Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
|---|---|---|
If $abc=1$ then $\sum\limits_{cyc}^{}{\frac{1}{b(a+b)}}\ge \frac{3}{2}$
If $abc=1$ for positive $a,b,c$, then $\sum\limits_{cyc}^{}{\dfrac{1}{b(a+b)}}\ge \dfrac{3}{2}$
I have tried the following,in decreasing order of success:
1)AM-GM:$a+b+c\ge 3$ and $ab+bc+ca\ge 3$
2)Substituting $1=abc$ yields nothing
3)Substituting $a=\frac{x}{y},b=\frac{y}{z},c=\frac{z}{x}$ yields something weird
4)Rearrangement inequality on the sequences $\frac{1}{a},\frac{1}{b},\frac{1}{c}$ and $\frac{1}{a+b},\frac{1}{b+c},\frac{1}{c+a}$
5)Titu's lemma doesn't help
A little nudge in the right direction would help.
| Look at $\frac{1}{b(a+b)}$ and $\frac{1}{a(a+b)}$. Adding together yields $\frac{1}{ab}$. So the sum is just
$$\frac{1}{ab} + \frac{1}{bc} + \frac{1}{ac} = \frac{a + b + c}{abc}$$
The denominator is 1 and the numerator is at least 3 by AM-GM for positive $a,b,c$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1141774",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
"answer_id": 2
} |
Finding number of integer solutions using Generating Functions This is a problem for a practice test my professor gave me.
$$\text{How many integer solutions are there to } x_1+x_2+x_3+x_4 \leq 50 \\ \text{with } x_i \geq 2 \text{ for all } i = 1,2,3,4 \text{ and } x_1,x_2 \leq 7 \text{?}$$
This is how I approached the problem, using generating functions:
$\text{Same as}$
$$x_1+x_2+x_3+x_4+x_5 = 50, \space x_5 \geq 0 $$
$\text{Find the coefficient of } x^{50} \text{ in}$
$$(x^2 + x^3 + x^4 + \dotso)^2 (x^2 + x^3 + x^4 + \dotso + x^7)^2 (1 + x + x^2 + x^3 + \dotso)$$
After some factoring, we'll have:
$$x^8(1+x^2+x^3+ \dotso + x^5)^2 (1+x+x^2+\dotso)^3$$
This is the same as:
$\text{Find the coefficient of } x^{42} \text{ in}$
$$(1+x^2+x^3+ \dotso + x^5)^2 (1+x+x^2+\dotso)^3$$
To simplify further:
$$(1-x^6)^2 \frac{1}{(1-x)^3}$$
So, this is where I'm confused. I was using the formula which is based off this answer on Math.SE, but I don't get the correct answer. According to my professor, the correct answer is:
$$\dbinom{30+5-1}{30} - 2\dbinom{36+5-1}{36} + \dbinom{42+5-1}{42} = 26,781$$
What I end up doing mirrors that of the linked question on Math.SE:
$$(1-x^6)^2 = 1-2x^6+x^{12} \\
\\
(1-2x^6+x^{12}) \frac{1}{(1-x)^3}$$
Using the formula from the linked question:
$\text{We do this three times, for } k=0, k=6, \text{ and } k=12$. The result is
$$(1-2x^6+x^{12})\frac{1}{(1-x)^3}={m-0+2 \choose 2}- 2{m-6+2 \choose 2} + {m-12+2 \choose 2}
\\
\
\\m = 42 \\
\
\\={42-0+2 \choose 2}- 2{42-6+2 \choose 2} + {42-12+2 \choose 2} = 36$$
As you can see, my answer differs greatly from what my professor said was correct. I don't understand why this formula I used doesn't work; I've used it for lots of other problems of this same type, and I calculated the correct number; for this one though, it doesn't seem to be working.
What am I doing wrong?
| Inclusion exclusion is easier than generating functions in this case. First you reduce to
$$ x_1 + x_2 + x_3 + x_4 + x_5 = 42 $$
where $x_i \geq 0$ and $x_1, x_2 \leq 5$. Then IE immediately gives
$$ {42+5-1 \choose 5-1} - 2{36+5-1 \choose 5-1} + {30+5-1 \choose 5-1} \\
= {46 \choose 4} - 2{40 \choose 4} + {34 \choose 4} = 26781 $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1143075",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
} |
Existence of two primes satisfying the given conditions I want to know whether
the equation
$x^a-x=y^b-y$
has a solution or not satisfying the conditions that $x$ and $y$ are distinct odd primes, $a$ and $b$ are integers both greater than $1$.
| Michael Bennett, On some exponential equations of S. S. Pillai, Canadian Journal of Math. 53 (2001) 897-922, reports eight solutions of $x^a-x=y^b-y$ in positive integers, with $a,b>1$:
$6=2^3-2=3^2-3$,
$30=2^5-2=6^2-6$,
$210=6^3-6=15^2-15$,
$240=3^5-3=16^2-16$,
$2184=3^7-3=13^3-13$,
$8190=2^{13}-2=91^2-91$,
$78120=5^7-5=280^2-280$,
$24299970=30^5-30=4930^2-4930$.
See the OEIS for links. I don't think any new ones have been discovered, and I don't think anyone has proved there are none. Of those eight solutions, the only one satisfying the (rather unmotivated) primality stipulations is the one Barry reported in his answer five years ago.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1143521",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
} |
Sum of $1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{6}+\frac{1}{8}+\frac{1}{9}+\frac{1}{12}+\cdots$ My problem is to find the sum of the series
$$
S = 1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{6}+\frac{1}{8}+\frac{1}{9}+\frac{1}{12}+\cdots
$$
where the terms are the reciprocals of the positive integers whose only prime factors are $2$s and $3$s.
I can see that every term in $S$ is of the form $\dfrac{1}{2^m3^n}$ where $m,n\geq 0$. I can also see that each term only occurs once, but I'm not really sure how to use this information effectively. Any ideas (apparently geometric series was a hint)?
| $$S = 1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{6}+\frac{1}{8}+\frac{1}{9}+\frac{1}{12}+\cdots$$
$$\frac{1}{2}S = \frac{1}{2}+\frac{1}{4}+\frac{1}{6}+\frac{1}{8}+\frac{1}{12}+\frac{1}{16}+\frac{1}{18}+\frac{1}{24}+\cdots$$
What terms are missing? We've lost all the terms of the form $\frac{1}{3^n}$, but have retained everything else.
$$S-\frac{1}{2}S = 1+\frac{1}{3}+\frac{1}{9}+\dots$$
$$\frac{1}{2}S = \frac{3}{2}$$
$$S = 3$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1146102",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 2
} |
How to prove this floor function equation? How can I prove the following equation?
$$
\lfloor nx \rfloor =
\lfloor x \rfloor +
\Big\lfloor x + \frac{1}{n} \Big\rfloor +
\Big\lfloor x + \frac{2}{n} \Big\rfloor +
\Big\lfloor x + \frac{3}{n} \Big\rfloor +
\Big\lfloor x + \frac{4}{n} \Big\rfloor +
\Big\lfloor x + \frac{5}{n} \Big\rfloor+ \dotsb +
\Big\lfloor x + \frac{n-1}{n} \Big\rfloor
$$
$n∈N$ and $x∈R$
| If you take a closer look, you will notice that the second term within each floored term will be less than one.
E.g in $$\left\lfloor x+\frac{4}{n}\right\rfloor$$As you can see $\frac{4}{n}$ is less than one.
So, we can conclude that each term will be reduced to $\lfloor x\rfloor$ if $n$ and $x$ are integers.
We will get
$$\lfloor nx\rfloor = n\lfloor x\rfloor$$
Since $n$ and $x$ are integers, we can remove the floor function.
Hence proved.
This identity will not be true for all values of $n$ and $x$ if they can be any number other than integers. Try putting any fractional value.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1147404",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 1
} |
Why does $\int\frac{1}{2x+1}dx=\frac{1}{2}\ln|2x+1|+C$? The way I am thinking is as follows:
$$\int\frac{1}{2x+1}\,dx = \int\frac{1}{2}\frac{1}{x+\frac{1}{2}}\,dx = \frac{1}{2}\int\frac{1}{x+\frac{1}{2}}\,dx = \frac{1}{2}\ln\left|x+\frac{1}{2}\right|+C$$
However, the textbook answer is $\frac{1}{2}\ln|2x+1|+C$. Where is my thinking wrong?
| You forgot to add a constant, $C$. This is important: your answer differs from the textbook's only by a constant.
$$\frac{1}{2}\ln\left|x + \frac{1}{2}\right| = \frac{1}{2}\ln\left|\frac{2x + 1}{2}\right| = \frac{1}{2}[\ln|2x + 1| - \ln(2)] = \frac{1}{2}\ln|2x + 1| - \frac{1}{2}\ln(2)$$
So
$$\int \frac{1}{2x+1}\, dx = \frac{1}{2}\ln\left|x + \frac{1}{2}\right| + C = \frac{1}{2}\ln|2x + 1| + C.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1148261",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
Finding the value of the infinite sum $1 -\frac{1}{4} + \frac{1}{7} - \frac{1}{10} + \frac{1}{13} - \frac{1}{16} + \frac{1}{19} + ... $ Can anyone help me to find what is the value of $1 -\frac{1}{4} + \frac{1}{7} - \frac{1}{10} + \frac{1}{13} - \frac{1}{16} + \frac{1}{19} + ... $ when it tends to infinity
The first i wanna find the pattern but it seems do not have any unique pattern can anyone help me?
| the series equal to
$$\sum_{n=0}^{\infty}\frac{1}{6n+1}-\frac{1}{6n+4}$$
the first term
$$\frac{1}{1-x^6}=\sum_{n=0}^{\infty} x^{6n}$$
$$\int_0^1\frac{1}{1-x^6}dx=\int_0^1\sum_{n=0}^{\infty}x^{6n}dx$$
$$\int_0^1\frac{1}{1-x^6}dx=\sum_{n=0}^{\infty}\frac{1}{6n+1}$$
the second term
$$\frac{1}{1-x^6}=\sum_{n=0}^{\infty}x^{6n}$$
mutiply by $x^3$
$$\frac{x^3}{1-x^6}=\sum_{n=0}^{\infty}x^{6n+3}$$
$$\int_0^1\frac{x^3}{1-x^6}dx=\int_0^1\sum_{n=0}^{\infty}x^{6n+3}dx$$
$$\int_0^1\frac{x^3}{1-x^6}dx=\sum_{n=0}^{\infty}\frac{1}{6n+4}$$
hence
$$\sum_{n=0}^{\infty}\frac{1}{6n+1}-\frac{1}{6n+4}=\int_0^1\frac{1}{1-x^6}dx-\int_0^1\frac{x^3}{1-x^6}dx=\int_0^1\frac{1}{1+x^3}dx$$
$$=\frac{1}{3}\log 2+\frac{\pi\sqrt{3}}{9}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1148862",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 3,
"answer_id": 0
} |
The Finishing Step to Showing that $1/(1-4x)^{1/2}$ generates the sequence $\binom{2r}{r}$ The Full Question:
Show that $(1-4x)^{-\frac{1}{2}}$ generates the sequence $\binom{2n}{n}$, $n\in
\mathbb N$
My Research
How to show that $1 \over \sqrt{1 - 4x} $ generate $\sum_{n=0}^\infty \binom{2n}{n}x^n $
Show $\sum\limits_{n=0}^{\infty}{2n \choose n}x^n=(1-4x)^{-1/2}$
Both these questions seem similar, but I found they used a very different approach (more calculus based) to find their answers whereas my question requires me to use more algebra and definitions to get the answer.
My Work
1) Let $y = -4x$
2) We know the $r^{th}$ term of$(1-y)^{-1/2}$ will have the coefficient of $\binom{-1/2}{r}(-4^r)$
3) $\binom{-1/2}{r}(-4^r) = \frac{(-1/2)(-1/2-1)(-1/2-2)\cdots(-1/2-r+1)}{r!}(-4)^r$
4) If we factor out $(-1/2)$ from the whole thing, we have: $\frac{(1)(1+2)(1+4)\cdots(1+2r-2)}{r!}(2^r)=\frac{(2r-1)(2r-3)\cdots(5)(3)(1)}{r!}2^r$
5) If we multiply top and bottom by $r!$ we get:$\frac{(2r-1)(2r-3)\cdots(5)(3)(1)2^rr!}{r!r!}$
6) All that remains is to show the numerator to equal $(2r)!$ and I'm golden, however I don't really know how to do this.
My Question
Can anyone help me with step 6? I feel I'm very close but haven't quite been able to see it.
|
Note: Avoiding a small calculation error in OPs work in 4.) and considering a small trick in 6.) will close the gap. Here's a calculation:
\begin{align*}
\binom{-\frac{1}{2}}{r}
&=\frac{\left(-\frac{1}{2}\right)\left(-\frac{1}{2}-1\right)\left(-\frac{1}{2}-2\right)\cdot\ldots\cdot
\left(-\frac{1}{2}-(r-1)\right)}{r!}\\
&=\frac{\left(-\frac{1}{2}\right)\left(-\frac{3}{2}\right)\left(-\frac{5}{2}\right)\cdot\ldots\cdot
\left(-\frac{2r-1}{2}\right)}{r!}\\
&=\left(-\frac{1}{2}\right)^r\cdot\frac{(2r-1)(2r-3)\cdot\ldots\cdot5\cdot3\cdot1}{r!}\tag{1}\\
&=\left(-\frac{1}{2}\right)^r\cdot\frac{(2r)(2r-1)(2r-2)(2r-3)\cdot\ldots\cdot5\cdot4\cdot3\cdot2\cdot1}
{r!(2r)(2r-2)(2r-4)\cdot\ldots\cdot4\cdot2}\tag{2}\\
&=\left(-\frac{1}{2}\right)^r\cdot\frac{(2r)!}
{r!2^r\cdot r(r-1)\cdot\ldots\cdot2\cdot1}\\
&=\left(-\frac{1}{4}\right)^r\frac{(2r)!}{r!r!}\\
&=\left(-\frac{1}{4}\right)^r\binom{2r}{r}\\
\end{align*}
Comment:
*
*In (1) we factor out $\left(-\frac{1}{2}\right)^r$. Please note that $2^r$ is part of the denominator (check corr. part (4) in your calculation)
*In (2) we fill in even factors to complete the numerator to $(2r)!$ and we also write these factors in the denominator as compensation.
Clever notation: In (1) it's convenient to use the compact notation of double factorials.We observe
$$(2r)!=(2r)!!(2r-1)!!=2^rr!(2r-1)!!$$
and using this notation in (1):
\begin{align*}
\binom{-\frac{1}{2}}{r}
&=\ldots\\
&=\left(-\frac{1}{2}\right)^r\frac{(2r-1)!!}{r!}\\
&=\left(-\frac{1}{2}\right)^r\frac{(2r)!}{r!(2r)!!}\\
&=\left(-\frac{1}{2}\right)^r\frac{(2r)!}{r!2^rr!}\\
&=\left(-\frac{1}{4}\right)^r\binom{2r}{r}\\
\end{align*}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1151233",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Prove that $a^2 + b^2 \geq 8$ if $ x^4 + ax^3 + 2x^2 + bx + 1 = 0 $ has at least one real root. If it is known that the equation
$$ x^4 + ax^3 + 2x^2 + bx + 1 = 0 $$
has a (real) root, prove the inequality
$$ a^2 + b^2 \geq 8. $$
I am stuck on this problem, though, it is a very easy problem for my math teacher. Anyway, I can't figure out.
| I assume that $a,b$ ar real.
Suppose that $a^2+b^2<8$ and the polynomial has the real root $\xi$. It follows that:
$$
0=\xi^4 + a\xi^3 + 2\xi^2 + b\xi + 1>\xi^4 + a\xi^3 + \frac{a^2+b^2}{4}\xi^2 + b\xi + 1=\xi^2\left(\xi+\frac{a}{2}\right)^2+\left(\frac{b}{2}\xi+1\right)^2
$$
But the sum of squares of real numbers is always positive, therefore $\xi$ has to be complex; contradiction. Thus we have $a^2+b^2\ge8$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1151480",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 4,
"answer_id": 2
} |
Deriving an expression for $\cos^4 x + \sin^4 x$
Derive the identity $\cos^4 x + \sin^4 x=\frac{1}{4} \cos (4x) +\frac{3}{4}$
I know $e^{i4x}=\cos (4x) + i \sin (4x)=(\cos x +i \sin x)^4$. Then I use the binomial theorem to expand this fourth power, and comparing real and imaginary parts, I conclude that $\cos^4 x + \sin^4 x = \cos (4x) + 6 \cos^2 (x) \sin^2 (x)$.
So now I need to show that $\cos (4x) + 6 \cos^2 (x) \sin^2 (x)=\frac{1}{4} \cos (4x) +\frac{3}{4}$, which has stumped me.
| $$\begin{align}
\cos^4x + \sin^4 x &= \left( \cos^2 x \right)^2 + \left( \sin^2 x \right)^2 \\[4pt]
&= \left( \frac{1 + \cos 2 x}{2}\right)^2 + \left( \frac{1-\cos 2x}{2} \right)^2 \\[4pt]
&= \frac{1}{2}\left( 1 + \cos^2 2 x \right) \\[4pt]
&= \frac{1}{2}\left( 1 + \frac{1+\cos 4 x}{2} \right) \\[4pt]
&= \frac{1}{4}\left(\; 3 + \cos 4 x \;\right)
\end{align}$$
From your particular stopping place, you could proceed thusly:
$$\begin{align}
\cos 4 x + 6 \sin^2 x \cos^2 x &= \cos 4 x + \frac32\cdot(2\sin x \cos x)^2 \\[4pt]
&= \cos 4 x + \frac32\cdot \sin^2 2 x \\[4pt]
&= \cos 4 x + \frac32\cdot \frac{1 - \cos 4 x}{2} \\[4pt]
&= \frac14\left(\;3 + \cos 4 x\;\right)
\end{align}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1153131",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 4,
"answer_id": 2
} |
Is there a simple expression for ${}_2F_1(a,a+\tfrac{1}{2};a+1;z)$? I have been searching through some books and also this
but I have not succeed.
I wonder if there is a simple equivalent form for ${}_2F_1(a,a+\tfrac{1}{2};a+1;z)$, in terms of elementary functions or other simpler functions.
In principle, $a>0$ but a solution for integer positive values of $a$ would be useful to me as well.
Thank you in advance.
| For the sake of simplicity, I'll assume $z$ is real and $0 < a \le 1/2$. First, suppose $a = 1/2$ and $z\neq 0,1$. Then
\begin{align}{}_2F_1\left(a, a+\frac{1}{2}; a + 1; z\right) &= {}_2F_1\left(\frac{1}{2},1;\frac{3}{2};z \right)\\
&= \frac{1}{i\sqrt{z}}\cdot i\sqrt{z}\,{}_2F_1\left(1,\frac{1}{2};\frac{3}{2};-(i\sqrt{z})^2\right)\\
&=\frac{1}{\sqrt{z}}\arctan(i\sqrt{z})\\
&= \frac{1}{2\sqrt{z}}\ln\frac{1 + \sqrt{z}}{1 - \sqrt{z}}.
\end{align}
Now suppose $a < 1/2$ and $z = 1$. Since $(a + 1) - a - (a + 1/2) = 1/2 - a > 0$, Gauss's summation formula gives
\begin{align}
{}_2F_1\left(a, a + \frac{1}{2};a + 1; z\right) &=
\frac{\Gamma(a+1)\Gamma\left(\frac{1}{2}-a\right)}{\Gamma(1)\Gamma\left(\frac{1}{2}\right)}\\
&= \frac{\Gamma(a+1)\Gamma\left(\frac{1}{2}-a\right)}{\sqrt{\pi}}.
\end{align}
In general, since $a + 1 > a > 0$, Euler's integral representation of ${}_2F_1$ gives
\begin{align}
{}_2F_1\left(a,a+\frac{1}{2};a+1;z\right) &= \frac{\Gamma(a+1)}{\Gamma(a)\Gamma(1)} \int_0^1 t^{a-1} (1 - zt)^{-a-1/2}\, dt\\
&= a\int_0^z u^{a-1}(1 - u)^{-a-1/2}\, \frac{du}{z^a}\\
&= az^{-a}B\left(a,\frac{1}{2}-a;z\right)
\end{align}
for $z \neq 0$, where the $B$ function in the last step is the incomplete Beta function.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1155081",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
Implicit differentiation of $(x^2 + y^2)^2 = (x-y)^2$ $$
(x^2 + y^2)^2 = (x-y)^2$$
Wolfram alpha yields this answer:
$$
y'(x) = \frac{(-2 x^3-2 x y^2+x-y)}{((2 x^2-1) y+x+2 y^3)}$$
But it's impossible to get $-y$ in the denominator
Actually, my answer is pretty much the same, except it's $+ y$ in the denominator
When differentiating the left side, you'll get:
$$(2x^2 + 2y^2) * (2x + 2yy') = 4x^3 + 4x^2 y y' + 4y^2x+4y^3 y'$$
dividing by 2 both sides and taking y' out, the left side would look:
$$y'(2x^2 y + 2y^3 + x + y)$$
as you can see, it's $+y$
so why does wolfram alpha says otherwise?
| You start off with
$$
(x^2+y^2)^2=(x-y)^2 \Longleftrightarrow x^4+2x^2y^2+y^4=x^2-2xy+y^2.
$$
Now pull everything with a $y$ in it to the LHS and terms only with $x$ to the RHS:
$$
y^4-y^2+2x^2y^2+2xy=x^2-x^4.
$$
Now implicitly differentiation everything (using the product rule twice):
$$
4y^3y'-2yy'+(4xy^2+4x^2yy')+(2y+2xy')=2x-4x^3.
$$
Now factor out the $y'$:
$$
y'(4y^3-2y+4x^2+2x) = 2x-4x^3-2y-4xy^2.
$$
Isolate $y'$ (and divide everything by $2$ once you have done this), and you will see that
$$
y'=\frac{x-2x^3-2xy^2-y}{2y^3-y+2x^2y+x},
$$
and this matches the output produced by Wolfram|Alpha.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1161323",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
How to factorise $x^4 - 3x^3 + 2$, so as to compute the limit of a quotient? Question:
Find the limit: $$\lim_{x \to 1}\frac{x^4 - 3x^3 + 2}{x^3 -5x^2+3x+1}$$
The denominator can be simplified to: $$(x-1)(x^2+x)$$
However, I am unable to factor the numerator in a proper manner (so that $(x-1)$ will cancel out)
I know upon graphing that the limit is $5\over4$. What should I do here?
Note: To be done without the use of L'Hospital Rule
| You may use Horner here. Since $x=1$ is a root of the nominator (you can check that easily) then you can deduce that down by applying a long division.
Hence your limit is deduced down to:
$$\lim_{x\rightarrow 1}\frac{x^4-3x^3+2}{x^3-5x^2+3x+1}=\lim_{x\rightarrow 1}\frac{(x-1)\left ( x^3-2x^2-2x-2 \right )}{\left ( x-1 \right )\left ( x^2-4x-1 \right )}=\frac{1-2-2-2}{1-4-1}=5/4$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1163127",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 0
} |
If $a\leq (b + c)/2$ with $a,b,c>0$, why $a^2\leq (b^2 + c^2)/2$? If $a\leq (b + c)/2$ with $a,b,c>0$, why $a^2\leq \frac{b^2 + c^2}{2}$? I can only see how to get $a^2\leq \frac{b^2+c^2 + 2ab}{4}$.
| because $(a+b)^2 \neq a^2+b^2.$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1165071",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 6,
"answer_id": 5
} |
Prove that $1^2+2^2+\cdots+n^2=\frac{n(n+1)(2n+1)}{6}$ for $n \in \mathbb{N}$. Problem: Prove that $1^2+2^2+\cdots+n^2=\frac{n(n+1)(2n+1)}{6}$ for $n \in \mathbb{N}$.
My work: So I think I have to do a proof by induction and I just wanted some help editing my proof.
My attempt:
Let $P(n)=1^2+2^2+\cdots+n^2=\frac{n(n+1)(2n+1)}{6}$ for $n \in \mathbb{N}$.
Then $$P(1)=1^2=\frac{1(1+1)(2+1)}{6}$$
$$1=\frac{6}{6}.$$
So $P(1)$ is true.
Next suppose that $P(k)=1^2+2^2+\cdots+k^2=\frac{k(k+1)(2k+1)}{6}$ for $k \in \mathbb{N}$. Then adding $(k+1)^2$ to both sides of $P(k)$ we obtain the following:
$$1^2+2^2+\cdots+k^2+(k+1)^2=\frac{k(k+1)(2k+1)}{6}+(k+1)^2$$
$$=\frac{2k^3+3k^2+k+6(k^2+2k+1)}{6}$$
$$=\frac{2k^3+9k^2+13k+6}{6}$$
$$=\frac{(k^2+3k+2)(2k+3)}{6}$$
$$=\frac{(k+1)(k+2)(2k+3)}{6}$$
$$=\frac{(k+1)((k+1)+1)(2(k+1)+1)}{6}$$
$$=P(k+1).$$
Thus $P(k)$ is true for $k \in \mathbb{N}$.
Hence by mathematical induction, $1^2+2^2+\cdots+n^2=\frac{n(n+1)(2n+1)}{6}$ is true for $n \in \mathbb{N}$.
| Your inductive assumption is such that the formula marked $\color{red}{\mathrm{red}}$ (several lines below) holds for $i=k$:
$$\sum^{i=k}_{i=1} i^2=\frac{k(k+1)(2k+1)}{6} $$
You need to prove that for $i=k+1$: $$\sum^{i=k+1}_{i=1} i^2=\color{blue}{\frac{(k+1)(k+2)(2k+3)}{6}}$$
To do this you cannot use: $$\sum^{i=k}_{i=n} i^2=\color{red}{\frac{n(n+1)(2n+1)}{6}} $$ as this is what you are trying to prove.
