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Taylor series at x = 1: arctan(x)

sweatingbear

Member
May 3, 2013
91
Hey forum. Is there any way one can take advantage of the Maclaurin series of \(\displaystyle \arctan (x)\) to obtain the Taylor series of \(\displaystyle \arctan (x)\) at \(\displaystyle x = 1\)? I attempted to obtain the series in the suggested manner but to no avail.

We have

\(\displaystyle \arctan (x) = \sum_{n=0}^\infty \frac { f^{(n)}(1) }{n!} (x - 1)^n \, . \)

Let \(\displaystyle z = x - 1 \iff x = z + 1\). Thus

\(\displaystyle \arctan (z + 1) = \sum_{n=0}^\infty \frac { f^{(n)}(0) }{n!} z^n \, . \)

We are taking the derivatives at \(\displaystyle z = 0\) in the sum in the right-hand side, because \(\displaystyle z = 0\) when \(\displaystyle x = 1\). But it seems we cannot use the Maclaurin series for \(\displaystyle \arctan (z + 1)\) since the argument is not around \(\displaystyle 0\) when \(\displaystyle z\) is around \(\displaystyle 0\).

It seems as if I will have to resort to brute-force derivative computations?
 

ZaidAlyafey

Well-known member
MHB Math Helper
Jan 17, 2013
1,667
How did \(\displaystyle f^{(n)}(1) \) changed to \(\displaystyle f^{(n)}(0)\) by substitution ?
 

ZaidAlyafey

Well-known member
MHB Math Helper
Jan 17, 2013
1,667
Let us start by the following

\(\displaystyle \frac{1}{1+(x-1)^2} = \sum^{\infty}_{n=0} (-1)^n(x-1)^{2n}\) converges \(\displaystyle |x-1|<1 \)

\(\displaystyle \arctan(x-1) = \sum^{\infty}_{n=0} \frac{(-1)^n (x-1)^{2n+1}}{2n+1}\) converges on the same disk .

[HR][/HR]

\(\displaystyle \frac{1}{1+x^2} = \sum^{\infty}_{n=0} (-1)^n x^{2n}\) converges \(\displaystyle |x|<1\)

\(\displaystyle \arctan(x) = \sum^{\infty}_{n=0} \frac{(-1)^n x^{2n+1}}{2n+1}\) converges on the same disk
 

sweatingbear

Member
May 3, 2013
91
How did \(\displaystyle f^{(n)}(1) \) changed to \(\displaystyle f^{(n)}(0)\) by substitution ?
Does not the point of differentiation change when you introduce \(\displaystyle z\)? We were first wanting to take derivatives at \(\displaystyle x = 1\). But after the variable substitution we have a polynomial of \(\displaystyle z\) and when \(\displaystyle x = 1\), \(\displaystyle z\) is equal to \(\displaystyle 0\). Hence the point of differentiation must be zero and not one? Isn't this really the whole reason behind why we are able to manipulate Maclaurin series for Taylor series about expansion points other than zero. I am not entirely confident though, correct me if I am wrong.
 

ZaidAlyafey

Well-known member
MHB Math Helper
Jan 17, 2013
1,667
If we started with the following :

\(\displaystyle f(z)= \arctan (z) = \sum_{n=0}^\infty \frac { f^{(n)}(1) }{n!} (z - 1)^n \, . \) around 1

\(\displaystyle g(z) = \arctan (z + 1) = \sum_{n=0}^\infty \frac { g^{(n)}(0) }{n!} z^n \, . \) around 0

but since \(\displaystyle g(z) = f(z+1) \)

\(\displaystyle \arctan (z + 1) = \sum_{n=0}^\infty \frac { f^{(n)}(1) }{n!} z^n \, . \) around 0
 

sweatingbear

Member
May 3, 2013
91
If we started with the following :

\(\displaystyle f(z)= \arctan (z) = \sum_{n=0}^\infty \frac { f^{(n)}(1) }{n!} (z - 1)^n \, . \) around 1

\(\displaystyle g(z) = \arctan (z + 1) = \sum_{n=0}^\infty \frac { g^{(n)}(0) }{n!} z^n \, . \) around 0
Hm, I am not quite sure I understand how it became an approximation around 0 just by increasing the argument by one. How we are suddenly taking the derivative at 1? Could you please elaborate?
 

