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The graphs of $x^2+y^2+6x-24y+72=0$ and $x^2-y^2+6x+16y-46=0$ intersect at four points.Find the sum of the distances of these four points The graphs of $x^2+y^2+6x-24y+72=0$ and $x^2-y^2+6x+16y-46=0$ intersect at four points.Compute the sum of the distances of these four points from the point $(-3,2).$
$x^2+y^2+6x-24y+72=0$ is a circle and $x^2-y^2+6x+16y-46=0$ is a hyperbola.These cut at four points.
$x^2+y^2+6x-24y+72=0$ becomes $(x+3)^2+(y-12)=9^2$.
$x^2-y^2+6x+16y-46=0$ becomes $(x+3)^2-(y-8)^2=-9$.
When i solved these two equations to find the points of intesection,$y=10\pm \sqrt{41}$ and $x$ is $-3\pm\sqrt{36\pm 4\sqrt{41}}$
Now it is very difficult to find the sum of distances of these points from $(-3,2)$.Answer is $40$ given in the book.
|
Rather than solving the problem by eliminating x we can eliminate y. We will get a parabola which contains all the roots having x axis as directrix and (3,-2) as focus. So, all we need to do is find the sum of ordinates to find the distance. And, then you can eliminate x from equation to get quadratic in y and find sum of roots which would be doubled because of repeated ordinates.
So, ans is 40
Upvote if you like the answer!!!
|
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?Where does the integral went wrong $$\int_{-a}^{a}\int_{0}^{b\cdot \sqrt{1-\frac{x^2}{a^2}}}2x^2ydydx=\int_{-a}^{a}2b^2(x^2-x^4/a^2) dx =2b^2(2/3a^2-2a^5/5a^2)=\frac{8}{15}a^3b^2$$
But the actually answer in $\frac{4}{15}a^3b^2$. Which step did I go wrong?
|
The first equality should have been $$\int_{-a}^{a}\int_{0}^{b\cdot \sqrt{1-\frac{x^2}{a^2}}}2x^2ydydx=\int_{-a}^{a}b^2(x^2-x^4/a^2)dx$$
since the antiderivative of $2y$ is $y^2$. That's why you get $2$ times the desired result.
|
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Series of squares of n integers - where is the mistake? Given the following two series:
$$1^3 + 2^3 + ... + n^3$$
$$0^3 + 1^3 + .... + (n-1)^3$$
I take the difference vertically of the two:
$$\left(1^3-0^3\right) + \left(2^3-1^3\right) + .... + \left(n^3-(n-1)^3\right)$$
This equals to $n^3$
If I now express this in sum notation:
$$\sum_{i=1}^n\left(i^3-(i-1)^3\right) = n^3$$
If I expand: $(i-1)^3 = i^3 - 3i^2 + 3i - 1$
Thus
$$\left(i^3 - (i-1)^3\right) = 3i^2 - 3i +1$$
My sum is now:
$$3\sum_{i=1}^n i^2 -3 \sum_{i=1}^n i + n = n^3$$
$$\sum_{i=1}^n i^2 = \frac{1}{3} \left(n^3 + 3 \frac{n(n-1)}{2} -n\right)$$
And the expression on the RHS above is not $\frac{1}{6} n (n+1) (2n+1)$
I don't want to solve the above using forward difference, I want to keep it backward.
|
Note that $\sum_{i=1}^n i = \frac 12n(n{\color{red}+}1)$, hence
\begin{align*}
\frac 13 \left(n^3 + 3\frac{n(n+1)}2 - n\right)
&= \frac 16 n(2n^2 + 3n + 3 - 2)\\
&= \frac 16 n(2n^2 + 3n + 1)\\
&= \frac 16 n(2n+1)(n+1)
\end{align*}
|
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An elegant way to solve $\frac {\sqrt3 - 1}{\sin x} + \frac {\sqrt3 + 1}{\cos x} = 4\sqrt2 $ The question is to find $x\in\left(0,\frac{\pi}{2}\right)$:
$$\frac {\sqrt3 - 1}{\sin x} + \frac {\sqrt3 + 1}{\cos x} = 4\sqrt2 $$
What I did was to take the $\cos x$ fraction to the right and try to simplify ;
But it looked very messy and trying to write $\sin x$ in terms of $\cos x$ didn't help.
Is there a more simple (elegant) way to do this.
|
Another approach is to write the equation as
$(\sqrt{3} - 1) \cos x +(\sqrt{3} + 1) \sin x = 4 \sqrt2 \sin x \cos x$
then rearranging gives
$\frac{\sqrt{3} - 1}{2 \sqrt{2}} \cos x +\frac{\sqrt{3} + 1}{2 \sqrt{2}} \sin x = 2 \sin x \cos x$.
Now note that $(\frac{\sqrt{3} - 1}{2 \sqrt{2}})^2 +(\frac{\sqrt{3} + 1}{2 \sqrt{2}})^2 = \frac{1}{8} ((3 -2 \sqrt{3} + 1) + (3 +2 \sqrt{3} + 1) = 1$ so that we can write $\frac{\sqrt{3} - 1}{2 \sqrt{2}}$ as the $\sin$ of some angle, say $\alpha$, and identify $\frac{\sqrt{3} + 1}{2 \sqrt{2}}$ as $\cos \alpha$.
Then you have to solve $\sin(\alpha + x) = \sin(2x)$, with $\tan \alpha = \frac{\sqrt{3} - 1}{\sqrt{3} + 1}$.
|
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Divisibility of $n^4 -n^2$ by 4 (induction proof) We have to show that $$ n^4 -n^2 $$ is divisible by 3 and 4 by mathematical induction
Proving the first case is easy however I do not know how what to do in the inductive step. Thank you.
|
Let's prove that
if $n^4-n^2$ is a multiple of both $3$ and $4$,
then so is $(n+1)^4-(n+1)^2$.
Consider
$
((n+1)^4-(n+1)^2)-(n^4-n^2)=4 n^3+6 n^2+2 n
$.
Ignoring $6n^2$, we have $4 n^3+2 n=3n^3+3n+n^3-n=3n^3+3n+6\binom{n+1}{3}$, which is always a multiple of $3$.
Ignoring $4n^3$, we have
$6 n^2+2 n=4n^2+2n(n+1)=4n^2+4\binom{n+1}{2}$, which is always a multiple of $4$.
We're done.
This problem is definitely one which is much easier not by induction but instead simply by arguing that $n^{4} - n^{2} = n^{2} (n + 1) (n - 1)$, as @eltonjohn suggested.
|
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Calculate the limit of: $x_n = \frac{\ln(1+\sqrt{n}+\sqrt[3]{n})}{\ln(1 + \sqrt[3]{n} + \sqrt[4]{n})}$, $n \rightarrow \infty$ Is it ok to solve the following problem this way? What I have done is to solve parts of the limit first (that converges to $0$), and then solve the remaining expression? Or is this flawed reasoning?
Question
Calculate the limit of:
$$x_n = \frac{\ln(1+\sqrt{n}+\sqrt[3]{n})}{\ln(1 + \sqrt[3]{n} + \sqrt[4]{n})}$$
when $n$ goes to infinity.
Answer
This can also be written as:
$$\lim_{n \to \infty} \frac{\ln(1+\sqrt{n}+\sqrt[3]{n})}{\ln(1 + \sqrt[3]{n} + \sqrt[4]{n})}$$
The denominator can be written as:
$$\ln(1 + \sqrt[3]{n} + \sqrt[4]{n}) = \ln(1 + \frac{1 + \sqrt[4]{n}}{\sqrt[3]{n}}) + \ln(\sqrt[3]{n})$$
From this we can see that:
$$\lim_{n \to \infty} \ln(1 + \frac{1 + \sqrt[4]{n}}{\sqrt[3]{n}}) \rightarrow 0$$
The numerator can be written as:
$$\ln(1 + \sqrt{n} + \sqrt[3]{n}) = \ln(1 + \frac{1 + \sqrt[3]{n}}{\sqrt{n}}) + \ln(\sqrt{n})$$
From this we can see that:
$$\lim_{n \to \infty} \ln(1 + \frac{1 + \sqrt[3]{n}}{\sqrt{n}}) \rightarrow 0$$
This means that we have the following limit:
$$\lim_{n \to \infty} \frac{\ln(\sqrt{n})}{\ln(\sqrt[3]{n})} = \lim_{n \to \infty} \frac{\ln(n^{\frac{1}{2}})}{\ln(n^{\frac{1}{3}})} = \lim_{n \to \infty} \frac{\frac{1}{2}\ln(n)}{\frac{1}{3}\ln(n)} = \lim_{n \to \infty} \frac{3 \ln(n)}{2\ln(n)} \rightarrow \frac{3}{2}$$
The limit converges towards $\frac{3}{2}.$
|
The answer is correct but I would say the working is not quite rigorous. In effect you have said that
$$\lim a=0\ ,\quad \lim c=0$$
implies
$$\lim\frac{a+b}{c+d}=\lim\frac bd\ .$$
It's true in this case since $b,d\to\infty$; it would also be true if $b,d$ have any finite non-zero limits; but it is not always true if $b,d\to0$.
For example, it is not true that
$$\lim\frac{\frac1n+\frac2n}{\frac1n+\frac3n}=\lim\frac{0+\frac2n}{0+\frac3n}\ .$$
|
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If possible, solve $7x^2 - 4x + 1 \equiv 0 $ (mod 11) If possible, solve $7x^2 - 4x + 1 \equiv 0 $ (mod 11)
Not sure if I'm approaching this problem correctly, any help is appreciated.
So far I have:
$7x^2 - 4x + 1 \equiv 0 $ (mod 11)
$21x^2 - 12x + 3 \equiv 0 $ (mod 11)
$-x^2 - x + 3 \equiv 0 $ (mod 11)
$-x^2 - x \equiv -3 $ (mod 11)
$-1(x^2 + x) \equiv -3 $ (mod 11)
$(x^2 + x) \equiv 3 $ (mod 11)
-Not sure if I am heading the right way, once again any help is appreciated.
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Multiply by $8$ (which is the inverse of $7$ mod $11$) and get: $x^2+x+8=0$ (mod $11$).
Then complete the square. "Half of 1" is the same as "half of 12" (mod 11) so we can rewrite our equation as $(x+6)^2-36+8=0$ (mod $11$) which simplifies to $(x+6)^2=6$ (mod $11$).
So your equation has a solution if and only if 6 is a square (mod 11).
It isn't, so there is no solution.
To show that 6 is not a square (directly) consider: $0^2=0$, $(\pm 1)^2=1$, $(\pm 2)^2=4$, $(\pm 3)^2=9$, $(\pm 4)^2=16=5$, and $(\pm 5)^2 = 25=3$ (all mod $11$).
|
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Induction for divisibility: $3\mid 12^n -7^n -4^n -1$ I must use mathematical induction to show that
$a_{n} = 12^n −7^n −4^n −1$ is divisible by 3 for all positive integers n.
Assume true for $n=k$
$a_{k} = 12^k -7^k -4^k -1$
Prove true for $n=k+1$
$a_{k} = 12^{k+1} -7^{k+1} -4^{k+1} -1$
$ = (12^k)(12) - (7^k)(7) - (4^k)(4) -1$
$ = (12^k)(12) - (7^k)(3+4) - (4^k)(3+1) -1$
I'm not really sure about the last step, as someone just told me to do it. Am I supposed to find the right addends to use and then distribute the exponent terms until I get a multiple of the original $a_{k}$? Because I can't get it to work out evenly, and the -1 at the end gives me trouble. Also, I know that $12^n$ is a multiple of three already, but I don't know how to implement that fact to my advantage. Can I prove that $7^{n}-4^{n}-1$ is also a multiple of three and go from there?
|
Hint: Instead of the last step, I would do this:
\begin{align*} a_{k+1} &= 12^{k+1} -7^{k+1} -4^{k+1} -1 \\
&= 12^k \cdot 12 - 7^k \cdot 7 - 4^k \cdot 4 - 4 + 3 \\
&= 12^k \cdot 8 - 7^k \cdot 3 + 4(12^k-7^k-4^k-1)+3 \end{align*}
|
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|
How to integrate $\int\limits_{0}^{\pi/2}\frac{dx}{\cos^3{x}+\sin^3{x}}$? I have$$\int\limits_{0}^{\pi/2}\frac{\text{d}x}{\cos^3{x}+\sin^3{x}}$$
Tangent half-angle substitution gives a fourth-degree polynomial in the denominator that is difficult to factor.
|
$$I = \int_{0}^{\pi/2}\frac{dx}{\sqrt{2}\cos\left(x-\frac{\pi}{4}\right)\left(1-\frac{1}{2}\sin(2x)\right)}=\int_{-\pi/4}^{\pi/4}\frac{dx}{\sqrt{2}\cos(x)\left(1-\frac{1}{2}\cos(2x)\right)} $$
hence, through the substitution $x=\arcsin t$:
$$ I = \sqrt{2}\int_{0}^{\pi/4}\frac{dx}{\cos(x)\left(\frac{3}{2}-\cos^2 x\right)}=\sqrt{2}\int_{0}^{\frac{1}{\sqrt{2}}}\frac{dt}{(1-t^2)\left(\frac{1}{2}+t^2\right)}$$
and the last integral is perfectly manageable through partial fraction decomposition.
The outcome is:
$$ \int_{0}^{\pi/2}\frac{d\theta}{\sin^3\theta+\cos^3\theta} = \color{red}{\frac{\pi}{3}+\frac{2\sqrt{2}}{3}\,\log\left(1+\sqrt{2}\right)}.$$
|
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Derive $\tan(3x)$ in terms of $\tan(x)$ using De Moivre's theorem
Derive the following identity: $$\tan(3x)=\frac{3\tan(x)-\tan^3(x)}{1-3\tan^2(x)}$$
The way I approached the questions is that I first derived $\sin(3x)$ and $\cos(3x)$ because $\tan(\theta)=\frac{\sin(\theta)}{\cos(\theta)}$. Then substituting:
$$\tan(3x)=\frac{\sin(3x)}{\cos(3x)}=\frac{-4\sin^3(x)+3\sin(x)}{4\cos^3(x)-3\cos(x)}$$
I transformed it into $$\tan(3x)=\frac{-4\sin^2(x)\tan(x)+3\tan(x)}{4\cos^2(x)-3}$$ and also to $$\tan(3x)=-\tan(x)\frac{4\sin^2(x)-3}{-4\sin^2(x)+1}$$ but I'm stuck on either of these forms. Any help?
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Don't “normalize” the triplication formulas to only sines and cosines.
We have
$$
\cos3x+i\sin3x=\cos^3x+3i\cos^2x\sin x-3\cos x\sin^2x-i\sin^3x
$$
so
\begin{align}
\sin3x&=3\cos^2x\sin x-\sin^3x \\[6px]
\cos3x&=\cos^3x-3\cos x\sin^2x \\[12px]
\tan3x&=\frac{3\cos^2x\sin x-\sin^3x}{\cos^3x-3\cos x\sin^2x}
\end{align}
Now divide numerator and denominator by $\cos^3x$.
|
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Prove the function Prove
$\forall n\in\mathbb{N}: n\ge 1 \rightarrow 2^n\le 2^{n+1}-2^{n-1}-1.$
I did that $2^n \le\ 2^{n+2} - 2^{n}$ and then $2^n < 2^{n+2} - 2^{n-1}$
but have no idea how to add $-1$ in the function and let the $2^n \le\ 2^{n+2} - 2^{n} - 1$
|
Use a geometric series to show that $1+2+\cdots +2^n = 2^{n+1}-1$.
Here is an alternative approach:
Write the equation as $2^{n+1} \ge 2^n + 2^{n-1} +1$.
For $n=1$ it is easy to check that $4 \ge 2 + 1 +1$.
Suppose the statement is true for $n$ and multiply across by $2$ to get
$2^{n+2} \ge 2^{n+1} + 2^{n} +2 \ge 2^{n+1} + 2^{n} +1$,
hence the statement is true for $n+1$.
A moment's reflection will show the connection between this approach and
my first approach above.
|
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What is the range of $f :R → R$, and $f(x) = x^2 + 6x − 8$ I have this discrete math question I have done completing the square but not sure how to continue. May I get some guide? Thanks!
What is the range of $f :R → R$, and $f(x) = x^2 + 6x − 8$
$f(x)=x^2+6x-8$
$f(x)=(x^2+6x+9)-8-9$
$f(x)=(x+3)^2-17$
|
\begin{align}
f(x) & = x^2 + 6x - 8 \\
& = (x^2 + 6x + 9) - 8 - 9 \\
& = (x + 3)^2 - 17 \\
\end{align}
Thus, the range is $[-17, \infty)$, which follows immediately from the fact that $(x + 3)^2 \ge 0$ and that $f(x)$ is not bounded from above.
While this is the general approach to finding ranges of quadratic functions, consider this instead if you do not get the idea above:
Suppose that $f(x) = k$ where $x, k \in \Bbb R$, then the range of $f(x)$ is just the range of $k$.
\begin{align}
f(x) = k & \Leftrightarrow f(x) - k = 0 \\
& \Leftrightarrow x^2 + 6x - (8 + k) = 0 \\
\end{align}
Since $x \in \Bbb R$, that is, $x^2 + 6x- (8 + k) = 0$ has real roots, which is equivalent to
\begin{align}
\Delta & = b^2 - 4ac \\
& = 6^2 + 4 \cdot 1 \cdot (8 + k) \\
& = 68 + 4k \\
& \ge 0 \\
\end{align}
Thus,
$$f(x) = k \ge -17$$
and $[-17, \infty)$ is the range of $f(x)$.
|
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Finding limit of $ \lim \limits_{x,y \to 0,0}{(1 + x^2 y^2)}^{-\frac{1}{x^2 + y^2}}$ Here is my limit:
$$ \lim \limits_{x,y \to 0,0}{(1 + x^2 y^2)}^{-\frac{1}{x^2 + y^2}}$$
I have learned two methods. One where we replace y with for example $y = kx $ (because $y = y_0 + k(x - x_0)$ and $y_0 = 0, x_0 = 0$). Or with $x = r *cos(\phi)$ and $x = r *sin(\phi)$ where $r \to 0$.
Neither seem to help me at the moment (or at least when I tried solving with both I didn't get a good answer.
It kind of seems like I could use $ \lim \limits_{x \to \infty}{(1 + \frac{1}{x})}^{x} = e$, but I tried and also couldn't get a decent answer.
Any ideas?
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We can use your idea of setting $x= r \cos \theta$, $y=r \sin \theta$ and the limit becomes
$$\lim_{r \rightarrow 0} \left(1+\frac{r^4 \sin^2 2\theta}{4} \right)^{-\frac{1}{r^2}}$$
For a given $r$, the maximum and minimum values of this function are
$1$ and $\left( 1+\frac{r^4}{4} \right)^{-\frac{1}{r^2}}$ obtained by setting $\theta =0$ and $\theta = \frac{\pi}{4}$ respectively.
The second limit as $r\rightarrow 0$ is
$\lim_{r\rightarrow 0}\left( 1+\frac{r^4}{4} \right)^{-\frac{1}{r^2}} = \lim_{x\rightarrow \infty} \left( 1+\frac{1}{x^2} \right)^{\frac{x}{2}} = 1$
where we have taken $x = \frac{2}{r^2}$
Because this the minimum value the function can take on the circle, we can say the following:
For any $\epsilon >0$ there exists $r$ such that $x^2 + y^2 < r^2 \Rightarrow |(1+x^2y^2)^{-\frac{1}{x^2+y^2}} -1| < \epsilon$ so the limit is $1$.
|
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Trigonometric substitution $\tan{\frac{x}{2}}=t$. What is $\cos{x}$ then? For example, the integral is:
$$\int \frac{\sin{x}}{3\sin{x}+4\cos{x}}dx$$
And we use the substitution: $\tan{\frac{x}{2}}=t$
Now, to get $\cos{x}$ in terms of $\tan\frac{x}{2}$, I first expressed $\cos^2\frac{x}{2}$ and $\sin^2\frac{x}{2}$ in temrs of $\tan\frac{x}{2}$:
$$\cos^2\frac{x}{2}=\frac{1}{\frac{1}{\cos^2\frac{x}{2}}}=\frac{1}{\frac{\sin^2\frac{x}{2}+\cos^2\frac{x}{2}}{\cos^2\frac{x}{2}}}=\frac{1}{1+\tan^2\frac{x}{2}}=\frac{1}{1+t^2}$$
$$\sin^2\frac{x}{2}=1-\cos^2\frac{x}{2}=1-\frac{1}{1+t^2}=\frac{t^2}{1+t^2}$$
Now, using trig idendity: $\cos{2x}=\cos^2{x}-\sin^2{x}$ we have:
$$\cos{x}=\bigg(\cos^2\frac{x}{2}-\sin^2\frac{x}{2}\bigg)=\frac{1}{1+t^2}-\frac{t^2}{1+t^2}=\frac{1-t^2}{1+t^2}$$
which is good, but, I don't know what is wrong with this next procedure (first, getting $\sin^2{x}$, then using trig identity $\sin^2{x}+\cos^2{x}=1$ getting $\cos{x}$).
So, first, expressing $\sin{x}$ in terms of $\tan\frac{x}{2}$ using trig identity $\sin{2x}=2\sin{x}\cos{x}$:
$$\sin^2{x}=4\sin^2\frac{x}{2}\cos^2\frac{x}{2}=4\cdot\frac{t^2}{1+t^2}\cdot\frac{1}{1+t^2}=\frac{4t^2}{(1+t^2)^2}$$
So:
$$\cos^2{x}=1-\sin^2{x}=1-\frac{4t^2}{(1+t^2)^2}=\frac{t^4-2t^2+1-4t^2}{(1+t^2)^2}=\frac{(t^2-1)^2}{(t^2+1)^2}$$
And, $\cos{x}$ is then:
$$\cos{x}=\pm\frac{t^2-1}{t^2+1}$$
Why do I get $\pm$ here? Did I make a mistake somewhere? Thank you for your time.
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These are well-known formulae :
$$\sin x =\frac{2t}{1+t^2},\qquad \cos x =\frac{1-t^2}{1+t^2},\qquad \tan x =\frac{2t}{1-t^2}\quad(t\not\equiv\pm\frac\pi4\mod\pi).$$
A geometric proof of the formulae:
Consider the unit circle and a line passing through the point $(-1,0)$which has equation $y=t(x+1)$. Let $M$ be the second interection point of the line with the circle. If $x$ is the polar angle of $M$, we have $t=\tan \frac x 2$, and the coordinates of $M$ are $\;x_M=\cos x,\enspace y_M=\sin x$. Now the equation for the abscissae of the intersection points is
$$(1+t^2)x_M^2+2t^2x_M+t^2-1=0,$$
of which one of the roots is $-1$, hence the other root, by Viète's relations, is $x_M=-\dfrac{t^2-1}{t^2+1}$. You obtain also
$\;\sin x=y_M=t\biggl(\dfrac{1-t^2}{t^2+1}+1\biggr)=\dfrac{2t}{t^2+1}$.
That said, Bioche's rules stipulate you should set $u=\tan x$. Indeed, $\mathrm d\mkern1mu x=\dfrac{\mathrm d\mkern1mu u}{1+u^2}$, hence
$$I=\int \frac{\sin{x}}{3\sin{x}+4\cos{x}}\mathrm d\mkern1mu x=\int\frac{u}{3+4u^2}\dfrac{\mathrm d\mkern1mu u}{1+u^2}.$$
Decomposing into partial fractions:
$$\frac u{(3+4u^2)(1+u^2)}=\frac{4u}{3+4u^2}-\frac{u}{1+u^2},$$
we get
\begin{align*}I&=\frac12\bigl(\ln(3+4u^2)-\ln(1+u^2)\bigr)\\&= \frac12\ln\biggl(\frac{3+4\tan^2x}{1+\tan^2x}\biggr)=\frac12\ln(4-\cos^2x).
\end{align*}
|
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|
integrate $\int \frac{\tan^4x}{4}\cos^3x$
$$\int \frac{\tan^4x}{4}\cos^3x$$
$$\int \frac{\tan^4x}{4}\cos^3x=\frac{1}{4}\int \frac{\sin^4x}{\cos^4x}\cos^3x=\frac{1}{4}\int\frac{\sin^4x}{\cos x}=\frac{1}{4}\int\frac{\sin^2x\cdot\sin^2x}{\cos x}=\frac{1}{4}\int\frac{\sin^2x\cdot(1-\cos^2x)}{\cos x}=\frac{1}{4}\int \frac{\sin^2x}{\cos x}-\frac{\cos^2x}{\cos x}=\frac{1}{4}\int \frac{1-\cos^2x}{\cos x}-\cos x=\frac{1}{4} \int \frac{1}{\cos x}-2\cos x=\frac{1}{4}(\ln(\tan x+\sec x)+2\sin x$$
Is it correct?
|
No, you've made a mistake in your fifth step! Solve it like this:
$$\int\frac{\tan^4(x)\cos^3(x)}{4}\space\text{d}x=\frac{1}{4}\int\tan^4(x)\cos^3(x)\space\text{d}x=\frac{1}{4}\int\tan(x)\sin^3(x)\space\text{d}x=$$
Substitute $u=\sin(x)$ and $\text{d}u=\cos(x)\space\text{d}x$:
$$\frac{1}{4}\int-\frac{u^4}{u^2-1}\space\text{d}u=-\frac{1}{4}\int\left[1+u^2+\frac{1}{2(u-1)}-\frac{1}{2(u+1)}\right]\space\text{d}u=$$
$$-\frac{1}{4}\left[\int1\space\text{d}u+\int u^2\space\text{d}u+\frac{1}{2}\int\frac{1}{u-1}\space\text{d}u-\frac{1}{2}\int\frac{1}{u+1}\space\text{d}u\right]=$$
Notice:
*
*$$\int 1\space\text{d}y=y+\text{C}$$
*$$\int y^n\space\text{d}y=\frac{y^{n+1}}{n+1}+\text{C}$$
$$-\frac{1}{4}\left[u+\frac{u^3}{3}+\frac{1}{2}\int\frac{1}{u-1}\space\text{d}u-\frac{1}{2}\int\frac{1}{u+1}\space\text{d}u\right]=$$
*
*Substitute $p=u-1$ and $\text{d}p=\text{d}u$
*Substitute $s=u+1$ and $\text{d}s=\text{d}u$
$$-\frac{1}{4}\left[u+\frac{u^3}{3}+\frac{1}{2}\int\frac{1}{p}\space\text{d}p-\frac{1}{2}\int\frac{1}{s}\space\text{d}s\right]=-\frac{1}{4}\left[u+\frac{u^3}{3}+\frac{\ln\left|p\right|}{2}-\frac{\ln\left|s\right|}{2}\right]+\text{C}=$$
$$-\frac{1}{4}\left[u+\frac{u^3}{3}+\frac{\ln\left|u-1\right|}{2}-\frac{\ln\left|u+1\right|}{2}\right]+\text{C}=$$
$$-\frac{1}{4}\left[\sin(x)+\frac{\sin^3(x)}{3}+\frac{\ln\left|\sin(x)-1\right|}{2}-\frac{\ln\left|\sin(x)+1\right|}{2}\right]+\text{C}=$$
$$-\frac{1}{4}\left[\frac{3\sin(x)+\sin^3(x)}{3}+\frac{\ln\left|\sin(x)-1\right|-\ln\left|\sin(x)+1\right|}{2}\right]+\text{C}$$
|
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|
How to compute such a limit? Knowing $f(x,y) = 2x^2 +3y^2 -7x +15y$, one simply proves $$|f(x,y)|\leq 5(x^2+y^2)+22 \sqrt{x^2 + y^2}$$ How can I use this info to compute
$$ \lim_{(x,y)\to(0,0)} \frac{f(x,y) - 2(x^2+y^2)^{1/4}}{(x^2+y^2)^{1/4}}\;\;\; ?$$
Thanks!
|
This can be done quite easily if you convert to polar coordinates.
