Why Does the Inequality Hold in This Taylor Series Expansion?

[evinda]

Hello! (Wasntme)

I want to find the Taylor series of the function $f(x)=\log(1+x), x \in (-1,+\infty)$. We take $\xi=0, I=(-1,1)$

It is: $$f'(x)=(1+x)^{-1}, f''(x)=-1 \cdot (1+x)^{-2}, f'''(x)=2 \cdot (1+x)^{-3} , f^{(4)}(x)=-6 \cdot (1+x)^{-4}, f^{(5)}(x)=24(1+x)^{-5}$$

So,we see that $\frac{d^n}{dx^n} f(x)=\frac{(-1)^{n-1} \cdot (n-1)!}{(1+x)^n}$

So,the possible Taylor series is: $\sum_{n=1}^{\infty} \frac{(-1)^{n-1}}{n!} x^n$

To check if it is the right Taylor series,we have to check if the radius of convergence converges to $0$.

$R_{n,\xi} (n,\zeta)=\frac{(-1)^n}{(n+1)(1+\zeta)^{n+1}}x^{n+1}, \zeta \in [x,0] \text{ or } \zeta \in [0,x]$

• $0<x<1$ :
  
  $$|R_{n,\xi} (n,\zeta)| \leq \frac{x^{n+1}}{(n+1)(1+\zeta)^{n+1} } \leq \frac{1}{n+1} \to 0$$
  
• $-1<x<0$:
  
  $$|R_{n,\xi} (n,\zeta)|=|\frac{1}{n!} \int_{0}^x \frac{ (x-t)^n (-1)^n n!}{(1+t)^{n+1}}|= \int_0^x \frac{(t-x)^n}{(1+t)^{n+1}} dt $$
  
  According to my notes, $$(\frac{t-x}{1+t})^n \leq |x^n|, \forall t \in [x,0] $$
  
  But... why does the inequality $(\frac{t-x}{1+t})^n \leq |x^n|, \forall t \in [x,0] $ stand? And,also why $t \in [x,0] $ ? (Thinking)

 

[I like Serena]

Hello! (Wasntme)

I want to find the Taylor series of the function $f(x)=\log(1+x), x \in (-1,+\infty)$. We take $\xi=0, I=(-1,1)$

It is: $$f'(x)=(1+x)^{-1}, f''(x)=-1 \cdot (1+x)^{-2}, f'''(x)=2 \cdot (1+x)^{-3} , f^{(4)}(x)=-6 \cdot (1+x)^{-4}, f^{(5)}(x)=24(1+x)^{-5}$$

So,we see that $\frac{d^n}{dx^n} f(x)=\frac{(-1)^{n-1} \cdot (n-1)!}{(1+x)^n}$

So,the possible Taylor series is: $\sum_{n=1}^{\infty} \frac{(-1)^{n-1}}{n!} x^n$

To check if it is the right Taylor series,we have to check if the radius of convergence converges to $0$.

$R_{n,\xi} (n,\zeta)=\frac{(-1)^n}{(n+1)(1+\zeta)^{n+1}}x^{n+1}, \zeta \in [x,0] \text{ or } \zeta \in [0,x]$

• $0<x<1$ :
  
  $$|R_{n,\xi} (n,\zeta)| \leq \frac{x^{n+1}}{(n+1)(1+\zeta)^{n+1} } \leq \frac{1}{n+1} \to 0$$
  
• $-1<x<0$:
  
  $$|R_{n,\xi} (n,\zeta)|=|\frac{1}{n!} \int_{0}^x \frac{ (x-t)^n (-1)^n n!}{(1+t)^{n+1}}|= \int_0^x \frac{(t-x)^n}{(1+t)^{n+1}} dt $$
  
  According to my notes, $$(\frac{t-x}{1+t})^n \leq |x^n|, \forall t \in [x,0] $$
  
  But... why does the inequality $(\frac{t-x}{1+t})^n \leq |x^n|, \forall t \in [x,0] $ stand? And,also why $t \in [x,0] $ ? (Thinking)

Hi hi! (Mmm)

We have $t \in [x,0]$ because the integral is taken with respect to t with boundaries x and 0.

With a little work we can show that
$$\left(\frac{t-x}{1+t}\right)^n$$
is a monotone increasing function of $t$.
It takes its maximum value for $t=0$ at $|x|^n$. (Wink)

 

[Prove It]

Hello! (Wasntme)

I want to find the Taylor series of the function $f(x)=\log(1+x), x \in (-1,+\infty)$. We take $\xi=0, I=(-1,1)$

It is: $$f'(x)=(1+x)^{-1}, f''(x)=-1 \cdot (1+x)^{-2}, f'''(x)=2 \cdot (1+x)^{-3} , f^{(4)}(x)=-6 \cdot (1+x)^{-4}, f^{(5)}(x)=24(1+x)^{-5}$$

So,we see that $\frac{d^n}{dx^n} f(x)=\frac{(-1)^{n-1} \cdot (n-1)!}{(1+x)^n}$

So,the possible Taylor series is: $\sum_{n=1}^{\infty} \frac{(-1)^{n-1}}{n!} x^n$

To check if it is the right Taylor series,we have to check if the radius of convergence converges to $0$.

