Taylor series of 1/ln(t+1) at t=0

[laurabon]

I have tried to use the natural Taylor expansion of ln(t+1)
and working with long division but I get the result 1/t+1/2+t/12−t^2/24+⋯
instead of 1/t+1/2-t/12+t^2/24+... that is the right result
I have tried to do this several time but still don't works. Is there a miscalculation in my long division?

 

[fresh_42]

I have tried to use the natural Taylor expansion of ln(t+1)
and working with long division but I get the result 1/t+1/2+t/12−t^2/24+⋯
instead of 1/t+1/2-t/12+t^2/24+... that is the right result
I have tried to do this several time but still don't works. Is there a miscalculation in my long division?

Hard to tell without seeing your calculation.

The Taylor series at a point $t_0=0$ is
$$
f(t)=\sum_{n=0}^\infty \dfrac{(t-t_0)^n}{n!} \left.\dfrac{d^{n}}{dt^n}\right|_{t_0}f(t)=\sum_{n=0}^\infty\dfrac{t^n}{n!}f^{(n)}(0)
$$
so we need primarily the derivatives of $f(t)=\log(t+1).$
\begin{align*}
f^{(0)}(t)&=\log(t+1)&\quad f^{(0)}(0)=\log(1)=0\\
f^{(1)}(t)&=\dfrac{1}{t+1}&\quad f^{(1)}(0)=\dfrac{1}{0+1}=1\\
f^{(2)}(t)&=-\dfrac{1}{(t+1)^2}&\quad f^{(2)}(0)=-\dfrac{1}{(0+1)^2}=-1\\
f^{(3)}(t)&=\dfrac{2}{(t+1)^3}&\quad f^{(3)}(0)=\dfrac{2}{(0+1)^3}=2\\
&\ldots &\ldots
\end{align*}
Now go on as long as you like and gather what you have.

 

[laurabon]

Hard to tell without seeing your calculation.

The Taylor series at a point $t_0=0$ is
$$
f(t)=\sum_{n=0}^\infty \dfrac{(t-t_0)^n}{n!} \left.\dfrac{d^{n}}{dt^n}\right|_{t_0}f(t)=\sum_{n=0}^\infty\dfrac{t^n}{n!}f^{(n)}(0)
$$
so we need primarily the derivatives of $f(t)=\log(t+1).$
\begin{align*}
f^{(0)}(t)&=\log(t+1)&\quad f^{(0)}(0)=\log(1)=0\\
f^{(1)}(t)&=\dfrac{1}{t+1}&\quad f^{(1)}(0)=\dfrac{1}{0+1}=1\\
f^{(2)}(t)&=-\dfrac{1}{(t+1)^2}&\quad f^{(2)}(0)=-\dfrac{1}{(0+1)^2}=-1\\
f^{(3)}(t)&=\dfrac{2}{(t+1)^3}&\quad f^{(3)}(0)=\dfrac{2}{(0+1)^3}=2\\
&\ldots &\ldots
\end{align*}
Now go on as long as you like and gather what you have.

I agree with you and mabe is the safest way . About my calculation : I have used also walphram alpha and it says that 1/(x-x^2/2+x^3/6) (I used only these terms for ln(1+t) = 1/t+1/2+t/12−t^2/24

instead 1/log(1+t)=1/t+1/2-t/12+t^2/24 why this happens?

 

[PeroK]

I have tried to use the natural Taylor expansion of ln(t+1)
and working with long division but I get the result 1/t+1/2+t/12−t^2/24+⋯
instead of 1/t+1/2-t/12+t^2/24+... that is the right result
I have tried to do this several time but still don't works. Is there a miscalculation in my long division?

So, what you did was:
$$\ln(1 + t) = t - \frac{t^2} 2 + \frac{t^3} 3 -\frac{t^4}{4} \dots$$$$\frac 1 {\ln(1+t)} = \frac 1 {t - \frac{t^2} 2 + \frac{t^3} 3 -\frac{t^4}{4} \dots} = \frac 1 t + \frac 1 2 + \frac t {12} - \frac {t^2}{24} \dots $$Using polynomial long division?

