Write a closed form expression for the approximation y(nC)

[IntegrateMe]

y(4C) ≈ 7.3 + C + \frac{C}{3^{10C}} + \frac{C}{3^{20C}} + \frac{C}{3^{30C}}

Would:

y(nC) ≈ 7.3 + C\sum_{n = 0}^{\infty}{\frac{1}{3^{10Cn}}}

Be an acceptable answer? If not, what am I doing wrong here?

 

[IntegrateMe]

Any help?

 

[Dick]

The sum part is a geometric series r^n where r=1/(3^(10C).

 

[IntegrateMe]

Oh, ok. So should I be relating this to a\frac{1-r^n}{1-r}
In particular, will my solution look something like:

y(nC) ≈ 7.3 + C\frac{1-{\frac{1}{3^{10C}}}^n}{1-\frac{1}{3^{10C}}}
?

 

[Dick]

Depends. What are you really trying to do here? Why are you writing approximately equal to? (1+r+r^2+...+r^n) is actually (1-r^(n+1))/(1-r).

 

[IntegrateMe]

Well, because y(nC) is an approximation of some function, I assume. I just realized that there should have been a "..." in the initial y(4C).

 

[Dick]

Well, because y(nC) is an approximation of some function, I assume. I just realized that there should have been a "..." in the initial y(4C).

Now why would you do that? Then the first sum is the same as second sum. You can sum a geometric series like that exactly. As you've basically already said. If n is large and r=1/(3^(10C) is small you can approximate it by the infinite sum. I don't see any other role for 'approximately equal' here.

 

[IntegrateMe]

I just assumed that I couldn't use the sigma notation since they specified a "closed form expression." But, then again, I'm not entirely what that means.