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kondor19780726 [428]
3 years ago
6

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.
Mathematics
1 answer:
amm18123 years ago
6 0

Answer:

To Prove: 9e^x is equal to the sum of its Maclaurin series.

Step-by-step explanation:

If f(x) = 9e^x, then f ^{(n + 1)(x)} =9e^x for all n. If d is any positive number and   |x| ≤ d, then |f^{(n + 1)(x)}| = 9e^x\leq  9e^d.

So Taylor's Inequality, with a = 0 and M = 9e^d, says that |R_n(x)| \leq \dfrac{9e^d}{(n+1)!} |x|^{n + 1} \:for\: |x| \leq  d.

Notice that the same constant M = 9e^d works for every value of n.

But, since lim_{n\to\infty}\dfrac{x^n}{n!} =0 $ for every real number x$,

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

It follows from the Squeeze Theorem that lim_{n\to\infty} |R_n(x)|=0 and therefore lim_{n\to\infty} R_n(x)=0 for all values of x.

THEOREM\\If f(x)=T_n(x)+R_n(x), $where $T_n $is the nth degree Taylor Polynomial of f at a and  $ lim_{n\to\infty} R_n(x)=0 \:  for \: |x-a|

By this theorem above, 9e^x is equal to the sum of its Maclaurin series, that is,

9e^x=\sum_{n=0}^{\infty}\frac{9x^n}{n!}  for all x.

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Need help with math probelm if do 5 stars and brainly points
PtichkaEL [24]

Step-by-step explanation:

let's first convert feet to inches

4 ft = 48 inches ( width )

6.5 ft = 78 inches ( length )

12.5 ft = 150 inches ( height )

each cube is 1/4 x 1/4 x 1/4

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78 ÷ 1/4 = 78 x 4 = 312 cubes

y-axis ( width )

48 x 4 = 192 cubes

z-axis ( height )

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the total number of cubes fit in the closet is :

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2 years ago
A random sample of 25 ACME employees showed the average number of vacation days taken during the year is 18.3 days with a standa
Norma-Jean [14]

Answer:

a) Null hypothesis:\mu \leq 15  

Alternative hypothesis:\mu > 15  

b) df=n-1=25-1=24  

For this case the p value is given p_v = 0.0392

If we compare the p value and the significance level given \alpha=0.05 we see that p_v so we can conclude that we have enough evidence to reject the null hypothesis, so we can conclude that the true mean is higher than 15 at 5% of signficance.  

c) Type I error, also known as a “false positive” is the error of rejecting a null  hypothesis when it is actually true. Can be interpreted as the error of no reject an  alternative hypothesis when the results can be  attributed not to the reality.

So for this case a type I of error would be reject the hypothesis that the true mean is less or equal than 15 and is actually true.

Step-by-step explanation:

Data given and notation  

\bar X=18.3 represent the sample mean

s=3.72 represent the sample standard deviation

n=25 sample size  

\mu_o =15 represent the value that we want to test

\alpha=0.05 represent the significance level for the hypothesis test.  

t would represent the statistic (variable of interest)  

p_v represent the p value for the test (variable of interest)  

Part a: State the null and alternative hypotheses.  

We need to conduct a hypothesis in order to check if the true mean for vacation days is higher than 15, the system of hypothesis would be:  

Null hypothesis:\mu \leq 15  

Alternative hypothesis:\mu > 15  

If we analyze the size for the sample is < 30 and we don't know the population deviation so is better apply a t test to compare the actual mean to the reference value, and the statistic is given by:  

t=\frac{\bar X-\mu_o}{\frac{s}{\sqrt{n}}}  (1)  

t-test: "Is used to compare group means. Is one of the most common tests and is used to determine if the mean is (higher, less or not equal) to an specified value".  

Part b: P-value  and conclusion

The first step is calculate the degrees of freedom, on this case:  

df=n-1=25-1=24  

For this case the p value is given p_v = 0.0392

Conclusion  

If we compare the p value and the significance level given \alpha=0.05 we see that p_v so we can conclude that we have enough evidence to reject the null hypothesis, so we can conclude that the true mean is higher than 15 at 5% of signficance.  

Part c

Type I error, also known as a “false positive” is the error of rejecting a null  hypothesis when it is actually true. Can be interpreted as the error of no reject an  alternative hypothesis when the results can be  attributed not to the reality.

So for this case a type I of error would be reject the hypothesis that the true mean is less or equal than 15 and is actually true.

3 0
3 years ago
Pls tell the answer in step by step
IceJOKER [234]

Answer:

5

Step-by-step explanation:

here,  the highest degree is 5 so the degree of polynomial is also 5.

4 0
3 years ago
Juan is 1 1/4 feet shorter than Maria. Maria is 1/3 foot taller than Luis. Luis is 62 inches tall, how tall are Maria and Juan
liraira [26]
It's really easy i think u can handle yourself
7 0
3 years ago
Read 2 more answers
Solve the inequality for x. 75+15x is greater than or equal to 140
Gre4nikov [31]
75+ 15x ≥ 140
⇒ 15x ≥ 140 -75
⇒ 15x ≥ 65
⇒ x ≥ 65/15
⇒ x ≥ 13/3
⇒ x ≥ 4 1/3

The final answer is x ≥ 4 1/3~
8 0
3 years ago
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