 

[laurabon]

So, what you did was:
$$\ln(1 + t) = 1 - t + t^2 - t^3 \dots$$$$\frac 1 {\ln(1+t)} = \frac 1 {1 - t + t^2 - t^3 \dots} = \frac 1 t + \frac 1 2 + \frac t {12} - \frac {t^2}{24} \dots $$Using polynomial long division?

yes. Is it wrong ? the coefficient for log are different because I have used the Talor series but the calculation is that

 

[fresh_42]

I agree with you and mabe is the safest way . About my calculation : I have used also walphram alpha and it says that 1/(x-x^2/2+x^3/6) (I used only these terms for ln(1+t) = 1/t+1/2+t/12−t^2/24

instead 1/log(1+t)=1/t+1/2-t/12+t^2/24 why this happens?

Sorry, I didn't see the inversion. It wasn't part of your post and was hidden in the title.
Please use https://www.physicsforums.com/help/latexhelp/ and show what you actually did.

You can easily make sign errors in a long division if you aren't careful enough. And to deal with the long denominator is a nightmare, too. You'd probably better solve
$$
1=\left(x-\dfrac{x^2}{2}+\dfrac{x^3}{3}-\dfrac{x^4}{4}\pm \ldots\right)\cdot (a_0+a_1x+a_2x^2+a_3x^4+\ldots)
$$

 

[laurabon]

Sorry, I didn't see the inversion. It wasn't part of your post and was hidden in the title.
Please use https://www.physicsforums.com/help/latexhelp/ and show what you actually did.

You can easily make sign errors in a long division if you aren't careful enough. And to deal with the long denominator is a nightmare, too. You'd probably better solve
$$
1=\left(x-\dfrac{x^2}{2}+\dfrac{x^3}{3}-\dfrac{x^4}{4}\pm \ldots\right)\cdot (a_0+a_1x+a_2x^2+a_3x^4+\ldots)
$$

Yes you are right , it was my error . I was considering the denominator of taylor expansionfor log(1+x) with factorial but... it doesn't hae factorials... about what you proposed in yyour post , doing multiplication shouldn't a_0 be 0? because on the right I have 1+0x+0x^2=a_0x

 

[fresh_42]

Yes you are right , it was my error . I was considering the denominator of taylor expansionfor log(1+x) with factorial but... it doesn't hae factorials... about what you proposed in yyour post , doing multiplication shouldn't a_0 be 0? because on the right I have 1+0x+0x^2=a_0x

If I look at the result, I should have started with $a_{-1}\dfrac{1}{x}+a_0+a_1x+\ldots$ Then $a_0$ turns out to be a half.

 

[vela]

If you only need a few of the first terms, I'd do it like this:
\begin{align*}
\frac{1}{\log(1+t)} &= \frac{1}{t-t^2/2+t^3/3-t^4/4+\cdots} \\
&= \frac{1}{t}\left[\frac{1}{1-\underbrace{\left(\frac{t}{2}-\frac{t^2}{3}+\frac{t^3}{4}+\cdots\right) }_x}\right] \\
&= \frac{1}{t}(1+x+x^2+x^3+\cdots)
\end{align*} Inside the parentheses, the term linear in $t$ only gets a contribution from $x$. The $t^2$ term has two contributions: the second term of $x$ and the first term of $x^2$, so you'd have
$$\frac{1}{\log(1+t)} = \frac{1}{t}\left[1+\frac{1}{2} t + \left(-\frac 13 + \frac 14\right) t^2 + \cdots\right]$$ And so on.

 

[pasmith]

1/\log(1 + t) doesn't have a Taylor series about t = 0, because the function is not defined (and therefore not differentiable) there: \log(1 + 0) = \log 1 = 0. It does have a Laurent series about t = 0.

 

[topsquark]

I have tried to use the natural Taylor expansion of ln(t+1)

What is "the natural Taylor expansion"? Is this is typo?

-Dan