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Finding the number of real roots of an unusual(!) equation How many real roots does the below equation have?
\begin{equation*}
\frac{x^{2000}}{2001}+2\sqrt{3}x^2-2\sqrt{5}x+\sqrt{3}=0
\end{equation*}
A) 0 B) 11 C) 12 D) 1 E) None of these
I could not come up with anything.
(Turkish Math Olympiads 2001)
|
We have that $x^{2000} \geq 0$, because squares are nonnegative.
Further, we have $(x-\frac{1}{2}\sqrt{\frac{5}{3}})^2 \geq 0$. This gives $x^2-\sqrt{\frac{5}{3}}x+\frac{5}{12} \geq 0$
Therefore $2 \sqrt{3}x^2-2\sqrt{5}x+\frac{10}{12}\sqrt{3} \geq 0$
Therefore $2 \sqrt{3}x^2-2\sqrt{5}x+\sqrt{3} > 0$
Therefore $\frac{x^{2000}}{2001} + 2 \sqrt{3}x^2-2\sqrt{5}x+\sqrt{3} > 0$
Therefore there are no real roots.
|
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"url": "https://math.stackexchange.com/questions/1270107",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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|
Determinant properties Prove without expanding:
\begin{equation}\begin{vmatrix}1&1&1\\a^2&b^2&c^2\\a^3&b^3 & c^3\end{vmatrix} = (ab + ac + bc)(b - a)(c - a)(c - b)\end{equation}
*I tried to zero some elements and expand until I reach the Right hand side.
*Also tried C1-C3, C2-C3 then decompose the determinant into two determinants and taking common factors. But I couldn't get (ab + ac + bc) part.
*I can only use the properties shown here http://www.vitutor.com/alg/determinants/properties_determinants.html
|
I found a long proof but that's the only way I could answer it.
\begin{gather*}
\begin{bmatrix}1&1&1\\a^2&b^2&c^2\\a^3&b^3&c^3\end{bmatrix} \\
= \begin{bmatrix}1&0&0\\a^2&b^2-a^2&c^2-a^2\\a^3&b^3-a^3&c^3 - a^3\end{bmatrix} \\
= (b-a)(c-a)\begin{bmatrix}1&0&0\\a^2&b+a&c+a\\a^3&b^2+ab+a^2&c^2+ac+a^2\end{bmatrix} \\
= (b-a)(c-a)\begin{bmatrix}1&0&0\\a^2&b+a&c-b\\a^3&b^2+ab+a^2&c^2-b^2+ac-ab\end{bmatrix} \\
= (b-a)(c-a)(c-b)\begin{bmatrix}1&0&0\\a^2&b+a&1\\a^3&b^2+ab+a^2&a+b+c\end{bmatrix}\\
= (b-a)(c-a)(c-b)[(b+a)(a+b+c)-(b^2+ab+a^2)] \\
= (ab+ac+bc)(b-a)(c-a)(c-b)\\
= R.H.S
\end{gather*}
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1273025",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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|
Find the smallest integer that is divisible by exactly $X$ perfect squares. Is there a method to find the smallest integer divisible by exactly $X$ perfect squares?
Example: find the smallest positive integer divisible by exactly 2015 perfect squares.
I've been trying to figure this out, but I haven't made much progress... Please help!
|
First of all, there won't be a simple formula.
If $X$ is prime, the smallest number is $2^{2X-2}$. So the smallest number divisible by example five perfect squares is $2^{8}=256$.
More generally, it will depend on factorizations of $X$.
In general, if $N$ is the smallest integer with exactly $X$ positive integer divisors, then $N^2$ will be the smallest integer with exactly $X$ perfect square divisors.
The number of divisors of $N$ is in terms of the prime factorization of $N$. So if $$N=p_1^{a_1}p_2^{a_2}\cdots p_n^{a_n}$$ then the number of divisors is:
$$\tau(N) = (a_1+1)(a_2+1)\cdots(a_n+1)$$
So lets try the example $X=15$. Then either $N=p_1^{14}$ or $N=p_1^{4}p_2^{2}$. In this case $2^43^2=144$ is the smallest number with exactly $15$ divisors, o $144^2$ is the smallest with number with exactly $15$ perfect square divisors.
It gets harder with even more facts. Try $X=12$. Then there are many different ways to factor
$$\begin{align}X&=24\\&=12\cdot 2
\\&=8\cdot 3
\\&=6\cdot 4
\\&=6\cdot 2\cdot 2
\\&=4\cdot 3\cdot 2
\\&= 3\cdot 2\cdot 2\cdot 2\end{align}$$
Which yields prospects:
$$N=2^{23},2^{11}3^1, 2^{7}3^2, 2^{5}3^{3}, 2^{5}3^15^1, 2^33^25^1, 2^23^15^17^1$$
Here, the best value is $2^33^25^1=360$, and $N^2=360^2$ is your answer.
|
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"timestamp": "2023-03-29T00:00:00",
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|
Can $T:P_3\rightarrow P_4$ be a function of two variables? I'm given the following question:
Denote by $P_n$ the set of polynomials in the variable $x$ of degree at most $n$. The usual basis for $P_n$ is given by $\left\{1,x,x^2,\ldots,x^n\right\}$.
Define the linear transformation $T:P_3\rightarrow P_4$ by $T(p(x))=3p'(x)+\int_0^xp(t)dt$. Find the image of $T$.
Why is there $\boldsymbol{\int_0^xp(t)dt}$ in $T$? Neither basis is expressed in terms of $t$. Why can't we just use the same variable $x$?
My professor is a smart guy, and it's been awhile since we saw this question; could this be a typo? Does it make any sense?
|
We can re-write the formula for $T$ as
\begin{align*}
T(a+bx+cx^2+dx^3)
&= \frac{d}{dx}(a+bx+cx^2+dx^3)+\int_0^x(a+bt+ct^2+dt^3)\,dt \\
&= b+2\,cx+3\,dx^2+\left[at+\frac{b}{2}t^2+\frac{c}{3}t^3+\frac{d}{4}t^4\right]_0^x \\
&= b+2\,cx+3\,dx^2+ax+\frac{b}{2}x^2+\frac{c}{3}x^3+\frac{d}{4}x^4-0 \\
&= b+(2\,c+a)x+\left(3\,d+\frac{b}{2}\right)x^2+\frac{c}{3}x^3+\frac{d}{4}x^4
\end{align*}
This confirms that the map $T$ is, in fact, well-defined.
Note that this formula implies that the matrix of $T$ relative to the standard basis is
$$
[T]=
\begin{bmatrix}
0 & 1 & 0 & 0\\
1 & 0 & 2 & 0\\
0 & 1/2 & 0 & 3\\
0 & 0 & 1/3 & 0\\
0 & 0 & 0 & 1/4
\end{bmatrix}
$$
The reduced row-echelon form of this matrix is
$$
\DeclareMathOperator{rref}{rref}\rref[T]
=
\begin{bmatrix}
1 & 0 & 0 & 0 \\
0& 1 & 0 & 0 \\
0&0&1&0\\
0&0&0&1\\
0&0&0&0
\end{bmatrix}
$$
This means that $T$ is rank four. That is, the dimension of the image of $T$ is four and a basis for the image corresponds to the span of the columns of [T]
$$
\DeclareMathOperator{Image}{Image}\Image T
=
\DeclareMathOperator{Span}{Span}\Span\left\{
x,
1+\frac{1}{2}\,x^2,
2\,x+\frac{1}{3}\,x^3,
3\,x^2+\frac{1}{4}\,x^4
\right\}
$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1275632",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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|
Solving equations including floor function. I got a little trouble solving equations that involve floor function in an efficient way.
For example :
$$
\left\lfloor\frac{x+3}{2}\right\rfloor = \frac{4x+5}{3}
$$
In the one above, I get that you basically let $$ \frac{4x+5}{3} = k $$ and then inserting $k$ in the left side, take $k = 8l, 8l+1$, and so on and test it.
If there's a better solution to the one above plese tell me.
My main problem is when it comes down to functions that have multiple floors such as :
$$
\left\lfloor \frac{x+1}{3}\right\rfloor + \left\lfloor\frac{2x+5}{6}\right\rfloor = \frac{3x-5}{2}
$$
Using the same method for each of them and then intersecting the solutions should give me the right answer but is there a faster way to solve equations like this ?
|
Since $\dfrac{4x+5}3 = \left\lfloor \dfrac{x+3}2 \right \rfloor$ is an integer, we need $3$ to divide $4x+5$, i.e.,
$$\dfrac{4x+5}3 = m \in \mathbb{Z} \implies x = \dfrac{3m-5}4 \text{ where }m \in \mathbb{Z}$$
Hence,
$$\left\lfloor \dfrac{x+3}2 \right\rfloor = m \implies \dfrac{x+3}2 = m + e \implies x+3 = 2m+2e \implies x = 2m-3+2e$$
where $e \in[0,1)$.
Hence, we need
$$\dfrac{3m-5}4 = 2m-3+2e \implies 3m-5 =8m-12+8e \implies 8e = -5m+7$$ This gives us $e = \dfrac{7-5m}8$. Since $e \in [0,1)$, we have $7-5m \in [0,8) \implies m = 0,1$. Hence, $$x=-\dfrac54,-\dfrac12$$
|
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"timestamp": "2023-03-29T00:00:00",
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|
Find all the right inverses of a matrix How do I find the right inverse of a non square matrix?
The matrix i have is
$$M =
\begin{bmatrix}
1 & 1 & 0 \\
2 & 3 & 1\\
\end{bmatrix}$$
Im really not sure how to even start this?
|
Right inverse means a matrix $A_{3 \times 2}$ such that $MA=I_{2 \times 2}$. So you are looking for a matrix $A=\begin{pmatrix}x&p\\y&q\\z&r\end{pmatrix}$ such that
$$MA =
\begin{pmatrix}
1 & 1 & 0 \\
2 & 3 & 1\\
\end{pmatrix}\begin{pmatrix}x&p\\y&q\\z&r\end{pmatrix}=\begin{pmatrix}1&0\\0&1\end{pmatrix}.$$
This gives the following system:
\begin{align*}
x+y & = 1\\
2x+3y+z & = 0\\
p+q & = 0\\
2p+3q+r & = 1.
\end{align*}
Solving this gives
$$A=\begin{pmatrix}3+z & r-1\\-2-z & 1-r\\z & r\end{pmatrix},$$
where $r,z \in \mathbb{R}$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1276049",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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|
Solving $ \sqrt{x - 4} + \sqrt{x - 7} = 1 $. I have the equation $ \sqrt{x - 4} + \sqrt{x - 7} = 1 $.
I tried to square both sides, but then I got a more difficult equation:
$$
2 x - 11 + 2 \sqrt{x^{2} - 11 x - 28} = 1.
$$
Can someone tell me what I should do next?
|
squaring gives
$$x-4+x-7+2\sqrt{x-4}\sqrt{x-7}=1$$
$$\sqrt{x-4}\sqrt{x-7}=12-2x$$
$$(x-4)(x-7)=36+x^2-12x$$
$$x=8$$
but $8$ fulfills not our equation.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1280676",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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|
Find the remainder when $3^{89}7^{86}$ is divided by $17$
Find the remainder when $3^{89}7^{86}$ is divided by $17$.
I guess the problem is to be solved by congruencies. But unfortunately, I have no clear conception about it.
Can someone please explain it.
Thank you.
|
You can check $3$ has order $16$ modulo $17$. This means $3^{16}\equiv 1\mod 17$, but $3^k\not\equiv 1$ for $1\le k<16$ Actually, Little Fermat ensures $a^{16}\equiv 1 \mod 16$ for any $a$ not divisible by $17$, and Lagrange's theorem says the order of $a$ is a divisor of $16$. Also $3^8\equiv -1$.Thus:
$$3^{89}=\bigl(3^{16}\bigr)^5\cdot 3^8\cdot 3\equiv 1\cdot(-1)\cdot3=-3\mod 17$$
Similarly $7^2\equiv-2\mod 17$, hence $7^4\equiv(-2)^2=4$, $7^8\equiv4^2\equiv -1$ and finally $17^{16}\equiv 1$. Thus
$$7^{86}\equiv 7^4\cdot7^2\equiv-8$$
whence $3^{89}7^{86}\equiv (-3)(-8)=24\equiv 7\mod 17.$
|
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"timestamp": "2023-03-29T00:00:00",
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|
Solving complex numbers equation $z^3 = \overline{z} $ We have the following equation:
$$z^3 = \overline{z} $$
I set z to be $z = a + ib$ and since I know that $ \overline{z} = a - ib$. I was trying to solve it by opening the left side of the equation.
$$ z^3 = (a+ib)^3 \Rightarrow $$
$$ [a^2+b^2+i(ab + ba)](a+ib) \Rightarrow $$
$$ a^3 - b^2a - 2b^2a + i (2a^2b + b^2a - b^3) $$
but this is where I got so far and I'm not sure how continue and if my solution so far is even the right way to solve it.
|
You have a good start. Rewrite equation as $z^3-\bar{z} =0$, now do $z=a+bi$, so we get $$a^3-3b^2a-a+i(3a^2b-b^3-b) = 0$$
Now both the imaginary and the real part must be equal to zero, so we get the following system of equations $$a^3-3b^2a-a=0 \wedge 3a^2b-b^3-b=0$$
Factoring gives:
$$a(a^2-3b^2-1)=0 \wedge b(3a^2-b^2-1)=0$$
So we have four possibilities:
*
*$a=0, b=0$
*$a=0, 3a^2-b^2-1=0$
*$a^2-3b^2-1=0, b=0$
*$a^2-3b^2-1=0, 3a^2-b^2-1=0$
First one clearly gives $z=0$.
Second one: Substitute $a=0$ in to get $b^2-1=0$, so $b=1$ or $b=-1$.
This gives $z=i$ and $z=-i$.
Third one: Substitute $b=0$ in to get $a^2-1=0$, so $a=1$ or $a=-1$.
This gives $z=1$ and $z=-1$.
Fourth one: Subtract the first equation trice form the second. This gives $8b^2+2=0$, so $b^2+\frac{1}{4}=0$, so $b=\pm\frac{1}{2}i$. This gives no solutions, since we defined $b = \Im(z)$ and it must be real.
Conclusion: The solutions are $z=0,1,-1,i,-i$.
|
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"timestamp": "2023-03-29T00:00:00",
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|
What is the biggest $n$ in $4^n$ that divides $7^{2048} - 1$? A few days ago I stumbled on the following question, it was used in the Museum of mathematics masters tournament:
What is the biggest integer $n$ in $4^n$, that divides $7^{2048} - 1$?
a) 1
b) 3
c) 5
d) 7
It was not phrased exactly like this, but it is similar enough.
Using ghci and a python terminal I guessed the answer is 7.
But I don't know how to go about it using only pen and paper.
|
You can write
\begin{align*}
7^{2048} - 1 &= 7^{2048} - 1^{2048} \\
&= (7^{1024} + 1)(7^{512} + 1)(7^{256} + 1)(7^{128} + 1)(7^{64} + 1)(7^{32} + 1)(7^{16} + 1)(7^{8} + 1)(7^{4} + 1)(7^{2} + 1)(7 + 1)(7-1)
\end{align*}
For all $k \geq 2$, $7^{2^k} + 1 \equiv (-1)^{2^k} + 1 \equiv 2 \mod 4$, so each term of the form $(7^{2^k} + 1)$ has only one factor of 2. The $(7+1)$ term has 3 factors of 2 and the $(7-1)$ has one factor of 2. So the highest power of 2 that divides $7^{2048} + 1$ is $1\cdot 10 + 3 + 1 =14$, which means for you $n=\lfloor 14/2 \rfloor = 7$.
|
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"timestamp": "2023-03-29T00:00:00",
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|
Solve the equation $\log_{2} x \log_{3} x = \log_{4} x$ Question:
Solve the equations
a)
$$\log_{2} x + \log_{3} x = \log_{4} x$$
b)
$$\log_{2} x \log_{3} x = \log_{4} x$$
Attempted solution:
The general idea I have been working on is to make them into the same base, combine them and take the inverse to get the solution for x.
a)
$$\log_{2} x + \log_{3} x = \log_{4} x \Leftrightarrow$$
$$\log_{2} x + \frac{\log_{2} x}{\log_{2} 3} = \frac{\log_{2} x}{\log_{2} 4} \Leftrightarrow$$
$$\log_{2} x + \frac{\log_{2} x}{\log_{2} 3} = \frac{\log_{2} x}{2} \Leftrightarrow$$
$$\frac{2 \log_{2} 3 \log_{2}x + 2\log_{2} x - \log_{2}3 \log_{2}x}{2 \log_{2}3} = 0 \Leftrightarrow $$
Moving the denominator over and solving for $\log_{2} x$
$$\frac{\log_{2} x (2\log_{2}3 + 1 - \log_{2}3)}{2\log_{2} 3} = 0 \Leftrightarrow$$
$$\log_{2} x = 0 \Rightarrow x = 2^{0} = 1$$
b)
$$\log_{2} x \log_{3} x = \log_{4} x$$
$$\log_{2} x \frac{\log_{2} x}{\log_{2} 3} - \frac{\log_{2} x}{\log_{2} 4} = 0 \Leftrightarrow$$
$$\frac{4 (\log_{2}x)^2 - \log_{2}3 \log_{2}x}{4 \log_{2}3} = 0 \Leftrightarrow $$
$$4 (\log_{2}x)^2 - \log_{2}3 \log_{2}x = 0 \Leftrightarrow $$
Substituting $t = \log_{2} x$ gives:
$$4t^2 - \log_{2}3t = 0 \Rightarrow$$
$$t^2 - \frac{\log_{2}3t}{4} = 0 \Rightarrow$$
$$t \left(t- \frac{\log_{2}3}{4}\right) = 0$$
$$t_{1} = 0 \Rightarrow x = 1$$
$$t_{2} = \frac{\log_{2}3}{4} \Rightarrow x = 2^{\frac{\log_{2}3}{4}}$$
However, the second solution here should be $\sqrt{3}$, so I must have made a mistake somewhere. Any suggestions?
|
rewrite the second equation as $$\frac{\ln(x)}{\ln(2)}\cdot\frac{\ln(x)}{\ln(3)}= \frac{\ln(x)}{2\ln(2)}$$
and the first equation as
$$\frac{\ln(x)}{\ln(2)}+\frac{\ln(x)}{\ln(3)}=\frac{\ln(x)}{2\ln(2)}$$
|
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"timestamp": "2023-03-29T00:00:00",
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|
Listing all the possible echelon forms of a $3\times 3$ matrix My approach.
Let $a$ denote a leading entry and $b$ be any value.
The possible echelon forms of a $3\times 3$ matrix are:
$$\begin{bmatrix} a & b & b \\ 0 & a & b \\ 0 & 0 & a\end{bmatrix}, \begin{bmatrix} a & b & b \\ 0 & a & b \\ 0 & 0 & 0\end{bmatrix}, \begin{bmatrix} a & b & b \\ 0 & 0 & 0 \\ 0 & 0 & 0\end{bmatrix}, \begin{bmatrix} 0 & a & b \\ 0 & 0 & 0 \\ 0 & 0 & 0\end{bmatrix}, \begin{bmatrix} 0 & 0 & a \\ 0 & 0 & 0 \\ 0 & 0 & 0\end{bmatrix}.$$
If these are correct, we can pretty much augment any matrix of the appropriate size to the right of these matrices and these matrices would remain in echelon form right? Even if suppose for the fifth matrix I augment it with something that will make the system inconsistent, i.e.
$$\begin{bmatrix} 0 & 0 & a \\ 0 & 0 & 0 \\ 0 & 0 & 0\end{bmatrix}\begin{bmatrix} 0 \\ 5 \\ 0\end{bmatrix}$$
is still in echelon form. Or am I mistaken somewhere?
|
here are the rrefs of $3 \times 3:$
of rank $3:$ $\quad \pmatrix{1&0&0\\0&1&0\\0&0&1}$
of rank $2: \quad\pmatrix{1&0&x\\0&1&x\\0&0&0} , \pmatrix{1&x&0\\0&0&1\\0&0&0}, \pmatrix{0&1&0\\0&0&1\\0&0&0}$
of rank $1: \quad\pmatrix{1&x&x\\0&0&0\\0&0&0}, \pmatrix{0&1&x\\0&0&0\\0&0&0}, \pmatrix{0&0&1\\0&0&0\\0&0&0}$
of rank $0: \quad \pmatrix{0&0&0\\0&0&0\\0&0&0}$
where $x$ represents an arbitrary number. hopefully this is complete.
|
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|
Find the integer x: $x \equiv 8^{38} \pmod {210}$ Find the integer x: $x \equiv 8^{38} \pmod {210}$
I broke the top into prime mods:
$$x \equiv 8^{38} \pmod 3$$
$$x \equiv 8^{38} \pmod {70}$$
But $x \equiv 8^{38} \pmod {70}$ can be broken up more:
$$x \equiv 8^{38} \pmod 7$$
$$x \equiv 8^{38} \pmod {10}$$
But $x \equiv 8^{38} \pmod {10}$ can be broken up more:
$$x \equiv 8^{38} \pmod 5$$
$$x \equiv 8^{38} \pmod 2$$
In the end,I am left with:
$$x \equiv 8^{38} \pmod 5$$
$$x \equiv 8^{38} \pmod 2$$
$$x \equiv 8^{38} \pmod 7$$
$$x \equiv 8^{38} \pmod 3$$
Solving each using fermat's theorem:
*
*$x \equiv 8^{38}\equiv8^{4(9)}8^2\equiv64 \equiv 4 \pmod 5$
*$x \equiv 8^{38} \equiv 8^{1(38)}\equiv 1 \pmod 2$
*$x \equiv 8^{38} \equiv 8^{6(6)}8^2\equiv 64 \equiv 1 \pmod 7$
*$x \equiv 8^{38} \equiv 8^{2(19)}\equiv 1 \pmod 3$
So now, I have four congruences. How can i solve them?
|
we have $$64\equiv 8^{38}\mod 210$$
|
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|
$ \lim_{x\rightarrow 0^{+}}\frac{\sin ^{2}x\tan x-x^{3}}{x^{7}}=\frac{1}{15} $ Can someone show me how is possible to prove that
\begin{equation*}
\lim_{x\rightarrow 0^{+}}\frac{\sin ^{2}x\tan x-x^{3}}{x^{7}}=\frac{1}{15}
\end{equation*}
but without Taylor series. One can use L'Hospital rule if necessary. I was not able.
|
@Paramanand. Thank you very much. Your method is powerfull. After reading
it, I am able now to combine it with some unpublished computations I did.
My favorite method is to do not use LHR nor Taylor expansion whenever
possible and to came back all the computations to the basic limits only. For
example, to compute the limit
\begin{equation*}
\lim_{x\rightarrow 0}\frac{\tan x-\sin x}{x^{3}}
\end{equation*}
one can make use the following basic limits
\begin{equation*}
\lim_{x\rightarrow 0}\frac{\tan x-x}{x^{3}}=\frac{1}{3},\ \ \ \ \ \ \text{and
}\ \ \ \ \ \ \ \lim_{x\rightarrow 0}\frac{x-\sin x}{x^{3}}=\frac{1}{6},
\end{equation*}
as follows
\begin{eqnarray*}
\lim_{x\rightarrow 0}\frac{\tan x-\sin x}{x^{3}} &=&\lim_{x\rightarrow 0}%
\frac{(\tan x-x)+(x-\sin x)}{x^{3}} \\
&=&\lim_{x\rightarrow 0}\frac{\tan x-x}{x^{3}}+\lim_{x\rightarrow 0}\frac{%
x-\sin x}{x^{3}} \\
&=&\frac{1}{3}+\frac{1}{6}=\frac{1}{2}.
\end{eqnarray*}
To compute the limit $A=\lim\limits_{x\rightarrow 0^{+}}\dfrac{\tan x+2\sin
x-3x}{x^{5}}$ I suggest the same way but to push further one more order. By
LHR one can use the following basic limits:
\begin{equation*}
\lim_{x\rightarrow 0^{+}}\frac{(\tan x-x-\frac{1}{3}x^{3})}{x^{5}}=\frac{2}{%
15},\ \ \ \ \ \ and\ \ \ \ \ \ \ \lim_{x\rightarrow 0^{+}}\frac{\sin x-x+%
\frac{1}{6}x^{3}}{x^{5}}=\frac{1}{120}.
\end{equation*}
Hence,
\begin{eqnarray*}
\lim_{x\rightarrow 0^{+}}\frac{\tan x+2\sin x-3x}{x^{5}} &=&\lim_{x%
\rightarrow 0^{+}}\frac{(\tan x-x-\frac{1}{3}x^{3})+2(\sin x-x+\frac{1}{6}%
x^{3})}{x^{5}} \\
&=&\lim_{x\rightarrow 0^{+}}\frac{(\tan x-x-\frac{1}{3}x^{3})}{x^{5}}%
+2\lim_{x\rightarrow 0^{+}}\frac{\sin x-x+\frac{1}{6}x^{3}}{x^{5}} \\
&=&\frac{2}{15}+2\left( \frac{1}{120}\right) \\
&=&\frac{3}{20}.
