How do I use induction more rigorously when making Taylor expansions?

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Discussion Overview

The discussion revolves around the use of mathematical induction in making Taylor expansions more rigorous. Participants explore the process of deriving coefficients for polynomial expansions and the relevance of specific examples and approaches to different types of functions.

Discussion Character

  • Exploratory
  • Technical explanation
  • Debate/contested
  • Mathematical reasoning

Main Points Raised

  • One participant describes their method of taking the first few derivatives of a function to induce a pattern for Taylor expansions, questioning the rigor of this process and mentioning "plugging in n+1."
  • Another participant expresses uncertainty about the use of induction in this context, suggesting that the Taylor series formula provides definitive powers.
  • A participant requests an example, noting that the approach may depend on the specific function, with periodic functions like sine and cosine being easier to handle.
  • A later reply provides an example involving the derivation of non-relativistic kinetic energy from a relativistic energy expression, illustrating a pattern in derivatives using induction.
  • One participant asks if the original poster understands the inductive step in the process of induction.
  • The original poster admits to not understanding the inductive step, leading to a suggestion to consult an external resource on writing proofs by induction.

Areas of Agreement / Disagreement

Participants express differing views on the application of induction in Taylor expansions, with some questioning its relevance while others seek to clarify and provide examples. The discussion remains unresolved regarding the rigor of the inductive approach.

Contextual Notes

Participants do not reach a consensus on the necessity or effectiveness of using induction for Taylor expansions, and there are varying levels of understanding regarding the inductive process itself.

Mayhem
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When I do Taylor expansions, I take the first 3 or 4 derivatives of a function and try to induce a pattern, and then evaluate it at some value a (often 0) to find the coefficients in the polynomial expansion.

This is how my textbook does it, and how several other online sources do it as well, but can I make this inductive process slightly more rigorous? I remember hearing about "plugging in n+1" from a YouTube video a long time ago, and I'm wondering if that's relevant.

Sorry for being so vague, I hope it is clear what I'm trying to achieve.
 
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Taylor series formula gives us each power definitely. I am not sure about you say it induction.
 
Mayhem said:
When I do Taylor expansions, I take the first 3 or 4 derivatives of a function and try to induce a pattern, and then evaluate it at some value a (often 0) to find the coefficients in the polynomial expansion.

This is how my textbook does it, and how several other online sources do it as well, but can I make this inductive process slightly more rigorous? I remember hearing about "plugging in n+1" from a YouTube video a long time ago, and I'm wondering if that's relevant.

Sorry for being so vague, I hope it is clear what I'm trying to achieve.
Can you give an example? The approach would tend to depend on the function in question. Sines and cosines are periodic, so they should be easier. But, a typical function requires a specific inductive argument.
 
PeroK said:
Can you give an example? The approach would tend to depend on the function in question. Sines and cosines are periodic, so they should be easier. But, a typical function requires a specific inductive argument.
Sure. This is from a recent assignment where I had to derive the non-relativistic KE from a relativistic energy expression, where we let ## v/c \rightarrow 0##

$$
\begin{align*}
f'(x) &= \frac{1}{2} \cdot \frac{1}{(1-x)^{3/2}} \\
f''(x) &= \frac{1}{2} \cdot \frac{3}{2} \cdot \frac{1}{(1-x)^{5/2}} \\
f'''(x) &= \frac{1}{2} \cdot\frac{3}{2} \cdot \frac{5}{2}\frac{1}{(1-x)^{7/2}} \\
f''''(x) &= \frac{1}{2}\cdot \frac{3}{2} \cdot\frac{5}{2}\cdot \frac{7}{2}\frac{1}{(1-x)^{9/2}} \\
&\Downarrow \mathrm{Induction} \\
f^{(n)}(x) &= \frac{1\cdot 3 \cdot 5 \cdot \cdot \cdot (2n-1)}{2^n} \frac{1}{(1-x)^{2n+1/2}}
\end{align*}
$$
 
Okay, so you can guess the formula (which must hold for ##n = 1##). Do you know how induction works, in terms of the inductive step?
 
PeroK said:
Okay, so you can guess the formula (which must hold for ##n = 1##). Do you know how induction works, in terms of the inductive step?
No, not really.
 

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