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Taylor series expansion for $f(x)=\sqrt{x}$ for $a=1$ I seem to be stuck defining an alternating sequence of terms in this series because $f^{(0)}(x)=f(x)$ is positive, as well as $f'(x)$, but then every other term starting with $f''(x)$ is negative. How can I define $f^{(n)}(x)$ given this?
\begin{array}{ll}
f(x)=x^{\frac{1}{2}} & f(1)=1 \\
f'(x)=\frac{1}{2}\cdot x^{-\frac{1}{2}} & f'(1)=\frac{1}{2} \\
f''(x)=(-1)^1\cdot\left(\frac{1}{2}\right)^{2}\cdot x^{-\frac{3}{2}} & f''(1)=(-1)^1\cdot\left(\frac{1}{2}\right)^{2} \\
f'''(x)=(-1)^2\cdot 3\cdot\left(\frac{1}{2}\right)^{3}\cdot x^{-\frac{5}{2}} & f'''(1)=(-1)^2\cdot3\cdot\left(\frac{1}{2}\right)^{3} \\
f^{(4)}(x)=(-1)^3\cdot3\cdot 5\cdot\left(\frac{1}{2}\right)^{4}\cdot x^{-\frac{7}{2}} & f^{(4)}(1)=(-1)^3\cdot3\cdot 5\cdot\left(\frac{1}{2}\right)^{4} \\
f^{(n)}(x)=(-1)^{n-1}\left(\frac{1}{2}\right)^{n}\cdot x^{\frac{1-2n}{2}} & f^{(n)}(1)=(-1)^{n-1}\left(\frac{1}{2}\right)^{n}
\end{array}
I thought I had the right answer until I realized that I'd be defining $f(x)$ to be negative.
|
This is a formula which won't display so well in a comment.
$$1\cdot3\cdot5\cdot7=\frac {1\cdot2\cdot3\cdot4\cdot5\cdot6\cdot7}{2\cdot4\cdot6}=\frac {7!}{2^33!}$$ You should be able to work out the general term from there.
Note also that there is no reason that every term of the sum has to fit the same neat formula. You can always write it as $$a_0+\sum_{r=1}^\infty a_rx^r$$ where $a_0$ is the term which does not fit the pattern, and $a_r$ has a general form.
|
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|
Evaluate $\int_2^\infty{\frac{3x-2}{x^2(x-1)}}$ To be shown that $\int_2^\infty{\dfrac{3x-2}{x^2(x-1)}}=1-\ln2$
My thought: $\dfrac{3x-2}{x^2(x-1)}=\dfrac{3x}{x^2(x-1)}-\dfrac{2}{x^2(x-1)}$
• $\dfrac{3x}{x^2(x-1)}=\dfrac{3}{x(x-1)}=\ldots=-\dfrac{3}{x}+\dfrac{3}{x-1}$
• $\dfrac{2}{x^2(x-1)}=\ldots=-\dfrac{2}{x^2}+\dfrac{2}{x-1}$
But developing the sum of the integrals of the above two gives ln of infinite in my results. I don't know what I am doing wrong. Can you please confirm the dots above? Although I have several times. Thanks a lot
|
HINT:
Using Partial Fraction Decomposition,
$$\dfrac{3x-2}{x^2(x-1)}=\frac A{x^2}+\frac Bx+\frac C{x-1}$$
$$\implies 3x-2=A(x-1)+Bx(x-1)+Cx^2=(B+C)x^2+x(A-B)-A$$
Comparing the constants in the above identity $A=2$
Comparing the coefficients of $x,A-B=3\implies B=A-3=-1$
Comparing the coefficients of $x^2,B+C=0\implies C=-B=1$
Alternatively, $$ 3x-2=A(x-1)+Bx(x-1)+Cx^2$$
Putting $x=1$ in the above identity, $C=3\cdot 1-2=1$
Putting $x=0,-2=A(0-1)\implies A=2$
Comparing the coefficients of $x^2,B+C=0\implies B=-C=-1$
|
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|
Finding indefinite integral $$\int\frac{\sqrt{a-x}}{\sqrt{a}-\sqrt{x}}\, dx$$
Can someone give a hint as to how to break this apart into a do-able integral?
Just spent ages trying to substitute $\sqrt{a-x}$ and $\sqrt{x}$ as $u$, without success.
Advice/help much appreciated.
|
Note that $$\frac{{\sqrt {a - x} }}{{\sqrt a - \sqrt x }} = \frac{{\sqrt {a - x} }}{{a - x}}\left( {\sqrt a + \sqrt x } \right) = \frac{{\sqrt a + \sqrt x }}{\sqrt{a - x}}$$
This allows to get an easy integral in the first summand. For the second one, we make a series of elementary yet (maybe) non-obvious manipulations
$$\begin{align}
\int {\sqrt {\frac{x}{{a - x}}} dx} &\mathop = \limits^{\left( 1 \right)} \int {\sqrt {\frac{a}{u} - 1} \left( { - du} \right)} \\
&\mathop = \limits^{\left( 2 \right)} \int {w\frac{{2wa}}{{{{\left( {{w^2} + 1} \right)}^2}}}dw} \\
&\mathop = \limits^{\left( 3 \right)} - \frac{{aw}}{{1 + {w^2}}} + a\int {\frac{{dw}}{{{w^2} + 1}}} \\
&\mathop = \limits^{\left( 4 \right)} - \frac{{aw}}{{1 + {w^2}}} + a\arctan w\end{align} $$
Explanation
$(1)$ Make $u=a-x$
$(2)$ Make $w^2=\dfrac ua-1$
$(3)$ Integrate by parts with $f=w,g'=\dfrac{2w}{(1+w^2)^2}$
$(4)$ Use the usual trigonometric integral $(\tan^{-1}x)'=(1+x^2)^{-1}$
The above gives $$\int {\sqrt {\frac{x}{{a - x}}} dx} = a\arctan \sqrt {\frac{x}{{a - x}}} - \sqrt {x\left( {a - x} \right)} + C$$
|
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|
Find the determinant of the following general matrix Let $A_r$ and $B_r$ be the $r\times r$ matrix blocks
$A_r=\left(
\begin{array}{A}
1-t & t^2 & 0 & 0 & 0 & \cdots & 0 \\
t^2 & 0 & 0 & 0 & 0 & \cdots & 0 \\
0 & 0 & t^2 & 0 & 0 & \cdots & 0 \\
0 & 0 & 0 & t^2 & 0 & \cdots & 0 \\
0 & 0 & 0 & 0 & t^2 & \cdots & 0 \\
\vdots & \vdots & \vdots & \vdots & \vdots & \ddots & 0 \\
0 & 0 & 0 & 0 & 0 & \cdots & t^2 \\
\end{array}
\right)$,
$B_r=\left(
\begin{array}{B}
0 & t^2 & 0 & 0 & 0 & \cdots & 0 & -1 \\
-1 & 0 & 0 & 0 & 0 & \cdots & 0 & 1-t \\
0 & -1 & 0 & 0 & 0 & \cdots & 0 & 1-t \\
0 & 0 & -1 & 0 & 0 & \cdots & 0 & 1-t \\
0 & 0 & 0 & -1 & 0 & \cdots & 0 & 1-t \\
\vdots & \vdots & \vdots & \vdots & \ddots & \ddots & \vdots & 1-t \\
0 & 0 & 0 & 0 & 0 & \ddots & 0 & 1-t \\
0 & 0 & 0 & 0 & 0 & \cdots & -1 & 1-t \\
\end{array}
\right)
$.
Find the determinant of the following $rs\times rs$ matrix $C$:
$$C = \left(
\begin{array}{C}
A_r & B_r & 0 & 0 & 0 & \cdots & 0 & 0 \\
0 & A_r & B_r & 0 & 0 & \cdots & 0 & 0 \\
0 & 0 & A_r & B_r & 0 & \cdots & 0 & 0 \\
0 & 0 & 0 & A_r & B_r & \cdots & 0 & 0 \\
0 & 0 & 0 & 0 & A_r & \ddots & 0 & 0 \\
\vdots & \vdots & \vdots & \vdots & \vdots & \ddots & \ddots & \vdots \\
0 & 0 & 0 & 0 & 0 & \cdots & A_r & B_r \\
B_r & 0 & 0 & 0 & 0 & \cdots & 0 & A_r \\
\end{array}
\right) $$
Following user1551's answer, I find the matrix $A^{-1}B$ to be
$$
-\frac{1}{t^4} \left(
\begin{array}{D}
t^2 & 0 & 0 & 0 & 0 & \cdots & 0 & -t^2 \\
-(1-t) & -t^4 & 0 & 0 & 0 & \cdots & 0 & 1-t+t^2 \\
0 & t^2 & 0 & 0 & 0 & \cdots & 0 & t^2(1-t) \\
0 & 0 & t^2 & 0 & 0 & \cdots & 0 & t^2(1-t) \\
0 & 0 & 0 & t^2 & 0 & \cdots & 0 & t^2(1-t) \\
\vdots & \vdots & \vdots & \vdots & \ddots & \ddots & \vdots & \vdots \\
0 & 0 & 0 & 0 & 0 & \ddots & 0 & t^2(1-t) \\
0 & 0 & 0 & 0 & 0 & \cdots & t^2 & t^2(1-t) \\
\end{array}
\right)
$$
But then I am stuck at calculating $(-A^{-1}B)^s$...
(By the way, from the SciLab calculations I have done, I guess that the final result of the determinant depends on $\mu$, $D$ and $N$ where $r$ = $\mu D$ and $s$ = $\mu N$ and $D$ and $N$ are relatively prime.)
|
Here is a partial answer. For convenience, let us drop the subscript $r$. Write $C=\pmatrix{A&R\\ S&T}$, where $T$ is $r(s-1)\times r(s-1)$. Note that
$$
T^{-1} =
\left[\begin{array}{rrrrr}
A^{-1}&-A^{-1}BA^{-1}&A^{-1}BA^{-1}BA^{-1}&\cdots&(-1)^{s-2}(A^{-1}B)^{s-2}A^{-1}\\
&A^{-1}&-A^{-1}BA^{-1}&\ddots&\ddots\\
&&\ddots&\ddots&\ddots\\
&&A^{-1}&-A^{-1}BA^{-1}&A^{-1}BA^{-1}BA^{-1}\\
&&&A^{-1}&-A^{-1}BA^{-1}\\
&&&&A^{-1}
\end{array}\right].
$$
Therefore, using Schur complement, we get
\begin{align*}
\det(C) &= \det(T)\det(A-RT^{-1}S)\\
&= \det(A)^{s-1} \det(A - (-1)^{s-2}B(A^{-1}B)^{s-2}A^{-1}B)\\
&= \det(A)^s \det(I - (-1)^{s-2}A^{-1}B(A^{-1}B)^{s-2}A^{-1}B)\\
&= (-t^{2r})^s \det(I - (-A^{-1}B)^s).
\end{align*}
So, the question boils down to finding $\det(I - (-A^{-1}B)^s)$.
|
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|
Maximize $(a-1)(b-1)(c-1)$ knowing that : $a+b+c=abc$. If : $a,b,c>0$, and : $a+b+c=abc$, then find the maximum of $(a-1)(b-1)(c-1)$.
I noted that : $a+b+c\geq 3\sqrt{3}$, I believe that the maximum is at : $a=b=c=\sqrt{3}$. (Can you give hints).
|
Let $a+b+c=3u$, $ab+ac+bc=3v^2$ and $abc=w^3$.
Hence, the condition does not depend on $v^2$ and we need to find a maximum of
$$abc-ab-ac-bc+a+b+c-1$$ or
$$6u-3v^2-1,$$
which is linear function of $v^2$,
which says that it's enough to find a maximum for an extremal value of $v^2$,
which happens for equality case of two variables.
Let $b=a$.
Hence, $c=\frac{2a}{a^2-1}$, where $a>1$ and we need to find $\max\limits_{a>1}f$, where
$$f(a)=(a-1)^2\left(\frac{2a}{a^2-1}-1\right)^2,$$
which gives that
$$\max\limits_{a>1}f=-10+6\sqrt3.$$
Done!
|
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|
Is it possible to calculate $3^{-1}\equiv ?\pmod{10}$? If I wanted to calculate $3^{-1}\equiv ?\pmod{10}$ would I first calcuate $3^1$ which is just
$3\equiv 3\pmod{10}$ and then divide both sides by $3^2$ which would get $3^{-1}\equiv 3^{-1} mod{10}$ Then im not sure what to do next. My book states that $3^{-1}\equiv 7\pmod{10}$ which I don't know how they get?
|
A number $x^{-1}$ is one such that $x \cdot x^{-1} = 1$ (here $\mod 10$). In this case it is easy, because there is only 10 possibilities: $0,1,2,3,4,5,6,7,8,9$, with product being $0,3,6,9,2,5,8,1,4,7$ respectively. That means there exists, and there exists only one number such that $$x \cdot x^{-1} = 1 \mod 10.$$ There are other solutions like $17, 27, 37$ but those are $\geq 10$ or $< 0$ (curiously $-3$ is such a number, that is $3\cdot -3 = -9 = 1 \mod 10$).
The general approach can be done via the Euclidean algorithm, which for any numbers $a,b$ lets you find $x,y$ such that
$$a x + b y = \gcd(a,b).$$
Happily $3$ and $10$ are coprime, that is $\gcd(3,10) = 1$, which allows as to run the algorithm and get $3 \cdot (-3) + 10 \cdot (1) = 1$, giving us $x = -3$, now it is enough to move $x$ to appropriate interval, that is $x+10k$ such that $0 \leq x+10k < 10$, and that gives us $7$.
I hope this helps ;-)
|
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|
Simplifying the expression $(1+\sqrt[4]3)/(1-\sqrt[4]3)+1/(1+\sqrt[4]3)+2/(1+\sqrt{3})$ Can anyone give provide me some help to simplify this expression?
The three denominators are pretty much different, and I can't find a common denominator.
$$\frac{1+\sqrt[4]3}{1-\sqrt[4]3}+\frac1{1+\sqrt[4]3}+\frac2{1+\sqrt{3}}$$
The calculator said it's equal to $-2$, but I don't get how a complicated expression like this would be equal to $-2$.
|
Putting $\sqrt[4]3=x$
$$\frac{1+\sqrt[4]3}{1-\sqrt[4]3}+\frac1{1+\sqrt[4]3}+\frac2{1+\sqrt{3}}$$
$$=\frac{1+x}{1-x}+\frac1{1+x}+\frac2{1+x^2}$$
$$=\frac x{1-x}+\frac1{1-x}+\frac1{1+x}+\frac2{1+x^2}$$
$$=\frac x{1-x}+\frac{1+x+1-x}{(1-x)(1+x)}+\frac2{1+x^2}$$
$$=\frac x{1-x}+2\left(\frac1{1-x^2}+\frac1{1+x^2}\right)$$
$$=\frac x{1-x}+2\left(\frac{1+x^2+1-x^2}{(1-x^2)(1+x^2)}\right)$$
$$=\frac x{1-x}+\frac4{1-x^4}$$
Now $x^4=3$ as $\sqrt[4]3=x$
|
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|
For what natural numbers is $n^3 < 2^n$? Prove by induction Problem
For what natural numbers is $n^3 < 2^n$?
Attempt @ Solution
*
*For $n=1$, $1 < 2$
*Suppose $n^3 < 2^n$ for some $n = k \ge 1$
*It looks like the inequality is true for $n = 0$, $n = 1$ and $n\ge10$
*But, how can I prove this through induction?
|
Claim: If $n \in \{0,1\} \cup \{n \in \Bbb{N} \mid n \ge 10\}$, then $n^3<2^n$.
You have verified the claim for $n=0$ and $n=1$. The next time the inequality holds is at $n=10$, since $10^3=1000<1024=2^{10}$. This is our base case.
Induction Hypothesis: Assume that $n^3<2^n$ holds true for $n=k\ge10$.
It remains to prove the claim true for $n=k+1$. Observe that since $10 \le k$, we have:
$$ \begin{align*}
(k+1)^3 &= k^3+3k^2+3k+1 \\
&< k^3+9k^2+9k+10 \\
&\le k^3+9k^2+9k+(k) \\
&= k^3+9k^2+10k \\
&\le k^3+9k^2+(k)k \\
&= k^3+10k^2 \\
&\le k^3+(k)k^2 \\
&= 2k^3 \\
&< 2(2^k) & \text{by the induction hypothesis}\\
&= 2^{k+1}
\end{align*} $$
as desired. This completes the induction.
|
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|
Let $f(x) = \int \frac{x}{1-x^{8}}dx\,$
*
*Let $f(x) = \int \frac{x}{1-x^{8}}dx\,$.
*
*Represent $I(x)$ by a power series $\sum^{\infty}a_{n}x^{n}$.(Find $a_{n}$)
*What is the radius of convergence of $I(x)$ ?
*Two curves are generated by polar equations $r=1+\sin\theta$ and $r=-\sin\theta$.
*
*Find the area of the region that lies inside both two curves.
*Find the length of the part of the curve $r=1+\sin\theta$ that lies inside the curve $r=-\sin\theta$.
|
HINT:
For the first problem, $$\frac x{1-x^8}=\frac x2\left(\frac1{1-x^4}+\frac1{1+x^4}\right)$$
$$=\frac x4\left(\frac1{1-x^2}+\frac1{1+x^2}\right)+\frac12\frac x{1+x^4}$$
$$=\frac x4\left(\frac1{1-x^2}\right)+\frac14\frac x{1+x^2}+\frac12\frac x{1+x^4}$$
$$=\frac 18\left(\frac{1+x-(1-x)}{1-x^2}\right)+\frac14\frac x{1+x^2}+\frac12\frac x{1+x^4}$$
$$=\frac 18\frac1{1-x}-\frac18\frac1{1+x}+\frac14\frac x{1+x^2}+\frac12\frac x{1+x^4}$$
For the last two terms put $x^2=u$
You will find all required the Series Formula here
In fact, we can make the substitution $x^2=u$ from the very start, but then we need substitute back $u$ with $x$ as we need the Power Series of $x$
|
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|
Why does $ \frac{2x}{2+x}$ provide a particularly tight lower bound for $\ln(1+x)$ for small positive values of $x$? EDIT:
My question was poorly worded.
I wasn't asking about showing $\ln(1+x) > \frac{2x}{2+x}$ for $x>0$.
What I wanted to know is why the lower bound provided by $ \frac{2x}{2+x}$ was so tight for small positive values of $x$. This is addressed in robjohn's answer.
|
As I mentioned in a comment, the power series near $x=0$ are
$$
\begin{align}
\log(1+x)&=x-\frac{x^2}{2}+\frac{x^3}{3}+O(x^4)\\
\frac{2x}{2+x}&=x-\frac{x^2}{2}+\frac{x^3}{4}+O(x^4)\\
\log(1+x)-\frac{2x}{2+x}&=\frac{x^3}{12}+O(x^4)
\end{align}
$$
So near $x=0$, the value, and the first and second derivatives match. That means the functions $\log(1+x)$ and $\frac{2x}{2+x}$ match to second order. Not quite as simple, $\frac{x(6+x)}{6+4x}$ matches $\log(1+x)$ to third order. Rational approximations to functions are called Padé Approximations.
Furthermore,
$$
\begin{align}
\frac{\mathrm{d}}{\mathrm{d}x}\left(\log(1+x)-\frac{2x}{2+x}\right)
&=\frac{\mathrm{d}}{\mathrm{d}x}\left(\log(1+x)+\frac{4}{2+x}-2\right)\\
&=\frac1{1+x}-\frac4{(2+x)^2}\\
&=\frac{x^2}{(1+x)(2+x)^2}
\end{align}
$$
So, for $x\gt-1$, $\log(1+x)-\frac{2x}{2+x}$ is an increasing function. At $x=0$, $\log(1+x)-\frac{2x}{2+x}=0$.
Therefore, for $x\gt0$,
$$
\log(1+x)\gt\frac{2x}{2+x}
$$
and for $-1\lt x\lt0$,
$$
\log(1+x)\lt\frac{2x}{2+x}
$$
|
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Prove that $2^n < \binom{2n}{n} < 2^{2n}$ Prove that $2^n < \binom{2n}{n} < 2^{2n}$. This is proven easily enough by splitting it up into two parts and then proving each part by induction.
First part: $2^n < \binom{2n}{n}$. The base $n = 1$ is trivial. Assume inductively that some $k$ satisfies our statement. The inductive step can be proved as follows.
$2^k < \binom{2k}{k} \implies 2^{k+1} < 2\binom{2k}{k} = \frac{2(2k)!}{k!k!} = \frac{2(2k)!(k + 1)}{k!k!(k + 1)} = \frac{2(2k)!(k+2)}{(k+1)!k!}<\frac{(2k)!(2k+2)(2k+1)}{(k+1)!k!(k+1)} = \binom{2(k+1)}{k+1}$
Second part: $2^{2n} > \binom{2n}{n}$. Again, the base is trivial. We can assume that for some $k$ our statement is satisfied and prove that inductive step as follows:
$2^{2k} > \binom{2k}{k} \implies 2^{2k + 2} > 2^2\binom{2k}{k} = \frac{2\cdot2(2k)!}{k!k!} = \frac{2\cdot2(2k)!(k+1)(k+1)}{k!k!(k+1)(k+1)} = \frac{(2k)!(2k+2)(2k+2)}{(k+1)!(k+1)!} > \frac{(2k)!(2k+1)(2k+2)}{(k+1)!(k+1)!} = \binom{2(k+1)}{k+1}$
Is there a non-inductive derivation for the inequality?
|
For the second part note that $$2^{2n}=(1+1)^{2n}=1+\binom {2n}1 +\dots + \binom {2n}n+\dots\gt \binom{2n}n$$
|
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|
Show the sequence $(1 - \frac{1}{n})^{-n}$ is decreasing. How do you show the sequence $(1 - \frac{1}{n})^{-n}$ is decreasing?
I understand that the binomial theorem should be used here but I don't see how we can use it to prove that $a_{n+1} < a_n$.
I will rewrite the sequence as,
\begin{align*}
(1 - \frac{1}{n})^{-n} &= (\frac{n-1}{n})^{-n} \\
&= (\frac{n}{n-1})^n \\
&= (1 + \frac{1}{n-1})^n
\end{align*}
Then I can apply binomial theorem to it.
This is as far as I got now.
|
Let
$$a_n=(1 - \frac{1}{n})^{-n}=\frac{n^n}{(n-1)^n}$$
Then
$$\frac{a_n}{a_{n+1}}=\frac{n^n}{(n-1)^n}\frac{n^{n+1}}{(n+1)^{n+1}}=\frac{n^{2n+2}}{(n-1)^{n+1}(n+1)^{n+1}}\frac{n-1}{n}$$
$$=\left( \frac{n^{2}}{n^2-1} \right)^{n+1} \frac{n-1}{n}=\left( 1+\frac{1}{n^2-1} \right)^{n+1} \frac{n-1}{n}$$
By Bernoully or Binomial Theorem
$$\frac{a_n}{a_{n+1}} \geq \left( 1+\frac{n+1}{n^2-1} \right) \frac{n+1}{n}=\frac{n}{n-1} \frac{n-1}{n}=1$$
|
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|
Finding $\frac{a+b}{a-b}$ such that $a^2+b^2=6ab$ For $a,b > 0$ such that $a^2+b^2=6ab$ .How to find $\frac{a+b}{a-b}$
|
We have $a^2+b^2=6ab$.
To both sides, add $2ab$ to obtain, $(a+b)^2 = 8ab$.
Similarly, subtract $2ab$ to obtain, $(a-b)^2=4ab$.
Thus, $\left(\dfrac{a+b}{a-b}\right)^2 = 2$.
So ultimately $\frac{a+b}{a-b}= \pm\sqrt2$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/453044",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 0
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|
Existence of linear mapping I am studying for an exam in linear algebra and I am having trouble solving the following:
Do linear mappings $\phi : \mathbb{R}^2 \rightarrow \mathbb{R}^2$ with the following properties exist?
$1)$ $\phi_1 \begin{pmatrix} 2 \\ 0 \end{pmatrix} = \begin{pmatrix} 0 \\ 1 \end{pmatrix} $, $\phi_1 \begin{pmatrix} 1 \\ 1 \end{pmatrix} = \begin{pmatrix} 5 \\ 2 \end{pmatrix} $, $\phi_1 \begin{pmatrix} 1 \\ 2 \end{pmatrix} = \begin{pmatrix} 2 \\3 \end{pmatrix} $
$2)$ $\phi_2 \begin{pmatrix} 2 \\ 3 \end{pmatrix} = \begin{pmatrix} 0 \\0\end{pmatrix} $, $\phi_2 \begin{pmatrix} 0 \\ 0 \end{pmatrix} = \begin{pmatrix} 2\\ 3 \end{pmatrix} $
$3)$ $\phi_3 \begin{pmatrix} 1 \\ -1 \end{pmatrix} = \begin{pmatrix} 1\\ 0 \end{pmatrix} $, $\phi_3 \begin{pmatrix} -2 \\ 5 \end{pmatrix} = \begin{pmatrix} -3 \\ 1 \end{pmatrix} $, $\phi_3 \begin{pmatrix} 2 \\ 1 \end{pmatrix} = \begin{pmatrix} 1 \\1 \end{pmatrix} $
I know that the following properties have to hold for a linear mapping:
*
*$f(\mathbf{x}+\mathbf{y}) = f(\mathbf{x})+f(\mathbf{y})$
*$f(\alpha \mathbf{x}) = \alpha f(\mathbf{x})$
*$f(0) = 0 $
I conclude that $2)$ is a not a linear mapping since $\phi(0)$ is not mapped to $0$.
