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Find the minimun of $MN+\frac{3}{5}MP$, $MN$ and $MP$ is two sides of a quadrilateral. In a quadrilateral $OPMN$ ，$\angle NOP=90^\circ$,$ON=1$,$OP=3$, and $M$ satisfy $\vec{MO}\cdot\vec{MP}=4$, find the minimum of
$MP+\frac{3}{5}MN$
I choose the vertex $O$ of $OPMN$ as the origin of the coordinate system, and ON as the $x$-axis, $OP$ as the $y$-axis.
so $N(1,0)$ and $P(0,-3)$ and assume $M(x,y)$ then $\vec{MO}=(-x,-y)$ $\vec{MP}=(-x,-3-y)$. then $$x^2+3y+y^2=4\Rightarrow x^2+\left(y+\frac{3}{2}\right)^2=\left(\frac{5}{2}\right)^2$$
I know trail of point $M$ is a circle centered at $(0,-\frac{3}{2})$.
but I have no idea to go to next. and I use $x=\frac{5}{2}\cos
\theta, y=\frac{5}{2}\sin\theta -\frac{3}{2}$ and we have $$MP+\frac{3}{5}MN =f(\theta)= \sqrt{\frac{17}{2}+\frac{15}{2}\sin
\theta} + \frac{3}{5}\sqrt{\frac{19}{2}-5\cos\theta-\frac{15}{2}\sin\theta}$$
I use Mathematica to find the numerical minimum of $f(\theta)$. It is about $$f_{\min}(\theta)\approx f(-1.45162)=3.45245$$
it seems when $M$ near the bottom of the circle, $MP+\frac{3}{5}MN$ take the minimum . Please help me .thanks very much. :)
|
from Maple:
$y = -\frac{969+85 \sqrt{273} }{596} = -3.982265922$
Minimum of $MP+\frac{3}{5}MN$ is $\frac{\sqrt{298}}{5} = 3.452535300$
|
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|
Counting with Recurrence Relations Find the recurrence relation for a(n) - number of ternary strings of length n, containing the number 2 odd times.
Some of these: 012,112,12,02,... .
|
We divide all that sequences into $4$ non-intersecting subclasses:
a) Odd count of $2$s, ends with $1$ or $0$,
b) Even count of $2$s, ends with $1$ or $0$,
c) Odd count of $2$s, ends with $2$,
d) Even count of $2$s, ends with $2$,
Let $a_n,\,b_n,\,c_n,\,d_n$ be the cardinality of a corresponding class of sequences of length $n$. For $n=1$ we have
$$a_1=0,\,b_1=2,\,c_1=1,\,d_1=0$$
Let $n>1$. We consider the last digit and the rest part of length $n-1$.
If a sequence is of classes a or b, then the rest part will have the same amount of $2$s, therefore
$$a_n=2(a_{n-1}+c_{n-1})\\b_n=2(b_{n-1}+d_{n-1})$$
Otherwise the rest part will have one less $2$, thus
$$c_n=b_{n-1}+d_{n-1}\\d_n=a_{n-1}+c_{n-1}$$
Let $\mathbf{x}_n=(a_n,b_n,c_n,d_n)^T$, we have $\mathbf{x}_n=A\mathbf{x}_{n-1}$ where
$$A=\begin{pmatrix}
2&0&2&0\\
0&2&0&2\\
0&1&0&1\\
1&0&1&0
\end{pmatrix}$$
Performing diagonalization of $A$ we have $A=SDS^{-1}$ where
$$S=\begin{pmatrix}
0 & -1 & 2 & 2\\
-1 & 0 & -2 & 2\\
0 & 1 & -1 & 1\\
1 & 0 & 1 & 1
\end{pmatrix}\\
D=\operatorname{diag}(0,0,1,3)$$
As $\mathbf{x}_n$ $=A^{n-1}\mathbf{x}_1$ $=SD^{n-1}S^{-1}\mathbf{x}_1$ the desired value will be $(1,0,1,0)\cdot \mathbf{x}_n$ $=
(1,0,1,0)\cdot SD^{n-1}S^{-1}\mathbf{x}_1$ $=\frac{3^n-1}{2}$
|
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|
Combinations series: $\frac{{n \choose 1}(n-1)^3+{n \choose 3}(n-3)^3+\ldots}{n^2(n+3)\cdot 2^n}$
Evaluate $\frac{{n \choose 1}(n-1)^3+{n \choose 3}(n-3)^3+\ldots}{n^2(n+3)\cdot 2^n}$ for $n=10$.
Attempt: I'll deal with the case n being even, as we need to evaluate for n=10.
the numerator is
$${n \choose 1}(n-1)^3+{n \choose 3}(n-3)^3+\ldots$$
$$=\sum_{r=odd} {n \choose r}(n-r)^3$$(not sure if this is a correct notation).
$$=\sum_{r=odd}{n \choose n-r}r^3=\sum_{r=odd} {n \choose r}r^3$$(parity being same as n is even and r is odd, although I don't think this matters much).
Using the identity ${n \choose r}=\frac{n}{r} {n-1 \choose r-1}$ repeatedly in following steps,
$$=n\sum_{r=even} {n-1 \choose r-1}r^2$$
$$=[n(n-1)](1+\sum_{r=odd} {n-2 \choose r-2}[(r-2+3)+\frac{1}{r-1}]$$
$$=[n(n-1)](1+(n-2)\sum_{r=even}{n-3 \choose r-3}+3\sum_{r=odd}{n-2 \choose r-2}+\frac{1}{n-1} \sum_{r=even}{n-1 \choose r-1} -1)$$
$$=[n(n-1)]((n-2)\cdot 2^{n-4} +3\cdot 2^{n-4}+\frac{2^{n-2}}{n-1}$$
This simplfies to $n \cdot 2^{n-4} (n^2+7n-4)$.
Which is incorrect. The answer for $n=10$ (numerator/denominator is given as $\frac{1}{16}$).
Where am I going wrong?
Also the hint given for this problem was "expand $\frac{(e^x+1)^n - (e^x-1)^n}{2}$ in two different ways". I didn't quite understand this approach?
Could someone please explain this approach and any other approach also?
|
Following the given hint, we have that
$$\begin{align}\sum_{r \text{ odd}} {10 \choose r}(10-r)^3&=\frac{1}{2}\left[\left((e^x+1)^{10}-(e^x-1)^{10}\right)'''\right]_{x=0}\\
&=\left[360(e^x+1)^7e^{3x}+135(e^x+1)^8e^{2x}+5(e^x+1)^9e^x\right.\\
&\quad \left.-360(e^x-1)^7e^{3x}-135(e^x-1)^8e^{2x}-5(e^x-1)^9e^x\right]_{x=0}\\
&=360\cdot 2^7+135\cdot 2^8+5\cdot 2^9.
\end{align}.$$
The same approach works for any integer $n\geq 4$:
$$\begin{align}\sum_{r \text{ odd}} {n \choose r}(n-r)^3
&=\frac{1}{2}\left[\left((e^x+1)^{n}-(e^x-1)^{n}\right)'''\right]_{x=0}\\
&=3\binom{n}{3}2^{n-3}+3\binom{n}{2}2^{n-2}+n2^{n-2}
=\frac{n^2(n+3)2^n}{16}.
\end{align}$$
P.S. We could also use the Taylor series of $e^x$: for $n\geq 4$
$$\begin{align}\frac{1}{2}\left[\left((e^x+1)^{n}-(e^x-1)^{n}\right)'''\right]_{x=0}
&=\frac{3!}{2}[x^3]\left((e^x+1)^{n}-(e^x-1)^{n}\right)\\
&=3[x^3]\left(2+x+\frac{x^2}{2}+\frac{x^3}{6}\right)^{n}\\
&=3[x^3]n2^{n-1}\left(x+\frac{x^2}{2}+\frac{x^3}{6}\right)\\
&\quad +3[x^3]\binom{n}{2}2^{n-2}\left(x+\frac{x^2}{2}+\frac{x^3}{6}\right)^2\\
&\quad +3[x^3]\binom{n}{3}2^{n-3}\left(x+\frac{x^2}{2}+\frac{x^3}{6}\right)^3\\
&=n2^{n-2}+3\binom{n}{2}2^{n-2}+3\binom{n}{3}2^{n-3}\\
&=\frac{n^2(n+3)2^n}{16}.
\end{align}$$
|
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|
Proving That $\sum^{n}_{k=0} \bigl(\frac{4}{5}\bigr)^k < 5$
Using induction, prove that $$\sum_{k=0}^n \biggl(\frac 4 5 \biggr)^k = 1+\frac{4}{5}+\bigg(\frac{4}{5}\bigg)^2+\bigg(\frac{4}{5}\bigg)^3+\cdots +\bigg(\frac{4}{5}\bigg)^n<5$$ for all natural numbers $n.$
What I have tried is as follows.
Consider the statement $$P(n):1+\frac{4}{5}+\bigg(\frac{4}{5}\bigg)^2+\cdots +\bigg(\frac{4}{5}\bigg)^n<5.$$
For $n=1$, we have that $\displaystyle P(1):1+\frac{4}{5}<5$ is true.
For $n=k$, we assume that $$\displaystyle P(k):1+\frac{4}{5}+\bigg(\frac{4}{5}\bigg)^2+\cdots +\bigg(\frac{4}{5}\bigg)^k<5.$$
$\displaystyle P(k+1):1+\frac{4}{5}+\bigg(\frac{4}{5}\bigg)^2+\cdots +\bigg(\frac{4}{5}\bigg)^k+\bigg(\frac{4}{5}\bigg)^{k+1}<5+\bigg(\frac{4}{5}\bigg)^{k+1}$
How can I prove that the sum on the left is $< 5?$ Help me please. Thanks.
|
$$1+\frac{4}{5}+\bigg(\frac{4}{5}\bigg)^2+\bigg(\frac{4}{5}\bigg)^3+\cdots +\bigg(\frac{4}{5}\bigg)^n = 5\left(1-\dfrac{4^{n+1}}{5^{n+1}} \right)$$
|
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|
Given a tetrahedron, whose sides are $AB=3,AC=4,BC=5,AD=6,BD=7,CD=8$ . Find the volume of the tetrahedral $ABCD$ .
Given a tetrahedron, whose sides are $AB= 3, AC= 4, BC= 5, AD= 6, BD= 7, CD= 8$ . Find the volume of the tetrahedral $ABCD$ .
Assume that the tetrahedral $ABCD$ has its height $DH$ , whose length I will find by using vectors and the following lemma:
Given three real numbers $f, t, u$ so that $f+ t+ u= 1$ and $H$ on the plane $BCD$ . We'll have
$$\overrightarrow{DH}= f\overrightarrow{DA}+ t\overrightarrow{DB}+ u\overrightarrow{DC}$$
From hypothesis
$$\overrightarrow{DH}\cdot \overrightarrow{AB}= 0\Rightarrow \left ( f\overrightarrow{DA}+ t\overrightarrow{DB}+ u\overrightarrow{DC} \right )\left ( \overrightarrow{DA}- \overrightarrow{DB} \right )= 0$$
$$\Rightarrow f\overrightarrow{DA}\cdot \overrightarrow{DB}+ 49t+ u\overrightarrow{DC}\cdot \overrightarrow{DB}- 36f- t\overrightarrow{DB}\cdot \overrightarrow{DA}- u\overrightarrow{DC}\cdot \overrightarrow{DA}= 0$$
$$49t- 36f+ (f- t)\overrightarrow{DA}\cdot \overrightarrow{DB}+ u\overrightarrow{DC}\cdot \overrightarrow{DB}- u\overrightarrow{DC}\cdot \overrightarrow{DA}= 0$$
On the other hand
$$\cos DCA= \frac{64+ 36- 16}{2\cdot 8\cdot 6}= \frac{7}{8}$$
$$\cos ADB= \frac{36+ 49- 9}{2\cdot 6\cdot 7}= \frac{19}{21}$$
$$\cos DBC= \frac{64+ 49- 25}{2\cdot 8\cdot 7}= \frac{11}{14}$$
Therefore
$$\overrightarrow{DA}\cdot \overrightarrow{DB}= 6\cdot 7\cdot \frac{19}{21}= 38$$
$$\overrightarrow{DC}\cdot \overrightarrow{DB}= 8\cdot 7\cdot \frac{11}{14}= 44$$
$$\overrightarrow{DC}\cdot \overrightarrow{DA}= 8\cdot 6\cdot \frac{7}{8}= 42$$
$$\therefore 49t- 36f+ 38(f- t)+ 44u- 42u= 0\Rightarrow 2f+ 11t+ 2u= 0$$
Similarly
$$\overrightarrow{DH}\cdot \overrightarrow{BC}= 0\Rightarrow \left ( f\overrightarrow{DA}+ t\overrightarrow{DB}+ u\overrightarrow{DC} \right )\left ( \overrightarrow{DC}- \overrightarrow{DB} \right )= 0$$
$$\Rightarrow f\left ( \overrightarrow{DA}\cdot \overrightarrow{DC}- \overrightarrow{DA}\cdot \overrightarrow{DB} \right )+ t\overrightarrow{DB}\cdot \overrightarrow{DC}- t\left | \overrightarrow{DB} \right |^{2}+ u\left | \overrightarrow{DC} \right |^{2}- u\overrightarrow{DC}\cdot \overrightarrow{DB}= 0$$
$$\Rightarrow 4f+ 44t- 49t+ 64u- 44u= 0\Rightarrow 4f- 5t+ 20u= 0$$
A solution to the system of linear equations given by
$$f= \frac{115}{72}, t= -\frac{2}{9}, u= -\frac{3}{8}$$
$$\Rightarrow \overrightarrow{DH}= \frac{115}{72}\overrightarrow{DA}- \frac{2}{9}\overrightarrow{DB}- \frac{3}{8}\overrightarrow{DC}$$
$$\Rightarrow \left | \overrightarrow{DH} \right |^{2}= \frac{115^{2}}{75^{2}}\cdot 36+ \frac{4}{81}\cdot 49+ \frac{9}{64}\cdot 64- \frac{115\cdot 4}{72\cdot 9}\cdot 38+ \frac{2\cdot 6}{9\cdot 8}\cdot 44- \frac{115\cdot 6}{72\cdot 8}\cdot 42= \frac{1199}{36}$$
Is there any way to find the length of $DH$ ? Thanks a real lot.
|
The volume of the pyramid in terms of its side lengths
can be found as
\begin{align}
V&=
\frac1{12}\,
\left(
4\, u^2\, v^2\, w^2+(u^2+v^2-c^2)\, (v^2+w^2-a^2)\, (u^2+w^2-b^2)
\right.
\\
&\phantom{=}
\left.
-u^2\, (v^2+w^2-a^2)^2-v^2\, (u^2+w^2-b^2)^2-w^2\, (u^2+v^2-c^2)^2
\right)^{1/2}
,
\end{align}
where
\begin{align}
a&=|BC|
,\quad
b=|AC|
,\quad
c=|AB|
,\\
u&=|AD|
,\quad
v=|BD|
,\quad
w=|CD|
.
\end{align}
For
$|AB|=3, |AC|=4, |BC|=5, |AD|=6, |BD|=7, |CD|=8$,
\begin{align}
V&=\tfrac13\,\sqrt{1199}
\approx 11.542
.
\end{align}
See also
Cayley-MengerDeterminant,
as advised in
this answer.
Check-in:
one of instances of such tetrahedron can be
presented with the coordinates of vertices as follows:
\begin{align}
A &= (0, 0, 0)
,\\
B &= (-3, 0, 0)
,\\
C &= (0, 4, 0)
,\\
D &= (\tfrac23, -\tfrac32, \tfrac16\,\sqrt{1199})
.
\end{align}
|
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|
Find maximum $k \in \mathbb{R}^{+}$ such that $ \frac{a^3}{(b-c)^2} + \frac{b^3}{(c-a)^2} + \frac{c^3}{(a-b)^2} \geq k (a+b+c) $ Find maximum $k \in \mathbb{R}^{+}$ such that $$ \frac{a^3}{(b-c)^2} + \frac{b^3}{(c-a)^2} + \frac{c^3}{(a-b)^2} \geq k (a+b+c) $$
for all $a, b, c$ that are distinct positive real numbers ( $a \neq b$, $b \neq c$, $a \neq c$)
Usually when I see this kind of cyclic, symmetrical inequality, the extreme values are taken at $a = b = c$， which is obviously not the case here. So I am not sure how to approach this one..
|
This is a partial answer when $a,b,c$ form a side of a triangle.
Letting $c\to 0^+$, the inequality is in the form of:
$$\dfrac{a^3}{b^2}+\dfrac{b^3}{a^2}-k(a+b)\geq o(c),$$
for $o(c)\geq 0.$
But the left hand side is:
$$(a+b)\left(\dfrac{(a-b)^2(a^2+ab+b^2)}{a^2b^2}\right)+(a+b)(1-k).$$
If $k>1,$ then the whole thing can be made negative by taking $a = n+\epsilon$ and $b = n:$
$$(2n+\epsilon)\left(\epsilon^2\cdot\dfrac{n^2+3n\epsilon+3\epsilon^2}{n^2(n+\epsilon)^2} + 1-k\right).$$
For $k=1,$ the inequality is equivalent to:
$$\sum\dfrac{a(a-b+c)(a+b-c)}{(b-c)^2}\geq 0.$$
This begs for the Ravi substitution: $a = y+z,...$ and so:
$$\sum\dfrac{yz(y+z)}{(y-z)^2}\geq 0\iff \sum yz(y+z)(x-y)^2(x-z)^2\geq 0.$$
But this one is a direct consequence of a generalized Schur or Vornicu-Schur inequality, which can be found in Theorem $4$ here.
For when $a,b,c$ do not form a triangle, exactly one of $x,y,z$ is negative and the Vornicu-Schur does not apply here, at least not directly. I have a feeling that this is already a solved problem in Vasc's famous inequality book if you happen to have access to it.
|
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|
A problem of definite integral inequality? $$
\text { Minimum odd value of $a$ such that }\left.\left|\int_{10}^{19} \frac{\sin x d x}{\left(1+x^{a}\right)}\right|<\frac{1}{9} \text { is (where } a \in N\right)
$$
I proceed this way
As $|\sin x|<1$
Integration $|\frac{\sin x}{1+x^a}|<|\frac{1}{1+x^a}|$
And using trial and error method to get $a=3$
Any other way ??
|
Continuing from where you stopped we have that $$\left| \frac{1}{1+x^a}\right| \le \left| \frac{1}{1+10^a}\right|$$ on the interval given. Multiplying by $|\sin x|$ tells us that then integral is less than $$\int_{10}^{19}\frac{1}{1+10^a}\mathrm dx=\frac{9}{1+10^a}.$$
For this quantity to be less than $1/9$ we must have $${1+10^a}{9}>9,$$ or $1+10^a>81,$ which gives $$10^a>80.$$ The first integer $a$ satisfying this condition is $2.$ Hence the first odd such integer is $3.$
|
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|
Solving a second-order recurrence relation with complex characteristic roots in polar form. I am self-studying this topic from a textbook and am stuck with trying work through one example.
Suppose we are solving the recurrence equation, $u_n = 2u_{n-1} - 2u_{n-2}$.
This has the characteristic equation $r^2 - 2r + 2 = 0$, which has two characteristic complex roots $1\pm i$. (assume this is correct)
The complex roots can be written in three forms (modulus = $\sqrt{2}$, arguments = $\pm\frac{\pi}{4}$).
\begin{align*}
\text{Rectangular: }&1 \pm i \\
\text{Polar: }&\sqrt{2}\left(\cos\frac{\pi}{4} \pm i\sin\frac{\pi}{4}\right) \\
\text{Exponential: }&\sqrt{2}e^{\pm\frac{\pi i}{4}}
\end{align*}
The general solution to this recurrence relation with two roots is: $u_n = A(\text{root}_1)^n + B(\text{root}_2)^n$, where $A$ and $B$ are arbitrary constants (assume this is correct).
I imagine the complex roots can be used in any of their three forms in the general solution, but am having particular trouble with the polar form.
\begin{align*}
u_n &= A\left(\sqrt{2}\left(\cos \frac{\pi}{4} + i\sin \frac{\pi}{4}\right)\right)^n + B\left(\sqrt{2}\left(\cos \frac{\pi}{4} - i\sin \frac{\pi}{4}\right)\right)^n && \\
&= A (\sqrt{2})^n\left( \cos\frac{n\pi}{4} + i\sin\frac{n\pi}{4} \right) + B (\sqrt{2})^n\left( \cos\frac{n\pi}{4} - i\sin\frac{n\pi}{4} \right) && \text{Using De Moivre's theorem} \\
&= (\sqrt{2})^n\left( (A+B)\cos\frac{n\pi}{4} + (A-B)i\sin\frac{n\pi}{4} \right) && \text{Switching $A\pm B$ for other arbitrary constants} \\
&= (\sqrt{2})^n\left( C\cos\frac{n\pi}{4} + D\begingroup\color{red}i\endgroup\sin\frac{n\pi}{4} \right)
\end{align*}
I am not sure where I went wrong, but this result does not agree with the result in the textbook I am studying from, in that the imaginary number (colored in red) is absent from the textbook result. I might have thought that is a texbook error, but it explicitly draws attention to the fact that using the general solution in polar form includes only real numbers!
The textbook example does not show the derivation in steps and seems to assume this is straightforward. Also it does not explicitly show that complex roots in exponential form can be used (can they?).
|
Solving the characteristic polynomial is just a way of providing linearly independent solutions... In fact, both the real and imaginary parts will be solutions to your equation. So, the general solution will be in fact
$$
(\sqrt{2})^n (C \cos \frac{n \pi}{4} + D \sin \frac{n \pi}{4}).
$$
Very simply speaking, you have space of dimension 2, and you come up with two linearly independent vectors, which will of course form a basis.
|
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|
Problem with $a\sin(x)+b\cos(x)=\pm\sqrt{a^2+b^2}\sin\left(\arctan\left(\frac{b}{a}\right)+x \right) $
Consider $f(x)=a\sin(x)+b\cos(x)$
where $a,b$ are some real constants.
Putting $f(x)=R\sin(\alpha+x)$, I got
$$f(x)=\pm\sqrt{a^2+b^2}\sin\left(\arctan\left(\frac{b}{a}\right)+x \right) \tag{1}$$
According to my Desmos graphs:
$$f(x)=+\sqrt{a^2+b^2}\sin\left(\arctan\left(\frac{b}{a}\right)+x \right) \tag{2}$$
for $a>0$
and
$$f(x)=-\sqrt{a^2+b^2}\sin\left(\arctan\left(\frac{b}{a}\right)+x \right) \tag{3}$$
for $a<0$.
So the sign $\pm$ taken is independent of the value of $b$.
I tried to prove this by discussing the possible values taken by each of $a$ and $b$: e.g. when $a,b>0$
$$0<\arctan\left(\frac{b}{a}\right)<\frac{\pi}{2} \tag{4}$$
But I struggled to proceed in discussing $$\sin\left(\arctan\left(\frac{b}{a}\right)+x \right) $$ because $\sin()$ can take any values regardless of what $a,b$ are.
Is there a way to determine which sign should be taken in the RHS of $(1)$?
|
Denote
$$
\alpha = \arctan \frac{b}{a}.
$$
Then evidently
$$
\tan \alpha = \frac{b}{a}.
$$
Let's start with identity
$$
\sin^2 \alpha + \cos^2 \alpha = 1
$$
and divide each term by $\cos^2 \alpha$ to get
$$
\tan^2 \alpha + 1 = \frac{1}{\cos^2 \alpha} \implies \cos^2 \alpha = \frac{1}{1+\tan^2 \alpha} = \frac{a^2}{a^2 + b^2}.
$$
This gives us also
$$
\sin^2 \alpha = \frac{b^2}{a^2 + b^2}.
$$
Now (looking at $\alpha$) you can decide when one should take positive/negative value for $\sin$ and $\cos$.
Suppose, for example
$$
0 \le \alpha < \frac{\pi}{2}.
$$
This means that both sine and cosine will be nonnegative and we have
$$
\sin \alpha = \frac{|b|}{\sqrt{a^2 + b^2}}, \; \cos \alpha = \frac{|a|}{\sqrt{a^2 + b^2}}.
$$
This will give us
$$
\sin(\alpha+x ) = \sin \alpha \cos x + \cos \alpha \sin x = \frac{|a|}{\sqrt{a^2 + b^2}} \sin x + \frac{|b|}{\sqrt{a^2 + b^2}} \cos x.
$$
Now let's consider the case
$$
0 > \alpha > -\frac{\pi}{2}.
$$
This means that sine will be negative and cosine will be positive. So
$$
\sin \alpha = -\frac{|b|}{\sqrt{a^2 + b^2}}, \; \cos \alpha = \frac{|a|}{\sqrt{a^2 + b^2}}.
$$
$$
\sin(\alpha+x ) = \sin \alpha \cos x + \cos \alpha \sin x = \frac{|a|}{\sqrt{a^2 + b^2}} \sin x - \frac{|b|}{\sqrt{a^2 + b^2}} \cos x.
$$
|
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|
Find the angle $x$ in this triangle
This image was doing the rounds on a popular text messaging application, so I decided to give it a try.
From sine rule in $\triangle ABP$:
$$\frac{AB}{\sin(150^\circ)} = \frac{AP}{\sin(10^\circ} \\ \implies AP = 2AB \sin(10^\circ)$$
Applying sine rule again in $\triangle APC$:
$$\frac{AP}{\sin(60^\circ + x)} = \frac{AC}{\sin(x)}$$
Manipulating the equation and using some properties gives us
$$x = \arctan\left(\frac{\sqrt 3}{4\sin(10^\circ) - 1}\right)$$
This gives $x = -80^\circ$, but since it's an arctan, $x = 100^\circ$. Also, since $\sin(x) = \sin(\pi - x)$, $x = 80^\circ$ as well.
My question is: Is there a way to solve this problem that does not require a calculator? I tried to chase angles but that did not work out in this case. This solution requires computing $\sin(10^\circ)$ as well as the $\arctan$ of that expression, which needs a calculator.
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Let $\angle ACP =z $. By trigonometric form of Ceva's theorem we have:
$$\frac {\sin60^\circ}{\sin20^\circ}\frac {\sin10^\circ}{\sin40^\circ}\frac {\sin(50^\circ-z)}{\sin z}=1.\tag1
$$
Further we have the following property for product of sines:
$$
\prod_{k=1}^{n-1}2\sin\frac{k\pi}n=n.
$$
Particularly for $n=9$ it gives
$$
(2^4\sin20^\circ\sin40^\circ\sin60^\circ\sin80^\circ)^2=9
\implies\sin20^\circ\sin40^\circ\sin60^\circ\sin80^\circ=\frac3{16}.\tag2
$$
Combining (1) and (2) we obtain:
$$
\frac {\sin z}{\sin(50^\circ-z)}=\frac{\sin^2 60^\circ\sin10^\circ\sin80^\circ}{\frac3{16}}
=4\sin10^\circ\cos10^\circ=\frac{\sin20^\circ}{\sin30^\circ}\implies z=20^\circ,
$$
and finally $$x=180^\circ-60^\circ-z=100^\circ.$$
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|
Find the minimum of $x^3+\frac{1}{x^2}$ for $x>0$ Finding this minimum must be only done using ineaqualities.
$x^3+\frac{1}{x^2}=\frac{1}{2}x^3+\frac{1}{2}x^3+\frac{1}{3x^2}+\frac{1}{3x^2}+\frac{1}{3x^2}$
Using inequalities of arithemtic and geometric means:
$\frac{\frac{1}{2}x^3+\frac{1}{2}x^3+\frac{1}{3x^2}+\frac{1}{3x^2}+\frac{1}{3x^2}+1}{6}\geqslant
\sqrt[6]{\frac{1}{2}x^3\frac{1}{2}x^3\frac{1}{3x^2}\frac{1}{3x^2}\frac{1}{3x^2}}=\sqrt[6]{\frac{1}{108}}\Rightarrow x^3+\frac{1}{x^2}\geqslant 6\sqrt[6]{\frac{1}{108}}-1 $
Sadly $\ 6\sqrt[6]{\frac{1}{108}}-1$ is not correct answer, it is not the minimum.
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We need to find a positive number $a$ such that
$$x^3+{1\over x^2}\ge a^3+{1\over a^2}$$
for all $x\gt0$. The existence of such an $a$ is not in doubt, since $x^3+{1\over x^2}\to\infty$ as $x\to0$ and $x\to\infty$. But
$$\begin{align}
x^3+{1\over x^2}\ge a^3+{1\over a^2}
&\iff(x^3-a^3)-\left({1\over a^2}-{1\over x^2}\right)\ge0\\
&\iff(x-a)\left((x^2+ax+a^2)-{x+a\over a^2x^2} \right)\ge0
\end{align}$$
Now $x-a$ changes sign at $x=a$, so in order for nonnegativity to be maintained, the other factor, $(x^2+ax+a^2)-{x+a\over a^2x^2}$, must do so as well. In particular that factor must also equal $0$ at $x=a$, so we must have
$$3a^2={a+a\over a^2a^2}={2\over a^3}$$
or $a=\sqrt[5]{2/3}$.
|
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|
How to Prove $\int_{0}^{\infty}\frac {1}{x^8+x^4+1}dx=\frac{π}{2\sqrt{3}}$ Question:- Prove that
$\int_{0}^{\infty}\frac {1}{x^8+x^4+1}dx=\frac{π}{2\sqrt{3}}$
On factoring the denominator we get,
$\int_{0}^{\infty}\frac {1}{(x^4+x^2+1)(x^4-x^2+1)}dx$
Partial fraction of the integrand contains big terms with their long integral.So i didn't proceed with partial fraction.I'm unable figure out any other method.I think that there might be some other method for evaluation of this definite integral since its value is
$\frac{π}{2\sqrt{3}}$.
