Conquer Club

Yeah, so I'm in AP Calculus BC and we just learned about Taylor and McLaurin Polynomials. I have this problem where I'm supposed to find all values x for which f(x)=e^x=1+x+x^2/2!+x^3/3!, x<0 where the error cannot exceed 0.001.

I have no idea how to do this and my friend refuses to help as he's doing College Writing stuff.

HELP!

That doesn't really make sense to my feeble little mind...

with problems that complicated the answer usually comes out to be 0. Put that down, and no one can blame you if you're wrong anyways.

haha, the back of the book actually says -.3967<x<0, but the point is showing my work. That's the bitch these advanced math classes.

/edit: -.3936<x<0

Just write down 7.

7 is always a good guess.

AndrewLC wrote:Just write down 7.

7 is always a good guess.

WRONG!

42 is the answer to everything.

muy_thaiguy wrote:

AndrewLC wrote:Just write down 7.

7 is always a good guess.

WRONG!

42 is the answer to everything.

Touche

I always do 7 though

I haven't done polynomials for ages! I don't have the slightest clue. Makes me want to take math classes again! What jobs are there for guys that are good at math?

DaGip wrote:I haven't done polynomials for ages! I don't have the slightest clue. Makes me want to take math classes again! What jobs are there for guys that are good at math?

Physicist, Engineering of any sort,... Math teacher... Any job at NASA... That's about it...

And i thought math was my best subject.

Your shit out of luck my friend

to find the zeros of a function, don't you take the derivative and set it equal to zero and solve for x?

i believe the derivative of e^x is e^x, so take e^x = 0 and sove for x?

or, show the derivative of the mclaurin series

1 + x/1! + x^2/2! + x^3/3! + x^4/4!...

take the derivative and it becomes

0 + 1 + x + x^2/2! + x^3/3! + ....

lo and behold the same as when you started! (we already said the d/dx of e^x is e^x)

clear it up for you?

The error R(x) of a Taylor series is less than or equal to the (n+1)th term of the series except the derivative is evaluated at the greatest value possible on the interval in question. Therefore the error of a three term expansion of the e^x taylor series is (e^z)(x^4)/4! < .001, here though z should equal x and as far as I can tell -.4393<x<.3597 but if I remember correctly one often approximated on the side of safety when dealing with error, epsilons, closure operators, etc. in mathematics. Surely, though, there is a better place to gain mathematical insight than conquer club?

Nobuo wrote:The error R(x) of a Taylor series is less than or equal to the (n+1)th term of the series except the derivative is evaluated at the greatest value possible on the interval in question. Therefore the error of a three term expansion of the e^x taylor series is (e^z)(x^4)/4! < .001, here though z should equal x and as far as I can tell -.4393<x<.3597 but if I remember correctly one often approximated on the side of safety when dealing with error, epsilons, closure operators, etc. in mathematics. Surely, though, there is a better place to gain mathematical insight than conquer club?

Haha, finally someone who knew what they were talking about. Luckily my friend finally finished his paper and helped me with the problem.

And no, sadly, this is the only real forum that I ever attend, so Conquer Club was my best choice for advanced math help late at night.

I'm so glad I'm all done with that shit.......