But how shall I proceed with the others?
|
Very fundamental theorem says that linerar mapping from a vector space is the same thing as any function from some basis. Any such function can be uniquelly extended to whole vector space. In your case for example vectors $(2, 0)$, $(1, 1)$ are lineary independent and so form a basis of $\mathbb{R}^2$. So there exists unique linear mapping $\phi_1$ such that first two conditions hold. You have to find out if this $\phi_1$ satisfies $\phi_1(1, 2) = (2, 3)$. You can do this by decomposing $(1, 2)$ to basis $(2, 0)$, $(1, 1)$. You should be also able to write down the matrix of $\phi_1$ wrto standard basis and see what value it takes on $(1, 2)$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/453988",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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|
Finding the limit $\lim \limits_{n \to \infty}\ (\cos \frac x 2 \cdot\cos \frac x 4\cdot \cos \frac x 8\cdots \cos \frac x {2^n}) $ This limit seemed quite unusual to me as there aren't any intermediate forms or series expansions which are generally used in limits. Stuck on this for a while now .Here's how it goes :
$$
\lim \limits_{n \to \infty}
\left[\cos\left(x \over 2\right)\cos\left(x \over 4\right)
\cos\left(x \over 8\right)\ \cdots\ \cos\left(x \over 2^{n}\right)\right]
$$
|
Setting
$$
u_n(x)=\cos\frac{x}{2}\cdot\cos\frac{x}{4}\cdot\cdots\cos\frac{x}{2^n},
$$
we have
$$
v_n(x):=u_n(x)\cdot\sin\frac{x}{2^n}=u_{n-1}(x)\cdot\cos\frac{x}{2^n}\cdot\sin\frac{x}{2^n}=\frac12u_{n-1}(x)\cdot\sin\frac{x}{2^{n-1}}=\frac12v_{n-1}(x).
$$
It follows that
$$
v_n(x)=\frac{1}{2^{n-1}}v_1(x)=\frac{1}{2^{n-1}}\cdot\cos\frac{x}{2}\cdot\sin\frac{x}{2}=\frac{1}{2^n}\sin x.
$$
Hence, provided $\sin\frac{x}{2^n} \ne 0$, we have
$$
u_n(x)=\frac{\sin x}{2^n\sin\frac{x}{2^n}}.
$$
Thus
$$
\lim_{n\to\infty}u_n(x)=\lim_{\xi\to0}\frac{\sin x}{x}\cdot\left(\frac{\sin\xi}{\xi}\right)^{-1}=\frac{\sin x}{x}.
$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/455995",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "24",
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|
$2+2 = 5$? error in proof $$\begin{align} 2+2 &= 4 - \frac92 +\frac92\\
&= \sqrt{\left(4-\frac92\right)^2} +\frac92\\
&= \sqrt{16 -2\times4\times\frac92 +\left(\frac92\right)^2} + \frac92\\
&= \sqrt{16 -36 + \left(\frac92\right)^2} +\frac92\\
&= \sqrt {-20 +\left(\frac92\right)^2} + \frac92\\
&= \sqrt{25-45 +\left(\frac92\right)^2} +\frac92\\
&= \sqrt {5^2 -2\times5\times\frac92 + \left(\frac92\right) ^2} + \frac92\\
&= \sqrt {\left(5-\frac92\right)^2} +\frac92\\
&= 5 + \frac92 - \frac92 \\
&= 5\end{align}$$
Where did I go wrong
|
Here's what your "proof" would look like correcting all the errors. As you can see, it's not nearly as impressive as a proof that 2+2=5.
$$\begin{align} 2+2 &= 4 - \frac92 +\frac92\\
&= -\sqrt{(4-\frac92)^2} +\frac92\\
&= -\sqrt{16 -2\times4\times\frac92 +(\frac92)^2} + \frac92\\
&= \left(-\sqrt{16 -36 + (\frac92)^2}\right) +\frac92\\
&= \left(-\sqrt {-20 +(\frac92)^2}\right) + \frac92\\
&= -\sqrt{25-45 +(\frac92)^2} +\frac92\\
&= -\sqrt {5^2 -2\times5\times\frac92 + (\frac92) ^2} + \frac92\\
&= -\sqrt {(5-\frac92)^2} +\frac92\\
&= -5 + \frac92 + \frac92 \\
&= -5+9\end{align}$$
For reference, the most serious mistake was in the 2nd line. In general, it's not true that $\sqrt{x^2} = x$, but rather $\sqrt{x^2} = |x|$. For $x=4-\frac92<0$, you need to keep track of the extra minus sign coming from the absolute value. Other than that, there were some obvious typos that I've corrected.
|
{
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"url": "https://math.stackexchange.com/questions/457490",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "16",
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|
Question involving approximation, taylor series and proving Question:
Consider the approximation $$\ln(2)\approx 2\left ( \frac{1}{3}+\frac{1}{3\times 3^{3}}+\frac{1}{5\times 3^{5}} \right )$$
Prove that the error in this approximation is less than $$\frac{1}{7\times 2^{2} \times 3^{5}}$$
Attempt:
It looks like the expression comes from the taylor series expansion so:
$\ln(1+x)=x-\frac{x^{2}}{2}+\frac{x^{3}}{3}-\frac{x^{4}}{4}+... \text{ for }\ -1< x< 1$
$\ln(1-x)=-x-\frac{x^{2}}{2}-\frac{x^{3}}{3}-\frac{x^{4}}{4}+...$
$\therefore \ln\left ( \frac{1+x}{1-x} \right )=2\left ( x+\frac{x^{3}}{3}+\frac{x^{5}}{5}+\frac{x^{7}}{7} \right )$
$\text{Now let}\ x=\frac{1}{3}$
$\therefore \ln(2)=2\left ( \frac{1}{3}+\frac{1}{3\times 3^{3}}+\frac{1}{5\times 3^{5}}+\frac{1}{7\times 3^{7}}\right )$
So we have to prove that:
$2\left (\frac{1}{7\times 3^{7}}+\frac{1}{9\times 3^{9}}+\frac{1}{11\times 3^{11}}\cdots\right) < \frac{1}{7\times 2^{2} \times 3^{5}}$
|
Since $9\gt 7$, and $11\gt 7$, and $13\gt 7$, and so on, the tail
$$\frac{2}{7\cdot 3^7}+\frac{2}{9\cdot 3^9}+\frac{2}{11\cdot 3^{11}}+\frac{2}{13\cdot 3^{13}}+\cdots$$
is less than the sum of the geometric series
$$\frac{2}{7\cdot 3^7}\left(1+\frac{1}{3^2}+\frac{1}{3^4}+\frac{1}{3^6}+\cdots\right).$$
But
$$1+\frac{1}{3^2}+\frac{1}{3^4}+\frac{1}{3^6}+\cdots=\frac{9}{8}.$$
|
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"language": "en",
"url": "https://math.stackexchange.com/questions/459146",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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|
How to check whether this sum converge or diverge? Consider the following sum:
$$\sum_{k=1}^\infty\prod_{j=1}^k\frac{1}{\sqrt{j+1}-\sqrt{j}+1}$$
How could I check whether this sum converge or diverge? Root and ratio tests are inconclusive...
|
A different solution, which is not mine, is the following:
Let $\alpha_{k}=\prod_{j=1}^{k}\frac{1}{\sqrt{j+1}-\sqrt{j}+1}\,,\; k\in\mathbb{N}$, then
\begin{align*}
1-\frac{\alpha_{k+1}}{\alpha_k}&=1-\frac{\prod_{j=1}^{k+1}\frac{1}{\sqrt{j+1}-\sqrt{j}+1}}{\prod_{j=1}^{k}\frac{1}{\sqrt{j+1}-\sqrt{j}+1}}\\
&=1-\frac{1}{\sqrt{k+2}-\sqrt{k+1}+1}\\
&=\frac{\sqrt{k+2}-\sqrt{k+1}}{\sqrt{k+2}-\sqrt{k+1}+1}\\
&=\frac{k+2-k-1}{\big(\sqrt{k+2}-\sqrt{k+1}+1\big)\big(\sqrt{k+2}+\sqrt{k+1}\big)}\\
&=\frac{1}{\sqrt{k+2}+\sqrt{k+1}+1}\,.
\end{align*}
Because
\begin{align*}
\mathop{\lim}\limits_{k\to+\infty}\frac{k}{\sqrt{k+2}+\sqrt{k+1}+1}&=\mathop{\lim}\limits_{k\to+\infty}\frac{\sqrt{k}}{\sqrt{1+\frac{2}{k}}+\sqrt{1+\frac{1}{k}}+\frac{1}{\sqrt{k}}}=+\infty\,,
\end{align*}
by Raabe's criterion, we have that the series $\sum_{{k}=1}^{\infty}\big(\prod_{j=1}^{k}\frac{1}{\sqrt{j+1}-\sqrt{j}+1}\big)$ converges.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/462174",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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|
$n\text{ odd}\implies n^2=8k+1$ for some $k\in \mathbb{Z}$ So this girl tells me "Did you know that if $n$ is odd, then $n^2=8k+1$ for some $k\in \mathbb{Z}$?" And so I was like, "Really?" She said, "Yeah!" So I wrote this down:
If $n$ is odd, then $n=2m+1$, and so by squaring $n$ we get
$$n^2=(2m+1)^2=8(\frac{1}{2}m^2+\frac{1}{4}m)+1,$$
which means that for any $m$ of the form $m=4t$ for some $t\in \mathbb{Z}$ we get that
$$8(\frac{1}{2}m^2+\frac{1}{4}m)+1=8(\frac{1}{2}(4t)^2+\frac{1}{4}(4t))+1=8(8t^2+t)+1,$$
so for any value of $t$ we can find a value $m$ which makes
$$8(\frac{1}{2}m^2+\frac{1}{4}m)+1=n^2,$$
for some $n$ that is odd, as desired.
My question here is, is this a good approach to proving what the girl said?
|
The problem is that your proof doesn't handle the case when $m$ is not a multiple of $4$ - but the claim is still true then (e.g. $n = 5$, so that $m = 2$).
A much easier way to proceed is as follows: If $n$ is odd, then $n$ leaves a remainder of either $1$ or $3$ upon division by $4$. If the remainder is $1$, we find $n = 4m + 1$, giving
$$n^2 = 16 m^2 + 16 m + 1 = 8(2m^2 + 2m ) + 1$$
On the other hand, if the remainder is $3$ and $n = 4m + 3$, we get
$$n^2 = 16 m^2 + 24 m + 9 = 8 (2m^2 + 3m + 1) + 1$$
as desired.
Alternatively, if $n = 2m + 1$, then
$$n^2 = 4m^2 + 4m + 1 = 4(m^2 + m) + 1$$
But $m^2 + m$ is always even, regardless of $m$; hence, $m^2 + m = 2k$ for some $k$ and $n^2 = 8k + 1$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/463258",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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|
can someone explain this limit i have,
$$\lim_{x\to 0} \frac {\sqrt{x+49}-7}{3-\sqrt{x+9}}$$ the correct answer is $-\frac{3}{7}$ and in my case the result is $\frac{7}{3}$ i don't understand.
i tried this
$\lim_{x\to 0} \frac {\sqrt{x+49}-7}{3-\sqrt{x+9}}$=$\lim_{x\to 0} \frac {\sqrt{x+49}-7}{3-\sqrt{x+9}}.\frac {\sqrt{x+49}+7}{\sqrt{x+49}+7}.\frac {3+\sqrt{x+9}}{3+\sqrt{x+9}}$=
$\lim_{x\to 0} \frac {x+49-49}{9-x+9}.\frac {\sqrt{x+49}+7}{3+\sqrt{x+9}}= \frac {x(\sqrt{x+49}-7)}{x(3-\sqrt{x+9})}=\frac{14}{3}=\frac{7}{3}$ i don't know what is the error!
|
You can also write your expression as
$$\frac{7}{3}\frac{\sqrt{1+x/49}-1}{1-\sqrt{1+x/9}}$$
If $x$ is small, then $\sqrt{1+x/49}=1+x/98, \sqrt{1+x/9}=1+x/18$
and finally we have $-3/7.$
|
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"language": "en",
"url": "https://math.stackexchange.com/questions/464866",
"timestamp": "2023-03-29T00:00:00",
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|
How to solve $\sin x \cdot\sin 2x\cdot\sin 3x + \cos x\cdot\cos 2x\cdot\cos 3x =1$ How to solve $\sin x \cdot\sin 2x\cdot\sin 3x + \cos x\cdot\cos 2x\cdot\cos 3x =1$
I don't know the solution for this.
Help me!
Thank all!
|
The use of the "multiple-angle" formulas is probably the surest way to establish the solutions to this equation; however, it requires a goodly amount of algebraic manipulations. We can also say something about the solutions by investigating the properties of the terms in the equation
$$( \ \sin x \cdot \sin 2x \cdot \sin 3x \ ) \ + \ ( \ \cos x \cdot \cos 2x \cdot \cos 3x \ ) \ = \ 1 \ .$$
Something to consider is that both product terms of the sum have the same period, since the individual factors have periods of $ \ 2 \pi \ , \ \pi \ , \ \text{and} \ \frac{2 \pi}{3} \ . $ The period of each product term is $ \ \pi \ $ , since both $ \ \sin x \ \ \text{and} \ \ \sin 3x \ $ (or $ \ \ \cos x \ \ \text{and} \ \ \cos 3x \ $ ) complete an odd number of half-cycles in intervals of that length, restoring each product to the same parity at $ \ x = \pi \ $ that it had at $ \ x = 0 \ . $ So we only need to focus for the present on the behavior of the terms in the interval $ \ [ \ 0 \ , \ \pi \ ) \ . $
The first term has zeroes at all multiples of $ \frac{\pi}{2} \ \text{and} \ \frac{\pi}{3} \ $ , while the second term has its zeroes at all odd multiples of $ \frac{\pi}{6} \ , \ \frac{\pi}{4} \ \text{and} \ \frac{\pi}{2} \ . $ Upon checking the sign of each factor in both product terms, we find that the "product of sines" is positive on $ \ ( \ 0 \ , \ \frac{\pi}{3} \ ) \ \ \text{and} \ \ ( \ \frac{\pi}{2} \ , \ \frac{2 \pi}{3} \ ) \ , $ while the "product of cosines" is positive on $ \ ( \ 0 \ , \ \frac{\pi}{6} \ ) \ , \ ( \ \frac{\pi}{4} \ , \ \frac{\pi}{2} \ ) \ , \ ( \ \frac{\pi}{2} \ , \ \frac{3 \pi}{4} \ ) \ , \ \text{and} \ \ ( \ \frac{5 \pi}{6} \ , \ \pi \ ) \ . $ Both terms, then, are positive only on $ \ ( \ 0 \ , \ \frac{\pi}{6} \ ) \ , \ ( \ \frac{\pi}{4} \ , \ \frac{\pi}{3} \ ) \ , \ \text{and} \ \ ( \ \frac{\pi}{2} \ , \ \frac{2 \pi}{3} \ ) \ . \ $ [There are, of course, corresponding "windows" in every other periodic interval.] This is important to what follows.
As the terms are products of sine and cosine functions, it is the case that
$$-1 \ \le \ \sin x \cdot \sin 2x \cdot \sin 3x \ \le \ +1 \ , $$
and similarly for the product-of-cosines. But we can be more specific about the range of these terms in the intervals where both are positive.
At $ \ x = 0 \ , $ $ \ \cos x \cdot \cos 2x \cdot \cos 3x \ $ plainly equals 1 and falls to zero at $ \ x = \frac{\pi}{6} \ $ , since all three factors are decreasing. In the same interval, all three factors of $ \ \sin x \cdot \sin 2x \cdot \sin 3x \ $ are increasing and
$$0 \ \le \ \sin x \cdot \sin 2x \cdot \sin 3x \ \le \ (\sin \frac{\pi}{6}) \cdot (\sin \frac{\pi}{3}) \cdot (\sin \frac{\pi}{2}) \ \le \ \frac{\sqrt{3}}{4} \ \approx \ 0.433 \ . $$
So we can be reasonably satisfied that the only solution to the equation in the interval $ \ [ \ 0 \ , \ \frac{\pi}{6} \ ] \ $ is $ \ x = 0 \ . \ $ [It may require a bit more work -- short of making a graph -- to be completely satisfied that this is so.]
In the interval $ \ ( \ \frac{\pi}{4} \ , \ \frac{\pi}{3} \ ) \ , \ $ we note that
$$0 \ \le \ \sin x \cdot \sin 2x \cdot \sin 3x \ \le \ (\sin \frac{\pi}{3}) \cdot (\sin \frac{\pi}{2}) \cdot (\sin \frac{3 \pi}{4}) \ \le \ \frac{\sqrt{6}}{4} \ \approx \ 0.612 \ $$
and
$$0 \ \le \ \cos x \cdot \cos 2x \cdot \cos 3x \ \le \ (\cos \frac{\pi}{4}) \cdot (\cos \frac{2 \pi}{3}) \cdot (\cos \pi) \ \le \ \frac{\sqrt{2}}{4} \ \approx \ 0.354 \ . $$
Thus, there is no prospect that the sum of these terms can equal 1 in this interval. (The fact that the individual terms change in opposite directions over the interval guarantees this. And, in any event, the products do not even attain the indicated upper bounds.)
Finally, on $ \ ( \ \frac{\pi}{2} \ , \ \frac{2 \pi}{3} \ ) \ , \ $ we may conclude that
$$0 \ \le \ \sin x \cdot \sin 2x \cdot \sin 3x \ \le \ (\sin \frac{\pi}{2}) \cdot (\sin \frac{4 \pi}{3}) \cdot (\sin \frac{3 \pi}{2}) \ \le \ \frac{\sqrt{3}}{2} \ \approx \ 0.866 \ $$
and
$$0 \ \le \ \cos x \cdot \cos 2x \cdot \cos 3x \ \le \ (\cos \frac{2 \pi}{3}) \cdot (\cos \frac{4 \pi}{3}) \cdot (\cos 2 \pi) \ \le \ \frac{1}{4} \ . $$
Somewhat closer checking is needed to show that the "product of sines" falls well short of the upper bound suggested here, and that this product reaches its maximum and then decreases before the "product of cosines" attains its upper bound. So, again, the sum of terms fails to reach 1 in this interval.
Therefore, in the entire interval $ \ [ \ 0 \ , \ \pi \ ) \ , \ $ the only solution found for the equation is $ \ x = 0 \ . $ This means that the set of all solutions to the equation contains only all integer multiples of $ \ \pi \ . $ A graph of each term and of their sum is presented below.
The orange curve represents the "product of sines", the green curve, the "product of cosines", and the blue curve, their sum.
|
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"timestamp": "2023-03-29T00:00:00",
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|
How can I calculate this integral I solve integral considerable number, but the following integral I could not solve. It is about this integral:
$$
\int\frac{\ln^2(a+bx)dx}{x^n}
$$
I want to make the solution through partial method: $\int udv=uv-\int vdu$
I've solved the integral $$\int\frac{\ln(a+bx)dx}{x^n}$$.
Please help, thank you for your help preliminarily
|
This is just an (failed) attempt to find a reduction formula, and as @mtiano noted, the close form requires some hypergeometric functions. Let
$$\begin{align}
u =& \frac{\ln(a+bx)}{x^n} \\
du =& \frac{\frac{bx^n}{a+bx}-n x^{n-1}\ln(a+bx)}{x^{2n}}dx
= \left[ \frac{b}{x^n(a+bx)}-\frac{n\ln(a+bx)}{x^{n+1}}\right]dx\\
dv =& \ln(a+bx) dx \\
v =& \frac{(a+bx)\ln(a+bx)}{b}-x
\end{align}$$
Also let (since OP has solved it)
$$I_n(x)=\int\frac{\ln(a+bx)dx}{x^n}$$
$$\begin{align}
J_n(x)=&\int\frac{\ln^2(a+bx)dx}{x^n}\\
=& \int\frac{\ln(a+bx)}{x^n}\ln(a+bx)dx\\
=& \frac{\ln(a+bx)}{x^n} \cdot \left[ \frac{(a+bx)\ln(a+bx)}{b}-x \right]- \\
& \int{\left[ \frac{(a+bx)\ln(a+bx)}{b}-x\right] \left[\frac{b}{x^n(a+bx)}-\frac{n\ln(a+bx)}{x^{n+1}}\right]dx}\\
=& \frac{(a+bx)\ln^2(a+bx)}{bx^n}-\frac{\ln(a+bx)}{x^{n-1}} -\\
& \int \left[ \frac{\ln(a+bx)}{x^n} - \frac{b}{x^{n-1}(a+bx)}- \frac{n(a+bx)\ln^2(a+bx)}{bx^{n+1}} +\frac{n\ln(a+bx)}{x^{n}}\right]dx \\
=& \frac{(a+bx)\ln^2(a+bx)}{bx^n}-\frac{\ln(a+bx)}{x^{n-1}} - (n+1) I_n(x) +\frac{na}{b} J_{n+1}(x) + nJ_n(x) +\\
& \int\frac{b\,dx}{x^{n-1}(a+bx)}\\
=& \frac{(a+bx)\ln^2(a+bx)}{bx^n}-\frac{\ln(a+bx)}{x^{n-1}} - (n+1) I_n(x) +\frac{na}{b} J_{n+1}(x) + nJ_n(x) +\\
& \left(-\frac{b}{a}\right)^{n-1} \int \frac{b\,dx}{a+bx}
- \sum_{k=1}^{n-1} \left(-\frac{b}{a}\right)^{n-k}\int\frac{dx}{x^k} &\text{(1)}\\
=& \frac{(a+bx)\ln^2(a+bx)}{bx^n}-\frac{\ln(a+bx)}{x^{n-1}} - (n+1) I_n(x) +\frac{na}{b} J_{n+1}(x) + nJ_n(x) +\\
& \left(-\frac{b}{a}\right)^{n-1} \ln(a+bx) -\left(-\frac{b}{a}\right)^{n-1} \ln x - \sum_{k=2}^{n-1} \left(-\frac{b}{a}\right)^{n-k}\frac{1}{(-k+1)x^{k-1}} &\text{(2)}\\
\end{align}$$
Notice, for line $\text{(1)}$, the final summation term exists only for $n \ge 2$. For line $\text{(2)}$, the second last term (involving $\ln x$) exists only for $n \ge 2$, and the final summation term exists only for $n \ge 3$.
Making $J_{n+1}(x)$ the main term,
$$\begin{align}
-\frac{na}{b} J_{n+1}(x)=& \frac{(a+bx)\ln^2(a+bx)}{bx^n}-\frac{\ln(a+bx)}{x^{n-1}} - (n+1) I_n(x) + (n-1)J_n(x) +\\
& \left(-\frac{b}{a}\right)^{n-1} \ln(a+bx) -\left(-\frac{b}{a}\right)^{n-1} \ln x - \sum_{k=2}^{n-1} \left(-\frac{b}{a}\right)^{n-k}\frac{1}{(-k+1)x^{k-1}}\\
J_{n+1}(x)=& -\frac{(a+bx)\ln^2(a+bx)}{nax^n}+\frac{b\ln(a+bx)}{nax^{n-1}} + \frac{b(n+1)}{na} I_n(x) - \frac{b(n-1)}{na}J_n(x) +\\
& \left(-\frac{b}{a}\right)^{n} \frac{\ln(a+bx)}{n} -\left(-\frac{b}{a}\right)^{n} \frac{\ln x}{n} - \frac{1}{n}\sum_{k=2}^{n-1} \left(-\frac{b}{a}\right)^{n-k+1}\frac{1}{(-k+1)x^{k-1}}\\
\end{align}$$
This gives a reduction formula for $J_{n+1}(x)$ for $n = 1, 2, 3, \ldots$. Same as above, the second last term (involving $\ln x$) exists only for $n \ge 2$, and the final summation term exists only for $n \ge 3$. Lastly, I could not solve the base case
$$J_1(x) = \int \frac{\ln^2 (a + bx)dx}{x}$$
yet, and reading from WolframAlpha the solution requires polylogarithm function.
|
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"timestamp": "2023-03-29T00:00:00",
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|
Compute $\int x^2 \cos \frac{x}{2} \mathrm{d}x$ I am trying to compute the following integral:
$$\int x^2 \cos \frac{x}{2} \mathrm{d}x$$
I know this requires integration by parts multiple times but I am having trouble figuring out what to do once you have integrated twice. This is what I have done:
Let $u = \cos \frac{x}{2}$ and $\mathrm{d}u = \frac{-\sin\left(\dfrac{x}{2}\right)}{2} \mathrm{d}x$ and $\mathrm{d}v = x^2$ and $v= \frac{x^3}{3}$.
\begin{align}
&\int x^2 \cos \frac{x}{2} \\
&\cos \frac{x}{2} \cdot \frac{x^3}{3} - \int \frac{x^3}{3} \cdot \frac{-\sin\left(\dfrac{x}{2}\right)}{2} \mathrm{d}x
\end{align}
So now I integrate $\int \frac{x^3}{3} \cdot \frac{-\sin\left(\dfrac{x}{2}\right)}{2} \mathrm{d}x$ to get:
\begin{align}
\frac{-\sin\left(\dfrac{x}{2}\right)}{2} \cdot \frac{x^4}{12} - \int \frac{x^4}{12} \cdot \frac{\cos x}{2}
\end{align}
Now, this is where I get stuck. I know if I continue, I will end up with $\frac{-\sin\left(\dfrac{x}{2}\right)}{2}$ again when I integrate $\cos \frac{x}{2}$. So, where do I go from here?
Thanks!
|
HINT
Go in the opposite direction. Let $u = x^2$, $dv = \cos \frac{x}{2}$.
|
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|
Prove that $\sum_{n=1}^{\infty} \frac{1}{2^n - 1}$ is convergent and $\sum_{n=1}^{\infty} \frac{1}{2^n - 1} < 2$. Prove that $\sum_{n=1}^{\infty} \frac{1}{2^n - 1}$ is convergent and $\sum_{n=1}^{\infty} \frac{1}{2^n - 1} < 2$.
