a problem for calculus enthusiasts

by Jeremiah Mburuburu Wed May 08, 2013 12:32 pm

http://sulekha.forumotion.com/t350-a-problem-for-calculus-enthusiasts#727

by Rishi Wed May 08, 2013 1:23 pm

g'(4) = 1

by Rishi Wed May 08, 2013 1:29 pm

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

by Jeremiah Mburuburu Wed May 08, 2013 1:44 pm

Rishi wrote:g'(4) = 1

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.

by Jeremiah Mburuburu Wed May 08, 2013 1:55 pm

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

check your h.

by Rishi Wed May 08, 2013 2:06 pm

Jeremiah Mburuburu wrote:
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
check your h.

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

by Jeremiah Mburuburu Wed May 08, 2013 2:12 pm

Rishi wrote:
Jeremiah Mburuburu wrote:
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
check your h.

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

yes! however, i'm disappointed that there was no controversy here.

by Rishi Wed May 08, 2013 2:20 pm

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?

by Seva Lamberdar Wed May 08, 2013 2:26 pm

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.

by Jeremiah Mburuburu Wed May 08, 2013 2:29 pm

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?

nearly everything was proved. we were rarely tested directly on the proofs, but the problems tested our understanding of the core concepts.

by Seva Lamberdar Thu May 09, 2013 9:49 am

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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