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Determining all the positive integers $n$ such that $n^4+n^3+n^2+n+1$ is a perfect square. I successfully thought of bounding our expression examining consecutive squares that attain values close to it, and this led to the solution I'll post as an answer, which was the one reported. However, before that, I had briefly tried manipulating $$n^4+n^3+n^2+n+1=m^2.$$Since $n=1$ is not a solution, I rewrote this as $$\frac{n^5-1}{n-1}=m^2 \\ n^5-1=nm^2-m^2 \\ m^2-1=n(m^2-n^4) \\ (m-1)(m+1)=n(m-n^2)(m+n^2),$$ but in vain. Does manipulation lead somewhere? Is there a different approach from both of mine?
|
Hint: notice that
$$ (2n^2+n)^2 = 4n^4+4n^3+n^2, $$
$$ (2n^2+n+1)^2 = 4n^4+4n^3+5n^2+2n+1 $$
are two consecutive squares and $4(n^4+n^3+n^2+n+1)$ is just between them, with very few exceptions. An old motto says "too close to a square to be a square, it is not a square".
|
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"timestamp": "2023-03-29T00:00:00",
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|
For which values of $a,b,c$ is the matrix $A$ invertible? $A=\begin{pmatrix}1&1&1\\a&b&c\\a^2&b^2&c^2\end{pmatrix}$
$$\Rightarrow\det(A)=\begin{vmatrix}b&c\\b^2&c^2\end{vmatrix}-\begin{vmatrix}a&c\\a^2&c^2\end{vmatrix}+\begin{vmatrix}a&b\\a^2&b^2\end{vmatrix}\\=ab^2-a^2b-ac^2+a^2c+bc^2-b^2c\\=a^2(c-b)+b^2(a-c)+c^2(b-a).$$
Clearly, $$\left\{\det(A)\neq0\left|\begin{matrix}c\neq b\\a\neq c\\b\neq a\\a,b,c\neq 0\end{matrix}\right.\right\}\\$$
Is it sufficient to say that the matrix is invertible provided that all 4 constraints are met? Would Cramer's rule yield more explicit results for $a,b,c$ such that $\det(A)\neq0$?
|
$$\det(A)= \det\begin{pmatrix}1&a&a^2\\1&b&b^2\\1&c&c^2\end{pmatrix}= \det\begin{pmatrix}1&a&a^2\\0&b-a&b^2-a^2\\0&c-a&c^2-a^2\end{pmatrix} \\
=\det\begin{pmatrix}b-a&(b-a)(b+a)\\c-a&(c-a)(c+a)\end{pmatrix} =(b-a)(c-a)\det\begin{pmatrix}1&b+a\\1&c+a\end{pmatrix}$$
|
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|
How do you add two series together How do you add the series
$$\frac{1}{2}\left(\sum_{n=0}^{\infty}\frac{2^{n}}{(z-3)^{n+1}} + \sum_{n=0}^{\infty}\frac{(z-3)^{n}}{4^{n+1}}\right)$$
?
is this right?
$$\begin{aligned}
&\frac{1}{2}\sum_{n=0}^{\infty}\left(\frac{2^{n}}{(z-3)^{n+1}} + \frac{(z-3)^{n}}{4^{n+1}}\right)\\
=\> & \frac{1}{2}\sum_{n=0}^{\infty}\left(\frac{(4^{n+1} \cdot 2^{n}) + (z-3)^{n} (z-3)^{n+1}}{(z-3)^{n+1} 4^{n+1}} \right)\\
=\>& \frac{1}{2}\sum_{n=0}^{\infty}\left(\frac{(4^{n+1} \cdot 2^{n}) + (z-3)^{n+1+n}}{(z-3)^{n+1} 4^{n+1}}\right)\\
=\>& \frac{1}{2}\sum_{n=0}^{\infty}\left(\frac{(4^{n+1} \cdot 2^{n}) + (z-3)^{2n+1}}{(z-3)^{n+1} 4^{n+1}}\right)\end{aligned}$$
|
This isn't wrong but if i were you i would write it as Laurent series. $\displaystyle\frac {1}{2}\left(\sum_{n=0}^\infty \frac{2^n}{(z-3)^{n+1}}+\sum_{n=0}^\infty \frac{(z-3)^n}{4^{n+1}}\right) = \frac {1}{2}\left(\sum_{n=-1}^{-\infty} \frac{(z-3)^n}{2^{n+1}}+\sum_{n=0}^\infty \frac{(z-3)^n}{4^{n+1}}\right)$
$\displaystyle = \frac 12 \sum_{n=-\infty}^\infty a_n (z-3)^n$
where $a_n := \begin{cases}\frac{1}{2^{n+1}}\quad\text{if n<0}\\\frac{1}{4^{n+1}}\quad\text{if }n\geq 0\end{cases}$.
If you want to evaluate the sum i suggest just leaving it as it is.
|
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|
Find the possible values of a in the cubic equation. Given that $(x-a)$ is a factor of $x^3-ax^2+2x^2-5x-3$, find the possible values of the constant $a$.
I believe you first have to find the $a$ in the cubic equation then the other $a$ in $(x-a)$, but I'm having problems finding the first $a$. So far all I have is:
$x^3-ax^2+2x^2-5x-3=0$
$x^3-2ax^2-5x=3$
|
Hint: If $x-a$ is a factor, then $a$ is a zero. Hence,
$$a^3-a^3+2a^2-5a-3=0\implies 2a^2-5a-3=0.$$
|
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|
Inequality with Algebra $\sup_{x,y,z\in A}\left(\left(13x+\frac{5}{x}\right)+\left(13y+\frac{5}{y}\right)+\left(13z+\frac{5}{z}\right)\right)= 63$ let $A=\{(x,y,z)|x,y,z\in [\frac{1}{2},2],xyz=1\}$
Maybe have
$$f=\sup_{x,y,z\in A}\left(\left(13x+\dfrac{5}{x}\right)+\left(13y+\dfrac{5}{y}\right)+\left(13z+\dfrac{5}{z}\right)\right)= 63?$$
Here the maximum of $f$ is $63$,reached at the point $(x,y,z)$ on the space surface $xyz=1$,and $x=2,y=\dfrac{1}{2},z=1$
I attempted a proof by AM-GM inequality,But I am not able to solve the upper part, this inequality when $x=2,y=\dfrac{1}{2},z=1$,then $f=63$
This problem can use from space Analysis geometry? ,take $y=\dfrac{15}{3x}$
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$p=x+y,xy=q, \dfrac{1}{2} \le z \le 2 \implies \dfrac{1}{2} \le q \le 2 $
$f=13p+\dfrac{5p+13}{q}+5q$
consider $\dfrac{A}{q}+Bq$ will get max at two ends, so $f(p,q)$ will get max at $q=2$ or $q= \dfrac{1}{2}$
when $q=2 ,y=\dfrac{2}{x}$, put it in $f$,it become $f(x)= \dfrac{A}{x}+Bx+C$,again, $f(x)$ will get max at two ends,so you will have $x=2$ or $x=\dfrac{1}{2}$ to check which one is bigger.
|
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|
Determinant properties Prove without expanding:
\begin{equation}\begin{vmatrix}1&1&1\\a^2&b^2&c^2\\a^3&b^3 & c^3\end{vmatrix} = (ab + ac + bc)(b - a)(c - a)(c - b)\end{equation}
*I tried to zero some elements and expand until I reach the Right hand side.
*Also tried C1-C3, C2-C3 then decompose the determinant into two determinants and taking common factors. But I couldn't get (ab + ac + bc) part.
*I can only use the properties shown here http://www.vitutor.com/alg/determinants/properties_determinants.html
|
\begin{gather*}
\begin{bmatrix}1&1&1\\a^2&b^2&c^2\\a^3&b^3&c^3\end{bmatrix} \\
= \begin{bmatrix}1&0&0\\a^2&b^2-a^2&c^2-a^2\\a^3&b^3-a^3&c^3 - a^3\end{bmatrix} \\
= \begin{bmatrix}b^2-a^2&c^2-a^2\\b^3-a^3&c^3 - a^3\end{bmatrix} \\
= (b^2-a^2)(c^3 - a^3)-(b^3-a^3)(c^2 - a^2)
= (b-a)(c-a)[(b+a)(c^2+ac+a^2)-(c+a)(b^2+ab+a^2)] \\
= (ab+ac+bc)(b-a)(c-a)(c-b)\\
= R.H.S
\end{gather*}
|
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|
Find all values of positive integers $x,y,z$ so that $4^x+4^y+4^z$ is a perfect square I have to solve the equation
$$4^x+4^y+4^z=k^2$$
I posted my solution but I don't know if there are other proofs. How can I demonstrate that this expression is a perfect square?
Thanks in advance.
|
The equation that I have to solve is:
$$4^x+4^y+4^z=k^2$$
Manipulating the equation it becomes: $$4^x(4^{|y-x|}+4^{|z-x|}+1)=k^2$$
Now $4^x$ is always a perfect square, therefore we have to find value of $x,y, z$ so that $4^{|y-x|}+4^{|z-x|}+1$ is a perfect square.
We put $u=|y-x|$ and $v=|z-x|$ and the equation becomes $$4^u+4^v+1=m^2(k^2|m^2)$$
and
$$4(4^{u-1}+4^{v-1})=(m+1)(m-1)$$
Now we analyse three systems of equations:
$1)$
$$
\left\{
\begin{array}{c} 4=m-1\\ 4^{u-1}+4^{v-1}=m+1\end{array}
\right.$$
$2)$ $$
\left\{
\begin{array}{c} 4=m+1\\ 4^{u-1}+4^{v-1}=m-1 \end{array}
\right.$$
$3)$
$$
\left\{
\begin{array}{c} 4=(m-1)(m+1)\\
4^{u-1}+4^{v-1}=1 \end{array}
\right.$$
From the first system, if $4=m-1$ then $m=5$, but $4^u+4^v+1=5$ hasn't solutions.
From the second system, if $4=m+1$ then $m=3$, therefore $4^u+4^v+1=9$ has as only solutions $u=1$ and $v=1$.
From the third system, if $4=(m+1)(m-1)$ then $m=\sqrt 5$ that isn't a integer.
If $u=1=|y-x|$ and $v=1=|z-x|$ the solutions are $z=y=x\pm 1$.
|
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|
Finding out this combination In how many ways three non-empty strings of length less than or equal to $N$ using $k$ different characters can be selected so that in each case, among the three strings, no string is prefix (not necessarily proper prefix) of one of the other two strings.
Example:
Let $N=1,k=3$.
Result will be $6$
$$
[a, b, c]\\
[a, c, b]\\
[b, a, c]\\
[b, c, a]\\
[c, a, b]\\
[c, b, a]\\
$$
I am getting stuck for the "prefix" part. Please help. I am looking for a way that is computationally cheap as much as possible as $N$ could be as big as $10^{9}$ ($k$ will be much smaller though - upto two digits) .
EDIT :
$$
1\le N\le 10^9\\
1\le k\le 50\\
$$
More Example : For $N=2,k=2$ answer is $36$
|
Too long for a comment:
If $N=1$, then your problem has $k(k-1)(k-2)$ solutions.
If $N=2$, then the string lengths are $(1,1,1),(1,1,2),(1,2,2)$ or $(2,2,2)$. The number of solutions is $$F(1,1,1)=k(k-1)(k-2)\\F(1,1,2)=k(k-1)(k-2)k\\F(1,2,2)=k(k-1)k[(k-1)k-1]\\F(2,2,2)=k^2(k^2-1)(k^2-2)$$
respectively. The total is
$$F(1,1,1)+3F(1,1,2)+3F(1,2,2)+F(2,2,2)\\=k^6+3k^5-6k^4-8k^3+8k^2+2k$$
In general, it is a polynomial in $k$ of degree $3N$.
It might be simpler to do the problem for two strings, which might give you insights for the three-string problem.
(edited to reflect the change in question)
If $a\leq b\leq c$ then $F(a,b,c)=k^a(k^b-k^{b-a})(k^c-k^{c-a}-k^{c-b})\\
=k^{a+b+c}-k^{b+c}-k^{a+c}-k^{b+c}+k^{b+c-a}+k^c$
$$X(k,N)=6\sum_{a<b<c}F(a,b,c)+3\sum_{a<b}(F(a,a,b)+F(a,b,b))+\sum_aF(a,a,a)$$
The sum of $k^{a+b+c}$ over all string-lengths $(1\leq a\leq N,1\leq b\leq N,1\leq c\leq N)$ is $[(k^{N+1}-k)/(k-1)]^3$. The sum for $a<b$ of $k^{a+c}$ equals the sum for $a<b$ of $k^{c+(b-a)}$. So
$$X(k,N)=\frac{(k^{N+1}-k)^3}{(k-1)^3}+6\sum_{b<c}(b-1)(-2k^{b+c}+k^c)\\
+3\sum_{a<b}(-3k^{a+b}+2k^b)\\
+3\sum_{a<b}(-k^{a+b}-2k^{2b}+k^{2b-a}+k^b)\\
+\sum_a(-3k^{2a}+2k^a)
$$
Some simplifying, including the sum of $k^{2b-a}$ equals the sum of $k^{a+b}$, gives
$$X(k,N)=\frac{(k^{N+1}-k)^3}{(k-1)^3}\\+\sum_{a<b}(-12a+3)k^{a+b}\\
+\sum_a(-6a+3)k^{2a}\\
+\sum_a(3a^2-1)k^a$$
WolframAlpha can handle each piece, (and gives an answer with a dozen or two terms, not $N$ or $N^2$ terms).
I think it comes to this, but I could easily have made an error:
$$\frac{1}{(k-1)^3(k+3)^2}\left[-2k^4-24k^3-6k^2-2k\right.\\
+k^N((3N^2-1)k^5-(6N+6)k^4+(-6N^2+6N-19)k^3+6Nk^2+(-3N^2+6N+2)k)\\
+k^{2N}(-6Nk^5+(-6N+6)k^4+(6N+12)k^3+(6N+6)k^2)\\
\left.+k^{3N}(k+1)^2\right]$$
|
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|
Limit with polylog How do you show the following limit?
$$\lim_{x\to\infty} x\log(-e^x + 1)+\operatorname{Li}_2(e^x)-\frac12x^2=\frac{\pi^2}3$$
Where $\operatorname{Li}_n(x)$ is the polylogarithm.
This question is inspired by a thread in the sagemath mailinglist.
|
The asymptotic expansion of dilog is :
$$Li_2(X)=-\frac{1}{2}\ln^2\left(\frac{1}{X}\right)+i\pi\ln\left(\frac{1}{X}\right)+\frac{\pi^2}{3}+O\left(\frac{1}{X}\right)$$
$$Li_2(e^x)=-\frac{1}{2}\ln^2\left(\frac{1}{e^x}\right)+i\pi\ln\left(\frac{1}{e^x}\right)+\frac{\pi^2}{3}+O\left(e^{-x}\right)=-\frac{x^2}{2}-i\pi x+\frac{\pi^2}{3}+O\left(e^{-x}\right)$$
The asymptotic expansion of $x\ln(1-e^x)$ is :
$$x\ln(1-e^x)=x\ln\left(-e^x(1-e^{-x})\right)=x\ln(-e^x)+x\ln(1-e^{-x})=x^2+i\pi x+O\left(xe^{-x}\right)$$
Then, comming back to $Li_2(e^x)+x\ln(1-e^x)-\frac{x^2}{2}$
$$Li_2(e^x)+x\ln(1-e^x)-\frac{x^2}{2}=-\frac{x^2}{2}-i\pi x+\frac{\pi^2}{3}+x^2+i\pi x-\frac{x^2}{2}+O\left(xe^{-x}\right)$$
After simplification :
$$Li_2(e^x)+x\ln(1-e^x)-\frac{x^2}{2}=\frac{\pi^2}{3}+O\left(xe^{-x}\right)$$
So, the limit for $x$ tending to infinity is $\frac{\pi^2}{3}$
|
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|
Proving that $12^n + 2(5^{n-1})$ is a multiple of 7 for $n\geq 1$ by induction
Prove by induction that $12^n + 2(5^{n-1})$ is a multiple of $7$.
Here's where I am right now:
Assume $n= k $ is correct:
$$12^k+2(5^{k-1}) = 7k.$$
Let $n= k+1 $:
$$12^{k+1} + 2(5^k)$$
$$12^k(12) + 2(5^k)$$
Any ideas on how I can proceed from here?
|
Basis Step:
$P(1) = 12^1 + 2\cdot5^{1-1} = 12 + 2 = 14 = 7\cdot 2$
Thus $P(1)$ is true.
Inductive Step:
Assume $P(k)$ holds for an arbitrary positive integer $k$. Under this assumption, let us prove that $P(k+1)$ is true, namely that
$$12^{k+1} + 2\cdot5^{k} $$ is also a multiple of 7.
\begin{align}
12^{k+1}+2\cdot 5^k
&=12\cdot 12^k+2\cdot 5^k\\
&=12\cdot(7q-2\cdot 5^{k-1})+2\cdot 5^k\\
&=7\cdot12q-24\cdot5^{k-1}+10\cdot5^{k-1}\\
&=7\cdot12q+(-24+10)\cdot5^{k-1}\\
&=7\cdot12q+(-14)\cdot5^{k-1}\\
&=7\cdot12q+(7\cdot(-2))\cdot5^{k-1}\\
\end{align}
$7\cdot12q$ is a multiple of 7. Furthermore, $7\cdot(-2))\cdot5^{k-1}$ is a multiple of 7, thus their sum is a multiple of 7, illustrated by the right hand side of the equation.
The last line shows that $P(k+1)$ is true under the assumption that $P(k)$ is true. This completes the inductive step.
Thus by the Principal of Mathematical Induction, the initial proposition is true for all positive integers n. This completes the proof.
|
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|
Solving equations including floor function. I got a little trouble solving equations that involve floor function in an efficient way.
For example :
$$
\left\lfloor\frac{x+3}{2}\right\rfloor = \frac{4x+5}{3}
$$
In the one above, I get that you basically let $$ \frac{4x+5}{3} = k $$ and then inserting $k$ in the left side, take $k = 8l, 8l+1$, and so on and test it.
If there's a better solution to the one above plese tell me.
My main problem is when it comes down to functions that have multiple floors such as :
$$
\left\lfloor \frac{x+1}{3}\right\rfloor + \left\lfloor\frac{2x+5}{6}\right\rfloor = \frac{3x-5}{2}
$$
Using the same method for each of them and then intersecting the solutions should give me the right answer but is there a faster way to solve equations like this ?
|
As the left hand side is integer, so should be $\dfrac{3x-5}2\iff2|3(x-1)\iff2|(x-1)\implies x$ is odd (assuming $x$ to be an integer)
Again as lcm$(3,6)$ we need to test for $x\equiv0,1,2,3,4,5\pmod6$
But as $x$ is odd, $x\equiv1,3,5\pmod6$
If $x=6b+1$
$$\left\lfloor\frac{x+1}{3}\right\rfloor + \left\lfloor\frac{2x+5}{6}\right\rfloor=\left\lfloor\frac{6b+1+1}{3}\right\rfloor + \left\lfloor\frac{2(6b+1)+5}{6}\right\rfloor=2b+(2b+1)=4b+1$$
and $$\dfrac{3x-5}2=\dfrac{3(6b+1)-5}2=9b-1$$
and so on
If $x$ is not necessarily an integer, $\dfrac{3x-5}2+I\iff x=\dfrac{5+2I}3$
Check for $I\equiv0,1,2\pmod3$
|
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|
number $e$, why $\sum_{k=0}^\infty \frac{1}{k!}=e$? The definition that I have for $e$ is
$$e:=\lim_{n\to\infty }\left(1+\frac{1}{n}\right)^n.$$
How can we get that
$$e=\sum_{k=0}^\infty \frac{1}{k!}\ \ ?$$
My try
By Newton formula
$$\left(1+\frac{1}{n}\right)^n=\sum_{k=0}^n\binom{n}{k}\frac{1}{n^k}=\sum_{k=0}^n\frac{n!}{k!(n-k)!}\frac{1}{n^k},$$
but I don't know how to prove that
$$\lim_{n\to\infty }\sum_{k=0}^n\frac{n!}{k!(n-k)!}\frac{1}{n^k}=\sum_{k=0}^\infty \frac{1}{k!}.$$
|
As you say, by the Binomial Theorem,
$$
\left(1+\frac1n\right)^n
=\sum_{k=0}^n\binom{n}{k}\frac1{n^k}\tag{1}
$$
For each $k\le n$, by Bernoulli's Inequality,
$$
\begin{align}
\frac{\binom{n+1}{k}\frac1{(n+1)^k}}{\binom{n}{k}\frac1{n^k}}
&=\frac{n+1}{n-k+1}\left(1-\frac1{n+1}\right)^k\\
&\ge\frac{n+1}{n-k+1}\left(1-\frac{k}{n+1}\right)\\
&=1\tag{2}
\end{align}
$$
and for each $k\gt n$, $\binom{n}{k}\frac1{n^k}=0$. Thus, for each $k$,
$$
\binom{n+1}{k}\frac1{(n+1)^k}\ge\binom{n}{k}\frac1{n^k}\tag{3}
$$
Furthermore, since
$$
\binom{n}{k}\frac1{n^k}=\frac1{k!}\cdot\overbrace{\frac{n}{n}\cdot\frac{n-1}{n}\cdot\frac{n-2}{n}\cdots\frac{n-k+1}{n}}^{\text{$k$ terms}}\tag{4}
$$
we have, for $n\ge k$,
$$
\frac1{k!}\left(\frac{n-k+1}{n}\right)^k\le\binom{n}{k}\frac1{n^k}\le\frac1{k!}\tag{5}
$$
Thus, by the Squeeze Theorem,
$$
\lim_{n\to\infty}\binom{n}{k}\frac1{n^k}=\frac1{k!}\tag{6}
$$
So, by $(3)$ and $(6)$, each term of the sum in $(1)$ increases to $\frac1{k!}$. Therefore, by the Monotone Convergence Theorem,
$$
\begin{align}
\lim_{n\to\infty}\left(1+\frac1n\right)^n
&=\lim_{n\to\infty}\sum_{k=0}^n\binom{n}{k}\frac1{n^k}\\
&=\sum_{k=0}^\infty\frac1{k!}\tag{7}
\end{align}
$$
|
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|
Integrate $\int \frac{x^5 dx}{\sqrt{1+x^3}}$ I took $1+x^3$ as $t^2$ . I also split $x^5$ as $x^2 .x^3$ . Then I subsituted the differentiated value in in $x^2$ . I put $x^3$ as $1- t^2$ . I am getting the last step as $2/9[\sqrt{1+x^3}x^3 ]$ but this is the wrong answer , i should get $2/9[\sqrt{1+x^3}(x^3 +2)]$.
Please help me. Thanks
|
Substitute $t=1+x^3$ and $dt=3x^2$:
\begin{align*}
\int \frac{x^5 dx}{\sqrt{1+x^3}}&=\frac{1}{3}\int \frac{(t-1) dt}{\sqrt{t}}\\
&=\frac{1}{3}\int \left( \sqrt{t}-\frac{1}{\sqrt{t}}\right)dt\\
&=\frac{1}{3}\left( \frac{2}{3}t^\frac{3}{2}-2\sqrt{t}\right)\\
&=\frac{2}{9}(x^3-2)\sqrt{1+x^2}
\end{align*}
And of course don't forget the constant of integration.
|
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"timestamp": "2023-03-29T00:00:00",
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|
Solving $ \sqrt{x - 4} + \sqrt{x - 7} = 1 $. I have the equation $ \sqrt{x - 4} + \sqrt{x - 7} = 1 $.
I tried to square both sides, but then I got a more difficult equation:
$$
2 x - 11 + 2 \sqrt{x^{2} - 11 x - 28} = 1.
$$
Can someone tell me what I should do next?
|
Suppose that $x$ is a solution. Flip things over. We get
$$\frac{1}{\sqrt{x-4}+\sqrt{x-7}}=1.$$
Rationalizing the denominator, and minor manipulation gives
$$\sqrt{x-4}-\sqrt{x-7}=3.$$
From this, addition gives $2\sqrt{x-4}=4$. The only possibility is $x=8$, which is not a solution.
|
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|
Seating arrangements with no 3 objects together.
Suppose that five $1$'s and six $0$'s need to be arranged in such a way that no three $0$'s are consecutive. How many different arrangements are possible?
This is a variation on a problem where some number of boys and girls need to be seated such that no two boys sit next to each other. What are the solution strategies if the number of boys is increased to $3$, $4$, etc?
Thank you.
|
user84413 has provided an elegant approach to solving this problem.
Here is another approach:
If we place five ones in a row, we create six spaces, four between successive ones and two at the ends, which we wish to fill with six zeros. If we let $x_k$ denote the number of zeros placed in the $k$th space from the left, then the number of ways we can arrange five ones and six zeros so that at most two zeros are consecutive is the number of nonnegative integer solutions of the equation
$$x_1 + x_2 + x_3 + x_4 + x_5 + x_6 = 6 \tag{1}$$
subject to the restrictions that $x_k \leq 2$ for $1 \leq k \leq 6$.
When there were no restrictions, then the number of solutions of equation 1 in the nonnegative integers is equal to the number of ways we can place five addition signs in a row of six ones, which is $\binom{6 + 5}{5} = \binom{11}{5}$. However, we must exclude those cases in which three or more zeros are consecutive.
Suppose that $x_1 \geq 3$. Let $y_1 = x_1 - 3$. Then $y_1$ is a nonnegative integer and
\begin{align*}
x_1 + x_2 + x_3 + x_4 + x_5 + x_6 & = 6\\
y_1 + 3 + x_2 + x_3 + x_4 + x_5 + x_6 & = 6\\
y_1 + x_2 + x_3 + x_4 + x_5 + x_6 & = 3 \tag{2}
\end{align*}
Equation 2 is an equation in the nonnegative integers, with $\binom{3 + 5}{5} = \binom{8}{5}$ solutions. Since there are six ways one of the variables could exceed $2$, we must subtract $\binom{6}{1}\binom{8}{5}$ from $\binom{11}{5}$. However, doing so subtracts those cases in which there are two distinct blocks of three zeros twice. There are $\binom{6}{2}$ ways to fill two of the six spaces with three zeros. Therefore, by the Inclusion-Exclusion Principle, the number of bit strings consisting of five ones and six zeros in which at most two zeros are consecutive is
$$\binom{11}{5} - \binom{6}{1}\binom{8}{5} + \binom{6}{2}$$
|
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|
Find the absolute maximum and minimum value of $f$ I have to find the absolute maximum and minimum value of $$f(x, y)=\sin x+\cos y$$ on the rectangle $[0, 2\pi] \times [0, 2\pi]$.