I'm not sure, but I suppose that $$\sum_{n=1}^{\infty} \frac{1}{2^n - 1} < \sum_{n=0}^{\infty} \frac{1}{2^n} = 2$$
I don't know how can I prove it exactly, but we have:
$$\sum_{n=1}^{\infty} \frac{1}{2^n - 1} = 1+ \frac{1}{3} + \frac{1}{7} + \frac{1}{15}+... \\
\sum_{n=0}^{\infty} \frac{1}{2^n} = 1+ \frac{1}{2} + \frac{1}{4} + \frac{1}{8}+...$$
Because
$$1 \le 1 \\ \frac{1}{3} < \frac{1}{2} \\ \frac{1}{7} < \frac{1}{4} \\ \frac{1}{15} < \frac{1}{8} \\ \mbox{and so on}$$
we have
$$\sum_{n=1}^{\infty} \frac{1}{2^n - 1} = 1+ \frac{1}{3} + \frac{1}{7} + \frac{1}{15}+... < 1+ \frac{1}{2} + \frac{1}{4} + \frac{1}{8}+... = \sum_{n=0}^{\infty} \frac{1}{2^n}$$
Could you tell me, does it work? Maybe exist more formal proof?
|
Notice that
$$\sum_{n=0}^\infty \frac{1}{2^n}=\sum_{n=1}^\infty \frac{1}{2^{n-1}}=2$$
and the result follows since
$$\frac{1}{2^{n-1}}\geq \frac{1}{2^{n}-1}\iff 2^{n-1}\geq1,\; \forall n\geq1\quad\text{which's true}$$
|
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"timestamp": "2023-03-29T00:00:00",
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|
How prove that $10(a^3+b^3+c^3)-9(a^5+b^5+c^5)\le\dfrac{9}{4}$ let $a,b,c\ge 0$, such that $a+b+c=1$, prove that
$$10(a^3+b^3+c^3)-9(a^5+b^5+c^5)\le\dfrac{9}{4}$$
This problem is simple as 2005, china west competition problem
$$10(a^3+b^3+c^3)-9(a^5+b^5+c^5)\ge 1$$
see:(http://www.artofproblemsolving.com/Forum/viewtopic.php?p=362838&sid=00aa42b316d41e251e24e658594fcc51#p362838)
for 2005 china west problem we have two methods (at least)
solution 1：
note
$$(a+b+c)^3=a^3+b^3+c^3+3(a+b)(b+c)(a+c)$$
$$(a+b+c)^5=a^5+b^5+c^5+5(a+b)(b+c)(a+c)(a^2+b^2+c^2+ab+bc+ac)$$
then
$$\Longleftrightarrow 10[1-3(a+b)(b+c)(a+c)]-9[1-5(a+b)(b+c)(a+c)(a^2+b^2+c^2+ab+bc+ac)]\ge 1$$
$$\Longleftrightarrow 3(a+b)(b+c)(a+c)(a^2+b^2+c^2+ab+bc+ac)-2(a+b)(b+c)(a+c)\ge 0$$
$$\Longleftrightarrow 3(a^2+b^2+c^2+ab+bc+ac)\ge 2=2(a+b+c)^2$$
$$ a^2+b^2+c^2-ab-bc-ac\ge 0$$
It's Obviously.
solution 2:
$$10(a+b+c)^2(a^3+b^3+c^3)-9(a^5+b^5+c^5)-(a+b+c)^5\ge0$$
it is equivalent to
$$15(a+b)(b+c)(c+a)(a^2+b^2+c^2-ab-bc-ca)\ge0$$
But for
$$10(a^3+b^3+c^3)-9(a^5+b^5+c^5)\le\dfrac{9}{4}$$
and for this equality I think
$$10a^3-9a^5\le p (a-1/3)+q$$
and let
$$f(x)=10x^3-9x^5\Longrightarrow f'(x)=30x^2-45x^4\Longrightarrow p=f'(1/3)=\dfrac{25}{9}$$
$$q=f(1/3)=\dfrac{1}{9}$$
so if we can prove this
$$10a^3-9a^5\le\dfrac{25}{9}(a-1/3)+\dfrac{1}{9}$$
These methods I can't work, can someone help deal it. Thank you
|
Set $c=1-a-b$ and find the three lines on which $\frac{\partial f(a,b)}{\partial a\partial b}=0$. Now derive $f(a,b(a))$ according to $a$ to find the extrema. This is a bit tedious, but you fill get the desired $9/4$.
|
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|
Find the value of $\textrm{cosec}^2\left(\frac\pi7\right) +\textrm{cosec}^2\left(\frac{2\pi}7\right)+\textrm{cosec}^2\left(\frac{4\pi}7\right)$ What is the value of $$\textrm{cosec}^2\left(\frac\pi7\right) +\textrm{cosec}^2\left(\frac{2\pi}7\right)+\textrm{cosec}^2\left(\frac{4\pi}7\right) \qquad\qquad ? $$
I tried to write $\textrm{cosec}^2\left(\frac{4\pi}7\right)$ as $\textrm{cosec}^2\left(\frac{3\pi}7\right)$. Then converted in $\sin$... But in vain.. Is there any other approach?
|
If $7x=\pi,4x=\pi-3x$
$\implies \sin4x=\sin(\pi-3x)=\sin3x$
$\implies 2\sin2x\cos2x=3\sin x-4\sin^3x$
$\implies 4\sin x\cos x\cos2x=3\sin x-4\sin^3x$
If $\sin x\ne0,$ we have $4\cos x\cos2x=3-4\sin^2x\implies 4\cos x(1-2\sin^2x)=3-4\sin^2x$
On squaring & rearrangement, $64(\sin^2x)^3-112(\sin^2x)^2+56\sin^2x-7=0$
which is a cubic equation in $\sin^2x$ with roots being $\sin^2\frac{r\pi}7$
where $r=(1$ or $6),(2$ or $5)$ and $(3$ or $4)$ as $\sin\frac{(7-r)\pi}7=\sin (\pi-\frac{r\pi}7)=\sin\frac{r\pi}7$
Using Vieta's Formula we have,
$\displaystyle\sin^2\frac{\pi}7\sin^2\frac{2\pi}7\sin^2\frac{4\pi}7=\frac7{64}$ and $\displaystyle\sin^2\frac{\pi}7\sin^2\frac{2\pi}7+\sin^2\frac{2\pi}7\sin^2\frac{4\pi}7+\sin^2\frac{4\pi}7\sin^2\frac{\pi}7=\frac{56}{64}$
We need to find
$\displaystyle\frac1{\sin^2\frac{\pi}7}+\frac1{\sin^2\frac{2\pi}7}+\frac1{\sin^2\frac{4\pi}7}=\displaystyle\frac{\sin^2\frac{\pi}7\sin^2\frac{2\pi}7+\sin^2\frac{2\pi}7\sin^2\frac{4\pi}7+\sin^2\frac{4\pi}7\sin^2\frac{\pi}7}{\sin^2\frac{\pi}7\sin^2\frac{2\pi}7\sin^2\frac{4\pi}7}=\frac{\frac{56}{64}}{\frac7{64}}=\frac{56}7=8$
|
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"timestamp": "2023-03-29T00:00:00",
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|
Prove by induction $\frac{n^n}{3^n}$\dfrac{n^n}{3^n}<n!<\dfrac{n^n}{2^n}$
The case $n!<\dfrac{n^n}{2^n}$ is easier.
|
1) Show that $n!<\frac{n^n}{2^n}$ for $n\ge6$
a) This is true for $n=6$, since $6!=720<729=3^6$.
b) Assume that $n!<\frac{n^n}{2^n}$ for some integer $n\ge6$.
Then $(n+1)!=(n+1)n!<(n+1)\cdot\frac{n^n}{2^n}$, and
$\frac{(n+1)^n}{n^n}=\big(\frac{n+1}{n}\big)^n=\big(1+\frac{1}{n}\big)^n\ge1+n(\frac{1}{n})=2$ by Bernoulli's inequality;
so $(n+1)\cdot\frac{n^n}{2^n}=\frac{(n+1)^{n+1}}{2^n}\cdot\frac{n^n}{(n+1)^n}\le\frac{(n+1)^{n+1}}{2^n}\cdot\frac{1}{2}=\frac{(n+1)^{n+1}}{2^{n+1}}$
and therefore $(n+1)!<\frac{(n+1)^{n+1}}{2^{n+1}}$.
2) Show that $\frac{n^n}{3^n}<n!$ for all integers $n\ge1$:
a) This is true for $n=1$, since $\frac{1}{3}<1$.
b) Assume that $\frac{n^n}{3^n}<n!$ for some integer $n\ge1$.
Then $(n+1)!=(n+1)n!>(n+1)\cdot\frac{n^n}{3^n}$, and
since $\frac{(n+1)^n}{n^n}=\big(1+\frac{1}{n}\big)^n<3$, as we will show below,
$(n+1)\cdot\frac{n^n}{3^n}=\frac{(n+1)^{n+1}}{3^n}\cdot\frac{n^n}{(n+1)^n}>\frac{(n+1)^{n+1}}{3^n}\cdot\frac{1}{3}=\frac{(n+1)^{n+1}}{3^{n+1}}$;
so $(n+1)!>\frac{(n+1)^{n+1}}{3^{n+1}}$.
To show that $\big(1+\frac{1}{n}\big)^n<3$ for all n, we have
$\big(1+\frac{1}{n}\big)^n=\sum_{k=0}^{n}\binom{n}{k}(\frac{1}{n})^k=\sum_{k=0}^{n}\frac{n(n-1)(n-2)\cdots(n-k+1)}{k!\cdot n^k}\le\sum_{k=0}^{n}\frac{1}{k!}$, so
$\big(1+\frac{1}{n}\big)^n\le1+1+\frac{1}{2!}+\frac{1}{3!}+\frac{1}{4!}+\cdots\frac{1}{n!}\le1+1+\frac{1}{2!}+\frac{1}{3\cdot 2!}+\frac{1}{3^{2}\cdot 2!}+\cdots\frac{1}{3^{n-2}\cdot2!}=$
$\;\;\;\;\;\;\;2+\frac{\frac{1}{2}}{1-\frac{1}{3}}\big(1-(\frac{1}{3})^{n-1}\big)=2+\frac{3}{4}\big(1-(\frac{1}{3})^{n-1}\big)<2+\frac{3}{4}<3.$
|
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"timestamp": "2023-03-29T00:00:00",
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|
Find $f(x)$ such that $f(x)+f\left(\frac{1}{x}\right)=f(x)\cdot f\left(\frac{1}{x}\right)$
Given a polynomial $f(x)$ of n degree such that
$$f(x)+f\left(\frac{1}{x}\right)=f(x)\cdot f\left(\frac{1}{x}\right)$$
Find the polynomial
I've tried considering $f(x)=\sum \limits_{i=1}^{n} a_{i}x^{i}$ and after a number of steps I finally arrived at the answer.However I am not convinced.Is there a better way to solve this functional equation.
Thanks in advanced!
|
$$f(y) = 1+y$$
$$f(x)+f\left(\frac{1}{x}\right) = 1+x+1+\frac{1}{x} = 2 + x + \frac{1}{x}$$
$$f(x) f\left(\frac{1}{x}\right) = (1+x) \left ( 1+\frac{1}{x}\right) = 1 + x + \frac{1}{x} + x \frac{1}{x} = 2 + x + \frac{1}{x}$$
In fact, $f(y) = 1+y^k$ for $k \in \mathbb{N}$ seems to work for the same reason.
|
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"url": "https://math.stackexchange.com/questions/472708",
"timestamp": "2023-03-29T00:00:00",
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|
Find the value of the the term The sequence $a_1,a_2,a_3,\ldots$ satisfies $a_1=1$, $a_2=2$, and $$a_{n+2}=\frac2{a_{n+1}}+a_n\;;$$ find the value of
$$\frac{a_{2012}2^{2009}}{2011}$$
|
HINT:
$$a_{n+2}=\frac2{a_{n+1}}+a_n\iff a_{n+2}a_{n+1}=a_{n+1}a_n+2$$
$$\implies a_{n+2}a_{n+1}=2n+a_2a_1=2n+2\ \ \ \ (1)$$
$$\text{Replacing }n\text{ with } n-1,\implies a_{n+1}a_n=2n$$
$$\implies a_{n+2}a_{n+1}\cdot2n=a_{n+1}a_n(2n+2)$$
$$\implies a_{n+2} =a_n\frac{(n+1)}n$$
$$n=2m-2\implies a_{2m} =a_{2m-2}\frac{(2m-1)}{2m-2}$$
$$ a_{2m} =\frac{(2m-1)(2m-3)\cdots3}{(2m-2)(2m-4)\cdots4}a_2$$
$$\text{Now,} \frac{(2m-1)(2m-3)\cdots3}{(2m-2)(2m-4)\cdots4}=2\frac{(2m-1)(2m-2)\cdots4\cdot3\cdot2\cdot1}{2^{m-1}\cdot (m-1)! ((2m-2)\cdot4\cdot2)}=\frac{(2m-1)!}{2^{2m-2} \{(m-1)!\}^2}$$ and $a_2=2$
|
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|
How to prove $\sin x+\frac{\sin 2x}{2}+\frac{\sin 3x}{3}+\cdots$ is positive? Let $0<x<\pi$. $n$ be a natural number.
How to prove
$$\sin x+\frac{\sin 2x}{2}+\frac{\sin 3x}{3}+\cdots+ \frac{\sin nx}{n}>0$$
|
Let $f_n(x)=\sum\limits_{k=1}^n\frac{\sin{kx}}{k}$.
For $n=1$ it's obvious.
Let $f_n(x)>0$ for any $0<x<\pi.$
It's enough to prove that $f_{n+1}(x)>0.$
Indeed, let $f_{n+1}(x)\leq0$ for some value of $x\in(0,\pi)$.
Since $f_{n+1}$ is a continuous function on $[0,\pi],$ $f_{n+1}(0)=f_{n+1}(\pi)=0,$
there is $x_0\in(0,\pi)$ for which $f_{n+1}$ gets a minimal value and we know that $f_{n+1}(x_0)\leq0.$
Now, $$f'_{n+1}(x_0)=0$$ or
$$\sum_{k=1}^{n+1}\cos{kx_0}=0$$ or
$$\sum_{k=1}^{n+1}2\sin\frac{x_0}{2}\cos{kx_0}=0$$ or
$$\sum_{k=1}^{n+1}\left(\sin\left(kx_0+\frac{x_0}{2}\right)-\sin\left(kx_0-\frac{x_0}{2}\right)\right)=0$$ or
$$\sin\left(nx_0+\frac{3x_0}{2}\right)=\sin\frac{x_0}{2},$$ which gives:
$$0\geq f_{n+1}(x_0)=f_n(x_0)+\frac{1}{n+1}\sin(n+1)x_0=$$
$$=f_n(x_0)+\frac{1}{n+1}\sin\left(\left(n+\frac{3}{2}\right)x_0-\frac{x_0}{2}\right)=$$
$$=f_n(x_0)+\frac{1}{n+1}\left(\sin\left(n+\frac{3}{2}\right)x_0\cos\frac{x_0}{2}-\cos\left(n+\frac{3}{2}\right)x_0\sin\frac{x_0}{2}\right)=$$
$$=f_n(x_0)+\frac{\sin\frac{x_0}{2}}{n+1}\left(\cos\frac{x_0}{2}-\cos\left(n+\frac{3}{2}\right)x_0\right)=$$
$$=f_n(x_0)+\frac{\sin\frac{x_0}{2}}{n+1}\left(\left|\cos\left(n+\frac{3}{2}\right)x_0\right|-\cos\left(n+\frac{3}{2}\right)x_0\right)\geq f_n(x_0),$$ which is a contradiction and by induction we are done!
|
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|
The integer $c_n$ in $(1+4\sqrt[3]2-4\sqrt[3]4)^n=a_n+b_n\sqrt[3]2+c_n\sqrt[3]4$ For non-negative integer $n$, write
$$(1+4\sqrt[3]2-4\sqrt[3]4)^n=a_n+b_n\sqrt[3]2+c_n\sqrt[3]4$$
where $a_n,b_n,c_n$ are integers. For any non-negative integer $m$, prove or disprove
$$2^{m+2}|c_n\iff2^m|n$$
So far I have
$$\left[\begin{array}{c}a_n\\b_n\\c_n\end{array}\right]=\left[\begin{array}{ccc}1&-8&8\\4&1&-8\\-4&4&1\end{array}\right]\left[\begin{array}{c}a_{n-1}\\b_{n-1}\\c_{n-1}\end{array}\right].$$
|
$$
x \equiv \sqrt[3]{2}
\quad\Longrightarrow\quad
x^{2} = \sqrt[3]{4},\ x^{3} = 2,\ x^{4} = 2x,\ x^{5} = 2x^{2}, x^{6} = 4,\ x^{7} = 4x\ldots
$$
$$\left(1 + 4\sqrt[3]{2} - 4\sqrt[3]{4}\right)^{n}
=
\left(1 + 4x - 4x^{2}\right)^{n}
=
a_{n} + b_{n}x + c_{n}x^{2}
$$
$$
{1
\over
1- z\left(1 + 4x - 4x^{2}\right)}
=
\overbrace{\sum_{n = 0}^{\infty}z^{n}a_{n}}^{\equiv\ {\rm A}\left(z\right)}\
+\
x\quad\overbrace{\sum_{n = 0}^{\infty}z^{n}b_{n}}^{\equiv\ {\rm B}\left(z\right)}\
+\
x^{2}\quad\overbrace{\quad\sum_{n = 0}^{\infty}z^{n}c_{n}}^{\equiv\ {\rm C}\left(z\right)}
$$
$$
1=
\left\lbrack\left(1- z\right) - 4zx + 4zx^{2}\right\rbrack
\left\lbrack
{\rm A}\left(z\right) + {\rm B}\left(z\right)x + {\rm C}\left(z\right)x^{2}
\right\rbrack
$$
$$
\left\lbrace%
\begin{array}{rcrcrcl}
\left(1 - z\right){\rm A}\left(z\right)
& + &
8z{\rm B}\left(z\right)
& - &
8z{\rm C}\left(z\right)
& = &
1
\\[1mm]
-4z{\rm A}\left(z\right)
& + &
\left(1 - z\right){\rm B}\left(z\right)
& + &
8z{\rm C}\left(z\right)
& = &
0
\\[1mm]
4z{\rm A}\left(z\right)
& - &
4z{\rm B}\left(z\right)
& + &
\left(1 - z\right){\rm C}\left(z\right)
& = &
0
\end{array}\right.
$$
$$
\left(1 - z{\bf M}\right)\left\vert\Psi\left(z\right)\right\rangle
=
\left\vert\Psi_{0}\right\rangle
$$
$$
{\bf M}
\equiv\left(%
\begin{array}{rrr}
1 & -8 & 8
\\
4 & 1 & -8
\\
-4 & 4 & 1
\end{array}\right)\,,
\qquad
\left\vert\Psi\left(z\right)\right\rangle
\equiv
\left(%
\begin{array}{c}
{\rm A}\left(z\right)
\\[1mm]
{\rm B}\left(z\right)
\\[1mm]
{\rm C}\left(z\right)
\end{array}\right)
$$
$$
\left\vert\Psi_{0}\right\rangle
\equiv
\left(%
\begin{array}{c}
1
\\[1mm]
0
\\[1mm]
0
\end{array}\right)\,,
\qquad
\left\vert\Psi_{1}\right\rangle
\equiv
\left(%
\begin{array}{c}
0
\\[1mm]
1
\\[1mm]
0
\end{array}\right)\,,
\qquad
\left\vert\Psi_{2}\right\rangle
\equiv
\left(%
\begin{array}{c}
0
\\[1mm]
0
\\[1mm]
1
\end{array}\right)\,,
$$
and
$$
\left(%
\begin{array}{c}
a_{n} \\ b_{n} \\ c_{n}
\end{array}\right)
=
\left.%
{1 \over n!}\,{{\rm d}^{n}\left\vert\Psi\left(z\right)\right\rangle \over {\rm d}z^{n}}
\right\vert_{z = 0}
$$
Since
\begin{align}
\left(1 - z{\bf M}\right)\left\vert\Psi\left(z\right)\right\rangle
=
\left\vert\Psi_{0}\right\rangle
&\quad\Longrightarrow&
\left\vert\Psi\left(0\right)\right\rangle
=
\left\vert\Psi_{0}\right\rangle
\\[1mm]
-{\bf M}\left\vert\Psi\left(z\right)\right\rangle
+
\left(1 - z{\bf M}\right)\left\vert\Psi\left(z\right)\right\rangle'
=
0
&\quad\Longrightarrow&
\left\vert\Psi\left(0\right)\right\rangle'
=
{\bf M}\left\vert\Psi\left(0\right)\right\rangle
=
{\bf M}\left\vert\Psi_{0}\right\rangle
\\[1mm]
-2{\bf M}\left\vert\Psi\left(z\right)\right\rangle'
+
\left(1 - z{\bf M}\right)\left\vert\Psi\left(z\right)\right\rangle''
=
0
&\quad\Longrightarrow&
\left\vert\Psi\left(0\right)\right\rangle''
=
2{\bf M}\left\vert\Psi\left(0\right)\right\rangle'
=
2{\bf M}^{2}\left\vert\Psi_{0}\right\rangle
\\[1mm]
-3{\bf M}\left\vert\Psi\left(z\right)\right\rangle''
+
\left(1 - z{\bf M}\right)\left\vert\Psi\left(z\right)\right\rangle'''
=
0
&\quad\Longrightarrow&
\left\vert\Psi\left(0\right)\right\rangle'''
=
3{\bf M}\left\vert\Psi\left(0\right)\right\rangle'
=
3!\,{\bf M}^{3}\left\vert\Psi_{0}\right\rangle
\\ \vdots&\vdots&\vdots
\end{align}
we'll get
$$
\left(%
\begin{array}{ccc}
a_{n} \\ b_{n} \\ c_{n}
\end{array}\right)
=
{1 \over n!}\,n!\,{\bf M}^{n}\left\vert\Psi_{0}\right\rangle
\quad\Longrightarrow\quad
c_{n}
=
\left\langle\Psi_{2}\left\vert{\bf M}^{n}\right\vert\Psi_{0}\right\rangle
$$
In addition, ${\bf M} = 1 + 4{\bf Q}$ where
$$
{\bf Q}
\equiv\left(%
\begin{array}{rrr}
0 & -2 & 2
\\
1 & 0 & -2
\\
-1 & 1 & 0
\end{array}\right)
\quad\mbox{such that}\quad
{\bf M}^{n}
=
\sum_{\ell = 0}^{n}{n \choose \ell}2^{2\ell}{\bf Q}^{\ell}
$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/476342",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 3,
"answer_id": 2
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|
A real solution to a cubic equation What is the easiest way to find the real solution of the equation $x^3-6x^2+6x-2=0$?
I know the solution to be $x=2+2^{2/3}+2^{1/3}$ (Mathematica) but I would like to find it analytically. If possible, not by plugging the coefficients in Cardano's or similar formula.
|
$$\begin{equation*}
x^{3}-6x^{2}+6x-2=0 \tag{1}
\end{equation*}$$
Set $x=t+2.$ Then
$$\begin{equation*}
t^{3}-6t-6=0\tag{2}
\end{equation*}$$
Set $t=u+v.$ Then
\begin{equation*}
( u+v) ^{3}-6\left( u+v\right) -6=0,
\end{equation*}
\begin{eqnarray*}
\left( u+v\right) ^{3}-6\left( u+v\right) -6 &=&( u^{3}+v^{3}-6)
+(3u^{2}v+3uv^{2}-6u-6v) \\
&=&( u^{3}+v^{3}-6) +( 3uv-6) ( u+v)
\end{eqnarray*}
If the auxiliary variables $u,v$ satisfy the following system, $t$ satisfies $(2)$.
\begin{equation*}
\left\{
\begin{array}{c}
u^{3}+v^{3}=6 \\
3uv=6
\end{array}
\right. \Leftrightarrow \left\{
\begin{array}{c}
u^{3}+v^{3}=6 \\
u^{3}v^{3}=8.
\end{array}
\right.
\end{equation*}
Set $U=u^{3},V=v^{3}$. Since we know the sum and the product of $U,V$, we have
\begin{equation*}
\left\{
\begin{array}{c}
U+V=6 \\
UV=8
\end{array}
\right. \Leftrightarrow \left\{
\begin{array}{c}
U=u^{3}=4, \\
V=v^{3}=2
\end{array}
\right. \;\vee \left\{
\begin{array}{c}
U=u^{3}=2, \\
V=v^{3}=4.
\end{array}
\right.
\end{equation*}
The pair $(u,v)=(2^{\frac{2}{3}},2^{\frac{1}{3}})$ leads to one of the solutions of $(2)$, the solution
\begin{equation*}
t=u+v=2^{\frac{2}{3}}+2^{\frac{1}{3}}.\tag{3}
\end{equation*}
The corresponding solution of $(1)$ is thus
\begin{equation*}
x=t+2=2^{\frac{2}{3}}+2^{\frac{1}{3}}+2.\tag{4}
\end{equation*}
Remark: This agrees with Old$\ $John's creative method, because $( 2^{
\frac{2}{3}}+2^{\frac{1}{3}}+2) ( 2^{\frac{1}{3}}-1) =2^{\frac{1}{3}}$.
ADDED. The other roots of $(1)$ are complex conjugates.
|
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "13",
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|
$\tan A + \tan B + \tan C = \tan A\tan B\tan C\,$ in a triangle I want to prove this (where each angle may be negative or greater than $180^\circ$):
When $A+B+C = 180^\circ$ \begin{equation*} \tan A + \tan B + \tan C = \tan A\:\tan B\:\tan C. \end{equation*}
We know that
\begin{equation*}\tan(A+B) = \frac{\tan A+\tan B}{1-\tan A\tan B}\end{equation*} and that \begin{equation*}\text{and that}~A+B = 180^\circ-C.\end{equation*}
Therefore $\tan(A+B) = -\tan C.$
From here, I got stuck.
|
Maybe a little bit obvious but what I would do is the following:
$\textrm{Let:}$
$$\omega + \phi + \psi = \pi$$
Then take the tangent function to both sides.
$$\tan \left ( \omega + \phi + \psi \right) = \tan \left ( \pi \right)$$
Since $\tan \pi = 0$, then:
$$\tan \left ( \omega + \phi + \psi \right) = \tan \left ( \pi \right)$$
Now group:
$$\tan \left ( \left ( \omega + \phi + \right) + \psi \right) = \tan \left ( \pi \right)$$
Resolving we have:
$$\frac{\tan \left (\omega + \phi \right) + \tan \left (\psi \right) }{1-\tan \left( \omega + \phi \right) \tan \left (\psi \right)} = 0$$
In order to make the whole equation to zero, the numerator has to be zero as well, therefore just replace:
$$ \tan \left (\omega + \phi \right) + \tan \left (\psi \right) = 0$$
By expanding it:
$$\tan \left (\omega + \phi \right) = - \tan \left (\psi \right )$$
$$\frac{\tan \omega + \tan \phi}{1- \tan \omega \tan \phi} = - \tan \psi$$
$$\tan \omega + \tan \phi = \left( 1- \tan \omega \tan \phi \right) \left ( - \tan \psi \right )$$
Don't worry, we're almost there:
$$\tan \omega + \tan \phi = - \tan \psi + \tan \omega \tan \phi \tan \psi$$
$$\tan \omega + \tan \phi + \tan \psi = \tan \omega \tan \phi \tan \psi$$
Therefore we have proved the identity!.
