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Simple factor of equation I have this polynomial: $5z^4-12z^3+30z^2-12z+5$
How do I factor it to get the following?: $(5z^2-2z+1)(z^2-2z+5)$
Can someone show me the procedure to perform whenever I encounter with a case like this? Thank you.
|
You should exploit the symmetry: write the polynomial as
$$
z^2\left(5z^2+\frac{5}{z^2}-12z-\frac{12}{z}-30\right)
$$
and observe that
$$
z^2+\frac{1}{z^2}=\left(z+\frac{1}{z}\right)^{\!2}-2
$$
so you can rewrite the expression as
$$
z^2\left(5\left(z+\frac{1}{z}\right)^{\!2}-12\left(z+\frac{1}{z}\right)-40\right)
$$
The polynomial
$$
5t^2-12t-40
$$
has roots
$$
a=\frac{6+\sqrt{236}}{5},\quad b=\frac{6-\sqrt{236}}{5}
$$
so we get
$$
5z^2\left(z+\frac{1}{z}-a\right)\left(z+\frac{1}{z}-b\right)
$$
that can be rewritten as
$$
5(z^2-az+1)(z^2-bz+1)
$$
If your polynomial is
$$
5z^4-12z^3\color{red}{+}30z^2-12z+5
$$
the same procedure would give a polynomial in $t$ without real roots. In particular, there is no real root for the polynomial.
In this case you know that if $\alpha$ is a root, also $\alpha^{-1}$ is root. Since a real factorization exists, the roots must have modulus $1$ and if we pair the conjugate pairs, we get a factorization in the form
$$
5z^4-12z^3+30z^2-12z+5=
(5z^2+az+b)(bz^2+az+5)
$$
(try seeing why). Now it's quite easy to find $a$ and $b$.
Or you can try finding the complex roots. The procedure reduces to the polynomial
$$
5t^2-12t+20
$$
whose roots are
$$
\frac{6+8i}{5},\qquad \frac{6-8i}{5}
$$
so you just have to solve the equations
$$
z+\frac{1}{z}=\frac{6+8i}{5}
\qquad
z+\frac{1}{z}=\frac{6-8i}{5}
$$
|
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"language": "en",
"url": "https://math.stackexchange.com/questions/1712115",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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|
Proving that $\int_0^1 \frac{\log^2(x)\tanh^{-1}(x)}{1+x^2}dx=\beta(4)-\frac{\pi^2}{12}G$ I am trying to prove that
$$I=\int_0^1 \frac{\log^2(x)\tanh^{-1}(x)}{1+x^2}dx=\beta(4)-\frac{\pi^2}{12}G$$
where $\beta(s)$ is the Dirichlet Beta function and $G$ is the Catalan's constant. I managed to derive the following series involving polygamma functions but it doesn't seem to be of much help.
$$
\begin{align*}
I &=\frac{1}{64}\sum_{n=0}^\infty \frac{\psi_2 \left(\frac{n}{2}+1 \right) -\psi_2\left(\frac{n+1}{2} \right)}{2n+1} \\
&= \frac{1}{8}\sum_{n=1}^\infty \frac{\psi_2(n)}{2n-1}-\frac{1}{32}\sum_{n=1}^\infty\frac{\psi_2\left(\frac{n}{2}\right)}{2n-1}
\end{align*}
$$
Numerical calculations show that $I \approx 0.235593$.
|
The generalization of the main integral follows easily by employing the same ideas used in this post and the previous one.
Let $n$ be a natural number. Then, we have
$$\int_0^1 \frac{\log^{2n}(x)\operatorname{arctanh}(x)}{1+x^2}\textrm{d}x$$
$$=\lim_{s\to0}\frac{d^{2n}}{ds^{2n}}\left(\frac{\pi}{16}\cot \left(\frac{\pi s}{2}\right) \left(\psi \left(\frac{3}{4}-\frac{s}{4}\right)-\psi\left(\frac{1}{4}-\frac{s}{4}\right)\right)-\frac{\pi ^2 }{16} \csc \left(\frac{\pi s}{2}\right)\right),$$
where $\psi$ represents the Digamma function.
Another similar generalization
Let $n$ be a natural number. Then, we get
$$\int_0^1 \frac{\log^{2n}(x)\arctan(x)}{1-x^2}\textrm{d}x$$
$$=\frac{\pi}{4} \left(1-2^{-2 n-1}\right) \zeta (2 n+1)(2 n)!$$
$$-\lim_{s\to0}\frac{d^{2n}}{ds^{2n}}\left(\frac{\pi}{16} \csc \left(\frac{\pi s}{2}\right) \left(\pi \cos \left(\frac{\pi s}{2}\right)+\psi\left(\frac{s+1}{4}\right)-\psi\left(\frac{s+3}{4}\right)\right)\right),$$
where $\zeta$ represents the Riemann zeta function and $\psi$ denotes the Digamma function.
A solution in large steps by Cornel I. Valean to the main integral
$$\int_0^1 \frac{\log^2(x)\operatorname{arctanh}(x)}{1+x^2}dx$$
We follow the strategy used to the auxiliary result from the previous post, and then we immediately arrive at
$$\int_0^1 \frac{\log^2(x)\operatorname{arctanh}(x)}{1+x^2}dx=\frac{1}{2}\Re\biggr\{ \int_0^{\infty } \frac{\log ^2(x) \operatorname{arctanh}(x)}{1+x^2} \textrm{d}x\biggr \}$$
$$=\frac{1}{2} \int_0^{\infty }\left(PV\int_0^1 \frac{x \log ^2(x)}{(1-y^2 x^2)(1+x^2)} \textrm{d}y\right)\textrm{d}x$$
$$=\frac{1}{2}\int_0^1\left(PV\int_0^{\infty} \frac{x \log ^2(x)}{(1-y^2 x^2)(1+x^2)} \textrm{d}x\right)\textrm{d}y$$
$$=\frac{\pi^2}{12}\int_0^1 \frac{\log(y)}{1+y^2}\textrm{d}y-\frac{1}{6}\int_0^1 \frac{\log^3(y)}{1+y^2}\textrm{d}y=\beta(4)-\frac{\pi^2}{12}G,$$
as desired.
End of story.
A note: Using the Cauchy product $\displaystyle \frac{\operatorname{arctanh}(x)}{1+x^2}=\sum _{n=1}^{\infty } \sum _{k=1}^n \frac{(-1)^{n+k} x^{2 n-1}}{2 k-1}$, an the value of the main integral, we immediately obtain the beautiful series
$$\sum _{n=1}^{\infty }\frac{(-1)^{n-1}}{n^3} \sum _{k=1}^n \frac{(-1)^{k-1}}{2 k-1}=4\beta(4)-\frac{\pi^2}{3}G.$$
Some kind of bonus: Using the integrals relation obained with integration by parts as shown in Shobhit Bhatnagar's post and combining it with the results obtained in this post and the previous one, we obtain the value of the other integral,
$$\int_0^1\frac{\log^2(x)\arctan(x)}{1-x^2}\textrm{d}x= -\beta(4)-\frac{\pi^2}{24}G+\frac{7\pi}{16}\zeta(3).$$
A note: It's clear the generalization $\displaystyle \int_0^1 \frac{\log^{2n}(x)\arctan(x)}{1-x^2}\textrm{d}x$ may be approached in the same way as $\displaystyle \int_0^1 \frac{\log^2(x)\operatorname{arctanh}(x)}{1+x^2}dx$.
|
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"language": "en",
"url": "https://math.stackexchange.com/questions/1712715",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "18",
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|
Prove: $2^k$ is the sum of two perfect squares If $k$ is a nonnegative integer, prove that $2^k$ can be represented as a sum of two perfect squares in exactly one way. (For example, the unique representation of $10$ is $3^2+1^2$; we do not count $1^2+3^2$ as different.)
I understand that $2^{2n}=0+2^{2n}$ and $2^{2n+1}=2^{2n}+2^{2n}$. But how can we prove that $2^k$ can be represented as two perfect squares in exactly one way?
|
Suppose
$2^{2n+1}$
is the smallest odd power of two
that can represented in
more than one way
as the sum of two squares.
Then
$2^{2n+1}
=(2^{n}+a)^2+(2^{n}-b)^2
$
where
$1 \le a < 2^n$
and
$1 \le b < 2^n$.
Then
$\begin{array}\\
2^{2n+1}
&=(2^{n}+a)^2+(2^{n}-b)^2
\qquad(0)\\
&=2^{2n}+a2^{n+1}+a^2+2^{2n}-b2^{n+1}+b^2\\
\text{or}\\
(b-a)2^{n+1}
&=a^2+b^2
\qquad(1)\\
\text{or}\\
b(2^{n+1}-b)
&=a(2^{n+1}+a)
\qquad(2)\\
\end{array}
$
From $(0)$,
$a$ and $b$ must be even,
otherwise the right side
is $\equiv 2 \bmod 4$.
Suppose
$2^k || a$,
so that
$2^k | a$
and
$2^{k+1}\not\mid a$.
Then
$2^{2k}||a(2^{n+1}+a)
$
so
$2^{2k}||b(2^{n+1}-b)
$.
If $2^j || b$,
reasoning as for $a$,
$2^{2j} ||b(2^{n+1}-b)
$.
Therefore
$j=k$,
so that
$a=2^kc$
and
$b = 2^kd$
where $c$ and $d$ are odd.
Substituting in $(2)$,
$2^kd(2^{n+1}-2^kd)
=2^kc(2^{n+1}+2^kc)
$
or
$d(2^{n+1-k}-d)
=c(2^{n+1-k}+c)
$
or
$0
=2c2^{n-k}+c^2-2d2^{n-k}+d^2
$
or
$\begin{array}\\
2^{2(n-k)+1}
&=2^{2(n-k)}+2c2^{n-k}+c^2+2^{2(n-k)}-2d2^{n-k}+d^2\\
&=(2^{(n-k)}+c)^2+(2^{(n-k)}-d)^2\\
\end{array}
$
Therefore,
a smaller odd power of $2$
can be represented
in more than one way
as the sum of two squares.
But this contradicts $2^{2n+1}$
being the smallest such odd power.
Therefore
$2^{2n+1}$
can only be represented
in one way
as the sum of two squares.
This feels good!
This is the first time
I have independently constructed
a proof by infinite descent.
Maybe I'll try
even powers,
but this is enough for now.
|
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"url": "https://math.stackexchange.com/questions/1713733",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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|
When is matrix $A$ diagonalizable? I got the following matrix:
$$ A =
\begin{pmatrix}
a & 0 & 0 \\
b & 0 & 0 \\
1 & 2 & 1 \\
\end{pmatrix}
$$
I need to answer when this matrix is diagonalizable.
Its characteristic polynomial is $ t(t-a)(t-1) $. So its 3 eigenvalues are 0, 1 and a. Both the algebraic and geometry multiplicities of those values are 1 (for all of them).
Let's look at the matrices for those eigenvalues:
$$ A - 0I =
\begin{pmatrix}
a & 0 & 0 \\
b & 0 & 0 \\
1 & 2 & 1 \\
\end{pmatrix}
$$
$$ A - I =
\begin{pmatrix}
a -1 & 0 & 0 \\
b & -1 & 0 \\
1 & 2 & 0 \\
\end{pmatrix}
$$
$$ A - aI =
\begin{pmatrix}
0 & 0 & 0 \\
b & -a & 0 \\
1 & 2 & 1 - a \\
\end{pmatrix}
$$
$\rho (A - 0I) = 2 $
$\rho (A - 1I) = 2 $
$\rho (A - aI) = 2 $
It seems that for every $a$ and $b$ this matrix would be diagonalizable.
But it's not. Where am I wrong?
|
1) If $a\ne 0, 1,\;$ then A is diagonalizable since it has 3 distinct eigenvalues.
2) If $a=0$, then A is diagonalizable $\iff$ $\text{nullity}(A-0I)=\text{nullity}(A)=2 \iff \text{rank}(A)=1$
$\hspace{2.3 in}\iff\text{rank}\begin{pmatrix} 0&0&0\\b&0&0\\1&2&1\end{pmatrix}=1\iff b=0$
3) If $a=1$, then A is diagonalizable $\iff$ $\text{nullity}(A-1I)=\text{nullity}(A-I)=2 \iff \text{rank}(A-I)=1$
$\hspace{2.3 in}\iff\text{rank}\begin{pmatrix} 0&0&0\\b&-1&0\\1&2&0\end{pmatrix}=1\iff b=-\frac{1}{2}$
|
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"language": "en",
"url": "https://math.stackexchange.com/questions/1714473",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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|
Three questions about the form $X^2 \pm 3Y^2 = Z^3$ and a related lemma In Ribenboim’s Fermat’s Last Theorem for Amateurs, he gives the following lemma [Lemma 4.7, pp. 30–31].
Lemma. Let $E$ be the set of all triples $(u, v, s)$ such that $s$ is odd, $\gcd(u,v) = 1$ and $s^3 = u^2 + 3v^2$. Let $F$ be the set of all pairs $(t,w)$ where $\gcd(t, w) = 1$ and $t \not\equiv w\!\pmod{2}$. The mapping $\Phi : F \rightarrow E$ given by $\Phi (t, w) = (u, v, s)$ with
\begin{cases}
\, u = t(t^2−9w^2), \\
\, v = 3w(t^2-w^2),\\
\, s = t^2 + 3w^2,
\end{cases}
is onto $E$.
Q1: Am I understanding “onto” correctly by interpreting this to mean that this is a complete integer parameterization of the form $X^2 + 3Y^2 = Z^3$?
Q2: Is there a similar solution for the form $X^2-3Y^2 = Z^3$? It appears that
\begin{cases}
\, x = t(t^2+9w^2), \\
\, y = 3w(t^2+w^2),\\
\, z = t^2 - 3w^2,
\end{cases}
satisfies, but I want to be sure it’s complete.
Q3: Is this generalizable to the form $X^2 \pm kY^2 = Z^3$ for any [possibly non-square or squarefree] integer $k$?
|
The formula quoted in Ribenboim has the general form,
$$u^2+dv^2 = (p^3 - 3 d p q^2)^2 + d(3 p^2 q - d q^3)^2 = (p^2+dq^2)^3\tag1$$
Assume $\gcd(u,v)=1$. For $d=3$, apparently it is integrally complete.
But we cannot necessarily extend the same conclusion to general $d$. For example, for $d=11$, the formula above becomes,
$$(p^3 - 33 p q^2)^2 + 11(3 p^2 q - 11 q^3)^2 = (\color{blue}{p^2+11q^2})^3\tag2$$
The rational substitution $p,\,q = \frac{x+4y}{2},\;\frac{x}{2}$ transforms $(2)$ to,
$$(-4 x^3 - 15 x^2 y + 6 x y^2 + 8 y^3)^2 + 11(-x^3 + 3 x^2 y + 6 x y^2)^2 = (\color{blue}{3 x^2 + 2 x y + 4 y^2})^3\tag3$$
As W. Jagy pointed out, for $d=11$, the forms $x^2+11y^2,\;3 x^2 + 2 x y + 4 y^2$ are the desired sums. But for $p$ to be an integer, then $x$ has to be even which results in $\gcd(u,v)\neq1$. For $d=11$, then $(1)$ is only rationally complete.
For other $d$, say $d=47$, Pepin gave,
$$(13p^3 + 30p^2q - 42p q^2 - 18q^3)^2 + 47(p^3 - 6p^2q - 6p q^2 + 2q^3)^2 =
2^3(3p^2 + p q + 4q^2)^3\tag4$$
But now there is no rational substitution to transform $(1)$ to $(4)$. Thus, for certain $d$, then $(1)$ is not even complete rationally. (See also related posts: this and this.)
$\color{green}{Update:}$
As requested by OP, for $d=-6$, then formula $(1)$ yields,
$$(p^3 + 18 p q^2)^2 - 6(-3 p^2 q - 6 q^3)^2 = (p^2 - 6 q^2)^3\tag5$$
while this other formula, using initial $a,b,c = 5,\,2,\,1$ yields,
$$(5 r^3 - 36 r^2 s + 90 r s^2 - 72 s^3)^2 -
6(2 r^3 - 15 r^2 s + 36 r s^2 - 30 s^3)^2 = (r^2-6 s^2)^3\tag6$$
and there is no rational transformation between $(5)$ and $(6)$. Thus, for $d=-6$, then $(1)$ is not rationally complete also.
|
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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|
Integer solutions for $n$ for $|{\sqrt{n} - \sqrt{2011}}| < 1$ $$|{\sqrt{n} - \sqrt{2011}}| < 1$$
What is the number of positive integer $n$ values, which satisfy the above inequality.
My effort:
$
({\sqrt{n} - \sqrt{2011}})^2 < 1 \\n + 2011 -2\sqrt{2011n} < 1\\ n+2010<2\sqrt{2011n}\\ n^2+2 \times 2010 \times n +2010^2<4 \times 2011n \\n^2 -4024n +2010^2 < 0 $
But it seems this won't lead me for desired answer.
|
$\sqrt{n}\in(\sqrt{2011}-1,\sqrt{2011}+1)\iff n\in ( (\sqrt{2011}-1)^2,(\sqrt{2011}+1)^2)=(2012-2\sqrt{2011},2012+2\sqrt{2011})$
There are $2\lfloor2\sqrt{2011} \rfloor + 1$ integers in this range.
We approximate $\sqrt{2011}$ as $44.8$ and we find the answer is $2\cdot 89 + 1=179$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1715920",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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|
Prove the function Prove
$\forall n\in\mathbb{N}: n\ge 1 \rightarrow 2^n\le 2^{n+1}-2^{n-1}-1.$
I did that $2^n \le\ 2^{n+2} - 2^{n}$ and then $2^n < 2^{n+2} - 2^{n-1}$
but have no idea how to add $-1$ in the function and let the $2^n \le\ 2^{n+2} - 2^{n} - 1$
|
$2^{n+1}-2^{n-1}-1=2^{n-1}*(4-1)-1=2^{n-1}*3 -1 \ge 2^{n-1}*3-2^{n-1}=2^{n-1}*2=2^n $
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1716073",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 3
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|
If $abcd=1$,where $a,b,c,d$ are positive reals,then find the minimum value of $a^2+b^2+c^2+d^2+ab+ac+ad+bc+bd+cd$. If $abcd=1$,where $a,b,c,d$ are positive reals,then find the minimum value of $a^2+b^2+c^2+d^2+ab+ac+ad+bc+bd+cd$.
Let $E=a^2+b^2+c^2+d^2+ab+ac+ad+bc+bd+cd=(a+b+c+d)^2-(ab+ac+ad+bc+bd+cd)$
I do not know how to use $abcd=1$ in this question to get the minimum value of the expression.
|
with AM-GM, we get
a²+b²+c²+d² ≥ 4√((abcd)²) = 4
ab+ac+ad+bc+bd+cd ≥ 6√((abcd)³) = 6
minimum value of a²+b²+c²+d²+ab+ac+ad+bc+bd+cd is 4+6 = 10 when a=b=c=d=1.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1719036",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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|
The only natural number $x$ for which $x+\sqrt{-2}$ is a cube in $\mathbb{Z}[\sqrt{-2}]$ is $x=5$ Let $A = \mathbb{Z}[\sqrt{-2}]= \{a+b\sqrt{-2} \ : a, b \in \mathbb{Z}\}$. Show that the only natural number $x$ for which $x+\sqrt{-2}$ is a cube in $A$ is $x=5$.
So I have to show that there exists $c,d \in \mathbb{Z}$ such that $(x+\sqrt{-2})^3= c + d \sqrt{-2}$. From the last equation, I obtain $x(x^2-6) = c \in \mathbb{Z}$ and $3x^2-2= b \in \mathbb{Z}$, and here I blocked. Could I conclude from both equations? If not, is there exists another way to proceed (hint)?
Thanks!
|
Directly
$$(a+b\sqrt{-2})^2(a+b\sqrt{-2})=(a^2-2b^2+2ab\sqrt{-2})(a+b\sqrt{-2})=$$
$$=(a^3-6ab^2)+(3a^2b-2b^2)\sqrt{-2}$$
We get that it must be
$$3a^2b-2b^2=1\iff b(3a^2-2b)=1\iff b,\,3a^2-2b=\pm 1$$
But $b=-1\implies 3a^2-2(-1)=-1\implies3a^2=-3$ , impossible, so it must
$$\;b=1\implies3a^2-2=1\iff 3a^2=3\implies a^2=1\iff a=\pm1\;$$
and thus we want $\;\pm1+\sqrt{-2}\;$ , so that for example
$$(1+\sqrt{-2})^3=1+3\sqrt{-2}-6-2\sqrt{-2}=-5+\sqrt{-2}$$
But $\;-5\notin\Bbb N\;$ , so let us try the other option with $\;a=-1\,,\,\,b=1\;$ :
$$(-1+\sqrt{-2})^3=-1+3\sqrt{-2}+6-2\sqrt{-2}=5+\sqrt{-2}$$
and this time $\;x=5\in\Bbb N\;$ .
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1719328",
"timestamp": "2023-03-29T00:00:00",
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|
Converging sequence $a_{{n+1}}=6\, \left( a_{{n}}+1 \right) ^{-1}$ I know the sequence is converging. But I find it difficult proving it, by induction. So far I have drawn a diagram and calculate the five first numbers. From the diagram I can se that the sequence can be split into two sequences, one that is increasing and one tha is decreasing. I need to show that
I need to show that
$a_{{1}}=1$,
$a_{{3}}=3/2$ and
$a_{{5}}={\frac {30}{17}}$ is an increasing sequence and that
$a_{{2}}=3$,
$a_{{4}}={\frac {12}{5}}$ is decreasing.
Any help would be appriciated.
$a_{{1}}=1, a_{{n+1}}=6\, \left( a_{{n}}+1 \right) ^{-1},1\leq n$
*
*$a_{{1}}=1$
*$a_{{2}}=3$
*$a_{{3}}=3/2$
*$a_{{4}}={\frac {12}{5}}$
*$a_{{5}}={\frac {30}{17}}$
|
(At the end,
I have added my form
of the explicit solution.)
$a_{1}=1, a_{n+1}=\dfrac{6}{a_{{n}}+1},1\leq n
$
If it has a limit $L$,
then
$L = \dfrac{6}{L+1}$
or
$L^2+L-6 = 0
$
or
$(L+3)(L-2) = 0
$.
Since
$a_n > 0$
for all $n$,
the only possibility
is $L=2$.
To get a sequence
whose terms should
go to $0$,
let
$a_n = b_n+2
$.
Then
$b_{n+1}+2=\dfrac{6}{b_{{n}}+3}
$
or
$b_{n+1}
=\dfrac{6-2(b_{n}+3)}{b_{n}+3}
=\dfrac{-2b_{n}}{b_{n}+3}
$.
Since
$b_1 = a_1-2
= -1
$,
$b_2 = \dfrac{2}{2}
= 1
$,
$b_3 = \dfrac{-2}{4}
= \dfrac12
$,
$b_4 = \dfrac{-1}{7/2}
= -\dfrac{2}{7}
$.
Since
$\left(-\dfrac{2 x}{x+3}\right)' = -\dfrac{6}{(x+3)^2}
$,
if
$-\dfrac12 \le b_n
\le \dfrac12$,
then
$b_{n+1}
\le \dfrac{1}{5/2}
=\dfrac25
\lt \dfrac12
$
and
$b_{n+1}
\ge \dfrac{-1}{7/2}
=-\dfrac27
\gt -\dfrac12
$.
Therefore
$|b_n|
\le \dfrac12
$
for $n \ge 3$.
Therefore,
for
$n \ge 3$,
$\begin{array}\\
\left|\dfrac{b_{n+1}}{b_n}\right|
&=\left|\dfrac{2}{b_n+3}\right|\\
&\le \dfrac{2}{5/2}\\
&= \dfrac45\\
&< 1\\
\end{array}
$
so that
$b_n
\to 0
$
as $n \to \infty$.
Therefore
$a_n
\to 2
$
as $n \to \infty$.
(added later)
I can show that
$b_m
= \frac{15}{8(-\frac32)^m-3}
$
so that
$b_{2m}
=\frac{15}{8(\frac94)^{m}-3}
$
is decreasing
and
$b_{2m+1}
=\frac{-5}{4(\frac94)^{m}+1}
$
is increasing,
both approaching $0$
as their limit.
|
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|
How to prove $1+\cos \left(\frac{2\pi}{7}\right)-4\cos^2 \left(\frac{2\pi}{7}\right)-8\cos^3 \left(\frac{2\pi}{7}\right) \neq 0$ The task is to prove the following non-equality by hand:
$$1+\cos \left(\frac{2\pi}{7}\right)-4\cos^2 \left(\frac{2\pi}{7}\right)-8\cos^3 \left(\frac{2\pi}{7}\right) \neq 0$$
Wolframalpha shows this, but I can't prove it.
http://www.wolframalpha.com/input/?i=1%2Bcos(2pi%2F7)-4cos%5E2(2pi%2F7)-8cos%5E3(2pi%2F7)%3D0
|
Putting $x=\cos(\frac{2\pi}{7})$ you have the polynomial $1+x-4x^2-8x^3\not= 0$ then $8x^3+4x^2-x-1\not=0$ from that
$$8x^3+4x^2-x-1=4x^2(2x+1)-x-1-x+x=4x^2(2x+1)-(2x+1)+x=(2x+1)(4x^2-1)+x=(2x+1)^2(2x-1)+x$$
Since $(2x+1)^2>0$ and $x>\frac{1}{2}$ since $\frac{2\pi}{7}<\frac{\pi}{3}$ we have that $(2x-1)>0$ and a sum of positive numbers isn't zero.
|
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|
Solving the following trigonometric equation: $\sin x + \cos x = \frac{1}{3} $ I have to solve the following equation:
$$\sin x + \cos x = \dfrac{1}{3} $$
I use the following substitution:
$$\sin^2 x + \cos^2 x = 1 \longrightarrow \sin x = \sqrt{1-\cos^2 x}$$
And by operating, I obtain:
$$ \sqrt{(1-\cos^2 x)} = \dfrac{1}{3}-\cos x$$
$$ 1 - \cos^2 x = \dfrac{1}{9} + \cos^2 x - \dfrac{2}{3}\cos x$$
$$ -2\cos^2 x + 2/3\cos x +\dfrac{8}{9}=0$$
$$ \boxed{\cos^2 x -\dfrac{1}{3}\cos x -\dfrac{4}{9} = 0}$$
Can I just substitute $\cos x$ by $z$ and solve as if it was a simple second degree equation and then obtain $x$ by taking the inverse cosine? I have tried to do this but I cannot get the right result. If I do this, I obtain the following results:
$$ z_1 = -0.520517 \longrightarrow x_1 = 121.4º\\
z_2= 0.8538509 \longrightarrow x_2 = 31.37º$$
I obtain $x$ from $z$ by taking the inverse cosine.
