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Set of linear equations Find eigenvalues and eigenvectors of the matrix:
$\begin{pmatrix} 1 & 0 & -2 \\ 1 & 3 & -1 \\ -1 & 0 & 2 \end{pmatrix}$
$\begin{pmatrix} 1-\lambda & 0 & -2 \\ 1 & 3-\lambda & -1 \\ -1 & 0 & 2-\lambda \end{pmatrix}$
I have found the eigenvalues, which are: 0 and 3.
For the 0-eig I have found their corresponding eigenvector (-6,1,-3)
and now I try to find the eigenvector for eig = 3:
$\begin{pmatrix} -2 & 0 & -2 \\ 1 & 0 & -1 \\ -1 & 0 & -1 \end{pmatrix}$~$\begin{pmatrix} 1 & 0 & -1 \\ 0 & 0 & 1 \end{pmatrix}$
and I am then supposed to find this solution: (0,1,0)
First, I don't understand how to get from the first matrix to the second matrix.
Secondly, how to find the solution.
May you help me please.
Inverse matrix:
First step:
$$
\left[
\begin{array}{ccc|ccc}
1&0&0&1&0&0\\
-1&2&1&0&1&0\\
0&0&1&0&0&1\\
\end{array}
\right]
$$
Second step:
$$\left[
\begin{array}{ccc|ccc}
1&0&0&1&0&0\\
0&2&1&1&1&0\\
0&0&1&0&0&1\\
\end{array}
\right]
$$
Last step:
$$\left[
\begin{array}{ccc|ccc}
1&0&0&1&0&0\\
0&1&0&1/2&1/2&-1/2\\
0&0&1&0&0&1\\
\end{array}
\right]
$$
|
First thing first. To find the eignvalues and the eigenvectors of a matrix you need to compute characteristic polynomial, which is given by the evaluation of
$$\mathcal{X}_A(\lambda) =det(\lambda - A) = det (A - \lambda) = 0$$
where $$ A=\begin{pmatrix} 1 & 0 & -2 \\ 1 & 3 & -1 \\ -1 & 0 & 2 \end{pmatrix}$$
Once you have done this you should get $\mathcal{X}(\lambda) = \lambda(\lambda -3)^2$. From this equation you can see that you have three eigenvalues: $\lambda_1 = 0, \lambda_{2,3} = 3$. Now you are interested in finding the eigenvectors corresponding to these eigenvalues and you can do that by evaluating the eigenspaces of the eigenvalues:
$$E(A,\lambda_1):=Kern(A-\lambda_1) = Kern (A)$$
To find out $Kern(A)$ you need to find the vector $v_1$ such that $A \cdot v_1 = 0$ you can do that by solving the system
$$\begin{pmatrix} 1 & 0 & -2 \\ 1 & 3 & -1 \\ -1 & 0 & 2 \end{pmatrix} \cdot v_1= \begin{pmatrix} 0 \\ 0 \\ 0\end{pmatrix}$$ You should be able to solve this system of linear equation with the help of the gaussian algorithm. In our case we subtract the first line to the second line of our matrix and this is what we get:
$$\begin{pmatrix} 1 & 0 & -2 \\ 1 & 3 & -1 \\ -1 & 0 & 2 \end{pmatrix} \cdot v_1= \begin{pmatrix} 0 \\ 0 \\ 0\end{pmatrix} \overset{II - I}{\leadsto} \begin{pmatrix} 1 & 0 & -2 \\ 0 & 3 & 1 \\ -1 & 0 & 2 \end{pmatrix} \cdot v_1= \begin{pmatrix} 0 \\ 0 \\ 0\end{pmatrix}$$
The next step of the guassian algorithm is for example this one: you add the third line to the first one and you see that the third line cancels out:
$$\begin{pmatrix} 1 & 0 & -2 \\ 0 & 3 & 1 \\ -1 & 0 & 2 \end{pmatrix} \cdot v_1= \begin{pmatrix} 0 \\ 0 \\ 0\end{pmatrix} \overset{III + I}{\leadsto} \begin{pmatrix} 1 & 0 & -2 \\ 0 & 3 & 1 \\ 0 & 0 & 0 \end{pmatrix} \cdot v_1= \begin{pmatrix} 0 \\ 0 \\ 0\end{pmatrix}$$
What you can see is that the third component of the solution vector $v_1$ is a parameter, we set it to be $k$. In fact you have infinitely many solution to this system of equation, and when you evaluate the eigenspaces of a matrix you will always have infinitely many solutions! Those solutions are in fact the span of the eigenvectors. However if you set the third component of $v_1$ to be $k$ you will see that
$$v_1 = \begin{pmatrix} 2k \\ \frac{k}{3} \\ k \end{pmatrix}= k \cdot \begin{pmatrix} 2 \\ \frac{-1}{3} \\ 1 \end{pmatrix}$$
Where $k \in \mathbb{R}\backslash \{0\}$ (you find your solution of $v_1$ by taking $k=3$). Hence $E(A,\lambda_1) = span \{ v_1 \}$
To find the other eigenvectors you do the same thing. You evaluate solution to the eigenspace
$$E(A, \lambda_{2,3}) = Kern (A-3\mathbb{1}) = Kern\left(\begin{pmatrix} -2 & 0 & -2 \\ 1 & 0 & -1 \\ -1 & 0 & -1 \end{pmatrix}\right)$$
I leave the calculation to you and the solution should be $v_2 = \begin{pmatrix}0 \\ 1 \\ 0\end{pmatrix}$, which is exactly the solution you are after. I hope that now that i've shown you how to find such solution you are also able to find it, if not just write it in the comment section below and i'll help you.
There are a few remarks that you should keep in consideration. As you can see the dimension of you matrix is $3$, but we have only $2$ eigenvectors. This is due to the fact that your matrix is not diagonalizable. In fact a matrix is diagonalizable if and only if the sum of the dimension of the eigenspaces is equal to the dimension of the matrix. Probably you will see very soon the famous Jordan Normal Form of a matrix and once you have seen it i recommend you to redo this exercise and try to understand why the matrix $A$ has only two eigenvectors and where is the third eigenvector disappeared. I hope that you understand my calculations, if not please ask me :-)
|
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|
Limits of $ \mathop {\lim }\limits_{n \to + \infty } \left( {1 + n + n\cos n} \right)^{\frac{1}{{2n + n\sin n}}} $ To calculate the limts
the First
$\mathop {\lim }\limits_{x \to 0^ + } \left( {2\sin \sqrt x + \sqrt x \sin \frac{1}{x}} \right)^x$
Suppose:
$\frac{1}{x} = n$
We find
$
\mathop {\lim }\limits_{x \to 0^ + } \left( {2\sin \sqrt x + \sqrt x \sin \frac{1}{x}} \right)^x = \mathop {\lim }\limits_{n \to + \infty } \left( {2\sin \frac{1}{{n^2 }} + \frac{1}{{n^2 }}\sin n} \right)^{\frac{1}{n}} = 0
$
The second
$\mathop {\lim }\limits_{n \to + \infty } \left( {1 + n + n\cos n} \right)^{\frac{1}{{2n + n\sin n}}}$
No idea to me
|
For $n$ sufficiently large:
$$0\le\frac{1}{2n+n\sin n}\log(1+n+n\cos n)\le\frac{\log(3n)}{n}=\frac{\log3+\log n}{n}\xrightarrow{n\to\infty}0$$
hence the desired limit is $e^0=1$.
|
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|
Simple algebra formula for which I can't find the right answer I have the formula $y + (z + 1) = \frac{1}{2} \cdot (z + 1) \cdot (z + 2)$, and I should work to $y = \frac{1}{2}\cdot z \cdot (z + 1)$.
Somebody showed me how it's done:
$y + (z + 1) = \frac{1}{2} \cdot (z + 1) \cdot (z + 2)$
$y + (z + 1) = \frac{1}{2} \cdot ((z + 1) \cdot (z + 2))$
$y + (z + 1) = \frac{1}{2} \cdot (z^2 + 3z + 2)$
$y + (z + 1) = \frac{1}{2}(z^2) + \frac{1}{2}(3z) + \frac{1}{2}(2)$
$y + (z + 1) = \frac{1}{2}(z^2) + 1\frac{1}{2}z + 1$
$y = \frac{1}{2}(z^2) + 1\frac{1}{2}z + 1$ - z - 1
$y = \frac{1}{2}(z^2) + \frac{1}{2}z$
$y = \frac{1}{2}z(z + 1)$
Great! But, my try went completely wrong, and I don't understand what I'm doing wrong:
$y + (z + 1) = \frac{1}{2} \cdot (z + 1) \cdot (z + 2)$
$y + (z + 1) = \frac{1}{2} \cdot ((z + 1) \cdot (z + 2))$
$y + (z + 1) = \frac{1}{2} \cdot (z^2 + 3z + 2)$
$y = \frac{1}{2} \cdot z^2 + 3z + 2 - z - 1$
$y = \frac{1}{2} \cdot z^2 + 2z + 1$
$y = \frac{1}{2} \cdot (z^2 + 2z + 1)$
$y = \frac{1}{2}(z^2) + \frac{1}{2}(2z) + \frac{1}{2}1$
$y = \frac{1}{2}(z^2) + z + \frac{1}{2}$
But from this last step, I can't get anywhere near $y = \frac{1}{2}z(z + 1)$, and I do not understand what I did wrong.
|
Your step $y=\frac12 z^2+3z+2-z-1$ is wrong because $k(a+b)=ka+kb\neq ka+b$. A faster way to tackle the question is this.
$y+z+1=\frac12 (z+1)(z+2)$
$y=\frac12 (z+1)(z+2)-(z+1)$
$y=(z+1)(\frac 12 z+1-1)$
$y=\frac12 z(z+1)$
|
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|
Linear algebra, power of matrices $P^{-1}AP =
\begin{pmatrix}
-1 & 1 & 0 & 0 \\
0 & -1 & 1 & 0 \\
0 & 0 & -1 & 0 \\
0 & 0 & 0 & 2 \\
\end{pmatrix}
$
with
$P=
\begin{pmatrix}
-1 & 1 & 0 & 0 \\
0 & -1 & 1 & 0 \\
0 & 0 & -1 & 0 \\
0 & 0 & 0 & 1
\end{pmatrix}
$
and $P^{-1}$ is the inverse of $P$
Find $A$, $A^{100}$?
I found $A$, $P^{-1}AP=B$ and multiplied by $P$ and $P^{-1}$. How can I find $A^{100}$? I have to use eigenvalues/vectors.
|
Note that $B=P^{-1}AP=(-I_3+N)\oplus2$. Therefore $B^{100}=P^{-1}A^{100}P=(-I_3+N)^{100}\oplus2^{100}$. Since $N^3=0$, by binomial theorem,
$$(-I+N)^{100}=(-I)^{100}+\binom{100}1(-I)^{99}N+\binom{100}2(-I)^{98}N^2.$$
|
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"language": "en",
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|
If $ab=3$ and $\frac1{a^2}+\frac1{b^2}=4,$ then $(a-b)^2=\;$? If $ab=3$ and $\frac1{a^2}+\frac1{b^2}=4$, what is the value of $(a-b)^2$? I think $a^2+b^2=36$, please confirm and is it possible to to figure out one of the variables?
|
You know that $(a-b)^2=30$. The same strategy tells you that $(a+b)^2=42$.
Thus
$$a+b=\pm \sqrt{42}\quad\text{and}\quad a-b=\pm\sqrt{30}.$$
Now by adding and subtracting, we can find $2a$ and $2b$. and hence $a$ and $b$. Note that there are $4$ combinations, though if we have found one solution $a=p$, $b=q$, the other three are $a=-p$, $b=-q$, and $a=q$, $b=p$, and $a=-q$, $b=-p$.
One of the solutions is $a=\frac{\sqrt{42}+\sqrt{30}}{2}$, $b=\frac{\sqrt{42}-\sqrt{30}}{2}$.
|
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|
Improper integral comparison test Having this integral
$$\int_1^{\infty}\frac{3x^2+2x +1}{x^3+6x^2+x+4}$$
In order to do the comparison test at some point it gets like
$$\frac{3x^2+2x +1}{x^3+6x^2+x+4}\geq \frac{1}{4x}$$
How is $\frac{1}{4x}$ found ? It doesnt seem obvious to me.
EDIT
Also how is this found
$$0\leq\frac{3x +1}{\sqrt{x^5+4x^3+5x}}\leq \frac{4}{x^{3/2}}$$
Where does 4 came from here ?Minimum value ? 3x1 +1 ? Why.
|
For example, and taking into account the fact that for large values $\;x\;$ only the summands with the largest exponent of $\;x\;$ are important (in this sense), we get
$$\frac{3x^2+2x +1}{x^3+6x^2+x+4}\ge\frac{3x^2}{x^3+6x^3+x^3+4x^3}=\frac1{4x}$$
|
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"language": "en",
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|
Solutions of $x^2-6x+\lfloor x \rfloor+7=0$
What are the roots of $x^2-6x+\lfloor x \rfloor+7=0$, where $\lfloor x \rfloor$ is the greatest integer function?
Is there some way to solve the equation without graphing?
|
$x-1 < [x] \le x$, so $x^2-5x+6<0$ while $x^2-5x+7\ge 0$, which furthur reduces to $2<x<3$. We know $[x]=2$, so $x^2-6x+9=0$ and $x=3$. But this contradicts $x<3$ and hence no solution.
|
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|
Two ways to show that $\sin x -x +\frac {x^3}{3!}-\frac {x^5}{5!}< 0$ Show that: $\large \sin x -x +\frac {x^3}{3!}-\frac {x^5}{5!}< 0$ on: $0<x<\frac {\pi}2$
I tried to solve it in two ways and got a little stuck:
One way is to use Cauchy's MVT, define $f,g$ such that $f(x)=\sin x -x +\frac {x^3}{3!}$ and $g(x)=\frac {x^5}{5!}$ so $\frac {f(x)}{g(x)}<1$ and both are continuous and differentiable on $x\in (0,\frac{\pi}2)$ so
$$\large \frac {f(x)}{g(x)}=\frac{f(x)-f(0)}{g(x)-g(0)}=\frac {f'(c)}{g'(c)}\Rightarrow
\frac{\sin x -x +\frac {x^3}{3!}} {\frac {x^5}{5!}}=\frac {\cos y -1 +\frac{y^2}{2}}{\frac{y^4}{24}}$$
Such that $y\in (0,x)$.
Now what ? I need to show that $\frac {f'(y)}{g'(y)}<1$ but I tried to input $\frac{\pi} 2$ as a value to $f'/g'$ but it was larger than 1.
The other way: With Taylor expansion.
$\large \sin x=x-\frac {x^3}{3!}+\frac {x^5}{5!}+R_{5,0}(x)\Rightarrow\\
\large \sin x-x+\frac {x^3}{3!}-\frac {x^5}{5!}=R_{5,0}(x)$
Now we're left with showing that $R_{5,0}(x)<0$
So $\large R_{5,0}(x)=\frac{-\sin(c)c^7}{7!}$ for $c\in (0,x)$, now can I say that because it's negative and $(\sin(c)c^7>0)$ for all $c\in(0,x)$, $R_{5,0}(x)<0$ ?
Also, would it be correct to say that: $\large R_{5,0}(x)=\frac{cos(c)c^6}{6!}$?
|
Repeat use of the Cauchy Mean Value theorem $3$ times we have to prove:
$\dfrac{sinx}{6x} < 1$. But since $x > 0$, we need to prove: $h(x) = sinx - 6x < 0$.
$h'(x) = cosx - 6 < 0$, so $h(x) < h(0) = 0$. Done.
|
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|
Divergence theorem integrating over a cylinder Problem: Calculate $\int \int_S \langle F,n \rangle dS$ where $S$ is the half cylinder $y^2+z^2=9$ above the $xy$-plane, and $F(x,y,z) = (x,y,z).$
My working: I did this using a surface integral and the divergence theorem and got different results. First, using a surface integral:
Write $z=h(x,y)=(9-y^2)^\frac{1}{2}$. So the normal is given by $N=(h_x, h_y, -1)$. Calculating the partial derivatives gives $h_x =0, h_y = \frac{-y}{(9-y^2)^\frac{1}{2}}$.
The unit normal is $n=\frac{(9-y^2)^\frac{1}{2}}{3}(0,\frac{-y}{(9-y^2)^\frac{1}{2}},-1) = (0, \frac{-y}{3}, \frac{-(9-y^2)^\frac{1}{2}}{3})$ .
So $\langle F,n \rangle = \frac{-y^2}{3} - \frac{9-y^2}{3}$ and $\int \int_S \langle F,n \rangle dS = \int_0^3 \int_{-3}^3 \frac{-y^2}{3} - \frac{9-y^2}{3} dy dx = 0$.
However, using the divergence theorem, we have $$\int \int_S \langle F,n \rangle dS = \int_{x=0}^{x=3} \int_{\theta=0}^{\theta=\pi} \int_{r=0}^{r=3} 3 r dr d\theta dx = \frac{81\pi}{2}$$
Where have I made a mistake?
|
For the surface $z=h(x,y)=(9-y^2)^{\frac1{2}}$ the outward unit normal vector is
$$n = \Big(0,\frac{y}{3},\frac{(9-y^2)^{\frac1{2}}}{3}\Big)$$
and the differential surface area element is
$$dS = \sqrt{1+h_x^2+h_y^2}dxdy= 3(9-y^2)^{\frac{-1}{2}}dxdy$$
so the surface integral over the top is
$$\int F \cdot n dS= \int_{0}^{3}\int_{-3}^{3}\Big[y\frac{y}{3}+(9-y^2)^{\frac1{2}}\frac{(9-y^2)^{\frac1{2}}}{3}\Big]3(9-y^2)^{\frac{-1}{2}}dxdy=3\int_{-3}^{3}9(9-y^2)^{\frac{-1}{2}}dy=27\pi$$
There is also a contribution of $\frac{27\pi}{2}$ from the surface integral over the end where $x=3$. The sum is $\frac{81\pi}{2}$ which matches the volume integral of the divergence.
|
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|
Evaluating the indefinite integral $ \int \sqrt{\tan x} ~ \mathrm{d}{x}. $ I have been having extreme difficulties with this integral. I would appreciate any and all help.
$$
\int \sqrt{\tan x} ~ \mathrm{d}{x}.
$$
|
Let $I = \int\sqrt{\tan x}\;\mathrm{d}x$ and $J = \int\sqrt{\cot x}\;\mathrm{d}x$.
Now $$\begin{align}I + J
&= \int\left(\sqrt{\tan x} + \sqrt{\cot x}\right) \;\mathrm{d}x \\
&= \sqrt{2} \int\frac{\sin x + \cos x}{\sqrt{\sin 2x}} \;\mathrm{d}x \\[5pt]
&= \sqrt{2} \int\frac{(\sin x - \cos x)'}{\sqrt{1-(\sin x - \cos x)^2}} \;\mathrm{d}x \\[5pt]
&= \sqrt{2} \sin^{-1}(\sin x - \cos x) + \mathbb{C_1} \tag{1} \\
\end{align}$$
and $$\begin{align}I - J
&= \int\left(\sqrt{\tan x} - \sqrt{\cot x}\right) \;\mathrm{d}x \\
&= \sqrt{2} \int\frac{(\sin x - \cos x)}{\sqrt{\sin 2x}} \;\mathrm{d}x \\
&= -\sqrt{2} \int\frac{(\sin x + \cos x)'}{\sqrt{(\sin x + \cos x)^2 - 1}} \;\mathrm{d}x \\
&= -\sqrt{2} \ln\left|(\sin x + \cos x) + \sqrt{(\sin x + \cos x)^2 - 1}\right| + \mathbb{C_2} \tag{2} \\
\end{align}$$
Now, adding $(1)$ and $(2)$:
$$I = \frac{1}{\sqrt{2}} \sin^{-1}(\sin x - \cos x) - \frac{1}{\sqrt{2}} \ln\left|\sin x + \cos x + \sqrt{\sin 2x} \vphantom{x^{x^x}} \right| + \mathbb{C}$$
|
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|
Applying Rouche's Theorem for an Annulus We hadn't really covered much of Rouche's Theorem within class, so I'm kind of asking a trial question to see if I understand it.
I've been asked to find out how many zeros of the function $z^5 + \frac{1}{8}\cdot e^z + 1$, counting multiplicity, that lie inside the annulus $\{ z : \frac{1}{2} < |z| < 2\}$
So, I started off by considering the circle $|z| = \frac{1}{2}$, and let $f(z) = 1$, and $g(z) = z^5 + \frac{1}{8} \cdot e^z$. Then;
$$|g(z)| = |z^5 + \frac{1}{8} \cdot e^z| = |z|^5 + \frac{1}{8} \cdot e^{|z|} = \frac{1}{2^5} + \frac{e^{\frac{1}{2}}}{8} < 1 = |f(z)|$$
Thus, there are no zeros of the function that exist within the circle $|z| \le \frac{1}{2}$.
Then, I considered the circle $|z| = 2$, and let $f(z) = z^5$, and $g(z) = \frac{1}{8} \cdot e^z + 1$. Then;
$$|g(z)| = |\frac{1}{8} \cdot e^z + 1| =\frac{1}{8} \cdot e^{|z|} + 1 =\frac{1}{8} \cdot e^{2} + 1 < 32 = 2^5 = |z|^5 = |f(z)|$$
Thus, we have five zeros within the circle $|z| \le 2$, and further, all of the zeros lie within the annulus.
Is this a proper application of Rouche's theorem, assuming that I placed "by Rouche's theorem...", etc?
|
Yes, this is a correct way to apply Rouché's theorem here.
However, the first of two applications could just as well be replaced by one-line argument with the reverse triangle inequality. Namely: when $|z|\le 1/2$,
$$\left|z^5 + \frac{1}{8}\cdot e^z + 1\right|\ge |1|-\left|z^5\right| - \left|\frac{1}{8}\cdot e^z \right| \ge 1-\frac{1}{32} - \frac{1}{8}e^{1/2}>0$$
so there are no zeros there.
You do need the theorem to count the zeros inside of $|z|=2$, as you did.
|
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|
Matrix Power Formula
Prove that for a fixed $a \in \mathbb{R}$ we have the matrix power formula for all $n \in \mathbb{Z}_+$:
$$\begin{pmatrix}a & 1\\0 & a\end{pmatrix}^n = \begin{pmatrix}a^n & na^{n-1}\\0 & a^n\end{pmatrix}$$
How would we prove this? Right now we are doing work with proofs by induction... but how would I prove this?
|
For $n = 1$ we should have (according to the rule)
$$
\begin{pmatrix}a & 1\\0 & a\end{pmatrix}^1 = \begin{pmatrix}a^1 & 1\cdot a^0\\0 & a^1\end{pmatrix}
$$
but of course
$$
\begin{pmatrix}a^1 & 1\cdot a^0\\0 & a^1\end{pmatrix} = \begin{pmatrix}a & 1\\0 & a\end{pmatrix}
$$
so it's correct (this is called the base case and it's the starting point of the induction).