By the way I used the angles $\omega$, $\phi$ and $\psi$ as I feel more comfortable working with them but in your case you may want them to be replaced by the letters $\textrm{A, B and C}$.
I hope this have helped you.
|
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "15",
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|
Prove that there are infinitely many perfect cubes of the form $p^2+3q^2$ Prove that there are infinitely many perfect cubes of the form $p^2+3q^2$ where $p$ and $q$ are integers.
Hint: one approach is to set $p^2+3q^2=(a^2+3b^2)^3$ and then find $(p,q)$ in terms of $a,b$.
Any different approach is very welcome!
|
Using the hint that OP stated,
$$ \begin{array} { l l}
(a^2 + 3b^2 )^3 & = a^6 + 9 a^4 b^2 + 27a^2 b^4 + 27b^6 \\
& = (a^6 + 6 a^4 b^2 + 9 a^2 b^4) + 3( a^4 b^2 + 6a^2 b^4 + 9b^6 ) \\
& = (a^3 + 3ab^2)^2 + 3 (a^2b + 3b^3)^2 \end{array}$$
|
{
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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|
Implicit differentiation question Given that $x^n + y^n = 1$, show that $$\frac{d^2y}{dx^2} = -\frac{(n-1)x^{n-2}}{y^{2n-1}}.$$
I found that $\displaystyle nx^{n-1}+ny^{n-1}\frac{dy}{dx} = 0$ so that $\displaystyle y'=\frac{-x^{n-1}}{y^{n-1}}$.
Then $$n(n-1)x^{n-2}+n(n-1)y^{n-2}\left(\frac{dy}{dx}\right)^2 + \frac{d^2y}{dx^2}ny^{n-1} = 0.$$
Therefore $$y'' = \displaystyle \frac{-n(n-1)x^{n-2}-n(n-1)y^{n-2}(y')^2}{ny^{n-1}}.$$
Substituting the first derivative:
$$y'' = \displaystyle\frac{-n(n-1)x^{n-2}-n(n-1)y^{n-2}\left(\dfrac{-x^{n-1}}{y^{n-1}}\right)^2}{ny^{n-1}}.$$
I've been trying tons of different steps and can't establish which way to eliminate the $y$ terms from the numerator. Could someone offer a hint on how to proceed. Thanks.
|
Expanding your final expression, we have
\begin{align*}
y'' &= \frac{-n(n-1)x^{n-2} -n(n-1)y^{n-2}\dfrac{x^{2n-2}}{y^{2n-2}}}{ny^{n-1}}\\
&= \frac{-n(n-1)x^{n-2} -n(n-1)y^{-n}x^{2n-2}}{ny^{n-1}}\\
&= \frac{-n(n-1)x^{n-2}(1 + x^{n}y^{-n})}{ny^{n-1}}\\
&= \frac{-(n-1)x^{n-2}(1 + x^{n}y^{-n})}{y^{n-1}}.
\end{align*}
Now note that $x^n + y^n = 1$, so $x^ny^{-n} + 1 = y^{-n}$. Therefore
\begin{align*}
y'' &= \frac{-(n-1)x^{n-2}(1 + x^{n}y^{-n})}{y^{n-1}}\\
&= \frac{-(n-1)x^{n-2}y^{-n}}{y^{n-1}}\\
&= \frac{-(n-1)x^{n-2}}{y^{2n-1}}.
\end{align*}
|
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"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
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|
Solutions of $x^2 + 7y^2 = 2^n$ where $x$ and $y$ are odd numbers Is it true that for any $n\geq 2$ the equation $x^2 + 7y^2 = 2^n$ has a solution with $x$ and $y$ odd ??
|
for $n=3$, they are solutions ($x_1=1,y_1=1$).
If you have odd solutions $x_n$, $y_n$ for $2^n$, then
$$\left(\frac{7y_n-x_n}{2}\right)^2+7\left(\frac{x_n+y_n}{2}\right)^2=2x_n^ 2+14y_n^2=2(x_n^2+7y_n^2)=2^{n+1}$$
$$\left(\frac{7y_n+x_n}{2}\right)^2+7\left(\frac{x_n-y_n}{2}\right)^2=2x_n^ 2+14y_n^2=2(x_n^2+7y_n^2)=2^{n+1}$$
Hence $x_{n+1}=\frac{7y_n-x_n}{2}$ and $y_{n+1}=\frac{x_n+y_n}{2}$
or $x_{n+1}=\frac{7y_n+x_n}{2}$ and $y_{n+1}=\frac{x_n-y_n}{2}$
This is, in fact, the same answer as wendy.
Note that always one of the solution is odd and the other is even. Take the odd one !
|
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"timestamp": "2023-03-29T00:00:00",
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|
Evaluation Of Maximum Value Without Calculus $x$,$y$,$z$ are non-negative numbers.
$x+y+z=3$
Find the maximum value of $~$ $x^{2}y+y^{2}z+z^{2}x$ $~$ without calculus.
|
in general
if $x,y,z\ge 0$,and such $x+y+z=3$, then we have
$$x^ky+y^kz+z^kx\le\max{\{3,\dfrac{3^{k+1}k^k}{(k+1)^{k+1}}\}}$$
For $k=2$ I have nice methods
with out loss of let $x=\max{(x,y,z)}$
then we use Benoulli inequality,we have
$$(1+\dfrac{z}{x})^2\ge 1+\dfrac{2z}{x}$$
so
$$(x+z)^2y=x^2y(1+\frac zx)^2\ge x^2y(1+\dfrac{2z}{x})=x^2y+xyz+xyz\ge x^2y+y\cdot yz+xz^2$$
so
$$x^2y+y^2z+z^2x\le (x+z)^2y=2^2\left(\dfrac{x+z}{2}\right)^2\cdot y\le 2^2\left(\dfrac{x+z+y}{3}\right)^3=4$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/489292",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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|
compute $1-\frac 12+ \frac15 - \frac 16+ \frac 19- \frac{1}{10}+ \cdots + \frac{1}{4n+1}-\frac{1}{4n+2}$ Knowing that $1 - \frac 12 + \frac 13 - \cdots = \ln 2$ and $1 - \frac 13 + \frac 15 - \cdots = \frac{\pi}{4}$, compute $1-\frac 12+ \frac15 - \frac 16+ \frac 19- \frac{1}{10}+ \cdots + \frac{1}{4n+1}-\frac{1}{4n+2}$. I programmed and found that $\frac{\ln 2}{4}+{\frac{\pi}{4}}/2$ is precise to the eighth digit.
|
$$\begin{aligned}\sum_{n\geq 0}\frac{1}{4n+1}-\frac{1}{4n+2} =\sum_{n\geq 0}\int_0^1 t^{4n}(1-t)\,dt=\int_0^1 \frac{t-1}{t^4-1}\,dt=\frac{\ln 2}{4}+\frac{\pi}{8}\end{aligned}$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/492105",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
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|
Prove that $(x^2-x^3)(x^4-x) = \sqrt{5}$, where $x= \cos(2\pi/5)+i\sin(2\pi/5)$
Prove $(x^2-x^3)(x^4-x) = \sqrt{5}$ if $x= \cos(2\pi/5)+i\sin(2\pi/5)$.
I have tried it by substituting $x = \exp(2i\pi/5)$
but it is getting complicated.
|
As was already mentioned, we have
$$0=x^5-1=(x-1)(x^4+x^3+x^2+x+1)$$
$$|x|=|e^{2\pi i/5}|=1\implies x^2=x^{-3}\;,\;x=x^{-4}\;,\;x^{-k}=\overline{x^k}\implies$$
$$(x^2-x^3)^2(x^4-x)^2=(x^2-x^{-2})^2(x^{-1}-x)^2=(x-x^{-1})^4(x+x^{-1})^2=$$
$$=\left(2i\sin\frac{2\pi}5\right)^4\left(2\cos\frac{2\pi}5\right)^2=64\sin^4\frac{2\pi}5\cos^2\frac{2\pi}5=$$
$$=64\left(\frac14\sqrt{10+2\sqrt5}\right)^4\left(\frac14\left(-1+\sqrt5\right)\right)^2=5$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/495174",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 5,
"answer_id": 3
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|
Solving a system of equation modul0 5 Consider the system of linear equations $$\begin{pmatrix} 6 & -3\\ 2 & 6 \end{pmatrix}\begin{pmatrix} x_1\\ x_2 \end{pmatrix}=\begin{pmatrix} 3\\ 1 \end{pmatrix} $$
a) Solve the system in $\mathbb{F}_5$
I just want to make sure my solution is correct:
We have: $$A=\begin{pmatrix} 6 & -3\\ 2 & 6 \end{pmatrix}\\ \Rightarrow A^{-1}= \frac{1}{42}\begin{pmatrix} 6 & 3\\ -2 & 6 \end{pmatrix}\\ \equiv \begin{pmatrix} 3 & \frac{3}{2}\\ -1 & 3 \end{pmatrix} \mbox{ } [5]$$
Therefore: $\begin{pmatrix} x_1\\ x_2 \end{pmatrix}=\begin{pmatrix} 3 & \frac{3}{2}\\ -1 & 3 \end{pmatrix}\begin{pmatrix} 3\\ 1 \end{pmatrix}= \begin{pmatrix} \frac{21}{2}\\ 0 \end{pmatrix} [5] $
|
Note that modulo $5$ one has
$$
A =
\begin{bmatrix} 6 & -3\\ 2 & 6 \end{bmatrix}
=
\begin{bmatrix} 1 & 2\\ 2 & 1 \end{bmatrix}.
$$
The latter matrix has determinant $-3 \equiv 2$, whose inverse is $3$, so
$$
A^{-1} =
\begin{bmatrix} 3 & -1\\ -1 & 3 \end{bmatrix}.
$$
Now
$$
\begin{bmatrix} x_{1}\\x_{2} \end{bmatrix}
=
A^{-1}
\begin{bmatrix} 3\\ 1 \end{bmatrix}
=
\begin{bmatrix} 3 & -1\\ -1 & 3 \end{bmatrix}.
\begin{bmatrix} 3\\ 1 \end{bmatrix}
=
\begin{bmatrix} 3\\ 0 \end{bmatrix},
$$
which is your result, as
$$
\frac{21}{2} \equiv 21 \cdot 3 \equiv 3 \pmod{5}.
$$
|
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|
How to evaluate $\int_{0}^{+\infty}\exp(-ax^2-\frac b{x^2})\,dx$ for $a,b>0$ How can I evaluate
$$I=\int_{0}^{+\infty}\!e^{-ax^2-\frac b{x^2}}\,dx$$
for $a,b>0$?
My methods:
Let $a,b > 0$ and let
$$I(b)=\int_{0}^{+\infty}e^{-ax^2-\frac b{x^2}}\,dx.$$
Then
$$I'(b)=\int_{0}^{\infty}-\frac{1}{x^2}e^{-ax^2-\frac b{x^2}}\,dx.$$
What the other methods that can I use to evaluate it? Thank you.
|
The integral can be evaluated as follows
$$
\begin{align}
\int_{x=0}^\infty \exp\left(-a\left(x^2+\frac{b}{ax^2}\right)\right)\,dx&=2\int_{x=0}^\infty \exp\left(-a\left(x^2-2\sqrt{\frac{b}{a}}+\frac{b}{ax^2}+2\sqrt{\frac{b}{a}}\right)\right)\,dx\\
&=\int_{x=0}^\infty \exp\left(-a\left(x-\frac{1}{x}\sqrt{\frac{b}{a}}\right)^2-2\sqrt{ab}\right)\,dx\\
&=e^{\large-2\sqrt{ab}}\int_{x=0}^\infty \exp\left(-a\left(x-\frac{1}{x}\sqrt{\frac{b}{a}}\right)^2\right)\,dx.
\end{align}
$$
The trick to solve the last integral is by setting
$$
I=\int_{x=0}^\infty \exp\left(-a\left(x-\frac{1}{x}\sqrt{\frac{b}{a}}\right)^2\right)\,dx.
$$
Let $t=-\frac{1}{x}\sqrt{\frac{b}{a}}\;\rightarrow\;x=-\frac{1}{t}\sqrt{\frac{b}{a}}\;\rightarrow\;dx=\frac{1}{t^2}\sqrt{\frac{b}{a}}\,dt$, then
$$
I_t=\sqrt{\frac{b}{a}}\int_{t=0}^\infty \frac{\exp\left(-a\left(-\frac{1}{t}\sqrt{\frac{b}{a}}+t\right)^2\right)}{t^2}\,dt.
$$
Let $t=x\;\rightarrow\;dt=dx$, then
$$
I_t=\int_{t=0}^\infty \exp\left(-a\left(t-\frac{1}{t}\sqrt{\frac{b}{a}}\right)^2\right)\,dt.
$$
Adding the two $I_t$s yields
$$
2I=I_t+I_t=\int_{t=0}^\infty\left(1+\frac{1}{t^2}\sqrt{\frac{b}{a}}\right)\exp\left(-a\left(t-\frac{1}{t}\sqrt{\frac{b}{a}}\right)^2\right)\,dt.
$$
Let $s=t-\frac{1}{t}\sqrt{\frac{b}{a}}\;\rightarrow\;ds=\left(1+\frac{1}{t^2}\sqrt{\frac{b}{a}}\right)dt$ and for $0<t<\infty$ is corresponding to $-\infty<s<\infty$, then
$$
I=\frac{1}{2}\int_{s=-\infty}^\infty e^{-as^2}\,ds=\frac{1}{2}\sqrt{\frac{\pi}{a}}.
$$
Thus
$$
\begin{align}
\int_{x=0}^\infty \exp\left(-a\left(x^2+\frac{b}{ax^2}\right)\right)\,dx&=e^{\large-2\sqrt{ab}}\int_{x=0}^\infty \exp\left(-a\left(x-\frac{1}{x}\sqrt{\frac{b}{a}}\right)^2\right)\,dx\\
&=\frac{1}{2}\sqrt{\frac{\pi}{a}}e^{\large-2\sqrt{ab}}.
\end{align}
$$
|
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|
Smallest number with specific number of divisors Is there a general method for finding smallest number of specific number of divisors? I am doing "Higher Algebra by Barnard JM Child" and came across a question that "find the smallest number with 24 divisors", that's how I tried to solve it, alert me if I am wrong:
Since $24$ can not have more than $4$ prime factors, the number can not have more than 4 prime factors.
As a single number it is :$2$^${23}$
as product of two numbers: $2^5*3^3$,$2^{11}*3$,$2^7*3^2$,
since $5+3<7+2<11+1$, so $2^5*3^3$ is the min of these numbers
as product of three numbers :$2^5*3*5$,$2^3*3^2*5$ since $3+2+1<5+1+1$, hence $2^3*3^2*5$ is the lesser of two
as product of 4 numbers $2^2*3*5*7$
$k=\min(2^{23},2^5*3^3,2^3*3^2*5,2^2*3*5*7)$
=$2^3*3^2*5=360$
The above method seems to be fishy and laborious, is there a general approach to find the smallest number with specific number of divisors?
|
Take any number $N=p_1^{m_1}...p_k^{m_k}$, where $p_i$ are its prime divisors, and compute how many divisors it has. Each divisor would be a product of the same primes in varying powers, $D=p_1^{d_1}...p_k^{d_k}$, where $0\leq d_i \leq m_i$. Different divisors have different collections of powers, so the number of divisors will be $(m_1+1)(m_2+1)...(m_k+1)$.
Now let's find the smallest $N$ such that $(m_1+1)(m_2+1)...(m_k+1)=24$. There are only few possibilities to break 24 into a product of decreasing numbers: $3*2*2*2=4*3*2=6*4=6*2*2=8*3=12*2=24$. The corresponding candidates with the smallest primes chosen for the smallest powers would be these ones:
$2^{3-1}*3^{2-1}*5^{2-1}*7^{2-1}=2^2*3*5*7=420$
$2^{4-1}*3^{3-1}*5^{2-1}=2^3*3^2*5=360$
$2^5*3^3=2592$
$2^5*3*5=480$
$2^{11}*3=6144$
$2^{23}=8388608$
Then just pick the smallest one, which happens to be 360.
|
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|
How to solve this matrix using gauss-elimination by hand I feel like i am having a brain fart. I have been given this $Ax=b$ system:
$A= \begin{pmatrix} 0.913 & 0.659 \\ 0.780 & 0.563 \end{pmatrix}$
$b= \begin{pmatrix} 0.254 \\ 0.217 \end{pmatrix}$
I know the answer is $x_1 = 1$ and $x_2 = -1$ but for some reason when i try to solve this using Gaussian elimination, the entire bottom row goes to $0$ giving me $1$ equation with $2$ variables.
Can someone get me out of this please?
|
Alright I worked it out step by step, so bear with me (I'll use A followed by b notation):
\begin{pmatrix} 0.913 & 0.659 \\ 0.780 & 0.563 \end{pmatrix} \begin{pmatrix} 0.254\\ 0.217 \end{pmatrix}
We multiply the top row by the bottom and vice versa:
\begin{pmatrix} 0.71214 & 0.51402 \\ 0.71214 & 0.514019 \end{pmatrix} \begin{pmatrix} 0.19812\\ 0.198121 \end{pmatrix}
Subtract the bottom from top:
\begin{pmatrix} 0 & 0.000001 \\ 0.71214 & 0.514019 \end{pmatrix} \begin{pmatrix} -0.000001\\ 0.198121 \end{pmatrix}
Divide the top row by 0.000001 to reduce it to:
\begin{pmatrix} 0 & 1 \\ 0.71214 & 0.514019 \end{pmatrix} \begin{pmatrix} -1\\ 0.198121 \end{pmatrix}
Multiply the top row by 0.514019 and subtract it from the bottom to get:
\begin{pmatrix} 0 & 1 \\ 0.71214 & 0\end{pmatrix} \begin{pmatrix} -1\\ 0.71214\end{pmatrix}
Diving the bottom row by 0.71214 we end with:
\begin{pmatrix} 0 & 1 \\ 1 & 0\end{pmatrix} \begin{pmatrix} -1\\ 1\end{pmatrix}
from which we can readily see x1 = 1 and x2 = -1
|
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"timestamp": "2023-03-29T00:00:00",
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|
Fermat's Combinatorial Identity: How to prove combinatorially? $$\binom{r}{r} + \binom{r+1}{r} + \binom{r+2}{r} + \dotsb + \binom{n}{r} = \binom{n+1}{r+1}$$
I don't have much experience with combinatorial proofs, so I'm grateful for all the hints.
(Presumptive) Source: Theoretical Exercise 1.11, P18, A First Course in Pr, 8th Ed, by S Ross
|
This is analogous to https://math.stackexchange.com/a/357087/53259. The RHS here imports the number of ways of picking $r+1$ numbers out of $\{1,2,...,\color{magenta}{r}, \color{green}{r +1},...,\underbrace{n}_{= \color{magenta}{r} + (n - r)},\color{green}{n +1}\} \qquad (*)$
Now for any given choice of $r+1$ numbers, call the highest number chosen $k$ and esteem it as the $(r + 1)$th number. Observe :
The "leftmost" subset of $(*)$ containing $r + 1$ elements $ = \{1, 2, ..., r, \color{green}{r +1}\},$
Also, the "rightmost" subset of $(*)$ containing $r + 1$ elements $ = \{\color{green}{n +1} - (r + 1), ..., n , \color{green}{n +1}\}.$
Thus, $\color{green}{r +1} \leq k \leq \color{green}{n +1}$.
For each $k \in [\color{green}{r +1}, \color{green}{n +1}]$. , we must select the remaining $r$ numbers to be chosen
from the $k-1$ numbers smaller than $k$.
For $k = r +1$, must pick $r$ numbers to the left of $r + 1$, out of $\{\color{ #0073CF}{1, 2, ..., r}, r+1\}$.
Since there are $ \color{#0073CF}{r}$ such numbers, so $\color{#0073CF}{r}$ possible choices for $r$.
Thus the total number of choices for $r$ numbers $= \binom{\color{#0073CF}{r}}{r}$.
For $k = r +2$, must pick $r$ numbers to the left of $r + 2$, out of $\{\color{ #0073CF}{1, 2, ..., r, r +1}, r+2\}$.
Since there are $ \color{#0073CF}{r + 1}$ such numbers, so $\color{#0073CF}{r + 1}$ possible choices for $r$.
Thus the total number of choices for $r$ numbers $= \binom{\color{#0073CF}{r + 1}}{r}$.
...
For $k = n$, must pick $r$ numbers to the left of $n$, out of $\{\color{ #0073CF}{1, 2, ..., r, r + 1, ..., n - 1}, n\}$.
Since there are $ \color{#0073CF}{n - 1}$ such numbers, so $\color{#0073CF}{n - 1}$ possible choices for $r$.
Thus the total number of choices for $m$ numbers $= \binom{\color{#0073CF}{n - 1}}{r}$.
For $k = n + 1$, must pick $r$ numbers to the left of $n + 1$, out of $\{\color{ #0073CF}{1, 2, ..., r, r + 1, ..., n}, n + 1\}$.
Since there are $ \color{#0073CF}{n}$ such numbers, so $\color{#0073CF}{n}$ possible choices for $r$.
Thus the total number of choices for $m$ numbers $= \binom{\color{#0073CF}{n}}{r}$.
Summing up the number of ways of doing this for $k=r+1,...,r+n+1$ yields the LHS.
Remark : (Presumptive) Source: Theoretical Exercise 1.11, P18, A First Course in Pr, 8th Ed, by S Ross, via: $\color{blue}{\mathsf{i \text{ there }}}$ $ = i + 1$ here, $k$ there $= r + 1$ here, $\color{blue}{\mathsf{n \text{ there }}}$ $ = n + 1$ here.
$$\begin{align} \binom{r}{r} + \binom{r+1}{r} + \binom{r+2}{r} + \dotsb + \binom{n}{r} &= \binom{n+1}{r+1} \\
\sum_{\Large{r \le i \le n \text{ or } i \in [r,n]}} \binom{i}{r}, n \ge r & =\end{align}$$
|
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How to solve the following equation $\sqrt[3]{x+3}+\sqrt[3]{x}=\sqrt[3]{3+8x}$ I am trying to solve this equation:
$$\sqrt[3]{x+3}+\sqrt[3]{x}=\sqrt[3]{3+8x}$$
I would like to get some advice, how to solve it.
Thanks.
|
First note that $x=0$ is a solution. Now we consider $x\neq 0$ and we divide by $\sqrt[3]{x}$. We get $\sqrt[3]{1+\frac{3}{x}}+1=\sqrt[3]{8+\frac{3}{x}}$, so lets define $y=\frac{3}{x}$ and write
$$1+\sqrt[3]{1+y} = \sqrt[3]{8+y}.$$
Now we cube both sides to obtain
$$1+3\sqrt[3]{1+y}+3\sqrt[3]{1+y}^2+1+y = 8+y$$
or
$$\sqrt[3]{1+y}+\sqrt[3]{1+y}^2 = 2.$$
We complete the square to obtain
$$\left(\sqrt[3]{1+y}+\frac{1}{2}\right)^2=\frac{9}{4}$$
or
$$\sqrt[3]{1+y} = \frac{-1\pm 3}{2}.$$
Hence $y = 0$ or $y=-9$.
There is no $x$ corresponding to $y=0$, but corresponding to $y=-9$ we have $x=-\frac{1}{3}$.
Concluding, $x\in\left\{-\frac{1}{3},0\right\}$.
|
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"timestamp": "2023-03-29T00:00:00",
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|
How can I solve this equation if I know the solution for a similar equation For equation
\begin{equation}
-y''-2a^2\operatorname{sech}^2(ax)~y=k^2y
\end{equation}
I know there is the solution
\begin{equation}
y(x)=C_1\frac{-a \tanh (a x)+i k}{a+i k}e^{i k x} +C_2\frac{-a \tanh (a x)-i k}{a-i k}e^{-i k x}
\end{equation}
How can I find the solution for equation
\begin{equation}
-y''-2a\operatorname{sech}^2(ax) y=k^2y \text{ ?}
\end{equation}
$a$ and $k$ are real numbers.
|
Your reference book is lying!