\end{eqnarray*}
Now I can say that taking into account the development of Paramanand Singh,
and the remark above, it is (finally) possible to compute the original
limit by making use of only the basic following limits
\begin{eqnarray*}
\lim_{x\rightarrow 0}\frac{\sin x}{x} &=&1 \\
\lim_{x\rightarrow 0}\frac{\tan x-x}{x^{3}} &=&\frac{1}{3} \\
\lim_{x\rightarrow 0}\frac{\sin x-x}{x^{3}} &=&-\frac{1}{6} \\
\lim_{x\rightarrow 0}\frac{\tan x-x-\frac{1}{3}x^{3}}{x^{5}} &=&\frac{2}{15}
\\
\lim_{x\rightarrow 0}\frac{\sin x-x+\frac{1}{6}x^{3}}{x^{5}} &=&\frac{1}{120}%
.
\end{eqnarray*}
The computations could be conducted as follows
\begin{eqnarray*}
\frac{\sin ^{2}x\tan x-x^{3}}{x^{7}} &=&\left( \frac{\sin x-x}{x^{3}}\right)
\left( \frac{\tan x-x}{x^{3}}\right) \left( \frac{\sin x}{x}\right) +\left(
\frac{\sin x\tan x-x\sin x+\sin ^{2}x-x^{2}}{x^{6}}\right) \\
&=&\left( \frac{\sin x-x}{x^{3}}\right) \left( \frac{\tan x-x}{x^{3}}\right)
\left( \frac{\sin x}{x}\right) +\left( \frac{\sin x-x}{x^{3}}\right) \left(
\frac{\tan x-x}{x^{3}}\right) \\
&&+\left( \frac{x\tan x+\sin ^{2}x-2x^{2}}{x^{6}}\right) \\
&=&\left( \frac{\sin x-x}{x^{3}}\right) \left( \frac{\tan x-x}{x^{3}}\right)
\left( \frac{\sin x}{x}\right) +\left( \frac{\sin x-x}{x^{3}}\right) \left(
\frac{\tan x-x}{x^{3}}\right) \\
&&+\left( \frac{\sin x-x}{x^{3}}\right) ^{2}+\left( \frac{\tan x+2\sin x-3x}{%
x^{5}}\right) \\
&=&\left( \frac{\sin x-x}{x^{3}}\right) \left( \frac{\tan x-x}{x^{3}}\right)
\left( \frac{\sin x}{x}\right) +\left( \frac{\sin x-x}{x^{3}}\right) \left(
\frac{\tan x-x}{x^{3}}\right) \\
&&+\left( \frac{\sin x-x}{x^{3}}\right) ^{2}+\left( \frac{\tan x-x-\frac{1}{3%
}x^{3}}{x^{5}}\right) +2\left( \frac{\sin x-x+\frac{1}{6}x^{3}}{x^{5}}%
\right)
\end{eqnarray*}
Therefore,
\begin{eqnarray*}
\lim_{x\rightarrow 0}\frac{\sin ^{2}x\tan x-x^{3}}{x^{7}} &=&\left( \frac{-1%
}{6}\right) \left( \frac{1}{3}\right) \left( \frac{1}{1}\right) +\left(
\frac{-1}{6}\right) \left( \frac{1}{3}\right) +\left( \frac{-1}{6}\right)
\left( \frac{-1}{6}\right) +\left( \frac{2}{15}\right) +2\left( \frac{1}{120}%
\right) \\
&=&\frac{1}{15}.
\end{eqnarray*}
It is like decomposing a non-prime number into a product of prime
numbers...Here, the basic limits are the prime numbers, and performing
'a' decomposition is a state of art.
Thanks again to Paramanand.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1289063",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
}
|
Find the limit of: $\lim_{x\to 1}\left(\frac{\sqrt{2x-1}-1}{x^2-1}\right)$
Find the limit of: $\lim\limits_{x\to 1}\left(\frac{\sqrt{2x-1}-1}{x^2-1}\right)$
I tried multiplying by the conjugate of $\sqrt{2x-1}-1$ ,but I obtain
$\frac{2x-2}{\left(x-1\right)\left(x+1\right)\left(\sqrt{2x-1}+1\right)}$,
which is again zero over zero indeterminate form.
How to approach this type of problems?
|
Another approach:
With the following substitution: $$\begin{aligned}\sqrt{2x-1}=t\implies x=&\ \frac{t^2+1}2\\\\\frac1{x+1}=&\ \frac2{t^2+3}\\\\\frac1{x-1}=&\ \frac2{t^2-1}=\frac2{(t-1)(t+1)}\ ,\end{aligned}$$
we have:
$$\lim_{x\to 1}\frac{\sqrt{2x-1}-1}{x^2-1}=\lim_{x\to 1}\frac{\sqrt{2x-1}-1}{(x-1)(x+1)}=\lim_{t\to 1}\left((t-1)\cdot\frac2{(t-1)(t+1)}\cdot\frac2{t^2+3}\right)=\frac12$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1289783",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 2
}
|
Probability of $ax^2 + bx + c = 0$ having real solutions
$a$, $b$, $c$ are random integer numbers between $1$ and $100$ (including $1$ and $100$, and uniformly distributed). What is the probability that the equation $ax^2 + bx + c = 0$ has real solutions?
This is from a final high school math exam, and I don't know how to get the answer without using computer.
|
Interesting question. What kind of class was this?
We can approximate this in the continuous case by letting $(a, b, c)$ reside in the open unit cube in the positive octant. Given any value of $b$ in the interval $(0, 1)$, the allowable values of $(a, c)$ are those in the region of the first-quadrant unit square "inside" the curve $4ac = b^2$. The area of this region is
$$
\frac{b^2}{4} + 2\int_{a=b/2}^1 \frac{b^2}{4a} \, da
= \frac{b^2}{4}-\frac{b^2}{2}\ln\frac{b}{2}
$$
We can then integrate this area over $b \in (0, 1)$:
$$
P = \int_{b=0}^1 \left( \frac{b^2}{4}-\frac{b^2}{2} \ln \frac{b}{2} \right)\,db
= \left. \frac{5b^3}{36}-\frac{b^3}{6}\ln\frac{b}{2} \right]_{b=0}^1
= \frac{5+6\ln 2}{36} \doteq 0.25441
$$
But I'm not sure how you get this in the discrete case without some extensive counting.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1291841",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "20",
"answer_count": 3,
"answer_id": 0
}
|
How do you find the $4\times 4$ matrix corresponding to the transformation T with respect to the basis? If the transformation $T$ acting on the vector space $A \in Mat_{2,2}$ is given by $T(A)=CA$, where
$
C= \left( \begin{array}{cc}
1 & 2 \\
3 & 4 \end{array} \right)
$ how would you find the $4\times 4$ matrix corresponding to $T$ with respect to the basis of $Mat_{2,2}$?
|
Hint:
If we take the standard basis $\Biggl\{\begin{pmatrix}1&0\\0&0\end{pmatrix}, \begin{pmatrix}0&1\\0&0\end{pmatrix}, \begin{pmatrix}0&0\\1&0\end{pmatrix}, \begin{pmatrix}0&0\\0&1\end{pmatrix}\Biggr\}$ for $M_{2,2}$,
then $T(v_1)=\begin{pmatrix}1&0\\3&0\end{pmatrix}, T(v_2)=\begin{pmatrix}0&1\\0&3\end{pmatrix}, T(v_3)=\begin{pmatrix}2&0\\4&0\end{pmatrix}, T(v_4)=\begin{pmatrix}0&2\\0&4\end{pmatrix}$.
Now write each of these matrices as linear combinations of $v_1,\cdots,v_4$, and use the coefficients to get the columns of the matrix of T with respect to this basis.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1293376",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
}
|
Let $A$ be a complex $2$ by $2$ matrix having distinct eigenvalues $a, b$. Show that $A^n =\frac{ a^n}{a - b}(A - bI) + \frac{b^n}{b - a}(A - aI)$.
Let $A\in\mathscr{M}_{2\times 2}(\mathbb{C})$ be a matrix having distinct eigenvalues $a\neq b$. Show that, for all $n > 0$,
\begin{equation*}
A^n =\frac{ a^n}{a - b}(A - bI) + \frac{b^n}{b - a}(A - aI).
\end{equation*} (Exercise 707 from Golan, The Linear Algebra a Beginning Graduate Student Ought to Know.)
I proved it by induction, under the assumption that it is true for $A,A^2$,
\begin{equation*}
A^n =\frac{ a^n}{a - b}(A - bI) + \frac{b^n}{b - a}(A - aI)=\frac{(a^n-b^n)A+(ab^n-a^nb)I}{a-b}\\\Rightarrow A^{n+1}=A\frac{(a^n-b^n)A+(ab^n-a^nb)I}{a-b}=\frac{(a^n-b^n)A^2+(ab^n-a^nb)A}{a-b}\cdots
\end{equation*}
but I can't prove it for $A^2$.
I'm also wondering if there is another way to prove it by induction just starting with $A$ or a straight way.
Thanks.
|
We have
$$A = X \begin{bmatrix}a & 0\\0 & b\end{bmatrix} X^{-1}$$
This means
$$A^n = X \begin{bmatrix}a^n & 0\\0 & b^n\end{bmatrix} X^{-1}$$
We have
$$A-bI = X \begin{bmatrix}a-b & 0\\0 & 0\end{bmatrix} X^{-1} \text{ and }A-aI = X \begin{bmatrix}0 & 0\\0 & b-a\end{bmatrix} X^{-1}$$
Hence,
$$\dfrac{a^n}{a-b}\left(A-bI\right) = X \begin{bmatrix}a^n & 0\\0 & 0\end{bmatrix} X^{-1} \text{ and }\dfrac{b^n}{b-a}\left(A-aI\right) = X \begin{bmatrix}0 & 0\\0 & b^n\end{bmatrix} X^{-1}$$
Hence,
\begin{align}
\dfrac{a^n}{a-b}\left(A-bI\right) + \dfrac{b^n}{b-a}\left(A-aI\right) & = X \begin{bmatrix}a^n & 0\\0 & 0\end{bmatrix} X^{-1} + X \begin{bmatrix}0 & 0\\0 & b^n\end{bmatrix} X^{-1}\\
& = X \begin{bmatrix}a^n & 0\\0 & b^n\end{bmatrix} X^{-1} = A^n
\end{align}
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1296807",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
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|
Finding the largest triangle inscribed in the unit circle Among all triangles inscribed in the unit circle, how can the one with the largest area be found?
|
Given any triangle $\triangle ABC$ of sides $a,b$ and $c$, let $R$ be its circumradius and $\mathcal{A}$ be its area. We have this interesting identity:
$$4 R \mathcal{A} = abc$$
When $ABC$ is inscribed inside the unit circle, $R = 1$ and by $GM \le AM$, we have
$$\mathcal{A} = \frac14 abc \le \frac14 \left(\frac{a^2+b^2+c^2}{3}\right)^{3/2}$$
Notice
$$\begin{align}a^2 + b^2 + c^2
&= |\vec{A} - \vec{B}|^2 + |\vec{B} - \vec{C}|^2 + |\vec{C}-\vec{A}|^2\\
&= 6 - 2\left(\vec{A}\cdot\vec{B} + \vec{B}\cdot\vec{C} + \vec{C}\cdot\vec{A}\right)\\
&= 9 - |\vec{A} + \vec{B} + \vec{C}|^2
\end{align}
$$
This leads to an upper bound for the area
$$\mathcal{A} \le \frac14 \left(\frac{9}{3}\right)^{3/2} = \frac{3\sqrt{3}}{4}$$
Since this upper bound is attained by an equilateral triangle of side $\sqrt{3}$, the maximum area is $\frac{3\sqrt{3}}{4}$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1298379",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "16",
"answer_count": 8,
"answer_id": 0
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|
What will be the equation of side $BC$.
The equation of two equal sides $AB$ and $AC$ of an isosceles triangle $ABC$ are $x+y=5$ and $7x-y=3$ respectively . What will be the equation of the side $BC$ if the area of the triangle $\triangle ABC$ is $5$ square units.
$a.)x+3y-1=0\\
b.)x-3y+1=0\\
c.)2x-y-5=0\\
\color{green}{d.)x+2y-5=0}\\$
$\quad\\~\\~\\$
Let the slope of the required line be $m$.
I used the slope formula between two lines
$\angle B=\angle C\\~\\\left|{\dfrac{(-1-m)}{(1-m)}}\right|=\left|{\dfrac{(7-m)}{(1+7m)}}\right|\\
\implies m=-3,\dfrac{1}{3}$
But the book is giving right answer as option $d.)$.
By finding $m$ i only found slope not the whole equation.
Also i would like to know if their is clean simple short way , and also using $Area=5$ .
I have studied maths upto $12th$ grade.
Update : by using Geogebra I found that $x-3y+1=0$ fits perfectly
|
Let the lines: $x+y=5$ & $7x-y=3$ represent AC & AB. Then the acute angle $\theta$ between the equal sides AB & AC is given as $$\tan \theta=\left|\frac{7-(-1)}{1+7(-1)}\right|=\left|\frac{8}{-6}\right|=\frac{4}{3} \implies \sin \theta=\frac{4}{5}$$ The length of two sides AB & AC of isosceles triangle are equal hence, the area of triangle is given as $$\frac{1}{2}(AB)(AC)\sin\theta=5 \implies \frac{1}{2}(AB)^2\left(\frac{4}{5}\right)=5 \implies AB=\frac{5}{\sqrt{2}}$$ The equations of lines bisecting the angle between the equal sides: $x+y=5$ & $7x-y=3$ are given as $$\frac{x+y-5}{\sqrt{1^2+1^2}}=\pm \frac{7x-y-3}{\sqrt{7^2+(-1)^2}}\implies 5x+5y-25=\pm(7x-y-3)$$$$ \implies x-3y+11=0 \quad \text{&} \quad 3x+y-7=0 $$ From above, equations it's clear that $3x+y-7=0$ is acute angle bisector having negative slope & passing through vertex A. We can also check it out by a rough sketch.
Let the equation of third (unknown) line i.e. side BC be $x-3y+c=0$ normal to the angle bisector: $3x+y-7=0$. Now, solving: $x-3y+c=0$ & any given line say $7x-y=3$, we get the intersection point $\left(\frac{c+9}{20}, \frac{7c+3}{20} \right)$. Similarly, solving both the equations of the given lines, we get intersection point $\left(1, 4 \right)$. Now, calculating the length $AB=\frac{5}{\sqrt{2}}$ of equal sides of isosceles triangle using point-to-point distance-formula as
$$AB=\sqrt{\left(1-\frac{c+9}{20}\right)^2+\left(4-\frac{7c+3}{20}\right)^2}=\frac{5}{\sqrt{2}}$$$$ (c-11)^2=100 \implies c-11=\pm 10 \implies c=21 \quad \text{&} \quad c=1$$ Thus by setting the values of $c$, we get two equations of parallel lines representing the third unknown side BC as: $\color{#0ae}{x-3y+1=0}$ & $ \color{#0ae} {x-3y+21=0}$ lying on either side of vertex A of given isosceles triangle ABC. Thus, we see that the option (b) $\color{#0b4}{x-3y+1=0}$ is correct.
According to the option (d) in your book the unknown side is: $x+2y-5=0$. Note that this line has slope $\frac{-1}{2}$ i.e. negative but see in the figure you drew the slope of unknown side must be only positive. Thus option provided in your book is absolutely wrong (may be due to some printing mistake).
|
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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|
Prove that $f=x^6+ax+5$ is reducible over $\mathbb{Z_7},\forall a\in\mathbb{Z_7}$ We have $f=x^6+ax+5\in\mathbb{Z_7}$ and we have to show that it is reducible on $\mathbb{Z_7}$, $\forall a\in\mathbb{Z_7}$. Here are all my steps:
For $a=0$ we'll get $f=x^6+5\in\mathbb{Z_7}$. But $\forall x\in\mathbb{Z_7}$ with $x\neq0\Rightarrow x^6=1$. Therefore $f\neq0$ and it means that $f$ doesn't have linear factors. How can we continue from here?
For $a\neq0$ and $x\neq0$ we'll get $f=ax+6$, $\forall x\in\mathbb{Z_7}$. Here the only solution for which $f$ is reducible over $\mathbb{Z_7}$ is for $x=a^{-1}$. But if $x\neq a^{-1}$ then $ f$ is not reducible over $\mathbb{Z_7}$.
How can I show that $f$ is reducible over $\mathbb{Z_7},\forall a\in\mathbb{Z_7}$ ? Where am I wrong?
|
Note that $f(a^{-1})=0$ as you noted before when $a$ is not $0$. Then we know that $(x-a^{-1})$ is a factor. We can do a "complete the sextic" approach on the function as follows:
$f(x) = (x-a^{-1})(x^5) = x^6-a^{-1}x^5$
$f(x) = (x-a^{-1})(x^5+a^{-1}x^4) = x^6 + a^{-2}x^4$
Verify that $f(x) = (x-a^{-1})(x^5+a^{-1}x^4+a^{-2}x^3+a^{-3}x^2 + a^{-4}x + 2a) = x^6-a^{-5}x + 2ax - 2= x^6-ax + 2ax +5 = x^6+ax+5$
The last $2a$ came from the observation that $a^{-1}$ was a root of both $x^6-ax$ and $x^6+ax+5$, meaning that the two functions were some multiple of $(x-a^{-1})$ from each other.
Also, when $a=0$, we need to factorize $f(x) = x^6+5 = x^6-9 = (x^3+3)(x^3-3)$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1301026",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 1
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|
First order differential equation: did i solve this equation right So i'm trying to solve:
$$x^2\frac{dy}{dx} + 2xy = y^3$$
I'm given this differential equation, that Bernoulli equation:
$$\frac{dy}{dx} + p(x)y = q(x)y^{n} $$
I think i've solved it and got
$$ u = \frac{2}{5x} +Cx^4$$
I'm just not sure i am right i will show you how i get there but firstly...
This was part of another question which i've already solved
Show that if $y$ is the solution of the above Bernoulli differential
equation and $u = y^{1−n}$, then $u$ satisfies the linear differential
equation:
$$\frac{du}{dx} + (n-1)p(x)u = (1-n)q(x)$$
Applying the chain rule to $u = y^{1-n}$ we obtain that
\begin{align}
\frac{d u}{dx}(x)&= \frac{du}{dy}\cdot\frac{dy}{dx}\\
&= (1-n)y^{-n}\cdot\frac{dy}{dx}
\end{align}
Futhermore using the Bernoulli equation we have
$$
\frac{dy}{dx}=q(x)y^n-p(x)y
$$
and
\begin{align}
\frac{d u}{dx}&= (1-n)y^{-n}\cdot\frac{dy}{dx}\\
&=(1-n)y^{-n}\cdot q(x)y^n - (1-n)y^{-n}\cdot p(x) y\\
&=(1-n)q(x) -(1-n)p(x)y^{1-n}\\
&=(1-n)q(x) -(1-n)p(x)u
\end{align}
Hence U satisfies the equation
$$
\frac{du}{dx}+(1-n)p(x)u = (1-n)q(x)
$$
$$x^2\frac{dy}{dx} + 2xy = y^3$$
Divide both sides by $x^2$
$$\frac{dy}{dx} + \frac{2}{x}y = x^{-2} y^3$$
Consider
$$\frac{du}{dx} + (n-1)p(x)u = (1-n)q(x)$$
We know that
*
*n = 3
*1- n = 1-3 = -2
*p(x) = $ \frac{2}{x}$
*q(x) = $x^{-2}$
*u = $y^{1-3} = y^{-2}$
Subbing these in...
$$
\frac{du}{dx} + (-2)\frac{2}{x}u = (-2)x^{-2}
$$
$$
\frac{du}{dx} + \left(-\frac{4}{x}\right)u = (-2)x^{-2}
$$
So...
$$ \text{integrating factor} = e^{\int p(x) \, dx} $$
- p(x) dx = $-\frac{4}{x}$
$$ -4 \int \frac{1}{x} = -4log(x) = log (x^{-4}) $$
$$ \text{integrating factor} = e^{log (x^{-4})}= x^{-4} = \frac{1}{x^4}$$
So multiply this to the equation
$$\frac{1}{x^4}\frac{du}{dx} + \left(\frac{-4}{x^5} \right)u = \frac{-2}{x^6}$$
So we want to solve
$$ \frac{d}{dx}\frac{1}{x^4}u = \frac{-2}{x^6} $$
$$ \int \frac{d}{dx}\frac{1}{x^4}u = \int \frac{-2}{x^6} $$
$$ \frac{1}{x^4}u = -2\int \frac{1}{x^6} $$
$$ \frac{1}{x^4}u = -2\frac{1}{-5x^5} + c $$
$$ \frac{1}{x^4}u = \frac{2}{5x^5} + c $$
$$ \therefore u= \frac{2}{5x} + cx^{4} $$
is this fine? Or do i need to somehow equate this y or sub $u=y^{1-n}$
As $$u=y^{-2}$$
$$\frac{1}{y^2}= \frac{2}{5x} + cx^{4} $$
$$y^2= \frac{5x}{2} + \frac{1}{cx^{4}} $$
$$y= \sqrt{\frac{5x}{2} + \frac{1}{cx^{4}}} $$
|
$$
x^2y' + 2xy = y^3
$$
first thing to notice is
$$
\dfrac{d}{dx}x^2y = x^2y' + 2xy
$$
so we have
$$
\dfrac{d}{dx}x^2y = y^3
$$
let $v = x^2y$
we then have
$$
\dfrac{dv}{dx} = \left(\frac{v}{x^2}\right)^3 = \frac{1}{x^6}v^3
$$
|
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"url": "https://math.stackexchange.com/questions/1302602",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 3
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|
why $ \sin \theta = \frac{7}{8} \cos \theta$? I have an example:
$$ \tan \theta = \frac{\sin \theta}{\cos \theta} = \frac{7}{8} $$
and then this equation is true? why there is cos multiplied?:
$$ \sin \theta = \frac{7}{8} \cos \theta$$
|
Your initial equation is $$\frac ab = \frac 78.$$ If you multiply this equation by $b$, you get
$$\frac ab \cdot b = \frac 78\cdot b\\
\frac{a\cdot b}{b} = \frac78\cdot b\\
a =\frac78\cdot b$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1303744",
"timestamp": "2023-03-29T00:00:00",
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|
Difficult inverse tangent identity
Prove that:
$$\arctan\left(\frac{\sqrt{1 + x} - \sqrt{1-x}}{\sqrt{1 + x} + \sqrt{1-x}} \right) = \frac{\pi}{4} - \frac{1}{2}\arccos(x), -\frac{1}{\sqrt{2}} \le x \le 1$$
I'd multiply the inside of $\arctan$ by the conjugate of the denominator.
I get:
$$\arctan\left(\frac{1 - 1\sqrt{1 - x^2}}{x} \right)$$
But that is still very difficult.
Any HINTS, no solutions?
|
Let $\dfrac12\arccos x=y\implies x=\cos2y$
and $-\dfrac1{\sqrt2}\le x\le1\implies-\dfrac1{\sqrt2}\le\cos2y\le1$
Using the definition of Principal values of $\arccos,0\le2y\le\dfrac{3\pi}4\iff0\le y\le\dfrac{3\pi}8 \ \ \ \ (1)$
$\implies\sin y,\cos y\ge0$
$\implies\sqrt{1-x}=\sqrt{1-\cos2y}=+\sqrt2\sin y$
$\implies\sqrt{1+x}=\sqrt{1+\cos2y}=+\sqrt2\cos y$
Now $$\arctan\left(\frac{\sqrt{1 + x} - \sqrt{1-x}}{\sqrt{1 + x} + \sqrt{1-x}} \right)=\cdots =\arctan\left[\tan\left(\dfrac\pi4-y\right)\right] $$
which is $\dfrac\pi4-y$
if $-\dfrac\pi2\le\dfrac\pi4-y\le\dfrac\pi2\iff\dfrac{3\pi}4\ge y\ge-\dfrac\pi4$ which is satisfied by $(1)$
|
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|
How do you solve this quadratic equation? The number of values of a for which
$$ (a^2 - 3a + 2)x^2 + (a^2-5a + 6)x + a^2-4 = 0$$
is an identity in x is?
Here's how much I was able to solve through:-
$$ (a^2 - 3a + 2)x^2 + (a^2-5a + 6)x + a^2-4 = 0$$
$$ ((a-1)(a-2))x^2 + ((a-3)(a-2))x + (a+2)(a-2) = 0$$
$$ (a-2)[(a-1)x^2 + (a-3)x + (a+2)] = 0$$
$$ so \hspace{5pt} (a-2) = 0 \hspace{5pt} or \hspace{5pt} [(a-1)x^2 + (a-3)x + (a+2)] = 0$$
$$ so \hspace{5pt} a = 2$$
$$ Now \hspace{5pt} [(a-1)x^2 + (a-3)x + (a+2)] = 0 $$
I don't know what to do next. Thanks in advance
|
Since
$$(a-1)x^2+(a-3)x+a+2$$
is a quadratic polynomial, it has, at most, two roots, then $$(a^2 - 3a + 2)x^2 + (a^2-5a + 6)x + a^2-4 = 0$$
is an identity in $x$ only if $a=2$.
|
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|
Evaluate $\lim_{n\to\infty}nI_n$ with $I_n=\int_0^1\frac{x^n}{x^2+3x+2}dx$ We have to evaluate: $$\lim_{n\to\infty}nI_n$$ with $$I_n=\int_0^1\frac{x^n}{x^2+3x+2}\:dx.$$
There is an elegant way to solve this problem?
