2. Let ƒ : R² → R where f(x, y) = x²y, c = (1, 2), and u = (1,0). Use the E-6 definition to prove L₂ =? (you need to find Lu in your scratchwork).

Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
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Real Analysis II Please find related sample as guide to solve above problem
fi
fa
f: R² →→ R², f(x, y) = (x²y (3x + 2y), U=(1,0), C = (1.1)
claim
Lu= (2,3)
ex.
Fix eso. Choose d
ff:
E
choose d= [2]. Assume 0</4/<d. Then
ок
11/t [f (c + tu) - f(c)] - Lu|| = || / f ( 1++, 1) - (1,5)]-(2,3)||
= || / + [ (1+4) ², 3 (1++) + 2) - (1,5)] = (2,3)||
[[ ' / E ((₁+2+ + + ²₁, 3+ +5) = (1,5)] - (2,3)||
2
work
=
11 / [(2+ + t²₁ 3+) ] - (2,3) || = || ( 2 ++, 3) - (2,3)||
~₁ = || (t,0) || = |t| 2 √ = E
5
(1
why is Lue = (2,3)?
[2f₁ 2f1
2x
24
242 212
L2x
L
=
Lu = 2xy x²
3
17
2
[2xy x²
11
3
2xy
3
2
1
Lul
faxy
3
3
Transcribed Image Text:fi fa f: R² →→ R², f(x, y) = (x²y (3x + 2y), U=(1,0), C = (1.1) claim Lu= (2,3) ex. Fix eso. Choose d ff: E choose d= [2]. Assume 0</4/<d. Then ок 11/t [f (c + tu) - f(c)] - Lu|| = || / f ( 1++, 1) - (1,5)]-(2,3)|| = || / + [ (1+4) ², 3 (1++) + 2) - (1,5)] = (2,3)|| [[ ' / E ((₁+2+ + + ²₁, 3+ +5) = (1,5)] - (2,3)|| 2 work = 11 / [(2+ + t²₁ 3+) ] - (2,3) || = || ( 2 ++, 3) - (2,3)|| ~₁ = || (t,0) || = |t| 2 √ = E 5 (1 why is Lue = (2,3)? [2f₁ 2f1 2x 24 242 212 L2x L = Lu = 2xy x² 3 17 2 [2xy x² 11 3 2xy 3 2 1 Lul faxy 3 3
definition to prove Lu=? (you need to find Lu in your scratchwork).
2. Let f: R² → R where f(x, y) = x²y, c = (1,2), and u = (1,0). Use the
E-6 definition to prove L₁ =? (you need to find Lu in your scratchwork).
3 Let f. 2
(²1 3r I
K R2 where f(rw)
(1.1) and
Transcribed Image Text:definition to prove Lu=? (you need to find Lu in your scratchwork). 2. Let f: R² → R where f(x, y) = x²y, c = (1,2), and u = (1,0). Use the E-6 definition to prove L₁ =? (you need to find Lu in your scratchwork). 3 Let f. 2 (²1 3r I K R2 where f(rw) (1.1) and
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