Let $u=-\tanh ax$ ,
Then $\dfrac{dy}{dx}=\dfrac{dy}{du}\dfrac{du}{dx}=-a(\text{sech}^2ax)\dfrac{dy}{du}$
$\dfrac{d^2y}{dx^2}=\dfrac{d}{dx}\left(-a(\text{sech}^2ax)\dfrac{dy}{du}\right)=-a(\text{sech}^2ax)\dfrac{d}{dx}\left(\dfrac{dy}{du}\right)-2a^2(\text{sech}^2ax\tanh ax)\dfrac{dy}{du}=-a(\text{sech}^2ax)\dfrac{d}{du}\left(\dfrac{dy}{du}\right)\dfrac{du}{dx}-2a^2(\text{sech}^2ax\tanh ax)\dfrac{dy}{du}=-a(\text{sech}^2ax)\dfrac{d^2y}{du^2}(-a~\text{sech}^2ax)-2a^2(\text{sech}^2ax\tanh ax)\dfrac{dy}{du}=a^2(\text{sech}^4ax)\dfrac{d^2y}{du^2}-2a^2(\text{sech}^2ax\tanh ax)\dfrac{dy}{du}$
For $-y''-2a^2(\text{sech}^2ax)y=k^2y$ , the ODE becomes
$-\left(a^2(\text{sech}^4ax)\dfrac{d^2y}{du^2}-2a^2(\text{sech}^2ax\tanh ax)\dfrac{dy}{du}\right)-2a^2(\text{sech}^2ax)y=k^2y$
$a^2(\text{sech}^4ax)\dfrac{d^2y}{du^2}-2a^2(\text{sech}^2ax\tanh ax)\dfrac{dy}{du}+(k^2+2a^2\text{sech}^2ax)y=0$
$a^2(1-\tanh^2ax)^2\dfrac{d^2y}{du^2}-2a^2(\tanh ax)(1-\tanh^2ax)\dfrac{dy}{du}+(k^2+2a^2(1-\tanh^2ax))y=0$
$a^2(1-u^2)^2\dfrac{d^2y}{du^2}-2a^2u(1-u^2)\dfrac{dy}{du}+(k^2+2a^2(1-u^2))y=0$
For $-y''-2a(\text{sech}^2ax)y=k^2y$ , the ODE becomes
$-\left(a^2(\text{sech}^4ax)\dfrac{d^2y}{du^2}-2a^2(\text{sech}^2ax\tanh ax)\dfrac{dy}{du}\right)-2a(\text{sech}^2ax)y=k^2y$
$a^2(\text{sech}^4ax)\dfrac{d^2y}{du^2}-2a^2(\text{sech}^2ax\tanh ax)\dfrac{dy}{du}+(k^2+2a~\text{sech}^2ax)y=0$
$a^2(1-\tanh^2ax)^2\dfrac{d^2y}{du^2}-2a^2(\tanh ax)(1-\tanh^2ax)\dfrac{dy}{du}+(k^2+2a(1-\tanh^2ax))y=0$
$a^2(1-u^2)^2\dfrac{d^2y}{du^2}-2a^2u(1-u^2)\dfrac{dy}{du}+(k^2+2a(1-u^2))y=0$
Which both relates to the associated Legendre differential equation
|
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|
Evaluating the area between the curves $r=2\sin\theta$ and $r=\sin\theta+\cos\theta$
So the problem asked me to find the area of the region that lies inside both of the circles
$$r=2\sin\theta, \quad r=\sin\theta +\cos\theta $$
I know that $r=2\sin\theta$ is $x^2+(y-1)^2=1,$but the second one is a little bit harder to me.
$$\begin{align*}
x&=r\cos\theta=\cos\theta(\sin\theta + \cos\theta)\\
y&=\sin\theta(\sin\theta+\cos\theta)
\end{align*}$$
so $x+y=1+2\cos\theta \sin\theta$....Which gives me no help.
I'm getting used to polar coordinates, but I need some help! Thank you
|
To clarify the common area, you need to plot two graphs. Both are circles.
$\cos{\theta} + \sin{\theta} = \sqrt{2}\sin(\theta + \pi/4)$. So draw a circle $r = \sqrt{2} \sin {\theta}$, and rotate it $\pi/4$ clockwise.
Then, it is the circle whose cernter is (1/2,1/2) and the radius is $\frac{\sqrt{2}}{2}$.
Purely algebraically, $r= \sqrt{x^2 + y^2}$, $\cos\theta = \frac{x}{\sqrt{x^2+y^2}}$, and $\sin\theta = \frac{y}{\sqrt{x^2+y^2}}$.
Then $\sqrt{x^2 + y^2 } = \frac{x}{\sqrt{x^2+y^2}} + \frac{y}{\sqrt{x^2+y^2}}$, i.e. $x^2 + y^2 = x + y$.
So $(x-\frac{1}{2})^2 + (y - \frac12)^2 = \frac12$.
|
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|
Express sequence in closed type Given the sequence $a_n = \sqrt{2+a_{n-1}}$. Is there anyway to find a closed form for this sequence?
Thank you for your time.
|
We are given
$a_n = \sqrt{2+a_{n-1}}$.
I will assime that
$0 \le a_0 < 2$.
If $a_{n-1} < 2$,
$a_n < \sqrt{2+2} = 2$,
so all subsequent $a_n < 2$.
Since $a_1 = \sqrt{2+a_0} < 2$,
all $a_n < 2$.
Let $d_n = 2-a_n$,
so $a_n = 2-d_n$.
$2-d_n = \sqrt{2+(2-d_{n-1})}
=\sqrt{4-d_{n-1}}
$
or,
since $1-x < \sqrt{1-x} < 1-x/2$
if $0 < x < 1$,
$d_n
= 2-\sqrt{4-d_{n-1}}
= 2-2\sqrt{1-d_{n-1}/4}
< 2-2(1-d_{n-1}/4)
= d_{n-1}/2
$
so
$d_n \to 0$
geometrically,
so $a_n \to 2$.
|
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|
Quantum Fourier Transform and roots of unity. I need to find $QFT_{6}$ for the state quantum state $\frac{1}{\sqrt2}(|0\rangle + |3\rangle)$. I received a very sufficient answer recently on simplifying nth roots of unity, but I am having a lot of trouble applying it to this problem. I have obtained the $QFT_6$ matrix below, and multiplied it by the shown state:
$$
\begin{pmatrix}
\frac{1}{\sqrt6} & \frac{1}{\sqrt6} & \frac{1}{\sqrt6} & \frac{1}{\sqrt6} & \frac{1}{\sqrt6} & \frac{1}{\sqrt6}\\
\frac{1}{\sqrt6} & \frac{1}{\sqrt6}\omega & \frac{1}{\sqrt6}\omega^2 & \frac{1}{\sqrt6}\omega^3 & \frac{1}{\sqrt6}\omega^4 & \frac{1}{\sqrt6}\omega^5\\
\frac{1}{\sqrt6} & \frac{1}{\sqrt6}\omega^2 & \frac{1}{\sqrt6}\omega^4 & \frac{1}{\sqrt6}\omega^6 & \frac{1}{\sqrt6}\omega^8 & \frac{1}{\sqrt6}\omega^{10}\\
\frac{1}{\sqrt6} & \frac{1}{\sqrt6}\omega^3 & \frac{1}{\sqrt6}\omega^6 & \frac{1}{\sqrt6}\omega^9 & \frac{1}{\sqrt6}\omega^{12} & \frac{1}{\sqrt6}\omega^{15}\\
\frac{1}{\sqrt6} & \frac{1}{\sqrt6}\omega^4 & \frac{1}{\sqrt6}\omega^8 & \frac{1}{\sqrt6}\omega^{12} & \frac{1}{\sqrt6}\omega^{16} & \frac{1}{\sqrt6}\omega^{20}\\
\frac{1}{\sqrt6} & \frac{1}{\sqrt6}\omega^5 & \frac{1}{\sqrt6}\omega^{10} & \frac{1}{\sqrt6}\omega^{15} & \frac{1}{\sqrt6}\omega^{20} & \frac{1}{\sqrt6}\omega^{25}\\
\end{pmatrix}
\begin{pmatrix}
\frac{1}{\sqrt2}\\
0 \\
0 \\
\frac{1}{\sqrt2}\\
0 \\
0
\end{pmatrix}
$$
Right, well now I don't understand how to just express all of the omegas in terms of $\omega$ and $\omega^2$. Help would be much appreciated.
|
Position a regular hexagon on the unit circle with vertices at $\pm1$. Let $\zeta$ be the first complex root of unity counterclockwise from $1$. Then $\zeta^2$ is a third full rotation away from $1$, in other words it is a third root of unity. And $\zeta^3=-1$ since it sweeps out exactly half a rotation. Algebraically speaking we know $(\zeta^3)^2=1$ and $\zeta^3\ne1$ (since $\zeta$ is a primitive $6$th root) so $\zeta^3=-1$.
This tells us that $\zeta^4=-\zeta$ and $\zeta^5=-\zeta^2$. We already know $\zeta^6=1$. Can $\zeta^2$ be reduced?
Indeed it can. The theory of cyclotomic polynomials tells us that
$$\Phi_6(x)=\frac{(x^6-1)(x-1)}{(x^3-1)(x^2-1)}=\frac{x^3+1}{x+1}=x^2-x+1.$$
So $\zeta^2=\zeta-1$. Thus all of $\zeta^2,\zeta^3,\zeta^4,\zeta^5,\zeta^6$ can be written in the form $a+b\zeta$ with these relations.
One can also argue $\zeta^2=\zeta-1$ with a more geometric toolkit. Looking at the hexagon it is clear that $\zeta^2$ bisects the vectors representing $-1$ and $\zeta$, so $\zeta^2=r(\zeta-1)$ already; at this stage one shows $\zeta-1$ has modulus $1$ so $r=1$ hence $\zeta^2=\zeta-1$.
Finally with the cycling $1=\zeta^6=\zeta^{12}=\zeta^{18}=\cdots$ you should have all you need.
|
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|
For what $n$ does $x^n \equiv 2\pmod{13}$ have a solution? I want to ask for what values of $n$ the congruence $$x^n \equiv 2 \pmod{13}$$ has a solution for $x$.
|
Hints: Note that $x^{12k+m}\equiv x^m\pmod{13}$. So you will "only" have to examine the cases $m=0$ to $11$.
If $m$ and $12$ are relatively prime, then the congruence $x^m\equiv a\pmod{13}$ always has a solution.
So we are down to $m=0$, $2$, $3$, $4$, $6$, $8$, $9$, $10$.
You can find shortcuts to deal with some of these. For example, once you find that $x^2\equiv 2$ has no solutions, that rules out the even $m$ in the list. (One can use a theorem to rule out $m=2$, or one can square the numbers $1$ to $6$.) There are other theorems one could quote to bypass calculations.
Remark: The following theorem lets us deal quickly with the problem,with minimal calculation.
Theorem: Let $a$ be relatively prime to the prime $p$. Then the congruence $x^k\equiv a\pmod{p}$ has a solution if and only if $a^{(p-1)/d}\equiv 1\pmod{p}$, where $d=\gcd(k,p-1)$.
Let $p=13$ and $a=2$. The possible $d$ are $1,2,3,4,6,12$. Since $2^4\not\equiv 1\pmod{13}$ and $2^6\not\equiv 1\pmod{13}$, it follows that $x^k\equiv 2\pmod{13}$ has a solution if and only if $\gcd(k,12)=1$.
|
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|
Independence of $\frac{1-\cos(x)+k\sin(x)}{\sin(x)+k(1+\cos(x))}$ from $k$. How to one can show that the value of the following expression
$$\frac{1-\cos(x)+k\sin(x)}{\sin(x)+k(1+\cos(x))}$$
doesn't depend to values of $k$?
|
We have $\frac{1-\cos x + k \sin x}{\sin x + k(1+\cos x)}$ and what bothers me at first sight is that $k$ multiplies $\sin$ in numerator but $\cos$ in denominator. So, let's turn $\cos$ into $\sin$:
$$
\begin{align}\frac{1-\cos x + k \sin x}{\sin x + k(1+\cos x)} &= \frac{1-\cos x + k \sin x}{\sin x + k(1+\cos x)}\cdot \frac{1-\cos x}{1-\cos x}\\
&= \frac{(1-\cos x)(1-\cos x + k \sin x)}{(1-\cos x)\sin x + k(1-\cos^2 x)}\\ &= \frac{(1-\cos x)(1-\cos x + k \sin x)}{(1-\cos x)\sin x + k\sin^2 x}\\
&= \frac{(1-\cos x)(1-\cos x + k \sin x)}{\sin x(1-\cos x + k\sin x)}\\
&= \frac{1-\cos x}{\sin x}
\end{align}
$$
EDIT: Exactly the same technique can be used to solve Narasimham's problem as well.
|
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|
Trigonometric Equation $\sin x=\tan\frac{\pi}{15}\tan\frac{4\pi}{15}\tan\frac{3\pi}{10}\tan\frac{6\pi}{15}$ How can I solve this trigonometric equation?
$$\sin x=\tan\frac{\pi}{15}\tan\frac{4\pi}{15}\tan\frac{3\pi}{10}\tan\frac{6\pi}{15}$$
|
Using this solution,
$$\tan x\tan(60^\circ-x)\tan(60^\circ+x)=\tan3x$$
Putting $x=12^\circ,$ $$\tan12^\circ\tan48^\circ\tan72^\circ=\tan36^\circ$$
$$\implies \tan 12^\circ \tan 48^\circ \tan 54^\circ \tan 72^\circ =\tan 54^\circ \tan36^\circ=\tan(90^\circ-36^\circ)\tan36^\circ=\cot36^\circ\tan36^\circ=1$$
|
{
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"timestamp": "2023-03-29T00:00:00",
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|
Finding the Limit $\lim_{x\to 0} \frac{\sin x(1 - \cos x)}{x^2}$ Here's the problem.
$$\lim_{x\to 0} \frac{\sin x(1 - \cos x)}{x^2}$$
I really don't know where to start with this. Please help.
|
I assume you know that $$\lim_{x \to 0} \frac{\sin x}{x}=1.$$
Now, $$\lim_{x \to 0} \frac{1-\cos x}{x^2} = \lim_{x \to 0} \frac{(1-\cos x)(1+\cos x)}{x^2 (1+\cos x)}=\lim_{x \to 0} \frac{\sin^2 x}{x^2(1+\cos x)},$$
and therefore $$\lim_{x \to 0} \frac{1-\cos x}{x^2}=\lim_{x \to 0} \left( \frac{\sin x}{x} \right)^2 \times\lim_{x \to 0}\frac{1}{1+\cos x}=\frac{1}{2}.$$
Finally, $$\lim_{x \to 0} \frac{\sin x (1-\cos x)}{x^2}=\lim_{x \to 0} \sin x \times \lim_{x \to 0} \frac{1-\cos x}{x^2}=0 \times \frac{1}{2}=0.$$
|
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|
Finding the curve whose asymptotes are given?
Find the equation of the cubic curve whose asymptotes are $x+a=0$,
$y-a=0$, $x+y+a=0$ and which touches the axis of X at origin and
passes through the point (-2a,-2a)?
My solution: The general equation of the curve must be $(x+a)(y-a)(x+y+a)+g(x,y)=0$
Its given that curve touches the x-axis at origin, which implies, x-axis $y=0$ is tangent at origin. So, the lowest degree terms in curve equation must have $y$ as a factor (because, we don't know if $y=0$ is going to be the only tangent at origin.If it was then the only lowest degree factor should have bee $y$). In either case, I don't know how to apply this concept and get $g(x,y)$.
I'm stuck here. I don't know how to proceed further. Final answer must be $xy(x+y)+x(y^2-x^2+xy)-6a^2y=0$
EDIT: By the way, can any one kindly suggest a good book that contains such problems?
|
Consider the most general form of a cubic planar curve $$Ax^3+By^3+Cx^2y+Dxy^2+Ex^2+Fxy+Gy^2+Hx+Ky+L=0.$$
Since we know that the curve must pass through the origin we can conclude that $L=0$. Now using the implicit differentiation we obtain $$y'_x=-\frac{3Ax^2+2Cxy+Dy^2+2Ex+Fy+H}{3By^2+Cx^2+2Dxy+Fx+2Gy+K}=\tag{1}$$ $$-\frac{3A+D\left(\frac{y}{x}\right)^2+2C\left(\frac{y}{x}\right)+\frac{1}{x}\left(2E+F\left(\frac{y}{x}\right)\right)+\frac{1}{x^2}H}{3B\left(\frac{y}{x}\right)^2+C+2D\left(\frac{y}{x}\right)+\frac{1}{x}\left(F+2G\left(\frac{y}{x}\right)\right)+\frac{1}{x^2}K}\tag{2}$$
Since we know that the tangent line at the origin is horizontal we have that $H=0$ from $y'(0)=0$. From the asymptotic behavior for $x\longrightarrow\infty$, $y\longrightarrow a$ (horizontal asymptote) we know $y'\longrightarrow 0$. Using these requirements we must conclude that $A=0$ from formula (2). Exactly the same argument will hold for the vertical asymptote $x=-a$. In this case we will need to compute $$x'_y=-\frac{3B+\ldots}{2X\frac{x}{y}+\ldots}\longrightarrow0\mbox{ as }y\longrightarrow\infty.$$ So conclude that $A=B=0$. Lets look at the slant asymptote $x+y=a$. When $x$ approaches infinity in a way that $y/x\longrightarrow-1$ we must have that $y'_x\longrightarrow -1$. Then we have from 1: $$-1=\frac{D+2C}{C+2D}\Longrightarrow C=D,$$ and we narrow down our search to the curves whose forms fit $$Cxy(x+y)+Ex^2+Fxy+Gy^2+Ky=0.$$ Without loosing generality we can set $C=1$ by rescaling. This is possible since $C=0$ the curve will no longer be cubic. Thus we have $$xy(x+y)+Ex^2+Fxy+Gy^2+Ky=0\tag{3}$$
This is a quadratic equation with respect $x$ or $y$. If we try to resolve (3) for $x$ and $y$ we will obtain $$(G+x)y^2+\mbox{lower order terms in }y=0\tag{4}$$ $$(E+y)x^2+\mbox{lower order terms in }x=0\tag{5}$$ If we want to have a horizontal asymptote at $y=a$ the leading coefficient of [5] must vanish for $y=a$ and we have $E=-a$. Similarly, the presence of the vertical asymptote $x=-a$ is only possible when the leading coefficient of [4] is $0$, i.e. $G=a$ and we obtain more narrow description of all such possible curves: $$xy(x+y)-ax^2+Fxy+ay^2+Ky=0\tag{6}$$ We have not used the fact that $(-2a,-2a)$ must belong to the curve. If we substitute these values into (6), another constrain is revealed: $$-8a^2+2Fa-K=0.\tag{7}$$ We know that $y=a-x$ gets closes and closer to the curve (6) if $x$ is sufficiently large. This means that if we substitute $y=a-x$ into (6) and let $x\longrightarrow\infty$ we should obtain $0$ (this follows from the continuity of the left hand-side of (6)): $$0=\lim_{x\to\infty}\left[x(a-x)(x+a-x)-ax^2-a(a-x)^2+Fx(a-x)+K(a-x)\right]=$$ $$=\lim_{x\to\infty}\left[x^2(-a-F)+x(\ldots)+\ldots\right]\Longrightarrow F=-a\Longrightarrow K=-10a^2\mbox{, follows from (7)}.$$ We arrive to the final form of the necessary cubic curve $$xy(x+y)+a(y^2+xy-x^2)-10a^2y=0.$$
W$_\alpha$ confirms our computations here and here.
|
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Homework - ERO with Unknown in Matrix I was having a problem with how to properly perform elementary row operation (ERO) on a matrix.
In the question, we were given an augmented matrix $\begin{bmatrix}1&1&-1&1\\2&3&a&3\\1&a&3&2\end{bmatrix}$ and require us to determine all possible values of $a$ such that the system is
*
*consistent with infinitely many solution
*consistent with one and only one solution
*inconsistent.
So I decided to make the matrix in row echelon form. Here are my steps:
1) $R_{3} - R_{1} \rightarrow R_{3}$
$\begin{bmatrix}1&1&-1&1\\2&3&a&3\\0&a-1&4&1\end{bmatrix}$
2) $R_{2} - 2R_{1} \rightarrow R_{2}$
$\begin{bmatrix}1&1&-1&1\\0&1&a+2&1\\0&a-1&4&1\end{bmatrix}$
3) $R_{3} - (a-1)R_{2} \rightarrow R_{3}$
$\begin{bmatrix}1&1&-1&1\\0&1&a+2&1\\0&0&4-(a+2)(a-1)&1-(a-1)\end{bmatrix}$
After I further simplify the matrix in step 3, I got $\begin{bmatrix}1 & 1 & -1 & 1 \\ 0 & 1 & a+2 & 1 \\ 0 & 0& (3+a)(2-a) & 2-a \end{bmatrix}$
With the simplified matrix, I have attempted to perform another ERO, $\frac{R_{3}}{2-a} \rightarrow R_{3}$ so that $\begin{bmatrix}1 & 1 & -1 & 1 \\ 0 & 1 & a+2 & 1 \\ 0 & 0& (3+a) & 1 \end{bmatrix}$. After this ERO, there will not be such a value that the system is consistent with infinitely many solutions, which is not same as the suggested answer. It appear that the ERO $\frac{R_{3}}{2-a} \rightarrow R_{3}$ is illegal.
I tried to check for the definition of ERO again, the textbook state that ERO include
*
*interchange any 2 equations of a system of linear equation
*multiply both sides of any equation in a system by a non-zero scalar
*add a multiple of one equation to another equation within the system
Under this definition, it seems that my step 3 is also illegal as it include a unknown $a$ in the ERO. However, using the result of step 3, I can manage to get the correct answer.
So I want to know that is that we can only perform ERO with a constant value (e.g. $R_{3} - 6R_{1} \rightarrow R_{3}$) or we can perform ERO with variable (e.g. $R_{3} - (a-8)R_{2} \rightarrow R_{3}$) but with certain restriction?
|
Hint: consider what you have after your step 3 and the simplification, that is,
$\qquad\begin{bmatrix}1 & 1 & -1 & 1 \\ 0 & 1 & a+2 & 1 \\ 0 & 0& (3+a)(2-a) & 2-a \end{bmatrix}$
*
*If $a=2$, then …
*If $a=-3$, then …
*If $a\ne2$ and $a\ne 3$ then …
Explanation of the error
You can't do $\frac{R_{3}}{2-a} \rightarrow R_{3}$ unless you know that $a\ne 2$. So if you perform that operation, you must also consider the case $a=2$ separately.
|
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|
Valid values of constant for probability mass function (PMF)
For what values of constant $c$ do the following functions define a valid PMF for random variable $X$ on supports $X=\{1,2,...\}$
(1) $f(x) = c/2^x$
(2) $f(x) = c2^x/x!$
I was thinking $\sum_{i=1}^\infty f(x) = 1$ so,
$$\begin{aligned}
\sum_{i=1}^\infty \frac{c}{2^x} &= 1 \\
c &= \sum_{i=1}^\infty 2^x
\end{aligned}$$
and $$\begin{aligned}
\sum_{i=1}^\infty \frac{c2^x}{x!} &= 1 \\
c &= \sum_{i=1}^\infty \frac{x!}{2^x}
\end{aligned}$$
Is these correct? (1) sums to infinity? Not sure about (2)
|
For the first, your start was correct, we want
$$c\left(\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+\cdots\right)=1.$$
The inner sum above is an infinite geometric series, with first term $\frac{1}{2}$ and common ratio $\frac{1}{2}$.
It has sum is $\dfrac{1/2}{1-1/2}$, which is $1$.
So we want $(c)(1)=1$, and therefore $c=1$.
For the second problem, again the start was correct, we want
$$c\sum_{n=1}^\infty \frac{2^n}{n!}=1.\tag{1}$$
You may recall that for any $t$, we have
$$\sum_{n=0}^\infty \frac{t^n}{n!}=e^t.\tag{2}$$
Comparing with the sum in (1), and noting that the first term of (2) is missing in (1), we find that
$$\sum_{n=1}^\infty \frac{2^n}{n!}=e^2-1.$$
So we want $c(e^2-1)=1$, making $c=\frac{1}{e^2-1}$.
Remark: What was going wrong in each calculation was the built-in assumption that the reciprocal of a sum (in this case an infinite "sum") is equal to the sum of the reciprocals. To simplify the mistake, you were assuming that the reciprocal $\frac{1}{a+b}$ of $a+b$ is equal to $\frac{1}{a}+\frac{1}{b}$. This is in general not true.
|
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How to find sum of 3 perpendiculars of a triangle? Q. ABC is an equilateral triangle with side 10cm and P is a point inside the triangle, at a distance of 2cm from AB. If PD, PE and PF are perpendiculars to the three sides, find sum PD+ PF+PE.
What I've done:
In triangles ADP and AFP,
$AP^2=x^2+PF^2$
$AP^2=y^2+4$
In DPB and PBE
$PB^2=4+(10-y)^2$
$PB^2=PE^2+(10-z)^2$
In PEC and PFC
$PC^2=PE^2+z^2$
$PC^2=PF^2+(10-z^2)$
|
From equilateral triangle and from $DB=FC$ we see $x=y$.
By using Carnot's theorem:
$$(10-z)^2+(10-y)^2+y^2=z^2+x^2+(10-y)^2$$
$$100-20z=x^2-y^2=0$$
$$5=z$$
From here we can say that the points $A,P,E$ are on a line and $AE=BE=EC=5$.
So $AP$ is bisector of $DAF$ and from there we can conclude that $FP=2$ and $AP=4$.
Hence $$PE+PD+PF=1+2+2=5$$
|
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|
Need to prove that $4\cdot 10^{2n} + 9\cdot 10^{2n-1} + 5$ is divisible by $99$ for all $n \in \mathbb{N} $, using induction. First, obviously, I figured out the base case. So I have $4\cdot 10^{2n} + 9\cdot 10^{2n-1} + 5 = 99k$ for some $k \in \mathbb{N} $. As for the inductive step, I was thinking about splitting it up into two parts; proving that the $4\cdot 10^{2n} + 9\cdot 10^{2n-1} + 5$ is divisible by $9$, and that it's divisible by $11$. Unfortunately, I factor it out and get $10(4\cdot10^{2n} + 9\cdot10^{2n-1} + 5)-45$, which is equal to $10(99k)-45$, but of $99$'s divisors, only $3$ divides $45$.
Is there a more ingenious way to factor this equation so that all of my problems are solved?
|
A number is a multiple of $9$ iff the sum of its decimal digits is a multiple of $9$. Your numbers satisfy this: $4+9+5=18$.
A number is a multiple of $11$ iff the alternating sum of its decimal digits is a multiple of $11$. Your numbers satify this: $4-9+5=0$.
Explicitly, we have $4900\cdots05=9\cdot 544\cdots45$ and $4900\cdots05=11\cdot 4454\cdots5455$.
