Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
|---|---|---|
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 mat... | 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 +... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2584789",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 7,
"answer_id": 6
} |
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 ... | 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 $$\frac{x^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2585274",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Calculate the determinant of $A-5I$ Question
Let $ A =
\begin{bmatrix}
1 & 2 & 3 & 4 & 5 \\
6 & 7 & 8 & 9 & 10 \\
11 & 12 & 13 & 14 & 15 \\
16 & 17 & 18 & 19 & 20 \\
21& 22 & 23 & 24 & 25
\end{bmatrix}
$.
Calculate the determinant of $A-5I$.
My approach
the nullity of $A$ is $3$, so the algebraic multiplicity of $\lam... | The calculation of $\lambda_4 \lambda_5$ (which in this case is equivalent to the cubic term of the characteristic polynomial) is indeed given by the sum of principal $2\times2$ subdeterminants
$$\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|.$$
I... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2585742",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 4,
"answer_id": 1
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How many $3$-digit numbers can be formed using the digits $ 2,3,4,5,6,8 $ such that the number contains the digits $5$ and repetitions are allowed? The solution I have is by counting the complement which gives an answer of $91$. But I think that it should be solved as follows-
Let the $3$-digit number be denoted by $3$... | The total numbers that can be formed using these $6$ digits is $6^3$. We find the number of numbers in which there is no $5$. Therefore numbers without the digit $ 5$ is $5^3$. Therefore the numbers with at least one $5$ in the number is $6^3-5^3=91$
Hence answer is $91$
Or you can do it the other way using cases:
1) ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2586564",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 2
} |
Solving $\cos x + \cos 2x - \cos 3x = 1$ with the substitution $z = \cos x + i \sin x$ I need to solve
$$\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$... | 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... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2586848",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 6,
"answer_id": 0
} |
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
bee... | For your calculation of $P(A\cap B)$, it appears you're missing some cases.
Here's one way to organize the count . . .
$$
\begin{array}
{|c|c|c|c|}
A&B&C&\text{count}&\text{times}\\ \hline
1&4&0&{\large{\binom{5}{1}}}{\large{\binom{4}{4}}}=5&2\\
2&3&0&{\large{\binom{5}{2}}}{\large{\binom{3}{3}}}=10&2\\ \hline
1&3&1&{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2588051",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Evaluate $\int (2x+3) \sqrt {3x+1} dx$ Evaluate $\int (2x+3) \sqrt {3x+1} dx$
My Attempt:
Let $u=\sqrt {3x+1}$
$$\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{... | 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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"answer_id": 2
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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:
$${\left(-1\right)}^nA_n+1=2^{n+... | The striking symmetry of this binomial identity might lead us to ask what are the roles of $2$ and $-1$ and are there similar identities like this one? Here is a slight generalisation using essentially the same approach as @MarkoRiedel did in his nice answer.
We consider $a\in\mathbb{C}$ instead of $2$ resp. $-1$ and ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2593862",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
Problem with factorials: find the least n for which (n + 16)!(n+20)! ends with a number of zeros divisible by 2016 I have been working on this problem during the last few days but I can't find a good solution.
| Number of zeros ended $n!$ gives formula
$$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]... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2594655",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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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} \Righ... | 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 pr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2595010",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 2
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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 dif... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2595273",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
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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 Cro... | Hint: $y = x^2-1$:
$3(x^2 - 1)^3 + 7(x^2 - 1)^2 +4(x^2 - 1) = 3y^3+7y^2+4y = y(3y^2+7y+4) = y(y+1)(3y+4)$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2595704",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 3
} |
$P(X=x, Y=y)= e^{-2}\binom{x}{y} \frac{3}{4}^{y}\frac{1}{4}^{x-y}\frac{2^{x}}{x!}$ ; 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 si... | If it's too hard to find the marginals, write
$$\begin{align}
\mathbb{E}[Y] &= \sum_{\text{all x}}\sum_{\text{all }y}y \cdot \mathbb{P}(X = x, Y = y)\\
&=\sum_{x=0}^{\infty}\sum_{y=0}^{x}y\cdot e^{-2}\binom{x}{y}\left(\dfrac{3}{4}\right)^y\left(\dfrac{1}{4}\right)^{x-y}\dfrac{2^x}{x!} \\
&= \sum_{x=0}^{\infty}e^{-2}\df... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2598323",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"answer_id": 0
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Prove by induction: $\sum\limits_{k=1}^n (-1)^{k+1}k^2 = \frac{(-1)^{n+1}n(n+1)}{2}$ The whole problem has been translated from German, so apologies if I made any mistakes. Thank you for taking the time to help!
