Q. \(1 + \frac{(a)_{1} (a-b+1)_{1}}{(-x)^{1}}\).
Answer
Since \((a)_1=a\) and \((a-b+1)_1=a-b+1\), the expression becomes:
\(1-\frac{a(a-b+1)}{x}\)
Final result: \(1-\frac{a(a-b+1)}{x}\)
Detailed Explanation
- Write the original expression.\(1 + \frac{(a)_1\,(a-b+1)_1}{(-x)^1}\)
- Recall the definition of the Pochhammer (rising factorial) for n = 1: by definition, for any symbol c, \((c)_1 = c\). Therefore:\((a)_1 = a\) and \((a-b+1)_1 = a-b+1\).
- Substitute these values into the numerator of the fraction:\( (a)_1\,(a-b+1)_1 = a\,(a-b+1).\)
- Simplify the denominator \((-x)^1\). Raising to the first power leaves the base unchanged, so:\((-x)^1 = -x.\)
- Put the simplified numerator and denominator into the fraction:\(1 + \dfrac{a\,(a-b+1)}{-x}.\)
- Use the sign in the denominator to move the negative sign in front of the fraction:\(1 – \dfrac{a\,(a-b+1)}{x}.\)
- If you prefer a single rational expression, combine the terms over the common denominator x. Write 1 as \( \dfrac{x}{x}\) and subtract the fraction:\(\dfrac{x}{x} – \dfrac{a\,(a-b+1)}{x} = \dfrac{x – a\,(a-b+1)}{x}.\)
- Optionally expand the product in the numerator for a fully expanded form. Expand \(a\,(a-b+1)\):\(a\,(a-b+1) = a^2 – a b + a.\)
Then the combined fraction becomes
\(\dfrac{x – (a^2 – a b + a)}{x} = \dfrac{x – a^2 + a b – a}{x}.\)
- Summary of equivalent simplified forms (choose the preferred one):
- \(1 – \dfrac{a\,(a-b+1)}{x}\)
- \(\dfrac{x – a\,(a-b+1)}{x}\)
- \(\dfrac{x – a^2 + a b – a}{x}\)
See full solution
Algebra FAQs
What does \( (a)_1 \) mean?.
\( (a)_1 \) is the Pochhammer (rising factorial) for \(n=1\), so \( (a)_1=a\). For general \(n\), \( (a)_n=a(a+1)\cdots(a+n-1)\).
How do I simplify \(1+\frac{(a)_1(a-b+1)_1}{(-x)^1}\)?
Use \( (a)_1=a\) and \( (a-b+1)_1=a-b+1\). The expression simplifies to \(1-\dfrac{a(a-b+1)}{x}\)..
Are there domain restrictions?
Yes: \(x\neq 0\) because of division by \(x\). There are no extra restrictions for \(n=1\); Pochhammer values are defined for all complex \(a\)..
How does this relate to the \(\Gamma\) function?
For integer \(n\), \( (a)_n=\dfrac{\Gamma(a+n)}{\Gamma(a)}\). For \(n=1\), \( (a)_1=\Gamma(a+1)/\Gamma(a)=a\).
Is \( (a)_n \) the same as a falling factorial?.\
No. \( (a)_n\) is the rising factorial. The falling factorial is \(a^{\underline{n}}=a(a-1)\cdots(a-n+1)\). Notation must be checked in context.
What is the limit as \(x\to\infty\)?.
What is the limit as \(x\to\infty\)?.
For which \(x\) does the expression equal zero?
Solve \(1-\dfrac{a(a-b+1)}{x}=0\). Thus \(x=a(a-b+1)\), assuming \(x\neq 0\).
Can you give a quick numeric example?
Let \(a=2,b=1,x=3\). Then \( (a)_1=2,\ (a-b+1)_1=2\), so \(1+\dfrac{4}{-3}=1-\tfrac{4}{3}=-\tfrac{1}{3}\). .
Where does this term appear in hypergeometric series?
It is the first-order term (n=1) of a hypergeometric coefficient like in \( {}_2F_1(a,a-b+1;\, \cdot \,; \, z) \) or similar series, arising from Pochhammer factors in the general term.
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