Here is all my steps:
*
*My first ideea was to find a recurrence relation such that:
$$I_{n+2}+3I_{n+1}+2I_n=\frac{1}{n+1},\forall x\in\mathbb{N}$$
*
*Next step I show that $\forall x\in[0,1]\Rightarrow I_{n}\ge I_{n+1}\ge I_{n+2}$
Therefore it involving that: $$6I_n\ge 4I_{n+1}+2I_n\ge\frac{1}{n+1},\forall x\in\mathbb{N}$$
As I said above $$6I_{n+2}\leq 4I_{n+2}+2I_n\leq\frac{1}{n+1}$$
$\Rightarrow \frac{n}{6(n+1)}\leq nI_n\leq\frac{n}{6(n-1)},\forall x\in\mathbb{N}$
Therefore by squeeze thereom:
$$nI_n\to\frac{1}{6}\:as\:n\to\infty$$
|
We may just integrate by parts,
$$
\begin{align}
I_n=\int_0^1\frac{x^n}{(x+1)(x+2)}dx&=\left. \frac{x^{n+1}}{(n+1)}\frac{1}{(x+1)(x+2)}\right|_0^1+\frac{1}{(n+1)}\int_0^1\frac{(2x+3)\:x^{n+1}}{(x+1)^2(x+2)^2}\:dx\\\\
&=\frac1{6(n+1)}+\frac{1}{n+1}\int_0^1\frac{(2x+3)}{(x+1)^2(x+2)^2}\:x^{n+1}dx\\\\
&=\frac1{6(n+1)}+\frac{1}{n+1}J_n \tag1
\end{align}
$$ and one may observe that
$$
0\leq \int_0^1\frac{(2x+3)}{(x+1)^2(x+2)^2}\:x^{n+1}dx\leq \frac{(2\times1+3)}{(0+1)^2(0+2)^2}\int_0^1x^{n+1}dx
$$ or
$$
0\leq J_n\leq \frac{5}{4}\frac{1}{(n+2)}. \tag2
$$
Then using $(1)$ and $(2)$ gives easily
$$ \lim_{n \to +\infty}nI_n=\frac16.$$
|
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|
Why cannot $(2x^2 + x)^2$ be simplified to $2x^4 + x^2$? So, I want to simplify an equation : $(2x^2 + x)^2$.
I thought this would simplify to $2x^4 + x^2$
But, if you input a value for $x$, the answers do not equal. For example, if you input $x = 3$, then:
$$(2x^2+x)^2
= 21^2
= 441$$
AND:
$$2x^4 + x^2
= 2(82) + 9
= 173$$
Can anyone explain why this is the case?
|
In general, $(a + b)^2 = (a+b)(a+b) = a^2 + 2ab + b^2$. So, in your case,
$$(2x^2 + x)^2 = (2x^2)^2 + 2(2x^2\cdot x) + x^2 = 4x^4 + 4x^3 + x^2$$
|
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|
Why does $e^{-(x^2/2)} \approx \cos[\frac{x}{\sqrt{n}}]^n$ hold for large $n$? Why does this hold:
$$
e^{-x^2/2} = \lim_{n \to \infty} \cos^n \left( \frac{x}{\sqrt{n}} \right)
$$
I am not sure how to solve this using the limit theorem.
|
For sure, this is not as elegant as previous answers but I love too much Taylor expansions !
Starting with $$ \cos \left( \frac{x}{\sqrt{n}} \right)=1-\frac{x^2}{2 n}+\frac{x^4}{24 n^2}-\frac{x^6}{720 n^3}+\frac{x^8}{40320
n^4}-\frac{x^{10}}{3628800 n^5}+O\left(x^{12}\right)$$ and raising to power $n$ (using binomial theorem) $$ \cos^n \left( \frac{x}{\sqrt{n}} \right)=1+\sum_{n=1}^\infty a_n x^{2n}$$ with $$a_1=-\frac 12$$ $$a_2=\frac{3n-2}{24n}\to \frac 18$$ $$a_3=-\frac{15 n^2-30 n+16}{720 n^2}\to -\frac 1{48} $$ $$a_4=\frac{105 n^3-420 n^2+588 n-272}{40320 n^3}\to \frac 1{384}$$ $$a_5=-\frac{945 n^4-6300 n^3+16380 n^2-18960 n+7936}{3628800 n^4}\to -\frac 1{3840}$$ $$a_6=\frac{10395 n^5-103950 n^4+429660 n^3-893640 n^2+911328 n-353792}{479001600 n^5}\to \frac 1{46080}$$
|
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|
simple 2 sides inequality
$$2<\frac{x}{x-1}\leq 3$$
Is the only way is to multiple both sides by $(x-1)^2$?
so we get $2x^2-4x+2<x^2-x $ and $x^2-x<3x^2-6x+3$ which is $-x^2+3x-2$ and $-2x^2+5x-3<0$ so the sloutions are:
$1<x\leq \frac{3}{2}$ and $1<x\leq 2$ so overall it is $1<x\leq\frac{3}{2}$
|
you can rewrite $$2 < \frac x{x-1} = 1 + \frac 1{x-1} \le 3 \to 1 < \frac1 {x-1} \le 2 \tag 1$$ from the graph of $y = \frac1{x-1},$ we see that there are no solutions $(-\infty, 1)$ and $x - 1 > 0$ is necessary so $(1)$ is equivalent to $$\frac 12 \le x - 1< 1\to \frac32 \le x < 2. $$
|
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|
What is wrong with this integral reasoning? $$\int\frac{x^{2}+1}{x\sqrt{x^{4}+1}}dx$$
We start by multiplying by $1=\frac{x}{x}$.
$$\int\frac{x^{2}+1}{x^{2}\sqrt{x^{4}+1}}xdx$$
Next, we use the substitution $u=x^{2}$;$\frac{du}{2}=xdx$.
$$\frac{1}{2}\int\frac{u+1}{u\sqrt{u^{2}+1}}du=\frac{1}{2}\int\frac{1}{\sqrt{u^{2}+1}}du+\frac{1}{2}\int\frac{1}{u\sqrt{u^{2}+1}}du=$$
$$=\frac{1}{2}\ln\left(u+\sqrt{u^{2}+1}\right)+\frac{1}{2}\int\frac{1}{u\sqrt{u^{2}+1}}du$$
The problem is now reduced to computing the integral $\frac{1}{2}\int\frac{1}{u\sqrt{u^{2}+1}}du$.
$$\frac{1}{2}\int\frac{1}{u\sqrt{u^{2}+1}}du=\frac{1}{2}\int\frac{1}{u^2\sqrt{1+\frac{1}{u^{2}}}}du$$
We use substitution again $v=\frac{1}{u}$;$-dv=\frac{1}{u^{2}}du$, then we have the next integral.
$$-\frac{1}{2}\int\frac{1}{\sqrt{1+v^{2}}}dv=-\frac{1}{2}\ln\left(v+\sqrt{1+v^{2}}\right)=-\frac{1}{2}\ln\left(\frac{1}{u}+\sqrt{1+\frac{1}{u^{2}}}\right)=$$
$$=-\frac{1}{2}\ln\left(\frac{1+\sqrt{u^{2}+1}}{u}\right)$$
Finally the solution is:
$$\int\frac{x^{2}+1}{x\sqrt{x^{4}+1}}dx=\frac{1}{2}\ln\left(u+\sqrt{1+u^{2}}\right)-\frac{1}{2}\ln\left(\frac{1+\sqrt{u^{2}+1}}{u}\right)=\frac{1}{2}\ln\left(\frac{x^{4}+x^{2}\sqrt{x^{4}+1}}{1+\sqrt{x^{4}+1}}\right).$$
But the problem is, the book has the following solution:
$$\int\frac{x^{2}+1}{x\sqrt{x^{4}+1}}dx=\ln\left(\frac{x^{2}-1+\sqrt{x^{4}+1}}{x}\right)$$
which is obviously different.
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The two expressions are equal up to a constant.
Take from your answer $$\frac{x^2(x^2+\sqrt{x^4+1})}{1+\sqrt{x^4+1}}$$
and multiply top and bottom by $\sqrt{x^4+1}-1$
you get
$$\frac{x^4-x^2+1+(x^2-1)\sqrt{x^4+1}}{x^2}=\frac{1}{2}\frac{(x^2-1+\sqrt{x^4-1})^2}{x^2}$$
So $$\frac{1}{2} \ln \frac{x^2(x^2+\sqrt{x^4+1})}{1+\sqrt{x^4+1}}=
\frac{1}{2} \ln \frac{1}{2}\frac{(x^2-1+\sqrt{x^4-1})^2}{x^2}=\frac{1}{2}\ln \frac{1}{2}+\frac{1}{2} \ln \frac{(x^2-1+\sqrt{x^4-1})^2}{x^2}=
\frac{1}{2}\ln \frac{1}{2}+\ln \frac{x^2-1+\sqrt{x^4-1}}{x}$$
|
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|
$(1+i)^6$ in polar form $re^{i\theta}$ I used De Moivre's formula and got
$$
\left(\frac{\sqrt{2}}{2}\right)^6 \times
\cos\left(6 \times \frac{1}{4\pi}\right) +
i\sin\left(6 \times \frac{1}{4\pi}\right) =
\frac{1}{8} e^{\frac{3}{2\pi}}.
$$
But the answer is $8e^{\frac{3}{2\pi}}$, can someone explain where the $8$ is from?
Thanks!
|
$1+i = \sqrt{2}\left(\cos (\frac{\pi}{4}) + i\sin (\frac{\pi}{4})\right)= \sqrt{2}e^{i\frac{\pi}{4}} \Rightarrow (1+i)^6 = (\sqrt{2})^6\cdot e^{i(3\frac{\pi}{2})}=-8i$
|
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|
How to do power series expansion What is the coefficient of $x^{11}$ in the power series expansion of $\frac 1{1-x-x^4}$?
How do I do power series expansions?
|
suppose $$\frac{1}{1-x-x^4} = 1 + x + a_2x^2 + a_3x^3 + \cdots. $$ then we have $$1 =(1-x-x^4)(1 + x + a_2x^2 + a_3x^3 +\cdots)=1+(a_2-1)x^2 + (a_3-a_2)x^3+\\(a_4-a_3-1)x^4+\cdots+(a_n-a_{n-1} -a_{n-4})x^n+\cdots $$ equating the coefficients of $x^2, x^3, \cdots$, we find that
$$ a_0 = 1, a_1 = 1, a_2 = 1, a_3 =1, a_4=2 \text{ and } a_n=a_{n-1} +a_{n-4} $$
$\begin{array}{|c|c|c|c|c|c|c|c|c|c|c|} \hline
n&0&1&2&3&4&5&6&7&8&9&10&11&12 \\ \hline
a_n&1&1&1&1&2&3&4&5&7&10&14&19&26 \\ \hline
\end{array}$
$$ \text{ i found }19 \text{ to be the coefficient of } x^{11}.$$
|
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|
Where am I wrong in $\int_0^1 x(1-x)^ndx$ We have $\int_0^1 x(1-x)^ndx=-\frac{x(1-x)^{n+1}}{n+1}|^{^1}_{_0}+\frac{(1-x)^{n+2}}{(n+1)(n+2)}|^{^1}_{_0}=-\frac{1}{(n+1)(n+2)}$
If I use substitution $u=1-x\Rightarrow \int_0^1 (1-u)u^n du=\frac{u^{n+1}}{n+1}|^{^1}_{_0}-\frac{u^{n+2}}{n+2}|^{^1}_{_0}=\frac{1}{(n+1)(n+2)}$
Where am I wrong ?
|
REVISED BY REQUEST
We begin with the integral $\int_0^1 x(1-x)^ndx$. Integrating by parts gives
$$\begin{align}
\int_0^1 x(1-x)^ndx&=\left.-\frac{x(1-x)^{n+1}}{n+1}\right|^{^1}_{_0}+\frac{1}{n+1}\int_0^1 (1-x)^{n+1}dx\\\\
&=\frac{1}{n+1}\int_0^1 (1-x)^{n+1}dx
\end{align}$$
Now, integrating the remaining term reveals that
$$\begin{align}
\frac{1}{n+1}\int_0^1 (1-x)^{n+1}dx&=-\frac{1}{n+1}\left.\frac{(1-x)^{n+2}}{n+2}\right|^{^1}_{_0}\\\\
&=\frac{1}{(n+1)(n+2)}
\end{align}$$
ORIGINAL ANSWER
$$\begin{align}
\int_0^1 x(1-x)^ndx&=\left.-\frac{x(1-x)^{n+1}}{n+1}\right|^{^1}_{_0}\color{red}{-}\left.\frac{(1-x)^{n+2}}{(n+1)(n+2)}\right|^{^1}_{_0}\\\\
&=\frac{1}{(n+1)(n+2)}
\end{align}$$
|
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|
Evaluate $\lim\limits_{n\to\infty}\prod\limits_{k=2}^{n}\frac{k^2+k-2}{k^2+k}$ I can't find the product of a sequence.
We have
$$\frac{(2+2)(2-1)}{2(2+1)}\frac{(3+2)(3-1)}{3(3+1)}...\frac{(k+2)(k-1)}{k(k+1)}$$
I am stuck with $$P=\frac{2(n+2)}{n^2(n-1)}$$ but that isn't correct. Can the squeeze theorem be used?
|
HINT:
We have
$$\begin{align}
\prod_{k=2}^n\frac{k^2+k-2}{k^2+k}&=\prod_{k=2}^{n} \left(\frac{k+2}{k+1}\frac{k-1}{k}\right)\\\\
&=\prod_{k=2}^{n} \left(\frac{k+2}{k+1}\right)\prod_{k=2}^{n} \left(\frac{k-1}{k}\right)
\end{align}$$
|
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|
Evaluate $\lim\limits_{x\to\ 0}(\frac{1}{\sin(x)\arctan(x)}-\frac{1}{\tan(x)\arcsin(x)})$
Evaluate $$\lim\limits_{x\to\ 0}\left(\frac{1}{\sin(x)\arctan(x)}-\frac{1}{\tan(x)\arcsin(x)}\right)$$
It is easy with L'Hospital's rule, but takes too much time to calculate derivatives. What are some shorter methods?
|
The series expansions are
$\sin x = x - x^3/6...,\,\,\tan x = x + x^3/3...,\,\, \arctan x=x - x^3/3...$
Equate numerators
$$\frac{(x+x^3/3)-(x-x^3/6)}{(x - x^3/6)(x + x^3/3)(x^3/3)}$$
$$\frac{x^3/2}{(x - x^3/6)(x + x^3/3)(x^3/3)}$$
We only care about the $x^3$ in the numerator
$$\frac{x^3/2}{x^3}$$
$$1/2$$
|
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|
Remembering that $\sin^2(\theta) = 1/2 - 1/2\cos(2\theta)$? How do you remember this for integrals?
It doesn't seem obvious and I can never remember it when I come across it in integrals.
|
$\cos^2 \theta + \sin^2 \theta = 1$
$\cos^2 \theta - \sin^2 \theta = \cos 2\theta$
Add and you get
$2\cos^2 \theta = 1 + \cos 2\theta$
$\cos^2 \theta = \frac 1 2 + \frac 1 2\cos 2\theta$
Subtract and you get
$2\sin^2 \theta = 1 - \cos 2\theta$
$\sin^2 \theta = \frac 1 2 - \frac 1 2\cos 2\theta$
|
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|
Find a unit tangent vector to a curve that is an intersection of two surfaces. The intersection of the two surfaces given by the Cartesian equations $2x^2+3y^2-z^2=25$ and $x^2+y^2=z^2$ contains a curve $C$ passing through the point $P=(\sqrt{7},3,4)$. These equations may be solved for $x$ and $y$ in terms of $z$ to give a parametric representation of $C$ with $z$ as a parameter.
(a) Find a unit tangent vector $T$ to $C$ at the point $P$ without using an explicit knowledge of the parametric representation.
(b) Check the result in part (a) by determining a parametric representation of $C$ with $z$ as a parameter.
For (b), I solved the two equations for the surfaces given to get $y^2=25-z^2$ and $x^2=2z^2-25$. Since we're looking for the curve that contains $P$, $x$, $y$ should be positive so we get $y=\sqrt{25-z^2}$ and $x=\sqrt{2z^2-25}$. So from this I get the parametric representation $(\sqrt{2z^2-25}, \sqrt{25-z^2},z)$ for the curve $C$.
Is this the correct way of finding the parametrization? Moreover, I do not know how to find the unit targent vector $T$ to $C$ at $P$, without getting a parametrization. How can I find this? The answer to $T$ is $\frac{1}{\sqrt{751}}(24,-4\sqrt{7},3\sqrt{7})$.
I would greatly appreciate any solutions, hints or suggestions.
|
Hint a:
The normal to $2x^2+3y^2-z^2=25$ at $(\sqrt7,3,4)$ is parallel to the gradient: $(4x,6y,-2z)=2(2\sqrt7,9,-4)$
The normal to $x^2+y^2-z^2=0$ at $(\sqrt7,3,4)$ is parallel to the gradient: $(2x,2y,-2z)=2(\sqrt7,3,-4)$
The common tangent to both surfaces would be perpendicular to both of these normals, which would make it parallel to their cross product
$$
(2\sqrt7,9,-4)\times(\sqrt7,3,-4)=(-24,4\sqrt7,-3\sqrt7)
$$
Hint b:
We are given that $2x^2+3y^2-z^2=25$ and $x^2+y^2-z^2=0$. Therefore, by subtracting the equations, we have that
$$
x^2+2y^2=25
$$
which can be parametrized by $(x,y)=\left(5\cos(\theta),\frac5{\sqrt2}\sin(\theta)\right)$. For each point $(x,y)$, the $z$ coordinate can be computed from the equation from either surface.
|
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|
Coefficient Problem (polynomial expansion)
Let $C$ be the coefficient of $x^2$ in the expansion of the product $(1 - x)(1 + 2x)(1 - 3x)\cdots(1 + 14x)(1 - 15x).$ Find $|C|.$
Just to begin,
$(1-x)(1+2x) = -2x^2 + x + 1$
$(1-x)(1+2x)(1-3x) = 6x^3 - 5x^2 - 2x + 1$
But expanding on like this take too long.
In the end the terms will be like:
$(1 - 15x)(a_0 + a_1x^1 + a_2x^2 + ....)$
Then just looking at the $x^2$, it will be:
$(a_2x^2 - 15a_1x^2) = x^2(a_2 - 15a_1)$
But it it still a very hard problem, any hints?
|
Somewhat more generally, let $$\prod_{j=1}^n (1 + (-1)^j j x) = 1 + c(n) x + d(n) x^2 + \ldots$$
so you want $d(15)$.
We have $c(0) = d(0) = 0$ with
$$ (1 + c(n-1) x + d(n-1) x^2)(1 + (-1)^n n x) = 1 + c(n) x + d(n) x^2 + \ldots $$
so that
$$ \eqalign{c(n) &= c(n-1) + (-1)^n n \cr
d(n) &= d(n-1) + (-1)^n n c(n-1) \cr} $$
Can you solve these recursions?
|
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}
|
A limit question with a parameter $\displaystyle\lim_{a\to0} \int^{1+a}_0 \frac{1}{1+x^2+a^2}\,dx$
How to solve it?
Can I solve it in this way?
\begin{align}
&\lim_{a\to0} \int^{1+a}_0 \frac{1}{1+x^2+a^2}\,dx\\
&=\int^{1}_0 \lim_{a\to0} \frac{1}{1+x^2+a^2}\,dx\\
&=\int^{1}_0 \frac{1}{1+x^2}\,dx\\
&=π/4
\end{align}
The answer is right, but I don't know whether the argument is correct.
|
$$1+a^2+x^2=(1+a^2)\left(1+\left(\frac{x}{\sqrt{1+a^2}}\right)^2\right)\implies$$
$$\int_0^{1+a}\frac{dx}{1+a^2+x^2}=\frac1{\sqrt{1+a^2}}\int_0^{1+a}\frac{\left(\frac1{\sqrt{1+a^2}}\right)dx}{1+\left(\frac{x}{\sqrt{1+a^2}}\right)^2}=$$
$$=\left.\frac1{\sqrt{1+a^2}}\arctan\frac x{\sqrt{1+a^2}}\right|_0^{1+a}=\frac1{\sqrt{1+a^2}}\arctan\frac{1+a}{\sqrt{1+a^2}}$$
Thus, in the limit you get $\;\dfrac\pi4\;$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1318271",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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|
Finding the Range of a Trigonometric function The range of $$f(x)=3\cos^2x-8\sqrt3 \cos x\cdot\sin x+5\sin^2x-7$$is given by:(1)$[-7,7]$(2)$[-10,4]$(3)$[-4,4]$(4)$[-10,7]$
ANS: (2)
My Solution
The equation can be written as: $$3\cos^2x-8\sqrt3 \cos x\cdot \sin x+16\sin^2x-11\sin^2x-7 \\\implies (\sqrt3\cos x-4\sin x)^2-11\sin^2x-7$$
So let $y=\sqrt3\cos x+4\sin x$
$$-\sqrt{(\sqrt3)^2+4^2} \le y\le \sqrt{(\sqrt3)^2+4^2} \implies 19 \le y^2 \le 19 \implies y^2\in[0,19]$$
CASE 1: When $y^2=0$
$$f(x)=0-11\sin^2x-7 \text{ ,taking } \sin^2 x=0\text{, minimum value of }f(x)= -7$$
CASE 2: When $y^2=19$
$$f(x)=19-11\sin^2x-7 \text{ ,taking } \sin^2 x=1\text{, minimum value of }f(x)= 1$$
So my range is $y\in[-7, 1]$
where is the problem?
|
The problem is when you say $y=0$ you fix the value of $sinx$ as $\sqrt3cosx-4sinx=0$ so $tanx=\frac{\sqrt3}{4}$ and $sin^2x= \frac{3}{19}$ not 0 or 1 . You should write it as the form of $asin2x+bcos2x+c$ as stated in the comment.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1319532",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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|
Calculate $\lim_{x \to \infty} (\sqrt{x^2 + 3x} - \sqrt{x^2 + 1})$ without using L'Hospital's rule Question:
Calculate
$$\lim_{x \to \infty} (\sqrt{x^2 + 3x} - \sqrt{x^2 + 1})$$
without using L'Hospital's rule.
Attempted solution:
First we multiply with the conjugate expression:
$$\lim_{x \to \infty} (\sqrt{x^2 + 3x} - \sqrt{x^2 + 1}) = \lim_{x \to \infty} \frac{x^2 + 3x - (x^2 + 1)}{\sqrt{x^2 + 3x} + \sqrt{x^2 + 1}}$$
Simplifying gives:
$$\lim_{x \to \infty} \frac{3x - 1}{\sqrt{x^2 + 3x} + \sqrt{x^2 + 1}}$$
Breaking out $\sqrt{x^2}$ from the denominator and $x$ from the numerator gives:
$$\lim_{x \to \infty} \frac{x(3 - \frac{1}{x})}{x(\sqrt{1 + 3x} + \sqrt{1 + 1})} = \lim_{x \to \infty} \frac{3 - \frac{1}{x}}{\sqrt{1 + 3x} + \sqrt{2}}$$
The result turns out to be $\frac{3}{2}$, but unsure how to proceed from here.
|
$$
\sqrt{x^2+3x} - \sqrt{x^2+1} = x\sqrt{1+\frac{3}{x}} - x\sqrt{1+\frac{1}{x^2}} = x\left(\sqrt{1+\frac{3}{x}} -\sqrt{1+\frac{1}{x^2}} \right)
$$
expand the radicals we find
$$
\sqrt{1+\frac{3}{x}} = 1 + \frac{3}{2x} + O(x^{-2})\\
\sqrt{1+\frac{1}{x^2}} = 1 + O(x^{-2})
$$
put it all together
$$
\lim_{x\to\infty}\left(\sqrt{x^2+3x} - \sqrt{x^2+1}\right) \to x\left(1 + \frac{3}{2x}-1\right) = \frac{3}{2}
$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1324388",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 3
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|
How does one calculate: $\left(\frac{z}{2!}-\frac{z^3}{4!}+\frac{z^5}{6!}-\cdots\right)^2$ How does one calculate:
$$\left(\frac{z}{2!}-\frac{z^3}{4!}+\frac{z^5}{6!}-\cdots\right)^2$$
Is the best way to just take the first term times the following two, and the second two times the next two to see the pattern?