Now for the induction step. Assume that the rule is correct for all $n \leq k$ for some $k$ (this is called the induction hypothesis). We need to check, on the basis of this alone, that it also holds for $n = k+1$. We know that
$$
\begin{pmatrix}a & 1\\0 & a\end{pmatrix}^{k+1} = \begin{pmatrix}a & 1\\0 & a\end{pmatrix}^k\cdot \begin{pmatrix}a & 1\\0 & a\end{pmatrix}
$$
Now, $\begin{pmatrix}a & 1\\0 & a\end{pmatrix}^k$ we can rewrite (in red below), according to the induction hypothesis (remember, we assumed that the rule was true for all $n \leq k$). Therefore we have
$$
\color{red}{\begin{pmatrix}a & 1\\0 & a\end{pmatrix}^k}\cdot \begin{pmatrix}a & 1\\0 & a\end{pmatrix} = \color{red}{\begin{pmatrix}a^k & ka^{k-1}\\0 & a^k\end{pmatrix}}\cdot \begin{pmatrix}a & 1\\0 & a\end{pmatrix}
$$
Now we just use matrix multiplication:
$$
\begin{pmatrix}a^k & ka^{k-1}\\0 & a^k\end{pmatrix}\cdot \begin{pmatrix}a & 1\\0 & a\end{pmatrix} = \begin{pmatrix}a^k \cdot a + ka^{k-1}\cdot 0 & a^k\cdot 1 + ka^{k-1}\cdot a\\ 0\cdot a + a^k \cdot 0& 0 \cdot 1 + a^k \cdot a\end{pmatrix}\\
= \begin{pmatrix}a^{k+1} & (k+1)a^k\\0 & a^{k+1}\end{pmatrix}
$$
which is exactly the rule for $n = k+1$, finishing the inductive step.
|
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Show $ f(x,y)= 2x\sin(\frac{1}{\sqrt{x^2 + y^2}}) - x\cos(\frac{1}{\sqrt{x^2 + y^2}}) \sqrt{x^2+y^2}^{-1}$ not continuous at $(0,0)$ $f(x,y) = 2x\sin(\frac{1}{\sqrt{x^2 + y^2}}) - \frac{x\cos(\frac{1}{\sqrt{x^2 + y^2}})}{ \sqrt{x^2+y^2}}
$
I'm a bit puzzled. The statement is obviously true if you plot the function.
Formal: $\quad f(a_n,b_n) \not\rightarrow (0,0)\quad for \quad (a_n,b_n) \rightarrow (0,0)$
I've tried something along the lines of $\sin(\frac{1}{a_n}) = \sin(\pi[4n+1]) = 1$, but this attempt remained fruitless.
|
Let $x =r \cos(\phi), y = r\sin(\phi)$ then you have
\begin{align*}
\lim_{(x,y)\to(0,0)}f(x,y) &= \lim_{(x,y)\to(0,0)}2x\sin(\frac{1}{\sqrt{x^2 + y^2}}) - \frac{x\cos(\frac{1}{\sqrt{x^2 + y^2}})}{ \sqrt{x^2+y^2}} \\
&= \lim_{r\to 0}2 r\cos(\phi) \sin\left(\dfrac{1}{r}\right)- \dfrac{r\cos(\phi)\cos\left(\dfrac{1}{r}\right)}{r}
\end{align*}
Now $\lim_{r\to0}2r\cos(\phi)\sin\left(\frac{1}{r}\right)$ converges to $0$ indipendently from the angle $\phi$ but $\lim_{r\to 0} \cos(\phi)\cos\left(\frac{1}{r}\right)$ goes to $0$ dependently on the angle $\phi$ hence your function is not continuous in $(0,0)$
|
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|
Problem about functional equations: $ f \left( x ^ 2 + y \right) = f ( x ) ^ 2 + \frac { f ( x y ) } { f ( x ) } $ This problem was taken from the Bulgarian selection team test for the 47th IMO and appeared in a Chinese magazine, I came across it in my own training.
http://www.math.ust.hk/excalibur/v10_n4.pdf
The problem is as follows:
Consider $ f : \mathbb R ^ * \to \mathbb R ^ * $ where $ \mathbb R ^ * $ is the set of non-zero real numbers. Find all such functions that satisfy the following functional equation for all $ x $ and $ y $ in $ \mathbb R ^ * $ with $ y \ne - x ^ 2$:
$$ f \left( x ^ 2 + y \right) = f ( x ) ^ 2 + \frac { f ( x y ) } { f ( x ) } $$
I suspect the only function is the identity, but the problem has proved itself to be too difficult for me, I've pondered about it for various hours. Any hint or solution is welcome! Thanks.
|
This might not be the optimal approach... but it attempts to avoid reasoning based on real analysis (well, apart from the very definition of irrational numbers).
Determine the value of $f(1)$
Let $f(1)=A$. Then, for any $y$ different from $0$ and $(-1)$ we have $$f(1+y)=f(1)^2 + f(y)/f(1) = A^2 + f(y)/A$$ which allows us to express the following quantities: $$\begin{eqnarray}
f(2) & = & A^2 + 1 \\
f(3) & = & A^2 + A + \frac{1}{A} \\
f(4) & = & A^2 + A + 1 + \frac{1}{A^2} \\
f(5) & = & A^2 + A + 1 + \frac{1}{A} + \frac{1}{A^3} \\
\end{eqnarray}$$
At the same time, we have $f(x^2+1)=f(x)^2+f(x)/f(x)=f(x)^2+1$. This gives us $$f(5)=f(2)^2+1$$ Equating the two expressions for $f(5)$ and substituting for $f(2)$ gives us
$$(A^2+1)^2 + 1 = A^2+A+1+\frac{1}{A} + \frac{1}{A^3}$$ which simplifies to $$\begin{eqnarray}0 & = & A^7 + A^5 - A^4 + A^3 - A^2 - 1 \\ & = & (A-1)(A^2-A+1)(A^2+A+1)^2\end{eqnarray}$$ whose only real root is $A=1$. Thus, $f(1)=1$ and we have $f(1+y)=1+f(y)$ for all $y$ other than $0$ and $(-1)$. In particular, this means that $f(k)=k$ for any positive integer $k$.
Show additive property for numbers of same sign
Now, consider real numbers $u,v$ of the same sign and these two equalities:
$$\begin{eqnarray}
f(v^2 + u/v) & = & f(v)^2 + f(u)/f(v) \\
f(v^2 + (u/v + 1)) & = & f(v)^2 + f(u+v)/f(v) \\
\end{eqnarray}$$
We can apply the equality $f(1+y)=1+f(y)$ since both $v^2$ and $u/v$ are positive to obtain
$$f(u+v)=f(u)+f(v)$$
Note: This equation is very similar to Cauchy's functional equation; a quite interesting problem of its own. The equality given in our problem is stronger than "just" plain additivity, though.
Determine values of $f$ on positive rationals
It's quite easy to see that this equality implies that $f\left(\frac{p}{q}\right)=\frac{p}{q}$ for any positive integers $p,q$; one simply expands $$p=f(p)=f\left(\frac{p}{q}\right)+f\left(\frac{p}{q}\right)+\ldots+f\left(\frac{p}{q}\right)=q\cdot f\left(\frac{p}{q}\right)$$
Show that $f$ is increasing on positive reals
For $y>0$, we can write $$\begin{eqnarray}
f(y) = f(y)+1-1 & = & f(y+1)-1 \\
& = & f\left((\sqrt{y})^2+1\right)-1 = \\
& = & f\left(\sqrt{y}\right)^2+1-1 \\
& = & f\left(\sqrt{y}\right)^2>0
\end{eqnarray}$$
Thus, for any positive reals $u>v>0$, we get $$f(u)=f(v+(u-v))=f(v)+f(u-v)>f(v)$$
Determine the values of $f$ for positive reals
Let's show that $f(x)=x$ for $x>0$. If that wasn't the case, we'd have either $f(x)>x$ or $0<f(x)<x$. In the first case, we could find rational number $z$ with $f(x)>z>x$. First inequality implies $f(x)>f(z)$ since $f(z)=z$, while the second one implies $f(z)>f(x)$, a contradiction. The other case can be treated in the same way. Thus, $f(x)=x$.
Alternate approach is to realize that the Dedekind cut corresponding to irrational $x>0$ must be the same as the Dedekind cut corresponding to $f(x)$.
Finish the proof by considering the negative reals
If $x<0$ is non-integer, we can simply use the equality $f(1+y)=1+f(y)$ to show that $f(x)=x$; just by adding $1$ until we reach a positive number. For negative integers, we can bootstrap the process by considering $f(-1)=f\left(-\frac{1}{2}\right)+f\left(-\frac{1}{2}\right)=(-1)$ and do the same by reaching $(-1)$.
The only function satisfying the given condition is $f(x)=x$ Q.E.D.
|
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|
Equations of lines tangent to an ellipse Determine the equations of the lines that are tangent to the ellipse $\displaystyle{\frac{1}{16}x^2 + \frac{1}{4}y^2 = 1}$
and pass through $(4,6)$.
I know one tangent should be $x = 4$ because it goes through $(4,6)$ and is tangent to the ellipse but I don't know how to find the other tangents. Any help is appreciated.
|
We have that $x^2+4y^2=16$, so using implicit differentiation gives $2x+8yy^{\prime}=0$ and therefore $\;\;\;\displaystyle y ^{\prime}=-\frac{x}{4y}$.
Since the slope of the line between $(x,y)$ and $(4,6)$ is given by $\frac{y-6}{x-4}$, you have that $\displaystyle -\frac{x}{4y}=\frac{y-6}{x-4}$.
This gives $-x^2+4x=4y^2-24y$, so now you can use this equation and the equation
$x^2+4y^2=16$ to find the other point of tangency.
|
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|
Solve $y^{2/3}+(y')^{2/3}=1$ other than the direct method? Is there any way to solve
$$y^{2/3}+(y')^{2/3}=1$$
other than just solving for $y'$ and then integrate?
|
$\cos^2 u + \sin^2 u = 1 \Rightarrow (\cos^3 u)^{2/3} + (\sin^3 u)^{2/3} = 1$
Let $y = \cos^3 u$ such that $y' = \sin^3 u$, if possible.
$y = \cos^3 u \Rightarrow\\
y' = -3 (\cos^2 u \sin u) u' \Rightarrow\\
-3 u'\cos^2 u \sin u = \sin^3 u \Rightarrow\\
3u' = -\tan^2 u \Rightarrow\\
\displaystyle \int -3 \cot^2 u\, du = x + c \Rightarrow\\
3 \displaystyle \int 1 - \csc^2 u\, dy = x + c \Rightarrow\\
3(u + \cot u) = x + c
$
Now, $y = \cos^3 u \Rightarrow u = \cos^{-1} y^{1/3} \Rightarrow \cot u = \dfrac{y^{1/3}}{\sqrt{1 - y^{2/3}}}$
Thus the solution is
$$\boxed{3\left( \cos^{-1} y^{1/3} + \dfrac{y^{1/3}}{\sqrt{1 - y^{2/3}}} \right) = x + c}$$
|
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|
How does Wolfram get from the first form to the second alternate form? So, I was trying to compute an integral but I couldn't actually manage getting anywhere with it in its initial form. So, I inserted the function in Wolfram Alpha and I really got a nicer form (second alternate form). But I want to understand how that was done, I really need some help here.
Input:
$\dfrac{1}{(e^t + 1)(e^t + 2)(e^t + 3)}$
Alternate forms:
$\dfrac{1}{11e^t + 6e^{2t} + e^{3t} + 6}\\
-\dfrac{1}{e^t + 2} + \dfrac{1}{2(e^t + 3)} + \dfrac{1}{2(e^t + 1)}
$
|
One trick to help deriving the $2^{nd}$ alternate form is following identity:
$$\frac{1}{(z+a)(z+b)} = \frac{1}{b-a}\left(\frac{1}{z+a} - \frac{1}{z+b}\right)$$
In particular, when $a,b$ differs by an integer, it reduces to a pattern relatively easy
to detect and apply this trick. e.g.
$$\begin{align}
\frac{1}{(z+1)(z+2)(z+3)}
&= \left(\frac{1}{z+1}-\frac{1}{z+2}\right)\frac{1}{z+3}\\
&= \frac{1}{(z+1)(z+3)} - \frac{1}{(z+2)(z+3)}\\
&= \frac12\left(\frac{1}{z+1}-\frac{1}{z+3}\right) - \left(\frac{1}{z+2} - \frac{1}
{z+3}\right)\\
&= \frac12\frac{1}{z+1} - \frac{1}{z+2} +\frac12\frac{1}{z+3}
\end{align}
$$
Substitute $z$ by $e^{t}$, you obtain the $2^{nd}$ alternate form from WA.
|
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|
finding the explicit function of a recursive sequence So I have the recursive sequence $f(0) = 0, f(n+1) = 2f(n)+ (n+1)^2$, and I'm not quite sure how to make it explicit. Substituting $n$ for $n+1$ cleans it up a little, yielding $f(n) = 2f(n-1) + n^2$, but I'm not quite sure what to do after that. It seems to be a lot harder than it looks. Any help would be appreciated.
|
Here is a generating functions solution.
Let
$$
F(x) = \sum_{n \ge 0} f(n) x^n.
$$
Then the recurrence $f(n) = 2 f(n-1) + n^2 \;\; (n \ge 1)$ becomes
$$
F(x) = 2xF(x) + \sum_{n \ge 1} n^2 x^n + f(0)
$$
or
$$
F(x) = 2xF(x) + \frac{x(1 + x)}{(1 - x)^3}.
$$
Solving,
$$
F(x) = \frac{x + x^2}{(1 - x)^3(1 - 2x)}
$$
The partial fraction decomposition of this is
$$
F(x) =
\frac{6}{1-2x}
- \frac{3}{1-x}
- \frac{1}{(1-x)^2} - \frac{2}{(1-x)^3}
$$
which gives us the general formula for $f(n)$:
$$
f(n)
= 6 \cdot 2^n - 3 - (n+1) - (n+2)(n+1)
= \boxed{6 \cdot 2^n - n^2 - 4n - 6}.
$$
You can check directly that this verifies the initial value $f(0) = 0$ and the recurrence.
|
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|
Area of a triangle in terms of areas of certain subtriangles In triangle $ABC$ , $X$ and $Y$ are points on sides $AC$ and $BC$ respectively . If $Z$ is on the segment $XY$ such that $\frac{AX}{XC} = \frac{CY}{YB} = \frac{XZ}{ZY}$ , then how to prove that the area of triangle $ABC$ is given by $[ABC]=([AXZ] ^{1/3} + [BYZ]^{1/3})^3.$
|
We can say that points $X$, $Y$, $Z$ separate respective segments $AC$, $CB$, $XY$ into sub-segments with ratios $b:r$ (think "blue : red"), where $b+r=1$.
Let's add a few perpendiculars, transfer our ratio-colors accordingly (as the reader may verify, using similar triangles), and compute the area of $\triangle AXZ$:
$$\begin{align}
|\triangle AXZ| &= \frac{1}{2}\;|\overline{AX}|\;|\overline{RZ}| \\[6pt]
&= \frac{1}{2}\cdot b\;|\overline{AC}|\cdot b\;|\overline{RS}| \\[6pt]
&= \frac{1}{2}\cdot b\;|\overline{AC}|\cdot b\cdot b\;|\overline{PB}| \\[6pt]
&= b^3\cdot \frac{1}{2}|\overline{AC}|\;|\overline{PB}| \\[6pt]
&= b^3\;|\triangle ABC| \\[8pt]
\text{Similarly, } |\triangle BYZ| &= r^3\;|\triangle ABC|
\end{align}$$
Therefore,
$$\frac{|\triangle AXZ|}{|\triangle ABC|} = b^3 \qquad\text{and}\qquad
\frac{|\triangle BYZ|}{|\triangle ABC|} = r^3$$
Recalling $b+r=1$, the result follows. $\square$
|
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|
Limit Set of All Real Numbers Not even sure how to start this one. Does anyone know how to do this?
Prove that there exists a sequence such that its limit set is the set
of all real numbers
Limit set is the set of all subsequence limits btw
|
Start the sequence with $0$.
Then write $-1,1$.
Then write $-\frac{4}{2},-\frac{3}{2},-\frac{1}{2}, \frac{1}{2},\frac{2}{2},\frac{3}{2},\frac{4}{2}$.
Then write $-\frac{16}{4},-\frac{15}{4}, -\frac{13}{4},\dots, -\frac{1}{4},\frac{1}{4},\dots, \frac{16}{4}$.
Then write $-\frac{64}{8}, \frac{63}{8},\dots, \frac{63}{8},\frac{64}{8}$, skipping $\frac{0}{8}$, though it really doesn't matter.
Then write everything from $-\frac{256}{16}$ to $\frac{256}{16}$, in steps of $\frac{1}{16}$, skipping \frac{0}{16}$.
Continue.
|
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Find length of $CD$ where $\measuredangle BCA=120^\circ$ and $CD$ is the bisector of $\measuredangle BCA$ meeting $AB$ at $D$ $ABC$ is a triangle with $BC=a,CA=b$ and $\measuredangle BCA=120^\circ$. $CD$ is the bisector of $\measuredangle BCA$ meeting $AB$ at $D$. Then the length of $CD$ is ____ ?
A)$\frac{a+b}{4}$
B)$\frac{ab}{a+b}$
C)$\frac{a^2+b^2}{2(a+b)}$
D)$\frac{a^2+ab+b^2}{3(a+b)}$
I have tried using $AD+DB=AB$ and using cosine rule on each term. However it gets lengthy.
|
Use the sine law. Let $x=BD$ and $y=AD$ and $z=CD$ which is what we want to find.Then we have in triangle $ABC$,
$$\frac{\sin 120}{x+y}=\frac{\sin A}{a}=\frac{\sin B}{b}$$ and so
$$\sin A=\frac{\sqrt{3}}{2}\frac{a}{x+y}$$
$$\sin B=\frac{\sqrt{3}}{2}\frac{b}{x+y}$$
Next in triangles $ACD$ and $BCD$ you have
$$\frac{\sin 60}{y}=\frac{\sin A}{z}$$
$$\frac{\sin 60}{x}=\frac{\sin B}{z}$$
So $$z=\frac{2}{\sqrt{3}} y \ \sin A = \frac{ay}{x+y}$$
$$z=\frac{2}{\sqrt{3}} x \sin B= \frac{bx}{x+y}$$
multiplying the first by $b$, the second by $a$ and adding,
$$(a+b)z=ab$$
So I think its B)
|
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For an odd prime $p$, $\;p =1 \pmod 4,\; \;x^2+1 = 0 \pmod p$ $$\text{ If } p \neq 2 \text{ is a prime, we know that: }$$
$$\left(\left(\frac{p-1}{2}\right)!\right)^2 \equiv (-1)^{\frac{p+1}{2}} \pmod p$$
According to this,prove that:
$$p \equiv 1 \pmod 4 \Rightarrow x^2+1\equiv 0 \pmod p \text{ has a solution}$$
$$\text{Which is the solution?}$$
Solve: $\displaystyle{ x^2+1 \equiv 0 \pmod{29} \text{ , } x^2+1 \equiv 0 \pmod{37}}$
Inversely,prove that $x^2+1 \equiv 0 \pmod p \text{ has a solution } \Rightarrow p \equiv 1 \pmod 4$
I have thought the following:
*
*$$\left(\left(\frac{p-1}{2}\right)!\right)^2-(-1)^{\frac{p+1}{2}} \equiv 0 \pmod p \Rightarrow \left(\left(\frac{p-1}{2}\right)!\right)^2+(-1)^{\frac{p+3}{2}} \equiv 0 \pmod p $$
As $p=4k+1$, $(-1)^{\frac{p+3}{2}}=(-1)^{\frac{4k+4}{2}}=1$
So,we have: $$\left(\left(\frac{p-1}{2}\right)!\right)^2+1 \equiv 0 \pmod p$$
So,we conclude that $x^2+1 \equiv 0 \pmod p \text{ has a solution , this one: } (\frac{p-1}{2})!$
Is it right or have I done something wrong?
*
*So,are the solutions of $x^2+1 \equiv 0 \pmod{29} \text{ and } x^2+1 \equiv 0 \pmod{37}$,these one: $\displaystyle{(14)! \text{ and } (18)! \text{ respectively}}$ ?
Are these the only solutions??
*How can I show the inverse?
|
That seems correct to me. The last part comes from the fact that the equation $x^2 +1 \equiv 0$ (mod $p$) has at most 2 distinct solutions because it is a degree 2 polynomial and $p$ is prime. And if $x$ is a solution, then so is $-x$.
|
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|
Infinite Sum of algebraic expression Prove that $$\sum_{i=1}^{\infty} \frac{1}{i(2i+3)} = \frac89 -\frac23\ln2$$
I tried using integration but failed miserably. Hints please.
|
You can do it by integration too.
$$\begin{aligned}
\frac{2}{3}\sum_{i=1}^{\infty} \left(\frac{1}{2i}-\frac{1}{2i+3}\right) &=\frac{2}{3}\int_0^1 \sum_{i=1}^{\infty} \left(x^{2i-1}-x^{2i+2}\right)\,dx\\
&=\frac{2}{3}\int_0^1 \left(\frac{1}{x}-x^2\right)\frac{x^2}{1-x^2}\,dx=\frac{2}{3}\int_0^1 \frac{(1-x)(1+x+x^2)x}{(1-x)(1+x)}\,dx\\
&=\frac{2}{3}\int_0^1 \left(x+\frac{x^3}{1+x}\right)\,dx=\frac{2}{3}\int_0^1 \left(x+\frac{x^3+1}{x+1}-\frac{1}{x+1}\right)\,dx\\
&=\frac{2}{3}\int_0^1 \left(x+1-x+x^2-\frac{1}{x+1}\right)\,dx \\
&= \boxed{\dfrac{8}{9}-\dfrac{2}{3}\ln 2}
\end{aligned}$$
|
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|
Square roots modulo powers of $2$ Experimentally, it seems like every $a\equiv1 \pmod 8$ has $4$ square roots mod $2^n$ for all $n \ge 3$ (i.e. solutions to $x^2\equiv a \pmod {2^n}$)
Is this true? If so, how can I prove it? If not, is it at least true that the maximum (over $a$) number of square roots of $a$ mod $2^n$ is bounded by some polynomial in $n$?
|
If $n\ge3$ and $a=8k+1$ then $a$ has exactly four distinct square roots modulo $2^n$.
Proof that there is at least one square root: use induction. The case $n=3$ is easy to check; if $x^2\equiv a\pmod{2^n}$ then $x^2=a+2^nk$ for some integer $k$; it is easy to see that $x$ is odd and so
$$(x+2^{n-1}k)^2=a+2^nk(1+x)+2^{n+1}(k^22^{n-3})\equiv a\pmod{2^{n+1}}\ .$$
Finding all square roots: since $a\equiv x^2\pmod{2^n}$ we need to solve $y^2=x^2\pmod{2^n}$. This gives
$$2^n\mid (y-x)(y+x)\ .$$
Each factor on the RHS is even; but not both are divisible by $4$ as then their sum $2y$ would be a multiple of $4$, which is impossible as $y$ is odd. So we have
$$2^{n-1}\mid y-x\quad\hbox{or}\quad 2^{n-1}\mid y+x\ .$$
This gives the four possibilities
$$y\equiv x\,,\ y\equiv x+2^{n-1}\,,\ y\equiv -x\,,\
y\equiv -x+2^{n-1}\pmod{2^n}\ ;$$
it is not hard to check that they are all different modulo $2^n$, and that they are in fact square roots of $a$.
|
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|
How to use mathematical induction with inequality? I am stuck with this question.
Given that $n$ is a positive integer where $n≥2$, prove by the method of mathematical induction that
(a) $$ \sum_{r=1}^{n-1} r^3 < \frac{n^4}{4} $$
(b) $$ \sum_{r=1}^{n} r^3 > \frac{n^4}{4} $$
|
Hint for the first question:
*
*Assume $\displaystyle\sum\limits_{r=1}^{n-1}r^3<\frac{n^4}{4}$
*Prove $\displaystyle\sum\limits_{r=1}^{n}r^3<\frac{(n+1)^4}{4}$
$\displaystyle\sum\limits_{r=1}^{n}r^3=$
$\displaystyle\sum\limits_{r=1}^{n-1}r^3+n^3<$
$\displaystyle\frac{n^4}{4}+n^3=$
$\displaystyle\frac{n^4+4n^3}{4}<$
$\displaystyle\frac{n^4+4n^3+6n^2+4n+1}{4}=$
$\displaystyle\frac{(n+1)^4}{4}$
Note that the assumption is used in order to infer $\displaystyle\sum\limits_{r=1}^{n-1}r^3+n^3<\frac{n^4}{4}+n^3$
Hint for the second question:
*
*Assume $\displaystyle\sum\limits_{r=1}^{n}r^3>\frac{n^4}{4}$
*Prove $\displaystyle\sum\limits_{r=1}^{n+1}r^3>\frac{(n+1)^4}{4}$
$\displaystyle\sum\limits_{r=1}^{n+1}r^3=$
$\displaystyle\sum\limits_{r=1}^{n}r^3+(n+1)^3>$
$\displaystyle\frac{n^4}{4}+(n+1)^3=$
$\displaystyle\frac{n^4+4n^3+12n^2+12n+4}{4}>$
$\displaystyle\frac{n^4+4n^3+6n^2+4n+1}{4}=$
$\displaystyle\frac{(n+1)^4}{4}$
Note that the assumption is used in order to infer $\displaystyle\sum\limits_{r=1}^{n}r^3+(n+1)^3>\frac{n^4}{4}+(n+1)^3$
|
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|
When is the sum of two squares the sum of two cubes When does $a^2+b^2 = c^3 +d^3$ for all integer values $(a, b, c, d) \ge 0$. I believe this only happens when: $a^2 = c^3 = e^6$ and $b^2 = d^3 = f^6$. With the following exception:
*
*$1^3+2^3 = 3^2 + 0^2$
Would that statement be correct?
