Solve equation without symbolic math toolbox

Hello ,

I'm trying to solve this equation : 0 = (U/r.^3)* sqrt(-2+((3*r.^3)*(Bx/U)))-B

U is a constant

Bx and B are column matrix

I would have as a result a value of r for each value of Bx and B

is it possible to do this without the Symbolic math toolbox ?

Thank you

Answer 1

[Chris C]

You can run this with two functions. The first function I have designated as myMain.m and it is written as..

r0 = rand(3,3);
fsolve(@myfun,r0)

The second function myfun.m is called by the first function by passing initial guesses of r as r0. I altered the way your equation above was written and wrote the function as...

function F = myfun(r)
Bx = rand(3,1);
B = rand(3,1);
U = rand(1);

F = (U)*sqrt(-2+((3*r.^3)*(Bx/U)))-r.^3*B;
end

Change the matrices to whatever numbers you need and it should work.

Good luck.

[Benoit]

Hello,

Thank you for reactivity, I've tried you're solution. Im new to fsolve function so maybe I am missing something but your answer gives me just one solution.

[Benoit]

Ok , i've got it.

Thank you

Answer 2

[glady]

Two fair dice are thrown and the product of the numbers on the dice is recorded. Given that one die lands on 5, find the probability that the product on the dice is

(a) Exactly 15 =

(b) More than 12 =

[Torsten]

5*1 =  5
5*2 = 10
5*3 = 15
5*4 = 20
5*5 = 25
5*6 = 30

[Sam Chak]

Hi @glady

For Question (a), are you inclined to answer, "If the first die is specified to be 5, and the conditional sample space consists of the six equally likely outcomes as shown by @Torsten, then since only the pair (5, 3) gives a product of 15, the probability is 1/6"?

For Question (b), are you inclined to answer, "Under the same conditional sample space, since only the pairs (5, 3), (5, 4), (5, 5), and (5, 6) give a product greater than 12, the probability is 4/6, which simplifies to 2/3"?

However, it seems to me that this may be a trick question. Many students might opt for the six ordered outcomes because they are linguistically simpler for the human mind to interpret, without clarifying the initial meaning of "Two fair dice are thrown."

Perhaps we could seek the opinions of @John D'Errico and @Walter Roberson.

[Torsten]

However, it seems to me that this may be a trick question. Many students might opt for the six ordered outcomes because they are linguistically simpler for the human mind to interpret, without clarifying the initial meaning of "Two fair dice are thrown."

It means that all 6 outcomes {1,2,3,4,5,6} are equally likely for both dice. If the second dice was biased, the probabilities for the product outcomes {5,10,15,20,25,30} were no longer 1/6 for each case, but would differ.

[Sam Chak]

Hi @Torsten

I do not have a definitive answer because I encountered a similar question previously, in which the math teacher listed all ordered pairs (die1, die2). In this case, if the condition states that "one die lands on 5," it implies that at least one die shows a 5. Therefore, the possible outcomes are as follows:

(5, 1) =  5, 
(5, 2) = 10, 
(5, 3) = 15, 
(5, 4) = 20, 
(5, 5) = 25, 
(5, 6) = 30, 
(1, 5) =  5, 
(2, 5) = 10, 
(3, 5) = 15, 
(4, 5) = 20, 
(6, 5) = 30.

with (5, 5) counted only once.

[Paul]

For me, the ambiguity in the problem is with "Given that one die lands on 5 ..."

Is that

(a) at least one die (Sam's assumption)

(b) one specific die (Torsten's assumption)

(c) exactly one die (another reasonable assumption)

This answer thread should be deleted insofar as it has nothing to do with the original question nor with Matlab, unless the ask is to show how to solve the problem with Matlab, in which case it should be a new question.

[Walter Roberson]

I would say "one die lands on 5" implies that the other one did not land on 5.

[Torsten]

Google's AI supports Sam Chak's last interpretation and gives probabilities of 2/11 and 7/11 :-)

[Paul]

Well, we can't really trust AI w/o verification, can we? :-)

[D1,D2] = deal((1:6).');
S = combinations(D1,D2);
N = sym(height(S));
S.A1 = prod(S{:,"D"+(1:2)},2) == 15;
S.A2 = prod(S{:,"D"+(1:2)},2)  > 12;
B1 = or(S.D1 == 5,S.D2 == 5);   % at least one
B2 = S.D1 == 5;                 % one specific
B3 = xor(S.D1 == 5,S.D2 == 5);  % exactly one
S.B = [B1,B2,B3];
PA1gB = (sum(S.A1 & S.B,1)/N) ./ (sum(S.B,1)/N)

PA1gB = 

PA2gB = (sum(S.A2 & S.B,1)/N) ./ (sum(S.B,1)/N)

PA2gB =