So what you do instead is notice that:
$$\sum^{i=k+1}_{i=1} i^2= \underbrace{\frac{k(k+1)(2k+1)}{6}}_{\text{sum of k terms}} + \underbrace{(k+1)^2}_{\text{(k+1)th term}}$$
$$\sum^{i=k+1}_{i=1} i^2= \frac{k(k+1)(2k+1)}{6}+\frac{6(k+1)^2}{6}$$
$$\sum^{i=k+1}_{i=1} i^2= \frac{(k+1)\left(k(2k+1)+6(k+1)\right)}{6}$$
$$\sum^{i=k+1}_{i=1} i^2= \frac{(k+1)(2k^2+\color{green}{7k}+6)}{6}=\frac{(k+1)(2k^2+\color{green}{4k+3k}+6)}{6}=\frac{(k+1)\left(2k(k+2)+3(k+2)\right)}{6}=\color{blue}{\frac{(k+1)(k+2)(2k+3)}{6}}\quad \forall \space k \in \mathbb{N}$$
Which is the relation we set out to prove. So the method is to substitute $i=k+1$ into the formula you are trying to prove and then use the inductive assumption to recover the $\color{blue}{\mathrm{blue}}$ equation at the end.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1166027",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
"answer_count": 3,
"answer_id": 2
} |
Modulo polynomial in ring theory Let $x^4-16$ be an element of the polynomial ring $E= \mathbb{Z}[x]$ and use the bar notation to denote passage to the quotient ring $\mathbb{Z}[x]/(x^4-16)$. Find a polynomial of degree $\leq 3$ that is congruent to $7x^{13} -11x^9 + 5x^5-2x^3+3$ modulo $x^4-16$. Can anyone help understand how to mod out polynomials.
| ${\rm mod}\,\ x^4-2^4\!:\,\ x^4\equiv 2^4\,\Rightarrow\, \color{#c00}{x^{4n}}\equiv 2^{4n}$
$\! \begin{align}{\rm therefore}\quad &\ \ 7\,\color{#c00}{x^{12}}x -11\,\color{#c00}{x^8}\,x + 5\,\color{#c00}{x^4}\,x-2x^3+3\\
\equiv &\ \ 7\, 2^{12}\,x - 11\, 2^8\, x + 5\, 2^4\, x -2\,x^3+3\quad {\bf QED}\end{align}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1166851",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 6,
"answer_id": 4
} |
Show that $PF.PG=b^2$ in a hyperbola If the normal at P to the hyperbola $\frac {x^2}{a^2}-\frac {y^2}{b^2}=1$ meets the transverse axis in G and the conjugate axis in G' and CF be the perpendicular to the normal from the center C then show that $$PF.PG=b^2\space and \space PF.PG'=a^2.$$ We know that the equation of the normal at parametric point $P\equiv(a\sec\theta,b\tan\theta)$ is given by
$$a\,x\cos\theta+b\,y\cot\theta=a^2+b^2$$
Equation of the transverse axis is $y=0$ and that of the conjugate axis is $x=0$.
Hence
$$G\equiv(\frac {a^2+b^2}{a}\sec\theta,0), \, G'\equiv(0,\frac {a^2+b^2}{b}\tan\theta)$$
C is the origin and CF is perpendicular to the normal. Hence
$$ CF\equiv b\,x\sec\theta-a\,y\tan\theta=0\implies y=\frac {b\,x\sec\theta}{a \tan\theta}$$
Substituting this in the equation of the normal we get
$$a\,x\cos\theta+b \,\cot\theta\,\frac {b\,x\sec\theta}{a\,\tan\theta}=a^2+b^2\ $$
implies
$$ (a^2 \cos\theta \sin^2\theta+ b^2\cos\theta)x=(a^2+b^2)\sin^2\theta$$
After finding the coordinates of F, the calculation becomes very complicated. So is there an easier way to approach the problem? I would love to see a pure geometric solution to this problem.
| $\rm PF.PG$ = Power of P wrt the circle with CG as diameter, the equation of whose is:
$$x(x-(1/a)(a^2+b^2)\sec\theta)+y^2=0$$
So, $${\rm PF.PG}=|a\sec\theta(a\sec\theta-(1/a)(a^2+b^2)\sec\theta)+b^2\tan^2\theta|=b^2$$
Similiarly:
$\rm PF.PG'$ = Power of P wrt the circle with CG' as diameter, the equation of whose is:
$$x^2+y(y-(1/b)(a^2+b^2)\tan\theta)=0$$
So, $${\rm PF.PG}=|a^2\sec^2\theta+b\tan\theta(b\tan\theta-(1/b)(a^2+b^2)\tan\theta)|=a^2$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1167958",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
} |
Use $\sum_{n=1}^\infty \frac{1}{n^4} = \frac{\pi^4}{90}$ to compute $\sum_{n=1}^\infty \frac{(-1)^n}{n^4}$ Is it possible to use the fact that $\sum_{n=1}^\infty \frac{1}{n^4} = \frac{\pi^4}{90}$ to compute $\sum_{n=1}^\infty \frac{(-1)^n}{n^4}$?
| Hint:
Let
$$E=\sum_{n=1, n=even}^\infty \frac{1}{n^4}$$
$$O=\sum_{n=1, n=odd}^\infty \frac{1}{n^4}$$
You are given that $E+O=\frac{\pi^4}{90}$ and want to find $E-O$.
To do this note that
$$E=\sum_{n=1, n=even}^\infty \frac{1}{n^4}=\sum_{k=1}^\infty \frac{1}{(2k)^4}=\frac{1}{16}\sum_{k=1}^\infty \frac{1}{k^4}=\frac{1}{16}\frac{\pi^4}{90}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1169861",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Finding square root of $-5-12i$ by formula and by De Moivre's Theorem I was trying to obtain the square root of $-5-12i$ by the formula for square root (given below) and also by De Moivre's theorem and verify that both give the same result. But the two results are somehow not matching for this complex number. I am writing my solution below in two cases for each method:
Case - I:
As given on pg - 3 of Complex Analysis - Newman and Bak, the equation $(x+iy)^2 = a+ib$ has the solution: $x=\pm\sqrt{\frac{a+\sqrt{a^2+b^2}}{2}}$ and $y=\pm\sqrt{\frac{-a+\sqrt{a^2+b^2}}{2}}.($sign $b)$
Putting $a=-5, b=-12$ & $($sign $b) = -ve$ in the above formula for $x$ & $y$, we get that the square roots of $-5-12i$ are $2-3i$ & $-2+3i$
Case - II:
By De Moivre's theorem, we know that given $z = r(cos \theta + i sin \theta)$; its $n$th root $z_k$ is given by $z_k = r^{1/n}(cos (\frac{\theta + 2k \pi}{n}) + i sin (\frac{\theta + 2k \pi}{n}))$, where $k=0,1,...,n-1$
Here, $z=-5-12 i = r(cos \theta + i sin \theta)$. Thus, $r=13$ and $\theta = atan(\frac{-12}{-5}) = 1.176005207$ (in radian)
Hence, $z_k = \sqrt{13}(cos (\frac{\theta + 2k \pi}{2}) + i sin (\frac{\theta + 2k \pi}{2}))$, where $k=0,1$
For $k=0$, $z_0 = \sqrt{13} (cos (\frac{\theta}{2}) + i sin (\frac{\theta}{2})) = \sqrt{13} (0.8320502943 + i 0.5547001962) = 3 + 2i$
For $k=1$, $z_1=\sqrt{13} (cos (\frac{\theta + 2 \pi}{2}) + i sin (\frac{\theta + 2 \pi}{2})) = \sqrt{13} (cos (\pi + \frac{\theta}{2}) + i sin (\pi + \frac{\theta}{2})) = - \sqrt{13} (cos (\frac{\theta}{2}) + i sin (\frac{\theta}{2})) = - \sqrt{13} (0.8320502943 + i 0.5547001962) = -3 - 2i$
Here, the square roots of $-5-12i$ are $3+2i$ and $-3-2i$
I think there must be some error in the solution because the square roots are coming out to be different. Thanks...
| Toby's answer has a minor miscue. Also, in the first chapter of Bruce Palka's "An Introduction To Complex Function Theory", an easier algorithm is given.
Let $\;z = -5 - 12i,\;$ and let $\alpha = \text{Arg}(z).$ This means that $\;\alpha \,\in (-\pi,\pi]\;$ and $\;z = |z|[\cos(\alpha) + i\sin(\alpha)].$
Further, let $\beta = \alpha/2 \Rightarrow \cos(\beta) \geq 0 \;\text{and}\; \sin(\beta) < 0 \Leftrightarrow \sin(\alpha) < 0.$
Then the principal $\sqrt{z} = +\sqrt{|z|}[\cos(\beta) + i\sin(\beta)],\;$ while the other square root will be $-1 \times \;$ the principal square root.
In this case, $|z| = 13, \cos(\alpha) = -5/13, \;\text{and}\; \sin(\alpha) < 0.\;$ Therefore, by a half-angle formula,
$\cos\beta = \sqrt{\frac{1 - 5/13}{2}} = \frac{2}{\sqrt{13}}.\;$
Further, $\;\sin(\beta) < 0 \Rightarrow \sin(\beta) = \frac{-3}{\sqrt{13}}.$
Thus, the principal sqrt is $\;\sqrt{13}[\frac{2}{\sqrt{13}} + i\frac{-3}{\sqrt{13}}] \;=\; (2 - 3i).$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1171492",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
How to prove $x^3-y^3 = (x-y)(x^2+xy+y^2)$ without expand the right side? I can prove that $x^3-y^3 = (x-y)(x^2+xy+y^2)$ by expanding the right side.
*
*$x^3-y^3 = (x-y)x^2 + (x-y)(xy) + (x-y)y^2$
*$\implies x^3 - x^2y + x^2y -xy^2 + xy^2 - y^3$
*$\implies x^3 - y^3$
I was wondering what are other ways to prove that $x^3-y^3 = (x-y)(x^2+xy+y^2)$
| $x^3-y^3=x^2(x-y)+yx^2-y^3=x^2(x-y)+yx(x-y)+xy^2-y^3=...$ (Simly insert $x^2y-yx^2$, then insert $y^2x-xy^2$,...)
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1172119",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 12,
"answer_id": 7
} |
Indefinite integral question: $\int \frac{1}{x\sqrt{x^2+x}}dx$ How can I solve this integral: $$\int \frac{1}{x\sqrt{x^2+x}}dx$$ I first completed the square and got: $$\int \frac{1}{x\sqrt{(x+\frac{1}{2})^2-\frac{1}{4}}}dx$$ Then I factored out 1/4 and got: $$2\int \frac{1}{x\sqrt{(2x+1)^2-1}}dx$$ Then I substituted $2x+1$ with $t$ and got: $$2\int \frac{1}{(t-1)\sqrt{t^2-1}}dt$$ I'm not sure what to do next. Please give me a hint ;)
| Here is an approach.
$$
\begin{align}
\int \frac{1}{x \sqrt{x^2+x}}dx &=\int \frac{1}{\sqrt{1+\frac{1}{x}}}\frac{dx}{x^2}\\\\
&=-\int \frac{1}{\sqrt{1+u}}\:du \quad (u=1/x)\\\\
&=-2\sqrt{u+1}+C\\\\
&=-2\sqrt{\frac{x+1}{x}}+C
\end{align}
$$ where $C$ is any constant.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1172599",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 4,
"answer_id": 1
} |
Critical numbers of the function: $x\sqrt{5-x}$ Let f(x) = $$\displaystyle f(x) = x\sqrt{5-x} $$
On the interval: [-6,4]
Critical numbers are the the values of x in the domain of f for which f'(x) = 0 or f'(x) is undefined.
Derivative of the function:
$$ \frac{1}{2} \cdot x (5-x)^{\frac{-1}{2}} \cdot -1$$
$$ \frac {\frac{-x}{2}}{\sqrt{5-x}}$$
$f'(x) = 0$, when $x = 0, $ and is undefined when x= 5
Plugging in the roots of the derivative function and the end points of the interval into the original function:
\begin{align*}
f(0) & = 0\\
f(5) & = 0\\
f(-6) & = -6\sqrt{11}\\
f(4) & = 4 \cdot 1 = 4
\end{align*}
So why is the 4 not the absolute maximum value?
p.s. I assumed the first term goes to zero when taking a derivative by the product rule. I confused d/dx x = 1, with any number d/dd = 0
| it would be much easier to work with $$y^2 = x^2(5-x), x \le 5.$$ taking the difference we find $$2y \, dy = (10x - 3x^2)\, dx $$ setting $\frac{dy}{dx} = 0$ gives you $$x = 0,\, x = \frac{10}3.$$ the spurious critical number $x = 0$ is an artifact of squaring. we know that $y = x \sqrt 5+\cdots$ for $x = 0+\cdots.$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1175845",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
Multiplication and Addition tables the following: What would be the addition and multiplication tables of $Z_2[x]/\langle x^2 + x\rangle$?
I know how to do the addition and multiplication tables for normal modular arithmetic, butam not sure about this.
| The tables are
$$
\begin{array}{r|c c c c}
+ & 0 & 1 & x & 1+x \\
\hline
0 & 0 & 1 & x & 1+x \\
1 & 1 & 0 & 1+x & x \\
x & x & 1+x & 0 & 1 \\
1 + x & 1+x & x & 1 & 0
\end{array}
$$
and
$$
\begin{array}{r|c c c c}
\times & 0 & 1 & x & 1+x \\
\hline
0 & 0 & 0 & 0 & 0 \\
1 & 0 & 1 & x & 1+x \\
x & 0 & x & x & 1 \\
1 + x & 0 & 1+x & 0 & 1+x
\end{array},
$$
where we have identified the elements $0,1,x,1+x$ with their cosets $0 + I$, etc, where $I$ is the ideal in question.
But how do we arrive at these tables? Well, the first step is to find manageable representatives for the cosets $p + I$ for any polynomial $p$. These representatives are $0,1,x,$ and $1+x$.
The way we find these representatives is by taking an arbitrary polynomial
$$
p(z) = a_nx^n + ... + a_0 \in \mathbb{Z}/2\mathbb{Z}[x].
$$
Notice however, that $x + x^2 \in I$, so $x + I = x + (x^2+x) I = x^2 + I.$ Similarly, we also have that $x^n + I = x^{n+1} + I$ for any $n \geq 1.$ This relation shows that
$$
p(x)+I = p(z) + a_nx^n + a_nx^{n-1} + I = (a_{n}+a_{n+1})x^{n-1} + a_{n-2}x^{n-2} +... +I.
$$
Now, apply this trick again with the polynomial on the right hand side of the above. We gradually reduce the degree of our representative of $p(x) + I$. When we can no longer apply this trick, we must have that the highest degree monomial in our representative $q(x)$ must be less than $2$. Otherwise, we could simply apply the trick again. Thus $q$ must be one of $0,1, x,$ or $1+x$, as these are the polynomials of degree less than $2$ in $\mathbb{Z}/2\mathbb{Z}[x]$. To compute the addition and multiplication tables, we now just take representatives of any two cosets and reduce them in this fashion.
For example, if we are computing what $(x+I)(1+x+I)$ is, we know by definition that this is equal to $x+x^2 + I$. Since $x+x^2 \in I$, $x+x^2 + I = I = 0 +I$. Thus $x(1+x) = 0$ in the ring $\mathbb{Z}/2\mathbb{Z}[x]/I$. Similarly, if we are computing $(x+I)(x+I)$ in the ring, we see by definition that this is $x^2 + I$. Since $x^2 + x \in I$ $x^2 + I = x^2 + x^2 + x + I = x + I$. Thus $x^2 = x$ in $\mathbb{Z}/2\mathbb{Z}[x]/I.$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1176045",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Prove by induction that $3^{3n+1} + 2^{n+1}$ is divisible by 5 How do I do this? I've tried using logarithms, factoring, but nothing seems to work.
| Other users have already outlined the proof by induction, but I think a direct proof is interesting as well.
By Fermat's little theorem (or by inspection), we know that
$$3^4 \equiv 2^4 \equiv 1 \pmod{5}$$
This means that powers of $3$, modulo $5$, are determined by the exponents remainder when divided by $4$.
$$3^{3n+1} \equiv \begin{cases}3 \pmod{5} &\mbox{if }\,\, n \equiv 0 \pmod{4}\\
1 \pmod{5} &\mbox{if }\,\, n \equiv 1 \pmod{4}\\
2 \pmod{5} &\mbox{if }\,\, n \equiv 2 \pmod{4}\\
4 \pmod{5} &\mbox{if }\,\, n \equiv 3 \pmod{4}\\
\end{cases}$$
$$2^{n+1} \equiv \begin{cases}2 \pmod{5} &\mbox{if }\,\, n \equiv 0 \pmod{4}\\
4 \pmod{5} &\mbox{if }\,\, n \equiv 1 \pmod{4}\\
3 \pmod{5} &\mbox{if }\,\, n \equiv 2 \pmod{4}\\
1 \pmod{5} &\mbox{if }\,\, n \equiv 3 \pmod{4}\\
\end{cases}$$
For each possible residue for $n$ modulo $4$, the sum is $0$ modulo $5$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1176215",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 8,
"answer_id": 4
} |
If ${a}$ is an arbitrary integer, then prove that ${360|a^2(a^2-1)(a^2-4)}$. I think I have solved the problem. I want to verify my proof, since I don't have a teacher to help me.
Proof:
Since, ${360=8*45}$ and ${gcd(45,8)=1}$, hence if we can prove that ${45|a^2(a^2-1)(a^2-4)}$ and ${8|a^2(a^2-1)(a^2-4)}$ , then we are done proving the question.
Now, to prove divisibility by ${45}$, we see that ${45=9*5}$, also ${gcd(9,5)=1}$ and thus we can prove divisibility by ${45}$ if we can prove ${9|a^2(a^2-1)(a^2-4)}$ and ${5|a^2(a^2-1)(a^2-4)}$.
Since ${a}$ is an arbitrary integer, it may be represented as ${2k}$ or ${2k+1}$.
So, ${a^2(a^2-1)(a^2-4)=a^2(a+1)(a-1)(a+2)(a-2)}$ may be represented as ${4k^2(2k+1)(2k-1)(2k+2)(2k-2)=16k^3(2K-1)(2k+1)(k+1)}$, which is divisible by ${8}$.
Similarly, for ${a=2k+1}$, we prove the divisibility by substituting $a$ by ${2k+1}$ to get ${(2k+1)^2(2k+2)(2k)(2k+3)(2k-1)=4k(k+1)(2k+1)^2(2k-1)(2k+3)}$ which is divisible by ${8}$.
Therefore ${8|a^2(a^2-1)(a^2-4)}$ .
Similarly, to prove divisibility by $9$, we represent $a$ by ${3k}$, ${3k+1}$ and ${3k+2}$ and prove divisibility for all of them in a similar manner.
Same method is used to prove divisibility by $5$ by represent $a$ by $5k,5k+1,5k+2,5k+3,5k+4$ and prove divisibility for all of these cases.
(While doing in paper, I actually prove these divisibilities.)
Hence, ${8|a^2(a^2-1)(a^2-4)}$ and ${45|a^2(a^2-1)(a^2-4)}$.
Since, $gcd(8,45)=1$, so ${(8*45)|a^2(a^2-1)(a^2-4)}$ .
Therefore, it is proved that ${360|a^2(a^2-1)(a^2-4)}$ .
Now, please tell me if my method is correct or not. If incorrect, then please show me how to solve it properly otherwise please show me how to improve my solution. Thank you! :)
| Perhaps not as formal, but easy to follow if you look at it like this:
$$360 = 2^3 \cdot 3^2 \cdot 5$$
and
$$a^2(a^2-1)(a^2-4) = (a-2)(a-1)a^2(a+1)(a+2)$$
Notice that this is a product of five consecutive integers (with the middle one squared), so:
*
*certainly one of the factors will be divisible by 5
*either $a$ is divisble by $3$ (so $a^2$ by $3^2$) and if not, both factors in either $(a-1)(a+2)$ or $(a-2)(a+1)$ are divisible by 3 (so the product by $3^2$)
*if $a-2$ is even, then...; if not, then... - can you finish the reasoning for the divisibility by $2^3$?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1176779",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 1,
"answer_id": 0
} |
Given the $ x+y+z =3$ Prove that $ x^2+y^2+z^2 \geq 3xyz$ Given the $ x+y+z =3$ and x, y and z are positive numbers. How to prove that $ x^2+y^2+z^2 \geq 3xyz$.
I tried many methods but I failed.
I did the AM-HM in-equality, but failed.
$\frac{\frac{x}{yz} + \frac{y}{zx} + \frac{z}{xy}}{3} \geq \frac{3}{\frac{xy}{z}+\frac{yz}{x}+\frac{zx}{y}} $.
The denominator of RHS of the equation must have, minimum value. So it boils down to
$\frac{x^2y^2+y^2z^2+z^2x^2}{xyz} \leq 3 $. Don't know how to proceed after that.
Taking partial derivative in range $(0,3)$ for the function $(x^2+y^2+z^2)$ also didn't give me the required insight.
Thankyou.
| LHS= $(x^2+y^2+z^2)(\frac {x+y+z} {3}) \geq 3(xyz)^{\frac {2} {3}}.(xyz)^{\frac {1} {3}}$=RHS. (I have used here AM-GM inequality twice.)
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1179131",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
Let $\theta=\frac{2 \pi}{67}$ consider the rotation matrix $A$. What is $A^{2010}$?
Let $\theta=\frac{2 \pi}{67}$. Consider the matrix
$$A = \begin{pmatrix}
\cos\theta & \sin\theta\\
-\sin \theta& \cos \theta
\end{pmatrix} $$
Then the matrix $A^{2010}$ is?
My approach
$$A^2 = \begin{pmatrix}
\cos2\theta & \sin2\theta\\
-\sin 2\theta& \cos 2\theta
\end{pmatrix} $$
$2010$ is a multiple of $67$. So I'm trying to convert in such format by taking $A$ matrix as power of $2$ and multiplying. But $2048$ is the nearest $2$ power term. So help please. I dont know if you can understand what I did!! Sorry for being not explaining properly
| Observe that when you multiply the matrices
$$\begin{pmatrix}
\cos\theta & \sin\theta\\
-\sin \theta& \cos \theta
\end{pmatrix} *
\begin{pmatrix}
\cos\phi & \sin\phi\\
-\sin \phi& \cos \phi
\end{pmatrix}
=
\begin{pmatrix}
\cos(\theta+\phi) & \sin(\theta+\phi)\\
-\sin (\theta+\phi)& \cos (\theta+\phi)
\end{pmatrix}.$$
(You can show this with the multiangle formulae for $\sin$ and $\cos$.) So we have
$$A^{67} = \begin{pmatrix}
\cos67\theta & \sin67\theta\\
-\sin 67\theta& \cos 67\theta
\end{pmatrix}
=
\begin{pmatrix}
\cos2\pi & \sin2\pi\\
-\sin 2\pi& \cos 2\pi
\end{pmatrix}
=
\begin{pmatrix}
1 & 0 \\
0 & 1
\end{pmatrix} = I.$$
$$\therefore A^{2010} = (A^{67})^{30} = I.$$
(I tried to put the end bit as a 'spoiler' which you have to mouse over to view, but I couldn't work out how to!)
As mentioned by other people, applying the matrix
$$M_\theta = \begin{pmatrix}
\cos\theta & \sin\theta\\
-\sin \theta& \cos \theta
\end{pmatrix}$$
has the action of rotation a vector (anticlockwise) by $\theta$. Thus doing it repeatedly simply means that they sum: $M_\theta * M_\phi = M_{\theta + \phi}$.
Hopefully this helps! :)
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1179407",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 0
} |
Integrals, arc length I found this on my test today, and i didn't manage to solve it.
My question is.
How to find arc length of $f(x) = x^3/3$ from $x=1$ to $x=2$?
If I use formula for arc length which is $\ell = \int_a^b \sqrt{1 + (f'(x))^2} \, dx$ where is $f'(x)=x^2$ I will get $\ell = \int_1^2 \sqrt{1+x^4} \, dx$
I used everything I could (know), partition and substitution, but nothing works...
