Undernetmath’s Weblog

Solutions_2008

heureka.jpg This page contains solutions of former potds. You are explicitly encouraged to post your own solutions to old or new potds as comments to this site. An administrator will incorporate them into this page later on (assuming they are correct and stated in a reasonably accessible manner). Please note that in the case of essentially identical solutions only one will be incorporated into the page (usually the first one).


2008-11-19 by kmh

Compute \displaystyle{\lim_{x\rightarrow 0} \frac{\int_0^x \ln(1+t^2)\,dt}{x^3}}
Solution by bor0
v = t
dv = dt
\ln(1 + t^2) = u

\dfrac{2t}{t^2+1}dt = du

Use integration by parts.

\int u dv = uv - \int v du

\int \ln(t^2 + 1) dt = t\ln(t^2 + 1) - 2\int\dfrac{t^2}{t^2+1} dt

\int\dfrac{t^2}{t^2+1} dt = \int\dfrac{t^2+1-1}{t^2+1} dt = \int\dfrac{t^2+1}{t^2+1} dt - \int\dfrac{dt}{t^2+1}

\int\dfrac{t^2+1}{t^2+1} dt - \int\dfrac{dt}{t^2+1} = t - \arctan(t)

\int(t^2 + 1) dt = t\ln(t^2 + 1) - 2t + 2\arctan(t)

\int(x^2 + 1) dx = x(\ln(x^2+1) - 2) + 2\arctan(x)

\lim_{x\to0} \dfrac{x(\ln(x^2+1) - 2) + 2\arctan(x)}{x^3} (Indeterminate form 0/0, apply l’Hospital)

\lim_{x\to0} \dfrac{\ln(x^2+1)}{3x^2} (Indeterminate form 0/0, apply l’Hospital)

\lim_{x\to0} \dfrac{\dfrac{2x}{x^2+1}}{6x} = \lim_{x\to0} \dfrac{1}{3x^2+3} = \dfrac{1}{3}


2008-10-30 by kmh

Show that if 100m+n is divisible by 7, then m+4n is divisble by 7 as well.
Solution by Magnus_RM and rafno
   100m+n\, mod\, 7=2m+n :
\Leftrightarrow (2m+n) \,mod\, 7=0
\Leftrightarrow  4\cdot (2m+n)\, mod\, =0
\Leftrightarrow 8m+4n\, mod\, 7=0
\Leftrightarrow m+4n\, mod\, 7=0

2008-7-17 by kmh
Show that for all power of 3 the 2nd last digit is even. For instance we have 3^5=729 and 2 is even.
(original source: de.rec.denksport)
Solution by rafno
Show that for all power of 3 the 2nd last digit is even. For instance we have 3^5=729 and 2 is even.
We prove by induction.
For k=1 the second last digit of 3^1=3=03 is 0. True.
Supose its true for k-1.
3^k=3*3^(k-1).
Then the second last digt of 3^(k-1) is even. So 3^(k-1)= ….+2n*10+a written in base 10.
So 3^k= … +(2*(3n)+b)*10+c
where 3*a=b*10+c. So we must verify that b is even.
but the last digit of any power of 3 is 1,3,9 or 7. and 3 times any of these gives an even last second digit. So b=2*r and then 3^k=…+2*(3n+k)*10+c.
Therefore all power of 3 has even second last digit.


2008-6-10 by CaosTheory

Compute \displaystyle{ \int_0^\infty \frac{\sin(x)^2}{x^2} dx}.
Solution by cst-link
\displaystyle{\int_0^\infty\frac{\sin(x)^2}{x^2}dx} \displaystyle{=\lim_{a\rightarrow 0} \lim_{b\rightarrow\infty} \int_a^b\frac{\sin(x)^2}{x^2}dx} \displaystyle{\stackrel{(1)}{=}\lim_{a\rightarrow 0} \lim_{b\rightarrow\infty} \left[-\frac{\sin(x)^2}{x}\right]_a^b+\int_a^b\frac{2\sin(x)\cos(x)}{x} dx} \displaystyle{\stackrel{(2)}{=}\lim_{a\rightarrow 0} \lim_{b\rightarrow\infty} \left[-\frac{\sin(x)^2}{x}\right]_a^b+\int_a^b\frac{\sin(2x)}{x} dx} \displaystyle{\stackrel{(3)}{=}\lim_{a\rightarrow 0} \lim_{b\rightarrow\infty} \left[-\frac{\sin(x)^2}{x}\right]_a^b+\int_\frac{a}{2}^\frac{b}{2}\frac{\sin(z)}{z} dz} \displaystyle{=\lim_{a\rightarrow 0} \underbrace{\frac{\sin(a)}{a}}_{\rightarrow 1}\cdot\sin(a)-0+\int_\frac{a}{2}^\infty\frac{\sin(z)}{z} dz} \displaystyle{=0-0+\int_0^\infty\frac{\sin(z)}{z} dz=\frac{\pi}{2}}