So this is a problem from my math book
Prove that $\sum\limits_{k=1}^n (-1)^{k+1}k^2 = \frac{(-1)^{n+1}n(n+... | This is obviously true for $n=1$. Assume it's true for any positive integer $n$
$$\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\limi... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2601559",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 2
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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... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2601731",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 3
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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/comm... | 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
$$a+b+c... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2602035",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "18",
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Prove that $ \frac{1}{1+n^2} < \ln(1+ \frac{1}{n} ) < \frac {1}{\sqrt{n} }$ For $n >0$ ,
Prove that $$ \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... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2604662",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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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$
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2605444",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 6,
"answer_id": 4
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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... | 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... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2612141",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 0
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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:$$f(12)=f(2)f(3)f(4)f(6)=f... | The polynomials in the exponent of $b$ can be written
$${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\cho... | {
"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
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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}{... | 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
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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:
$$\lim_{n \to \infty} \left(\frac{2\sin\left(\frac{\pi}{3 \cdot 3^n} \rig... | If you want to stick with your approach using the ratio test, you can (although Olivier's answer is more direct). Caveat: I'll detail every step of the derivation, which is not actually necessary for a proof.
We will only rely on elementary arguments, specifically the fact that $\sin'(0)=\cos 0 = 1$ — which is equivale... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2615300",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 2
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Evaluate limit containing $\sum{n^6}$ Evaluate:
$$\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}$$
$$... | 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",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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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.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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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{... | 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 th... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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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,
$~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~$$\sum_{k=1}^nk^2=\frac{n(n+1)(2n+1)}{6}$
The following "trick" is given below, to u... | You need to know also that $1+2+...+n=\frac{n(n+1)}{2}$ and use your work:
$$n^3+3n^2+3n=3x+3\cdot\frac{n(n+1)}{2}+n,$$ which gives which you wish.
| {
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Proving that the sequence $\{\frac{3n+5}{2n+6}\}$ is Cauchy. I'm not quite sure how to tackle these kinds of questions in general, but I tried something that I thought could be right. Hoping to be steered in the right direction here!
Let $\{\frac{3n+5}{2n+6}\}$ be a sequence of real numbers. Prove that this sequence i... | $|\frac{3n+5}{2n+6}-\frac{3m+5}{2m+6}|=\frac{|n-m|}{(2n+6)(2m+6)}\le\frac{8|n-m|}{nm} =8|\frac{1}{n}-\frac{1}{m}|$.
Can you proceed ?
| {
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Proof explanation about matrix for block. Prove:
$$\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$$
be QR decompositions of $A$ and $B$. Then
\begin{align*}
\det \begin{bmatrix} A & C \\ 0 & B \end{bmatrix} &= \det \begin{bmatrix} Q_A R_A & Q_A Q_A^T C... | Determinants multiply.
With $A$ square $m$ by $m,$ and $B$ square $n$ by $n,$ also $C$ $m$ by $n,$
$$
\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)
$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Find the area of underneath the curve $ \ f(x)=4-3x \ $ over the interval $ \ [2,4] \ $ by dividing the interval into $ \ n \ $ Find the area of underneath the curve $ \ f(x)=4-3x \ $ over the interval $ \ [2,4] \ $ by dividing the interval into $ \ n \ $ equal subintervals and taking limit.
