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Counting Possible Sums in a Group of Integers Say you have 5 buckets.
Each bucket contains a distinct number $[1, 10]$.
So why can it be guaranteed that the total sum of 1 or more numbers from the 5 buckets will equal the total sum of 1 or more OTHER numbers from the 5 buckets?
Example:
6, 7, 8, 9, 10.
We know that $6+9=15$ and $7+8= 15$. Regardless of the distinct numbers we fill the buckets with, we will always be able to find cases where this condition holds.
Another Example:
1,2,3,5,10 We know $1+2= 3$ and $3 = 3$
I find this extremely interesting but I'm not quite sure how to explain this observation.
In my reasoning, I know the lowest value that can be obtained is 1, and the highest, $10+9+8+7=34$ (assuming we must leave one number for the other group).
What principles explain simply why this works?
|
This is the pigeonhole principle at work, although it takes a little digging to get it to work.
Firstly we can always assume that the numbers are different, or we could just match two that are identical. So although you gave me that the numbers were distinct, that's actually not needed to make this work.
So given the smallest value chosen, $s$, the largest sum possible is $s+7+8+9+10 = s+34$, giving a range of $35$ possible sums, although the all-five-numbers can't possibly match anything.
Now the analysis has to devolve into cases, to get the range small enough that the pigeonhole principle applies, of more possible subset sums than the range that contains them. Discarding the empty set and the five-number set, there are 30 subset sums of interest.
Looking at the top four numbers, we review cases to drive them below $30$ by attempting to avoid a common sum [noting that $s$ cannot equal a difference in the proposed top-$4$ set]:
Sum: $34$, combination: $7,8,9,10$ - $7+10=8+9$
Sum: $33$, combination: $6,8,9,10$ - $s$ cannot be any of $1,2,3,4$ and $s=5$ gives $5+10=9+6$
Sum: $32$, combination (1): $5,8,9,10$ - $s$ cannot be any of $1,2,3,4$
Sum: $32$, combination (2): $6,7,9,10$ - $6+10=7+9$
Sum: $31$, combination (1): $4,8,9,10$ - $s$ cannot be any of $1,2$ and $s=3$ gives $3+10=4+9$
Sum: $31$, combination (2): $5,7,9,10$ - $s$ cannot be any of $1,2,3,4$
Sum: $31$, combination (3): $6,7,8,10$ - $s$ cannot be any of $1,2,3,4$ and $s=5$ gives $5+8=6+7$
Sum: $30$, combination (1): $3,8,9,10$ - $s$ cannot be any of $1,2$
Sum: $30$, combination (2): $4,7,9,10$ - $s$ cannot be any of $1,2,3$
Sum: $30$, combination (3): $5,6,9,10$ - $5+10=6+9$
Sum: $30$, combination (4): $5,7,8,10$ - $5+10=7+8$
Sum: $30$, combination (5): $6,7,8,9$ - $6+9=7+8$
And otherwise the top numbers sum below $30$ and the pigeonhole principle guarantees matching sums.
|
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|
Solve $\lim_{x \rightarrow 3} \frac{\ln(\sqrt{x-2})}{x^2-9}$ without using L'Hôpital's rule $\lim_{x \rightarrow 3} \frac{\ln(\sqrt{x-2})}{x^2-9}$
To do this I tried 2 approaches:
1: If $\lim_{x \rightarrow 0} \frac{\ln(x+1)}{x} = 1$, $\lim_{x\to1}\frac{\ln(x)}{x-1}=1$ and $\lim_{x\to0}\frac{\ln(x+1)}x=\lim_{u\to1}\frac{\ln(u)}{u-1}$, then I infer that $\frac{\ln(x)}{y} = 1$ where $x \rightarrow 1$ and $y \rightarrow 0$, then I have:
$$\sqrt{3-2} \rightarrow 1
\\ 3^2-9 \rightarrow 0
$$
and so
$$\lim_{x \rightarrow 3} \frac{\ln(\sqrt{x-2})}{x^2-9} = 1$$
But this is wrong. So I tried another method:
2:
$$\lim_{x \rightarrow 3} \frac{\ln(\sqrt{x-2})}{x^2-9} = \frac{\ln(\sqrt{x-2})}{(x-3)(x+3)} = \lim_{x \rightarrow 3} \frac{\ln(\sqrt{x-2})}{(x-3)} \cdot \lim_{x \rightarrow 3} \frac{1}{x+3} = \frac{1}{6}$$
Which is also wrong.
My questions are: What did I do wrong in each method and how do I solve this?
EDIT: If $\frac{\ln(x)}{y} = 1$ where $x \rightarrow 1$ and $y \rightarrow 0$, and $\ln(x) \rightarrow 0^+$, does this mean that $\frac{0^+}{0^+} \rightarrow 1$?
|
$$\lim_{x\to3}\frac{\ln \sqrt {x-2}}{x^2-9} = \lim_{x\to3}\frac{\ln (1+\sqrt{x-2}-1)}{\sqrt {x-2}-1}\cdot\frac{\sqrt {x-2}-1}{x^2-9} = \lim_{x\to3}\frac{x-3}{1+\sqrt {x-2}}\cdot\frac{1}{x^2-9} = \frac{1}{12}$$
|
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|
Find $a, b\in \mathbb Z$ where $(a^2+b)(a+b^2)=(a-b)^3$ Find all non-zero $a, b\in \mathbb Z$ where
$$(a^2+b)(a+b^2)=(a-b)^3$$
I actually had no clue on what to try. Thanks for your help.
I believe I've already tried but per the 1st comment let me expand both sides and see what I cancel.
$$a^3+a^2b^2+ab+b^3=a^3-3a^2b+3ab^2-b^3$$
$$b(a^2b+2b^2+3a^2-3ab+a)=0$$
And then..?
|
Let me post it as an answer to mark this answered.
Thanks again to астон вілла олоф мэллбэрг!
$$a^3+a^2b^2+ab+b^3=a^3-3a^2b+3ab^2-b^3$$
$$b(a^2b+2b^2+3a^2-3ab+a)=0$$
$$b=0\quad or \quad 2b^2+(a^2-3a)b+3a^2+a=0$$
Applying quadratic formula on b,
$$b=\frac{a(3-a)\pm\sqrt{a^2(a-3)^2-24a^2-8a}}{4}=\frac{a(3-a)\pm (a+1)\sqrt{a(a-8)}}{4}$$
So $a(a-8)$ should be a square, let's say $n^2$.
$$a^2-8a=n^2$$
$$(a-4)^2:=m^2=n^2+16$$
$$(m,n)=(\pm4,0),(\pm5,\pm3)$$
$$(a,b)=(8,-10),(-1,-1),(9,-6),(9,-21)$$
|
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|
Formal way of showing inequality? At the end of my class today, there was a proof that involved using the fact that given $f(x)=3+4x-x^2-x^4$
$f(x)\leqslant 0 $ if $|x|>2 $
this seems obvious but I was wondering if anyone could show me a more formal way of proving it?
|
For $x\geq2$ we have $$x^4+x^2-4x-3=x^4-2x^3+2x^3-4x^2+3x^2-6x+2x-4+1=$$
$$=(x-2)(x^3+2x^2+3x+2)+1>0$$
For $x\leq-2$ we obtain:
$$x^4+x^2-4x-3=x^4+3x^3+4x^2-4x^3-16x^2-16x+13x^2+52x+52-40x-80+25=$$
$$=(x+2)^2(x^2-4x+13)-40(x+2)+25>0$$
|
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|
How to take the integral $\int\frac{dx}{\sin^3x}$ There's
$$\int\frac{\mathrm dx}{\sin^3x}.$$
I tried to write it like
$$\int\frac{(\sin^2x+\cos^2x)}{\sin^3x}\,\mathrm dx,$$
and then made partial fractions from it, but it didn't help much, the answer is still incorrect.
|
$$
\begin{aligned}
\int \frac{1}{\sin^3x}dx
& = \int \frac{\left(t^2+1\right)^2}{4t^3}dt
\\& =\frac{1}{4}\int \:\frac{1}{t^3}+t+\frac{2}{t}dt
\\& = \color{red}{\frac{1}{4}\left(\frac{1}{2}\tan ^2\left(\frac{x}{2}\right)-\frac{1}{2}\cot ^2\left(\frac{x}{2}\right)+2\ln \left|\tan \left(\frac{x}{2}\right)\right|\right)+C}
\end{aligned}
$$
Solved by substitution $\color{blue}{t=\tan \left(\frac{x}{2}\right)}$
|
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|
Which of the following numbers is greater? Which of the following numbers is greater?
Without using a calculator and logarithm.
$$7^{55} ,5^{72}$$
My try
$$A=\frac{7^{55} }{5^{55}×5^{17}}=\frac{ 7^{55}}{5^{55}}×\frac{1}{5^{17}}= \left(\frac{7}{5}\right)^{55} \left(\frac{1}{5}\right)^{17}$$
What now?
|
Observe that $7^2/5^2 = 49/25 < 2$. Cubing both sides, $(7^2/5^2)^3 < 8$. Since $7/5^3 = 7/125 < 1/8$ we have $7^7/5^9 < 8 \times 1/8 = 1$.
Then $7^{55}/5^{72} = 1/7 \times (7^7/5^9)^8 < 1/7 < 1$ and so $7^{55} < 5^{72}$.
|
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|
How to solve $p^2 - p + 1 = q^3$ over primes?
How to solve $p^2 - p + 1 = q^3$ over primes?
I checked that $p$ and $q$ must be odd and I don't know what to do next. I don't have any experience with such equations.
|
COMMENT.-Primes $p$ of the form $6x-1$ cannot be solution: in fact
$$(6x-1)^2-(6x-1)+1=36x^2-18x+3=3(12x^2-6x+1)$$ so $q=3$ which is not compatible with $12x^2-6x+1=3^2$.
It follows $p$ must be of the form $6x+1$ so we have
$$(6x+1)^2-(6x+1)+1=36x^2+6x+1=q^3$$ Solving the equation, where $X=6x$,
$$X^2+X+1-q^3=0$$ we have $$2X=-1+\sqrt{4q^3-3}$$ and the equation $$4q^3-3=z^2$$ seems to have $(z,q)=(37,7)$ as only solution which gives $\boxed{(p,q)=(19,7)}$.
Is it $(p,q)=(19,7)$ the only solution?
|
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|
If the points $(0,0),(a,11),(b,37)$ are vertices of equilateral triangle what is the value of ab? If the points $(0,0),(a,11),(b,37)$ are vertices of equilateral triangle, find the value of $ab$
|
Pretty lengthy equations (and probably bad method to solve). So here are the equations :
$$a^2+11^2=b^2+37^2=(b-a)^2+(37-11)^2$$
$$\implies b^2+a^2-2ab +676 = a^2+121\implies b^2-2ab+555=0\,\,\,\,(1)$$
$$b^2 = a^2-1278 \implies b=\sqrt{a^2-1278}\,\,\,\,\,\,(2)$$
substitute $b$ in $(1)$$$\implies (a^2-1278)+555 = 2a(\sqrt{a^2-1278})$$
Now this is ugly step :- square both sides=>
$$\implies a^4 + 693^2 -2(693)a^2 = 4a^2(a^2-1278)$$
And if you assign $a^2$ to some $t$ it is in quadraric form and should be solvable.
|
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|
Determinant of a matrix that is almost lower triangular Calculate the determinant of
$$ \left[
\begin{array}{cccc}
1 & 0 & 0 & 0 & \cdots & 1\\
1 & a_1 & 0 & 0 & \cdots & 0 \\
1 & 1 & a_2 & 0 & \cdots & 0 \\
1 & 0 & 1 & a_3 & \cdots & 0 \\
\vdots & \vdots & \vdots & \vdots & \ddots & \vdots \\
1 & 0 & 0 & \cdots& 1 & a_{n-1} \\
\end{array}
\right]$$
I tried to develop at the first line, but got stuck. Any helps or hints appreciated.
|
Expanding from the first one will force a product along the leading diagonal; the one at the end of the first row will leave you with another determinant
\begin{eqnarray*}
\left[
\begin{array}{cccc}
1 & 0 & 0 & 0 & \cdots & 1\\
1 & a_1 & 0 & 0 & \cdots & 0 \\
1 & 1 & a_2 & 0 & \cdots & 0 \\
1 & 0 & 1 & a_3 & \cdots & 0 \\
\vdots & \vdots & \vdots & \ddots & \ddots & \vdots \\
1 & 0 & 0 & \cdots& 1 & a_{n-1} \\
\end{array}
\right]=a_1 \cdots a_{n-1}+(-1)^{n+1}\left[
\begin{array}{cccc}
1 & a_1 & 0 & 0 & \cdots & 0 \\
1 & 1 & a_2 & 0 & \cdots & 0 \\
1 & 0 & 1 & a_3 & \cdots & 0 \\
\vdots & \vdots & \vdots & \ddots & \ddots & \vdots \\
1 & 0 & 0 & \cdots& 1 & a_{n-2} \\
1 & 0 & 0 & \cdots& 0 & 1 \\
\end{array}
\right]
\end{eqnarray*}
Now take the last row of this determinant & make it the first row & shift all the other rows down by one; this will introduce another $(-1)^{n+1}$ but will be essentially the same determinant that you started with but of size one smaller
\begin{eqnarray*}
-(-1)^{n+1}(-1)^{n+1}\left[
\begin{array}{cccc}
1 & 0 & 0 & 0 & \cdots & 1\\
1 & a_1 & 0 & 0 & \cdots & 0 \\
1 & 1 & a_2 & 0 & \cdots & 0 \\
1 & 0 & 1 & a_3 & \cdots & 0 \\
\vdots & \vdots & \vdots & \ddots & \ddots & \vdots \\
1 & 0 & 0 & \cdots& 1 & a_{n-2} \\
\end{array}
\right]
\end{eqnarray*}
So now we can use the above arguement recursively & we have
\begin{eqnarray*}
D=a_1 \cdots a_{n-1}-a_1 \cdots a_{n-2} +\cdots (-1)^{n-2}a_1+ (-1)^{n-1}.
\end{eqnarray*}
Tidy this expression up a little bit & we have the answer (exactly as Arden states in his comment)
\begin{eqnarray*}
D=\sum_{k=0}^{n-1}(-1)^{n-1-k}\prod_{i=1}^ka_i.
\end{eqnarray*}
|
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|
Solving: $(b+c)^2=2011+bc$
Solve $$(b+c)^2=2011+bc$$ for integers $b$ and $c$.
My tiny thoughts:
$(b+c)^2=2011+bc\implies b^2+c^2+bc-2011=0\implies b^2+bc+c^2-2011=0$
Solving in $b$ as Quadratic.$$\implies b=\frac{-c\pm \sqrt{8044-3c^2}} {2}.$$
So $8044-3c^2=k^2$, as $b$ and $c$ are integers. We also have inequalities: $8044>3c^2,8044>3b^2\\ \ \ \ \ 51>c\ \ \ \ , \ \ \ \ 51>b$
How to proceed further. Help.
|
Decided to see what I could do in my head.
$2011 \equiv 1 \pmod 3$ is a prime. This means it has both expressions as $u^2 + uv + v^2$ and $x^2 + 3 y^2.$ Note that $45^2 = (4 \cdot 5) \cdot 100 + 25 = 2025$ is too large for $x^2.$ Then $44^2 = 45^4 - 2 \cdot 45 + 1 = 2025 - 90 + 1 = 1935 + 1 = 1936$ is small enough. Then $2011 - 1936 = 11 + 64 = 75 = 3 \cdot 25 = 3 \cdot 5^2.$
So far we have $$ 2011 = 44^2 + 3 \cdot 5^2 $$
The order I can remember is $r^2 + rs + k s^2$ when $r = p - q, s = 2q$ gives $p^2 - 2pq + q^2 + 2 p q - 2 q^2 + 4kq^2 = p^2 + (4k-1)q^2. $ This time $k=1.$ That is, $p^2 + 3 q^2 = r^2 + rs + r^2$ when $r = p-q, s = 2 q.$ we got $p = 44, q = 5.$ So $r = 44 - 5, s = 2 \cdot 5,$ or
$$ r^2 + rs + s^2 = 2011, \; \; r = 39, \; \; s = 10. $$
It happens that $x^2 + xy + y^2$ and $u^2 + 3 v^2$ integrally represent exactly the same numbers. This is the only time this happens with positive forms. The other time is $x^2 + xy - y^2$ and $u^2 -5 v^2.$
As soon as we get to $x^2 + xy + k y^2$ with $|k| > 1$ and $1-4k$ not a square, there is a difference. For example, $x^2 + xy + 2 y^2$ represents more numbers (namely those $2 \pmod 4$) than $u^2 + 7 v^2,$ although both are very well behaved. Both represent all odd primes $p \equiv 1,2,4 \pmod 7.$ Hmmm, probable correction for negative $k.$ It appears $x^2 + xy - 3 y^2$ and $u^2 - 13 v^2$ represent the same numbers. Good to know.
|
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|
Quadratic equation find all the real values of $x$ Find all real values of $x$ such that
$\sqrt{x - \frac{1}{x}} + \sqrt{1 - \frac{1}{x}} = x$
I tried sq both sides by taking 1 in RHS but it didn't worked out well...
|
Divide by $x$:
$$\sqrt{\frac1x-\frac1{x^3}}+\sqrt{\frac1{x^2}-\frac1{x^3}}=1$$
Change $t=1/x$:
$$\sqrt{t-t^3}+\sqrt{t^2-t^3}=1$$
Square and rearrange:
$$2\sqrt{t^3-t^4-t^5+t^6}=t+t^2-2t^3-1$$
Square and rearrange again:
$$4t^6-4t^5-4t^4+4t^3=4t^6-4t^5-3t^4+6t^3-t^2-2t+1$$
Finally we get
$$t^4+2t^3-t^2-2t+1=0$$
That is,
$$(t^2+t-1)^2=0$$
|
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|
Multiple choice question of indefinite integral, $\int \frac{x + 9}{x^3 + 9x} dx$. If $\int \frac{x + 9}{x^3 + 9x} dx = k\arctan(mx) + n\ln (x) + p \ln (x^2 + 9) + c$, then $(m+n)/(k+p) = $
(A) 6
(B) -8
(C) -3
(D) 4
I tried solving it by differentiating the R.H.S. but couldn't arrive at the answer.
|
Try to separate the denominator like this
$$\frac{x+9}{x^3+9x}=\frac{x+9}{x(x^2+9)}$$
$$\qquad\qquad\ \,=\frac Ax+\frac{Bx+C}{x^2+9}$$
Then construct an equality for $A,B,C$ and we have
$$Ax^2+9A+Bx^2+Cx=x+9$$
Here we get $A=1,\ B=-1,\ C=1$, and that gives us
$$\int\frac{x+9}{x^3+9x}\,dx\ =\ \int\frac1x+\frac1{x^2+9}-\frac x{x^2+9}\,dx$$
Could you continue with this?
$----------$
Note that
\begin{align*}
&\int\frac{x}{x^2+9}\,dx\\
=\ \frac12&\int\frac{du}{u+9}\qquad(u=x^2,\ du=2x\,dx)\\
=\ \frac12&\ln(u+9)+c\\
=\ \frac12&\ln(x^2+9)+c
\end{align*}
|
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|
Show that $\sqrt{2-\sqrt{3}}=\frac{1}{2}(\sqrt{6}-\sqrt{2})$
Prove the following:
$\sqrt{2-\sqrt{3}}=\frac{1}{2}(\sqrt{6}-\sqrt{2})$
How do you go about proving this?
|
Previous comments and other answers show you the path to the solution.
I would like to propose another formula of the same kind, thats illustrates the fact that cancellation of nested square roots may sometimes occur ... but not necessarily in any obvious way !
Question Find a simpler form for $x=\sqrt{2+\sqrt{3}}+\sqrt{4-\sqrt{7}}$
Answer
We have :
$$x\sqrt{2}=\sqrt{4+2\sqrt{3}}+\sqrt{8-2\sqrt{7}}=\sqrt{\left(1+\sqrt{3}\right)^{2}}+\sqrt{\left(1-\sqrt{7}\right)^{2}}=1+\sqrt{3}+\sqrt{7}-1=\sqrt{3}+\sqrt{7}$$
Hence : $2x^{2}=10+2\sqrt{21}$ and finally (since $x\ge0$) :
$$\boxed{x=\sqrt{5+\sqrt{21}}}$$
A slightly more general formula could be :
For all $(p,q)\in\mathbb{N}^2$ :
$$\sqrt{p+1+\sqrt{2p+1}}+\sqrt{q+1-\sqrt{2q+1}}=\sqrt{p+q+1+\sqrt{\left(2p+1\right)\left(2q+1\right)}}$$
|
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|
Regarding complex roots of a polynomial I am having difficulty in finding the roots of the following polynomial when it is given that all the roots are complex:-
$$f(x)= x^4+4x^3+8x^2+8x+4$$
How can I factorize the polynomial to get its roots?
|
Using binomial theorem;
$$\begin{align}f(x)&=x^4+4x^3+8x^2+8x+4\\
&=\left(x^4+4x^3+6x^2+4x+1\right)+2\left(x^2+2x+1\right)+1\\
&=(x+1)^4+2(x+1)^2+1\\
&=\left((x+1)^2+1\right)^2\\
&=\left(x^2+2x+2\right)^2\end{align}$$
|
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|
If $\sin A+\sin^2 A=1$ and $a\cos^{12} A+b\cos^{8} A+c\cos^{6} A-1=0$ If $\sin A+\sin^2 A=1$ and $a\cos^{12} A+b\cos^{8} A+c\cos^{6} A-1=0$.
Find the value of $2b + \dfrac {c}{a}$.
My Attempt:
$$\sin A+\sin^2 A=1$$
$$\sin A + 1 - \cos^2 A=1$$
$$\sin A=\cos^2 A$$
Now,
$$a(\cos^2 A)^{6}+b(\cos^2 A)^{4}+c(\cos^2 A)^{3}=1$$
$$a\sin^6 A+ b\sin^4 A+c\sin^3 A=1$$
How do I proceed further?
|
@MyGlasses .. In the third line from below you forgot that there is $(-a)$ twice in your attempt to prove the required identity, resulting in a mistake in your answer.
|
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"language": "en",
"url": "https://math.stackexchange.com/questions/2166927",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
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|
How many integers between $10000$ and $99999$ ( inclusive) are divisible by $3$ or $5$ or $7$? How many integers between $10000$ and $99999$ (inclusive) are divisible by $3$ or $5$ or $7$ ?
My Try :
Total Integers between $10000$ and $99999$ are $89999$.
$\left\lfloor\frac{89999}{3}\right\rfloor+\left\lfloor\frac{89999}{5}\right\rfloor+\left\lfloor\frac{89999}{7}\right\rfloor$ - $\left\lfloor\frac{89999}{3\times5}\right\rfloor-\left\lfloor\frac{89999}{3\times7}\right\rfloor-\left\lfloor\frac{89999}{5\times7}\right\rfloor$ + $\left\lfloor\frac{89999}{3\times5\times7}\right\rfloor$ =
$48857$
I don't have an answer for this. Am I right here ?
|
Another way:
$10000=3\times 3333+1\implies$ First integer $>10000$ divisible by $3$ is $10002$.