Does anyone have nice way to solve it!
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Since your function is even, your integral is$$\require{cancel}\frac12\int_{-\infty}^\infty\frac{\mathrm dx}{x^8+x^4+1}.$$On the other hand, $x^8+x^4+1=\dfrac{x^{12}-1}{x^4-1}$ and therefore the roots of $x^8+x^4+1$ are the roots of order $12$ of $1$ which are not fourth roots of $1$. Among these, those with imaginary part greater than $0$ are $e^{\pi i/6}$, $e^{\pi i/3}$, $e^{2\pi i/3}$ and $e^{5\pi i/6}$. The residue of $\dfrac1{z^8+z^4+1}$ at these points is, respectively, $-\dfrac1{4\sqrt3}$, $-\dfrac i{4\sqrt3}$, $-\dfrac i{4\sqrt3}$, and $\dfrac1{4\sqrt3}$. Therefore, if $R>1$, then, by the residue theorem\begin{multline}\int_{-R}^R\frac{\mathrm dx}{x^8+x^4+1}+\int_{|z|=R,\ \operatorname{Im}z\geqslant0}\frac{\mathrm dz}{z^8+z^4+1}=\\=2\pi i\left(\cancel{-\frac1{4\sqrt3}}-\frac i{4\sqrt3}-\frac i{4\sqrt3}+\cancel{\frac1{4\sqrt3}}\right)=\frac\pi{\sqrt3}\end{multline}and so, since$$\lim_{R\to\infty}\int_{|z|=R,\ \operatorname{Im}z\geqslant0}\frac{\mathrm dz}{z^8+z^4+1}=0,$$we have\begin{align}\int_{-\infty}^\infty\frac{\mathrm dx}{x^8+x^4+1}&=\lim_{R\to\infty}\int_{-R}^R\frac{\mathrm dx}{x^8+x^4+1}\\&=\frac\pi{\sqrt3}.\end{align}
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|
Diophantine equation : $6^m+2^n+2=x^2$
Find $m,n,x\in\mathbb{N}$ such that $6^m+2^n+2=x^2$.
My first approach is to show that for $m,n\geq2$, there exist no solution for $x$ by using modulo $4$.
Case $1$ : $m=1$, $x^2=2^n+8$.
As $n\geq1\implies2\mid RHS\implies2\mid x^2\implies4\mid x^2\implies4\mid LHS\implies 4\mid 2^n\implies n\geq 2$.
The equation can be reduced into $2+2^{n-2}=\bar x^2$ where $2\bar x=x$.
If $n-2\geq2$, $LHS\equiv2$ and $RHS\equiv0,1\mod4$. Therefore $n-2<2\implies n\leq3$.
Checking for $2\leq n\leq3$, we have $m=1,n=3,x=4$ as a solution.
Case $2$ : $n=1$, $x^2=6^m+4$.
$m=1$ is not a solution, therefore $m\geq2\implies 4\mid LHS\implies2\mid x$.
The equation can be reduced into $2^{m-2}3^m+1=\bar x^2$ where $2\bar x=x$.
I do not know how to solve the problem after this step. Any hints or solution is appreciated.
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Assume that $m$ and $n$ are both greater than $1$. Then, we have:
$$x^2 \equiv 6^m+2^n+2 \equiv 0+0+2 \equiv 2 \pmod{4}$$
which is impossible. Thus, we either have $m=1$ or $n=1$.
Case $1$ : $m=1$
Substituting yields:
$$2^n+8=x^2$$
If $n>3$, then $8 \mid x^2$ but $16 \nmid x^2$ which would be a contradiction. Thus, $n \leqslant 3$. It is clear by plugging in values $n \leqslant 3$, that the only solution is:
$$(m,n,x)=(1,3,4)$$
Case $2$ : $n=1$
Substituting yields:
$$6^m+4=x^2 \implies6^m=x^2-4=(x-2)(x+2)$$
Thus, the values $x-2$ and $x+2$ must contain only powers of $2$ and $3$. Clearly, only one of the factors is divisible by $3$, and is thus divisible by $3^m$. The other factor is atleast $3^m-4$, which yields:
$$6^m \geqslant 3^m(3^m-4) \implies 2^m \geqslant 3^m-4 \implies m \leqslant 2$$
Plugging in both $m=1$ and $m=2$ shows that no such solutions exist.
Thus, the only solution is $(m,n,x)=(1,3,4)$.
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|
Prove that $a^2bc + ab^2c + abc^2 \le a^3b+ac^3+b^3c$ Note: $a,b$ and $c$ are positive real numbers.
I tried to use excel and I believe that, after going through a bunch of numbers, this preposition is true. However, I do not know how to prove it mathematically. Can someone help me to prove this question.
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You have to show
$$abc(a+b+c)\leq a^3b+ac^3+b^3c$$
or (since $a,b,c > 0)$
$$a+b+c \leq \frac{a^2}{c} + \frac{c^2}{b} + \frac{b^2}{a}$$
Now, Cauchy-Schwarz inequality helps as follows
$$(a+b+c)^2 = \left(\sqrt c\frac a{\sqrt c}+\sqrt a \frac b{\sqrt a }+\sqrt b \frac c{\sqrt b } \right)^2\stackrel{C.-S.}{\leq} (c+a+b)\cdot \left(\frac{a^2}{c}+ \frac{b^2}{a} + \frac{c^2}{b} \right)$$
Hence,
$$a+b+c \leq \frac{a^2}{c} + \frac{c^2}{b} + \frac{b^2}{a}$$
|
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|
Integrate $\int \frac{x^2-1}{x^3 \sqrt{2x^4-2x^2+1}} \mathop{dx}$ Evaluate $$\int \frac{x^2-1}{x^3 \sqrt{2x^4-2x^2+1}} \mathop{dx}$$
I tried $u=\sqrt{2x^4-2x^2+1}$, $u=\dfrac{1}{x}$ and $u=\sqrt{x}$ but none of these worked.
A friend gave this and said it's from IIT JEE and answer is $$\dfrac{\sqrt{2x^4-2x^2+1}}{2x^2}+C$$
I would like a hint or suggestion.
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$$I = \int \frac{x^2-1}{x^3\sqrt{2x^4-2x^2+1}}\,dx$$
Take $x^4$ factor out from the square root.
$$ = \int\frac{x^2-1}{x^5\sqrt{2-2x^{-2}+x^{-4}}}$$
Now substitute $\sqrt{2-2x^2+x^4}=t$
We have,
$$\,dt = \frac{4(x^{-3}-x^{-5})}{2\sqrt{2-2x^{-2}+x^{-4}}}\,dx$$
$$I = \frac{1}{2}\int dt$$
$$I = \frac{1}{2}\sqrt{2-2x^{-2}+x^{-4}}+C = \frac{\sqrt{2x^4-2x^2+1}}{2x^2}+C$$
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|
On the joint numerical range of a pair of symmetric matrices Proposition 13.4 of Alexander Barvinok's A Course in Convexity shows the existence of the following result:
Let $n\ge 3$. For two $n\times n$ symmetric matrices $A$ and $B$, and a PSD matrix $X$ with $\mbox{trace}(X) = 1$, there exists a unit vector $x$ such that $x^T A x = \mbox{trace}(AX)$ and $x^T B x = \mbox{trace}(BX)$.
This book does not show how to construct such a vector!
In Boyd & Vandenberghe's Convex Optimization, on page 656, there exists a constructive method for a looser version of above (basically, no constraints on the trace of $X$ and consequently, the magnitude of $x$).
I could not so far use their proof for the above stronger version result and basically construct such $x$. Any proof, idea, or help?
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Some thoughts
Let $\alpha = \mathrm{Tr}(AX)$ and $\beta = \mathrm{Tr}(BX)$.
We need to find $x$ such that
\begin{align}
x^\mathsf{T} A x &= \alpha, \\
x^\mathsf{T} B x &= \beta, \\
x^\mathsf{T} x &= 1.
\end{align}
We use the Newton-Raphson method to solve the above system of three quadratic equations.
The Newton-Raphson map is
$$x \mapsto x - J^\mathsf{T}(JJ^\mathsf{T})^{-1} \left(
\begin{array}{l}
x^\mathsf{T} A x - \alpha \\
x^\mathsf{T} B x - \beta \\
x^\mathsf{T} x - 1
\end{array}
\right)
$$
where $J$ is the Jacobian matrix given by
$$J = \left(
\begin{array}{l}
2x^\mathsf{T}A \\
2x^\mathsf{T}B \\
2x^\mathsf{T}
\end{array}
\right).
$$
Example. Let $n=5$,
$$A = \left(\begin{array}{rrrrr} -6 & -2 & 0 & -1 & 3\\ -2 & 2 & 2 & -2 & -4\\ 0 & 2 & 2 & -5 & 2\\ -1 & -2 & -5 & 4 & -2\\ 3 & -4 & 2 & -2 & 0 \end{array}\right),$$
$$B = \left(\begin{array}{rrrrr} 2 & -1 & -3 & -2 & 2\\ -1 & -6 & -4 & -1 & 0\\ -3 & -4 & 0 & -5 & -1\\ -2 & -1 & -5 & -6 & -4\\ 2 & 0 & -1 & -4 & 6 \end{array}\right)$$
and
$$X = \frac{1}{127}\left(\begin{array}{rrrrr} 29 & -9 & -20 & 16 & -1\\ -9 & 35 & 9 & -20 & 6\\ -20 & 9 & 26 & -22 & -8\\ 16 & -20 & -22 & 24 & 5\\ -1 & 6 & -8 & 5 & 13 \end{array}\right).$$
Randomly choose an initial $x_0$. After a few iterations, we get $x \approx [0.5829, 0.5326, 0.2344, -0.4843, 0.2951]^\mathsf{T}$.
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|
Integer solutions to $2^m - 3^n = p \cdot C$ where $m, n , p$ are positive integer variables and $C$ is a positive integer constant. How many solutions are there for equation $2^m - 3^n = p \cdot C$ where $m, n, p$ are positive integer variables and $C$ is an odd positive integer constant greater than $3$?
Can we say that if there will be none, one, finitely many, or infinite number of solutions?
What happens if $C$ is prime? We believe that for prime $C$, there is at least one solution. Can we prove it?
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There are infinitely many solutions. Let $C=11$, $n=1$ and $m=10l+8,\exists l\in\mathbb{N}$ then $$2^{m}-3^n\equiv 2^{10l+8}-3\equiv 2^8-3\equiv 0\pmod{11}$$
and thus, we can let natural number $p=\dfrac{2^{10l+8}-3}{11}$ so that $(m,n,p,C)=(10l+8,1,1,11)$ are the solutions to the equation $2^m-3^n=11k$ $\Box$
And when $C$ is prime we can let $m= (C-1)q_1$ , $ n=(C-1)q_2$ for some natural number $q_1,q_2$ so that, by Fermat's little theorem
$$2^m-3^n\equiv1-1\equiv 0\pmod {C}$$
and choose $p=\dfrac{2^{(C-1)q_1}-3^{(C-1)q_2}}{C}$ that are the solutions to $2^m-3^n=pC$ where $C$ is prime, Hence, again there is infinitely many solution for prime $C$.
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|
Rate of convegence of $\frac{E^2[X^n]}{ E[X^{n-1}] E[X^{n+1}]}$ as $n \to \infty$ Let $X$ be a non-negative discrete random variable such that $0\le X\le B$ and$$f(n)=\frac{E^2[X^n]}{E[X^{n-1}]E[X^{n+1}]}$$for $n \ge 1$.
I am interested in the rate of growth of $f(n)$ as $n \to \infty$. In other words, is there a way to quantify how fast $f(n)$ approaches its limit as a function of $n$ and some “property” of $X$?
A few observations:
*
*Upper bound: By Cauchy-Schwarz we have that $f(n) \le 1$.
*Exact limit: It can be shown that $\lim\limits_{n \to \infty} f(n)=1$. So, we are trying to refine this limit.
*Trivial example: If $X$ is supported on $\{0,B\}$, then $f(n)= 1$ for all $n \ge 1$.
*Connection to Strong Cauchy-Schwarz: Essentially, this is a question about the sharper versions of Cauchy-Schwartz inequality. Therefore, using the expression for the correction term in Cauchy-Schwartz we have
$$
f(n)=1-\frac{1}{2} E \left[ \left| \frac{X^\frac{n-1}{2}}{ \sqrt{E[X^{n-1}]}} -\frac{X^\frac{n+1}{2}}{ \sqrt{E[X^{n+1}]}}\right|^2 \right].
$$
I would like to answer this with minimal assumptions, but, if needed, we can assume there is a mass at $B$.
|
Some idea
Let $m\ge 2$. Let $0 \le a_1 < a_2 < \cdots < a_m$ and $\mathrm{Pr}(a_1) = p_1, \mathrm{Pr}(a_2) = p_2, \cdots, \mathrm{Pr}(a_m) = p_m$ with $p_1+p_2+\cdots + p_m = 1$. We have
$$f(n) = \frac{(1+A)^2}{(1+B)(1+C)}$$
where
\begin{align}
A = \frac{\sum_{j=1}^{m-1} p_j a_j^n}{p_m a_m^n}, \
B = \frac{\sum_{j=1}^{m-1} p_j a_j^{n-1}}{p_m a_m^{n-1}}, \
C = \frac{\sum_{j=1}^{m-1} p_j a_j^{n+1}}{p_m a_m^{n+1}}.
\end{align}
To proceed, we need the following result. The proof is given at the end.
Fact 1: Let $x, y, z \ge 0$ and $yz \ge x^2$. Then
$$\frac{(1+x)^2}{(1+y)(1+z)} \ge 1 - (y + z - 2x).$$
Let us proceed. By Cauchy-Bunyakovsky-Schwarz inequality, we have $BC \ge A^2$. By Fact 1, we have
$$1 - (B + C - 2A) \le f(n) \le 1$$
that is
$$1 - \frac{\sum_{j=1}^{m-1} p_j a_j^{n-1}(a_m - a_j)^2 }{p_m a_m^{n+1}}
\le f(n) \le 1. \tag{1}$$
Remark: Actually, $1 - (B + C - 2A)$ is the Taylor expansion of $\frac{(1+A)^2}{(1+B)(1+C)}$ around $A=0, B=0, C=0$.
We have
\begin{align}
1 - \frac{\sum_{j=1}^{m-1} p_j a_j^{n-1}(a_m - a_j)^2 }{p_m a_m^{n+1}}
&\ge 1 - \frac{\sum_{j=1}^{m-1} p_j a_{m-1}^{n-1}(a_m - 0)^2 }{p_m a_m^{n+1}}\\
&= 1 - \frac{1-p_m}{p_m} \left(\frac{a_{m-1}}{a_m}\right)^{n-1}.
\end{align}
Thus, we have
$$1 - \frac{1-p_m}{p_m} \left(\frac{a_{m-1}}{a_m}\right)^{n-1} \le f(n) \le 1. \tag{2}$$
$\phantom{2}$
Proof of Fact 1: After clearing the denominators, it suffices to prove that
$$(y+z + yz - x)^2 - yz (1+y)(1+z) \ge 0.$$
From $yz \ge x^2$, we have $y + z \ge 2\sqrt{yz} \ge 2x$ and thus $0 \le y + z + yz - \sqrt{yz} \le y + z + yz - x$.
It suffices to prove that
$$(y+z + yz - \sqrt{yz})^2 - yz (1+y)(1+z) \ge 0$$
or
$$(y+z + yz)(\sqrt{y} - \sqrt{z})^2 \ge 0.$$
It is true. We are done.
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|
Evaluate integral: $\int_0^{\frac{\pi}{2}}\ln(a^2\cos^2 x +b^2\sin^2x)dx$? For $a,b>0$, show that
$$\int_0^{\frac{\pi}{2}}\ln(a^2\cos^2 x+b^2\sin^2 x) dx=\pi\ln\left(\frac{a+b}{2}\right)$$
I came across to this problem in the book Table of integrals, series and product. So here is my try to prove the closed form.
For $a>b$ then, we can write the $$ a^2\cos^2x +b^2\sin^2x =b^2\cos^2x+b^2\sin^2x + k\cos^2x = b^2 +k\cos^2x $$ where $ k= a^2-b^2$ and similarly, for $ b>a$ we can write $a^2\cos^2x +b^2\sin^2x = a^2+l\sin^2x$ where $l= b^2-a^2$. Hence we have $$ \int_0^{\frac{\pi}{2}}\ln(b^2+k\cos^2x )dx=\\
\int_0^{\frac{\pi}{2}}\left(\ln(b^2) + \ln\left(1+\frac{k}{b^2}\cos^2x\right)\right)dx\cdots(1)$$ and $$\int_0^{\frac{\pi}{2}}\ln(a^2+l\sin^2x )dx \\= \int_0^{\frac{\pi}{2}}\left(\ln(a^2) + \ln\left(1+\frac{l}{a^2}\sin^2x\right)\right)dx, \; \; b>a$$ Since $\left| \frac{k}{b^2}\cos^2 x\right|\leq 1$ so we use the Machlaurin series of $\ln(1+x)$ for $|x|<1$ giving us $$ \pi \ln(b)+\sum_{m\geq 1} \frac{(-1)^{m-1}}{m}\left(\frac{k}{4b^2}\right)^m\int_0^{\frac{\pi}{2}}\cos^{2m}xdx$$ The latter integral is well know result know as Wallis integral thus gives us $$ \frac{\pi}{2}\sum_{m\geq 1} \frac{(-1)^{m+1}}{m}\left(\frac{k}{4b^2}\right)^m{2m\choose m}\cdots(2)$$ Recalling the generating function of Central binomial coefficients for $|y| <\frac{1}{4}$, ie $$\sum_{m\geq 1} {2m\choose m} y^{m}=\frac{1}{\sqrt{1-4y}}-1.$$ Dividing by $y$ and an integrating from $0$ to $z$ we have $$ \sum_{m\geq 1}\frac{1}{m}{2m\choose m} z^{m} =-2\ln\left(\frac{\sqrt{1-4z}+1}{2}\right) $$ multiplying thoroughly by $-1$ and set $z=-\frac{k}{4b^2}=-\frac{a^2-b^2}{4b^2}$ gives us $$\sum_{m\geq 1} \frac{(-1)^{m+1}}{m}\left(\frac{k}{4b^2}\right)^m{2m\choose m}\\ = 2\ln\left(\frac{{a+b}}{2}\right)-2\ln b $$ From $(1)$ and $(2)$ we have $$\pi\ln(b)+\pi\ln\left(\frac{a+b}{2}\right)-\pi\ln b \\= \pi\ln\left(\frac{a+b}{2}\right)$$ Similarly, for the case of $b>a$ we replace $ \cos^2x$ by $\sin^2x$, $k$ by $ l$. We obtained the same desired result. Moreover, we can also have the following result.
$$\int_0^{\frac{\pi}{2}}\ln(a^2\cos^2 x+b^2\sin^2 x) dx=\frac{1}{2}\int_0^{\pi}\ln(a^2\cos^2x+b^2\sin^2x)dx=\pi\ln\left(\frac{a+b}{2}\right)$$
Now I'm wishing to know other different approaches/proofs for the aforementioned integral.
Thank you.
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Let
$$ I(a)=\int_{0}^{\frac{\pi}{2}} \ln(a^2 \cos^2 x+b^2 \sin^2 x) dx $$
Thus
$$ I(b)=\int_{0}^{\frac{\pi}{2}} \ln(b^2 \cos^2 x+b^2 \sin^2 x) dx=\pi \ln b $$
\begin{align}
I'(a)&=\int_{0}^{\frac{\pi}{2}} \frac{2a\cos^2 x}{a^2 \cos^2 x+b^2 \sin^2 x}dx \\
&=2a \int_{0}^{\frac{\pi}{2}} \frac{d(\tan x)}{(a^2+b^2 \tan^2 x)(1+\tan^2 x)} \\
&=\frac{\pi}{a+b} \\
\end{align}
Thus
$$ \int_b^a I'(a) da=\int_b^a \frac{\pi}{a+b} da $$
$$ I(a)-I(b)=\pi (\ln(a+b)-\ln(2b)) $$
$$ I(a)=\pi \ln \frac{a+b}{2} $$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3733804",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "13",
"answer_count": 2,
"answer_id": 1
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|
Solve for $y$ in $\frac{dy}{dx}-\frac{3y}{2x+1}=3x^2$ I saw a challenge problem on social media by a friend, solve for $y$ in $$\frac{dy}{dx}-\frac{3y}{2x+1}=3x^2$$
I think this is an integration factor ODE
$$\frac{1}{{(2x+1)}^{\frac{3}{2}}} \cdot \frac{dy}{dx}-\frac{3y}{{(2x+1)}^{\frac{5}{2}}}=\frac{3x^2}{{(2x+1)}^{\frac{3}{2}}}$$
Is this correct?
$$\left(\frac{y}{{(2x+1)}^{\frac{3}{2}}} \right)'=\frac{3x^2}{{(2x+1)}^{\frac{3}{2}}}$$
$$\left(\frac{y}{{(2x+1)}^{\frac{3}{2}}} \right)=\int \frac{3x^2}{{(2x+1)}^{\frac{3}{2}}} \mathop{dx}$$
|
$$\frac{dy}{dx} -\frac{3}{2x + 1}y = 3x^2$$
For an integrating factor we have
$$\frac{dy}{dx} + Py = Q, \quad I = \exp(\int P \; dx)$$
then $$Iy = \int IQ\;dx$$
For our method
$$I = \exp(\int -\frac{3}{2x + 1} \; dx) = \exp(-\frac{3}{2}ln(2x + 1)) = (2x + 1)^{-\frac{3}{2}}$$
so
$$y = (2x + 1)^\frac{3}{2} \int \frac{3x^2}{(2x + 1)^{\frac{3}{2}}} \; dx= C_1(2x + 1)^\frac{3}{2} + (2x + 1)(x^2 - 2x -2)$$
since the integral gives
$$\frac{x^2 - 2x - 2}{(2x + 1)^\frac{1}{2}}$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3734196",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 1
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|
$\int \frac{x^3+3x+2}{(x^2+1)^2 (x+1)} \ dx$ without using partial fraction decomposition One way to evaluate the integral $$\int \frac{x^3+3x+2}{(x^2+1)^2(x+1)} \ dx $$ is to rewrite it as $$ \int \frac{x^3+x+2x+2}{(x^2+1)^2(x+1)} dx \\=\int\frac{x(x^2+1) +2(x+1)}{(x^2+1)^2(x+1)}dx\\=\int\frac{x}{(x^2+1)(x+1)}dx+\int\frac{2}{(x^2+1)^2}dx$$ and then proceed by using partial fraction decomposition on the first integral. The second integral could be dealt with by substituting $x=\tan \theta$.
Is there a way to evaluate this integral without using partial fractions, and preferably without splitting it into two integrals as I did here?
|
$$\operatorname*{Res}_{z=-1}\frac{z^3+3z+2}{(z^2+1)^2(z+1)}=\lim_{z\to -1}\frac{z^3+3z+2}{(z^2+1)^2}=-\frac{1}{2}$$
so we know in advance that the integrand function plus $\frac{1}{2(x+1)}$ can be written as $\frac{p(x)}{(x^2+1)^2}$:
$$ \frac{x^3+3x+2}{(x^2+1)^2(x+1)}+\frac{1}{2(x+1)} = \frac{x^3+x^2+x+5}{2(x^2+1)^2}=\frac{x}{2(x^2+1)}+\frac{x^2+5}{2(x^2+1)^2}. $$
This leads to the decomposition
$$ \frac{x^3+3x+2}{(x^2+1)^2(x+1)}=-\frac{1}{2(x+1)}+\frac{x}{2(x^2+1)}+\frac{1}{2(x^2+1)}+\frac{2}{(x^2+1)^2} $$
and also to
$$\int\frac{x^3+3x+2}{(x^2+1)^2(x+1)}\,dx= -\frac{1}{2}\log(x+1)+\frac{1}{4}\log(x^2+1)+\frac{1}{2}\arctan(x)+2\int\frac{dx}{(x^2+1)^2} $$
where the last integral is immediately solved by $x\mapsto\tan\theta$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3735451",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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|
Equations via Lambert's W function I'm studying Lambert's W function and I came across the equation $2^x = 2x$.
Upon inspection it is easy to see that $x = 1$ and $x = 2$ are the real solutions to the equation.
Solving for Lambert's W function, we have:
$$
\begin{split}
&2^x = 2x \ \Rightarrow \ x2^{-x} = \frac{1}{2} \ \Rightarrow \ -x\log 2 \ e^{-x\log 2} = -\frac{\log 2}{2} \ \Rightarrow \ W(-x \log 2\ e^{-x\log 2}) = W\biggl(-\frac{\log 2}{2}\biggr) \ \Rightarrow\\
&-x\log 2 = W\biggl(-\frac{\log 2}{2}\biggr) \ \Rightarrow \ x = -\frac{1}{\log 2}W\biggl(-\frac{\log 2}{2}\biggr)
\end{split}
$$
But, $-\frac{1}{e} < -\frac{\log 2}{2} < 0$. Thus,
$$
x =
\begin{cases}
-\frac{1}{\log 2}W_0\biggl(-\frac{\log 2}{2}\biggr) = -\frac{1}{\log 2}W_0[2^{-1}\log(2^{-1})] = -\frac{1}{\log 2}\cdot \log(2^{-1}) = 1\\
-\frac{1}{\log 2}W_-1\biggl(-\frac{\log 2}{2}\biggr) = -\frac{1}{\log 2}W_{-1}[2^{-1}\log(2^{-1})] = -\frac{1}{\log 2}\cdot 2\log(2^{-1}) = 2
\end{cases}
$$
Why $W_{-1}[2^{-1}\log(2^{-1})] = -2\log 2$? Where did factor $2$ come from?
|
$\require{begingroup} \begingroup$
$\def\e{\mathrm{e}}\def\W{\operatorname{W}}\def\Wp{\operatorname{W_0}}\def\Wm{\operatorname{W_{-1}}}$
When the argument of $\W(u)$ is in the range $(-\tfrac1\e,0)$,
the number $u$ always can be expressed in two equivalent forms:
\begin{align}
u&=\ln\left(a^{\tfrac 1{1-a}} \right)\cdot a^{\tfrac 1{1-a}}
\tag{1}\label{1}
\\
&=\ln\left(a^{\tfrac a{1-a}} \right)\cdot a^{\tfrac a{1-a}}
\tag{2}\label{2}
.
\end{align}
Indeed, multiplying \eqref{1} by $1=a\cdot\frac1a$, we have
\begin{align}
u&=
a\cdot\tfrac 1a\cdot\ln\left(a^{\tfrac 1{1-a}} \right)\cdot a^{\tfrac 1{1-a}}
\tag{3}\label{3}
\\
&=
a\cdot\ln\left(a^{\tfrac 1{1-a}} \right)\cdot a^{\tfrac 1{1-a}}\cdot \tfrac 1a
\tag{4}\label{4}
\\
&=
\ln\left(a^{\tfrac a{1-a}} \right)\cdot a^{\tfrac 1{1-a}-1}
=
\ln\left(a^{\tfrac a{1-a}} \right)\cdot a^{\tfrac 1{1-a}-\tfrac{1-a}{1-a}}
=
\ln\left(a^{\tfrac a{1-a}} \right)\cdot a^{\tfrac a{1-a}}
,
\tag{5}\label{5}
\end{align}
that is, \eqref{1}$\equiv$\eqref{2}.
Recall that the expression of the form $\W(t\exp{t})=t$,
just by the definition of $\W$, ignoring any questions about branches of $\W$.
So, applying $\W$ to both forms of $u$ in \eqref{1}, \eqref{2},
we have two distinct results,
\begin{align}
\W\left( \ln\left(a^{\tfrac 1{1-a}} \right)\cdot a^{\tfrac 1{1-a}} \right)
&=
\ln\left(a^{\tfrac 1{1-a}} \right)
=\frac{\ln a}{1-a}
\tag{6}\label{6}
\\
\text{and }
\W\left( \ln\left(a^{\tfrac a{1-a}} \right)\cdot a^{\tfrac a{1-a}} \right)
&=
\ln\left(a^{\tfrac a{1-a}} \right)
=\frac{a\ln a}{1-a}
\tag{7}\label{7}
.
\end{align}
The last two expressions are known as
Parametric representation of the real branches of the Lambert W function.
Note that both values obtained from \eqref{6}, \eqref{7}
are negative, and one of them
is always greater than $-1$
(belongs to the branch $\Wp$),
while the other is less than $-1$
(belongs to the branch $\Wm$).
