# Tag Info

Accepted

### Different evaluation control inside of Integrate than outside of it

This is exactly what Inactivate was designed for: int = Inactivate[Integrate[x^n/(x + 1)^(n + 1), {x, 0, 1}], Integrate] ...
• 23.3k

### How to remember evaluation across sessions?

Strategy: Use memoization and then save the relevant definition in a file at the end of your session. Next session you can recover the "memoized" definition. Memoization Look at memoization ...
• 28.7k

### HoldForm is not working as I would expect

Why it happens When you enter an expression and evaluate it in Mathematica, MakeExpression is used to convert the box expression (i.e. the description of the exact ...
• 23.3k
Accepted

### HoldForm is not working as I would expect

(The basic idea is from this answer by xzczd.) Try this: ...
• 55.5k
Accepted

### How can I efficiently define functions with different names but doing the same stuff?

To define identical functions with SetDelayed use (#[x_] := x^2 + x) & /@ {f, g, h}; This approach is also useful when ...
• 119k

### Different evaluation control inside of Integrate than outside of it

You can use Trott-Strzebonski or RuleCondition or controlled evaluation; see Replacement inside held expression, which might be considered a duplicate. Variations: <...
• 213k

### Different evaluation control inside of Integrate than outside of it

I found a way, but does the code really have to be so ridiculous for such a simple task? ...
• 2,503

### HoldForm is not working as I would expect

It is just enough to wrap a/b. Try this: Have fun!
• 34.8k
Accepted

### Wierd problem with Table and Evaluate

You need to use v on the correct part of the solution as follows ...
• 108k
1 vote

### How can I efficiently define functions with different names but doing the same stuff?

If I understand your question correctly, this is how I usually implement this. ...
• 1,180
1 vote

### How can I efficiently define functions with different names but doing the same stuff?

Just to show another perspective that could also work on multiple definitions. Define your base function (which could have multiple definitions) ...
• 4,979

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