Maclaurin and Taylor Series

[warwick]

1. Obtain the Maclaurin Series for the function by making an appropriate substitution in the Maclaurin Series for 1 / (1 - x). Include general term in your answer.

(d) 1 / (2 - x)

3. Obtain the first four nonzero terms of the Maclaurin series for the function by making an appropriate substitution in one of the following binomial series: 1 / (1 + x)^ 2 1 / (1 + x)^(1/2)

(a) (2 + x) ^ (-1/2)

(b) (1 - x^2) ^ (-2)

 

[Deleted member 4993]

1. Obtain the Maclaurin Series for the function by making an appropriate substitution in the Maclaurin Series for 1 / (1 - x). Include general term in your answer.

(d) 1 / (2 - x)

Hint:

\(\displaystyle \frac {1}{2-x}\ = \frac {1}{1 - (x - 1)}\)


3. Obtain the first four nonzero terms of the Maclaurin series for the function by making an appropriate substitution in one of the following binomial series: 1 / (1 + x)^ 2 1 / (1 + x)^(1/2)

(a) (2 + x) ^ (-1/2)

(b) (1 - x^2) ^ (-2)

For the problems above - exactly where are you stuck?

 

[soroban]

Hello, warwick!

I'll get you started by doing #1 . . .

\(\displaystyle \text{1. Obtain the Maclaurin Series for the function by making an appropriate substitution}\)
\(\displaystyle \text{in the Maclaurin Series for: }\:\frac{1}{1 - x}.\;\;\text{Include the general term in your answer.}\)

. . . \(\displaystyle (d)\;\frac{1}{2 - x}\)

\(\displaystyle \text{We know that: }\frac{1}{1-u} \;=\;1 + u + u^2 + u^3 + u^4 + \cdots\)


\(\displaystyle \text{We are given: }\;\frac{1}{2-x} \;=\;\frac{1}{2\left(1 - \frac{x}{2}\right)} \;=\;\frac{1}{2}\left(\frac{1}{1-\frac{x}{2}}\right)\)

\(\displaystyle \text{This is one-half the above Maclaurin Series with }u = \frac{x}{2}\)


\(\displaystyle \text{Therefore: }\;\frac{1}{2-x} \;=\;\frac{1}{2}\bigg[1 + \frac{x}{2} + \left(\frac{x}{2}\right)^2 + \left(\frac{x}{2}\right)^3 + \cdots + \left(\frac{x}{2}\right)^n + \cdots \bigg]\)

 

[warwick]

Hello, warwick!

I'll get you started by doing #1 . . .

\(\displaystyle \text{1. Obtain the Maclaurin Series for the function by making an appropriate substitution}\)
\(\displaystyle \text{in the Maclaurin Series for: }\:\frac{1}{1 - x}.\;\;\text{Include the general term in your answer.}\)

. . . \(\displaystyle (d)\;\frac{1}{2 - x}\)

\(\displaystyle \text{We know that: }\frac{1}{1-u} \;=\;1 + u + u^2 + u^3 + u^4 + \cdots\)


\(\displaystyle \text{We are given: }\;\frac{1}{2-x} \;=\;\frac{1}{2\left(1 - \frac{x}{2}\right)} \;=\;\frac{1}{2}\left(\frac{1}{1-\frac{x}{2}}\right)\)

\(\displaystyle \text{This is one-half the above Maclaurin Series with }u = \frac{x}{2}\)


\(\displaystyle \text{Therefore: }\;\frac{1}{2-x} \;=\;\frac{1}{2}\bigg[1 + \frac{x}{2} + \left(\frac{x}{2}\right)^2 + \left(\frac{x}{2}\right)^3 + \cdots + \left(\frac{x}{2}\right)^n + \cdots \bigg]\)

Ok. Originally, I got as far as the second equal bar on the "we are given..." line. I just didn't make the u = x/2 connection. I also found 1/1-(x-1), but how would that help? Would plugging in u = x-1 be a correct but different looking answer?

