Find the Taylor polynomials of degree n approximating the functions for x near 0. (Assume p is a constant) tan⁡x,n=3, 4

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Find the Taylor polynomials of degree n approximating the functions for x near 0. (Assume p is a constant) tan⁡x,n=3, 4

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Function f(x) is given asf(x)=tan⁡x AndNow we need to calculate the derivatives of the above function at x= 0f′(x)=sec2xf′(0)=sec20=1f″(zx)=2tan⁡xsec2xf″(0)=2tan⁡0sec20=0f(3)(x)=4tan2xsec2x+2sec4xf(3)(0)=4tan20sec20+2sec40=2f(4)(x)=8tan3xsec2x+16tan⁡xsec4xf(4)(0)=8tan30sec20+16tan⁡0sec40=0Since the Taylor Polynomial of Degree n Approximating f(a) for x near 0 is given asf(x)≈Pn(x)=f(0)+f′(0)(x)+f′(0)2!(x)2+…+f(n)(0)n!(x)nWe call Pn(x) the Taylor polynomial of degree n centered at x = 0 or the Taylor polynomial about x = 0n−3f(x)≈P3(x)=f(0)+f′(0)(x)+f′(0)2!(x)2+…+f(n)(0)n!(x)n=0=(1)x+(0)x22!+(2)x33!=x+x33

Witionsion · Accepted Answer

Recall the useful identities that sin⁡(0)=0 and cos⁡(0)=1.f(x)=tan⁡(x) f(0)=0f′(x)=(cos⁡x)2 f′(0)=1f′(x)=−2(cos⁡x)−3(−sin⁡x)=2(cos⁡x)−3(sin⁡x) f′(0)=0f(3)(x)=−6(cos⁡x)−4(−sin⁡x)(sin⁡x)+2(cos⁡x)−3(sin⁡x)=6(cos⁡x)−4(sin⁡x)2+2(cos⁡x)−2 f(3)(0)=2f(4)(x)=−24(cos⁡x)−5)−(sin⁡x)(sin⁡x)2+12(cos⁡x))−4(sin⁡x)(cos⁡x)−4(cos⁡x)−3(−sin⁡x)=24(cos⁡x)−5(sin⁡x)3+16(cos⁡x)}−3}(sin⁡x) f(4)(0)=0Because the fourth derivative is zero, both P3(x) and P4(x) will be equal:P4(x)=0+1×x+012x2+213!x3+014!x4=x+13x3

Find the Taylor polynomials of degree n approximating the functions

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