Is there a general formula for when this happens?
|
Let $c=x^2,d=y^2,i=\sqrt{-1}$, then $$c^3+d^3=x^6+y^6=(x^3-(yi)^3)(x^3+(yi)^3)\\
=(x-yi)(x^2+xyi-y^2)(x+yi)(x^2-xyi-y^2)$$
Let $a+bi=(x-yi)(x^2-xyi-y^2),a-bi=(x+yi)(x^2+xyi-y^2)$, then $a^2+b^2=c^3+d^3$.
We get $a=x^3-2xy^2,b=y^3-2x^2y,c=x^2,d=y^2.$
If $a<0$ or $b<0$, we can take the absolute value.
For example, let $x=1,y=2$ we get $a=-7,b=4,c=1,d=4$ hence $7^2+4^2=1^3+4^3,$
let $x=3,y=2,$ we get $a=3,b=-28,c=9,d=4$ hence $3^2+28^2=9^3+4^3.$
|
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|
Limit with cube root conjugate Calculate the limit as $n -> \infty$ of $$n(\sqrt[3]{(1+1/n)} -1)$$
Summary of my work:
*
*Put n into the denominator in the form of $1/n$
*Multiplied by $n/n$
*Multiplied by conjugate of numerator: $[\sqrt[3]{(n^3 + n^2)^2} + 2\sqrt[3]{n^3 +n^2} + n^2]/[\sqrt[3]{(n^3 + n^2)^2} + 2\sqrt[3]{n^3 +n^2} + n^2]$
*Multiplied by $((1/n^2)/(1/n^2))$
Ended up with a final answer of 1/2
|
$$n\left(\sqrt[3]{1+\frac{1}{n}}-1\right)=n\left(\frac{\frac{1}{n}}{\sqrt[3]{(1+\frac{1}{n})^{2}}+\sqrt[3]{1+\frac{1}{n}}+1}\right)$$
which tends to $\frac{1}{3}$ as $n\to\infty$.
|
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|
Simple Differentiation Problem Involving Area Radius and Circumference A stone is dropped into a pool of water, and the area covered by the spreading ripple increases at a rate of $4 m^2 s^{-1} $.
Calculate the rate at which the circumference of the circle formed is increasing 3 seconds after the stone is dropped.
My method:
$$ \begin{align}
& \cfrac{dA}{dt}=4\\
& A = \pi \times r \times r \ \text{so} \ \cfrac{dA}{dr} = 2 \times \pi \times r \\
& C = 2 \times \pi \times r \ \text{so} \ \cfrac{dC}{dr} =2 \times \pi \\
& \cfrac{dA}{dt} = \cfrac{dA}{dr} \times \cfrac{dr}{dt} \\
& 4 = 2 \times \pi \times (3) \cfrac{ dr}{dt} \\
& \text{so} \ \cfrac{dr}{dt} = \cfrac{2}{(3 \times \pi)} \\
& \text{now} \ \cfrac{dC}{dt} = \cfrac{dC}{dr} \times \cfrac{dr}{dt} \\
& \text{so} \ \cfrac{dC}{dt} = 2 \times \pi \times \cfrac{2}{(3 \times \pi)}
\end{align} $$
I'm getting $1.3333...$
But the answer is $2.05m/s$ , I don't understand how to get it...
|
$$\begin{align} & \cfrac{dA}{dr} = 2 \pi r \\ & \cfrac{dC}{dt} = 2\pi r \\ & \cfrac{dA}{dt} = \cfrac{dA}{dr} \times \cfrac{dr}{dt} \\ & 4 = 2 \pi r \times \cfrac{dr}{dt} \\ & \cfrac{dr}{dt} = \cfrac{4}{2\pi r} \\ & \cfrac{dr}{dt} = \cfrac{2} {\pi r} \\ & r dr = \cfrac{2}{\pi } dt \\ & \int r dr = \int \cfrac{2}{\pi } dt \\ & \cfrac{r^2}{2} = \cfrac{2}{\pi} \times t \\ & r^2 = \cfrac{4}{\pi} \times t \\ & \text{Since t = 3 seconds} \ r^2 = \cfrac{4}{\pi} \times 3 = \cfrac{12}{\pi} \\ & r = \sqrt{\cfrac{12}{\pi} } \\ & \text{Since :} \ \cfrac{dC}{dt} = \cfrac{dC}{dr} \times \cfrac{dr}{dt} = 2 \pi \times \cfrac{2}{\pi r} \\ & \cfrac{dC}{dt} = \require{cancel}{2 \cancel{\pi} \times \cfrac{2}{\cancel{\pi}r}} \\ & \text{Put the value of r } \ \cfrac{dC}{dt} = \cfrac{4}{r} = \cfrac{4}{\sqrt{\cfrac{12}{\pi}}} \\ & \cfrac{dC}{dt} = \cfrac{4 \times \sqrt{\pi}}{2\sqrt{3}} \end{align} $$
This is approximately equal to : $2.046 m/s $
|
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|
If $ x^2+y^2+z^2 =1$ for $x,y,z \in \mathbb{R}$, then find maximum value of $ x^3+y^3+z^3-3xyz $. If $ x^2+y^2+z^2 =1$, for $x,y,z \in \mathbb{R}$, what is the maximum of
$ x^3+y^3+z^3-3xyz $ ?
I factorize it... Then put the maximum values of $x+y+z$ and min value of $xy+yz+zx$...
But it is wrong as they don't hold simultaneously!
Also can it be solved using partial differentiation ?
And plz provide a solution without it.!
|
Let $f(t)=\frac{t}{2}(3-t^2)$. It is straightforward to check that $f$ attains its maximum on the interval $[-\sqrt{3},\sqrt{3}]$ at $t=1$, with $f(1)=1$.
Now, if $x^2+y^2+z^2=1$ then
$$\eqalign{
x^3+y^3+z^3-3xyz&=(x+y+z)(x^2+y^2+z^2-xy-yz-zx)\cr
&=\frac{ x+y+z }{2}\left(3(x^2+y^2+z^2)-(x+y+z)^2\right)\cr
&=f(x+y+z)
}
$$
But by Cauchy-Schwarz inequality we have
$$|x+y+z|\leq\sqrt{3}\sqrt{x^2+y^2+z^2}=\sqrt{3}$$
So,
$$
x^3+y^3+z^3-3xyz\leq f(1)=1
$$
with equality if and only if $x+y+z=1$. This corresponds to the points of the circle
of intersection of the sphere of equation $x^2+y^2+z^2=1$ with the plane of equation
$x+y+z=1$.
|
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|
Proof of the inequality $\frac{a}{b+c}+\frac{b}{a+c}+\frac{c}{a+b} \geq \frac{3}{2}$ I am currently attempting to prove the following inequality
$\dfrac{a}{b+c}+\dfrac{b}{a+c}+\dfrac{c}{a+b} \geq \dfrac{3}{2}$ for all $ a,b,c>0$
My instinctive plan of attack is to use the AM/GM inequality with $x_1=\dfrac{a}{b+c}$ etc.
Using that I get this
$\dfrac{a}{b+c}+\dfrac{b}{a+c}+\dfrac{c}{a+b} \geq 3\times \sqrt[3]{\dfrac{abc}{(a+b)(b+c)(a+c)}}$
From here, I used the fact that $(a+b)(b+c)(a+c)\geq 8abc$, which can be easily proven by considering that $a+b\geq 2\sqrt{ab}$
But by using this, I get the following...
$\dfrac{a}{b+c}+\dfrac{b}{a+c}+\dfrac{c}{a+b} \geq 3\times \sqrt[3]{\dfrac{abc}{(a+b)(b+c)(a+c)}} \leq 3 \times \sqrt[3]{\dfrac{abc}{8abc}} = \dfrac{3}{2}$
Everything seems so perfect because I get the value $\dfrac{3}{2}$ as required, but this method isn't valid due to the change in direction! What is going on?
Is there a way of proving this inequality otherwise then?
|
By Cauchy-Schwarz $$\begin{align}\sum_{cyc}\frac{a}{b+c}&=\sum_{cyc}\left(\frac{a}{b+c}+1\right)-3\\
&=\sum_{cyc}a\sum_{cyc}\frac{1}{b+c}-3\\
&=\frac{1}{2}\sum_{cyc}(b+c)\sum_{cyc}\frac{1}{b+c}-3\\
&\geq\frac{1}{2}\cdot9-3=\frac{3}{2}.\end{align}$$
Done!
Another way.
By C-S $$\sum\limits_{cyc}\frac{a}{b+c}=\sum\limits_{cyc}\frac{a^2}{ab+ac}\geq\frac{(a+b+c)^2}{\sum\limits_{cyc}(ab+ac)}=\frac{(a+b+c)^2}{2(ab+ac+bc)}\geq\frac{3}{2},$$
where the last inequality it's just $\sum\limits_{cyc}(a-b)^2\geq0$.
Done!
Another way.
$$\sum\limits_{cyc}\frac{a}{b+c}-\frac{3}{2}=\sum\limits_{cyc}\left(\frac{a}{b+c}-\frac{1}{2}\right)=\sum\limits_{cyc}\frac{2a-b-c}{2(b+c)}=\sum\limits_{cyc}\frac{a-b-(c-a)}{2(b+c)}=$$
$$=\sum\limits_{cyc}(a-b)\left(\frac{1}{2(b+c)}-\frac{1}{2(a+c)}\right)=\sum\limits_{cyc}\frac{(a-b)^2}{2(a+c)(b+c)}\geq0.$$
Done!
|
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|
Proof of $\sin2x+x\sin^2x \lt\dfrac{1}{4}x^2+2$ How can be proven the following inequality?
$$\forall{x\in\mathbb{R}},\left[\sin(2x)+x\sin(x)^2\right]\lt\dfrac{1}{4}x^2+2$$
Thanks
|
For $x<0$ inequality is obvious.
$\displaystyle(\frac{x}{2}-1)^2=\frac{x^2}{4}-x+1 \geq 0$, so
$\displaystyle \frac{x^2}{4}+2 \geq x+1$, but for $x \geq 0$
$x \geq x\ (\sin x)^2$ and $1 \geq \sin 2x$
If you want to have $\left[\sin(2x)+x\sin(x)^2\right]\lt\dfrac{1}{4}x^2+2$ instead of $\left[\sin(2x)+x\sin(x)^2\right]\le\dfrac{1}{4}x^2+2$ you must have $\sin x=\pm 1$ and $\sin 2x=1$, but it's impossible.
|
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|
System of quadratic equations with three variables The problem is as follows:
For $x,y,z \in R$,
$$
\left\{
\begin{array}{l}
x^{2} -yz-8x+7=0 \\
y^{2}+z^{2}+yz-6x+6=0
\end{array}
\right.
$$
What is the domain of $x$?
One way to solve this is to use another variable. This is shown in this answer. What other ways are there to solve this?
Addition:
The answer to this question is $1 \leq x \leq 9$.
|
We have that $yz=x^2-8x+7$ and, substituting into the 2nd equation, $y^2+z^2=6x-6-yz=6x-6-(x^2-8x+7)=-x^2+14x-13$.
Since the system of equations $y^2+z^2=a$ and $yz=b$ has a solution iff $a\ge2b$ and $a\ge-2b$,
as shown below,
the given system has a solution iff
1) $-x^2+14x-13\ge2(x^2-8x+7)$ and 2) $-x^2+14x-13\ge -2(x^2-8x+7)$.
Since 1) $-x^2+14x-13\ge2(x^2-8x+7) \iff -3x^2+30x-27\ge0 \iff x^2-10x+9\le0 \iff
(x-1)(x-9)\le0 \iff 1\le x\le9$
and
2) $-x^2+14x-13\ge -2(x^2-8x+7) \iff x^2-2x+1\ge0 \iff (x-1)^2\ge0$, which is true $\;\;\;\;\;\;\;\;\;$for all $x\in\mathbb{R}$,
the values of $x$ for which this system has a solution are the values of $x$ in $[1,9]$.
$----------------------------------------$
$\Longrightarrow$ If $y^2+z^2=a$ and $yz=b$ has a solution, then $y^2-2yz+z^2=(y-z)^2\ge0\implies a\ge2b$, and $y^2+2yz+z^2=(y+z)^2\ge0\implies a\ge-2b$.
$\Longleftarrow$ If $a\ge2b$ and $a\ge-2b$, then $a\ge2|b|\ge0$ and $a^2\ge4b^2$.
If $a=0$, then $b=0$ and the system has the solution $y=0, z=0$.
If $a>0$, then $y=\big(\frac{a+\sqrt{a^2-4b^2}}{2}\big)^{1/2}$ satisfies $y^4-ay^2+b^2=0$, so $y^2-a+\frac{b^2}{y^2}=0$ and therefore letting $z=\frac{b}{y}$ gives a solution of $y^2+z^2=a$ and $yz=b$.
|
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|
Is $\mathbb{Z}[\sqrt{2} + \sqrt{3}]$ integrally closed?
Is $\mathbb{Z}[\sqrt{2} + \sqrt{3}]$ integrally closed? (Or could it have a relation to another domain like $\mathbb{Z}[\sqrt{-3}]$ does with $\mathbb{Z}[\omega]$?) Also, is it UFD? What are its units?
I have never read about this domain in any book, though i did read try to read one book with very general techniques that could theoretically be applied here.
reason I even known about this domain is because their was a question on this website last month asking if it was closed under multiplication. A lot of people fell into the trap of thinking that its quadratic but its not, myself included. Their was a whole bunch of snarky comments from people who know its quartic (is that the right term? not quadratic but only one person explained the difference -- he did a great job answering the question and explaining the misconception, and maybe from his answer one can derive the answers to the questions I'm asking today, but I'm not sure how if that's the case.
|
Interestingly enough this is not integrally closed. (spelling not corrected)
We determine the rings of integers of
$\mathbb{Z}[\sqrt{2}+\sqrt{3}]$ and $\mathbb{Q}[\sqrt{2},\sqrt{3}]$
We prove first that
$$a+b\sqrt{2}+c\sqrt{3}+d\sqrt{6} \in \mathbb{Z}[\sqrt{2}+\sqrt{3}]$$ if and only if $b$ and $c$ have the same parity and $d$ is even.
An additive basis is given by,
$\{1,\sqrt{2}+\sqrt{3}, 2\sqrt{2}, 2\sqrt{6}\}$.
We show this as follows,
$(\sqrt{2}+\sqrt{3})^2=5+2\sqrt{6}$
so $2\sqrt{6} \in\mathbb{Z}[\sqrt{2}+\sqrt{3}]$
this gives on multiplying $2\sqrt{6}$ by the original expression we have
$$6\sqrt{2} +4\sqrt{3}\in\mathbb{Z}[\sqrt{2}+\sqrt{3}]$$ from which we can easily get $2\sqrt{2}, 2\sqrt{3} \in \mathbb{Z}[\sqrt{2}+\sqrt{3}]$.
Next,
$$(\sqrt{2}+\sqrt{3})^3=11\sqrt{2}+9\sqrt{3}.$$
Now noting that $2\sqrt{3}=2(\sqrt{2}+\sqrt{3})-2\sqrt{2}$ we see that
$1,(\sqrt{2}+\sqrt{3}),(\sqrt{2}+\sqrt{3})^2,(\sqrt{2}+\sqrt{3})^3$ are all expressed in terms of these our basis.
From which we easily deduce the claim.
Now let us determine the ring of integers of $\mathbb{Q}(\sqrt{2},\sqrt{3})$
Since we have automorphisms such as $\sqrt{2},\sqrt{3} \mapsto -\sqrt{2},\sqrt{3}$
and $\sqrt{2},\sqrt{3} \mapsto \sqrt{2},-\sqrt{3}$
and $\sqrt{2},\sqrt{3} \mapsto -\sqrt{2},-\sqrt{3}$
It is easily seen that any algebraic integer has the form
$$\frac{a+b\sqrt{2}+c\sqrt{3}+d\sqrt{6}}{2}.$$
If we write this as $\frac{\alpha+\beta\sqrt{3}}{2}$ and take its norm over $\mathbb{Q}(\sqrt{2})$ we get an algebraic integer in $\mathbb{Q}(\sqrt{2})$ and we know the integers here are of the form $n+m\sqrt{2}$, $n,m \in \mathbb{Z}$.
Skipping
the details of the calculation we find $4|a^2+c^2+2(b^2+d^2)$ and $2|ab+cd$ from which a little simple number theory gives that $a$ and $c$ are even and $b$ and $d$ have the same parity.
Thus we see that the integers in $\mathbb{Q}(\sqrt{2},\sqrt{3})$
have the form
$$a+b\sqrt{2}+c\sqrt{3}+d\frac{\sqrt{2}+\sqrt{6}}{2}.$$
It remains to check that this last is in fact algebraic integer.
So let $$\alpha=\frac{\sqrt{2}+\sqrt{6}}{2}\in \mathbb{Q}[\sqrt{2},\sqrt{3}]$$ then
$$\alpha^2=2+\sqrt{3}$$ therefore
$$(\alpha-2)^2=3$$ so
$$\alpha^4-4\alpha^2+1=0$$
so $\alpha$ is an algebraic integer.
Note that this shows that $\mathbb{Z}[\sqrt{2},\sqrt{3}]$ is not integrally closed.
|
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|
Finding the conditions of (x,y,z,t) for them to belong to the span of a set of vectors So I got this math exercise, and I don't know how to go about it:
In $\mathbb{R}^4$, $S$ is the subspace spanned by the following set of vectors: $(1, 1, 1, 0) , (1, 2, 1, 1) , (2, 0, 1, 1) , (3, 0, 4, 2)$
Find the condition (or a series of conditions) that $(x, y, z, t)$ must meet to so they belong to $S$.
|
If a vector is in the subspace spanned by those four vectors, then we have:
$$\begin{pmatrix} x\\ y\\ z\\ t\\
\end{pmatrix} =\alpha \begin{pmatrix}1\\1\\1\\0\\ \end{pmatrix}+\beta \begin{pmatrix}1\\2\\1\\1\\ \end{pmatrix} + \gamma \begin{pmatrix}2\\0\\1\\1\\\end{pmatrix} + \delta \begin{pmatrix}3\\0\\4\\2\\\end{pmatrix}$$
For some coefficients $\alpha, \beta, \gamma, \delta$. This is just saying that it can be expressed as a linear combination of these vectors.
In the context of linear algebra, this can be expressed as
$$ \begin{pmatrix}1&1&2&3\\1&2&0&0\\1&1&1&4\\0&1&1&2\\ \end{pmatrix} \begin{pmatrix} \alpha\\ \beta\\ \gamma\\ \delta\\
\end{pmatrix}=\begin{pmatrix} x\\ y\\ z\\ t\\
\end{pmatrix} $$
This should allow you to find your answer.
|
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|
Finding the derivative of sinus and cosinus. Trigonometric identities How can we see that $$\sin(x+h)-\sin(x)=2\sin\left(\frac h2\right)\cos\left(x+\frac h2\right)$$
How can we see that $$\cos(x+h)-\cos(x)=-2\sin\left(\frac h2\right)\sin\left(x+\frac h2\right)$$
Do these identities have a name?
|
For any $a,b$, we have the well known "addition theorems"
$$ \sin(a+b) = \sin a \cos b + \sin b \cos a$$
and $$ \sin(a-b) = \sin a \cos b - \sin b \cos a $$
Subtracting these two equations, we get
$$ \sin(a+b) - \sin(a-b) = 2\sin b \cos a $$
For the cosine, we have
$$ \cos(a+b) = \cos a \cos b - \sin a \sin b $$
and $$ \cos(a-b) = \cos a \cos b + \sin a \sin b $$
Subtraction again
$$ \cos(a+b) - \cos(a-b) = -2\sin a\sin b$$
Now let $b = \frac h2$, $a = x+\frac h2$.
|
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|
If $\alpha, \beta \in [0,\pi]$ then the minimum value of $\sin(\frac{\alpha +\beta}{2})$ is... Problem : If $\alpha, \beta \in [0,\pi]$ then the minimum value of $\sin(\frac{\alpha +\beta}{2})$ is
a) $\frac{\sin\alpha +\sin\beta}{2}$
b) $|\sin\alpha -\sin\beta|$
c) $\frac{\cos\alpha +\cos\beta}{2}$
d) $|\cos\alpha -\cos\beta|$
Solution :
Using $$\sin(A+B)= \sin A \cos B +\sin B \cos A $$ we can write $$\sin(\frac{\alpha+ \beta}{2}) =\sin\alpha/2 \cos\beta /2 +\sin\beta/2 \cos\alpha/2 \qquad(i)$$
Now using Arithemtic Mean $\geq $ Geometric Mean for $(i)$ we get :
$$\Rightarrow \sin(\frac{\alpha +\beta}{2}) \geq 2\sqrt{\frac{1}{4}\sin\alpha \sin\beta}\quad \Rightarrow \quad \sin(\frac{\alpha +\beta}{2}) \geq \sqrt{\sin\alpha \sin\beta}$$
Now how to proceed further in this I am not getting the clue please guide thanks..
|
Using Werner Formula,
$$2\sin\frac{A+B}2\cos\frac{A-B}2=\sin A+\sin B$$
But,
$$2\sin\frac{A+B}2\cos\frac{A-B}2\le2\sin\frac{A+B}2$$ as for $\displaystyle A,B\in[0,\pi]; 0\le \cos\frac{A-B}2\le1$(why?)
$$\implies2\sin\frac{A+B}2\ge\sin A+\sin B$$
Please try Werner formula, with $\displaystyle2\sin\frac{A+B}2\sin\frac{A-B}2$
|
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|
Proving 7n+5 is never a cubic number? This is from a question that starts with:
An arithmetic progression of integers an is one in which $a_n=a_0+nd$, where $a_0$ and $d$ are integers and n takes successive values $0, 1, 2, \cdots$ Prove that if one term in the progression is the cube of an integer there will be an infinite number of such terms.
I have done this part (with help) but I am now stuck on proving that $7n+5$ can never be a cubic number (i.e. $x^3$ where $x$ is an integer) and $n$ is a positive integer. My first plan was proof by induction, trying this with both $x$ (from $x^3=7n+5$) and then $n$. Nether of these worked as I got formula that seemed impossible to solve. I have also tried manipulating $x^3=7n_1+5$ and $y^3=7n_2+5$ since if $x$ is an integer then there must also be a $y$ which this statement is also true. Any hints on where to start would be great thanks!
|
This is quite a quick thing to prove with modular arithmetic, but I will avoid that and use first principles. Suppose $m^3 = 7n+5$. There are seven cases:
*
*$m=7k$
*$m=7k+1$
*$m=7k+2$
*$m=7k+3$
*$m=7k+4$
*$m=7k+5$
*$m=7k+6$
So basically, $m=7k+i$ for $i$ in $\{0,1,\ldots,6\}$.
