In the picture above we have the expression of stretch with respect to strain tensor, but in case of infinitesimal strain theory, we take stretch equal to 1+T.E.T

I don't understand why by taking Taylor series expansion, the answer would be 1+ x, with x= T.E.T, with my calculation, the first derivative is 1/radical(1+2x) not x

  • $\begingroup$ Did you ever learn about the Binomial Theorem? This is just expanding the square root as a series, and ignoring the higher order terms (which is what "infinitesimal" means). $\endgroup$ – alephzero Mar 19 at 15:29
  • $\begingroup$ I know its a taylor series expansion, but when I calculated it, the first derivative gave 1/radical(1+2x) , and not x. So this is why I'm not convinced with the answer. I have edited my question $\endgroup$ – user134613 Mar 19 at 16:53
  • $\begingroup$ Because derivative of radical U is U'/2*radical U $\endgroup$ – user134613 Mar 19 at 16:55
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    $\begingroup$ You're expanding the series at $x = 0$. That means the denominator of the derivative is equal to 1. See emathzone.com/tutorials/calculus/… $\endgroup$ – Biswajit Banerjee Mar 19 at 22:44
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    $\begingroup$ Taylor series is doing it the hard way. The Binomial theorem says $(1 + a)^{1/2} = 1 + \frac 1 2a + (\frac1 2)(\frac 1 2-1)(\frac 1 {2!})a^2 + \dots$. $\endgroup$ – alephzero Mar 19 at 23:46

Taylor is straightforward: $$ \sqrt{1+2x} =\left.\sqrt{1+2x}\right|_0 +\left.{d \over dx}\sqrt{1+2x}\right|_0x +O(x^2) \\ =1 +\left.{d \over dx}{1 \over \sqrt{1+2x}}\right|_0x +O(x^2) \\ =1 +x +O(x^2) \\ $$

Note that Taylor holds for matrices variables under some conditions.

ps.If you can handle it, you also have the Generalized Binomial Expansion for complex numbers (not matrices) $$ (x+y)^r=\sum_{k=0}^\infty\left(\array{r\\k}\right)x^{r-k} y^k \ , \ \left(\array{r\\k}\right)={(r)\cdots (r-k+1)\over k!}\\ \sqrt{1+2x}=(1+2x)^{1/2} =1+\left(\array{1/2\\1}\right)(2x)+O(x^2)\\ =1+{1/2 \over 1!}2x+O(x^2)=1+x+O(x^2) $$

  • $\begingroup$ Thank you very much $\endgroup$ – user134613 Mar 20 at 6:48
  • $\begingroup$ The OP's notation isn't very clear, but by any reasonable engineering definition "stretch" is a (real) number, not a matrix. Presumably $T.E.T$ with the underbrace $X$ is a component of some sort of triple product... $\endgroup$ – alephzero Mar 20 at 20:14

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