I have done the following:
$$\nabla f=(\cos x, -\sin y)$$
$$\nabla f=0 \Rightarrow \cos x=0 \text{ and } -\sin y=0 \Rightarrow x=\frac{\pi}{2}, \frac{3 \pi}{2} \text{ and } y=0, \pi, 2\pi$$
The critical points are $$\left (\frac{\pi}{2}, 0\right ), \left (\frac{\pi}{2}, \pi\right ), \left (\frac{\pi}{2}, 2 \pi\right ), \left (\frac{3\pi}{2}, 0\right ), \left (\frac{3\pi}{2}, \pi\right ), \left (\frac{3\pi}{2}, 2\pi\right )$$
The second derivatives are:
$$\frac{\partial^2{f}}{\partial{x^2}}=-\sin x \ \ ,\ \ \frac{\partial^2{f}}{\partial{y^2}}=-\cos y \ \ ,\ \ \frac{\partial^2{f}}{\partial{x}\partial{y}}=0$$
*
*$$\left (\frac{\pi}{2}, 0\right ): \ \ \frac{\partial^2{f}}{\partial{x^2}}\left (\frac{\pi}{2}, 0\right )=-1<0$$
*$$\left (\frac{\pi}{2}, \pi\right ): \ \ \frac{\partial^2{f}}{\partial{x^2}}\left (\frac{\pi}{2}, \pi\right )=-1<0$$
*$$\left (\frac{\pi}{2}, 2 \pi\right ): \ \ \frac{\partial^2{f}}{\partial{x^2}}\left (\frac{\pi}{2}, 2 \pi\right )=-1<0$$
*$$\left (\frac{3\pi}{2}, 0\right ): \ \ \frac{\partial^2{f}}{\partial{x^2}}=1>0 \\ D=(\frac{\partial^2{f}}{\partial{x^2}})(\frac{\partial^2{f}}{\partial{y^2}})-(\frac{\partial^2{f}}{\partial{x}\partial{y}})^2=-1<0$$
*$$\left (\frac{3\pi}{2}, \pi\right ): \ \ \frac{\partial^2{f}}{\partial{x^2}}\left (\frac{3\pi}{2}, \pi\right )=1>0 \\ D=(\frac{\partial^2{f}}{\partial{x^2}})(\frac{\partial^2{f}}{\partial{y^2}})-(\frac{\partial^2{f}}{\partial{x}\partial{y}})^2=1>0 \Rightarrow \left (\frac{3\pi}{2}, \pi\right ) \text{ is a local minima } $$
*$$\left (\frac{3\pi}{2}, 2\pi\right ): \ \ \frac{\partial^2{f}}{\partial{x^2}}\left (\frac{3\pi}{2}, 2\pi\right )=1>0 \\ D=(\frac{\partial^2{f}}{\partial{x^2}})(\frac{\partial^2{f}}{\partial{y^2}})-(\frac{\partial^2{f}}{\partial{x}\partial{y}})^2=1>0 \Rightarrow \left (\frac{3\pi}{2}, 0\right ) \text{ is a local minima }$$
That means that the minimum value of $f$ is $$f\left (\frac{3\pi}{2}, 0\right )=f\left (\frac{3\pi}{2}, 2\pi\right )=0$$
Is this correct??
$$$$
EDIT1:
We have to do the following steps:
*
*Find all the critical points of $f$ in $U$.
*Find the critical points of $f$, when we look at it as a function that is defined only at $\partial U$.
*Calculate the value of $f$ at all the critical points.
*Compare all these values and pick the greatest one and the smallest one.
$U$ is the open rectangle. How can we do the second step?
$$$$
EDIT2:
First step:
$$\nabla f=(\cos x, -\sin y)$$
$U=(0, 2 \pi ) \times(0, 2\pi)$
$$\nabla f=0 \Rightarrow \cos x=0 \text{ and } -\sin y=0 \Rightarrow x=\frac{\pi}{2}, \frac{3 \pi}{2} \text{ and } y= \pi$$
The critical points in $U$ are $$\left (\frac{\pi}{2}, \pi\right ), \left (\frac{3\pi}{2}, \pi\right )$$
Second step:
Do we suppose that $x$ and $y$ can take only the values $0$ and $2 \pi$??
I have to find the absolute maximum and minimum value of $$f(x, y)=xy$$ on the rectangle $[-1, 1] \times [-1, 1]$.
I have done the following:
We have to do the following steps:
*
*Find all the critical points of $f$ in $U=(-1, 1) \times (-1, 1)$.
*Find the critical points of $f$, when we look at it as a function that is defined only at $\partial U$.
*Calculate the value of $f$ at all the critical points.
*Compare all these values and pick the greatest one and the smallest one.
First step:
$$\nabla f=(y, x)$$
$$\nabla f=0 \Rightarrow y=0 \text{ and } x=0$$
The only critical point in $U$ is $$(0, 0)$$
Second step:
$\partial U=A\cup B\cup C\cup D$ with
\begin{eqnarray*}
A &=&\left\{ (x,-1):-1\leq x\leq 1 \right\} \\
B &=&\left\{ (1 ,y):-1\leq y\leq 1 \right\} \\
C &=&\left\{ (x,1 ):-1\leq x\leq 1 \right\} \\
D &=&\left\{ (-1,y):-1\leq y\leq 1 \right\} .
\end{eqnarray*}
$$A: f(x, -1)=-x=g(x) \\ g'(x)=0 \Rightarrow -1=0$$
$$B: f(1, y)=y=g(y) \\ g'(y)=0 \Rightarrow 1=0$$
$$C: f(x, 1)=x=g(x) \\ g'(x)=0 \Rightarrow 1=0$$
$$D: f(-1, y)=-y=g(y) \\ g'(y)=0 \Rightarrow -1=0$$
That means that there are no critical points on the boundary $\partial{U}$.
Third step:
The value of $f$ at the critical point $(0, 0)$ is $f(0, 0)=0$.
Is it correct so far??
Forth step:
How can $f$ have an absolute maximum and minimum value if there is only one critical point?? Or have I done something wrong??
|
Answer to question 2. (only)
$f(x,y)=\sin x+\cos y.$ One has $\partial U=A\cup B\cup C\cup D$ with
\begin{eqnarray*}
A &=&\left\{ (x,0):0\leq x\leq 2\pi \right\} \\
B &=&\left\{ (2\pi ,y):0\leq y\leq 2\pi \right\} \\
C &=&\left\{ (x,2\pi ):0\leq x\leq 2\pi \right\} \\
D &=&\left\{ (0,y):0\leq y\leq 2\pi \right\} .
\end{eqnarray*}
We study the restriction of $f$ to each set $A,\ B,$ $C,$ and $D.$
\begin{eqnarray*}
f_{\mid A}(x,y) &=&f(x,0)=\sin x+\cos 0=\sin x+1,\ \ \ with\ \ \ 0\leq x\leq
2\pi . \\
\max_{A}f(x,y) &=&\max_{0\leq x\leq 2\pi }\sin x+1=1+1=2.
\end{eqnarray*}
\begin{eqnarray*}
f_{\mid B}(x,y) &=&f(2\pi ,y)=\sin 2\pi +\cos y=\cos y,\ \ \ with\ \ \ 0\leq
y\leq 2\pi . \\
\max_{B}f(x,y) &=&\max_{0\leq y\leq 2\pi }\cos y=1.
\end{eqnarray*}
\begin{eqnarray*}
f_{\mid C}(x,y) &=&f(x,2\pi )=\sin x+\cos 2\pi =\sin x+1,\ \ \ with\ \ \
0\leq x\leq 2\pi . \\
\max_{C}f(x,y) &=&\max_{0\leq x\leq 2\pi }\sin x+1=1+1=2.
\end{eqnarray*}
\begin{eqnarray*}
f_{\mid D}(x,y) &=&f(0,y)=\sin 0+\cos y=\cos y,\ \ \ with\ \ \ 0\leq y\leq
2\pi . \\
\max_{D}f(x,y) &=&\max_{0\leq y\leq 2\pi }\cos y=1.
\end{eqnarray*}
Now we have
\begin{eqnarray*}
\max_{\partial U}f(x,y) &=&\max \{\max_{A}f(x,y),\ \max_{B}f(x,y),\
\max_{C}f(x,y),\ \max_{D}f(x,y)\} \\
&=&\max \{2,\ 1,\ 2,\ 1\} \\
&=&2.
\end{eqnarray*}
|
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|
Solve the equation $\log_{2} x \log_{3} x = \log_{4} x$ Question:
Solve the equations
a)
$$\log_{2} x + \log_{3} x = \log_{4} x$$
b)
$$\log_{2} x \log_{3} x = \log_{4} x$$
Attempted solution:
The general idea I have been working on is to make them into the same base, combine them and take the inverse to get the solution for x.
a)
$$\log_{2} x + \log_{3} x = \log_{4} x \Leftrightarrow$$
$$\log_{2} x + \frac{\log_{2} x}{\log_{2} 3} = \frac{\log_{2} x}{\log_{2} 4} \Leftrightarrow$$
$$\log_{2} x + \frac{\log_{2} x}{\log_{2} 3} = \frac{\log_{2} x}{2} \Leftrightarrow$$
$$\frac{2 \log_{2} 3 \log_{2}x + 2\log_{2} x - \log_{2}3 \log_{2}x}{2 \log_{2}3} = 0 \Leftrightarrow $$
Moving the denominator over and solving for $\log_{2} x$
$$\frac{\log_{2} x (2\log_{2}3 + 1 - \log_{2}3)}{2\log_{2} 3} = 0 \Leftrightarrow$$
$$\log_{2} x = 0 \Rightarrow x = 2^{0} = 1$$
b)
$$\log_{2} x \log_{3} x = \log_{4} x$$
$$\log_{2} x \frac{\log_{2} x}{\log_{2} 3} - \frac{\log_{2} x}{\log_{2} 4} = 0 \Leftrightarrow$$
$$\frac{4 (\log_{2}x)^2 - \log_{2}3 \log_{2}x}{4 \log_{2}3} = 0 \Leftrightarrow $$
$$4 (\log_{2}x)^2 - \log_{2}3 \log_{2}x = 0 \Leftrightarrow $$
Substituting $t = \log_{2} x$ gives:
$$4t^2 - \log_{2}3t = 0 \Rightarrow$$
$$t^2 - \frac{\log_{2}3t}{4} = 0 \Rightarrow$$
$$t \left(t- \frac{\log_{2}3}{4}\right) = 0$$
$$t_{1} = 0 \Rightarrow x = 1$$
$$t_{2} = \frac{\log_{2}3}{4} \Rightarrow x = 2^{\frac{\log_{2}3}{4}}$$
However, the second solution here should be $\sqrt{3}$, so I must have made a mistake somewhere. Any suggestions?
|
Your denominator of $4$ in $(b)$ is wrong - $\log_2(4)=2$, not $4$. Then you'd get:
$$2^{\frac{\log_2(3)}{2}} = \left(2^{\log_2(3)}\right)^{1/2}=3^{1/2}$$
It might be easier to just use that $\log_4{x}=\frac{\log_2(x)}{2}$ and solve:
$$\log_2(x)\log_3(x)=\frac{\log_2(x)}{2}$$
This is only true when $\log_2(x)=0$ or when $\log_3(x)=\frac{1}{2}$.
|
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|
Find the integer x: $x \equiv 8^{38} \pmod {210}$ Find the integer x: $x \equiv 8^{38} \pmod {210}$
I broke the top into prime mods:
$$x \equiv 8^{38} \pmod 3$$
$$x \equiv 8^{38} \pmod {70}$$
But $x \equiv 8^{38} \pmod {70}$ can be broken up more:
$$x \equiv 8^{38} \pmod 7$$
$$x \equiv 8^{38} \pmod {10}$$
But $x \equiv 8^{38} \pmod {10}$ can be broken up more:
$$x \equiv 8^{38} \pmod 5$$
$$x \equiv 8^{38} \pmod 2$$
In the end,I am left with:
$$x \equiv 8^{38} \pmod 5$$
$$x \equiv 8^{38} \pmod 2$$
$$x \equiv 8^{38} \pmod 7$$
$$x \equiv 8^{38} \pmod 3$$
Solving each using fermat's theorem:
*
*$x \equiv 8^{38}\equiv8^{4(9)}8^2\equiv64 \equiv 4 \pmod 5$
*$x \equiv 8^{38} \equiv 8^{1(38)}\equiv 1 \pmod 2$
*$x \equiv 8^{38} \equiv 8^{6(6)}8^2\equiv 64 \equiv 1 \pmod 7$
*$x \equiv 8^{38} \equiv 8^{2(19)}\equiv 1 \pmod 3$
So now, I have four congruences. How can i solve them?
|
Notice
$$8^5=210\cdot{156}+8$$
So
$$8^5\equiv8\pmod{210}--(1)$$
Also
$$8^2\equiv64\pmod{210}--(2)$$
By (1),(2) we have
$$8^7\equiv8\cdot{64}\equiv92\pmod{210}$$
Futhermore
$$8^{35}=(8^5)^7\equiv8^7\equiv92\pmod{210}--(3)$$
Morever
$$8^3\equiv512\equiv92\pmod{210}--(4)$$
Therefore, by (3),(4),
$$8^{38}=8^{35}\cdot8^3\equiv92^2\equiv64\pmod{210}$$
|
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|
What is the area of shaded region which is lies between outer and inner circle. There is a outer circle with radius 2r and another inner circle with radius r whose center is the middle of big circle.As depicted in the following figure.
Foo graph Image
There is a sector of 120 degree in inner circle which leads to shaded part between outer and inner circle. What is the area of shaded circle in terms of r ?
i.e. What is the value of K ?
|
You can also easily solve the problem by rotating the figure by $90^o$ clockwise so that the dotted line coincides with the x-axis & center of outer circle at the origin. Thus, equation of the outer circle: $x^2+y^2=4r^2$ & the equation radius above x-axis: $y-0=\tan60^o(x-r)$ or $y=\sqrt{3}(x-r)$ solving the equation for intersection point of line & outer circle as follows $$x^2+(\sqrt{3}(x-r))^2=4r^2 => 4x^2-6rx-r^2$$ now, solving for x-coordinate, we get $$x=\frac{6r\pm\sqrt{36r^2+16r^2}}{8}=\frac{r(3\pm \sqrt{13})}{4}$$ Consider a right triangle above x-axis obtained by joining the intersection point (in I-quadrant) to the center of outer circle, we get semi-aperture angle $\alpha$ at the center of outer circle as follows $$\alpha=\cos^{-1}\left(\frac{\frac{r(3+\sqrt{13})}{4}}{2r}\right)=\cos^{-1}\left(\frac{3+\sqrt{13}}{8}\right)=\sin^{-1}\left(\frac{\sqrt{42-6\sqrt{13}}}{8}\right)$$
Hence, the shaded area is $$=\frac{1}{2}(2\alpha)(2r)^2-2\left(\frac{1}{2}(r)(2r)(\sin\alpha)\right)-\frac{1}{2}\left(\frac{2\pi}{3}\right)(r^2)=r^2\left(4\cos^{-1}\left(\frac{3+\sqrt{13}}{8}\right)-\frac{\pi}{3}-\left(\frac{\sqrt{42-6\sqrt{13}}}{4}\right)\right)$$ Hence, we get $$K=4\cos^{-1}\left(\frac{3+\sqrt{13}}{8}\right)-\frac{\pi}{3}-\frac{\sqrt{42-6\sqrt{13}}}{4}\approx0.222026268$$
|
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|
Question regarding Proof by Contradiction This is taken out of Apostol's Calculus I. It's part of his retelling of Archimedes method for finding the integral of $x^2$.
At one point, the conclusion is reached
Therefore any number $A$ must also satisfy
$$\frac{b^3}{3} - \frac{b^3}{n} < A < \frac{b^3}{3} + \frac{b^3}{n}$$
for every integer $n \geq 1$. Now there are only thee possibilities:
$$A > \frac{b^3}{3}, A < \frac{b^3}{3}, A = \frac{b^3}{3}$$
Afterwards Apostle goes on to prove that the first two possibilities lead to a contradiction and that the third option has to be correct. But this is where I'm confused. Why must it hold for all $n$ in the third case if it doesn't hold for all $n$ in the first two? It's not exactly the same as when you have 3 excluding cases. We're assuming that because it only holds for some $n$ in the first two cases that it therefore must hold for all $n$ in the third. Where am I going wrong here?
|
Note that "for all integers $n \geq 1$'' goes with the first sentence of the quote, not the second. The alternatives in the second sentence do not depend on $n$, but rather compare two constants.
In his version of Archimedes' proof, Apostol assumes $A > \frac{b^3}{3}$ and then uses the first statement of the quote -- that for all integers $n \geq 1$, $A$ is between $\frac{b^3}{3}-\frac{b^3}{n}$ and $\frac{b^3}{3}+\frac{b^3}{n}$ -- to show that all positive integers $n$ must be less than a finite constant $\frac{b^3}{A-\frac{b^3}{3}}$. There are infinitely many $n$ greater than that constant, so we have infinitely many contradictions and the original assumption must be false. The original assumption $A > \frac{b^3}{3}$ doesn't depend on $n$. The second alternative $A < \frac{b^3}{3}$ can be disposed of with a similar argument. Thus, since $A$ isn't greater than $\frac{b^3}{3}$ and $A$ isn't less than $\frac{b^3}{3}$, $A$ must be equal to $\frac{b^3}{3}$.
|
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Minimum number of circular segments. Let K be any natural number. Consider the unit square, and the circle of diameter 1 inside of the square. We then consider circular segments of area $\frac{1}{2K}$ and claim that there exists a constant $c$ such that $\#$(mutually disjoint circular segments of area $\frac{1}{2K}$)$\geq cK^{\frac{1}{3}}$.
Why is this correct? I tried explaining to myself by geometry but not with the best of results. Any hint would be appreciated.
|
Forget about the unit square, it's irrelevant.
Let K be a positive real. The maximum area of disjoint segments arises if the segments bound a polygon inscribed in the circle, and since the areas of a set of segments for a given K are equal that polygon is regular.
Let $n$ be the number of segments. Let $\theta$ be the angle subtended at the centre of the circle by each segment, so $\theta = 2\pi/n$.
The area of a segment is $\pi r^2/n - \frac{1}{2}r^2 \sin\theta$; $r = \frac{1}{2}$, so
$\frac{1}{2K} = \pi /4n - \sin\theta / 8$
$\frac{4}{K} = 2\pi /n - \sin\theta$
$\frac{4}{K} = \theta - \sin\theta$
Now $\sin \theta = \theta - \theta^3/3! + \theta^5/5! - ...$
For small $\theta$ (large $n$) we can ignore the 3rd & subsequent terms. So
$$\begin{align}
\frac{4}{K} & \approx \theta^3/6\\
\frac{24}{K} & \approx \theta^3\\
\left(\frac{24}{K}\right)^\frac{1}{3} & \approx \theta = 2\pi/n\\
\left(\frac{K}{24}\right)^\frac{1}{3} & \approx n / 2\pi\\
n & \approx 2\pi\left(\frac{K}{24}\right)^\frac{1}{3}\\
n & \approx \left(\frac{2\pi}{24^\frac{1}{3}}\right) K^\frac{1}{3}
\end{align}$$
So $n$ is proportional to $K^\frac{1}{3}$ for sufficiently large $n$. I'll let you clean up the fine details relating to bounds. :)
|
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|
Find the length of tangent $x$.
Two circles $C_1$ and $C_2$ of radius $2$ and $3$ respectively touch each other as shown in the figure .If $AD$ and $BD$ are tangents then the length of $BD$ is
$a.)3\sqrt6\\
b.)5\sqrt6\\
\color{green}{c.)\dfrac{7\sqrt6}{3}}\\
d.)6\\$
I did a construction of $CE$
And with help of Pythagorus found $AE=2\sqrt{10}$
And with again pythagorus i applied
$10^2+x^2=(x+2\sqrt{10})^2\implies x=3 \sqrt{\dfrac52}$
But the book is giving option $c.$ what is the mistake ?
|
As shown in the second diagram, we have in right $\Delta ABD$ we have $$AD=\sqrt{x^2+10^2}=\sqrt{x^2+100}$$ Now, in right triangles $\Delta ABD$ & $\Delta AEC_{2}$, we have $$\sin \angle DAB=\sin \angle EAC_{2} \implies \frac{BD}{AD}=\frac{EC_{2}}{AC_{2}} \implies \frac{x}{\sqrt{x^2+100}}=\frac{3}{7}$$ $$49x^2-9x^2=900 \implies x=\sqrt{\frac{900}{40}}=\sqrt{\frac{45}{2}}=3\sqrt{\frac{5}{2}}$$
The result is obviously same as you have obtained hence, there is certainly some printing mistake in the options provided in your book.
|
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"url": "https://math.stackexchange.com/questions/1291650",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
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|
Calculate the sum of three series which may be telescoping Let $$\sum_{n=1}^\infty \frac{n-2}{n!}$$
$$\sum_{n=1}^\infty \frac{n+1}{n!}$$
$$\sum_{n=1}^\infty \frac{\sqrt{n+1} -\sqrt n}{\sqrt{n+n^2}}$$
I have to calculate their sums. So I guess they are telescoping. However, I've no idea about how to make the telescopy emerge.
|
Lets begin with the first two since they represent a very simple class. Consider the function
$$ e^x = 1 + x + \frac{1}{2}x^2 + \frac{1}{3!}x^3 + ... = \sum_{n=0}^{\infty}{\frac{1}{n!}{x^n}} $$
From here it suffices to consider
$$ \frac{e^x}{x^2} = \sum_{n=0}^{\infty}{\frac{1}{n!}{x^{n-2}}} $$
And therefore
$$ \frac{d}{dx}\left[\frac{e^x}{x^2} \right] = \frac{e^x}{x^2}-2\frac{e^x}{x^3} = \sum_{n=0}^{\infty}{\frac{n-2}{n!}{x^{n-3}}} $$
Let $x= 1$
$$ \frac{e^1}{1^2}-2\frac{e^1}{1^3} = -e = \sum_{n=0}^{\infty}{\frac{n-2}{n!}{1^{n-3}}} = \sum_{n=0}^{\infty}{\frac{n-2}{n!}}$$
But your sum starts from one so,
$$ -e - \frac{0-2}{0!} = \sum_{n=1}^{\infty}{\frac{n-2}{n!}} $$
Thus:
$$ 2 - e = \sum_{n=1}^{\infty}{\frac{n-2}{n!}}$$
For the second one we pull the same technique instead by multiplying by $x$
$$\frac{d}{dx}[xe^x] = \sum_{n=0}^{\infty}{\frac{n+1}{n!}{x^{n}}} $$
$$e^x(1+x) = \sum_{n=0}^{\infty}{\frac{n+1}{n!}{x^{n}}} $$
$$2e = \sum_{n=0}^{\infty}{\frac{n+1}{n!}{1^{n}}} $$
$$2e-1 = \sum_{n=1}^{\infty}{\frac{n+1}{n!} }$$
|
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"language": "en",
"url": "https://math.stackexchange.com/questions/1297395",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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|
Multiplicative Inverse Element in $\mathbb{Q}[\sqrt[3]{2}]$ So elements of this ring look like
$$a+b\sqrt[3]{2}+c\sqrt[3]{4}$$
If I want to find the multiplicative inverse element for the above general element, then I'm trying to find $x,y,z\in\mathbb{Q}$ such that
$$(a+b\sqrt[3]{2}+c\sqrt[3]{4})(x+y\sqrt[3]{2}+z\sqrt[3]{4})=1$$
I can see that expanding gives me the system
$$ax+2cy+2bz=1$$
$$bx+ay+2cz=0$$
$$cx+by+az=0$$
I don't want to solve this using matrices because I know it will turn ugly. Is there a more elegant way to approach the inverse calculation to avoid the ugly calculation? The only thing I thought of was setting the bottom two equations equal to each other
$$bx+ay+2cz=cx+by+az$$
Which seems to indicate that
$$a=b, b=c, a=2c$$ but this would make me think $a=b=c=0$ and thus a multiplicative inverse does not exist.
|
First let me remark that there is a general abstract argument which shows that if $L/K$ is a field extension and $a \in L$ is algebraic over $K$, then the $K$-algebra $K[a]$ is a field. Namely, $K[a]$ is an integral domain which is also a finite-dimensional vector space over $K$. This implies for $0 \neq x \in K[a]$ that the linear map $K[a] \to K[a]$, $y \mapsto x \cdot y$ is surjective, since it is injective, which means that $x$ is invertible. To get a constructive proof, we just have to sit down and make the linear algebra argument here explicit.
The underlying vector space of $\mathbb{Q}[\sqrt[3]{2}]$ has basis $1,\sqrt[3]{2},\sqrt[3]{4}$. For some fixed non-zero element $x=a + b \sqrt[3]{2} + c \sqrt[3]{4}$, let's write down the linear map $y \mapsto x \cdot y$ in terms of this basis:
$~~~\,1 \mapsto a + b \sqrt[3]{2} + c \sqrt[3]{4}$
$\sqrt[3]{2} \mapsto 2c + a \sqrt[3]{2} + b \sqrt[3]{4} $
$\sqrt[3]{4} \mapsto 2b+2c\sqrt[3]{2} + a \sqrt[3]{4} $
The corresponding matrix is:
$$\begin{pmatrix} a & 2c & 2b \\ b & a & 2c \\ c & b & a \end{pmatrix}$$
From linear algebra we know how to invert matrices, for example via Cramer's rule. In this case, we get:
$$\frac{1}{a^3 - 6abc + 2b^3 + 4c^3} \cdot \begin{pmatrix} a^2 - 2bc & 2b^2 - 2ac & 4c^2 - 2ab \\ 2c^2 - ab & a^2 - 2bc & 2b^2 - 2ac \\ b^2 - ac & 2c^2 - ab & a^2 - 2bc \end{pmatrix}$$
The determinant $a^3 - 6abc + 2b^3 + 4c^3$ has been computed via the Rule of Sarrus and the cofactors have been computed by the usual formula for $2 \times 2$-determinants.
