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Fast/smart way to write polar curve in cartesian Is there a fast way to write the curve:
$$r=\frac{a}{1-\frac{1}{\sqrt{2}}\cos(\theta)}$$
as a cartesian curve $f(x,y)=0$?
It seems I can take
$$r(1-\frac{1}{\sqrt{2}}\cos(\theta)) = a$$
$$r-\frac{x}{\sqrt{2}}=a$$
$$\sqrt{x^2+y^2}-\frac{x}{\sqrt{2}}=a$$
$$x^2+y^2=(a+\frac{x}{\sqrt{2}})^2$$
and then expand out, and complete the square. But I seem to get an error. Perhaps there is a smart way to do this?
|
\begin{align*}
(a + \frac{x}{\sqrt{2}})^2 = a^2 + ax\sqrt{2} + \frac{x^2}{2}
\end{align*}
And then you can rewrite the equation
\begin{equation*}
\frac{x^2}{2} - ax\sqrt{2} + y^2 = a^2
\end{equation*}
as
\begin{equation*}
\left( \frac{x}{\sqrt{2}} - a \right)^2 + y^2 = \left( \frac{x - a\sqrt{2}}{\sqrt{2}} \right)^2 + y^2 = 2a^2
\end{equation*}
And then you can divide through by $2a^2$ to get the equation for a shifted ellipse:
\begin{equation*}
\frac{(x-a\sqrt{2})^2}{4a^2} + \frac{y^2}{2a^2} = 1
\end{equation*}
Unfortunately as far as I can tell there is no "smart" way to do it -- it's just crunching numbers.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2389057",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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"answer_id": 1
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|
What is the result of $\lim_{x\to0_+} \frac{e^x - x e^x - 1}{\left(e^x - 1 \right)^2}$ without L'Hôpital's rule. I have the limit
$$
\lim_{x\to0_+} \frac{e^x - x e^x - 1}{\left(e^x - 1 \right)^2}
$$
which I need to compute without L'Hôpital's rule.
(The result is $-\frac{1}{2}$ with L'Hôpital's rule).
Thanks.
|
It is also simple to look at Taylor expansions around $0$ up to at least second order
$$e^x = 1 + x + \frac{1}{2}x^2... $$ then $$\frac{e^x - xe^x - 1}{(e^x-1)^2} \approx \frac{1 + x + \frac{1}{2}x^2 - x - x^2 - \frac{1}{2}x^3 - 1} {(x+ \frac{1}{2}x^2)^2} = \frac{ - \frac{1}{2}x^2 - \frac{1}{2}x^3}{x^2 + \frac{1}{4}x^4 + x^3} $$ and now by looking at the lowest order terms, the conclusion that $$\lim_{x \to 0^+} = -\frac{1}{2}$$ follows quick
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2390227",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"answer_id": 2
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|
Isosceles triangle inscribed in an ellipse.
Find the maximum area of an isosceles triangle, which is inscribed inside an ellipse $\dfrac {x^2} {a^2} + \dfrac {y^2} {b^2} = 1$ with its (unique) vertex lying at one of the ends of the major axis of the ellipse.
The three vertices of the triangle would be $(a,0), (x,y), (x, -y)$.
The area of the triangle by Heron's formula is $$A^2 = (x-a)^2y^2 = (x-a)^2b^2\left( 1-
\dfrac{x^2}{a^2}\right) \tag{1}.$$
Hence $$\dfrac{dA}{dx} = 0 \implies (x-a)^2 \left( x + \dfrac{a}2\right) = 0.$$
We have minimum at $x = a$ and maximum at $x = -\frac{a}{2}$.
Substituting back in $(1)$ and taking square roots on both the sides gives $$A = \dfrac{\sqrt{3}ab}{4}$$.
The given answer is $3A$.
What mistake I made ? Is it possible to find the answer to this question if the triangle was scalene ?
|
By AM-GM
$$S_{\Delta}=(a-x)y=(a-x)b\sqrt{1-\frac{x^2}{a^2}}=ab\left(1-\frac{x}{a}\right)\sqrt{1-\frac{x^2}{a^2}}=$$
$$=\frac{ab}{\sqrt3}\cdot\sqrt{\left(1-\frac{x}{a}\right)^3\left(3+\frac{3x}{a}\right)}\leq\frac{ab}{\sqrt3}\cdot\sqrt{\left(\frac{3\left(1-\frac{x}{a}\right)+3+\frac{3x}{a}}{4}\right)^4}=\frac{3\sqrt3ab}{4}.$$
The equality occurs for $1-\frac{x}{a}=3+\frac{3x}{a}$, which says that the equality indeed occurs,
which says that $\frac{3\sqrt3ab}{4}$ is a maximal value.
Done!
|
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"timestamp": "2023-03-29T00:00:00",
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|
When is this integer a perfect square: $n^2 + 20n + 12$
Let $S_n = n^2 + 20n + 12$, where $n$ a positive integer. What is the sum of all possible values of $n$, for which $S_n$ is a perfect square?
Help me to solve it.
|
Use $$(n+4)^2<n^2+20n+12<(n+10)^2.$$
I got $n^2+20n+12=(n+8)^2$, which gives $n=13$ or
$n^2+20n+12=(n+6)^2$, which gives $n=3$.
Id est, the answer is $16$
because in cases $n^2+20n+12=(n+5)^2$, $n^2+20n+12=(n+7)^2$ and $n^2+20n+12=(n+9)^2$ we can not get solutions.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2391025",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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|
Given prime $p$, for any integer $a$, exist integer $x,y$ such that $p\mid y^2+x^3+a$. Let $p$ be prime number and $a$ an integer. Show that the equation
$$y^2+x^3+a\equiv 0\pmod p$$has integer solutions.
It seem a famous problem? But I can't prove it.
|
The solutions to problem N8 of the 53rd IMO, as mentioned by Gerry Myerson, do indeed apply to this problem. In fact it can be generalised. For any strictly positive integer $n$ and prime power $q=p^m$, then for any element $r$ of the finite field $F_q$ of size $q$, the equation
$$
y^2+x^n=r\qquad{\rm(*)}
$$
has solutions for $x,y\in F_q$, so long as $q$ is large enough. I'll show that $q\ge 2n^2-3n+2$ is enough.
In the question above, we have $n=3$ and, using mod-$p$ arithmetic is the same as working in the field of size $p$, so there are solutions whenever $p\ge 11$. This only leaves $p=2,3,5,7$ to be checked by hand.
As the case with $n=1$ is trivial, I'll assume that $n>1$. Also, if $q$ is a power of $2$ then squaring is injective in $F_q$, so (*) has solutions with $x=0$. I will therefore assume that $q$ is odd. Let $N$ be the number of solutions in $F_q$ to
$$
a^2+b^n= c^2+d^n.\qquad{\rm(**)}
$$
Upper bound for $N$: First, let $k$ be the number of solutions to $b^n=d^n$. There is one solution with $b=d=0$. Then, for each $d\not=0$ there are at most $n$ possibilities for $b$. So,
$$
k\le 1+n(q-1).
$$
Similarly, replacing $n$ by $2$, there are at most $1+2(q-1)=2q-1$ solutions to $a^2=c^2$. Putting these together, the number of simultaneous solutions in $F_q$ to $a^2=c^2$ and $b^n=d^n$ is at most $(2q-1)k$.
Now looking at each of the $q^2-k$ solutions to $b^n\not=d^n$, we can write (**) as $(a+c)(a-c)=d^n-b^n$. There are at most $q-1$ possibilities for $a-c$, which uniquely determines $a+c$ and, hence, both $a$ and $c$ (this uses the fact that $q$ is odd, so we can divide by $2$ in $F_q$).
\begin{align}
N&\le(2q-1)k+(q^2-k)(q-1)=q(q^2-q+k)\\
&\le q(q^2+(n-1)q+1-n)\qquad{\rm(1)}
\end{align}
Lower bound for $N$: Letting $s_r$ be the number of solutions to $y^2+x^n=r$ in $F_q$, by counting the number of solutions to $a^2+b^n=r=c^2+d^n$ for each $r$, we have
$$
N=\sum_{r\in F_q}s_r^2.\qquad{\rm(2)}
$$
I'll calculate a lower bound for $N$ in the situation that there exists some $r\in F_q$ for (*) has no solution -- i.e., that $s_r=0$. Letting $(a,b)$ be a solution to $a^2+b^n=r$ and $z=t^{2n}\not=0$ be a $2n$'th power in $F_q$,
$$
(at^n)^2+(bt^2)^n=rz,
$$
from which we see that $s_{rz}=s_r$. As there are at least $(q-1)/(2n)$ nonzero powers of $2n$ in $F_q$, we get $s_r=0$ for at least $(q-1)/(2n)$ values of $r$. Applying Cauchy-Schwartz to (2),
$$
N\ge\frac1{q-(q-1)/2n}\left(\sum_{r\in F_q}s_r\right)^2.
$$
However, $\sum s_r$ is just the number of solutions to $a^2+b^n=r$, which is $q^2$. So,
$$
N\ge\frac{q^4}{q-(q-1)/2n}\qquad{\rm(3)}
$$
Finally, in the situation that (*) has no solution for some $r\in F_q$, we can combine inequalities (1) and (3) to get
$$
q(q^2+(n-1)q+1-n)\ge\frac{q^4}{q-(q-1)/2n}.
$$
Rearranging,
$$
q^3+2(n-1)^2q+n-1\le (2n^2-3n+2)q^2.
$$
As the final two term on the left are positive, this gives $q <2n^2-3n+2$.
|
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"timestamp": "2023-03-29T00:00:00",
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|
$\lim_{(x,y)\rightarrow (0,0)}\frac{x^5-\sin(y^3)}{4x^4+3y^2}$ Are the next computations correct?
$$
0\leq \left|\frac{x^5-\sin(y^3)}{4x^4+3y^2}\right|
\leq\frac{|x^5|}{4x^4}+\frac{|y^3|}{3y^2}
=\frac{|x|}{4}+\frac{|y|}{3}
$$
Then, by squeeze theorem, the limit is zero.
|
$$0\leq |\frac{x^5-\sin(y^3)}{4x^4+3y^2}|\leq\frac{|x^5|}{4x^4+3y^2}+\frac{|y^3|}{3y^2+4x^4} \leqslant\frac{|x|}{4}+\frac{|y|}{3} \leqslant \frac{|x|+|y|}{3}=\frac{\sqrt{x^2}+\sqrt{y^2}}{3}\leqslant \frac{2(\sqrt{x^2+y^2})}{3}=\frac{2 \delta}{3}$$
If you take $\delta=\frac{3 \epsilon}{2}$ you have the $\epsilon-\delta$ proof of your limit.
Also as for a second solution:
$$0\leq |\frac{x^5-\sin(y^3)}{4x^4+3y^2}|\leq\frac{|x^5|}{4x^4+3y^2}+\frac{|y^3|}{3y^2+4x^4}$$
Compute the limits $$ \lim_{(x,y) \rightarrow (0,0)}\frac{|x^5|}{4x^4+3y^2}$$ $$\lim_{(x,y) \rightarrow (0,0)}\frac{|y^3|}{3y^2+4x^4}$$
by taking polar coordinates:
$$x=r\cos{\theta}$$ $$y=r\sin{\theta}$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2392352",
"timestamp": "2023-03-29T00:00:00",
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|
If $450^\circ<\alpha<540^\circ$ and $\cot\alpha=-\frac{7}{24},$ calculate $\cos\frac{\alpha}{2}$. Why is my solution wrong? The problem says:
If $450^\circ<\alpha<540^\circ$ and $\cot\alpha=-\frac{7}{24},$ calculate $\cos\frac{\alpha}{2}$
I solved it in the following way: $$\begin{align} -\frac{7}{24}&=-\sqrt{\frac{1+\cos2\alpha}{1-\cos2\alpha}}\\ \frac{49}{576}&=\frac{1+\cos2\alpha}{1-\cos2\alpha}\\ 625\cos2\alpha&=527\\ 2\cos^2\alpha-1&=\frac{527}{625}\\ \cos\alpha&=-\frac{24}{25}, \end{align}$$ therefore, $$\begin{align} \cos\frac{\alpha}{2}&=\sqrt{\frac{1-\frac{24}{25}}{2}}\\ &=\sqrt{\frac{1}{50}}\\ &=\frac{\sqrt{2}}{10}. \end{align}$$ But there is not such an answer:
A) $0.6$
B) $\frac{4}{5}$
C) $-\frac{4}{5}$
D) $-0.6$
E) $0.96$
I have checked the evaluating process several times. While I believe that my answer is correct and there is a mistake in the choices, I want to hear from you.
|
Hint:
$$\cot\alpha=\frac{1-\tan^2\frac\alpha2}{2\tan\frac\alpha2}=\frac{1-t^2}{2t}$$ or
$$t^2+2\cot\alpha\ t-1=0,$$
or
$$t=-\cot\alpha\pm\sqrt{1-\cot^2\alpha}=-\frac34,\frac43.$$
Then $$\cos\frac\alpha2=\pm\frac1{\sqrt{1+t^2}}=\pm\frac1{\sqrt{2-2\cot\alpha\ t}}=\pm\frac45,\pm\frac35.$$
Then given range of $\alpha$ tells you that
$$-\frac1{\sqrt2}<\cos\frac\alpha2<0.$$
|
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"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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|
If $x^3+y^3+(x+y)^3+33 xy=2662$, $x,y\in \Bbb R$, find $S=x+y$.
If $x^3+y^3+(x+y)^3+33 xy=2662$ and $\{x,y\}\subset \Bbb R$, find $S=x+y$.
This question from an olympiad contest. Answer stated: $S=x+y=11$
Tried to develop $(x+y)^3$ to find something useful for the situation, but without success.
|
Let $x+y=S$
$$S^3+(S-x)^3+33 x (S-x)+x^3$$
has a derivation
$$ \frac{d}{dx}(x^3+(S-x)^3+S^3+33 x (S-x) )= -3 (-11 + S) (S - 2 x) $$
So, sollution $ \frac{d}{dx}=0$ leads to $S=11$ or $ S=2x$, i.e. $x=y$
For $x=y$, $x^3+33 x y+(x+y)^3+y^3=10x^3+33x^2=2662$ has a unique real solution $x(=y)=11/2$
For $y=c-x$, function is a ploynomial $2 S^3 + (33 S - 3 S^2) x + (-33 + 3 S) x^2$ with determinant
$$ -351384 + 31944 S + 1089 S^2 + 66 S^3 - 15 S^4 $$
which is negative except $S=11$
|
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"timestamp": "2023-03-29T00:00:00",
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|
Finding the exterma of $x^2+y^2+z^2-yz-zx-xy$ s.t. $x^2+y^2+z^2-2x+2y+6z+9=0$ using Lagrange's multiplier,
Using Lagrange's multiplier method, obtain the maxima and minima of $$x^2+y^2+z^2-yz-zx-xy$$ subject to the condition $$x^2+y^2+z^2-2x+2y+6z+9=0$$
My attempt:
I formed the expression
$$F=x^2+y^2+z^2-yz-zx-xy+\lambda(x^2+y^2+z^2-2x+2y+6z+9)=0$$
Differentiated partially wrt $x$, $y$ and $z$, and equated to $0$.
I get the following equations
$$(2\lambda+2)x-y-z=2\lambda$$
$$-x+(2\lambda+2)y-z=-2\lambda$$
$$-x-y+(2\lambda+2)z=6\lambda$$
In matrix form:
$$\begin{bmatrix} 2\lambda+2 & -1 & -1 \\-1 & 2\lambda+2 & -1 \\ -1 & -1 & 2\lambda+2 \end{bmatrix} \begin{bmatrix} x \\ y \\ z \end{bmatrix}=\lambda \begin{bmatrix}2 \\ -2 \\ -6 \end{bmatrix} $$
When I try to find $x$, $y$ and $z$ from these equations, it gets complex. Is there an easier way to ascertain their values?
|
You solve the system and get
$\left[x= -\dfrac{3-2 \lambda}{2 \lambda+3},\;y= -1,\;z= -\dfrac{3 (2 \lambda+1)}{2 \lambda+3}\right]$
then you use the constraint
$x^2+y^2+z^2-2x+2y+6z+9=0$
and substitute. Simplify
$(2 \lambda-3) (2 \lambda+9)=0$
$\lambda=\dfrac{3}{2};\;\lambda=-\dfrac{9}{2}$
Substitute again in the solutions and get
$x= 2,\;y= -1,\;z= -4$ we have a maximum $27$
and
$x = 0,\; y = -1,\; z = -2$ we have a minimum $3$
hope this helps
|
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|
Evaluate $(\sqrt{3}-3i)^6$
Evaluate $$(\sqrt{3}-3i)^6.$$
So I assume that we should write the following in polar form
$r=\sqrt{(\sqrt{3})^2+(-3)^2}=\sqrt{3+9}=\sqrt{12}=2\sqrt{3},$
$\theta=\arctan{\frac{-3}{\sqrt{3}}}=-\frac{\pi}{3}.$
So $$(\sqrt{3}-3i)^6=[2\sqrt{3}e^{-i\frac{\pi}{3}+2\pi k}]^6=1728e^{-2\pi i+2\pi m},$$ where $m\in \mathbb{Z}.$
So $$1728e^{-2\pi i+2\pi m}=1728cos(-2\pi)=1728.$$
Is it correct?
|
You are right. It's
$$27\cdot2^6\cos^6(-60^{\circ})=1728.$$
|
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|
Find an invertible matrix $S$ and a matrix $J$ in Jordan form, such that $S^{-1}AS = J$ $A = \begin{pmatrix}
0 & -1 & 0 & 0 \\
2 & 0 & -1 & 0 \\
3 & -1 & -2 & -1 \\
-1 & 0 & 1 & 1 \\
\end{pmatrix}$
We need to find an invertible matrix $S$ and a matrix in Jordan form $J$, such that $S^{-1}AS=J$.
I've found that the characteristic polynomial is $P_A(\lambda) = \lambda^3(\lambda+1)$. Therefore, $\lambda = 0,-1$.
The minimal polynomial is $m_A(\lambda) = P_A(\lambda).$
At this stage, we know that $J = \begin{pmatrix}
-1 & 0 & 0 & 0 \\
0 & 0 & 1 & 0 \\
0 & 0 & 0 & 1 \\
0 & 0 & 0 & 0 \\
\end{pmatrix}$ (up to blocks' order).
Next, I've found the Null Spaces:
For $\lambda=0$:
$N(A) = N \begin{pmatrix}
0 & -1 & 0 & 0 \\
2 & 0 & -1 & 0 \\
3 & -1 & -2 & -1 \\
-1 & 0 & 1 & 1 \\
\end{pmatrix} = span \begin{Bmatrix} \begin{pmatrix}
-1 \\ 0 \\ -2 \\ 1
\end{pmatrix} \end{Bmatrix}$
$N(A^2) = N \begin{pmatrix}
-2 & 0 & 1 & 0 \\
-3 & -1 & 2 & 1 \\
-7 & -1 & 4 & 1 \\
2 & 0 & -1 & 0 \\
\end{pmatrix} = span \begin{Bmatrix}
\begin{pmatrix}
1 \\ 1 \\ 2 \\ 0
\end{pmatrix},
\begin{pmatrix}
0 \\ 1 \\ 0 \\ 1
\end{pmatrix}
\end{Bmatrix}$
$N(A^3) = N \begin{pmatrix}
3 & 1 & -2 & -1 \\
3 & 1 & -2 & -1 \\
9 & 3 & -6 & -3 \\
3 & -1 & 2 & 1 \\
\end{pmatrix} = span \begin{Bmatrix}
\begin{pmatrix}
-1 \\ 3 \\ 0 \\ 0
\end{pmatrix},
\begin{pmatrix}
2 \\ 0 \\ 3 \\ 0
\end{pmatrix},
\begin{pmatrix}
1 \\ 0 \\ 0 \\ 3
\end{pmatrix}
\end{Bmatrix}$
For $\lambda=-1$:
$N(A+I) = N \begin{pmatrix}
1 & -1 & 0 & 0 \\
2 & 1 & -1 & 0 \\
3 & -1 & -1 & -1 \\
-1 & 0 & 1 & 2 \\
\end{pmatrix} = span \begin{Bmatrix} \begin{pmatrix}
1 \\ 1 \\ 3 \\ -1
\end{pmatrix} \end{Bmatrix}$
I'm not sure how to continue from here...
|
Once you have found $J$, finding $S$ is very easy. Just write $S=(s_{ij})$ and solve the linear equations in these coefficients given by the equation
$$
AS=SJ.
$$
There are many solutions for $S$ here, and you have to pick one which satisfies $\det(S)\neq 0$.
Edit: Say, $S=(x_1,x_2,x_3,x_4)$ with column vectors $x_1=(s_1,s_2,s_3,s_4)^T$ until $x_4=(s_{13},s_{14},s_{15},s_{16})^T$. Then
$AS=SJ$ is equivalent to $16$ easy equations
$$
s_1-s_2 = 0,\; s_6 =0, \; \cdots ,\; - s_{12} - s_{13} + s_{15}
+ s_{16} = 0.
$$
This can be solved even by hand in a few minutes, e.g.,
$$
S=\begin{pmatrix} 1 & 1 & -1 & 0 \cr 1 & 0 & -1 & 1 \cr
3 & 2 & -2 & 1 \cr -1 & -1 & 0 & -1\end{pmatrix}
$$
|
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"timestamp": "2023-03-29T00:00:00",
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|
Pattern recognition for $S_n=2+5+13+35+...+n^{th}_{term}$ I want to find sum of the first $n^{th}$ term of this sqequence .
$$2,5,13,35,97,275,793,...\\s_n=2+5+13+35+97+...$$
What is the closed form formula for $s_n$?
|
Alternatively, the sequence is:
$$2,3\cdot 2-1,3\cdot 5-2,3\cdot 13-4,3\cdot 35-8,...$$
It is the recurrence relation:
$$a_n=3a_{n-1}-2^{n-2},a_1=2.$$
Divide it by $2^{n}$:
$$\frac{a_n}{2^{n}}=\frac{3a_{n-1}}{2^{n}}-\frac14.$$
Denote: $b_n=\frac{a_n}{2^n}$ to get:
$$b_n=\frac{3}{2}b_{n-1}-\frac{1}{4},b_1=1.$$
Solution is:
$$b_n=\frac{1}{3}\left(\frac{3}{2}\right)^n+\frac12.$$
Hence:
$$a_n=2^nb_n=2^{n-1}+3^{n-1}.$$
Now the sum $S_n$ is calculated in the same way as in previous solution.
|
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|
Functional equation : $ f(1)^3 + f(2)^3 + \ldots + f(n)^3 = (f(1) + f(2) + \ldots + f(n))^2$ Find all function $f:\mathbb{N}\rightarrow\mathbb{N}\cup\{0\}$ satisfying
$$ f(1)^3 + f(2)^3 + \ldots + f(n)^3 = (f(1) + f(2) + \ldots + f(n))^2$$
$\forall n \in \mathbb{N}$.
Thank you, Batominovski and Guy Fabrice..
Is my understanding correct ? Please let me know if there is any mistake.
Substitute $n=1$, $f(1)^3 =f(1)^2$, so $f(1)=0$ or $1$
$f(1)=0$ : substitute $n=2$, $f(2)^3 =f(2)^2$, so $f(2)=0$ or $1$
If $f(2) = 0$, let $l$ be the maximal value such that $f(l) =0$.
so $ f(1)^3 + f(2)^3 + \ldots + f(l+1)^3 = (f(1) + f(2) + \ldots + f(l+1))^2$
then $f(l+1)^3=(f(l+1))^2$ so $f(l+1) = 1$
If $f(2) = 1$, then $0+1+f(3)^3=(f(1)+f(2)+f(3))^2$ , so $f(3) = 0$ or $2$
Since $ f(1)^3 + f(2)^3 + \ldots + f(n)^3 = (f(1) + f(2) + \ldots + f(n))^2$
and $ f(1)^3 + f(2)^3 + \ldots + f(n+1)^3 = (f(1) + f(2) + \ldots + f(n+1))^2$
so $f(n+1)^3 = 2(f(1)+f(2)+\ldots+f(n))f(n+1)+f(n+1))^2$
so $f(n+1)^2-f(n+1)=2(f(1)+f(2)+\ldots+f(n))$
i.e., if $f(n) \not= 0$, then $f(n)=n, \forall n \in \mathbb{N}$
so if $f(x)=0, \forall x \in \mathbb{N}$ then $f(n) = 0, \;\forall n \in \mathbb{N}$
If $f(x)$ be the value such that Max$\{f(1), f(2), \ldots ,f(x)\}=k$ then $f(x+1) = k+1$ or $0$ ---[1]
By induction, let $P(n)$ denotes [1]
Basic step , it's obvious that if $f(1) = 0$, then $f(l_i)=1$, $1\leq l_i\leq x$
so $1+f(l_i+1)^3 = (f(l_i+1) +1)^2$ so $f(l_i+1)=2$ or $0$
Inductive step, suppose that $P(k)$ is true.
$1^3+2^3+\ldots+k^3+f(x+1)^3=(1+2+\ldots+k+f(x+1))^2$,
so $f(x+1)=k+1$ or $0$, so $P(n)$ is true.
The sequence of $f$ is $\underbrace{0\ldots0}_{\text{$k_1$}}1\underbrace{0\ldots0}_{\text{$k_2$}}2\ldots$, where $k_1, k_2, \ldots \in \mathbb{N}$
Answer check :
$ f(1)^3 + f(2)^3 + \ldots + f(n)^3 = 1^3+2^3+\ldots+m^3=\left(\frac{m(m+1)}{2}\right)^2=(f(1) + f(2) + \ldots + f(n))^2$
$\blacksquare$
|
Hint: Prove by induction on $n$ that, if $f(n)\neq 0$, then
(1) we have $f(n)=M_n+1$, where $M_n:=\max\big\{f(1),f(2),\ldots,f(n-1)\big\}$ for $n>1$, and $M_1:=0$;
(2) for $n>1$, each the numbers $1,2,\ldots,M_n$ appears only once among $f(1),f(2),\ldots,f(n-1)$ in an increasing order (i.e., $i$ appears before $i+1$).