2008-5-30 by kmh

Determine the size of the dark area.
ellipse.jpg
Solution by pisagor
drawing.jpg


2008-5-19 by kmh
Show that \frac{\binom{2n}{n}}{n+1} \in \mathbb{N} for all n\in\mathbb{N} .
(Original source: The Red Book of Mathematical Problems)
Solution by cst-link
First note that \frac{1}{n+1}=1-\frac{n}{n+1} and \binom{2n}{n}=\frac{(2n)!}{n!n!}
Now we have
\frac{\binom{2n}{n}}{n+1} = \frac{(2n)!}{n!n!}\cdot(1-\frac{n}{n+1})=\frac{(2n)!}{n!n!}-\frac{(2n)!}{(n-1)!(n+1)!} = \binom{2n}{n}-\binom{2n}{n-1} \in \mathbb{N}


2008-4-29 by kmh
Find all values of \sin(x^8-x^6-x^4+x^2) for x\in\mathbb{N} and x^8-x^6-x^4+x^2 denoting an angle in degrees.
Solution by cst-link
Because´sin(u) = sin(u mod 360)
for u expressed in degrees, this problem is suitable for brute force.
For:
f(x) = x^8 – x^6 – x^4 + x^2
and x natural, one may easily see that
f(x) mod 360 = f(x mod 360) mod 360
and, subsequently
sin(f(x)) = sin(f(x) mod 360) = sin(f(x mod 360) mod 360)
Thus it suffices to find by means of computer/code the set
{f(x mod 360) mod 360 | x in N} =
= {f(x) mod 360 | x in {0, 1,…, 359}} =
= {0, 180}
which gives:
{sin(f(x)) | x in N} = {0}


2008-4-29 by kmh

Compute \displaystyle \lim_{n\rightarrow\infty} \sum_{k=n}^{2n} \frac{1}{k}
( original source: Internet Math Olympiad Israel 2008 )
Solution by cst-link
For f(x) = 1/x and each x in [k, k+1] and k>1 one has:
f(x) \leq f(k) \leq f(x-1)
Subsequently:
\displaystyle \int_k^{k+1} f(x)dx \leq \int_k^{k+1} f(k)dx
ln(k+1) - ln(k) \leq (k+1 - k)/k
ln(k+1) - ln(k) \leq 1/k

Since the left sum is telescoping we have:
\displaystyle \sum_{k=n}^{2n} (ln(k+1) - ln(k)) \leq \sum_{k=n}^{2n} 1/k
\displaystyle ln((2n+1)/n) \leq \sum_{k=n}^{2n} 1/k
So with n\rightarrow\infty and an analog argument for f(k-1) you’ll get:
\displaystyle  ln(2) \leq \lim_{n\rightarrow\infty} \sum_{k=n}^{2n} 1/k \leq ln(2)


2008-3-31 by Karlo

Construct the foci of arbitrary ellipse given as a graph using ruler and compass only.

ellipse.jpg

Solution by kmh

Construct 2 non collinear pairs of parallel chords. For each pair of parallel chords construct the midpoints of the chords and a line through the 2 midpoints. Due to the conjugated diameter property of the ellipse those 2 lines will intersect in the center C . Now draw a circle around C such that it has 4 intersection points with the ellipse. Construct the bisecting perpendiculars through those intersection points, they are the major and minor axis of the ellipse. The major intersects the ellipse in Q and the minor in R. Draw a circle with radius |CQ| around R this circle with will intersect the major in the foci.

2008-2-5 by karlo

Three friends are taking me out for my birthday. The product of their ages is 2450. The sum of their ages is my cousin’s age. I could tell you my cousin’s age, but to find the ages of my friends, you’d also need to know that each of the three is younger than I am. How old am I?

