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Novay_Z [31]
3 years ago
5

Consider the following function. f(x) = 1/x, a = 1, n = 2, 0.6 ≤ x ≤ 1.4 (a) Approximate f by a Taylor polynomial with degree n

at the number a. T2(x) = (b) Use Taylor's Inequality to estimate the accuracy of the approximation f(x) ≈ Tn(x) when x lies in the given interval. (Round your answer to eight decimal places.) |R2(x)| ≤ 7.71604938 Incorrect: Your answer is incorrect. (c) Check your result in part (b) by graphing |Rn(x)|.
Mathematics
1 answer:
Xelga [282]3 years ago
7 0

Answer:

y - 1 = 0

Step-by-step explanation:

move constant to the left by adding its opposite to both sides y - 1 = 1 - 1

the sum two opposites equals 0

y =1

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Answer:

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Step-by-step explanation:

An infinite geometric series is defined as limit of partial sum of geometric sequences. Therefore, to find the infinite sum, we have to find the partial sum first then input limit approaches infinity.

However, fortunately, the infinite geometric series has already set up for you. It’s got the formula for itself which is:

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We can also write in summation notation rather S-term as:

\displaystyle \large{\sum_{n=1}^\infty a_1r^{n-1} = \dfrac{a_1}{1-r}}

Keep in mind that these only work for when |r| < 1 or else it will diverge.

Also, how fortunately, the given summation fits the formula pattern so we do not have to do anything but simply apply the formula in.

\displaystyle \large{\sum_{n=1}^\infty 4(0.5)^{n-1} = \dfrac{4}{1-0.5}}\\\\\displaystyle \large{\sum_{n=1}^\infty 4(0.5)^{n-1} = \dfrac{4}{0.5}}\\\\\displaystyle \large{\sum_{n=1}^\infty 4(0.5)^{n-1} = 8}

Therefore, the sum will converge to 8.

Please let me know if you have any questions!

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