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Oct
8
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.
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
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Oct
1
revised Beginner Calculus — Finding the Derivative Before Evaluating
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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
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Sep
29
awarded  Popular Question
Sep
26
revised Is there a function to *reverse* a number
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Sep
26
revised Is there a function to *reverse* a number
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Sep
25
revised Is there a function to *reverse* a number
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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.
Sep
24
reviewed Approve Find complex numbers in polynomial