The question is asking to approximate (1.01)^1/5 to 3 decimal places using power series. I tried thunds of ways to represent this as a power series, but can't seem to find the right one. Can you help and please explain how you got it.
Thanks
THANKS ALL!
Finally got it!
??????
Favorite Answer
x^y = exp(y * ln(x))
and
exp(x) = 1 + x + x²/2! + x³/3! + x^4/4! + ...
with n! = n*(n-1)*(n-2)*...*2*1 (factorials)
and
ln(1+x) = x - x²/2 + x³/3 - x^4/4 + ....
so (1.01)^0.2
= exp(0.2 ln(1.01))
= exp(0.2 (0.01 - 0.01²/2 + 0.01³/3 - ....) )
= exp(0.0019901)
= 1 + 0.0019901 + 0.0019901²/2 + 0.0019901³/6 + ...
= 1.002
Anonymous
Most calculus books include the example of the binomial series:
(1+x)^p = 1 + px + (p(p-1)/2)x^2 + (p(p-1)(p-2)/6)x^3 + ...
Set x=.01 and p=1/5.
SS4
It's asking for the binomial expansion
1.01 = 1 + 0.01
(1+0.01)^(1/5) = (1.01)^(1/5)
The expansion is
(1+x)^n = 1 + nx + [n(n-1)/2!]*x² + [n(n-1)(n-2)/3!]x³ + ... + n!/(n-r)!r! * x^n
statman
Start with (1 + x)^n
The first few terms are
1^n + n(1)^(n - 1)x + [n(n - 1)/2](1)^(n - 2)x^2 + . . .
= 1 + nx + [n(n - 1)/2]x^2 + . . .
Now x = 0.01 and n = 1/5 = 0.2, so the series is
1 + (0.01)(0.2) + [(0.2)(-0.8)/2](0.01)^2 + . . .
= 1 + 0.002 - 0.08(0.0001) + . . .
= 1 + 0.002 - 0.000008 + . . .
= 1.001992
Compare this to the exact value, which is 1.00199204766653...
Since you needed three digits, the first two terms would have sufficed.
The first two terms are 1 + nx, so we have
1 + (0.2)(0.01) = 1 + 0.002 = 1.002, which is good to three digits.