Or directly, $4900\cdots05=99\cdot 4949\cdots495$, which is the formula given by Martin.
|
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|
Prove by induction that $1^3 + \dots + n^3 = (1 + \dots + n)^2$ I'm suppose to prove by induction:
$1^3 + \dots + n^3 = (1 + \dots + n)^2$
This is my attempt; I'm stuck on the problem of factoring dots.
|
Your attempt looks OK as far as it goes (except for a missing superscript $2$ at one point, but that's not causing any further errors).
To prove that something is equal to $1^3+2^3+3^3+4^3+\cdots+n^3$, one must show that at each step the amount that gets added is the next cube.
So how much has to be added to $(1+2+\cdots+n)^2$ to get $(1+2+\cdots+n+[n+1])^2$? We need to show that that is the next cube, $(n+1)^3$. The amount that gets added is
$$
\begin{align}
& \phantom{={}} (1+2+\cdots+n+[n+1])^2 - (1+2+\cdots+n)^2 \\[6pt]
& = (A + [n+1])^2 - (A)^2 \\[6pt]
& = A^2 + 2A[n+1]+[n+1]^2 - A^2 \\[6pt]
& = 2A[n+1] + [n+1]^2.
\end{align}
$$
Now here it helps to know that
$$
A = 1 + 2 + 3 + \cdots+n = \frac{n(n+1)}{2}.
$$
Thus the difference above is
$$
2A[n+1]+[n+1]^2 = n(n+1)[n+1]+[n+1]^2
$$
and this simplifies to
$$
n^3+3n^2+3n+1
$$
and finally to
$$
(n+1)^3.
$$
Next, there's the problem of organizing that into a presentable proof by induction.
|
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|
If $17! = 355687\underline{ab}8096000$. Then value of $(a,b)$ is
*
*If $17! = 355687\underline{ab}8096000$, what is the value of $(a,b)$?
*If $34! = 295232799\underline{cd}9604140847618609643\underline{ab}0000000$, what is the value of $(a,b,c,d)$?
My Attempt:
We know that
$$
17! = 1\times 2 \times 3 \times 4 \times 5\times 6\times 7\times 8 \times 9\times 10 \times 11\times 12\times 13\times 14\times 15\times 16\times 17
$$
Thus the RHS must be divisible by $3$,$7$, or $11$.
Step 1: Divisibility by $3$
We know that for some $K\in\mathbb{N}$
$$
\begin{align}
K&=\frac{3+5+5+6+8+7+a+b+8+0+9+6+0+0+0}{3}\\
&=\frac{57+a+b}{3}\\
&= 19+\frac{a+b}{3}
\end{align}
$$
Therefore, $0\leq a+b\leq 18$, which implies that $(a+b) \in\{0,3,6,9,12,15,18\}$.
Step 2: Divisibility by $11$
We know that for some $L\in\mathbb{N}$
$$
\begin{align}
L&=\frac{(3+5+8+a+8+9+0+0)-(5+6+7+b+0+6+0)}{11}\\
&=\frac{9+a-b}{11}\\
\end{align}
$$
This implies that $a-b=2$ or $b-a=9$, and thus we have to solve for $a-b = 2$ and $(a+b) \in\{6,12\}$. Therefore $(a,b) \in\lbrace{(4,2),(7,5)\rbrace}$. Again we have to solve for $b-a=9$ and $b+a\in\{3,9\}$, giving $(a,b)\in\{ (3,6),(0,9)\}$.
So we end up with $(a,b) \in\{(4,2),(7,5),(0,9)\}$. But answer given is $(a,b) = (4,2)$. What did I do wrong?
|
To avoid unanswered questions, I will compile the tips from the comments here, along with a bit more.
Instead of using divisibility by $3$, use divisibility by $9,$ so that we can actually conclude that $\frac{57+a+b}{9}=6+\frac{3+a+b}9$ is an integer. Since $0\le a+b\le 18,$ then it follows that either $a+b=6$ or $a+b=15$.
As with your work, we can also conclude that $a-b=2$ or $b-a=9$. The latter, though, is impossible, since $b\le 9$ and $a\ge 0,$ so that $b-a=9$ if and only if $b=9$ and $a=0,$ in which case, $a+b\ne 6,15.$ Therefore, we have $a-b=2.$ Since $2a=(a+b)+(a-b)=a+b+2,$ then $2a=8$ or $2a=17,$ but the latter is clearly false, since $a$ is an integer. Thus, $a=4,$ so since $a-b=2,$ then solving the system gives us $(a,b)=(4,2).$
Your second question was previously answered here. Calvin Lin's answer, in particular, seems very accessible.
|
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Given that $gcd(a,b)=1$, prove that $gcd(a+b,a^2-ab+b^2)=1$ or $3$, also when will it equal $1$? It is an exercise on the lecture that i am unable to prove.
Given that $gcd(a,b)=1$, prove that $gcd(a+b,a^2-ab+b^2)=1$ or $3$, also when will it equal $1$?
|
HINT:
Let prime $p$ divides $a+b, a^2-ab+b^2$
$\implies p$ divides $\{(a+b)^2-(a^2-ab+b^2)\}=3ab$
If $p|a,$ as $p|(a+b),p$ must divide $(a+b)-a=b\implies p|(a,b)$
But as $(a,b)=1,p$ can not divide $a$
Similarly, $p$ can not divide $b$
$\implies p|3$
$\implies (a+b,a^2-ab+b^2)|3$
and $(a+b,a^2-ab+b^2)=3$ if $3|(a+b)\iff 3|(a^2-ab+b^2)$
|
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|
Evaluating $\int_{-2}^{2} 4-x^2 dx$ with a Riemann sum I'm having problems with a Riemann sum ... I need to find the integral:$$\int_{-2}^2 (4-x^2)\;dx$$Clearly we have $$\int_{-2}^{2}(4-x^2)\;dx=4x-\frac{x^3}{3}\mid_{-2}^{2}=(4\cdot2-\frac{2^3}{3})-(4\cdot(-2)-\frac{(-2)^3}{3})=\frac{32}{3}$$OK.
On the other hand, we have
$$\Delta x=\frac{b-a}{n}=\frac{4}{n}$$
and
$$\xi_1=-2+\frac{4}{n};\;\;\xi_2=-2+2\frac{4}{n};\;\;\ldots\;\;;\xi_n=-2+n\frac{4}{n}$$
then
$$\xi_i=-2+\frac{4i}{n}=\frac{4i-2n}{n},$$
so that
$$ \begin{align} \int_{-2}^2 {4-x^2}\;dx=&\lim_{n\to+\infty} \sum_{i=1}^n\left(4-\left( \frac{4i-2n}{n}\right)^2\right)\frac{4}{n}\\
=&\lim_{n\to+\infty} \sum_{i=1}^n\left(4-\left( \frac{4i-2n}{n}\right)^2\right)\cdot\frac{4}{n}\\
=&\lim_{n\to+\infty} \sum_{i=1}^n\left(4-\left( \frac{16i^2-16ni+4n^2}{n^2}\right)\right)\cdot\frac{4}{n}\\
=&\lim_{n\to+\infty} \sum_{i=1}^n \left(\frac{4n^2-16i^2+16ni-4n^2}{n^2} \right)\cdot\frac{4}{n}\\
=&\lim_{n\to+\infty} \sum_{i=1}^n \frac{-16i^2+16ni}{n^2}\cdot\frac{4}{n} \\
=&\lim_{n\to+\infty}\sum_{i=1}^{n} \frac{-64i^2+64ni}{n^3}\\
=&\lim_{n\to+\infty} \frac{64}{n^3}\left(-\sum_{i=1}^{n} i^2+n\sum_{i=1}^{n} i\right) \\
=&\lim_{n\to+\infty} \frac{64}{n^3} \left(-\frac{n(n+1)(2n+1)}{6}+n\cdot \frac{n(n+1)}{2} \right)\\
=&\lim_{n\to+\infty} \frac{64}{n^3} \left(-\frac{\color{#ff0000}{2}n^3+3n^2+n}{6}+\frac{n^3+n^2}{2} \right) \\
=&\lim_{n\to+\infty} \frac{64}{n^3} \left(\frac{-\color{#ff0000}{2}n^3-3n^2-n+3n^3+3n^2}{6}\right)\\
=&\lim_{n\to+\infty}\frac{32}{3} \left(\frac{\color{#ff0000}{1\times}n^3-n}{n^3}\right) \\
=&\lim_{n\to+\infty}\frac{64}{\color{#ff0000}{2\times}3}-\frac{32}{3n^2}=\frac{64}{\color{#ff0000}{2\times}3} \color{#ff0000}{=}\frac{32}{3}=\int_{-2}^{2} {4-x^2}\;dx
\end{align}$$
Where is the mistake?
|
The problem is an accidental algebra mistake, as I pointed out in comments: $n(n+1)(2n+1)=2n^3+3n^2+n\neq n^3+3n^2+n$, and this should do it.
|
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|
Identity proofs I'm strugling with these two combinatorial identities:
$$ \binom{n+2}{3}=\sum_{i=1}^{n} i(n+1-i) $$
and
$$\binom{n+1}{2}^2=\sum_{i=1}^{n} i^3$$
Please give me some footholds and hints
|
For the first problem split the RHS:
$$\sum_{i=1}^n i(n+1-i) = \sum_{i=1}^n in +i - i^2 = n\sum_{i=1}^n i + \sum_{i=1}^n i - \sum_{i=1}^n i^2$$
Now we know that the some for consecutive integer from $1$ to $n$ is $\frac{n(n+1)}{2}$ and for the squares of consecutive integers from $1$ to $n$ is $\frac{n(n+1)(2n+1)}{6}$. So we substitute and we have:
$$n\sum_{i=1}^n i + \sum_{i=1}^n i - \sum_{i=1}^n i^2 = n \times \frac{n(n+1)}{2} + \frac{n(n+1)}{2} - \frac{n(n+1)(2n+1)}{6}$$
$$n \times \frac{n(n+1)}{2} + \frac{n(n+1)}{2} - \frac{n(n+1)(2n+1)}{6} = \frac{n(n+1)}{6} \times (3n + 3 - 2n - 1) = \frac{n(n+1)(n+2)}{3!} = \frac{(n+2)!}{3!(n-1)!} = \binom{n+2}{3}$$
And for the second problem it's a direct consequence from:
$$\sum_{i=1}^n i^3 = \left(\frac{n(n+1)}{2}\right)^2 = \binom{n+1}{2}^2$$
Note that this substitutions for the summations are well known, but they are proven using induction.
|
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Is there a way to calculate $\int \limits_0^1\frac{x^3}{\sqrt{x^2-1}}\frac{1}{1-a^2x^2}\frac{1}{1-b^2x^2}\frac{1}{c-x}\mathrm dx$ I want to calculate
$\displaystyle \int \limits_0^1\dfrac{x^3}{\sqrt{x^2-1}}\dfrac{1}{1-a^2x^2}\dfrac{1}{1-b^2x^2}\dfrac{1}{c-x}\mathrm dx$
$a$ and $b$ are real parameters, c could be complex and is the solution of a cubic equation.
I tried to find an appropriate contour in the complex plane but failed because it seems impossible to go around the cut (integration from $-1$ to $1$ would be easier). Have anyone an idea?
Or is any other integration technique better suited?
|
Let us denote$A:=\arccos(a)$, $B:=\arccos(b)$ and $C:=\arccos(1/c)$.
By firstly substituting $x=\sin(y)$ and then $u=\tan(y/2)$ we get:
\begin{eqnarray}
&&\int\limits_0^1 \frac{x^3}{\sqrt{1-x^2}} \frac{1}{1-a^2 x^2} \frac{1}{1-b^2 x^2} \frac{1}{c-x} dx= \\
&&\frac{16}{c} \int\limits_0^{\tan(\pi/4)} \frac{u^3+u^5}{\prod\limits_{\xi_2=\pm}(u-\exp(\xi_2 \imath C))\prod\limits_{\xi_1=\pm}(u+\xi_1 \exp(\xi_2 \imath A))(u+\xi_1 \exp(\xi_2 \imath B))} du
\end{eqnarray}
Now the only thing we need to do is to use the following partial fraction decomposition:
\begin{eqnarray}
\frac{u^p}{\prod\limits_{j=1}^d (u-u_j)} = \sum\limits_{i=1}^d \frac{u_i^p}{(u-u_i)} \cdot \prod\limits_{j\ne i} \frac{1}{u_i-u_j}
\end{eqnarray}
which is valid for $d\ge 1$ and $p=0,1,2,\cdots,d-1$ and the computation is completed.
|
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Help with exact value of: $\tan (\sin^{-1}(-1/2) - \tan^{-1}(3/4))$ Ok, so I used the tan formula of difference of angles, and so far I've got to:
I found that $$\tan \alpha = \frac{-1/2}{\sqrt{3}/2} = \pm \frac{\sqrt3}{3}$$
so
$$\tan \left(\sin^{-1}\left(\frac{-1}{2}\right) - \tan^{-1}\left(\frac{3}{4}\right)\right) = \frac{\sqrt3/3 - 3/4}{1+(\sqrt3/3)(3/4)}$$
I went a little past this but keep running into trouble. Can someone show me a step by step way to solve this from beginning to end so I can see how this is done? Also note it's not a proof, and I can't use exact values in form of decimals. It needs to be rationalized square-roots and fractions. Much appreciated
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Let $a=\sin^{-1}\left(-\frac{1}{2}\right)$ and $b=\tan^{-1}\left(\frac{3}{4}\right)$. We have
$$\sin^2a+\cos^2a=1 \Rightarrow \frac{1}{4}+\cos^2a=1 \Rightarrow \cos a=\frac{\sqrt 3}{2} \Rightarrow \tan a = -\frac{\sqrt 3}{3}$$
since $\sin^{-1}x$ has $\left[-\frac{\pi}{2},\frac{\pi}{2}\right]$ as image.
Also,
$$\frac{\sin b}{\cos b}=\frac{3}{4} \Rightarrow \frac{9}{16}\cos^2 b+\cos^2b=1 \Rightarrow \cos b=\frac{4}{5} \Rightarrow \sin b=\frac{3}{5}$$
since $\tan^{-1}x$ has $\left(-\frac{\pi}{2},\frac{\pi}{2}\right)$ as image.
We then have
$$\tan(a-b)=\frac{\tan a-\tan b}{1+\tan a \tan b} = \frac{-\frac{\sqrt 3}{3}-\frac{3}{4}}{1-\frac{\sqrt 3}{3} \frac{3}{4}}=-\frac{4\sqrt3 + 9}{12-3\sqrt 3}=-\frac{4\sqrt3 + 9}{12-3\sqrt 3}\frac{12+3\sqrt 3}{12+3\sqrt 3}=$$
$$-\frac{108 + 48\sqrt 3 + 27\sqrt 3 +36}{144-27}=-\frac{144+75\sqrt 3}{117}=-\frac{48+25\sqrt 3}{39}$$
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If $\sqrt[3]{a} + \sqrt[3]{b}$ is rational then prove $\sqrt[3]{a}$ and $\sqrt[3]{b}$ are rational Assume there exist some rationals $a, b$ such that $\sqrt[3]{a}, \sqrt[3]{b}$ are irrationals, but:
$$\sqrt[3]{a} + \sqrt[3]{b} = \frac{m}{n}$$
for some integers $m, n$
$$\implies \left(\sqrt[3]{a} + \sqrt[3]{b}\right)^3 = \frac{m^3}{n^3}$$
$$\implies a + b + 3 \cdot \sqrt[3]{ab}\left(\sqrt[3]{a} + \sqrt[3]{b}\right) = \frac{m^3}{n^3}$$
Since $a +b$, $\sqrt[3]{a} + \sqrt[3]{b}$ are rational, $\sqrt[3]{ab}$ must be rational as well.
For convenience let us say $\sqrt[3]{a} = p, \sqrt[3]{b} = q \implies pq$ is rational. This means, for all $i$, $p^iq^i$ is rational.
$$\implies (p + q)^2 = \frac{m^2}{n^2}$$
$$ = p^2 + q^2 + 2pq = \frac{m^2}{n^2}$$
Since $pq$ is rational, $2pq$ is rational and so is $p^2 + q^2$.
Assume, for some $i$ that $p^i + q^i$ and $p^{i-1} + q^{i-1}$ is rational.
$$\implies (p^i + q^i)(p + q) - pq(p^{i - 1} + q^{i - 1}) = p^{i+1} + q^{i+1}$$
is rational as well.
So for all $i$,
$$a^{\frac{i}{3}} + b^{\frac{i}{3}}$$
and
$$a^{\frac{i}{3}}b^{\frac{i}{3}}$$
are rational.
I know I'm really close to the answer, but it somehow just keeps slipping through my fingers.
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Statement for the general case (for any $n>1$ in place of $3$)
Let $a,b$ be two positive rational numbers such that $\sqrt[n]{a}+\sqrt[n]{b}$ is a rational number for some natural number $n>1$. Then both $\sqrt[n]{a}$, $\sqrt[n]{b}$ are rational numbers.
Proof
Let $s=\sqrt[n]{a}+\sqrt[n]{b}\in\mathbb{Q}$. Then $\sqrt[n]{a}$ is a root of the polynomial $p(X)=X^n-a$ (If $\sqrt[n]{a}$ is not rational then $p$ is in fact the minimal polynomial). Again $\sqrt[n]{a}$ is also a root of $q(X)=(s-X)^n-b$. Now we show that $\sqrt[n]{a}$ is the unique common root of $p$ and $q$. Let $z$ be a common root of $p$ and $q$. Then $z=\sqrt[n]{a}\omega_1$ and $s-z=\sqrt[n]{b}\omega_2$ for some $n^{th}$ roots of unity $\omega_1,\omega_2$. Then $$\sqrt[n]{a}+\sqrt[n]{b}=\sqrt[n]{a}\omega_1+\sqrt[n]{b}\omega_2\tag{1}$$ Since $|\omega_1|=|\omega_2|=1$ therefore $|\mathrm{Re}(\omega_i)|\leq1$ with equality if and only if $\omega_i=\pm1$ for $i\in\{1,2\}$. Now comparing real parts in the equation $(1)$ we can easily conclude that $\omega_1=\omega_2=1$ implying that $\sqrt[n]{a}$ is the unique common root.
Then $$\gcd(p(X),q(X))=X-\sqrt[n]{a}$$ But $p(X),q(X)\in\mathbb{Q}[X]$ implies $X-\sqrt[n]{a}$ is in $\mathbb{Q}[X]$ then $\sqrt[n]{a}\in\mathbb{Q}$. Similarly one can prove that $\sqrt[n]{b}$ is also rational.
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How prove this $x+y=0$ if $\left(\sqrt{y^2-x^3}-x\right)\left(\sqrt{x^2+y^3}-y\right)=y^3$ Question:
let $x,y$ are real numbers,and such
$$\left(\sqrt{y^2-x^3}-x\right)\left(\sqrt{x^2+y^3}-y\right)=y^3$$
show that
$$x+y=0\tag{1}$$
before I have solve following problem:
if
$$(x+\sqrt{x^2+1})(y+\sqrt{y^2+1})=1,\Longrightarrow x+y=0\tag{2}$$
solution:we have
$$y+\sqrt{y^2+1}=\sqrt{x^2+1}-x\tag{3}$$
$$x+\sqrt{x^2+1}=\sqrt{y^2+1}-y\tag{4}$$
$(3)+(4)$
then $x+y=0$
But for $(1)$ we can't use this methods to solve it.maybe can have other nice methods, Thank you
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Below some characteristics of the function $f(x,y) = \left(\sqrt{y^2-x^3}-x\right)\left(\sqrt{x^2+y^3}-y\right)-y^3$ are shown. The area where $f(x,y)$ is undefined is in gray: $y^2 - x^3 < 0$ or $x^2 + y^3 < 0 $. Therefore the domain of the function is mainly restricted to $x\le 0$ and $y\ge 0$. The line $x+y=0$ is in red, some contour lines of $f(x,y)$ are in blue. The function is somewhat symmetrical around $x+y=0$, but not quite. Shouldn't the restriction $x\le 0$ and $y\ge 0$ be added to the problem in the first place?It's easy to prove that $x+y=0$ is a solution of $f(x,y)=0$, but proving the reverse seems to be tedious; at least I see no "nice" way to do it.
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Integers whose sum and product are integers Let $a$, $b$ be real numbers such that $a + b$ and $ab$ are integers.
a. Prove that $a^n + b^n$ is an integer for every natural number $n$.
b. Suppose that $a$ does not equal $b$. Prove that $\dfrac{a^n - b^n}{a - b}$ is an integer for every positive integer $n$.
Attempts
I am thinking mathematical induction can be applied to part a.
1) Base Case: n = 1
$a^1$ + $b^1$ = $a + b$ which is an integer.
2) Assume $a^{n-1}$ + $b^{n-1}$ is an integer. We want to show that $a^n$ + $b^n$ is also an integer.
3) $a^n$ + $b^n$ = ... Not sure what to do next
Mathematical induction seems to also apply to part b.
1) Base Case: n = 1
$\dfrac{a^1 - b^1}{a - b}$ = $\dfrac{a-b}{a - b}$ = 1 which is an integer.
2) Assume $\dfrac{a^{n-1} - b^{n-1}}{a - b}$ is an integer. We want to show that $\dfrac{a^{n} - b^{n}}{a - b}$ is also an integer.
3) $\dfrac{a^{n} - b^{n}}{a - b}$ = ... Not sure what to do next
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For a)
$\\ { a }^{ 2 }+{ b }^{ 2 }={ \left( a+b \right) \left( a+b \right) }-2ab,\quad integer\\ { a }^{ 3 }+{ b }^{ 3 }=\left( a+b \right) { \left( { a }^{ 2 }+{ b }^{ 2 } \right) }-ab\left( a+b \right) ,\quad integer\\ { a }^{ 4 }+{ b }^{ 4 }=\left( a+b \right) { \left( { a }^{ 3 }+{ b }^{ 3 } \right) }-ab\left( { a }^{ 2 }+{ b }^{ 2 } \right) ,\quad integer\\ \vdots \\ { a }^{ n }+{ b }^{ n }=\left( a+b \right) { \left( { a }^{ n-1 }+{ b }^{ n-1 } \right) }-ab\left( { a }^{ n-2 }+{ b }^{ n-2 } \right) ,\quad integer$
|
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Existence of positive integer k that are both squares Is there a positive integer k such that $4k+1$ and $9k+1$ are both squares?
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The answer is no for k>0.
Let $b>a$, $a^2=4k+1$ and $b^2=9k+1$. Note that,
$$
5k=b^2-a^2=(b+a)(b-a)=\left(\sqrt{9k+1}+\sqrt{4k+1}\right)\cdot \left(\sqrt{9k+1}-\sqrt{4k+1} \right)
$$
Case 1:set $k$ a prime number. Then
*
*$k=\left(\sqrt{9k+1}+\sqrt{4k+1}\right)$ and $5=\left(\sqrt{9k+1}-\sqrt{4k+1}\right)$ or
*$k=\left(\sqrt{9k+1}-\sqrt{4k+1}\right)$ and $5=\left(\sqrt{9k+1}+\sqrt{4k+1}\right)$.
In both cases,
$
k+5=2\sqrt{9k+1}.
$
And this quadratic equation has no integer solution to $ k $.
Case 2:set $k=\alpha\cdot\beta\cdot p$ whit $\beta\geq \alpha$ and $p>1$ natural numbers. By a procedure completely analogous to the previous case we obtain,
*
*$\beta\cdot p =\left(\sqrt{9k+1}+\sqrt{4k+1}\right)$ and $5\alpha=\left(\sqrt{9k+1}-\sqrt{4k+1}\right)$ or
*$\beta\cdot p=\left(\sqrt{9k+1}-\sqrt{4k+1}\right)$ and $5\alpha=\left(\sqrt{9k+1}+\sqrt{4k+1}\right)$.
In both cases, $p\beta +5\alpha=2\sqrt{9\alpha\beta p+1}.$ Also in this case you can also check that there is no real solution to $ p $ for all $\alpha,\beta\in\mathbb{N}^*$.
|
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How do I prove inequalities and one-to-one function? Can anyone please help me with these questions?
1.Given
x + 1 < 0
Prove that:
i) $2x - 1 < 0 $
ii) ${2x-1\over x+1} > 2$
2.For $g(x) = {kx + 8\over 4x - 5}$
i) Find k if gg(x) = x
Is it fine if I just let any value of x for this question?
ii) Find the value of k so that g(x) is not a one-to-one function.
Thank you very much.
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(1 i) $x+1<0\Rightarrow 2\left( x+1 \right)<2.0\Rightarrow 2x+2<0\Rightarrow 2x+2-3<-3\Rightarrow 2x-1<-3<0$
so $2x-1<0.$
(1 ii)$$\frac{2x-1}{x+1}=\frac{2\left( x+1 \right)-3}{x+1}=2-\frac{3}{x+1}$$
because $x+1<0$ given we can write $$-\frac{3}{x+1}>0\Rightarrow 2-\frac{3}{x+1}>2 \Rightarrow\frac{2x-1}{x+1}>2$$
(2 i) we know if $g\left( x \right)=\frac{ax+b}{cx+d}$ then ${{g}^{-1}}\left( x \right)=\frac{-dx+b}{cx-a}$.
$g\left( g\left( x \right) \right)=x\Rightarrow g\left( x \right)={{g}^{-1}}\left( x \right)$ this means $g(x)=\frac{kx+8}{4x-5}=\frac{5x+8}{4x-k}={{g}^{-1}}\left( x \right)$ so $k=5.$
(2 ii) similar way
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Writing a number as power of two Number of powers of two needed to be added to the given powers of two to write the (sum of given powers and added powers of two)number of the form $2^k-1$ where $k$ is any integer.
Okay, let me explain the question.
Suppose you are given: $2^0 , 2^1 , 2^2$
Here sum of the numbers $=1+2+4=7=2^3-1$. So, there is no need to add any powers of two :)
Suppose you are given: $2^2 , 2^4 , 2^5$
Here sum of the numbers $=4+16+32=52$ which cannot be written as $2^k-1$ for any $k$.
So, we are adding here $2^0,2^1,2^3$ to the above sum which makes it $63 = 2^6 -1$.
