $4a^2+\sqrt 2 a-\sqrt 2=0$, that is
$a=\frac{-\sqrt{2}+\sqrt{\Delta }}{2\times 4}=\frac{-\sqrt{2}+\sqrt{2+16\sqrt{2}}}{8}$
However obviously if you replace it directly to calculate $S$, it would be extremely time-consuming.
Is there another way? I think the first equation can be changed (without needing to solve it) to calculate $S$, as when I use brute force (with calculator), I found out that $S\approx 1.414213562\approx\sqrt 2$.
| Rewrite using the value of $a^2 = \frac{1-a}{2\sqrt2}$
$$\frac{a+1}{\sqrt{\tfrac{(a-1)^2}{8}+a+1}+\tfrac{(a-1)}{2\sqrt{2}}} = \frac{(a+1)2\sqrt2}{\sqrt{a^2+6a+9} + a-1} = \frac{(a+1)2 \sqrt 2}{(a+3) + a-1} \\= \sqrt{2}$$
| {
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"url": "https://math.stackexchange.com/questions/2669096",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 0
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Prove $\sum\limits_{cyc}\sqrt[3]{\frac{a^2+bc}{b+c}}\ge\sqrt[3]{9(a+b+c)}$ Let $a,b,c$ be positive real numbers. Show that
$$\sum_{cyc}\sqrt[3]{\dfrac{a^2+bc}{b+c}}\ge\sqrt[3]{9(a+b+c)}$$
I have tried C-S and Holder inequalities, without success. How to solve it?
| Just a little hint
By AM GM
$$\sum \sqrt[3] {\frac {a^2+bc}{b+c}}\ge 3\sqrt[3] {\prod \left (\frac {a^2}{b+c}+ \frac {bc}{b+c}\right)^{\frac {1}{3}}}$$
Now by Hölder's inequality
$$3\sqrt[3] {\prod \left(\frac {a^2}{b+c}+ \frac {bc}{b+c}\right)^{\frac {1}{3}}}\ge 3\sqrt[3] {\left (\frac {a^2b^2c^2}{(a+b)(b+c)(a+c)}\right)^{\frac {1}{3}}+\left (\frac {a^2b^2c^2}{(a+b)(b+c)(a+c)}\right)^{\frac {1}{3}}}=3\sqrt[3] {2\sqrt[3] {\frac {a^2b^2c^2}{(a+b)(b+c)(a+c)}}}$$
Now again by AM GM on denominator
$$3\sqrt[3] {2\sqrt[3] {\frac {a^2b^2c^2}{(a+b)(b+c)(a+c)}}}\ge 3\sqrt[3] {\frac {3}{(a+b+c)}\sqrt[3] {a^2b^2c^2}}$$
Hope it helped any way
| {
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"timestamp": "2023-03-29T00:00:00",
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Showing $x^4-x^3+x^2-x+1>\frac{1}{2}$ for all $x \in \mathbb R$
Show that
$$x^4-x^3+x^2-x+1>\frac{1}{2}. \quad \forall x \in \mathbb{R}$$
Let $x \in \mathbb{R}$,
\begin{align*}
&\mathrel{\phantom{=}}x^4-x^3+x^2-x+1-\frac{1}{2}=x^4-x^3+x^2-x+\dfrac{1}{2}\\
&=x^2(x^2-x)+(x^2-x)+\dfrac{1}{2}=(x^2-x)(x^2+1)+\dfrac{1}{2}\\
&=x(x-1)(x^2+1)+\dfrac{1}{2}.
\end{align*}
Is there any way to solve this question?
| You can note that the inequality holds for $x=-1$. For $x\ne-1$ it is equivalent to
$$
\frac{x^5+1}{x+1}>\frac{1}{2}
$$
that can be rewritten as
$$
\frac{2x^5-x+1}{x+1}>0
$$
Let's analyze $f(x)=2x^5-x+1$, with $f'(x)=10x^4-1$, which vanishes for $x=\pm10^{-1/4}$. So $f$ has a relative maximum at $-10^{-1/4}$ and a relative minimum at $10^{-1/4}$. Let $a=10^{-1/4}$, for simplicity; then $-1<-a$, so we just need to analyze
$$
f(a)=2a^5-a+1=\frac{1}{5}a-a+1=1-\frac{4}{5}a>0
$$
(prove it).
Since $f(-1)=0$, we conclude that $f(x)<0$ for $x<-1$ and $f(x)>0$ for $x>-1$, so
$$
\frac{f(x)}{x+1}>0
$$
for every $x\ne-1$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2670433",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 7,
"answer_id": 5
} |
Evaluate the limit $\lim_{n\to \infty}\sqrt{n^2 +2} - \sqrt{n^2 +1}$ I know that
$$\lim_{n\to \infty}(\sqrt{n^2+2} - \sqrt{n^2+1})=0.$$
But how can I prove this?
I only know that $(n^2+2)^{0.5} - \sqrt{n^2}$ is smaller than $\sqrt{n^2+2} - \sqrt{n^2}$ = $\sqrt{n^2+2} - n$.
Edit:
Thank Y'all for the nice and fast answers!
| Hint:
Try to multiply by $$1=\frac{\sqrt{n^2+2}+\sqrt{n^2+1}}{\sqrt{n^2+2}+\sqrt{n^2+1}}$$
And use the face that
$$a^2-b^2=(a-b)(a+b)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2670779",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
Tangent vector of curve $ \Psi(t)= (2t^3 - 2t, 4t^2, t^3+t )^T $ expressed in spherical coordinates I have a curve where $ t\in R^{+}_0$
$$
\Psi(t)= (2t^3 - 2t, 4t^2, t^3+t )^T
$$
I need to express tangent vector T in standard spherical coordinates in terms of normalized 'frame vectors' $\hat h_i $.
I started from denoting spherical cordinates and comparing with the given data $$x=r \sin\Phi \cos\lambda=2t^3 - 2t=2t(t^2-1)$$
$$y=r \sin\Phi \sin\lambda=4t^2$$
$$z=r \cos\Phi=t^3+t=t(t+1) $$
Tangent vector is defined as:
$$T=\dot r |h_r| \hat h_r +\dot \phi |h_\phi| \hat h_\phi +\dot \lambda |h_\lambda| \hat h_\lambda $$
I don't know how to determine $r, \phi, \lambda$. Any tips?
| \begin{align}
\mathbf{r} &=
\begin{pmatrix} 2t^3-2t \\ 4t^2 \\ t^3+t \end{pmatrix} \\
r &= \sqrt{(2t^3-2t)^2+(4t^2)^2+(t^3+t)^2} \\
&= t(t^2+1)\sqrt{5} \tag{$t>0$} \\
\dot{r} &= (3t^2+1)\sqrt{5} \\
\cos \theta &= \frac{z}{r} \\
&= \frac{1}{\sqrt{5}} \\
\sin \theta &= \frac{2}{\sqrt{5}} \\
\dot{\theta} &= 0 \\
\tan \phi &= \frac{y}{x} \\
\phi &= \tan^{-1} \frac{2t}{t^2-1} \\
\dot{\phi} &= -\frac{2}{t^2+1} \\
\mathbf{v} &=
\dot{r} \, \mathbf{e}_r+
r\dot{\theta} \, \mathbf{e}_{\theta}+
r\dot{\phi} \sin \theta \, \mathbf{e}_{\phi} \\
&= \sqrt{5}(3t^2+1) \, \mathbf{e}_r-4t\, \mathbf{e}_{\phi} \\
\mathbf{T} &=
\frac{(3t^2+1)\sqrt{5} \, \mathbf{e}_r-
4t\, \mathbf{e}_{\phi}}{\sqrt{45t^4+46t^2+5}}
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2672604",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
To check whether a function is unbounded
Show that $f:[-1,1]\to \mathbb{R}$ defined by
$f(x)=2x\sin\frac{1}{x^2}-\frac{2}{x}\cos\frac{1}{x^2}, ~~x\neq 0$ and $f(0)=0$ is unbounded on every neighbourhood of $0$.
Attempt:
$|f(x)|=|2x\sin\frac{1}{x^2}-\frac{2}{x}\cos\frac{1}{x^2}|\leq |2x\sin\frac{1}{x^2}|+|\frac{2}{x}\cos\frac{1}{x^2}|\leq 2+|\frac{2}{x}|$. At the neighbourhood
of $x=0$, $\frac{1}{x}$ is unbounded, so $f(x)$ cannot be written as $|f(x)|\leq K$ for some positive $K$. Thus $f(x)$ is unbounded.
