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[SOLVED] Taylor Series expansions

fionamb83

New member
May 4, 2012
4
Hi! I'm taking a course on Perturbation theory and as it's quite advanced the lecturer assumes everyone has a good level of maths. One of the parts is expanding roots of a quadratic equation about 0, I can understand how simple ones of the form $(1 + x)^2$ but I don't know where the answers are coming from for a couple I have here. This is the positive root of an equation:

$\frac{1}{2x}(1 + \sqrt{1 + 4x})$

The expansion for this is :

$1 + x + 2x^2 + 5x^3 + ...$

If anyone could explain this do me or do out the method for doing expansions of this type I would be forever grateful!!!
 
Last edited by a moderator:

Sudharaka

Well-known member
MHB Math Helper
Feb 5, 2012
1,621
Hi! I'm taking a course on Perturbation theory and as it's quite advanced the lecturer assumes everyone has a good level of maths. One of the parts is expanding roots of a quadratic equation about 0, I can understand how simple ones of the form \(\displaystyle $(1 + x)^2$\(\displaystyle but I don't know where the answers are coming from for a couple I have here. This is the positive root of an equation:

\(\displaystyle $\frac{1}{2x}(1 + \sqrt{1 + 4x})$ \(\displaystyle

The expansion for this is :

\(\displaystyle $1 + x + 2x^2 + 5x^3 + ...$ \(\displaystyle

If anyone could explain this do me or do out the method for doing expansions of this type I would be forever grateful!!!\)\)\)\)\)\)
\(\displaystyle \(\displaystyle \(\displaystyle \(\displaystyle \(\displaystyle \(\displaystyle

Hi fionamb83,

For the expansion of, \(\frac{1}{2x}(1 + \sqrt{1 + 4x})\) you can use the Taylor Series. Briefly put it states that, if \(f(x)\) is an infinitely differentiable function around a neighborhood of \(a\) then,

\[f(x)=\sum_{n=0} ^ {\infty} \frac {f^{(n)}(a)}{n!} \, (x-a)^{n}=f(a)+\frac {f'(a)}{1!} (x-a)+ \frac{f''(a)}{2!} (x-a)^2+\frac{f^{(3)}(a)}{3!}(x-a)^3+ \cdots\]

Try to use this to expand \(\sqrt{1 + 4x}\) around \(0\). First you have to find an expression for the derivatives of \(\sqrt{1 + 4x}\) at \(x=0\).\)\)\)\)\)\)
 

CaptainBlack

Well-known member
Jan 26, 2012
890
Hi! I'm taking a course on Perturbation theory and as it's quite advanced the lecturer assumes everyone has a good level of maths. One of the parts is expanding roots of a quadratic equation about 0, I can understand how simple ones of the form $(1 + x)^2$ but I don't know where the answers are coming from for a couple I have here. This is the positive root of an equation:

$\frac{1}{2x}(1 + \sqrt{1 + 4x})$

The expansion for this is :

$1 + x + 2x^2 + 5x^3 + ...$

If anyone could explain this do me or do out the method for doing expansions of this type I would be forever grateful!!!
You do not need \(\displaystyle tags here \$ or \$\$ work and are sufficient.

CB\)
 

CaptainBlack

Well-known member
Jan 26, 2012
890
Hi! I'm taking a course on Perturbation theory and as it's quite advanced the lecturer assumes everyone has a good level of maths. One of the parts is expanding roots of a quadratic equation about 0, I can understand how simple ones of the form $(1 + x)^2$ but I don't know where the answers are coming from for a couple I have here. This is the positive root of an equation:

$\frac{1}{2x}(1 + \sqrt{1 + 4x})$

The expansion for this is :

$1 + x + 2x^2 + 5x^3 + ...$

If anyone could explain this do me or do out the method for doing expansions of this type I would be forever grateful!!!
I think you will have to be more explicit about the equation that you are trying to get an expansion for the roots of.

Also the variable in the perturbation series should not be the same as the unknown in the equation but should be in terms of a parameter in the equation, should it not?

CB
 

Opalg

MHB Oldtimer
Staff member
Feb 7, 2012
2,702
$\frac{1}{2x}(1 + \sqrt{1 + 4x})$

The expansion for this is :

$1 + x + 2x^2 + 5x^3 + ...$
Please check whether you have written this correctly. In fact, $\frac{1}{2x}(1 + \sqrt{1 + 4x})$ goes to infinity as $x\to0$, so it cannot have a power series expansion at all.

For expressions like $\sqrt{1 + 4x}$, you should use the binomial expansion $$(1+t)^{1/2} = 1+\tfrac12t+\dfrac{\bigl(\frac12\bigr) \bigl(-\frac12\bigr)}{2!}t^2 + \dfrac{\bigl(\frac12\bigr) \bigl(-\frac12\bigr) \bigl(-\frac32\bigr)}{3!}t^3 + \ldots = 1+\tfrac12t -\tfrac18t^2 +\tfrac1{16}t^3 +\ldots,$$
substituting $t=4x$.
 

fionamb83

New member
May 4, 2012
4
Ok, sorry for the confusion. The actual example of this is found in Simon Mallams notes at the end of page 15, example 2.2, the second result if you google "simon malham introduction to asymptotic analysis" (sorry I can't post links yet).

