Lagrange Error Bound
It’s also called the
Lagrange Error Theorem
, orTaylor's Remainder Theorem
.
To approximate a function more precisely, we’d like to express the function as a sum of a Taylor Polynomial & a Remainder.
(▲ For T
is the Taylor polynomial with n terms, and R
is the Remainder after n terms.)
▲Jump back to review the note on Error estimation Theorem.
►Jump over to have practice at Khan academy: Lagrange Error Bound.
The tricky part of that expression is to “preset” the accuracy of the Error
, aka. the Remainder
.
For bounding the Error
, out strategy is to apply the Lagrange Error Bound
theorem.
Simply saying, the theorem is:
- If a function’s ALL DERIVATIVES are bounded by a number over the interval
(C, x)
:
(▲ for C
is the centre of approximation)
- where the max value of all derivatives of the function is:
(▲ for z
is any value between C
and x
makes the derivative to the max)
(▲ and note that: the input has to be n+1
)
- then the function’s Remainder MUST satisfy this theorem:
Example
Solve:
- This problem is to approximate a function with Taylor Polynomial.
- To do so, we’re to set use the
Error estimation
method, which first to set up the equation:
- In this case, it is:
- Since it’s only asking for the
error bound
, so we only focus on the ErrorRn
. - We want to apply the
Lagrange Error Bound Theorem
, and bound it to0.001
:
- For those unknowns variables in the theorem, we know that:
- The approximation is centred at
1.5π
, soC = 1.5π
. - The input of function is
1.3π
, sox = 1.3π
. - For The
M
value, because all the derivatives of the functioncos(x)
, are bounded to 1 even without an interval , so let's say the max valueM = 1
. - Therefore, the formula of this theorem becomes:
- Unfortunately, at this moment we don’t have easier method to solve for
n
except trying some numbers in:
- We could see that, with the degree gets larger and larger, the
Error
becomes smaller and smaller. - Only until
n=4
, which means the4th derivative
, theError
is less than0.001
. - So the answer is
4th derivative
.
Example
Solve:
- Same with the problem above, we want to apply the
Lagrange Error Bound Theorem
, and bound it to0.001
:
- For those unknowns variables in the theorem, we know that:
- The approximation is centred at
0
because it's told as aMaclaurin Series
, soC = 0
. - The input of function is
-0.95
, sox = -0.95
. - The interval is (C, x) or (x, C), which is
(-0.95, 0)
in this case. - For The
M
value, since all the derivatives ofeˣ
is justeˣ
, andeˣ
is unbounded at all, so we're to examine the Max value over the interval(-0.95, 0)
- With the help from
Desmos Calculator
, we know that over the interval(-0.95, 0)
, the max value ofeˣ
ise⁰ = 1
:
- So boundary is
M = 1
. - Therefore, the formula of this theorem becomes:
- In this case, we need to try some numbers for
n
to get the desired value:
- After tried
n=5
andn=6
, we could see that only untiln=6
, which means the6th derivative
, theError
is less than0.001
. - So the answer is
6th derivative
.
Example
Solve:
- Same with the problem above, we want to apply the
Lagrange Error Bound Theorem
, and bound it to0.001
:
- For those unknowns variables in the theorem, we know that:
- The approximation is centred at
2
, soC = 2
. - The input of function is
2.5
, sox = 2.5
. - The interval then is
(2, 2.5)
. - For the
M
value, it's not easy to figure out, but we've been told the formula for derivative.
- So the expression for
M
would be: - Let’s directly plug in the
M
expression into the Remainder:
- With the help from
Desmos grapher
we know that when within the interval2≤ z ≤ 2.5
, thatz=2
makes the formula to the max:
- So let’s set the inequality:
- After trying out some number for
n
, we get thatn ≥ 3
.