anonymous
  • anonymous
Why is ∑cos(pi*x)/(x^3/4) convergent by the alternating series test?
Mathematics
  • Stacey Warren - Expert brainly.com
Hey! We 've verified this expert answer for you, click below to unlock the details :)
SOLVED
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.
chestercat
  • chestercat
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
anonymous
  • anonymous
\[\sum_{1}^{\infty}\cos(Pi x)/x ^{3/4}\] I know that cos(pi x ) becomes (-1)^x, but doesn't the function no decrease and therefore not converge?
anonymous
  • anonymous
*not
anonymous
  • anonymous
When 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.

Looking for something else?

Not the answer you are looking for? Search for more explanations.

More answers

anonymous
  • anonymous
But the derivative of 1/x^(3/4) is -3/(4*x^(7/4)). As x approaches infinity doesn't it become zero?
anonymous
  • anonymous
If 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
  • anonymous
Wait, Isnt the derivative zero as it approaches infinity? Or is the derivative always negative because x is restricted by the x^(7/4)?
anonymous
  • anonymous
hmmm, but still. it would be -3/infinity.
anonymous
  • anonymous
Yes as it approaches infinity, the derivative is 0.
anonymous
  • anonymous
So wouldn't it not satisfy the requirement of the test: "An+1 < An" in other words, decreasing?
anonymous
  • anonymous
It doesn't say the derivative decreases, it says the original function decreases.
anonymous
  • anonymous
well, if the slope is negative, the function is decreasing.
anonymous
  • anonymous
Yeah, 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
  • anonymous
My work points to no, but the answer key says yes. So...
anonymous
  • anonymous
Is there another way to test out this requirement? Derivative was the only way I could think of.
anonymous
  • anonymous
So, are you stuck because of the An+1 < An part of the alternating series test?
anonymous
  • anonymous
Yeah
anonymous
  • anonymous
Okay, 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
  • anonymous
Seems 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 non-calc, 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
  • anonymous
Actually a better question would be, how would you test this An+1 < An? Using derivative, plugging in, or something else?
anonymous
  • anonymous
oh, plugging in n+1 for n and comparing would work right?
anonymous
  • anonymous
How 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 x-intercept. 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 x-intercepts. 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 x-axis.
anonymous
  • anonymous
Ok, 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!

Looking for something else?

Not the answer you are looking for? Search for more explanations.