Answer here is $3$ because we are adding here three powers of two ($2^0,2^1,2^3$).
I know that any number can be represented as sum of powers of two. But,can somebody give me an insight how to derive solution for this problem?
Thank you!
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Suppose that you’re given $2^{a_1},2^{a_2},\ldots,2^{a_n}$, where $a_1\le a_2\le\ldots\le a_n$. If $a_1=a_2$, erase $2^{a_1}$ and $2^{a_2}$ and write down $2^{a_1+1}$; then sort the powers of $2$ into non-decreasing order again. Repeat until the two smallest powers are different. Then do the same thing with the second-smallest power, and so on up the line.
Example: You’re given $2^2,2^2,2^2,2^2,2^3,2^4,2^7$. The stages are shown below: $$\begin{align*}&\color{brown}{2^2,2^2},2^2,2^2,2^3,2^4,2^7\\&\color{brown}{2^3},2^2,2^2,2^3,2^4,2^7\\&2^2,2^2,2^3,2^3,2^4,2^7\\&\color{brown}{2^2,2^2},2^3,2^3,2^4,2^7\\&\color{brown}{2^3},2^3,2^3,2^4,2^7\\&2^3,2^3,2^3,2^4,2^7\\&\color{brown}{2^4},2^3,2^4,2^7\\&2^3,2^4,2^4,2^7\\&2^3,\color{brown}{2^4,2^4,}2^7\\&2^3,\color{brown}{2^5},2^7\\&2^4,2^5,2^7\end{align*}$$
When the process is complete, you’ll have at most one copy of each power of $2$. Now fill in all missing powers below the largest; in the example that means adding $2^0,2^1,2^2,2^3$, and $2^6$. If the highest power that you have after the consolidation step is $2^a$, the sum will be $2^{a+1}-1$.
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$x^2+xy+y^2$ and $x^2-xy+y^2$ are not both perfect squares
Prove that $x^2+xy+y^2$ and $x^2-xy+y^2$ cannot be both perfect squares.
Surely $x$ and $y$ are natural numbers. If $x^2+xy+y^2 =a^2$ and $x^2-xy+y^2=b^2$ simultaneously then we have to show that there are no such integers $a$ and $b$.
I have tried that:
Suppose $x^2+xy+y^2=a^2$ and $x^2-xy+y^2=b^2$. Then $2xy=a^2-b^2=(a+b)(a-b)$, so one of $x$ and $y$ is even, but then I am stuck.
|
This is not yet a full answer, but I need to get ready for work now and I'm not sure how to proceed anyway.
This statement clearly doesn't hold if $x$ or $y$ can be $0$, and negative numbers are easily flipped about, so assume that $x$ and $y$ are positive integers.
If $x$ and $y$ are both even, then dividing each by $2$ will yield another solution. Thus if there is any solution, there must be one for which $x$ or $y$ is odd, so we will assume, without loss of generality, that $y$ is odd.
Since $2xy=(a+b)(a-b)$, either $a+b$ or $a-b$ must be even. But $a+b = (a-b)+2b$, so in fact $a+b$ and $a-b$ are both even.
Thus $4\mid (a+b)(a-b)=2xy$, so $x$ or $y$ is even. Since $y$ is assumed to be odd, $x$ must be even.
Thus $a$ and $b$ are both odd.
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Can $(1-\frac{1}{2})(1-\frac{1}{2^2})(1-\frac{1}{2^3})...(1-\frac{1}{2^{n-1}})(\frac{1}{2^n})$ be simplified? Can $(1-\frac{1}{2})(1-\frac{1}{2^2})(1-\frac{1}{2^3})...(1-\frac{1}{2^{n-1}})(\frac{1}{2^n})$ be simplified? It seems like an expression from a simple induction proof problem that's missing its result.
|
You could write
$$\frac{1}{2} \cdot \frac{3}{4} \cdot \frac{7}{8} \cdot \frac{15}{16} \cdot \dots \cdot \frac{2^{n - 1} - 1}{2^{n - 1}} \cdot \frac{1}{2^n}$$
and then the denominator is
$$2 \cdot 4 \cdot \dots 2^n = 2^{1 + 2 + 3 + \dots + n} = 2^{n(n + 1)/2}$$
while the numerator is
$$(2 - 1) (2^2 - 1) (2^3 - 1) \cdots (2^{n - 1} - 1) = \prod_{k = 1}^n (2^k - 1) = (-1)^n (2; 2)_n$$
where $(2; 2)_n$ is a $q$-Pochhammer symbol. So the final "simplification" would be
$$\frac{(-1)^n (2; 2)_n}{2^{n(n + 1)/2}}$$
|
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I have an abstract algebra proof I am a little iffy on. Help please! I was independently studying abstract algebra and found the following problem:
The problem says to show $\mathbb{Q}(\sqrt{3},\sqrt{7}) = \mathbb{Q}(\sqrt{3} + \sqrt{7})$.
The solution given was as follows:
Everything made since except line 4. Where does the $-3(\sqrt{3} + \sqrt{3})$ come from and why does it equal $4\sqrt{3}$?
Would someone be able to show be how this works for any case? In other words,
$\mathbb{Q}(\sqrt{a}, \sqrt{b}) = \mathbb{Q}(\sqrt{a} + \sqrt{b})$ where $gcd(a,b) = 1$.
I wanted to use a similar argument as the above proof.
|
$(\sqrt{a} + \sqrt{b})^2 = a + b + 2\sqrt{ab} \in \mathbb{Q}(\sqrt{a}+\sqrt{b}) \Rightarrow \sqrt{ab} \in \mathbb{Q}(\sqrt{a}+\sqrt{b})$.
Then, $\sqrt{ab}(\sqrt{a}+\sqrt{b}) = a\sqrt{b} + b\sqrt{a} \in \mathbb{Q}(\sqrt{a}+\sqrt{b})$
$\Rightarrow a\sqrt{b} + b\sqrt{a} - a(\sqrt{b}+\sqrt{a}) = b\sqrt{a}-a\sqrt{a} \in \mathbb{Q}(\sqrt{3}+\sqrt{7})$
$\Rightarrow \displaystyle\frac{1}{b-a} \cdot \sqrt{a}(b-a) = \sqrt{a} \in \mathbb{Q}(\sqrt{a} + \sqrt{b})$
Since $\sqrt{a} \in \mathbb{Q}(\sqrt{a} + \sqrt{b}), \sqrt{b} = (\sqrt{a} + \sqrt{b}) - \sqrt{a} \in \mathbb{Q}(\sqrt{a},\sqrt{b})$.
|
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Proper method for solving quadratic equations with exponents $(\sqrt {x^2-5x+6}+\sqrt{x^2-5x+4})^{x/2}$ + $(\sqrt {x^2-5x+6}-\sqrt{x^2-5x+4})^{x/2}$ = $2^{(x+4)/4}$
I have found out, by trial and error method, that $x=0$ and $x=4$ satisfy this equation. But is there a proper way to solve this equation and get the solutions?
|
Put
$$
a=(\sqrt {x^2-5x+6}+\sqrt{x^2-5x+4})^{x/2},\\
b=(\sqrt {x^2-5x+6}-\sqrt{x^2-5x+4})^{x/2}
$$
Then
$$
a+b=2^{(x+4)/4},\\
a \cdot b=2^{x/2}
$$
By solving it we get that $a=b,$ or
$$
(\sqrt {x^2-5x+6}+\sqrt{x^2-5x+4})^{x/2}=(\sqrt {x^2-5x+6}-\sqrt{x^2-5x+4})^{x/2}.
$$
It is a simple equation.
|
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|
Finding the limit of $(\root 3 \of {{n^3} + {n^2}} - \root 3 \of {{n^3} + 1} )$ Any Ideas/Hints?
$$\lim\limits_{n \to \infty } (\root 3 \of {{n^3} + {n^2}} - \root 3 \of {{n^3} + 1} )$$
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Hint: $a^3-b^3=(a-b)(a^2+ab+b^2)$. Take $a=\sqrt[3]{n^3+n^2}$ and $b=\sqrt[3]{n^3+1}$.
|
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When is the quadratic congruence $ax^2 + bx +c \equiv 0 \pmod p$ solvable? I am learning about quadratic congruences and I don't now how to decide, for which $a, b, c$ and $p$ there is a solution of the congruence. Is it sufficient if the discrminant $b^2-4ac$ has a solution in $\Bbb Z_p^*$?
|
If $a = 0$, this reduces to a linear congruence which has a unique solution (provided $b \neq 0$); I assume you know how to find the solution in this case.
Now suppose $a \neq 0$.
If $\underline{p \neq 2}$, multiply the congruence by $4a$ to obtain $4a^2x^2 + 4abx + 4ac \equiv 0 \pmod p$. Completing the square gives $(2ax + b)^2 + 4ac - b^2 \equiv 0 \pmod p$. Now let $y = 2ax + b$, then the congruence becomes $y^2 \equiv b^2 - 4ac \pmod p$. Therefore the discriminant $b^2 - 4ac$ must be a square modulo $p$ (i.e. a square in $\mathbb{Z}_p$); we call such an element a quadratic residue modulo $p$.
If $b^2 - 4ac$ is not a quadratic residue modulo $p$, then the congruence has no solutions.
If $b^2 - 4ac$ is a quadratic residue modulo $p$, then there are two values of $y$, call them $y_1$, $y_2$, which solve $y^2 \equiv b^2 - 4ac$. Furthermore, these two values are negatives of each other, i.e. $y_2 = - y_1$. Note that it is possible that $y_1 = y_2$, namely when $b^2 - 4ac = 0$ in which case $y_1 = y_2 = 0$. For each value $y_i$, we obtain a value $x_i$ by solving the linear congruence $2ax_i + b \equiv y_i \pmod p$ which has a solution because $a \neq 0$. Note that $x_1$ and $x_2$ are distinct if and only if $y_1$ and $y_2$ are distinct.
In summary, the equivalence $ax^2 + bx + c \equiv 0 \pmod p$ has
\begin{align*}
\text{no solutions}\ &\text{if}\ b^2 - 4ac\ \text{is not a quadratic residue}\\
\text{one solution}, x = 0\ &\text{if}\ b^2 - 4ac = 0\\
\text{two distinct solutions}\ &\text{if}\ b^2 - 4ac\ \text{is a non-zero quadratic residue modulo}\ p.
\end{align*}
The problem of determining whether or not $b^2 - 4ac$ is a quadratic residue modulo $p$ can be solved by using the law of quadratic reciprocity.
If $\underline{p = 2}$, then as $a \neq 0$, $a = 1$ so the congruence becomes $x^2 + bx + c \equiv 0 \pmod 2$. If $x = 0$, then we obtain the congruence $c \equiv 0 \pmod 2$; if $x = 1$, then we obtain the congruence $b + c \equiv 1 \pmod 2$. With these conditions at hand, we can determine the solutions for all four possible congruences:
\begin{align*}
x^2 &\equiv 0 \pmod 2\ \text{has one solution,}\ x = 0\\
x^2 + 1 &\equiv 0 \pmod 2\ \text{has one solution,}\ x = 1\\
x^2 + x &\equiv 0 \pmod 2\ \text{has two solutions,}\ x= 0, 1\\
x^2 + x + 1 &\equiv 0 \pmod 2\ \text{has no solutions.}
\end{align*}
|
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|
Show that if $x>0$, then $\ln(x)\geq 1-\frac{1}{x} $ Show that if $x>0$, then
$$ \ln(x)\geq 1-\dfrac{1}{x}. $$
I tried a few things but so far nothing has worked, I could use a hint.
|
You can use mean value theorem for $\ln$.
*
*$x = 1$, then equality holds.
*$0 < x < 1$:
$$\ln x \ge \frac{x-1}{x} \\
\frac{\ln x - \ln 1}{x- 1} \le \frac{1}{x} \\
\frac{1}{c} \le \frac{1}{x} \\
c \ge x$$
Which indeed is true, because $c \in (x,1)$. Change from $\ge$ to $\le$ is due to division by $(x-1)$ which is smaller than $0$.
*$x>1$ similarly as in 2.:
$$\ln x \ge \frac{x-1}{x} \\
\frac{\ln x - \ln 1}{x- 1} \ge \frac{1}{x} \\
\frac{1}{c} \ge \frac{1}{x} \\
c \le x$$
here $c \in (1, x)$ so the inequality is true as well.
|
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|
trigonometric equation opening Solve:
$$ \sin x + \sin 3x + \sin 5x = 0 . $$
Attempt at a solution: applying formulas for summation of sine we get after a series of operations: $ \sin x(8 \cos x \cos 2x \cos x + 1) = 0$ equaling sine to $0$ we get one solution $180k$. comparing the other factor we eventually get $\cos^2 x = 0.25 $ this in turn produces the solutions plus/minus $120 + 180 k$ and plus/minus $60 + 180 k$ , giving us $5$ solutions in total. ( $K \in Z $ )
Is this ok?
|
Using $e^{xi}=\cos x+i\sin x$, we can easily derive a canonical form for $\sin(ax)$, for $a\in\mathbb{N}$.
For example, $e^{3xi}=(\cos x+i\sin x)^3=\cos 3x+i\sin 3x$. Equating the imaginary parts of each expression, we obtain $$3\cos^2x \sin x-\sin^3x=\sin 3x$$ If we reuse this technique for 5, $e^{5xi}=(\cos x+i\sin x)^5=\cos 5x+i\sin 5x$. Equating the imaginary parts again,
$$ 5\cos^4x\sin x-10\cos^2x\sin^3x+\sin^5x=\sin 5x $$
So, your question reduces to
$$ \begin{align}
\sin x+3\cos^2x \sin x-\sin^3x+5\cos^4x\sin x-10\cos^2x\sin^3x+\sin^5x&=0 \\
\sin x \left(1+3(1-\sin^2x)-\sin^2x+5(1-\sin^2x)^2-10\sin^2x(1-\sin^2x)+\sin^4x\right)&=0 \\
\sin x(9-24\sin^2x+16\sin^4x)&=0
\end{align}$$
Now, we have $\sin x=0$ or $9-24\sin^2x+16\sin^4x=0$. Using the quadratic formula,
$$ \sin^2x=\frac{24\pm\sqrt{24^2-4\cdot9\cdot16}}{32}=\frac{3}{4} $$
So your solutions are $\sin x=0,\ \pm\frac{\sqrt{3}}{2}$, or when $x$ is an integer multiple of $\frac{\pi}{3}$
|
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|
converting decimals to base negative-10 I have a decimal (base $10$) number, $44$, and would like to convert it to base $-10$. I know how to convert
$$
164_{-10} \mapsto 44_{10},
$$
but not the other way around.
|
The simplest way is to simply carry and 'borrow' as needed, keeping track of odd vs. even digit positions. For your example, $44 = 4\times 10^1+4\times 10^0$. The ones' digit is clearly the same, so now we can concentrate on the tens' digit. Since $4=10-6$, we can express $4\times 10^1=40$ by borrowing from (or carrying into) the hundreds' digit; $40 = 4\times 10^1$ $= 10\times 10^1 - 6\times 10^1$ $= 1\times 10^2 - 6\times 10^1$ $= 1\times (-10)^2 + 6\times (-10)^1$; in other words a digit $d\gt 0$ in any 'odd' position (10s, 1000s, etc.) converts to a digit $d'=10-d$ in that position and a carry of $1$ into the next digit up. This can sometimes lead to a cascading effect of borrows and carries; for instance, imagine trying to convert $N=944_{10} = 9\times 10^2 + 4\times 10^1 + 4\times 10^0$. The ones' place is trivial, as before; and as before, we borrow/carry from the hundreds' place to invert the teens; $N = 9\times 10^2 + 1\times 10^2 - 6\times 10^1 + 4\times 10^0 = 10\times 10^2 - 6\times 10^1 + 4\times 10^0$. Now, this would 'carry' into the thousands' place as a positive digit: $N=1\times 10^3-6\times 10^1 + 4\times 10^0$, forcing us to do one more round of borrowing, turning $d=1$ into $10-d'=10-9$ : $N=1\times 10^4 - 9\times 10^3 + 0\times 10^2 - 6\times 10^1 + 4\times 10^0$ $=1\times (-10)^4 + 9\times (-10)^3 + 0\times (-10)^2 + 6\times (-10)^1 + 4\times 10^0$ $= 19064_{-10}$.
|
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|
How do you find the smallest possible value of aan equation with two unknowns? I'm solving a list of problems where I'm given an equation and I find the smallest possible value by comparing the equation to a quadratic equation and completing the square, however the next one involves another unknown $y$:
$$ x^2 - 3x + 2y^2 + 4y + 2. $$
I've been thinking about maybe somehow making $c$ equal $2y^2 + 4y + 2$?
The answer from the answerbook is: $$ x^2 - 3x + 2y^2 + 4y + 2 = \left( x - \dfrac {3}{2} \right)^2 + 2 \left( y + 1 \right)^2 - \dfrac {9}{4}. $$
|
EDIT:
$$ x^2 - 3x + 2y^2 + 4y + 2$$
can be re-written as
$$ x^2-2*x\frac{3}{2}+\frac{9}{4}+2(y^2+2y+1)-\frac{9}{4}$$
Obeserve that the first two terms in the expression are of the form
$$a^2+2a.b+b^2=(a+b)^2$$
Hence we get
$$ x^2 - 3x + 2y^2 + 4y + 2=x^2-2x\frac{3}{2}+\frac{9}{4}+2(y^2+2y+1)-\frac{9}{4}$$
$$\Rightarrow x^2 - 3x + 2y^2 + 4y + 2=(x-\frac{3}{2})^2+2(y+1)^2-\frac{9}{4}$$
On the RHS we have the sum of two squares is is . The minimum value of each square term will be 0 as a square term cannot be negative in $IR$. Hence the minimum value of the given expression is -$\frac{9}{4}$ when the two square terms are 0(at x=$\frac{3}{2}$ and y=-1).
|
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|
which is larger number? $\sqrt{7}-\sqrt{6}$ or $\sqrt{6}-\sqrt{5}$ Which is larger number? $\sqrt{7}-\sqrt{6}$ or $\sqrt{6}-\sqrt{5}$?
Squaring both sides will give me something but I could not go any further.
|
You can also use the inequality between the arithmetic mean and the quadratic mean: $$ \frac{x+y}{2} \leq \sqrt{\frac{x^2 + y^2}{2}}. $$ Setting $x = \sqrt{5}$ and $y = \sqrt{7}$ gives you $$\frac{\sqrt{5} + \sqrt{7}}{2} \leq \sqrt{6},$$ which is equivalent to $$ \sqrt{7} - \sqrt{6} \leq \sqrt{6} - \sqrt{5}.$$
|
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|
Prove by induction that $3\mid n^3 - n$
Prove by induction that $3\mid n^3 - n$.
I'm having an argument with my professor whether my exam was right or not. Before I sign a formal complain to get a review on my exam, I'd like to be sure it's correct. My answer:
Proof by induction:
Proposition: $\forall n \in\mathbb{ N}$ (belongs to natural numbers), $3 | (n^3 - n)$;
Basic Step: $P(0) = 0^3 - 0$ is divisible by 3.
Inductive Step:
$p(n+1) = (n+1)^3 - (n+1)$
$= n^3 + 3n^2 + 3n + 1 - n - 1$
$= n^3 + 3n^2 + 2n$
$= n^3 + 3n^2 + 2n - (n^3 - n)$
$= 3n^2 + 3n$
By induction, $p(n+1)$ is true if:
$3\mid n$ and $3\mid m$ and $3 | n - m$
to prove that, assume that $n$ is $3p$ and $m$ is $3q$, therefore $3 | 3p - 3q$.
My professor wanted:
$= (n^3 -n) + 3n^2 + (2n + n)$
|
The trouble comes from saying that $n^3+3n^2+2n=n^3+3n^2+2n-(n^3-n)$. Instead we can consider $(n+1)^3-(n+1)=n^3+3n^2+3n+1-n-1$. We rewrite this as $(n^3-n)+(3n^2+3n)=(n^3-n)+3(n^2+n)$. The first term is our induction hypothesis which we can rewrite as $3k$ where $k\in\mathbb{Z}$ and the last term is divisible by three. So $3(k+n^2+n)$ is divisible by three for all integers $(k+n^2+n)$. Thus by the Principle of Mathematical Induction $3|(n^3-n)$ for all $n \in \mathbb{N}$.
|
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|
Congruence-What is remainder when $a^2+2b$ is divided 3? Suppose $a\equiv 7 \pmod 9 $, $b\equiv 1\pmod 6$. What is remainder when $a^2+2b$ is divided by 3 ?----- So, since $a=9m+7$, $b=6n+1$, $m,n\in \mathbb{Z}$, $a^2+2b=(9m+7)^2+2(6n+1)$,=$81m^2+12n+177$=$3(27m^2+4n+59)$....I am not sure what I am getting here.. Can I say the remainder is $0$?
|
Another solution using the properties of modular arithmetic.
$a \equiv 7 \pmod 9 \Rightarrow a^2 \equiv 49 \equiv 4 \pmod 9$
$b \equiv 1 \pmod 6 \Rightarrow 2b \equiv 2 \pmod 6$
So $a^2 = 9n + 4$ and $2b = 6m + 2$, thus $a^2 + 2b = 9n + 6m + 6 = 3(3n + 2m + 2)$ which is definitely divisible by 3.
|
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|
Questions about $(\mathbb{Z}/2^n\mathbb{Z})^{\times}$ We have $(\mathbb{Z}/2^n\mathbb{Z})^{\times}$ is a group with $\varphi(2^n)=2^{n-1}$ elements. Prove that $x^2=1$ has exactly four solutions in $\mathbb{Z}/2^n\mathbb{Z}$. Moreover, can we show that $q$ is a perfect square in $(\mathbb{Z}/2^n\mathbb{Z})^{\times}$ if and only if $q=8k+1$ for some integer $k$. It is well known that $o_{2^n}(5)=2^{n-2}$, but can we strenghten this to $o_{2^n}(3)=o_{2^n}(5)=2^{n-2}$? Furthermore which elements of $(\mathbb{Z}/2^n\mathbb{Z})^{\times}$ are expressible as a power of $3$? What about a power of $5$. Any help is greatly appreciated.
|
For the first question, we need $n\ge 3$. We show that there are precisely $4$ integers $x$ between $0$ and $2^n-1$ such that $x^2\equiv 1\pmod{2^n}$, that is, such that $2^n$ divides $(x-1)(x+1)$.
Note that if $a$ and $b$ are two consecutive even integers, then one of them is congruent to $2$ modulo $4$.
Suppose that $2^n$ divides $(x-1)(x+1)$, and $x+1$ is of the shape $4k+2$. Then $2^{n-1}$ must divide $x-1$. Since we can take $0\le x\le 2^n-1$, it follows that $x=1$ or $x=2^{n-1}+1$.
Suppose that $2^n$ divides $(x-1)(x+1)$ and $x-1$ is of the shape $4k+2$. Then $2^{n-1}$ divides $x+1$, and therefore the only possibilities between $0$ and $2^{n}-1$ are $2^n-1$ and $2^{n-1}-1$.
If $n\ge 3$, the $4$ possibilities we have listed are distinct.
Now we look at the second question, and sketch a proof of the fact that (again when $n\ge 3$), $a$ is a quadratic residue modulo $2^n$ if and only if $a$ is of the form $8k+1$.
Since the quadratic residues modulo $8$ are $0$, $1$, and $4$ any odd square is of the form $8k+1$, so numbers of the form $8k+3$, $8k+5$, and $8k+7$ cannot be quadratic residues of $2^n$.
Define the function $f(x)$ on the odd positive integers $\lt 2^n$ as the remainder when $x^2$ is divided by $2^n$. By our count of the solutions of $x^2\equiv 1\pmod{2^n}$, $f(x)$ is $4$ to $1$. Thus there must be $2^{n-3}$ quadratic residues. But the only candidates have shape $8k+1$. Since there are only $2^{n-3}$ of these, they must all be quadratic residues.
|
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For every integer $n$, the remainder when $n^4$ is divided by $8$ is either $0$ or $1$. I am trying to prove the following statement:
For every integer $n$, the remainder when $n^4$ is divided by $8$ is either $0$ or $1$.
So far I have figured out that $n^4 = 8m$ or $n^4 = 8m + 1$.
Any help or hints are appreciated! Thanks!
|
An easier way is by using the powerful tool "$mod\ 8$" for any $ n \in \mathbb{N}: n \equiv 0,1, 2, 3 ,4 , 5 , 6 \ or\ 7 \ mod\ 8 \equiv 0, 1, 2, 3, 4, -3, -2 \ or\ -1 \ mod \ 8$
so for ${n^4}$ we have
${n^4}={n^2}^2 = n^3 n \equiv 0^4, 1^4, 2^3 \times 2 , {3^2}^2, {4^2}^2, {-3^2}^2, -2^3 \times 2 \ or\ -1^4 \ mod \ 8 \equiv 0, 1, 0, 1,0,1,0,1 \ mod\ 8 $
I hope I didn't make it more complicated .
|
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Solve for: $8\log_4\sqrt{x^2-9}+3\sqrt{2\log_4\left(x+3\right)^2}=10+\log_2\left(x-3\right)^2$ Solve for: $$8\log_4\sqrt{x^2-9}+3\sqrt{2\log_4\left(x+3\right)^2}=10+\log_2\left(x-3\right)^2$$
My try:
$8\log_4\sqrt{x^2-9}+3\sqrt{2\log_4\left(x+3\right)^2}=10+\log_2\left(x-3\right)^2\\\Leftrightarrow \log_2\left(x^2-9\right)^2+3\sqrt{\log_2\left(x+3\right)^2}=10+\log_2\left(x-3\right)^2\\\Leftrightarrow \log_2\left(x-3\right)^2+\log_2\left(x+3\right)^2+3\sqrt{\log_2\left(x+3\right)^2}-10-\log_2\left(x-3\right)^2=0\\\Leftrightarrow \log_2\left(x+3\right)^2+3\sqrt{\log_2\left(x+3\right)^2}-10=0$
But I don't know Conditions defined for this math? Could you help me please?
|
$$\log_2(x+3)=y$$
$$y^2+3y-10=0$$
$$y_{1,2}=\frac{-3\pm7}{2},y_1=2,y_2=-5$$
$$\log_2(x+3)=2,x+3=4,x=1$$
$$\log_2(x+3)=-5,x+3=2^{-5},x=2^{-5}-3$$
|
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|
How find this sum $\sum\limits_{n=0}^{\infty}\frac{1}{(3n+1)(3n+2)(3n+3)}$ Find this sum
$$I=\sum_{n=0}^{\infty}\dfrac{1}{(3n+1)(3n+2)(3n+3)}$$
My try: let
$$f(x)=\sum_{n=0}^{\infty}\dfrac{x^{3n+3}}{(3n+1)(3n+2)(3n+3)},|x|\le 1$$
then we have
$$f^{(3)}(x)=\sum_{n=0}^{\infty}x^{3n}=\dfrac{1}{1-x^3}$$
then we find the $f(x)$,Following is very ugly(can you someone can post your follow solution,) have other simple methods? Thank you very much.
|
I undeleted my answer and fixed it up since one of the other answers refers to it.