We convert
$$lim_{x, y \to (0, 0)} \frac{2x^2 + 3y^2 - 7x + 15y - 2(x^2 + y^2)^{\frac{1}{4}}}{(x^2 + y^2)^{\frac{1}{4}}}$$
turns into
$$lim_{r \to 0} \>\> \frac{2r^2\cos^2 \theta + 3r^2 \sin^2 \theta - 7r \cos \theta + 15 r \sin \theta - 2 \sqrt{r}}{\sqrt{r}}$$
$$lim_{r \to 0} \> \>2 r^{\frac{3}{2}}\cos^2 \theta + 3 r^{\frac{3}{2}} \sin^2 \theta -7 \sqrt{r} \cos \theta + 15 \sqrt{r} \sin \theta - 2 $$
$$= -2$$
|
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|
Let $F=\mathbf Z_2$ and let $f(x)=x^3+x+1 \in F[x]$ and let a be a zero of $f(x)$ in some extension of $F$ a) How many elements does $F(a)$ have?
b) Express $a^5, a^{-2}, a^{100}$ in the form $c_2a^2+c_1a+c_0$
I know that $F(a)$ has 8 elements since $f(x)$ has no zeros in $\mathbf{Z_2}$ and by a theorem that states :
$$F(a)\cong F[x]/<p(x)>$$ for $p(x)$ irreducible over F, with a being a zero of p(x) on some extension E of F, there are two choices for each coefficient. Theorem then states that all elements of can be written in the form:
$$c_{n-1}a^{n-1}+c_{n-2}a^{n-2}+...+c_1a+c_0,$$ where $c_0, c_1,..., c_{n-1} \in F$
I need to know how to do b). I have the answer for it and I don't understand how is it that $a^5=a^2+a+1$ for example. I need an explanation of how this is done.
|
If $a$ is a zero of $x^3+x+1$, then $a^3=-a-1=a+1$, since we're working mod $2$.
Then, $a^4=a\cdot a^3=a(a+1)=a^2+a$.
Moreover, $a(a^2+1)=1$ and so $a^{-1}=a^2+1$ and $a^{-2}=(a^2+1)^2=a^4+1=a^2+a+1$.
Now, $a$ is a non-zero element in a field of $8$ elements. Hence, by Lagrange's theorem applied to the multiplicative group, we have $a^7=1$.
Therefore, $a^5=a^{-2}=a^2+a+1$ and $a^{100}=a^{100 \bmod 7}=a^2$.
You can avoid Lagrange's theorem by computing $a^5=a\cdot a^4=\cdots=a^2+a+1=a^{-2}$, which implies $a^7=1$.
|
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|
Exponential equation on the set of real numbers Solve the following equation on the set of real numbers:
$8^x+27^x+2·30^x+54^x+60^x=12^x+18^x+20^x+24^x+45^x+90^x$
$x=1; x=0; x=-1$ are trivial solutions, but I'm stuck with proving that there are no others...
|
Assuming we are looking for real valued solutions for $x$:
Start by writing all the terms on the left hand side, with zero on the other side, and write $a=2^x$, $b=3^x$ and $c=5^x$
The equation then becomes $$a^3+b^3+2abc+ab^3+a^2bc-a^2b-ab^2-a^2c-a^3b-b^2c-ab^2c=0$$
When $b$ is replaced by $a$ the polynomial is identically zero, so $(b-a)$ is a factor.
So we have one solution already, given by $$b-a=0\Rightarrow 2^x=3^x\Rightarrow x=0$$
After some long division, the remaining factor is $$a^2b+ab^2-abc-a^2+ac+b^2-bc$$
This factorises as $$(ab-a+b)(a+b-c)$$
So there are two more possible solutions to consider:
firstly,$$2^x3^x-2^x+3^x=0\Rightarrow 6^x+3^x=2^x,$$ which has only one solution, namely $x=-1$
And secondly, $$2^x+3^x-5^x=0$$ which again has only one solution, namely $x=1$
So the three solutions you identified in the first place are indeed the only solutions.
|
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|
Sums involving floor function I am looking for a direct formula for this sum
$$\sum_{k=0}^n \lfloor{\sqrt{n+k}}\rfloor\lfloor{\sqrt{k}}\rfloor$$
Or a method to efficiently compute the sum for large n
|
Here's an exact $O(\sqrt{n})$ formula:
$$\left\lfloor \sqrt{n}\right\rfloor \left((n+1) \left\lfloor \sqrt{n}\right\rfloor +\frac{1}{3} \left(-\left(\left\lfloor \sqrt{n}\right\rfloor +1\right)^3+\frac{3}{2} \left(\left\lfloor \sqrt{n}\right\rfloor +1\right)^2+\frac{1}{2} \left(-\left\lfloor \sqrt{n}\right\rfloor -1\right)\right)\right)+\frac{1}{6} \left\lfloor \sqrt{n}\right\rfloor \left(-\left\lfloor \sqrt{n}\right\rfloor \left(2 \left\lfloor \sqrt{n}\right\rfloor +3\right)+6 n+5\right) \left\lfloor \sqrt{2} \sqrt{n}-\left\lfloor \sqrt{n}\right\rfloor \right\rfloor +\sum _{j=\left\lfloor \sqrt{n}\right\rfloor +1}^{\left\lfloor \sqrt{2n} \right\rfloor } \left(\left\lfloor \sqrt{j^2-n}\right\rfloor ^3-\left(j^2-n\right) \left\lfloor \sqrt{j^2-n}\right\rfloor -\frac{1}{6} \left(\left\lfloor \sqrt{j^2-n}\right\rfloor -1\right) \left(4 \left\lfloor \sqrt{j^2-n}\right\rfloor +1\right) \left\lfloor \sqrt{j^2-n}\right\rfloor \right)$$
It can be derived via the following steps:
*the $\left\lfloor\sqrt{k+n}\right\rfloor$ factor can be ignored at first; substitute in $\sqrt{n}$ and use the known formula for summing $\left\lfloor\sqrt{k}\right\rfloor$
*now add in the deficit: each $k$ where $\left\lfloor\sqrt{k+n}\right\rfloor$ increases by $1$ (there are $O(\sqrt{n})$ such points), sum $\left\lfloor\sqrt{k}\right\rfloor$ from this $k$ to $n$ (use the known formula again, with suitable modifications).
*Clean up: shift the summation index to make the sum formula maybe nicer looking; pull factors which only depend on $n$ outside the sum, etc.
|
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|
Find formula for $\frac{1}{\sqrt 1}+ \frac{1}{\sqrt 2}+\cdots+\frac{1}{\sqrt n}$ I have the series:
$$\frac{1}{\sqrt 1}+ \frac{1}{\sqrt 2}+\cdots+\frac{1}{\sqrt n}$$
I find hard to generalize into one formula, any explanation would be helpful.
|
Analogous to Euler-Mascheroni Constant we have:
$$1+\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{3}}+\ldots+\frac{1}{\sqrt{n}}=
2\sqrt{n}-1.4603545088\ldots+\frac{1}{2\sqrt{n}}-\frac{1}{24\sqrt{n^{3}}}+O\left( \frac{1}{\sqrt{n^{7}}} \right)$$
where $\displaystyle
\lim_{n\to \infty} \left( 2\sqrt{n}-1-\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{3}}-
\ldots-\frac{1}{\sqrt{n}} \right) = 1.4603545088\ldots$
|
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|
Prove that $\bigg\lfloor\frac{m}{p^k}\bigg\rfloor-\bigg\lfloor\frac{n}{p^k}\bigg\rfloor-\bigg\lfloor\frac{m-n}{p^k}\bigg\rfloor$ equal to zero or one
Prove that $\bigg\lfloor\frac{m}{p^k}\bigg\rfloor-\bigg\lfloor\frac{n}{p^k}\bigg\rfloor-\bigg\lfloor\frac{m-n}{p^k}\bigg\rfloor$ equal to zero or one for all $k,m,n\in \mathbb N$. where $m\geqslant n$ and $p$ is a prime.
What I did:
$\bigg\lfloor\frac{m}{p^k}\bigg\rfloor-\bigg\lfloor\frac{n}{p^k}\bigg\rfloor-\bigg\lfloor\frac{m}{p^k}-\frac{n}{p^k}\bigg\rfloor$
Let $x:=\frac{m}{p^k}$ and let $y:=\frac{n}{p^k}$
we get $\lfloor x\rfloor-\lfloor y\rfloor-\lfloor x-y\rfloor$
What should I do now?
|
Note that
$$
\bigg\lfloor\frac{m}{p^k}\bigg\rfloor\le \frac{m}{p^k}<\bigg\lfloor\frac{m}{p^k}\bigg\rfloor+1\quad\text{and}\quad \bigg\lfloor\frac{n}{p^k}\bigg\rfloor\le \frac{n}{p^k}<\bigg\lfloor\frac{n}{p^k}\bigg\rfloor+1.
$$
Hence,
$$
\frac{m}{p^k}-1-\frac{n}{p^k}<\bigg\lfloor\frac{m}{p^k}\bigg\rfloor-\bigg\lfloor\frac{n}{p^k}\bigg\rfloor<\frac{m}{p^k}-\frac{n}{p^k}+1.
$$
Now you can do the same for the third term in your expression.
|
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|
Integrate $\int_0^4 \frac{x}{\sqrt{16 + 5x}} \, dx$ I'm trying to solve
$$
\int_0^4 \frac{x}{\sqrt{16 + 5x}} \, dx
$$
using the substitution rule. The substitution rule, as far as I know it, reads
Let $g^\prime$ be a continuous function on $[a,b]$ whose range is an interval $I$, and let $f$ be continuous on $I$. Then
$$
\int_a^b f(g(x))g^\prime(x)\, dx = \int_{g(a)}^{g(b)} f(u)\, du
$$
The solutions manual gives the solution as follows. Let $u = 16 + 5x$, so $x = \frac{1}{5}(u - 16)$ and $du = 5dx$. When $x = 0$, $u = 16$ and when $x = 4$, $u = 36$. So,
\begin{align*}
\int_0^4 \frac{x}{\sqrt{16 + 5x}} \, dx & = \int_{16}^{36} \frac{\frac{1}{5}(u - 16)}{\sqrt{u}} \, \frac{du}{5} = \ldots
\end{align*}
I'm comfortable solving from there, but I don't see exactly how I used the substitution rule as stated above. It seems like we are saying $g(x) = 16 + 5x$, so $g^\prime(x) = 5$, and $f(u) = u^{-1/2}$. Then $x = \frac{1}{5}(g(x) - 16)$, and we have the form
$$
\int_0^4 \frac{x}{\sqrt{16 + 5x}} \, dx = \int_{0}^{4}\frac{1}{5}(g(x) - 16)f(g(x))\, dx,
$$
which is not what we need for the substitution rule. How can I write the original integral in terms of $f$, $g$ and $g^\prime$ so it is apparent how to apply the substitution rule?
|
$$\int_{0}^{4}\frac{x}{\sqrt{16+5x}}\space\text{d}x=$$
Substitute $u=16+5x$ and $\text{d}u=5\space\text{d}x$.
This gives a new lower bound $u=16+5\cdot0=16$ and upper bound $u=16+5\cdot4=36$:
$$\frac{1}{5}\int_{16}^{36}\frac{u-16}{5\sqrt{u}}\space\text{d}u=$$
$$\frac{1}{5}\int_{16}^{36}\left[\frac{\sqrt{u}}{5}-\frac{16}{5\sqrt{u}}\right]\space\text{d}u=$$
$$\frac{1}{5}\left[\int_{16}^{36}\frac{\sqrt{u}}{5}\space\text{d}u-\int_{16}^{36}\frac{16}{5\sqrt{u}}\space\text{d}u\right]=$$
$$\frac{1}{5}\left[\frac{1}{5}\int_{16}^{36}\sqrt{u}\space\text{d}u-\frac{16}{5}\int_{16}^{36}\frac{1}{\sqrt{u}}\space\text{d}u\right]=$$
$$\frac{1}{5}\left[\frac{1}{5}\int_{16}^{36}u^{\frac{1}{2}}\space\text{d}u-\frac{16}{5}\int_{16}^{36}u^{-\frac{1}{2}}\space\text{d}u\right]=$$
Use:
$$\int y^b\space\text{d}y=\frac{y^{b+1}}{b+1}+\text{C}$$
$$\frac{1}{5}\left[\frac{1}{5}\left[\frac{2u^{\frac{3}{2}}}{3}\right]_{16}{36}-\frac{16}{5}\left[2\sqrt{u}\right]_{16}{36}\right]=$$
$$\int y^b\space\text{d}y=\frac{y^{b+1}}{b+1}+\text{C}$$
$$\frac{1}{5}\left[\frac{2}{15}\left[u^{\frac{3}{2}}\right]_{16}{36}-\frac{32}{5}\left[\sqrt{u}\right]_{16}{36}\right]=$$
$$\frac{1}{5}\left[\frac{2}{15}\left(36^{\frac{3}{2}}-16^{\frac{3}{2}}\right)-\frac{32}{5}\left(\sqrt{36}-\sqrt{16}\right)\right]=$$
$$\frac{1}{5}\left[\frac{2}{15}\left(216-64\right)-\frac{32}{5}\left(6-4\right)\right]=$$
$$\frac{1}{5}\left[\frac{2}{15}\left(152\right)-\frac{32}{5}\left(2\right)\right]=$$
$$\frac{1}{5}\left[\frac{304}{15}-\frac{64}{5}\right]=$$
$$\frac{1}{5}\left[\frac{112}{15}\right]=$$
$$\frac{112}{75}$$
|
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|
Prove the convergence and find the sum of series $\sum\limits_{n=1}^{\infty}\left(n^3\sin\frac{\pi}{3^n}\right)$. We know that $0<\sin\frac{\pi}{3^n}\le\frac{\sqrt 3}{2},\forall n\ge 1$. How to find the boundary for $n^3\sin\frac{\pi}{3^n}$ (how to use comparison test here)?
I tried using the ratio test, but the limit $$\lim\limits_{n\to\infty}\left((n+1)^3\sin\frac{\pi}{3^{n+1}}\cdot \frac{1}{n^3\sin\frac{\pi}{3^n}}\right)$$ isn't that easy to evaluate.
|
Since $\sin x \le x$ for $x\ge 0$ and $0\le \frac{\pi}{3^n}\le \pi$,
$$
0\le n^3 \sin\frac{\pi}{3^n} \le \frac{n^3 \pi}{3^n}
$$
and $\sum_{n=1}^{\infty}\frac{n^3 \pi}{3^n}$ converges by ratio test. By comparison test, given series converges.
|
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|
Is there integral or series for $\sqrt{10}-\frac{4^4}{3^4}$ (to prove the inequality)? Both of these numbers are bad approximations for $\pi$, but they turn out to be much closer together:
$$\sqrt{10}-\frac{4^4}{3^4}=0.00178$$
Since there is a lot of questions here about integrals and series which prove such close inequalities, I wanted to know if something exists for this inequality as well:
$$\sqrt{10}>\frac{4^4}{3^4}$$
If we use the continued fraction for $\sqrt{10}$:
$$\sqrt{10}=3+\cfrac{1}{6+\cfrac{1}{6+\cdots}}$$
One of the approximants will be:
$$\sqrt{10} \approx \frac{117}{37}=\frac{234}{74}$$
$$\frac{4^4}{3^4}=\frac{256}{81}$$
This is just a curiosity for me, there is no other context for the question.
|
Well, this is the same as $\sqrt{2}\sqrt{5} - \frac{2^8}{3^4}$. So, you can use tools from calculus to show that this function is positive at x = 2:
$f(x) = \sqrt{x}\sqrt{5} - \frac{x^8}{3^4}$
First of all, its zero is at:
$\frac{x^8}{\sqrt{x}} = 3^4 \sqrt{5} $
This gives us:
$x= 3^{(8/15)} 5^{(1/15)}$
This is slightly larger than 2.
Also, the derivative of f(x) is:
$f'(x) = \frac{\sqrt{5}}{2 \sqrt{x}} - \frac{8x^7}{3^4}$
This is negative for $x > (\frac{3}{2})^{(8/15)} 5^{(1/15)}$ . So, $f(x)$ must be positive before $x= 3^{(8/15)} 5^{(1/15)}$.
|
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|
how do I prove this by induction? (recursion) The terms are given recursively:
$P_0=3$
$P_1=7$
and $P_n = 3P_{n-1}-2P_{n-2}$ for $n\ge2$
What should I assume and what step proves that $P_n=2^{n+2}-1$ is a closed form of the sequence.
Suppose $n_0=1$ and the base cases are $0$ and $1$.
I think this book has a mistake
problem from the book
|
Note that $2^{0+2}-1=2^{2}-1=4-1=3$ and $2^{1+2}-1=2^{3}-1=8-1=7$,
so that the base case holds. Suppose now that the statement holds
for $P_{n-1}$ and $P_{n-2}$ and note that
\begin{align*}
P_{n} & =3P_{n-1}-2P_{n-2}\\
& =3\left(2^{n+1}-1\right)-2\left(2^{n}-1\right)\\
& =3\cdot2^{n+1}-3-2^{n+1}+2\\
& =2\cdot2^{n+1}-1\\
& =2^{n+2}-1.
\end{align*}
|
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|
Evaluate $\int_0^1 \sqrt{x^2+4x+1}\,dx$
Evaluate $\displaystyle\int_0^1 \sqrt{x^2+4x+1}\,dx$
I tried simplify by doing this:
$$\int_0^1 \sqrt{x^2+4x+1}\:dx=\displaystyle\int_0^1 \sqrt{(x+2)^2-3}\:dx$$
Then, by letting $t=x+2$, $dt=dx$
$$\int_0^1 \sqrt{t^2-3}\:dt=\displaystyle\int_0^1 \sqrt{(t-\sqrt 3)(t+\sqrt 3)}\:dt$$
I tried solving $\displaystyle\int_0^1 \sqrt{t^2-3}\:dt$ by substituting: $$k=t^2-3, dt=\frac{dk}{2\sqrt{|k+3|}}$$
But that leads me nowhere:
$$\frac{1}{2}\displaystyle\int_0^1 \frac{\sqrt{k}}{\sqrt{|k+3|}}\:dk$$
Any hints?
|
Using the Euler substitution $\sqrt{x^2+4x+1}=t-x$, we obtain
$\displaystyle x=\frac{t^2-1}{2(t+2)}$ and $\displaystyle dx=\frac{t^2+4t+1}{2(t+2)^2}dt$,
so $\displaystyle\int_0^1\sqrt{x^2+4x+1}dx=\int_1^{1+\sqrt{6}}\left(t-\frac{t^2-1}{2(t+2)}\right)\frac{t^2+4t+1}{2(t+2)^2}dt=\frac{1}{4}\int_1^{1+\sqrt{6}}\frac{(t^2+4t+1)^2}{(t+2)^3}dt$.
Now let $u=t+2, t=u-2, dt=du$ to get
$\displaystyle\frac{1}{4}\int_3^{3+\sqrt{6}}\frac{(u^2-3)^2}{u^3}du=\frac{1}{4}\int_3^{3+\sqrt{6}}\left(u-\frac{6}{u}+9u^{-3}\right)du=\frac{1}{4}\left[\frac{u^2}{2}-6\ln u-\frac{9}{2}u^{-2}\right]_3^{3+\sqrt{6}}$
$\displaystyle=\frac{3}{4}(\sqrt{6}+1)-\frac{3}{2}\ln\left(\frac{3+\sqrt{6}}{3}\right)-\frac{9}{8}\left(\frac{-6(1+\sqrt{6})}{9(15+6\sqrt{6})}\right)=\frac{3}{2}\sqrt{6}-1-\frac{3}{2}\ln\left(\frac{3+\sqrt{6}}{3}\right)$
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|
Integration of a $\int \frac{v^2+1}{v^3-v^2+v+1}\,dv$ While solving the DE $$\frac{dy}{dx}=\frac{y^2-x^2}{y^2+x^2}$$ with the initial substitution $y=vx$ I got stuck in the integration of :
$$\int \frac{v^2+1}{v^3-v^2+v+1}\,dv$$
I don't know how to proceed further. Kindly help.
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The main difficulty is that your cubic doesn't have nice roots. Lets call them $v_r$, $v_c$ and $\bar{v_c}$ and let $v_c=a+bi$. I.e.
$$v^3-v^2+v+1=(v-v_r)(v-v_c)(v-\bar{v_c})$$
$$=(v-v_r)(v^2-2av+a^2+b^2)=(v-v_r)((v-a)^2+b^2)$$
Hence we can write:
$$\frac{v^2+1}{v^3-v^2+v+1}=\frac{A}{v-v_r}+\frac{Bv+C}{(v-a)^2+b^2}$$
$$v^2+1=A((v-a)^2+b^2)+(Bv+C)(v-v_r)$$
Next work out the coefficients $A$, $B$, and $C$.
If $v=v_r$ then:
$$v_r^2+1=A((v_r-a)^2+b^2)$$
$$A=\frac{v_r^2+1}{(v_r-a)^2+b^2}$$
If $v=0$ then:
$$1=A(a^2+b^2)+C(-v_r)$$
$$C=\frac{A(a^2+b^2)-1}{v_r}=\frac{2a+v_r(a^2+b^2-1)}{(v_r-a)^2+b^2}$$
Rearranging the equation for $B$ gives:
$$B=\frac{1+v^2-A((v-a)^2+b^2)-C(v-v_r)}{v(v-v_r)}$$
Subbing $A$ and $C$ back in and solving for $B$ gives (note the $v$ cancels out):
$$B=\frac{a^2+b^2-1-2av_r}{(v_r-a)^2+b^2}$$
You can then carry out the integral and finally sub back in the $A$, $B$ and $C$.
$$\int\frac{A}{v-v_r}+\frac{Bv+C}{(v_r-a)^2+b^2)}\ dv$$
$$=A\log{|v-v_r|}+\frac{aB+C}{b}\arctan\left(\frac{v-a}{b}\right)+\frac{B}{2}\log((v-a)^2+b^2)$$
Needless to say it doesn't end up pretty and doesn't have a nice closed form as you also need to substitute in the actual values for $v_r$, $a$ and $b$ which are:
$$v_r=\frac{(3\sqrt{33}-17)^{\frac23}+(3\sqrt{33}-17)^{\frac13}-2}{3(3\sqrt{33}-17)^{\frac13}}$$
$$a=\frac{-(3\sqrt{33}-17)^{\frac23}+2(3\sqrt{33}-17)^{\frac13}+2}{6(3\sqrt{33}-17)^{\frac13}}$$
$$b=\frac{\sqrt{3}(3\sqrt{33}-17)^{\frac13}+2\sqrt{3}}{6(3\sqrt{33}-17)^{\frac13}}$$
Of note Wolframalpha also struggles to express this integral nicely.
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|
Algebraic shortcut to find $a^n + b ^n$? Recently, I found this problem online:
Given $a+b=1$ and $a^2+b^2=2$, find $a^7+b^7$.
Although I could've solved it by substituting the first equation into the second and then using the quadratic formula; the way the question was set up, I suspected that there was a shortcut. However, I couldn't find the solution to the problem on that website, so I'm asking here:
If you are given $a+b$ and $a^2+b^2$, is there a shortcut to finding $a^n+b^n$?
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Note that
$$a^{n+1}+b^{n+1}=(a^n+b^n)(a+b)-ab(a^{n-1}+b^{n-1}).\tag{1}$$
Since $(a+b)^2=1$, and $a^2+b^2=2$, we have $2ab=-1$ and therefore $ab=-\frac{1}{2}$.
It follows from (1) that
$$a^{n+1}+b^{n+1}=(a^n+b^n)(1)+\frac{1}{2}(a^{n-1}+b^{n-1}).$$
Let $f(k)=a^k+b^k$. We have obtained the recurrence
$$f(n+1)=f(n)+\frac{1}{2}f(n-1).$$
Using this recurrence, we can compute our way to $f(7)$ fairly quickly.
Remarks: $1.$ The same strategy can be used for any given $a+b$ and $a^2+b^2$.
$2.$ To go a little faster, we could use $a^{n+2}+b^{n+2}=(a^n+b^n)(a^2+b^2)-a^2b^2(a^{n-2}+b^{n-2})$.
