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hram777 [196]
3 years ago
7

What is the minimum number of weighings on a balance scale need to find a counterfeit coin among 8 coins if the counterfeit coin

is either lighter or heavier that the other true coins (which are all the same weight.) describe the algorithm to find the counterfeit coin in this minimum number of weighings?
Mathematics
1 answer:
olga nikolaevna [1]3 years ago
5 0
It would take only 3 weighings to find the counterfeit coin (at the most). 

First you would weigh coins against each other in pairs. 

-Coins 1 and 2 vs Coins 3 and 4
-Coins 5 and 6 vs Coins 7 and 8

One of the 4 pairs would weigh more or less than the other. Once you have determined that pair and whether it is more or less, then weigh the coins in that pair against each other. Use whether the pairing was more or less than the other pairing to determine which of the coins you are looking for. 
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Answer:

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Step-by-step explanation:

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Let p0, p1, and p2 be the orthogonal polynomials described below, where the inner product on P4 is given by evaluation at -2, -1
Mamont248 [21]

Answer:

$\frac{51}{5}t$

Step-by-step explanation:

Let W = $(p_0, p_1, p_2)$  be orthogonal polynomials which is equal to $(4, 3t, t^2 -2)$, which defines the inner products as

$(f,g)=f(-2)g(-2)+f(-1)g(-1)+f(0)g(0)+f(1)g(1)+f(2)g(2)$

Now, we find the orthogonal projection of $p=3t^3$ on W.

So the projection is

$Proj_W p = \frac{(p_0,p)}{(p_0,p_0)}p_0+\frac{(p_1,p)}{(p_1,p_1)}p_1+\frac{(p_2,p)}{(p_2,p_2)}p_2$

$(p_0,p)=p_0(-2)p(-2)+p_0(-1)p(-1)+p_0(0)p(0)+p_0(1)p(1)+p_0(2)p(2)$

          $=4(-24)+4(-3)+4(0)+4(3)+4(24)=0$

$(p_0,p_0)=p_0(-2)p_0(-2)+p_0(-1)p_0(-1)+p_0(0)p_0(0)+p_0(1)p_0(1)+p_0(2)p_0(2)$

            $=4(4)+4(4)+4(4)+4(4)+4(4)=80$

$(p_1,p)=p_1(-2)p(-2)+p_1(-1)p(-1)+p_1(0)p(0)+p_1(1)p(1)+p_1(2)p(2)$

          $=(-6)(-24)+(-3)(-3)+0(0)+3(3)+6(24)=306$

$(p_1,p_1)=p_1(-2)p_1(-2)+p_1(-1)p_1(-1)+p_1(0)p_1(0)+p_1(1)p_1(1)+p_1(2)p_1(2)$

            $=(-6)(-6)+(-3)(-3)+0(0)+3(3)+6(6)=90$

$(p_2,p)=p_2(-2)p(-2)+p_2(-1)p(-1)+p_2(0)p(0)+p_2(1)p(1)+p_2(2)p(2)$

         $=2(-24)+(-1)(-3)+(-2)(0)+(-1)(3)+2(24)=0$

$(p_2,p_2)=p_2(-2)p_2(-2)+p_2(-1)p_2(-1)+p_2(0)p_2(0)+p_2(1)p_2(1)+p_2(2)p_2(2)$

            $=(2)(2)+(-1)(-1)+(-2)(-2)+(-1)(-1)+2(2)=14$

Therefore,

$Proj_W p = \frac{(p_0,p)}{(p_0,p_0)}p_0+\frac{(p_1,p)}{(p_1,p_1)}p_1+\frac{(p_2,p)}{(p_2,p_2)}p_2$

              $=\frac{0}{80}(4)+\frac{306}{90}(3t)+\frac{0}{14}(t^2-2)$

              $=\frac{51}{5}t$

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