Note also, that the positive parameter $a$ can be either less, or greater than $1$.
When $a<1$, the expression \eqref{7} is recognized as $\Wp(u)$
while the expression \eqref{6} is recognized as $\Wm(u)$
and $a=\tfrac{\Wp(u)}{\Wm(u)}$.
In context of the original question, we have
for $a=\tfrac12$,
\begin{align}
u&=\tfrac12\cdot\ln\tfrac12
\tag{8}\label{8}
\\
&=\ln\left(a^{\tfrac a{1-a}} \right)\cdot a^{\tfrac a{1-a}}
=\ln\left(a^{\tfrac 1{1-a}} \right)\cdot a^{\tfrac 1{1-a}}
\tag{9}\label{9}
\\
&=\tfrac12\cdot\ln\tfrac12
=\tfrac14\cdot\ln\tfrac14
\tag{10}\label{10}
.
\end{align}
\begin{align}
\W(u)&=\W(\tfrac12\cdot\ln\tfrac12)=\ln\tfrac12
=-\ln2
=\Wp(u)\approx -0.69314718>-1
\tag{11}\label{11}
,\\
\W(u)&=\W(\tfrac14\cdot\ln\tfrac14)=\ln\tfrac14
=-2\ln2
=\Wm(u)\approx -1.38629436<-1
\tag{12}\label{12}
.
\end{align}
$\endgroup$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3735711",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
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|
Recursive squaring algorithm by squaring five numbers From Erickson (PDF):
**(c) Describe a recursive algorithm that squares any $n$-digit number in $O(n^{\log_3{5}})time$, by reducing to squaring only five $\left( n/3+O(1)\right)$-digit numbers. [Hint: What is $(a+b+c)^2+(a−b+c)^2$?]
Solving the hint,
$$\begin{align}
&(a+b+c)^2+(a−b+c)^2 \\
&= \left(a^2+b^2+c^2+2ab+2ac+2bc\right) + \left(a^2+b^2+c^2-2ab+2ac-2bc\right) \\
&= 2\left(a^2+b^2+c^2+2ac\right) \\
&= 2\left(b^2+(a+c)^2\right) \\
\end{align}$$
Let the given $n$-digit number take the form of $10^{2m}x+10^{m}y+z$.
Then
$$\begin{align} &(10^{2m}x+10^{m}y+z)^2 \\
&= 10^{4m}x^2 + 10^{2m}y^2 + z^2 + 10^{3m}2xy + 10^{2m}2xz + 10^{m}2yz \\
&= 10^{4m}x^2 + z^2 + 10^{3m}2xy + 10^{m}2yz + 10^{2m}(y^2 + 2xz) \\
&= 10^{4m}x^2 + z^2 + 10^{3m}2xy + 10^{m}2yz + 10^{2m}(y^2 + (x+z)^2 - x^2 - z^2) \\
&= (10^{4m} - 10^{2m})x^2 + (1 - 10^{2m})z^2 + 10^{3m}2xy + 10^{m}2yz + 10^{2m}(y^2 + (x+z)^2) \\
&= (10^{4m} - 10^{2m})\bbox[yellow]{x^2} + (1 - 10^{2m})\bbox[yellow]{z^2} + \bbox[pink]{10^{m}2y\left(10^{2m} x + z\right)} + \frac{10^{2m}}{2}\left(\bbox[yellow]{(x+y+z)^2} + \bbox[yellow]{(x-y+z)^2}\right) \\
\end{align}
$$
At this point I've gotten 4 out of the 5 squares required but I'm stuck at converting the last term into some kind of square. I've tried expanding $(a+b+c)^2-(a−b+c)^2$ but it leads to $2b(a+c)$ which somewhat resembles the last term in pink but not quite.
Am I doing this right?
|
Apparently the algorithm that the question is looking for simply Toom-Cook, and that the missing last square is $\left(x(-2)^2 + y(-2) + z\right)^2 = (4x-2y+z)^2$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3736216",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
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|
Simplify $\frac{4\cos ^2\left(2x\right)-4\cos ^2\left(x\right)+3\sin ^2\left(x\right)}{4\sin ^2\left(x\right)-\sin ^2\left(2x\right)}$
Simplify:
$$\frac{4\cos ^2\left(2x\right)-4\cos ^2\left(x\right)+3\sin ^2\left(x\right)}{4\sin ^2\left(x\right)-\sin ^2\left(2x\right)}$$
After the substitution as $\cos(x)=a$ and $\sin(x)=b$, $(a^2+b^2=1)$, the expression becomes
$$\frac{4(a^2-b^2)^2-4a^2+3b^2}{4b^2-4a^2b^2}=\frac{4a^4-8a^2b^2+4b^4-4a^2+3b^2}{4b^4}=\bigg(\frac{a^2}{b^2}-1\bigg)^2-\frac{4a^2-3b^2}{4b^4}$$But I don't think I got anything useful... Any help is appreciated.
|
Another way is to express everything in terms of $\cos 2x$: $$\frac{4\cos^22x -3(\cos^2x-\sin^2x)-\cos^2x}{4\cdot\frac{1-\cos2x}{2} -(1-\cos^22x)} \\=\frac{4\cos^22x-3\cos 2x -\frac{1+\cos 2x}{2}}{2(1-\cos 2x)+\cos^22x-1}\\=\frac{8\cos^22x-7\cos 2x-1}{2(1-\cos 2x)^2}\\=\frac{(8\cos 2x+1)(\cos 2x-1)}{2(\cos 2x -1)^2}\\=\frac{8\cos 2x+1}{2\cos 2x-2}$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3736352",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 2
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|
The derivative of $f(x)=\frac{3 \sin x}{2+\cos x}$ My solution:
The background
$$\begin{align} &3\frac{d}{dx}\left(\frac{\sin(x)}{2+\cos(x)}\right)\\
&=3\frac{\frac{d}{dx}(\sin(x))(2+\cos(x))-\frac{d}{dx}(2+\cos(x))\sin(x)}{(2+\cos(x))^2}\\
&=3\frac{\cos(x)(2+\cos(x))-(-\sin(x))\sin(x)}{(2+\cos(x))^2}\\
&=\frac{3+6\cos(x)}{(2+\cos(x))^2}\end{align}$$
However, I plugged $f(x)=\dfrac{3 \sin x}{2+\cos x}$ into the derivative calculator in wolfram alpha and received the following calculation:
$$\frac{3\left(\sin^2(x)+\cos^2(x)+2\cos(x)\right)}{(2+\cos(x))^2}$$
Is my solution incorrect and the one from wolfram alpha correct? If so, where did I go wrong?
|
Your answer is correct. Use trig. identity $\sin^2(x)+\cos^2(x)=1$
$$\therefore \frac{3(\sin^2(x)+\cos^2(x)+2\cos(x))}{(2+\cos(x))^2}=\frac{3(1+2\cos(x))}{(2+\cos(x))^2}=\frac{3+6\cos(x)}{(2+\cos(x))^2}$$
Above matches your solution
|
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 0
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|
Find the sum of all positive integers $n$ such that when $1^3+2^3+3^3 +\dots+ n^3$ is divided by $n+5$ the remainder is $17.$
Find the sum of all positive integers $n$ such that when $1^3+2^3+3^3 +\dots+ n^3$ is divided by $n+5$ the remainder is $17.$
Letting $k= n+5$ we get that $1^3+2^3+3^3 +\dots+ (k-5)^3 \equiv 17 \text{ (mod $k$)}$.
Knowing the sum of cubes formula we get that ($\frac{(k-5)(k-4)}{2})^2\equiv 17 \text{ (mod $k$)}$.
From here I'm not sure how I should continue. What would be my options?
|
From $$n\equiv -5 \pmod{n+5} \Rightarrow \\
n^2\equiv 25 \pmod{n+5} \Rightarrow\\
n^2+n\equiv 20 \pmod{n+5} \Rightarrow \\
(n^2+n)^2\equiv 400 \pmod{n+5} \Rightarrow \\
n+5 \mid 4\cdot\left(\frac{n^2(n+1)^2}{4}-100\right)$$
but we want
$$\sum\limits_{k=1}^n k^3=\frac{n^2(n+1)^2}{4}=Q\cdot (n+5)+17$$
or
$$n+5 \mid 4\cdot\left(Q\cdot (n+5)+17-100\right)=
4\left(Q\cdot (n+5)-83\right)\Rightarrow\\
n+5\mid 2^2\cdot 83$$
which yields the following options to check
$$n+5=83$$
$$n+5=2\cdot83$$
$$n+5=2^2\cdot83$$
The first two will satisfy the initial condition, so $n=78, n=161$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3737447",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 5,
"answer_id": 1
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|
prove $\sum\cos^3{A}+64\prod\cos^3{A}\ge\frac{1}{2}$ In every acute-angled triangle $ABC$,show that
$$(\cos{A})^3+(\cos{B})^3+(\cos{C})^3+64(\cos{A})^3(\cos{B})^3(\cos{C})^3\ge\dfrac{1}{2}$$
I want use Schur inequality
$$x^3+y^3+z^3+3xyz\ge xy(y+z)+yz(y+z)+zx(z+x)$$
then we have
$$x^3+y^3+z^3+6xyz\ge (x+y+z)(xy+yz+zx)$$
But I can't use this to prove my question
and I use this post methods links also can't solve my problem,use $AM-GM $ inequality$$\cos^3{A}+\dfrac{\cos{A}}{4}\ge\cos^2{A}$$
so
$$LHS\ge \sum_{cyc}\cos^2{A}-\dfrac{1}{4}\sum_{cyc}\cos{A}+64\prod_{cyc}\cos^3{A}$$use
$$\cos^2{A}+\cos^2{B}+\cos^2{C}+2\cos{A}\cos{B}\cos{C}=1$$
it must to prove
$$\frac{1}{2}+64\cos^3{A}\cos^3{B}\cos^3{C}\ge 2\cos{A}\cos{B}\cos{C}+\dfrac{1}{4}(\cos{A}+\cos{B}+\cos{C})$$
|
The Euler's identity states that for all $x,y,z$ reals we have
$$x^3+y^3+z^3=3xyz+(x+y+z)(x^2+y^2+z^2-xy-yz-zx)$$
But if $A,B,C$ angles of an acute triangle $ABC$ and $x=\cos A$, $y=\cos B$, $z=\cos C$, we have
$$
x+y+z=1+\frac{\rho}{R},\tag 1
$$
$$
xyz=\frac{\rho_0}{2R}\tag 2
$$
$$
x^2+y^2+z^2=1-\frac{\rho_0}{R}\tag 3
$$
$$
xy+yz+zx=\frac{1}{2}\left(\frac{\rho}{R}\right)^2+\frac{\rho}{R}+\frac{1}{2}\frac{\rho_0}{R}\tag 4
$$
The symbols $R,\rho$ are the circumradius and inradius of $ABC$ resp. The symbol $\rho_0$ is the inradius of triangle $A'B'C'$. The $A'B'C'$ (orthic triangle) is formed by the intersections of heights of $ABC$ with its sides. Hence
$$
\Pi=\sum_{cyc}x^3+64(xyz)^3=
$$
$$
=3\frac{\rho_0}{2R}+\left(1+\frac{\rho}{R}\right)\left(1-\frac{\rho_0}{R}-\frac{1}{2}\left(\frac{\rho}{R}\right)^2-\frac{\rho}{R}-\frac{1}{2}\frac{\rho_0}{R}\right)+64\left(\frac{\rho_0}{2R}\right)^3=
$$
$$
=1-\frac{1}{2}\left(\frac{\rho}{R}\right)^3+8\left(\frac{\rho_0}{R}\right)^3-\frac{3}{2}\left(\frac{\rho}{R}\right)^2-\frac{3}{2}\frac{\rho}{R}\cdot\frac{\rho_0}{R}=
$$
$$
=f(t,t_0):=1-\frac{3t^2}{2}-\frac{t^3}{2}-\frac{3tt_0}{2}+8t_0^3\textrm{, }(t,t_0)\in D=\left[0,\frac{1}{2}\right]\times \left[0,\frac{1}{4}\right],
$$
where (Euler's theorem)
$$
t=\frac{\rho}{R}\leq\frac{1}{2}\textrm{ and }t_0=\frac{\rho_0}{R}\leq\frac{1}{4}
$$
and $0<t\leq\frac{1}{2}$, $0<t_0\leq \frac{1}{4}$. One can see easily that
$$
f(t,t_0)\geq\frac{1}{2}=f\left(\frac{1}{2},\frac{1}{4}\right)\textrm{, }\forall (x,x_0)\in D
$$
This can be done easily and the proof is complete.
|
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"language": "en",
"url": "https://math.stackexchange.com/questions/3737759",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
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|
Suppose that $z = x + iy$, with $y > 0$. Show that there are positive real numbers $u$ and $v$ with $2u^{2} = |z| + x$ and $2v^{2} = |z| - x$ Suppose that $z = x + iy$, with $y > 0$. Show that there are positive real numbers $u$ and $v$ with $2u^{2} = |z| + x$ and $2v^{2} = |z| - x$. Calculate $(u+iv)^{2}$. Show that $z$ has exactly two complex square roots. Show that the same holds when $y < 0$.
MY ATTEMPT
Let us solve this problem backwards:
\begin{align*}
\begin{cases}
2u^{2} = |z| + x\\\\
2v^{2} = |z| - x
\end{cases} \Longleftrightarrow
\begin{cases}
|z| = u^{2} + v^{2}\\\\
x = u^{2} - v^{2}
\end{cases}
\end{align*}
Consequently, one has that
\begin{align*}
|z|^{2} = (u^{2} + v^{2})^{2} = x^{2} + y^{2} = (u^{2} - v^{2})^{2} + y^{2} & \Longleftrightarrow u^{4} + 2u^{2}v^{2} + v^{4} = u^{4} - 2u^{2}v^{2} + v^{4} + y^{2}\\\\
& \Longleftrightarrow y^{2} = 4u^{2}v^{2}
\end{align*}
Thus it suffices to choose $x = u^{2} - v^{2}$ and $y = 2uv$, where $u,v\in\textbf{R}_{\geq0}$.
Thus we conclude that
\begin{align*}
(u+iv)^{2} = u^{2} - v^{2} + 2uvi = x + yi = z
\end{align*}
If $y < 0$, it suffices to take $y = -2uv$.
Finally, I did not get what it means to show that $z$ has exactly two complex roots. Any help on this?
If there is a nice way to approach it, I'd be grateful to know. Any contribution is appreciated.
|
I think you accidentally solved for $x$ and $y$ instead of $u$ and $v$. Remember that $z = x+iy$ is fixed, so we're looking for $u$ and $v$ in terms of $x$ and $y$. I think you can probably reuse most of the work you already have for that, though.
All they mean by showing $z$ has two complex roots is that since you showed that $(u+iv)^2 = z$, we can also write $\sqrt{z} = u+iv$. For the second complex root, you can just take use that $(-(u+iv))^2 = (u+iv)^2 = z$.
|
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"timestamp": "2023-03-29T00:00:00",
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|
If $a,b,c$ are sides of a triangle, then find range of $\frac{ab+bc+ac}{a^2+b^2+c^2}$ $$\frac{ab+bc+ac}{a^2+b^2+c^2}$$
$$=\frac{\frac 12 ((a+b+c)^2-(a^2+b^2+c^2))}{a^2+b^2+c^2}$$
$$=\frac 12 \left(\frac{(a+b+c)^2}{a^2+b^2+c^2}-1\right)$$
For max value, $a=b=c$
Max =$1$
How do I find the minimum value
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In $\Delta ABC$, $a^2=b^2+c^2-2bc \cos A > b^2+c^2-2bc$.
Adding up the similar inequalities gives $ab+bc+ca > \frac{1}{2} (a^2+b^2+c^2)$
|
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|
Can someone explain the limit $\lim _{n \rightarrow \infty} \left(\frac{1}{\sqrt{n^2+1}}+\frac{1}{\sqrt{n^2+2}}+\cdots+\frac{1}{\sqrt{n^2+n}}\right)$? $$\begin{aligned}
&\text { Find the following limit: } \lim _{n \rightarrow \infty} \left(\frac{1}{\sqrt{n^2+1}}+\frac{1}{\sqrt{n^2+2}}+\cdots+\frac{1}{\sqrt{n^2+n}}\right)\\
&\text { **In a solution I got** Ans:- Let } u_n=\frac{n}{\sqrt{n^2+n}}\\
&\therefore \lim _{n \rightarrow \infty} u_n=\lim _{n \rightarrow \infty} \frac{n}{\sqrt{n^{2}+n}}=\lim _{n \rightarrow \infty} \frac{1}{\sqrt{1+\frac{1}{n}}}=1\\
&\text { By Cauchy's first theorem:- } \lim _{n \rightarrow \infty} \left(\frac{u_1+\cdots+u_n}{n}\right)=1\\
&\text{So, } \lim _{n \rightarrow \infty}\left(\frac{1}{\sqrt{n^2+1}}+\frac{1}{\sqrt{n^2+2}}+\cdots+\frac{1}{\sqrt{n^2+n}}\right)=1
\end{aligned}$$
I am unable to understand this. How the general term comes like this! Any such explanation to the whole problem would be greatly appreciated.
|
$$\frac{1}{\sqrt{n^2+n}} \leq\frac{1}{\sqrt{n^2+1}}\leq \frac{1}{\sqrt{n^2}}\\
\frac{1}{\sqrt{n^2+n}} \leq\frac{1}{\sqrt{n^2+2}}\leq \frac{1}{\sqrt{n^2}}\\
\frac{1}{\sqrt{n^2+n}} \leq\frac{1}{\sqrt{n^2+3}}\leq \frac{1}{\sqrt{n^2}}\\
\vdots\\
\frac{1}{\sqrt{n^2+n}} \leq\frac{1}{\sqrt{n^2+n}}\leq \frac{1}{\sqrt{n^2}}.$$ Now look at sum of them
$$\frac{n}{\sqrt{n^2+n}} \leq(\frac{1}{\sqrt{n^2+1}}+...+\frac{1}{\sqrt{n^2+n}})\leq \frac{n}{\sqrt{n^2}}.$$ Now take limit.
|
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|
If $ 1+ \frac{1}{2}+\frac{1}{3}+.....+\frac{1}{100}=\frac{A}{B}$ where $A$ and $B$ are coprime positive integers, then $5\nmid A$ and $5\nmid B$.
Let the sum $$1+ \frac{1}{2}+\frac{1}{3}+.....+\frac{1}{100}=\frac{A}{B}$$ where $A,B\in \mathbb{N}$ and $\gcd(A,B)=1$. Show that neither $A $ nor $B $ is divisible by $5$.
My attempt: $$\begin{align}1+ \frac{1}{2}+\frac{1}{3}+\ldots+\frac{1}{100}&=(1+\frac{1}{100})+(\frac{1}{2}+\frac{1}{99})+\cdots +(\frac{1}{50}+\frac{1}{51})\\&=101(\frac{1}{100}+\frac{1}{2\times99}+\frac{1}{3\times 98}+\ldots+\frac{1}{51 \times 50})\,.\end{align}$$ denominator is $100!$ and numerator is multiple of $101$. Now how to prove the statement? Please help me. Thanks in advance.
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Let $S:=\displaystyle\sum_{k=1}^{100}\,\dfrac1{k}$, and write $[n]:=\{1,2,\ldots,n\}$ for each positive integer $n$. Note that $B$ is not divisible by $5$ because
$$\begin{align}S&=\sum_{\substack{k\in[100]\\ 5\nmid k}}\,\dfrac1k +\frac15\,\sum_{\substack{k\in[20]\\5\nmid k}}\frac1k+\frac1{25}\,\sum_{k=1}^4\,\frac1k\tag{1}
\\&=\sum_{\substack{k\in[100]\\ 5\nmid k}}\,\dfrac1k +\frac15\,\sum_{t=0}^{3}\,\sum_{r=1}^4\frac1{5t+r}+\frac{1}{25}\cdot\frac{25}{12}\tag{2}
\\&=\sum_{\substack{k\in[100]\\ 5\nmid k}}\,\dfrac1k +\frac15\,\sum_{t=0}^{3}\,\left(\frac{10t+5}{(5t+1)(5t+4)}+\frac{10t+5}{(5t+2)(5t+3)}\right)+\frac{1}{12}\tag{3}
\\&=\sum_{t=0}^{19}\,\sum_{r=1}^4\,\frac{1}{5t+r} +\sum_{t=0}^{3}\,(2t+1)\left(\frac{1}{(5t+1)(5t+4)}+\frac{1}{(5t+2)(5t+3)}\right)+\frac{1}{12}\tag{4}\,.\end{align}$$
Now, $A$ is not divisible by $5$ because, from the equation above,
$$\begin{align}S&\equiv \sum_{t=0}^{19}\,\sum_{r=1}^4\,\frac{1}{5t+r}+\sum_{t=0}^3\,(2t+1)\left(\frac{1}{1\cdot 4}+\frac{1}{2\cdot 3}\right)+\frac{1}{12}\pmod{5}\tag{5}
\\&\equiv \sum_{t=0}^{19}\,\sum_{r=1}^4\,\frac{1}{r}+\left(\frac{1-5}{1\cdot 4}+\frac{1+5}{2\cdot 3}\right)\,\sum_{t=0}^3\,(2t+1)+\frac{1-25}{12}\pmod{5}\tag{6}
\\&\equiv \sum_{t=0}^{19}\,\frac{25}{12}+\big((-1)+1\big)(1+3+5+7)-2\pmod{5}\tag{7}
\\&\equiv 0+0\cdot 16-2 =-2\not\equiv 0\pmod{5}\tag{8}\,.\end{align}$$
Exercise for the Reader. Let $\dfrac{A}{B}=\sum\limits_{k=1}^{20}\,\dfrac{1}{k}$, where $A$ and $B$ are relatively prime positive integers. Prove that $5$ divides $A$ but $25$ does not divide $A$. Prove also that $5$ does not divide $B$.
|
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|
Given matrix $A^2$, how to find matrix $A$?
Let $$A^2 = \begin{pmatrix} 3 & 1 \\ 2 & 2 \end{pmatrix}$$ Knowing that $A$ has positive eigenvalues, what is $A$?
What I did was the following:
$$A = \begin{pmatrix}
a & b \\
c & d
\end{pmatrix}$$
so
$$A^2 = \begin{pmatrix}
a^2 + bc & ab+bd \\
ac+cd & bc+d^2
\end{pmatrix}$$
I got stuck here after trying to solve the 4 equations. Can someone help, please?
|
Computing matrix powers can be done with diagonalization.
The eigenvalues of $A^2$ have sum $5$ (trace) and product $4$ (determinant), so they are $1$ and $4$.
The corresponding eigenvectors of $A^2$ are $\pmatrix{1\\-2}$ and $\pmatrix{1\\1}$, respectively.
Therefore, $A^2$ is diagonalized as follows: $\pmatrix{1&1\\-2&1}\pmatrix{1&0\\0&4}\pmatrix{1&1\\-2&1}^{-1}=\pmatrix{1&1\\-2&1}\pmatrix{1&0\\0&4}\dfrac{\pmatrix{1&-1\\2&1}}3.$
Therefore, we can take $A=\pmatrix{1&1\\-2&1}\pmatrix{1&0\\0&2}\dfrac{\pmatrix{1&-1\\2&1}}3=\dfrac{\pmatrix{5&1\\2&4}}3$.
|
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|
Finding the equation of the normal to the parabola $y^2=4x$ that passes through $(9,6)$
Let $L$ be a normal to the parabola $y^2 = 4x$. If $L$ passes through the point $(9, 6)$, then $L$ is given by
(A) $\;y − x + 3 = 0$
(B) $\;y + 3x − 33 = 0$
(C) $\;y + x − 15 = 0$
(D) $\;y − 2x + 12 = 0$
My attempt: Let $(h,k)$ be the point on parabola where normal is to be found out. Taking derivative, I get the slope of the normal to be $\frac{-k}{2}$. Since the normal passes through $(9,6)$, so, the equation of the normal becomes:$$y-6=\frac{-k}{2}(x-9)$$$$\implies \frac{kx}{2}+y=\frac{9k}{2}+6$$
By putting $k$ as $2,-2,-4$ and $6$, I get normals mentioned in $A,B,C$ and $D$ above (not in that order).
But the answer is given as $A,B$ and $D$. What am I doing wrong?
|
The equation of the normal is $\color{red}{y=\frac{-kx}{2}+\frac{9k}{2}+6}$. The point $(h,k)$ lies on it so we get
$$k=\frac{-kh}{2}+\frac{9k}{2}+6.$$
But $k^2=4h$ (since $(h,k)$ lies on the parabola as well), so we get
$$k^3-28k-48=0 \implies (k-6)(k+2)(k+4)=0 \implies k=6,-2,-4.$$
These value can be used to find the equation of the normal as:
\begin{align*}
y&=-3x+33\\
y &=x-3\\
y &=2x-12
\end{align*}
|
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|
Evaluate $\lim_{x\to+\infty} \frac{3x^3+x\cos{\sqrt{x}}}{x^4\sin{\frac{1}{x}}+1}$ Evaluate $$\lim_{x\to+\infty} \frac{3x^3+x\cos{\sqrt{x}}}{x^4\sin{\frac{1}{x}}+1}$$
My attempt: $$\lim_{x\to+\infty} \frac{3x^3+x\cos{\sqrt{x}}}{x^4\sin{\frac{1}{x}}+1}=\lim_{x\to+\infty} \frac{3x^2+\cos{\sqrt{x}}}{x^3\sin{\frac{1}{x}}+\frac{1}{x^3}}$$ Let $t=\frac{1}{x}$, then
$$\lim_{x\to+\infty} \frac{3x^2+\cos{\sqrt{x}}}{x^3\sin{\frac{1}{x}}+\frac{1}{x^3}}=\lim_{t\to0^+} \frac{\frac{3}{t^2}+\cos{\frac{1}{\sqrt{t}}}}{\frac{1}{t^3}\sin{t}+t^3}$$
That's where I got stuck. I think this substitution didn't help much. Maybe there's a way to apply the squeeze theorem, but it's not so obvious. Hint, please?
|
$$\lim_{x\to+\infty} \frac{3x^3+x\cos{\sqrt{x}}}{x^4\sin{\frac{1}{x}}+1}=\lim_{x\to+\infty} \frac{3x^3(1+\frac{x\cos{\sqrt{x}}}{3x^3}) }{x^3 \left( \frac{\sin{\frac{1}{x}}}{\frac{1}{x}} +\frac{1}{x^3} \right)} = 3$$
|
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|
Functions of several variables If $f(x,y) = x^2 + xy + y^2 - 3x + 4y - 5$. I know the domain is $\mathbb R^2$. How to determine the image of f is my issue.
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Have a look at my answer here: https://math.stackexchange.com/a/3619647/399263
You can always translate conics to cancel terms in $x,y$.
$f(x+a,y+b)=x^2+(-3+b+2a)x+xy+y^2+(2b+4+a)y+[\cdots]$
Solve $\begin{cases}2a+b-3=0\\2b+4+a=0\end{cases}\iff \begin{cases}a=\frac{10}3\\b=-\frac {11}3\end{cases}$
$$f(x+a,y+b)=\overbrace{(x^2+xy+y^2)}^{\ge 0}-\dfrac{52}3$$
In this case $x^2+xy+y^2\ge 0$ is always positive (discriminant $-3y^2<0$) with equality for $x=y=0$.
Thus $f$ has a minimum for $(x+a,y+b)=(0,0)\iff (x=-a,y=-b)$ of value $-\dfrac{52}3$.
Also $f$ is unbounded above since for instance $f(x,0)=x^2-3x-5$ is unbounded above.
|
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|
If $\alpha,\beta,\gamma$ are the roots of $x^3+x+1=0$, then find the equation whose roots are: $(\alpha-\beta)^2,(\beta-\gamma)^2,(\gamma-\alpha)^2$ Question:
If $\alpha,\beta,\gamma$ are the roots of the equation, $x^3+x+1=0$, then find the equation whose roots are: $({\alpha}-{\beta})^2,({\beta}-{\gamma})^2,({\gamma}-{\alpha})^2$
Now, the normal way to solve this question would be to use the theory of equations and find the sum of roots taken one at a time, two at a time and three at a time. Using this approach, we get the answer as $(x+1)^3+3(x+1)^2+27=0$. However, I feel that this is a very lengthy approach to this problem. Is there an easier way of doing it?
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Let $a,b,c$ be the roots of $x^3+x+1=0$ so we have $a+b+c=0, ab+bc+ca=1,abc=-1$, so $a^2+b^2+c^2=-2$ and $c^3=-c-1$
We would explore a transformation from $x$ to $y$ to get the required cubic equation of $y$.
