Given $f(x),f(x+h),f(x+3h),f(x-5h)$, approximate $f''(x)$.

Book's solution: From Taylor' series, $$ \\ f(x+h)=f(x)+f'(x)h+0.5f''(x)h^2+{1\over6}f'''(x)h^3+O(h^4) \\ f(x+3h)=f(x)+3f'(x)h+4.5f''(x)h^2+4.5f'''(x)h^3+O(h^4) \\ f(x-5h)=f(x)-5f'(x)h+12.5f''(x)h^2-{125\over6}f'''(x)h^3+O(h^4) $$ Thus, $$ \\ af(x+h)+bf(x+3h)+cf(x-5h)+df(x)= \\(a+b+c+d)f(x)+(a+3b-5c)f'(x)h+(0.5a+4.5b+12.5c)f''(x)h^2+({a\over6}+{27\over6}-{125\over 6})f'''(x)h^3+O(h^4) $$ Now we want to set the cofficients of $f'(x)$ and of $f'''(x)$ because they are not given. So we need to calculate the linear equations system $$ \\ a+b+c+d=0 \\a+3b-5b=0 \\0.5a+4.5b+12.5c=a \\{a\over6}+{27b\over6}-{125c\over6}=0 $$

My question is why did they set the third equation they on $a$?


They want to find a linear combinaison so that only the third order term remains. So they need to cancel the terms of other orders and have a non-zero number otherwise. They could have picked any non-zero value, it just turns out that it easier if you pick a.

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    $\begingroup$ I don't understand your answer. Are you saying that the third equation could be $0.5a+4.5b+12.5c=1000a?$ $\endgroup$ – mfl Feb 21 at 20:19
  • $\begingroup$ I would have chosen $1$, to get $f''$ directly. Notice that if you set it to $1000a$ you get $f''(x)=\frac 1{1000 a}(af(x+h)+bf(x+3h)+cf(x-5h)+df(x))$. $\endgroup$ – Andrei Feb 21 at 21:42
  • $\begingroup$ Yes @mfl Exactly $\endgroup$ – PackSciences Feb 22 at 4:44

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