Both statements should be proven simultaneously (i.e., not with separate inductive proofs).
|
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|
How prove infinitely many postive integers triples $(x,y,z)$ such $(x+y+z)^2+2(x+y+z)=5(xy+yz+zx)$
show that there exsit infinitely many postive integers triples $(x,y,z)$
such
$$(x+y+z)^2+2(x+y+z)=5(xy+yz+zx)$$
May try it is clear $(x,y,z)=(1,1,1)$ is one solution，and
$$(x+y+z+1)^2=5(xy+yz+xz)+1$$
|
Here's a way to more constructively get the answer user399601 provided:
Note that since $(x + y + z)^2 = x^2 + y^2 + z^2 + 2(xy + yz + xz)$, the equation
\begin{equation}\label{eqn:constraint}\tag{1}
(x + y + z)^2 + 2(x + y + z) = 5(xy + yz + xz)
\end{equation}
is equivalent to
\begin{align*}
x^2 + y^2 + z^2 + 2(x + y + z) &= (x + 1)^2 + (y + 1)^2 + (z + 1)^2 - 3\\
&= 3xy + 3yz + 3xz
\end{align*}
Now take $z = 1$ to get
\begin{equation*}
(x + 1)^2 + (y + 1)^3 + 1 = 3(x + 1)(y + 1)
\end{equation*}
Making the substitutions $u = x+1$ and $v = y + 1$, the equation becomes
\begin{equation}\label{eqn:simplified}\tag{2}
u^2 + v^2 + 1 = 3uv
\end{equation}
and
\begin{equation*}
(u - v)^2 + 1 = uv
\end{equation*}
Now suppose there's some function $c:\mathbb{Z}^2 \to \mathbb{Z}$ such that $(u + c, v)$ is a solution whenever $(u, v)$ is a solution. Then
\begin{align*}
(u - v + c)^2 + 1 &= (u + c)v\\
(u - v)^2 + 2c(u-v) + c^2 + 1 &= uv + cv \\
c^2 + 2(u - v)c &= cv
\end{align*}
So $$c(u, v) = 2(v - u) + v = 3v - 2u$$
Now note that if $(u, v)$ solves equation \eqref{eqn:simplified}, then $(u-1, v-1, 1)$ solves equation \eqref{eqn:constraint}. Thus the solution $(3v - u, v)$ maps to $(3y - x + 1, y, 1)$.
|
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|
Determining if a vector is in the row space I am just trying to determine if the vector $[0, 7, 4]$ belongs in the row space of the matrix $$A = \begin{bmatrix}
1 & 2 & 0 \\
3 & -1 & 4 \\
1 & -5 & 4 \\
\end{bmatrix}
$$
What I have done so far is created an augmented matrix like so
$$ \left[
\begin{array}{ccc|c}
1 & 3 & 1 & 0 \\
2 & -1 & -5 & 7 \\
0 & 4 & 4 & 4 \\
\end{array}
\right]
$$
(naive gaussian) reducing to
$$ \left[
\begin{array}{ccc|c}
1 & 3 & 1 & 0 \\
0 & -7 & -7 & 7 \\
0 & 0 & 0 & 32/7 \\
\end{array}
\right]
$$
returning inconsistent, i.e. not existing in the row space. However apparently it does in fact belong in the row space, so clearly I have gone about this the wrong way. Is someone able to correct this?
|
A vector $\vec b$ is in the row space of $A$ if and only if $\vec b$ is in the column space of $A^\top$. Thus, to determine if the vector $\vec b=\left[\begin{array}{r}
0 \\
7 \\
4
\end{array}\right]$ is in the row space of $A = \left[\begin{array}{rrr}
1 & 2 & 0 \\
3 & -1 & 4 \\
1 & -5 & 4
\end{array}\right]$, form the augmented matrix
$$
\left[\begin{array}{rrr|r}
1 & 3 & 1 & 0 \\
2 & -1 & -5 & 7 \\
0 & 4 & 4 & 4
\end{array}\right]
$$
Row reducing gives
$$
\DeclareMathOperator{rref}{rref}\rref\left[\begin{array}{rrr|r}
1 & 3 & 1 & 0 \\
2 & -1 & -5 & 7 \\
0 & 4 & 4 & 4
\end{array}\right]=
\left[\begin{array}{rrr|r}
1 & 0 & -2 & 0 \\
0 & 1 & 1 & 0 \\
0 & 0 & 0 & 1
\end{array}\right]
$$
This gives an inconsistent system. Hence $\vec b$ is not in the row space of $A$.
|
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|
$\sin(40^\circ)<\sqrt{\frac{3}7}$
Prove without using of calculator, that $\sin40^\circ<\sqrt{\frac{3}7}$.
My attempt.
Since $$\sin(40^\circ)=2\sin(20^\circ)\cos(20^\circ)<2\sin(20^\circ)$$
$$=2\sin(60^\circ-40^\circ)=\sqrt{3} \cos(40^\circ)-\sin(40^\circ),$$
$$2\sin(40^\circ)<\sqrt{3} \cos(40^\circ).$$
Hence, $$4\sin^2(40^\circ)<3\cos^2(40^\circ)=3(1-\sin^2(40^\circ))$$
$$7\sin^2(40^\circ)<3$$
$$\sin(40^\circ)<\sqrt{\frac{3}7}$$
Is there another way to prove this inequality?
|
Applying the identity $\sin 3x=3\sin x-4\sin^3 x$ we get
$$\frac{\sqrt 3}2=\sin 120^\circ=3\sin 40^\circ-4\sin^3 40^\circ$$
Now, consider the polynonial
$$P(x)=8x^3-6x+\sqrt 3$$
We know that $\sin40^\circ$ is a root of $P$. Also, we know that $\sin 40^\circ>\sin30^\circ=1/2$. Differentiating $P$, we see that $P$ is increasing in $[1/2,\infty)$. Furthermore,
$$P\left(\sqrt{\frac 37}\right)=\frac{24}{49}\sqrt{21}-\frac67\sqrt{21}+\sqrt 3=\sqrt {21}\left(-\frac{18}{49}+\sqrt{\frac17}\right)\stackrel*>0$$
Since $\sqrt{3/7}>1/2$ and $P(\sqrt{3/7})>P(\sin 40^\circ)$, then the inequality follows.
$*$: Perhaps this inequality needs a proof.
Since $18^2=324<343=7^3$,
$$\frac{18^2}{7^4}<\frac17$$
|
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|
Convergence of sequence $\lim_{n \rightarrow \inf }n\sqrt{n}(\sqrt{n+1}-a\sqrt{n}+\sqrt{n-1})$ I am able to show that the sequence converge only if $a=2$, I also did a numeric analysis of the sequence, and I am sure that the value of this limit is $-\frac{1}{4}$. However, I do not know how to prove that
$\lim_{n \rightarrow \inf }n\sqrt{n}(\sqrt{n+1}-2\sqrt{n}+\sqrt{n-1})=-\frac{1}{4}$
|
You can consider $n=1/x$ and study the limit of the function at $0$ from the right:
$$
\lim_{x\to0^+}
\frac{1}{x}\frac{1}{\sqrt{x}}
\left(\sqrt{\frac{1}{x}+1}-a\sqrt{\frac{1}{x}}+\sqrt{\frac{1}{x}-1}\right)
=
\lim_{x\to0^+}\frac{\sqrt{1+x}-a+\sqrt{1-x}}{x^2}
$$
If $a\ne2$, the limit is infinite, because the limit of the numerator is $2-a$ (and this decides between $\infty$ and $-\infty$).
Thus we can assume $a=2$.
Now you can rationalize; first multiply by $\sqrt{1+x}-2-\sqrt{1-x}$:
$$
\lim_{x\to0^+}\frac{(1+x+4-4\sqrt{1+x})-(1-x)}{x^2(\sqrt{1+x}-2-\sqrt{1-x})}
$$
Isolating the part that has limit $-2$ and simplifying the $-2$ in the numerator, we reduce to
$$
\lim_{x\to0^+}\frac{2\sqrt{1+x}-x-2}{x^2}
\cdot
\lim_{x\to0^+}\frac{-2}{\sqrt{1+x}-2-\sqrt{1-x}}
$$
The second factor has limit $1$ and can be disregarded. Now you can rationalize again:
$$
\lim_{x\to0^+}\frac{2\sqrt{1+x}-x-2}{x^2}=
\lim_{x\to0^+}\frac{4(1+x)-(x^2+4x+4)}{x^2(2\sqrt{1+x}+x+2)}
$$
and finish up.
With a Taylor expansion, recalling that
$$
\sqrt{1+t}=1+\frac{1}{2}t-\frac{1}{8}t^2+o(t^2)
$$
we have
$$
\lim_{x\to0^+}\frac{\sqrt{1+x}-2+\sqrt{1-x}}{x^2}=
\lim_{x\to0^+}\frac{(1+x/2-x^2/8)-2+(1-x/2-x^2/8)+o(x^2)}{x^2}=-\frac{1}{4}
$$
|
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|
Polynomial : $P(a)P(a^2)P(a^3)P(a^4)$ Let $P(x) = x^3 + 2x^2+3x+4$ and $a$ be the root of equation $x^4+x^3+x^2+x+1=0$.
Find the value of $P(a)P(a^2)P(a^3)P(a^4)$
Is my answer correct ?
Since root of equation $x^4+x^3+x^2+x+1=0$ is the $5^{th}$ primitive root of 1,
so $a, a^2, a^3, a^4$ are roots of $x^4+x^3+x^2+x+1=0$
$P(a)P(a^4)=(a^3+2a^2+3a+4)(\frac{1}{a^3}+\frac{2}{a^2}+\frac{3}{a}+4)= 15+5a^4+5a$
Similarly, $P(a^2)P(a^3)=15+5a^3+5a^2$
$P(a)P(a^2)P(a^3)P(a^4)=(15+5a^4+5a)(15+5a^3+5a^2)=125$
|
Here is another way. Call the product that you want Q.
Note that $xP(x)-P(x)=x^4+x^3+x^2+x-4$
As you correctly observe the values you are substituting are all roots of $x^4+x^3+x^2+x+1=0$ and hence for these values you get $(x-1)P(x)=-5$ whence $$(a-1)(a^2-1)(a^3-1)(a^4-1)Q=(-5)^4=625$$
Now pairing the first and last factor and the middle two factors and noting that $a^5=1$ we have $$(a-1)(a^2-1)(a^3-1)(a^4-1)=(2-a-a^4)(2-a^2-a^3) =$$$$=4-2(a+a^2+a^3+a^4)+a^3+a^4+a+a^2$$ and we know that $$a^4+a^3+a^2+a=-1$$ So $5Q=625$ and $Q=125$
|
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|
take $\int_\lambda f \ ds$ (Stieltjes) where $f(x,y) = \sqrt{x^2+y^2}$ and $\lambda(t) = (a\cos t, b\sin t)$
Calculate $\int_\lambda f \ ds$, where $f:\mathbb{R}^2\to\mathbb{R}$
is given by $f(x,y) = \sqrt{x^2+y^2}$ and $\lambda:[0,
2\pi]\to\mathbb{R^2}$ is the path $\lambda(t) = (a\cos t, b\sin t)$,
which describes an ellipse.
When $\lambda$ is $C^1$ and $w$ is a continuous $1-form$, we have:
$$\int_\lambda w =\int_a^b w(\lambda(t))\cdot \lambda'(t) dt$$
so:
$$\int_\lambda f\ ds = \int_0^{2\pi} \sqrt{(a\cos t)^2 + (b\sin t)^2}\cdot (-a\sin t, b\cos t) \ dt = \\ \int_0^{2\pi} \sqrt{a^2\cos^2 t + b^2\sin^2 t} \cdot (-a\sin t, b\cos t) \ dt $$
but I can't even take the dot product... I don't hink this is the way to solve.
|
I studied analysis about twenty years ago, so my answer should be considered not as a rigor proof but as an idea.
$$\int_\lambda f\ ds = \int_0^{2\pi} f(t)\frac{ds}{dt}dt.$$
Let $\lambda(t)=(\lambda_x(t), \lambda_y(t))$. Then
$ds=\sqrt{\lambda_x’^2+\lambda_y’^2}dt=\sqrt{a^2\sin^2t+b^2\cos^2 t}dt$,
$$\int_\lambda f\ ds=\int_0^{2\pi} \sqrt{a^2\cos^2 t + b^2\sin^2 t}\cdot
\sqrt{a^2\sin^2t+b^2\cos^2 t}dt=$$
Now let’s try to find
$$I=\int \sqrt{a^2\cos^2 t + b^2\sin^2 t}\cdot \sqrt{a^2\sin^2t+b^2\cos^2 t}dt$$
To get rid of trigonometric functions, make a substitution $u=\tan t$. Then $\cos^2 t=\frac{1}{1+\tan^2 t}=\frac{1}{1+u^2}$, $\sin^2 t=\frac{\tan^2 t}{1+\tan^2 t}=\frac{u^2}{1+u^2}$, $t=\operatorname{atan} u$, and $dt=\frac {du}{1+u^2}$. Hence
$$I=\int \frac{\sqrt{a^2+b^2u^2}\cdot \sqrt{a^2u^2+b^2}}{(1+u^2)^2}du.$$
Then I used for references a fundamental and huge (it has more than two thousand pages in three volumes) book “Differential and Integral Calculus” by Grigorii Fichtenholz. There is written that such integrals are called (pseudo)elliptic and usually cannot be expressed in a finite form by elementary functions even with the extended meaning of this term. Mathcad failed to calculate $I$ too.
Update. If we interpret the given integral as
$$\int_\lambda f\ dr =\int_\lambda f\ d\lambda= \int_0^{2\pi} f(t)\frac{d\lambda}{dt}dt.$$
then the calculations become simple. Indeed, let
$\lambda(t)=(\lambda_x(t), \lambda_y(t))$. Then
$d\lambda=(\lambda_x’,\lambda_y’)dt=(-a\sin t, b\cos t)dt$,
$$\int_\lambda f\ dr=\int_0^{2\pi} \sqrt{a^2\cos^2 t + b^2\sin^2 t}\cdot
(-a\sin t, b\cos t)dt.$$
Now let’s try to find
$$I=\int \sqrt{a^2\cos^2 t + b^2\sin^2 t}\cdot (-a\sin t, b\cos t)dt=(I_1,I_2),$$
where
$$I_1=\int \sqrt{a^2\cos^2 t + b^2\sin^2 t}(-a\sin t)dt,$$
and $$I_2=\int \sqrt{a^2\cos^2 t + b^2\sin^2 t}(b\cos t)dt.$$
But
$$I_1=\int \sqrt{a^2\cos^2 t + b^2\sin^2 t}\cdot a\cdot d\cos t=
a\int \sqrt{(a^2-b^2)u^2 + b^2}du,$$
and
$$I_2=\int \sqrt{a^2\cos^2 t + b^2\sin^2 t}\cdot b\cdot d\sin t=
b\int \sqrt{(b^2-a^2)u^2 + a^2}du$$
At last I recall the standard integrals
$$\int \sqrt{\alpha t^2+\beta}\cdot dt=\frac 12t\sqrt{\alpha t^2+\beta}+\frac{\beta}2\int \frac{dt}{\sqrt{\alpha t^2+\beta}},$$
and
$$\int \frac{dt}{\sqrt{\alpha t^2+\beta}}=\cases{
\frac 1{\sqrt{\alpha}}\ln\left|t+\sqrt{t^2+\frac{\beta}{\alpha}}\right|+C,\mbox{ if }\alpha>0 \\ \frac{1}{\sqrt{|\alpha|}}\operatorname{asin} \left(\sqrt{\frac{|\alpha|}{\beta}}t \right)+C,\mbox{ if }\alpha<0}.$$
|
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|
Find all values of $x$ at which $P(x)=x^4-4x^3+22x^2-36x+18$ is a perfect square Find all values of positive integer $x$ at which the following expression is perfect square
$$P(x)=x^4-4x^3+22x^2-36x+18$$
I tried to assume $P= (x^2+ax+b)^2$ ; and comparing the cofactors , get that $a= -2 ; b= 9$
, but when expand the $x^2-2x+9$, I didn't get the same as $P$
What is wrong in my work?
|
If $x$ is an integer, then
$$x^4-4x^3-22x^2-36x+18$$
$\qquad$is congruent to $2$ mod $4$ if $x$ is even,
$\qquad$and is congruent to $5$ mod $8$ if $x$ is odd,
hence can't be a perfect square.
(Even squares are congruent to $0$ mod $4$, and odd squares are congruent to $1$ mod $8$.)
Some details . . .
If $x$ is even, all the terms are multiples of $4$ except $18$, so
\begin{align*}
&x^4-4x^3-22x^2-36x+18\\[4pt]
\equiv\;&18\;(\text{mod}\;4)\\[4pt]
\equiv\;&2\;(\text{mod}\;4)\\[4pt]
\end{align*}
If $x$ is odd, then $x^2 \equiv 1\;(\text{mod}\;8)$, hence
\begin{align*}
&x^4-4x^3-22x^2-36x+18\\[4pt]
\equiv\;&x^4-4x^3-6x^2-4x+2\;(\text{mod}\;8)\\[4pt]
\equiv\;&(x^2)^2-4(x^2)x-6x^2-4x+2\;(\text{mod}\;8)\\[4pt]
\equiv\;&(1)^2-4(1)x-6(1)-4x+2\;(\text{mod}\;8)\\[4pt]
\equiv\;&-3-8x\;(\text{mod}\;8)\\[4pt]
\equiv\;&5\;(\text{mod}\;8)\\[4pt]
\end{align*}
|
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|
Area of square as a function of $\hat{a}$? Suppose $ABCD $ is square ,and $AM=DN=QB=PC$ so $$A'B'C'D'$$
is a square too.
Can someone help me to find area of $\bf{smaller -square}$(or $\color{red} {\Box A'B'C'D'}
$) as a function of angle $\hat{a}$ .It seems to be easy ,but I get stuck on this problem.Thanks in advance.
|
Label the points $A, B, C, D, E, F, G, H, I, J, K, L$ as in this diagram:
Now, let $AD = x$ and let $\angle DAF = \alpha$.
We wish to express the area of square $IJKL$ in terms of $x$ and $\alpha$.
First, let $DF = AE = y$. The tangent of $\alpha$ can then be written
$$\tan \alpha = \frac{y}{x} \,\,\implies\,\, y = x\tan\alpha$$
Triangle $ADF$ is right, so we can apply the Pythagorean theorem to find that $$AF = \sqrt{x^2+y^2}$$
To find $IL$, we will find $AI$ and $LF$ and subtract both from $AF$.
To find $AI$, note that triangles $AIE$ and $ADF$ are similar. This means that
$$\frac{AE}{AI} = \frac{AF}{AD}$$
$$\frac{y}{AI}=\frac{\sqrt{x^2+y^2}}{x}$$
$$AI = \frac{xy}{\sqrt{x^2+y^2}}$$
To find $LF$, note that triangles $DLF$ and $ADF$ are similar. This means that
$$\frac{DF}{LF} = \frac{AF}{DF}$$
$$\frac{y}{LF} = \frac{\sqrt{x^2+y^2}}{y}$$
$$LF = \frac{y^2}{\sqrt{x^2+y^2}}$$
Finally we can find $IL$:
$$IL = AF - AI - LF$$
$$IL = \sqrt{x^2+y^2} - \frac{y^2+xy}{\sqrt{x^2+y^2}}$$
This means that our final answer for the area of square $IJKL$ is:
$$(IL)^2 = \boxed{\left(\sqrt{x^2+y^2} - \frac{y^2+xy}{\sqrt{x^2+y^2}}\right)^2\,}$$
where $x = AF$ and $y = x\tan \alpha$, with $\alpha = \angle DAF$.
|
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|
Finding $a + b + c$ for the $101$st number of the form $3^a + 3^b + 3^c$ when these numbers are listed in increasing order $a,b,c\neq0$ & $a,b,c$ are distinct & $a,b,c\in \mathbb{N}$
numbers in form of
$3^a+3^b+3^c$
If we order them in increasing order, what is the sum of
$a+b+c=?$
for the $101$st such number?
There are some things I need to clarify about the question a number is represented by $3^a+3^b+3^c$, not $3^a,3^b,3^c$. These do not represent an individual number.
That being said we could accept $a>b>c$ and then acknowledge that $a,b,c\in \left\{ 0,1,2,3,\cdots,n \right\}$. There are $\dbinom{n+1}{3}$ choices possible for $n$. Therefore, $n$ must give a number greater than $101$. I have chosen $\dbinom{10}{3}=120$, so $n=9$. But I also used $\dbinom{9}{3}=84$ to know which number is $3^9+3^1+3^0$ (it is the $84$th). I couldn't spot how I would proceed in getting the $101$st number? What do you propose?
|
At the top of your question you say $a,b,c\neq 0$ but then the rest of what you say seems to include the possibility that one of them can be $0$. I'll assume that $0$ is not allowed in what follows. Obviously if you include $3^0$ as a possible power the answer will be different, but the method is the same.
As you say, $\binom 93=84$ and $\binom{10}3=120$. So there are $84$ numbers of the form $3^a+3^b+3^c$ where $a,b,c$ are distinct elements of $\{1,2,...,9\}$, and another $36$ if we allow $10$. All of these are less than $3\times3^{10}=3^{11}$, so these are the first $120$ numbers of the form $3^a+3^b+3^c$ where $a,b,c$ are distinct positive integers.
Any choice of $a,b,c$ where the largest is $10$ will be bigger than any choice where they are all less than $10$, so the $84$ where all powers are at most $9$ come first, and then you want the $17$th number of the form $3^{10}+3^b+3^c$, where $b,c$ are distinct positive integers less than $10$.
Hint for finishing from here: how many choices for $b,c$ are there if $b,c\in\{1,2,...,6\}$?
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Let $a = \frac{9+\sqrt{45}}{2}$. Find $\frac{1}{a}$ I've been wrapping my head around this question lately:
Let
$$a = \frac{9+\sqrt{45}}{2}$$
Find the value of
$$\frac{1}{a}$$
I've done it like this:
$$\frac{1}{a}=\frac{2}{9+\sqrt{45}}$$
I rationalize the denominator like this:
$$\frac{1}{a}=\frac{2}{9+\sqrt{45}} \times (\frac{9-\sqrt{45}}{9-\sqrt{45}})$$
This is what I should get:
$$\frac{1}{a} = \frac{2(9-\sqrt{45})}{81-45} \rightarrow \frac{1}{a}=\frac{18-2\sqrt{45}}{36})$$
Which I can simplify to:
$$\frac{1}{a}=\frac{2(9-\sqrt{45})}{36}\rightarrow\frac{1}{a}=\frac{9-\sqrt{45}}{18}$$
However, this answer can't be found in my multiple choice question here:
Any hints on what I'm doing wrong?
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Since the $ 45 $ inside the radical is expressed a product containing a perfect square ($ 9 $), $$ \sqrt{45} = \sqrt{9 \times 5} = \sqrt{9}\sqrt{5} = 3\sqrt{5} $$
Thus, $$ \frac{1}{a} = \frac{9 - \sqrt{45}}{18} = \frac{9 - 3\sqrt{5}}{18} $$
Dividing the numerator and denominator by $ 3 $ (the greatest common factor of the coefficients $ 3 $ and $ 9 $ in the numerator and $ 18 $ in the denominator):
$$ \frac{(9 - 3\sqrt{5}) / 3}{18 / 3} = \frac{3 - \sqrt{5}}{6} $$ which is one of the choices (choice (b)).
|
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|
What are the number of solutions of $11^x+13^x+17^x-19^x$? My attempt: $ 11^x+13^x+17^x-19^x=0\implies (\frac{11}{19})^x+(\frac{13}{19})^x+(\frac{17}{19})^x=1 $.
Now taking limit as $x\rightarrow\infty$,we get $0=1$,which is absurd.Hence,the equation has no solution.
Is it correct??
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Let $f(x)= 11^x+13^x+17^x-19^x=19^x[(\frac{11}{19})^x+(\frac{13}{19})^x+(\frac{17}{19})^x-1]$.
Clearly,$g(x)=(\frac{11}{19})^x+(\frac{13}{19})^x+(\frac{17}{19})^x-1$ is a strictly decreasing function.Also,$\lim_{x\to \infty}g(x)=-1 $ & $\lim_{x\to -\infty}g(x)=\infty$.Thus $f(x)=0$ has exactly one solution.
|
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Proving two binomial coefficient identities based on the expansion of $(1 + x)^{2n}$ This is a very interesting combinatorics problem that I came across in an old textbook of mine. So I know its got something to do with permutations and combinations, which yields the shortest, simplest proofs, but other than that, the textbook gave no hints really and I'm really not sure about how to approach it. Any guidance hints or help would be truly greatly appreciated. Thanks in advance :) So anyway, here the problem goes:
Use counting arguments to prove these identities:
$$1. \quad\begin{pmatrix}
n \\
0\\
\end{pmatrix}^2 + \begin{pmatrix} n \\ 1\\ \end{pmatrix}^2 +\begin{pmatrix} n \\ 2\\ \end{pmatrix}^2 + \ldots+\begin{pmatrix} n \\ n\\ \end{pmatrix}^2= \begin{pmatrix} 2n \\ n\\ \end{pmatrix}$$
$$2. \quad\ \begin{pmatrix}
n \\
1\\
\end{pmatrix}^2 + 2\begin{pmatrix} n \\ 2\\ \end{pmatrix}^2 +3\begin{pmatrix} n \\ 3\\ \end{pmatrix}^2 + \ldots+n\begin{pmatrix} n \\ n\\ \end{pmatrix}^2= \frac n2\begin{pmatrix} 2n \\ n\\ \end{pmatrix}$$
Use $(1+x)^{2n}$ to prove Identity $1$.
Use Identity $1$ to prove Identity $2$.
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Consider a group of $n$ males and $n$ females from which a committee of $n$ need to be chosen.
There are $n + 1$ possible cases:
Case $1$: $0$ males and $n$ females are chosen- $\binom{n}{0} \cdot \binom{n}{n} = \binom{n}{0}^2 ways$
Case $2$: $1$ male and $n -1$ females are chosen- $\binom{n}{1} \cdot \binom{n}{n - 1} = \binom{n}{1}^2 ways$
..
Case $n + 1$: $n$ males and $0$ females are chosen- $\binom{n}{n} \cdot \binom{n}{0} = \binom{n}{n}^2 ways$.
But using a more direct approach, the selection can be made using $\binom{2n}{n}$ ways.
Thus, $\binom{2n}{n} = \sum_{k = 0}^{n} \binom{n}{k}^2$.
