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

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  • #1
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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  • #2
anuttarasammyak
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Taylor series formula gives us each power definitely. I am not sure about you say it induction.
 
  • #3
PeroK
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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.
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.
 
  • #4
Mayhem
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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##

$$
\begin{align*}
f'(x) &= \frac{1}{2} \cdot \frac{1}{(1-x)^{3/2}} \\
f''(x) &= \frac{1}{2} \cdot \frac{3}{2} \cdot \frac{1}{(1-x)^{5/2}} \\
f'''(x) &= \frac{1}{2} \cdot\frac{3}{2} \cdot \frac{5}{2}\frac{1}{(1-x)^{7/2}} \\
f''''(x) &= \frac{1}{2}\cdot \frac{3}{2} \cdot\frac{5}{2}\cdot \frac{7}{2}\frac{1}{(1-x)^{9/2}} \\
&\Downarrow \mathrm{Induction} \\
f^{(n)}(x) &= \frac{1\cdot 3 \cdot 5 \cdot \cdot \cdot (2n-1)}{2^n} \frac{1}{(1-x)^{2n+1/2}}
\end{align*}
$$
 
  • #5
PeroK
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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?
 
  • #6
Mayhem
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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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