So if we count a number of integers in sequence $10002,10005,\cdots,99999$ we have our answer. the common difference is $3$ first term is $10002$ hence, $a_n=10002+(n-1)3=99999\implies n=30000$. Similarly for others too.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2168060",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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|
How do I determine a basis of the vector space of polynomials degree 3 or less that satisfy $\int_{0}^{1}[xp'(x)-p(x)]=p(1)-2p(0)$? I have proven that the set of polynomials of degree 3 or less that satisfy $\int_{0}^{1}[xp'(x)-p(x)]=p(1)-2p(0)$ is indeed a subspace of $\mathscr{P}_{3}(\mathbb{R})$.
Now I need to find a basis. I found that all degree zero polynomials work. I've tried just using the monomials of each degree $(\lambda,0,0,0),(0,\lambda x,0,0),...,(0,0,0,\lambda x^{3})$ and I've found that none of the rest work...
It doesn't seem right to say that my basis is just one vector, but I'm at a loss on how to find anything else.
|
Let $p(x) = a + bx + cx^2 + dx^3$. Then the equation reads
$$
\int_0^1 x(b + 2cx + 3d x^2) - (a + bx + cx^2 + dx^3) \mathrm{d} x = a+b+c+d - 2a
$$
Evaluating the left-hand side gives
$$
\frac b2 + \frac{2c}{3} + \frac{3d}{4} - a - \frac{b}{2} - \frac{c}{3} - \frac{d}{4} = a + b + c + d - 2a,
$$
or
$$
b = - \frac{2}{3} c - \frac{1}{2} d.
$$
Thus the general form is
$$
p(x) = a + \left(- \frac{2}{3} c - \frac{1}{2} d \right)x + cx^2 + dx^3 = a + c \left( x^2 - \frac{2}{3} x \right) + d \left( x^3 - \frac{1}{2} x \right)
$$
and we can identify a basis $\left\{ 1, x^2 - \frac{2x}{3}, x^3 - \frac{x}{2} \right\}$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2169563",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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|
Sum of reciprocal of the positive divisors of $1800.$ Sum of reciprocal of the positive divisors of $1800.$
Attempt: divisors of $1800 = 2^3\times 3^2 \times 5^2$
so sum of divisors of $1800$ is $\displaystyle (1+2+2^2+2^3)\times (1+3+3^2)\times (1+5+5^2)$
$ \displaystyle = \bigg(\frac{2^4-1}{2-1}\bigg)\times \bigg(\frac{3^3-1}{3-1}\bigg)\times \bigg(\frac{5^3-1}{5-1}\bigg)$
but answer is different, could some help me to solve it. Thanks.
|
HINT:
$$\sum_{d=1,d|n}^n\dfrac1d=\dfrac1n\sum_{d=1,d|n}^n\dfrac nd=\dfrac1n\sum_{d=1,d|n}^nd$$
Now use this
See also: Is there a formula to calculate the sum of all proper divisors of a number?
|
{
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"timestamp": "2023-03-29T00:00:00",
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|
Determine: $S = \frac{2^2}{2}{n \choose 1} + \frac{2^3}{3}{n \choose 2} + \frac{2^4}{4}{n \choose 3} + \cdots + \frac{2^{n+1}}{n+1}{n \choose n}$ We are given two hints: consider $(n+1)S$; and use the Binomial Theorem. But we are not to use calculus.
My consideration of $(n+1)S$ goes like this:
\begin{align*}
\sum\limits_{k=1}^{n}\frac{2^{k+1}}{k+1}{n \choose k}
&= \frac{1}{n+1}\sum\limits_{k=1}^{n}(n+1)\frac{2^{k+1}}{k+1}{n \choose k} \\
&= 2\frac{1}{n+1}\sum\limits_{k=1}^{n}2^k{n+1 \choose k+1} \\
&= 2\frac{1}{n+1}\sum\limits_{k=1}^{n}(1+1)^k{n+1 \choose k+1} \\
\end{align*}
Now I think I'm in a position to use the Binomial Theorem, giving
\begin{equation*}
2\frac{1}{n+1}\sum\limits_{k=1}^{n}\sum\limits_{i=0}^{k}{k \choose i}{n+1 \choose k+1}
\end{equation*}
I don't know if I am on the right track, but I do know that I'm stuck. Can anyone offer any advice on how to proceed?
|
The trick is to get rid of the factors $1/(k+1)$.
You do so by absorbing $k+1$ in $k!$, and
$$\frac{\binom nk}{k+1}=\frac{n!}{(k+1)k!(n-k)!}=\frac{n!}{(k+1)!(n+1-k-1)!}=\frac{\binom{n+1}{k+1}}{n+1}.$$
Now,
$$S=\sum\limits_{k=1}^{n}\frac{2^{k+1}}{k+1}{n \choose k}=\frac1{n+1}\sum\limits_{k=1}^{n}2^{k+1}{n+1 \choose k+1}.$$
The summation is $3^{n+1}$ from wich the terms $k=-1$ and $0$ have been removed. Hence
$$S=\frac{3^{n+1}-1-2(n+1)}{n+1}.$$
|
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"timestamp": "2023-03-29T00:00:00",
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|
why is $2^{\log^{2}\frac{1}{2}} = 1/2$? I have no understanding of $\log^2n$, wouldn't $2^{\log^{2}\frac{1}{2}} = 1/4$ since $2^{\log\frac{1}{2}} = 1/2$?
$\log$ here has the base of $2$.
|
Using a combination of exponent rules and the one log rule you mentioned,
$$2^{\log^2\frac{1}{2}} = 2^{\log\frac{1}{2} \cdot \log\frac{1}{2}} = (2^{\log\frac{1}{2}})^ {\log\frac{1}{2}} = \left(\frac{1}{2}\right)^{\log\frac{1}{2}} = \frac{1}{2^{\log\frac{1}{2}}} = \frac{1}{\frac{1}{2}} = 2$$
|
{
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 1
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|
Fisher information of exponential distribution using the generic formula. I am trying to calculate Fisher information when $X_1...X_n$ are IID and follow the exponential distribution:
$$f(x)=\frac{1}{\theta}e^{-\frac{x}{\theta}}$$
I use the following formula of Fisher information to confirm that the result is indeed the same as with the other formulas:
$$I(\theta) = E[(\frac{\partial}{\partial \theta }logf(X;\theta))^2]$$
I have calculated that for $\frac{\partial}{\partial \theta }logf(X;\theta)$ we have:
$$\frac {-n}{\theta}+ \frac{\sum_1^nX_i}{\theta^2}$$
Squaring that:
$$\frac{n^2}{\theta^2}+\frac{(\sum_1^n X_i)^2}{\theta^4}-\frac{2n\sum_1^nX_i}{\theta^3}$$
And finally taking the expected value and letting it pass through the constants::
$$\frac{n^2}{\theta^2}+\frac{E(\sum_1^n X_i)^2}{\theta^4}-\frac{2nE(\sum_1^nX_i)}{\theta^3}$$
Recalling that $E(X)= \theta$ and $E(X^2)=2\theta^2$ it seems as though the 2nd and 3rd term cancel out leaving $\frac{n^2}{\theta^2}$, but the correct answer is $\frac{n}{\theta^2}$. Where does this go wrong?
|
Ok using that parameterisation I agree your likelihood is correct! So method one we differentiate again to get
$$
\ell_{\theta \theta} = -\frac{2 \sum x_i }{\theta^3} + \frac{n}{\theta^2}
$$
now since $\mathbb{E} \sum_i x_i = n \theta$ we get
$$
\begin{align}
\mathcal{I} &= - \mathbb{E}\left[ \ell_{\theta \theta} \right] \\
&= 2 \frac{n\theta}{\theta^3} - \frac{n}{\theta^2} \\\
&= \frac{n}{\theta^2}
\end{align}
$$
Alternatively noting that
$$
\begin{align}
\mathbb{E}(\sum_i X_i )^2 &= n\mbox{Var}(X_1) + (n\theta)^2 \\
&= n \theta^2 + n^2 \theta^2
\end{align}
$$
we have
$$
\begin{align}
\mathbb{E} \left[ \ell_{\theta} ^2 \right] &= \mathbb{E} \left[ \left(\frac{1}{\theta^2} \sum_i x_i - \frac{n}{\theta} \right)^2\right] \\
&= \frac{1}{\theta^4}\mathbb{E} \left[\left(\sum_i X_i\right)^2 \right]- \frac{2n}{\theta^3}\mathbb{E}\left[\sum X_i \right]+ \frac{n^2}{\theta^2} \\
&= \frac{n\theta^2}{\theta^4} + \frac{n^2 \theta^2}{\theta^4} - \frac{2 n^2 \theta}{\theta^3} + \frac{n^2}{\theta^2} \\
&=\frac{n}{\theta^2}+\frac{2 n^2}{\theta^2} - \frac{2n^2}{\theta^2} \\
&= \frac{n}{\theta^2}.
\end{align}
$$
|
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"timestamp": "2023-03-29T00:00:00",
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|
Find minimum of $y=\sqrt{-x^2+4x+12}-\sqrt{-x^2+2x+3}$
Find minimum of $y=\sqrt{-x^2+4x+12}-\sqrt{-x^2+2x+3}$
My work so far:
1) $$y =\frac{2x+9}{\sqrt{-x^2+4x+12}+\sqrt{-x^2+2x+3}} $$
2) I used a derivative and found the answer ($y=\sqrt3$ at $x=0$). Is there any other way?
|
Domain gives $-1\leq x\leq3$.
If $x=0$ so $y=\sqrt3$.
We'll prove that it's a minimal value of $y$.
Id est, we need to prove that
$$\sqrt{-x^2+2x+3}+\sqrt3\leq\sqrt{-x^2+4x+12}$$ or after squaring of the both sides we need to prove that
$$\sqrt{3(-x^2+2x+3)}\leq x+3$$ or since $x+3>0$, we need to prove that
$$3(-x^2+2x+3)\leq(x+3)^2,$$
which is $x^2\geq0$.
Done!
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2180644",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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|
Prove that $b_n = b_{n+100}$
Let $b_n$ denote the units digit of $\displaystyle\sum_{a=1}^n a^a$. Prove that $b_n = b_{n+100}$.
I tried rewriting the sum, but didn't see how to prove the equality. For example, if $n = 178$ we have $$\displaystyle\sum_{a=1}^{178} a^a = (1^1+2^2+3^3+\cdots+78^{78})+(79^{79}+80^{80}+81^{81}+\cdots+178^{178})$$ and so we need to show that $79^{79}+80^{80}+81^{81}+\cdots+178^{178} \equiv 0 \pmod{10}$. Is there an easier way to prove the general result?
|
It is sufficient to prove that for each integer $k$
$$\sum_{a=t+1}^{t+100} a^a \equiv 0 \mod10$$
Consider a subsequence with indices $k,k+10,\dots,k+90$. We have
$$
b(k) = k^k + (k+10)^{k+10} +\dots+(k+90)^{k+90}
$$
Obviously, the following holds $$b(k) \equiv k^k + k^{k+10} + \dots+k^{k+90}\mod10$$
The right side could be rewritten
$$k^k + k^{k+10} +\dots+k^{k+90} = k^k(1 + k^{10} +\dots+k^{90})$$
Lemma 1$$1 +k^{10} + k^{20}+\dots+k^{90} + k^{100} \equiv 1 \mod10$$
Proof. Here we use Fermat's little theorem, which states that $b$ coprime with $10$ sutisfies the condition $b^\varphi(10) = b^4\equiv 1\mod 10$. And for $b$ coprime with $5$ we have $b^\varphi(5) = b^4\equiv 1\mod 10$.
So, in the proof it is sufficient to check only the integers from $[0,9]$.
*
*For the coprimes with $10$ we have $$1 +k^{10} + k^{20}+\dots+k^{100}\equiv 1 + 5(1+k^2)\equiv 1\mod10.$$
*For $k = 5$ we have $5^{10i}(1-5^{100-10i})\equiv 1\mod10$, where the expression inside the brackets is even.
*For an even $k$ we get
$$1 + k^{10}(1 + k^{10}+\dots+k^{90})\equiv 1 \mod10,$$
where the expression in the brackets is divisible by $5$ because
$$1 + k^{10}+\dots+k^{90}\equiv 5(1+k^2)\equiv 0\mod5,$$
Then
$$k^k + k^{k+10} + k^{k+20}+\dots+k^{k+90} \equiv k^k(1-k^{100})\mod10$$
Lemma 2 $$a^a(1 - a^{100})\equiv 0\mod10,$$ for an integer $a$.
Proof.
*
*For the comprimes with $10$, it is straightforward from Fermat's little theorem.
*For $a \equiv 5\mod10$ we have even value for $(1-a^{100})$.
*For an even number $a$ comprime with $5$, the quantity $(1-a^{100})$ is divisible by $5$ according to Fermat's little theorem.
*For $a\equiv 0 \mod10$ it is obvious.
Finally, we can return to the initial sum and rewrite it
\begin{align}
\sum_{a=k}^{k+100}a^a &\equiv
k^k(1 - k^{100}) + (k+1)^{k+1}(1 - (k+1)^{100}) +\dots+(k+9)^{k+9}(1 - (k+9)^{100})\mod10
\end{align}
Where each term is divisible by $10$ according to Lemma 2.
Therefore, the whole last sum is actually divisible by $10$.
|
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"timestamp": "2023-03-29T00:00:00",
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|
Evaluate the sum of series $\sum_{k=1}^{+\infty}(-1)^{k-1}\frac{1}{k(4k^2-1)}$ Evaluate the sum of series
$$\sum_{k=1}^{+\infty}(-1)^{k-1}\frac{1}{k(4k^2-1)}$$
I have tried two methods:
1) using power series
2) using partial sums
but I can't find the sum.
1) Using power series:
$$\sum_{k=1}^{+\infty}(-1)^{k-1}\frac{1}{k(4k^2-1)}x^{k(4k^2-1)}$$
$$f(x)=\sum_{k=1}^{+\infty}(-1)^{k-1}\frac{1}{k(4k^2-1)}x^{k(4k^2-1)}$$
After derivation:
$$f'(x)=\sum_{k=1}^{+\infty}(-1)^{k-1}x^{4k^3-k-1}$$
The problem here is that:
$$\sum_{k=1}^{+\infty}(-1)^{k-1}x^{4k^3-k-1}=x^2-x^{29}+x^{104}-...$$
Is it possible to find the closed form for the last series?
2) Using partial sums:
$$S_n=\sum_{k=0}^{n}(-1)^{k}\frac{1}{(k+1)(4(k+1)^2-1)}$$
Now, using the formula:
$$S_n+a_{n+1}=a_0+\sum_{k=1}^{n+1}(-1)^{k}\frac{1}{(k+1)(4(k+1)^2-1)}\Rightarrow$$
$$S_n+(-1)^{n+1}\frac{1}{(k+2)(4(k+2)^2-1)}=\frac{1}{3}+\sum_{k=1}^{n+1}(-1)^{k}\frac{1}{(k+1)(4(k+1)^2-1)}$$
$$S_n=\frac{1}{3}-\frac{1}{30}+...+(-1)^{n}\frac{1}{(n+1)(4(n+1)^2-1)}$$
$$\sum_{k=1}^{n+1}(-1)^{k}\frac{1}{(k+1)(4(k+1)^2-1)}=T_n=-\frac{1}{30}+...+(-1)^{n+1}\frac{1}{(n+2)(4(n+2)^2-1)}$$
$$T_n=S_n-\frac{1}{3}+(-1)^{n+1}\frac{1}{(n+2)(4(n+2)^2-1)}$$
Going back to the formula $$S_n+a_{n+1}=a_0+\sum_{k=1}^{n+1}(-1)^{k}\frac{1}{(k+1)(4(k+1)^2-1)}$$
we have that $S_n$ cancels, so we can't determine partial sums using this method?
$$S_n+(-1)^{n+1}\frac{1}{(k+2)(4(k+2)^2-1)}=\frac{1}{3}+T_n$$
$$S_n+(-1)^{n+1}\frac{1}{(k+2)(4(k+2)^2-1)}=\frac{1}{3}+S_n-\frac{1}{3}+(-1)^{n+1}\frac{1}{(n+2)(4(n+2)^2-1)}$$
Question: How to find the sum of this series?
|
With telescoping we obtain
\begin{align*}
\sum_{k=1}^\infty& (-1)^{k-1}\frac{1}{k(4k^2-1)}\\
&=\lim_{N\to\infty}\sum_{k=1}^N(-1)^{k-1}\frac{1}{k(4k^2-1)}\\
&=\lim_{N\to\infty}\sum_{k=1}^N(-1)^{k-1}\left(\frac{1}{2k-1}+\frac{1}{2k+1}-\frac{1}{k}\right)\\
&=1-\sum_{k=1}^\infty(-1)^{k-1}\frac{1}{k}\\
&=1-\ln 2
\end{align*}
|
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"timestamp": "2023-03-29T00:00:00",
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|
Variation of Coupon Collectors Problem for the case where $k\le n$. I was working on this problem which says,
Suppose one draws balls with replacement from an urn containing $n$ unique balls and records its number. Then what is the expected number of draws required for getting
(a) the set of balls numbered $\{1,2,\cdots,k\}$ recorded? Where $k\leq n$ a fixed number.
(b) the set of any $k$ distinct balls?
N.B I have solved the part (a) which comes out to be $nH_k$; $H_k=\sum_{j=1}^{k}\frac{1}{j}$. But I'm stuck at the second part. Any help is much appreciated.
|
By way of enrichment here is a generating function approach to the
question of when the first $k$ coupons where $k\le n$ have been seen.
Using the notation from the following MSE
link we get from
first principles for the probability of $m$ draws that
$$P[T=m] = \frac{1}{n^m} \times {k\choose k-1} \times
\sum_{q=0}^{n-k} {n-k\choose q} {m-1\brace q+k-1}
(q+k-1)!.$$
What happens here is that we choose the $k-1$ of the $k$ values that
go into the prefix, which also determines the value that will complete
the set with the last draw. We then choose a set of $q$ values not
from the $k$ initial ones and partition the first $m-1$ draws or slots
into $q+k-1$ sets, one for each value.
We verify that this is a probability distribution, getting
$$\sum_{m\ge 1} P[T=m] \\ =
\sum_{m\ge 1} \frac{1}{n^m} \times {k\choose k-1} \times
\sum_{q=0}^{n-k} {n-k\choose q} (m-1)! [z^{m-1}]
(\exp(z)-1)^{q+k-1}
\\ = k \sum_{m\ge 1} \frac{1}{n^m} \times
(m-1)! [z^{m-1}] \sum_{q=0}^{n-k} {n-k\choose q}
(\exp(z)-1)^{q+k-1}
\\ = k \sum_{m\ge 1} \frac{1}{n^m} \times
(m-1)! [z^{m-1}] (\exp(z)-1)^{k-1}\exp(z(n-k))
\\ = k! \sum_{m\ge 1} \frac{1}{n^m}
\sum_{q=0}^{m-1} {m-1\choose q} {q\brace k-1} (n-k)^{m-1-q}
\\ = k! \sum_{q\ge 0} {q\brace k-1}
\sum_{m\ge q+1} {m-1\choose q} \frac{1}{n^m} (n-k)^{m-1-q}
\\ = k! \sum_{q\ge 0} {q\brace k-1} \frac{1}{n^{q+1}}
\sum_{m\ge 0} {m+q\choose q} \frac{1}{n^m} (n-k)^m
\\ = k! \sum_{q\ge 0} {q\brace k-1} \frac{1}{n^{q+1}}
\frac{1}{(1-(n-k)/n)^{q+1}}
= k! \sum_{q\ge 0} {q\brace k-1} \frac{1}{k^{q+1}}.$$
Recall the OGF of the Stirling numbers of the second kind which says
that
$${n\brace k} = [z^n] \prod_{p=1}^k \frac{z}{1-pz}.$$
We obtain
$$k! \sum_{q\ge 0} \frac{1}{k^{q+1}}
[z^q] \prod_{p=1}^{k-1} \frac{z}{1-pz}
= (k-1)! \prod_{p=1}^{k-1} \frac{1/k}{1-p/k}
\\ = (k-1)! \prod_{p=1}^{k-1} \frac{1}{k-p} = 1$$
and the sanity check goes through. For the expectation of when the
first $k$ have been seen we recycle the above, inserting a factor of
$m$, starting from
$$k! \sum_{q\ge 0} {q\brace k-1}
\sum_{m\ge q+1} {m-1\choose q} \frac{m}{n^m} (n-k)^{m-1-q}
\\ = k! \sum_{q\ge 0} {q\brace k-1}
\sum_{m\ge q+1} {m\choose q+1} \frac{q+1}{m} \frac{m}{n^m} (n-k)^{m-1-q}
\\ = k! \sum_{q\ge 0} {q\brace k-1} (q+1)
\sum_{m\ge q+1} {m\choose q+1} \frac{1}{n^m} (n-k)^{m-1-q}
\\ = k! \sum_{q\ge 0} {q\brace k-1} (q+1) \frac{1}{n^{q+1}}
\sum_{m\ge 0} {m+q+1\choose q+1} \frac{1}{n^m} (n-k)^{m}
\\ = k! \sum_{q\ge 0} {q\brace k-1} (q+1) \frac{1}{n^{q+1}}
\frac{1}{(1-(n-k)/n)^{q+2}}
\\ = n \times k! \sum_{q\ge 0} {q\brace k-1} (q+1) \frac{1}{k^{q+2}}
= \frac{n}{k^2} \times k!
\sum_{q\ge 0} {q\brace k-1} (q+1) \frac{1}{k^{q}}.$$
Activating the OGF produces
$$\frac{n}{k^2} \times k!
\sum_{q\ge 0} \frac{1}{k^{q}}
[z^q] \left(\prod_{p=0}^{k-1} \frac{z}{1-pz}\right)'
\\ = \frac{n}{k^2} \times k!
\sum_{q\ge 0} \frac{1}{k^{q}}
[z^q] \prod_{p=0}^{k-1} \frac{z}{1-pz}
\sum_{p=0}^{k-1} \frac{1}{z(1-pz)}
\\ = \frac{n}{k^2} \times k!
\prod_{p=0}^{k-1} \frac{1/k}{1-p/k}
\sum_{p=0}^{k-1} \frac{1}{1/k(1-p/k)}
\\ = n \times k!