$3.$ If we are in a real hurry, and $n$ is not small, there are various tricks to speed up computation. For example
$$a^{2k}+b^{2k}=(a^k+b^k)^2-2(ab)^k.$$
So we can take "giant" steps.
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|
Trigonometric Indentities $\dfrac{1}{\cos x}+ \dfrac{2\cos x}{\cos 2x}$
If $x=\dfrac{\pi}{7}$ Prove that above expression is equal to $4$.
I worked a few steps and reached here:
$\dfrac{(4\cos^2 x -1)}{\cos x (2\cos^2 x-1)}$
Not able to proceed after this!!!
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$$F=\dfrac{\cos2x+2\cos^2x}{\cos x\cos2x}=\dfrac{3-4\sin^2x}{\cos x\cos2x}$$
If $\sin x\ne0,$
$$F=\dfrac{\sin x(3-4\sin^2x)}{\sin x\cos x\cos2x}=\frac{2\sin3x}{\sin2x\cos2x}=4\cdot\dfrac{\sin3x}{\sin4x}$$
Now $\sin4x=\sin3x$
if $4x=n\pi+(-1)^n3x$ where $n$ is any integer
If $n$ is odd $=2m+1,$(say), $7x=(2n+1)\pi, x=\dfrac{(2n+1)\pi}7$ where $n\equiv0,\pm1,\pm2,\pm3\pmod7$
But $n\equiv3\pmod7\implies\sin x=0$
So, $\sin x\ne0\implies n\equiv0,\pm1,\pm2,-3\pmod7$
If $n$ is even $=2m$(say), $x=2m\pi\implies\sin x=?$
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|
Determine the real numbers $x, y, z, t$ satisfying an equation Be non-zero real numbers $a, b, c$ so $a + b + c \neq 0$ and $ab + bc + ca = 0$.
Determine the real numbers $x, y, z, t$ knowing that
$$\frac{\frac{a^2 + b^2 + c^2}{2}+x^2+y^2+z^2}{a+b+c}= t+ \sqrt{ax+by+cz-t(a+b+c)}.$$
I tried converting equality data but managed to find the numbers.
I wonder: is even possible to determine the numbers $x, y, z, t$ ?
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is even possible to determine the numbers x, y, z, t?
Yes, it's possible.
Let $x^2+y^2+z^2=P,a+b+c=Q$.
Since
$$(a+b+c)^2=a^2+b^2+c^2+2(ab+bc+ca)=a^2+b^2+c^2$$
we have
$$\frac{\frac{a^2 + b^2 + c^2}{2}+x^2+y^2+z^2}{a+b+c}= t+ \sqrt{ax+by+cz-t(a+b+c)}$$
$$\Rightarrow\quad \frac{\frac{Q^2}{2}+P}{Q}= t+ \sqrt{ax+by+cz-Qt}$$
$$\Rightarrow\quad Q^2+2P-2Qt=2Q\sqrt{ax+by+cz-Qt}$$
Squaring the both sides gives
$$\Rightarrow\quad (Q^2+2P-2Qt)^2=4Q^2(ax+by+cz-Qt)$$
$$\Rightarrow\quad Q^4+4P^2+4Q^2t^2+4PQ^2-4Q^3t-8PQt=4Q^2(ax+by+cz)-4Q^3t$$
$$\Rightarrow\quad Q^4+4P^2+4Q^2t^2+4PQ^2-8PQt=4Q^2(ax+by+cz)$$
$$\Rightarrow\quad 4Q^2t^2-8PQt+Q^4+4P^2+4PQ^2-4Q^2(ax+by+cz)=0$$
$$\Rightarrow\quad t=\frac{2P\pm Q\sqrt{-(a-2x)^2-(b-2y)^2-(c-2z)^2}}{2Q}$$
Hence, in order for $t\in\mathbb R$ to exist, we have to have
$$a-2x=b-2y=c-2z=0,$$
i.e.
$$\color{red}{x=\frac a2,\quad y=\frac b2,\quad z=\frac c2}$$
and so $$t=\frac{P}{Q}=\frac{(a/2)^2+(b/2)^2+(c/2)^2}{a+b+c}=\frac{(a+b+c)^2}{4(a+b+c)}=\color{red}{\frac{a+b+c}{4}=t}$$
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$\forall x\in R,$find the range of the function $f(x)=\cos x(\sin x+\sqrt{\sin^2x+\sin^2\alpha});\alpha\in[0,\pi]$ $\forall x\in R,$find the range of the function $f(x)=\cos x(\sin x+\sqrt{\sin^2x+\sin^2\alpha});\alpha\in[0,\pi]$
$f(x)=\cos x(\sin x+\sqrt{\sin^2x+\sin^2\alpha});\alpha\in[0,\pi]$
$f'(x)=\cos x(\cos x+\frac{\sin x\cos x}{\sqrt{\sin^2x+\sin^2\alpha}})-\sin x(\sin x+\sqrt{\sin^2x+\sin^2\alpha})$
I am stuck here and could not find the minimum and maximum values of $f(x),$The answer given is $-\sqrt{1+\sin^2\alpha}\leq f(x)\leq\sqrt{1+\sin^2\alpha}$.
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Let $$y=\cos x\left[\sin x+\sqrt{\sin^2 x+\sin^2 a}\right] = \sin x\cdot \cos x+\cos x\cdot \sqrt{\sin^2 x+\sin^2 a}$$
Now Using $\bf{Cauchy\; Schwartz\; Inequality}$
We get $$(\sin^2 x+\cos ^2 x)\cdot \left[\cos^2 x+\sin^2 x+\sin^2 a\right]\geq \left(\sin x\cdot \cos x+\cos x\cdot \sqrt{\sin^2 x+\sin^2 a}\right)^2$$
So we get $$y^2\leq (1+\sin^2 a)\Rightarrow |y| \leq\sqrt{1+\sin^2 a}$$
So we get $$-\sqrt{1+\sin^2 a}\leq y \leq \sqrt{1+\sin^2 a}.$$
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curve of $(x^2+y^2)^2=2(x^2-y^2)$ The diagram shows the curve $(x^2+y^2)^2=2(x^2-y^2)$ and one of its maximum points $M$. Find the coordinates of $M$.
My attempt.
Differentiate the equation and I got
$\frac{dy}{dx}=-\frac{x(x^2+y^2-1)}{y(x^2+y^2+1)}$
How should I proceed?
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I think you take the top part set it to zero $x(x^2+y^2-1)=0$, since thats how you get the critical numbers. Then you have :
$x=0$ or $y=\sqrt{1-x^2}$
plug y back into original formula and solve to get something like
$x=\pm \sqrt{\frac{3}{4}}$ so max would be at one of those $x$ values.
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find the rate of change $f(x) = 4\sin^3 x$ when $x = \frac{5\pi}{6}$ find the rate of change $f(x) = 4\sin^3 x$ when $x = \frac{5\pi}{6}$
To find the rate of change I need to find $\frac{dy}{dx}$ using the chain rule $h'(x) = g'(f(x)).f'(x)$
$g'(f(x)) = 12\sin^2 x$
$f'(x) = \cos x$
$h'(x) = 12\sin^2 x \cos x$
After this, I am completely stuck, this comes from a paper that does not allow calculators and am I supposed to know what $\frac{5\pi}6$ means and be able to apply it to the derivative function? If so could somebody offer a link or something for me to learn becasue I do not know this.
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Trig by Reference Triangles:
The angle $\frac{5 \pi}{6}$ in radians will be given by $150^o$. This simple conversion can be done by remembering that $180^o = \pi$ $rad $.
To solve for
$h'(x)=12 sin^2x cosx$
$h'( \frac{ 5 \pi}{6} )=12 sin^2 ( \frac{5 \pi}{6}) cos( \frac{5 \pi}{6} )$
we can use right-angled triangles of the standard angles. E.g. here our reference triangle is the $(90, 60, 30^o)$ triangle in the secondant quadrant of the cartesian plane. However, we are more interested in the radian form of the triangle ($ \frac{\pi}{2}, \frac{ \pi}{3}, \frac{\pi}{6}) $. The respective side lengths of this triangle are therefore $(2, - \sqrt{3}, 1)$.
$sin (\frac{5 \pi}{6} ) = \frac{ - \sqrt {3}}{2}$
and,
$-cos (\frac{5 \pi}{6} ) = \frac{1}{2}$
(See link:Reference Triangles)
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Multiple Angle formulas, alternate forms Relatively simple question, that might not be simple to answer:
I have noticed that there are ways of expressing every double angle formula of a given trigonometric function using only that function except for $\sin$ and $\csc$. That is,
$\sin2\theta=2\sin\theta\cos\theta=?$
$\cos2\theta=2\cos^2\theta-1$
$\tan2\theta=\dfrac{2\tan\theta}{1-\tan^2\theta}$
$\csc2\theta=\dfrac{1}{2}\csc\theta\sec\theta=?$
$\sec2\theta=\dfrac{\sec^2\theta}{2-\sec^2\theta}$
$\cot2\theta=\dfrac{\cot^2\theta-1}{2\cot\theta}$
For triple angle formulas, all 6 trig functions have expressions using only the given trig function. They are
$\sin3\theta=3\sin\theta-4\sin^3\theta$
$\cos3\theta=4\cos^{3}\theta-3\cos\theta$
$\tan3\theta=\dfrac{3\tan\theta-\tan^3\theta}{1-3\tan^2\theta}$
$\csc3\theta=\dfrac{\csc^3\theta}{3\csc^2\theta-4}$
$\sec3\theta=\dfrac{\sec^3\theta}{4-3\sec^2\theta}$
$\cot3\theta=\dfrac{3\cot\theta-\cot^3\theta}{1-3\cot^2\theta}$
My question is: What are the the formulas, provided they exist, for $\sin2\theta$ and $\csc2\theta$ in terms of $\sin\theta$ and $\csc\theta$ respectively. If they do not exist, then some explanation as to why it is not possible would be most insightful.
Edit: I appreciate the answers so far, but what I really wanting to know is: Is there a known closed form (no piecewise-defined function) expression for the given expressions. Or if not, how to show that there is no such expression?
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A formula using no trig functions other than the sine, and
without the ambiguity of symbols such as $\pm$, is
$$
\sin(2\theta) = \begin{cases}
2\sin\theta \sqrt{1-\sin^2\theta} &
\text{if $2n\pi-\frac\pi2 \leq \theta \leq 2n\pi+\frac\pi2$
for some integer $n$,} \\
-2\sin\theta \sqrt{1-\sin^2\theta} & \text{otherwise}.
\end{cases}
$$
The reason this works is that
$\sqrt{1-\sin^2\theta} = \lvert \cos\theta\rvert$,
which is $\cos\theta$ when $\cos\theta \geq 0$, which occurs
whenever $2n\pi-\frac\pi2 \leq \theta \leq 2n\pi+\frac\pi2$
for some integer $n$;
but for any other $\theta$, $\cos\theta < 0$ and therefore
$\lvert\cos\theta\rvert = -\cos\theta$.
In fact, this formula is really just a combination of the identity
$\sin(2\theta) = 2\sin\theta\cos\theta$ with the identity
$$
\cos\theta = \begin{cases}
\sqrt{1-\sin^2\theta} &
\text{if $2n\pi-\frac\pi2 \leq \theta \leq 2n\pi+\frac\pi2$
for some integer $n$,} \\
-\sqrt{1-\sin^2\theta} & \text{otherwise}.
\end{cases}
$$
Not surprisingly, few people choose to write out such a complicated
formula merely to have a formula for $\sin(2\theta)$ that involves
no trig functions other than $\sin\theta$.
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Prove that this system of linear equations generates $\left| \left( \begin{smallmatrix} 1/2 \\ n \end{smallmatrix} \right) \right|$ as a solution? This infinite system of linear equations:
$$ \begin{array}( 2x_1=1 \\ 3x_1+4x_2=2 \\ 4x_1+5x_2+6x_3=3 \\ \cdots \end{array} $$
In other words, this is particular case of a system:
$$ \begin{array}( a_{11}x_1=b_1 \\ a_{21}x_1+a_{22}x_2=b_2 \\ a_{31}x_1+a_{32}x_2+a_{33}x_3=b_3 \\ \cdots \\ a_{m1}x_1+a_{m2}x_2+ \cdots + a_{mn}x_n + \cdots=b_m \\ \cdots \end{array} $$
With the coefficients given by:
$$a_{mn}=m + n~~~~~~~~~b_m=m$$
The systems seems to give the following sequence of solutions:
$$x_n= \left| \left( \begin{matrix} \frac{1}{2} \\ n \end{matrix} \right) \right|=\left(\frac{1}{2}, \frac{1}{8}, \frac{1}{16}, \frac{5}{128}, \cdots \right)$$
How can I prove that it's true?
Why does it happen? Can other fractional binomial coefficients be obtained this way (and how)? By 'this way' I mean a triangular system with simple regular sequence of integer coefficients.
I had some thoughts, and that's what I tried. Let's assume, that $x_n$ I found are really binomial coefficients. Then for $x_m$ we will have:
$$x_m= \left| \left( \begin{matrix} \frac{1}{2} \\ m \end{matrix} \right) \right|=\frac{1/2 (1-1/2)(2-1/2)\cdots (m-1-1/2)}{m!}=\frac{m-3/2}{m}x_{m-1}$$
$$x_m= \left(1- \frac{3}{2m} \right)x_{m-1}=x_1-\frac{3}{2}\left(\frac{x_1}{2}+\frac{x_2}{3}+\cdots+ \frac{x_{m-1}}{m} \right)$$
$$x_m=\frac{1}{2}-\frac{3}{2}\left(\frac{x_1}{2}+\frac{x_2}{3}+\cdots+ \frac{x_{m-1}}{m} \right)$$
On the other hand, from the $m^{th}$ equation we can derive:
$$x_m= \frac{b_m}{a_{mm}}-\frac{1}{a_{mm}}\left(a_{m1}x_1+a_{m2}x_2+\cdots+ a_{m,m-1}x_{m-1} \right)$$
$$x_m= \frac{1}{2}-\frac{1}{2m}\left((m+1)x_1+(m+2)x_2+\cdots+ (m+m-1)x_{m-1} \right)$$
I don't see how to get this relation into the same form as the previous one. Maybe I should modify it somehow.
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Alright, I figured it out. First, we write the recurrence relation for the absolute value of the binomial coefficients:
$$x_m= \left| \left( \begin{matrix} \frac{1}{2} \\ m \end{matrix} \right) \right|=\frac{1/2 (1-1/2)(2-1/2)\cdots (m-1-1/2)}{m!}=\frac{m-3/2}{m}x_{m-1}$$
$$x_1=\frac{1}{2}$$
Next, we work with our system.
We subtract the equation number $m-2$ from the equation number $m-1$ to show:
$$x_1+x_2+\cdots + x_{m-2}=1-2(m-1)x_{m-1}$$
Now we add this equation to the equation number $m-1$ and obtain:
$$(m+1)x_1+(m+2)x_2+\cdots + (2m-2)x_{m-2}+(2m-2)x_{m-1}=m-2(m-1)x_{m-1}$$
$$(m+1)x_1+(m+2)x_2+\cdots + (2m-2)x_{m-2}=m-(4m-4)x_{m-1}$$
Now we use the RHS from this equation in the equation number $m$:
$$m-(4m-4)x_{m-1}+(2m-1)x_{m-1}+2mx_m=m$$
Finally we obtain:
$$x_m=\frac{2m-3}{2m}x_{m-1}$$
$$x_1=\frac{1}{2}$$
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Limit $\lim_{n\to\infty}\sum_{k=0}^n\frac{n}{n^2+k}$ I came across this problem that I'm supposed to be able to solve in under 5 minutes (for a competition).
$$\lim_{n\to \infty} \sum_{k=0}^n \frac {n}{n^2+k}$$
I tried solving this for small sums, $\sum_{k=0}^2 \frac {n}{n^2+k}$, $\sum_{k=0}^3 \frac {n}{n^2+k} $ and made this:
$$\sum_{k=0}^2 \frac {n}{n^2+k} = \frac {n}{n^2+1} + \frac{n}{n^2+2} = \frac {n(n^2+2)+n(n^2+1)}{(n^2+1)(n^2+2)} = \frac {2n^3+...}{n^4+...}$$
And it's limit is $0$, for $k$ up to $3$ is $0$ too, so I assumed that the original limit must be $0$. Wrong, my book says that is $1$, I graphed the function and indeed that is correct. How should I solve it?
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$\displaystyle \sum_{k=0}^n \frac {n}{n^2+k} = \frac {n}{n^2+0} + \frac{n}{n^2+1} +.......... +\frac{n}{n^2+n} $
$\displaystyle = \frac {n(n^2+1)(n^2+2)\cdots(n^2+n)+n(n^2)(n^2+2)......}{n^2(n^2+1)(n^2+2)....(n^2+n)} = \frac {n.n^{2n}(n+1)+}{n^2\cdot n^{2n}+}$
$\displaystyle \lim_{n\rightarrow \infty} \sum_{k=0}^n \frac {n}{n^2+k}= \lim_{n\rightarrow \infty}\frac {n.n^{2n}(n+1)+....}{n^2\cdot n^{2n}+.....} = 1$
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Pade approximant for the function $\sqrt{1+x}$ I'm doing the followiwng exercise:
The objective is to obtain an approximation for the square root of any given number using the expression
$$\sqrt{1+x}=f(x)\cdot\sqrt{1+g(x)}$$
where g(x) is an infinitesimal. If we choose $f(x)$ as an approximation of $\sqrt{1+x}$, then we can calculate $g(x)$:
$$g(x)=\frac{1+x}{f^2(x)}-1$$
$f(x)$ can be chosen as a rational function $p(x)/q(x)$, such that $p$ and $q$ have the same degree and it's Mclaurin series is equal to the Mclaurin series of the function $\sqrt{1+x}$ until some degree. Find a rational function $f(x):=p(x)/q(x)$, quotient of two linear polynomials, such that the McLaurin series of $p(x)-\sqrt{1+x}\cdot q(x)$ have the three first terms equal to $0$.
How can I do this? Has something to be with the Pade approximant?
Any hint would be really appreciated. Thanks for your time.
|
Using the fact that $\sqrt{1+x}=1+\frac x{1+\sqrt{1+x}}$, we get
$$
\begin{align}
\sqrt{1+x}
&=1+\cfrac x{2+\cfrac x{2+\cfrac x{2+\cfrac x{2+\cdots}}}}
\end{align}
$$
Which gives the approximants
$$
\begin{align}
\color{#C00000}{1+\frac12x}&=1+\frac12x\\
\color{#00A000}{\frac{1+\frac34x}{1+\frac14x}}&=1+\frac12x-\frac18x^2+\frac1{32}x^3-\frac1{128}x^4+O\left(x^5\right)\\
\color{#5555FF}{\frac{1+x+\frac18x^2}{1+\frac12x}}&=1+\frac12x-\frac18x^2+\frac1{16}x^3-\frac1{32}x^4+O\left(x^5\right)\\
\color{#C0A000}{\frac{1+\frac54x+\frac5{16}x^2}{1+\frac34x+\frac1{16}x^2}}&=1+\frac12x-\frac18x^2+\frac1{16}x^3-\frac5{128}x^4+O\left(x^5\right)\\
\sqrt{1+x}&=1+\frac12x-\frac18x^2+\frac1{16}x^3-\frac5{128}x^4+O\left(x^5\right)
\end{align}
$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1750968",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 5,
"answer_id": 0
}
|
How to evaluate this limit using Taylor expansions? I am trying to evaluate this limit:
$\lim_{x \to 0} \dfrac{(\sin x -x)(\cos x -1)}{x(e^x-1)^4}$
I know that I need to use Taylor expansions for $\sin x -x$, $\cos x -1$ and $e^x-1$. I also realise that all of these are just their regular Taylor expansions with their first term removed so the series starts at $n=1$ rather than at 0.
However, when I actually try to evaluate I get stuck at:
$\lim_{x \to 0} \dfrac{(-\frac{1}{3!}+\frac{x^2}{5!}...)(-\frac{9x^2}{2!}+\frac{81x^4}{4!}...)}{(\frac{1}{x}+\frac{1}{2!}+\frac{x}{3!}+\frac{x^2}{4!}...)^4}$
I don't know how to proceed from here.
|
Turns out to just be a foolish mistake. When I factored $x$ out of the denominator I should have raised it to 4. That would be an $x^5$ on the bottom:
$\lim_{x \to 0} \dfrac{x^3(-\frac{1}{3!}+\frac{x^2}{5!}...)x^2(-\frac{9}{2!}+\frac{81x^2}{4!}...)}{x^5({1}+\frac{x}{2!}+\frac{x^2}{3!}+\frac{x^3}{4!}...)^4}$
Now I can simply cancel out the $x^5$s and then substitute remaining $x$s for zeroes.
$\lim_{x \to 0} \dfrac{(-\frac{1}{3!})(-\frac{9}{2!})}{({1})^4} = \dfrac{3}{4} $
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1752176",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 2
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|
Find the limit $\lim_{x\rightarrow 0} \frac{\sin x - x}{\tan x - x}$ Find the limit if it exists. (this exercise is taken from Calculus - The Classic Edition by Swokowski Chapter 10, section 1, no. 9, p.498)
$$\lim_{x\rightarrow 0} \frac{\sin x - x}{\tan x - x}$$
since the limit is $0/0$ therefore, we use L'Hopital's rule, that is,
$$\lim_{x\rightarrow 0} \frac{\sin x - x}{\tan x - x} = \lim_{x\rightarrow 0}\frac{\cos x - 1}{\sec^2 x - 1 }$$
since $\lim_{x\rightarrow 0}\frac{\cos x - 1}{\sec^2 x - 1 } = \frac {1 - 1}{ 1- 1} = \frac00$. Thus, we use the L'Hopital's rule again. that is,
$$\lim_{x\rightarrow 0}\frac{\cos x - 1}{\sec^2 x - 1 } = \lim_{x\rightarrow 0}\frac{-\sin x}{2 \sec^2 x \sec x \tan x}$$
since $\lim_{x\rightarrow 0} \frac{-\sin x}{2 \sec^2 x \sec x \tan x}= \frac{0}{2(1)(0)} = \frac00$. Thus, it always goes to zero by zero. But the answer for this question is $\frac{-1}2$. How is that?!! there must be something missing that either I forget or misunderstand.
|
For $x\to 0$ we have $\sin x\sim x-\frac{x^3}{3!} $ and $\tan x\sim x+\frac{x^3}{3} $
then
$$
\frac{\sin x- x}{\tan x-x}\sim \frac{-\frac{x^3}{3!}}{+\frac{x^3}{3}}=-\frac{1}{2}
$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1752715",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 7,
"answer_id": 4
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|
n tends to infinity $\displaystyle\lim_{n\to\infty}
{\left(
\frac{\sqrt {n^2+n} - 1}{n}
\right)}^{
\left(2 \sqrt{n^2+n} - 1
\right)}$
Sorry for the bad format of the question I don't know much latex
I tried rationalization of the power and the base but was not able to get the answer.
|
Let $y = \displaystyle\lim_{n\to\infty}
{\left(
\frac{\sqrt {n^2+n} - 1}{n}
\right)}^{2 \sqrt{n^2+n} - 1}$
and let $g(n) = \sqrt{n^2 + n}$
Then $g'(n) = \dfrac{2n+1}{2g(n)}$ and
\begin{align}
\ln(y)
&= (2\sqrt{n^2+n} - 1) \ln{ \left( \frac{\sqrt{n^2+n} - 1}{n} \right)} \\
&= (2g(n) - 1) \ln{ \left( \frac{g(n) - 1}{n} \right)} \\
&= \dfrac
{ln(g(n) - 1) -\ln n}
{\left( \dfrac{1}{2 g(n) + 1} \right)}
\end{align}
Using L'Hospital,
\begin{align}
\lim_{n\to\infty}\dfrac
{ln(g(n) - 1) -\ln n}
{\left( \dfrac{1}{2 g(n) + 1} \right)}
&=-\lim_{n\to\infty}\dfrac
{\left( \dfrac{g'(n)}{g(n)-1} \right) - \dfrac 1n}
{\left( \dfrac{2g'(n)}{(2g(n)+1)^2} \right)} \\
&=-\lim_{n\to\infty}\dfrac
{\left( \dfrac{1}{g(n)-1} \right) - \dfrac{2g(n)}{n(2n+1)}}
{\left( \dfrac{2}{4n^2 + 4n + 4g(n) + 1} \right)} \\
&=-\lim_{n\to\infty}\dfrac
{\left( \dfrac{2n^2 + n - 2n^2 - 2n + 2g(n)}{(2n^2 + n)(g(n)-1)} \right)}
{\left( \dfrac{2}{4n^2 + 4n + 4g(n) + 1} \right)} \\
&=-\lim_{n\to\infty}
\dfrac
{(-n + 2g(n))(4n^2 + 4n + 4g(n) + 1)}
{2(2n^2 + n)(g(n)-1)} \\
&=-\lim_{n\to\infty}
\dfrac
{2\sqrt{n^2 + n}-n}
{\sqrt{n^2 + n}-1}
\cdot \lim_{n\to\infty} \dfrac
{(4n^2 + 4n + 4g(n) + 1)}
{4n^2 + 4n} \\
&= -1
\end{align}
So $\ln y = -1$ and $y = \dfrac 1e$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1754965",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 1
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|
Ellipse and chord length There is a analytic geometry problem:
In the ellipse $\frac{x^2}{4}+y^2=1$, segment $AB$ is a chord and $AB=\sqrt{3}$, find the maximum and minimum area of $\triangle AOB$.