Is this proof right?
| Let $x_k = \frac{1}{\sqrt{2k\pi}}$ where $k\in \mathbb{Z}, k > 0$.
Then $$f \left( \frac{1}{\sqrt{2k\pi}} \right)=-2\sqrt{2k\pi}$$
As $k\to \infty, x_k \to 0, |f(x_k)| = 2\sqrt{2k\pi}\to \infty$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2677042",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
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} |
Does the following limit exists? $$
\lim_{a\to 0}\left\{a\int_{-R}^{R}\int_{-R}^{R}
\frac{\mathrm{d}x\,\mathrm{d}y}
{\,\sqrt{\,\left[\left(x + a\right)^{2} + y^{2}\right]
\left[\left(x - a\right)^{2} + y^{2}\right]\,}\,}\right\}
\quad\mbox{where}\ R\ \mbox{is a}\ positive\ \mbox{number.}
$$
The integral exists, since you can take small disks around singularities and make a change of variables. It seems that the limit should be $0$, as $\lim_{a \to 0}\left[a\log\left(a\right)\right] = 0$.
| This is not an answer yet but just a few manipulations that might hopefully simplify things a bit. I might have made some calculation errors.
$$\int\limits_{-R}^{R}\int\limits_{-R}^{R}\frac{1}{\sqrt{\left[(x+a)^2+y^2\right]\left[(x-a)^2+y^2\right]}}\,dx\,dy \leq$$
$$\int\limits_{0}^{\sqrt{2}R}\int\limits_{0}^{2\pi}\frac{1}{\sqrt{\left(r^2+2ar\cos\theta+a^2\right)\left(r^2-2ar\cos\theta+a^2\right)}}r\,d\theta\,dr=$$
$$\int\limits_{0}^{\sqrt{2}R}\int\limits_{0}^{2\pi}\frac{1}{\sqrt{r^4-2a^2r^2\cos2\theta+a^4}}r\,d\theta\,dr=$$
$$\frac{1}{2}\int\limits_{0}^{2\pi}\int\limits_{0}^{2R^2}\frac{1}{\sqrt{\rho^2-2a^2\rho\cos2\theta+a^4}}\,d\rho\,d\theta=\frac{1}{2}\int\limits_{0}^{2\pi}\int\limits_{0}^{2R^2}\frac{1}{\sqrt{\left(\rho-a^2\cos2\theta\right)^2+a^4\sin^22\theta}}\,d\rho\,d\theta=$$
$$\frac{1}{2}\int\limits_{0}^{2\pi}\int\limits_{-a^2\cos2\theta}^{2R^2-a^2\cos2\theta}\frac{1}{\sqrt{\rho^2+a^4\sin^22\theta}}\,d\rho\,d\theta=$$
$$\frac{1}{2}\int\limits_{0}^{2\pi}\log\left(\frac{\sqrt{4R^4 - 4R^2a^2\cos2\theta+a^4} + 2R^2-a^2\cos2\theta}{2a^2\sin^2\theta}\right)\,d\theta \leq$$
$$\frac{1}{2}\int\limits_{0}^{2\pi}\log\left(\frac{4R^2+2a^2\sin^2\theta}{2a^2\sin^2\theta}\right)\,d\theta=\frac{1}{2}\int\limits_{0}^{2\pi}\log\left(1+2\left(\frac{R}{a\sin\theta}\right)^2\right)\,d\theta$$
The last integral looks more easy to bound hopefully.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2678757",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
finding the values of a such that the system $AB\cdot x=0$ has infinitely many solutions the following matrix are given:
$$A=
\begin{bmatrix}
a & 1 & 1 \\
a & a & 2 \\
a & a & a \\
\end{bmatrix}
$$
$$B=
\begin{bmatrix}
a-3 & 1 & 1 \\
a-3 & a-3 & 2 \\
a-3 & a-3 & a-3 \\
\end{bmatrix}
$$
find the values of a such that the system $AB\cdot x=0$ has infinitely many solutions.
$$$$I know that a matix has infinitely many solutions $\iff det(AB)=0$, so this is what I did :
$$A'=
\begin{bmatrix}
a & 1 & 1 \\
0 & a-1 & 2 \\
0 & 0 & a-2 \\
\end{bmatrix}
$$
and $det(A) =0$ if $a\cdot(a-1)\cdot(a-2)=0$ , so if $a=0,1,2.$ A has infinitely many solutions.
$$B=
\begin{bmatrix}
a-3 & 1 & 1 \\
0 & a-4 & 2 \\
0 & 0 & a-5 \\
\end{bmatrix}
$$
and $det(B) =0$ if $(a-3)\cdot(a-4)\cdot(a-5)=0$ , so if $a=3,4,5.$ B has infinitely many solutions. $$$$So $AB\cdot x=0$ , if $a=0,1,2,3,4,5$, BUT the textbook answer is $a=1,2,3$. What am I doing wrong?
| Your answer is right (I checked it) and therefore your textbook is wrong.
| {
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"url": "https://math.stackexchange.com/questions/2680706",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
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} |
Find the sum to $n$ terms of the series: $1^2.1+2^2.3+3^2.5+....$ Find the sum to $n$ terms of the series:
$$1^2.1+2^2.3+3^2.5+.....$$
My Attempt:
Here, $n^{th}$ term of $1,2,3,....=n$
$n^{th}$ term of $1^2,2^2,3^2,....=n^2$
Also, $n^{th}$ term of $1,3,5,....=2n-1$
Hence, $n^{th}$ term of the given series is $t_n=n^2(2n-1)$
| $$\sum_{k=1}^nk^2(2k-1)=\sum_{k=1}^n\left[12\binom{k}3+10\binom{k}2+\binom{k}1\right]=12\binom{n+1}4+10\binom{n+1}3+\binom{n+1}2$$
This under the convention that $\binom{k}{r}=0$ if $r\notin\{0,\dots,k\}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2680816",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 3
} |
Given that $P(x)$ is a polynomial such that $P(x^2+1) = x^4+5x^2+3$, what is $P(x^2-1)$? How would I go about solving this? I can't find a clear relation between $x^2+1$ and $x^4+5x^2+3$ to solve $P(x^2-1)$.
| Here is another way. Let $t :=\sqrt{x-1}$. Then note that
$$\begin{align}
t&=\sqrt{x-1};\\
t^2&=x-1;\\
t^2+1&=x;\\
t^4&=x^2-2x+1.
\end{align}$$
So that $$\begin{align}
P(x)=P(t^2+1)&=t^4+5t^2+3\\
&=(x^2-2x+1)+5(x-1)+3\\
&=x^2+3x-1.
\end{align}.$$
Lastly, evaluate $P(x^2-1)=\ldots=x^4+x^2-3$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2683178",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 3
} |
How to deal with $(-1)^{k-1}$ It's a problem on mathematical induction.
$$1^2-2^2+3^2-.....+(-1)^{n-1}n^2=(-1)^{n-1}\frac{n.(n+1)}{2}$$
I have proved it for values of $n=1,2$.
