How are $c_0$, $c_1$, and $c_2$ found for the quadratic approximation of $e^x$ at $x=0$?

Conditions

• The approximation is centered at $x=0$.
• The target function is $e^x$.
• The approximating polynomial is $c_0+c_1x+c_2x^2$.

Reasoning, step by step

1. Match the function values at $x=0$: $c_0+c_1(0)+c_2(0)^2=e^0=1$, which simplifies to $c_0=1$.
2. Differentiate both sides to match slopes: $d/dx(e^x)=e^x$ and $d/dx(1+c_1x+c_2x^2)=c_1+2c_2x$.
3. Evaluate the derivative equality at $x=0$: $e^0=c_1+2c_2(0)$, which simplifies to $c_1=1$.
4. Differentiate again to match concavities: $d^2/dx^2(e^x)=e^x$ and $d^2/dx^2(1+1x+c_2x^2)=2c_2$.
5. Evaluate the second derivative equality at $x=0$: $e^0=2c_2$, which simplifies to $1=2c_2$, so $c_2=1/2$.
6. Substitute the coefficients back to get $e^x$ ≈ $1+x+1/2x^2$.

Example

The board shows the sequence from $e^x$≈$c_0+c_1x+c_2x^2$ to $1+c_1x+c_2x^2$, then $1+1x+c_2x^2$, and finally $1+x+1/2x^2$. The speaker explicitly plugs in $x=0$, differentiates, plugs in zero again, differentiates again, and substitutes the found coefficients.

Common misconceptions

• Substituting $x=0$ before differentiating, which would erase the information needed to determine the higher coefficients.
• Confusing the first derivative (slope) with the second derivative (concavity) when setting up the equations for $c_1$ and $c_2$.

Watch the explanation

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• What is the quadratic approximation of $e^x$ at 0?
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