My progress
Assume $A(x_1,y_1),B(x_2,y_2)$, we have
$$ \left\{
\begin{array}{}
(x_1-x_2)^2+(y_1-y_2)^2=3 \\
\frac{x_1^2}{4}+y_1^2=1 \\
\frac{x_2^2}{4}+y_2^2=1
\end{array}
\right.$$
And
$$ S_{\triangle AOB}=\frac{1}{2}(x_1y_2-x_2y_1)$$
After a complicated simplify, I got
$$ \frac{3}{4}x_1^2+\frac{3}{4}x_2^2-2x_1x_2-\frac{\sqrt{16-4x_1^2-4x_2^2+x_1^2x_2^2}}{2}=1$$
and the target $\frac{1}{2}(x_1y_2-x_2y_1)$ turned to
$$ \frac{\sqrt{4x_1^2-x_1^2x_2^2}-\sqrt{4x_2^2-x_1^2x_2^2}}{2}$$
It likely is a conditional extremum problem and I don't know what to do next.
|
It likely is a conditional extremum problem and I don't know what to do next.
I don't know what to do next either.
So, let us take another approach.
If the line $AB$ is parallel to $y$-axis, we can easily see that the area of the triangle is $\sqrt 3/2$.
If the line $AB$ is not parallel to $y$-axis, we can set $AB : y=ax+b$.
Then,
$$\frac{x^2}{4}+(ax+b)^2=1\iff (4a^2+1)x^2+8abx+4b^2-4=0$$
Let $\alpha,\beta$ be the $x$ coordinates of $A,B$. Then,
$$\alpha+\beta=\frac{-8ab}{4a^2+1},\quad\alpha\beta=\frac{4b^2-4}{4a^2+1}$$
Now, we can write
$$\sqrt 3=AB=\sqrt{(\alpha-\beta)^2+((a\alpha+b)-(a\beta+b))^2}$$
Squaring the both sides gives
$$3={(a^2+1)((\alpha+\beta)^2-4\alpha\beta)}={(a^2+1)\left(\left(\frac{-8ab}{4a^2+1}\right)^2-4\frac{4b^2-4}{4a^2+1}\right)}$$
and so
$$b^2=\frac{16a^4+56a^2+13}{16(a^2+1)}$$
Let $S$ be the area of the triangle. Then,
$$S^2=\left(\frac{\sqrt 3}{2}\cdot\frac{|a\cdot 0-0+b|}{\sqrt{a^2+1}}\right)^2=\frac{3b^2}{4(a^2+1)}=\frac{3(16a^4+56a^2+13)}{64(a^2+1)^2}:=f(a)$$
$$f'(a)=\frac{9a(5-4a^2)}{16(a^2+1)^3}=0\iff a=0,\pm\frac{\sqrt 5}{2}$$
Hence, the maximum of $f(a)$ is $f(\pm\sqrt 5/2)=1$, and the minimum of $f(a)$ is $f(0)=39/64$.
Therefore, the maximum area of the triangle is $\color{red}{1}$, and the minimum area is $\color{red}{\sqrt{39}/8}$.
|
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"url": "https://math.stackexchange.com/questions/1756245",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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|
Test for the convergence of the sequence $S_n =\frac1n \left(1 + \frac{1}{2} + \frac{1}{3} + \cdots+ \frac{1}{n}\right)$ $$S_n =\frac1n \left(1 + \frac{1}{2} + \frac{1}{3} + \cdots+ \frac{1}{n}\right)$$
Show the convergence of $S_n$ (the method of difference more preferably)
I just began treating sequences in school, and our teacher taught that monotone increasing sequence, bounded above and monotone decreasing sequences, bounded below converge.
and so using that theorem here..
I found the $$(n+1)_{th} term$$,
$$S_{n+1} = \frac1{n+1} \left(1 + \frac{1}{2} + \frac{1}{3} + \cdots+ \frac{1}{n+1}\right)$$
and then subtracted the (n)th term from it
What I was able to get was...
$$S_{n+1}-S_n = \frac1{(n+1)^2} - \frac{\left(1+\frac12+\dots+\frac1n\right)}{n(n+1)}.$$
...but then this is where I get stucked, but i'm trying to prove that the sequence > 0(i.e Converges) or < 0 (i.e diverges).
|
Avoiding appeals to logarithms:
Writing $H_n = 1 +\frac{1}{2}+\cdots + \frac{1}{n}$, note that for $n>M$:
$$H_n=1 +\frac{1}{2}+\cdots + \frac{1}{n}=1 +\frac{1}{2}+\cdots + \frac{1}{m}+\cdots+\frac{1}{n}=H_M+\frac{1}{M+1}+\cdots+\frac{1}{n}$$
Now, each of the terms after $H_M$ is $<\frac{1}{M}$, whence $H_n<H_M+\frac{n-M}{M}$.
Thus, $\frac{H_n}{n}<\frac{H_M}{n}+\frac{n-M}{nM}=\frac{H_m-1}{n}+\frac{1}{M}$ for any positive integer $M$, as long as $n$ is larger than $M$.
From this, we deduce that $\limsup\frac{H_n}{n}\le\frac{1}{M}$ for all positive integers $M$, and thus that $\limsup\frac{H_n}{n}=0$.
As $\frac{H_n}{n}>0$, it is thus trivial to deduce that the sequence converges to $0$.
|
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"language": "en",
"url": "https://math.stackexchange.com/questions/1757859",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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|
Can this puzzle be solved without brute force?
Consider positive integers $a$ and $b$, where $a \ge b$ and the sum $\frac{a+1}{b}+\frac{b+1}{a}$ is also an integer. Find the sum of all $a$ values less than $1000$ that meet this criteria.
For example, if $a=3$ and $b=2$, then
$$\frac{a+1}{b}+\frac{b+1}{a} = \frac{3+1}{2}+\frac{2+1}{3} = 3$$
which is an integer and provides one legal value of $a$. The problem asks for all such values of $a \lt 1000$. I’m assuming that a particular $a$ suffices if there exists even one $b \le a$ that satisfies the condition.
This problem can be solved by simply writing a computer program that double-loops through the positive integers less than 1000, testing each case, however I’m wondering if there is a more general/abstract way to solve it.
Here are some thoughts … I’m not sure how helpful they are:
*
*$\frac{a+1}{b}+\frac{b+1}{a}$, which is an integer, can be rewritten as $\frac{a}{b}+\frac{b}{a}+\frac{1}{b}+\frac{1}{a}$.
*either the quotients $(a+1) \div b$ and $(b+1) \div a$ are each also integers, or their fractional parts add up to $1$.
|
We use a now-standard technique called Vieta jumping, which has a fascinating and surprisingly recent history.
Suppose $a^2 + a + b^2 + b = kab$ for some integer $k$. Then, writing this as a polynomial in $a$ with integer coefficients:
$$a^2 - (kb-1) a + (b^2+b) = 0,$$
we see that if $(a,b)$ is one of the roots of the above quadratic, then the other one is $c = (b^2+b)/a$, which is clearly positive. Furthermore, since the two roots sum to an integer, namely $kb-1$, then $c$ is necessarily a positive integer.
Now let's consider sizes. If $a>b$, then $a \ge b+1$ so $c = b(b+1)/a \le b$. So if $(a,b)$ is one solution to the problem, then either $a=b$ or $(b,c)$ is a smaller solution, with the same value of $k$. Since all values remain positive, they can only get smaller finitely many times, so eventually we must reach a solution where $a=b$.
The case where $a=b$ is very easy: we must have $k = 2 + 2/a$, so the only solutions of this type are $(1,1)$ with $k=4$ and $(2,2)$ with $k=3$.
Finally, since every solution eventually reduces to either $(1,1)$ or $(2,2)$ by the reduction process, we can recover all solutions by running the process in reverse, starting from $(1,1)$ and $(2,2)$:
$$(1,1) \leftarrow (2,1) \leftarrow (6,2) \leftarrow (21,6) \leftarrow \cdots $$
$$(2,2) \leftarrow (3,2) \leftarrow (6,3) \leftarrow (14,6) \leftarrow \cdots $$
The values grow exponentially, so it doesn't take long for either sequence to exceed $1000$.
|
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"language": "en",
"url": "https://math.stackexchange.com/questions/1758896",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
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|
is it possible to choose points on the graph of $y = x^2$ to form vertices of an equilateral triangle? is it possible to choose points on the graph of $y = x^2$ to form vertices of an equilateral triangle $\Delta ABC$?where three ponit not $(0,0)$ .and find $(S_{\Delta ABC})_{min}$
|
Summary
*
*Are there other points which can form equilateral triangle?
The answer is yes. In fact, there are infinitely many of them.
The centroids of the triangles lie on another parabola $y = 9x^2 + 2$.
*The minimal area is $3\sqrt{3}$. achieved by the equilateral triangle with vertices $(0,0)$, $(\pm\sqrt{3},3)$.
Identify the euclidean plane $\mathbb{R}^2$ with complex plane $\mathbb{C}$.
Let $z = x + iy$ and $\bar{z} = x - iy$ be its complex conjugate, the equation of the parabola becomes
$$y = x^2 \iff \frac{z - \bar{z}}{2i} = \left(\frac{z+\bar{z}}{2}\right)^2 \iff
(z + \bar{z})^2 + 2i(z-\bar{z}) = 0$$
Let $\omega = e^{2\pi i/3}$. Given any equilateral triangle $T$, we can always find two complex numbers $\rho = u+iv$ and $a$ such that the vertices of $T$ are $\rho + a\omega^k$ for $k = 0, \pm 1$.
If they all lie on the parabola above, then for $k = 0, \pm 1$, we have
$$
\begin{align}
&\; (\rho + \bar{\rho} + a\omega^k + \bar{a}\omega^{-k})^2
+ 2i(\rho - \bar{\rho} + a\omega^k - \bar{a}\omega^{-k})\\
= &\; (\rho+\bar{\rho})^2
+ 2(\rho+\bar{\rho})(a\omega^k + \bar{a}\omega^{-k})
+ (a^2\omega^{-k} + 2|a|^2 + \bar{a}^2\omega^k)
+ 2i(\rho - \bar{\rho} + a\omega^k - \bar{a}\omega^{-k})\\
= &\; 0
\end{align}
$$
Comparing coefficients of different powers of $\omega$, we get
$$
\begin{cases}
(\rho + \bar{\rho})^2 + 2i(\rho - \bar{\rho}) + 2|a|^2 &= 0\\
2(\rho + \bar{\rho} + i)a + \bar{a}^2 &= 0\\
2(\rho + \bar{\rho} - i)\bar{a} + a^2 &= 0
\end{cases}
$$
Multiply the $2^{nd}$ equation by $4a^2$ and simplify it using $1^{st}$ equation,
we get
$$8(\rho + \bar{\rho} +i )a^3 + \left[(\rho + \bar{\rho})^2 + 2i(\rho - \bar{\rho})\right]^2 = 0\tag{*1}$$
Since $\rho + \bar{\rho} + i = 2u + i\ne 0$, we can use this to determine $a$ up to a factor $\omega^k$.
For this $a$ to be compatible with the $1^{st}$ equation, the condition is
$$\begin{align}
& (\rho + \bar{\rho})^2 + 2i(\rho - \bar{\rho}) = -2|a|^2 = -8|\rho + \bar{\rho} + i|^2
= -8 ((\rho+\bar{\rho})^2 + 1)\\
\iff & 9(\rho + \bar{\rho})^2 + 2i(\rho - \bar{\rho}) + 8 = 36u^2 - 4v + 8 = 0\\
\iff & v = 9u^2 + 2
\end{align}
$$
Base on this, we see start from any point $(u,v)$ from the parabola $y = 9x^2 + 2$, if one define $\rho = u + iv$ and use $(*1)$ to compute $a$, the 3 points
$\rho + a\omega^k$ will lie on the original parabola
$y = x^2$ and form an equilateral triangle.
Back to question of minimization of area. It is clear it is equivalent to
minimization of $|a|$. Since
$$|a|^2 = -\frac12 \left[ (\rho + \bar{\rho})^2 + 2i(\rho - \bar{\rho}) \right]
= 2(v-u^2) = 2(2+8u^2)$$
the minimal value $|a|$ is achieved at $(u,v) = (0,2)$ with value $2$.
The corresponding triangle has vertices at $(0,0)$, $(\pm\sqrt{3},3)$
with side length $2\sqrt{3}$ and area $3\sqrt{3}$.
|
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "3",
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|
Evaluate the $\lim_{x \to \ -\infty} (x + \sqrt{x^2 + 2x})$ Evaluate : $$\lim_{x \to \ -\infty} (x + \sqrt{x^2 + 2x})$$
I've tried some basic algebraic manipulation to get it into a form where I can apply L'Hopital's Rule, but it's still going to be indeterminate form.
This is what I've done so far
\begin{align}
\lim_{x \to \ -\infty} (x + \sqrt{x^2 + 2x}) &= \lim_{x \to \ -\infty} (x + \sqrt{x^2 + 2x})\left(\frac{x-\sqrt{x^2 + 2x}}{x-\sqrt{x^2 + 2x}}\right)\\ \\
&= \lim_{x \to \ -\infty} \left(\frac{x^2 - (x^2 + 2x)}{x-\sqrt{x^2 + 2x}}\right)\\ \\
&= \lim_{x \to \ -\infty} \left(\frac{-2x}{x-\sqrt{x^2 + 2x}}\right)\\
\\
\end{align}
And that's as far as I've gotten. I've tried applying L'Hopitals Rule, but it still results in an indeterminate form.
Plugging it into WolframAlpha shows that the correct answer is $-1$
Any suggestions on what to do next?
|
This is based on @Battani's answer but with a more in-depth explanation
\begin{align}
\lim _{ x\to -\infty } \left( \frac { -2x }{ x-\sqrt { x^{ 2 }+2x } } \right) &= \lim _{ x\rightarrow -\infty }\left( \frac { -2x }{ x-\sqrt { { x }^{ 2 }\left( 1+\frac { 2 }{ x } \right) } } \right) \\ \\
&\text{Now because $\sqrt{x^2}$ = $|x|$} \\ \\
&= \lim _{ x\rightarrow -\infty } \frac { -2x }{ x-\left| x \right| \sqrt { 1+\frac { 2 }{ x } } } \\ \\
\text{Recall that } & \ |x| = \begin{cases}
x &\text{ if } \ \ x \geq 0\\
- x &\text{ if } \ \ x < 0\\
\end{cases}
\\ \\
&\text{$x$ is approaching $-\infty$} \\ \\
&\therefore \ \ \ \ |x| = -x
\\ \\
&= \lim _{ x\rightarrow -\infty } \frac { -2x }{ x- (-x)\sqrt { 1+\frac { 2 }{ x } } } \\ \\
&= \lim _{ x\rightarrow -\infty } \frac { -2x }{ x + x\sqrt { 1+\frac { 2 }{ x } } } \\ \\
&= \lim _{ x\rightarrow -\infty } \frac { -2x }{ x\left( 1+\sqrt { 1+\frac { 2 }{ x } } \right) } \\ \\
&= \lim _{ x\rightarrow -\infty } \frac { -2 }{ 1+\sqrt { 1+\frac { 2 }{ x } } } \\\\
&= -1
\end{align}
|
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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|
How to evaluate $\lim_{c \rightarrow \infty} \int_{-c}^c f(x)dx$ I'm trying to evaluate:
$$\lim_{c \rightarrow \infty} \int_{-c}^c \frac{1+x}{1+x^2}dx$$
but I don't understand how to evaluate
$$\lim_{c \rightarrow \infty} \int_{-c}^c f(x)dx$$
How?
|
Hint. Observe that
$$
\int_{-c}^c \frac{1+x}{1+x^2}dx=\int_{-c}^c \frac{1}{1+x^2}dx=2\int_0^c \frac{1}{1+x^2}dx=2\arctan(c).
$$ Can you take it from here?
Some details. We have
$$
\begin{align}
\color{blue}{\int_{-c}^c \frac{x}{1+x^2}dx}&=\int_{-c}^0 \frac{x}{1+x^2}dx+\int_0^c \frac{x}{1+x^2}dx
\\\\&=\int_{c}^0 \frac{-x}{1+(-x)^2}(-dx)+\int_0^c \frac{x}{1+x^2}dx
\\\\&=-\int_0^c \frac{x}{1+x^2}dx+\int_0^c \frac{x}{1+x^2}dx
\\\\&=\color{blue}{0}
\end{align}
$$ giving
$$
\int_{-c}^c \frac{1+x}{1+x^2}dx=\int_{-c}^c \frac{1}{1+x^2}dx+\color{blue}{\int_{-c}^c \frac{x}{1+x^2}dx}=\int_{-c}^c \frac{1}{1+x^2}dx+\color{blue}{0}.
$$
|
{
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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|
How do I solve the following differential equation (It's not seperable)? I'm trying to Solve the following equation: find the solution $y:(-1,1) \rightarrow \mathbb{R}$ of $y'=\dfrac{y}{1-x^2}+x$?
It is not separable and I have no other Tools to solve it.
|
Subtract $ \frac {y(x)}{1-x^2} $ from both sides :
$ \frac {dy(x)}{dx} + \frac {y(x)}{x^2-1} = - \frac {-x^3 + x}{x^2-1}$
Let $ m(x) = e^{\int \frac {1}{x^2-1} dx } = \frac {\sqrt{1-x}}{\sqrt{x+1}} $
Then multiplay both sides by m(x) and substitute : $ \frac {\sqrt{1-x}}{\sqrt{x+1}(x^2-1)} = \frac {d}{dx} \frac {\sqrt{-x+1}}{\sqrt{x+1}} $. Apply the reverse product rule : $ g \frac{df}{dx} + f \frac{dg}{dx} = \frac{d}{dx}(fg) $ to the left side of the equation above and integrate both sides :
$\int \frac {\sqrt{-x+1}y(x)}{\sqrt{x+1}} dx= \int - \frac {\sqrt{-x+1}(-x^3 + x)}{\sqrt{x+1}(x^2-1)}dx $
You will then get :
$ \frac {\sqrt{-x+1}y(x)}{\sqrt{x+1}} = - sin^1(\frac{\sqrt{x+1}}{\sqrt{2}}) + \frac{1}{2}(x-2)\sqrt{-x^2 +1} + c $
Divide both sides by m(x) that we defined above and you get :
$ y(x) = \frac{\sqrt{x+1}(- sin^1(\frac{\sqrt{x+1}}{\sqrt{2}}) + \frac{1}{2}(x-2)\sqrt{-x^2 +1} + c)}{\sqrt{-x+1}} $
This is the solution to your differential equation, I hope I didn't do any typing mistake while writing it down !
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"timestamp": "2023-03-29T00:00:00",
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|
Solve $ 1 + \frac{\sqrt{x+3}}{1+\sqrt{1-x}} = x + \frac{\sqrt{2x+2}}{1+\sqrt{2-2x}} $
Solve for $x \in \mathbb{R}$
$$ 1 + \dfrac{\sqrt{x+3}}{1+\sqrt{1-x}} = x + \dfrac{\sqrt{2x+2}}{1+\sqrt{2-2x}} $$
I tried some substitutions and squaring but that didn't help. I also tried to use inequalities as done in my previous problem, but that too didn't help.
|
Given : $ 1 + \dfrac{\sqrt{x+3}}{1+\sqrt{1-x}} = x + \dfrac{\sqrt{2x+2}} {1+\sqrt{2-2x}} $
Let $\alpha = x+3 $ ; $\beta = 2x+2$ ; $\left(\beta - \alpha\right) = x-1$
$\implies 1+ \dfrac{\sqrt{\alpha}}{1+\sqrt{4-\alpha}} = x + \dfrac{\sqrt{\beta}} {1+\sqrt{4-\beta}} $
$\implies \alpha + \dfrac{\sqrt{\alpha}}{1+\sqrt{4-\alpha}} = \beta + \dfrac{\sqrt{\beta}} {1+\sqrt{4-\beta}} $
Let $f(x) = x + \dfrac{\sqrt{x}}{1+\sqrt{4-x}}$, the given equation becomes,
$f(\alpha) = f(\beta)$
Note that $f(x)$ is monotonic increasing in its domain,
$\therefore \alpha = \beta$
$\implies x+3 = 2x +2$
$\implies x=\boxed{1}$
|
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"timestamp": "2023-03-29T00:00:00",
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|
Solving a mixed radical and quadratic equation
Solve for $x \in \mathbb{R}$
$$4x^2(x+2) +3(2x^2-4x-3)\sqrt{4x+3} +6x = 0$$
I tried taking square by isolating the radical, but the resultant equation couldn't be solved.
Any help will be appreciated.
Thanks.
|
Let $X=4x+3$. Then,
$$3(2x^2-X)\sqrt{X}=-2x(2x^2+X)$$
Squaring the both sides gives
$$9(2x^2-X)^2X=4x^2(2x^2+X)^2,$$
i.e.
$$4x^4X-4x^2X^2+9X^3-16x^6+16x^4X-36x^2X^2=0$$
$$X(4x^4-4x^2X+9X^2)-4x^2(4x^4-4x^2X+9X^2)=0$$
$$(X-4x^2)(4x^4-4x^2X+9X^2)=0$$
since $9X^2=X^2+8X^2$
$$(X-4x^2)(4x^4-4x^2X+X^2+8X^2)=0$$
to have
$$(X-4x^2)((2x^2-X)^2+8X^2)=0$$
finally,
$$(-4x^2+4x+3)((2 x^2-4 x-3)^2+8 (4 x+3)^2)=0$$
and so
$$x=\frac 32,-\frac 12.$$
The former is sufficient while the latter isn't. So, $\color{red}{x=\frac 32}$ is the only solution.
|
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"timestamp": "2023-03-29T00:00:00",
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|
if $x^y=y^x$ show that $x+y>2e$
Let $0<x<y$, such that
$$x^y=y^x$$
show that
$$x+y>2e$$
Since $$y\ln{x}=x\ln{y}\Longrightarrow \dfrac{\ln{y}}{y}=\dfrac{\ln{x}}{x}$$
Let
$$f(x)=\dfrac{\ln{x}}{x}\Longrightarrow f'(x)=\dfrac{1-\ln{x}}{x^2}$$
If $0<x<e$ then $f'(x)>0$, if $x>e$,then $f'(x)<0$
$\Longleftrightarrow y-e+x-e>0$,maybe consider
$f(x-e)$ and $f(y-e)$,wihch bigger?
I just do it now
Thanks in advance!
|
As MathematicianByMistake answer,we want prove
$$g(r)=r^{\frac{1}{r-1}}+r^{\frac{r}{r-1}}=(x+1)^{\frac{1}{x}}+(x+1)^{\frac{x+1}{x}}=(x+1)^{\frac{1}{x}}(x+2),x>0$$
since
$$g'(x)=(x+1)^{\frac{1}{x}-1}\cdot\dfrac{x(x^2+2x+2)-(x^2+3x+2)\ln{(x+1)}}{x^2}$$
we only prove
$$h(x)=x(x^2+2x+2)-(x^2+3x+2)\ln{(x+1)}>0,x>0$$
since
$$h'(x)=3x(x+1)-(3x+2)\ln{(x+1)}>3x(x+1)-x(3x+2)=x>0$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1771348",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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|
Use induction to prove that $2^n \gt n^3$ for every integer $n \ge 10$. Use induction to prove that $2^n \gt n^3$ for every integer $n \ge 10$.
My method:
If $n = 10$, $2^n \gt n^3$ where $2^{10} \gt 10^3$ which is equivalent to $1024 \gt 1000$, which holds for $n = 10$. $2^k \gt k^3$.
$2^{k + 1} \gt (k + 1)^3$
$2^{k + 1} \gt (k + 1)(k + 1)(k + 1)$
$2^{k + 1} \gt (k^2 + 2k + 1)(k + 1)$
$2^{k + 1} \gt (k^3 + 3k^2 + 3k + 1)$
$(k + 1) \cdot 2^k \gt k^3 + 3k^2 + 3k + 1$
Since $2^{10} \gt 10^3$, the inequality holds for $n = 10$. Assume $2^k \gt k^3$ for $k \ge 10$. We show that $2^{k + 1} \gt (k + 1)^3$. Hence, $2^{k + 1} = 2 \cdot 2^k \gt 2k^3$
$= k^3 + k^3 \ge k^3 + 10k^2$
$= k^3 + 10k^2 = k^3 + 4k^2 + 6k^2 \ge k^3 + 4k^2 + 6 \cdot 10 = k^3 + 4k^2 +60$
$k^3 + 4k^2 + 60 \gt k^3 + 3k^2 + 3k + 1$
Therefore, $2^{k + 1} \gt (k + 1)^3$. Hence, $2^n \gt n^3$ for every integer $n \ge 10$.
I was trying to fix this but I am not sure how to go about doing so.
|
We do the induction step, perhaps somewhat along the lines that you had considered.
Let $k\ge 10$, and suppose that $2^k\gt k^3$. We want to show that $2^{k+1}\gt (k+1)^3$.
We have
$$2^{k+1}=2\cdot 2^k\gt k^3+k^3.$$
It will now be enough to show that $k^3+k^3\gt (k+1)^3$, or equivalently that $(k+1)^3\lt k^3+k^3$.
We have
$$(k+1)^3=k^3+3k^2+3k+1\le k^3+3k^2+3k^2+k^2=k^3+7k^2.$$
But $7k^2\lt k^3$, and therefore $(k+1)^3\lt k^3+k^3$.
|
{
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"timestamp": "2023-03-29T00:00:00",
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|
How to find the orthogonal projection of the given vector on the given subspace $W$ of the inner product space $V$. $V=P(R)$ with the inner product $\langle f(x),g(x) \rangle$=$\int_0^1 f(t)g(t )dt$, $h(x)=4+3x-2x^2$ and $W=P_1(R)={\{1,x}\}$.
I don't know how to do this the question. All I know is that it has something do with the Gram-Schmidt process.
Please provide full details ,if possible as I don't know much about about orthogonalization projection.
|
The inner product structure of your vector space $V$ is $$\langle f | g
\rangle = \int_0^1 f(x)g(x) dx$$
To project a vector
$$h(x) = 4 + 3x - 2x^2$$
on the subspace $W$ of the vector space $V$, you just add the projections of $h$ on each of the basis vectors of the subspace.