For verifying the result using the binomial theorem,
*
*What is the coefficient of $x^n$ in $(1 + x)^{2n}$?
*What is the coefficient of $x^n$ in $(1 + x)^n \cdot (1 + x)^n$?
*For the second question, you should be able to find the derivative for the identity involving the binomial expansion of $(1 + x)^{2n}$ and proceed in a similar manner.
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|
Determine whether the following series is convergent or divergent, and find its sum if it is convergent. PROBLEM
Is the following complex series convergent, and if so, what is its sum?
$$\sum_{n=0}^{\infty}{\frac{\cos(n\theta)}{3^n}}, \text{ } \theta \in \mathbb{R}$$
MY ATTEMPT #1
Let $$a_n = \frac{\cos(n\theta)}{3^n}.$$
By the Ratio Test
$$\left|\frac{a_{n+1}}{a_n}\right| = \left|\frac{\cos((n+1)\theta)}{3^{n+1}}\cdot\frac{3^n}{\cos(n\theta)}\right| = \frac{1}{3}\cdot\left|\frac{\cos((n+1)\theta)}{\cos(n\theta)}\right|.$$
Let $x = (n+1)\theta, y = n\theta$. Then we have
$$\frac{x}{y} = \frac{n+1}{n} \to 1 \text{ as } n \to \infty$$
$$L = \lim_{n \to \infty}{\frac{1}{3}\cdot\left|\frac{\cos((n+1)\theta)}{\cos(n\theta)}\right|} = \frac{1}{3}\cdot\lim_{n \to \infty}{\left|\frac{\frac{\cos(x)}{x}}{\frac{\cos(y)}{y}}\right|}\cdot\lim_{n \to \infty}{\left|\frac{x}{y}\right|} = \frac{1}{3}\cdot{\frac{\lim_{x \to \infty}{\left|\frac{\cos(x)}{x}\right|}}{\lim_{y \to \infty}{\left|\frac{\cos(y)}{y}\right|}}} \to \frac{1}{3}\cdot\frac{0}{0}.$$
Applying L'Hopital's Rule, we get
$$\frac{1}{3}\cdot{\frac{\lim_{x \to \infty}{\left|\frac{\cos(x)}{x}\right|}}{\lim_{y \to \infty}{\left|\frac{\cos(y)}{y}\right|}}} = \frac{1}{3}\cdot{\frac{\lim_{x \to \infty}{\left|\frac{-x\sin(x) - \cos(x)}{x^2}\right|}}{\lim_{y \to \infty}{\left|\frac{-y\sin(y) - \cos(y))}{y^2}\right|}}}$$
But
$$\lim_{z \to \infty}{\left|\frac{-z\sin(z) - \cos(z)}{z^2}\right|} = \lim_{z \to \infty}{\left|\frac{z\sin(z) + \cos(z)}{z^2}\right|} = \left|\lim_{z \to \infty}{\frac{\sin(z)}{z}}\right| + \left|\lim_{z \to \infty}{\frac{\cos(z)}{z^2}}\right| = 0 + 0 = 0.$$
HENCE THIS IS NOT THE CORRECT APPROACH.
MY ATTEMPT #2
Still by the Ratio Test,
$$\left|\frac{a_{n+1}}{a_n}\right| = \left|\frac{\cos((n+1)\theta)}{3^{n+1}}\cdot\frac{3^n}{\cos(n\theta)}\right| = \frac{1}{3}\cdot\left|\frac{\cos((n+1)\theta)}{\cos(n\theta)}\right|.$$
Applying the trigonometric identity
$$\cos(A+B) = \cos(A)\cos(B) - \sin(A)\sin(B)$$
to $\cos((n+1)\theta)$ does not appear to help either.
QUESTION
Does this mean that the Ratio Test is not an appropriate test of convergence for this series? If so, what test should I use, and how?
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As hinted in the comments, the series
$$\sum_{n=0}^{\infty}{\frac{\cos(n\theta)}{3^n}}$$
is convergent (by the Comparison Test) since $\theta \in \mathbb{R}$ implies that
$$\left|a_n\right| \leq \frac{1}{3^n},$$
where $\left(\frac{1}{3}\right)^n$ is a convergent geometric series.
Now, to compute the sum, we consider the geometric series
$$\sum_{n=0}^{\infty}{\bigg(\frac{e^{i\theta}}{3}\bigg)^n}$$
which (since $\left|e^{i\theta}/3\right| = 1/3 < 1$) has sum
$$\frac{1}{1 - \frac{e^{i\theta}}{3}} = \frac{3}{3 - e^{i\theta}} = \frac{\bigg(9 - 3\cos\theta\bigg)+i\bigg(3\sin\theta\bigg)}{\sin^2\theta + \bigg(3 - \cos\theta\bigg)^2}.$$
But by De Moivre's Theorem,
$$(e^{i\theta})^n = e^{i(n\theta)} = \cos(n\theta)+i\sin(n\theta).$$
Hence,
$$\sum_{n=0}^{\infty}{\frac{\cos(n\theta)}{3^n}} = \Re\Bigg(\sum_{n=0}^{\infty}{\bigg(\frac{e^{i\theta}}{3}\bigg)^n}\Bigg) = \frac{9 - 3\cos\theta}{\sin^2\theta + \bigg(3 - \cos\theta\bigg)^2} = \frac{9 - 3\cos\theta}{10 - 6\cos\theta},$$
where $\Re(z)$ denotes the real part of $z \in \mathbb{C}$.
|
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Prove that the expression $5^{2n+1} \cdot 2^{n+2} + 3^{n+2} \cdot 2^{2n+1}$ is divisible by 19.
Prove that the expression
$$5^{2n+1} * 2^{n+2} + 3^{n+2} * 2^{2n+1}$$
is divisible by $19$.
I'll skip the basis step (as I have done last time) but I can conclude that it's only divisible by 19 for integers n ≥ 0 (or whole numbers).
II.
Assume that
$$5^{2k+1} * 2^{k+2} + 3^{k+2} * 2^{2k+1}$$
is divisible by 19. Then,
$$5^{2k+3} * 2^{k+3} + 3^{k+3} * 2^{2k+3}$$
is divisible by 19.
Now this is where I get lost, I try to "dismember" the expression to get
$$5^{2k}* 5^3 * 2^k * 2^3 + 3^k * 3^3 * 2^{2k} * 2^3$$
I also try to get it similar to to the assumption to make use of the said assumption yielding
$$5^{2k}* 5 * 5^2 * 2^k * 2^2 * 2 + 3^k * 3^2 * 3 * 2^{2k} * 2 * 2^2$$
$$5^{2k+1} * 5^2 * 2^{k+2} * 2 + 3^{k+2} * 3 * 2^{2k+1} * 2^2$$
$$5^{2k+1} * 25 * 2^{k+2} * 2 + 3^{k+2} * 3 * 2^{2k+1} * 2^2$$
$$50 * 5^{2k+1} * 2^{k+2} + 12 * 3^{k+2} * 2^{2k+1}$$
And this is where I get lost.. : (
Am I missing out something? Had I done it wrong? The number 19 is prime which makes it hard to handle for me. Thanks!
EDIT : After some pondering, I answered it this way :
$$50 * 5^{2k+1} * 2^{k+2} + 12 * 3^{k+2} * 2^{2k+1}$$
I realized that 50 can be written as 38 + 12 (and 38 is a multiple of 19) Hence,
$$ 38 + 12 * 5^{2k+1} * 2^{k+2} + 12 * 3^{k+2} * 2^{2k+1} $$
Factoring out 12, I get :
$$ 38 + 12(5^{2k+1} * 2^{k+2} + 3^{k+2} * 2^{2k+1}) $$
38 is divisible by 19 and the long expression is divisible by 19 (per the assumption) and qed.
Is this correct ?
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Without using induction
$$5^{2n+1}2^{n+2}+3^{n+2}2^{2n+1}=20\cdot50^n+18\cdot12^n$$
$$\equiv1\cdot12^n+(-1)\cdot12^n\pmod{19}$$ as $20\equiv1,18\equiv-1,50\equiv12\pmod{19}$
|
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true false combinatorics questions about arrangements I've solved some exercises regarding combinatorics. The questions are true\false questions, but I'm not sure that I've done them right. Please take a look and indicate any corrections I need to make.
1) The number of different arrangements of the string 'AAABBC' is larger than the number of arrangements of the string 'AABBCC'
2) The number of different arrangements of the string 'AABBCC' is equal to the number of possible 'cuttings' (subsets) with $3$ classes that include $2$ elements each of the set $\{1,2,3,4,5,6\}$.
3) The number of different arrangements of the string 'AABBCC' is equal to the number of ways in which $6$ students can submit $3$ different tasks in pairs.
4) The number of ways in which you can put $3$ identical apples in $4$ different baskets is equal to the number of ways in which you can put $4$ identical apples in $3$ different baskets.
5) The number of ways in which you can put $4$ different apples in $4$ different baskets is $16$ times the number of ways in which you can put $4$ identical apples in $4$ different baskets.
6) The number of different separations of the set $a= \{1,2,3,4,5,6\}$ to classes with $3$ elements each is equal to the number of possible combinations of a choosing a set with $3$ elements from $a$.
What I did:
1) False. If $a$ is the number of As, and $b$ is the number of Bs and $c$ is the number of Cs, then $a = 3$, $b=2$ and $c=1$, and the possible ways to arrange the string'AAABBC' is:$\frac {6!}{3!2!}$. regarding 'AABBCC': $\frac {6!}{2!2!2!}$, which means that the latter is larger.
2) False. Using the same method, the different arrangements of 'AABBCC' is equal to $\frac{6!}{2!2!2!}$, and the number of possible mappings from $\{1,2,3,4,5,6\}$ to subsets with $3$ classes with $2$ elements each is: $\frac{\binom {6}{2} \binom {4}{2} \binom {2}{2}}{2}$ (I'm sure I've done a mistake here).
3) False. Arranging 'AABBCC' is equal to $\frac{6!}{2!2!2!}$, and the number of possibilities in which $6$ students can submit $3$ different tasks in pairs is $\frac{6 \cdot 3}{2}$.
4) True. Seems to be equal, since if you have $4$ apples and $3$ baskets, you can only put $3$ apples in the baskets.
5) False. The number of ways in which you can put $4$ different apples in $4$ different baskets is: $\binom {4+3}{4}$, and the number of ways in which you can put $4$ identical apples in $4$ different baskets is not $16$ times less than that. If I made a mistake in calculation, please correct me if possible.
6) True. From what I understand, it's basically doing the same.
Please correct me if I've made a mistake, so I can learn and improve. Thank you very much.
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The number of different arrangements of the string AAABBB is larger than the number of different arrangements of the string AABBCC.
Your work is correct.
The number of different arrangements of the string AABBCC is equal to the number of ways of cutting the set $\{1, 2, 3, 4, 5, 6\}$ into three subsets with two elements each.
You are correct that the number of distinguishable arrangements of AABBCC is
$$\frac{6!}{2!2!2!}$$
However, your count of the number of ways to divide the set $\{1, 2, 3, 4, 5, 6\}$ into three subsets with two elements each is incorrect.
Method 1: The number of ways to divide the set $\{1, 2, 3, 4, 5, 6\}$ into three labeled subsets with two elements each is
$$\binom{6}{2}\binom{4}{2}\binom{2}{2}$$
since there are $\binom{6}{2}$ ways to select two of the six elements for the subset with the first label, $\binom{4}{2}$ ways to select two of the four remaining elements for the subset with the second label, and one way to place the remaining elements in the subset with the third label. However, the subsets are not labeled. Since the order in which we list the subsets does not matter and there are $3!$ ways we could order the subsets, the number of distinguishable ways of dividing the set $\{1, 2, 3, 4, 5, 6\}$ into three two-element subsets is
$$\frac{1}{3!}\binom{6}{2}\binom{4}{2}\binom{2}{2} = \frac{1}{6} \cdot \frac{6!}{4!2!} \cdot \frac{4!}{2!2!} \cdot \frac{2!}{2!0!} = \frac{1}{6} \cdot \frac{6!}{2!2!2!}$$
Hence, the statement is still false.
Method 2: There are five choices for the element that is placed in the same subset as $1$. That leaves four elements. There are three ways of placing a number in the same subset as the smallest remaining element. The remaining two elements must be placed in the third subset. Hence, there are
$$5 \cdot 3 \cdot 1$$
ways of dividing the set $\{1, 2, 3, 4, 5, 6\}$ into three two-element subsets.
The number of different arrangements of the string AABBCC is equal to the number of ways in which six students can submit three different tasks in pairs.
There are $\binom{6}{2}$ ways to select which two of the six students submit task 1, $\binom{4}{2}$ ways to select which two of the four remaining students select task 2, and one way for the remaining pair of students to submit task three. Hence, the number of ways in which six students can submit three different tasks in pairs is
$$\binom{6}{2}\binom{4}{2}\binom{2}{2} = \frac{6!}{2!2!2!}$$
which is equal to the number of distinguishable arrangements of the string
AABBCC.
The number of ways you can put three identical apples in four different baskets is equal to the number of ways you can put four identical apples in three different baskets.
Let $x_j$ be the number of apples placed in the $j$th basket.
The number of ways you can put three identical apples in four different baskets is the number of solutions of the equation
$$x_1 + x_2 + x_3 + x_4 = 3$$
in the nonnegative integers. A particular solution corresponds to the placement of three addition signs in a row of three ones. For instance,
$$+ 1 + 1 1 +$$
corresponds to the solution $x_1 = 0$, $x_2 = 1$, $x_3 = 2$, and $x_4 = 0$. There are
$$\binom{3 + 3}{3} = \binom{6}{3}$$
such solutions since we must choose which three of the six positions for the six symbols (three ones and three addition signs) will be filled with addition signs.
In general, the number of solutions of the equation
$$x_1 + x_2 + x_3 + \ldots + x_k = n$$
in the nonnegative integers is
$$\binom{n + k - 1}{k - 1}$$
since we must choose which $k - 1$ positions of the $n + k - 1$ positions needed for the $n$ ones and $k - 1$ addition signs will be filled with addition signs.
The number of ways four identical apples can be placed in three different baskets is the number of solutions of the equation
$$x_1 + x_2 + x_3 = 4$$
in the nonnegative integers, which is
$$\binom{4 + 3 - 1}{3 - 1} = \binom{6}{2}$$
Since $\binom{6}{3} > \binom{6}{2}$, there are more ways to distribute three identical apples to four different baskets than there are ways to distribute four identical apples to three different baskets.
The number of different ways you can put four different apples in four different baskets is $16$ times the number of ways you can put four identical apples in four different baskets.
As you concluded in the comments, there are $4^4$ ways to put four different apples in four different baskets and $\binom{4 + 4 - 1}{4 - 1} = \binom{7}{4}$ ways to place four identical apples in four different baskets. Since
$$4^4 = 256 \neq 16 \cdot 35 = 16\binom{7}{3}$$
the statement is false.
The number of different separations of the set $A = \{1, 2, 3, 4, 5, 6\}$ is equal to the number of ways of choosing a subset of three elements from $A$.
The statement is false.
There are $$\binom{6}{3}$$ ways to select a subset with three elements from a six-element set.
We will count the number of ways to separate set $A$ into two sets with three elements in two ways.
Method 1: Since choosing a subset of three elements of $A$ also determines its complement, choosing a subset of three elements counts each separation twice, once when you count the subset and once when you count its complement. Thus, there are
$$\frac{1}{2}\binom{6}{3}$$
ways to separate set $A$ into two three-element subsets.
Method 2: In any separation of set $A$ into two sets with three elements, there are $$\binom{5}{2}$$ ways to choose which two of the other five elements will be in the same subset as $1$.
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What will be the $n^{th}$ term of this given series? Given series is:
$$1+\frac{1\times x^2}{2\times 4}+\frac{1\times 3\times 5\times x^4}{2\times 4\times 6\times 8}+\frac{1\times 3\times 5\times 7\times 9\times x^6}{2\times 4\times 6\times 8\times 10\times 12}+.....\infty$$
I need to find it's $n^{th}$ term but am having trouble in dealing with increasing number of multiples in each step. Kindly Guide in How to proceed
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Hint: Your $n$th term (counting the first non-$1$ term as the first term and excepting the $0$th term) looks to be
$$x^{2n}\frac{1\times 3 \times \cdots \times (4n-3)}{2\times 4\times \cdots \times (4n)}.$$
This equals
$$x^{2n}\frac{1\times 2 \times \cdots \times (4n-2)}{\left(2\times 4\times \cdots \times (4n-2)\right)\left(2\times 4\times \cdots \times (4n)\right)}.$$
How can we represent this in terms of factorials?
|
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|
About using the lagrange multiplier. Today I have seen a question like the following;
$x+y+z=5$ and $xy+yz+xz=3$ and $x,y,z \in \mathbb{R}^+$
What is the maximum value $x$ can get?
Now it is pretty obvious that question is solvable using many simple elementary methods like say $(x+y+z)^2=x^2+y^2+z^2+2(xy+yz+xz)=25$
$25=x^2+y^2+z^2+6 \rightarrow x^2+y^2+z^2=19$
Applying Cauchy-Schwarz $(y^2+z^2)(1+1)\geq (z+y)^2$
$2(19-x^2)\ge(5-x)^2$
$x_{max}=\dfrac{13}{3}$
However I would also like to solve it by the lagrange multiplier, here are my efforts
$f(x,y,z,k)=x+5-(y+z)+k(x+y+z-5)$
$f_k'=0$
$f_x'=k+1$
$f_y'=k-1$
$f_z'=k-1$
$k=1,-1$ and $x$ goes nowhere near $\dfrac{13}{3}$
So my question is where am I wrong and how can I fix it?
Besides are there maxima minima questions that can't be solved with lagrange,
is this one of them?
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Your function $f(x,y,z,k)=x+5-(y+z)+k(x+y+z-5)$ is not the right function fot Lagrange. The right function is given by
$f(x,y,z,k)=x+5-(y+z)+k(xy+yz+xz-3)$:
|
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Find $\tan^{-1} (i\sqrt{2})$. Problem: Find $\tan^{-1} (i\sqrt{2})$.
My attempt: We must find $z \in \mathbb{C}$ such that $\tan z = i\sqrt{2}$.
$$\tan z = \frac{\sin z}{\cos z} = i\frac{e^{-iz} - e^{iz}}{e^{iz} + e^{-iz}}$$
Let $u = e^{iz}$. Then $$i\frac{u^{-1}-u}{u+u^{-1}} = i\sqrt{2}$$ so $$1-u^2 = \sqrt{2}(u^2 + 1)$$
and
$$u^2 = \frac{1-\sqrt{2}}{1+\sqrt{2}}$$
Then we have
$$e^{iz} = \sqrt{\frac{1-\sqrt{2}}{1+\sqrt{2}}}$$
Taking the natural logarithm of both sides gives
$$z = \frac{1}{2i}\ln\frac{1-\sqrt{2}}{1+\sqrt{2}}$$
Since $\ln$ is multi-valued and $\frac{1-\sqrt{2}}{1+\sqrt{2}}+0i=\frac{1-\sqrt{2}}{1+\sqrt{2}}e^{i2n\pi},$
we have
$$\frac{1}{2i}\ln{\frac{1-\sqrt{2}}{1+\sqrt{2}}}=\frac{-i}{2}\texttt{Ln}\frac{1-\sqrt{2}}{1+\sqrt{2}}+ n\pi$$
That's how far I got. The book lists the final answer as $$\pi/2 + n\pi -i\texttt{Ln}(\sqrt{2}-1)$$
|
Let $u=e^{iz}$ just as you did.
$$\begin{array}{rcl}
u^2 &=& \dfrac{1-\sqrt{2}}{1+\sqrt{2}} \\
u^2 &=& \dfrac{(1-\sqrt{2})^2}{(1+\sqrt{2})(1-\sqrt{2})} \\
u^2 &=& \dfrac{(1-\sqrt{2})^2}{-1} \\
e^{2iz} &=& -(1-\sqrt{2})^2 \\
e^{2iz} &=& (1-\sqrt{2})^2 e^{i\pi} \\
2iz &=& \ln[(1-\sqrt{2})^2e^{i\pi}] + 2ni\pi \\
2iz &=& 2\ln[\sqrt{2}-1] + i\pi + 2ni\pi \\
z &=& -i\ln[\sqrt{2}-1] + \dfrac\pi2 + n\pi \\
\end{array}$$
|
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|
If $8\sin x - \cos x=4$, then find possible values of $x$ I am not understanding what exactly can watch do here. First I thought that if I could square it but it was in vain. Please help me.
|
$8\sin x -\cos x =4$
solve for $\cos x$
$\cos x = 8\sin x -4$
plug in the fundamental identity $\sin^2 x + \cos^2 x=1$
$\sin^2 x + \left ( 8\sin x -4\right)^2=1$
$65 \sin ^2 x -64 \sin x +15=0$
$\sin x = \dfrac{64\pm \sqrt{64^2-4\cdot 65 \cdot 15}}{130}$
$\sin x = \dfrac{3}{5} \to x = \arcsin \left(\dfrac{3}{5}\right)+2k \pi \lor x=\pi-\arcsin \left(\dfrac{3}{5}\right)+2k \pi$
$\sin x = \dfrac{5}{13} \to x = \arcsin \left(\dfrac{5}{13}\right)+2k \pi \lor x=\pi-\arcsin \left(\dfrac{5}{13}\right)+2k \pi$
Hope this helps
|
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|
Show that ${\sum^{n}_{k=1}} \frac{1}{n+k} \le \frac{3}{4}$
Prove that $\displaystyle{\sum^{n}_{k=1}} \frac{1}{n+k} \le \frac{3}{4}$ for each positive integer $n$.
My work. I think that i have to use induction, but i can't see how... What i did:
$$f(n)=\displaystyle{\sum^{n}_{k=1}} \frac{1}{n+k} \implies k(n)=f(n+1)-f(n)=\frac{1}{2n+1}-\frac{1}{2n+2}.$$
Now we have
$$f(n+1)=\displaystyle{\sum^{n}_{k=1}} \left ( \frac{1}{2k+1} \right )- \displaystyle{\sum^{n}_{k=1}} \left ( \frac{1}{2k+2} \right )$$
But now I have no more ideas.
|
$f(x) = 1/(n+x)$ is convex in $x$. So we have that the value for $f(x)$ for any $x \in (0 \; n]$ is less or equal to (line equation between the two interval boundaries) $1/n + x (1/(2n)- 1/n)/n $. Summing the values gives
$$
\sum^{n}_{k=1} \frac{1}{n+k} \le \sum^{n}_{k=1} 1/n + k (1/(2n)- 1/n)/n = \\1 + \frac12 n (n+1) (1/(2n)- 1/n)/n = 1 - \frac14 \frac{n+1}{n} = \frac{3}{4} - \frac{1}{4 n} \le \frac{3}{4}
$$
|
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|
Let $(x, y, z)\in \Bbb R$, prove that $\big(\frac{2x-y}{x-y}\big)^2+\big(\frac{2y-z}{y-z}\big)^2 +\big(\frac{2z-x}{z-x}\big)^2 \geqslant 5$
Olympiad Inequation
Let $x$, $y$ and $c$ be distinct real numbers. Prove that:
$$\left(\frac{2x-y}{x-y}\right)^2+\left(\frac{2y-z}{y-z}\right)^2+\left(\frac{2z-x}{z-x}\right)^2 \geqslant 5.$$
This is an assignment I got from a teacher. I've been baffled by the negative sign in the denominator.
Could you please provide insight as to how would one resolve this inequality please? (Do not solve yet, I've yet to turn in this assignment, I just want to get some tips)
|
This is a special case of the following inequality: $$\sum_{a,b,c}\Big(\dfrac{ma-nb}{a-b}\Big)^2\geq m^2+n^2,$$
where $m,n\in\mathbb{R}$ and $a,b,c$ are distinct reals.
The proof is simply completing the square: start by writing $$\dfrac{ma-nb}{a-b} = m+(m-n)\dfrac{b}{a-b}$$ and $$\dfrac{mc-na}{c-a} = n+(m-n)\dfrac{c}{c-a}$$ and leave the middle term untouched first.
Let me know if still need further assistance.
