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This is the problem I'm trying to solve:

Let $f(x,y)=ay+sin(bx)+c$. Evaluate the Taylor polynomial at $P(0,0)$ and find the values for $a$, $b$ and $c$ if $P(x,y)=-1+2x-y$

I do know the linearization formula, but it requires me to use the value of $f$ at the given point. However, since there are 3 unknown constants, I don't know how to work this out... Thanks for your help.

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The linear Taylor-Polynomial of $f$ with respect to $(0,0)$ is given by:

$$f(x,y) \approx P(x,y;(0,0))= f(0,0) + f'_x(0,0)(x-0) + f'_y(0,0)(y-0)$$

Can you take it from here? The spoiler contains the next step - try to do it yourself, first.

$$P(x,y) = [c] + [b\cos(b0)x] + [ay] = c + bx + ay\overset{!}{=}-1+2x-y$$

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  • $\begingroup$ Thank you :) What I'm stuck with is, how do you relate $c$ with $f(0,0)$? I know the function and the linearization have the same value at the evaluated point, so I got $-1 + [b\cos(b0)x] + [ay]$ and I got that $-1$ from evaluating the given $P(x,y)$ at the given $(0,0)$. But how can I say that "c = -1"? $\endgroup$ – Floella Dec 14 '17 at 15:36
  • $\begingroup$ $c=f(0, 0)=P(0, 0) = -1$. This should follow directly from the associativity of equality ($a=b$ and $b=c$ implies $a=c$). $\endgroup$ – Austin Weaver Dec 14 '17 at 15:41
  • $\begingroup$ Thank you. Actually, I understand that $f(0, 0)=P(0, 0)$. What I don't get is why those two are the same as $c$. Probably a silly question, but I haven't been able to figure it out. $\endgroup$ – Floella Dec 14 '17 at 15:48
  • $\begingroup$ In the spoiler we calculated the general Taylor-Polynomial. This has to be ($\overset{!}{=}$) equal to the polynomial from the Task. It holds: Two polynomials $f(x)=a_0 + a_1x+…$ and $g(x)=b_0 + b_1x+ …$ are equal if their coefficients are equal: $$f=g ⇔ a_0=b_0, a_1=b_1 ...$$ So we can "equate coefficients". (Small remark: This statement is not always correct in the way written here. But for the sake of this answer it is true.) $\endgroup$ – P. Siehr Dec 14 '17 at 15:58
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The second order taylor polynomial of $f(x, y)$ at $(0, 0)$ is $$f_{approx}(x, y) = c + b\cos(bx_0)x+ay+0xy = c+bx+ay.$$

But we are given that $f_{approx}(x, y) = -1+2x-y$, so we set the expressions equal to solve for $a, b,$ and $c$. This yields:

$$-1+2x-y=c+bx+ay\\\implies a = -1, b = 2, c = -1.$$

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