In this case, since
$$W = P_1 = \left\{ 1, x \right\}$$
and the vector we wish to project is $h$, we need to find
$$
w = 1 \times\langle h |1\rangle + x \times\langle h|x\rangle
$$
Where $w$ is the projection of $h$ in $W$
Let's now compute $w$
$$
w = 1 \times\langle h |1\rangle + x \times\langle h|x\rangle \\
= 1 \times \int_0^1h \cdot 1dx + x \times \int_0^1 h \cdot x dx \\
= \int_0^1 (4 + 3x - 2x^2)dx + x\int_0^1 (4 + 3x - 2x^2)x dx \\
= \int_0^1 (4 + 3x - 2x^2)dx + x\int_0^1 (4x + 3x^2 - 2x^3) dx \\
= 4x + \frac{3x^2}{2} - \frac{2x^3}{3} \bigg \vert_0^1 +
x \left( \frac{4x^2}{2} + \frac{3x^3}{3} - \frac{2x^4}{4} \bigg \vert_0^1 \right) \\
= \left(4 + \frac{3}{2} - \frac{2}{3} \right) + x \left(\frac{4}{2} - \frac{3}{3} - \frac{2}{4}\right) \\
= \frac{12 + 9 - 4}{6} + x \left(2 - 1 - \frac{1}{2} \right )\\
= \frac{17}{6} + \frac{x}{2}
$$
Hence, the projection of $h$ on $W$, or
$$
w = \langle h | W \rangle = \frac{17}{6} + \frac{x}{2}
$$
|
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"timestamp": "2023-03-29T00:00:00",
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|
Prove that $a^2b+b^2c+c^2a \leqslant 3$ for $a,b,c >0$ with $a^ab^bc^c=1$
Let $a,b,c >0$ and $a^ab^bc^c=1$. Prove that
$$a^2b+b^2c+c^2a \leqslant 3.$$
I don't even know what to do with the condition $a^ab^bc^c=1$. At first I think $x^x>1$, but I was wrong. This inequality is true, following by the verification from Mathematica
|
Remarks: @Wiley gave a very nice proof. Here we give an alternative proof.
Using the well-known inequality $a^2b + b^2c + c^2a \le \frac{4}{27}(a + b + c)^3 - abc$, it suffices to prove that
$$\frac{4}{27}(a + b + c)^3 - abc \le 3. \tag{1}$$
The condition $a^a b^b c^c = 1$ is written as $$a\ln a + b\ln b + c\ln c = 0. \tag{2}$$
Note that $x\mapsto x\ln x$ is convex on $x > 0$. By Jensen's inequality and (2), we have
$$0 = a\ln a + b\ln b + c\ln c
\ge 3 \cdot \frac{a + b + c}{3}\ln \frac{a + b + c}{3}$$
which results in
$$a + b + c \le 3. \tag{3}$$
Fact 1: It holds that
$x\ln x \ge x - 1 + \frac14(x - 1)^2$ for all $x \in (0, 3]$.
(The proof is given at the end.)
Using Fact 1 and (2) and (3), we have
$$
0 = a\ln a + b\ln b + c\ln c \ge a + b + c - 3 + \frac14(a-1)^2 + \frac14(b-1)^2 + \frac14(c-1)^2
$$
or
$$a^2 + b^2 + c^2 + 2a + 2b + 2c - 9 \le 0. \tag{4}$$
We use the pqr method.
Let $p = a + b + c, q = ab + bc + ca, r = abc$.
We have
\begin{align*}
&\frac{4}{27}(a+b+c)^3 - abc - 3\\
=\,& \frac{4}{27}p^3 - r - 3\\
\le\,& \frac{4}{27}p^3 - \frac{4pq - p^3}{9} - 3 \tag{5}\\
=\,& \frac{7}{27}p^3 - \frac49 pq - 3\\
\le\,& \frac{7}{27}p^3 - \frac49 p \cdot \frac{p^2 + 2p - 9}{2} - 3 \tag{6}\\
=\,& \frac{1}{27}(p - 3)(p^2 - 9p + 27)\\
\le\,& 0. \tag{7}
\end{align*}
Explanations:
(5): Degree three Schur inequality yields $r \ge \frac{4pq - p^3}{9}$.
(6): From (4), we have
$p^2 - 2q + 2p - 9\le 0$
which results in
$q \ge \frac{p^2 + 2p - 9}{2}$.
(7): From (3), we have $p \le 3$.
We are done.
Proof of Fact 1:
It suffices to prove that
$$\ln x \ge \frac{1}{x}\left(x - 1 + \frac14(x - 1)^2\right).$$
Let $F(x) := \mathrm{LHS} - \mathrm{RHS}$. We have
$$F'(x) = \frac{(x-1)(3-x)}{4x^2}.$$
We have $F'(x) < 0$ on $(0, 1)$,
and $F'(x) > 0$ on $(1, 3)$,
and $F'(1) = 0$, and $F'(3) = 0$.
Thus, $F(x) \ge F(1) = 0$ for all $x\in (0, 3]$.
We are done.
|
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|
A rational function integration Evaluate $$\int \frac{3x^2+1}{(x^2-1)^3}dx$$ I tried breaking the numerator in terms of the denominator but it didn't help much. I also tried a few substitutions buy most of them were useless. Please give some hints. Thanks.
|
Let $$I = \int\frac{3x^2+1}{(x^2-1)^3}dx = \int\frac{3x^2+1}{x^{\frac{3}{2}}\left(x^{\frac{3}{2}}-x^{-\frac{1}{2}}\right)^3}dx$$
So $$I = \int\frac{3x^{\frac{1}{2}}+x^{-\frac{3}{2}}}{\left(x^{\frac{3}{2}}-x^{-\frac{1}{2}}\right)^3}dx\;,$$ Now Put $\left(x^{\frac{3}{2}}-x^{-\frac{1}{2}}\right)=t\;,$ Then $\left(3x^{\frac{1}{2}}+x^{-\frac{3}{2}}\right)dx = 2dt$
So we get $$I = 2\int t^{-3}dt = -\frac{1}{t^2}+\mathcal{C} = -\frac{x}{(x^2-1)^2}+\mathcal{C}$$
|
{
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"timestamp": "2023-03-29T00:00:00",
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|
How to differentiate this fraction $\frac{2}{x^2+3^3}$? $\frac{2}{(x^2+3)^3}$.
I have ${dy}/{dx}$ x 2 x ${x^2+3^3}$ - 2 x ${dy}/{dx}$ x ${x^2+3^3}$ over $({x^2+3)^6}$
And then simplifying to $-12x^5 + 36x^2$ over $({x^2+3)^6}$
I'm not sure if this is right.
|
$$\frac {d}{dx} \frac 2{(x^2+3)^3}=\frac {d}{dx} (2(x^2+3)^{-3})=2\frac {d}{dx} (x^2+3)^{-3}=2\times(-3)(x^2+3)^{-4}2x=-\frac {12x}{(x^2+3)^{4}}$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1776669",
"timestamp": "2023-03-29T00:00:00",
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|
If $x \geq C$, where $C > 0$ is a constant, then what is the least upper bound for $\dfrac{2x}{x + 1}$? The title says it all.
Since
$$f(x) = \dfrac{2x}{x + 1} = 2\left(1 - \dfrac{1}{x + 1}\right),$$
then because $x \geq C$ where $C > 0$, an upper bound is given by
$$\dfrac{2x}{x + 1} < 2.$$
Note that the lower bound is equivalent to
$$x \geq C \iff x + 1 \geq C + 1 \iff \dfrac{1}{x + 1} \leq \dfrac{1}{C + 1}$$ $$\iff 2\left(1 - \dfrac{1}{x + 1}\right) \geq 2\left(1 - \dfrac{1}{C + 1}\right),$$
so that if $2 - \varepsilon$ is an upper bound for $f(x)$ ($\varepsilon > 0$), then $\varepsilon$ satisfies the inequality
$$2 - \varepsilon \geq f(x) \geq 2\left(1 - \dfrac{1}{C + 1}\right) \implies 0 < \varepsilon < \dfrac{2}{C + 1}.$$
My question is essentially whether we can do better than this.
|
\begin{align*}
x&\ge C \\
f(x) = \dfrac{2x}{x + 1} &= 2\left(1 - \dfrac{1}{x + 1}\right)\\
x&\in [C,\infty)\\
x+1&\in[C+1, \infty)\\
\frac{1}{x+1}&\in\Bigg(0,\frac{1}{C+1}\Bigg]\\
1-\frac{1}{x+1}&\in \Bigg[1-\frac{1}{C+1}, 1\Bigg)\\
\dfrac{2x}{x + 1} = 2\left(1 - \dfrac{1}{x + 1}\right)&\in\Bigg[\frac{2C}{C+1}, 2\Bigg)\\
\end{align*}
So you have $$\frac{2C}{C+1}\le f(x) < 2$$
Alternative:
Since you have already (and correctly) found the lower bound we'll focus on the upper bound. We know that the function is monotonic, since
$$f'(x) = \frac{2}{(x+1)^2} > 0$$
Hence, we know that the function attains it's maximum (if any) as $x\to \infty$.
Hence we find $$\lim_{x\to\infty} f(x) = \lim_{x\to\infty} \frac{2x}{x+1} = \lim_{x\to\infty} \frac{2}{1+\frac{1}{x}} = \fbox{2}$$
Hence, $f(x)<2$.
|
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|
How to prove that $(x^2+y)(y^2+z)(z^2+x)+2xyz \leqslant 10$, where $x,y,z \in \mathbb{R}^+$ and $x^2+y^2+z^2+xyz=4$ $x,y,z \in \mathbb{R}^+$ and $x^2+y^2+z^2+xyz=4$, prove
$$(x^2+y)(y^2+z)(z^2+x)+2xyz \leqslant 10$$
I try several trig substitutions but feel hopeless with the cyclic term here. The condition $x^2+y^2+z^2+xyz=4$ made it too difficult to homogenize the inequality. I don't even know how to do brutal force either.
|
$$ x^2+y^2+z^2+xyz=4;\space\space x,y,z>0\qquad (1)\\ (x^2+y)(y^2+z)(z^2+x)+2xyz \le 10\qquad (2)$$
$(1)\Rightarrow 0<x,y,z\le 2 \text{ and}\space x=2\iff y=z=0$
If $x=2$ then $(2)$ is trivially verified.
Put $x=2-\epsilon$ where $0\le \epsilon\le 2$ so $(1)$ becomes
$$y^2+z^2+(2-\epsilon)yz=\epsilon(4-\epsilon)\qquad (1')$$
$(1’)$ is the equation of an ellipse whose axes are contained in the diagonals. In fact, changing coordinates by a rotation of $45^{\circ}$, straightforward calculation in $(1’)$ can transforms $(y,z)$ in $(y_1,z_1)$ giving $$\left(\frac{y_1}{\sqrt{2\epsilon}}\right)^2+\left(\frac{z_1}{2(4-\epsilon)}\right)^2=1$$
The concerned points $(y,z)$ are those of the red arc in the figure below corresponding to the value $\epsilon = 0.8$.
From $(2)$ we get
$$F(\epsilon,y,z)=((2-\epsilon)^2+y)(y^2+z)(z^2+2-\epsilon)+(4-2\epsilon)yz\qquad (2’)$$ where $$\begin{cases}0\le \epsilon\le 2\\ 0<y,z\le \sqrt{\epsilon(4-\epsilon)}\end{cases}$$
For $\epsilon$ fixed, $(2’)$ is maximum when $y=z$ (@HN_NH exercise for) and this occurs when $y^2=\epsilon$ (easily get from $(1’)$ or from the drawn ellipse). It follows
$$F_1(\epsilon)=(2-\epsilon)^3(\epsilon+\sqrt{\epsilon})+(2-\epsilon)^2(\epsilon^2+\epsilon\sqrt{\epsilon})+(2-\epsilon)(3\epsilon+\epsilon\sqrt{\epsilon})+\epsilon^2\sqrt{\epsilon}+\epsilon^2$$
$$F_1(\epsilon)=(2\epsilon^2-6\epsilon+8)\sqrt{\epsilon}+2\epsilon^3-10\epsilon^2+14\epsilon$$
Now, $F_1(\epsilon)$ has in its domain a maximum at $\epsilon=1$ in whose case $F_1(1)=10$. This corresponds to $(x,y,z)=(1,1,1)$ in $(2)$; for the other allowed values the proposed inequality becomes $$(x^2+y)(y^2+z)(z^2+x)+2xyz \lt 10$$
|
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|
When is $\frac{a}{n} = n^a$ true? Title says it all really, I am trying to figure out if theres a situation where
$\frac{a}{n} = n^a$ is true or if this is impossible.
This is not realy from somewhere, just for the sake of curiosity.
|
$\frac an = n^a \implies$
$a = n^{a+1} \implies $
$1 = \log_a a = \log_a n^{a+1} = (a +1 ) \log_a n \implies$
$\frac 1{a+1} = \log_a n \implies$
$n = a^{\frac 1{a+1}}$
So for all positive $a$, $n$ is solvable. As $0/0$ (as well as $0^0$) are undefined $n = a = 0$ is not a solution (despite $0 = 0^1$).
For $a < 0$ it gets a little more complicated.
If $n > 0$ then $a/n < 0$ while $n^a > 0$ so $n < 0$.
But $n <0$ implies $a/n > 0$ so $(-|n|)^a > 0$ so $a = -p/q$ must be that $p$ is even. (Remember: negative numbers to irrational powers are not defined.) And as $\gcd (p,q) =1$, $q$ must be odd.
Then $n = a^{1/(a+ 1)} = a^{q/(q-p)}$.
Example: $a = -2/3; n = (-2/3)^{3}$. And $a/n = \frac 1{(2/3)^2} = (-3/2)^2 = ((-3/2)^3)^{2/3} = ((-2/3)^3)^{-2/3} = n^a$
Or $a = -2; n=-1/2$ and $a/n = 4 = (-2)^2 = (-1/2)^{-2}$.
Or $a = -4/3; n = (-4/3)^{-1/3} = -\sqrt[3]{3/4}$.
etc.
|
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For real numbers $a,b,c$ calculate the value of: $\frac{c}{a+b}+\frac{a}{b+c}+\frac{b}{c+a}$ if we have... For real numbers $a,b,c$ we have: $a+b+c=11$ and $\frac1{a+b}+\frac1{b+c}+\frac1{c+a}=\frac{13}{17}$, calculate the value of: $\frac{c}{a+b}+\frac{a}{b+c}+\frac{b}{c+a}$
I think we should use a trick to solve this,because doing algebraic operations on this problem are too tedius!
|
Hint:
As $(a+b+c) ( \frac1{a+b}+\frac1{b+c}+\frac1{c+a})= 3+\frac{a}{b+c} +\frac{b}{c+a} + \frac{c}{a+b}$
Now applying the values given in question we get $\frac{a}{b+c} +\frac{b}{c+a} + \frac{c}{a+b}=(11 \cdot \frac{13}{17})- 3=\frac{92}{17}$
|
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|
Prove that $4n+2=x^2+y^2+z^2$ for some odd $x,y$ and even $z$ Show that for all $n\in \mathbb{N}$, exists $x,y,z \in \mathbb{N}$, such that $x,y$ are odd and $z$ is even, such that $4n+2=x^2+y^2+z^2$.
I started by using the fact that every natural number has a decomposition to a sum of 4 square, and tried to prove that one of the numbers is zero, basically $4n+2=x^2+y^2+z^2+w^2$.
Firstly, I falsely assumed that all of those numbers weren't 0, and then I proved that at least one of those numbers has to be odd, because if they were all even then $4n+2=x^2+y^2+z^2+w^2 \equiv 0 (mod 4)$, contradicting the fact that $4n+2\equiv 2 (mod 4)$. Then I showed that since $4n+2$ is even then there have to be 2 even numbers exactly and 2 odd numbers exactly among $x,y,z,w$.
From here I assumed that $x,y$ are odd, and that $z,w$ are even and tried to prove that at least one of $z,w$ is 0, but I didn't really have ideas in this direction, so this didn't work out.
|
If x and y are odd while z is even, then you can set $x = 2a+1, y = 2b+1, z = 2c$
$4n+2 = (2a+1)^2+(2b+1)^2+(2c)^2 = 4a^2 + 4a+1+4b^2+4b+1+4c^2 = 4(a^2+a+b^2+b+c^2)+2$
Now you're trying to prove that every possible $n$ can be described as $n = a^2+a+b^2+b+c^2 = a(a+1) + b(b+1) + c^2$
Note that $a(a+1)+b(b+1) \equiv 0 (mod 2)$
Does this help?
|
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|
How does one solve this recurrence relation? We have the following recursive system:
$$
\begin{cases}
& a_{n+1}=-2a_n -4b_n\\
& b_{n+1}=4a_n +6b_n\\
& a_0=1, b_0=0
\end{cases}
$$
and the 2005 mid-exam wants me to calculate answer of $ \frac{a_{20}}{a_{20}+b_{20}} $.
Do you have any idea how to solve this recursive equation to reach a numerical value?
|
We can write the recurrence relation in matrix form
$$\begin{bmatrix} a_{k+1}\\ b_{k+1}\end{bmatrix} = \begin{bmatrix}-2 & -4\\ 4 & 6\end{bmatrix} \begin{bmatrix} a_{k}\\ b_{k}\end{bmatrix}$$
Hence,
$$\begin{bmatrix} a_{n}\\ b_{n}\end{bmatrix} = \begin{bmatrix}-2 & -4\\ 4 & 6\end{bmatrix}^n \begin{bmatrix} a_{0}\\ b_{0}\end{bmatrix}$$
Unfortunately, the matrix is not diagonalizable. Its Jordan decomposition gives us
$$\begin{array}{rl}\begin{bmatrix} a_{n}\\ b_{n}\end{bmatrix} &= \begin{bmatrix}-1 & \frac{1}{4}\\ 1 & 0\end{bmatrix} \begin{bmatrix} 2 & 1\\ 0 & 2\end{bmatrix}^n \begin{bmatrix} 0 & 1\\ 4 & 4\end{bmatrix} \begin{bmatrix} a_{0}\\ b_{0}\end{bmatrix}\\\\ &= \begin{bmatrix}-1 & \frac{1}{4}\\ 1 & 0\end{bmatrix} \begin{bmatrix} 2^n & n \, 2^{n-1}\\ 0 & 2^n\end{bmatrix} \begin{bmatrix} b_{0}\\ 4 a_{0} + 4 b_{0}\end{bmatrix}\\\\ &= \begin{bmatrix} -2^n & (1 - 2n) \, 2^{n-2}\\ 2^n & n \, 2^{n-1}\end{bmatrix} \begin{bmatrix} b_{0}\\ 4 a_{0} + 4 b_{0}\end{bmatrix}\end{array}$$
If $a_0 = 1$, $b_0 = 0$ and $n = 20$,
$$\begin{bmatrix} a_{20}\\ b_{20}\end{bmatrix} = \begin{bmatrix} -2^{20} & -39 \cdot 2^{18}\\ 2^{20} & 20 \cdot 2^{19}\end{bmatrix} \begin{bmatrix} 0\\ 2^2\end{bmatrix} = 2^{20} \begin{bmatrix} -39\\ 40\end{bmatrix}$$
Thus,
$$\dfrac{a_{20}}{a_{20} + b_{20}} = \dfrac{-39}{-39 + 40} = -39$$
|
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|
Determine all integral solutions of $a^2+b^2+c^2=a^2b^2$ I started like this :
$a^2+c^2=b^2(a^2-1)\\c^2 +1=(a^2-1)(b^2-1)$
but it's leads to nowhere.
can you help please ?
|
Suppose that this Diophantine equation has an integral solution $(a,b,c)\neq(0,0,0)$. Then, $a\neq 0$ and $b\neq 0$, whence $a^2-1\geq 0$ and $b^2-1\geq 0$. If $a$ or $b$ is even, then $a^2-1\equiv 3\pmod{4}$ or $b^2-1\equiv3\pmod{4}$. Hence, $\left(a^2-1\right)\left(b^2-1\right)=c^2+1$ is divisible by a prime natural number $p\equiv 3\pmod{4}$. Thus, $c^2\equiv -1\pmod{p}$, which is a contradiction as $\left(\frac{-1}{p}\right)=(-1)^{\frac{p-1}{2}}=-1$. Therefore, $a$ and $b$ must be odd. Thus, $$c^2=a^2b^2-a^2-b^2\equiv1-1-1=-1\pmod{8}\,,$$ which is a contradiction.
|
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Edwards Differential Calculus for Beginners, 1896. Chapter 4, Question 82 This question asks for an evaluation of an infinite series assuming only knowledge of basic differential calculus. I couldn't figure it out, but user 'Dr. MV' gave me a hint which was sufficient for me to find the answer. I hope this makes the question seem more relevant, I'm new around these parts and have yet to learn what makes for adequate structure in regards to asking questions.
Here is the question.
Prove that if $x <1$, then
$$\sum_{k=0}^\infty\frac{2^kx^{2^k-1}-2^{k+1}x^{2^{k+1}-1}}{1-x^{2^k}+x^{2^{k+1}}}=\frac{1-2x}{1-x+x^2} + \frac{2x-4x^3}{1-x^2+x^4} + \frac{4x^3-8x^7}{1-x^4+x^8} + ... =\frac{1+2x}{1+x+x^2}$$
|
We have:$$\frac{2^nx^{2^n-1}+2^{n+1}x^{2^{n+1}-1}}{1+x^{2^n}+x^{2^{n+1}}}-\frac{2^nx^{2^n-1}-2^{n+1}x^{2^{n+1}-1}}{1-x^{2^n}+x^{2^{n+1}}}=\frac{2^{n+1}x^{2^{n+1}-1}+2^{n+2}x^{2^{n+2}-1}}{1+x^{2^{n+1}}+x^{2^{n+2}}}$$
So if we add and substract $$\frac{1+2x}{1+x+x^2}$$ at the lefthand side of the sum, the subtracted term and the $\frac{1-2x}{1-x+x^2}$ term give $$-\frac{2x+4x^3}{1-x^2+x^4}$$ That combines with the next term and so on.
So we are left with just $$\frac{1+2x}{1+2x+x^2}$$
|
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|
Inverse of almost quadratic functions Hi all first question here,
I'm currently trying to invert the following two (two since the $\pm$ sign) functions:
$y = x^2 \pm \sqrt{(ax + b \sin{\theta})^2 + (b\cos{\theta})^2}$
Where a,b, $\theta$ are real and positive.
I tried banging my head against the wall and using Maple, but Maple gives me answers in terms of RootOf and _Z(yes lower case _ and then Z) which I interpreted as "go away". Then again I'm bad at Maple.
I'm aware that the global inverse doesn't exist but locally I should be able to find something right?
For case $\theta = 0$ I currently have two non-global inverses in the form of:
$x = \frac{1}{2} \sqrt{2} \sqrt{a^2 + 2y \pm \sqrt{a^4 +4 a^2 y + 4 b^2}}$
(Note that this only inverts to positive x but for $\theta = 0$ the function is symmetric in 0.)
Any help, direction or solution is appreciated.
Cheers,
Piotr
|
$$y-x^2=\pm\sqrt{(ax+b\sin\theta)^2+(b\cos\theta)^2}$$
$$(y-x^2)^2=(ax+b\sin\theta)^2+(b\cos\theta)^2$$
$$y^2-2x^2y+x^4=ax^2+2ab\sin\theta x+b^2(\sin^2\theta+\cos^2\theta)$$
$sin^2\theta+\cos^2\theta=1$
$$y^2-2x^2y+x^4=ax^2+2ab\sin\theta x+b^2$$
$$x^4-2(y+a)x^2-2ab\sin\theta x+y^2-b^2=0$$
$$x^4+kx^2+l=mx^2+2ab\sin\theta x+n$$
Where $m-k=2(y+a)$ and $l-n=y^2-b^2$.
To make the LHS a perfect square, I also want the relationship $k^2=4l$,
$$x^4+kx^2+l=(x^2+\sqrt{l})^2$$
To make the RHS a perfect square, I want the relationship $\sqrt{mn}=ab\sin\theta$,
$$mx^2+2ab\sin\theta x+n=(\sqrt mx+\sqrt n)^2$$
This allows us to set the two equal and square root both sides, followed by an application of the quadratic formula.
To solve for $k,l,m,n$, we must solve cubic polynomials, which is doable.
After some work, we can get
$$\frac14k^3+\frac{y+a}2k^2-(y^2-b^2)k-2(y+a)(y^2-b^2)-(ab\sin\theta)^2=0$$
Using the cubic formula, we get
$$k=\left(\frac{\sqrt3}2-\frac12i\right)^{1,2,3}\left(\sqrt[3]{q+\sqrt{q^2+(r-p^2)^3}}+\sqrt[3]{q-\sqrt{q^2+(r-p^2)^3}}\right)+p$$
Where $p=-\frac{2(y+a)}{3}$, $q=p^3+\frac{-4(y+a)2(y^2-b^2)+6(2(y+a)(y^2-b^2)-(ab\sin\theta)^2)}{3}$, and $r=\frac{4(b^2-y^2)}3$.
You can quickly see how this becomes excruciatingly painful? I don't even write the value of $k$ outright, I use extra variables all over the place! And there are three solutions for $k$, where we have $1,2,3$ in the exponent. (Kind of like the plus minus of the quadratic formula but we have three different solutions)
If you really wanted to, you could try and solve for the other variables now that you have $k$. Then you can set the two binomials equal and cancel the square roots (adding in $\pm$ when you cancel them) and then apply quadratic formula.
It is too tedious to do, which is why I won't do it. I hope I left a good enough explanation that you could do it, if you really really wanted to try, and feel free to leave a comment if you are confused on anything.
|
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|
Prove $\sum\limits_{cyc}\left(\frac{a^4}{a^3+b^3}\right)^{\frac34} \geqslant \frac{a^{\frac34}+b^{\frac34}+c^{\frac34}}{2^{\frac34}}$ When $a,b,c > 0$, prove
$$\left(\frac{a^4}{a^3+b^3}\right)^{\frac34}+\left(\frac{b^4}{b^3+c^3}\right)^ {\frac34}+\left(\frac{c^4}{c^3+a^3}\right)^{\frac34} \geqslant \frac{a^{\frac34}+b^{\frac34}+c ^{\frac34}}{2^{\frac34}}$$
I tried the substitution $x=a^4,\ldots$ but I have no idea how to deal with the left- hand side. I tried some C-S but it goes nowhere. I think Bernoulli's inequality may be the only way to prove this inequality.
|
It's not a proof, but I think it can help.
