Taylor Series

[merry kestrel]

Does anyone know if this is correct?

[thorn night]

@merry kestrel The last part, if you’re taking a derivative then you don’t put in an extra constant

That’s only done for indefinite integrals to account for the zeroth order term that gets eliminated by the power rule

[merry kestrel]

Ok, thank you, the important thing is whether the result of the series is good

[thorn night]

I’ll have to work it out on paper and get back to you. Can you explain to me how you applied the power rule to get the first derivative? I can’t read your work very well

[merry kestrel]

is thishttps://www.youtube.com/watch?v=Jmn8gx3eziI&t=188s&ab_channel=MateFacil

[thorn night]

Here's my work for the first derivative. I am not sure that video you looked at works exactly as expected; I just used the Fundamental Theorem of Calculus (specifically the part that says that for a function G(x) which outputs the integral of a function from 0 to x of a function f(t) has the derivative G'(x) = f(x).)

@merry kestrel

[merry kestrel]

If so, use the fundamental theorem of calculus

you have a error here

you don't need to grab x as u

[thorn night]

You do, because you have two functions of x so you have to use the product rule

that u is just code for u(x)

[merry kestrel]

Oh I think I made a mistake

[thorn night]

It's definitely better than doing it on a test

[merry kestrel]

ok I think so, now this is the result

use product rule and ftc

[thorn night]

the second line, second term on the RHS should have a coefficient of x by the product rule

[merry kestrel]

oh forget the x you are multiplying, ok so this answer is definitely fine

[merry kestrel]

this is correct

[thorn night]

hmm, alright then

[merry kestrel]

The question I have is, what function am I trying to approximate, the F(x)?

[thorn night]

Yes, which is why you are taking its derivatives and evaluating them at x = 0.

The rest of your solution is fine then

[merry kestrel]

But I don't think this is a good approximation xdddd

[thorn night]

They're usually not great approximations when the order is low. Your g(x) is wrong as well, should be x(1-e^3x)

[merry kestrel]

oh

I still don't think it's a good approximation, but it must be as you say, because of the degree.

Thanks for the help bro

[thorn night]

No problem. One last thing - your taylor series coefficient for the x^1 term should not be there

[merry kestrel]

where?

[thorn night]

the term in the taylor series, -x is actually x * 0 = 0. Evaluate the 1st derivative again at x = 0

[merry kestrel]

oh right, it's 0

so the answer is -3x^2

yesss

thanks