The correct result should be around 329º which corresponds to 4.165 rad. My question is if what I am doing is wrong because I have tried multiple times and I obtain the same result (or in the worst case, I have done the same mistake multiple times).
|
we have after squaring $$\sin(x)^2+\cos(x)^2+\sin(2x)=\frac{1}{9}$$ or $$\sin(2x)=-\frac{8}{9}$$
the answer is $$c_1\in \mathbb{Z}\land \left(x=2 \pi c_1+2 \tan ^{-1}\left(\frac{1}{10}
\left(9-\sqrt{161}\right)\right)\lor x=2 \pi c_1+2 \tan
^{-1}\left(\frac{1}{10} \left(9+\sqrt{161}\right)\right)\right)$$
|
{
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|
integrate $\int \frac{\tan^4x}{4}\cos^3x$
$$\int \frac{\tan^4x}{4}\cos^3x$$
$$\int \frac{\tan^4x}{4}\cos^3x=\frac{1}{4}\int \frac{\sin^4x}{\cos^4x}\cos^3x=\frac{1}{4}\int\frac{\sin^4x}{\cos x}=\frac{1}{4}\int\frac{\sin^2x\cdot\sin^2x}{\cos x}=\frac{1}{4}\int\frac{\sin^2x\cdot(1-\cos^2x)}{\cos x}=\frac{1}{4}\int \frac{\sin^2x}{\cos x}-\frac{\cos^2x}{\cos x}=\frac{1}{4}\int \frac{1-\cos^2x}{\cos x}-\cos x=\frac{1}{4} \int \frac{1}{\cos x}-2\cos x=\frac{1}{4}(\ln(\tan x+\sec x)+2\sin x$$
Is it correct?
|
More easy way $tan(x).cos(x)=sin(x)$ so integral becomes $sin^3(x).tan(x)=sin^4(x)/cos(x)$ Now $sinx=u$ so integral becomes $$\int\frac{u^4}{1-u^2}$$ which on simplification becomes $$\int \frac{2}{(1/u^2)(1/u^2-1)}du$$ now let $1/u=t$ thus the common $ u^2$ cancels to give $\int\frac{1}{1-t^2}dt$ whose integral is $\frac{1}{2}log(\frac{1+t}{1-t})+c$ and the resubstitution gives the answer .
|
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|
In $\triangle ABC$, if $\cos A\cos B\cos C=\frac{1}{3}$, then $\tan A\tan B+\tan B \tan C+\tan C\tan A =\text{???}$
In $\triangle ABC$, if
$$\cos A \cos B \cos C=\frac{1}{3}$$
then can we find value of
$$\tan A\tan B+\tan B \tan C+\tan C\tan A\ ?$$
Please give some hint. I am not sure if $\tan A \tan B+\tan B \tan C+\tan C \tan A$ will be constant under given condition.
|
Hint: For a triangle ABC
$A+B=\pi-C$
and
$1-2\cos A \cos B \cos C=\cos^2A+\cos^2B+\cos^2C$
Edit $1$:
$A+B=\pi-C$
Apply $\cos$ on both sides
Divide each term by $\cos A.\cos B$
We get $\tan A.\tan B=1+\frac{\cos C}{\cos A. \cos B}$
Similarly write $2$ more equations and add three equations.
Now use $1-2\cos A \cos B \cos C=\cos^2A+\cos^2B+\cos^2C$
|
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|
Last Digit of $x^0 + x^1 + x^2 + \cdots + x^{p-1} + x^p$
Given $x$ and $p$. Find the last digit of $x^0 + x^1 + x^2 + \cdots + x^{p-1} + x^p$
I need a general formula. I can find that the sum is equal to
$\dfrac{x^{p+1}-1}{x-1}$
But how to find the last digit.
P.S: $x\leq 999999$ and $p \leq 10^{15}$
|
We'll work it out mod 5 and mod 2. We'll assume $p > 2$.
Mod 2, we have $x^i \equiv x$ for all $i$, except for $x^0$. Therefore the parity of the sum is just the same as the opposite of that of $x$ [unless $p=2$].
Mod 5: if $x \equiv 0 \pmod{5}$ then the result is certain to be $1$ mod $5$. Otherwise, $x^4 \equiv 1 \pmod{5}$, so we just have lots of expressions $1+x+x^2+x^3$ all added together; this is $4$ if $x \equiv 1 \pmod{5}$, and $0$ otherwise.
Hence:
*
*If $x$ is $0$ mod 5, the result is 1 mod 5.
*If $x$ isn't 1 mod 5, we'll get $1+x \pmod{5}$ if $p$ is $1$ mod 4, and $1+x+x^2+x^3 = 0$ if $p$ is $3$ mod 4.
*If $x$ is 1 mod 5, we'll get $-(p-1)/4 + (1+x) \pmod{5}$ if $p$ is $1$ mod 4, and $-(p+1)/4 \pmod{5}$ if $p$ is $3$ mod 4.
These together are enough to assemble the answer.
|
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|
How to solve $a = \cos x - b\sin x$ where $a$ and $b$ are real numbers? I found this equation when solving a physics problem related to finding an angle when entering a river, that has a known current, and trying to get to a specific point on the other side. I'm not sure how to solve this equation explicitly for the angle $x$.
|
Method 1:
*
*$a=\cos x - b\sin x$
*$\cos x = a+b\sin x$
*$\cos^2x = a^2+2ab\sin x+b^2\sin^2x$
*$1-\sin^2x = a^2+2ab\sin x+b^2\sin^2x$
*$(a^2-1)+2ab\sin x+(b^2+1)\sin^2x=0$
Then quadratic.
Method 2:
$$\sin(\alpha-x)=\sin\alpha\cos x-\cos\alpha\sin x$$
$$\frac{\sin(\alpha-x)}{\sin\alpha}=\cos x-\frac1{\tan\alpha}\sin x$$
Let $\frac1{\tan\alpha}=b$, then $\frac1{\sin\alpha}=\sqrt{b^2+1}$.
The equation becomes:
*
*$a=\sqrt{b^2+1}\sin(\alpha-x)$
*$\sin(\alpha-x)=\dfrac{a}{\sqrt{b^2+1}}$
|
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|
Find sum of power series. Having a small mistake. Find the sum of the series. My answer is $-\frac{3}{4}$, but it should be $\frac{3}{4}$. Where did i make a mistake?
$$ \sum_{n=1}^{\infty} \frac{n}{3^n} $$
$$ \frac{d}{dx} (\frac{1}{1-x}) = \sum_{n=0}^{\infty} \frac{d}{dx} (x^n) $$
$$ \frac{-1}{(1-x)^2} = \sum_{n=0}^{\infty} n x^{n-1} $$
$$ \frac{-x}{(1-x)^2} = \sum_{n=0}^{\infty} n x^{n} $$
Substitute $\frac{1}{3} $ into x
$$ \frac{-\frac{1}{3}}{(1-\frac{1}{3})^2} = \frac{-3}{4} $$
|
Since $(1-x)'=-1$,
$$
\frac{d}{dx}\frac{1}{1-x}=-\frac{(1-x)'}{(1-x)^2}=\frac{1}{(1-x)^2}
$$
by chain rule. Now the correct answer will appear.
|
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|
Find remainder when $1^{5} + 2^{5} \cdots +100^{5}$ divided by 4 I'm studding D.M Burton & want to solve: Find remainder when $1^{5} + 2^{5} \cdots +100^{5}$ divided by $4$. . Please help me by giving your solution to it. I'm new comer to number theory so please don't use theorems above Theory of Congruence.
|
Another way is to use the "paired element" approach... Since $a^5+b^5=(a+b)(a^4-a^3b+a^2b^2-ab^3+b^4)$, we have
$$1^5+99^5+2^5+98^5+\dots+ 50^5+100^5\\
=100(1-99+99^2-99^3+99^4)+100(2^4-\dots)+\dots+100(49^4-\dots)+2^5\cdot(25)^5+100^5\\
\equiv 0\pmod 4$$
|
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|
Find all $x,y$ so that $\dfrac{x+y+2}{xy-1}$ is an integer. I am trying to find the integers $x,y$ so that
$\dfrac{x+y+2}{xy-1}$ is an integer.
What I have done:
I suppose there exists $t$ such that $$t=\dfrac{x+y+2}{xy-1}$$ where $xy\neq 1$ then consider the following scenarios:
$$x=y$$ $$x>y>0$$ $$x>0>y$$ ,etc.
This approach helps me find the solutions but it is very long. Any simpler method?
|
Wlog $|x|\le |y|$.
$x=0$ leads to $\frac{y+2}{-1}$, which is always an integer.
$x=1$ leads to $\frac{y+3}{y-1}=1+\frac4{y-1}$ which is an integer iff $y-1$ is a (positive or negative) divisor of $4$, so $y\in\{-3,-1,2,3,5\}$ ($y=0$ is excluded by $|y|\ge |x|$)
$x=-1$ leads to $\frac{y+1}{-y-1}=-1$, always an integer
$|x|\ge2$ leads to $|xy-1|\ge 2|y|-1\ge 3$ and $|x+y+2|\le 2|y|+2$, so that $\left|\frac{x+y+2}{xy-1}\right|\le\frac{2|y|+2}{2|y|-1}=1+\frac{3}{2|y|-1}\le 2$ with equality only for $|x|=|y|=2$. So we need only check the special case $|x|=|y|=2$ and otherwise if the quotient can be $-1$ or $0$ or $1$.
*
*$(2,2)$ leads to quotient $2$; $(2,-2)$ and $(-2,2)$ lead to $-\frac 25$; $(-2,-2)$ leads to $-\frac 23$
*The quotient is $0$ iff $y=-x-2$.
*The quotient is $1$ iff $x+y+2=xy-1$, i.e., $(x-1)(y-1)=4$; the finitely many factorizations give us finitely many solutions $x=y=3$ or $x=2,y=5$ (all else is exclude by $|y|\ge |x|\ge 2$)
*The quotient is $-1$ iff $x+y+2=1-xy$, i.e., $(x+1)(y+1)=0$ - contradicting $|y|\ge|x|\ge 2$.
So in summary, as solutions $(x,y;\frac{x+y+2}{xy-1})$ we have the families
$$ (0,t;-t-2), (t,0;-t-2), (-1,t;-1),(t,-1;-1),(t,-2-t;0)$$
and the special solutions
$$(1,2;5),(2,1;5),(1,3;3),(3,1;3),(1,5;2),(5,1;2),(2,2;2),(3,3;1),(2,5;1),(5,2;1).$$
|
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|
solve $\sin 2x + \sin x = 0 $ using addition formula
$\sin 2x + \sin x = 0 $
Using the addition formula, I know that
$\sin 2x = 2\sin x \cos x$
=> $2\sin x \cos x + \sin x = 0$
=> $\sin x(2\cos x + 1) = 0$
=> $\sin x = 0$ and $\cos x = -\frac{1}2 $
I know that $\sin x = 0$ in first and second quadrant so $x = 0$ and $x = 180$
What I do not know is what to do with $\cos x = -\frac{1}2$ and which quadrants this applies to.
The book I got the question from gives the following answer which does not make sense to me:
0, 120, 180, 240, 360
|
We know that $\cos 60^\circ = \frac{1}{2}$.
The cosine of an angle is defined to be the $x$-coordinate of the point where the terminal side of an angle in standard position (initial side on the positive $x$-axis and vertex at the origin) intersects the unit circle. Therefore, the cosine function is positive if the terminal side of the angle lies in the first quadrant, fourth quadrant, or on the positive $x$-axis; $0$ if the terminal side of the angle lies on the $y$-axis; and negative if the terminal side of the angle lies in the second quadrant, third quadrant, or on the negative $x$-axis.
Now consider the diagram below.
Observe that $\cos(\pi - \theta) = \cos(180^\circ - \theta) = -\cos\theta$. Hence,
$$\cos(180^\circ - 60^\circ) = \cos(120^\circ) = -\cos(60^\circ) = -\frac{1}{2}$$
Also, observe that $\cos(\pi + \theta) = \cos(180^\circ - \theta) = -\cos\theta)$. Hence,
$$\cos(180^\circ + 60^\circ) = \cos(240^\circ) = -\cos(60^\circ) = -\frac{1}{2}$$
|
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|
Find the Min of P(x,y) Find the Minimum of the following function :
$$P(x,y) = \frac{(x-y)}{(x^4+y^4+6)}.$$
This is a math problem I found in an internet math competition but it is really complex to me !!!
|
Hint: Armed with your guess, it is not hard to complete the square to get
$$\frac{x-y}{x^4+y^4+6}+\frac14= \frac{x^4+y^4+4x-4y+6}{4(x^4+y^4+6)}=\frac{(x^2-1)^2+(y^2-1)^2+2(x+1)^2+2(y-1)^2}{4(x^4+y^4+6)} \geqslant 0$$
with equality possible iff $x=-1, y=1$.
|
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|
In a $\triangle ABC,a^2+b^2+c^2=ac+ab\sqrt3$,then the triangle is In a $\triangle ABC,a^2+b^2+c^2=ac+ab\sqrt3$,then the triangle is
$(A)$equilateral
$(B)$isosceles
$(C)$right angled
$(D)$none of these
The given condition is $a^2+b^2+c^2=ac+ab\sqrt3$.
Using sine rule,
$a=2R\sin A,b=2R\sin B,c=2R\sin C$,we get
$\sin^2A+\sin^2B+\sin^2C=\sin A\sin C+\sin A\sin B\sqrt3$
I am stuck here.
|
$$a^2+b^2+c^2=ac+ab\sqrt3$$ The above equation can be re-written as
$$\frac{a^2}{4}-ac+c^2+\frac{3a^2}{4}-ab\sqrt3+b^2=0$$
which is
$$(\frac{a}{2}-c)^2+(\frac{\sqrt3 a}{2}-b)^2 = 0$$
which implies that
$\frac{a}{2} = c$ and $\frac{\sqrt3 a}{2}= b$
.Based on this it can be concluded that the ratio of sides is 1:$\sqrt{3}$:2, which is a right angled-triangle.I hope this was helpful.
|
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|
Proof of $\sum_{j=1}^{p-1} \lfloor jq/p \rfloor = \frac{1}{2}(q-1)(p-1)$ Involving Pairing of Summands I've seen the proof of the identity
$$\sum_{j=1}^{p-1} \lfloor jq/p \rfloor = \frac{1}{2}(q-1)(p-1)$$
where $p$ and $q$ are coprime positive integers. This involves counting the remainders $r_{j}/p$ of the summands $\lfloor jq/p \rfloor = jq/p - r_{j}/p$. But when I attempted at showing the identity, I used the fact that if $kq/p = n + \alpha$ for some $\alpha < 1$ (which is nonzero since $p$ and $q$ are coprime) and some integer $n$, then we can pair up the summands
$$\big\lfloor kq/p \big\rfloor + \big\lfloor \dfrac{(p-k)q}{p} \big\rfloor = n + q - (n + 1) = q-1.$$
There are $(p-1)/2$ such pairs in the sum if $p$ is odd, and thus we get $(p-1)(q-1)/2$. If $p$ is even, then $q$ is odd and there are $(p-2)/2$ such pairs, with an extra term $\lfloor (p/2)(q/p)\rfloor = (q-1)/2$. Thus we have
$$\dfrac{(q-1)(p-2)}{2} + \dfrac{q-1}{2} = \dfrac{(p-1)(q-1)}{2}$$
as desired.
My question is: Is there something I might be missing in this argument? I can't find a similar one anywhere else, which makes me think that there is a hole somewhere. But I cannot find anything in my view, so maybe others' views will help.
|
Your procedure is correct, in fact your steps can be reformulated as follows:
$$
\eqalign{
& \sum\limits_{j = 1}^{p - 1} {\left\lfloor {j\,q/p} \right\rfloor } = {1 \over 2}\left( {\sum\limits_{j = 1}^{p - 1} {\left\lfloor {j\,q/p} \right\rfloor + \sum\limits_{j = 1}^{p - 1} {\left\lfloor {\left( {p - j} \right)\,q/p} \right\rfloor } } } \right) = \cr
& = {1 \over 2}\sum\limits_{j = 1}^{p - 1} {\left( {\left\lfloor {j\,\left\lfloor {q/p} \right\rfloor + j\left\{ {q/p} \right\}} \right\rfloor + \left\lfloor {\left( {p - j} \right)\,\left\lfloor {q/p} \right\rfloor + \left( {p - j} \right)\left\{ {q/p} \right\}} \right\rfloor } \right)} = \cr
& = {1 \over 2}\sum\limits_{j = 1}^{p - 1} {\left( {p\,\left\lfloor {q/p} \right\rfloor + \left\lfloor {j\left\{ {q/p} \right\}} \right\rfloor + \left\lfloor {\left( {p - j} \right)\left\{ {q/p} \right\}} \right\rfloor } \right)} = \cr
& = {1 \over 2}\sum\limits_{j = 1}^{p - 1} {\left( {p\,\left\lfloor {q/p} \right\rfloor + p\left\{ {q/p} \right\} + \left\lfloor {j\left\{ {q/p} \right\}} \right\rfloor + \left\lfloor { - j\left\{ {q/p} \right\}} \right\rfloor } \right)} \cr}
$$
and since:$$
\left\lfloor x \right\rfloor + \left\lfloor { - x} \right\rfloor = \left\lfloor x \right\rfloor - \left\lceil x \right\rceil = - \left[ {0 < \left\{ x \right\}} \right]
$$
where the square brackets are the Iverson bracket, and because:$$
\gcd (q,p) = 1\quad \Rightarrow \quad \gcd \left( {q\bmod p,p} \right) = 1\quad \Rightarrow \quad 0 < \left\{ {j\left\{ {q/p} \right\}} \right\}\,\;\left| {\,1 \le j \le p - 1} \right.
$$
then the above converts to:$$
= {1 \over 2}\sum\limits_{j = 1}^{p - 1} {\left( {q - 1} \right)} = {{\left( {p - 1} \right)\left( {q - 1} \right)} \over 2}
$$
---- addendum (in reply to om joglekar) ---
We can write the duplicated summand as
$$
\begin{array}{l}
\left\lfloor {\frac{{jq}}{p}} \right\rfloor + \left\lfloor {\frac{{\left( {p - j} \right)q}}{p}} \right\rfloor =
\left\lfloor {\frac{{jq}}{p}} \right\rfloor + \left\lfloor {q - \frac{{jq}}{p}} \right\rfloor = \\
= q + \left\lfloor {\frac{{jq}}{p}} \right\rfloor + \left\lfloor { - \frac{{jq}}{p}} \right\rfloor =
q + \left\lfloor {\frac{{jq}}{p}} \right\rfloor + \left\lfloor { - \left( {\left\lfloor {\frac{{jq}}{p}} \right\rfloor
+ \left\{ {\frac{{jq}}{p}} \right\}} \right)} \right\rfloor = \\
= q + \left\lfloor {\frac{{jq}}{p}} \right\rfloor - \left\lfloor {\frac{{jq}}{p}} \right\rfloor
+ \left\lfloor { - \left\{ {\frac{{jq}}{p}} \right\}} \right\rfloor =
q + \left\lfloor { - \left\{ {\frac{{jq}}{p}} \right\}} \right\rfloor \\
\end{array}
$$
Now the fractional part is limited to be
$$
0 \le \left\{ x \right\} < 1
$$
Since $\gcd(q,p)=1$ and $1 \le j \le p-1$ the fraction $j q / p$ cannot be an integer, hence its fractional part will never be zero.
Therefore
$$
0 < \left\{ {\frac{{jq}}{p}} \right\} < 1\quad \Leftrightarrow \quad - 1 < - \left\{ {\frac{{jq}}{p}} \right\} < 0
$$
and the floor of that will be constant at $-1$.
|
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|
Limit of tan function This is a question from an old tutorial for a basic mathematical analysis module.
Show that $$\lim\limits_{n\to \infty}\tan^n(\frac{\pi}{4}+\frac{1}{n}) = e^2$$
My tutor has already gone through this in class but I am still confused. Is there anything wrong with the following reasoning?
Since $\frac{\pi}{4}+\frac{1}{n} \to \frac{\pi}{4}$ as $n \to \infty$,
it seems to me that $\tan(\frac{\pi}{4}+\frac{1}{n}) \to \tan(\frac{\pi}{4}) = {1}$,
and hence $\tan^n(\frac{\pi}{4}+\frac{1}{n}) \to 1^n = 1$.
Additionally, my tutor has given a hint, to use Squeeze theorem along with the definition $e = \lim\limits_{n \to \infty }(1 + {1\over n})^n$, but I can't see how these are to be used.
Edit: Here's another attempt I've made.
After using addition formula for tangent, we get
$$\frac{1 + \tan{\frac{1}{n}}}{1- \tan{\frac{1}{n}}} = 1 - \frac{2\tan{\frac{1}{n}}}{1-\tan{\frac{1}{n}}} = 1 - \frac{1}{\frac{1-\tan{\frac{1}{n}}}{2\tan{\frac{1}{n}}}} = [(1 - \frac{1}{\frac{1-\tan{\frac{1}{n}}}{2\tan{\frac{1}{n}}}})^\frac{1}{\frac{1-\tan{\frac{1}{n}}}{2\tan{\frac{1}{n}}}}]^\frac{2\tan{\frac{1}{n}}}{1-\tan{\frac{1}{n}}}$$
so given that $\lim\limits_{n \to \infty}{\frac{\tan{\frac{1}{n}}}{\frac{1}{n}}}=1$, we have
$$\lim\limits_{n\to \infty}\tan^n(\frac{\pi}{4}+\frac{1}{n}) = \lim\limits_{n \to \infty}[(1 - \frac{1}{\frac{1-\tan{\frac{1}{n}}}{2\tan{\frac{1}{n}}}})^\frac{1}{\frac{1-\tan{\frac{1}{n}}}{2\tan{\frac{1}{n}}}}]^\frac{2n\tan{\frac{1}{n}}}{1-\tan{\frac{1}{n}}} = e^\frac{2*1}{1-0} = e^2$$
I'm really hoping that this method works as well! So sorry for the ugly formatting, I couldn't figure out some parts.
|
Solution
Applying $\textbf{L'Hospital's Rule}$, we have $$\begin{align*}\lim_{x \to +\infty} \left[x\ln \tan \left(\dfrac{\pi}{4}+\dfrac{1}{x}\right)\right]&=\lim_{x \to +\infty}\dfrac{\ln \tan \left(\dfrac{\pi}{4}+\dfrac{1}{x}\right)}{\dfrac{1}{x}}\\&=\lim_{x \to +\infty}\dfrac{1}{\sin\left(\dfrac{\pi}{4}+\dfrac{1}{x}\right)\cos \left(\dfrac{\pi}{4}+\dfrac{1}{x}\right)}\\&=\lim_{x \to +\infty}\dfrac{2}{ \sin\left(\dfrac{\pi}{2}+\dfrac{2}{x}\right)}\\&=\lim_{x \to +\infty}\dfrac{2}{ \cos\left(\dfrac{2}{x}\right)}\\&=2. \end{align*}$$ Hence, $$\begin{align*}\lim\limits_{n\to \infty}\tan^n\left(\frac{\pi}{4}+\frac{1}{n}\right)&=\lim\limits_{x\to +\infty}\tan^x\left(\frac{\pi}{4}+\frac{1}{x}\right)\\&=\lim\limits_{x\to +\infty} \exp\left[x\ln \tan \left(\dfrac{\pi}{4}+\dfrac{1}{x}\right)\right]\\&=\exp\left[\lim_{x \to \infty}x\ln \tan \left(\dfrac{\pi}{4}+\dfrac{1}{x}\right)\right]\\&=\mathbb{e}^2.\end{align*}$$
|
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|
Calculating eigenvalues and eigenvectors Question: Calculate the eigenvectors and eigenvalues of the following matrix $$\begin{pmatrix}3&-3\\0&-2\\ \end{pmatrix}$$
My attempt:
I have calculated the eigenvalues to be $\lambda = 3$ and $\lambda =-2$ and I have managed to get the eigen vector for $\lambda = 3$ to be \begin{pmatrix}1\\0\\ \end{pmatrix}
However I get \begin{pmatrix}1\\\frac{5}{3}\\ \end{pmatrix}
but the correct answer seems to be
\begin{pmatrix}0\\1\\ \end{pmatrix}
Where am I going wrong?
|
The eigenvalues of a triangular matrix are simply the diagonal entries, so your eigenvalues $3$ and $-2$ are correct. As
$$\begin{pmatrix}
3 & -3 \\
0 & -2 \\
\end{pmatrix}
\begin{pmatrix}
1 \\ 0
\end{pmatrix}
= \begin{pmatrix}
3 \\ 0
\end{pmatrix}
= 3
\begin{pmatrix}
1 \\ 0
\end{pmatrix},$$
the vector $\begin{pmatrix}1 \\ 0 \end{pmatrix}$ is an eigenvector corresponding to $\lambda = 3$.
As
$$\begin{pmatrix}
3 & -3 \\
0 & -2 \\
\end{pmatrix}
\begin{pmatrix}
1 \\ 5/3
\end{pmatrix}
= \begin{pmatrix}
-2 \\ -10/3
\end{pmatrix}
= -2
\begin{pmatrix}
1 \\ 5/3
\end{pmatrix},$$
the vector $\begin{pmatrix}1 \\ 5/3\end{pmatrix}$ is an eigenvector corresponding to $\lambda = -2$.