We can rewrite this inequality in the following form.
$$\sum_{cyc}\frac{a^4}{\sqrt[4]{\left(\frac{a^4+b^4}{2}\right)^3}}\geq a+b+c,$$
where $a$, $b$ and $c$ are positives.
Now, since for $(a,b,c)=(1.98,0.89,1.38)$
$$\sum_{cyc}\frac{a^4}{\sqrt[4]{\left(\frac{a^4+b^4}{2}\right)^3}}-a-b-c=0.0080...,$$
we can use the Holder's inequality.
I checked that the following Holder does not help.
$$\left(\sum_{cyc}\frac{a^4}{\sqrt[4]{(a^4+b^4)^3}}\right)^4\sum_{cyc}a^4(a^4+b^4)^3(a+mb+nc)^5\geq$$
$$\geq\left(\sum_{cyc}a^4(a+mb+nc)\right)^5$$
because the inequality
$$8\left(\sum_{cyc}a^4(a+mb+nc)\right)^5\geq(a+b+c)^4\sum_{cyc}a^4(a^4+b^4)^3(a+mb+nc)^5$$ is wrong for all $m\geq0$ and $n\geq0$.
By the way, I think the following Holder can help.
$$\left(\sum_{cyc}\frac{a^4}{\sqrt[4]{(a^4+b^4)^3}}\right)^4\sum_{cyc}a^4(a^4+b^4)^3(a^2+kb^2+mc^2+lab+nac+pbc)^5\geq$$
$$\geq\left(\sum_{cyc}a^4(a^2+kb^2+mc^2+lab+nac+pbc)\right)^5,$$
where $a^2+kb^2+mc^2+lab+nac+pbc>0$ for all positives $a$, $b$ and $c$.
It's enough to prove that
$$8\left(\sum_{cyc}a^4(a^2+kb^2+mc^2+lab+nac+pbc)\right)^5\geq$$
$$\geq(a+b+c)^4\sum_{cyc}a^4(a^4+b^4)^3(a^2+kb^2+mc^2+lab+nac+pbc)^5.$$
Now, we can substitute in the last inequality $(a,b,c)=(1.98,0.89,1.38)$
and we can choose parameteres $k,$ $l$, $m$, $n$ and $p$ such that the inequality is true.
I hope that it's possible! I think it's possible even for one or more of these parameters equal to zero. For this we need some software, which I have no.
After choosing of parameters we can try to prove the inequality by BW, for which we need software again.
About BW see here: https://artofproblemsolving.com/community/c6h522084
After proving of the starting inequality we can say that this inequality we can prove by hand, but after some days of idiotic computations.
|
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|
How do I find the cdf of $X_1 + X_2$? $X_1$ uniform $(0,1)$ and $X_2$ uniform $(0,2)$
$$
\begin{cases}
f(x_1,x_2) = \frac{1}{2}, &\quad \mbox{for} \ 0<x_1<1, 0<x_2<2 \\
0, & \quad \mbox{otherwise}
\end{cases}
$$
The density of $X_1+X_2$:
$$
f(x_1,x_2) =
\begin{cases}
\displaystyle \frac{z}{2} & \quad 0<z<1\\
\displaystyle \frac{1}{2} & \quad 1<z<2\\
\displaystyle \frac{3-z}{2} & \quad 2<z<3\\
\end{cases}
$$
The question is how do I find the cdf of $X_1+X_2$?
I would do:
$$
f(x_1,x_2) =
\begin{cases}
\displaystyle \int_0^z \frac{z}{2} dx =\frac{z^2}{4}, & \quad 0<z<1\\
\displaystyle \int_0^1 \frac{1}{2} dx =\frac{1}{2}, & \quad 1<z<2\\
\end{cases}
$$
That just seems wrong.. Help
|
The PDF of $Y = X_1 + X_2$ is the convolution of the PDFs of $X_1$ and $X_2$
$$f_Y (y) = \begin{cases} \dfrac{y}{2} & \text{if } y \in [0,1]\\\\ \dfrac{1}{2} & \text{if } y \in [1,2]\\\\ \dfrac{3-y}{2} & \text{if } y \in [2,3]\\\\ 0 & \text{otherwise}\end{cases}$$
Integrating, we obtain the CDF
$$F_Y (y) = \begin{cases} 0 & \text{if } y < 0\\\\ \dfrac{y^2}{4} & \text{if } y \in [0,1]\\\\ \dfrac{1}{4} + \dfrac{y-1}{2} & \text{if } y \in [1,2]\\\\ \dfrac{3}{4} + \dfrac{3}{2} (y-2) - \dfrac{1}{4} (y^2 - 2^2) & \text{if } y \in [2,3]\\\\ 1 & \text{if } y \geq 3\end{cases}$$
|
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Sum of Specific Convergent Series Let L be the sum of the following alternating convergent series $$L = \sum_{n=1}^{\infty}\frac{(-1)^{n-1}}{n}$$
Now consider the rearrangement
$$1+\frac{1}{3}-\frac{1}{2}+\frac{1}{5}+\frac{1}{7}-\frac{1}{4}+\frac{1}{9}+\frac{1}{11}-\frac{1}{6}+... = \sum_{n=1}^{\infty}\left(\frac{1}{4n-3}+\frac{1}{4n-1}-\frac{1}{2n}\right)$$
Show that it sums to $\frac{3L}{2}$.
We are given 2 extra hints:
*
*the difference between $\frac{1}{2n+1}+\frac{1}{2n+3}+...+\frac{1}{4n-1}$ and $\frac{1}{2n+2}+\frac{1}{2n+4}+...+\frac{1}{4n}$ is small;
*the sum $A(n) = \frac{1}{n+1}+\frac{1}{n+2}+...+\frac{1}{2n}$ can be rewritten when considering the expression $A(n)-A(n-1)$.
I need help starting... what can I conclude from evaluating the hinted upon difference? And rewriting $A(n)$ leads to a recursion relation, with solution in the form of a harmonic partial sum, but again I can't see how it helps in any way. Thanks.
|
Multiply the given relation by a factor of $\frac{1}{2}$ and add it to the original sequence. Now we get:
$$L = 1 - \frac{1}{2} + \frac 13 - \frac 14 + \frac 15 - \frac 16 + \frac 17 - \frac 18 + ...$$
$$\frac 12 L = 0 + \frac 12 + 0 - \frac 14 + 0 + \frac 16 + 0 - \frac 18+ ...$$
$$\frac 32 L = 1 + 0 + \frac 13 - \frac 12 + \frac 15 + 0 + \frac 17 - \frac 14 +...$$
$$\frac 32 L = 1 + \frac 13 - \frac 12 + \frac 15 + \frac 17 - \frac 14 +... = \sum_{n=1}^{\infty}\left(\frac{1}{4n-3}+\frac{1}{4n-1}-\frac{1}{2n}\right)$$
Additionally it's a well-known fact that $L = \ln(2)$. Also this kind of an rearragements are possible due to the fact that the series is conditionally convergent. Actually due to the Riemann Series Theorem we can rearrange any conditionally sequence in such a manner that it will converge to any giver number in $\mathbb{R}$
|
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|
Prove $x+y^2+z^3 \geqslant x^2y+y^2z+z^2x$ for $xy+yz+zx=1$
Let $x,y,z \geqslant 0$ and $xy+yz+zx=1$. Prove that
$$x+y^2+z^3 \geqslant x^2y+y^2z+z^2x.$$
What I try:
$$x+y^2+z^3 \geqslant x^2y+y^2z+z^2x$$
$$\Leftrightarrow x(xy+yz+zx)+y^2+z^3- x^2y-y^2z-z^2x \geqslant 0$$
$$\Leftrightarrow \left(x^2z+\frac{z^3}{4}-z^2x \right)+ \left(y^2+\frac{3z^3}{4}+xyz-y^2z\right) \geqslant 0$$
I strongly believe that $\left(y^2+\frac{3z^3}{4}+xyz-y^2z\right) \geqslant 0$, but I have no proof.
|
Alternative proof:
Using $xy + zx \le 1$ and $\sqrt{yz} \le 1$, we have
\begin{align*}
&x + y^2 + z^3 - x^2 y - y^2z - z^2x \\
\ge\,& x(xy + zx) + y^2\sqrt{yz} + z^3 - x^2 y - y^2z - z^2x\\
=\,& x^2z + y^2\sqrt{yz} + z^3 - y^2z - z^2x \\
=\,& \left(x^2z - z^2x + \frac14 z^3\right) + \frac34 z^3 + y^2\sqrt{yz} - y^2z\\
=\,& \frac14 z(2x - z)^2 + \frac34 z^3 + y^2\sqrt{yz} - y^2z \\
\ge\,& \frac14 z(2x - z)^2 + \frac34 z^3 + y^2\cdot \frac{2yz}{y + z} - y^2z \\
=\,& \frac14 z(2x - z)^2 + \frac{z}{4(y+z)}(4y^3 + 3yz^2 - 4y^2z + 3z^3) \\
\ge\,& 0
\end{align*}
where we have used GM-HM to get
$\sqrt{yz} \ge \frac{2yz}{y + z}$ (note: $y + z > 0$),
and AM-GM to get
$4y^3 + 3yz^2 - 4y^2z \ge 2\sqrt{4y^3 \cdot 3yz^2} - 4y^2z = (4\sqrt 3 - 4)y^2z \ge 0$.
We are done.
|
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|
What am I doing wrong in calculating the following limit? $$\lim_{x\to-2} \frac{x+2}{\sqrt{6+x}-2}=\lim_{x\to-2} \frac{1+2/x}{\sqrt{(6/x^2)+(1/x)}-2/x^2}$$ Dividing numerator and denominator by $x \neq0$
$$\frac{1+2/-2}{\sqrt{(6/4)+(1/-2)}-2/4}=\frac{0}{1/2}=0$$ but the limit is $4$ according to Wolfram Alpha?
|
You say you divide by $x$, but that's not what you do in the denominator; it would be:
$$\lim_{x\to-2} \frac{x+2}{\sqrt{6+x}-2}
=\lim_{x\to-2} \frac{1+\tfrac{2}{x}}{\tfrac{\sqrt{6+x}}{x}-\tfrac{2}{x}} =\lim_{x\to-2} \frac{1+\tfrac{2}{x}}{-\sqrt{\tfrac{6}{x^2}+\tfrac{1}{x}}-\tfrac{2}{x}} $$
A better approach:
$$\begin{array}{rl}
\displaystyle \lim_{x\to-2} \frac{x+2}{\sqrt{6+x}-2}
& \displaystyle = \lim_{x\to-2} \frac{\left(x+2\right)\color{blue}{\left(\sqrt{6+x}+2\right)}}{\left(\sqrt{6+x}-2\right)\color{blue}{\left(\sqrt{6+x}+2\right)}} \\[7pt]
& \displaystyle = \lim_{x\to-2} \frac{\left(x+2\right)\left(\sqrt{6+x}+2\right)}{x+2} \\[7pt]
& \displaystyle = \lim_{x\to-2} \left(\sqrt{6+x}+2\right) \\
& = 4
\end{array}$$
|
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|
Which exponents r>0 is the limit finite I am trying to find values of $r>0$ such that $\lim\limits_{n\rightarrow \infty} \sum\limits_{k=1}^{n^2}\frac{n^{r-1}}{n^r+k^r}$ is finite. I have tried to use integral methods for this limit such as $\sum\limits_{k=1}^{n^2}\frac{n^{r-1}}{n^r+k^r}=\sum\limits_{k=1}^{n}\frac{n^{r-1}}{n^r+k^r}+\sum\limits_{k=n}^{n^2}\frac{n^{r-1}}{n^r+k^r} = \frac1n\sum\limits_{k=1}^{n}\frac{1}{1+(\frac kn)^r}+\sum\limits_{k=n}^{n^2}\frac{n^{r-1}}{n^r+k^r}$. So the first term, I could use the integral method to estimate. However, the second one is hard to evaluate. Could you please help me with this? Thanks
|
In short: diverges for $r\in [0,1]$, converges for $r>1$.
*
*First case (simpler): $r\in(0,1)$:
$$
\sum_{k=1}^{n^2}\frac{n^{r-1}}{n^r+k^r} = \frac{1}{n}\sum_{k=1}^{n^2}\frac{n^{r}}{n^r+k^r}
\geq \frac{1}{n}\sum_{k=1}^{n^2}\frac{n^{r}}{n^r+n^{2r}}
\geq \frac{1}{n}\sum_{k=1}^{n^2}\frac{n^{r}}{n^{2r}+n^{2r}}
= \frac{n^2}{n}\frac{1}{2n^r} = \frac{n^{1-r}}{2}
$$
so for $r\in[0,1)$ the sequence will diverge by comparison.
*
*Second case: for $r=1$,
$$\begin{align}
\sum_{k=1}^{n^2}\frac{n^{r-1}}{n^r+k^r} &=
\sum_{k=1}^{n^2}\frac{1}{n + k} =
\sum_{k=1}^{n}\frac{1}{n + k} + \sum_{k=n+1}^{n^2}\frac{1}{n + k}\\
&= \underbrace{\frac{1}{n}\sum_{k=1}^{n}\frac{1}{1 + \frac{k}{n}}}_{\text{Converges (Riemann sum)}} + \sum_{k=n+1}^{n^2}\frac{1}{n + k}
\end{align}$$
but
$$
\sum_{k=n+1}^{n^2}\frac{1}{n + k} =
\sum_{\ell=1}^{n-1} \sum_{k=\ell n+1}^{(\ell+1)n}\frac{1}{n + k}
\geq \sum_{\ell=1}^{n-1} \sum_{k=\ell n+1}^{(\ell+1)n}\frac{1}{(\ell+2)n}
= \sum_{\ell=1}^{n-1} \frac{1}{\ell+2} \sim_{n\to\infty} \ln n
$$
so
$$
\sum_{k=1}^{n^2}\frac{1}{n+k} \xrightarrow[n\to\infty]{} \infty
$$
*Last case: $r>1$. Same technique, and as before we only need to analyze the second term.
$$\begin{align}
\sum_{k=n+1}^{n^2}\frac{n^{r-1}}{n^r+k^r}
&= \frac{1}{n}\sum_{k=n+1}^{n^2}\frac{1}{1+\left(\frac{k}{n}\right)^r}
= \frac{1}{n}\sum_{\ell=1}^{n-1} \sum_{k=\ell n+1}^{(\ell+1)n} \frac{1}{1+\left(\frac{k}{n}\right)^r}
\leq
\frac{1}{n}\sum_{\ell=1}^{n-1} \sum_{k=\ell n+1}^{(\ell+1)n} \frac{1}{1+\ell^r}
\\
&=
\sum_{\ell=1}^{n-1} \frac{1}{1+\ell^r}
\end{align}$$
which therefore converges by comparison ($p$-series with exponent $r>1$).
$$
\exists L_r <\infty,\qquad \sum_{k=1}^{n^2}\frac{n^{r-1}}{n^r+k^r} \xrightarrow[n\to\infty]{} L_r
$$
|
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|
Show $x^2 + y^2 + 1 = 0 \pmod m$, iff $\,m \pmod 4 \ne 0$.
Show that $x^2 + y^2 + 1 = 0$ $\pmod m$ has solutions iff $\,m \pmod 4
\ne 0$.
I know hot to show that this equation has solutions if m = p It's easy to show "$=>$", but I'm completery stucked with the opposite direction.
|
If $m \equiv 0\pmod 4$ it is simple to prove there is no solution, test all possible pairs $(x,y)$. If $2m \equiv 2\pmod 4$ then a solution can be constructed by the chinese remainder theorem from the solution of the odd $m$ and the solution $x=0,y=1$ modulo $2$. So we assume that $m$ is odd.
If $m$ is a prime then $\frac{m+1}{2}$ of the numbers $0,\ldots,m-1$ are squares. Therefore both sets $\{-x^2\mid x=0,\ldots\ m$ and $\{y^2+1\mid y=0,\ldots\ m$} contain $\frac{m+1}{2}$ elements. So the intersection of these sets is not empty. So there is a $x$ and a $y$ such that $y^2+1=-x^2$ if $m$ is an odd prime.
Now by induction we show that the equation has a solution for every power of an odd prime.
If $(x,y)=(x_0,y_0)$ is a solution of
$$x^2+y^2+1 \equiv 0\pmod{p^n}$$
then
$$x_0^2+y_0^2+1=q_0p^n$$
Therefore
$$ \begin{align}(x_0+ap^n)^2+(y_0+bp^n)^2+1&=x_0^2+y_0^2+1+2p^n(ax_0+by_0)+(a^2+b^2) p^{2n}\\&=p^n(q_0+2ax_0+2by_0)+p^{n+1}(a^2+b^2)p^{n-1}\\
&=p^n(pt)+p^{n+1}(a^2+b^2)p^{n-1}\\
&\equiv 0 \pmod {p^{n+1}}\end{align}$$
because the equation
$$q_0+2ax_0+2by_0 \equiv 0 \pmod{p}$$
has a solution $(a,b) \pmod p$ and therefore
$$q_0+2ax_0+2by_0=tp$$.
Here is the proof:
At leas oen of $x_0$ or $y_0$ is not equal to $0$, so let's assume $x_0 \ne 0$. So we select an arbitrary b and get
$$a=\frac{2by_0-q_0}{2x_0} \pmod{p}$$
We have shown that there is a solution if $m$ is a power of an odd prime. By the chinese remainder theorem we can construct a solution for $m$ that is a product of odd prime powers.
|
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"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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|
Show that the series converges uniformly Show that $\sum_{n=1}^{\infty} \sin \left(\dfrac{x}{n}\right)^2$ converges uniformly on $[a,-a]$ , $a\in \mathbb{R}$.
My attempt:
Using Taylor's formula, we have:$$ \sin\left(\dfrac{x}{n}\right)^2 = \dfrac{x^2}{n^2} + o\left( \dfrac{x^2}{n^2}\right)$$
Therefore,
$$\sin\left(\dfrac{x^2}{n^2}\right) \leq \dfrac{x^2}{n^2} \leq \dfrac{a^2}{n^2}$$
We then have $\| \sin\dfrac{x^2}{n^2}\| \leq \dfrac{a^2}{n^2} \leq \infty$
Therefore, according to Weierstrass's criteria:$\sum_{n=1}^{\infty} \sin \left(\dfrac{x}{2}\right)^2$ converges uniformly on $[a,-a]$
Is that the right approach to the problem?
|
Since your $x$ is bounded, there exists an $N$ such that
\begin{equation}
0<\frac{x^2}{n^2}\leq \frac{\pi}{2}
\end{equation}
for all $n\geq N$ and $x\in[-a,a]$. Now you can use the inequality $0\leq \sin x\leq x $ for $0\leq x\leq \frac{\pi}{2}$.
|
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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|
proof that $\sum_{n=1}^{\infty}\frac{n}{n^2+2n+1}$ diverges, by comparsion I need to prove that
$$\sum_{n=1}^{\infty}\frac{n}{n^2+2n+1}$$
diverges by comparsion.
The way I did was to use
$$\frac{n}{n^2+2n+1}>\frac{n}{n^2+2n^2+n^2} = \frac{n}{4n^2} = \frac{1}{4n}$$
which diverges. Can I do that? Because $$2n+1<2n^2+n^2\implies n^2+2n+1<n^2+2n^2+n^2\implies $$
$$\frac{1}{n^2+2n+1}>\frac{1}{n^2+2n^2+n^2}$$
for large $n$.
I also tried seeing
$$\sum_{n=1}^{\infty}\frac{n}{n^2+2n+1} = \sum_{n=1}^{\infty}\frac{n}{(n+1)^2}$$
but I couldn't find any comparsion for that. Am I right in the first demonstration? Can you find a comparsion for the second attempt I did?
|
$n/(n^2+2 n+1)=1/(n+2 +1/n^2)\geq 1/(n+3)$, and $\sum_n 1/(n+3)$ diverges.
|
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"timestamp": "2023-03-29T00:00:00",
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|
Solution of an initial value problem (MCQ) (CSIR DEC 2015) The solution of the initial value problem
$ (x-y) u_{x} + (y-x-u) u_{y} = u $
with the initial condition $u(x,0) = 1$ satisfies
*
*$ u^2(x-y+u) + (y-x-u) = 0$
*$ u^2(x+y+u) + (y-x-u) = 0$
*$ u^2(x-y+u) - (x+y+u) = 0$
*$ u^2(y-x+u) + (x+y-u) = 0$
This is what I am able to do
The characteristic equations are :
$\frac{dx}{x-y} = \frac{dy}{y-x-u} = \frac{du}{u} $
From this we get
$dx + dy + du = 0$
Integrating we get a characteristic curve
$x+y+u =C_{1}$
I am unable to get a second characteristic curve.
Please help.
The correct answer is 2.
Thanks in advance!
|
Second characteristic curve :
$\frac{dx}{x-y} = \frac{dy}{y-x-u} = \frac{du}{u} =\frac{dx-dy+du}{(x-y)-(y-x-u)+u} = \frac{d(x-y+u)}{2(x-y+u)} = \frac{du}{u} $
$\frac{1}{2}\ln|x-y+u|-\ln|u|=$constant $\quad\to\quad \frac{x-y+u}{u^2}=C_2$
The implicit solution is $\quad x+y+u=F\left(\frac{x-y+u}{u^2}\right) \quad $any differentiable function $F$.
The condition $u(x,0)=1$ determines the function $F(x+1)=x+1 \quad\to\quad F(X)=X$.
$F\left(\frac{x-y+u}{u^2}\right)=\frac{x-y+u}{u^2}$
$$x+y+u=\frac{x-y+u}{u^2}$$
|
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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|
The best notation for this identity involving pentagonal numbers $\omega(n)$ and the $3x+1$ map Let the $3x+1$ map
$$ f(n) =
\begin{cases}
3n+1 & \text {if $n$ is odd} \\
\frac{n}{2} & \text {if $n$ is even}
\end{cases} .$$
Now we read the Wikipedia's page for the Collatz problem, also known by several names.
Example. Since $f(1)=4$, $f(4)=2$ and $f(2)=1$, the integer $n=1$ is not a counterexample for the Collatz conjecture (in other words, is the easier example of an integer satisfying the conjecture).
In the other hand we know that the pentagonal numbers of the form $$\omega(n)=\frac{3n^2-n}{2}$$ (related with partitions) are defined as $$\omega(n)=\sum_{k=0}^{n-1}(3k+1).$$
Thus if $n-1$ is odd (this is the first case of two, the second case with $n-1$ even) one can deduce easily combining previous and the definition of the $3x+1$ map that $$\omega(n)=1+\sum_{k\in \left\{ 1,3,5,\ldots,n-1 \right\}}f(k)+6\left(\sum_{k\in \left\{ 2,4,6,\ldots,n-2 \right\}}f(k)\right)+\sum_{k\in \left\{ 2,4,6,\ldots,n-2 \right\}}1.$$
Question. I am stuck to get the best concise formula ( I say the previous last identity, if it is neccesary by cases to get also the case $n-1$ is even) with the right notation. I don't know if I should be to use counting funtions to count the number of odd positive integers $\leq n-1$ (respectively even positive integers) or ceil and floor functions. Can you help to get this simple proposition, both cases, with a good notation Thanks in advance.
|
At first some notes: We usually regard the closed formula
\begin{align*}
\omega(n)=\frac{3n^2+n}{2}\qquad\qquad\qquad\qquad\qquad n\geq 1\tag{1}
\end{align*}
as simpler than the summation formula
\begin{align*}
\omega(n)=\sum_{k=0}^{n-1}(3k+1)\qquad\qquad\qquad\qquad\quad n\geq 1\tag{2}
\end{align*}
So, if we want to represent $\omega(n)$ by the function
\begin{align*}
f(n)=\begin{cases}
3n+1\qquad & n \text{ odd}\\
\frac{n}{2}\qquad & n \text{ even}
\end{cases}\qquad\qquad n\geq 1
\end{align*}
we would rather try to avoid sums and start from (1). Another aspect is there is no best notation. An appropriate notation is usually application dependent and may vary according to different needs. But we can try to do some simplifications which are typically useful.
First variant: Expressing $f(n)$ by (2)
We start similar as OP did.
We obtain
\begin{align*}
\omega(n)&=\sum_{k=0}^{n-1}(3k+1)\\
&=\sum_{{k=0}\atop{k \text{ odd}}}^{n-1}(3k+1)+\sum_{{k=0}\atop{k \text{ even}}}^{n-1}(6\cdot\frac{k}{2}+1)\tag{3}\\
&=\sum_{{k=0}\atop{k \text{ odd}}}^{n-1}f(k)+\sum_{{k=0}\atop{k \text{ even}}}^{n-1}(6f(k)+1)\tag{4}\\
&=\sum_{k=1}^{\left\lfloor\frac{n}{2}\right\rfloor}f(2k-1)
+\sum_{k=0}^{\left\lfloor\frac{n-1}{2}\right\rfloor}(6f(2k)+1)\tag{5}\\
&=\sum_{k=0}^{\left\lfloor\frac{n}{2}\right\rfloor-1}f(2k+1)
+6\sum_{k=0}^{\left\lfloor\frac{n-1}{2}\right\rfloor}f(2k)+\left\lfloor\frac{n-1}{2}\right\rfloor+1\tag{6}\\
&=\sum_{k=0}^{\left\lfloor\frac{n-1}{2}\right\rfloor}\left(f(2k+1)+6f(2k)\right)\\
&\qquad\qquad+\left\lfloor\frac{n-1}{2}\right\rfloor+1-f(2\left\lfloor\frac{n-1}{2}\right\rfloor+1)\cdot\frac{1-(-1)^n}{2}\tag{7}\\
\end{align*}
Comment:
*
*In (3) we split the sum in odd and event parts of the index $k$
*In (4) we use the function $f(n)$
*In (5) we use the floor function to set the upper limit of $k$. For the odd part we use the substitution $k \rightarrow 2k-1$ and for the even part we use the substitution $k \rightarrow 2k$.