The proper equation is $\epsilon x^2 + x - 1 = 0$ One of its roots I gave above (only I substituted x for $\epsilon$ because I was lazy I had been putting it in to wolfram alpha and matlab to try and understand what was going on). So modified (and fixing a slight mistake, I forgot a minus sign on the one) the positive root is $\frac{1}{2\epsilon}(-1 + \sqrt{1 + 4\epsilon})$

I tried the binomial expansion $$(1+t)^{1/2} = 1+\tfrac12t+\dfrac{\bigl(\frac12\bigr) \bigl(-\frac12\bigr)}{2!}t^2 + \dfrac{\bigl(\frac12\bigr) \bigl(-\frac12\bigr) \bigl(-\frac32\bigr)}{3!}t^3 + \ldots = 1+\tfrac12t -\tfrac18t^2 +\tfrac1{16}t^3 +\ldots,$$
substituting $t=4\epsilon$.

But the using that I am getting $1 + 2\epsilon - 2\epsilon^2 + 4\epsilon^3 + ...$
So I can't figure out where the answer $1 - {\epsilon} +2{\epsilon}^2 - 5{\epsilon}^3 + ...$ is coming from.

I'm sorry if all this seems a bit daft. I moved from physics to simulation science and they hadn't felt the need to teach us any maths other than O.D.E.'s and P.D.E.'s in my Maths Physics course, so I have big gaps in my knowledge unfortunately. It's just the expansion that's stumping me, which is really a major part if I'm going to be able to do it in my exam.

In the meantime I will try some other things. I just wanted to clarify the above.
 
Last edited:

Opalg

MHB Oldtimer
Staff member
Feb 7, 2012
2,702
So modified (and fixing a slight mistake, I forgot a minus sign on the one) the positive root is $\frac{1}{2\epsilon}(-1 + \sqrt{1 + 4\epsilon})$

I tried the binomial expansion $$(1+t)^{1/2} = 1+\tfrac12t+\dfrac{\bigl(\frac12\bigr) \bigl(-\frac12\bigr)}{2!}t^2 + \dfrac{\bigl(\frac12\bigr) \bigl(-\frac12\bigr) \bigl(-\frac32\bigr)}{3!}t^3 + \ldots = 1+\tfrac12t -\tfrac18t^2 +\tfrac1{16}t^3 +\ldots,$$
substituting $t=4\epsilon$.

But the using that I am getting $1 + 2\epsilon - 2\epsilon^2 + 4\epsilon^3 + ...$
So I can't figure out where the answer $1 - {\epsilon} +2{\epsilon}^2 - 5{\epsilon}^3 + ...$ is coming from.
You needed to take one more term in the binomial series: $$(1+t)^{1/2} = 1+\tfrac12t+\dfrac{\bigl(\frac12\bigr) \bigl(-\frac12\bigr)}{2!}t^2 + \dfrac{\bigl(\frac12\bigr) \bigl(-\frac12\bigr) \bigl(-\frac32\bigr)}{3!}t^3 + \dfrac{\bigl(\frac12\bigr) \bigl(-\frac12\bigr) \bigl(-\frac32\bigr) \bigl(-\frac52\bigr)}{4!}t^4 + \ldots = 1+\tfrac12t -\tfrac18t^2 +\tfrac1{16}t^3 -\tfrac5{128}t^4 +\ldots.$$ Now substitute $t=4\varepsilon$ to get $(1+4\varepsilon)^{1/2} = 1+2\varepsilon -2\varepsilon^2 + 4\varepsilon^3 -10\varepsilon^4 + \ldots.$ Finally (which you are forgetting to do) subtract 1 and divide by $2\varepsilon$ to get the approximation for $\frac{1}{2\epsilon}(-1 + \sqrt{1 + 4\epsilon}).$
 

fionamb83

New member
May 4, 2012
4
You needed to take one more term in the binomial series: $$(1+t)^{1/2} = 1+\tfrac12t+\dfrac{\bigl(\frac12\bigr) \bigl(-\frac12\bigr)}{2!}t^2 + \dfrac{\bigl(\frac12\bigr) \bigl(-\frac12\bigr) \bigl(-\frac32\bigr)}{3!}t^3 + \dfrac{\bigl(\frac12\bigr) \bigl(-\frac12\bigr) \bigl(-\frac32\bigr) \bigl(-\frac52\bigr)}{4!}t^4 + \ldots = 1+\tfrac12t -\tfrac18t^2 +\tfrac1{16}t^3 -\tfrac5{128}t^4 +\ldots.$$ Now substitute $t=4\varepsilon$ to get $(1+4\varepsilon)^{1/2} = 1+2\varepsilon -2\varepsilon^2 + 4\varepsilon^3 -10\varepsilon^4 + \ldots.$ Finally (which you are forgetting to do) subtract 1 and divide by $2\varepsilon$ to get the approximation for $\frac{1}{2\epsilon}(-1 + \sqrt{1 + 4\epsilon}).$

Ah right! It's clear to me now. Thanks, that's a great help!