Klaas van Aarsen

MHB Seeker
Staff member
Mar 5, 2012
8,864
I believe you have to brute-force the evaluations of $\arctan^{(n)}(1)$ to get your Taylor expansion.
 

sweatingbear

Member
May 3, 2013
91
I believe you have to brute-force the evaluations of $\arctan^{(n)}(1)$ to get your Taylor expansion.
Hm, all right thanks. But I would really appreciate it if somebody could help me understand this better:

Does not the point of differentiation change when you introduce \(\displaystyle z\)? We were first wanting to take derivatives at \(\displaystyle x = 1\). But after the variable substitution we have a polynomial of \(\displaystyle z\) and when \(\displaystyle x = 1\), \(\displaystyle z\) is equal to \(\displaystyle 0\). Hence the point of differentiation must be zero and not one? Isn't this really the whole reason behind why we are able to manipulate Maclaurin series for Taylor series about expansion points other than zero. I am not entirely confident though, correct me if I am wrong.
 

Klaas van Aarsen

MHB Seeker
Staff member
Mar 5, 2012
8,864
Does not the point of differentiation change when you introduce \(\displaystyle z\)? We were first wanting to take derivatives at \(\displaystyle x = 1\). But after the variable substitution we have a polynomial of \(\displaystyle z\) and when \(\displaystyle x = 1\), \(\displaystyle z\) is equal to \(\displaystyle 0\). Hence the point of differentiation must be zero and not one? Isn't this really the whole reason behind why we are able to manipulate Maclaurin series for Taylor series about expansion points other than zero. I am not entirely confident though, correct me if I am wrong.
Yes. The point of differentiation changes.

You're talking a Taylor expansion of either $\arctan(x)$ around $x=1$, or of $\arctan(1+z)$ around $z=0$.
But I'm afraid that in both cases the Taylor expansion requires you to evaluate $\arctan^{(n)}(1)$.
 

sweatingbear

Member
May 3, 2013
91
Yes. The point of differentiation changes.
Ok thank you! So would you say my line of argument is valid here:

We have

\(\displaystyle \arctan (x) = \sum_{n=0}^\infty \frac { f^{(n)}(1) }{n!} (x - 1)^n \, . \)

Let \(\displaystyle z = x - 1 \iff x = z + 1\). Thus

\(\displaystyle \arctan (z + 1) = \sum_{n=0}^\infty \frac { f^{(n)}(0) }{n!} z^n \, . \)

We are taking the derivatives at \(\displaystyle z = 0\) in the sum in the right-hand side, because \(\displaystyle z = 0\) when \(\displaystyle x = 1\).
Namely that I shifted the \(\displaystyle n\):th derivative from \(\displaystyle f^{(n)}(1)\) to \(\displaystyle f^{(n)}(0) \)? Zaid seemingly objected this procedure, so I am a tad bit dubious...

____________________________________

You're talking a Taylor expansion of either $\arctan(x)$ around $x=1$, or of $\arctan(1+z)$ around $z=0$.
But I'm afraid that in both cases the Taylor expansion requires you to evaluate $\arctan^{(n)}(1)$.
All right, derivatives it is. Thank you!
 
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Klaas van Aarsen

MHB Seeker
Staff member
Mar 5, 2012
8,864
Ok thank you! So I my line of argument is valid here:
Erm... there's a small inconsistency there that is confusing.

You have the function $f$ in both formulas.
But $f$ also changes.
In the first formula you have $f(x)=\arctan(x)$, while in the second formula you should have $f^*(z)=\arctan(1+z)$.
Note that I have added a star * to indicate it's a different function.

In particular ${f^*}^{(n)}(0)=\arctan^{(n)}(1+0)=\arctan^{(n)}(1)$.
 

sweatingbear

Member
May 3, 2013
91
You have the function $f$ in both formulas.
But $f$ also changes.
In the first formula you have $f(x)=\arctan(x)$, while in the second formula you should have $f^*(z)=\arctan(1+z)$.
Note that I have added a star * to indicate it's a different function.

In particular ${f^*}^{(n)}(0)=\arctan^{(n)}(1+0)=\arctan^{(n)}(1)$.
All right, but look at Zaid's post:

If we started with the following :

\(\displaystyle f(z)= \arctan (z) = \sum_{n=0}^\infty \frac { f^{(n)}(1) }{n!} (z - 1)^n \, . \) around 1

\(\displaystyle g(z) = \arctan (z + 1) = \sum_{n=0}^\infty \frac { g^{(n)}(0) }{n!} z^n \, . \) around 0

but since \(\displaystyle g(z) = f(z+1) \)

\(\displaystyle \arctan (z + 1) = \sum_{n=0}^\infty \frac { f^{(n)}(1) }{n!} z^n \, . \) around 0
He seemingly disagrees. What is going on?
 