Now I assume for $n=k$
$$P(k):1^2-2^2+3^2-.....+(-1)^{k-1}k^2=(-1)^{k-1}\frac{k.(k+1)}{2}$$.
$$P(k+1):1^2-2^2+3^2-.....+(-1)^{k-1}k^2+(k+1)^2=(-1)^{k-1}\frac{k.(k+1)}{2}+(k+1)^2\\=\frac{(k+1)}{2} [(-1)^{k-1}.k+2k+2]$$
I need suggestion to deal with the $(-1)^{k-1}$ so that I can prove the whole. Any help is appreciated.
| If$$1^2-2^2+\cdots+(-1)^{k-1}k^2=(-1)^{k-1}\frac{k(k+1)}2,$$then\begin{align}1^2-2^2+\cdots+(-1)^k(k+1)^2&=(-1)^{k-1}\frac{k(k+1)}2+(-1)^k(k+1)^2\\&=(-1)^k\left(-\frac{k(k+1)}2+(k+1)^2\right)\\&=(-1)^k(k+1)\left(k+1-\frac k2\right)\\&=(-1)^k(k+1)\frac{k+2}2\\&=(-1)^k\frac{(k+1)(k+2)}2.\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2688026",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
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} |
Proving $1+\tan^2(x)=\sec^2(x)$ If one is asked to prove $1+\tan^2(x)=\sec^2(x)$, this is how I would prove it. $$\frac{d}{dx} \frac{\sin(x)}{\cos(x)}=\frac{\cos^2(x)+\sin^2(x)}{\cos^2(x)}=\frac{1}{\cos^2(x)}=\sec^2(x)$$ and $$\frac{d}{dx} \frac{\sin(x)}{\cos(x)}=\frac{\cos^2(x)+\sin^2(x)}{\cos^2(x)}=1+\frac{\sin^2(x)}{\cos^2(x)}=1+\left(\frac{\sin(x)}{\cos(x)}\right)^2=1+\tan^2(x)$$ Therefore $1+\tan^2(x)=\sec^2(x)$ So is my prove valid, and are there any easier ways to prove this?
| Note simply that
$$1+\tan^2(x)=1+\frac{\sin^2 x}{\cos^2 x}=\frac{\cos^2x+\sin^2 x}{\cos^2 x}=\frac{1}{\cos^2 x}=\sec^2(x)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2688627",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Number of binary strings not having even-length runs of $0$ or $1$ I was solving this algorithmic problem. This problem boils down to finding the finding the number of binary strings of given length $n$ such that those strings don't have any even-length run of $0$ or $1$. The post says that the answer is twice the $n$th term of the Fibonacci series, where the term numbering starts at $0$.
I have tried myself to arrive at the same solution but without success. Any help/hint is appreciated.
| Using $z$ for ones and $w$ for zeroes we get the generating function
$$f(z, w) = (1+z+z^3+z^5+\cdots) \\ \times
\left(\sum_{q\ge 0} (w+w^3+w^5+\cdots)^q (z+z^3+z^5+\cdots)^q\right)
\\ \times (1+w+w^3+w^5+\cdots).$$
This simplifies to
$$f(z, w) = \left(1+\frac{z}{1-z^2}\right)
\left(\sum_{q\ge 0} w^q \frac{1}{(1-w^2)^q}
z^q \frac{1}{(1-z^2)^q}\right)
\left(1+\frac{w}{1-w^2}\right)
\\ = \left(1+\frac{z}{1-z^2}\right)
\frac{1}{1-wz/(1-w^2)/(1-z^2)}
\left(1+\frac{w}{1-w^2}\right)
\\ = \left(1+z-z^2\right)
\frac{1}{(1-w^2)(1-z^2)-wz}
\left(1+w-w^2\right).$$
Since we are only interested in the count we may end the distinction
between zeroes and ones to obtain
$$g(z) = \frac{(1+z-z^2)^2}{(1-z^2)^2-z^2}
= \frac{(1+z-z^2)^2}{1-3z^2+z^4}
= \frac{1+z-z^2}{1-z-z^2}.$$
Hence
$$(1-z-z^2) g(z) = 1 + z - z^2.$$
Extracting coefficients we get
$$[z^0] (1-z-z^2) g(z) = [z^0] g(z) = 1.$$
so that $g_0 = 1.$ We also have
$$[z^1] (1-z-z^2) g(z) = g_1-g_0 = 1$$
so that $g_1 = 2.$ Furthermore
$$[z^2] (1-z-z^2) g(z) = g_2-g_1-g_0 = -1$$
so that $g_2 = 2.$ Finally for $n\ge 3$ we find
$$[z^n] (1-z-z^2) g(z) = 0$$
so that $g_n - g_{n-1} - g_{n-2} = 0$ or
$$g_n = g_{n-1} + g_{n-2}.$$
With $g_1 = 2 F_1$ and $g_2 = 2 F_2$ and the induction hypothesis $g_n
= 2 F_n$ we get from the recurrence that $g_n = 2 F_{n-1} + 2 F_{n-2}
= 2 F_n$ so that indeed $g_n = 2 F_n$ for $n\ge 1$ as claimed.
Remark. We may also observe that
$$g(z) = 1 + 2 \frac{z}{1-z-z^2}$$
where we recognize the Fibonacci number OGF so that we may conclude by
inspection.
| {
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"url": "https://math.stackexchange.com/questions/2688983",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 2
} |
Number of divisors of the number $2079000$ which are even and divisible by $15$
Find the number of divisors of $2079000$ which are even and divisible by $15$?
My Attempt: Since they are divisible by $15$ and are even, $2$ and $5$ have to included from the numbers prime factors.
$2079000 = 2^3 \cdot 3^3 \cdot 5^3 \cdot 7 \cdot 11$
Therefore, the number of divisors should be
$2 \cdot 2 \cdot (3+1) \cdot (1+1) \cdot (1+1)$
But however this answer is wrong.
Any help would be appreciated.
| Any factors that are even and divisible by $15$ are divisible by $30$.
Effectively you need to find the number of factors of $2079000 /30 = 69300 = 2^2 \cdot 3^2\cdot 5^2\cdot 7\cdot 11$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2690113",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 8,
"answer_id": 0
} |
How to show that $x^2 - 37y^2 =2$ does not have integer solutions We need to prove that $x^2 - 37y^2 =2$ does not have integer solutions.
I have two angles I thought about approaching it from:
*
*Since 37 is prime, I can show that for $x$ not divisible by $37$, we have $x^{36} ≡ 1mod(37)$ but I don't see how that's useful
*I could manipulate the equation and make it to: $x^2 - 4 = 37y^2 - 2$
$\implies (x-2)(x+2) = 37y^2 - 2$
Then if the RHS is even, then $y^2$ is even $\implies$ $y^2$ ends with $0, 4,$ or $6$ $\implies$ $37y^2$ ends with $0, 8,$ or $2$
$\implies 37y^2 -2$ ends with $0, 6,$ or $8$
But then I reach a dead end here too
Any suggestions or ideas?
| $$x^2-37y^2=2\Rightarrow x^2-y^2\equiv2\mod(4)\Rightarrow x^2\equiv 2+y^2\mod(4)$$
but $2+y^2\equiv 2\mod(4)$ or $2+y^2\equiv 3\mod(4)$ since any square has remainder $0$ or $1$ when divided by $4$. Either case is impossible since $x^2$ is a square itself.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2691514",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 2
} |
binomial limit when limit approches to infinity $\displaystyle \lim_{n\rightarrow\infty}\binom{n}{x}\left(\frac{m}{n}\right)^x\left(1-\frac{m}{n}\right)^{n-x}$
solution i try
$\displaystyle \lim_{n\rightarrow\infty}\left(\frac{m}{n}\right)^x\left(1-\frac{m}{n}\right)^{n-x}=\lim_{n\rightarrow\infty}(\frac{m}{n}+1-\frac{m}{n})^n=1$
I have edited my post
This is wrong how i find right answer. Help me
| Assuming $x$ is a natural number less than $n$, then we see that
\begin{align}
\binom{n}{x} \left(\frac{m}{n}\right)^x\left(1-\frac{m}{n}\right)^{n-x} =&\ \frac{n!}{x!(n-x)!}\left(\frac{m}{n}\right)^x\left(1-\frac{m}{n}\right)^{n-x}\\
\text{Stirling Approximation } \sim&\ \frac{\sqrt{2\pi n} \left(\frac{n}{e} \right)^n}{x! \sqrt{2\pi(n-x)}(\frac{n-x}{e})^{n-x}}\left(\frac{m}{n}\right)^x\left(1-\frac{m}{n}\right)^{n-x}\\
=&\ \frac{1}{x!}\sqrt{\frac{n}{n-x}}\left(1-\frac{m-x}{n-x} \right)^{n-x}\left( \frac{m}{e}\right)^x\\
\rightarrow &\ \frac{1}{x!} \left( \frac{m}{e}\right)^xe^{-(m-x)}= \frac{m^x}{x!}e^{-m}
\end{align}
| {
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"url": "https://math.stackexchange.com/questions/2691740",
"timestamp": "2023-03-29T00:00:00",
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"question_score": "2",
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Solving $2^x \cdot 5^y = 0,128$ $x,y \in \mathbb Z$
$2^x \cdot 5^y = 0,128$
$x+y = ?$
My attempt:
I know that
$$0,128 = \frac{128}{1000}$$
$$5^3 = 125$$
$$2^{-3} = \frac{1}{8}$$
EDIT:
$2^7 = 128$
Then we need to get
$0,128$
| You know that $128=2^7$, and $1000=10^3=(2\cdot 5)^3=2^3\cdot 5^3$. Therefore
$$0.128=\frac{128}{1000}=\frac{2^7}{2^3\cdot 5^3}$$
Can you do the rest?
| {
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"timestamp": "2023-03-29T00:00:00",
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Solve $1+a^c+b^c=\text{lcm}(a^c,b^c)$ Question: Find all positive integer solutions to $1+a^c+b^c=\text{lcm}(a^c,b^c)$ where LCM denotes the least common multiple.