Let $$y=(a-b)^2=a^2+b^2-2ab\implies y=-2-c^2+2/c \implies c=\frac{3}{1+y}$$
Replacing $c$ by $x$ we get the required transformation $x=\frac{3}{1+y}$, putting it in the given $x$ equation, we get:
$$\frac{27}{(1+y)^3}+\frac{3}{(1+y)}+1=0 \implies y^3+6y^2+9y+31=0,$$
which is the required cubic equation.
|
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|
Optimizing a quadratic in one variable with parameterized coefficients Recently I was doing a physics problem and I ended up with this quadratic in the middle of the steps:
$$ 0= X \tan \theta - \frac{g}{2} \frac{ X^2 \sec^2 \theta }{ (110)^2 } - 105$$
I want to find $0 < \theta < \frac{\pi}2$ for which I can later take the largest $X$ value that solves this equation, i.e. optimize the implicit curve to maximize $X$.
I tried solving this by implicit differentiation (assuming $X$ can be written as a function of $\theta$) with respect to $\theta$ and then by setting $\frac{dX}{d\theta} = 0$:
\begin{align}
0 &= X \sec^2 \theta + \frac{ d X}{ d \theta} \tan \theta - \frac{g}{2} \frac{ 2 \left( X \sec \theta \right) \left[ \frac{dX}{d \theta} \sec \theta + X \sec \theta \tan \theta \right]}{ (110)^2 } - 105 \\
0 &= X \sec^2 \theta - \frac{g}{2} \frac{ 2( X \sec \theta) \left[ X \sec \theta \tan \theta\right] }{ (110)^2 } \\
0 &= 1 - \frac{ Xg \tan \theta}{(110)^2} \\
\frac{ (110)^2}{ g \tan \theta} &= X
\end{align}
This is still not an easy equation to solve. However, one of my friends told we could just take the discriminant of the quadratic in terms of $X$, and solve for $\theta$ such that $D=0$.
Taking discriminant and equating to 0, I get
$$ \sin \theta = \frac{ \sqrt{2 \cdot 10 \cdot 105} }{110}$$
and, the angle from it is, 24.45 degrees
I tried the discriminant method, but it gave me a different answer from the implicit differentiation method. I ended up with two solutions with the same maximum value of $X$ but different angles: $\theta =24.45^\text{o}$ and $X=1123.54$ (from discriminant method), and
$\theta = 47^\text{o}$ and $X=1123.54$ (from implicit differentiation).
I later realized the original quadratic can only have solutions if $D(\theta) > 0$, where $D$ is the discriminant. Using the discriminant, I can find a lower bound on the angle. Once I have the lower bound, if I can prove that $X$ decreases monotonically as a function of $\theta$, then I can use the lower bound for further calculations of $\theta$.
So then I used the implicit function theorem and got
$$ \frac{dX}{ d \theta} =- \frac{X \sec^2 \theta -\frac{g X^2}{2 (110)^2} 2 \sec \theta ( \sec \theta \tan \theta) } {\tan \theta - \frac{g \sec^2 \theta}{2 (110)^2} 2X }$$
Now the problem here is that I can't prove this function is in monotonic in terms of $\theta$ as the implicit derivative is a function of both $\theta$ and $X$.
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This is a similar but simple approach which gives the same result. Seeing that $y=\tan(\theta)$ can take any positive value, we go for maximizing $x$ and get $\max(x) = 1123$.
We have: $105 = x\tan\theta-\frac{g}{2} \frac{x^2\sec^2\theta}{(110)^2}$
Let $\ a=105,\ y=\tan\theta,\ b = \frac{g}{2} \frac{1}{(110)^2}.\quad$ Then, $\ a=xy-bx^2(y^2+1)$
Taking $xy=c,\; x^2=\frac{c-bc^2-a}{b}$ (Note that since $y$ can take any positive value, so can $c$)
So, for $\,x_{\max},\quad \frac{d}{dc}\ (\frac{c-bc^2-a}{b}) = 0$ which gives $c=\frac{1}{2b}$
Now, $x_{\max} = \sqrt{\frac{c-bc^2-a}{b}}\quad at\, c=\frac{1}{2b},$ i.e. $\,x_{\max} \approx 1123$ by plugging in values of $a$ and $b.$
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Proof of inequality by Muirhead We have to prove:
$$\frac{\sqrt{pq}}{p+q+2r}+\frac{\sqrt{pr}}{p+r+2q}+\frac{\sqrt{pr}}{p+r+2q}\leq\frac{3}{4}$$
By multiplying it all out we get the following equivalent:
\begin{align*}
4\sum_{cyc}{\sqrt{pq}(p+r+2q)(q+r+2p)}\leq \\
3(p+r+2q)(p+q+2r)(r+q+2p)
\end{align*}
Let us put:
\begin{align*}
x=\sqrt{p}\\
y=\sqrt{q}\\
z=\sqrt{r}\\
\end{align*}
Now, with help of Wolfram Alpha we rewrite the equation in Muirhead-Notation:
\begin{align*}
4(2[5,1,0]+\frac{1}{2}[4,1,1]+3[3,2,1]+\frac{5}{2}[3,3,0])\\
=8[5,1,0]+2[4,1,1]+12[3,2,1]+5[3,3,0]\\
\leq 3[6,0,0]+21[4,2,0]+12[2,2,2]
\end{align*}
We have:
\begin{align*}
[6,0,0]+[2,2,2]\geq 2[4,1,1]\\
2([6,0,0]+[4,2,0])\geq 4[5,1,0]\\
6([4,2,0]+[2,2,2])\geq 12[3,2,1]\\
10 [4,2,0]\geq 10[3,3,0]\\
\end{align*}
Where we use $\frac{[p]+[q]}{2}\geq [\frac{p+q}{2}]$ multiple times.
Now we still have to prove:
$4[5,1,0]\leq 3[4,2,0]+6[2,2,2]$
Now the LHS of the inequality is homogenous in $x,y,z$ so we scale them so that $\max (x,y,z)\leq \frac{3}{4}$. Now:
$[5,1,0]\leq [5,2,0]\leq \frac{3}{4}[4,2,0]\leq \frac{3}{4}[4,2,0]+6[2,2,2]$.
The entirety of this proof seems rather fishy to me. Is it correct? If no, where did I go wrong?
|
Two last lines in your proof are wrong.
Since $$(5,1,0)\succ(4,2,0),$$ by Muirhead $$\sum_{sym}x^5y\geq\sum_{sym}x^4y^2,$$
but you wrote a reversed inequality.
By the way, there is a proof of your inequality without expanding.
Indeed, we need to prove that
$$\sum_{cyc}\frac{yz}{2x^2+y^2+z^2}\leq\frac{3}{4}$$ or
$$\sum_{cyc}\left(\frac{1}{4}-\frac{yz}{2x^2+y^2+z^2}\right)\geq0$$ or
$$\sum_{cyc}\frac{2x^2+y^2+z^2-4yz}{2x^2+y^2+z^2}\geq0$$ or
$$\sum_{cyc}\frac{(x-y)(x+2z-y)-(z-x)(x+2y-z)}{2x^2+y^2+z^2}\geq0$$ or
$$\sum_{cyc}(x-y)\left(\frac{x+2z-y}{2x^2+y^2+z^2}-\frac{y+2z-x}{2y^2+x^2+z^2}\right)\geq0$$ or
$$\sum_{cyc}(x-y)^2(2z^2-2(x+y)z+3(x^2+y^2))(2z^2+x^2+y^2)\geq0,$$ which is obvious.
|
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|
Iterates of $\frac{\sqrt{2}x}{\sqrt{x^2 +1}}$ converge to $\text{sign}(x)$. In this post, a comment states that if $f(x):= \dfrac{\sqrt{2}x}{\sqrt{x^2 +1}}$ and $F_n:=\underbrace{f\circ \dots\circ f}_{n\text{ times}}$, then the pointwise limit $\lim\limits_{n \to \infty} F_n$ is equal to the sign function
$$
\text{sign}(x):=\begin{cases}
1 & : x>0\\
0 & : x=0\\
-1 & : x<0
\end{cases}.
$$
*
*When $x=0$, the limit is clear since $f(0)=0$ is a fixed point.
*When $x>0$, I have the bound $0<F_n(x)<f(x)^{-2^n}\to 1$.
I guess the way to go about showing that the (pointwise) limit holds is by establishing a lower limit for $x>0$ (and arguing similarly for $x<0$). However, I can't manage to figure what that lower limit should look like...
|
You can find the $n^{th}$ iterate of $f$ explicitly.
$$f(x)=\dfrac{\sqrt2x}{\sqrt{x^2+1}},$$
$$f(f(x))=\frac{\sqrt2\dfrac{\sqrt2x}{\sqrt{x^2+1}}}{\sqrt{\dfrac{2x^2}{x^2+1}+1}}=\frac{2x}{\sqrt{3x^2+1}},$$
$$f(f(f(x)))=\frac{\sqrt2\dfrac{2x}{\sqrt{x^2+1}}}{\sqrt{\dfrac{4x^2}{3x^2+1}+1}}=\frac{2\sqrt2x}{\sqrt{7x^2+1}}$$
and more generally,
$$f^{(n)}(x)=\frac{2^{n/2}x}{\sqrt{(2^n-1)x^2+1}}=\frac x{\sqrt{(1-2^{-n})x^2+2^{-n/2}}}.$$
Then as $n$ tends to infinity,
$$f^{(\infty)}(x)=\frac x{|x|}.$$
|
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|
Integer exponent equation
Show that $$(2^a-1)(2^b-1)=2^{2^c}+1$$ doesn't have solution in positive integers $a$, $b$, and $c$.
After expansion I got
$$2^{a+b}-2^a-2^b=2^{2^c}\,.$$
Any hint will be appreciated.
|
$$2^{a+b}=2^a+2^b+2^{2^c}$$
Case $1$:
If $a=b$, then $$2^{2a}=2^{a+1}+2^{2^c}$$
If $a+1$ and $2^c$ are distinct, then surely, the hamming weight on the RHS is $2$ but the hamming weight on the left is $1$.
Hence we must have $a+1=2^c$
$$2^{2a}=2^{a+2}$$
Hence $2a=a+2$, hence $a=2$, but since $a+1=2^c$, we have $3=2^c$ which is a contradiction.
Case $2: a \ne b$, if $2^c$ is not equal to $a$ or $b$, then the hamming weight on the right is $3$ but the hamming weight on the left is $1$. Also, let $a> b$. In order to have the hamming weight of $2^a+2^b$ to drop to $1$ upon adding $2^{2^c}$. We need $a=b+1$ and $b=2^c$.
$$2^{2b+1}=2^{b+1}+2^b+2^b=2^{b+2}$$
$$2b+1=b+2$$
$$b=1$$
Hence $c=0$.
Hence, we do not have positive solution.
|
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Prove that $\tan^{-1}\frac{\sqrt{1+x^2}+\sqrt{1-x^2}}{\sqrt{1+x^2}-\sqrt{1-x^2}}=\frac{\pi}{4}+\frac 12 \cos^{-1}x^2$ Let the above expression be equal to $\phi$
$$\frac{\tan \phi +1}{\tan \phi-1}=\sqrt{\frac{1+x^2}{1-x^2}}$$
$$\frac{1+\tan^2\phi +2\tan \phi}{1+\tan^2 \phi-2\tan \phi}=\frac{1+x^2}{1-x^2}$$
$$\frac{1+\tan^2\phi}{2\tan \phi }=\frac{1}{x^2}$$
$$\sin 2\phi=x^2$$
$$\phi=\frac{\pi}{4}-\frac 12 \cos^{-1}x^2$$
Where am I going wrong?
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A bit late answer but I thought worth mentioning it.
First note that we can substitute $y=x^2$ and consider $0<y\leq 1$. Furthermore, the argument of $\arctan$ can be simplified as follows:
$$\frac{\sqrt{1+y}+\sqrt{1-y}}{\sqrt{1+y}-\sqrt{1-y}}=\frac{1+\sqrt{1-y^2}}{y}$$
Now, setting $y = \cos t$ for $t \in \left[0,\frac{\pi}2\right)$, to show is only
$$\arctan \frac{1+\sin t}{\cos t} = \frac{\pi}{4}+\frac t2$$
At this point half-angle formulas come into mind:
$$\frac{1+\sin t}{\cos t} = \frac{(\cos \frac t2 + \sin \frac t2)^2}{\cos^2 \frac t2 - \sin^2 \frac t2} = \frac{\cos \frac t2 + \sin \frac t2}{\cos \frac t2 - \sin \frac t2}$$ $$ = \frac{1+\tan \frac t2}{1-\tan \frac t2} = \tan\left(\frac{\pi}{4}+\frac t2\right)$$.
Done.
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Let $a_2,a_3,\cdots,a_n$ be positive real numbers and $s=a_2+a_3+\cdots+a_n$. Show that $\sum\limits_{k=2}^na_k^{1-\frac{1}{k}}I tried this problem in this way:-
Let the real numbers are $b_2,b_3,\cdots,b_n$. such that $(b_2,b_3,\cdots,b_n)$is a permutation of the numbers given $(a_2,a_3,\cdots,a_n)$. Hence $s=b_2+b_3+\cdots+b_n$
Now denote $a_2=b_2,\ a_3=b_3,\cdots ,\ a_n=b_n$
Hence$$\sum\limits_{k=2}^na_k^{1-\frac{1}{k}}=b_2^{1-\frac{1}{2}} + b_3^{1-\frac{1}{3}} + \cdots +b_n^{1-\frac{1}{n}}$$
Now denote $a_2=b_n,\ a_3=b_2,\ a_4=b_3,\cdots ,\ a_n=b_{n-1}$
Hence $$\sum\limits_{k=2}^na_k^{1-\frac{1}{k}}=b_n^{1-\frac{1}{2}} + b_2^{1-\frac{1}{3}} + b_3^{1-\frac{1}{4}}+ \cdots +b_{n-1}^{1-\frac{1}{n}}$$
Now denote $a_2=b_{n-1},\ a_3=b_n,\ a_4=b_2,\cdots ,\ a_n=b_{n-2}$
Hence $$\sum\limits_{k=2}^na_k^{1-\frac{1}{k}}=b_{n-1}^{1-\frac{1}{2}} + b_n^{1-\frac{1}{3}} + b_2^{1-\frac{1}{4}}+ \cdots +b_{n-2}^{1-\frac{1}{n}}$$
$$\vdots$$ $$\vdots$$
Now denote $a_2=b_{3},\ a_3=b_4,\ a_4=b_5,\cdots ,\ a_{n-1}=b_n,\ a_n=b_{2}$
Hence $$\sum\limits_{k=2}^na_k^{1-\frac{1}{k}}=b_{3}^{1-\frac{1}{2}} + b_4^{1-\frac{1}{3}} + \cdots +b_{n}^{1-\frac{1}{n-1}}+b_{2}^{1-\frac{1}{n}}$$
Add all these and we get:- $$\sum\limits_{k=2}^n \left(\sum\limits_{i=2}^n b_k^{1-\frac{1}{i}} \right)$$
Hence we need to prove $$\sum\limits_{k=2}^n \left(\sum\limits_{i=2}^n b_k^{1-\frac{1}{i}} \right)<(n-1)\left(s+\sqrt{s}\right)$$
Now $$\sum\limits_{k=2}^n \left(\sum\limits_{i=2}^n b_k^{1-\frac{1}{i}}\right)=\sum\limits_{k=2}^n \left(\sum\limits_{i=2}^n b_k\left(b_k^{-\frac{1}{i}}\right)\right)$$
Now let a positive real number $m$ and a positve integer $p$. then $$m^{\frac{1}{p}}\leq \frac{\frac{1}{m}+\overbrace{1+1+\cdots}^{(p-1)\ \text{times}}}{p}$$
Hence $$\sum\limits_{k=2}^n \left(\sum\limits_{i=2}^n b_k\left(b_k^{-\frac{1}{i}}\right)\right)\leq\sum\limits_{k=2}^n \left(\sum\limits_{i=2}^n b_k\left(\frac{\frac{1}{b_k}+\overbrace{1+1+\cdots+1}^{(i-1)\ \text{times}}}{i}\right)\right)= \sum\limits_{k=2}^n \left(\sum\limits_{i=2}^n \left(\frac{1}{i}\right) + b_k\sum\limits_{i=2}^n\left(\frac{i-1}{i}\right)\right)=
(n-1)\sum\limits_{i=2}^n \left(\frac{1}{i}\right) + \sum\limits_{k=2}^n \left(b_k\sum\limits_{i=2}^n\left(1-\frac{1}{i}\right)\right)=
(n-1)\sum\limits_{i=2}^n \left(\frac{1}{i}\right) + (n-1)\sum\limits_{k=2}^n b_k -\sum\limits_{k=2}^n\left(b_k\sum\limits_{i=2}^n\left(\frac{1}{i}\right)\right)=(n-1)s+(n-1)\sum\limits_{i=2}^n \left(\frac{1}{i}\right)-\sum\limits_{k=2}^n\left(b_k\sum\limits_{i=2}^n\left(\frac{1}{i}\right)\right)$$
Now i am stuck. Can anyone help me? If you have any other process please mention that also. This problem was in the Old and New Inequalities Volume 2. So please try to be limited in AM-GM and Cauchy-Schwarz.
This problem was proposed by George Tsintsifas in American Mathematical Monthly. So if anyone gives the original solution to this problem i will gladly welcome that.
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Inspired by the solution published in the American Mathematical Monthly:
We can suppose that $\ \forall k \ , \ 0<a_k<1 $.
$\displaystyle \sum_{k=2}^n a_k^{1-\frac{1}{k}} - s = \sum_{k=2}^n \left( a_k^{1-\frac{1}{k}}-a_k\right) = \sum_{k=2}^n a_k^{\frac{1}{2}}\left( a_k^{\frac{1}{2}-\frac{1}{k}}-a_k^{\frac{1}{2}}\right)$
By using the Cauchy-Schwarz inequality:
$\displaystyle \left(\sum_{k=2}^n a_k^{1-\frac{1}{k}} - s\right)^2 \leqslant \sum_{k=2}^n a_k \ \sum_{k=2}^n \left( a_k^{\frac{1}{2}-\frac{1}{k}}-a_k^{\frac{1}{2}}\right)^2 = s\ \sum_{k=2}^n \left( a_k^{\frac{1}{2}-\frac{1}{k}}-a_k^{\frac{1}{2}}\right)^2 $
We have, for all $k\geqslant 3$:
$\left( a_k^{\frac{1}{2}-\frac{1}{k}}-a_k^{\frac{1}{2}}\right)^2= a_k^{1-\frac{2}{k}}\left( 1-\exp\left( \dfrac{\ln a_k}{k}\right)\right)^2 \leqslant a_k^{\frac{1}{3}} \ln^2(a_k)\dfrac{1}{k^2} \leq \dfrac{36}{k^2e^2}$
So:
$\displaystyle \sum_{k=2}^n \left( a_k^{\frac{1}{2}-\frac{1}{k}}-a_k^{\frac{1}{2}}\right)^2 \leqslant 1+\dfrac{36}{e^2}\sum_{k=3}^n\dfrac{1}{k^2} \leqslant 1+\dfrac{36}{e^2}\left(\dfrac{\pi^2}{6}-\dfrac{5}{4}\right) < 4$
Eventually:
$\displaystyle \left(\sum_{k=2}^n a_k^{1-\frac{1}{k}} - s\right)^2 \leqslant 4s$
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How can $\sin(2nx)$ be expressed as a polynomial? Let $P_n (u)$ be a polynomial in $u$ of degree $n$. Then, prove that for every positive integer $n$, $\sin(2nx)$ is expressible as
$(\cos x)(P_{2n-1} (\sin x))$ for some $P_{2n-1}$.
I tried by starting as
$\sin(2nx)=2\sin(nx)\cos(nx),$ and ta-da, it satisfies for $n=1,2,3$. But I can't figure it in general.
P.S. Complex exponent is not allowed.
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Hint:
$$ \sin(2x)=(\cos x)(2 \sin x)$$
$$ \sin(4x) = 2 \sin(2x) \cos (2x) = 2(\cos x)(2 \sin x)(1- 2 \sin^2x) = (\cos x)(4\sin x - 8 \sin^3x)$$
$$ \sin(6x) = \sin(4x + 2x)= \sin(4x)\cos(2x) +\sin(2x)\cos(4x)$$
$$= \sin(4x) (1-2\sin^2 x) + \sin(2x) ( 1- \sin^2(2x))$$
$$ = \sin(4x) (1- 2\sin^2 x) + \sin(2x) - \sin^3(2x) $$
We then use our decompositions for $\sin(2x)$ and $\sin(4x)$.
$$ = (\cos x) P_3(\sin x) (1 - 2\sin^2 x) + (\cos x) P_1(\sin x) - \cos^3 x P_1(\sin x)^3 $$
$$ = (\cos x) P_3(\sin x) (1 - 2\sin^2 x)) + (\cos x) P_1(\sin x) - (\cos x)\left(1- \frac{1}{4}P_1(\sin x)^2\right) P_1(\sin x)^3 $$
Note: Last step above uses
$$ \cos^{2m+1} y = \cos y (1-\sin^2y)^m = \cos y \left(1-\frac{1}{4}P_1(\sin y)^2\right)^m $$
Next level, $8x$, gets treated as $8x = 4x +4x$ (to avoid having to deal with $\cos(6x)$):
$$ \sin(8x) = \sin(4x + 4x)= 2\sin(4x)\cos(4x) = 2\sin(4x)(1- 2\sin^2(2x)) $$
$$ = 2(\cos x) P_3(\sin x) - 2(\cos x) P_3(\sin x) 8 \cos^2 x P_1(\sin x)^2 $$
$$= 2(\cos x) P_3(\sin x) - 2(\cos x) P_3(\sin x) 8 (1 -\sin^2 x) P_1(\sin x)^2 $$
This has all ingredients of a more formal induction based proof of the decomposition (existence of $P_{2n-1}$ for all $n\geq 1$).
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How to evaluate $\int _0^{\infty }\frac{\ln \left(x^3+1\right)}{\left(x^2+1\right)^2}\:dx$ without complex analysis This particular integral evaluates to,
$$\int _0^{\infty }\frac{\ln \left(x^3+1\right)}{\left(x^2+1\right)^2}\:dx=\frac{\pi }{8}\ln \left(2\right)-\frac{3\pi }{8}+\frac{\pi }{3}\ln \left(2+\sqrt{3}\right)-\frac{G}{6}$$
And its been proven here.
But i'd like to know how to evaluate this without complex analysis.
One of the answers uses differentiation under the integral sign directly and partial fraction decomposition on a similar integral, but doing it that way doesnt help me with this case here
I tried to evaluate this way but got stuck,
$$\int _0^{\infty }\frac{\ln \left(x^3+1\right)}{\left(x^2+1\right)^2}\:dx=\int _0^1\frac{\ln \left(x^3+1\right)}{\left(x^2+1\right)^2}\:dx+\int _1^{\infty }\frac{\ln \left(x^3+1\right)}{\left(x^2+1\right)^2}\:dx\:\:\:\:\:\: \text{then sub}\:\:x=\frac{1}{t}\:\:\text{for the 2nd integral}$$
$$=\int _0^1\frac{\ln \left(t^3+1\right)}{\left(t^2+1\right)^2}\:dt+\int _0^1\frac{t^2\ln \left(t^3+1\right)}{\left(t^2+1\right)^2}\:dt-3\int _0^1\frac{t^2\ln \left(t\right)}{\left(t^2+1\right)^2}\:dt$$
$$=\int _0^1\frac{\ln \left(t^3+1\right)}{t^2+1}\:dt+3G+3\int _0^1\frac{\ln \left(t\right)}{\left(t^2+1\right)^2}\:dt$$
I managed to evaluate the last integral expanding the denominator but i cant think of a way to evaluate the 1st integral, please help me.
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As said in comments, consider
$$I(a)=\int _0^{\infty }\frac{\log \left(ax^3+1\right)}{\left(x^2+1\right)^2}\,dx$$
$$I'(a)=\int _0^{\infty }\frac{x^3}{\left(x^2+1\right)^2 \left(a x^3+1\right)}\,dx$$
Using partial fraction decomposition, the integrand is
$$\dfrac{ a(2a^2-1)x^2+3a^2x-a(a^2+2)}{(a^2+1)^2\left(ax^3+1\right)}-\dfrac{\left(2a^2-1\right)x+3a}{(a^2+1)^2\left(x^2+1\right)}-\dfrac{x-a}{\left(a^2+1\right)\left(x^2+1\right)^2}$$ which is not too bad.
This leads to
$$I'(a)=\frac{-8 \sqrt{3} \pi a^{8/3}+24 \sqrt{3} \pi a^{4/3}+16 \sqrt{3} \pi
a^{2/3}+9 \pi a^3-18 a^2+12 \left(2 a^2-1\right) \log (a)-45 \pi a-18}{36
\left(a^2+1\right)^2}$$ This is not the most pleasant part but it leads to the result.
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How to evaluate $\int \frac{\sqrt{1+x}+\sqrt{1-x}}{\sqrt{1+x}-\sqrt{1-x}} dx$? I am trying to evaluate
$$\int \frac{\sqrt{1+x}+\sqrt{1-x}}{\sqrt{1+x}-\sqrt{1-x}} dx \quad (1)$$
The typical way to confront this kind of integrals are the conjugates i.e:
$$\int \frac{\sqrt{1+x}+\sqrt{1-x}}{\sqrt{1+x}-\sqrt{1-x}} dx = $$
$$ \int \left(\frac{\sqrt{1+x}+\sqrt{1-x}}{\sqrt{1+x}-\sqrt{1-x}}\right)\left(\frac{\sqrt{1+x}+\sqrt{1-x}}{\sqrt{1+x}+\sqrt{1-x}}\right)\left(\frac{\sqrt{1+x}-\sqrt{1-x}}{\sqrt{1+x}-\sqrt{1-x}}\right) dx = $$
$$\int \left(\frac{(\sqrt{1+x})^2-(\sqrt{1-x})^2}{(\sqrt{1+x})^2-(\sqrt{1-x})^2}\right)\left(\frac{\sqrt{1+x}-\sqrt{1-x}}{\sqrt{1+x}+\sqrt{1-x}}\right)dx = $$
$$\int 1*\left(\frac{\sqrt{1+x}-\sqrt{1-x}}{\sqrt{1+x}+\sqrt{1-x}}\right)dx $$
That's a dead end.
I also tried other conjugate approaches (only the numerator, only the denominator etc) with no better luck.
Any ideas?
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Let $\sqrt{\frac{1+x}{1-x}}=t.$
Thus, $$x=\frac{t^2-1}{t^2+1},$$ $$dx=\frac{2t(t^2+1)-(t^2-1)2t}{(t^2+1)^2}dt=\frac{4tdt}{(t^2+1)^2}$$ and we need to calculate $$\int\frac{4t(t+1)}{(t-1)(t^2+1)^2}dt.$$
Now, easy to show that:
$$\int\frac{4t(t+1)}{(t-1)(t^2+1)^2}dt=\frac{2t}{t^2+1}+\ln\frac{(t-1)^2}{t^2+1}+C.$$
Can you end it now?
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Other way to evaluate $\int \frac{1}{\cos 2x+3}\ dx$? I am evaluating
$$\int \frac{1}{\cos 2x+3} dx \quad (1)$$
Using Weierstrass substitution:
$$ (1)=\int \frac{1}{\frac{1-v^2}{1+v^2}+3}\cdot \frac{2}{1+v^2}dv =\int \frac{1}{v^2+2}dv \quad (2) $$
And then $\:v=\sqrt{2}w$
$$ (2) = \int \frac{1}{\left(\sqrt{2}w\right)^2+2}\sqrt{2} dw$$$$= \frac{1}{2} \int \frac{1}{\sqrt{2}\left(w^2+1\right)}dw$$$$ = \frac{1}{2\sqrt{2}}\arctan \left(w\right) + C$$$$= \frac{1}{2\sqrt{2}}\arctan \left(\frac{\tan \left(x\right)}{\sqrt{2}}\right)+C$$
Therefore,
$$\int \frac{1}{\cos 2x+3} dx = \frac{1}{2\sqrt{2}}\arctan \left(\frac{\tan \left(x\right)}{\sqrt{2}}\right)+C $$
That's a decent solution but I am wondering if there are any other simpler ways to solve this (besides Weierstass). Can you come up with one?
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$$\cos(2x)=\cos (x+x) =\cos x \cos x-\sin x \sin x=\cos^2x-\sin^2x$$
$$I =\int \frac{1}{\cos 2x+3}dx= \int \frac{\sec^2x}{\sec^2x(\cos^2x-\sin^2x+3)}dx = \int \frac{\sec^2x}{1-\tan^2x+3\sec^2x}dx $$ Substitute $t=\tan x$
so that $dt=\sec^2x dx$
$$I=\int \frac{1}{1-t^2+3(1+t^2)}dt=\int \frac{1}{4+2t^2} dt=\frac{1} {2\sqrt 2} \tan^{-1}\frac{\sqrt 2 t} {2}+c$$ Now substitute back $t=\tan x$
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How to solve this ODE: $x^3dx+(y+2)^2dy=0$? I am trying to solve $$ x^3dx+(y+2)^2dy=0 \quad( 1)$$
Dividing by $dx$, we can reduce the ODE to seperate variable form, i.e
$$ (1) \to (y+2)^2y'=-x^3 $$
Hence,
$$ \int (y(x)+2)^2y'(x) dy = \int -x^3dx = - \frac{x^4}{4} + c_1$$
This LHE seems to be easy to solve using integration by parts:
$$ \int (y(x)+2)^2y'(x) dy = y(x)(y(x)+2)^2 - \int y^3dy- \int 4y^2dy + \int4ydy \iff$$
$$ \iff - \frac{y^4}{4} + -\frac13 y^3 + 4y^2 + 8y + c_2$$
But then solving the ODE for $y(x)$ is a struggle:
$$ \iff - \frac{y^4}{4} + -\frac13 y^3 + 4y^2 + 8y = - \frac{3}{4}x + C$$
Any ideas on how I can solve this?