*In (6) we shift the index of the left sum by one to start with $k=0$ and we extract the constant $1$ from the right sum and get $\left\lfloor\frac{n-1}{2}\right\rfloor+1$.
*In (7) we can further reduce the number of summation symbols (if we like) by observing that
\begin{align*}
\left\lfloor\frac{n-1}{2}\right\rfloor=\begin{cases}
\left\lfloor\frac{n}{2}\right\rfloor &\qquad n \text{ odd}\\
\left\lfloor\frac{n}{2}\right\rfloor-1 &\qquad n \text{ even}\\
\end{cases}
\end{align*}
So, we can use one summation symbol and a common upper limit $\left\lfloor\frac{n-1}{2}\right\rfloor$. Since in the odd case we have one more than in the even case we have to subtract the last element. We do this by subtracting $f(2\left\lfloor\frac{n-1}{2}\right\rfloor+1)$ and multiplying it with
\begin{align*}
\frac{1-(-1)^n}{2}=\begin{cases}
1\qquad& n\text{ odd}\\
0\qquad& n\text{ even}
\end{cases}
\end{align*}
Second variant: Expressing $f(n)$ by (1)
Since it is usually more convenient to avoid summation symbols if possible at all we start now from (1). Here is one from many different possibilities.
We obtain
\begin{align*}
\omega(n)&=\frac{3n^2-n}{2}\\
&=(3n+1)\frac{n}{2}-n\\
&=\begin{cases}
f(n)\frac{n}{2}-n\qquad & n\text{ odd}\\
(3n+1)f(n)-2f(n)\qquad& n\text{ even}
\end{cases}\tag{8}\\
&=\begin{cases}
f(n)\frac{n}{2}-n\qquad\qquad\qquad & n\text{ odd}\\
(3n-1)f(n)\qquad\qquad\qquad& n\text{ even}
\end{cases}\\
&=(3n-1)f(n)\cdot\frac{1+(-1)^n}{2}+\frac{n}{2}\left(f(n)-2\right)\frac{1-(-1)^n}{2}\tag{9}\\
&=\frac{1}{4}\left(7n-2+(5n-2)(-1)^n\right)f(n)-\frac{n}{2}\left(1-(-1)^n\right)\tag{10}
\end{align*}
Comment:
*
*In (8) we use $f(n)$ according to odd and even $n$
*In (9) we put odd and even cases together using a technique as we did in (7)
*In (10) we do some final rearrangement
|
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|
Prove $\sqrt[2]{x+y}+\sqrt[3]{y+z}+\sqrt[4]{z+x} <4$
$x,y,z \geqslant 0$ and $x^2+y^2+z^2+xyz=4$, prove
$$\sqrt[2]{x+y}+\sqrt[3]{y+z}+\sqrt[4]{z+x} <4$$
A natural though is that from the condition $x^2+y^2+z^2+xyz=4$, I tried a trig substitutions $x=2\cdot \cos A$, $y=2\cdot \cos B$ and $z=2\cdot \cos C$, where $A,B,C$ are angles of an acute triangle.
Our inequality becomes
$$\sqrt[2]{2(\cos A+ \cos B)}+\sqrt[3]{2(\cos B+\cos C)}+\sqrt[4]{2(\cos C+\cos A)} <4$$
I used formula $\cos(A)+\cos(B) < 2$ and estimate that $LHS <5$ but not $4$.
I do not know how to proceed.
|
Remark: My second proof is given in https://artofproblemsolving.com/community/c6h2483725p20868007.
There, KaiRain@AoPS gave a nice proof.
Proof without using derivative
We split into three cases:
*
*$y+z \le 1$ and $z+x \le 1$: We have $x + y \le 2$.
Thus, we have $\mathrm{LHS} \le \sqrt{2} + 1 + 1 < 4$. The inequality is true.
*$y + z > 1$ and $z + x \le 1$: Since $\sqrt[3]{y+z} < \sqrt{y+z}$ and $\sqrt[4]{z+x} \le 1$, it suffices to prove that
$$\sqrt{x+y} + \sqrt{y+z} \le 3.$$
By AM-QM inequality, we have $\sqrt{x+y} + \sqrt{y+z} \le \sqrt{2(x+y + y + z)}$. Thus,
It suffices to prove that $$x+2y+z \le \frac{9}{2}.$$
From $x^2+y^2+z^2 + xyz = 4$, we have
$y^2 + (\frac{1}{2} + \frac{y}{4})(x + z)^2 + (\frac{1}{2} - \frac{y}{4})(x - z)^2 = 4$
which results in $y^2 + (\frac{1}{2} + \frac{y}{4})(x + z)^2 \le 4$
and $x + z \le \sqrt{\frac{4(4 - y^2)}{2 + y}}$.
Thus, it suffices to prove that $2y + \sqrt{\frac{4(4 - y^2)}{2 + y}} \le \frac{9}{2}$.
It suffices to prove that $(\frac{9}{2} - 2y)^2 \ge \frac{4(4 - y^2)}{2 + y}$
that is $\frac{1}{4}(4y - 7)^2 \ge 0$. The inequality is true.
*$z + x > 1$: Since $\sqrt[3]{z+x} > \sqrt[4]{z+x}$, it suffices to prove that
$$
\sqrt{x+y} + \sqrt[3]{y+z} + \sqrt[3]{z+x} < 4.
$$
Note that $u\mapsto \sqrt[3]{u}$ is concave on $u > 0$.
It suffices to prove that
$$
\sqrt{x+y} + 2\sqrt[3]{\frac{y+z + z+x}{2}} < 4.
$$
From $x^2+y^2+z^2 + xyz = 4$, we have $(z + \frac{1}{2}xy)^2 = \frac{1}{4}(4-x^2)(4-y^2)$
which results in $$z = \frac{1}{2}\sqrt{(4-x^2)(4-y^2)}-\frac{1}{2}xy \le \frac{1}{2}\frac{4-x^2+4-y^2}{2} - \frac{1}{2}xy = 2- \frac{(x+y)^2}{4}.$$
Thus, it suffices to prove that
$$
\sqrt{x+y} + 2\sqrt[3]{\frac{y+x}{2} + 2- \frac{(x+y)^2}{4}} < 4.
$$
Let $v = \sqrt{x+y}$. Since $x+y \le \sqrt{2(x^2+y^2)} \le \sqrt{2 \cdot 4} < 4$, we have $v\in (0, 2)$.
It suffices to prove that $$v + 2\sqrt[3]{\frac{v^2}{2} + 2- \frac{v^4}{4}} < 4$$
or $$\frac{v^2}{2} + 2- \frac{v^4}{4} < \frac{1}{8}(4 - v)^3$$ or $$\frac{1}{8}(2v^4-v^3+8v^2-48v+48) > 0$$ for $v\in (0, 2)$.
It is easy to prove that $2v^4-v^3+8v^2-48v+48 > 0$ for $v\in (0, 2)$.
We are done.
|
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|
Choosing Functions for the Squeeze Theorem
Evaluate $$\lim_{n\to \infty}\dfrac{1}{\sqrt{n^2}}+\dfrac{1}{\sqrt{n^2+1}}+...+\dfrac{1}{\sqrt{n^2+2n}}$$
$$$$
I came across the the question on this site itself but had a few doubts on the given solution. As I do not yet have 50 reputation points, I cannot comment over there. Could somebody please help me?
$$$$
From what I understand of the Squeeze Theorem, the three functions are related as
$$g(x)\le f(x)\le h(x)$$
$$$$
Now in the selection of terms, the following inequality has been used:
$$\displaystyle \frac{1}{n+1} \leq \frac{1}{\sqrt{n^2+k}} \leq \frac{1}{n} $$when $0 \leq k \leq 2n$
$$$$
This inequality lead to the one used as the three functions for the application of the Squeeze theorem:
$$\frac{2n+1}{n+1} \leq S(n) \leq \frac{2n+1}{n}$$
where $S(n)=\dfrac{1}{\sqrt{n^2}}+\dfrac{1}{\sqrt{n^2+1}}+...+\dfrac{1}{\sqrt{n^2+2n}}$
$$$$
I don't understand how $$\displaystyle \frac{1}{n+1} \leq \frac{1}{\sqrt{n^2+k}} $$
$$$$
Isn't $$n^2+k\le n^2+2n<(n+1)^2 \Rightarrow n^2+k<(n+1)^2$$
$$\Rightarrow \sqrt{n^2+k}< (n+1)$$$$\Rightarrow \displaystyle \frac{1}{n+1} < \frac{1}{\sqrt{n^2+k}}$$
$$$$
Thus shouldn't the resulting set of inequalities be $$\displaystyle \frac{1}{n+1} < \frac{1}{\sqrt{n^2+k}} \leq \frac{1}{n} $$
$$\Longrightarrow \frac{2n+1}{n+1} < S(n) \leq \frac{2n+1}{n}$$
$$$$
In this case there is a $<$ sign instead of the $\le$ sign. How then can the Squeeze Theorem be applied?
Many thanks in advance.
$$$$
EDIT: Also, since Limits preserve Inequalities, how can $$\lim_{n\to \infty} \frac{2n+1}{n+1} = \lim_{n\to \infty}\frac{2n+1}{n}$$ when $$\frac{2n+1}{n+1} < \frac{2n+1}{n}$$
|
Remember that $a<b\implies a\le b$. This is because $a\le b$ means $a<b$ or $a=b$ as you already noted in the comment.
If you are still confused, recall that "$p\implies q$" means $q$ is true whenever $p$ is true. And when $a<b$ is true, $a<b$ or $a=b$ is true. Hence, $a<b\implies a\le b$.
For example, is "$0\le 1$" true?
|
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Solve the differential equation $\frac{dx}{dt}=-\frac{1}{5}\sqrt{(x+1)}$ Solve the differential equation $$\frac{dx}{dt}=-\frac{1}{5}\sqrt{(x+1)}$$ given that $x=80$ when $t=0$. Give your answer in the form $x=f(t)$.
\begin{align}
\frac{dx}{dt} & = -\frac{1}{5}\sqrt{(x+1)} \\
\frac{dt}{dx} & = \frac{-5}{\sqrt{(x+1)}} \\
\int-\frac{1}{5}dt & = \int\frac{1}{\sqrt{(x+1)}}dx \\
-\frac{1}{5}t & = 2\sqrt{(x+1)} \\
-\frac{t}{10} & = \sqrt{(x+1)} \\
\left(-\frac{t}{10}\right)^2 & = x+1 \\
x & = \left(-\frac{t}{10}\right)^2-1+C \\
\end{align}
Let $x=80$ and $t=0$
\begin{align}
80 & = -1+C \\
C & = 81 \\
\end{align}
Answer in the Mark Scheme is $C=18$
$$x=\left(9-\frac{t}{10}\right)^2-1$$
|
You didn't put $+C$ in the right place. You should have $C-\frac{t}{10}=\sqrt{x+1}$. Solve for $C$ here, then isolate $x$.
|
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How to find $L = \int_0^1 \frac{dx}{1+{x^8}}$ Let $L = \displaystyle \int_0^1 \frac{dx}{1+{x^8}}$ . Then
*
*$L < 1$
*$L > 1$
*$L < \frac{\pi}{4}$
*$L > \frac{\pi}{4}$
I got some idea from this video link. But got stuck while evaluating the second integral.
Please help!!
Thanks in advance!
|
$\forall x \in [0,1], \frac{1}{1+x^8}<1$, so $L<1$.
$\forall x \in [0,1], \frac{1}{1+x^8}>\frac{1}{1+x^2}$, so $L>\int_0^1 \frac{dx}{1+{x^2}}=\frac{\pi}{4}$
How to evaluate this integral :
$$\begin{align} \int \frac{1}{1+x^8}dx &= \int (1+x^8)^{-1}dx \\&= \int_0^x(1+t^8)^{-1}dt +a \\&= \int_0^{x^8} (1+t)^{-1}d(t^{1/8})+b \\ &=\frac{1}{8}\int_0^{x^8}(1+t)^{-1}t^{1/8 -1}dt+c \\ &= \frac{1}{8}\int_0^{1} (1+x^8t)^{-1} (x^8t)^{1/8-1} d(x^8t)+d \tag{1} \\ &= \frac{x}{8} \int_0^1(1+x^8t)^{-1}x^{1-8}t^{1/8-1}x^8dt+e \tag{2} \\ &= x\,{}_2F_1 \left( 1,\frac{1}{8},\frac{9}{8},-x^8 \right) +f\end{align}$$
So $L={}_2F_1 \left( 1,\frac{1}{8},\frac{9}{8},-1 \right)$.
$(1)$ I substitute $t$ with $x^8t$
$(2)$ Here $x$ is just a constant
|
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Please remind me of how this technique works ... We had a high school mathematics teacher who taught us a cool technique that I've forgotten. It can be used, for example, for developing a formula for the sum of squares for the first "n" integers. You start by making a column for Sn, and then determine the differences until you get a constant. See the picture.
(sorry about the rotated picture)
How do you proceed from here to the formula?
|
*
*backgrounded on the fact that the difference of two consecutive squares is an odd number:
$(n+1)^2-n^2=2n+1$
the difference of two consecutive odd numbers is always $2$ thus the self-saying technique.
So when we proceed to find a square $n^2$ we sum up all odd numbers from 1 to $2n-1$ which is an arithmetic sequence of distance 2.
$n^2=\frac{n(2n)}{2}=\frac{\frac{(2n)(2n)}{2}}{2}=\frac{\frac{(2n-1)(2n)}{2}+n}{2}=\frac{\binom{2n}{2}+n}{2}$
Also we note that
$n^2=\frac{\frac{(2n)(2n)}{2}}{2}=\frac{\frac{(2n+1)(2n)}{2}-n}{2}=\frac{\binom{2n+1}{2}-n}{2}$
Summing up the two sums of two formulas $$\sum_n{\frac{\binom{2i}{2}+n}{2}}+\sum_n{\frac{\binom{2i+1}{2}-n}{2}}=\sum_n{\frac{\binom{2i}{2}+\binom{2i+1}{2}}{2}}=\sum_{2n+1}{\frac{\binom{i}{2}}{2}}=2\sum{i^2}$$ is just the sum of values in second column of pascal triangle (even and odd binomials) until $2n+1$, divided by 2.
1
1 1
1 2 |1 |
1 3 |3 | 1
1 4 |6 | 4 1
1 5 |10| 10 5 1
1 6 |15| 20 15 6 1
. . \ \
. . \ \
. . \ \
1 7 21 \35\ . . . .
Summing up the second column will result in the next row/column situated pascal value, which is known to be $\binom{2n+2}{3}$
So ... $$2\sum_n{i^2}=\frac{\binom{2n+2}{3}}{2}=\frac{\frac{2n(2n+1)(2n+2)}{6}}{2}$$
|
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|
How to find $\max\int_{a}^{b}\left (\frac{3}{4}-x-x^2 \right )\,dx$ over all possible values of $a$ and $b$, $(aI tried finding the maxima of $f(x)=\frac{3}{4}-x-x^2$ by taking the derivative and so on and use the fact that $\displaystyle\int_{a}^{b}f(x)\,dx \leq M(b-a)$ where $M$ is the global maximum, but then the maximum value depends on the values of $a$ and $b$.
|
Given $$f(a,b) = \int_{a}^{b}\left(\frac{3}{4}-x-x^2\right)dx$$
Do the integration to get:
$$f(a,b) = \frac{a^3}{3}+\frac{a^2}{2}-\frac{3 a}{4}-\frac{b^3}{3}-\frac{b^2}{2}+\frac{3 b}{4} $$
Diff $f(a,b)$ with w.r.t $a$ and set to $0$
$$a^2+a-\frac{3}{4} =0$$
Solve. We get these $2$ critical points:
$$a = - \frac{3}{2} \ \ a=\frac{1}{2} $$
Diff $f(a,b)$ with wrt $b$ and set to $0$
$$-b^2-b+\frac{3}{4} = 0$$
We get these $2$ critical points:
$$ b = - \frac{3}{2} \ \ b=\frac{1}{2} $$
only $\displaystyle a = -\frac{3}{2} \ \ b = \frac{1}{2} $ need to be tried. $( a < b )$
$$f\left( - \frac{3}{2} , \frac{1}{2}\right) = \frac{4}{3} .$$
|
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|
Show that $1^7+7^7+13^7+19^7+23^7\equiv{0}\pmod{63}$ Show that $1^7+7^7+13^7+19^7+23^7\equiv{0}\pmod{63}$
According to Fermat's theorem:
$$1^7+7^7+13^7+19^7+23^7\equiv{1+7+13+19+23}\pmod{7}\equiv{63}\pmod{7}\equiv{0}\pmod{7}$$
Now we need to show: $1^7+7^7+13^7+19^7+23^7\equiv{0}\pmod{9}$ , but how??
|
Since OP demands a proof without Euler's theorem, below is one:
First we denote the sum as $S$ and modulo $9:$
$$S\equiv1^7+(-2)^7+4^7+1^7+5^7\pmod9.$$
Since $(-2)^3\equiv1\pmod9,$ we find that $(-2)^7\equiv-2\pmod9,$ so that $$S\equiv4^7+5^7\pmod9.$$
But $5\equiv-4\pmod9,$ so $$S\equiv4^7+(-4)^7\equiv0\pmod9.$$
Hope this helps.
|
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|
Using a Taylor polynomial to approximate $\cos(\frac14)$ with an error no more than $10^{-12}$ So the Lagrange remainder is given by:
$$R_n(x)=\frac{f^{n+1}(c)}{(n+1)!}(x-a)^{n+1}.$$
We want $\cos(\frac14)$ and we can do it around $a=0$. We know that $f^{n+1}$ is either $\pm\cos x$ or $\pm\sin x$, which are both bounded with an absolute value of $1$:
$$|\cos x|\le1,\qquad |\sin x|\le 1.$$
So we can say the following is true:
$$|R_n(x)|=\frac{f^{n+1}(c)}{(n+1)!}\left(\frac14\right)^{n+1}\le 1\cdot\frac{1}{(n+1)!}\left(\frac14\right)^{n+1} $$
And we get:
$$|R_n(x)|\le \frac{1}{(n+1)!}\left(\frac14\right)^{n+1} $$
Let's try $n=6$:
$$\frac{1}{10^{12}}\le \frac{1}{7!}\left(\frac14\right)^{7} =\frac{1}{5040}\cdot\frac{1}{16384}=\frac{1}{82575360} $$
$$\frac{1}{10^{12}}\le \frac{1}{82575360} $$
Is $n=6$ enough? Or do I have to continue?
|
You want the error to be less than $10^{-12}$, so you want $|R_n(x)|<\frac{1}{10^{12}}$. You know that
$$|R_n(x)|\leq\frac{1}{(n+1)!}\left(\frac{1}{4}\right)^n,$$
so you want to find $n$ such that
$$\frac{1}{(n+1)!}\left(\frac{1}{4}\right)^n<\frac{1}{10^{12}},$$
so that you can combine the inequalities to get $|R_n(x)|<\tfrac{1}{10^{12}}$. The bound decreases as $n$ increases, and you have already calculated the bound for $n=7$, which was too big. So try $n=8$, or $n=9$, or...
|
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If $\sin \theta+\cos\theta+\tan\theta+\cot\theta+\sec\theta+\csc\theta=7$, then $\sin 2\theta$ is a root of $x^2 -44x +36=0$ My own bonafide attempt.
$$ 0<\theta<\pi/2$$
and $$\sin\theta+\cos\theta+\tan\theta+\cot\theta+\sec\theta+\csc\theta=7$$
then show that $\sin 2\theta$ is a root of the equation $$x^2 -44x +36=0$$
I tried to use the above given equation of all the trigonometric ratios, though I ended up with an expression of $\sin\theta\cos\theta$.
But this too is in the form of $\sin\theta$ and $\cos\theta$ which was $\sin\theta\cos\theta=\frac{1}{6-\sin\theta-\cos\theta}$.
When I put that in the quadratic equation, it would again transform into the form of $\sin\theta$ and $\cos\theta$, and hence at last I couldn't prove the thing.
Even hints would work, as I would like to solve the question myself.
|
$$\sin(\theta)+\cos(\theta)+\tan(\theta)+\cot(\theta)+\sec(\theta)+\csc(\theta)=7$$
$$\sin(\theta)+\cos(\theta)+\frac{\sin(\theta)}{\cos(\theta)}+\frac{\cos(\theta)}{\sin(\theta)}+\frac{1}{\cos(\theta)}+\frac{1}{\sin(\theta)}=7$$
$$\sin^2\theta\cos\theta+\sin\theta\cos^2\theta+\sin^2\theta+\cos^2\theta+\sin\theta+\cos\theta=7\sin\theta\cos\theta$$
Let $\sin\theta+\cos\theta=u; \sin\theta\cos\theta=v$
$$uv+1+u=7v$$
$$u(1+v)=7v-1$$
$$u=\frac{7v-1}{v+1}$$
$$u^2=\left(\frac{7v-1}{v+1}\right)^2$$
$u^2=(\sin\theta+\cos\theta)^2=1+2\sin\theta\cos\theta=1+2v$
$$1+2v=\left(\frac{7v-1}{v+1}\right)^2$$
where $v=\sin\theta\cos\theta=\frac12 \sin2\theta$
Let $\sin2\theta=x$. Then $$1+x=\left(\frac{\frac72x-1}{\frac12x+1}\right)^2$$
$$1+x=\left(\frac{7x-2}{x+2}\right)^2$$
$$\color{red}{x^2-44x+36=0}$$
|
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Integration - Find the area under the curve. Im not sure how to do this at all. Need help walking through the steps on how to get the answer. So below is the question. I need any and all help.
Find the area under the curve:
$$ y=2\sin(3x-\pi/3)$$
between $x=0$ and $x=\pi/2$, and give your answer to $4$ decimal places.
|
I assume by "area under the curve," you mean area between the graph of $y=f(x)$ and the $x$-axis. The area under the curve between $x=x_1$ and $x=x_2$ is the definite integral of the function $f(x)$ with bounds $x_1$ and $x_2$.
Here is a picture of this area from Wolfram Alpha (note that when the function goes below the $x$-axis, that area counts as negative):
The reason that this is the area in question is that this integral is defined as the limit as $\Delta x \rightarrow 0$ of the area under the curve approximated by rectangles of width $\Delta x$ and height $f(x)$ (the smaller $\Delta x$, the better the approximation).
In this case, the requested area is: $$
\begin{align}
&\int_{0}^{\frac{\pi}{2}}{2 \sin (3x - \dfrac{\pi}{3}) \ dx} \\
&= -\dfrac{2}{3} \cos (3x - \dfrac{\pi}{3}) |_{0}^{\frac{\pi}{2}} \\
&= -\dfrac{2}{3} \cos (\dfrac{3\pi}{2} - \dfrac{\pi}{3}) + \dfrac{2}{3} \cos (0 - \dfrac{\pi}{3}) \\
&= -\dfrac{2}{3} \cos (\dfrac{7\pi}{6}) + \dfrac{2}{3} \cos ( - \dfrac{\pi}{3}) \\
&\approx -\dfrac{2}{3} \cdot (-0.8660) + \dfrac{2}{3} \cdot(\dfrac{1}{2}) \\
&\approx 0.9107
\end{align}
$$
|
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For $f(x)=\frac {1}{x}$ and $g(x)=\sqrt{x-4}$, find the domain of the composite function $g\circ f(x)$.
For $f(x)=\frac {1}{x}$ and $g(x)=\sqrt{x-4}$, find the domain of the composite function $g\circ f(x)$.
My Attempt
Here, $$f(x)=\frac {1}{x}$$
$$g(x)=\sqrt{x-4}$$
Now,
$$g\circ f(x)=\sqrt{\frac {1-4x}{x}}$$
How can I proceed further?
|
We have $$g\circ f(x)=\sqrt{f(x)-4}=\sqrt{\frac{1}{x}-4}$$
This function is defined when $x\neq 0$ because of the $\frac{1}{x}$ and when $\frac{1}{x}-4\geq 0$ because of the $\sqrt{\quad}$. Now, if $x<0$ then $\frac{1}{x}<0$ and so $\frac{1}{x}-4<0$. Hence, we need $x>0$. Finally if $x>0$, then
$$\frac{1}{x}-4\geq 0 \iff \frac{1}{x}\geq 4 \iff 1 \geq 4x \iff \frac{1}{4}\geq x.$$
This finally shows that the domain of $g\circ f$ is $D=(0,\frac{1}{4}]$.
|
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First integrals for solving system of ODEs Assume a problem $$
\begin{cases}
\frac{\mathrm{dx}}{\mathrm{dt}} = \frac{y}{x-y}, \\[2ex]
\frac{\mathrm{dy}}{\mathrm{dt}} = \frac{x}{x-y}.
\end{cases}$$
Additionally, $x = x(t)$ and $y=y(t)$.
Attempt
Multiply the first equation by $x$ and the second one by $y$. Subtract the latter from the former.$$xx'-yy'=0$$
Integrate.
$$\frac{x^2}{2}-\frac{y^2}{2}=C_0$$
Next, let us simply substract the two original equations from one another.
$$\frac{\mathrm{d(x-y)}}{\mathrm{dt}}=-1$$
Multiply both sides by $\mathrm{dt}$ and integrate.
$$x-y=-t+C$$
Therefore, we have reached the solutions
$$
\begin{cases}
\frac{x^2}{2}-\frac{y^2}{2}=C_0, \\[2ex]
x-y=-t+C
\end{cases}\Rightarrow
\begin{cases}
x^2-y^2+C_1=0=\Psi_1(x,y) + C_1, \\[2ex]
x-y+t+C_2=0=\Psi_2(x,y) + C_2.
\end{cases}$$
Question
*
*How can I show that $\Psi_1(x,y)$ and $\Psi_2(x,y)$ are both first integrals?
*How would one demonstrate that these are independent first integrals (assuming 1. stands)?