First term will contribute:
$$\left(\frac{z^2}{2!2!}-\frac{2z^4}{2!4!}+\frac{2z^6}{2!6!}-\cdots\right)$$
$$\left(\frac{z^2}{4}-\frac{z^4}{4!}+\frac{z^6}{6!}-\cdots\right)$$
Second term will contribute:
$$\left(-\frac{z^4}{4!}+\cdots\right)$$
This already seems wrong, so expanding via the first term seems good, and I hazard a guess that
$$\left(\frac{z}{2!}-\frac{z^3}{4!}+\frac{z^5}{6!}-\cdots\right)^2=\left(\frac{z^2}{4}-\frac{z^4}{4!}+\frac{z^6}{6!}-\cdots\right)$$
|
Your series in parentheses is
$$\frac{1-\cos z} {z}. $$
Expand the binomial. The series for $\cos^2z$ is a bit complicated, but can be written using binomial coefficients.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1324472",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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|
Calculating determinant of a Vandermonde type matrix of order n Det$
\begin{bmatrix}
1 & 2 & 3 &\ldots &n\\
1& 2^3& 3^3& \ldots & n^3\\
1 &2^5& 3^5& \ldots & n^5\\
\vdots & \vdots& & \vdots \\
1&2^{2n-1}& 3^{2n-1}& \ldots &n^{2n-1}
\end{bmatrix}
$
If the powers were consecutively increasing down the rows than we could use the Vandermonde’s identity. However I am quite uncertain about how to tackle this. I am convinced that elementary row operations will be to no avail. Any hints?
|
If you multiply the second column by $2$, the third by $3$ and so on, you get
$$
\det
\begin{bmatrix}
1 & 2 & 3 &\ldots &n\\
1& 2^3& 3^3& \ldots & n^3\\
1 &2^5& 3^5& \ldots & n^5\\
\vdots & \vdots& & \vdots \\
1&2^{2n-1}& 3^{2n-1}& \ldots &n^{2n-1}
\end{bmatrix}
=
\frac{1}{n!}\det
\begin{bmatrix}
1 & 2^2 & 3^2 &\ldots &n^2\\
1& 2^4& 3^4& \ldots & n^4\\
1 &2^6& 3^6& \ldots & n^6\\
\vdots & \vdots& & \vdots \\
1&2^{2n}& 3^{2n}& \ldots &n^{2n}
\end{bmatrix}
$$
and the matrix is now Vandermonde.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1326026",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
}
|
PDF of $Z=\frac{X^2+Y^2}{2}$ where $X\sim N(0,1)$ and $Y\sim N(0,1)$ Say $X \sim N(0,1)$ and $Y\sim N(0,1)$ are independent random variables. So: $f_X(x) = \frac{1}{\sqrt{2\pi}}e^{\frac{-1}{2}x^2}$ and $f_Y(y) = \frac{1}{\sqrt{2\pi}}e^{\frac{-1}{2}y^2}$. Now I am interested in the probability density function (PDF) $f_Z(z)$ when $Z=\frac{1}{2}(X^2+Y^2)$.
I know some things about the sum of two normally distributed random variables, for example: $Z_1 = X +Y$ gives that $Z_1 \sim N(0,2)$ by the use of convolution.
How to obtain (efficiently) the PDF for $Z$, again by the use of convolution?
|
$$\mathbb{P}[X^2+Y^2\leq R^2]=\frac{1}{2\pi}\iint_{x^2+y^2\leq r^2}e^{-\frac{x^2+y^2}{2}}\,dx\,dy=\frac{1}{2\pi}\int_{0}^{2\pi}\int_{0}^{R}\rho e^{-\rho^2/2}\,d\rho\,d\theta$$
so:
$$\mathbb{P}[X^2+Y^2\leq R^2]=1-e^{-R^2/2}$$
and $X^2+Y^2$ has an exponential distribution with parameter $\lambda=\frac{1}{2}$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1326167",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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|
Sum of digits of $\dfrac{360}{2^n}$ always 9 - proper explanation? What is the exact explanation behind this "series"?
*
*$\dfrac{360} {2^1} = 180 => 1+8+0 = 9$
*$\dfrac{360}{2^2} = 90 => 9+0 = 9$
..
*$\dfrac{360}{2^{10}} => 0.3515625 => 0+3+5+1+5+6+2+5 = 27 => 2+7 = 9$
and so forth..
how is this kind of "recursive series" explained?
|
Note that multiplying by $10$ doesn't change the sum of the digits, even if the digits extend beyond the decimal point, so the sum digits in $\frac{360}{2^n}$ is the same as the sum of digits of $\frac{360}{2^n} \times 10^n$, which is equal to $360 \times 5^n$. But
$$360 \times 5^n = 9 \times 40 \times 5^n$$
so this number is always divisible by $9$, and you can apply the sum-of-digits trick.
For what it's worth, the same result holds for any number of the form
$$\frac{9k}{2^m \times 5^n}$$
where $k,m,n \in \mathbb{N}$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1326644",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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|
One more confusing factoring question. The question is:
$x^6 + 5x^3 + 8$
Please can someone help me in factorising this. I saw some solutions but they are not meant for a IX grade student.
Thanks for the help.
|
We can rearrange the polynomial as follows and use a difference of cubes:
\begin{align*}
x^6+5x^3+8&=(x^2+2)^3-6x^4+5x^3-12x^2\\
&=(x^2+2)^3-x^3-x^2(6x^2-6x+12)\\
&=\left[(x^2+2)-x\right]\left[(x^2+2)^2+x(x^2+2)+x^2\right]-6x^2(x^2-x+2)\\
&=(x^2-x+2)\left[(x^2+2)^2+x(x^2+2)-5x^2\right]\\
&=(x^2-x+2)\left(x^4+x^3-x^2+2x+4\right)\\
\end{align*}
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1328499",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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|
What is $\lim_{x\to 0} \sum_{n=2}^\infty \frac{\sqrt{x}\ln n}{1+n^2 x}$?
What is $\displaystyle\lim_{x\to 0} \sum_{n=2}^\infty \frac{\sqrt{x}\ln n}{1+n^2 x}$ ?
Find an asymptotic expansion of $\displaystyle \sum_{n=2}^\infty \frac{\sqrt{x}\ln n}{1+n^2 x}$ as $x\to 0$
One the one hand, $\sqrt{x}\to 0$, but $\displaystyle\frac{\ln n}{1+n^2 x}\sim \ln n$ which suggests that $\sum_{n=2}^\infty \frac{\ln n}{1+n^2 x}$ "diverges" to $\infty$.
I haven't been able to tell which term dominates here, let alone the asymptotic expansion.
Here's the graph of $x\to \sum_{n=2}^\infty \frac{\sqrt{x}\ln n}{1+n^2 x}$ for $x\in (0,0.5)$
This suggests the limit is $\infty$ and the estimate should be $1/x^\alpha$ for some positive $\alpha$.
|
If we look at the corresponding integral, we get
$$\int_2^\infty \frac{\sqrt{x}\ln t}{1 + t^2x}\,dt = \int_{2\sqrt{x}}^\infty \frac{\ln u - \ln \sqrt{x}}{1+u^2}\,du.$$
Both functions,
$$\frac{\ln u}{1+u^2} \quad\text{and}\quad \frac{1}{1+u^2}$$
are Lebesgue integrable over $(0,\infty)$, so for the integral, we have the asymptotic development
\begin{align}
I(x) &= \int_0^\infty \frac{\ln u}{1+u^2}\,du - \frac{\pi}{4}\ln x - \int_0^{2\sqrt{x}}\frac{\ln u}{1+u^2}\,du + \frac{1}{2}\ln x\int_0^{2\sqrt{x}} \frac{1}{1+u^2}\,du\\
&= C - \frac{\pi}{4}\ln x + \frac{1}{2}\arctan (2\sqrt{x})\ln x - 2\sqrt{x}\ln (2\sqrt{x}) + 2\sqrt{x} + O(x)\\
&= C - \frac{\pi}{4}\ln x + 2(1-\ln 2)\sqrt{x} + O(x).
\end{align}
Now let's look at the difference between the series and the integral.
$$\frac{\sqrt{x}\ln n}{1+n^2x} - \int_{n}^{n+1} \frac{\sqrt{x}\ln t}{1+t^2x}\,dt = \sqrt{x}\int_n^{n+1} \frac{(t^2-n^2)x\ln n}{(1+n^2x)(1+t^2x)} - \frac{\ln \frac{t}{n}}{1+t^2x}\,dt.$$
We always have $0 \leqslant \ln \frac{t}{n} \leqslant \ln \frac{3}{2}$ on $[n,n+1]$, so we can estimate
$$0 < \int_2^\infty \frac{\sqrt{x}\ln \frac{t}{\lfloor t\rfloor}}{1+t^2x}\,dt < \ln \frac{3}{2} \int_0^\infty \frac{du}{1+u^2}.$$
For the other term, noting that $n \geqslant \frac{2}{3}t$ and $t^2 - n^2 < 2t+1 < 3t$, we can estimate
\begin{align}
0 &< \int_2^\infty \frac{x^{3/2}(t^2-\lfloor t\rfloor^2)\ln \lfloor t\rfloor}{(1+\lfloor t\rfloor^2x)(1+t^2x)}\,dt\\
&< \int_2^\infty \frac{x^{3/2}3t\ln t}{\bigl(1+\frac{4}{9}t^2x\bigr)(1+t^2x)}\,dt\\
&= 3\sqrt{x}\int_{2\sqrt{x}}^\infty \frac{u\ln \frac{u}{\sqrt{x}}}{\bigl(1+\frac{4}{9}u^2\bigr)(1+u^2)}\,du.
\end{align}
The latter tends to $0$ when $x\to 0$, due to the $\sqrt{x}$ factor, so the difference between the series and the integral remains bounded.
Overall
$$\sum_{n=2}^\infty \frac{\sqrt{x}\ln n}{1+n^2x} = -\frac{\pi}{4}\ln x + O(1).$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1331503",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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|
Finalising proof from Humphreys´ "Introduction to Lie Algebras and Representation Theyory" $L=\mathfrak{sl}(2, \mathbb{F})$ with standard Chevalley basis $(x, \ y, \ h)$ and $a, \ c\in \mathbb{Z}^{+}$. Humphrey gives a Lemma in chapter 26.2 saying:
$\frac{x^{c}}{c!}\frac{y^{a}}{a!}=\sum\nolimits_{k=0}^{min\{a,c\}} \frac{y^{a-k}}{(a-k)!} \begin{pmatrix} h-a-c+2k \\ k \end{pmatrix} \frac{x^{c-k}}{(c-k)!}$
But only proves half of it.
He showes via induction that it works for $a$, when $c=1$, by proveing $\frac{xy^{a}}{a!}=\frac{y^{a}x}{a!}+\frac{y^{a-1}}{(a-1)!}(h-a+1)$.
The rest he only describes:
One has to use induction on $c$ utilising that $xf(h)=f(h-2)x$ for $f(T)$ polynomal and $\begin{pmatrix} n+1 \\ k \end{pmatrix} - \begin{pmatrix} n \\ k \end{pmatrix} = \begin{pmatrix} n \\ k-1 \end{pmatrix}$ to finish the proof. I am stuck with the induction:
The intitial step was given, with proving the $a$ part.
I tried the following for the induction step: $c \to c+1$
$\frac{x^{c+1}}{(c+1)!}\frac{y^{a}}{a!} \\
= \frac{x}{c+1}(\frac{x^{c}y^{a}}{c!a!}) \\
= \frac{x}{c+1}(\sum\nolimits_{k=0}^{min\{a,c\}} \frac{y^{(a-k)}}{(a-k)!}\begin{pmatrix} h-a-c+2k \\ k \end{pmatrix} \frac{x^{(c-k)}}{(c-k)!}) \\
= \frac{1}{c+1} (\sum\nolimits_{k=0}^{min\{a,c\}} \frac{y^{(a-k)}}{(a-k)!} \begin{pmatrix} h - a - c + 2(k-1) \\ k \end{pmatrix} \frac{x^{(c-(k-1))}}{(c-k)!}) \\
= \dots$
But I have no clue how to continue.
By now I have calculeted some more:
First of all instead of the last step it is:
$\frac{1}{c+1}(\sum\nolimits_{k=0}^{min(c,a)} \frac{y^{a-k}}{(a-k)!} \begin{pmatrix} h-a-c+2(k-1) \\ k \end{pmatrix} \frac{x^{c-(k-1)}}{(c-k)!})+\frac{1}{c+1}(\sum\nolimits_{k=0}^{min(c,a)}\frac{y^{a-(k+1)}}{(a-(k+1))!}\begin{pmatrix}h-a+k+1\end{pmatrix} \begin{pmatrix} h-a-c+2k \\ k \end{pmatrix} \frac{x^{c-k}}{(c-k)!})$
Also I have tried doing things back to front:
$\sum_{k=0}^{min(a,c+1)}\frac{y^{a-k}}{(a-k)!}\begin{pmatrix} h-a-c+2k-1 \\ k \end{pmatrix}\frac{x^{c-(k-1)}}{(c-(k-1))!} \\
=\sum_{k=0}^{min(a,c+1)}\frac{y^{a-k}}{(a-k)!}\left( \begin{pmatrix} h-a-c+2(k-1) \\ (k-1)\end{pmatrix} - \begin{pmatrix} h-a-c+2(k-1) \\ k \end{pmatrix} \right) \frac{x^{c-(k-1)}}{(c-(k-1))!} \\
=\left(\sum_{k=0}^{min(a,c+)} \frac{y^{a-k}}{(a-k)!}\begin{pmatrix} h-a-c+2(k-1) \\ (k-1)\end{pmatrix}\frac{x^{c-(k-1)}}{(c-(k-1))!}\right) + \left(\sum_{k=0}^{min(a,c+1)}\frac{y^{a-k}}{(a-k)!} \begin{pmatrix} h-a-c+2(k-1) \\ k \end{pmatrix} \frac{x^{c-(k-1)}}{(c-(k-1))!}\right) \\
=\left(\sum_{k=1}^{min(a,c+1)} \frac{y^{a-k}}{(a-k)!}\begin{pmatrix} h-a-c+2(k-1) \\ (k-1)\end{pmatrix}\frac{x^{c-(k-1)}}{(c-(k-1))!}\right) + \left(\sum_{k=0}^{min(a,c+1)}\frac{y^{a-k}}{(a-k)!} \begin{pmatrix} h-a-c+2(k-1) \\ k \end{pmatrix} \frac{x^{c-(k-1)}}{(c-(k-1))!}\right) \\
=\left(\sum_{k=0}^{min(a,c+1)-1} \frac{y^{a-1-k}}{(a-1-k)!}\begin{pmatrix} h-a-c+2k \\ k\end{pmatrix}\frac{x^{c-k}}{(c-k))!}\right) + \left(\sum_{k=0}^{min(a,c+1)}\frac{y^{a-k}}{(a-k)!} \begin{pmatrix} h-a-c+2(k-1) \\ k \end{pmatrix} \frac{x^{c-(k-1)}}{(c-(k-1))!}\right) \\
=\dots$
Still I don´t know how to bring this together.
Thank you for helping me!
|
You do need to keep careful track of the terms for the induction to go smoothly. In addition to the listed steps my calculation needed the formula (proof is straightforward)
$$
k\binom n k=(n-k+1)\binom n {k-1}.\qquad(*)
$$
I abbreviate
$$
y^{[k]}=\frac{y^k}{k!}
$$
and similarly with $x^{[k]}$. Things are a bit simpler if we declare $x^{[-1]}=y^{[-1]}=0$. So we have the obvious formula
$$
x\cdot x^{[k]}=\frac1{k+1}x^{[k+1]}.
$$
With all that the inductive step $c\to c+1$ proceeds as follows
$$
\begin{array}{lcl}
x^{[c+1]}y^{[a]}&=&\dfrac x{c+1}x^{[c]}y^{[a]}\\
&\overset{\text{ind.hyp.}}=
&\displaystyle\frac1{c+1}\sum_{k=0}^{\min(a,c)}xy^{[a-k]}\binom{h-a-c+2k}kx^{[c-k]}\\
&\overset{c=1}=&
\displaystyle\sum_{k=0}^{\min(a,c)}\frac1{c+1}
\left\{y^{[a-k]}x+y^{[a-k-1]}(h-a+k+1)\right\}\binom{h-a-c+2k}kx^{[c-k]}\\
&=&\displaystyle\sum_{k=0}^{\min(a,c)}\frac1{c+1}y^{[a-k]}x\binom{h-a-c+2k}kx^{[c-k]}\\
&+&\displaystyle\sum_{k=0}^{\min(a,c)}\frac1{c+1}y^{[a-k-1]}(h-a+k+1)\binom{h-a-c+2k}kx^{[c-k]}\\
&=&\displaystyle\sum_{k=0}^{\min(a,c)}\frac1{c+1}y^{[a-k]}\binom{h-2-a-c+2k}kx\cdot x^{[c-k]}\\
&+&\displaystyle\sum_{k=1}^{\min(a,c+1)}\frac1{c+1}y^{[a-k]}(h-a+k)\binom{h-a-c+2k-2}{k-1}x^{[c-k+1]}\\
&=&\displaystyle\sum_{k=0}^{\min(a,c)}y^{[a-k]}\frac{c+1-k}{c+1}\binom{h-2-a-c+2k}k x^{[c+1-k]}\\
&+&\displaystyle\sum_{k=1}^{\min(a,c+1)}y^{[a-k]}\frac{h-a+k}{c+1}\binom{h-a-c+2k-2}{k-1}x^{[c-k+1]}.\\
\end{array}
$$
Comparing this with what we want we see that it suffices (modulo the two terms at the ends that are easy to check) to verify the identity
$$
\frac{c-k+1}{c+1}\binom{h-2-a-c+2k}k+\frac{h-a+k}{c+1}\binom{h-a-c+2k-2}{k-1}=\binom{h-a-c-1+2k}k
$$
To do that I rewrote the fractional multipliers of the binomial coefficients in the form
$$
\frac{c-k+1}{c+1}=1-\frac k{c+1}
$$
and
$$
\frac{h-a+k}{c+1}=1+\frac{h-a-c+k-1}{c+1}.
$$
With the $1$s you use the Pascal triangle identity to get what you want. The extra terms cancel by identity $(*)$.
|
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"timestamp": "2023-03-29T00:00:00",
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|
Find the value of $x$ such that $(3-\log_3x)\log _{3x}3=1$. Find the value of $x$ such that $(3-\log_3x)\log _{3x}3=1$.
Is there another way to solve other than this attempt?
My attempt,
$(3-\log_3x)\log _{3x}3=1$
$\frac{\log(3)\left(3-\frac{\log (x)}{\log (3)}\right)}{\log (3x)}=1$
$\log (3)\left(3-\frac{\log (x)}{\log (3)}\right)=\log (3x)$
$3\log (3)-\log(x)=\log (3x)$
$3\log (3)-\log (x)-\log(3x)=0$
$\log (27)+\log (\frac{1}{x})+\log (\frac{1}{3x})=0$
$\log \left(\frac{27}{x(3x)}\right)=0$
$\log \left(\frac{9}{x^2}\right)=0$
$\frac{9}{x^2}=1$
$9=x^2$
$x=\pm 3$
$-3$ is rejected. So the only solution is $3$
|
Using another approach, recall that $\log_b x=\frac{\log_c x}{\log_cb}$.
Now, for $b=3x$ and $c=3$ we have
$$\log_{3x}3=\frac{\log_33}{\log_3 3x}=\frac{\log_33}{\log_3 3+\log_3x}=\frac{1}{1+\log_3x}$$
Let $y=\log_3 x$. Then, the equation
$$\left(3-\log_3x\right)\log_{3x}3=1\implies\frac{3-y}{1+y}=1\implies y=1\implies x=3$$
|
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "2",
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|
Solve recurcion using generating function I have got a problem with solving this equation using generating functions.
$$ P_{n}=2nP_{n-1}-10n+5 $$
$$ P_{0}=5 $$
I started like that:
$$
f(x)=\sum_{n=0}^{\infty}P_{n}x^{n}=5+\sum_{n=1}^{\infty}P_{n}x^{n}=5+\sum_{n=1}^{\infty}(2nP_{n-1}-10n+5)x^{n}=5+\sum_{n=1}^{\infty}2nP_{n-1}x^{n}-\sum_{n=1}^{\infty}10nx^{n}+\sum_{n=1}^{\infty}5x^{n}
$$
So:
$$\sum_{n=1}^{\infty}10nx^{n}=10x\sum_{n=1}^{\infty}(x^{n})'=\frac{10x}{(1-x)^{2}} $$
Also:
$$
5\sum_{n=1}^{\infty}x^{n}=\frac{5x}{1-x}
$$
But I don't know what to do with:
$$
2\sum_{n=1}^{\infty}nP_{n-1}x^{n}
$$
Can you help me to solve this equations?
|
Method 1: Given $P_{n}=2nP_{n-1}-10n+5$ and $P_{0}=5$ then
\begin{align}
\sum_{n=0}^{\infty} P_{n+1} \, t^{n} &= 2 \, \sum_{n=0}^{\infty} (n+1) \, P_{n} \, t^{n} - 10 \, \sum_{n=0}^{\infty} n \, t^{n} - 5 \, \sum_{n=0}^{\infty} t^{n} \\
\sum_{n=1}^{\infty} P_{n} \, t^{n-1} &= 2 \, \frac{d}{dt} \, \sum_{n=0}^{\infty} P_{n} \, t^{n+1} - 10 \frac{d}{dt}\left(t \, \sum_{n=0}^{\infty} t^{n} \right) - \frac{5}{1-t} \\
\frac{1}{t} \left( - P_{0} + P(t) \right) &= 2 \, \frac{d}{dt} \left( t P(t) \right) - 10 \frac{d}{dt} \left( \frac{t}{1-t} \right) - \frac{5}{1-t} \\
\frac{1}{t} \left( -5 + P(t) \right) &= 2t P'(t) + 2P(t) - \frac{5(1+t)}{(1-t)^{2}}
\end{align}
which becomes
\begin{align}
2 t^{2} \, P'(t) - (1-2t) \, P(t) = \frac{5(1-3t)}{(1-t)^{2}}
\end{align}
where $P(t) = \sum_{n=0}^{\infty} P_{n} \, t^{n}$. This result leads to more complications.
Method 2:
Consider the exponential generating function method.
\begin{align}
P(t) = \sum_{n=0}^{\infty} P_{n} \, \frac{t^{n}}{n!} &= 2 \, \sum_{n=0}^{\infty} P_{n-1} \, \frac{t^{n}}{(n-1)!} - 10 \, \sum_{n=0}^{\infty} \frac{t^{n}}{(n-1)!} + 5 \sum_{n=0}^{\infty} \frac{t^{n}}{n!} \\
&= 2 \sum_{n=1}^{\infty} P_{n-1} \, \frac{t^{n}}{(n-1)!} - 10 \, \sum_{n=1}^{\infty} \frac{t^{n}}{(n-1)!} + 5 \, e^{t} \\
&= 2 t P(t) - 10 t \, e^{t} + 5 \, e^{t} = 2t P(t) + 5(1-2t) \, e^{t}
\end{align}
This leads to $P(t) = 5 e^{t}$ or $P_{n} = 5$. A verification can be made and is indeed a solution.
Method 3: From the difference equation it is readily found that $P_{n} = 5 \cdot 2^{n} \, n!$ for $n \geq 0$. A quick check verifies this is also a valid solution.
|
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"timestamp": "2023-03-29T00:00:00",
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|
Finding the limit of $(1-\cos x)/x^2$ $$\lim _{x \to 0}{1-\cos x\over x^2}={2\sin^2\left(\frac{x}{2}\right)\over x^2}={\frac{2}{x^2}\cdot {\sin^2\left(\frac{x}{2}\right)\over \left(\frac{x}{2}\right)^2}}\cdot\left(\frac{x}{2}\right)^2$$
now $$\lim_{x \to 0}{\sin^2\left(\frac{x}{2}\right)\over \left(\frac{x}{2}\right)^2}=\left({\sin\left(\frac{x}{2}\right)\over \left(\frac{x}{2}\right)}\right)^2=1^2$$
So we have $$\frac{2}{x^2}\cdot \left(\frac{x}{2}\right)^2=\frac{2}{x^2}\cdot \left(\frac{x^2}{4}\right)=\frac{1}{2}$$
Are the moves right?
|
I'm surprised nobody used L'Hospital's Rule:
$$\lim _{x \to 0}{1-\cos x\over x^2}\equiv \lim _{x \to 0}{\sin x\over 2x}\equiv\lim _{x \to 0}{\cos x\over 2}=\frac{1}{2}
$$
|
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"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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|
$a^2+b$ and $a+b^2$ prime implies $\gcd(ab+1,a+b)=1$ Let $a,b>1$ be integers such that $a^2+b$ and $a+b^2$ are prime. Prove that $\gcd(ab+1,a+b)=1$.
Clearly $a$ and $b$ are of different parities; suppose $a$ is odd and $b$ even. If a prime $p\neq 2$ divides $ab+1$ and $a+b$, then it also divides $(ab+1)+(a+b)=(a+1)(b+1)$ and $(ab+1)-(a+b)=(a-1)(b-1)$. So either $p$ divides both $a+1,b-1$ or $p$ divides both $a-1,b+1$.
|
Continuing where you left off, suppose without loss of generality that $p$ divides $a+1$ and $b-1$. Then
$$
a+b^2 \equiv -1 + 1^2 \equiv 0 \pmod{p}
$$
Since $a+b^2$ is prime, it follows that $p=a+b^2$. But by assumption $p$ divides $a+b$, and $b > 1$...
|
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|
$xy + yz + zx + 2xyz = 1$ implies $4x+y+z\geq 2$ Let $x,y,z>0$ satisfy $$xy + yz + zx + 2xyz = 1.$$ Prove that $4x+y+z\geq 2$.