\prod_{p=0}^{k-1} \frac{1}{k-p}
\sum_{p=0}^{k-1} \frac{1}{k-p}
= n \times k! \times \frac{1}{k!} \times H_k.$$
This yields the answer
$$\bbox[5px,border:2px solid #00A000]{ n H_k.}$$
What we have here are in fact two annihilated coefficient extractors
(ACE) more of which may be found at this MSE
link.
|
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"url": "https://math.stackexchange.com/questions/2181969",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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|
How can I derive the conditions of Positive semidefinite cone in $2\times2$ matrix. By the definition, in order for $X$ to be positive semidefinite cone in $S^2$, it should satisfy that
\begin{equation}
X=\left[
\begin{array}{cc}
x & y \\
y & z
\end{array}
\right]\in S_+^2 \quad\Longleftrightarrow\quad x\ge0,\quad z\ge0, \quad xz\ge y^2,\tag{1}
\end{equation}
where
$$
S_+^2 = \left\{X\in S^2 | X \succeq 0\right\}.
$$
I have failed to prove the $(1)$.
$$
\begin{align}
\forall \alpha,\beta,\quad
\left[
\begin{array}{cc}
\alpha & \beta
\end{array}
\right]
\left[
\begin{array}{cc}
x & y \\
y & z
\end{array}
\right]
\left[
\begin{array}{cc}
\alpha \\
\beta
\end{array}
\right] &\ge 0\\
\alpha^2x + 2\alpha\beta y + \beta^2z &\ge 0\\
\left(\alpha\sqrt{x} + \beta\sqrt{z}\right)^2 + 2\alpha\beta(y-\sqrt{xz}) &\ge 0
\\\therefore \alpha\beta(y-\sqrt{xz})\ge0\\
\text{if}\quad\alpha\beta\ge0,\quad \text{then} \quad y^2\ge xz\\
\text{if}\quad\alpha\beta\le0,\quad \text{then} \quad y^2\le xz
\end{align}
$$
How can I reach to
$$
x\ge0,\quad z\ge0,\quad xz\ge y^2\qquad?
$$
|
For the 2x2 case, you can form the characteristic polynomial, obtaining an expression for the eigenvalues using the quadratic formula. Setting the eigenvalues greater than or equal to zero yields the inequalities:
$2 \lambda_{+} = (x + z) + \sqrt{(x+z)^2 + 4(y^2-xz)} \ge 0$
$2 \lambda_{-} = (x + z) - \sqrt{(x+z)^2 + 4(y^2-xz)} \ge 0$
Observe that $(x+z)$ less than zero implies that $\lambda_{-}$ will always be negative. Since we require both eigenvalues to be positive, $x+z \ge 0$.
$x+z \ge 0 \implies \lambda_{+} \ge 0$
(the square roots will not yield complex numbers, because real symmetric matrices always have real eigenvalues)
$\lambda_{-} \ge 0 \implies (x+z) - \sqrt{(x+z)^2 + 4(y^2-xz)} \ge 0$
$\implies (x+z) \ge \sqrt{(x+z)^2 + 4(y^2-xz)}$
$\implies (x+z)^2 \ge (x+z)^2 + 4(y^2-xz)$
$\implies 0 \ge y^2-xz$
$\implies xz \ge y^2$
We thus have $x+z \ge 0$ and $xz \ge y^2$ being the conditions which ensure that $X$ is positive semidefinite.
Now observe that $x$ and $z$ cannot have opposing signs if their product is greater than or equal to the (always positive) $y^2$. It follows that $x \ge 0$ and $z \ge 0$.
Putting it altogether, $X$ is positive semidefinite if and only if $x \ge 0$, $z \ge 0$ and $xz \ge y^2$.
|
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"language": "en",
"url": "https://math.stackexchange.com/questions/2183078",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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|
How many $2 \times 2$ matrices are there with entries from the set ${\{0,1,2,...,i}\}$ in which there are no zeros rows and no zero columns? How many $2 \times 2$ matrices are there with entries from the set ${\{0,1,2,...,i}\}$ in which there are no zeros rows and no zero columns?
attempt:
Suppose we have a matrix $ \begin{pmatrix}
a & b \\
c & d
\end{pmatrix}$. Then we don't want $ \begin{pmatrix}
0 & 0 \\
0 & 0
\end{pmatrix}$ or $ \begin{pmatrix}
0 & 0 \\
c & d
\end{pmatrix}$ or $ \begin{pmatrix}
a & b \\
0 & 0
\end{pmatrix}$.
Similarly for the columns.
Then we have the first case:
that all of the $a,b,c,d \neq 0$,and so we have $i^4$ options to choose such a matrix.
second case: we have that $ \begin{pmatrix}
0 & b \\
c & 0
\end{pmatrix}$ or $ \begin{pmatrix}
a & 0 \\
0 & d
\end{pmatrix}$. so we have $i^2$ or $i^2$ on both,so $2i^2$ ways to choose a matrix.
third case : either the $a,b,c,d$ is zero and the rest nonzero.
I am not really sure. I would add the cases to get the final answer.
Can someone please help me? Any feedback would really help.
Thanks
|
You may count the number of two tuples that can be formed from the elements of your set. clearly there are $(i+1)^2$ two tuples. Now the first entry should not be the vector $(0,0)$ so you have $(i+1)^2-1$ options for first row of marix. You have number of options for second row but then you have to subtract matrices of form $$ \begin{pmatrix}
0 & b \\
0& d
\end{pmatrix}$$ and $$ \begin{pmatrix}
a& 0\\
c & 0
\end{pmatrix}$$ (where $a$, $b$, $c$ and $d$ are non zero)
which are $2i^2$ in number.
so your answer should be $${[(i+1)^2-1]}^2-2i^2$$
|
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"language": "en",
"url": "https://math.stackexchange.com/questions/2184101",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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"answer_id": 0
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|
Evaluate: $\int_0^1 \frac{\left(1-x^4\right)^{3/4}}{\left(x^4+1\right)^2} \, dx$ Here is what I did:
$\int_0^1 \frac{\left(1-x^4\right)^{3/4}}{\left(x^4+1\right)^2} \, dx$
Put $x^2=\tan (\theta )$
Then: $x=\sqrt{\tan (\theta )}$ and $dx=\frac{1}{2} \tan ^{-\frac{1}{2}}(\theta ) \sec ^2(\theta )d\theta$
So the integral becomes $\int_0^{\frac{\pi }{4}} \frac{\left(1-\tan ^2(x)\right)^{3/4} \tan ^{-\frac{1}{2}}(x) \sec ^2(x)}{2 \left(\tan ^2(x)+1\right)^2} \, dx$
= $\frac{1}{2} \int_0^{\frac{\pi }{4}} \frac{\sin ^{-\frac{1}{2}}(x) {\cos^{\frac{1}{2}} (x)} \sec ^2(x) \left(\frac{\cos ^2(x)-\sin ^2(x)}{\cos ^2(x)}\right)^{3/4}}{\sec ^4(x)} \, dx$
= $\frac{1}{2} \int_0^{\frac{\pi }{4}} \sin ^{-\frac{1}{2}}(x) \cos (x) \cos ^{\frac{3}{4}}(2 x)\, dx$
Now how do I proceed?
I tried converting it to the form $\sin ^p(2 x) \cos ^q(2 x)$ but that proved impossible.
The textbook gives the answer as $\frac{1}{2^{9/2}}B\left(\frac{7}{4},\frac{1}{4}\right)$ which WolframAlpha tells me is 0.1472... but if ask it to evaluate the integral, I get the answer as 0.700... Who is right?
|
These are three successive substitutions that will get you to the answer:
First, use: $ a) \, x^2 = \tanh{\theta}$
Then, use: $ b) s^2 = \sinh{\theta}$
Finally, use: $ c) \, u = 2s^4$
Then use the following definition of the Beta Function:
$\displaystyle \text{B}(p,q) = \int_0^\infty \frac{x^{p-1}}{(1+x)^{p+q}} \text{d}t$
The answer is $\displaystyle \frac{1}{4\sqrt[4]{2}}\text{B}\left(\frac{1}{4},\frac{7}{4}\right) = \frac{3\pi \sqrt[4]{2}}{16}$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2184406",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
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|
Differentiation of a function We need to find out the derivative $\frac{dy}{dx}$ of the following:
$x^m$$y^n$=$(x+y)^{m+n}$
*
*I know how to differentiate the function and on solving we get $\frac{dy}{dx}$=$\frac{y}{x}$. But we notice that $\frac{dy}{dx}$ is independent of values of $m$ and $n$. So what I did was again starting the problem from start but this time substituting any arbitrary values of $m$ and $n$, like $m$=$n$=$1$ (lets say). That is we get $xy$=$(x+y)^2$. Now opening the square bracket and solving we get
$x^2+y^2+xy=0$
So $\frac{dy}{dx}$=$\frac{-2x-y}{2y+x}$
But the answer should be $\frac{y}{x}$.Whats wrong in assuming arbitrary values of $m$ and $n$ if the answer dosen't depend upon values of $m$ and $n$.Please help.
How I got $\frac{dy}{dx}$=$\frac{y}{x}$?
$x+y$=$x^\frac{m}{m+n}$$y^\frac{m}{m+n}$
$ln(x+y)$=$\frac{m(ln(x))}{(m+n)}$+$\frac{n(ln(y))}{(m+n)}$
Differentiating we get
$\frac{1}{x+y}$.$(1+\frac{dy}{dx})$=$\frac{m}{(m+n)x}$+$\frac{n}{(m+n)y}\frac{dy}{dx}$
$\frac{dy}{dx}(\frac{1}{x+y}-\frac{n}{(m+n)y})$=$\frac{m}{(m+n)x}-\frac{1}{x+y}$
Just rearrange and you are done $\frac{dy}{dx}$=$\frac{y}{x}$
|
I'm not quite sure how you got $\frac{dy}{dx}=\frac yx$. You should instead have
$$\frac d{dx}x^my^n=mx^{m-1}y^n+nx^my^{n-1}\frac{dy}{dx}$$
$$\frac d{dx}(x+y)^{m+n}=(m+n)(x+y)^{m+n-1}\left(1+\frac{dy}{dx}\right)$$
Put these two together and we get
$$mx^{m-1}y^n+nx^my^{n-1}\frac{dy}{dx}=(m+n)(x+y)^{m+n-1}\left(1+\frac{dy}{dx}\right)$$
Expand a bit:
$$mx^{m-1}y^n+nx^my^{n-1}\frac{dy}{dx}=(m+n)(x+y)^{m+n-1}+(m+n)(x+y)^{m+n-1}\frac{dy}{dx}$$
Group the $\frac{dy}{dx}$ terms:
$$\left(nx^my^{n-1}-(m+n)(x+y)^{m+n-1}\right)\frac{dy}{dx}=(m+n)(x+y)^{m+n-1}-mx^{m-1}y^n$$
Divide both sides:
$$\frac{dy}{dx}=\frac{(m+n)(x+y)^{m+n-1}-mx^{m-1}y^n}{nx^my^{n-1}-(m+n)(x+y)^{m+n-1}}$$
and I'm fairly certain this does not equal $\frac yx$. It does simplify a little though:
$$(x+y)^{m+n-1}=\frac{x^my^n}{x+y}$$
$$\frac{dy}{dx}=\frac{(m+n)\frac{x^my^n}{x+y}-mx^{m-1}y^n}{nx^my^{n-1}-(m+n)\frac{x^my^n}{x+y}}$$
Divide the numerator and denominator by $x^my^n$ to get
$$\frac{dy}{dx}=\frac{\frac{m+n}{x+y}-mx^{-1}}{ny^{-1}-\frac{m+n}{x+y}}$$
Now simplify the fractions:
$$\frac{dy}{dx}=\frac{(m+n)xy-my(x+y)}{nx(x+y)-(m+n)xy}$$
Expand and cancel like terms:
$$\frac{dy}{dx}=\frac{nxy-my^2}{nx^2-mxy}$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2186338",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
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|
The probability that $a$ and $b$ satisfy the inequality $a-2b+10>0$? Nine identical balls are numbered $1,2,3,.........,9$ are put in a bag.$A$ draws a ball and gets the number $a$ and puts back in the bag. Next $B$ draws a ball and gets the number $b$.
The probability that $a$ and $b$ satisfy the inequality $a-2b+10>0$ ?
My Try :-
Total pairs of $(a,b)$ possible are $81$ .
*
*If $a=1$, then $b = 1,2,3,4,5$. Similarly for $a=2$.
*If $a=3$, then $b = 1,2,3,4,5,6$. Similarly for $a=4$.
*If $a=5$, then $b = 1,2,3,4,5,6,7$. Similarly for $a=6$.
*If $a=7$, then $b = 1,2,3,4,5,6,7,8$. Similarly for $a=8$.
*If $a=9$, then $b = 1,2,3,4,5,6,7,8,9$.
Total favourable pairs are $61$.
Hence, Total Probability = $\frac{61}{81}$
However, I don't have an answer for this. Am I right or missing something ?
|
$\newcommand{\bbx}[1]{\,\bbox[8px,border:1px groove navy]{\displaystyle{#1}}\,}
\newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace}
\newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
\newcommand{\dd}{\mathrm{d}}
\newcommand{\ds}[1]{\displaystyle{#1}}
\newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
\newcommand{\ic}{\mathrm{i}}
\newcommand{\mc}[1]{\mathcal{#1}}
\newcommand{\mrm}[1]{\mathrm{#1}}
\newcommand{\pars}[1]{\left(\,{#1}\,\right)}
\newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
\newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
\newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
\newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$
\begin{align}
&\sum_{a = 1}^{9}{1 \over 9}\sum_{b = 1}^{9}{1 \over 9}\bracks{a - 2b + 10 > 0} =
{1 \over 81}\sum_{b = 1}^{9}\sum_{a = 1}^{9}\bracks{a > 2b - 10}
\\[5mm] = &\
{1 \over 81}\sum_{b = 1}^{9}\braces{%
\bracks{2b - 10 < 1}\sum_{a = 1}^{9}1 +
\bracks{1 \leq 2b - 10 \leq 9}\sum_{a = 2b - 9}^{9}1}
\\[5mm] = &\
{1 \over 81}\sum_{b = 1}^{9}\braces{%
\bracks{b < {11 \over 2}}9 +
\bracks{{11 \over 2} \leq b \leq {19 \over 2}}\pars{19 - 2b}} =
{1 \over 9}\sum_{b = 1}^{5}1 + {1 \over 81}\sum_{b = 6}^{9}\pars{19 - 2b}
\\[5mm] = &\
{5 \over 9} + {1 \over 81}\pars{4 \times 19 - 2 \times 30} =
\bbx{\ds{61 \over 81}}
\end{align}
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2187290",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
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|
Prove that there exist integers $p$ and $q$ such that $\det(A^3+B^3) = p^3+q^3$
Let $A$ and $B$ be $3 \times 3$ matrices with integer entries so that $AB = BA$ and $\det(A) = \det(B) = 0$. Prove that there exist integers $p$ and $q$ such that $\det(A^3+B^3) = p^3+q^3$.
I thought about factorizing $A^3+B^3$. We have $$A^3+B^3 = (A+B)(A^2-AB+B^2).$$ Then we have $$\det(A^3+B^3) = \det(A+B)\det(A^2-AB+B^2).$$ How do we use the fact that $A$ and $B$ are $3 \times 3$? For $3 \times 3$ matrices is it true that $\det(A+xB) = x(a+bx)$ where $a,b$ are integers? If so, then we have since $$A^2-AB+B^2 = (A-e^{2\pi i/3}B)(A-e^{-2\pi i/3}B),$$ that $$\det(A^3+B^3) = (a+b)(a-be^{2\pi i/3})(a-be^{-2 \pi i/3}).$$
|
In general it is not true that $\det(A+xB) = x(a+bx)$, but in this case you can prove it.
Consider the polynomial
$$P(X)=\det(A+XB)$$
This is a polynomial of degree at most $3$ with integer coefficients (this is an easy exercise, just write out the formula for the determinants). Write it as
$$
P(x)=aX^3+bX^2+cX+d
$$
You know that $P(0)=\det(A)=0$. Hence, $d=0$.
Moreover, for $x \neq 0$.
$$a+bX+cX^2+dX^3=X^3 P(\frac{1}{X})= X^3 \det(A+\frac{1}{X}B)=\det(XA+B)$$
Since those polynomials agree for $x \neq 0$, it follows they also agree for $x=0$. Plugging in $x=0$ you get
$$a=\det(B)=0$$
Therefore, $\det(A)=0$ and $\det(B)=0$ implies $a=d=0$ and hence
$$\det(A+XB)=cX^2+dX$$
Now to complete the proof, use your last formula:
$$\det(A^3+ B^3)=\det(A+B)\det(A+\omega B)\det(A+\omega^2B)\\=(c+d)(c\omega^2+d\omega)(c\omega^4+d\omega^2)=c^3+d^3$$
where $\omega$ is one of the complex roots to
$$\omega^3+1=0$$
|
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"language": "en",
"url": "https://math.stackexchange.com/questions/2189377",
"timestamp": "2023-03-29T00:00:00",
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"question_score": "3",
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|
Three Distinct Points and Their Normal Lines
Suppose That three points on the graph of $y=x^2$ have the property that their normal lines intersect at a common point. Show that the sum of their $x$-coordinates is $0$.
I have a lot going but can not finish it.
Proof:
Let $(a,a^2)$, $(b,b^2)$, and $(c,c^2)$ be three distinct points on $y=x^2$ such that $a\not=b\not=c$.
Find the tangent slope and the slope of their normal lines.
$$(a,a^2) \hspace{4mm}m_{tan}=2a, \hspace{4mm} m_{norm}=\frac{-1}{2a}$$
$$(b,b^2) \hspace{4mm}m_{tan}=2b, \hspace{4mm} m_{norm}=\frac{-1}{2b}$$
$$(c,c^2) \hspace{4mm}m_{tan}=2c, \hspace{4mm} m_{norm}=\frac{-1}{2c}$$
Normal Line $(a,a^2)$
$y-a^2=-\frac{1}{2a}(x-a) \implies y=-\frac{1}{2a}x+\frac{1}{2}+a^2$
Normal Line $(b,b^2)$
$y-b^2=-\frac{1}{2b}(x-b) \implies y=-\frac{1}{2b}x+\frac{1}{2}+b^2$
Normal Line $(c,c^2)$
$y-c^2=-\frac{1}{2c}(x-c) \implies y=-\frac{1}{2c}x+\frac{1}{2}+c^2$
Find the $x$-interception points between the normal lines of $(a,a^2)$ and $(b,b^2)$.
$-\frac{1}{2a}x+\frac{1}{2}+a^2=-\frac{1}{2b}x+\frac{1}{2}+b^2 \implies x=-(b+a)2ab$
Find the $x$-interception points between the normal lines of $(a,a^2)$ and $(c,c^2)$.
$-\frac{1}{2a}x+\frac{1}{2}+a^2=-\frac{1}{2c}x+\frac{1}{2}+c^2 \implies x=-(c+a)2ac$
Find the $x$-interception points between the normal lines of $(b,b^2)$ and $(c,c^2)$.
$-\frac{1}{2b}x+\frac{1}{2}+b^2=-\frac{1}{2c}x+\frac{1}{2}+c^2 \implies x=-(c+b)2bc$
Show that $a+b+c=0$
$$\begin{align}
....\\
....\\
....\\
\end{align}$$
I do not know how to show that $a+b+c=0$. Any advice on how to continue? Thanks in advance!
|
Michele Maschio’s answer picks up nicely where you left off, although neither your partial solution nor that answer deals with the case that one of the points is the origin, where the slope of the normal is undefined. This is a slightly different approach that doesn’t need to consider that case separately.
If $y=f(x)$ is differentiable at $x=x_0$, then an equation of the line normal to the curve at the point $(x_0,f(x_0))$ is $(x-x_0)+f'(x_0)(y-f(x_0))=0$. In contrast to the slope-intercept form of the equation for a line, this form is also valid when $f'(x_0)=0$. For this problem the equation is $$(x-x_0)+2x_0(y-x_0^2)=0$$ or, after rearranging a bit, $$x+2x_0y=x_0(1+2x_0^2).$$ The condition that the normals through three distinct points with $x$-coordinates $a$, $b$ and $c$ gives us the system $$\begin{align}x+2ay&=a(1+2a^2)\\x+2by&=b(1+2b^2)\\x+2cy&=c(1+2c^2).\end{align}$$ Subtract the first equation from the other two to eliminate $x$: $$\begin{align}2(b-a)y&=2b^3-2a^2+b-a\\2(c-a)y&=2c^3-2a^3+c-a.\end{align}$$ Cross-multiply and subtract to eliminate $y$, producing $$\begin{align}0&=(c-a)(2b^3-2a^3+b-a)-(b-a)(2c^3-2a^3+c-a) \\ &=2a^3b-2ab^3+2b^3c-2bc^3+2c^3a-2ca^3 \\ &= 2(b-a)(c-b)(a-c)(a+b+c).\end{align}$$ The last two expressions have a satisfying symmetry that reflects the symmetry of the roles of the three points in the problem. Since the three points are distinct, we must have $a+b+c=0$ for our system of equations to be consistent.
|
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"timestamp": "2023-03-29T00:00:00",
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|
Solving Ax=b Using the Basis of the Nullspace Let B = \begin{bmatrix}2&3&1&-1\\1&2&1&2\\3&5&2&1\\4&7&3&3\end{bmatrix}
Find the complete solution to the nonhomogenous system Bx=\begin{bmatrix}6\\-4\\2\\-2\end{bmatrix} by first computing a basis for the nullspace of B.
x = $c_1$ + $sc_2$ + $tc_3$
Where $c_1$, $c_2$ and $c_3$ are vectors.
I know how to normally solve an Ax=b equation, by taking the inverse of A and multiplying by b, but I'm really not understanding the method that's being used here.
|
Start with the reduced row echelon form:
$$
\begin{align}
\mathbf{B} &\mapsto \mathbf{E}_{\mathbf{B}} \\
%
\left[
\begin{array}{rrrr}
2 & 3 & 1 & -1 \\
1 & 2 & 1 & 2 \\
3 & 5 & 2 & 1 \\
4 & 7 & 3 & 3 \\
\end{array}
\right]
%
&\mapsto
%
\left[
\begin{array}{rrrr}
\color{blue}{1} & 0 & -1 & -8 \\
0 & \color{blue}{1} & 1 & 5 \\
0 & 0 & 0 & 0 \\
0 & 0 & 0 & 0 \\
\end{array}
\right]
\end{align}
$$
The message is that the fundamental columns are the first two, highlighted by blue coloring.
Your question is look for the values of $\alpha$ and $\beta$ which solve
$$
\left[
\begin{array}{r}
6 \\ -4 \\ 2 \\ -2
\end{array}
\right]
%
=
%
\alpha
\left[
\begin{array}{r}
2 \\ 1 \\ 3 \\ 4
\end{array}
\right]
%
+
%
\beta
\left[
\begin{array}{r}
3 \\ 2 \\ 5 \\ 7
\end{array}
\right]
$$
We are look for two unknowns, so solve using two rows. Let's pick rows 1 and 3, liberating us from the chance of sign errors. The problem is
now
$$
%
\left[
\begin{array}{rr}
2 & 3 \\
3 & 5
\end{array}
\right]
%
\left[
\begin{array}{c}
\alpha \\
\beta
\end{array}
\right]
%
=
\left[
\begin{array}{c}
6 \\
2
\end{array}
\right]
%%
\quad \Rightarrow \quad
%
\left[
\begin{array}{c}
\alpha \\
\beta
\end{array}
\right]
=
%
\left[
\begin{array}{rr}
5 & -3 \\
-3 & 2
\end{array}
\right]
%
\left[
\begin{array}{c}
6 \\
2
\end{array}
\right]
%
=
%
\left[
\begin{array}{r}
24 \\
-14
\end{array}
\right]
$$
It doesn't seem that we need to resolve the nullspace. But it is a byproduct of the augmented reduction producing $\mathbf{E}_{\mathbf{B}}$:
$$
\mathcal{N} \left( \mathbf{B}^{\mathrm{T}} \right) =
\text{span } \left\{ \,
%
\left[
\begin{array}{r}
8 \\
-5 \\
0 \\
1
\end{array}
\right], \,
%
\left[
\begin{array}{r}
-1 \\
-1 \\
1 \\
0
\end{array}
\right]
%
\, \right\}
$$
Solution
The solution to
$$
\begin{align}
\mathbf{B} x &= y \\
%%
\left[
\begin{array}{rrrr}
2 & 3 & 1 & -1 \\
1 & 2 & 1 & 2 \\
3 & 5 & 2 & 1 \\
4 & 7 & 3 & 3 \\
\end{array}
\right]
%
\left[
\begin{array}{c}
x_{1} \\
x_{2} \\
x_{3} \\
x_{4}
\end{array}
\right]
%
&=
%
\left[
\begin{array}{r}
6 \\
-4 \\
2 \\
-2
\end{array}
\right]
%
\end{align}
$$
is
$$
%
\left[
\begin{array}{c}
x_{1} \\
x_{2} \\
x_{3} \\
x_{4}
\end{array}
\right]
%
=
\left[
\begin{array}{r}
24 \\
-14 \\
0 \\
0
\end{array}
\right]
%
+
%
s
\left[
\begin{array}{r}
8 \\
-5 \\
0 \\
1
\end{array}
\right]
%
+
%
t
\left[
\begin{array}{r}
-1 \\
-1 \\
1 \\
0
\end{array}
\right]
$$
where the constants $s$ and $t$ are arbitrary.