Attempt: Substitute $a^c=A; b^c=B$. We know that $\text{lcm}(A,B) \cdot \gcd(A,B)=AB$, so if we denote by $d$ the $\gcd$ of A and B, we get that $\text{lcm}(A,B)=\frac{AB}{\gcd(A,B)}=\frac{AB}{D}; \frac{AB}{D}=1+A+B; AB=D(1+A+B)$
$AB=D(1+A+B)$ // $+DAB$
$DAB+AB=D(1+A+B)+DAB$
$(D+1)AB=D(1+A+B+AB)$
$(D+1)AB=D(A+1)(B+1)$
But I am stuck here. I don't even know if any of this is useful.
| Let $a^c=A$ and $b^c=B$. Therefore since $A,B|lcm(A,B)$ we have:$$1+B=lcm(A,B)-A\\1+A=lcm(A,B)-B$$which concludes that $$A|1+B\\B|1+A$$therfore $\gcd(A,B)=1$ and we have $lcm(A,B)=AB$. Embarking on this:$$1+A+B=AB$$ or $$(A-1)(B-1)=2$$which yields to $$(A,B)=(1,2)\\(A,B)=(2,1)$$or equivalently $$a^c=1\\b^c=2$$(the other answer has been ignored hence of symmetry) the only positive integer answer is $a=c=1$ and $b=2$
| {
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Let $x_n=\frac{1}{3}+(\frac{2}{5})^2+(\frac{3}{7})^3+....(\frac{n}{2n+1})^2$ Is $(x_n) $cauchy sequence? Let $$x_n=\frac{1}{3}+\left(\frac{2}{5}\right)^2+\left(\frac{3}{7}\right)^3+\cdots+\left(\frac{n}{2n+1}\right)^n$$ Is $(x_n)$ Cauchy sequence ?
My work
for $n>m:$
$$\begin{align}|x_n-x_m|&=\left|\frac{1}{3}+\left(\frac{2}{5}\right)^2+\cdots+\left(\frac{n}{2n+1}\right)^n-\left(\frac{1}{3}+\left(\frac{2}{5}\right)^2+\cdots+\left(\frac{m}{2m+1}\right)^m\right)\right|\\
&=\left|\left(\frac{m+1}{2(m+1)+1}\right)^{m+1}+\cdots+\left(\frac{n}{2n+!}\right)^n\right|\end{align}$$ how to proceed from here
| For $n > m$ we have
\begin{align}
|x_n - x_m| &= \left(\underbrace{\frac{m+1}{2(m+1)+1}}_{\le \frac12}\right)^{m+1}+\cdots +\left(\underbrace{\frac{n}{2n+1}}_{\le \frac12}\right)^n \\
&\le \frac{1}{2^{m+1}} + \cdots + \frac{1}{2^n} \\
&= \frac{1}{2^{m+1}}\left( 1 + \frac12 + \cdots + \frac1{2^{n-m-1}}\right)\\
&= \frac{1}{2^{m+1}} \frac{2^{n-m}-1}{2-1}\\
&= \frac{2^{n-m}-1}{2^{m+1}} \xrightarrow{m\to\infty} 0
\end{align}
Therefore your sequence $(x_n)_n$ is Cauchy.
| {
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Inequality $\frac{a^3+b^3+c^3}{3}\geq\sqrt{\frac{a^4+b^4+c^4}{3}}$ Let $a$, $b$ and $c$ be non-negative numbers such that $(a+b)(a+c)(b+c)=8$. Prove that:
$$\frac{a^3+b^3+c^3}{3}\geq\sqrt{\frac{a^4+b^4+c^4}{3}}$$
Some attempts:
*
*From the condition follows $a^3+b^3+c^3 = (a+b+c)^3 -24$
*It is known (see here)
$$\frac{a+b+c}{3}\geq\sqrt[27]{\frac{a^3+b^3+c^3}{3}}$$
*Setting $2x=a+b$, $2y = b+c$, $2z = a+c$, we can express $a =x+z-y$ etc. The condition then becomes $xyz = 1$ which can be parametrized with free variables $0\leq q \leq 2 \pi /3 $ and arbitrary $r$ by
$$
x = \exp(r \cos q) \; ; \; y = \exp(r \cos (q + 2 \pi /3)) \; ; \; z = \exp(r \cos (q - 2 \pi /3))
$$
Using that, the condition can be removed and then calculus may be used.
*The question may be interpreted geometrically. Expressions such as $a^3+b^3+c^3 = $const. and $a^4+b^4+c^4 =$ const. can be interpreted as hypersurfaces of what has been called an N(3)-dimensional ball in p-norm, see here. A nice visualization is given in here. Then properties such as extrema, convexity etc. of these surfaces can be used.
I couldn't put the pieces together.
| We start with two identities :
$$\frac{1}{9}[((a+b+c)^2-2(ab+bc+ca))^2+(a^2-b^2)^2+(a^2-c^2)^+(b^2-c^2)^2]=\frac{a^4+b^4+c^4}{3}$$
Furthermore we have :
$$\sum_{cyc}^{}(x+y)(y+z)=\sum_{cyc}^{}3xy+y^2$$
And:
$$\sum_{cyc}^{}(x-y)(y-z)=\sum_{cyc}^{}xy-y^2$$
Finally we have :
$$\frac{\sum_{cyc}^{}(x-y)(y-z)+\sum_{cyc}^{}(x+y)(y+z)}{4}=xy+yz+zx$$
And
$$\frac{1}{3}[(a+b+c)^3-3(a+b)(b+c)(c+a)]=\frac{a^3+b^3+c^3}{3}$$
So the fist identity becomes :
$$\frac{1}{9}[((a+b+c)^2-2(\frac{\sum_{cyc}^{}(a-b)(b-c)+\sum_{cyc}^{}(a+b)(b+c)}{4}))^2+(a^2-b^2)^2+(a^2-c^2)^+(b^2-c^2)^2]=\frac{a^4+b^4+c^4}{3}$$
So if we put the following substitution we get :
$x=a+b$
$y=b+c$
$z=a+c$
We get :
$$\frac{1}{9}[((\frac{x+y+z}{2})^2-2(0.25(xy+zx+zy+(x-y)(y-z)+(y-z)(x-z)+(x-z)(x-y)))^2+x^2(z-y)^2+y^2(x-z)^2+z^2(x-y)^2]=\frac{a^4+b^4+c^4}{3}$$
And
$$\frac{1}{3}[(\frac{x+y+z}{2})^3-3xyz]=\frac{a^3+b^3+c^3}{3}$$
The condition becomes :
$xyz=8$
Futhermore we now that :
$$x^2(z-y)^2+y^2(x-z)^2+z^2(x-y)^2=2(xy+yz+zx)^2-6xyz(x+y+z)$$
So we have to prove this :
$$\sqrt{[((\frac{x+y+z}{2})^2-2(0.25(2xy+2zx+2zy-(x^2+y^2+z^2)))^2+2(xy+yz+zx)^2-48(x+y+z)]}\leq [(\frac{x+y+z}{2})^3-24] $$
We put $x+y+z=cst=\beta$ and we can maximizing the LHS and the RHS becomes constant .