Now
$$\begin{align}
(7k+i)^3 &= 7n+5\\
7^3k^3+3\cdot7^2k^2i+3\cdot7ki^2+i^3&=7n+5\\
7^3k^3+3\cdot7^2k^2i+3\cdot7ki^2&=7n+5-i^3\\
7^3k^3+3\cdot7^2k^2i+3\cdot7ki^2-7n&=5-i^3\\
7\left(7^2k^3+3\cdot7k^2i+3\cdot k i^2-n\right)&=5-i^3
\end{align}$$
And this means $7$ is a divisor of $5-i^3$. If we check what $5-i^3$ is for all seven cases, we see that this is impossible. So the original premise that $m^3 = 7n+5$ cannot be true.
|
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|
Closed form for the sum: $\sum_{n=1}^{\infty}\frac{1}{n(n + 1/3)}$ I tried using partial fractions to compute the sum of the series
$$
\sum_{n=1}^{\infty}\frac{1}{n(n + 1/3)}
$$
Another technique is to turn this series into a definite integral of 0 to 1. but do not know how to do.
Thanks for any help.
|
Note that $\displaystyle\sum_{n=1}^{\infty}\dfrac{1}{n(n+1/3)} = \sum_{n=1}^{\infty}\dfrac{9}{3n(3n+1)} = 9\sum_{n=1}^{\infty}\left(\dfrac{1}{3n}-\dfrac{1}{3n+1}\right)$.
Now, let $f(x) = \displaystyle\sum_{n=1}^{\infty}\left(\dfrac{x^{3n}}{3n}-\dfrac{x^{3n+1}}{3n+1}\right)$. Clearly, $f(0) = 0$ and our sum is $9f(1)$.
Also, termwise differentiation yields $f'(x) = \displaystyle\sum_{n=1}^{\infty}\left(x^{3n-1}-x^{3n}\right) = \dfrac{x^2-x^3}{1-x^3} = \dfrac{x^2}{1+x+x^2}$.
Using partial fractions, we get $f(1) = f(0) + \displaystyle\int_{0}^{1}f'(x)\,dx$ $= \displaystyle\int_{0}^{1}\dfrac{x^2}{1+x+x^2}\,dx = \left[x - \dfrac{1}{2}\ln(1+x+x^2) - \dfrac{1}{\sqrt{3}}\tan^{-1}\left(\dfrac{2x+1}{\sqrt{3}}\right) \right]_0^1$ $= 1 - \dfrac{1}{2}\ln 3 - \dfrac{\pi}{6\sqrt{3}}$
Therefore, our sum is $9f(1) = 9 - \dfrac{9}{2}\ln 3 - \dfrac{\pi\sqrt{3}}{2}$.
|
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|
Perfect squares and modularly congruency in mod 5 There is not any perfect square $k$ such that $k \equiv 3\ (\textrm{mod}\ 5)$? Why? How can I prove it?
|
You can check the 5 cases:
If $n \equiv 0 \pmod{5} \Rightarrow n^2 \equiv 0 \pmod{5}$
If $n \equiv 1 \pmod{5} \Rightarrow n^2 \equiv 1 \pmod{5}$
If $n \equiv 2 \pmod{5} \Rightarrow n^2 \equiv 4 \pmod{5}$
If $n \equiv 3 \pmod{5} \Rightarrow n^2 \equiv 4 \pmod{5}$
If $n \equiv 4 \pmod{5} \Rightarrow n^2 \equiv 1 \pmod{5}$
|
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|
Prove that if $n$ is not divisible by $5$, then $n^4 \equiv 1 \pmod{5}$ Suppose $n$ is an integer which is not divisible by $5$. Prove that $n^4 \equiv 1 \pmod{5}$.
|
$n^{4}-1 = (n^{2}-1)(n^{2}+1)$ and (mod $5$) we have $n^{2}+1 \equiv n^{2} - 5n +6 = (n-2)(n-3),$
while also (mod $5$), we have $n^{2}-1 = (n-1)(n+1) \equiv (n-1)(n-4).$ Hence we have $(n^{4}-1) \equiv (n-1)(n-2)(n-3)(n-4) $ (mod $5$), and as long as $n$ is not divisible by $5$, one of $n-1,n-2,n-3,n-4$ is divisible by $5$.
|
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|
How to show that $(a+b)^p\le 2^p (a^p+b^p)$ If I may ask, how can we derive that $$(a+b)^p\le 2^p (a^p+b^p)$$ where $a,b,p\ge 0$ is an integer?
|
By induction.
Assume without loss of generalization that $A \leq B$
Note that $A^{n-1}(B-A) \leq B^{n-1}(B-A) \Rightarrow A^{n-1}B - A^n \leq B^n - AB^{n-1} \Rightarrow AB^{n-1} + A^{n-1}B) \leq A^n + B^n$
For $p = 1$ this is obvious. Supose true for all $p < n$.
$(A+B)^n = (A+B)^{n-1}(A+B) \leq 2^{n-1}(A^{n-1}+B^{n-1})(A+B) = 2^{n-1}(A^n + B^n + AB^{n-1} + A^{n-1}B) \leq 2^{n-1}(A^n + B^n + A^n + B^n) = 2^n(A^n + B^n)$
|
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|
If the circle $x^2+y^2+4x+22y+c=0$ bisects the circmuference of the circle $x^2+y^2-2x+8y-d=0$ the... Problem :
If the circle $x^2+y^2+4x+22y+c=0$ bisects the circmuference of the circle $x^2+y^2-2x+8y-d=0$ then c +d equals
(a) 60
(b) 50
(c) 40
(d) 30
Solution :
Equation of common chord of the circles is given by $S-S'=0$ where $S = x^2+y^2+4x+22y+c=0 ; S' = x^2+y^2-2x+8y-d=0$
$\Rightarrow 4x+22y+c-(2x+8y-d)=0$
$\Rightarrow 4x+22y+c-2x-8y+d =0$
$\Rightarrow c+d = -(2x+14y)$
Now how to get the value of c +d , please suggest thanks..
|
The Common chord has equation $6x+14y+c+d=0$ and it must pass through the center $(1,-4)$ of $S'$. Thus $c+d=50$.
|
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|
Can you show me where the "1/2" comes from? I'm struggling with a math assignment:
$$\frac12 \cos(x)·(3+2\sin(2x))−\cos(x)=0 ⇔ \cos(x)\left(\color{red}{\frac12}+\sin(2x)\right)=0$$
According to my knowledge it needs to be:
$$\frac12 \cos(x)·(3+2\sin(2x))−\cos(x)=0 ⇔ \cos(x)\left(\color{red}{\frac32}+\sin(2x)\right)=0$$
But why is it $\frac12$ and not $\frac32$?
|
\begin{align*}
\frac{1}{2} \cos(x)( 3 + 2 \sin(2x)) - \cos(x)
&= \color{blue}{\frac{3}{2} \cos(x)} + \cos(x) \sin (2x) \color{blue}{-\cos(x)} \\
&= \color{blue}{\frac{1}{2} \cos(x)} + \cos(x) \sin (2x) \\
&= \cos(x)(\frac{1}{2} + \sin(2x))
\end{align*}
|
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|
How prove this $3^{\frac{5^{2^n}-1}{2^{n+2}}}\equiv (-5)^{\frac{3^{2^n}-1}{2^{n+2}}}\pmod {2^{n+4}}$ Question:
show that:
$$3^{\frac{5^{2^n}-1}{2^{n+2}}}\equiv (-5)^{\frac{3^{2^n}-1}{2^{n+2}}}\pmod {2^{n+4}},n\geq 1$$
My idea: since
I have prove
$$5^{2^n}-1\equiv 0\pmod {2^{n+2}}$$
$$3^{2^n}-1\equiv 0\pmod {2^{n+2}}$$
so let
$$\frac{5^{2^n}-1}{2^{n+2}}=k_{1},\frac{3^{2^n}-1}{2^{n+2}}=k_{2},k_{1},k_{2}\in N$$
so we must prove
$$3^{k_{1}}-(-5)^{k_{2}}\equiv 0\pmod {2^{n+4}}$$
and How prove it?
Thank you
|
This is not even true. Just try with n=0
$3^1 = (-5)^{1/2}$ (mod 16) makes no sense.
|
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|
Find $\tan x $ if $\sin x+\cos x=\frac12$ It is given that $0 < x < 180^\circ$ and $\sin x+\cos x=\frac12$, Find $\tan x $.
I tried all identities I know but I have no idea how to proceed. Any help would be appreciated.
|
$\sin x+\cos x=\frac12$ $\Rightarrow$
$(\sin x+\cos x)^2=\frac14$ $\Rightarrow$
$\sin^2 x+2\sin x\cos x+\cos^2 x=\frac14$ $\Rightarrow$
$1+2\sin x\cos x=\frac14$ $\Rightarrow$
$2\sin x\cos x = -\frac34$ $\Rightarrow$
$\sin x\cos x=-\frac38$.
If $a=\sin x$, $b=\cos x$, we have the system of equations
$$
\begin{align}
a+b&=\frac12\\
ab&=-\frac38
\end{align}
$$
which leads to
$$a-\frac3{8a}=\frac12\\
a^2-\frac12a-\frac38=0$$
with the solutions $a=\frac{1\pm\sqrt7}4$.
So we get
$$
\sin x=\frac{1+\sqrt7}4, \qquad \cos x= \frac{1-\sqrt7}4 \qquad \Rightarrow \qquad \tan x = \frac{1+\sqrt7}{1-\sqrt7}\\
\sin x=\frac{1-\sqrt7}4, \qquad \cos x= \frac{1+\sqrt7}4 \qquad \Rightarrow \qquad \tan x = \frac{1-\sqrt7}{1+\sqrt7}
$$
You could also simplify the result:
$$\frac{1+\sqrt7}{\frac1-\sqrt7} = \frac{(1+\sqrt7)^2}{1-7} = -\frac{8+2\sqrt7}6= -\frac{4+\sqrt7}3$$
|
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|
taylor series of $\ln(1+x)$? Compute the taylor series of $\ln(1+x)$
I've first computed derivatives (up to the 4th) of ln(1+x)
$f^{'}(x)$ = $\frac{1}{1+x}$
$f^{''}(x) = \frac{-1}{(1+x)^2}$
$f^{'''}(x) = \frac{2}{(1+x)^3}$
$f^{''''}(x) = \frac{-6}{(1+x)^4}$
Therefore the series:
$\ln(1+x) = f(a) + \frac{1}{1+a}\frac{x-a}{1!} - \frac{1}{(1+a)^2}\frac{(x-a)^2}{2!} + \frac{2}{(1+a)^3}\frac{(x-a)^3}{3!} - \frac{6}{(1+a)^4}\frac{(x-a)^4}{4!} + ...$
But this doesn't seem to be correct. Can anyone please explain why this doesn't work?
The supposed correct answers are: $$\ln(1+x) = \int \left(\frac{1}{1+x}\right)dx$$
$$\ln(1+x) = \sum_{k=0}^{\infty} \int (-x)^k dx$$
|
Start with the Taylor series of the derivative
$$\frac 1{1+x}=\frac 1{(1+a)+(x-a)}=\frac 1{1+a}\,\,\frac 1{ 1+\frac {x-a}{1+a}}$$ Let $t=\frac {x-a}{1+a}$
$$\frac 1{1+t}=\sum_{n=0}^\infty (-1)^n t^n$$ So
$$\frac 1{1+x}=\frac 1{1+a}\,\,\sum_{n=0}^\infty (-1)^n \left(\frac{x-a}{a+1}\right)^n$$
Now, integrate
$$\log(1+x)=\frac 1{1+a}\,\,\sum_{n=0}^\infty (-1)^n \int\left(\frac{x-a}{a+1}\right)^n \,dx=\frac 1{1+a}\,\,\sum_{n=0}^\infty (-1)^n \frac{1+a}{n+1} \left(\frac{x-a}{a+1}\right)^{n+1}+C$$ Simplify
$$\log(1+x)=\sum_{n=0}^\infty (-1)^n \frac{1}{n+1} \left(\frac{x-a}{a+1}\right)^{n+1}+C$$ and using $x=a$, then $C=\log(1+a)$
|
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|
Prove that $\sin(12^\circ)\sin(48^\circ)\sin(54^\circ)=\frac18$ Prove that $$\sin(12^\circ)\sin(48^\circ)\sin(54^\circ)=\frac18.$$ Without using a calculator. I tried all identities I know but I have no idea how to proceed. I always get stuck on finding $\sin36^\circ$.
|
$2\sin 12\cdot \sin 48 = \cos (48 - 12) - \cos (48 + 12) = \cos 36 - \dfrac{1}{2}$. So the problem is to find $\cos 36$. Use $1 - 2x^2 = 3x - 4x^3$ to solve for $\sin 18$ ( not hard ), then find $\cos 36$, and $\sin 54$. In fact, the equation is: $4x^3 - 2x^2 - 3x + 1 = 0 \to (x-1)(4x^2 + 2x - 1) = 0$, so $\sin 18 = x = \dfrac{\sqrt{5} - 1}{4}$. You can take it from here.
|
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|
How prove $x^3+y^3+z^3-3xyz\ge C|(x-y)(y-z)(z-x)|$
let $x,y,z\ge 0$,and such
$$x^3+y^3+z^3-3xyz\ge C|(x-y)(y-z)(z-x)|$$
Find the maximum of the $C$
witout loss of we assume that $$x+y+z=1$$
I think
$$x^3+y^3+z^3-3xyz=(x+y+z)(x^2+y^2+z^2-xy-yz-xz)=(x+y+z)^3-3(xy+yz+xz)(x+y+z)=1-3(yz+xz+xy)$$
then $$(x-y)(y-z)(x-z)=?$$
so I can't，It is said $$C_{max}=\sqrt{9+6\sqrt{3}}$$
|
this might be helpful
$x^3+y^3+z^3-3xyz = \frac{1}{2}(x+y+z)((x-y)^2+(y-z)^2+(z-x)^2)$
|
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|
Is there also an other way to show the equality: $\left\lfloor \frac{n}{2}\right\rfloor + \left\lceil \frac{n}{2} \right\rceil=n$? I want to show that:
$$ \left\lfloor \frac{n}{2}\right\rfloor + \left\lceil \frac{n}{2} \right\rceil=n$$
That's what I have tried:
*
*$ \left\lfloor \frac{n}{2}\right\rfloor=\max \{ m \in \mathbb{Z}: m \leq \frac{n}{2}\}$
*$\left\lceil \frac{n}{2} \right\rceil= \min \{ m \in \mathbb{Z}: m \geq \frac{n}{2}\}$
If $n=2k,k \in \mathbb{Z}$,then: $\frac{n}{2} \mathbb{Z}$,so
$$\left\lfloor \frac{n}{2}\right\rfloor=\max \left\{ \frac{n}{2}, \frac{n-2}{2}, \dots \right\}=\frac{n}{2} \\ \left\lceil \frac{n}{2} \right\rceil=\min \left\{ \frac{n}{2}, \frac{n+2}{2}, \dots \right\}=\frac{n}{2}$$
Therefore, $ \left\lfloor \frac{n}{2}\right\rfloor + \left\lceil \frac{n}{2} \right\rceil=n$.
If $n=2k+1, k \in \mathbb{Z}$,then $\frac{n}{2} \notin \mathbb{Z}$.So:
$$\left\lfloor \frac{n}{2}\right\rfloor=\max \left\{ \frac{n-1}{2}, \frac{n-3}{2}, \dots \right\}=\frac{n-1}{2}\\ \left\lceil \frac{n}{2} \right\rceil=\min \left\{ \frac{n+1}{2}, \frac{n+3}{2}, \dots \right\}=\frac{n+1}{2}$$
Therefore, $ \lfloor \frac{n}{2}\rfloor + \lceil \frac{n}{2} \rceil=\frac{n-1}{2}+\frac{n+1}{2}=n$
Is there also an other way to show the equality or is it the only one?
|
Since the function $f(x)=\{x\}=x-\lfloor x\rfloor$ is periodic with period one, the function:
$$ g(n) = \left\lfloor\frac{n}{2}\right\rfloor+\left\lceil\frac{n}{2}\right\rceil -n$$
is periodic with period $2$. Since $g(0)=g(1)=0$, $g(n)=0$ for every $n\in\mathbb{N}$.
|
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|
The limit of $((1+x)^{1/x} - e+ ex/2)/x^2$ as $x\to 0$ $$\lim_{x\rightarrow 0}\frac{(1+x)^{1/x}-e+\frac{ex}{2}}{x^2}=\,?$$
*
*by directly substituting $x=0$ i got $\infty$
*by using L-H's rule, i got $-1/8$
the given options are
$a)\frac{24e}{11}$
$b)\frac{11e}{24}$
$c)\frac{e}{11}$
$d)\frac{e}{24}$
may be the question would be
$$\lim_{x\rightarrow 0}\frac{(1+x)^{1/x}-e+\frac{ex}{2}}{x^2}=\,?$$
|
We will use three limits derived using L'Hospital:
$$
\lim_{x\to0}\frac{1-e^x}{x}=-1\tag1
$$
and
$$
\lim_{x\to0}\frac{1+x-e^x}{x^2}=-\frac12\tag2
$$
and
$$
\lim_{x\to0}\frac{1+x+\frac12x^2-e^x}{x^3}=-\frac16\tag3
$$
Now
$$
\begin{align}
&\frac{(1+x)^{1/x}-e+\frac{ex}2}{x^2}\\[9pt]
&=\frac{e\left(1+\frac{1+x-e^x}{e^x}\right)^{1/x}-e+\frac{ex}2}{x^2}\tag{4a}\\[6pt]
&=\frac{e}{x^2}\left(\color{#C00}{1}\color{#090}{{+}\frac{1{+}x{-}e^x}{e^x}\frac{\frac1x}1}\color{#00F}{{+}\left(\frac{1{+}x{-}e^x}{e^x}\right)^2\frac{\frac1x\left(\frac1x{-}1\right)}2}{+}O\!\left(x^3\right)\color{#C00}{{-}1}\color{#090}{{+}\frac{x}2}\right)\tag{4b}\\
&=\frac{e}{x^2}\left(\color{#090}{\frac{1+x-e^x+\frac12x^2e^x}{xe^x}}\color{#00F}{+\left(\frac{1+x-e^x}{e^x}\right)^2\frac{\frac1x\left(\frac1x-1\right)}2}+O\!\left(x^3\right)\right)\tag{4c}\\
&=\frac{e}{x^2}\left(\color{#090}{\frac{1{+}x{+}\frac12x^2{-}e^x{+}\frac12x^2\left(e^x{-}1\right)}{xe^x}}\color{#00F}{+\left(\frac{1{+}x{-}e^x}{e^x}\right)^2\frac{\frac1x\left(\frac1x{-}1\right)}2}+O\!\left(x^3\right)\right)\tag{4d}\\
&=e\left(\color{#090}{\frac{1+x+\frac12x^2-e^x}{x^3e^x}+\frac{e^x-1}{2xe^x}}\color{#00F}{+\left(\frac{1+x-e^x}{x^2e^x}\right)^2\frac{1-x}2}+O(x)\right)\tag{4e}\\[3pt]
&\!\!\overset{x\to0}=e\left(\color{#090}{-\frac16+\frac12}\color{#00F}{+\frac18}\right)\tag{4f}\\[6pt]
&=\frac{11}{24}e\tag{4g}
\end{align}
$$
Explanation:
$\text{(4a)}$: $\frac{1+x}{e^x}=1+\frac{1+x-e^x}{e^x}$
$\text{(4b)}$: factor $\frac{e}{x^2}$ out and use the first $3$ terms of the Binomial Theorem
$\text{(4c)}$: cancel the red terms and combine the green terms
$\text{(4d)}$: add and subtract $\frac12x^2$ to the green numerator
$\text{(4e)}$: distribute $\frac1{x^2}$
$\text{(4f)}$: evaluate the limits using $(1)$, $(2)$, and $(3)$
$\text{(4g)}$: simplify
|
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|
Coordinate Geometry of circles; Radical Axis question
If one of the diameters of the circle $x^2+y^2-2x-6y+6=0$ is a chord to the circle with center at $(2, 1)$, then the radius of the second circle is?
Apparently the solution is $3$, with the cryptic reasoning that $r^2= 2^2+ (3−1)^2+ (2−1)^2 ⇒ r =3$.
I have no idea what this means and here is how I solved this problem. The equation of the second circle is $(x-2)^2+ (y-1)^2 = r^2$. Therefore, we can subtract the equation of first circle with that of the second circle to get the equation to the radical axis (which in this case is also the chord they are talking about).
Equation of chord (radical axis) is $2x -3y +1-r^2=0$
The condition we haven't used so far is that, the center of the first circle $(2,3)$ obtained from the equation lies on the radical axis, because it says the diameter of the first circle is a chord to the second. substituting $(2,3)$ in this equation and solving for $r$, I get $r=2$. Which doesn't match the solution.
(and common sense says if if the diameter is the chord, then the second circle likely has a radius greater than $2$)
What am I missing? Can someone explain the the cryptic solution?
|
Explanation of cryptic solution.
First let us rewrite the given equation in standard form:
$$x^2+y^2-2x-6y+6=0\implies (x-1)^2+(y-3)^2=4$$
Now we know the original circle has radius $r=2$ and is centered at $(1,3)$
Now we can make a convenient right triangle with $r$ the radius of the original circle, $d$ the distance between the centers, and $R$ the radius of the unknown circle:
Hence we have $r^2+d^2=R^2$. The distance between two points $(x_1,y_1)$ and $(x_2,y_2)$ is $\sqrt{(x_1-x_2)^2+(y_1-y_2)^2}$. Hence $d^2=(1-2)^2+(3-1)^2$, which gives us the final step
$$R^2=r^2+d^2=2^2+(1-2)^2+(3-1)^2=9\implies R=3$$
|
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|
A Fixed Field of $\mathbb{Q}(\sqrt{2},\sqrt{3},\sqrt{5})$ I was working on a review problem from Dummit and Foote and came across the following issue. It is clear that the Galois group of the splitting field for the polynomial $(x^2-2)(x^2-3)(x^2-5)$ has order $8$ and one of the automorphisms is
$$
a+b\sqrt{2}+c\sqrt{3}+d\sqrt{5} \mapsto a-b\sqrt{2}-c\sqrt{3}+d\sqrt{5}
$$
which I would think has associated fixed field $\mathbb{Q}(\sqrt{5})$. However, after completing the problem I found in the solutions that this automorphism has fixed field $\mathbb{Q}(\sqrt{6},\sqrt{5})$. Why is the $\sqrt{6}$ there? I would think that by setting
$$
a+b\sqrt{2}+c\sqrt{3}+d\sqrt{5} =a-b\sqrt{2}-c\sqrt{3}+d\sqrt{5}
$$
we find
$$
b\sqrt{2}+c\sqrt{3}=-b\sqrt{2}-c\sqrt{3}
$$
so that $b\sqrt{2}+c\sqrt{3}=0$ so that $(b/c)^2=3/2$, impossible as $b,c \in \mathbb{Q}$. How how is it that this automorphism is fixing $\mathbb{Q}(\sqrt{6})$?
|
Under this automorphism, $\sqrt{2}\mapsto -\sqrt{2}$ and $\sqrt{3}\mapsto -\sqrt{3}$, so that $\sqrt{6}\mapsto\sqrt{6}$.
|
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|
Show that $\frac{\Gamma(\frac 1 3)^2}{\Gamma(\frac 1 6)}=\frac{\sqrt {\pi}\sqrt[3] 2}{\sqrt 3}$. Show that $$\frac{\Gamma(\frac 1 3)^2}{\Gamma(\frac 1 6)}=\frac{\sqrt {\pi}\sqrt[3] 2}{\sqrt 3}$$
Since there's $\sqrt {\pi}$, I suspect I have to related it to $\Gamma(1/2)$. Please give me some idea.
|
It's the Legendre duplication formula
$$\Gamma (2z) = \frac{2^{2z-1}}{\sqrt{\pi}}\Gamma(z)\Gamma\left(z+\tfrac{1}{2}\right),$$
and Euler's reflection formula
$$\Gamma(z)\Gamma(1-z) = \frac{\pi}{\sin (\pi z)}.$$
Thus we have
$$\Gamma\left(\tfrac{1}{3}\right)^2 = \underbrace{\frac{2^{-2/3}}{\sqrt{\pi}}\Gamma\left(\tfrac{1}{6}\right)\Gamma\left(\tfrac{4}{6}\right)}_{\text{duplication } \Gamma\left(\tfrac{1}{3}\right)}\cdot\Gamma\left(\tfrac{1}{3}\right) = \frac{2^{-2/3}}{\sqrt{\pi}}\Gamma\left(\tfrac{1}{6}\right) \frac{\pi}{\sin \frac{\pi}{3}} = \frac{2^{1/3}\sqrt{\pi}}{\sqrt{3}} \cdot\Gamma\left(\tfrac{1}{6}\right).$$
|
{
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|
Line parallel to a plane and have 45 degrees between another I need to find a direction vector for a line parallel to a plane $x+y+z = 0$ and that have $45$ degrees with the plane $x-y = 0$
So, i've assumed the vector $\vec V_r = (a,b,c)$ and since it is parallel to the first plane, the product:
$$(a,b,c)\cdot(1,1,1) = 0$$
(where $(1,1,1)$ is the vector normal to the first plane).
