VIPSolutionsAct like a helpful tutor and exlain me : Give step-by-step solution with explanation and final answer:1. Using Taylor series expansion to derive the following Eq. up to O(At*). Note that the high-order terms must be included in your answer. Qu(w.y +A) at
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Act like a helpful tutor and exlain me :
Give step-by-step solution with explanation and final answer:
1. Using Taylor series expansion to derive the following Eq. up to O(At*). Note that the
high-order terms must be included in your answer.
Qu(w.y +A)
at
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Asked by: manish kumar
Created at: 2025-10-05 23:22:55
Asked by: manish kumar
Created at: 2025-10-05 23:22:55
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## Step-by-Step Solution
Let's break down the problem:
**We are asked to use the Taylor series to expand \( u(x, y, t + \Delta t) \) about \( t \) up to order \( O(\Delta t^4) \) and then derive the time derivative \( \frac{\partial u(x, y, t + \Delta t)}{\partial t} \) including all higher order terms up to \( O(\Delta t^4) \).**
---
### **Step 1: Taylor Series Expansion**
The Taylor series for a general function \( u(x, y, t) \) expanded about \( t \) is:
\[
u(x, y, t + \Delta t) =
u(x, y, t)
+ \Delta t \frac{\partial u}{\partial t}
+ \frac{(\Delta t)^2}{2!} \frac{\partial^2 u}{\partial t^2}
+ \frac{(\Delta t)^3}{3!} \frac{\partial^3 u}{\partial t^3}
+ \frac{(\Delta t)^4}{4!} \frac{\partial^4 u}{\partial t^4}
+ O(\Delta t^5)
\]
---
### **Step 2: Find the Time Derivative**
Now, let's differentiate \( u(x, y, t + \Delta t) \) with respect to \( t \):
\[
\frac{\partial u(x, y, t + \Delta t)}{\partial t}
\]
Since \( u(x, y, t + \Delta t) \) is a function of \( t \), we differentiate each term:
- The first term \( u(x, y, t) \) gives \( \frac{\partial u}{\partial t} \).
- The second term \( \Delta t \frac{\partial u}{\partial t} \) gives \( \Delta t \frac{\partial^2 u}{\partial t^2} \).
- The third term \( \frac{(\Delta t)^2}{2!} \frac{\partial^2 u}{\partial t^2} \) gives \( \frac{(\Delta t)^2}{2} \frac{\partial^3 u}{\partial t^3} \).
- The fourth term \( \frac{(\Delta t)^3}{3!} \frac{\partial^3 u}{\partial t^3} \) gives \( \frac{(\Delta t)^3}{6} \frac{\partial^4 u}{\partial t^4} \).
- The fifth term \( \frac{(\Delta t)^4}{4!} \frac{\partial^4 u}{\partial t^4} \) gives \( \frac{(\Delta t)^4}{24} \frac{\partial^5 u}{\partial t^5} \).
So,
\[
\frac{\partial u(x, y, t + \Delta t)}{\partial t} =
\frac{\partial u}{\partial t} +
\Delta t \frac{\partial^2 u}{\partial t^2} +
\frac{(\Delta t)^2}{2} \frac{\partial^3 u}{\partial t^3} +
\frac{(\Delta t)^3}{6} \frac{\partial^4 u}{\partial t^4} +
\frac{(\Delta t)^4}{24} \frac{\partial^5 u}{\partial t^5} +
O(\Delta t^5)
\]
---
### **Step 3: Final Answer (Explicit Form)**
\[
\boxed{
\frac{\partial u(x, y, t + \Delta t)}{\partial t} =
\frac{\partial u}{\partial t}
+ \Delta t \frac{\partial^2 u}{\partial t^2}
+ \frac{(\Delta t)^2}{2} \frac{\partial^3 u}{\partial t^3}
+ \frac{(\Delta t)^3}{6} \frac{\partial^4 u}{\partial t^4}
+ O(\Delta t^4)
}
\]
> **Note:** The \( O(\Delta t^4) \) notation means we are keeping all terms up to \( (\Delta t)^3 \), and the next term (with \( (\Delta t)^4 \)) and higher are included in the error term.
---
### **Summary Table**
| Term Order | Expression |
|----------------|-------------------------------------------------|
| \( O(1) \) | \( \frac{\partial u}{\partial t} \) |
| \( O(\Delta t) \) | \( \Delta t \frac{\partial^2 u}{\partial t^2} \) |
| \( O(\Delta t^2) \) | \( \frac{(\Delta t)^2}{2} \frac{\partial^3 u}{\partial t^3} \) |
| \( O(\Delta t^3) \) | \( \frac{(\Delta t)^3}{6} \frac{\partial^4 u}{\partial t^4} \) |
| \( O(\Delta t^4) \) | Higher order (in error term) |
---
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