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The coefficient of ${\rm d}x$ depends only on $x$ and the coefficient of $y$ depends only on ${\rm d}y$. So $$0 = x^3\,{\rm d}x + (y+2)^2\,{\rm d}y = {\rm d}\left(\frac{x^4}{4} + \frac{(y+2)^3}{3}\right),$$which says that your solutions are described by $$\frac{x^4}{4} + \frac{(y+2)^3}{3} = c,$$for some $c \in \Bbb R$. Thus $$y = -2+\left(\frac{c-3x^4}{4}\right)^{1/3},$$after letting $c$ absorb some constants.
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Evaluating $\int _0^1\frac{\ln \left(x^3+1\right)}{x+1}\:dx$ What methods would work best to find $\displaystyle \int _0^1\frac{\ln \left(x^3+1\right)}{x+1}\:dx$
As usual with this kind of integral i tried to differentiate with the respect of a parameter
$$\int _0^1\frac{\ln \left(ax^3+1\right)}{x+1}\:dx$$
$$\int _0^1\frac{x^3}{\left(x+1\right)\left(ax^3+1\right)}\:dx=\frac{1}{a-1}\int _0^1\left(\frac{ax^2-ax+a}{ax^3+1}-\frac{1}{x+1}\right)\:dx$$
in the end $2$ of these integrals are nice but the other $2$ are not
$\displaystyle \int _0^1\frac{-ax}{ax^3+1}\:dx$ and $\displaystyle \int _0^1\frac{a}{ax^3+1}\:dx$
Is there a better approach to this?
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Instead of using Feynman's trick at once use the following substitution first
$$\underbrace{\int _0^1\frac{\ln \left(1+x^3\right)}{1+x}\:dx}_{x=\frac{1-t}{1+t}}$$
$$=\ln \left(2\right)\underbrace{\int _0^1\frac{1}{1+x}\:dx}_{\ln \left(2\right)}+\int _0^1\frac{\ln \left(1+3x^2\right)}{1+x}\:dx-3\underbrace{\int _0^1\frac{\ln \left(1+x\right)}{1+x}\:dx}_{\frac{1}{2}\ln ^2\left(2\right)}$$
Now make use of the general result proved here.
$$\int _0^1\frac{\ln \left(b+ax^2\right)}{1+x}\:dx$$
$$=-\frac{\ln ^2\left(b\right)}{4}-\frac{\text{Li}_2\left(-\frac{a}{b}\right)}{2}-\frac{\ln ^2\left(a+b\right)}{4}+\frac{\ln \left(b\right)\ln \left(a+b\right)}{2}-\arctan ^2\left(\sqrt{\frac{a}{b}}\right)+\ln \left(2\right)\ln \left(a+b\right)$$
So by setting $a=3$ and $b=1$ we have
$$-\frac{\text{Li}_2\left(-3\right)}{2}-\arctan ^2\left(\sqrt{3}\right)+\ln ^2\left(2\right)-\frac{1}{2}\ln ^2\left(2\right)$$
Thus,
$$\int _0^1\frac{\ln \left(1+x^3\right)}{1+x}\:dx=-\frac{\text{Li}_2\left(-3\right)}{2}-\frac{\pi ^2}{9}+\frac{1}{2}\ln ^2\left(2\right)$$
Where $\text{Li}_2\left(z\right)$ is the Dilogarithm function.
As suggested by others one could also turn the integral into
$$\int _0^1\frac{\ln \left(1+x^3\right)}{1+x}\:dx=\int _0^1\frac{\ln \left(1-x+x^2\right)}{1+x}\:dx+\int _0^1\frac{\ln \left(1+x\right)}{1+x}\:dx$$
From the result above we obtain the value of that other integral which is
$$\int _0^1\frac{\ln \left(1-x+x^2\right)}{1+x}\:dx=-\frac{\text{Li}_2\left(-3\right)}{2}-\frac{\pi ^2}{9}$$
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|
How to integrate $\int {2\over (x^2+2)\sqrt{x^2+4}}dx$?
Solve the following indefinite integral:
$$\int \frac{2}{(x^2+2)\sqrt{x^2+4}} dx$$
My approach:
I used the substitution: $x=2\tan t$, $dx=2\sec^2t dt$
$$\int \frac{2}{(x^2+2)\sqrt{x^2+4}} dx=\int \frac{2}{(4\tan^2t+2)\sqrt{4\tan^2t+4}}\cdot 2\sec^2t\ dt$$
$$=\int \frac{4\sec^2t }{2(2\tan^2t+1)2\sec t} dt$$
$$=\int \frac{\sec t}{2\tan^2t+1}dt$$
In numerator I have $\sec t$ but not $\sec^2t$ therefore I can't see a way to take it further. Please help me solve this integral. Thanks in advance.
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Alternatively, let $ x = 2\sinh t \Rightarrow \frac{dx}{dt} = 2\cosh t$.
The integral becomes
$$\begin{array} {r c l }
\displaystyle \int \frac2{(4\sinh^2 t + 2)(2\cosh t)} \cdot 2\cosh t \, dt
&=& \displaystyle \frac12 \int \frac1{2\sinh^2 t + 1} \, dt \\
&=& \displaystyle \frac12 \tan^{-1} (\tanh t) + C \\
&=& \displaystyle \frac12 \tan^{-1} \left (\tanh \left (\sinh^{-1} \tfrac x2 \right )\right) + C \\
\end{array} $$
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|
Why is the convergence point of $ \sum _{n=1}^{\infty }\frac{1}{2^n}-\frac{1}{2^{n+1}} $ negative? I am trying to evaluate $\frac1{2^1} - \frac1{2^2} + \frac1{2^3} - \frac1{2^4} + \cdots$
I re-wrote the sum using sigma notation as:
$$ \sum _{n=1}^{\infty } \left( \frac{1}{2^n}-\frac{1}{2^{n+1}} \right) \quad (1) $$
Hence,
$$ (1) = \sum _{n=1}^{\infty } \frac{2^{n+1} - 2^n }{2^n2^{n+1}} = \sum _{n=1}^{\infty } \frac{2^n(2-1) }{2^n2^{n+1}} = - \sum _{n=1}^{\infty } \frac1{2^{n+1}} = - \sum _{n=2}^{\infty } \frac1{2^{n}} = -\left( \sum _{n=0}^{\infty } \left(\frac12\right)^n -1-\frac12\right) = -2+\frac32 =-\frac12 $$
Therefore,
$$ \boxed{\sum _{n=1}^{\infty }\frac{1}{2^n}-\frac{1}{2^{n+1}} =-\frac12} $$
I can't spot any algebraic mistakes, thus I assume the sum is correct. But I don't understand the result. How can a real valued sum converge to a negative number?
|
You wrote, $\frac{1}{2^1} - \frac{1}{2^2} + \frac{1}{2^3} - \frac{1}{2^4} + \cdots = \sum _{n=1}^{\infty } \left( \frac{1}{2^n}-\frac{1}{2^{n+1}} \right) $ ,
This is not correct,
$\frac{1}{2^1} - \frac{1}{2^2} + \frac{1}{2^3} - \frac1{2^4} + \cdots = \sum_{n=1}^{\infty} (-1)^{n+1} \frac{1}{2^n} = \sum_{n=1}^{\infty} \frac{1}{2^{2n-1}} -
\sum_{n=1}^{\infty} \frac{1}{2^{2n}} = \frac{1}{2} \frac{1}{1-\frac{1}{2^2}} - \frac{1}{2^2} \frac{1}{1-\frac{1}{2^2}} = \frac{2}{3} - \frac{1}{3} = \frac{1}{3} $ .
|
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"timestamp": "2023-03-29T00:00:00",
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|
Prove complex numbers $a$ and $b$ are antipodal under stereographic projection $\iff a \overline{b} = -1$ I'm trying to prove the following statement:
Given $a, b \in \mathbb{C}$, prove that $a$ and $b$ correspond to antipodal points on the Riemann sphere under stereographic projection if and only if $a \overline{b} = -1$
My attempt
I wanted to make a proof where all my implications were reversible to avoid making a proof of each implication separately. As previous knowledge, I know that if a have a point $a \in \mathbb{C}$, then the stereographic projection $f: \mathbb{C} \to S^2$ is given by
$$
f(a) = \left(\frac{a + \overline{a}}{1 + |a|^2},\frac{a - \overline{a}}{i\left(1 + |a|^2\right)},\frac{|a|^2-1}{|a|^2+1}\right)
$$
Now, given that $P,Q\in S^2$ are antipodal if and only if $P =-Q$, I get the following:
\begin{align}
f(a) = -f(b) &\iff
\begin{cases}
\frac{a + \overline{a}}{1 + |a|^2} = \frac{-b - \overline{b}}{1 + |b|^2} \\
\frac{a - \overline{a}}{i\left(1 + |a|^2\right)} = \frac{\overline{b}-b}{i\left(1 + |b|^2\right)} \\
\frac{|a|^2-1}{|a|^2+1} = \frac{1-|b|^2}{|b|^2+1} \\
\end{cases}\\
&\iff\begin{cases}
a + \overline{a}+a|b|^2 +\overline{a}|b|^2 = -b - \overline{b}-b|a|^2 -\overline{b}|a|^2 \\
a - \overline{a}+a|b|^2 -\overline{a}|b|^2 = -b + \overline{b}-b|a|^2 +\overline{b}|a|^2 \\
|ab|^2+|a|^2-|b|^2-1 =-|ab|^2+|a|^2-|b|^2+1 \\
\end{cases}\\
&\iff\begin{cases}
a +a|b|^2 = -b -b|a|^2 \\
\overline{a} +\overline{a}|b|^2 = -\overline{b} -\overline{b}|a|^2 \\
|ab|^2=1 \\
\end{cases}\\
&\iff\begin{cases}
a +b +a|b|^2+b|a|^2 =0 \\
|a||b|=1 \\
\end{cases}\\
\end{align}
Where here I use brackets to indicate that all those equations are true simultaneously. On this last step is where I ran into trouble because I couldn't find a way to show that both conditions in the last step are equivalent to $b =- \frac{1}{\overline{a}}$.
Is my attempt correct (up to what I have already written)? And if so, does somebody know how I could conclude the proof of equivalence? Any help would be greatly appreciated. Thank you!
|
For the direct implication, one could also use the inverse function of $f$, $\phi$:
$$ \phi (x,y,u) = \frac{x+iy}{1-u}$$
for $(x,y,u)\not= (0,0,1)$, $x^2+y^2+u^2=1.$
If $ P = (x,y,u)$ and $Q=(-x,-y,-u)$, then
$$ \phi(P)\overline{\phi(Q)} = \frac{x+iy}{1-u} \cdot \frac{-x+iy}{1+u} = -\frac{x^2+y^2}{1-u^2} = -1$$
The indirect implication is straightforward. For example:
$$\frac{a + \bar{a}}{1+|a|^2} = \frac{-\bar{b}^{-1} -b^{-1}}{1+|b|^{-2}} = -\frac{b + \bar{b}}{1+|b|^2}.$$
Edit: Note that
$$ a + b + a|b|^2 + b|a|^2 = 0$$
is equivalent to
$$ a(1+|b|^2) = - b(1+|a|^2) $$
Multiplying by $\bar{b}$, we get:
$$ a\bar{b}(1+|b|^2) = - |b|^2(1+|a|^2) $$
which implies that $ a\bar{b}$ is real negative.
|
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|
Find $\lim _{x \to \infty}\frac{5^x+7^{x+1}}{7^x+5^{x+1}}$
I just started studying and in our first maths-exam we had to find the following limit:
$\lim \limits_{x \to \infty}\frac{5^x+7^{x+1}}{7^x+5^{x+1}}$
The other tasks where rather easy, so I feel like I'm missing something obvious here, but I don't even really know how to start with this one.
A walkthrough would be great, but hints in the right direction would also help.
Thank you!
|
I would divide both numerator and denominator by $7^{x+1}$:
\begin{align*}
\frac{5^x+7^{x+1}}{7^x+5^{x+1}} &= \frac{\frac{5^x}{7^{x+1}} + \frac{7^{x+1}}{7^{x+1}}}{\frac{7^x}{7^{x+1}} + \frac{5^{x+1}}{7^{x+1}}}\\
&= \frac{\frac{1}{7}\left(\frac{5}{7}\right)^x + 1}{\frac{1}{7} + \left(\frac{5}{7}\right)^{x+1}}
\end{align*}
Since $\frac{5}{7}<1$, you know what $\left(\frac{5}{7}\right)^x$ tend to as $x\to\infty$.
|
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|
How to factorize $a^2-2ab+a^2b-2b^2$? I have been stuck on factorizing this:
$$a^2-2ab+a^2b-2b^2$$
I thought I could solve it by making $(a+b)$ as one factor but it didn't work then I tried to add and deduct some terms which that didn't lead me to anything either.
I don't really know what to do next.
|
$a^2-2ab+a^2b-2b^2=$
$=-2b^2+a(a-2)b+a^2.$
In order to factorize the last polynomial, we have to solve the following quadratic equation:
$-2b^2+a(a-2)b+a^2=0$
that is equivalent to
$2b^2-a(a-2)b-a^2=0.$
$\Delta=a^2(a-2)^2+8a^2=a^2(a^2-4a+12)\ge0,$
$b=\frac{a(a-2)\pm\sqrt{a^2(a^2-4a+12)}}{4}=\frac{a(a-2)\pm a\sqrt{a^2-4a+12}}{4}.$
Now we can factorize the polynomial:
$-2b^2+a(a-2)b+a^2=$
$=-2\left(b-\frac{a(a-2)-a\sqrt{a^2-4a+12}}{4}\right)\cdot\left(b-\frac{a(a-2)+a\sqrt{a^2-4a+12}}{4}\right)=$
$\begin{align*}
&=-\frac{1}{8}\left(4b-a^2+2a+a\sqrt{a^2-4a+12}\right)\cdot\left(4b-a^2+2a-a\sqrt{a^2-4a+12}\right).\\
\end{align*}$
Therefore we get that
$\begin{align}
&a^2-2ab+a^2b-2b^2=\\
&=-\frac{1}{8}\left(4b-a^2+2a+a\sqrt{a^2-4a+12}\right)\cdot\left(4b-a^2+2a-a\sqrt{a^2-4a+12}\right).\\
\end{align}$
|
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|
Find the remainder when $(x - 1)^{100} + (x - 2)^{200}$ is divided by $x^2 - 3x + 2$ .
Find the remainder when $(x - 1)^{100} + (x - 2)^{200}$ is divided by $x^2 - 3x + 2$ .
What I tried: In some step I messed up with this problem and so I think I am getting my answer wrong, so please correct me.
We have $x^2 - 3x + 2$ = $(x - 1)(x - 2)$
and I can see $(x - 1)^2 \equiv 1$ $($mod $x - 2)$ . We also have :-
$$\frac{(x - 1)^{100}}{(x - 1)(x - 2)} = \frac{(x - 1)^{99}}{(x - 2)}.$$
We have :- $(x - 1)^{98} \equiv 1$ $($mod $x - 2).$ $\rightarrow (x - 1)^{99} \equiv (x - 1)$ $($mod $x - 2)$. Now for the case of $(x - 2)^{200}$ we have :-
$$\frac{(x - 2)^{200}}{(x - 1)(x - 2)} = \frac{(x - 2)^{199}}{(x - 1)}.$$
We have :- $(x - 2) \equiv (-1)$ $($mod $x - 1)$ $\rightarrow (x - 2)^{199} \equiv (-1)$ $($mod $x - 1)$.
Adding all these up we have :- $(x - 1)^{100} + (x - 2)^{200} \equiv (x - 2)$ $($mod $x² - 3x + 2)$ .
On checking my answer with wolfram alpha , I found the remainder to be $1$, so I messed up in some step .
Can anyone help me?
|
Write $$(x - 1)^{100} + (x - 2)^{200}=k(x)(x-2)(x-1)+ax+b$$
SInce this is valid for all $x$ it is valid also for $$x=1: \;\;\; 1=a+b$$ and $$x=2: \;\;\; 1=a2+b$$
So $a=0$ and $b=1$.
|
{
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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|
If $x\sin A+y\sin B+z\sin C=x^2\sin2A+y^2\sin2B+z^2\sin 2C=0$, show $x^3\sin3A+y^3\sin3B+z^3\sin 3C=0$ I'm having trouble with a question that came in one of my exam and is about complex numbers and trigonometry:
If $$x \sin A +y \sin B+z\sin C=0$$ and $$x^2 \sin 2A + y^2 \sin 2B + z^2 \sin 2C=0$$ then prove that $$x^3 \sin 3A+y^3 \sin 3B +z^3 \sin 3C=0$$ where $x$, $y$, $z$ belong to $\mathbb{R}$ and $A+B+C=\pi$.
What I tried: I tried maybe doing it as imaginary part of $xe^{iA} + ye^{iB} +ze^{iC} =0$ and similarly, but I'm reaching a dead end in that approach
|
If $\sin{C}=0$, so $$x\sin{A}+y\sin(180^{\circ}k-A)=0,$$ where $k$ is an integer number.
Thus, easy to show (just consider two cases: $k$ is odd and $k$ is even) that $$x^3\sin3A+y^3\sin(540^{\circ}k-3A)=0.$$
Now, let $\prod\limits_{cyc}\sin{A}\neq0.$
Thus, the first condition gives $$z=-\frac{x\sin{A}+y\sin{B}}{\sin(A+B)}$$ and with the second we obtain:
$$(x^2\sin2A+y^2\sin2B)\sin(A+B)=2(x\sin{A}+y\sin{B})^2\cos(A+B)$$ or
$$(\sin{A}\cos{A}\sin(A+B)-\sin^2A\cos(A+B))x^2-2\sin{A}\sin{B}\cos(A+B)xy+$$
$$+(\sin{B}\cos{B}\sin(A+B)-\sin^2B\cos(A+B))=0$$ or
$$\sin{A}\sin{B}(x^2-2xy\cos(A+B)+y^2)=0$$ or
$$x^2-2xy\cos(A+B)+y^2=0,$$ which since $|\cos(A+B)|\neq1,$ gives $$x=y=0.$$
Can you end it now?
|
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|
How to compute $\int_{0}^{\frac{\pi}{2}}\frac{\sin(x)}{\sin^3(x)+ \cos^3(x)}\,\mathrm{d}x$?
How to compute the following integral?
$$\int_{0}^{\frac{\pi}{2}}\frac{\sin(x)}{\sin^3(x)+ \cos^3(x)}\,\mathrm{d}x$$
So what I did is to change $\sin(x)$ to $\cos(x)$ with cofunction identity, which is $\sin(\frac{\pi}{2} -x) = \cos(x)$. The integral changes into easier:
$$\int_{0}^{\frac{\pi}{2}} \frac{\cos(x)}{\sin^3(x)+ \cos^3(x)} \, \mathrm{d}x$$
And then I divided by $\cos^3(x)$. It will turn everything to
$$\int_{0}^{\frac{\pi}{2}} \frac{\sec^2(x)}{\tan^3(x)+1} \, \mathrm{d}x.$$
And I used $u$-substitution setting $u = \tan(x)$ $\Rightarrow$ $\mathrm{d}u = \sec^2(x) \, \mathrm{d}x$ and the bounds $u = \tan(\frac{\pi}{2}) = \infty$ and $u =\tan(0) = 0$ and the integral changed into integral to
$$ \int_{0}^{\infty} \frac{\mathrm{d}u}{1+u^3} $$
This is where I used partial fraction decomposition and my answer is divergent and my answer is wrong:
\begin{align*}
\int_{0}^{\infty} \frac{1}{3(u+1)} + \frac{-u+2}{3(u^2-u+1)} \, \mathrm{d}u
\end{align*}
by this it looks like it will divergent.
The correct answer is $\frac{2\pi}{3\sqrt{3}}$. So, what do I do, next?
Then I did the another method which gives me divergent again,
\begin{align*}
\int_{0}^{\frac{\pi}{2}} \frac{\sin(x)}{\sin^3(x)+\cos^3(x)} \, \mathrm{d}x
&= \int_{0}^{\frac{\pi}{2}} \frac{\tan(x)\sec^2(x)}{1+\tan^3(x)} \, \mathrm{d}x
\end{align*}
and $u = \tan(x)$ $\Rightarrow$ $\mathrm{d}u = \sec^2(x) \, \mathrm{d}x$, the integral
$$\int_{0}^{\frac{\pi}{2}} \frac{u}{1+u^3} \, \mathrm{d}u$$
\begin{align*}
\int_{0}^{\frac{\pi}{2}} \frac{-1}{3(u+1)} + \frac{u+1}{3(u^2-u+1)} \, \mathrm{d}u
\end{align*}
and did partial fraction and this gives divergent.
I have no idea what to do next for the first method of work, or the second method of work.
|
As proved here
For all $n\geq 2$ we shall show that
$$I(n)=\int_{0}^{\frac{\pi}{2}} \frac{\sin x}{\sin^{2n-1}x+\cos^{2n-1}x}dx=\frac{\pi}{2n-1}\sum_{k=0}^{n-2}{n-2\choose k}\operatorname{csc}\left(\frac{(2\pi(n-k-1)}{2n-1}\right)$$. Just set $n=2$ we have $$I(1)=\int_0^{\frac{\pi}{2}}\frac{\sin x}{\sin^3x+\cos^3x}dx =\frac{2\pi}{3\sqrt 3}$$.
Call original integral as $I_1$ After we were done with $x=\frac{\pi}{2}-x$ we have $$I_1=\int_0^{\frac{\pi}{2}}\frac{\cos x}{\sin^3x +\cos^3x }dx $$ adding the $I_1$ and $I_2$ we have $$I= \frac{1}{2}\int_0^{\frac{\pi}{2}}\frac{\sin x+\cos x }{\sin^3 x+\cos^3 x}dx = \frac{1}{2}\int_0^{\frac{\pi}{2}}\frac{\sec^2x}{\tan^2x-\tan x +1}dx\underbrace {=}_{\tan x=y} \frac{1}{2}\int_0^{\infty}\frac{dy}{y^2-y+1}=\frac{1}{\sqrt 3} \tan^{-1}\left(\frac{2y-1}{\sqrt3}\right)\bigg|_0^{\infty}=\frac{2\pi}{3\sqrt 3} $$
|
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|
CDF in Probability $$f(x)=\left\{\begin{array}{ll}
C & \text { for }-3 \leq x<3 \\
Dx & \text { for } 3 \leq x<5 \\
0 & \text { otherwise }
\end{array}\right.$$
Given that $P(-3 \leq x < 3) = \frac{2}{3}P(3 \leq x < 5)$, What is the cumulative distribution function $CDF(x)$.
So far I have tried to solve the following:
$P(-3 \leq x < 3) = \frac{2}{3}P(3 \leq x < 5)$
$\int_{-3}^x f(x)dx$ = $\frac{2}{3} \int_{3}^x f(x)dx$
$\int_{-3}^x (C)dx$ = $\frac{2}{3} \int_{3}^x (Dx)dx$
$\int_{-3}^x (C)dx$ = $ \frac{2}{3} \int_{3}^x (Dx)dx$
$ \left[Ax\right]_-3^x $ = $\frac{2}{2*3} \left[D* x^{2}\right]_3^x $
C [3 - (x)] = $ \frac{1}{3} \left [D * (x^{2} - 3^{2}\right] $
(3-x)* C = $ \frac{D*(x^{2} - 3^{2})}{3} $
C = $ -\frac{D*(x + 3)}{3} $
How do I find the values for C and D and then calculate the CDF?
|
$\int_{-3}^{x}f(t)dt \neq \frac{2}{3} \int_{3}^{x}f(t)dt $, this need not be true
but
$\int_{-3}^{3}f(x)dx = \frac{2}{3} \int_{3}^{5}f(x)dx $
other equation will be
$\int_{-\infty }^{\infty }f(x)dx = 1 $
Get $C,D$ from above two equations
Finally CDF = $\int_{-\infty }^{x }f(t)dt$
|
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|
The value of the following product is?
Evaluate the following product:
$$\newcommand{\T}[1]{\frac{\sin\frac{\theta}{#1}}{\tan^2\frac{\theta}{#1}\tan\frac{2\theta}{#1} + \tan\frac{\theta}{#1}}} \\
P(\theta) = \T{2} \times \T{2^2} \times \T{2^3} \times .... \infty$$
For $\theta = \frac \pi 4$
Simplified, $P(\theta)$ is
$$P(\theta) = \lim_{n \to \infty}\prod_{r=1}^n T(\theta,r)= \lim_{n \to \infty}\prod_{r=1}^n\T{2^r}$$
The denominator can be simplified as follows:
$$D = \tan\frac{\theta}{2^r}\left( \tan\frac{\theta}{2^r}\tan\frac{\theta}{2^{r-1}} + 1\right) \\
= \tan\frac{\theta}{2^{r-1}} - \tan\frac{\theta}{2^{r}}$$
After this, $P(\theta)$ becomes
$$P(\theta) = \lim_{n \to \infty}\prod_{r=1}^n \frac{\sin\frac{\theta}{2^r}}{\tan\frac{\theta}{2^{r-1}}- \tan\frac{\theta}{2^r}}$$
One more detail I found out is that $\lim_{n \to \infty} T(\theta,n) = 1$, but I couldn't proceed further from here. Any hints/solutions are appreciated.
EDIT: After the hints in the comments, $T(\theta, r)$ resolves to $\cos \frac \theta {2^{r-1}} \cos \frac \theta {2^r}$ as follows (assuming $\frac \theta {2^r} = t$)
$$\begin{gather}
T(\theta, n) = \frac{\sin t}{\tan^2t\tan 2t + \tan t} \\
= \frac{\cos t}{\tan t \tan 2t + 1} \\
= \frac{\cos t(1-\tan^2t)}{1+\tan^2t} \\
= \cos t \cos 2t \\
= \cos \frac \theta {2^{r-1}} \cos \frac \theta {2^r}
\end{gather}$$
Now,
$$P(\theta) = \lim_{n \to \infty} \frac{ \left( \cos\theta\cos\frac\theta2... \cos \frac{\theta}{2^n} \right)^2 }{\cos\theta} = \frac{\sin^2\theta}{2^{2n}\sin^2 \frac \theta {2^n}\cos \theta} = \frac{\sin^2 \theta}{\theta^2 \cos \theta}$$
Therefore,
$$\boxed{P(\pi/4) = \frac{8\sqrt2}{\pi^2}}$$
However, the answer mentioned in the textbook is $\frac{2}{\pi}$. Where am I going wrong? (I think there's a silly mistake somewhere here; just not able to find it :(
|
Carrying on from the answer:
$$\begin{gather}
T(\theta, n) = \frac{\sin t}{\tan^2t\tan 2t + \tan t} \\
= \frac{\cos t}{\tan t \tan 2t + 1} \\
= \frac{\cos t(1-\tan^2t)}{1+\tan^2t} \\
= \cos t \cos 2t \\
= \cos \frac \theta {2^{r-1}} \cos \frac \theta {2^r}
\end{gather}$$
Now,
$$P(\theta) = \cos\theta \cos \frac \theta 2 \cdot \cos\frac\theta2 \cos\frac\theta{2^2}\cdot ... = \left( \cos\theta \cos\frac\theta2...\right) \left( \cos\frac\theta2\cos\frac\theta{2^2}...\right)$$
Let
$$\begin{gather}
S = \lim_{n \to \infty}\cos\frac\theta2...\cos\frac\theta{2^n}\\
S\sin\frac\theta{2^n} = \lim_{n \to \infty} \frac{\sin\theta}{2^n} \\
S = \lim_{n \to \infty} \frac{\sin\theta}{2^n \sin\frac{\theta}{2^n}} = \lim_{t \to 0} \frac{t\sin\theta}{\sin(\theta t)}\\
S = \frac{\sin\theta}{\theta}
\end{gather}$$
Therefore, $P$ reduces to
$$P(\theta) = \frac{\sin2\theta\sin\theta}{2\theta^2}$$
And the value of $P(\pi/4)$ would be
$$P(\pi/4) = \frac{16}{2\sqrt2\pi^2}\\
\boxed{P(\pi/4) = \frac{4\sqrt2}{\pi^2}}$$
|
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|
Convergence of series with $a_{n + 1} = \sin{a_n}$ Given the sequence $a_n \mspace{10mu},\mspace{10mu}a_{n+1}=\sin(a_n)\mspace{10mu} and \mspace{10mu} a_1=1. $
Find if the series $\sum_{k=1}^{\infty}a_k$ converges or diverges.
