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# Towards Maclaurin

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Age 16 to 18

Challenge Level

(1) We know $\cos x \leq 1$ for all $x$. By considering the derivative of the function $$f(x) = x - \sin x$$ prove that $\sin x \leq x$ for $x \geq 0$.

(2) By considering the derivative of the function $$f(x) = \cos x - \left(1 - {x^2\over 2}\right)$$ prove that $\cos x \geq 1 - {x^2\over 2}$ for $x \geq 0$.

(3) By considering the derivative of the function $$f(x) = \left(x - {x^3 \over 3!}\right) - \sin x $$ prove that $\sin x \geq (x - {x^3 \over 3!})$ for $x \geq 0$.

(4) By considering the derivative of the function $$f(x) = \cos x - \left(1 - {x^2 \over 2!} + {x^4\over 4!}\right) $$ prove that $\cos x \leq \left(1 - {x^2\over 2!} + {x^4 \over 4!}\right)$ for $x \geq 0$.

(5) What can you say about continuing this process?

(2) By considering the derivative of the function $$f(x) = \cos x - \left(1 - {x^2\over 2}\right)$$ prove that $\cos x \geq 1 - {x^2\over 2}$ for $x \geq 0$.

(3) By considering the derivative of the function $$f(x) = \left(x - {x^3 \over 3!}\right) - \sin x $$ prove that $\sin x \geq (x - {x^3 \over 3!})$ for $x \geq 0$.

(4) By considering the derivative of the function $$f(x) = \cos x - \left(1 - {x^2 \over 2!} + {x^4\over 4!}\right) $$ prove that $\cos x \leq \left(1 - {x^2\over 2!} + {x^4 \over 4!}\right)$ for $x \geq 0$.

(5) What can you say about continuing this process?