And also, using the formula for the angle between a line and a plane, where $n$ is the normal vector for the second plane:
$$\sin(\alpha) = \frac{|\vec V_r\cdot\vec n|}{\|\vec V_r\| \|\vec n\|}$$
so:
$$\sin\Bigl(\frac{\pi}{4}\Bigr) = \frac{\bigl|(a,b,c)\cdot(1,-1,0)\bigr|}{\sqrt{a^2+b^2+c^2}\sqrt{2}}$$
Then I end up with two equations:
$$a + b + c = 0$$
$$\frac{1}{\sqrt{2}} = \frac {|a-b|}{\sqrt{a^2+b^2+c^2}\sqrt{2}}$$
But i'm not able to solve for $(a,b,c)$. Could somebody help me? Thanks :)
|
After "cross multiplying" and canceling the factor of $\sqrt{2}$ on each side, we have
$$
\sqrt{a^2 + b^2 + c^2} = |a - b|.
$$
Now, square both sides and remove $a^2$ and $b^2$ terms that appear on both sides, yielding
$$
c^2 = -2ab.
$$
From the original equation, we know that $a + b + c = 0$, so
$$
b = -a - c.
$$
At this point, we can make the simplifying assumption that $a = 1$. Why? (Hover to reveal the answer after you've tried to answer this yourself.)
As long as $a \ne 0$ and we're looking for a direction vector, we may choose the scale of one coordinate. How do we know that $a \ne 0$? If it were true that $a = 0$, then the quadratic equation would give $c^2 = 0$ and so $c = 0$. From there, the linear equation would give $b = 0$. Now, $(a, b, c) = (0, 0, 0)$. No good.
Substitute $b = -1 - c$ into $c^2 = -2b$, producing $c^2 = 2 + 2c$ or $(c - 1)^2 = 3$. Therefore,
$$
c = 1 \pm \sqrt{3}.
$$
Now,
$$
b = -1 - (1 \pm \sqrt{3}) = -2 \mp \sqrt{3}.
$$
Putting this all together, their are two possible direction vectors:
$$
\vec V_r =
\begin{bmatrix}
a \\ b \\ c
\end{bmatrix}
=
\begin{bmatrix}
1 \\ -2 \mp \sqrt{3} \\ 1 \pm \sqrt{3}
\end{bmatrix}.
$$
|
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|
Show that the inequality holds $\frac{1}{n}+\frac{1}{n+1}+...+\frac{1}{2n}\ge\frac{7}{12}$ We have to show that:
$\displaystyle\frac{1}{n}+\frac{1}{n+1}+...+\frac{1}{2n}\ge\frac{7}{12}$
To be honest I don't have idea how to deal with it. I only suspect there will be need to consider two cases for $n=2k$ and $n=2k+1$
|
Hint By the monotonicity of $x \mapsto \frac{1}{x}$, we have
$$\frac{1}{n} + \dots + \frac{1}{2n} = \sum_{k=n}^{2n} \frac{1}{k} \geq \int_n^{2n} \frac{1}{x} \, dx.$$
A direct calculation of the latter integral yields an even sharper bound:
$$\frac{1}{n} + \dots + \frac{1}{2n} \geq \ln 2 > \frac{7}{12}.$$
For the last inequality note that it follows from $e \leq 3$ that
$$e^{3/5} \leq 3^{3/5} <2;$$
hence $\ln 2>\frac{3}{5} > \frac{7}{12}$.
|
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|
How do I go from this $\frac{x^2-3}{x^2+1}$ to $1-\frac{4}{x^2+1}$? So I am doing $\int\frac{x^2-3}{x^2+1}dx$ and on wolfram alpha it says the first step is to do "long division" and goes from $\frac{x^2-3}{x^2+1}$ to $1-\frac{4}{x^2+1}$. That made the integral much easier, so how would I go about doing that in a clear manner? Thanks in advance for the help!
|
$$
\begin{array}{ccccccccc}
& & 1 \\ \\
x^2 + 1 & ) & x^2 & - & 3 \\
& & x^2 & + & 1 \\ \\
& & & & -4 & \longleftarrow\text{remainder}
\end{array}
$$
The quotient is $1$ and the remainder is $-4$.
So $\displaystyle \frac{x^2-3}{x^2+1} = 1 - \frac{4}{x^2+1}$
|
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|
How to prove this $(n+1)^n < n^{n+1}$ for $\space n \ge 3$ I'm having some more trouble with induction
I know how to prove this using $\ln$, but I need to use induction only.
prove that:
$(n+1)^n < n^{n+1}$
for any $ n\ge 3$
|
Hypothesis
\[ n^{n+1}>(n+1)^{n},\ \mbox{for}\ n\ge 3\]
Basis
\[ 3^{3+1} > (3+1)^{3} \]
\[ 3^{4} > 4^{3} \]
\[ 81 > 64\]
Induction
\[
(n+1)^{n+2}=n^{n+1}\left[\frac{(n+1)^{n+2}}{n^{n+1}}\right]
\]
Now via the induction hypothesis, we have
\[
n^{n+1}\left[\frac{(n+1)^{n+2}}{n^{n+1}}\right] > (n+1)^{n}\left[\frac{(n+1)^{n+2}}{n^{n+1}}\right]
\]
$$
\begin{align}
(n+1)^{n}\left[\frac{(n+1)^{n+2}}{n^{n+1}}\right] &= (n+2)^{n+1}\left[\frac{(n+1)^{2n+2}}{n^{n+1}(n+2)^{n+1}}\right]\\&= (n+2)^{n+1}\left[\frac{(n^{2}+2n+1)^{n+1}}{(n^{2}+2n)^{n+1}}\right]
\end{align}
$$
Since $n$ is positive and the numerator is larger than the denominator, then the fraction is greater than $1$. Hence,
\[
(n+2)^{n+1}\left[\frac{(n^{2}+2n+1)^{n+1}}{(n^{2}+2n)^{n+1}}\right] > (n+2)^{n+1}
\]
Thus,
\[ n^{n+1}>(n+1)^{n},\ \mbox{for}\ n\ge 3\]
I hope this helps you understand.
|
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|
Prove that $ \frac {12^{x-2}.4^{x}} {6^{x-2}} = 2^{3x-2} $ Can someone please help me with this question?
$ \large \frac {12^{x-2}.4^{x}} {6^{x-2}} = 2^{3x-2} $
My steps so far:
$ \large \frac {4^{x-2}.3^{x-2}.4^{x}}{3^{x-2}.2^{x-2}} = 2^{3x-2} $
$ \large \frac {4^{x-2}.4^{x}}{2^{x-2}} = 2^{3x-2} $
$ \large \frac {4^{2x-2}}{2^{x-2}} = 2^{3x-2}$
I get stuck here but I am assuming I need to get that 4 to a 2 somehow so I can combine them... so:
$ \large \frac {(2^2)^{2x-2}}{2^{x-2}} = 2^{3x-2}$
I feel like I am not on the right track here as I have no idea where to go now. Could anyone help please? Thank you!
|
You are doing the right thing. Just be careful at the end. Remember that: $$(a^b)^c = a^{bc}$$
So $$4^{2x - 2} = (4^2)^{x-1} = (2^4)^{x - 1} = 2^{4x - 4}$$This happens because $2^4 = 4^2 = 16$. In general $a^b \neq b^a$. And you can write $\frac{1}{2^{x - 2}} = 2^{2 - x}$. This gives: $$2^{4x - 4} \cdot 2^{2 -x} = 2^{3x - 2}$$
Seeing your attempt, I'm sure you can go on your own now. But if you need a little more help, say.
|
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|
Trigonometric Proof: Question:
If $m\cos\alpha-n\sin\alpha=p$ then prove that $m\sin\alpha+n\cos\alpha=\pm \sqrt{m^2+n^2-p^2}$
My Efforts:
$(m\cos\alpha-n\sin\alpha)^{2}=p^2$
$m^2\cos^2\alpha+n^2\sin^2\alpha-2mn\cos\alpha\ \sin\alpha=p^2$
Now i think we need to add something on both side but i can't figure out what to add
|
From
$$m^2\cos^2\alpha+n^2\sin^2\alpha-2mn\cos\alpha\ \sin\alpha=p^2$$
we can add $m^2\sin^2\alpha+n^2\cos^2\alpha$ to both sides (to make use of the $\sin^2x+\cos^2x=1$ identity) to achieve
$$m^2(\sin^2\alpha+\cos^2\alpha)+n^2(\sin^2\alpha+\cos^2\alpha)-2mn\cos\alpha\ \sin\alpha=p^2+m^2\sin^2\alpha+n^2\cos^2\alpha
\\\iff m^2+n^2-p^2=m^2\sin^2\alpha+n^2\cos^2\alpha+2mn\cos\alpha\ \sin\alpha=(m\sin\alpha+n\cos\alpha)^2$$
The result follows.
|
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|
Prove that $\int_0^1\frac{\ln(1-x)\ln^2x}{x-1}dx=\frac{\pi^4}{180}$ Prove that (please)
$$\int_0^1\frac{\ln(1-x)\ln^2x}{x-1}dx=\frac{\pi^4}{180}$$
I've tried using Taylor series and I ended up with
$$-\sum_{m=0}^\infty\sum_{n=1}^\infty\frac{2}{n(m+n+1)^3}$$
I am stuck there and I couldn't continue it using partial fraction to get any familiar sum of series. Could anyone here please help me to prove the integral preferably (if possible) with elementary ways (high school methods)? Any help would be greatly appreciated. Thank you.
|
The answer by user111187 is the cleanest way to approach such problem. I will elaborate on the sums (which he doesnt show).
$$I = (-)\cdot\int_{0}^{1} \frac{\log(1-x)\log^2(x) dx}{1-x}$$
Let $u = 1-x \implies du = -dx$
$$I = -\int_{0}^{1} \frac{\log(u)\log^2(1-u) du}{u}$$
$$\sum_{n=1}^{\infty} H_n u^n = -\frac{\log(1-u)}{1-u}$$
$$\sum_{n=1}^{\infty} \frac{H_nu^{n+1}}{n+1} = \frac{\log^2(1-u)}{2}$$
$$(2)\cdot\sum_{n=1}^{\infty} \frac{H_n\log(u)\cdot u^{n}}{n+1} = \frac{\log^2(1-u)\log(u)}{u}$$
$$(-2)\cdot \sum_{n=1}^{\infty}\frac{H_n}{n+1}\cdot \int_{0}^{1} \log(u) \cdot u^{n} du = \int_{0}^{1} \frac{\log^2(1-u)\log(u)}{u} du = I$$
For the integral in the LHS, we simply use Leibniz's rule considering $\displaystyle J(n) = \int_{0}^{1} u^n du$ then differentiate.
$$(2)\cdot \sum_{n=1}^{\infty}\frac{H_n}{(n+1)^3} = \int_{0}^{1} \frac{\log^2(1-u)\log(u)}{u} du = I$$
$$H_n = H_{n-1} + \frac{1}{n} \implies H_{n+1} = H_{n} + \frac{1}{n+1} \implies H_n = H_{n+1} - \frac{1}{n+1}$$
$$ = (2)\cdot \sum_{n=1}^{\infty}\frac{H_{n+1}}{(n+1)^3} - (2)\cdot\sum_{n=1}^{\infty} \frac{1}{(n+1)^4}$$
$$ = (2)\cdot \sum_{n=2}^{\infty} \frac{H_{n}}{n^3} - (2)\cdot\sum_{n=2}^{\infty} \frac{1}{n^4}$$
$$ = (2)\cdot \left( \sum_{n=1}^{\infty} \frac{H_{n}}{n^3} - 1 \right) - (2) \cdot \left( \sum_{n=1}^{\infty} \frac{1}{n^4} - 1 \right)$$
From Generalised Euler Sum
$$ = (2)\cdot \left( \left(1 + \frac{3}{2} \right)\zeta(4) - \frac{1}{2}\zeta^2(2) - 1 \right) - 2\zeta(4) + 2$$
$$ = 5\zeta(4) - \zeta^2(2) - 2 - 2\zeta(4) + 2 = 3\zeta(4) - \zeta^2(2) = \frac{\pi^4}{180}$$
Finally,
$$I = \int_{0}^{1} \frac{\log(1-x)\log^2(x) \space dx}{x-1} = \frac{\pi^4}{180}$$
|
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|
condition for tuple of non-zero integers $(a,b,c,d)$ such that $ad+bc=ac-bd=ab+cd=a^2-b^2+c^2-d^2=0$ As the title says, what would be the condition for tuple of non-zero integers $(a,b,c,d)$ such that $ad+bc=ac-bd=ab+cd=a^2-b^2+c^2-d^2=0$? Would there be infinitely many tuples that satisfy the condition above?
|
Suppose that there is a set of non-zero integers $(a,b,c,d)$.
First, since $b\not=0$, we have
$$ac=bd\Rightarrow d=\frac{ac}{b}\tag1$$
Moreover, we have
$$ad+bc=ac-bd\Rightarrow (a+b)d=c(a-b).$$
If $a+b=0$, then we have $a-b=0$ because $c\not=0$. This leads $a=b=0$. This is a contradiction. Hence, we have $a+b\not =0$.
Hence, we have $$d=\frac{a-b}{a+b}c\tag2$$
Hence, since $c\not=0$, from $(1),(2)$ we have
$$\frac{ac}{b}=\frac{a-b}{a+b}c\Rightarrow ac(a+b)=bc(a-b)\Rightarrow a^2+b^2=0\Rightarrow a=b=0.$$
This is a contradiction. Hence, we know that there is no sets of non-zero integers $(a,b,c,d)$.
|
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|
Evaluate $\frac{1}{a}\int_0^\infty{x^2}e^{-\frac{x^2}{2a}}\,dx$ Evaluate the following integral:
$$\frac{1}{a}\int_0^\infty{x^2}e^{-\large\frac{x^2}{2a}}\,dx.$$
|
$$\frac{1}{a}\int_{0}^{\infty}x^{2}e^{-\frac{x^{2}}{2a}}dx=-\int_{0}^{\infty}x(e^{-\frac{x^{2}}{2a}})'dx$$
Use integration by parts and the value of Gaussian integral to evaluate:
$$=-(xe^{-\frac{x^{2}}{2a}})\vert_{0}^{\infty}+\int_{0}^{\infty}e^{-\frac{x^{2}}{2a}}dx=\sqrt{2a}\int_{0}^{\infty}e^{-(\frac{x}{\sqrt{2a}})^{2}}\frac{1}{\sqrt{2a}}dx=\sqrt{2a}\cdot\frac{\sqrt{\pi}}{2}$$
Alternatively notice that:
$$-\frac{1}{2}\sqrt{\frac{a\pi}{2}}=\frac{d}{db}(\sqrt{\frac{2a}{b}}\cdot\frac{\sqrt{\pi}}{2})\vert_{b=1}=\frac{d}{db}(\int_{0}^{\infty}e^{-\frac{bx^{2}}{2a}}dx)\vert_{b=1}=-\frac{1}{2a}\int_{0}^{\infty}x^{2}e^{-\frac{x^{2}}{2a}}dx$$
|
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|
Dealing with absolute values after trigonometric substitution in $\int \frac{\sqrt{1+x^2}}{x} \text{ d}x$. I was doing this integral and wondered if the signum function would be a viable method for approaching such an integral. I can't seem to find any other way to help integrate the $|\sec \theta|$ term in the numerator of the integrand.
$$ \begin{aligned} \int \dfrac{\sqrt{x^2 + 1}}{x} \text{ d}x \ \ & \overset{x=\tan \theta}= \int \dfrac{\sqrt{\sec^2 \theta} \sec^2 \theta}{\tan \theta} \text{ d}\theta \\ & \ \ = \int \dfrac{|\sec \theta| \sec^2 \theta}{\tan \theta} \text{ d}\theta \\ & \ \ = \int \dfrac{\text{sgn} (\sec \theta) \sec^3 \theta}{\tan \theta} \text{ d}\theta \\ & \ \ = \text{sgn} (\sqrt{1+x^2}) \left( - \log \left| \dfrac{\sqrt{1+x^2} + 1}{x} \right| + \sqrt{1+x^2} \right) + \mathcal{C} \end{aligned} $$
It's clear that $\text{sgn} (\sqrt{1+x^2}) = 1$ since the sign of the argument of that function is always positive and the signum function extracts the sign. So I'd leave the integral as: $$ \begin{aligned} \int \dfrac{\sqrt{x^2 + 1}}{x} \text{ d}x = \log \left| x \right| - \log \left( \sqrt{1+x^2} + 1 \right) + \sqrt{1+x^2} + \mathcal{C} \end{aligned} $$
Would that be ok?
Also, apparently Wolfram has suggested that my final result should have $x$ as opposed to $|x|$ in the argument of the first logarithm. Why is that?
Any help would be greatly appreciated!
|
I think you are correct. The Wolfram answer is only defined for $x>0$, whereas $\dfrac{\sqrt{x^2+1}}{x}$ has an antiderivative for $x<0$ as well.
|
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|
Divisor function asymptotics Define $\tau_{r}(n) = \sum_{d_1...d_r = n}1$.
One exercise in a book on sieve theory asked for an elementary proof by induction of the fact that
$$\sum_{n\le x}\tau_r(n) = \frac{1}{(r - 1)!}x(\ln x)^{r - 1} + O\left(x(\ln x)^{r - 2}\right)$$
The base case $r = 2$ is easy with reversing the order of summation.
The only other progress that I made was the fact that $\sum_{n\le x}\tau_{r}(n) = \sum_{d\le x}\left\lfloor\frac{x}{d}\right\rfloor\tau_{r- 1}(d)$, but I don't know how to proceed.
|
We can write
$$\tau_{r+1}(n)= \sum_{d\mid n} \tau_r(d),$$
which leads to your
$$\sum_{n\leqslant x} \tau_{r+1}(n) = \sum_{d\leqslant x} \left\lfloor \frac{x}{d}\right\rfloor \tau_r(d),$$
but we can also write
$$\tau_{r+1}(n) = \sum_{d\mid n} \tau_r\left(\frac{n}{d}\right),$$
and that gives us
$$\begin{align}
\sum_{n\leqslant x} \tau_{r+1}(n)
&= \sum_{n\leqslant x} \sum_{d\mid n} \tau_r\left(\frac{n}{d}\right)\\
&= \sum_{d\leqslant x} \sum_{k\leqslant \frac{x}{d}} \tau_r(k)\\
&= \sum_{d\leqslant x} C_r\frac{x}{d}\left(\ln \frac{x}{d}\right)^{r-1} + O\left(\frac{x}{d}\left(\ln \frac{x}{d}\right)^{r-2}\right).
\end{align}$$
Now we can estimate the lower-order part by $K\cdot \frac{x}{d}(\ln x)^{r-2}$, and since $\sum_{d\leqslant x} \frac{1}{d} = \ln x + O(1)$, the sum of these terms is, as it should be, $O(x(\ln x)^{r-1})$.
For the dominant terms, we find
$$\begin{align}
\sum_{d\leqslant x} \frac{x}{d}\left(\ln \frac{x}{d}\right)^{r-1}
&= \sum_{d\leqslant x} \frac{x}{d}\left(\ln x - \ln d\right)^{r-1}\\
&= \sum_{d\leqslant x} \frac{x}{d} \sum_{k=0}^{r-1} (-1)^k\binom{r-1}{k} (\ln x)^{r-1-k}(\ln d)^k\\
&= \sum_{k=0}^{r-1}(-1)^k x(\ln x)^{r-1-k}\binom{r-1}{k} \sum_{d\leqslant x} \frac{(\ln d)^k}{d}\\
&= \sum_{k=0}^{r-1}(-1)^k x(\ln x)^{r-1-k}\binom{r-1}{k} \left(\frac{(\ln x)^{k+1}}{k+1} + O(1)\right)\\
&= x(\ln x)^r\sum_{k=0}^{r-1}(-1)^k\binom{r-1}{k}\frac{1}{k+1} + O\left(x(\ln x)^{r-1}\right),
\end{align}$$
and since
$$\sum_{k=0}^{r-1} (-1)^k\binom{r-1}{k}\frac{1}{k+1} = -\frac{1}{r} \sum_{k=0}^{r-1} (-1)^{k+1}\binom{r}{k+1} = \frac{1}{r},$$
we get
$$\sum_{n\leqslant x} \tau_{r+1}(n) = \frac{C_r}{r} x(\ln x)^r + O\left(x(\ln x)^{r-1}\right).$$
Since $\tau_1(n) = 1$ for all $n$, we immediately have
$$\sum_{n\leqslant x} \tau_1(n) = \lfloor x\rfloor = 1\cdot x(\ln x)^0 + O\left(x(\ln x)^{-1}\right),$$
hence $C_1 = 1 = \frac{1}{0!}$, and the recurrence $C_{r+1} = C_r/r$ yields $C_r = \frac{1}{(r-1)!}$, as desired.
|
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|
Prove $\frac{(xy)^7}{x^8+(xy)^7+y^8}+\frac{(yz)^7}{y^8+(yz)^7+z^8}+\frac{(zx)^7}{z^8+(zx)^7+x^8}\leq1$
If $x,y,z$ are positive real numbers that $xyz=1$ , Prove
a) $\frac{xy}{x^8+xy+y^8}+\frac{yz}{y^8+yz+z^8}+\frac{zx}{z^8+zx+x^8}\leq1$
b)$\frac{(xy)^7}{x^8+(xy)^7+y^8}+\frac{(yz)^7}{y^8+(yz)^7+z^8}+\frac{(zx)^7}{z^8+(zx)^7+x^8}\leq1$
Additional info: We must use the following inequality for proving$$x^8+y^8\geq x^7y+y^7x\geq x^6y^2+y^6x^2\geq...\geq2x^4y^4$$
and more general this inequality $$a^{m+n}+b^{n+m}\geq a^nb^m+a^mb^n$$
Things I have tried so far: For Proving a) i can write $$\frac{xy}{x^8+xy+y^8}\leq\frac{xy}{x^7y+xy+y^7x}=\frac{1}{x^6+y^6+1}$$
And continuing this$$\frac{1}{x^6+y^6+1}\leq \frac{1}{x^4y^2+y^4x^2+x^2y^2z^2}=\frac{1}{x^2y^2(x^2+y^2+z^2)}=\frac{1}{\frac{1}{z^2}(x^2+y^2+z^2)}=\frac{z^2}{x^2+y^2+z^2}$$
By doing these for two other fractions we can prove inequality.
for part b, i tried to do same thing but I was unsuccessful. for example i end up with these inequality $\frac{(xy)^7}{x^8+(xy)^7+y^8}\leq \frac{x^3y^3}{2+x^2y^2}$ which is not useful for proving the problem.
|
After the first step it will become
$\frac{(xy)^6}{x^6+y^6+(xy)^6}=\frac{1}{\frac{1}{y^6}+\frac{1}{x^6}+1}$
After set as I said it will be
$\frac{1}{a^6+b^6+1}$
Now you just need to continue what you do in a)
|
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|
Simplify the following compound fraction: $$\frac{2x+1}{\frac{3}{x^2}+\frac{2x+1}{x}}$$
My calculator says the final answer is $$\frac{x^2(2x+1)}{2x^2+x+3}$$
Please show the work. Thanks.
|
First recall that
\[ \frac{a}{b} \pm \frac{c}{d} =\frac{ad \pm cb}{bd} \]
And
\[ \frac{a}{\left(\frac{c}{d}\right)} =\frac{ad}{c} \]
Then
\[
\frac{2x+1}{\frac{3}{x^{2}}+\frac{2x+1}{x}}= \frac{2x+1}{\frac{3x+x^{2}(2x+1)}{x^{3}}}= \frac{x^{3}(2x+1)}{3x+x^{2}(2x+1)}= \frac{x^{3}(2x+1)}{2x^{3}+x^{2}+3x}= \frac{x^{3}(2x+1)}{x(2x^{2}+x+3)}= \frac{x^{2}(2x+1)}{2x^{2}+x+3}
\]
|
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|
Any idea how to linearize this equation? $X^2-Y^2=aZ+bZ^2$ The intention is to linearize this equation $X^2-Y^2=aZ+bZ^2$ into something which looks like $Z=mX+nY+c$ so that a graph of $Z$ against $X$ or $Y$ can be plotted.