To emphasize:
$$
\left[
\begin{array}{rrrr}
2 & 3 & 1 & -1 \\
1 & 2 & 1 & 2 \\
3 & 5 & 2 & 1 \\
4 & 7 & 3 & 3 \\
\end{array}
\right]
\left(
\left[
\begin{array}{r}
24 \\
-14 \\
0 \\
0
\end{array}
\right]
%
+
%
s
\left[
\begin{array}{r}
8 \\
-5 \\
0 \\
1
\end{array}
\right]
%
+
%
t
\left[
\begin{array}{r}
-1 \\
-1 \\
1 \\
0
\end{array}
\right]
\right)
%
=
%
\left[
\begin{array}{r}
6 \\
-4 \\
2 \\
-2
\end{array}
\right]
$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2192954",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
}
|
Deducing a supremum from a given property Consider $$ S = \bigg\{ { \frac{x^2}{1+2x^2} } :x\in\mathbb{R}\bigg\}$$
we may guess that $\sup A=\frac{1}{2}$
But how does one prove this without taking limits?
|
$\frac{1}{2}$ is an upper bound for $S$. In fact
$$\frac{x^2}{1+2x^2} =\frac{1}{2} \frac{2x^2}{1+2x^2}\le\frac{1}{2}\frac{1+2x^2}{1+2x^2}= \frac{1}{2} $$
for all $x \in \mathbb R$. Moreover,
$$\lim_{x \to +\infty}\frac{x^2}{1+2x^2} =\frac{1}{2} $$
and so for any $\varepsilon >0$
$$\left|\frac{x^2}{1+2x^2}-\frac{1}{2} \right|<\varepsilon $$
for $x$ large enough. In particular, for $x$ large enough, we have
$$\frac{x^2}{1+2x^2} > \frac{1}{2}-\varepsilon, $$
so the desired supremu is indeed $\frac{1}{2}$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2193499",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 1
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|
When is $\frac{x^2+xy+y^2}{49}$ an integer?
Find the number of distinct ordered pairs $(x, y)$ of positive integers such that $ 1\leq x, y \leq 49$ and $\frac{x^2+xy+y^2}{49}$ is an integer.
Multiplying the given equation by $(x-y)$ gives $x^3 \equiv y^3 \pmod{49}$. Thus we can't have $7 \mid x$ while $7 \nmid y$ or vice-versa. Rearranging the given equation gives $$(x+y)^2 \equiv xy \pmod{49}.$$
If $7 \mid x,y$ then there are $49$ solutions. Now suppose that $7 \nmid x,y$. Then we have $$(x+y)^2(xy)^{-1} \equiv xy^{-1}+2+xy^{-1} \equiv 1 \pmod{49}.$$ Thus, $xy^{-1}+yx^{-1} \equiv xy^{-1}(1+(yx^{-1})^2) \equiv -1 \pmod{49}$.
I didn't see how to continue from here. The answer is $$2\varphi(49)+49,$$ and $\varphi(49)$ is the number of units modulo $49$, so maybe we can use that to solve this question.
|
The square roots of $-3 \pmod 7$ are $2,5.$ We ask when
$$ (2 + 7t)^2 \equiv -3 \pmod {49}, $$
with only solution $t =5$ and the square root being $37 \pmod{49}.$
We ask when
$$ (5 + 7t)^2 \equiv -3 \pmod {49}, $$
with only solution $t =1$ and the square root being $12 \pmod {49}.$
This little procedure is the beginning of Hensel's Lemma. Indeed, given two solutions $r_i$ to $$ r_i^2 \equiv -3 \pmod {7^k}, $$
we arrive at exactly two solutions to
$$ s_i^2 \equiv -3 \pmod {7^{k+1}} $$ such that
$$s_i \equiv r_i \pmod{7^k}. $$
Take $x \equiv 1 \pmod {49}$ and solve
$$ x^2 + xy + y^2 \equiv 0 \pmod {49}. $$
Two and four are invertible $\pmod {49},$ so we have
$$ 4 x^2 + 4xy + 4 y^2 \equiv 0 \pmod {49} $$
W took $x=1,$ so we now have
$$ 4 + 4y + 4 y^2 \equiv 0 \pmod{49}. $$
$$ 3 + (2y+1)^2 \equiv 0 \pmod {49}. $$
$$ (2y+1)^2 \equiv -3 \pmod {49}. $$
$$ 2y+1 \equiv 12, 37 \pmod {49}. $$
$$ 2y \equiv 11,36 \pmod{49}. $$
$$ 2y \equiv 60,36 \pmod {49}. $$
$$ y \equiv 30, 18 \pmod {49}. $$
Each of the two solutions
$$ (1,30) \; \; \; (1,18) \pmod {49} $$
can be multiplied by any invertible element $t$ to give all
$$ (t,30t) \; \; \; (t,18t) \pmod {49} $$
There are $42$ values of $t,$ giving $84 $ solutions with both $x,y \neq 0 \pmod 7.$ In turn, given any $(x,y),$ there is an element $1/x \pmod {49}$
and we arrive at a fixed solution $(1, y/x) \pmod {42}.$
If $x \equiv 0 \pmod 7,$ it follows that $y \equiv 0 \pmod 7.$ Furthermore these are all solutions, giving another $7 \cdot 7 = 49.$
Together $84 + 49.$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2195685",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
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|
How can I prove this trigonometric equation with squares of sines? Here is the equation:
$$\sin^2(a+b)+\sin^2(a-b)=1-\cos(2a)\cos(2b)$$
Following from comment help,
$${\left(\sin a \cos b + \cos a \sin b\right)}^2 + {\left(\sin a \cos b - \cos a \sin b\right)}^2$$
$$=\sin^2 a \cos^2b + \cos^2 a \sin^2 b + \sin^2 a \cos^2 b + \cos^2 a \sin^2 b$$
I am stuck here, how do I proceed from here?
Edit: from answers I understand how to prove,but how to prove from where I am stuck?
|
We have $$
\begin{align}
\sin^2(a+b)+\sin^2(a-b)&=\frac{1-\cos(2(a+b))}{2}+\frac{1-\cos(2(a-b))}{2}\\
&=1-\frac{\color{red}{\cos(2a+2b)}+\color{blue}{\cos(2a-2b)}}{2}\\
&=1-\frac{\color{red}{\cos 2a\cos 2b-\sin 2a\sin 2b}+ \color{blue}{\cos 2a\cos 2b+\sin 2a\sin 2b}}{2}\\
&=1-\frac{2\cos 2a\cos 2b}{2}\\
&=1-\cos 2a\cos 2b.
\end{align}
$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2198091",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 8,
"answer_id": 2
}
|
Solve $\lim_{x \rightarrow \infty}\frac{\ln(x+1)-\ln(x)}{x}$ without using L'Hopital or Taylor Series
Solve $$\lim_{x \rightarrow \infty}\frac{\ln(x+1)-\ln(x)}{x}$$ without using L'Hopital or Taylor Series
I tried:
\begin{align}\lim_{x \rightarrow \infty}\frac{\ln(x+1)-\ln(x)}{x} &=
\frac{\ln(x+1)}{x}-\frac{\ln(x)}{x} \\
&=\lim \frac{\ln(x+1)}{x}-\lim\frac{\ln(x)}{x} \\
&=\lim \frac{\ln(x+1)}{x}-0 \\
&=\lim \frac{\ln(x+1)}{x} \\
&=\lim_{x \rightarrow 0} \frac{\ln\left(\frac{1}{x}+1\right)}{\frac{1}{x}} \\
&=\lim_{x \rightarrow 0} x\ln\left(\frac{1}{x}+1\right) \\
&=\lim_{x \rightarrow 0} \ln\left(\left(1+\frac{1}{x}\right)^x\right) \\
&= ???\end{align}
How do I solve this? What do I do next?
|
Recall that
$$ \ln x := \int_1^x \frac{1}{t} ~dt.$$
Thus
\begin{align}
\lim_{x \to \infty} \frac{\ln (x+1) - \ln x}{x} &= \lim_{x \to \infty} \frac{1}{x} \left[ \int_1^{x+1} \frac{1}{t} ~dt - \int_1^x \frac{1}{t} ~dt \right] \\
&= \lim_{x \to \infty} \frac{1}{x} \int_x^{x+1} \frac{1}{t} ~dt
\end{align}
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2199268",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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|
Show that $\lim_{n \rightarrow \infty} \frac{\Sigma_{i=0}^k \binom{n}{3i}}{2^n} = \frac{1}{3}$ Let $n = 3k$ . Show that $\displaystyle{\lim\limits_{n \rightarrow \infty} \frac{\Sigma_{i=0}^k \binom{n}{3i}}{2^n} = \frac{1}{3}}$ . In other words, the sum of every third element of the nth row of the Pascal triangle is roughly one third of the sum of all elements of that row; after this, generalize the result.
attempt:
$\displaystyle{\begin{array}{ll}
\lim\limits_{n \rightarrow \infty} \displaystyle{\frac{\Sigma_{i=0}^k \binom{3k}{3i}}{2^n}} & = \\\\
\lim\limits_{n \rightarrow \infty} \displaystyle{\frac{\binom{3k}{0} + \binom{3k}{3} + \binom{3k}{6} + \cdot \cdot \cdot + \binom{3k}{k} }{2^n}} & = \\\\
\lim_\limits{n \rightarrow \infty} \displaystyle{\frac{\frac{3k!}{0! 3k!} + \frac{3k!}{3!(3k-3)!} + \frac{3k!}{6!(3k-6)!} + \cdot \cdot \cdot + \frac{3k!}{3k!(3k - 3k)!} }{2^n}} & = \\\\
\lim_\limits{n \rightarrow \infty} \displaystyle{\frac{\frac{1}{1} + \frac{3k (3k-1)(3k-2)(3k-3)....2 \cdot 1}{3!(3k-3)!} + \frac{3k(3k-1)(3k-2)...(3k-6)...2.1}{6!(3k-6)!} + \cdot \cdot \cdot + \frac{1}{1} }{2^n}} & = \\\\
\lim_\limits{n \rightarrow \infty} \displaystyle{\frac{\frac{1}{1} + \frac{3k (3k-1)(3k-2)(3k-3)}{3!} + \frac{3k(3k-1)(3k-2)...(3k-5)}{6!} + \cdot \cdot \cdot + \frac{1}{1} }{2^n}}
\end{array}}$
I am stuck. If $n = 3k$ approaches infinity then it seems that we would have $0$ on the numerator and so the expression would go to $0$ and not $1/3$ .
Could someone please help me ? and give me some feedback. Thank you
|
Let $\omega=\exp\left(\frac{2\pi i}{3}\right)$. The characteristic function of the integers $n$ of the form $3k$ can be written as $\frac{1+\omega^n+\omega^{2n}}{3}$ (this is the principle behind the discrete Fourier transform) hence
$$ \sum_{i\geq 0}\binom{n}{3i} = \frac{1}{3}\sum_{k=0}^{n}\binom{n}{k}(1+\omega^k+\omega^{2k})=\frac{2^n+(1+\omega)^n+(1+\omega^2)^n}{3}$$
and accidentally both $1+\omega$ and $1+\omega^2$ have unit modulus, hence the term $(1+\omega)^n+(1+\omega^2)^n$ is bounded by $2$ in modulus, regardless of $n$. It follows that
$$ \lim_{n\to +\infty}\frac{1}{2^n}\sum_{i\geq 0}\binom{n}{3i}=\color{red}{\frac{1}{3}} $$
as wanted. In other terms,
If we consider a random subset of $\{1,\ldots,n\}$, its cardinality equals $3k$ for some $k\in\mathbb{N}$ with a probability closer and closer to $\frac{1}{3}$ as $n$ increases.
That holds also if $3$ is replaced by $4,5,6,7,\ldots$, but the speed of convergence is quite different: to prove it (through the DFT) is an interesting exercise I leave to the reader / to the OP.
|
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"url": "https://math.stackexchange.com/questions/2200539",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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|
Calculate the following determinant How do I calculate the following determinant?
$$
\begin{vmatrix}
a_n & -1 & 0 & \dots & 0 & 0 \\
a_{n-1} & x & -1 & \dots & 0 & 0 \\
a_{n-2} & 0 & x & \dots & 0 & 0 \\
\vdots & \vdots & \vdots & \ddots & \vdots & \vdots \\
a_1 & 0 & 0 & \dots & x & -1 \\
a_0 & 0 & 0 & \dots & 0 & x
\end{vmatrix}
$$
|
Let $F(a_0, a_1, \dots, a_n)$ be the determinant in your question. Expanding by minors across the first row, we get that $F(a_0, a_1, \dots, a_n)$ is equal to
$$
a_n
\begin{vmatrix}
x & -1 & \dots & 0 & 0 \\
0 & x & \dots & 0 & 0 \\
\vdots & \vdots & \ddots & \vdots & \vdots \\
0 & 0 & \dots & x & -1 \\
0 & 0 & \dots & 0 & x
\end{vmatrix}
-(-1)
\begin{vmatrix}
a_{n-1} & -1 & \dots & 0 & 0 \\
a_{n-2} & x & \dots & 0 & 0 \\
\vdots & \vdots & \ddots & \vdots & \vdots \\
a_1 & 0 & \dots & x & -1 \\
a_0 & 0 & \dots & 0 & x
\end{vmatrix}
$$
or $a_n x^n + F(a_0, a_1, \dots, a_{n-1}).$
From there, you can get the answer by induction on $n$.
|
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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|
Solving binomial theorem via induction I'm trying to prove binomial theorem by induction, but I'm a little stuck. I would look at online resources as this problem has been done many times, but the version I am trying to prove the binomial theorem in a different form.
$$(1 + x)^n = \sum_{k = 0}^{n} \binom{n}{k} x^k$$
I'm mostly confused as to how I can make the left side be equivalent to a summation, any help is appreciated. Try to hint me along!
|
We show by induction the following is valid for $n\geq 0$
\begin{align*}
(1 + x)^n = \sum_{k = 0}^{n} \binom{n}{k} x^k
\end{align*}
Base step: $n=0$
We have to show
\begin{align*}
(1 + x)^0 = \sum_{k = 0}^{0} \binom{0}{k} x^k
\end{align*}
Since the left-hand side is $$(1+x)^0=1$$
and the right-hand side is $$\sum_{k = 0}^{0} \binom{0}{k} x^k=\binom{0}{0}x^0=1,$$ both sides are equal and the claim is valid for $n=0$.
Induction hypothesis: $n=N$
We assume the validity of
\begin{align*}
(1 + x)^N = \sum_{k = 0}^{N} \binom{N}{k} x^k\tag{1}
\end{align*}
Induction step: $n=N+1$
We have to show
\begin{align*}
(1 + x)^{N+1} = \sum_{k = 0}^{N+1} \binom{N+1}{k} x^k
\end{align*}
We obtain
\begin{align*}
(1 + x)^{N+1} &= (1+x)(1+x)^N\tag{2}\\
&=(1+x)\sum_{k = 0}^{N} \binom{N}{k} x^k\tag{3}\\
&=\sum_{k = 0}^{N} \binom{N}{k} x^k+\sum_{k = 0}^{N} \binom{N}{k} x^{k+1}\tag{4}\\
&=\binom{N}{0}x^0+\sum_{k=1}^N\binom{N}{k}x^k+\sum_{k=0}^{N-1}\binom{N}{k}x^{k+1}+\binom{N}{N}x^{N+1}\tag{5}\\
&=\binom{N+1}{0}x^0+\sum_{k=1}^N\binom{N}{k}x^k+\sum_{k=1}^{N}\binom{N}{k-1}x^{k}+\binom{N+1}{N+1}x^{N+1}\tag{6}\\
&=\sum_{k=0}^{N+1}\binom{N+1}{k}x^k\tag{7}
\end{align*}
and the claim follows.
Comment:
*
*In (2) we split the product, since we want to apply the induction hypothesis.
*In (3) we apply the induction hypothesis (1).
*In (4) we multiply out.
*In (5) we separate the first summand $\binom{N}{0}$ from the left sum and the last summand $\binom{N}{N}x^{N+1}$ from the right sum.
*In (6) we use the binomial identities
\begin{align*}
\binom{N}{0}=\binom{N+1}{0}=1\qquad\text{and}\qquad\binom{N}{N}=\binom{N+1}{N+1}=1
\end{align*}
We also shift the index $k$ of the right sum by one to start from $k=1$. This all is a preparation for the next step to easily collect all the terms in one sum.
*In (7) we apply the binomial identity
\begin{align*}
\binom{N}{k}+\binom{N}{k-1}=\binom{N+1}{k}
\end{align*}
and the two sums can be merged into one sum. We also see the left-most term $\binom{N+1}{0}$ and the right-most term $\binom{N+1}{N+1}$ can be made part of the sum using index $k=0$ and $k=N+1$.
|
{
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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|
Show: $\cos \left( \frac{ 3\pi }{ 8 } \right) = \frac{1}{\sqrt{ 4 + 2 \sqrt{2} }}$ I'm having trouble showing that:
$$\cos\left(\frac{3\pi}{8}\right)=\frac{1}{\sqrt{4+2\sqrt2}}$$
The previous parts of the question required me to find the modulus and argument of $z+i$ where $z=\operatorname{cis{\theta}}$. Hence, I found the modulus to be $2\cos{\left(\frac{\pi}{4}-\frac{\theta}{2}\right)}$ units and that the argument would be $\operatorname{arg}(z+i)=\frac{\pi}{4}+\frac{\theta}{2}$.
Now, the next step that I took was that I replaced every theta with $\frac{3\pi}{8}$ in the polar form of the complex number $z+i$. So now it would look like this:
$$z+i=\left[2\cos{\left(\frac{\pi}{8}\right)}\right]\operatorname{cis}{\left(\frac{3\pi}{8}\right)}$$
Then, I expanded the $\operatorname{cis}{\left(\frac{3\pi}{8}\right)}$ part to become $\cos{\left(\frac{3\pi}{8}\right)}+i\sin{\left({\frac{3\pi}{8}}\right)}$. So now I've got the $\cos\left({\frac{3\pi}{8}}\right)$ part but I don't really know what to do next. I've tried to split the angle up so that there would be two angles so I can use an identity, however, it would end up with a difficult fraction instead. So if the rest of the answer or a hint would be given to finish the question, that would be great!!
Thanks!!
|
Hint: by the double angle formula:
$$-\,\frac{1}{\sqrt{2}}=\cos\left(\frac{3\pi}{4}\right)=2\,\cos^2\left(\frac{3\pi}{8}\right)-1 \;\;\implies\;\; \cos\left(\frac{3\pi}{8}\right) = \sqrt{\frac{1}{2}\left({1-\frac{1}{\sqrt{2}}}\right)}=\frac{1}{2}\sqrt{2-\sqrt{2}}$$
The above is equivalent to the posted form since $\sqrt{2-\sqrt{2}} \cdot \sqrt{4+2\sqrt2} = 2\,$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2206391",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 0
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|
How to prove this condition if three vectors are colinear? So I was given this problem:
Let $\vec{a} = \begin{pmatrix}x_a\\y_a\\z_a\end{pmatrix}$, $\vec{b} = \begin{pmatrix}x_b\\y_b\\z_b\end{pmatrix}$, and $\vec{c} = \begin{pmatrix}x_c\\y_c\\z_c\end{pmatrix}$.
Show that $(x_a,y_a,z_a)$, $(x_b,y_b,z_b)$, and $(x_c,y_c,z_c)$ are collinear if and only if
$$\vec{a} \times \vec{b} + \vec{b} \times \vec{c} + \vec{c} \times \vec{a} = \mathbf{0}.$$
I tried plugging it in and it looks like it is true:
Let $\vec{a} = \begin{pmatrix}2\\4\\6\end{pmatrix}$, $\vec{b} = \begin{pmatrix}4\\8\\12\end{pmatrix}$, and $\vec{c} = \begin{pmatrix}8\\16\\24\end{pmatrix}$.
$$\vec{a} \times \vec{b} + \vec{b} \times \vec{c} + \vec{c} \times \vec{a}
=\begin{pmatrix}0\\0\\0\end{pmatrix}
+\begin{pmatrix}0\\0\\0\end{pmatrix}
+\begin{pmatrix}0\\0\\0\end{pmatrix}\\
=
\begin{pmatrix}0\\0\\0\end{pmatrix}\\
$$
How could I prove this to be true? I am not entirely sure where to begin.
|
We have $|v \times w| = |v||w|\sin \theta$, where $\theta$ is the angle between $u$ and $v$. Therefore three points $a$, $b$ and $c$ are collinear if and only
$$(b - a) \times (c - a) = 0$$
Expanding
$$b \times c - b\times a - a \times c + a \times a = 0$$
Since $a \times a = 0$ and anticommutivity
$$a\times b + b \times c + c \times a = 0$$
|
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"url": "https://math.stackexchange.com/questions/2211230",
"timestamp": "2023-03-29T00:00:00",
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|
$a+b+c = 13$; if $b/a=c/b$, find the maximum and minimum values of $a$ and the corresponding $b$ and $c$ Question :
The sum of $3$ integers $a,b$ and $c$ is $13$. If $\dfrac{b}{a}=\dfrac{c}{b}$, find the maximum and minimum values of $a$ and the corresponding $b$ and $c$.
To tackle this problem I let $x=\dfrac{b}{a}=\dfrac{c}{b}$ because I wanted to create a quadratic equation in order to use the discriminant theorem.
From the equation above I can deduce that $b=ax$ and $c=ax^2$.
Because $a+b+c=13$.
Therefore; $$a+ax+ax^2=13$$
$$\implies 1+x+x^2-\frac{13}{a} = 0 $$ (where $a \ne 0$, $b \ne 0$, $c \ne 0$)
I can only work up to here. I do not know how to use the discriminant theorem to work out the maximum and minimum of $a$, $b$ and $c$.
|
∵ $x=b/a$ is a solution of (1)
∴ (1) has at least one solution.
$△=1-4(1-(13/9))≥0$
$52/a≥3...(2)$
From (2) $a>0$ and $52/3≥a$
$∴0<a≤52/3$
$1<a≤17$
Minimum of $a=1$ and maximum of $a =17$.