We get a one variable inequality this is the following :
$$\sqrt{\frac{11}{48}\beta^4-48\beta}\leq \frac{\beta^3}{8}-24$$
Finally the condition becomes :
$$xyz=8\implies x+y+z\geq 6$$
So the one variable inequality is true for $\beta\geq 6$
Done !
| {
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What is the best notation to write ``equations with reasons''? I'm learning my analytic number theory course. When I make notes, I find that it is effective to use ``equations with reasons'' as follow
$$\begin{array}{rll}
\vartheta(x)&\displaystyle=\sum_{p\leq x} \log p& \textrm{partial sum $\begin{cases}
1 & \textrm{$n$ is prime}\\
0 & \textrm{otherwise}
\end{cases}$ and $\log x$}\\
&\displaystyle =\pi(x)\log x-\int_1^x \frac{\pi(t)}{t}\mathrm{d} t & \textrm{Assume that $\displaystyle\pi(x)=\frac{x}{\log x}+O\left(\frac{x}{\log x}\right)$}\\
&\displaystyle=x-\int_1^x \frac{1}{\log t}\mathrm{d} t+O\left(x-\int_2^x \frac{1}{\log t}\mathrm{d} t\right)
& \displaystyle\because\int_2^x \frac{1}{\log t}\mathrm{d} t\leq \frac{x}{\log x}\leq x \\
&\displaystyle=x+O(x)\\
\end{array}$$
My question is are there any standard on this notations? For example, the reason should put into which line, before the result of the result? And, is it acceptable to use the symbol $\because$?
| Here are two examples of the notation I use, although I don't think there is just one way to do it.
$$\begin{align*}
c^2 & = a^2 + b^2 & \text{(by Pythagorean Theorem)} \\
& = 2a^2 & \text{(since } a = b \text{)} \\
& = 2 & \text{(since } a = 1 \text{)}
\end{align*}$$
and it's code:
\begin{align*}
c^2 & = a^2 + b^2 & \text{(by Pythagorean Theorem)} \\
& = 2a^2 & \text{(since } a = b \text{)} \\
& = 2 & \text{(since } a = 1 \text{)}
\end{align*}
And the other example:
$$\begin{align*}
c^2 = a^2 + b^2 & \Rightarrow c^4 = (a^2 + b^2)^2 & \text{(square both sides)} \\
& \Rightarrow c^4 = a^4 + 2a^2b^2 + b^4 & \\
& \Rightarrow c^4 = 2a^4 + 2a^2a^2 & \text{(since } a = b \text{)} \\
& \Rightarrow c^4 = 4a^4 & \\
& \Rightarrow a^4 = \frac{c^4}{4} & \\
\end{align*}$$
and it's code:
\begin{align*}
c^2 = a^2 + b^2 & \Rightarrow c^4 = (a^2 + b^2)^2 & \text{(square both sides)} \\
& \Rightarrow c^4 = a^4 + 2a^2b^2 + b^4 & \\
& \Rightarrow c^4 = 2a^4 + 2a^2a^2 & \text{(since } a = b \text{)} \\
& \Rightarrow c^4 = 4a^4 & \\
& \Rightarrow a^4 = \frac{c^4}{4} & \\
\end{align*}
| {
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"timestamp": "2023-03-29T00:00:00",
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GCD of two numbers $4a^2 + 3,4ab-1$ I came across this problem while solving some contest math. Please help me out with this.
What are all the possible GCD for two numbers of the form $4a^2 + 3,4ab-1$?
| We show below that there are infinitely many primes which can divide the gcd.
We also characterize all of them.
Claim 1: If $p$ is any prime
$$p|gcd( 4a^2 + 3,4ab-1) .$$
then $p \neq 2,3$.
Proof $4a^2 + 3$ is odd hence $p \neq 2$.
Assume by contradiction that $p=3$. Then $3|4a^2 + 3 \Rightarrow 3|a \Rightarrow 3|4ab \Rightarrow 3 \nmid 4ab-1$.
Claim 2: If $p\neq 3$ any odd prime such that $-3$ is a quadratic residue modulo $p$, then there exists $a,b$ such that
$$p|gcd( 4a^2 + 3,4ab-1) .$$
Note that Legendre symbol immediately tells you what the prime needs to be modulo 12.
Proof of claim 1.
The equation
$$ 4a^2 + 3 \equiv 0 \pmod{p}$$
has solutions.
Pick one such $a$, and then the equation
$$3b=-a \pmod{p}$$
has solutions. Pick such $b$.
Then
$$4a^2\equiv -3 \pmod{p}$$
and
$$12ab \equiv 4a (3b) \equiv -4a^2 \equiv 3 \pmod{p}$$
which yields
$$4ab \equiv 1 \pmod{p}$$
Claim 3: If $p$ is any prime
$$p|gcd( 4a^2 + 3,4ab-1) .$$
then $-3$ is a quadratic residue $\pmod{p}$.
Proof
$$p |gcd( 4a^2 + 3,4ab-1) \Rightarrow p|4a^2+3 \Rightarrow \\
-3=(2a)^2 \pmod{p}$$
Conclusion Let $p$ be a prime. Then there exists $a,b$ such that $p|gcd( 4a^2 + 3,4ab-1)$ if and only if $p \neq 2,3$ and
$$ \left(\frac{-3}{p} \right)=1$$
if and only if $$p\equiv 1,7 \pmod{12}$$
| {
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Find the value of $\frac{\tan A}{\tan B}$, given $\frac{\sin A}{\sin B}=5$
If $\displaystyle \frac{\sin A}{\sin B}=5$, then find the value of $\displaystyle \frac{\tan A}{\tan B}$
Try using the Componendo and Dividendo formula:
$$\frac{\sin A+\sin B}{\sin A-\sin B}=\frac{3}{2}$$
$$\frac{\tan(A+B)/2}{\tan(A-B)/2}=\frac{3}{2}$$
Can someone help me find: $$\frac{\tan A}{\tan B}$$
Thanks
| This is impossible to answer for general $A$ and $B$ satisfying the condition.
If $\displaystyle \frac{\tan A}{\tan B}=k$, then $\displaystyle k=\frac{\sin A\cos B}{\cos A\sin B}=\frac{5\cos B}{\cos A}$.
$$5\cos B=k\cos A$$
$$(5\sin B)^2+(5\cos B)^2=\sin^2A+k^2\cos^2A$$
$$k^2=\frac{25-\sin^2A}{\cos^2A}$$
So $k$ depends on $A$ and is not a constant.
| {
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Solve $\sqrt {x^2-3}=x-3$ in $\mathbb R$
Solve $\sqrt {x^2-3}=x-3$ in $\mathbb R$
My attempt:
$|x^2-3|=(x-3)^2$
So $-(x^2-3)=(x-3)^2$ or $(x^2-3)=(x-3)^2$
If $-(x^2-3)=(x-3)^2=x^2+9-6x$
So no solutions in $\mathbb R$
And if $(x^2-3)=(x-3)^2$
So $x^2-3=x^2+9-6x$
Now, can I delete $x^2$ with $x^2$ ? Like this
$x^2-x^2-3-9+6x=0$
$6x=12$
$x=2$
But $f(2)$ isn’t equal to $0$?
| From
$$\sqrt {x^2-3}=x-3$$
what we need stricktly as condition is that $x^2-3\ge 0$, then we can square both sides
$$\sqrt {x^2-3}=x-3\iff x^2-3=(x-3)^2$$
and for $x\neq 3$, noting that $x=3$ is not a solution, we obtain
$$\iff x^2-3=x^2-6x+9 \iff 6x=12\iff x=2$$
but since $x=2$ doesn't satisfy the equation it is to reject.
This is a legit way to solve, the extra solution come in from squaring both sides but we can exclude it at the end by checking directly on the original equation.
As an alternative we can observe that since LHS is $\ge 0$ also RHS must be then $x-3\ge0 $ is required by the original equation.
| {
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How to prove divergent sequences? For this question, I know that the sequence diverges to infinity, but I'm not sure if I am doing it right. Here is what I have so far. Can anyone please help me out?
Determine whether the following sequence is convergent or divergent
$a_n = \{8n^3 + n^2 -2\}$
$\lim_{n \to \infty} a_n = \infty$
Wts for any $M>0$, there exists some $N>0$, st if $n>N$, then $a_n>M$.
$n^3 > N^3 > M$
$n^3(8 + \frac{1}{n} - \frac{2}{n^3}) > N^3(8 + \frac{1}{N} - \frac{2}{N^3}) > M$
$n^3(8 + \frac{1}{n} - \frac{2}{n^3}) > N > (\frac{M}{8+\frac{1}{N} -\frac{2}{N^3}})^\frac{1}{3}$
| Let $M > 0$.