Klaas van Aarsen

MHB Seeker
Staff member
Mar 5, 2012
8,864
All right, but look at Zaid's post:

He seemingly disagrees. What is going on?
It's the same thing.
Zaid name the other function $g$, where I named it $f^*$.
He introduced $g$ to write the Taylor expansion around z=0.
Then he showed that you still have to evaluate the $\arctan$ derivatives at 1.
 

sweatingbear

Member
May 3, 2013
91
It's the same thing.
Zaid name the other function $g$, where I named it $f^*$.
He introduced $g$ to write the Taylor expansion around z=0.
Then he showed that you still have to evaluate the $\arctan$ derivatives at 1.
Hm, all right. Great, good to know that I was thinking correct in terms of changing the point of differentation.

How did \(\displaystyle f^{(n)}(1) \) changed to \(\displaystyle f^{(n)}(0)\) by substitution ?
ZaidAlyafey, is it wrong to do that according to you or were you thinking of something else? I would really like to know what you had in mind!

Let us start by the following

\(\displaystyle \frac{1}{1+(x-1)^2} = \sum^{\infty}_{n=0} (-1)^n(x-1)^{2n}\) converges \(\displaystyle |x-1|<1 \)

\(\displaystyle \arctan(x-1) = \sum^{\infty}_{n=0} \frac{(-1)^n (x-1)^{2n+1}}{2n+1}\) converges on the same disk .

[HR][/HR]

\(\displaystyle \frac{1}{1+x^2} = \sum^{\infty}_{n=0} (-1)^n x^{2n}\) converges \(\displaystyle |x|<1\)

\(\displaystyle \arctan(x) = \sum^{\infty}_{n=0} \frac{(-1)^n x^{2n+1}}{2n+1}\) converges on the same disk
I understand where every series comes from, but I am afraid I don't understand what you wanted to put across. Could you please elaborate?
 

sweatingbear

Member
May 3, 2013
91
Would really appreciate it if somebody could help me see things clearer!
 

ZaidAlyafey

Well-known member
MHB Math Helper
Jan 17, 2013
1,667
Let us start by the following

\(\displaystyle \frac{1}{1+(x-1)^2} = \sum^{\infty}_{n=0} (-1)^n(x-1)^{2n}\) converges \(\displaystyle |x-1|<1 \)

\(\displaystyle \arctan(x-1) = \sum^{\infty}_{n=0} \frac{(-1)^n (x-1)^{2n+1}}{2n+1}\) converges on the same disk .

[HR][/HR]

\(\displaystyle \frac{1}{1+x^2} = \sum^{\infty}_{n=0} (-1)^n x^{2n}\) converges \(\displaystyle |x|<1\)

\(\displaystyle \arctan(x) = \sum^{\infty}_{n=0} \frac{(-1)^n x^{2n+1}}{2n+1}\) converges on the same disk
what is the difference between the expansion of \(\displaystyle \arctan(x)\) and \(\displaystyle \arctan(x-1)\) ?
 

sweatingbear

Member
May 3, 2013
91
what is the difference between the expansion of \(\displaystyle \arctan(x)\) and \(\displaystyle \arctan(x-1)\) ?
The first one is about \(\displaystyle x = 0\) and the other one about \(\displaystyle x=1\), right?
 

ZaidAlyafey

Well-known member
MHB Math Helper
Jan 17, 2013
1,667
The first one is about \(\displaystyle x = 0\) and the other one about \(\displaystyle x=1\), right?
I meant in the expansion , you realize that we only needed to put \(\displaystyle x-1\) instead of \(\displaystyle x\) to find the expansion of \(\displaystyle \arctan(x-1)\) using \(\displaystyle \arctan(x)\) but the coefficients are still the same right ?
 

sweatingbear

Member
May 3, 2013
91
I meant in the expansion , you realize that we only needed to put \(\displaystyle x-1\) instead of \(\displaystyle x\) to find the expansion of \(\displaystyle \arctan(x-1)\) using \(\displaystyle \arctan(x)\) but the coefficients are still the same right ?
Yes the coefficients remain the same but simply changing the argument from \(\displaystyle x\) to \(\displaystyle x-1\) didn't really change the point of expansion, right? About \(\displaystyle x = 0\), we have

\(\displaystyle \arctan (x) = \sum_{n=0}^\infty (-1)^n \frac{x^{2n+1}}{2n+1} \, .\)

If we suddenly change the argument to \(\displaystyle x-1\), we get

\(\displaystyle \arctan (x - 1) = \sum_{n=0}^\infty (-1)^n \frac{(x-1)^{2n+1}}{2n+1} \, ,\)

but we are still expanding about \(\displaystyle x = 0\), not \(\displaystyle x = 1\). The problem is that we want an expansion about \(\displaystyle x=1\), hence my confusion.
 