How to solve this definite integral?
$$\ell =\int_1^2 \sqrt{1+x^4} \, dx$$
| We have:
$$\ell = \int_{1}^{2}x\,\sqrt{x^2+\frac{1}{x^2}}\,dx=\int_{1}^{2}x\,\sqrt{\left(x-\frac{1}{x}\right)^2+2}\,dx$$
and by replacing $x-\frac{1}{x}$ with $t$ we have:
$$\ell = \frac{1}{4}\int_{0}^{3/2}\left(t+\sqrt{4+t^2}\right)^2\sqrt{\frac{2+t^2}{4+t^2}}\,dt$$
hence:
$$\ell = \frac{17\sqrt{17}}{48}-\frac{\sqrt{2}}{3}+\frac{1}{2}\int_{0}^{3/2}(2+t^2)^{3/2}(4+t^2)^{-1/2}\,dt $$
but:
$$\begin{eqnarray*} \frac{1}{2}\int_{0}^{3/2}(2+t^2)^{3/2}(4+t^2)^{-1/2}\,dt &=& \frac{1}{4}\int_{0}^{9/4}(2+t)^{3/2}(t(4+t))^{-1/2}\,dt\\&=&\sqrt{2}\int_{0}^{9/8}(1+t)^{3/2}((t+1)^2-1)^{-1/2}\\ &=& \frac{1}{4}\int_{0}^{9/4}(2+t)^{3/2}(t(4+t))^{-1/2}\,dt\\&=&\sqrt{2}\int_{1}^{17/8}t^{3/2}(t^2-1)^{-1/2}\,dt\end{eqnarray*}$$
that can be written as an incomplete beta function if we replace $t$ with $\frac{1}{u}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1179639",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
What are the ways to solve this integral? $$\int _{0}^{\frac{\pi}{2}}\frac{1}{5\sin x+3} dx.$$
I've tried the way of let $u=\tan \left( \frac{x}{2}\right) $ , but it's very complicated.
| No it's not very complicated, proceed carefully:
$$\scriptsize I=\int\frac{{\rm d}x}{3+5\sin x}=\int\frac{{\rm d}x}{3+5\left(\frac{2\tan(x/2)}{1+\tan^2(x/2)}\right)}=\int\frac{(1+\tan^2(x/2)){\rm d}x}{3(1+\tan^2(x/2))+10\tan (x/2)}=\int\frac{\sec^2(x/2){\rm d}x}{3\tan^2(x/2)+10\sin x+3}$$
Now, take $y=\tan(x/2)\implies {\rm d}y=\sec^2(x/2).(1/2).{\rm d}x$ using Chain Rule.
$$I=\int\frac{2{\rm d}y}{3y^2+10y+3}=\int\frac{2{\rm d}y}{3(y+5/3)^2-16/3}$$
Now take $z=\sqrt3(y+5/3),{\rm d}z=\sqrt3{\rm d}y$:
$$I=\frac2{\sqrt3}\int\frac{{\rm d}z}{z^2-(4/\sqrt3)^2}=\frac2{\sqrt3}.\frac1{2(4/\sqrt3)}\ln\left|\frac{z-4/\sqrt3}{z+4/\sqrt3}\right|+{\rm C}$$
So:
$$\int\frac{{\rm d}x}{3+5\sin x}=\frac14\ln\left|\frac{3(\tan (x/2)+5/3)-4}{3(\tan (x/2)+5/3)+4}\right|+{\rm C}=\frac14\ln\left|\frac{3\tan (x/2)+1}{3\tan (x/2)+9}\right|+{\rm C}$$
Now just put the limits:
$$\frac14\ln\left|\frac{3\tan (x/2)+1}{3\tan (x/2)+9}\right|\Bigg|_0^{\pi/2}=\frac14\ln\left|\frac{3+1}{3+9}\right|-\frac14\ln\left|\frac{0+1}{0+9}\right|=\frac14\ln\frac{4/12}{1/9}=\frac14\ln3\approx0.27465$$
Moral:
$$\text{You are just lazy.}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1183550",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 2
} |
Start from $259^2 + 1^2 = 34 \cdot 1973$ and use the descent procedure to write the prime 1973 as a sum of two squares.
Start from $259^2 + 1^2 = 34 \cdot 1973$ and use the descent procedure to write the prime 1973 as a sum of two squares.
How to solve it using fermat descent method?
| $34 \cdot 1973 = 259^2 + 1^2; \\ (21^2 + 1^2)(259^2 + 1^2)=(21 \cdot 259 +1)^2+(259-21)^2 \implies 13 \cdot 1973 = 7^2 + 160^2. $
$(7^2 + 4^2)(7^2 + 160^2)=(7 \cdot 7 +160 \cdot 4)^2+(160 \cdot 7-4 \cdot 7)^2 \implies 5 \cdot 1973 = 53^2 + 84^2. $
$(2^2 + 1^2)(53^2 + 84^2)=(2 \cdot 53 + 84 \cdot 1 )^2+(84 \cdot 2 -53 \cdot 1 )^2 \implies 1973 = 38^2 + 23^2.$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1183726",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Show that $\int_1^{\infty } \frac{(\ln x)^2}{x^2+x+1} \, dx = \frac{8 \pi ^3}{81 \sqrt{3}}$ I have found myself faced with evaluating the following integral: $$\int_1^{\infty } \frac{(\ln x)^2}{x^2+x+1} \, dx. $$
Mathematica gives a closed form of $8 \pi ^3/(81 \sqrt{3})$, but I have no idea how to arrive at this closed form. I've tried playing around with some methods from complex analysis, but I haven't had much luck (it has been a while). Does anyone have any ideas? Thanks in advance!
| Shocked, shocked! that there is no contour integration yet. So, without further ado...
Note that
$$f(x) = \frac{\log^2{x}}{x^2+x+1} \implies f \left ( \frac1{x} \right ) = x^2 f(x) $$
Thus,
$$\int_1^{\infty} dx \frac{\log^2{x}}{x^2+x+1} = \int_0^{1} \frac{\log^2{x}}{x^2+x+1} = \frac12 \int_0^{\infty} dx \frac{\log^2{x}}{x^2+x+1} $$
Now consider
$$\oint_C dz \frac{\log^3{z}}{z^2+z+1} $$
where $C$ is a keyhole contour of outer radius $R$ and inner radius $\epsilon$. Taking the limit as $R \to \infty$ and $\epsilon \to 0$, we get that the contour integral is equal to
$$\int_0^{\infty} dx \frac{\log^3{x} - (\log{x}+i 2 \pi)^3}{x^2+x+1} $$
or
$$-i 6 \pi \int_0^{\infty} dx \frac{\log^2{x}}{x^2+x+1} + 12 \pi^2 \int_0^{\infty} dx \frac{\log{x}}{x^2+x+1} +i 8 \pi^3 \int_0^{\infty} dx \frac{1}{x^2+x+1} $$
Note that the first integral is what we seek, the second integral is zero (by the same trick we applied above), and the third integral is relatively easy to find:
$$\int_0^{\infty} \frac{dx}{x^2+x+1} = \int_0^{\infty} \frac{dx}{(x+1/2)^2+3/4} = \frac{2}{\sqrt{3}} \left [\arctan{\frac{2}{\sqrt{3}} \left ( x+\frac12 \right )} \right ]_0^{\infty} = \frac{2 \pi}{3 \sqrt{3}}$$
The contour integral is also equal to $i 2 \pi$ times the sum of the residues at the poles of the integrand, which are at $z_+ = e^{i 2 \pi/3}$ and $z_- = e^{i 4 \pi/3}$. The sum of the residues is
$$\frac{-i 8 \pi^3/27}{i \sqrt{3}} + \frac{-i 64 \pi^3/27}{-i \sqrt{3}} = \frac{56 \pi^3}{27 \sqrt{3}}$$
Then
$$-i 6 \pi \int_0^{\infty} dx \frac{\log^2{x}}{x^2+x+1} = i 2 \pi \frac{56 \pi^3}{27 \sqrt{3}} - i 8 \pi^3 \frac{2 \pi}{3 \sqrt{3}} = -i \frac{32 \pi^4}{27 \sqrt{3}}$$
Thus,
$$\int_1^{\infty} dx \frac{\log^2{x}}{x^2+x+1} = \frac12 \int_0^{\infty} dx \frac{\log^2{x}}{x^2+x+1} = \frac{8 \pi^3}{81 \sqrt{3}} $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1186002",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 5,
"answer_id": 0
} |
Evaluation of $\int\frac{1}{x^4-5x^2+16}dx$
Evaluation of $\displaystyle \int\frac{1}{x^4-5x^2+16}\,dx$
$\bf{My\; Try::}$ Given $$\displaystyle \int\frac{1}{x^4-5x^2+16}dx = \frac{1}{8}\int\frac{\left(x^2+4\right)-\left(x^2-4\right)}{x^4-5x^2+16}\,dx$$
So We get $$\displaystyle = \frac{1}{8}\int\frac{x^2+4}{x^4-5x^2+16}\,dx-\frac{1}{8}\int\frac{x^2-4}{x^4-5x^2+16}\,dx$$
So We get $$\displaystyle = \frac{1}{8}\int\frac{1+\frac{4}{x^2}}{\left(x-\frac{4}{x}\right)^2+\left(\sqrt{3}\right)^2}dx-\frac{1}{8}\int\frac{1-\frac{4}{x^2}}{\left(x+\frac{4}{x}\right)^2-\left(\sqrt{13}\right)^2}\,dx$$
Now Using $$\displaystyle \left(x-\frac{4}{x}\right)=t$$ and $$\displaystyle \left(1+\frac{4}{x^2}\right)dx = dt$$ in First Integral and
Using $$\displaystyle \left(x+\frac{4}{x}\right)=u$$ and $$\displaystyle \left(1-\frac{4}{x^2}\right)\,dx = du$$ in Second Integral.
So We Get $$\displaystyle = \frac{1}{8}\int\frac{1}{t^2+\left(\sqrt{3}\right)^2}dt-\frac{1}{8}\int\frac{1}{u^2-\left(\sqrt{13}\right)^2}\,du$$
So We Get $$\displaystyle = \frac{1}{8\sqrt{3}}\tan^{-1}\left(\frac{x^2-4}{\sqrt{3}x}\right)-\frac{1}{16\sqrt{13}}\ln \left|\frac{x^2-\sqrt{13}x+4}{x^2+\sqrt{13}x+4}\right|+{C}$$
My question is can we solve the above question any other Method, If yes then
please explain here. Thanks
| Factor the denominator and expand the rational function into partial fractions:
\begin{eqnarray*}
\frac{1}{x^{4}-5x^{2}+16} &=&\frac{1}{\left( x^{2}-\sqrt{13}x+4\right)
\left( x^{2}+\sqrt{13}x+4\right) } \\
&=&\frac{1}{104}\frac{\sqrt{13}x+13}{x^{2}+\sqrt{13}x+4}-\frac{1}{104}\frac{
\sqrt{13}x-13}{x^{2}-\sqrt{13}x+4}.
\end{eqnarray*}
Then complete the square in the denominators
\begin{equation*}
\frac{1}{x^{4}-5x^{2}+16}=\frac{\sqrt{13}}{104}\frac{x+13/\sqrt{13}}{\left(
x+\sqrt{13}/2\right) ^{2}+3/4}-\frac{\sqrt{13}}{104}\frac{x-13/\sqrt{13}}{
\left( x-\sqrt{13}/2\right) ^{2}+3/4},
\end{equation*}
and use, respectivelly, the substitutions $$u=x+\sqrt{13}/2,\qquad \text{and}\qquad v=x-\sqrt{13}/2.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1186356",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 1
} |
Center of mass of Right angle Trapezoid Given bases $a$, $b$, and the height $h$.
Get the $M(x,y)$ coordinates formula from point $O(0,0)$, where $M$ is center of mass. Wiki has a formula for $M(y) = \frac{h}{3}\frac{2a+b}{a+b}$. And I'm interested in how to find that formula (also for $M(x)$).
The median $c$ is dividing the Trapezoid into 2 equal surfaces. I need to draw $m$ in right place $y$, so all 4 surfaces stay equal each other.
Here's my try:
$y(\frac{a+m}{2})=(h-y)(\frac{m+b}{2})$
$\frac{a+m}{2}=(\frac{h}{y}-1)(\frac{m+b}{2})$
$y=\frac{h(m+b)}{a+2m+b}$
on the other look
$2y(\frac{a+m}{2})=h(\frac{a+b}{2})$
$y=\frac{h(a+b)}{2(a+m)} = \frac{h(m+b)}{a+2m+b}$
From the last equality we have: $m^2 = \frac{a^2+b^2}{2}$
But I still cant get that formula back for $y$
| 1190922
$x=\frac{{ \begin{vmatrix}0&0&1\\a&0&1\\0&h&1\\\end{vmatrix} · (0+a+0) }+{ \begin{vmatrix}b&h&1\\0&h&1\\a&0&1\\\end{vmatrix} · (b+0+a) }}{ 3\left(\begin{vmatrix}0&0&1\\a&0&1\\0&h&1\\\end{vmatrix} + \begin{vmatrix}b&h&1\\0&h&1\\a&0&1\\\end{vmatrix}\right) }$
$y=\frac{{ \begin{vmatrix}0&0&1\\a&0&1\\0&h&1\\\end{vmatrix} · (0+0+h) }+{ \begin{vmatrix}b&h&1\\0&h&1\\a&0&1\\\end{vmatrix} · (h+h+0) }}{ 3\left(\begin{vmatrix}0&0&1\\a&0&1\\0&h&1\\\end{vmatrix} + \begin{vmatrix}b&h&1\\0&h&1\\a&0&1\\\end{vmatrix}\right) }$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1190922",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
eigenvectors of a matrix with known eigenvalues $$
\begin{matrix}
1 & 1 & 0 & 0 \\
0&1&0&0\\
0&0&0&2\\
0&0&2&0
\end{matrix}
$$
This matrix has eigenvalues 1, 2, & -2, I've solved the eigenvectors for 1 & 2, but not for -2. Subtracting (-2) from the diagonal produces:
$$
\begin{matrix}
3 & 1 & 0 & 0 \\
0&3&0&0\\
0&0&2&2\\
0&0&2&2
\end{matrix}
$$
Solving the augmented matrix (not shown) gives me
$$
\begin{matrix}
0 \\
0\\
-1\\
1\\
\end{matrix}
$$
but the answer is
$$
\begin{matrix}
0 \\
0\\
1\\
-1\\
\end{matrix}
$$
I've checked multiple sources and I'm wrong but I don't understand why.
| you can take advantage of the block structure of the matrix. you have two $2 \times 2$ matrices $B$ and $C$ where $B = \pmatrix{1&1\\0&1}, C = \pmatrix{0&2\\2&0}.$ $B$ has eigenvalues $1,1$ but only one eigenvector $(1,0)^T$ and $C$ has eigenvalues $-2, 2.$ the eigenvector corresponding to $-2$ is $(1,-1)^T$ and an eigenvector corresponding to $2$ is $(1, 1)^T.$
the eigenvalues of the mattrix $\pmatrix{B&0\\0&C}$ are $1,1, -2, 2.$ an eigenvector corresponding to the eigenvalue $1$ is $\pmatrix{1&0&0&0}^T,$ an eigenvector corresponding to the eigenvalue $-2$ is $\pmatrix{0&0&1&-1}^T$ and an
eigenvector corresponding to the eigenvalue $2$ is $\pmatrix{0&0&1&1}^T$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1191247",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Proving $2(\sqrt{n} - 1) < \sum\limits_{i=1}^n\frac{1}{\sqrt i}$ I'm trying to prove $$2(\sqrt{n} - 1) < \sum_{i=1}^n\frac{1}{\sqrt i}$$ (Which is the opposite pretty much of Prove by induction that $\sum_{i = 1}^{n} \frac{1}{\sqrt{i}} \leq 2\sqrt{n} - 1$)
And I'm encountering some troubles I fail to see through.
My attempt is by induction, and for $n=1$ it's clear, then if I assume for $k$ I get that for $k+1$:
$$2(\sqrt{k+1} - 1) < \sum_{i=1}^{k+1}(\frac{1}{\sqrt i})$$
$$\sum_{i=1}^{k+1}(\frac{1}{\sqrt i}) > 2(\sqrt{k}-1)+\frac{1}{\sqrt{k+1}}$$
Which would mean I would like to find that:
$$2(\sqrt{k}-1)+\frac{1}{\sqrt{k+1}} - 2(\sqrt{k+1} - 1) > 0$$
But this is what I get:
$$\frac{1}{\sqrt{k+1}}+2\sqrt{k}-2\sqrt{k+1}=\frac{1}{\sqrt{k+1}}+2\frac{\left(\sqrt{k}-\sqrt{k+1}\right)\left(\sqrt{k}+\sqrt{k+1}\right)}{\sqrt{k}+\sqrt{k+1}}=$$
$$ \frac{1}{\sqrt{k+1}}+2\frac{k-k-1}{\sqrt{k}+\sqrt{k+1}}=\frac{1}{\sqrt{k+1}}-\frac{2}{\sqrt{k}+\sqrt{k+1}}=$$
$$ \frac{\sqrt{k}+\sqrt{k+1}-2\sqrt{k+1}}{\sqrt{k}\left(\sqrt{k}+\sqrt{k+1}\right)}<0)
$$
So that is not true, how can I find that $\sum_{i=1}^{k+1}(\frac{1}{\sqrt i}) > 2(\sqrt{k+1} - 1)$ if what I get from my induction hypothesis doesn't help me since it's already smaller than what I'm searching for?
| This is just to illustrate a non-inductive method to proving the inequality:
\begin{align}\sum_{i = 1}^n \frac{1}{\sqrt{i}} &= 2\sum_{i = 1}^n \frac{1}{2\sqrt{i}} \\ & =2\sum_{i = 1}^n \frac{1}{\sqrt{i} + \sqrt{i}} \\
&> 2\sum_{i = 1}^n \frac{1}{\sqrt{i+1} + \sqrt{i}} \\
&= 2\sum_{i = 1}^n (\sqrt{i+1} - \sqrt{i})\\
&= 2[(\sqrt{2} - \sqrt{1}) + (\sqrt{3} - \sqrt{2}) + \cdots + (\sqrt{n+1} - \sqrt{n})]\\
&= 2(\sqrt{n+1} - 1),
\end{align}
and $2(\sqrt{n+1} - 1) > 2(\sqrt{n} - 1)$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1192428",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 3,
"answer_id": 1
} |
Express $\sqrt{3}\sin\theta - \cos\theta$ as: $a\cos (\theta + \alpha) $ Express $\sqrt{3}\sin\theta - \cos\theta$ as: $a\cos (\theta + \alpha) $
Can someone please explain to me how to go about doing this?
| If
\begin{align*}
\cos\varphi & = \frac{b}{\sqrt{a^2 + b^2}}\\
\sin\varphi & = \frac{a}{\sqrt{a^2 + b^2}}
\end{align*}
then
$$a\sin\theta - b\cos\theta = -\sqrt{a^2 + b^2}\cos(\theta + \varphi)$$
since
\begin{align*}
-\sqrt{a^2 + b^2}\cos(\theta + \varphi) & = -\sqrt{a^2 + b^2}(\cos\theta\cos\varphi - \sin\theta\sin\varphi)\\
& = -\sqrt{a^2 + b^2}\left(\cos\theta \cdot \frac{b}{\sqrt{a^2 + b^2}} - \sin\theta \cdot \frac{a}{\sqrt{a^2 + b^2}}\right)\\
& = -b\cos\theta + a\sin\theta\\
& = a\sin\theta - b\cos\theta
\end{align*}
In the expression, $\sqrt{3}\sin\theta - \cos\theta$, $a = \sqrt{3}$ and $b = 1$, so $$a^2 + b^2 = \sqrt{3}^2 + 1^2 = 3 + 1 = 4$$ Thus, $\sqrt{a^2 + b^2} = \sqrt{4} = 2$. Hence,
\begin{align*}
\cos\varphi & = \frac{b}{\sqrt{a^2 + b^2}} = \frac{1}{2}\\
\sin\varphi & = \frac{a}{\sqrt{a^2 + b^2}} = \frac{\sqrt{3}}{2}
\end{align*}
Therefore,
$$\varphi = \arccos\left(\frac{1}{2}\right) = \arcsin\left(\frac{\sqrt{3}}{2}\right) = \frac{\pi}{3}$$
so we obtain
$$\sqrt{3}\sin\theta - \cos\theta = -2\cos\left(\theta + \frac{\pi}{3}\right)$$
Note: We could also express the answer in the form
$$\sqrt{3}\sin\theta - \cos\theta = \sin\left(\theta - \frac{\pi}{6}\right)$$
if we use the formula
$$a\sin\theta - b\cos\theta = \sqrt{a^2 + b^2}\sin(\theta - \varphi)$$
where
\begin{align*}
\cos\varphi & = \frac{a}{\sqrt{a^2 + b^2}}\\
\sin\varphi & = \frac{b}{\sqrt{a^2 + b^2}}
\end{align*}
with $a = \sqrt{3}$, $b = 1$, and
$$\varphi = \arcsin\left(\frac{1}{2}\right) = \arccos\left(\frac{\sqrt{3}}{2}\right) = \frac{\pi}{6}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1192729",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 2
} |
Taylor series of $x/(x^2-4x+5)$ I'm supposed to find the Taylor series of this function (I can choose to center it at any A I want):
$$f(x)= x/(x^2-4x+5)$$
When I derivate, it only gets more and more confusing. How can I make any sense out of this?
| The first step is to complete the square: $x^2-4x+5 = (x-2)^2 + 1$. This suggests centering the expansion at $x-2$, since then the formula for the sum of a geometric series gives
$$
\frac{1}{1 + (x-2)^2} = \sum_{i=0}^\infty (-1)^i (x-2)^{2i}.
$$
Writing $x = (x-2) + 2$ we deduce
$$
\frac{x}{1 + (x-2)^2} = \sum_{i=0}^\infty (-1)^i [2(x-2)^{2i} + (x-2)^{2i+1}].
$$
The first few terms are
$$
\frac{x}{1 + (x-2)^2} = 2 + (x-2) - 2(x-2)^2 - (x-2)^3 + 2(x-2)^4 + (x-2)^5 - 2(x-2)^6 - (x-2)^7 + \cdots.
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1193204",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
The order of accuracy of the implicit Euler method is equal to $1$ I want to show that the order of accuracy of the implicit Euler method is equal to $1$.
That's what I have tried:
We have the initial value problem
$\left\{\begin{matrix}
y'(t)=f(t,y(t)) &, a \leq t \leq b \\
y(a)=y_0 &
\end{matrix}\right.$
Using Taylor's expasion, we have $y(t^{n+1})=y(t^n)+hy'(t^n)+\frac{h^2}{2}y''(\xi_n)$.
From the implicit Euler method we have $y^{n+1}=y^n+hf(t^{n+1},y^{n+1})$.
Thus:
$$|y(t^{n+1})-y^{n+1}|=|y(t^n)-y^n+h(f(t^n,y(t^n))-f(t^{n+1},y^{n+1}))+ \frac{h^2}{2}y''(\xi_n)| \leq |y(t^n)-y^n|+h|y(t^n)-y^{n+1}|+ \frac{h^2}{2}|y''(\xi_n)|=|y(t^n)-y^n|+h|y^{n+1}-y(t^n)|+ \frac{h^2}{2}|y''(\xi_n)| \leq |y(t^n)-y^n|+h|y^{n}-y(t^n)|+ \frac{h^2}{2}|y''(\xi_n)|=(1+h)|y(t^n)-y^n|+ \frac{h^2}{2}|y''(\xi_n)| \\\leq (1+h)^{n+1} |y(t^0)-y(0)|+ \sum_{i=0}^n (1+h)^i \frac{h^2}{2}|y''(\xi_{n-i})|=\frac{h^2}{2} \sum_{i=0}^n (1+h)^i |y''(\xi_{n-i})| \overset{M:=\max_{i=0, \dots, n} |y''(\xi_i)|}{\leq } \frac{Mh^2}{2} \sum_{i=0}^n (1+h)^i= \frac{Mh^2}{2} \left( \frac{(1+h)^{n+1}-1}{h}\right)=\frac{Mh}{2}((1+h)^{n+1}-1) \leq \frac{Mh}{2}(e^{(n+1)h}-1) \overset{(\star)}{\leq } \frac{Mh}{2} (e^{b-a}-1)=CM \frac{h}{2}$$
$(\star): t^{n+1}=a+(n+1)h \\ t^{n+1} \leq b \\ \Rightarrow a+(n+1)h \leq b \Rightarrow (n+1)h \leq b-a$
Therefore, the order of accuracy is at least $1$.
We consider the initial value problem:
$\left\{\begin{matrix}
y'(t)=2t &, 0 \leq t \leq 1 \\
y(0)=0 &
\end{matrix}\right.$
Its real solution is $y(t)=t^2, 0 \leq t \leq 1$.
$y^{n+1}=y^n+2ht^{n+1}=y^n+2h (n+1)h=y^n+ 2h^2(n+1) $
$y^1=2h^2=1 \cdot 2 \cdot h^2$
$y^2=2h^2+4h^2=6h^2= 3 \cdot 2 \cdot h^2$
$y^3=6h^2+6h^2=12h^2=3 \cdot 4 \cdot h^2$
$y^4= 12h^2+8h^2=20h^2=4 \cdot 5 \cdot h^2$
We see that $y^n=n(n+1)h^2$
For $n=N$ we have $y^N=N(N+1)h^2=(N+1)h=1+h$.
$\epsilon^N=|y(t^N)-y^N|=|y(1)-(1+h)|=|1-1-h|=h$.
So the order of accuracy is exactly $1$.
Could you tell me if it is right?
| Your first line should be, using $y(t^n)=y(t^{n+1})-h·f(t^{n+1}, y(t^{n+1}))+\frac{h^2}{2}·y''(\xi_n)$ as the appropriate Taylor expansion,
$$
|y(t^{n+1})-y^{n+1}|
=\left|y(t^n)-y^n+h·\bigl(f(t^{n+1},y(t^{n+1}))-f(t^{n+1},y^{n+1})\bigr)+ \frac{h^2}{2}·y''(\xi_n)\right|
\\
\leq |y(t^n)-y^n|+h·L·|y(t^{n+1})-y^{n+1}|+ \frac{h^2}{2}·|y''(\xi_n)|
$$
where $L$ is the Lipschitz constant in $y$-direction of $f$. So that
$$
|y(t^{n+1})-y^{n+1}|\leq \frac{1}{1-L·h}·\left(|y(t^n)-y^n|+ \frac{h^2}{2}·|y''(\xi_n)|\right)
$$
Everything else looks correct at least in principle.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1194443",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Find all prime numbers p such that both numbers $4p^2+1$ and $6p^2+1$ are prime numbers? I tried $p$ for $2, 3$ and $5$ and they are not primes for both cases. How can I find all these prime numbers that satisfy those conditions?
| You might want to double-check your numbers: $4 \times 5^2 + 1 = 4 \times 25 + 1 = 101$, which is prime, and $6 \times 5^2 + 1 = 6 \times 25 + 1 = 151$, which is also prime.
