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anonymous
 5 years ago
Why is ∑cos(pi*x)/(x^3/4) convergent by the alternating series test?
anonymous
 5 years ago
Why is ∑cos(pi*x)/(x^3/4) convergent by the alternating series test?

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anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0\[\sum_{1}^{\infty}\cos(Pi x)/x ^{3/4}\] I know that cos(pi x ) becomes (1)^x, but doesn't the function no decrease and therefore not converge?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0When determining the decreasing function, you have to disregard the component that makes the series alternating. So, you want to look at 1/x^(3/4) only. The limit as x approaches infinity of this function is 0 and it is a decreasing function.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0But the derivative of 1/x^(3/4) is 3/(4*x^(7/4)). As x approaches infinity doesn't it become zero?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0If we have \[\sum_{1}^{\infty}\cos \pi x * a _{n}\] Then \[\lim_{n \rightarrow \infty}a _{n}=0\] and An is a decreasing function. We agree that the limit is zero as n approaches infinity right? And the derivative is always negative, so it is a decreasing function.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Wait, Isnt the derivative zero as it approaches infinity? Or is the derivative always negative because x is restricted by the x^(7/4)?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0hmmm, but still. it would be 3/infinity.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Yes as it approaches infinity, the derivative is 0.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0So wouldn't it not satisfy the requirement of the test: "An+1 < An" in other words, decreasing?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0It doesn't say the derivative decreases, it says the original function decreases.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0well, if the slope is negative, the function is decreasing.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Yeah, I think I may not be on the same page as you... haha. Do you agree that it converges by the alternating series test?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0My work points to no, but the answer key says yes. So...

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Is there another way to test out this requirement? Derivative was the only way I could think of.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0So, are you stuck because of the An+1 < An part of the alternating series test?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Okay, then. An = 1/n^(3/4), so we look at the function f(x) = 1/x^(3/4). If we take the derivative of this function, we get 3/(4*x^(7/4)). This derivative is always negative. Since f'(x) is always negative, we can conclude that the function f(x) is a decreasing function (or that the slope of f(x) is always negative). Being that the slope of f(x) is always negative and f(x) is always decreasing, An is a decreasing sequence. Therefore, An+1 < An for all n. If you want to check without using the derivative. Choose values for x (i.e. 1, 2, 3, 4, 5) and put them into f(x) = 1/x^(3/4). You'll find that the values will get lower and lower (1, 0.59, 0.44, .35, .299), so you could conclude that the function is decreasing and the sequence An is decreasing, and An+1 < An.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Seems like the problem I'm having is proving that the derivative is negative. I just graphed it and sure enough when x>0, it's negative. But the limit does approach 0, which is throwing me off. I forgot the fact that although the limit approaches zero, the y values could still be negative. Since most of these questions are noncalc, Is there a way to look at the derivative and just know that it's negative like you did? Is that just a matter of knowing what it looks like?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Actually a better question would be, how would you test this An+1 < An? Using derivative, plugging in, or something else?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0oh, plugging in n+1 for n and comparing would work right?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0How would you test An+1 < An? Pick a value for n, say n = 1. Find A1 by plugging 1 into the function f(x). n+1 = 2, so find A2 by plugging 2 into the function f(x). You should find that f(1) is greater than f(2). Now, just plugging in numbers doesn't effectively prove the derivative is negative. If you look at the derivative, f'(x) = 3/(4*x^(7/4)), there's no way you can get a positive value for f'(x) because there are some negative values (you can find f'(1) and f'(2) for example) and there is no xintercept. I suppose that's how I'd prove the derivative is negative: That f'(x) is continuous, has some verifiable negative values, and has no xintercepts. When talking about the limit approaching infinity, don't worry about that when talking about whether a function is totally negative, because the function's not actually going to hit the xaxis.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Ok, that makes sense. Although it's a bit long, it does get rid of the derivative method issue. Well, Thanks for all the help, it really helped clear stuff up. I really appreciate it!
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