Taylor Expansions in 2d

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1 Taylor Expansions in 2 In your irst year Calculus course you evelope a amily o ormulae or approximating a unction F(t) or t near any ixe point t 0. The cruest approximation was just a constant. F(t 0 + t) F(t 0 ) The next better approximation inclue a correction that is linear in t. F(t 0 + t) F(t 0 )+F (t 0 ) t The next better approximation inclue a correction that is quaratic in t. F(t 0 + t) F(t 0 )+F (t 0 ) t+ 2 F (t 0 ) t 2 An so on. The approximation that inclues all corrections up to orer t n is F(t 0 + t) F(t 0 )+F (t 0 ) t+ 2! F (t 0 ) t 2 + 3! F(3) (t 0 ) t n! F(n) (t 0 ) t n You may have also oun a ormula or the error introuce in making this approximation. The error E n ( t) is eine by an obeys F(t 0 + t) = F(t 0 )+F (t 0 ) t+ 2! F (t 0 ) t n! F(n) (t 0 ) t n +E n ( t) E n ( t) = (n+)! F(n+) (t ) t n+ or some (unknown) t between t 0 an t 0 + t. Even though we o not know what t is, we can still learn a lot rom this ormula or E n. I we know that F (n+) (t) Mn+ or all t between t 0 an t 0 + t, then E n ( t) M n+ (n+)! tn+, which tells us that E n ( t) goes to zero like the (n+) st power o t as t tens to zero. We now generalize all this to unctions o more than one variable. To be concrete an to save writing, we ll just look at unctions o two variables, but the same strategy works or any number o variables. We ll also assume that all partial erivatives exist an are continuous. Suppose that we wish to approximate 0 + x,y 0 + y) or x an y near zero. The trick is to write 0 + x,y 0 + y) = F() with F(t) = 0 +t x,y 0 +t y) an think o x 0, y 0, x an y as constants so that F is a unction o the single variable t. Then we can apply our single variable ormulae with t 0 = 0 an t =. To o so we nee to compute various erivatives o F(t) at t = 0, by applying the chain rule to F(t) = ( ) with x(t) = x 0 +t x, y(t) = y 0 +t y c Joel Felman. 20. All rights reserve. October 28, 20 Taylor Expansions in 2

2 Since t x(t) = x an y(t) = y, the chain rule gives t F t (t) = 2 F t 2 (t) = ( ) ) tx(t)+ y( t y(t) = x 0 +t x,y 0 +t y) x+ y 0 +t x,y 0 +t y) y ( ) t x(t)+ ( ) ] x y t y(t) x + y ( ) t x(t)+ y y ( ) t y(t) ] y = 2 2 x y y x+ 2 y x y + 2 y 2 y 2 = 2 2 x y x y + 2 y 2 y 2 an so on. It s not har to prove by inuction that, in general, or any natural number k, where ( ) k l = k! l!(k l)! Subbing these into gives F (k) (t) = k ( k k l) l y k l 0 +t x,y 0 +t y) x l y k l l=0 is the stanar binomial coeicient. So when t = 0, F(0) = 0,y 0 ) F (0) = t 0,y 0 ) x+ y 0,y 0 ) y 2 F(0) = 2 2! t ,y 0 ) x y 0,y 0 ) x y y 2 0,y 0 ) y 2 k F k! (0) = t k 0 l,m k l+m=k k l y m 0,y 0 ) x l y m 0 + x,y 0 + y) = F(t 0 + t) t0 =0, t= = n k=0 k! F(k) (0)+E n () 0 + x,y 0 + y) = 0 l,m n l+m n = 0,y 0 ) l+m l y m 0,y 0 ) x l y m +E n + x 0,y 0 ) x+ y 0,y 0 ) y ] xx 0,y 0 ) x 2 +2 xy 0,y 0 ) x y + yy 0,y 0 ) y E 2 I all partial erivatives o o orer n+ are boune in magnitue by M, then En = (n+)! F (n+) (t ) n+ ( n+ ) (n+)! l M x l y n+ l ( ) = M n+ (n+)! x + y l=0 c Joel Felman. 20. All rights reserve. October 28, 20 Taylor Expansions in 2 2