But you're right about $2$ and $3$: $4 \times 2^2 + 1 = 17$, which is prime, but $6 \times 2^2 + 1 = 25$ which is obviously composite; and $4 \times 3^2 + 1 = 37$, which is prime, but $6 \times 3^2 + 1 = 55$, which is also composite.
Maybe if $p \neq 5$ then either $5 \mid (4p^2 + 1)$ or $5 \mid (6p^2 + 1)$? Well, if $p$ is a prime other than $5$, then $p \equiv 2, 3 \textrm{ or } 4 \pmod 5$. If $p \equiv 4 \pmod 5$ then $p^2 \equiv 1 \pmod 5$ and $4p^2 + 1 \equiv 0 \pmod 5$. But if $p \equiv 2 \textrm{ or } 3 \pmod 5$, then $p^2 \equiv 4 \pmod 5$ and $6p^2 + 1 \equiv 0 \pmod 5$.
Ergo, no other prime can satisfy both conditions.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1195009",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
} |
Various evaluations of the series $\sum_{n=0}^{\infty} \frac{(-1)^n}{(2n+1)^3}$ I recently ran into this series:
$$\sum_{n=0}^{\infty} \frac{(-1)^n}{(2n+1)^3}$$
Of course this is just a special case of the Beta Dirichlet Function , for $s=3$.
I had given the following solution:
$$\begin{aligned}
1-\frac{1}{3^3}+\frac{1}{5^3}-\cdots &=\sum_{n=0}^{\infty}\frac{(-1)^n}{\left ( 2n+1 \right )^3} \\
&\overset{(*)}{=} \left ( 1+\frac{1}{5^3}+\frac{1}{9^3}+\cdots \right )-\left ( \frac{1}{3^3}+\frac{1}{7^3}+\frac{1}{11^3}+\cdots \right )\\
&=\sum_{n=0}^{\infty}\frac{1}{\left ( 4n+1 \right )^3} \; -\sum_{n=0}^{\infty}\frac{1}{\left ( 4n+3 \right )^3} \\
&= -\frac{1}{2\cdot 4^3}\psi^{(2)}\left ( \frac{1}{4} \right )+\frac{1}{2\cdot 4^3}\psi^{(2)}\left ( \frac{3}{4} \right )=\frac{1}{2\cdot 4^3}\left [ \psi^{(2)}\left ( 1-\frac{1}{4} \right )-\psi^{(2)}\left ( \frac{1}{4} \right ) \right ]\\
&=\frac{1}{2\cdot 4^3}\left [ 2\pi^3 \cot \frac{\pi}{4} \csc^2 \frac{\pi}{4} \right ] \\
&=\frac{\pi^3 \cot \frac{\pi}{4}\csc^2 \frac{\pi}{4}}{4^3}=\frac{\pi^3}{32}
\end{aligned}$$
where I used polygamma identities and made use of the absolute convergence of the series at $(*)$ in order to re-arrange the terms.
Any other approach using Fourier Series, or contour integration around a square, if that is possible?
| Differentiating twice the logarithm of the Weierstrass representation of sine gives
$$ \sum\limits_{n=-\infty}^{\infty} {1\over (z+n)^2}=\frac{\pi^{2}}{\sin^{2}(\pi z)} $$
(as i've been answered in here.)
Now differentiate once more and consider $z=\frac{1}{4}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1195285",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 5,
"answer_id": 1
} |
Find solution of $(1-x^2)y''-xy'+p^2y=0, p \in \mathbb{R}$ The following differential equation is given:
$$(1-x^2)y''-xy'+p^2y=0, p \in \mathbb{R}$$
*
*Find the general solution of the differential equation at the interval $(-1,1)$ (with the method of power series).
*Are there solutions of the differential equation that are polynomials?
That's what I have tried:
We are looking for a solution of the form $y(x)= \sum_{n=0}^{\infty} a_n x^n$ with radius of convergence $R>0$.
$$y'(x)= \sum_{n=1}^{\infty} n a_n x^{n-1}= \sum_{n=0}^{\infty} (n+1) a_{n+1} x^n$$
$$y''(x)= \sum_{n=1}^{\infty} (n+1)n a_{n+1}x^{n-1}= \sum_{n=0}^{\infty} (n+2)(n+1) a_{n+2} x^n$$
$$-x^2y''(x)= \sum_{n=0}^{\infty} -n(n-1)a_n x^n$$
We have:
$$\sum_{n=0}^{\infty} \left[ (n+2)(n+1) a_{n+2}-n(n-1) a_n-na_n+p^2a_n\right]x^n=0, \forall x \in (-R,R)$$
It has to hold:
$$(n+2)(n+1) a_{n+2}+ \left[ -n(n-1)-n+p^2\right] a_n=0 \\ \Rightarrow (n+2)(n+1)a_{n+2}+ \left[ -n(n-1+1)+p^2 \right]a_n=0 \\ \Rightarrow a_{n+2}=\frac{n^2-p^2}{(n+2)(n+1)} a_n \forall n=0,1,2, \dots$$
For $n=0: a_2=-\frac{p^2}{2}a_0$
For $n=1: a_3= \frac{1-p^2}{2 \cdot 3} a_1$
For $n=2: a_4= \frac{(2^2-p^2)(-p^2)}{2 \cdot 3 \cdot 4 } a_0$
For $n=3: a_5=\frac{(3^2-p^2)(1-p^2)}{2 \cdot 3 \cdot 4 \cdot 5}a_1$
For $n=4: a_6=\frac{(4^2-p^2)(2^2-p^2)(-p^2)}{2 \cdot 3 \cdot 4 \cdot 5 \cdot 6}a_0$
For $n=5: a_7=\frac{(5^2-p^2)(3^2-p^2)(1-p^2)}{2 \cdot 3 \cdot 4 \cdot 5 \cdot 6 \cdot 7}$
So:
$$a_{2n}=\frac{\prod_{i=0}^{n-1} (2i)^2-p^2}{(2n)!} a_0$$
$$a_{2n+1}=\frac{\prod_{i=0}^{n-1} (2i+1)^2-p^2}{(2n+1)!} a_1$$
So the solution $y$ can be written as follows:
$$y(x)=\sum_{n=0}^{\infty} a_{2n}x^{2n}+ \sum_{n=0}^{\infty}a_{2n+1}x^{2n+1}$$
Is it right so far? If so, then do we have to find for the power series $\sum_{n=0}^{\infty} a_{2n}x^{2n}$ and $\sum_{n=0}^{\infty}a_{2n+1}x^{2n+1}$ the radius of convergence and show that they define functions that are infinitely many times differentiable?
If so, then do we have to take the ratio
$$\left| \frac{a_{2(n+1)} x^{2(n+1)}}{a_{2n} x^{2n}}\right|$$
? If yes, could we deduce something from the latter about the radius of convergence?
| Those ratios simplify rather nicely, so you can indeed deduce something about the radius of convergence. But in fact there is a general theory that says there should be analytic solutions in any region (in the complex plane) where the coefficients are analytic and the leading coefficient is nonzero. In this case the only bad points are $x = \pm 1$ where the leading coefficient is $0$, so the radius of convergence should be (at least) 1.
Also note what happens for integer values of $p$.
By the way (though this is not really connected to the power-series method), the substitution $x = \sin(z)$ produces something nice (equivalently, substitute $y(x) = u(\arcsin(x))$ in to the differential equation and simplify).
EDIT: For the ratio test, note that as $n \to \infty$, if $p$ is not an integer
$$ \dfrac{a_{2(n+1)} x^{2(n+1)}}{a_{2n} x^{2n}} = \dfrac{((2n)^2-p^2) x^2}{(2n+2)(2n+1)} \to x^2 $$
and similarly for $a_{2(n+1)+1} x^{2(n+1)+1}/(a_{2n+1} x^{2n+1})$, so the radii of convergence of the series are $1$.
If $p$ is an integer, one of the factors $(2i)^2 - p^2$ or $(2i+1)^2 - p^2$ is $0$, which makes either $\sum_{n} a_{2n} x^{2n}$ or $\sum_n a_{2n+1} x^{2n+1}$ a polynomial.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1197701",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Finf $f(x)$ which is a second degree polynomial, such that $f(1)=0$ and $f(x) = f(x-1)$ I must find a function $f(x) = ax^2+bx+c$ such that:
$$f(1) = a+b+c=0\\f(x)=f(x-1)\implies ax^2+bx+c = a(x-1)^2+b(x-1)+c\implies\\ax^2+bx+c = ax^2+(-2a+b)x+a-b+c\implies\\a = a, b = -2a+b, c = a-b+c$$
but this results fo $a=b=c=0$. What am I doing wrong?
| You simply get that
$$\{f(x)=ax^2+bx+c\mid f(1)=0, f(x)=f(x-1)\}=\{0\}$$
which is actually correct.
But in other way, you can say that $$\{f(x)\in \mathbb R_2[x]\mid \deg f=2,\ f(1)=0,\ f(x)=f(x-1)\}=\emptyset$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1200463",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 2
} |
Finding the limit $\lim_{x\rightarrow \infty} \sqrt[3]{x+1}-\sqrt[3]{x}$ I am trying to find this limit
$$\lim_{x\rightarrow \infty} \sqrt[3]{x+1}-\sqrt[3]{x}$$
My so far method is this
*
*$f(x)>0.$
*$f^{\prime}(x)=\frac{1}{3\sqrt[3]{(x+1)^2}}-\frac{1}{3\sqrt[3]{x^2}}<0.$
*For every $0<y<1$ the equation $f(x)=y$ has a unique solution(found in maple):
$$x=\frac{1}{3}\cdot \frac{\frac{1}{3}y^2(3y^2+\sqrt{-3y^4+12y})+\frac{1}{6}\frac{3y^2+\sqrt{-3y^4+12y}}{y}-y^4-2y}{y}$$
The previous facts implies that the limit is $0$.
I am wondering if there is any way easier than this to find the limite.
thanks in advance.
| Hint:-
$$0<\left(\sqrt[3]{x+1}-\sqrt[3]{x}\right)=\displaystyle\frac{1}{\left(\sqrt[3]{(x+1)^2}+\sqrt[3]{x(x+1)}+\sqrt[3]{x^2}\right)}<\displaystyle\frac{1}{3\sqrt[3]{x^2}}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1200544",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 6,
"answer_id": 4
} |
Solve this system of equation Solve this system of equations for real $x$ and $y$:
*
*$5x\left(1+\dfrac{1}{x^2+y^2}\right)=12$
*$5y\left(1-\dfrac{1}{x^2+y^2}\right)= 4$
I juggled with those equations and got $x-y+\dfrac{x+y}{x^2+y^2}=\dfrac{8}{5}$, from where I guessed a solution $(2,1)$.
But I don't know how to approach mathematically.
Please help me.
| $$\left\{ {\begin{array}{*{20}{c}}{5x\left( {1 + \frac{1}{{{x^2} + {y^2}}}} \right) = 12}\\{5y\left( {1 - \frac{1}{{{x^2} + {y^2}}}} \right) = 4}\end{array}} \right.$$
If you sum the above equations, you can find formula of a line all the points on it are solutions to that system of equations;
$$\left\{ {\begin{array}{*{20}{c}}{1 + \frac{1}{{{x^2} + {y^2}}} = \frac{{12}}{{5x}}}\\{1 - \frac{1}{{{x^2} + {y^2}}} = \frac{4}{{5y}}}\end{array}} \right.$$
Here is the formula of the line:
$$1 = \frac{6}{{5x}} + \frac{2}{{5y}}$$
Obviously, zero value for $x$ or $y$ is not in the domain of solutions to the equations. As verification, all the results above can be found by this formula: $(x,y)=(\frac{2}{5},-\frac{1}{5})$, $(2,1)$ and $(1,-2)$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1200638",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
How come, in this problem, the maximum product is always achieved using only $2$s and $3$s? Consider the following problem.
Given a number $N$, write it as a sum $n = n_1 + n_2 + \cdots + n_k$, such that the product $p = n_1 \times n_2 \times \cdots \times n_k$ is maximized.
For example, $11$ can be written as $2 + 3 + 3 + 3$ or as $7 + 4$, yielding the products $2 \times 3 \times 3 \times 3 = 54$ and $7 \times 4 = 28$.
So in this case, the answer is is $2\times 3\times 3\times 3$.
For $17$ we have the same situation: $2 \times 3 \times 3 \times 3 \times 3 \times 3 = 486$.
So my question is, why is the maximum product always achieved by using only $2$s and $3$s?
| Any factor $1$ can be merged with some other factor, and the product will get larger. When $n>4$ then $(n-3)\cdot 3>n$, furthermore $4=2\cdot2$. Therefore the largest possible product can be written with $3$s and $2$s only.
As $2\cdot 2\cdot 2<3\cdot 3$ it is advantageous to replace three $2$s by two $3$s as often as possible. At the end there remain only $3$s and zero, one, or two $2$s, depending on the remainder of the given $n$ modulo $3$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1200745",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Prove that the algebraic expressions are equivalent.
$$\frac{3^{k+1}-1}{2} + 3^{k+1} = \frac{3^{k+2}-1}{2}$$
with steps make left hand side = right hand side by modifying one or both expressions
Thanks for your help guys, I solved it like this:
$$\frac{3^{k+1}-1}{2} + 3^{k+1} = \frac{3^{k+1}-1}{2} + \frac{2*3^{k+1}}{2} = \frac{3^{k+2}-1}{2} $$
| $${3^{k+1}-1\over 2}+3^{k+1}={3^{k+1}\over 2}-{1\over 2}+{2(3^{k+1})\over 2}=({3\over 2})3^{k+1}-{1\over 2}={3^{k+2}-1\over 2}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1201612",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Find integers $a,b,c,d$ such that is it possible for someone to run a computer search for me to turn up integers $a,b,c,d$ such that $$1+\sqrt{2}+\sqrt{3}+\sqrt{6}=\sqrt{a+\sqrt{b+\sqrt{c+\sqrt{d}}}}$$? Note: I know they exist, I just can't find them.
| Problem statement:
$$1+\sqrt{2}+\sqrt{3}+\sqrt{6}=\sqrt{a+\sqrt{b+\sqrt{c+\sqrt{d}}}}$$
Let's try this identity that you posted:
$$\sqrt{x}+\sqrt{y}=\sqrt{x+y+\sqrt{4xy}}$$
Use your variables:
$$\sqrt{3}+\sqrt{6}$$
$$=\sqrt{3+6+\sqrt{4\cdot 3\cdot 6}}$$
$$=\sqrt{9+\sqrt{72}}$$
$$\sqrt{1}+\sqrt{2}$$
$$=\sqrt{1+2+\sqrt{4\cdot 1 \cdot 2}}$$
$$=\sqrt{3+\sqrt{8}}$$
$$\sqrt{9+\sqrt{72}}+\sqrt{3+\sqrt{8}}$$
$$=\sqrt{\left(9+\sqrt{72}\right)+\left(3+\sqrt{8}\right)+\sqrt{4\cdot \left(9+\sqrt{72}\right)\cdot\left(3+\sqrt{8}\right)}}$$
$$=\sqrt{12+\sqrt{72}+\sqrt{8}+\sqrt{\left(36+\sqrt{1152}\right)\cdot\left(3+\sqrt{8}\right)}}$$
$$=\sqrt{12+\sqrt{72}+\sqrt{8}+\sqrt{108+3\sqrt{1152}+36\sqrt{8}+\sqrt{9216}}}$$
$$=\sqrt{12+8\sqrt{2}+\sqrt{108+144\sqrt{2}+\sqrt{9216}}}$$
$$=\sqrt{12+8\sqrt{2}+\sqrt{108+144\sqrt{2}+96}}$$
$$=\sqrt{12+8\sqrt{2}+\sqrt{204+144\sqrt{2}}}$$
$$=\sqrt{12+\sqrt{128}+\sqrt{204+\sqrt{41472}}}$$
Almost there!
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1201810",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 1,
"answer_id": 0
} |
Given $\frac{dy}{dx}=x^2+y^2$ and initial condition $\varphi (0)=1$, find the first 6 terms in the Taylor expansion solution $y=\varphi (x)$ Given $\frac{dy}{dx}=x^2+y^2$ and initial condition $\varphi (0)=1$, use the method of reduction to an integral equation and successive approximation to find the first 6 terms in the Taylor expansion solution $y=\varphi (x)$.
We have that if $f:D \rightarrow \mathbb{R}$ is continuous, $\varphi$ is defined and continuous on $I=\{x|x_0-h<x<x_0+h\}$ to $\mathbb{R}$, and $(x_0,y_0) \in D$ with $\varphi (x_0)=y_0$, then $\varphi$ is a solution of $\frac{d\varphi}{dx}=f[x,\varphi (x)]$ on $I$ only if $\varphi (x)=y_0+\int_{x_0}^{x} f[t,\varphi (t)] dt$ for $x \in I$. This is what I mean by reduction to an integral equation.
EDIT:
In case it helps anyone refine their answer, the answer should be $1+x+x^2+\frac{4}{3}x^3+\frac{7}{6}x^4+\frac{6}{5}x^5$. This comes out of the back of my book. Of course, my problem is that I cannot get to this answer.
| $$y = 1 + \int_0^x y^2 + t^2 \, dt$$ we will define $$y_{n+1} = 1 + \int_0^x (t^2 + y_n^2)\, dt = 1 + \frac 13 x^3 + \int_0^x y_n^2 \, dt,\, y_0 = 1.$$ so $$y_1 = 1+ \frac 13 x^3 + \int_0^x \,dt = 1 + x + \frac 13 x^3 \tag 1\\ y_2 = 1 + \frac 13 x^3 + \int_0^x \left(1 + t + \frac 13 t^3\right)^2\, dt = 1 + \frac 13 x^3 + \int_0^x \left(1 + 2t + t^2 + \frac 23 t^3 + \frac 23 t^4 + \frac 19 t^6\right)$$
$$y_2 = 1 + x+ x^2 + \frac23 x^3+\frac 1{6}x^4+\frac 2{15}x^5+\frac 1{63}x^7\tag 2\\
y_2^2 = 1 + 2x + 3x^2 + \frac{10} 3 x^3+\frac 83 x^4 + \cdots\\
y_3 = 1 + \frac 13 x^3 + \int_0^x \left(1 + 2t + 3t^2 + \frac{10} 3 t^3+\frac 83 t^4 + \cdots \right)dt $$
$$y_3=1 + x+x^2 + \frac 43 x^3+\frac 5{17}x^4 + \frac 8{15}x^5+\cdots\tag 3\\
y_3^2 = 1 + 2x + 3x^2 + \frac{14}3x^3+\cdots$$
$$y_4 = 1 + \frac 13 x^3 + \int_0^x \left(1 + 2t + 3t^2 + \frac{14} 3 t^3+ \cdots \right)dt$$
$$
\\y_4 = 1 + x + x^2 + \frac43x^3 + \frac 76 x^4 + \cdots \tag 4
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1205397",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
Prove the logarithmic inequality Prove that:
$(\log_{24}{48})^2+(\log_{12}{54})^2>4$
I tried to put $t=\log_23$ and get the equation $6t^4+32t^3+22t^2-84t-74>0$.
But I can't do anything with it...
| Let $l=\log_{2}3$, so $\log_{3}2=\frac{1}{l}$ and $l<\frac{8}{5}$ since $2^8>3^5$, and $l>\frac{11}{7}$ since $2^{11}<3^7$.
Then $\displaystyle\log_{24}48=\frac{\log_{2}48}{\log_{2}24}=\frac{4+l}{3+l}>\frac{4+8/5}{3+8/5}=\frac{28}{23}\;\;\;$ and $\hspace{.36 in}\displaystyle\log_{12}54=\frac{\log_{3}54}{\log_{3}12}=\frac{3+\log_{3}2}{1+2\log_{3}2}=\frac{3+1/l}{1+2/l}=\frac{3l+1}{l+2}>\frac{3(\frac{11}{7})+1}{\frac{11}{7}+2}=\frac{8}{5}$,
so $\displaystyle \left(\log_{24}48\right)^2+\left(\log_{12}54\right)^2>\left(\frac{28}{23}\right)^2+\left(\frac{8}{5}\right)^2>4$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1206452",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Find a vector whose image under $T$ is the vector $b$
Hey everyone, I'm having some trouble solving this problem. To find the vector can I multiply the matrix $A$ by the column vector $[a, b, c]$ and set that equal to the vector $b$? or do I have to row reduce to solve this problem? Thanks for the help
| An alternative way, using advanced methods: Scalar product and cross product.
Setting the columns from $A$
$$u = \left( {\begin{array}{*{20}{c}}
4 \\
{ - 3} \\
5
\end{array}} \right),v = \left( {\begin{array}{*{20}{c}}
4 \\
{ - 6} \\
{ - 3}
\end{array}} \right),w = \left( {\begin{array}{*{20}{c}}
3 \\
2 \\
6
\end{array}} \right)$$
then I do first check, if $\{ u,v,w\}$ is a set of linear independent vectors.
Because of
$$v \times w = \left( {\begin{array}{*{20}{c}}
4 \\
{ - 6} \\
{ - 3}
\end{array}} \right) \times \left( {\begin{array}{*{20}{c}}
3 \\
2 \\
6
\end{array}} \right) = \left( {\begin{array}{*{20}{c}}
{ - 36 + 6} \\
{ - 9 - 24} \\
{8 + 18}
\end{array}} \right) = \left( {\begin{array}{*{20}{c}}
{ - 30} \\
{ - 33} \\
{26}
\end{array}} \right)$$
and
$$u \cdot (v \times w) = \left( {\begin{array}{*{20}{c}}
4 \\
{ - 3} \\
5
\end{array}} \right) \cdot \left( {\begin{array}{*{20}{c}}
{ - 30} \\
{ - 33} \\
{26}
\end{array}} \right) = - 120 + 99 + 130 = 109 \ne 0$$
it is the case: $\{ u,v,w\} $ is a set of linear independent vectors.
Can be used as basis. Therefore $\{ u,v,w,b\}$ is a set of linear dependent
vectors. There exists a unique linear combination:
$$xu + yv + zw = b$$
These numbers are computed by the formulae:
$$\begin{gathered}
x = \frac{{b \cdot (v \times w)}}{{u \cdot (v \times w)}} \hfill \\
y = \frac{{b \cdot (u \times w)}}{{v \cdot (u \times w)}} \hfill \\
z = \frac{{b \cdot (u \times v)}}{{w \cdot (u \times v)}} \hfill \\
\end{gathered} $$
Known as Cramer's rule, written in an old fashion style.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1209197",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Elementary proof that $\sum_{n \geq 1} \frac{(-1)^n}{\sqrt{n}}+i \frac{1}{n^2}$ is conditionally convergent. I must prove that the complex series $$\sum_{n \geq 1} \frac{(-1)^n}{\sqrt{n}}+i \frac{1}{n^2}$$ is conditionally convergent. The catch is, I'm not supposed to use anything "too advanced", meaning I only have in hands the ratio test and the root test, which are pretty much useless, since the failure of the ratio test implies the failure of the root test. We can use the comparison test too.
I managed to prove it using Raabe's test, here goes:
$$\sum_{n \geq 1} \left| \frac{(-1)^n}{\sqrt{n}}+i \frac{1}{n^2}\right| = \sum_{n \geq 1} \sqrt{\frac{1}{n}+\frac{1}{n^4}},$$ the ratio test and root test both fail for $\sqrt{\frac{1}{n}+\frac{1}{n^4}} = \frac{\sqrt{n^3+1}}{n^2}$, but by Raabe's test: $$\lim_{n \to +\infty} n\left(1 - \frac{n^2}{(n+1)^2}\sqrt{\frac{(n+1)^3+1}{n^3+1}}\right) = \frac{1}{2} < 1,$$ the series diverges (actually, I used Wolfram Alpha to get this limit, cheated again).
Now, we know that the complex series converges if and only if $\sum_{n \geq 1}\frac{(-1)^n}{\sqrt{n}}$ and $\sum_{n \geq 1}\frac{1}{n^2}$ both converge. The second one obviously converge. And for the first one, both the ratio test and the root test fail, but by Raabe's test: $$\lim_{n \to +\infty} n\left(1+\sqrt{\frac{n}{n+1}}\right) = +\infty > 1,$$and the series converges.
Question: Is there an elementary approach to this?
| Since $$\sqrt{\frac{1}{n} + \frac{1}{n^4}} \ge \sqrt{\frac{1}{n}} = \frac{1}{\sqrt{n}}$$ and $\sum_{n = 1}^\infty \frac{1}{\sqrt{n}}$ diverges, by direct comparison the series
$$\sum_{n = 1}^\infty \left|\frac{(-1)^n}{\sqrt{n}} + \frac{i}{n^2}\right| = \sum_{n = 1}^\infty\sqrt{\frac{1}{n} + \frac{1}{n^4}}$$ diverges.
To see $\sum_{n = 1}^\infty (-1)^n/\sqrt{n}$ converges, let $s_n$ be the sequence of partial sums of the series $\sum_{n = 1}^\infty (-1)^{n-1}/\sqrt{n}$. Then
$$s_{2n} = s_{2n-2} + \frac{1}{\sqrt{2n-1}} - \frac{1}{\sqrt{2n}} \ge s_{2n-2}$$
and
$$s_{2n} = 1 - \left(\frac{1}{\sqrt{2}} - \frac{1}{\sqrt{3}}\right) - \cdots - \left(\frac{1}{\sqrt{2n-2}} - \frac{1}{\sqrt{2n-1}}\right) - \frac{1}{\sqrt{2n}} \le 1.$$
for all $n\in \Bbb N$. So the sequence $\{s_{2n}\}$ is increasing and bounded above by $1$. By the monotone convergence theorem, $s_{2n}$ converges. Let $s = \lim_{n\to \infty} s_{2n}$. Then
$$\lim_{n\to \infty} s_{2n+1} = \lim_{n\to \infty} \left(s_{2n} + \frac{1}{\sqrt{2n+1}}\right) = \lim_{n\to \infty} s_{2n} = s.$$
Since $\lim_{n\to \infty} s_{2n} = s = \lim_{n\to \infty} s_{2n+1}$, $\{s_n\}$ converges (with limit $s$). So $\sum_{n = 1}^\infty (-1)^n/\sqrt{n}$ converges (to $-s$).
