63. Representation of functions as power series Consider a power series. ( 1) n x 2n for all 1 < x < 1


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1 3 9. SEQUENCES AND SERIES 63. Representtion of functions s power series Consider power series x 2 + x 4 x 6 + x 8 + = ( ) n x 2n It is geometric series with q = x 2 nd therefore it converges for ll q = x 2 < or x (, ). Using the formul for the sum of geometric series, one infers tht + x 2 = x2 + x 4 + = ( ) n x 2n for ll < x < This shows tht the function /( + x 2 ) cn be represented s power series in the open intervl (, ). Note well tht the found representtion is vlid only in the intervl of convergence of the power series despite tht the function /( + x 2 ) is defined on the entire rel line. In generl, one cn construct representtion of function by power series in (x ) for some. The intervl of vlidity of this representtion depends on the choice of. Exmple 2. Find representtion of /x s power series in (x ), >, nd determine the intervl of its vlidity. Solution: Put y = x. The function cn rewritten in the form tht resembles the sum of geometric series: x = ( + y/) = ( y ) n ( ) n = (x n+ )n, x (, 2) The geometric series converges if q = y/ = y / < nd, hence, this representtion is vlid only if < y < or < x < or < x < Differentition nd integrtion of power series. A formul for the sum of power series x n is often complicted nd, in most cses, cnnot even be found explicitly. How cn functions defined by power series be differentited nd integrted? If function is finite sum f(x) = u (x)+ +u n (x), then the derivtive is the sum of derivtives f = u + + u n nd, similrly, the integrl is the sum of integrls fdx = u dx + + u n dx. This is not generlly true for infinite sums. As n exmple, consider function defined by the series f(x) = u n (x) = n= n= sin(nx) n 2
2 63. REPRESENTATION OF FUNCTIONS AS POWER SERIES 3 By comprison with p series: u n (x) = sin(nx) /n 2 /n 2, this series converges for ll x becuse /n 2 converges. If the series is differentited just like finite sum, i.e. termbyterm, u n (x) = cos(nx)/n, then the series u n (x) diverges for x = 2πk for ny integer k s the hrmonic series /n. So, f (2πk) does not exist. Thus, lthough the terms u n (x) re differentible functions in the intervl of convergence of the series u n, the series of derivtives u n my not converge nd, hence, f = u n my not be differentible everywhere in its domin. It ppers tht if u n (x) = (x ) n, tht is, u n (x) is power series, then the termbyterm differentition or integrtion is justified. A proof of this ssertion goes beyond the scope of this course. Theorem 46. (Differentition nd integrtion of power series) If the power series (x ) n hs nonzero rdius of convergence R >, then the function f defined by f(x) = c + c (x ) + c 2 (x ) 2 + = (x ) n is differentible (nd therefore continuous) on the intervl ( R, +R) nd f (x) = c + 2c 2 (x ) + 3c 3 (x ) 2 + = n (x ) n (x ) 2 f(x)dx = C + c (x ) + c + = C + 2 n= The rdii of convergence of these power series re both R. (x ) n+ n + Thus, for power series the differentition or integrtion nd the summtion cn be crried out in ny order: d cn (x ) n = d dx dx [(x ) n ] ( ) cn (x ) n dx = [ (x ) n ]dx Remrk. Theorem 46 sttes the rdius of convergence of power series does not chnge fter differentition or integrtion of the series. This does not men tht the intervl of convergence does not chnge. It my hppen tht the originl series converges t n endpoint, wheres the differentited series diverges there. Exmple 3. Find the intervls of convergence for f, f, nd f if f(x) = n= xn /n 2
