SME
Calculus I
Linear approximation, Newton's method and L'Hôpital's rule
Use this for “Use Newton’s method with x₀ = 2 to approximate ∛10: find x₁ and x₂.”
Chart
Newton: each tangent meets the axis at the next x: x₀ = 2, x₁ = 2.167, x₂ = 2.155
This is an example. Type your own values to start a new problem.
f(x)
=
x
−
N
x₁
=
x₀
−
f(x₀)
f′(x₀)
=
x₀
−
f(x₀)
f′(x₀)
=
x₁
x₂
=
x₂
ⁿ√N
True root, to compare
worked out (not typed)
Formulas
f
(
x₀
)
=
x₀
n
−
N
−2
=
2
3
−
10
f
′(
x₀
)
=
n
×
x₀
n
− 1
12
=
3
×
2
3 − 1
x₁
=
x₀
−
f
(
x₀
)
f
′(
x₀
)
2.166667
=
2
−
−2
12
x₂
=
x₁
−
x₁
n
−
N
n
×
x₁
n
− 1
2.154504
=
2.166667
−
2.166667
3
− 10
3 ×
2.166667
3 − 1
ⁿ√
N
=
N
1 ÷
n
2.154435
=
10
1 ÷ 3
Assumptions
•
ⁿ√N is the positive root of f(x) = xⁿ − N, and the power rule gives f′(x) = nxⁿ⁻¹.
•
Each step follows the tangent at the guess down to where it crosses the x-axis: xₖ₊₁ = xₖ − f(xₖ) ÷ f′(xₖ).
•
Near a simple root the number of correct digits roughly doubles with each step.
•
A guess far from the root takes many steps, and a guess where f′ = 0 (x = 0) fails: the tangent is flat.
Step-by-step
WE KNOW
N = 10, n = 3, x₀ = 2
FIND
value of f at the first guess (f(x₀)), slope at the first guess (f′(x₀)), guess after one step (x₁), guess after two steps (x₂), true root, to compare (ⁿ√N)
Step 1 · Find value of f at the first guess (f(x₀))
Use f(x₀) = x₀ⁿ − N
Put
the
guess
into
f:
how
far
xⁿ
is
from
N
shows
how
far
the
guess
is
from
the
root.
f
(
x₀
)
=
2
3
−
10
f
(
x₀
)
=
−2
Step 2 · Find slope at the first guess (f′(x₀))
Use f′(x₀) = n × x₀ⁿ⁻¹
Power
rule
on
xⁿ:
bring
the
power
down
in
front
and
lower
it
by
1;
N
has
slope
0.
f
′(
x₀
)
=
3
×
2
2
f
′(
x₀
)
=
12
Step 3 · Find guess after one step (x₁)
Use x₁ = x₀ − f(x₀) ÷ f′(x₀)
The
tangent
at
x₀
has
slope
f
′(
x₀
),
so
it
meets
the
x-axis
f
(
x₀
)
f
′(
x₀
)
to
the
side
of
x₀
.
x₁
=
2
−
−2
12
x₁
=
2
−
(−0.1667)
x₁
=
2.166667
Step 4 · Find guess after two steps (x₂)
Use x₂ = x₁ − (x₁ⁿ − N) ÷ (n × x₁ⁿ⁻¹)
Repeat
the
step
from
x₁
:
work
out
f
and
its
slope
there,
then
follow
that
tangent
down
to
the
x-axis.
x₂
=
2.166667
−
f
(
x₁
)
f
′(
x₁
)
f
(
x₁
)
=
2.166667
3
−
10
=
10.171296
−
10
=
0.171296
f
′(
x₁
)
=
3
×
2.166667
2
=
14.08333
x₂
=
2.166667
−
0.171296
14.08333
=
2.154504
x₂
=
2.154504
Step 5 · Find true root, to compare (ⁿ√N)
Use ⁿ√N = N1 ÷ n
The
true
root,
worked
out
to
compare
with
x₁
and
x₂
.
ⁿ√
N
=
10
1 ÷ 3
ⁿ√
N
=
2.154435
Check
Put the numbers back in. Both sides should match.
−2
=
2
3
−
10
✓
12
=
3
×
2
3 − 1
✓
2.166667
=
2
−
−2
12
✓
2.154504
=
2.166667
−
2.166667
3
− 10
3 ×
2.166667
3 − 1
✓
2.154435
=
10
1 ÷ 3
✓
Topic check
See what you know: 5 questions on this topic
New questions each time. Pass to mark it complete.