This matrix represents the linear map $y \mapsto x^{-1} \cdot y$ with respect to our basis. Thus, to get $x^{-1}$, we just have to evaluate at $1$, and we get:
$$x^{-1} = \frac{1}{a^3 - 6abc + 2b^3 + 4c^3} \cdot ((a^2 - 2bc) + (2c^2 - ab) \sqrt[3]{2} + (b^2 - ac) \sqrt[3]{4})$$
Of course, this method works quite generally. For example, for $x=a + b \sqrt[3]{p} + c \sqrt[3]{p}^2$ we have:
$$x^{-1} = \frac{1}{a^3 - 3pabc + pb^3 + p^2 c^3} \cdot ((a^2 - pbc) + (pc^2 - ab) \sqrt[3]{p} + (b^2 - ac) \sqrt[3]{p}^2)$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1298030",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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|
If $a_1,a_2,a_3$ are roots $x^3+7x^2-8x+3,$ find the polynomial with roots $a_1^2,a_2^2,a_3^2$ If $a_1,a_2,a_3$ are the roots of the cubic $x^3+7x^2-8x +3,$ find the cubic polynomial whose roots are:
$a_1^2,a_2^2,a_3^2$ and the polynomial whose roots are $\frac{1}{a_1}, \frac{1}{a_2}, \frac{1}{a_3}.$ Not really sure where to go. Hints appreciated.
|
For the second,
if
$f(x)
= x^3+ux^2+vx+w
=0
$
then,
if
$g(x)
=x^3f(1/x)
= x^3(1/x^3+u/x^2+v/x+w)
= 1+ux+vx^2+wx^3
$
$g(x)$
has the same roots as
$f(1/x)$
as long as none of the roots
are zero.
Therefore,
just reverse the order of the coefficients:
$3x^3-8x^2+7x+1$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1299134",
"timestamp": "2023-03-29T00:00:00",
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|
Decompose $\frac{x^2-2x+3}{(x-1)^2(x^2+4)}$, partial fraction braindead decompose $\frac{x^2-2x+3}{(x-1)^2(x^2+4)}$
the way my teacher wants us to solve is by substitution values for x,
I set it up like this:
(after setting the variables to the common denominator and getting rid of the denominator in the original equation)
$x^2-2x+3= \frac{A}{x-1}+\frac{B}{(x-1)^2}+\frac{Cx+D}{x^2+4}$
1)Let $x=1$, after plugging in for $x$ I got $2=5B$, or $B=\frac{2}{5}$
2)let the $x=0$, I get $3=-4A+4B+D$, and if I substitute B and simplify I get $D=4A-\frac{7}{5}$. Furthermore, no matter what value I plug in I still end up with two variables and cant seem to find a way to eliminate one. It seems like I have to set it up as the triple system of equations but I just dont know how to apply it here. Would really appreciate if you guys could give me a hint on how to go about it.
|
After $B=\frac25$, you may obtain $C$ and $D$ similarly with
$$\frac{x^2-2x+3}{(x-1)^2}=\frac{A(x^2+4)}{x-1}+\frac{B(x^2+4)}{(x-1)^2}+{Cx+D}$$
by setting $x^2=-4$ to get $1=-8C+3D$, $2=3C+2D$, yielding
$C=\frac4{25}$ and $D=\frac{19}{25}$. On the other hand, $A$ can be obtained by examining the limit $x\to \infty$ of above equation, which leads to $A=-C=-\frac4{25}$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1301773",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
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|
First order differential equation: did i solve this equation right So i'm trying to solve:
$$x^2\frac{dy}{dx} + 2xy = y^3$$
I'm given this differential equation, that Bernoulli equation:
$$\frac{dy}{dx} + p(x)y = q(x)y^{n} $$
I think i've solved it and got
$$ u = \frac{2}{5x} +Cx^4$$
I'm just not sure i am right i will show you how i get there but firstly...
This was part of another question which i've already solved
Show that if $y$ is the solution of the above Bernoulli differential
equation and $u = y^{1−n}$, then $u$ satisfies the linear differential
equation:
$$\frac{du}{dx} + (n-1)p(x)u = (1-n)q(x)$$
Applying the chain rule to $u = y^{1-n}$ we obtain that
\begin{align}
\frac{d u}{dx}(x)&= \frac{du}{dy}\cdot\frac{dy}{dx}\\
&= (1-n)y^{-n}\cdot\frac{dy}{dx}
\end{align}
Futhermore using the Bernoulli equation we have
$$
\frac{dy}{dx}=q(x)y^n-p(x)y
$$
and
\begin{align}
\frac{d u}{dx}&= (1-n)y^{-n}\cdot\frac{dy}{dx}\\
&=(1-n)y^{-n}\cdot q(x)y^n - (1-n)y^{-n}\cdot p(x) y\\
&=(1-n)q(x) -(1-n)p(x)y^{1-n}\\
&=(1-n)q(x) -(1-n)p(x)u
\end{align}
Hence U satisfies the equation
$$
\frac{du}{dx}+(1-n)p(x)u = (1-n)q(x)
$$
$$x^2\frac{dy}{dx} + 2xy = y^3$$
Divide both sides by $x^2$
$$\frac{dy}{dx} + \frac{2}{x}y = x^{-2} y^3$$
Consider
$$\frac{du}{dx} + (n-1)p(x)u = (1-n)q(x)$$
We know that
*
*n = 3
*1- n = 1-3 = -2
*p(x) = $ \frac{2}{x}$
*q(x) = $x^{-2}$
*u = $y^{1-3} = y^{-2}$
Subbing these in...
$$
\frac{du}{dx} + (-2)\frac{2}{x}u = (-2)x^{-2}
$$
$$
\frac{du}{dx} + \left(-\frac{4}{x}\right)u = (-2)x^{-2}
$$
So...
$$ \text{integrating factor} = e^{\int p(x) \, dx} $$
- p(x) dx = $-\frac{4}{x}$
$$ -4 \int \frac{1}{x} = -4log(x) = log (x^{-4}) $$
$$ \text{integrating factor} = e^{log (x^{-4})}= x^{-4} = \frac{1}{x^4}$$
So multiply this to the equation
$$\frac{1}{x^4}\frac{du}{dx} + \left(\frac{-4}{x^5} \right)u = \frac{-2}{x^6}$$
So we want to solve
$$ \frac{d}{dx}\frac{1}{x^4}u = \frac{-2}{x^6} $$
$$ \int \frac{d}{dx}\frac{1}{x^4}u = \int \frac{-2}{x^6} $$
$$ \frac{1}{x^4}u = -2\int \frac{1}{x^6} $$
$$ \frac{1}{x^4}u = -2\frac{1}{-5x^5} + c $$
$$ \frac{1}{x^4}u = \frac{2}{5x^5} + c $$
$$ \therefore u= \frac{2}{5x} + cx^{4} $$
is this fine? Or do i need to somehow equate this y or sub $u=y^{1-n}$
As $$u=y^{-2}$$
$$\frac{1}{y^2}= \frac{2}{5x} + cx^{4} $$
$$y^2= \frac{5x}{2} + \frac{1}{cx^{4}} $$
$$y= \sqrt{\frac{5x}{2} + \frac{1}{cx^{4}}} $$
|
How's about...
$$
\frac{du}{dx} + \left(-\frac{4}{x}\right)u = (-2)x^{-2}
$$
$$
\left(-\frac{4}{x}\right)u = (-2)x^{-1} -\frac{du}{dx}
$$
$$
u = \frac{2}{4x} + \frac{x}{4}\frac{du}{dx}
$$
As $u=y^{1-n} = y^{-2}$
$$\frac{1}{y^2} = \frac{2}{4x} + \frac{x}{4}\frac{du}{dx} $$
$$ y^2 = = \frac{4x}{2} + \frac{4}{x}\frac{dx}{du} $$
$$ y = \sqrt{\frac{4x}{2} + \frac{4}{x}\frac{dx}{du}} $$
|
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"language": "en",
"url": "https://math.stackexchange.com/questions/1302602",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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|
What are the facts used in each step of this proof? What are the facts used in each step of this proof ?
Suppose that $A\in F^{nm}$ and $B\in F^{ml}$
$$\begin{align}rank A + rank B &= rank\begin{bmatrix}0 & A\\B & 0\\ \end{bmatrix}\\
&\le rank\begin{bmatrix}0 & A\\B & I\\ \end{bmatrix}\\ &= rank\begin{bmatrix}I & A\\0 & I\\ \end{bmatrix}\begin{bmatrix}-AB & 0\\B & I\\ \end{bmatrix}\\&\le rank\begin{bmatrix}-AB & 0\\B & I\\ \end{bmatrix}\\&\le rank\begin{bmatrix}-AB&0\\ \end{bmatrix}+rank\begin{bmatrix}B&I\\ \end{bmatrix}\\&=rankAB+m\end{align}$$
I especially din't understand why we can write this?
$$\begin{align}rank\begin{bmatrix}-AB & 0\\B & I\\ \end{bmatrix}\le rank\begin{bmatrix}-AB&0\\ \end{bmatrix}+rank\begin{bmatrix}B&I\\ \end{bmatrix}\end{align}$$
and why
$$\begin{align}rank\begin{bmatrix}-AB&0\\ \end{bmatrix}+rank\begin{bmatrix}B&I\\ \end{bmatrix}=rankAB+m\end{align}$$
also I cannot understand the part
$$\begin{align}rank A + rank B &= rank\begin{bmatrix}0 & A\\B & 0\\ \end{bmatrix}\\
&\le rank\begin{bmatrix}0 & A\\B & I\\ \end{bmatrix}\end{align}$$
|
Pick rank-many independent rows from $\begin{pmatrix}-AB&0\\B&1\end{pmatrix}$. Some of these rows live in $\begin{pmatrix}-AB&0\end{pmatrix}$ and some in $\begin{pmatrix}B&1\end{pmatrix}$, where they are still linearly independent hence certainly not more than $\operatorname{rank}\begin{pmatrix}-AB&0\end{pmatrix}$ and $\operatorname{rank}\begin{pmatrix}B&1\end{pmatrix}$, respectively.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1302768",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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|
Solving for a three dimensional vector. Let
$a = \begin{pmatrix} 2 \\ 5 \\ -1 \end{pmatrix}$ and $b = \begin{pmatrix} -6 \\ 4 \\ -3 \end{pmatrix}$
There exists two nonzero three-dimensional vectors
${v} = \begin{pmatrix} x \\ y \\ z \end{pmatrix}$
that are orthogonal to both ${a}$ and ${b}$, such that its entries $x$, $y$, and $z$ are integers, that satisfy $\gcd(x,y,z) = 1$. Find either vector.
I have tried to write out the equations, but there aren't enough equations and too many variables. I still can't figure out how to use $\gcd(x,y,z) = 1$. Any help is appreciated.
|
\begin{align}
2x + 5y - z = 0 \rightarrow x + \frac{5}{2}y - \frac{1}{2}z = 0\\
-6x + 4y - 3z = 0 \rightarrow x - \frac{2}{3}y + \frac{1}{2}z = 0\\
\rightarrow \\
\left(\frac{5}{2} + \frac{2}{3}\right)y - z = 0
\end{align}
This gives that $z = \frac{19}{6}y$ which gives that $x = -\frac{5}{2}y + \frac{1}{2}\frac{19}{6}y = -\frac{11}{12}y$. This means that $y$ must be divisible by $12$ (to produce an integer). this gives that $x = -11$, $y = 12$, and $z = 38$. The gcd of these is $1$ because one of the values is a prime and none of the other values have $11$ as a factor (there is only one value divisible by $3$ and only one divisible by $19$ but there are two values divisible by $2$).
...the other one is just the negation of this one: $x = 11$, $y = -12$, $z = -38$.
|
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"url": "https://math.stackexchange.com/questions/1303456",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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|
Is it possible to make a dataset given these following mean, range, and standard variance? Given that the:
$Mean = 30$
Range $= X_n - X_1 = 10$
$S^2 = Variance = 40$
is it possible to construct a dataset with those values?
So, to be honest I have no idea how to properly approach this and I have tried "brute forcing" this to find a set of numbers to make this work, but no matter what I do I can't seem to do it. Thus I believe that the answer is "no", however, I'm not sure about this.
Is there a standard procedure I can follow to solve this logically?
|
No, it is impossible.
Given any list of $n$ numbers $X_k$ without any constraint of its order. Let
$$\begin{cases}
\overline{X} &= \frac{1}{n}\sum_{k=1}^n X_k\\
\overline{X^2} &= \frac{1}{n}\sum_{k=1}^n X_k^2
\end{cases}
\quad\text{ and }\quad
\begin{cases}
P &= \max\{ X_k : 1 \le k \le n \}\\
Q &= \min\{ X_k : 1 \le k \le n \}
\end{cases}
$$
The (biased) sample variances and unbiased sample variance for the numbers $X_k$ are given by the formulas
$$\begin{cases}
\sigma_X^2 &= \frac{1}{n}\sum\limits_{k=1}^n (X_k - \overline{X})^2 = \overline{X^2} - \overline{X}^2\\
s_X^2 &= \frac{1}{n-1}\sum\limits_{k=1}^n (X_k - \overline{X})^2 = \frac{n}{n-1}\sigma_X^2
\end{cases}$$
Since $(P - X_k)(X_k - Q) \ge 0$ for all $1 \le k \le n$, we have
$$\frac{1}{n}\sum_{k=1}^n (P - X_k)(X_k - Q) \ge 0
\iff -PQ + (P+Q)\overline{X} - \overline{X^2} \ge 0\\
\implies \sigma_X^2 = (\overline{X^2} - \overline{X}^2) \le (P - \overline{X})(\overline{X} - Q)$$
This inequality is a special case of the Bhatia-Davis inequality.
Since $P \ge \overline{X} \ge Q$, this leads to
the Popoviciu's inequality on variances:
$$\sigma_X^2 \le \frac14 (P-Q)^2
\quad\iff\quad
s_X^2 = \frac{n}{n-1}\sigma_X^2 \le \frac{n}{4(n-1)}(P-Q)^2$$
Apply this to our problem where $P - Q = 10$, it is clear there is no way for the (biased) sample variances $\sigma_X^2$ to be equal to $40$. For the unbiased sample variances, if we want $s_X^2 = 40$, we need
$$40 \le \frac{n}{4(n-1)}10^2 = 25\frac{n}{n-1}\quad\implies\quad n \le \frac{8}{3}$$
Since $n$ is an integer $\ge 2$, $n$ can only be $2$. However when $n = 2$,
$$\begin{cases}
\overline{X} &= \frac{P+Q}{2}\\
\overline{X^2} &= \frac{P^2+Q^2}{2}
\end{cases}
\quad\implies\quad
\begin{align}
s_X^2 &= \frac{2}{2-1}\left(\overline{X^2} - \overline{X}^2\right)
= P^2 + Q^2 - \frac{(P+Q)^2}{2}\\
&= \frac12(P-Q)^2 = 50
\end{align}
$$
This doesn't match the given number $40$ again. As a result, there is no data-set which can produces the desired mean, range and sample variances (for both biased and unbiased one).
|
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|
For which $n$ does $\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=\frac{1}{a+b+c}$ imply $\frac{1}{a^n}+\frac{1}{b^n}+\frac{1}{c^n}=\frac{1}{a^n+b^n+c^n}$ I'm having trouble finishing a problem on an old national competition.
As the title states, the question says asks:
Given $a,b,c \neq 0,a+b=c$ such that $\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=\frac{1}{a+b+c}$,
Find all integers $n$ such that $\frac{1}{a^n}+\frac{1}{b^n}+\frac{1}{c^n}=\frac{1}{a^n+b^n+c^n}$.
I know that if it is true for $n$, it is also true for $-n$ since $(a^n+b^n+c^n)(\frac{1}{a^n+b^n+c^n})=1$ and $(a^n+b^n+c^n)(\frac{1}{a^n}+\frac{1}{b^n}+\frac{1}{c^n})=1$ and the statment follows for $-n$ given that we can make the substitution for $n$. Thus I found out that $n=-1$ satisfies the equality.
I also know that it fails for all even $n$. If $n$ were even, $(a^n,b^n,c^n)$ would all be positive, and Titu's Lemma asserts that $\frac{1}{a^n}+\frac{1}{b^n}+\frac{1}{c^n}\geq \frac{9}{a^n+b^n+c^n}$. Thus $n$ must be odd.
I also observed that $(2,-1,1)$ was a solution, and from here noted that this particular solution satisfies all odd $n$.
I have been trying to prove that all other solutions satisfy all odd $n$ with no success.
Any help would be greatly appreciated. Thanks!
|
Lemma: if $\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}=\dfrac{1}{a+b+c}$,then we have
$$\dfrac{1}{a^n}+\dfrac{1}{b^n}+\dfrac{1}{c^n}=\dfrac{1}{a^n+b^n+c^n}$$
where $n$ is odd.
proof:
$$\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}=\dfrac{1}{a+b+c}
\Longrightarrow \dfrac{ab+bc+ac}{abc}=\dfrac{1}{a+b+c}$$
$$\Longrightarrow (a+b+c)(ab+bc+ac)=abc$$
since
$$(a+b+c)(ab+bc+ac)=(a+b)(b+c)(a+c)+abc\Longrightarrow (a+b)(b+c)(a+c)=0$$
WOLG, let $a=-b$,and Note $n$ is odd.so
$$a^n+b^n=0,\dfrac{1}{a^n}+\dfrac{1}{b^n}=0$$
so
$$\dfrac{1}{a^n}+\dfrac{1}{b^n}+\dfrac{1}{c^n}=\dfrac{1}{a^n+b^n+c^n}$$
|
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"timestamp": "2023-03-29T00:00:00",
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|
The biggest and smallest integer solution of $\sqrt{(5+2\sqrt6)^{2x}}+\sqrt{(5-2\sqrt6)^{2x}}\le98$ are? $$\sqrt{(5+2\sqrt6)^{2x}}+\sqrt{(5-2\sqrt6)^{2x}}\le98$$
I noticed that $5+2\sqrt6=(\sqrt2+\sqrt3)^2$ but that hardly helps.
After cancelling the square root and the power of two, I tried multiplying the whole thing with $(5+2\sqrt6)^{x}$ and replacing $(5+2\sqrt6)^{x}=t$ would get a quadratic equation that wouldn't help me without some serious approximation which I don't trust myself enough to do.
Is there an easier way to go around this?
|
Firstly note that
$$(5+2\sqrt{6})(5-2\sqrt{6}) = 25-24=1$$
so $(5+2\sqrt{6})^{-1}= (5-2\sqrt{6})$.
Putting $t=(5+2\sqrt{6})^x$ you get the inequality
$$t + \frac{1}{t} \leq 98$$
which has solution
$$(5+2\sqrt{6})^{-2} = 49 - 20 \sqrt{6} \le t \le 49 + 20 \sqrt{6} = (5+2\sqrt{6})^2$$
which means exactly
$$-2 \le x \le 2$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1305885",
"timestamp": "2023-03-29T00:00:00",
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|
Finding the Laurent representation of a complex function How can i find the Laurent representation fot the function:
$$f(z)=\dfrac{1}{1-z^2}+\dfrac{1}{3-z}$$
In the region of:
a) $\{z\in\mathbb C:1<|z|<3\}$
b) $\{z\in\mathbb C:1<|z-2|<3\}$
|
$\frac{1}{1 - z} = \sum z^n$--the simplest function. You have the following:
$$
\frac{1}{2}\frac{1}{1 - (-z)} + \frac{1}{2}\frac{1}{1 - z} + \frac{1}{3}\frac{1}{1 - \frac{z}{3}}
$$
When $1 < |z| < 3$, we need the Laurent series for $\frac{1}{1 \pm z}$ but the Taylor series for $\frac{1}{3 - z}$:
$$
\frac{1}{1 - z} = \frac{1}{z}\frac{1}{z^{-1} - 1} = -z^{-1}\frac{1}{1 - z^{-1}} = -z^{-1}\sum_0^\infty z^{-n} = -\sum_1^\infty z^{-n}
$$
Which gives:
\begin{align}
\frac{1}{1 - z} =&\ -\sum_1^\infty z^{-n} \\
\frac{1}{1 - (-z)} =&\ -\sum_1^\infty (-1)^n * z^{-n} \\
\frac{1}{1 - \frac{z}{3}} =&\ \sum_0^\infty \left(\frac{z}{3}\right)^n
\end{align}
Adding this gives something to the affect of:
$$
a_n = \begin{cases}
\frac{1}{3} & n = 0 \\
-z^{-n} + \frac{1}{3}\frac{z^n}{3^n} & n \text{ even} \\
\frac{1}{3}\frac{z^n}{3^n} & n \text{ odd}
\end{cases}
$$
I'm not quite clear on the domain of the second part.
p.s. You actually don't need the partial fraction decomposition (and I believe there was a previous answer which did not use partial fractions).
We know that $\frac{1}{1 - z^2}$'s Taylor series has a radius of convergencce of $|z| < 1$ (and is clearly undefined for $z = \pm 1$), and finally since the Laurent series for $\frac{1}{1 - z} = -\sum_1^\infty z^{-n}$, meaning:
$$
\frac{1}{1 - z^2} = -\sum_1^\infty z^{-2n} \text{, for } |z| > 1
$$
Adding to the Taylor series we get:
$$
-\sum_1^\infty z^{-2n} + \frac{1}{3}\sum_0^\infty \left(\frac{z}{3}\right)^n
$$
...but the exponents don't match up (which would be nice)...so match them up by splitting the Taylor series into two sums:
$$
-\sum_1^\infty z^{-2n} + \frac{1}{3}\left(\sum_0^\infty \left(\frac{z}{3}\right)^{2n} + \sum_0^\infty \left(\frac{z}{3}\right)^{2n + 1} \right)
$$
Which can be written as:
$$
\frac{1}{3} + \sum_1^\infty\left(\frac{1}{3}\left(\frac{z}{3}\right)^{2n} - z^{-2n}\right) + \frac{1}{3}\sum_0^\infty\left(\frac{z}{3}\right)^{2n + 1}\\
\text{or} \\
\frac{1}{3} + \sum_1^\infty\left(\frac{1}{3}\left(\frac{z}{3}\right)^{2n} - z^{-2n}\right) + \frac{1}{3}\sum_1^\infty\left(\frac{z}{3}\right)^{2n - 1} \\
\text{or}\\
\frac{1}{3} + \sum_1^\infty\left(\frac{1}{3}\left(\frac{z}{3}\right)^{2n} - z^{-2n} + \frac{1}{3}\left(\frac{z}{3}\right)^{2n - 1}\right)
$$
|
{
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"timestamp": "2023-03-29T00:00:00",
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|
Prove that $\frac{ab}{c}+\frac{bc}{a}+\frac{ac}{b}\ge a+b+c$ How to prove that
\begin{equation*}\frac{ab}{c}+\frac{bc}{a}+\frac{ac}{b}\ge a+b+c,\ where \ a,b,c>0\end{equation*}
I tried the following:
\begin{equation*}abc(\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2})\ge a+b+c\end{equation*}
Using Chebyshev's inequality
\begin{equation*}(\frac{1}{a}+\frac{1}{b}+\frac{1}{c})(\frac{1}{a}+\frac{1}{b}+\frac{1}{c})\le3(\frac{1}{a}\frac{1}{a}+\frac{1}{b}\frac{1}{b}+\frac{1}{c}\frac{1}{c})\end{equation*}
from first inequality follows
\begin{equation*}\frac{1}{3}(\frac{1}{a}+\frac{1}{b}+\frac{1}{c})^2abc\ge a+b+c\end{equation*}
equivalent to
\begin{equation*}\frac{abc}{3(a+b+c)}\ge (\frac{1}{\frac{1}{a}+\frac{1}{b}+\frac{1}{c}})^2\end{equation*}
and by amplifying both members by 9
\begin{equation*}\frac{3abc}{a+b+c}\ge (\frac{3}{\frac{1}{a}+\frac{1}{b}+\frac{1}{c}})^2\end{equation*}
now using mean inequality
\begin{equation*}\sqrt[3]{abc}\ge \frac{3}{\frac{1}{a}+\frac{1}{b}+\frac{1}{c}}\end{equation*}
the inequality in question becomes
\begin{equation*}3abc\ge (a+b+c)(\sqrt[3]{abc})^2\end{equation*}
which yields
\begin{equation*}3\sqrt[3]{abc}\ge a+b+c\end{equation*}
not what I wanted.
|
We can do a slick AM-GM "pairwise" token that I picked up from the "Cauchy masters":
$\dfrac{ab}{c} + \dfrac{bc}{a} \geq 2\sqrt{\dfrac{ab}{c}\cdot\dfrac{bc}{a}}= 2b$, and similarly: $\dfrac{bc}{a}+\dfrac{ca}{b} \geq 2c$, and $\dfrac{ca}{b} + \dfrac{ab}{c} \geq 2a$. Add them up and divide by $2$ to get the answer.
|
{
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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|
Why cannot $(2x^2 + x)^2$ be simplified to $2x^4 + x^2$? So, I want to simplify an equation : $(2x^2 + x)^2$.
I thought this would simplify to $2x^4 + x^2$
But, if you input a value for $x$, the answers do not equal. For example, if you input $x = 3$, then:
$$(2x^2+x)^2
= 21^2
= 441$$
AND:
$$2x^4 + x^2
= 2(82) + 9
= 173$$
Can anyone explain why this is the case?
|
Hint:
$$
(A+B)^2= (A+B)(A+B)=A^2+AB+BA+B^2=A^2+B^2+2AB
$$
Use: $A=2x^2$ and $B=x$ and you find the right result.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1309780",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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"answer_id": 1
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|
Homogenous equation to higher order ODE Hello I have a quick question in regard to general form of the solution to
$$y^{(4)}-2y^{(3)}+y''=0$$
I had thought to find this solution we would consider
$r^{4}-2r^{3}+r^{2}=0$
which factors as $r^2(r-1)^{2}$
that is we have $r_{1,2}=0$ and $r_{3,4}=1$
So what from what I had thought I knew, I thought this implied a solution of form
$$y=c_1e^{0x}+c_2xe^{0x}+c_3e^{x}+c_4xe^{x}=c_1+c_2x+c_{3}e^{x}+c_{4}xe^{x}$$ with $c \in \mathbb{R}$
But wolfram says it is $$y(x)=e^x(c_{2}x+c_{1}-2c_{2})+c_{4}x+c_{3}$$
So where am I going wrong? Thanks all
PS: I hear it is correct, and I would by wondering about solving
$y^{(4)}-2y^{(3)}+y''=x$
Would I be able to use method of undetermined coefficients for this?