EDIT: Completing the hint: $$\sum_{a,b,c}\Big(\dfrac{ma-nb}{a-b}\Big)^2-m^2-n^2 = \Big(\dfrac{mb-nc}{b-c}\Big)^2+(m-n)^2\Big(\dfrac{b^2}{(a-b)^2}+\dfrac{c^2}{(c-a)^2}\Big)+2(m-n)\Big(\dfrac{mb}{a-b}+\dfrac{nc}{c-a}\Big)=$$
$$=\Big(\dfrac{mb-nc}{b-c}\Big)^2+(m-n)^2\Big(\dfrac{b^2}{(a-b)^2}+\dfrac{c^2}{(c-a)^2}\Big)+2(m-n)\dfrac{bc(m-n)-a(mb-nc)}{(a-b)(c-a)}=$$
$$=\Big(\dfrac{mb-nc}{b-c}\Big)^2+(m-n)^2\Big(\dfrac{b^2}{(a-b)^2}+\dfrac{c^2}{(c-a)^2}+\dfrac{2bc}{(a-b)(c-a)}\Big)-\dfrac{2a(mb-nc)(m-n)}{(a-b)(c-a)}=$$
$$=\Big(\dfrac{mb-nc}{b-c}\Big)^2+(m-n)^2\Big(\dfrac{b}{a-b}+\dfrac{c}{c-a}\Big)^2 - \dfrac{2a(m-n)(mb-nc)}{(a-b)(c-a)}=$$
$$\Big(\dfrac{mb-nc}{b-c}\Big)^2+(m-n)^2\Big(\dfrac{a(c-b)}{(a-b)(c-a)}\Big)^2 - \dfrac{2a(m-n)(mb-nc)}{(a-b)(c-a)}=$$ $$=\Big(\dfrac{mb-nc}{b-c} - \dfrac{a(m-n)(c-b)}{(a-b)(c-a)}\Big)^2\geq 0.$$
|
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|
Logarithmic problem If $\log_{16} 15 =a$ and $\log_{12} 18 =b$, then show that $$\log_{25} 24 = \frac{5-b}{16a-8ab-4b+2}.$$
|
Let $l_2= \ln 2$,$l_3= \ln 3$ and $l_5= \ln 5$. Now change base & we have
\begin{eqnarray*}
\frac{5-b}{16a-8ab-4b+2} = \frac{5-\frac{l_2+2l_3}{2l_2+l_3}}{16 \frac{l_3+l_5}{4l_2}-8 \frac{l_3+l_5}{4l_2} \frac{l_2+2l_3}{2l_2+l_3} -4 \frac{l_2+2l_3}{2l_2+l_3} +2} \\
\end{eqnarray*}
\begin{eqnarray*}
= \frac{(5(2l_2+l_3)-(l_2+2l_3))l_2} {4 (l_3+l_5)(2l_2+l_3) -2 (l_3+l_5)(l_2+2l_3) -4 (l_2+2l_3)l_2 +2(2l_2+l_3)l_2} \\
\end{eqnarray*}
\begin{eqnarray*}
= \frac{(9l_2+3l_3)l_2} {(l_3+l_5)6l_2-6l_2l_3} \\
\end{eqnarray*}
\begin{eqnarray*}
= \frac{(3l_2+l_3)} {2 l_5} = \log_{25}(24) \\
\end{eqnarray*}
|
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|
Prove that $\int \frac{x^m dx}{\ln x}=\ln(\ln x)+\frac{(m+1)\ln x}{1}+\frac{(m+1)^2 \ln ^2 x}{1 \times 2^2}+\cdots $ Prove that
$$\int \frac{x^m dx}{\ln x}=\ln(\ln x)+\frac{(m+1)\ln x}{1}+\frac{(m+1)^2 \ln ^2 x}{1 \times 2^2}+\frac{(m+1)^3 \ln^3 x}{1 \times 2 \times 3^2}+\cdots \infty $$
My Try:
I started with $$I=\int \frac{x^m}{\ln x}=\int \frac{x^{m+1} dx}{x \ln x}$$
Now by using parts taking $u=x^{m+1}$ and $v=\frac{1}{x \ln x}$ we get
$$I=x^{m+1} \times \ln(\ln x)-(m+1)\int x^m \ln (\ln x)dx$$
can i have any clue to proceed
|
Since differentiating is usually easier than finding primitives, and you are given a candidate, the simplest way is to differentiate the right hand side. Write that as
$$f(x) = \ln (\ln x) + \sum_{n = 1}^{\infty} \frac{(m+1)^n(\ln x)^n}{n\cdot n!}.\tag{1}$$
The series converges locally uniformly on $(0,+\infty)$, so it can be differentiated term-by-term. This yields
\begin{align}
f'(x) &= \frac{1}{x\ln x} + \sum_{n = 1}^{\infty} \frac{(m+1)^n(\ln x)^{n-1}}{n!\cdot x} \\
&= \frac{1}{x\ln x} + \frac{1}{x\ln x}\sum_{n = 1}^{\infty} \frac{(m+1)^n(\ln x)^n}{n!} \\
&= \frac{1}{x\ln x} \sum_{n = 0}^{\infty} \frac{(m+1)^n(\ln x)^n}{n!} \\
&= \frac{1}{x\ln x} e^{(m+1)\ln x} \\
&= \frac{x^{m+1}}{x\ln x} \\
&= \frac{x^m}{\ln x}
\end{align}
for $x > 1$, so the right hand side is indeed a primitive of the integrand.
|
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|
Proof $(1+i)^{n}$ = $2^{n/2}(\cos(nπ/4)+i\sin(nπ/4))$ While binomial problem I struck here
$$(1+i)^{n} = 2^{\frac{n}{2}}\left(\cos(\frac{n\pi}{4})+i\sin(\frac{n\pi}{4})\right).$$
Please proof this equation .
Any help will be appreciated.
And Thanks for help
|
This is not a binomial question,
Using Taylor series,
$$\sin x=x-\frac{x^3}{3!}+\frac{x^5}{5!}-\cdots$$
$$\cos x=1-\frac{x^2}{2!} +\frac{x^4}{4!} -\cdots $$
$$e^x=1+x+\frac{x^2}{2!}+\frac{x^3}{3!} +\cdots$$
$$e^{ix}=1+ix+\frac{(ix)^2}{2!}+\frac{(ix)^3}{3!} +\cdots$$
$$e^{ix}=1+ix-\frac{(x)^2}{2!}+i\frac{(x)^3}{3!} +\cdots$$
$$e^{ix}=(1-\frac{x^2}{2!}+\frac{x^4}{4!}-\cdots)+i(x-\frac{x^3}{3!}+\frac{x^5}{5!}-\cdots)$$
$$e^{ix}=\cos x+i\sin x$$
And,
We can write
$$a+ib=r\cos t+ir\sin t$$
$$a+ib=r e^{it}$$
Hint
$$r=|a+ib|$$
$$t=\tan^{-1} (\frac{b}{a})$$
If you want answer
So(wait a minute...)\begin{align}(1+i)=\sqrt{1+1}e^{\frac{iπ}{4}}\end{align} \begin{align} (1+i)^n = 2^{\frac{n}{2}} e^{\frac{inπ}{4}}\end{align} \begin{align} (1+i)^n= 2^{\frac{n}{2}} (\cos {\frac{nπ}{4}} +i\sin {\frac{nπ}{4}})\end{align}
|
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|
Is there a general theorem to manipulate recursive functions? I was messing with some recursive functions and realized it was equivalent to the Fibonacci sequence, but I couldn't figure out why. I then played a little further, and discovered some other interesting patterns. Let's consider functions $A,B$.
Let $A(1) = 0$, $B(1) = 1$, $A(n) = B(n-1)$, $B(n) = A(n-1) + B(n-1)$. Then, $B(n) = B(n-2) + B(n-1)$, which causes $A(n) = A(n-1) + A(n-2)$. Of course, now both $A$ and $B$ are the Fibonacci sequence.
I was trying to figure out how this happened precisely, so I went back along the
chain. I then discovered $B(n) = 2B(n-2) + B(n-3) = 2B(n-1) - B(n-3)$ and several other similar relations. What causes these relations to happen? Is there a general method of manipulating functions to get equalities?
|
(Not a complete answer, still maybe relevant.) The recurrence can be written in matrix form as:
$$
\begin{pmatrix}
B_{n} \\
A_{n}
\end{pmatrix}
=
\begin{pmatrix}
1 & 1 \\
1 & 0
\end{pmatrix}
\cdot
\begin{pmatrix}
B_{n-1} \\
A_{n-1}
\end{pmatrix}
\quad\quad \text{with}\;\;
\begin{pmatrix}
B_{1} \\
A_{1}
\end{pmatrix}
=
\begin{pmatrix}
1 \\
0
\end{pmatrix}
$$
It follows by telescoping (or induction) that:
$$
\begin{pmatrix}
B_{n} \\
A_{n}
\end{pmatrix}
=
\begin{pmatrix}
1 & 1 \\
1 & 0
\end{pmatrix}^{n-1} \cdot
\begin{pmatrix}
B_{1} \\
A_{1}
\end{pmatrix}
$$
But $\displaystyle\; \begin{pmatrix}
1 & 1 \\
1 & 0
\end{pmatrix}^{n-1}
=
\begin{pmatrix}
F_n & F_{n-1} \\
F_{n-1} & F_{n-2}
\end{pmatrix}
\;$ (see this for example), so in the end:
$$
\begin{pmatrix}
B_{n} \\
A_{n}
\end{pmatrix}
=
\begin{pmatrix}
F_n & F_{n-1} \\
F_{n-1} & F_{n-2}
\end{pmatrix}
\cdot
\begin{pmatrix}
1 \\
0
\end{pmatrix}
=
\begin{pmatrix}
F_n \\
F_{n-1}
\end{pmatrix}
$$
|
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|
Solve the following nonlinear system of equations:$ xy+xz+yz=12 , xyz=2+x+y+z$
Solve the following system of equations in $\Bbb R^+$:
$$
\left\{
\begin{array}{l}
xy+yz+xz=12 \\
xyz=2+x+y+z\\
\end{array}
\right.
$$
I did as follows.
First equation yields to: $(x+y+z)^2-x^2-y^2-z^2=24$. Now, using second equation we get: $x^2y^2z^2-4xyz=20+x^2+y^2+z^2$ . Here I stopped...
|
Using AM-GM:
$$12=xy+yz+xz\ge 3\sqrt[3]{(xyz)^2} \Rightarrow xyz\le 8 \ \ \ \ (1)$$
$$xyz=2+x+y+z\le8 \Rightarrow x+y+z\le 6 \ \ \ \ \ (2)$$
The equality occurs when $x=y=z=2$.
Using C-S:
$$(x^2+y^2+z^2)(y^2+z^2+x^2)\ge (xy+yz+zx)^2 \Rightarrow x^2+y^2+z^2\ge 12 \ \ \ \ (3)$$
The equality occurs when $x=y=z=2$.
From the first equation:
$$(x+y+z)^2=24+x^2+y^2+z^2\ge 36 \Rightarrow x+y+z\ge 6 \ \ \ \ \ (4)$$
From $(2)$ and $(4)$, and then from the second equation we get:
$$x+y+z=6, \\ xyz=8.$$
So, we get:
$$x=y=z=2.$$
|
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|
What is the probability we need k rolls of a fair sided die to get at least one five and one six? I HAVE checked the questions and none of them seem to specify any method to compensate for at least one 5 AND one 6.
Here is how I tried it.
$$ P(k\ Rolls \ and \ last \ is \ a \ six)= P(\ k -1 \ rolls \ have \ at \ least\ one \ 5 )*P(\ k \ is \ a \ six)$$
$$ = \sum_{j=1}^{k-1} {k-1\choose j}(\frac{1}{6})^j (\frac{4}{6})^j $$
$$ = (\frac{5}{6})^{k-1}- {k-1\choose0}(\frac{1}{6})^0(\frac{4}{6})^{k-1} $$
$$ =(\frac{5}{6})^4 - (\frac{4}{6})^4$$
|
It's not clear to me if the question is whether $k$ rolls suffice or whether exactly $k$ rolls are required. I try to cover both possibilities below.
Let's say the first roll on which we have at least one 5 and at least one 6 is roll $K$.
Let's also say $A$ is the event of rolling no 5 in $k$ rolls and $B$ is the event of rolling no 6 in $k$ rolls. Then the probability that we have at least one 5 and at least one 6 by roll $k$ is
$$\begin{align}
\Pr(K \le k) &=\Pr(A^c \cap B^c) \\
&= 1 - \Pr(A \cup B) \\
&= 1 - \Pr(A) - \Pr(B) + \Pr(A \cap B) \\
&= 1 - 2 \left( \frac{5}{6} \right)^k + \left( \frac{4}{6} \right)^k \\
\end{align}$$
So the probability that the first roll on which we have at least one 5 and at least one 6 is roll $k$ is
$$\Pr(K = k) = \Pr(K \le k) - \Pr(K \le k-1) = \frac{1}{3} \left( \frac{5}{6} \right)^{k-1} - \frac{1}{3} \left( \frac{4}{6} \right)^{k-1}$$
|
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|
Turkevicius inequality From Zdravko Cvetkovski's Inequalities — Theorems, Techniques and Selected:
Let $a$, $b$, $c$ and $d$ be positive real numbers. Prove that:
$$a^4+b^4+c^4+d^4+2abcd \geq a^2b^2+ a^2c^2+ a^2d^2+ b^2c^2+ b^2d^2+ c^2d^2.$$
Please suggest how to prove this inequality using basic methods. No logarithms.
|
Assume $a\geq b\geq c\geq d.$ Let $f(a) = a^4+b^4+c^4+d^4+2abcd - a^2b^2-a^2c^2-a^2d^2-b^2c^2-b^2d^2-c^2d^2.$ Then, $f'(a) = 4a^3 -2a(b^2+c^2+d^2)+2bcd$ and $f''(a) = 12a^2 - 2b^2-2c^2-2d^2\geq 0.$ Hence, $f'$ is increasing and $f'(b) = 2(b^3-b(c^2+d^2)+bcd) \geq 2b(b-c)(b-d)\geq 0.$ Thus, $f(a)$ is increasing and therefore:
$$f(a)\geq f(b) = b^4+c^4+d^4+2b^2cd-2b^2c^2-2b^2d^2-c^2d^2 = $$$$b^4-2b^2(c^2+d^2-cd)+c^4+d^4-c^2d^2 = (b^2 - c^2-d^2+cd)^2+2cd(c-d)^2\geq 0.$$
|
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|
Sum of terms $1^2-2^2+3^2-...$ Prove that: $$1^2-2^2+3^2-4^2+...+(-1)^{n-1}n^2=\frac{(-1)^{n-1}n(n+1)}{2}$$
I proved it by induction but is there any other way to solve it?
If it was not a proof but rather a question like find the term,how to solve it?
I realized that alternate terms were under same sign but can't understand whether to take $\frac{n}{2}$ odd and even terms[if n is even] or $\frac{n+1}{2}$ odd terms and $\frac{n-1}{2}$ even terms[if n is odd].
I thought of this $$1^2-2^2+3^2-4^2+5^2=1^2+{(1+2)}^2+{(1+4)}^2-2^2-4^2$$
$$=1^2+1^2+1^2+2(0+2+4)+2^2+4^2-2^2-4^2$$
But then how to generalize??
|
$$\sum_{r=1}^{2n}(-1)^{r-1}r^2=\sum_{r=1}^n\{(2r-1)^2-(2r)^2\}=\sum_{r=1}^n(1-4r)$$
$$=\dfrac n2(1-4+1-4n)=\cdots=(-1)^{2n-1}\dfrac{2n(2n+1)}2$$
Now $$\sum_{r=1}^{2n+1}(-1)^{r-1}r^2=(2n+1)^2+\sum_{r=1}^{2n}(-1)^{r-1}r^2=?$$
|
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|
How to compute $S_{2016}=\sum\limits_{k=1}^{2016}\left(\sum\limits_{n=k}^{2016}\frac1n\right)^2+\sum\limits_{k=1}^{2016}\frac1k$? I came across a question asking the value of the following sum:
\begin{align}
\left(1+ \frac{1}{2}+\frac{1}{3}+\cdots +\frac 1{2015}+\frac{1}{2016}\right)^2 \\
+\left(\frac{1}{2}+\frac{1}{3}+\cdots +\frac 1{2015}+\frac{1}{2016}\right)^2 \\
+ \left(\frac{1}{3}+\cdots +\frac 1{2015}+\frac{1}{2016}\right)^2 \\[5 pt]
+\cdots \qquad\quad \vdots\qquad\qquad\\[5 pt]
+ \left(\frac{1}{2015}+\frac{1}{2016}\right)^2 \\
+ \left(\frac{1}{2016}\right)^2\\
+ \left(1+ \frac{1}{2}+\frac{1}{3}+\cdots +\frac 1{2015}+\frac{1}{2016}\right)\;
\end{align}
I can not find a good way to solve it. Any ideas?
Edit: That is, with no dots, $$S_{2016}=\sum_{k=1}^{2016}\left(\sum_{n=k}^{2016}\frac1n\right)^2+\sum_{k=1}^{2016}\frac1k$$
|
If you expand all:
$$S=1+2\times \frac{1}{2^2}+3\times \frac{1}{3^2}+\cdots +2016 \cdot \frac{1}{2016^2}+$$
$$2\cdot 1\left(1\cdot \frac12+1\cdot \frac13+\cdots 1\cdot \frac{1}{2016}\right)+$$
$$2\cdot 2\left(\frac12 \cdot \frac13+\frac12 \cdot \frac14+\cdots +\frac12 \cdot \frac{1}{2016}\right)+$$
$$\vdots$$
$$2\cdot 2015\left(\frac{1}{2015}\cdot \frac{1}{2016}\right)+$$
$$1+\frac12+\cdots +\frac{1}{2016}=$$
$$2\left(1+\frac12+\cdots+\frac{1}{2016}\right)+$$
$$2\left(\frac12+\frac13+\cdots+\frac{1}{2016}\right)+$$
$$\vdots$$
$$2\left(\frac{1}{2016}\right)=$$
$$2\cdot (2016)=4032.$$
|
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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|
If $z = 2 - 3i$, show that $z^2 - 4z + 13 = 0$ and hence find the value of $4z^3 - 3z^2 + 169$ This is an example given in my book, and it seems to involve division by 0, so I'm rather confused
Here is how it's done in the book:
We have $z = 2 - 3i $
So, $z - 2 = -3i$
Now, ${(z-2)}^2 = {(-3i)}^2$
Or, $z^2 + 4 - 4z = 9i^2$
Or $z^2 -4z + 13 = 0$
Now, $4z^3 - 3z^2 + 169 = 4z(z^2 - 4z + 13) + 13(z^2 -4z + 13)$
In the above step, clearly $4z^3 - 3z^2 + 169$ has been divided by $z^2 - 4z + 13$ to obtain the factors. But $z^2 - 4z + 13 $ is equal to $0$, so wouldn't the division step give an undefined answer ?
|
Alt. hint: if $z = 2 - 3i$, then $z+\bar z = 2 \operatorname{Re}(z) = 4$ and $ z \bar z = |z|^2 = 2^2+3^2=13$. It follows by Vieta's formulas that $z$ and $\bar z$ are the roots of $z^2-4z+13=0\,$.
But $z^2-4z+13=0 \iff z^2 = \color{blue}{4z -13}\,$, then multiplying by $z \ne 0\,$ it follows that $z^3 = 4z^2-13 z = 4(4z-13)-13z=\color{green}{3z - 52}\,$
So in the end $\,4z^3 - 3z^2 + 169 = 4\color{green}{(3z-52)}-3\color{blue}{(4z-13)}+169=0\,$.
|
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|
Induction proof of $F(n)^2+F(n+1)^2=F(2n+1)$, where $F(n)$ is the $n$th Fibonacci number.
Let $F(n)$ denotes the $n$th number in Fibonacci sequence. Then for all $n\in\mathbb{N}$,
$$F(n)^2+F(n+1)^2=F(2n+1).$$
I know how to prove it by using the formula
$$F(n)=\frac{\left(\frac{1+\sqrt5}{2}\right)^n-\left(\frac{1-\sqrt5}{2}\right)^n}{\sqrt{5}},$$
but is there a way to prove it by induction?
I am year 12 standard so please don't go too deep.
|
The base cases $n=0$ and $n=1$ are easy to verify.
Note that
\begin{align*}
F(2n+3)&=F(2n+2)+F(2n+1)=(F(2n)+F(2n+1))+F(2n+1)\\
&=F(2n)+2F(2n+1)\\
&=(F(2n+1)-F(2n-1))+2F(2n+1)\\&=
3F(2n+1)-F(2n-1).
\end{align*}
Hence, by the inductive hypothesis,
\begin{align*}
F(2n+3)&=3F(2n+1)-F(2n-1)=3(F(n)^2+F(n+1)^2)-(F(n-1)^2+F(n)^2)\\
&=2F(n)^2+3F(n+1)^2-F(n-1)^2\stackrel{?}{=}F(n+1)^2+F(n+2)^2.
\end{align*}
So it remains to show that
$$F(n+2)^2+F(n-1)^2\stackrel{?}{=}2F(n)^2+2F(n+1)^2$$
which holds because
\begin{align*}
F(n+2)^2+F(n-1)^2&=(F(n+1)+F(n))^2+(F(n+1)-F(n))^2\\&=2F(n)^2+2F(n+1)^2.
\end{align*}
|
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|
Algebraic solution of complex equation For solving algebraically any complex equation involves two components for the real & imaginary parts. Let the real part be - $a$, imaginary part - $b$.
For the complex equation $$x^3 = 1-i $$
Substituting $x = a +bi$, we get: $$(a+bi)^3 = 1 - i.$$
Expanding the l.h.s. :
$$a^3 -(b^3)i + 3(a^2)bi -3(b^2)a$$
Equating real and imaginary parts, we get:
$$ a^3 -3(b^2)a = 1 \tag{1}$$
$$ b^3 -3(a^2)b = 1 \tag{2}$$
Factoring $(1)$, $(2)$ to get some roots in the process:
$$a(a^2 - 3(b^2)) = 1 \tag{1'}$$
$$b(b^2 - 3(a^2)) = 1 \tag{2'}$$
From $(1')$, the roots are possibly given by equations below:
$$a = 1 \tag{3}$$
$$ - \text{or} - $$ // 'or' is used in logical or sense, i.e. either or both
$$ a^2 - 3(b^2) = 1 \tag{4}$$
Substituting $a = 1$ in $(4)$, we get:
$b = 0$; which cannot be a possible solution, as the right side of
question has $b = -1$. Hence, $a \ne 1$ also.
Next, trying for the possible value from $(2')$.
$$ b = 1 \tag{5} $$
$$ -\text{or}- $$ // 'or' is used in logical or sense, i.e. either or both
$$ b^2 - 3(a^2) = 1 \tag{6} $$
Substituting $b = 1$ in $(6)$, we get:
$$ a = 0; $$
which cannot be a possible solution, as the right side of question
has $a = 1$. Hence, $b \ne 1$ also.
I am unable to solve it further, as no solution emerges from the
two equations - $(1')$, $(2')$.
|
Someone has pointed out that one solution of $x^3 - (1-i) =0$ is $\dfrac{-1-i}{\sqrt[3]2}.$
Therefore $x+\dfrac{1+i}{\sqrt[3]2}$ is a factor of $x^3 - (1-i).$ So we can do long division:
$$
\begin{array}{cccccccccc}
& & & & x^2 & - & \dfrac{1+i}{\sqrt[3]2}x & + & i\sqrt[3]2 \\ \\
x + \dfrac{1+i}{\sqrt[3]2} & \Big) & x^3 & + & 0x^2 & + & 0x & - & (1-i) \\
& & x^3 & + & \dfrac{1+i}{\sqrt[3]2} x^2 \\ \\ \\
& & & & -\dfrac{1+i}{\sqrt[3]2} x^2 & & & - & (1-i) \\
& & & & -\dfrac{1+i}{\sqrt[3]2} x^2 & - & i x \sqrt[3]2 \\ \\ \\
& & & & & & ix\sqrt[3]2 & - & (1-i) \\
& & & & & & ix\sqrt[3]2 & - & (1-i) \\ \\ \\
& & & & & & & & 0 \\
\hline
\end{array}
$$
Thus we have found that
$$
x^3-(1-i) = \left( x + \frac{1+i}{\sqrt[3]2} \right) \left( x^2 - \frac{1+i}{\sqrt[3]2} x + i\sqrt[3]2 \right).
$$
The other two solutions are now solutions of a quadratic equation. If we write the equation as $ax^2+bx+c=0$ then we have $a=1,$ $b= - \dfrac{1+i}{\sqrt[3]2},$ and $c=i\sqrt[3]2,$ and
$$
b^2-4ac = -3i\sqrt[3]2.
$$
So what is $\pm\sqrt{b^2-4ac~} \text{ ?}$ We just saw that $(1+i)^2 = 2i,$ i.e. $\pm\sqrt{2i} = \pm(1+i),$ so that gives us $\pm\sqrt{b^2-4ac~}.$
|
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|
Show that $\lim_{n\to\infty}\frac{1}{n^4}\sum_{j=1}^{n}\left((2j-1)\sum_{k=1}^{n+1-j}k\right)=\frac{1}{12}$ Show that$$\lim_{n\rightarrow \infty} \frac{1}{n^4} \left(1\left(\sum_{k=1}^{n}k\right)+ 3\left(\sum_{k=1}^{n-1}k\right)+5\left(\sum_{k=1}^{n-2}k\right)+\cdots+(2n-1)\cdot1\right)=\frac{1}{12}$$
This is getting more and more complicated as number of terms are decreasing.
|
$$\begin{align}
\sum_{r=1}^n\left[(2r-1)\sum_{k=1}^{n-r+1}k\right]
&=\sum_{r=1}^n\left[(2r-1)\sum_{s=r}^n (n-s+1)\right]
&&(s=n-k+1)\\
&=\sum_{r=1}^n\sum_{s=r}^n\sum_{t=s}^n(2r-1)\\
&=\sum_{t=1}^n\sum_{s=1}^t\sum_{r=1}^s (2r-1)
&&(1\le r\le s\le t\le n)\\
&=\sum_{t=1}^n \sum_{s=1}^t s^2\\
&=\sum_{t=1}^n \sum_{s=1}^t \binom s2+\binom {s+1}2\\
&=\sum_{t=1}^n \binom {t+1}3+\binom {t+2}3\\
&=\binom {n+2}4+\binom {n+3}4\\
&=\frac 14\big[(n-1)+(n+3)\big]\binom {n+2}3\\
&=\frac 12(n+1)\binom {n+2}3\\
&=\frac {(n+2)(n+1)^2n}{12}\\
\lim_{n\to \infty}\frac 1{n^4}\sum_{r=1}^n\left[(2r-1)\sum_{k=1}^{n-r+1}k\right]
&=\lim_{n\to\infty}\frac 1{n^4}\cdot\frac {(n+2)(n+1)^2n}{12}\\
&=\lim_{n\to\infty}\frac 1{n^4}\cdot \frac {n^4+O(n^3)}{12}\\
&=\lim_{n\to\infty}\frac {1+O(\frac 1n)}{12}\\
&=\frac 1{12}
\end{align}$$
|
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|
How to calculate $\lim_{n\to\infty}f(n)\sin(\frac{1}{n})$ where $f(x)=\int_{x}^{x^2}(1+\frac{1}{2t})^t\sin{\frac{1}{\sqrt{t}}}dt(x>0)$
$$f(x)=\int_{x}^{x^2}\left(1+\frac{1}{2t}\right)^t\sin\left(\frac{1}{\sqrt{t}}\right)dt\hspace{1cm}(x>0)$$
try to find $$\lim_{n\to\infty}f(n)\sin\left(\frac{1}{n}\right)$$
I found this problem from a problem book and with a hint which tells me to apply L'Hospital's rule.
But when $n\to\infty$ then $\frac{1}{\sin(1/n)}\to\infty$ and $f(n)\text{ seems}\to0$.