So, your answers are correct. Note that $\begin{pmatrix}0 \\ 1\end{pmatrix}$ is not an eigenvector of this matrix, because
$$\begin{pmatrix}
3 & -3 \\
0 & -2 \\
\end{pmatrix}
\begin{pmatrix}0 \\ 1\end{pmatrix}
= \begin{pmatrix}-3 \\ -2\end{pmatrix}
$$
which is not a scalar multiple of $\begin{pmatrix}0 \\ 1\end{pmatrix}$.
|
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|
Is the following solution correct? Question: $ \sqrt{x^2 + 1} + \frac{8}{\sqrt{x^2 + 1}} = \sqrt{x^2 + 9}$
My solution: $(x^2 + 1) + 8 = \sqrt{x^2 + 9} \sqrt{x^2 + 1}$
$=> (x^2 + 9) = \sqrt{x^2 + 9} \sqrt{x^2 + 1}$
$=> (x^2 + 9) - \sqrt{x^2 + 9} \sqrt{x^2 + 1} = 0$
$=> \sqrt{x^2 + 9} (\sqrt{x^2 + 9} - \sqrt{x^2 + 1}) = 0$
So, either $\sqrt{x^2 + 9} = 0$ or $(\sqrt{x^2 + 9} - \sqrt{x^2 + 1}) = 0$
From the first expression, I get $x = \pm 3 i$ and from the second expression, I get nothing.
Now, notice how in the 2nd step, I could've divided both the sides by $\sqrt{x^2 + 9}$, but I didn't because I learned here that we must never do that and that we should always factor: Why one should never divide by an expression that contains a variable.
So, my question is: is the solution above correct? Would it have been any harm had I divided both the sides by $\sqrt{x^2 + 9}$?
|
A suggested simplification. You should always look for simplifications to the algebra if they're easy to find.
Let $y = x^2 + 1$
Then you're solving $\sqrt y + \frac{8}{\sqrt y} = \sqrt{y + 8}$
$\frac{y + 8}{\sqrt y} = \sqrt{y + 8}$
$(y+8)^2 = y(y+8)$
$8(y+8) = 0$
From this, it should be obvious that $y = -8$ is the only root, giving $x^2 + 1 = -8$ or $x = \pm 3i$.
|
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|
Show that $\int_{-\infty}^\infty {{x^2-3x+2}\over {x^4+10x^2+9}}dx={5\pi\over 12}$ Show that $$\int_{-\infty}^\infty {{x^2-3x+2}\over {x^4+10x^2+9}}dx={5\pi\over 12}.$$
Any solutions or hints are greatly appreciated.
I know I can rewrite the integral as $$\int_{-\infty}^\infty {(x-1)(x-2)\over {(x^2+1)(x^2+9)}}dx.$$ but I'm not sure how to proceed.
|
Well, $\int_{-\infty}^{+\infty}\frac{-3x}{x^4+10x^2+9}\,dx = 0$, hence the problem boils down to computing:
$$ \int_{-\infty}^{+\infty}\frac{(x^2+2)}{(x^2+1)(x^2+9)}\,dx=\frac{1}{8}\int_{-\infty}^{+\infty}\frac{1}{x^2+1}\,dx+\frac{7}{8}\int_{-\infty}^{+\infty}\frac{1}{x^2+9} $$
that is trivially equal to $\left(\frac{1}{8}+\frac{7}{24}\right)\pi = \color{red}{\large\frac{5\pi}{12}}$.
Footnote: how to find the coefficients $\frac{1}{8}$ and $\frac{7}{8}$ very fast. We know that for some $A,B$
$$ g(z)=\frac{z+2}{(z+1)(z+9)}=\frac{A}{z+1}+\frac{B}{z+9} $$
must hold. On the other hand, $A=\lim_{z\to -1}g(z)(z+1)$ as well as $B=\lim_{z\to -9}g(z)(z+9)$, so:
$$ A = \lim_{z\to -1}\frac{z+2}{z+9},\qquad B=\lim_{z\to -9}\frac{z+2}{z+1}.$$
|
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|
Extract coefficients for a formal power series using Lagrange Inversion Formula Given $f(x)$ is a formal power series that satisfies $f(0) = 0$
$(f(x))^{3} + 2(f(x))^{2} + f(x) - x = 0$
I know that the Lagrange inversion formula states given f(u) & $\varphi(u)$ are formal power series with respect to u, and $\varphi(0) = 1$ then the following is true.
$[x^{n}](f(u(x))) = \frac{1}{n}[u^{n-1}](f'(u)\varphi(u)^{n})$
How do I find the coefficient of $x^{n}$ in $f(x)$ using Lagrange inversion formula?
|
Suppose we have
$$f(z)^3 + 2f(z)^2 + f(z) = z$$
and we seek $[z^n] f(z).$
While we wait for a contribution from LIF experts in the meantime we
can use Poor Man's Lagrange Inversion which is the Cauchy Residue
Theorem. We have
$$[z^n] f(z)
= \frac{1}{2\pi i}
\int_{|z|=\epsilon} \frac{1}{z^{n+1}} f(z) \; dz.$$
Now put $w=f(z)$ so that $w^3+2w^2+w = z$ and
$3w^2 + 4w + 1 \; dw = dz$ to get
$$\frac{1}{2\pi i}
\int_{|w|=\epsilon} \frac{1}{w^{n+1} (1+w)^{2n+2}}
\times w \times (3w^2 + 4w + 1) \; dw.$$
Now $$w \times (3w^2 + 4w + 1)
= 3 (1+w)^3 - 5 (1+w)^2 + 2 (w+1).$$
Extracting coefficients we thus have
$$3 (-1)^n {n+2n-2\choose n}
- 5 (-1)^n {n+2n-1\choose n}
+ 2 (-1)^n {n+2n\choose n}
\\ = (-1)^n \times
\left(3{3n-2\choose n}
- 5 {3n-1\choose n}
+ 2{3n\choose n}\right).$$
This is
$$(-1)^n {3n-2\choose n}
\left(3 - 5\frac{3n-1}{2n-1}
+ 2 \frac{(3n-1)3n}{(2n-1)2n}\right)
\\ = \frac{(-1)^n}{1-2n} {3n-2\choose n}.$$
Remark. The fact that we are given $f(0) = 0$ determines the
choice of branch so that $z=0$ corresponds to $w=0$ (which is needed
when we make the substitution). Also, if we choose $\epsilon$
sufficiently small we also get a small circle for $w$ (with a
different $\epsilon.$)
Addendum, five years later. A better choice of integral
is
$$[z^n] f(z) = \frac{1}{n} [z^{n-1}] f'(z)
= \frac{1}{n} \frac{1}{2\pi i}
\int_{|z|=\varepsilon}
\frac{1}{z^n} f'(z) \; dz.$$
We put $w = f(z)$ and obtain
$$\frac{1}{n} \frac{1}{2\pi i}
\int_{|w|=\gamma}
\frac{1}{w^n (1+w)^{2n}} \; dw
= \frac{(-1)^{n-1}}{n} {n-1+2n-1\choose n-1}
\\ = \frac{(-1)^{n-1}}{n} {3n-2\choose n-1}.$$
|
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|
n tends to infinity $\displaystyle\lim_{n\to\infty}
{\left(
\frac{\sqrt {n^2+n} - 1}{n}
\right)}^{
\left(2 \sqrt{n^2+n} - 1
\right)}$
Sorry for the bad format of the question I don't know much latex
I tried rationalization of the power and the base but was not able to get the answer.
|
When we see both the base and exponent as variable then the best approach is to take logs. Thus if $L$ is the desired limit then
\begin{align}
\log L &= \log\left\{\lim_{n \to \infty}\left(\frac{\sqrt{n^{2} + n} - 1}{n}\right)^{(2\sqrt{n^{2} + n} - 1)}\right\}\notag\\
&= \lim_{n \to \infty}\log\left(\frac{\sqrt{n^{2} + n} - 1}{n}\right)^{(2\sqrt{n^{2} + n} - 1)}\text{ (via continuity of log)}\notag\\
&= \lim_{n \to \infty}(2\sqrt{n^{2} + n} - 1)\log\left(\frac{\sqrt{n^{2} + n} - 1}{n}\right)\notag\\
&= \lim_{n \to \infty}(2\sqrt{n^{2} + n} - 1)\log\left(1 + \frac{\sqrt{n^{2} + n} - 1 - n}{n}\right)\notag\\
&= \lim_{n \to \infty}(2\sqrt{n^{2} + n} - 1)\cdot\dfrac{\sqrt{n^{2} + n} - 1 - n}{n}\cdot\dfrac{\log\left(1 + \dfrac{\sqrt{n^{2} + n} - 1 - n}{n}\right)}{\dfrac{\sqrt{n^{2} + n} - 1 - n}{n}}\notag\\
&= \lim_{n \to \infty}(2\sqrt{n^{2} + n} - 1)\cdot\dfrac{\sqrt{n^{2} + n} - 1 - n}{n}\cdot 1\notag\\
&= \lim_{n \to \infty}\left(2\sqrt{1 + \frac{1}{n}} - \frac{1}{n}\right)\cdot\dfrac{(n^{2} + n) - (1 + n)^{2}}{\sqrt{n^{2} + n} + 1 + n}\notag\\
&= \lim_{n \to \infty}2\cdot\dfrac{-1 - n}{\sqrt{n^{2} + n} + 1 + n}\notag\\
&= -2\lim_{n \to \infty}\dfrac{1 + \dfrac{1}{n}}{\sqrt{1 + \dfrac{1}{n}} + 1 + \dfrac{1}{n}}\notag\\
&= -1\notag
\end{align}
Hence $L = 1/e$. We have used the fact that the expression $$\frac{\sqrt{n^{2} + n} - 1 - n}{n}$$ tends to $0$ as $n \to \infty$ and $(1/x)\log(1 + x) \to 1$ as $x \to 0$.
|
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|
Show that $\sum_{k=1}^{\infty}\frac{F_{2^{k-1}}}{L_{2^k}-2}=\frac{15-\sqrt{5}}{10}$
Show that$$\sum_{k=1}^{\infty}\frac{F_{2^{k-1}}}{L_{2^k}-2}=\frac{15-\sqrt{5}}{10},$$
where $F_n$ is a Fibonacci number and $L_n$ is a Lucas number.$^1$
Motivation: For example, when calculating Millin series
$$
\sum_{n=1}^{\infty} \frac{1}{F_{2^n}}=\frac{7-\sqrt{5}}{2},
$$
we can show that
$$
\sum_{k=1}^n \frac{1}{F_{2^k}} =3-\frac{F_{2^n-1}}{F_{2^n}}.
$$
and finish the calculation by sending $n$ to $\infty$.
Also, when proving
$$
\sum_{n=1}^{\infty}\frac{F_{2^{n-1}}}{L_{2^n}+1}=\frac{1}{\sqrt{5}},
$$
solution first shows that
$$
\sum_{k=1}^n \frac{F_{2^{k-1}}}{L_{2^k}+1}=\frac{F_{2^n}}{L_{2^n}+1}
$$
After deducting such formulas, it is easy to calculate limits. Then, how to find a formula about $$\sum_{k=1}^n \frac{F_{2^{k-1}}}{L_{2^k}-2}=?$$
My attempt: Since
$$
\frac{15-\sqrt{5}}{10} = \frac{3}{2}-\frac{1}{2}\cdot\frac{1}{\sqrt{5}},
$$
I made a conjecture
$$
\sum_{k=1}^n \frac{F_{2^{k-1}}}{L_{2^k}-2}=\frac{3}{2}-\frac{1}{2}\frac{F_{2^n}}{L_{2^n}-2}.
$$
For $n=1$,
$$\frac{F_1}{L_2-2}=1=\frac{3}{2}-\frac{1}{2}=\frac{3}{2}-\frac{1}{2}\frac{F_2}{L_2-2}.$$
Suppose that the conjecture holds for some $n$. Then
\begin{align}
\sum_{k=1}^{n+1} \frac{F_{2^{k-1}}}{L_{2^k}-2}&=\frac{3}{2}-\frac{1}{2}\frac{F_{2^n}}{L_{2^n}-2}+\frac{F_{2^n}}{L_{2^{n+1}}-2}\\
&=\dots?
\end{align}
My question: How to complete the induction step? If the conjecture is wrong, then how to find other ways?
$^1$ Elementary Problems and Solutions (May 2015). Fibonacci Quarterly. Volume 53. Number 2.
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Using
$$F_{2N}=F_NL_N$$
$$L_{2N}=L_N^2-2(-1)^N$$
we have
$$F_{2^{n+1}}=F_{2^n}L_{2^n}$$
$$L_{2^{n+1}}=L_{2^n}^2-2$$
and so
$$\begin{align}\frac 32-\frac 12\cdot\frac{F_{2^n}}{L_{2^n}-2}+\frac{F_{2^n}}{L_{2^{n+1}}-2}&=\frac 32-\frac 12\cdot\frac{F_{2^n}}{L_{2^n}-2}+\frac{F_{2^n}}{L_{2^n}^2-4}\\\\&=\frac 32-\frac{F_{2^n}(L_{2^n}+2)-2F_{2^n}}{2(L_{2^n}-2)(L_{2^n}+2)}\\\\&=\frac 32-\frac{F_{2^n}L_{2^n}}{2(L_{2^{n}}^2-4)}\\\\&=\frac 32-\frac 12\cdot\frac{F_{2^{n+1}}}{L_{2^{n+1}}-2}\end{align}$$
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|
Is $\frac{1}{11}+\frac{1}{111}+\frac{1}{1111}+\cdots$ an irrational number? Obviously:
$$\frac{1}{10}+\frac{1}{100}+\frac{1}{1000}+\cdots=0.1111\dots=\frac{1}{9}$$
is a rational number.
Now, if we make terms with demoninators in the form:
$$q_n=\sum_{k=0}^{n} 10^k$$
Then the sum will be:
$$\sum_{n=1}^{\infty}\frac{1}{q_n}=\frac{1}{11}+\frac{1}{111}+\frac{1}{1111}+\cdots=0.1009181908362007\dots$$
The decimal expansion of this number appears to be non-periodic.
How can we prove/disprove that this number is irrational?
Edit
This number is at OEIS: http://oeis.org/A065444
|
This is not the answer to the question of irrationality. This a piece of information about an interesting closed form of the series.
$$q_n=\sum_{k=0}^n 10^k = \frac{10^{n+1}-1}{9}$$
$q_0=1\quad;\quad q_1=11\quad;\quad q_2=111\quad...$
$\sum_{n=0}^m \frac{1}{q_n}=\frac{1}{1}+\frac{1}{11}+\frac{1}{111}+... \quad$ ($m+1$ terms).
$$\sum_{n=0}^m \frac{1}{q_n}=\sum_{n=1}^m \frac{9}{10^{n+1}-1}=
\frac{9}{\ln(10)}\left(\psi_{10}(m+2)-\psi_{10}(1)\right)-9(m+1)$$
$\psi_q(x)$ is the q-digamma function. The leading terms of the asymptotic series are:
$$\psi_{10}(x)\sim (x-\frac{1}{2})\ln(10)-\ln(9)$$
$$\sum_{n=0}^{m\to\infty} \frac{1}{q_n}\sim
\frac{9}{\ln(10)}\left( (m+2-\frac{1}{2})\ln(10)-\ln(9)-\psi_{10}(1)\right)-9(m+1)$$
After simplification :
$$\sum_{n=0}^{\infty} \frac{1}{q_n}=\frac{9}{2}
-\frac{9}{\ln(10)}\left( \ln(9)+\psi_{10}(1)\right)$$
$\psi_{10}(1)\simeq-1.32759401026424207 $
$\frac{9}{2}
-\frac{9}{\ln(10)}\left( \ln(9)+\psi_{10}(1)\right)\simeq 1.100918190836200736 $
Note that $\sum_{n=0}^{\infty} \frac{1}{q_n}$ includes the first term$=1$
|
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|
Contour Integral of $\int\limits_0^{2\pi}\frac{d\theta}{1+a\cos\theta}$ for $a^2<1$ (textbook wrong?) My book is telling me that the answer is $\frac{2\pi}{\sqrt{1-a^2}}$. I'm getting an extra a on the numerator. Could somebody verify if I'm wrong, or if it's my book (it has been wrong numerous times).
|
Contour Integration
$$
\begin{align}
\int_0^{2\pi}\frac{\mathrm{d}\theta}{1+a\cos(\theta)}
&=\oint\frac{2\,\mathrm{d}z}{iz\left(2+az+\frac az\right)}\tag{1}\\
&=-\frac{2i}a\oint\frac{\mathrm{d}z}{z^2+\frac2az+1}\tag{2}\\
&=-\frac{2i}a\oint\frac{\mathrm{d}z}{\left(z+\frac1a+\sqrt{\frac1{a^2}-1}\right)\left(z+\frac1a-\sqrt{\frac1{a^2}-1}\right)}\tag{3}\\
&=\frac{2\pi}{\sqrt{1-a^2}}\tag{4}
\end{align}
$$
Explanation:
$(1)$: $z=e^{i\theta}$
$(2)$: algebra
$(3)$: factor the denominator
$(4)$: the residue of the integrand at $z=-\frac1a+\sqrt{\frac1{a^2}-1}$ is $\frac1{2\sqrt{\frac1{a^2}-1}}$
$\phantom{(4)\text{:}}$ the singularity at $z=-\frac1a-\sqrt{\frac1{a^2}-1}$ is outside the unit circle
Weierstrass Substitution
$$
\begin{align}
\int_0^{2\pi}\frac{\mathrm{d}\theta}{1+a\cos(\theta)}
&=2\int_0^{\pi}\frac{\mathrm{d}\theta}{1+a\cos(\theta)}\tag{5}\\
&=2\int_0^\infty\frac{\frac{2\,\mathrm{d}z}{1+z^2}}{1+a\frac{1-z^2}{1+z^2}}\tag{6}\\
&=4\int_0^\infty\frac{\mathrm{d}z}{(1+a)+(1-a)z^2}\tag{7}\\
&=\frac4{1+a}\sqrt{\frac{1+a}{1-a}}\int_0^\infty\frac{\mathrm{d}z}{1+z^2}\tag{8}\\
&=\frac{2\pi}{\sqrt{1-a^2}}\tag{9}
\end{align}
$$
Explanation:
$(5)$: periodicity and evenness of $\cos(x)$
$(6)$: Weierstrass substitution
$(7)$: algebra
$(8)$: substitute $z\mapsto\sqrt{\frac{1+a}{1-a}}\,z$
|
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|
How to find continuities with square root? I don't understand how to find $$\frac{4-x^2}{3-\sqrt{x^2+5}}$$
The book says to multiply the equation by $\frac{3 + \sqrt {x^2+5}}{3 + \sqrt {x^2+5}}$. I don't understand where that comes from. It says the multiplication simplifies to "$3 + \sqrt {x^2+5}$" - I don't see how that's possible. Is the book wrong?
|
For this simplification, you need to know that $(a+b)(a-b)=a^2-b^2$. It is important to recognize this, as this is something you need very often. You also need this when simplifying expressions with square roots in the denominator. In this case, this gives us the following simplification: $(3+\sqrt{x^2+5})(3-\sqrt{x^2+5})=3^2-\left(\sqrt{x^2+5}\right)^2=9-(x^2+5)=4-x^2$.
For this, we subsitute $a=3$ and $b=\sqrt{x^2+5}$ in the identity $(a+b)(a-b)=a^2-b^2$.
This means that \begin{align*} \frac{4-x^2}{3-\sqrt{x^2+5}} &=\frac{4-x^2}{3-\sqrt{x^2+5}} \cdot \frac{3+\sqrt{x^2+5}}{3+\sqrt{x^2+5}} \\ &= \frac{(4-x^2)\left(3+\sqrt{x^2+5}\right)}{\left(3-\sqrt{x^2+5}\right)\left(3+\sqrt{x^2+5}\right)} \\ &= \frac{(4-x^2)\left(3+\sqrt{x^2+5}\right)}{4-x^2} \\ &= 3+\sqrt{x^2+5}\end{align*}
However, this is not valid for all $x$, since the former expression is not defined for all $x$, while the latter expression is. This is where the question is actually about: Finding the discontinuities of the function. For this, the following is useful:
Fact. If $f(x)$ and $g(x)$ are continuous, then $\frac{f(x)}{g(x)}$ is continuous where $g(x) \neq 0$.
So we need to find where $3-\sqrt{x^2+5}=0$. So $\sqrt{x^2+5}=3$, so $x^2+5=9$ and $x^2=4$ which means that $x=2$ or $x=-2$. These both satisfy the given equation.
|
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finding the smallest number $n$ such that $n!=n(n+1)(n+2)(n+3)$ What is the smallest number $n$ such that $n!=n(n+1)(n+2)(n+3)$?
How will I solve this type of problems?
|
Okay... Is there only one solution? Well, yeah... because $n!$ increases by a factor of $(n+1)!/n! = n+1$ as $n$ increases by one, while $n(n+1)(n+2)(n+3)$ only increases by $(n+1)(n+2)(n+3)(n+4)/n(n+1)(n+2)(n+3) = (n+4)/n = 1 + 4/n$. As $n!$ increases faster for $n > 2$ there is at most one solution.
(Unless there is a solution for $n = 1$ or $0$ which obviously there isn't.)
So..... Let's start.
$n! = n(n+1)(n+2)(n+3) \iff (n-1)! = (n+1)(n+2)(n+3)$
So $(n-4)!(n-3)(n-2)(n-1) = (n+1)(n+2)(n+3)$
$(n+1)(n+2)(n+3) = n^3 + 6n^2 + 11n + 6$and $(n-1)(n-2)(n-3) = n^3 - 6n^2 + 11n - 6$
So $(n-4)!(n^3 - 6n^2 + 11n-6) = n^3 + 6n^2 + 11n + 6$
So $[(n-4)!-1](n^3 + 11n) =[(n-4)!+1](6n^2 + 6)$
$\frac{(n-4)! - 1}{(n-4)! + 1} = 6\frac{n^2+1}{n^3+11n}$
$\frac{(n-4)!+1}{(n-4)! + 1}- \frac{2}{(n-4)! + 1} = 6(\frac{n^2+11}{n^3+11n}- \frac{10}{n^3+11n})$.
$1 - idgybit = 6\frac 1 n - oodgybit$
So $n \approx 6$.
Iff $n=6$
$1 - \frac 2 3 = 1 - \frac{10}{36+11}$... Mmm. not quite.
Iff $n = 7$
$1 - \frac 2{3! + 1} = 6/7 - \frac{60}{7^3 + 77}$
$1 - 2/7 = 6/7[1 - 10/(49 + 11)]$
$5/7 = 6/7[1 - 1/6] = 6/7*5/6 = 5/7$. So that does it.
$n = 7$ is the only solution.
|
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Solving system of non-linear equations. So I'm trying to find the stationary points for $$f(x,y,z) = 4x^2 + y^2 +2z^2 -8xyz$$
Setting the partial derivatives to zero leads to:
$$x-yz=0 \\ y-4xz=0\\z-2xy=0$$
Substiting $z=2xy$ into the first two equations to get $y(1-8x^2)=0$ and $x(1-2y^2)=0$. Which have the solution $y=0,x=\frac{1}{\sqrt{8}}, x=-\frac{1}{\sqrt{8}}$ and $x=0, y=\frac{1}{\sqrt{2}},y=-\frac{1}{\sqrt{2}}$
Now how do I go about pairing my $x,y$ solutions?
|
You have two conditions:
*
*$y=0$ or $x=\frac{1}{\sqrt{8}}$ or $x=-\frac{1}{\sqrt{8}}$
*$x=0$ or $y=\frac{1}{\sqrt{2}}$ or $y=-\frac{1}{\sqrt{2}}$
Both of these must be true. That means that one of the conditions gives the value for $x$, while the other must give the value for $y$. For example, if $y=0$ then the first statement is true. Then $y\neq\frac{1}{\sqrt{2}}$ and $y\neq-\frac{1}{\sqrt{2}}$, so $x=0$ must be true to make the second statement true.
So $(x,y)=(0,0)$ is possible and gives you a possible value for $z$. In a similiar way, $(x,y)=(\frac{1}{\sqrt{8}},\frac{1}{\sqrt{2}})$,$(x,y)=(\frac{1}{\sqrt{8}},-\frac{1}{\sqrt{2}})$, $(x,y)=(-\frac{1}{\sqrt{8}},\frac{1}{\sqrt{2}})$ and $(x,y)=(-\frac{1}{\sqrt{8}},-\frac{1}{\sqrt{2}})$ are possible. Each of them gives you a value for $xy$ and hence for $z$.
Don't forget to check whether every solution indeed satisfies all equations in the system. You can do this by just calculating the given expressions with values for $x,y,z$.
|
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Find the sum using The question is as follows:
Find the sum:
$1\cdot2 + 2\cdot3 + ... + (n-1)n$
What I have tried so far:
We can write $(n-1)n$ as $\frac{(n+1)!}{(n-1)!}$ which we can also write as $2\cdot\dbinom{n+1}{2}$
I believe it is possible to use the binomial theorem here, setting $a = b = 1$ in $(a+b)^n$. I am not sure how to proceed however.
|
$$1\cdot2 + 2\cdot3 + ... + (n-1)n+\color{red}{n(n+1)-n(n+1)}=$$
$$=1(1+1)+2(2+1)+...+(n-1)n+\color{red}{n(n+1)-n(n+1)}=$$
$$=1^2+2^2+...+n^2+1+2+...+n-n(n+1)=$$
$$=\frac{n(n+1)(2n+1)}{6}+\frac{n(n+1)}{2}-n(n+1)=$$
$$=\frac{n(n+1)(n+2)}{3}-n(n+1)=\frac{n(n^2-1)}{3}$$
|
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Find the minimum $k$
Find the minimum $k$, which $\exists a,b,c>0$, satisfies
$$ \frac{kabc}{a+b+c}\geq (a+b)^2+(a+b+4c)^2$$
My Progress
With the help of Mathematica, I found that when $k=100$, we can take $a=1,b=1,c=1/2$. And I'm pretty sure that $k=100$ is the answer, but I couldn't prove it.
|
Effectively, you want to show
$$\frac{(a+b)^2+(a+b+4c)^2}{abc}(a+b+c) \geqslant 100$$
and you already have a case of equality.