And if so, because r=0 is a double root of the equations, would by assumed form for a particular solution need to be $x^2({A_{o}x+A_{1}})$? If I did it with this, could I get a correct answer? Im not sure exactly
|
For the equation $y^{(4)} - a y^{(3)} + y^{(2)} = c_{0}x$ the solution is as follows. Integrate twice to obtain $y^{(2)} - a y^{(1)} + y = c_{0} \frac{x^{3}}{6} + c_{1} x + c_{2}$. Now let $y(x) = f(x) + b_{1} x^{3} + b_{2} x^{2} + b_{3} x + b_{4}$ to obtain
\begin{align}
f'' - a f' + f + [b_{1} x^{3} + (b_{2}-3 a b_{1}) x^{2} +(6 b_{1} - 2 a b_{2} + b_{3}) x + (2 b_{2} - a b_{3} + b_{4}) ] = c_{0} \frac{x^{3}}{6} + c_{2}
\end{align}
which leads to $b_{1} = c_{0}/6$, $b_{2} = a c_{0}/2$, $b_{3} = c_{1} - (a^{2} + 1) c_{0}$, $b_{4} = c_{2} + a c_{1} - (2 a + a^{3}) c_{0} $ and
$$y(x) = f(x) + \frac{c_{0} x^{3}}{6} + \frac{a c_{0} x^{2}}{2} + (c_{1} - (a^{2} + 1) c_{0}) x + c_{2} +a c_{1} -a(2 + a^{2})c_{0}$$.
The equation for $f(x)$ is $f'' - a f + f = 0$ which has the solution
\begin{align}
f(x) = e^{ax/2} \left( A e^{\frac{x}{2} \sqrt{a^{2} - 4}} + B e^{- \frac{x}{2} \sqrt{a^{2} - 4}} \right).
\end{align}
Collecting all the parts together leads to
\begin{align}
y(x) = e^{ax/2} \left( A e^{\frac{x}{2} \sqrt{a^{2} - 4}} + B e^{- \frac{x}{2} \sqrt{a^{2} - 4}} \right) + \frac{c_{0} x^{3}}{6} + \frac{a c_{0} x^{2}}{2} + (c_{1} - (a^{2} + 1) c_{0}) x + c_{2} +a c_{1} -a(2 + a^{2})c_{0}.
\end{align}
When $a=2$ this reduces to $y = (A + Bx) e^{x} + \frac{c_{0} x^{3}}{6} + c_{0} x^{2} + (c_{1} - 5 c_{0}) x + c_{2} + 2 c_{1} - 12 c_{0}$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1310881",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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|
simple 2 sides inequality
$$2<\frac{x}{x-1}\leq 3$$
Is the only way is to multiple both sides by $(x-1)^2$?
so we get $2x^2-4x+2<x^2-x $ and $x^2-x<3x^2-6x+3$ which is $-x^2+3x-2$ and $-2x^2+5x-3<0$ so the sloutions are:
$1<x\leq \frac{3}{2}$ and $1<x\leq 2$ so overall it is $1<x\leq\frac{3}{2}$
|
Multiplying $(x-1)^2\gt 0$ is a good idea, but you made mistakes after that.
$$2(x-1)^2\lt x(x-1)\iff x^2-3x+2\lt 0$$$$\iff (x-2)(x-1)\lt 0\iff 1\lt x\lt 2$$
and
$$x(x-1)\le 3(x-1)^2\iff 2x^2-5x+3\ge 0$$$$\iff (x-1)(2x-3)\ge 0\iff x\le 1\ \text{or}\ x\ge\frac 32.$$
Hence, the answer is
$$\frac 32\le x\lt 2.$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1310939",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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|
$(1+i)^6$ in polar form $re^{i\theta}$ I used De Moivre's formula and got
$$
\left(\frac{\sqrt{2}}{2}\right)^6 \times
\cos\left(6 \times \frac{1}{4\pi}\right) +
i\sin\left(6 \times \frac{1}{4\pi}\right) =
\frac{1}{8} e^{\frac{3}{2\pi}}.
$$
But the answer is $8e^{\frac{3}{2\pi}}$, can someone explain where the $8$ is from?
Thanks!
|
We have that $|1+i| = \sqrt{2}$ and ${\rm Arg}(1+i) = \pi/4$. So: $$1+i = \sqrt{2}\left(\cos \frac{\pi}{4}+i \sin\frac{\pi}{4}\right).$$By De Moivre's formula: $$(1+i)^6 = (\sqrt{2})^6 \left(\cos \frac{6\pi}{4}+i\sin\frac{6\pi}{4}\right).$$But $\sqrt{2}^6 = (2^{1/2})^6 = 2^3=8$. Simplifying: $$(1+i)^6 = 8\left(\cos \frac{3\pi}{2}+i\sin\frac{3\pi}{2}\right) = 8e^{3i\pi/2}$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1311507",
"timestamp": "2023-03-29T00:00:00",
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|
Find all Integral solutions to $x+y+z=3$, $x^3+y^3+z^3=3$. Suppose that $x^3+y^3+z^3=3$ and $x+y+z=3$.
What are all integral solutions of this equation?
I can only find $x=y=z=1$.
|
Using the Power Mean Inequality we have that:
$$ 1=\frac{1}{3}(x^3 + y^3 + z^3) \leq \left[\frac{1}{3}(x+y+z)\right]^3 =1$$
Equality only holds when $x=y=z=1$.
Let's replace $z \mapsto -z$ and try to solve the system of equations for $x,y,z \geq 0$:
$$ x^3 + y^3 = 3 + z^3 \text{ and } x + y = z + 3 $$
We can still use the power mean inequality (in only two of the varaibles) and use the linear relation
$$ \frac{1}{3}z^3 < \frac{1}{3}(3 + z^3 ) = \frac{1}{3}(x^3 + y^3 ) \leq \left[\frac{1}{3}(x+y)\right]^3 = \left[\frac{1}{3}(z+3)\right]^3
< \frac{1}{24}(z+3)^3 $$
Maybe if I take cube roots the left and right sides are easier to compare. I also rounded $3^3 \approx 27$
$$ z < \frac{1}{2}(z + 3)$$
This means $0 \leq z < 3$. Can it be that $0^3 + 3 = 3, 1^3 + 3 = 4, 2^3 + 3 = 11$ are the sums of two cubes?
If two of the 3 numbers are negative, the starting point should be:
$$ x^3 + y^3 = -3 + z^3 \text{ and } x + y = z - 3 $$
Then using the power mean inequality we can attempt to deduce a range on $z$
$$ \frac{1}{3}(-3 + z^3 ) < \frac{1}{24}(z-3)^3 $$
This one is harder to estimate. After fiddling around, we see that $z \leq 0$:
$$ z < \frac{1}{2}\sqrt[3]{(z-3)^3 + 24}
< \frac{1}{2}z $$
So there are definitely no solutiuons of this kind.
|
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"timestamp": "2023-03-29T00:00:00",
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|
Difficulty proving $x_n$ satisfaction in the inequality I am trying to prove that $$x_n = \int_{0}^{1}\left(\frac {n}{x}\right)^n \,dx$$ satisfies the inequalities
$$\frac {n}{n+1}\left(\frac {\pi^2}{6}-\sum_{k=1}^{n}\frac {1}{k^2}\right)<x_n<\frac {n}{n+1}\left(\frac {\pi^2}{6}-\sum_{k=1}^{n-1}\frac {1}{k^2}\right)$$
$$\frac {n^2}{n+1}\left(\frac {\pi^2}{6}-\sum_{k=1}^{n}\frac {1}{k^2}\right)<x_n<\frac {n^2}{n+1}\left(\frac {\pi^2}{6}-\sum_{k=1}^{n-1}\frac {1}{k^2}\right)$$
I tried to make the substitution $\frac {n}{x} =t$ but I couldn't get the desired result.
|
$$x_n = n \int_{n}^{\infty} \frac {t^n}{t^2}\,dt = n \sum_{k=n}^{\infty}\int_{k}^{k+1} \frac {t^n}{t^2} \,dt = n \sum_{k=n}^{\infty}\int_{k}^{k+1} \frac {(t-k)^n}{t^2} \,dy = n \sum_{k=n}^{\infty} \int_{0}^{1} \frac {y^n}{(k+y)^2}\,dy = n \int_{0}^{1} y^n \left(\sum_{k=n}^{\infty} \frac {1}{(k+y)^2}\right) \,dy $$
Since $\frac {1}{(k+1)^2} < \frac {1}{(k+y)^2} < \frac {1}{k^2}$, for positive integers $k$, and $y \in (0,1)$, it follows that
$$n \int_{0}^{1} y^n \left(\sum_{k=n}^{\infty} \frac {1}{(k+1)^2}\right) \,dy < x_n < n \int_{0}^{1} y^n \left(\sum_{k=n}^{\infty} \frac {1}{k^2}\right) \,dy$$
Thus,
$$\frac {n}{n+1}\left(\frac {\pi^2}{6}-\sum_{k=1}^{n}\frac {1}{k^2}\right)<x_n<\frac {n}{n+1}\left(\frac {\pi^2}{6}-\sum_{k=1}^{n-1}\frac {1}{k^2}\right)$$
Which implies that $\lim\limits_{n\to \infty} x_n = 0$.
The second limit equals $1$. We have, based on the preceding inequalities, that
$$\frac {n^2}{n+1}\left(\frac {\pi^2}{6}-\sum_{k=1}^{n}\frac {1}{k^2}\right)<x_n<\frac {n^2}{n+1}\left(\frac {\pi^2}{6}-\sum_{k=1}^{n-1}\frac {1}{k^2}\right)$$
and the result follows since $\displaystyle\lim\limits_{n \to \infty} n\left(\frac {\pi^2}{6}-\sum_{k=1}^{n} \frac {1}{k^2}\right) = 1$
|
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"timestamp": "2023-03-29T00:00:00",
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|
How can one evaluate the integral $\int\sqrt{\frac{1-ax}{1-x}}\ dx$? How can I evaluate the following integral:
$$\int\sqrt{\frac{1-ax}{1-x}}\ dx?$$
Here $a$ is a positive constant. I know that the function
$$\sqrt{\frac{1-ax}{1-x}}$$
has a primitive, but I don't know how to find it.
|
The standard way for these integrals is to use substitution: set
$$y^2=\frac{1-ax}{1-x}=a+\frac{1-a}{1-x},\enspace y>0,\enspace\text{whence}\enspace \mathrm 2y\, d\mkern1mu y=\frac{1-a}{(1-x)^2}\mathrm d\mkern1mu x$$
so that
$$\mathrm d\mkern1mu x =2y\frac{(1-x)^2}{1-a}y=\frac{2(1-a)y}{(y^2-a)^2}\mathrm d\mkern1mu y$$
and finally
\begin{align*}\int\sqrt{\frac{1-ax}{1-x}}\,\mathrm d\mkern1mu x&=2(1-a)\int \frac{y^2}{(y^2-a)^2}\,\mathrm d\mkern1mu y=2(1-a)\int \frac{\mathrm d\mkern1mu y}{y^2-a}+2a(1-a)\int \frac{\mathrm d\mkern1mu y}{(y^2-a)^2}\\
&=\frac{1-a}{\sqrt a}\,\ln\biggl(\frac{y-\sqrt a}{y-\sqrt a}\biggr)\end{align*}
Now use partial fractions:$$\frac{y^2}{(y^2-a)^2}=\frac A{y-\sqrt a}+\frac B{y+\sqrt a}+\frac C{(y-\sqrt a)^2}+\frac D{(y+\sqrt a)^2}$$
*
*For symmetry reasons, one has $A=-B$, $C=D$.
*Multiplying both sides by $(y-\sqrt a)^2$ and setting $y=\sqrt a$ shows $\,C=D=\dfrac14$.
*Setting $y=0\,$ leads to $A=-B=\dfrac1{4\sqrt a}$
Thus
\begin{align*}\int \frac{y^2}{(y^2-a)^2}\,\mathrm d\mkern1mu y&=\dfrac1{4\sqrt a}\,\ln\biggl(\frac{y-\sqrt a}{y+\sqrt a}\bigg)-\frac14\biggl(\frac1{y-\sqrt a}+\frac1{y+\sqrt a}\biggr)\\
&=\dfrac1{4\sqrt a}\,\ln\biggl(\frac{y-\sqrt a}{y+\sqrt a}\bigg)-\frac y{2(y^2-a)}\\
&=\dfrac1{4\sqrt a}\,\ln\biggl(\frac{\sqrt{1-ax}-\sqrt{a(1+ax)}}{\sqrt{1-ax}+\sqrt{a(1+ax)}}\biggr)-\frac{\sqrt{(1-x)(1-ax)}}{2(1-a)}.
\end{align*}
|
{
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"timestamp": "2023-03-29T00:00:00",
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|
Integral of Trigonometric Identities $$\int(\sin(x))^3(\cos(2x))^2dx$$
I can write $$\sin^3(x)=\sin(x)(1-\cos^2(x)=\sin(x)-\sin(x)\cos^2(x)$$
for $$\cos^2(2x)=(1-\sin^2(x))^2=1-4\sin^2(x)+4\sin^4(x)$$
after simplifying the Trig identities i get:
$$\int(sin^3(x)-4sin^5(x)+4sin^7(x))dx$$
so i need to know how to go further :)
|
$$\int\left(\sin^3(x)\cos^2(2x)\right)dx=$$
$$\int \left(\frac{7}{16}\sin(x)-\frac{5}{16}\sin(3x)+\frac{3}{16}\sin(5x)-\frac{1}{16}\sin(7x)\right)dx=$$
$$\int \frac{7}{16}\sin(x)dx-\int \frac{5}{16}\sin(3x)dx+\int \frac{3}{16}\sin(5x)dx-\int \frac{1}{16}\sin(7x)dx =$$
$$\frac{7}{16}\int \sin(x)dx-\frac{5}{16}\int \sin(3x)dx+\frac{3}{16}\int \sin(5x)dx-\frac{1}{16}\int \sin(7x)dx =$$
$$-\frac{7}{16}\cos(x)-\frac{5}{16}\left(-\frac{1}{3}\cos(3x)\right)+\frac{3}{16}\left(-\frac{1}{5}\cos(5x)\right)-\frac{1}{16}\left(-\frac{1}{7}\cos(7x)\right) =$$
$$-\frac{7}{16}\cos(x)+\frac{5}{48}\cos(3x)-\frac{3}{80}\cos(5x)+\frac{1}{112}\cos(7x)$$
|
{
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"url": "https://math.stackexchange.com/questions/1315757",
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|
If $T(n)= T(n-1) + 2T(n-2)$ If $T(n)= T(n-1) + 2T(n-2)$ with $T(0)=0$ and $T(1) = 1$
What is $T(n)$ (in $Θ$–notation) in terms of $n$?
I am trying to solve by substitution, but I am not sure if I am doing this right, as I get stuck. Can anybody tell me where to go from here?
$T(n)= T(n-1) + 2T(n-2)$
$T(n-1)= T(n-2) + 2T(n-3)$
$T(n-2)= T(n-3) + 2T(n-4)$
$T(n)= T(n-k) + 2T(n-(k+1))$
Substitute $n$ for $k$
$T(n)= T(n-n) + 2T(n-(n+1))$
$T(n)= T(0) + 2T(1)$
I am not sure where to go from here to find $T(n)$ (in $Θ$–notation) in terms of $n$.
|
Generating functions for the win. Let $T(x) = \sum_{n=0}^{\infty}T_nx^n = \sum_{n=1}^{\infty}T_nx^n$, since $T_0 = 0$.
Then:
$$\begin{split}
T(x) &= x + \sum_{n=2}^{\infty}T_nx^n \\
&= x + \sum_{n=2}^{\infty}(T_{n-1} + 2T_{n-2})x^n \\
&= x + \sum_{n=2}^{\infty}T_{n-1}x^n + \sum_{n=2}^{\infty}2T_{n-2}x^n
\end{split}$$
Let's rewrite those sums. The first one:
$$\begin{split}
\sum_{n=2}^{\infty}T_{n-1}x^n &= x\sum_{n=2}^{\infty}T_{n-1}x^{n-1} \\
&= x\sum_{n=1}^{\infty}T_nx^n \\
&= xT(x)
\end{split}$$
Likewise, the second sum works out to $2x^2T(x)$. Thus:
$$T(x) = x + xT(x) + 2x^2T(x)$$
So:
$$\begin{split}
T(x) &= \frac{x}{1 -x - 2x^2} = \frac{x}{(1-2x)(1+x)} \\
&= \frac{\frac{1}{3}}{1-2x} + \frac{-\frac{1}{3}}{1+x} \\
&= \frac13\sum_{n=0}^{\infty}2^nx^n - \frac13\sum_{n=0}^{\infty}(-1)^nx^n \\
&= \sum_{n=0}^{\infty}\left(\frac{2^n - (-1)^n}{3}\right)x^n
\end{split}$$
Which gives us our recurrence: $T_n = \left(\frac{2^n - (-1)^n}{3}\right) = \Theta(2^n)$
|
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Solving for n in the equation $\left ( \frac{1}{2} \right )^{n}+\left ( \frac{1}{4} \right )^{n}+\left ( \frac{3}{4} \right )^{n}=1$
Solving for $n$ in the equation $$\left ( \frac{1}{2} \right )^{n}+\left ( \frac{1}{4} \right )^{n}+\left ( \frac{3}{4} \right )^{n}=1$$
Can anyone show me a numerical method step-by-step to solve this?
Thanks
|
$\left ( \frac{1}{2} \right )^{n}+\left ( \frac{1}{4} \right )^{n}+\left ( \frac{3}{4} \right )^{n}= \frac{2^n}{4^n} +\frac{1^n}{4^n} +\frac{3^n}{4^n} = \frac{1^n+2^n+3^n}{4^n}=1,\: \therefore 1^n+2^n+3^n=4^n $
I don't think this can be solved analytically,
you just have to use trial and error with n
|
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How many strings of $8$ English letters are there (repetition allowed)? a) at least one vowel
b) start with $x$ and at least one vowel
c) start and end with $x$ and at least one vowel
I can solve them easily by considering $total-no$ $vowel$. So,
a) $26^8 -21^8$
b) $26^7 -21^7$
c) $26^6 -21^6$
But, can I try
a) choose $1$ vowel out of $5$ in $5$ ways and put in one of $8$ places, and then the remaining places $26^7$, which gives $5\cdot 8 \cdot 26^7$? What should be the logic thinking this way?
|
Your calculation has a lot of overcounting, whenever there is more than one vowel present.
If you really want to avoid (or, say, cross-check) the "negative space" method of simply excluding options with no vowels, you could perhaps sum through the possibilities of where the first vowel is:
*
*Vowel in first place: $5\cdot 26^7 $ ways
*First vowel in second place: $21\cdot 5\cdot 26^6 $ ways
*First vowel in third place: $21^2\cdot 5\cdot 26^5 $ ways
*etc.
Total
$$5\cdot 26^7 + 21\cdot 5\cdot 26^6 + 21^2\cdot 5\cdot 26^5 + 21^3\cdot 5\cdot 26^4 \\
\quad\quad+ 21^4\cdot 5\cdot 26^3 + 21^5\cdot 5\cdot 26^2 + 21^6\cdot 5\cdot 26 + 21^7\cdot 5 \\
= 5 \sum_{k=0}^7 21^k \,26^{7-k}$$
Not easy.
Another method is to calculate options for an exact number of vowels. This can be calculated by setting the pattern in one step, eg for three vowels:
$$BBABBAAB$$
which is ${8 \choose 3}$, then multiplying by the options for consonants and vowels respectively, so
$${8 \choose 3}5^3\,21^5$$
for exactly three vowels. Then add up all options of interest (or, if simpler, add up the non-qualifying options and subtract from total).
|
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Is there a way to find expansion of powers of multinomials without any coefficients? For example, $(a + b + c)^3 = a^3 + b^3 + c^3 + 3ab^2 + 3ac^2 + 3a^2b + 3a^2c + 3bc^2 + 3b^2c + 6abc$
Knowing the value of a, b and c, is there a way to find this without the coefficients i.e. $a^3 + b^3 + c^3 + ab^2 + ac^2 + a^2b + a^2c + bc^2 + b^2c + abc$ using previous values or some other way?
|
The answer to this question gives the formula you are looking for ($m=3$).
$$\frac{a^{m+2}}{(a-b)(a-c)}+\frac{b^{m+2}}{(b-a)(b-c)}+\frac{c^{m+2}}{(c-a)(c-b)}$$
|
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with inequality $\frac{1}{3a+5b+7c}+\frac{1}{3b+5c+7a}+\frac{1}{3c+5a+7b}\le\frac{\sqrt{3}}{4}$
let $a,b,c>0$, such $ab+bc+ac=1$,show that
$$\dfrac{1}{3a+5b+7c}+\dfrac{1}{3b+5c+7a}+\dfrac{1}{3c+5a+7b}\le\dfrac{\sqrt{3}}{4}$$
by Macavity C-S:with inequality $\frac{y}{xy+2y+1}+\frac{z}{yz+2z+1}+\frac{x}{zx+2x+1}\le\frac{3}{4}$
$$\dfrac{1}{3a+5b+7a}\le\dfrac{1}{2}\left(\dfrac{1}{6a+9b}+\dfrac{1}{b+7c}\right)$$
It suffices to show
$$\sum_{cyc}\left(\dfrac{1}{6a+9b}+\dfrac{1}{b+7c}\right)\le\dfrac{\sqrt{3}}{2}$$
|
Using Cauchy-Schwarz in a different way:
$$\begin{align}
2(3a+5b+7c) &= 15(a+b+c)-(9a+5b+c) \\
&= \sqrt{(118+107)(2+a^2+b^2+c^2)} - (9a+5b+c)\\
&\ge \sqrt{118\cdot2}+\sqrt{(9^2+5^2+1^2)(a^2+b^2+c^2)} -(9a+5b+c)\\
&\ge 2\sqrt{59}
\end{align}
$$
$$\implies \sum_{cyc} \frac1{3a+5b+7c} \le \frac3{\sqrt{59}} < \frac{\sqrt3}4$$
P.S. The maximum is in fact $\dfrac{\sqrt3}5$, though a simple way to show that eludes me for now.
|
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Finding the Range of a Trigonometric function The range of $$f(x)=3\cos^2x-8\sqrt3 \cos x\cdot\sin x+5\sin^2x-7$$is given by:(1)$[-7,7]$(2)$[-10,4]$(3)$[-4,4]$(4)$[-10,7]$
ANS: (2)
My Solution
The equation can be written as: $$3\cos^2x-8\sqrt3 \cos x\cdot \sin x+16\sin^2x-11\sin^2x-7 \\\implies (\sqrt3\cos x-4\sin x)^2-11\sin^2x-7$$
So let $y=\sqrt3\cos x+4\sin x$
$$-\sqrt{(\sqrt3)^2+4^2} \le y\le \sqrt{(\sqrt3)^2+4^2} \implies 19 \le y^2 \le 19 \implies y^2\in[0,19]$$
CASE 1: When $y^2=0$
$$f(x)=0-11\sin^2x-7 \text{ ,taking } \sin^2 x=0\text{, minimum value of }f(x)= -7$$
CASE 2: When $y^2=19$
$$f(x)=19-11\sin^2x-7 \text{ ,taking } \sin^2 x=1\text{, minimum value of }f(x)= 1$$
So my range is $y\in[-7, 1]$
where is the problem?
|
Let's do this.
\begin{align}
f(x)&=3\cos^2 x−8\sqrt3 \cos x\sin x + 5\sin^2x−7 =\\
&= 3(\cos^2x + \sin^2 x) - 4\sqrt3\cdot 2\sin x\cos x + 2\sin^2 x - 7=\\
&=3 - 4\sqrt 3\sin 2x + 2\sin^2 x - 7 = \\
&=-4 -4\sqrt 3\sin 2x + 1 - \cos 2x=\\
&=-\cos2x -4\sqrt 3\sin 2x-3.
\end{align}
Now we have
$$1\cdot\cos2x +4\sqrt 3\cdot\sin 2x = \sqrt{1^2 + (4\sqrt 3)^2}\sin(2x+\varphi)=7\sin(2x+\varphi),$$
where $\sin\varphi=1/7$, $\cos\varphi=4\sqrt3/7$. So you have
$$f(x)=-7\sin(2x+\varphi)-3,$$
and range is $[-7-3,7-3] = [-10, 4]$.
|
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Calculate $\lim_{x \to \infty} (\sqrt{x^2 + 3x} - \sqrt{x^2 + 1})$ without using L'Hospital's rule Question:
Calculate
$$\lim_{x \to \infty} (\sqrt{x^2 + 3x} - \sqrt{x^2 + 1})$$
without using L'Hospital's rule.
