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3 years ago

Is there a general rule to convert a function to a power series? Examples would be greatly appreciated:)

Mathematics

1 answer:

Maslowich3 years ago

5 0

Step-by-step explanation:

Use the Maclaurin expansion:

f(x) = ∑ₙ₌₀°° f⁽ⁿ⁾(0) xⁿ / n!

For example, for f(x) = eˣ:

f(x) = ∑ₙ₌₀°° e⁰ xⁿ / n!

f(x) = ∑ₙ₌₀°° xⁿ / n!

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84% of Astrid's books are fiction. What fraction of Astrid's books are fiction?

Fed [463]

<h2>Greetings!</h2>

Answer:

$\frac{21}{25}$

Step-by-step explanation:

If 84% are fiction, then the fraction of these are simply the percentage over 100:

$\frac{84}{100}$

Simplified down by dividing by 4:

$\frac{21}{25}$

<h2>Hope this helps!</h2>

5 0

3 years ago

Read 2 more answers

Two linear functions of x are shown. Which Statement about the functions is true?

kolbaska11 [484]

Answer:

OPTION D: The rate of change of Function 2 is less than that of Function 1.

Step-by-step explanation:

From the table, we will determine the equation of function 2.

From the table, g(1) = 14 & g(-3) = - 86

Since, it is a linear function it should have been of the form, y = a + bx.

We will denote Function 2 as g(x).

So, g(1) = 14 = a + b . . . (1)

And g(-3) = -86 = -3a + b . . . (2)

From (1), a = 14 - b

Substituting this in (2), we get:

-86 = -3(14 - b) + b

⇒ - 86 = - 42 + 4b

⇒ - 44 = 4b

⇒ b = -11

So, a = 14 - (-11)

⇒ a = 25

So, equation of function 2 is: y = 25 - 11x.

Hence, Option A and B are eliminated.

In a linear equation of the form, y = ax + b, y - intercept = b.

So, in function 1, y - intercept = 19

And in function 2, y - intercept = -11

Clearly, OPTION C can be eliminated.

Now, rate of change of Function 1 = $\frac{dy}{dx} = 30$

and rate of change of Function 2 = $\frac{dy}{dx} = 25$

Therefore, we can say that the rate of change if function 2 is less than that of function 1.

Option D is our answer.

5 0

3 years ago

You are investing $4000 with a 2.5% nominal interest; compare what happens if the interest is compounded annually, quarterly, or

Dominik [7]

Every year he pay like 100$
Compounded gon to be half
Quarter 1.3%

7 0

3 years ago

Use mathematical induction to prove the statement is true for all positive integers n. 1^2 + 3^2 + 5^2 + ... + (2n-1)^2 = (n(2n-

Charra [1.4K]

Answer:

The statement is true is for any $n\in \mathbb{N}$ .

Step-by-step explanation:

First, we check the identity for $n = 1$ :

$(2\cdot 1 - 1)^{2} = \frac{2\cdot (2\cdot 1 - 1)\cdot (2\cdot 1 + 1)}{3}$

$1 = \frac{1\cdot 1\cdot 3}{3}$

$1 = 1$

The statement is true for $n = 1$ .

Then, we have to check that identity is true for $n = k+1$ , under the assumption that $n = k$ is true:

$(1^{2}+2^{2}+3^{2}+...+k^{2}) + [2\cdot (k+1)-1]^{2} = \frac{(k+1)\cdot [2\cdot (k+1)-1]\cdot [2\cdot (k+1)+1]}{3}$

$\frac{k\cdot (2\cdot k -1)\cdot (2\cdot k +1)}{3} +[2\cdot (k+1)-1]^{2} = \frac{(k+1)\cdot [2\cdot (k+1)-1]\cdot [2\cdot (k+1)+1]}{3}$

$\frac{k\cdot (2\cdot k -1)\cdot (2\cdot k +1)+3\cdot [2\cdot (k+1)-1]^{2}}{3} = \frac{(k+1)\cdot [2\cdot (k+1)-1]\cdot [2\cdot (k+1)+1]}{3}$

$k\cdot (2\cdot k -1)\cdot (2\cdot k +1)+3\cdot (2\cdot k +1)^{2} = (k+1)\cdot (2\cdot k +1)\cdot (2\cdot k +3)$

$(2\cdot k +1)\cdot [k\cdot (2\cdot k -1)+3\cdot (2\cdot k +1)] = (k+1) \cdot (2\cdot k +1)\cdot (2\cdot k +3)$

$k\cdot (2\cdot k - 1)+3\cdot (2\cdot k +1) = (k + 1)\cdot (2\cdot k +3)$

$2\cdot k^{2}+5\cdot k +3 = (k+1)\cdot (2\cdot k + 3)$

$(k+1)\cdot (2\cdot k + 3) = (k+1)\cdot (2\cdot k + 3)$

Therefore, the statement is true for any $n\in \mathbb{N}$ .

4 0

3 years ago

when you flip a biased coin the probability of getting tails is 0.53 what is the probability of getting heads?

garri49 [273]

Im sorry im
just answering to get points to ask questions

5 0

4 years ago