Using homogeneity, we may set $a+b+c=5$, to equivalently show
$$(5-c)^2+(5+3c)^2 \geqslant 20 abc$$
Now $a+b = 5-c$, so for any $c$, we have $ab$ maximized when $a=b$. Thus it is enough to show
$$(5-c)^2+(5+3c)^2 \geqslant 20 \left(\frac{5-c}2\right)^2c$$
which leads us to the polynomial inequality $5(c-1)^2(10-c) \geqslant 0$ which is obvious.
|
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|
Solving a 6th degree polynomial equation I have a polynomial equation that arose from a problem I was solving. The equation is as follows:
$$-x^6+x^5+2x^4-2x^3+x^2+2x-1=0 .$$
I need to find $x$, and specifically there should be a real value where $\sqrt3<x<\sqrt{2+\sqrt2}$, in accordance to the problem I am solving. I know that it would be possible for me to find approximations of the roots of the equation, but I would prefer to know the exact value of this specific root (i.e. with the answer as a surd, with nested surds if required). I am unable to do this as I do not know any method of solving polynomials of degree $> 4$.
If this cannot be done, could you tell me an approximate decimal value of $x$, or at least check that a solution exists within the range I have given (it is possible that I made an error earlier in my algebra).
|
Using a complicated computer algebra system (Mathematica 10.4), we can get all the roots to this equation as radicals. Two roots are complex and the rest are all real (surprisingly).
Module[{roots},
roots = Solve[-x^6 + x^5 + 2 x^4 - 2 x^3 + x^2 + 2 x - 1 == 0, x];
Transpose[{
N[x /. roots],
FullSimplify[Element[x, Reals] /. roots],
x /. roots
}]
] // TableForm
*
*There are two complex roots at $\frac{1}{2} \pm \mathrm{i}\sqrt{3}$.
*There is a real root at $-1$.
*The other three are complicated and real: \begin{align}
1.80194\dots{} &= \frac{1}{3} \left(1+\frac{7^{2/3}}{\sqrt[3]{\frac{1}{2} \left(-1+3 \mathrm{i}
\sqrt{3}\right)}}+\sqrt[3]{\frac{7}{2} \left(-1+3 \mathrm{i}
\sqrt{3}\right)}\right), \\
-1.24698\dots{} &= \frac{1}{3}-\frac{7^{2/3} \left(1+\mathrm{i}
\sqrt{3}\right)}{3\ 2^{2/3} \sqrt[3]{-1+3 \mathrm{i} \sqrt{3}}}-\frac{1}{6}
\left(1-\mathrm{i} \sqrt{3}\right) \sqrt[3]{\frac{7}{2} \left(-1+3 \mathrm{i}
\sqrt{3}\right)}, \\
0.445042\dots{} &= \frac{1}{3}-\frac{7^{2/3} \left(1-\mathrm{i} \sqrt{3}\right)}{3\
2^{2/3} \sqrt[3]{-1+3 \mathrm{i} \sqrt{3}}}-\frac{1}{6} \left(1+\mathrm{i} \sqrt{3}\right)
\sqrt[3]{\frac{7}{2} \left(-1+3 \mathrm{i} \sqrt{3}\right)} \text{.}
\end{align}
Conveniently, the first one is in the interval you require.
These three complicated roots are the roots of the same polynomial @Travis gets.
Looking at the Galois group structure, I believe we cannot dispense with complex numbers in these expressions.
|
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$\sqrt{x+938^2} - 938 + \sqrt{x + 140^2} - 140 = 38$ - I keep getting imaginary numbers $$\sqrt{x+938^2} - 938 + \sqrt{x + 140^2} - 140 = 38$$
My attempt
$\sqrt{x+938^2} + \sqrt{x + 140^2} = 1116$
$(\sqrt{x+938^2} + \sqrt{x + 140^2})^2 = (1116)^2$
$x+938^2 + 2*\sqrt{x+938^2}*\sqrt{x + 140^2} + x + 140^2 = 1116^2$
$2x + 2*\sqrt{x+938^2}*\sqrt{x + 140^2} = 1116^2 - 938^2 - 140^2$
$x + \sqrt{x^2 + 2(938^2 + 140^2)x+(938*140)^2} = 1116^2 - 938^2 - 140^2$
At this point trying to solve for x inside the sqrt in the quadratic gives me an imaginary number. How is it possible to solve this?
|
You received good answers and solutions. However, it seems that you made a few mistakes in your calculations. The line $$2x + 2\sqrt{x+938^2}\sqrt{x + 140^2} = 1116^2 - 938^2 - 140^2$$ is correct. What is not correct is in the next line since $$(x+938^2)(x + 140^2)=x^2+(938^2+140^2)x+938^2\times 140^2$$ I do not know how appeared the factor $2$ in the radical and disappeared in other places.
Using this correction, we then have $$2x+2\sqrt{x^2+(938^2+140^2)x+938^2\times 140^2}=346012$$ Dividing both sides by $2$ and moving the $x$ term to the rhs then gives $$\sqrt{x^2+899444 x+17244942400}=173006-x$$ Squaring again, the $x^2$ disappear and what is left is $$1245456 x-12686133636=0\implies x=\frac{352392601}{34596}$$
But, as Achille Hui answered, replace your numbers by letters to avoid these enormous intermediate numbers.
|
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How can we see that $ \sum_{n=0}^{\infty}\frac{2^n(1-n)^3}{(n+1)(2n+1){2n \choose n}}=(\pi-1)(\pi-3) $? I wonder will it help me so prove it if I was to decompose it into partial fractions?
Mathematica approves of the identity; it is converges. can anyone help me to prove it?
$$
\sum_{n=0}^{\infty}\frac{2^n(1-n)^3}{(n+1)(2n+1){2n \choose n}}=(\pi-1)(\pi-3)
$$
|
Hint. One may observe that, for $n\geq 0$, using the Euler beta function,
$$
\frac{2^n(1-n)^3}{(n+1)(2n+1){2n \choose n}}=-\frac{(n-1)^3}{n+1}\int_0^1(2x(1-x))^ndx. \tag1
$$ Then, one may write
$$\begin{align}
&\sum_0^\infty\frac{2^n(1-n)^3}{(n+1)(2n+1){2n \choose n}}
\\\\&=-\sum_0^\infty\frac{(n-1)^3}{n+1}\int_0^1(2x(1-x))^ndx\\\\
&=-\int_0^1\sum_0^\infty\frac{(n-1)^3}{n+1}(2x(1-x))^n \:dx\\\\
&=\int_0^1\left(\frac{-7+34 x-82 x^2+96 x^3-48 x^4}{(1-2 x+2 x^2)^3}-\frac{4 \log(1-2 x+2 x^2)}{(1-x) x}\right)dx\\\\
&=\underbrace{\int_0^1\frac{-7+34 x-82 x^2+96 x^3-48 x^4}{(1-2 x+2 x^2)^3}\,dx}_{\large \color{blue}{3-4\pi}}+\underbrace{4\int_0^1-\frac{\log(1-2 x+2 x^2)}{(1-x) x}\,dx}_{\large \color{red}{\pi^2}}
\\\\&=\color{blue}{3-4\pi}+\color{red}{\pi^2}
\\\\&=(\pi-1)(\pi-3)
\end{align}
$$
as announced.
|
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Find the infinite simple continued fractions for ... Find the infinite simple continued fractions for $\sqrt{2};\sqrt{5};\sqrt{6};\sqrt{7};\sqrt{8}$.
*
*I have solved similar equations for continued fractions but only using a fraction, if someone could please demonstrate how to do this to ANY of these values I will be good from there just need an example to work off of. I realize I can just get these values off of Wolfram Alpha but I need to know how to actually work through them. Thank you in advance!
|
Just an example: $\sqrt{7}$. Since $4<7<9$, $\left\lfloor \sqrt{7}\right\rfloor = 2$, so:
$$ \sqrt{7} = 2+(\sqrt{7}-2) = \color{blue}{2}+\frac{1}{\frac{\sqrt{7}+2}{3}}\tag{1}.$$
Since $2+\sqrt{7}\in (4,5)$, $\left\lfloor\frac{\sqrt{7}+2}{3}\right\rfloor =1$, so:
$$ \frac{\sqrt{7}+2}{3} = 1+\frac{\sqrt{7}-1}{3} = 1+\frac{1}{\frac{\sqrt{7}+1}{2}}$$
and by plugging this identity back into $(1)$ we get:
$$\sqrt{7}=\color{blue}{2}+\frac{1}{\color{blue}{1}+\frac{1}{\frac{\sqrt{7}+1}{2}}}.\tag{2}$$
Now $\frac{1+\sqrt{7}}{2}\in(1,2)$, hence:
$$ \frac{\sqrt{7}+1}{2}=1+\frac{\sqrt{7}-1}{2} = 1+\frac{1}{\frac{\sqrt{7}+1}{3}}$$
and by plugging it back into $(2)$ we get:
$$\sqrt{7}=\color{blue}{2}+\frac{1}{\color{blue}{1}+\frac{1}{\color{blue}{1}+\frac{1}{\frac{\sqrt{7}+1}{3}}}}.\tag{3}$$
Continuing that way, we have:
$$\frac{\sqrt{7}+1}{3} = 1+\frac{\sqrt{7}-2}{3} = 1+\frac{1}{\sqrt{7}+2}$$
hence:
$$ \color{red}{\sqrt{7}}=[2;1,1,1,2+\sqrt{7}]=\color{red}{[2;1,1,1,\overline{4,1,1,1}]}.\tag{4}$$
Now you just have to check that the same algorithm leads to:
$$ \sqrt{2} = [1;\overline{2}],\quad \sqrt{5}=[2;\overline{4}],\quad \sqrt{6}=[2;\overline{2,4}],\quad \sqrt{8}=[2;\overline{1,4}]$$
(yes, I took the most complex example on purpose).
|
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Can I apply Chinese remainder theorem here?
A number when divided by a divisor leaves $27$ remainder. Twice the number when divided by the same divisor leaves a remainder $3$. Find the divisor.
My attempt:
Let, the number be=$n$ and the divisor be=$d$.
Thus, we have $n\equiv 27\pmod d$ and $2n\equiv 3\pmod d$.
Thus we have two congruence equations. Can we, by any chance, apply the Chinese Remainder theorem to find $d$?
|
Let the number be $n$, and the divisor be $d$.
$\frac{n}{d} = X + \frac{27}{d}$ and also $\frac{2n}{d} = Y + \frac{3}{d}$
Where $X$ and $Y$ are different or same positive integers, comparing the two
$$2X + \frac{54}{d} = Y + \frac{3}{d}$$
$$2X = Y + \frac{3}{d} - \frac{54}{d}$$
$$2X = Y - \frac{51}{d}$$
Now the value of $d$, must be a factor of $51$, the factors are $1, 3, 17$ or $51$.
It could also be the negative of any of those values.
If the question had stated that $n$ and $d$ are in their lowest term
$Y$ is an odd number, because $d$ is also odd
$n > 27$ and $n > X*d$
because $\frac{n}{d} = X + \frac{27}{d}$, if $n$ and $d$ were in their lowest term then $n > d$, this is a proper fraction, then $\frac{27}{d}$ must be the improper part, so that $d > 27$
Therefore d would be $51$
If $X$ is odd, $n$ would be even and vice versa.
|
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Roll 6 dice, find the number of outcomes with 3 distinct numbers. Suppose you roll six dice, how many outcomes are there with 3 distinct numbers.
My attempt:
First there are ${6 \choose 3}$ ways to choose these 3 distinct numbers.
We consider 3 cases;
Case 1: 3 pars of repeated numbers e.g. $223344$. There are ${3\choose 3}$ choices for the values, and for the ordering there are ${6\choose 2} \times {4\choose 2}$.
Case 2: One number reapeated 3 times, one number repeated twice, one number appears once. There are ${3\choose 1}$ choices for the thrice repeated number and ${2\choose 1}$ choices for the twice repeated number. Also there are ${6\choose 3}\times {3\choose 2}$ orderings.
Case 3: One number repeated 4 times and 2 numbers appearing once. There are again ${3\choose 1}$ choices for the number appearing $4$ times and the number of orderings is ${6\choose 4}\times2\times 1$.
Hence in total there are $${6\choose 3}\bigg({6\choose 2} \times {4\choose 2}+3\times2\times {6\choose 3}\times {3\choose 2}+2\times 1\times{6\choose 4}\bigg)$$
Is this correct? I am most uncertain about case 2, is it correct to choose the value for and then still order it?
|
This is correct, except you forgot to include the factor $\binom31$ for case $3$ in the final result. We can double-check the result using inclusion-exclusion: There are $\binom63$ ways to choose the $3$ numbers, and by inclusion-exclusion there are
$$\sum_{k=0}^3(-1)^k\binom3k(3-k)^6=3^6-3\cdot2^6+3\cdot1^6=540$$
ways to form strings of $6$ that use all $3$ of them, for a total of $10800$, in agreement with your result.
|
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Show $3\cos 2x + 1 = 4\cos^2 x - 2\sin^2 x$ Show $3\cos 2x + 1 = 4\cos^2 x - 2\sin^2 x$
Using the formula $\cos 2x = \cos x - \sin^2 x$
I can say $3\cos 2x + 1 = 3(\cos^2 x - \sin^2 x) + 1$
$\Rightarrow 3\cos x^2 - 3\sin^2 x + 1$
But from there I don't see how I can get the answer.
|
You are very close. All you need to do is use the identity $\sin^2{x} + \cos^2{x} = 1$:
$\begin{eqnarray}3\cos{2x} + 1 & = & 3(\cos^2{x}-\sin^2{x}) + 1 \\
& = & 3(\cos^2{x}-\sin^2{x}) + \sin^2{x} + \cos^2{x} \\
& = & 4\cos^2{x} - 2\sin^2{x}
\end{eqnarray}$
QED.
Alternatively, if you don't feel comfortable turning the $1$ into $\sin^2{x}+\cos^2{x}$ (and trust me, it's a perfectly valid thing to do), you can equivalently break out one of the existing $\sin^2{x}$s and replace it with $1 - \cos^2{x}$, and as long as you get all the signs correct it amounts to the same thing.
|
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Finding $a, b, c$ values of a polynomial from a graph. I was doing my homework and I am now stuck on question number 7 which is:
The diagram shows the curve with the equation $y = (x + a)(x - b)^2$ where $a$ and $b$ are positive integers.
(i) Write down the values of $a$ and $b$, and also of $c$, given that the curve crosses the $y$-axis at $(0, c)$.
I have attached a picture of my text book here:
I have so far found $b$ like this:
$(x - b)^2 = x^2 - 2bx + b^2$
Since $y = 0$ at $x = 1$,
$0 = 1 - 2b + b^2$
Solving that quadratic equation gives us $b = 1$.
I need to find $a$ and $c$ and I am totally puzzled on how to find them.
|
From the graph it is evident that the roots are $-2$ and $1$ ($1$ being a repeated root),
So the equation becomes:
$$(x+2)(x-1)^2=0$$
Comparing it with $$(x+a)(x-b)^2=0$$ We find $a=2,b=1$
Also for finding the point $(0,c)$ we can plug in $x=0$ in the equation $y=(x+2)(x-1)^2$ and we get $c=2$
|
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Finding lim${_{n \rightarrow \infty}}\left( \frac{n^3}{2^n} \right)$ For a class of mine we were given a midterm review; however, I just cannot figure out how to finish this one:
Find the limit $$\lim_{n \rightarrow \infty}\left( \dfrac{n^3}{2^n} \right)$$
My attempt so far:
Let $s_n=\dfrac{n^3}{2^n}$.
Note that $2^n=(1+1)^n$. Thus by the Binomial Theorem we have that,
$
(1+1)^n=\sum_{k=0}^{n} {n \choose k}(1)^{n-k}(1)^n
$
Evaluating some of the first couple I get the following terms:
$1+n+\dfrac{n(n-1)}{2}+\dfrac{n(n-1)(n-2)}{6}+\dfrac{n(n-1)(n-2)(n-3)}{24}+...$
I have noticed that for $n<4$, $n^3 \geq 2^n$. However, it seems that when $n\geq 4, n^3 < 2^n$. Thus, would it be possible to make an argument that for $n \geq 4, s_{4}>s_{5}>s_{6}>...$? Therefore, by evaluating for $n=4$ and using some algebra you get the following:
$s_{4}=\dfrac{24n^3}{24+24n+4n(n-1)(n-2)+n(n-1)(n-2)(n-3)} \leq \dfrac{24n^3}{n^4} = \dfrac{24}{n}$
and we know that lim$_{n \rightarrow \infty} \left( \dfrac{1}{n} \right)=0$ and lim$_{n \rightarrow \infty} \left( -\dfrac{1}{n} \right)=0$
Therefore, since $\dfrac{1}{n} \geq s_{4} > s_{5} >s_{6} > ...s_{n}\geq-\dfrac{1}{n}$, by the Squeeze Theorem $s_{n}$ converges to zero as well?
|
Using your first idea, what about using the fact that $$(1+1)^n = \sum_{k=1}^n \binom{n}{k} \geq \binom{n}{4} = \frac{n(n-1)(n-2)(n-3)}{24} \geq \frac{(n-3)^4}{24}$$and
concluding by the squeeze theorem? (as $\frac{24n^3}{(n-3)^4} \xrightarrow[n\to\infty]{} 0$).
|
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|
Evaluate the double integral.
To find:
$$\iint_R(x^2-xy)dA$$
enclosed by
$$y=x, y=3x-x^2,$$
$$x=3x-x^2,$$
$$x=0, x=2,$$
$$y=0, y=2,$$
$$R=((x,y): 0\le x\le 2), (x \le y \le 3x-x^2)$$
$$\int_0^2\int_x^{3x-x^2}(x^2-xy)\,dy\,dx$$
$$\int_0^2 \left. \left(x^2y-x\frac{y^2}{2}\right) \right|_x^{3x-x^2}\,dx$$
where to from here?
|
Given your domain, let's first make a small correction:
\begin{align}
R &= \left\{(x,y) \in \mathbb{R}^2: 0 \leq x\leq 2, x \leq y \leq 3x-x^2 \wedge 2 \right\} \\
&= \left\{(x,y) \in \mathbb{R}^2: 0 \leq x\leq 1, x \leq y \leq 3x-x^2 \right\} \\
&\qquad\cup \left\{(x,y) \in \mathbb{R}^2: 1 < x \leq 2, x \leq y \leq 2 \right\}
\end{align}
So,
\begin{align}
I
&= \int_R x^2 - xy \;\mathrm{d}A\\
& = \int_0^2 \int_x^{3x-x^2 \wedge 2} x^2 - xy \;\mathrm{d}y\,\mathrm{d}x \\
%
%
&= \int_0^1 \int_x^{3x-x^2} x^2 - xy \;\mathrm{d}y\,\mathrm{d}x +
\int_1^2 \int_x^{2} x^2 - xy \;\mathrm{d}y\,\mathrm{d}x \\
%
%
&= \int_0^1 x \left[x y - \frac{y^2}{2} \right]_{x}^{3 x-x^2} \;\mathrm{d}x +
\int_1^2 x \left[x y - \frac{y^2}{2} \right]_{x}^{2} \; \mathrm{d}x \\
%
%
&= \int_0^1 \left(- 2 x^3 + 2 x^4 -\frac{x^5}{2} \right) \;\mathrm{d}x +
\int_1^2 \left(-2 x +2 x^2 -\frac{x^3}{2} \right) \;\mathrm{d}x \\
%
%
&= \left[-\frac{x^4}{2}+\frac{2 x^5}{5}-\frac{x^6}{12} \right]_0^1 +
\left[-x+\frac{2 x^3}{3}-\frac{x^4}{8} \right]_1^2 \\
%
%
&= \left(-\frac{11}{60} \right) +\left(-\frac{5}{24}\right)\\
%
%
&= -\frac{47}{120}
\end{align}
|
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|
Proving that a function maintain a certain equation I need to prove that the function $$g(x,y,z)=f(\frac{1}{y}-\frac{1}{x},xye^{\frac{-z^2}{2}})$$ maintain the equation $$x^2g_x+y^2g_y=-\frac{x+y}{z}g_z$$ while $y\neq0$ and $f(u,v)$ differentiable for all $(u,v)$
How do i prove it without knowing what is $f$?
|
You start by writing $g_x$ as function of $f_u$ and $f_v$, thus
$$g_x = \frac{ \partial g}{\partial x} = \frac{ \partial u}{\partial x} \frac{\partial f}{\partial u} + \frac{ \partial v}{\partial x} \frac{\partial f}{\partial v} = \frac{ \partial \frac{1}{y} -\frac{1}{x}}{\partial x} f_u + \frac{ \partial xy \exp\left(- \frac{z^2}{2} \right)}{\partial x} f_v =x^{-2}f_u +y \exp\left(- \frac{z^2}{2} \right) f_v.$$
Likewise you find
$$g_y = - \frac{1}{y^2} f_u +x \exp\left(- \frac{z^2}{2} \right) f_v \textrm{ and } g_z = -zxy\exp\left(- \frac{z^2}{2} \right)f_v.$$
Then replace $g_x$, $g_y$ and $g_z$ in $$x^2 g_x + y^2 g_y + \frac{x+y}{z} g_z=0$$
by above formulas, and you see that the LHS is indeed $0$.
|
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|
Proof of $1^3+1^3+2^3+3^3+5^3+\cdots +F_n^3=\frac{F_nF_{n+1}^2+(-1)^{n+1}[F_{n-1}+(-1)^{n+1}]}{2}$ Fibonacci series
$F_0=0$, $F_1=1$; $F_{n+1}=F_n+F_{n-1}$
This is a well known identity
$1^2+1^2+2^2+3^2+5^2+\cdots +F_n^2=F_nF_{n+1}$
I was curious and look every websites for a closed form of
$1^3+1^3+2^3+3^3+5^3+\cdots +F_n^3=S_n$, but couldn't find one, so I went and look for it.
It takes me a lot of time through experimental mathematics and deduced a closed form for the cube of the Fibonacci series.
$$1^3+1^3+2^3+3^3+5^3+\cdots +F_n^3=\frac{F_nF_{n+1}^2+(-1)^{n+1}[F_{n-1}+(-1)^{n+1}]}{2}$$
I need someone to verify it by proof. I try next to determine for the 4th power but fail, so I just wonder is there a closed form for nth power Fibonacci series.
|
For a (somewhat complicated) combinatorial proof see Identity 5 in Benjamin, Carnes, Cloitre. Recounting the Sums of Cubes of Fibonacci
Numbers.
|
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|
Derivation of Spherical Law of Cosines I am trying to get a derivation of the spherical law of cosines. The Wikipedia page [https://en.wikipedia.org/wiki/Spherical_law_of_cosines ] contains a proof that I don't understand because there are not enough intermediate steps shown.
The Wikipedia page says that for unit vectors $\mathbf{u},\mathbf{v},\mathbf{w}$ :
$$
\begin{align}
\cos(a) = \mathbf{u}\cdot\mathbf{v}
\end{align}
$$
\begin{align}
\cos(b) = \mathbf{u}\cdot\mathbf{w}
\end{align}
$$
\begin{align}
\cos(c) = \mathbf{v}\cdot\mathbf{w}
\end{align}
$$
The unit vector $\mathbf{t}_a$ is defined as the unit vector perpendicular to $\mathbf{u}$ in the u-v plane, whose direction is given by the component of v perpendicular to u. Wikipedia explains that
$$
\begin{align}
\mathbf{t}_a =
{
{\mathbf{v}-\mathbf{u}(\mathbf{u \cdot v})}
\over
{
\left\lVert
{\mathbf{v}-\mathbf{u}(\mathbf{u \cdot v})}
\right\rVert
}}
=
{{{\mathbf{v}-\mathbf{u}\cos(a)}}
\over
{\sin(a)}}
\end{align}
$$
Similarly,
$$
\begin{align}
{\mathbf{t}_b}
=
{{{\mathbf{w}-\mathbf{u}\cos(b)}}
\over
{\sin(b)}}
\end{align}
$$
Without any further justification, their proof ends by the claim:
$$
\begin{align}
{\mathbf{t}_a} \cdot {\mathbf{t}_b}
=
{{\cos(c)-\cos(a)\cos(b)}
\over
{\sin(a)\sin(b)}}
\end{align}
$$
It is the final step, which yeilds the formula for $\mathbf{t}_a\cdot\mathbf{t}_b$, that I do not comprehend. Can somebody please explain to me the justification for this expression?
|
It looks like you just follow the angle definitions given and distribute.