Attempted solution:
First we multiply with the conjugate expression:
$$\lim_{x \to \infty} (\sqrt{x^2 + 3x} - \sqrt{x^2 + 1}) = \lim_{x \to \infty} \frac{x^2 + 3x - (x^2 + 1)}{\sqrt{x^2 + 3x} + \sqrt{x^2 + 1}}$$
Simplifying gives:
$$\lim_{x \to \infty} \frac{3x - 1}{\sqrt{x^2 + 3x} + \sqrt{x^2 + 1}}$$
Breaking out $\sqrt{x^2}$ from the denominator and $x$ from the numerator gives:
$$\lim_{x \to \infty} \frac{x(3 - \frac{1}{x})}{x(\sqrt{1 + 3x} + \sqrt{1 + 1})} = \lim_{x \to \infty} \frac{3 - \frac{1}{x}}{\sqrt{1 + 3x} + \sqrt{2}}$$
The result turns out to be $\frac{3}{2}$, but unsure how to proceed from here.
|
I do not agree with your result after factorizing in the last step. I get
$$\frac{x(3-1/x)}{x\left(\sqrt{1+\frac{3}{x}}+\sqrt{1+\frac{1}{x^2}}\right)}=\frac{3-1/x}{\sqrt{1+\frac{3}{x}}+\sqrt{1+\frac{1}{x^2}}}\rightarrow 3/2$$
|
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How does one calculate: $\left(\frac{z}{2!}-\frac{z^3}{4!}+\frac{z^5}{6!}-\cdots\right)^2$ How does one calculate:
$$\left(\frac{z}{2!}-\frac{z^3}{4!}+\frac{z^5}{6!}-\cdots\right)^2$$
Is the best way to just take the first term times the following two, and the second two times the next two to see the pattern?
First term will contribute:
$$\left(\frac{z^2}{2!2!}-\frac{2z^4}{2!4!}+\frac{2z^6}{2!6!}-\cdots\right)$$
$$\left(\frac{z^2}{4}-\frac{z^4}{4!}+\frac{z^6}{6!}-\cdots\right)$$
Second term will contribute:
$$\left(-\frac{z^4}{4!}+\cdots\right)$$
This already seems wrong, so expanding via the first term seems good, and I hazard a guess that
$$\left(\frac{z}{2!}-\frac{z^3}{4!}+\frac{z^5}{6!}-\cdots\right)^2=\left(\frac{z^2}{4}-\frac{z^4}{4!}+\frac{z^6}{6!}-\cdots\right)$$
|
Here is another way to square the series without multiplying out the terms.
Since $\cos z = 1 - \dfrac{z^2}{2!} + \dfrac{z^4}{4!} - \dfrac{z^6}{6!} + \cdots$, we have $\dfrac{1-\cos z}{z} = \dfrac{z}{2!} - \dfrac{z^3}{4!} + \dfrac{z^5}{6!} - \cdots$.
Thus, $\left(\dfrac{z}{2!} - \dfrac{z^3}{4!} + \dfrac{z^5}{6!} - \cdots\right)^2 = \dfrac{(1-\cos z)^2}{z^2}$.
Now, note that $(1-\cos z)^2 = 1-2\cos z + \cos^2 z = 1-2\cos z + \dfrac{1+\cos 2z}{2} = \dfrac{3}{2} - 2\cos z + \dfrac{1}{2}\cos 2z$.
The Taylor series for $(1-\cos z)^2 = \dfrac{3}{2} - 2\cos z + \dfrac{1}{2}\cos 2z$ is given by:
$\dfrac{3}{2} - 2\displaystyle\sum_{n = 0}^{\infty}\dfrac{(-1)^nz^{2n}}{(2n)!} + \dfrac{1}{2}\sum_{n = 0}^{\infty}\dfrac{(-1)^n(2z)^{2n}}{(2n)!}$ $= \displaystyle\sum_{n = 1}^{\infty}\dfrac{(-1)^n(-2+2^{2n-1})z^{2n}}{(2n)!}$.
Now, divide by $z^2$ to get the series for $\dfrac{(1-\cos z)^2}{z^2}$.
EDIT: If you only need the first few terms of the series, then carefully multiplying out the first few terms of $\left(\dfrac{z}{2!} - \dfrac{z^3}{4!} + \dfrac{z^5}{6!} - \cdots\right)\left(\dfrac{z}{2!} - \dfrac{z^3}{4!} + \dfrac{z^5}{6!} - \cdots\right)$ will be easier. However, if you need all the terms of the series, the above method will be easier.
|
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How to retrieve the expression of $a^3+b^3+c^3$ in terms of symmetric polynomials? I recently had to find without any resources the expression of $a^3+b^3+c^3$ in terms of $a+b+c$, $ab+ac+bc$ and $abc$.
Although it's easy to see that $a^2+b^2+c^2=(a+b+c)^2-2(ab+ac+bc)$, I couldn't come up myself with $a^3+b^3+c^3=(a+b+c)^3-3(a+b+c)(ab+ac+bc)+3abc$.
Can someone explain to me how to retrieve this formula with only pen and paper, or how to memorize it for good ?
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Let $E$ be the expression equal to $a^3+b^3+c^3$ in terms of symmetric polynomials. Since the reduced expression has no mixed terms, you must have $E = (a+b+c)^3 + E'$, where
$$ E' = a^3+b^3+c^3 - (a+b+c)^3 = -3(a^2b +ab^2 +a^2 c +2abc +b^2c +ac^2 +bc^2) $$
To get all of the terms like $a^2b$, we must have $(a+b+c)(a^2+b^2+c^2)$ somewhere in $E'$; after calculating this, the final answer should be clear.
|
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Prove that $a^7 + b^7 + c^7 \geq a^4b^3 + b^4c^3 + c^4a^3$
Prove that $a^7 + b^7 + c^7 \geq a^4b^3 + b^4c^3 + c^4a^3$
Values $a,b,c$ are all positive reals. I tried Muirhead and a few AM $\geq$ GM.
This problem is equivalent to proving $a^4b^3 + b^4c^3 + c^4a^3 \geq c^2a^5 + b^2c^5+a^2b^5$.
|
Earth lended me it's energy. Without any assumption on the size of $a,b,c$:
$a^7 + a^7 + a^7 + a^7 + b^7 + b^7 + b^7 \geq 7a^4b^3$
$b^7 + b^7 + b^7 + b^7 + c^7 + c^7 + c^7 \geq 7b^4c^3$
$c^7 + c^7 + c^7 + c^7 + a^7 + a^7 + a^7 \geq 7c^4a^3$
adding up I get what I wanted
|
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Integral ${\large\int}_0^\infty\frac{dx}{\sqrt[4]{7+\cosh x}}$ How to prove the following conjectured identity?
$$\int_0^\infty\frac{dx}{\sqrt[4]{7+\cosh x}}\stackrel{\color{#a0a0a0}?}=\frac{\sqrt[4]6}{3\sqrt\pi}\Gamma^2\big(\tfrac14\big)\tag1$$
It holds numerically with precision of at least $1000$ decimal digits.
Are there any other integers under the radical except $7$ and $1$ that result in a nice closed form?
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Take the integral in the form $$I=\frac{1}{2^{1/4}}\int_{0}^{1}\frac{du}{\left(3u^{4}+u^{2}\right)^{1/4}\left(1-u^{2}\right)^{1/2}}=\frac{1}{2^{1/4}}\int_{0}^{1}\frac{du}{\left(3u^{2}+1\right)^{1/4}\left(1-u^{2}\right)^{1/2}\left(u^{2}\right)^{1/4}}
$$ then put $u^{2}=s
$ $$=\frac{1}{2^{5/4}}\int_{0}^{1}\frac{ds}{\left(3s+1\right)^{1/4}\left(1-s\right)^{1/2}s^{3/4}}
$$ and now put $s=1-t
$ $$=\frac{1}{2^{7/4}}\int_{0}^{1}\frac{dt}{\left(1-3t/4\right)^{1/4}\left(1-t\right)^{3/4}t^{1/2}}.
$$ Now recalling the identity $$\,_{2}F_{1}\left(a,b;c;z\right)=\frac{\Gamma\left(c\right)}{\Gamma\left(b\right)\Gamma\left(c-b\right)}\int_{0}^{1}\frac{t^{b-1}\left(1-t\right)^{c-b-1}}{\left(1-tz\right)^{a}}dt
$$ we have $$I=\frac{1}{2^{7/4}}\frac{\Gamma\left(\frac{1}{2}\right)\Gamma\left(\frac{1}{4}\right)}{\Gamma\left(\frac{3}{4}\right)}\,_{2}F_{1}\left(\frac{1}{4},\frac{1}{2};\frac{3}{4};\frac{3}{4}\right)
$$ and in this case it is possible calculate the exact value of the hypergeometric function (see the update in the Jack D'Aurizio's answer for reference) $$\,_{2}F_{1}\left(\frac{1}{4},\frac{1}{2};\frac{3}{4};\frac{3}{4}\right)=\frac{2\sqrt{2}}{3^{3/4}}
$$ and so $$I=\frac{\Gamma^{2}\left(\frac{1}{4}\right)}{6^{3/4}\sqrt{\pi}}.
$$ The result is not equal to $\sqrt[4]{6}\Gamma^{2}\left(\frac{1}{4}\right)/\left(2\sqrt{\pi}\right)
$ but I haven't found a mistake in my calculations.
|
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Points of intersection of two functions Through the following steps I found the x-coordinates of the intersection points of two functions:
$(x)= -x^{2}+3x+1\: and\: g(x)=3/x $
The numbers I found are x=(1, 2, 3)
But on the graph, one of the points has a negative x value, could you guys point me to anything I have missed in my calculations.
$-x^{2}+3x+1=3/x
\\x(-x^{2}+3x+1)=3
\\x=3
\\and
\\-x^{2}+3x+1=3
\\x^{2}-3x-1=-3
\\x^{2}-3x-1+3=-3+3
\\x^{2}-3x+2=0
\\(x-1)(x-2)=0
\\So
\\x=1 \:and\: x=2
$
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The first equation is a cubic, $ -(x+1)(x-1) (x-3) = 0, $
with roots $ x = \pm 1, 3 $ at points of intersection.
|
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Mathematical Reflections J339: Solving $\frac{x-1}{y+1} + \frac{y-1}{z+1} + \frac{z-1}{x+1} = 1$ in positive integers I have been trying to solve $\frac{x-1}{y+1} + \frac{y-1}{z+1} + \frac{z-1}{x+1} = 1$ in positive integers and I'm shaky on one specific step. I solved the problem through case-by-case analysis and I wonder if someone else has a better, more natural solution.
Observe that (2,2,2) is a trivial solution.
Assume that one variable, WLOG $x=1$. Then we need to solve $\frac{y-1}{z+1} + \frac{z-1}{2} = 1$ which after clearing denominators is equivalent to $2(y-1) + z^2-1 = 2z+2$ or $(z-3)(z+1)=-2(y-1)$. The quadratic is even whenever $z$ is odd so testing $z=3$ and $z=1$ shows that $(1,1,3)$ (and its permutations) are solutions.
This is my shaky step: Besides $(1,1,3)$, all other solutions have $x,y,z\geq2$.
I then verify that other than $(2,2,2)$ there are no solutions with all three of $x,y,z\geq2$.
Is this logic correct? Can I say that no other solutions involve 1 now that I have done casework on WLOG $x=1$?
Thanks in advance!
|
observe that the quadratic $(z-3)(z+1)$ has a minimum where $z=1$ and its value there is $-4$, so we get $-2(y-1)\ge -4$, that is $y\le 3$. since $y\ge 1$, we get $(z-3)(z+1)\le 0$, so $z\le 3$ too.
|
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Proving “The sum of $n$ consecutive cubes is equal to the square of the sum of the first $n$ numbers.” This site
states:
Example $\boldsymbol 3$. The sum of consecutive cubes. Prove this remarkable fact of arithmetic: $$1^3 +2^3 +3^3 +\ldots +n^3 =(1 +2 +3 +\ldots +n)^2.$$
“The sum of $n$ consecutive cubes is equal to the square of the sum of the first $n$ numbers.”
In other words, according to Example $1$: $$1^3 +2^3 +3^3 +\ldots +n^3 = \frac{n^2 (n+1)^2}{4}.$$
Should:
$$1^3 +2^3 +3^3 +\ldots +n^3 = \frac{n^2 (n+1)^2}{4}$$
not be:
$$1^3 +2^3 +3^3 +\ldots +n^3 = \frac{n^3 +(n + 1)^3}{2^3}$$
as everything in the left-hand side is cubed?
|
Well you have:
$$1^3+\ldots+n^3=\frac{n^2(n+1)^2}{4} $$
You can prove it easily by induction. Now your proposition could not work without even knowing the formula for the following reason:
$$\frac{n^3+(n+1)^3}{2}\text{ is equivalent to } n^3$$
Whereas I claim that:
$$\int_0^{n}x^3dx \leq \sum_{k=1}^nk^3\leq \int_0^{n+1}x^3dx$$
From which it follows that:
$$\sum_{k=1}^nk^3\text{ is equivalent to } \frac{n^4}{4}$$
|
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Solving the definite integral: $\int_0^1 x\sqrt{px + 1}\,dx$ with $p>0$?
How to integrate $$I = \int_0^1 x\sqrt{px + 1}\,dx$$ , $p>0, p\in\mathbb{R}$ ?
I tried this:
$$ t = \sqrt{px + 1} \implies x = \frac{t^2 - 1}{p}$$
$$ x = 0 \implies t = 1$$
$$ x = 1 \implies t = \sqrt{p+1}$$
so I have:
$$ I = \int_1^\sqrt{p+1} \frac{t^2 - 1}{p}t \,dt = \frac{1}{p}\int_1^\sqrt{p+1} t^3 - t \,dt = \frac{1}{p}\bigg[\frac{t^4}{4}\bigg\vert_1^\sqrt{p+1} - \frac{t^2}{2}\bigg\vert_1^\sqrt{p+1}\bigg]$$
$$ I = \frac{1}{p} \bigg[\frac{1}{4}(p^2 + 2p + 1 - 1) - \frac{1}{2}(p + 1 - 1) \bigg] = \frac{1}{p}\frac{1}{4}p^2 = \frac{1}{4}p $$
So for $p=4$, the integral should have the value $1$, but WA doesn't agree: http://wolfr.am/5p-i9uOA
I looked at it for quite a while but I can't locate the mistake. Since $p>0$, everything seems fine to me but apparently it isn't.
|
One can also integrate by parts. To that end, we have
$$\begin{align}
I=\int_0^1x\sqrt{1+px}dx&=\left.\left(\frac{2x}{3p}(1+px)^{3/2}\right)\right|_0^1-\frac{2}{3p}\int_0^1(1+px)^{3/2}dx\\\\
&=\frac{2}{3p}(1+p)^{3/2}-\frac{2}{3p}\left.\left(\frac{2}{5p}(1+px)^{5/2}\right)\right|_0^1\\\\
&=\frac{2}{3p}(1+p)^{3/2}-\frac{4}{15p^2}\left((1+p)^{5/2}-1\right)\\\\
&=\frac{2(3p-2)(1+p)^{3/2}+4}{15p^2}
\end{align}$$
|
{
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"url": "https://math.stackexchange.com/questions/1330334",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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|
How to prove that a root $z_2=\frac{-a-\sqrt{a^2+b^2}}{2b} <1$ I want to prove that that the integral $I= \int_{0}^{2π} \frac{1}{a+b \cos\theta}\,d \theta$, where $a>0$, $b>0$ and $a>b$, is equal to $ I=\dfrac{2πb}{\sqrt{a^2+b^2}}$ with the method of residues.
My approach is:
Let $z=e^{i\theta}$ be the parametrization of the circle; I will convert the complex integral to a complex integral on the circle $\gamma : \lvert z \lvert=1$
We have $\dfrac{dz}{d\theta}= ie^{i\theta}\Rightarrow dz=iz\,d \theta \Rightarrow \frac{1}{iz}\,dz=d\theta$
Also $\cos\theta = \dfrac{e^{i\theta}+e^{-i\theta}}{z} =\dfrac{z+ \bar{z}}{2}= \dfrac{z(z+ \bar{z})}{2z}=\dfrac{z^2+1}{2z}$
If I replace this quantity in the integral $Ι$, with easy calculations I have:
$$I = \int_{0}^{2π} \frac{2}{bz^2+2az+1}\,dz$$
So, with the factorization of denominator we have $\displaystyle I = \int_{0}^{2π} \frac{2}{(z-z_1)(z-z_2)}\,dz$ where $z_1=\dfrac{-a+\sqrt{a^2+b^2}}{2b}$ and $z_2=\dfrac{-a-\sqrt{a^2+b^2}}{2b}$
The discriminant is positive since we have $a>0$ and $b>0$ .
To continue with the method of residues and finally find the result, clearly I have to cancel a root from $z_1$ and $z_2$ which is not contained in the circle $\lvert z \lvert=1$.
Intuitively this root is $z_2$. How can I prove that this root is outside this circle? Any obvious condition, I think doesn't show anything.
|
Without loss of generality, we can divide everything by $b$, and so look at $c=a/b$:
$$ -\frac{a/b}{2} \pm \frac{\sqrt{(a/b)^2+1}}{2} = -\frac{c}{2} \pm \frac{\sqrt{1+c^2}}{2}. $$
The conditions give $c>1$. Now, $\sqrt{1+c^2}>\sqrt{c^2}=c$, so
$$ -\frac{c}{2} - \frac{\sqrt{1+c^2}}{2} < -\frac{c}{2}-\frac{c}{2} = -c < -1. $$
Meanwhile, to check the other one is inside the circle,
$$ \frac{\sqrt{1+c^2}-c}{2} = \frac{(\sqrt{1+c^2}-c)(\sqrt{1+c^2}+c)}{2(\sqrt{1+c^2}+c)} = \frac{1}{2(\sqrt{1+c^2}+c)}, $$
and this is obviously positive, but also $\sqrt{1+c^2}+c >1+c$, and so
$$ \frac{1}{2(\sqrt{1+c^2}+c)}<\frac{1}{2(1+c)}<1, $$
certainly.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1332518",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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|
If real numbers $x$ and $y$ satisfy the equation $\frac {2x+i}{y+i}= \frac {1+i\sin{\alpha}}{1-i\sin{3\alpha}}$ then quotient $\frac xy$ is equal to?
If real numbers $x$ and $y$ satisfy the equation $\frac {2x+i}{y+i}= \frac {1+i\sin{\alpha}}{1-i\sin{3\alpha}}$, then quotient $\frac xy$ is equal to?
Other conditions are ($\alpha \neq k\pi,\ \alpha \neq \frac \pi2+ k\pi,\ k\in\mathbb Z,\ i^2 = -1$). I tried this by making nominator the difference of squares but it does not lead me anywhere. Solution for this task is $-2-4\cos{2\alpha}$
|
Lethal weapon #2: Multiply and divide by the same thing:
$$
\frac{1+i\sin{\alpha}}{1-i\sin{3\alpha}}=\frac{i}{i}\frac{1+i\sin{\alpha}}{1-i\sin{3\alpha}}=\frac{i-\sin\alpha}{i+\sin 3\alpha}.
$$
Thus $2x=-\sin\alpha$ and $y=\sin3\alpha$, so
$$
\frac{x}{y}=-\frac{\sin\alpha}{2\sin3\alpha}=-\frac{1}{2+4\cos2\alpha}.
$$
|
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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|
Find $\lim\limits_{x \to 0}\frac {1}{x^3}\int_0^x \frac{t\ln (1+t)}{t^4+4}\mathrm{d}t $ without using L'hopital's rule Find $$\lim\limits_{x \to 0}\frac {1}{x^3}\int_0^x \frac{t\ln (1+t)}{t^4+4}\mathrm{d}t $$ without using L'hopital's rule. Using the rule, I got it as 1/12. What's a way to do it without L'hopital?
|
Let's first analyze the limit when $x \to 0^{+}$. If $0 < t < 1$ then we know that $$t - t^{2} = t(1 - t) < \frac{t}{1 + t} < \log(1 + t) < t$$ Multiplying by $t$ we get $$t^{2} - t^{3} < t\log(1 + t) < t^{2}\tag{1}$$ Also we can easily see that $$\frac{1}{4}\left(1 - \frac{t}{4}\right) < \frac{1}{t + 4} < \frac{1}{t^{4} + 4} < \frac{1}{4}\tag{2}$$ If $0 < t < 1$ then all the terms involved in the inequalities $(1)$ and $(2)$ are positive and hence we can multiply them together to get $$\left(t^{2} - t^{3}\right)\left(\frac{1}{4} - \frac{t}{16}\right) < \frac{t\log(1 + t)}{t^{4} + 4} < \frac{t^{2}}{4}$$ or $$\frac{t^{2}}{4} + o(t^{2}) < \frac{t\log(1 + t)}{t^{4} + 4} < \frac{t^{2}}{4}$$ Integrating the above between $0$ and $x$ where $0 < x < 1$ we get $$\frac{x^{3}}{12} + o(x^{3}) < \int_{0}^{x}\frac{t\log(1 + t)}{t^{4} + 4}\,dt < \frac{x^{3}}{12}$$ and dividing by $x^{3}$ we get $$\frac{1}{12} + o(1) < \frac{1}{x^{3}}\int_{0}^{x}\frac{t\log(1 + t)}{t^{4} + 4}\,dt < \frac{1}{12}$$ and thus by Squeeze theorem we get $$\lim_{x \to 0^{+}}\frac{1}{x^{3}}\int_{0}^{x}\frac{t\log(1 + t)}{t^{4} + 4}\,dt = \frac{1}{12}$$ A similar argument can be made for the case when $x \to 0^{-}$.
We don't need to use any advanced tools like Taylor series or L'Hospital's Rule and job is done via very simple inequalities and Squeeze theorem.
|
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"url": "https://math.stackexchange.com/questions/1334221",
"timestamp": "2023-03-29T00:00:00",
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|
How to solve this type of exercises $\sqrt{x^6+x^5-2x^3+O(x^2)}$ I have a simulation test with this type of exercise, asymptotic expansion:
$$\sqrt{x^6+x^5-2x^3+O(x^2)}$$
with
$$ x\rightarrow \infty$$
I have studied the theory of Landau's symbols but I have no idea about how to solve.
Can someone please explain me how to do that?
|
Write
$$\sqrt{x^6+x^5-2x^3+O(x^2)}=$$
$$=\sqrt{x^6+x^5-2x^3}+\sqrt{x^6+x^5-2x^3+O(x^2)}-\sqrt{x^6+x^5-2x^3}=$$
$$=\sqrt{x^6+x^5-2x^3}+\frac{O(x^2)}{\sqrt{x^6+x^5-2x^3+O(x^2)}+\sqrt{x^6+x^5-2x^3}}.$$
But $\sqrt{x^6+x^5-2x^3+O(x^2)}+\sqrt{x^6+x^5-2x^3}\sim 2x^3$, hence
$$\frac{O(x^2)}{\sqrt{x^6+x^5-2x^3+O(x^2)}+\sqrt{x^6+x^5-2x^3}}=O(x^{-1})$$
thus
$$\sqrt{x^6+x^5-2x^3+O(x^2)}=\sqrt{x^6+x^5-2x^3}+O(x^{-1})$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1334816",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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|
Understanding the solution to a basic annuity problem involving an unknown interest rate The following is the problem and the solution:
Before looking at the solution, here is how I approached the problem:
Let $X$ be the amount that each child receives. (i) and (ii) imply that $Xa_{20} = v^{20}\cdot{}2X\cdot\frac{1}{i},$ where $a_{20}$ is the present value of an annuity immediate and $\frac{1}{i}$ is the present value of a perpetuity immediate.
Let $T$ be the value of the estate. Then $Xa_{20} = \frac{1}{3}T$ and $v^{20}\cdot{}2X\cdot\frac{1}{i} = \frac{1}{3}T.$
We also know that $Xa_{20} + Xa_{20} + v^{20}\cdot{}2X\cdot\frac{1}{i} = T$.
At this point I didn't know how to solve for either the interest rate or the value of the estate.
The solution does not even appear to define the value of the estate. At the same time, I don't understand the sentence:
"Payment to each child for 20 years = $\frac{1}{3a_{20}}$." Could someone walk me through the solution and specifically the idea of taking the reciprocal?
Also, is there a way to make my solution attempt work?
|
First of all it is assumed that john´s estate is 1. It works also with other values. Thus all three, children and charity, receive $\frac{1}{3}$.
Now you use the formula for the present value. The present value have to be 1/3 for both children.
The present value for one child is
$C_0=r\cdot v\cdot \frac{1-v^n}{1-v}\Rightarrow\frac{1}{3}= r \cdot v\cdot\frac{1-v^n}{1-v}$
r are the annual payments.
$v=\frac{1}{1+i}\ $. i is the interest rate.
Solving the equation for r gives:
$r=\frac{1}{3v}\cdot\frac{1-v}{1-v^n}$
Simplifying $\frac{1-v}{v}$
$\frac{1-v}{v}=\frac{1-\frac{1}{1+i}}{\frac{1}{1+i}}=(1+i)\cdot (1-\frac{1}{1+i})=1+i-1=i $
Therefore it is $r=\frac{i}{3\cdot(1-v^n)}$
$n=20$
And for both children it is $\frac{2i}{3\cdot(1-v^{20})}$
The present value for a perpetuity is $C_0=\frac{r}{i}$.
$C_0=\frac{r^c}{i}$
The index c is for charity.
But it has to be discounted for 20 years, because it starts 20 years later.
$\frac{1}{3}=\frac{r^c}{i}\cdot v^{20}$
Solving for $r^c$
$r^c=C_0\cdot i=\frac{1}{3\cdot v^{20}}\cdot i$
Setting both terms equal:
$\frac{1}{3\cdot v^{20}}\cdot i=\frac{2i}{3\cdot(1-v^{20})}$
Multiplying both sides by 3 and dividing both sides by i:
$\frac{1}{ v^{20}}=\frac{2}{(1-v^{20})}$
Taking the reciprocals of both sides
$v^{20}=\frac{(1-v^{20})}{2}$
$v^{20}=\frac{1}{2}-\frac{v^{20}}{2} \quad |+\frac{v^{20}}{2}$
$\frac{v^{20}}{2}+v^{20}=\frac{1}{2}$
$\frac{3}{2}v^{20}=\frac{1}{2}$
$3\cdot v^{20}=1$
$v^{20}=\frac{1}{3}$
$v=\left( \frac{1}{3} \right)^{1/20}$
$v\approx 0.9465=\frac{1}{1+i}$
$\Rightarrow \frac{1}{0.9465}=1+i$
$i=\frac{1}{0.9465}-1\approx 0.0565=5.65\%$
|
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"timestamp": "2023-03-29T00:00:00",
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|
How to proceed with evaluating $\int\frac{dx}{\sqrt{9+4x^2}}$ and $\int\tan^2(3x)dx$
*
*$\displaystyle\int\frac{dx}{\sqrt{9+4x^2}}$
*$\displaystyle\int\tan^2(3x)dx$
For the first one i'm not sure if I did it correctly, here is what I did:
Let $2x=3\tan(t)$, so $x=\frac{3}{2}\tan(t)$ and $dx=\frac{3}{2}\sec^2(t)dt$. So by substitution,
$$\begin{align}\displaystyle\int\frac{dx}{\sqrt{9+4x^2}}&=\int\frac{\frac{3}{2}\sec^2(t)}{3\sec(t)}dt\\
&=\frac{1}{2}\int\sec(t)dt\\
&=\frac{1}{2}\left(\ln|\sec(t)+\tan(t)| + C\right)\\
&=\frac{1}{2}\left(\ln\bigg|\frac{\sqrt{9+4x^2}}{3}+\frac{2x}{3}\bigg|+C\right)\\
\end{align}$$
For the second one, i'm unable to proceed, what I did was
$\displaystyle\int\tan^2(3x)dx=\int\frac{\sin^2(3x)}{\cos^2(3x)}dx=\int\frac{\frac{1}{2}(1-\cos(6x)}{\frac{1}{2}(1+\cos(6x)}dx=\int\frac{(1-\cos(6x)}{(1+\cos(6x)}dx$
is this the right way to proceed?