Would you help me with this problem? Best regards!
|
First prove that $f(n)\to\infty$ as $n\to\infty$ then apply L'Hopital rule
$$\lim_{n\to\infty}f(n)\sin\left(\frac{1}{n}\right)=\lim_{n\to\infty}\frac{f(n)}{\frac{1}{\sin\left(\frac{1}{n}\right)}}=\lim_{n\to\infty}\frac{f'(n)}{\left(\dfrac{1}{\sin\left(\frac{1}{n}\right)}\right)'}$$
$$f'(n)=2 \left(\frac{1}{2 n^2}+1\right)^{n^2} n \sin \left(\frac{1}{\sqrt{n^2}}\right)-\left(\frac{1}{2 n}+1\right)^n \sin \left(\frac{1}{\sqrt{n}}\right)$$
and
$$\left(\dfrac{1}{\sin\left(\frac{1}{n}\right)}\right)'=\frac{2 \cos \left(\frac{1}{n}\right)}{n^2 \left(1-\cos \left(\frac{2}{n}\right)\right)}$$
$$\lim_{n\to\infty} \frac{2 \left(\frac{1}{2 n^2}+1\right)^{n^2} n \sin \left(\frac{1}{\sqrt{n^2}}\right)-\left(\frac{1}{2 n}+1\right)^n \sin \left(\frac{1}{\sqrt{n}}\right)}{\frac{2 \cos \left(\frac{1}{n}\right)}{n^2 \left(\cos \left(1-\frac{2}{n}\right)\right)}}=$$
$$=\lim_{n\to\infty}\left[n^2 \sin \left(\frac{1}{n}\right) \left(2 \left(\frac{1}{2 n^2}+1\right)^{n^2} n \sin \left(\frac{1}{\sqrt{n^2}}\right)-\left(\frac{1}{2 n}+1\right)^n \sin \left(\frac{1}{\sqrt{n}}\right)\right) \tan \left(\frac{1}{n}\right)\right]$$
Now recalling that $$\lim_{t\to\infty}\left(1+\frac{1}{t}\right)^t=e;\;\lim_{z\to 0}\frac{\sin z}{z}=1;\;\lim_{z\to 0}\frac{\tan z}{z}=1$$
you should arrive to the result $2 \sqrt{e}$
|
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"timestamp": "2023-03-29T00:00:00",
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|
Let $a$, $b$ and $c$ be integers. Prove that if $4|(a+bc)$ and $6|(b+ac)$, then $2|(a^2-b^2)$. This is my work so far.
How can I isolate the $(a^2-b^2)$?
I am not asking for a full solution, just a hint and next step.
All done! Thanks Michael.
|
$4|(a+bc)$ then $a+b c = 4k\to c=\dfrac{4 k-a}{b}$
$6|(b+ac)$ then $b+ac=6h$ plug $c$
$b + a \,\dfrac{4 k-a}{b}=6h$
$b^2+4ak-a^2=6bh$
$a^2-b^2=2 (2 a k-3 b h)$
Which means that $2|(a^2-b^2)$
QED
|
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|
A system of two nonlinear algebraic equations Is it possible to determine real solutions $x,y$ for the following system of nonlinear equations in terms of the real constants $a,b,k$?
$$
a=kx(1-y^2)\qquad\mbox{and}\qquad b=ky(1-x^2).
$$
If so, what is (are) the solution(s)?
|
Square the first equation $a^2 = k^2x^2(1-y^2)^2$. Now $k^2(1-y^2)^2$ minus this equation gives
\begin{eqnarray*}
k^2(1-y^2)^2-a^2 = k^2(1-y^2)^2(1-x^2)
\end{eqnarray*}
Multiply this by $y$ and substitute for $ky(1-x^2)=b$ and we have
\begin{eqnarray*}
\color{red}{y(k^2(1-y^2)^2-a^2) = kb(1-y^2)^2}.
\end{eqnarray*}
This is a quintic (order $5$) in $y$ and will not be easy to solve.
|
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|
how to solve $\int e^{7x}\cos(2x)dx$ I'm having problems to solve this integral, It looks like it will never end because every time I do integration by parts I need to do it again and again...
$$\int e^{7x}\cos(2x)dx$$
$$u=\cos(2x) \ \ \ dv = e^{7x}$$
$$du = -2\sin(2x)dx \ \ \ v = \frac{1}{7}e^{7x}$$
$$\frac {\cos(2x)e^{7x}}{7}-\int \frac{1}{7}e^{7x}(-2\sin(2x))dx$$
$$\frac {\cos(2x)e^{7x}}{7}+\frac{2}{7}\int e^{7x}(-2\sin(2x))dx$$
doing integration by parts again
$\int e^{7x}(-2\sin(2x))dx$
$u = \sin(2x)$ $dv = e^{7x}$
$du = \cos(2x)$ $v =\frac{1}{7}e^{7x}$
$\frac {\sin(2x)e^{7x}}{7} - \frac{2}{7}\int e^{7x}\cos(2x)dx$
|
You're almost there:
\begin{align}
I &= \int e^{7x}\cos(2x)\,dx \\
&= \frac {\cos(2x)e^{7x}}{7}+\frac{2}{7}\int e^{7x}(-2\sin(2x))\,dx \\
&= \frac {\cos(2x)e^{7x}}{7}+\frac{2}{7}\left(\frac {\sin(2x)e^{7x}}{7} - \frac{2}{7}\int e^{7x}\cos(2x)\,dx\right)\\
&= \frac {\cos(2x)e^{7x}}{7}+\frac {2\sin(2x)e^{7x}}{49} - \frac{4}{49}\int e^{7x}\cos(2x)\,dx\\
&= \frac {7\cos(2x)+2\sin(2x)}{49}e^{7x} - \frac{4}{49}I\\
\end{align}
Thus, by solving for $I$ we get:
$$I = \frac {7\cos(2x)+2\sin(2x)}{53}e^{7x}$$
|
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|
Find a and b, that there are two roots of $x^3 - 5x^2+ 7x = a$ and these numbers are roots of $x^3 - 8x + b = 0$ Find a and b, that there are two roots of $x^3 - 5x^2+ 7x = a$, that they are roots of $x^3 - 8x + b = 0$
I find that $a+b=5x(-x+3) $
Also I tried to solve second equation to get roots, depending on b, but my approach was unsuccessful.
Any help is appreciated!
|
They are roots of the equation
$$x^2=3x-\frac{a+b}{5},$$ which you got.
Thus, since $x^3-8x+b=0$ has these two roots, we see that the equation
$$x\left(3x-\frac{a+b}{5}\right)-8x+b=0$$ or
$$x^2-\frac{1}{3}\left(\frac{a+b}{5}+8\right)x+\frac{b}{3}=0$$
has these two roots.
Thus, we have the following system
$$\frac{1}{3}\left(\frac{a+b}{5}+8\right)=3$$ and
$$\frac{b}{3}=\frac{a+b}{5},$$
which gives $a=2$ and $b=3$.
|
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|
Proof that $\binom{n + 3}{4} = n + 3 \binom{n + 2}{4} - 3 \binom{n + 1}{4} + \binom{n}{4}$. I was trying to count the number of equilateral triangles with vertices in an regular triangular array of points with n rows. After putting the first few rows into OEIS, I saw that this was described by A000332: $\binom{n}{4} = n(n-1)(n-2)(n-3)/24$.
A000332 has the comment:
Also the number of equilateral triangles with vertices in an equilateral triangular array of points with n rows (offset 1), with any orientation.
The linked solution is insightful, but the proof is very mechanical.
I tried to write an inductive proof, but I'm unable to come up with a nice way to prove the final identity (in particular for $n \geq 4$):
$$\binom{n + 3}{4} = n + 3 \binom{n + 2}{4} - 3 \binom{n + 1}{4} + \binom{n}{4}$$.
Wolfram|Alpha admits that these are indeed equal, and this could certainly be shown by writing everything as a polynomial.
However, I was hoping to find some nice combinatorial identities that give some intuition as to why this equality holds.
|
Here is a combinatorial proof, with generalization. Let $m$, $n$, and $r$ be non-negative integers such that $m\leq n$. Consider $(m+r)$-subsets of the set $S=\{1,2,\ldots,n,n+1,n+2,\ldots,n+r\}$ which contain all numbers $n+1$, $n+2$, $\ldots$, $n+r$. Such a subset must be of the form $V\cup \{n+1,n+2,\ldots,n+r\}$ with $V\subseteq \{1,2,\ldots,n\}$ having $m$ elements. So there are $\binom{n}{m}$ possible subsets of this form. We now count the number of these subsets in a different way, using PIE.
Let $T_s$ denote the set of $s$-subsets of $T=\{n+1,n+2,\ldots,n+r\}$. Let $E_A$ denote the set of $(m+r)$-subsets $U$ of $S$ such that $A$ does not intersect $U$. Let $\mathcal{C}$ be the set of all $(m+r)$-subsets of $S$. Observe that $\mathcal{C}\setminus \bigcup_{k=1}^r E_{\{n+k\}}$ is the set composed by all $(m+r)$-subsets of $S$ which contain $n+1$, $n+2$, $\ldots$, $n+r$. By PIE,
$$\left|\mathcal{C}\setminus \bigcup_{k=1}^rE_{\{n+k\}}\right|=\sum_{s=0}^r(-1)^s\sum_{A\in T_s}\left|E_A\right|.$$
Since $|E_A|=\binom{n+r-s}{m+r}$ and $|T_s|=\binom{r}{s}$, we get
$$\binom{n}{m}=\left|\mathcal{C}\setminus \bigcup_{k=1}^rE_{\{n+k\}}\right|=\sum_{s=0}^r(-1)^s\binom{r}{s}\binom{n+r-s}{m+r}.$$
In particular, when $m=1$ and $r=3$, we have the identity
\begin{align}n&=\binom{3}{0}\binom{n+3}{4}-\binom{3}{1}\binom{n+2}{4}+\binom{3}{2}\binom{n+1}{4}-\binom{3}{3}\binom{n}{4}\\&=\binom{n+3}{4}-3\binom{n+2}{4}+3\binom{n+1}{4}-\binom{n}{4}.\end{align}
This is equivalent to
$$\binom{n+3}{4}=n+3\binom{n+2}{4}-3\binom{n+1}{4}+\binom{n}{4}.$$
|
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|
Value of $1/e$ with the partial sum of $e$ How to show $$1/e=1-1+\frac1{2!}-\frac1{3!}+\frac1{4!}+\dots$$
I have consider the product of the nth partial sums of the expansions for $e$ and $1/e$ but still can't prove there are equal to 1
with the definition $e=1+1+\frac1{2!}+\frac1{3!}+\dots$
|
I assume you know that
$$e = 1 + 1 + \frac{1}{2!} + \frac{1}{3!} + \ldots = \sum_{k=0}^{\infty} \frac{1}{k!}.$$
Now let
$$f = 1 - 1 + \frac{1}{2!} - \frac{1}{3!} + \ldots = \sum_{k=0}^{\infty} \frac{(-1)^k}{k!}.$$
By the Mertens theorem and the binomial formula
$$\begin{align*}
e \cdot f & = \left( \sum_{k=0}^{\infty} \frac{1}{k!} \right) \cdot \left( \sum_{k=0}^{\infty} \frac{(-1)^k}{k!} \right) = \sum_{n=0}^{\infty} \sum_{k=0}^n \frac{(-1)^{n-k}}{k! (n-k)!} \\[1ex]
& = \sum_{n=0}^{\infty} \frac{1}{n!} \sum_{k=0}^n \binom{n}{k} \cdot 1^k \cdot (-1)^{n-k} = \sum_{n=0}^{\infty} \frac{(1-1)^n}{n!} = \frac{1}{0!} = 1
\end{align*}$$
so $f = \frac{1}{e}$. In the end we used the fact that
$$0^n = \begin{cases} 1 & \text{ if } n = 0 \\ 0 & \text{ if } n > 0 \end{cases}$$
|
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|
A Geometry Question Concerning Six Equilateral Triangles The question is as follows:
The figure below is built by joining six equilateral triangles ABC, ACD, CDE, DEF, EFG, and FGH, all of whose edges are 1 unit long. It is given that HIJKLMB is straight.
(a) There are five triangles in the figure that are similar to CMB. List them, making sure that you match corresponding vertices.
(b) Find the lengths of CM and EK.
(c) List the five triangles that are similar to AMB.
(d) Find the lengths of CL, HI, IJ, and JK.
I believe that we need to use a concept based on similar triangles, but I am unsure as to what I should be doing. Any help will be greatly appreciated.
|
Since $\Delta CMB\sim\Delta AMH$, we obtain:
$$\frac{CM}{AM}=\frac{BC}{AH}$$ or
$$\frac{CM}{AC-CM}=\frac{BC}{AH},$$
which says $$CM=\frac{1}{4}.$$
$EK=\frac{1}{2}$ by symmetry.
Now, $\Delta BLC\sim\Delta HLD.$
Thus, $$\frac{CL}{LD}=\frac{BC}{HD}$$ or
$$\frac{CL}{1-CL}=\frac{1}{2},$$
which gives $$CL=\frac{1}{3}.$$
Now, since $FI=CM=\frac{1}{4}$, by the law of cosines we obtain:
$$HI=\sqrt{1^2+\left(\frac{1}{4}\right)^2-2\cdot1\cdot\frac{1}{4}\cdot\frac{1}{2}}=\frac{\sqrt{13}}{4}.$$
Also, since $FI=\frac{1}{2}$ and $AJ=\frac{1}{3}$, we obtain:
$$IJ=\sqrt{\left(\frac{1}{4}\right)^2+\left(\frac{1}{3}\right)^2-2\cdot\frac{1}{4}\cdot\frac{1}{3}\cdot\frac{1}{2}}=\frac{\sqrt{13}}{12}$$ and since $EJ=1-\frac{1}{3}=\frac{2}{3},$ we obtain:
$$JK=\sqrt{\left(\frac{2}{3}\right)^2+\left(\frac{1}{2}\right)^2-2\cdot\frac{2}{3}\cdot\frac{1}{2}\cdot\frac{1}{2}}=\frac{\sqrt{13}}{6}.$$
Done!
|
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An integral of a rational function with high degrees: Evaluate $\int_{-\infty}^{\infty}\frac{(x^4 + 1)^2}{x^{12} + 1}dx$.
Calculate: $$\int_{-\infty}^{\infty}\frac{(x^4 + 1)^2}{x^{12} + 1}dx.$$
What I have tried is to divide both numerator and denominator with $x^4 + 1$ and then get two following integrals, because of parity of the integrand (nothing else worked for me): $$2\int_0^\infty \frac{x^4 + 1}{(x^4 - \sqrt3x^2 + 1)(x^4 + \sqrt3x^2 + 1)}dx = $$ $$ = \int_0^\infty \frac1{x^4 - \sqrt3x^2 + 1}dx + \int_0^\infty\frac1{x^4 + \sqrt3x^2 + 1}dx = $$ $$ = \int_0^\infty \frac1{(x^2 - \frac{\sqrt3}2)^2 + \frac14}dx + \int_0^\infty \frac1{(x^2 + \frac{\sqrt3}2)^2 + \frac14}dx.$$
I don't see what would be continuation of this. Any help is appreciated.
Thank you for any help. Appreciate it.
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Hint: For integrals of rational functions of even polynomials over the real line, the following substitution can come in handy:
$$\begin{align}
\mathcal{I}
&=\int_{-\infty}^{\infty}\mathrm{d}x\,\frac{\left(x^{4}+1\right)^{2}}{x^{12}+1}\\
&=\int_{-\infty}^{\infty}\mathrm{d}x\,\frac{\left(x^{4}+1\right)^{2}}{\left(x^{4}+1\right)\left(x^{8}-x^{4}+1\right)}\\
&=\int_{-\infty}^{\infty}\mathrm{d}x\,\frac{x^{4}+1}{x^{8}-x^{4}+1}\\
&=2\int_{0}^{\infty}\mathrm{d}x\,\frac{x^{4}+1}{x^{8}-x^{4}+1}\\
&=2\int_{0}^{\infty}\mathrm{d}x\,\frac{\left(x^{2}+x^{-2}\right)}{x^{2}\left(x^{4}+x^{-4}-1\right)}\\
&=2\int_{0}^{\infty}\mathrm{d}x\,\frac{\left(x^{2}+x^{-2}\right)}{\left(x^{4}+x^{-4}-1\right)};~~~\small{\left[x\mapsto\frac{1}{x}\right]}\\
&=\int_{0}^{\infty}\mathrm{d}x\,\frac{\left(x^{2}+x^{-2}\right)}{x^{2}\left(x^{4}+x^{-4}-1\right)}+\int_{0}^{\infty}\mathrm{d}x\,\frac{\left(x^{2}+x^{-2}\right)}{\left(x^{4}+x^{-4}-1\right)}\\
&=\int_{0}^{\infty}\mathrm{d}x\,\frac{x^{2}+1}{x^{2}}\cdot\frac{\left(x^{2}+x^{-2}\right)}{\left(x^{4}+x^{-4}-1\right)}\\
&=\int_{-\infty}^{\infty}\mathrm{d}y\,\frac{2\left(4y^{2}+2\right)}{\left(16y^{4}+16y^{2}+1\right)};~~~\small{\left[\frac{x^{-1}-x}{2}=y\iff x=\sqrt{y^{2}+1}-y\right]}.\\
\end{align}$$
Note that we have cut the degrees of the polynomials in half and still have all integer coefficients. This could be considered an advantage over the brute-force partial fraction method if you're looking to avoid messy factorizations.
|
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In how many ways can a student score exactly $100$ points on four $50$ point exams? greater than $100$? A student takes up $4$ exams with $50$ points each. In how many ways can he score exactly $100$? Similarly, in how many ways can he score greater than $100$?
My try:
I cannot seem to fill in the $4$ possibilities. The range depends on the constraint of equating to $100$.
Thanks for any help!
|
To score exactly 100, first suppose there are no constraints other than having nonnegative numbers. Then the number of solutions to $a+b+c+d = 100$ is simply $\binom{103}{3}$ by a standard stars and bars argument. Now we need only remove solutions where one of the values is greater than $50.$ Suppose $ a > 50,$ then we ask for solutions to $(a-51) + b + c + d = 49,$ where $a,b,c,d$ are nonnegative, which has $\binom{52}{3}$ solutions. Since we could have had any of the four be the one larger than $50,$ there are $4\cdot \binom{52}{3}$ such "bad" solutions. Hence the required total is $N_{100} = \binom{103}{3} - 4\binom{52}{3}.$
For the second part, any solution to $a + b + c +d > 100$ corresponds with $(50-a) + (50-b) + (50-c) + (50-d) < 100.$ Hence the number of such solutions is just $(51^4 - N_{100})/2,$ where $N_{100}$ is the answer to the first part
|
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Limit of $\sqrt{x^2+3x}+x$ when $x\to-\infty$ Limit of $ \lim_{x\to -\infty}(\sqrt{x^2+3x}+x)$, I know that the final answer is $-3/2$, my question is about Wolfram Alpha step by step solution:
$$x+\sqrt{x^2+3x}=\frac{(x+\sqrt{x^2+3x})(x-\sqrt{x^2+3x})}{x-\sqrt{x^2+3x}}$$
$$=-\frac{3x}{x-\sqrt{x^2+3x}}$$
$$\lim_{x\to-\infty}-\frac{3x}{x-\sqrt{x^2+3x}}$$
$$\lim_{x\to-\infty}-\frac{3x}{x-\sqrt{x^2+3x}}=-3$$
$$\lim_{x\to-\infty}\frac{x}{x-\sqrt{x^2+3x}}=-3\lim_{x\to-\infty}\frac{x}{x-\sqrt{x^2+3x}}$$
$$\frac{x}{x-\sqrt{x^2+3x}}=\frac{1}{1-\frac{\sqrt{x^2+3x}}{x}}$$
$$-3\lim_{x\to-\infty}\frac{x}{x\left(1-\frac{\sqrt{x^2+3x}}{x}\right)}$$
To prepare the product $\frac{1}{1-\frac{\sqrt{x^2+3x}}{x}}$ for solution by l'Hopital's rule, write it as $\frac{x}{x\left(1-\frac{\sqrt{x^2+3x}}{x}\right)}$
$$-3\lim_{x\to-\infty}\frac{x}{x\left(1-\frac{\sqrt{x^2+3x}}{x}\right)}$$
Is it correct to use L'Hopital here like Wolfram did?
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$$
x+\sqrt{x^2+3x}=\frac{x^2-({x^2+3x})}{x-\sqrt{x^2+3x}}=\frac{-3x}{x-\sqrt{x^2+3x}}=\frac{-3}{1+\sqrt{1+\frac3x}}\stackrel{\tiny x\to-\infty}{\longrightarrow} -\frac 32
$$
|
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|
Find the arclength of the function $ x = \frac{1}{3} y^{\frac{-3}{2}} + \frac{1}{7}y^{\frac{7}{2}}$ for $1\leq y \leq 25$. I'm trying to find the arclength of the function $ x = \frac{1}{3} y^{\frac{-3}{2}} + \frac{1}{7}y^{\frac{7}{2}}$ for $1\leq y \leq 25$.
I can find the equation for the length pretty easily but I'm looking at thow to solve for the actual length. It looks like a very complex integral so I'm assuming I made a mistep or theres some easy reduction I can make.
After determining the area of an incredibly small section of the function:
$$ds = \sqrt{\left(\frac{dx}{dy}\right)^2 + 1}$$
$$\frac{dx}{dy} = \frac{-y^{\frac{-5}{2}}}{2} + \frac{y^{\frac{5}{2}}}{2} $$
$$ds = \sqrt{\left(\frac{-y^{\frac{-5}{2}}}{2} + \frac{y^{\frac{5}{2}}}{2}\right)^2 + 1}$$
This leaves me with the integral
$$\int{\sqrt{\left(\frac{-y^{\frac{-5}{2}}}{2} + \frac{y^{\frac{5}{2}}}{2}\right)^2 + 1}} dy$$
I do still have to calculate from 1 - 25 but I like to plug in after I solve my integral. Anyway, I can't tell how to solve this, but I have a feeling I need to play with the squared term. Perhaps a substitution or maybe the reciprocals simplify into something. If anyone has any tips I appreciate it!
EDIT: Problem solved! (I think)
In the answer to the question I can make the simplification
$$\int{\sqrt{\left(\frac{y^{\frac{5}{2}} + y^{\frac{-5}{2}}}{2}\right)^2}} dy$$
$$\int{\left(\frac{y^{\frac{5}{2}} + y^{\frac{-5}{2}}}{2}\right)} dy$$
$$\frac{1}{2}\left(\int{y^{\frac{5}{2}}}dy + \int{y^{\frac{-5}{2}}}dy\right)$$
$$\frac{y^{\frac{7}{2}}}{7} + \frac{1}{-3y^{\frac{3}{2}}}$$
We plug in our bounds here and the answer is $F(25) - F(1)$
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HINT:
$$1+\left(\frac{y^{5/2}-y^{-5/2}}{2}\right)^2=\left(\frac{y^{5/2}+y^{-5/2}}2\right)^2$$
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A difficult integral $I=\int_0^1\sqrt{1+\sqrt{1-x^2}}\frac{dx}{1+x^2}$ How to prove
$$I=\int_0^1\sqrt{1+\sqrt{1-x^2}}\frac{dx}{1+x^2}=\sqrt{\sqrt{2}+1}\arctan\sqrt{\sqrt{2}+1}-\frac{1}{2}\sqrt{\sqrt{2}-1}\ln(1+\sqrt{2}+\sqrt{2+2\sqrt{2}})$$
$$ I=\int_0^{\pi/4}\sqrt{1+\sqrt{1-\tan^2y}}dy=\int_0^{\pi/4}\sqrt{{cosy}+\sqrt{\cos2y}}\frac{dy}{\sqrt{cosy}} $$
put $$x=tany$$
But how to calculate this integral?
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By enforcing the substitution $x=\sin\theta$ we are left with
$$ I = \int_{0}^{\pi/2}\sqrt{1+\cos\theta}\frac{\cos\theta}{1+\sin^2\theta}\,d\theta = \sqrt{2}\int_{0}^{\pi/2}\frac{\cos\theta\cos\frac{\theta}{2}}{1+\sin^2\theta}\,d\theta$$
and by enforcing the substitution $\theta=2\varphi$ we get:
$$ I = 2 \sqrt{2}\int_{0}^{\pi/4}\frac{\cos(\varphi)(2\cos^2(\varphi)-1)}{1+4\sin^2(\varphi)\cos^2(\varphi)}\,d\varphi. $$
By setting $\varphi=2\arctan t$, the original integral is converted into the integral of a rational function over the interval $J=\left(0,\tan\frac{\pi}{8}\right)=\left(0,\sqrt{2}-1\right)$, namely
$$ I = 4\sqrt{2}\int_J \frac{1-7 t^2+7 t^4-t^6}{1+20 t^2-26 t^4+20 t^6+t^8}\,dt. $$
Now the closed form of $I$ just depends on the partial fraction decomposition of the last integrand function, i.e. on the roots of $p(t)=1+20 t^2-26 t^4+20 t^6+t^8$. $p(t)$ is an even and palindromic polynomial, so its roots are given by triple-nested square roots, but finding them is pretty easy, since the whole problem boils down to solving a quadratic equation. I will let you fill the missing details.
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Given $\frac{(a-b)(b-c)(a-c)}{(a+b)(b+c)(c+a)}=\frac{1}{11}$. Find $\frac{a}{a+b}+\frac{b}{b+c}+\frac{c}{c+a}.$
Given: $\{a,b,c\}\subset \Bbb R$, $\frac{(a-b)(b-c)(a-c)}{(a+b)(b+c)(c+a)}=\frac{1}{11}\ \ (1)$. Find the value of $\frac{a}{a+b}+\frac{b}{b+c}+\frac{c}{c+a}\ \ (2).$
This question appeared in OBM 2005, the Brazilian Math Olympiad. The answer is $\frac{17}{11}$. But my solution is, perhaps, unnecessarily complicated as involves a lot of algebra. Is there an easy way to solve?
My attempt (outline): First, I developed expression (1), getting
$$abc=5a^2c+5ab^2+5bc^2-6a^2b-6b^2c-6c^2a\ \ (3)$$
Then, developed (2) to get
$$(2)=\frac{{3abc + 2a^2 b + 2ac^2 + 2b^2 c + a^2 c + ab^2 + bc^2 }}{{2abc + a^2 b + ac^2 + a^2 c +b^2 c + ab^2 + bc^2 }}\ (4)$$
Now, replace $abc$ definition from (3) into (4), and after some additional algebra, get
$$\frac{{17a^2 c + 17ab^2 + 17bc^2 - 17a^2 b - 17ac^2 - 17b^2 c}}{{11a^2 c + 11ab^2 + 11bc^2 - 11a^2 b - 11ac^2 - 11b^2 c}}=\frac{17}{11}$$
Question: is there a simpler way to get to the result? Simpler answers/helpful hints are welcomed.
