Taylor Series for Square Root Function

[nuuskur]

Homework Statement

Expand $f(x) = \sqrt{2x+1}$ into a Taylor series around point $c=1$. Find the interval of convergence.

Homework Equations

The Attempt at a Solution

I do know that $f(x) = \sum\frac{1}{n!}f^{(n)}(c)(x-c)^n$ assuming the function is representable as a Taylor series. How do I figure out the series for this particular problem?
I have calculated some of the first derivatives of $f(1)$:
f(x) = 3^\frac{1}{2} + 3^{-\frac{1}{2}}(x-1) - \frac{1}{2!}3^{-\frac{3}{2}}(x-1)^2+\frac{1}{3!}3\cdot 3^{-\frac{5}{2}}+ ... = \sum\limits_{n=0}^\infty \frac{1}{n!} ? (x-1)^n
What do I do to come up with the general term for $?$

 

[Ray Vickson]

Homework Statement

Expand $f(x) = \sqrt{2x+1}$ into a Taylor series around point $c=1$. Find the interval of convergence.

Homework Equations

The Attempt at a Solution

I do know that $f(x) = \sum\frac{1}{n!}f^{(n)}(c)(x-c)^n$ assuming the function is representable as a Taylor series. How do I figure out the series for this particular problem?
I have calculated some of the first derivatives of $f(1)$:
f(x) = 3^\frac{1}{2} + 3^{-\frac{1}{2}}(x-1) - \frac{1}{2!}3^{-\frac{3}{2}}(x-1)^2+\frac{1}{3!}3\cdot 3^{-\frac{5}{2}}+ ... = \sum\limits_{n=0}^\infty \frac{1}{n!} ? (x-1)^n
What do I do to come up with the general term for $?$

I think the easiest way is to write $x = 1 + t/2$ and expand around $t = 0$.

 

[nuuskur]

I think the easiest way is to write $x = 1 + t/2$ and expand around $t = 0$.

What is the idea behind this substitution? What is your reason for picking so-and-so substitution? What are you trying to simplify?

 

[Ray Vickson]

What is the idea behind this substitution? What is your reason for picking so-and-so substitution? What are you trying to simplify?

The expansion of $(1+u)^p$ about $u = 0$ is well known, even if $p$ is fractional and/or negative. If you are not familiar with it, you should learn it; basically, it is the binomial expansion extended to arbitrary powers, with an appropriate definition of $C(p,k)$ ("p choose k") for integer $k \geq 0$ and arbitrary real $p \neq 0$ (Newton, circa 1665).

Of course, you could use $x = 1 + t$ instead of $x = 1 + t/2$. Try these and see what you get.

 

[nuuskur]

Well for $(1+x)^p = 1 + px +\frac{p(p-1)}{2!}x^2 + \frac{p(p-1)(p-2)}{3!}x^3 + ...$ Then for this problem it would become:

(1+2x)^{\frac{1}{2}} = 1 + \frac{1}{2}(2x) + \frac{1}{2!}\cdot\frac{1}{2}\cdot-\frac{1}{2}(2x)^2 + \frac{1}{3!}\cdot\frac{1}{2}\cdot -\frac{1}{2}\cdot-\frac{3}{2}(2x)^3 + \frac{1}{4!}\cdot\frac{1}{2}\cdot -\frac{1}{2}\cdot -\frac{3}{2}\cdot -\frac{5}{2}(2x)^4 + ... = 1 + x -\frac{1}{2}x^2 + \frac{1}{2}x^3 - \frac{5}{8}x^4
If I substitute $x = t+\frac{1}{2}$ then
\sqrt{2}(1+t)^{\frac{1}{2}} = \sqrt{2}\left (1 + \frac{1}{2}t -\frac{1}{8} + \frac{1}{16}t^3 -\frac{5}{128}t^4 + ...\right ) , |t|<1
up until the 5th summand everything looks ok and then I get a factor of $\frac{5}{128}$. How should I proceed?

I also tried $x = t+1$ the problem is the same, at the 5th summand the logic breaks down.

Can I say that if $|t|< 1$ then substituting back: $|x-\frac{1}{2}|< 1$?

 

[Noctisdark]

(1+t)^p , expand that then replace p by 1/2 and t = 2x
(1+t)^p = Σ(p)n*(t^n)/n! Where (p)n is the falling factorial so just change t with 2x
(P)n = p(p-1)(p-2)...