For any $n \ge \sqrt[3]{M+2}$ we have
$$8n^3 + n^2 - 2 \ge n^3 - 2 \ge (M+2) - 2 = M$$
Hence your sequence is unbounded.
| {
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Spivak's Calculus: Chapter 1, Problem 18b (Quadratic determinant less than zero) The problem in question is as follows:
18b Suppose that $b^2 -4c \lt 0$. Show that there are no numbers $x$ that satisfy $x^2 + bx + c = 0$; in fact, $x^2 + bx + c \gt 0$ for all $x$. Hint: complete the square.
Trying to apply the hint, I began by constructing $b^2 - 4c < 0 \therefore (b-\frac{2c}{b})^2 - \frac{4c^2}{b^2} \lt 0$, but manipulating this ultimately just leads you to $b^2 \lt 4c$ which you didn't need to complete the square to get anyway.
The only other idea I had was that one could construct the quadratic equation beginning from the assumption that $x^2 + bx + c = 0$ and then go for proof by contradiction e.g.
$x^2 + bx + c =0$
$x^2 + bx = -c$
$x^2 + bx + (\frac{b}{2})^2 = -c + (\frac{b}{2})^2$
$(x + \frac{b}{2})^2 = \frac{b^2 - 4c}{4}$
$\therefore$ Given that for all real values of $x$ and $b$, $(x + \frac{b}{2})^2 \gt 0$, by transitivity of equality, $\frac{b^2 - 4c}{4} \gt 0$
$\therefore 4(\frac{b^2 - 4c}{4}) \gt 4(0)$
$\therefore b^2 - 4c \gt 0$ for all x such that $x^2 + bx + c = 0$
But that still leaves the statement "in fact, $x^2 + bx + c \gt 0$ for all $x$" unproven, unless it's supposed to obviously follow, in which case I'm not seeing how.
| Essentially the same algebraic manipulations used in the second part of your question will give a proof without aiming for a contradiction.
Indeed, these manipulations will give you that $x^2 + bx + c = (x + \frac{b}{2})^2 - \frac{b^2 - 4c}{4}$. This is strictly positive since $(x + \frac{b}{2})^2 \geq 0$ and $\frac{b^2 - 4c}{4} < 0$. In particular, $x^2 + bx + c > 0$ for every $x$ which completes the proof.
| {
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Evaluate $\int \frac{1}{ax^2-bx}\,dx$ with substitution Evaluate $\int\frac{1}{ax^2-bx}\,dx$
First try:
$$\int\frac{1}{ax^2-bx}\,dx = \int\frac{1}{bx^2(\frac{a}{b}-\frac{1}{x})}\,dx$$
By substituting $u=\frac{a}{b}-\frac{1}{x}$ so $du=\frac{1}{x^2}\,dx$ we have,
$$\int\frac{1}{bx^2(\frac{a}{b}-\frac{1}{x})} \, dx = \frac{1}{b} \int\frac{1}{u} \, du = \frac{\ln|u|}{b}+C = \frac{\ln|\frac{a}{b}-\frac{1}{x}|}{b}+C=\frac{\ln|\frac{ax-b}{bx}|}{b}+C$$
Second try:
$$\int\frac{1}{ax^2-bx} \, dx = \frac{1}{b}\int\frac{b}{x^2(a-\frac{b}{x})} \, dx$$
By substituting $u=a-\frac{b}{x}$ so $du=\frac{b}{x^2}dx$ we have,
$$\frac{1}{b}\int\frac{b}{x^2(a-\frac{b}{x})}dx = \frac{1}{b}\int\frac{1}{u}\,du = \frac{\ln|u|}{b}+C = \frac{\ln|a-\frac{b}{x}|}{b}+C=\frac{\ln|\frac{ax-b}{x}|}{b} + C$$
It seems to me that $\frac{\ln|\frac{ax-b}{bx}|}{b}+C\neq\frac{\ln|\frac{ax-b}{x}|}{b}+C$, so, whats the matter ?!
| $$\frac{1}{b}\ln\left|\frac{ax-b}{bx}\right|=\frac{1}{b}\left(\ln\left|\frac{ax-b}{x}\right|-\ln|b|\right)=\frac{1}{b}\ln\left|\frac{ax-b}{x}\right|+\frac{1}{b}\ln|b|$$
And $\frac{\ln|b|}{b}$ is just a constant.
The primitive function is not just a function - it's a set of functions:
$$\int f :=\{g\mid g'=f\}$$
And we usually denote this set as one of its element + a constant. So, for example, we can say that $\int x = \frac{x^2}{2}+C$, or $\int x = \frac{x^2}{2}+2-e^\pi+C$, they denote the same set.
Note: Our calculus teacher was not using the $+C$. You can also leave it out, but it might cause some problem later, for example, when you will deal with differential equations.
| {
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Solve the differential equation 2y'=yx/(x^2 + 1) - 2x/y I have to solve the equation:
$$ 2y'= {\frac{xy}{x^2+1}} - {\frac{2x}{y}} $$
I know the first step is to divide by y, which gives the following equation:
$$ {\frac{2y'}{y}} ={\frac{x}{x^2+1}} - {\frac{2x}{y^2}}$$
According to my notes I get that I should make a substitution:
$$ z = {\frac{1}{y^2}} $$
and the derivative of z:$$ z' = {\frac{y'}{y^3}}$$
But I don't know how to proceed after this... Any help is appreciated!
| write your equation in the form
$$2\frac{dy(x)}{dx}y(x)-\frac{xy(x)^2}{x^2+1}=-2x$$
substituting
$$v(x)=y(x)^2$$ and $$\mu(x)=e^{\int\frac{-x}{\sqrt{x^2+1}}dx}=\frac{1}{\sqrt{x^2+1}}$$
we get
$$\frac{\frac{dv(x)}{dx}}{\sqrt{x^2+1}}-\frac{xv(x)}{\sqrt{x^2+1}^{3/2}}=\frac{2x}{\sqrt{x^2+1}}$$
and we get
$$\frac{\frac{d v(x)}{dx}}{\sqrt{x^2+1}}+\frac{d}{dx}\left(\frac{1}{\sqrt{x^2+1}}\right)v(x)=-\frac{2x}{\sqrt{x^2+1}}$$
and now $$\int\frac{d}{dx}\left(\frac{v(x)}{\sqrt{x^2+1}}\right)dx=\int-\frac{2x}{\sqrt{x^2+1}}dx$$
| {
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A partition of 186 into five parts under divisibility constraints The sum of 5 positive natural numbers, not necessarily distinct, is 186. If placed appropriately on the vertices of the following graph, two of them will be joined by an edge if and only if they have a common divisor greater than 1 (that is, they are not relatively prime).
What, in non-decreasing order, are those 5 numbers? The answer is unique.
| We cannot have at least three even numbers, for then there would be a triangle. So, we have four odd numbers and one even.
None of the numbers in the square can be a prime number, for else the two numbers it is connected two would have that prime number as a common divisor, and share an edge.
So, each number in the square must share one prime factor with one neighbor, and a different prime factor with another, and since it does not share en edge with the third, those two numbers have different prime factors altogether, meaning that there are at least four odd prime factors involved among the four numbers in the square.
One of them could be multiplied by $2$, but maybe the singleton is the even number ... well, let's see what is possible:
With those four odd prime numbers being $3,5,7,11$, we could form:
$3\cdot 5$, $3 \cdot 7$, $5 \cdot 11$, and $7 \cdot 11$ ... sum is $168$, so leftover number is $18$, which has prime factor $3$ ... does not work.
We can try to multiply one of the numbers by $2$, but not going over $186$ that can only be done for $3 \cdot 5$, meaning a sum of $183$ with a leftover of $5$ ... also does not work.
$3 \cdot 5$, $5 \cdot 7$, $3 \cdot 11$, $7 \cdot 11$ ... sum is $160$, so leftover is $26$ ... works!
Multiplying $3 \cdot 5$ by $2$ gives leftover $11$ .. does not work
$3 \cdot 7$, $3 \cdot 11$, $5 \cdot 7$, $5 \cdot 11$ ... sum is $144$ ... leftover $42$ ... does not work
Multiplying by $2$ gives leftovers $21$, $9$, and $7$ ... none work
Taking our four odd primes as $3,5,7,13$:
$3\cdot 5$, $3 \cdot 7$, $5 \cdot 13$, $7 \cdot 13$ ... sum is $193$ ... too high
$3 \cdot 5$, $3 \cdot 13$, $5\cdot 7$, $7 \cdot 13$ ... sum $180$ ... leftover $6$ ... does not work
$3 \cdot 7$, $3 \cdot 13$, $5 \cdot 7$, $5 \cdot 13$ ... sum $160$ ... leftover $26$ ... does not work
multiplying by $2$: only $3 \cdot 7$ possible, but gives leftover of $5$ does not work.