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1209397",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Help to resolve a Double Integral I'm doing a workout guide about double integrals and I came across an exercise that I could not resolve for a while.
$$\int_0^2\int_1^2 \frac{x}{\sqrt{1+x^2+y^2}} \,\mathrm dx\,\mathrm dy$$
I guess that the easier order of integration is $dxdy$, because if I try to integrate respect to $y$ first, I would have to deal with the integral of a root of $1+x^2+y^2$, while outside of the root there's no $y$.
I tried Integration by substitution (with $u = 1 + x^2 + y^2$ and $du = 2xdx$) to solve the inner integral, but then I felt I couldn't solve the outside integral.
Any hints you can give me?.
| The antiderivative may be deduced, but it takes a little patience. I prefer to integrate by parts rather than do a trig sub right away.
$$\begin{align}\int dx \sqrt{x^2+a^2} &= x \sqrt{x^2+a^2} - \int dx \frac{x^2}{\sqrt{x^2+a^2}}\\ &= x \sqrt{x^2+a^2} - \int dx \sqrt{x^2+a^2} + a^2 \int \frac{dx}{\sqrt{x^2+a^2}} \end{align}$$
Now a trig sub is less work:
$$\implies 2 \int dx \sqrt{x^2+a^2} = x \sqrt{x^2+a^2} + a^2 \int dt \, \sec{t} $$
or
$$\int dx \sqrt{x^2+a^2} = \frac12 x \sqrt{x^2+a^2} + \frac12 a^2 \log{\left (x+\sqrt{a^2+x^2}\right )} + C$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1209636",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 3,
"answer_id": 0
} |
Let $a,b,c$ be the nonnegative real numbers such that $a+b+c=1$. Prove that $\sqrt{a+\frac{(b-c)^2}4}+\sqrt b+\sqrt c\le\sqrt3$
Let $a,b,c$ be the nonnegative real numbers such that $a+b+c=1$. Prove that
$$\sqrt{a+\frac{(b-c)^2}4}+\sqrt b+\sqrt c\le\sqrt3$$
I first wrote $a$ as $1-b-c$ and substituted it in main inequality
$$\sqrt{4(1-b-c)+(b-c)^2}+2\sqrt b+2\sqrt c\le2\sqrt 3$$
I tried to find some relation between this inequality and QAGH inequalities, but I couldn't. Then I come up with an idea to square main inequality several times to cancel square roots and then to reduce it to sum of squares, but I am sure there is a better way to prove it. What is the best way to prove it?
| I use Cauchy-Schwarz inequality we have
$$\left(\sqrt{a+\dfrac{(b-c)^2}{4}}+\sqrt{b}+\sqrt{c}\right)^2
\le \left(a+\dfrac{(b-c)^2}{4}+\dfrac{(\sqrt{b}+\sqrt{c})^2}{2}\right)(1+2)
$$
$$\Longleftrightarrow \left(a+\dfrac{(b-c)^2}{4}+\dfrac{(\sqrt{b}+\sqrt{c})^2}{2}\right)\le 1$$
since $1-a=b+c$
$$\Longleftrightarrow \dfrac{(b-c)^2}{4}+\dfrac{(\sqrt{b}+\sqrt{c})^2}{2}\le b+c$$
$$\Longleftrightarrow \dfrac{(b-c)^2}{4}\le\dfrac{(\sqrt{b}-\sqrt{c})^2}{2}$$
$$\Longleftrightarrow (\sqrt{b}+\sqrt{c})^2\le 2$$
It is clear,Because use Cauchy-schwarz inequality
$$b+c\le 1\Longrightarrow (\sqrt{b}+\sqrt{c})^2\le[1+1](b+c)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1209896",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Sum of Fibonacci numbers While trying to find find a formula to calculate the length of the golden spiral I came across the sum of the Fibonacci numbers.
I noticed that
$$\text{Fibonacci numbers: }1,1,2,3,5,8,13,21,34...$$
$$1+1+2= 5-1$$
$$1+1+2+3= 8-1$$
and that
$$2+3+5+= 13-2$$
$$3+5+8=21-5$$
so generalized that writing:
$$\sum^n_{i=k} F_i=F_{n+2}-F_{k+1}$$
where $F_n$ is the nth Fibonacci number.
But the sentence "I noticed that" is not a sufficient demonstration; I tried a lot but I couldn't find a correct demonstration.
How can I find it?
| Fibonacci numbers can be written as a matrix using:
$$\begin{bmatrix} 1 & 1 \\ 1 & 0 \end{bmatrix}^n = \begin{bmatrix} F_{n+1} & F_{n} \\ F_{n} & F_{n-1}\end{bmatrix}$$
So that any sum, using $X= \begin{bmatrix} 1 & 1 \\ 1 & 0 \end{bmatrix}$, is :
$$\sum_{k=a}^b F_n = \left( \sum_{k=a}^b X^n \right)_{2,1}$$
which is a geometric sum. So you can use geometric sum formula:
$$\begin{align} \sum_{k=a}^b X^n &= \sum_{k=0}^b X^n - \sum_{k=0}^{a-1} X^n \\
&= (X^{b+1} - I)(X-I)^{-1}- (X^{a} - I)(X - I)^{-1} \\
&= \left(X^{b+1} - X^{a}\right)(X - I)^{-1} \\
\end{align}$$
Now $(X - I)^{-1} = X$ for this particular matrix (property of Fibonacci recursion):
$$\begin{align} \sum_{k=a}^b X^n
&= \left(X^{b+1} - X^{a}\right)X \\
&= \left(X^{b+2} - X^{a+1}\right) \\
\end{align}$$
$$\sum_{k=a}^b F_n = F_{b+2} - F_{a+1}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1211909",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 0
} |
Probability that minimum of two numbers is less than 4 Suppose I have to choose two numbers from set $$S=\{1,2,3,4,5,6 \}$$ without a replacement , then what is the probability that minimum of two is less than $4$?
I made two groups for this problem $A= \{1,2,3 \}$ and $B=\{4,5,6 \}$
.There are two possibilities , either both are from $A$ or one from $A$ and one from $B$ to satisfy our requirement. Hence
$$P(E)=\frac{3}{6}.\frac{2}{5} + \frac{3}{6}.\frac{3}{5}$$
but answer is incorrect. Help?
| The error with your approach is that "One from $A$ and one from $B$" has two ways of happening: $A$ then $B$, or $B$ then $A$. So you need to multiply your calculation of this case by 2.
$$P(E) = \frac{3}{6} \cdot \frac{2}{5} + 2 \cdot \frac{3}{6} \cdot \frac{3}{5} = \frac{4}{5}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1213148",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 3
} |
Lowering powers of $\cos^2x \sin^4x$ First, I will be straight up, this is a homework question.
I need to write $\cos^2 x \sin^4 x$ in terms of cosine to the first power. I know that $\sin^4x$ =
$$ \frac{3-4\cos 2x+\cos 4x}{8}$$
from there I go:
$$ \frac{1+\cos 2x}{2} \cdot \frac{3-4\cos 2x+\cos 4x}{8}$$
$$ \frac{3(1 + \cos 2x) - 4\cos 2x(1 + \cos 2x) + \cos 4x (1 + \cos 2x)} {16} $$
$$ \frac{3 + 3\cos 2x - 4\cos 2x \color{red}{+} 4 \cos^2 2x + \cos 4x + \cos 4x \cos 2x}{16}$$
$$ \frac{3 - \cos 2x + \cos 4x + 4 \frac{1 + \cos 4x}{2} \cos 4x \cos 2x}{16}$$
Then $$ 4 \frac{1 + \cos 4x}{2}$$ simplifies to $$ 2 + 2\cos 4x$$
adding this into the rest of the fraction gives me
$$ \frac{5 - \cos 2x + 3\cos 4x + \cos 4x \cos 2x}{16}$$
However, the answer sheet that I have says the answer is
$$ \frac{1 - \cos 2x - \cos 4x + \cos 2x \cos 4x }{16}$$
This is problem $13$ from section $7.3$ of Stewart, Redlin and Watson precalculus $6$th edition
Where is my mistake?
| There is a systematic approach to solve such questions, based on complex numbers.
We have $\cos kx=\dfrac{e^{ikx}+e^{-ikx}}2=\dfrac{z^k+z^{-k}}2$ and $\sin kx=\dfrac{e^{ikx}-e^{-ikx}}{2i}=\dfrac{z^k-z^{-k}}{2i}$, where $z=e^{ix}$.
The expression can be rewritten
$$\left(\frac{z+z^{-1}}{2}\right)^2\left(\frac{z-z^{-1}}{2i}\right)^4=\frac{z^6+z^{-6}-2z^4-2z^{-4}-z^2-z^{-2}+4}{64}=\frac{\cos6x-2\cos4x-\cos2x+2}{32}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1216162",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
Proving $\frac{1}{\sqrt{2}}=1-\frac{1}{2^2}-\frac{1}{2!2^4}-\frac{3!!}{3!2^6}-\frac{5!!}{4!2^8}-\cdots$ How can I prove
$$\frac{1}{\sqrt{2}}=1-\frac{1}{2^2}-\frac{1}{2!2^4}-\frac{3!!}{3!2^6}-\frac{5!!}{4!2^8}-\cdots$$
I wanted to prove it by using the Taylor series of $\sqrt{2}$, but I couldnt do.
| As Lucian pointed out, we can write
$\displaystyle\frac{1}{\sqrt{2}}=\left(1-\frac{1}{2}\right)^{1/2}=\sum_{k=0}^{\infty}\binom{1/2}{k}\left(-\frac{1}{2}\right)^k=1+\sum_{k=1}^{\infty}\frac{\frac{1}{2}(\frac{1}{2}-1)(\frac{1}{2}-2)\cdots(\frac{1}{2}-(k-1))}{k!}\cdot\frac{(-1)^k}{2^k}$
$\displaystyle=1+\sum_{k=1}^{\infty}\frac{1(-1)(-3)(-5)\cdots(-(2k-1))}{k!2^k}\cdot\frac{(-1)^k}{2^k}=1-\sum_{k=1}^{\infty}\frac{(2k-1)!!}{k!2^{2k}}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1217847",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
find taylor series to fourth term I'm wondering if there is faster method than just calculating derivatives with finding taylor series up to 4 term of function $\displaystyle f(x)=\frac{(1+x^4)}{(1+2x)^3(1-2x)^2}$
| The idea is to use the formula
$$
\frac{1}{(1-x)^t} = \sum_{n=0}^\infty \binom{n+t-1}{n} x^n,
$$
which can be proved by induction on $t$.
Using this we get
$$
\frac{1+x^4}{(1+2x)^3(1-2x)^2} = \\
(1+O(x^4))(1 - 6x + 24x^2 - 80x^3+O(x^4))(1 + 4x + 12x^2 + 32x^3 + O(x^4)) = \\
1 + (-6+4)x + (24-6\cdot 4+12)x^2 + (-80 + 24\cdot 4 - 6\cdot 12 + 32)x^3 +
O(x^4) = \\
1 - 2x + 12x^2 - 24x^3 + O(x^4).
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1219248",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
How to differentiate $y=(x+1)^3/x^{3/2}$ and $y=2x^4/(b^2-x^2)$ I need to solve a list of derivatives to help me on an exam; however, I'm in doubt when they use another variable (constant) or when I have a fraction with functions that use the power rule.
For example:
$$y = \frac{(x + 1)^3}{x^{3/2}}.$$
This one I tried having $u = x + 1$, deriving it and having $v = x^{3/2}$, but I couldn't get the right answer.
And:
$$y = \frac{2x^4}{b^2 - x^2}.$$
And this one I tried to separate the constant, like $y = \frac{1}{b^2} \frac{dx}{dy} \frac{2x^4}{-x^2}$, but also coudn't find the answer.
Can you give a tip on how to find which rule to apply to these kind of derivatives?
| When the expression contains products and ratioé, logarithmic differentiation make life much easier.
Let us consider the first problem $$y = \frac{(x + 1)^3}{x^{3/2}}$$ Taking logarithms $$\log(y)=3\log(x+1)-\frac32\log(x)$$ Now, differentiate both sides $$\frac{y'}y=\frac3{x+1}-\frac3{2x}=\frac{3 (x-1)}{2 x (x+1)}$$ $$y'=\frac{3 (x-1)}{2 x (x+1)}y=\frac{3 (x-1)}{2 x (x+1)} \frac{(x + 1)^3}{x^{3/2}}=\frac{3 (x-1) (x+1)^2}{2 x^{5/2}}$$
Do the same for the second problem $$y = \frac{2x^4}{b^2 - x^2}$$ $$\log(y)=\log(2)+4\log(x)-\log(b^2-x^2)$$ $$\frac{y'}y=0+\frac 4x+\frac{2x}{b^2-x^2}=\frac{2(2b^2-x^2)}{x(b^2-x^2)}$$ $$y'=\frac{2(2b^2-x^2)}{x(b^2-x^2)}y=\frac{2(2b^2-x^2)}{x(b^2-x^2)}\frac{2x^4}{b^2 - x^2}=\frac{4(2b^2-x^2)x^3}{(b^2-x^2)^2}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1219363",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Definite integral with trigonometric functions I have problem finding, how to solve this integral
$$
\int _0 ^{\frac{\pi}{4}} \frac{3 \sin x + 2 \cos x}{2 \sin x + 3 \cos x}dx
$$
This I can rewrite as
$$
\int _0 ^{\frac{\pi}{4}} \frac{12 \sin ^2 x - 5 \cos x \sin x - 6}{4-13 \cos ^2 x}dx =
$$
$$
= \int _0 ^{\frac{\pi}{4}} \frac{12 \sin ^2 x}{4-13 \cos ^2 x}dx - \int _0 ^{\frac{\pi}{4}} \frac{5 \cos x \sin x}{4-13 \cos ^2 x}dx - \int _0 ^{\frac{\pi}{4}} \frac{6}{4-13 \cos ^2 x}dx
$$
But from this point, I can't get any closer to the answer.
Can you, please, help me?
| Whenever you have to integrate a fraction involving odd powers of $\sin$ and $\cos$, the preferred algorithmic approach is the substitution $t = \tan \frac x 2$. This leads to $\mathrm{d}x = \frac 2 {1 + t^2} \mathrm{d}t$ and then, using that $\frac {\sin x} {\cos x} = \tan x = \tan (\frac x 2 + \frac x 2) = \frac {2t} {1-t^2}$ and $\sin^2 x + \cos^2 x = 1$ you get two equations with two unknowns ($\sin$ and $\cos$) that have for solutions $\cos x = \frac {1-t^2} {1+t^2}$ and $\sin x = \frac {2t} {1+t^2}$ (the signs were dictaded by the fact that on $[0,\frac {\pi} 4]$ $\sin$ and $\cos$ are positive). Next, use partial fraction expansion (i.e. split your integrand into as many simple fractions as possible, by studying the real roots of its denominator); if the degree of the numerator is larger than the one of the denominator, first divide the polynomials. The whole solution is too long to write here.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1222861",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
How to deduce that $1\cdot 1 + 2\cdot 1 + 2\cdot 2 + 3\cdot 1+3\cdot 2+3\cdot 3 +...+(n\cdot n) = n(n+1)(n+2)(3n+1)/24$ I know how to reason $$1\cdot2 + 2\cdot3 + 3\cdot4 + n(n-1) = \frac{1}{3}n(n-1)(n+1)$$
However, I'm stuck on proving $$1\cdot1 + (2\cdot1 + 2\cdot2) + (3\cdot1+3\cdot2+3\cdot3) + \cdots +(n\cdot 1+...+n\cdot n) = \frac{1}{24}n(n+1)(n+2)(3n+1).$$
Could somebody shed some light on me?
| You can do as the following :$$\begin{align}\sum_{k=1}^{n}k(1+2+\cdots+k)&=\sum_{k=1}^{n}k\cdot\frac{k(k+1)}{2}\\&=\frac 12\sum_{k=1}^{n}k^3+\frac 12\sum_{k=1}^{n}k^2\\&=\frac 12\left(\frac{n(n+1)}{2}\right)^2+\frac 12\cdot\frac{n(n+1)(2n+1)}{6}.\end{align}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1225311",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 0
} |
Trig function integral I'm trying to solve
$$\int_{0}^{\pi}\frac{dx}{cos^2(x)-a^2}, \hspace{5mm} 0<a<1$$
There are numerous examples of similar integrals but non with the condiction that $0<a<1$, say $a = 0.5$. Since the function is even one can expand the domain of integration to $2\pi$ and use the residuum theorem, but zeros lie on the unit circle making that method useless. Numerical integration in Mathematica or Fortran is also problematic and filled with errors.
So, does anyone have a clue how to solve this? I encountered this integral while calculating renormalization of energy dispersion in graphene due to electron-phonon interaction.
| The Cauchy principal value of the integral is $0$.
We have
$$F(a) = 2\int_0^{\pi/2} \dfrac{dx}{\cos^2(x)-a^2} = 2 \int_0^{\pi/2} \dfrac{\sec^2(x)dx}{1-a^2\sec^2(x)}$$
Setting $\tan(x) = t$, we obtain
\begin{align}
F(a) & = 2 \int_0^{\infty} \dfrac{dt}{1-a^2-a^2t^2} = \dfrac2{a^2} \int_0^{\infty} \dfrac{dt}{\left(\dfrac{1-a^2}{a^2}\right)-t^2}
\end{align}
Note that since $a^2 < 1$, we set $b^2 = \dfrac{1-a^2}{a^2}$, where $b \in \mathbb{R}^+$. We have
\begin{align}
F(a) & = \dfrac2{a^2} \int_0^{\infty} \dfrac{dt}{b^2-t^2}
\end{align}
We will interpret $F(a)$ as the Cauchy principal value of the integral, i.e.,
\begin{align}
\dfrac{a^2F(a)}2 & = \lim_{\delta \to 0^+} \left(\int_0^{b-\delta} \dfrac{dt}{b^2-t^2} + \int_{b+\delta}^{\infty} \dfrac{dt}{b^2-t^2} \right)
\end{align}
We have
\begin{align}
\int_0^{b-\delta} \dfrac{dt}{b^2-t^2} + \int_{b+\delta}^{\infty} \dfrac{dt}{b^2-t^2} & = \dfrac1{2b}\ln\left(\dfrac{b+t}{b-t}\right)_0^{b-\delta} + \left.\dfrac1{2b}\ln\left(\dfrac{t+b}{t-b}\right) \right \vert_{b+\delta}^{\infty}\\
& = \dfrac1{2b}\left(\ln\left(\dfrac{2b-\delta}{\delta}\right)-\ln(1) + \ln(1) - \ln\left(\dfrac{2b+\delta}{\delta}\right)\right)\\
& = \dfrac1{2b}\ln\left(\dfrac{2b-\delta}{2b+\delta}\right)
\end{align}
Hence, we obtain
$$\dfrac{a^2F(a)}2 = \lim_{\delta \to 0^+} \dfrac1{2b}\ln\left(\dfrac{2b-\delta}{2b+\delta}\right)=0$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1227237",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Number in tens place Can you figure out the digit in the tens place of the solution to $4^{2015} \cdot 9^{2016}$ without using a calculator?
By tens place I mean, for example if you have number $2451$, the number in tens place here is $5$.
I know the answer, but I don't know how to get it, so if you got any ideas, please share. :)
| You have
$$4^{11}=4194304\equiv 4 \pmod{100}$$
And since $2015=183\times 11+2$, and $183=16\times 11+7$, you have
$$4^{2015}=4^{11\times183+2}=4^2\times(4^{11})^{183}\equiv 4^2\times 4^{183} \pmod{100}$$
$$=4^2\times4^{11\times16+7}=4^9\times(4^{11})^{16}\equiv 4^9\times4^{16}=4^{25}=4^3\times(4^{11})^2\equiv4^5 \pmod{100}$$
$$=1024\equiv 24\pmod{100}$$
Likewise,
$$9^{10}=3486784401\equiv1\pmod{100}$$
Thus
$$9^{2016}=9^6\times (9^{10})^{201}\equiv 9^6=531441\equiv 41\pmod{100}$$
Therefore
$$4^{2015}\times9^{2016}\equiv 24\times41=984\equiv 84\pmod{100}$$
And the digit in tens place is $8$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1229260",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
limit using taylor series I keep getting an error in the expansion
$$\lim_{x\to 0 }\frac{2\exp(\sin(x))-2-x-x^2-\arctan (x) }{x^3}$$
The numerator works out as
$$\approx 2 (1+\sin(x) + 1/2 \sin^2(x)) -2-x-x^2 - (x-x^3/3+x^5/5)$$
$$\approx 2(1+x-x^3/3!+1/2(x-x^3/3!)^2 -2-2x+x^3/3-x^2 -x^5/5 $$
$$ = o(x^4)$$
so that the limit is zero. But it is supposed to be $1/3$. Where is my mistake?
| Starting from $$e^y=1+y+\frac{y^2}{2}+\frac{y^3}{6}+\frac{y^4}{24}+O\left(y^5\right)$$ Replace $y$ by the expansion of $\sin(x)$ to get $$e^{\sin(x)}=1+x+\frac{x^2}{2}-\frac{x^4}{8}-\frac{x^5}{15}+O\left(x^6\right)$$ Just as quid answered while I was typing, one term was missing in the expansion.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1229968",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 3
} |
Simplify $ \frac{b}{a-b}+ \frac{a}{b-a}$ I have a question regarding simplification of an algebraic expression.
Here is the problem:
$ \frac{b}{ a - b} + \frac{a}{b-a} $
The outcome is $ -1 $
Here is how I try to simplify it:
*
*Add fractions: $ \frac{ b(-a+b) + a(a-b)}{(a-b)(-a+b)} $
*FOIL: $ \frac{a^2 - 2ab + b^2}{(a-b)(-a+b)} $
*Apply Binomial Decomposition: $ \frac{(a-b)^2}{(a-b)(-a+b)} $
*Reduce fractions: $ \frac{\color{red}{(a-b)}(a-b)}{\color{red}{(a-b)}(-a+b)}$
The answer is: $ \frac{a-b}{-a+b} $
Now in the book the answer is, as previously mentioned is $-1$
How do I get my answer $ \frac{a-b}{-a+b} $ to become $-1$ ?
Could you, please, explain the steps necessary to get that answer?
Thank you
| Notice that $-a+b=-(a-b)$. Thus: $\frac{a-b}{-a+b}=\frac{a-b}{-(a-b)}=-\frac{a-b}{a-b}=-1$.
You could also have done this in a much easier way:
$\frac{b}{a-b}+\frac{a}{b-a}=\frac{b}{a-b}-\frac{a}{-(b-a)}=\frac{b}{a-b}-\frac{a}{a-b}=\frac{b-a}{a-b}=-1$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1231556",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Prove the following trigonometric identity $$\frac{\tan{(\frac{\pi}{4}+x)}-\tan{(\frac{\pi}{4}-x)}}{\tan{(\frac{\pi}{4}+x)}+\tan{(\frac{\pi}{4}-x)}} = 2\sin{x}\cos{x}$$
==============
On L.H.S, I've tried to write it using the sum and difference formula so it becomes
$$\frac{\dfrac{1+\tan x}{1-\tan x}-\dfrac{1-\tan x}{1+\tan x}}{\dfrac{1+\tan x}{1-\tan x}+\dfrac{1-\tan x}{1+\tan x}}$$
Then I try to rationalize and it got really messy
$$\frac{\dfrac{(1+\tan x)(1+ \tan x)}{(1-\tan x)(1+\tan x)} - \dfrac{(1+\tan x)(1+\tan x)}{(1-\tan x)(1+\tan x)}}{\dfrac{(1+\tan x)(1+\tan x)}{(1-\tan x)(1+\tan x)} + \dfrac{(1+\tan x)(1+\tan x)}{(1-\tan x)(1+\tan x)}}$$
Somehow I distributed the terms and try to simplify and I got stuck.
Any help would be appreciated. Thank you.
| You may write
$$
\begin{align}
\dfrac{\dfrac{1+\tan x}{1-\tan x}-\dfrac{1-\tan x}{1+\tan x}}{\dfrac{1+\tan x}{1-\tan x}+\dfrac{1-\tan x}{1+\tan x}}=\dfrac{(1+\tan x)^2-(1-\tan x)^2}{(1+\tan x)^2+(1-\tan x)^2}
&=\dfrac{4\tan x}{2(1+\tan^2x)}\\\\&=2\tan x \cos^2x\\\\&=2\sin x\cos x
\end{align}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1232129",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
} |
If the distance between two lines is $ \frac{1}{\sqrt{3}} $, then $ \alpha $ is...
If the shortest distance between the lines $\displaystyle \frac{x-1}{\alpha}=\frac{y+1}{-1}=\frac{z}{1}\;, (\alpha \neq - 1)$ and
$x+y+z+1=0 = 2x-y+z+3 = 0$ is $\displaystyle \frac{1}{\sqrt{3}}\;,$ Then $\alpha = $
$\bf{My\; Solution:: }$ Here $x+y+z+1=0$ and $2x-y+z+3=0$ represent $2$ plane and
we know that locus of point of Intersection of $2$ plane is also a Line.