*Is going through points 1 and 2 sufficient to conclude that these form the general solution of the original system?
|
Don't forget $x^2-y^2=(x+y)(x-y)$,
\begin{align*}
\frac{(x+y)(C-t)}{2} &= C_{0} \\
x+y &= \frac{2C_{0}}{C-t}
\end{align*}
Hence,
$$\left \{
\begin{align*}
x &= \frac{C_{0}}{C-t}+\frac{C-t}{2} \\
y &= \frac{C_{0}}{C-t}+\frac{t-C}{2} \\
\end{align*} \right.$$
|
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Prove $\sum_{n=0}^{\infty}{2^n(n^2-n\pi+1)(n^2+n-1)\over (2n+1)(2n+3){2n\choose n}}=1$ $$S=\sum_{n=0}^{\infty}{2^n(n^2-n\pi+1)(n^2+n-1)\over (2n+1)(2n+3){2n\choose n}}=1$$
Beta function
$${1\over {2n\choose n}}=n\int_{0}^{1}x^{n-1}(1-x)^ndx$$
$$\sum_{n=0}^{\infty}{2^n\over {2n\choose n}}=\int_{0}^{1}{2(1-t)\over [1-2t(1-t)]^2}dt$$
We can split out the sum
$$\sum_{n=0}^{\infty}{2^{n-1}(n^2-n\pi+1)\over (2n+1){2n\choose n}}-\sum_{n=0}^{\infty}{2^{n-1}(n^2+n-1)\over (2n+3){2n\choose n}}=1$$
With out the denominator $2n+1$ and $2n+3$ We can apply the beta function directly and integrate. I don't how to get the factors $2n+1$ and $2n+3$ with the beta function. Any hints on how to prove S?
|
Through Euler's beta function you may prove that:
$$ \forall x\in(0,4), \qquad f(x)=\sum_{n\geq 0}\frac{x^n}{(2n+1)\binom{2n}{n}}=\frac{4\arcsin\left(\frac{\sqrt{x}}{2}\right)}{\sqrt{x(4-x)}} \tag{1}$$
(have also a look at this question), then by computing $\int x^2 f(x^2)\,dx$ we get:
$$ \forall x\in(0,2),\quad g(x)=\sum_{n\geq 0}\frac{x^{2n}}{(2n+1)(2n+3)\binom{2n}{n}}=\frac{4 \left(x \sqrt{4-x^2}-(4-x^2)\arcsin\frac{x}{2}\right)}{x^3 \sqrt{4-x^2}}\tag{2}$$
that is more than enough to prove your claim.
|
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Find the equation of the sphere which touches the sphere $x^2+y^2+z^2+2x-6y+1=0$ at $(1,2,-2)$ and pass through $(1,-1,0)$ Find the equation of the sphere which touches the sphere $x^2+y^2+z^2+2x-6y+1=0$ at $(1,2,-2)$ and pass through $(1,-1,0)$
My Attempt:
Let the equation of the sphere be $x^2+y^2+z^2+2ux+2vy+2wz+d=0$.As this sphere touches the sphere $x^2+y^2+z^2+2x-6y+1=0$ at $(1,2,-2)$.
So $1+4+4+2u+4v-4w+d=0\implies2u+4v-4w+d=-9.......(1)$
Also the sphere $x^2+y^2+z^2+2ux+2vy+2wz+d=0$ passes through $(1,-1,0)$
So $1+1+2u-2v+d=0\implies2u-2v+d=-2......(2)$
I am stuck here.Please help me.
|
As sphere touch each other at $(1,2,-2)$ so distance between their centres=$r_1+r_2$ and distance between centre of sphere and point $(1,-1,0)$ is the radius of sphere(at that point we have a tangent). Now you have $4$ equations and $4$ unknowns you can now solve them
|
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Using AM-GM inequality to prove prove that $$x^4 + y^4 + z^4 \geq xyz(x+y+z)$$
This AM-GM inequalities are seriously stumping me. I'd appreciate a full proof and explanation and hints for proving other inequalities like this. Thanks.
|
The inequality is equivalent to ( by expanding RHS)
$$ x^4 + y^4 +z^4 \ge x^2yz + y^2xz + z^2xy$$
By AM-GM
$$ x^4+x^4 +y^4 +z^4 \ge 4(x^8y^4z^4)^\frac{1}{4} = 4x^2yz$$
$$ y^4+y^4 +x^4 +z^4 \ge 4(y^8x^4z^4)^\frac{1}{4} = 4y^2xz$$
$$ z^4+z^4 +x^4 +y^4 \ge 4(z^8x^4y^4)^\frac{1}{4} = 4z^2xy$$
Adding them up we get
$$4(x^4 +y^4 +z^4) \ge 4(x^2yz+y^2xz + z^2xy)$$
$$ x^4 +y^4 +z^4 \ge x^2yz+y^2xz + z^2xy $$
|
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|
USA $2011$ contest inequality problem, proving $\frac{ab+1}{(a+b)^2}+\frac{bc+1}{(b+c)^2}+\frac{ca+1}{(c+a)^2}\ge 3$, under given condition.
If $a^2+b^2+c^2+(a+b+c)^2\le4$, then $$\frac{ab+1}{(a+b)^2}+\frac{bc+1}{(b+c)^2}+\frac{ca+1}{(c+a)^2}\ge 3.$$
My attempt:
From the given criteria, one can easily obtain that $$(a+b)^2+(b+c)^2+(c+a)^2\le 4\tag{$\clubsuit$}$$
Now, I tried Cauchy-Scwartz on the sets $$\left\{\frac{\sqrt{ab+1}}{a+b},\frac{\sqrt{bc+1}}{b+c},\frac{\sqrt{ca+1}}{c+a}\right\}\text{and}\left\{\sqrt{ab+1}(a+b),\sqrt{bc+1}(b+c),\sqrt{ca+1}(c+a)\right\},$$
and got $$\frac{ab+1}{(a+b)^2}+\frac{bc+1}{(b+c)^2}+\frac{ca+1}{(c+a)^2}\ge\frac{(ab+bc+ca+3)^2}{\sum(ab+1)(a+b)^2}\\
\ge\frac{(ab+bc+ca+3)^2}{\sum ab(a+b)^2+4}[\text{see}\; (\clubsuit)\text{ marked inequality}]$$
Now, what is left is just proving $$\frac{(ab+bc+ca+3)^2}{\sum ab(a+b)^2+4}\ge 3,$$ i.e. (after some simple arithmetic), $$6ab+6bc+6ca+2a^bc+2ab^2c+2abc^2\ge3+a^3b+ab^3+b^3c+bc^3+c^3a+ca^3+a^2b^2+b^2c^2+c^2a^2.$$
Now, I got totally stuck! Even Muirhead fails to majorise the powers of LHS, so, any idea......
P.S. Though I am tagging this (algebra-precalculus), I won't mind, if someone uses calculus. Any kind of solutions are welcome, but still, I'd like precalculus more, since it is a contest problem.
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For positives $a$, $b$ and $c$ by AM-GM we obtain:
$2\sum\limits_{cyc}\frac{1+ab}{(a+b)^2}\geq\sum\limits_{cyc}\frac{a^2+b^2+c^2+ab+ac+bc+2ab}{(a+b)^2}=3+\sum\limits_{cyc}\frac{(c+a)(c+b)}{(a+b)^2}\geq3+3=6$
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|
Showing $\int_{0}^{1}{x^{2n}-x\over 1+x}\cdot{dx\over \ln{x}}=\ln\left({2\over \pi}\cdot{(2n)!!\over (2n-1)!!}\right)$ Integrate
$$I=\int_{0}^{1}{x^{2n}-x\over 1+x}\cdot{dx\over \ln{x}}=\ln\left({2\over \pi}\cdot{(2n)!!\over (2n-1)!!}\right)\tag1$$
$${x^{2n}-x\over 1+x}=\sum_{k=0}^{\infty}(-1)^k(x^{2n}-x)x^k\tag2$$
Sub $(2)$ into $(1)\rightarrow (3)$
$$I=\sum_{n=0}^{\infty}\int_{0}^{1}(x^{2n+k}-x^{k+1})\cdot{dx\over \ln{x}}\tag3$$
Frullani's theorem
$$\int_{0}^{1}(x^m-x^n)\cdot{dx\over \ln{x}}=\ln{m+1 \over n+1}\tag4$$
Apply $(4)$ to $(3)\rightarrow (5)$
$$I=\sum_{n=0}^{\infty}(-1)^n\ln\left({2n+k+1\over k+2}\right)\tag5$$
$${2\over \pi}={(2n-1)!!\over (2n)!}\prod_{k=0}^{\infty}\left({2n+k+1\over k+2}\right)^{(-1)^k}\tag6$$
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Enforce the substitution $x\to e^{-x}$ to write
$$\begin{align}
I(n)&=\int_0^1 \frac{x^{2n}-x}{1+x}\frac{1}{\log(x)}\,dx\\\\
&=\int_0^\infty \frac{e^{-x}-e^{-2nx}}{x}\frac{e^{-x}}{1+e^{-x}}\,dx\\\\
&=\sum_{k=0}^\infty (-1)^k \int_0^\infty \frac{e^{-(k+2)x}-e^{-(k+2n+1)x}}{x}\\\\
&=\sum_{k=0}^\infty (-1)^k \log\left(\frac{k+2n+1}{k+2}\right)\\\\
&=\sum_{k=1}^\infty (-1)^{k-1} \log\left(\frac{k+2n}{k+1}\right)\\\\
\end{align}$$
Now, note that we can write the partial sum
$$\begin{align}
\sum_{k=1}^{2N} (-1)^{k-1} \log\left(\frac{k+2n}{k+1}\right)&=\sum_{k=1}^N \log\left(\frac{2k-1+2n}{2k}\right)-\sum_{k=1}^N \log\left(\frac{2k+2n}{2k+1}\right)\\\\
&=\sum_{k=1}^N \log\left(\frac{2k-1+2n}{2k+2n}\right)+\sum_{k=1}^N \log\left(\frac{2k+1}{2k}\right)\\\\
&=\sum_{k=n+1}^{n+N} \log\left(\frac{2k-1}{2k}\right)+\sum_{k=1}^N \log\left(\frac{2k+1}{2k}\right)\\\\
&=\sum_{k=1}^{n+N} \log\left(\frac{2k-1}{2k}\right)+\sum_{k=1}^{n+N} \log\left(\frac{2k+1}{2k}\right)\\\\
&-\sum_{k=1}^{n} \log\left(\frac{2k-1}{2k}\right)-\sum_{k=N+1}^{n+N} \log\left(\frac{2k+1}{2k}\right)\\\\
&=\sum_{k=1}^{n+N} \log\left(\frac{(2k-1)(2k+1)}{(2k)(2k)}\right)\\\\
&+\log\left(\frac{(2n)!!}{(2n-1)!!}\right)-\sum_{k=N+1}^{n+N} \log\left(\frac{2k+1}{2k}\right)\\\\
\end{align}$$
Recalling Wallis' Product, we see that
$$\lim_{N\to \infty}\sum_{k=1}^{n+N} \log\left(\frac{(2k-1)(2k+1)}{(2k)(2k)}\right)=-\log(\pi/2)$$
And since $\lim_{N\to \infty}\sum_{k=N+1}^{n+N} \log\left(\frac{2k+1}{2k}\right)=0$, we find
$$I(n)=\log\left(\frac{2}{\pi}\frac{(2n)!!}{(2n-1)!!}\right)$$
as was to be shown!!
APPENDIX:
In THIS ANSWER, I evaluated the integral
$$J(n)=\int_0^1 \frac{x^{2n+1}-x}{1+x}\frac{1}{\log(x)}\,dx=\log\left(\frac{(2n+1)!!}{(2n)!!}\right)$$
by making use of the integral evaluated herein.
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How to put derivative of composition in Jacobian matrix? Here are two functions:
$f\left(u,v\right)=u^{2}+3v^{2}$
$g\left(x,y\right)=\begin{pmatrix} e^{x}\cos y \\ e^{x}\sin y \end{pmatrix} $
I need to make Jacobian matrix of $f\circ g$. I found derivative of their composition:
$\frac{d\left(f\circ g\right) }{d\left(x,y\right) }=2e^{2x}\cos^{2}{y}+4e^{2x}\sin{y}\cos{y}+6e^{2x}sin^{2}{y} $
How do I put that in Jacobian matrix?
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Using the chain rule instead:
\begin{align*}D(f\circ g)(x,y)& =\color{red}{Df(g(x,y))}\cdot\color{blue}{ Dg(x,y)}\\
& = \color{red}{\begin{pmatrix} 2u&6v \end{pmatrix}\circ(g(x,y))}\cdot \color{blue}{ \begin{pmatrix}e^x\cos y & -e^x\sin y \\ e^x\sin y & e^x\cos y\end{pmatrix}}\\
& =\color{red}{ \begin{pmatrix} 2e^x\cos y&6e^x\sin y \end{pmatrix}}\cdot \color{blue}{ \begin{pmatrix}e^x\cos y & -e^x\sin y \\ e^x\sin y & e^x\cos y\end{pmatrix}}\\
\phantom{asd} \\
& = \begin{pmatrix}2e^{2x}\cos^2y + 6e^{2x}\sin^2y & -2e^{2x}\cos y \sin y + 6e^{2x}\sin y\cos y \end{pmatrix}\\
\phantom{asd} \\
& = 2e^{2x}\begin{pmatrix}1 + 2\sin^2y & 2\sin y\cos y \end{pmatrix}
\end{align*}
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|
Equation involving the Jacobi symbol: $\left( \frac {-6} p \right) = 1$? I have to determine the values of $p \in \{0, \dots, 23 \}$ such that $\left( \frac {-6} p \right) = 1$.
I have that:
$$\left( \frac {-6} p \right) = \left( \frac 2 p \right) \left( \frac {-3} p \right)$$
and I know that $\left( \frac 2 p \right) = 1$ if $p \equiv 1,7 \pmod 8$ and $\left( \frac 2 p \right) = -1$ if $p \equiv 3,5 \pmod 8$, and also that $\left( \frac {-3} p \right) = 1$ if $p\equiv 1 \pmod 3$.
Using the Chinese Remainder Theorem, I find that $p\equiv 1,?,7,?\pmod{24}$.
Thanks in advance!
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I saw this the other day as well. Call this a side note, the primes $1,7 \pmod {24}$ are all represented by $x^2 + 6 y^2.$ It is easy to see that if $p = x^2 + 6 y^2$ in integers, then $(-6|p) = 1.$
$$ 1, 7, 31, 73, 79, 97, 103, 127, 151, 193,
199,
$$
the primes $5,11 \pmod {24}$ are all represented by $2x^2 + 3 y^2.$
$$ 2, 3, 5, 11, 29, 53, 59, 83, 101, 107,
131, 149, 173, 179, 197,
$$
Of course, $2x^2 + 3 y^2$ also represents $2$ and $3,$ but those divide $6.$
Meanwhile, IF $(-6|p)= 1,$ it is easy enough to construct a (positive) binary quadratic form $f(x,y) = p x^2 + Bxy + C y^2$ of discriminant $-24,$ and this form reduces to one of the two indicated forms above; that process tells us how to represent $p.$ For part of that, prove if n - natural number divide number $34x^2-42xy+13y^2$ then n is sum of two square number
Why not: if $(-6|p)= 1,$ then $(-24|p)= 1,$ we have some integer solution to $\beta^2 \equiv -24 \pmod p.$ If the original $\beta$ we found was odd, replace it by $B= p - \beta$ so that $B$ is even, and $B^2 \equiv -24 \pmod {4p}.$ That is, $B^2 = -24 + 4pC.$ Well, $B^2 - 4pC = -24,$ and the positive binary quadratic form $\langle p, B, C \rangle$ has discriminant $-24.$ The notation $\langle p, B, C \rangle$ means the form $f(x,y) = p x^2 + Bxy + C y^2$
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Find the area of $S=\{(x,y)|\rm{\exists ~}\theta,\beta,x=\sin^2{\theta}+\sin{\beta},y=\cos^2{\theta}+\cos{\beta}\}$ Let $S$ be the domain defined by $$S=\{(x,y)|\rm{\exists ~}\theta,\beta,x=\sin^2{\theta}+\sin{\beta},y=\cos^2{\theta}+\cos{\beta}\}$$
find the area of $S$
This is middle school problem,so I think it can be solved without integral methods?
$$(x-\sin^2{\theta})^2+(y-\cos^2{\theta})^2=1$$
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From $(x - \sin^2\theta)^2 + (y - \cos^2\theta)^2 = 1$ we get that $S$ is a set of circles with radius $1$ and centers on the points of $(\sin^2\theta, \cos^2\theta)$, i.e. on the segmet from point $(1,0)$ to point $(0,1)$ (see figure below).
So the area of $S$ is sum of areas of two semicircles and area of rectngle with sides lengths $2$ and $\sqrt{2}$:
$$
\frac{1}{2}\pi\cdot 1^2 + 2\cdot\sqrt{2} + \frac{1}{2}\pi\cdot 1^2 = \pi + 2\sqrt{2}.
$$
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|
Prove that $\lim\limits_{x\to \infty} a\sqrt{x+1}+b\sqrt{x+2}+c\sqrt{x+3}=0$ if and only if $ a+b+c=0$ Prove that
$$ \displaystyle \lim_{x\to\infty } \left({a\sqrt{x+1}+b\sqrt{x+2}+c\sqrt{x+3}}\right)=0$$
$$\text{if and only if}$$
$$ a+b+c=0.$$. I tried to prove that if $a+b+c=0$, the limit is $0$ first, but after getting here i got stuck
$$\lim_{x\to\infty } \left({\sqrt{x+1}\left(a+b\sqrt{1+\frac{1}{x+1}}+c\sqrt{1+\frac{2}{x+1}}\right)}\right)$$
Got here by substituting $\sqrt{x+2}$ with $\sqrt{(x+1)(1+\dfrac{1}{x+1})}$
Edit: x tends to infinity, not to 0. I transcribed wrongly.
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The limit is bounded if a+b+c = 0
verify:
$\lim_\limits{x \to \infty} \sqrt{x+k} - \sqrt {x} = 0$
$\lim_\limits{x \to \infty} a\sqrt{x+1} -a\sqrt{x}+ b\sqrt{x+2} -b\sqrt{x} + c\sqrt{x+3} -c\sqrt{x}= 0$
$\lim_\limits{x \to \infty} (a\sqrt{x+1} + b\sqrt{x+2} + c\sqrt{x+3}) = \lim_\limits{x \to \infty}(a+b+c) \sqrt{x}$
$\lim_\limits{x \to \infty}(a+b+c) \sqrt{x} = 0 \implies (a+b+c = 0)$
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|
Is function invertible? Reflection on the unit circle:
Let $E=\mathbb R ^{2} - \left\{0,0\right\} $ be perforated plane and $f: E \mapsto E$ defined by $f\left(x,y\right)=\left(\frac{x}{x^{2}+y^{2} } , \frac{y}{x^{2}+y^{2} } \right)
$
Show with Jacobian matrix that $f$ is in all points local invertible. Show that $f$ is also global invertible. Find $f^{-1}$ and explain mapping geometrically.
What I did:
I started with determinante of Jacobian matrix to show that function is local invertible, but I got the following result.
Is that enough for showing that function is local invertible? How do I do rest?
$Df=\frac{-x^{4}-y^{4}-2x^{2}y^{2} }{\left(x^{2}+y^{2} \right) ^{2} } $
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See the comment on the question for local invertibility.
Now, we prove that $f$ is invertible by showing that $f$ is one-one and onto. Given $(x_1, y_1), (x_2,y_2) \in \Bbb R^2 - \{(0,0)\}$, we have:
$$f(x_1,y_1) = f(x_2, y_2) \implies \begin{cases} \frac{x_1}{x_1 ^2 + y_1^2} = \frac{x_2}{x_2 ^2 + y_2^2} \\ \frac{y_1}{x_1 ^2 + y_1^2} = \frac{y_2}{x_2 ^2 + y_2^2}\end{cases}$$
Squaring the equations and adding them to each other, we get $x_1^2 + y_1^2 = x_2^2 + y_2^2$, which then given $x_1 = x_2$ and $y_1 = y_2$.
Next, given $(X,Y) \in \Bbb R^2 - \{(0,0)\}$, we want to find $(x,y)$ such that $f(x,y) = (X,Y)$, so we are solving for $x$ and $y$:
$$\begin{cases} X = \frac{x}{x^2 + y^2} \\ Y = \frac{y}{x^2 + y^2} \end{cases}$$
Notice then that $X^2 + Y^2 = \frac1{x^2 + y^2}$
Edit
Now, on one hand, $X = \frac{x}{x^2 + y^2}$ gives (after cross multiplying) $x = X(x^2 + y^2)$, but $x^2 + y^2 = \frac1{X^2 + Y^2}$ (by taking reciprocals in the equation right above "Edit"), therefore:
$$x = X \times \frac{1}{X^2 + Y^2} = \frac{X}{X^2 + Y^2}$$
Similarly, we get:
$$y = \frac{Y}{X^2 + Y^2}$$
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|
The solution of equation $4+6+8+10+\cdots +x =270$ is 15. The solution of equation
$4+6+8+10+\cdots +x =270$
is $x=15$.
How can I prove it?
I ve tried the geometric sequence but I cannot figure out the pattern.
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$$4+6+8+10+\cdots +x=270\\ \frac { 4+x }{ 2 } \left( \frac { x-4 }{ 2 } +1 \right) =270\\ \frac { \left( 4+x \right) }{ 2 } \frac { \left( x-2 \right) }{ 2 } =270\\ \left( x-2 \right) \left( x+4 \right) =1080\\ { x }^{ 2 }+2x-1088=0\\ x=-1\pm \sqrt { 1089 } \\ x=32$$
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problem proving: $(1+q)(1+q^2)(1+q^4)...(1+q^{{2}^{n}}) = \frac{1-q^{{2}^{n+1}}}{1-q}$ I'm trying to prove this, and it is really frustrating, because it seems a really easy problem to prove, however, I'm having a little problem with exponents:
$$(1+q)(1+q^2)(1+q^4)...(1+q^{{2}^{n}}) = \frac{1-q^{{2}^{n+1}}}{1-q}$$
Hypothesis
$F(x)=(1+q)(1+q^2)(1+q^4)...(1+q^{{2}^{x}}) = \frac{1-q^{{2}^{x+1}}}{1-q}$
The format can be a little problematic here, so, to clarify:
$(1+q^{{2}^{x}})$ = 1+(q^(2^x))
and
$1-q^{{2}^{x+1}}$ = 1 - (q^2^(x+1))
Proof:
$P1 | F(x) = (\frac{1-q^{{2}^{x+1}}}{1-q})(1+q^{{2}^{x+1}}) = \frac{1-q^{{2}^{x+2}}}{1-q} $
$P2| \frac{[(1)-(q^{{2}^{x+1}})][(1)+(q^{{2}^{x+1}})]}{(1-q)} = \frac{1-q^{{2}^{x+2}}}{1-q}$
Here I don't know if I should:
$P3 | \frac{(1-q^{{2}^{x+1}})^2}{1-q} = \frac{1-q^{{2}^{x+2}}}{1-q}| $ applying $ (a+b)(a-b) = a^2 - b^2 $
or
$P3 | \frac{1-q^{{2}^{x+1}+{2}^{x+1}}}{1-q} = \frac{1-q^{{2}^{x+2}}}{1-q} $
This arises because, suposedly, the property goes:
$(a^n)^m = a^{n*m}$
But, it seems, in this problem, the proponderance is different, kinda like this:
$a^{(n^{m})}$
Because, for example, when q=2:
$ F(0) = 1+2^{({2}^{0})} = 1 + 2^{1} = 3 $
which seems to be true, since when evaluating RHS:
$ F(0) = \frac{1-2^{{2}^{0+1}}}{1-2} = 3 $
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Define $f(q)=(1+q)(1+q^2)(1+q⁴)...(1+q^{{2}^{n}})$,
We calculate
\begin{align*}
(1-q)f(q)&= (1-q)(1+q)(1+q^2)(1+q⁴)...(1+q^{{2}^{n}}) \\
&=(1-q^2)(1+q^2)(1+q^{2^2})...(1+q^{{2}^{n}})\\
&=(1-q^{2^2})(1+q^{2^2})...(1+q^{{2}^{n}})\\
&\, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \vdots \\
&=1-q^{2^{n+1}}
\end{align*}
Now divide by $1-q$ and you get the identity.
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|
induction clarification about the step $n+1$ Suppose i need to prove that $\frac{1}{2^2}+\frac{1}{3^2}...+\frac{1}{n^2}<1-\frac{1}{n}$
So in the step of $n+1$, the right side becomes $<1-\frac{1}{n+1}$ or is it: $<1-\frac{1}{n}-\frac{1}{n+1}$? i guess it's the first one but why? i mean in the left side it's in addition to the first term $(\frac{1}{n^2})$ which is $\frac{1}{(n+1)^2}$.
*This question is mostly because i saw answers for this where they add $\frac{1}{(k+1)^2}$ to both sides, which is complex than just to write : $1-\frac{1}{n}=\frac{n-1}{n}$ and in the step of $n+1$ it just becomes: $\frac{n}{n+1}$ from there it's simple so i didn't understand why to add that term unless i'm missing something...
Thank you!
Thanks you.
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Your induction should be as follows: assume that this property holds for all $k≤n$. Then your inductive step for $n+1$ is: $$\frac{1}{3^2} +\ldots +\frac{1}{n^2}+\frac{1}{(n+1)^2}<\frac{1}{2^2}+\frac{1}{3^2}+\ldots +\frac{1}{n^2}$$ This holds as you can verify for yourself that $\frac{1}{2^2}>\frac{1}{(n+1)^2}$ for any $n≥2$. Since $\frac{1}{n}>\frac{1}{n+1}$, and by our assumption, we now have that the equation is: $$\frac{1}{3^2} +\ldots +\frac{1}{(n+1)^2}< \frac{1}{2^2}+\ldots +\frac{1}{n^2}<1-\frac{1}{n}<1-\frac{1}{n+1}$$ And our inductive step is shown.
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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.