# \$P_3(x)\$ and \$R_3(x)\$ of \$f(x)=e^{-3x}+3 sin (x)-1\$ Code Answer

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the maclarin series for \$sin(x)\$

is \$\$ sin(x)=sum^{infty}_{k=0} frac{(-1)^k*x^{2k+1}}{(2k+1)!}\$\$

so summation to \$k=3\$

\$\$
begin{aligned}
sum^{k=3}_{k=0} frac{(-1)^k*x^{2k+1}}{(2k+1)!}
&= frac{(-1)^0*x^{2*0+1}}{(2*0+1)!}
+ frac{(-1)^1*x^{2*1+1}}{(2*1+1)!}
+frac{(-1)^2*x^{2*2+1}}{(2*2+1)!}
+ frac{(-1)^3*x^{2*3+1}}{(2*3+1)!}
\&=x-frac{x^3}{3!}+frac{x^5}{5!}-frac{x^7}{7!}
end{aligned}
\$\$

the maclarin series for \$e^{-3x}\$ using \$e^x=sum frac{x^k}{k!}\$

\$\$begin{aligned}
e^{-3x} &= sum^{infty}_{k=0} frac{(-3*x)^k}{k!}= frac{(-3*x)^0}{0!}+frac{(-3*x)^1}{1!}+frac{(-3*x)^2}{2!}+frac{(-3*x)^3}{3!}
\ &=1-3x+frac{3^2x^2}{2!}-frac{3^3*x^3}{3!}
end{aligned}
\$\$

So, the third degree polynomial of the function is nothing more than

\$\$
begin{aligned}
P_3(x)&=sum^3_0 frac{(-3x)^k}{k!} +3*sum^3_0 frac{(-1)^k*x^{2k+1}}{(2k+1)!}-1
\ &=1-3x+frac{3^2x^2}{2!}-frac{3^3*x^3}{3!}+ 3(x-frac{x^3}{3!}+frac{x^5}{5!}-frac{x^7}{7!})-1
\ &=1-3x+frac{3^2x^2}{2!}-frac{3^3*x^3}{3!}+ (3x-3frac{x^3}{3!}+3frac{x^5}{5!}-3frac{x^7}{7!})-1
\& =-3x+frac{3^2x^2}{2!}-frac{3^3*x^3}{3!}+ (3x-3frac{x^3}{3!}+3frac{x^5}{5!}-3frac{x^7}{7!})
\& =frac{3^2x^2}{2!}-frac{3^3*x^3}{3!}+ (-3frac{x^3}{3!}+3frac{x^5}{5!}-3frac{x^7}{7!})
\&=frac{3^2x^2}{2!}
+(-frac{3^3*x^3}{3!}+ -3frac{x^3}{3!})+ 3frac{x^5}{5!}-3frac{x^7}{7!}
\&=frac{3^2x^2}{2!}+
(-3-3^3)frac{x^3}{3!}+ 3frac{x^5}{5!}-3frac{x^7}{7!}
\&=frac{9*x^2}{4}+
(-30)frac{x^3}{3*2}+ 3frac{x^5}{5*4*3*2*1}-3frac{x^7}{7*6*5**4*3*2*1}
\&=frac{9*x^2}{4}+
(-5)frac{x^3}{1}+ frac{x^5}{5*4*2*1}-frac{x^7}{7*6*5**4*2*1}
\&=frac{9x^2}{4}
-5x^3+ frac{x^5}{40}-frac{x^7}{1680}
end{aligned}
\$\$

finding the primes of \$f\$

\$\$begin{aligned}
f^0(x)&=e^{-3x}+3sin(x)-1
\f^1(x)&= -3*e^{-3x}+3cos(x)
\ f^2(x)&=(-3)^2*e^{-3x}-3*sin(x)
\ f^3(x)&=(-3)^3*e^{-3x}+3cos(x)
\ f^4(x)&=(-3)^4e^{-3x}+3sin(x)
end{aligned} \$\$

The remainder of the \$P_n(x)\$ is \$\$R_n(x)=f^{n+1}(c)/(n+1)! (x-a)^{n+1} text{ for some } xin[a,x]\$\$

In our case \$n=3\$ and \$cin[0,x]\$

\$\$R_3(x)=frac{f^4(c)}{4!}(x)^{4} =left ((-3)^4e^{-3c}+3sin(c) right)*frac{x^4}{4!} \$\$

Anything wrong????? If no I want to delete it Or even better vote to close it if its ok

I think that you made a few mistakes in your calculations (in particular \$2!=2\$ and not \$4\$).
\$\$e^{-3x}=1-3 x+frac{9 x^2}{2}-frac{9 x^3}{2}+frac{27 x^4}{8}-frac{81 x^5}{40}+frac{81
x^6}{80}-frac{243 x^7}{560}+Oleft(x^8right)\$\$
\$\$sin(x)=x-frac{x^3}{6}+frac{x^5}{120}-frac{x^7}{5040}+Oleft(x^8right)\$\$ These make
\$\$e^{-3x}+3sin(x)-1=frac{9 }{2}x^2-5 x^3+frac{27 }{8}x^4-2 x^5+frac{81 }{80}x^6-frac{73
}{168}x^7+Oleft(x^8right)\$\$