Know Here we have to calculate the equation of Common line of these $2$ planes.
For Calculation of equation of line....
Add these $2$ equations... $\displaystyle 3x+2z+4=0\Rightarrow z= - \frac{\left(4+3x\right)}{2}$ and Put into $x+y+z+1=0$
$\displaystyle 2x+2y+2z+2=0\Rightarrow 2x+2y-4-3x+2=0\Rightarrow -x+2y=2$
So we get $\displaystyle x=2y-2\Rightarrow \frac{x-0}{1} = \frac{y-1}{\frac{1}{2}}$ and above we get $\displaystyle \frac{x-0}{1}=\frac{z+2}{-\frac{3}{2}}.$
So We get equation of line is $\displaystyle \frac{x-0}{1}=\frac{y-1}{\frac{1}{2}}=\frac{z+2}{-\frac{3}{2}}.$
Now we have to Calculate Distance b/w two lines $\displaystyle \frac{x-1}{\alpha}=\frac{y+1}{-1}=\frac{z}{1}\;, (\alpha \neq - 1)$
and $\displaystyle \frac{x-0}{1}=\frac{y-1}{\frac{1}{2}}=\frac{z+2}{-\frac{3}{2}}.$
Is my equation of line is Right or not, If not then how can i calculate common
equation of line and value of $\alpha.$
Help me , Thanks
| I like to use Plücker coordinates for lines
*
*Line along $(\alpha,-1,1)$, through Point $(1,-1,0)$
$$ L_1 = \left\{ \begin{array}{c} \vec{e} \\ \vec{r}\times\vec{e} \end{array} \right\} = \left\{ \begin{array}{c} \begin{pmatrix} \alpha\\-1\\1\end{pmatrix} \\
\begin{pmatrix} 1\\-1\\0\end{pmatrix} \times \begin{pmatrix} \alpha\\-1\\1\end{pmatrix}
\end{array} \right\} = \left\{ \begin{array}{c} \begin{pmatrix} \alpha\\-1\\1\end{pmatrix} \\ \begin{pmatrix} -1\\-1\\ \alpha-1 \end{pmatrix} \end{array} \right\} =\left\{ \begin{array}{c} \vec{\ell}_1 \\ \vec{m}_1 \end{array} \right\} $$
*
*Line where Plane $\vec{n_A}=(-1,-1,-1)$, $d_A=1$ and Plane $\vec{n_B}=(-2,1,-1)$, $d_B=3$ meet
$$ L_2 = \left\{ \begin{array}{c} \vec{n}_A \times \vec{n}_B \\ d_B \vec{n}_A-d_A \vec{n}_A \end{array} \right\} = \left\{ \begin{array}{c} \begin{pmatrix}-1\\-1\\-1\end{pmatrix} \times \begin{pmatrix}-2\\1\\-1\end{pmatrix} \\ 3 \begin{pmatrix}-1\\-1\\-1\end{pmatrix}-1 \begin{pmatrix}-2\\1\\-1\end{pmatrix} \end{array} \right\}=\left\{ \begin{array}{c} \begin{pmatrix}2\\1\\-3\end{pmatrix} \\ \begin{pmatrix}-1\\-4\\-2\end{pmatrix} \end{array}\right\} =\left\{ \begin{array}{c} \vec{\ell}_2 \\ \vec{m}_2 \end{array}\right\} $$
*Minimum distance between lines $L_1$ and $L_2$
$$ \rho = \frac{\vec{m}_1 \cdot \vec{\ell}_2 + \vec{m}_2 \cdot \vec{\ell}_1} {\| \vec{\ell}_1 \times \vec{\ell}_2 \| } = \frac{\begin{pmatrix} -1 \\ -1 \\ \alpha-1 \end{pmatrix}\cdot \begin{pmatrix} 2 \\ 1 \\-3 \end{pmatrix}+\begin{pmatrix}-1\\-4\\-2 \end{pmatrix}\cdot\begin{pmatrix}\alpha\\-1\\1 \end{pmatrix}}{\| \begin{pmatrix} \alpha\\-1\\1\end{pmatrix}\times \begin{pmatrix}2\\1\\-3 \end{pmatrix} \|} = \frac{2-4 \alpha}{\sqrt{10 \alpha^2+16 \alpha+12}} $$
We know that $\rho=1/\sqrt{3}$ and thus
$$\left. \frac{1}{\sqrt{3}} = \frac{2-4 \alpha}{\sqrt{10 \alpha^2+16 \alpha+12}} \right\} \boxed{ \alpha=0 \;\&\; \alpha=\frac{32}{19}}$$
From the two solutions we pick $\alpha=0$ (simplest solution) to get an additional level of detail:
*
*Line normal to both lines $L_1$ and $L_2$ through the shortest distance (for reference)
$$ L_3 =\left\{ \begin{array}{c} \vec{\ell}_1 \times \vec{\ell}_2 \\ \vec{m}_1 \times \vec{\ell}_2 - \vec{m}_2 \times \vec{\ell}_1 \end{array} \right\} =
\left\{ \begin{array}{c} \begin{pmatrix}2\\2\\ 2\end{pmatrix} \\ \begin{pmatrix} 10 \\ -6 \\ 0 \end{pmatrix} \end{array}\right\} =\left\{ \begin{array}{c} \vec{\ell}_3 \\ \vec{m}_3 \end{array} \right\}$$
*
*The direction $\vec{e}_3$ and position $\vec{r}_3$ of the cross line is
$$ \vec{e}_3 = \frac{ \vec{\ell}_3 }{ \| \vec{\ell}_3 \| } = \begin{pmatrix}\frac{1}{\sqrt{3}}\\ \frac{1}{\sqrt{3}} \\ \frac{1}{\sqrt{3}} \end{pmatrix} $$
$$ \vec{r}_3 = \frac{ \vec{\ell}_3 \times \vec{m}_3 }{\| \vec{\ell}_3 \|^2} = \begin{pmatrix}1\\ \frac{5}{3} \\ -\frac{8}{3} \end{pmatrix}$$
*
*Confirm results with GeoGebra 3D
Distance AB is $1/\sqrt{3} = 0.5774$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1232568",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Trigonometric integrals How do I evaluate this indefinite integral ? Integral $$\int\frac{x^2+n(n-1)}{(x\sin(x)+n\cos(x))^2}dx$$ What type of integral is it ? Is there any intuition involved in the approach to solve it?
Edit: The complete term in denominator has 2 as exponent. There was a typo previously.
| $\bf{Another\; Solution::}$ Given $$\displaystyle \int\frac{x^2+n(n-1)}{\left(x\sin x+n\cos x\right)^2}dx\;,$$ Now Multiply both $\bf{N_{r}}$ and $\bf{D_{r}}$
by $x^{2n-2};,$ We Get $$\displaystyle \int\frac{\left[x^{2}+n(n-1)\right]\cdot x^{2n-2}}{\left(x^{n}\sin x+nx^{n-1}\cos x\right)^2}dx$$
Now Let $$x^n\sin x+nx^{n-1}\cos x = t\;,$$ Then
$$\left[nx^{n-1}\sin x+\cos x\cdot x^{n}-nx^{n-1}\sin x+\cos x\cdot n(n-1)x^{n-2}\right]dx = dt$$
So We get $$\displaystyle x^{n-2}\cdot \cos x \cdot \left[x^2+n(n-1)\right]dx = dt.$$
So Our Integral Convert into $$\displaystyle \int \underbrace{x^{n}\sec x }_{\bf{1^{st}\; fun.}}\cdot \underbrace{\frac{x^{n-2}\cdot\cos x\cdot [x^2+n(n-1)] }{\left(x^2\sin x+nx^{n-1}\cos x\right)^2}}_{\bf{II^{nd}\; fun.}}dx$$
Now Using Integration by Parts, We Get
$$\displaystyle = -x^2\sec x\cdot \frac{1}{\left(x^n\sin x+nx^{n-1}\cos x\right)}+\int \frac{x^n\sec x\cdot \tan x+\sec x\cdot nx^{n-1}}{\left(x^n\sin x+nx^{n-1}\cos x \right)}dx$$
After Simplifying The second Integral, We Get
$$\displaystyle = -\frac{x^2\sec x}{\left(x^n\sin x+nx^{n-1}\cos x \right)}+\int \sec^2 xdx$$
So We Get $$\displaystyle \int\frac{x^2+n(n-1)}{(x\sin x+n\cos x)^2}dx = -\frac{x^n\sec x}{\left(x^n\sin x+nx^{n-1}\cos x \right)}+\tan x+\mathcal{C}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1233433",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Find the largest $k$ such that $3^k$ divides the product of the first $100$ odd integers
Let $P$ be the product of the first 100 positive odd integers. Find the largest integer $k$ such that $P$ is divisible by $3^k$.
There are $50$ odd numbers and $50$ even numbers between $0$ and $100$, $99$ being the $50$th odd. So it makes sense that, $199$ is the $100$th odd number.
I got that
$$P = 1 \cdot 3 \cdots 199 = \frac{200!}{2 \cdot 4 \cdot 6 \cdots 198},$$
but that doesn't really help.
I am after
$$\prod_{n = 0}^{99} (2n + 1),$$
but I don't see a general formula.
| evidently the highest power of $3$ dividing the first $100$ odd numbers is the same as the highest power of $3$ which divides $\frac{200!}{100!}$
$$
200 = 2.3^4 +1.3^3 + 1.3^2+ 0.3^1+2.3^0
$$
so the sum-of-digits in the 3-ary represention of 200 is 6. similarly for 100 the sum of 3-ary digits is 4.
hence the number required is:
$$
\frac{200-6}{3-1} - \frac{100-4}{3-1} = 97-48 = 49
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1234316",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
} |
Find all solutions to $4x^2+6x+1 \equiv 0 \pmod {13}$ Find all solutions to $4x^2+6x+1 \equiv 0 \pmod {13}$
I think it has no solutions but I am not sure how to show this.
| We have $4x^2+6x+1 = (2x+3/2)^2-5/4$. Hence,
\begin{align}
4x^2+6x+1 \equiv 0 \pmod{13} \implies (2x+3/2)^2-5/4 \equiv 0 \pmod{13}
\end{align}
This simplifies into
\begin{align}
(4x+3)^2 \equiv 5 \pmod{13}
\end{align}
This has no solutions, since
$$5^{(13-1)/2}\pmod{13} \equiv 5^6\pmod{13} \equiv (25)^3 \pmod{13} \equiv-1\pmod{13}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1235156",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 1
} |
What is that function? Polynomial? Is it a polynomial or rational polynomial or else?
$y = \dfrac{a}{x^4} + \dfrac {b}{x^2} + c$
I need to fit a curve to a discrete data of that form, so I need to know what fitting to use.
| We have $$\frac{a}{x^4}+\frac{b}{x^2}+c=\frac{a}{x^4}+\frac{bx^2}{x^4}+\frac{cx^4}{x^4}=\frac{a+bx^2+cx^4}{x^4}$$
This is a rational function.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1239102",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
Prove by mathematical induction: $\frac{1}{n}+\frac{1}{n+1}+\dots+\frac{1}{n^2}>1$ Could anybody help me by checking this solution and maybe giving me a cleaner one.
Prove by mathematical induction:
$$\frac{1}{n}+\frac{1}{n+1}+\dots+\frac{1}{n^2}>1; n\geq2$$.
So after I check special cases for $n=2,3$, I have to prove that given inequality holds for $n+1$ case by using the given $n$ case. Ok, so this is what I've got by now:
$$\frac{1}{n+1}+\frac{1}{n+2}+\dots+\frac{1}{n^2}+\frac{1}{n^2+1}+\dots+\frac{1}{(n+1)^2}\overset{?}{>}1$$
$$\frac{1}{n+1}+\frac{1}{n+2}+\dots+\frac{1}{n^2}+\frac{1}{n^2+1}+\dots+\frac{1}{n^2+2n+1}\overset{?}{>}1$$
$$\frac{1}{n}+\frac{1}{n+1}+\frac{1}{n+2}+\dots+\frac{1}{n^2}+\frac{1}{n^2+1}+\dots+\frac{1}{n^2+2n+1}\overset{?}{>}1+\frac{1}{n}$$
From the $n$ case we know that:
$$\frac{1}{n}+\frac{1}{n+1}+\frac{1}{n+2}+\dots+\frac{1}{n^2}+\frac{1}{n^2+1}+\dots+\frac{1}{n^2+2n+1}>1+\frac{1}{n^2+1}+\dots+\frac{1}{n^2+2n+1}$$
So we basically have to prove that:
$$1+\frac{1}{n^2+1}+\dots+\frac{1}{n^2+2n+1}\overset{?}{>}1+\frac{1}{n}$$
$$\frac{1}{n^2+1}+\dots+\frac{1}{n^2+2n+1}\overset{?}{>}\frac{1}{n}$$
Since
$$n^2+1<n^2+2<\dots<n^2+2n+1<2n^2+n$$
for $n\geq2$, then also:
$$\frac{1}{n^2+1}>\frac{1}{2n^2+n}$$
$$\frac{1}{n^2+2}>\frac{1}{2n^2+n}$$
.
.
$$\frac{1}{n^2+2n+1}>\frac{1}{2n^2+n}$$
so then we have:
$$\frac{1}{n^2+1}+\dots+\frac{1}{n^2+2n+1}>\frac{1}{2n^2+n}+\dots+\frac{1}{2n^2+n}=(2n+1)\frac{1}{2n^2+n}=\frac{1}{n}$$
| Your argument is fine and quite clearly presented. You can shorten the presentation considerably, though, by doing something like this:
For $n\ge 2$ let $a_n=\sum_{k=n}^{n^2}\frac1k$. Since $a_2=\frac12+\frac13+\frac14>1$, it suffices to show that $a_{n+1}\ge a_n$ for $n\ge 2$. Since $n^2+2n+1<2n^2+n$ for $n\ge 2$, we have
$$a_{n+1}-a_n=\sum_{k=1}^{2n+1}\frac1{n^2+k}-\frac1n\ge\sum_{k=1}^{2n+1}\frac1{2n^2+n}-\frac1n=\frac{2n+1}{2n^2+n}-\frac1n=0\;,$$
so $a_{n+1}\ge a_n$, and the result follows by induction.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1239518",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 6,
"answer_id": 0
} |
Antiderivative of $ (x^2 + c)^{-3/2} $ What method should be used to determine the antiderivative of this expression?
Edit: I have $ c > 0 $ in the problem I'm working on.
| $$\int(x^2+c)^{-3/2}dx=\int\frac{1}{\sqrt{(x^2+c)^3}}$$
Let us use a new variable $t$ such that $\sqrt{x^2+c}=x+t$. This is the first Euler substitution.
From this formula we can solve for $x$ to get $x=\frac{c-t^2}{2t}$, $dx=\frac{-c-t^2}{2t^2}dt$, $\sqrt{x^2+c}=x+t=\frac{c+t^2}{2t}$, and $\sqrt{(x^2+c)^3}=\frac{(c+t^2)^3}{8t^3}$.
We put these into the integral and get
$$\int\frac{8t^3}{(c+t^2)^3}\frac{-(c+t^2)}{2t^2}dt=-2\int\frac{2t}{(c+t^2)^2}dt=2\frac{1}{c+t^2}$$
We can return to the variable $x$ since $t=\sqrt{x^2+c}-x$. Therefore
$$\int(x^2+c)^{-3/2}=\frac{2}{c+\left(\sqrt{x^2+c}-x\right)^2}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1244354",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Fermat numbers of the form of $b^2$ For n > 1 Let $F_n = 2^{2^n} + 1$ be a fermat number and b = $2^{2^{n - 2}}$ * ($2^{2^{n - 1}}$ - 1 ).
Then $b^2$ $\equiv$ 2 (mod $F_n$)
I tried to square the original expression I got something ugly that I couldn't simplify further.
I got $b^2$ = $2^{2^{n - 1}}$ * ($2^{2^n}$ - $2 * 2^{2^{n - 1}}$ + 1) $\equiv$ - ($2 * 2^{2^n} * 2^{2^{n-1}}$) (mod $F_n$). I stopped here I can't simplify it anymore.
| You're almost there! Just notice that $2^{2^n}\equiv -1\pmod{2^{2^n}+1}$ :
$$\begin{align}b^2&=2^{2^{n-1}}(2^{2^n}-2*2^{2^{n-1}}+1)\\&=2^{2^{n-1}}(2^{2^n}+1-2*2^{2^{n-1}})\\&\equiv2^{2^{n-1}}(0-2*2^{2^{n-1}})\pmod{2^{2^n}+1}\\&\equiv-2*2^{2^{n-1}}*2^{2^{n-1}}\pmod{2^{2^n}+1}\\&\equiv-2*2^{2^{n-1}+2^{n-1}}\pmod{2^{2^n}+1}\\&\equiv-2*2^{2^n}\pmod{2^{2^n}+1}\\&\equiv -2*(-1)\pmod{2^{2^n}+1}\end{align}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1244564",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Computing $\int_3^5 \frac{x^2\,dx}{\sqrt{(x-3)(5-x)}}$
$$
\int_3^5 \frac{x^2\,dx}{\sqrt{(x-3)(5-x)}}
$$
how? $x^2/\sqrt{8x-x^2-15}$ and what to do then?
| $$
-(x^2 - 8x + 15) = 1-(x^2 - 8x + 16) = 1 -(x-4)^2 = 1-\sin^2\theta
$$
So the square root of this is $\cos\theta$.
$$
dx = \cos\theta\,d\theta
$$
$$
x^2 = (4+\sin\theta)^2
$$
As $x$ goes from $3$ to $5$, $\sin\theta=x-4$ goes from $-1$ to $1$, so $\theta$ goes from $-\pi/2$ to $\pi/2$. We get
$$
\int_{-\pi/2}^{\pi/2} \frac{(4+\sin\theta)^2}{\cos\theta}\,d\theta = \int_{-\pi/2}^{\pi/2} \frac{(4+\sin\theta)^2}{1-\sin^2\theta}\,(\cos\theta\,d\theta) = \int_{-1}^1 \frac{(4+u)^2}{1-u^2} \, du.
$$
Then partial fractions should do it.
Or you might try a rationalizing substitution rather than a trigonometric substitution.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1245223",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
How to evaluate volume of oblique frustum of right circular cone with elliptical section?
Let there be an oblique frustum, with an elliptical section, of a right circular cone with apex point O & cone angle $2\alpha=60^{o}$. It is obtained by cutting the cone by a plane at a normal distance $OM=h=20 cm$ & making an angle $\theta=60^{o}$ with the axis OO' of the frustum (cone) (as shown in the above diagram). How to evaluate the volume of this oblique frustum with elliptical section?
Any help is greatly appreciated!
| First, assign an $xyz$ coordinate system. For simplicity, let $M$ be the origin, and let the positive $x$-axis point to the right, and the positive $y$-axis point into the screen, while the $z$-axis points upward.
Next, find the unit vector in this coordinate system of the axis of the cone. This can be found quite easily to be
$ a = ( - \cos 60^\circ, 0 , \sin 60^\circ ) = \left( - \dfrac{1}{2}, 0, \dfrac{\sqrt{3}}{2} \right) $
Now the equation of the surface of the cone is
$ (r - O)^T Q (r - O) = 0 $
where $r = [x, y, z]^T$, and $O$ is the tip of the cone, and
$ Q = \cos^2 \alpha I - a a^T = \dfrac{3}{4} I - \begin{bmatrix} -\dfrac{1}{2} \\ 0 \\ \dfrac{\sqrt{3}}{2} \end{bmatrix} \begin{bmatrix} -\dfrac{1}{2} && 0 && \dfrac{\sqrt{3}}{2} \end{bmatrix} $
Upon evaluting the above expression, one finds that
$ Q = \begin{bmatrix} \dfrac{1}{2} && 0 && \dfrac{\sqrt{3}}{4} \\ 0 && \dfrac{3}{4} && 0 \\ \dfrac{\sqrt{3}}{4} && 0 && 0 \end{bmatrix} $
Since we want to find the area of the elliptical base, we set $z=0$ in the equation of the cone, and this leads to the following equation in $x$ and $y$ (Note that the coordinates of point $O$ are $(0, 0, 20) $
$\dfrac{1}{2} x^2 + \dfrac{3}{4} y^2 + \dfrac{\sqrt{3}}{2} (-20 x) = 0 $
Multiplying through by $4$,
$ 2 x^2 + 3 y^2 - 40\sqrt{3} x = 0 $
Completing the square in $x$,
$ 2 (x - 10\sqrt{3})^2 + 3 y^2 = 600 $
Dividing through by $600$
$ \dfrac{(x-10\sqrt{3})^2}{300} + \dfrac{ y^2 }{200} = 1 $
Hence the elliptical base has a semi-major axis along the $x$-axis of length $ = \sqrt{300} = 10\sqrt{3} \text{ cm} $ and semi-minor axis along the $y$-axis of length $=\sqrt{200} = 10\sqrt{2} \text{ cm} $.
The area of the base is therefore $A = \pi a b = 100 \sqrt{6} \pi \text{ cm}^2 $
And the volume is $ V = \dfrac{1}{3} A h = 2000 \sqrt{\dfrac{2}{3}} \pi \text{ cm}^3 $
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1246580",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 1
} |
Solve for $x:1 + \tan^2(x) = 8\sin^2(x)$ I have a tricky problem , I tried several methods and I can't seem to get a definite answer.
$1 + \tan^2(x) = 8\sin^2(x), x \in [\frac{\pi}{6} , \frac{\pi}{2}]$
I got to $8\cos^4(x)-8\cos^2(x)+1=0$ and found that $\cos^2(x) = \frac{1}{4}[2-\sqrt{2}]$ but that is not too useful.
I need to find the angle "x" which is:
a)$\frac{\pi}{8}\quad$ b)$\frac{\pi}{6}\quad$ c)$\frac{\pi}{4}\quad$ d)$\frac{5\pi}{6}\quad$ e)$\frac{3\pi}{4}\quad$ f)$\frac{3\pi}{8}$
| Left side is $ \sec^2 x $ and exploiting $2\sin x \cos x =\sin 2x $ should ring a bell.
$$2\sin^2 2 x = 1 \implies \sin 2x = \pm \dfrac1{\sqrt2} $$
where $ x = \frac12 $ of $\pi/4 = \pi/8 $ in the first quadrant.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1248525",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 6,
"answer_id": 2
} |
"Simpler" geometrical description So i was asked to find:
Find the matrix that represents the linear transformation of the plane obtained by:
*
*reflecting in the line y = x, $\begin{bmatrix} 0&1 \\ 1&0 \end{bmatrix}$
*then rotating anticlockwise through an angle of 45 degrees, $\begin{bmatrix} \frac{1}{\sqrt{2}}&-\frac{1}{\sqrt{2}} \\ \frac{1}{\sqrt{2}}&\frac{1}{\sqrt{2}} \end{bmatrix}$
*and finally reflecting in the y axis. $\begin{bmatrix} -1&0 \\ 0&1 \end{bmatrix}$
*Give a simpler geometrical description of what this transformation does.
Which i guess is just multiplying all the steps
$$\begin{bmatrix} 0&1 \\ 1&0 \end{bmatrix} \begin{bmatrix} \frac{1}{\sqrt{2}}&-\frac{1}{\sqrt{2}} \\ \frac{1}{\sqrt{2}}&\frac{1}{\sqrt{2}} \end{bmatrix} \begin{bmatrix} -1&0 \\ 0&1 \end{bmatrix} = \begin{bmatrix} -\frac{1}{\sqrt{2}}&\frac{1}{\sqrt{2}} \\ \frac{1}{-\sqrt{2}}&-\frac{1}{\sqrt{2}} \end{bmatrix} $$
How do i give a simpler geometrical description of this other than what's already stated in the question: reflect in the line y=x, rotate through angle of 45 degree and reflect by the y axis.
Is this some kind of inverse project it feels a lot like the geometric aspect of an inverse to its main function.
| there is a result from geometry that says any rotation by angle $2\theta$ is equivalent to two reflections on mirrors separated by an angle $\theta.$ the two mirrors can be placed on ant line through the point of rotation.
we will also use the fact that reflections are involuntary; that they are their own inverses.
let $R_{x = 0}, R_{y= \tan(3\pi/8) x},$ and $ R_{y = x}$ represent the reflections on the lines $x = 0, y= \tan(3\pi/8) x$ and $y = x.$ by the geometry argument we have $$Rot_{\pi/4} = R_{y= \tan(3\pi/8) x} R_{y=x}$$ where $Rot_{\pi/4}$ represents the rotation by $45^\circ$ counterclockwise about the origin.
we can compute $$\begin{align}R_{x = 0} Rot_{\pi/4} R_{y=x} &= R_{x = 0} R_{y= \tan(3\pi/8) x} R_{y=x} R_{y=x}\\
&= R_{x = 0} R_{y= \tan(3\pi/8) x}\\
&=Rot_{\pi/4}\end{align}$$
the there transformations result in a rotation by $45^\circ$ counter clockwise. this is represented by the matrix $\pmatrix{1/\sqrt 2& -1/\sqrt 2\\1/\sqrt 2&1\sqrt 2}.$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1248791",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
$a^2 = 2b^3 = 3c^5$ Find the smallest value of $abc$. We have following equation:
$a^2 = 2b^3 = 3c^5$
Where $a, b, c$ are natural numbers.
Find the smallest possible value of product $abc$.
| Since $6\mid c$ we may write $c=2^r3^s$ with $r,s\ge 1$, so that $a^2=3c^5=2^{5r}3^{5s+1}$. Since $a^2$ is a square, $5r$ and $5s+1$ must be even. The minimal $r,s$ are then $r=2$ and $s=1$, so that $c=12$ and $a=864$. It follows that $b=72$, and
$$
abc=a^2=3b^3=2c^5.
$$
This value is minimal, because $c$ is minimal (see the comments above).
