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Let $f(x)=x^4$ be approximated by a polynomial of degree less or equal to 2, which interpolates $x^4$ at x = -1,0,1then the maximum absolute interpolation error over the interval[-1,1] is equal to?

I use the result of error term in lagrange interpolation that is $Mh^nh/(n+1)$ and get the bound 2 but in the book they provide the answer is 0.22 to 0.28. please help me

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    I may be misintepreting the problem statement, but: The approximation is quite obviously $g(x)=x^2$ because we are required to have $g(0)=0$, $g(\pm1)=1$. We have to find the maximum of $|f(x)-g(x)|$ for $x\in[-1,1]$. As the derivative of $f(x)-g(x)$ is $4x^3-2x$, the candidates are $x=0$ and $x=\pm\frac1{\sqrt 2}$. The error is $0$ at $x=0$, but it is $\frac14$ at $x=\pm\frac1{\sqrt 2}$. – 2017-01-24
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    @Hagen von Eitzen thanks but this function $x^4−x^2$ has minimum at x=1/sqrt(2) not maximum,how can u say that maximum is 1/4? kindly explain it – 2017-01-24
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    Because we actually look for the maximum for $|f(x)-g(x)|$ – 2017-01-24

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So there is a known error bound for Lagrange interpolation on equally spaced nodes on $[a, b]$, namely (from Wikipedia: https://en.wikipedia.org/wiki/Polynomial_interpolation#For_equally_spaced_intervals)

$$|R_n(x)| \le {h^{n+1} \over 4(n+1)} \max_{\xi \in [a, b]} |f^{(n+1)}(\xi)|$$

where $n$ is the degree of the polynomial and $h$ is the spacing between nodes.

In your case $h = 1, n = 2$ and so you get

$$|R_3(x)| \le {1 \over 12} \max_{\xi \in [-1, 1]} |f^{(3)} (\xi)|$$

Now $f^{(3)}(x) = 24x$, and the maximum of this function on $[-1, 1]$ is 24, so you get giving $|R_3(x)| \le 1/12 \times 24 = 2$, as you specified.

But now let's think about what this polynomial actually is. The interpolating polynomial is $g(x) = x^2$ (since this has the same value as $f(x) = x^4$ at -1, 0, 1), and so you're looking for the maximum of $|x^4 - x^2|$ over $[-1, 1]$. This can be seen to be $1/4$ by elementary calculus. This doesn't contradict the bound given, because this bound applies to any function $f$, while it turns out that $x^4$ "looks like" a quadratic on $[-1, 1]$ and therefore can be approximated relatively well by a quadratic.

As to why your text gives the range $[0.22, 0.28]$, is it possible that they are doing some numerical approximation?

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    thanks,but this fuction $x^4-x^2$ has minimum at $x=1/sqrt(2)$ not maximum,how can u say that is 1/4 – 2017-01-24
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    thanks,but this function $x^4−x^2$ has minimum at x=1/sqrt(2) not maximum,how can u say that maximum is 1/4? kindly explain it – 2017-01-24
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    For $x^4-x^2$ , points of inflection are $\frac{1}{\sqrt{2}}$ and $\frac{-1}{\sqrt{2}}$ . Taking the second derivative, we will see that for both the points, $f''(x) > 0$ implying a minimum. So how did you conclude? – 2017-10-05