The condition invites the factoring $(1+x)(1+y)(1+z)+xyz-2=x+y+z$, but having the factor $4$ in the desired inequality makes things more difficult.
|
since
$$xy+yz+xz+2xyz=1$$
Note this following indentity
$$\sum_{cyc}\dfrac{ab}{(b+c)(c+a)}+\dfrac{2abc}{(a+b)(b+c)(c+a)}=1$$
so we Let $$x=\dfrac{a}{b+c},y=\dfrac{b}{c+a},z=\dfrac{c}{a+b},a,b,c>0$$
$$\Longleftrightarrow \dfrac{4a}{b+c}+\dfrac{b}{c+a}+\dfrac{c}{a+b}\ge 2\tag{1}$$
Use Cauchy-Schwarz inequality we have
$$\left(\dfrac{4a}{b+c}+\dfrac{b}{c+a}+\dfrac{c}{a+b}\right)\left(a(b+c)+b(c+a)+c(a+b)\right)\ge (2a+b+c)^2$$
then it suffices prove
$$(2a+b+c)^2\ge 4(ab+bc+ac)$$
$$4a^2+b^2+c^2-2bc\ge 0$$
$$4a^2+(b-c)^2\ge 0$$
It is clear.
Proof $(1)$ other methods
let $$b+c=x,c+a=y,a+b=z$$
then
$$a=\dfrac{y+z-x}{2},b=\dfrac{x+z-y}{2},c=\dfrac{x+y-z}{2}$$
then inequality $(1)$ can write
$$\Longleftrightarrow \dfrac{2(y+z-x)}{x}+\dfrac{x+z-y}{2y}+\dfrac{x+y-z}{2z}\ge 2$$
$$\Longleftrightarrow \left(\dfrac{2y}{x}+\dfrac{x}{2y}\right)+\left(\dfrac{2z}{x}+\dfrac{x}{2z}\right)+\left(\dfrac{z}{2y}+\dfrac{y}{2z}\right)-3\ge 2$$
Use AM-GM inequality By Done
|
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|
Solve the system of equations by variable estimation Solve the system of equations: $\left\{\begin{array}{l}(x-1)\sqrt{x-y^2}=y(x-2y+1)\\y\sqrt{x-1}+3\sqrt{x-y^2}=2x+y-1\end{array}\right.$
I guess there is only one solution $(x;y)=(2;1)$.
This is my try
Condition: $x\ge 1;x\ge y^2$.
We have: $x-2y+1\ge x-2|y|+1\ge x-(y^2+1)+1=x-y^2\ge 0 \Rightarrow y\ge 0$
Applying AM-GM inequality, we have:
$2(2x+y-1)=2y\sqrt{x-1}+6\sqrt{x-y^2}\le(y^2+x-1)+3(1+x-y^2)=4x-2y^2+2$
$\Rightarrow y^2+y-2\le0 \Rightarrow 0\le y\le 1$
Squaring both side of first equation, we get: $(x-2y^2+2y-1)(x^2-x+2y^2-2xy)=0$
If $x-2y^2+2y-1=0$, we get $\dfrac{1}{2}\le x\le 1$. Thus, we get $x=1$ and $y=0$ or $y=1$. Two pairs of $(x,y)$ don't satisfy the system of equations.
If $x^2-x+2y^2-2xy=0$, ...
Who can help me finish it?
|
since you have
$$x^2-x+2y^2-2xy=0\Longrightarrow x-y^2=x^2-2xy+y^2=(x-y)^2$$
Note $x>y$,so have
$$\sqrt{x-y^2}=x-y$$
take the second equation we have
$$y\sqrt{x-1}+3(x-y)=2x+y-1$$
so we have
$$y=\dfrac{x+1}{4-\sqrt{x-1}}\in[0,1]\Longrightarrow 1\le x\le 2$$
take
$$x^2-x+2y^2-2xy=0\Longrightarrow x^2-x+2\left(\dfrac{x+1}{4-\sqrt{x-1}}\right)^2-2x\dfrac{x+1}{4-\sqrt{x-1}}=0,1\le x\le 2$$
$$\Longrightarrow \dfrac{x^3-6\sqrt{x-1}x^2+8x^2+10\sqrt{x-1}x-19x+2}{(\sqrt{x-1}-4)^2}=0$$
take $\sqrt{x-1}=t$,then we have
$$\Longrightarrow (t^2+1)^3-6t(t^2+1)^2+8(t^2+1)^2+10t(t^2+1)-19(t^2+1)+2=0$$
$$\Longrightarrow (t-1)(t^5-5t^4+6t^3+4t^2+4t+8)=0,0\le t\le 1$$
since
$$f(t)=t^5-5t^4+6t^3+4t^2+4t+8=t^5+6t^3+4t^2+4t+(8-5t^4)>0,0\le t\le 1$$
so $t=1$
then you have $x=2$
|
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|
How to proceed with evaluating $\int\frac{dx}{\sqrt{9+4x^2}}$ and $\int\tan^2(3x)dx$
*
*$\displaystyle\int\frac{dx}{\sqrt{9+4x^2}}$
*$\displaystyle\int\tan^2(3x)dx$
For the first one i'm not sure if I did it correctly, here is what I did:
Let $2x=3\tan(t)$, so $x=\frac{3}{2}\tan(t)$ and $dx=\frac{3}{2}\sec^2(t)dt$. So by substitution,
$$\begin{align}\displaystyle\int\frac{dx}{\sqrt{9+4x^2}}&=\int\frac{\frac{3}{2}\sec^2(t)}{3\sec(t)}dt\\
&=\frac{1}{2}\int\sec(t)dt\\
&=\frac{1}{2}\left(\ln|\sec(t)+\tan(t)| + C\right)\\
&=\frac{1}{2}\left(\ln\bigg|\frac{\sqrt{9+4x^2}}{3}+\frac{2x}{3}\bigg|+C\right)\\
\end{align}$$
For the second one, i'm unable to proceed, what I did was
$\displaystyle\int\tan^2(3x)dx=\int\frac{\sin^2(3x)}{\cos^2(3x)}dx=\int\frac{\frac{1}{2}(1-\cos(6x)}{\frac{1}{2}(1+\cos(6x)}dx=\int\frac{(1-\cos(6x)}{(1+\cos(6x)}dx$
is this the right way to proceed?
Thanks
|
The first one is correct. For the second one, you can use the hopefully well-known derivative:
$$
D\tan x=1+\tan^2x.
$$
Hence,
$$
\int \tan^2(3x)\,dx=\int 1+\tan^2(3x)-1\,dx=\frac{1}{3}\tan(3x)-x+C.
$$
|
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"timestamp": "2023-03-29T00:00:00",
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|
calculate the $1/6+1/12+1/24+1/48 \ldots $. Wolfram is wrong? I am trying to calculate the following sum
$$
S = \frac{1}{6}+\frac{1}{12}+\frac{1}{24}+\frac{1}{48} + \cdots
$$
so
$$
S+\frac{1}{3} = \frac{1}{3} + \frac{1}{6}+\frac{1}{12}+\frac{1}{24}+\frac{1}{48} + \cdots = \frac{1}{3} \cdot \sum_{k=0}^{\infty} \frac{1}{2^k}
$$
This is equal to
$$
\frac{1}{3} \cdot \frac{1}{1-0.5} = \frac{2}{3}
$$
so
$$
S + \frac{1}{3} = \frac{2}{3}
$$
and thus $S = \frac{1}{3}$.
So why in Wolfram Alpha when I do that [EDIT: new link ]
I get the following answer:
$$
0.3450320298895586027335724702689612099836962897927387
$$
|
Here's a link. Wolfram Alpha looks right to me. Your link is rather unclear, likely truncated: what exactly did you look up on Alpha?
|
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|
Substituting a value of sine function in a trigonometric equation I am trying to really understand trigonometric equations and I've stumbled upon a rather confusing example. Solve the following equation:
$\sin x= 2|\sin x|+ {\sqrt{3}}\cos x$
First step is to define the absolute $\sin x$:
$$|\sin x| =
\begin{cases}
\sin x & \sin x \in [0, 1]\\
-\sin x & \sin x \in [-1, 0)
\end{cases}
$$
Let's see what happens in the first case:
$$\sin x \in [0,1] \Rightarrow\, x \in {[2k{\pi}, (2k+1){\pi}]}$$
which means x is in the first or second quadrant.
Bringing everything to the right side:
$$\sin x+{\sqrt{3}}\cos x=0$$
Now we notice terms ${\sqrt{3}}$ with $\cos x$ and $1$ with $\sin x$ and do the following:
\begin{align*}
\sin x+{\sqrt{3}}\cos x& =0 \quad /:2\\
\frac{1}{2}\sin x+{\frac{\sqrt{3}}{2}}\cos x& = 0
\end{align*}
Now, the goal is to try and substitute integer terms with trigonometric functions they are solutions to, in order to apply one of the trigonometric identities for solving the problem. Let's aim at the following formula: $\sin(x+y)=\sin x\cos y + \sin y\cos x$ .We have several options:
*
*$\frac {1}{2} $is value of $\cos$ for either $x=\frac {\pi}{3} + 2k\pi$ OR $x=\frac {5\pi}{3} + 2k\pi$. Of all of these angles, only $x=\frac {\pi}{3} $ falls under our domain.
*${\frac {\sqrt{3}}{2}} $is value of $\sin$ for either $x=\frac {\pi}{3} + 2k\pi$ OR $x=\frac {2\pi}{3} + 2k\pi$. Out of all of these angles, there are two of them that fall under our domain. $x=\frac {\pi}{3}$ and $x=\frac {2\pi}{3}$.
This is where the problem is. I feel I should consider both of the $\sin$ values and solve two different versions of the equation. The first one is by substituting $\frac {1}{2} = \cos\frac {\pi}{3}$ and $\frac {\sqrt{3}}{2} = \sin\frac {\pi}{3}$, which is easily solved using the identity mentioned above. The second case is when substituting $\frac {\sqrt{3}}{2} = \sin\frac {2\pi}{3}$. This does not conform to any trig. identity. I tried treating $\sin\frac {2\pi}{3}$ as $\sin$ of double angle, but ended up at the beginning. I typed this equation into Symbolab and examined their step-by-step solution. It turns out that at some point they divided the equation with $\cos x$. Even when $\cos x$ is definitely not zero, I was taught dividing a trigonometric equation with anyting other than integers was a risky move very likely to result in a loss of solutions. So, my questions are:
Should you just ignore possible angles that won't make you a simple trigonometric identity?
Can you divide trigonometric equation with some trigonometric function and when?
I hope I explained myself clearly enough. Thank you in advance.
|
Hint:
note that if $y=\dfrac{2 \pi}{3}$ than $\cos y = -\dfrac{1}{2}$, so you cannot use $\dfrac{1}{2}\sin x + \dfrac{\sqrt{3}}{2}\cos x=0$.
|
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|
Why doesn't using the approximation $\sin x\approx x$ near $0$ work for computing this limit? The limit is
$$\lim_{x\to0}\left(\frac{1}{\sin^2x}-\frac{1}{x^2}\right)$$
which I'm aware can be rearranged to obtain the indeterminate $\dfrac{0}{0}$, but in an attempt to avoid L'Hopital's rule (just for fun) I tried using the fact that $\sin x\approx x$ near $x=0$. However, the actual limit is $\dfrac{1}{3}$, not $0$.
In this similar limit, the approximation reasoning works out.
|
If we take one more term in the Taylor expansion:
\begin{align}
\sin x&\approx x-\frac{x^3}6+\cdots\\
\sin^2 x&\approx x^2-2x\frac{x^3}6+\cdots\\
\frac 1{\sin^2 x}&\approx\frac 1{x^2-x^4/3}\\
&=\frac 1{x^2}\cdot\frac 1{1-x^2/3}\\
\lim_{x\to 0}\left[\frac 1{\sin^2 x}-\frac 1{x^2}\right]&=\lim_{x\to 0}\left[\frac 1{x^2}\left(\frac 1{1-x^2/3}-1\right)\right]\\
&=\lim_{x\to 0}\left[\frac 1{x^2}\cdot\frac{1-1+x^2/3}{1-x^2/3}\right]\\
&=\lim_{x\to 0}\frac 1{3-x^2}\\
&=\frac 1 3
\end{align}
To see where the first-order expansion went wrong, it's necessary to keep track of where the error term goes:
\begin{align}
\sin x&= x+\text{O}(x^3)\\
\sin^2 x&=x^2+2x\text{O}(x^3)+\text{O}(x^3)^2\\
&=x^2+\text{O}(x^4)+\text{O}(x^6)\\
&=x^2+\text{O}(x^4)\\
\frac 1{\sin^2 x}&=\frac 1{x^2+\text{O}(x^4)}\\
&=\frac 1{x^2}\cdot\frac 1{1+\text{O}(x^2)}\\
\frac 1{\sin^2 x}-\frac 1{x^2}&=\frac 1{x^2}\left[\frac 1{1+\text{O}(x^2)}-1\right]\\
&=\frac 1{x^2}\cdot\frac{1-1+\text{O}(x^2)}{1+\text{O}(x^2)}\\
&=\frac{\text{O}(x^2)}{x^2}\cdot\frac 1{1+\text{O}(x^2)}\\
&=\text{O}(1)
\end{align}
Thus the $\sin x\approx x$ approximation is not accurate enough to estimate even the constant term of the expression in the limit. (Note that it does allow us to say that there are no $\text{O}(n^{-1})$ or bigger terms, so the limit probably won't diverge.)
|
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|
Help to solve this geometry problem
How to find the length of $BG$ and $EC$ using $\alpha$ and $\beta$ ?
|
Writing
$$a := |\overline{BC}| \qquad b := |\overline{CA}| \qquad c := |\overline{AB}|$$
$$y := |\overline{CE}| = |\overline{ED}| \qquad z := |\overline{BG}| = |\overline{GF}|$$
we have, in right triangles $\triangle ADG$ and $\triangle AFE$,
$$\frac{b-2y}{c-z} = \cos A = \frac{c-2z}{b-y}$$
Solving for $y$ and $z$ gives
$$\begin{align}
y &= \frac{2 b - c \cos A - b \cos A^2}{(2 - \cos A) (2 + \cos A)} =
\frac{(b - c \cos A) + b (1-\cos A^2)}{(2 - \cos A) (2 + \cos A)} =
\frac{a \cos C + b \sin A^2}{(2 - \cos A) (2 + \cos A)} \\
z &= \frac{2 c - b \cos A - c \cos A^2}{(2 - \cos A) (2 + \cos A)} = \frac{(c - b \cos A) + c (1-\cos A^2)}{(2 - \cos A) (2 + \cos A)} = \frac{a \cos B + c \sin A^2}{(2 - \cos A) (2 + \cos A)}
\end{align}$$
By the Law of Sines, we can write $b = a \sin B/\sin A$ and $c = a \sin C/\sin A$, so that
$$y = a\;\frac{\cos C + \sin A \sin B}{(2 - \cos A) (2 + \cos A)} \qquad z = a\;\frac{\cos B + \sin C \sin A}{(2 - \cos A) (2 + \cos A)}$$
With $A+B+C=\pi$, so that $\sin A = \sin(B+C)$ and $\cos A = - \cos(B+C)$, we get an answer that agrees with @Jack's (noting that his expressions are for $|\overline{BF}| = 2y$ and $|\overline{CD}| =2z$):
$$y = a\;\frac{3\cos C - \cos(2B+C)}{2(4 - \cos^2(B+C))} \qquad z = a\;\frac{ 3\cos B - \cos(B+2C)}{2(4-\cos^2(B+C))}$$
|
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|
$\int \limits_0^{\infty} x^2 \exp(-2x^2) dx$ How to evaluate this integral?
$$\int \limits_0^{\infty} x^2 \exp(-2x^2) dx$$
I found similar problem, but don't know how to apply them here.
What do I have to substitute?
|
There are often several ways to evaluate some integrals. Here is one of the less standard ways.
Consider the integral
\begin{align}
I(a) = \int_{0}^{\infty} x^2 \, e^{- a x^2} \, dx
\end{align}
for which
\begin{align}
I(a) &= - \partial_{a} \, \int_{0}^{\infty} e^{-a x^2} \, dx
= - \frac{1}{2} \, \partial_{a} \left( \sqrt{\frac{\pi}{a}} \right)
= \frac{1}{4a} \, \sqrt{\frac{\pi}{a}}.
\end{align}
Where $a=2$ the result desired is obtained.
Integral evaluation for clarity:
\begin{align}
J(a) &= \int_{0}^{\infty} e^{-a x^{2}} \, dx \\
&= \frac{1}{2 \sqrt{a}} \, \int_{0}^{\infty} e^{-t} \, t^{-1/2} \, dt \hspace{10mm} t= a x^{2} \\
&= \frac{1}{2\sqrt{a}} \, \sqrt{\frac{\pi}{a}}.
\end{align}
Further calculations:
\begin{align}
\int_{0}^{\infty} x^{2n} \, e^{-a x^{2}} \, dx &= (-1)^{n} \partial_{a}^{n} \left( \frac{1}{2} \sqrt{\frac{\pi}{a}} \right) \\
&= \frac{(2n)!}{2^{2n+1} \, n! \, a^{n}} \, \sqrt{\frac{\pi}{a}}
\end{align}
|
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"timestamp": "2023-03-29T00:00:00",
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|
Proof:Taylor expansion of inverse trigonometric functions I find it quite difficult to remember the Taylor expansion of inverse trigonometric functions.Actually in school we have been just taught the series (for finding limits in calculus without teaching us the proof).
Can someone provide a short proof of these expansions (mainly for inverse trigonometric functions) so that I can derive them even if I forget.
|
You can remember the generalized binomial formula
$$(1+x)^p=1+px+\frac{p(p-1)x^2}2+\frac{p(p-1)(p-2)x^3}{3!}+\frac{p(p-1)(p-2)(p-3)x^4}{4!}\cdots,$$
and apply it for $p=-1$ (fairly easy)
$$\frac1{1+x}=1-x+x^2-x^3+x^4\cdots,$$
or $p=-1/2$,
$$\frac1{\sqrt{1+x}}=1-
\frac{x}2+
\frac{3\,x^2}{2^2\cdot 2}-
\frac{3\cdot5\,x^3}{2^3\cdot3!}+
\frac{3\cdot5\cdot7\,x^4}{2^4\cdot4!}\cdots.$$
Then for the $\arctan$, substitute $x^2$ for $x$ and integrate.
$$\int\frac{dx}{1+x^2}=x-\frac{x^3}3+\frac{x^5}5-\frac{x^7}7+\frac{x^9}9\cdots,$$
and for the $\arcsin$ substitute $-x^2$ for $x$ and integrate
$$\int\frac{dx}{\sqrt{1-x^2}}=x+
\frac{x^3}{2\cdot3}+
\frac{3\,x^5}{2^2\cdot 2\cdot5}+
\frac{3\cdot5\,x^7}{2^3\cdot3!\cdot7}+
\frac{3\cdot5\cdot7\,x^9}{2^4\cdot4!\cdot9}\cdots.$$
For the same "price", you also get $\ln$, the natural logarithm,
$$\int\frac{dx}{1+x}=x-\frac{x^2}2+\frac{x^3}3-\frac{x^4}4+\frac{x^5}5\cdots.$$
|
{
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"timestamp": "2023-03-29T00:00:00",
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|
Minimum value of cosA+cosB+cosC in a triangle ABC I have used jensen's inequality but couldn't move on.
|
$\cos A+\cos B+\cos C$
$=2\cos\dfrac{A+B}2\cos\dfrac{A-B}2+1-2\sin^2\dfrac C2$
$=2\sin\dfrac C2\cos\dfrac{A-B}2+1-2\sin\dfrac C2\cos\dfrac{A+B}2$ as $A+B=\pi-C,\dfrac{A+B}2=\dfrac\pi2-\dfrac C2$
$=1+2\sin\dfrac C2\left[\cos\dfrac{A-B}2-\cos\dfrac{A+B}2\right]$
$=1+4\sin\dfrac A2\sin\dfrac B2\sin\dfrac C2$
Now $\sin\dfrac A2\sin\dfrac B2\sin\dfrac C2$ can be made arbitrarily small, but$>0$ by setting $A,B\to0,C\to\pi$
|
{
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"timestamp": "2023-03-29T00:00:00",
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|
Alternative Quadratic Formula Well the formula for solving a Quadratic equation is :
$$\text{If }\space ax^2+bx+c=0$$
then
$$x=\dfrac{-b \pm \sqrt{b^2 -4ac} }{2a}$$
But looking at this : [Wolfram Mathworld] (And also in other places)
They give An Alternate Formula:
$$x=\dfrac{2c}{-b \pm \sqrt{b^2 -4ac}}$$
How does one get this?
Also in the first formula (the one we know) , $a \neq 0$ ... but here is it still the case?
Please help, Thanks!
|
\begin{align}
\dfrac{-b \pm \sqrt{b^2 -4ac} }{2a}
&= \dfrac{-b \pm \sqrt{b^2 -4ac} }{2a}
\cdot \frac{-b \mp \sqrt{b^2 -4ac}}{-b \mp \sqrt{b^2 -4ac}} \\
&= \dfrac{b^2-(b^2-4ac)}{2a(-b \mp \sqrt{b^2 -4ac})} \\
&= \dfrac{4ac}{2a(-b \mp \sqrt{b^2 -4ac})} \\
&= \dfrac{2c}{-b \pm \sqrt{b^2 -4ac}} \\
\end{align}
|
{
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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|
The value of $x$ for which function attains max value
At what value of $x,\ x\in \mathbb{Z}$ will the function $\dfrac{x^2+3x+1}{x^2-3x+1}$ attain its maximum value .
$\color{green}{a.)\ 3 }\\
b.)\ 4 \\
c.) -3 \\
d.)\ \text{none of these} \\ $
$\dfrac{x^2+3x+1}{x^2-3x+1}\\
=1+\dfrac{6x}{x^2-3x+1}\\
=1+\dfrac{6}{x-3+\frac{1}{x}}\\
$
here i thought to use $\text{AM-GM}$ inequality for the part $x+\dfrac{1}{x}$
and concluded that $x+\dfrac{1}{x}=2 \implies x=1$
But after inspecting some values i came up with
$$\begin{array}{|c|c|} \hline
x & k \\ \hline
-4 & \dfrac{5}{29} \\ \hline
-3 & \dfrac{1}{19} \\ \hline
-2 & -\dfrac{1}{11} \\ \hline
-1 & -\dfrac{1}{5} \\ \hline
0 & 1 \\ \hline
1 & -5 \\ \hline
2 & -11 \\ \hline
\color{green}{3 }& \color{red}{19} \\ \hline
4 &\dfrac{29}{5}\\ \hline
\end{array}$$
I look for a short and simple way .
also i don't want to use calculus.
I have studied maths up to $12$th grade.
|
Your last two formulas for $f(x)$ in your first edit are incorrect: they should be
$$f(x)=1+\frac{6x}{x^2-3x+1}=1+\frac{6}{x-3+\frac{1}{x}}$$
(I see you corrected that formula in your question by a later edit.)
You maximize that by making the denominator positive but as small as possible. The $-3$ in that denominator means that you make $x=3$. Any smaller, and the denominator becomes negative. Any bigger, and the denominator becomes bigger and reduces the value of the function.
Nothing fancy was needed here. The AM-GM inequality shows that $x+\frac 1x$ has a minimum of $2$, but that makes the denominator negative and is thus not relevant here. You want to make $x+\frac 1x$ just above $3$, and that is done at $x=3$.
Here is another way to look at it. If $x$ can be any real number, your function $f(x)$ has no maximum. It tends to infinity as the denominator tends to zero. We can solve the quadratic equation of the denominator equaling zero and get
$$x=\frac{3\pm\sqrt 5}{2}\approx 0.38,\ 2.62$$
We get a positive denominator for $x<0.38$ and $x>2.62$ (approximations used here).
We still want a small positive denominator, so if we limit $x$ to integers we now look at $x=0$ (below 0.38) and $x=3$ (above 2.62). We see that $x=3$ gives us the larger value, so that gives us the maximum for all integers.
|
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"timestamp": "2023-03-29T00:00:00",
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|
Given primitive solution to $\frac{1}{a}+\frac{1}{b}=\frac{1}{c}$, show $a+b$ is a perfect square If $a,b,c$ are positive integers and
$\gcd(a,b,c)$ is $1$. Given that $\frac{1}{a} + \frac{1}{b} = \frac{1}{c}$ then prove that $a+b$ is a perfect square.
I was trying to get something useful from the information given in the question but was unable to get that.
|
Ah well, since Jack deleted his answer, I'll fix it for him.