Sub these values of $a$ into (1) to find the corresponding maximum and minimum of b and c.
|
{
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"url": "https://math.stackexchange.com/questions/2214092",
"timestamp": "2023-03-29T00:00:00",
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"question_score": "4",
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|
Calculating value of a variable in combination? Using the formula (bionomial theorem):
$${}_nC_r \cdot a^{n-r} \cdot b^{r}$$
i was trying to find out the term with the coefficient $400$ from $(2x + 5y)^5$? Then i came up with this equation which i have no idea to solve beside trying numbers between 1 and 5. Is there any other way?
$$2^{5-r} \cdot 5^{r} \cdot {}_5C_r = 400$$
can we find the value of r without brute forcing?
|
$\frac{2^5}{2^r} \cdot 5^r \cdot \binom 5r = 400$
$\frac{32}{2^r} \cdot 5^r \cdot \binom 5r = 400$
$\frac{1}{2^r} \cdot 5^r \cdot \binom 5r = \frac{25}{2}$
$\frac{1}{2^r} \cdot 5^r \cdot \binom 5r = \frac{5 \cdot 5}{2}$
$\frac{1}{2^r} \cdot 5^r \cdot \binom 5r = \frac{5}{2} \cdot \binom 51$
On comparing both sides,
$2^r = 2^1$
$r = 1$
|
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|
Evaluating the indefinite integral $\int \frac{x^3}{\sqrt{4x^2 -1}}~dx$.
Find $$\int {x^3\over \sqrt{4x^2 -1}}\,dx.$$
Let
$2x = \sec u$, $2 =\sec (u) \tan(u) u^{'}(x).$ Then
$$
\begin{align*}
\int {x^3\over \sqrt{4x^2 -1}}\,dx &= \frac1{16}\int {\sec^3 u\over \tan u}\tan u \sec u \, du\\
&= \frac1{16}\int {\sec^4 u} \, du\\
&= \frac1{16}\left(\tan u + {\tan^3 u\over 3} \right) + C\\
&= \frac1{16}\left(\tan (\sec^{-1} 2x) + {\tan^3 (\sec^{-1} 2x)\over 3} \right) + C.
\end{align*}$$
Given answer : $$\dfrac{\left(2x^2+1\right)\sqrt{4x^2-1}}{24}+C$$
Why is my answer incorrect ?
|
Hint:
$$\tan(\sec^{-1}(x))=\pm\sqrt{x^2-1}$$
(that is, your answer simplifies a bit further)
|
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|
Proving $\cos(4\theta) - 4\cos(2\theta) \equiv 8\sin^4\theta - 3$.
Prove the following identity:
$$\cos(4\theta) - 4\cos(2\theta) \equiv 8\sin^4\theta - 3$$
How can I express $\cos(4\theta) $ in other terms?
|
Since $\cos(4\theta)=2\cos^2(2\theta)-1$ and $\cos(2\theta)=1-\sin^2(\theta)$, we have
$$\begin{align}
\cos(4\theta) - 4\cos(2\theta)&=2\cos^2(2\theta)- 4\cos(2\theta)-1 \\
&=2(1-2\sin^2(\theta))^2-4(1-2\sin^2(\theta))-1\\
&=2-8\sin^2(\theta)+8\sin^4(\theta)-4+8\sin^2(\theta)-1\\
&=8\sin^4(\theta)-3.
\end{align}$$
|
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|
Need direction for solving integral
Evaluate: $$\int \frac{\sin^2x}{\cos^2x +4}dx.$$
I tried to this things:
First
$$\tan\left(\frac{x}{2}\right) = t$$
$$dx = \frac{2~dt}{1+t^2}$$
$$\sin x= \frac{2t}{1+t^2}$$
$$\cos x= \frac{1-t^2}{1+t^2}$$
Tried also but this is same thing?
$$\tan\left(\frac{x}{2}\right) = t$$
There I don't have to place $2$ left to integral ( same thing )
Second
I tried trigonometric transformations like
$$\sin^2x \equiv 1-\cos^2x$$
$$\int \frac{1-\cos^2x}{\cos^2x +4}dx$$
$$-1\cdot \int \frac{\cos^2x-1}{\cos^2x +4}~dx$$
$$-1\cdot \int \frac{\cos^2x+4-5}{\cos^2x +4}~dx$$
But I get stucked in this part
$$5\int \frac{1}{\cos^2x +4}dx.$$
And mix something from first step
Third
I tried to divide everything with $\sin^2x$, but didn't succeed to solve + mix something from first step
Can anyone give me direction how to solve this?
sorry, for late update....
My goal is to complete this task without trigonometry function sec
|
$$\int \frac{1}{\cos^2x +4}dx$$
Divide top and bottom of integrand by $\cos^2 x$,
$$=\int \frac{\sec^2 x}{1+4\sec^2 x} dx $$
Now use $\tan^2 x+1=\sec^2 x$ to get,
$$=\int \frac{\sec^2 x}{1+4(\tan^2 x+1)} dx$$
$$=\int \frac{\sec^2 x}{5+4 \tan^2 x} dx$$
Let $u=\tan x$.
$$=\int \frac{1}{5+4u^2} du$$
$$=\frac{1}{5} \int \frac{1}{1+\frac{4}{5}u^2} du$$
$$=\frac{1}{5} \int \frac{1}{1+(\frac{2}{\sqrt{5}}u)^2} du$$
$$=\frac{1}{5} \frac{\sqrt{5}}{2} \int \frac{\frac{2}{\sqrt{5}}}{1+(\frac{2}{\sqrt{5}}u)^2} du$$
I think you can deal with this.
|
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|
Sum of Sequence with Squares of Fibonacci Numbers in Denominator Find the sum of this sequence:
$$\frac{1}{1^2+1}-\frac{1}{2^2-1}+\frac{1}{3^2+1}-\frac{1}{5^2-1}+\frac{1}{8^2+1}-...$$
So, alternating series, but I've got nothing. I tried regrouping by pairs and got $$\frac{1}{6}+\frac{7}{120}+\frac{103}{10920}$$
which, helped me none.
|
Note: most of this is the identity
$$ F_{n+1} F_{n-1} - F_n^2 = (-1)^n $$ which is how I how I saw the two parts telescope.
The $+$ part is
$$ \frac{1}{1 \cdot 2} + \frac{1}{2 \cdot 5} + \frac{1}{5 \cdot 13} + \frac{1}{13 \cdot 34} + $$
Do the $\pm$ parts separately. The partial sums for the $+$ part are
$$ \frac{1}{2}, \frac{3}{5}, \frac{8}{13}, \frac{21}{34}, \cdots $$
which are ratios of Fibonacci numbers.
The $-$ part is
$$ \frac{1}{1 \cdot 3} + \frac{1}{3 \cdot 8} + \frac{1}{8 \cdot 21} + \frac{1}{21 \cdot 55} + $$
Try this for the $-$ parts.
$$ \frac{1}{3}, \frac{3}{8}, \frac{8}{21}, \frac{21}{55}, \cdots $$
Needs a bit of precision to get the limit of the difference.
I see
$$ \frac{1}{\phi} - \frac{1}{\phi^2} = \frac{\phi - 1}{\phi^2} = \frac{1}{\phi^3} $$
|
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|
Prove that $2^p+p^2$ is prime for $p=3$ only I do know that all prime numbers larger than $3$ can be expressed as $3k + 1$ and or $3k + 2$. Plugging those in I still see no solution.
EDIT: $p$ can only be a prime number.
|
From Fermat's little theorem, it follows that $2^p \equiv 2 \pmod p$. Then $2^p + p^2 \equiv 2 \pmod p$... oops, that doesn't help, never mind.
Oh, wait a minute: since $2 = 3 - 1$, then $p^2 \equiv 1 \pmod 3$ by Fermat's little theorem. That means $p^2$ is of the form $3k + 1$.
So as long as $2^p \not \equiv 2 \pmod 3$, the number $2^p + p^2$ has a shot at being prime. However, since $2^2 \equiv 1 \pmod 3$ but $2^3 \equiv 2 \pmod 3$, it follows that the powers of $2$ alternate $2, 1 \pmod n$ according to whether $n$ is odd or even. But since $p > 2$ is odd, it follows that $2^p \equiv 2 \pmod 3$. Therefore, with $p > 3$ odd, $2^p + p^2$ must be a multiple of $3$.
And the only prime multiples of $3$ are $3$ itself and $-3$. So it is only with $p = 3$ that we have $2^3 + 3^2 = 17$.
Just as an amusing aside: given $p = -3$, we have $$2^p + p^2 = \frac{73}{8}.$$
|
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|
How do I solve this system of equations (with squares and square roots)? Consider the system of equations:
\begin{align*}
x+y+z&=6\\
x^2+y^2+z^2&=18
\\\sqrt{x}+\sqrt{y}+\sqrt{z}&=4.
\end{align*}
How do I solve this? I've tried squaring, adding equations side by side, substituting, etc., but without success, e.g.
$$x^2+y^2+z^2+2(xy+yz+xz)=36\implies xy+yz+xz=9,$$
but then I don't know what to do next. Please help me solve this.
|
Let $s = \sqrt{x}$, $t = \sqrt{y}$, $u = \sqrt{z}$. Then your system is
$$ \eqalign{s^2 + t^2 + u^2 &= 6\cr
s^4 + t^4 + u^4 &= 18\cr
s + t + u &= 4\cr}$$
Solving the third equation for $u$ and substituting in the others gives
$$ \eqalign{s^2+t^2+(4-s-t)^2-6 &= 0\cr s^4+t^4+(4-s-t)^4-18 &= 0\cr}$$
The resultant of the left sides with respect to $t$ is
$$ 4096 (s-2)^2 (s-1)^4$$
So the only solutions have $s = 1$ or $s = 2$ (and by symmetry, also $t=1$ or $t=2$ and $u=1$ or $u=2$). In fact the solutions all have two of the variables $=1$ and one $=2$. In terms of $x,y,z$, two are $1$ and one is $4$.
|
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|
How to find all possible solutions to $a^2 + b^2 = 2^k$? I'm working on the following problem, this is for an introductory discrete mathematics class.
Find all possible solutions to the equation $a^2 + b^2 = 2^k, k\geq1$ and $a$ and $b$ positive integers.
I've observed that the following are answers:
$2^1 = 1^2 + 1^2$
$2^3 = 2^2 + 2^2$
$2^5 = 4^2 + 4^2$ or $2^5 = 2^4 + 2^4$
From that, there seems to be a pattern such as $2^{k+1}=2^k + 2^k$ for some even $k \geq 0$
I've also been able to observe that $k$ has to be odd. I've tried the following:
$k = 2l + 1$ for some $l$
$2^{2l+1} = a^2 + b^2$
$2^l.2 = a^2 + b^2$ Now, if I set $a=b$
$2.2^l = 2a^2$ <=> $2^l = a^2$
But I don't know how to prove that $a$ has to be equal to $b$
Also, in the case where $k$ is even:
$k=2l$ for some $l$
$2^{2l}$ = $a^2 + b^2$ Again, setting $a=b$
$2^{2l+1} = a^2$
$2^{(2l-1)/2} = a^2$ where $(2l-1)/2$ is not an integer.
But again, I don't know how to go about it when $a \neq b$.
Am I on the right path? Any tips or suggestions on how to go forward?
|
easier than you think. If $a^2 + b^2$ is divisible by $4,$ then $a,b$ are both even. Easy to prove. Induction says if $a^2 + b^2$ is divisible by $16,$ then $a,b$ are both divisible by $4.$ If if $a^2 + b^2$ is divisible by $64,$ then $a,b$ are both divisible by $8.$ And so on.
Which means that there are just two cases to write up, with $n \geq 0,$
$$ a^2 + b^2 = 4^n, $$
$$ a^2 + b^2 = 2 \cdot 4^n. $$
In both cases we can write
$$ a = 2^n t, \; \; \; b = 2^n u $$ and proceed
|
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|
Will there always be integer solutions for $x$ and $y$ in $3^y = 2^x - 1$? Is there a way to find the solutions of $3^y=2^x-1$ where $(x,y)$ are integer coordinates (maybe within a certain interval)? Are there an infinite number of integer coordinate pairs for this equation? Is it possible to even determine this? I'm not quite sure the difficulty of this problem, but substituting $x$ or $y$ values into this equation to find integer pairs is time consuming and not very promising.
|
There is an entirely elementary way to show that $(x,y)=(1,0)$ and $(2,1)$ are the only integer solutions to $3^y=2^x-1$.
If $y\ge1$, $x$ must be even since $2^{2u+1}-1\equiv1$ mod $3$. Writing $x=2u$, we have
$$3^y=2^{2u}-1=(2^u-1)(2^u+1)$$
which implies $2^u-1$ and $2^u+1$ are each a power of $3$, say $2^u-1=3^r$ and $2^u+1=3^{r+s}$ with $s\ge1$ (since $2^u+1$ is clearly greater than $2^u-1$). This gives us
$$2=(2^u+1)-(2^u-1)=3^r(3^s-1)$$
the only solution to which is $r=0$ and $s=1$, leading to $u=1$, which is to say $(x,y)=(2,1)$.
|
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|
A determinant from analytic geometry? I have a question regarding the following determinant:
$\begin{vmatrix}
+ax - by - cz & bx+ay & cx+az \\
bx+ay& -ax+by-cz & bz+cy \\
cx+az & bz+cy & -ax-by+cz
\end{vmatrix}
=
(a^2 + b^2 + c^2)(x^2 + y^2 +z^2)(ax+by+cz).$
I can prove the above equality by performing row operations and column operations. However the above equation has a lot of geometrical terms and it seems to me that this equation could have other interpretations that I am missing.
So I have two questions:
1) Can we write the given determinant as product of two determinants (or three even)?
2) Is there a conceptual proof of the above equality through linear algebra or analytic geometry?
|
Now I can answer the first question. We can write the matrix as a product of two rectangular matrices.
$$\left[ \begin{matrix}
+ax - by - cz & bx+ay & cx+az \\
bx+ay& -ax+by-cz & bz+cy \\
cx+az & bz+cy & -ax-by+cz
\end{matrix}\right] = \left[ \begin{matrix}
a & -b & c & 0\\
b & a & 0 & -c \\
c & 0 & -a & b \\
\end{matrix}\right] \left[ \begin{matrix}
x & y & z\\
y & -x & 0\\
-z & 0 & x\\
0 & z & -y\\
\end{matrix}\right] = A^TB$$
The determinant can be evaluated using Cauchy-Binet formula:
$$\begin{vmatrix}
a & -b & c & 0\\
b & a & 0 & -c \\
c & 0 & -a & b \\
\end{vmatrix}\begin{vmatrix}
x & y & z\\
y & -x & 0\\
-z & 0 & x\\
0 & z & -y\\
\end{vmatrix}= |A^TB|= A_{123}B_{123}+A_{124}B_{124}+A_{134}B_{134}+A_{234}B_{234}.$$
where $A_{ijk}$ stands for the determinant of the matrix formed by rows $i,j,k$.
Now just verify that
$$A_{123}B_{123}=ax(a^2+b^2+c^2)(x^2+y^2+z^2).$$
$$A_{124}B_{124}=by(a^2+b^2+c^2)(x^2+y^2+z^2).$$
$$A_{134}B_{134}=cz(a^2+b^2+c^2)(x^2+y^2+z^2).$$
$$A_{234}B_{234}=0.$$
|
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|
Need help in an elementary number theory problem from Higher Algebra by S. BARNARD Can someone help me solve this question:
If a and b are coprime and n is a prime,prove that $$\frac{(a^n+b^n)}{(a+b)}$$ and $$a+b$$ have no common factor,unless $a+b$ is a multiple of $n$.
I've tried it out but not sure about the solution. So can someone help enlighten me. Thanks in advance. It's from Higher Algebra by S. BARNARD.
|
We have $$\frac{a^n+b^n}{a+b}=\sum_{j=0}^{n-1}a^j(-b)^{n-1-j}\equiv \sum_{j=0}^{n-1}a^ja^{n-1-j}=na^{n-1}\mod a+b$$ for every odd $n$
Hence if $n$ is an odd prime and $q\ne n$ a prime dividing both $a+b$ and $\frac{a^n+b^n}{a+b}$, we get $$0\equiv \frac{a^n+b^n}{a+b}\equiv na^{n-1}\mod q$$ hence $q$ must didvide $a$ and because $q$ divides $a+b$, $q$ must also divide $b$. But $a$ and $b$ are coprime, we get a contradiction.
In the case $n=2$, we have $a^2+b^2=(a+b)^2-2ab$, hence if $q\ne 2$ divides both $\frac{a^2+b^2}{a+b}$ (and therfore $a^2+b^2$) and $a+b$, it divides $2ab$ and therefore $a$ or $b$ and therefore $a$ and $b$. We get a contradiction again.
|
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|
How can I solve $5^{2x} + 4(5^x) - 5 = 0$? This is a math problem I'm currently working on.
$$5^{2x} + 4(5^x) - 5 = 0$$
I've used logarithm to try solve the problem. Here's what I've done so far:
\begin{align}5^{2x} + 4(5^x) - 5 &= 0\\
5^{2x} + 5^x &= \frac{5}{4}\\
\log_5{2x} + \log_5{x} &= \log_5\left(\frac{5}{4}\right)\\
\log_5{2x^2} &= \log_5\left(\frac{5}{4}\right)\\
5^{2x^2} &= 5^\frac{5}{4}\\
(2x^2)\log5 &= \frac{5}{4}\log5\\
2x^2 &= \frac{5}{4}\\
2x^2 &= 1.25\\
x^2 &= 0.625\\
\sqrt {x^2} &= \sqrt {0.625}\\
x &= \frac{\sqrt {10}}{4}\end{align}
Substituting the value for $x$ into the equation doesn't equate it to zero. I've tried several different ways but I have still not come up with a correct answer.
|
Hint:
A useful trick is to substitute $u=5^x$ to obtain a quadratic equation:
$$u^2+4u-5=0$$
Can you solve for $u$, and then solve for $x$?
|
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|
Answering a question containing $\sqrt{1.5}$ without using calculator. Of the following which is the best approximation of $\sqrt{1.5}(266)^{\frac{3}{2}}$?
(A)1,000
(B)2,700
(C)3,200
(D)4,100
(E)5,300
How can I answer this without using a calculator and in about 2.5 minutes?
|
$\sqrt{1.5}(266)^{\frac{3}{2}}
= 266 \times\sqrt{1.5 \times 266}
= 266 \times\sqrt{266 + 133}
= 266 \times\sqrt{399}$
Note that $\sqrt{399}$ is very close to $\sqrt{400} = 20$.
So $266 \times\sqrt{399} \approx 266 \times 20 = 5320.$
By the way, if you want a bit more accurracy, let $f(x) = \sqrt{x}$. Then
$f'(x) = \dfrac{1}{2 \sqrt x}$
\begin{align}
f(400+\delta) &\approx f(400) + \delta f'(400) \\
f(400-1) &\approx \sqrt{400} - \dfrac{1}{2 \sqrt{400}} \\
\sqrt{399} &\approx 20 - \dfrac{1}{40} \\
\hline
266 \times\sqrt{399}
&\approx 266 \times \left(20-\dfrac{1}{40} \right) \\
&\approx 5320 - 6.65 \\
&\approx 5313.35
\end{align}
To ten digits, the correct answer is $5313.345839$.
|
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|
If $f(x-3) = x^2$, $f(x) = (x+3)^2$? I was wondering suppose you have $f(x-3) = x^2$. Is it correct to say that $f(x) = f(x-3 +3 ) = (x+3)^2$. If so, why or why not? Why can we say that if $g(x) = x^2 + x$, then $g(x+2) = (x+2)^2 + (x+2)$ but not the other way around?
|
I would interpret
$f(x-3) = x^2$
as composition
$f \circ u$
where $u(x) = x - 3$.
In this case $x = u + 3$ and $f(u) = x^2 = (u+3)^2$ or $f(x) = (x + 3)^2$.
|
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|
Computing $\int_0^\pi (4 + \sin^2 \theta)^{-1} d\theta$ with the residue theorem It's asked to find the value of the real integral
$$
I = \int\limits_0^{\pi}\frac{\rm{d}\theta}{4+\sin^2{\theta}}
$$
I don't understand how to apply the theorem, basically because I cannot parametrise $z = e^{i\theta}$ as I'm not on a circumference.
|
Note that $\sin^2(\theta)$ is an even function. Hence, we assert that
$$\int_0^\pi \frac{1}{4+\sin^2(\theta)}\,d\theta=\frac12\int_{-\pi}^\pi \frac{1}{4+\sin^2(\theta)}\,d\theta \tag 1$$
And now one can proceed with the substitution $z=e^{i\theta}$ in $(1)$ such that
$$\int_0^\pi \frac{1}{4+\sin^2(\theta)}\,d\theta=\frac12\oint_{|z|=1}\frac{1}{4+\left(\frac{z-z^{-1}}{2i}\right)^2}\,\frac{1}{iz}\,dz\tag 2$$
We can make things a bit easier by noting that
$$\begin{align}
\int_{0}^\pi \frac{1}{4+\sin^2(\theta)}\,d\theta&=\int_{0}^{\pi} \frac{2}{9-\cos(2\theta)}\,d\theta\\\\
&=\int_0^{2\pi}\frac{1}{9-\cos(\theta)}\,d\theta\\\\
&=\oint_{|z|=1}\frac{1}{9-\left(\frac{z+z^{-1}}{2}\right)}\,\frac{1}{iz}\,dz\tag 3\\\\
&=\oint_{|z|=1}\frac{2i}{z^2-18z+1}\,dz\\\\
&=\oint_{|z|=1}\frac{2i}{(z-9-4\sqrt{5})(z-9+4\sqrt{5})}\,dz\\\\
&=2\pi i \text{Res}\left(\frac{2i}{(z-9-4\sqrt{5})(z-9+4\sqrt{5})}, z=9-8\sqrt{5}\right)\\\\
&=\frac{\pi}{2\sqrt 5}
\end{align}$$
The integrand of $(3)$ is of simpler form than that of $(2)$ and facilitates evaluating the residue from the pole inside $|z|=1$.
|
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|
Ellipse tangent to two circles We're given two circles with radii $p$ and $q$, one centered at the origin, and one centered at the point $(w,0)$.
I want to construct an ellipse of the form
$$
\frac{(x-c)^2}{a^2} + \frac{y^2}{b^2} = 1
$$
that is tangent to the two circles, as shown here:
If $b$ is known, can we obtain closed form expressions for $a$ and $c$ as functions of $p$, $q$, $w$ and $b$.
|
For ellipse $$\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$$
Equation of normal at $(x_1,y_1)$:
$$a^2y_1(x-x_1)=b^2x_1(y-y_1)$$
Put $y=0$,
$$x=\frac{a^2-b^2}{a^2}x_1=e^2x_1 \tag{$y_1 \ne 0$}$$
which is the $x$-intercept.