Answer:
The length of the i... | There is a mistake in one of your steps:
$$ \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... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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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
B... | 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 $$\theta =\cos^{-1} (\frac {\sqrt 5}{5})$$
| {
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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
$$\frac{a+b+c}{d+e+f}=-\frac{5}{6}$$
| {
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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\... | {
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"timestamp": "2023-03-29T00:00:00",
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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:
$$bc=\dfrac {a+c}{2} \dfrac... | Since $a,b,c$ are in AP we have $a=b-x$ and $c=b+x$ for some $x$ and since $b,c,a$ are in HP we have $$(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$$
| {
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Number of answers : $f(x)=f^{-1}(x)$
let $f(x)= 1+\sqrt{x+k+1}-\sqrt{x+k} \ \ k \in \mathbb{R}$
Number of answers :
$$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)$$
... | Let $y=f(x)=f^{-1}(x)$ and then
\begin{eqnarray}
y=1+\sqrt{x+k+1}-\sqrt{x+k},\tag{1}\\
x=1+\sqrt{y+k+1}-\sqrt{y+k}.\tag{2}
\end{eqnarray}
Subtracting (1) from (2) gives
\begin{eqnarray}
x-y&=&(\sqrt{y+k+1}-\sqrt{x+k+1})-(\sqrt{y+k}-\sqrt{x+k})\\
&=&\frac{y-x}{\sqrt{y+k+1}+\sqrt{x+k+1}}-\frac{y-x}{\sqrt{y+k}+\sqrt{x+k}}... | {
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Recursion of polynomials I have the following equation on the functions $f_i$ for $i>1$, defined in $[0,1/4)$
$$f_i(x) = f_{i-1}(x) + \frac{x}{1-4x}f_{i-2}(x)$$
And assume that $f_0$ equals $c(x)$, the generating function of the Catalan numbers, and $f_1(x)=(c(x)-1)/x$.
Any idea on how can I get a closed form if it exi... | Testing the form $f_i = g^i$, we get $g^2 = g + \frac{x}{1-4x} \implies g_{\pm} = \frac12 \pm \frac1{2\sqrt{1-4x}}$
So letting $f_i =a g_+^i + b g_-^i$, we get $c = a+b$ and
$$\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 = \fra... | {
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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 inde... | 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",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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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
$$\lim_{n\rightarrow \infty } \frac{1^4 + 2^4 + \ldots + n^4}{n^5}$$
$$\lim_{n\rightarrow \infty } \frac{n^4( \frac{1^4}{n^4} + \frac{2^4}{n^4} +\ldots + 1 )}{n... | If
$m > 1$ then
$k^m
\lt \int_k^{k+1} x^m dx
\lt (k+1)^m
$.
Summing the left inequality
from $1$ to $n-1$,
$\begin{array}\\
\sum_{k=1}^{n-1}k^m
&\lt \sum_{k=1}^{n-1}\int_k^{k+1} x^m dx\\
&= \int_1^{n} x^m dx\\
&= \dfrac{x^{m+1}}{m+1}\big|_1^{n}\\
&< \dfrac{n^{m+1}-1}{m+1}+(n+1)^m\\
&< \dfrac{n^{m+1}}{m+1}\\
\text{so}\\... | {
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Calculate floor of $\frac{1}{x_1 + 1} + \frac{1}{x_2 + 1} + ... + \frac{1}{x_{100} + 1}$ There is a recurrence sequence $x_1 = \frac{1}{2}$, $x_{n+1} = x_n^2 + x_n$. How much is floor of $\frac{1}{x_1 + 1} + \frac{1}{x_2 + 1} + ... + \frac{1}{x_{100} + 1}$? Floor is an integer part of a real number.
| The instructive hints of @AchilleHui deserve an answer by its own.
