I ask members here kindly for their assistance. I'm having some confusion over the process of integrating inequalities, in particular for obtaining the series expansion for the exponential function by integration. The text by Backhouse and Holdsworth (Pure Mathematics 2), shows the expansion of the exponential function by integration with the assumption that (x^n)→0 as n→∞ when |x|<1; Let the variable x lie in the range of values from 0 to c, where c is any positive constant, thus 0 < x < c since e^0 = 1, ∴ 1 < (e^x) < e^c Integrating from 0 to x gives, x < (e^x - 1) < x(e^c) The problem i'm having here is the integration step, where does the negative one come from in the middle of the inequality? Thanks. Lukka
Thanks for popping by Voko. That's the definite integral from zero to x.. i.e. the indefinite integral evaluated at x minus the indefinite integral evaluated at 0 right? So if we evaluate at x we have; ∫ e^x dx = e^x +c1, and at zero; ∫ e^0 dx = ∫ (1) dx = x +c2 Thus the definite integral of e^x from 0 to x; [e^x +c1] - [x +c2] = [e^x - x] +c1 + c2
It would be better to write this as ##\int_0^x e^t dt ## I.e., use a dummy variable that is different from the variable in the limits of integration. No, this isn't right. An antiderivative of e^{x} is e^{x}. For a definite integral you can choose the antiderivative where the arbitrary constant is zero. So, $$ \int_0^x e^t~dt = \left. e^t\right|_0^x = e^x - e^0 = e^x - 1$$ Where you went wrong was in integrating e^{0}. You don't need to (and shouldn't) do two integrations. Once you have your antiderivative, just evaluate it at the two limits of integration.
I agree personally, but this unfortunately tends to confuse students, in part because the ambiguous notation is well entrenched in textbooks.
Hey Mark, thanks for sharing your insight, i guess i will have to review the topic a little and practice a bunch of examples. Much more legible with the use of dummy variable too.. with regards to the integral when i evaluate at zero, I think i understand where i'm going wrong here.. I should just write the integral of e^0 as; ∫ e^0 dx = e^0 +c = 1 +c Is that right? ie. i cant just substitute 1 in for e^0 before i integrate? Thank you both for sharing your expertise, your help is greatly appreciated :)
No, this is not correct at all. As Mark44 said, you do not compute the indefinite integral two times. You do it just once, and then you plug in the upper and lower limits.
Yes. i understand what you are saying. i just integrate the function e^t like an indefinite case and than plug in the upper and lower limits of integration and obtain the difference. sorry for the confusion there and thank you for pointing this out. Where can i find the tex commands for the definite integral on the edit page?
Here: https://www.physicsforums.com/showpost.php?p=3977517&postcount=3. There's a table of frequently used symbols about halfway down the page. Integrals are shown near the end of the table. The table doesn't show definite integrals, so maybe I can add some to the table. Anyway, here's what I did. I put this in HTML code tags so that it won't render in the browser, letting you see what I wrote. Note that in the table they use [ tex ] and [ itex ] tags (without the extra spaces). You can also put $$ before and after your expression (same as tex and \tex) or ## before and after your expression (same as itex and \itex). Code (Text): $$\int_0^x e^t~dt = \left.e^t\right|_0^x = e^x - e^0$$ The above renders as $$\int_0^x e^t~dt = \left.e^t\right|_0^x = e^x - e^0$$
Brilliant :) thanks for the link. It doesn't look too complicated i shall begin using it right away. I found some more latex under the sigma tab on the advanced page too. thanks again Mark ;)
It's actually \left.e^t\right|_0^x This is to get the vertical bar off to the right with the two limits of integration (\right|_0^x). The other end, \left. doesn't display anything, but a \right command can't appear without a \left command. This string displays like this: $$\left.e^t\right|_0^x$$
Yes okay i understand. That's pretty cool. So the bar is inserted at the end with the underscore to indicate the limits themselves? I have bookmarked the Latex page for future reference. The sandbox is very helpful for getting started. Might take some getting used to but i'm happy to learn it, after all it helps us all better communicate and understand each other. Is Latex a universal code? i mean can i use the same tags when visiting other forums?
The underscore and caret (or circumflex) are pretty standard in LaTeX for subscripts and exponents, respectively. The same notation is used in integrals and summations and elsewhere, and limits use a similar notation. Code (Text): $$ \int_0^1 f(x)dx$$ $$ \int_0^1 f(x)dx$$ Code (Text): $$ \sum_{n = 1}^{\infty} \frac 1 n$$ $$ \sum_{n = 1}^{\infty} \frac 1 n$$ Code (Text): $$ \lim_{x \to \infty} f(x)$$ $$ \lim_{x \to \infty} f(x)$$
This is a bit of a weird case, the more common use is with brackets. E.g. compare Code (Text): $$( \frac{1}{x} [ \sum_{n = 1}^N n ] )$$ $$( \frac{1}{x} [ \sum_{n = 1}^N n ] )$$ and Code (Text): $$ [u]\left([/u] \frac{1}{x} [u]\left[[/u] \sum_{n = 1}^N n [u]\right][/u] [u]\right)[/u] $$ $$\left( \frac{1}{x} \left[ \sum_{n = 1}^N n \right] \right)$$ The nice trick is the brackets don't have to match, e.g. you can use Code (Text): $$\left[ \sum_{n = 1}^N n \right\}$$ $$\left[ \sum_{n = 1}^N n \right\}$$ And if you put a dot, you will get no bracket, e.g. Code (Text): $$\left\{ \sum_{n = 1}^N n \right.$$ $$\left\{ \sum_{n = 1}^N n \right.$$ and you can combine with superscripts and subscripts Code (Text): $$\left. \left( \sum_{n = 1}^N n \right)^2 \right|_{x = 0}$$ $$\left. \left( \sum_{n = 1}^N n \right)^2 \right|_{x = 0}$$