Friends please if you can help me with this exercise, this came into my calculus test 1: parameters found for "w" and "n" the equation cos (wx) - x / n has a real solution and calculate how many second solutions has exactly ..... please still do not know how to solve it thanks
Stacey Warren - Expert brainly.com
Hey! We 've verified this expert answer for you, click below to unlock the details :)
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga.
Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus.
Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
The directions are somewhat unclear. Are you quoting word for word?
Is this the equation :
\( \Large \cos (wx) - \frac xn = 0 \)
Not the answer you are looking for? Search for more explanations.
search parameters of "w" and "n" for the equation has a real solution
but also sought to find the exact number of solutions of the equation
That's harder :)
We can try using calculus
That confused me
This exercise was in my calculus test 1
this problem is within the subject of calculus 1
we can see that as n gets large
\(\Large \cos (wx) - \frac xn \to \cos(wx) \)
We could modify the equation
\Large n\cdot \cos (wx) -x = 0
The derivative is
\( \Large -n \sin(wx)w -1 = 0\)
However I don't see how this helps
Again just to be clear you want to know exactly how many real solutions are there to the equation:
$$\Large \cos (wx) - \frac xn = 0$$
Does n have to be an integer? Clearly n=0 won't work.
I don't see anything that pops out.
They call these things transcendental equations, don't they? Graphing is the way to go about these.
yes . not an algebraic equation.
i dont see any other way than by checking for sign change, using intermediate value theorem . but this does not give you a nice result
theorem of Bolzano or intermediate value theorem that thought but how to use it with the values of "w" and "n"
but forget to mention
\[w >0,n \ge1\]
we can assume n are integers.
I think it has infinitely many solutions. As n goes larger, the continuity of the cosine (regarding W values) will ensure the existence of number k= cos (c) for some c in R such that k between (-1,1). Now we can vary n in such a way that x/n is between (-1,1) for any value of x (since n is independent).
That is to say, for any x in the domain, there exist some n such that their ratio is between (-1,1). on the other hand, we can change w as we want such that the equality still holds . in short, infinitely many solutions
That's just what I think