Given the recurrence relation
\begin{align*}
x_{n+1}&=x_n(x_n+1)\tag{1}\\
x_1&=\frac{1}{2}\\
\end{align*}
we derive from (1) a telescoping representation of the reciprocal values via
\begin{align*}
\frac{1}{x_{n+1}}&=\frac{1}{x_n(x_n+1)}
=\frac{1}... | {
"language": "en",
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"source": "stackexchange",
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Evaluating a limit using Taylor series How do I evaluate the limit
$$ \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)}$$
depending on $\alpha, $using Taylor series? I know I consider
$e^t = 1 +t +\frac{t^2}{2} + o(t^2)$
$ \log(1+t) = t - \frac{t^2}{2} + \frac{t^3}{3} ... | Note that
$$\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 + ... | {
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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)^... | 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, $$(k+3)^{k+2}\leq(k+3)(k+1)(k+2)^{k+1}=$$
$$=(k^2+4... | {
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Integral of $\int{\frac{\cos^4x + \sin^4x}{\sqrt{1 + \cos 4x}}dx}$ How do we integrate the following?
$\int{\frac{\cos^4x + \sin^4x}{\sqrt{1 + \cos 4x}}dx}$ given that $\cos 2x \gt 0$
I tried to simplify this, but I cannot seem to proceed further than the below form:
$\int{\frac{\sec2x}{\sqrt{2}}dx + \sqrt{2}\int{\f... | If $\cos 2x > 0$ and the integrand is wrong as Rudr Pratap Singh points out, then:
\begin{align}
\frac{\cos^4 x-\sin^4 x}{\sqrt{1+\cos 4x}}
&= \frac{(\cos^2x+\sin^2x)(\cos^2x-\sin^2x)}{\sqrt{2\cos^2 2x}} \\
&= \frac{(1)(\cos 2x)}{\sqrt2|\cos2x|} \\
&= \frac 1 {\sqrt2}.
\end{align}
Hence,
$$\int \frac{\cos^4x-\sin^4x}{\... | {
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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:
*
*$$y = (2x^3 + 5)^4$$
$$ so \frac{dy}{dx} = 4(2x^3+5)^3 * 6x^2$$
$$ = 24x^2(2x^3 + 5)^3$$
*$$f(x) = (5x^6 + 2x^3)^4$$
$$f'(x) = 4(5x^6 ... | My comment was getting too long so I'll post a short answer.
For number $5$, as mentioned in the comments you should use the product rule. Recall that:
$$\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}$$
and differentiate both (u... | {
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Closed form of $u_n$ with $u_{0} =1$ and $u_n=\frac{1}{3}u_{n-1}-1$? Let $u_n$ be a sequence defined by recurrence relation as : $u_{0} =1$ and $u_n=\frac{1}{3}u_{n-1}-1$, I'm not familiar to look for the closed form of $u_n $ , I have tried to find $u_n $ with a function of $n$ but I don't succeed because it's not def... | You can express $u_n$ through $u_0$ sequentially through $u_{n-1}$, $u_{n-2}$ and so on
$$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... | {
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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... | {
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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}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2636083",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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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 $x,y\in \mathbb Z$.
... | $abx=(a+b)$, 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=... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2636366",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "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 ... | 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.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2637276",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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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$$\left| z^3+\frac{1}{z^3} \right| = \le... | $$2\ge|a^3+b^3|=|(a+b)^3-3ab(a+b)|$$
$$\ge|a+b|^3-3|ab(a+b)|$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2638606",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Find Area bounded by Inverse of $f(x)=x^3+3x+2$ between $x=-2$ and $x=6$ Find Area bounded by Inverse of $f(x)=x^3+3x+2$ between $x=-2$ and $x=6$
My Try:
we need $$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'(... | 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... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2640310",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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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... | For equation $a+b=c$ $$i(1+i)\;+\;(2+3i)^4\;=\;i(3+2i)^4$$
we have
$$Q = \frac{\log\left(13^2\right)}{\log\left(13\sqrt{2}\right)}\approx 1.761929.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2642142",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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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 div... | 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.