Any other combination of four odd primes gives sum above $186$.
So: only one combination possible: the square is $15, 33, 77, 35$, and the singleton is $26$
| {
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In how many ways can $3$ girls and $5$ boys be placed in three distinct cars that each hold at most three children?
There are $3$ cars of different makes available to transport $3$ girls and $5$ boys on a field trip. Each car can hold up to $3$ children. Find the number of ways in which they can be accommodated, if the internal arrangement of children inside the car is considered to be immaterial.
I considered three cases:
*
*$3$ G in car 1, $3$ B in car 2, and $2$ B in car 3.This can be done in $\binom{3}{3}\binom{5}{3}\binom{2}{2}3!=60$ ways;
*$2$ G, $1$ B in car 1; $1$ G, $2$ B in car 2; and $2$ B in car 3. This can be done in $\binom{3}{2}\binom{5}{1}\binom{1}{1}\binom{4}{2}\binom{2}{2}3!=540$ ways;
*$1$ G, $2$ B in car 1; $1$ G, $2$ B in car 2; and $1$ G, $1$ B in car 3. This can be done in $\binom{3}{1}\binom{5}{2}\binom{2}{1}\binom{3}{2}\binom{1}{1}\binom{1}{1}\frac{3!}{2!}=540$ ways.
The total number of ways thus equals $540+540+60=1140,$ but the answer given is $1680$. Where did I go wrong?
| Christian Blatter has provided an elegant solution. The argument below shows you the two cases you omitted and a different way of counting the cases you handled.
The number $3$ can be partitioned into three parts as follows:
\begin{align*}
3 & = 3 + 0 + 0\\
& = 2 + 1 + 0\\
& = 1 + 1 + 1
\end{align*}
The number $5$ can be partitioned into three parts not larger than $3$ as follows:
\begin{align*}
5 & = 3 + 2 + 0\\
& = 3 + 1 + 1\\
& = 2 + 2 + 1
\end{align*}
With that in mind, we consider cases:
Case 1: Three girls in one car, three boys in another car, and the remaining two boys in the remaining car.
Since the cars are different, there are $\binom{3}{1}$ ways to select the car for the three girls and one way to place all three girls in that car, $\binom{2}{1}$ ways to select which of the remaining cars will receive three boys, and $\binom{5}{3}$ ways to select which three boys are placed in that car. The remaining two boys must be placed in the remaining car. Hence, there are
$$\binom{3}{1}\binom{2}{1}\binom{5}{3} = 60$$
such distributions as you found.
Case 2: Three boys in one car, one boy in each of the other cars, with two girls being placed in one of the cars with one boy and the other girl being placed in the other car with one boy.
There are $\binom{3}{1}$ ways to select the car which will receive three boys and $\binom{5}{3}$ ways to select the boys who will be placed in that car. There are $2!$ ways to place one boy in each of the two remaining cars. There are $\binom{2}{1}$ ways to select which of the cars with one boy will receive two girls and $\binom{3}{2}$ ways to select which two of the girls will be placed in that car. The remaining girl must be placed in the other car with one boy. Hence, there are
$$\binom{3}{1}\binom{5}{3}2!\binom{2}{1}\binom{3}{2} = 360$$
such distributions. You omitted this case.
Case 3: Two girls and one boy are placed in one car, one girl and two boys are placed in another car, and the remaining two boys are placed in the remaining car.
There are $\binom{3}{1}$ ways of selecting the car that receives two girls and $\binom{3}{2}$ ways to select which two of the girls that car receives. There are $\binom{2}{1}$ ways to select which of the remaining cars receives the remaining girl and one way to place her there. There are $\binom{5}{1}$ ways to select the boy who is placed in the same car as the two girls and $\binom{4}{2}$ ways to select the boys who are placed in the same car as the single girl. The remaining two boys must be placed in the empty car. There are
$$\binom{3}{1}\binom{3}{2}\binom{2}{1}\binom{5}{1}\binom{4}{2} = 540$$
such distributions, which you counted correctly.
Case 4: Two girls are placed in one car, two boys and a girl are placed in a different car, and three boys are placed in the remaining car.
There are $\binom{3}{1}$ ways to select the car that receives two girls and $\binom{3}{2}$ ways to select which two of the three girls that car receives. There are $\binom{2}{1}$ ways to select which of the remaining cars receives one girl and one way to place her there. There are $\binom{5}{2}$ ways to select which two boys are placed in the car with the single girl. The remaining boys must be placed in the remaining car. There are
$$\binom{3}{1}\binom{3}{2}\binom{2}{1}\binom{5}{2} = 180$$
such distributions. You omitted this case.
Case 5: One girl is placed in each of the three cars, with two boys placed in each of two cars, and the remaining boy placed in the other car.
There are $3!$ ways to place one girl in each of the three cars. There are $\binom{3}{1}$ ways to select the car which receives one boy and $\binom{5}{1}$ ways to select that boy. There are $\binom{4}{2}$ ways to select which two boys will be placed in the same as the older of the two girls who does not yet have a boy in her car. The remaining two boys must be placed in the remaining car. There are
$$3!\binom{3}{1}\binom{5}{1}\binom{4}{2} = 540$$
such distributions. You counted this case correctly.
Total: Since these cases are mutually exclusive and exhaustive, the answer is found by adding up the cases.
| {
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Derivative of the function $\sin^{-1}\Big(\frac{2^{x+1}}{1+4^x}\Big)$
Find y' if $y=\sin^{-1}\Big(\frac{2^{x+1}}{1+4^x}\Big)$
In my reference $y'$ is given as $\frac{2^{x+1}\log2}{(1+4^x)}$. But is it a complete solution ?
Attempt 1
Let $2^x=\tan\alpha$
$$
\begin{align}
y=\sin^{-1}\Big(\frac{2\tan\alpha}{1+\tan^2\alpha}\Big)=\sin^{-1}(\sin2\alpha)&\implies \sin y=\sin2\alpha=\sin\big(2\tan^{-1}2^x\big)\\
&\implies y=n\pi+(-1)^n(2\tan^{-1}2^x)
\end{align}
$$
$$
\begin{align}
y'&=\pm\frac{2.2^x.\log2}{1+4^x}=\pm\frac{2^{x+1}.\log2}{1+4^x}\\
&=\color{blue}{\begin{cases}
\frac{2^{x+1}.\log2}{1+4^x}\text{ if }-n\pi-\frac{\pi}{2}\leq2\tan^{-1}2^x\leq -n\pi+\frac{\pi}{2}\\
-\frac{2^{x+1}.\log2}{1+4^x}\text{ if }n\pi-\frac{\pi}{2}\leq2\tan^{-1}2^x\leq n\pi+\frac{\pi}{2}
\end{cases}}
\end{align}
$$
Attempt 2
$$
\begin{align}
y'&=\frac{1}{\sqrt{1-\frac{(2^{x+1})^2}{(1+4^x)^2}}}.\frac{d}{dx}\frac{2^{x+1}}{1+4^x}\\
&=\frac{1+4^x}{\sqrt{1+4^{2x}+2.4^x-4^x.4}}.\frac{(1+4^x)\frac{d}{dx}2^{x+1}-2^{x+1}\frac{d}{dx}(1+4^x)}{(1+4^x)^2}\\
&=\frac{(1+4^x).2^{x+1}.\log2-2^{x+1}.4^x.\log2.2}{\sqrt{1+4^{2x}-2.4^x}.(1+4^x)}\\&=\frac{2^{x+1}\log2\big[1+4^x-2.4^x\big]}{\sqrt{(1-4^x)^2}.(1+4^x)}\\&=\frac{2^{x+1}\log2\big[1-4^x\big]}{|{(1-4^x)}|.(1+4^x)}\\
&=\color{blue}{\begin{cases}\frac{2^{x+1}\log2}{(1+4^x)}\text{ if }1>4^x>0\\
-\frac{2^{x+1}\log2}{(1+4^x)}\text{ if }1<4^x
\end{cases}}
\end{align}
$$
In both my attempts i am getting both +ve and -ve solutions. Is it the right way to find the derivative?