First I found that $a_n$ is monotone decreasing sequence. If $a_n\in[0;\pi/2]$ then $\sin(a_n)\leqslant a_n$. And $a_n\leq 0$.
$\lim_{n\rightarrow \infty}a_n=0.$
Ratio test gives $\lim_{n\rightarrow \infty}\frac{a_{n+1}}{a_n}=1$.
More investigation of the series is required.
|
As you mentioned, $(a_n)$ decreases to $0$. Therefore, you have
$$a_{n+1}^{-2} - a_n^{-2} = \sin(a_n)^{-2} - a_n^{-2} = \left(a_n - \frac{a_n^3}{6} + o(a_n^4)\right)^{-2} - a_n^{-2}$$ $$=a_n^{-2} \left[\left(1 - \frac{a_n^2}{6} + o(a_n^3)\right)^{-2} - 1 \right] = a_n^{-2} \left(1 + \frac{a_n^2}{3} + o(a_n^2) - 1 \right) = \frac{1}{3} + o(1)$$
So the sequence $a_{n+1}^{-2} - a_n^{-2}$ converges to $\frac{1}{3}$. You can now use Cesaro to see that
$$a_n^{-2} \sim \frac{n}{3}$$
Therefore
$$a_n \sim \sqrt{\frac{3}{n}}$$
so by comparison, the series diverges.
|
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Find the sum of all possible values of $a$ such that the following equation $(x - a)^2 + (x^2 - 3x + 2)^2 = 0$ has real root in $x$ :-
Find the sum of all possible values of $a$ such that the following equation $(x - a)^2 + (x^2 - 3x + 2)^2 = 0$ has real root in $x$ :-
What I Tried :- I know $(x^2 - 3x + 2) = (x - 1)(x - 2)$ .
So :- $(x - a)^2 = - (x^2 - 3x + 2)^2$
=> $(x - a) = -(x^2 - 3x + 2)$
=> $(x - a) = -(x - 1)(x - 2) = (1 - x)(x - 2)$ .
From here I don't have a good hint or a clue to move forward . Can anyone help ?
|
Hint:sum of two non negative terms can be zero if they are both zero.
|
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If $x$, $y$ and $z$ satisfy $xy=1$, $yz=2$ and $xz=3$, what is the value of $x^2+y^2+z^2$? Express your answer as a common fraction.
If $x$, $y$ and $z$ satisfy $xy=1$, $yz=2$ and $xz=3$, what is the value of $x^2+y^2+z^2$? Express your answer as a common fraction.
I tried going along the path of computing $(x+y+z)^2$, which expands to $(x^2+y^2+z^2) + 2\cdot (xy+yz+xz)$, but I couldn't go anywhere from there (other than substituting xy, yz, xz, but that's not enough information.
|
Hint: Calculate $${(xy)(yz)\over xz}$$
|
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|
Finding $\lim_{x \to \infty} (x + \frac{2x^{3}}{3} - \frac{2(x^2+1)^{\frac{3}{2}}}{3})$ $\lim_{x \to \infty} (x + \frac{2x^{3}}{3} - \frac{2(x^2+1)^{\frac{3}{2}}}{3})$ this limit according to wolframalpha is equal to $0$.
So this is my work thus far
$\lim_{x \to \infty} (x + \frac{2x^{3}}{3} - \frac{2(x^2+1)^{\frac{3}{2}}}{3})$ output is $\infty - \infty$ which is indeterminate form.
So next I basically but it on the same denominator: $\frac{1}{3}$ $((3x + 2x^3 - 2(x^2+1)^{\frac{3}{2}})$ and turned $2(x^2+1)^{\frac{3}{2}}$ into something easier to work with $2\sqrt{x^2+1}+2x^{2}\sqrt{x^2+1}$
now the limit is $\frac{1}{3} \lim_{x \to \infty} ((3x + 2x^3-2\sqrt{x^2+1} -2x^{2}\sqrt{x^2+1})$ and this is where I am stuck to do next and lost.
|
$$x + \frac{2x^{3}}{3} - \frac{2(x^2+1)^{\frac{3}{2}}}{3}=\frac{\left(x+\frac{2}{3}x^3\right)^2-\frac{4}{9}(x^2+1)^3}{x + \frac{2x^{3}}{3}+ \frac{2(x^2+1)^{\frac{3}{2}}}{3}}=$$
$$=\frac{-\frac{1}{3}x^2-\frac{4}{9}}{x + \frac{2x^{3}}{3}+ \frac{2(x^2+1)^{\frac{3}{2}}}{3}}=\frac{-\frac{1}{3x}-\frac{4}{9x^3}}{\frac{1}{x^2} + \frac{2}{3}+ \frac{2(1+\frac{1}{x^2})^{\frac{3}{2}}}{3}}\rightarrow0$$ for $x\rightarrow+\infty.$
|
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|
condition on complex numbers to form cyclic quadrilateral. Consider complex numbers $z,z^2,z^3,z^4$ in that order which form a cyclic quadrilateral . If $\arg z=\alpha$ and $\alpha$ lies in $[0,2\pi]$.Find the values $\alpha$ can take.
I encountered this question in one competitive exam.i tried using the property of cyclic quadrilateral to get $$\arg\left(\frac{z^3-z^4}{z-z^4}\right)+\arg\left(\frac{z-z^2}{z^3-z^2}\right)=\pi$$ This can be simplfied furher but does not help.
I also tried using coni theorem but of no use. Answer given is alpha lies in $(0,\frac{2\pi}{3})and(\frac{4\pi}{3},2\pi)$
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By Ptolemy we obtain:
$$|z-z^2|\cdot|z^3-z^4|+|z-z^4|\cdot|z^2-z^3|=|z-z^3|\cdot|z^2-z^4|$$ or
$$|z|+|z^2+z+1|=|(z+1)^2|.$$
Now, we can use a triangle inequality.
Id est, for $|z|=r$ we obtain:
$$(\cos\alpha,\sin\alpha)||(r^2\cos2\alpha+r\cos\alpha+1,r^2\sin2\alpha+r\sin\alpha),$$ which gives
$$\sin\alpha(r^2\cos2\alpha+r\cos\alpha+1)=\cos\alpha(r^2\sin2\alpha+r\sin\alpha)$$ or
$$\sin\alpha=r^2\sin\alpha$$ and since $\sin\alpha\neq0$, we obtain $r=1$.
|
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|
Five roots of $x^5+x+1=0$ and the value of $\prod_{k=1}^{5} (2+x_k^2)$
*
*Here, $x_{k}$ are five roots of $x^{5} + x + 1 = 0$.
*I know two roots are $\omega, \omega^{2}$ and next I can find a cubic dividing it by $x^{2} + x + 1$ and using the connection of $3$ roots with the coefficients of this cubic( Vieta's formulas ).
*But the calculation becomes very tedious, where I do not get the required value of $\prod_{k = 1}^{5}\left(2 + x_{k}^{2}\right) = 51$.
Can there be a simpler way of doing this ?.
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Note that $\prod_{k=1}^5 (2+x_k^2) = \prod_{k=1}^5 (\sqrt{2}+ix_k)(\sqrt 2 - ix_k)$.
This is the product of all roots of a polynomial, whose roots are exactly $\sqrt{2} \pm ix_k$ for $k=1,...,5$.
Note that if $x^5+x+1$ has roots $x_1,...,x_5$, then $p(y) = (-iy+\sqrt 2i)^5 + (-iy+\sqrt 2i) + 1$ has roots $\sqrt 2 + ix_k$, $k=1,...,5$. The conjugate of this polynomial $\bar{p}$ has roots $\sqrt 2 - ix_k$.
Which means that the polynomial which has roots exactly equal to those we want, is $p\bar p$, and we need just the constant term of this whole polynomial, because by Vieta that is the product of all the roots. The constant of $p$ is $\sqrt 2^5i^5 + \sqrt 2i^5 +1 = 5\sqrt 2i + 1$, similarly of $\bar{p}$ is $1-5\sqrt 2 i$. Multiply these to get $1+(5\sqrt 2)^2 = 1+50=51$ and we are done.
|
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|
Find the inequality with the best possible $k= constant$ (with the condition $x^{2}+ y^{2}\leq k$).
Find the inequality with the best possible $constant$
*
*Given two non-negative numbers $x, y$ so that $x^{2}+ y^{2}\leq \frac{2}{7}$. Prove that
$$\frac{1}{1+ x^{2}}+ \frac{1}{1+ y^{2}}+ \frac{1}{1+ xy}\leq \frac{3}{1+ \left ( \frac{x+ y}{2} \right )^{2}}$$
where $constant= \frac{2}{7}$ is the best possible.
*Given two non-negative numbers $x, y$ so that $x^{2}+ y^{2}\leq \frac{2}{5}$. Prove that
$$\frac{1}{\sqrt{1+ x^{2}}}+ \frac{1}{\sqrt{1+ y^{2}}}+ \frac{1}{\sqrt{1+ xy}}\leq \frac{3}{\sqrt{1+ \left ( \frac{x+ y}{2} \right )^{2}}}$$
where $constant= \frac{2}{5}$ is the best possible.
They are my two examples. I'm looking forward to seeing more many inequalities alike. Thanks for all your nice comments.
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For example.
Let $x$ and $y$ be non-negative numbers such that $x^2+y^2\leq\frac{2}{15}.$ Prove that:
$$\frac{1}{1+ x^{2}}+ \frac{1}{1+ y^{2}}+ \frac{1.5}{1+ xy}\leq \frac{3.5}{1+ \left ( \frac{x+ y}{2} \right )^{2}},$$
where $\frac{2}{15}$ is a best such constant.
It's interesting that:
For any non-negatives $x$ and $y$ the following inequality is true.
$$\frac{1}{1+ x^{2}}+ \frac{1}{1+ y^{2}}+ \frac{2}{1+ xy}\geq \frac{4}{1+ \left ( \frac{x+ y}{2} \right )^{2}}$$
By the way, the last inequality is true for any reals $x$ and $y$ such that $xy+1>0.$
|
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|
If $\tan\theta =\cos2\alpha\tan\phi$ then prove that $\tan(\phi-\theta)=\frac{\tan^2\alpha\sin2\phi}{1+\tan^2\alpha\cos2\phi}$ If $\tan\theta =\cos2\alpha\tan\phi$ then prove that $\tan(\phi-\theta)=\frac{\tan^2\alpha\sin2\phi}{1+\tan^2\alpha\cos2\phi}$
I have tried in a few different ways applying the formula $\tan(\phi - \theta) = \frac{\tan \phi - \tan \theta} {1+\tan \phi \tan \theta}$ then substituting $\tan \theta = \cos 2\alpha \tan \phi$. But all in vain. Assuming the problem to be correct, my highest achievement was getting the numerator correct, but the denominator really gives me a lot of headache. Please guide me. Thanks in advance.
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$\tan(\phi-\theta)=\frac{\tan\phi-\tan\theta}{1+\tan\phi\tan\theta}$.
If we replace $\tan\theta=\cos 2\alpha \tan \phi$ we get
\begin{align*}
\frac{\tan\phi-\cos 2\alpha\tan\phi}{1+\tan\phi\cos2\alpha\tan\phi}&=\frac{\tan\phi-(\cos^2\alpha-\sin^2 \alpha)\tan\phi}{1+\tan^2\phi(\cos^2\alpha-\sin^2\alpha)}\\
&=\frac{2\sin^2\alpha\tan\phi\cos^2\phi}{\cos^2\phi+\sin^2\phi(\cos^2\alpha-\sin^2\alpha)}\\
&=\frac{2\sin^2\alpha \sin\phi\cos\phi}{\cos^2\phi+\sin^2\phi(1-2\sin^2\alpha)}\\
&=\frac{\sin^2\alpha \sin 2\phi}{1-2\sin^2\alpha\sin^2\phi}\\
&=\frac{\sin^2\alpha \sin 2\phi}{\cos^2\alpha+\sin^2\alpha-2\sin^2\alpha\sin^2\phi}\\
&=\frac{\sin^2\alpha\sin 2\phi}{\cos^2\alpha+\sin^2\alpha(1-2\sin^2\phi)}\\&=\frac{\sin^2\alpha\sin2\phi}{\cos^2\alpha(1+\tan^2\alpha \cos2\phi)}\\
&=\frac{\tan^2\alpha \sin 2\phi }{1+\tan^2\alpha \cos2\phi}.
\end{align*}
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Sum of Telescopic Series Question :
$$\sum_{r=1}^n\frac{(r^2-r-1)}{(r+1)!}=?$$
I tried but I got $(-1)!$ in the end. This is what I tried:
$$
\require{cancel}
\frac{r^2-1}{(r+1)!}-\frac{r}{(r+1)!}\\
\frac{\cancel{(r+1)}\cancel{(r-1)}}{\cancel{(r+1)}r\cancel{(r-1)}(r-2)!}-\frac{\cancel{r}}{(r+1)\cancel{r}(r-1)!}\\
T_r=\frac{1}{r(r-2)!}-\frac{1}{(r+1)(r-1)!}\\
\sum_{r=1}^{n} T_r = \frac{1}{1(1-2)!}-\frac{1}{(n+1)(n-1)!}\\
= \frac{1}{(-1)!}-\frac{1}{(n+1)(n-1)!}\\
$$
Please point out my mistake.
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$$\begin{eqnarray*}\frac{r^2-r-1}{(r+1)!}&=&\frac{r^2-1}{(r+1)!}-\frac{r}{(r+1)!}\\&=&\frac{(r-1)(r+1)}{(r+1)!}-\frac{r}{(r+1)!}\\&=&\frac{r-1}{r!}-\frac{r}{(r+1)!}\end{eqnarray*}$$
Therefore:
$$\begin{eqnarray*}\sum_{r=1}^n\frac{(r^2-r-1)}{(r+1)!}&=&\sum_{r=1}^n\left[\frac{r-1}{r!}-\frac{r}{(r+1)!}\right]\\&=&\frac{0}{1!}-\frac{1}{2!}+\frac{1}{2!}-\frac{2}{3!}+\frac{2}{3!}-\frac{3}{4!}+...+\frac{n-1}{n!}-\frac{n}{(n+1)!}\\&=&\color{blue}{-\frac{n}{(n+1)!}}\end{eqnarray*}$$
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|
System of Three Equations - Prove that at least two numbers' absolute value is equal. Question :
Prove that at least two of the numbers $a,b,c$ must have the absolute values equal in order that the system of following equations in $x,y$ may be consistent
$$ a^2x+b^2y+c^2=0$$
$$a^4x+b^4y+c^4 =0$$
$$x+y+1=0$$
I found $x=\frac{c^2(b^2-c^2)}{a^2(a^2-b^2)}$ and $y=\frac{c^2(c^2-a^2)}{b^2(a^2-b^2)}$ from first two equation.
After putting $x$ and $y$ in $3rd$ equation.
$$\frac{c^2(b^2-c^2)}{a^2(a^2-b^2)}+\frac{c^2(c^2-a^2)}{b^2(a^2-b^2)}+1=0 $$
I did prove it but I am not satisfied with the process as I have to use an identity(never seen and found in the simplification)$$a^2b^2(a^2-b^2)+b^2c^2(b^2-c^2)+c^2a^2(c^2-a^2)=-(a^2-b^2)(b^2-c^2)(c^2-a^2)$$ and the simplification nearly took 30 minutes(which is not good in exam)
Is there any other fast and efficient way to prove or simplify this ?
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$$a^2x+a^2y+a^2=0$$
$$\implies (a^2-b^2)y+(a^2-c^2)=0$$
$$a^4x+a^4y+a^4=0$$
$$\implies (a^4-b^4)y+(a^4-c^4)=0$$
If $|a|\neq |b|$ and $|a|\neq |c|$, then ${a^4-b^4\over a^2-b^2}={a^4-c^4\over a^2-c^2}$ which imlies $a^2+b^2=a^2+c^2$ which is $|b|=|c|$
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Evaluating two integrals involving $\tan^{-1}\left(\frac{\sqrt{x(1-x)}}{x+\frac12}\right)$
I want to show
$$I:=\int_0^{1}\tan^{-1}\left(\frac{\sqrt{x(1-x)}}{x+\frac12}\right)dx=\frac{\pi}{8}$$
and
$$J:=\int_0^{1}\frac{1}{1-x}\tan^{-1}\left(\frac{\sqrt{x(1-x)}}{x+\frac12}\right)dx=\pi\log \frac{3}{2}$$
My work:
Let us try to do the first one. Note that by making a change of variable $x\mapsto 1-x$ we have
$$I=\int_0^1 \tan^{-1}\left(\frac{\sqrt{x(1-x)}}{\frac32-x}\right)dx$$
Adding the two expression of $I$ and doing some simple algebra lead us to
$$2I=\int_0^1 \tan^{-1}\left(\frac{8}{3}\sqrt{x(1-x)}\right)dx$$
We can obtain similar expression for $2J$ as well. But I am not sure how to deal with this integral. The fractor $\frac{3}{8}$ looks really odd here. However, if you plug it in wolframalpha they indeed give you the desired result. I also tried substituting $x=\sin^2\theta$ or $x=\cos^2\theta$. The expression didn't simplify. Perhaps there is some clever way to do it.
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We can use an Euler substitution to simplify both integrals, namely: $\frac{\sqrt{x(1-x)}}{x}=t\Rightarrow x=\frac{1}{1+t^2}$.
$$I=\int_0^1\arctan\left(\frac{\sqrt{x(1-x)}}{\frac{1}{2}+x}\right)dx=\int_0^\infty\arctan\left(\frac{2t}{3+t^2}\right)\left(\frac{1}{1+t^2}\right)'dt$$
$$\overset{IBP}=\int_0^\infty \left(\frac{1}{1+t^2}-\frac{3}{9+t^2}\right)\frac{1}{1+t^2}dt=\color{blue}{\int_0^\infty \frac{1}{(1+t^2)^2}dt}-\int_0^\infty \frac{1}{1+t^2}\frac{3}{9+t^2}dt$$
$$\overset{\color{blue}{t\to \frac{1}{t}}}=\color{blue}{\frac12\int_0^\infty \frac{1}{1+t^2}dt}+\frac{1}{8}\int_0^\infty \frac{3}{9+t^2}dt-\frac{3}{8}\int_0^\infty \frac{1}{1+t^2}dt=\frac{\pi}{8}$$
Similarly for the second integral we obtain:
$$J=\int_0^1\arctan\left(\frac{\sqrt{x(1-x)}}{\frac{1}{2}+x}\right)\frac{dx}{1-x}=2\int_0^\infty \arctan\left(\frac{2t}{3+t^2}\right) \frac{1}{t(1+t^2)}dt$$
$$\overset{IBP}=\int_0^\infty \left(\frac{1}{1+t^2}-\frac{3}{9+t^2}\right)(\ln(1+t^2)-2\ln t)dt$$
Now we can differentiate under the integral sign considering the following integral:
$$J(a)=\int_0^\infty \left(\frac{1}{1+t^2}-\frac{3}{9+t^2}\right)(\ln(a^2+t^2)-2\ln t)dt$$
$$\Rightarrow J'(a)=\int_0^\infty \left(\frac{1}{1+t^2}-\frac{3}{9+t^2}\right)\frac{2a}{a^2+t^2}dt$$
$$=\int_0^\infty \left(\frac{1}{1-a^2}\frac{2a}{a^2+t^2}-\frac{2a}{1-a^2}\frac{1}{1+t^2}+\frac{2a}{9-a^2}\frac{3}{9+t^2}-\frac{3}{9-a^2}\frac{2a}{a^2+t^2}\right)dt$$
$$=\pi\left(\frac{1}{1-a^2}-\frac{a}{1-a^2}+\frac{a}{9-a^2}-\frac{3}{9-a^2}\right)=\pi\left(\frac{1}{1+a}-\frac{1}{3+a}\right)$$
$$J(0)=0\Rightarrow J=\pi\int_0^1 \left(\frac{1}{1+a}-\frac{1}{3+a}\right)da=\pi\ln\left(\frac{3}{2}\right)$$
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Sum of the roots of $x^2-6[x]+6=0$, where $[.]$ is GIF Sum of the roots of $x^2-6[x]+6=0$, where $[.]$ is GIF
I have done this problem by inspection as $$\frac{x^2+6}{6}=[x] \implies x>0.$$
Let [x]=0, then $x$ is non real. Let $[x]=1$, then $x=0$ which contradicts. Let $[x]=2$, it gives $x=\sqrt{6}$, in agrrement. Similarly assuming $[x]=3,4$; we get correct roots as $\sqrt{12}$ and $\sqrt{18}$. But if er let $[x]=5$, it gives $x=\sqrt{24}$. which contradicts. So I get the sum of roots as $\sqrt{6}(1+\sqrt{2}+\sqrt{3}).$
The question is: Have I found all the real roots. In any case, what is(are) more appropriate method(s) of doing it.
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$$\left\lfloor x \right\rfloor\le x$$
$$-6\left\lfloor x \right\rfloor\ge -6x$$
$$x^2-6\left\lfloor x \right\rfloor+6\ge x^2-6x+6$$
$$0\ge x^2-6x+6$$
$$x\in[3-\sqrt{3},3+\sqrt{3}] \ \ \ \& \ \ \left\lfloor x \right\rfloor\in\{1,2,3,4\}$$
so let's consider the cases:
$1^{\circ}$ $\left\lfloor x \right\rfloor=1\Rightarrow x=0$ which is not a root of $x^2-6\left\lfloor x \right\rfloor+6$
$2^{\circ}$ $\left\lfloor x \right\rfloor=2\Rightarrow x=\sqrt{6} $ which is a root of $x^2-6\left\lfloor x \right\rfloor+6$
$3^{\circ}$ $\left\lfloor x \right\rfloor=3\Rightarrow x=\sqrt{12} $ which is a root of $x^2-6\left\lfloor x \right\rfloor+6$
$4^{\circ}$ $\left\lfloor x \right\rfloor=4\Rightarrow x=\sqrt{18} $ which is a root of $x^2-6\left\lfloor x \right\rfloor+6$
$$\sum \text{roots}=\sqrt{6}+\sqrt{12}+\sqrt{18} $$
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Show that $\frac{\left(2n\right)!}{2^{n}n!} \ge \frac{\left(2n\right)!}{\left(n+1\right)^{n}}$. I am attempting to solve this by induction, yet I am stuck on the last part. Below is my attempted solution.
Show that $\frac{\left(2n\right)!}{2^{n}n!} \ge \frac{\left(2n\right)!}{\left(n+1\right)^{n}}, \forall n \in \mathbf{N} $.
My attempted solution, by induction:
For $n=1$ we have,
$\frac{\left(2\right)!}{2^{1} \cdot 1!} \ge \frac{\left(2\right)!}{\left(1+1\right)^{1}} $
$2 \ge 1$.
Thus, the inequality holds for $n=1$.
Now assume that it holds $\forall k \in \mathbf{N}$.
Then, for $k+1$,
$\left(k+1\right) \frac{\left(2k\right)!}{2^{k}k!} \ge \frac{\left(2k\right)!}{\left(k+1\right)^{k}} \left(k+1\right) $.
This is where I am stuck.
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By the AM-GM inequality you have
$$\sqrt{1 \cdot n} \leq \frac{n+1}{2} \\
\sqrt{2 \cdot (n-1)} \leq\frac{n+1}{2} \\
...\\
\sqrt{n \cdot 1} \leq \frac{n+1}{2} $$
Multiplying everything together you get
$$n! \leq \left( \frac{n+1}{2}\right)^n$$
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Solve system of equations $\begin{align} \begin{cases} |y|=|x-3| \\(2\sqrt{z}+2-y)y=1+4y \\ x^2+z-4x=0\end{cases} \end{align}$ Solve system of equations$$\begin{align}
\begin{cases}
|y|=|x-3| \\
(2\sqrt{z}+2-y)y=1+4y \\
x^2+z-4x=0
\end{cases}
\end{align}$$
*
*$$y=3-x;z=4x-x^2$$
Then the second equation is equivalent with $$2\sqrt {4x-x^2}=\frac{x^2-4x+4}{x-3}(0\le x\le 4)$$
This equation has no root but i do not know how to prove it. In 2nd case $y=3-x$ i have done but i am stuck on this case.
Help me, ty.
|
In this case $4x-x^2\geq0,$ which gives $0\leq x\leq4.$
Now, after substitution $z=4x-x^2$ rewrite the second equation in the form:
$$2(3-x)\sqrt{4x-x^2}=(x-4)^2,$$ which gives $x=4$ or
$$2(3-x)\sqrt{x}=\sqrt{(4-x)^3},$$ which gives also $0<x<3$ and after squaring of the both sides we obtain:
$$4x(3-x)^2=(4-x)^3$$ or $$(x-2)^2(16-5x)=0,$$ which gives also $x=2.$
In the second case we'll prove that the equation $$2(x-3)\sqrt{4x-x^2}=(x-2)^2$$ has no real roots for $3<x<4.$
For which we'll prove that:
$$(x-2)^4>4(x-3)^2(4x-x^2)$$ or
$$5x^4-48x^3+156x^2-176x+16>0$$ for which it's enough to prove that
$$5x^4-48x^3+156x^2-176x+15>0$$ or
$$(x-3)(5x^3-33x^2+57x-5)>0$$ or
$$(5x^3-33x^2+55x)+(2x-5)>0,$$ which is obvious for any $x>3.$
|
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|
Why are the functions $y=x^{\sin^2(x)}+x^{\cos^2(x)}$ and $y=(x-2\sqrt{x}+1)\cos^2(2x)+2\sqrt{x}$ so similar? I was just playing around with functions and plotting them in Desmos and I found that these functions are remarkably similar despite their totally different expressions. The functions are
$$y=x^{\sin^2(x)}+x^{\cos^2(x)}$$ and $$y=(x-2\sqrt{x}+1)\cos^2(2x)+2\sqrt{x}$$
Link to Desmos page:
https://www.desmos.com/calculator/meyktbtz0i
Edit: They aren't exactly the same if you zoom in closer as pointed out to me.
|
We have that
$$f(x)=x^{\sin^2(x)}+x^{\cos^2(x)}=x^{1-\cos^2(x)}+x^{\cos^2(x)}$$
$$g(x)=(x-2\sqrt{x}+1)\cos^2(2x)+2\sqrt{x}=$$
$$=(x-2\sqrt{x}+1)\cos^2(2x)+2\sqrt{x}(\cos^2(2x)+\sin^2(2x))=$$
$$=(x+1)\cos^2(2x)+2\sqrt{x}\sin^2(2x)$$
with
$$2\sqrt x \le f(x), g(x) \le x+1$$
and since, as noticed in the graph linked to your question, we have that
*
*at $x=\frac \pi 4 + \frac12k\pi$
$$f(x)=x^{\sin^2(x)}+x^{\cos^2(x)}=2\sqrt x \quad f'(x)=\frac1{\sqrt x}$$
$$g(x)=(x+1)\cos^2(2x)+2\sqrt{x}\sin^2(2x)=2\sqrt x \quad g'(x)=\frac1{\sqrt x}$$
*
*at $x=\frac \pi 2 + \frac12k\pi$
$$f(x)=x^{\sin^2(x)}+x^{\cos^2(x)}=x+1 \quad f'(x)=1$$
$$g(x)=(x+1)\cos^2(2x)+2\sqrt{x}\sin^2(2x)=x+1 \quad g'(x)=1$$
the two functions coincide in infinitely many points with the same slope at these points thus they appear very close one to each other.
|
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|
Intuitive explanation for $\lim\limits_{n\to\infty}\left(\frac{\left(n!\right)^{2}}{\left(n-x\right)!\left(n+x\right)!}\right)^{n}=e^{-x^2}$ In this post I noticed (at first numerically) that:
$$\lim\limits_{n\to\infty}\left(\frac{\left(n!\right)^{2}}{\left(n-x\right)!\left(n+x\right)!}\right)^{n}=e^{-x^2}$$
This can be proved by looking at the Taylor expansion
$$n\ln\left(\frac{\left(n!\right)^{2}}{\left(n-x\right)!\left(n+x\right)!}\right)=-2n\sum_{k=1}^{\infty}\frac{\psi^{(2k-1)}\left(n+1\right)}{\left(2k\right)!}x^{2k}$$
and the asymptotic expansion
$$\psi^{(m)}(n+1)=\left(-1\right)^{\left(m+1\right)}\sum_{k=0}^{\infty}\frac{\left(k+m-1\right)!}{k!}\frac{B_k}{n^{k+m}}$$
where we have chosen $B_1=-\frac12$.