(*) whe... | {
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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.
$$K^n = \left\{\begin{pmatrix}
a_{1} \\
a_{2... | It is a convention, which you can take as a definition.
Since $K^n$ is an $n$-dimensional vector space when $n$ is a positive integer, we would like to have $K^0$ to denote a zero-dimensional vector space, which would be $\{0\}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2648273",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Trigonometric Inequality $\sin{A}+\sin{B}-\cos{C}\le\frac32$ To prove
$$\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... | 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 $\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2648521",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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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... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2649304",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 1
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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 s... | 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
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2652008",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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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... | 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}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2652742",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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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,
$$u_1 = ae-bf-cg-dh\\
u_2 = af+be+ch-dg\\
u_3 = ag-bh+ce+df\\
u_4 = ah+bg-... | Summarizing some comments and adding a bit on top:
The identity is a special case of
$$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{al... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
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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... | 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.
$$ \tan \frac x2
= \frac{\sin \frac x2}{\cos \frac x2}
= \frac{2 \sin \frac x2 \ \cos \frac x2}{1 + 2\cos^2 \frac x2 - 1}
= \frac{\sin x}{1 + \cos x}
= \frac{1 - \... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2655018",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Why limit for this function exists as $x$ approaches $0$? $$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}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2658222",
"timestamp": "2023-03-29T00:00:00",
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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: $$\left(\frac{1}{2}+\cos \frac{\pi}{20}\right)\left(\frac{1}{2}+\cos \frac{3\pi}{... | Let $\xi=\exp\left(\frac{2\pi i}{40}\right)$. The given product equals
$$ \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}-... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2659008",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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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
$$\sqrt{2+\tan (x)}=u^2 \implies x=-\tan ^{-1}\left(2-u^2\right)\implies dx=\frac{2 u}{1+\left(2-u^2\right)^2}\,du$$
$$\int \sqrt{2+\tan (x)}\, dx=\int \frac{2 u^2}{1+\left(2-u^2\right)^2}\,du$$
$$1+\left(2-u^2\right)^2=u^4-4 u^2+5=(u^2-(2-i))(u^2-(2+i))$$ N... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2661168",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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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 $$\frac14=x^4... | Use exponential form: since $|z|=1$, $z=\mathrm e^{i\theta}$, so the second equation can be rewritten as
$$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\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2662090",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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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 functio... | 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 ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2663047",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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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",
