#[Taylor Series] How can f(x) possibly be equal to T_n(x) + R_n(x)?

49 messages · Page 1 of 1 (latest)

fluid ingot
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T_n lasts n terms while f(x) has infinite terms. Shouldn't R_n be the remaining terms - I thought, but it's merely one term n + 1. I don't understand why?

placid basaltBOT
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slim comet
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context?

slim comet
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@fluid ingot can you send what formula of the taylor series are you seeing

fluid ingot
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Oh, it's in another language

slim comet
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the math will be the same innit? 💀

fluid ingot
slim comet
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bro

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R_n contains f

fluid ingot
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Yeah

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I thought

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$f(x) = \sum_{k = 0}^{\infty} \frac{f^{k}(a)}{k!} (x-a)^k$

slim comet
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yes

pliant shoreBOT
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domini

fluid ingot
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But T_n only reaches k = n

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And R_n extends it to n + 1

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What about [n+2, inf)?

slim comet
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as i said

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R_n contains f

brittle steeple
fluid ingot
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I don't understand

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Oh

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T_n has a and R_n has theta

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Is that the difference you're referring to

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Oh I see

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It has a little f in there

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But what does that mean

zenith smelt
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By explicit definition, $R_n(x):=f(x)-T_n(x)$
You then prove the equivalent forms of $R_n$

pliant shoreBOT
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Omegabet_

zenith smelt
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The proof in essence proves the integral form of $R_n$ first, then you can determine $\xi$ so that the remaining integral is equivalent to $f^{(n+1)}(\xi)$

pliant shoreBOT
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Omegabet_

fluid ingot
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Oh, it's a freaking theorem

zenith smelt
fluid ingot
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Thanks

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I thought it was something that intuitively followed

zenith smelt
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no

fluid ingot
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And was extremely confused

zenith smelt
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it is extremely not intuitive that R_n should be writable just as those

fluid ingot
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Yeah

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Thanks so much

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!close

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?close

keen grailBOT
fluid ingot
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&close

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;-; lol

hasty lichen
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+close