You have:
$$
\begin{align}
\cos(a) = \mathbf{u}\cdot\mathbf{v}
\end{align}
$$
\begin{align}
\cos(b) = \mathbf{u}\cdot\mathbf{w}
\end{align}
$$
\begin{align}
\cos(c) = \mathbf{v}\cdot\mathbf{w}
\end{align}
$$
$$
\begin{align}
\mathbf{t}_a =
{{{\mathbf{v}-\mathbf{u}\cos(a)}}
\over
{\sin(a)}}
\end{align}
$$
$$
\begin{align}
{\mathbf{t}_b}
=
{{{\mathbf{w}-\mathbf{u}\cos(b)}}
\over
{\sin(b)}}
\end{align}
$$
So:
$$
\begin{align}
{\mathbf{t}_a} \cdot {\mathbf{t}_b}
=
\frac{1}
{\sin(a)\sin(b)}(\mathbf{v}\cdot\mathbf{w}-\mathbf{u}\cdot\mathbf{w}\cos(a)-\mathbf{v}\cdot\mathbf{u}\cos(b)+\mathbf{u}\cdot\mathbf{u}\cos(a)\cos(b))
\end{align}
$$
And simplify.
|
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|
Factor $6x^2 −7x−5=0$ I'm trying to factor
$$6x^2 −7x−5=0$$
but I have no clue about how to do it. I would be able to factor this:
$$x^2-14x+40=0$$
$$a+b=-14$$
$$ab=40$$
But $6x^2 −7x−5=0$ looks like it's not following the rules because of the coefficient of $x$. Any hints?
|
Consider the product
\begin{align*}
(4x + 7)(3x - 8) & = 4x(3x - 8) + 7(3x - 8)\\
& = \color{blue}{12}x^2 \color{green}{- 32}x + \color{green}{21}x \color{blue}{-56}\\
& = \color{blue}{12}x^2 \color{green}{- 11}x \color{blue}{- 56}
\end{align*}
Observe that the product of the quadratic and constant coefficients is equal to the product of the two linear coefficients in the expression $\color{blue}{12}x^2 \color{green}{- 32}x + \color{green}{21}x \color{blue}{-56}$, that is,
$$(\color{blue}{12})(\color{blue}{-56}) = (\color{green}{-32})(\color{green}{21}) = -672$$
Now suppose that $ax^2 + bx + c$ is a quadratic polynomial with integer coefficients that has factorization
\begin{align*}
ax^2 + bx + c & = (rx + s)(tx + u)\\
& = rx(tx + u) + s(tx + u)\\
& = \color{blue}{rt}x^2 + \color{green}{ru}x + \color{green}{st}x + \color{blue}{su}\\
& = \color{blue}{rt}x^2 + (\color{green}{ru + st})x + \color{blue}{su}\\
\end{align*}
Equating coefficients yields
\begin{align*}
a & = \color{blue}{rt}\\
b & = \color{green}{ru + st}\\
c & = \color{blue}{su}
\end{align*}
Note that the product of the quadratic and linear coefficients is equal to the product of the two linear coefficients that have sum $b$, that is,
$$(\color{blue}{rt})(\color{blue}{su}) = (\color{green}{ru})(\color{green}{st})$$
Hence, if a quadratic polynomial $ax^2 + bx + c$ factors with respect to the rational numbers, we can split the linear term into two linear terms whose coefficients have product $ac$ and sum $b$.
To split the linear term of $6x^2 - 7x - 5$, we must find two numbers with product $6 \cdot (-5) = -30$ and sum $-7$. They are $-10$ and $3$. Hence,
\begin{align*}
6x^2 - 7x - 5 & = 0\\
6x^2 - 10x + 3x - 5 & = 0 && \text{split the linear term}\\
2x(3x - 5) + 1(3x - 5) & = 0 && \text{factor by grouping}\\
(2x + 1)(3x - 5) & = 0 && \text{extract the common factor}
\end{align*}
To solve the equation, use the zero product property $ab = 0 \iff a = 0~\text{or}~b = 0$.
|
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|
Prove: $\cos^2 x (\sec x - 1)(\sec x + 1) = (1 - \cos x)(1 + \cos x)$
Prove the trigonometric identity
$$\cos^2 x (\sec x - 1)(\sec x + 1) = (1 - \cos x)(1 + \cos x)$$
I've searched high and low on the net and cannot find identities where there is $+$ or $- 1$'s in the equation. Any help is appreciated.
Edit after reviewing comments.
$\cos^2 {x} (\sec {x} -1)(\sec {x} +1) = (1 - \cos {x})(1 + \cos {x})\quad$
$\cos^2 {x} (\sec^2 {x}- 1) = 1 - \cos^2 {x}\quad$
$\cos^2 {x}·tan^2 {x} = \sin^2 {x} \quad$
$\cos^2 {x} ·\frac{\sin^2 {x}}{\cos^2 {x}} = \sin^2 {x} \quad$
$\frac{\sin^2 {x}}{\cos^2 {x} }=\frac{\sin^2 {x}}{\cos^2 {x} }\quad$
How's that? I tried the $\sec {x }= \frac1{\cos {x}}\quad$ method. However I was unsuccessful.
|
Substitute $\frac1{\cos{x}}\;$ for $\sec{x}\;$ in the identity, and it will shout its truth to you.
|
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|
How do I show that $\ln(2) = \sum \limits_{n=1}^\infty \frac{1}{n2^n}$? My task is this:
Show that $$\ln(2) = \sum \limits_{n=1}^\infty \frac{1}{n2^n}$$
My work so far:
If we approximate $\ln(x)$ around $x = 1$, we get:
$\ln(x) = (x-1) - \frac{(x-1)^2}{2} + \frac{(x-1)^3}{3} - \frac{(x-1)^4}{4} + ...$
Substituting $x = 2$ then gives us:
$\ln(2) = 1 - \frac{1}{2} + \frac{1}{3} - \frac{1}{4} + \frac{1}{5} + ...$
No suprise there, we should always get alternating series for $\ln(x)$ when we are doing the taylor expansion. By using Euler transform which is shown at the middle of this page on natural logarithm one can obtain the wanted result, but how do you derive it? I need someone to actually show and explain in detail how one starts from the left side of the equation and ends up on the other.
Thanks in advance
|
Note that
\begin{align}
\frac{1}{1-x}&=1+x+x^2+\cdots\qquad(-1<x<1).
\end{align}
Thus
\begin{align}
-\ln(1-x)&=\int\frac{1}{1-x}{\rm d}x\\
&=x+\frac{1}{2}x^2+\frac{1}{3}x^3+\cdots\\
&=\sum_{n=1}^\infty\frac{1}{n}x^n.
\end{align}
By taking $x=\frac{1}{2}$ into the above equation, we have
$$\sum_{n=1}^\infty\frac{1}{n2^n}=-\ln\left(\frac{1}{2}\right)=\ln(2).$$
|
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|
is $y = \sqrt{x^2 + 1}− x$ a injective (one-to-one) function? I have a function $y = \sqrt{x^2 + 1}− x$ and I need to prove if it's a Injective function (one-to-one). The function f is injective if and only if for all a and b in A, if f(a) = f(b), then a = b
$\sqrt{a^2 + 1} − a = \sqrt{b^2 + 1} − b$
$\sqrt{a^2 + 1} + b = \sqrt{b^2 + 1} + a$
$(\sqrt{a^2 + 1} + b)^2 = (\sqrt{b^2 + 1} + a)^2$
$a^2 + 1 + b^2 + 2b\sqrt{a^2 + 1} = b^2 + 1 + a^2 + 2a\sqrt{b^2 + 1}$
$2b\sqrt{a^2 + 1} = 2a\sqrt{b^2 + 1}$
$b\sqrt{a^2 + 1} = a\sqrt{b^2 + 1}$
$(b\sqrt{a^2 + 1})^2 = (a\sqrt{b^2 + 1})^2$
$b^2(a^2 + 1) = a^2(b^2 + 1)$
$b^2a^2 + b^2 = a^2b^2 + a^2$
$b^2 = a^2$
$\sqrt{b^2} = \sqrt{a^2}$
$b = \pm a$
then the function is not injective because $b$ in not equal to $a$, but using online solver the function is injective.
|
Hint:
$$y=\sqrt{x^2+1}-x=\frac1{\sqrt{x^2+1}+x}$$
Then $$f(a)=f(b) \Leftrightarrow \frac1{\sqrt{a^2+1}+a}=\frac1{\sqrt{b^2+1}+b}\Leftrightarrow$$
$$\Leftrightarrow \sqrt{a^2+1}+a=\sqrt{b^2+1}+b \Leftrightarrow a=b$$
$f(t)=\sqrt{t^2+1}+t -$ strictly increasing
|
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|
sum of all non real roots of the equation in a bi-quadratic equation
Consider the equation $8x^4-16x^3+16x^2-8x+a=0\;\left(a\in \mathbb{R}\right)\;,$ Then the sum of
all non real roots of the equation can be
$\bf{OPTIONS::}\;\; (a)\;\; 1\;\;\;\;\;\; (b)\;\; 2\;\;\;\;\;\; (c)\; \displaystyle \frac{1}{2}\;\; \;\;\;\; (d)\;\; None$
$\bf{My\; Try::}$ Let $f(x)=8x^4-16x^3+16x^2-8x+a\;,$ Then $f'(x)=32x^3-48x^2+32x-8$
And $\displaystyle f''(x) = 96x^2-96x+32 = 96\left[x^2-x+\frac{1}{3}\right]=96\left[\left(x-\frac{1}{2}\right)^2+\frac{1}{12}\right]>0\;\forall x \in \mathbb{R}$
So Using $\bf{LMVT\;,}$ We get $f'(x)=0$ has at most $1$ real roots and
$f(x)=0$ has at most $2$ real roots
Now How can i solve it after that, Help me
Thanks
|
Now How can i solve it after that
Noting that $f'(1/2)=0$ is a key.
We have
$$f'(x)=8(4x^3-6x^2+4x-1)=8(2x-1)\left(2\left(x-\frac 12\right)^2+\frac 12\right)$$
So, $f(x)$ is decreasing for $x\lt 1/2$ and is increasing for $x\gt 1/2$.
By the way,
$$f\left(\frac 12+s\right)=8s^4+4s^2-\frac 32+a$$
and so $f(1/2-\alpha)=0$ is equivalent to $f(1/2+\alpha)=0$.
Also, by Vieta's formula, the sum of roots is $-(-16)/8=2$.
If $a\le 3/2$, then $f(1/2)\le 0$, and the answer is
$$2-(1/2-\alpha)-(1/2+\alpha)=\color{red}{1}$$
If $a\gt 3/2$, then $f(1/2)\gt 0$, and the answer is $\color{red}{2}$.
|
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|
Solution to a 2D recurrence equation I am seeking an explicit solution to this 2D recurrence equation:
\begin{eqnarray}
f(0,b) & = & b\\
f(a,0) & = & a\\
f(a,b) & = & f(a-1,b) - f(a,b-1)
\end{eqnarray}
So, for example, for $a=3$,
$$f(3,0)=3 \;,$$
$$f(3,1)=f(2,1)-f(3,0)=-2-3=-5 \;,$$
etc. Here is $f(3,b)$ for $b=0,\ldots,10$:
$$
3, -5, 9, -12, 18, -22, 30, -35, 45, -51, 63 \;.
$$
Note the sums of adjacent terms of this list follow a clear pattern:
$$
-2, 4, -3, 6, -4, 8, -5, 10, -6, 12 \;.
$$
Plot cropped above and below. $f(10,10)=146760$.
Can anyone see or calculate an explicit solution as a function of $(a,b)$?
|
Consider the generating function $g_n(x) = \sum_{k=0}^n f(k,n-k) x^k$. Thus $g_0(x) = f(0,0) = 0$ and
$$g_n(x) = n + n x^n + \sum_{k=1}^{n-1} (f(k-1,n-k) - f(k,n-1-k)) x^k
= 2n-1 + x^n + (x-1) g_{n-1}(x)$$
which has solution (according to Maple)
$$ g_n(x) = -2\,{\frac {n+1}{x-2}}-{\frac {-x+4}{ \left( x-2 \right) ^{2}}}+{x}^{
n+1}-{\frac {x \left( -3+x \right) \left( x-1
\right) ^{n+1}}{ \left( x-2 \right) ^{2}}}
$$
Despite appearances, this is in fact a polynomial in $x$, i.e. the singularity at $x=2$ is removable.
Then $f(k,m)$ is the coefficient of $x^k$ in $g_{m+k}(x)$.
|
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|
Indicate on an Argand Diagram the region of the complex plane in which $ 0 \leq \arg (z+1) \leq \frac{2\pi}{3} $
Question: Indicate on an Argand Diagram the region of the complex plane in which $$ 0 \leq \arg (z+1) \leq \frac{2\pi}{3} $$
I've tried this
Consider $$ 0 \leq \arg (z+1) \leq \frac{2\pi}{3} $$
Split it into two inequalities : $ 0 \leq \arg (z+1) $ and $\arg (z+1) \leq \frac{2\pi}{3} $
Then:
$$ z+1 \Leftrightarrow x+iy+1 \Leftrightarrow (x+1)+i(y) $$
$$ \tan^{-1}\left(\frac{\Im(z)}{\Re(z)} \right) = \arg(z+1)$$
$$ \Leftrightarrow \tan^{-1}\left(\frac{y}{x+1} \right) = \arg(z+1)$$
So for the first inequality
$$ 0 \leq \arg (z+1) $$
$$ 0 \leq \tan^{-1}\left(\frac{y}{x+1} \right) $$
$$ 0 \leq \frac{y}{x+1} $$
$$ 0 \leq y ~~ x \neq -1 $$
Second Inequality:
$$\tan^{-1}\left(\frac{y}{x+1} \right) \leq \frac{2\pi}{3} $$
$$ \frac{y}{x+1} \leq -\sqrt{3} $$
$$ y \leq -\sqrt{3} (x+1) $$
So we have two conditions
$$ 0 \leq y ~~ x \neq -1 $$
and $$ y \leq -\sqrt{3} (x+1) $$
How do I draw these?
|
You switched the sign of the inequality somewhere. It should be $y\ge-\sqrt3(x+1)$. So you plot the two lines $y=0$ and $y=-\sqrt3(x+1)$ and mark the sector between them as shown.
|
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"language": "en",
"url": "https://math.stackexchange.com/questions/1784619",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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|
How to find the integral of a quotient of rational functions? How do I compute the following integral: $$\int \dfrac{x^4+1}{x^3+x^2}\,dx$$
My attempt:
We can write $$\dfrac{x^4+1}{x^3+x^2} = \dfrac{A}{x^2} + \dfrac{B}{x} + \dfrac{C}{x+1}$$
It is easy to find that
$A=1$,
$B=2$, and
$C=-1$.
Therefore
$$\frac{x^4+1}{x^3+x^2} = \frac{1}{x^2} + \frac{2}{x} - \frac{1}{x+1}$$
Therefore:
$$\int \frac{x^4+1}{x^3+x^2}\,dx = \int \frac{dx}{x^2} + \int \dfrac{2\,dx}{x} - \int \frac{dx}{x+1} = -\frac{1}{x} +2\log \vert x\vert - \log \vert x+1 \vert + C$$
The problem is I was supposed to find: $$\int \frac{x^4+1}{x^3+x^2}\,dx = \frac{x^2}{2} - x - \frac{1}{2} - \log \vert x \vert + 2 \log \vert x+1 \vert + C$$
Where is my mistake?
|
Your long division is wrong, because:
$$\frac{x^4+1}{x^3+x^2}=\frac{x^4+1}{x^2(x+1)}=\frac{1}{x^2}+x+\frac{2}{x+1}-\frac{1}{x}-1$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1784930",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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|
Find $\int_{\pi /4}^{65\pi /4} \frac{dx}{(1+2^{\cos x})(1+2^{\sin x})}$ Find the value of: $$I=\int_{\pi /4}^{65\pi /4} \frac{dx}{(1+2^{\cos x})(1+2^{\sin x})}$$
First, I rewrote the limits as the function goes from $\frac{\pi}{4}$ to $\frac{9\pi}{4}$. Now the integral with the reduced limits has the value $\frac{I}{8}$. I used standard properties of definite integrals like converting $f(x)$ to $f(a+b-x)$ where $a,b$ are the lower and upper limits of the integral. Considering symmetry of the trigonometric functions, I believe I can even change the limits to $0$ to $2\pi$. However, this does not simplify the problem as much as I would like. Please advice.
|
\begin{align*}
I&=\int_{\pi /4}^{65\pi /4} \frac{dx}{(1+2^{\cos x})(1+2^{\sin x})} \\
&= 8\int_{-\pi}^{\pi} \frac{dx}{(1+2^{\cos x})(1+2^{\sin x})} \tag{1}\\
&= 8\int_{-\pi}^{\pi}\frac{dx}{\left(1+2^{\cos (-x)}\right)\left(1+2^{\sin (-x)}\right)} \tag{2}\\
&=8\int_{-\pi}^{\pi} \frac{dx}{\left(1+2^{\cos x}\right)\left(1+2^{-\sin x}\right)} \tag{3}\\
&= 8\int_{-\pi}^{\pi} \frac{2^{\sin x} dx}{\left(1+2^{\cos x}\right)\left(1+2^{\sin x}\right)} \tag{4}\\
\implies 2I &= 8\int_{-\pi}^{\pi} \frac{1+2^{\sin x} dx}{\left(1+2^{\cos x}\right)\left(1+2^{\sin x}\right)} \tag{5}\\
&= 8 \int_{-\pi}^{\pi} \frac{dx}{1+2^{\cos x}} \\
\implies I &= 4 \int_{-\pi}^{\pi} \frac{dx}{1+2^{\cos x}} \\
&= 8 \int_{0}^{\pi} \frac{dx}{1+2^{\cos x}} \tag{6} \\
&= 8 \int_{0}^{\pi} \frac{dx}{1+2^{\cos (\pi-x)}} \tag{7} \\
&= 8 \int_{0}^{\pi} \frac{dx}{1+2^{-\cos x}} \tag{8} \\
&= 8 \int_{0}^{\pi} \frac{2^{\cos x} dx}{1+2^{\cos x}} \tag{9} \\
\implies 2I &= 8\int_{0}^{\pi} \frac{1+2^{\cos x} dx}{1+2^{\cos x}} \tag{10} \\
\implies I &= 4 \int_{0}^{\pi} dx \\
&= \color{red}{4 \pi}
\end{align*}
In $(1)$, we have used the periodicity of $\sin x$ and $\cos x$ and brought the limits down to more familiar territory.
In $(2)$, we have used the $a+b-x$ trick.
In $(3)$ we used that $\sin x$ is odd and $\cos x$ is even.
In $(5)$ we have added $(1)$ to $(4)$
In $(6)$ we use that $\cos x$ is even.
In $(7)$, $a+b-x$ again.
In $(8)$, $\cos(\pi-x) = -\cos x$
In $(10)$, we do $(6) + (9)$
|
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"timestamp": "2023-03-29T00:00:00",
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|
Derivatives using matrices good $$\left|\begin{matrix}
(1+x)^{a_1b_1} & (1+x)^{a_1b_2} & (1+x)^{a_1b_3} \\
(1+x)^{a_2b_1} & (1+x)^{a_2b_2} & (1+x)^{a_2b_3} \\
(1+x)^{a_3b_1} & (1+x)^{a_3b_2} & (1+x)^{a_3b_3} \\
\end{matrix}\right |$$
Find the coefficient of $x$ in the above determinant.
Try using differentiability.
|
Denote the determinant by $f(x)$. Then the $x$ coefficient is given by $f'(0)$. Now the derivative of a determinant equals the sum of determinants of the matrices obtained by taking the derivative of each row separately. Hence
$$
\begin{align}
f'(0)
&=
\frac{\mathrm{d}}{\mathrm{d}x}
\left|
\begin{matrix}
(1+x)^{a_1b_1} & (1+x)^{a_1b_2} & (1+x)^{a_1b_3} \\
(1+x)^{a_2b_1} & (1+x)^{a_2b_2} & (1+x)^{a_2b_3} \\
(1+x)^{a_3b_1} & (1+x)^{a_3b_2} & (1+x)^{a_3b_3} \\
\end{matrix}
\right|_{x=0} \\
&=
\left|
\begin{matrix}
a_1b_1 & a_1b_2 & a_1b_3 \\
1 & 1 & 1 \\
1 & 1 & 1 \\
\end{matrix}
\right|
+
\left|
\begin{matrix}
1 & 1 & 1 \\
a_2b_1 & a_2b_2 & a_2b_3 \\
1 & 1 & 1 \\
\end{matrix}
\right|
+
\left|
\begin{matrix}
1 & 1 & 1 \\
1 & 1 & 1 \\
a_3b_1 & a_3b_2 & a_3b_3 \\
\end{matrix}
\right| \\
&= 0
\end{align}
$$
|
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"timestamp": "2023-03-29T00:00:00",
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|
Integrate $ \int \frac{1}{1 + x^3}dx $ $$ \int \frac{1}{1 + x^3}dx $$
Attempt:
I added and subtracted $x^3$ in the numerator but after a little solving I can't get through.
|
$$\begin{align}
\int\frac{1}{1+x^3}\,dx
&=\int\frac{1-x^2+x^2}{1+x^3}\,dx\\
&=\int\frac{1-x^2}{1+x^3}\,dx+\int\frac{x^2}{1+x^3}\,dx\\
&=\int\frac{1-x}{1-x+x^2}\,dx+\frac13\ln\left(1+x^3\right)+C\\
&=\int\frac{1-(u+1/2)}{1-(u+1/2)+(u+1/2)^2}\,du+\frac13\ln\left(1+x^3\right)+C\\
&=\int\frac{1/2-u}{3/4+u^2}\,du+\frac13\ln\left(1+x^3\right)+C\\
&=\frac12\int\frac{1}{3/4+u^2}\,du-\int\frac{u}{3/4+u^2}\,du+\frac13\ln\left(1+x^3\right)+C\\
&=\frac12\frac{1}{\sqrt{3/4}}\arctan\left(\frac{u}{\sqrt{3/4}}\right)-\frac12\ln\left(3/4+u^2\right)+\frac13\ln\left(1+x^3\right)+C\\
&=\frac{1}{\sqrt{3}}\arctan\left(\frac{x-1/2}{\sqrt{3/4}}\right)-\frac12\ln\left(1-x+x^2\right)+\frac13\ln\left(1+x^3\right)+C\\
\end{align}$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1788443",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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|
What is a sumset? Say we have sets $X = \{0,2,3\}$ and $Y = \{1,2,5\}$.
Is the sumset defined to be
$X + Y = \{0 + 1,2 +2,3+5\} = \{1,4,8\}$
or summing every element pairwise
$X + Y = \{1,2,5,3,4,7,8\} $
?
|
$$ A + B = \{a+b : a \in A, b \in B\}. $$
If $X = \{0,2,3\}$ and $Y = \{1,2,5\}$.
$$X+Y=\{0+1;0+2;0+5;2+1;2+2;2+5;3+1;3+2;3+5\}$$
|
{
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"timestamp": "2023-03-29T00:00:00",
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|
538.com's Puzzle of the Overflowing Martini Glass - How to compute the minor and major axis of an elliptical cross-section of a cone FiveThirtyEight.com Riddler Puzzle / May 13
The puzzle goes like this;
"It’s Friday. You’ve kicked your feet up and have drunk enough of your martini that, when the conical glass () is upright, the drink reaches some fraction p of the way up its side. When tipped down on one side, just to the point of overflowing, how far does the drink reach up the opposite side?"
I came up with an an answer of $ p^2 $, based on the following (probably incorrect) solution;
$
\begin{aligned}
\overline{AE} &= \overline{AD} &&= 1 \dots \text{length of one side}\\
\overline{AC} &= p &&=\text{initial distance of liquid along one side from the bottom}\\
\overline{AB} &= q &&= \text{distance when glass is tipped over to the other side}\\
h_p &= p \cos\theta &&= \text{height of upright triangle} \\
b_p &= 2p \sin\theta &&= \text{base of upright triangle} \\
h_q &= \overline{AE} \sin(2\theta) &&= \text{height of tipped-over triangle} \\
&= 2\sin\theta\cos\theta\\
b_q &= q &&= \text{base of tipped-over triangle} \\
&&&\\
\triangle(ACF) &= \left( \displaystyle \frac{h_p b_p}{2} \right) &&= p^2\sin\theta \cos\theta\\
\triangle(ABE) &= \left(\displaystyle\frac{h_q b_q}{2}\right) &&= q\sin\theta\cos\theta&&\\
\end{aligned}
$
$\begin{aligned}
\text{As } & \text{volume of cone AEB} &&= \text{volume of cone ACF,}\\
& \triangle(ACF) &&= \triangle(ABE) \dots \text{(leap of faith?)}\\
& q &&= p^2 \\
\end{aligned}
$
So the questions are;
a) Is it valid to assume that equal volumes imply equal vertical cross-sections?
b) How to derive the major and minor axes of the elliptical cross section $BE$ (so we can state the equation for the tilted volume in terms of $\overline{AB}$, and ultimately state $\overline{AB}$ as a function of $p$)?
|
Diagram below shows side and aerial views of upright and tilted martini glass.
For the upright glass, the shape of the martini liquid is an inverted right circular cone, with height $p\sin\theta$ and circular base of area $A_1$. Volume of martini is given by:
$$\begin{align}
V&=\frac 13\cdot A_1 \cdot p\cos\theta\\
&=\frac 13\cdot \pi (p\sin\theta)^2\cdot p\cos\theta\\
&=\frac 13\cdot \pi p\sin\theta\cdot \underbrace{\left(\frac 12 \cdot p^2\cdot \sin 2\theta\right)}_{\text{area of sideview cross-section}}
&&\cdots (1)
\end{align}$$
For the tilted glass, the shape of the martini liquid is an inverted ellipsoidal cone, with height $h$ and ellipse base with area $A_2$ and semi-major and semi-minor axes of $a$ and $b$ respectively. Volume of martini is given by:
$$\begin{align}
V&=\frac 13\cdot A_2 \cdot h\\
&=\frac 13\cdot \pi ab\cdot h\\
&=\frac 13 \cdot \pi b \cdot \underbrace{\left(\frac 12 \cdot 2a\cdot h\right)}_{\text{area of sideview cross-section}}\\
&=\frac 13 \cdot \pi b \cdot \underbrace{\left(\frac 12 \cdot q\cdot 1\cdot \sin 2\theta\right)}_{\text{area of sideview cross-section}}\\
\end{align}$$
Also, when glass is tilted, top surface of martini liquid changes from a circle to an ellipse, elongating along direction of tilt, with width remaining unchanged, i.e semi-minor axis of ellipse is equal to radius of original circle, or $b=p\sin\theta$. Hence
$$V=\frac 13 \cdot \pi p\sin\theta \cdot \left(\frac 12 \cdot q\cdot 1\cdot \sin 2\theta\right)\qquad\cdots (2)$$
Volume remains the same:
$(1)=(2)$ :
$$\begin{align}
\frac 13\cdot \pi p\sin\theta\cdot \left(\frac 12 \cdot p^2\cdot \sin 2\theta\right)
&=\frac 13 \cdot \pi p\sin\theta \cdot \left(\frac 12 \cdot q\cdot 1\cdot \sin 2\theta\right)\\
\underbrace{\frac 12 \cdot p^2\cdot \sin 2\theta}_{\text{area of upright sideview cross-section}}&=\underbrace{\frac 12 \cdot q\cdot 1\cdot \sin 2\theta}_{\text{area of tilted sideview cross-section}}\\\\
\color{red}{q}&\color{red}{=p^2}\qquad\blacksquare
\end{align}$$
The above also shows that the area of the sideview cross-section remains the same before and after tilting.
