How to change variables elegantly?

How do you solve this Discrete Math question? Change of Variables?

  • Hello All, I have a change of variables question on products, similar to that of summation. The given variable is t= k+1, which can be arranged to k= t-1. Now, the original problem is Greek Capital Pi (k/(k^2+4) starting at k=1 till the n term. So here are the first four terms. 1/5, 2/8, 3/13, 4/20. From this point, I need help plugging this in to get an equivalent product. Any help would be greatly appreciated! Thank you.

  • Answer:

    If k goes from 1 to n, then t = k+1 goes from 1+1 to 1+n, or from 2 to n+1. That tells you what the bounds of the product will be in terms of t. To get the terms of the product, you just replace every occurrence of k in the product with the corresponding expression of k in terms of t. In this case, k = t - 1, so the product, from k = 1 to n, of k/(k^2 + 4) = the product, from t = 2 to n+1, of (t-1)/((t-1)^2 + 4). Of course, if you liked you could expand out (t - 1)^2 + 4 = t^2 - 2t + 1 + 4 = t^2 - 2t + 5 and write it as the product from t = 2 to n + 1 of (t - 1)/(t^2 - 2t + 5). [This doesn't do much for you, as the polynomial t^2 - 2t + 5 doesn't factor in any nice way, so the fraction doesn't simplify much; but it's certainly another valid way of writing the product.]

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If k goes from 1 to n, then t = k+1 goes from 1+1 to 1+n, or from 2 to n+1. That tells you what the bounds of the product will be in terms of t. To get the terms of the product, you just replace every occurrence of k in the product with the corresponding expression of k in terms of t. In this case, k = t - 1, so the product, from k = 1 to n, of k/(k^2 + 4) = the product, from t = 2 to n+1, of (t-1)/((t-1)^2 + 4). Of course, if you liked you could expand out (t - 1)^2 + 4 = t^2 - 2t + 1 + 4 = t^2 - 2t + 5 and write it as the product from t = 2 to n + 1 of (t - 1)/(t^2 - 2t + 5). [This doesn't do much for you, as the polynomial t^2 - 2t + 5 doesn't factor in any nice way, so the fraction doesn't simplify much; but it's certainly another valid way of writing the product.]

mcbengt

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