 

[warwick]

1. Obtain the Maclaurin Series for the function by making an appropriate substitution in the Maclaurin Series for 1 / (1 - x). Include general term in your answer.

(d) 1 / (2 - x)

Hint:

\(\displaystyle \frac {1}{2-x}\ = \frac {1}{1 - (x - 1)}\)


3. Obtain the first four nonzero terms of the Maclaurin series for the function by making an appropriate substitution in one of the following binomial series: 1 / (1 + x)^ 2 1 / (1 + x)^(1/2)

(a) (2 + x) ^ (-1/2)

(b) (1 - x^2) ^ (-2)

For the problems above - exactly where are you stuck?

I suppose rewriting them as the following is a start.

1 / (2 + x)^ (1/2)

1 / (1 - x)^ 2

 

[Deleted member 4993]

I suppose rewriting them as the following is a start.

1 / (2 + x)^ (1/2)

1 / (1 - x^2)^ 2

That's a good supposition - now continue by writing out McLauren series of the original functions and make appropriate substitution.

 

[warwick]

I suppose rewriting them as the following is a start.

1 / (2 + x)^ (1/2)

1 / (1 - x^2)^ 2

That's a good supposition - now continue by writing out McLauren series of the original functions and make appropriate substitution.

Ok. I got the second one.

1 / (1 - x^2)^ 2 = 1 + 2x^2 + 3x^3 + 4x^6 + ...

I have no idea how the book did the first one.

 

[warwick]

13) Find the first four nonzero terms of the Maclaurin series for the function by multiplying the Maclaurin series of the factors.

a. e^x sin x = (1 + x + x^2/2! + x^3/3! + ...) (x - x^3/3! + x^5/5! - x^7/7! + ...) = x + x^2 + x^3/3

So far my answer is right, but I can't get next term right. I'd rather not type out all the products...

(1 + x)^(1/2) ln (1 + x) = (1 - (1/2)x + (3/8)x^2 - (12/48)x^3 ) (x - x^2/2 + x^3/3 - x^4/4 + ...)

 

[galactus]

For the first one, rewrite as \(\displaystyle (2+x)^{\frac{-1}{2}}\)

a=2, b=x and n = -1/2.

Now, apply the binomial theorem by plugging in your values.

\(\displaystyle (a+b)^{n}=a^{n}+na^{n-1}b+\frac{n(n-1)}{2!}a^{n-2}b^{2}+.......+\frac{n(n-1)(n-2).....(n-k+1)}{k!}a^{n-k}b^{k}}\)\(\displaystyle +..........+nab^{n-1}+b^{n}\)

I know it looks menacing, but plug in your values for the first few to get several terms.

You can also take n derivatives using \(\displaystyle f^{n}(0)\)

 

[Deleted member 4993]

I suppose rewriting them as the following is a start.

1 / (2 + x)^ (1/2)


Ok. I got the second one.

1 / (1 - x^2)^ 2 = 1 + 2x^2 + 3x^3 + 4x^6 + ... <-- How did you get these terms

I have no idea how the book did the first one.

 

[warwick]

1 / (1 - x^2)^ 2 = 1 + 2x^2 + 3x^3 + 4x^6 + ... <-- How did you get these terms

u = -x^2 I plugged that into the appropriate binomial series.

 

[warwick]

For the first one, rewrite as \(\displaystyle (2+x)^{\frac{-1}{2}}\)

a=2, b=x and n = -1/2.

Now, apply the binomial theorem by plugging in your values.

\(\displaystyle (a+b)^{n}=a^{n}+na^{n-1}b+\frac{n(n-1)}{2!}a^{n-2}b^{2}+.......+\frac{n(n-1)(n-2).....(n-k+1)}{k!}a^{n-k}b^{k}}\)\(\displaystyle +..........+nab^{n-1}+b^{n}\)

I know it looks menacing, but plug in your values for the first few to get several terms.

You can also take n derivatives using \(\displaystyle f^{n}(0)\)

We didn't go over the binomial series in class. Binomial Series is in the book, but your formula is easier to apply They don't have a or b or n.