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miskamm [114]
3 years ago
14

Find the Taylor series for f(x)=sin(x) centered at c=π/2.sin(x)=∑ n=0 [infinity]On what interval is the expansion valid? Give yo

ur answer using interval notation.
Mathematics
1 answer:
agasfer [191]3 years ago
3 0

Answer:

sinx =1-\frac{1}{2} (x-\frac{\pi}{2} )^2+\frac{1}{24} (x-\frac{\pi}{2} )^4-\frac{1}{720} (x-\frac{\pi}{2} )^6+\frac{1}{40320} (x-\frac{\pi}{2} )^8+...

Step-by-step explanation:

given that f(x) = sin x

we have to find the Taylor series for that

f(x) = sin x   : f( = 1\\f'(x) = cos x :(f'\frac{\pi}{2})=0\\f"(x) = -sinx :f" (\frac{\pi}{2}) =-1\\f^4 (x) = -cosx : f^4 (\frac{\pi}{2}) =0

and so on.

i.e. 2nd, 4th, 6th terms would be 0

and also 1st, 5th, 9th terms would be positive for f value and 3rd, 9th,... would be negative

Using the above we can write Taylor series as

f(x) = f(a)+\frac{f'(a)}{1!} (x-\frac{\pi}{2}) +...+f^n(a) /n! (x- \frac{\pi}{2})^n+...

sinx =1-\frac{1}{2} (x-\frac{\pi}{2} )^2+\frac{1}{24} (x-\frac{\pi}{2} )^4-\frac{1}{720} (x-\frac{\pi}{2} )^6+\frac{1}{40320} (x-\frac{\pi}{2} )^8+...

This is valid for all real values of x.

x ∈(-\infty, infty)

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Answer:

The expected loss is $275 million.

Step-by-step explanation:

Expected loss can be determined as the sum of the product of each possible loss by the its probability of occurence. In this situation, there are only two possible losses listed since the probability of no loss doesn't add any value to the expected loss and should be disregarded.

Expected loss (in millions) = EL

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The expected loss is $275 million.

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3 years ago
Renae likes to make pizza dough on the weekends. She has 8 1/3 cups of flour. She needs 5/6 of a cup for each pizza.How many piz
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Answer:

She can make 10 pizzas.

Step-by-step explanation:

8 1/3=25/3

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3 years ago
Please help me answer this question its urgent
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Answer:

Step-by-step explanation:

1). \frac{2}{3}n-\frac{3}{4}n+\frac{1}{6}n+2\frac{2}{9}n

= \frac{2}{3}n-\frac{3}{4}n+\frac{1}{6}n+(2+\frac{2}{9})n

= (\frac{2}{3}-\frac{3}{4}+\frac{1}{6})n+(2+\frac{2}{9})n

= (\frac{8}{12}-\frac{9}{12}+\frac{2}{12})n+(2+\frac{2}{9})n

= \frac{(8-9+2)}{12}n+(2+\frac{2}{9})n

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= (\frac{3}{36}+\frac{72}{36}+\frac{8}{36})n

= \frac{83}{36}n

= 2\frac{11}{36}n

2). \frac{2}{5}g-\frac{1}{6}-g+\frac{3}{10}g-\frac{4}{5}

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= (\frac{4}{10}-\frac{10}{10}+\frac{3}{10})g-(\frac{5}{30}+ \frac{24}{30})

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= 7i-\frac{3}{7}i+\frac{1}{2}i+\frac{1}{3}h-h+\frac{1}{4}h

= (7-\frac{3}{7}+\frac{1}{2})i+(\frac{1}{3}-1+\frac{1}{4})h

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= (\frac{98-6+7}{14})i+(\frac{4-12+3}{12})h

= \frac{99}{14}i-\frac{5}{12}h

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Answer:

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