Thanks
|
The first integral is correct.
The second is easier! Substute $3x=t$ and note that $\tan^2 t= \dfrac{1}{\cos^2t}-1$
|
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|
Substituting a value of sine function in a trigonometric equation I am trying to really understand trigonometric equations and I've stumbled upon a rather confusing example. Solve the following equation:
$\sin x= 2|\sin x|+ {\sqrt{3}}\cos x$
First step is to define the absolute $\sin x$:
$$|\sin x| =
\begin{cases}
\sin x & \sin x \in [0, 1]\\
-\sin x & \sin x \in [-1, 0)
\end{cases}
$$
Let's see what happens in the first case:
$$\sin x \in [0,1] \Rightarrow\, x \in {[2k{\pi}, (2k+1){\pi}]}$$
which means x is in the first or second quadrant.
Bringing everything to the right side:
$$\sin x+{\sqrt{3}}\cos x=0$$
Now we notice terms ${\sqrt{3}}$ with $\cos x$ and $1$ with $\sin x$ and do the following:
\begin{align*}
\sin x+{\sqrt{3}}\cos x& =0 \quad /:2\\
\frac{1}{2}\sin x+{\frac{\sqrt{3}}{2}}\cos x& = 0
\end{align*}
Now, the goal is to try and substitute integer terms with trigonometric functions they are solutions to, in order to apply one of the trigonometric identities for solving the problem. Let's aim at the following formula: $\sin(x+y)=\sin x\cos y + \sin y\cos x$ .We have several options:
*
*$\frac {1}{2} $is value of $\cos$ for either $x=\frac {\pi}{3} + 2k\pi$ OR $x=\frac {5\pi}{3} + 2k\pi$. Of all of these angles, only $x=\frac {\pi}{3} $ falls under our domain.
*${\frac {\sqrt{3}}{2}} $is value of $\sin$ for either $x=\frac {\pi}{3} + 2k\pi$ OR $x=\frac {2\pi}{3} + 2k\pi$. Out of all of these angles, there are two of them that fall under our domain. $x=\frac {\pi}{3}$ and $x=\frac {2\pi}{3}$.
This is where the problem is. I feel I should consider both of the $\sin$ values and solve two different versions of the equation. The first one is by substituting $\frac {1}{2} = \cos\frac {\pi}{3}$ and $\frac {\sqrt{3}}{2} = \sin\frac {\pi}{3}$, which is easily solved using the identity mentioned above. The second case is when substituting $\frac {\sqrt{3}}{2} = \sin\frac {2\pi}{3}$. This does not conform to any trig. identity. I tried treating $\sin\frac {2\pi}{3}$ as $\sin$ of double angle, but ended up at the beginning. I typed this equation into Symbolab and examined their step-by-step solution. It turns out that at some point they divided the equation with $\cos x$. Even when $\cos x$ is definitely not zero, I was taught dividing a trigonometric equation with anyting other than integers was a risky move very likely to result in a loss of solutions. So, my questions are:
Should you just ignore possible angles that won't make you a simple trigonometric identity?
Can you divide trigonometric equation with some trigonometric function and when?
I hope I explained myself clearly enough. Thank you in advance.
|
For simplicity, let's find the solutions in $[0,2\pi)$.
1) If $\sin x\ge0$, we get $\sin x=2\sin x+\sqrt{3}\cos x$, which yields $-\sin x=\sqrt{3}\cos x$.
$\hspace{.2 in}$Dividing by $-\cos x$ gives $\tan x=-\sqrt{3}$, so $x=\pi-\frac{\pi}{3}=\color{red}{\frac{2\pi}{3}}$
$\hspace{.2 in}$(since $\sin x>0, \tan x<0\implies x$ is in Quadrant II, with reference angle $\frac{\pi}{3})$.
2) If $\sin x<0$, we get $\sin x=-2\sin x+\sqrt{3}\cos x$, which yields $3\sin x=\sqrt{3}\cos x$.
$\hspace{.2 in}$Dividing by $3\cos x$ gives $\tan x=\frac{1}{\sqrt{3}}$, so $x=\pi+\frac{\pi}{6}=\color{red}{\frac{7\pi}{6}}$
$\hspace{.2 in}$(since $\sin x<0, \tan x>0\implies x$ is in Quadrant III, with reference angle $\frac{\pi}{6})$.
|
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|
Why doesn't using the approximation $\sin x\approx x$ near $0$ work for computing this limit? The limit is
$$\lim_{x\to0}\left(\frac{1}{\sin^2x}-\frac{1}{x^2}\right)$$
which I'm aware can be rearranged to obtain the indeterminate $\dfrac{0}{0}$, but in an attempt to avoid L'Hopital's rule (just for fun) I tried using the fact that $\sin x\approx x$ near $x=0$. However, the actual limit is $\dfrac{1}{3}$, not $0$.
In this similar limit, the approximation reasoning works out.
|
Since you already received good answers, let me add another point of view.
Rewriting the expression
$$A=\frac1{\sin^2(x)}-\frac1{x^2}=\frac{x^2-\sin^2(x)}{x^2\sin^2(x)}$$ and knowing that, close to $0$, $\sin(x)\approx x$, then the denominator looks like $x^4$ that is to say that, if the limit exist the numerator should also be developed at least up to order $x^4$; again, since $\sin(x)\approx x$, then $\sin(x)$ should be developed for at least one more term.
So, using $\sin(x)=x-\frac{x^3}{6}+O\left(x^4\right)$, after simplication, the numerator is just $\frac{x^4}{3}+\cdots$ and the denominator is $x^4+\cdots$; so the limit $\frac 13$.
If we needed more than the limit (say for example how it is approached), more terms would be required. Using $\sin(x)=x-\frac{x^3}{6}+\cdots$ and using it for both numerator and denominator, we should get $$A\approx \frac{x^2-\left(x-\frac{x^3}{6}\right)^2}{x^2 \left(x-\frac{x^3}{6}\right)^2}$$ which, after simplication and long division would give $$A\approx \frac{1}{3}+\frac{x^2}{12}+\cdots$$
Using instead $\sin(x)=x-\frac{x^3}{6}+\frac{x^5}{120}+\cdots$ and using it for both numerator and denominator, and doing the same as before, we should get $$A\approx \frac{1}{3}+\frac{x^2}{15}+\cdots$$ which is slightly different.
To illustrate the above, I suggest you plot on the same graph the three functions $y_1=\frac1{\sin^2(x)}-\frac1{x^2}$ , $y_2=\frac{1}{3}+\frac{x^2} {12}$, $y_3=\frac{1}{3}+\frac{x^2}{15}$ for $-\frac 12 \leq x\leq \frac 12$ . This would be better than a long and tedious speech.
|
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|
Volume of the intersection of two cylinders I have two infinite cylinders of unit radius
in $\mathbb{R}^3$, whose axes are skew lines.
Say that the axis of one is centered on the $x$-axis, and the axis of the
other is determined by the two points $a$ and $b$. Is there a formula for the
volume of their intersection, as a function of $a$ and $b$?
|
Let
*
*$r = 1$ be the common radii of the two cylinders.
*$2d$ be the nearest distance between the two axis of the cylinders.
*$\alpha = 2\beta$ be the angle between the two axis of the cylinders.
Choose the coordinate system so that the axis of the cylinders pass through
$(0, \pm d, 0)$ with tangent vectors $( \pm\sin\beta, 0, \cos\beta)$ respectively. It is not hard to see the two cylinders are give by.
$$\mathcal{C}_{\pm} = \left\{ (x,y,z) \in \mathbb{R}^3 : ( x\cos\beta \mp z\sin\beta )^2 + (y \mp d)^2 \le r^2 \right\}$$
If we intersect $\mathcal{C}_{\pm}$ with a plane of constant $z$, the intersections will be two axis aligned ellipses with semi major axis $\frac{r}{\cos\beta}$ in the $x$-direction and semi minor axis $r$ in the $y$-direction.
Introduce new coordinates
$$(u,v,w) = (x\cos\beta, y, z\sin\beta) \quad\iff\quad
(x,y,z) = \left(\frac{u}{\cos\beta}, v, \frac{w}{\sin\beta}\right)
$$
The intersections of $\mathcal{C}_{\pm}$ with a plane of constant $w$ become two circles of radii $r$.
$$(u \mp w)^2 + (v \mp d)^2 \le r^2$$
and these two circles intersect when and only when $d^2 + w^2 \le r^2$.
When they intersect, let $\theta = \cos^{-1}\left(\frac{\sqrt{d^2 + w^2}}{r}\right)$. In the diagram below, the intersection is the one bounded by the two circular arc in red. Its area is equal to
$$\begin{align}
& \verb/Area/(\text{sector } ABD) + \verb/Area/(\text{sector } CDB)
- \verb/Area/(\text{rhombus } ABCD)\\
= & 2\verb/Area/(\text{sector } ABD) - 4\verb/Area/(\text{triangle } OAB)\\
= & 2\left(\frac12 r^2(2\theta)\right) - 4\left(\frac12 (r\sin\theta)(r\cos\theta)\right)\\
= & r^2(2\theta - \sin(2\theta))
\end{align}
$$
$\hspace1.25in$
Let $\displaystyle\;\theta_0 = \cos^{-1}\left(\frac{d}{r}\right)\;$,
the volume we want becomes
$$\begin{align}\verb/Volume/
&= \overbrace{\frac{1}{\cos\beta\sin\beta}}^{\color{blue}{\text{Jacobian}}}
\int_{-\sqrt{r^2 - d^2}}^{\sqrt{r^2 - d^2}} r^2 (2\theta - \sin(2\theta)) d w\\
&= \frac{-4r^3}{\sin\alpha} \int_{0}^{\theta_0} (2\theta - \sin(2\theta)) d\sqrt{\cos^2\theta - \cos^2\theta_0}\\
\color{blue}{\text{ integrate by part } \rightarrow}
&= \frac{16 r^3}{\sin\alpha}\int_0^{\theta_0} \sin^2\theta \sqrt{\cos^2\theta - \cos^2\theta_0} d\theta\\
\color{blue}{ m = \sin^2\theta_0 \rightarrow}
&= \frac{16 r^3}{\sin\alpha}\int_0^\sqrt{m} \frac{\sin^2\theta}{\cos\theta}\sqrt{m - \sin\theta^2} d\sin\theta\\
\color{blue}{ \sin\theta = \sqrt{m} t \rightarrow}
&= \frac{16 r^3 m^2}{\sin\alpha}\int_0^1 t^2\sqrt{\frac{1-t^2}{1-mt^2}} d t\\
&= \frac{16 r^3 m^2}{\sin\alpha}\int_0^1 \frac{t^2 - t^4}{\sqrt{(1-t^2)(1-mt^2)}} dt
\end{align}
$$
Let $Q(t) = (1-t^2)(1-mt^2)$, it is easy to check
$$\frac{d}{dt}\left(t \sqrt{Q(t)}\right)
= \frac{1}{\sqrt{Q(t)}}\left(3m t^4 - 2(m+1) t^2 + 1\right)$$
We can use this to transform the last integral as
$$\begin{align}\verb/Volume/
&= \frac{16 r^3 m}{3\sin\alpha}\int_0^1 \frac{3m t^2 - 2(m+1) t^2 + 1}{\sqrt{Q(t)}} dt\\
&= \frac{16 r^3}{3\sin\alpha}\int_0^1 \frac{(2-m)(1-mt^2) - 2(1-m)}{\sqrt{Q(t)}} dt
\end{align}
$$
Compare this with the definition of the complete
elliptic integrals of the $1^{st}$ and $2^{nd}$ kind.
$$\begin{align}
K(m)
&= \int_0^{\pi/2} \frac{d\theta}{\sqrt{1-m\sin^2\theta}}
= \int_0^1 \frac{dt}{\sqrt{(1-mt^2)(1-t^2)}}\\
E(m) &= \int_0^{\pi/2} \sqrt{1-m\sin^2\theta} d\theta
= \int_0^1 \sqrt{\frac{1-mt^2}{1-t^2}} dt
\end{align}
$$
We get
$$\verb/Volume/ = \frac{16 r^3}{3\sin\alpha}\left((2-m)E(m) - 2(1-m)K(m)\right)$$
As a sanity check, when $d \to 0$, $m \to 1^{-}$, we have
$$(1-m)K(m) \to 0\quad\text{ and }\quad (2-m)E(m) \to E(1) = 1$$
This leads to a familiar result for the volume of intersection when the axis intersect.
$$\verb/Volume/_{d = 0} = \frac{16r^3}{3\sin\alpha}$$
|
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"language": "en",
"url": "https://math.stackexchange.com/questions/1338708",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "12",
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|
Minimum value of cosA+cosB+cosC in a triangle ABC I have used jensen's inequality but couldn't move on.
|
$$
\begin{align}
\sqrt[3]{\sin(\frac{A}{2}) \sin(\frac{B}{2}) \sin(\frac{C}{2})}
&
\le \frac{\sin(\frac{A}{2}) + \sin(\frac{B}{2}) + \sin(\frac{C}{2}) }{3} \text{ (AM-GM inequality)}\\
&
\le \sin \left( \frac{ \frac{A}{2} + \frac{B}{2} + \frac{C}{2}}{3} \right) \text{ (Jensen's inequality)}\\
&
= \sin(\frac{\pi}{6})\\
&
= \frac{1}{2}
\end{align}
$$
$$
0 \lt \sin(\frac{A}{2}) \sin(\frac{B}{2}) \sin(\frac{C}{2}) \le \frac{1}{8}
$$
since
$$
\cos A+\cos B+\cos C=1+4\sin(\frac{A}{2}) \sin(\frac{B}{2}) \sin(\frac{C}{2})
$$
so we have:
$$
1 \lt \cos A+\cos B+\cos C \le \frac{3}{2}
$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1340174",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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|
Alternative Quadratic Formula Well the formula for solving a Quadratic equation is :
$$\text{If }\space ax^2+bx+c=0$$
then
$$x=\dfrac{-b \pm \sqrt{b^2 -4ac} }{2a}$$
But looking at this : [Wolfram Mathworld] (And also in other places)
They give An Alternate Formula:
$$x=\dfrac{2c}{-b \pm \sqrt{b^2 -4ac}}$$
How does one get this?
Also in the first formula (the one we know) , $a \neq 0$ ... but here is it still the case?
Please help, Thanks!
|
I would like to add some point to answer posted user @mixedmath.
For equation
$$
ax^2 + bx + c = 0
$$
roots are
$$
x = \frac {-b\ \pm \ \sqrt {b^2 - 4ac} }{2a}
$$
Lets put
$$
y = \frac {1}{x}
$$
Then we get converted equation as,
$$
cy^2 + by + a = 0
$$
Equivalent roots are,
$$
y = \frac {-b\ \mp \ \sqrt {b^2 - 4ac} }{2c}
$$
And thus if you invert to get values of $x$ then those are,
$$
x = \frac{1}{y} = \frac{2c}{-b \mp \sqrt{b^2 -4ac}}
$$
Do note the difference between the signs $\pm$ and $\mp$.
You can verify that by trying to equate both sides as follows:
$$
\frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{2c}{-b \mp \sqrt{b^2 - 4ac}}
$$
Lets take one sign at a time. Lets choose $+$ from left side, this $-$ from right. It becomes,
$$
\frac{-b + \sqrt{b^2 - 4ac}}{2a} = \frac{2c}{-b - \sqrt{b^2 - 4ac}}
$$
$$
(-b)^2 - (\sqrt{b^2 -4ac})^2 = 4ac
$$
$$
b^2 - b^2 + 4ac = 4ac
$$
$$
4ac = 4ac
$$
Thus things are in correct place.
Lets verify with one quick example.
Consider a quadratic equation as follows,
$$
f(x) = (x-3)(2x-5) = 2x^2 - 11 x + 15 = 0
$$
According to first formula we get values as follows,
$$
x = \frac{11 \pm 1}{4}
$$
$$
\{x_+, x_-\} = \{3, \frac{5}{2}\}
$$
And with second formula we get,
$$
x = \frac{30}{11 \mp 1}
$$
$$
\{x_+, x_-\} = \{\frac{5}{2}, 3\}
$$
|
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"language": "en",
"url": "https://math.stackexchange.com/questions/1340267",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"answer_id": 2
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|
Given primitive solution to $\frac{1}{a}+\frac{1}{b}=\frac{1}{c}$, show $a+b$ is a perfect square If $a,b,c$ are positive integers and
$\gcd(a,b,c)$ is $1$. Given that $\frac{1}{a} + \frac{1}{b} = \frac{1}{c}$ then prove that $a+b$ is a perfect square.
I was trying to get something useful from the information given in the question but was unable to get that.
|
$$
\frac1a+\frac1b=\frac1c\implies ac+bc=ab
$$
$a\mid bc$. Therefore, each prime that divides $a$ must divide one and only one of $b$ or $c$ (since $(a,b,c)=1$). Therefore, $a=(a,b)(a,c)$. Similarly, $b=(a,b)(b,c)$ and $c=(a,c)(b,c)$.
Thus,
$$
a+b=\frac{ab}c=\frac{(a,b)(a,c)(a,b)(b,c)}{(a,c)(b,c)}=(a,b)^2
$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1341162",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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|
Need hint to solve a nasty integral.
Let $f(x)=\frac{x+2}{2x+3}$, $x>0$. If
$$\int \left( \frac{f(x)}{x^2} \right)^{1/2}dx=\frac{1}{\sqrt{2}}g \left(\frac{1+\sqrt{2f(x)}}{ 1-\sqrt{2f(x)}} \right) -\sqrt{\frac{2}{3}}h \left(\frac{\sqrt{3f(x)}+\sqrt{2}}{\sqrt{3f(x)}-\sqrt{2} } \right)+C$$
where $C$ is the constant of integration, then find $g(x)$ and $h(x)$.
I tried solving it like
$$\int \left( \frac{f(x)}{x^2} \right)^{1/2}dx=\int \sqrt{\frac{x+2}{2x^3+3x^2}}dx$$ but it leads us to nowhere.
Any hint will be of great help.
|
if you find this integral, you can
Hint: Let
$$\sqrt{\dfrac{x+2}{2x+3}}=t,\Longrightarrow x=\dfrac{3t^2-2}{1-2t^2}$$
$$-\int \dfrac{1-2t^2}{3t^2-2}\cdot t\cdot\dfrac{2t}{(2t^2-1)^2}dt=\int\dfrac{2t^2}{(3t^2-2)(2t^2-1)}dt$$
and Note
$$\dfrac{2t^2}{(3t^2-2)(2t^2-1)}=\dfrac{4}{3t^2-2}-\dfrac{2}{2t^2-1}$$
maybe as this result you can some calculation have your form
|
{
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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|
Find the last two digits of $33^{100}$
Find the last two digits of $33^{100}$
By Euler's theorem, since $\gcd(33, 100)=1$, then $33^{\phi(100)}\equiv 1 \pmod{100}$. But $\phi(100)=\phi(5^2\times2^2)=40.$
So $33^{40}\equiv 1 \pmod{100}$
Then how to proceed?
With the suggestion of @Lucian:
$33^2\equiv-11 \pmod{100}$ then $33^{100}\equiv(-11)^{50}\pmod{100}\equiv (10+1)^{50}\pmod{100}$
By using the binomial expansion, we have:
$33^{100}\equiv (10^{50}+50\cdot 10^{49}+ \cdots + 50\cdot 10+1)\pmod{100}$
$\implies 33^{100}\equiv (50\cdot 10+1)\pmod{100}\equiv 01 \pmod{100}$
|
$$(33)^{100}\equiv(33)^{5\cdot5\cdot4}\equiv(93)^{5\cdot4}\equiv(93)^4\equiv 01\pmod{100}$$
|
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"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "6",
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"answer_id": 3
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|
Compute $(1 - \omega + \omega^2)(1 + \omega - \omega^2)$ where $\omega^3 = 1$ If $\omega^3 = 1$ and $\omega \neq 1$, then compute $(1 - \omega + \omega^2)(1 + \omega - \omega^2)$
I'm pretty lost, I don't really know where to start.
Thanks
|
$$
\begin{align}
(1-\omega+\omega^2)(1+\omega-\omega^2)
&=1-\omega^2+2\omega^3-\omega^4\\
&=1-\omega^2+2-\omega\\
&=4-(1+\omega+\omega^2)\\
&=4
\end{align}
$$
The last step is true since $(1+\omega+\omega^2)\overbrace{(1-\omega)}^{\text{not $0$}}=1-\omega^3=0$.
|
{
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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|
Why are the coefficients always equal? Take the equation $ax^{2} + bx + c = 3x^{2} + 4x + 53$.
Why is it always true that $a = 3, b = 4$ and $c = 53$?
I've seen many examples like this where the coefficients are equated, and was just wondering why that is always true.
|
Suppose that $ax^{2} + bx + c = 3x^{2} + 4x + 53$ for all $x$, or $ax^{2} + bx + c - (3x^{2} + 4x + 53) = 0$ for all $x$.
$ax^{2} + bx + c - (3x^{2} + 4x + 53) = 0$ is a polynomial with at most degree $2$ so by the fundamental theorem of algebra it has at most $2$ roots if it is not the zero polynomial.
But $ax^{2} + bx + c - (3x^{2} + 4x + 53) = 0$ has infintely many roots (it is zero for every value of $x$), so it must be the zero polynomial. Then we have $a=3$, $b=4$ and $c=53$.
$$ax^{2} + bx + c - (3x^{2} + 4x + 53) = (a-3)x^2+(b-4)x+(c-53)$$
Every coefficient must be equal to zero for it to be the zero polynomial, thus $a-3=0$, and $a=3$. Similiarly, one has $b=4$ and $c=53$.
|
{
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"url": "https://math.stackexchange.com/questions/1344347",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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|
Complex numbers - roots of unity
Let $\omega$ be a complex number such that $\omega^5 = 1$ and $\omega \neq 1$. Find
$$\frac{\omega}{1 - \omega^2} + \frac{\omega^2}{1 - \omega^4} + \frac{\omega^3}{1 - \omega} + \frac{\omega^4}{1 - \omega^3}.$$
I have tried adding the first two and the second two separately, then adding those sums but how do I get a numerical value as the answer?
Thanks
|
HINT : $$\frac{\omega}{1-\omega^2}+\frac{\omega^2}{1-\omega^4}+\frac{\omega^3}{1-\omega}+\frac{\omega^4}{1-\omega^3}$$
$$=\frac{\omega}{1-\omega^2}+\frac{\omega^2}{1-\omega^4}+\frac{\omega^7}{\omega^4-\omega^5}+\frac{\omega^6}{\omega^2-\omega^5}$$
|
{
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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|
Definite integral with logarithm and arctangent inside of arctangent
How to prove $$\int_0^1 \left[ \frac{2}{\pi }\arctan \left(\frac 2 \pi \arctan \frac{1}{x} + \frac{1}{\pi }\ln \frac{1 + x}{1 - x}\right) - \frac{1}{2} \right]\frac{\mathrm{d}x} x = \frac{1}{2} \ln \left( \frac \pi {2\sqrt 2 } \right).$$
I have tried let $t=\frac1x$, but it seems no use! Could you help me to solve it?
|
One may show that this integral is equivalent to the integral in this problem as follows:
First, rewrite
$$\frac12 \log{\left ( \frac{1+x}{1-x} \right )} = \tanh^{-1}{x}$$
Then note that
$$\tan^{-1}{\left ( \frac1{x} \right )} = \frac{\pi}{2} - \tan^{-1}{x} $$
The integrand is then equal to
$$\left [\frac{2}{\pi} \tan^{-1}{\left (\frac{2}{\pi} \left (\frac{\pi}{2} - \tan^{-1}{x} + \tanh^{-1}{x} \right ) \right )}-\frac12 \right ] \frac1{x}$$
which is equal to
$$\left [\frac{2}{\pi} \tan^{-1}{\left (1+\frac{2}{\pi} \left ( \tanh^{-1}{x} - \tan^{-1}{x} \right ) \right )}-\frac12 \right ] \frac1{x}$$
Now, let
$$\tan^{-1}{(1+y)} = \frac{\pi}{4} + \tan^{-1}{w} $$
Then it is straight forward to show that
$$y = \frac{1+w}{1-w} - 1 = \frac{2 w}{1-w} \implies w=\frac{y}{y+2} $$
With $y=(2/\pi) (\tanh^{-1}{x} - \tan^{-1}{x} )$, we have
$$\frac{2}{\pi} \tan^{-1}{\left (1+\frac{2}{\pi} \left ( \tanh^{-1}{x} - \tan^{-1}{x} \right ) \right )}-\frac12 = \frac{2}{\pi} \tan^{-1}{\left (\frac{\tanh^{-1}{x} - \tan^{-1}{x} }{\tanh^{-1}{x} - \tan^{-1}{x} +\pi} \right )} $$
Thus, the integral in question is equal to
$$\frac{2}{\pi} \int_0^1 \frac{dx}{x} \tan^{-1}{\left (\frac{\tanh^{-1}{x} - \tan^{-1}{x} }{\tanh^{-1}{x} - \tan^{-1}{x} +\pi} \right )} $$
The evaluation of this integral is detailed in this arXiv paper.
|
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|
Closed form for a binomial series I am wondering if any knows how to compute a closed form for the following two series.