 

[nuuskur]

Can you just write the series as (1+2x)^{\frac{1}{2}} = \sum_{n=0}^\infty \frac{P(n)}{n!}(2x)^n

 

[Noctisdark]

Can you just write the series as (1+2x)^{\frac{1}{2}} = \sum_{n=0}^\infty \frac{P(n)}{n!}(2x)^n

(P)n is a falling factorial, (P)n = P(P-1)(P-2) ... for more on (P)n
http://mathworld.wolfram.com/FallingFactorial.html

 

[nuuskur]

(1+2x)^{\frac{1}{2}} = \sum_{n=0}^\infty \frac{(P)_{\frac{1}{2}}}{n!}(2x)^n? Does the index represent the power?

Anyhow, if $|t|<1$, this is around point $c=0$. If I substitute back $|x-\frac{1}{2}|<1$ would this be around point $c=1$?
If so then \begin{cases}x&gt;-\frac{1}{2}, &amp;-1&lt; x-\frac{1}{2}&lt;0\\x&lt;\frac{3}{2}, &amp;0\leq x-\frac{1}{2}&lt;1 \end{cases}

 

[Noctisdark]

In your case p = 1/2 so (1/2)n = ((-1)^(n+1))*(2n-1)!/2^n for n>1 when n = 0 (1/2)n = 1, i think you can expand it now,1+ Σ[((-1)^n+1)*(2n-1)!*x^n]/n!

 

[Noctisdark]

(1+2x)^{\frac{1}{2}} = \sum_{n=0}^\infty \frac{(P)_{\frac{1}{2}}}{n!}(2x)^n? Does the index represent the power?

Anyhow, if $|t|<1$, this is around point $c=0$. If I substitute back $|x-\frac{1}{2}|<1$ would this be around point $c=1$?
If so then \begin{cases}x&gt;-\frac{1}{2}, &amp;-1\leq x-\frac{1}{2}&lt;0\\x&lt;\frac{3}{2}, &amp;0\leq x-\frac{1}{2}&lt;1 \end{cases}

Read wha you've written !
Can absolute value be less than zero ?
Edit, if you have evaluated (p)n you'd find that -1<=x<=1 because the 2^n goes by the falling factorial, so -1/2 <=x<=1

 

[nuuskur]

I am attempting to read what you have written :D (highly recommend TeXing mathematical text, I put more emphasis in deciphering your post compared to actually focusing on its meaning in the context of the current problem)
EDIT: nonsensical

 

[Ray Vickson]

I am attempting to read what you have written :D (highly recommend TeXing mathematical text, I put more emphasis in deciphering your post compared to actually focusing on its meaning in the context of the current problem)
If $p=\frac{1}{2}$ then it would be (1+2x)^{\frac{1}{2}} = 1 + \sum_{n=1}^\infty \frac{(-1)^{n+1}(2n-1)!}{n!}x^n

Unless I misunderstood your original question, I thought you were asked to find the expansion of $f(x) = \sqrt{2x+1}$ about the point $x = 1$, which means you want a series involving $(x-1)^n$, not $x^n$. That is the reason I suggested writing $x = 1 + t$, so $f(x) = \sqrt{3+2t}$. You want the expansion of that in $t^n$ terms.

 

[nuuskur]

Right you are, I am getting side-tracked and confused :S.
If $x = t+1$ then \sqrt{3}(1+\frac{2}{3}t)^{\frac{1}{2}} = 1 + \left [\frac{1}{2}\left (\frac{2}{3}t\right )\right ] + \left [\frac{1}{2!}\frac{1}{2}\left (-\frac{1}{2}\right )\left (\frac{2}{3}t\right )^2\right ] + \left [\frac{1}{3!}\frac{1}{2}\left (-\frac{1}{2}\right )\left (-\frac{3}{2}\right )\left (\frac{2}{3}t\right )^3\right ] + \left [\frac{1}{4!}\frac{1}{2}\left (-\frac{1}{2}\right )\left (-\frac{3}{2}\right )\left (-\frac{5}{2}\right )\left (\frac{2}{3}t\right )^4\right ]

However, to come up with the general term. I understand I can express this in the form $1 + \sum_{n=1}^\infty(-1)^{n+1} \frac{1}{n!} \left (\frac{2}{3}\right )^nt^n$. Problem is what comes in between there? The falling factorial seems viable, but it isn't mentioned in my textbook, I suspect I am expected to solve this problem differently.