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"answer_id": 2
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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 alway... | 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
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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
$a=\frac{-\sqrt{2}+\sqrt{\Delta }}{2\times 4}=\frac{-\sqrt{2}+\sqrt{2+16\sqrt{2}}}{8}$... | Rewrite using the value of $a^2 = \frac{1-a}{2\sqrt2}$
$$\frac{a+1}{\sqrt{\tfrac{(a-1)^2}{8}+a+1}+\tfrac{(a-1)}{2\sqrt{2}}} = \frac{(a+1)2\sqrt2}{\sqrt{a^2+6a+9} + a-1} = \frac{(a+1)2 \sqrt 2}{(a+3) + a-1} \\= \sqrt{2}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2669096",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"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
$$3\sqrt[3] {\prod \left(\frac {a^2}{b+c}+ \frac {bc}{b+c}\right)^{\frac {1}{3}}}\ge 3\sqrt[3] {\left (\frac {a^2b^2c^2}{(a+b)(b+c)(a+c)}\r... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Showing $x^4-x^3+x^2-x+1>\frac{1}{2}$ for all $x \in \mathbb R$
Show that
$$x^4-x^3+x^2-x+1>\frac{1}{2}. \quad \forall x \in \mathbb{R}$$
Let $x \in \mathbb{R}$,
\begin{align*}
&\mathrel{\phantom{=}}x^4-x^3+x^2-x+1-\frac{1}{2}=x^4-x^3+x^2-x+\dfrac{1}{2}\\
&=x^2(x^2-x)+(x^2-x)+\dfrac{1}{2}=(x^2-x)(x^2+1)+\dfrac{1}{2... | You can note that the inequality holds for $x=-1$. For $x\ne-1$ it is equivalent to
$$
\frac{x^5+1}{x+1}>\frac{1}{2}
$$
that can be rewritten as
$$
\frac{2x^5-x+1}{x+1}>0
$$
Let's analyze $f(x)=2x^5-x+1$, with $f'(x)=10x^4-1$, which vanishes for $x=\pm10^{-1/4}$. So $f$ has a relative maximum at $-10^{-1/4}$ and a rela... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2670433",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 7,
"answer_id": 5
} |
Evaluate the limit $\lim_{n\to \infty}\sqrt{n^2 +2} - \sqrt{n^2 +1}$ I know that
$$\lim_{n\to \infty}(\sqrt{n^2+2} - \sqrt{n^2+1})=0.$$
But how can I prove this?
I only know that $(n^2+2)^{0.5} - \sqrt{n^2}$ is smaller than $\sqrt{n^2+2} - \sqrt{n^2}$ = $\sqrt{n^2+2} - n$.
Edit:
Thank Y'all for the nice and fast... | Hint:
Try to multiply by $$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)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2670779",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
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Tangent vector of curve $ \Psi(t)= (2t^3 - 2t, 4t^2, t^3+t )^T $ expressed in spherical coordinates I have a curve where $ t\in R^{+}_0$
$$
\Psi(t)= (2t^3 - 2t, 4t^2, t^3+t )^T
$$
I need to express tangent vector T in standard spherical coordinates in terms of normalized 'frame vectors' $\hat h_i $.
I started from d... | \begin{align}
\mathbf{r} &=
\begin{pmatrix} 2t^3-2t \\ 4t^2 \\ t^3+t \end{pmatrix} \\
r &= \sqrt{(2t^3-2t)^2+(4t^2)^2+(t^3+t)^2} \\
&= t(t^2+1)\sqrt{5} \tag{$t>0$} \\
\dot{r} &= (3t^2+1)\sqrt{5} \\
\cos \theta &= \frac{z}{r} \\
&= \frac{1}{\sqrt{5}} \\
\sin \theta &= \frac{2}{\sqrt{5}} \\
\dot{\theta}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2672604",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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To check whether a function is unbounded
Show that $f:[-1,1]\to \mathbb{R}$ defined by
$f(x)=2x\sin\frac{1}{x^2}-\frac{2}{x}\cos\frac{1}{x^2}, ~~x\neq 0$ and $f(0)=0$ is unbounded on every neighbourhood of $0$.
Attempt:
$|f(x)|=|2x\sin\frac{1}{x^2}-\frac{2}{x}\cos\frac{1}{x^2}|\leq |2x\sin\frac{1}{x^2}|+|\frac{2}{x... | Let $x_k = \frac{1}{\sqrt{2k\pi}}$ where $k\in \mathbb{Z}, k > 0$.
Then $$f \left( \frac{1}{\sqrt{2k\pi}} \right)=-2\sqrt{2k\pi}$$
As $k\to \infty, x_k \to 0, |f(x_k)| = 2\sqrt{2k\pi}\to \infty$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2677042",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
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Does the following limit exists? $$
\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.}
$$
The integral exists, since ... | This is not an answer yet but just a few manipulations that might hopefully simplify things a bit. I might have made some calculation errors.
