Estimating errors in partial sums For each of the following convergent alternating series, evaluate the nth partial sum for the given value of n . Then use Theorem 10.18 to find an upper bound for the error | S − S n | in using the nth partial sum S n to estimate the value of the series S . 31. ∑ k = 0 ∞ ( − 1 ) k k 4 + k + 2 1 ; n = 5
Estimating errors in partial sums For each of the following convergent alternating series, evaluate the nth partial sum for the given value of n . Then use Theorem 10.18 to find an upper bound for the error | S − S n | in using the nth partial sum S n to estimate the value of the series S . 31. ∑ k = 0 ∞ ( − 1 ) k k 4 + k + 2 1 ; n = 5
Solution Summary: The author evaluates the nth partial sum and finds an upper bound for the error left|S-S_nright|.
Estimating errors in partial sums For each of the following convergent alternating series, evaluate the nth partial sum for the given value of n. Then use Theorem 10.18 to find an upper bound for the error |S − Sn| in using the nth partial sum Sn to estimate the value of the series S.
Given lim x-4 f (x) = 1,limx-49 (x) = 10, and lim→-4 h (x) = -7 use the limit properties
to find lim→-4
1
[2h (x) — h(x) + 7 f(x)] :
-
h(x)+7f(x)
3
O DNE
17. Suppose we know that the graph below is the graph of a solution to dy/dt = f(t).
(a) How much of the slope field can
you sketch from this information?
[Hint: Note that the differential
equation depends only on t.]
(b) What can you say about the solu-
tion with y(0) = 2? (For example,
can you sketch the graph of this so-
lution?)
y(0) = 1
y
AN
(b) Find the (instantaneous) rate of change of y at x = 5.
In the previous part, we found the average rate of change for several intervals of decreasing size starting at x = 5. The instantaneous rate of
change of fat x = 5 is the limit of the average rate of change over the interval [x, x + h] as h approaches 0. This is given by the derivative in the
following limit.
lim
h→0
-
f(x + h) − f(x)
h
The first step to find this limit is to compute f(x + h). Recall that this means replacing the input variable x with the expression x + h in the rule
defining f.
f(x + h) = (x + h)² - 5(x+ h)
=
2xh+h2_
x² + 2xh + h² 5✔
-
5
)x - 5h
Step 4
-
The second step for finding the derivative of fat x is to find the difference f(x + h) − f(x).
-
f(x + h) f(x) =
= (x²
x² + 2xh + h² -
])-
=
2x
+ h² - 5h
])x-5h) - (x² - 5x)
=
]) (2x + h - 5)
Macbook Pro
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