I am having difficulties with interpolating polynomials to approximate. I am familiar when the problem involves 2 points but this particular problem involves 3 points. Can you explain the process to me? For the given function $f(x)=\sin(\pi x)$ let $x_0=1,x_1=1.25,x_2=1.6$. Construct interpolation polynomials of degree at most 1 and at most 2 to approximate $f(1.4)$ and find the absolute error.
Interpolating polynomials for $f(x)=\sin(\pi x)$
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1Presumably, for the interpolation of degree one, you only need $x_1$ and $x_2$. – 2012-09-12
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0So I would interpolate 2 polynomials in this case, one for x_1 and x_2 and the other for x_0 and x_1 ? – 2012-09-12
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0So you chose those 2 points since we need to approximate f(1.4)? – 2012-09-12
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0To your last question: yes, indeed! – 2012-09-12
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0Yup that was correct, I got the right answer. Thanks :)))) – 2012-09-12
1 Answers
Interpolation of degree at most $1$ is linear interpolation. You only need the value of $f$ evaluated in $x_1$ and $x_2$ since $x_1<1.4
You can also interpolate using a piecewise polynomial as well but in this case you get fixed degree of smoothness. For example cubic splines, which are piecewise cubic polynomials define a twice differentiable function, and not more ! The nice thing about these interpolation polynomials is that between two knots they have degree at most 3, preventing Runge's phenomenon that Lagrange interpolation polynomial present when increasing the number of knots.
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0Thanks your answer is very clear. I was getting confused there. I was just noticing your equation for the linear interpolation. I was taught a more confusing formula and find yours to be neat and clean. Thanks – 2012-09-12
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0First construct three polynomials $l_i(x)$, $i=1,2,3$ such that $l_i(x_j)=0$ if $i\ne j$ and $l_i(x_i)=1$. Then set $$ P(x) = l_0(x) f(x_0) + l_1(x) f(x_1) + l_2(x) f(x_2) $$ Now to construct the $l_i$'s just do as such (example with $i=0$) : $$ l_0(x) = \frac{(x-x_1)(x-x_2)}{(x_0-x_1)(x_0-x_2)} $$ – 2012-09-12
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0What am I doing incorrectly? $L_0(x)= \frac{3} {20}(x^2-2.85x+2)\cdot\sin( \pi \cdot1) = 0, L_1(x) = \frac{-7} {20}(x^2-2.6x+1.6)\cdot\sin( \pi *1.25), L_2 = \frac{21} {100}(x^2-2.25x+1.25)\cdot\sin( \pi \cdot1.6)$ – 2012-09-12
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0$L_0$ is fine but it is $\frac{20}{3}$ not its inverse. I didn't check the other ones. – 2012-09-12
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0well either way it becomes $0$ so $L_0$ is kinda out of the question – 2012-09-12
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0Whoops $L_1(x) = \frac{-7} {80}(x^2-2.6x+1.6)\cdot\sin( \pi *1.25)$ – 2012-09-12
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1@math101: $$ f(1)\frac{20}{3}(x^2-2.85x+2) -f(1.25)\frac{80}{7}(x^2-2.6x+1.6) +f(1.6)\frac{100}{21}(x^2-2.25x+1.25) $$ – 2012-09-12