pbs
Reputation
3,401
Top tag
Next privilege 5,000 Rep.
Approve tag wiki edits
 1d comment I think I found a flaw in Riemann Zeta Function Regularization In general I get a feeling these "weird" identities are lacking some information... in particular to do with modulo some number... but that's another story. Getting back to the point, one problem might be: $\sum n^3 = \left(\sum n\right)^2$ is only true for finite number of terms. 1d comment How to prove $\sum_{i=2}^\infty { ({ 1 \over x })^i \over i!} \leq O({1 \over x^2})$? Where does the $n$ come from ? Maybe you mean $x$ not $n$... If so, not sure about a full proof, but it's clear to see that $1/x^i$ appears in the left and $i\geq 2$. Also - what are your permitted values of $x$ ? Oct 1 comment Beginner Calculus — Finding the Derivative Before Evaluating Yes, that's correct, but you may be able to use the limit definition of differentiation to substitute directly. You will have to try that out (similar to as I did in my answer above). But - for the questions you are answering it sounds like you should just differentiate and then substitute at the end. This is the simplest way to do things. Very loosely speaking, the "calculus" kind of abstracts the limit definitions into a neat package which is easier to manipulate for most applications. I liken it to abstraction in object oriented programming, but that's just the way I see it. Good luck. Oct 1 comment Beginner Calculus — Finding the Derivative Before Evaluating No. The product rule gives this: $g'(x)=xf'(x)+f(x)$. Since $x=1$, $f'(1)=-1$, and $f(1)=3$, then we get $g'(1)=1\cdot f'(1)+f(1)=1\cdot(-1)+3=2$. Oct 1 comment Beginner Calculus — Finding the Derivative Before Evaluating You've made a mistake in your calculation: the correct substitutions give $g'(x)=xf'(x)+f(x)$, so that $g'(1)=1\cdot(-1)+3=2,$ as required. Oct 1 comment Beginner Calculus — Finding the Derivative Before Evaluating We know that $g'(x)=xf'(x)+f(x)$, so if you know for sure the values of $f'(x)$ and $f(x)$ for a specific $x$ then yes you can substitute to get the answer - assuming those values are actually correct with respect to $f(x)$. Oct 1 revised Beginner Calculus — Finding the Derivative Before Evaluating added 26 characters in body Oct 1 revised Beginner Calculus — Finding the Derivative Before Evaluating added 255 characters in body Oct 1 answered Beginner Calculus — Finding the Derivative Before Evaluating Sep 29 comment How to Prove limit does not Exists This question and answers may help you: math.stackexchange.com/questions/1212672/… Sep 29 revised Is there a function to *reverse* a number deleted 28 characters in body Sep 29 awarded Popular Question Sep 26 revised Is there a function to *reverse* a number added 9 characters in body Sep 26 revised Is there a function to *reverse* a number added 21 characters in body Sep 25 revised Is there a function to *reverse* a number added 77 characters in body Sep 25 answered Is there a function to *reverse* a number Sep 25 comment Is there a function to *reverse* a number Something like: $$r(a)=\sum_{n=0}^N 10^n \left( \left\lfloor 10^na\right\rfloor -10\left\lfloor 10^{n-1}a \right\rfloor \right).$$ Assuming this formula is correct, then you may be able to find a closed form from it... if one such exists ! Sep 25 comment Is there a function to *reverse* a number Use the summation symbol $\Sigma$. A function can include summations. Sep 25 comment Is there a function to *reverse* a number Use your extractor function. Sep 25 comment Is there a function to *reverse* a number The number $1234$ has decimal digits $a=(1,2,3,4)$, with value $\sum_{k=0}^3 10^{3-k}a_k$, so the reverse is $\sum_{k=0}^3 10^k a_k$. Not exactly what you're looking for, but you can build a function using this and the number extractor function.