Solution by …

Let a1, a2 and a3 be the 3 ages of the 3 friends. “The product of their ages is 2450” => a1 * a2 * a3 = 2450 = 2 * 5^2 * 7^2. It is fairly simple (by hand, or via a computer) to find all distinct triplets that satisfy the product rule. (Note: throughout thist solution, I call (t1, t2, t3) and (q1, q2, q3) –distinct triplets– if and only if {t1, t2, t3} != {q1, q2, q3}. This means that, for example, (t1, t2, t3) and (t2, t3, t1) are -Not- distinct triplets). “I could tell you my cousin’s age, but to find the ages of my friends, you’d also..” => knowing the sum a1+a2+a3 does Not suffice in finding a1, a2, a3. Picture the triplets “determined” above. Each triplet generates a sum (of its elements). Assume one of these triplets, Triplet_X, generates Sum_X and that no other triplet generates Sum_X. If the sum of the ages (cousin’s age) would be equal to Sum_X, we’d know for sure that the friends age triplet is Triplet_X, since it is the only one that satisfies the “friends_ages_sum = cousins_age” condition. Again: this can not happen, the sum (cousin’s age) must not be sufficient and therefore the sum, whichever it is, must satisfy the following condition: (at least) two distinct triplets must exist so that each triplet has the sum of its elements equal to the cousin’s age. Go back to the distinct triplets “determined” at step1. For each, compute the sum of their elements. If a sum appears only once, this means that the triplet that generated it is invalid. Find any sum which appears at least twice. Luckly, there is only one sum with this proprety, and mainly: 64, generated by distinct triplets (7, 7, 50) and (5, 10, 49). Therefore, it is certain, that the friends age triplet (a1, a2, a3) is equal to one of the two determined previously. The final condition holds the key. “you’d also need to know that each of the three is younger than I am.”. This means that knowing that all 3 of the friends are younger than Karlo MUST invalidate one of the two triplets. If Karlo is older than 50 (> 50), both triplets are “valid”. If Karlo is younger than 49 (<= 49) both triplets are “invalid”. One possibility remains: Karlo is 50, which makes the first triplet: (7, 7, 50) invalid and the second triplet (5, 10, 49) valid. To conclude: the friends are aged 5, 10, 49; the cousin is 64 and Karlo is 50. Happy birthday to Karlo!

2008-1-17 by yfk

Let x_1, x_2,\ldots, x_n \in \{0,1\} and \overline{x}= \frac{1}{n} \displaystyle \sum_{i=1}^n x_i. Show that \frac{1}{n} \displaystyle \sum_{i=1}^n (x_i-\overline{x})^2 \leq\frac{1}{4}.

Solution by kmh

First note that due to x_i \in \{0,1\} we have x_i=x_i^2 for i=1,\ldots,n. This yields us \frac{1}{n} \displaystyle \sum_{i=1}^n (x_i-\overline{x})^2=\frac{1}{n} \displaystyle \sum_{i=1}^n (x_i^2 -2x_i\overline{x}+\overline{x}^2)=\frac{1}{n} \displaystyle \sum_{i=1}^n x_i -2\overline{x}\frac{1}{n} \displaystyle \sum_{i=1}^n x_i+\frac{1}{n} \displaystyle \sum_{i=1}^n \overline{x}^2 =\overline{x}-2\overline{x}^2+\overline{x}^2=\overline{x}-\overline{x}^2\stackrel{(\star)}{\leq}\frac{1}{4} (*): If we consider the last expression as a function of \overline{x}, then we have an upside down parabola with a maximum of \frac{1}{4}. This can be determined completing the square or finding the the root of the derivative. Completing the square:

\overline{x}-\overline{x}^2=-(\overline{x}^2-\overline{x})=-((\overline{x}-\frac{1}{2})^2-\frac{1}{4} )=-(\overline{x}-\frac{1}{2})^2+\frac{1}{4}

2008-1-8 by lhrrwcc

Let A be a noetherian ring. Prove that if f:A\rightarrow A is a surjective ring homomorphism then f is bijective.

Solution by lhrrwcc

A being noetherian means that every increasing chain of ideals (a_1) \subset (a_2) \subset (a_3) \subset \ldots eventually becomes stationary, ie (a_k) = (a_{k+1}) = \ldots. Now in order to prove that any surjection is an inyection we can prove that the kernel is trivial. So assume that f: A \rightarrow A is a surjective ring homomorphism and that Ker f is not trivial. Let m \in Ker f. There is, because f is surjective, an element t such that f(t)=m, so f(f(t)) = f(m) = 0 so t in Ker f^2. Now, there is an element s in A such that f(s) = t, and again f(f(f(s))) = f(f(t)) =f(m) = 0 so s in Ker f^3. If we keep doing will we will get an increasing chain of ideals Ker f  \subset Ker f^2 \subset Ker f^3 \subset \ldots contradicting the hypothesis that A is noetherian, therefore the kernel is trivial, and f is an injection

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