-2
$\begingroup$

Let $T_n$ = $\sqrt{2000^2 - n^2}$

The sum of this series to n terms is $S_n$

Then find $\frac{2}{2000^2}(2000 + 2S_{2000})$ upto $2$ decimal places.

I know in homework questions we do have to write what approach we have tried, but trust me I have tried $\mathbf {EVERYTHING}$ that I could, and can't possibly list them all here. Squaring and summing up $n^2$,the method of differences, a method whose name I forgot in which you multiply and divide by the difference of the two terms and using which you can calculate sum of almost any series except $\mathbf {HP}$(what was it called, pls tell) and on and on and on. Apparently, like Edison I have found 2000 ways how NOT to solve this question,which ironically contains 2000.(I HATE 2000 xD). I asked all my friends. None of them could solve it. So basically this is a question approximately 32 people want to ask.

Thank you.

  • 0
    Are you allowed to use technology to find $S_n$? – 2017-02-19
  • 0
    hmmm something to do with $\pi$? :) – 2017-02-19
  • 0
    @TimThayer Nope no technology. Cant just tell my teacher that wolfram did it, I know not how. Gotta show the full working. – 2017-02-19
  • 0
    @hypergeometric No idea. Are you telling me or asking me. Btw, I fail to see how $\pi$ could possibly come up here?? – 2017-02-19
  • 0
    What level of math are you taking? – 2017-02-19
  • 0
    @TimThayer Well anything till 12th grade would do. Though I am still in 10th, advanced maths is taught to us. – 2017-02-19
  • 2
    Have you learned about approximating series? As far as I know, summing square roots doesn't have a closed-form solution. – 2017-02-19
  • 0
    @TimThayer Well no the series I have handled till date always tend to have an exact answer. However, the question DID say that answer approximately till 2 decimal places, which I thought was due to the fact that there was $2000^2$ in the denominator, so the expression mentioned divided by $10^6$ gotta have some decimals. At least thats what I thought. Approximating series?? Could you please post a solution and explain in detail .Maybe I can understand it. – 2017-02-19

1 Answers 1

3

Note that the Riemann sum for the unit circle gives $$\lim_{n\to\infty}\frac 4{n^2}\sum_{r=1}^n\sqrt{n^2-r^2}=\pi$$ Setting $n=2000$ gives $$\frac 4{2000^2}\sum_{r=1}^{2000}\sqrt{2000^2-r^2}$$ which is the second term of the second expression in the question.

Hence it would appear to be some approximation of $\pi$.

There doesn't appear to be a simple way to compute the summation in the question without using mechanical means.

Wolframalpha returns the first as $3.14058\times 10^6$ and the second as $3.14158$.

  • 0
    "the Riemann sum for the unit circle". Umm what the hell is this??(No offence). I guess this is above 12th grade. But I seriously doubt that something like this would come up in question of series and summations for 10th grade, with difficulty uptil 12th acceptable. Are u SURE that it is the only method?? – 2017-02-19
  • 0
    Very nice observation! – 2017-02-19
  • 0
    @TimThayer - Thanks. Glad you like it :) – 2017-02-19
  • 0
    Also, isn't $2000$ a bit small when n tending to infinity gives us $\pi$. I think that it can't be even **approximated** as $\pi$. If n was like 100000 , wouldn't that have been a reasonable approximation?? – 2017-02-19
  • 0
    @TimThayer Is this what you meant by "approximating series"?? – 2017-02-19
  • 0
    There are techniques to rewrite the square root in a way that allows us to approximate the answer. You could use a Taylor polynomial, for instance. You could search on the topic "numerical analysis" for more information. – 2017-02-19
  • 0
    @Pranav - Draw a circle with radius of $1$. Its area is $\pi\cdot 1^2=\pi$. Take one quadrant of the circle. Slice it into $n$ vertical slices of equal width of $\frac 1n$. The $r$-th slice has height $\sqrt{1-\left(\frac rn\right)^2}$. Multiply these to get the area of the $r$-th slice. Sum from $r=1$ to $n$ to get approx area of one quadrant. Multiply by $4$ to get approx area for whole circle, i.e. an approximation for $\pi$. – 2017-02-19
  • 0
    @Pranav - The solution makes an observation that it *seems* to be *some* approximation of $\pi$, not that it *is* a *good* approximation of $\pi$. Just trying to make sense of the question. – 2017-02-19
  • 0
    @hypergeometric Thanks. I think that this question might have crept in accidentally. Not much series and summation in here. Btw, do you know what that method is called?? – 2017-02-19
  • 0
    @TimThayer Thanks. Do you know what that method is called – 2017-02-19
  • 0
    @hypergeometric's method is called a Riemann sum. Riemann sums are the basis for definite integration in Calculus. – 2017-02-19