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Set $$X = \frac{a}{a+b} + \frac{b}{b+c} + \frac{c}{a+c}$$ and $$Y = \frac{b}{a+b} + \frac{c}{b+c} + \frac{a}{a+c}.$$ Then evidently $X+Y = 3$ and we may compute that $$X - Y = \frac{a-b}{a+b} + \frac{b-c}{b+c} + \frac{c-a}{a+c} = \frac{(a-b)(b-c)(a-c)}{(a+b)(b+c)(a+c)} = \frac{1}{11}$$ Solving this system of linear equations, it follows that $$ (X,Y) = \left(\frac{17}{11}, \frac{16}{11}\right)$$
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Solve $\frac{\left(x-7\right)}{\sqrt{x-3}+2}+\frac{\left(x-5\right)}{\sqrt{x-4}+1}=\sqrt{10}$ The answer in wolfram is 13. Any easier technique to solve this equation?
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$$\frac{x-7}{\sqrt{x-3}+2}+\frac{x-5}{\sqrt{x-4}+1}=\sqrt{10}$$
Substitute $x=y+4$
$$\frac{y-3}{\sqrt{y+1}+2}+\frac{y-1}{\sqrt{y}+1}=\sqrt{10}$$
$$\frac{y-3}{\sqrt{y+1}+2}+\frac{(\sqrt{y}+1)(\sqrt{y}-1)}{\sqrt{y}+1}=\sqrt{10}$$
$$\frac{y-3}{\sqrt{y+1}+2}+\sqrt{y}-1=\sqrt{10}$$
Substitute $y=z-1$
$$\frac{z-4}{\sqrt{z}+2}+\sqrt{z-1}-1=\sqrt{10}$$
$$\frac{(\sqrt{z}-2)(\sqrt{z}+2)}{\sqrt{z}+2}+\sqrt{z-1}-1=\sqrt{10}$$
$$\sqrt{z}-2+\sqrt{z-1}-1=\sqrt{10}$$
$$\sqrt{z-1}=3+\sqrt{10}-\sqrt{z}$$
$$(\sqrt{z-1})^2=\left(-\sqrt{z}+\sqrt{10}+3\right)^2$$
$$z-1=z-2 \sqrt{10 z}-6 \sqrt{z}+6 \sqrt{10}+19$$
$$z-\left(z-6 \sqrt{z}-2 \sqrt{10 z}+6 \sqrt{10}+19\right)-1=0$$
$$2 \sqrt{10} \sqrt{z}+6 \sqrt{z}-6 \sqrt{10}-20=0$$
Substitute $\sqrt{z}=w$
$$2 \sqrt{10} w+6 w-6 \sqrt{10}-20=0$$
$$\left(6+2 \sqrt{10}\right) w-6 \sqrt{10}-20=0$$
$$w=\frac{10+3 \sqrt{10}}{3+\sqrt{10}}\to w=\sqrt{10}$$
$z=w^2$ so $z=10$ and $y=z-1=9$ and $x=y+4\to \color{red}{x=13}$
Hope this is useful
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How do you simplify an expression involving fourth and higher order trigonometric functions? The problem is as follows:
Which value of $K$ has to be in order that $R$ becomes independent from $\alpha$?.
$$R=\sin^6\alpha +\cos^6\alpha +K(\sin^4\alpha +\cos^4\alpha )$$
So far I've only come up with the idea that the solution may involve $R=0$, therefore
$$\sin^6\alpha +\cos^6\alpha +K(\sin^4\alpha +\cos^4\alpha)=0$$
as a result the expression becomes $0$ thus independent from $\alpha$, however the result is like this
$$-K=\frac{\sin^6\alpha +\cos^6\alpha}{\sin^4\alpha +\cos^4\alpha}$$
I am not sure if this is the right way.
Moreover, how can I simplify this expression, as it has order four and six?
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$$1=(\sin^2x+\cos^2x)^3= \sin^6x+ \cos^6x+ 3\sin^2x\cos^2x(\sin^2x+\cos^2x)= \sin^6x+ \cos^6x+ 3\sin^2x\cos^2x$$
$$3\sin^2x\cos^2x=\frac{3}{2}((\sin^2x+\cos^2x)^2 - \sin^4x-\cos^4x)=\frac{3}{2}((1- \sin^4x-\cos^4x))$$
Substitute and compare with the equation you have
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Find all natural number(s) $n$ such that $55\mid n^2 + 3n + 1$
Find all natural number(s) $n$ such that $55\mid n^2 + 3n + 1$
I see that $n^2+3n+1 =n^2-2n+1+5n\equiv n^2-2n+1=(n-1)^2 \pmod{5}$
I also see that $n^2+3n+1=n^2+3n-10+11=(n-2)(n+5)+11\equiv(n-2)(n+5) \pmod{11}$
After that what?
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There are infinitely many solutions, in particular you have the identity
$$(55t-119)^2+3(55t-119)+1=55(55t^2-235t+251)$$which shows solutions $x=55t-119$ for each integer $t$ you choose. There are maybe other solutions out of this infinity of them.
|
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surface area of $x^2+y^2+z^2-14z=0$, where $0\leq x^2+y^2\leq3z$ What is the simplest way to compute the surface area of
$$x^2+y^2+z^2-14z=0,$$
where
$$0\leq x^2+y^2\leq3z\tag*{?}$$
I am having problems with various parameterizations.
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Well, $z^2-14z+3z\geq x^2+y^2+z^2-14z\geq 0$ combined with $z\geq 0$ gives $z\geq 11$.
Another form of the equation is $x^2+y^2+(z-7)^2=49$, which is a sphere centered at $(0, 0, 7)$ with radius 7.
So we need to calculate a part of the area of sphere, from $z=11$ to $z=14$, or the area of sphere $x^2+y^2+z^2 = 49$ from $z=4$ to $z=7$. We even don't need to parameters in this calculation: notice that a sphere is a surface of revolution.
$S=\int_{4}^{7}2\pi\sqrt{49-z^2}\sqrt{1+(\sqrt{49-z^2})^\prime} \mathrm{d}z= 14\pi z|_4^7=42\pi$
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If $a+b+c=0$ prove that $ (a^2+b^2+c^2) \times (a^2+b^2+c^2) = 2(a^4+b^4+c^4)$
If $a+b+c=0$, for $a,b,c \in\mathbb R$, prove
$$(a^2+b^2+c^2) \times (a^2+b^2+c^2) = 2(a^4+b^4+c^4)$$
What is a good way to do this?
This question came from answering this slightly harder question. Those answers were somewhat hard to understand for me. To get something easier to digest I made a very similar but easier (lower exponent) question.
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Put $c=-a-b$. Then
$$
(a^2+b^2+c^2)^2=4(a^4 + 2a^3b + 3a^2b^2 + 2ab^3 + b^4)=2(a^4+b^4+c^4).
$$
A very similar calculation was given here.
|
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Asymptotic Analysis for two variables i have a function $ f(x,y)= \sqrt{x^2+2xy+3y^2}$ $\space$ , and $g(x,y)= \sqrt{x^2+y^2}$
I have to prove that $f$ is bounded both above and below by $g$ asymptotically $\Rightarrow f(x,y)=\Theta(g(x,y)) \Rightarrow \text{there is c and C $\in {R_{>0}}$ so that $c g(x,y)\leq f(x,y) \leq C g(x,y)$}$
I was only able to find the big C , but i am still not sure how to find the little c
$$ {x^{2}+2xy+3y^{2}} \leq 3x^{2}+6xy+3y^{2}= 3(x+y)^2 \leq 6(x^2+y^2)$$
$$\Rightarrow \sqrt{x^{2}+2xy+3y^{2}} \leq \sqrt{6}\sqrt{x^{2}+y^{2}}$$
which means that C =$\sqrt{6}$
Any hints how can i find c
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$$c\sqrt{x^2+y^2}\leq \sqrt{x^2+2 x y+3 y^2}$$
square both sides
$$c^2 \left(x^2+y^2\right)\leq x^2+2 x y+3 y^2$$
expand and rearrange
$$\left(c^2-1\right) x^2-2 x y+\left(c^2-3\right) y^2\leq 0\quad(*)$$
which is true for any $x,y\in\mathbb{R}$ if
$$\left(1-\left(c^2-1\right) \left(c^2-3\right)\right) y^2<0\land c^2-1<0$$
that is
$$-c^4+4 c^2-2<0\land c^2-1<0$$
$c^2-1<0\to -1<c<1\quad(*)$
$-c^4+4 c^2-2<0\to c^4-4c^2+2>0$
set $c^2=d$ so the inequality becomes
$d^2-4d+2>0\to d<2-\sqrt{2}\lor d>2+\sqrt{2}$
reset $d=c^2$
$c^2<2-\sqrt{2}\lor c^2>2+\sqrt{2}$
$c<-\sqrt{2+\sqrt{2}}\lor -\sqrt{2-\sqrt{2}}<c<\sqrt{2-\sqrt{2}}\lor c>\sqrt{2+\sqrt{2}}$
recalling $(*)$ and the constraint given that $c>0$ we get $0<c<\sqrt{2-\sqrt{2}}$
so we can take $\color{red}{c=\sqrt{2-\sqrt{2}}}$
hope this helps
$(*)$
Discriminant $\Delta=(2y)^2-4(c^2-1)(c^2-3)y^2=4y^2(1-(c^2-1)(c^2-3))$
$$
.
$$
Edit
Using the same technique for the right bound I found
$C=\sqrt{2+\sqrt{2}}$
which is a bit smaller than $\sqrt 6$
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Check solution of recurrence relation $a_0 = 3$, $a_1 = 7$, $a_n = 3a_{n-1} - 2a_{n-2}$ for $n \geq 2$ $a_0 = 3, a_1 = 7, a_n = 3a_{n-1} - 2a_{n-2}$ for $n \geq 2$
I know I've got the wrong answer because my outputs don't match. So if someone could show me where I'm going wrong I would be super grateful.
Let $ y = \sum\limits_{n = 0}^{\infty} a_nx^n$
multiplying original formula through by $x^n$ I get
$a_nx^n = 3a_{n-1}x^n - 2a_{n-2}x^n$
sum over $n \geq 2$
$\sum\limits_{n \geq 2} a_nx^n = \sum\limits_{n \geq 2} 3a_{n-1}x^n - \sum\limits_{n \geq 2}2a_{n-2}x^n$
The left hand side is equal to $y - a_1 - a_0$
The first part of the right hand side
$\sum\limits_{n \geq 2} 3a_{n-1}x^n = 3x \sum\limits_{n \geq 2} a_{n-1}x^{n-1} = 3x(y-a_0)$
The second part of the right hand side
$\sum\limits_{n \geq 2}2a_{n-2}x^n = 2x^2\sum\limits_{n \geq 2}a_{n-2}x^{n-2} = 2x^2y$
So putting it all together and substituting for $a_0$ and $a_1$
$y - 10 = 3x(y-3) - 2x^2y$
all the $y$'s to one side
$y - 3xy + 2x^2y = -9x + 10 \Rightarrow y = \cfrac{-9x + 10}{(1-2x)(1-x)}$
after fraction decomposition I get
$y = \cfrac{11}{(1-2x)}-\cfrac{1}{(1-x)}$
Which then gives me
$a_n = 11 \cdot 2^n -1$
which is definitely wrong :(
|
If you are interested in solving only this particular relation, then you can do it with induction.
First, it is not difficult to gues some starting values:
$$a_1 = 3 = 2^2-1$$
$$a_2 = 7= 2^3-1$$
$$ a_3 = 3\cdot 7-2\cdot 3 = 15 = 2^4-1$$
$$ a_4 = 3\cdot 15-2\cdot 7 = 31 = 2^5-1$$
So it is reasonably to believe that $a_n = 2^n-1$ for each $n$.
Do the induction step from $n-1,n\to n+1$:
$$a_{n+1} = 3\cdot (2^n-1)-2\cdot (2^{n-1}-1) = 2^{n+1}-1$$
and we are done.
|
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|
Let $r$ be a root of the polynomial $p(x) = (\sqrt{5} - 2\sqrt{3})x^3 + \sqrt{3}x - \sqrt{5} + 1$. Find another polynomial $q(x)$ with integer coefficients such that $q(r) = 0$.
I have no clue how to do this question. Can't use rational root theorem and I see no feasible way to get the roots of $p(x)$. Any help would be appreciated.
|
$p(x)=(\sqrt{5}-2\sqrt{3})x^3+\sqrt{3}x-\sqrt{5}+1=\sqrt{5}(x^3-1)+\sqrt{3}(x-2x^3)+1$
Let's set : $\begin{cases}a=x^3-1\\b=x-2x^3\\p(x)=\sqrt{5}a+\sqrt{3}b+1\end{cases}$
For $p(x)=0$ we have $5a^2+3b^2+2\sqrt{15}ab=(-1)^2=1$
And then $60a^2b^2=(1-5a^2-3b^2)^2$
Finally substitute $a,b$ with their values, I find the same polynomial as M.Rozenberg.
$49x^{12}−168x^{10}+140x^9+186x^8−240x^7−76x^6+60x^5+153x^4−80x^3−36x^2+16=0$
|
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|
How to Simplify $\sqrt {-X} \times \sqrt {-Y}$ For example : $\sqrt {-18} \times \sqrt {-12}$
Would I start by multiplying the 2 numbers under a square root, In which case the double negatives cancel out?
$ \sqrt {-18\times-12} = \sqrt {216} = 6\sqrt {6}$
Or get the i out of the square roots in which case I get a $i^2$?
$$ \sqrt {18}i \times \sqrt {12}i = 6\sqrt {6}i^2 = - 6\sqrt {6}$$
|
Simply
$$i\sqrt{18}\cdot i\sqrt{12}$$
$$-1\sqrt{12\cdot 18}$$
$$-1\sqrt{216}$$
$$-6\sqrt{6}$$
ALWAYS start with $i$
|
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|
How to find the maximum and minimum values of $\frac{8x(x^2-1)}{(x^2+1)^2}$ algebraically? The function is $f(x) = \frac{8x(x^2-1)}{(x^2+1)^2}$.
I have tried using calculus, only to fail.
|
$$f(x) = \frac{8x(x^2-1)}{(x^2+1)^2}={8(x^3-x)}{(x^2+1)^{-2}}\\f'(x)=8\left(\left(x^3-x\right)\left(-\dfrac{4x}{\left(x^2+1\right)^3}\right)+\left(3x^2-1\right){(x^2+1)^{-2}}\right)\\=
8\left(-\dfrac{4x^4-4x^2}{\left(x^2+1\right)^3}+\frac{3x^2-1}{(x^2+1)^{2}}\right)
\\=
8\left(\dfrac{4x^2-4x^4}{\left(x^2+1\right)^3}+\frac{(3x^2-1)(x^2+1)}{(x^2+1)^{2}(x^2+1)}\right)
\\=
8\left(\dfrac{4x^2-4x^4+(3x^2-1)(x^2+1)}{\left(x^2+1\right)^3}\right)
\\=
\frac{-8 (x^4 - 6 x^2 + 1)}{(x^2 + 1)^3}$$
now just solve for $0=x^4 - 6 x^2 + 1$
$$0=x^4 - 6 x^2 + 1,u=x^2\\0=u^2-6u+1\\u_{1,2}={\frac {6\pm {\sqrt {36-4\ }}}{2}}=\begin{cases}3+2\sqrt2\\3-2\sqrt2\end{cases}$$
so you get $4$ answers:$$x_{1,2,3,4}=\begin{cases}\sqrt{3+2\sqrt2}&=&1+\sqrt2\\[2ex]
-\sqrt{3+2\sqrt2}&=&-1-\sqrt2\\[2ex]
\sqrt{3-2\sqrt2}&=&\sqrt2-1\\[2ex]
-\sqrt{3-2\sqrt2}&=&1-\sqrt2\end{cases}$$
|
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|
How many solutions does the equation $a + b + c = 10$ if $(1, 4, 5)$ and $(1, 5, 4)$ are not considered distinct?
Question: What if we consider (1,4,5) and (1,5,4) as non-distinct possibilities, then what should we do?
$${{9}\choose{2}}-2\cdot\frac{{9}\choose{2}}{3}$$
|
Since solutions obtained by a permutation are considered the same we may assume $1\leq a\leq b\leq c$ to begin with. We therefore put
$$a=1+x_1,\quad b=a+x_2=1+x_1+x_2,\quad c=b+x_3=1+x_1+x_2+x_3\ ,$$
whereby $x_i\geq0$ $\>(1\leq i\leq3)$ and
$a+b+c=3+3x_1+2x_2+x_3=10$,
or $$3x_1+2x_2+x_3=7\ .$$
If $x_1=0$ then $2x_2+x_3=7$, which leads to the four solutions $(0,0,7)$, $(0,1,5)$, $(0,2,3)$, $(0,3,1)$.
If $x_1=1$ then $2x_2+x_3=4$, which leads to the three solutions $(1,0,4)$, $(1,1,2)$, $(1,2,0)$.
If $x_1=2$ we obtain the single solution $(2,0,1)$.
In terms of $a$ $b$, $c$ we get the $8$ solutions listed by N.F. Taussig. It remains unclear how your text arrived at $36$ solutions.
|
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|
Let $\alpha $, $\beta $ are the roots of $3x^2+x+5=0$ Let $\alpha $, $\beta $ are the roots of $3x^2+x+5=0$ then find a quadratic equation with roots as $\dfrac {\alpha +1}{\alpha -3}$ and $\dfrac {\beta +1}{\beta -3}$.
I got this solution but didn't understand what procedure is used, what concepts are used and how's it working?
|
For a different route, try this.
Let $a=\dfrac {\alpha +1}{\alpha -3}$ and $b=\dfrac {\beta +1}{\beta -3}$.
An equation with roots $a$ and $b$ is $(x-a)(x-b)=0$, or $x^2-(a+b)+ab=0$.
Now
$$
a+b=\frac{2 (\alpha \beta - \alpha - \beta - 3)}{\alpha \beta - 3 \alpha - 3 \beta + 9},
\quad
ab = \frac{\alpha \beta + \alpha + \beta + 1}{\alpha \beta - 3 \alpha - 3 \beta + 9}
$$
Finally,
$$
\alpha + \beta = -\frac{1}{3},
\quad
\alpha \beta = \frac{5}{3}
$$
|
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|
Number of sequences that have at least five consecutive positions in which the numbers are in increasing order? A sequence of numbers is formed from the numbers $1, 2, 3, 4, 5, 6, 7$ where all $7!$ permutations are equally likely. What is the probability that anywhere in the sequence there will be, at least, five consecutive positions in which the numbers are in increasing order?
I approached this problem in the following way, but I am wondering if there is a better way, since my approach is quite complicated.
My Approach: There are three possibilities: a sequence have $7$ consecutive positions in which numbers increase, have $6$ consecutive positions in which numbers increase, and $5$ consecutive positions in which numbers increase.
There is only $1$ sequence that have $7$ consecutive positions. Namely, the sequence $(1,2,3,4,5,6,7)$.
There are $12$ sequences that have $6$ consecutive positions. Namely, we choose $1$ number from $(1,2,3,4,5,6,7)$, and move it to either sides. As an illustration, if we choose $3$, then we can get $(3,1,2,4,5,6,7)$ or $(1,2,4,5,6,7,3)$.
Now consider when there are $5$ consecutive positions in which numbers increase. We choose $2$ numbers that are not in the increasing subsequence.
If $1$ and $7$ are not chosen, we can place them in front of the subsequence, of after. For example, if we choose $(2,5)$, then we will have $(2,5,1,3,4,6,7)$,$(5,2,1,3,4,6,7)$, $(1,3,4,6,7,2,5)$ and $(1,3,4,6,7,5,2)$. This is $\binom{5}{2}\times4$.
Then I'm not sure how to proceed when we choose $1$ and/or $7$?
|
This solution does not differ in any essential way from that of N. Shales. I am posting this here so N. Shales can compare our approaches.
Since the sequence contains seven numbers, any block of five consecutive increasing numbers must start in the first, second, or third positions. Let $A_1$, $A_2$, and $A_3$ denote, respectively, the set of sequences containing five consecutive increasing numbers that begin in the first, second, and third positions.
By the Inclusion-Exclusion Principle, the number of sequences containing a block of at least five consecutive increasing numbers is
$$|A_1 \cup A_2 \cup A_3| = |A_1| + |A_2| + |A_3| - |A_1 \cap A_2| - |A_1 \cap A_3| - |A_2 \cap A_3| + |A_1 \cap A_2 \cap A_3|$$
$|A_1|$: There are $\binom{7}{5}$ ways to select five of the seven numbers to be in a block of five consecutive increasing numbers and one way to arrange the numbers within that block so that they are increasing. There are $2!$ ways to arrange the remaining two numbers in the remaining two positions. Hence,
$$|A_1| = \binom{7}{5}2!$$
By symmetry, $|A_1| = |A_2| = |A_3|$.
$|A_1 \cap A_2|$: If both the first five and second five numbers are increasing, then the first six numbers must form an increasing sequence. There are $\binom{7}{6}$ ways to select six of the seven numbers to be in the block of six consecutive increasing numbers and one way to arrange the numbers within the block so that they are increasing. There is one way to place the remaining number in the remaining position. Hence,
$$|A_1 \cap A_2| = \binom{7}{6}1!$$
By symmetry, $|A_1 \cap A_2| = |A_2 \cap A_3|$.
$|A_1 \cap A_3|$: If the first five and last five of the seven numbers are increasing, then all seven numbers must be increasing. There is only one way to arrange all seven numbers in an increasing sequence. Hence,
$$|A_1 \cap A_3| = \binom{7}{7}$$
$|A_1 \cap A_2 \cap A_3|$: There is only one way for all seven numbers to be arranged in increasing order. Hence,
$$|A_1 \cap A_2 \cap A_3| = \binom{7}{7}$$
Therefore,
$$|A_1 \cup A_2 \cup A_3| = 3\binom{7}{5}2! - 2\binom{7}{6}1! - \binom{7}{7} + \binom{7}{7} = 3\binom{7}{5}2! - 2\binom{7}{6}$$
Hence, the probability that a sequence of seven numbers contains a block of at least five consecutive increasing numbers is
$$\frac{3\dbinom{7}{5}2! - 2\dbinom{7}{6}1!}{7!}$$
|
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|
Does in plane exist $22$ points and $22$ such circles that each circle contains at least $7$ points and each point is on at least $7$ circles.
Does in plane exist $22$ points and $22$ such circles that each circle contains at least $7$ points and each point is on at least $7$ circles.
I have solved this one but now I can't remember how I did it. I just remember that I used some linear algebra and double counting.
Suppose each point $P_i\in \{P_1,P_2,...P_{22}\}$ is on $p_i\geq 7$ circles among circles $C_1,C_2,...,C_{22}$. Since each pair $\{C_i,C_j\}$ share at most $2$ points and each point is on $\displaystyle{p_i\choose 2}$ pair of circles, we have:
$$ 2\cdot {22\choose 2} \geq \sum _{i=1}^{22} {p_i\choose 2} \geq 22 {7\choose 2} $$
Since we must have all equalities we deduce that $p_i = 7$ for all $i$ and each pair of circles intersect in exactly two points. Now since $$ 22\cdot 7 \leq \sum _{i=1}^{22} |C_i| = \sum _{i=1}^{22}p_i = 22\cdot 7$$
so each circle contains exactly $7$ points.
|
As we already see in the introduction each point is on $7$ circles and each circle contains $7$ points. Also every two circles meet at $2$ points.
Let $A$ be an incident matrix, so $a_{ij} = 1$ if $P_j \in C_i$, else $a_{ij} =0$. Let $M:= A\cdot A^T$, then determinant of
$M$ is a perfect square: $\det (M) =\det (A\cdot A^T)
=\det A \cdot \det A^T =(\det A)^2$. But $$ M=
\begin{bmatrix}
7 & 2 & 2 & \cdots & 2 & 2 \\
2 & 7 & 2 & \cdots & 2 & 2 \\
2 & 2 & 7 & \cdots & 2 & 2 \\
\vdots & \vdots & \vdots & \cdots & \vdots & \vdots \\
2 & 2 & 2 & \cdots& 7 & 2 \\
2 & 2 & 2 & \cdots & 2 & 7 \\
\end{bmatrix} \sim
\begin{bmatrix}
5 & 0 & 0 & \cdots & 0 & -5 \\
0 & 5 & 0 & \cdots & 0 & -5 \\
0 & 0 & 5 & \cdots & 0 & -5 \\
\vdots & \vdots & \vdots & \cdots & \vdots & \vdots \\
0 & 0 & 0 & \cdots& 5 & -5 \\
2 & 2 & 2 & \cdots & 2 & 7 \\
\end{bmatrix} \sim
\begin{bmatrix}
5 & 0 & 0 & \cdots & 0 & 0 \\
0 & 5 & 0 & \cdots & 0 & 0 \\
0 & 0 & 5 & \cdots & 0 & 0 \\
\vdots & \vdots & \vdots & \cdots & \vdots & \vdots \\
0 & 0 & 0 & \cdots& 5 & 0 \\
2 & 2 & 2 & \cdots & 2 & 49 \\
\end{bmatrix}
$$
so $\det (M)= 5^{21}\cdot 49$ which is not a perfect square. Thus such a configuration does not exist.
|
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|
Find $\lim_{x \to 0} (\frac {\sin x}{x})^{\frac{1}{\sin^2x}}$ Is there any other way to compute the problem $$\lim_{x \to 0} (\frac {\sin x}{x})^{\frac{1}{\sin^2(x)}}$$
I try to use L'Hospital rule but it's too complicated.
|
In the same spirit as other answers but assuming that you want to go beyond the limit.
$$A=\left(\frac {\sin (x)}{x}\right)^{\frac{1}{\sin^2(x)}}\implies \log(A)={\frac{1}{\sin^2(x)}}\log\left(\frac {\sin (x)}{x}\right)$$ Using Taylor for one more term
$$\frac {\sin (x)}{x}=1-\frac{x^2}{6}+\frac{x^4}{120}+O\left(x^6\right)$$
$$\log\left(\frac {\sin (x)}{x}\right)=-\frac{x^2}{6}-\frac{x^4}{180}+O\left(x^6\right)$$
$$\sin^2(x)=x^2-\frac{x^4}{3}+O\left(x^6\right)$$
$$\log(A)=\frac{-\frac{x^2}{6}-\frac{x^4}{180}+O\left(x^6\right) } { x^2-\frac{x^4}{3}+O\left(x^6\right)}=\frac{-\frac{1}{6}-\frac{x^2}{180}+O\left(x^4\right) } { 1-\frac{x^2}{3}+O\left(x^4\right)}$$ Now, long division
$$\log(A)=-\frac{1}{6}-\frac{11 x^2}{180}+O\left(x^4\right)$$ Finally, Taylor again
$$A=e^{\log(A)}=\frac{1}{\sqrt[6]{e}}-\frac{11 x^2}{180 \sqrt[6]{e}}+O\left(x^4\right)$$
This is even a good approximation : try $x=\frac \pi 6$ (quite far away from $0$). The exact formula would give $$A=\frac{81}{\pi ^4}\approx 0.831545$$ while the above approximation would give $$\frac{1}{\sqrt[6]{e}}\left(1-\frac{11 \pi ^2}{6480}\right)\approx 0.832300$$
|
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|
Evaluate $\lim_{n\rightarrow \infty}4\sqrt{n}\sin (\pi\sqrt{4n^2+\sqrt{n}})$ Evaluate the limit
$$\lim_{n\rightarrow \infty}4\sqrt{n}\sin (\pi\sqrt{4n^2+\sqrt{n}})$$
We know that :
$$\sin (\pi-x)=\sin x$$
So we have :
$$\lim_{n\rightarrow \infty}4\sqrt{n}\sin (\pi-\pi\sqrt{4n^2+\sqrt{n}})$$
Now :
$$(\pi-\pi\sqrt{4n^2+\sqrt{n}})\cdot \frac{\pi+\pi\sqrt{4n^2+\sqrt{n}}}{\pi+\pi\sqrt{4n^2+\sqrt{n}}}=\frac{\pi^2-\pi^2(4n^2+\sqrt{n})}{\pi+\pi\sqrt{4n^2+\sqrt{n}}}$$
$$\lim_{n\rightarrow \infty}4\sqrt{n}\sin \left(\frac{\pi^2-\pi^2(4n^2+\sqrt{n})}{\pi+\pi\sqrt{4n^2+\sqrt{n}}}\right)$$
Now what ?
