Error estimate for Taylor polynomials

[emc92]

Use the error estimate for Taylor polynomials to find an n such that

| e - (1 + (1/1!) + (1/2!) + (1/3!) + ... + (1/n!) | < 0.000005

all i have right now is the individual components...

f(x) = e^x
T_n (x) = 1/ (n-1)!

k/(n-1)! |x-a|ⁿ⁺¹ = 0.000005
a = 0
x = 1

I don't know where to go from here

 

[LCKurtz]

Use the error estimate for Taylor polynomials to find an n such that

| e - (1 + (1/1!) + (1/2!) + (1/3!) + ... + (1/n!) | < 0.000005




all i have right now is the individual components...

f(x) = e^x
T_n (x) = 1/ (n-1)!

k/(n-1)! |x-a|ⁿ⁺¹ = 0.000005
a = 0
x = 1

I don't know where to go from here

The max error if your last term is$$
\frac {f^{(n)}(a)(x-a)^n}{n!}$$ is$$
\left | \frac {f^{(n+1)}(c)(x-a)^{n+1}}{(n+1)!}\right |$$As you have already observed $a=0$ and $x = 1$. How big can that derivative term be for $0\le c\le 1$? Once you have that, figure out how large $n$ needs to be to make it small enough.

 

[emc92]

how do i figure out c?

 

[emc92]

i just looked over my notes again. we started this problem in class, and our professor told us to pick a value for k above the value of e.

so if k is 3,
3/(n+1)! = 0.000005

so now i have to find n?

 

[LCKurtz]

The max error if your last term is$$
\frac {f^{(n)}(a)(x-a)^n}{n!}$$ is$$
\left | \frac {f^{(n+1)}(c)(x-a)^{n+1}}{(n+1)!}\right |$$As you have already observed $a=0$ and $x = 1$. How big can that derivative term be for $0\le c\le 1$? Once you have that, figure out how large $n$ needs to be to make it small enough.

how do i figure out c?

i just looked over my notes again. we started this problem in class, and our professor told us to pick a value for k above the value of e.

so if k is 3,
3/(n+1)! = 0.000005

so now i have to find n?

You don't have to "figure out c". You have the nth derivative of your function evaluated at c, and c is in the interval [0,1]. You don't know the exact value of c so you ask yourself, "how big can $f^{(n)}(c)= e^c$ be for c in [0,1]. Do you understand why your instructor says to pick k > e?

To answer your last question, you don't want 3/(n+1)! = 0.000005. You would be very lucky to find an integer n giving equailty. You want 3/(n+1)! < 0.000005. Writing it a different way, you want$$
(n+1)! > \frac 3 {.000005}$$Factorials grow very quickly. Is shouldn't be difficult to check by hand how big n needs to be.