Calc 2 series

[joku1234]

I don't even know to approach these problems! Please help!

Question 2: Suppose \(\displaystyle \dfrac{2}{1\, +\, 3x}\, =\, \)\(\displaystyle \displaystyle{ \sum_{n\, =\, 0}^{\infty}\, a_n\, x^n}\)

Find a₃.

Question 3: Suppose \(\displaystyle \dfrac{x}{(1\, +\, 4x)^2}\, =\, \)\(\displaystyle \displaystyle{ \sum_{n\, =\, 0}^{\infty}\, a_n\, x^n}\)

Find a₂.

Question 4: Suppose \(\displaystyle \int\, x^2\, \ln(1\, +\, x)\, dx\, =\, C\, +\,\)\(\displaystyle \displaystyle{ \sum_{n\, =\, 0}^{\infty}\, a_n\, x^n}\)

Find a₄ and express your answer as a decimal.

 

[Steven G]

I don't even know to approach these problems! Please help!

1/(1-r) =summation of r^2.
so 1/(1+r) = 1/(1-(-r)) = summation of (-r)^2.
This should help you with the 1st question.

 

[Deleted member 4993]

I don't even know to approach these problems! Please help!

I don't even know to approach these problems! Please help!

Question 2: Suppose \(\displaystyle \dfrac{2}{1\, +\, 3x}\, =\, \)\(\displaystyle \displaystyle{ \sum_{n\, =\, 0}^{\infty}\, a_n\, x^n}\)

Find a₃.

Question 3: Suppose \(\displaystyle \dfrac{x}{(1\, +\, 4x)^2}\, =\, \)\(\displaystyle \displaystyle{ \sum_{n\, =\, 0}^{\infty}\, a_n\, x^n}\)

Find a₂.

Question 4: Suppose \(\displaystyle \int\, x^2\, \ln(1\, +\, x)\, dx\, =\, C\, +\,\)\(\displaystyle \displaystyle{ \sum_{n\, =\, 0}^{\infty}\, a_n\, x^n}\)

Find a₄ and express your answer as a decimal.

The answers depend on:

what do you know?

Do you know Taylor series expansion of those functions - i.e. - (1+3x)^-1, (1+4x)^-2 & ln(1+x)?

 

[Ishuda]

Suppose
\(\displaystyle \frac{17}{1+12x}\, =\, \underset{n=0}{\overset{\infty}{\Sigma}}\, a_n\, x^n\)
then
17 = (1 + 12x) (a₀ + a₁ x + a₂ x² + a₃ x³ + ...)
= a₀ + a₁ x + a₂ x² + a₃ x³ + ... + 12 a₀ x + 12 a₁ x² + 12 a₂ x³ + ...
= a₀ + (a₁ + 12 a₀) x + (a₂ + 12 a₁) x² + (a₃ + 12 a₂) x³ + ...
Thus
a₀ = 17
a_j = -12 a_j-1; j = 1, 2, 3, ...
which easily leads to
a_j = (-1)^j 17 12^j; j = 1, 2, 3, ...

Of course that 'proof' holds only for the region of absolute convergence but it generally suffices in the region of convergence.

 

[daon2]

Another way to do this is with derivatives.

If

\(\displaystyle \displaystyle f(x) = \sum_{n=0}^{\infty} a_n(x-c)^n\)

then

\(\displaystyle a_n = \dfrac{f^{(n)}(c)}{n!}\)

This is particularly useful if you have the aid of software, otherwise might be cumbersome. For example, and I'll use Ishuda's example, if \(\displaystyle f(x) = \dfrac{17}{1+12x}\) then

\(\displaystyle f^{(0)}(x) = f(x)= \dfrac{17}{1+12x}\)

\(\displaystyle f^{(3)}(x) = \dfrac{-176256}{(1+12 x)^4}\)

so \(\displaystyle a_0 = f(0) = 17\) and \(\displaystyle a_3 = \dfrac{-176256}{3!\cdot (1+12 (0))^4} = -29376\).

 

[Ishuda]

Another way to do this is with derivatives.
... and \(\displaystyle a_3 = \dfrac{-176256}{3!\cdot (1+12 (0))^4} = -29376\).

Just for grins and giggles, you might note that
-29376 = (-1)³ 17 12³

EDIT: And I certainly agree with Jomo that if you have software to make life easier, like computing the derivatives and their values, it is nice to use them if you need the answer in a hurry or as part of a larger problem you are working on.

 

[juragan]

nice thread,
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