MATH 1172 Midterm: MATH 1172 Ohio State University Math 1172 10.3 & 10.4 Solutions Fa 2014
31 Jan 2019
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Davis Buenger Math 1172 10.3 & 10.4 October 21, 2015
1. (Test Problem) True or false: Suppose
f(x) = x3+ 2x2. Then the 3rd-order Taylor
polynomial for fcentered at a= 1 is
p3= 3 + 7(x−1) + 10(x−1)2+ 6(x−1)3.
False: Observe that:
f(x) = x3+ 2x2f(1) = 3
f′(x) = 3x2+ 4x f′(1) = 7
f′′(x) = 6x+ 4 f′′(1) = 10
f′′′(x) = 6 f′′′(1) = 6
Thus the third order polynomial for f(x) cen-
tered at a= 1 is
f(1)+f′(1)(x−1)+ f′′ (1)
2! (x−1)2+f′′′(1)
3! (x−1)3
= 3 + 7(x−1) + 10
2! (x−1)2+6
3!(x−1)3.
2. (Test Problem) True or false:
0 = π−π3
3! +π5
5! −π7
7! +. . . [Hint: The Taylor
series for sin(x) centered at 0 may be useful.]
True: Recall that the taylor series for sin(x) is
x−x3
3! +x5
5! −x7
7! +. . . , and the series converges
for all x. By plugging in πfor the Taylor series,
we have
0 = sin(π) = π−π3
3! +π5
5! −π7
7! +. . . .
3. Find the first four nonzero terms of the
taylor series centered at 0 for the function
f(x) = 4
√1 + xand use these terms to
approximate 4
√1.09.
Solution: Let us find first derivatives of f(x)
and evaluate the derivatives at 0.
f′(x) = 1
4(1 + x)
−3
4f′(0) = 1
4
f′′(x) = −3
16 (1 + x)
−7
4f′′(0) = −3
16
f′′′(x) = 21
64(1 + x)
−11
4f′′′(0) = 21
64
Additionally f(0) = 1.Thus the first four terms
of the taylor series are
1 + 1
4·x+−3
16 ·x2
2! +21
64 ·x3
3! .
Thus 4
√1.09 = f(0.09) and
f(0.09) ≈1 + 1
4·0.09 + −3
16 ·0.092
2! +21
64 ·0.093
3! .
4. Consider the function f(x) = sin(3x) and the
remainder in the taylor series centered at the
point 0 for f(x). Show that limn→∞ Rn(x) = 0
for all xin R.
Solution: Recall that we defined Rn(x) as
Rn(x) = f(x)−Pn(x),
where Pn(x) is the nth Taylor polynomial for
f(x). Furthermore, by Taylor’s Theorem, for
ever x∈Rthere exists zin between 0 and x
such that
sin(3x)−Pn(x) = f(n+1)(z)xn+1
(n+ 1)!.
Observe that every derivative of sin(3x) is of
the form 3ksin(3x), −3ksin(3x), 3kcos(3x),or
−3kcos(3x), where kis equal to the order of
the derivative. In any case,
f(n+1) = 3n+1g(x)
where g(x) is one of sin(3x), −sin(3x), cos(3x),
or −cos(3x). Thus for any value of z,
|f(n+1)(z)| ≤ 3n+1.
So for any value of x
|Rn(x)| ≤ 3n+1 |x|n+1
(n+ 1)!.
As factorials grow faster than exponentials, this
tends to 0 for every value of x.
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Document Summary
October 21, 2015: (test problem) true or false: suppose f (x) = x3 + 2x2. Then the 3rd-order taylor polynomial for f centered at a = 1 is p3 = 3 + 7(x 1) + 10(x 1)2 + 6(x 1)3. Thus the third order polynomial for f (x) cen- tered at a = 1 is f (1)+f (1)(x 1)+ f (1) = 3 + 7(x 1) + (x 1)2 : (test problem) true or false: [hint: the taylor series for sin(x) centered at 0 may be useful. ] (x 1)3. True: recall that the taylor series for sin(x) is x x3. , and the series converges for all x. By plugging in for the taylor series, we have. 7: find the rst four nonzero terms of the taylor series centered at 0 for the function f (x) = 4 1 + x and use these terms to approximate 4 1. 09. Thus the rst four terms of the taylor series are.