However, this limit seems so beautiful and interesting that it produces the Gaussian function. It makes me wonder if there is a more intuitive way to understand this limit, possibly in a context of probabilities.
|
Consider first
$$ \begin{align}
\frac{(n+5)!}{n!} &=(n+5)(n+4)\cdots (n+1)\\
&= n^5 (1+5/n)(1+4/n) \cdots (1+1/n) \\
& \approx n^5 \left(1 + \frac{5+4+\cdots+1}{n}\right) \tag1\\
& = n^5 \left(1 + \frac{5\times6}{2n}\right) \\
\end{align}$$
Or, in general
$$ \frac{(n+x)!}{n!} \approx n^x \left(1 + \frac{x(x+1)}{2n}\right) \tag2$$
Similarly:
$$ \frac{n!}{(n-x)!} \approx n^x \left(1 - \frac{x(x-1)}{2n}\right) \tag3$$
Then
$$
\begin{align}
\frac{\left(n!\right)^{2}}{\left(n-x\right)!\left(n+x\right)!}
&\approx \frac{\left(1 - \frac{x(x-1)}{2n}\right)}{\left(1 + \frac{x(x+1)}{2n}\right)}\\
&\approx \left(1 - \frac{x(x-1)}{2n}\right) \left(1 - \frac{x(x+1)}{2n}\right)\\
&\approx \left(1 - \frac{x^2}{n}\right) \tag4
\end{align}
$$
And, using the standard exponential limit:
$$\lim_{n\to \infty} \left(1 - \frac{x^2}{n}\right)^n =e^{-x^2} \tag5$$
Edit Result $(4)$ can also be obtained by a probabilistic reasoning: Imagine the following scenario: from a bag with $n$ white balls and $x$ black balls we pick $x$ balls. Which is the probability that all picked balls are white?
This is $$ \frac{\binom{n}{x}}{\binom{n+x}{x}}=\frac{\left(n!\right)^{2}}{\left(n-x\right)!\left(n+x\right)!}$$
precisely the LHS in eq $(4)$.
Now, if $n \gg x$, the probability that more than one of the picked balls are black is negligible, hence we can approximate this by $P \approx 1 - x \frac{x}{n}= 1- x^2/n$
BTW: The complete term in the original limit, $\left(\frac{\left(n!\right)^{2}}{\left(n-x\right)!\left(n+x\right)!}\right)^{n}$, can then be regarded as the probability of getting no black balls (after adding $n$ white balls) in $n$ tries.
I don't see how this probability could be associated with a gaussian density.
|
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|
Find a matrix $A$ such that $X$ generates the subspace $W$ (the solution space of the system $AX=0$.) Consider $W$ a subspace of $\mathbb{R}^{5}$ generated by:
\begin{align*}
X=\left \{(1,-1,0,5,1), (1,0,1,0,-2),(-2,0,-1,0,-1)\right \}
\end{align*}
Find a system of linear equations $AX=0$ such that W be the solution space of the system.
I understand that what I've to do is to find a matrix $A_{5x5}$ such that:
\begin{align*}
A \begin{pmatrix}
1\\
-1\\
0\\
5\\
1
\end{pmatrix}=\begin{pmatrix}
0\\
0\\
0\\
0\\
0
\end{pmatrix} \ \text{ , } \ A \begin{pmatrix}
1\\
0\\
1\\
0\\
-2
\end{pmatrix}=\begin{pmatrix}
0\\
0\\
0\\
0\\
0
\end{pmatrix} \ \text{ and, } \ A \begin{pmatrix}
-2\\
0\\
-1\\
0\\
-1
\end{pmatrix}=\begin{pmatrix}
0\\
0\\
0\\
0\\
0
\end{pmatrix}
\end{align*}
So I think that the solution can be to propose a new system of linear equations but I'm not sure how can I do it. How can I find the matrix $A$?
|
Hint: The kernel is the orthogonal complement of the row space. So $A$ needs row space equal to $W^\perp$.
|
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|
Proving $\cos(4x)=8\cos^4(x)-8\cos^2(x)+1$ without using LHS and RHS
Prove that
$$\cos(4x)=8\cos^4(x)-8\cos^2(x)+1$$
My Attempt
$$\Rightarrow \cos^2(2x)-\sin^2(2x) =8\cos^4(x)-8\cos^2(x)+1$$
Add 1 to both sides
$$\cos^2(2x)-\sin^2(2x) +1=8\cos^4(x)-8\cos^2(x)+2$$$$\Rightarrow \cos^2(2x)=4\cos^4(x)-4\cos^2(x)+1$$
$$\Rightarrow \cos^2(2x)=(2\cos^2(x)-1)^2$$ $$\Rightarrow \cos^2(2x)=(\cos^2(x)-\sin^2(x))^2=(\cos(2x))^2$$
$$\therefore \cos(4x)=8\cos^4(x)-8\cos^2(x)+1$$
My professor says that this is an invalid proof as it was not proven using LHS and RHS and assumed that they were equivalent to add 1 to each side. My question: are there any proofs that would not work if you assumed that the two side were equivalent? Any examples will be much appreciated!
|
For example: $$8\cos^4x-8\cos^2x+1=1-8\cos^2x\sin^2x=1-2\sin^22x=\cos4x.$$
Also, we can use your idea:
$$\cos4x=2\cos^22x-1=2(2\cos^2x-1)^2-1=8\cos^4x-8\cos^2x+1.$$
|
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|
Finding $x$ when $\sqrt{3x - 4} + \sqrt[3]{5 - 3x} = 1$ Find $x$ if $\sqrt{3x - 4} + \sqrt[3]{5 - 3x} = 1.$
I was thinking of trying to substitute some number $y$ written in terms of $x$ than solving for $y$ to solve for $x.$ However, I'm not sure what $y$ to input, so can someone give me a hint?
|
$$\sqrt{3x - 4} + \sqrt[3]{5 - 3x} = 1$$
$$ \sqrt[3]{5 - 3x} = 1-\sqrt{3x - 4}$$
$$ 5 - 3x = (1-\sqrt{3x - 4})^3$$
$$ 5 - 3x = (1-3x)\sqrt{3x-4}+9x-11$$
$$ -4(3x-4)= (1-3x)\sqrt{3x-4}$$
$$ -4(3x-4)= (1-3x)\sqrt{3x-4}$$
$$ -4\sqrt{3x-4}\cdot \sqrt{3x-4}= (1-3x)\sqrt{3x-4}$$
so $x=4/3$ or: $$ -4 \sqrt{3x-4}=1-3x$$
which is easier. You can finish?
|
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|
Proving $\frac{1+\cos\theta}{\sec\theta-\tan\ \theta}+\frac{\cos\theta-1}{\sec\theta+\tan\ \theta}=2+2\tan\ \theta$
Prove this trigonometric identity:
$$\frac{1+\cos\theta}{\sec\theta-\tan\ \theta}+\frac{\cos\theta-1}{\sec\theta+\tan\ \theta}=2+2\tan\ \theta$$
I've simplified it until
$$\frac{2\cos^2\theta}{1-\sin\theta}$$
but couldn't get $2+2\tan\theta$ from it.
|
First, let's make a common denominator of $\frac{1+\cos\theta}{\sec\theta-\tan\theta}+\frac{\cos\theta-1}{\sec\theta+\tan\theta}$.
Ready, set, go!
$$\require{cancel}\begin{aligned}\color{blue}{\frac{1+\cos\theta}{\sec\theta-\tan\theta}}+\color{red}{\frac{\cos\theta-1}{\sec\theta+\tan\theta}}&=\frac{\color{blue}{\left(1+\cos\theta\right)\left(\sec\theta+\tan\theta\right)}+\color{red}{\left(\cos\theta-1\right)\left(\sec\theta-\tan\theta\right)}}{\color{blue}{\left(\sec\theta+\tan\theta\right)}\color{red}{\left(\sec\theta-\tan\theta\right)}}\\&=\frac{\color{blue}{\cancel{\sec x}+\tan x+\cos{\theta}\sec{\theta}\cancel{+\cos{\theta}}\tan{\theta}}+\color{red}{\cos{\theta}\sec{\theta}\cancel{-\cos{\theta}\tan{\theta}}\cancel{-\sec x}+\tan x}}{\color{blue}{\left(\sec x+\tan x\right)}\color{red}{\left(\sec x-\tan x\right)}}\\&=\frac{2\tan x+2\cos{\theta}\sec{\theta}}{\sec^{2}x-\tan^{2}x}&&\text{The denominator is equal to 1.}\\&=\frac{2\tan x+2\cos{\theta}\sec{\theta}}{1}\\&=2\tan x+2\cos{\theta}\sec{\theta}\\&=2\tan\theta+2\cancel{\cos\theta}\cdot\frac{1}{\cancel{\cos\theta}}\\&=2+2\tan\theta\end{aligned}$$
I hope this helps.
|
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|
Minimum of a function without calculus. $a=\frac{{(1+t^2)}^3}{t^4}$
Find the the minimum value of $a$. .$$a=\frac{{(1+t^2)}^3}{t^4}$$.
Instead of calculus i tried using the AM-GM inequality.,as follows:we have $$3+\frac{1}{t^4}+\frac{3}{t^2}+3\left( \frac{t^2}{3}\right)\ge 3+{\left(\frac{1}{9}\right)}^{1/5}$$ which is not the minimum i got by using calculus.What mistake am i making using this inequality? And also it would be very helpul if anyone could arrive at the minima using basic inequalities.(withoutv calculus)
using calculus minima occured at ,$t=\sqrt{2}$.
|
Using the AM-GM inequality, we have
$$1+t^2 = 1+\frac{t^2}{2}+\frac{t^2}{2} \geqslant 3\sqrt[3]{1 \cdot \frac{t^2}{2} \cdot \frac{t^2}{2}} = 3\sqrt[3]{\frac{t^4}{4}},$$
therefore
$$(1+t^2)^3 \geqslant \frac{27}{4}t^4,$$
or
$$\frac{(1+t^2)^3}{t^4}\geqslant \frac{27}{4}.$$
Equality occur when $\frac{t^2}{2} = 1,$ or $t = \pm \sqrt{2}.$
|
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|
Determine a in the system such that the system is consistent. The system is:
$x_1 + 2x_2 - x_3 = 2, $
$2x_1 - x_2 + x_3 = 1 $,
$-x_1 + 4x_2 -2x_3 = a $
To begin solving the system, I did Row 2 - 2(Row 1). Row 3 + Row 1, then Row 3 + Row 2.
This left me with
\begin{bmatrix}1&2&-1&2\\0&-5&3&-3\\0&0&0&a-1\end{bmatrix}
If all I am trying to find is if the system is inconsistent, do I need to continue reducing it completely to reduced echelon form? Or is it that since I solved the row with a, I can stop there leaving the system as consistent when a = 1.
|
Generally you don't need to continue reducing to reduced echelon form since you can see that for the system to be consistent you need $a-1=0$ so $a=1$. But there is a a mistake.
Row 3 + Row 1 gives
$$\begin{bmatrix}1 & 2 & -1 & 2\\0 & -5 & 3 & -3\\0 & 6 & -3 & a+2\end{bmatrix}$$
so Row 3 + Row 2 will give
$$\begin{bmatrix}1 & 2 & -1 & 2\\0 & -5 & 3 & -3\\0 & 1 & 0 & a-1\end{bmatrix}$$
Then you can continue to do Row 2 + 5$\times$Row3 to give
$$\begin{bmatrix}1 & 2 & -1 & 2\\0 & 0 & 3 & 5a-8\\0 & 1 & 0 & a-1\end{bmatrix}$$
and similarly Row 1 -2$\times$Row 3
$$\begin{bmatrix}1 & 0 & -1 & 4-2a\\0 & 0 & 3 & 5a-8\\0 & 1 & 0 & a-1\end{bmatrix}$$
and finally Row 1 +$\frac{1}{3}\times$ Row 2 gives
$$\begin{bmatrix}1 & 0 & 0 & \frac{4}{3}-\frac{a}{3}\\0 & 0 & 3 & 5a-8\\0 & 1 & 0 & a-1\end{bmatrix}$$
So you can see that the solutions are $x_{1}=\frac{4}{3}-\frac{a}{3},x_{2}=a-1$ and $x_{3}=\frac{5a}{3}-\frac{8}{3}$. Thus the system is consistent for all a.
Indeed we can check:
*
*$x_{1}+2x_{2}-x_{3}=\frac{4}{3}-\frac{a}{3}+2a-2-\frac{5a}{3}+\frac{8}{3}=2$
*$2x_{1}-x_{2}+x_{3}=\frac{8}{3}-\frac{2a}{3}-a+1+\frac{5a}{3}-\frac{8}{3}=1$
*$-x_{1}+4x_{2}-2x_{3}=-\frac{4}{3}+\frac{a}{3}+4a-4-\frac{10a}{3}+\frac{16}{3}=a$
as required.
|
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|
There are triads of perfect squares that are consecutive terms of arithmetic progression? Prove that there exist infinitely many triples of positive integers $ x , y , z $ for which the numbers $ x(x+1) , y(y+1) , z(z+1) $ form an increasing arithmetic progression.
$ \bigg( $ It is equivalent to find all triples of $ 4x(x+1)+1=(2x+1)^{2} , 4y(y+1)+1=(2y+1)^{2} , 4z(z+1)+1=(2z+1)^{2} $ $ \bigg) $
Note : I know $ \big( 1^{2} , 5^{2} , 7^{2} \big) $ , $ \big( 7^{2} , 13^{2} , 17^{2} \big) $ , $ \big( 7^{2} , 17^{2} , 23^{2} \big) $ , $ \big( 17^{2} , 25^{2} , 31^{2} \big) $ , but how i can found all triples ?
|
We have $$x^2+x+z^2+z=2(y^2+y)$$ or $$(x+z+1)^2+(x-z)^2=(2y+1)^2$$
We got Pythagorean triples.
Let $x+z+1=m^2-n^2$ and $x-z=2mn$, where $m>n$ and $m$ and $n$ have a different parity.
Thus, $$(x,y,z)=\left(\frac{m^2-n^2+2mn-1}{2},\frac{m^2+n^2-1}{2},\frac{m^2-n^2-2mn-1}{2}\right).$$
Also, we can always take $\frac{m^2-n^2-2mn-1}{2}>0.$
For example, for $m=6k$ and $n=2k-1$, where $k$ is a positive integer, we obtain:
$$(x,y,z)=(28k^2-4k-1,20k^2-2k,4k^2+8k-1).$$
|
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|
To Prove $\frac{1}{b}+\frac{1}{c}+\frac{1}{a} > \sqrt{a}+\sqrt{b}+\sqrt{c}$ The three sides of a triangle are $a,b,c$, the area of the triangle is $0.25$, the radius of the circumcircle is $1$.
Prove that $1/b+1/c+1/a > \sqrt{a}+\sqrt{b}+\sqrt{c}$.
what I've tried:
$$\frac{1}{4} = \frac{1}{2}ab\sin C \Rightarrow ab=\frac{1}{2}\sin C \\c=2\sin C \Rightarrow \frac{1}{c}=\frac{1}{2}*\sin C $$
so, $$\frac{1}{c}=ab \Rightarrow abc=1 \Rightarrow \sqrt{abc}=1$$
now the problem becomes
$$ab+bc+ac > \frac{1}{\sqrt{ab}}+\frac{1}{\sqrt{bc}}+\frac{1}{\sqrt{ac}},$$
with $0<a\leq b\leq c\leq 2$, and $a+b>c$.
But even so I don't know how to prove it.
Any help or hint is appreciated. Thank you.:)
|
Area of triangle ABC $=\frac{abc}{4R}=0.25$ where $R=1$ and a>0, b>0, c>0
$abc=1$
$\frac{1}{a}+\frac{1}{b}\geq\frac{2}{\sqrt{ab}}=\sqrt c$
$\frac{1}{b}+\frac{1}{c}\geq\frac{2}{\sqrt{bc}}=\sqrt a$
$\frac{1}{a}+\frac{1}{c}\geq\frac{2}{\sqrt{ac}}=\sqrt a$
$\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\geq \sqrt a+ \sqrt b + \sqrt c$
Since equality holds when $a=b=c=1$ but Area of triangle $ABC=\frac{\sqrt3}{4}\not=0.25$
Then
$\frac{1}{a}+\frac{1}{b}+\frac{1}{c}> \sqrt a+ \sqrt b + \sqrt c$
|
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|
Calculate the Taylor polynomial of a definite integral I want to determine the Taylor polynomial of degree 1 to $f$ where $x=0$
$$f(x)=\int_0^{\sin x} \frac{e^t}{1+t^3} \,dt$$
This is my attempt:
$$f'(x)=\frac{e^{\sin x}}{1+\sin^3 x}$$
$$P(x)=f(0)+f'(0)x$$
$$f(0)=\int_0^0 \frac{e^t}{1+t^3}\,dt=0$$
$$f'(0)=\frac{e^{\sin 0}}{1+\sin^3 0}=\frac{1}{1}=1$$
$$P(x)=0+1\cdot x=x$$
Is this correct if not, can you help me where I calculated wrong? :)
Many thanks
|
You have
$$f(x)=\int_0^{a(x)} g(t) \, dt \implies f'(x)=g(a(x))\, a'(x)$$ by the fundamental theorem of calculus.
So, for your case
$$f'(x)=\frac{e^{\sin (x)}\cos (x)}{1+\sin^3(x)} $$
Now, it is a nice problem of series composition
$$\sin(x)=x-\frac{x^3}{6}+\frac{x^5}{120}-\frac{x^7}{5040}+O\left(x^9\right)$$
$$e^{\sin (x)}=1+x+\frac{x^2}{2}-\frac{x^4}{8}-\frac{x^5}{15}-\frac{x^6}{240}+\frac{x^7}{90}+\frac
{31 x^8}{5760}+O\left(x^9\right)$$
$$\cos(x)=1-\frac{x^2}{2}+\frac{x^4}{24}-\frac{x^6}{720}+\frac{x^8}{40320}+O\left(x^{10}\right)$$
$$e^{\sin (x)}\cos (x)=1+x-\frac{x^3}{2}-\frac{x^4}{3}-\frac{x^5}{40}+\frac{7 x^6}{90}+\frac{31
x^7}{720}+\frac{x^8}{630}+O\left(x^9\right)$$
$$1+\sin^3(x)=1+x^3-\frac{x^5}{2}+\frac{13 x^7}{120}-\frac{41 x^9}{3024}+O\left(x^{11}\right)$$
Long division to get
$$f'(x)=1+x-\frac{3 x^3}{2}-\frac{4 x^4}{3}+\frac{19 x^5}{40}+\frac{187 x^6}{90}+\frac{913
x^7}{720}-\frac{839 x^8}{630}+O\left(x^9\right)$$ Integrate termwise
$$f(x)=x+\frac{x^2}{2}-\frac{3 x^4}{8}-\frac{4 x^5}{15}+\frac{19 x^6}{240}+\frac{187
x^7}{630}+\frac{913 x^8}{5760}-\frac{839 x^9}{5670}+O\left(x^{10}\right)$$
|
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|
Count all positive integers less than 100000 with at least two consecutive 1s? Okay, aforementioned was the problem:
We need to count all the positive integers with at least two consecutive ones.
The way I thought about it was:
There are $10^{5}$ ways to have numbers that are less than 100 000. Now from these, I need to subtract the number of numbers which have either no consecutive ones or one pair of consecutive ones. And we can do that by saying:
Number of no consecutive ones: ....
But this doesn't work. How to think about the problem in a easy way?Can someone help?
|
Exactly 2 1's?
What is the number of 5 digit numbers with 2 1's?
$1^2\cdot9^3\cdot{5\choose 2}$
How many of those have the 2 1's next to each other? How many locations are there for the first 1?
$1^2\cdot9^3\cdot{4\choose 1}$
3 ones
$1^3\cdot9^2\cdot{5\choose 3}$
And one of ${5\choose 3}$ has the 1's all separated by an intervening digit.
$1^3\cdot 9^2\cdot({5\choose 3}-1)$
4 1's and 5 1's
$1^4*9^1*{5\choose 4}$ and $1^5*9^0*{5\choose 5},$ respectively.
$4\cdot 9^3 + 9\cdot 9^2 + 5\cdot 9 + 1 = 3691$
|
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"url": "https://math.stackexchange.com/questions/3838060",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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|
if $x^5=1$ with $x\neq 1$ then find value of $\frac{x}{1+x^2}+\frac{x^2}{1+x^4}+\frac{x^3}{1+x}+\frac{x^4}{1+x^3}$
if $x^5=1$ with $x\neq 1$ then find value of $$\frac{x}{1+x^2}+\frac{x^2}{1+x^4}+\frac{x^3}{1+x}+\frac{x^4}{1+x^3}$$
So my first observation was x is a non real fifth root of unity. Also $$x^5-1=(x-1)(1+x+x^2+x^3+x^4)=0$$
Thus $$1+x+x^2+x^3+x^4=0$$ I tried using this condition to simplify the above expression but nothing interesting simplified. Please note i am looking for hints rather than complete solutions.
EDIT:I came to know its a duplicate ,but i feel that the answers given below are different from those in the original.
|
Note that
$$
\frac1{1+x^n}=\frac12\frac{1+x^{5n}}{1+x^n}=\frac{1-x^n+x^{2n}-x^{3n}+x^{4n}}2\tag1
$$
Applying $(1)$ gives
$$
\begin{align}
\frac{x}{1+x^2}
&=\frac{x-x^3+x^5-x^7+x^9}2\\
&=\frac{x-x^3+1-x^2+x^4}2\tag2
\end{align}
$$
$$
\begin{align}
\frac{x^2}{1+x^4}
&=\frac{x^2-x^6+x^{10}-x^{14}+x^{18}}2\\
&=\frac{x^2-x+1-x^4+x^3}2\tag3
\end{align}
$$
$$
\begin{align}
\frac{x^3}{1+x}
&=\frac{x^3-x^4+x^5-x^6+x^7}2\\
&=\frac{x^3-x^4+1-x+x^2}2\tag4
\end{align}
$$
$$
\begin{align}
\frac{x^4}{1+x^3}
&=\frac{x^4-x^7+x^{10}-x^{13}+x^{16}}2\\
&=\frac{x^4-x^2+1-x^3+x}2\tag5
\end{align}
$$
Each power of $x$ appears twice positive and twice negative, except for $1$ which is always positive. Therefore,
$$
\begin{align}
\frac{x}{1+x^2}+\frac{x^2}{1+x^4}+\frac{x^3}{1+x}+\frac{x^4}{1+x^3}
&=\frac{1+1+1+1}2\\
&=2\tag6
\end{align}
$$
|
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"timestamp": "2023-03-29T00:00:00",
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|
Where is the mistake when finding $\int x \sqrt{4+5 x} \ dx$? We have to find the integral of $\frac{dy}{dx}=x \sqrt{4+5x}$. This is from Morris Kline's book, Chapter $7$, exercise $5$, question $1$m.
I try to solve like following:
Let $u=4+5x$. Then $x=\frac{u-4}{5}$. Hence $\frac{d y}{d x}=\left(\frac{u-4}{5}\right) \sqrt{u}$
Now $$\int(x \sqrt{4+5 x}) \cdot d x=\int\left(\frac{u-4}{5}\right)
\sqrt{u} \cdot d x$$
$$ = \int\left(\frac{u^{3 / 2}-4 u^{1 / 2} }{5}\right)
\cdot d x$$
$$=\frac{\int u^{3 / 2}-4 \int u^{1 / 2}}{5} \cdot d x$$
$$=\frac{\frac{u^{5 / 2}}{5 / 2} -\frac{4 u^{3 / 2}}{3 /
2}}{5} + C$$
$$=\frac{2 u^{5 / 2}}{25}-\frac{8 u^{3 / 2}}{15}+C$$
$$\Rightarrow y=\frac{2(4+5 x)^{5 / 2}}{25} - \frac{8(4+5 x)^{3 /
2}}{15}+C$$
Where $C$ is a constant.
But the answer given in the book is: $$y=\left(\frac{2(4+5 x)^{5 /
2}}{125}\right)-\left(\frac{8(4+5 x)^{3 / 2}}{75}\right)+C.$$
Where is the mistake?
|
Answer:
$I=\int_{}^{} x\sqrt{5x+4}$
$\Rightarrow $ $I=\frac{1} {5}\int_{}^{} 5x\sqrt{5x+4}$
$\Rightarrow $ $I=\frac{1} {5}\int_{}^{} (5x+4-4)\sqrt{5x+4}$
$\Rightarrow $ $I=\frac{1} {5}\int_{}^{} (5x+4)^{\frac{3}{2}} -\frac{4}{25}\int_{}^{} 5\sqrt{5x+4} $
$\Rightarrow $$I= \frac{1}{5} [\frac{2}{25} (5x+4)^{\frac{5}{2}}] - \frac{4}{25} [\frac{2} {3} (5x+4)^{\frac{3}{2}} ] +c$
I=$\frac{2}{125} [(5x+4)^{\frac{5}{2}}] - \frac{8}{75} [ (5x+4)^{\frac{3}{2}} ]+c$.
|
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|
Show that $a_{n+1} = \frac{1}{2}(a_n+b_n), $ $b_{n+1} = \frac{1}{2}(b_n+c_n) $and $c_{n+1} = \frac{1}{2}(c_n+a_n)$ are convergent,
Question: Let $a_0,b_0,c_0$ be real numbers.
Define the sequences $(a_n)_n,(b_n)_n, (c_n)_n$ recursively by
$$a_{n+1} = \frac{1}{2}(a_n+b_n), \quad b_{n+1} = \frac{1}{2}(b_n+c_n) \quad \text{and}\quad c_{n+1} = \frac{1}{2}(c_n+a_n).$$
Prove that the sequences are convergent and find their limits.
My attempt: I try to find the closed-form solution for $a_n,b_n,c_n$.
Note that
$$
\begin{pmatrix}
a_{n+1} \\
b_{n+1} \\
c_{n+1} \\
\end{pmatrix} = \begin{pmatrix}
\frac{1}{2} & \frac{1}{2} & 0 \\
0 & \frac{1}{2} & \frac{1}{2} \\
\frac{1}{2} & 0 & \frac{1}{2}
\end{pmatrix} \begin{pmatrix}
a_n \\
b_n \\
c_n
\end{pmatrix} = A\begin{pmatrix}
a_n \\
b_n \\
c_n
\end{pmatrix}.
$$
Clearly
$$
\begin{pmatrix}
a_{n} \\
b_{n} \\
c_{n} \\
\end{pmatrix} = A^{n}\begin{pmatrix}
a_0 \\
b_0 \\
c_0 \\
\end{pmatrix}.
$$
To compute $A^n$, I try to diagonalize $A$ over complex numbers.
From Wolfram alpha, $A$ has eigenvalues $1, \frac{1}{4}(1+i\sqrt3)$ and $\frac{1}{4}(1-i\sqrt3)$.
Therefore,
$$
\begin{pmatrix}
a_{n} \\
b_{n} \\
c_{n} \\
\end{pmatrix} = X \begin{pmatrix}
1 & 0 & 0 \\
0 & \left( \frac14 (1+i\sqrt 3) \right)^n & 0 \\
0 & 0 & \left( \frac14 (1-i\sqrt 3) \right)^n
\end{pmatrix} X^{-1} \begin{pmatrix}
a_0 \\
b_0 \\
c_0 \\
\end{pmatrix}
$$
for some invertible matrix $X$ (in fact, columns of $X$ are eigenvectors of $A$).
However, I stuck at here.
I do not know how to proceed to show that the sequences $(a_n), (b_n), (c_n)$ convergent.
It would be good if someone can solve this problem by continuing my method above.
EDIT: After reading comments below, I attempted the following.
From Wolfram alpha, we have
$$\lim_{n\to\infty} A^n = \frac13 J_3$$
where $J_3$ is the $3\times 3$ matrix with all entries $1$.
Therefore,
$$\lim_{n\to\infty} \begin{pmatrix}
a_{n} \\
b_{n} \\
c_{n} \\
\end{pmatrix} = \frac13 J_3 \begin{pmatrix}
a_{0} \\
b_{0} \\
c_{0} \\
\end{pmatrix} = \frac13 \begin{pmatrix}
a_{0} + b_{0} + c_{0} \\
a_{0} + b_{0} + c_{0} \\
a_{0} + b_{0} + c_{0}
\end{pmatrix}.$$
Hence, the sequences $(a_n),(b_),(c_n)$ are convergent and their limits are $\frac13 (a_0 + b_0 + c_0)$.
|
Hint: $$a_{n+1}+b_{n+1}+c_{n+1}=a_n+b_n+c_n=\ldots=a_0+b_0+c_0$$
$$|a_{n+1}-b_{n+1}|+|b_{n+1}-c_{n+1}|+|c_{n+1}-a_{n+1}|=\frac12\,(|a_n-b_n|+|b_n-c_n|+|c_n-a_n|)$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3852072",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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|
A Stirling number identity Let $s(n,j)$ denote the signed Stirling numbers of the first kind and $S(n,j)$ the Stirling numbers of the second kind.
