The Student Room Group

Product of two finite summations

I was wondering what the product of two finite summations was and I stumbled upon this formula:
3a1a45c272d0924b8f0de37a9f697679.png
(I would have re-written it in LaTex but I can't seem to get the upper and lower bound above and beneath the sigma).

I don't really understand the formula so if someone could show me how it works with an example, it would be much appreciated.
Thank you in advance!
Reply 1
Its so faint! I can't make out most of the notation
Reply 2
Original post by Haesights
Its so faint! I can't make out most of the notation


Sorry! It was fine for me. Here's the link: http://en.wikipedia.org/wiki/Cauchy_product#Finite_summations
Reply 3
Original post by Xilian Empires
I was wondering what the product of two finite summations was and I stumbled upon this formula:
3a1a45c272d0924b8f0de37a9f697679.png
(I would have re-written it in LaTex but I can't seem to get the upper and lower bound above and beneath the sigma).

I don't really understand the formula so if someone could show me how it works with an example, it would be much appreciated.
Thank you in advance!


Yeah... that thing came up at uni... we have to 'Be aware of it' and be able to do it by hand up to 'n' terms for n<10 or so - there isn't really an easy way without a short program on a computer as far as I know
Reply 4
Original post by Xilian Empires

(I would have re-written it in LaTex but I can't seem to get the upper and lower bound above and beneath the sigma).





01\displaystyle \sum_0^1
Reply 5
Original post by natninja
Yeah... that thing came up at uni... we have to 'Be aware of it' and be able to do it by hand up to 'n' terms for n<10 or so - there isn't really an easy way without a short program on a computer as far as I know


But surely you understand the formula? Because I literally don't understand the notation.

Original post by BabyMaths
01\displaystyle \sum_0^1


Thank you, I'll keep that in mind. :smile:
Reply 6
Original post by Xilian Empires
But surely you understand the formula? Because I literally don't understand the notation.


means (a1+a2+a3+...+an)(b1+b2+b3+...+bm) where n=/=m
Reply 7
Original post by natninja
means (a1+a2+a3+...+an)(b1+b2+b3+...+bm) where n=/=m


Wouldnt it be equal to (a0+a1+a2+...+an)(b0+b1+b2+...+bn) (a_0+a_1+a_2+...+a_n)(b_0+b_1+b_2+...+b_n) ?
And what's mm?
Reply 8
Original post by Xilian Empires
Wouldnt it be equal to (a0+a1+a2+...+an)(b0+b1+b2+...+bn) (a_0+a_1+a_2+...+a_n)(b_0+b_1+b_2+...+b_n) ?
And what's mm?


Looking at the website I think it deals specifically with the upper limit being n for both series.

Where are you struggling exactly - have you used the sigma notation before, or is it all totally unfamiliar to you?
Reply 9
Original post by Xilian Empires
Wouldnt it be equal to (a0+a1+a2+...+an)(b0+b1+b2+...+bn) (a_0+a_1+a_2+...+a_n)(b_0+b_1+b_2+...+b_n) ?
And what's mm?



yup, but the series don't have to have the same number of terms - hence n and m numbers of terms
Original post by Xilian Empires
I was wondering what the product of two finite summations was and I stumbled upon this formula:
3a1a45c272d0924b8f0de37a9f697679.png
(I would have re-written it in LaTex but I can't seem to get the upper and lower bound above and beneath the sigma).

I don't really understand the formula so if someone could show me how it works with an example, it would be much appreciated.
Thank you in advance!


can't read it use abx2\sum\limits_a^b x^2
\sum\limits_a^b x^2

Use the latex wikibook. http://en.wikibooks.org/wiki/LaTeX/Advanced_Mathematics
Original post by davros
Looking at the website I think it deals specifically with the upper limit being n for both series.

Where are you struggling exactly - have you used the sigma notation before, or is it all totally unfamiliar to you?

I understand the sigma notation, just not why this formula doesn't work when I apply to something. For example,

A=k=022k A= \sum\limits_{k=0}^{2} 2k
and
B=k=02n+1B= \sum\limits_{k=0}^{2} n+1

We can easily ascertain that AB=36AB=36.
Using the formula...

k=02ni=0kaibkik=0n1(aki=n+12nkbi+bki=n+12nkai) \sum\limits_{k=0}^{2n} \sum\limits_{i=0}^{k} a_i b_{k-i} - \sum\limits_{k=0}^{n-1}(a_k \sum\limits_{i=n+1}^{2n-k} b_i + b_k \sum\limits_{i=n+1}^{2n-k} a_i )

If you plug in the numbers n=2;ak=2k;bk=k+1n=2; a_k=2k; b_k=k+1

k=04i=0k(2i)(ki+1)k=01(2ki=34k(i+1)+(k+1)i=34k(2i))[br] \sum\limits_{k=0}^{4} \sum\limits_{i=0}^{k} (2i)(k-i+1) - \sum\limits_{k=0}^{1}(2k \sum\limits_{i=3}^{4-k} (i+1) + (k+1) \sum\limits_{i=3}^{4-k} (2i) ) [br]

Okay I just triple-checked by adding all the parentheses and it actually works :colondollar: :
https://www.desmos.com/calculator/volag6sdvt
Although if you could give me some ideas on how to prove it that would be great!
Original post by natninja
yup, but the series don't have to have the same number of terms - hence n and m numbers of terms


Sure, that makes perfect sense.

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