Question

let p be some fixed real number. find the linearization of f(x) = (1 + x)^p at 0

Answer 1

y = (1+x)^p
ln(y) = p*ln(1+x)
Say now Y = ln(y) and X = ln(1+x), then:
Y = p*X
That is a linear function now

Answer 2

is that kind of linearization you want?

Answer 3

ohh i guess you want the linearization by taylor series, right?

Answer 4

\[f(x) \approx \sum_{n=0}^{\infty} \frac{f^n(0) \cdot (x)^n}{n!}\]

Answer 5

that is the Maclaurin series, which is an particular case of taylor series, just apply the formula till the first order

Answer 6

Okay, let's make the linearization of f(x) at x=a:
f(x) = f(a) + f'(a)(x-a), for any "a" in the range of f(x)
f'(a) is the derivative of f(x) avaluated at x=a

Answer 7

yes, the tangent line as in linear approximation

Answer 8

I have L(x) = f(0) + p(1+x)^(p-1) (0-p)

Answer 9

good catch

Answer 10

it should be:
L(x) = f(0) + p(1+0)^(p-1) (x-0)

Answer 11

simplifying:
L(x) = f(0) + px