$$\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}\fra... | {
"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
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finding the values of a such that the system $AB\cdot x=0$ has infinitely many solutions the following matrix are given:
$$A=
\begin{bmatrix}
a & 1 & 1 \\
a & a & 2 \\
a & a & a \\
\end{bmatrix}
$$
$$B=
\begin{bmatrix}
a-3 & 1 & 1 \\
a-3 & a-3 & 2 \\
a-3 & a-3 & a-3 \\
\end{bmatr... | Your answer is right (I checked it) and therefore your textbook is wrong.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2680706",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Find the sum to $n$ terms of the series: $1^2.1+2^2.3+3^2.5+....$ Find the sum to $n$ terms of the series:
$$1^2.1+2^2.3+3^2.5+.....$$
My Attempt:
Here, $n^{th}$ term of $1,2,3,....=n$
$n^{th}$ term of $1^2,2^2,3^2,....=n^2$
Also, $n^{th}$ term of $1,3,5,....=2n-1$
Hence, $n^{th}$ term of the given series is $t_n=n^2(2... | $$\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}.$$
Lastly, evaluate $P(x^2-1)=\ldots=x^4+x^2-3$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2683178",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 3
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How to deal with $(-1)^{k-1}$ It's a problem on mathematical induction.
$$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^... | 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,
"answer_id": 2
} |
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 $$\frac{d}{dx} \frac{\sin(x)}{\cos(x)}=\frac{\cos^2(x)+\sin^2(x)}{\cos^2(x)}=1+\frac{\sin^2(x)}... | Note simply that
$$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)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2688627",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Number of binary strings not having even-length runs of $0$ or $1$ I was solving this algorithmic problem. This problem boils down to finding the finding the number of binary strings of given length $n$ such that those strings don't have any even-length run of $0$ or $1$. The post says that the answer is twice the $n$t... | Using $z$ for ones and $w$ for zeroes we get the generating function
$$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}... | {
"language": "en",
"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 \cdo... | 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 ... | $$x^2-37y^2=2\Rightarrow x^2-y^2\equiv2\mod(4)\Rightarrow x^2\equiv 2+y^2\mod(4)$$
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... | 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... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2691740",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 1
} |
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?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2697396",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 1
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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}... | 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 othe... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2698842",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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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 $n>m:$
$$\begin{align}|x_n-x_m|&=\left|\frac{1}{3}+\left(\... | For $n > m$ we have
\begin{align}
|x_n - x_m| &= \left(\underbrace{\frac{m+1}{2(m+1)+1}}_{\le \frac12}\right)^{m+1}+\cdots +\left(\underbrace{\frac{n}{2n+1}}_{\le \frac12}\right)^n \\
&\le \frac{1}{2^{m+1}} + \cdots + \frac{1}{2^n} \\
&= \frac{1}{2^{m+1}}\left( 1 + \frac12 + \cdots + \frac1{2^{n-m-1}}\right)\\
&= \frac... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2698972",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Inequality $\frac{a^3+b^3+c^3}{3}\geq\sqrt{\frac{a^4+b^4+c^4}{3}}$ Let $a$, $b$ and $c$ be non-negative numbers such that $(a+b)(a+c)(b+c)=8$. Prove that:
$$\frac{a^3+b^3+c^3}{3}\geq\sqrt{\frac{a^4+b^4+c^4}{3}}$$
Some attempts:
*
*From the condition follows $a^3+b^3+c^3 = (a+b+c)^3 -24$
*It is known (see here)
$$\... | 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 :
$$\frac{\sum_{cyc}^{}(x-y)(y-z)+\sum_{cyc}^{}(x... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2699175",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 4,
"answer_id": 2
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What is the best notation to write ``equations with reasons''? I'm learning my analytic number theory course. When I make notes, I find that it is effective to use ``equations with reasons'' as follow
$$\begin{array}{rll}
\vartheta(x)&\displaystyle=\sum_{p\leq x} \log p& \textrm{partial sum $\begin{cases}
1 & \textrm{... | 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... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2701717",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
} |
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 \R... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2702035",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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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:
$$\frac{\sin A+\sin B}{\sin A-\sin B}=\frac{3}{2}$$
$$\frac{\tan(A+B)/2}{\tan(A-B)/2}=\fra... | This is impossible to answer for general $A$ and $B$ satisfying the condition.
