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Taylor Expansions
Can anyone help me with these Taylor expansions:
1. Using an expansion around 0, and replacing h with x, construct a fourth order Taylor expansion of
i) f (x) = e^(cx)
2. Economists sometimes measure inflation as ln(pt/pt−1), where pt is an index of retail prices. By writing pt = pt − p(t−1) + p(t−1), and using the Taylor Series expansion for ln(1 + x) show that
ln(pt/pt−1) ≈ (pt − p(t−1)) / (p(t-1))
How good is the approximation if a)
pt/p(t−1) = 1.01, or b) pt/pt−1 = 1.10?
Any ideas, especially the second one, cheers,
AK
1. Using an expansion around 0, and replacing h with x, construct a fourth order Taylor expansion of
i) f (x) = e^(cx)
2. Economists sometimes measure inflation as ln(pt/pt−1), where pt is an index of retail prices. By writing pt = pt − p(t−1) + p(t−1), and using the Taylor Series expansion for ln(1 + x) show that
ln(pt/pt−1) ≈ (pt − p(t−1)) / (p(t-1))
How good is the approximation if a)
pt/p(t−1) = 1.01, or b) pt/pt−1 = 1.10?
Any ideas, especially the second one, cheers,
AK
(1)
Replace h by cx in
e^h = 1 + h + (1/2)h^2 + (1/6)h^3 + (1/24)h^4 + ...
(2)
For h close to 0, ln(1 + h) ~ h. The closer h is to 0, the better the approximation.
ln(p(t)/p(t - 1))
= ln[(p(t) - p(t - 1) + p(t - 1))/p(t - 1)]
= ln[1 + (p(t) - p(t - 1))/p(t - 1)]
~ (p(t) - p(t - 1))/p(t - 1)
The closer (p(t) - p(t - 1))/p(t - 1) is to 0, the better the approximation.
Ie, the closer p(t)/p(t - 1) is to 1, the better the approximation.
Check on your calculator:
ln(1.01) = 0.00995 -------> 0.01 is a good approximation (0.5% error)
ln(1.1) = 0.0953 -------> 0.1 is a less good approximation (5% error)
Replace h by cx in
e^h = 1 + h + (1/2)h^2 + (1/6)h^3 + (1/24)h^4 + ...
(2)
For h close to 0, ln(1 + h) ~ h. The closer h is to 0, the better the approximation.
ln(p(t)/p(t - 1))
= ln[(p(t) - p(t - 1) + p(t - 1))/p(t - 1)]
= ln[1 + (p(t) - p(t - 1))/p(t - 1)]
~ (p(t) - p(t - 1))/p(t - 1)
The closer (p(t) - p(t - 1))/p(t - 1) is to 0, the better the approximation.
Ie, the closer p(t)/p(t - 1) is to 1, the better the approximation.
Check on your calculator:
ln(1.01) = 0.00995 -------> 0.01 is a good approximation (0.5% error)
ln(1.1) = 0.0953 -------> 0.1 is a less good approximation (5% error)
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