Answer:
The four first terms are:
[tex]7x,7x^{2},\frac{21x^{3}}{6},\frac{7x^{4}}{6}[/tex]
Step-by-step explanation:
The function is:
[tex]f(x)=7xe^{x}[/tex]
The Taylor series around a is given by:
[tex]F(x)=\sum^{\infty}_{n=0} \frac{f^{n}(a)(x-a)^{n}}{n!}[/tex]
The first 4 terms will be:
[tex]F(x)=f(0)+\frac{f^{'}(0)(x)}{1}+\frac{f^{''}(0)(x)^{2}}{2}+\frac{f^{'''}(0)(x)^{3}}{6}[/tex]
Let's find first the derivatives:
[tex]f'(x)=7(xe^{x}+e^{x})[/tex]
[tex]f'(0)=7(0e^{0}+e^{0})=7[/tex]
[tex]f''(x)=7xe^{x}+7e^{x}+7e^{x}=7xe^{x}+14e^{x}[/tex]
[tex]f''(0)=14[/tex]
[tex]f'''(0)=21[/tex]
[tex]f''''(0)=28[/tex]
[tex]F(x)=0+\frac{7(x)}{1}+\frac{14(x)^{2}}{2}+\frac{21(x)^{3}}{6}+\frac{28(x)^{4}}{24}[/tex]
Therefore, the four first terms are:
[tex]7x,7x^{2},\frac{21x^{3}}{6},\frac{7x^{4}}{6}[/tex]
I hope it helps you!
So the Taylor series for the function informed will be:
[tex]7x; 7x^2; \frac{21x^3}{6} ; \frac{7x^4}{6}[/tex]
The function is:
[tex]f(x)= 7e^xx[/tex]
The Taylor series around a is given by:
[tex]f(x)= \sum \frac{f^n (a) (x-a)^n }{n!}[/tex]
The first four terms will be:
[tex]F(x)=f(0)+\frac{f'(0)(x)}{1}+\frac{f''(0)(x)^2}{2} + \frac{f'''(0)(x)^3}{6}[/tex]
Let's find first the derivaties:
[tex]f'(x)= 7(e^xx+e^x)\\f'(0)= 7\\f''(x)= 7e^xx+14e^xx\\f''(0)=14\\f'''(0)= 21\\f''''(0)= 28[/tex]
[tex]F(x)= 0+\frac{7x}{1}+\frac{14x^2}{2}+\frac{21x^3}{6}+\frac{28x^4}{24}[/tex]
Therefore, the four first terms are:
[tex]7x; 7x^2; \frac{21x^3}{6} ; \frac{7x^4}{6}[/tex]
See more about Taylor series at : brainly.com/question/6953942