*
*$$\sum_{m=1}^{n}\frac{(-1)^m}{m^2}\binom{2n}{n+m}$$
*$$\sum_{m=1}^{n}\frac{(-1)^m}{m^4}\binom{2n}{n+m}$$
Mathematica gave a result in terms of the Hypergeometric function, which I don't understand. I was wondering if anyone can illustrate how to express these two series in a nice closed form, without using the Hypergeomeyric function. I was thinking along the line using a combinatorial identity or method, but I having problems figuring it out.
|
The first expression:
$\begin{align}\sum\limits_{m=1}^{n}\frac{(-1)^m}{m^2}\binom{2n}{n+m}
&= \binom{2n}{n}\sum\limits_{m=1}^{n}\frac{(-1)^m}{m^2}\binom{n}{m}\frac{m!n!}{(m+n)!}\\
&= \binom{2n}{n}\sum\limits_{m=1}^{n}\frac{(-1)^m}{m}\binom{n}{m}\frac{(m-1)!n!}{(m+n)!}\\
&= \binom{2n}{n}\sum\limits_{m=1}^{n}\frac{(-1)^m}{m}\binom{n}{m}\int_0^1 x^{m-1}(1-x)^n\,dx\\
&= \binom{2n}{n}\int_0^1\int_0^{x}\sum\limits_{m=1}^{n}\binom{n}{m}(-1)^my^{m-1} \frac{(1-x)^n}{x}\,dy\,dx\\
&= \binom{2n}{n}\int_0^1\int_0^{x}\frac{(1-y)^n -1 }{y}. \frac{(1-x)^n}{x}\,dy\,dx\\
&= \binom{2n}{n}\int_0^1 \frac{(1-x)^n}{x}\int_0^{x}\frac{(1-y)^n -1 }{y} \,dy\,dx\\
&= \binom{2n}{n}\int_0^1 \frac{x^n}{1-x}\int_0^{1-x}\frac{(1-y)^n -1 }{y} \,dy\,dx\\
&= -\binom{2n}{n}\int_0^1 \frac{x^n}{1-x}\int_0^{1-x}\frac{1-(x+y)^n}{1-(x+y)} \,dy\,dx\\
&= -\binom{2n}{n}\int_0^1 \frac{x^n}{1-x}\left[\sum\limits_{m=1}^{n}\frac{(x+y)^m}{m}\right]_0^{1-x}\,dx\\
&= -\binom{2n}{n}\int_0^1 \frac{x^n}{1-x}\sum\limits_{m=1}^{n}\frac{1-x^m}{m}\,dx\\
&= -\binom{2n}{n}\sum\limits_{m=1}^{n}\frac{1}{m}\sum\limits_{k=1}^{m}\frac{1}{n+k}= -\binom{2n}{n}\left(\sum\limits_{m=1}^{n}\frac{H_{m+n}}{m} - H_n^2\right)\end{align}$
The second one looks like a troublesome calculation to make.
$$\sum\limits_{m=1}^{n}\frac{(-1)^m}{m}\binom{2n}{n+m} = -\binom{2n}{n}(H_{2n} - H_n)$$
With @Chris's sis's help in chat I figured out how to simplify the expression further:
$$\begin{align}\sum\limits_{m=1}^{n}\frac{H_{m+n}}{m} &= \sum\limits_{m=1}^{n}\frac{H_m}{m}+\sum\limits_{m=1}^{n}\sum\limits_{k=1}^{n} \frac{1}{m(k+m)}\\&= \frac{1}{2}(H_n^{(2)}+H_n^2) + \frac{1}{2}\sum\limits_{m=1}^{n}\sum\limits_{k=1}^{n} \left(\frac{1}{m(k+m)}+\frac{1}{k(k+m)}\right)\\&= \frac{1}{2}(H_n^{(2)}+H_n^2) + \frac{1}{2}\sum\limits_{m=1}^{n}\sum\limits_{k=1}^{n} \frac{1}{mk}\\&= H_n^2 + \frac{1}{2}H_n^{(2)} \tag{1} \end{align}$$
Hence, $\displaystyle \sum_{m=1}^{n}\frac{(-1)^m}{m^2}\binom{2n}{n+m} = -\frac{1}{2}\binom{2n}{n}H_n^{(2)}$
Edit: For the second one, a short computation similar to one above shows:
$$\sum\limits_{m=1}^{n}\frac{(-1)^m}{m^4}\binom{2n}{n+m} = -\sum\limits_{m=1}^{n}\sum\limits_{j=1}^{m}\sum\limits_{k=1}^{j} \frac{H_{k+n} - H_n}{mjk}$$
Couldn't figure out a more direct line of computation, used differencing to complete the proof:
Call the triple sum $\displaystyle F(n) = \sum\limits_{m=1}^{n}\sum\limits_{j=1}^{m}\sum\limits_{k=1}^{j} \frac{H_{k+n} - H_n}{mjk}$
Then,
$\begin{align}F(n+1) - F(n) &= \sum\limits_{m=1}^{n+1}\sum\limits_{j=1}^{m}\sum\limits_{k=1}^{j} \frac{H_{k+n+1} - H_{n+1}}{mjk} - \sum\limits_{m=1}^{n}\sum\limits_{j=1}^{m}\sum\limits_{k=1}^{j} \frac{H_{k+n} - H_n}{mjk}\\
&= \sum\limits_{m=1}^{n+1}\sum\limits_{j=1}^{m}\sum\limits_{k=1}^{j} \frac{H_{k+n} - H_n - \frac{k}{(n+1)(n+k+1)} }{mjk} - \sum\limits_{m=1}^{n}\sum\limits_{j=1}^{m}\sum\limits_{k=1}^{j} \frac{H_{k+n} - H_n}{mjk}\\
&= \frac{1}{n+1}\sum\limits_{j=1}^{n+1}\sum\limits_{k=1}^{j} \frac{H_{k+n+1} - H_{n+1}}{jk} - \sum\limits_{m=1}^{n}\sum\limits_{j=1}^{m}\sum\limits_{k=1}^{j} \frac{k}{mjk(n+1)(n+k+1)}\\
&= \frac{1}{n+1}\left(\sum\limits_{j=1}^{n+1}\sum\limits_{k=1}^{j} \frac{H_{k+n+1} - H_{n+1}}{jk} - \sum\limits_{m=1}^{n}\sum\limits_{j=1}^{m} \frac{H_{n+j+1} - H_{n+1}}{mj}\right)\\&= \frac{1}{(n+1)^2}\sum\limits_{j=1}^{n+1} \frac{H_{n+j+1} - H_{n+1}}{j} = \frac{H_{n+1}^{(2)}}{2(n+1)^2}\end{align}$
In the last line I used the previous result $(1)$.
Hence, $\displaystyle F(n) = \frac{1}{2}\sum\limits_{k=1}^{n} \frac{H_{k}^{(2)}}{k^2} = \frac{\left(H_{n}^{(2)}\right)^2 + H_{n}^{(4)}}{4}$
|
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"question_score": "7",
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|
What are the ways to solve trig equations of the form $\sin(f(x)) = \cos(g(x))$? if I have the following trig equation:
$$\sin(10x) = \cos(2x)$$
I take the following steps to solve it:
*
*I rewrite $\cos(2x)$ as $\sin\left(\frac{\pi}{2} + 2x\right)$ or as $\sin\left(\frac{\pi}{2} - 2x\right)$ cause $\sin\left(\frac{\pi}{2} - a\right) = \sin\left(\frac{\pi}{2} + a\right) = \cos(a)$;
*Let's say that I have chosen $\sin\left(\frac{\pi}{2} + 2x\right)$, the equation becomes:
$$\sin(10x) = \sin\left(\frac{\pi}{2} + 2x\right).$$
*
*Then, I know that:
$$\sin(f(x)) = \sin(g(x)) \Leftrightarrow f(x) = g(x) + 2\pi n, n \in \mathbb{Z} \lor f(x) = (\pi - g(x)) + 2\pi k, k \in \mathbb{Z}$$
This means that (for $f(x) = 10x$ and $g(x) = \frac{\pi}{2} + 2x$):
$$\sin(10x) = \cos(2x) \Leftrightarrow 10x = \frac{\pi}{2} + 2x + 2\pi n, n \in \mathbb{Z} \lor 10x = (\pi - (\frac{\pi}{2} + 2x)) + 2\pi k, k \in \mathbb{Z}$$
Solving, I get the following results:
$$x_{1} = \frac{\pi}{16} + \frac{\pi}{4}n,\,\,x_{2} = \frac{\pi}{24} + \frac{\pi}{6}k,\,\,\,\,n,k \in \mathbb{Z}$$
Now, are there any other methods for solving such equations or could this one be just fine?
|
Otherwise as you ask could be as follows (although it is almost the same).
You have $$sin (f(x))-sin (\frac\pi2-g(x))=0$$ so from the identity
$$sin(a)-sin(b)=2sin\frac {a-b}{2}cos\frac{a+b}{2}$$
it follows $$2sin\frac{(f(x)+g(x)- \frac{\pi}{2})}{2}cos\frac{(f(x)-g(x)+\frac{\pi}{2})}{2}=0$$ hence $$f(x)+g(x)=2n\pi+\frac\pi2$$ and $$f(x)-g(x)=n\pi-\frac\pi2$$
|
{
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|
How to find this infinite product How to find this infinite product ?
$$\prod_{n=0}^\infty \left(1-\dfrac{2}{4(2n+1)^2+1}\right)$$
I try to use infinite product of $\cos{x}$ but it doesn't work.
Thank you.
|
$$\begin{eqnarray*}\prod_{n\geq 0}\left(1-\frac{2}{4(2n+1)^2+1}\right) &=& \prod_{n\geq 0}\left(\frac{4(2n+1)^2-1}{4(2n+1)^2+1}\right)\\&=&\color{purple}{\prod_{n\geq 0}\left(1-\frac{1}{4(2n+1)^2}\right)}\color{blue}{\prod_{n\geq 0}\left(1+\frac{1}{4(2n+1)^2}\right)^{-1}}\end{eqnarray*}$$
hence by exploiting the Weierstrass products for the $\color{purple}{\cos}$ and $\color{blue}{\cosh}$ function we get:
$$\prod_{n\geq 0}\left(1-\frac{2}{4(2n+1)^2+1}\right) = \frac{1}{\sqrt{2}\cosh\frac{\pi}{4}}=\color{red}{\frac{\sqrt{2}}{e^{\pi/4}+e^{-\pi/4}}}.$$
|
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"timestamp": "2023-03-29T00:00:00",
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|
Solve $x+y+z=1; x^2+y^2+z^2=35; x^3+y^3+z^3=97$ It may be surprising that I can't get any analytical way of verifying that one of the solutions of $$x+y+z=1$$ $$x^2+y^2+z^2=35$$ $$x^3+y^3+z^3=97$$ is $x=-1, y=-3$ and $z=5$. Although it may be possible that I have to revisit my elementary arithmetic.
|
First note that
$$x^{2}+y^{2}+z^{2}=(x+y+z)^{2}-2(xy+yz+zx)=35$$
Now substituting $x+y+z=1$, we have $1^{2}-2(xy+yz+zx)=35$, and thus $xy+yz+zx=-17$.
Then, $$x^{3}+y^{3}+z^{3}=(x+y+z)^{3}-3(x+y+z)(xy+yz+zx)+3xyz=97$$ and substituting with $1$ and our above result once again leads us to have $xyz=15$. Now we need a product of $15$ and sum of $1$ for $(x,y,z)$, giving us $(-1, -3, 5)$ as a solution, amongst others.
|
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|
Using equation to find value of $1/x - 1/y$ $$\left(\frac{48}{10}\right)^x=\left(\frac{8}{10}\right)^y=1000$$
What is the value of $\frac{1}{x}-\frac{1}{y}$?
I have already used that when $48$ divided by $10$ then it becomes $4.8$ and when $8$ divided by $10$ then it becomes $0.8$ by getting $10^x$ and $10^y$ to make it simpler but I cannot proceed further.
|
HINT :
$$4.8=1000^{\frac 1x},\ \ \ 0.8=1000^{\frac 1y}$$
So, $$1000^{\frac 1x-\frac 1y}=\frac{1000^{1/x}}{1000^{1/y}}=\frac{4.8}{0.8}$$
|
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"timestamp": "2023-03-29T00:00:00",
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|
Limit and factorization I have the following very interesting homework exercise:
Let $$f(x)=\frac{x^2-4}{2-x\cdot \sqrt {x+2}+\sqrt{x+2}}$$ Find the
following limit, if it exists: $$\lim_{x\to 2}f(x)$$
I understand that I need to "delete" the $x-2$ factor from both nominator and denominator and then evaluate the limit. It is obvious that $x^2-4=(x-2)\cdot (x+2)$, so the nominator is done.
So, my main problem is how to factor the denominator of $f(x)$. Any help with the factorization, without the final solution, would be appreciated. Then, after I understand how to factor it, anyone can post the full solution, even I.
|
$$\begin{align}\lim_{x\to 2}\frac{x^2-4}{2-x\sqrt{x+2}+\sqrt{x+2}}&=\lim_{x\to 2}\frac{(x-2)(x+2)}{2-(x-1)\sqrt{x+2}}\\&=\lim_{x\to 2}\frac{(x-2)(x+2)}{2-(x-1)\sqrt{x+2}}\cdot\frac{2+(x-1)\sqrt{x+2}}{2+(x-1)\sqrt{x+2}}\\&=\lim_{x\to 2}\frac{(x-2)(x+2)(2+(x-1)\sqrt{x+2})}{4-(x-1)^2(x+2)}\\&=\lim_{x\to 2}\frac{(x-2)(x+2)(2+(x-1)\sqrt{x+2})}{-(x-2)(x+1)^2}\\&=\lim_{x\to 2}\frac{(x+2)(2+(x-1)\sqrt{x+2})}{-(x+1)^2}\\&=\frac{4\cdot (2+2)}{-3^2}\end{align}$$
|
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|
proof by induction $2^n \leq 2^{n+1}-2^{n−1}-1$ My question is prove by induction for all $n\in\mathbb{N}$, $2^n \leq 2 ^{n+1}-2^{n−1}-1$
My proof
$1+2+3+4+....+2^n \leq 2^{n+1}-2^{n−1}-1$
Assume $n=1$,$1 ≤ 2$
Induction step
Assume statement is true for $n=k$, show true for $n=k + 1$
$1+2+3+4+....+2^k+2^k+1 ≤ 2 ^{k+1} −2^{k−1} - 1$
$2^{k+1} - 2^{k-1} -1 + 2^{k+1} ≤ 2^{k+1} - 2^{k-1} -1$
$4^{k+1} -2^{k-1} -1 ≤ 2^{k+1} - 2^{k-1} -1$
I do not know how to proceed from here, and i am confused because it seems to me this is not true.
|
In the case that $n=1$ we have $$2 \leq 2^2 - 2^0 - 1 \iff 2 \leq 2$$ which is true. Now, assuming that $$2^k \leq 2^{k+1} - 2^{k-1} - 1$$ holds. We proceed by multiplying our inductive hypothesis above by $2$ to get $$2 \cdot 2^{k} \leq 2(2^{k+1} - 2^{k-1} - 1)$$ Which simplifies to $$2^{k+1} \leq 2^{k+2} - 2^{k} - 2 \leq 2^{k+2} - 2^{k} - 1$$
So we get by transitivity that $$2^{k+1} \leq 2^{k+1 + 1} - 2^{k + 1 -1} - 1$$
So your statement is true by the principle of Mathematical Induction.
|
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|
Quadratic approximation of $9^{1/3}$ Find a quadratic approximation of the cube root of $9$ by using the equation $9=8(1+\frac 18)$, and estimate the difference between the exact value and the approximation.
How am I supposed to find the quadratic approximation of this formula? If there is no $x$, and the formula for quad approx is $Qa(x) = f(a) + f '(a)(x-a) + f ''(a)(x-a)^2/2$, so does it mean that my quadratic approx will be equal just to $f(a)$?
Could anyone help please, it looks easy, but I can't get it.
|
Since $9=8(1+\frac{1}{8})$,
$\;\;\sqrt[3]9=2\big(1+\frac{1}{8}\big)^{1/3}=2\big(1+\frac{1}{3}(\frac{1}{8})-\frac{1}{9}(\frac{1}{8})^2+\frac{5}{81}(\frac{1}{8})^3+\cdots\big)$
since $\displaystyle(1+x)^{1/3}=1+\frac{1}{3}x+\frac{(\frac{1}{3})(\frac{1}{3}-1)}{2!}x^2+\frac{(\frac{1}{3})(\frac{1}{3}-1)(\frac{1}{3}-2)}{3!}+\cdots$
Since the series is alternating (starting with the second term) and has terms decreasing in absolute value ,
the error in using this approximation is at most $2(\frac{5}{81})(\frac{1}{8})^3$, the absolute value of the first omitted term.
|
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|
Simplifying $ 2\cos^2(x)\sin^2(x) + \cos^4(x) + \sin^4(x)$ who can simplify the following term in a simplest way?
$$ 2\cos^2(x)\sin^2(x) + \cos^4(x) + \sin^4(x)$$
(The answer is 1). Thanks for any suggestions.
|
$$ 2\cos^2(x)\sin^2(x) + \cos^4(x) + \sin^4(x)$$
$$(a+b)^2=2ab+a^2+b^2$$
$$=(\cos^2x+\sin^2x)^2$$
$$=1^2$$
$$=\boxed{\color{red}1}$$
|
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|
The expression $(1+q)(1+q^2)(1+q^4)(1+q^8)(1+q^{16})(1+q^{32})(1+q^{64})$ where $q\ne 1$, equals The expression
$(1+q)(1+q^2)(1+q^4)(1+q^8)(1+q^{16})(1+q^{32})(1+q^{64})$
where $q\ne 1$, equals
(A) $\frac{1-q^{128}}{1-q}$
(B) $\frac{1-q^{64}}{1-q}$
(C) $\frac{1-q^{2^{1+2+\dots +6}}}{1-q}$
(D) none of the foregoing expressions
What I have done
$(1+q)(1+q^2)(1+q^4)(1+q^8)(1+q^{16})(1+q^{32})(1+q^{64})$ is a polynomial of degree $127$. Now the highest degree of the polynomial in option (A), (B) and (C) is $127$, $63$ and $41$ respectively. And therefore (A) is the correct answer.
I get to the correct answer but I don't think that my way of doing is correct. I mean what if, if option (B) was $\frac{1+q^{128}}{1-q}$.
Please show how should I approach to the problem to get to the correct answer without any confusion.
|
$a^n – b^n = (a – b)(a^{n–1} +a^{n–2}b + a^{n–3}b^2 + ··· + ab^{n–2} + b^{n–1})$
If $n$ is odd,$a^n+b^n = (a+b)(a^{n–1}-a^{n–2}b + a^{n–3}b^2 - ··· - ab^{n–2} + b^{n–1})$
|
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|
Wanted : for more formulas to find the area of a triangle? I know some formulas to find a triangle's area, like the ones below.
*
*Is there any reference containing most triangle area formulas?
*If you know more, please add them as an answer
$$s=\sqrt{p(p-a)(p-b)(p-c)} ,p=\frac{a+b+c}{2}\\s=\frac{h_a*a}{2}\\s=\frac{1}{2}bc\sin(A)\\s=2R^2\sin A \sin B \sin C$$
Another symmetrical form is given by :$$(4s)^2=\begin{bmatrix}
a^2 & b^2 & c^2
\end{bmatrix}\begin{bmatrix}
-1 & 1 & 1\\
1 & -1 & 1\\
1 & 1 & -1
\end{bmatrix} \begin{bmatrix}
a^2\\
b^2\\
c^2
\end{bmatrix}$$
Expressing the side lengths $a$, $b$ & $c$ in terms of the radii $a'$, $b'$ & $c'$ of the mutually tangent circles centered on the triangle's vertices (which define the Soddy circles)
$$a=b'+c'\\b=a'+c'\\c=a'+b'$$gives the paticularly pretty form $$s=\sqrt{a'b'c'(a'+b'+c')}$$
If the triangle is embedded in three dimensional space with the coordinates of the vertices given by $(x_i,y_i,z_i)$ then $$s=\frac{1}{2}\sqrt{\begin{vmatrix}
y_1 &z_1 &1 \\
y_2&z_2 &1 \\
y_3 &z_3 &1
\end{vmatrix}^2+\begin{vmatrix}
z_1 &x_1 &1 \\
z_2&x_2 &1 \\
z_3 &x_3 &1
\end{vmatrix}^2+\begin{vmatrix}
x_1 &y_1 &1 \\
x_2&y_2 &1 \\
x_3 &y_3 &1
\end{vmatrix}^2}$$
When we have 2-d coordinate $$ s=\frac{1}{2}\begin{vmatrix}
x_a &y_a &1 \\
x_b &y_b &1 \\
x_c &y_c & 1
\end{vmatrix}$$
In the above figure, let the circumcircle passing through a triangle's vertices have radius $R$, and denote the central angles from the first point to the second $q$, and to the third point by $p$ then the area of the triangle is given by:
$$ s=2R^2|\sin(\frac{p}{2})\sin(\frac{q}{2})\sin(\frac{p-q}{2})|$$
|
I have a little more:)
$$
S = 4R^2(\sin A + \sin B + \sin C)\sin\frac A2\sin\frac B2\sin\frac C2\\
S = \frac{r^2}{4}\frac{\sin A + \sin B + \sin C}{\sin\dfrac A2\sin\dfrac B2\sin\dfrac C2}\\
S = r^2\left(\cot\frac A2+\cot\frac B2 + \cot\frac C2\right)\\
S = r^2\cot\frac A2 \cot\frac B2 \cot\frac C2\\
S = 2p^2 \frac{\sin A\sin B\sin C}{(\sin A+\sin B+\sin C)^2}\\
S = 4p^2 \frac{\sin\dfrac A2 \sin\dfrac B2 \sin\dfrac C2}{\sin A+\sin B+\sin C}
$$
|
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|
$ x^2 + \frac {x^2}{(x-1)^2} = 2010 $ I found this question from last year's maths competition in my country. I've tried any possible way to find it, but it is just way too hard.
Given $$ x^2 + \frac {x^2}{(x-1)^2} = 2010,$$
find $\dfrac {x^2} {x-1}.$
(A) $1+\sqrt {2011}$ (B) $ 1-\sqrt {2011}$ (C) $1\pm \sqrt{2011} $ (D) $\sqrt {2011}$
So I multiply them with $(x-1)$
$$x^2(x-1) + \frac {x^2} {x-1} = 2010(x-1)$$
$$\frac {x^2} {x-1} = (x-1)(2010-x^2)$$
and I stuck in here, dont know how to remove $x$ in there
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$${ x }^{ 2 }+\frac { { x }^{ 2 } }{ { \left( x-1 \right) }^{ 2 } } =2010\\ { \left( x+\frac { x }{ x-1 } \right) }^{ 2 }-2\frac { { x }^{ 2 } }{ x-1 } =2010\\ { \left( \frac { { x }^{ 2 } }{ x-1 } \right) }^{ 2 }-2\left( \frac { { x }^{ 2 } }{ x-1 } \right) -2010=0\\ \left( \frac { { x }^{ 2 } }{ x-1 } \right) =t\\ { t }^{ 2 }-2t-2010=0\\ t=1\pm \sqrt { 2011 } \\ \\ $$
|
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|
Find $x$ if $\frac {1} {x} + \frac {1} {y+z} = \frac {1} {2}$ I found this question from past year's maths competition in my country. I've tried any possible way to find it, but it is just way too hard.
Find $x$ if \begin{align}\frac {1} {x} + \frac {1} {y+z} &= \frac {1} {2}\\ \frac {1} {y} + \frac {1} {x+z} &= \frac{1}{3}\\ \frac {1} {z} + \frac {1} {x+y} &= \frac {1} {4}\end{align}
$(A)\;\frac 32$
$(B)\;\frac {17}{10}$
$(C)\;\frac {19}{10}$
$(D)\;\frac {21}{10}$
$(E)\;\frac {23}{10}$
EDIT: I'm very sorry guys, it should be $\frac {1} {x+y}$ not $xy$, I'm sorry for the typos (idk what is wrong with me)
|
\begin{align}
\frac {1} {x} + \frac {1} {y+z} &= \frac {1} {2}\\ \frac {1} {y} + \frac {1} {x+z} &= \frac{1}{3}\\ \frac {1} {z} + \frac {1} {x+y} &= \frac {1} {4}
\end{align}
Is equivalent to the system
\begin{align}
xy+xz&=2(x+y+z)
\\
xy+yz&=3(x+y+z)
\\
xz+yz&=4(x+y+z),
\end{align}
Solving this system as a linear system of
equations in terms of $xy,xz$ and $yz$ and $x+y+z$,
we arrive at
\begin{align}
xy&=\tfrac12 (x+y+z), \quad(1)
\\
yz&=\tfrac52 (x+y+z),
\\
xz&=\tfrac32 (x+y+z).
\end{align}
Dividing these equations pairwise, we get
\begin{align}
z&=5x
\\
y&=\tfrac53 x,
\end{align}
combining with (1) we have
\begin{align}
\tfrac{5}{3}x^2-\tfrac{23}{6}x&=0
\end{align}
and since $x\ne0$, the answer is $x=\tfrac{23}{10}$.
|
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|
Using complex variables to find sums of Fourier series Use complex variables to find the sum of the Fourier Series:
$$\sin(\theta) + \frac{\sin(2\theta)}{2^{2}} + \frac{\sin(3\theta)}{2^{3}}+\cdots$$
where $\theta$ is a real variable.
|
We have
$$\begin{align}
f(\theta)&=\sin\theta + \sum_{k=2}^{\infty}\frac{\sin k\theta}{2^k} \tag 1\\\\
&=\sin\theta+\text{Im}\left( \sum_{k=2}^{\infty}\left(\frac{e^{i\theta}}{2}\right)^k\right) \tag 2\\\\
&=\sin\theta+\text{Im}\left(\frac{e^{i2\theta}}{2-e^{i\theta}}\right) \tag 3\\\\
&=\sin\theta+\frac12 \frac{2\sin 2 \theta-\sin \theta}{5-4\cos \theta} \tag 4\\\\
&=\frac12 \frac{9 \sin \theta-2\sin 2\theta}{5-4\cos \theta}\tag 5
\end{align}$$
In going from $(1)$ to $(2)$, we note that the imaginary part of $e^{i\theta}=\cos \theta+i\sin \theta$ is $\sin \theta$ and thus write $\sin \theta=\text{Im}(e^{ \theta})$. We also make use of the fact that $e^{ik\theta}=\left(e^{i\theta}\right)^k$.
