Skip to content

Commit

Permalink
more
Browse files Browse the repository at this point in the history
  • Loading branch information
@tkoz0
tkoz0 committed Mar 2, 2025
1 parent c630884 commit 2de976d
Showing 1 changed file with 168 additions and 0 deletions.
168 changes: 168 additions & 0 deletions academic/misc/2025-math-calendar/01-jan.html
View file
Original file line number Diff line number Diff line change
Expand Up @@ -289,26 +289,194 @@ <h3 id="jan04">Jan 04</h3>

<h3 id="jan05">Jan 05</h3>

<svg version="1.1" xmlns="http://www.w3.org/2000/svg" viewBox="0 2 22 12" width="330" height="180" class="img_center">
<g stroke="black" stroke-width="0.1" fill="none">
<rect x="1" y="10" width="2" height="1" />
<rect x="3" y="10" width="2" height="1" />
<rect x="5" y="10" width="2" height="1" />
<rect x="7" y="10" width="2" height="1" />
<rect x="9" y="10" width="2" height="1" />
<rect x="11" y="10" width="2" height="1" />
<rect x="13" y="10" width="2" height="1" />
<rect x="15" y="10" width="2" height="1" />
<rect x="17" y="10" width="2" height="1" />
<rect x="19" y="10" width="2" height="1" />
<polygon points="11 11 13 13 9 13" />
<rect x="5" y="5" width="2" height="5" />
<rect x="7" y="5" width="2" height="5" />
<rect x="15" y="5" width="2" height="5" />
<rect x="19" y="5" width="2" height="5" />
</g>
<g fill="black" font-size="0.07em">
<text x="4.75" y="4">7kg</text>
<text x="6.75" y="4">5kg</text>
<text x="14.75" y="4">1kg</text>
<text x="18.75" y="4">xkg</text>
</g>
</svg>

<p>
The torque applied is the mass times the distance from the center. For each of
these blocks, we use the midpoint. The values to balance form the equation
\[7\cdot2.5+5\cdot1.5=1\cdot2.5+x\cdot4.5\Rightarrow17.5+7.5=2.5+4.5x
\Rightarrow22.5=4.5x\Rightarrow x=5\]
</p>

<h3 id="jan06">Jan 06</h3>

\[\sqrt{7+\sqrt{13}}\cdot\sqrt{7-\sqrt{13}}=\sqrt{7^2-13}=\sqrt{36}=6\]

<h3 id="jan07">Jan 07</h3>

<p>
Find the sum of the roots of \(x^8-7x^7+8\)
</p>

<p>
Using Vieta's formula, is is \(-a_7/a_8=-(-7)/1=7\).
</p>

<h3 id="jan08">Jan 08</h3>

<p>
\(y\) is a positive integer and \(x=1+y+y^2+\ldots+y^7\) is a power of a prime.
</p>

<p>
By finite geometric sum, \(x={y^8-1\over y-1}\) when \(y&gt;1\). This can be
further factored:
\[\begin{align}&x={y^8-1\over y-1}={(y^4+1)(y^4-1)\over y-1}
={(y^4+1)(y^2+1)(y^2-1)\over y-1}\\&
={(y^4+1)(y^2+1)(y+1)(y-1)\over y-1}=(y^4+1)(y^2+1)(y+1)\end{align}\]
From here, \(y+1\) is a power of a prime. Suppose \(y+1=p^a\) for some prime
\(p\) and integer \(a&gt;0\). Then
\[y=p^a-1\Rightarrow y^2=p^{2a}-2p^a+1\Rightarrow y^2+1=p^{2a}-2p^a+2\]
This quantity must be equal to some other power of the same prime so suppose for
some integer \(b&gt;0\)
\[p^{2a}-2p^a+2=p^b\Rightarrow 2=p^b-p^{2a}+2p^a\]
We see that 2 is a multiple of \(p\) therefore our prime must be \(p=2\). This
means \(y+1=2^a\). Then
\[y^2=(2^a-1)^2=2^{2a}-2\cdot2^a+1\Rightarrow y^2+1=2^{2a}-2\cdot2^a+2=2^b\]
The only time this is true is when \(a=1\). We can see this by factoring it to
\[y^2+1=2\cdot(2^{2a-1}-2^a+1)\]
which is twice an odd number whenever \(a&gt;1\). Therefore, the only
possibility is \(y+1=2\Rightarrow y=1\). In this case, \(x=8=2^3\) which is the
solution.
</p>

<h3 id="jan09">Jan 09</h3>

<p>
Find the number of factors of 36.
</p>

<p>
Factor it to \(36=2^2\cdot3^2\). For each prime, there are 3 power choices to
form a factor (0,1,2). Therefore the number of factors is \(3\times3=9\).
</p>

<h3 id="jan10">Jan 10</h3>

\[{14\over15}x-{2\over3}x+{3\over5}x-{7\over9}x={8\over9}\]

<p>
\[x{42-30+27-35\over45}={40\over45}\Rightarrow 4x=40\Rightarrow x=10\]
</p>

<h3 id="jan11">Jan 11</h3>

\[\begin{align}&3\sin\theta+4\cos\theta=5\\
&5\sin\theta+5\cos\theta+3\cot\theta=x\end{align}\]