The equation is equivalent to $ac+bc=ab$ or $(a-c)(b-c)=c^2$.
Claim: $\gcd(a-c,b-c)=1$.
Proof: $\gcd(a,b,c)=1$ implies $\gcd(a-c,b-c,c)=1$ which implies $\gcd(a-c,b-c,c^2)=1$. But $a-c\mid c^2$, so $\gcd(a-c,b-c,c^2)=\gcd(a-c,b-c)$.
When the product of two relatively prime positive integers is a perfect square, the two numbers are perfect squares. So:
$$a-c=d^2$$ for some $d\mid c$, and $b-c=\frac{c^2}{d^2}$.
So $$a+b = c+d^2 + c+\frac{c^2}{d^2} = \left(d+\frac{c}{d}\right)^2$$
[ In particular, Jack's assumption that $d=1$ or $d=c$ was wrong because in the case:
$$\frac{1}{10} + \frac{1}{15}=\frac{1}{6}$$
we have $d=2$.]
|
{
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"timestamp": "2023-03-29T00:00:00",
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|
Find the last two digits of $33^{100}$
Find the last two digits of $33^{100}$
By Euler's theorem, since $\gcd(33, 100)=1$, then $33^{\phi(100)}\equiv 1 \pmod{100}$. But $\phi(100)=\phi(5^2\times2^2)=40.$
So $33^{40}\equiv 1 \pmod{100}$
Then how to proceed?
With the suggestion of @Lucian:
$33^2\equiv-11 \pmod{100}$ then $33^{100}\equiv(-11)^{50}\pmod{100}\equiv (10+1)^{50}\pmod{100}$
By using the binomial expansion, we have:
$33^{100}\equiv (10^{50}+50\cdot 10^{49}+ \cdots + 50\cdot 10+1)\pmod{100}$
$\implies 33^{100}\equiv (50\cdot 10+1)\pmod{100}\equiv 01 \pmod{100}$
|
$$\begin{align}
33^{100}&=9^{50}\cdot11^{100}\\
&=(1-10)^{50}(1+10)^{100}\\
&=(1-50\cdot10+\cdots)(1+100\cdot10+\cdots)\\
&\equiv1\mod100
\end{align}$$
|
{
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"timestamp": "2023-03-29T00:00:00",
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|
Compute $(1 - \omega + \omega^2)(1 + \omega - \omega^2)$ where $\omega^3 = 1$ If $\omega^3 = 1$ and $\omega \neq 1$, then compute $(1 - \omega + \omega^2)(1 + \omega - \omega^2)$
I'm pretty lost, I don't really know where to start.
Thanks
|
Note that $x^3-1=(x-1)(x^2+x+1)$ and $\omega\neq 1$ so that $\omega^2+\omega+1=0$
This means that $1+\omega-\omega^2=-2\omega^2$ and $1-\omega+\omega^2=-2\omega$
The product then reduces to $4\omega^3=4$
|
{
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|
Complex numbers - roots of unity
Let $\omega$ be a complex number such that $\omega^5 = 1$ and $\omega \neq 1$. Find
$$\frac{\omega}{1 - \omega^2} + \frac{\omega^2}{1 - \omega^4} + \frac{\omega^3}{1 - \omega} + \frac{\omega^4}{1 - \omega^3}.$$
I have tried adding the first two and the second two separately, then adding those sums but how do I get a numerical value as the answer?
Thanks
|
A less efficient answer with a more general approach.
You may notice that $\{\omega,\omega^2,\omega^3,\omega^4\}$ are the roots of $\frac{x^5-1}{x-1}$. If we set $Z=\{\omega,\omega^2,\omega^3,\omega^4\}$
we have
$$ \sum_{z\in Z}\frac{z}{1-z^2}=\sum_{z\in Z}\frac{z^3}{1-z}=\sum_{z\in Z}\frac{1}{1-z}-\sum_{z\in Z}(1+z+z^2)=-2+\sum_{z\in Z}\frac{1}{1-z}.$$
If $z\in Z$, $1-z$ is a root of $\frac{1-(1-x)^5}{x}=x^4-5x^3+10x^2-10x+5$.
By Vieta's theorem it follows that
$$ \sum_{z\in Z}\frac{1}{1-z} = \frac{10}{5} = 2$$
hence:
$$ \sum_{z\in Z}\frac{z}{1-z^2} = \color{red}{0}.$$
Key steps:
*
*$z\mapsto z^3$ is a bijection on $Z$
*for any $k\in[1,4]$ we have $\sum_{z\in Z}z^k = -1$.
|
{
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|
Matrix multiplication and determinant question Show that if $\det(\begin{bmatrix}b & c\\a & b\end{bmatrix})=0$ with $A=\begin{bmatrix}a & a\\b & b\end{bmatrix}$ and $B=\begin{bmatrix}b & b\\c & c\end{bmatrix}$ then $AB=BA.$
How do I go by solving this? I tried finding the determent, $AB$, and $BA$ but it didn't work.
|
$$
\begin{bmatrix} a & a \\ b & b \end{bmatrix} \begin{bmatrix} b & b \\ c & c \end{bmatrix} = \begin{bmatrix} ab+ac & ab+ac \\ c^2+bc & b^2 + bc \end{bmatrix}
$$
$$
\begin{bmatrix} b & b \\ c & c \end{bmatrix} \begin{bmatrix} a & a \\ b & b \end{bmatrix} = \begin{bmatrix} ab+b^2 & ab+b^2 \\ ac+bc & ac + bc \end{bmatrix}
$$
So we ask whether $ab+ac$ is the same as $ab+b^2$. They're the same if $ac=b^2$. That happens if $b^2-ac=0$. And $b^2-ac$ is the determinant that you mentioned. So check the other three entries in the same way.
|
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|
Find the coefficient of $x^9$ in $(1+x)(1+x^2)(1+x^3)\cdots(1+x^{100})$ This question had come in jee advanced 2015. Give a hint to solve it.
|
You have $100$ factors, each a sum of a $1$ and a positive power of $x$. When the product is expanded as a sum, every term results from picking a $1$ from some factors and a positive power of $x$ from others. If the positive powers of $x$ that you pick in one case are $x^3$, $x^{20}$, and $x^{35}$, then the term is $x^{3+20+35} = x^{58}$.
If they are $x^2$, $x^3$, and $x^4$, then the term is $x^{2+3+4}=x^9$. ${}\qquad 2+3+4=9$.
If they are $x^1$, $x^2$, and $x^6$, then the term is $x^{1+2+6} = x^9$. ${}\qquad 1+2+6=9$.
So the question is: In how many ways can you write $9$ as a sum of distinct positive integers? That will be the coefficient of $x^9$.
|
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|
Find the residue of $e^{\frac{1}{z^2-1}}\sin(\pi z)$ at $z=1$ I'm dealing with the following problem (from an old qualifying exam):
Let $\gamma$ be a closed curve in the right half-plane that has index $N$ with respect to the point 1. Find
$$
\int_{\gamma}e^{\frac{1}{z^2-1}}\sin(\pi z)\,dz.
$$
I take this problem as, "Find $\text{Res}\left(f,1\right)$," where $f$ is the integrand. For an essential singularity like I believe $f$ has at $z=1$, you have to expand into the Laurent series. This is the closest I got: By partial fraction decomposition,
$$
\frac{1}{z^2-1}=\frac{1}{2(z-1)}-\frac{1}{2(z+1)},
$$
So
$$
f(z)=e^{\frac{1}{2(z-1)}}e^{-\frac{1}{2(z+1)}}\sin(\pi z)=\frac{1}{2i}e^{\frac{1}{2(z-1)}}\sin\left(\frac{z\pi}{2(z+1)}\right).
$$
If I could magically find a Taylor expansion about $z=1$ of $\sin\left(\frac{z\pi}{2(z+1)}\right)$ I'd be all set, but it seems quite disgusting to work out. I've also tried integrating over a simpler closed curve containing $z=1$ but I can't seem to find something that turns out integrable. Any help is greatly appreciated. Thanks.
|
Expanding the functions $\exp$ and $\sin$ in Laurent series we get
[
\begin{align}
\exp \left( {\frac{1}{{z^2 - 1}}} \right) = \sum {\frac{1}{{\left( {z^2 - 1} \right)^n n!}}} = \sum {\frac{1}{{\left( {z + 1} \right)^n \left( {z - 1} \right)^n n!}}} = \sum {\frac{{\frac{1}{{\left( {z + 1} \right)^n }}}}{{\left( {z - 1} \right)^n n!}}}
\end{align}
Also,
\begin{align}
\sin \pi z = \sin \pi \left( {z + 1 - 1} \right)=\sin \left( {\pi \left( {z - 1} \right) + \pi } \right) = \sin \pi \left( {z - 1} \right)\cos \pi + \sin \pi \cos \pi \left( {z - 1} \right) \\
= - \sin \pi \left( {z - 1} \right)
\end{align}
and since
\begin{align}
\sin z = \sum {\frac{{\left( { - 1} \right)^n }}{{\left( {2n + 1} \right)!}}z^{2n + 1} } \Rightarrow \sin \pi \left( {z - 1} \right) = \sum {\frac{{\left( { - 1} \right)^n }}{{\left( {2n + 1} \right)!}}\pi ^{2n + 1} \left( {z - 1} \right)^{2n + 1} }
\end{align}
Thus,
\begin{align}
&\exp \left( {\frac{1}{{z^2 - 1}}} \right)\sin \pi \left( {z - 1} \right)
\\
&= - \left( {\sum {\frac{1}{{\left( {z + 1} \right)^n \left( {z - 1} \right)^n n!}}} } \right)\left( {\sum {\frac{{\left( { - 1} \right)^n }}{{\left( {2n + 1} \right)!}}\pi ^{2n + 1} \left( {z - 1} \right)^{2n + 1} } } \right) \
\\
&=- \left( {1 + \frac{{\frac{1}{{\left( {z + 1} \right)}}}}{{\left( {z - 1} \right)}} + \frac{{\frac{1}{{\left( {z + 1} \right)^2 }}}}{{\left( {z - 1} \right)^2 2!}} + \frac{{\frac{1}{{\left( {z + 1} \right)^3 }}}}{{\left( {z - 1} \right)^3 3!}} + \frac{{\frac{1}{{\left( {z + 1} \right)^4 }}}}{{\left( {z - 1} \right)^4 4!}} + \frac{{\frac{1}{{\left( {z + 1} \right)^5 }}}}{{\left( {z - 1} \right)^5 5!}} + \cdots } \right) \\
&\times \left( {\pi \left( {z - 1} \right) - \frac{{\pi ^3 }}{{3!}}\left( {z - 1} \right)^3 + \frac{{\pi ^5 }}{{5!}}\left( {z - 1} \right)^5 + \cdots + \cdots } \right) \\
&\Rightarrow \text{The coefficient of }\frac{1}{z-1}=-
\left( {\frac{\pi }{{1!2!}}\frac{1}{{\left( {z + 1} \right)^2 }} - \frac{{\pi ^3 }}{{3!4!}}\frac{1}{{\left( {z + 1} \right)^4 }} + \frac{{\pi ^5 }}{{5!6!}}\frac{1}{{\left( {z + 1} \right)^6 }} + \ldots } \right)
\\
&=
- \sum {\frac{{\left( { - 1} \right)^n \pi ^{2n + 1} }}{{n!\left( {n + 1} \right)!}}\frac{1}{{\left( {z + 1} \right)^{2n + 2} }}}
= g\left( z \right)
\end{align}
\begin{align}
{\mathop{\rm Res}\nolimits} \left\{ {f\left( z \right),1} \right\} = g\left( 1 \right)
=
- \sum {\frac{{\left( { - 1} \right)^n \pi ^{2n + 1} }}{{n!\left( {n + 1} \right)!}}\frac{1}{{2^{2n + 2} }}}
\end{align}
|
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|
Find the inverse Laplace transform of: $\frac{1}{(s^2+a^2)(s^2+b^2)}$ I'm having trouble doing this homework problem because I'm not sure how to deal with the $a$ and $b$. I did it the usual way we were taught - use partial fraction decomposition and then try to solve for the coefficients. When I solved for them, this is the conclusion I came to:
$$
\frac{A}{B}=\frac{-a^2}{b^2}
$$
Thus my original problem becomes
$$
\frac{-a^2}{(s^2+a^2)}+\frac{b^2}{(s^2+b^2)}
$$
But I'm stuck here. I know that the inverse of $\frac{a}{(s^2+a^2)}$ is $sin(at)$ but I'm not sure if I can use that here....
Also, is there another way of solving this without using partial fraction decomposition?
|
Just note $$\frac{1}{s^2+a^2}-\frac{1}{s^2+b^2}=\frac{b^2-a^2}{(s^2+a^2)(s^2+b^2)}$$
Then, if $a^2\neq b^2$ and $ab\neq 0$, we have
\begin{align*}
\frac{1}{(s^2+a^2)(s^2+b^2)}&=\frac{\frac{1}{b^2-a^2}}{s^2+a^2}-\frac{\frac{1}{b^2-a^2}}{s^2+b^2}\\
\mathcal{L}^{-1}\left\{\frac{1}{(s^2+a^2)(s^2+b^2)}\right\}&=\frac{1}{a(b^2-a^2)}\sin(at)-\frac{1}{b(b^2-a^2)}\sin(bt)
\end{align*}
|
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|
Find the value of $x$ such that $\sqrt{4+\sqrt{4-\sqrt{4+\sqrt{4-x}}}}=x$ Find the value of $x$, $$\sqrt{4+\sqrt{4-\sqrt{4+\sqrt{4-x}}}}=x$$
Help guys please, I have tried and I got, $x=-2, x=1$, and I think it's wrong
|
$$\begin{align}\sqrt{4+\sqrt{4-\sqrt{4+\sqrt{4-x}}}}&=x&\Longleftrightarrow \\
\left(\sqrt{4+\sqrt{4-\sqrt{4+\sqrt{4-x}}}}\right)^2&=x^2&\Longleftrightarrow\\
4+\sqrt{4-\sqrt{4+\sqrt{4-x}}}&=x^2&\Longleftrightarrow\\
\left(\sqrt{4-\sqrt{4+\sqrt{4-x}}}\right)^2&=\left(x^2-4\right)^2&\Longleftrightarrow\\
4-\sqrt{4+\sqrt{4-x}}&=\left(x^2-4\right)^2&\Longleftrightarrow\\
-\sqrt{4+\sqrt{4-x}}&=\left(x^2-4\right)^2-4&\Longleftrightarrow\\
\sqrt{4+\sqrt{4-x}}&=-\left(x^2-4\right)^2+4&\Longleftrightarrow\\
\left(\sqrt{4+\sqrt{4-x}}\right)^2&=\left(-\left(x^2-4\right)^2+4\right)^2&\Longleftrightarrow\\
4+\sqrt{4-x}&=\left(-\left(x^2-4\right)^2+4\right)^2&\Longleftrightarrow\\
\sqrt{4-x}&=\left(x^4-8x^2+12\right)^2-4&\Longleftrightarrow\\
\left(\sqrt{4-x}\right)^2&=\left(\left(x^4-8x^2+12\right)^2-4\right)^2&\Longleftrightarrow\\
4-x &=\left(x^4-8x^2+10\right)^2\left(x^4-8x^2+14\right)^2 &&
\end{align}
$$
Solving $x$ gives us $\frac{1}{2}+\frac{\sqrt{13}}{2}$
|
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|
Finding the roots of a different Quadratic equation from the roots of a Given Quadratic equation The Question:
If $\alpha$ and $\beta$ are the roots of the equation $ax^2+bx+c=0$...
Then find the roots of the equation $ax^2-bx(x-1)+c(x-1)^2=0$
My Attempt:
The new equation can be made into a quadratic as:
$$(a-b+c)x^2+(b-2c)x+c=0$$
Now $$\text{Sum of roots}=\dfrac{-b}{a} = \dfrac{2c-b}{a-b+c}$$
And $$\text{Product of roots}=\dfrac{c}{a}=\dfrac{c}{a-b+c}$$
But I don't seem to be going anywhere with the way I'm proceeding
Please Help! Thanks!
|
hint: $\dfrac{2c-b}{a-b+c} = \dfrac{2\dfrac{c}{a}-\dfrac{b}{a}}{1-\dfrac{b}{a}+\dfrac{c}{a}}=\dfrac{2\alpha\cdot \beta+(\alpha+\beta)}{1+\alpha+\beta+\alpha\cdot \beta}$, can you continue?
|
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|
Proving basic floor function inequality: $-1 \lt \lfloor 2x \rfloor - 2 \lfloor x \rfloor \lt 2$ As a direct consequence of the definition of $\lfloor x \rfloor $ I know that $$2x-1 \lt \lfloor 2x \rfloor \le 2x$$ and $$2x-2 \lt 2\lfloor x\rfloor \le 2x$$
How can I use these to show that $-1 \lt \lfloor 2x \rfloor - 2 \lfloor x \rfloor \lt 2$? I'm a bit rusty with inequalities.
Any help is appreciated.
|
Let $n = \lfloor x\rfloor$ and $f=x-n = x-\lfloor x\rfloor$, i.e. $x=n+f$. Then
$$\begin{align*}
\lfloor 2x\rfloor-2\lfloor x\rfloor&= \lfloor 2n+2f\rfloor - 2\lfloor n+f\rfloor\\
&=2n+\lfloor 2f\rfloor - 2n\\
&= \lfloor 2f\rfloor
\end{align*}$$
Since $0\le f<1$, $0\le 2f<2$ and then $\lfloor 2f\rfloor$ can only be $0$ or $1$.
Hence we have a stronger result:
$$0\le\lfloor 2x\rfloor-2\lfloor x\rfloor\le1$$
|
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|
Side limits of the derivative of this function $f:\mathbb{R}\to \mathbb{R}$ with $f\left(x\right)=\left(x^3+3x^2-4\right)^{\frac{1}{3}}$
Calculate side limits of this function's derivative, $f'_s\:and\:f'_d$, in $x_o=-2$
The answer key says I should get $\infty $ and $-\infty$ but I'm not getting that. The derivative I get is $\frac{x\left(x+2\right)}{\left(\left(x-2\right)^2\left(x+2\right)^4\right)^{\frac{1}{3}}}$ and by doing the multiplication from the denominator I would get something with $x^2$.
|
You have the wrong expression for the derivative.
$$\begin{align}
\frac d{dx}\left[(x^3+3x^2-4)^{1/3}\right]
&= \frac 13(x^3+3x^2-4)^{-2/3}(3x^2+6x) \\[2ex]
&= \frac {x^2+2x}{[(x+2)^2(x-1)]^{2/3}} \\[2ex]
&= \frac {x(x+2)}{(x+2)^{4/3}(x-1)^{2/3}} \\[2ex]
&= \frac {x}{(x+2)^{1/3}(x-1)^{2/3}} \\[2ex]
\end{align}$$
That last expression's denominator tends to zero as $x\to-2$ but the numerator does not tend to zero, which means an infinite limit on both sides of $-2$. As $x\to-2$ from the left, both numerator and denominator are negative, so the expression tends to $+\infty$. As $x\to-2$ from the right, the numerator is negative but the denominator is positive, so the expression tends to $-\infty$.
|
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|
Find eigenvalues and eigenvectors of this matrix Problem: Let \begin{align*} A = \begin{pmatrix} 1 & 1 & 1 & 1 \\ 1 & 1 & 1 & 1 \\ 1 & 1 & 1 & 1 \\ 1 & 1 & 1 & 1 \end{pmatrix}. \end{align*} Compute all the eigenvalues and eigenvectors of $A$.
Attempt at solution: I found the eigenvalues by computing the characteristic polynomial. This gives me \begin{align*} \det(A - x \mathbb{I}_4) = \det \begin{pmatrix} 1-x & 1 & 1 & 1 \\ 1 & 1-x & 1 & 1 \\ 1 & 1 & 1-x & 1 \\ 1 & 1 & 1 & 1-x \end{pmatrix} = -x^3 (x-4) = 0 \end{align*} after many steps. So the eigenvalues are $\lambda_1 = 0$ with multiplicity $3$ and $ \lambda_2 = 4$ with multiplicity $1$.
Now I was trying to figure out what the eigenvectors are corresponding to these eigenvalues. Per definition we have $Av = \lambda v$, where $v$ is an eigenvector with the corresponding eigenvalue. So I did for $\lambda_2 = 4$: \begin{align*} \begin{pmatrix} 1 & 1 & 1 & 1 \\ 1 & 1 & 1 & 1 \\ 1 & 1 & 1 & 1 \\ 1 & 1 & 1 & 1 \end{pmatrix} \begin{pmatrix} x_1 \\ x_2 \\ x_3 \\ x_4 \end{pmatrix} = 4 \begin{pmatrix} x_1 \\ x_2 \\ x_3 \\ x_4 \end{pmatrix} \end{align*} I think this is only possible when $x_1 = x_2 = x_3 = x_4 = 1$. So am I right in stating that all the eigenvectors corresponding to $\lambda_2$ are of the form $t \begin{pmatrix} 1 \\ 1 \\ 1 \\ 1 \end{pmatrix}$ with $t$ some number $\neq 0$? For $\lambda_1 = 0$, I'm not sure how to find the eigenvectors. The zero vector is never an eigenvector. This means $x_1, x_2, x_3$ and $x_4$ can be anything aslong as they add to zero?
|
You are exactly right! $x_1,x_2,x_3$ and $x_4$ can be anything as long as they sum to zero. This is already a complete solution in some sense (you have "found" all of the eigenvectors). If you would like, you could find 3 linearly independent such vectors, and then you would be sure that these three span the whole space.
|
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|
Find $LK_1^2 + LK_2^2 + \dots + LK_{11}^2$. $K_1 K_2 \dotsb K_{11}$ is a regular $11$-gon inscribed in a circle, which has a radius of $2$. Let $L$ be a point, where the distance from $L$ to the circle's center is $3$. Find
$LK_1^2 + LK_2^2 + \dots + LK_{11}^2$.
Any suggestions as to how to solve this problem? I'm unsure what method to use.
|
Let $\omega = e^{2 \pi i/11}$, a primitive $11^{\text{th}}$ root of unity. We can assume that the circle is centered at the origin. We can also assume that $A_k$ is associated with the complex number $2 \omega^k$
Let $p$ be complex number associated with the point $P$. Then
$PA_1^2 + PA_2^2 + \dots + PA_{11}^2 = \sum_{k = 0}^{10} |p - 2 \omega^k|^2.$
From the identity $z \cdot \overline{z} = |z|^2$,
\begin{align*}
\sum_{k = 0}^{10} |p - 2 \omega^k|^2 &= \sum_{k = 0}^{10} (p - 2 \omega^k)(\overline{p} - 2 \overline{\omega}^k) \\
&= \sum_{k = 0}^{10} (p \overline{p} - 2 \overline{\omega}^k p - 2 \omega^k \overline{p} + 4 \omega^k \overline{\omega}^k) \\
&= 11 p \overline{p} - 2p \sum_{k = 0}^{10} \overline{\omega}^k - 2 \overline{p} \sum_{k = 0}^{10} \omega^k + 4 \sum_{k = 0}^{10} \omega^k \overline{\omega}^k.
\end{align*}
The distance from $P$ to the origin is 3, so $11p \overline{p} = 11 \cdot |p|^2 = 11 \cdot 9 = 99$.
Since $\omega$ is a primitive $11^{\text{th}}$ root of unity, $\omega^{11} - 1 = 0$, which factors as
$(\omega - 1)(\omega^{10} + \omega^9 + \dots + \omega + 1) = 0.$
Since $\omega \neq 1$, we have $\omega^{10} + \omega^9 + \dots + \omega + 1 = 0$. Therefore,
$2 \overline{p} \sum_{k = 0}^{10} \omega^k = 0.$
Also, $|\omega| = 1$, so $\overline{\omega} = 1/\omega$, which means
$\sum_{k = 0}^{10} \overline{\omega}^k = 1 + \frac{1}{\omega} + \dots + \frac{1}{\omega^9} + \frac{1}{\omega^{10}} = \frac{\omega^{10} + \omega^9 + \dots + \omega + 1}{\omega^{10}} = 0.$
Finally, $\omega^k \overline{\omega}^k = \omega^k/\omega^k = 1$, so
$4 \sum_{k = 0}^{10} \omega^k \overline{\omega}^k = 4 \cdot 11 = 44.$
Therefore,
$PA_1^2 + PA_2^2 + \dots + PA_{11}^2 = 99 + 44 = \boxed{143}.$
|
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|
Way to make this homogoneous ODE seperable? Is my algebra correct, turning this:
$$\frac{\mathrm{d}y}{\mathrm{d}x} = \frac{4y-3x}{2x-y}$$
Into this:
I split into the difference of the two fractions,
then factored x out of the left fraction, and factored y
out of the right fraction, getting:
$$\frac{4\frac{y}{x}}{x(1-\frac{y}{x})} - \frac{3\frac{x}{y}}{2\frac{x}{y}-1}$$
Now I can substitute $v = \frac{y}{x}, v^{-1}= \frac{x}{y}$, right?
