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3 years ago

80 Superscript one-fourth x

Mathematics

1 answer:

ANTONII [103]3 years ago

6 0

Can you please help me with this I’ll help you too

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Use the Taylor series you just found for sinc(x) to find the Taylor series for f(x) = (integral from 0 to x) of sinc(t)dt based

Marina CMI [18]

In this question (brainly.com/question/12792658) I derived the Taylor series for $\mathrm{sinc}\,x$ about $x=0$ :

$\mathrm{sinc}\,x=\displaystyle\sum_{n=0}^\infty\frac{(-1)^nx^{2n}}{(2n+1)!}$

Then the Taylor series for

$f(x)=\displaystyle\int_0^x\mathrm{sinc}\,t\,\mathrm dt$

is obtained by integrating the series above:

$f(x)=\displaystyle\int\sum_{n=0}^\infty\frac{(-1)^nx^{2n}}{(2n+1)!}\,\mathrm dx=C+\sum_{n=0}^\infty\frac{(-1)^nx^{2n+1}}{(2n+1)^2(2n)!}$

We have $f(0)=0$ , so $C=0$ and so

$f(x)=\displaystyle\sum_{n=0}^\infty\frac{(-1)^nx^{2n+1}}{(2n+1)^2(2n)!}$

which converges by the ratio test if the following limit is less than 1:

$\displaystyle\lim_{n\to\infty}\left|\frac{\frac{(-1)^{n+1}x^{2n+3}}{(2n+3)^2(2n+2)!}}{\frac{(-1)^nx^{2n+1}}{(2n+1)^2(2n)!}}\right|=|x^2|\lim_{n\to\infty}\frac{(2n+1)^2(2n)!}{(2n+3)^2(2n+2)!}$

Like in the linked problem, the limit is 0 so the series for $f(x)$ converges everywhere.

7 0

3 years ago

Given An = 5a(n-1) where a1= 2. What are the first 4 terms

nexus9112 [7]

Answer:

2, 10, 50, 250

Step-by-step explanation:

Using the formula with a₁ = 2 , then

a₂ = 5a₁ = 5 × 2 = 10

a₃ = 5a₂ = 5 × 10 = 50

a₄ = 5a₃ = 5 × 50 = 250

The first 4 terms are 2, 10, 50, 250

8 0

3 years ago

after taking math online . what would you say are some pross (good) and cons (bad) about taking classes . online? what would mak

Elden [556K]

Answer:

Cons: Could be harder, Difficult to solve, hard to see problems online, Things to help: Post assignments in portions Pros: hand wont hurt if writing, Wont loose paper if it’s on paper

Step-by-step explanation:

5 0

3 years ago

Read 2 more answers

What is 2 6/7 as an improper fraction

Makovka662 [10]

Answer:

20/7

Step-by-step explanation:

Check the picture

6 0

2 years ago

Kurt drew a rectangular maze with the length of 3/4 foot and a width of 5/12 foot.What is the area,in square feet,of Kurt's maze

nata0808 [166]

Answer: $\cfrac{5}{16} \textnormal { ft}^2$

Explanation:

$\textnormal {Area} = \cfrac{3}{4} \times \cfrac{5}{12}$

$\textnormal {Area} = \cfrac{5}{16} \textnormal { ft}^2$

4 0

3 years ago