Turn a Solved DE into a Taylor Series

In summary: Differential Equations.In summary, Eric is working on a problem for his differential equations class, but is stuck on the last step. He has tried to solve the problem using techniques he learned in class, but has run into trouble. He has been told by his teacher that he needs to find a similar Taylor Series to solve for C, but is not sure how to do that. He has also been told that he needs to find an exponential function to relate to his original equation. He believes he has found the correct equation, but is unsure if he has turned it into a taylor series correctly. He has also been told that he needs to find an e in his equation to make it correct.
  • #1
epheterson
22
0
Many of you have probably used the book Differential Equations by Lomen & Lovelock.
For my class I'm working on Example 2, Page 153.

You don't need to see the book, though, to help me out. It's a four-part problem and I'm on the last step not knowing where to take it.

In Part B, we created a DE:

dP/dt = (397/4169) P + (6580236/4169)

and solved it:

P(t) = (397/8338) t^2 + (6580236/4169) t + (1225953/4169)

Now in Part D they ask:
"Expand the explicit solution you obtained in Part B in a Taylor Series about the origin."

I'm not so sure I've ever done anything like this, so I'm not clear on what steps I should take. I believe my teacher said I should find a similar Taylor Series and manipulate it to fit my needs but that still doesn't get me very far.

Absolutely any insight is appreciated, hope you can help me!

Thanks,
Eric
 
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  • #2
If your differential equation really is
dP/dt = (397/4169) P + (6580236/4169)
then dP/dt = (397/4169) P + (6580236/4169)
is NOT the solution. Your P(t) satisfies dP/dt = (397/4169) t + (6580236/4169).

The solution to dP/dt = (397/4169) P + (6580236/4169)
will involve an exponential function.
 
  • #3
I think I'm going to have to look at it in a different light. I could've sworn I put it correctly into maple, that's what it threw back at me.

Now that I try again, I get P(t) = -315680/19+exp((19/200)*t)*_C1, which I have checked by hand. I'm not entirely sure it's right though, I solved for C using a point from my original data table and got funny results when trying to use it.

I'll try again tomorrow. But now that I have an e in it, I believe I'll just be modifying the taylor series for e.
 
  • #4
Okay, I've figured out the problem. The one I posted was from a different part of the question.

I believe my correct equation is
[tex]P(t)=e^{.095t}16543 - 16571[/tex]

Now I know that E is
[tex]e^{x} = \Sigma\frac{x^{n}}{n!}[/tex]

So, correct me if I'm wrong, but do I manipulate the series as so to match my equation?

[tex]\Sigma[\frac{(0.095t)^n}{n!}16543-16571][/tex]

To be proper, all the summations here are from n=0 to [tex]\infty[/tex]

In the question they ask me to relate it to the solution I got in part A, which is similar to what I originally posted:
[tex]P(t) = a t^2 + b t + c[/tex]

I don't think how I have it set up results in an answer in that form. Did I turn it into a taylor series properly?
 
Last edited:
  • #5
epheterson said:
Okay, I've figured out the problem. The one I posted was from a different part of the question.

I believe my correct equation is
[tex]P(t)=e^{.095t}16543 - 16571[/tex]

Now I know that E is
[tex]e^{x} = \Sigma\frac{x^{n}}{n!}[/tex]

So, correct me if I'm wrong, but do I manipulate the series as so to match my equation?

[tex]\Sigma[\frac{(0.095t)^n}{n!}16543-16571][/tex]
No, the "-16571" is NOT part of the exponential and is NOT summed an infinite number of times!

To be proper, all the summations here are from n=0 to [tex]\infty[/tex]

In the question they ask me to relate it to the solution I got in part A, which is similar to what I originally posted:
[tex]P(t) = a t^2 + b t + c[/tex]

I don't think how I have it set up results in an answer in that form. Did I turn it into a taylor series properly?
And, again, that quadratic is WRONG.
 
  • #6
Well stay with me here...

I was unsure and that's what I was asking about, now as I understand it my new equation should be:
[tex]P(t) = (16543)\Sigma[\frac{(0.095t)^n}{n!}]-16571[/tex]

True, the [tex]P(t) = a t^2 + b t + c[/tex] is incorrect for the original DE I gave you, but it is correct for a different part of the problem. This step is relating the quadratic to our new equation which we're making with summations.


I think I did this right, to relate them, I calculated out the summation to three terms and plugged them into get:
[tex]P(t) = 74.7t^2+1571.6t-28[/tex]

which I can relate to my quadratic using similarities.
 
  • #7
epheterson said:
Well stay with me here...

I was unsure and that's what I was asking about, now as I understand it my new equation should be:
[tex]P(t) = (16543)\Sigma[\frac{(0.095t)^n}{n!}]-16571[/tex]
Good! This is now correct.

True, the [tex]P(t) = a t^2 + b t + c[/tex] is incorrect for the original DE I gave you, but it is correct for a different part of the problem. This step is relating the quadratic to our new equation which we're making with summations.


I think I did this right, to relate them, I calculated out the summation to three terms and plugged them into get:
[tex]P(t) = 74.7t^2+1571.6t-28[/tex]

which I can relate to my quadratic using similarities.
Okay, I didn't do the calculations myself but assuming the arithmetic is correct, this is the "2nd order Taylor Polynomial" for your function.
 

1. What is a Taylor Series?

A Taylor Series is a mathematical representation of a function as an infinite sum of terms that are calculated from the values of the function's derivatives at a single point.

2. How do you turn a solved differential equation into a Taylor Series?

To turn a solved differential equation into a Taylor Series, you first need to find the derivatives of the function at a specific point. Then, plug these derivatives into the formula for a Taylor Series and simplify to get the infinite sum representation.

3. Why would I want to turn a solved differential equation into a Taylor Series?

Turning a solved differential equation into a Taylor Series can be useful for approximating the function in a specific range of values. It can also help with understanding the behavior of the function and making predictions about its future values.

4. Are there any limitations to using a Taylor Series to represent a function?

Yes, there are limitations to using a Taylor Series. The series may not converge for certain functions or may only converge within a certain range. Additionally, the accuracy of the series may decrease as the number of terms in the infinite sum increases.

5. Can a Taylor Series be used for non-polynomial functions?

Yes, a Taylor Series can be used for non-polynomial functions as long as the derivatives of the function exist at the chosen point. However, the resulting series may not always converge or may only converge within a limited range of values.

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