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I think the following is better.
$$\lim_{n\rightarrow+\infty}4\sqrt{n}\sin (\pi\sqrt{4n^2+\sqrt{n}})=-\lim_{n\rightarrow+\infty}4\sqrt{n}\sin \left(2\pi n-\pi\sqrt{4n^2+\sqrt{n}}\right)=$$
$$=\lim_{n\rightarrow+\infty}4\sqrt{n}\sin\frac{\pi\sqrt{n}}{2 n+\sqrt{4n^2+\sqrt{n}}}=\lim_{n\rightarrow+\infty}4\sqrt{n}\sin\frac{\pi}{\sqrt{n}\left(2 +\sqrt{4+\frac{1}{n\sqrt{n}}}\right)}=$$
$$=\lim_{n\rightarrow+\infty}\left(\frac{\sin\frac{\pi}{\sqrt{n}\left(2 +\sqrt{4+\frac{1}{n\sqrt{n}}}\right)}}{\frac{\pi}{\sqrt{n}\left(2 +\sqrt{4+\frac{1}{n\sqrt{n}}}\right)}}\cdot\frac{4\pi}{2 +\sqrt{4+\frac{1}{n\sqrt{n}}}}\right)=\pi.$$
|
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|
Factoring a quartic expression
We have a function $$f(x)=x^{\frac{2}{3}}-\frac{2}{x^{\frac{1}{3}}}+1$$ and let line segment $AB$ be represented by $g(x)$ such that $$g(x)=x-4$$
Find the greatest positive real solution representing the intersection between the two lines.
The first step is obviously to let them equal each other, and the other steps will follow logically, they are demonstrated here
$$x-4=x^{\frac{2}{3}}-\frac{2}{x^{\frac{1}{3}}}+1$$ $$x-5=\frac{x-2}{x^{\frac{1}{3}}}$$ $$x^{\frac{1}{3}}=\frac{x-2}{x-5}$$ $$x=\frac{x^3-6x^2+12x-8}{x^3-15x^2+75x-125}$$ $$x^4-15x^3+75x^2-125x=x^3-6x^2+12x-8$$ $$x^4-16x^3+81x^2-137x+8=0$$
and here I'm very much stuck. By the factor theorem we know that for $(x-a)$ to a factor, $a$ must be a factor of $8$, i.e. $1,2,4,-1,-2$ or $-4$, but I've tried every one of these values and haven't found a value of $a$ such that $(x-a)$ is a factor of the latest expression.
The graph of the two equations is given here, and the point of intersection clearly falls at the intersection of two grid lines, so I know the solution is an integer, but under examination conditions I obviously won't have a graphing tool this sophisticated, so I'm eager to know how to proceed from here algebraically.
Any help is appreciated, thank you.
|
Perhaps it would be easier if you write $t=x^{1\over 3}$. Then the equation you get is $$t^2-{2\over t}+1 = t^3-4 \;\;\;\Longrightarrow \;\;\; t^4-t^3-5 t+2=0$$
Last one can be factor like this $$(t-2)(t^3+t^2+2t-1)=0$$
So $t=2$ and for $t>2$ we get $$t^3+t^2+2t-1 \geq 8+4+4-1 > 0$$
So $t=2$ and $x=8$ is a maximum solution.
|
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|
Summation over roots of unity Find the value of $\displaystyle\sum_{r=1}^{4} \frac{1}{2-\alpha^r} $
where $ \alpha^k (k=0,1,2,3,4,5) $ are fifth roots of unity.
My approach:-
As we know that $ \alpha^k (k=0,1,2,3,4,5) $ are fifth roots of unity, then $ \alpha^k - 1$ should be equal to zero. Therefore the final answer to the summation $\displaystyle\sum_{r=1}^{4} \frac{1}{2-\alpha^r} $ should be $\displaystyle\sum_{r=1}^{4} 1 = 4 $
But the answer given is $ \dfrac{ 49}{31} $
Any help or hint will be much appreciated!
|
Since $\alpha_0,\alpha_1,\alpha_2 \dots \alpha_{4}$ are roots of the equation
$$x^5-1=0 \tag1$$
You can apply Transformation of Roots to find a equation whose roots are$$\frac{1}{2-\alpha_0} , \frac{1}{2-\alpha_1},\dots \frac{1}{2-\alpha_{4}}$$
Let $P(y)$ represent the polynomial whose roots are $\frac{1}{2-\alpha_k}$
$$y=\frac{1}{2-\alpha_k}=\frac{1}{2-x} \implies x=\frac{2y-1}{y}$$
Put in $(1)$
$$\Bigg(\frac{2y-1}{y}\Bigg)^5-1=0 \implies (2y-1)^{5}-y^{5}=0$$
Use Binomial Theorem to find coefficient of $y^5$ and $y^{4}$.You will get sum of the roots using Vieta's Formulas.
Hope it helps!
|
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|
Solve $\lfloor \sqrt x +\sqrt{x+1}+\sqrt{x+2}\rfloor=x$ $$\lfloor \sqrt x +\sqrt{x+1}+\sqrt{x+2}\rfloor=x$$
I tried to solve this equation.
First thing is $\lfloor \sqrt x +\sqrt{x+1}+\sqrt{x+2}\rfloor \in \mathbb{Z} $ so $x \in \mathbb{Z}$
second $$\sqrt x +\sqrt{x+1}+\sqrt{x+2} \geq \sqrt 0 +\sqrt{0+1}+\sqrt{0+2} \\\to x \in \mathbb{N}$$ so we can check $x=1,2,3,4,5,6,7,8,9,\ldots$ by a MATLAB program. I checked the natural numbers to find solution. I found $x=8,9$ worked here.
Now my question is about somehow an analytical solving of the equation, or another idea. Can any one help me? Thanks in advance.
|
This is a little plodding, but $x\in\mathbb{Z}$ and $x\ge0$ (required in order for $\sqrt x$ to be real) tells us
$$x\ge\lfloor\sqrt0+\sqrt1+\sqrt2\rfloor=\lfloor2.414\rfloor=2$$
which tells us
$$x\ge\lfloor\sqrt2+\sqrt3+\sqrt4\rfloor=\lfloor5.145\rfloor=5$$
which tells us
$$x\ge\lfloor\sqrt5+\sqrt6+\sqrt7\rfloor=\lfloor7.331\rfloor=7$$
which tells us
$$x\ge\lfloor\sqrt7+\sqrt8+\sqrt9\rfloor=\lfloor8.474\rfloor=8$$
Finally, we see that if $x\ge10$, then the function $f(x)=\sqrt x+\sqrt{x+1}+\sqrt{x+2}-x$ is negative, since $f(10)=-0.056997...$ and
$$f'(x)={1\over2\sqrt x}+{1\over2\sqrt{x+1}}+{1\over2\sqrt{x+2}}-1\lt{3\over2\sqrt 9}-1=-{1\over2}\lt0$$
so we cannot have $x\le\sqrt x+\sqrt{x+1}+\sqrt{x+2}$, which is required in order to have $x=\lfloor\sqrt x+\sqrt{x+1}+\sqrt{x+2}\rfloor$, for $x\ge10$. This leaves the two possibilities $x=8$ and $9$, which do solve the equation.
|
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|
Prove that $1^2 + 2^2 + \cdots + n^2 = \frac{n(n+1)(2n+1)}{6}$ without using induction. I have to deduce the following formula $$1^2 + 2^2 + \cdots + n^2 = \frac{n(n+1)(2n+1)}{6},$$ while using the given formula $$\binom{k}{0}+\binom{k+1}{1}+\cdots+\binom{k+r}{r}=\binom{k+r+1}{r}$$
I tried to find values for $k$, such that $\binom{k}{0}=1^2$ etc. but that didn't work. Does anybody have a push in the right direction? Thanks!
|
Using your identity, we can write
$$\binom{k}{k} + \binom{k+1}{k} + \cdots + \binom{k+r}{k} = \binom{k+r+1}{k+1}$$
and take $k=2$ to give
$$\binom{2}{2} + \binom{3}{2} + \cdots + \binom{r+2}{2} = \binom{r+3}{3}$$ but we know that $\binom{k}{2} = \frac{k(k-1)}{2}$, so
$$\begin{align}\binom{r+3}{3} &= \sum_{i=0}^r \binom{i+2}{2} \\
\frac{(r+3)(r+2)(r+1)}{6} &= \sum_{i=0}^r \frac{(i+1)(i+2)}{2} \\
&= \sum_{i=0}^r \left(\frac{1}{2}i^2 + \frac{3}{2}i+1\right) \\
&= \frac{1}{2}\sum_{i=1}^ri^2 + \frac{3}{2}\frac{r(r+1)}{2}+r+1
\end{align}$$
from which the desired identity easily follows.
|
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|
Compute $\zeta(6)$ using Fourier series How can i prove that $\sum_{n=1}^\infty \frac{1}{n^6} = \frac{\pi^6}{945}$ knowing that $f(t)=t^2 = \frac{\pi^2}{3} + 4\sum_{n=1}^\infty \frac{(-1)^n}{n^2}\cos(nt)$ on $(-\pi, \pi)$. I've integrated four times but i don't get anywhere. Thank's for helping.
|
Starting with
$$\sum_{n=1}^{\infty} \frac{(-1)^{n}}{n^2} \, \cos(n t) = \frac{t^{2}}{4} - \frac{\zeta(2)}{2}$$
then integrate from $0$ to $t$ to obtain
$$\sum_{n=1}^{\infty} \frac{(-1)^{n}}{n^3} \, \sin(n t) = \frac{t^{3}}{2 \cdot 3!} - \frac{\zeta(2) \, t}{2} + c_{0}.$$
When $t=0$ it is determined that $c_{0} = 0$. Integrate several more times evaluating each series at $t=0$ to determine the constants of integration to obtain:
\begin{align}
\sum_{n=1}^{\infty} \frac{(-1)^{n}}{n^3} \, \sin(n t) &= \frac{t^{3}}{2 \cdot 3!} - \frac{\zeta(2) \, t}{2} \\
\sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{n^4} \, \cos(n t) &= \frac{t^{4}}{2 \cdot 4!} - \frac{\zeta(2) \, t^{2}}{4} + \frac{7}{8} \, \zeta(4) \\
\sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{n^5} \, \sin(n t) &= \frac{t^{5}}{2 \cdot 5!} - \frac{\zeta(2) \, t^{3}}{12} + \frac{7}{8} \, \zeta(4) \, t \\
\sum_{n=1}^{\infty} \frac{(-1)^{n}}{n^6} \, \cos(n t) &= \frac{t^{6}}{2 \cdot 6!} - \frac{\zeta(2) \, t^{4}}{48} + \frac{7}{16} \, \zeta(4) \, t^{2} - \frac{31}{32} \, \zeta(6).
\end{align}
In the last series let $t = \pi$ to obtain
$$\zeta(6) = \frac{\pi^{6}}{945}.$$
|
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|
Prove by induction that $\sum_{i=1}^{n} i \cdot 2^i = (n-1) \cdot 2^{n+1} +2$. Help finding my mistake
Prove by induction:
$$\sum_{i=1}^{n} i \cdot 2^i = (n-1) \cdot 2^{n+1} +2$$
Basis: let $p(n)$ be the predicate.
Let $n=1$ this gives $(1-1) \cdot 2^1+1+2 = 2$ and $1 \cdot 2^1 = 2$ so its true for $p(1)$
Induction: assume $n=k$ thus $$\sum_{i=1}^{k} i \cdot 2^i = (k-1) \cdot 2^{k+1} +2$$
when $n=k+1$:
$$\sum_{i=1}^{k+1} i \cdot 2^i = \sum_{i=1}^{k} i \cdot 2^i + ((k+1)-1) \cdot 2^{(k+1)+1} +2$$
$$ =(k-1) \cdot 2^{k+1} +2 + ((k+1)-1) \cdot 2^{k+2} +2$$
$$= 2^{k+1}k - 2^{k+1} +2 +2^{k+2}(k+1) - 2^{k+2}+2$$
$$=\frac{2^{k+1}k-2^{k+1}+2 \cdot 2^{k+1}(k+1)-2 \cdot 2^{k+1} +4}{2}$$
$$=2^{k+1}k-2^{k+1}+2^{k+1}(k+1)-2^{k+1}+2$$
$$= (k-1) \cdot 2^{k+1}+((k+1)-1) \cdot 2^{k+1} +2$$
I've been messing around with the arithmetic part of this for a while now and it's getting frustrating, I'm thinking that I've possibly made a mistake at the beginning of the induction step and that's why I cant get this to work, or maybe I'm just missing something in the algebra, either way can someone please help.
|
The following is strange:
when $n=k+1$:
$$\sum_{i=1}^{k+1} i \cdot 2^i = \sum_{i=1}^{k} i \cdot 2^i + \color{red}{((k+1)-1) \cdot 2^{(k+1)+1} +2}$$
$$\sum_{i=1}^{k+1} i \cdot 2^i = \left(\sum_{i=1}^{k} i \cdot 2^i \right)+\color{blue}{(k+1)2^{k+1}}$$
|
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|
Can someone explain the steps of this Partial fraction decomposition?
My thoughts:
$$\frac{(Ax + B)}{(x^2+1)} + \frac{(Cx + D)}{(x^2+4)} = \frac{x}{(x^2+1)(x^2+4)}$$
I combined the left terms
Set the numerator of the combined left term to "x" which is the numerator of the right term
I got
$$4Ax + Cx = x$$
I am not sure what to do next
|
You have only found the equation for $x$. The full set of equations is derived as follows:
$$(Ax+B)(x^2+4)+(Cx+D)(x^2+1)\equiv x$$
So compare coefficients of each of the powers of $x$:
$$(A+C)x^3\equiv 0\\(B+D)x^2\equiv 0\\(4A+C)x\equiv x\\(4B+D)1\equiv 0$$
Solving this gives $A=-C,B=-D$. So $$4A-A=1\implies A=\frac13\implies C=-\frac13$$ and $$4B-B=0\implies B=0\implies D=0$$
And this gives the left hand side as
$$\frac{\frac13 x}{x^2+1}-\frac{\frac 13 x}{x^2+4}$$ as required.
|
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|
Representing a unit speed curve on a sphere in terms of its Frenet Frame Let $\alpha$ be a unit speed curve with positive curvature $\kappa \gt 0$ and non-zero torsion $\tau \ne 0$, lying on a sphere of radius $r$ centred at $c \in \Bbb{R}^3$.
Show that $\alpha - c = -\frac{1}{\kappa}N - (\frac{1}{\kappa})'\frac{1}{\tau}B$.
Where N is the principle normal vector field and B is the binormal vector field to $\alpha$. I'm not sure how it is possible to show this as all the components on the right hand side of the equation involve derivatives of the curve on the left hand side, any help would be much appreciated.
Then, deduce a formula for the radius of the sphere in terms of $\kappa,\tau$.
|
Well, we have
$(\alpha - c) \cdot (\alpha - c) = r^2, \tag 1$
since $\alpha$ lies on the sphere of radius $r$ centered at $c$; this is just what equation (1) affirms. If we differentiate (1) with respect to $s$, the arc-length along $\alpha$, we obtain
$2\alpha' \cdot (\alpha - c) = \alpha' \cdot (\alpha - c) + (\alpha - c) \cdot \alpha' = 0, \tag 2$
whence
$\alpha' \cdot (\alpha - c) = 0; \tag 3$
since $\alpha$ is a unit-speed curve, we have
$\alpha' = T, \; \Vert T \Vert = 1, \tag 4$
the unit tangent vector to $\alpha$; thus (3) becomes
$T \cdot (\alpha - c) = 0, \tag 5$
which is nearly self-evident, since $T$ is tangent to the sphere (1), hence normal to the radial vector $\alpha - c$; in any event, we may differentiate (5) once again to obtain
$T' \cdot (\alpha - c) + T \cdot T = T' \cdot (\alpha - c) + T \cdot \alpha' = 0, \tag 6$
or
$T' \cdot (\alpha -c) + \Vert T \Vert^2 = 0; \tag 7$
we now use the Frenet-Serret equation
$T' = \kappa N, \tag 8$
where $N$ is the normal vector to $\alpha$, and (4) to re-write (7) as
$\kappa N \cdot (\alpha - c) + 1 = 0, \tag 9$
which since $\kappa > 0$ implies
$N \cdot (\alpha - c) = -\dfrac{1}{\kappa}; \tag {10}$
we may now differentiate (9) with respect to $s$ to find
$\kappa' N \cdot (\alpha - c) + \kappa N' \cdot (\alpha - c) + kN \cdot \alpha'= 0; \tag{11}$
we have
$N \cdot \alpha' = N \cdot T = 0, \tag{12}$
and also the Frenet-Serret equation
$N' = -\kappa T + \tau B, \tag{13}$
where $\tau$ is the torsion and $B = T \times N$ the binormal vector of $\alpha$; when (10), (12) and (13) are substituted into (11) we obtain
$-\dfrac{\kappa'}{\kappa} + \kappa (-\kappa T + \tau B) \cdot (\alpha - c) = 0, \tag{14}$
or
$-\dfrac{\kappa'}{\kappa} -\kappa^2 T \cdot (\alpha - c) + \kappa \tau B \cdot (\alpha - c) = 0, \tag{15}$
which by virtue of (5) reduces to
$-\dfrac{\kappa'}{\kappa} + \kappa \tau B \cdot (\alpha - c) = 0, \tag{16}$
and since $\kappa > 0 \ne \tau$ we have
$-\dfrac{\kappa'}{\kappa^2 \tau} + B \cdot (\alpha - c) = 0, \tag{17}$
whence
$B \cdot (\alpha - c) = \dfrac{\kappa'}{\kappa^2 \tau} = -\left ( \dfrac{1}{\kappa} \right )' \dfrac{1}{\tau}; \tag {18}$
(5), (10) and (18) express the components of the radial vector $\alpha - c$ in the orthonormal frame $T$, $N$, $B$, whence
$\alpha - c = -\dfrac{1}{\kappa}N -\left ( \dfrac{1}{\kappa} \right )' \dfrac{1}{\tau} B, \tag{19}$
the requisite result.
We may find $r$ in terms of $\kappa$ and $\tau$ by inserting (19) into (1):
$r^2 = (\alpha - c) \cdot (\alpha - c)$
$= \left (-\dfrac{1}{\kappa}N -\left ( \dfrac{1}{\kappa} \right )' \dfrac{1}{\tau} B \right ) \cdot \left (-\dfrac{1}{\kappa}N -\left ( \dfrac{1}{\kappa} \right )' \dfrac{1}{\tau} B \right ); \tag{20}$
since
$\Vert N \Vert = \Vert B \Vert = 1, \; N \cdot B = 0, \tag{21}$
(20) reduces to
$r^2 = \dfrac{1}{\kappa^2} + \left ( \left ( \dfrac{1}{\kappa} \right )' \dfrac{1}{\tau} \right )^2 = \dfrac{1}{\kappa^2} + \left ( -\dfrac{\kappa'}{\kappa^2} \right )^2 \dfrac{1}{\tau^2} = \dfrac{1}{\kappa^2} + \dfrac{(\kappa')^2}{\kappa^4 \tau^2} = \dfrac{(\kappa')^2 + \kappa^2 \tau^2}{\kappa^4 \tau^2}, \tag{22}$
whence
$r = \sqrt{\dfrac{(\kappa')^2 + \kappa^2 \tau^2}{\kappa^4 \tau^2}} = \dfrac{1}{\kappa^2 \tau} \sqrt{(\kappa')^2 + \kappa^2 \tau^2}, \tag{23}$
as the desired formula for $r$.
|
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|
Equivalent method for solving lyapunov equality I am trying to show that the lyapunov equality:
$A^{T}P + PA = -Q$
Where $Q =
\begin{bmatrix}
1 & 0 \\
0 & 1 \\
\end{bmatrix}
$
is equivalent to solving:
$M
\begin{bmatrix}
P_{11}\\
P_{12}\\
P_{22}\\
\end{bmatrix}
= v
$
The stated solution is:
$M:=
\begin{bmatrix}
-4 & -2 & 0\\
1 & -2 & -1\\
0 & 2 & 0\\
\end{bmatrix}
$
and
$ v:=
\begin{bmatrix}
-1 \\
0 \\
-1 \\
\end{bmatrix}
$
I can see that the values for the vector $v$ come from the corresponding $Q_{11},Q_{12},Q_{22}$ values. However, I do not know how the M values are defined. If I multiply out the left side of the equation I get:
$
\begin{bmatrix}
-4P_{11} - P_{12} - P_{21} & P_{11} - 2P_{12} - P_{22}\\
P_{11} - 2P_{21} - P_{22} & P_{12} + P_{21}\\
\end{bmatrix}
$
Which if i try to map this to an M matrix I would get:
$M =
\begin{bmatrix}
-4 & -1 & 0\\
1 & -2 & -1\\
0 & 1 & 0\\
\end{bmatrix}
$
Which is close, but not the same as the solution.
|
If in your case the Lyapunov equality can be written as
$$
\begin{bmatrix}
-4\,P_{11} - P_{12} - P_{21} & P_{11} - 2\,P_{12} - P_{22} \\
P_{11} - 2\,P_{21} - P_{22} & P_{12} + P_{21}
\end{bmatrix} = -
\begin{bmatrix}
Q_{11} & Q_{12} \\ Q_{21} & Q_{22}
\end{bmatrix} \tag{1}
$$
then by stacking the columns of $(1)$ on top of each other, then it can also be written as
$$
\begin{bmatrix}
-4\,P_{11} - P_{12} - P_{21} \\ P_{11} - 2\,P_{21} - P_{22} \\ P_{11} - 2\,P_{12} - P_{22} \\ P_{12} + P_{21}
\end{bmatrix} = -
\begin{bmatrix}
Q_{11} \\ Q_{21} \\ Q_{12} \\ Q_{22}
\end{bmatrix}. \tag{2}
$$
Factoring out $P$ gives
$$
\begin{bmatrix}
-4 & -1 & -1 & 0 \\
1 & -2 & 0 & -1 \\
1 & 0 & -2 & -1 \\
0 & 1 & 1 & 0
\end{bmatrix}
\begin{bmatrix}
P_{11} \\ P_{21} \\ P_{12} \\ P_{22}
\end{bmatrix} = -
\begin{bmatrix}
Q_{11} \\ Q_{21} \\ Q_{12} \\ Q_{22}
\end{bmatrix}. \tag{3}
$$
Now if we assume that $P = P^\top$ and $Q = Q^\top$, which implies $P_{21} = P_{12}$ and $Q_{21} = Q_{12}$, then the second and third equation from $(2)$ and $(3)$ are identical, so one of them can be omitted. Also the second column of the matrix, which gets multiplied by the $P$ vector, gets multiplied by $P_{21}$ which is identical to $P_{12}$. But the third column of the matrix gets multiplied by $P_{12}$. So this can be simplified down to that the sum of the second and third column get multiplied by $P_{12}$. Applying this to $(3)$ allows it to be rewritten as
$$
\begin{bmatrix}
-4 & -2 & 0 \\
1 & -2 & -1 \\
0 & 2 & 0
\end{bmatrix}
\begin{bmatrix}
P_{11} \\ P_{12} \\ P_{22}
\end{bmatrix} = -
\begin{bmatrix}
Q_{11} \\ Q_{12} \\ Q_{22}
\end{bmatrix}. \tag{4}
$$
So from this it can be seen that this indeed matches the stated solution for $M$. You probably just forgot to add the second and third column.