X,Y,Z are variables while a,b,c are constants
|
EDIT: Thanks to dylan37 for pointing out that the this is probably not the sort of "linearization" that curiousfurious (the OP) intended.
Maybe you could use the trick that's employed in deriving the Quadratic Formula.
Starting from $ X^{2}-Y^{2}=aZ+bZ^{2}$, first divide both sides by $b$ to get
$\frac{1}{b}X^{2}-\frac{1}{b}Y^{2}=\frac{a}{b}Z+Z^{2}=2\left(\frac{a}{2b}\right)Z+Z^{2}.$
Now, complete the square:
$
\frac{1}{b}X^{2}-\frac{1}{b}Y^{2}+\frac{a^{2}}{4b^{2}}=\frac{a^{2}}{4b^{2}}+2\left(\frac{a}{2b}\right)Z+Z^{2}=(Z+\frac{a}{2b})^{2}
$
Finally, let$\;\;U=X^{2};\;\;V=Y^{2};\;\;W=(Z+\frac{a}{2b})^{2}.$
In this way,
$X^{2}-Y^{2}=aZ+bZ^{2}$ becomes $\frac{1}{b}U-\frac{1}{b}V+\frac{a^{2}}{4b^{2}}=W$, or $W=\frac{1}{b}U-\frac{1}{b}V+\frac{a^{2}}{4b^{2}}$.
|
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|
Find the kernel of linear transformation Linear transformation $l:\mathbb{R}^3 \mapsto \mathbb{R}$ is determined as follows: $l(1,0,0)=1$; $l(1,4,0)=-1$; $l(0,0,1)=1$.
I need to find $\text{Ker}(l)$.
Answer should be $\text{Span}\{(0,2,1),(1,2,0)\}$.
But my answer is $\text{Span}\{(1,4,1),(2,4,0)\}$.
I tried to solve the problem the following way:
$$l(x) = x_1 l(1,0,0) + x_2 l(1,4,0) + x_3 l(0,0,1).$$
Then $l(x) = x_1 - x_2 +x_3$. After than solve the system $x_1 - x_2 + x_3 = 0$. I got $$(x_1,x_2,x_3) = \text{Span}\{(0,1,1) (1,1,0)\}.$$
Express to standart coordinates I got: $x = b(1,0,0) + (a+b)(1,4,0) + a(0,0,1)$. So kernel is $\text{Span}\{(1,4,1),(2,4,0)\}$. Can't find the mistake.
|
Both answers are correct. That is $\mathcal{A}=((0,2,1)^T,(1,2,0)^T)$ and $\mathcal{B}=((1,4,1)^T,(2,4,0)^T)$ are both basis of the Kernel. Indeed for the basis $\mathcal{A}$ we have
$$\begin{pmatrix}0\\2\\1\end{pmatrix} = -\frac12\begin{pmatrix}1\\0\\0\end{pmatrix}+\frac12\begin{pmatrix}1\\4\\0\end{pmatrix}+\begin{pmatrix}0\\0\\1\end{pmatrix}$$
Applying the function $l$ and using the properties of a linear map we get
\begin{align*}l\left( \begin{pmatrix}0\\2\\1\end{pmatrix}\right) &=l\left(-\frac12\begin{pmatrix}1\\0\\0\end{pmatrix}+\frac12\begin{pmatrix}1\\4\\0\end{pmatrix}+\begin{pmatrix}0\\0\\1\end{pmatrix}\right)\\
&=-\frac12l\left(\begin{pmatrix}1\\0\\0\end{pmatrix}\right)+\frac12l\left(\begin{pmatrix}1\\4\\0\end{pmatrix}\right)+l\left(\begin{pmatrix}0\\0\\1\end{pmatrix}\right)\\
&=-\frac12\cdot 1 + \frac12\cdot (-1)+ 1 =0\end{align*}
And
$$\begin{pmatrix}1\\2\\0\end{pmatrix} = \frac12\begin{pmatrix}1\\0\\0\end{pmatrix}+\frac12\begin{pmatrix}1\\4\\0\end{pmatrix}+0\begin{pmatrix}0\\0\\1\end{pmatrix}$$
Applying the function $l$ and using as above the properties for linear maps:
\begin{align*}
l\left(\begin{pmatrix}1\\2\\0\end{pmatrix} \right) &= \frac12 l\left(\begin{pmatrix}1\\0\\0\end{pmatrix}\right) + \frac12 l\left(\begin{pmatrix}1\\4\\0\end{pmatrix}\right) \\
&=\frac12 \cdot 1 + \frac12 \cdot (-1)
\end{align*}
While for the basis $\mathcal{B}$ you can see that $(0,2,4)^T = 2(0,2,1)^T$ and for
$$\begin{pmatrix}1\\4\\1\end{pmatrix}=0\cdot\begin{pmatrix}1\\0\\0\end{pmatrix}+\begin{pmatrix}1\\4\\0\end{pmatrix}+\begin{pmatrix}0\\0\\1\end{pmatrix}$$
And again applying $l$
\begin{align*}
l\left(\begin{pmatrix}1\\4\\1\end{pmatrix} \right) &= l\left(\begin{pmatrix}1\\4\\0\end{pmatrix} \right) + l\left( \begin{pmatrix}0\\0\\1\end{pmatrix}\right) \\
&= -1 + 1 = 0
\end{align*}
Hope this helps :-)
|
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|
Number of solutions of $x_1 + x_2 + x_3 + x_4 = 14$ such that $x_i \le 6$
Let $x_1, x_2, x_3, x_4$ be nonnegative integers.
(a) Find the number of solutions to the following equation:
$$ x_1 + x_2 + x_3 + x_4 = 14 $$
I got $17 \choose 3$ for this. Is that correct?
(b) Find the number of solutions if we add the restriction that $x_i \le 6$ for
$1 \le i \le 4$
|
Use generating functions. The ways to pick each $x_i$ is represented by:
$$
1 + z + z^2 + z^3 + z^4 + z^5 + z^6 = \frac{1 - z^7}{1 - z}
$$
Now you want four of those, and have them add up to $n$:
$\begin{align}
[z^n] \frac{(1 - z^7)^4}{(1 - z)^4}
&= [z^n] (1 - 4 z^7 + 6 z^{14} - 4 z^{21} + z^{28})
\cdot \sum_{k \ge 0} (-1)^k \binom{-4}{k} z^k \\
&= [z^n] (1 - 4 z^7 + 6 z^{14} - 4 z^{21} + z^{28})
\cdot \sum_{k \ge 0} \binom{k + 4 - 1}{4 - 1} z^k \\
&= \binom{n + 3}{3}
- 4 \binom{n - 4}{3}
+ 6 \binom{n - 11}{3}
- 4 \binom{n - 18}{3}
+ \binom{n - 25}{3} \\
\end{align}$
Here you have to be careful, the terms with negative upper index aren't present (no such terms are present in the sum, the first factor essentially picks out some of them).
In the particular case $n = 14$:
$$
\binom{14 + 3}{3} - 4 \binom{14 - 4}{3} + 6 \binom{14 - 11}{3}
= 206
$$
|
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|
Find the sum of the multiples of $3$ and $5$ below $709$? I just cant figure this question out:
Find the sum of the multiples of $3$ or $5$ under $709$
For example, if we list all the natural numbers below $10$ that are multiples of $3$ or $5$, we get $3$, $5$, $6$ and $9$. The sum of these multiples is $23$.
|
If inc/exclusion is unknown then, note $\,3,5\mid n\!+\color{blue}{\!15k}\iff 3,5\mid n,\,$ so the multiples of $\,3,5\,$ have periodicity $15,\,$ so we can split the sum into chunks from each period as follows
$$\begin{eqnarray}
\color{blue}{0}+\overbrace{\{0,3,5,6,9,10,12\}}^{\large \rm sum\, =\, \color{#c00}{45}}\\
\color{blue}{15}+\{0,3,5,6,9,10,12\}\\
\color{blue}{30}+\{0,3,5,6,9,10,12\}\\
\cdots\qquad\qquad\\
\color{blue}{15\cdot 46}+\{0,3,5,6,9,10,12\}\\
\underbrace{15\cdot 47}_{\large\color{#0a0}{705}} + \underbrace{15\cdot 47+3}_{\large\color{#0a0}{708}}\qquad\quad\ \ \, \\
\hline
\end{eqnarray}\qquad\qquad$$
with total sum $\, =\, \color{blue}{7\cdot 15\, (47\cdot 46/2)} + 47\cdot \color{#c00}{45}\color{#0a0}{ + 705 + 708} = 117033$
|
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|
Double Integral $\int_{0}^{4} \int_{\sqrt{x}}^{2} \frac{1}{1+y^3} \mathrm{d}y\;\mathrm{d}x$ I am having trouble computing the double integral:
$$
\int_{0}^{4} \int_{\sqrt{x}}^{2} \frac{1}{1+y^3} \mathrm{d}y\,\mathrm{d}x
$$
I computed the inner integral:
$$
\left [ \frac{1}{3}\ln|y + 1| - \frac{1}{6}\ln(y^2 - y + 1) + \sqrt\frac{1}{3}\tan^{-1}[\frac{(2y-1)}{\sqrt(3)}] \right ]_{\sqrt{x}}^{2}
$$
So now would I just continue as:
$$
\left [ \frac{1}{3}\ln|2 + 1| - \frac{1}{6}\ln(2^2 - 2 + 1) + \sqrt\frac{1}{3}\tan^{-1}[\frac{(2(2)-1)}{\sqrt{3}}] \right ] - \left[\frac{1}{3}\ln|\sqrt{x} + 1| - \frac{1}{6}\ln(\sqrt{x}^2 - \sqrt{x} + 1) + \sqrt\frac{1}{3}\tan^{-1}[\frac{(2\sqrt{x}-1)}{\sqrt 3}]\right]
$$
To finish the inner integral?
|
If you draw the region of integration, the integral you have is equal to this integral:
$$\int_0^2\int_0^{y^2} \frac{1}{1+y^3} \operatorname{d}x\operatorname{d}y.$$
This is a bit easier to integrate. *After integrating with respect to $x$, you can use a $u$ substitution, $u=1+y^3$.
|
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|
Evaluation of $ \int_{0}^{2a}x\cdot \sin^{-1}\left(\frac{1}{2}\sqrt{\frac{2a-x}{a}}\right)dx$
$(1)$ Evaluation of $\displaystyle \int_{0}^{\frac{\pi}{2}}\frac{a\sin x+b\cos x}{\sin \left(x+\frac{\pi}{4}\right)}dx$
$(2)$ Evaluation of $\displaystyle \int_{-1}^{1}\ln\left(\frac{1+x}{1-x}\right)\cdot \frac{x^3}{\sqrt{1-x^2}}dx$
$(3)$ Evaluation of $\displaystyle \int_{0}^{2a}x\cdot \sin^{-1}\left(\frac{1}{2}\sqrt{\frac{2a-x}{a}}\right)dx$
$\bf{My\; Try::}$ For $(1)$ one
Let $\displaystyle I = \int_{0}^{\frac{\pi}{2}}\frac{a\sin x}{\sin (x+\frac{\pi}{4})}dx+\int_{0}^{\frac{\pi}{2}}\frac{b\cos x}{\sin (x+\frac{\pi}{4})}dx$
Now Let $\displaystyle J = \int_{0}^{\frac{\pi}{2}}\frac{a\sin x}{\sin (x+\frac{\pi}{4})}dx$ and $\displaystyle K = \int_{0}^{\frac{\pi}{2}}\frac{b\cos x}{\sin (x+\frac{\pi}{4})}dx$
Now We will Calculate $\displaystyle J = \int_{0}^{\frac{\pi}{2}}\frac{a\sin x}{\sin (x+\frac{\pi}{4})}dx$
Using $\displaystyle \left(x+\frac{\pi}{4}\right)=t\;,$ Then $dx = dt$.
So $\displaystyle J = a\int_{0}^{\frac{\pi}{4}}\frac{\sin \left(t-\frac{\pi}{4}\right)}{\sin t}dt = a\cdot \frac{1}{\sqrt{2}}\cdot \frac{\pi}{2}$
Similarly we will Calculate $\displaystyle K = \int_{0}^{\frac{\pi}{2}}\frac{b\cos x}{\sin (x+\frac{\pi}{4})}dx$
Using $\displaystyle \left(x+\frac{\pi}{4}\right)=t\;,$ Then $dx = dt$.
So $\displaystyle K = b\int_{0}^{\frac{\pi}{4}}\frac{\cos \left(t-\frac{\pi}{4}\right)}{\sin t}dt = b\cdot \frac{1}{\sqrt{2}}\cdot \frac{\pi}{2}$
So $\displaystyle I = J+K = \frac{\pi}{2\sqrt{2}}\cdot (a+b)$
Is there is any Shorter Solution for $(1)$ one and How can I calcultae $(2)$ and $(3)$ one
Help me
Thanks
|
(1)
We have:
$$\begin{eqnarray*}\int_{0}^{\pi/4}\frac{\sin x}{\sin(x+\pi/4)}\,dx &=& \int_{\pi/4}^{\pi/2}\frac{\sin(x-\pi/4)}{\sin x}\,dx = \frac{1}{\sqrt{2}}\int_{\pi/4}^{\pi/2}\left(1-\cot x\right)\,dx\\&=&\frac{1}{4\sqrt{2}}\left(\pi-\log 4\right),\end{eqnarray*}$$
$$\begin{eqnarray*}\int_{0}^{\pi/4}\frac{\cos x}{\sin(x+\pi/4)}\,dx &=& \int_{\pi/4}^{\pi/2}\frac{\cos(x-\pi/4)}{\sin x}\,dx = \frac{1}{\sqrt{2}}\int_{\pi/4}^{\pi/2}\left(\cot x+1\right)\,dx\\&=&\frac{1}{4\sqrt{2}}\left(\pi+\log 4\right).\end{eqnarray*}$$
(2) We have:
$$\begin{eqnarray*}\int_{-1}^{1}\log\left(\frac{1+x}{1-x}\right)\frac{x^3}{\sqrt{1-x^2}}=2\int_{0}^{\pi/2}\cos^3\theta\cdot\log\left(\frac{1+\cos\theta}{1-\cos\theta}\right)d\theta\end{eqnarray*}$$
and integrating by parts we get:
$$\begin{eqnarray*}\int_{-1}^{1}\log\left(\frac{1+x}{1-x}\right)\frac{x^3}{\sqrt{1-x^2}}=\int_{0}^{\pi/2}\left(3+\frac{\sin(3\theta)}{3\sin\theta}\right)d\theta=\frac{5\pi}{3}.\end{eqnarray*}$$
(3) We have:
$$\begin{eqnarray*}\int_{0}^{2a}x\cdot\arcsin\left(\frac{1}{2}\sqrt{\frac{2a-x}{a}}\right)dx&=&a^2\cdot\int_{0}^{2}x\cdot\arcsin\left(\frac{\sqrt{2-x}}{2}\right)dx\\&=&4a^2\cdot\int_{0}^{1}x\cdot\arcsin\sqrt{\frac{1-x}{2}}\,dx\\&=&2a^2\int_{0}^{\pi/2}\theta \sin\theta\cos\theta\,d\theta\\&=&\frac{\pi a^2}{4}.\end{eqnarray*}$$
|
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|
The inverse of $2x^2+2$ in $\mathbb{Z}_3[x]/( x^3+2x^2+2)$ What is the independent coefficient in the inverse of $2x^2+2$ in $\mathbb{Z}_3[x]/(x^3+2x^2+2)$ ?
I have been calculating some combinations, but I don't know how I can calculate the inverse.
|
This is in no way a standard solution, but I just want to include it for fun.
Note that $x^3=-(2x^2+2)$, we will find $x^{-3}$. Now
$$x^3-x^2=-2=1$$
implies
$$x(x^2-x)=x^2(x-1)=1.$$
Thus
$$x^{-1}=x(x-1)\text{ and }x^{-2}=x-1.$$
Multiplying the two we get
$$x^{-3} = x(x-1)^2=x(x^2+x+1)=x^3+x^2+x.$$
Using $x^3=x^2+1$ on the RHS, we see that
$$x^{-3}=2x^2+x+1.$$
Thus
$$(2x^2+2)^{-1} = -x^{-3} = -2x^2-x-1 = x^2+2x+2.$$
|
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|
Infinite Series $\sum_{n=1}^\infty\frac{H_n}{n^32^n}$ I'm trying to find a closed form for the following sum
$$\sum_{n=1}^\infty\frac{H_n}{n^3\,2^n},$$
where $H_n=\displaystyle\sum_{k=1}^n\frac{1}{k}$ is a harmonic number.
Could you help me with it?
|
By first finding the following integral by using the algebraic identity
$a^2b=\frac{1}{6}\left(a+b\right)^3-\frac{1}{6}\left(a-b\right)^3-\frac{1}{3}b^3$ one can easily prove avoiding Euler sums that:
$$\int _0^1\frac{\ln ^2\left(1-x\right)\ln \left(1+x\right)}{1+x}\:dx=-\frac{1}{4}\zeta \left(4\right)+2\ln \left(2\right)\zeta \left(3\right)-\ln ^2\left(2\right)\zeta \left(2\right)+\frac{1}{4}\ln ^4\left(2\right)$$
Now:
$$\int _0^1\frac{\ln ^2\left(1-x\right)\ln \left(1+x\right)}{1+x}\:dx=\frac{1}{2}\ln \left(2\right)\int _0^1\frac{\ln ^2\left(x\right)}{1-\frac{x}{2}}\:dx+\frac{1}{2}\int _0^1\frac{\ln ^2\left(x\right)\ln \left(1-\frac{x}{2}\right)}{1-\frac{x}{2}}\:dx$$
$$=2\ln \left(2\right)\sum _{k=1}^{\infty }\frac{1}{k^3\:2^k}-2\sum _{k=1}^{\infty }\frac{H_k}{k^3\:2^k}+2\sum _{k=1}^{\infty }\frac{1}{k^4\:2^k}$$
$$=2\ln \left(2\right)\operatorname{Li}_3\left(\frac{1}{2}\right)-2\sum _{k=1}^{\infty }\frac{H_k}{k^3\:2^k}+2\operatorname{Li}_4\left(\frac{1}{2}\right)$$
$$=\frac{7}{4}\ln \left(2\right)\zeta \left(3\right)-\ln ^2\left(2\right)\zeta \left(2\right)+\frac{1}{3}\ln ^4\left(2\right)-2\sum _{k=1}^{\infty }\frac{H_k}{k^3\:2^k}+2\operatorname{Li}_4\left(\frac{1}{2}\right)$$
By making use of the result we find:
$$\sum _{k=1}^{\infty }\frac{H_k}{k^3\:2^k}=\frac{1}{8}\zeta \left(4\right)+\operatorname{Li}_4\left(\frac{1}{2}\right)-\frac{1}{8}\ln \left(2\right)\zeta \left(3\right)+\frac{1}{24}\ln ^4\left(2\right)$$
|
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|
What is the greatest common divisor of $3^{3^{333}}+1$ and $3^{3^{334}}+1$? What is the greatest common divisor of $3^{3^{333}}+1$ and $3^{3^{334}}+1$ ? Could somebody please help me ?
|
Write $T_n=3^{3^n}+1$. Then
$$\eqalign{T_n^3
&=(3^{3^n})^3+3(3^{3^n})^2+3(3^{3^n})+1\cr
&=3^{3^{n+1}}+1+3(3^{3^n})(3^{3^n}+1)\cr
&=T_{n+1}+3^{3^n+1}T_n\ .\cr}$$
From this you should be able to see an impotant relationship between $T_n$ and $T_{n+1}$.
|
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|
If $ab+bc+ca=0$ then the value of $1/(a^2-bc)+1/(b^2-ac)+1/(c^2-ab)$ is If $ab+bc+ca=0$
then the value of
$1/(a^2-bc)+1/(b^2-ac)+1/(c^2-ab)$ is...
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$$\frac{1}{a^2+ab+ca}+\frac{1}{b^2+ab+bc}+\frac{1}{c^2+ca+bc}$$
$$=\frac{1}{a(a+b+c)}+\frac{1}{b(a+b+c)}+\frac{1}{c(a+b+c)}=\frac{ab+bc+ca}{abc(a+b+c)}=0.$$
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Prove that if $\frac{x+y}{3a-b}=\frac{y+z}{3b-c}=\frac{z+x}{3c-a}$ then $\frac{x+y+z}{a+b+c}=\frac{ax+by+cz}{a^2+b^2+c^2}$ If $\displaystyle\frac{x+y}{3a-b}=\frac{y+z}{3b-c}=\frac{z+x}{3c-a}$ then prove that $\displaystyle\frac{x+y+z}{a+b+c}=\frac{ax+by+cz}{a^2+b^2+c^2}$
I tried to prove this in many ways. First, I tried to multiply the first equality by $\displaystyle(3a-b)(3b-c)(3c-a)$, but then it seems too complicated. Is there any easy method?
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If we set
$$k=\frac{x+y}{3a-b}=\frac{y+z}{3b-c}=\frac{z+x}{3c-a},$$
then we have
$$x+y=(3a-b)k,\ \ y+z=(3b-c)k,\ \ z+x=(3c-a)k.$$
Hence, adding these three gives us
$$2(x+y+z)=(3a-a+3b-b+3c-c)k\iff \frac{x+y+z}{a+b+c}=k\tag 1$$
On the other hand, we have
$$a(x+y)=a(3a-b)k,\ \ b(y+z)=b(3b-c)k,\ \ c(z+x)=c(3c-a)k$$
$$\Rightarrow (ax+by+cz)+(ay+bz+cx)=\{3(a^2+b^2+c^2)-(ab+bc+ca)\}k\tag2$$
$$b(x+y)=b(3a-b)k,\ \ c(y+z)=c(3b-c)k,\ \ a(z+x)=a(3c-a)k$$
$$\Rightarrow (ax+by+cz)+(az+bx+cy)=\{3(ab+bc+ca)-(a^2+b^2+c^2)\}k\tag3$$
$$c(x+y)=c(3a-b)k,\ \ a(y+z)=a(3b-c)k,\ \ b(z+x)=b(3c-a)k$$
$$\Rightarrow (ay+bz+cx)+(az+bx+cy)=\{3(ab+bc+ca)-(ab+bc+ca)\}k\tag4$$
Hence, calculating $(2)+(3)-(4)$ gives us
$$2(ax+by+cz)=2(a^2+b^2+c^2)k\iff \frac{ax+by+cz}{a^2+b^2+c^2}=k\tag5$$
From $(1)$ and $(5)$, we have the conclusion.
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The length of the smallest possible ladder to change the bulb.
In the drawing, the point P, a bit located under the bulb, has coordinates (a, b), where a and b are two parameters. You want to change the bulb, and for this it is necessary to install a ladder such that
*
*It relies on the ground (the $Ox$) axis at the point $S$.
