Shouldn't ##m^2-24n \gt 0## rather than ##m^2-24n \geq 0##, since the last term after completing the squares is ##\frac {3}{4} B^2## which will always be positive because ##b \approx 212##?
I think the reason its under Calculus is because the first chapter in our Calculus course is Functions where the concept of ##f(x)## is explained. Here we need to use that concept since we need to know what ##f(1)##,##f(2)## and ##f(3)## mean.
After understanding the question based on your post, I tried to go about it, but its far too complex since another requirement we have is that no calculator is allowed.
My analysis is as below.
Let ##a=f(1)##, ##b=f(2)## and ##c=f(3)##.
I can easily see that ##a= 102 + 7\sin{1} \approx 102##...
The angle in these trigonometric ratios of sin 1, sin 2 and sin 3 must be treated as radians.
1 radian is about 57 degrees, 2 radians is about 114 degrees and 3 radians is about 171 degrees.
So, we cannot use the assumption that sin x = x since the angle x in these cases are nowhere close to...
Thanks for your answer.
I was assuming that y needs to be substituted by f(x) resulting in an equation involving x, which was what was making it impossible to solve. The reason why I assumed this is because in Calculus we generally take it for granted that y=f(x).
When I look at the left hand side of the equation in above question then I can see that the highest degree of x would be 6 after the denominators are eliminated.
I know that a polynomial of degree n will have n roots, but this one is not a pure polynomial since there is also a trigonometric...
Yes. The loops considered must be independent loops to get independent equations i.e. no considered loop can be formed from a combination of loops being considered.
I would either ignore any one of the first two loops or any one of the last two loops from the list of loops you provided, in order to get 4 independent equations. And the justification would be the first paragraph in post#23 .i.e. "When you add the first two loops you get the same loop as when...
When you add the first two loops you get the same loop as when you add the last two loops. So clearly, there are overlapping loops which are going to yield dependent equations. I'm not sure if that's a valid way of looking at it.
So, I would either ignore any one of the first two loops or any...
I meant the following when I said "fully contained".
If you take the loop 1624351, then it will contain the loops 16351 and 62436. So to get independent equations, either take the loop 1624351 or take the two loops 16351 and 62436 which are fully contained in the loop 1624351.
In your list of...
Would you say that the approach followed by the book's solution (that is mentioned under UPDATE 2 in OP) is a good one for limiting the number of unknowns? In the book's solution, only 3 unknowns are there, whereas in my approach I had 5 unknowns. It would be quicker to solve for 3 unknowns...
Wow, that is great knowledge. I never knew these facts. Thanks for clearing such an important and often missed concept.
So, the key is to apply Kirchhoff's Loop rule to loops such that no considered loop is fully contained in another considered loop, in order to get independent equations.
One...
Yes, you're right. Point 2 is not true. As @kuruman pointed out in post#8 that if 2 smaller loops add up to another bigger loop, then there would be only two independent equations coming from applying Loop rule to the 2 smaller loops and the 1 bigger loop.