$R_{n,\xi} (n,\zeta)=\frac{(-1)^n}{(n+1)(1+\zeta)^{n+1}}x^{n+1}, \zeta \in [x,0] \text{ or } \zeta \in [0,x]$

• $0<x<1$ :
  
  $$|R_{n,\xi} (n,\zeta)| \leq \frac{x^{n+1}}{(n+1)(1+\zeta)^{n+1} } \leq \frac{1}{n+1} \to 0$$
  
• $-1<x<0$:
  
  $$|R_{n,\xi} (n,\zeta)|=|\frac{1}{n!} \int_{0}^x \frac{ (x-t)^n (-1)^n n!}{(1+t)^{n+1}}|= \int_0^x \frac{(t-x)^n}{(1+t)^{n+1}} dt $$
  
  According to my notes, $$(\frac{t-x}{1+t})^n \leq |x^n|, \forall t \in [x,0] $$
  
  But... why does the inequality $(\frac{t-x}{1+t})^n \leq |x^n|, \forall t \in [x,0] $ stand? And,also why $t \in [x,0] $ ? (Thinking)

The easiest way to get a Taylor (MacLaurin) series for $\displaystyle \begin{align*} \log{(1 + x)} \end{align*}$ is to recognise that $\displaystyle \begin{align*} \frac{d}{dx} \left[ \log{(1 + x)} \right] = \frac{1}{1 + x} = \frac{1}{1 - \left( -x \right) } \end{align*}$, which is easily recognised as the closed form of the geometric series $\displaystyle \begin{align*} \sum_{n = 0}^{\infty} \left( -x \right)^n = \sum_{n = 0}^{\infty} (-1)^n \,x^n \end{align*}$, which is convergent for $\displaystyle \begin{align*} \left| -x \right| < 1 \implies \left| x \right| <1 \end{align*}$.

So if you integrate the series for $\displaystyle \begin{align*} \frac{1}{1 + x} \end{align*}$, you will get a series for $\displaystyle \begin{align*} \log{(1 + x)} \end{align*}$, which will have the same radius of convergence (though may also converge on the endpoints which you must check).

 

[evinda]

Hi hi! (Mmm)

We have $t \in [x,0]$ because the integral is taken with respect to t with boundaries x and 0.

With a little work we can show that
$$\left(\frac{t-x}{1+t}\right)^n$$
is a monotone increasing function of $t$.
It takes its maximum value for $t=0$ at $|x|^n$. (Wink)

So,is it like that?

We set $g(t)=(\frac{t-x}{1+t})^n$

$$g'(t)=\frac{n(t-x)^n}{(1+t)^{n+2}}$$

$$g'(t)=0 \Rightarrow t=x$$

$$g'(\frac{x}{2})=\frac{n(-x)^n}{2^n (1+\frac{x}{2})^{n+2}}>0$$

So,we conclude that at the interval $[\xi,x], g'>0$,so $g$ achieves its maximum at the point $x$? But..then the maximum would be $g(t)=0$ ..(Worried)
Or have I done something wrong? (Thinking)(Thinking)

 

[evinda]

The easiest way to get a Taylor (MacLaurin) series for $\displaystyle \begin{align*} \log{(1 + x)} \end{align*}$ is to recognise that $\displaystyle \begin{align*} \frac{d}{dx} \left[ \log{(1 + x)} \right] = \frac{1}{1 + x} = \frac{1}{1 - \left( -x \right) } \end{align*}$, which is easily recognised as the closed form of the geometric series $\displaystyle \begin{align*} \sum_{n = 0}^{\infty} \left( -x \right)^n = \sum_{n = 0}^{\infty} (-1)^n \,x^n \end{align*}$, which is convergent for $\displaystyle \begin{align*} \left| -x \right| < 1 \implies \left| x \right| <1 \end{align*}$.

So if you integrate the series for $\displaystyle \begin{align*} \frac{1}{1 + x} \end{align*}$, you will get a series for $\displaystyle \begin{align*} \log{(1 + x)} \end{align*}$, which will have the same radius of convergence (though may also converge on the endpoints which you must check).

Could you explain me further how I can find the Taylor series in that way?? (Thinking)

I have tried something,but it should be wrong,because I don't get the right Taylor series.. That's what I have tried:

$$\sum_{n=0}^{\infty} (-x)^n=\frac{1}{1+x} \Rightarrow \int_0^x \sum_{n=0}^{\infty} (-x)^n=\int_0^x \frac{1}{1+x} \Rightarrow \sum_{n=0}^{\infty} (\frac{-x^{n+1}}{n+1})=\ln(1+x)$$

Could you tell me what I have done wrong??

