Waldronjw

2022-07-15

Let $y=f(x)=({x}_{1}^{2}+2{x}_{2},{x}_{1}{x}_{2}-3{x}_{1})$

Is the linear approximation just $f(y)=f(x)+A(y-x)$ whenever y is approximately near $x$?

Is the linear approximation just $f(y)=f(x)+A(y-x)$ whenever y is approximately near $x$?

Wade Atkinson

Beginner2022-07-16Added 12 answers

A differentiable function $f:{\mathbb{R}}^{2}\to {\mathbb{R}}^{2}$ can be approximated by it's derivative in the sense that given a point, say, $(1,1)$, you can write:

$f(1+\delta x,1+\delta y)\approx f(1,1)+Df(1,1)(\delta x,\delta y)$

where $Df(1,1)$ is a matrix applied to the vector $(\delta x,\delta y)$. So, yes, if $y$ is near $x$ (as ($(1+\delta x,1+\delta y)$) is near ($(1,1)$)), then you can approximate the difference $f(y)-f(x)$ by the derivative evaluated at $x$, applied to the difference $y-x$. That is:

$f(y)\approx f(x)+Df(x)(y-x)$

Compare this to the one dimensional case, where the tangent line to the graph of a differentiable function locally approximates the graph, in the sense that:

$f(y)\approx f(x)+{f}^{\prime}(x)(y-x)$

$f(1+\delta x,1+\delta y)\approx f(1,1)+Df(1,1)(\delta x,\delta y)$

where $Df(1,1)$ is a matrix applied to the vector $(\delta x,\delta y)$. So, yes, if $y$ is near $x$ (as ($(1+\delta x,1+\delta y)$) is near ($(1,1)$)), then you can approximate the difference $f(y)-f(x)$ by the derivative evaluated at $x$, applied to the difference $y-x$. That is:

$f(y)\approx f(x)+Df(x)(y-x)$

Compare this to the one dimensional case, where the tangent line to the graph of a differentiable function locally approximates the graph, in the sense that:

$f(y)\approx f(x)+{f}^{\prime}(x)(y-x)$

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B) $\frac{5}{9}$ degree on fahrenheit scale

C) 1 degree on fahrenheit scale

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What's the correct way to go about computing the Inverse Laplace transform of this?

$\frac{-2s+1}{({s}^{2}+2s+5)}$

I Completed the square on the bottom but what do you do now?

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$X(s)=\frac{s}{{s}^{4}+1}$

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inverse laplace transform - with symbolic variables:

$F(s)=\frac{2{s}^{2}+(a-6b)s+{a}^{2}-4ab}{({s}^{2}-{a}^{2})(s-2b)}$

My steps:

$F(s)=\frac{2{s}^{2}+(a-6b)s+{a}^{2}-4ab}{(s+a)(s-a)(s-2b)}$

$=\frac{A}{s+a}+\frac{B}{s-a}+\frac{C}{s-2b}+K$

$K=0$

$A=F(s)\ast (s+a)$