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1249775",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
Am I misinterpreting this matrix determinant property? I was reading matrix determinant properties from wikipedia.
The property reads
$\det(cA) = c^n \det(A)$ for $n \times n$ matrix.
However I am not able to realize it. What I find is $\det(cA) = c\det(A)$
For example, multiplying matrix by 2 and then taking the determinant of the resultant matrix:
$
2\begin{bmatrix}
4 & 5 & 6 \\
6 & 5 & 4 \\
4 & 6 & 5 \\
\end{bmatrix}=
\begin{bmatrix}
8 & 10 & 12 \\
6 & 5 & 4 \\
4 & 6 & 5
\end{bmatrix}
$
and
$
\begin{vmatrix}
8 & 10 & 12 \\
6 & 5 & 4 \\
4 & 6 & 5
\end{vmatrix}=60
$
Now first taking the determinant and then multiplying by 2 yields the same result:
$$
2\begin{vmatrix}
4 & 5 & 6 \\
6 & 5 & 4 \\
4 & 6 & 5 \\
\end{vmatrix}
= 2 \cdot 30 = 60
$$
Where I am mistaking?
| You have $\det (cB) = \det (cI) \det B $ and you can see from the formula for $\det$ that $\det (cI) = c^n$.
Another way is to notice that $\det$ is a multilinear function of the columns (or rows), that is, we can write
$\det(A) = f(a_1,...,a_n)$ where $a_k$ is the $k$th row of $A$, and $f$ is
linear in each argument separately.
Then $\det(cA) = f(c a_1,...,c a_n) = c f(a_1, c a_2,...,c a_n) = c^2 f(a_1, a_2,...,c a_n)= \cdots = c^n f(a_1,...,a_n) = c^n \det A$.
Addendum:
You multiplied the matrix incorrectly.
$A=\begin{bmatrix}
4 & 5 & 6 \\
6 & 5 & 4 \\
4 & 6 & 5
\end{bmatrix}$, $2A = \begin{bmatrix}
8 & 10 & 12 \\
12 & 10 & 8 \\
8 & 12 & 10
\end{bmatrix}$.
$\det A = 30, \det (2A) = 240$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1250186",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 4,
"answer_id": 0
} |
In how many ways can $1000000$ be expressed as a product of five distinct positive integers? I'm trying to solve the following problem:
"In how many ways can the number $1000000$ be expressed as a product of five distinct positive integers?"
Here is my attempt:
Since $1000000 = 2^6 \cdot 5^6$, each of its divisors has the form $2^a \cdot 5^b$, and a decomposition of 1000000 into a product of five factors has the form
$$
1000000 = (2^{a_1} \cdot 5^{b_1})(2^{a_2} \cdot 5^{b_2})(2^{a_3} \cdot 5^{b_3})(2^{a_4} \cdot 5^{b_4})(2^{a_5} \cdot 5^{b_5})
$$
where $a_i$ and $b_i$ are nonnegative integers which satisfy the
conditions
$$
a_1 + a_2 + a_3 + a_4 + a_5 = 6, b_1 + b_2 + b_3 + b_4 + b_5 = 6
$$
The total number of systems of $a_i$ which satisfy the first equation is $210$ and the same number is for $b_i$. Thus the total number of decompositions is $210 \cdot 210 = 44100$.
However, in this enumeration, factorizations which differ only in the brder of the factors have been counted separately; that is, some factorizations are counted several times each.
To get the number of distinct unordered decompositions I must, at first, substract the number of ordered decompositions with at least two identical factors and, at second, divide resulting number by $5!$ to leave only unordered ones.
And I'm stuck at the step of counting the number of ordered decompositions with at least two identical factors.
The number of decompositions with $k$ identical factors is $(\lfloor\dfrac{a}{k}\rfloor + 1)(\lfloor\dfrac{b}{k}\rfloor+1){5 \choose k}$, that is, the number of decompositions with two identical factors is $16 \cdot {5 \choose 2} = 160$, three identical factors - $9 \cdot {5 \choose 3} = 90$, four ones - $4 \cdot {5 \choose 4} = 20$ and five ones - $4 \cdot {5 \choose 5} = 4$. Thus the number of distinct decompositions must be equal $\dfrac{44100-160-90-20-4}{5!}$, but this number is not integral. I suppose that the numbers of decompositions with $k$ identical factors overlap and I misuse inclusion-exclusion principle. But I have no idea how I can count that overlapped decompositions.
Please, help!
Thanks!
| For $n{\ge}0$, $\;$ let $F(n)$ be the ways in which the number $10^n$ can be expressed as a product of five distinct positive integers.
Then
$$F(n)=\left(\frac{1}{5!}\right)\left(24\left(\left\lfloor\frac{n}{5}\right\rfloor-\left\lfloor\frac{n-1}{5}\right\rfloor\right)^{2}
+(-30)\left(\left\lfloor\frac{n}{4}\right\rfloor+1\right)^{2}
+(-20)\left(\left\lfloor\frac{n+2}{2}\right\rfloor-\left\lfloor\frac{n+2}{3}\right\rfloor\right)^{2}
+20\left(\left\lfloor\frac{(n+2)(n+3)}{6}\right\rfloor\right)^{2}
+15\left(\frac{2n^{2}+10n+11+(2n+5)(-1)^n}{16}\right)^{2}
+(-10)\left(\frac{4n^{3}+30n^{2}+68n+45+3(-1)^{n}}{48}\right)^{2}
+\left(\frac{(n+1)(n+2)(n+3)(n+4)}{24}\right)^{2}
\right)$$
The answer for this question is $F(6)=194$.
$$F(10)=6355,\quad F(100)=175519523577,\quad F(1000)=14758828112205481602.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1251256",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "22",
"answer_count": 4,
"answer_id": 2
} |
Finding triplets $(a,b,c)$ such that $\sqrt{abc}\in\mathbb N$ divides $(a-1)(b-1)(c-1)$ When I was playing with numbers, I found that there are many triplets of three positive integers $(a,b,c)$ such that
*
*$\color{red}{2\le} a\le b\le c$
*$\sqrt{abc}\in\mathbb N$
*$\sqrt{abc}$ divides $(a-1)(b-1)(c-1)$
Examples : The followings are positive integers.
$$\frac{(2-1)(8-1)(49-1)}{\sqrt{2\cdot 8\cdot 49}},\ \frac{(6-1)(24-1)(529-1)}{\sqrt{6\cdot 24\cdot 529}},\frac{(7-1)(63-1)(3844-1)}{\sqrt{7\cdot 63\cdot 3844}}$$
Then, I began to try to find every such triplet. Then, I found
$$(a,b,c)=(k,km^2,(km^2-1)^2)$$
where $k,m$ are positive integers such that $k\ge 2$ and $km^2\ge 3$, so I knew that there are infinitely many such triplets. However, I can neither find the other triplets nor prove that there are no other triplets. So, here is my question.
Question : How can we find every such triplet $(a,b,c)$?
Added : There are other triplets : $(a,b,c)=(k,k,(k-1)^4)\ (k\ge 3)$ by a user user84413, $(6,24,25),(15,15,16)$ by a user Théophile. Also, from the first example by Théophile, I got $(2k,8k,(2k-1)^2)\ (k\ge 3)$.
Added : $(a,b,c)=(k^2,(k+1)^2,(k+2)^2)\ (k\ge 2)$ found by a user coffeemath. From this example, I got $(k^2,(k+1)^2,(k-1)^2(k+2)^2)\ (k\ge 2)$.
Added : I got $(a,b,c)=(2(2k-1),32(2k-1),(4k-3)^2)\ (k\ge 5)$.
Added : I got $(a,b,c)=(k,(k-1)^2,k(k-2)^2)\ (k\ge 4)$.
Added : A squarefree triplet $(6,10,15)$ and $(4,k^2,(k+1)^2)\ (k\ge 2)$ found by a user martin.
Added : user52733 shows that $(6,10,15)$ is the only squarefree solution.
| We study solutions, if any, of the shape $(a,b,c) = (x^2 P + 1,~ y^2 P + 1,~ z^2)$. We need
$$z^2 \left(x^2 y^2 P^2 + (x^2 + y^2) P + 1 ) \right)$$
be a square. Namely, we need
\begin{align}
2 x^2 y^2 P &= -(x^2 + y^2) + \sqrt{ (x^2 + y^2)^2 + 4(w^2 - 1)x^2 y^2} \\
\implies P &= \frac{ -(x^2 + y^2) + u }{ 2 x^2 y^2}
\end{align}
where we have
$$(x^2 + y^2)^2 + (2wxy)^2 = (2xy)^2 + u^2$$
The equation $X^2 + Y^2 = V^2 + W^2$ like the Pythagorean one has also infinitely many solutions. For arbitrary integer $M,N,r,s$ we have
\begin{align}
2X &= r M + s N & 2Y &= r N – s M \\
2V &= r N + s M & 2W &= r M – s N
\end{align}
making
\begin{align}
2(x^2 + y^2) &= r M + s N & 4wxy &= r N – s M \\
4xy &= rN + sM & 2u &= rM – sN
\end{align}
The main equation
$$(z^2 - 1) x^2 y^2 P^2 = Zzw \quad (?)$$
with $Z$ integer becomes
$$4(z^2 - 1)s^2 N^2 = Zz (r^2 N^2 - s^2 M^2)$$
and we have to choose conveniently, if possible, among the arbitrary parameters in order this equation be verified, in particular $z$ must divide $4s^2 N^2$ for instance. No time for me now. If someone wants to continue go on.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1251576",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "56",
"answer_count": 3,
"answer_id": 2
} |
Finding a matrix from equation we've got the following 4x4 Matrix
$$\begin{pmatrix}
4 & -2 & 3 & 2\\
3 & 5 & 1 & -4\\
-1 & 6 & -4 & -7\\
-2 & 0 & -2 & 4
\end{pmatrix}$$
and I need to find $B$ from the equation: $(A-3I)B=0$.
i started to solve it by finding first $A-3I$. and I got:
$$\begin{pmatrix}
1 & -2 & 3 & 2\\
3 & 2 & 1 & -4\\
-1 & 6 & -7 & -7\\
-2 & 0 & -2 & 1
\end{pmatrix}$$
now I know that every column $[AB]^j$ [$j$ represents column number] can be calculated by $A[B]^j$ [$j$ represents column number].
and I was trying to solve it by multiplying $A$ with a specific column in $B$ but I wasn't able to reach the zero matrix.
EDIT:
almost forgot to mention that B needs to be 4x4 matrix and different form 0!
| $$\begin{pmatrix}
1 & -2 & 3 & 2\\
3 & 2 & 1 & -4\\
-1 & 6 & -7 & -7\\
-2 & 0 & -2 & 1
\end{pmatrix}$$
is a zero divisor since : column$(3)=$column$(1)-$column$(2)$.
So the matrix:
$$B=\begin{pmatrix}
0 & 0 & -a & 0\\
0 & 0 & a & 0\\
0 & 0 & a & 0\\
0 & 0 & 0 & 0
\end{pmatrix}$$
gives $AB=0$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1252645",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Poisson Integral relation If $$ I_n(r) = \int_0^\pi \frac{\cos nx}{r^2-2r\cos x+1} \, dx $$
How to prove that
$$ I_{n-1}(r)+I_{n+1}(r)= \left(r+\frac{1}{r}\right)I_n(r)\text{ ?}$$
I only find that $$I_{n-1}(r)+I_{n+1}(r)= \int_0^\pi \frac{2\cos nx\cos x}{r^2-2r\cos x+1} \, dx$$
| $$\begin{aligned}
I_{n-1}(r)+I_{n+1}(r) &=\int_0^{\pi}\frac{2\cos(nx) \cos x}{r^2-2r\cos x+1}\,dx \\
&=-\frac{1}{r}\int_0^{\pi} \frac{\cos(nx)(r^2-2r\cos x+1-r^2-1)}{r^2-2r\cos x+1}\,dx \\
&=-\frac{1}{r}\int_0^{\pi}\cos(nx)\,dx+\frac{r^2+1}{r}\int_0^{\pi} \frac{\cos nx}{r^2-2r\cos x+1}\,dx\\
&=\left(r+\frac{1}{r}\right)I_n(r) \\
\end{aligned}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1252896",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 1,
"answer_id": 0
} |
Factorize Trigonometric Equation: $ 3\sin(x)^2 - 2\sin(x)\cos(x) - \cos(x)^2 = 0 $ I have a problem with the following trigonometric equation:
$$ 3\sin(x)^2 - 2\sin(x)\cos(x) - \cos(x)^2 = 0 $$
It's from the book Engineering Mathematics 7th edition by Stroud.
The book is giving the answer, but I can't seem to be able to find out how to factorize it. I can't figure it out.
Consider the following solution:
This equation can be written as $3\sin ^2x-\cos ^2x = 2\sin x \cos x.$
That is: $3\sin ^2x-2\sin x \cos x-\cos ^2 x = 0.$
That is: $(3\sin x + \cos x)(\sin x - \cos x) = 0.$
So that $3 \sin x \cos x = 0$ or $\sin x - \cos x = 0.$
If $3 \sin x + \cos x = 0,$ then $\tan x = \frac{-1}{3},$ and so $x = -0.3218 ± n \pi,$ and if $\sin x - \cos x = 0,$ then $\tan x = 1,$ and so $x = \frac{\pi}{4}.$
If anyone could help me understand how to factorize this equation to get the one shown in the image it would help me very much.
Thank you in advance.
| If you put $\sin x = a$, and $\cos x = b$, then you might be able to see the "structure" of the equation:
$$\begin{align} 3\sin^2x - 2\sin x \cos x - \cos^2x & = 3a^2 - 2ab - b^2 \\
&= 3a^2-3ab+ab-b^2\\ & =3a(a-b)+b(a-b)\\ & =(a-b)(3a+b)\\ & = (3a+ b)(a-b) \\ & = (3\sin x + \cos x)(\sin x - \cos x) = 0\end{align}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1254334",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
} |
How can a binomial coefficient can be approximated by using Stirling's formula? I've met some difficulties with such question:
How can we approximate a binomial coefficient by using a Stirling's factorial approximation.
I've evaluate a little bit and got this
How can I transform the right side of this equation for getting estimation like (1 + ?/n + O(1/n^2))
| Another case is when
$k$ is a constant times $n$.
Let
$k = a n $
where $0 < a < 1$,
so $k/n = a$.
Then
$\begin{align}
\binom{n}{k}
&=\frac{n!}{k!(n-k)!}\\\\
&\approx. \frac{\sqrt{2\pi n}(\frac{n}{e})^n} {\sqrt{2\pi k}(\frac{k}{e})^k \sqrt{2\pi (n-k)}(\frac{n-k}{e})^{n-k}}\\\\
&=\frac{1}{\sqrt{2\pi k}(k/n)^k \sqrt{1-(k/n)}(1-(k/n))^{n-k}}\\
&=\frac{1}{\sqrt{2\pi k(1-k/n)}a^{an} (1-a)^{n-an}}\\
&=\frac{1}{\sqrt{2\pi n a(1-a)}(a^a(1-a)^{1-a})^n}\\
\end{align}
$
For example,
if $a = 1/2$,
since
$a(1-a) = 1/4$
and
$a^a (1-a)^{1-a}
=(1/2)^{1/2}(1/2)^{1/2}
=1/2
$,
this becomes
$\binom{n}{n/2}
\approx \frac{1}{\sqrt{2n\pi/4}(1/2)^n}
= \frac{2^n}{\sqrt{n\pi/2}}
$.
For another example,
if $a = 1/3$,
since
$a(1-a) = 2/9$
and
$a^a (1-a)^{1-a}
=(1/3)^{1/3}(2/3)^{2/3}
=\frac{(2^2)^{1/3}}{3}
=\frac{4^{1/3}}{3}
=(4/27)^{1/3}
$,
this becomes
$\binom{n}{n/3}
\approx \frac{1}{\sqrt{2n\pi(2/9)}(4/27)^{n/3}}
= \frac{3(27/4)^{n/3}}{2\sqrt{n\pi}}
$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1256545",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 3,
"answer_id": 0
} |
Diophantine equation $x^2 + xy + y^2 = \left({{x+y}\over{3}} + 1\right)^3$. Solve in integers the equation$$x^2 + xy + y^2 = \left({{x+y}\over3} + 1\right)^3.$$
| Setting $x+y=3t$ and $xy=s$, we obtain that
$$9t^2-s = (t+1)^3 \implies s = -t^3+6t^2-3t-1$$
$x$ and $y$ satisfying the quadratic $a^2 -3ta + s =0$. This means $9t^2-4s = k^2$. Eliminating $s$, we obtain
$$4t^3-15t^2+12t+4 = k^2 \implies 64t^3 - 240t^2 + 192t + 64 = (4k)^2$$
$$(4t-5)^3 - 108t + 189 = (8k)^2 \implies (4t-5)^3 - 27(4t-5) + 54 = (4k)^2$$
Hence, it boils down to solving for integer points on the elliptic curve
$$Y^2 = X^3 - 27X+54 = (X-3)^2(X+6)$$
Hence, we need $X+6=m^2$, i.e., $4t+1=m^2 \implies t = \dfrac{m^2-1}4$. This means $m$ has to be odd. Choosing $m=2p+1$, we obtain
$$t = \dfrac{4p^2+4p}4 = p^2 + p$$
This gives us that
$$Y = m(m^2-9) = (2p+1)(4p^2+4p-8) \implies k=(2p+1)(p+2)(p-1)$$
This gives us $x,y = \dfrac{3t \pm k}2 = t + \dfrac{t \pm k}2$.
Hence, to summarize,
*
*Pick any $p \in \mathbb{Z}$.
*Set $t=p(p+1)$ and $k = (2p+1)(p+2)(p-1)$
*This gives us $\boxed{\color{blue}{x,y = t + \dfrac{t \pm k}2}}$
If we only want unordered pairs, it suffice to vary $p$ just over $\mathbb{N}$. The first few values by varying $p$ from $0$ to $5$ are
\begin{array}{|c|c|c|}
\hline
p & x & y\\
\hline
0 & -1 & 1\\
1 & 3 & 3\\
2 & 19 & -1\\
3 & 53 & -17\\
4 & 111 & -51\\
5 & 199 & -109\\
\vdots & \vdots & \vdots\\
\hline
\end{array}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1256663",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
} |
Prove that $n^2(n^2+1)(n^2-1)$ is a multiple of $5$ for any integer $n$. Prove that $n^2(n^2+1)(n^2-1)$ is a multiple of $5$ for any integer $n$.
I was thinking of using induction, but wasn't really sure how to do it.
| If one wishes to use induction:
The statement holds for $n=0$, so let us assume it's true for $n=k$: $$k^2(k^2+1)(k^2-1)=(k-1)k^2(k+1)(k^2+1)=5m $$ and consider it for $n=k+1$. We have $$(k+1)^2((k+1)^2+1)((k+1)^2-1)=\\ (k+1)^2(k^2+2k+2)(k^2+2k)=\\ k(k+1)^2(k+2)(k^2+2k+2).$$ Suppose this isn't a multiple of $5$. Then none of its factor is, so our inductive hypothesis yields that for some integer $a,b$ either $k-1=5a$ or $k^2+1=5b$, or both:
*
*if both, then $$(k-1)+(k^2+1)=k(k+1)=5(a+b),$$ contradiction;
*if only the former, we have $$k+4=5(a+1),$$ whence $$ k^2+2k+2=25(a+1)^2-6k-14=25(a+1)^2-2(3k+7)=\\25(a+1)^2-2(3(k-1)+10)=5\left(5(a+1)^2-2\left(\frac{3(k-1)}{5}+2\right)\right),$$ contradiction;
*if only the latter, we also have for some integer $c$ $$k+3=5c,$$ whence $$k^2+2k+2=k^2+2k+7-5=(k^2+1)+2(k+3)-5=5(b+2c)-5=5(b+2c-1),$$ contradiction.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1257632",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 10,
"answer_id": 4
} |
Help with solving mathematical induction problem I need help with the following:
Use mathematical induction to prove that for every $n\in N$,
$$
\sum_{k=1}^n\frac{1}{\cos kx \cos(k+1)x}=\frac{\tan(n+1)x-\tan x}{\sin x}
$$
For $n=1$, the statement is true.
Suppose that the statement is true for $n=m\in N$, and prove that it is true for $n=m+1$.
$$\sum_{k=1}^{m+1} \frac{1}{\cos kx \cos(k+1)x}=\frac{\tan(m+2)x-\tan x}{\sin x}(*)$$
Proof:
$$\frac{\tan(m+2)x-\tan x}{\sin x}=\frac{\tan(m+1)x-\tan x}{\sin x}+\frac{1}{\cos(k+1)x \cos(k+2)x}$$
Using trigonometry rule $\cos x \cos y$, the right side of equation is
$$\frac{(\cos x+\cos(2k+3)x)(\tan(m+1)x-\tan x)+2\sin x}{\sin x(\cos x+\cos(2k+3)x)}$$
How to further transform this expression to become $(*)$
Thanks for replies.
| $$\frac{\tan(m+2)x-\tan x}{\sin x}-\frac{\tan(m+1)x-\tan x}{\sin x}$$
$$=\frac{\tan(m+2)x-\tan(m+1)x}{\sin x}$$
$$=\dfrac{\sin[(m+2)x-\sin(m+1)x]}{\cos(m+2)x\cdot\cos(m+1)x\cdot \sin x}$$
$$=\dfrac1{\cos(m+2)x\cdot\cos(m+1)x}$$ if $\sin x\ne0$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1262046",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
need some help with a power series convergence test problem Find the interval of convergence for the given power series:
$$\sum\limits_{n=1}^\infty \frac{(x - 1)^n }{n(-4)^n}$$
First I applied the generalized ratio test, came out with $\frac{(1-x)}{4}$
Solved the inequality $|1-x| \lt 4$ and got $-3 \lt x \lt 5$.
But webwork refuses to accept my answer. Am I doing something wrong?
| $$\sum\limits_{n=1}^{\infty} \frac{(x - 1)^n }{n(-4)^n}$$
Using the ratio test, we have
$$ \lim\limits_{n\to\infty} \left|\frac{\frac{(x - 1)^{n+1}}{(n+1)(-4)^{n+1}}}{\frac{(x - 1)^n }{n(-4)^n}}\right|$$
$$ =\lim\limits_{n\to\infty} \left|\frac{(x - 1)^{n+1}n(-4)^n}{(x-1)^n(n+1)(-4)^{n+1}}\right|$$
$$ =\lim\limits_{n\to\infty} \left|\frac{(x - 1)n}{(n+1)(-4)}\right|$$
$$ =\frac14\left|x-1\right|\lim\limits_{n\to\infty} \left|\frac{n}{n+1}\right|$$
$$ =\frac14\left|x-1\right|\lim\limits_{n\to\infty} \left|\frac{1}{1+\frac1n}\right|$$
$$ =\frac14\left|x-1\right|$$
Now let's find the interval of convergence
$$ \frac14\left|x-1\right|\lt 1 $$
$$ \left|x-1\right|\lt 4 $$
$$ -4\lt x-1\lt 4 $$
$$ -3\lt x\lt 5 $$
At $x=-3$, we have
$$\sum\limits_{n=1}^{\infty} \frac{(-4)^n }{n(-4)^n}= \sum\limits_{n=1}^{\infty} \frac{1}{n}\Rightarrow \mbox{diverges} $$
At $x=5$, we have
$$\sum\limits_{n=1}^{\infty} \frac{4^n }{n(-4)^n}= \sum\limits_{n=1}^{\infty} \frac{(-1)^n}{n}\Rightarrow \mbox{converges} $$
Therefore, the interval of convergence is
$$ -3\lt x\leq 5 $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1264123",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 3
} |
A question on the Lagrange Inversion Formula I have to use the L.I.F. for
\begin{align*}
s\left(x,y\right)=\frac{1}{2}\left(1-x-y-\sqrt{1-2x-2y-2xy+x^2+y^2}\right)
\end{align*}
to obtain that
\begin{align*}
s\left(x,y\right) = \sum_{p,q\geq1}\frac{1}{p+q-1}\binom{p+q-1}{p}\binom{p+q-1}{q}x^py^q
\end{align*}
I've tried finding some resources on examples on how to do this but so far I am quite clueless. Any help or suggestions are appreciated.
| We seek to use Lagrange Inversion to show that
$$s(x,y) = \frac{1}{2}
\left(1-x-y-\sqrt{1-2x-2y-2xy+x^2+y^2}\right)$$
has the series expansion
$$\sum_{p,q\ge 1} \frac{1}{p+q-1}
{p+q-1\choose p} {p+q-1\choose q} x^p y^q.$$
On squaring we obtain
$$4 s(x,y)^2 = (1-x-y)^2 + 1-2x-2y-2xy+x^2+y^2
\\ - 2(1-x-y) (1-x-y-2s(x,y))
\\ = 2 (1-x-y)^2 - 4xy - 2(1-x-y) (1-x-y-2s(x,y))
\\ = - 4xy + 4 (1-x-y) s(x, y).$$
We finally get
$$s(x,y)^2 = -xy + (1-x-y) s(x,y).$$
This implies
$$x = \frac{s(x,y) (1 - y - s(x,y))}{y + s(x,y)}.$$
We get with $p\ge 1$
$$[x^p] s(x, y) = \frac{1}{p} [x^{p-1}] \frac{d}{dx} s(x, y)
= \frac{1}{p} \; \underset{x}{\mathrm{res}} \;
\frac{1}{x^p} \frac{d}{dx} s(x,y).$$
Now put $s(x,y) = u$ so that $\frac{d}{dx} s(x, y) \; dx = du$
$$\frac{1}{p} \; \underset{u}{\mathrm{res}} \;
\frac{(y+u)^p}{u^p (1-y-u)^p}
\\ = \frac{1}{p} \frac{1}{(1-y)^p} \; \underset{u}{\mathrm{res}} \;
\frac{(y+u)^p}{u^p (1-u/(1-y))^p}.$$
This is
$$\frac{1}{p} \frac{1}{(1-y)^p}
\sum_{r=0}^{p-1} {p\choose r} y^{p-r}
{2p-2-r\choose p-1} \frac{1}{(1-y)^{p-1-r}}.$$
Extracting the coefficient on $[y^q]$ where we see that $q\ge 1$:
$$\frac{1}{p}
\sum_{r=0}^{p-1} {p\choose r} [y^q] y^{p-r}
{2p-2-r\choose p-1} \frac{1}{(1-y)^{2p-1-r}}
\\ = \frac{1}{p}
\sum_{r=0}^{p-1} {p\choose r}
{2p-2-r\choose p-1} {q+p-2\choose 2p-2-r}.$$
Next observe that
$${2p-2-r\choose p-1} {q+p-2\choose 2p-2-r}
= \frac{(q+p-2)!}{(p-1)! \times (p-1-r)! \times (q+r-p)!}
\\ = {p+q-2\choose p-1} {q-1\choose p-1-r}.$$
We get for our sum
$$\frac{1}{p} {p+q-2\choose p-1}
\sum_{r=0}^{p-1} {p\choose r} {q-1\choose p-1-r}
\\ = \frac{1}{p} {p+q-2\choose p-1} {p+q-1\choose p-1}.$$
This was Vandermonde. Some binomial coefficient algebra now yields
$$\bbox[5px,border:2px solid #00A000]{
\frac{1}{p+q-1} {p+q-1\choose p} {p+q-1\choose q}}$$
as claimed.
