Question

(5 pts) Consider the function f(x) = 8e7x. We want to find the Taylor series of f(x) at x = -5. (a) The nth derivative of f(x) is f(n)(x) = At r = -5, we get f(n)(-5) = (c) The Taylor series at r = -5 is +00 T(x) = { (3+5)" n=0 = (d) To find the radius of convergence, we use the ratio test. an+1 L= lim n+too an and so its radius of convergence is R= |x +51 Enter INF if the radius is infinite.

Answer 1

Given functions f()=8279 taylor llor series centered at 2=65) We know that, Taylor series for any function f(x) centered at a is given by; f(x) = f(a) + f (a)(2-a) +f'ta) 6-a33 fla, l-ain t no -se' -35 here; a=-5; (-35) flan e7ae f(-5) = 8 è #f(-5) = (8964)(2] similarly, fW=8670e7a Illa)= f 867030 f. ܀ f'(X) = 8603320 %f"(-5)=() c7@33 '"(-5) =(87(7038 35 f%) = 8(709 yf substituting above values in the known formula to obtain nta 81706835 Taylor sentes 1-35 fQ= 81@5357810e (a+s'+2676235 21 3 + 8(7234-35 to n- ni ni no (275) (x+5)+. 31 + Equation Therefore, from above; Ta fhol) 8 cone and f(-5) = 8606835) also from equation (1); Taylor series at 2=5 's; TOW 8109050 (475) (ory T(a) 8(70007 475) nzo (d) From Ratio Test, for conven convergence nce of seres; Rim antil ell antil (picomezos (4753 ni = lim 1835 Gets (n+121 8 Conless ots)" ni - dim ntly (75) nicht) = lim 1(7)(4+5)*+ 6+1) as nyo; Into also intito + Radlus =) = infinite ..R=INF (infinite) here, an= nal ilim, antilalim nya an nato gitizin nyo h! nyo Scanned with CamScanner