Recall that our discussion of the function log(x) was motivated by the interest in having a function f with f'(x) = 1/x. Continuing this line of reasoning, let fo(x) = log(x) and suppose that for integers n ≥ 1, we would like to have functions fn with (fn)' = fn-1. It turns out that fn can be expressed in term of polynomials and logarithms, i.e., no new functions are required. (i) Find an expression for fn. (ii) Show that the polynomial part of fn has degree n. Determine the leading coefficient Cn. Hint: The n-th harmonic number is defined by 1 1 H₂ = 1 + + + n 2 3 + 1 n
Recall that our discussion of the function log(x) was motivated by the interest in having a function f with f'(x) = 1/x. Continuing this line of reasoning, let fo(x) = log(x) and suppose that for integers n ≥ 1, we would like to have functions fn with (fn)' = fn-1. It turns out that fn can be expressed in term of polynomials and logarithms, i.e., no new functions are required. (i) Find an expression for fn. (ii) Show that the polynomial part of fn has degree n. Determine the leading coefficient Cn. Hint: The n-th harmonic number is defined by 1 1 H₂ = 1 + + + n 2 3 + 1 n
Chapter6: Exponential And Logarithmic Functions
Section6.8: Fitting Exponential Models To Data
Problem 60SE: Use the result from the previous exercise to graph the logistic model P(t)=201+4e0.5t along with its...
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![Recall that our discussion of the function log(x) was motivated by the interest in having
a function f with f'(x) = 1/x. Continuing this line of reasoning, let fo(x) = log(x) and
suppose that for integers n ≥ 1, we would like to have functions fn with
(fn)' = fn-1.
It turns out that fn can be expressed in term of polynomials and logarithms, i.e., no new
functions are required.
(i) Find an expression for fn.
(ii) Show that the polynomial part of fn has degree n. Determine the leading coefficient Cn.
Hint: The n-th harmonic number is defined by
Hn
1 1
= = 1 + + +
2 3
+
1
n](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F257d13f8-7f6f-4222-87ae-f2dbc3da1604%2F6e5656fe-be40-4c12-95b9-d9bc2fdca250%2F9jdqs3_processed.jpeg&w=3840&q=75)
Transcribed Image Text:Recall that our discussion of the function log(x) was motivated by the interest in having
a function f with f'(x) = 1/x. Continuing this line of reasoning, let fo(x) = log(x) and
suppose that for integers n ≥ 1, we would like to have functions fn with
(fn)' = fn-1.
It turns out that fn can be expressed in term of polynomials and logarithms, i.e., no new
functions are required.
(i) Find an expression for fn.
(ii) Show that the polynomial part of fn has degree n. Determine the leading coefficient Cn.
Hint: The n-th harmonic number is defined by
Hn
1 1
= = 1 + + +
2 3
+
1
n
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