Opalg

MHB Oldtimer
Staff member
Feb 7, 2012
2,725
Hey forum. Is there any way one can take advantage of the Maclaurin series of \(\displaystyle \arctan (x)\) to obtain the Taylor series of \(\displaystyle \arctan (x)\) at \(\displaystyle x = 1\)?
The blunt answer is NO. For any sufficiently smooth function $f$, the Taylor series for $f$ at the point $a$ is given by \(\displaystyle f(a+x) = \sum_{n=0}^\infty \frac{f^{(n)}(a)}{n!}x^n.\) In particular, the first term in the series, the constant term, is $f(a)$. For the function $f(x) = \arctan x$ and the point $a=1$, the series starts with the constant term $\arctan(1) = \pi/4$. So any answer that does not start with the constant term $\pi/4$ must be wrong. Of course, the Maclaurin series for $\arctan x$ does not contain any terms involving $\pi$, so you cannot expect to get the Taylor series at $x=1$ from that.

What we want to do here is to find a power series for $\arctan(1+x)$. One way to try to do that would be to notice that the first derivative of $\arctan(1+x)$ is \(\displaystyle \frac1{1+(1+x)^2}.\) So a good strategy would be to try to find a power series for that function and then integrate it term by term, adding the constant term $\pi/4$ at the start. That way, you can at least find the first few terms of the series: $$ \frac1{1+(1+x)^2} = \frac1{2+2x+x^2} = \frac12\,\frac1{1 + x\bigl(1+\frac x2\bigr)}.$$ Now use the series $(1+t)^{-1} = 1-t +t^2-t^3 + \ldots$, with $t = x\bigl(1+\frac x2\bigr)$, to get $$ \frac12\,\frac1{1 + x\bigl(1+\frac x2\bigr)} = \frac12\Bigl(1 - x\bigl(1+\tfrac x2\bigr) + x^2\bigl(1+\tfrac x2\bigr)^2 - x^3\bigl(1+\tfrac x2\bigr)^3 + \ldots\Bigr).$$ You can pick out the coefficients of the first few powers of $x$ to get a series starting $\frac12 - \frac12x + \frac14x^2 +\ldots$. (I think that the next term is $0$ but I haven't checked that.) Now integrate each term, add on the $\pi/4$ at the start, and you have the beginnings of the Taylor series for $\arctan x$ around $x=1$. But I don't believe that there is any simple formula for the general term of the series.
 

ZaidAlyafey

Well-known member
MHB Math Helper
Jan 17, 2013
1,667
Yes the coefficients remain the same but simply changing the argument from \(\displaystyle x\) to \(\displaystyle x-1\) didn't really change the point of expansion, right? About \(\displaystyle x = 0\), we have

\(\displaystyle \arctan (x) = \sum_{n=0}^\infty (-1)^n \frac{x^{2n+1}}{2n+1} \, .\)

If we suddenly change the argument to \(\displaystyle x-1\), we get

\(\displaystyle \arctan (x - 1) = \sum_{n=0}^\infty (-1)^n \frac{(x-1)^{2n+1}}{2n+1} \, ,\)

but we are still expanding about \(\displaystyle x = 0\), not \(\displaystyle x = 1\). The problem is that we want an expansion about \(\displaystyle x=1\), hence my confusion.
Exactly . I just tried to point out that the substitution you did was not correct in the first post. It is not easy to find the Taylor expansion using the Maclaurin By substitution because you are transforming the function into different function .
 

sweatingbear

Member
May 3, 2013
91
Thanks a bunch for the answers. I understand now that one cannot take advantage of the Maclaurin series of \(\displaystyle \arctan(x)\) in order to find the Taylor series about \(\displaystyle x=1\) for the function.

I will start a new thread regarding a question the discussion in this thread gave rise to (in order to avoid serious digressions).
 
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