3 32 9. SEQUENCES AND SERIES Solution: Here = /n 2 n nd, hence, = / n n 2 = (/ n n) 2 = α. So the rdius of convergence is R = /α =. For x = ±, the series is p series /n 2 which converges (p = 2 > ). Thus, f(x) is defined on the closed intervl x [, ]. By Theorem.27, the derivtives f (x) = n= xn /n nd f (x) = n=2 (n )x n 2 /n hve the sme rdius of convergence R =. For x =, the series f ( ) = ( ) n /n is the lternting hrmonic series which converges, wheres the series f ( ) = ( ) n (n )/n diverges becuse the sequence of its terms does not converge to zero: ( ) n (n )/n = /n. For x =, the series f () = /n is the hrmonic series nd, hence, diverges. The series f () = (n )/n lso diverges ((n )/n does not converge to zero). Thus, the intervls of convergence for f, f, nd f re, respectively, [, ], [, ), nd (, ). The termbyterm integrtion of power series cn be used to obtin power series representtion of ntiderivtives. Exmple 4. Find power series representtion for tn (x) Solution: tn (x) = dx + x 2 = ( ( x 2 ) )dx n = C + ( ) n+ x2n+ 2n + Since tn () =, the integrtion constnt C stisfies the condition = C + or C =. The geometric series with q = x 2 converges if q <. Hence, the rdius of convergence of the series for tn (x) is R = (the power series representtion is vlid for x (, )). In prticulr, the number / 3 is less thn the rdius of convergence of the power series for tn (x). So, the number tn (/ 3) = π/6 cn be written s the numericl series by substituting x = / 3 into the power series for tn (x). This leds to the following representtion of the number π: π = 2 ( ) n 3 (2n + )3 n Power series nd differentil equtions. A power series representtion is often used to solve differentil equtions. A reltion between function f(x), its rgument x, nd its derivtives f (x), f (x) etc. is clled differentil eqution. A function f(x) tht stisfies differentil eqution is generlly difficult to find in closed form. A power series representtion turns out to be helpful. Since in this representtion function is defined by sequence { }, f(x) = x n, nd
4 63. REPRESENTATION OF FUNCTIONS AS POWER SERIES 33 so re its derivtives f (k) (x), differentil eqution imposes conditions on which re solved recursively. Exmple 5. Find power series representtion of the solution of the eqution f (x) = f(x) nd determine its rdius of convergence. Solution: Put f(x) = x n nd, hence, f (x) = n x n. Then the eqution f = f gives: c + 2c 2 x + 3c 3 x 2 + 4c 4 x 3 + = c + c x + c 2 x 2 + c 3 x 3 + By mtching the coefficients t the monomil terms, x, x 2, x 3, etc., one finds: c = c c 2 = c 2, c 3 = c 2 3,..., = n Using the ltter reltion recursively: = n = n(n ) 2 = n(n )(n 2) 3 = = c So, f(x) = c n= xn / where c is constnt (the eqution is stisfied for ny choice of c ). By the rtio test, the series converges for ll x (so R = ). Indeed, = / nd + / = /(n + ) = α nd, hence, R = /α =. For this simple differentil eqution, it is not difficult to find f(x) = c e x by reclling the properties of the exponentil function: (e x ) = e x. The condition f() = e = determines the constnt c =. Thus, the exponentil function hs the following power series representtion (6) e x = + x! + x2 2! + x3 3! + = The series converges on the entire rel line. In prticulr, the number e hs the following series representtion e = +! + 2! + 3! + = Approximtion of definite integrls. If n indefinite integrl of f(x) is difficult to obtin, then the evlution of the integrl b f(x)dx poses problem. A power series representtion offers simple wy to pproximte the vlue of the integrl. Suppose tht f(x) = x n for R < x < R. By Theorem 46, for ny x n
5 34 9. SEQUENCES AND SERIES R < < b < R, b f(x)dx = b x n dx = b n+ n + n+ n + n b k+ n c k k + c k k+ k + Errors of the pproximtion of the series sum by finite sums hve been discussed erlier. Exmple 6. How mny terms does one need in the power series pproximtion of the integrl of f(x) = e x2 over the intevl [, ] to mke the bsolute error smller thn 5? Solution: Note first tht the indefinite integrl e x2 dx cnnot be expressed in elementry functions! So, direct use of the fundmentl theorem of clculus becomes problemtic. However, e x2 dx cn be represented s power series tht converges on the entire rel line by replcing x in Eq. (6) by ( x 2 ). One hs e x2 dx = ( ) k k! x 2k dx = ( ) k k!(2k + ) n ( ) k k!(2k + ) To determine n in the finite sum pproximtion of the series, recll the lternting series estimtion theorem (Theorem 37) where b n = /((2n + )): e x2 dx n ( ) k b n+ = k!(2k + ) (n + )!(2n + 3) < 5 A direct clcultion shows tht b nd b So n = 7 is sufficient to pproximte the integrl with the required ccurcy.
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