Construction and Uniqueness of a Periodic Interpolating Function with Special Contraints
$begingroup$
Consider the following points: $(t_{1},y_{1})$ $(t_{2},y_{2})$ … $(t_{n},y_{n})$
where $t_{1}<t_{2}<...t_{n}<t_{1}+P$ for some $P in mathbb{R}$
Define $G(t)$ as a function $mathbb{R}mapstomathbb{R}$ with the following properties:
- $G(t_{i})=y_{i}$
$G(t_{i}+nP)=y_{i}$ $forall ninmathbb{N}$
$frac{dG(t_{i}+nP)}{dt}=0$ $forall ninmathbb{N}$
- $G(t)in mathcal{C}^{infty}$
Is $G(t)$ unique? Is there an algorithm for constructing $G(t)$? Trigonometric interpolation doesn't work, to my knowledge, because it doesn't require that the given points are local extrema. I imagine there is some solution involving an infinite sum of trigonometric functions, although I am not sure how to begin proving constructing a general algorithm.
In trivial cases, it's simple to find counterexamples showing that these functions are not unique. Consider in the case:
$(t_{i},y_{i})$=$(0,1) (pi,-1)$
In this case, there are an infinite set of functions that satisfy the four stated requirements.
$G(t)=cos(t)$ or $G(t)=ncos(t) forall n in mathbb{N}$
I'm pretty sure I don't have to prove that these functions satisfy the four previous requirements. The more interesting question is if there are any other smooth functions (which we may call other non-trivial solutions) that would solve this problem.
For simplicity we can first consider this case for just the two points $(0,1) (pi,-1)$, and then expand to an arbitrary number of arbitrary points.
interpolation
asked Dec 9 '18 at 7:44
Daniel AgramonteDaniel Agramonte
11
$endgroup$
add a comment |
$begingroup$
Consider the following points: $(t_{1},y_{1})$ $(t_{2},y_{2})$ … $(t_{n},y_{n})$
where $t_{1}<t_{2}<...t_{n}<t_{1}+P$ for some $P in mathbb{R}$
Define $G(t)$ as a function $mathbb{R}mapstomathbb{R}$ with the following properties:
- $G(t_{i})=y_{i}$
$G(t_{i}+nP)=y_{i}$ $forall ninmathbb{N}$
$frac{dG(t_{i}+nP)}{dt}=0$ $forall ninmathbb{N}$
- $G(t)in mathcal{C}^{infty}$
Is $G(t)$ unique? Is there an algorithm for constructing $G(t)$? Trigonometric interpolation doesn't work, to my knowledge, because it doesn't require that the given points are local extrema. I imagine there is some solution involving an infinite sum of trigonometric functions, although I am not sure how to begin proving constructing a general algorithm.
In trivial cases, it's simple to find counterexamples showing that these functions are not unique. Consider in the case:
$(t_{i},y_{i})$=$(0,1) (pi,-1)$
In this case, there are an infinite set of functions that satisfy the four stated requirements.
$G(t)=cos(t)$ or $G(t)=ncos(t) forall n in mathbb{N}$
I'm pretty sure I don't have to prove that these functions satisfy the four previous requirements. The more interesting question is if there are any other smooth functions (which we may call other non-trivial solutions) that would solve this problem.
For simplicity we can first consider this case for just the two points $(0,1) (pi,-1)$, and then expand to an arbitrary number of arbitrary points.
interpolation
asked Dec 9 '18 at 7:44
Daniel AgramonteDaniel Agramonte
11
$endgroup$
add a comment |
$begingroup$
Consider the following points: $(t_{1},y_{1})$ $(t_{2},y_{2})$ … $(t_{n},y_{n})$
where $t_{1}<t_{2}<...t_{n}<t_{1}+P$ for some $P in mathbb{R}$
Define $G(t)$ as a function $mathbb{R}mapstomathbb{R}$ with the following properties:
- $G(t_{i})=y_{i}$
$G(t_{i}+nP)=y_{i}$ $forall ninmathbb{N}$
$frac{dG(t_{i}+nP)}{dt}=0$ $forall ninmathbb{N}$
- $G(t)in mathcal{C}^{infty}$
Is $G(t)$ unique? Is there an algorithm for constructing $G(t)$? Trigonometric interpolation doesn't work, to my knowledge, because it doesn't require that the given points are local extrema. I imagine there is some solution involving an infinite sum of trigonometric functions, although I am not sure how to begin proving constructing a general algorithm.
In trivial cases, it's simple to find counterexamples showing that these functions are not unique. Consider in the case:
$(t_{i},y_{i})$=$(0,1) (pi,-1)$
In this case, there are an infinite set of functions that satisfy the four stated requirements.
$G(t)=cos(t)$ or $G(t)=ncos(t) forall n in mathbb{N}$
I'm pretty sure I don't have to prove that these functions satisfy the four previous requirements. The more interesting question is if there are any other smooth functions (which we may call other non-trivial solutions) that would solve this problem.
For simplicity we can first consider this case for just the two points $(0,1) (pi,-1)$, and then expand to an arbitrary number of arbitrary points.
interpolation
asked Dec 9 '18 at 7:44
Daniel AgramonteDaniel Agramonte
11
$endgroup$
Consider the following points: $(t_{1},y_{1})$ $(t_{2},y_{2})$ … $(t_{n},y_{n})$
where $t_{1}<t_{2}<...t_{n}<t_{1}+P$ for some $P in mathbb{R}$
Define $G(t)$ as a function $mathbb{R}mapstomathbb{R}$ with the following properties:
- $G(t_{i})=y_{i}$
$G(t_{i}+nP)=y_{i}$ $forall ninmathbb{N}$
$frac{dG(t_{i}+nP)}{dt}=0$ $forall ninmathbb{N}$
- $G(t)in mathcal{C}^{infty}$
Is $G(t)$ unique? Is there an algorithm for constructing $G(t)$? Trigonometric interpolation doesn't work, to my knowledge, because it doesn't require that the given points are local extrema. I imagine there is some solution involving an infinite sum of trigonometric functions, although I am not sure how to begin proving constructing a general algorithm.
In trivial cases, it's simple to find counterexamples showing that these functions are not unique. Consider in the case:
$(t_{i},y_{i})$=$(0,1) (pi,-1)$
In this case, there are an infinite set of functions that satisfy the four stated requirements.
$G(t)=cos(t)$ or $G(t)=ncos(t) forall n in mathbb{N}$
I'm pretty sure I don't have to prove that these functions satisfy the four previous requirements. The more interesting question is if there are any other smooth functions (which we may call other non-trivial solutions) that would solve this problem.
For simplicity we can first consider this case for just the two points $(0,1) (pi,-1)$, and then expand to an arbitrary number of arbitrary points.
interpolation
interpolation
asked Dec 9 '18 at 7:44
Daniel AgramonteDaniel Agramonte
11
asked Dec 9 '18 at 7:44
Daniel AgramonteDaniel Agramonte
11
asked Dec 9 '18 at 7:44
Daniel AgramonteDaniel Agramonte
11
asked Dec 9 '18 at 7:44
Daniel AgramonteDaniel Agramonte
11
asked Dec 9 '18 at 7:44
Daniel AgramonteDaniel Agramonte
11
11
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