If $\displaystyle \frac{\tan A}{\tan B}=k$, then $\displaystyle k=\frac{\sin A\cos B}{\cos A\sin B}=\frac{5\cos B}{\cos A}$.
$$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}$$
So $k$ depends... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2704183",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Solve $\sqrt {x^2-3}=x-3$ in $\mathbb R$
Solve $\sqrt {x^2-3}=x-3$ in $\mathbb R$
My attempt:
$|x^2-3|=(x-3)^2$
So $-(x^2-3)=(x-3)^2$ or $(x^2-3)=(x-3)^2$
If $-(x^2-3)=(x-3)^2=x^2+9-6x$
So no solutions in $\mathbb R$
And if $(x^2-3)=(x-3)^2$
So $x^2-3=x^2+9-6x$
Now, can I delete $x^2$ with $x^2$ ? Like this
$x^2-x... | From
$$\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$$
but since $x=2$ doesn't satisfy the equation it is to re... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2704837",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 3
} |
How to prove divergent sequences? For this question, I know that the sequence diverges to infinity, but I'm not sure if I am doing it right. Here is what I have so far. Can anyone please help me out?
Determine whether the following sequence is convergent or divergent
$a_n = \{8n^3 + n^2 -2\}$
$\lim_{n \to \infty} a_n =... | Let $M > 0$.
For any $n \ge \sqrt[3]{M+2}$ we have
$$8n^3 + n^2 - 2 \ge n^3 - 2 \ge (M+2) - 2 = M$$
Hence your sequence is unbounded.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2705509",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 0
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Spivak's Calculus: Chapter 1, Problem 18b (Quadratic determinant less than zero) The problem in question is as follows:
18b Suppose that $b^2 -4c \lt 0$. Show that there are no numbers $x$ that satisfy $x^2 + bx + c = 0$; in fact, $x^2 + bx + c \gt 0$ for all $x$. Hint: complete the square.
Trying to apply the hint, I ... | Essentially the same algebraic manipulations used in the second part of your question will give a proof without aiming for a contradiction.
Indeed, these manipulations will give you that $x^2 + bx + c = (x + \frac{b}{2})^2 - \frac{b^2 - 4c}{4}$. This is strictly positive since $(x + \frac{b}{2})^2 \geq 0$ and $\frac{b... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2706487",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 7,
"answer_id": 3
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Evaluate $\int \frac{1}{ax^2-bx}\,dx$ with substitution Evaluate $\int\frac{1}{ax^2-bx}\,dx$
First try:
$$\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 = \fr... | $$\frac{1}{b}\ln\left|\frac{ax-b}{bx}\right|=\frac{1}{b}\left(\ln\left|\frac{ax-b}{x}\right|-\ln|b|\right)=\frac{1}{b}\ln\left|\frac{ax-b}{x}\right|+\frac{1}{b}\ln|b|$$
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 w... | {
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"url": "https://math.stackexchange.com/questions/2707839",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"answer_id": 1
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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 su... | 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(... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2708655",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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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... | 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 sq... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2710554",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 1
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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 t... | 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$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2713476",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
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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\alp... | $$y=\sin^{-1} \Big(\frac{2^{x+1}}{1+4^x}\Big)$$ 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 b... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2714111",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
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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 $... | 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| ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2715630",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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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 num... | 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{a... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2718015",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 0
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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 $... | 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 $(ac^2)^2+(bc)^4=(c^2)^3$.
$3^2+2^4=5^2$ $\implie... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2722096",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Subsets and Splits
Fractions in Questions and Answers
The query retrieves a sample of questions and answers containing the LaTeX fraction symbol, which provides basic filtering of mathematical content but limited analytical insight.