I liked this simple calculus exercise
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I liked this simple calculus exercise
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Re: I liked this simple calculus exercise
#2This reminds me of an exercise I'll never forget from my Math Methods course: finding the derivative of arcsin(x).
It seems almost impossible because, just looking at it, there seems to be nothing you can do to simplify it. Then, out of sheer nothing-else-to-do-ism, you take the sin() of it and realize sin(arcsin(x)) = x. Take the derivative of both sides, apply chain rule and draw a right triangle and you have the answer.
Like the words the author uses for the integral, it's all valuable technique.
Re: I liked this simple calculus exercise
#3This reminds me of an exercise I'll never forget from my Math Methods course: finding the derivative of arcsin(x). It seems almost impossible because, just looking at it, there seems to be nothing you can do to simplify it. Then, out of sheer nothing-else-to-do-ism, you take the sin() of it and realize sin(arcsin(x)) = x. Take the derivative of both sides, apply chain rule and draw a right triangle and you have the a…
Alternatively, x = sin(theta) and arcsin(sin(theta)) = theta, depending on the limits. It’s a cool technique, thanks for the reminder!
Re: I liked this simple calculus exercise
#4This reminds me of an exercise I'll never forget from my Math Methods course: finding the derivative of arcsin(x). It seems almost impossible because, just looking at it, there seems to be nothing you can do to simplify it. Then, out of sheer nothing-else-to-do-ism, you take the sin() of it and realize sin(arcsin(x)) = x. Take the derivative of both sides, apply chain rule and draw a right triangle and you have the a…
Dude. That's just a direct application of the inverse function theorem.
Even geometrically you can see that swapping the axis (x and y) gives you the desired result.
Re: I liked this simple calculus exercise
#5It's easy to fall into the trap of relying on rote memorization of integration rules, but problems like (⋆) force students to truly understand the concepts behind the math.
Re: I liked this simple calculus exercise
#6Current Calc 2 student here. I would be braindead approaching this problem honestly, I don't think I'd even know how to begin; I'm hoping that's normal.
Why would the exponent be equal to x/2 - floor(x/2) be equal to x/2 on the interval [0, 2)? And how does the graph of x/2 - floor(x/2) imply anything about the behavior of e^(x/2 - floor(x/2))? I'm hoping I just haven't learned enough yet?
Re: I liked this simple calculus exercise
#7Current Calc 2 student here. I would be braindead approaching this problem honestly, I don't think I'd even know how to begin; I'm hoping that's normal. Why would the exponent be equal to x/2 - floor(x/2) be equal to x/2 on the interval [0, 2)? And how does the graph of x/2 - floor(x/2) imply anything about the behavior of e^(x/2 - floor(x/2))? I'm hoping I just haven't learned enough yet?
> Why would the exponent be equal to x/2 - floor(x/2) be equal to x/2 on the interval [0, 2)?
floor(x/2) = 0 on the interval [0, 2), so the expression reduces to x/2.
> And how does the graph of x/2 - floor(x/2) imply anything about the behavior of e^(x/2 - floor(x/2))?
If y = x/2 - floor(x/2) is periodic, then e^y = e^(x/2 - floor(x/2)) must be periodic as well, with the same period.
Re: I liked this simple calculus exercise
#8Current Calc 2 student here. I would be braindead approaching this problem honestly, I don't think I'd even know how to begin; I'm hoping that's normal. Why would the exponent be equal to x/2 - floor(x/2) be equal to x/2 on the interval [0, 2)? And how does the graph of x/2 - floor(x/2) imply anything about the behavior of e^(x/2 - floor(x/2))? I'm hoping I just haven't learned enough yet?
> Why would the exponent be equal to x/2 - floor(x/2) be equal to x/2 on the interval [0, 2)? floor(x/2) = 0 on the interval [0, 2), so the expression reduces to x/2. > And how does the graph of x/2 - floor(x/2) imply anything about the behavior of e^(x/2 - floor(x/2))? If y = x/2 - floor(x/2) is periodic, then e^y = e^(x/2 - floor(x/2)) must be periodic as well, with the same period.
Thank you!
Re: I liked this simple calculus exercise
#9Current Calc 2 student here. I would be braindead approaching this problem honestly, I don't think I'd even know how to begin; I'm hoping that's normal. Why would the exponent be equal to x/2 - floor(x/2) be equal to x/2 on the interval [0, 2)? And how does the graph of x/2 - floor(x/2) imply anything about the behavior of e^(x/2 - floor(x/2))? I'm hoping I just haven't learned enough yet?
I think this kind of problem is less about maths and more about how one might approach an unfamiliar problem. Not knowing where to begin is normal. What you're looking to do is to build up the intuition for how you can break down the problem into smaller pieces so that you can investigate its properties.
x/2 - floor(x/2) is the natural place to start because it's the smallest independent piece of the equation. Take a couple of minutes to plot this on a graph for a small range of values, like 0 With this, you can calculate and sketch out e^(above result) on a graph. Finally, knowing the principle that a definite integral calculates the area under the curve, you should be able to use your sketch to reason out how to calculate the entire original integral.
Hopefully you can see how solving this kind of problem isn't about knowing anything about this particular problem, but simply investigating it without any prior expectations, which is why the author thinks this is an interesting exercise for students.
Re: I liked this simple calculus exercise
#10This reminds me of an exercise I'll never forget from my Math Methods course: finding the derivative of arcsin(x). It seems almost impossible because, just looking at it, there seems to be nothing you can do to simplify it. Then, out of sheer nothing-else-to-do-ism, you take the sin() of it and realize sin(arcsin(x)) = x. Take the derivative of both sides, apply chain rule and draw a right triangle and you have the a…
Dude. That's just a direct application of the inverse function theorem. Even geometrically you can see that swapping the axis (x and y) gives you the desired result.
GP just happened to be one of the lucky 10,000 on that day in math class. https://xkcd.com/1053/