Special Note:
In the name of science and mathematics, I ordered myself a martini!
Pics shown below. Toothpick measures diameter of circle and minor axis of ellipse, and shows that they are both the same. Olive marks centre of circle/ellipse respectively.
Upright martini glass
Tilted martini glass
Edit
This section has been added to address the point raised by @DavidK about the assumption on the ellipse semi-minor axis.
Recap:
By equating volumes, we have
$$\begin{align}
V_\text{upright}&=V_\text{tilted}\\
\frac 13\cdot \pi (p\sin\theta)^2 \cdot (p\cos\theta)&=\frac 13\cdot \pi ab \cdot h\\
p\sin\theta\cdot \left(\frac 12\cdot 2p\sin\theta\cdot p\cos\theta\right)&=b\cdot \left(\frac 12 \cdot 2a\cdot h\right)\\
p\sin\theta\cdot \left(\frac 12 p^2 \sin 2\theta\right)&=b\cdot \left(\frac 12\cdot q\cdot 1\cdot \sin 2\theta\right)\\
p^3\sin\theta&=bq&&(1)
\end{align}$$
Now consider the side view of the tilted glass (see diagram below).
From the diagram it is clear that
$$\begin{align}
r&=\frac 12 (1+q)\sin\theta\\
d&=\frac 12 (1-q)\sin\theta\\
b&=\sqrt{r^2-d^2}\\
&=\sqrt {q} \sin\theta&&(2)
\end{align}$$
From $(1), (2)$,
$$\begin{align}
p^3\sin\theta&=\sqrt{q}\sin\theta\cdot q\\
&=q^{3/2}\sin\theta\\
\color{red}q&\color{red}{=p^2}\qquad\blacksquare\end{align}$$
Substuting the result in $(2)$ gives
$$b=p\sin\theta$$.
Additional Note:
From additional analysis it can be seen that, for a conical glass,
constant volume implies constant lateral width, i.e. $b=p\sin\theta$, and vice versa.
|
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"timestamp": "2023-03-29T00:00:00",
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|
If $x^4 + 3\cos(ax^2 + bx +c) = 2(x^2-2) $ has two solutions with $a,b,c \in (2,5)$, then find the maximum value of $\frac{ac}{b^2} $ The answer given is 1. i tried like this $3\cos(ax^2 + bx +c) = -x^4 +2x^2-4 = -(x^2 -1)^2 -3 $. The maximum value of $-x^4 +2x^2-4$ is $-3$ so $3\cos(ax^2 + bx +c) =-3$ and the two values of x are $1$ and $-1$. Now $\cos(a + b +c) =-1$ and $\cos(a - b +c) =-1$. How do i find maximum value of $ \frac{ac}{b^2} $
|
You are on the right track. The next step is to utilize the condition $a, b, c \in (2, 5)$.
Since $\cos(a + b + c) = \cos(a - b + c)$ and $a + b + c > a - b + c$, we have
$$a + b + c = a - b + c + 2k\pi$$
for some positive integer $k$, implying $b = k\pi$. Since $b \in (2, 5)$, it can be seen that
$k = 1$, and $b = \pi$. Additionally, $\cos(a + b + c) = -1$, $a \in (2, 5)$, $c \in (2, 5)$ and $b = \pi$ imply that
$$a + b + c = (2\ell + 1)\pi$$
with $(2\ell + 1)\pi \in (4 + \pi, 10 + \pi)$, hence $\ell \in (2/\pi, 5/\pi)$, that is $\ell = 1$. It then follows that $a + c = 2\pi$.
By AM-GM inequality, when $a = c = \pi$:
$$ac \leq \left(\frac{a + c}{2}\right)^2 = \pi^2.$$
Recall $b = \pi$, thus
$$\frac{ac}{b^2} \leq \frac{\pi^2}{\pi^2} = 1.$$
|
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|
Prove $\left(\frac{a+1}{a+b} \right)^{\frac25}+\left(\frac{b+1}{b+c} \right)^{\frac25}+\left(\frac{c+1}{c+a} \right)^{\frac25} \geqslant 3$ $a,b,c >0$ and $a+b+c=3$, prove
$$\left(\frac{a+1}{a+b} \right)^{\frac25}+\left(\frac{b+1}{b+c} \right)^{\frac25}+\left(\frac{c+1}{c+a} \right)^{\frac25} \geqslant 3$$
I try to apply AM-GM
$$\left(\frac{a+1}{a+b} \right)^{\frac25}+\left(\frac{b+1}{b+c} \right)^{\frac25}+\left(\frac{c+1}{c+a} \right)^{\frac25} \geqslant 3\cdot \sqrt[3]{\left(\frac{a+1}{a+b} \right)^{\frac25}\left(\frac{b+1}{b+c} \right)^{\frac25}\left(\frac{c+1}{c+a} \right)^{\frac25}}$$
Thus it remains to prove
$$\left(\frac{a+1}{a+b} \right)\left(\frac{b+1}{b+c}\right)\left(\frac{c+1}{c+a} \right) \geqslant 1 $$ with the condition $a+b+c=3.$
But I found the counter example for $$\left(\frac{a+1}{a+b} \right)\left(\frac{b+1}{b+c}\right)\left(\frac{c+1}{c+a} \right) \geqslant 1 $$ :(
|
Partial Hint :
With your work one can show that :
$$\left(\frac{a+1}{3-x-a+a} \right)\left(\frac{3-x-a+1}{3-x-a+x}\right)\left(\frac{x+1}{x+a} \right) \geqslant 1 $$
With $1\leq a\leq 2$ and $x\in[2-a,1]$
I cannot prove it but using Bernoulli's inequality we have :
$$ \left(\frac{a+1}{x+a} \right)^{\frac25}+\frac{1}{1+\frac{2}{5}\left(\frac{3-x-a+a}{3-x-a+1}-1\right)}+\left(\frac{x+1}{3-x-a+x} \right)^{\frac25}\geq 3$$
With $x\in[0,1]$ and $1\leq a\leq 1.5$
|
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "24",
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|
How to prove $\sum_{n=1}^{\infty} \frac{3}{\sqrt[3]{n^2+2}}$ diverges? $$\sum_{n=1}^{\infty} \frac{3}{\sqrt[3]{n^2+2}}$$
It seems clear to me that this series diverges because the dominant temr is $1/n^{2/3}$, a p-series with $p < 1$
However I need to prove divergence using something like the comparison test, integral test, or similar.
I can't work out a suitable comparison to make to prove divergence. Suggestions?
|
Note that for all positive integers $n$,
$$n^2 + 2 < n^2 + 4n + 4 = (n+2)^2.$$ Therefore, $$\frac{3}{\sqrt[3]{n^2 + 2}} > \frac{3}{\sqrt[3]{(n+2)^2}} > \frac{1}{(n+2)^{2/3}},$$ hence $$\sum_{n=1}^\infty \frac{3}{\sqrt[3]{n^2+2}} > \sum_{n=1}^\infty \frac{1}{(n+2)^{2/3}} = - 1^{2/3} - \frac{1}{2^{2/}} + \sum_{n=1}^\infty \frac{1}{n^{2/3}},$$ the last sum of which is a divergent $p$-series, hence the given sum also diverges.
|
{
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"timestamp": "2023-03-29T00:00:00",
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|
Finding the area of a circle that is formed by cutting a sphere. Say I have a sphere $x^2+y^2+z^2=a^2$ and a plane $x+y+z=b.$ How do I find the surface area of the circle cut by the sphere on the plane? I think I would use the surface integral and for graphs the surface "element" is $\sqrt{1+f_x(x,y)^2+f_y(x,y)^2}$ where $f_x$ and $f_y$ are the derivatives of $x$ and $y$ accordingly. What would $f(x,y)$ be here exactly? Or should I project this circle onto $xOy$ plane, have $f(x,y)=b-x-y$. Either way I don't know how to project the surface onto the $xOy$ plane, where it is most likely a ellipse, where then I could parameterize the domain and go on from there.
I would really like to know how to project this onto $xOy$, because I can use this for other problems.
|
By symmetry the center of the circle is $c=\left(\frac b3,\frac b3,\frac b3\right)$, as noted in the comments. By the Pythagoras's theorem the radius of the circle is
$$r=\sqrt{a^2-|c|^2}=\sqrt{a^2-\frac{1}{3}b^2}$$
whenever the expression is defined, otherwise there is no intersection between the sphere and the plane, and the area is zero. Therefore the area of the circle is
$$A_C=\pi r^2=\pi\left(a^2-\frac{1}{3}b^2\right).$$
You can obtain the same result projecting on the $xOy$ plane. Substitute for $z=b-x-y$ in the sphere equation to obtain
$$x^2+y^2+(b-x-y)^2=a^2. $$
Now we know that the projection will be an ellypse (every affine image of a circle is), with principal axes in direction parallel to $(1,1)$ and $(1,-1)$ (again by symmetry). Therefore we make the orthonormal change of variable $$\begin{cases}w=\frac{1}{\sqrt 2}(x+y)\\z=\frac{1}{\sqrt2}(x-y)\end{cases}$$
that makes the coordinate axes parallel to the principal axes of the ellypse, preserving areas, in order to obtain an equation of an ellypse in the normal form. The equation becomes
\begin{align}
&\frac12(w+z)^2+\frac12(w-z)^2+(b-\sqrt2 w)^2=a^2\\
& z^2+3w^2-2\sqrt2 bw+b^2=a^2\\
& z^2+3\left(w-\frac{\sqrt2}{3}b\right)^2=a^2-\frac13 b^2
\end{align}
which is the equation of an ellypse with center $(w,z)=\left(\frac{\sqrt 2}{3}b,0\right)$ and semiaxes $\ell_1=\frac{1}{\sqrt3}\sqrt{a^2-\frac13 b^2}$ and $\ell_2=\sqrt{a^2-\frac13 b^2}$, with area $A_E=\pi \ell_1\ell_2=\frac{\pi}{\sqrt 3}\left(a^2-\frac13 b^2\right)$, again under the condition that $a^2-\frac13 b^2$ be non negative.
Now the surface element of $f(w,z)=b-x-y=b-\sqrt 2 w$ is
$$\sqrt{1+f_w(w,z)^2+f_z(w,z)^2}=\sqrt{1+(-\sqrt2)^2}=\sqrt3$$
which integrated on the ellypse gives again
$$A_C=\sqrt3 A_E=\pi\left(a^2-\frac13 b^2\right).$$
|
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|
Let $z \in\Bbb{C}^{\times}$ such that $|z^3+\frac{1}{z^3}|\leq 2$. Prove that $|z+\frac{1}{z}|\leq 2$. Problem :
Let $z \in\Bbb{C}^{\times}$ such that $|z^3+\frac{1}{z^3}|\leq 2$. Prove that $|z+\frac{1}{z}|\leq 2$.
My approach :
Since :
$(a^3+b^3)=(a+b)^3-3ab(a+b)$
$\Rightarrow z^3+\frac{1}{z^3}=(z+\frac{1}{z})^3-3(z+\frac{1}{z})$
Now I don't know further about this problem whether or not this will help here. Please guide to solve this problem will be of great help, thanks.
|
You are on the right path; you have shown that
$$\left(z+\tfrac{1}{z}\right)^3=z^3+\tfrac{1}{z^3}+3\left(z+\tfrac{1}{z}\right).$$
Taking norms it follows from the triangle inequality that
$$\left|\left(z+\tfrac{1}{z}\right)^3\right|=\left|z^3+\tfrac{1}{z^3}+3\left(z+\tfrac{1}{z}\right)\right|\leq\left|z^3+\tfrac{1}{z^3}\right|+\left|3\left(z+\tfrac{1}{z}\right)\right|,$$
and rewriting and rearranging the terms a bit yields
$$\left|z+\tfrac{1}{z}\right|\cdot\left(\left|z+\tfrac{1}{z}\right|^2-3\right)\leq\left|z^3+\tfrac{1}{z^3}\right|\leq2.$$
Let $x:=\left|z+\tfrac{1}{z}\right|$ so that we above becomes $x\cdot(x^2-3)\leq2$, or equivalently $x^3-3x-2\leq0$.
Using the rational root test if you like, it's easy to see that $x^3-3x-2=(x-2)(x+1)^2$, from which it follows that $x\leq2$ as desired.
|
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|
Limit of ($\sqrt{x^2+8x}-\sqrt{x^2+7x}$) as $x$ approaches infinity I've been stuck on this one problem for 3 days now, I don't know how to proceed. Any help would be appreciated.
The problem is asking for the $$\lim_{x\to\infty} (\sqrt{x^2+8x}-\sqrt{x^2+7x}) $$
Every time I attempt this problem, I can never get rid of infinity in the numerator without making the denominator zero. How can I do this problem?
|
You want
$\lim_{x\to\infty} (\sqrt{x^2+8x}-\sqrt{x^2+7x})
$.
For reals $a$ and $b$,
$\begin{array}\\
\sqrt{x^2+ax}-\sqrt{x^2+bx}
&=(\sqrt{x^2+ax}-\sqrt{x^2+bx})\dfrac{\sqrt{x^2+ax}+\sqrt{x^2+bx}}{\sqrt{x^2+ax}+\sqrt{x^2+bx}}\\
&=\dfrac{(\sqrt{x^2+ax}-\sqrt{x^2+bx})(\sqrt{x^2+ax}+\sqrt{x^2+bx})}{\sqrt{x^2+ax}+\sqrt{x^2+bx}}\\
&=\dfrac{(x^2+ax)-(x^2+bx)}{\sqrt{x^2+ax}+\sqrt{x^2+bx}}\\
&=\dfrac{ax-bx}{\sqrt{x^2+ax}+\sqrt{x^2+bx}}\\
&=\dfrac{x(a-b)}{\sqrt{x^2+ax}+\sqrt{x^2+bx}}\\
&=\dfrac{a-b}{\sqrt{1+a/x}+\sqrt{1+b/x}}\\
&\to \dfrac{a-b}{2} \text{ as } x \to \infty\\
\end{array}
$
The result is the same
if the expression is
$\sqrt{x^2+ax+c}-\sqrt{x^2+bx+d}
$.
|
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|
Generalized eigenvector space with $\lambda_1=\lambda_2=\lambda_3=2$ Let be $$M:=\begin{pmatrix}
1 & 1 & 1 \\
-1 & 3 & 1 \\
0 & 0 & 2
\end{pmatrix}$$
The characteristic polynomial would then be $\chi_M(\lambda)=\lambda^3-6\lambda^2+12\lambda-8$, and his zeros points are $\lambda_1=\lambda_2=\lambda_3=2$. Now we have the matrix:
$$N:=\lambda_1I_3-M=\begin{pmatrix}
1 & -1 & -1 \\
1 & -1 & -1 \\
0 & 0 & 0
\end{pmatrix}$$
From there we calculate that there are two eigenvectors: $v_1=\begin{pmatrix} 1 \\ 1 \\ 0 \end{pmatrix}$ and $v_2=\begin{pmatrix} 1 \\ 0 \\ 1 \end{pmatrix}$. If we want to calculate the generalized eigenvectors we need to find a vector $v_3$ such that $N^2v_3=0$, but we get that $N^2=0$, a nullmatrix 3x3. How can we calculate the third vector? As I know there should be a minimum of $\dim{M}=3$ generalized eigenvectors.
|
Let $$v_3=\begin{pmatrix}-1\\0\\0\end{pmatrix}.$$ Then $$(M-2I)v_3 = \begin{pmatrix}-1&1&1\\-1&1&1\\0&0&0\end{pmatrix}\begin{pmatrix}-1\\0\\0\end{pmatrix} = \begin{bmatrix}1\\1\\0\end{bmatrix}=v_1, $$
so
$$(M-2I)^2v_3 = (M-2I)(M-2I)v_3=(M-2I)v_1=0, $$
and hence $v_3$ is a generalized eigenvector of $M$.
|
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|
Find the range of $a$ for $a^2 - bc - 8a + 7 = 0$ and $b^2 + c^2 + bc - 6 a + b = 0$. Let $a, b, c$ be real numbers such that
\begin{cases}
a^2 - bc - 8a + 7 = 0\\
b^2 + c^2 + bc - 6 a + b = 0\\
\end{cases}
Find the range of $a$.
By adding the two equations I have $$(a - 7)^2 + (b + \frac 12)^2 + c^2 = \frac {169}4,$$
thus $a \in [\frac 12, \frac {27}2]$. Then I found that $a$ cannot obtain the boundaries so $a \in (\frac 12, \frac {27}2)$. But this is only a rough restriction. Exactly how do I solve this?
Edit: When I got this question from my friend, it was a $b$, but after I saw the official solution, it should be a $6$ instead. So the official answer I posted is actually not the answer to the question here but the answer to the original question. I'm really sorry for this mistake.
|
This is the "official" solution which I found later:
\begin{cases}
a^2 - bc - 8a + 7 = 0\\
b^2 + c^2 + bc - 6 a + \mathbf6= 0\\
\end{cases}
is equivalent to
\begin{cases}
bc=a^2 - 8a + 7\\
b+c=\pm(a-1)\\
\end{cases}
So $b,c$ are the roots of the equation
$$x^2\pm(a-1)x+(a^2-8a+7)=0$$
A necessary and sufficient condition that $b,c\in\mathbb{R}$ is
$$\Delta=(a-1)^2-4(a^2-8a+7)\ge0$$ Solving this gives the answer $a\in[1,9]$.
|
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|
Prove $\frac{x}{y+z}+\frac{y}{z+x}+\frac{z}{x+y} \geqslant \frac32+ \frac{27}{16}\frac{(y-z)^2}{(x+y+z)^2}$
$x,y,z >0$, prove
$$\frac{x}{y+z}+\frac{y}{z+x}+\frac{z}{x+y} \geqslant \frac32+ \frac{27}{16}\frac{(y-z)^2}{(x+y+z)^2}$$
This inequality is easier than the other one. Previously, I learned from Jack D'Aurizio at this post, so my first step is to use the Lagrange's identity. I get
$$\sum_{cyc}\frac{x}{y+z}=\frac32+\frac12\sum_{cyc}\frac{(x-y)^2}{(x+z)(y+z)}$$
Thus it remains to prove
$$\tag{1}\sum_{cyc}\frac{(x-y)^2}{(x+z)(y+z)} \geqslant \frac{27}{8} \frac{(y-z)^2}{(x+y+z)^2}$$
I spent several hours to prove (1) but not success. Please help.
|
Here's a proof.
Following the task description, we want to establish
$$
\sum_{cyc}\frac{(x-y)^2}{(x+z)(y+z)} \geqslant \frac{27}{8} \frac{(y-z)^2}{(x+y+z)^2}
$$
We will follow two paths for separate cases.
Path 1:
Clearing denominators, we obtain
$$
(x-y)^2 (x+y) + (y-z)^2 (y+z)+ (z-x)^2 (z+x) \geq \frac{27}{8} \frac{(y-z)^2}{(x+y+z)^2} (x+y) (y+z) (z+x)
$$
Now we use the general formula (by AM-GM):
$$
(a+b+c)^3 \geq 27 a b c
$$
Applying this to $a = (x+y); b = (y+z); c = (z+x)$ gives
$$
(x+y+z)^3 \geq \frac{27}{8} (x+y) (y+z) (z+x)
$$
It then suffices to show
$$
(x-y)^2 (x+y) + (y-z)^2 (y+z) + (z-x)^2 (z+x) \geq \frac{(y-z)^2}{(x+y+z)^2} (x+y+z)^3
$$
which is
$$
(x-y)^2 (x+y) + (z-x)^2 (z+x) \geq (y-z)^2 x
$$
Since
$$
(y-z)^2 = (y-x + x- z)^2 = (y-x)^2 + (x- z)^2 + 2 (y-x)(x-z)
$$
this translates into
$$
(y-x)^2 y + (x-z)^2 z \geq 2 x(y-x)(x-z)
$$
For the two cases $y\geq x ; z\geq x $ and $y\leq x ; z\leq x $ the RHS $\leq 0$ so we are done. For the other two cases, by symmetry, it remains to show the case $y> x ; z < x $.
Rearranging terms, we can also write
$$
(y-x)^3 - (x-z)^3 +x ((y - x) + (z-x))^2 \geq 0
$$
This holds true at least for $(y-x)^3 \geq (x-z)^3$ or $y+z\geq 2 x$.
So the proof is complete other than for the case $y+z < 2x$ and [ $y> x ; z < x $ or $z> x ; y < x $ ].
Path 2.
For the remaining case $y+z < 2 x $ and [$y> x ; z < x$ or $z> x ; y < x $] we will follow a different path. Again, by symmetry, we must inspect only $y+z < 2x$ and $y> x > z$.
Clearing all denominators gives
$$
8 (x+y+z)^2 \sum_{cyc} {(x-y)^2}{(x+y)} - 27 (y-z)^2 (x+y)(y+z) (z+x) \geqslant 0
$$
By homogeneity, we set $y=1+z$.
The condition $y+z < 2x$ then translates into $1+2z < 2x$, hence we further set $x = z + (1 +q)/2$ where $0\leq q \leq 1$ since also $x = z + (1 +q)/2 < y = 1 +z$.
This gives a lengthy expression
$$
8 (3 z + 3/2 + q/2)^2 \left\{ ((q-1)/2)^2 (2z + (3 +q)/2) + ((1 +q)/2 )^2(2z +(1 +q)/2 ) + (1 + 2z) \right\} - 27 (2z + (3 +q)/2) (2z +(1 +q)/2 )(1 + 2z)\geqslant 0
$$
The LHS is a third order expression in $z$ with leading (in $z^3$ ) term $72 q^2 z^3$, hence for large enough $z$ it is rising with $z$. A remarkable feature of this expression is that for the considered range $0\leq q \leq 1$ it is actually monotonously rising for all $z$. To see this, consider whether there are points with zero slope. The first derivative of the expression with respect to $z$ is
$$
216 q^2 z^2 + (84 q^3 + 216 q^2 - 108 q) z + (81 q^2)/2 + 42 q^3 + 8 q^4 - 54 q + 27/2
$$
Equating this to zero gives
$$
z_{1,2} = -(18 q \pm q \sqrt{q^2 - 45} + 7 q^2 - 9)/(36 q)
$$
however there are no real roots in the considered range $0\leq q \leq 1$, so monotonicity (rising with $z$) is established.
Hence, to show the inequality it suffices to inspect the LHS at the smallest $z=0$. This gives
$$
q^5/2 + 4 q^4 + 10 q^3 + (9 q^2)/4 - (27 q)/2 + 27/4 \geq 0
$$
An even stronger requirement is
$$
f(q) =
10 q^3 + (9 q^2)/4 - (27 q)/2 + 27/4 \geq 0
$$
In the considered range $0\leq q \leq 1$, $f(q)$ has a minimum which is obtained by taking the first derivative,
$$
30 q^2 + (9 q)/2 - 27/2
$$
and equating to zero, which gives $q = 3/5$, and the above $f(q)$ then gives
$$
f(q = 3/5) = 81/50
$$
This establishes the inequality. $ \qquad \Box$
|
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|
How should I try to evaluate the integral $\int_a^b \sqrt{1 + \frac{x^2}{r^2 - x^2}} \; dx$ I've tried to evaluate $\displaystyle\int_{-r}^r \sqrt{1 + \frac{x^2}{r^2 - x^2}} \; dx$ on my own, but I have encountered a problem I cannot get around.