Also the matrix in the formulation in $(3)$ can easily be constructed for any $A \in \mathbb{R}^{2 \times 2}$ using
$$
\hat{M} =
\begin{bmatrix}
A + I\,A_{11} & I\,A_{12} \\
I\,A_{21} & A + I\,A_{22}
\end{bmatrix}. \tag{5}
$$
So from this it can also be deduced that your $A$ matrix should equal
$$
A =
\begin{bmatrix}
-2 & -1 \\ 1 & 0
\end{bmatrix}.
$$
In general when $A \in \mathbb{R}^{n \times n}$ then $\hat{M} \in \mathbb{R}^{n^2 \times n^2}$, such that it consists out of a $n$ by $n$ grid of $\mathbb{R}^{n \times n}$ sub-matrices. By denoting $\hat{M}_{ij}$ as the sub-matrix on the $i$th place in the vertical direction and of the $j$th place in the horizontal direction, then each sub-matrix can be constructed using
$$
\hat{M}_{ij} = \left\{
\begin{array}{ll}
A + A_{ij}\,I, & \text{if}\ i = j \\
A_{ij}\,I, & \text{otherwise}
\end{array}\right. \tag{6}
$$
|
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|
Arithmetic and geometric progression - 3 terms Find $3$ numbers $x, y, z$ which are consecutive terms of a geometric series, if $xy$, $yz$, $zx$ and $xyz$ are consecutive terms of an arithmetic series.
OK $y=xa$ and $z=xa^2$.
Also $yz=xy+b$
$zx=xy+2b$ and
$xyz=xy+3b$
So by substituting we get: $a=0$ (rejected) and $a=-\frac 1 2 $. But then how do I get $b$?
|
If $x = a$, $y = ar$ and $z = ar^2$, $(a,r \ne 0)$, then
$xy = a^2r$, $yz = a^2r^3$, $zx = a^2 r^2$ and $xyz = a^3r^3$
Since $xy$, $yz$ and $zx$ are in AP.
$$yz = \dfrac{xy + zx}{2} \iff a^2r^3 = \dfrac{a^2r + a^2r^2}{2} \implies 2r^3 = r+ r^2 \implies r(2r^2 - r - 1) = 0$$
So $r = 1, -\dfrac12, 0$.
Also, $$zx=\dfrac{xyz + yz}{2}\iff a^2r^2 = \dfrac{a^3r^3 + a^2r^3}{2} \implies 2 = ar + r \implies a = \dfrac{2 - r}{r}$$
$a = 1 (r = 1), -5(r = -1/2)$
|
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|
How do I simplify $\frac{bH}{a\sqrt{a^2+H^2}}$ to $\frac{b}{a\sqrt{a^2H^{-2}+1}}$?
I have trouble following the algebraic simplification between step 2 and 3.
How do I simplify $\frac{bH}{a\sqrt{a^2+H^2}}$ to $\frac{b}{a\sqrt{a^2H^{-2}+1}}$?
|
There you go
$$\frac{bH}{a\sqrt{a^2+H^2}}$$
$$=\frac{bH}{a\sqrt{(H^2)(\frac{a^2}{H^2}+1)}}$$
$$=\frac{bH}{aH\sqrt{\frac{a^2}{H^2}+1}}$$
$$=\frac{b}{a\sqrt{a^2H^{-2}+1}}$$
|
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|
Find the number of real roots for $x+\sqrt{a^2-x^2}=b$, $a>0$, $b>0$, as a function of $a$ and $b$
Given: (1) $x+\sqrt{a^2-x^2}=b$, $(a,b,x)\subset \mathbb R$, $a>0$, $b>0$.
Find: number of roots for (1), given possible values for $a$ and $b$.
This is a question from a book for the preparation for math contests.
It states as final answer: (a) 1 root if $b<a$; and (b) 2 roots if $a<b<a\sqrt{2}$.
I'm having difficulties on finding this answer. I don't know whether it is correct or perhaps I'm not finding the right approach.
I started moving $x$ in (1) to the left, to get
$$\sqrt{a^2-x^2}=b-x$$
Before proceeding with squaring both sides, I saved 2 needed conditions for checking the final solution (c1) $a^2-x^2\ge 0$ and (c2) $b-x\ge 0$. Then squaring both sides, we get:
$$a^2-x^2=b^2+x^2-2bx\Leftrightarrow 2x^2-2bx+(b^2-a^2)=0$$
with discriminant $\triangle$ defined by:
$$\triangle=4(2a^2-b^2)$$
From this it is easy to see that a condition for 2 roots is (c3) $\sqrt{2}a>b,$ and for 1 root is (c4) $\sqrt{2}a=b,$ as $a>0$ and $b>0$. Then I find the roots as $$x=\frac{2b\pm \sqrt{\triangle}}{4}=\frac{b\pm \sqrt{2a^2-b^2}}{2}$$
From this point, I can't see a way to reach the stated answer, if it is right.
Full solutions or helpful hints are welcome. Sorry if it is a duplicate.
|
As you wrote, we have
$$2x^2-2bx+b^2-a^2=0$$
with$$a^2-x^2\ge 0\quad\text{and}\quad b-x\ge 0,$$
i.e.
$$-a\le x\le a\quad\text{and}\quad x\le b\tag1$$
Here, let us separate it into cases :
*
*Case 1 : If $(a\sqrt 2\gt)\ a\gt b$, then $(1)\iff -a\le x\le b$. We have $\frac{b+\sqrt{2a^2-b^2}}{2}\gt \frac{b+\sqrt{2b^2-b^2}}{2}=b$ and $b\gt 0\gt \frac{b-\sqrt{2a^2-b^2}}{2}\gt \frac{b-2a}{2}=\frac b2-a\gt -a$. So, in this case, 1 root.
*Case 2 : If $a\le b\lt a\sqrt 2$, then $(1)\iff -a\le x\le a$, and $$\small\frac{b+\sqrt{2a^2-b^2}}{2}\le a\iff \sqrt{2a^2-b^2}\le 2a-b\iff 2a^2-b^2\le (2a-b)^2\iff (a-b)^2\ge 0$$which indeed holds, and $$a\gt \frac{b+\sqrt{2a^2-b^2}}{2}\gt\frac{b-\sqrt{2a^2-b^2}}{2}\gt \frac{b-\sqrt{2b^2-b^2}}{2}=0$$So, in this case, 2 roots.
*Case 3 : If $b=a\sqrt 2$, then $(1)\iff -a\le x\le a$, and $x=\frac{a}{\sqrt 2}$ satisfies this. So, in this case, 1 root.
*Case 4 : If $b\gt a\sqrt 2$, the equation has no real solutions.
Therefore,
1 root if $a\gt b$ or $b=a\sqrt 2$
2 roots if $a\le b\lt a\sqrt 2$
0 root if $b\gt a\sqrt 2$
|
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|
Evaluate the following line integral: $\int_l\sqrt{x^2+y^2}dl$ Evaluate the following line integral:
$$\int_l\sqrt{x^2+y^2}dl$$ where $$l:x^2+y^2=ax$$
What I've already done is:
$$x^2+y^2=ax \Rightarrow \left( x-\frac{a}{2} \right)^2+y^2=\left(\frac{a}{2}\right)^2$$
Spherical coordinates:
$$
\begin{cases}
x-\frac{a}{2}=a \cos t \\
y=a \sin t
\end{cases}$$
$$dl=\sqrt{(-a \sin t)^2+(a \cos t)^2}dt=a\space dt$$
$$\int_l\sqrt{x^2+y^2}dl=4 \int_0^{\frac{\pi}{2}} \sqrt{a^2 \cos^2{t}+a^2\cos t+\frac{a^2}{4}+a^2 \sin^2{t}} \space dt=4 a^2 \int_0^{\frac{\pi}{2}} \sqrt{\cos t+\frac{5}{4}} \space dt$$
and it seems to me that this integral doesn't have a nice answer.
The answer should be $2a^2$
|
Your parameterization is making it very hard for you. I would recommend using straight polar coordinates. You should try that on your own before reading this solution.
Letting our path $l=r$, we then have
$$\int_l\sqrt{x^2+y^2}\,dl=\int_rr\,dr$$
Note that the path $x^2+y^2=ax$ becomes
$$r^2=ar\cos\theta$$
which gives us
$$r=a\cos\theta$$
Therefore,
$$dr=a\sin\theta\,d\theta$$
The line integral is then
$$4\int_0^{\frac{\pi}{2}}(a\cos\theta)(a\sin\theta)\,d\theta=4a^2\int_0^{\frac{\pi}{2}}\cos\theta\sin\theta\,d\theta=2a^2$$
Picking the wrong parameterization can really mess up your integrals! If it doesn't work out at first, try a different parameterization and see what happens.
|
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|
How can find all solution of $x^2(y-1)+y^2(x-1)=1$ such that $x,y \in \mathbb{Z}$?
How can find all solution of $x^2(y-1)+y^2(x-1)=1$ such that $x,y \in \mathbb{Z}$?
I have no clue to find solution .
I tried to change into quadratic equation $$y^2(x-1)+yx^2-(x^2+1)=0\\y=\frac{-y\pm\sqrt{x^4+4(x-1)(x^2+1)(-1)}}{2(x-1)}$$
but stop here ...
Please help to me to find a method of solving . Thanks in advance.
Remark : by graphing that I found $$\quad{(x,y)=\\(2,1)\\(1,2)\\(2,-5)\\(-5,2)}$$
|
Write like this:
$$ xy (x+y)-(x+y)^2+2xy =1$$
If we put $a=x+y$ and $b=xy$ we get $$ b ={1+a^2\over a+2}\;\;\;\Longrightarrow \;\;\;a+2\mid a^2+1$$
Now since $a+2\mid a^2-4$ we have $a+2\mid 5$ so $a+2\in\{\pm 1,\pm 5\}$ and thus $a\in\{-3,-1,-7,3\}$
\begin{array}{c|c|c|c}
a & b & x&y\\\hline
-7 & -10 &/&/\\
-3 & -10 &-5(2)&2(-5)\\
-1 & 2 &/&/\\
3& 2 &2(1)&1(2)\\
\end{array}
|
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|
Limit of $f(x)$ when $x$ goes to zero Let $f(x) = \frac{1 + \tan x + \sin x - \cos x}{\sin^2 x + x^3}$ . Find value of $\lim_{x \to 0} f(x)$ if it exists . I can solve it using L'Hospital's Rule and Taylor series but I'm looking for another way suing trigonometric identities .
|
$$\lim_{x\rightarrow0}\frac{1 + \tan x + \sin x - \cos x}{\sin^2 x + x^3}=\lim_{x\rightarrow0}\frac{2\sin^2\frac{x}{2} +2\sin\frac{x}{2}\cos\frac{x}{2}\left(\frac{1}{\cos{x}}+1\right) }{\sin^2 x + x^3}=$$
$$=\lim_{x\rightarrow0}\left(\frac{\sin\frac{x}{2}}{\frac{x}{2}}\cdot\frac{\sin\frac{x}{2} +\cos\frac{x}{2}\left(\frac{1}{\cos{x}}+1\right) }{\frac{\sin x}{x}\cdot\sin{x} + x^2}\right)=\infty$$
because $\frac{\sin{x}}{x}\rightarrow1$.
|
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|
GCD of $n^3+3n+1$ and $7n^3+18n^2-n-2$ is always $1$ A problem I found yesterday says to prove $\gcd(n^3+3n+1, 7n^3+18n^2-n-2)=1$ for all integers $n\ge 1$. To begin, I used the Euclidean algorithm to observe that $7n^3+18n^2-n-2=7\left(n^3+3n+1\right)+\left(18n^2-22n-9\right)$, so $$\gcd(n^3+3n+1, 7n^3+18n^2-n-2)=\gcd(n^3+3n+1, 18n^2-22n-9).$$ From here, I got stuck however, since the next step of polynomial division involves rational numbers. I observed that the GCD cannot be a multiple of $3$, since if $3$ divides the term on the right, then $3\mid n$, however, then $3\nmid n^3+3n+1$. Similarly, by a parity argument, both terms are odd. Any suggestions for how to finish the problem from here?
|
HINT: Because both terms are odd, you know that
$$\gcd(n^3+3n+1,18n^2-22n-9)=\gcd(2(n^3+3n+1),18n^2-22n-9).$$
A similar argument works for the prime $3$, and perhaps for $3^2$. Can you take it from here?
|
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|
Find value of $a_{2012}$ A sequence $\left\{a_n\right\}$ is defined as:
$a_1=1$, $a_2=2$ and
$$a_{n+1}=\frac{2}{a_n}+a_{n-1}$$ $\forall$ $n \ge 2$
Find $a_{2012}$
My Try:
we have
$$a_{n+1}-a_{n-1}=\frac{2}{a_n}$$
$$a_n a_{n+1}-a_{n-1}a_n=2 \tag{1}$$
Replacing $n$ with $n-1$ we get
$$a_{n-1} a_{n}-a_{n-2}a_{n-1}=2 \tag{2}$$ adding $(1)$ and $(2)$
we get
$$a_n a_{n+1}-a_{n-2}a_{n-1}=4 \tag{3}$$
Again replace $n$ with $n-1$ in $(3)$ and adding with $(1)$ we get
$$a_n a_{n+1}-a_{n-2}a_{n-3}=6 \tag{4}$$ Again replace $n$ with $n-1$ in $(4)$ and adding with $(1)$ we get
$$a_n a_{n+1}-a_{n-3}a_{n-4}=8 \tag{5}$$
Continuing the process we get
$$a_na_{n+1}-a_{n-2010}a_{n-2011}=4022$$
Now in above equation put $n=2012$ we get
$$a_{2012}a_{2013}-a_1a_2=4022$$ $\implies$
$$a_{2012}a_{2013}=4024$$
Any further clue?
|
By your work we obtain: $$a_{n+2}a_{n+1}=a_{n+1}{a_n}+2,$$
which gives $$a_{n+1}a_n=2+(n-1)2=2n.$$
Thus, $$a_{n}=\frac{2n-2}{\frac{2n-4}{a_{n-2}}}=\frac{n-1}{n-2}a_{n-2},$$
which for even $n$ gives
$$a_n=\frac{(n-1)(n-3)...1}{(n-2)(n-4)...2}a_2=\frac{2(n-1)!!}{(n-2)!!}.$$
Id est, $$a_{2012}=\frac{2\cdot2011!!}{2010!!}.$$
|
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|
Solution to a cubic-like polynomial with fractional powers I wish to solve the following equation for x, but am not certain how:
$\frac{(1+x)^{1/3}}{1+x^{1/3}} = (\frac{3^{1/3}}{.691})(R/a)(p/h)^{-1/3}$
I suspect that there is some way to manipulate this into a cubic equation and solve by the usual methods, but despite much work I have not succeeded in finding any solution for x. Any help would be appreciated.
|
Set $x^{1/3}=z\to x=z^3$
and $k=(\frac{3^{1/3}}{.691})(R/a)(p/h)^{-1/3}$
the equation becomes
$$\frac{\sqrt[3]{z^3+1}}{z+1}=k$$
$$\frac{z^3+1}{(z+1)^3}=k^3$$
$$\frac{(z+1)(z^2-z+1)}{(z+1)^3}=k^3$$
$$z^2-z+1=k^3(z^2+2z+1)$$
$$z=\frac{-2 k^3-1\pm\sqrt{12 k^3-3}}{2 \left(k^3-1\right)}$$
$$x=z^3\to x=\left(\frac{-2 k^3-1\pm\sqrt{12 k^3-3}}{2 \left(k^3-1\right)}\right)^3$$
set back the value of $k$
$$x_1=\frac{0.125 \left(a^3 p \left(\sqrt{\frac{36.3703 h R^3}{a^3 p}-3}+1\right)+6.06172 h R^3\right)^3}{\left(a^3 p-3.03086 h R^3\right)^3}$$
$$x_2=-\frac{0.125 \left(a^3 p \left(\sqrt{\frac{36.3703 h R^3}{a^3 p}-3}-1\right)-6.06172 h R^3\right)^3}{\left(a^3 p-3.03086 h R^3\right)^3}$$
|
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|
Calculating infinite sums of power series. This past week I have been studying power series, first I studied how to determinate the intervals of convergence and I have no problem doing that (usually I just have to apply the root or ratio test of convergence). However, now I am asked to calculate the sums of:
I know some results of infinite series, like the geometric or telescopic series, however this is not enough to calculate any of those infinite sums. Is there any general procedure to calculate this sums? Or any differentation/integration theorems of power series I could use?
|
\begin{align}
\sum_{n=1}^\infty \frac{n(x+3)^n}{2^n} & = (x+3)\sum_{n=1}^\infty \frac{n(x+3)^{n-1}}{2^n} = (x+3)\sum_{n=1}^\infty \frac d {dx} \, \frac{(x+3)^n}{2^n} \\[10pt]
& = (x+3) \frac d {dx} \sum_{n=1}^\infty \frac{(x+3)^n}{2^n} \qquad \text{See the comment on this step below.} \\[10pt]
& = (x+3) \frac d {dx} \, \frac{\text{first term}}{1 - \text{common ratio}} = \text{etc.}
\end{align}
Is the sum of the derivatives equal to the derivative of the sum? In first-semester calculus you see it proved that that is true when there are only finitely many terms. It doesn't entirely generally work with inifinitely many terms, but it does with power series in the interior of the interval of convergence.
\begin{align}
n^3 x^n & = x^3 \cdot\Big( n(n-1)(n-2)x^{n-3}\Big) + 3x^2 \cdot \Big( n(n-1) x^{n-2} \Big) + 3x \cdot\Big( nx^{n-1}\Big) \\[10pt]
& = x^3 \frac {d^3} {dx^3} x^n + 3x^2 \frac{d^2}{dx^2} x^n + 3x \frac d {dx} x^n \\[12pt]
\text{So } \sum_{n=0}^\infty \frac{n^3 x^n}{n!} & = x^3 \frac{d^3}{dx^3} \sum_{n=0}^\infty \frac{x^n}{n!} + 3x^2 \frac{d^2}{dx^2} \sum_{n=0}^\infty \frac{x^n}{n!} + 3x \frac d {dx} \sum_{n=0}^\infty \frac{x^n} {n!}
\end{align}
And presumably you know how to sum the series in the last line above.
|
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|
How to simplify an expression I have an expression:
$$\frac{\sqrt{6} + 1}{6} - \frac{\sqrt{10-4\sqrt{6}}}{6}$$
and it seems like it must be equal to $\frac{1}{2}$. How could i simplify this?
|
It is : $(\sqrt{6}-2)^2=10-4\sqrt{6}$ so you get :
$$\frac{\sqrt{6} + 1}{6} - \frac{\sqrt{10-4\sqrt{6}}}{6} = \frac{\sqrt{6} + 1}{6}-\frac{\sqrt{(\sqrt6 - 2)^2}}{6}= \frac{\sqrt{6} + 1}{6}- \frac{\sqrt6-2}{6} = \frac{3}{6} =\frac{1}{2}$$
Note that : $\sqrt{6}-2 > 0 $ and that's why you remove the root simply enough.
|
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Given $ \frac{1}{r}+\frac{1}{s}=a; \frac{1}{r}\times\frac{1}{s}=b; a+b=r; a\times b=s$, find $a$. (Brazilian Math Olympics, 2016)
Given: $\{a,b,r,s\}\subset \mathbb R$, $a>0$,
$\frac{1}{r}$, $\frac{1}{s}$ are roots for $x^2-ax+b=0$, and
$a$,$~b$ are roots for $x^2-rx+s$.
Find: the numeric value of $a$.
This is question 3, level 2, phase 3, Brazilian Math Olympics (OBM, 2016). No answer provided.
The first step is easy, using Girard relations, we can get the system
$$
\left\{
\begin{array}{c}
\frac{1}{r}+\frac{1}{s}=a \\
\frac{1}{r}\times\frac{1}{s}=b\\
a+b=r\\
a\times b=s
\end{array}
\right.
$$
I'm having a hard time on solving this system of equations. All tricks I know seems to lead to nowhere. It was considered a hard question in the contest (level 2 in OBM is for students up to 9th grade).
Hints and solutions are appreciated. Sorry if this is a duplicate.
|
$$\frac{1}{r}+\frac{1}{s}=a\tag1$$
$$\frac{1}{r}\times\frac{1}{s}=b\tag2$$
$$a+b=r\tag3$$
$$a\times b=s\tag4$$
Substituting $(3)(4)$ into $(1)(2)$ gives
$$(1)\implies\frac{1}{a+b}+\frac{1}{ab}=a\implies ab+a+b=a^2b(a+b)=a^3b+a^2b^2\tag5$$
$$(2)\implies \frac{1}{a+b}\times\frac{1}{ab}=b\implies 1=ab^2(a+b)\tag6$$
From $(5)(6)$, we have
$$(a^2b^2(a+b)=)\ \ b(ab+a+b)=a\implies (a+1)b^2=a-ab\tag7$$
Multiplying the both sides of $(5)$ by $a+1$ gives
$$(a+1)(ab+a+b)=(a+1)a^3b+a^2(a+1)b^2\tag8$$
Substituting $(7)$ into $(8)$ gives
$$(a+1)(ab+a+b)=(a+1)a^3b+a^2(a-ab),$$
i.e.
$$-(a^4-a^2-2a-1)b=a(a^2-a-1),$$
i.e.
$$-(a^2-(a+1)^2)b=a(a^2-a-1),$$
i.e.
$$-(a^2-a-1)(a^2+a+1)b=a(a^2-a-1)$$
If $a^2-a-1\not=0$, then $$(a+1)b^2=-ab-a^2b^2\tag9$$
From $(7)(9)$, we get
$$-ab-a^2b^2=a-ab\implies -ab^2=1$$
This is impossible since the LHS is negative while the RHS is positive.
So, we have $a^2-a-1=0\implies a=\frac{1+\sqrt 5}{2}$.
Therefore, the answer is $$\color{red}{a=\frac{1+\sqrt 5}{2}}$$
|
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|
Set of solutions to a sum I want to know the set of integer solutions to the following sum:
$$\sum\limits_{n=1}^{k}\frac{n}{a_n}=1$$
For instance, the series
$a_n=\frac{1}{nk}$ satisfies this for all integer $k$. So far, this is the only solution that I have come up with for a general $k$.
Here are some solutions that I have for specific values of $k$:
If $k=2$, $a_1=a_2=\frac{1}{3}$ works.
If $k=3$, $a_1=a_2=\frac{1}{4}, a_3=\frac{1}{12}$ works.
If $k=4$, $a_1=a_2=\frac{1}{5}, a_3=\frac{1}{15}, a_4=\frac{1}{20}$ works.
Are there any generalizations or ways to generate new solutions?
|
Yet another generic solution:
since $$\sum_{n=1}^{k-2} \frac{n}{2^{n+1}}=1-\frac{k}{2^{k-1}},$$
we can write the fraction $\dfrac{k}{2^{k-1}}$ in the form
$$\frac{k}{2^{k-1}}=\frac{k-1}{2^{k-1}}+\frac{1}{2^{k-1}} = \color{red}{\frac{k-1}{2^{k-1}}}+\color{darkviolet}{\frac{k}{k\cdot 2^{k-1}}}.$$
Then for given $k$ ($k\ge 3$) we have solution
$$(a_1,a_2,\ldots,a_{k-2},a_{k-1},a_k) = \left(4,8,\ldots, 2^{k-1}, \color{red}{2^{k-1}}, \color{darkviolet}{k\cdot 2^{k-1}}\right).$$
Examples:
$k=3$: $(4,\color{red}{4},\color{darkviolet}{12})$;
$k=4$: $(4,8,\color{red}{8},\color{darkviolet}{32})$;
$k=5$: $(4,8,16,\color{red}{16},\color{darkviolet}{80})$;
$k=6$: $(4,8,16,32,\color{red}{32},\color{darkviolet}{192})$;
$\cdots$ .
In general, the set of solutions is rather wide:
$k=3$:
(2,5,30)
(2,6,18)
(2,7,14)
(2,8,12)
(2,10,10)
(2,12,9)
(2,16,8)
(2,28,7)
(3,4,18)
(3,6,9)
(3,12,6)
(3,30,5)
(4,3,36)
(4,4,12)
(4,8,6)
(5,4,10)
(5,10,5)
(5,40,4)
(6,3,18)
(6,4,9)
(6,6,6)
(6,24,4)
(8,4,8)
(8,16,4)
(10,5,6)
(12,3,12)
(12,12,4)
(14,4,7)
(15,6,5)
(20,10,4)
(30,3,10)
(36,9,4)
|
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>Does $\{f_n\}$ converge pointwise to a function on $[0,\infty)?$ Let $g_n(x)=\sin^2(x+\frac{1}{n}),x\in (0,\infty)$
and $f_n(x)=\int _0^x g_n(t)\, dt$.
Does $\{f_n\}$ converge pointwise to a function on $[0,\infty)?$
I try to show that $$\int_0^x \sin^2\left(t +\frac1n\right) \, dt = \frac{x}{2} - \frac{1}{4}\sin\left(2x+\frac2n\right) + \frac{1}{4} \sin \left(\frac2n\right),$$
and
$$\int_0^x \sin^2(t) \, dt = \frac{x}{2} - \frac{1}{4}\sin(2x) .$$
Hence,
$$|f_n(x) - f(x)| \leqslant \frac{1}{4}\left|\sin\left(2x+\frac2n\right) - \sin(2x) \right|+ \frac{1}{4} \left|\sin\left(\frac2n\right)\right| \leqslant \frac{1}{2n} + \frac{1}{4}\left|\sin\left(\frac2n\right)\right|$$
We have uniform convergence on $[0,\infty)$.
From this proof i can conclude that $\{f_n\}$ doesnot converge pointwise to a
function $f$ on $[0,\infty)$. so My answer is no.......IS it correct ?
|
You have correctly calculated $f_n$. Now just let $n\to\infty$:
$$f_n(x) = \frac{x}{2} - \frac{1}{4}\sin\left(2x+\frac{2}n\right) + \frac{1}{4} \sin \left(\frac2n\right)= \frac{x}2 - \frac{1}2\cos\left(x+\frac2n\right)\sin x \xrightarrow{n\to\infty} \frac{x}2 - \frac12\cos x\sin x$$
So,
$$\lim_{n\to\infty} f_n(x) = \frac{x}2 - \frac12\cos x\sin x = \frac{x}2 - \frac14\sin(2x) = \int_0^x \sin^2 t\,dt$$
is the poinwise limit of the sequence $(f_n)_{n=1}^\infty$.
Another way to see this is to use the Lebesgue Dominated Convergence Theorem. Functions $f_n$ are bounded by the integrable function $1$ so we can interchange the limit and the integral:
$$\lim_{n\to\infty} f_n = \lim_{n\to\infty}\int_0^x \sin^2 \left(t+\frac1n\right)\ dt =\int_0^x \lim_{n\to\infty} \sin^2 \left(t+\frac1n\right)\ dt = \int_0^x \sin^2 t \ dt$$
|
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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.