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a problem for calculus enthusiasts
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a problem for calculus enthusiasts
http://sulekha.forumotion.com/t350-a-problem-for-calculus-enthusiasts#727
Jeremiah Mburuburu- Posts : 1251
Join date : 2011-09-09
Re: a problem for calculus enthusiasts
The ordered pairs for g will be
(6.00, 1.92); (5.00, 1.94); (4.40, 1.96); (4.10, 1.98); (4.00, 2.00)
Using the linear approximation technique
g(a+h) = g(a) + h*g'(a)
here h=-0.02
g(4.10) = g(4.00) + (-0.02) * g'(4.00)
1.98=2.00 +(-0.02) * g'(4.00)
-0.02 = -0.02 * g'(4.00)
g'(4.00) = 1
(6.00, 1.92); (5.00, 1.94); (4.40, 1.96); (4.10, 1.98); (4.00, 2.00)
Using the linear approximation technique
g(a+h) = g(a) + h*g'(a)
here h=-0.02
g(4.10) = g(4.00) + (-0.02) * g'(4.00)
1.98=2.00 +(-0.02) * g'(4.00)
-0.02 = -0.02 * g'(4.00)
g'(4.00) = 1
Rishi- Posts : 5129
Join date : 2011-09-02
Re: a problem for calculus enthusiasts
no. you might remember that while f describes how the output of f varies with the input, g, the inverse of f describes how the input to f varies with the output, i.e., the inverse of f tells you the input to f required for a specified output. interpret g', the derivative of the inverse of f similarly.Rishi wrote:g'(4) = 1
Jeremiah Mburuburu- Posts : 1251
Join date : 2011-09-09
Re: a problem for calculus enthusiasts
check your h.Rishi wrote:The ordered pairs for g will be
(6.00, 1.92); (5.00, 1.94); (4.40, 1.96); (4.10, 1.98); (4.00, 2.00)
Using the linear approximation technique
g(a+h) = g(a) + h*g'(a)
here h=-0.02
g(4.10) = g(4.00) + (-0.02) * g'(4.00)
1.98=2.00 +(-0.02) * g'(4.00)
-0.02 = -0.02 * g'(4.00)
g'(4.00) = 1
Jeremiah Mburuburu- Posts : 1251
Join date : 2011-09-09
Re: a problem for calculus enthusiasts
Jeremiah Mburuburu wrote:check your h.Rishi wrote:The ordered pairs for g will be
(6.00, 1.92); (5.00, 1.94); (4.40, 1.96); (4.10, 1.98); (4.00, 2.00)
Using the linear approximation technique
g(a+h) = g(a) + h*g'(a)
here h=-0.02
g(4.10) = g(4.00) + (-0.02) * g'(4.00)
1.98=2.00 +(-0.02) * g'(4.00)
-0.02 = -0.02 * g'(4.00)
g'(4.00) = 1
Thanks
g'(4.00) = -0.2
g(a+h) = g(a) + h*g'(a)
here h=0.10
g(4.10) = g(4.00) + 0.10 * g'(4.00)
1.98=2.00 + (0.10) * g'(4.00)
-0.02 = 0.10 * g'(4.00)
g'(4.00) = -0.02/0.10 = -0.2
Rishi- Posts : 5129
Join date : 2011-09-02
Re: a problem for calculus enthusiasts
yes! however, i'm disappointed that there was no controversy here.Rishi wrote:Jeremiah Mburuburu wrote:check your h.Rishi wrote:The ordered pairs for g will be
(6.00, 1.92); (5.00, 1.94); (4.40, 1.96); (4.10, 1.98); (4.00, 2.00)
Using the linear approximation technique
g(a+h) = g(a) + h*g'(a)
here h=-0.02
g(4.10) = g(4.00) + (-0.02) * g'(4.00)
1.98=2.00 +(-0.02) * g'(4.00)
-0.02 = -0.02 * g'(4.00)
g'(4.00) = 1
Thanks
g'(4.00) = -0.2
g(a+h) = g(a) + h*g'(a)
here h=0.10
g(4.10) = g(4.00) + 0.10 * g'(4.00)
1.98=2.00 + (0.10) * g'(4.00)
-0.02 = 0.10 * g'(4.00)
g'(4.00) = -0.02/0.10 = -0.2
Jeremiah Mburuburu- Posts : 1251
Join date : 2011-09-09
Re: a problem for calculus enthusiasts
It looks like I am beginning to understand calculus.
Btw when you took calculus 1, were you expected to prove theorems like mean value theorem on the exam or at least be able to reproduce steps involved in the proof?
Btw when you took calculus 1, were you expected to prove theorems like mean value theorem on the exam or at least be able to reproduce steps involved in the proof?
Rishi- Posts : 5129
Join date : 2011-09-02
Re: a problem for calculus enthusiasts
Rishi, the problem you and JM solved above is a typical example of the Finite Difference Method used in numerical evaluation / solution of differential equations. Before the computer packages (using FEA etc.) became available to solve differential equations in last 30 yrs. or so, finite difference method was quite popular using the earlier main frame computers to solve differential equations.
Re: a problem for calculus enthusiasts
nearly everything was proved. we were rarely tested directly on the proofs, but the problems tested our understanding of the core concepts.Rishi wrote:It looks like I am beginning to understand calculus.
Btw when you took calculus 1, were you expected to prove theorems like mean value theorem on the exam or at least be able to reproduce steps involved in the proof?
Jeremiah Mburuburu- Posts : 1251
Join date : 2011-09-09
Re: a problem for calculus enthusiasts
Seva Lamberdar wrote:Rishi, the problem you and JM solved above is a typical example of the Finite Difference Method used in numerical evaluation / solution of differential equations. Before the computer packages (using FEA etc.) became available to solve differential equations in last 30 yrs. or so, finite difference method was quite popular using the earlier main frame computers to solve differential equations.
One of the interesting things about the finite difference method is that it very closely demonstrates the originally intended use / application of calculus (as thought by Newton et al.).
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