The indefinite integral $\sqrt{\frac{r^2}{r^2-x^2}} \sqrt{r^2-x^2} \tan ^{-1}\left(\frac{x}{\sqrt{r^2-x^2}}\right) + C$ is undefined at both limits of integration
(the fractions $\frac{r^2}{r^2-r^2} = \frac{r^2}0 = \text{indetermined}$ and $\frac{({-r})^2}{r^2-r^2} = \frac{r^2}0 = \text{indetermined}$)
so I really don't know what to do. Apparently, the value of this integral should be $\pi r.$ Any help would be appreciated.
|
$$\displaystyle\int_{-r}^r \sqrt{1 + \frac{x^2}{r^2 - x^2}} \; dx=2r\int_{0}^r \frac{1}{\sqrt{r^2-x^2}}dx=2r\,\left. {{\sin }^{-1}}\left( \frac{x}{r} \right) \right|_{0}^{r}=r\pi$$
|
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|
Evaluation of $\sum^{\infty}_{n=0}\frac{1}{16^n}\binom{2n}{n}.$
Prove that $\displaystyle \int_{0}^{\frac{\pi}{2}}\sin^{2n}xdx = \frac{\pi}{2}\frac{1}{4^{n}}\binom{2n}{n}$ and also find value of $\displaystyle \sum^{\infty}_{n=0}\frac{1}{16^n}\binom{2n}{n}.$
$\bf{My\; Try::}$ Let $$\displaystyle I_{n} = \int_{0}^{\frac{\pi}{2}}\sin^{2n}xdx = \int_{0}^{\frac{\pi}{2}}\sin^{2n-2}x\cdot \sin^2 xdx = \int_{0}^{\frac{\pi}{2}}\sin^{2n-2}x\cdot (1-\cos^2 x)dx$$
$$I_{n} =I_{n-1}-\int_{0}^{\frac{\pi}{2}}\cos x\cdot \sin^{2n-2}\cdot \cos xdx$$
Now Using Integration by parts, We get $$I_{n} = I_{n-1}-\frac{I_{n}}{2n-1}\Rightarrow I_{n} = \frac{2n-1}{2n}I_{n-1}$$
Now Using Recursively, We get $$I_{n} = \frac{2n-1}{2n}\cdot \frac{2n-3}{2n-2}I_{n-2} =\frac{2n-1}{2n}\cdot \frac{2n-3}{2n-2}\cdot \frac{2n-5}{2n-4}I_{n-3}$$
So we get $$I_{n} = \frac{2n-1}{2n}\cdot \frac{2n-3}{2n-2}\cdot \frac{2n-5}{2n-4}\cdot \frac{2n-7}{2n-6}\cdot \cdot \cdot \cdot \cdot \cdot \cdot\cdot \frac{3}{2}I_{0}$$
and we get $\displaystyle I_{0} = \frac{\pi}{2}$
So we get $$I_{n} = \frac{(2n)!}{4^n\cdot n!\cdot n!}\cdot \frac{\pi}{2}$$
Now I did not understand How can I calculate value of $\displaystyle \sum^{\infty}_{n=0}\frac{1}{16^n}\binom{2n}{n}.$
Help Required, Thanks.
|
Hint. From what you have proved, one may deduce that
$$
\frac{\pi}{2}\sum^{\infty}_{n=0}\frac{1}{16^n}\binom{2n}{n}=\int_{0}^{\large \frac{\pi}{2}}\sum^{\infty}_{n=0}\frac{1}{4^n}\sin^{2n}xdx=\int_{0}^{\large \frac{\pi}{2}}\frac{4}{4-\sin^2 x}dx
$$ the latter integral being easy to evaluate.
|
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|
How does $\sqrt{x^2+y^2}+\sqrt{y^2+z^2}+\sqrt{z^2+x^2}$ relate to $\sqrt{x^2+y^2+z^2}$? Another possible 'mean' for three positive real numbers $x,y,z$ is made of pairwise quadratic means:
$$\frac{\sqrt{x^2+y^2}+\sqrt{y^2+z^2}+\sqrt{z^2+x^2}}{3 \sqrt{2}}$$
By QM-AM inequality it is greater than or equal to arithmetic mean of the three numbers:
$$\frac{\sqrt{x^2+y^2}+\sqrt{y^2+z^2}+\sqrt{z^2+x^2}}{3 \sqrt{2}} \geq \frac{x+y+z}{3}$$
Now what is its relationship to the quadratic mean of the three numbers:
$$\frac{\sqrt{x^2+y^2}+\sqrt{y^2+z^2}+\sqrt{z^2+x^2}}{3 \sqrt{2}} \text{ ? } \sqrt{\frac{x^2+y^2+z^2}{3}}$$
Using the obvious inequality $x^2+y^2+z^2 \geq x^2+y^2$, we obtain the following relationship:
$$\sqrt{\frac{x^2+y^2+z^2}{3}} \geq \sqrt{\frac{2}{3}} \frac{\sqrt{x^2+y^2}+\sqrt{y^2+z^2}+\sqrt{z^2+x^2}}{3 \sqrt{2}} $$
Can we make this bound any tighter? How?
|
Never mind, it's easy to prove that
$$\sqrt{\frac{x^2+y^2+z^2}{3}} \geq \frac{\sqrt{x^2+y^2}+\sqrt{y^2+z^2}+\sqrt{z^2+x^2}}{3 \sqrt{2}}$$
Square the RHS:
$$(\sqrt{x^2+y^2}+\sqrt{y^2+z^2}+\sqrt{z^2+x^2})^2=2(x^2+y^2+z^2)+2 \sum_{cyc} \sqrt{(x^2+y^2)(y^2+z^2)}$$
Using AM-GM we obtain:
$$\sqrt{(x^2+y^2)(y^2+z^2)} \leq \frac{x^2+2y^2+z^2}{2}$$
So it follows that:
$$(\sqrt{x^2+y^2}+\sqrt{y^2+z^2}+\sqrt{z^2+x^2})^2 \leq 6(x^2+y^2+z^2)$$
Or:
$$\sqrt{x^2+y^2}+\sqrt{y^2+z^2}+\sqrt{z^2+x^2} \leq \sqrt{6(x^2+y^2+z^2)}$$
Now we just divide by $3\sqrt{2}$.
|
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|
Compute the minimum value of $ \underline{A}\ \underline{B}\ \underline{C} - (A^2 + B^2 + C^2). $ Let $\underline{A}\ \underline{B}\ \underline{C}$ represent a three-digit base 10 number whose digits are $A$, $B$, and $C$ with $A \ge 1$. Compute the minimum value of $$ \underline{A}\ \underline{B}\ \underline{C} - (A^2 + B^2 + C^2). $$
I simplified the expression given above as $100a-a^2+10b-b^2+c-c^2$. This expression contains three quadratic expressions,namely $100a-a^2$,$10b-b^2$,$c-c^2$. I tried using parabolas of these quadratics and the fact that the maximum value attained by a quadratic equation is in its vertex,so the minimum value should be distant to the vertex of the parabola. I get the answer as 99 but it is wrong. Any help is appreciated.
|
Your attempts are good; you want to minimize the expression
$$100a-a^2+10b-b^2+c-c^2,$$
which is the sum of three quadratics. It is minimal when the three quadratics
$$100a-a^2,\qquad 10b-b^2,\qquad c-c^2,$$
are minimal. These are parabolas with their maxima at $50$, $5$ and $\tfrac{1}{2}$, respectively. They are minimal when we are as far away from the maximum as possible. This gives us $a=1$, $b=0$ and $c=9$.
|
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|
$\int\frac{\sin x}{\sqrt{1-\sin x}}dx=?$ Calculate this integral $\displaystyle\int\dfrac{\sin x}{\sqrt{1-\sin x}}dx=?$
Effort;
$1-\sin x=t^2\Rightarrow \sin x=1-t^2\Rightarrow \cos x=\sqrt{2t^2-t^4}$
$1-\sin x=t^2\Rightarrow-\cos x dx=2tdt\Rightarrow dx=\frac{2t}{\sqrt{t^4-2t^2}}dt$
$\displaystyle\int\frac{1-t^2}{t}\cdot\frac{2t}{\sqrt{t^4-2t^2}}dt=2\int\frac{1-t^2}{\sqrt{t^4-2t^2}}dt$
$\ = 2\displaystyle\int\frac{1}{\sqrt{t^4-2t^2}}dt-2\displaystyle\int\frac{t}{\sqrt{t^2-2t}}dt$
$\ = 2\displaystyle\int t^{-1}(t^2-2)^{-\frac{1}{2}}dt-2\displaystyle\int t(t^2-2t)^{-\frac{1}{2}}dt$
But after that I don't know how to continue.
|
Slightly different attempt to get rid of the trig functions (assuming a domain where no sign issues cause trouble):
$$\begin{array}{rl}
\displaystyle \frac{\sin x}{\sqrt{1-\sin x}}
& \displaystyle
= \frac{\sin x\sqrt{1+\sin x}}{\sqrt{1-\sin x}\sqrt{1+\sin x}}\\[7 pt]
& \displaystyle
= \frac{\sin x\sqrt{1+\sin x}}{\sqrt{1-\sin^2 x}}\\[7 pt]
& \displaystyle
= \frac{\cos x \sin x\sqrt{1+\sin x}}{\cos^2 x}\\[7 pt]
& \displaystyle
= \frac{\cos x \sin x\sqrt{1+\sin x}}{1-\sin^2 x}
\end{array}$$
Now let $t = \sin x$:
$$\int \frac{\sin x}{\sqrt{1-\sin x}} \,\mbox{d}x
=\int \frac{\cos x \sin x\sqrt{1+\sin x}}{1-\sin^2 x}\,\mbox{d}x \to
\int \frac{t\sqrt{1+t}}{1-t^2} \,\mbox{d}t$$
This can be rationalized with $u^2 = 1+t$ to get (after simplifying):
$$\int \left( -2 - \frac{2}{u^2-2} \right) \, \mbox{d}u$$
This appears to be a longer route than the suggestions given in some other answers ;o).
|
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|
derivative with square root I have been trying to figure this equation for some time now, but have come up empty. I have tried multiple ways on solving it. Whether by using the Quotient Rule or some other method, I can't seem to figure it out. Any help would be appreciated.
Find the derivative of the function
$
y = \frac{x^{2} + 8x + 3 }{\sqrt{x}}
$
The answer is:
$
y' = \frac{3x^{2}+ 8x - 3 }{2x^\frac{3}{2}}
$
I'm having trouble getting to that answer. So if someone could point me in the right direction by showing the steps, I would be grateful. Thanks!
|
For differentiating $\sqrt x$ one should rewrite it as $x^{\frac{1}{2}}$ and use the power rule. Given $y=\frac{x^2+8x+3}{\sqrt x}$ we have:
$$y'=\frac{\left(\sqrt x\right)\left(2x+8\right)-\left(\frac{1}{2}x^{-\frac{1}{2}}\right)\left(x^2+8x+3\right)}{\left(\sqrt x\right)^2}
\\=x^{-\frac{3}{2}}\left(2x^2+8x-\frac{1}{2}x^2-4x-\frac{3}{2}\right)
\\=\frac{1}{2}x^{-\frac{3}{2}}\left(3x^2+8x-3\right)$$
|
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|
Integration by Parts on $\sqrt{1+x^2}dx$ I have been asked to show that $$\int\sqrt{1+x^2}dx=\frac{1}{2}x\sqrt{1+x^2}+\frac{1}{2}\int\frac{1}{{\sqrt{1+x^2}}}dx$$
I'm aware of how to do this with trig substitution, but the question is specifically regarding integration by parts, and the only examples I can find keep using trig subs.
So far I have integrated twice. The first time was with $u=\sqrt{1+x^2},v'=1$. Resulting in the following:
$$\int\sqrt{1+x^2}dx=x\sqrt{1+x^2}-\int\frac{x}{\sqrt{1+x^2}}dx$$
The first term is almost correct. Integrating this new integral, with $u=\frac{x}{\sqrt{1+x^2}}, v'=1$. Gave me the following:
$$\int\sqrt{1+x^2}dx=x\sqrt{1+x^2}-\frac{x^2}{\sqrt{1+x^2}}+\frac{1}{2}\int\frac{x}{\sqrt{1+x^2}}dx$$
From here I'm pretty much stuck. I can't seem to figure out how I am supposed to get rid of the middle term. The final integral can be put into the right form if I divide by $x$, but then the first term loses it's $x$ scalar and I still can't drop the middle term.
I feel as though I've made a mistake on one of the integrations, but I've been looking over it for that long that the whole thing is blurry.
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Start with integration by parts
$$v=x, \ dv=dx$$
$$u=\sqrt{1+x^2}, \ du=\frac{x}{\sqrt{1+x^2}}dx$$
Therefore
\begin{align*}\int \sqrt{1+x^2}dx&=x\sqrt{1+x^2}-\int\frac{x^2}{\sqrt{1+x^2}}dx \\
&=x\sqrt{1+x^2}-\int\frac{(1+x^2)-1}{\sqrt{1+x^2}}dx\\
&=x\sqrt{1+x^2}-\int\frac{1+x^2}{\sqrt{1+x^2}}dx+\int\frac{dx}{\sqrt{1+x^2}}\\
&=x\sqrt{1+x^2}-\int\sqrt{1+x^2}dx+\frac{dx}{\sqrt{1+x^2}}dx.\end{align*}
Just add $\int\sqrt{1+x^2}dx$ to both sides. This means that
$$2\int\sqrt{1+x^2}dx=x\sqrt{1+x^2}+\int\frac{dx}{\sqrt{1+x^2}}$$
or
$$\int\sqrt{1+x^2}dx=\frac12x\sqrt{1+x^2}+\frac12\int\frac{dx}{\sqrt{1+x^2}}.$$
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|
Problem solving a word problem using a generating function
How many ways are there to hand out 24 cookies to 3 children so that they each get an even number, and they each get at least 2 and no more than 10? Use generating functions.
So the first couple steps are easy.
The coefficient is $x^{24}$
$g(x) = x^6(1+x^2+x^4+x^6+x^8)^3$ or what I got was $x^6 (1 + (x^2)^1 +...+ (x^2)^4)^3$
now finding the closed formula is where I am having problems
My answer: using the fact that $\dfrac{1-x^{n+1}}{1- x}$
I get $x^6\left(\dfrac{1-x^9}{1-x}\right)$ which is wrong
The correct answer: $x^6\left(\dfrac{1-x^{10}}{1-x^2}\right)$
If someone could explain in some detail on how to get the correct formula would be much appreciated. Thanks!
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Your expression $g(x) = x^6 (1+x^2+\cdots+x^8)^3$ is correct, but the geometric series formula needs to be used correctly. Recall the geometric series formula for the sum of 5 terms: $a+ar+\cdots+ar^4 = \frac{a(1-r^5)}{1-r}$. In our case, $r=x^2$ and $a=1$. So $1+x^2+\cdots+x^8$ simplifies to $\frac{1-(x^2)^5}{1-x^2} = \frac{ 1-x^{10}}{1-x^2}$.
Hence, the number of ways in question is the coefficient of $x^{24}$ in $x^6 \left(\frac{1-x^{10}}{1-x^2} \right)^3$. Notice some differences with your answer: the denominator has $x^2$ (not $x$) since the common ratio of the geometric series is $x^2$; the geometric series formula with $n$ terms has an $n$ (not $n+1$) in the exponent. Also, don't forget to take the third power.
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Find the rightmost digit of: $1^n+2^n+3^n+4^n+5^n+6^n+7^n+8^n+9^n$ Find the rightmost digit of: $1^n+2^n+3^n+4^n+5^n+6^n+7^n+8^n+9^n(n$ arbitrary positive integer)
First of all I checked a few cases for small $n$'s and in all cases the rightmost digit was $5$, so maybe this is the case for all values of $n$.
Then I thought maybe it's better to consider odd and even cases for $n$ but there's no unified rule here, because for example: $8^2\equiv{4}\pmod{10}$ and $8^4\equiv{6}\pmod{10}$. Any ideas??
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Hint: This is equivalent to asking what $$1^n+2^n+3^n+4^n+5^n+6^n+7^n+8^n+9^n$$ is modulo $10$. But $m=m-10$ mod $10$, so modulo 10 the above is the same as
$$1^n+2^n+3^n+4^n+5^n+(-4)^n+(-3)^n+(-2)^n+(-1)^n.$$
What does this equal if $n$ is odd versus even?
|
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|
expansion of a generating function I found this formula in a book
$$\sqrt{1-4x}=1-2\sum_{n=1}^{\infty}\frac{1}{n} {{2n-2}\choose {n-1}} x^n$$
How can I prove that?
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There are quite a few ways to prove it.
You could note that $\frac1n\binom{2n-2}{n-1}$ is $C_{n-1}$, the $(n-1)$-st Catalan number, so that the formula can be written
$$\sqrt{1-4x}=1-2\sum_{n\ge 0}\frac1{n+1}\binom{2n}nx^{n+1}=1-2\sum_{n\ge 0}C_nx^{n+1}\;.$$
Now you can simply manipulate the known generating function for the Catalan numbers:
$$\sum_{n\ge 0}C_nx^n=\frac{1-\sqrt{1-4x}}{2x}\;,$$
so
$$\begin{align*}
\sqrt{1-4x}&=1-2x\sum_{n\ge 0}C_nx^n\\
&=1-2\sum_{n\ge 0}C_nx^{n+1}\;,
\end{align*}$$
as desired.
Or you can use the generalized binomial theorem:
$$\begin{align*}
(1-4x)^{1/2}&=\sum_{n\ge 0}\binom{1/2}n(-4x)^n\\
&=1+\sum_{n\ge 1}\binom{1/2}n(-1)^n4^nx^n\\
&\overset{*}=1+\sum_{n\ge 1}\frac{(-1)^{n-1}}{2^{2n-1}n}\binom{2n-2}{n-1}(-1)^n4^nx^n\\
&=1-\sum_{n\ge 1}\frac{2^{2n}}{2^{2n-1}n}\binom{2n-2}{n-1}x^n\\
&=1-2\sum_{n\ge 1}\frac1n\binom{2n-2}{n-1}x^n\;,
\end{align*}$$
where the starred step is carried out in detail here.
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proof of $\frac{\frac{1}{n}}{\sqrt{x^2+\frac{1}{n}}+x}\leq\frac{1}{\sqrt{n}}$ for large $n$ I'm trying to conclude this question in which I just need to prove that:
$$0<\frac{\frac{1}{n}}{\sqrt{x^2+\frac{1}{n}}+x}\leq\frac{1}{\sqrt{n}}$$
and then I'll have that, for large $n$:
$$\begin{align}
\left\lvert\cos\left(\sqrt{x^2+\frac{1}{n}}\right)-\cos(x)\right\rvert &= \left\lvert2\sin\frac{\sqrt{x^2+\frac{1}{n}}-x}{2}\sin\frac{\frac{\frac{1}{n}}{\sqrt{x^2+\frac{1}{n}}+x}}{2}\right\rvert\\
&\leq
2\sin\frac{\sqrt{x^2+\frac{1}{n}}-x}{2}\sin\frac{1}{2\sqrt{n}}\xrightarrow[n\to\infty]{} 0
\end{align}$$
because $\sin\frac{1}{2\sqrt{n}}\to 0$ and $\sin\frac{\sqrt{x^2+\frac{1}{n}}-x}{2}$ is bounded.
I liked the answers given in the question but I really need to solve it this way, I'm trying to investigate the inequation. I know that, the worst case is for $n$ so large that $\frac{1}{n}$ in the LHS of the inequality goes to $0$, and the best case is when $n=1$ and therefore we have:
$$\sqrt{x^2}+x<\sqrt{x^2+\frac{1}{n}}+x<\sqrt{x^2+1}+x\le\sqrt{2}+1$$
*for $0\le x\le 1$
but it doesn't help, I think.
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Well, as $x$ is positive, by monotonicity of the square root
$$\sqrt{x^2+\frac1n}+x \geq \sqrt{x^2+\frac1n}+\geq \sqrt{\frac1n}=\frac1{\sqrt{n}}$$
Thus
$$\frac{\frac{1}{n}}{\sqrt{x^2+\frac1n}+x}\leq \frac{\frac{1}{n}}{\sqrt{x^2+\frac1n}}\leq \frac{\frac{1}{n}}{\frac1{\sqrt{n}}}=\frac{1}{\sqrt{n}}$$
|
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Positive integers $a$ and $b$ are such that $ab+1$ is a factor of $a^2+b^2$. Prove that $\frac{a^2+b^2}{ab+1}$ is a perfect square. IMO 1988 question No. 6
I have a confusion in the question.The question is as follows:
$a$ and $b$ are positive integers and $ab+1$ is a factor of $a^2+b^2$. Prove that $\displaystyle\frac{a^2+b^2}{ab+1}$ is a perfect square.
My confusion: According to me, $a$ and $b$ have only one value ie $a=b=1$. What are the other possible values of $a$ and $b$?
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Let us assume $(a^2 + b^2)/(ab + 1) = k \in \mathbb{N}$. We then have $a^2 - kab + b^2 = k$. Let us assume that $k$ is not an integer square, which implies $k \ge 2$. Now, we observe the minimal pair $(a, b)$ such that $a^2 - kab + b^2 = k$ holds. We may assume, without loss of generality, that $a \ge b$. For $a = b$, we get $k = (2 - k)a^2 \le 0$, so we must have $a > b$.
Let us observe the quadratic equation $x^2 - kbx + b^2 - k = 0$, which has solutions $a$ and $a_1$. Since $a+ a_1 = kb$, it follows that $a_1 \in \mathbb{Z}$. Since $a > kb$ implies $k > a + b^2 > kb$ and $a = kb$ implies $k = b^2$, it follows that $a < kb$ and thus $b^2 > k$. Since $aa_1 = b^2 - k > 0$ and $a>0$, it follows that $a_1 \in \mathbb{N}$ and $a_1 = (b^2 - k)/a < (a^2 - 1)/a < a$. We have thus found an integer pair $(a_1, b)$ with $0 < a_1 < a$ satisfies the original equation. This is a contradiction of the initial assumption that $(a, b)$ is minimal. Hence, $k$ must be an integer square.
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|
number of real triplets $(x,y,z)$ in system of equations
Total number of real triplets of $(x,y,z)$ in $x^3+y^3+z^3=x+y+z$ and $x^2+y^2+z^2=xyz$
$\bf{My\; Try::}$ Let $x+y+z=a$ and $xy+yz+zx=b\;,$
Then $(x+y+z)^2-2(xy+yz+zx)=xyz\implies a^2-2b=xyz$
And using $\displaystyle x^3+y^3+z^3-3xyz=(x+y+z)\left[x^2+y^2+z^2-(xy+yz+zx)\right]$
So we get $a-3xyz=a[xyz-b]\implies a+ab=(a+3)xyz$
So we get $\displaystyle \frac{a+ab}{a+3} = a^2-2b$
Now how can I solve it after that, Help required, Thanks.
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Assume that $x+y+z=a$ and $xyz=b$. We have:
$$ e_2 = xy+xz+yz = \frac{a^2-b}{2} $$
and we may consider that:
$$ p(t) = (t-x)(t-y)(t-z) = t^3 - e_1 t^2 + e_2 t^2 - e_3 = t^3 - at^2 + \frac{a^2-b}{2}t - b. $$
On the other hand, for any $w\in\{x,y,z\}$ we have $p(w)=0$ or
$$ w^3 = e_1 w^2 - e_2 w + e_3,\tag{1} $$
hence by summing $(1)$ over $w\in\{x,y,z\}$ we get:
$$ b=x^3+y^3+z^3 = a b - \frac{a^2-b}{2}a+3b \tag{2}$$
from which $b=\frac{a^3}{4+3a}$. The problem boils down to understanding which values of $a$ grant that
$$ p(t) = t^3 - at^2 + \frac{(a+2)a^2}{4+3a}\,t-\frac{a^3}{4+3a}\tag{3} $$
has three real roots, i.e. a non-negative discriminant. That cannot happen if $a>-3$, but happens for sure if $a$ belongs to some sub-interval of $\mathbb{R}^-$, so we have an infinite number of triplets fulfilling the given constraints.
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|
Integral of $\int{(x^2+2x)\over \sqrt{x^3+3x^2+1}} dx$
Find the integral of the following:
$$\int{(x^2+2x)\over \sqrt{x^3+3x^2+1}} dx$$
Do set $u=x^3+3x^2+1$?
So, $du=(3x^2+6x)dx$?
And, $x^2+2x={u-1-x^2\over x}$?
So then,
$$\int{({u-1-x^2\over x})\over \sqrt u} du$$
This seems very complicated, is there any easier way to do this? or if this is correct, could anyone show me how to go on with this? Thanks!
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$du = (3x^2+6x)dx = 3(x^2+2x)dx$, so your answer changes as $\frac{1}{3}\int \frac{du}{\sqrt{u}} = \frac{2}{3}\sqrt{u}$
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Prove if $n \in \mathbb N$, then $\frac{1}{2!}+\cdots+\frac{n}{(n+1)!}=1-\frac{1}{(n+1)!}$
Prove if $n \in \mathbb N$, then $\frac{1}{2!}+\frac{2}{3!}+\frac{3}{4!}+\cdots+\frac{n}{\left(n+1\right)!} = 1-\frac{1}{\left(n+1\right)!}$
So I proved the base case where $n=1$ and got $\frac{1}{2}$
Then since $n=k$ implies $n=k+1$ I setup the problem like so:
$\frac{k}{(k+1)!}+\frac{(k+1)}{(k+2)!}=1-\frac{1}{(k+2)!}$
After trying to simplify it I got the following:
$\frac{k(k+2)!+(k+1)(k+1)!}{(k+1)!(k+2)!}=1-\frac{1}{(k+2)!}$
However, I'm having trouble simplifying it to match the RHS. Hints?
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You have your base case.
Explicitly state your inductive hypothesis.
Suppose:
$\sum_\limits{n=1}^k\frac{n}{(n+1)!} = 1-\frac{1}{(k+1)!}$
We will show that:
$\sum_\limits{n=1}^{k+1}\frac{n}{(n+1)!} = 1-\frac{1}{(k+2)!}$
$\sum_\limits{n=1}^k\frac{n}{(n+1)!}+\frac{(k+1)}{(k+2)!}$
$1-\frac{1}{(k+1)!}+\frac{(k+1)}{(k+2)!}$(By the inductive hypothesis.)
$1-\frac{1}{(k+2)!}$
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Use integration by parts $\int^{\infty}_{0} \frac{x \cdot \ln x}{(1+x^2)^2}dx$ $$I=\int^{\infty}_{0} \frac{x \cdot \ln x}{(1+x^2)^2}dx$$
Clearly $$-2I=\int^{\infty}_{0} \ln x \cdot \frac{-2x }{(1+x^2)^2} dx$$
My attempt :
$$-2I=\left[ \ln x \cdot \left(\frac{1}{1+x^2}\right)\right]^\infty_0 - \int^{\infty}_{0} \left(\frac{1}{1+x^2}\right) \cdot \frac{1}{x} dx$$
$$-2I=\left[ \ln x \cdot \left(\frac{1}{1+x^2}\right)\right]^\infty_0 - \int^{\infty}_{0} \frac{1}{x(1+x^2)} dx$$
$$-2I=\left[ \ln x \cdot \left(\frac{1}{1+x^2}\right)\right]^\infty_0 - \int^{\infty}_{0} \frac{1+x^2-x^2}{x(1+x^2)} dx$$
$$-2I=\left[ \ln x \cdot \left(\frac{1}{1+x^2}\right)\right]^\infty_0 - \int^{\infty}_{0} \frac{1}{x}+ \frac{1}{2}\int^{\infty}_{0} \frac{2x}{1+x^2} dx$$
$$-2I=\left[ \ln x \cdot \left(\frac{1}{1+x^2}\right)\right]^\infty_0 -\left[ \ln x -\frac{1}{2}\cdot \ln(1+x^2) \right]^\infty_0 $$
$$-2I=\left[ \frac{\ln x}{1+x^2}\right]^\infty_0 -\left[\ln \left (\frac{x}{\sqrt{1+x^2}} \right) \right]^\infty_0 $$
How can I evaluate the last limits ?
