Taylor's Theorem .... Loring W. Tu, Lemma 1.4 .... ....

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In summary, the conversation discusses Loring W.Tu's book "An Introduction to Manifolds" and focuses on Lemma 1.4 which is Taylor's Theorem with Remainder. The conversation covers questions related to the application of the theorem and its proof, specifically the change from $\mathbb{R}^{n}$ to $\mathbb{R}$ and the use of summation in the equation. The overall strategy of the proof is to repeatedly apply the lemma in the case of $\mathbb{R}^{1}$ and obtain the $i+1$st order Taylor polynomial from single-variable calculus.
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I am reading Loring W.Tu's book: "An Introduction to Manifolds" (Second Edition) ...

I need help in order to fully understand the proof of Tu's Lemma 1.4: Taylor's Theorem with Remainder ...

Lemma 1.4 reads as follows:View attachment 8631
View attachment 8632
My questions are as follows:Question 1

In the above text from Tu we read the following:

" ... ... In case \(\displaystyle n = 1\) and \(\displaystyle p = 0\), this lemma says that

\(\displaystyle f(x) = f(0) + x g_1(x)\) ... ... "Now Tu seems to put \(\displaystyle n= 1\) in the equation in the lemma but does not change \(\displaystyle \mathbb{R}^n\) to \(\displaystyle \mathbb{R}^1\) and does not change \(\displaystyle x = (x^1, x^2, \ ... \ ... \ x^n)\) to \(\displaystyle x = (x^1)\) ... ... How can this be valid?Question 2In the above text from Tu we read the following:

" ... ... Applying the lemma repeatedly gives \(\displaystyle g_i(x) = g_i(0) + x g_{ i + 1 } (x)\) ... ... "How exactly does Tu arrive at the above equation ... I take it he puts \(\displaystyle f = g_i\) and he pits p = 0 ... but how does he get \(\displaystyle x g_{ i + 1 } (x)\) out of the summation term .. ? ( ... note that it is the i + 1 term in g_{ i + 1 } that I find puzzling ... )
Question 3I must say that generally I am having trouble following the overall 'strategy' of the proof ... can it be summarised as transforming the equations of the lemma into a valid Taylor series ...?

... ... but mind you he only seems to show this for \(\displaystyle p= 0\)?
Hope someone can help ...?

Peter
 

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  • #2
Hi Peter,

Before answering your questions directly, I think it is worth noting the difference between when $x^{i}$ represents the $i$th coordinate of the vector $x=(x^{1}, x^{2}, \ldots, x^{n})$ versus when $x^{i}$ is taken to mean the single variable $x$ raised to the $i$th power. In the proof of the lemma it is the former, in the justification of the result after the lemma (e.g., see equation (1.2)), it is the latter.

Peter said:
Question 1
Now Tu seems to put \(\displaystyle n= 1\) in the equation in the lemma but does not change \(\displaystyle \mathbb{R}^n\) to \(\displaystyle \mathbb{R}^1\) and does not change \(\displaystyle x = (x^1, x^2, \ ... \ ... \ x^n)\) to \(\displaystyle x = (x^1)\) ... ...

How can this be valid?

The change from $\mathbb{R}^{n}$ to $\mathbb{R}$ has occurred (the superscript $x^{1}$ is dropped and he just writes $x$ when in $\mathbb{R}$) as evidenced by the equation $$f(x)=f(0)+xg_{1}(x).$$ Had he been working in $\mathbb{R}^{n}$ this would instead be written as $$f(x) = f(0) + x^{1}g_{1}(x) + x^{2}g_{2}(x) +\cdots + x^{n}g_{n}(x),$$ where $x=(x^{1}, x^{2},\ldots, x^{n}).$
Peter said:
Question 2
How exactly does Tu arrive at the above equation ... I take it he puts \(\displaystyle f = g_i\) and he pits p = 0 ... but how does he get \(\displaystyle x g_{ i + 1 } (x)\) out of the summation term .. ? ( ... note that it is the i + 1 term in g_{ i + 1 } that I find puzzling ... )

You are correct here by taking $f=g_{i}$ and reapplying the lemma in the case of $\mathbb{R}^{1},$ nicely done. There is no summation because, in this case, $g_{i}(x)$ is a function of a single variable. As mentioned above, the powers of $x$ in equation (1.2) are exponents on the single variable $x$ and do not represent coordinates of a vector.

Peter said:
Question 3
I must say that generally I am having trouble following the overall 'strategy' of the proof ... can it be summarised as transforming the equations of the lemma into a valid Taylor series ...?

... ... but mind you he only seems to show this for \(\displaystyle p= 0\)?

I would say that it is the direct application of the lemma to the case where $n=1$, $p=0$ repeated over and over again on the sequence of functions $g_{i}(x)$ to obtain the $i+1$st order Taylor polynomial from single-variable calculus.

Let me know if anything is still unclear.
 
  • #3
GJA said:
Hi Peter,

Before answering your questions directly, I think it is worth noting the difference between when $x^{i}$ represents the $i$th coordinate of the vector $x=(x^{1}, x^{2}, \ldots, x^{n})$ versus when $x^{i}$ is taken to mean the single variable $x$ raised to the $i$th power. In the proof of the lemma it is the former, in the justification of the result after the lemma (e.g., see equation (1.2)), it is the latter.
The change from $\mathbb{R}^{n}$ to $\mathbb{R}$ has occurred (the superscript $x^{1}$ is dropped and he just writes $x$ when in $\mathbb{R}$) as evidenced by the equation $$f(x)=f(0)+xg_{1}(x).$$ Had he been working in $\mathbb{R}^{n}$ this would instead be written as $$f(x) = f(0) + x^{1}g_{1}(x) + x^{2}g_{2}(x) +\cdots + x^{n}g_{n}(x),$$ where $x=(x^{1}, x^{2},\ldots, x^{n}).$

You are correct here by taking $f=g_{i}$ and reapplying the lemma in the case of $\mathbb{R}^{1},$ nicely done. There is no summation because, in this case, $g_{i}(x)$ is a function of a single variable. As mentioned above, the powers of $x$ in equation (1.2) are exponents on the single variable $x$ and do not represent coordinates of a vector.
I would say that it is the direct application of the lemma to the case where $n=1$, $p=0$ repeated over and over again on the sequence of functions $g_{i}(x)$ to obtain the $i+1$st order Taylor polynomial from single-variable calculus.

Let me know if anything is still unclear.

All clear now, thanks GJA ...

I appreciate your most helpful post ...

Peter
 

FAQ: Taylor's Theorem .... Loring W. Tu, Lemma 1.4 .... ....

What is Taylor's Theorem?

Taylor's Theorem is a mathematical theorem that gives an expression for a function in terms of its derivatives at a single point.

Who is Loring W. Tu?

Loring W. Tu is a mathematician and professor at Tufts University, known for his work in differential geometry and topology.

What is Lemma 1.4 in Loring W. Tu's work?

Lemma 1.4 is a lemma, or a small, self-contained theorem, in Loring W. Tu's work. It may be used as a building block in proving larger theorems.

What is the significance of Loring W. Tu's work?

Loring W. Tu's work is significant in the fields of differential geometry and topology, as he has made important contributions to these areas through his research and teaching.

How is Taylor's Theorem used in mathematics?

Taylor's Theorem is used in mathematics to approximate a function with a polynomial, providing a tool for analyzing and understanding functions and their behavior around a specific point.

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