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22.04.2020 •
Mathematics
EXAMPLE 2 Prove that 9ex is equal to the sum of its Maclaurin series. SOLUTION If f(x) = 9ex, then f (n + 1)(x) = for all n. If d is any positive number and |x| ≤ d, then |f (n + 1)(x)| = ≤ 9ed. So Taylor's Inequality, with a = 0 and M = 9ed, says that |Rn(x)| ≤ (n + 1)! |x|n + 1 for |x| ≤ d. Notice that the same constant M = 9ed works for every value of n. But, from this equation, we have lim n → [infinity] 9ed (n + 1)! |x|n + 1 = 9ed lim n → [infinity] |x|n + 1 (n + 1)! = . It follows from the Squeeze Theorem that lim n → [infinity] |Rn(x)| = 0 and therefore lim n → [infinity] Rn(x) = for all values of x. By this theorem, 9ex is equal to the sum of its Maclaurin series, that is, 9ex = [infinity] 9xn n! n = 0 for all x.
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Ответ:
To Prove:
is equal to the sum of its Maclaurin series.
Step-by-step explanation:
If
, then
for all n. If d is any positive number and |x| ≤ d, then ![|f^{(n + 1)(x)}| = 9e^x\leq 9e^d.](/tpl/images/0617/0695/ef926.png)
So Taylor's Inequality, with a = 0 and M =
, says that ![|R_n(x)| \leq \dfrac{9e^d}{(n+1)!} |x|^{n + 1} \:for\: |x| \leq d.](/tpl/images/0617/0695/9478b.png)
Notice that the same constant
works for every value of n.
But, since
,
We have![lim_{n\to\infty} \dfrac{9e^d}{(n+1)!} |x|^{n + 1} =9e^d lim_{n\to\infty} \dfrac{|x|^{n + 1}}{(n+1)!} =0](/tpl/images/0617/0695/cebce.png)
It follows from the Squeeze Theorem that
and therefore
for all values of x.
By this theorem above,
is equal to the sum of its Maclaurin series, that is,
Ответ:
Step-by-step explanation:
y = 40x + 25
where x = 1 class
y = total cost
y = 40 (1) + 25
= 65