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Put $x=\dfrac1y$
$$I=\int_\infty^0\dfrac{y^4\cdot -\ln(y)}{y(y^2+1)^2}\cdot-\dfrac{dy}{y^2}=\int_\infty^0\dfrac{y\ln y}{(1+y^2)^2}dy=-\int_0^\infty\dfrac{y\ln y}{(1+y^2)^2}dy=-I$$
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first orderr non linear ODE I came along this first order non linear ODE, and cannot solve it.
$$\frac{dv}{dt}=\frac{-b}{(vt)^2}+k$$ (where b and k are constants)
The question asked to express v as a function of t.
Thank you very much, any help is welcome!
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Assume $b\neq0$ and $k\neq0$ for the key case:
Hint:
Approach $1$:
$\dfrac{dv}{dt}=-\dfrac{b}{(vt)^2}+k$
$\dfrac{dv}{dt}-k=-\dfrac{b}{v^2t^2}$
$\dfrac{d(v-kt)}{dt}=-\dfrac{b}{v^2t^2}$
Let $x=v-kt$ ,
Then $\dfrac{dx}{dt}=-\dfrac{b}{(x+kt)^2t^2}$
$\dfrac{dt}{dx}=-\dfrac{k^2t^4+2kxt^3+x^2t^2}{b}$
This belongs to a "Chini-like" equation.
Approach $2$:
Let $x=vt$ ,
Then $v=\dfrac{x}{t}$
$\dfrac{dv}{dt}=\dfrac{1}{t}\dfrac{dx}{dt}-\dfrac{x}{t^2}$
$\therefore\dfrac{1}{t}\dfrac{dx}{dt}-\dfrac{x}{t^2}=-\dfrac{b}{x^2}+k$
$\dfrac{1}{t}\dfrac{dx}{dt}=k-\dfrac{b}{x^2}+\dfrac{x}{t^2}$
$\dfrac{1}{t}\dfrac{dx}{dt}=\dfrac{(kx^2-b)t^2+x^3}{x^2t^2}$
$((kx^2-b)t^2+x^3)\dfrac{dt}{dx}=x^2t$
Let $u=\dfrac{1}{t^2}$ ,
Then $\dfrac{du}{dx}=-\dfrac{2}{t^3}\dfrac{dt}{dx}$
$\therefore-\dfrac{((kx^2-b)t^2+x^3)t^3}{2}\dfrac{du}{dx}=x^2t$
$\left(\dfrac{1}{t^2}+\dfrac{k}{x}-\dfrac{b}{x^3}\right)\dfrac{du}{dx}=-\dfrac{2}{xt^4}$
$\left(u+\dfrac{k}{x}-\dfrac{b}{x^3}\right)\dfrac{du}{dx}=-\dfrac{2u^2}{x}$
This belongs to an Abel equation of the second kind.
Let $v=u+\dfrac{k}{x}-\dfrac{b}{x^3}$ ,
Then $u=v-\dfrac{k}{x}+\dfrac{b}{x^3}$
$\dfrac{du}{dx}=\dfrac{dv}{dx}+\dfrac{k}{x^2}-\dfrac{3b}{x^4}$
$\therefore v\left(\dfrac{dv}{dx}+\dfrac{k}{x^2}-\dfrac{3b}{x^4}\right)=-\dfrac{2}{x}\left(v-\dfrac{k}{x}+\dfrac{b}{x^3}\right)^2$
$v\dfrac{dv}{dx}+\left(\dfrac{k}{x^2}-\dfrac{3b}{x^4}\right)v=-\dfrac{2}{x}\left(v^2-\left(\dfrac{2k}{x}-\dfrac{2b}{x^3}\right)v+\dfrac{(kx^2-b)^2}{x^6}\right)$
$v\dfrac{dv}{dx}+\left(\dfrac{k}{x^2}-\dfrac{3b}{x^4}\right)v=-\dfrac{2v^2}{x}+\left(\dfrac{4k}{x^2}-\dfrac{4b}{x^4}\right)v-\dfrac{2(kx^2-b)^2}{x^7}$
$v\dfrac{dv}{dx}=-\dfrac{2v^2}{x}+\left(\dfrac{3k}{x^2}-\dfrac{b}{x^4}\right)v-\dfrac{2(kx^2-b)^2}{x^7}$
Let $v=\dfrac{w}{x^2}$ ,
Then $\dfrac{dv}{dx}=\dfrac{1}{x^2}\dfrac{dw}{dx}-\dfrac{2w}{x^3}$
$\therefore\dfrac{w}{x^2}\left(\dfrac{1}{x^2}\dfrac{dw}{dx}-\dfrac{2w}{x^3}\right)=-\dfrac{2w^2}{x^5}+\left(\dfrac{3k}{x^2}-\dfrac{b}{x^4}\right)\dfrac{w}{x^2}-\dfrac{2(kx^2-b)^2}{x^7}$
$\dfrac{w}{x^4}\dfrac{dw}{dx}-\dfrac{2w^2}{x^5}=-\dfrac{2w^2}{x^5}+\left(\dfrac{3k}{x^2}-\dfrac{b}{x^4}\right)\dfrac{w}{x^2}-\dfrac{2(kx^2-b)^2}{x^7}$
$\dfrac{w}{x^4}\dfrac{dw}{dx}=\left(\dfrac{3k}{x^2}-\dfrac{b}{x^4}\right)\dfrac{w}{x^2}-\dfrac{2(kx^2-b)^2}{x^7}$
$w\dfrac{dw}{dx}=\left(3k-\dfrac{b}{x^2}\right)w-\dfrac{2(kx^2-b)^2}{x^3}$
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Integral solution and odd prime Here is my question:
Let $p\neq 7$ be an odd prime. Suppose that only one of the two equations,
$$
x^2 + 7y^2 = p, \quad x^2 - 7y^2 = p
$$
has integral solution $(x,y)$. Prove that $p \equiv 3\;(\mathrm{mod}\; 4)$.
What I have done is that in modulo $7$, $x^2$ can be $1, 2$ or $4$.
And in modulo $4$, $x^2$ can be $0$ or $1$.
But I don't know how to prove that $p$ is not congruent $1$ modulo $4$ when only one of two equation has an integral solution.
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First we show that there are integers $x$ and $y$ such that $x^2+7y^2=p$, and $p$ is a prime not equal to $7$, then $-7$ is a quadratic residue of $p$.
If $x^2+7y^2=p$, then $y$ is not divisible by $p$. For suppose to the contrary that $p$ divides $y$. Then $p$ divides $x$, so $p^2$ divides $x^2+7y^2$, which is impossible.
If $x^2+7y^2=p$, then $-7y^2\equiv x^2\pmod{p}$. If we multiply both sides by $u^2$, where $u$ is the inverse of $y$ modulo $p$, we find that $(xu)^2\equiv-7$, so is a quadratic residue of $p$.
A similar argument shows that if $x^2-7y^2=p$, then $7$ is a quadratic residue of $p$.
Now suppose that $p\equiv 1\pmod{4}$. Then $-1$ is a QR of $p$. It follows that $-7$ is a QR of $p$ if and only if $7$ is a QR of $p$.
But since exactly one the equations $x^2+7y^2=p$ and $x^2-7y^2=p$ has a solution, exactly one of $-7$ and $7$ is a QR of $p$. Thus $p$ cannot be of the form $4k+1$, so it is of the form $4k+3$.
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|
If $x$ and $y$ are non-negative integers for which $(xy-7)^2=x^2+y^2$. Find the sum of all possible values of $x$. I am not able to reach to the answer. I have used discriminant as $x$ and $y$ are both integers but it didn't give any hint to reach to answer. I am not able to understand how should I deal with these type of question.
|
We have $$(xy)^2-12xy+49=(x+y)^2\iff(xy-6)^2+13=(x+y)^2$$
hence $$13=(x+y+xy-6)(x+y-xy+6)$$ It follows the only possibility (after discard$-13$ and $-1$)
$$x+y+xy-6=13\\x+y-xy+6=1$$ This implies clearly $$x+y=7$$ which gives the candidates $(0,7),(1,6),(2,5),(3,4)$ and symmetrics and it is verified that the only to be accepted are $(0,7)$ and $(4,3)$.
Consequently, taking $(x,y)=(0,7),(7,0),(4,3),(3,4)$ $$S=0+7+4+3=\color{red}{14}$$
|
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|
Integrating $\int\frac{x^3}{\sqrt{9-x^2}}dx$ via trig substitution What I have done so far:
Substituting $$x=3\sin(t)\Rightarrow dx=3\cos(t)dt$$ converting our integral to
$$I=\int\frac{x^3}{\sqrt{9-x^2}}dx=\int \frac{27\sin^3(t) dt}{3\sqrt{\cos^2(t)}}3\cos(t)dt\\
\Rightarrow \frac{I}{27}=\int \sin^3(t)dt=-\frac{\sin^2(t)\cos(t)}{3}+\frac{2}{3}\int\sin(t)dt\\
\Rightarrow I=18\cos(t)-9\sin^2(t)\cos(t)+c$$
Which is where I'm stuck. I can substitute back in for the $\sin(t)$ terms but don't know hot to deal with the $\cos(t)$ terms and back out information about $x$.
edited: because I miswrote the problem, I am sorry.
|
A nice trick to do in this situation is to write
$$
\sin^3t = (1-\cos^2t)\sin t.
$$
You can now do a secondary $u$-substitution where you can take $u = \cos t$ and $du = -\sin tdt$.
Edit
Where you left off you had
$$
\frac{I}{27} \;\; =\;\; \int \sin^3t dt \;\; =\;\; \int (1-\cos^2t)\sin t dt
$$
If we now take the $u$-substitution I suggested above we find
$$
\frac{I}{27} \;\; =\;\;- \int (1 - u^2) du \;\; =\;\; \frac{1}{3}u^3 - u + C
$$
for some constant $C$. We can do the first re-substitution
$$
\frac{I}{27} \;\; =\;\; \frac{1}{3} \cos^3t - \cos t+C
$$
and now you can do the other substitution by noting that $\cos t = \sqrt{1 - \sin^2t} = \sqrt{1 - \frac{x^2}{9}} = \frac{1}{3} \sqrt{9 - x^2}$.
Additional Edit
If we plug back in the above expressions we get
\begin{eqnarray*}
\frac{I}{27} & = & \cos t\left (\frac{1}{3}\cos^2t - 1\right ) + C \\
& = & \frac{1}{3} \sqrt{9-x^2}\left (\frac{1}{27}(9-x^2) - 1\right ) + C\\
I & = & \frac{1}{3} \sqrt{9-x^2} (9- x^2 - 27) + C \\
& = & -\frac{1}{3} \sqrt{9-x^2}(x^2 + 18) + C.
\end{eqnarray*}
The answer you saw on Wolfram Alpha was the result of taking through the calculation as far as it could go.
|
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|
Probability that at least 2 edges of $\Gamma_{n,N}$ shall have a point in common In the classic paper of Erdos,Renyi On the evolution of random graphs[page 7] ,it is argued that the probability that at least 2 edges of $\Gamma_{n,N}$ shall have a point in common is given by $1-\frac{{n\choose 2N}(2N)!}{2^N N!{{n \choose 2}\choose N}} = \mathcal{O}(\frac{N^2}{n})$.
Here $n$ is order and $N$ is size of the random graph.
Can anyone please describe how the asymptotic formula is derived?
IN general, how to derive such results involving factorials and powers?
|
An important condition on $N$ in that paper is that $N = o(n^{1/2}).$ This allows us to more easily deal with the binomial coefficient terms. Namely, since $N = o(n^{1/2})$, we have that
$$
{n \choose 2N} = (1+o(1)) \frac{n^{2N}}{(2N)!},
$$
and
$$
{{n \choose 2} \choose N} = (1+o(1)) \frac{ {n \choose 2}^N}{N!}.
$$
But we will need to know more about these $o(1)$ terms and so we have to go deeper. To do this, let's expand the first binomial expression:
\begin{align*}
{n \choose 2N} &= \frac{1}{(2N)!} \prod_{i=0}^{2N-1} (n - i) \\&= \frac{n^{2N}}{(2N)!} \prod_{i=0}^{2N-1} \left( 1 - \frac{i}{n} \right).
\end{align*}
Further, note that
\begin{align*}
\prod_{i=0}^{2N-1} \left( 1 - \frac{i}{n} \right) &= \exp \sum_{i=0}^{2N-1} \log \left(1- \frac{i}{n}\right) \\ & = \exp \sum_{i=0}^{2N-1} \left( - \frac{i}{n} + O\left(\frac{N^2}{n^2}\right) \right) \\&= \exp \left( \frac{-2N(2N-1)}{n} + O\left( \frac{N^3}{n^2}\right)\right) \\&= \exp \left(O \left( \frac{N^2}{n} \right)\right) = 1+ O\left( \frac{N^2}{n} \right).
\end{align*}
Thus
$$
{n \choose 2N} = \left(1+O\left(\frac{N^2}{n}\right)\right)\frac{n^{2N}}{(2N)!}.
$$
The asymptotic expression for the other binomial expression can be shown using a similar method to get
$$
{{n \choose 2} \choose N} = \left(1+O\left(\frac{N^2}{{n\choose 2}}\right)\right)\frac{{n \choose 2}^{N}}{N!}.
$$
Putting these all together, we have that
\begin{align*}
\frac{ { n \choose 2N} (2N)!} {2^N N! {{n \choose 2} \choose N}} &= \left(1+O\left(\frac{N^2}{n}\right)\right) \frac{ n^{2N}}{2^N {n \choose 2}^N} \\&= \left(1+O\left(\frac{N^2}{n}\right)\right) \frac{n^{2N}}{2^N n^N (n-1)^N/2^N} \\&= \left(1+O\left(\frac{N^2}{n}\right)\right) \frac{1}{\left(1-\frac{1}{n}\right)^N}.
\end{align*}
Now note that $\left(1-\frac{1}{n}\right)^N = 1+O\left(\frac{N}{n}\right).$
As a consequence,
$$
\frac{ { n \choose 2N} (2N)!} {2^N N! {{n \choose 2} \choose N}} = 1+O\left(\frac{N^2}{n}\right).
$$
In this case, we did not need to approximate the factorials since each factorial was divided out. If you want an asymptotic formula for a factorial, you would use Stirling's approximation. Namely,
$$
n! = (1+o(1)) \sqrt{2\pi \, n} \, \left(\frac{n}{e}\right)^n.
$$
If you wanted to approximate further, that $o(1)$ error term above is actually on the order of $1/n$.
|
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|
Prove that $2+2\sqrt{12n^2+1}$ is perfect square Problem:
Let $n \in \mathbb{N}$ such that $2+2\sqrt{12n^2+1}$ is the integer. Prove that $2+2\sqrt{12n^2+1}$ is perfect square.
I tried to found $n$ such that $\sqrt{12n^2+1}$ is integer, i.e. $12n^2 + 1 = k^2$. It is a Pell equation, and it has solution: $(n_0,k_0) = (0,1)$,
$$\begin{cases} n_{i+1} &=& 7n_i + 2k_i \\ k_{i+1} &=& 24n_i + 7k_i.\end{cases}$$ You can see this link. But I don't know how the next step. Or maybe you have another approach.
|
Let $$12n^2+1=4k^2+4k+1 \Rightarrow 3n^2=k^2+k=k(k+1)$$
This implies that $k=3a^2, k+1=b^2$ or $k=b^2, k+1=3a^2$. However, the second case is impossible as $$b^2 =3a^2-1 \equiv -1 \pmod 3$$
Thus, we have that $k+1$ is a perfect square.
Note $$2+2\sqrt{12n^2+1}=4k+4=(2b)^2$$ Our proof is done.
|
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|
Prove that $\int_{0}^{\infty}{1\over x^4+x^2+1}dx=\int_{0}^{\infty}{1\over x^8+x^4+1}dx$ Let
$$I=\int_{0}^{\infty}{1\over x^4+x^2+1}dx\tag1$$
$$J=\int_{0}^{\infty}{1\over x^8+x^4+1}dx\tag2$$
Prove that $I=J={\pi \over 2\sqrt3}$
Sub: $x=\tan{u}\rightarrow dx=\sec^2{u}du$
$x=\infty \rightarrow u={\pi\over 2}$, $x=0\rightarrow u=0$
Rewrite $(1)$ as
$$I=\int_{0}^{\infty}{1\over (1+x^2)^2-x^2}dx$$
then
$$\int_{0}^{\pi/2}{\sec^2{u}\over \sec^4{u}-\tan^2{u}}du\tag3$$
Simplified to
$$I=\int_{0}^{\pi/2}{1\over \sec^2{u}-\sin^2{u}}du\tag4$$
Then to
$$I=2\int_{0}^{\pi/2}{1+\cos{2u}\over (2+\sin{2u})(2-\sin{2u})}du\tag5$$
Any hints on what to do next?
Re-edit (Hint from Marco)
$${1\over x^8+x^4+1}={1\over 2}\left({x^2+1\over x^4+x^2+1}-{x^2-1\over x^4-x^2+1}\right)$$
$$M=\int_{0}^{\infty}{x^2+1\over x^4+x^2+1}dx=\int_{0}^{\infty}{x^2\over x^4+x^2+1}dx+\int_{0}^{\infty}{1\over x^4+x^2+1}dx={\pi\over \sqrt3}$$
$$N=\int_{0}^{\infty}{x^2-1\over x^4-x^2+1}dx=0$$
$$J=\int_{0}^{\infty}{1\over x^8+x^4+1}dx={1\over 2}\left({\pi\over \sqrt3}-0\right)={\pi\over 2\sqrt3}.$$
|
Both integrands are even, so the integrals can be extended to negative infinity. The integrands decay quickly enough for the contour to be closed with a semicircle at infinity in the upper half plane without chaning the integral. Then the integrals can be determined with the residue theorem.
Multiplying the denominator of the first integrand by $x^2-1$ yields $x^6-1$, so there are simple poles at $x=\omega_k=\exp\left(\frac{2\pi\mathrm i k}6\right)$ for $k=1,2,4,5$. The ones for $k=1,2$ lie in the upper half plane, and the corresponding residues are
$$
r_k=\lim_{x\to\omega_k}\frac{\left(x^2-1\right)\left(x-\omega_k\right)}{x^6-1}=\lim_{x\to\omega_k}\frac{(x-\omega_0)(x-\omega_3)(x-\omega_k)}{\prod_{j=0}^5(x-\omega_j)}\;,
$$
with
$$
r_1=\frac1{(\omega_1-\omega_2)(\omega_1-\omega_4)(\omega_1-\omega_5)}=-\frac14-\frac1{4\sqrt3}\mathrm i
$$
and
$$
r_2=\frac1{(\omega_2-\omega_1)(\omega_2-\omega_4)(\omega_2-\omega_5)}=\frac14-\frac1{4\sqrt3}\mathrm i
$$
Thus the contour integral is
$$
2\pi\mathrm i\left(\left(-\frac14-\frac1{4\sqrt3}\mathrm i\right)+\left(\frac14-\frac1{4\sqrt3}\mathrm i\right)\right)=\frac\pi{\sqrt3}\;,
$$
and your real integral is half of that.
For the second integral, multiplying the denominator by $x^4-1$ yields $x^{12}-1$, so there are simple poles at $\nu_k=\exp\left(\frac{2\pi\mathrm ik}{12}\right)$ for $k=1,2,4,5,7,8,10,11$, of which the ones for $k=1,2,4,5$ lie in the upper half plane. I'll leave it to you to find and add their residues as above.
|
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|
Prove that $\frac{2^a+1}{2^b-1}$ is not an integer
Let $a$ and $b$ be positive integers with $a>b>2$. Prove that $\frac{2^a+1}{2^b-1}$ is not an integer.
This is equivalent to showing there always exists some power of a prime $p$ such that $2^a+1 \not \equiv 0 \pmod{p^a}$ but $2^b-1 \equiv 0 \pmod{p^a}$. How do we prove the statement from this or is there an easier way?
|
Assume that $2^b-1$ is a divisor of $2^a+1$ and write $a$ as $kb+r$ with $0\leq r<b$. Then:
$$2^a+1\equiv (2^b)^k\cdot 2^r+1 \equiv 2^r+1 \pmod{(2^b-1)} $$
but since $r<b$ and $b>2$, $2^r+1$ is too small to be $\equiv 0\pmod{2^b-1}$.
|
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|
Given: 2 lines containing the diameter of a circle and a point lying on this circle; Find: the equation of this circle The lines $ y = \frac{4}{3}x - \frac{5}{3} $ and $ y = \frac{-4}{3}x - \frac{13}{3} $ each contain diameters of a circle. and the point $ (-5, 0) $ is also on that circle.
Find the equation of this circle.
|
Prerequisites
Given
*
*We have 2 lines with given equations (see below).
*
*$ y = \frac{4}{3}x - \frac{5}{3} $
*$ y = \frac{-4}{3}x - \frac{13}{3} $
*These lines each contain a diameter of the circle-in-question.
*We have a point with the given coordinate (see below).
*
*$ P = (-5, 0) $
*This point is on the circle in question.
Problem
Find the equation of the circle-in-question.
Here is our progress:
$ (x - h)^2 + (y - k)^2 = r^2 $
Solution
Step 1
Let us find the center of the circle-in-question.
Note that diameters of circles go through the center of a circle.
We can find the intersection of the two given lines.
Let us find the x-component of the center of the circle-in-question.
$ \frac{4}{3}x - \frac{5}{3} = \frac{-4}{3}x - \frac{13}{3} $
$ 4x - 5 = -4x - 13 $
$ 8x = -8 $
$ x = -1 $
Let us find the y-component of the center of the circle-in-question.
$
y
= \frac{4}{3}x - \frac{5}{3}
= \frac{4}{3}(-1) - \frac{5}{3}
= \frac{-4}{3} - \frac{5}{3}
= \frac{-9}{3}
= -3
$
So...
The center of the circle-in-question is $ Q = (-1, -3) $.
Here is our progress:
$ (x - (-1))^2 + (y - (-3))^2 = r^2 $
$ (x + 1)^2 + (y + 3)^2 = r^2 $
Step 2
Let us find the radius of the circle-in-question.
We know that the radius of the circle-in-question is the distance between the center of the circle-in-question and the given point that lies on the circle-in-question.
Let us find the distance between such points.
$
l
= \sqrt{((-1)-(-5))^2+((-3)-(0))^2}
= \sqrt{(4)^2+(-3)^2}
= \sqrt{16+9}
= \sqrt{25}
= 5
$
So...
The length of the radius is $ r = 5 $
Here is our progress:
$ (x + 1)^2 + (y + 3)^2 = (5)^2 $
$ (x + 1)^2 + (y + 3)^2 = 25 $
Answer
The equation of the circle-in-question is:
$$ (x + 1)^2 + (y + 3)^2 = 25 $$
|
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Find the values of a and b so that $x^4+x^3+8x^2+ax+b$ is exactly divisible by $x^2+1$ I have been trying this question for a long time but I am not getting it. So please help me and try to make it as fast as possible
|
Method 1.
Subtracting $x^2(x^2+1)$ from $x^4+x^3+8x^2+a x+b$ gives $x^3+7 x^2 +a x+b.$
Subtracting $x(x^2+1)$ from $x^3+7 x^2+a x+b$ gives $7 x^2+(a-1)x+b.$
Subtracting $7(x^2+1)$ from $7 x^2+(a-1)x+b$ gives $(a-1)x+(b-7).$
Observe that $(a-1)x+(b-7)=0$ for every $x$ if and only if $a=1$ and $b=7.$
Method 2.
$f(x)=x^4+x^3+8 x^2+a x+b$ is divisible by $x^2+1$ if $f(i)=f(-i)=0,$ where $i^2+1=0.$
We have $i^3=i(i^2)=i(-1)=-i$ , and $i^4=(i^2)^2=(-1)^2=1.$
So $f(i)=(b-7)+i(a-1)$ and $f(-i)=(b-7)-i(a-1).$
These are both $0$ if and only if $b-7=a-1=0.$
|
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|
The roots of the equation $x^2 - 6x + 7 = 0$ are $α$ and $β$. Find the equation with roots $α + 1/β$ and $β + 1/α$. Quadratic equation question, as specified in the title.
The roots of the equation $x^2 - 6x + 7 = 0$ are $α$ and $β$. Find the equation with roots $α + \frac{1}{β}$ and $β + \frac{1}{α}$.
I gather that $α + β = -\frac{b}{a} = \frac{6}{1} = 6$ and that $αβ = \frac{c}{a} = \frac{7}{1} = 7$. Do I need to convert $α + \frac{1}{β}$ and $β + \frac{1}{α}$ into a format whereby I can sub in the values for adding together or multiplying $α$ and $β$ ? If so, how ?
|
As $\alpha\beta=\dfrac71,$
let $y=\dfrac{\alpha\beta+1}\alpha=\dfrac{7+1}\alpha\iff\alpha=\dfrac8y$
Put the value of $\alpha$ in $$x^2-6x+7=0$$ and rearrange.
|
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|
Prove that $\sum_{x=1}^{n} \frac{1}{x (x+1)(x+2)} = \frac{1}{4} - \frac{1}{2 (n+1) (n+2)}$
Prove that $\displaystyle \sum_{x=1}^{n} \frac{1}{x (x+1)(x+2)} = \frac{1}{4} - \frac{1}{2 (n+1) (n+2)}$.
I tried using the partial fraction decomposition $a_j = \frac{1}{2j} - \frac{1}{j+1} + \frac{1}{2(j+2)}$, but I don't see how that helps.
|
Hint. From what you wrote one may observe that
$$
\frac{1}{j (j+1)(j+2)}=\frac{1}{2j} - \frac{1}{j+1} + \frac{1}{2(j+2)}=\frac12\left(\frac1j - \frac1{j+1}\right)+ \frac12\left(\frac1{j+2} - \frac1{j+1}\right)
$$ then noticing that terms telescope.
Remark. One may also write
$$
\frac{1}{j (j+1)(j+2)}=\frac12\frac1{j(j+1)}-\frac12\frac1{(j+1)(j+2)}
$$ then by summing terms telescope giving
$$
\sum_{j=1}^n\frac{1}{j (j+1)(j+2)}=\frac12\frac1{1\times(1+1)}-\frac12\frac1{(n+1)(n+2)}= \frac{1}{4} - \frac{1}{2 (n+1) (n+2)}
$$
as announced.
|
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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.