I need the following (probably trivial) identity
$\sum\limits_{j = 0}^n {s(n,j)S(m + j,k)} = 0$ for $k < n$ and $\sum\limits_{j = 0}^n {s(n,j)S(m + j,n)} = {n^m},$ but don't see how to prove it-
|
We seek to evaluate (note that this is zero by inspection when $k\gt
n+m$):
$$\sum_{j=0}^n (-1)^{n+j} {n\brack j} {m+j\brace k}$$
where $k\le n.$ Using standard EGFs this becomes
$$n! [z^n] \sum_{j=0}^n (-1)^{n+j}
\frac{1}{j!} \left(\log\frac{1}{1-z}\right)^j
(m+j)! [w^{m+j}] \frac{(\exp(w)-1)^k}{k!}
\\ = (-1)^n n! m! [z^n] \sum_{j=0}^n (-1)^{j}
{m+j\choose j} \left(\log\frac{1}{1-z}\right)^j
\\ \times \frac{1}{2\pi i} \int_{|w|=\gamma}
\frac{1}{w^{m+j+1}} \frac{(\exp(w)-1)^k}{k!} \; dw
\\ = (-1)^n n! m! [z^n]
\frac{1}{2\pi i} \int_{|w|=\gamma}
\frac{1}{w^{m+1}} \frac{(\exp(w)-1)^k}{k!}
\\ \times \sum_{j=0}^n (-1)^{j}
{m+j\choose j} \left(\log\frac{1}{1-z}\right)^j \frac{1}{w^j} \; dw.$$
Now $\left(\log \frac{1}{1-z}\right)^j = z^j+\cdots$ so the coefficient
extractor $[z^n]$ enforces the upper limit of the sum:
$$(-1)^n n! m! [z^n]
\frac{1}{2\pi i} \int_{|w|=\gamma}
\frac{1}{w^{m+1}} \frac{(\exp(w)-1)^k}{k!}
\\ \times \sum_{j\ge 0} (-1)^{j}
{m+j\choose j} \left(\log\frac{1}{1-z}\right)^j
\frac{1}{w^j} \; dw
\\ = (-1)^n n! m! \frac{1}{2\pi i} \int_{|z|=\epsilon}
\frac{1}{z^{n+1}}
\frac{1}{2\pi i} \int_{|w|=\gamma}
\frac{1}{w^{m+1}} \frac{(\exp(w)-1)^k}{k!}
\\ \times \sum_{j\ge 0} (-1)^{j}
{m+j\choose j} \left(\log\frac{1}{1-z}\right)^j
\frac{1}{w^j} \; dw \; dz
\\ = (-1)^n n! m! \frac{1}{2\pi i} \int_{|z|=\epsilon}
\frac{1}{z^{n+1}}
\\ \times \frac{1}{2\pi i} \int_{|w|=\gamma}
\frac{1}{w^{m+1}} \frac{(\exp(w)-1)^k}{k!}
\frac{1}{(1+\frac{1}{w}\log\frac{1}{1-z})^{m+1}} \; dw \; dz
\\ = (-1)^n n! m! \frac{1}{2\pi i} \int_{|z|=\epsilon}
\frac{1}{z^{n+1}}
\\ \times \frac{1}{2\pi i} \int_{|w|=\gamma}
\frac{(\exp(w)-1)^k}{k!}
\frac{1}{(w+\log\frac{1}{1-z})^{m+1}} \; dw \; dz.$$
Now observe that for the geometric series in $j$ to converge we must have
$|\log\frac{1}{1-z}| \lt |w|.$ Note that with $\log \frac{1}{1-z} = z +
\cdots$ the image of $|z|=\epsilon$ makes one turn around the origin, a
circle of radius $\epsilon$ plus additional lower order fluctuations. We
therefore choose $\epsilon$ to shrink this pseudo-circle to be entirely
contained in $|w|=\gamma.$ With this choice the pole at
$-\log\frac{1}{1-z}$ is inside the contour in $w.$ We thus require
$$\frac{1}{k! \times m!}
\left(\sum_{q=0}^k {k\choose q} (-1)^{k-q} \exp(qw)\right)^{(m)}
= \frac{1}{k! \times m!}
\sum_{q=0}^k {k\choose q} (-1)^{k-q} q^m \exp(qw).$$
Evaluating the integral in $w$ we find
$$(-1)^n \frac{n!}{k!} \frac{1}{2\pi i} \int_{|z|=\epsilon}
\frac{1}{z^{n+1}}
\sum_{q=0}^k {k\choose q} (-1)^{k-q} q^m (1-z)^q \; dz$$
which is
$$\bbox[5px,border:2px solid #00A000]{
\frac{n!}{k!}
\sum_{q=0}^k {k\choose q} {q\choose n} (-1)^{k-q} q^m.}$$
Now when $k\lt n$ we have ${q\choose n}= 0$ so the entire sum vanishes
as claimed. We get just one term when $k=n$ namely
$$\frac{n!}{n!} {n\choose n} {n\choose n} (-1)^{n-n} n^m = n^m$$
also as claimed. This concludes the argument.
|
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "3",
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|
The tangent at $(1,7)$ to the curve $x^2=y-6$ touches the circle $x^2+y^2+16x+12y+c=0$ at...
The tangent at $(1,7)$ to the curve $x^2=y-6$ touches the circle $x^2+y^2+16x+12y+c=0$ at...
What I tried...
The equation $x^2=y-6$ is of a parabola. To find the slope of the tangent to the parabola at the point $(1,7)$,
$$\frac{dy}{dx}\Bigg|_{(1,7)}=2\tag{Slope of the line tangent to the parabola}$$
So the equation of the line is $2x-y+5=0\implies y=2x+5$
Substituting this in the equation of circle to find the point of intersection of the line with the circle, we get,
$$x^2+(2x+5)^2+16x+12(2x+5)+c=0$$
Solving this, I get a complicated equation and then the answer comes out in terms of $c$ but the actual answer does not contain $c$ at all.
I would prefer a more analytical/geometrical approach if possible
|
By rearranging the equation of the circle, we quickly find that the centre of the circle is $(-8,-6)$ (as you have illustrated).
Let $O$ be the centre of the circle, and $P$ be the point where the line $y=2x+5$ and circle touch. Since the line and circle touch at $P,$ the line through $O$ and $P$ must be the line perpendicular to $y=2x+5$ at $P$ (that is, the radius is perpendicular to the tangent).
So now we're looking for a line perpendicular to $y=2x+5$ and passing through $O=(-8,-6).$ By vector geometry, this line has parametric form $r(t) = (-8,-6) + t(-2,1),$ i.e., $r(t) = (-2t-8,t-6).$ Now set $x=-2t-8$ and $y=t-6$ in $y=2x+5$ and solve for $t$: we find $t=1,$ i.e., $P=(-6,-7).$
|
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|
Prove that $\frac{a}{1+a^2}+\frac{b}{1+a^2+b^2}+\frac{c}{1+a^2+b^2+c^2}+\frac{d}{1+a^2+b^2+c^2+d^2}\leq\frac{3}{2}$
For any reals $a$, $b$, $c$ and $d$ prove that:
$$\frac{a}{1+a^2}+\frac{b}{1+a^2+b^2}+\frac{c}{1+a^2+b^2+c^2}+\frac{d}{1+a^2+b^2+c^2+d^2}\leq\frac{3}{2}$$
C-S in the IMO 2001 stile does not help here:
\begin{align}
&\frac{a}{1+a^2}+\frac{b}{1+a^2+b^2}+\frac{c}{1+a^2+b^2+c^2}+\frac{d}{1+a^2+b^2+c^2+d^2} \\
\leq&\sqrt{4\left(\tfrac{a^2}{(1+a^2)^2}+\tfrac{b^2}{(1+a^2+b^2)^2}+\tfrac{c^2}{(1+a^2+b^2+c^2)^2}+\tfrac{d^2}{(1+a^2+b^2+c^2+d^2)^2}\right)}\\
\leq&2\sqrt{\tfrac{a^2}{1+a^2}+\tfrac{b^2}{(1+a^2)(1+a^2+b^2)}+\tfrac{c^2}{(1+a^2+b^2)(1+a^2+b^2+c^2)}+\tfrac{d^2}{(1+a^2+b^2+c^2)(1+a^2+b^2+c^2+d^2)}} \\
=&2\sqrt{1-\tfrac1{1+a^2}+\tfrac1{1+a^2}-\tfrac1{1+a^2+b^2}+\tfrac1{1+a^2+b^2}-\tfrac1{1+a^2+b^2+c^2}+\tfrac1{1+a^2+b^2+c^2}-\tfrac1{1+a^2+b^2+c^2+d^2}} \\
<&2
\end{align}
We can assume that our variables are non-negative, of course.
For two variables we can get a best estimation here:
$$\frac{a}{1+a^2}+\frac{b}{1+a^2+b^2}\leq\sqrt{\frac{207+33\sqrt{33}}{512}}\approx0.88...$$
There is also the following Ji Chen's estimation:
$$\frac{x_1}{1+x_1^2}+\frac{x_2}{1+x_1^2+x_2^2}+\dotsb+\frac{x_n}{1+x_1^2+x_2^2+\dotsb+x_n^2}<\sqrt{n}-\dfrac{\ln{n}}{2\sqrt{n}},$$
but it does not help.
Thank you!
|
First, we give some auxiliary results (Facts 1 through 3). The proofs are easy and thus omitted.
Fact 1: Let $a, b$ be reals. Then $\frac{a}{1+a^2}+\frac{b}{1+a^2+b^2}\le \sqrt{\frac{207+33\sqrt{33}}{512}}$.
Fact 2: Let $\gamma$ be real. Then $\frac{\gamma}{1 + \gamma^2} + \sqrt{\frac{207+33\sqrt{33}}{512}}\frac{1}{\sqrt{1+\gamma^2}} < \frac{6}{5}$.
Fact 3: Let $a$ be real. Then $\frac{a}{1+a^2} + \frac{6}{5}\frac{1}{\sqrt{1+a^2}} < \frac{3}{2}$.
$\phantom{2}$
Now, let $\alpha = \frac{c}{\sqrt{a^2 + b^2 + 1}}$ and $\beta = \frac{d}{ \sqrt{a^2 + b^2 + 1}}$.
We have
\begin{align}
&\frac{c}{1+a^2+b^2+c^2}+\frac{d}{1+a^2+b^2+c^2+d^2} \\
=\ & \frac{\alpha \sqrt{a^2 + b^2 + 1} }{1+a^2+b^2+\alpha^2(a^2 + b^2 + 1)}\\
&\quad + \frac{\beta \sqrt{a^2 + b^2 + 1}}{1+a^2+b^2+\alpha^2(a^2 + b^2 + 1) +\beta^2(a^2 + b^2 + 1)}\\
=\ & \left(\frac{\alpha}{1 + \alpha^2} + \frac{\beta}{1 + \alpha^2 + \beta^2}\right)\frac{1}{\sqrt{a^2+b^2+1}}\\
\le\ & \sqrt{\frac{207+33\sqrt{33}}{512}}\frac{1}{\sqrt{a^2+b^2+1}}
\end{align}
where we have used Fact 1.
Let $\gamma = \frac{b}{\sqrt{1+a^2}}$. We have
\begin{align}
&\frac{a}{1+a^2} + \frac{b}{1+a^2+b^2} + \sqrt{\frac{207+33\sqrt{33}}{512}}\frac{1}{\sqrt{a^2+b^2+1}}\\
=\ & \frac{a}{1+a^2} + \frac{\gamma \sqrt{1+a^2}}{1+a^2+\gamma^2(1+a^2)} + \sqrt{\frac{207+33\sqrt{33}}{512}}\frac{1}{\sqrt{a^2+\gamma^2(1+a^2)+1}}\\
=\ & \frac{a}{1+a^2} + \left(\frac{\gamma}{1 + \gamma^2} + \sqrt{\frac{207+33\sqrt{33}}{512}}\frac{1}{\sqrt{1+\gamma^2}}\right)\frac{1}{\sqrt{1+a^2}}\\
<\ & \frac{a}{1+a^2} + \frac{6}{5}\frac{1}{\sqrt{1+a^2}}\\
<\ & \frac{3}{2}
\end{align}
where we have used Facts 2 and 3.
We are done.
|
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|
Differentiation: $y = 9x + \frac{3}{x}$ Differentiate $y = 9x + \frac{3}{x}$
I think the first step is to turn $\frac{3}{x}$ into a more "friendly" format, so $x$ to the power of something maybe? How do I get $x$ with an index from $\frac{3}{x}$?
|
You can also apply the definition (whenever $x\neq 0$):
\begin{align*}
f(x) = 9x + \frac{3}{x} \Rightarrow f'(x) & = \lim_{h\to 0}\frac{1}{h}\left(9(x + h) - 9x + \frac{3}{x + h} - \frac{3}{x}\right)\\\\
& = \lim_{h\to 0}\frac{1}{h}\left(9h - \frac{3h}{x(x + h)}\right)\\\\
& = \lim_{h\to 0}\left(9 - \frac{3}{x(x+h)}\right)\\\\
& = 9 - \frac{3}{x^{2}}
\end{align*}
Hopefully this helps !
|
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|
can't solve $a-2b=3$ & $5b=4a+1$ $$a-2b=3\tag1\label1$$
$$5b=4a+1\tag2\label2$$
Coming to \eqref{1}, $a=2b+3$. Substituting that in \eqref{2}, we get
$5b=4(2b+3)+1$.
$5b=8b+12+1$.
$3b=-13$.
$b=-13/3$.
Substituting value of $b$ in \eqref{1}.
$a-2(-13/3)=3$.
$a-(-26/3)=3$.
Adding $26/3$ to both sides.
$a-0=3+26/3$.
$a=35/3$.
Substituting $a$ and $b$ in \eqref{2}.
$5b=4a+1$.
$5(-13/3)=4(35/3)+1$.
$-65/3=4(35/3)+1$.
$-65/3=140/3+1$.
$-65/3=143/3$.
$-65=143$.
But this is not true.
|
No, you messed up where you said "Adding 26/3 to both sides." If you check again, you will see that this should actually be "subtracting 26/3 to both sides" since $$a-\left(-\frac{26}{3}\right)=a+\frac{26}{3}=3\implies a=-\frac{17}{3}$$.
Your answer and algebra for $b$ are correct, giving $a=-\frac{17}{3}, b=-\frac{13}{3}$.
|
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|
What is the sum of first $n$ terms of the series $1+\frac{1}{4}+3+1+\frac{1}{4}+3+1+\frac{1}{4}+3+\ldots \ldots $ Question: What is the sum of the first $n$ terms of the divergent series :
$$
1+\frac{1}{4}+3+1+\frac{1}{4}+3+1+\frac{1}{4}+3+\ldots \ldots
$$.
My trials: $(1+1+1+\dots)+(\frac{1}{4}+\frac{1}{4}+\frac{1}{4}+\dots)+(3+3+3+\dots)$. $n$th term of fist, second, third series are respectively, $a_n=1, b_n=\frac{1}{4}, c_n=3$. What can I do now?
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The series is divergent
$S_n=\frac {n}{3}\times \frac{17}{4}$, if $n=3k,k\in Z^+$
$S_n=\frac {n+1}{3}\times \frac{17}{4}-3$, if $n=3k-1,k\in Z^+$
$S_n=\frac {n+2}{3}\times \frac{17}{4}-\frac{13}{4}$, if $n=3k-2,k\in Z^+$
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A box contains 7 white and 5 black balls. A box contains 7 white and 5 black balls. If 3 balls are drawn simultaneously at random, what is the probability that they are not all of the same colour? Calculate the probability of the same event for the case where the balls are drawn in succession with replacement between drawings.
Probability that they are not all of the same colour $$= \frac{^7C_2\times ^5C_1}{^{12}C_3}+\frac{^5C_2\times ^7C_1}{^{12}C_3}=\frac{35}{44}$$
For the second case, I did it like:
Probability that they are not all of the same colour where the balls are drawn in succession with replacement between drawings $$= \frac{7^2\times 5}{12^3}+\frac{5^2\times 7}{12^3}=\frac{35}{144}$$
But in my book the answer is $\frac{35}{48}$.
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How about counting the complementary event--i.e., the three balls drawn are of the same color? This seems easier. There are two mutually exclusive cases: either all the balls drawn are white, or they are all black. In the first case, there are clearly $$\binom{7}{3} = \frac{7!}{3! 4!} = 35$$ ways to pick three white balls. In the second case, there are $$\binom{5}{3} = \frac{5!}{3! 2!} = 10$$ ways to pick three black balls. Since there are $$\binom{5 + 7}{3} = \frac{12!}{3! 9!} = 220$$ ways to pick any three balls, the complementary probability of getting all the same color is $$\frac{35 + 10}{220} = \frac{9}{44},$$ thus the desired probability of different colors is $$1 - \frac{9}{44} = \frac{35}{44}.$$ This matches your computation.
When balls are drawn with replacement, we again count the complementary outcomes, but the computation is different because the outcome of each draw is independent and identically distributed. In each of the three draws, the probability of obtaining a white ball is $7/12$. So the probability of getting three white balls is $$(7/12)^3 = \frac{343}{1728}.$$ Similarly, the probability of getting three black balls is $$(5/12)^3 = \frac{125}{1728}.$$ So the total probability of getting the same color in three draws is $$\frac{343 + 125}{1728} = \frac{13}{48},$$ and the desired probability of getting both colors in three draws is $$1 - \frac{13}{48} = \frac{35}{48}.$$ So your book is correct for this second scenario.
Where did you go wrong? The issue is that the probability of getting, say, two white balls and one black ball is not simply $$\frac{7^2 \cdot 5}{12^3}.$$ The actual probability is three times this, because the outcomes can be ordered; e.g., $$\{w, w, b\}, \{w, b, w\}, \{b, w, w\}$$ are all distinct outcomes. Therefore, you will find that if you multiply your answer by $3$, you get the book's answer: $$\frac{35}{144} \cdot 3 = \frac{35}{48}.$$
Another way to reason about this is to note that when the number of draws is fixed--in this case, $n = 3$ draws--then the number of white balls drawn determines the number of black balls drawn. For instance, if you draw three balls with replacement, saying you got exactly two white balls is the same as saying you got exactly one black ball. Or if you got zero white balls, this is the same as saying you got three black balls.
So, to say that you got balls of both colors is equivalent to saying you got either $1$ or $2$ white balls, no more, no less. So the random number $X$ of white balls is a binomial random variable with $n = 3$ and probability of drawing a white ball $p = 7/12$; i.e., $$\Pr[X = x] = \binom{n}{x} p^x (1-p)^{n-x} = \binom{3}{x} (7/12)^x (5/12)^{3-x}, \quad x \in \{0, 1, 2, 3\}.$$ Thus we have $$\Pr[X = 1] + \Pr[X = 2] = \binom{3}{1} \frac{7^1 5^2}{12^3} + \binom{3}{2} \frac{7^2 5^1}{12^3} = \frac{35}{48}.$$
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Simplify $\frac{x^3+1}{x+\sqrt{x-1}}$
Simplify $$A=\dfrac{x^3+1}{x+\sqrt{x-1}}.$$
Firstly, $x-1\ge0$ and $x+\sqrt{x-1}\ne0:\begin{cases}x-1\ge0\\x+\sqrt{x-1}\ne0\end{cases}.$ The first inequality is equivalent to $x\ge1$. Can we use that in the second inequality? I mean can we say that $x+\sqrt{x-1}>0$ because $x>0$ and $\sqrt{x-1}\ge0.$ (I am asking if we can use in the inequalities after $x\ge1$ that $x$ is actually greater than or equal to $1$.)
After that we have $A=\dfrac{(x+1)(x^2-x+1)}{x+\sqrt{x-1}}.$ What from here? Thank you in advance!
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The answer to your first question is yes, you can assume that $x\geq 1$ because of $\sqrt{x-1}$, as you said.
The second aim is also true: $x+\sqrt{x-1}\geq 0 $ because $x> 0$ and $\sqrt{x-1}\geq 0$.
So, the domain of your equation is $[1,+\infty)$.
Then, one thing you can do to simplify $A$ is to multiply and divide by the conjugate of the divisor:
$$
A=\dfrac{(x+1)(x^2-x+1)}{x+\sqrt{x-1}}\dfrac{x-\sqrt{x-1}}{x-\sqrt{x-1}}
=\dfrac{(x+1)(x^2-x+1)(x+\sqrt{x-1})}{(x-\sqrt{x-1})(x+\sqrt{x-1})}
$$
Using the Difference of squares:
$$
A=\dfrac{(x+1)(x^2-x+1)(x-\sqrt{x-1})}{x^2-(\sqrt{x-1})^2)}
=\dfrac{(x+1)(x^2-x+1)(x-\sqrt{x-1})}{x^2-x+1}
=(x+1)(x-\sqrt{x-1})
$$
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Rationalize the denominator of $2\over{2 - \sqrt[4]{2}}$?
Rationalize the denominator of $2\over{2 - \sqrt[4]{2}}$.
Here's my progress. Let $x = \sqrt[4]{2}$. Then our expression can be written as $x^4/(x^4 - x)$, which simplifies to $x^3/(x^3 - 1)$. Multiply top and bottom by $(x^3 + 1)$ to get $x^3(x^3 + 1)/(x^6 - 1)$. Multiply top and bottom by $(x^6 + 1)$ to get$$x^3(x^3 + 1)(x^6 + 1)/(x^{12} - 1) = x^3(x^3 + 1)(x^6 + 1)/7 = {1\over7}(8 + 4\sqrt[4]{2} + 2 \sqrt{2} + 2^{3/4}).$$However, Wolfram Alpha also tells me that we can write this as$${1\over{14}}\Big(16 + 4\sqrt{2} + 7\sqrt{{{64}\over{49}} + {{72{\sqrt2}}\over{49}}}\Big)$$But how do I derive that? Seems impossible!
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The systematic way is to compute the inverse of $2-x$ mod $x^4-2$ using the extended Euclidean algorithm:
$$
1= \frac{1}{14}(x^4-2) + \frac{1}{14} (x^3 + 2 x^2 + 4 x + 8)(2-x)
$$
(courtesy of WA).
Therefore,
$$
\frac{1}{2 - \alpha} = \frac{1}{14} (\alpha^3 + 2 \alpha^2 + 4 \alpha + 8)
$$
where $\alpha=\sqrt[4]{2}$.
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$\lim_{x \rightarrow 1} \frac{ x^2-1 + \sqrt{x^3+1} - \sqrt{x^4+1} }{ x-1 + \sqrt{x+1} - \sqrt{x^2+1} } $ Calculate below limit
$$\lim_{x \rightarrow 1} \frac{ x^2-1 + \sqrt{x^3+1} - \sqrt{x^4+1} }{ x-1 + \sqrt{x+1} - \sqrt{x^2+1} } $$
Using L'Hôpital's rule might be too tedious. I wonder if there is a trick given the resemblence of numerator and denominator?
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Using binomial first order expansion we have
*
*$\sqrt{x^3+1}=\sqrt{2+(x^3-1)}=\sqrt 2\sqrt{1+\frac{x^3-1}2}\sim \sqrt 2+\sqrt 2\frac{x^3-1}4$
*$\sqrt{x^4+1}\sim \sqrt 2+\sqrt 2\frac{x^4-1}4$
*$\sqrt{x+1}\sim \sqrt 2+\sqrt 2\frac{x-1}4$
*$\sqrt{x^2+1}\sim \sqrt 2+\sqrt 2\frac{x^2-1}4$
then
$$\frac{x^2-1 + \sqrt{x^3+1} - \sqrt{x^4+1}}{x-1 + \sqrt{x+1} - \sqrt{x^2+1}}\sim \frac{x^2-1+\sqrt 2\frac{x^3-1}4-\sqrt 2\frac{x^4-1}4}{x-1+\sqrt 2\frac{x-1}4-\sqrt 2\frac{x^2-1}4}=$$
$$=\frac{x+1+\sqrt 2\frac{x^2+x+1}4-\sqrt 2\frac{x^3+x^2+x+1}4}{1+\sqrt 2\frac{1}4-\sqrt 2\frac{x+1}4}\to \frac{2-\frac14 \sqrt 2}{1-\frac14 \sqrt 2}=\frac{15+2\sqrt2}{7}$$
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Simplify expression involving fractions for mathematical induction proof. I have this expression
$1-\frac{1}{k+1}+\frac{1}{(k+1)(k+2)}$
that I need to manipulate into looking like
$1-\frac{1}{k+2}$
as the final step in my mathematical induction proof process.
My idea was to go something like this:
\begin{align*}
1-\frac{1}{k+1}+\frac{1}{(k+1)(k+2)} &= 1-\frac{k+2}{(k+1)(k+2)}+\frac{1}{(k+1)(k+2)}\\
&= 1-\frac{(k+2)+1}{(k+1)(k+2)}\\
&= 1-\frac{(k+2)+1}{(k+1)(k+2)} \cdot \frac{k+1}{k+2}\\
&= 1-\frac{1}{k+2}
\end{align*}
Is this algebraically correct?
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The first part is right, but:
$$1-\frac{k+2}{(k+1)(k+2)}+\frac{1}{(k+1)(k+2)}$$
$$= 1 - \left(\frac{k+2}{(k+1)(k+2)} -\frac{1}{(k+1)(k+2)} \right)$$
$$= 1 - \frac{k+2 \color{red}{-} 1}{(k+1)(k+2)}$$
and you can continue from here.
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We have sides $a,b,c$ of triangle $ABC$ which are equal to $\sqrt{6}, \sqrt{3}, 1$, Prove that angles $A=90+B$ We have sides $a,b,c$ of triangle $ABC$ which are equal to $\sqrt{6}, \sqrt{3}, 1$, Prove that angles $A=90+B$
I just did the question above in the following way:
$t=\frac{\sqrt{6}+\sqrt{3}+1}{2}$
From Heron we have that:
$=\sqrt{\frac{\sqrt{6}+\sqrt{3}+1}{2}*\frac{\sqrt{3}+1-\sqrt{6}}{2}*\frac{\sqrt{6}-\sqrt{3}+1}{2}*\frac{\sqrt{6}+\sqrt{3}-1}{2}}$
$=\frac{1}{4}\sqrt{8}$
I state that AK is a height of the triangle.
We have that it is enough if $sin(90+B)=sin(A)$
However we also have that $sin(90+B)=sin(BAK)$ and $sin(BAK)=\frac{BK}{1}=BK$ (BAK is a right angled triangle)
Also:
$\frac{1}{2}*sin(A)*\sqrt{3}=\frac{1}{4}\sqrt{8}$
$sin(A)=\frac{\frac{1}{2}*\sqrt{8}}{\sqrt{3}}$
So we just have to prove that $BK=\frac{\frac{1}{2}*\sqrt{8}}{\sqrt{3}}$
We have that:
$\frac{AK*BC}{2}=\frac{1}{4}\sqrt{8}$
$AK=\frac{\sqrt{4}}{2\sqrt{3}}$
Hence from Pythagoras we have that:
$BK=\sqrt{\frac{8}{12}}=\frac{\frac{1}{2}*\sqrt{8}}{\sqrt{3}}=sin(A)$
Hence it is proved.
I realize that my method of proving it is complex. Could you please show some simple solutions to this problem?
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We are to make use of $$ \cos (90+\theta) = -\sin \theta$$
Let $a=\sqrt{6}$, $b=\sqrt{3}$, $c=1$. So we assumed $a>b>c$ or, $A>B>C$.
Now use cosine-law,
$$ \cos A = \dfrac{b^2+c^2-a^2}{2bc}$$
and show by calculation that
$$ \cos A = -\sin B $$
where, $\sin B = \sqrt{1-\cos^2 B}$.
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Present value of deferred annuity with varying amounts An annual annuity pays the amount 1, 2, 3, 4, 5, 6, 5, 4, 3, 2, 1 (in dollars), the first payment occurring at the end of the second year. The present value at $=0$ of this annuity is 25 dollars at an annual effective rate $i$.
Another annual annuity pays the amount 1, 2, 3, 4, 5, 6, 6, 5, 4, 3, 2, 1 (in dollars), the first payment occurring at the end of the first year. The present value at $=0$ of this annuity is $V$ dollars at an annual effective rate $i$.
What is the value of $V$?
I have been trying to do this problem for two days, but I do not know where to start. Do I need to find the annual effective rate first using the present value given? My issue with this is that since the payments are not all equal, do I need to calculate them all individually? Or is there a formula?
Thank you !!
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Let $$p(v)=v^2+2v^3+3v^4+4v^5+5v^6+6v^7+5v^8+4v^9+3v^{10}+2v^{11}+v^{12}$$
Since $p(0)=0$, $p$ is strictly increasing on $v\geq0$, and $p(v)\to\infty$ as $v\to\infty$, there is a unique $r>0$ such that $p(r)=25$.
Now, $$\begin{align}
p(v)&=v^2(1+2v+3v^2+4v^3+5v^4+6v^5+5v^6+4v^7+3v^8+2v^9+v^{10})\\
&=v^2(1+v+v^2+v^3+v^4+v^5)^2\\
&=\left(v\frac{v^6-1}{v-1}\right)^2
\end{align}$$
where the second equation is proved in the accepted answer to this question, so that $$\frac{r^7-r}{r-1}=5$$
Let $$q(v)=v+2v^2+3v^3+4v^4+5v^5+6v^6+6v^7+5v^8+4v^9+3v^{10}+2v^{11}+v^{12}$$
Then $$\begin{align}
q(v)-p(v)&=v+2v^2+3v^3+4v^4+5v^5+6v^6-(v^2+2v^3+3v^4+4v^5+5v^6)\\
&=v+v^2+v^3+v^4+v^5+v^6\\&=\frac{v^7-v}{v-1}
\end{align}$$ so that
$$q(r)-p(r)=\frac{r^7-r}{r-1}=5$$ and $$q(r)=25+5=\boxed{30}$$
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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.