This was from Egorychev's Combinatorial sums.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1266250",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Evaluate $\int_0^1 \int_\sqrt{y}^1 \int_0^{x^2+y^2} dz dx dy$.
Evaluate $\int_0^1 \int_\sqrt{y}^1 \int_0^{x^2+y^2} dz dx dy$.
Attempt:
$$ \int_0^1 \int_\sqrt{y}^1 \int_0^{x^2+y^2} dz dx dy
= \int_0^1 \int_\sqrt{y}^1 x^2 + y^2 dx dy
= 1/3 + 1/3 - 2/15 - 2/7 = \frac{26}{105}.$$
However, the solution should be $26/35$ so it appears that I'm off by a factor of $3$ but I don't see where this fits in. Where did I go wrong?
| There is nothing wrong with your calculations.
$$\int_0^1 \int_{y^{1/2}}^1 x^2 + y^2 \ dx \ dy = \int_0^1 y^2 - y^{5/2} + \frac13(1-y^{3/2}) \ dy \\
= \frac 13 - \frac 27 + \frac 13 - \frac 2{15} = \frac{26}{105}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1268849",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Find solutions to $\cot(x)+\csc(x)=\sqrt3$ in range $[0,2\pi]$ What is the best way to do the above? Are there any tricks I should be aware of.
I know how to simplify it to $\dfrac{\cos(x)}{\sin(x)} + \dfrac{1}{\sin(x)} = \sqrt{3}$
so multiplying both sides by $\sin(x)$, we get $\cos(x)+1=\sqrt{3}\sin(x)$.
But I'm stuck from there.
| Let $x = 2t$, then we can start at where you left off:
$\cos x + 1 = \sqrt{3}\sin x\Rightarrow \cos (2t)+1 = \sqrt{3}\sin (2t) \Rightarrow 2\cos^2 t = 2\sqrt{3}\sin t\cos t \Rightarrow 2\cos t(\cos t - \sqrt{3}\sin t) = 0\Rightarrow \cos t = 0 \Rightarrow t = (2n+1)\dfrac{\pi}{2} \Rightarrow x = 2t = (2n+1)\pi, n \in \mathbb{Z}$ or $\cos t - \sqrt{3}\sin t = 0 \Rightarrow 2\cos (t+\frac{\pi}{3})=0 \Rightarrow t+\dfrac{\pi}{3} = (2n+1)\dfrac{\pi}{2} \Rightarrow t = n\pi + \dfrac{\pi}{6} \Rightarrow x = 2t = 2n\pi + \dfrac{\pi}{3}, n \in \mathbb{Z}$. Let $n = 0$ we obtain $x = \pi, \dfrac{\pi}{3}$ are the only $2$ solutions that belong to the range $[0,2\pi]$, and they are the only solutions.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1269194",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 6,
"answer_id": 4
} |
Continued product in $\sin$ series Find the value of the product $$(\sin 1°)(\sin 3°)(\sin 5°)\ldots(\sin 89°)$$ I tried multiplying and dividing by $2$ and then combining and then converting into cosine, but doesn't work out.
| Notice for any integer $N > 0$, we have
$$\begin{align}
z^{2N} + 1
&= \prod_{k=-N}^{N-1} \left( z - e^{\frac{2k+1}{2N}\pi i} \right)
= \prod_{k=0}^{N-1}\left(z - e^{\frac{2k+1}{2N}\pi i}\right)\left(z - e^{-\frac{2k+1}{2N}\pi i}\right)\\
&= \prod_{k=0}^{N-1}\left[ z^2+1 - 2z\cos\left(\frac{2k+1}{2N}\pi\right)\right]
\end{align}
$$
Set $z = 1$, we find
$$2
= 2^N \prod_{k=0}^{N-1} \left[1 - \cos\left(\frac{2k+1}{2N}\pi\right)\right]
= 2^{2N} \prod_{k=0}^{N-1} \sin^2\left(\frac{2k+1}{4N}\pi\right)\\
$$
Since all the $\sin(\cdots)$ involved are positive, this leads to
$$\prod_{k=0}^{N-1} \sin\left(\frac{2k+1}{4N}\pi\right) = 2^{\frac12 - N}$$
Substitute $N = 45$, we get
$$\sin(1^\circ)\sin(3^\circ)\cdots\sin(89^\circ)
= \prod_{k=0}^{44}\sin\left(\frac{2k+1}{180}\pi\right) =
2^{-89/2}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1270286",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
} |
$f(x)$ is a polynomial satisfying $2 + f(x)f(y)=f(x)+f(y)+f(xy)$, find $f(f(2)$), given $f(2)=5.$
If f(x) is a polynomial satisfying $2 + f(x)f(y)=f(x)+f(y)+f(xy)$, find $f(f(2))$, given $f(2)=5.$
ATTEMPT:- $f(f(2))=f(5)$, We can find $f(0)$,$f(1)$ and $f(1/2)$ to be $1,2$ and $5/4$ respectively.
we can change the function to the form $g(x)*g(y)=g(xy)$ by basic transformation.
but how to get $f(5)$ without using transformations.
| Notice that you can get values for $f(2n) = (2n)^2 + 1$. Observing this, I suppose that $f$ is a quadratic polynomial with form $f(t) = at^2 + bt + 1$ ($+1$ since we know that $f(0) = 1$).
Writing the functional equation as $(f(x) - 1)(f(y) - 1) = f(xy) - 1$, we get $(ax^2 + bx)(ay^2 + by) = (ax^2 y^2 + bxy)$, so comparing coefficients we get the following system:
$a^2 = a$
$ab = 0$
$b^2 = b$
So this gives answers of $(1, 0), (0,1),$ and $(0,0)$. This gives answers of $f(x) = 1, f(x) = x+1, f(x) = x^2 + 1$. Using our initial condition $f(2) = 5$, we know that only the last equation can be correct.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1270658",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
How to interpret relation of variables as eigenvector? I am trying to calculate the eigenvectors of a square matrix $A \in \mathbb{R}^{4x4}$.
$$A =
\begin{pmatrix}
a & 1 & 0 & 0 \\
0 & a & 1 & 0 \\
0 & 0 & a & 0 \\
0 & 0 & 0 & b
\end{pmatrix}
$$
To receive the eigenvalues I did:
$$\chi_A = \begin{vmatrix}
a-\lambda & 1 & 0 & 0 \\
0 & a-\lambda & 1 & 0 \\
0 & 0 & a-\lambda & 0 \\
0 & 0 & 0 & b-\lambda
\end{vmatrix} = (\lambda-a)^3(\lambda-b)$$
This means we have $\lambda_1 = a$ with multiplicity $m=3$ and $\lambda_2 = b$ with multiplicity $m=1$.
Now I try to calculate the eigenvectors first for $\lambda_1 = a$
$$\begin{bmatrix}
0 & 1 & 0 & 0 \\
0 & 0 & 1 & 0 \\
0 & 0 & 0 & 0 \\
0 & 0 & 0 & b-a
\end{bmatrix} = \begin{bmatrix}
0 \\
0 \\
0 \\
0
\end{bmatrix}$$
which gives us $x_1 \in \mathbb{R}$, $x_2 = 0$, $x_3 = 0$ and $(b-a)x_4 = 0$.
When I try to convert this result to the eigenvector for eigenvalue $a$ I am stuck with
$$Eig(\lambda_1) = <\begin{pmatrix} 1 \\ 0 \\ 0 \\ ?\end{pmatrix}>$$
I do not know what value (or variables) I could use for the fourth entry as their value depends on the statement of $(b-a)x_4 = 0$ being true.
| There are two cases to separate:
*
*If $a = b$, then you 'only' get one eigenvalue $a$ with multiplicity 4. In this case
$$\begin{bmatrix}
0 & 1 & 0 & 0 \\
0 & 0 & 1 & 0 \\
0 & 0 & 0 & 0 \\
0 & 0 & 0 & b-a
\end{bmatrix} = \begin{bmatrix}
0 & 1 & 0 & 0 \\
0 & 0 & 1 & 0 \\
0 & 0 & 0 & 0 \\
0 & 0 & 0 & 0
\end{bmatrix} = \begin{bmatrix}
0 \\
0 \\
0 \\
0
\end{bmatrix}$$
implies $x_2 = x_3 = 0$ and $x_1,x_4 \in \mathbb{R}$ arbitrary. Hence,
$$Eig(\lambda_1 = a) = < \begin{pmatrix} 1 \\ 0 \\ 0 \\ 0 \end{pmatrix}, \begin{pmatrix} 0 \\ 0 \\ 0 \\ 1 \end{pmatrix} >$$
2. If $a \neq b$, then $b-a \neq 0$ and we can divide by it. Hence, it follows that $x_4 = 0$ and we have
$$Eig(\lambda_1 = a) = < \begin{pmatrix} 1 \\ 0 \\ 0 \\ 0 \end{pmatrix} >$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1271090",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Let $a,b$ be relative integers such that $2a+3b$ is divisible by $11$. Prove that $a^2-5b^2$ is also divisible by $11$. The divisibility for $11$ of $a^2 - 5b^2$ can be easily verified; in fact: $$a \equiv \frac {-3}{2}b \pmod {11}$$ therefore $$\frac {9}{4}\cdot b^2 - 5b^2 = 11(-\frac{b^2}{4}) \equiv 0 \pmod {11}.$$
The solution doesn't need the use of the rules of modular-arithmetic. How can I demonstrate it?
| We have for some integer $c$ $$2a+3b=11c,$$ and this is equivalent to $$a=\frac{11c-3b}{2}.$$ Now, squaring both sides we get $$a^2=\frac{11^2c^2-66bc+9b^2}{4} $$ and substracting $5b^2$: $$a^2-5b^2=\frac{11^2c^2-66bc-11b^2}{4} $$ $$ a^2-5b^2=11\left(\frac{11c^2-6bc-b^2}{4}\right).\tag{$\star$}$$ Note that the quantity in brackets in the RHS of $(\star)$ is an integer because $11$ and $4$ are coprime, so the numerator of the fraction must be divisible by $4$ in order to make the RHS itself an integer, just as the LHS is.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1271972",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 0
} |
$\sum \limits_{n \geq 0}a_n \frac{x^n}{n!}=e^{x+x^2/2}$ implies $a_n \sim \frac1{\sqrt2} n^{\frac n2}e^{ -\frac n2+\sqrt n -\frac14 }$
Prove the following asymptotic formula for the exponential generating function coefficients of $e^{x+x^2/2}$: $\; \; a_n \sim \frac1{\sqrt2} n^{\frac n2}e^{ -\frac n2+\sqrt n -\frac14 }$
Stanley Richard writes in equation 1.12 of his book Enumerative Combinatorics the preceding formula without a proof, claiming it is 'routine with complex variable theory'.
It was already deduced that $a_n=\sum\limits_{j \geq 0} \binom{n}{2j}\frac{(2j)!}{2^jj!}$, so I tried to use Stirling's approximation on it and ended with a horrendous sum involving $(n-2j)^{n-2j}$, which seems clearly off-the-path.
| Note: This answer follows section 5.4, example 2 of H.Wilf's Generatingfunctionology
Observe that $f(x)=e^{x+\frac{x^2}{2}}$ is an entire function which can therefore be written as power series
\begin{align*}
f(x)=\sum_{n\geq 0}a_nx^n
\end{align*}
converging everywhere in the complex plane. For this type of function we can check if Hayman's method is applicable.
$$$$
Theorem (Hayman): Let $f(x)=\sum_{n\geq 0}a_nx^n$ be an admissible function.
Let $r_n$ be the positive real root of the equation $a(r_n) = n$, for each $n = 1,2,\ldots$ where $a(r)$ is given by
\begin{align*}
a(r)=r\frac{f^{\prime}(r)}{f(r)}.
\end{align*}
Then
\begin{align*}
a_n\sim \frac{f(r_n)}{r_n^n\sqrt{2\pi b(r_n)}}\qquad\qquad (n\rightarrow \infty)\tag{1}
\end{align*}
where $b(r)$ is given by
\begin{align*}
b(r)=ra^{\prime}(r)=r\frac{f^{\prime}(r)}{f(r)}+r^2\frac{f^{\prime\prime}(r)}{f(r)}-r^2\left(\frac{f^{\prime}(r)}{f(r)}\right)^2.
\end{align*}
We have to check if our function $f(x)$ is admissable in order to apply Hayman's method. One criterion for admissability is according to H.Wilf, section 5.4 (E):
Let $P(x)$ be a nonconstant polynomial with real coeffcients, and let $f(x) = e^{P(x)}$. If $a_n > 0$ for all suffciently large $n$, then $f(x)$ is admissible in the plane.
Indeed, since $f(x)=e^{x+\frac{x^2}{2}}$ has this shape with $P(x)=x+\frac{x^2}{2}$ and the coefficients $a_n=\sum\limits_{j \geq 0} \binom{n}{2j}\frac{(2j)!}{2^jj!}>0$ for all $n$, we can apply this method.
We start with
Calculation of $a(r)$ and $b(r)$
We find that
\begin{align*}
a(r)=r\frac{f^{\prime}(r)}{f(r)}=r+r^2\qquad\text{ and }\qquad b(r)=ra^{\prime}(r)=r+2r^2
\end{align*}
Calculation of $r_n$
According to Hayman's theorem we have to solve the quadratic equation
\begin{align*}
a(r)=r+r^2=n\tag{2}
\end{align*}
and get
\begin{align*}
r=-\frac{1}{2}\pm\frac{1}{2}\sqrt{4n+1}
\end{align*}
Since we need the positive real root $r_n$, we obtain using the binomial series representation of the square root of $n$ following asymptotic estimation for $r_n$
\begin{align*}
r_n&=-\frac{1}{2}+\sqrt{n+\frac{1}{4}}\\
&=-\frac{1}{2}+\sqrt{n}\left(1+\frac{1}{4n}\right)^{\frac{1}{2}}\\
&=-\frac{1}{2}+\sqrt{n}\sum_{k\geq 0}\binom{\frac{1}{2}}{k}\frac{1}{(4n)^{k}}\\
&=-\frac{1}{2}+\sqrt{n}\left(1+\frac{1}{8n}+\mathcal{O}(n^{-2})\right)\tag{3}\\
&=\sqrt{n}-\frac{1}{2}+\frac{1}{8\sqrt{n}}+\mathcal{O}(n^{-\frac{3}{2}})\tag{4}
\end{align*}
Comment:
*
*In (3) we take the summands with $k=0,1$, since every other summand with $k>1$ has already order of $\mathcal{O}(n^{-2})$
To calculate the asymptotic estimation for $a_n$ we need the asymptotic estimations for $r_n^n, f(r_n), b(r_n)$. We start with
Asymptotic estimation of $f(r_n)$
We obtain
\begin{align*}
f(r_n)&=e^{r_n+\frac{1}{2}r_n^2}\\
&=e^{\frac{1}{2}(r_n+n)}\tag{5}\\
&=e^{\frac{n}{2}}e^{\frac{r_n}{2}}\\
&=e^{\frac{n}{2}}\exp\left(\frac{1}{2}\sqrt{n}-\frac{1}{4}+\mathcal{O}(n^{-\frac{1}{2}})\right)\tag{6}\\
&=\exp\left(\frac{n}{2}+\frac{1}{2}\sqrt{n}-\frac{1}{4}+\mathcal{O}(n^{-\frac{1}{2}})\right)\\
&\sim e^{\frac{n}{2}+\frac{1}{2}\sqrt{n}-\frac{1}{4}}\qquad\qquad\qquad\qquad\qquad\qquad (n\rightarrow \infty)
\end{align*}
Comment:
*
*In (5) we use the relationship $r_n+r_n^2=n$ from (2)
*In (6) we use the expansion of $r_n$ in (4) up to the order of $n^{-\frac{1}{2}}$
Asymptotic estimation of $b(r_n)$
We get
\begin{align*}
b(r_n)&=r_n+2r_n^2\\
&=n+r_n^2\tag{7}\\
&=n+\left(\sqrt{n}+\mathcal{O}(1)\right)^2\tag{8}\\
&\sim 2n\qquad\qquad\qquad\qquad\qquad\qquad (n\rightarrow \infty)
\end{align*}
Comment:
*
*In (7) we use the relationship $r_n+r_n^2=n$ from (2)
*In (8) we use the expansion of $r_n$ in (4) up to the order of constants
Asymptotic estimation of $r_n^n$
We use $a^b=\exp(b\ln(a))$ and the Taylor series for $\ln$ to get
\begin{align*}
r_n^n&=\left(\sqrt{n}-\frac{1}{2}+\mathcal{O}(n^{-\frac{1}{2}})\right)^n\\
&=n^{\frac{n}{2}}\left(1-\frac{1}{2\sqrt{n}}+\mathcal{O}(n^{-1})\right)^n\\
&=n^{\frac{n}{2}}\exp\left(n\ln\left(1-\frac{1}{2\sqrt{n}}+\mathcal{O}(n^{-1})\right)\right)\\
&=n^{\frac{n}{2}}\exp\left(-n\sum_{k\geq 1}\left(\frac{1}{2\sqrt{n}}+\mathcal{O}(n^{-1})\right)^k\frac{1}{k}\right)\tag{9}\\
&=n^{\frac{n}{2}}\exp\left(-n\left(\frac{1}{2\sqrt{n}}+\mathcal{O}(n^{-1})\right)\right)\tag{10}\\
&=n^{\frac{n}{2}}\exp\left(-\frac{1}{2}\sqrt{n}+\mathcal{O}(1)\right)\\
&\sim n^{\frac{n}{2}}e^{-\frac{\sqrt{n}}{2}}\qquad\qquad\qquad\qquad\qquad\qquad (n\rightarrow \infty)
\end{align*}
Comment:
*
*In (9) we use $\mathcal{O}(n^{-1})=-\mathcal{O}(n^{-1})$
*In (10) we take the first summand with $k=1$, since every other summand with $k>1$ has already order of $\mathcal{O}(n^{-1})$
Final step: Asymptotic estimation of $a_n$
We can now put everything together to calculate the asymptotic expansion of $a_n$ according to (1). One final aspect we have to consider is, that the admissability of $e^{x+\frac{1}{2}x^2}$ was based upon the representation
$$e^{P(x)}=\sum_{n\geq 0}\frac{t_n}{n!}$$
So, we have the situation
\begin{align*}
a_n=\frac{t_n}{n!}\sim \frac{f(r_n)}{r_n^n\sqrt{2\pi b(r_n)}}\qquad\qquad (n\rightarrow \infty)
\end{align*}
and we will therefore also take the Stirling approximation of $n!$
$$n!\sim \sqrt{2\pi}n^{n+\frac{1}{2}}e^{-n}\qquad\qquad\qquad\qquad (n\rightarrow \infty)$$
We finally obtain
\begin{align*}
a_n &\sim n!\frac{f(r_n)}{r_n^n\sqrt{2\pi b(r_n)}}\\
&=\sqrt{2\pi}n^{n+\frac{1}{2}}e^{-n}
\frac{e^{\frac{n}{2}+\frac{1}{2}\sqrt{n}-\frac{1}{4}}}{n^{\frac{n}{2}}e^{-\frac{\sqrt{n}}{2}}\sqrt{4\pi n}}\\
&=\frac{1}{\sqrt{2}}n^{\frac{n}{2}}e^{-\frac{n}{2}+\sqrt{n}-\frac{1}{4}}
\end{align*}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1272904",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Probability Roulette Problem Suppose the Roulette table has 37 numbers (European Roulette table). During 37 spins, I always do the same bet: 35 numbers straight (35 chips in 35 different numbers).
Then:
*
*the probability of winning the 37 consecutive spins is $(\frac{35}{37})^{37}\approx 0.1279$,
*the probability of losing at least once during the 37 spins is $1-(\frac{35}{37})^{37}\approx 0.8720$,
*the probability of losing exactly once during the 37 spins is $\binom{37}{1}(\frac{2}{37})(\frac{35}{37})^{36}\approx 0.2705$
*the probability of losing exactly twice during the 37 spins is $\binom{37}{2}(\frac{2}{37})^2(\frac{35}{37})^{35}\approx 0.2782$
*the probability of losing exactly 3 times during the 37 spins is $\binom{37}{3}(\frac{2}{37})^3(\frac{35}{37})^{34}\approx 0.1855$
*the probability of losing exactly $m$ times during the 37 spins is $\binom{37}{m}(\frac{2}{37})^m(\frac{35}{37})^{37-m}$, being decreasing when $2\leq m\leq 37$, reaching the lowest value $(\frac{2}{35})^{37}$ (losing the 37 spins).
Why the probability of losing exactly twice during the 37 spins is greater than the probability of losing exactly once during the 37 spins? I expected to get a decreasing sequence of probabilities starting from $m=1$, not from $m=2$.
This tells us that 37 is a number of spins big enough to lose exactly twice being more probably than losing exactly once, but 37 isn't a number of spins big enough to lose exactly more than 3 times.
For which numbers $n$ of spins, we have a decreasing sequence of probabilities
$\binom{n}{m}(\frac{2}{37})^m(\frac{35}{37})^{n-m}$ ($1\leq m\leq n$)? How to find such $n$? What is special about these $n$'s?
| Second question:
It has to be $Bin(n,\frac{2}{37}|1) > Bin(n,\frac{2}{37}|2)$.
${n \choose 1} \left( \frac{2}{37} \right)^1\cdot \left( \frac{35}{37} \right)^{n-1} >{n \choose 2} \left( \frac{2}{37} \right)^2\cdot \left( \frac{35}{37} \right)^{n-2}$
$n\cdot \frac{2}{37} \cdot \frac{37}{35} \cdot \left( \frac{35}{37} \right)^n > \frac{n\cdot (n-1)}{2} \cdot \left( \frac{2}{37} \right) ^2\cdot \left( \frac{37}{35} \right) ^2\cdot \left( \frac{35}{37} \right)^{n}$
After dividing the equation by $\left( \frac{35}{37} \right)^{n},n, \frac{37}{35} $ and $\frac{2}{37}$ and cancelling out 37, we get
$1 >\frac{n-1}{2} \cdot \frac{2}{35} \Rightarrow 35 > n-1 \Rightarrow 36 > n $
$\Rightarrow 1 \leq n \leq 35 $
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1273353",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Given $f(1)=10,f(2)=20,f(3)=30$ find $f(12)+f(-8)$ for a 4-th degree monic polynomial If $f(x)=x^4+ax^3+bx^2+cx+d$.
Given $f(1)=10,f(2)=20,f(3)=30$ find $f(12)+f(-8)$.
This problem has troubled me a lot.The more I try to solve it,it becomes lengthier.
My problem is that there are four unknowns and only three equations.
Please help me.
| The fact that there are three equiations for four unknowns states that the solution will have one unconstrained unknown. Since you're asked for a concrete answer, the $f(12) + f(-8)$ is a specially crafted expression that doesn't depend that particular unconstrained unknown.
Let's simplifiy things by introducing $y = x - 2$. Let $g(y)$ be $f(x)$ for this new variable:
$$
g(y) = f(y + 2) = (y + 2)^4 + a (y+2)^3 + \dots =
y^4 + A y^3 + B y^2 + C y + D.
$$
$g(y)$ has similar constraints,
$$
g(-1) = f(1) = 10\\
g(0) = f(2) = 20\\
g(1) = f(3) = 30
$$
and we're asked about $f(12) + f(-8) = g(10) + g(-10)$. Note the symmetry in constraints and the value we are interesting in.
Now,
$$
40 = g(-1) + g(1) = 2\cdot 1^4 + 2B\cdot 1^2 + 2D = 2 + 2B + 2D\\
40 = 2g(0) = g(0) + g(-0) = 2\cdot 0^4 + 2B\cdot 0^2 + 2D = 2D\\
g(10) + g(-10) = 2\cdot 10^4 + 2B\cdot 10^2 + 2D = 20000 - 200 + 40 = 19840
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1274133",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
"answer_id": 1
} |
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.