# Stretch in Infinitesimal strain theory 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

• 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). – alephzero Mar 19 at 15:29
• 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 – user134613 Mar 19 at 16:53
• Because derivative of radical U is U'/2*radical U – user134613 Mar 19 at 16:55
• You're expanding the series at $x = 0$. That means the denominator of the derivative is equal to 1. See emathzone.com/tutorials/calculus/… – Biswajit Banerjee Mar 19 at 22:44
• Taylor series is doing it the hard way. The Binomial theorem says $(1 + a)^{1/2} = 1 + \frac 1 2a + (\frac1 2)(\frac 1 2-1)(\frac 1 {2!})a^2 + \dots$. – alephzero Mar 19 at 23:46

Taylor is straightforward: $$\sqrt{1+2x} =\left.\sqrt{1+2x}\right|_0 +\left.{d \over dx}\sqrt{1+2x}\right|_0x +O(x^2) \\ =1 +\left.{d \over dx}{1 \over \sqrt{1+2x}}\right|_0x +O(x^2) \\ =1 +x +O(x^2) \\$$
ps.If you can handle it, you also have the Generalized Binomial Expansion for complex numbers (not matrices) $$(x+y)^r=\sum_{k=0}^\infty\left(\array{r\\k}\right)x^{r-k} y^k \ , \ \left(\array{r\\k}\right)={(r)\cdots (r-k+1)\over k!}\\ \sqrt{1+2x}=(1+2x)^{1/2} =1+\left(\array{1/2\\1}\right)(2x)+O(x^2)\\ =1+{1/2 \over 1!}2x+O(x^2)=1+x+O(x^2)$$
• 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... – alephzero Mar 20 at 20:14