Radius $p$:
\begin{align*}
p^2 &= (x_1-e^2x_1)^2+y_1^2 \\
&= \frac{b^4x_1^2}{a^4}+y_1^2 \\
&= \frac{b^4x_1^2}{a^4}+b^2\left( 1-\frac{x_1^2}{a^2} \right) \\
&= b^2-\frac{b^2e^2x_1^2}{a^2} \tag{1}
\end{align*}
Similarly,
$$q^2=b^2-\frac{b^2e^2x_2^2}{a^2} \tag{2} $$
$(1)-(2)$,
$$p^2-q^2=\frac{b^2e^2}{a^2}(x_2^2-x_1^2) \tag{3}$$
Also $$w=e^2(x_2-x_1) \tag{4}$$
On solving,
$$x_1=\frac{a^2(p^2-q^2)}{2b^2w}-\frac{a^2w}{2(a^2-b^2)}$$
$$x_2=\frac{a^2(p^2-q^2)}{2b^2w}+\frac{a^2w}{2(a^2-b^2)}$$
and from $(1)$,
$$a=\frac{b\sqrt{(p^2-q^2)^2+w^2
\left( \sqrt{b^2-p^2}+\sqrt{b^2-q^2} \right)^2}}{p^2-q^2}$$
Note that $$c=-e^2x_1$$
The required ellipse is
$$\frac{1}{a^2}
\left[
x-\frac{w}{2}+\frac{(a^2-b^2)(p^2-q^2)}{2b^2w}
\right]^2+\frac{y^2}{b^2}=1$$
provided $p,q \le b$
A good combination of $p$, $q$, $w$ and $b$:
A bad combination of $p$, $q$, $w$ and $b$:
A special combination of $p$, $q$, $w$ and $b$:
|
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|
General form of multiplication of a matrix by its transpose When multiplying $A^T*A$, does it usually end up in the form $\begin{pmatrix} a&b\\b&-2b\end{pmatrix}$? If not, are there any special cases where it does? If so, why?
|
Per the comment, we generally have
$$
\begin{pmatrix} a&b\\c&d\end{pmatrix} \begin{pmatrix} a&c\\b&d\end{pmatrix} = \begin{pmatrix} a^2+b^2 & ac+bd \\ ac+bd & c^2+d^2\end{pmatrix} \overset{?} = \pmatrix{\alpha & -\frac 12 \beta\\ - \frac 12 \beta & \beta}
$$
Your question amounts to asking when $a,b,c,d$ are such that
$$
ac + bd = - \frac 12 (c^2 + d^2)
$$
Given any $c,d$, there is a line of solutions $(a,b)$ to this equation.
|
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|
Prove inequality $\sum\frac{{{a^3} - {b^3}}}{{{{\left( {a - b} \right)}^3}}}\ge \frac{9}{4}$ Given $a,b,c$ are positive number. Prove that $$\frac{a^3-b^3}{\left(a-b\right)^3}+\frac{b^3-c^3}{\left(b-c\right)^3}+\frac{c^3-a^3}{\left(c-a\right)^3}\ge \frac{9}{4}$$
$$\Leftrightarrow \sum \frac{3(a+b)^2+(a-b)^2}{(a-b)^2} \ge 9$$
$$\Leftrightarrow \frac{(a+b)^2}{(a-b)^2}+\frac{(b+c)^2}{(b-c)^2}+ \frac{(c+a)^2}{(c-a)^2}\ge 2$$
Which $$\frac{a+b}{a-b}.\frac{b+c}{b-c}+\frac{b+c}{b-c}.\frac{c+a}{c-a}+\frac{c+a}{c-a}.\frac{a+b}{a-b} =-1$$
Use $x^2+y^2+z^2 \ge -2(xy+yz+zx)$ then inequality right
I don't know why use $x^2+y^2+z^2 \ge -2(xy+yz+zx)$ then inequality right?
P/s: Sorry i knowed, let $x=\frac{a+b}{a-b}$
We have $(x+1)(y+1)(z+1)=(x-1)(y-1)(z-1) => xy+yz+zx=-1 $
|
$$(x+y+z)^2\ge 0\\x^2+y^2+z^2+2xy+2xz+2yz\ge 0\\x^2+y^2+z^2\ge -2(xy+yz+zx)$$
|
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|
solve for $x$ and $y$ in the following equation $x^2 +y^2 -3=3xy$ and $2x^2 +y^2 =6$ Solve for $x$ and $y$ in the following equations: $x^2 +y^2 -3=3xy$ and $2x^2 +y^2 =6$. I made $y^2$ the subject of the formula in eqn 2.
This gives $y^2 = 6 -2x^2$.
I substitute this into the first eqn.
This gives $x^2+3xy -3$. There's where am stuck.
Can someone pull me out.
|
multiplying the first equation by $-1$ and adding to the second we get
$$x^2+3xy=3$$ thus $$y=\frac{3-x^2}{3x}$$ if $$x\ne 0$$ and you can eliminate $y$
|
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|
Find the rational canonical form of a matrix from its minimal and characteristic polynomials
What is the rational canonical form of $A$?
$$A=\begin{bmatrix}
1 & 1 & 0 & 0 \\
0 & 1 & 0 & 0 \\
2 & 3 & -1 & 4\\
1 & 1 & -1 & 3\\
\end{bmatrix}$$
I found that the minimal polynomial $m_A(x)=(x-1)^2$ and the characteristic polynomial $c_A(x)=(x-1)^4$. Therefore the invariant factors can be
$$x-1,x-1,(x-1)^2$$
or
$$(x-1)^2,(x-1)^2$$
Therefore the rational canonical form may be
$$\begin{bmatrix}
1 & 0 & 0 & 0 \\
0 & 1 & 0 & 0 \\
0 & 0 & 0 & -1\\
0 & 0 & 1 & 2\\
\end{bmatrix}$$
or
$$\begin{bmatrix}
0 & -1 & 0 & 0 \\
1 & 2 & 0 & 0 \\
0 & 0 & 0 & -1\\
0 & 0 & 1 & 2\\
\end{bmatrix}$$
How do I quickly figure out which one is the correct one?
|
The "first" $2\times 2$ principal block is clearly not an $x-1,x-1$ block as it is not the identity. Nor is the "other" $2\times 2$ principal block since it is not the identity. So we have two $(x-1)^2$ blocks.
|
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|
Finding $\int x \sqrt{1- x^2 \over 1 + x^2}dx$.
$$J =\int x \sqrt{1- x^2 \over 1 + x^2}dx$$
Substituting $u = \sqrt{1 + x^2}$
$$J = \int \sqrt{2 - u^2} du = \sqrt{2}\int \sqrt{1 - \left({u \over \sqrt{2}}\right)^2} du$$
Now substituting $\sin z = u/\sqrt{2}$
$$J = 2\int \cos^2 z dz = z + (\sin 2z)/2 + C = \arcsin\left(\sqrt{1+ x^2}\over 2\right) + {\sqrt{1-x^4}\over 2} + C $$.
The give answer is $\displaystyle \arcsin(x^2) + {\sqrt{1-x^4}\over 2} + C$.
What went wrong in my attempt ?
|
HINT:
Let $x^2=\cos2y$ as $x^2\ge0,0\le2y\le\dfrac\pi2\ \ \ \ (1)$
$\implies x\ dx=-\sin2y\ dy,\sin2y=+\sqrt{1-x^4}$
Now $\sin2y=2\sin y\cos y$
and $\sqrt{\dfrac{1-x^2}{1+x^2}}=+\tan y\text{ by } (1)$
Can you take it from here?
|
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|
For a given condition is it true that $a+b+c=3$. Suppose a, b, c are positive real numbers such that
$$(1+a+b+c)\left(1+\frac 1a+\frac 1b+\frac 1c\right)=16$$
Then is it true that we must have $a+b+c=3$ ?
Please help me to solve this. Thanks in advance.
|
By Cauchy-Schwarz $$(1+a+b+c)\left(1+\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\geq\left(1\cdot1+\sqrt{a}\cdot\frac{1}{\sqrt{a}}+\sqrt{b}\cdot\frac{1}{\sqrt{b}}+\sqrt{c}\cdot\frac{1}{\sqrt{c}}\right)^2= 16$$
The equality occurs for $a=b=c=1$, which says that $a+b+c=3$.
|
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|
Prove by induction that $\forall n\geq 1,\ 7\mid 3^{2n+1} + 2^{n-1}$ Prove by induction that $$7 \mid 3^{2n+1} + 2^{n-1},\ \forall n\geq 1$$
Base case $n=1$:
$$3^{2 × 1+1} + 2^{1-1} = 28.$$
Induction:
$$P(k): 3^{2k+1} + 2^{k-1},\ P(k+1): 3^{2(k+1)+1} + 2^{(k+1)-1}.$$
$$3^{2k+3} + 2^k = 9 \times 3^{2k+1} + 2^{k-1} \times 2 = 7 \times 3^{2k+1} + 2 \times 3^{2k+1} + 2^{k-1} \times 2.$$
Where do I go from here?
|
The correct expression is obviously $$3^{2n+1}+ 2^{n-1}$$.
$n=1: 3^3 +2^0 = 27 +1 =28$
$n=2: 3^5 +2^1 = 245 = (35)(7)$
$n=3: 3^7 +2^2 = 2191 = (313)(7)$
So
3^[2(n+1)+1] +2^(n+1-1) = [3^(2n+1)](9) + 2^n
= [7k-2^(n-1)](9) +2^n
= 63k -2^(n-1)(9-2)
= (7)[9k - 2^(n-1)]
|
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|
How $10 +2\sqrt{2} \sqrt{10}+2 > 17$ becomes $2\sqrt{2} \sqrt{10} > 5$ due to theorem $3$? I'm reading Kazarinoff's Analytic inequalities. He gives the theorem 3 and 6:
*
*If $a>b$ and $c>d$ then $a+c> b+d$.
*If $a>b>0$, then $a^{\frac{p}{q}}> b^{\frac{p}{q}}$.
There is an exercise a little bit further: Show that $\sqrt{10}+\sqrt{2}> \sqrt{17}$.
And the solution goes as following:
$$10 +2\sqrt{2} \sqrt{10}+2 > 17\tag{$\text{Theorem 6}$}$$
$$2\sqrt{2} \sqrt{10} > 5\tag{$\text{Theorem 3}$}$$
How is subtracting from both sides a consequence of theorem $3$ (Supposing that this is what is being done)? I took:
$$12 +2\sqrt{2} \sqrt{10} > 17$$
Then $12$ or $2\sqrt{2}\sqrt{10}$ might be $a$ or $c$ and writing $17$ as $12+5$, I'd have:
$$12 +2\sqrt{2} \sqrt{10} > 12+5$$
If $a=12$, then by theorem $3$, $b=5$ to achieve $12>5$ according to the theorem, $c=2\sqrt{2} \sqrt{10}$, $d=12$ but to use the theorem, shouldn't $2\sqrt{2}\sqrt{10}>12$? Also, I could choose $a=2\sqrt{2}\sqrt{10}$ but the only reasonable choice for $c$ would be $5$ and I'd have to have $12>12$.
I looked at Wikipedia and found the following:
Which seems natural (and I am acquainted with its existence), but I'm a bit confused at how he deduced that from theorem $3$.
I am aware that perhaps, he deduced something that looks like a subtraction from both sides but is quite different, but I couldn't imagine more and all my experiments failed.
|
It does not follow directly from Theorem 3, but one can make a corollary:
$a>b$ $\implies$ $a+c>b+c$
Proof: By Definition 1, p. 2, $a>b$ means $a=b+h$, $h>0$. As $h/2>0$ too (by Theorem 4 that says $a>b$, $c>0$ $\implies$ $a/c>b/c$), the same definition gives that $a>b+h/2$. Now take $c$ and $d=c-h/2$. We have $c>d$, then by Theorem 3
$$
a+c>b+h/2+d=b+c.
$$
|
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|
Knowing that for any set of real numbers $x,y,z$, such that $x+y+z = 1$ the inequality $x^2+y^2+z^2 \ge \frac{1}{3}$ holds.
Knowing that for any set of real numbers $x,y,z$, such that $x+y+z = 1$ the inequality $x^2+y^2+z^2 \ge \frac{1}{3}$ holds.
I spent a lot of time trying to solve this and, having consulted some books, I came to this:
$$2x^2+2y^2+2z^2 \ge 2xy + 2xz + 2yz$$ $$2xy+2yz+2xz = 1-(x^2+y^2+z^2) $$ $$2x^2+2y^2+2z^2 \ge 1 - x^2 -y^2 - z^2 $$ $$x^2+y^2+z^2 \ge \frac{1}{3}$$
But this method is very unintuitive to me and I don't think this is the best way to solve this. Any remarks and hints will be most appreciated.
|
We have an instance of the least-norm problem
$$\begin{array}{ll} \text{minimize} & \| {\rm x} \|_2^2 \\ \text{subject to} & {\Bbb 1}_3^\top {\rm x} = 1\end{array}$$
The minimizer is
$${\rm x}_{\min} := {\Bbb 1}_3 \left( {\Bbb 1}_3^\top {\Bbb 1}_3 \right)^{-1} = \frac13 {\Bbb 1}_3$$
and, thus, the minimum is $\| {\rm x}_{\min} \|_2^2 = \frac13$. Hence, $\color{blue}{ \| {\rm x} \|_2^2 \geq \frac13 }$.
|
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|
Showing $\frac{1}{x^3+x}$ is continuous at $x=1$ I've been attempting to use the definition of continuity and felt a little uncertain about my working on this question.
We define $\epsilon>0$ and choose $\delta $ s.t .....
Let $\lvert x-1\rvert < 1 $:
$\lvert \frac{1}{x^3+x}-\frac{1}{2} \rvert = \lvert \frac{2-x^3-x}{2x^3+2x}\rvert = \lvert (x-1)\rvert\lvert\frac{-x^2-x-2}{2x^3+2x}\rvert$
Then we restrict $\delta < 1$ so now $ 0<x<2$ which tells us that
$\lvert\frac{-x^2-x-2}{2x^3+2x}\rvert<-\frac{2}{5}$
Hence $\lvert (x-1)\rvert\lvert\frac{-x^2-x-2}{2x^3+2x}\rvert < -\frac{2\delta}{5}<\epsilon$.
So we must choose $\delta > -\frac{2\epsilon}{5}$.
However I am used to finding something of the form $0<\delta<$min{$a,b$}
Does this mean the solution is $-\frac{2\epsilon}{5}<\delta<1$ ?
Apologies if this is really trivial/simple, I have just never seen a solution like that so wondered if perhaps I made an error somewhere.
|
I like to my limits
go to zero,
so,
let $x = 1+y$.
Then
$\begin{array}\\
\lvert \frac{1}{x^3+x}-\frac{1}{2} \rvert
&=\lvert \frac{1}{(1+y)^3+(1+y)}-\frac{1}{2} \rvert\\
&=\lvert \frac{1}{y^3+3y^2+4y+2}-\frac{1}{2} \rvert\\
&=\lvert \frac{2-(y^3+3y^2+4y+2)}{2(y^3+3y^2+4y+2)}\rvert\\
&=\lvert \frac{-(y^3+3y^2+4y)}{2(y^3+3y^2+4y+2)}\rvert\\
&=\lvert \frac{-y(y^2+3y+4)}{2(y^3+3y^2+4y+2)}\rvert\\
\end{array}
$
As $y \to 0$,
$\frac{(y^2+3y+4)}{2(y^3+3y^2+4y+2)}
\to 1
$,
so
$\frac{-y(y^2+3y+4)}{2(y^3+3y^2+4y+2)}
\to 0$.
|
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|
why$2 (-1 + 2^{1 + n})$ is the answer to the recurrence relation $a_{n}=2a_{n-1}+2$? $a_{0}=2$
$a_{1}=2(2)+2$
$a_{2}=2(2(2)+2)+2$
$a_{3}=2(2(2(2)+2)+2)+2$
$a_{4}=2(2(2(2(2)+2)+2)+2)+2$
$a_{5}=2(2(2(2(2(2)+2)+2)+2)+2)+2$
To simplifiy
$a_{6}=2^{6}+2^{5}...2^{1}$
so my answer is
$a_{n}=2^{n+1}+2^{n}+...2^{1}$
The correct answer is $2 (-1 + 2^{1 + n})$
How do I make this transition?
|
You can also do this in base 2 fairly easily:
$$
\begin{align*}
a_0 &= 10_2 \\
a_1 &= 2a_0 + 2 = 100_2 + 10_2 = 110_2 \\
a_2 &= 1100_2 + 10_2 = 1110_2.
\end{align*}
$$
We can see that by induction we will have.
$$a_n = \underbrace{11\cdots1}_{n+1}0_2 = 2(\underbrace{11\cdots1_2}_{n+1}).$$
Now note that
$$\underbrace{11\cdots1_2}_{n+1} + 1 = 1\underbrace{00\cdots0_2}_{n+1} = 2^{n + 1}.$$
Therefore
$$a_n = 2(2^{n + 1} - 1).$$
|
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|
Limit of trigonometric function $\lim_{x\to\pi/3} \frac{1 - 2\cos(x)}{\sin(x - \frac{\pi}{3})}$ I want to compute this limit:
$\displaystyle \lim_{x\to\pi/3} \frac{1 - 2\cos(x)}{\sin(x - \frac{\pi}{3})}$
Using L'Hopital, is easy to get the result, which is $\sqrt{3}$
I tried using linear approximation (making $u = x - \displaystyle \frac{\pi}{3}$)
$\displaystyle \lim_{u\to 0} \frac{1 - 2\cos(u + \frac{\pi}{3})}{\sin(u)} = \lim_{u\to 0} \frac{1 - \cos(u) + \sqrt{3}\sin(u)}{u} \approx \lim_{u\to 0} \frac{1 - 1 + \sqrt{3}u}{u} = \sqrt{3}$
But it bothers me using that sort of linear approximation, I want to get the result in a more formal way. I have tried using double angle properties
$$\cos(2x) = \cos^2(x) - \sin^2(x)$$
$$\sin(2x) = 2\sin(x)\cos(x)$$
But I reach to a point of nowhere, I cannot come up with a way of simplifying expressions to get the results:
$\displaystyle\lim_{x\to\pi/3} 2\frac{3\sin^2(\frac{x}{2}) - \cos^2(\frac{x}{2})}{\sqrt{3}\sin(x) - \cos(x)}$
Is there a way of computing this limit without approximations and without L'Hopital?
|
HINT:
Using Prosthaphaeresis & Double angle Formula as $\cos\dfrac\pi3=?$
$$2\cdot\dfrac{\cos\dfrac\pi3-\cos x}{\sin\left(x-\dfrac\pi3\right)}=2\cdot\dfrac{2\sin\dfrac{\pi+3x}6\sin\dfrac{3x-\pi}6}{2\sin\dfrac{3x-\pi}6\cos\dfrac{3x-\pi}6}$$
Now as $x\to\dfrac\pi3\iff3x\to\pi,3x\ne\pi\implies\sin\dfrac{3x-\pi}6\ne0$, so it can be cancelled safely.
|
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|
For an even and $2\pi$ periodic function, why does $\int_{0}^{2\pi}f(x)dx = 2\int_{0}^{\pi}f(x)dx $ How does one derive the fact that $$\int_{0}^{2\pi}f(x)dx = 2\int_{0}^{\pi}f(x)dx $$ whenever $f(x)$ is an even and $2\pi$ periodic function. I do know the result that for even functions, we have $$\int_{-a}^{a}f(x)dx = 2\int_{0}^{a}f(x)dx$$
But for the case where the end points are not symmetrical wrt the origin i do not know how to do.
|
Hint:
We have,
$$\int_{0}^{2\pi}\frac{1}{a+b\cos x}dx$$
$$=\int_{0}^{\pi} \frac{1}{a+b \cos x} dx+\int_{\pi}^{2\pi} \frac{1}{a+b \cos (x-2\pi)} dx$$
Can you see why? What happens if you let $x-2\pi=u$ on the second part?
\begin{align} \int_{0}^{2\pi} \frac{1}{a+b\cos x}dx \ =\int_{0}^{\pi} \frac{1}{a+b\cos x}dx+\int_{\pi}^{2\pi} \frac{1}{a+b\cos x}dx \ =\int_{0}^{\pi} \frac{1}{a+b\cos x}dx+\int_{\pi}^{2\pi} \frac{1}{a+b\cos (x-2\pi)}dx \ =\int_{0}^{\pi} \frac{1}{a+b\cos x}dx+\int_{-\pi}^{0} \frac{1}{a+b\cos x} dx \ =\int_{-\pi}^{\pi} \frac{1}{a+b\cos x}dx \ =2 \int_{0}^{\pi} \frac{1}{a+b\cos x}dx \end{align}
|
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|
Geometry Problem Concerning Lengths with a Square and Two Right Triangles Square $ABCD$ has side length 13, and points $E$ and $F$ are exterior to the square such that $BE = DF = 5$ and $AE = CF = 12$. Find $EF^2$.
Click Here for the Attached Diagram
|
Solution:
First, by Pythagoras, triangles $ABE$ and $CDF$ are right triangles. We extend lines $\overline{EA}$, $\overline{EB}$, $\overline{FC}$, and $\overline{FD}$ to construct quadrilateral $EGFH$.
Right triangles $ABE$ and $CDF$ are congruent, so $\angle DCF = \angle EAB$, which means $\angle GDA = \angle EAB$. Also, $AD = AB$, so triangles $EAB$ and $GDA$ are congruent. Hence, $\angle AGD = \angle AEB = 90^\circ$.
Also, $GE = AG + AE = 5 + 12 = 17$, and $GF = DF + DG = 5 + 12 = 17$, so triangle $EGF$ is a right isosceles triangle. Then $EF = GE \sqrt{2} = 17 \sqrt{2}$, so $EF^2 = (17 \sqrt{2})^2 = \boxed{578}$.
|
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"timestamp": "2023-03-29T00:00:00",
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|
Suppose $ \sum_{n=1}^{\infty} b_n x^n= \frac{x^3}{(x^4-1)^2}$. What could be an expression of $b_n$? A practice problem reads:
Suppose $$\sum_{n=1}^{\infty} b_n x^n = \frac{x^3}{(x^4-1)^2}.$$
What could be an expression of $b_n$?
Some of the possible answers read $ 2^{3n}nx^{3n-1}, nx^{3n-1}, nx^{4n-1}$
There are a few other answers. I was just not sure how to start this problem?
|
Another variation is based upon the binomial series expansion.
\begin{align*}
(1+x)^\alpha=\sum_{n=0}^\infty\binom{\alpha}{n}x^n\qquad\qquad |x|<1
\end{align*}
We obtain
\begin{align*}
\frac{x^3}{(x^4-1)^2}&=x^3\sum_{n=0}^\infty\binom{-2}{n}(-x^4)^n\tag{1}\\
&=x^3\sum_{n=0}^\infty\binom{n+1}{1}x^{4n}\\
&=\sum_{n=0}^\infty (n+1)x^{4n+3}\tag{2}\\
&=x^3 + 2 x^7 + 3 x^{11} + 4 x^{15} + 5 x^{19} + 6 x^{23}+\cdots
\end{align*}
In (1) we use the binomial identity $\binom{-p}{q}=\binom{p+q-1}{q}(-1)^q$.