*It rests against the vertical wall (the $Oy$ axis) at the point $M$.
*She passes exactly through the point $P$.
Question : What, in function of $a$ and $b$ only, the length of the smallest possible ladder fulfilling the three conditions above?
My try:
I tried to compute the coordinate of $M(0,y_M)$ in function of $a,b$ but I am stuck with some ugly calculs.
I wrote $\Vert\overrightarrow{MS}\Vert=\Vert\overrightarrow{MP}\Vert+\Vert\overrightarrow{PS}\Vert$ and square it. It looks like
$$
2by_M=a^2-b^2+b^2+(c-a)^2+2\sqrt{a^2(c-a)^2+(ab)^2+b^2(b-y_M)^2+(b-y_M)^2(c-a)^2}
$$
As usual, I think I miss something.
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Let $x$ be the base of the triangle and let $y$ be the height.
To minimize the length $L$ of the ladder, we can minimize $L^2=x^2+y^2$.
By similar triangles, $\displaystyle\frac{y}{x}=\frac{b}{x-a}$, so $\displaystyle y=\frac{bx}{x-a}$ and $\displaystyle f(x)=L^2=x^2+\frac{b^2x^2}{(x-a)^2}$.
Then $f^{\prime}(x)=2x+\frac{(x-a)^2(2b^{2}x)-b^{2}x^2(2(x-a))}{(x-a)^4}=2x-\frac{2ab^{2}x}{(x-a)^3}=0$ if $2x(x-a)^3=2ab^{2}x$,
so $(x-a)^3=ab^2$ since $x\ne0$. Therefore $x=a+a^{1/3}b^{2/3}=a^{1/3}(a^{2/3}+b^{2/3})$ and
$\displaystyle \;\;\;\;y=\frac{bx}{x-a}=\frac{ba^{1/3}(a^{2/3}+b^{2/3})}{a^{1/3}b^{2/3}}=b^{1/3}(a^{2/3}+b^{2/3}).$
Now find $L=\sqrt{x^2+y^2}$.
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How find the minimum $\frac{(5y+2)(2z+5)(x+3y)(3x+z)}{xyz}$,if $x,y,z>0$
let $x,y,z>0$, find the minimum of the value
$$\dfrac{(5y+2)(2z+5)(x+3y)(3x+z)}{xyz}$$
I think we can use AM-GM inequality to find it.
$$5y+2=y+y+y+y+y+1+1\ge 7\sqrt[7]{y^5}$$
$$2x+5=x+x+1+1+1+1+1\ge 7\sqrt[7]{x^2}$$
$$x+3y=x+y+y+y\ge 4\sqrt[4]{xy^3}$$
$$3x+z=x+x+x+z\ge 4\sqrt[4]{x^3z}$$
but this is not true,because not all four equalities can hold at once.
This problem is from china middle school test,so I think have without Lagrange methods,so I think this inequality have AM-GM inequality
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Here is yet another way, using (weighted) AM-GM. I would still prefer Holder (the solution posted earlier) for its simplicity, this is just to illustrate it can be done - especially if you know the point of equality.
Let $a = \sqrt{\frac2{15}}$. Then using weighted AM-GM, we can write the following inequalities. Note that they are constructed so that the point of equality is easily maintained:
$$5y+2=5a \cdot \frac{y}a+2 \cdot 1 \ge (5a+2)\left(\frac ya \right)^{\frac{5a}{5a+2}}$$
$$\parallel ly, \quad 2z+5=\frac2a \cdot az+5\cdot1 \ge (\tfrac2a+5)(az)^{\frac2{5a+2}}$$
$$x+3y = 1\cdot x + 3a\cdot\frac{y}a \ge (1+3a)x^{\frac1{3a+1}}\left(\frac ya\right)^{\frac{3a}{3a+1}}$$
$$3x+z = 3\cdot x + \frac1a \cdot az \ge (3+\tfrac1a)x^{\frac{3a}{3a+1}}(az)^{\frac1{3a+1}}$$
Multiplying the lot, and substituting the value of $a$, we get
$$\dfrac{(5y+2)(2z+5)(x+3y)(3x+z)}{xyz} \ge 241+44\sqrt{30}$$
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How prove this inequality $a^3b+b^3c+c^3a+a^3b^3+b^3c^3+c^3a^3\le 6$ let $a,b,c>0$, and such
$$a^2+b^2+c^2=3$$
show that
$$a^3b+b^3c+c^3a+a^3b^3+b^3c^3+c^3a^3\le 6\tag{2}$$
I know this famous inequality( creat by valsie)
$$(a^2+b^2+c^2)^2\ge 3(a^3b+b^3c+c^3a)$$ this inequality if and only if
$$a=b=c, or,a:b:c=\sin^2{\dfrac{\pi}{7}}:\sin^2{\dfrac{2\pi}{7}}:\sin^2{\dfrac{3\pi}{7}}$$
can see:
Vasc inequality solution
so
$$(a^3b+b^3c+c^3a)\le \dfrac{1}{3}(a^2+b^2+c^2)^2=3$$
so we only prove
$$a^3b^3+b^3c^3+a^3c^3\le 3 \tag{1}$$
but this (1) inequality is not true.such as
$$a=\dfrac{5}{\sqrt{17}},b=\dfrac{5}{\sqrt{17}},c=\dfrac{1}{\sqrt{17}}$$
but
$$a^3b^3+c^3a^3+b^3c^3\approx 3.23$$
see :inequality
so How prove this inequality (2),this (2) inequality is true, because I have use Maple test it.
Ps: when I join in the stack exchange,it is always appear
"Mathematics Stack Exchange requires external JavaScript from another domain, which is blocked or failed to load",That's why? then I can't Can't comment
Hello,chenbai,why it is clear $A,B,C,D,E,F\ge 0$, and this is not me download it. Thank you
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both side multiply $a^2+b^2+c^2=3$, we have:
$(a^2+b^2+c^2)(a^3b+b^3c+c^3a)+3(a^3b^3+b^3c^3+c^3a^3)\le \dfrac{2}{3}(a^2+b^2+c^2)^3 \iff 2(a^2+b^2+c^2)^3-3((a^2+b^2+c^2)(a^3b+b^3c+c^3a)+3(a^3b^3+b^3c^3+c^3a^3)) \ge 0 $
let $x= $min{$a,b,c$}, the other two are $x+u,x+v$, we will always have
$Ax^4+Bx^3+Cx^2+Dx+F \ge 0 $
$A=12u^2-12uv+12v^2 \\ B=31u^3-36u^2v-9uv^2+31v^3 \\C=27u^4-21u^3v-33u^2v^2+6uv^3+27v^4 \\D=9u^5-3u^4v-12u^3v^2-15u^2v^3+12uv^4+9v^4 \\ F=2v^6+6u^2v^4-12u^3v^3+6u^4v^2-3u^5v+2u^6$
$u,v$ may swap each other when $x=a$ or $b$ or $c$
it is not difficult to prove $A,B,C,D,F \ge 0 $ and when and only when $u=v=0$ to get $LHS=0$
QED
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When $n$ is divided by $14$, the remainder is $10$. What is the remainder when $n$ is divided by $7$? I need to explain this to someone who hasn't taken a math course for 5 years. She is good with her algebra.
This was my attempt:
Here's how this question works. To motivate what I'll be doing,
consider \begin{equation*} \dfrac{5}{3} = 1 + \dfrac{2}{3}\text{.}
\end{equation*} This is because when 5 is divided by 3, 3 goes into 5
once (hence the $1$ term) and there is a remainder of $2$ (hence the
$\dfrac{2}{3}$ term). Note the following: every division problem can
be decomposed into an integer (the $1$ in this case) plus a fraction,
with the denominator being what you divide by (the $3$ in this
case).
So, when $n$ is divided by 14, the remainder is 10. This can be
written as \begin{equation*} \dfrac{n}{14} = a + \dfrac{10}{14}
\end{equation*} where $a$ is an integer.
We want to find the remainder when $n$ is divided by 7, which I'll
call $r$. So \begin{equation*} \dfrac{n}{7} = b + \dfrac{r}{7}\text{,}
\end{equation*} where $b$ is an integer.
Here's the key point to notice: notice that \begin{equation*}
\dfrac{n}{7} = \dfrac{2n}{14} = 2\left(\dfrac{n}{14}\right)\text{.}
\end{equation*} This is because $\dfrac{1}{7} = \dfrac{2}{14}$.
Thus, \begin{equation*} \dfrac{n}{7} = 2\left(\dfrac{n}{14}\right) =
2\left(a + \dfrac{10}{14}\right) = 2a + 2\left(\dfrac{10}{14}\right) =
2a + \dfrac{10}{7} = 2a + \dfrac{7}{7} + \dfrac{3}{7} = (2a+1) +
\dfrac{3}{7}\text{.} \end{equation*} So, since $a$ is an integer, $2a
+ 1$ is an integer, which is our $b$ from the original equation. Thus, $r = 3$.
To her, this method was not very intuitive. She did understand the explanation. Are there any suggestions for how I can explain this in another way?
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$n=14k+10=7\cdot2\cdot k+7+3=(2k+1)7+3$, the remainder is $3$
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Trig Substitution Problem - Integration Suppose I want to integrate this (I chose an easy one):
$$\int \frac {dx}{\sqrt{x^2-4}}$$
Method 1: (Trig Substitution) $x=2\sec(\theta)$
$$\int \frac {dx}{\sqrt{x^2-4}}=\int \frac{\sec(\theta)\tan{(\theta)}}{|\tan(\theta)|}d\theta$$
Now removing the absolute value I find is tough. The inverse secant function splits out values of $\theta$ between $0$ and $\pi/2$, and $\pi/2$ to $\pi$. How can I resolve this? Tangent is positive from $0$ and $\pi/2$ - first quadrant, and negative from $\pi/2$ to $\pi$. I know most books overlook this, but I want to be most accurate.
I say "Method 1" because I also have another question right after this.
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Notice that the integrand $\dfrac{1}{\sqrt{x^2-4}}$ is only defined on $(-\infty,-2)\cup(2,\infty)$.
Hence, we can consider the antiderivative on $x > 2$ and on $x < -2$ seperately.
If $x > 2$, then we substitute $x = 2\sec \theta$ where $0 < \theta < \tfrac{\pi}{2}$.
Over this range, $\tan \theta > 0$. So, $\sqrt{x^2-4} = 2\tan \theta$, and thus,
$\displaystyle\int\dfrac{dx}{\sqrt{x^2-4}} = \int \dfrac{2\sec \theta \tan \theta}{|2\tan \theta|} \,d\theta = \int \dfrac{\sec \theta \tan \theta}{\tan \theta} \,d\theta = \int \sec \theta \,d\theta$
$= \ln|\sec\theta + \tan \theta|+C = \ln\left|\frac{x}{2}+\frac{\sqrt{x^2-4}}{2}\right|+C = \ln\left(\frac{x}{2}+\frac{\sqrt{x^2-4}}{2}\right)+C$
If $x < -2$, then we substitute $x = 2\sec \theta$ where $\tfrac{\pi}{2} < \theta < \pi$.
Over this range, $\tan \theta < 0$. So, $\sqrt{x^2-4} = -2\tan \theta$, and thus,
$\displaystyle\int\dfrac{dx}{\sqrt{x^2-4}} = \int \dfrac{2\sec \theta \tan \theta}{|2\tan \theta|} \,d\theta = \int \dfrac{\sec \theta \tan \theta}{-\tan \theta} \,d\theta = -\int \sec \theta \,d\theta$
$= -\ln|\sec\theta + \tan \theta|+C = -\ln\left|\frac{x}{2}-\frac{\sqrt{x^2-4}}{2}\right|+C = \ln\left|\dfrac{1}{\frac{x}{2}-\frac{\sqrt{x^2-4}}{2}}\right|+C$.
$= \ln\left|\dfrac{\frac{x}{2}+\frac{\sqrt{x^2-4}}{2}}{\left(\frac{x}{2}-\frac{\sqrt{x^2-4}}{2}\right)\left(\frac{x}{2}+\frac{\sqrt{x^2-4}}{2}\right)}\right|+C = \ln\left|\dfrac{\frac{x}{2}+\frac{\sqrt{x^2-4}}{2}}{\frac{x^2}{4}-\frac{x^2-4}{4}}\right|+C = \ln\left|\frac{x}{2}+\frac{\sqrt{x^2-4}}{2}\right|+C$
As you can see, the two expressions are the same. So, in this case, we don't lose anything by carelessly writing $\sqrt{x^2-4} = 2\tan \theta$ instead of $|2\tan \theta|$.
Alternatively, if $x < -2$, then substitute $x = 2\sec \theta$ where $\pi < \theta < \frac{3\pi}{2}$. Then $\tan \theta > 0$ over this range, and so, we don't need to worry about the absolute value signs.
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Closed Form for the Imaginary Part of $\text{Li}_3\Big(\frac{1+i}2\Big)$
$\qquad\qquad$ Is there any closed form expression for the imaginary part of $~\text{Li}_3\bigg(\dfrac{1+i}2\bigg)$ ?
Motivation: We already know that $~\Re\bigg[\text{Li}_3\bigg(\dfrac{1+i}2\bigg)\bigg]=\dfrac{\ln^32}{48}-\dfrac5{192}~\pi^2~\ln2+\dfrac{35}{64}~\zeta(3)$,
so I was wondering whether a similar closed form expression might also exist for its
imaginary part as well. Thank you !
Apparently, $~\Im~\text{Li}_3\bigg(\dfrac{1+i}2\bigg)~+~\Im~\text{Li}_3(1+i)~=~\dfrac7{128}\cdot\pi^3~+~\dfrac3{32}\cdot\pi\cdot\ln^22,~$ so the question
is equivalent to asking for the closed form of the imaginary part of $~\Im~\text{Li}_3(1+i).$
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Using the relation between 3 trilogarithms given here:
\begin{equation}
\operatorname{Li}_3(z)=-\operatorname{Li}_3(\frac{z}{z-1})-\operatorname{Li}_3(1-z)+\frac{1}{6}\log^3(1-z)-\frac{1}{2}\log(z)\log^2(1-z)+\frac{\pi^2}{6}\log(1-z)+\zeta(3)
\end{equation}
with $z=(1+i)/2$, we have $z/(z-1)=-i$ and $1-z=(1-i)/2$. With the known specific value (see here)
\begin{align}
\operatorname{Li}_3(-i)&=-\frac{3}{32}\zeta(3)-i\frac{\pi^3}{32}
\end{align}and remarking that $\operatorname{Li}_n(z)=\overline{\operatorname{Li}_n(\overline{z})}$,
\begin{equation}
\operatorname{Li}_3(z)+\operatorname{Li}_3(1-z)=2\Re\operatorname{Li}_3(\frac{1+i}{2})
\end{equation}
and
\begin{align}
\log\left( z \right)&=-\frac{\log 2}{2}+i\frac{\pi}{4}\\
\log\left(1- z \right)&=-\frac{\log 2}{2}-i\frac{\pi}{4}
\end{align}
we deduce
\begin{align}
\Re\operatorname{Li}_3(\frac{1+i}{2})&=\frac{1}{2}\left[
\frac{3}{32}\zeta(3)+i\frac{\pi^3}{32}
+ \frac{1}{6}\left( -\frac{\log 2}{2}-i\frac{\pi}{4} \right)^3-\right.\\
&\left.\frac{1}{2}\left( -\frac{\log 2}{2}+i\frac{\pi}{4} \right))\left( -\frac{\log 2}{2}-i\frac{\pi}{4} \right)^2+\frac{\pi^2}{6}\left( -\frac{\log 2}{2}-i\frac{\pi}{4} \right)+\zeta(3)
\right]\\
&=\frac{35}{64}\zeta(3)+\frac{\log^32}{48}-\frac{5}{192}\pi^2\ln 2
\end{align}
as expected.
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|
Find $ y''$ and $ y'''$ if $b^2x^2+a^2y^2=a^2b^2$ answers :- $$y''=-\frac{b^4}{a^2y^3}$$ and $$ y'''=-\frac{3b^6x}{a^4y^5}$$i have tried solve in this manner:-$$b^2x^2+a^2y^2=a^2b^2 -1.)$$diff wrt x :$$2b^2x+2a^2y'=0$$ $$a^2y'=-b^2x$$$$\frac{dy}{dx}=\frac{-b^2x}{a^2y}$$on substituting eq -1 we get :-$$\frac{dy}{dx}=\frac{-b^2x}{a \sqrt{a^2b^2-b^2x^2}}$$$$\frac{dy}{dx}=-a^{-1}.b^2.x.(a^2b^2-b^2x^2)^{-\frac{1}{2}}$$on diff wrtx :- $$y''=\frac{-b}{a}[(a^2b^2-b^2x^2)^{-\frac{1}{2}}+{-\frac{1}{2}}(a^2b^2-b^2x^2)^{-\frac{3}{2}}.-2b^2.x.x]$$$$y''=\frac{-b}{a}[(a^2b^2-b^2x^2)^{-\frac{1}{2}}+(a^2b^2-b^2x^2)^{-\frac{3}{2}}.b^2.x^2]$$now substituting$(a^2b^2-b^2x^2)$as $a^2y^2$ from eq -1.):- $$y''=b^2\frac{(a^2y^2)^3+ax^2}{a(a^2y^2)^{\frac{3}{2}}}$$**This was all i could have done someone please help**
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The question is not very clear but it seems that you are writing $y_1$ for the derivative of $y$ and $y_2$ for the second derivative.
If I am correct in this, then
$$\eqalign{b^2x^2+a^2y^2=a^2b^2\quad
&\Rightarrow\quad 2b^2x+2a^2yy'=0\cr
&\Rightarrow\quad 2b^2+2a^2(yy''+(y')^2)=0\qquad(*)\cr
&\Rightarrow\quad y''=-\frac{b^2}{a^2y}-\frac{(a^2yy')^2}{a^4y^3}\cr
&\Rightarrow\quad y''=-\frac{a^2b^2y^2+b^4x^2}{a^4y^3}\cr
&\Rightarrow\quad y''=-\frac{b^2(a^2b^2)}{a^4y^3}=-\frac{b^4}{a^2y^3}\ .\cr}$$
You can use similar ideas to find $y'''$: it will probably be easiest to begin by differentiating $(*)$.
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Generating functions to solve number of integer solution problem If I have $x_1 + x_2 + x_3 =10$ with $1\leq x_1 \leq 5, \; 2 \leq x_2 \leq 6, \;3 \leq x_3 \leq 9$
I know that I compute $(t^1+\dots + t^5)(t^2 +\dots + t^6)(t^3+\dots +t^9)$ and look at the coefficient of $t^{10}$ to find number of integer solutions.
But below I have $3a_2,5a_3$
Say I have $a_1+3a_2+5a_3=33$ with $1\leq a_1 \leq 11,\;3\leq a_2 \leq 18,\; 5 \leq a_3 \leq 13$.
Do I simply change it to $9 \leq 3a_2 \leq 54$ etc
and compute $(t^1+t^2+\dots +t^{10}+ t^{11})(t^9+t^{12}+\dots +t^{51}+t^{54})(t^{25}+t^{30}+\dots +t^{60}+t^{65})$ and look at coefficient $33$?
Meaning do I count the exponents up in the increased rate?
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You are looking for ($\dotsb$ at the end of a factor covers terms that don't affect the result):
$\begin{align}
[z^{10}] (z + \dotsb + z^5) &(z^2 + \dotsb + z^6) (z^3 + \dotsb + z^9) \\
&= [z^{10 - 1 - 2 - 3}] (1 + \dotsb + z^4) (1 + \dotsb + z^4) (1 + \dotsb + z^6) \\
&= [z^6] \frac{(1 - z^5)^2 (1 - z^7)}{(1 - z)^3} \\
&= [z^6] (1 - 2 z^5 - \dotsb) \sum_{k \ge 0} (-1)^k \binom{-3}{k} z^k \\
&= [z^6] (1 - 2 z^5) \sum_{k \ge 0} \binom{k + 3 - 1}{3 - 1} z^k \\
&= \binom{6 + 2}{2} - 2 \binom{6 - 5 + 2}{2} \\
&= 22
\end{align}$
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|
Integrate $\int\frac{\cos x dx}{\cos^{3/2}2x}=\frac{\sin x}{\sqrt{\cos 2x}}+C$ Wolfram gives this nice result:
$$\int\frac{\cos x dx}{\cos^{3/2}2x}=\frac{\sin x}{\sqrt{\cos 2x}}+\text{constant}$$
I have tried writing $\cos 2x = \cos^2x - \sin^2x $ and doing Weierstrass substitution $\tan (x/2) = t$ but its getting very complicated. Any help/hints ?
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$$\int\dfrac{\cos xdx}{\cos^{3/2}2x}=\int\dfrac{\cos xdx}{(\cos^2x-\sin^2x)^{3/2}}=$$
$$\dfrac{\cos xdx}{(1-\tan^2x)^{3/2}\cos^3x}=\int\dfrac{\sec^2xdx}{(1-\tan^2x)^{3/2}}$$
$$u=\tan x,du=\sec^2xdx$$
$$\int\frac{du}{(1-u^2)^{3/2}}$$
$$u=\sin\theta,du=\cos\theta d\theta$$
$$\int\dfrac{\cos\theta d\theta}{\cos^3\theta}=\tan\theta+C=\frac{u}{\sqrt{1-u^2}}+C=\frac{\tan x}{\sqrt{1-\tan^2x}}+C=$$
$$\dfrac{\sin x}{\sqrt{\cos^2x-\sin^2x}}+C=\dfrac{\sin x}{\sqrt{\cos2x}}+C$$
|
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|
Show that two expressions are equivalent I am trying to prove a hyperbolic trigonometric identity and I ran into the following expression:
$$\frac{\left (\sqrt{x^2+1}+x \right )^2+1}{2\left ( \sqrt{x^2+1} + x \right )} \quad.$$
This expression is supposed to be equivalent to
$$\sqrt{x^2+1} \quad.$$
I tried to algebraically manipulate the original expression to get the required expression and got to:
$$\frac{\left (\sqrt{x^2+1}+x \right )^2+1}{2\left ( \sqrt{x^2+1} + x \right )} = x+\frac{1}{\sqrt{x^2+1} + x} \quad.$$
but I don't know how I can show that
$$x+\frac{1}{\sqrt{x^2+1} + x} = \sqrt{x^2+1}\quad .$$
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For $b\ne 0$, $\frac{a}{b}=c\Leftrightarrow bc=a$.
So a strategy would be to try to show that
$$(\sqrt{x^2+1}+x)^2+1=2(\sqrt{x^2+1}+x)\cdot \sqrt{x^2+1}$$
|
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|
Prove of Nesbitt's inequality in 6 variables I was just reading "Pham kim hung secrets in inequalities,Volume 1" book and there was an interesting problem on it's Cauchy-Schwarz and Holder section that caught my eye.
Prove that for all positive real numbers $a,b,c,d,e,f$,we always have $$\frac{a}{b+c}+\frac{b}{c+d}+\frac{c}{d+e}+\frac{d}{e+f}+\frac{e}{f+a}+\frac{f}{a+b}\ge 3$$
The writer of the book proved it by Cauchy-Schwarz,But there was an another method in the start of the book for proving original Nesbitt inequality.
Prove that for all non-negative real numbers $a,b,c$ $$\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}\ge\frac{3}{2}$$
Solution: set $S=\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}$,$M=\frac{b}{b+c}+\frac{c}{c+a}+\frac{a}{a+b}$,$N=\frac{c}{b+c}+\frac{a}{c+a}+\frac{b}{a+b}$.
obliviously $M+N=3$.and by AM-GM we get the $$M+S\ge3,N+S\ge3$$
So $M+N+2S\ge6$ and $M+N=3$ we get $S\ge\frac{3}{2}$.
Like this method (calling it $S,M,N$),He proves $$\frac{a}{b+c}+\frac{b}{c+d}+\frac{c}{d+a}+\frac{d}{a+b}\ge2$$
for all non-negative $a,b,c,d$.