3 Example. In this example, we in the secon orer Taylor expansion o,y) = +4x 2 +y 2 about 0,y 0 ) = (,2) an use it to compute approximately (., 2.05). We irst compute all partial erivatives up to orer 2 at 0,y 0 ).,y) = +4x 2 +y 2 0,y 0 ) = 3 x,y) = 4x x 0,y 0 ) = 4 +4x2 +y 2 3 y y,y) = y 0,y 0 ) = 2 +4x 2 +y xx,y) = 6x 2 +4x 2 +y 2 +4x 2 +y 2 ] 3/2 xx 0,y 0 ) = = xy xy,y) = +4x 2 +y 2 ] 3/2 xy 0,y 0 ) = 8 27 yy,y) = y 2 +4x 2 +y 2 +4x 2 +y 2 ] 3/2 yy 0,y 0 ) = = 5 27 So the quaratic approximation to about 0,y 0 ) is ( x 0 + x, y 0 + y ) 0,y 0 )+ x 0,y 0 ) x+ y 0,y 0 ) y ] xx 0,y 0 ) x 2 +2 xy 0,y 0 ) x y + yy 0,y 0 ) y In particular, with x = 0. an y = 0.05, = x y + 27 x x y y2 (., 2.05) (0.) (0.05)+ 27 (0.0) 8 27 (0.005) (0.0025) = 3.69 The actual value, to our ecimal places, is Example. In this example, we in the thir orer Taylor expansion o,y) = e 2x sin(3y) about 0,y 0 ) = (0,0) in two ierent ways. The irst way uses the canne ormula. We irst compute all partial erivatives up to orer 3 at 0,y 0 ).,y) = e 2x sin(3y) 0,y 0 ) = 0 x,y) = 2e 2x sin(3y) x 0,y 0 ) = 0 y,y) = 3e 2x cos(3y) y 0,y 0 ) = 3 xx,y) = 4e 2x sin(3y) xx 0,y 0 ) = 0 xy,y) = 6e 2x cos(3y) xy 0,y 0 ) = 6 yy,y) = 9e 2x sin(3y) yy 0,y 0 ) = 0 xxx,y) = 8e 2x sin(3y) xx 0,y 0 ) = 0 xxy,y) = 2e 2x cos(3y) xxy 0,y 0 ) = 2 xyy,y) = 8e 2x sin(3y) xyy 0,y 0 ) = 0 yyy,y) = 27e 2x cos(3y) yyy 0,y 0 ) = 27 c Joel Felman. 20. All rights reserve. October 28, 20 Taylor Expansions in 2 3

4 So the Taylor expansion, about (0,0) to orer three is,y) = 0 l,m 3 l+m 3 l+m l y m (0,0) x l y m +E 3,y) = 0!! 3y +!! 6xy + 2!! 2x2 y 0!3! 27y3 +E 3,y) = 3y +6xy +6x 2 y 9 2 y3 +E 3,y) A secon way to get the same result exploits the single variable Taylor expansions e x = +x+ 2! x2 + 3! x3 + siny = y 3! y3 + Replacing x by 2x in the irst an y by 3y in the secon an multiplying the two together, keeping track only o terms o egree at most three, gives,y) = e 2x sin(3y) ] ] = +(2x)+ 2! (2x)2 + 3! (2x)3 + (3y) 3! (3y)3 + ] ] = +2x+2x x3 + 3y 9 2 y3 + just as in the irst computation. = 3y 9 2 y3 +6xy +6x 2 y + Problem Derive the Taylor expansion, to orer 2 in powers o x, y an z, o ( x 0 + x,y 0 + y,z 0 + z ) Answer: In general, the expansion to orer N is 0 + x,y 0 + y,z 0 + z) = 0 l,m,n 2 l+m+n N n! l+m+n l y m z n 0,y 0,z 0 ) ( x) l ( y) m ( z) n +E n Problem 2 Fin theseconorertaylor expansion o,y) = +x 2 +y 2 about 0,y 0 ) = (, ). Answer: (+ x, + y) = x+ 2 9 y + 27 x x y + 27 y2 + Problem 3 FintheseconorerTaylor expansiono,y,z) = +x 2 +y 2 +z 2 about 0,y 0,z 0 ) = (0,0,0). Answer:,y,z) = x 2 y 2 z 2 + c Joel Felman. 20. All rights reserve. October 28, 20 Taylor Expansions in 2 4

5 Problem 4 Fin the secon orer Taylor expansion o,y) = e x+2y about 0,y 0 ) = (0,0). Boun the error o this approximation i x 0. an y 0.. Answer:,y) = + x + 2y + 2 x2 + 2xy + 2y 2 + E 2,y) with E2,y) (0.3) 3 3! e 0.3 when x 0. an y 0.. (O course there are many other correct bouns on E 2,y).) Problem 5 The unction z =,y) solves the equation x+2y +z +e 2z =. Fin the Taylor polynomial o egree two or,y) in powers o x an y. Answer:,y) = 3 x 2 3 y 2 27 x xy 8 27 y2 + Problem 6 Suppose that,y) is a polynomial o egree n. Prove that or all real numbers x 0, y 0, x an y, 0 + x,y 0 + y) = 0 l,m n l+m n Note that this ormula is exact. There is no error term E n. l+m l y m 0,y 0 ) ( x) l ( y) m c Joel Felman. 20. All rights reserve. October 28, 20 Taylor Expansions in 2 5