<p>
We should focus on the first equation to find what \(\theta\) is. This might
make us think of the 3-4-5 right triangle which is related. That can be used to
find \(\sin(\theta)=3/5,\cos(\theta)=4/5\). These values can be used to find
\(x\) in the second equation
\[5\sin\theta+5\cos\theta+3\cot\theta=5(3/5)+5(4/5)+3{4/5\over3/5}=3+4+4=11\]
</p>

<h3 id="jan12">Jan 12</h3>

\[\sum_{x=0}^\infty{2x+1\over2^{x-1}}\]

<p>
Multiply by \(2/2\) so the denominator becomes \(2^x\)
\[2\sum_{x=0}^\infty{2x+1\over2^x}=2\left[2\sum_{x=0}^\infty{x\over2^x}
+\sum_{x=0}^\infty2^{-x}\right]\]
The simpler sum is
\[\sum_{x=0}^\infty2^-x={1\over1-1/2}={1\over1/2}=2\]
The other sum can be rewritten as a double summation by changing the order of
summation thinking about the following table
\[\begin{array}{cccc}1/2&&&\\1/4&1/4&&\\1/8&1/8&1/8&\\\vdots&&&\ddots\\
\end{array}\]
Each term corresponds to a row so now if we sum column by column
\[\begin{align}&\sum_{x=1}^\infty{x\over2^x}=\sum_{x=1}^\infty
\sum_{y=x}^\infty{2^{-y}}=\sum_{x=1}^\infty\left[\sum_{y=0}^\infty2^{-y}
-\sum_{y=0}^{x-1}2^{-y}\right]=\sum_{x=1}^\infty\left[
{1\over1-1/2}-{1-2^{-x}\over1-1/2}\right]\\&
=\sum_{x=1}^\infty(2-2+2\cdot2^{-x})=2\sum_{x=1}^\infty2^{-x}
=2\left[{1\over1-1/2}-2^0\right]=2\end{align}\]
So it turns out both of these summations are 2 and if we substitute them into
the expression we found earlier, the answer is \(6\).
</p>

<h3 id="jan13">Jan 13</h3>

<p>
Find the largest solution of \((25-2y)^{(y^2-25)}=1\).
</p>

<p>
If \(|25-2y|\neq1\) then we need \(y^2-25=0\Rightarrow y=\pm5\).
</p>

<p>
If \(|25-2y|=1\) then \(y=12,13\). The only requirement is the exponent is
positive to avoid the division by zero problem, which is true for these values.
Therefore, the largest solution is \(y=13\).
</p>

<h3 id="jan14">Jan 14</h3>

<p>
Two natural numbers differ by 4. The sum of their squares is 106. Find their
sum.
</p>

<p>
Without loss of generality, let \(a,b\) be the 2 numbers with \(a&gt;b\). Then
\(a-b=4\) and \(a^2+b^2=106\). By substitution
\[(b+4)^2+b^2=106\Rightarrow2b^2+8b+16=106\Rightarrow b^2+4b-45=0
\Rightarrow(b+9)(b-5)=0\]
The solutions are \(b=5,-9\) so the only natural number solution is \(b=5\).
Then \(a=b+4=9\) so the sum is \(a+b=14\).
</p>

<h3 id="jan15">Jan 15</h3>

\[{\cot^3(\pi/12)-\tan^3(\pi/12)\over2\sqrt{3}}\]

<p>
The easiest way to solve this is probably by first finding \(\tan(\pi/12)\)
which can be done with the angle difference identity
\[\tan(\pi/3-\pi/4)={\tan(\pi/3)-\tan(\pi/4)\over1+\tan(\pi/3)\tan(\pi/4)}
={\sqrt{3}-1\over1+\sqrt{3}}\]
Then from this we can easily find
\[\cot(\pi/12)={1+\sqrt{3}\over\sqrt{3}-1}\]
These can be substituted in with some gross algebra
\[\begin{align}&{\cot^3(\pi/12)-\tan^3(\pi/12)\over2\sqrt{3}}
={\left({1+\sqrt{3}\over\sqrt{3}-1}\right)^3
-\left({\sqrt{3}-1\over1+\sqrt{3}}\right)^3\over2\sqrt{3}}
={{6\sqrt{3}+10\over6\sqrt{3}-10}-{6\sqrt{3}-10\over6\sqrt{3}+10}
\over2\sqrt{3}}\\&
={\left(6\sqrt{3}+10\right)^2-\left(6\sqrt{3}-10\right)^2
\over2\sqrt{3}\left(6\sqrt{3}-10\right)\left(6\sqrt{3}+10\right)}
={36\cdot3+100+120\sqrt{3}-36\cdot3-100+120\sqrt{3}
\over2\sqrt{3}\left(36\cdot3-100\right)}
={240\sqrt{3}\over16\sqrt{3}}=15\end{align}\]
</p>

<h3 id="jan16">Jan 16</h3>

<h3 id="jan17">Jan 17</h3>
Expand Down

0 comments on commit 2de976d

Please sign in to comment.