We conclude according to (2) the representation
\begin{align*}
\sum_{n=1}^{\infty} b_n x^n=\sum_{n=0}^\infty (n+1)x^{4n+3}
\end{align*}
implies for $n\geq 0$
\begin{align*}
b_{4n+k}=
\begin{cases}
n+1&\qquad k=3\\
0&\qquad k\neq 3
\end{cases}
\end{align*}
|
{
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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|
Find a quadratic integer polynomial that annihilates a given $2\times2$ integer matrix Suppose $A=\begin{bmatrix}3&2\\2&3\end{bmatrix}$. How can we find integers $b$ and $c$ such that $A^2+bA+cI_2=0$?
|
$A=\begin{pmatrix}3&2\\ 2&3\end{pmatrix}\implies A^2= \begin{pmatrix}3&2\\ 2&3\end{pmatrix}\begin{pmatrix}3&2\\ 2&3\end{pmatrix} =\begin{pmatrix}13&12\\ 12&13\end{pmatrix}$
Then $$A^2+bA+cI_2=0\implies \begin{pmatrix}13&12\\ 12&13\end{pmatrix}+b\begin{pmatrix}3&2\\ 2&3\end{pmatrix}+c\begin{pmatrix}1&0\\ 0&1\end{pmatrix} =\begin{pmatrix}0&0\\ 0&0\end{pmatrix}$$
Then we get the following equations:
$$13+3b+c=0\tag{1}$$
$$12+2b=0\tag{2}$$
$(2)\implies b=-6\quad\text{then}\quad (1)\implies 13+3\cdot -6+c=0\implies c=5$
|
{
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|
Proving a variable always returns true in an inequality I have the following inequality
$$(1+a)^n\leq 1+(2^n−1)a$$ for $0 ≤ a ≤ 1$
What's the approach to prove that any value of the variable returns true ?
|
We will prove this by induction on $n$, assuming $n > 0$. First, the base for the induction, $n=1$:
$$
(1 + a)^1 \leq 1 + (2^1 - 1)a
\iff 1 + a \leq 1 + a
$$
which is true. Now, the induction hypothesis is $(1 + a)^k \leq 1 + (2^k - 1)a$. We can add $2^ka$ to both sides to get
$$\begin{align}
1 + (2^{k+1} - 1)a &\geq (1 + a)^k + 2^ka \\
&\geq (1 + a)^k + (1 + a)^ka \\
&= (1 + a)^{k+1}
\end{align}$$
and we are done. Note that we used the fact that $a \leq 1$ when we wrote $2^k \geq (1 + a)^k$ and the fact that $0 \leq a$ when we multiplied this by $a$.
For negative $n$, the statement becomes $\frac{1}{(1 + a)^n} \leq 1 + \big(\frac{1}{2^n} - 1\big)a$. The proof proceeds almost exactly as before; for the base, at $n = 1$,
$$\begin{align}
\frac{1}{(1 + a)^1} &\leq 1 + \bigg(\frac{1}{2^1} - 1\bigg)a \\
\iff \frac{1}{1 + a} &\leq 1 - \frac{a}{2} = \frac{2 - a}{2} \\
\iff 2 &\leq -a^2 + a + 2 \\
\iff a &\leq 1
\end{align}$$
Assuming $\frac{1}{(1 + a)^k} \leq 1 + \big(\frac{1}{2^k} - 1\big)a$ and subtracting $\frac{a}{2^{k+1}}$ from both sides , we have
$$\begin{align}
1 + \big(\frac{1}{2^{k+1}} - 1\big)a &\geq \frac{1}{(1 + a)^k}-\frac{a}{2^{k+1}} \\
&\geq \frac{1}{(1 + a)^k}-\frac{a}{(1 + a)^{k+1}} \\
&= \frac{1}{(1 + a)^{k+1}}
\end{align}$$
as desired. We used $0 \leq a \leq 1$ in the base case as well as in the inductive step, so that should not be an issue. Note that the double inversion of a negative and a reciprocal allows us to use $2 \geq 1 + a$ as before.
This proves the result for $n \geq 1$ and $n \leq -1$. For $n = 0$, the result is trivial and boring, but completes the proof for all integers $n$.
|
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|
Finding minimum of trigonometric function Let $$v=\sqrt{a^2\cos^2(x)+b^2\sin^2(x)}+\sqrt{b^2\cos^2(x)+a^2\sin^2(x)}$$
Then find difference between maximum and minimum of $v^2$.
I understand both of them are distance of a point on ellipse from origin, but how do we find maximum and minimum?
I tried guessing, and got maximum $v$ when $x=45^{o}$ and minimum when $x=0$, but how do we justify this?
|
Let $$p=\sqrt{a^2\cos^2(x)+b^2\sin^2(x)},q=\sqrt{b^2\cos^2(x)+a^2\sin^2(x)}$$
Now $p^2+q^2=a^2+b^2$
and $$2(p^2+q^2)-(p+q)^2=(p-q)^2\ge0\iff(p+q)^2\le2(p^2+q^2)=?$$
|
{
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"timestamp": "2023-03-29T00:00:00",
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|
Evaluate $\displaystyle \lim_{n \to \infty} n \int_0^1 (\cos x - \sin x)^n dx$ Evaluate $$\lim_{n \to \infty} n \int_0^1 (\cos x - \sin x)^n dx$$
The answer should be $1$.
I tried to solve it similar to how one user solved this integral Limit of integral with cos and sin
but it seems like it doesn't work because the upper bound will become $1 - \pi/4$.
|
If you know how to deal with
$$n \int_0^{\frac{\pi}{4}} (\cos x - \sin x)^n\,dx,$$
just split the integral at $\frac{\pi}{4}$ and see whether you can find something useful for the other part.
Since $x \mapsto \cos x - \sin x$ is strictly decreasing on $\bigl[0, \frac{\pi}{2}\bigr]$, and $\cos \frac{\pi}{4} = \sin \frac{\pi}{4}$, we have
$$\lvert \cos x - \sin x\rvert \leqslant c := \sin 1 - \cos 1$$
for $\frac{\pi}{4} \leqslant x \leqslant 1$. Hence
$$\Biggl\lvert \int_{\frac{\pi}{4}}^1 (\cos x - \sin x)^n\,dx\Biggr\rvert \leqslant \int_{\frac{\pi}{4}}^1 c^n\,dx < c^n.$$
Since $0 < c < 1$, it follows that
$$n \int_{\frac{\pi}{4}}^1 (\cos x - \sin x)^n\,dx \to 0.$$
That is useful indeed. Thus we need only consider
$$\int_0^{\frac{\pi}{4}} (\cos x - \sin x)^n\,dx = 2^{n/2} \int_0^{\frac{\pi}{4}} \sin^n \biggl(\frac{\pi}{4} - x\biggr)\,dx = 2^{n/2}\int_0^{\frac{\pi}{4}} \sin^n x\,dx.$$
Substituting $u = \sin x$, then $v = \sqrt{2}\cdot u$, and afterwards integrating by parts, we find (for $n \geqslant 1$)
\begin{align}
n\int_0^{\frac{\pi}{4}} (\cos x - \sin x)^n\,dx
&= n 2^{n/2} \int_0^{\frac{\pi}{4}} \sin^n x\,dx \\
&= n 2^{n/2} \int_0^{\frac{1}{\sqrt{2}}} \frac{u^n}{\sqrt{1-u^2}}\,du \\
&= n \int_0^1 \frac{v^n}{\sqrt{2-v^2}}\,dv \\
&= v^n\cdot \frac{v}{\sqrt{2-v^2}}\biggr\rvert_0^1 - \int_0^1 v^n\biggl(\frac{1}{\sqrt{2-v^2}} + \frac{v^2}{(2-v^2)^{3/2}}\biggr)\,dv \\
&= 1 - \int_0^1 \frac{2v^n}{(2-v^2)^{3/2}}\,dv.
\end{align}
Since $\frac{1}{\sqrt{2}} < \frac{2}{(2-v^2)^{3/2}} < 2$ for $0 < v < 1$, we thus have
$$1 - \frac{2}{n+1} < n\int_0^{\frac{\pi}{4}} (\cos x - \sin x)^n\,dx < 1 - \frac{1}{\sqrt{2}\,(n+1)}.$$
|
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|
Is there a possible explanation in plain English how to use the Chinese Reminder Theorem? For example, if it is the problem of
Find the smallest integer that leave a remainder of 3 when divided by
5, a remainder of 5 when divided by 7, and a remainder of 7 when
divided by 11
In some explanation such as in this article it started to use mod, and the whole explanation seemed to very difficult to understand. Is there a more plain English explanation of how this is solved?
|
Find the smallest integer, $n$, that leaves
$\text{A remainder of $3$ when divided by $5$.}\tag{1}$
$\text{A remainder of $5$ when divided by $7$.}\tag{2}$
$\text{A remainder of $7$ when divided by $11$.}\tag{3}$
Without using moduli, the Chinese Remainder theorem says that the integers $n$ that meet the above three requirements must be of the form $385A + r$ for a unique $0 \le r < 385$; where $385 = 5 \cdot 7 \cdot 11$.
Looking at the first two conditions, there must exists integers $u$ and $v$ such that
$$\text{$n=5u+3\quad$ and $\quad n = 7v + 5$}$$
So
$$\text{$7n=35u+21\quad$ and $\quad 5n = 35v + 25$}$$
We find (Bezout's Identity) that $-2 \cdot 7n + 3 \cdot 5n = n$. So
\begin{align}
n &= -2 \cdot 7n + 3 \cdot 5n \\
n &= -2(35u+21) + 3(35v + 25) \\
n &= 35(-2u+3v) + 33 \\
n &= 35w + 33 \\
\end{align}
What have we gained from this? The CRT guarantees that every integer $n$ that satisfies conditions $(1)$ and $(2)$ must be of the form $35w + 33$ for some integer $w$.
Condition $(3)$ tells us that there must exists some integer, $x$, such that
$n = 11x + 7$. So
$$\text{$11n=385w+363\quad$ and $\quad 35n = 385x + 245$}$$
We find that $16\cdot 11n - 5 \cdot 35n = n$ So
\begin{align}
n &= 16\cdot 11n - 5 \cdot 35n \\
n &= 16(385w+363) - 5(385x + 245) \\
n &= 385(16w-5x) + 4583\\
n &= 385(16w-5x) + 11\cdot 385 + 348\\
n &= 385(16w-5x+11) + 348\\
n &= 385y + 348\\
\end{align}
A quick check shows that 348 satisfies conditions $(1), (2)$, and $(3)$.
|
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|
Locating a line perpendicular to an ellipse At any given point on an ellipse, the line of the perpendicular is easy to calculate; what about the inverse? Given a line of known orientation, at what point on the circumference of an ellipse would it be perpendicular?
|
Given the perpendicular line, you can find the slope of the tangent line by taking the negative reciprocal of the slope of the perpendicular line. Then, once you have the slope of the tangent at a point on the ellipse, find the first derivative of the ellipse and set it equal to that quantity, and then solve. There will be two solutions.
For example, let the axes of the ellipse be $a$ and $b$ and let the line have slope $m$. Then the slope of the tangent at the point is perpendicular to the perpendicular line, and its slope is $-\frac{1}{m}$. Now we must find the first derivative of the equation of an ellipse. The equation of an ellipse takes the form
$$\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$$
If we solve for $y$, we get
$$y=\pm b\sqrt{1-\frac{x^2}{a^2}}$$
So that
$$y'=\pm \frac{bx}{a\sqrt{a^2-x^2}}$$
We set this equal to our tangent slope and solve the equation
$$-\frac{1}{m}=\pm \frac{bx}{a\sqrt{a^2-x^2}}$$
Or, because of the "plus or minus", we get
$$\frac{1}{m}=\pm \frac{bx}{a\sqrt{a^2-x^2}}$$
Now square both sides to get
$$\frac{1}{m^2}=\frac{b^2x^2}{a^2(a^2-x^2)}$$
Multiply both sides by $(a^2-x^2)$ to get
$$\frac{1}{m^2}(a^2-x^2)=\frac{b^2}{a^2}x^2$$
$$\frac{a^2}{m^2}-\frac{1}{m^2}x^2=\frac{b^2}{a^2}x^2$$
$$\frac{a^2}{m^2}=(\frac{b^2}{a^2}+\frac{1}{m^2})x^2$$
$$\frac{a^2}{m^2}=(\frac{a^2+b^2m^2}{a^2m^2})x^2$$
$$x=\pm\sqrt{\frac{a^4}{a^2+b^2m^2}}$$
Then once you have these $x$ values, plug it in and solve for the $y$ values, and you should have your points. This is a lot of algebra, so please let me know in the comments if I made a stupid mistake, and I'll fix it.
|
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|
$A$ and $B$ are the positive acute angles satisfying the equations $3\cos^2 A + 2\cos^2 B=4$ $A$ and $B$ are the positive acute angles satisfying the equations $3\cos^2 A + 2\cos^2 B=4$ and $\dfrac {3\sin A}{\sin B}=\dfrac {2\cos B}{\cos A}$. Then $A+2B$ is equal to:
$1$. $\dfrac {\pi}{4}$
$2$. $\dfrac {\pi}{3}$
$3$. $\dfrac {\pi}{6}$
$4$. $\dfrac {\pi}{2}$.
My Attempt
$$\dfrac {3\sin A}{\sin B}=\dfrac {2\cos B}{\cos A}$$
$$3\sin A.\cos A= 2\cos B.\sin B$$
$$\dfrac {3}{2} \sin 2A=2\sin 2B$$.
How do I proceed further?
|
Hint: Square both sides of the second equation, and and replace $\sin^2 $ by $1 - \cos^2$, and coupled with the first solve a system of equations for $\cos A, \cos B$. Can you manager to proceed?
|
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|
Find the shortest distance between the curve $y^2=x$ to the line $y=x+1$
Find the shortest distance between the curve $y^2=x$
to the line $y=x+1$.
Im not sure how to attack a problem like this.
|
Let the parabola and the line be parameterized as follows
$$\mathcal P := \{ (t_1^2, t_1) \mid t_1 \in \mathbb R \} \qquad \qquad \qquad \mathcal L := \{ (t_2, t_2+1) \mid t_2 \in \mathbb R \}$$
The squared distance between $\mathcal P$ and $\mathcal L$ is given by the minimum of the quartic objective function
$$f (t_1, t_2) := \left( t_1^2 - t_2 \right)^2 + \left( t_1 - t_2 - 1 \right)^2$$
Differentiating $f$ with respect to $t_1$ and $t_2$ and finding where the partial derivatives vanish, we obtain a system of two polynomial equations
$$\begin{array}{rl} 4 t_{1} \left( t_{1}^{2} - t_{2} \right) + 2 t_{1} - 2 t_{2} - 2 &= 0\\ - 2 t_{1}^{2} - 2 t_{1} + 4 t_{2} + 2 &= 0\end{array}$$
whose solution is
$$(t_1, t_2) = \left( \frac 12, -\frac 18 \right)$$
Thus, the distance between the parabola and the line is
$$\sqrt{\left( \left( \frac 12 \right)^2 + \frac 18 \right)^2 + \left( \frac 12 + \frac 18 - 1 \right)^2} = \sqrt{\frac{9}{32}} = \color{blue}{\frac{3}{4 \sqrt{2}}}$$
Plot
SymPy code
>>> from sympy import *
>>> t1, t2 = symbols('t1 t2')
>>> squareddistance = (t1**2 - t2)**2 + (t1 - t2 - 1)**2
>>> solve([diff(squareddistance, t1), diff(squareddistance, t2)], t1, t2)
[(1/2, -1/8), (1/2 - sqrt(3)*I/2, -1/2 - sqrt(3)*I/2), (1/2 + sqrt(3)*I/2, -1/2 + sqrt(3)*I/2)]
Thus, the unique real solution is (1/2, -1/8).
|
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|
Finding $\gcd(x^3, x^3+x+1)$ Suppose $f(x) = x^3$ and $g(x) = x^3+x+1$. I am trying to show that the $\gcd(f,g)= 1$ but am running into some trouble...
My attempt: (Division algorithm)
$x^3+x+1 = (1)(x^3)+(x+1)$
$x^3 = x^2(x+1)-x^2$
But I cannot go further than this, what is going wrong here?
|
The gcd of $x^3+x+1$ and $x^3$ is the same as the gcd of $x^3$ and $(x^3+x+1)-x^3=x+1$.
Now $x+1$ is irreducible and doesn't divide $x^3$. End.
By the way, the remainder of the division of $f(x)=x^3$ by $x+1$ is $f(-1)=-1$ and, indeed, $x^3+1=(x+1)(x^2-x+1)$, but it's not needed.
|
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|
Summation of the Sine Function I was messing around on Wolfram Alpha's summation calculator and when I plugged in the summation
$$\sum_{i=1}^n\sin\frac{i\pi}{180}$$
and it gave me the value
$$\frac12\left(\cot\frac\pi{360}-\csc\frac\pi{360}\cos\frac{(2n+1)\pi}{360}\right)$$
I don't understand... how does it arrive at this formula? And how do I verify this? If I wanted to, how could I find similar formulas in the future?
|
We have
$$ 2\sin{A}\sin{B} = -\cos{(A+B)}+\cos{(A-B)} $$
by using the angle addition formulae. Putting $A=ak+b$, $B=a/2$ gives
$$ 2\sin{\left(\frac{a}{2}\right)}\sin{(ak+b)} = -\cos{\left(a\left(k+\frac{1}{2}\right)+b\right)} + \cos{\left(a\left(k-\frac{1}{2}\right)+b\right)}. $$
Summing from $k=1$ to $n$ gives
$$ 2\sin{\left(\frac{a}{2}\right)}\sum_{k=1}^n \sin{(ak+b)} = \sum_{k=0}^n \left( - \cos{\left(a\left(k+\frac{1}{2}\right)+b\right)} + \cos{\left(a\left(k+\frac{1}{2}\right)+b\right)} \right). $$
The sum on the right telescopes, and the only remaining terms give
$$ \sum_{k=1}^n \sin{(ak+b)} = \frac{1}{2}\csc{\left(\frac{a}{2}\right)} \left( \cos{\left(\frac{a}{2}+b\right)} - \cos{\left(a\left(n+\frac{1}{2}\right)+b\right)} \right) $$
This is essentially the most general formula of its kind: choosing different values for $a$ and $b$ gives a number of useful trigonometric sums. In your case, $a=\pi/180$, $b=0$, which gives
$$ \sum_{k=1}^n \sin{\frac{k\pi}{180}} = \frac{1}{2}\csc{\left(\frac{\pi}{360}\right)} \left( \cos{\left(\frac{\pi}{360}\right)} - \cos{\left(\frac{(2n+1)\pi}{360}\right)} \right), $$
as expected.
These formulae may also be used in integrating the trigonometric functions from first principles.
|
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|
For $x \ge 1$, is $2^{\sqrt{x}} \ge x$ For $x \ge 1$, is $2^{\sqrt{x}} \ge x$
The answer appears to me to be yes since:
$$\sqrt{x} \ge \log_2 x$$
Here's my reasoning:
(1) For $x > 16$, $5\sqrt{x} > 4\sqrt{x} + 4$
(2) For $x > 25$, $x > 5\sqrt{x}$
(3) So, $x > 25$, $2x > x + 4\sqrt{x} + 4$ and it follows that $\sqrt{2x} > \sqrt{x} +2$
(4) So, as $x$ doubles, $\sqrt{2x} > \sqrt{x} + 2$ but $\log_2(2x) = \log_2(x) +1$
Is my reasoning correct? Is there a simpler way to analyze this?
|
The inequality doesn't always hold, as can be seen by taking $x=9$.
Notice that your question is about comparing $a^b$ and $b^a$ with $a=2,b=\sqrt{x}$. It is well known that $a^b \leq b^a$ if and only if $$a^{1/a} \leq b^{1/b} $$
which in our case is
$$\sqrt{2} \leq b^{1/b}.$$
This inequality holds iff $$2 \leq b \leq 4 $$
which translates to $$ 4 \leq x \leq 16 .$$
Thus your inequality holds for $x \in [1,4] \cup [16,\infty)$.
|
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|
If $\sec \theta + \tan \theta =x$, then find the value of $\sin \theta$. If $\sec \theta + \tan \theta =x$, then find the value of $\sin \theta$.
$$\sec \theta + \tan \theta = x$$
$$\dfrac {1}{\cos \theta }+\dfrac {\sin \theta }{\cos \theta }=x$$
$$\dfrac {1+\sin \theta }{\sqrt {1-\sin^2 \theta }}=x$$
$$1+\sin \theta =x\sqrt {1-\sin^2 \theta }$$
$$1+2\sin \theta + \sin^2 \theta = x^2-x^2 \sin^2 \theta $$
$$x^2 \sin^2 \theta + \sin^2 \theta + 2\sin \theta = x^2-1$$
$$\sin^2 \theta (x^2+1) + 2\sin \theta =x^2-1$$
|
The equation becomes
$$
1+\sin\theta=x\cos\theta
$$
Set $X=\cos\theta$ and $Y=\sin\theta$, so the equation becomes
$$
\begin{cases}
X^2+Y^2=1 \\[4px]
1+Y=xX
\end{cases}
$$
Note that $x\ne0$ and substitute $X=x^{-1}(1+Y)$ in the first equation getting
$$
(1+Y)^2+x^2Y^2=x^2
$$
that simplifies to
$$
(1+x^2)Y^2+2Y+1-x^2=0
$$
that yields
$$
Y=-1 \qquad\text{or}\qquad Y=\frac{x^2-1}{x^2+1}
$$
Is $Y=-1$ a solution for the problem?
By the way, you also get $\cos\theta$, since
$$
X=\frac{1}{x}(1+Y)=\frac{1}{x}\frac{x^2+1+x^2-1}{x^2+1}=\frac{2x}{x^2+1}
$$
|
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|
Solving $|x-1|^{\log^2(x)-\log(x^2)}=|x-1|^3$
Solve the equation:$$|x-1|^{\log^2(x)-\log(x^2)}=|x-1|^3.$$
There are three solutions of $x$: $10^{-1}$, $10^3$ and $2$. I obtained the first two solutions but I have been unsuccessful in getting $2$ as a solution. Please help.
|
First, the presence of $\ln x$ in the equation, i.e. in $\ln^2 x - \ln(x^2)$, dictates that $x > 0$. Next, if $|x - 1| = 0$, then $x = 1$ and $\ln^2 x - \ln(x^2) = 0$, but $0^0$ is undefined and thus $x = 1$ is excluded.
Now the equation reduces to$$
|x - 1|^{\ln^2 x - \ln(x^2) - 3} = 1,
$$
which is equivalent to$$
|x - 1| = 1 \text{ or } \ln^2 x - \ln(x^2) - 3 = 0.
$$
Case 1: $|x - 1| = 1$.
In this case, $x = 0$ or $2$ and $x > 0$ implies that $x = 2$.
Case 2: $\ln^2 x - \ln(x^2) - 3 = 0$.
Note that $\ln(x^2) = 2\ln x$ for $x > 0$, so this equation becomes $(\ln x)^2 - 2 \ln x - 3 = 0$, or$$
(\ln x + 1)(\ln x - 3) = 0,
$$
which implies $\ln x = -1$ or $3$, i.e. $x = \dfrac{1}{10}$ or $1000$.
Therefore, the solutions are $\dfrac{1}{10}$, $2$, $1000$.
|
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|
Inverse element of
Given the extension $ \dfrac{\mathbb Q[x]}{\langle x^3-5\rangle}: \mathbb Q$ and $\alpha =1+x^2+\langle x^3-5\rangle \in \dfrac{\mathbb Q[x]}{\langle x^3-5 \rangle}$, find $\alpha^{-1}$.
Can somebody give a hint? or an answer? please?
|
Since $x^3-5$ is a polynomial of degree $3$, every element in $k:= \dfrac{\mathbb{Q}[x]}{\langle x^3-5\rangle}$ has a representative of degree at most $2$. Thus we are looking for a polynomial $\beta:=a + bx + cx^2 + \langle x^3 - 5\rangle$ where $a,b,c\in \mathbb{Q}$ and such that $\alpha\beta = 1 + \langle x^3 + 5\rangle$.
