Taylor Series Question

[MDBFS]

This is the cubic polyn of the Taylor series



I am trying to figure out what is required for A. ii.

 

[MDBFS]

this is what i got for the error term

 

[Steven G]

What is f'(0)?

 

[MDBFS]

What is f'(0)?

f'(0) is 1/4

 

[blamocur]

f'(0) is 1/4

But [imath]f(x) = \sqrt[4]{1-x}[/imath] is a decreasing function. I.e., it cannot have a positive derivative.

 

[BigBeachBanana]

f'(0) is 1/4

You forgot your chain rules.

 

[blamocur]

This is the cubic polyn of the Taylor series
View attachment 32543


I am trying to figure out what is required for A. ii.
View attachment 32544

You have errors in 3 out of 4 terms of your Taylor expansion.

 

[MDBFS]

You have errors in 3 out of 4 terms of your Taylor expansion.

rookie mistake in forgetting to differenciate the bracket, my bad

 

[MDBFS]

You forgot your chain rules.

ahahaha my bad, i fixed it, could you guide me on how I go about doing part ii

 

[Deleted member 4993]

ahahaha my bad, i fixed it, could you guide me on how I go about doing part ii

Part (ii) is fairly simple - guess and try. Let's see some of your efforts in that regard.

 

[MDBFS]

Part (ii) is fairly simple - guess and try. Let's see some of your efforts in that regard.

alright ill give it a shot, is k the f'''' term as in x^4/4! and epsilon 231/256?

 

[Deleted member 4993]

alright ill give it a shot, is k the f'''' term as in x^4/4! and epsilon 231/256?

Show how do those satisfy the given equation.

 

[MDBFS]

Show how do those satisfy the given equation.

i got 0.9 using the value of f''''(0)

 

[blamocur]

View attachment 32549

rookie mistake in forgetting to differenciate the bracket, my bad

Almost there, but the sign for [imath]f^{\prime\prime}[/imath] is wrong.

 

[MDBFS]

Almost there, but the sign for [imath]f^{\prime\prime}[/imath] is wrong.

how? wouldn't the minus signs cancel off?

 

[Deleted member 4993]

View attachment 32552

i got 0.9 using the value of f''''(0)

That does not satisfy the condition of 'K' being an integer. What number has an integer fourth root around 600 ?

 

[MDBFS]

That does not satisfy the condition of 'K' being an integer. What number has an integer fourth root around 600 ?

5?

 

[blamocur]

how? wouldn't the minus signs cancel off?

You have three minuses there -- only two of them cancel off

 

[Deleted member 4993]

5?

5 is incorrect - but I think you meant 625 (=5⁴) and that is CORRECT. Now guess ε and finish the rest.

 

[MDBFS]

5 is incorrect - but I think you meant 625 (=5⁴) and that is CORRECT. Now guess ε and finish

yes k^4 where k = 5
so 5^4
in this case epsilon is 0.04?
so 600 = 5^4 (1 - 0.04)