- #1
raxAdaam
- 32
- 0
Hi there,
I was hammering out the coefficients for the Taylor Series expansion of f(x) = \frac{1}{\sqrt{1-x^2}}, which proved to be quite unsatisfying, so decide to have a look around online for alt. approaches.
What I found (in addition to the method that uses the binomial theorem) was an old post here claiming that a trig sub could be used (here, post #5). However, the sub. - at least as written there, definitely doesn't line up:
x = sin(u), which yields f(x) = sec(u). However the post seems to mix up the differentials, claiming: \frac{df}{dx} = \frac{df}{du}\cdot\frac{dx}{du}, which is - unless I'm entirely missing something - not correct, right? We should actually have:
\frac{df}{dx} = \frac{df}{du}\cdot\frac{du}{dx} = sec(u)tan(u)\cdot \frac{1}{\sqrt{1-x^2}} = sec(u)tan(u)\cdot\frac{1}{cos(u)} ...
I'm hoping that I've missed something here, because the method would be stellar and super insightful if it was valid, but it seems the differentials are not handled properly - can anyone confirm this?
Cheers,
Rax
I was hammering out the coefficients for the Taylor Series expansion of f(x) = \frac{1}{\sqrt{1-x^2}}, which proved to be quite unsatisfying, so decide to have a look around online for alt. approaches.
What I found (in addition to the method that uses the binomial theorem) was an old post here claiming that a trig sub could be used (here, post #5). However, the sub. - at least as written there, definitely doesn't line up:
x = sin(u), which yields f(x) = sec(u). However the post seems to mix up the differentials, claiming: \frac{df}{dx} = \frac{df}{du}\cdot\frac{dx}{du}, which is - unless I'm entirely missing something - not correct, right? We should actually have:
\frac{df}{dx} = \frac{df}{du}\cdot\frac{du}{dx} = sec(u)tan(u)\cdot \frac{1}{\sqrt{1-x^2}} = sec(u)tan(u)\cdot\frac{1}{cos(u)} ...
I'm hoping that I've missed something here, because the method would be stellar and super insightful if it was valid, but it seems the differentials are not handled properly - can anyone confirm this?
Cheers,
Rax