In going from $(2)$ to $(3)$, we made use of the sum for a geometric series $\sum_{k=m}^{\infty}r^k=\frac{r^m}{1-r}$ with $r=\frac{e^{i\theta}}{2}$.
In going from $(3)$ to $(4)$, we multiplied both numerator and denominator by the complex conjugate of the denominator, thereby rendering the denominator real and equal to the square of the magnitude of $2-e^{i\theta}$.
NOTE:
We remark that if the first term of the series were $\frac{\sin \theta}{2}$ rather than $\sin \theta$, then $(5)$ would be
$$\begin{align}
\frac12 \frac{9 \sin \theta-2\sin 2\theta}{5-4\cos \theta}-\frac 12 \sin \theta &=\frac12 \frac{9\sin \theta-2\sin 2 \theta-(5\sin \theta -2 \sin 2\theta)}{5-4\cos \theta}\\\\
&=\frac{2\sin \theta}{5-4\cos \theta}
\end{align}$$
|
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|
$f(x) =ax^6 +bx^5+cx^4+dx^3+ex^2+gx+h $, find $f(7)$. Problem :
Given a polynomial $f(x) =ax^6 +bx^5+cx^4+dx^3+ex^2+gx+h$
such that $f(1)= 1, f(2) =2 , f(3) = 3, f(4) =4, f(5)=5, f(6) =6$.
Find $f(7)$ in terms of $h$.
My approach:
We can put the values of $f(1) = 1$ in the given equation and $f(2) = 2$, etc. But this is quite time consuming by making six different equations and then solve them to get the values of $a,b,c,d,e,g,h$. Please suggest some alternate solution for this.
|
let $g(x) = f(x) - x$
Then:
$$g(1)=g(2)=g(3)=g(4)=g(5)=g(6)=0$$
So:
$$g(x) = A(x-1)(x-2)(x-3)(x-4)(x-5)(x-6)$$
We also know that $g(0)=f(0)-0=f(0)=h$ so we can find $A$.
$h=A\times 1\times 2\times 3\times 4\times 5\times 6$ so $A=\frac{h}{6!}$
$f(7) = g(7) + 7 = 6\times 5\times 4\times 3\times 2\times 1\times \frac{h}{6!} + 7 = h + 7$
|
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|
Proving $\sum_{i=1}^n\frac{1}{i(i+1)(i+2)}=\frac{n(n+3)}{4(n+1)(n+2)}$ for $n\geq 1$ by mathematical induction Prove using mathematical induction that
$$\frac{1}{1\cdot 2\cdot 3} + \frac{1}{2\cdot 3\cdot 4} + \cdots + \frac{1}{n(n+1)(n+2)} = \frac{n(n+3)}{4(n+1)(n+2)}.$$
I tried taking $n=k$,
so it makes
$$\frac{1}{1\cdot 2\cdot 3} + \frac{1}{2\cdot 3\cdot 4} + \cdots + \frac{1}{n(n+1)(n+2)} = \frac{k(k+3)}{4(k+1)(k+2)}.$$
Then proving the statement for $n=k+1$:
$$\frac{k(k+3)}{4(k+1)(k+2)}+ \frac{1}{(k+1)(k+2)(k+3)} = \frac{(k+1)(k+1+3)}{4(k+1+1)(k+1+2) }.$$
|
You actually copied it over incorrectly; the general term should be $\frac{1}{n(n+1)(n+2)}$, not $\frac{1}{n(n+1)(n+3)}$. That in mind, try to see if you can follow this argument:
For each $n\geq 1$, let $S(n)$ denote the statement
$$
\frac{1}{1\cdot2\cdot3}+\frac{1}{2\cdot3\cdot4}+\frac{1}{3\cdot4\cdot5}+\cdots+\frac{1}{n(n+1)(n+2)}=\frac{n(n+3)}{4(n+1)(n+2)}.
$$
Base step ($n=1$): $S(1)$ says that $\frac{1}{1\cdot2\cdot3}=\frac{1\cdot4}{4\cdot2\cdot3}$, and this is correct since both sides equal $\frac{1}{6}$.
Inductive step ($S(k)\to S(k+1)$): For some fixed $k\geq 1$, assume the inductive hypothesis
$$
S(k) : \sum_{i=1}^k\frac{1}{i(i+1)(i+2)}=\frac{k(k+3)}{4(k+1)(k+2)}
$$
to be true. It remains to show that
$$
S(k+1) : \sum_{i=1}^{k+1}\frac{1}{i(i+1)(i+2)}=\frac{(k+1)(k+4)}{4(k+2)(k+3)}
$$
follows. Starting with the left-hand side of $S(k+1)$,
\begin{align}
\sum_{i=1}^{k+1}\frac{1}{i(i+1)(i+2)} &= \sum_{i=1}^k\frac{1}{i(i+1)(i+2)}+\frac{1}{(k+1)(k+2)(k+3)}\\[1em]
&= \frac{k(k+3)}{4(k+1)(k+2)}+\frac{4}{4(k+1)(k+2)(k+3)}\tag{by $S(k)$}\\[1em]
&= \frac{k(k+3)(k+3)}{4(k+1)(k+2)(k+3)}+\frac{4}{4(k+1)(k+2)(k+3)}\\[1em]
&= \frac{k^3+6k^2+9k+4}{4(k+1)(k+2)(k+3)}\\[1em]
&= \frac{(k+1)^2(k+4)}{4(k+1)(k+2)(k+3)}\\[1em]
&= \frac{(k+1)(k+4)}{4(k+2)(k+3)},
\end{align}
one arrives at the right-hand side of $S(k+1)$, completing the inductive step.
By mathematical induction, the statement $S(n)$ is true for all $n\geq 1$. $\blacksquare$
|
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"url": "https://math.stackexchange.com/questions/1375015",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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|
Infinite limit of trigonometric series The value of $\displaystyle\lim_{n\to\infty}(\sin^4x+\frac{1}{4}\sin^4(2x)+\cdots+\frac{1}{4^n}\sin^4(2^nx))$ is
(A) $\sin^4x$
(B) $\sin^2x$
(C) $\cos^2x$
(D) does not exist
My attempt:
$$\lim_{n\to\infty}(\sin^4x+\frac{1}{4}\sin^4(2x)+\cdots+\frac{1}{4^n}\sin^4(2^nx))=$$
$$=(\sin^4x+\frac{1}{4}16\sin^4x\cos^4 x+\cdots+\frac{1}{4^n}\sin^4(2^{n-1}x)\cos^4(2^{n-1}x)$$
i could not solve further.Any hint will be useful.
|
HINT:
$$\sin^2y-\sin^4y=\sin^2y(1-\sin^2y)=\dfrac{\sin^22y}4$$
$y=x\implies$ $$\sin^2x-\sin^4x=\dfrac{\sin^22x}4$$
$y=2x\implies$ $$\dfrac{\sin^22x-\sin^42x}{4^1}=\dfrac{\sin^22x}{4^2}$$
Set $y=4x,2^rx$ and add to get
$$\sin^2x-\sum_{r=0}^n\dfrac{\sin^4(2^rx)}{4^r}=\dfrac{\sin^4(2^{n+1}x)}{4^{n+1}}$$
$$\lim_{n\to\infty}\dfrac{\sin^4(2^{n+1}x)}{4^{n+1}}=0$$
|
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"timestamp": "2023-03-29T00:00:00",
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|
$\lim_{x\to 0}\frac{e^x-1}{\sin x}$ equal to $\lim_{x\to 0}\frac{e^x-1}{x}$ because $x$ and $\sin x$ tend both to $0$ for ${x\to 0}$ I'm stuck in this limit:
$$\lim_{x\to 0}\frac{x(e^x-1)}{\cos x-1}$$
I tried to solve it using special limits, so:
$$\lim_{x\to 0}\frac{x(e^x-1)}{\cos x-1}=$$
$$=\lim_{x\to 0}(e^x-1)\frac{x(\cos x+1)}{(\cos x-1)(\cos x+1)}=$$
$$=\lim_{x\to 0}-\frac{x}{\sin^2 x}(e^x-1)(\cos x+1)=$$
$$=\lim_{x\to 0}-\frac{x}{\sin x}\cdot\frac{e^x-1}{\sin x}\cdot(\cos x+1)$$
I know that $\lim_{x\to 0}\left(\frac{x}{\sin x}\right)=1$
I know also the special limit $\lim_{x\to 0}\left(\frac{e^x-1}{x}\right)=1$
My question is: can I intend $\lim_{x\to 0}\left(\frac{e^x-1}{\sin x}\right)$ equal to $\lim_{x\to 0}\left(\frac{e^x-1}{x}\right)$ because $x$ and $\sin x$ tend both to $0$ for ${x\to 0}$?
If not, how can I solve my limit, without series or something difficult?
Thank you!
|
$$\lim _{ x\rightarrow 0 }{ \frac { x\left( { e }^{ x }-1 \right) }{ \cos { x } -1 } = } \lim _{ x\rightarrow 0 }{ \frac { { x }^{ 2 }\left( { e }^{ x }-1 \right) }{ x\left( \cos { x } -1 \right) } = } \lim _{ x\rightarrow 0 }{ \frac { \left( { e }^{ x }-1 \right) }{ x } \lim _{ x\rightarrow 0 }{ \frac { { x }^{ 2 } }{ -2\sin ^{ 2 }{ \frac { x }{ 2 } } } = } } \\ =\lim _{ x\rightarrow 0 }{ \frac { \left( { e }^{ x }-1 \right) }{ x } \lim _{ x\rightarrow 0 }{ \frac { { x }^{ 2 } }{ -2\sin ^{ 2 }{ \frac { x }{ 2 } } } = } } -\frac { 1 }{ 2 } \lim _{ x\rightarrow 0 }{ \frac { \left( { e }^{ x }-1 \right) }{ x } \lim _{ x\rightarrow 0 }{ \frac { 4 }{ { \left( \frac { \sin { \frac { x }{ 2 } } }{ \frac { x }{ 2 } } \right) }^{ 2 } } = } } -2$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1375382",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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|
evaluate $\frac 1{1+\sqrt2+\sqrt3} + \frac 1{1-\sqrt2+\sqrt3} + \frac 1{1+\sqrt2-\sqrt3} + \frac 1{1-\sqrt2-\sqrt3}$
Evaluate $\frac 1{1+\sqrt2+\sqrt3} + \frac 1{1-\sqrt2+\sqrt3} + \frac 1{1+\sqrt2-\sqrt3} + \frac 1{1-\sqrt2-\sqrt3}$
How to evalute this equation without using calculator?
|
from here, separating it so that they have same thing
$$\frac 1{1+(\sqrt2+\sqrt3)} + \frac 1{1-(\sqrt2-\sqrt3)} + \frac 1{1+(\sqrt2-\sqrt3)} + \frac 1{1-(\sqrt2+\sqrt3)}$$
let $x = \sqrt 2 + \sqrt 3$, $y = \sqrt 2 - \sqrt 3$
$$\frac 1{1+x} + \frac 1{1-y} + \frac 1{1+y} + \frac 1{1-x}$$
combine fraction with x into one fraction, same thing as y
$$=\frac {(1+x)+(1-x)}{(1+x)(1-x)} + \frac {(1+y)+(1-y)}{(1+y)(1-y)}$$
$$=\frac {2}{1-x^2} + \frac {2}{1-y^2}$$
let them become one fraction
$$=\frac {2(1-x^2)+2(1-y^2)}{(1-x^2)(1-y^2)}$$
$$=\frac {4-2x^2-2y^2}{1-x^2-y^2+x^2y^2}$$
calculate $x^2$ and $y^2$, $x^2=(\sqrt 2 + \sqrt 3)^2= 2 + 2(\sqrt 6) + 3 = 2 + \sqrt {6\times4} + 3 = 5 + \sqrt {24}$, $y^2=(\sqrt 2 - \sqrt 3)^2= 2 - 2(\sqrt 6) + 3 = 2 - \sqrt {6\times4} + 3 = 5 - \sqrt {24}$
now replace $x^2$ and $y^2$
$$=\frac {4-2(5 + \sqrt {24}) - 2(5-\sqrt {24})}{1-(5+\sqrt {24})-(5-\sqrt {24})+((5+\sqrt {24})(5-\sqrt {24}))}$$
$$=\frac {4-10 - 2\sqrt {24} - 10+2\sqrt {24}}{1-5-\sqrt {24}-5+\sqrt {24}+(25-24)}$$
$$=\frac {-16}{-8} = 2$$
|
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"language": "en",
"url": "https://math.stackexchange.com/questions/1375531",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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|
Show that $B$ is invertible if $B=A^2-2A+2I$ and $A^3=2I$ If $A$ is $40\times 40$ matrix such that $A^3=2I$ show that $B$ is invertible where $B=A^2-2A+2I$.
I tried to evaluate $B(A-I)$ , $B(A+I)$ , $B(A-2I)$ ... but I couldn't find anything.
|
A few manipulations:
$$B=A^2-2A+2I$$
$$A^3 = 2I$$
then,
$$AB = A^3 -2A^2+2A \\= 2I -2A^2+2A \\= -2(A^2-2A+2I)-2A+6I \\= -2B -2A+6I$$
Now, the fact that $A^3 = 2I$ means $A$ is invertible, because $\det(A) \neq 0$
Thus,
$$B = -2A^{-1}B-2I+6A^{-1} $$
Take $B$ to one side:
$$B(I+2A^{-1}) = -2I+6A^{-1}$$
Multiply both side with $A$:
$$B(A+2I) = -2A+6I$$
I suppose those quantities will be similar.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1378132",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "13",
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|
Finding the equation of a circle. A circle of radius $2$ lies in the first quadrant touching both axis. Find the equation of the circle centered at $(6,5)$ and touching the above circle externally.
Let me share how I answered this question with your suggestions.
Since the radius of the first circle is $2$ and touches both axes. It follows that $h,k$ is $(2,2)$. The formula for this circle is
$$(x-2)^2 + (y-2)^2 = 4 .$$
Given that the center of the second circle is $(6,5)$, then we can get the distance to another circle. The distance between $(6,5)$ and $(2,2)$ is $5$.
We can subtract and get the radius of the second circle. $r=5-2$.
Then, the radius is $3$.
The equation of the second circle is
$$(x-6)^2 + (y-5)^5 = 9 .$$
Thank you everyone! :)
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The first circle is centered at $(2,2)$, so the distance between two centers of both the circles would be $\sqrt{(6-2)^2 +(5-2)^2} =5$. Hence the radius of circle centred at(6,5) is 3 units. So the final equation of circle is $(x-6)^2+(y-5)^2 = 9$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1378941",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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|
Angle between medians in right triangle In a right angled triangle,medians are drawn from the acute-angles to the opposite sides.If maximum acute angle between these medians can be expressed as $tan^{-1}(\frac{p}{q})$ where p and q are relatively prime positive integers.Find $p+q$.
Let triangle is right angled at B.Let us take B (0,0),C(a,0),A(0,c).Let AD is median from A to opposite side BC.Let CE is median from C to side AB.Coordinates of D are$(\frac{a}{2},0)$,coordinates of E are $(0,\frac{c}{2})$.Slope of AD is $\frac{-2a}{c}$.Slope of CE is $\frac{-a}{2c}.$
Angle between medians$=\frac{\frac{a}{2c}-\frac{2a}{c}}{1+\frac{a^2}{c^2}}$,I am not getting answer,beacause these variables are not cancelling out.
Can someone help me getting the answer?
|
I think the slope of $AD$ is $\frac{2c}{-a}$ and that the slope of $CE$ is $\frac{-c}{2a}$.
Then, the acute angle between the medians can be expressed as
$$\arctan\frac{\frac{-c}{2a}-\frac{2c}{-a}}{1+\frac{2c}{-a}\cdot\frac{-c}{2a}}=\arctan\frac{3ac}{2a^2+2c^2}=\arctan\frac{\frac{3ac}{a^2}}{\frac{2a^2}{a^2}+\frac{2c^2}{a^2}}=\arctan\frac{3s}{2+2s^2}$$
where $\frac ca=s\gt 0$.
Here, let $f(s)=\frac{3s}{2+2s^2}$. Then, we have
$$f'(s)=\frac{6(1-s)(1+s)}{(2+2s)^2}.$$
So, we have $f(s)\le f(1)=\frac 34$.
Hence, the maximum acute angle is $\arctan\frac 34$.
|
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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|
Does $\frac{x+y}{2}>\frac{a+b}{2}$ hold? $a$ and $b$ are two real positive numbers. Given that $x=\sqrt{ab}$ and $y=\sqrt{\frac{a^2+b^2}{2}}$, which one has a higher value, $\frac{x+y}{2}$ or $\frac{a+b}{2}$?
We know that $y=\sqrt{\frac{a^2+b^2}{2}}>\frac{a+b}{2}>x=\sqrt{ab}$ by inequality, and at this point I'm stuck.
|
We can drop the factor of $\frac{1}{2}$, so we have $\sqrt{ab}+\sqrt{\frac{a^2+b^2}{2}}$ and $a+b$. Squaring gives $ab + \frac{a^2+b^2}{2}+\sqrt{ab\frac{a^2+b^2}{2}}$ and $a^2+b^2+ab$. They both have a factor $ab$ and $\frac{a^2+b^2}{2}$, so we have to compare $\sqrt{ab\frac{a^2+b^2}{2}}$ and $\frac{a^2+b^2}{2}$. Rewriting we find $\sqrt{ab}\sqrt{\frac{a^2+b^2}{2}}$ and $\sqrt{\frac{a^2+b^2}{2}}\sqrt{\frac{a^2+b^2}{2}}$, so dropping out the common factor and squaring, we get $ab$ and $\frac{a^2+b^2}{2}$. We have that $\frac{a^2+b^2}{2}\geq ab$ implies that $\frac{a^2+b^2}{2} -ab = (a-b)^2 \geq 0$, which holds.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1381476",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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|
Finding $P$ such that $P^TAP$ is a diagonal matrix
Let $$A = \left(\begin{array}{cc} 2&3 \\ 3&4 \end{array}\right) \in
M_2(\mathbb{C})$$
Find $P$ such that $P^TAP = D$ where $D$ is a diagonal matrix.
So here's the solution:
$$A = \left(\begin{array}{cc|cc} 2&3&1&0\\ 3&4&0&1 \end{array}\right) \sim \left(\begin{array}{cc|cc} 2&0&1&-3/2\\ 0&-1/2&0&1 \end{array}\right)$$
Therefore, $$P = \left(\begin{array}{cc} 1&-3/2\\ 0&1 \end{array}\right) \\ P^TAP = \left(\begin{array}{cc} 2&0\\ 0&-1/2 \end{array}\right) $$
What was done here exactly? I'd be glad elaborate about the process.
Thanks.
|
When you parameterize conics with rotation, its usual make the variable change $\mathbf {X=P^TY}$ so that $P^T\mathbf{A}P=D$ where $D$ is a diagonal matrix and $\mathbf{A}$ is the matrix associate to the conics. The goal is eliminate cross terms $Bxy$.
The change of variable $\mathbf {X=P^TY}$ is deduced by completing the square (with respect to $x$ and $y$) in the conic of equation
$Ax^2+Bxy+Cy^2+Dx+Ey+F=0$
If $B \neq 0 $ and $C \neq 0, $ the matrix $\mathbf {A} =
\begin{pmatrix} A & B/2 \\ B/
2 & A\end{pmatrix} $ is "diagonalized" by $\mathbf {P} = \
\begin {pmatrix} 1 & 0 \\ -
B/(2 C) & 1\end {pmatrix} $ because $$\mathbf {P}^
T\mathbf {A}\mathbf {P} = \left(
\begin{array}{cc}
A-\frac{B^2}{4 C} & 0 \\
0 & C \\
\end{array}
\right)$$
If $B \neq 0 $ and $A \neq 0, $ the matrix $\mathbf {A} =
\begin{pmatrix} A & B/2 \\ B/
2 & A\end{pmatrix} $ is "diagonalized" by $\mathbf {P} = \
\begin {pmatrix} 1 & -
B/(2 C) \\ 0 & 1\end {pmatrix} $ because $$\mathbf {P}^
T\mathbf {A}\mathbf {P} = \left(
\begin{array}{cc}
A & 0 \\
0 & C-\frac{B^2}{4 A} \\
\end{array}
\right)
$$
As special case, If $A=C$ then $\mathbf{P}=\left(
\begin{array}{cc}
1 & 1 \\
1 & -1 \\
\end{array}
\right)$.
$$\mathbf {P}^
T\mathbf {A}\mathbf {P}=\left(
\begin{array}{cc}
2 A+B & 0 \\
0 & 2 A-B \\
\end{array}
\right)$$
In this last case, if you take $\frac{1}{\sqrt{2}}\mathbf P$, you get eingevalues in diagonal
Edit: For a general method you can see: "Linear Algebra
Done Wrong" by Sergei Treil
).
|
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "5",
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|
Matrix representation with non-standard bases. In chaprer 2.2 of Fiedberg's Linear Algebra is wroten about matrix representation. But all examples are only with standard ordered bases. I made a task to understand it. Please, could you show me matrix representation on this task or your own example?
Let $$ T\begin{pmatrix}x\\y\end{pmatrix}=\begin{pmatrix}x+y\\x-y\end{pmatrix}$$ is transformation from vector space V with a basis $$ v= \{ \begin{pmatrix}1\\2\end{pmatrix}, \begin{pmatrix}3\\4\end{pmatrix}\} $$ to vector space W with a basis $$ w= \{ \begin{pmatrix}1\\4\end{pmatrix}, \begin{pmatrix}2\\3\end{pmatrix}\} $$.
How to find the matrix representation $$ [T]_v^w $$?
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Problem: Let
$$ T\begin{pmatrix}x\\y\end{pmatrix}=\begin{pmatrix}x+y\\x-y\end{pmatrix}$$
is transformation from vector space V with a basis
$$ v= \{ \begin{pmatrix}1\\2\end{pmatrix}, \begin{pmatrix}3\\4\end{pmatrix}\} $$ to vector space W with a basis
$$ w= \{ \begin{pmatrix}1\\4\end{pmatrix}, \begin{pmatrix}2\\3\end{pmatrix}\} $$. Find $[T]_v^w$ for $T: \mathbb{R}^2 \rightarrow \mathbb{R}^2$ as given.
Solution: there is a simple way to calculate for given examples. Basically, for each domain basis element (in $v$) we need to express it as a linear combination of the basis of the codomain ($w$). Let's begin:
$$ T(v_1)=T(1,2) = (1+2,1-2) = (3,-1) = A_{11}w_1+A_{21}w_2 \qquad \star$$
where $w_1 = (1,4)$ and $w_2 = (2,3)$. Likewise,
$$ T(v_2)=T(3,4) = (3+4,3-4) = (7,-1) = A_{12}w_1+A_{22}w_2 \qquad \star^2$$
So we must solve $\star$ and $\star^2$ for $A_{ij}$. That is,
$$ (3,-1)=A_{11}(1,4)+A_{21}(2,3) \ \ \& \ \ (7,-1)=A_{21}(1,4)+A_{22}(2,3)$$
These can be solved in many ways, for example:
$$ \left[ \begin{array}{c} 3 \\ -1 \end{array}\right] = \left[ \begin{array}{cc} 1 & 2 \\ 4 & 3 \end{array}\right]\left[ \begin{array}{c} A_{11} \\ A_{21} \end{array}\right] \ \ \& \ \
\left[ \begin{array}{c} 7 \\ -1 \end{array}\right] = \left[ \begin{array}{cc} 1 & 2 \\ 4 & 3 \end{array}\right]\left[ \begin{array}{c} A_{21} \\ A_{22} \end{array}\right]$$
reduce to a single matrix equation
$$ \left[ \begin{array}{cc} 3 & 7 \\ -1 & -1 \end{array}\right] = \left[ \begin{array}{cc} 1 & 2 \\ 4 & 3 \end{array}\right]\left[ \begin{array}{cc} A_{11} & A_{12} \\ A_{21} & A_{22} \end{array}\right]$$
which I solve by multiplication by inverse
$$ \left[ \begin{array}{cc} A_{11} & A_{12} \\ A_{21} & A_{22} \end{array}\right] =
\frac{1}{-5}\left[ \begin{array}{cc} 3 & -2 \\ -4 & 1 \end{array}\right]\left[ \begin{array}{cc} 3 & 7 \\ -1 & -1 \end{array}\right] = \frac{1}{-5}\left[ \begin{array}{cc} 11 & 23 \\ -13 & -29 \end{array}\right] = [T]_v^w.$$
Let $\Phi_v: \mathbb{R}^2 \rightarrow \mathbb{R}^2$ be the coordinate map defined by linearly extending $\Phi_v(v_i)=e_i$ and likewise $\Phi_w (w_i)=e_i$ for the standard basis $e_1=(1,0)$ and $e_2 = (0,1)$. Then I also denote $\Phi_v(x) = [x]_v$ and $\Phi_w(x) = [x]_w$; the coordinate charts with respect to bases $v$ and $w$ respective. Given all this notation the meaning of $[T]_v^w$ is simply that it is the matrix which acts as $T$ when we swap vectors in the domain and codomain for their corresponding coordinate vectors. That is:
$$ [T]_v^w[x]_v = [T(x)]_w $$
To actually calculate the coordinate charts (and hence the matrix of $T$) it generally requires we solve some systems of equations because it it not immediately obvious how to express $T(x)$ in the $w$-basis. I have examples and more discussion in a notation pretty close to the one I use here at:my linear algebra lectures on You Tube.
|
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"timestamp": "2023-03-29T00:00:00",
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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.