 

[I like Serena]

So,we conclude that at the interval $[\xi,x], g'>0$,so $g$ achieves its maximum at the point $x$? But..then the maximum would be $g(t)=0$ ..(Worried)
Or have I done something wrong? (Thinking)(Thinking)

Uhh... the maximum would be $g(0)=|x|^n$... (Sadface)

Lemme explain.

Since we have that $-1 < x \le t < 0$, it follows that both $t-x \ge 0$ and $1+t>0$.
Therefore $g'(t)>0$, meaning that g is monotonously increasing.
The function $g$ takes it highest value when $t$ takes it highest value, which is when $t$ approaches $0$.
So the supremum is $g(0) = (\frac{0-x}{1+0})^n = |x|^n$.

Could you explain me further how I can find the Taylor series in that way?? (Thinking)

I have tried something,but it should be wrong,because I don't get the right Taylor series.. That's what I have tried:

$$\sum_{n=0}^{\infty} (-x)^n=\frac{1}{1+x} \Rightarrow \int_0^x \sum_{n=0}^{\infty} (-x)^n=\int_0^x \frac{1}{1+x} \Rightarrow \sum_{n=0}^{\infty} (\frac{-x^{n+1}}{n+1})=\ln(1+x)$$

Could you tell me what I have done wrong??

Only that you suddenly left out a couple of parentheses... (Crying)
And you didn't really simplify...

And I'm suddenly noticing that your original Taylor series is wrong!
You put $n!$ in the denominator but neglected to put the $(n-1)!$ that comes along with the derivatives in the numerator!

 

[evinda]

Uhh... the maximum would be $g(0)=|x|^n$... (Sadface)

Lemme explain.

Since we have that $-1 < x \le t < 0$, it follows that both $t-x \ge 0$ and $1+t>0$.
Therefore $g'(t)>0$, meaning that g is monotonously increasing.
The function $g$ takes it highest value when $t$ takes it highest value, which is when $t$ approaches $0$.
So the supremum is $g(0) = (\frac{0-x}{1+0})^n = |x|^n$.

A ok! I understand! (Mmm)

Only that you suddenly left out a couple of parentheses... (Crying)
And you didn't really simplify...
And I'm suddenly noticing that your original Taylor series is wrong!
You put $n!$ in the denominator but neglected to put the $(n-1)!$ that comes along with the derivatives in the numerator!

So,it is like that:

$$f(x)=\log(1+x)$$

$$f'(x)=\frac{1}{1+x}=\frac{1}{1-(-x)}=\sum_{n=0}^{\infty} (-x)^n$$
$$\int_0^x \frac{1}{1+x}=f(0)+ \int_0^x f'(t)dt \Rightarrow \log(1+x)=\int_0^x \sum_{n=0}^{\infty} (-t)^n dt \Rightarrow \log(1+x)=\sum_{n=0}^{\infty} (-1)^n \int_0^x t^n dt= \sum_{n=0}^{\infty} (-1)^n \frac{x^{n+1}}{(n+1)}$$

Is it right so far or have I done something wrong??

 

[I like Serena]

$$f'(x)=\frac{1}{1+x}=\frac{1}{1-(-x)}=\sum_{n=0}^{\infty} (-x)^n$$
$$\int_0^x \frac{1}{1+x}=f(0)+ \int_0^x f'(t)dt \Rightarrow \log(1+x)=\int_0^x \sum_{n=0}^{\infty} (-t)^n dt \Rightarrow \log(1+x)=\sum_{n=0}^{\infty} (-1)^n \int_0^x t^n dt= \sum_{n=0}^{\infty} (-1)^n \frac{x^{n+1}}{(n+1)}$$

Is it right so far or have I done something wrong??

Looks fine... you only needed to add a couple parentheses back after which you could split off the $(-1)^n$. (Sweating)

 

[evinda]

Looks fine... you only needed to add a couple parentheses back after which you could split off the $(-1)^n$. (Sweating)

And how can I find now the Lagrange remainder to prove that it is actually the power series,showing that it converges to $0$ ? (Thinking)

 

[I like Serena]

And how can I find now the Lagrange remainder to prove that it is actually the power series,showing that it converges to $0$ ? (Thinking)

Well... with your new method you're not doing a Taylor expansion, so you don't get a Lagrange remainder. (Doh)
What you do get is that $\frac{1}{1+x}$ is the result of a geometric series, which has a convergence radius of 1.
Its antiderivative will also have a convergence radius of 1.

If you want to use the Lagrange remainder, you should go back to your original expansion and correct the formulas. Or else you should do the Taylor expansion on $\frac{1}{1+x}$ to prove it has convergence radius 1, after which you can calculate its antiderivative. (Mmm)