To expand exlog(1+y) up to the second-order terms around (0,0) using a Taylor series, we follow these steps:
1. Function Definition
Let f(x,y)=exlog(1+y).
We want to expand this function around (x,y)=(0,0), including terms up to second order.
2. Taylor Expansion Formula
The Taylor expansion of a function of two variables f(x,y) around (0,0) is given by:
f(x,y)≈f(0,0)+∂x∂f(0,0)x+∂y∂f(0,0)y+21(∂x2∂2f(0,0)x2+2∂x∂y∂2f(0,0)xy+∂y2∂2f(0,0)y2)
3. Compute Function and Derivatives at (0,0)
f(0,0):
f(0,0)=e0log(1+0)=1⋅0=0
First-order partial derivatives:
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∂x∂f=exlog(1+y), so at (0,0):
∂x∂f(0,0)=e0log(1+0)=1⋅0=0
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∂y∂f=1+yex, so at (0,0):
∂y∂f(0,0)=1+0e0=1
Second-order partial derivatives:
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∂x2∂2f=exlog(1+y), so at (0,0):
∂x2∂2f(0,0)=e0log(1+0)=1⋅0=0
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∂y2∂2f=−(1+y)2ex, so at (0,0):
∂y2∂2f(0,0)=−(1+0)2e0=−1
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∂x∂y∂2f=1+yex, so at (0,0):
∂x∂y∂2f(0,0)=1+0e0=1
4. Combine the Terms
Now, using the Taylor expansion formula, we plug in the values we've computed:
f(x,y)≈0+0⋅x+1⋅y+21(0⋅x2+2⋅1⋅xy+(−1)⋅y2)
Simplifying:
f(x,y)≈y+xy−21y2
Final Answer:
The second-order Taylor expansion of exlog(1+y) around (0,0) is:
f(x,y)≈y+xy−21y2
Would you like to explore the details further or ask any questions about the steps?
Here are 5 questions to expand on this topic:
- How do we generalize Taylor series for higher dimensions?
- What is the significance of second-order terms in a Taylor expansion?
- Can you expand exlog(1+y) to third order?
- How does the mixed partial derivative ∂x∂y∂2f contribute to the expansion?
- Why is the expansion centered around (0,0) instead of another point?
Tip: When expanding around a point, ensure you calculate all partial derivatives correctly up to the desired order.