Comparison Test Problem & Estimitaing it's error

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SUMMARY

The discussion focuses on estimating the error in approximating the sum of the series defined by the function \(\frac{(\sin n)^2}{n^3}\) using the first 10 terms. Participants clarify that instead of integrating, one should consider the sum of the series and bound the error using the inequality \(\frac{(\sin k)^2}{k^3} \leq \frac{1}{k^3}\). The use of the non-elementary function Ci(x) is mentioned but deemed unnecessary for this problem. The key takeaway is to focus on bounding the error rather than attempting to compute an integral.

PREREQUISITES
  • Understanding of infinite series and convergence
  • Familiarity with the function \(\frac{(\sin n)^2}{n^3}\)
  • Basic knowledge of error estimation techniques in series
  • Awareness of the non-elementary function Ci(x) and its implications
NEXT STEPS
  • Learn about bounding errors in series using inequalities
  • Study the properties and applications of the Ci(x) function
  • Explore techniques for estimating the sum of series
  • Investigate convergence tests for infinite series
USEFUL FOR

Students studying calculus, mathematicians interested in series convergence, and anyone involved in numerical analysis or error estimation techniques.

JRangel42
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Homework Statement



The question in the book is: "Use the sum of the first 10 terms to approximate the sum of the series. Estimate the error.

My problem is estimating the error I'm looking for. I just need help with finding the integral.

Homework Equations



((sin n)^2)/(n^3)

The Attempt at a Solution



∞∫n ((sin n)^2)/(n^3)

u = ((sin n)^2) du = 2 ((sin n)^2)cos n dv = n^3 dx v = 1/4 x^4

This is as far as I got.
 
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JRangel42 said:

Homework Statement



The question in the book is: "Use the sum of the first 10 terms to approximate the sum of the series. Estimate the error.

My problem is estimating the error I'm looking for. I just need help with finding the integral.

Homework Equations



((sin n)^2)/(n^3)

The Attempt at a Solution



∞∫n ((sin n)^2)/(n^3)

u = ((sin n)^2) du = 2 ((sin n)^2)cos n dv = n^3 dx v = 1/4 x^4

This is as far as I got.

What integral? Aren't you just given an infinite series, and are asked to estimate the error when you stop at 10 terms? Anyway, if you did have to integrate the function you would need to use the non-elementary function Ci(x). Maple 11 gets:
f:=sin(x)^2/x^3;
> Jt:=Int(f,x=1..t);
\int_1^\infty \frac{\sin^2(x)}{x^3} \, dx
> value(Jt) assuming t>1;
<br /> \frac{1}{4} \frac{t-t^2 \cos(2)+2t^2 \sin(2)-2\mbox{Ci}(2) t^2 - 2 + 2\cos^2(t)<br /> - 4t \cos(t) \sin(t) + 4\mbox{Ci}(2t) t^2}{t^2}

RGV
 
Last edited:
Oh. It seems that I have been doing this the wrong way since I haven't been introduced to Ci before. I'll make sure to double check my steps this time.
 
JRangel42 said:
Oh. It seems that I have been doing this the wrong way since I haven't been introduced to Ci before. I'll make sure to double check my steps this time.

Your error is the sum of (sin(k))^2/k^3 over all of the terms you didn't include. Don't sweat the Ci function. All you need is an estimate to bound the error. (sin(k))^2/k^3<=1/k^3. Estimate that instead.
 
JRangel42 said:
Oh. It seems that I have been doing this the wrong way since I haven't been introduced to Ci before. I'll make sure to double check my steps this time.

You still have not answered my question: what integral? Are you not just dealing with a sum? (The Ci stuff was just to show you that the integral is not 'doable', so you should not waste time on it!)

RGV
 
Alright guys. Thanks for the tip. I think get it now.
 
Question: A clock's minute hand has length 4 and its hour hand has length 3. What is the distance between the tips at the moment when it is increasing most rapidly?(Putnam Exam Question) Answer: Making assumption that both the hands moves at constant angular velocities, the answer is ## \sqrt{7} .## But don't you think this assumption is somewhat doubtful and wrong?

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