As I Liked this method, I started to proof the 6 variable variation using it. $$S=\sum\limits_{cyc}\frac{a}{b+c}$$ $$M=\sum\limits_{cyc}\frac{b}{b+c}$$ $$N=\sum\limits_{cyc}\frac{c}{b+c}$$
it easy too see that $M+N=6$.and using AM-GM it is easy to reach that $M+S\ge6$ but for proving $N+S \ge 6$ $$N+S=\frac{a+c}{b+c}+\frac{b+d}{c+d}+\frac{c+e}{d+e}+\frac{d+f}{e+f}+\frac{e+a}{f+a}+\frac{f+b}{a+b}\ge 6$$
I don't see any way to change this to something easy to work with it.By the way it is strange that why it is called Nesbitt's 6 variables inequality in book because Nesbitt Generalization is $$\sum_{i=1}^{n}\frac{a_i}{s-a_i}\ge\frac{n}{n-1}$$
Where $\sum_{i=1}^{n}a_i = s$ for positive $a_1,\ldots a_n$.
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$$\sum\limits_{i=1}^{6}\frac{x_i}{x_{i+1}+x_{i+2}} \ge 3$$ where, the indices are taken cyclically. Wlog, assume $\sum\limits_{i=1}^{6} x_i = 1$.
We start with the fact that $f(s) = \frac{1}{1-s}$ is convex on the interval $[0,1)$. Applying Jensen Inequality, $$\sum\limits_{i=1}^{6} \frac{x_i}{1-(x_i + x_{i-1}+x_{i-2}+x_{i-3})} \ge \frac{1}{1-\sum\limits_{i=1}^{6}x_i(x_i + x_{i-1}+x_{i-2}+x_{i-3})}$$
$$\iff \sum\limits_{i=1}^{6}\frac{x_i}{x_{i+1}+x_{i+2}} \ge \frac{1}{\left(\sum\limits_{i=1}^6 x_i\right)^2-\sum\limits_{i=1}^{6}x_i(x_i + x_{i-1}+x_{i-2}+x_{i-3})}$$
Now, $\displaystyle \left(\sum\limits_{i=1}^6 x_i\right)^2-\sum\limits_{i=1}^{6}x_i(x_i + x_{i-1}+x_{i-2}+x_{i-3}) = (x_1+x_4)(x_3+x_6)+(x_1+x_4)(x_2+x_5)+(x_2+x_5)(x_3+x_6)$
Since, $\displaystyle 1 = \left((x_1+x_4)+(x_3+x_6)+(x_2+x_5)\right)^2 \ge 3\left((x_1+x_4)(x_3+x_6)+(x_1+x_4)(x_2+x_5)+(x_2+x_5)(x_3+x_6)\right)$
the desired inequality follows.
As for the $N+S \ge 6$ inequality, is not always true.
Take the values $(a,b,c,d,e,f) = (1,0,1.02,0.02,1.01,0.02)$.
Then, $N+S = 5.99991989925 < 6$ (when I say $b = 0$ I mean to take a positive value as close to $0$ as possible)
|
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|
Closed form for $1 + 3 + 5 + \cdots +(2n-1)$ What is the closed summation form for $1 + 3 + 5 + \cdots + (2n-1)$ ?
I know that the closed form for $1 + 2 + 3+\cdots + n = n(n+1)/2$ and I tried plugging in $(2n-1)$ for $n$ in that expression, but it didn't produce a correct result:
$(2n-1)((2n-1)+1)/2$
plug in 3
$(2n-1)((2n-1)+1)/2 = 6 != 1 + 2 + 3 = 9$
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Sum of Arithmetic Progression $=\frac{N}{2}(a_1+a_n)$, where N is the nunber of terms
$N = \frac{(a_n-a_1)}{d} + 1 = \frac{(2n-2)}{2} + 1 = n-1+1 = n$
Thus the sum is $ = \frac{n}{2}(1+2n-1) = n^2$
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|
The remainder of $1^1+2^2+3^3+\dots+98^{98}$ mod $4$ How can I solve this problem:
If the sum $S=(1^1+2^2+3^3+4^4+5^5+6^6...+98^{98})$ is divided by $4$ then what is the remainder?
I know that all the even terms I can ignore since $(2n)^{2n}=4^nn^{2n})$ which is divisible by $4$,but i dont know what to do next it. Also I know we can write each of them as $(99-98)^1+(99-97)^2+(99-96)^3...+(99-1)^{98}$.
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In mod $4$, note that for any positive integer $m$,
$$(2m)^{2m}=(4m^2)^m\equiv 0$$$$(4m-1)^{4m-1}\equiv (-1)^{4m-1}\equiv -1$$$$(4m-3)^{4m-3}\equiv 1^{4m-3}\equiv 1$$
and that $97=4\cdot 25-3,\ 95=4\cdot 24-1$.
Hence, we have
$$S\equiv 24\cdot (-1)+25\cdot 1\equiv 1.$$
|
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|
Solve $x+3y=4y^3,y+3z=4z^3 ,z+3x=4x^3$ in reals
Find answers of this system of equations in reals$$
\left\{
\begin{array}{c}
x+3y=4y^3 \\
y+3z=4z^3 \\
z+3x=4x^3
\end{array}
\right.
$$
Things O have done: summing these 3 equations give $$4(x+y+z)=4(x^3+y^3+z^3)$$
$$x+y+z=x^3+y^3+z^3$$
I've also tried to show that $x,y,z$ should be between $1$ and $-1$(As $(-1,-1,-1)$ and $(1,1,1)$ are answers of this system of equations) but I was not successful.
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Consider $f(x) = 4x^3 - 3x$. We have: $f'(x) = 12x^2 - 3 = 0 \iff x = \pm \dfrac{1}{2}$. Thus:
If $x \in \left(-\dfrac{1}{2}, \dfrac{1}{2}\right)$ then $f'(x) < 0$, and $f$ decreases. This means:
If $x > y > z$, then: $x = f(y) < f(z) = y$, a contradiction. And we can obtain a contradiction for any other inequalities of $x,y,z$ using the same argument. The same argument works for $x \leq -\dfrac{1}{2}$ or $x \geq \dfrac{1}{2}$ since then $f$ increases.
For if $x > y = z$, then: $y = f(z) > f(x) = z$, contradiction again. Thus we must have:
$x = y = z$, and we have: $x = y = z = \pm 1$ and $0$
|
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|
Solving $y^3=x^3+8x^2-6x+8$ Solve for the equation $y^3=x^3+8x^2-6x+8$ for positive integers x and y.
My attempt- $$y^3=x^3+8x^2-6x+8$$
$$\implies y^3-x^3=8x^2-6x+8$$
$$\implies (y-x)(y^2+x^2+xy)=8x^2-6x+8$$
Now if we are able to factorise $8x^2-6x+8$ then we can compare LHS with RHS.Am I on the right track?Please help.
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We have
$y^3 − (x + 1)^3 = x^3 + 8x^2 − 6x + 8 − (x^3 + 3x^2 + 3x + 1) = 5x^2 − 9x + 7$.
Consider the quadratic equation
$5x^2 − 9x + 7 = 0$.
The discriminant of this equation is
$D = 92 − 4 × 5 × 7 = −59 < 0$ and hence the expression $5x^2 − 9x + 7$ is positive for all real values of x.
We conclude that $(x + 1)^3 < y^3$ and hence $x + 1 < y$.
On the other hand we have
$(x + 3)^3 − y^3 = x^3 + 9x^2 + 27x + 27 − (x^3 + 8x^2 − 6x + 8) = x^2 + 33x + 19 > 0$
for all positive $x$.
We conclude that $y < x + 3$.
Thus we must have $y = x + 2$. Putting this value of y, we get $0 = y^3 − (x + 2)^3 = x^3 + 8x^2 − 6x + 8 − (x^3 + 6x^2 + 12x + 8) = 2x^2 − 18x$.
We conclude that $x = 0$ and $y = 2$ or $x = 9$ and $y = 11$
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|
What is the sum of $\sum_{n=1}^\infty (n^2+n^3)x^{n-1}$? Consider the power sequence
$$\sum_{n=1}^\infty (n^2+n^3)x^{n-1}$$
What is the function to which it sums to?
My reasoning is to differntiate the sum with respect to $x$, then to integrate with respect to x from $0$ to $x$ after variation of the sum into a Taylor series form.
I mean:
$$\int _0 \frac{d}{dx}\sum_{n=1}^\infty (n^2+n^3)x^{n-1}dx=\int_0 \sum_{n=1}^\infty \frac{n^2+n^3}{n-1}x^{n-2}dx$$
But it doesn't seem to work in this exercise.
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This is a very general approach.
$$n^3+n^2 = n(n-1)(n-2) + 4n(n-1) + 2n$$
So $$(n^3+n^2)x^n = \left(x^3\left(\frac{d}{dx}\right)^3 + 4x^2\left(\frac{d}{dx}\right)^2 + 2x\frac{d}{dx}\right)x^n$$
So $$\sum (n^3+n^2)x^n =\left(x^3\left(\frac{d}{dx}\right)^3 + 4x^2\left(\frac{d}{dx}\right)^2 + 2x\frac{d}{dx}\right)\frac{1}{1-x}$$
In general, if $(n)_k = n(n-1)\cdots(n-k+1)$ is the falling factorial, then these are a basis for all polynomials, so if $p(n)$ is a polynomial of degree $d$, then we can write:
$$p(n)=\sum_{k=0}^{d} a_k(n)_k$$
Then $$\sum_{n=0}^\infty p(n)x^n = \left(\sum_{k=0}^d a_kx^k\left(\frac{d}{dx}\right)^k\right)\frac{1}{1-z}$$
Now $$\left(\frac{d}{dx}\right)^k\frac{1}{1-x} = \frac{k!}{(1-x)^{k+1}}$$
So that gives:
$$\sum_{n=0}^\infty p(n)x^n = \sum_{k=0}^d \frac{k!a_kx^k}{(1-x)^{k+1}}=\frac{\sum_{k=0}^d k!a_kx^k(1-x)^{d-k}}{(1-x)^{d+1}}$$
This shows that $(1-x)^{d+1}\sum_{n=0}^\infty p(n)x^n$ is a polynomial of degree at most $d$. The beauty of this is that you can just multiply polynomials:
$$(1-x)^{d+1}\sum_{n=0}^d p(n)x^n$$ and ignore the terms of degree bigger than $d$ to figure out what the numerator polynomial is.
So if $p(n)=n^3+n^2$ then
$$\begin{align}(1-x)^4(2x+12x^2+36x^3) &= 2x(1-4x+6x^2-4x^3+x^4)(1+6x+18x^2) \\
&= 2x(1+2x +0x^2 +\dots)
\end{align}$$
And your series is:
$$\sum_{n=1}^\infty (n^3+n^2)x^{n-1} = \frac{1}{x}\sum_{n=0}^\infty p(n)x^n=\frac{2(1+2x)}{(1-x)^4}$$
Interesting to note that $(1-x)^4\sum_{n=0}^d p(n)x^d$ is essentially the same as the approach in Semiclassical's answer - multiplying by $(1-x)$ repeatedly is essentially the same as doing his finite differences.
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|
Prove that $2|ab| \leq a^2 + b^2$ and $|a|+|b| \leq \sqrt {2}(a^2+b^2)^{1/2}$ I am having issues solving these problems:
I tried using $(|a|+|b|)^2 \geq 0$ and $(|a|-|b|)^2 \geq 0$ but I am having problems constructing the proof. I need to gain some intuition on how to proceed.
Show that for $a,b\in (-\infty,+\infty)$, the following inequalities apply:
a) $2|ab| \leq a^2 + b^2$
b) $|a|+|b| \leq \sqrt {2}(a^2+b^2)^{1/2}$
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(a) $0 \leq (|a|-|b|)^2 = |a|^2 -2|a||b|+|b|^2 = a^2 - 2|ab|+ b^2\Rightarrow 2|ab| \leq a^2+b^2 $
(b) by (a) we have $(|a| + |b|)^2 = a^2 + 2|ab| + b^2 \leq a^2+b^2+a^2 +b^2 = 2(a^2+b^2) \Rightarrow |a|+|b| \leq \sqrt {2(a^2+b^2)}$
|
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|
How to Solve $ \int \frac{dx}{x^3-1} $ I am having quite a difficult time integrating
$$
\int \frac{\mathrm{d}x}{x^3-1}
$$
My first approach was to apply a partial fraction decomposition
$$
\int \frac{\mathrm{d}x}{x^3-1} = \int \frac{\mathrm{d}x}{(x-1)(x^2+x+1)} = \frac{1}{3} \int \frac{\mathrm{d}x}{x-1} - \frac{1}{3} \int \frac{x+2}{x^2+x+1}\mathrm{d}x$$
The first term is simple enough to integrate but I can't figure out the second. I've tried each of the integration tricks in my arsenal, integration of parts gives a more complex equation, a trigonometric substitution doesn't seem to get me anywhere, and lastly I'm not sure what to do with a u-substitution since if $ u = x^2+x+1 $ then $ \mathrm{d}u = (2x+1)\mathrm{d}x \neq (x+2)\mathrm{d}x $. And further breaking up the formula further doesn't seem to eliminate these problems.
My last effort was to get a solution from Wolfram and try to reverse engineer it by taking the derivative
$$
\frac{\mathrm{d}}{\mathrm{d}x} \left( - \frac{1}{6} \ln (x^2+x+1) + \frac{1}{3} \ln (1-x) - \frac{\sqrt{3}}{3} \tan ^{ - 1} \left( \frac{2x+1}{\sqrt{3}} \right) \right)
$$
from which I got
$$
\frac{1}{3(1-x)} - \frac{1}{2(x^2 + x + 1)} - \frac{2x+1}{6x^2 + 6x + 6}
$$
However I can't seem to figure out the method or motivation for the fraction decomposition. I'm I missing something really obvious or am I missing a tool of integration? Thank you very much for your help!
From the first answer suggesting completing the square I believe I was able to figure the rest out. One then gets a difference of squares and can then use a trigonometric substitution where
$$ x + \frac{1}{2} = \frac{\sqrt{3}}{2} \tan \theta $$
the answer I got is
$$
\frac{1}{3} \ln |x-1| - \frac{1}{6} \ln(x^2+x+1) - \frac{\sqrt{3}}{3} \tan ^{ - 1} \left( \frac{2x+1}{\sqrt{3}} \right) + C
$$
My last confusion is the discrepancy in the solutions. Wolfram's second term seems to be undefined over $ (1, \infty) $, whereas I believe that $$
\int \frac{\mathrm{d}x}{x^3-1}
$$ should be defined over that interval.
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You can write $\displaystyle\int\frac{x+2}{x^2+x+1} dx=\frac{1}{2}\int\frac{2x+1}{x^2+x+1} dx + \frac{3}{2}\int\frac{1}{x^2+x+1} dx$
$\displaystyle=\frac{1}{2}\ln(x^2+x+1)+\frac{3}{2}\int\frac{1}{(x+\frac{1}{2})^2+\frac{3}{4}} dx$, and now let $u=x+\frac{1}{2}$.
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|
Solve the inequality: $|x^2 − 4| < 2$ This is a question on a calculus assignment our class received, I am a little confused on a few parts to the solution, can someone clear a few things up with it?
Since $x^2-4 = 0$ that means $x = 2$ and $x = -2$ are turning points.
When $x < -2$ :
$x^2-4<2$
$x^2<6$
$x < \sqrt{6}$ and $x < - \sqrt{6}$
But the solution says that $x > -\sqrt{6}$, what am I not getting?
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$$|x^2-4|<2 \\ \implies -2<x^2-4<2 \\ \implies 2<x^2<6 \\ \implies \sqrt{2}<\pm x<\sqrt{6} \\ \implies \sqrt{2}<x<\sqrt{6},\ \ -\sqrt{2}>x>-\sqrt{6}$$
|
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|
Find the exact length of the curve $y=\frac 12 x^2- \frac 12 \ln(x)$
Find the exact length of the curve $y = \frac 12 x^2- \frac 12 \ln(x)$, for $2 \le x \le 4$.
My attempt:
\begin{align}
L&= \int_2^4 \sqrt{1+\left[x-\frac 1{2x} \right]^2} \, dx \\
&= \int_2^4 \sqrt{1+x^2+\frac 1{4x^2}-1} \, dx \\
&= \int_2^4 \sqrt{x^2+\frac 1{4x^2}} \, dx
\end{align}
But I am stuck here. How can I proceed from this step onwards?
|
It is possible that there is a typo in the question: often in arclength problems, the constants are chosen so that $\sqrt{1+\left(\frac{dy}{dx}\right)^2}$ miraculously turns out to be the square of something nice. However, let us assume there is no typo.
We want to integrate $\frac{1}{2x}\sqrt{4x^4+1}$. Multiply top and bottom by $x^3$, and make the change of variable $4x^4+1=u^2$. We arrive at a standard integral.
|
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|
Two methods to integrate? Are both methods to solve this equation correct?
$$\int \frac{x}{\sqrt{1 + 2x^2}} dx$$
Method One:
$$u=2x^2$$
$$\frac{1}{4}\int \frac{1}{\sqrt{1^2 + \sqrt{u^2}}} du$$
$$\frac{1}{4}log(\sqrt{u}+\sqrt{{u} +1})+C$$
$$\frac{1}{4}log(\sqrt{2}x+\sqrt{2x^2+1})+C$$
Method Two
$$u=1+2x^2$$
$$\frac{1}{4}\int\frac{du}{\sqrt{u}}$$
$$\frac{1}{2}\sqrt{u}+C$$
$$\frac{1}{2}\sqrt{1+2x^2}+C$$
I am confused why I get two different answers.
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In first you have:
$$\frac{1}{4}\int \frac{1}{\sqrt{1+u}}du=\frac{1}{4}log(\sqrt{u}+\sqrt{{u} +1})+C$$
It should be:
$$\frac{1}{4}\int \frac{1}{\sqrt{1+u}}du=\frac{1}{4}\int(1+u)^{-\frac{1}{2}}du=\frac{1}{4}\cdot 2 \sqrt{1+u}+C$$
|
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|
Calculate the binomial sum $ I_n=\sum_{i=0}^n (-1)^i { 2n+1-i \choose i} $ I need any hint with calculating of the sum
$$
I_n=\sum_{i=0}^n (-1)^i { 2n+1-i \choose i}.
$$
Maple give the strange unsimplified result
$$
I_n={\frac {1/12\,i\sqrt {3} \left( - \left( \left( 1+i\sqrt {3} \right)
^{2\,{\it n}+2} \right) ^{2}+16\, \left( -1 \right) ^{2\,{\it n}}
\left( {2}^{2\,{\it n}} \right) ^{2} \right) }{{2}^{2\,{\it n}}
\left( 1+i\sqrt {3} \right) ^{2\,{\it n}+2}}},
$$
Сalculation for small $n$ are as follows $I_1=-1, I_2=0,I_3=1, I_4=-1, \ldots$
and leads to a hypothese:
$
I_n= -1 \text{ for } n=3k+1, =0, \text{ for } n=3k+2,=1, \text{ for } n=3k.
$
But how to prove it?
|
Start by restating the problem: we seek to evaluate
$$\sum_{q=0}^n (-1)^q {2n+1-q\choose q}
= \sum_{q=0}^n (-1)^q {2n+1-q\choose 2n+1-2q}.$$
Introduce the integral representation
$${2n+1-q\choose 2n+1-2q}
= \frac{1}{2\pi i}
\int_{|z|=\epsilon} \frac{(1+z)^{2n+1-q}}{z^{2n+2-2q}} \; dz.$$
This gives the following for the sum
(note that the integral correctly represents the fact of the binomial coefficient being zero for $q>n$, so we may extend the sum to infinity):
$$\frac{1}{2\pi i}
\int_{|z|=\epsilon} \frac{(1+z)^{2n+1}}{z^{2n+2}}
\sum_{q\ge 0} (-1)^q \frac{z^{2q}}{(1+z)^q} \; dz
\\ = \frac{1}{2\pi i}
\int_{|z|=\epsilon} \frac{(1+z)^{2n+1}}{z^{2n+2}}
\frac{1}{1+z^2/(1+z)} \; dz
\\ = \frac{1}{2\pi i}
\int_{|z|=\epsilon} \frac{(1+z)^{2n+2}}{z^{2n+2}}
\frac{1}{1+z+z^2} \; dz.$$
Extracting coefficients we find
$$\sum_{q=0}^{2n+1} {2n+2\choose 2n+1-q}
[z^q] \frac{1}{1+z+z^2}
= \sum_{q=0}^{2n+1} {2n+2\choose q+1}
[z^q] \frac{1}{1+z+z^2}.$$
Introduce the generating function
$$Q(w) = \sum_{n\ge 0} w^{2n+1}
\sum_{q=0}^{2n+1} {2n+2\choose q+1}
[z^q] \frac{1}{1+z+z^2},$$
which is
$$\sum_{q\ge 0} w^q [z^q] \frac{1}{1+z+z^2}
\sum_{2n+1\ge q} {2n+2\choose q+1} w^{2n+1-q}
\\ = \sum_{q\ge 0} w^q [z^q] \frac{1}{1+z+z^2}
\sum_{p\ge 0} {p+q+1\choose q+1} w^p
\\ = \sum_{q\ge 0} w^q [z^q] \frac{1}{1+z+z^2}
\frac{1}{(1-w)^{q+2}}
\\ = \frac{1}{(1-w)^2} \sum_{q\ge 0}
\left(\frac{w}{1-w}\right)^q [z^q] \frac{1}{1+z+z^2}.$$
What we have here is an annihilated coefficient extractor and the
sum simplifies to
$$Q(w) = \frac{1}{(1-w)^2}
\frac{1}{1+w/(1-w) + w^2/(1-w)^2}
\\ = \frac{1}{(1-w)^2+w(1-w) + w^2}
= \frac{1}{1-w+w^2}.$$
This is the ordinary generating function of a sequence with recurrence
$$a_{n+2} = a_{n+1} - a_n.$$
Since $a_0 = 1$ and $a_1 = 1$ we obtain
$$1, 1, 0, -1, -1, 0, 1, 1, 0, -1, -1, 0\ldots$$
thereby proving periodicity with period six.
Selecting the odd-index values preserves periodicity and we get the sequence
$$1, -1, 0, 1, -1, 0, \ldots$$
Post Scriptum.
If we introduce $$\rho_{1,2} = \frac{1\pm i\sqrt{3}}{2}
\quad\text{and set}\quad
c_{1,2} = \frac{3\pm i\sqrt{3}}{6}$$
we have the closed form
$$[w^n] \frac{1}{1-w+w^2}
= c_1 \rho_1^{-n} + c_2 \rho_2^{-n}$$
from which $[w^{2n+1}] Q(w)$ may be extracted.
The technique of annihilated coefficient extractors (ACE) is also employed at this
MSE link I and this MSE link II.
A trace as to when this method appeared on MSE and by whom starts at this
MSE link.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/936864",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 6,
"answer_id": 2
}
|
How to show this equals 1 without "calculations" We have $$ \sqrt[3]{2 +\sqrt{5}} + \sqrt[3]{2-\sqrt{5}} = 1 $$
Is there any way we can get this results through algebraic manipulations rather than just plugging it into a calculator?
Of course, $(2 +\sqrt{5}) + (2-\sqrt{5}) = 4 $, maybe this can in some way help?
|
I'll expand on Guillermo's comment: let
$$
a=\sqrt[3]{2+\sqrt{5}},\quad b=\sqrt[3]{2-\sqrt{5}}.
$$
You want to show that $a+b=1$. Consider
$$
(a+b)^3=a^3+3a^2b+3ab^3+b^3=(a^3+b^3)+3ab(a+b)=4+3\times(-1)(a+b)
$$
so if you let $x=(a+b)$ , then $x$ is a real number satisfying
$$
0=x^3+3x-4=x^3-x^2+x^2-x+4x-4\\
=(x-1)(x^2+x+4)=(x-1)\left[(x+0.5)^2+3.75\right].
$$
We infer that $x=1$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/937073",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
}
|
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