Then separate? Then integrate?
|
I solved it maybe it will help you:$$\frac { dy }{ dx } =\frac { 4y-3x }{ 2x-y } =\frac { 4\frac { y }{ x } -3 }{ 2-\frac { y }{ x } } \\ \frac { y }{ x } =t$$
$$ y=xt\\ y^{ \prime }=t+x{ t }^{ \prime }\\ t+x{ t }^{ \prime }=\frac { 4t-3 }{ 2-t } $$ $$ x{ t }^{ \prime }=\frac { { t }^{ 2 }+2t-3 }{ 2-t } =\frac { \left( t+3 \right) \left( t-1 \right) }{ 2-t } \\ \int { \frac { 2-t }{ \left( t+3 \right) \left( t-1 \right) } dt } =\int { \frac { dx }{ x } } \\ \frac { 2-t }{ \left( t+3 \right) \left( t-1 \right) } =\frac { A }{ t+3 } +\frac { B }{ t-1 } $$ $$2-t=A\left( t-1 \right) +B\left( t+3 \right) =\left( A+B \right) t+3B-A$$
$A=-\frac { 5 }{ 4 } $,$B=\frac { 1 }{ 4 } $
$$ -\frac { 5 }{ 4 } \int { \frac { dt }{ t+3 } } +\frac { 1 }{ 4 } \int { \frac { dt }{ t-1 } = } \int { \frac { dx }{ x } } $$
$$-\frac { 5 }{ 4 } \ln { \left| t+3 \right| } +\frac { 1 }{ 4 } \ln { \left| t-1 \right| =\ln { Cx } } $$ $$\ln { \sqrt [ 4 ]{ \left| \frac { t-1 }{ \left( t+3 \right) ^{ 5 } } \right| } } =\ln { Cx } $$ $$ \frac { t-1 }{ \left( t+3 \right) ^{ 5 } } =Cx^{ 4 }$$ $$ \frac { \frac { y }{ x } -1 }{ { \left( \frac { y }{ x } +3 \right) }^{ 5 } } =Cx^{ 4 }$$
|
{
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"timestamp": "2023-03-29T00:00:00",
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|
Using equation to find value of $1/x - 1/y$ $$\left(\frac{48}{10}\right)^x=\left(\frac{8}{10}\right)^y=1000$$
What is the value of $\frac{1}{x}-\frac{1}{y}$?
I have already used that when $48$ divided by $10$ then it becomes $4.8$ and when $8$ divided by $10$ then it becomes $0.8$ by getting $10^x$ and $10^y$ to make it simpler but I cannot proceed further.
|
From $4.8^x = 1000$ we get $4.8 = 1000^{1/x}$ and similarly for the $y$ term we get $0.8 = 1000^{1/y}$ by taking the $x$ and $y$-root of both sides respectively. Hence we get $$\frac{4.8}{0.8} = \frac{1000^{1/x}}{1000^{1/y}}$$
This yields $$6 = 1000^{1/x - 1/y}$$ so $$\bbox[10px, border: solid red 2px]{\frac{1}{x} - \frac{1}{y} = \log_{1000} 6}$$ by applying the $1000$-base logarithm of both sides of the equation.
|
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|
Limit and factorization I have the following very interesting homework exercise:
Let $$f(x)=\frac{x^2-4}{2-x\cdot \sqrt {x+2}+\sqrt{x+2}}$$ Find the
following limit, if it exists: $$\lim_{x\to 2}f(x)$$
I understand that I need to "delete" the $x-2$ factor from both nominator and denominator and then evaluate the limit. It is obvious that $x^2-4=(x-2)\cdot (x+2)$, so the nominator is done.
So, my main problem is how to factor the denominator of $f(x)$. Any help with the factorization, without the final solution, would be appreciated. Then, after I understand how to factor it, anyone can post the full solution, even I.
|
$$f(x)=\frac{x^2-4}{2-x\cdot \sqrt {x+2}+\sqrt{x+2}}\Longrightarrow$$
$$\lim_{x\to2}f(x)=\lim_{x\to2}\left(\frac{x^2-4}{2-x\cdot \sqrt {x+2}+\sqrt{x+2}}\right)=$$
$$\lim_{x\to2}\left(\frac{\frac{d}{dx}\left(x^2-4\right)}{\frac{d}{dx}\left(2-x\cdot \sqrt {x+2}+\sqrt{x+2}\right)}\right)=$$
$$\lim_{x\to2}\left(\frac{2x}{\frac{1}{2\sqrt{x+2}}-\frac{x}{2\sqrt{x+2}}-\sqrt{x+1}}\right)=$$
$$\lim_{x\to2}\left(-\frac{4x\sqrt{x+2}}{3(x+1)}\right)=$$
$$\frac{\lim_{x\to2}\left(-4x\sqrt{x+2}\right)}{\lim_{x\to2}\left(3(x+1)\right)}=$$
($3(x+1)$ is a polynomial and thus everywhere continuous):
$$\frac{\lim_{x\to2}\left(-4x\sqrt{x+2}\right)}{3(2+1)}=$$
$$\frac{\lim_{x\to2}\left(-4x\sqrt{x+2}\right)}{3(3)}=$$
$$\frac{\lim_{x\to2}\left(-4x\sqrt{x+2}\right)}{9}=$$
$$\frac{-4\lim_{x\to2}x\sqrt{\lim_{x\to2}(x+2)}}{9}=$$
($x$ is a polynomial and thus everywhere continuous):
$$\frac{-4\cdot2\sqrt{\lim_{x\to2}(x+2)}}{9}=$$
($x+2$ is a polynomial and thus everywhere continuous):
$$\frac{-4\cdot2\sqrt{(2+2)}}{9}=\frac{-4\cdot2\sqrt{4}}{9}=\frac{-4\cdot2\cdot2}{9}=\frac{-4\cdot4}{9}=\frac{-16}{9}=-\frac{16}{9}$$
So:
$$\lim_{x\to2}f(x)=\lim_{x\to2}\left(\frac{x^2-4}{2-x\cdot \sqrt {x+2}+\sqrt{x+2}}\right)=-\frac{16}{9}$$
|
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|
Verify that $\binom{n+1}{4} = \frac{\left(\substack{\binom{n}{2}\\{\displaystyle2}}\right)}{3}$ for $n \geq 4$
Verify that for $n \geq 4$
$$\dbinom{n+1}{4} = \frac{\left(\substack{\binom{n}{2}\\{\displaystyle2}}\right)}{3}$$
Now present a combinatoric argument for the above.
First, by verify does it mean check for some n > 3? if so, then n = 4 gives both sides value 5. I have tried expanding both sides and It just gets messy, but i'm sure that would be attempting to prove it?
I cant quite move ahead with this one, I have said that $\left(\substack{\binom{n}{2}\\{\displaystyle2}}\right)$ is the number of ways of choosing 2 pairs of objects from n objects. my reasoning for this is that $\dbinom{n}{2}$ is the number of ways of choosing 2 objects from n and hence $\left(\substack{\binom{n}{2}\\{\displaystyle2}}\right)$ is the number of ways of choosing 2 of these ways... What i am struggling to do is understand how the three comes into it.
|
Using
\begin{align}
\binom{n}{2} = \frac{n(n-1)}{2}
\end{align}
then
\begin{align}
\frac{1}{3} \, \binom{\binom{n}{2}}{2} &= \frac{1}{2} \binom{n}{2} \, \left(\frac{n}{2} - 1\right) \\
&= \frac{n(n-1)}{4!} \left( n(n-1)-2 \right) = \frac{(n+1)(n)(n-1)(n-2)}{4!} \\
&= \binom{n+1}{4}.
\end{align}
As to the modified component of the question: The comments seem to be helpful
|
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|
Does $(a^p + b^p)^{p-1} \equiv 1 \pmod {p^2}$ have any solutions where $a$ and $b$ are co-primes less than $p$? How will you prove that $(a^p + b^p)^{p-1} \equiv 1 \pmod {p^2}$ has no solution where $p$ is a prime number and $a$, $b$ are two co-primes less than $p$? If this equation has a solution, then what it is it?
Edit- Some solutions considered taking $a$ or $b$ as 1. Are there any solutions where neither $a$ nor $b$ is 1?
|
For $p=2$ it says $a^2 + b^2 \equiv 1 \mod 4$. This can only be achieved if one of $a$ and $b$ is divisible by 2, so this one doesn't have any solutions.
For $p=3$, one has $(a^3+b^3)^2 \equiv 1 \mod 9$, so $a^3+b^3 \equiv 1 \mod 9$ or $a^3+b^3 \equiv 8 \mod 9$. The former has no solutions, same argument as for $p=2$. But third powers are only congruent to 0,1,8 mod 9, so it won't work without using a 0.
For $p=5$ one can verify in the same way that there are no solutions.
For $p=7$ one has $a=1$, $b=2$ as a solution as well as 97 other pairs.
|
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|
Find the value of $2xy$ .
If $13x+17y=643$ ,$\{x,y\}\in \mathbb{N}$, then what is the value of two times the product of
$x$ and $y$ ?
Options
$a.)\ 744\quad \quad \quad \quad \quad
b.)\ 844\\
\color{green}{c.)\ 924}\quad \quad \quad \quad \quad
d.)\ 884\\$
I tried,
$13x+17y \pmod{13}\equiv 0\\
\implies 2y \pmod{13}\equiv 3 \\
\implies y=8
\implies y=8, x=39$
$2xy=624$
I look for a short and simple way .
I have studied maths up to $12$th grade.
|
When applying mod $13$, the equation $13x+17y=643$ becomes
$$4y\equiv 6\pmod{13}$$
or
$$40y\equiv 60\pmod {13}$$
that is, $y\equiv 8\pmod {13}$.
Now, to find $x$, apply mod $17$:
$$13x\equiv 14\pmod{17}$$
or
$$4\cdot 13x\equiv 56\pmod {17}$$
thus, $x\equiv 5\pmod{17}$.
Now we are to find the concrete values of $x$ and $y$:
$$13(17u+5)+17(13v+8)=643$$
which yields
$$221(u+v)+201=643$$
therefore, $u+v=2$. Since $u$ and $v$ must not be negative, we have three possibilities:
*
*$u=2$, $v=0$. Then $x=39$, $y=8$, so $2xy=624$.
*$u=v=1$. Then $x=22$, $y=21$. Then $2xy=924$.
*$u=0$, $v=2$. Then $x=5$, $y=34$. Then $2xy=340$.
|
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|
Why does this sum converge? I know that the following sum converges to 2 via WolframAlpha, but I am not sure why.
$$\sum_{k=1}^\infty k \left[\frac{2}{k} - \frac{4}{k+1} + \frac{2}{k+2}\right] = 2$$
WolframAlpha gives the following partial sum formula:
$$\sum_{k=1}^n k \left[\frac{2}{k} - \frac{4}{k+1} + \frac{2}{k+2}\right] = \frac{2n}{n+2}$$
I would intuitively guess that the result of the partial sum formula is 2 for $n = \infty$. Where did that partial sum formula come from? Can someone help me build some intuition here?
|
$$\begin{align}k\left(\frac 2k-\frac{4}{k+1}+\frac{2}{k+2}\right)&=2-\frac{4k}{k+1}+\frac{2k}{k+2}\\&=2-4\left(1-\frac{1}{k+1}\right)+2\left(1-\frac{2}{k+2}\right)\\&=4\left(\frac{1}{k+1}-\frac{1}{k+2}\right)\end{align}$$
|
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|
Derive Cartesian cubic Möbius strip from parametric The following link:
http://mathworld.wolfram.com/MoebiusStrip.html
shows the Möbius strip parametrized as
\begin{eqnarray}
x = [ R + s \cos \left ( \frac{1}{2} t \right ) ] \cos t \\
y = [ R + s \cos \left ( \frac{1}{2} t \right ) ] \sin t \\
z = s \sin \left ( \frac12 t \right )
\end{eqnarray}
The symbols for $R$ and $s$ and angle $t$ are explained there.
Then they say that from this parametrization we can derive the cubic.
\begin{equation}
-R^2 y + x^2 y + y^3 - 2 R x z - 2 x^2 z + y z^2 = 0.
\end{equation}
Any ideas about how to do this? I have tried with no success.
Thanks.
This is what I have done so far:
Square the first two equations above and add to find
\begin{equation}
x^2 + y^2 = \left ( R + s \left ( \cos \frac{t}{2} \right ) \right )^2
\end{equation}
Take the square root of this
\begin{equation}
\sqrt{x^2 + y^2 }= R + s \left ( \cos \frac{t}{2} \right ) \Longrightarrow
\sqrt{x^2 + y^2 } - R = s \left ( \cos \frac{t}{2} \right )
\end{equation}
Square this and the third equation (for $z$) and add to find
\begin{equation}
s^2 = \left ( \sqrt{x^2+y^2}- R \right )^2 + z^2
\end{equation}
Now, let us divide the second by the first equation
That is
\begin{equation}
\frac{y}{x} = \tan t = \frac{2 \tan (t/2)}{1 - \tan^2 (t/2)}
\end{equation}
multiply numerator and denominator by $\cos^2 (t/2)$
\begin{equation}
\frac{y}{x} = \frac{2 \sin(t/2) (\sqrt{1-\sin^2(t/2)} }{\cos^2 (t/2) - \sin^2(t/2)}
= \frac{2 \sin(t/2) \sqrt{1 - \sin^2(t/2)}}{1 - 2 \sin^2(t/2)}.
\end{equation}
Multiply numerator and denominator by $s^2$, then
\begin{equation}
\frac{y}{x} = \frac{2 z \sqrt{s^2 - z^2}}{s^2 - 2 z^2}
\end{equation}
That is
\begin{equation}
\frac{y}{x} = \frac{2 z (\sqrt{x^2 + y^2} - R)}{ ( \sqrt{x^2 + y^2} - R)^2 - z^2}
\end{equation}
or
\begin{equation}
\frac{y}{x} = \frac{2 z (\sqrt{x^2 + y^2} - R)}{ (x^2 + y^2) + R^2 - 2 R \sqrt{x^2+y^2}
- z^2}
\end{equation}
Or
\begin{equation}
y (x^2 + y^2) + y R^2 - 2 R y \sqrt{x^2+y^2} - z^2 y =
2 z x \sqrt{x^2 + y^2} -2 R x z
\end{equation}
|
Consider the Moebius strip with midline the circle of radius $R$ in the $(x,y)$-plane and having width $2a>0$:
$$M:\quad(\phi,s)\mapsto\left\{\eqalign{x&=\bigl(R+s\cos{\textstyle{\phi\over2}}\bigr)\cos\phi \cr
y&=\bigl(R+s\cos{\textstyle{\phi\over2}}\bigr)\sin\phi \cr
z&=s\sin{\textstyle{\phi\over2}}\ .\cr}\right.\tag{1}$$
The parameter domain is $-\pi<\phi<\pi$, $\ -a<s<a$. For "technical reasons" we have excluded $\phi=\pm\pi$. This guarantees $\cos{\phi\over2}>0$ over the whole parameter domain.
It is claimed that $M$ is part of a certain cubic surface $S\subset{\mathbb R}^3$. This surface results from revoking the condition $-a<s<a$ in $(1)$, so that now the parameter $s$ runs from $-\infty$ to $\infty$. Consider a fixed value $\phi\in\ ]{-\pi},\pi[\ $. Then
$$g_\phi:\quad s\mapsto\left(\bigl(R+s\cos{\textstyle{\phi\over2}}\bigr)\cos\phi, \ \bigl(R+s\cos{\textstyle{\phi\over2}}\bigr)\sin\phi , \ s\sin{\textstyle{\phi\over2}}\right)\tag{2}$$
describes a straight line lying on $S$. We now replace the parameter $s$ in $(2)$ by the new parameter $u:=R+s\cos{\phi\over2}$. In this way $g_\phi$ appears in the form
$$g_\phi:\quad u\mapsto\bigl(u\cos\phi, u\sin\phi, \tan{\textstyle{\phi\over2}}(u-R)\bigr)\qquad(-\infty<u<\infty)\ .\tag{3}$$
When we let $\phi$ vary as well in $(3)$ we obtain another parametrization of our surface $S$. In order to get rid of the trigonometric functions we introduce the new parameter $t:=\tan{\phi\over2}$, which runs from $-\infty$ to $\infty$. In this way we obtain the following rational representation of $S$:
$$S:\quad(t,u)\mapsto\left\{\eqalign{x&=u\ {1-t^2\over 1+t^2} \cr
y&=u\ {2t\over1+t^2}\cr
z&=t\ (u-R)\cr}\right.\qquad\qquad\bigl((t,u)\in{\mathbb R}^2\bigr)\ .$$
One has $2(z+Rt)=2ut=y(1+t^2)$ and $$2tx=y(1-t^2)\ .\tag{4}$$ Adding these two equations leads to
$$t={y-z\over R+x}\ .$$
We insert this value of $t$ into $(4)$ and obtain after clearing denominators
$$2x(R+x)(y-z)=y\bigl((R+x)^2-(y-z)^2\bigr)\ .$$
This equation is valid for all points $(x,y,z)\in S$, and as well on the line $g_{\pm \pi}$ omitted from consideration. It expands to
$$-R^2 y + x^2 y + y^3 - 2 R x z - 2 x^2 z - 2 y^2 z + y z^2=0\ ,$$
as given in the quoted Wikipedia link (in your question you have forgotten one term).
|
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|
Find the principal solutions of the trigonometric equation $\cos x-\sin x+\sin 2x+3\cos2x+1=0$ I am unable to simplify the expression. If I simplify the double angles, it leaves me with a nasty expression,
$\cos x-\sin x+2\sin x\cos x+6\cos^2 x-2=0$. What do I do next. Some hints, please. Also, is there some elegant solution? Thanks!
|
Observe:
$$
\cos x-\sin x+2\sin x\cos x+6\cos^2 x-2=(\sin x)(2\cos x-1)+(2\cos x-1)(3\cos x+2)=(\sin x+3\cos x-2)(2\cos x-1).
$$
Therefore, $\cos x=\frac{1}{2}$ is a solution.
The solutions to $\sin x+3\cos x+2=0$ are less pleasant, but one can solve $\sin x=-2-3\cos x$, square both sides and solve, using the quadratic formula for $\cos x$.
|
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|
Volume of Tetrahedron $ABCD$ is a regular tetrahedron of volume $1$. Maria glues regular tetrahedra $A'BCD$, $AB'CD$, $ABC'D$, and $ABCD'$
to the faces of $ABCD$. What is the volume of the tetrahedron $A'B'C'D'$?
|
Let the edge length of (original) regular tetrahedron $ABCD$ (centered at $O$) be $a$ then the volume of the regular tetrahedron $ABCD$ $$=\color{red}{\frac{a^3}{6\sqrt{2}}}$$ but the volume is $1$ hence, we have $$\frac{a^3}{6\sqrt{2}}=1$$ $$\implies \color{blue}{a=(6\sqrt{2})^{1/3}}$$ Distance of each face of regular tetrahedron $ABCD$ from the center $O$ is $$=\frac{a}{2\sqrt{6}}$$ Normal height of regular tetrahedron $ABCD$ is $$=a\sqrt{\frac{2}{3}}$$ Now, the distance of each of vertices $A'$, $B'$, $C'$ & $D'$ from the center $O$ $$=\frac{a}{2\sqrt{6}}+a\sqrt{\frac{2}{3}}=\frac{5a}{2\sqrt{6}}$$Now, let $a'$ be the edge length of new regular tetrahedron $A'B'C'D'$ then the distance of its each vertex from the center $O$ $$=\frac{a'}{2}\sqrt{\frac{3}{2}}$$ $$\implies \frac{a'}{2}\sqrt{\frac{3}{2}}=\frac{5a}{2\sqrt{6}}$$ $$\implies a'=\frac{5a}{3}=\color{red}{\frac{5(6\sqrt{2})^{1/3}}{3}}$$
Hence, the volume of new regular tetrahedron $A'B'C'D'$ $$=\frac{(a')^3}{6\sqrt{2}}=\frac{\left(\frac{5(6\sqrt{2})^{1/3}}{3}\right)^3}{6\sqrt{2}}$$ $$=\frac{125(6\sqrt{2})}{27(6\sqrt{2})}$$ $$=\color{blue}{\frac{125}{27}\approx 4.62962963\dots}$$
Note: All the formula of regular tetrahedron have been taken from Table of platonic solids
|
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|
Solving $6 \cos x - 5 \sin x = 8$ My attempt:
Using the formula for linear combinations of sine and cosine:
$$A \cos x+B \sin x=C \sin (x+\phi)$$
$$
\sqrt{51} \left(\frac{6}{\sqrt{51}} \cos x - \frac{5}{\sqrt{51}}\sin x\right) = 8
$$
$$
\frac{6}{\sqrt{51}} \cos x - \frac{5}{\sqrt{51}}\sin x = \frac{8}{\sqrt{51}}
$$
And then assume:
$$
\frac{6}{\sqrt{51}}= \cos \psi ; \frac{5}{\sqrt{51}}= \sin\psi ;
$$
$$
\cos \psi \cos x - \sin \psi \sin x = \cos (x+ \psi) = \cos(x + \arccos ( \frac{6}{\sqrt{51}}))
$$
$$
x + \arccos\left(\frac{6}{\sqrt{51}}\right) = \arcsin\left( \frac{8}{\sqrt{51}}\right)
$$
$$
x \approx 12^\circ
$$
But answer is:
$$
-\frac{\pi}{4} + (-1)^n \frac{\pi}{4} + \pi n , n\in\Bbb Z
$$
|
We have $$6 \cos x - 5\cos x = \cos x$$
But, a quick look at the graph of the cosine function shows us that it is bounded between $-1$ and $1$, so $\cos x = 8$ has no solutions. $\square$
Assuming your original equation was $6 \cos x - 5\sin x = 8$ as your body would suggest, we can represent this in the form $$\sqrt{61} \sin \left(x - \arctan\left(\frac{5}{6}\right)\right) = 8$$
using the same method you did in your question, can you take it from there?
Edit: This still doesn't make sense, we have $$\max(6 \cos x - 5\sin x) = \sqrt{61} < 8$$ So there are still no solutions.
|
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|
Wanted : for more formulas to find the area of a triangle? I know some formulas to find a triangle's area, like the ones below.
*
*Is there any reference containing most triangle area formulas?
*If you know more, please add them as an answer
$$s=\sqrt{p(p-a)(p-b)(p-c)} ,p=\frac{a+b+c}{2}\\s=\frac{h_a*a}{2}\\s=\frac{1}{2}bc\sin(A)\\s=2R^2\sin A \sin B \sin C$$
Another symmetrical form is given by :$$(4s)^2=\begin{bmatrix}
a^2 & b^2 & c^2
\end{bmatrix}\begin{bmatrix}
-1 & 1 & 1\\
1 & -1 & 1\\
1 & 1 & -1
\end{bmatrix} \begin{bmatrix}
a^2\\
b^2\\
c^2
\end{bmatrix}$$
Expressing the side lengths $a$, $b$ & $c$ in terms of the radii $a'$, $b'$ & $c'$ of the mutually tangent circles centered on the triangle's vertices (which define the Soddy circles)
$$a=b'+c'\\b=a'+c'\\c=a'+b'$$gives the paticularly pretty form $$s=\sqrt{a'b'c'(a'+b'+c')}$$
If the triangle is embedded in three dimensional space with the coordinates of the vertices given by $(x_i,y_i,z_i)$ then $$s=\frac{1}{2}\sqrt{\begin{vmatrix}
y_1 &z_1 &1 \\
y_2&z_2 &1 \\
y_3 &z_3 &1
\end{vmatrix}^2+\begin{vmatrix}
z_1 &x_1 &1 \\
z_2&x_2 &1 \\
z_3 &x_3 &1
\end{vmatrix}^2+\begin{vmatrix}
x_1 &y_1 &1 \\
x_2&y_2 &1 \\
x_3 &y_3 &1
\end{vmatrix}^2}$$
When we have 2-d coordinate $$ s=\frac{1}{2}\begin{vmatrix}
x_a &y_a &1 \\
x_b &y_b &1 \\
x_c &y_c & 1
\end{vmatrix}$$
In the above figure, let the circumcircle passing through a triangle's vertices have radius $R$, and denote the central angles from the first point to the second $q$, and to the third point by $p$ then the area of the triangle is given by:
$$ s=2R^2|\sin(\frac{p}{2})\sin(\frac{q}{2})\sin(\frac{p-q}{2})|$$
|
If $W$ is the "hypotenuse-face" of a right-corner tetrahedron, and $X$, $Y$, $Z$ are the (right-triangular) "leg-faces", then
$$W^2 = X^2 + Y^2 + Z^2$$
where, yes, we are squaring areas. (This fact is actually equivalent to Heron's formula for non-obtuse triangles. You can extend it to include obtuse triangles by allowing the tetrahedron to have imaginary(!) edge-lengths at its right corner.)