 

[Ray Vickson]

Right you are, I am getting side-tracked and confused :S.
If $x = t+1$ then \sqrt{3}(1+\frac{2}{3}t)^{\frac{1}{2}} = 1 + \left [\frac{1}{2}\left (\frac{2}{3}t\right )\right ] + \left [\frac{1}{2!}\frac{1}{2}\left (-\frac{1}{2}\right )\left (\frac{2}{3}t\right )^2\right ] + \left [\frac{1}{3!}\frac{1}{2}\left (-\frac{1}{2}\right )\left (-\frac{3}{2}\right )\left (\frac{2}{3}t\right )^3\right ] + \left [\frac{1}{4!}\frac{1}{2}\left (-\frac{1}{2}\right )\left (-\frac{3}{2}\right )\left (-\frac{5}{2}\right )\left (\frac{2}{3}t\right )^4\right ]

However, to come up with the general term. I understand I can express this in the form $1 + \sum_{n=1}^\infty(-1)^{n+1} \frac{1}{n!} \left (\frac{2}{3}\right )^nt^n$. Problem is what comes in between there? The falling factorial seems viable, but it isn't mentioned in my textbook, I suspect I am expected to solve this problem differently.

Are you not allowed to go to the library and look things up, learning about and using tools not in your textbook? Back when I was attending classes that would have been praiseworthy behavior, unless specifically forbidden for some reason----and that never happened. The only requirement was to cite sources. Of course, nowadays going on-line replaces a walk to the library.

Alternatively, you could just go ahead and compute $\left. d^n f(x)/dx^n \right|_{x=1}$ to get the Taylor expansion coefficients directly.

 

[vela]

If $x = t+1$ then \sqrt{3}(1+\frac{2}{3}t)^{\frac{1}{2}} = 1 + \left [\frac{1}{2}\left (\frac{2}{3}t\right )\right ] + \left [\frac{1}{2!}\frac{1}{2}\left (-\frac{1}{2}\right )\left (\frac{2}{3}t\right )^2\right ] + \left [\frac{1}{3!}\frac{1}{2}\left (-\frac{1}{2}\right )\left (-\frac{3}{2}\right )\left (\frac{2}{3}t\right )^3\right ] + \left [\frac{1}{4!}\frac{1}{2}\left (-\frac{1}{2}\right )\left (-\frac{3}{2}\right )\left (-\frac{5}{2}\right )\left (\frac{2}{3}t\right )^4\right ]

However, to come up with the general term. I understand I can express this in the form $1 + \sum_{n=1}^\infty(-1)^{n+1} \frac{1}{n!} \left (\frac{2}{3}\right )^nt^n$.

The second form you wrote isn't correct. Perhaps that's why it's not making sense to you.

Problem is what comes in between there? The falling factorial seems viable, but it isn't mentioned in my textbook, I suspect I am expected to solve this problem differently.

Forget about the falling factorial. Just look for patterns and simplify. For example, how many powers of 2 do you get in each term in the denominator? How many in the numerator? Do they follow a predictable pattern so you can cancel them out? What about the sign of each term?

 

[nuuskur]

Yes, the falling factorial is forgotten. This is what I have come up with.

1 + \frac{1}{1!}\frac{1}{3^1}t^1 -\frac{1}{2!}\frac{1}{3^2}t^2 + \frac{1}{3!}\frac{3}{3^3}t^3 - \frac{1}{4!}\frac{3\cdot 5}{3^4}t^4 + \frac{1}{5!}\frac{3\cdot 5\cdot 7}{3^5}t^5
I know what the pattern is, but I don't know how to express the $3\cdot 5\cdot 7\cdot 9 ...$ as a general term.

 

[vela]

Notation-wise, that's sometimes written using the double-factorial, e.g., $7! = 1\cdot 3\cdot 5\cdot 7$. If you define (-1)! = 1, you can write your series as
$$1 + \sum_{k=1}^\infty (-1)^{k+1}\frac{(2k-3)!}{k!3^k} t^k.$$ That's just notation though.

Another way, which is perhaps more useful to know, is to use the fact that, for example,
$$ 1\cdot 3\cdot 5\cdot 7 = \frac{1 \cdot 2 \cdot 3 \cdot 4 \cdot 5 \cdot 6 \cdot 7}{2 \cdot 4 \cdot 6} = \frac{7!}{2^3 3!}$$

 

[nuuskur]

Notation-wise, that's sometimes written using the double-factorial, e.g., $7! = 1\cdot 3\cdot 5\cdot 7$. If you define (-1)! = 1, you can write your series as
$$1 + \sum_{k=1}^\infty (-1)^{k+1}\frac{(2k-3)!}{k!3^k} t^k.$$ That's just notation though.

Another way, which is perhaps more useful to know, is to use the fact that, for example,
$$ 1\cdot 3\cdot 5\cdot 7 = \frac{1 \cdot 2 \cdot 3 \cdot 4 \cdot 5 \cdot 6 \cdot 7}{2 \cdot 4 \cdot 6} = \frac{7!}{2^3 3!}$$

Yes, I was experimenting around with that, too. I would like to avoid unnecessarely adding new concepts to the solution. Thank you for pointing me in the right direction.