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lora16 [44]
3 years ago
10

By recognizing each series below as a Taylor series evaluated at a particular value of x, find the sum of each convergent series

.
A. 1 + 1/5 + (1/5)^2 + (1/5)^3 + (1/5)^4 +.....+ (1/5)^n + .... = _____.
B. 1 + 5 + 5^2/2! + 5^3/3! + 5^4/4! +....+ 5^n/n! +....= _____.
Mathematics
1 answer:
goldfiish [28.3K]3 years ago
4 0

The first sum is a geometric series:

1+\dfrac15+\dfrac1{5^2}+\dfrac1{5^3}+\cdots+\dfrac1{5^n}+\cdots=\displaystyle\sum_{n=0}^\infty\frac1{5^n}

Recall that for |<em>x</em>| < 1, we have

\dfrac1{1-x}=\displaystyle\sum_{n=0}^\infty x^n

Here we have |<em>x</em>| = |1/5| = 1/5 < 1, so the first sum converges to 1/(1 - 1/5) = 5/4.

The second sum is exponential:

1+5+\dfrac{5^2}{2!}+\dfrac{5^3}{3!}+\cdots+\dfrac{5^n}{n!}+\cdots=\displaystyle\sum_{n=0}^\infty \frac{5^n}{n!}

Recall that

\exp(x)=\displaystyle\sum_{n=0}^\infty\frac{x^n}{n!}

which converges everywhere, so the second sum converges to exp(5) or <em>e</em>⁵.

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m=2/3, b=4

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Write the sentence as an absolute value inequality, then solve the inequality: A number “n” is more than 9 units from 3
Helga [31]

Answer:

absolute vlaue inequality: |x-3| > 9; solved: x<-6 and x>12

Step-by-step explanation:

I’m going to start this off by saying I learned all this right now by just searching up how to solve an absolute inequality equation and watching one video, so this might not be an accurate explanation. (I’m pretty sure the answer’s right though)

So an absolute value inequality must be written like this:  

| x - a | *inequality* b

a is going to be the number that the inequality is centered around, in this case, 3. b will be how far you can deviate from that number, which in this case is 9.  

Now, you will have this:

|x - 3| *inequality* 9

Now, to find the inequality, you need to understand the wording. If it says “more than” as it does here, then you would have the greater-than symbol (>). If you have “less than” then you’d have the less-than symbol (<). If the problem says “at least b away” then it would be greater-than-or-equal to (≥), and likewise, if it says “at most b away” then it would be less-than-or-equal-to (≤).

So now we're at:

|x - 3| > 9

To solve the equation, you just need to subtract 9(b) from 3(a) and add 9(a) to 3(b) to get -6 and 12. Since x must be more than 9 units away, you would get:

x<-6 and x>12

Hope this is helpful!

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4 years ago
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