How to calculate cos x from its nth term in Maclaurin series??
The Maclaurin series of "f(x)=\\cos x" is
"f(x)=\\sum_{n=0}^{\\infty}(-1)^{n} \\frac{x^{2 n}}{(2 n) !}"
Let us look at some details.
The Maclaurin series for f(x) in general can be found by
"f(x)=\\sum_{n=0}^{\\infty} \\frac{f^{(n)}(0)}{n !} x^{n}"
Let us find the Maclaurin series for "f(x)=\\cos x."
By taking the derivatives,
"\\begin{aligned}\n\n&f(x)=\\cos x \\Rightarrow f(0)=\\cos (0)=1 \\\\\n\n&f^{\\prime}(x)=-\\sin x \\Rightarrow f^{\\prime}(0)=-\\sin (0)=0 \\\\\n\n&f^{\\prime \\prime}(x)=-\\cos x \\Rightarrow f^{\\prime \\prime}(0)=-\\cos (0)=-1 \\\\\n\n&f^{\\prime \\prime \\prime}(x)=\\sin x \\Rightarrow f^{\\prime \\prime \\prime}(0)=\\sin (0)=0 \\\\\n\n&f^{(4)}(x)=\\cos x \\Rightarrow f^{(4)}(0)=\\cos (0)=1\n\n\\end{aligned}"
Since "f(x)=f^{(4)}(x)," the cycle of "\\{1,0,-1,0\\}" repeats itself.
So, we have the series
"f(x)=1-\\frac{x^{2}}{2 !}+\\frac{x^{4}}{4 !}-\\cdots=\\sum_{n=0}^{\\infty}(-1)^{n} \\frac{x^{2 n}}{(2 n) !}"
Comments
Leave a comment