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On page 7 of Loring Tu's book called "An Introduction to Manifolds", he makes a jump and a probable mistake. It pertains to his proof of Taylor's remainder Theorem.

https://www.math.toronto.edu/~jeffrey/matd67/tu.pdf

Tu states "Applying the lemma repeatedly gives gi(x) = gi(0) +xgi+1(x)."

Can anyone tell me where Tu got this result? Is Tu even right? Any help is appreciated. (I have tried to tackle this part of the proof on and off since 2010).

Jason Broadway
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    Tu is correct, there is no mistake. The equation comes from applying the lemma on $f(x)=g_2(x)$ then $f(x)=g_3(x)$ etc. Thus $f(x)=f(0)+x\cdot g_1(x)$ becomes $g_n(x)=g_n(0)+x \cdot g_n(0).$ Note that the $g_i(x)$ are smooth, i.e. infinitely often differentiable. – Marius S.L. Jan 01 '23 at 03:05
  • Thank you, Marius. Your time is appreciated. I am chewing on your answer, but I am still lost. Something is hazy to me. Is x a number or a vector? How did you get gn(x)=gn(0)+x*gn(0). I promise you that I am not trolling you. – Jason Broadway Jan 01 '23 at 03:17
  • Still lost. Can you all give me one more try? I am not a student; I teach at a college. I have been trying to crack this riddle since 2010. – Jason Broadway Jan 01 '23 at 03:30
  • Do you understand the previous line $f(x)=f(0)+xg_1(x)$? If so, just replace $f$ with $g_i$ and $g_1$ with $g_{i+1}$. In other words, $f=g_0$ and you get the result for every $i=0,1,2,....$ – Mr. Brown Jan 01 '23 at 04:03
  • Also note that that this example occurs after the proof ends, which happens right before "In case $n=1$ and $p=0$ [...]". – Brian Moehring Jan 01 '23 at 04:23
  • Zack, yes, I got that part. I thought that g(2) had a life of its own without g(1), and I thought that x was (in general) more than a 1D variable. These assumptions were pushing me in the ditch. – Jason Broadway Jan 01 '23 at 04:35
  • Brian...dually noted... I did not know that Tu was keeping it that way as he kept on talking. I thought (wrongly) that Tu had to reverted back to treating x as a vector after that sentence. – Jason Broadway Jan 01 '23 at 04:37

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Lemma 1.4. says for $n=1$ and $p=0$ that a smooth function can be written as
$$ f(x)=f(0)+(x-0)\cdot g_1(x) \text{ with } g_1(0)=\left. \dfrac{\partial f}{\partial x}\right|_{x=0}=\left. \dfrac{df}{dx}\right|_{x=0} =f'(0) $$ During the proof, we saw that $g_1(x)$ is smooth again. Now we can therefore apply Lemma 1.4 again, this time not on $f(x)$ but on $g_1(x).$ This gives us an equation $$ g_1(x)=g_1(0)+(x-0)\cdot g_2(x) \text{ with } g_2(0)=\left. \dfrac{\partial g_1}{\partial x}\right|_{x=0}=\left. \dfrac{dg_1}{dx}\right|_{x=0} =g_1'(0). $$ If we go on and on, then we get increasing indices for ever new functions $g_i,$ but the equation remains: $$ g_i(x)=g_i(0)+(x-0)\cdot g_{i+1}(x) \text{ with } g_{i+1}(0)=\left. \dfrac{\partial g_i}{\partial x}\right|_{x=0}=\left. \dfrac{dg_i}{dx}\right|_{x=0} =g_i'(0) $$ Here we have $0$ as the point $p,$ $g_i$ as the function $f$, and $x$ as the variable. Since the dimension $n=1$ is only one, partials turn into common derivatives of a real function.

It is important to note that in general $f'(x)\neq g_1(x)$ and $g'_i(x)\neq g_{i+1}(x).$ They are only identical at $p=0,$ so $f'(0)= g_1(0)$ and $g'_i(0)= g_{i+1}(0).$ This is important to remember when the accuracy, i.e. the discrepancy between $f'(x)$ and $g_1(x)$ will be investigated.

Marius S.L.
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  • ok ok...I follow you... your proof is plain as day....Tu was just doing single variable x here, not (x1, x2, x3, ... , xn). Also, the next g(i+1) is defined from the g(i) before. That is what was getting me here... ...I thought that the g(i+1) had an existence independent of the g(i) before... ohhhh...I am grateful to you! – Jason Broadway Jan 01 '23 at 04:33
  • as plain as day...I got excited and sacrificed my grammar. Thanks again. – Jason Broadway Jan 01 '23 at 04:45