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VCE Stuff => VCE Mathematics => VCE Mathematics/Science/Technology => VCE Subjects + Help => VCE Mathematical Methods CAS => Topic started by: vcesuccess on November 23, 2009, 08:51:02 pm

Title: This should be correct?
Post by: vcesuccess on November 23, 2009, 08:51:02 pm: Write the summation notation of
$4{x}^2+2{x}+1$

I wrote
$\sum_{i=1}^3 \frac{4}{i}\cdot{x}^{3-i}$
Title: Re: This should be correct?
Post by: TrueTears on November 23, 2009, 09:05:22 pm: That wouldn't work since $\frac{4}{3}\left(x^{3-3}\right) = \frac{4}{3}$

I'd do this:

$\left[\sum_{i=1}^2 \frac{4}{i}x^{3-i}\right] + 1$
Title: Re: This should be correct?
Post by: vcesuccess on November 23, 2009, 09:20:21 pm: Yours is correct?
Title: Re: This should be correct?
Post by: TrueTears on November 23, 2009, 10:03:12 pm: Well when you expand my one you get $4x^2+2x+1$ .
Title: Re: This should be correct?
Post by: QuantumJG on November 23, 2009, 11:52:59 pm: Quote from: TrueTears on November 23, 2009, 09:05:22 pm
That wouldn't work since $\frac{4}{3}\left(x^{3-3}\right) = \frac{4}{3}$

I'd do this:

$\left[\sum_{i=1}^2 \frac{4}{i}x^{3-i}\right] + 1$

Yep TT is right! It would be great if you could find the whole polynomial from the summation operator. If vcesuccess had of subtracted 1/3 from his summation it would yield the quadratic.

I can't wait until I do real analysis and look at:

convergence:

$[t[\sum_{I=0}^\infty \ f(I) = a$

divergence:

$[\sum_{I=0}^\infty \ f(I) = \infty$
Title: Re: This should be correct?
Post by: TrueTears on November 23, 2009, 11:53:46 pm: Quote from: QuantumJG on November 23, 2009, 11:52:59 pm
Quote from: TrueTears on November 23, 2009, 09:05:22 pm
That wouldn't work since $\frac{4}{3}\left(x^{3-3}\right) = \frac{4}{3}$

I'd do this:

$\left[\sum_{i=1}^2 \frac{4}{i}x^{3-i}\right] + 1$

Yep TT is right! It would be great if you could find the whole polynomial from the summation operator. If vcesuccess had of subtracted 1/3 from his summation it would yield the quadratic.

I can't wait until I do real analysis and look at:

convergence:

$[t[\sum_{I=0}^\infty \ f(I) = a$

divergence:

$[\sum_{I=0}^\infty \ f(I) = \infty$

Haha I've been doing quite a bit of convergence and divergence, they're really fun!
Title: Re: This should be correct?
Post by: enwiabe on November 24, 2009, 05:59:13 pm: Quote from: vcesuccess on November 23, 2009, 08:51:02 pm
Write the summation notation of
$4{x}^2+2{x}+1$

I wrote
$\sum_{i=1}^3 \frac{4}{i}\cdot{x}^{3-i}$

Oh so close!

Should be:

$\sum_{i=0}^2 \frac{4}{2^{i}}\cdot{x}^{2-i}$

The way I approached it was with the insight that the geometric series of co-efficients was 1, 2, 4 (I.E. multiplying/dividing by 2 each time), and that these would increase by 1 power each time, which was the same increase as the powers of the polynomials.

EDIT: *whacks head* or more simply:

$\sum_{i=0}^2 \ 2^{i}\cdot{x}^{i}$
Title: Re: This should be correct?
Post by: Collin Li on November 24, 2009, 06:04:05 pm: Are there an infinite number of ways to write a summation notation for that, or something? Or, is there always a way to write something in summation notation (without being cheap and just going i=1 to 1, and writing the expression)?
Title: Re: This should be correct?
Post by: kyzoo on November 25, 2009, 03:41:45 pm: Is this even part of Methods?
Title: Re: This should be correct?
Post by: humph on November 25, 2009, 11:53:25 pm: Quote from: Collin Li on November 24, 2009, 06:04:05 pm
Are there an infinite number of ways to write a summation notation for that, or something? Or, is there always a way to write something in summation notation (without being cheap and just going i=1 to 1, and writing the expression)?

Well this
Quote from: TrueTears on November 23, 2009, 09:05:22 pm
$\left[\sum_{i=1}^2 \frac{4}{i}x^{3-i}\right] + 1$
isn't really summation notation, because there's an extra term (a constant at the end). And this
Quote from: enwiabe on November 24, 2009, 05:59:13 pm
$\sum_{i=0}^2 \frac{4}{2^{i}}\cdot{x}^{2-i}$
is just this
Quote from: enwiabe on November 24, 2009, 05:59:13 pm
$\sum_{i=0}^2 \ 2^{i}\cdot{x}^{i}$
after reversing the direction of summation.
For a general series, you often can't say more than just $\sum^{\infty}_{i=0}{a_i x^i}$ for some sequence $(a_i)$ , as there might not be a suitably "nice" pattern that allows us to give some closed form expression for $(a_i)$ . For example, if $(a_i)$ is just some random sequence of real numbers, then with probability 1 there's going to be no closed-form way of writing it (as a combination of geometric series, arithmetic series, in terms of some elementary function, etc.).