And how do I connect the domain for each cases in attempt 1 and attempt 2 ?
| $$y=\sin^{-1} \Big(\frac{2^{x+1}}{1+4^x}\Big)$$ the right side of the equation will be called $b$ it will help later. Take sine of both sides and the derivative $$\cos(y) y' =\frac{ln(2)2^{x+1}}{4^x+1}-\frac{ln(4) 4^x 2^{x+1}}{(4^x+1)^2}$$ the right side of the equation will be called $b'$ so $$ y' cos(y)=b'$$ divide both sides by cosine $$y'=\frac{b'}{\cos(y)}$$ substitute $y$ for $b$ and you have $$y'=\frac{b'}{cos(b)}$$
| {
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Proving a function is continuous using $\varepsilon -\delta$ definition I am trying to prove the function $f(x)$ = $\frac{x^2+1}{x^2+3}$ is continous at $x = 1$.
Here is my proof so far, but I am having difficulty showing the function is bounded.
Proof:
$\forall \varepsilon$ > 0, wants $\exists \delta$ > 0, such that $\left|\frac{x^{2} + 1}{x^{3} + 3} - \frac{1}{2}\right| < \varepsilon \Leftarrow |x-1|$.
Analysis:
$$\begin{split}
\left|\frac{x^{2} + 1}{x^{3} + 3} - \frac{1}{2}\right| < \varepsilon &\Leftarrow \\
\left|\frac{2(x^{2} +1) - (x^2+1)}{2(x^{3} + 3)}\right| < \varepsilon &\Leftarrow \\
\left|\frac{2x^2 + 2 - x^2 -1}{2x^2 + 6}\right| < \varepsilon &\Leftarrow \\ \left|\frac{x^2+1}{2x^2 + 6}\right|<\varepsilon &\Leftarrow \\
|x^2+1|\frac{1}{|2x^2+6|} < \varepsilon &\Leftarrow
\end{split}$$
That is as far as I got on the analysis, I'm having trouble with the rest
| From
$$\left| \frac{x^2+1}{x^2+3}-\frac12\right| < \epsilon$$
we have
$$\left| \frac{2(x^2+1)-x^2-3}{x^2+3}\right| < \epsilon$$
$$\left| \frac{(x-1)(x+1)}{x^2+3}\right| < \epsilon$$
Let $|x-1|< 1$, then $0<x<2$.
Hence we have $1 < x+1 < 3$ and $3< x^2+3 < 7$
Hence $\left|\frac{x+1}{x^2+3} \right|< 1$
Hence if $|x-1| < 1$ then $\left| \frac{(x-1)(x+1)}{x^2+3}\right|< |x-1|$
Can you complete the task?
Your previous mistake:
$$\left| \frac{2(x^2+1)-(x^2+1)}{x^2+3}\right| < \epsilon$$
the second $x^2+1$ in the numerator should be $x^2+3$.
| {
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Combinatorics: Find a recursive formula for partitions of integers into three partitions. I need to find a recursive formula for $p_n$ the number of ways to partition $n$ into three partitions. For example if we look for the partitions of $6$ then they are $1+1+4$, $1+2+3$, and $2+2+2$. Intuitively I look for the number of ways of partitioning $5$, which are $1+1+3$ and $1+2+2$, and try to relate this to the number of partitions of $6$. Of course, for every partition of $n-1$ there is a corresponding partition of $n$ in which the (or a) maximal element is increased by $1$. However, I'm having trouble finding a good characterization for the remaining partitions which are not formed in this way. I could think about also increasing a minimal element, but then I would have to find a way to count the number of double-countings, which would mean counting the number of ways of partitioning $n=a+b+c$ such that $a\leq b\leq c$ and $a+1=b$ or $a+1=c+1$. And I'm not even fully confidence that this description captures all of the ways that elements could get double-counted.
In the case of partitioning $5$ and $6$, increasing the maximal element takes the partitions $1+1+3$ and $1+2+2$ and yields $1+1+4$ and $1+2+3$. Increasing the minimal element yields $1+2+3$ and $2+2+2$, so the double-counted element is $1+2+3$.
| I think it would be easiest to start with the generating function for partitions into $3$ parts.
The generating function will be used to build the partition in layers.
e.g. A partition of $15$ in to $3$ parts is formed in layers: $3$ layers of width $3$, $2$ layers of width $2$ and $2$ layers of width $1$.
$$\begin{array}{c|ccc}
\text{layer $7$}&\Large{*} & &\\
\text{layer $6$}&\Large{*} & &\\
\text{layer $5$}&\Large{*} &\Large{*} &\\
\text{layer $4$}&\Large{*} &\Large{*} &\\
\text{layer $3$}&\Large{*} &\Large{*} &\Large{*}\\
\text{layer $2$}&\Large{*} &\Large{*} &\Large{*}\\
\text{layer $1$}&\Large{*} &\Large{*} &\Large{*}\\\hline
\textbf{part no.}& 1 & 2 & 3
\end{array}$$
This is the partition $(7,5,3)$.
If we build partitions in layers, then we can have an arbitrary number ($\ge 1$) of layers of width $3$. These possibilities can be represented by:
$$x^{1\cdot 3}+x^{2\cdot 3}+x^{3\cdot 3}+\cdots=\frac{x^3}{1-x^3}\tag{1}$$
then an arbitrary number ($\ge 0$) of layers of width $2$. These possibilities can be represented by:
$$x^{0\cdot 2}+x^{1\cdot 2}+x^{2\cdot 2}+\cdots=\frac{1}{1-x^2}\tag{2}$$
then an arbitrary number ($\ge 0$) of of width $1$. These possibilities can be represented by:
$$x^{0\cdot 1}+x^{1\cdot 1}+x^{2\cdot 1}+\cdots=\frac{1}{1-x}\tag{3}$$
The generating function for partitions into exactly $3$ parts is
$$P(x)=\sum_{k\ge 0}p_kx^k\, , \tag{4}$$
where $p_k$ is the number of partitions of $k$ into exactly $3$ parts. $P(x)$ is therefore given by the product of the factors $(1)$, $(2)$ and $(3)$:
$$P(x)=\frac{x^3}{(1-x)(1-x^2)(1-x^3)}=\frac{x^3}{1-x-x^2+x^4+x^5-x^6}\tag{5}$$
So, our example partition of $15$ above is represented by the term receiving contributing factors $x^{3\cdot 3}x^{2\cdot 2}x^{2\cdot 1}$ in this expansion of $(5)$.
Rearranging $(5)$:
$$P(x)(1-x-x^2+x^4+x^5-x^6)=x^3$$
$$\implies P(x)=x^3+(x+x^2-x^4-x^5+x^6)P(x)\, .\tag{6}$$
Using $(4)$ in $(6)$ and equating $x^k$ coefficients:
$$p_k=p_{k-1}+p_{k-2}-p_{k-4}-p_{k-5}+p_{k-6}\tag{Answer}$$
with $p_3=1$ is our starting value as shown by $(6)$.
| {
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$a^n+b^{n+2}=c^{n+1}$ (any solutions, other than this one?) I just made up an equation $a^n+b^{n+2}=c^{n+1}$ and chosen an exponents $n$ and $n+2$ and $n+1$ to be right where they are so that we have one solution, that is, a solution $(1,2,3)$ for $n=1$.
Are there any other solutions for $(a,b,c) \in \mathbb N^3$ and $n \in \mathbb N$?
| If $n=1$, we can take arbitrary integers $b$ and $c$ and set $a=c^2-b^3$.
For any $n$, $(2^{n+2})^n+(2^n)^{n+2}=2^{n^2+2n+1}=(2^{n+1})^{n+1}$.
When $n=2$, $(46,3,13)$, $(75,10,25)$ and $(88,4,20)$ are also solutions (found with Excel).
For $n=2$, if $a^2+b^4=c^2$, then $(ac^2)^2+(bc)^4=(c^2)^3$.
$3^2+2^4=5^2$ $\implies$ $(75,10,25)$ is a solution.
$77^2+6^4=85^2$ $\implies$ $(556325,510,7225)$ is a solution.
$17^2+12^4=145^2$ $\implies$ $(357425,1740,21025)$ is a solution.
If $u^2-v^2$ or $2uv$ is a perfect square, the Pythagorean triple $(u^2-v^2,2uv,u^2+v^2)$ will lead to a solution for $n=2$.
| {
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