Now
\begin{align*}\alpha\beta &= a + bx + (a+c)x^2 + bx^3 + cx^4 + \langle x^3 -5\rangle \\ &= (a+5b) + (b+5c)x + (a+c)x^2 + \langle x^3 -5\rangle
\end{align*}
Therefore
$$
\begin{cases}
a+5b = 1 & \\ b + 5c = 0 & \\ a + c = 0.
\end{cases}
$$
Solving this we get
$$a = \dfrac{1}{26} \qquad b = \dfrac{5}{26} \qquad c = \dfrac{-1}{26}$$
So $\alpha^{-1} = \dfrac{1}{26}(1 + 5x - x^2) + \langle x^3 - 5\rangle$.
|
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|
Prove this inequality $(2+x)\ln{x}-e(x-1)>0,x>1$
Let $x>1$,show that
$$(2+x)\ln{x}-e(x-1)>0$$
The simaler problem:[How prove $(2+5x)\ln{x}-6(x-1)>0.\forall x>1$
this problem idea: $f(x)=\ln{x}-\dfrac{e(x-1)}{x+2}$
then
$$f'(x)=\dfrac{x^2+(4-3e)x+4}{x(x+2)^2}=\dfrac{\left(x-\dfrac{3e-4}{2}\right)^2+4-\dfrac{(3e-4)^2}{4}}{x(x+2)^2}$$
but $f'(x)>0,\forall x>1$ is not always hold,because
$$4-\dfrac{(3e-4)^2}{4}<0$$
|
It's easier to deal with $f(x)=(2+x)\ln x-e(x-1)$ directly. Note that $f(1)=0$.
Since $f'(x)=\ln x+\frac2{x}+1-e$ and $f''(x)=\frac{x-2}{x^2}$, $f'(x)$ is monotone decreasing in $1<x<2$ and monotone increasing in $x>2$.
From $f'(2)=\ln 2+2-e<0$ and $f'(2.6)=\ln 2.6+\frac2{2.6}+1-e>0$, there is a constant $2<c<2.6$ for which $f'(c)=0$. (Finding $2.6$ is a bit tricky. I used WolframAlpha.)
Then you can show that
\begin{align*}
f(c)&=(2+c)\ln c-e(c-1)\\
&=(2+c)(-\frac2c-1+e)-e(c-1)\\
&=-c-\frac4c+3e-4\\
&>-2.6-\frac4{2.6}+3e-4\\
&>0.\ \text{(Very subtle.)}
\end{align*}
The above argument is sufficient to show $f(x)>0$ for $1<x<2$, $2\le x<2.6$ and $2.6\le x$.
|
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|
$a,b,c,d$ are real numbers such that....
Suppost that $a,b,c,d$ are real numbers such that $$a^2+b^2=1$$ $$c^2+d^2=1$$ $$ac+bd=0$$
I've to show that $$a^2+c^2=1$$ $$b^2+d^2=1$$ $$ab+cd=0$$
Basically,I've no any idea or tactics to tackle this problem. Any methods? Thanks in advance.
EDITED.
The given hint in the books is $S:=(a^2+c^2-1)^2+(b^2+d^2-1)^2=(ac+bd)^2$
|
Note that
$$0=(ac+bd)^2 = a^2c^2+b^2d^2+2abcd = a^2c^2+(1-a^2)(1-c^2) +2ac(-ac)$$
$$ = a^2c^2+1-a^2-c^2+a^2c^2-2a^2c^2 = 1-a^2-c^2$$
And so $a^2+c^2=1$. A similar argument shows that $b^2+d^2=1$. Further, now we can say
$$(ac+bd)^2-(ab+cd)^2 =a^2c^2+b^2d^2-a^2b^2-c^2d^2$$
$$ = a^2c^2+b^2d^2-a^2(1-d^2)-c^2(1-b^2)$$
$$=a^2c^2+b^2d^2+a^2d^2+b^2c^2-a^2-c^2$$
$$=(a^2+b^2)(c^2+d^2)-(a^2+c^2)=0$$
And so $$ab+cd=0$$.
|
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|
How can we prove the equality of the following determinants? Assume that $A$ is an $n\times n$ real matrix whose entries are all $1$. Then how can we show the following determinant equality for any $x$?
$\det(A-xI)$=\begin{vmatrix}
1 -x & 1 & 1 & \cdots & 1 \\
1 & 1 -x & 1 & \cdots & 1 \\
1 & 1 & 1 -x & \cdots & 1\\
\vdots & \vdots & \vdots & \ddots & \vdots\\
1 & 1 & 1 & \cdots & 1 -x
\end{vmatrix}=
\begin{vmatrix}
n-x & n-x & n-x & \cdots & n -x \\
1 & 1 -x & 1 & \cdots & 1 \\
1 & 1 & 1 -x & \cdots & 1\\
\vdots & \vdots & \vdots & \ddots & \vdots\\
1 & 1 & 1 & \cdots & 1 -x
\end{vmatrix}
Thanks.
|
Note that
$$
\begin{bmatrix}
n-x & n-x & n-x & \cdots & n -x \\
1 & 1 -x & 1 & \cdots & 1 \\
1 & 1 & 1 -x & \cdots & 1\\
\vdots & \vdots & \vdots & \ddots & \vdots\\
1 & 1 & 1 & \cdots & 1 -x
\end{bmatrix} = \\
\begin{bmatrix}
1&1&1&\cdots&1\\
&1&0&\cdots&0\\
&&\ddots&\ddots&\vdots\\
&&&&0\\
&&&&1
\end{bmatrix}
\begin{bmatrix}
1 -x & 1 & 1 & \cdots & 1 \\
1 & 1 -x & 1 & \cdots & 1 \\
1 & 1 & 1 -x & \cdots & 1\\
\vdots & \vdots & \vdots & \ddots & \vdots\\
1 & 1 & 1 & \cdots & 1 -x
\end{bmatrix}
$$
|
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|
Generating functions ( recurrence relations ) Find $a_n$ using Generating Functions : $a_n = -a_{n-1} + 2a_{n−2}$, $n\ge2$ and $a_0 = 1$, $a_1 = 2$.
Approach : So I will form a characteristic equation $ r^2 + r - 2 = 0$ whose roots are $r_1 = -2$, $r_2 = 1$.
So my general solution is $a_n = α_1r_1^n + α_2r_2^n$.
$a_n = α_1(-2)^n + α_2(1)^n$
When $a_0 = 1$, then $1 = α_1(-2)^0 + α_2(1)^0$, then $α_2 = 1 - α_1 $.
When $a_1 = 2$, then $2 = α_1(-2)^1 + α_2(1)^1$, then $-2α_1 + 1 - α_1 = 2$.
$α_1 = -1/3$ and $α_2 = 4/3 $
So $a_n = -1/3r_1^n + 4/3r_2^n$.
Can anyone tell me if it is correct or not and any help will be appreciated :) .
Also, if I have $a_{n+2}=a_{n+1}+2a_n$. Can someone tell me if its char. equation should be like $r^2-r-2 = 0$? Just asking because of the addition symbol rather than subtraction.
|
The standard generating function approach yields
$$GF=\frac{(1+3x)}{1+x-2x^2}$$
which has a denominator which can be factorised
$$GF=\frac{1+3x}{(1+2x)(1-x)}$$
Applying the theory of partial fractions gives
$$GF=-\frac{1}{3(1+2x)}+\frac{4}{3(1-x)}$$
which are all recognizable contributions to a formula for the $n$th term
$$a_n=-\frac{1}{3}(-2)^n+\frac{4}{3}$$
which matches the answer by Cye Waldman
|
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|
Conditional Probability with dice
Two dice are rolled and their number is added. If the sum equal 7, the game is lost. If the sum equal 9, the game is won. If the sum is any other number, the dice are rolled again, until the sum equals 7 or 9. What is the probability that someone wins 1 out of 5 games?
My approach:
Sum of 9: $2 \times (6,3) + 2 \times (5,4) \implies p(9) = \frac{1}{9}$
Sum of 7: $2 \times (6,1) + 2 \times (5,2) + 2 \times (4,3) \implies p(7) = \frac{1}{6}$
$$P(x=1) = {5 \choose 1} \bigg(\frac{1}{9}\bigg)^1 \bigg(\frac{1}{6}\bigg)^{4} = 4.29 \times 10^{-4}$$
|
Let us consider $(a, b)$ a pair of values for the two dice, with $a$ the value for die 1 and $b$ the value for die 2. There are six equiprobable cases in which you lose:
$$(1, 6), (2, 5), (3, 4), (4, 3), (5, 2), (6, 1)$$
Similarly, there are four equiprobable cases in which you win:
$$(3, 6), (4, 5), (5, 4), (6, 3)$$
All other combinations result in a new dice roll, and thus need not be considered. As such, the probability of winning a game equals $\frac{4}{10} = 0.4$. Based on the binomial distribution, the probability of winning exactly 1 out of 5 games now equals:
$${5 \choose 1} \bigg(\frac{4}{10}\bigg)^1 \bigg(\frac{6}{10}\bigg)^4 = 0.2592$$
|
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|
Show that $\int_0^\infty \frac{\ln x}{(x^2+1)(x^2-1)}dx=\frac{\pi^2}{8}$ How do I show that: $\int_0^\infty \frac{\ln x}{(x^2+1)(x^2-1)}dx=\frac{\pi^2}{8}$ using contours and residues
My attempt:
I know that the singular points are $i,-i,-1,1,0$
consider $f(z)= \frac{\ln z}{(z^2+1)(z^2-1)}$
and the branch $|z|>0$, $0<\theta<2\pi$
$u: z=r, \rho\le r \le R$ (u is the upper edge)
$-l: z=r, \rho\le r \le R$ (lower edge)
$\int_ufdz-\int_{-l}fdz=\int_\rho^R \frac{\ln r + i0}{(z^2+1)(z^2-1)}-\int_\rho^R \frac{\ln r + i2\pi}{(z^2+1)(z^2-1)}$
How do I continue from here?
|
First, enforcing the substitution $x\to 1/x$ reveals
$$\int_0^\infty \frac{\log(x)}{(x^2+1)(x^2-1)}\,dx=\int_0^\infty \frac{x^2\log(x)}{(x^2+1)(x^2-1)}\,dx\tag 1$$
From $(1)$ it is evident that
$$\int_0^\infty \frac{\log(x)}{(x^2+1)(x^2-1)}\,dx=\frac12\int_0^\infty \frac{\log(x)}{x^2-1}\,dx\tag 2$$
We evaluate the integral $J$ defined by
$$J=\oint_C \frac{\log^2(z)}{z^2-1}\,dz$$
where $C$ is the classical keyhole contour with (i) the branch cut along the non-negative real axis and (ii) with deformations around $z=1$. Applying the residue theorem, it is easy to see that $J=i\pi^3$. Therefore, we find that
$$\begin{align}
J&=i\pi^3\\\\
&=\int_{0}^{\infty}\frac{\log^2(x)}{x^2-1}\,dx-\text{PV}\int_0^\infty \frac{\left(\log(x)+i2\pi\right)^2}{x^2-1}\,dx\\\\
&=-i4\pi\int_0^\infty \frac{\log(x)}{x^2-1}\,dx\\\\
&+\color{blue}{(4\pi^2)\text{PV}\left(\int_0^\infty \frac{1}{x^2-1}\,dx\right)}\\\\
&+\color{red}{(4\pi^2)\lim_{\epsilon \to 0^+}\int_{\pi}^{2\pi} \frac{1}{(1+\epsilon e^{i\phi})^2-1}\,(i\epsilon e^{i\phi})\,d\phi}\\\\
&=-i4\pi\int_0^\infty \frac{\log(x)}{x^2-1}\,dx+\color{blue}{0}+\color{red}{i2\pi^3}\tag 4
\end{align}$$
Finally, solving $(4)$ for the integral of interest yields
$$\frac12\int_0^\infty \frac{\log(x)}{x^2-1}\,dx=\frac{\pi^2}{8}$$
We now present an approach herein that relies on real analysis only.
Writing $\int_0^\infty \frac{\log(x)}{x^2-1}\,dx=\int_0^1 \frac{\log(x)}{x^2-1}\,dx+\int_1^\infty \frac{\log(x)}{x^2-1}\,dx$ and enforcing the substitution $x\to 1/x$ in the second integral, we find
$$\begin{align}
\int_0^\infty \frac{\log(x)}{(x^2+1)(x^2-1)}\,dx=\int_0^1 \frac{\log(x)}{x^2-1}\,dx\end{align}$$
We can use partial fraction expansion to write
$$\begin{align}
\int_0^1 \frac{\log(x)}{x^2-1}\,dx&=-\frac12\int_0^1 \frac{\log(x)}{1+x}\,dx-\frac12\int_0^1 \frac{\log(x)}{1-x}\,dx\\\\
&=\frac12 \int_0^1 \frac{\log(1+x)}{x}\,dx-\frac12\int_0^1 \frac{\log(1-x)}{x}\,dx\\\\
&=\frac12 \sum_{n=1}^\infty \frac{(-1)^{n-1}}{n} \int_0^1 x^{n-1}\,dx+\frac12 \sum_{n=1}^\infty \frac{1}{n}\int_0^1 x^{n-1}\,dx\\\\
&=\frac12 \sum_{n=1}^\infty \frac{1-(-1)^n}{n^2}\\\\
&= \sum_{n=1}^\infty \frac{1}{(2n-1)^2}\\\\
&=\frac{\pi^2}{8}
\end{align}$$
Where we used $\sum_{n=1}^\infty\frac1{n^2}=\frac{\pi^2}{6}$ along with $\sum_{n=1}^\infty \frac{1}{(2n-1)^2}=\sum_{n=1}^\infty \frac{1}{n^2}-\sum_{n=1}^\infty \frac{1}{(2n)^2}=\frac34 \sum_{n=1}^\infty\frac1{n^2}=\frac{\pi^2}{8}$
And in THIS ANSWER, I showed using only real analysis that $\sum_{n=1}\frac1{n^2}=\frac{\pi^2}{6}$.
|
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|
Help with finding the length of one of the sides of a triangle
\begin{align}\frac z{\sin E}&=\frac y{\sin D}\tag{1}\\
z^2 &= x^2 + y^2 - 2xy\cos E\tag{2}\\
y^2 &= z^2 + x^2 - 2zx\cos D\tag{3}\end{align}
Solve for $z$
My solution is as follow.
I stopped at the equation:
$$(2xy)^2 - (x^2 + y^2 - z^2)^2 = (2zx)^2 - (z^2 + x^2 - y^2)^2$$
Please help me solve for $z$ in the above equation.
Thank you
|
If you expand your final equation $$(2xy)^2 - (x^2 + y^2 - z^2)^2 = (2zx)^2 - (z^2 + x^2 - y^2)^2$$ you will get an equation which has equal terms in $x^4$, $y^4$, and $z^4$ on both sides. Removing these gets you:
$$2x^2y^2 + 2x^2z^2 + 2y^2z^2 = 2x^2y^2 + 2x^2z^2 + 2y^2z^2$$
which as you can see is always true.
The problem is that all your algebraic manipulations cannot add information, they can just rearrange what you already know. Look at your diagram. You have two sides of a triangle, and nothing else. There are an infinite number of triangles you can make that have two sides of given lengths. You can see this physically - get two sticks, attach their ends, and pivot. Every position gives you a new triangle, with a different third side length.
|
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Prove that the following are real numbers
*
*$$\frac{1}{z}+\frac{1}{\overline{z}}$$
*$$z^3\cdot\overline{z}+z\cdot\overline{z}^3$$
1.$$\frac{1}{z}+\frac{1}{\overline{z}}$$
$$\frac{\overline{z}}{z\cdot \overline{z}}+\frac{z}{z\cdot\overline{z}}$$
$$\frac{\overline{z}+z}{z\cdot \overline{z}}$$
$$\frac{2Re(z)}{|z|}$$ which is real number as a division of two real numbers
*$$z^3\cdot\overline{z}+z\cdot\overline{z}^3$$
$$\overline{z^3\cdot\overline{z}+z\cdot\overline{z}^3}$$
Due to: $\overline{z_{1}+z_{2}}=\overline{z_{1}}+\overline{z_{2}}$ and $\overline{z_{1}\cdot z_{2}}=\overline{z_{1}}\cdot\overline{z_{2}}$
we get:
$$\overline{z}^3\cdot{z}+\overline{z}\cdot{z}^3$$
Now we can say that we got the same elements and $(z=\overline{z})$ therefore it is real number?
Is the reasoning in both of these correct?
|
Alternatively:
*
*$\cfrac{1}{z}+\cfrac{1}{\overline{z}}=\left(\cfrac{1}{z}\right)+\overline{\left(\cfrac{1}{z}\right)}= 2 \operatorname{Re}\left(\cfrac{1}{z}\right)$
*$z^3\cdot\overline{z}+z\cdot\overline{z}^3 = z \bar z(z^2 + \bar z ^2) = |z|^2\cdot 2 \operatorname{Re}(z^2)$
|
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|
How to convert the given set of equations into a symmetric form How do I convert the equations
$$\left\{\begin{array}{} 3a +2b+ c & = & 0\\ a+4b+4c & = & 0 \end{array}\right.$$
to the form
$$\frac{a}{4}=\frac{b}{-11}=\frac{c}{10}=k.$$
Thank you
|
One way of doing this is to relate $\frac a4$ to $b$ and $c$, separately.
First, multiplying the first equation by $2$ gives
$2\times (3a+2b+c)=2\times0 \Leftrightarrow 6a+4b+2c=0$.
Subtracting the second equation from this result gives
$5a-2c=0$
$5a=2c$
$a=\frac{2}{5}c$
$\frac{a}{4}=\frac{c}{10}$.
Similarly, to relate $\frac{a}{4}$ to $b$, you can start by multiplying the second equation by $3$, then subtracting it from the first equation to get
$(3a+2b+c)-(3a+12b+12c)=0$
$-10b-11c=0$
$-11c=10b$
$c=\frac{10b}{-11}$.
Dividing everything by $10$ gives
$\frac{c}{10}=\frac{10b}{-110}=\frac{b}{-11}$.
But we know, $\frac{a}{4}=\frac{c}{10}$, so $\frac{a}{4}=\frac{-b}{11}$.
This simplifies to
$\frac{a}{4}=\frac{b}{-11}=\frac{c}{10}$,
as required.
You can google "linear simultaneous equations" for more details on how to solve this kind of questions.
|
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|
How to prove an inequality I didn't know where can I start because it doesn't fit to any theories or formulas.
If $a>0$, $b>0$, $c>0$ and $a+b+c=1$, prove that:
$$\frac{a^3}{a^2+b^2}+\frac{b^3}{b^2+c^2}+\frac{c^3}{c^2+a^2}\geq \frac{1}{2}$$
|
We need to prove that
$$\sum_{cyc}\frac{a^3}{a^2+b^2}\geq\frac{1}{2}$$ or
$$\sum_{cyc}\frac{a^3}{a^2+b^2}\geq\frac{a+b+c}{2}$$ or
$$\sum_{cyc}\left(\frac{a^3}{a^2+b^2}-\frac{a}{2}\right)\geq0$$ or
$$\sum_{cyc}\frac{a(a^2-b^2)}{a^2+b^2}\geq0$$ or
$$\sum_{cyc}\left(\frac{a(a^2-b^2)}{a^2+b^2}-(a-b)\right)\geq0$$ or
$$\sum_{cyc}\frac{b(a-b)^2}{a^2+b^2}\geq0.$$
Done!
|
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|
General Solution for a PDE I would like to know if someone could provide me the solutions for the next PDE:
\begin{align*}
f_{x}(x,y) - f_{y}(x,y) = \frac{y-x}{x^{2}+y^{2}}f(x,y)
\end{align*}
I empirically found the following particular solution given by $f(x,y) = \displaystyle\frac{x+y}{\sqrt{x^{2}+y^{2}}}$. Any contribution is appreciated. Thank you in advance.
|
$$f_{x}(x,y) - f_{y}(x,y) = \frac{y-x}{x^{2}+y^{2}}f(x,y)$$
The set of characteristic ODEs is:
$$\frac{dx}{1}=\frac{dy}{-1}=\frac{df}{\frac{y-x}{x^{2}+y^{2}}f}$$
The equation of a first family of characteristic curves comes from :
$$\frac{dx}{1}=\frac{dy}{-1} \quad\to\quad x+y=c_1$$
The equation of a second family of characteristic curves comes from :
$$\frac{dx}{1}=\frac{dy}{-1}=\frac{xdx+ydy}{x-y}=\frac{df}{\frac{y-x}{x^{2}+y^{2}}f}$$
$$\frac{df}{f}=-\frac{d(x^2+y^2)}{2(x^2+y^2)}\quad\to\quad \sqrt{x^2+y^2}f=c_2$$
The general solution of the PDE can be expressed on the form of an implicit equation :
$$\Phi\left(x+y)\:,\:\sqrt{x^2+y^2}f\right)=0$$
where $\Phi$ is any differentiable function of two variables.
Or, on the equivalent explicit form : $\quad \sqrt{x^2+y^2}f=F(x+y)$
$$f(x,y)=\frac{F(x+y)}{\sqrt{x^2+y^2}}$$
where $F$ is any differentiable function.
The function $F$ has to be determined according to the boundary condition (not specified in the wording of the question).
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"language": "en",
"url": "https://math.stackexchange.com/questions/2298285",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
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Finding solutions to $2^x+17=y^2$
Find all positive integer solutions $(x,y)$ of the following equation:
$$2^x+17=y^2.$$
If $x = 2k$, then we can rewrite the equation as $(y - 2^k)(y + 2^k) = 17$, so the factors must be $1$ and $17$, and we must have $x = 6, y = 9$.
However, this approach doesn't work when $x$ is odd.
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$x=3,5,9$ work with $y=5,7,23$ but fail nor $x=1,7$. If $x $ works for any other value then $x>4$.
$2^x+17=y^2$
$2^4 (2^{x-4}+1)=y^2-1=(y+1)(y-1) $
$y\pm 1$ must be even and $\gcd(x+1,x-1)=2$ and one is a multiple of $4$ while the other isn't.
That means::
$y\pm 1=8m;y\mp 1=2n;mn=2^{x-4}+1$
Also means
$2^x+8=y^2-9$
$2^3 (2^{x-3}+1)=(y-3)(y+3) $
So $y\mp 3 =4k=2n\mp 2;y\pm 3=2j=8m\pm 2;jk=2^{x-3}+1$
We have strict limitation here it seems. In that $m,n $ are relatively close to $j,k $ but their product is about half.
I'm too lazy to work out the details but there must be an upper limit to that being possible.
Also $2^x-8=2^3 (2^{x-3}-1)=y^2-25=(y-5)(y+5) $ and so $y\pm 2 =4r;y\mp 2=2s;rs=jk-2$
So.... it's a mess but we have clear limits.
I strongly suspect $x=9$ is the highest possible $x $.
Oh, and we have $2^x+1=(y-4)(y+4)\approx 16mn $.
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{
"language": "en",
"url": "https://math.stackexchange.com/questions/2298402",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "13",
"answer_count": 5,
"answer_id": 4
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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.