compute function f(x)

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janeann

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Compute the function f(x) when d^3f/dx^3=cos(x) and f(0)=1, f'(0)=-2, and f''=1/2.

I know somehow i have to work backwards on this function maybe cos(1)=1, sin(1)=0 ahh i have no clue where to begin! please help!
 
janeann said:
I know somehow i have to work backwards on this function maybe cos(1)=1, sin(1)=0 ahh i have no clue where to begin!
Click to expand...
This makes no sense. WORK BACKWARDS! You told us that. This means you DO have a clue.

\(\displaystyle \frac{d}{dx}\sin(x) = \cos(x)\)

\(\displaystyle \frac{d}{dx}\cos(x) = -\sin(x)\)

That's about all you need.
 
After what? I just repeated formulae. You have to do the "work backwards" part.
 
yea i dont get how to work backwords from that:
for f(0) if you plug it in sin(0)=cos(0)
0=1?
then f'(0) you get 1=0
That makes no sence to me....sorry
 


If you find it easier to differentiate forwards versus backwards, you can do that to answer this exercise. I mean, there are only four possibilities to consider.

So, I will give you a hint, and you can try differentiating forwards instead of backwards, if you like.

We are told that the third derivative of some unknown function f is f```(x) = cos(x).

This means that f(x) must be one of the following:

Case I: f(x) = cos(x)

Case II: f(x) = sin(x)

Case III: f(x) = -cos(x)

Case IV: f(x) = -sin(x)

So, for each case, you determine the first derivative f`(x), the second derivative f``(x), and the third derivative f```(x).

When you find the case that ends up with f```(x) = cos(x), then you will know which case represents the correct function f.

By then, perhaps, you will also realize how you could have worked this exercise backwards just once. 8-)

Cheers ~ Mark

 
janeann said:
Compute the function f(x) when d^3f/dx^3=cos(x) and f(0)=1, f'(0)=-2, and f''=1/2. Is that correct?
I know somehow i have to work backwards on this function maybe cos(1)=1, sin(1)=0 ahh i have no clue where to begin! please help!
Click to expand...

Are you doing Differential equations (DE)?

This is a third order DE with three initial conditions given.

Doing this problem assumes you know how to find anti-derivatives (like TK suggested).

antiderivative of d[sup:1r78lof4]3[/sup:1r78lof4][f(x)]/dx[sup:1r78lof4]3[/sup:1r78lof4] is d[sup:1r78lof4]2[/sup:1r78lof4][f(x)]/dx[sup:1r78lof4]2[/sup:1r78lof4]

and the antiderivative cos(x) is sin(x) + A

What is the anti-derivative of d[sup:1r78lof4]2[/sup:1r78lof4][f(x)]/dx[sup:1r78lof4]2[/sup:1r78lof4] ?

What is the anti-derivative of sin(x) + A ?

You have to repeat it till you get f(x) on the left-hand-side.

Then apply the initial conditions to solve for the constants of integration.
 
What do you mean get f(x) on the left side.... i have the antiderivative of sin(x)+a=ax-cos(x)+c, and the antiderivative of d^2[f(x)]/dx^2 = f'(x) and then antiderivative is f(x) of that?
 


janeann said:
What do you mean get f(x) on the left side

We start with f```(x) = stuff

We have f```(x) on the lefthand side

We differentiate in reverse (i.e., antidifferentiation) to get f``(x) = stuff

Now we have f``(x) on the lefthand side

We go backwards again to get f`(x) = stuff

Now we have f`(x) on the lefthand side

We continue on to get f(x) = stuff

At this point, f(x) is on the lefthand side, and we are finished.


antiderivative of sin(x) + a = ax - cos(x) + c

Yes, but I would solve for the constant at each step, rather than carry symbolic constants from one step backwards to the next.
Click to expand...

f```(x) = cos(x)

f``(x) = sin(x) + a

Solve for the constant at this point, using the initial condition that f``(0) = 1/2

1/2 = sin(0) + a

a = 1/2

Now continue backwards from:

f``(x) = sin(x) + 1/2

to get your:

f`(x) = -cos(x) + 1/2 x + c

The initial condition says f`(0) = -2

Use this to solve for c.

Continue antidifferentiating, determining the constants at each step, until you obtain f(x) on the lefthand side.

 
Re:

mmm4444bot said:



f```(x) = cos(x)

f``(x) = sin(x) + a

Solve for the constant at this point, using the initial condition that f``(0) = 1/2 <<< The original post indicates f" = 1/2 - commonly meant to say f"(x) = 1/2 for (-infnty,+infnty). The poster did not respond to my question.

1/2 = sin(0) + a

a = 1/2

Now continue backwards from:

f``(x) = sin(x) + 1/2

to get your:

f`(x) = -cos(x) + 1/2 x + c

The initial condition says f`(0) = -2

Use this to solve for c.

Continue antidifferentiating, determining the constants at each step, until you obtain f(x) on the lefthand side.
Click to expand...
 
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