In the comments, Lucian mentioned the generalization $$\sum_{n=0}^{\infty} \frac{1}{(n+a)(n+b)(n+c)} = \frac{(b-c) \, \psi(a)+ (c-a) \, \psi(b)+(a-b) \, \psi(c)}{(a-b)(b-c)(c-a)} \tag{1}$$ where $\psi(x)$ is the digamma function; $a,b,c >0$; $a \ne b$; $b \ne c$; and $a \ne c$.
We can prove $(1)$ by using the series representation $$\psi(x) = - \gamma + \sum_{n=0}^{\infty} \left(\frac{1}{n+1}-\frac{1}{n+x} \right) \, , \quad x>0.$$
Starting from the right side of $(1)$, we get
$$ \begin{align} &\frac{(b-c) \, \psi(a)+ (c-a) \, \psi(b)+(a-b) \, \psi(c)}{(a-b)(b-c)(c-a)} \\ &= \small\left[ (b-c)\left(-\gamma + \sum_{n=0}^{\infty} \left( \frac{1}{n+1} - \frac{1}{n+a} \right) \right)+ (c-a)\left(- \gamma + \sum_{n=0}^{\infty} \left( \frac{1}{n+1} - \frac{1}{n+b} \right) \right) + (a-b) \left( -\gamma +\sum_{n=0}^{\infty} \left( \frac{1}{n+1} - \frac{1}{n+c} \right) \right) \right] /{\left((a-b)(b-c)(c-a) \right)} \\ &= \frac{1}{(a-b)(b-c)(c-a)}\sum_{n=0}^{\infty} \left( \frac{c-b}{n+a}+\frac{a-c}{n+b} + \frac{b-a}{n+c} \right) \\ &=\frac{1}{(a-b)(b-c)(c-a)} \sum_{n=0}^{\infty} \ \frac{bc^{2}-b^{2}c+a^{2}c-ac^{2}+ab^{2}-a^{2}b}{(n+a)(n+b)(n+c)} \\ &= \frac{1}{(a-b)(b-c)(c-a)}\sum_{n=0}^{\infty} \frac{(a-b)(ac-c^{2}-ab+bc)}{(n+a)(n+b)(n+c)} \\ &=\frac{1}{(a-b)(b-c)(c-a)}\sum_{n=0}^{\infty} \frac{(a-b)(b-c)(c-a)}{(n+a)(n+b)(n+c)} \\ &= \sum_{n=0}^{\infty} \frac{1}{(n+a)(n+b)(n+c)} \end{align}$$
|
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|
How to prove(or disprove) $\begin{vmatrix} A&B\\ B&A \end{vmatrix}=|A^2-B^2|$ Let $A$ and $B$ be square matrices of the same size.
(1) If $f$ is not invertible and $AB=BA$, show that
$$\begin{vmatrix}
A&B\\
B&A
\end{vmatrix}=|A^2-B^2|.$$
(2) If $A$ is invertible and $AB\neq BA$, then do we have
$$\begin{vmatrix}
A&B\\
B&A
\end{vmatrix}=|A^2-B^2|?$$
My try: I have only proved this: if $A$ is invertible and $AB=BA$, then
$$\begin{vmatrix}
A&B\\
B&A
\end{vmatrix}=|A^2-B^2|.$$
For $(1)$ and $(2)$, I can't prove. Thank you.
|
Let
$$
J=\left(\begin{array}{cc}I&0\\0&-I\end{array}\right)
$$
and
$$
M(A,B)=\left(\begin{array}{cc}A&B\\B&A\end{array}\right).
$$
Then $\det J^2=\det I=1$, so $\det(JM(A,B)J)=\det M(A,B)$. But (check this)
$$JM(A,B)J=M(A,-B),$$
so $\det M(A,B)=\det M(A,-B)$.
On the other hand using $AB=BA$ we get
$$
M(A,B)M(A,-B)=\left(\begin{array}{cc}A^2-B^2&-AB+BA\\BA-AB&A^2-B^2\end{array}\right)=\left(\begin{array}{cc}A^2-B^2&0\\0&A^2-B^2\end{array}\right).
$$
Putting these together we get
$$
\det M(A,B)^2=\left|\begin{array}{cc}A^2-B^2&0\\0&A^2-B^2\end{array}\right|.
$$
Can you take it from here?
|
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stuck on a Cartesian question we have a circle $(x-1)^2+(y-2)^2=9$
Point $P=(5,2)$ lies outside the circle.
Solve the equation of the line which passes through $P$ and intersects the circle at two points whose mutual distance is $d=2$. Find the coordinates of the intersection points.
I sketched the circle and $P$ but could not find the relationship between these.
Please help me to get the intersection points.
The answer below is quite confused and is not correct. Could someone help me ?
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HINT:
Let the equation of the line be $$\frac{y-2}{x-2}=m\implies y=mx+2-2m\ \ \ \ (1)$$ where $m$ is the gradient
Replacing the value of $y$ in $$(x-1)^2+(y-2)^2=9$$
$$(x-1)^2+(mx-m)^2=9\iff x^2(1+m^2)-2x(1+m^2)+m^2-8=0\ \ \ \ (2)$$
If $(x_1,y_1);(x_2,y_2)$ are the points of intersection, we have $\displaystyle x_1+x_2=\frac{2(1+m^2)}{1+m^2}=2$ and $\displaystyle x_1x_2=\frac{m^2-8}{1+m^2}$
So, if $d$ is the distance between the two points,
$$d^2=(x_1-x_2)^2+(y_1-y_2)^2=(1+m^2)(x_1-x_2)^2=(1+m^2)\{x_1+x_2)^2-4x_1x_2\}$$
$$=(1+m^2)\left(2^2-4\frac{(m^2-8)}{1+m^2}\right)$$
Find $m$ from the given condition , then the two values of $x$ from $(2)$ and those of $y$s from $(1)$
|
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|
Boolean Algebra simplify minterms I have this equation
$$\bar{A}\cdot\bar{B}\cdot\bar{C} + A\cdot\bar{B}\cdot C + A\cdot B\cdot \bar{C} + A \cdot B\cdot C$$
and need to simplify it. I have got as far as I can and spent a good 2 hours at it. I've realized I now need to use De Morgan's law to continue however I am baffled as to which rule to use. If someone could send me in the right direction that would be great!
$$\bar{A}\cdot\bar{B}\cdot\bar{C} + A\cdot\bar{B}\cdot C + A\cdot B\cdot \bar{C} + A \cdot B\cdot C$$
$$\bar{A}\cdot\bar{B}\cdot\bar{C} + A(\bar{B}\cdot\bar{C} + B\cdot\bar{C} + B\cdot C$$
$$\bar{A}\cdot\bar{B}\cdot\bar{C} + A(\bar{B}\cdot C + B(\bar{C} + C)) $$
$$\bar{A}\cdot\bar{B}\cdot\bar{C} + A(\bar{B}\cdot C + B) $$
$$\bar{A}\cdot\bar{B}\cdot\bar{C} + A(C + B) $$
Thanks
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Picking up where you (initially) left off (your work thus far is correct, save for the last line were you remove $\overline B$)...
$$\begin{align} &\overline{A}\cdot\overline{B}\cdot\overline{C} + A\cdot\overline{B}\cdot C + A\cdot B\cdot \overline{C} + A \cdot B\cdot C \\ \\
&\vdots \\ \\
&=\overline A \cdot \overline B \cdot \overline C + A(\overline B\cdot C+B) \\ \\
& = \overline A \cdot \overline B \cdot \overline C + A\cdot \overline B \cdot C + AB\tag{as good as it gets!}\\ \\
& = \overline B(\overline A \cdot \overline C + A\cdot C) + AB\tag{doesn't help any}\\ \\
\end{align}$$
|
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|
Series: $ \sum_{n = 1}^{\infty} \left( \prod_{i = 1}^n (2i - 1) \middle/ \prod_{i = 1}^n 2i \right)^k $ In Section 10.16 of Apostol's Calculus (Problem 18), the reader is asked to prove that the series
$$ \sum_{n = 1}^{\infty} \left( \prod_{i = 1}^n (2i - 1) \middle/ \prod_{i = 1}^n 2i \right)^k $$
converges if $k > 2$ and diverges if $k \leq 2$ using Gauss's test. In the text, Gauss's test is given as follows. If there is an $N \geq 1$, an $s > 1$, and an $M > 0$ such that $a_{n + 1}/a_n = 1 - A/n + f(n)/n^s$ for all $n \geq N$, where $|f(n)| \leq M$ for all $n > 0$, then $\sum a_n$ converges if $A > 1$ and diverges if $A \leq 1$. Letting $a_n$ be the terms of the indicated sequence, we use the binomial series to obtain
$$\begin{align}
\frac{a_{n + 1}}{a_n} ~ = ~ & \left( \frac{2n + 1}{2n + 2} \right)^k \\
~ = ~ & \left( 1 - \frac{1}{2n + 2} \right)^k && \\
~ = ~ & \sum_{i = 0}^{\infty} {k \choose i} \left(\frac{-1}{2n + 2} \right)^i \\
~ = ~ & 1 - \frac{k}{2}\left(\frac{1}{n + 1}\right) + \sum_{i = 2}^{\infty}{k \choose i}\left(\frac{-1}{2n + 2} \right)^i \\
~ = ~ & 1 - \frac{k}{2}\left(\frac{1}{n}\right) + \left( \sum_{i = 2}^{\infty}{k \choose i}\left(\frac{-1}{2n + 2} \right)^i + \frac{k}{2}\left(\frac{1}{n(n + 1)}\right)\right) \\
~ = ~ & 1 - \frac{k}{2}\left(\frac{1}{n}\right) + \left(\left\{ n^2 \sum_{i = 2}^{\infty}{k \choose i}\left(\frac{-1}{2n + 2} \right)^i + \frac{k}{2}\left(\frac{n}{n + 1}\right)\right\} \middle/ n^2 \right)
\end{align}$$
This has the form required for Gauss's test, with $A = k/2$ and $s = 2$. It remains to determine whether the absolute value of the bracketed expression is bounded by some positive number. We see that it is, since the absolute value of $kn/2(n + 1)$ is bounded by $|k|/2$. In addition, the absolute value of $n^2$ times the above series indexed at $2$ is bounded by some positive constant $\phi$. Thus, the absolute value of the function parameter to Gauss's test is bounded above by $\phi + |k|/2$ (work omitted). So the series converges if $k/2 > 1$, $i$.$e$., if $k > 2$, and diverges if $k/2 \leq 1$, $i$.$e$., if $k \leq 2$.
Here is my question. I am trying to work through this text using only those results previously introduced, and Apostol hasn't yet introduced the generalized binomial series. My conclusion is that this problem was supposed to be solved without using the binomial series, or facts about its convergence and divergence. So how would one go about solving this problem without expanding $(1 - 1/(2n + 2))^k$ using the binomial series, assuming that $k$ is an arbitrary real number? My only thought is that I should decompose $k$ into its integer and fractional parts, then use the binomial theorem for integral values. But then it's difficult to apply Gauss's test with negative values of $k$.
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By the generalized mean value theorem, for $0<x<1$ we have
$$
\ln(1-x) = - x - \frac1{(1-\xi)^2} \frac{x^2}2 \quad\text{and}\quad e^{-x} = 1 - x + e^{-\eta} \frac{x^2}2
$$
for some $0\le\xi,\eta\le x$. In particular, for $0<x\le\frac12$ we have
$$
-x-2x^2 \le \ln (1-x) \le -x \quad\text{and}\quad 1-x \le e^{-x} \le 1 - x + x^2.
$$
Therefore for $0<x\le\frac1{2k}$ (we can assume $k\ge2$ and use the comparison test for smaller $k$),
$$
(1-x)^k = \exp(k \ln (1-x)) \le \exp(-kx) \le 1 - kx + k^2x^2
$$
and
$$
(1-x)^k = \exp(k \ln (1-x)) \ge \exp(-kx-2kx^2) \ge 1-kx-2kx^2.
$$
We conclude that
$$
\bigg( 1 - \frac1{2n+1} \bigg)^k = 1 - \frac k{2n+1} + \frac{g(n)}{(2n+1)^2}
$$
where $|g(n)| \le k^2$. One can easily convert this into the exact form required by Gauss's test.
|
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|
Limit at infinity involving $e$ I am to find the limit of $$\lim_{x \to \infty} \left(1+\frac{x}{5x^3+x^2+8}\right)^ {\dfrac{x^3+8}{x}}$$
I could not find the proper substitution here. I would be happy if someone could shed some light. Thanks.
|
$$\lim_{x\to\infty}\left(1+\frac{x}{5x^3+x^2+8}\right)^ {x^2+8/x}=\lim_{x\to\infty}\left(1+\frac{1}{5x^2+x+8/x}\right)^ {x^2+8/x}=$$
$$=\lim_{x\to\infty}\left(\left(1+\frac{1}{5x^2+x+8/x}\right)^{5x^2+x+8/x} \right)^{\frac{x^2+8/x}{5x^2+x+8/x}}=e^{1/5}$$
because
$$\lim_{x\to\infty}\left(1+\frac{1}{5x^2+x+8/x}\right)^{5x^2+x+8/x}=e$$
and
$$\lim_{x\to\infty}{\frac{x^2+8/x}{5x^2+x+8/x}}=\frac{1}{5}$$
|
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|
How to integrate $\frac{4x+4}{x^4+x^3+2x^2}$? Please could anyone help me to integrate
$\quad\displaystyle{4x + 4 \over x^4 + x^3 + 2x^2}.\quad$
I know how to use partial fraction and I did this:
$$
x^{4} + x^{3} + 2x^{2} = x^{2}\left(x^{2} + x + 2\right)
$$
And then ?.$\quad$ Thanks all.
|
We have:
$$ \int{\frac{4x+4}{x^4+x^3+2x^2}} dx $$
We need to decompose this fraction into pieces and then integrate each one separately. We start by simplifying the denominator:
$$ \frac{4x+4}{x^2(x^2+x+2)} $$
Since we have two quadratic terms in the denominator, we can guess the form:
$$ \frac{4x+4}{x^2(x^2+x+2)} = \frac{Ax+B}{x^2} + \frac{Cx+D}{x^2+x+2} $$
Simplifying this further yields:
$$ 4x+4 = (Ax+B)(x^2+x+2) + x^2(Cx+D) $$
Let's expand everything out so that we can solve for these constants A, B, C, and D:
$$ 4x+4 = (Ax^3+Ax^2+2Ax+Bx^2+Bx+2B)+(Cx^3+Dx^2) $$
$$ 4x+4 = (A+C)x^3+(A+B+D)x^2+(2A+B)x+2B $$
We now have a system of equations:
$$ A+C=0, A+B+D=0, 2A+B=4, 2B=4 $$
Solving (by working with the 2B=4 term first), we get:
$$ A=1, B=2, C=-1, D=-3$$
Plugging back our constants into our "guess form":
$$\frac{4x+4}{x^2(x^2+x+2)}=\frac{(1)x+(2)}{x^2}+\frac{(-1)x+(-3)}{x^2+x+2}$$
$$=\frac{x+2}{x^2} + \frac{-x-3}{x^2+x+2}$$
$$=\frac{1}{x}+\frac{2}{x^2}-\frac{x-3}{x^2+x+2} $$
That's much better, now you should be able to integrate each piece separately.
Can you finish?
|
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|
The indefinite integration of $\frac{1}{\sqrt{n^4-1}}$ I need the indefinite integral:
$$\int\frac{1}{\sqrt{n^4-1}}dn $$
I know it has a relation with the $\tanh^{-1}$ function, but can't find a proper substitution.
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Case $1$: $|n^4|\geq1$
Then $\int\dfrac{1}{\sqrt{n^4-1}}dn$
$=\int\dfrac{1}{n^2\sqrt{1-\dfrac{1}{n^4}}}dn$
$=\int\dfrac{1}{n^2}\sum\limits_{k=0}^\infty\dfrac{(2k)!n^{-4k}}{4^k(k!)^2}dn$
$=\int\sum\limits_{k=0}^\infty\dfrac{(2k)!n^{-4k-2}}{4^k(k!)^2}dn$
$=\sum\limits_{k=0}^\infty\dfrac{(2k)!n^{-4k-1}}{4^k(k!)^2(-4k-1)}+C$
$=-\sum\limits_{k=0}^\infty\dfrac{(2k)!}{4^k(k!)^2(4k+1)n^{4k+1}}+C$
Case $2$: $|n^4|\leq1$
Then $\int\dfrac{1}{\sqrt{n^4-1}}dn$
$=\int\dfrac{1}{i\sqrt{1-n^4}}dn$
$=-\int\sum\limits_{k=0}^\infty\dfrac{i(2k)!n^{4k}}{4^k(k!)^2}dn$
$=-\sum\limits_{k=0}^\infty\dfrac{i(2k)!n^{4k+1}}{4^k(k!)^2(4k+1)}+C$
|
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|
Evaluate $\lim\limits_{x \to a} \frac{x^m-a^m} {x-a}$ How to evaluate this:
$$\lim\limits_{x \to a} \frac{x^m-a^m} {x-a} ; m\in \mathbb{N}$$
if i take:
$ m = 1 $
$$\lim\limits_{x \to a} \frac{x^1-a^1} {x-a} =1 $$ Is this correct?
but how evaluate limit where $ m = 2 $
$$\lim\limits_{x \to a} \frac{x^2-a^2} {x-a} = ???$$
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so if i take $ m = 3 $
$$\lim\limits_{x \to a} \frac{x^3-a^3} {x-a} = \lim\limits_{x \to a} \frac{(x-a)(x^2+ax+a^2)} {x-a}= \lim\limits_{x \to a} (x^2+ax+a^2)= 3a^2 $$
so if i understood
$$\lim\limits_{x \to a} \frac{x^m-a^m} {x-a} = m \cdot a^{m-1} $$
|
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|
Trigonometric Limit: $\lim_{x\to 0}\left(\frac{1}{x^2}-\frac{1}{\tan^2x}\right)$ I cannot figure out how to solve this trigonometric limit:
$$\lim_{x\to 0} \left(\frac{1}{x^2}-\frac{1}{\tan^2x} \right)$$
I tried to obtain $\frac{x^2}{\tan^2x}$, $\frac{\cos^2x}{\sin^2x}$ and simplify, and so on. The problem is that I always go back to the indeterminate $\infty-\infty$
Has someone a different approach to solve this limit?
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$$\lim_{x\to 0}\left(\frac{1}{x^2}-\frac{1}{\tan^2 x}\right)=\lim_{x\to 0}\frac{\tan x-x}{x^3}\cdot\frac{x+\tan x}{x}\cdot\left(\frac{x}{\tan x}\right)^2=
2\cdot\lim_{x\to 0}\frac{\sec^2 x-1}{3x^2}=$$ $$=\frac{2}{3}\cdot\lim_{x\to 0}\left(\frac{\tan x}{x}\right)^2=\frac{2}{3}.$$
|
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|
How to find algebraic simplification for recurrence relation with closed-form solution, specifically for the Lucas-Lehmer primality test I have a question based on the section Proof of correctness in the article Lucas-Lehmer primality test, see following link.
https://en.wikipedia.org/wiki/Lucas%E2%80%93Lehmer_primality_test#Proof_of_correctness
"We begin by noting that $\langle s_i\rangle$ is a recurrence relation with a closed-form solution. Define $\omega = 2 + \sqrt{3}$ and $\bar{\omega} = 2 - \sqrt{3}; ...$"
How does $\omega$ and $\bar{\omega}$ come in to the picture? I have no problem following the prof ones $\omega$ and $\bar{\omega}$ is defined but I don't understand how they where defined in the first place.
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The whole setup began with Lucas. For given integers $P,Q$ we get Lucas sequences $U_n, V_n,$ where $V_n = a^n + b^n.$
In your case, $P=14, \; Q=1.$
Among the relations we find $$ V_{2n} = V_n^2 - 2 Q^n, $$ the fact that $Q=1$ takes us to
$$ V_{2n} = V_n^2 - 2; $$ furthermore $V_0 = 2, V_1 = P.$
For your application, define
$$ s_j = V_{2^{j-1}}, $$ so that $$ s_1 = V_1 = P, \; s_2 = V_2 = P^2 - 2, \; s_3 = V_4 = P^4 - 4P^2 +2. $$
$$ s_{j+1} = s_j^2 - 2. $$
In case $P+2$ is a square, we may also define $s_0 = \sqrt{P+2}$ as another integer. With $P=14$ this works.
So, with $Q=1,$ we get the roots $$ a = \frac{P + \sqrt{P^2 - 4}}{2}, \; b = \frac{P - \sqrt{P^2 - 4}}{2} $$
$$ s_j = a^{\left( 2^{(j-1)} \right)} + b^{\left( 2^{(j-1)} \right)}. $$
Go Figure.
EDIT: Well, I like this, and want to remember it in a simple format.
Given any real number $t_0 > 2,$ define the sequence $t_1 = t_0^2 - 2$ and, generally, $t_{n+1} = t_n^2 - 2.$ Then,
taking
$$ \color{magenta}{ A = \frac{t_0 + \sqrt {t_0^2 - 4}}{2}, \; \; \; B = \frac{t_0 - \sqrt {t_0^2 - 4}}{2} } $$
One may check quickly that $$ AB = 1. $$
Indeed, both are positive, $A$ is the larger, and
$$ A + B = t_0. $$
So, think about it, using $AB= 1,$ also $A^m B^m = 1$ for any positive integer $m,$
$$ t_1 = t_0^2 - 2 = A^2 + 2 A B + B^2 - 2 = A^2 + 2 A B + B^2 - 2 AB = A^2 + B^2. $$
Then $$ t_2 = t_1^2 - 2 = A^4 + 2 A^2 B^2 + B^4 - 2 = A^4 + 2 A^2 B^2 + B^4 - 2 A^2 B^2 = A^4 + B^4. $$
Then $$ t_3 = t_2^2 - 2 = A^8 + 2 A^4 B^4 + B^8 - 2 = A^8 + 2 A^4 B^4 + B^8 - 2 A^4 B^4 = A^8 + B^8. $$
Pattern holds true forever, by induction: given
$$ \color{magenta}{ t_n = A^{\left( 2^n \right)} + B^{\left( 2^n \right)},} $$
we get
$$ t_{n+1} = t_n^2 - 2 = A^{\left( 2^{n+1} \right)} + 2 A^{\left( 2^n \right)} B^{\left( 2^n \right)}+ B^{\left( 2^{n+1} \right)} -2 = A^{\left( 2^{n+1} \right)} + 2 A^{\left( 2^n \right)} B^{\left( 2^n \right)}+ B^{\left( 2^{n+1} \right)} -2 A^{\left( 2^n \right)} B^{\left( 2^n \right)}, $$
$$ t_{n+1} = A^{\left( 2^{n+1} \right)} + B^{\left( 2^{n+1} \right)}. $$
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|
What is the smallest Finite Field in which the following polynomial is factorizable to irreducible factors? What is the smallest Finite Field in which the following polynomial is factorizable to irreducible factors?
$$(x^2+x+1)(x^5+x^4+1)(x^7+x^6+x^3+1) $$
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There is no finite field in which all given three polynomials are irreducible. The last
polynomial, i.e., $x^7+x^6+x^3+1$ has always $-1$ as a root, hence is reducible over all finite fields.
On the other hand, $\mathbb{F}_3$ is the smallest finite field, in which all three polynomials are reducible: $x^2+x+1=(x+2)^2$, $x^5+x^4+1=(x^3 + 2x + 1)(x + 2)^2$, and
$x^7+x^6+x^3+1=(x^6 + x^2 + 2x + 1)(x + 1)$.
|
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function question
$ f(x) = \begin{cases} -1 && \text{for}-2\le x \le0\\ x-1 && \text{for } 0<x\le2
\end{cases}\\\text{Domain of }f(x) = [-2,2]\\
\text{Let there be be another function g(x),such that}\\
g(x) = f(|x|) + |f(x)| \\
$
$\text{What is function g(x)?}$
I worked the question and got $g(x) = \begin{cases}
0 & \text{for } 0<x<1 \\
-x & \text{for }-2\le x \le 0 \\
2(x-1) & \text{for } 1\le x \le 2
\end{cases} $
Is there anything wrong with my answer?
|
There's nothing wrong with your answer.
Case 1. $-2\leq x \leq 0$
\begin{array}{rl}
g(x)&=f(-x) + |f(x)| \\
&=(-x)-1 +|-1| & \\
&=-x
\end{array}
Case 2. $0< x < 1$
\begin{array}{rl}
g(x)&=f(x) + |x-1| \\
&=x-1 + [-(x-1)] & \\
&=x-1-x+1\\
&=0
\end{array}
Case 3. $1 \leq x \leq 2$
\begin{array}{rl}
g(x)&=f(x) + |x-1| \\
&= x-1 + (x-1) & \\
&=2(x-1).
\end{array}
|
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Subsets and Splits
Fractions in Questions and Answers
The query retrieves a sample of questions and answers containing the LaTeX fraction symbol, which provides basic filtering of mathematical content but limited analytical insight.