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# Use iteration to guess an explicit formula for the sequence? Topic: Types of parenthesis in math
July 19, 2019 / By Chonsie
Question: e(k) = 4e(k-1) + 5, for all integers k >= 1 e(0) = 2 **All terms in parenthesis are subscripts ## Best Answers: Use iteration to guess an explicit formula for the sequence? Aubrey | 4 days ago
e(k) = 4e(k-1) + 5 = 4(4e(k-2) + 5) + 5 = 4^2 e(k-2) + 25 = 4^2 (4e(k-3) + 5) + 25 = 4^3 e(k-3) + 105 = ... It's obvious that the this will be of the form 4^j e(k-j) + c(j) for some constant c(j). We have the first few values of c(j): c(1) = 5 c(2) = 4*5 + 5 = 25 c(3) = 4*4*5 + 4*5 + 5 = 105 ... From this, it's pretty clear that c(j) = 4^(j-1)*5 + 4^(j-2)*5 + ... + 4^(j-j)*5 This can be evaluated more easily using the geometric series formula. Here, it gives c(j) = 5/3 * (4^j - 1) So, we have e(k) = 4^j e(k-j) + 5/3 * (4^j - 1) Setting j=k and using e(0) = 2 gives the final formula, e(k) = 2 * 4^k + 5/3 * (4^k - 1) = (11 * 4^k - 5) / 3 Of course this can be proven with an easy induction. P.S. Often, subscripts are typed like this: e_k, or e_k-1, or e_{k-1}. The first and third options are valid LaTeX code, which is the typesetting system that's used very heavily in math and related fields.
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We found more questions related to the topic: Types of parenthesis in math Abaigael
Could you help me with a problem similar to this? I can figure out the first few answers but not the explicit formula.
👍 50 | 👎 -2 Originally Answered: What series/sequence is this? 1,2,3,4,5,7,24.6.? (reasking)?
The following polynomial p(x) works: (I have edited these since first posting due to a small error. The actual values don't really matter though, only that the technique works.) p(x) = R(x) + i * i(x) where R(x) ≈ -0.0043663095238095*x^9 + 0.20872845238095*x^8 - 4.261379523809527*x^7 + 48.57609277777789*x^6 - 339.1883991666676*x^5 + 1496.427338611115*x^4 - 4142.960431190484*x^3 + 6878.457840158739*x^2 - 6135.313423809527*x + 2199.058 i(x) ≈ .00044137557870370375*x^9 - 0.020268647073413*x^8 + 0.39863961061508*x^7 - 4.393082552083342*x^6 + 29.77215699131946*x^5 - 128.0045194010416*x^4 + 346.7821787962959*x^3 - 565.647609399801*x^2 + 497.6784332261902*x - 176.56637 These are approximate as π was approximated as well as certain coefficients. [ i(x) has zeros at x = 1,2,3,...,8 ] Simply create two polynomials, one real, one imaginary, and using whatever technique you wish subject to: for x=1,2,3,...,10; R(x) has values: 1, 2, 3, 4, 5, 7, 24.6, -0.1124, 1, -2 for x=1,2,3,...,10; i(x) has values: 0, 0, 0, 0, 0, 0, 0, 0, -16.4, 4π No discontinuities anywhere. Originally Answered: What series/sequence is this? 1,2,3,4,5,7,24.6.? (reasking)?
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