Taylor-MacLaurin series for (e^(-sqrt[x]) - 1) / sqrt[x]

[tonycashflow]

I have to find the MacLaurin series for the following function

.

I would like to have the solution for this problem and of course the answer. I would also like someone to give me the step to do when you want the MacLaurin series for a given function f(x) because I am having a hard time understanding this. That would be much appreciated.

Thanks

 

[tkhunny]

Re: Taylor-MacLaurin series

Try an antiderivative. That might lead to a simpler expression. The term-wise derivative could be what you seek.

or...

Try \(\displaystyle e^{x}\)
After that, \(\displaystyle e^{-x}\) (Change the sign on all the odd exponents.)
After that, \(\displaystyle e^{-\sqrt{x}}\) (Cut all the exponents in half.)
After that,...

While doing either of these, yoiu should at least think about Uniform Convergence.

 

[soroban]

Re: Taylor-MacLaurin series

Hello, tonycashflow!

\(\displaystyle \text{Find the MacLaurin series for: }\;f(x) \:=\:\frac{e^{\text{-}\sqrt{x}}-1}{\sqrt{x}}\)

\(\displaystyle \text{We know that: }\;e^z \;=\;1 + z + \frac{z^2}{2!} + \frac{z^3}{3!} + \frac{z^4}{4!} + \frac{z^5}{5!} + \hdots\)


\(\displaystyle \text{Let }z \,=\,\text{-}x^{\frac{1}{2}}\)
. . \(\displaystyle e^{\text{-}x^{\frac{1}{2}}} \;=\;1 + \left(\text{-}x^{\frac{1}{2}}\right) + \frac{\left(\text{-}x^{\frac{1}{2}}\right)^2}{2!} +\frac{\left(\text{-}x^{\frac{1}{2}}\right)^3}{3!} +\frac{\left(\text{-}x^{\frac{1}{2}}\right)^4}{4!} + \frac{\left(\text{-}x^{\frac{1}{2}}\right)^5}{5!} + \hdots\)

. . \(\displaystyle e^{\text{-}\sqrt{x}}\;= \;1 - x^{\frac{1}{2}} + \frac{x}{2!} - \frac{x^{\frac{3}{2}}}{3!} + \frac{x^3}{4!} - \frac{x^{\frac{5}{2}}}{5!} + \hdots\)


\(\displaystyle \text{Then: }\;e^{\text{-}\sqrt{x}} - 1 \;=\;-x^{\frac{1}{2}} + \frac{x}{2!} - \frac{x^{\frac{3}{2}}}{3!} + \frac{x^2}{4!} - \frac{x^{\frac{5}{2}}}{5!} + \hdots\)


\(\displaystyle \text{Divide by }\sqrt{x}\!:\quad\frac{e^{\text{-}\sqrt{x}} - 1}{\sqrt{x}} \;=\;\frac{\text{-}x^{\frac{1}{2}} + \frac{x}{2!} - \frac{x^{\frac{3}{2}}}{3!} + \frac{x^2}{4!} - \frac{x^{\frac{5}{2}}}{5!} + \hdots}{x^{\frac{1}{2}}}\)


\(\displaystyle \text{Therefore: }\;\frac{e^{\text{-}\sqrt{x}}-1}{\sqrt{x}} \;=\;-1 + \frac{x^{\frac{1}{2}}}{2!} - \frac{x}{3!} + \frac{x^{\frac{3}{2}}}{4!} - \frac{x^2}{5!} + \hdots + (-1)^n\frac{x^{\frac{n-1}{2}}}{n^!} + \hdots\)

 

[tonycashflow]

Re: Taylor-MacLaurin series

So what I understand from your solution, is that, in most cases, you need to know the 'basic' MacLaurin series (e^x, 1/(1-x), ln(1+x), etc.) and then 'adjust' them so they work for your function. Is that right ?

By the way thanks a lot to both of you.

 

[tonycashflow]

Re: Taylor-MacLaurin series

I just made another problem and I want to know if I am on the right track.

I had to find the MacLaurin series for f(x)=1/(x+x^3)

Then, (1/x) * 1/(1+x^2)

I then put it into this form : 1/(1-(-x^2))
to have the basic form of 1/(1-x) which MacLaurin series is (x^n)

I then get this : -x^2n

I multiply it by 1/x which equals to x^-1 so then i get the MacLaurin series for my function as this:

-x^(2n-1)

Is that correct ?

 

[skeeter]

Re: Taylor-MacLaurin series

\(\displaystyle \frac{1}{1-(-x^2)} = 1 - x^2 + x^4 - x^6 + x^8 - x^{10} + ...\)

\(\displaystyle \frac{1}{x} \cdot \frac{1}{1 - (-x^2)} = x^{-1} - x + x^3 - x^5 + x^7 - x^9 + ... = \sum_{n=0}^{\infty} (-1)^n x^{2n-1}\)

 

[tonycashflow]

Ok thanks, I should have checked the 1/(1+x) form...

But one last thing about MacLaurin series :

when you have this : 1/(1+x)^2, what do you do to get it to a 'standard' form, if you can't, how do you get a MacLaurin series from this ?

Thanks

 

[skeeter]

this one you have to "build" from scratch to see the pattern ...

\(\displaystyle f(x) = \frac{1}{(x+1)^2}\) ... \(\displaystyle f(0) = 1\)

\(\displaystyle f'(x) = \frac{-2}{(x+1)^3}\) ... \(\displaystyle f'(0) = -2\)

\(\displaystyle f''(x) = \frac{6}{(x+1)^4}\) ... \(\displaystyle f''(0) = 6\)

\(\displaystyle f'''(x) = \frac{-24}{(x+1)^5}\) ... \(\displaystyle f''(0) = -24\)

\(\displaystyle f(x) = 1 - \frac{2x}{1!} + \frac{6x^2}{2!} - \frac{24x^3}{3!} + ...\)

\(\displaystyle f(x) = 1 - 2x + 3x^2 - 4x^3 + 5x^4 - 6x^5 + ... = \sum_{n=0}^{\infty} (-1)^n(n+1)x^n\)

or ... you can look at it as the product of

\(\displaystyle \frac{1}{1 - (-x)} \cdot \frac{1}{1 - (-x)} = (1 - x + x^2 - x^3 + x^4 - x^5 + ...)(1 - x + x^2 - x^3 + x^4 - x^5 + ...) =\)

stack 'em and multiply ... you'll see the product comes out as

\(\displaystyle 1 - 2x + 3x^2 - 4x^3 + 5x^4 - 6x^5 + ...\)

 

[tkhunny]

...or

\(\displaystyle \int\frac{1}{(1+x)^2}dx = -\frac{1}{1+x} + C = -\frac{1}{1-(-x)} + C = -(1-x+x^{2}-x^{3}+...) + C\)

Find the term-wise first derivative of that last expression.

Are you THINKING about Uniform Convergence?

 

[tonycashflow]

Are you THINKING about Uniform Convergence?

No sorry I have not learned that in my course. But thanks for your answer.

 

[tkhunny]

No sorry I have not learned that in my course.

Tsk, tsk, tsk...

Generally, one does not want to be in the habit of just doing things willy-nilly. Feel free to ask your teacher or professor or course leader if the things suggested here are ALWAYS valid. It is such a marvelously useful process that it is almost magic. I should hope that we could prove that it is valid.