More generally, if $W$, $X$, $Y$, $Z$ are the faces of a tetrahedron, and $\angle XY$ (etc) represents the dihedral angle between faces $X$ and $Y$, then we have a familiar-looking Law of Cosines:
$$W^2 = X^2 + Y^2 + Z^2 - 2 X Y \cos \angle XY - 2 Y Z \cos \angle YZ - 2 Z X \cos \angle ZX$$
(This is easily proven with vectors.) The above further implies another, more-familiar-looking Law:
$$\begin{align}
W^2 + X^2 - 2 WX \cos\angle WX \;&=\; Y^2 + Z^2 - 2 YZ \cos\angle YZ \\
W^2 + Y^2 - 2 WY \cos\angle WY \;&=\; Z^2 + X^2 - 2 ZX \cos\angle ZX \\
W^2 + Z^2 - 2 WZ \cos\angle WZ \;&=\; X^2 + Y^2 - 2 XY \cos\angle XY
\end{align}$$
(At the point where you say to yourself, "If there's any justice, each of these expressions should equal the square of the area of some face!", you will have inferred the existence of the tetrahedron's "pseudo-faces". But I digress ...)
|
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|
$ x^2 + \frac {x^2}{(x-1)^2} = 2010 $ I found this question from last year's maths competition in my country. I've tried any possible way to find it, but it is just way too hard.
Given $$ x^2 + \frac {x^2}{(x-1)^2} = 2010,$$
find $\dfrac {x^2} {x-1}.$
(A) $1+\sqrt {2011}$ (B) $ 1-\sqrt {2011}$ (C) $1\pm \sqrt{2011} $ (D) $\sqrt {2011}$
So I multiply them with $(x-1)$
$$x^2(x-1) + \frac {x^2} {x-1} = 2010(x-1)$$
$$\frac {x^2} {x-1} = (x-1)(2010-x^2)$$
and I stuck in here, dont know how to remove $x$ in there
|
A straightforward approach not requiring recognition of an identity runs:
Let $y=\frac {x^2}{x-1}$ so that $x^2=y(x-1)$, which means we can reduce $x^2$ wherever it occurs and see what happens. So $$x^2+\frac {x^2}{(x-1)^2}=y\left(x-1+\frac 1{x-1}\right)=y\cdot\frac {x^2-2x+2}{x-1}=y^2-2y$$And it is easy from there (solve the quadratic)
|
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|
Find $x$ if $\frac {1} {x} + \frac {1} {y+z} = \frac {1} {2}$ I found this question from past year's maths competition in my country. I've tried any possible way to find it, but it is just way too hard.
Find $x$ if \begin{align}\frac {1} {x} + \frac {1} {y+z} &= \frac {1} {2}\\ \frac {1} {y} + \frac {1} {x+z} &= \frac{1}{3}\\ \frac {1} {z} + \frac {1} {x+y} &= \frac {1} {4}\end{align}
$(A)\;\frac 32$
$(B)\;\frac {17}{10}$
$(C)\;\frac {19}{10}$
$(D)\;\frac {21}{10}$
$(E)\;\frac {23}{10}$
EDIT: I'm very sorry guys, it should be $\frac {1} {x+y}$ not $xy$, I'm sorry for the typos (idk what is wrong with me)
|
Let $x=1/p$, $y=1/q$, and $z=1/r$. The three equations become
$$\begin{align}
{1\over2}&=p+{qr\over q+r}={s\over q+r}\\
{1\over3}&=q+{rp\over r+p}={s\over r+p}\\
{1\over4}&=r+{pq\over p+q}={s\over p+q}\\
\end{align}$$
where
$$s=pq+qr+rp$$
Thus
$$\begin{align}
2s&=q+r\\
3s&=r+p\\
4s&=p+q\\
\end{align}$$
From this it follows that
$$(q+r)+(p+q)=2(r+p)\implies2q=r+p=3s\implies q={3\over2}s$$
and the others follow:
$$r=2s-q={1\over2}s\quad\text{and}\quad p=4s-q={5\over2}s$$
But now we have
$$s=pq+qr+rp={15\over4}s^2+{3\over4}s^2+{5\over4}s^2={23\over4}s^2$$
hence $s=4/23$, so $p=10/23$, and thus $x=23/10$.
|
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|
Factorisation over $\Bbb C$ of $z^2 -10z+30$ I haven't done these questions in a long time, so I am just wondering if my approach and answer is correct.
When asked to $z^2-10z+30$ over $\Bbb C$,
My approach: I complete the square of the equation, and would get $(z-5+\sqrt 5)(z-5-\sqrt 5)$
My question/concern: I think that could be the answer, but I'm not sure if you have to include the "$i$" symbol to show that it's an imaginary factor.
If incorrect, please correct me.
Thanks ahead of time.
|
The "$i$" should be included, thusly:
$z^2 - 10z + 30 = (z - 5 + i\sqrt{5})(z - 5 - i\sqrt{5}) ; \tag{1}$
without it, we have
$(z - 5 + \sqrt{5})(z - 5 - \sqrt{5}) = ((z - 5) + \sqrt{5})((z - 5) -\sqrt{5})$
$= (z - 5)^2 - (\sqrt {5})^2 = z^2 - 10z + 25 - 5$
$= z^2 - 10z + 20; \tag{2}$
on the other hand, with it:
$(z - 5 + i\sqrt{5})(z - 5 - i\sqrt{5}) = ((z - 5)+ i\sqrt{5})((z - 5) - i\sqrt{5})$
$= (z - 5)^2 - (i\sqrt{5})^2 = z^2 - 10z + 25 + 5$
$= z^2 - 10 z + 30, \tag{3}$
since $(i\sqrt{5})^2 = - 5$; in each of the above calculations we used the identity
$(a + b)(a - b) = a^2 - b^2; a, b \in \Bbb C. \tag{4}$
To complete the square for $z^2 - 10z + 30$, we proceed as follows:
$z^2 - 10z + 30 = 0;\tag{5}$
subtract $30$:
$z^2 - 10z = -30; \tag{6}$
add $25 = (-10/2)^2$:
$z^2 - 10z + 25 = -5; \tag{7}$
then
$(z - 5)^2 = z^2 -10z + 25 = -5 = (i\sqrt{5})^2, \tag{8}$
so
$(z - 5)^2 - (i\sqrt{5})^2 = 0, \tag{9}$
whence, via (4),
$((z - 5) + i\sqrt{5})((z - 5) - i\sqrt{5}) = 0; \tag{10}$
this shows the roots of $z^2 -10z + 30$ are $z - 5 \pm i\sqrt{5}$, so the factorization (1) binds.
|
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|
Simultaneous equations, $\frac{1}{x}+\frac{1}{y}=1$,$x+y=a$,$\frac{y}{x}=m$ By eliminating $x$ and $y$ from the following equations, I need to find the relation between $m$ and $a$.
\begin{align*}
\frac{1}{x}+\frac{1}{y}=1 \\
x+y=a \\
\frac{y}{x}=m
\end{align*}
I tried different ways, but cannot arrive at the answer. I end up with a quadratic.
This is what I have.
\begin{align*}
\frac{1}{x}=1-\frac{1}{y} \\
x(y-1)=y \\
x=\frac{y}{y-1}
\end{align*}
Now using second equation:
\begin{align*}
\frac{y}{y-1}+\frac{y(y-1)}{y-1}=a \\
\frac{y^2}{y-1}=a \\
y^2-ay+a=0
\end{align*}
Now roots are: $y_{1,2}=\displaystyle{\frac{a\pm\sqrt{a^2-4a}}{2}}$
Then it follows that: $x_{1,2}=\displaystyle{\frac{a\pm\sqrt{a^2-4a}}{2}}$
Hence there are three scenarios for the relation between $a$ and $m$.
First: $\frac{4a}{4}=m$, $a=m$
Second: $\displaystyle{\frac{(a+\sqrt{a^2-4a})^2}{4}=m}$
Third: $\displaystyle{\frac{(a-\sqrt{a^2-4a})^2}{4}=m}$
Is this correct? Thank you
|
No, since for each value of $y$ there is at most one $x$ satisfying $x+y=a$. You can now reexamine your solution.
Here is a different approach:
From the last two equations, we can derive that $x=\frac{a}{1+m}$ and $y=\frac{ma}{1+m}$. Substitute into the first equation, and we have $(1+m)^2=ma$. It is clear $m\ne0$. Hence, $$a=\frac{(1+m)^2}{m}$$ is what you want.
|
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|
$f(x) =ax^6 +bx^5+cx^4+dx^3+ex^2+gx+h $, find $f(7)$. Problem :
Given a polynomial $f(x) =ax^6 +bx^5+cx^4+dx^3+ex^2+gx+h$
such that $f(1)= 1, f(2) =2 , f(3) = 3, f(4) =4, f(5)=5, f(6) =6$.
Find $f(7)$ in terms of $h$.
My approach:
We can put the values of $f(1) = 1$ in the given equation and $f(2) = 2$, etc. But this is quite time consuming by making six different equations and then solve them to get the values of $a,b,c,d,e,g,h$. Please suggest some alternate solution for this.
|
\begin{align}
f(x) &=ax^6 +bx^5+cx^4+dx^3+ex^2+fx+3
\end{align}
Given that :
$f(1)= 1$, $f(2) =2$ , $f(3) = 3$, $f(4) =4$, $f(5)=5$, $f(6) =6$,
find $f(7)$.
It is also useful to add that $f(0)=3$.
Application of successive differences
lowers the degree of the polynomial on every step
down to $0$, and this useful formula shows up:
\begin{align}
\sum_{k=0}^7 (-1)^k\binom{7}{k}f(x-k)
&=0
\end{align}
\begin{align}
f(7)&=
-\sum_{k=1}^7 (-1)^k\binom{7}{k}f(7-k)
\\
&=
7 f(6)-21 f(5)+35 f(4)-35 f(3)+21 f(2)-7 f(1)+f(0)
\\
&=
7\cdot 6-21 \cdot 5+35 \cdot 4-35 \cdot 3 +21 \cdot 2 -7\cdot 1+3
\\
f(7)&=10.
\end{align}
|
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|
Proving $\sum_{i=1}^n\frac{1}{i(i+1)(i+2)}=\frac{n(n+3)}{4(n+1)(n+2)}$ for $n\geq 1$ by mathematical induction Prove using mathematical induction that
$$\frac{1}{1\cdot 2\cdot 3} + \frac{1}{2\cdot 3\cdot 4} + \cdots + \frac{1}{n(n+1)(n+2)} = \frac{n(n+3)}{4(n+1)(n+2)}.$$
I tried taking $n=k$,
so it makes
$$\frac{1}{1\cdot 2\cdot 3} + \frac{1}{2\cdot 3\cdot 4} + \cdots + \frac{1}{n(n+1)(n+2)} = \frac{k(k+3)}{4(k+1)(k+2)}.$$
Then proving the statement for $n=k+1$:
$$\frac{k(k+3)}{4(k+1)(k+2)}+ \frac{1}{(k+1)(k+2)(k+3)} = \frac{(k+1)(k+1+3)}{4(k+1+1)(k+1+2) }.$$
|
Given that $$\frac{1}{1\times 2\times 3}+\frac{1}{2\times 3\times 4}+\frac{1}{3\times 4\times 5}+\ldots +\frac{1}{n(n+1)(n+2)}=\frac{n(n+3)}{4(n+1)(n+2)}$$
1. Substituting $\color{blue}{n=1}$ in the given equality, we get $$\frac{1}{1\times 2\times 3}=\frac{(1)(1+3)}{4(1+1)(1+2)}$$ $$\implies \frac{1}{6}=\frac{4}{24}\implies \frac{1}{6}=\frac{1}{6}$$ Hence, it holds for $n=1$
*Assuming that it holds for $\color{blue}{n=k}$ then we have $$\color{blue}{\frac{1}{1\times 2\times 3}+\frac{1}{2\times 3\times 4}+\frac{1}{3\times 4\times 5}+\ldots +\frac{1}{k(k+1)(k+2)}=\frac{k(k+3)}{4(k+1)(k+2)}}$$
*Now, substituting $\color{blue}{n=k+1}$ in the given equality, we have $$\frac{1}{1\times 2\times 3}+\frac{1}{2\times 3\times 4}+\frac{1}{3\times 4\times 5}+\ldots +\frac{1}{k(k+1)(k+2)}+\frac{1}{(k+1)(k+2)(k+3)}=\frac{(k+1)(k+4)}{4(k+2)(k+3)}$$
$$\implies \frac{1}{1\times 2\times 3}+\frac{1}{2\times 3\times 4}+\frac{1}{3\times 4\times 5}+\ldots +\frac{1}{k(k+1)(k+2)}=\frac{(k+1)(k+4)}{4(k+2)(k+3)}-\frac{1}{(k+1)(k+2)(k+3)}$$ $$=\frac{1}{(k+2)(k+3)}\left(\frac{(k+1)^2(k+4)-4}{4(k+1)}\right)$$ $$=\frac{k^3+2k^2+k+4k^2+8k+4-4}{4(k+1)(k+2)(k+3)}$$ $$=\frac{k(k^2+6k+9)}{4(k+1)(k+2)(k+3)}$$ $$=\frac{k(k+3)^2}{4(k+1)(k+2)(k+3)}$$ $$=\frac{k(k+3)}{4(k+1)(k+2)}$$ Which is true from (2) thus it also holds for $n=k+1$
Hence, it is clear from (1), (2) & (3), the given equality holds for all $\color{blue}{n\geq 1}$
|
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|
Infinite limit of trigonometric series The value of $\displaystyle\lim_{n\to\infty}(\sin^4x+\frac{1}{4}\sin^4(2x)+\cdots+\frac{1}{4^n}\sin^4(2^nx))$ is
(A) $\sin^4x$
(B) $\sin^2x$
(C) $\cos^2x$
(D) does not exist
My attempt:
$$\lim_{n\to\infty}(\sin^4x+\frac{1}{4}\sin^4(2x)+\cdots+\frac{1}{4^n}\sin^4(2^nx))=$$
$$=(\sin^4x+\frac{1}{4}16\sin^4x\cos^4 x+\cdots+\frac{1}{4^n}\sin^4(2^{n-1}x)\cos^4(2^{n-1}x)$$
i could not solve further.Any hint will be useful.
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$$\begin{align} \sin^2 (x) - \sin^4(x) &= \sin^2 (x) \cos^2(x) = \frac14 \sin^2(2x), \\
\frac14 \sin^2 (2x) - \frac14 \sin^4(2x) &= \sin^2 (2x) \cos^2(2x) = \frac{1}{4^2} \sin^2(4x),\\
&\cdots \\
\frac{1}{4^{n}} \sin^2 (2^{n}x) - \frac{1}{4^{n}} \sin^4(2^{n}x) &= \sin^2 (2^{n}x) \cos^2 (2^{n}x)= \frac{1}{4^{n+1}} \sin^2(2^{n+1} x).
\end{align}
$$
So , the limit
$$ \begin{align}
\lim_{n\to\infty}(\sin^4x+\frac{1}{4}\sin^4(2x)+\cdots+\frac{1}{4^n}\sin^4(2^nx)) &= \lim_{n\to\infty}\left( \sin^2 (x) - \frac{1}{4^{n+1}} \sin^2(2^{n+1} x) \right) \\
&= \sin^2 (x) .
\end{align}$$
If you do not know this. For your multiple choice question:
*
*Set $x = 0$. the limit (if exists) is $0$, so C is false.
*For all $n$, $|\frac{1}{4^n}\sin(2^n x)| \leqslant \frac{1}{4^n}$ , thus by dominated convergence theorem, the sequence uniformly converges to a continues function, say $g(x)$. D is false.
*Choose an $x_0 \in (0,\frac{\pi}{2})$ such that $\frac13 <\sin^4(2x_0) < 1$, then $\frac14 \sin^4(2x_0)$ is strictly superior to the sum of the absolute value of the rest of the terms (which exists since the series is absolutely convergent and is inferior to $\frac{1}{12}$ by a simple comparison). So $|g(x_0) - \sin^4(x_0)| \geqslant \frac14 \sin^4(2x_0) - \sum|\mbox{rest of the terms}| > 0$. So A is false.
So B is the only correct choice.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1375214",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
"answer_id": 2
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|
evaluate $\frac 1{1+\sqrt2+\sqrt3} + \frac 1{1-\sqrt2+\sqrt3} + \frac 1{1+\sqrt2-\sqrt3} + \frac 1{1-\sqrt2-\sqrt3}$
Evaluate $\frac 1{1+\sqrt2+\sqrt3} + \frac 1{1-\sqrt2+\sqrt3} + \frac 1{1+\sqrt2-\sqrt3} + \frac 1{1-\sqrt2-\sqrt3}$
How to evalute this equation without using calculator?
|
Rearranging, we have
$$\frac{1}{1+(a+b)}+\frac{1}{1-(a+b)}+\frac{1}{1-(a-b)}+\frac{1}{1+(a-b)},$$
giving
$$\frac{2}{1-(a+b)^2}+\frac{2}{1-(a-b)^2}.$$
Then,
$$\frac{4-2(a-b)^2-2(a+b)^2}{1-(a-b)^2-(a+b)^2+(a+b)^2(a-b)^2}.$$
Expanding,
$$\frac{4(1-a^2-b^2)}{1-2a^2-2b^2+(a^2-b^2)^2}.$$
In your case
$a=\sqrt{2}$ and $b=\sqrt{3}$, so that your sum is
$$\frac{4(1-2-3)}{1-4-6+1}=\frac{-16}{-8}=2.$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1375531",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 6,
"answer_id": 5
}
|
Summation of a logarithmic series for $\ln(2(r^2 - 1)/r^2)$ Given that
$$\sum_{r=2}^{n}\ln\frac{r^2-1}{r^2}=\ln\frac{n+1}{2n}$$
for $n >1$.
Express $$\sum_{r=32}^{62}{\ln\frac{2(r^2-1)}{r^2}}$$ as $$A\ln 2 + B\ln3 + C\ln7$$ where $A$, $B$, $C$ are positive integers that can be found.
The result is actually $\ln 2113929216$, which is $2^{25}$ multiplied by $3^2 \times 7 (=63)$, hence the $\ln 2$, $\ln3$ and $\ln 7$.
How would you show this without knowing the result?
The upper limit is $62$, which is one away from the $3^2 \times 7$ for the $63$, leaving $2^{25}$ for the first term, with $A$.
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You don't need to deal with huge numbers.$$\begin{align}\sum_{r=32}^{62}\ln\left(\frac{2(r^2-1)}{r^2}\right)&=\sum_{r=32}^{62}\left(\ln 2+\ln\left(\frac{r^2-1}{r^2}\right)\right)\\&=\sum_{r=32}^{62}\ln 2+\sum_{r=32}^{62}\ln\left(\frac{r^2-1}{r^2}\right)\\&=(62-32+1)\ln 2+\sum_{r=2}^{62}\ln\left(\frac{r^2-1}{r^2}\right)-\sum_{r=2}^{31}\ln\left(\frac{r^2-1}{r^2}\right)\\&=31\ln 2+\ln\frac{62+1}{2\cdot 62}-\ln\frac{31+1}{2\cdot 31}\\&=31\ln 2+\ln(3^2\cdot 7)-\ln(2^2\cdot 31)-\ln(2^5)+\ln(2\cdot 31)\\&=31\ln 2+2\ln 3+\ln 7-2\ln 2-\ln(31)-5\ln 2+\ln 2+\ln(31)\\&=25\ln 2+2\ln 3+\ln 7\end{align}$$
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{
"language": "en",
"url": "https://math.stackexchange.com/questions/1375798",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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|
Finite difference : relationship involving gamma Given the following PDE,
$$
\frac{\partial V}{\partial t} + \frac{1}{2}\sigma^2S^2\frac{\partial^2V}{\partial S^2}=0
$$
and its finite difference approximation,
$$
\frac{V_n^{m+1}-V_n^m}{\Delta t} + \frac{1}{2}\sigma^2S_n^2\Gamma_n^m=0 \qquad (*)
$$
where
$$
\Gamma_n^m=\frac{V_{n+1}^{m}-2V_n^m + V_{n-1}^m}{\Delta S^2}
$$
How do i show by considering suitable linear combination of (*) evaluated at mesh points $S_{n-1}$, $S_n$, $S_{n+1}$ that,
$$
\frac{\Gamma_n^{m+1}-\Gamma_n^m}{\Delta t} + \frac{1}{2}\sigma^2S_n^2\frac{\Gamma_{n+1}^m-2\Gamma_n^m+\Gamma_{n-1}^m}{\Delta S^2}+2\sigma^2S_n\frac{\Gamma_{n+1}^m-\Gamma_{n-1}^m}{2\Delta S}+\sigma^2\frac{\Gamma_{n+1}^m+\Gamma_{n-1}^m}{2}=0
$$
|
Basically we need to show that
$$
D (S_n^2 \Gamma_n) = S_n^2 D \Gamma_n + 4S_n \frac{\Gamma_{n+1} - \Gamma_{n-1}}{2\Delta S} + \Gamma_{n+1} + \Gamma_{n-1}
$$
where $D$ is second order finite difference derivative operator
$$
D u_n \equiv \frac{u_{n+1} -2 u_n+ u_{n-1}}{\Delta S^2}
$$
Note, that $D$ can be written as
$$
D = D_+ D_- = D_- D_+
$$
where
$$
D_+ u_n = \frac{u_{n+1} - u_n}{\Delta S}\\
D_- u_n = \frac{u_{n} - u_{n-1}}{\Delta S}\\
$$
are one-sided finite difference first order derivative operators. Really,
$$
D_+D_- u_n = D_+\left( \frac{u_{n} - u_{n-1}}{\Delta S}\right) =
\frac{u_{n+1} - u_{n}}{\Delta S^2} - \frac{u_{n} - u_{n-1}}{\Delta S^2}
= D u_n.
$$
Same for the inverse order.
Next,
$$
D_{\pm} (a_n b_n) = \pm\frac{a_{n\pm 1}b_{n \pm 1} -a_nb_n}{\Delta S}
= \pm\frac{a_{n\pm 1}b_{n \pm 1} -a_{n\pm 1}b_n+a_{n\pm 1}b_n-a_nb_n}{\Delta S} =
a_{n \pm 1} D_\pm b_n + b_n D_\pm a_n.
$$
Thus
$$D_\pm S_n^2 = S_{n \pm 1} D_\pm S_n + S_n D_\pm S_n = S_n + S_{n \pm 1}\\
D_\pm \Gamma_n S_n^2 = \Gamma_{n \pm 1} (S_n + S_{n\pm 1}) + S_n^2 D_\pm \Gamma_n.
$$
This far we found that
$$
D_+ \Gamma_n S_n^2 = \Gamma_{n + 1} (S_n + S_{n+1}) + S_n^2 D_+ \Gamma_n.
$$
Let's apply $D_-$ to that
$$
D_- D_+ \Gamma_n S_n^2 = D_-\Gamma_{n + 1} (S_n + S_{n+1})
+ D_- S_n^2 D_+ \Gamma_n.
$$
Starting with the $D_- S_n^2 D_+ \Gamma_n$ we have
$$
D_- S_n^2 D_+ \Gamma_n = S_{n-1}^2 D_-D_+ \Gamma_n + (S_n + S_{n-1})D_+\Gamma_n =
S_{n-1}^2 D \Gamma_n + (S_n + S_{n-1})D_+ \Gamma_n.
$$
Next
$$
D_-\Gamma_{n + 1} (S_n + S_{n+1}) = \Gamma_n D_-(S_n+S_{n+1}) + (S_n+S_{n+2}) D_-\Gamma_{n+1} =
2\Gamma_n + (S_n + S_{n+1}) D_+\Gamma_n.
$$
So far we have
$$
D S_n^2 \Gamma_n = 2\Gamma_n + (S_{n-1} + 2S_n + S_{n+1}) D_+ \Gamma_n + S_{n-1}^2 D\Gamma_n.
$$
Computing the same expression by reversing the $D_+D_-$ order, we have
$$
D S_n^2 \Gamma_n = 2\Gamma_n + (S_{n-1} + 2S_n + S_{n+1}) D_- \Gamma_n + S_{n+1}^2 D\Gamma_n.
$$
Also, since $S_{n\pm 1} = S_{n}\pm \Delta S$,
$$
D S_n^2 \Gamma_n = 2\Gamma_n + 4S_n D_+ \Gamma_n + (S_{n}-\Delta S)^2 D\Gamma_n\\
D S_n^2 \Gamma_n = 2\Gamma_n + 4S_n D_- \Gamma_n + (S_{n}+\Delta S)^2 D\Gamma_n.
$$
Halfsumming the equations finally we have
$$
D S_n^2 \Gamma_n = 2\Gamma_n + 4S_n \frac{\Gamma_{n+1} - \Gamma_{n-1}}{2\Delta S} + (S_n^2 + \Delta S^2)D\Gamma_n = \\ =
S_n^2 D\Gamma_n + 4S_n \frac{\Gamma_{n+1} - \Gamma_{n-1}}{2\Delta S} +
2 \Gamma_n + \Delta S^2\frac{\Gamma_{n-1} - 2\Gamma_n + \Gamma_{n+1}}{\Delta S^2}.
$$
Rest should be obvious now.
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"language": "en",
"url": "https://math.stackexchange.com/questions/1378578",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 0
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|
Subsets and Splits
Fractions in Questions and Answers
The query retrieves a sample of questions and answers containing the LaTeX fraction symbol, which provides basic filtering of mathematical content but limited analytical insight.